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190500	https://www.ahajournals.org/doi/10.1161/01.STR.30.6.1230	The Natural History of CADASIL \| Stroke Skip to main content Advertisement Become a member Volunteer Donate Journals BrowseCollectionsSubjectsAHA Journal PodcastsTrend Watch ResourcesCMEAHA Journals @ MeetingsJournal MetricsEarly Career Resources InformationFor AuthorsFor ReviewersFor SubscribersFor International Users Alerts 0 Cart Search Sign inREGISTER Quick Search in Journals Enter search term Quick Search anywhere Enter search term Quick search in Citations Journal Year Volume Issue Page Searching: This Journal This JournalAnywhereCitation Advanced SearchSearch navigate the sidebar menu Sign inREGISTER Quick Search anywhere Enter search term Publications Arteriosclerosis, Thrombosis, and Vascular Biology Circulation Circulation Research Hypertension Stroke Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Journal of the American Heart Association Circulation: Arrhythmia and Electrophysiology Circulation: Cardiovascular Imaging Circulation: Cardiovascular Interventions Circulation: Cardiovascular Quality & Outcomes Circulation: Genomic and Precision Medicine Circulation: Heart Failure Stroke: Vascular and Interventional Neurology Annals of Internal Medicine: Clinical Cases Information For Authors For Reviewers For Subscribers For International Users Submit & Publish Submit to the AHA Editorial Policies Open Science Value of Many Voices Publishing with the AHA Open Access Information Resources AHA Journals CME AHA Journals @ Meetings Metrics AHA Journals Podcast Network Early Career Resources Trend Watch Professional Heart Daily AHA Newsroom Current Issue Archive Journal Information About Stroke Author Instructions Editorial Board Information for Advertisers Features Stroke Alert Podcast Blogging Stroke Early Career Program Journal Awards Submit Reference #1 Research Article Originally Published 1 June 1999 Free Access The Natural History of CADASIL : A Pooled Analysis of Previously Published Cases David W.Desmond, Joan T.Moroney, Timothy Lynch, Stephen Chan, Steven S.Chin, and J. P.MohrAuthor Info & Affiliations Stroke Volume 30, Number 6 18,944 194 Metrics Total Downloads 18,946 Last 12 Months 866 Total Citations 194 Last 12 Months 2 View all metrics Track CitationsAdd to favorites PDF/EPUB Contents Abstract Subjects and Methods Results Discussion Acknowledgments Footnote References eLetters Information & Authors Metrics & Citations View Options References Figures Tables Media Share Abstract Background and Purpose—Although numerous families with cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy (CADASIL) have been reported, our understanding of the disease remains incomplete. Thus, we performed this study to investigate the phenotypic range and natural history of CADASIL. Methods—We performed a pooled analysis of previously published cases. Results—We identified 105 symptomatic patients in 33 families. Vascular risk factors were uncommon, with hypertension reported in only 8 patients. The mean age of symptom onset was 36.7±12.9 years. Stroke or transient ischemic attack was an initial symptom in 45 patients, with a mean age of onset of 41.2±9.2 years. Migraine was also a common initial symptom, reported by 42 patients at a younger mean age of 28.3±11.7 years. Other initial symptoms included depression in 9 patients, cognitive impairment in 6 patients, and seizures in 3 patients. Regarding clinical course, 71 patients experienced a stroke or transient ischemic attack, and 52 of those patients had 1 or more recurrent ischemic events. Dementia was reported in 44 patients. Only 3 additional patients experienced migraine at a later time, while 13 additional patients developed depression. Six patients had seizures. Twenty-two of the 105 patients had died, with a mean age of death of 54.8±10.6 years. Nineteen of those 22 patients had experienced a stroke or transient ischemic attack and 19 patients were demented. Conclusions—CADASIL typically becomes evident in early or middle adulthood with migraine or an ischemic event, later manifests itself through recurrent subcortical ischemic strokes leading to a stepwise decline and dementia, and results in reduced survival. Early reports of the disease that was later termed “cerebral autosomal dominant arteriopathy with subcortical infarcts and leukoencephalopathy” (CADASIL) were sporadic. Families with an apparently hereditary form of vascular dementia were described by Van Bogaert in 19551 and Sourander and Wålinder in 1977,2 but that disorder received little additional attention until 1991,34 when the current resurgence of interest in the disease now known as CADASIL began. Despite the numerous case reports that have since been published, however, our understanding of the phenotypic range and natural history of CADASIL remains incomplete. Thus, we performed a pooled analysis of the clinical characteristics of previously reported well-studied symptomatic CADASIL patients to provide a broader perspective on that disorder, with an emphasis on the age and nature of symptom onset and the association between stroke and dementia. Subjects and Methods To identify all previously reported well-studied symptomatic patients who were likely to have had CADASIL, we first performed a computerized review of the literature using MEDLINE and a broad group of search terms. We then performed a manual review of the literature that was guided by the reference lists of relevant articles. Finally, we performed a manual review of abstracts that had been presented at major neurological meetings. In certain cases, follow-up publications provided further information regarding patients who had been described previously. In those instances, we incorporated that information into our database, but we did not cite those follow-up publications unless new cases had been reported in order to maintain the clarity of our reference list. It should be noted that we did not extract any additional patients from the cohorts described by Chabriat et al5 and Dichgans et al.6 We constructed a database consisting of families with an autosomal dominant pattern of inheritance of disorders that are typical of CADASIL (eg, stroke, migraine). We first required that 1 or more family members had undergone pathological study with results consistent with CADASIL. Alternatively, families were required to have undergone a genetic analysis with results supportive of linkage to the CADASIL locus. Individual members of each of those families were then required to have brain imaging features7 or pathological findings8 typical of CADASIL as well as an adequate clinical description (eg, age of symptom onset, nature of first symptom) to be eligible for inclusion in our final database. Results Background Characteristics We identified 105 eligible patients in 33 families.2349101112131415161718192021222324252627282930313233343536 Although we did not extract any additional patients from the cohort studies of Chabriat et al5 and Dichgans et al,6 we did include patients whom they had presented in other publications. Unfortunately, Chabriat et al and Dichgans et al failed to state explicitly whether any of the patients that they presented in those other publications had been included in their cohort studies. After careful review of their descriptions of those cohorts, however, we infer that no more than 11 of our 105 patients (10.5%) overlap with the cohort of Chabriat et al and that no more than 7 of our 105 patients (6.7%) overlap with the cohort of Dichgans et al. Regarding our eligibility criteria, pathological findings consistent with CADASIL were reported in 1 or more members of 24 of the 33 families, the results of genetic testing were consistent with CADASIL in 17 families, and both forms of supportive information were available in 8 families. Brain imaging findings consistent with CADASIL were reported in 100 patients, pathological findings consistent with CADASIL were reported in 34 patients, and both forms of supportive information were available in 29 patients. It should be noted that pathological findings consistent with CADASIL were reported in each of the 5 patients for whom brain imaging was not available. Thirty-four of the 105 patients were from France, 16 were from Italy, 39 were from other European countries, 13 were from the United States, 2 were from El Salvador, and 1 was from Japan. The sample was evenly divided by sex, with 55 males and 50 females. Vascular risk factors were uncommon, with hypertension reported in only 8 patients. Initial Symptoms The mean age of symptom onset was 36.7±12.9 years, with a range of 10 to 59 years. Stroke or transient ischemic attack (TIA) was an initial symptom in 45 patients, with a mean age of onset of 41.2±9.2 years (range, 20 to 58 years). Migraine was also a common initial symptom, reported by 42 patients at a significantly younger mean age of 28.3±11.7 years (range, 10 to 54 years; P<0.001). Thirty of those 42 patients had experienced migraine with aura, 6 had experienced migraine without aura, and 6 had experienced unspecified migraine. Other initial symptoms included depression in 9 patients, cognitive impairment in 6 patients, and seizures in 3 patients. Given that our report of learning disorders as the initial symptom of 3 patients in another publication34 was novel, it should be noted that we considered depression to be the initial symptom of the patients coded as III-1 and IV-5 in that pedigree and migraine to be the initial symptom of patient IV-3 for the purposes of the preceding analysis. Initial symptoms stratified by decade of age of onset are presented graphically in the Figure. Clinical Course Seventy-one of the 105 patients experienced a stroke or TIA, and 52 of those 71 patients went on to experience 1 or more recurrent ischemic events. Forty-four of the 105 patients were reported to be demented, and 16 of those 44 patients exhibited characteristics clearly typical of frontal lobe involvement (eg, disinhibition, perseverative behavior). An additional 19 patients exhibited cognitive impairment but were not reported to have frank dementia, suggesting that 63 of the patients exhibited at least mild cognitive deficits. Thirty-six of the 44 demented patients (81.8%) had experienced a stroke or TIA, and 31 of those patients (70.5%) had 1 or more recurrent ischemic events, while 35 of the 61 nondemented patients (57.4%) had experienced a stroke or TIA, and 21 of those patients (34.4%) had 1 or more recurrent ischemic events. We determined by χ 2 analyses that patients reported to have dementia had significantly more frequently experienced a stroke or TIA (P=0.008) and 1 or more recurrent ischemic events (P<0.001) than nondemented patients. Only 3 additional patients experienced the new onset of migraine later in their clinical course, suggesting that migraine is unlikely to occur if it is not the initial symptom. Thirteen additional patients went on to experience at least 1 episode of depression, however, suggesting that depression is more likely to occur later in the course of the disease than as an early manifestation. Six patients had seizures. Among patients who were living at the time of their case report, the mean maximum age reported was 47.6±12.6 years (range, 19 to 76 years). Twenty-two of the 105 patients were reported to have died, with a mean age of death of 54.8±10.6 years (range, 30 to 75 years), and the mean time from symptom onset to death was 12.8±13.1 years (range, <1 to 65 years). Nineteen of those 22 patients (86.4%) had experienced at least 1 stroke or TIA and 19 patients were demented. Discussion Our findings are consistent with the results of the cohort studies of Chabriat et al5 and Dichgans et al6 in suggesting that CADASIL becomes evident in young or middle adulthood with migraine or an ischemic event. Later, it manifests itself through recurrent subcortical ischemic strokes leading to a stepwise decline and a dementia syndrome with frontal lobe features, thus exemplifying the stereotypic course of vascular dementia. Finally, although we noted wide variability in the time from symptom onset to death, CADASIL typically results in reduced survival. Our presentation is unique because it is based on a large multinational sample of symptomatic patients, but we recognize that it suffers from the methodological weaknesses of the studies from which those patients were drawn. Those methodological weaknesses include the incomplete reporting of background characteristics (eg, vascular risk factors other than hypertension), leading to our inability to use an accurate “denominator” for the calculation of the true frequency of those characteristics in patients with CADASIL, as well as the underrecognition of certain subtle clinical manifestations (eg, mild cognitive impairment, mild depression). In addition, many of the studies of CADASIL that have thus far been published have been compelled to rely on clinical and pathological findings for case identification because of the unavailability of testing for Notch3 mutations, the molecular cause of CADASIL,37 potentially resulting in diagnostic errors. Although numerous families with CADASIL have been reported, particularly during the last decade, its prevalence remains unclear, in part because of the challenges of differential diagnosis and case identification. In particular, questions persist regarding the apparent difference in prevalence between Europe and North America. Thus, methods for the epidemiological study of CADASIL are worthy of discussion. Our pooled analysis found that affected individuals typically present with migraine or an ischemic event between the ages of 10 and 58 years, suggesting that the brain imaging studies of such patients should be examined for abnormalities typical of CADASIL when they are performed.7 When such abnormalities are identified, each patient’s pedigree should be reviewed and conventional laboratory studies should be conducted to exclude alternative etiologies. Patients with an autosomal dominant pattern of transmission of clinical features characteristic of CADASIL should undergo a skin biopsy with electron microscopy.8 Whenever possible, DNA should be examined for the determination of Notch3 mutation status,37 but it should be noted that this test is not yet commercially available. Alternatively, linkage analysis could be informative if DNA is available from many family members. Finally, given that our pooled analysis suggested that dementia is an essentially universal end-state for patients with CADASIL, neuropsychological testing should be performed to detect the subtle cognitive disturbances that may be present at the onset of the disease38 and gain information relevant to patient prognosis.39 Although our analysis of the clinical characteristics of a pooled sample of previously published cases is instructive, the formation of a patient cohort by these diagnostic methods would permit more accurate and definitive conclusions to be drawn regarding the prevalence, phenotypic range, and natural history of CADASIL. Open in Viewer Figure 1. Initial symptoms of CADASIL stratified by decade of age of onset. Acknowledgments This study was supported by grant R01-NS26179 from the National Institutes of Health (NIH). Dr Moroney was supported by grant K08-NS02051 from the NIH and the Taub Center for Alzheimer’s Disease Research at Columbia-Presbyterian Medical Center. Dr Lynch was supported by grant K08-NS01966 from the NIH; the Irving Scholar Award; and funding from the National Alliance for Research on Schizophrenia and Depression (NARSAD), the Parkinson’s Disease Foundation, and the Lowenstein Foundation. Dr Chin was supported by grant K08-AG00656 from the NIH. Footnote Presented in part at the 50th Annual Meeting of the American Academy of Neurology, Minneapolis, Minn, May 1, 1998. References 1. Van Bogaert L. Encéphalopathie sous-corticale progressive (Binswanger) à évolution rapide chez deux soeurs. Med Hellen . 1955;24:961–972. Go to Citation Google Scholar 2. Sourander P, Wålinder J. Hereditary multi-infarct dementia: morphological and clinical studies of a new disease. Acta Neuropathol (Berl) . 1977;39:247–254. Crossref PubMed Google Scholar a [...] and Sourander and Wålinder in 1977, b [...] 105 eligible patients in 33 families. 3. Davous P, Fallet-Bianco C. 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Authors of the article cited in the comment will be invited to reply, as appropriate. Comments and feedback on AHA/ASA Scientific Statements and Guidelines should be directed to the AHA/ASA Manuscript Oversight Committee via its Correspondence page. Sign In to Submit a Response to This Article Information & Authors Information Authors Information Published In Stroke Volume 30 • Number 6 • June 1999 Pages: 1230 - 1233 PubMed: 10356105 Copyright Copyright © 1999 by American Heart Association. History Received: 28 January 1999 Revision received: 24 February 1999 Accepted: 22 March 1999 Published online: 1 June 1999 Published in print: June 1999 Permissions Request permissions for this article. Request permissions Keywords cerebral artery diseases dementia genetics migraine stroke Authors Affiliations Expand All David W.Desmond PhD From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author Joan T.Moroney MD, MRCPI From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author Timothy Lynch MD, MRCPI From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author Stephen Chan MD From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author Steven S.Chin MD From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author J. P.Mohr MD From the Departments of Neurology (D.W.D., J.T.M., T.L., J.P.M.), Radiology (S.C.), and Pathology (S.S.C.), Columbia University, College of Physicians and Surgeons, New York, NY. View all articles by this author Notes Correspondence to Dr David W. Desmond, Neurological Institute, 710 W 168th St, New York, NY 10032. E-mail dwd2@columbia.edu Metrics & Citations Metrics Citations 194 Metrics Article Metrics View all metrics Downloads Citations No data available. 18,905 193 Total 6 Months 12 Months Total number of downloads and citations See more details Referenced in 2 patents 109 readers on Mendeley 1 readers on CiteULike Citations Download Citations If you have the appropriate software installed, you can download article citation data to the citation manager of your choice. Select your manager software from the list below and click Download. Please select your download format: [x] Direct Import Amy A. Jolly, Success Anyanwu, Fatemeh Koohi, Robin G. Morris, Hugh S. Markus, Prevalence of Fatigue and Associations With Depression and Cognitive Impairment in Patients With CADASIL, Neurology, 104, 3, (2025). Crossref Nontapat Sukhonpanich, Fatemeh Koohi, Amy A Jolly, Hugh S Markus, Changes in the prognosis of CADASIL over time: a 23-year study in 555 individuals, Journal of Neurology, Neurosurgery & Psychiatry, 96, 7, (690-696), (2024). Crossref Mengru Qi, Mengke Ban, Zhaoyang Lu, Shuwen Cao, Ranran Yang, Ping Zhang, Jinggui Song, Repetitive transcranial magnetic stimulation alleviates depression in a young patient with CADASIL: A case report, Journal of Neurorestoratology, 12, 1, (100093), (2024). Crossref Camille Wicker, Charles-Joris Roux, Louise Goujon, Yvan de Feraudy, Marie Hully, Anais Brassier, Claire-Marine Bérat, Nicole Chemaly, Arnaud Wiedemann, Lena Damaj, Marie-Thérèse Abi-Warde, Dries Dobbelaere, Agathe Roubertie, Aline Cano, Alina Arion, Anna Kaminska, Sabrina Da Costa, Arnaud Bruneel, Sandrine Vuillaumier-Barrot, Nathalie Boddaert, Tiffany Pascreau, Delphine Borgel, Manoelle Kossorotoff, Annie Harroche, P. de Lonlay, Association between acute complications in PMM2-CDG patients and haemostasis anomalies: Data from a multicentric study and suggestions for acute management, Molecular Genetics and Metabolism, 140, 3, (107674), (2023). Crossref Oliwia Szymanowicz, Izabela Korczowska-Łącka, Bartosz Słowikowski, Małgorzata Wiszniewska, Ada Piotrowska, Ulyana Goutor, Paweł Jagodziński, Wojciech Kozubski, Jolanta Dorszewska, Headache and NOTCH3 Gene Variants in Patients with CADASIL, Neurology International, 15, 4, (1238-1252), (2023). Crossref Jing W Goh, Satyaki Kundu, Ragunath Durairajan, Cerebral Autosomal Dominant Arteriopathy With Subcortical Infarcts and Leukoencephalopathy (CADASIL): A Diagnosis to Consider in Atypical Stroke Presentations, Cureus, (2023). Crossref James F. Meschia, Bradford B. Worrall, Fanny M. Elahi, Owen A. Ross, Michael M. Wang, Eric D. Goldstein, Natalia S. Rost, Jennifer J. Majersik, José Gutierrez, Management of Inherited CNS Small Vessel Diseases: The CADASIL Example: A Scientific Statement From the American Heart Association, Stroke, 54, 10, (e452-e464), (2023)./doi/10.1161/STR.0000000000000444 Abstract Kholoud Aljaberi, Amna Ahli, Sudhir Kumar Palat Chirakkara, Ahmed Shatila, A Case Report of Cerebral Autosomal Dominant Arteriopathy With Subcortical Infarcts and Leukoencephalopathy Misdiagnosed as Multiple Sclerosis, Cureus, (2023). Crossref Wenjun Zhang, Xiangjun Zhao, Xuewei Qi, Susan J. Kimber, Nigel M. Hooper, Tao Wang, Induced pluripotent stem cell model revealed impaired neurovascular interaction in genetic small vessel disease Cerebral Autosomal Dominant Arteriopathy with Subcortical Infarcts and Leukoencephalopathy, Frontiers in Cellular Neuroscience, 17, (2023). Crossref Yumi Yamamoto, Yi-Chu Liao, Yi-Chung Lee, Masafumi Ihara, Jay Chol Choi, Update on the Epidemiology, Pathogenesis, and Biomarkers of Cerebral Autosomal Dominant Arteriopathy With Subcortical Infarcts and Leukoencephalopathy, Journal of Clinical Neurology, 19, 1, (12), (2023). Crossref See more Loading... View Options View options PDF and All Supplements Download PDF and All Supplements Download is in progress PDF/EPUB View PDF/EPUB Figures Open all in viewer Figure 1. Initial symptoms of CADASIL stratified by decade of age of onset. Go to FigureOpen in Viewer Tables Media Share Share Share article link Copy Link Copied! Copying failed. Share FacebookLinkedInX (formerly Twitter)emailWeChatBluesky References References 1. Van Bogaert L. Encéphalopathie sous-corticale progressive (Binswanger) à évolution rapide chez deux soeurs. Med Hellen . 1955;24:961–972. Go to Citation Google Scholar 2. Sourander P, Wålinder J. Hereditary multi-infarct dementia: morphological and clinical studies of a new disease. Acta Neuropathol (Berl) . 1977;39:247–254. Crossref PubMed Google Scholar a [...] and Sourander and Wålinder in 1977, b [...] 105 eligible patients in 33 families. 3. Davous P, Fallet-Bianco C. Démence sous-corticale familiale avec leucoencéphalopathie artériopathique: observation clinico-pathologique. Rev Neurol (Paris) . 1991;147:376–384. PubMed Google Scholar a [...] little additional attention until 1991, b [...] 105 eligible patients in 33 families. 4. Tournier-Lasserve E, Iba-Zizen MT, Romero N, Bousser MG. Autosomal dominant syndrome with strokelike episodes and leukoencephalopathy. 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Go to Citation Crossref PubMed Google Scholar Advertisement Recommended October 1999 The Natural History of CADASIL [...] Patrick Davous +0 authors October 2011 CADASIL [...] Philip B. Gorelick +0 authors October 2011 Effects of Gender on the Phenotype of CADASIL Bence Gunda, Dominique Hervé, Ophelia Godin, Martina Bruno, Sonia Reyes, Nassira Alili, Christian Opherk, Eric Jouvent, Marco Düring, Marie-Germaine Bousser, Martin Dichgans, and [...] Hugues Chabriat +8 authors Advertisement Submit a Response to This Article Close Compose eLetter Title: Comment text: Contributors (all fields are required) Remove Contributor First Name: Last Name: Email: Affiliation: Add Another Contributor Statement of Competing Interests Competing Interests? 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190501	https://pollution.sustainability-directory.com/term/radioactive-decay-chains/	Radioactive Decay Chains → Term Skip to main content search Menu Expertise Founders Purpose Services Reuse Question Terms Contact search Close Search Term Radioactive Decay Chains BySustainability Directory28 April 2025No Comments14 min Fundamentals↓Intermediate↓Academic↓ Fundamentals Radioactive decay chains represent the sequential series of transformations that certain unstable atomic nuclei undergo as they emit particles and energy, ultimately leading to a stable nuclear state. This process constitutes the fundamental mechanism by which many heavy radioactive isotopes lose their inherent instability. The initial unstable atom initiating this sequence is designated as the parent nuclide. Each decay event within the chain produces a new nucleus, termed a daughter nuclide. These daughters may themselves be radioactive, continuing the chain until a stable isotope is finally formed. The meaning behind understanding these chains, especially for those new to nuclear concepts or sustainability, lies in recognizing that the presence of one radioactive substance often signifies the future Meaning → Exploring future scenarios and possibilities is the core focus of the Prism sustainability satellite. appearance of others. Consider the definition of a decay chain as a lineage, tracing the radioactive progeny from ancestor to stable descendant. This continuous cascade of nuclear transformations holds significant significance in predicting the suite of radionuclides that will exist within a sample over time. For instance, the ubiquitous Uranium-238 (238 U) undergoes a protracted decay series. Its first decay step produces Thorium-234 (234 Th), which is also radioactive. Thorium-234 subsequently decays, leading to Protactinium-234 (234 Pa), then Uranium-234 (234 U), and the sequence continues through isotopes of Thorium, Radium, Radon, Polonium, Astatine, and Bismuth, before finally concluding with the stable Lead-206 (206 Pb). This long cascade illustrates the complexity inherent in characterizing radiological hazards; addressing the parent nuclide is only the initial step in managing the long-term radioactive inventory. A radioactive decay chain illustrates the stepwise transformation of an unstable atomic nucleus through a series of decays until it reaches a stable form. The Simple Process Explained The explanation of each step is rooted in nuclear physics. An unstable nucleus contains an imbalance of neutrons and protons or possesses excess energy. To attain a more stable configuration, it releases energy Meaning → The Energy domain delves into the transition from fossil fuels to renewable sources and the future of power generation. or particles. The two most common types of decay in these chains are alpha decay and beta decay. Alpha Decay → This involves the emission of an alpha particle, which is essentially a helium nucleus (two protons and two neutrons). This event reduces the atomic number by two and the mass number by four. Such decays are common in the initial steps of heavy element decay chains. Beta Decay → This involves the transformation of a neutron into a proton (or vice versa) within the nucleus, accompanied by the emission of a beta particle (an electron or positron) and an antineutrino or neutrino. Beta decay changes the atomic number by one but leaves the mass number essentially unchanged. Understanding these basic decay modes is fundamental to grasping the substance of how one element transforms into another within a chain. The specific sequence and type of decay for any given nuclide within a chain is predictable, though the exact moment of decay for an individual atom remains a random event. Visualizing a Simple Chain Consider a very short, hypothetical decay chain to solidify this fundamental designation of parent-daughter relationships: Nuclide A (Parent) decays via alpha emission. Nuclide B (Daughter of A, Parent of C) is formed, which is also radioactive. Nuclide B decays via beta emission. Nuclide C (Daughter of B) is formed, which is stable. This simplified model provides a basic clarification of the chain concept. While real-world chains are often longer and may involve branching (where a nuclide can decay via more than one path, though typically one is dominant), this fundamental idea of sequential transformation remains constant. The denotation of each step as a specific decay type (alpha, beta) provides the necessary detail to follow the transformation process. Intermediate Transitioning from the fundamental description of radioactive decay Meaning → Radioactive decay is the process where unstable atomic nuclei spontaneously transform into more stable forms, emitting energy. chains requires a deeper understanding of the temporal aspects governing these processes and their cumulative impact. The interpretation of chain behavior moves beyond simple sequential steps to encompass the rate at which these transformations occur and the resultant mixture of radionuclides present at any given time. This temporal dynamic introduces considerable complexity for environmental safety assessments and waste management strategies. The rate of decay for any given radionuclide is characterized by its half-life, which represents the time required for half of the atomic nuclei in a sample to undergo radioactive decay. Half-lives vary enormously, from fractions of a second to billions of years. Within a decay chain, each nuclide possesses its own characteristic half-life. The interplay between the half-lives of the parent and its daughters dictates the accumulation and depletion rates of the daughter nuclides within the chain. A key concept in understanding decay chain kinetics is secular equilibrium. This condition arises in a decay chain where a very long-lived parent nuclide produces a much shorter-lived daughter nuclide (and subsequent daughters). After a period significantly longer than the daughter’s half-life (but shorter than the parent’s), the activity (decay rate) of the daughter nuclide becomes equal to the activity of the parent. The daughter nuclei are decaying just as quickly as they are being produced from the parent. This state of equilibrium has profound implications for managing long-lived radioactive waste originating from isotopes like Uranium or Thorium; even if the long-lived parent has low initial activity, the build-up of potentially more hazardous, shorter-lived daughters in equilibrium can pose a sustained long-term risk. The interplay of differing half-lives within a chain dictates the changing radiological hazard profile over time, a critical consideration for long-term management. Analyzing Chain Kinetics The elucidation of chain behavior involves applying mathematical models based on differential equations (Bateman equations) to predict the activity of each nuclide in a chain over time, given initial activities and half-lives. This predictive capability is paramount in disciplines such as nuclear engineering, health physics, and environmental science, allowing for the forecasting of dose rates and material inventories over geological timescales. The specification of initial conditions (what isotopes are present, and how much) is therefore a critical first step in any such analysis. Consider the Uranium-238 chain again. While 238 U has a half-life of 4.5 billion years, one of its daughters, Radium-226 (226 Ra), has a half-life of 1600 years, and another, Radon-222 (222 Rn), is a gas with a half-life of only 3.8 days. In uranium ore deposits, secular equilibrium is typically established, meaning the activity of 238 U, 226 Ra, and 222 Rn are approximately equal. This highlights why controlling radon gas is a concern in areas with high natural uranium levels; it’s a short-lived, highly active daughter produced continuously by its longer-lived parents within the chain. The connotation of ‘chain’ here shifts from a simple sequence to a dynamic system where the concentrations and activities of linked species are constantly evolving relative to one another, governed by their intrinsic decay rates. This kinetic perspective provides a more sophisticated sense of the challenge in managing radioactive materials, recognizing that the problem isn’t static but changes its radiological character over time. Intermediate Considerations in Waste Management For industries managing radioactive waste, particularly nuclear power, understanding decay chains is not merely academic; it has direct operational and strategic import. The waste streams contain not only fission products (many of which are single, shorter-lived isotopes) but also actinides and their decay chain daughters. For instance, spent nuclear fuel contains Plutonium isotopes, including 239 Pu, which decays via a chain leading to long-lived 235 U and ultimately stable Lead-207 (207 Pb). Understanding this chain is essential for predicting the long-term hazard of plutonium waste. Selected Isotopes and Half-lives in Major Decay Chains\| Isotope \| Decay Chain \| Half-life \| Primary Decay Mode \| --- --- \| \| Uranium-238 (238 U) \| Uranium Series \| 4.5 x 10 9 years \| Alpha \| \| Radium-226 (226 Ra) \| Uranium Series (daughter of 238 U) \| 1600 years \| Alpha \| \| Radon-222 (222 Rn) \| Uranium Series (daughter of 226 Ra) \| 3.8 days \| Alpha \| \| Thorium-232 (232 Th) \| Thorium Series \| 1.4 x 10 10 years \| Alpha \| \| Plutonium-239 (239 Pu) \| Neptunium Series (parent is 243 Cm, often discussed as starting point for hazard) \| 24,110 years \| Alpha \| The table illustrates the vast range of half-lives encountered. Waste isolation strategies must account for the presence and potential build-up of daughters with different chemical properties and half-lives than their parents. For example, if a waste form is designed to immobilize a parent, but a gaseous or highly soluble daughter accumulates and escapes, the containment strategy could be compromised. This prompts intermediate-level discussions on waste form durability, container integrity, and the geological environment’s role as a barrier, all viewed through the lens of how decay chains evolve within that system over millennia. Academic At the academic and expert level, the study of radioactive decay chains transcends mere kinetic description; it delves into the profound complexities of predicting and managing the long-term environmental and health consequences arising from the continuous generation of novel radionuclides within defined material inventories. The meaning of radioactive decay chains from this vantage point encompasses not only the nuclear transformations themselves but also their intrinsic link to radiochemical speciation, environmental transport, and ultimately, dose assessment over geological timescales. This advanced interpretation recognizes that decay chains represent dynamic radionuclide sources whose characteristics evolve dramatically over time, posing unique challenges for permanent disposal solutions like deep geological repositories. The academic definition of a radioactive decay chain can be stated as a series of sequential radionuclide transformations governed by fundamental quantum mechanical probabilities, linking parent nuclides through a cascade of decay events (primarily alpha, beta, and sometimes spontaneous fission or electron capture) to ultimately form stable isotopes, with the temporal evolution of isotopic abundances within the chain dictated by their respective decay constants and branching ratios, subject to the Bateman equations. The significance at this level lies in the necessity to accurately model this evolution under diverse and uncertain future conditions. The substance of this analysis extends beyond simple decay physics to incorporate how these transforming radionuclides interact with their containment environment. A key, often controversial, aspect at the academic frontier involves the predictability of long-term chain behavior within engineered and natural barrier systems designed for waste isolation. While the decay constants of individual isotopes are well-established, the actual delivery of dose to the biosphere depends on the rate at which radionuclides migrate from the waste package through geological media. This migration is governed by chemical form (speciation), solubility, adsorption onto rock surfaces, and groundwater flow rates → all factors influenced by environmental conditions that may change over hundreds of thousands to millions of years. The implication is that predicting the maximum dose requires coupling intricate decay chain kinetics with complex, uncertain hydrogeological and geochemical transport models. For example, the Neptunium-237 (237 Np) decay chain, stemming from 241 Am and other actinides in spent fuel, is particularly problematic. 237 Np itself has a half-life of 2.14 million years, and its chain includes relatively mobile daughters like Protactinium-233 (233 Pa) and potentially dose-significant isotopes like Thorium-229 (229 Th) and Radium-225 (225 Ra), before reaching stable Bismuth-209 (209 Bi). Understanding the speciation and mobility of each member of this chain under expected repository conditions, and critically, unexpected future conditions, is a substantial research undertaking. Academic scrutiny of decay chains focuses on coupling nuclear transformations with environmental transport processes to predict dose accumulation over immense timescales. Challenges in Long-Term Safety Assessment The academic explication of challenges in managing decay chains in waste highlights the integration of multiple scientific disciplines. Nuclear physicists provide decay data, chemists study speciation and solubility, geochemists investigate rock-water interactions and adsorption, and hydrologists model groundwater movement. Engineering principles dictate waste form and barrier design, while computational scientists develop the integrated assessment models. The delineation of future performance for a repository thus rests on the robustness of data and models across this interdisciplinary landscape. A major area of ongoing research is the interpretation of radionuclide source term evolution within a degraded waste package. As containers corrode and waste forms leach, radionuclides are released. However, the rate and form of release are not constant but depend on the changing inventory dictated by the decay chain. For instance, Americium-241 (241 Am) decays to Neptunium-237 (237 Np). 241 Am is less soluble and less mobile under many conditions than 237 Np. As 241 Am decays, the inventory of the more mobile 237 Np increases, potentially altering the overall source term behavior over hundreds of thousands of years. This dynamic source term requires sophisticated modeling, considering the interplay of decay kinetics and geochemical processes occurring simultaneously. Integrated Modeling Approaches Academic endeavors in this field increasingly rely on integrated modeling platforms that couple nuclear decay calculations with reactive transport simulations. These models aim for a comprehensive specification of radionuclide migration. For example, research detailed in journals like Applied Geochemistry or Nuclear Technology often presents complex simulations predicting the migration of key decay chain members (like 226 Ra from the 238 U chain or 237 Np and its daughters) through specific host rock formations, accounting for varying groundwater chemistry (pH, redox conditions) and mineralogy. The purport of such studies is to provide the technical basis for safety arguments required by regulatory bodies worldwide, such as the U.S. Nuclear Regulatory Commission (NRC) or the International Atomic Energy Agency (IAEA). Key Decay Chain Isotopes and Environmental Relevance\| Isotope \| Chain Origin \| Key Concern in Environment \| Relevant Research Area \| --- --- \| \| Radium-226 (226 Ra) \| Uranium-238 \| Bone seeker, groundwater contaminant \| Geochemical transport, adsorption on minerals \| \| Radon-222 (222 Rn) \| Uranium-238 \| Noble gas, inhalation hazard \| Gas transport in porous media \| \| Protactinium-233 (233 Pa) \| Neptunium-237 \| Variable mobility depending on pH/redox \| Actinide speciation, colloid formation \| \| Thorium-229 (229 Th) \| Neptunium-237 \| Alpha emitter, challenging chemistry \| Thorium geochemistry, complexation \| The table highlights just a few examples illustrating that the environmental behavior is isotope-specific, necessitating detailed investigation of each relevant member of a decay chain. The long-term sense of responsibility in managing these materials prompts continuous academic effort to reduce uncertainties in these predictions, often requiring multi-decadal or even multi-century natural analogue studies examining the behavior of naturally occurring radionuclides in similar geological settings. The essence of the academic challenge lies in bridging the gap between fundamental nuclear physics and the incredibly complex, long-term behavior of radionuclides in the geosphere. Institutions like the Waste Isolation Pilot Plant (WIPP) in the USA, or research programs associated with planned repositories in Finland (Onkalo) or Sweden, represent real-world implementations where these academic principles are tested and applied. The intention behind this rigorous research is to build confidence that disposal systems can safely isolate radioactive materials, including their evolving decay chain inventories, for the periods required for the hazard to diminish to insignificant levels → a timespan far exceeding recorded human history. Furthermore, the academic discussion often touches upon the ethical and societal dimensions of managing hazards that persist for hundreds of thousands or even millions of years. How do we communicate risk across such timescales? What constitutes sufficient safety proof? The import of decay chain knowledge extends into policy debates and public acceptance challenges for nuclear technologies and their legacy waste. The sheer timescale involved imparts a unique emotional weight to this field, demanding not just technical prowess but a recognition of the unprecedented stewardship required. Academic research, therefore, also contributes to methodologies for long-term societal management and knowledge preservation regarding these enduring hazards. Tags: Nuclear Waste ManagementRadioactive Decay KineticsRadionuclide Transport Modeling Discover More Radioactive Isotopes Meaning → Unstable atomic variants emitting energy, used in medicine, industry, and research, posing sustainability challenges. Radioactive Series Meaning → Chains of unstable atomic nuclei transforming step-by-step until stability, releasing radiation. Radioactive Contamination Meaning → Unintended presence of radioactive substances requiring isolation to prevent radiation exposure and environmental harm over potentially vast timescales. Radioactive Decay Series Meaning → Sequential transformations of unstable atomic nuclei into stable ones through particle emission. What Role Does Nuclear Energy Play?Nuclear energy significantly reduces air pollution and carbon emissions, yet waste and safety remain key sustainability challenges. Nuclear Instability Meaning → Nuclear instability is the condition of an atomic nucleus prone to spontaneous transformation, leading to radioactive decay. Sustainability Satellites Dedicated Glossaries Just as satellites orbit the Earth, providing a focused and detailed view of specific regions, our “Sustainability Satellites” offer an in-depth exploration of theSustainability Directory. Each satellite represents a vital aspect of this broader field, and each is presented as a domain. Glossaries illuminate the key concepts, terminology, and challenges within each area, allowing to build a more complete and nuanced understanding of the interconnected world. 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190502	https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:remainder-theorem/v/polynomial-remainder-theorem-example	Remainder theorem: finding remainder from equation (video) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. 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Skip to lesson content Algebra 2 Course: Algebra 2>Unit 4 Lesson 4: Polynomial Remainder Theorem Intro to the Polynomial Remainder Theorem Remainder theorem: finding remainder from equation Remainder theorem examples Remainder theorem Remainder theorem: checking factors Remainder theorem: finding coefficients Remainder theorem and factors Proof of the Polynomial Remainder Theorem Polynomial division: FAQ Math> Algebra 2> Polynomial division> Polynomial Remainder Theorem © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Remainder theorem: finding remainder from equation CCSS.Math: HSA.APR.B, HSA.APR.B.2 Google Classroom Microsoft Teams About About this video Transcript The Polynomial Remainder Theorem simplifies the process of finding the remainder when dividing a polynomial by [x - a]. Instead of long division, you just evaluate the polynomial at [a]. This method saves time and space, making polynomial division more manageable. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted freshsneezes 8 years ago Posted 8 years ago. Direct link to freshsneezes's post “How can a remainder be ne...” more How can a remainder be negative? Answer Button navigates to signup page •1 comment Comment on freshsneezes's post “How can a remainder be ne...” (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “With polynomial division,...” more With polynomial division, you can get remainders that are negative. In numeric long division, you would not have a negative remainder. 2 comments Comment on Kim Seidel's post “With polynomial division,...” (11 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more jffrystwrt 10 years ago Posted 10 years ago. Direct link to jffrystwrt's post “This question was asked a...” more This question was asked a few comments down but I did not find the answer very satisfying. What is the significance of the remainder? In what situation would I need the remainder of a polynomial instead of both its quotient and remainder? Answer Button navigates to signup page •1 comment Comment on jffrystwrt's post “This question was asked a...” (11 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer emma 3 years ago Posted 3 years ago. Direct link to emma's post “when you have a negative ...” more when you have a negative remainder what does that mean? is negative and positive remainder mean the same thing? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Venkata 3 years ago Posted 3 years ago. Direct link to Venkata's post “There's not much of a dif...” more There's not much of a difference. The existence of a remainder essentially implies that if you added that value to the original equation (irrespective of sign), the resulting equation would be perfectly divisible by the divisor. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Dev sapariya 8 years ago Posted 8 years ago. Direct link to Dev sapariya's post “From where 8 come from? I...” more From where 8 come from? In"38-44+20-7". Answer Button navigates to signup page •1 comment Comment on Dev sapariya's post “From where 8 come from? I...” (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 8 years ago Posted 8 years ago. Direct link to Kim Seidel's post “The polynomial has "-3x^3...” more The polynomial has "-3x^3" as its first term. Sal is finding p(2). This means x=2. Thus, the 1st term becomes: -3(2)^3 = -38 because 2^3 = 222 = 8 Comment Button navigates to signup page (7 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more hwhwhw 8 months ago Posted 8 months ago. Direct link to hwhwhw's post “Why does the polynomial r...” more Why does the polynomial remainder theorem work? It almost seems like magic. Answer Button navigates to signup page •1 comment Comment on hwhwhw's post “Why does the polynomial r...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Mohammad a year ago Posted a year ago. Direct link to Mohammad's post “Is it positive 2 cause we...” more Is it positive 2 cause we are subtracting 2 and not -2 cause we are not adding? Answer Button navigates to signup page •Comment Button navigates to signup page (4 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Jayanthika a year ago Posted a year ago. Direct link to Jayanthika's post “It's because the form is ...” more It's because the form is (x-a) so in (x-2), 2 is a. But if it was (x+2), a would be -2. Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Calculus noob🤔 2 years ago Posted 2 years ago. Direct link to Calculus noob🤔's post “Why is there no recent qu...” more Why is there no recent questions? All of the questions are more than 5 months old(as the time of writing this) Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “The most recent question ...” more The most recent question prior to your was 2 months ago. There may not be more questions because people are finding the answers they need in the questions and answers already posted. Comment Button navigates to signup page (8 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Young In Kim 5 years ago Posted 5 years ago. Direct link to Young In Kim's post “Would the resulting remai...” more Would the resulting remainder be just -27 or would it be (-27/x-2)? I'm really confused about this. Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Hecretary Bird 5 years ago Posted 5 years ago. Direct link to Hecretary Bird's post “Think about when you're d...” more Think about when you're dividing normal numbers that don't go into each other easily, like 9 / 4. In this problem, the remainder would only be 1. However, if you were to write the quotient out, it would be 2 + 1/4. It's the same way with polynomials. When talking about the quotient, you have to write the remainder as a fraction, but when just asked to identify the remainder, you can just give the remainder itself. Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Samuel Rodriguez 7 years ago Posted 7 years ago. Direct link to Samuel Rodriguez's post “How can I interpret a neg...” more How can I interpret a negative remainder? Answer Button navigates to signup page •Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer E Z 9 years ago Posted 9 years ago. Direct link to E Z's post “um, what would happen if ...” more um, what would happen if "a" is a positive number? like "x + 2" instead of "x - 2"? Should I then treat "a" like a negative number? Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer JJ Sutton 9 years ago Posted 9 years ago. Direct link to JJ Sutton's post “If the sign is positive, ...” more If the sign is positive, like in "x +2", then a is a negative number. The formula is "x - a". So, if a is positive, the sign just stays negative like in "x - 2". But if a is negative, then it becomes "x - (-2)" which is the same as "x + 2". So, yes. When plugging it in, a is a negative number in the case of "x + 2" 1 comment Comment on JJ Sutton's post “If the sign is positive, ...” (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Video transcript So we have a polynomial here. What I'm curious about is what is the remainder if I were to divide this polynomial by, let's just say, x minus, I want the remainder when I divide this polynomial by x minus two? You could do this. You could figure this out with algebraic long division, but I'll give you a hint. It is much simpler and much less computation intensive and takes much less space on your paper if you use the polynomial remainder theorem. If that's unfamiliar to you, there's other videos that actually cover that. So why don't you have a go at it. All right, so now let's work through this together. The polynomial remainder theorem tells us that when I take a polynomial, p of x, and if I were to divide it by an x minus a, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of a. So in this case, our p of x is this. What is our a? Well our a is going to be positive two. Remember it's x minus a. So let me do this. Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder in this case is going to be equal to p of two, which is equal to, so let's see, it's going to be, I'll just do it all in magenta, negative three times eight minus, let's see, minus four times four plus 20 minus seven. So let's see, this is -24 minus 16 plus 20 minus seven. So that gives us, let's see, -24 minus 16, this is -40. All right, I'm just doing it step by step. This is equal to negative, actually I can do this in my head. All right, here we go. So this is -40 plus 20 is -20 minus seven is -27. That was pretty neat because if we attempted to do this without the polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the quotient and all of that, but we don't need the quotient, we don't need to know. So if we did all the algebraic long division, you know, we would have taken our p of x and then we would have divided the x minus a into it, and we would have gotten a quotient here, q of x, and we would have done all this business down here, all this algebraic long division. Probably wouldn't have even fit on the page. But eventually we would have gotten to a point where we got an expression that has a lower degree than this. It would have to be a constant because this is a 1st degree, so it would have be essentially a zero degree. So we would have eventually gotten to our -27. But this was much, much, much, much easier then having to go through this entire exercise. 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190503	https://simple.wikipedia.org/wiki/Hockey_puck	Hockey puck - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Create account Log in Hockey puck [x] 40 languages Afrikaans العربية Беларуская Български Bosanski Català Čeština Deutsch Eesti Ελληνικά English Español Euskara فارسی Français 한국어 Hrvatski Bahasa Indonesia Italiano עברית Қазақша Latina Latviešu Lëtzebuergesch Nederlands 日本語 Norsk bokmål Norsk nynorsk Олык марий Polski Română Русский Slovenčina Slovenščina Српски / srpski Suomi Svenska Türkçe Українська 中文 Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia Two standard hockey pucks. A hockey puck is a disk used in some games serving the same use as a ball in ball games. The best-known use of pucks is in ice hockey, a major international sport. Hockey pucks are hard disks of rubber that are used to play ice hockey. A standard ice hockey puck is black, 1 inch thick (25.4 mm), 3 inches in diameter (76.2 mm), and weighs between 5.5 and 6 ounces (156-170 g). Pucks are often marked with silkscreened team or league logos on one or both faces. This short article about sports can be made longer. You can help Wikipedia by adding to it. Retrieved from " Category: Ice hockey words Hidden category: Sports stubs This page was last changed on 31 July 2024, at 14:51. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search Hockey puck 40 languagesAdd topic
190504	https://adamdhalla.medium.com/1-3-a-complete-guide-to-gaussian-elimination-dd51f7b4a300	[1/3] A Complete Guide to Gaussian Elimination \| by adam dhalla \| Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Press enter or click to view image in full size [1/3] A Complete Guide to Gaussian Elimination Part 1 of 3: What it means, and how we [humans] compute it, pivots, singular cases, row exchanges, and a geometric understanding using span and linear combinations. adam dhalla Follow 18 min read · Jan 3, 2021 116 3 Listen Share This series can act as a kind of summary of the lectures 1–4 of prof. Gilbert Strang’s class 18.06 on MIT OCW, as well as some extra tidbits of intuition I was able to gather. Continuing on my own personal journey learning linear algebra, I’d like to share, mostly out of my own selfish reasons (writing about concepts does make them stick) intuitions on a fundamental concept in linear algebra: using matrix notation to solve a system of equations. Gaussian Elimination is a starter concept in many first-year Linear Algebra courses. This quick explanation is meant to help anyone in one of those courses, learning about gaussian elimination, and more importantly (we will spend more time on this) factorizing this system into the form A = LU. I’d recommend a solid understanding of basic gaussian elimination (and all concepts below that) although I’ll review it anyways. Contents: This is a comprehensive guide, but if you’re looking for specific tidbits of information, utilize these links: Why Our Classic Way of Solving Linear Systems Doesn’t Work Motivation for Gaussian Elimination The Upper Triangular Gaussian Elimination in a (2 x 2) System Gaussian Elimination in a (3 x 3) System Pivots Temporary breakdown and row exchanges (non-singular) Permanent Breakdown, i.e, the singular case Singular System Case #1: Infinite Solutions Singular System Case #2: No Solutions Geometric View of the Singular Case Why Our Classic Way of Solving Linear Systems Doesn’t Work We’ve been pretty familiar with linear systems of equations since middle school maybe — not later than the sophomore year of high school. There were a few ways to solve these — you could use one equation to solve for some unknown, and then replace that unknown with it’s new value in the other equation. This is called substitution. Press enter or click to view image in full size You could also eliminate entire equations by adding and subtracting the equations themselves. And while gaussian elimination is a little closer to the latter, there’s still a problem with either of these methods. Both of these work in small systems of 2 or 3 equations and unknowns, but when you get to large systems, you need a methodical and reliable formula that works with all systems. When you get to solving large systems, this is done by a computer. It is quite hard to tell a computer to do either elimination or substitution the ways we were taught classically, as the methods differ from equation. This need for a methodical formula is why Gaussian Elimination exists. Motivation for Gaussian Elimination Gaussian Elimination is a way of solving a system of equations in a methodical, predictable fashion using matrices. Let’s look at an example of a system, and solve it using elimination. We don’t need linear algebra to solve this, obviously. Heck, we can solve it at a glance. The answer is quite obviously x= y = 1. But things get exponentially harder with the more unknowns and equations we have. We can store all this information in one matrix and two vectors, a matrix which stores all our coefficients A,a vector of the unknowns xand our answers b. If we multiply this out we get the same system of equations. More importantly, we can actually get rid of the x and y vector during elimination, since it doesn’t contribute any information. We know that the system is 2 x 2, and thus has two unknowns. Thus, we can create an augmented matrix of Aand b. On the right of the dashed line is our coefficient matrix, on the left is our answers. We can read the above as two equations: 1x + 1y = 2, and 2x + 1y = 3 (if you imagine the x and y vector still being there — or, easier, just imagine the corresponding unknown sitting next to each coefficient. We can do elimination in this matrix form in a similar way. What we want to do is turn the coefficient matrix A into U — what we call the upper triangular. The Upper Triangular The upper triangular(U)is a special variant of matrix with only zeros below the diagonal. Upper triangular patterns only occur in square matrices. Below are (2 x 2), (3 x 3) and (4 x 4) upper triangular matrices. Press enter or click to view image in full size The reason why upper triangular matrices are favored when solving systems of equations is that they make them easy to solve. Let’s momentarily imagine a system of equations with three equations and three unknowns. Press enter or click to view image in full size on the left is the Ax + b format, and on the right the augmented style. Both are equivalent. If the coefficient matrix is already in upper triangular form, as it is above, the system becomes very easy to solve. You can solve it from the bottom up. The bottom element simply states what it is. In an upper triangular, this is what your system becomes. We can now solve easily from the bottom up. We can easily calculate what z must be, and we can substitute this value of z in the equation above it to solve for d easily. This is called back-substitution. This might be easier to understand with a concrete example. Let’s view another upper triangular in the format Ax = band augmented. Press enter or click to view image in full size on the left is the Ax + b format, and on the right the augmented style. Both are equivalent. The associated system of equations (you should have it in your head as you see this) looks like this. We can easily figure this out bottom-up. We can see at a glance that z is equal to 3, and we can then substitute this in the equations above (back-substitution)and find out each variable one at a time, making it quite easy. We can find out that y must = 1 and x = 2. Gaussian Elimination in a (2 x 2) System Let us now return to our example with a better idea of where we want to head. We want to turn the coefficient matrix on the left, somehow, into a upper triangular matrix that is easy to solve. Our (2 x 2) example. The way we do this is reminiscent of our classic elimination strategy. To create a (2 x 2) upper triangular matrix, we want the place occupied by our 2 to be occupied by a 0. We don’t care what the numbers are — remember, an upper triangular has only 0’s below the diagonal. That’s the only criteria. We do our first elimination to get this 2 to a 0. We multiply the first row of our coefficient matrix [ 1 1 ] and subtract it from the bottom row [ 2 1 ], choosing the multiplier that cancels out the 2. If it helps, think of this just like old-school elimination — we’re doing the exact same thing. The correct multiplier to multiply the first row is 2.After multiplying our first row by 2,we get [ 2 2 ], which we then subtract from the second row to get our updated matrix. But, you may point out, we didn’t change our right hand side! And you’d be completely right. Just like in classic elimination, we do the same multiply-subtract action on the left hand side. So, we multiply the first number 2 by our multiplier 2, and subtract it from 3, to get -1. Notice our top row doesn’t change when we multiply it and subtract it from the next row, just the row below changes. This is also reminiscent of how we do classic elimination — we use the equation as a tool to simplify the other equations — we don’t alter the original equation itself. And now look! We have a simplified system with a upper triangular as our coefficient matrix. Writing this out again in full form, we get: Where U is the upper triangular,x the unknown matrix,and c, our modified ‘right hand side’ (formerly b). Just to recap. We went from Ax = b to Ux = c by transforming our matrix A (elimination) into the upper triangular U while applying the same transformation to b to keep things consistent on both sides of the equality. So, these two completely equivalent systems. Gaussian Elimination in a (3 x 3) System Let’s get a little more practice with Gaussian Elimination in a less trivial form, in a (3 x 3) matrix. Let’s again use a concrete example. Say we wanted to solve this system. That looks rather intimidating. Note that the blank space in the first row corresponds to the first equation not including z. First, let’s convert it into the equivalent matrix form and augmented matrix Ax = b and A \| b. Press enter or click to view image in full size Now, we can convert our coefficient matrix into the upper triangular in a series of steps similar to the ones carried out on the (2 x 2) matrix. To make our matrix into the upper triangular, we want the three spots in the bottom left corner (currently occupied by 4, 2, and 2) to be 0. Again, we do not care about the numbers on and above the diagonal — they can be anything. Our first step is to make the 4 in the second row, first column a zero in a similar elimination manner. The multiplier we multiply the first row before subtracting is 2, so the 2 in the first row cancels out the 4 in the second row. We then subtract this 2 x [ 2 3 0 ] = [ 4 6 0 ] from the second row. We then do the same calculation on the right-hand side, multiplying 8 by 2 and subtracting it from 9, getting -7. One step closer getting from our A to U! Our next step is to eliminate the 2 in the bottom left hand corner. We will do this by multiplying the first row again, but this time, subtracting it from the third row, choosing a multiplier which will multiply with the 2 from the first row and cancel out the 2 in the third row. This multiplier is only 1. We multiply our first row by 1 x [ 2 3 0 ] = [ 2 3 0 ], which we then subtract from [ 2 2 3 ]. [ 2 2 3 ] - [ 2 3 0 ]= [ 0 -1 3]. And don’t forget to do the same update to the right hand side — taking 8, multiplying it by 1, and subtracting it from 9 to get 1. Notice now that we’ve cleared out all the numbers below the diagonal in the first column. We’re done with that column. Before we move onto clearing that -1 in the second column, I’d like to point out something about the eliminations we’ve been doing. Pivots So far, we’ve based our choice of multiplier on what multiplies with our 2 (first row first column) and eliminates with the numbers below it. It is how we decide our multiplier. This makes it significant, in this column anyways. We call this number the pivot. More specifically, the first pivot, since it is the first of the three pivots in our equation. Press enter or click to view image in full size These are the three pivots in our matrix right now. I stress the importance of that right now part — we don’t really consider pivots until we get to them. You will see how our third pivot, 3, will no longer be 3 once we do another step of elimination. Thus, we don’t really think about pivots until we get to them, since steps of elimination often alter the rows below it, and along with them, the pivots. Get adam dhalla’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe Pivots will come into more play later. For now let’s return to our example. We now have to get rid of the -1 in the third row, second column. (3, 2 position). To do this, we use our second pivot. Since the first column is all zeros and won’t impact our calculations, we can almost think of it as now finding the upper triangular of a 2 x 2 system . Just like before, we look at our second row, and see what number we should multiply it by for the -1 in the second row to cancel out with the -1 in the third row. This multiplier is 1, since when multiplied it’ll make -1 (-1 x 1), and when subtracted from the next row, it will add to 0 (-1- -1 = -1 + 1 = 0). Again, we do the same calculation with the right hand side, multiplying our by -7 by 1 and subtracting it from 1, to get 8 (1- -7 = 1 + 7 = 8). Now, our original matrix A has become the upper triangular U, and our right hand side has gone from b, the original right hand side, to c, the modified. It can now be easily solved, bottom up, using back-substitution. After starting from the bottom with z = 1, and expanding upwards, we get the answer z = 1,y = 2, and x = 2. To better understand Gaussian Elimination we must ask a critical question: When does Gaussian Elimination break down? There are two cases that can happen. Temporary failure and permanent failure. Zeros can never exist in the position of a pivot. When this happens, elimination either breaks down temporarily or permanently. Let’s see how. Temporary breakdown and row exchanges (non-singular) When a zero exists in anywhere but the last pivot, we have a temporary breakdown. i.e, they are non-singular and have a single solution. When zero exists in the pivot place, we cannot eliminate the 2 or 5 in the first column because there is nothing we can multiply 0 by to eliminate the 2 or 5, since 0 x any number = 0. What do we do to solve this? A row exchange. Since the number in the first pivot place is 0, we can switch the first and second rows, so that the zero does not cause us trouble any more. Now, we can multiply our pivot. As an added bonus, we don’t need to eliminate the number in the (2, 1) position since it’s already 0. This same problem can happen in the second column as well. Say we start of with a rather unassuming (3 x 3) matrix like this: After we use elimination to get rid of the first column (the 4) we’re left with a problem. The second pivot has turned into a 0! We can do the exact same thing to fix this, but this time, a row exchange between rows 2 and 1. This fixes the system, and also solves it. Remember, when doing a row exchange in an augmented matrix, we must also exchange the right hand side b. Press enter or click to view image in full size This makes a lot of sense if you think of it in terms of equations. Switching the order of equations around quite obviously requires you to switch the answers around. Taking a look at the equation picture of this will make that quite obvious. Okay, so when is breakdown permanent? Permanent breakdown, i.e, the singular case When we encounter permanent breakdown, it is because we either have not enough information in the system, or conflicting information in the system. The former creates infinite solutions, and the latter creates no solutions. Cutting to the chase: breakdown is permanent when the last pivot is zero. Logically, this makes sense. When you’re at the last row and encounter a zero, there is no more rows below it to switch with, and you can’t switch with a row above (it would ruin the upper triangular form you’re building to). Singular system case #1: Infinite Solutions Let’s look at these in (2 x 2) matrices since it’s quite easy to understand that way. Here is a system that is singular. Notice that when we do elimination, we get: Oh no — the last pivot is a 0, meaning it is singular. Let’s look at what this means as a system of equations. We can clearly see that this system of equations has infinite solutions. Since the second equation evaluates to 0 = 0, which is true no matter what the x and y values are, we’re just left with one equation to describe two unknowns, giving an infinite amount of examples. Looking back at the original system, it’s clear to see why we have an infinite amount of examples. The second equation adds no information to our system. It is simply the first equation multiplied by two. We are left with the same amount of information as if we we only had one equation, and are left with not enough restrictions to come up with a single value of x and y. This occurs in a multitude of situations. In a nutshell, this infinite-solutions-problem occurs when one (or more) of our equations are made up of some combination of other equations in the system. Here are two less obvious examples of this similar type of infinite-answer case. Press enter or click to view image in full size In both of these cases, we end up with two equations to describe 3 unknowns. This is not enough restrictions, and we thus get an infinite amount of solutions. Singular system case #2: No solutions The second type of permanent breakdown, where we end up with no solutions, is quite similar. In essence, it is when the left hand side of our system is some combination of each other, but the right hand side contains conflicting information. Let’s see what this means in a simple 2 x 2 example. This is identical to the system we explored in the infinite solutions case, except that I have changed the 12 in b to an 13. You might already see where this is going. Let’s use gaussian elimination to simplify the system of equations. We end up with this: If we write this out in equation form, we can see that it contains conflicting information. 0 does not equal 1. Thus this equation is untrue for any unknowns x or y. This system has no solutions. This type of no-solution singular case occurs when the left hand side is some combination of each other, as it is here — but the right hand sides do not match up to the same combination. Before, we had infinite solutions because the right hand side matched up and we ended up with 0 = 0. But since we changed the left hand side and changed the right hand side differently, in a sense, we have no solutions instead of an infinite amount. Our two other examples in the infinite case become examples for no solution cases if we just change the right hand sides to make them inconsistent. Press enter or click to view image in full size Really, changing 9 and 10 to any other number that does not = 9 or 10 would result in no solution. But how do we tell if a system is singular without going all the way through elimination? This, like many concepts in linear algebra, requires geometric understanding. Geometric view of the singular case This part requires a good understanding of span and linear combinations. If you’re iffy on those, this video explains it all. As well as that, we must look at the both the row picture but also the column view of a matrix. This involves looking at a matrix no longer as a system of equations, but vectors in space. If you’re unfamiliar with both of those, view, this lecture covers it all superbly. Here’s a super quick one-picture refresher on row picture and column picture. Press enter or click to view image in full size When using the row view to analyze a (2 x 2) system, a singular case can mean two different things. Let’s use the same examples as last time. Press enter or click to view image in full size In higher dimensions, there become more and more ways for the planes to intersect in ways that either produce infinite or no solutions. For example, in 3D, two planes can intersect for a solution, but the 3rd plane misses the intersection, leading there to be no solutions for the system as a whole — or the planes could intersect, but they could all be straight vertical — causing an entire vertical line of infinite solutions. But now let’s look at this in terms of linear combinations and span in the column form of our system to get some more intuition. Press enter or click to view image in full size Notice that the source of the problem in both of these singular cases is that the vectors are colinear, and thus can only describe a single line as a combination: their span is one-dimensional and exists only on the line they share. The 3D geometric counterpart to this would be two of three 3D vectors (a, b, c) and (e, f, g) existing on the same plane, causing the span of the three vectors in total to be only on the plane they share — if the answer, also a 3-dimensional vector, does not exist on the plane (which, probability-wise, is likely), there is no solution. And if the answer does exist on the plane, there are an infinite amount of combinations of the three vectors to make that answer. Back to 2D systems — we’ve identified that the problem is that they have colinear vectors. How do we identify this in advance? It’s a simple test. Take the columns of your matrix A and put them in the form of a linear combination. Is there any way for a combination of vectors to combine to form the zero vector where x≠ y ≠ 0? Press enter or click to view image in full size Why does this work as an appropriate litmus test for if a system contains dependent vectors? Well, if there is a way that we can find to cancel out the vectors using combinations of other vectors in the system, it means that one of the vectors must be made up of some combination of the others. And obviously, we don’t count x = y = 0 since that would work for any linear combination. Here, there is an answer that is nonzero, and that is: In summary, we can solve a system of equations by using using matrix notation to create a coefficient matrix A.We put this in either Ax = bor augmentedform to solve. We convert this matrix A to an easily-solvable upper triangular U using multiplication-subtraction elimination on the rows on both sides of the equality sign (turning binto con the other side of the equality sign). We then use back-substitutionto solve the system bottom-up. That’s all the information you need to know to carry out and understand elimination. There are many more efficient and interesting ways to carry out this elimination in which I will cover in the following two articles. The next article will deal with viewing these transformations, eliminations and row exchanges as matrix multiplications — essentially, finding matrices that we can multiply our A by to carry out elimination in the same way we just did manually. Adam Dhalla is a high school student out of Vancouver, British Columbia, currently in the STEM and business fellowshipTKS. He is fascinated with the outdoor world, and is currently learning about emerging technologies for an environmental purpose. To keep up, Follow his Instagram, and hisLinkedIn. For more, similar content,subscribe to hisnewsletter here. Mathematics Gaussian Elimination Linear Algebra System Of Equations Math 116 116 3 Follow Written by adam dhalla ---------------------- 995 followers ·101 following 17 y old learning about machine learning, as well as a lifelong naturalist. Climate activist in Vancouver. Writer. Visit me @ adamdhalla.com Follow Responses (3) Write a response What are your thoughts? Cancel Respond Acephoton Jan 11 Even o1 for confused in permanent failure, beautifully explained -- Reply Pavan Cheruvu Dec 15, 2024 Nicely explained for newbies to Linear Algebra. Thanks very much for the post. -- Reply Ononiwu Maureen (Chinwendum) Aug 28, 2024 Such a life saver! 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190505	https://books.google.com/books/about/Engineering_Mechanics.html?id=oB7zDwAAQBAJ	Engineering Mechanics: Dynamics - James L. Meriam, L. G. Kraige, J. N. Bolton - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Wiley.com Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» My library My History Engineering Mechanics: Dynamics =============================== James L. Meriam, L. G. Kraige, J. N. Bolton John Wiley & Sons, Jul 28, 2020 - Technology & Engineering - 624 pages Engineering Mechanics: Dynamics provides a solid foundation of mechanics principles and helps students develop their problem-solving skills with an extensive variety of engaging problems related to engineering design. More than 50% of the homework problems are new, and there are also a number of new sample problems. To help students build necessary visualization and problem-solving skills, this product strongly emphasizes drawing free–body diagrams, the most important skill needed to solve mechanics problems. More » Preview this book » Selected pages Table of Contents Index Contents Dynamics of Particles 3 Kinematics of Particles 16 Kinetics of Particles 56 Section B Work and Energy 71 Impulse and Momentum 87 Section A Force Mass and Acceleration 186 Bodies 184 226 APPENDIX B Mass Moments of Inertia 288 Selected Topics of Mathematics 302 Section A Kinematics 227 309 Problems 1 Chapter 4 46 Chapter 5 118 P195 195 FMTOC indd 14 6 Copyright Other editions - View all Engineering Mechanics: Dynamics James L. Meriam,L. G. Kraige,J. N. Bolton Limited preview - 2018 Engineering Mechanics: Dynamics. Volume 2 James L. Meriam,L. Glenn Kraige,Jeff N. Bolton No preview available - 2018 Common terms and phrases a₁angleangular accelerationangular momentumangular velocityappliedaxesaxisballCalculatecenter Gcircularcoefficientcoefficient of restitutioncomponentsconstant speedcoordinatescylinderdampingDetermine the angularDetermine the velocitydirectiondiskdisplacementdynamicsearthequation of motionexpressionfigurefixedforce Ffree-body diagramft/secHELPFUL HINTShorizontalimpactinitialinstantKinematicskinetic energykm/hlinear momentumm/s²magnitudemass centermi/hrmoment of inertiamovingNeglectNewton's second laworbitparticleplane motionposition shownprecessionProbProblems for Articlepulleyrad/srad/s²rad/secradius of gyrationrelativereleased from restrev/minrigid bodyrotationrotorSAMPLE PROBLEMscalarshaftsliderSolutionspacecraftspringsurfacetionU.S. customary unitsv₁vectorvelocity and accelerationverticalvibrationvrelwheelwork-energyz-axiszeroνο About the author(2020) Dr. James L. Meriam has contributed to the field of engineering mechanics as one of the premier engineering educators during the second half of the twentieth century. Dr. Meriam earned his B.E., M. Eng., and Ph.D. degrees from Yale University. He had early industrial experience with Pratt and Whitney Aircraft and the General Electric Company. During the Second World War, he served in the U.S. Coast Guard. He was a member of the faculty of the University of California-Berkeley, Dean of Engineering at Duke University, a faculty member at the California Polytechnic State University, and visiting professor at the University of California-Santa Barbara. He retired in 1990. Professor Meriam always placed great emphasis on teaching, and this trait was recognized by his students wherever he taught. At Berkeley in 1963, he was the first recipient of the Outstanding Faculty Award of Tau Beta Pi, given primarily for excellence in teaching. In 1978, he received the Distinguished Educator Award for Outstanding Service to Engineering Mechanics Education from the American Society for Engineering Education, and in 1992 was the Society's recipient of the Benjamin Garver Lamme Award, which is ASEE's highest annual national award. Dr. L. G. Kraige, coauthor of the Engineering Mechanics series since the early 1980s, has also made significant contributions to mechanics education. Dr. Kraige earned his B.S., M.S., and Ph.D. degrees at the University of Virginia, principally in aerospace engineering, and he currently serves as Professor of Engineering Science and Mechanics at Virginia Polytechnic Institute and State University. In addition to his widely recognized research and publications in the field of spacecraft dynamics. Professor Kraige has devoted his attention to the teaching of mechanics at both introductory and advanced levels. His outstanding teaching has been widely recognized and has earned him teaching awards at the departmental, college, university, state, regional, and national levels. Bibliographic information Title Engineering Mechanics: Dynamics AuthorsJames L. Meriam, L. G. Kraige, J. N. Bolton Edition 9 Publisher John Wiley & Sons, 2020 ISBN 1119724171, 9781119724179 Length 624 pages SubjectsTechnology & Engineering › General Technology & Engineering / General Technology & Engineering / Manufacturing Technology & Engineering / Mechanical Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
190506	https://math.stackexchange.com/questions/4202060/background-on-the-p-adic-valuation	group theory - Background on the $p$-adic valuation - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Background on the p p-adic valuation Ask Question Asked 4 years, 1 month ago Modified4 years, 1 month ago Viewed 657 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. The following is described in Exercises 5.30 & 5.31 in Atiyah & Macdonald's Introduction to Commutative Algebra text: Let A A be a valuation ring of a field K K. The group U U of units of A A is a subgroup of the multiplicative group K∗K∗ of K K. Let Γ=K∗/U Γ=K∗/U. If ξ,η∈Γ ξ,η∈Γ are represented by x,y∈K∗x,y∈K∗, define ξ≥η ξ≥η to mean x y−1∈A x y−1∈A. This defines a total ordering on Γ Γ which is compatible with the group structure (i.e., ξ≥η⇒ξ ω≥η ω ξ≥η⇒ξ ω≥η ω for all ω∈Γ ω∈Γ). In other words, Γ Γ is a totally ordered abelian group. It is called the value group of A A. The canonical homomorphism v:K∗→Γ v:K∗→Γ satisfies v(x+y)≥v(x+y)≥ min(v(x),v(y))(v(x),v(y)) and v(x y)=v(x)+v(y)v(x y)=v(x)+v(y) for all x,y∈K∗x,y∈K∗, and is known as the valuation of K K with values in Γ Γ. I'd like to apply this result to a particular case to show that the resulting valuation coincides with the p p-adic valuation. Below is what I've gathered so far towards this, but I have a few questions. Let (p)(p) be the principal ideal generated by a prime number p p. Then Z(p)Z(p), the integers localized at the prime ideal (p)=p Z(p)=p Z, is a valuation ring of Q Q. The group of units of Z(p)Z(p) is given by U={a b:a,b∈Z,p∤a,p∤b}U={a b:a,b∈Z,p∤a,p∤b}. U U is a subgroup of the multiplicative group of units Q∗Q∗ of Q Q, where Q∗Q∗ consists of all non-zero rational numbers. Forming the quotient group Γ=Q∗/U Γ=Q∗/U, then, we have Γ≅Z Γ≅Z, which is a totally ordered abelian group via the relation c≤d c≤d for c,d∈Z c,d∈Z if and only if d−c∈N d−c∈N. Every non-zero rational number x x (that is, every x∈Q∗)x∈Q∗) can be written uniquely as x=p n e f x=p n e f for integers e e and f f indivisible by p p and an integer n n (that is, x=p n⋅u x=p n⋅u for some u∈U u∈U). The resulting valuation v p:Q∗→Z v p:Q∗→Z, then, is the correspondence x↦n x↦n (where we also define v p(0)=∞v p(0)=∞). This is known as the p-adic valuation or p-adic order, and it satisfies the two properties given in the result above. Here are the questions that I have about this: Why is the group Γ=Q∗/U Γ=Q∗/U above isomorphic to Z Z? What exactly does the isomorphism look like? Why is it true that every non-zero rational number x x can be written uniquely as x=p n e f x=p n e f for integers e e and f f indivisible by p p and an integer n n? Thank you! group-theory commutative-algebra p-adic-number-theory localization valuation-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Jul 19, 2021 at 6:11 michiganbiker898michiganbiker898 1,583 6 6 silver badges 23 23 bronze badges 4 1 Consider the fraction 1/p 1/p. If you were to have 1/p=p n e/f 1/p=p n e/f for some e,f∈Z e,f∈Z, then f=p n+1 e∈p Z f=p n+1 e∈p Z is divisible by p p. Perhaps you mean that only e e is not a multiple of p p?qualcuno –qualcuno 2021-07-19 06:27:46 +00:00 Commented Jul 19, 2021 at 6:27 1 @guidoar notice they are not asserting n≥0 n≥0, one allows n n to be negative as well.TY Mathers –TY Mathers 2021-07-19 06:36:42 +00:00 Commented Jul 19, 2021 at 6:36 1 If you test this by writing 1/p=p−1(1/1)1/p=p−1(1/1) then you see that you are getting p 0⋅1=1 p 0⋅1=1 TY Mathers –TY Mathers 2021-07-19 06:41:47 +00:00 Commented Jul 19, 2021 at 6:41 1 Yeah, got it. Thanks qualcuno –qualcuno 2021-07-19 06:42:50 +00:00 Commented Jul 19, 2021 at 6:42 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Let's answer the second question first. We need to be a bit more precise for the statement to be true: Claim: any x∈Q∗x∈Q∗ can be written uniquely in the form x=p n e f x=p n e f for n∈Z n∈Z and coprime integers e,f e,f such that p∤e p∤e, p∤f p∤f and f∈N f∈N. Proof: For existence write x=a b x=a b for a,b∈Z a,b∈Z. Using prime factorization in Z Z write a=p k e a=p k e and b=p ℓ f b=p ℓ f for k,ℓ∈Z≥0 k,ℓ∈Z≥0 and e,f∈Z e,f∈Z. Then x=p k−ℓ e f x=p k−ℓ e f and if necessary you divide any common factors out of e f e f so that without loss of generality (e,f)=1(e,f)=1. For uniqueness suppose x=p n e f=p n′e′f′x=p n e f=p n′e′f′ for n,n′,e,e′∈Z n,n′,e,e′∈Z and f,f′∈N f,f′∈N with (e,f)=(e′,f′)=1(e,f)=(e′,f′)=1 and p p not dividing e,e′,f,f′e,e′,f,f′. Suppose without loss of generality n≥n′n≥n′; multiplying you get p n−n′e f′=e′f p n−n′e f′=e′f, and then p p does not divide the right side so it cannot divide the left either, and deduce n=n′n=n′. Now you have e f′=e′f e f′=e′f. From here you see e∣e′f e∣e′f, but because (e,f)=1(e,f)=1 you can deduce e∣e′e∣e′. By symmetry you can also deduce e′∣e e′∣e, so e=±e′e=±e′, and because f,f′f,f′ are both positive you can deduce e e and e′e′ must have the same sign so e=e′e=e′. From here you easily get f=f′f=f′ and you are done. Now for the second statement, if x∈Q∗x∈Q∗ you are going to write x=p n u x=p n u for u∈U u∈U and n∈Z n∈Z the uniquely determined integer of our previous claim. You map Q∗→Z Q∗→Z by sending x↦n x↦n. Uniqueness of n n means this is well-defined, and you can check that it is a homomorphism. Finally you notice that U U is exactly the kernel of this map and surjectivity is clear, so you conclude with the first isomorphism theorem. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jul 19, 2021 at 6:36 TY MathersTY Mathers 19.9k 2 2 gold badges 34 34 silver badges 84 84 bronze badges 1 Thanks a lot! It makes sense why the p p-adic valuation described really yields the isomorphism I was looking for in my first question.michiganbiker898 –michiganbiker898 2021-07-19 06:54:38 +00:00 Commented Jul 19, 2021 at 6:54 Add a comment\| You must log in to answer this question. 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190507	https://study.com/skill/learn/differentiating-constant-functions-to-get-0-explanation.html	Differentiating Constant Functions to Get 0 \| Calculus \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Differentiating Constant Functions to Get 0 AP Calculus AB Skills Practice Click for sound 2:22 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 Differentiating… 00:55 Differentiating… 01:47 Differentiating… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Kathryn Boddie Instructors Kathryn Boddie Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison. View bio Example SolutionsPractice Questions Differentiating Constant Functions to Get Zero Step 1: Verify that the function is a constant function. A constant function is of the form f(x)=k where k is a fixed, real number. Step 2: Take the derivative using the constant rule of differentiation: d d x k=0. Differentiating Constant Functions to Get Zero: Vocabulary and Equations Constant Function: A constant function is any function that can be written as f(x)=k where k is a real number. This real number can be an integer, rational number, or irrational number. Constant Rule of Differentiation: The constant rule of differentiation states that d d x k=0, where k is a constant. This follows from the definition of the derivative f′(x)=lim h→0 f(x+h)−f(x)h applied to a constant function f(x)=k: f′(x)=lim h→0 f(x+h)−f(x)h=lim h→0 k−k h=lim h→0 0 h=lim h→0 0=0 We will use these steps, definitions, and equations to differentiate constant functions to get zero in the following two examples. Differentiating Constant Functions to Get Zero: Example Problem 1 Find the derivative of the function f(x)=217. Step 1: Verify that the function is a constant function. A constant function is of the form f(x)=k where k is a fixed, real number. Since 217 is a constant and a real number, this function is a constant function. Step 2: Take the derivative using the constant rule of differentiation: d d x k=0. The constant rule of differentiation says that the derivative of a constant is equal to zero. The derivative of the function is then calculated as follows. f(x)=217 f′(x)=0 Differentiating Constant Functions to Get Zero: Example Problem 2 Find the derivative of the function f(x)=π 12. Step 1: Verify that the function is a constant function. A constant function is of the form f(x)=k where k is a fixed, real number. This function is a constant function, since π 12 is a constant, and raising a constant by a constant retains this property. Step 2: Take the derivative using the constant rule of differentiation: d d x k=0. f(x)=π 12 f′(x)=0 Get access to thousands of practice questions and explanations! 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190508	https://openstax.org/books/calculus-volume-1/pages/chapter-1	Skip to ContentGo to accessibility pageKeyboard shortcuts menu Calculus Volume 1 Chapter 1 Calculus Volume 1Chapter 1 Search for key terms or text. Checkpoint 1.1 and 1.2 Domain = range = 1.3 1.4 The domain is 1.5 1.6 1.7 is odd. 1.8 Domain = range = 1.9 The point-slope form is The slope-intercept form is 1.10 The zeros are The parabola opens upward. 1.11 The domain is the set of real numbers such that The range is the set 1.12 The domain of is The domain of is 1.13 1.15 1.16 Shift the graph to the left 1 unit, reflect about the -axis, then shift down 4 units. 1.17 330° 1.18 1.19 ft 1.20 for 1.21 1.22 To graph the graph of needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function will have a period of and an amplitude of 3. 1.23 No. 1.24 The domain of is The range of is 1.25 1.26 The domain of is The range of is The inverse function is given by the formula 1.27 1.28 1.29 After years, there will be approximately 1.30 1.31 1.32 1.33 The magnitude earthquake is roughly times as severe as the magnitude earthquake. 1.34 1.35 Section 1.1 Exercises 1. a. Domain = range = b. Yes, a function 3. a. Domain = range = b. No, not a function 5. a. Domain = range = b. Yes, a function 7. a. b. 3 c. 13 d. e. f. 9. a. Undefined b. 2 c. d. e f. 11. a. b. c. d. e. f. 13. a. 9 b. 9 c. 9 d. 9 e. 9 f. 9 15. no y-intercept 17. 19. no x-intercept; 21. no intercepts 23. 25. 27. 29. Function; a. Domain: all real numbers, range: b. c. d. and e. and f. Not constant g. y-axis h. Even 31. Function; a. Domain: all real numbers, range: b. c. d. e. None f. Not constant g. Origin h. Odd 33. Function; a. Domain: range: b. c. d. e. Not decreasing f. and g. Origin h. Odd 35. Function; a. Domain: range: b. c. d. Not increasing e. f. g. No Symmetry h. Neither 37. a. all real numbers b. all real numbers c. all real numbers d. 39. a. all real numbers b. all real numbers c. all real numbers d. 41. a. b. 6; c. d. 43. a. all real numbers b. all real numbers 45. a. all real numbers b. all real numbers 47. a. b. 49. a. Yes, because there is only one winner for each year. b. No, because there are three teams that won more than once during the years 2001 to 2012. 51. a. b. a cube of side length 11.8 each has a volume of approximately 1643 cubic units. 53. a. b. i. therefore, the vehicle can travel 300 mi on a full tank of gas. Ii. therefore, the vehicle can travel 225 mi on 3/4 of a tank of gas. c. Domain: range: d. The driver had to stop at least once, given that it takes approximately 39 gal of gas to drive a total of 578 mi. 55. a. b. Exact: approximately 95 cm2 c. d. Exact: approximately 35 cm 57. a. b. $962.50, $1090, $1217.50 c. 77 skateboards Section 1.2 Exercises 59. a. −1 b. Decreasing 61. a. 3/4 b. Increasing 63. a. 4/3 b. Increasing 65. a. 0 b. Horizontal 67. 69. 71. 73. 75. a. b. 77. a. b. 79. a. b. 81. a. b. 83. a. 2 b. c. −5 d. Both ends rise e. Neither 85. a. 2 b. c. −1 d. Both ends rise e. Even 87. a. 3 b. 0, c. 0 d. Left end rises, right end falls e. Odd 89. 91. 93. 95. a. b. 97. a. b. 99. True; 101. False; where is a real-valued constant, is a power function 103. a. b. means that the initial purchase price of the equipment is $20,500; means that in 7.5 years the computer equipment has no value. c. $6835 d. In approximately 6.4 years 105. a. b. $245 c. 167 cupcakes 107. a. b. In 4 years, the value of the car is $20,000. 109. $30,337.50 111. 96% of the total capacity Section 1.3 Exercises 113. 115. 117. 119. 121. 123. 125. 127. 129. a. b. 131. a. b. 133. a. b. 135. a. b. 137. a. b. 139. 141. 143. 145. 155. 157. 159. 161. 163. 165. 167. a. 1 b. c. units to the right 169. a. b. c. No phase shift 171. a. 3 b. c. units to the left 173. Approximately 42 in. 175. a. 0.550 rad/sec b. 0.236 rad/sec c. 0.698 rad/min d. 1.697 rad/min 177. 179. a. π/184; the voltage repeats every π/184 sec b. Approximately 59 periods 181. a. Amplitude = b. c. 14 hours later, or 2 p.m. d. Section 1.4 Exercises 183. Not one-to-one 185. Not one-to-one 187. One-to-one 189. a. b. Domain 191. a. b. Domain: all real numbers, range: all real numbers 193. a. b. Domain: range: 195. 197. 199. These are inverses. 201. These are not inverses. 203. These are inverses. 205. These are inverses. 207. 209. 211. 213. 215. 217. a. b. The inverse function determines the distance from the center of the artery at which blood is flowing with velocity V. c. 0.1 cm; 0.14 cm; 0.17 cm 219. a. $31,250, $66,667, $107,143 b. c. 34 ppb 221. a. b. c. 223. so, the temperature occurs on June 21 and August 15 225. 227. the expression does not equal 2.1 since —in other words, it is not in the restricted domain of since 2.1 is in the restricted domain of Section 1.5 Exercises 229. a. 125 b. 2.24 c. 9.74 231. a. 0.01 b. 10,000 c. 46.42 233. d 235. b 237. e 239. Domain: all real numbers, range: 241. Domain: all real numbers, range: 243. Domain: all real numbers, range: 245. Domain: all real numbers, range: 247. 249. 251. 253. 255. 257. 259. 261. 263. 265. Domain: range: 267. Domain: range: 269. Domain: range: 271. 273. 275. 277. 279. 281. 283. 285. 287. 289. 291. 293. 295. 297. 299. 301. Approximately $131,653 is accumulated in 5 years. 303. i. a. pH = 8 b. Base ii. a. pH = 3 b. Acid iii. a. pH = 4 b. Acid 305. a. million b. 94 years from 2013, or in 2107 307. a. b. hours 309. The San Francisco earthquake was times more intense than the Japanese earthquake. Review Exercises 311. 313. False 315. Domain: range: all real numbers 317. Domain: and range: all real numbers 319. Degree of 3, -intercept: 0, zeros: 0, 321. 323. 325. 327. One-to-one; yes, the function has an inverse; inverse: 329. 331. a. b. 100 shirts 333. The population is less than 20,000 from December 8 through January 23 and more than 140,000 from May 29 through August 2 335. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution-NonCommercial-ShareAlike License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Gilbert Strang, Edwin “Jed” Herman Publisher/website: OpenStax Book title: Calculus Volume 1 Publication date: Mar 30, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 24, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . 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190509	https://insight.jci.org/articles/view/163422	JCI Insight - Vascular changes in the cycling and early pregnant uterus Go to The Journal of Clinical Investigation About Editors Consulting Editors For authors Publication ethics Publication alerts by email Transfers Advertising Job board Contact Physician-Scientist Development Current issue Past issues By specialty Back COVID-19 Cardiology Immunology Metabolism Nephrology Oncology Pulmonology All ... 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Email the journal Go to Top Abstract Introduction Human uterus anatomy and the menstrual cycle Types of blood vessels in the human endometrium Factors regulating the human endometrial vasculature Vascular remodeling in the human menstrual cycle The rodent estrous cycle Types of blood vessels in the rodent uterus Events in early rodent pregnancy Vascular remodeling in rodent pregnancy Eye to the future Conclusions Acknowledgments Footnotes References Version history PDF Metrics See more details Posted by 2 X users 58 readers on Mendeley Article usage Citations to this article (21) Advertisement Review Open Access \| 10.1172/jci.insight.163422 ShareXFacebookLinkedInWeChatBlueskyEmailCopy Link Vascular changes in the cycling and early pregnant uterus Noura Massri,1,2Rachel Loia,3Jennifer L. Sones,4Ripla Arora,2,5 andNataki C. Douglas 6,7 1 Cell and Molecular Biology Graduate Program and 2 Institute for Quantitative Health Science and Engineering, Michigan State University, East Lansing, Michigan, USA. 3 School of Graduate Studies, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. 4 Veterinary Clinical Sciences, School of Veterinary Medicine, Louisiana State University, Baton Rouge, Louisiana, USA. 5 Department of Obstetrics, Gynecology and Reproductive Biology, Michigan State University, East Lansing, Michigan, USA. 6 Department of Obstetrics, Gynecology and Reproductive Health and 7 Center for Immunity and Inflammation, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Authorship note: RA and NCD are co–senior authors. Find articles by Massri, N. in: PubMed \| Google Scholar \| 1 Cell and Molecular Biology Graduate Program and 2 Institute for Quantitative Health Science and Engineering, Michigan State University, East Lansing, Michigan, USA. 3 School of Graduate Studies, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. 4 Veterinary Clinical Sciences, School of Veterinary Medicine, Louisiana State University, Baton Rouge, Louisiana, USA. 5 Department of Obstetrics, Gynecology and Reproductive Biology, Michigan State University, East Lansing, Michigan, USA. 6 Department of Obstetrics, Gynecology and Reproductive Health and 7 Center for Immunity and Inflammation, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Authorship note: RA and NCD are co–senior authors. Find articles by Loia, R. in: PubMed \| Google Scholar 1 Cell and Molecular Biology Graduate Program and 2 Institute for Quantitative Health Science and Engineering, Michigan State University, East Lansing, Michigan, USA. 3 School of Graduate Studies, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. 4 Veterinary Clinical Sciences, School of Veterinary Medicine, Louisiana State University, Baton Rouge, Louisiana, USA. 5 Department of Obstetrics, Gynecology and Reproductive Biology, Michigan State University, East Lansing, Michigan, USA. 6 Department of Obstetrics, Gynecology and Reproductive Health and 7 Center for Immunity and Inflammation, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Authorship note: RA and NCD are co–senior authors. Find articles by Sones, J. in: PubMed \| Google Scholar \| 1 Cell and Molecular Biology Graduate Program and 2 Institute for Quantitative Health Science and Engineering, Michigan State University, East Lansing, Michigan, USA. 3 School of Graduate Studies, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. 4 Veterinary Clinical Sciences, School of Veterinary Medicine, Louisiana State University, Baton Rouge, Louisiana, USA. 5 Department of Obstetrics, Gynecology and Reproductive Biology, Michigan State University, East Lansing, Michigan, USA. 6 Department of Obstetrics, Gynecology and Reproductive Health and 7 Center for Immunity and Inflammation, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Authorship note: RA and NCD are co–senior authors. Find articles by Arora, R. in: PubMed \| Google Scholar \| 1 Cell and Molecular Biology Graduate Program and 2 Institute for Quantitative Health Science and Engineering, Michigan State University, East Lansing, Michigan, USA. 3 School of Graduate Studies, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. 4 Veterinary Clinical Sciences, School of Veterinary Medicine, Louisiana State University, Baton Rouge, Louisiana, USA. 5 Department of Obstetrics, Gynecology and Reproductive Biology, Michigan State University, East Lansing, Michigan, USA. 6 Department of Obstetrics, Gynecology and Reproductive Health and 7 Center for Immunity and Inflammation, Rutgers Biomedical and Health Sciences, Newark, New Jersey, USA. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Authorship note: RA and NCD are co–senior authors. Find articles by Douglas, N. in: PubMed \| Google Scholar \| Authorship note: RA and NCD are co–senior authors. Published June 8, 2023 - More info Published in Volume 8, Issue 11 on June 8, 2023 JCI Insight. 2023;8(11):e163422. © 2023 Massri et al. This work is licensed under the Creative Commons Attribution 4.0 International License. To view a copy of this license, visit Published June 8, 2023 - Version history View PDF Abstract Uterine vascular remodeling is intrinsic to the cycling and early pregnant endometrium. Maternal regulatory factors such as ovarian hormones, VEGF, angiopoietins, Notch, and uterine natural killer cells significantly mediate these vascular changes. In the absence of pregnancy, changes in uterine vessel morphology and function correlate with different stages of the human menstrual cycle. During early pregnancy, vascular remodeling in rodents and humans results in decreased uterine vascular resistance and increased vascular permeability necessary for pregnancy success. Aberrations in these adaptive vascular processes contribute to increased risk of infertility, abnormal fetal growth, and/or preeclampsia. This Review comprehensively summarizes uterine vascular remodeling in the human menstrual cycle, and in the peri- and post-implantation stages in rodent species (mice and rats). Introduction Under physiologic conditions, most vessels in the body are highly stable once formed. Blood vessels in the mammalian uterus are an exception, with characteristic structural and functional changes in both the non-pregnant and pregnant states (1, 2). In humans, in the absence of pregnancy, ovarian-derived estrogen and progesterone mediate cyclic vascular remodeling in the endometrium (1–6). In early pregnancy in rodents and humans, uterine vascular remodeling supports embryo growth and development and subsequent placenta formation (7, 8). Abnormalities in the uterine vasculature increase the risk of infertility, miscarriage, and pregnancy complications such as fetal growth restriction and preeclampsia. Thus, the normal physiologic vascular changes that occur during pre-pregnancy and early pregnancy are likely evolutionarily selected to support pregnancy success. Due to ease of visualization and relationship to the placenta, most documented changes in uterine vasculature have extensively focused on post-placental stages of pregnancy, and a discussion of uterine vasculature prior to and in early stages of pregnancy is missing (7, 9). Here, we describe uterine vasculature anatomy in both humans and rodents and factors that regulate vascular remodeling during the menstrual cycle for humans and the pre-placentation stages of rodent pregnancy. Uterine spiral arterioles undergo clinically significant adaptations during rodent and human pregnancy and have been more extensively investigated than uterine veins; these vessels and their derivatives are the focus of this Review. Human uterus anatomy and the menstrual cycle The human uterus is composed of the myometrium, a smooth muscle layer; the endometrium, an inner mucosal lining surrounding mesenchymal stroma; immune cells; and an extensive vascular network (10). The endometrium is further divided into the deep stratum basalis, adjacent to the myometrium, and the superficial stratum functionalis. The normal human menstrual cycle lasts 21 to 35 days and is divided into the estrogen-dominated proliferative phase, which has marked endometrial layer thickening, and the progesterone-dominated secretory phase (Figure 1A) (10). Ovulation, with oocyte release and corpus luteum formation, marks the beginning of the secretory phase and progesterone-mediated endometrial differentiation in preparation for embryo implantation. In the absence of pregnancy, the endometrium and associated vasculature are shed in response to the demise of the ovarian corpus luteum and withdrawal of ovarian hormones. Thus, with each cycle, the functionalis layer responds to estrogens and progesterone, is shed with menstruation, and then regenerates from the basalis layer (10, 11). Endometrial hypoxia and vasoconstriction of spiral arterioles are thought to limit the amount of blood lost during menstruation (12). Figure 1 Comparison of the human menstrual cycle and rodent estrous cycle. (A) The human menstrual cycle is divided into the proliferative (follicular) phase and the secretory (luteal) phase. Under the influence of ovarian estrogen (blue line) in the follicular phase, the endometrium regenerates, endothelial cells proliferate, and vascular permeability increases. Mid-cycle, a surge in luteinizing hormone results in follicle rupture, ovulation of the oocyte, and formation of the corpus luteum. In the secretory phase, progesterone (red dashed line) regulates stromal cell decidualization and increases endometrial edema in preparation for embryo implantation. In the absence of embryo implantation, the corpus luteum demises, and estrogen and progesterone levels fall. This is followed by menstruation and the start of another menstrual cycle. (B) The rodent estrous cycle is composed of four different phases: proestrus, estrus, metestrus, and diestrus. In the proestrus phase, which lasts between 21 and 32 hours, estrogen increases. The estrus phase and ovulation follow the proestrus phase. The estrus phase lasts 12–48 hours. The formation of the corpus luteum following ovulation marks the shift from an estrogen- to a progesterone-dominated stage. The metestrus and diestrus phases represent the secretory phases of the cycle. The metestrus phase lasts 8–24 hours, while the diestrus phase lasts 48–72 hours. Types of blood vessels in the human endometrium Uterine vasculature comprising both blood and lymphatic vessels has been consistently observed in the human myometrium (13). In contrast, in the endometrium, there are conflicting reports about the location of lymphatic vessels, while blood vessels have been well characterized (13–16). Blood vessels localize to three regions of the human endometrium: the subepithelial zone, functionalis layer, and basalis layer. The subepithelial zone extends to less than 200 μm below the surface epithelium, the basalis layer extends to less than 300 μm above the myometrium, and the functionalis is between (17). Uterine arteries are located within the myometrium and serve as the major blood supply to the uterus, branching into the arcuate and then the radial arteries. At the myometrial-endometrial junction, the radial arteries split to form basal arteries, which supply the basalis layer, and spiral arterioles, which supply the functionalis layer. Arteries in the basalis consist of endothelial cells associated with vascular smooth muscle cells (VSMCs) that express α-smooth muscle actin (α-SMA), γ-SMA, and myosin heavy chain. As the vessels course toward the luminal epithelium, VSMC content is reduced. Capillaries branching off from the spiral arterioles are present throughout the stroma and around the glandular epithelium. Beneath the luminal epithelial lining of the endometrium, capillaries arising from the spiral arterioles contribute to the subepithelial capillary plexus (3, 18). Microvessels at the subepithelial luminal surface are devoid of mural cells, including VSMCs and pericytes, and consist only of endothelial cells. Factors regulating the human endometrial vasculature Cyclic changes in the endometrial vasculature in response to ovarian steroid hormones are associated with angiogenic factor activation. Steroid hormones Changes in the endometrial vasculature during the menstrual cycle are thought to be directly regulated by estrogens and progesterone. Localized expression of estrogen and progesterone receptors in the functionalis layer has been described in endometrium from hysterectomy specimens. Estrogen receptor 1 (ESR1) and ESR2, previously described as estrogen receptors (ERs) α and β, respectively, and progesterone receptor (PR) are expressed in VSMCs (19). ESR1 is the predominant ESR during the late follicular phase, and ESR2 is predominant in the late secretory phase, suggesting hormonal regulation of their expression (19, 20). Whereas both ESR and PR are expressed in isolated human endometrial endothelial cells (21), endothelial cell expression of ESR2 has been detected in both proliferative- and secretory-phase endometrium, whereas PR expression has been variably detected (20–22). Thus, vascular remodeling could be mediated by acutely hormone-sensitive hormone receptor–expressing endothelial cells and VSMCs. Angiogenic factors The endometrial expression of angiogenic factors, VEGF, angiopoietins (ANGPTs), FGF, and their receptors implicates involvement in regulating vascular changes in the proliferative and secretory phases (21, 23). VEGF is known to promote de novo blood vessel growth, angiogenesis, and vascular permeability via tyrosine kinase family receptors, including VEGFR1 (FLT1), VEGFR2 (KDR/FLK1), and VEGFR3 (FLT4) (24). In both the proliferative and secretory phases, VEGFA is expressed in neutrophils associated with the subepithelial capillary plexus and in microvessels in the functionalis layer (25). In the proliferative phase, VEGFA expression also localizes to glandular epithelial and stromal cells (23, 26), and VEGFR2 is found in endothelial cells (23). In the secretory phase, uterine natural killer (uNK) cells express high levels of the angiogenic factors VEGFC, placental growth factor (PGF), and VEGFR3. VEGFA expression is higher in the epithelium than in the stroma. In the secretory phase, endothelial VEGFR2 levels decline, whereas VEGFR1 levels increase (23, 26). In proliferative and secretory endometrium, ANGPT1 is predominantly expressed in stromal cells, while ANGPT2 and the angiopoietin receptor TEK receptor tyrosine kinase (TIE2) are highly expressed in endothelial cells (27). FGF1 and FGF2 are expressed in the epithelium and stroma, with higher expression in the epithelium throughout the cycle (23). Isolated human endometrial endothelial cells proliferate in response to VEGF or FGF2, and addition of 17β-estradiol increases, while progesterone decreases, endothelial cell proliferation in response to these angiogenic growth factors (21). Vascular remodeling in the human menstrual cycle Mechanisms of angiogenesis Angiogenesis can occur via three mechanisms: sprouting, elongation, or intussusception, each requiring unique changes in endothelial cells (Table 1). Endothelial cell expression of α v β 3 integrin, a marker of vascular sprouts in non-endometrial tissues (28), has been detected only inside existing vessels and not in any sprout-like vascular structures. This observation suggests that elongation and intussusception and not sprouting angiogenesis are the primary mechanisms of angiogenesis in the human endometrium (20). The following sections highlight the roles that VEGF signaling might play in the cyclic remodeling of the human endometrial vasculature. Table 1 Definitions of terms related to uterine vasculature Angiogenesis during proliferative phase In the proliferative phase, estradiol, the primary ovarian estrogen, promotes rapid endometrial growth and angiogenesis. Stereologic examination of vessel length and branch points showed that vascular length density in the subepithelial capillary plexus is greatest during the mid to late proliferative phase. These observations are consistent with vessel elongation as the major mechanism of endometrial angiogenesis in the estradiol-dominated proliferative phase (17). Fine spatial analysis revealed a greater percentage of proliferating endothelial cells in the basalis layer during the proliferative phase compared with the secretory phase. In contrast, the proportion of proliferating vessels in both the subepithelial capillary plexus and functionalis layer was relatively constant across the menstrual cycle (25). Estradiol increases endothelial cell proliferative response to VEGF and promotes elongation angiogenesis and vascular permeability (20). A significantly greater percentage of VEGF-expressing vessels and more proliferating vessels with VEGF-expressing cells were observed in proliferative compared with the secretory endometrium for all endothelial layers evaluated (subepithelial capillary plexus, functionalis, and basalis). VEGF displayed higher expression in the subepithelial capillary plexus compared with the functionalis and basalis layers. Finally, the observation of focal VEGF in marginating and adherent neutrophils associated with microvessels implicates neutrophils as the primary intravascular VEGF source for microvessels undergoing angiogenesis (25). Secretory-phase angiogenesis Vascular development in the progesterone-dominated secretory phase is thought to occur via intussusceptive angiogenesis (3). Structurally, spiral arterioles increase in size and tortuosity, and the subepithelial capillaries dilate during the secretory phase (29). Microvessel concentration is thought to significantly increase in the mid-secretory phase (29). After ovulation, VEGFR2 levels decline, whereas VEGFR1 levels increase. While VEGFR2 signaling in the proliferative phase promotes endothelial cell proliferation, VEGFR1 signaling is thought to increase endothelial cell migration in the secretory phase (23). Unlike endothelial cells, mural cells proliferate under the control of progesterone during the secretory phase and wrap around endothelial cells, regulating angiogenesis and coordinating the function of blood vessels. In the late secretory phase, perivascular mural cells form a thick layer and surround the spiral arterioles (26). uNK cells increase in number during the mid-secretory phase and are located close to the endometrial glands and spiral arterioles, suggesting a role in the vascular remodeling that occurs in this phase (30). Vascular breakdown during menses At the end of the secretory phase, right before menstruation, there is a rapid regression of the endometrium that induces increased coiling of spiral arterioles. Menstruation is associated with changes in the endometrial balance between matrix metalloproteinase (MMP) activators and tissue inhibitors of MMPs (TIMPs). Progesterone withdrawal seems to stimulate MMP activity, resulting in extracellular matrix (ECM) breakdown and bleeding. The transition from menses to the proliferative phase initiates with estradiol-independent repair of vascular beds during menstruation (29). The rodent estrous cycle Like the human menstrual cycle, the rodent estrous cycle is orchestrated by ovarian-derived hormones that prepare the uterus for embryo implantation and pregnancy. In rodents, cyclic uterine remodeling occurs over 4–5 days (Figure 1B). Remodeling includes cell shape changes in the luminal and glandular epithelium and stromal cells, changes in uterine gland secretions, and changes in uterine lumen size and shape based on the volume of uterine luminal fluid (31). Although mice do not menstruate, there are mouse models of menstruation where endometrial shedding is induced after decidualization by withdrawal of progesterone (32). Despite these models, while a lot is known about the vascular remodeling events during the human menstrual cycle, little is known about vascular changes in the rodent estrous cycle. Types of blood vessels in the rodent uterus Like in humans, in rodents, both blood and lymphatic vessels have been consistently described in the uterine myometrium, while the endometrium and decidua of pregnancy appear to be devoid of lymphatic vessels (33). The rodent uterus is radial but asymmetric with large uterine arteries present along the mesometrial pole (Figure 2A). The uterine arteries, formed from the anastomoses of the uterine and ovarian arteries, are referred to as utero-ovarian or arcuate arteries. These arteries supply blood to segmental arteries in the mesometrium that branch into circumferential arterioles that pass through the uterine smooth muscle layers and wrap around the uterus. Circumferential arterioles further branch into spiral arterioles that supply blood to the inner muscle layers, the endometrium, and subepithelial capillaries present in proximity to the luminal epithelium (Figure 2A) (4). Figure 2 Vascular remodeling from pre-implantation through decidualization. (A and B) Longitudinal cross sections of the GD3.5 (A) and the GD4.5 (B) mouse uterus illustrating arteries (red), veins (blue), and the uterine lumen (outlined in gray). The utero-ovarian artery (UA) and vein (UV) give rise to the segmental arteries and veins, respectively (SA and SV). The segmental arteries divide into circumferential arterioles (CA) that then branch to form the spiral arterioles (SpA). Subepithelial capillary plexus (SCP) arise from the spiral arterioles and supply blood to the endometrium. Following embryo attachment at GD4.5 (B), the vessels remodel and dilate in proximity to the implantation region. (C and D) Transverse cross sections at GD5.5 (C) and GD7.5 (D). At GD5.5 (C) the newly formed decidual capillaries are readily apparent. By GD7.5 (D), the vasculature surrounding the implantation chamber can be divided into different regions — the central region surrounding the embryo (CTR), the mesometrial region (MR), and the anti–mesometrial region (AMR) — each with a unique composition of vessels. The MR is composed of spiral arterioles and capillaries. The spiral arterioles contain endothelial cells and mural cells (pericytes and VSMCs). The decidual capillaries in the AMR are smaller vessels composed of endothelial cells associated with pericytes, while the decidual capillaries in the CTR contain very few pericytes. AM, anti-mesometrial pole; Cx, cervix; E, embryo; M, mesometrial pole; Myo, myometrium; O, ovary; UF, uterine fold. Events in early rodent pregnancy Mating events in rodents are easily documented by detection of a copulation plug, termed gestational day (GD) 0.5. After mating, fertilization takes place in the oviduct (the rodent fallopian tube) followed by a series of cell divisions to form the blastocyst-stage embryo. In the mouse, multiple blastocysts enter the uterus early on GD3.0, followed by embryo spacing and embryo attachment on GD4.0 (34). Embryos attach to the uterine lumen, forming an implantation chamber (35), followed by a rise in vessel permeability around the site of attachment at GD4.5 (36). The implantation chamber then invades into the underlying stroma as the luminal epithelium around the embryo undergoes cell death and the stromal cells undergo a mesenchymal-to-epithelial differentiation process termed decidualization (36, 37). Between GD5.5 and GD7.5, the embryo continues to develop in the decidua, and embryonic umbilical vasculature fuses with maternal circulation by GD8.5 to form the placenta by GD9.5 (36, 37). Rat gestation is about 2 days longer than that of the mouse, and all events described above are shifted accordingly (38). Limited information is available on early human pregnancy owing to lack of noninvasive methodologies to study the uterine microenvironment. Therefore, rodent models have been extensively used to examine endometrial changes required for successful embryo implantation and pregnancy progression. In the following sections, we describe several such rodent models and highlight human studies where applicable. Tables 2, 3, and 4 summarize additional factors not discussed in the text. Table 2 GD0.5–GD4.5: factors that regulate angiogenesis and vascular remodeling Table 3 GD5.0–GD8.5: factors that regulate angiogenesis and vascular remodeling Table 4 GD9.0–GD10.5: factors that regulate angiogenesis and vascular remodeling Vascular remodeling in rodent pregnancy Vascular modifications in rodents during peri-implantation (GD0.5–GD4.5) In rat pregnancy, endothelial cell proliferation is increased in the entire endometrium in the first few days following ovulation (39, 40). In rats and mice, as embryos enter the uterus, there is a substantial increase in the diameter of subepithelial capillaries in the endometrium (Figure 2B) (5, 41). Embryos attach to the anti-mesometrial pole and subepithelial capillaries in proximity to the implantation region dilate, leading to increased vessel permeability following embryo attachment (36, 42, 43). It is thought that the loss of vascular endothelial cadherin (VE-cadherin) facilitates this increase in permeability (44). In rodents, intravenous blue dye injection prior to euthanasia aids in the identification of embryo implantation sites owing to vascular leakage through permeable vessels (36, 43). Increased vascular permeability at implantation sites across several mammalian species, including sheep and pigs, underscores the importance of this vascular change for successful implantation and later pregnancy events (45–47). Factors contributing to increased vascular permeability (GD0.5–GD4.5) During pre-implantation stages, the factors that contribute to increased endothelial cell proliferation and the exact timing of this process are not well described (39, 40). In contrast, increased vascular permeability around implantation sites has been heavily studied, and several factors have been implicated in this process (Table 2), as discussed below. Steroid hormones. In rodents, ovulatory estrogen induces epithelial proliferation at GD1–GD2, and a small rise in estrogen levels along with high levels of progesterone supports implantation by GD4.0 (48, 49). Experiments conducted in rodents to assess estrogen effects on the vasculature are typically performed with high estrogen levels, which mimic pre-ovulation conditions. Treatment of ovariectomized rats with estradiol or the estrogen agonist tamoxifen induced VEGF transcription (24, 50). Similarly, 17β-estradiol treatment of ovariectomized mice induced VEGF and VEGFR2 expression in the stroma within 6 hours of treatment (51). Further, a single injection of high-dose 17β-estradiol stimulated VEGF expression in the stroma within 2 hours after treatment (52). These studies imply a role for ovulatory estrogen in promoting expression of VEGF family genes. Like estrogen, progesterone alone induced stromal VEGFR2 expression in ovariectomized mice, although expression occurred within 12 hours of treatment. Thus, increase in serum progesterone levels may mediate the changes in uterine VEGF signaling that accompany embryo implantation (24, 51, 53, 54). VEGF signaling. The VEGF receptors, VEGFR1, VEGFR2, VEGFR3, and neuropilin 1 (NRP1) are expressed in the peri-implantation mouse uterus (24, 55), and endothelial VEGF signaling in the peri-implantation period is required for embryonic growth and pregnancy progression. VEGF164, the predominant VEGF isoform in the mouse uterus, is expressed in epithelial cells from GD0 to GD3 and in subepithelial stroma on GD2 and GD3. Following embryo implantation, VEGF164 is localized to the luminal epithelium and stromal cells surrounding the blastocyst (24). Interestingly, VEGF expression was elevated around embryo sites in early pregnancy but not during pseudopregnancy, suggesting that the presence of an embryo is necessary to induce expression between GD4.5 and GD6.5 (56). Systemic administration of a VEGF-blocking antibody in early rat pregnancy (GD2) lowered vascular permeability, reducing the number of implantation sites at GD4 without significant effect on microvascular density or endothelial cell proliferation (57). Thus, VEGF signaling plays a crucial role in pre-implantation vascular remodeling events associated with embryo attachment (57). Prostaglandins. Prostaglandin (PG) synthesis is necessary for the increased vascular permeability and uterine angiogenesis associated with early pregnancy. Prostaglandin synthase 2–derived (PTGS2-derived) PGs (PGE and PGI2) are elevated in implantation sites of several mammalian species (58–63). In rodents, depletion of PG function during pre-implantation is associated with abnormal post-implantation events, including implantation failure, decreased vascular permeability, and/or impaired decidualization, resulting in embryo growth restriction between GD5 and GD7 (45–47, 60, 62, 64–66). Ptgs2–/– mice also show defects in implantation and vascular permeability. However, daily intraperitoneal administration of the stable PGI2 analog carbaprostacyclin to Ptgs2–/– mice starting at GD3.75 improves implantation and restores VEGF and VEGFR2 expression, along with blood vessel numbers (61, 62, 65), suggesting that PG-depletion phenotypes are partly associated with disrupted VEGF signaling. Despite the association of both PG and VEGF signaling with increased vascular permeability in early pregnancy, the impact of PGs and VEGFs on vessel structure around the site of embryo attachment is not understood. Vascular modifications in rodents during decidualization (GD5.0–GD8.5) From GD4.5 to GD7.5, decidual angiogenesis creates a rich, new capillary network (Figure 2C) (6). At GD5.5, the decidual transcriptome shows upregulation of endothelial cell–associated genes (67), with peak endothelial cell proliferation at GD6.5 (68). Thus, it is not surprising that differences in the decidual vasculature are readily apparent by GD7.5. The mesometrial region (MR) contains spiral arterioles covered with VSMCs and pericytes. The central region (CTR) contains capillaries with little or no pericyte coverage, while the capillaries in the anti-MR (AMR) are closely associated with pericytes (Figure 2D) (6, 69). Comparison of the GD7.5 vasculature to the estrous-stage non-pregnant vasculature revealed a 14.9-fold increase in the vascular sprouts in the CTR and a 7.4-fold increase in the vascular sprouts in the AMR. During early pregnancy, the intussusceptive blood vessels also increase 2.8-fold and 2.4-fold in the CTR and AMR, respectively. Thus, both sprouting and intussusceptive angiogenesis contribute to formation and remodeling of decidual vessels in the mouse. Factors contributing to increased vascular permeability and endothelial cell proliferation (GD5.0–GD8.5) Post-implantation decidual angiogenesis and vascular remodeling, which includes the formation of new vascular networks, are regulated by key angiogenic signaling pathways (Table 3), including VEGF, ANGPT, and Notch (6, 67, 70, 71), and immune cells (72, 73), such as uNK cells (6, 74, 75), as discussed below. Steroid hormones. Similar to the pre-implantation/implantation stages, both estrogen and progesterone influence angiogenic factor expression in the post-implantation uterus. De novo synthesis of aromatase-derived estrogen from the stromal cells promotes angiogenesis during decidualization. Treatment with the aromatase inhibitor letrozole reduced expression of the endothelial cell–specific marker CD31 in the stromal compartment. Further, letrozole promoted downregulation of angiogenesis-related factors, such as endothelial PAS domain protein 1 (EPAS1), ANGPT2, and ANGPT4. These observations indicate that local estrogen production from the stromal cells induces secretion of angiogenic factors that contribute to endothelial cell network formation during decidualization (76). Post-implantation treatment with a PR antagonist, RU486, decreased VEGF (6) and Angpt2 expression at GD6.5 and GD8.5 and impaired uterine blood vessel remodeling at GD8.5 (70, 77). Spatiotemporal colocalization of PR and VEGFA was observed in decidual stromal cells at GD4.5. Further, while progesterone treatment at GD4.5 and GD5.5 increased blood vessel density in the CTR and AMR and increased vascular sinus fold numbers at GD6.5 in control mice, this progesterone-induced increase was not observed in VEGFA-deficient mice. Thus, progesterone-mediated changes in vascular remodeling primarily occur through VEGF and ANGPT2 signaling (6, 70). Angiogenic factors. After implantation, Vegf mRNA is expressed in decidual cells in the MR and AMR of implantation sites from GD5 to GD7 (24, 55). Angpt2, Angpt3, and Tie2 are expressed in the endometrium starting after implantation at GD5 and reach maximum expression after decidualization at GD8.5 (61, 70). Another ANGPT family member, ANGPT4, is expressed in endometrial fibroblasts and endothelial cell populations during decidualization (78). VEGF signaling via VEGFR1, VEGFR2, and VEGFR3 contributes to post-implantation vascular remodeling events (24, 55, 56, 68). Intraperitoneal administration of a VEGFR2-blocking antibody at GD3.75 reduced uterine vasculature at GD5.5, with subsequent embryo loss by GD10.5. VEGFR2 blockade in a pseudopregnant mouse model with oil stimulus also decreased uterine vasculature at GD6.5 (68), suggesting that a physical stimulus is sufficient to induce VEGF-dependent vascular changes. In contrast, although pre-conception and early pregnancy administration of VEGFR1- or VEGFR3-blocking antibodies reduced vascular density at GD5.5, embryo growth and pregnancy outcomes were not compromised. The phenotypes observed with VEGFA signaling blockade and conditional loss of one copy of Vegfa suggest that VEGF/VEGFR2 signaling promotes an increase in vascular density, enlargement of vascular sinus folds, and endothelial cell proliferation in the CTR and MR on GD4.5–GD8.5 (6). Together, these data suggest that VEGF/VEGFR2 is the only decidual VEGF signaling pathway essential for progression of pregnancy prior to placenta formation (6, 68, 79). Studies in mice and women implicate aberrant VEGF expression in the decidua as the cause of later-stage pregnancy complications. The pregnant blood pressure high/5 (BPH/5) mouse is a superimposed preeclampsia model that exhibits placental defects, including shallow trophoblast invasion, inadequate spiral artery remodeling, and compromised fetal growth (80, 81). The decreased decidual vasculature observed in BPH/5 pregnancies at GD7.5 is associated with altered VEGF164, Vegfr1, Vegfr2, and Plgf expression (80). In the decidua of women with a history of recurrent pregnancy loss (RPL), VEGF mRNA and protein levels were significantly decreased in comparison with those without a history of RPL, who were undergoing voluntary termination of pregnancy (82). This observation suggests a link between disrupted VEGF signaling in early pregnancy and increased risk of miscarriage. Together, both mouse and human studies highlight the importance of VEGF signaling in decidual angiogenesis, placentation, and pregnancy success. Notch signaling. Like VEGF proteins, the distinct expression pattern of Notch proteins and ligands in the peri-implantation uterus reflects unique roles for Notch signaling in decidual angiogenesis and vascular remodeling (33, 69). Blockade of the Notch ligand delta-like ligand 4 (DLL4) at GD4.5 altered vascular tip and stalk cell identity, leading to excessive decidual angiogenesis and compromised embryonic growth by GD9.5 (71). Endothelial-specific deletion of the Notch regulator ADAM10 resulted in abnormal vascular patterning, impaired decidualization, and pregnancy loss by GD6.5 (67). In this model, there was increased diameter of blood vessels, a characteristic honeycomb structure of vessels near the implantation site, and increased endothelial cells in the AMR at GD5.5. In contrast, endothelial cell–specific loss of the Notch ligand Jagged1 at GD4.5 resulted in aberrant Dll4 expression and increased Notch signaling without a change in the decidual vasculature or interruption of pregnancy at GD7.5 (69). These studies suggest a role for Notch signaling in decidual angiogenesis. uNK cells. uNK cells are immune cells that contribute to uterine vessel remodeling (83, 84). uNK cell precursors are present in the blood vessels and the stromal compartment in the pre-implantation mouse uterus (GD2–GD3). After implantation, uNK cells are abundant in the decidua and undergo notable maturation and proliferation from GD5.5 to GD7.5 (85). In addition, uNK cells produce soluble factors, including VEGF, PGF, and Notch ligand DLL1, that contribute to decidual angiogenesis and vascular remodeling during early pregnancy (73, 74, 85, 86). Between GD4.5 and GD6.5, uNK cells and VEGF/VEGFR3 signaling coordinately regulate enlargement and elongation of uterine vascular sinus folds in the MR (6). Even though uNK cells do not express estrogen and progesterone receptors in both mice (87–89) and humans (90), they are regulated by ovarian hormone signaling (72, 91). This is likely through steroid hormone regulation of stromal cells that surround the uNK cells (90). Consequently, ovarian stimulation (superovulation) with pregnant mare serum gonadotropin and human chorionic gonadotropin results in decreased uNK density near the MR of the decidua, deficient cytokine secretion, compromised placental angiogenesis, and low embryonic weight at GD8.5 (92). Antibody-mediated depletion of uNK cells at GD6.5 and GD7.5 resulted in reduced blood vessel density and vascular sinus folding in the decidua CTR at GD8.5 (6, 75). There are conflicting data regarding the vascular phenotypes in uNK cell–deficient mouse models. IL-15 is a key cytokine for uNK cell maturation and is expressed in the decidual stromal cells (85). Il15−/− mice lack uNK cells and display expression of angiogenesis genes without abnormalities in vascular density or endothelial cell proliferation at GD7.5 (93). Conversely, a uNK cell–deficient model created by depletion of immune cells (Rag2–/– Il2rg–/–) exhibits delayed mesometrial angiogenesis and maturation of decidual vessels at GD6.5, which compromises fetal development at GD9.5 (75). This study suggests involvement of uNK cells in promoting vascular branching and vessel pruning following implantation (75). However, in addition to uNK cell depletion, Rag2–/– Il2rg–/– mice lack B and T cells (alymphoid). This may explain the differences in the two uNK cell–deficient models. In humans, uNK cells are present in the endometrium and decidua of early pregnancy. These uNK cells express VEGFA and VEGFC, suggesting a role in decidual angiogenesis (26, 84). In vitro studies revealed that human uNK cells produce angiogenic factors (94) and that estrogen promotes uNK cell motility and uNK cell–mediated angiogenesis (95). After exposure to asoprisnil, a PR modulator, uNK cell numbers and IL-15 expression were significantly reduced in human secretory-phase endometrium. Impaired spiral artery remodeling, characterized by increased VMSC content and thickened arteriole walls, was also observed (96, 97). Increased numbers of uNK cells as well as increased vascular density and vascular wall thickness were observed in RPL compared with normal pregnancy (98). Gap junction protein connexin 43. Connexin 43 (Cx43), also known as Gap junction alpha-1 protein (Gjα1), is expressed in decidual stromal cells on GD4 and GD6. Mice with a dominant-negative loss of Cx43 function exhibited an increased and irregular pattern of decidual sinusoids at the mesometrial pole, decreased uNK cells at GD6.5 and GD7.5, upregulation of angiogenic factors such as VEGFA and FGF2, and fetal growth restriction (99). Global Cx43 loss results in postnatal death, preventing any analysis of fertility in the adult mouse (100). However, conditional deletion of Cx43 in uterine stromal cells, epithelium, and the circular muscle via PR-Cre (PR Cre) reduced expression of ANGPT2 and ANGPT4 at GD6 and GD7. Uterine-specific Cx43-deficient mice also displayed reduced endothelial proliferation and a rudimentary vascular network on GD6 and GD7, resulting in poor decidualization and embryo development (101). These studies reveal the importance of intercellular communication via gap junctions in establishing the uterine vascular network that is necessary for decidual angiogenesis, embryo growth, and pregnancy success (82, 102, 103). Vascular remodeling in rodents during placentation (GD8.5–GD10.5) Uterine vascular remodeling during placenta formation is mediated by decidual stromal cells, local uterine immune cells, including uNK cells, and embryo-derived trophoblast cells. When trophoblast cells in the trophectoderm layer of the blastocyst contact the endometrium, the mural trophectoderm differentiates into invasive trophoblast giant cells (TGCs), and the polar trophectoderm gives rise to the extraembryonic ectoderm and ectoplacental cone (EPC) (104). The EPC differentiates into several types of trophoblasts, including endovascular TGCs, interstitial glycogen trophoblast cells, and cytotrophoblasts, that serve unique functions in placental development in rodents (105). From GD8.5 to GD10.5, the major vascular changes are spiral arteriole remodeling and formation of the placenta. Spiral arteriole remodeling. Spiral arteriole remodeling in rodents and women occurs via trophoblast-independent and trophoblast-dependent mechanisms. Trophoblast-independent, decidua-associated vascular remodeling involves endothelial vacuolation and VSMC swelling, leading to apoptosis. Trophoblast-dependent vascular remodeling involves the action of endovascular trophoblasts, the subtype of invasive trophoblasts that invades the arteriole wall, resulting in a chimeric vessel composed of both trophoblast and endothelial cells (Figure 3). Figure 3 Remodeling of spiral arterioles during pregnancy. Spiral arterioles are lined by endothelial cells and are wrapped on the outside by pericytes and VSMCs. From GD7.5 to GD9.5, the maternal spiral arterioles increase in diameter after embryo-derived trophoblast cells (green) integrate into the lining of the vessel wall. As a result, the maternal vasculature in the pregnant uterus is one of decreased resistance and increased blood flow to nourish the fetus. Interstitial and endovascular trophoblast invasion is required for placenta formation in rodents and women (106). In both species, interstitial trophoblasts degrade ECM proteins and invade the decidua to arrive at the spiral arteriole adventitial layer, whereas endovascular trophoblasts percolate through the vessel wall, restructure smooth muscle, and displace the endothelium (106). For rodent models of placentation, an important difference between rats and mice is the depth of trophoblast invasion. In rats, endovascular and interstitial trophoblasts invade beyond the decidua into the mesometrial triangle (or metrial gland) of the myometrium, while in mice, endovascular trophoblast invasion is limited, and interstitial trophoblasts do not invade into the myometrium (106). Furthermore, invasive trophoblasts in rats are intraluminal, replacing the endothelium from within the vessel, while those in mice are perivascular, replacing the endothelium from outside the vessel (106). Vascular changes in rodents and humans are similar. However, the depth of trophoblast invasion into the uterus and uterine vasculature and the intraluminal approach to vessel wall intercalation are most similar between rats and humans. Although the rat may be a more appropriate model of trophoblast invasion, the lack of transgenic rat models has limited investigations of molecular signaling underlying vascular changes. The mechanistic studies described below largely rely on mouse models, for which transgenics are more advanced. Studies of spiral arteriole remodeling in mice use immunohistochemical staining of endothelial cells with vWF and VSMCs with SMA (107) to identify these key vasculature components. Remodeled spiral arteries have similarities to lymphatic vessels, including reduced SMA, lack of a basement membrane, and a large, dilated lumen (108). Spiral arterioles that were once high-resistance and low-capacitance are thus remodeled to low-resistance high-capacitance vessels that allow for adequate delivery of nutrients and oxygen from the mother to the developing fetus (Figure 3). The placenta is fully formed by GD10.5 and is composed of three layers, the decidua, junctional zone, and labyrinth, each with unique vasculature. Within the maternal-derived decidua, arterioles undergo angiogenesis and trophoblast invasion after GD8.5 (73). The junctional zone, composed of spongiotrophoblast and glycogen trophoblast cells, is the confluence of maternal and fetal cells of the placenta. The labyrinth portion of the placenta is generally considered the fetal compartment, with fetal vascular endothelium and cytotrophoblast cells in direct contact with maternal blood (109). This type of placentation in rodents and humans, with intimacy of maternal and fetal cells for nutrient exchange, is classified as hemochorial (110). Non-trophoblast factors contributing to spiral arteriole remodeling (GD9.0–GD10.5) Trophoblast-independent rodent spiral arteriole remodeling involves apoptosis and is mediated by angiogenic factors and uNK cells, as described below. Apoptosis. The process of endovascular and interstitial trophoblast invasion requires endothelial cell and VSMC apoptosis, as evidenced by the uterine expression of cleaved caspase-3 (111). Studies using mice deficient in uterine Corin, a serine transmembrane protease necessary for atrial natriuretic peptide (ANP) synthesis, showed that ANP is essential for decidualization and spiral arteriole remodeling (111). ANP-deficient mice demonstrated low expression of uterine TNF-related apoptosis-inducing ligand (TRAIL), which is required for VSMC apoptosis and cyclophilin B–mediated TRAIL receptor upregulation in endothelial cells within spiral arterioles (111). Gene mutations and reduced expression and activity of Corin have been associated with human preeclamptic pregnancy (112), suggesting that these mouse models could help develop a better understanding of the process of spiral arteriole remodeling. uNK cells. uNK cells are the predominant CD45+ immune cells in the mouse decidua until GD7.5 (73, 113, 114). Following IL-15 stimulation, uNK cells begin secreting key angiogenic and vasoactive factors, including VEGF, PGF, and IFN-γ, that are required for vascular remodeling in pregnancy. Retention of decidual vessel SMA in uNK cell–deficient Rag2–/– IL2rg–/– mice results in constricted spiral arterioles at GD9.5 (115). At mid-gestation, mature uNK cells are essential for decidual maintenance. uNK cell–deficient mice have hypocellular decidual basalis at GD9.5–GD10 that becomes hypogranular at GD12 with spiral arterioles resembling those in non-pregnant mice (116). Angiogenic factors. A balance of pro- and anti-angiogenic factors in the placenta and maternal circulation is required for pregnancy success (Table 4). VEGF interacts directly with maternal uterine endothelial cells to promote angiogenesis and uteroplacental blood flow necessary for fetal circulation by GD9.5. Soluble VEGFR1 (sFlt1) is produced with pathologic placental hypoxia and scavenges free VEGF, preventing adequate placental angiogenesis (117). Paradoxically, VEGF overexpression in the endometrium of mice results in parallel upregulation of anti-angiogenic sFlt1 (118). Ultimately, endometrial VEGF overexpression in early pregnancy results in pathologically enlarged maternal blood spaces within the junctional zone while sparing labyrinth formation before GD11.5 (118). Although early pregnancy studies are lacking in rats, a transgenic rat model of human preeclampsia was created by breeding of female rats containing a human angiotensinogen transgene with male rats harboring a human renin transgene (119). Uterine arterioles displaying a lower stretch response at GD7.5 correlated with a hypertensive pregnancy resembling the human preeclampsia condition. Trophoblast factors contributing to spiral arteriole remodeling (GD9.0–GD10.5) Endovascular trophoblast cells replace endothelial cells of the spiral arterioles, transforming them from small resistant vessels to flaccid, large-caliber vessels that are unresponsive to vasoconstrictive agents. This complex process is thought to occur in rodents and women. At the inception, endovascular trophoblast cells form plugs in maternal uterine vessels. This is thought to protect the conceptus from excessively high oxygen levels, and thus placenta formation begins in a hypoxic state (73, 120). Once the plugs dissociate, maternal blood flow to the placenta increases. Notch is involved in trophoblast-dependent spiral arteriole remodeling prior to GD10.5. Notch signaling. Canonical Notch signaling is active in trophoblasts during early pregnancy, and trophoblast subtypes express Notch proteins, including DLL4, Notch2, and Notch4 (121). Notch2 deletion using the trophoblast-specific driver Tpbpa-Cre resulted in inadequate remodeling of spiral arterioles and a decline in placental perfusion at GD10.5 and GD14.5 (122). Eye to the future There are clear gaps in our understanding of uterine vascular remodeling during early pregnancy. Most studies describe vascular structure, gene expression, and, to some extent, functional changes in rodents (3, 24, 39–42, 57, 123). Very few studies simultaneously investigate signaling pathways and vessel architecture (6, 70). Future studies that combine genetic rodent models and advanced methodologies, such as organoids, organ-on-chip, and omics, with novel imaging techniques will allow integration of gene expression with spatial information and vessel function at different stages of the estrous/menstrual cycle and in early pregnancy. Endometrial organoids and organ-on-chip represent promising systems that can facilitate study of tissue function in physiologic and pathophysiologic contexts. These systems may provide a suitable approach to bridge the gap between 2D cultures and in vivo models (124). Endometrial organoids have been generated using epithelial cells alone or epithelial and stromal cells together but have not yet integrated endothelial cells (125). An implantation-on-chip model has also been constructed to understand the interaction of primary trophoblast cells, which were derived from first-trimester placenta, with uterine stromal and uNK cells. In this model, trophoblast invasion and migration directionality were drastically reduced in the absence of uterine endothelial cells. Further, non-uterine sources of endothelial cells promoted trophoblast invasion to a much lesser extent compared with uterine endothelial cells, supporting the unique properties of organ-specific endothelium (77). Transcriptomic profiling of the uterine microenvironment with RNA sequencing during different stages of pregnancy is emerging as a powerful tool to determine the expression patterns of angiogenesis genes that are necessary for implantation and pregnancy success (126–128). Single-cell RNA sequencing has the advantage of interrogating the transcriptome of specific cells within the heterogeneous placental tissue composed of maternal- and fetal-derived cells. Information on the cell-to-cell crosstalk between maternal and fetal cells can be investigated more elegantly with spatial transcriptomics (126). Equally important are new imaging advances, such as confocal and light sheet microscopy, that will allow connections between gene expression changes and vascular structural and functional changes during implantation, decidualization, and early pregnancy (125, 129). Conclusions While a lot has been uncovered about uterine vascular changes during early pregnancy, many unanswered questions remain. The principal molecular signals in uterine (maternal) vascular remodeling during early pregnancy and how vessels are modified during clinical conditions such as infertility and miscarriage are not known. The downstream vasoactive and angiogenic pathways directly regulated by ovarian steroid hormones that result in vessel remodeling during the estrous cycle and during pre-implantation stages of pregnancy have not been well studied. Finally, although implied by increased vascular permability during early pregnancy, how the changes in blood vessel structure relate to function needs further attention. Addressing these gaps in knowledge will provide novel perspectives regarding early pregnancy and allow design of therapeutics to address clinical conditions related to infertility and miscarriage. Acknowledgments RA receives support from March of Dimes grant 5-FY20-209 and NIH R01HD109152; NM receives support from National Institute of Child Health and Human Development T32HD087166; NCD receives support from NIH R01AI148695; and JS receives support from NIH P20GM135002. Address correspondence to: Ripla Arora, 775 Woodlot Dr., Rm 3312, East Lansing, Michigan 48824, USA. Email: ripla@msu.edu. Or to: Nataki C. Douglas, 185 South Orange Ave, MSB E569, Newark, New Jersey 07103, USA. Email: nd537@njms.rutgers.edu. Footnotes Conflict of interest: NCD is chair of the Modern Fertility Medical Advisory Board. 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Email the journal Metrics Article usage Citations to this article (21) Go to Top Abstract Introduction Human uterus anatomy and the menstrual cycle Types of blood vessels in the human endometrium Factors regulating the human endometrial vasculature Vascular remodeling in the human menstrual cycle The rodent estrous cycle Types of blood vessels in the rodent uterus Events in early rodent pregnancy Vascular remodeling in rodent pregnancy Eye to the future Conclusions Acknowledgments Footnotes References Version history Advertisement Advertisement Copyright © 2025 American Society for Clinical Investigation ISSN 2379-3708 Sign up for email alerts ✓ Thanks for sharing! AddToAny More… Posted by 2 X users 58 readers on Mendeley See more details
190510	https://artofproblemsolving.com/wiki/index.php/Combinatorial_identity?srsltid=AfmBOopEFKGBBrnLqAXpmrf8-QEp9uUT9I8g6V8NdIPxJUkfo2c7hc1q	Art of Problem Solving Combinatorial identity - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Combinatorial identity Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Combinatorial identity Contents [hide] 1 Pascal's Identity 1.1 Proof 1.2 Alternate Proofs 2 Vandermonde's Identity 2.1 Video Proof 2.2 Combinatorial Proof 2.3 Algebraic proof 3 Hockey-Stick Identity 3.1 Proof 4 Another Identity 4.1 Hat Proof 4.2 Proof 2 5 Even Odd Identity 6 Examples 7 See also Pascal's Identity Pascal's Identity is a very important formula for olympiad math and it states that for any positive integers and . Here, is the binomial coefficient . This result can be interpreted combinatorially as follows: the number of ways to choose things from things is equal to the number of ways to choose things from things added to the number of ways to choose things from things. Proof If then and so the result is trivial. So assume . Then Alternate Proofs Here, we prove this using committee forming. Consider picking one fixed object out of objects. Then, we can choose objects including that one in ways. Because our final group of objects either contains the specified one or doesn't, we can choose the group in ways. But we already know they can be picked in ways, so Also, we can look at Pascal's Triangle to see why this is. If we were to extend Pascal's Triangle to row n, we would see the term . Above that, we would see the terms and . Due to the definition of Pascal's Triangle, . Vandermonde's Identity Vandermonde's Identity states that , which can be proven combinatorially by noting that any combination of objects from a group of objects must have some objects from group and the remaining from group . Video Proof Combinatorial Proof Think of the right hand side as picking people from men and women. Think of the left hand side as picking men from the total men and picking women from the total women. The left hand side and right hand side are the same, thus Vandermonde's identity must be true. ~avn Algebraic proof For all The coefficients of on both sides must be the same, so using the Binomial Theorem, Hockey-Stick Identity For . This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed. We can also flip the hockey stick because pascal's triangle is symmetrical. Proof Inductive Proof This identity can be proven by induction on . Base Case Let . . Inductive Step Suppose, for some , . Then . Algebraic Proof It can also be proven algebraically with Pascal's Identity, . Note that , which is equivalent to the desired result. Combinatorial Proof 1 Imagine that we are distributing indistinguishable candies to distinguishable children. By a direct application of Balls and Urns, there are ways to do this. Alternatively, we can first give candies to the oldest child so that we are essentially giving candies to kids and again, with Balls and Urns, , which simplifies to the desired result. Combinatorial Proof 2 We can form a committee of size from a group of people in ways. Now we hand out the numbers to of the people. We can divide this into disjoint cases. In general, in case , , person is on the committee and persons are not on the committee. This can be done in ways. Now we can sum the values of these disjoint cases, getting Algebraic Proof 2 Apply the finite geometric series formula to : Then expand with the Binomial Theorem and simplify: Finally, equate coefficients of on both sides: Since for , , this simplifies to the hockey stick identity. Algebraic Proof 3 Consider the number of solutions to the equation ++++++.......+ ≤ N where each is a non-negative integer for 1≤i≤r. METHOD 1 We know since all numbers on LHS are non-negative therefore 0≤N and N is a integer. Therfore, ++++++.......+ = 0,1,2......N. Consider each case seperately. ++++++.......+ =0 by Stars-and-bars the equation has solutions. ++++++.......+ =1 by Stars-and-bars the equation has solutions. ++++++.......+ =2 by Stars-and-bars the equation has solutions. ........... ++++++.......+ =N by Stars-and-bars the equation has solutions. Hence, the equation has + + +.... = (where k=N+r-1) SOLUTIONS. METHOD 2 Since, ++++++.......+ ≤ N Therefore we may say ++++++.......+ = N -m where m is another non-negative integer. 0 ≤++++++.......+ ⇒ 0≤ N-m ⇒ m≤ N So, we need not count this as an extra restriction. Now, ++++++.......+ +m = N. Again by Stars-and-bars this has solutions. Therefore, the equation has = solutions(As N+r-1 =k). Since, both methods yeild the same answer ⇒ = . Taking r-1= p redirects to the honeystick identity. ~SANSGANKRSNGUPTA Another Identity Hat Proof We have different hats. We split them into two groups, each with k hats: then we choose hats from the first group and hats from the second group. This may be done in ways. Evidently, to generate all possible choices of hats from the hats, we must choose hats from the first and the remaining hats from the second ; the sum over all such is the number of ways of choosing hats from . Therefore , as desired. Proof 2 This is a special case of Vandermonde's identity, in which we set . Even Odd Identity Examples 1986 AIME Problem 11 2000 AIME II Problem 7 2013 AIME I Problem 6 (Solution 2) 2015 AIME I Problem 12 2018 AIME I Problem 10 2020 AIME I Problem 7 2016 AMC 10A Problem 20 2021 AMC 12A Problem 15 1981 IMO Problem 2 2022 AIME I Problem 12 See also Pascal's Triangle Combinatorics Pascal's Identity Retrieved from " Categories: Theorems Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190511	https://ajc.maths.uq.edu.au/pdf/89/ajc_v89_p024.pdf	AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 89(1) (2024), Pages 24–31 On round-robin tournaments with a unique maximum score Yaakov Malinovsky∗ Department of Mathematics and Statistics University of Maryland, Baltimore County Baltimore, MD 21250 U.S.A. yaakovm@umbc.edu John W. Moon Department of Mathematical and Statistical Sciences University of Alberta Edmonton, AB T6G 2G1 Canada jwmoon@ualberta.ca Abstract Richard Arnold Epstein (1927–2016) published the ﬁrst edition of “The Theory of Gambling and Statistical Logic” in 1967. He introduced some material on round-robin tournaments (complete oriented graphs) with n labeled vertices in Chapter 9; in particular, he stated, without proof, that the probability that there is a unique vertex with the maximum score tends to 1 as n tends to inﬁnity. Our goal here is to give a proof of this result along with some historical remarks and comments. 1 Introduction In a classical round-robin tournament, each of n players wins or loses a game against each of the other n −1 players (see Moon ). Let Xij equal 1 or 0 according to whether player i wins or loses the game played against player j, for 1 ≤i, j ≤n, i ̸= j, where Xij +Xji = 1. We assume that all n 2 pairs (Xij, Xji) are independently distributed with P(Xij = 1) = P(Xji = 0) = 1/2. Let si := si(n) = n j=1,j̸=i Xij ∗Supported in part by BSF Grant 2020063. ISSN: 2202-3518 c ⃝The author(s). Released under the CC BY 4.0 International License Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 25 denote the score of player i, 1 ≤i ≤n, after playing against all the other n −1 players. We refer to (s1, s2, . . . , sn) as the score sequence of the tournament. When the identity of the players with a given score in a tournament is of no particular signiﬁcance, we often rearrange the elements of the score sequence in nondecreasing order and refer to this rearranged sequence as the nondecreasing score sequence of the tournament. These tournaments can be represented by complete oriented graphs in which the vertices represent the players and each pair of distinct vertices i and j is joined by an edge oriented from i to j or from j to i according to whether Xij = 1 or Xji = 1. The frequency of a given n-vertex nondecreasing score sequence is the sum of the frequencies (or the total number) of all labelled n-vertex tournaments that contain the same number of vertices of score k as the number of elements of value k in the given nondecreasing sequence, for 0 ≤k ≤n −1. Round-robin tournaments can be considered as a model of paired comparison experiments used in an attempt to rank a number of objects with respect to some criterion–or at least to determine if there is any signiﬁcant diﬀerence between the objects—when it is impracticable to compare all the objects simultaneously: see, e.g., Zermelo , David , David and Edwards , and Aldous and Kolesnik . In particular, David generated the score sequences of tournaments with n players for 3 ≤n ≤8 and their frequencies by expanding products of the form F(n) = 1≤i<j≤n 1 2wi + 1 2wj . For example, F(3) = 1 23 w2 1w2 + w1w2 2 + w2 1w3 + w1w2 3 + w2 2w3 + w2w2 3 + 2w1w2w3 ; so the joint probability mass function of the scores are P (s1 = 2, s2 = 1, s3 = 0) = · · · = P (s1 = 0, s2 = 1, s3 = 2) = 1 8, and P (s1 = 1, s2 = 1, s3 = 1) = 2 8; and the frequencies of the nondecreasing score sequences (0, 1, 2) and (1, 1, 1) are six and two, respectively. David used this information to develop, among other things, tests for deciding whether the maximum in a given outcome was signiﬁcantly larger than the expected value (n −1)/2 of a given score. Let rn denote the probability that an ordinary tournament with n labeled ver-tices has a unique vertex with maximum score. Epstein [8, p. 353] gave the values r4 = .5, r5 = .586, r6 = .627, r7 = .581, and r8 = .634 without further explanation. However, a reference to David is given a few pages later, so presumably he de-duced these values for ri from Table 1 in , except for one error: the value for r8 should have been 160, 241, 152/228 = .5969 · · ·. Epstein also stated, without a proof or reference, that as n increases indeﬁnitely, rn approaches unity. Some later editions Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 26 of his book contain more material on tournaments but the material on rn remains unchanged. A survey paper by Guy , on various unsolved problems, mentions Epstein’s problem on rn as being still unsolved. Alon referred us to a paper by Erd˝ os and Wilson that considered the analogous problem for the vertices of max-imum degree in a random labelled graph in which pairs of distinct vertices are joined by an edge with probability 1/2. Stockmeyer has recently pointed out that MacMahon generated the score sequences and their frequencies for tournaments with up to 9 vertices and his results agree with David’s for n = 8. It follows from MacMahon’s data that r9 = 42, 129, 744, 768/236 = .6130 · · ·. As a partial check, we obtained the same value for r9, as given earlier, by determining the number of ways of constructing 9-vertex tournaments with a unique vertex v of maximum score by adjoining v to an 8-vertex tournament with any given 8-vertex nondecreasing score sequence. We then tried to determine the value of r10, which was unknown to us at the time, in the same way from information about the 9-vertex case. In doing this we discov-ered that MacMahon’s values of 361, 297, 520 for the nondecreasing score sequence (2, 2, 3, 3, 4, 4, 6, 6, 6) and its complement were incorrect; these frequencies should have been divisible by 9, since there are 9 choices for the label of the winner of the match between the two vertices of score 4 (in both cases). These two frequency values should each be increased by 10, 000; and, using these corrected values we found that r10 = 21, 293, 228, 876, 800/245 = .6051 · · ·. Later, we also discovered that Doron Zeilberger had extended MacMahon’s work and had generated the nondecreasing score sequences and their frequencies for tournaments with up to 15 vertices using the Maple program. (We remark that Zeilberger’s frequencies for the two sequences mentioned earlier agree with the corrections we gave.) This additional data as well as Monte-Carlo simulations up to n = 100, 000 (see Malinovsky and Moon ) strongly suggest that Epstein’s Conjecture is correct. In the next section, we show that this is indeed the case. 2 The Uniqueness of the Maximum Score 2.1 Useful Facts and Notation Let (pij) denote a probability matrix such that pij + pji = 1, and pij = P(Xij = 1) for 1 ≤i ≤j ≤n, pii = 0 for 1 ≤i ≤n; and where the variables Xij and Xji are as deﬁned in Section 1. Huber used a coupling argument to establish the following inequality for the joint distribution function of the scores s1, . . . , sn in a round-robin tournament: P (s1 < k1, . . . , sm < km) ≤P (s1 < k1) · · · P (sm < km) , (2.1) where m ≤n, for any such probability matrix (pij), and any numbers (k1, . . . , km); the inequality also holds if the < sign is replaced by the ≤sign throughout. We assume here that pij = 1/2 for all i ≤j. As we shall see presently, Huber’s inequality Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 27 has implications for the maximum scores in tournaments. For expository convenience we introduce some notation and relations that we shall need later. Let b(n −1, j) = P(si = j) = n −1 j 1 2n−1 and B(n −1, j) = P(si > j) = k>j b(n −1, k) for 0 ≤j ≤n −1 and 1 ≤i ≤n. Next, let tn−1 = ⌈(n −1)/2 + xn−1((n −1)/4)1/2⌉ (2.2) where xn−1 = (2 log(n −1) −(1 + ϵ) log(log(n −1)))1/2 (2.3) for an arbitrary constant ϵ between 0 and 1, say. It is not diﬃcult to see that xn−1 ≤((tn−1 −(n −1)/2) ((n −1)/4)−1/2 ≤xn−1 + ((n −1)/4)−1/2. (2.4) It follows from (2.4) and deﬁnition (2.3) that xn−1 →∞and xn−1 = o(n1/6) as n →∞, and the same conclusion holds when xn−1 is replaced by ((tn−1 −(n −1)/2) ((n −1)/4)−1/2 . Consequently, we may appeal to relation (4.5.1) in R´ enyi [19, p. 204] and relations (2.7) and (6.7) in Feller [10, pp. 180 & 193] to conclude that b(n −1, tn−1) ∼ 2 π(n −1) 1/2 e− x2 n−1 2 ∼ √ 2(log(n −1))(1+ϵ)/2 π(n −1)3 (2.5) and B(n −1, tn−1) ∼ 1 √ 2π 1 xn−1 e− x2 n−1 2 ∼(log(n −1))ϵ/2 √ 4π(n −1) . (2.6) 2.2 Main Result Theorem. The probability that a random n-vertex tournament Tn has a unique vertex of maximum score tends to 1 as n tends to inﬁnity. In particular, if tn−1 is deﬁned as in (2.2) and (2.3) and s⋆denotes the maximum value of the scores s1, . . . , sn in Tn, then the following statements hold: (i) [Huber ] P(s⋆> tn−1) →1 as n →∞. Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 28 (ii) If Wn = Wn(Tn) denotes the number of ordered pairs of distinct vertices u and v in Tn such that su = sv = h for some integer h such that tn−1 < h ≤n −1, then P(Wn > 0) →0 as n →∞. Proof: Proof of (i) [Huber ] Huber observed that the required conclusion follows from the facts that P (s⋆< tn−1) ≤(1 −B(n −1, tn−1))n ≤e−nB(n−1,tn−1) ≤(1 + o(1))e−(log(n−1))ϵ/2 √ 4π →0, (2.7) as n →∞, appealing to the deﬁnition of B (n −1, tn−1), inequality (2.1), the in-equality 1 −c ≤e−c , and relation (2.6). Proof of (ii) We now turn to conclusion (ii). In view of conclusion (i), we may restrict our at-tention to tournaments Tn in which the maximum value s⋆of the scores realized by the vertices is at least as large as t = tn−1. Recall that Wn = Wn(Tn) denotes the number of ordered pairs of distinct vertices u and v of Tn such that t < su = sv where t ≤n −1, i.e. Wn = 1≤v<u≤n I(t < su = sv). Let s′ u and s′ v denote the scores of two such vertices u and v in their matches with the remaining n−2 players and note that s′ u and s′ v are independent variables. Then it follows that P (su = h, sv = h) = 1/2P(s′ u = h −1)P(s′ v = h) + 1/2P(s′ u = h)P(s′ v = h −1) = n −2 h −1 (1/2)n−2 n −2 h (1/2)n−2 = 4 h n −1 1 − h n −1 n −1 h (1/2)n−1 n −1 h (1/2)n−1 ≤ (b(n −1, h))2 . (2.8) Hence, E(Wn) = E 1≤v<u≤n I(t < su = sv) = n(n −1)E (I(t < s1 = s2)) = n(n −1)P (t < s1 = s2) = n(n −1) n−1 h=t+1 P(s1 = h, s2 = h) ≤ n(n −1) n−1 h=t+1 b(n −1, h)2 ≤n(n −1)b(n −1, t + 1)B(n −1, t) ≤ n(n −1)b(n −1, t)B(n −1, t) ∼(log(n −1))1/2+ϵ π 2(n −1) →0, Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 29 as n →∞. Consequently, appealing to (2.5), (2.6), and to the fact that Wn = WnI(Wn > 0) ≥I(Wn > 0), we ﬁnd that 1 −P(Wn = 0) = P(Wn > 0) ≤E(Wn) →0, as required. 2 3 Remarks Remark 3.1 For the sake of completeness, we mention an upper bound, found by Huber , for the maximum score of all tournaments Tn, except for a fraction that tends to zero as n →∞. Let t′ = t′ n−1 be deﬁned as t = tn−1 was deﬁned earlier except that the ϵ in relation (2.3) is replaced by −ϵ and without the ceiling function, it turns out that a relation corresponding to (2.6) is B(n −1, t′ n−1) ∼(log(n −1))−ϵ/2 √ 4π(n −1) . Hence, it follows from Boole’s inequality that P (s⋆> t′) ≤nB(n −1, t′) = O (log(n −1))−ϵ/2 , (3.1) as n →∞. From (2.7) and (3.1) Huber concluded that s⋆−n −1 2 − n −1 4 2 log(n −1) →0 in probability as n →∞. Remark 3.2 Malinovsky and Moon and Malinovsky and Rinott have ex-tended Huber’s inequality to a more general round-robin tournament model and to other tournaments and games models, respectively. Remark 3.3 Malinovsky and Moon gave another proof of (i) by the 2nd Mo-ment Method frequently applied to probabilistic problems in Graph Theory (see Alon and Spencer [3, Chapter 4]). Bollob´ as and Frieze and Karo´ nski (see also references therein) have, among other things, derived numerous results on the distribution of the degrees of vertices in ordinary graphs in which edges are present with probability p. Acknowledgements We thank Noga Alon for referring us to the work of Paul Erd˝ os and Robin J. Wilson. We also thank Boris Alemi for executing the Maple program, mentioned toward the end of Section 1, on a powerful computer and obtaining the data in the required format. We would like to thank the Editor, Managing Editor and two referees for the careful reading and helpful comments and suggestions that led to improvements in the paper. Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 30 References D. J. Aldous and B. Kolesnik, To stay discovered: On tournament mean score sequences and the Bradley-Terry model, Stochastic Process. Appl. 150 (2022), 844–852. N. Alon, Personal communication, 2022. N. Alon and J. H. Spencer, The probabilistic method, fourth edition, John Wiley & Sons, Inc., Hoboken, NJ, 2016. B. Bollob´ as, Random graphs, second edition, Cambridge University Press, Cam-bridge, 2001. H. A. David, Tournaments and paired comparisons, Biometrika 46 (1959), 139– 149. H. A. David, The method of paired comparisons, Second edition, Charles Griﬃn & Co., Ltd., London; The Clarendon Press, Oxford University Press, New York, 1988. H. A. David and A. W. F. Edwards, The Evaluation of Tournament Outcomes: Comments on Zermelo (1929), Annotated readings in the history of statistics (2001), 161–166, Springer-Verlag, New York. R. A. Epstein, The theory of gambling and statistical logic, Academic Press, New York, 1967. P. Erd˝ os and R. J. Wilson, On the chromatic index of almost all graphs, J. Combinatorial Theory Ser. B 23 (1977), 255–257. W. Feller, An introduction to probabilty theory and its applications, Vol. I., third edition, New York-London-Sydney: Wiley, 1968. A. Frieze and M. Karo´ nski, Introduction to random graphs, Cambridge Univer-sity Press, Cambridge, 2016. R. K. Guy, A pentagonal pot-pourri of perplexing problems, primarily proba-bilistic, Amer. Math. Monthly 91 (1984), 559–563. P. J. Huber, A remark on a paper of Trawinski and David entitled: Selection of the best treatment in a paired comparison experiment, Ann. Math. Statist. 34 (1963), 92–94. P. A. MacMahon, An American tournament treated by the calculus of symmetric functions, Quart. J. Pure Appl. Math. XLIX no. 193 (1923), 1–36. Reprinted in Percy Alexander MacMahon Collected Papers, Vol. I, George E. Andrews, ed., MIT Press, 1978, pp. 308–343. Y. MALINOVSKY AND J.W. MOON / AUSTRALAS. J. COMBIN. 89 (1) (2024), 24–31 31 Y. Malinovsky and J. W. Moon, On the negative dependence inequalities and maximal score in round-robin tournament, Statist. Probab. Lett. 185 (2022), 109432. Y. Malinovsky and J. W. Moon, On round-robin tournaments with a unique maximum score, (2023). Y. Malinovsky and Y. Rinott, On tournaments and negative dependence, J. Appl. Probab. 60 (2023), 945–954. J. W. Moon, Topics on tournaments, Holt, Rinehart and Winston, New York, 1968. (A version (2013) with some corrections is publicly available at https:// www.gutenberg.org/ebooks/42833.) A. R´ enyi, Foundations of probability, Holden-Day, Inc., San Francisco, Calif.-London-Amsterdam, 1970. P. K. Stockmeyer, Counting various classes of tournament score sequences, J. Integer Seq. 26 (2023), Article 23.5.2. D. Zeilberger, On the most commonly-occurring score vectors of American tour-naments of n-players, and their corresponding records, rutgers.edu/~zeilberg/mamarim/mamarimhtml/Percy.html, 2016. E. Zermelo, Die Berechnung der Turnier-Ergebnisse als ein Maximumproblem der Wahrscheinlichkeitsrechnung, Math. Z. 29 (1929), 436–460. (Received 14 Apr 2023; revised 3 Jan 2024)
190512	https://mathoverflow.net/questions/260063/positive-root-of-a-polynomial	Positive root of a polynomial - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Positive root of a polynomial Ask Question Asked 8 years, 8 months ago Modified7 years, 4 months ago Viewed 499 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Is anything in general known about the positive root x+x+ of the following polynomial of degree (n+1)(n+1), as a function of the coefficients α 1,α 2,...,α n>0 α 1,α 2,...,α n>0: ∑i=1 n(1−x 2 α i)∏j=1 j≠i n(1+x α j)=0.∑i=1 n(1−x 2 α i)∏j=1 j≠i n(1+x α j)=0. Is this polynomial well-known? At this point, I know that x+x+ is unique, but would like to know its dependence on α i>0,i=1,...,n α i>0,i=1,...,n. Of course, I can employ root-finding algorithms, but currently my focus is understanding the qualitative nature of x+x+. I could not even tackle n=2 n=2 case. Please tag as appropriate. polynomials Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:19 CommunityBot 1 2 2 silver badges 3 3 bronze badges asked Jan 20, 2017 at 2:43 Abhishek HalderAbhishek Halder 1,538 9 9 silver badges 21 21 bronze badges 2 1 How do you know it is unique?Igor Rivin –Igor Rivin 2017-01-20 03:00:09 +00:00 Commented Jan 20, 2017 at 3:00 5 I know it two ways. First: the above polynomial is the first order optimality condition for a convex optimization problem over x>0 x>0. Second: if you break down the products in the above polynomial and use the fact that α i>0 α i>0, then you will see that the only change in sign occurs from the x n x n term to the x n−1 x n−1 th term. Then by Descartes' rule of sign, the number of positive roots is exactly one.Abhishek Halder –Abhishek Halder 2017-01-20 03:24:11 +00:00 Commented Jan 20, 2017 at 3:24 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Not a full answer, but some comments and an algorithm that the OP may find relevant. Assuming x≠−1/α 1,…,−1/α n x≠−1/α 1,…,−1/α n, we can rewrite the equation as ∏j(1+x α j)∑i(1−x 2 α i)1+x α i=0.∏j(1+x α j)∑i(1−x 2 α i)1+x α i=0. Under our assumption, this further simplifies to the OP's older equation: f(x):=∑i 1−x 2 α i 1+x α i=0.f(x):=∑i 1−x 2 α i 1+x α i=0. Perhaps at this point, you could try to: (i) bracket the root, obtaining points a<x∗<b a<x∗<b such that f(a)<0 f(a)<0, f(x∗)=0 f(x∗)=0 and f(b)>0 f(b)>0, where the bounds are reasonable functions of the α i α i; (ii) run (by hand) an iteration or two of Newton's method to get an approximate x∗x∗ (seems hard); (iii) try Lagrange inversion; or (iv) try a fixed-point iteration. Example of (iv). Rewrite the equation as follows x 2∑i α i 1+x α i=∑i 1 1+x α i.x 2∑i α i 1+x α i=∑i 1 1+x α i. Using the notation a(x)=∑i α i/(1+x α i)a(x)=∑i α i/(1+x α i) and b(x)=∑i 1/(1+x α i)b(x)=∑i 1/(1+x α i), we then solve for x x by running the iteration x k+1=[b(x k)/a(x k)]1/2,k=0,1,….x k+1=[b(x k)/a(x k)]1/2,k=0,1,…. An experiment suggests that started from x 0=0 x 0=0, this iteration monotonically increases x x (some calculation required, but one can show this easily). Since b(x)/a(x)b(x)/a(x) is bounded above, the sequence (x k)(x k) must converge. Using a couple of steps of this iteration, one can get a "reasonable" approximation to the the desired solution x∗x∗ (this seems to converge quite fast, so could be a reasonable method to solve for x x numerically too). ~~EDIT: It seems that this Fixed-Point Iteration (FPI) is used in the preprint authored by the OP (albeit, without attribution)!~~ EDIT 2: Meanwhile (as noted in the comments below) the OP has fixed the attribution; please check the link the his comment for the latest version. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 10, 2018 at 21:55 answered Jan 20, 2017 at 5:06 SuvritSuvrit 28.7k 7 7 gold badges 83 83 silver badges 152 152 bronze badges 14 1 Not really; I just wrote it back in this form as it seems more amenable to points (i) and (iii) above.Suvrit –Suvrit 2017-01-20 05:44:37 +00:00 Commented Jan 20, 2017 at 5:44 1 @AbhishekHalder Since the rational equation has poles, it is easier to bracket the roots.Federico Poloni –Federico Poloni 2017-01-20 07:38:51 +00:00 Commented Jan 20, 2017 at 7:38 1 If I recall correctly, you can directly show it by choosing x<y x<y and computing b(x)/a(x)−b(y)/a(y)b(x)/a(x)−b(y)/a(y), which is rather close of course to computing derivatives.Suvrit –Suvrit 2017-08-30 01:04:39 +00:00 Commented Aug 30, 2017 at 1:04 2 @AbhishekHalder I saw that you wrote a preprint arxiv.org/pdf/1803.08157.pdf which seems to use the above FPI, but your preprint contains no link to MO nor any acknowledgment? am I missing something?Suvrit –Suvrit 2018-03-23 01:58:48 +00:00 Commented Mar 23, 2018 at 1:58 1 FPI stands for fixed-point iteration?Gerry Myerson –Gerry Myerson 2018-03-23 06:20:28 +00:00 Commented Mar 23, 2018 at 6:20 \|Show 9 more comments You must log in to answer this question. 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190513	https://brilliant.org/wiki/applying-boundary-conditions-to-standing-waves/	Applying Boundary Conditions to Standing Waves Sign up with Facebook or Sign up manually Already have an account? Log in here. Matt DeCross and Christopher Williams contributed Standing waves on a string with fixed endpoint boundary conditions. . Boundary conditions for the wave equation describe the behavior of solutions at certain points in space. For instance, the strings of a harp are fixed on both ends to the frame of the harp. If the string is plucked, it oscillates according to a solution of the wave equation, where the boundary conditions are that the endpoints of the string have zero displacement at all times. In general, there are two major types of boundary conditions: fixed-endpoint or Dirichlet boundary conditions, and free-endpoint or Neumann boundary conditions, corresponding to holding the end of a string or allowing it to freely oscillate, respectively. Both types of boundary conditions can lead to standing waves, where certain points called nodes have zero displacement at all times and the maximum amplitude of the wave at all points does not change in time. Contents Standing Waves and the Wave Equation Fundamental Frequency and Harmonics References Standing Waves and the Wave Equation The wave equation is a partial differential equation which is second order in both time and space which models small oscillations in some quantity y: v21∂t2∂2y=∂x2∂2y, where v is the velocity of the wave. Depending on the medium in which the wave travels, the velocity v can take several different forms: Strings: v=μT where T is the tension of the string and μ is the mass density per unit length of the string. Sound/Pressure waves: v=ρK where ρ is the density of the gas through which the sound travels and K is the elastic bulk modulus given by K=γp, where p is the pressure of the gas and γ is the adiabatic index of the gas (equal to 5/3 for a monatomic ideal gas). Light: Light waves are the propagation of oscillations in the electromagnetic field, which travel at the speed of light v=c. 2 1 4 1/2 Sound waves propagate in a room filled with a monatomic ideal gas. Suppose more gas is flooded into the room, so that the new density of gas in the room is twice what it was previously. The room is climate-controlled so that the temperature stays fixed. By what factor does the speed of the sound waves change? The correct answer is: 1 Any solution of the wave equation can be expressed as the sum of left-traveling and right-traveling waves, which can be any functions of the form f(x−vt) and g(x+vt). Show that any solution to the wave equation can be expressed as a sum of left-traveling and right-traveling waves f(x−vt) and g(x+vt). Solution: The two (left-traveling and right-traveling) solutions can be verified by direct substitution. For the left-traveling solution: dx2d2f(x−vt)dt2d2f(x−vt)=f′′(x−vt)=v2f′′(x−vt) Therefore, dt2d2f(x−vt)=v2dx2d2f(x−vt), so f(x−vt) solves the wave equation. Substituting the right-traveling waves gets the identical result. Since the left-traveling and right-traveling waves are linearly independent, any solution of the wave equation can be expressed as a linear combination of them. If the wave is traveling in some one-dimensional region where both endpoints are fixed, the simplest examples of solutions are sinusoids: y(x,t)=y0sin(x−vt)+y0sin(x+vt). where y0 is the amplitude of the wave. Using trigonometric identities, this solution may be expanded: y(x,t)=y0(sin(x−vt)+sin(x+vt))=y0(sin(x)cos(vt)−cos(x)sin(vt)+sin(x)cos(vt)+cos(x)sin(vt))=2y0sin(x)cos(vt) This solution has constant amplitude and the spatial part sin(x) has no time dependence. This is the mathematical description of a standing wave, visualized below. A superposition of left-propagating and right-propagating traveling waves creates a standing wave when the endpoints are fixed . Fundamental Frequency and Harmonics Solutions to the wave equation can also be written in the form f(kx−ωt) rather than f(x−vt), where k is the wave number and ω is the frequency, using the fact that ω=vk for non-dispersive waves. In this form, the solution for the amplitude of harmonic (sinusoidal) standing waves on a string fixed at both ends described above is: y(x,t)=2y0sin(kx)cos(ωt) Since the amplitude y(x,t) must vanish at the endpoints for standing waves at all times t, the allowed values of k are highly constrained. Suppose the string is length L, and label the endpoints of the string as the coordinates x=0 and x=L. Plugging in above, since sin(0)=0, y(0,t)=0 regardless of time t. However, at the endpoint at x=L, y(L,t)=0 if and only if: kL=nπ, for n any integer. The wave number k is thus constrained to the set of values: k=nLπ. Using the fact that the wave number is related to the wavelength λ by: k=λ2π, the possible allowed wavelengths are similarly discrete: λ=n2L, as are the frequencies: f=n2Lv=n2kLω. The allowed frequencies of standing waves (and in general, the allowed frequencies of any oscillatory system) are typically called normal modes, with the n=1 frequency called the fundamental frequency and frequencies corresponding to higher n called harmonics, where the nth harmonic corresponds to the integer n. To find the allowed wave numbers or frequencies, it is typically useful to first draw some cases of allowed wave patterns, from which the wavelength can be extracted first, as illustrated in the diagram below: Standing wave patterns with increasing n and the corresponding wavelengths and frequencies . 4001 1201 2001 2401 An oscillating string of tension 100 N and mass density per unit length μ=1 kg/m fixed at both ends has fundamental frequency 400 Hz. What is the difference in meters between the wavelengths corresponding to the second and third harmonics? The correct answer is: 2401 This analysis applies not only to transverse waves on a string but in fact generic sinusoidal solutions to the wave equation. Another common example concerns longitudinal pressure/sound waves traveling in pipes. A pipe closed at both ends represents Dirichlet boundary conditions for the sound waves in the pipe, since the longitudinal displacement at the ends of the pipe must be zero. If one end of the pipe is opened, the boundary condition becomes Neumann: the pressure at the end should be atmospheric pressure, so there is no change in pressure as waves leave the pipe, and the slope of the longitudinal displacement is flat. What are the allowed frequencies of longitudinal pressure waves in a half-open pipe of length L? Solution: First, analyze the three longest allowed wavelengths to identify a pattern. The longitudinal displacement over the length of the pipe is illustrated in the following diagram: Amplitudes of longitudinal displacement of pressure waves in a pipe for the three lowest-frequency harmonics. The lowest-frequency harmonic completes a quarter wavelength in length L. The next highest harmonic completes three-quarters of a wavelength, and the harmonic after that completes five-fourths of a wavelength. Observing the pattern, the allowed wavelengths must be: L=42n−1λ⟹λ=2n−14L. The corresponding frequencies are found from the relation λf=v, which yields: f=4L(2n−1)v, where v is the velocity of the pressure waves, and n∈{1,2,3…}. The same length Eight times as long Four times as long Twice as long Compared to a half-open pipe, how much longer would a pipe open on both ends need to be in order to have the same fundamental frequency for longitudinal pressure waves? Image from under Creative Commons licensing for reuse and modification. Image from under Creative Commons licensing for reuse and modification. Image from lecture notes from San Francisco State University, August 2015. Accessed February 10, 2016. Cite as: Applying Boundary Conditions to Standing Waves. Brilliant.org. Retrieved 12:46, August 25, 2025, from
190514	https://www.youtube.com/watch?v=XesyFuxanew	The Equation of a Circle - Expanded Form KeysToMaths1 6380 subscribers 29 likes Description 5530 views Posted: 7 Jan 2013 The equation of a circle is developed using the notation (h,k) as centre and r as radius. The notation using (-g,-f) as centre is ignored and is unnessary (and confusing). Transcript: in the first video in this playlist We derived the general equation of a circle with Center h k and radius R basically we use Pythagoras's Theorem to find the relation between X and Y where X comma Y is any point on the circle in this video we're going to expand this out so we can write this in a different form it's just an expanded form basically so we just Square this out we get x^2 - 2 HX + h^2 when we Square out this we get y^ 2 - 2K y + k^ 2 and on the right hand side we have r squared the next thing that we will do is bring the square terms together then write down the X term and then the Y term finally we write down the numbers h^2 K2 bring R 2 over becomes minus r s we have zero on the right hand side now h k and R are numbers they are given we're given the center normally h k and the radius of a circle and then we usually go and get the equation so if we put these three numbers together we get some new number that we will call C so now this is the expanded form of the equation of a circle we have to keep in mind what C is C is H 2 + k^ 2 - R 2 you can see in this equation H and K we can actually get H and K directly from this by looking at the coefficients of X and Y however the other important piece of information for a circle is the radius of the circle R well if we know what c is and if we know what H and KR then we can find out what R is so we just rearrange this equation to make R the subject so we have - r^ 2 = c - h^2 - k^2 changing signs and getting the square root you see that R is h^2 + k^ 2us C let's take an example we are going to get the fully expanded form for the equation of a circle with Center minus one 2 and radius six so I'll use the original form x - h^2 + y - k^2 = R 2 and plug in these values so we have x - H well h k is the center so we have x - -1 or x + 1^ squared plus y - k^2 where K is 2 = R 2 where R is 6 R2 is 36 now let's expand this out and get it into this form up here uh we get X x^ 2 + 2x + 1 + y^ 2 - 4 y + 4 = 36 so we get x^2 + y^2 + 2x - 4 y we have 1 + 4 is 5 5 - 36 is - 31 equals z so now we have it in this form up here x² term y s term X term y term constant equals zero now let's consider a new question suppose that we are given this equation of a circle and that we are asked to find the center and radius how do we reverse all the steps that we just did of course we know what the answer is we know that the center of the circle is minus1 2 and its radius is six well we can just refer to the general form of the equation of a circle so imagine that we are given this equation we compare to the general form so we can see that - 2 H is the coefficient of x so we see that - 2 H = + 2 for this particular circle from this equation we can solve for h h = 2 divid min-2 which is minus one so now we found H the x value of the center now let's look at the y- values very similar the coefficient of Y in the general form is minus 2 K for this particular Circle - 2 K is -4 so we see that K is -4 ID min-2 which is + two so now we found the center we found that the center is -1a 2 next how do we get the radius well we use this formula here to get the radius I'll do it over here we have to get the square root of h^2 + k^2 minus c h is minus one if we Square minus one we get plus one k^2 is 2^2 here which is four minus C well C is the term at the end of the equation for this particular Circle C isus 31 so our formula is H2 + k^2 minus C so we minus - 31 that's + 31 so we have the square of 36 which is six so R is equal to 6 so that's how we work backwards when we're given the equation in this form let's see that again basically to get the center we just look at the coefficient of x half it and change the sign half of plus two is plus one change the sign to get minus one then we look at the coefficient of Y half it and change the sign because you see - 2 K is-4 so we're dividing by minus two basically so that's the same as halfing this and changing the sign to get plus two to find out what the radius is we use this formula here so you have to memorize this formula or you can derive it so we just plug the values we got for H and K into this formula C is this number here C is - 31 find the center and radius of the circle x^2 + y^2 - 4x - 4 y + 4 = 0 and show that the circle touches both axes okay so we compare the circle to the general form which is up here basically - 2 H the coefficient of x equals -4 so to find H we have the coefficient of x and change change the sign so the center is going to be two for the h value to get the K value we have the coefficient of Y and change the sign half of-4 is min-2 change the sign we get plus two so that's HK the center to get the radius we use this formula so the radius is root h^2 + k^ 2us C so h is two k is two and we have to subtract C where C is this number here okay if we compare to the general form of our equation the number at the end is C so for this example C is four so we have 2^ 2 is four and four is 8 8 - 4 is two is 4un 4 is 2 so the circle is centered at 22 and its radius is two so here is a very rough sketch the center is is 2 two radius is two so you can see if we draw a very rough sketch of this circle that it will touch both axes it'll touch the y axis and the xaxis because the horizontal distance of the 22 to the Y AIS is two which is equal to the radius and the vertical distance of the point 22 to the xaxis is also two which is equal to the radius so as a matter of fact we have two points on this circle this point here has coordinates 2 0 and this point here has coordinates 02 so because these two points are on the circle they should satisfy the equation of the circle so I will just show that for the 02 so we plug 02 into this equation so we replace x with zero and Y with two so we - 4 X which is 0 - 4 Y which is 2 + 4 and all of this should work out to zero because 02 is on this circle okay so we have 2^ 2 is 4 this is 0o uh this is- 8 8 + 4 so we get - 4 + 4 which is zero so it does check out similarly for this point if you like you can plug this point into the equation and it should also satisfy it as a matter of fact we can get other points in this circle we can get this point here we see that the coordinates of this point is 42 okay because the radius is two so two and two is four the diameter of this circle is four coordinates of this point up here are two four so you can check if you sub any of these four points into the equation of the circle you will see that they will satisfy it
190515	https://www.omnicalculator.com/physics/critical-damping	Critical Damping Calculator Welcome to our critical damping calculator, which can help you estimate the critical damping coefficient of a damped oscillator. If you used our simple pendulum calculator, you might have learned that a simple pendulum's motion is oscillatory, and the force describing that motion is proportional to the displacement of the bob from its mean position. However, in real situations, friction forces like air resistance also influence this motion and, in this case, oppose the pendulum's movement. Read on to find more about these opposing frictional forces — the definition of damping force, degrees of damping, damping coefficient, and formula for estimating critical damping coefficient. We also explain underdamped, overdamped, and critically damped oscillations using figures and examples. After reading, you'll never be in doubt about how to calculate any critically damped system. 🙋 If you want to learn more about friction, the free fall with air resistance calculator is another clear example of friction opposing a movement. What is the definition of damping? Damping is the process of dissipating energy from a vibrating structure, resulting in the reduction of the vibration's amplitude. Generally, it involves the conversion of the mechanical energy of the vibrating structure into thermal energy. We know that a rolling ball comes to a stop because of air resistance and friction due to the surface on which it rolls. This way, any movement of an object is resisted by its surrounding objects. The resistance reduces the energy of the moving object with time, eventually slowing it down to a stop. Now, let's apply this idea to a vibrating structure. The air surrounding such an object (or any medium in contact with it) tries to resist the vibrational motion and reduce its vibrational energy (mechanical energy). The higher the energy, the higher the vibration's amplitude. Thus, reducing the energy with time reduces the vibration's amplitude too. Now that we've covered the mechanical theory let's jump to the next section to learn what is the damping coefficient. What is the damping coefficient? The damping coefficient of the damping agent or damper tells us how effective the damper is in resisting an object's harmonic motion. We calculate the resistive force exerted by a damper or the damping force as follows: F=−cdtdx, where: c is the damping coefficient; x is the displacement of the oscillating object from its mean position; and dtdx denotes the change in displacement (x) with time (t) — this is also known as the instantaneous velocity. The damping coefficient c helps estimate the energy dissipated due to frictional forces, which slow down oscillation. Now it's high time to find out what is the critical damping coefficient. Want to learn more about damping and damping coefficient? Take a look at our damping ratio calculator. What are the three degrees of damping? There are three degrees of damping that depend on how quickly the amplitude of the oscillations decreases: Underdamping (light damping); Critical damping; and Overdamping (heavy damping). In underdamped oscillations, the object comes to rest after a few oscillations with amplitude decreasing exponentially with time. Let's go back to the first oscillating object we have learned — a simple pendulum! It's an underdamped system. In overdamping, the damper exerts a force that is what is required to prevent the oscillations. Hence, the object returns to its equilibrium position without oscillating. Have you noticed that the toilet flush handle gradually returns to its resting position without oscillating? It's an overdamped system. Another example is an automated door — it's overdamped by design to prevent damage to the glass. In critical damping, the object reaches the equilibrium point as quickly as possible without oscillating. An automobile shock absorber is a great example of the practical application of critical damping. What is the critical damping coefficient? The damping coefficient formula The value of the damping coefficient at which critical damping occurs is the critical damping coefficient of a system. It's also equal to the undamped resonant frequency of the oscillator. In terms of stiffness and mass of the oscillating object, we calculate the critical damping coefficient as follows: cc=2k⋅m, where: k is the stiffness of the oscillated body; and m is the mass of the oscillating body. How to use the critical damping calculator It's simple and straightforward! Select the units for each variable from the drop-down list beside them. Enter the oscillator's mass in the input box for "Mass (m)." Type the oscillator's stiffness in the input for "Stiffness (k)." The calculator displays the natural frequency (ωn) and critical damping coefficient (cc) of the oscillator. ✅ You can enter the mass and natural frequency to find out the other two variables, or you can enter the critical damping coefficient and one other variable to estimate the remainders. In short, enter any two variables and obtain the other two variables. FAQs What is simple harmonic motion? An oscillating body executes simple harmonic motion (SHM) if: The magnitude of the forces acting on it is directly proportional to the magnitude of its displacement from the mean position; and The restoring force always acts towards the mean position. What are the differences between overdamping, underdamping, and critical damping? There are three damping systems: An overdamped system returns to the equilibrium position slowly without oscillating. An underdamped system moves to the equilibrium position quickly and oscillates before coming to rest. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium. How do I find the critical damping coefficient using the natural frequency? The critical damping coefficient (cc) is twice the product of the mass (m) and natural frequency (ωn) of the oscillating object: cc = 2mωn. So, find the product of the mass and natural frequency and multiply it by 2 to obtain the critical damping coefficient. How do I calculate the stiffness of an oscillator with mass and natural frequency? To find the stiffness of an oscillator, equate the two formulas to estimate the critical damping coefficient: cc = 2mωn, where m is the mass and ωn is the oscillating object's natural frequency; and cc = 2√(k×m), where k is the stiffness. Thus, stiffness is the product of the oscillator's mass and square of the natural frequency: k = mωn2. How do I find the critical damping coefficient of an oscillator of mass 2 kg and stiffness 2 N/m? The formula for calculating the critical damping coefficient (cc) using the oscillator's mass (m) and stiffness (k) is: cc = 2√(k×m). So, the critical damping coefficient of an oscillator of mass 2 kg and stiffness 2 N/m is 2√(2 × 2) = 4 Ns/m. How do I calculate the stiffness of an oscillator of mass 5 kg and critical damping coefficient 1 Ns/m? The formula to calculate stiffness is: k = cc2/2m, where k is the stiffness, m is the mass, and ωn is the natural frequency of the oscillating object. Thus, an oscillator of mass 5 kg and critical damping coefficient 1 Ns/m has a stiffness of 12 / (2 × 5) = 0.05 N/m. Did we solve your problem today? Check out 15 similar acoustic waves calculators 🔊 Acoustic impedance Alfvén velocity Beat frequency
190516	https://unacademy.com/content/question-answer/mathematics/what-is-the-derivative-of-1-x/	What is The Derivative of 1/x Access free live classes and tests on the app Download + IIT JEE AFCAT AP EAMCET Bank Exam BPSC CA Foundation CAPF CAT CBSE Class 11 CBSE Class 12 CDS CLAT CSIR UGC GATE IIT JAM JEE Karnataka CET Karnataka PSC Kerala PSC MHT CET MPPSC NDA NEET PG NEET UG NTA UGC Railway Exam SSC TS EAMCET UPSC WBPSC CFA Search for: LoginJoin for Free Profile Settings Refer your friends Sign out Terms & conditions • Privacy policy About • Careers • Blog © 2023 Sorting Hat Technologies Pvt Ltd Question & Answer » Mathematics Questions » What is The Derivative of 1/x What is The Derivative of 1/x Share Answer:We can express 1 / x as x-1. We know that (as per the power rule), If x is a variable and n is the power, then the derivative of x raised to the power n can be written as: d / dx(x n)=n x n-1 Hence, d / dx(x-1)=-1 (x-1-1) d / dx(x-1)=-1 (x-2) d / dx(x-1)= –1 / x² Therefore, the derivative of 1 / x is –1 / x². Unacademy is India’s largest online learning platform. Download our apps to start learning Starting your preparation? Call us and we will answer all your questions about learning on Unacademy ##### Call +91 8585858585 Company About usShikshodayaCareers we're hiringBlogsPrivacy PolicyTerms and Conditions Help & support User GuidelinesSite MapRefund PolicyTakedown PolicyGrievance Redressal Products Learner appEducator appParent app Popular goals IIT JEEUPSCSSCCSIR UGC NETNEET UG Trending exams GATECATCANTA UGC NETBank Exams Study material UPSC Study MaterialNEET UG Study MaterialCA Foundation Study MaterialJEE Study MaterialSSC Study Material © 2025 Sorting Hat Technologies Pvt Ltd Goals AFCAT AP EAMCET Bank Exam BPSC CA Foundation CAPF CAT CBSE Class 11 CBSE Class 12 CDS CLAT CSIR UGC GATE IIT JAM JEE Karnataka CET Karnataka PSC Kerala PSC MHT CET MPPSC NDA NEET PG NEET UG NTA UGC Railway Exam SSC TS EAMCET UPSC WBPSC CFA Share via COPY
190517	https://www.cuemath.com/perfect-square-formula/	Perfect Square Formula The perfect square formula is used to find the square of the addition or subtraction of two terms, (a ± b)2 and is known as the perfect square formula. Let's learn more about the perfect square formula in detail in the following section. What is the Perfect Square Formula? We apply the perfect square formula when we have to calculate the square of any binomial. It calculates the square of sum or difference of two terms or can be used in factorization. The perfect square formula is: (a ± b)2 = (a2± 2ab + b2) perfect square formula Book a Free Trial Class Examples on Perfect Square Formula Let us consider few illustrations based on the (a+b)^2 formula in this solved examples section. Example1: Find the square of 6x+ 4y using the perfect formulas. Solution: To find: Square of 6x+ 4y, Using the perfect square formula. (a + b)2 = (a2+ 2ab + b2) Put the values, (6x + 4y)2 = ((6x)2+ 2 × 6x × 4y + (4y)2) (6x + 4y)2= (36x2+ 48x + 16y2) Answer: The the square of 6x+ 4y is (36x2+ 48x + 16y2). Example 2: Using the perfect square formula, find if x2 + 25 - 10x is perfect square or not. Solution: To find: x2 + 25 - 10x is perfect square or not. Rearranging the terms: x2 + 25 - 10x = x2 + 5 × 5 - 2 × 5 × x = x2 - 2 × 5 × x + 5 × 5 Using the perfect square formula. (a - b)2 = (a2- 2ab + b2) Comparing the values, x2 - 2 × 5 × x + 5 × 5 = (x - 5)2 Answer: x2 + 25 - 10x is perfect square. Example 3: Simplify the following using the perfect square formula. (7x - 2y)2 Solution: a = 7x and b = 2y Using perfect square formula (a - b)2 = a2 - 2ab + b2 (7x)2 - 2(7x)(2y) + (2y)2 49x2 - 28xy + 4y2 Answer: (7x - 2y)2 = 49x2 - 28xy + 4y2. FAQs on Perfect Square Formula What Is the Expansion of Perfect Square Formula? The expansion of the perfect square formula is expressed as (a + b)2 = a2 + 2ab + b2. What Is Are the Two Perfect Squares Formula in Algebra? The two perfect squares formula in algebra are (a + b)2 and (a - b)2. These two can be read as a plus b whole square or a minus b whole square. These two perfect squares formulas are expressed as (a + b)2 = a2 + 2ab + b2. How To Represent the Perfect Square Formula? The perfect square formula is represented in form of two terms such as (a + b)2 . The expansion of the perfect square formula is expressed as (a + b)2 = a2 + 2ab + b2. How To Use the Perfect Square Formula? The following steps are followed while using the perfect square formula.
190518	https://math.dartmouth.edu/~carlp/PDF/lambda030205.pdf	THE ITERATED CARMICHAEL λ-FUNCTION AND THE NUMBER OF CYCLES OF THE POWER GENERATOR GREG MARTIN AND CARL POMERANCE 1. INTRODUCTION A common pseudorandom number generator is the power generator: x 7→xℓ(mod n). Here, ℓ, n are ﬁxed integers at least 2, and one constructs a pseudorandom sequence by starting at some residue mod n and iterating this ℓth power map. (Because it is the easiest to compute, one often takes ℓ= 2; this case is known as the BBS generator, for Blum, Blum, and Shub.) To be a good generator, the period should be large. Of course, the pe-riod depends somewhat on the number chosen for the initial value. However, a universal upper bound for this period is λ(λ(n)) where λ is Carmichael’s function. Here, λ(m) is deﬁned as the order of the largest cyclic subgroup of the multiplicative group (Z/mZ)×. It may be computed via the identity λ(lcm{a, b}) = lcm{λ(a), λ(b)} and its values at prime powers: with φ being Euler’s function, λ(pa) = φ(pa) = (p −1)pa−1 for every odd prime power pa and for 2 and 4, and λ(2a) = φ(2a)/2 = 2a−2 for a ≥3. Statistical properties of λ(n) were studied by Erd˝ os, Schmutz, and the second author in , and in particular, they showed that λ(n) = n/ exp((1 + o(1)) log log n log log log n) as n →∞through a certain set of integers of asymptotic density 1. This does not quite pinpoint the normal order of λ(n) (even the sharper version of this theorem from falls short in this regard), but it is certainly a step in this direction, and does give the normal order of the function log(n/λ(n)). In this paper we prove a result of similar quality for the function λ(λ(n)), which we have seen arises in connection with the period of the power generator. We obtain the same expression as with λ(n), except that the log log n is squared. That is, λ(λ(n)) = n/ exp((1 + o(1))(log log n)2 log log log n) almost always. We are able to use this result to say something nontrivial about the number of cycles for the power generator. This problem has been considered in several papers, including , , and . We show that for almost all integers n, the number of cycles for the ℓth power map modulo n is at least exp((1 + o(1))(log log n)2 log log log n), and we conjecture that this lower bound is actually the truth. Under the assumption of the Generalized Riemann Hypothesis (GRH), and using a new result of Kurlberg and the second author , we prove our conjecture. (By the GRH, we mean the Riemann Hypothesis for Kummerian ﬁelds as used by Hooley in his celebrated conditional proof of the Artin conjecture.) For an arithmetic function f(n) whose values are in the natural numbers, let fk(n) de-note the kth iterate of f evaluated at n. One might ask about the normal behavior of λk(n) for k ≥3. Here we make a conjecture for each ﬁxed k. We also brieﬂy consider the func-tion L(n) deﬁned as the least k such that λk(n) = 1. A similar undertaking was made G.M. is supported in part by the National Sciences and Engineering Research Council of Canada. C.P. is supported in part by the National Science Foundation. 1 2 GREG MARTIN AND CARL POMERANCE by Erd˝ os, Granville, Spiro, and the second author in for the function F(n) deﬁned as the least k with φk(n) = 1. Though λ is very similar to φ, the behavior of L(n) and F(n) seem markedly different. We know that F(n) is always of order of magnitude log n, and it is shown in , assuming the Elliott–Halberstam conjecture on the average distribution of primes in arithmetic progressions with large moduli, that in fact F(n) ∼α log n on a set of asymptotic density 1 for a particular positive constant α. We know far less about L(n), not even its typical order of magnitude. We raise the possibility that it is normally of order log log n and show that it is bounded by this order inﬁnitely often. A more formal statement of our results follows. Theorem 1. The normal order of log n/λ(λ(n)) is (log log n)2 log log log n. That is, λ(λ(n)) = n exp −(1 + o(1))(log log n)2 log log log n as n →∞through a set of integers of asymptotic density 1. We actually prove the slightly stronger result: given any function ψ(n) going to inﬁnity arbitrarily slowly, we have λ(λ(n)) = n exp −(log log n)2(log log log n + O(ψ(n))) for almost all n. Given integers ℓ, n ≥2, let C(ℓ, n) denote the number of cycles when iterating the modular power map x 7→xℓ(mod n). Theorem 2. Given any ﬁxed integer ℓ≥2, there is a set of integers of asymptotic density 1 such that as n →∞through this set, C(ℓ, n) ≥exp (1 + o(1))(log log n)2 log log log n . (1) Further, if ε(n) tends to 0 arbitrarily slowly, we have C(ℓ, n) ≤n1/2−ε(n) for almost all n. More-over, for a positive proportion of integers n we have C(ℓ, n) ≤n.409. Finally, if the Generalized Riemann Hypothesis (GRH) is true, we have equality in (1) on a set of integers n of asymptotic density 1. Conjecture 3. The normal order of log(n/λk(n)) is (1/(k −1)!)(log log n)k log log log n. That is, for each ﬁxed integer k ≥1, λk(n) = n exp − 1 (k −1)! + o(1) (log log n)k(log log log n) for almost all n. Deﬁne L(n) to be the number of iterations of λ required to take n to 1, that is, L(n) equals the smallest nonnegative integer k such that λk(n) = 1. Theorem 4. There are inﬁnitely many integers n such that L(n) < (1/ log 2 + o(1)) log log n. 2. NOTATION, STRATEGY, AND PRELIMINARIES The proof of Theorem 1, our principal result, proceeds by comparing the prime divi-sors of λ(λ(n)) with those of φ(φ(n)). The primes dividing φ(m) and λ(m) are always the same. However, this is not always true for φ(φ(m)) and λ(λ(m)). The prime 2 clearly THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 3 causes problems; for example, we have φ(φ(8)) = 2 but λ(λ(8)) = 1. However this prob-lem also arises from the interaction between different primes, for example, φ(φ(91)) = 24 but λ(λ(91)) = 2. We shall use the following notation throughout the paper. The letters p, q, r will always denote primes. Let vq(n) denote the exponent on q in the prime factorization of n, so that n = ∏ q qvq(n) for every positive integer n. We let Pn = {p: p ≡1 (mod n)}. We let x > eee be a real number and y = y(x) = log log x. By ψ(x) we denote a function tending to inﬁnity but more slowly than log log log x = log y. In Sections 2–5, the phrase “for almost all n” always means “for all but O(x/ψ(x)) integers n ≤x”. First we argue that the “large” prime divisors typically do not contribute signiﬁcantly: Proposition 5. For almost all n ≤x, the prime divisors of φ(φ(n)) and λ(λ(n)) that exceed y2 are identical. Proposition 6. For almost all n ≤x, ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) log q ≪y2ψ(x). (2) Next we argue that the contribution of “small” primes to λ(λ(n)) is typically small: Proposition 7. For almost all n ≤x, we have ∑ q≤y2 vq(λ(λ(n))) log q ≪y2ψ(x). Finally, we develop an understanding of the typical contribution of small primes to φ(φ(n)) by comparing it to the additive function h(n) deﬁned by h(n) = ∑ p\|n ∑ r\|p−1 ∑ q≤y2 vq(r −1) log q. (3) Proposition 8. For almost all n ≤x, ∑ q≤y2 vq(φ(φ(n))) log q = h(n) + O(y log y · ψ(x)). Proposition 9. For almost all n ≤x, we have h(n) = y2 log y + O(y2). Proof of Theorem 1. Let x be a sufﬁciently large real number. For any positive integer n ≤x we may write log n λ(λ(n)) = log n φ(n) + log φ(n) φ(φ(n)) + log φ(φ(n)) λ(λ(n)) . Recall that n/φ(n) ≪log log n, and so the ﬁrst two terms are both O(log log log x). Thus, it sufﬁces to show that log φ(φ(n)) λ(λ(n)) = (log log x)2(log log log x + O(ψ(x))) = y2 log y + O(y2ψ(x)) (4) 4 GREG MARTIN AND CARL POMERANCE for almost all n ≤x. We write log φ(φ(n)) λ(λ(n)) = ∑ q vq(φ(φ(n))) −vq(λ(λ(n))) log q = ∑ q≤y2 vq(φ(φ(n))) log q −∑ q≤y2 vq(λ(λ(n))) log q (5) + ∑ q>y2 vq(φ(φ(n))) −vq(λ(λ(n))) log q. Since λ(λ(n)) always divides φ(φ(n)), the coefﬁcients of log q in this last sum are all nonnegative. On the other hand, Proposition 5 tells us that for almost all n ≤x, whenever vq(φ(φ(n))) > 0 we have vq(λ(λ(n))) > 0 as well. Therefore the primes q for which vq(φ(φ(n))) ≤1 do not contribute to this last sum at all, that is, 0 ≤∑ q>y2 vq(φ(φ(n))) −vq(λ(λ(n))) log q = ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) −vq(λ(λ(n))) log q ≤ ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) log q ≪y2ψ(x) for almost all n ≤x by Propositions 5 and 6. Moreover, Proposition 7 tells us that the second sum on the right-hand side of equation (5) is O(y2ψ(x)) for almost all n ≤x. Therefore equation (5) becomes log φ(φ(n)) λ(λ(n)) = ∑ q≤y2 vq(φ(φ(n))) log q + O(y2ψ(x)) for almost all n ≤x. By Proposition 8, the sum on the right-hand side can be replaced by h(n) for almost all n ≤x, the error O(y log y · ψ(x)) in that proposition being absorbed into the existing error O(y2ψ(x)). Finally, Proposition 9 tells us that h(n) = y2 log y + O(y2) for almost all n ≤x. We conclude that equation (4) is satisﬁed for almost all n ≤x, which establishes the theorem. □ Given integers a and n, recall that π(t; n, a) denotes the number of primes up to t that are congruent to a (mod n). The Brun–Titchmarsh inequality (see [10, Theorem 3.7]) states that π(t; n, a) ≪ t φ(n) log(t/n) (6) for all t > n. We use repeatedly a weak form of this inequality, valid for all t > ee, ∑ p≤t p∈Pn 1 p ≪log log t φ(n) , (7) THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 5 which follows from the estimate (6) with a = 1 by partial summation. When n/φ(n) is bounded, this estimate simpliﬁes to ∑ p≤t p∈Pn 1 p ≪log log t n . (8) For example, we shall employ this last estimate when n is a prime or a prime power and when n is the product of two primes or prime powers; in these cases we have n/φ(n) ≤3. We also quote the fact (see Norton or the paper of the second author) that ∑ p∈Pn p≤t 1 p = log log t φ(n) + O log n φ(n) . (9) This readily implies that ∑ p∈Pn p≤t 1 p −1 = log log t φ(n) + O log n φ(n) (10) as well, since (noting that the smallest possible term in the sum is p = n + 1) the difference equals ∑ p∈Pn p≤t 1 (p −1)p ≤ ∞ ∑ i=1 1 in(in + 1) ≪1 n2 . We occasionally use the Chebyshev upper bound ∑ p≤z log p ≤∑ n≤z Λ(n) ≪z, (11) where Λ(n) is the von Mangoldt function, as well as the weaker versions ∑ p≤z log p p ≪log z, ∑ p≤z log2 p p ≪log2 z (12) and the tail estimates ∑ p>z log p p2 ≪1 z, ∑ p>z 1 p2 ≪ 1 z log z, (13) each of which can be derived from the estimate (11) by partial summation. We shall also need at one point a weak form of the asymptotic formula of Mertens, ∑ p≤z log p p = log z + O(1). (14) For any polynomial P(x), we also note the series estimate ∞ ∑ a=0 P(a) ma ≪P 1 6 GREG MARTIN AND CARL POMERANCE uniformly for m ≥2, valid since the series ∑∞ a=0 P(a)za converges uniformly for \|z\| ≤1 2. The estimates ∑ a∈N P(a) ma ≪P 1 m, ∑ a∈N ma>z P(a) ma ≪P 1 z, (15) valid uniformly for any integer m ≥2, follow easily by factoring out the ﬁrst denominator occurring in each sum. 3. LARGE PRIMES DIVIDING φ(φ(n)) AND λ(λ(n)) Proof of Proposition 5. If q is any prime, then q divides φ(φ(n)) if and only if at least one of the following criteria holds: • q3 \| n, • there exists p ∈Pq2 with p \| n, • there exists p ∈Pq with p2 \| n, • there exist r ∈Pq and p ∈Pr with p \| n, • q2 \| n and there exists p ∈Pq with p \| n, • there exist distinct p1, p2 ∈Pq with p1p2 \| n. In the ﬁrst four of these six cases, it is easily checked that q \| λ(λ(n)) as well. (This is not quite true for q = 2, but in this proof we shall only consider primes q > y2.) Therefore we can estimate the number of integers n ≤x for which q divides φ(φ(n)) but not λ(λ(n)) as follows: ∑ n≤x q\|φ(φ(n)) q∤λ(λ(n)) 1 ≤∑ p∈Pq ∑ n≤x q2p\|n 1 + ∑ p1∈Pq ∑ p2∈Pq p2̸=p1 ∑ n≤x p1p2\|n 1 ≤∑ p∈Pq x q2p + ∑ p1∈Pq ∑ p2∈Pq x p1p2 . Using three applications of the Brun–Titchmarsh inequality (8), we conclude that for any odd prime q, ∑ n≤x q\|φ(φ(n)) q∤λ(λ(n)) 1 ≪xy q3 + xy2 q2 ≪xy2 q2 . Consequently, by the tail estimate (13) and the condition ψ(x) = o(log y), ∑ q>y2 ∑ n≤x q\|φ(φ(n)) q∤λ(λ(n)) 1 ≪xy2 ∑ q>y2 1 q2 ≪ xy2 y2 log y2 < x log y ≪ x ψ(x). Therefore for almost all n ≤x, every prime q > y2 dividing φ(φ(n)) also divides λ(λ(n)), as asserted. □ Lemma 10. Given a real number x ≥3 and a prime q > y2, deﬁne Sq = Sq(x) to be the set of all integers n ≤x for which at least one of the following criteria holds: • q2 \| n, • there exists p ∈Pq2 with p \| n, • there exist distinct p1, p2 ∈Pq with p1p2 \| n, • there exist r ∈Pq2 and p ∈Pr with p \| n, THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 7 • there exist distinct r1, r2, r3 ∈Pq and p ∈Pr1r2r3 with p \| n, • there exist distinct r1, r2, r3, r4 ∈Pq, p1 ∈Pr1r2, and p2 ∈Pr3r4 with p1p2 \| n. Then the cardinality of Sq is O(xy2/q2). Note that if q2 \| φ(n), then at least one of the ﬁrst three of the six conditions in the statement of the lemma must be satisﬁed. Proof. The number of integers up to x for which any particular one of the six criteria holds is easily shown to be O(xy2/q2). For the sake of conciseness, we show the details of this calculation only for the last criterion, which is the most complicated. The number of integers n up to x for which there exist distinct r1, r2, r3, r4 ∈Pq, p1 ∈Pr1r2, and p2 ∈Pr3r4 with p1p2 \| n is at most ∑ r1,r2,r3,r4∈Pq ∑ p1∈Pr1r2 p2∈Pr3r4 ∑ n≤x p1p2\|n 1 ≤ ∑ r1,r2,r3,r4∈Pq ∑ p1∈Pr1r2 p2∈Pr3r4 x p1p2 . Using six applications of the Brun–Titchmarsh estimate (8), we have ∑ r1,r2,r3,r4∈Pq ∑ p1∈Pr1r2 p2∈Pr3r4 x p1p2 ≪ ∑ r1,r2,r3,r4∈Pq xy2 r1r2r3r4 ≪xy6 q4 < xy2 q2 , the last inequality being valid due to the hypothesis q > y2. □ Proof of Proposition 6. Deﬁne S = S(x) to be the union of Sq over all primes q > y2, where Sq is deﬁned as in the statement of Lemma 10. Using #A to denote the cardinality of a set A, Lemma 10 implies that #S ≤∑ q>y2 #Sq ≪∑ q>y2 xy2 q2 ≪ xy2 y2 log y2 ≪ x ψ(x) by the tail estimate (13) and the condition ψ(x) = o(log y). Therefore to prove that the estimate (2) holds for almost all integers n ≤x, it sufﬁces to prove that it holds for almost all integers n ≤x that are not in the set S. This in turn is implied by the upper bound ∑ n≤x n/ ∈S ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) log q ≪xy2, (16) which we proceed now to establish. Fix a prime q > y2 and an integer a ≥2 for the moment. In general, there are many ways in which qa could divide φ(φ(n)), depending on the power to which q divides n itself, the power to which q divides numbers of the form p −1 with p \| n, and so forth. However, for integers n / ∈S, most of these various possibilities are ruled out by one of the six criteria deﬁning the sets Sq. In fact, for n / ∈S, there are only two ways for qa to divide φ(φ(n)): • there are distinct r1, . . . , ra ⊂Pq and distinct p1 ∈Pr1, . . . , pa ∈Pra with p1 . . . pa\|n, • there are distinct r1, . . . , ra ⊂Pq, distinct p1 ∈Pr1, . . . , pa−2 ∈Pra−2, and p ∈Pra−1ra with p1 . . . pa−2p\|n. 8 GREG MARTIN AND CARL POMERANCE (We refer to the former case as the “supersquarefree” case.) Still considering q and a ﬁxed, the number of integers n up to x satisfying each of these two conditions is at most ∑ r1,...,ra∈Pq 1 a! ∑ p1∈Pr1 ... pa∈Pra ∑ n≤x p1...pa\|n 1 ≤ ∑ r1,...,ra∈Pq 1 a! ∑ p1∈Pr1 ... pa∈Pra x p1 . . . pa and ∑ r1,...,ra∈Pq 1 2!(a −2)! ∑ p1∈Pr1 ... pa−2∈Pra−2 p∈Pra−1ra ∑ n≤x p1...pa−2p\|n 1 ≤ ∑ r1,...,ra∈Pq 1 (a −2)! ∑ p1∈Pr1 ... pa−2∈Pra−2 p∈Pra−1ra x p1 . . . pa−2p, respectively, the factors 1/a! and 1/2!(a −2)! coming from the various possible permu-tations of the primes ri. Letting c ≥1 be the constant implied in the Brun–Titchmarsh inequality (8) as applied to moduli n that are divisible by at most two distinct primes, we see that ∑ r1,...,ra∈Pq 1 a! ∑ p1∈Pr1 ... pa∈Pra x p1 . . . pa ≤ ∑ r1,...,ra∈Pq 1 a! x(cy)a r1 . . . ra ≤x(cy)2a a!qa and ∑ r1,...,ra∈Pq 1 (a −2)! ∑ p1∈Pr1 ... pa−2∈Pra−2 p∈Pra−1ra x p1 . . . pa−2p ≤ ∑ r1,...,ra∈Pq 1 (a −2)! x(cy)a−1 r1 . . . ra ≤x(cy)2a−1 (a −2)!qa . Therefore the number of integers n ≤x such that n / ∈S and qa \| φ(φ(n)) is ≤x(cy)2a a!qa + x(cy)2a−1 (a −2)!qa < c2axy4 (a −2)!q2 , (17) where we have used the assumption q > y2. We now establish the estimate (16). Note that ∑ n≤x n/ ∈S ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) log q ≤2 ∑ n≤x n/ ∈S ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) −1 log q = 2 ∑ q>y2 log q ∑ a≥2 ∑ n≤x n/ ∈S qa\|φ(φ(n)) 1. Therefore, using the bound (17) for each pair q and a, ∑ n≤x n/ ∈S ∑ q>y2 vq(φ(φ(n)))≥2 vq(φ(φ(n))) log q ≤2 ∑ q>y2 log q ∑ a≥2 c2axy4 (a −2)!q2 = 2c4ec2xy4 ∑ q>y2 log q q2 ≪xy4 y2 = xy2 THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 9 by the tail estimate (13). This establishes the estimate (16) and hence the proposition. □ 4. SMALL PRIMES AND THE REDUCTION TO h(n) Lemma 11. For any prime power qa, the number of positive integers n ≤x for which qa divides λ(λ(n)) is O(xy2/qa). Proof. When q is an odd prime, the prime power qa divides λ(λ(n)) if and only if at least one of the following criteria holds: • qa+2 \| n, • there exists p ∈Pqa+1 with p \| n, • there exists p ∈Pqa with p2 \| n, • there exist r ∈Pqa and p ∈Pr with p \| n. Even when q = 2, at least one of these four conditions must hold for qa to divide λ(λ(n)), although they are not quite sufﬁcient. In either case, we still have the upper bound ∑ n≤x qa\|λ(λ(n)) 1 ≤∑ n≤x qa+2\|n 1 + ∑ p∈Pqa+1 ∑ n≤x p\|n 1 + ∑ p∈Pqa ∑ n≤x p2\|n 1 + ∑ r∈Pqa ∑ p∈Pr ∑ n≤x p\|n 1 ≤ x qa+2 + ∑ p∈Pqa+1 p≤x x p + ∑ p∈Pqa p≤√x x p2 + ∑ r∈Pqa ∑ p∈Pr p≤x x p. (18) In the second of these three sums, it is sufﬁcient to notice that any p ∈Pqa must exceed qa, which leads to the estimate ∑ p∈Pqa p≤√x x p2 < ∑ m>qa x m2 < x qa . To bound the ﬁrst and third sums in (18), we invoke the Brun–Titchmarsh estimate (8) a total of three times: ∑ p∈Pqa+1 p≤x x p ≪ xy qa+1 ∑ r∈Pqa ∑ p∈Pr p≤x x p ≪∑ r∈Pqa r≤x xy r ≪xy2 qa . Using these three estimates, (18) gives ∑ n≤x qa\|λ(λ(n)) 1 ≪ x qa+2 + xy qa+1 + x qa + xy2 qa ≪xy2 qa , which establishes the lemma. □ Proof of Proposition 7. We have ∑ q≤y2 vq(λ(λ(n))) log q = ∑ q≤y2 log q ∑ a∈N qa\|λ(λ(n)) 1 ≤∑ q≤y2 log q ∑ a∈N qa≤y2 1 + ∑ q≤y2 log q ∑ a∈N qa>y2 qa\|λ(λ(n)) 1. 10 GREG MARTIN AND CARL POMERANCE Since the ﬁrst sum is simply ∑ q≤y2 log q ∑ a∈N qa≤y2 1 = ∑ m≤y2 Λ(m) ≪y2 by the Chebyshev estimate (11), we have uniformly for n ≤x, ∑ q≤y2 vq(λ(λ(n))) log q ≪y2 + ∑ q≤y2 log q ∑ a∈N qa>y2 qa\|λ(λ(n)) 1. (19) To show that this quantity is usually small, we sum this last double sum over n and apply Lemma 11, yielding ∑ n≤x ∑ q≤y2 log q ∑ a∈N qa>y2 qa\|λ(λ(n)) 1 = ∑ q≤y2 log q ∑ a∈N qa>y2 ∑ n≤x qa\|λ(λ(n)) 1 ≪∑ q≤y2 log q ∑ a∈N qa>y2 xy2 qa . Using the geometric series sum (15) and the Chebyshev estimate (11), this becomes ∑ n≤x ∑ q≤y2 log q ∑ a∈N qa>y2 qa\|λ(λ(n)) 1 ≪∑ q≤y2 log q · xy2 y2 ≪xy2. Therefore if we sum both sides of (19) over n, we obtain ∑ n≤x ∑ q≤y2 vq(λ(λ(n))) log q ≪xy2. This implies that for almost all n ≤x, we have ∑ q≤y2 vq(λ(λ(n))) log q ≪y2ψ(x), as desired. □ Proof of Proposition 8. Fix a prime q for the moment. For any positive integer m, the usual formula for φ(m) readily implies vq(φ(m)) = max{0, vq(m) −1} + ∑ p\|m vq(p −1), which we use in the form ∑ p\|m vq(p −1) ≤vq(φ(m)) ≤∑ p\|m vq(p −1) + vq(m). Using these inequalities twice, ﬁrst with m = φ(n) and then with m = n, we see that ∑ p\|φ(n) vq(p −1) ≤vq(φ(φ(n))) ≤∑ p\|φ(n) vq(p −1) + vq(φ(n)) ≤∑ p\|φ(n) vq(p −1) + ∑ p\|n vq(p −1) + vq(n). (20) THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 11 Now a prime r divides φ(n) if and only if either r2 \| n or there exists a prime p \| n such that r \| p −1. Therefore ∑ p\|n ∑ r\|p−1 vq(r −1) ≤∑ r\|φ(n) vq(r −1) ≤∑ p\|n ∑ r\|p−1 vq(r −1) + ∑ r: r2\|n vq(r −1), the latter inequality accounting for the possibility that both criteria hold for some prime r. When we combine these inequalities with those in equation (20) and subtract the double sum over p and r throughout, we obtain 0 ≤vq(φ(φ(n))) −∑ p\|n ∑ r\|p−1 vq(r −1) ≤∑ r: r2\|n vq(r −1) + ∑ p\|n vq(p −1) + vq(n) ≤2∑ p\|n vq(p −1) + vq(n). Now we multiply through by log q and sum over all primes q ≤y2 to conclude that for any positive integer n, 0 ≤∑ q≤y2 vq(φ(φ(n))) log q −h(n) ≤2 ∑ q≤y2∑ p\|n vq(p −1) log q + ∑ q≤y2 vq(n) log q. It remains to show that the right-hand side of this last inequality is O(y log y · ψ(x)) for almost all n ≤x, which we accomplish by establishing the estimate ∑ n≤x ∑ q≤y2∑ p\|n vq(p −1) log q + ∑ n≤x ∑ q≤y2 vq(n) log q ≪xy log y. (21) We may rewrite the ﬁrst term on the left-hand side as ∑ n≤x ∑ q≤y2∑ p\|n vq(p −1) log q = ∑ n≤x ∑ q≤y2∑ p\|n ∑ a∈N qa\|p−1 log q = ∑ q≤y2 log q ∑ a∈N ∑ p∈Pqa ∑ n≤x p\|n 1 ≤∑ q≤y2 log q ∑ a∈N ∑ p∈Pqa x p. Using the Brun–Titchmarsh inequality (8) and the geometric series estimate (15), we ob-tain ∑ n≤x ∑ q≤y2∑ p\|n vq(p −1) log q ≪x ∑ q≤y2 log q ∑ a∈N y qa ≪xy ∑ q≤y2 log q q ≪xy log y2. The second term on the left-hand side of (21) is even simpler: we have ∑ n≤x ∑ q≤y2 vq(n) log q = ∑ q≤y2 log q ∑ a∈N ∑ n≤x qa\|n 1 ≤∑ q≤y2 log q ∑ a∈N x qa , and using the geometric series bound (15) and the weak Chebyshev estimate (12) yields ∑ n≤x ∑ q≤y2 vq(n) log q ≪x ∑ q≤y2 log q q ≪x log y2. The last two estimates therefore establish (21) and hence the proposition. □ 12 GREG MARTIN AND CARL POMERANCE 5. THE NORMAL ORDER OF h(n) Recall the deﬁnition (3): h(n) = ∑p\|n ∑r\|p−1 ∑q≤y2 vq(r −1) log q. We now calculate the normal order of the additive function h(n) via the Tur´ an–Kubilius inequality (see , Lemma 3.1). If we deﬁne M1(x) = ∑ p≤x h(p) p , M2(x) = ∑ p≤x h(p)2 p , then the Tur´ an-Kubilius inequality asserts that ∑ n≤x (h(n) −M1(x))2 ≪xM2(x). (22) Proposition 12. We have M1(x) = y2 log y + O(y2) for all x > eee. Proposition 13. We have M2(x) ≪y3 log2 y for all x > eee. Proof of Proposition 9. Let N denote the number of n ≤x for which \|h(n) −M1(x)\| > y2. The contribution of such n to the sum in (22) is at least y4N. Thus, Proposition 13 implies that N ≪x(log y)2/y. Hence, Proposition 12 implies that h(n) = y2 log y + O(y2) for all n ≤x but for a set of size O(x(log y)2)/y). This proves Proposition 9. □ To calculate M1(x) and M2(x) we shall ﬁrst calculate ∑p≤t h(p) and ∑p≤t h(p)2 and then account for the weights 1/p using partial summation. We begin the evaluation of ∑p≤t h(p) with a lemma. Lemma 14. Let b be a positive integer and t > ee a real number. (a) If b > t1/4 then ∑ r∈Pb π(t; r, 1) ≪t log t b . (b) If b ≤t1/4 then ∑ r∈Pb r>t1/3 π(t; r, 1) ≪ bt φ(b)2 log t. and ∑ r∈Pb π(t; r, 1) ≪t log log t φ(b) log t Remark. The exponents 1 4 and 1 3 are rather arbitrary and chosen only for simplicity; any two exponents 0 < α < β < 1 2 would do equally well. Proof. Notice that in all three sums, the only contributing terms are those with r > b and r < t. If b > t1/4, then the trivial bound π(t; r, 1) ≤t/r gives ∑ r∈Pb π(t; r, 1) ≤ ∑ r∈Pb t1/4t1/3 π(t; r, 1) = #{(m, r): r ≡1 (mod b), r > t1/3, mr + 1 ≤t, mr + 1 and r both prime} ≤∑ m<t2/3 #{r < t m : r ≡1 (mod b), mr + 1 and r both prime} ≪∑ mt1/3 π(t; r, 1) ≪∑ m eee and t > ee, we have ∑ p≤t h(p) = 2t log log t log y log t + O t log log t log t + t log2 y log t + t3/4 log t · y2 . Remark. In particular, we have ∑p≤x h(p) ≪x log log x log y/ log x = xy log y/ log x. Proof. We may rewrite ∑ p≤t h(p) = ∑ p≤t ∑ r\|p−1 ∑ q≤y2 vq(r −1) log q = ∑ p≤t ∑ r\|p−1 ∑ q≤y2 ∑ a∈N qa\|r−1 log q = ∑ q≤y2 log q ∑ a∈N ∑ r: qa\|r−1 ∑ p≤t r\|p−1 1 = ∑ q≤y2 log q ∑ a∈N ∑ r∈Pqa π(t; r, 1). (24) The main contribution to this triple sum comes from the terms with qa ≤t1/4 and r ≤t1/3. In fact, using Lemma 14(a) we can bound the contribution from the terms with qa large 14 GREG MARTIN AND CARL POMERANCE by ∑ q≤y2 log q ∑ a∈N qa>t1/4 ∑ r∈Pqa π(t; r, 1) ≪∑ q≤y2 log q ∑ a∈N qa>t1/4 t log t qa ≪t log t ∑ q≤y2 log q t1/4 ≪t3/4 log t · y2, where the last two estimates are due to the geometric series bound (15) and the Chebyshev bound (11). Similarly, using the ﬁrst half of Lemma 14(b) we can bound the contribution from the terms with qa small and r large by ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r>t1/3 π(t; r, 1) ≪∑ q≤y2 log q ∑ a∈N qa≤t1/4 t qa log t ≪ t log t ∑ q≤y2 log q q ≪t log y log t , where again the last two estimates are due to the geometric series bound (15) and the weak Chebyshev bound (12). In light of these two estimates, equation (24) becomes ∑ p≤t h(p) = ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 π(t; r, 1) + O t3/4 log t · y2 + t log y log t . (25) Deﬁne E(t; r, 1) = π(t; r, 1) −li(t)/(r −1). We have ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 π(t; r, 1) = ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 li(t) r −1 + E(t; r, 1) = ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 li(t) r −1 + O ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 \|E(t; r, 1)\| . (26) Let Ω(m) denote the number of divisors of m that are primes or prime powers. Using the estimate Ω(m) ≪log m, we quickly dispose of ∑ q≤y2 log q ∑ a∈N qa≤t1/4 ∑ r∈Pqa r≤t1/3 \|E(t; r, 1)\| ≤2 log y ∑ r≤t1/3 \|E(t; r, 1)\| ∑ q≤y2 ∑ a∈N qa\|r−1 1 ≤2 log y ∑ r≤t1/3 \|E(t; r, 1)\| Ω(r −1) ≪log y log t ∑ r≤t1/3 \|E(t; r, 1)\| ≪t log y log t by the Bombieri–Vinogradov theorem (we could equally well put any power of log t in the denominator of the ﬁnal expression if we needed). Inserting this estimate into equation (26), we see that equation (25) becomes ∑ p≤t h(p) = li(t) ∑ q≤y2 log q ∑ a∈N ∑ r∈Pqa r≤t1/3 1 r −1 + O t3/4 log t · y2 + t log y log t . (27) THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 15 We have by equation (10) ∑ q≤y2 log q ∑ a∈N ∑ r∈Pqa r≤t1/3 1 r −1 = ∑ q≤y2 log q ∑ a∈N log log t1/3 φ(qa) + O log qa qa = (log log t + O(1)) ∑ q≤y2 log q ∑ a∈N 1 qa + O 1 qa+1 + O ∑ q≤y2 log2 q ∑ a∈N a qa = (log log t + O(1)) ∑ q≤y2 log q q + O log q q2 + O ∑ q≤y2 log2 q q , using the geometric series estimate (15). Using the Mertens formula (14) to evaluate the main term and the weak Chebyshev estimates (12) to bound the error terms, we see that ∑ q≤y2 log q ∑ a∈N ∑ r∈Pqa r≤t1/3 1 r −1 = log log t log y2 + O(log y + log log t + log2 y). We conclude from equation (27) and the fact that li(t) = t/ log t + O(t/ log2 t) that ∑ p≤t h(p) = li(t) log log t log y2 + O(log y + log log t + log2 y) + O t3/4 log t · y2 + t log y log t = 2t log log t log y log t + O t log log t log t + t log2 y log t + t3/4 log t · y2 , as asserted. □ Proof of Proposition 12. In an explicit example of the technique of partial summation, we write M1(x) = ∑ p≤x h(p) p = ∑ p≤ee h(p) p + ∑ ee eee and t > ee, we have ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N ∑ r∈Pqa1 1 ∩Pqa2 2 ∑ p≤t p≡1 (mod r) 1 ≪t7/8 log t · y2 log y + t log log t · log2 y log t . Proof. Since the exact form of Pq a1 1 ∩Pq a2 2 depends on whether or not q1 = q2, we split the expression in question into two separate sums: ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N ∑ r∈Pqa1 1 ∩Pqa2 2 ∑ p≤t p≡1 (mod r) 1 (28) = ∑ q≤y2 log2 q ∑ a1,a2∈N ∑ r∈Pqmax{a1,a2} π(t; r, 1) + ∑ q1,q2≤y2 q1̸=q2 log q1 log q2 ∑ a1,a2∈N ∑ r∈Pqa1 1 qa2 2 π(t; r, 1). Noting that there are exactly 2a −1 ordered pairs (a1, a2) for which max{a1, a2} = a, we have ∑ q≤y2 log2 q ∑ a1,a2∈N ∑ r∈Pqmax{a1,a2} π(t; r, 1) = ∑ q≤y2 log2 q ∑ a∈N (2a −1) ∑ r∈Pqa π(t; r, 1) ≪∑ q≤y2 log2 q ∑ a∈N qa>t1/4 at log t qa + ∑ q≤y2 log2 q ∑ a∈N qa≤t1/4 at log log t qa log t by Lemma 14. Since ∑ q≤y2 log2 q ∑ a∈N qa>t1/4 at log t qa ≪t log t log y2 ∑ q≤y2 log q t1/4 ≪t3/4 log t · y2 log y THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 17 by the Chebyshev bound (11), and ∑ q≤y2 log2 q ∑ a∈N qa≤t1/4 at log log t qa log t ≪t log log t log t ∑ q≤y2 log2 q q ≪t log log t · log2 y log t by (11) and its weaker version (12), the ﬁrst term on the right-hand side of equation (28) is bounded by the estimate asserted in the statement of the lemma. It remains to satisfactorily bound the second term on the right-hand side of equation (28). Again dividing the sum so that Lemma 14 can be applied, we have ∑ q1,q2≤y2 q1̸=q2 log q1 log q2 ∑ a1,a2∈N ∑ r∈Pqa1 1 qa2 2 π(t; r, 1) ≪ ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 q a2 2 >t1/4 t log t qa1 1 qa2 2 + ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 q a2 2 ≤t1/4 t log log t qa1 1 qa2 2 log t. In the ﬁrst of these two terms, at least one of the qai i must exceed t1/8, and so using the estimates (15), (11), and (12) we see that ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 q a2 2 >t1/4 t log t qa1 1 qa2 2 ≤2t log t ∑ q1≤y2 log q1 ∑ a1∈N q a1 1 >t1/8 1 qa1 1 ∑ q2≤y2 log q2 ∑ a2∈N 1 qa2 2 ≪t log t ∑ q1≤y2 log q1 t1/8 ∑ q2≤y2 log q2 q2 ≪t7/8 log t · y2 log y. In the second, we simply ignore the restriction qa1 1 qa2 2 ≤t1/4 and use the estimates (15) and (12), obtaining ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N t log log t qa1 1 qa2 2 log t = t log log t log t ∑ q≤y2 log q ∑ a∈N 1 qa 2 ≪t log log t log t ∑ q≤y2 log q q 2 ≪t log log t · log2 y log t . This concludes the proof of the lemma. □ The following lemma is similar in spirit to Lemma 14 but is a bit more complicated to state and prove. Lemma 17. Let b1 and b2 be positive integers and t > ee a real number. (a) If b1 > t1/8 or b2 > t1/8 then ∑ r1∈Pb1 ∑ r2∈Pb2 π(t; r1r2, 1) ≪t log2 t b1b2 . 18 GREG MARTIN AND CARL POMERANCE (b) If neither b1 nor b2 exceeds t1/8 then ∑ r1∈Pb1 ∑ r2∈Pb2 r1r2>t1/3 π(t; r1r2, 1) ≪ b2t log log t φ(b1)φ(b2)2 log t and ∑ r1∈Pb1 ∑ r2∈Pb2 π(t; r1r2, 1) ≪ t(log log t)2 φ(b1)φ(b2) log t. Remark. Again, the values 1/8 and 1/3 for the exponents are rather arbitrary. Proof. The bound in part (a) follows from the trivial estimate π(t; r1r2, 1) ≪t/r1r2, just as in the proof of Lemma 14(a). For the ﬁrst estimate in part (b), we my assume that r1 ≤r2 by symmetry. We use Brun’s method again: ∑ r1∈Pb1 ∑ r2∈Pb2 r1≤r2 r1r2>t1/3 π(t; r1r2, 1) = #{(m, r1, r2): r1 ≡1 (mod b1), r2 ≡1 (mod b2), r1 ≤r2, r1r2 > t1/3, mr1r2 + 1 ≤t, and r1, r2, and mr1r2 + 1 are all prime} ≤∑ m<t2/3 ∑ r1<√ t/m r1∈Pb1 ∑ r2<t/mr1 r2∈Pb2 mr1r2+1 prime 1 ≪∑ m<t2/3 ∑ r1<√ t/m r1∈Pb1 mr1b2 φ(b2)φ(mr1b2) · t/mr1 log2(t/mr1b2) . Notice that t/mr1b2 > ( p t/m)/b2 > t1/6/t1/8 = t1/24, and so ∑ r1∈Pb1 ∑ r2∈Pb2 r1≤r2 r1r2>t1/3 π(t; r1r2, 1) ≪ t log2 t ∑ m<t2/3 ∑ r1<√ t/m r1∈Pb1 b2 φ(b2)2φ(m)φ(r1) ≪ b2t log log t φ(b1)φ(b2)2 log2 t ∑ m t1/8 is ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 >t1/8 ∑ r1∈Pqa1 1 r2∈Pqa2 2 π(t; r1r2, 1) ≪ ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 >t1/8 t log2 t qa1 1 qa2 2 ≪t log2 t ∑ q1≤y2 ∑ a1∈N q a1 1 >t1/8 log q1 qa1 1 ∑ q2≤y2 ∑ a2∈N log q2 qa2 2 ≪t log2 t ∑ q1≤y2 log q1 t1/8 ∑ q2≤y2 log q2 q2 ≪t7/8 log2 t · y2 log y 20 GREG MARTIN AND CARL POMERANCE by Lemma 17(a) and the estimates (15), (11), and (12); the contribution from the terms for which qa2 2 > t1/8 is bounded likewise. The remaining contribution is ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 ,q a2 2 ≤t1/8 ∑ r1∈Pqa1 1 r2∈Pqa2 2 π(t; r1r2, 1) ≪ ∑ q1,q2≤y2 log q1 log q2 ∑ a1,a2∈N q a1 1 ,q a2 2 ≤t1/8 t(log log t)2 qa1 1 qa2 2 log t ≪t(log log t)2 log t ∑ q≤y2 ∑ a∈N log q qa 2 ≪t(log log t)2 log2 y log t by Lemma 17(b) and the estimates (15) and (12). Using both these bounds in equation (29), we conclude that ∑ p≤t h(p)2 ≪t7/8 log t · y2 log y + t(log log t)2 log2 y log t . We now evaluate M2(x) using partial summation. We have M2(x) = ∑ p≤x h(p)2 p = ∑ p≤ee h(p)2 p + 1 x ∑ ee 1, the sequence ui (mod n) is eventually periodic, and we denote the length of the eventual cycle by ord(u, n). So, letting n(u) denote the largest divisor of n coprime to u, we have ord(u, n) = ord(u, n(u)). For example, let u = 2, n = 24. The sequence ui (mod n) is 2, 4, 8, 16, 8, 16, . . . with cycle length 2, and so ord(2, 24) = ord(2, 3) = 2. THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 21 When iterating the ℓth power map modulo n, the length of the eventual cycle starting with x = u is given by ord(ℓ, ord(u, n)). We would like to have a criterion for when a residue is part of some cycle, that is, for when a residue is eventually sent back to itself when iterating x 7→xℓ(mod n). Lemma 18. A residue u is part of some cycle under iteration of the map x 7→xℓ(mod n) if and only if (ℓ, ord(u, n)) = 1 and, with d = (u, n), we have (d, n/d) = 1. Proof. If (u, n) = d, then high powers of u will be ≡0 (mod n/n(d)). Thus, for u to be in a cycle it is necessary that n/n(d) = d, that is, (d, n/d) = 1. Further, it is necessary that (ℓ, ord(u, n)) = 1. Indeed, if σ = ord(u, n), we would need ℓi (mod σ) to be purely periodic, which is equivalent to (ℓ,σ) = 1. This proves the necessity of the condition. For the sufﬁciency, we have just noted that (ℓ,σ) = 1 implies that ℓi (mod σ) is purely periodic. This implies in turn that the sequence uℓi (mod n(u)) is purely periodic. But the condition (d, n/d) = 1 implies that n(u) = n/d, and as each uℓi ≡0 (mod d), we have that uℓi (mod n) is purely periodic. □ For d\|n with (d, n/d) = 1, let Cd(ℓ, n) denote the number of cycles in the ℓth power map mod n that involve residues u with (u, n) = d. For the lower bound in Theorem 2 we shall deal only with C1(ℓ, n), that is, cycles involving numbers coprime to n. Lemma 19. We have C1(ℓ, n) ≥φ(n)(ℓ)/λ(λ(n)). Proof. It is easy to see that the subgroup of (Z/nZ)× of residues u with (ℓ, ord(u, n)) = 1 has size φ(n)(ℓ). (In fact, this is true for any ﬁnite abelian group G: the size of the subgroup of elements with order coprime to ℓis \|G\|(ℓ).) As the length of any cycle in the ℓth power map is bounded above by λ(λ(n)), the lemma follows immediately. □ To investigate the normal size of φ(n)(ℓ), we introduce the function fℓ(n) = ∑ p\|ℓ vp(φ(n)) log p. We also make use of the notation qa∥n, which means that qa is the exact power of q divid-ing n, that is, qa divides n but qa+1 does not. Proposition 20. For any ﬁxed ℓ, we have fℓ(n) ≤(log log n)2 for almost all n, in fact for all but Oℓ(x/ log log x) integers n ≤x. Proof. We have ∑ n≤x fℓ(n) = ∑ p\|ℓ∑ n≤x ∑ qa∥n vp(φ(qa)) log p ≤x∑ p\|ℓ log p ∑ qa≤x vp(φ(qa)) qa ≤x∑ p\|ℓ log p ∑ pa≤x a −1 pa + x∑ p\|ℓ log p ∑ q≤x vp(q −1) q . Now x∑ p\|ℓ log p ∑ pa≤x a −1 pa ≪ℓx 22 GREG MARTIN AND CARL POMERANCE and, by (8), x∑ p\|ℓ log p ∑ q≤x vp(q −1) q = x∑ p\|ℓ log p ∑ a≥1 ∑ q∈Ppa, q≤x 1 q ≪x∑ p\|ℓ log p ∑ a≥1 log log x pa ≪ℓx log log x. Hence, ∑ n≤x fℓ(n) ≪ℓx log log x, so that the number of n ≤x with fℓ(n) > (log log n)2 is Oℓ(x/ log log x). □ It is interesting that one can prove an Erd˝ os–Kac theorem for fℓ(n) using as a tool the criterion of Kubilius–Shapiro (see , ). Proof of the lower bound in Theorem 2. Noting that φ(n)(ℓ) = φ(n)/e fℓ(n), we have φ(n)(ℓ) ≥ φ(n)/ exp((log log n)2) for almost all n by Proposition 20. Of course, n ≥φ(n) ≫ n/ log log n for all n ≥3. Therefore, using Lemma 19 and Theorem 2, we have C(ℓ, n) ≥C1(ℓ, n) ≥φ(n)(ℓ) λ(λ(n)) ≥ φ(n) exp((log log n)2)λ(λ(n)) = φ(n)/n exp((log log n)2) n λ(λ(n)) = exp((1 + o(1))(log log n)2 log log log n) for almost all n. This completes the proof of the lower bound in Theorem 2. □ We now consider the upper bounds in Theorem 2, ﬁrst establishing a lemma. Lemma 21. Suppose m is a positive integer and (d, m) = 1. For any integer j \| λ(m), the number of integers u ∈[1, m] with (u, m) = 1 and ord(du, m) \| λ(m)/j is at most φ(m)/j. Proof. In fact, we prove a more general statement for any ﬁnite abelian group G: let λ(G) denote the exponent of G, that is, the order of the largest cyclic subgroup of G, or equiv-alently the least common multiple of the orders of the elements of G. Then for any d ∈G and any j \| λ(G), the number of elements u ∈G for which the order of du divides λ(G)/j is at most #G/j. It is clear that the lemma follows immediately from this statement upon taking G to be (Z/mZ)×. It is also clear that in this statement, the element d plays no role whatsoever except to shufﬂe the elements of G around, and so we assume without loss of generality that d is the identity of G. Let p be any prime dividing λ(G), and choose a ≤b so that pa∥j and pb∥λ(G). When we write G canonically as isomorphic to the direct product of cyclic groups of prime-power order, at least one of the factors must be isomorphic to Z/pbZ. In every such factor, only one out of every pa elements has order dividing λ(G)/j, since all but pb−a elements of the factor have order divisible by pb−a+1. Since there is at least one such factor for every pa∥j, we conclude that at most one out of every j elements of G has order dividing λ(G)/j, as claimed. □ THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 23 Note that this result in the case d = 1 is Lemma 1 in . The above proof, while similar in spirit to the proof in , is simpler. Let τ(m) denote the number of positive divisors of m. Proposition 22. For any integers ℓ, n ≥2 we have C(ℓ, n) ≤nτ(λ(n))τ(n)/ ord(ℓ, λ(n)). Proof. It is sufﬁcient to show that for each ℓ, n ≥2 and each d \| n with (d, n/d) = 1, we have Cd(ℓ, n) ≤ nτ(λ(n)) ord(ℓ, λ(n)) . (30) Let d \| n with (d, n/d) = 1. We have seen in Lemma 18 that for a residue u (mod n) with (u, n) = d to be involved in a cycle, it is necessary and sufﬁcient that (ℓ, ord(u, n/d)) = 1. For each integer j \| λ(n/d), let Cd,j(ℓ, n) denote the number of cycles corresponding to residues u with (u, n) = d and ord(u, n/d) = λ(n/d)/j. Writing such a residue u as du1, we have u1 ∈[1, n/d] and (u1, n/d) = 1. Thus, by Lemma 21, we have that the number of such residues u is at most φ(n/d)/j ≤n/dj. Hence we have Cd,j(ℓ, n) ≤ n/dj ord(ℓ, λ(n/d)/j) . Now λ(n/d) = λ(n)/d1 for some integer d1 ≤d. It is shown in (15) of that for k \| m we have ord(a, m/k) ≥ord(a, m)/k for any nonzero integer a. Hence ord(ℓ, λ(n/d)/j) = ord(ℓ, λ(n)/d1 j) ≥ord(ℓ, λ(n))/d1 j, so that Cd,j(ℓ, n) ≤ n/dj ord(ℓ, λ(n))/d1 j ≤ n ord(ℓ, λ(n)) . Letting j range over all divisors of λ(n/d), we get that Cd(ℓ, n) ≤ nτ(λ(n/d)) ord(ℓ, λ(n)) , which immediately gives (30). □ Proof of the upper bounds in Theorem 2. Note that from [6, Theorem 4.1], we have τ(λ(n)) < exp((log log n)2) for almost all n. Furthermore, letting Ω(n) denote the number of prime factors of n counted with multiplicity, we know that the normal order of Ω(n) is log log n; in particular, we have Ω(n) < log log n/ log 2 for almost all n. Since the inequality τ(n) ≤ 2Ω(n) is elementary, this implies that τ(n) < log n for almost all n. We conclude from Proposition 22 that C(ℓ, n) < n exp(2(log log n)2)/ ord(ℓ, λ(n)) for almost all n. The three upper bounds in Theorem 2 therefore follow respectively from three results in the new paper of Kurlberg and the second author : Theorem 4 (1), which states that for any function ε(n) →0, we have ord(ℓ, λ(n)) ≥n1/2+ε(n) almost always; Theorem 22, which states that a positive proportion of integers n have ord(ℓ, λ(n)) ≥n.592; and Theorem 28, which states that if the GRH is true, then ord(ℓ, λ(n)) = n/ exp((1 + o(1))(log log n)2 log log log n) 24 GREG MARTIN AND CARL POMERANCE on a set of asymptotic density 1. (Note that the proof of this result uses Theorem 1 of the current paper.) □ 7. HIGHER ITERATES Here we sketch what we believe to be a viable strategy for establishing an analogue of Theorem 1 for the higher iterates λk where k ≥3. As in the case of k = 2, we have generally that n λk(n) = n φk(n) φk(n) λk(n) . We always have n/φk(n) ≤(c log log n)k, which is already a good enough estimate for our purposes. Even better, however, it is known that for each ﬁxed k, we have n/φk(n) ≪(log log log n)k for almost all n. The problem therefore reduces to comparing λk(n) to φk(n). Probably it is not hard to get analogs of Propositions 5 and 6, where we replace y2 with yk. The problem comes in with the proliferation of cases needed to deal with small prime factors. As with the second iterate, we expect the main contribution to come from the “supersquarefree” case. In particular, let hk(n) = ∑ p1\|n ∑ p2\|p1−1 · · · ∑ pk\|pk−1−1 ∑ q≤yk vq(pk −1) log q. We expect hk(n) to be the dominant contribution to log(φk(n)/λk(n)) almost always. But it seems hard not only to prove this in general but also to establish the normal order of hk(n). It would seem useful in this endeavor to have a uniform estimate of the shape ∑ p∈Pm, p≤x 1 p ∼log log x −log log m φ(m) for x ≥m1+ε. (31) Even under the assumption of the Riemann Hypothesis for Dirichlet L-functions, (31) seems difﬁcult, and maybe it is false. It implies with x = m2 that the sum is ≪1/φ(m), when all we seem to be able to prove, via sieve methods, is that it is ≪(log log m)/φ(m). Assuming uniformity in (31), it seems that on average hk(n) ∼ 1 (k −1)!(log log n)k log log log n, supporting Conjecture 3. It would be a worthwhile enterprise to try to verify or disprove the Conjecture in the case k = 3, which may be tractable. Going out even further on a limb, it may be instructive to think of what Conjecture 3 has to say about the normal order of L(n), the minimum value of k with λk(n) = 1. The expression (1/(k −1)!)(log log n)k log log log n reaches its maximum value when k ≈ log log n. Is this formula then trying to tell us that we have L(n) ≪log log n almost always? Perhaps so. There is a second argument supporting the thought that L(n) ≪log log n almost al-ways. Let P(n) denote the largest prime factor of an integer n > 1, and let ℓ(n) = P(n) −1 for n > 1, ℓ(1) = 1. Clearly, ℓ(n) \| λ(n) for all n, so that if L0(n) is the least k with ℓk(n) = 1, then L0(n) ≤L(n). It may be that the difference L(n) −L0(n) is usually not large. In any event, it seems safe to conjecture that L0(n) is usually of order of magni-tude log log n, due to the following argument. For an odd prime p, consider the quantity THE ITERATED CARMICHAEL λ-FUNCTION AND CYCLES OF THE POWER GENERATOR 25 log ℓ(p)/ log p ≈log P(p −1)/ log(p −1). It may be that this quantity is distributed as p varies through the primes in the same way that log P(n)/ log n is distributed as n varies through the integers, namely the Dickman distribution. Such a conjecture has been made in various papers. If so, it may be that the sequence log ℓ(p) log p , log ℓ2(p) log ℓ(p) , . . . behaves like a sequence of independent random variables, each with the Dickman distri-bution. And if so, it may then be reasonable to assume that almost always we get down to small numbers and terminate in about log log n steps. A similar probabilistic model is considered in , but for the simpler experiment of ﬁnding the joint distribution of logarithmic sizes of the various prime factors of a given number n. At the very least, we can prove that L(n) ≪log log n inﬁnitely often. Proof of Theorem 4. Notice that the deﬁnition of λ(n) as a least common multiple, together with the fact that λ(pa) \| λ(pa+1) always, implies that λ lcm{m1, . . . , mj} = lcm λ(m1), . . . , λ(m j) for any positive integers m1, . . . , mj. A trivial induction then shows that λk lcm{m1, . . . , mj} = lcm λk(m1), . . . , λk(mj) for any k ≥0. Since the least common multiple of a set of numbers equals 1 precisely when each number in the set equals 1, we deduce that L lcm{m1, . . . , mj} = max L(m1), . . . , L(m j) . We apply this identity with mi = i. Let nj = lcm{1, 2, . . . , j}. We have log n j = ∑i≤j Λ(i), which is asymptotic to j by the prime number theorem. On the other hand, it is trivial that for any number n we have L(n) ≤1 + (1/ log 2) log n, as λi+1(n) ≤(1/2)λi(n) for 1 ≤i < L(n). Therefore L(nj) = max{L(1), . . . , L(j)} ≤1 + max log 1 log 2, . . . , log j log 2 = 1 + log j log 2 = 1 log 2 + o(1) log log nj. □ We can improve on the estimate in Theorem 4, but not by much. Say we let Nj be the product of all primes p ≤j3.29 with p −1 \| nj, with nj as in the above proof. It follows from Friedlander that a positive proportion of the primes p ≤j3.29 have the required property. Thus, Nj > exp(cj3.29) for some positive constant c and all sufﬁciently large values of j. But λ(Nj) \| nj, so that L(Nj) ≤2 + j/ log 2. Hence L(Nj) < .439 log log Nj for j sufﬁciently large. (This result can be improved by a very small margin using a more recent result of Baker and Harman , but the argument is a bit more difﬁcult, since they do not get a positive proportion of the primes with the required property.) It is likely that L(n) ≪log log log n inﬁnitely often, possibly even that L(n) ≪k logk n inﬁnitely often for arbitrary k-fold-iterated logarithms. 26 GREG MARTIN AND CARL POMERANCE One may also study the maximal order of L(n). The analogous problem for the iter-ated φ-function is relatively trivial, but not so for λ. If there can exist very long “Sophie Germain chains”, that is, sequences of primes p1, p2, . . . , pk where each pi = 2pi−1 + 1, for i > 1, then we might have L(pk) ∼(1/ log 2) log pk. We might even perturb such a chain by a small amount and keep the asymptotic relation, say by occasionally having pi = 4pi−1 + 1. It seems hard to prove that long enough chains to get the the asymptotic for L(pk) do not exist, but probably they don’t on probabilistic grounds. We can at least say that L(n) ≥1 + (1/ log 3) log n inﬁnitely often, since this inequality is attained when n is a power of 3. REFERENCES E. Bach, Analytic methods in the analysis and design of number-theoretic algorithms, MIT Press, Cam-bridge, MA, 1985. R. Baker and G. Harman, Shifted primes without large prime factors, Acta Arith. 83 (1998), 331–361. E. Blanton, S. Hurd, J. McCranie, On the digraph deﬁned by squaring mod m, when m has primitive roots, Cong. Numerantium 82 (1992), 167–177. J. J. Brennan and B. Geist, Analysis of iterated modular exponentiation: the orbit of xα mod N, Designs, Codes, and Cryptography 13 (1998), 229–245. P. Erd˝ os, A. Granville, C. Pomerance, and C. Spiro, On the normal behavior of the iterates of some arithmetic functions, in Analytic number theory (Allerton Park, IL, 1989), 165–204, Progr. Math., 85, Birkh¨ auser Boston, Boston, MA, 1990. P. Erd˝ os and C. Pomerance, On the normal number of prime factors of ϕ(n), Rocky Mountain J. Math., 15 (1985), 343–352. Corrigendum in . P. Erd˝ os, C. Pomerance, and E. Schmutz, Carmichael’s lambda function, Acta Arith., 58 (1991), 363–385. J. B. Friedlander, Shifted primes without large prime factors, in Number theory and applications (Banff, AB, 1988), 393–401, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 265, Kluwer Acad. Publ., Dordrecht, 1989. J. B. Friedlander, C. Pomerance, and I. E. Shparlinski, Period of the power generator and small values of Carmichael’s function, Math. Comp., 70 (2001), 1591–1605. Corrigendum. Math. Comp., 71 (2002), 1803–1806. H. Halberstam and H.-E. Richert, Sieve methods, Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], London-New York, 1974. London Mathematical Society Monographs, No. 4. J. P. Kubilius, Probabilistic methods in the theory of numbers, Translations of Mathematical Mono-graphs, Vol. 11, American Math. Soc., Providence, 1964. P. Kurlberg and C. Pomerance, On the period of the linear congruential and power generators, to appear. K. K. Norton, On the number of restricted prime factors of an integer. I, Illinois J. Math., 20 (1976), 681–705. C. Pomerance, On the distribution of amicable numbers, J. Reine Angew. Math., 293/294 (1977), 217– 222. T. D. Rogers, The graph of the square mapping on the prime ﬁelds, Discrete Math., 148 (1996), 317–324. H. N. Shapiro, Distribution functions of additive arithmetic functions, Proc. Nat. Acad. Sci. USA, 42 (1956), 426–430. DEPARTMENT OF MATHEMATICS, UNIVERSITY OF BRITISH COLUMBIA, ROOM 121, 1984 MATHEMATICS ROAD, VANCOUVER, BC V6T 1Z2, CANADA E-mail address: gerg@math.ubc.ca MATHEMATICS DEPARTMENT, DARTMOUTH COLLEGE, HANOVER, NH 03755-3551, U.S.A. E-mail address: carlp@math.dartmouth.edu
190519	https://www.youtube.com/watch?v=9cN8LrojDo8	Solving a System of Equations with LCM and GCD SyberMath 155000 subscribers 486 likes Description 12529 views Posted: 12 Apr 2021 Join this channel to get access to perks:→ My merch → Follow me → Subscribe → Suggest → If you need to post a picture of your solution or idea: NumberTheoryProblems #GcdLcm EXPLORE: Solving a Functional Equation: Inscribing a circle between two equilateral triangles: Solving a Quadratic System: 111 comments Transcript: hello everyone in this video we're going to be solving a system of equations so we do have x plus y is equal to 84 and we have the ratio lcm to gcd and that is equal to 12. now let's just go ahead and review what these mean lcm stands for the least common multiple of x and y and g c d stands for the greatest common divisor of x and y so we're given the ratio of the lcm and the gcd and we're given the sum of the two numbers and we're supposed to solve for x and y which are positive integers all right great so we're going to be using a lot of good algebra here let's go ahead and get started so the first fact we're going to use here is which is obviously very important number theory if you know the least common multiple and the greatest common divisor of two numbers then you can multiply them and the product is going to equal the product of the numbers x and y which is a very well known fact and you can definitely prove this the proof is not that complicated you can look it up or i can include some links in the description as well now so that's one of the facts we're going to use so if l stands for the lcm and g stands for the gcd so i can go ahead and write down an equation like this x y is equal to l g okay it's not a commercial we're not talking about the lg lg just the product l times g now this is one fact that we can derive from here but we didn't use any of the givens yet right so we're also given that the l to g ratio is equal to 12 which is kind of nice right so that that means that i can write the l as 12 g that's great obviously we already know the fact that i think we used this in a previous video there was another video for gcd lcm and the fact that gcd divides lcm is something that we've used before so that's an important thing and you can see that here too now we're going to go ahead and substitute this for l so let's go ahead and replace l with 12 g that's going to give us x y is equal to 12 g multiplied by g which is 12 g squared where g stands for the greatest common divisor of x and y all right so that's the first thing or equation that we're going to be coming up with but of course we do have other equations as well for example we knew we do know that x plus y is equal to 84. so we're talking about two numbers whose sum is 84. now at this point you may just say hey why don't we just guess right guess and check well could could it be 1 in 83 2 and 82 so on and so forth but there are so many cases to check and it's also it's not an elegant solution it's just trial and error i mean sometimes we have to use it but in this case we have a more elegant solution all right so what am i going to do well i do know that x plus y is equal to 84. so let me go ahead and write that down here as well and let's talk about what this means so this fact is actually super duper important why we're talking about two numbers here x and y and obviously x and y are interchangeable so because of the symmetry at the end when we write our solutions as an ordered pair x and y if they're not equal of course they can be switched around if they're equal then it doesn't really matter right so that's something we can do at the end but for now try to focus on the system we do have a system here right we do have x y and x plus y but the the problem here is that we have two equations but there are three variables but notice that we have the x times y which is a product x plus y is a sum well so what does that tell you well it should tell you to use the vieta's formulas right so this calls for vieta and let me tell you what vietnamese briefly so vietnam basically gives us relationships between the roots of a polynomial and its coefficients and this goes for all degrees obviously starting with the quadratic you can go to any equation that is polynomial and degree corticointic you know nonic whatever doesn't really matter vigentic and you can use vieta's formulas they're awesome so in this case we have a quadratic because we're talking about two roots here x and y so let's go ahead and make up an equation but here's the million dollar question how do you write the equation whose roots are given right so i'm giving that i'm given that the roots are x and y so how do you write this equation well obviously not to get confused here and to make it a little easier on ourselves let's use a different variable you could use a t we could use a u but u is kind of fun to use so let's go ahead and use u so if i use u as my main variable how can i write the equation whose roots are x and y and this is how i can write it by vieta it's written as u squared minus the quantity x plus y times u plus x y is equal to zero now vietnam's formulas give us the sum of the roots as negative b over a and the product as c over a in a quadratic equation and this is pretty much the same thing now i do know these so from here i can basically replace the x plus y with 84 so it's going to give me u squared minus 84 u plus and x times y now i do not know the numerical value of x y but at least i have something right i mean that will relate it to g which is the greatest common divisor of these two numbers and that's definitely going to help us find x and y so let's go ahead and replace x y with 12 g squared and the whole thing is equal to zero so this is my quadratic whose roots are x and y so basically u represents x and y here and again they're interchangeable so it doesn't matter which one is which okay so how do you solve this equation though well this is a quadratic so what do you think since it's a quadratic uh it's either factorable or we're going to use a quadratic formula well in this case it's it doesn't seem factorable because of the g but we can use completing square or just the formula i'm going to use the formula because formula is more fun let's go ahead and use it so the formula says u is equal to negative b which is 84 plus minus the square root of b squared in this case i'm going to write it as 84 squared that's a large number but don't worry i'm going to simplify it minus 4ac and in this case a is 1. so we're going to focus on c rather so it's going to be 4 times 12 g squared which is 48 g squared again these are large numbers what i'm going to simplify them and all over 2 times a which is 2. great now before i proceed i'd like to simplify what's inside the radical because that's going to help me later on so how do you simplify this expression 84 squared something okay i'm thinking about this 84 can be written as 4 times 21 so that is equal to 4 squared times 21 squared and 48 is 16 times 4 so it is like 4 squared times 3g squared great so that means that 4 squared can actually be taken out but when you square root the 4 squared it's going to become a 4. so our expression is going to look like this 84 plus minus 4 times the square root of now inside the parentheses i have 2 and 1 squared which is equal to 441 minus 3g squared is what remains and this is cool because the numbers are much smaller and definitely i don't have a fraction anymore because i can just go ahead and divide everything by 2. so from here u equals 42 plus minus 2 times the quantity square root of 441 minus 3 g squared okay great now it doesn't look that great i know that u is a radical i mean we have a radical on the right hand side u represents x and y but how am i going to solve this right i have nothing about g besides this well here's the thing g is an integer positive integer so you and u is an integer because that represents x and y so whatever is inside the radical needs to be a perfect square otherwise the answer is not going to be an integer it's not even going to be a rational number right so what do i need well this requires that 441 minus 3g squared b a perfect square how about writing it as h squared alphabetically next letter after g right that's all i can come up with so now let's go ahead and put the g and h on the same side and we're going to be getting a really cool diophantine equation of course g and h are both integers here positive integers and we're talking about kind of like a linear because if you replace g squared with something in h squared with something you get a linear differential but this is fairly easy to solve so let's not get into too much depth here let's divide both sides by 3 and here we're going to get something interesting because 441 is obviously divisible by 3 and that is equal to 4 147 right okay now what am i seeing well g squared is an integer 147 is an integer therefore h squared over 3 also needs to be an integer which means h squared is a multiple of 3 but that means h is a multiple of 3 therefore we can write h as 3 k isn't that awesome we're using a lot of number theory and algebra here though so that's kind of cool i think so let's replace h with 3 k that's going to give us 9 k squared divided by 3 is going to give us 3 k squared finally this is the type of equation you want to get at because from this point on it can easily dissolve now if you look at equation carefully we're talking about a perfect square plus three times a perfect square so you kind of go into modular arithmetic here like three mod three mod five so on and so forth but you don't really need to complicate things here because think about it if k is equal to zero obviously g squared equals 147 is not a perfect square uh if k is equal to one you get g squared is equal to 144 which means that g is equal to 12. great so this gives us a solution are there any other solutions well well for k equals two we don't get an integer k equals three k equals four k equals five and you don't have to go that far because for example if k is equal to 6 6 squared is going to be 36 and then it's going to be 108 you're going to get 39. if k is equal to 7 you're going to get 49 and you're going to get g equals zero obviously you don't want the greatest common divisor to be zero because we're talking about positive integers here therefore this is the only solution that we get from here okay great so we got k equals one we don't care about k but we do care about the g g is equal to 12. now what am i going to do with the g though i do need to find x and y but remember we solved an equation a quadratic in u right and u represents x and y so if i go ahead and plug in the g value into this i'll be getting the x and y values so let's go ahead and do that right now so u is equal to 42 plus minus 2 times the square root of 441 minus 3 g squared right that should be our equation and now what am i going to do i'm going to replace g with 12. so if you replace g with 12 u is going to be 42 plus minus 2 times the square root of 441 minus 3 times 12 squared which is equal to 144 and if you divide i mean multiply you know what i'm talking about right 3 times 144 is going to be 441 and this is actually going to be let me see looks like i made a mistake here okay so g is equal to 12 and 3 times 140 um 144 is going to be i'm sorry apologize this is going to be 432 therefore okay there you go so the difference is going to be 9 so u is going to equal 42 plus minus okay this is going to the whole thing here is going to be the square root of 9 which is 3 2 times 3 is equal to 6. so from from here we're going to be getting two values u is either 42 plus 6 or 42 minus 6 but that represents remember the x and y values so this means that our solutions are going to be made up of two pairs and they're going to be 48 comma 36 and 36 comma 48 so those are the x y values that work and those are the only values that work well this brings us to the end of this video i hope you enjoyed this video please let me know don't forget to comment like and subscribe and i'll see you tomorrow with another video until then be safe take care and bye bye
190520	https://im.kendallhunt.com/HS/students/2/2/8/index.html	Illustrative Mathematics \| Kendall Hunt Illustrative Mathematics \| Kendall Hunt Skip to main content IM CurriculumAbout UsContact UsGoogle ClassroomEducators Register/Log in Math Tools Four-Function CalculatorScientific CalculatorGraphing CalculatorGeometrySpreadsheetProbability CalculatorConstructions Geometry Algebra & GeometryAlgebra 1GeometryAlgebra 2Algebra 1 Supports Unit 2 GeometryUnit 1Unit 2Unit 3Unit 4Unit 5Unit 6Unit 7Unit 8 Lesson 123456789101112131415 Lesson 8 The Perpendicular Bisector Theorem Let’s convince ourselves that what we’ve conjectured about perpendicular bisectors must be true. LessonPractice 8.1: Which One Doesn’t Belong: Intersecting Lines Which one doesn’t belong? A Caption: A B Caption: B C Caption: C D Caption: D 8.2: Lots of Lines Diego, Jada, and Noah were given the following task: Prove that if a point is the same distance from as it is from , then must be on the perpendicular bisector of . At first they were really stuck. Noah asked, “How do you prove a point is on a line?” Their teacher gave them the hint, “Another way to think about it is to draw a line that you know is on, and prove that line has to be the perpendicular bisector.” They each drew a line and thought about their pictures. Here are their rough drafts. Diego’s approach: “I drew a line through that was perpendicular to and through the midpoint of . That line is the perpendicular bisector of and is on it, so that proves is on the perpendicular bisector.” Jada’s approach: “I thought the line through would probably go through the midpoint of so I drew that and labeled the midpoint . Triangle is isosceles, so angles and are congruent, and and are congruent. And and are congruent because is a midpoint. That made two congruent triangles by the Side-Angle-Side Triangle Congruence Theorem. So I know angle and angle are congruent, but I still don’t know if is the perpendicular bisector of .” Noah’s approach: “In the Isosceles Triangle Theorem proof, Mai and Kiran drew an angle bisector in their isosceles triangle, so I’ll try that. I’ll draw the angle bisector of angle . The point where the angle bisector hits will be . So triangles and are congruent, which means and are congruent, so is a midpoint and is the perpendicular bisector.” With your partner, discuss each student’s approach. What do you notice that this student understands about the problem? What question would you ask them to help them move forward? Using the ideas you heard and the ways you think each student could make their explanation better, write your own explanation for why must be on the perpendicular bisector of and . Are you ready for more? Elena has another approach: “I drew the line of reflection. If you reflect across, then and will switch places, meaning coincides with ,and coincides with. will stay in its place, so the triangles will be congruent.” What feedback would you give Elena? Write your own explanation based on Elena‘s idea. 8.3: Not Too Close, Not Too Far Work on your own to make a diagram and write a rough draft of a proof for the statement: If is a point on the perpendicular bisector of , prove that the distance from to is the same as the distance from to . With your partner, discuss each other’s drafts. Record your partner‘s feedback for your proof. What do you notice that your partner understands about the problem? What question would you ask them to help them move forward? Summary The perpendicular bisector of a line segment is exactly those points that are the same distance from both endpoints of the line segment. This idea can be broken down into 2 statements: If a point is on the perpendicular bisector of a segment, then it must be the same distance from both endpoints of the line segment. If a point is the same distance from both endpoints of a line segment, then it must be on the perpendicular bisector of the segment. These statements are converses of one another. Two statements are converses if the “if” part and the “then” part are swapped. The converse of a true statement isn‘t always true, but in this case, both statements are true parts of the Perpendicular Bisector Theorem. A line of reflection is the perpendicular bisector of segments connecting points in the original figure with corresponding points in the image. Therefore, these 3 lines are all the same: The perpendicular bisector of a segment. The set of points equidistant from the 2 endpoints of a segment. The line of reflection that takes the 2 endpoints of the segment onto each other, and the segment onto itself. Description:A figure: A large quadrilateral with a small triangle inside. The two upward facing sides of the quadrilateral each have two tick marks, the downward facing sides each have three tick marks. Two sides of the downward facing small triangle each have one tick mark. A perpendicular bisector is drawn through the figure, intersecting at three points. A horizontal line segment is drawn in the center of the figure. It is useful to know that the perpendicular bisector of a line segment is also all the points which are the same distance from both endpoints of the line segment, because then: If 2 points are both equidistant from the endpoints of a segment, then the line through those points must be the perpendicular bisector of the segment (because 2 points define a unique line). If 2 points are both equidistant from the endpoints of a segment, then the line through those must be the line of reflection that takes the segment to itself and swaps the endpoints. If a point is on the line of reflection, then it is the same distance from that point to a point in the original figure and to its corresponding point in the image. If a point is on the perpendicular bisector of a segment, then it is the same distance from that point to both endpoints of the segment. Glossary Entries auxiliary line An extra line drawn in a figure to reveal hidden structure. For example, the line shown in the isosceles triangle is a line of symmetry, and the lines shown in the parallelogram suggest a way of rearranging it into a rectangle. converse The converse of an if-then statement is the statement that interchanges the hypothesis and the conclusion. For example, the converse of "if it's Tuesday, then this must be Belgium" is "if this is Belgium, then it must be Tuesday." corresponding For a rigid transformation that takes one figure onto another, a part of the first figure and its image in the second figure are called corresponding parts. We also talk about corresponding parts when we are trying to prove two figures are congruent and set up a correspondence between the parts to see if the parts are congruent. In the figure, segment corresponds to segment , and angle corresponds to angle . parallelogram A quadrilateral in which pairs of opposite sides are parallel. Privacy Policy \| Accessibility Information © 2019 Illustrative Mathematics®. Licensed under the Creative Commons Attribution 4.0 license. The Illustrative Mathematics name and logo are not subject to the Creative Commons license and may not be used without the prior and express written consent of Illustrative Mathematics. This book includes public domain images or openly licensed images that are copyrighted by their respective owners. Openly licensed images remain under the terms of their respective licenses. See the image attribution section for more information.
190521	https://math.stackexchange.com/questions/4088748/explanation-of-formula-given-on-wikipedia-for-sign-in-half-angle-formula-for-si	Skip to main content Explanation of formula given on Wikipedia (for sign in half-angle formula for sine) Ask Question Asked Modified 4 years, 3 months ago Viewed 241 times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. This Wikipedia article listing trigonometric identities, states the following identity under half angles: sinθ2=sgn(2π−θ+4π⌊θ4π⌋)1−cosθ2−−−−−−−−√ where sgn(x) is a function that returns +1 if x>0 and −1 if x<0. I know and understand the part of the formula that says 1−cosθ2−−−−−√ and understand the fact that the remaining part is to decide the ± sign in the formula. From where, however, does the sign-deciding part of the expression come from? And why should it make sense? trigonometry Share CC BY-SA 4.0 Follow this question to receive notifications edited Apr 4, 2021 at 16:38 Peter LeFanu Lumsdaine 11.2k77 gold badges3232 silver badges6565 bronze badges asked Apr 4, 2021 at 7:58 Aadi PrasadAadi Prasad 10399 bronze badges 5 1 The sign thing is just a shorthand to express that the value is positive when θ∈[4kπ,(4k+2)π] for some integer k, and negative otherwise. It's just a notation trick, not something derived from anywhere. – user3733558 Commented Apr 4, 2021 at 8:29 @user3733558 Why 4π, though? – soupless Commented Apr 4, 2021 at 8:34 @user3733558 Won't any θ lie in that range? (put k=0 and the range becomes [0,2π]) Edit: I understand what you are saying now. But I still do not understand how the notational trick incorporates the same. – Aadi Prasad Commented Apr 4, 2021 at 8:39 Thanks @AadiPrasad, and a last note (because I feel the tone of my comments and answers don't reflect it): I want to applaud your curiosity, and your questioning of details like this one. That's a proper attitude towards anything mathematical. Well done. – user3733558 Commented Apr 4, 2021 at 9:06 1 Thank you very much to you too, @user3733558. And no, your answer or your comments do not reflect the 'it is how it is' tone at all. Not everything has to have a very profound explanation. – Aadi Prasad Commented Apr 4, 2021 at 10:26 Add a comment \| 4 Answers 4 Reset to default This answer is useful 6 Save this answer. Show activity on this post. I imagine it happened this way. In the past, someone saw sinθ2=±1−cosθ2−−−−−−−−√ and thought "Wait a minute, this isn't really satisfactory. There shouldn't be uncertainties here. We know what sign the expression should be for a given θ". And indeed, sinθ2⩾0 for 0⩽θ⩽2πsinθ2⩽0 for 2π⩽θ⩽4π and this pattern is periodic of period 4π. So that someone sat down and thought "What function can I cook up that would give me +1 for inputs in [0,2π] and −1 for inputs in [2π,4π], and would also be 4π periodic?". And he found the expression you can see on Wikipedia. This is not the only way to achieve the same goal. This is not a forced (derived) result. This is an artificial invention, made to achieve a specific goal, and it works. But don't look for a deeper meaning behind it, there isn't any. Share CC BY-SA 4.0 Follow this answer to receive notifications edited May 2, 2021 at 9:15 answered Apr 4, 2021 at 8:53 user3733558user3733558 1,25822 gold badges1010 silver badges2121 bronze badges Add a comment \| This answer is useful 2 Save this answer. Show activity on this post. Personally, I would go with (−1)⌊θ2π⌋={1−1when θ2π∈[0,1)(mod2)when θ2π∈[1,2)(mod2)(1) However, what was done on Wikipedia was to create a sawtooth wave, which again, I would give as 12−{θ4π}{>0<0when θ4π∈(0,12)(mod1)when θ4π∈(12,1)(mod1)(2) However, since {x}=x−⌊x⌋, (2) can be written as 12−{θ4π}=12−(θ4π−⌊θ4π⌋)(3) Note that all three functions have the same sign; negative in the gray regions and positive in the white regions. The sign in (3) is preserved upon multiplication by 4π: 2π−θ+4π⌊θ4π⌋(5) which has the same sign as (2) and is what appears in Wikipedia. Share CC BY-SA 4.0 Follow this answer to receive notifications edited Apr 4, 2021 at 21:44 answered Apr 4, 2021 at 18:39 robjohn♦robjohn 354k3838 gold badges498498 silver badges891891 bronze badges 5 Can you elaborate a bit more? How did you come up with (1), or (2)? All of it just looks Greek to me. – Aadi Prasad Commented Apr 5, 2021 at 3:20 Ok. We need to establish some context. Have you encountered the floor function, ⌊x⌋? {x} is the (periodic) fractional part function, defined as x−⌊x⌋. It goes from 0 to just less than 1 as x goes from n to just less than n+1 for n∈Z. – robjohn ♦ Commented Apr 5, 2021 at 4:25 I know all the definitions you mention. But why would you 'go with (1)'? From the way it is written, it seems as if equation (1) should be as obvious and natural as 2+2=4 is. It isn't so to me. (I am a high schooler who will be learning (single variable) calculus this year.) – Aadi Prasad Commented Apr 5, 2021 at 10:37 Look at ⌊θ2π⌋. It jumps up 1 every 2π. That means it goes from even (0) on [0,2π), to odd (1) on [2π,4π), to even (2) on [4π,6π), and so on. Thus, (−1)⌊θ2π⌋ goes from 1 on [0,2π) to −1 on [2π,4π), to 1 on [4π,6π). This is the same pattern as the sign of sin(θ2) (see the graph). – robjohn ♦ Commented Apr 5, 2021 at 12:54 Ooh! Now I get it — both, (1) and (2). Thanks a lot! – Aadi Prasad Commented Apr 6, 2021 at 9:19 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. The function inside the sgn function has zeroes only when: θ=2π±4kπ and the sign alternates between them, which gives the same pattern as the sinθ2 function. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 4, 2021 at 8:41 JMPJMP 22.7k5151 gold badges3737 silver badges5555 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. It is from the process of deriving the identity itself. First, let us use the double-angle formula cos2θ=2cos2θ−1. Solving for cos2θ, we get cos2θ=1+cos2θ2. Replacing θ by θ2, we get cos2(θ2)cos(θ2)=1+cosθ2=±1+cosθ2−−−−−−−−√ Using the identity sin2θ+cos2θ=1 to get sinθ2, sin2(θ2)+cos2(θ2)sin2(θ2)+1+cosθ2sin2(θ2)sin2(θ2)sin2(θ2)sin(θ2)=1=1=1−1+cosθ2=2−1−cosθ2=1−cosθ2=±1−cosθ2−−−−−−−−√ As you can see, the squares on the equation force it to take the plus-minus sign. Edit: I don't have an idea about the sgn(2π−θ+4π⌊θ4π⌋). I just used the basic ones. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Apr 4, 2021 at 8:12 souplesssoupless 2,38222 gold badges1313 silver badges3737 bronze badges 1 1 Thank you very much for writing out the entire derivation. But I am already aware of it — the issue I have is with what is inside the sgn function. – Aadi Prasad Commented Apr 4, 2021 at 8:32 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions trigonometry See similar questions with these tags. 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190522	https://www.youtube.com/watch?v=jUJEFAi3mP8	Equation of Parabola Given Focus and Directrix Mario's Math Tutoring 458000 subscribers 755 likes Description 119079 views Posted: 4 Oct 2017 Learn how to find the equation of a parabola given the focus and directrix in this free math video tutorial by Mario's Math Tutoring. Timestamps: 00:00 Intro 0:10 What is a Parabola 0:53 Formula for a Parabola 1:20 How to Know Which Direction the Parabola is Opening 1:42 Write the Equation of the Parabola with Directrix x=-1 and Focus (3,1) 3:12 How Wide is the Parabola. Finding 2 More Points on the Parabola Related Videos: How to Graph Parabolas and Write the Equation Organized List of My Video Lessons to Help You Raise Your Scores & Pass Your Class. Videos Arranged by Math Subject as well as by Chapter/Topic. (Bookmark the Link Below) ➡️JOIN the channel as a CHANNEL MEMBER at the "ADDITIONAL VIDEOS" level to get access to my math video courses(Algebra 1, Algebra 2/College Algebra, Geometry, PreCalculus), midterm & final exam reviews, ACT and SAT prep videos and more! (Over 390+ videos) 35 comments Transcript: Intro basically is a parabola. Well, a parabola is the set of all points. Okay, such that if you pick a point on that parabola and you measure the distance to the focus and you measure the distance What is a Parabola to a line called the directx, those distances are always going to be the same. So, the idea is if you're to go further out on this parabola here, you can see the distance to that focus is getting longer, but so is that perpendicular uh distance to the directx is also getting longer. But those points on the parabola, this U-shaped graph, okay, they're always going to be equidistant from a given point called the focus and a given line called the directx. The other component of parabas is this distance between the vertex and the directx and the vertex and the focus. We call that distance p. And the basic equation of a parabola is in the form x^2 = 4 py or y^2 = 4 px. Now the thing to realize is that Formula for a Parabola you know these parabas can be picked up and shifted left and right up and down. The way I've written the equation here this is where the vertex or that bending point that turning point is at the origin. But you know you can see the h and k can shift the graph left and right up and down. But the key thing to remember is that if the x term is squared that means the parabola is going to be opening either up or down. If the p is positive it's opening up. If the p is negative it's opening down. And then How to Know Which Direction the Parabola is Opening for the y^2, if you have a y square term but not an x squared term, that means it's opening to the right or the left. Left if p is negative, right if p is positive. So let's get into this problem and see if we can uh graph it and write the equation. So the directx here is at x=1. So let's go ahead and sketch that in. So that's just an imaginary, you know, invisible line. Okay? And we'll just write that x=1 right there. And Write the Equation of the Parabola with Directrix x=-1 and Focus (3,1) then the focus is at 31, which is right there. I'll just label that the focus. Now the thing to realize is that the vertex is halfway in between the focus and the directx. So if we find that midpoint you can see that's going to be right here at 1 one. So let me just write that down. The vertex is at uh right one up one. Okay. So now the distance from the vertex to the focus that distance is our p distance or the vertex to the directx that is also p. And you can see that p is equal to two. All right. So that's important when we go to write our equation. So it looks like the parabola is going to be opening to the right. It opens away from the directx towards the focus. Okay. So you can see it's going to be opening something like this to the right. So that means this is going to be an x² I'm sorry a y^2 variety. Okay. So this is going to be y - 1^ 2. Okay. Uh equ= 4 p. p is 2 here. So 4 2 and then uh x -1. So if we simplify this, this is 8 x -1 and this is y - 1^ 2. And there you can see our vertex is at 1 1. These are the opposite sign. Okay, 1 one. And then this is positive, which tells us opening to the right. And the other thing that I didn't mention in in the intro here is that the parabas, they're actually at the level of focus, they're 4 p wide. Okay? So that means 4 p in this case is eight. So at the level of focus, it's going to be eight wide. From here, I'm How Wide is the Parabola. Finding 2 More Points on the Parabola going to go up four. 1 2 3 4. That's a point on our graph. and down four 1 2 3 4 and so that gives us an idea about how wide the parabola is. Okay, like that. So the equation is what we were after and we've got it right there. So I hope that helped you to make uh you know finding the equation of ellipse a little bit easier when you're given the focus and the directx. Subscribe to the channel, check out more math tutoring videos on my YouTube channel, Mario's Math Tutoring, and I look forward to helping you in the future videos. I'll talk to you soon.
190523	https://www.youtube.com/watch?v=x6Pnf57wRAs	How To Calculate The Standard Deviation of Grouped Data The Organic Chemistry Tutor 9690000 subscribers 17493 likes Description 1528420 views Posted: 30 Jan 2019 This statistics video explains how to calculate the standard deviation of grouped data. It discusses how to calculate the mean and standard deviation given a frequency distribution table. Statistics - Free Formula Sheet: Statistics - Video Lessons: Introduction to Statistics: Arithmetic, Geometric, & Harmonic Mean: Cumulative Relative Frequency Table: Quartiles, Deciles, & Percentiles: Interquartile Range & Outliers: Skewness & Symmetric Distribution: Sample Mean and Population Mean: Mean, Median, & Mode - Grouped Data: Standard Deviation: Variance: Mean Absolute Deviation: Standard Deviation of Grouped Data: Introduction to Probability: Probability Formulas: Final Exams and Video Playlists: 434 comments Transcript: in this video we're going to talk about how to calculate the standard deviation of group data so here we have a class of students and in a table on the left side we have different classes of grades 50 to 59 we'll say that's an f 60 to 69 is the d this is the c and so forth and to the right of that we have a column that tells us the frequency of those values so this means that five students had a score between 60 and 69 and 12 students had a score between 80 and 89 so with this information we were given the class ranges and the frequencies how can we calculate the standard deviation of the scores that the students had let's say on an exam how can we calculate the standard deviation given this information the first thing we need to do is determine the midpoint of the class ranges so what is the midpoint between 50 and 59 the lower boundary and the upper boundary of that first range to find the midpoint we basically need to average the two numbers so we need to add them up and divide by two 109 divided by 2 is going to be 54.5 now if we average 60 and 69 for the second range that's going to give us 64.5 and if we follow the pattern we can see that the next one is going to be 74.5 and then 84.5 by the way one thing i want to mention is that for this particular problem the student scores that they received on an exam we're going to say that all of those scores have been rounded to a whole number because 69.5 is neither in this range or in that range so we're going to say that all of the scores are whole number values or integer values now we need to determine the midpoint between those two it's not going to be 94.5 so we're going to add the lower boundary which is 90 and the upper boundary which is 100 divided by 2. so this is 190 divided by 2 and that turns out to be 95 so that's the first step you want to calculate the midpoint of all of the ranges that you see here now the next thing we need to do is multiply the frequency column by the midpoint column so this is going to be f times m and let me give you the formula that we're going to use to calculate the standard deviation the standard deviation is going to be the square root of the sum of the square of the deviations the deviations being the data values minus the mean divided by n minus 1 where n is the number of data values in the list now we're going to modify this equation for this particular type of problem so whenever you're dealing with group data you can calculate the sum of the square of the deviations by multiplying the frequency by the difference between the midpoint and the mean and then you have to square those results divided by n minus 1. so we're going to be using this formula in this video so just keep that in mind now before we calculate the standard deviation we need to calculate the mean the mean is basically the sum divided by the number of numbers in a list for group data the sum is going to be the frequency times the midpoint values which is what we're going to get in a second i mean in the fourth column and then we're going to divide that by the frequency values which is what we're going to get here so notice that these two are the same the sum of the frequency values is equal to the number of data values in the list so let's get rid of this now let's multiply the frequency by the midpoint so first we have 3 times 54.5 and so that's going to be 163.5 next we're going to multiply those two so 5 times 64.5 that's 322. and then 9 times 74.5 i got 670.5 and then just repeat the pattern so for the last one 8 times 95 that's going to be 760. now we don't need the sum of this column but we do need the sum of the frequency column and the frequency times the midpoint values so let's begin by adding up these numbers so we have 3 plus 5 plus 9 plus 12 plus 8 and so that's going to give us 37. let me just put a line right here now let's take the sum of these values as well so i got a big number i'm going to see if i can fit it here it's 2930.5 so now we can calculate the mean so it's going to be the sum of the frequency times the midpoint values divided by the frequency values so it's 2930.5 divided by 37 and so that's about 79.2 now does this answer make sense do you think the mean is around 79 well notice that the bulk of the data is in this range so 79 as the mean makes sense because most students they scored between 70 and 89. so it makes sense that the mean is like around 79.80 so now in the next column we're going to put the mean which is 79.2 and that's of course a rounded value now for the next column we're going to take the difference between the mean i mean not the mean but the midpoint and the mean x bar is the mean so let's subtract 54.5 by 79.2 and so we're going to get negative 24.7 now let's subtract these two numbers so 64.5 minus 79.2 now let's subtract these two so that's going to be 74.5 minus 79.2 and so that's negative 4.7 and then we have 84.5 minus 79.2 and so that's positive 5.3 and then for the last one 95 minus 79.2 that's going to be 15.8 now we don't need to add up those values for the next column we're simply going to square the previous column so basically we're going to take a negative 24.7 and square it so it's going to turn into a positive value it's going to be 610.09 14.7 squared that's going to be 216.09 and then 4.7 squared is 22.09 next we have 5.3 squared this is going to be 28.09 and for the last one it's 249.64 now the last thing we're going to do is we're going to multiply the frequency by the square of the deviations so this is going to be f times m minus x bar squared so it's going to be 3 times this number and so that's going to be 1830.27 now let's multiply 5 by 216.09 and so you should get 1080.45 now this work is pretty tedious but this is how you would do it if you're not going to use excel or a calculator for the next one we have 12 times 28.09 and so that's going to be 337.08 and for the last one it's eight times two forty nine point six four and so that's nineteen ninety seven point twelve now we do need to take the sum of that column so let's add those five numbers and so you should get 5443.73 now that we have the sum of the frequency times the deviation squared we can calculate the standard deviation using this formula so it's going to be the sum of the frequency times the square of the deviations divided by n minus one so this number represents everything that you see here so we're going to replace this term with 5443.73 and is this number 37 so we have n minus 1 or 37 minus 1. so it's going to be 5443.73 divided by 36 because that's 37 minus 1. and so that's going to give you 151.2147 with some extra numbers and then when we take the square root of that result our rounded answer so it's going to be approximately 12.3 so that's the standard deviation for this grouped frequency data and that's how you could find it for this type of problem thanks for watching
190524	https://www.jrheum.org/content/jrheum/30/12/2521.full.pdf	Cooke, et al: Editorial 2521 Osteoarthritis (OA) is believed to arise from the interplay of many factors, some of them mechanical (Figure 1). Typically there is deterioration or failure of the joints’ carti-laginous bearing surfaces, and to a lesser extent of the collagenous supporting structures. In line with a multifacto-rial etiology for OA, current approaches to gonarthrosis tend to be multidisciplinary, in which the knee is viewed as an organ. Mechanical Factors We have focused on one aspect of the mechanical influences — the association between the disease and certain patterns of limb alignment (“static mechanical” factors). For example, it is well known that patients with knee OA are commonly bow-legged, meaning that they have a varus alignment of their limbs. In this condition (“varus OA”) there is almost always a loss of joint space (JSL) focused in the medial tibiofemoral compartment. In contrast, the less common “knock-knee” deformity (“valgus OA”) generally has the JSL focused in the lateral compartment. Both varus and valgus alignment abnormalities increase the odds of JSL occurring, and the progression of the JSL has been shown to correlate with the severity of the deformity1. Malalignment of the limbs is typical in OA, and JSL accounts for some but not all of this; important contributions are also made by femoral and tibial geometry2. Contributions of the tibia to varus limb alignment are well recognized in the developmental conditions of tibia vara, Blount’s disease, and in adults with the so-called “malicious mal-alignment.” However, in varus knee OA the tibia appears not to be the major contributing factor. This observation arises from the studies of 2 large populations, one Caucasian and the other Middle-Eastern3,4. Unexpec-tedly, we found that their mean tibial plateau varus angles were not significantly different from the mean values in healthy adults. (We noted tibia vara in some Middle-Eastern cases, but this was largely in advanced disease, where severe erosion of the proximal medial plateau had occurred4. We suspect these cases relate to the high prevalence of meta-bolic bone diseases (osteoporosis and osteomalacia) in the region, and possibly to lifestyles as well5.) What we found in both Caucasian and Middle-Eastern groups was a contri-bution of femoral geometry to the varus alignment in OA — in particular a reduced valgus angle at the distal femur. This femoral abnormality was also associated with disease advancement, especially in the Middle-Eastern group, where both the abnormality and the disease itself tended to be more pronounced than in the Caucasian group. These findings may implicate “reduced femoral valgus” as a predisposing factor for varus OA. On the other hand, it could be that the altered femoral geometry arose from significant remodeling of the femur during the disease process itself6. At present one cannot say which of these alternatives is the most influential in the course of OA. It is relevant, however, that normal aging may cause increased varus alignment, even in the absence of disease (i.e., no symptoms, no JSL)7. So if our bones tend to bow during natural aging, we may at some critical stage become vulner-able — first to onset of OA, and then to progressive joint deterioration as our bones continue to change. This needs to be considered in addition to all the many other factors that may predispose to OA. In osteotomy the tibia is usually chosen for varus correc-tion to unload the medial compartment. But the functional outcome and survival may be limited8,9 and other options have been used with benefit, including femoral and selective double osteotomy10,11. These choices were based on the principle of lessening compartmental overload medially, by correction of the most deformed bone(s) identified by analysis of bone and joint loading contributions. Directions for Future Research Among the many questions raised from these observations are the following: • Is “reduced femoral valgus” a developmental or a devel-oping condition; does it predispose to varus OA? OR • Does this condition arise as part of the pathogenesis of the disease? If so, how? • In varus OA cases that lack bone abnormalities, what is it that drives the disease? To what extent are various genetic, biological, dynamic mechanical, or constitutional factors involved? • In individual cases how does one define or differentiate between primary tissue weakness (biological) and overload (mechanical) as factors favoring the loss of cartilage? • When surgery is required in varus knee OA, why should Editorial Varus Knee Osteoarthritis: Whence the Varus? Personal, non-commercial use only. The Journal of Rheumatology Copyright © 2003. All rights reserved. www.jrheum.org Downloaded on September 28, 2025 from tibial osteotomy be the logical first choice if femoral abnor-malities are so frequent? AND • In which circumstances would femoral realignment, if applied in the early stages of the disease, serve to ameliorate the progression? • What nonsurgical approaches, such as bracing and selec-tive gait training, may be effectively applied to modify the disease process? Finally, to address some of these issues we propose the following as priorities for further evaluation: 1. Conduct longitudinal studies of limb alignment and bone geometry in populations that may be at risk, before symp-toms occur 2. Develop measures or protocols assigning weight to the many risk factors for OA, e.g., bone weakness (osteoporosis and osteomalacia); mechanical overload (static, dynamic, prior trauma); genetic and racial factors; body mass; lifestyle (exercise, drinking, smoking, etc); and history of medication. These may be applicable either to study popula-tions or individual cases, for developing or established OA 3. Establish relationships between alignment changes and biological markers of cartilage degradation, using suitable study populations 4. Re-examine studies of tibial osteotomy for varus knee OA, assessing differences in outcome in cases with mainly tibial versus femoral deformity. We hope these remarks will provoke discussion and stim-ulate study in some of these areas. T. DEREK V. COOKE, MB, BChir, FRCSC, Orthopaedic Consultant, Suite 406B, 797 Princess Street, Kingston, Ontario K7L 1G1. E-mail: derek@cookes.ca ALLAN SCUDAMORE, MA, PhD, Scientist, Department of Orthopaedics, King Faisal Specialist Hospital and Research Centre; currently: 730 Parkside Crescent, Kingston, ON Canada K7M 8A8. E-mail: scud@kingston.net WILLIAM GREER, BSc, PhD, Senior Scientist, Head of the Scientific Computing Research Unit, Department of Biostatistics, Epidemiology, and Scientific Computing, King Faisal Specialist Hospital and Research Centre, Riyadh, Saudi Arabia. E-mail: greer@kfshrc.edu.sa Address reprint requests to Dr. Cooke. REFERENCES 1. Sharma L, Song J, Felson DT, Cahue S, Shamiyeh E, Dunlop DD. The role of knee alignment in disease progression and functional decline in knee osteoarthritis. JAMA 2001;286:188-95. 2. Cooke TDV, Bryant JT, Scudamore RA. Biomechanical factors in alignment and arthritic disorders of the knee. In: Fu FH, Harner CD, Vince KG, editors. Knee surgery. Baltimore: Williams and Wilkins; 1994:1061-78. 3. Cooke TDV, Li J, Scudamore A, Wyss U, Bryant T, Costigan P. Axial lower-limb alignment: comparison of knee geometry in normal volunteers and osteoarthritis patients. Osteoarthr Cart 1997;5:39-47. 4. Cooke TDV, Harrison L, Khan B, Scudamore RA, Chaudhry MA. Analysis of limb alignment in the pathogenesis of osteoarthritis: a The Journal of Rheumatology 2003; 30:12 2522 Figure 1. Mechanical and other factors that contribute to osteoarthritis. Personal, non-commercial use only. The Journal of Rheumatology Copyright © 2003. All rights reserved. www.jrheum.org Downloaded on September 28, 2025 from comparison of Saudi Arabian and Canadian cases. Rheumatol Int 2002;22:160-4. 5. El-Desouki M. Bone mineral density of the spine and femur in the normal Saudi population. Saudi Med J 1995;16:30-5. 6. Nishiyama T, Hashimura M, Kurosaka M, Mizuno K. Radiographic and computed tomographic analysis of lower extremities of varus osteoarthritic knees [abstract]. Trans Orthop Res Soc 1997;22:647. 7. Sanfridsson J, Ryd L, Svahn G, Friden T, Jonsson K. Radiographic measurement of femorotibial rotation in weight-bearing. Acta Radiologica 2001;42:204-17. 8. Oberg U, Oberg T. Functional outcome after high tibial osteotomy: a study using individual goal achievement as primary outcome variable. J Rehab Res Dev 2000;37:501-10. 9. Weale AE, Newman JH. Unicompartmental arthroplasty and high tibial osteotomy for osteoarthrosis of the knee. A comparative study with a 12- to 17-year follow-up period. Clin Orthop 1994;302:134-7. 10. Babis GC, An K-A, Chao EYS, Rand JA, Sim FH. Double level osteotomy of the knee: a method to retain joint-line obliquity. J Bone Joint Surg Am 2002;84:1380-8. 11. Cooke TDV, Pichora D, Siu D, Scudamore RA, Bryant JT. Surgical implications of varus deformity of the knee with obliquity of joint surfaces. J Bone Joint Surg Br 1989;71:560-5. Cooke, et al: Editorial 2523 Personal, non-commercial use only. The Journal of Rheumatology Copyright © 2003. All rights reserved. www.jrheum.org Downloaded on September 28, 2025 from
190525	https://homepages.cwi.nl/~aeb/games/sudoku/solving28.html	Prev Up Next Digit patterns and jigsaw puzzles For a given digit, there are 9.6.3.6.4.2.3.2.1 = 66 = 46656 possible patterns. When a partial grid has been given, the number of possible patterns is much smaller. For example, in the example used to illustrate backtrack, the numbers of patterns consistent with the givens is 1: 88 patterns 2: 43 patterns 3: 62 patterns 4: 92 patterns 5: 62 patterns 6: 107 patterns 7: 70 patterns 8: 43 patterns 9: 48 patterns Now one can turn a Sudoku puzzle into a jigsaw puzzle: make pieces for all possible digit patterns, and find one piece for each digit such that these pieces fit together and form the 9x9 grid. Note that generating all possible patterns for each digit involves nine backtrack searches. Backtrack and fun There is debate on the rules of the game. If all one wants is the solution for some Sudoku puzzle, then one can feed it to a computer and obtain the answer in a fraction of a millisecond. Without a computer one can employ backtrack if needed and obtain the answer within an hour. But maybe that is no fun. The fun part is to apply logical reasoning to find out where the digits should go, and to avoid boring, mechanical work. The "digit pattern approach" is not really about solving Sudokus, it is about solving POM puzzles. One starts with a Sudoku puzzle, does some boring, mechanical work to find all possible digit patterns, and then has an interesting puzzle: given these jigsaw pieces, how to select pieces that fit together. Example This example was given by Myth Jellies, the inventor of the POM (Pattern Overlay Method, that is, jigsaw) technique. Consider the Sudoku The numbers of possible patterns are smaller than in the example above (because there are more givens) but is still a bit large: 1: 32 patterns 2: 28 patterns 3: 9 patterns 4: 90 patterns 5: 3 patterns 6: 9 patterns 7: 12 patterns 8: 5 patterns 9: 16 patterns Do first some easy work, to get to And now: 1: 7 patterns 2: 6 patterns 3: 1 pattern 4: 7 patterns 5: 2 patterns 6: 4 patterns 7: 2 patterns 8: 1 pattern 9: 10 patterns That looks manageable, 40 jigsaw pieces. How to represent them? Myth Jellies proposes the notation where the jigsaw pieces are named 1a,1b,...,1g,2a,...,2f,3,4a,...,9j and drawn in one grid: \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| 4ab 6a \| 8 \| 1ab 4cd 9abcdefgh \| 3 \| 4efg 6bc \| 2 \| 5a 9ij \| 1cde 7a \| 1fg 5b 6d 7b \| \| 1c 2ab 6bd \| 5 \| 1defg 2cdef \| 7 \| 9 \| 8 \| 4 \| 3 \| 1ab 6ac \| \| 3 \| 4efg 6c 9ij \| 7 \| 1 \| 4abcd 6ad \| 5 \| 8 \| 2acd 9abcd \| 2bef 6b 9efgh \| \| 1adf 5b 9aei \| 1bceg 9bf \| 8 \| 6 \| 2 \| 7 \| 3 \| 4acef 9ghj \| 4bdg 5a 9cd \| \| 2ce 4cde 5a \| 7 \| 2ab 4abfg \| 9 \| 3 \| 1 \| 2df 5b \| 6 \| 8 \| \| 2df 6c 9cgj \| 6abd 9dh \| 3 \| 5 \| 8 \| 4 \| 1 \| 2be 7b 9efi \| 2ac 7a 9ab \| \| 1beg 4fg 9bdfh \| 1adf 4abcd 9aceg \| 5 \| 2 \| 7 \| 3 \| 6 \| 8 \| 1c 4e 9ij \| \| 8 \| 2 \| 6 \| 4 \| 1 \| 9 \| 7 \| 5 \| 3 \| \| 7 \| 3 \| 1c 4e 9ij \| 8 \| 5 \| 6 \| 2abce 9abcdefgh \| 1abfg 2f 4bdg \| 1de 2d 4acf \| The original Sudoku is a graph colouring problem. Here we have a clique-finding problem: find the unique 9-clique in the graph where compatible jigsaw pieces are joined. Two pieces are compatible when they do not overlap. (And both graph colouring and clique-finding are difficult in general.) Vulnerable pairs If there are two fields such that every pattern for some digit d contains at least one of those, then all patterns for other digits that cover both positions can be discarded (since they overlap every possible d-pattern). This is called a vulnerable d-pair. In the above example, positions (6,1) and (9,7) cover all 2-patterns (that is, form a vulnerable 2-pair) and hence patterns 9c and 9g are impossible. Positions (1,7) and (4,1) form a vulnerable 5-pair, showing that 9i is impossible. Equivalences Use a string like 2be9f as abbreviation for the statement "precisely one of 2b and 2e and 9f holds". Of course, every string in an open grid square is such a (true) statement. Use = for the equivalence <=>. (We have now turned the problem into a satisfiability problem for the formula that is the conjunction of the statements found in the grid squares, together with the formulae 1abcdefg etc. describing that every digit has some pattern.) Since 2abcdef and 2abce9abdefh (by (9,7)), we have 2df = 9abdefh. By (6,1) it follows that 2df6c9j, that is, 6c9abdefhj. Since 9abdefhj it follows that 6c is impossible. Since 6abd and 6abd9dh (by (6,2)), 9d and 9h are impossible. By (3,8) we have 2acd9ab. Now 2df = 9abef and 2bef = 9ab, so that 2df includes 2bef, and 2b and 2e are impossible, and we have 2f = 9ab and 2d = 9ef and 9j = 2ac. From 2ac9abef and 2ac7a9ab (by (6,9)) it follows that 7a = 9ef and 7b = 2ac9ab. From 5ab and 5b9abef (by (5,6)) it follows that 5a = 9abef and 5b = 2ac. But 1fg5b6d7b by (1,9), that is, 1fg2aacc6d9ab. By the 'precisely one' clause, this means that 2a and 2c (and 9j) are impossible. And hence 5a is the right 5-pattern. From 4abcdefg and 4efg (by (3,2)) and 4acef (by (4,8)) and 4abfg (by (5,3)) it follows that 4f is the right 4-pattern. From 1beg4f9bf (by (7,1)) and 1de4f9ef (by (9,9)) it follows that 1b, 1d, 1e, 1g, 9b, 9e, 9f are all impossible. Examining what is left, we see 9a, 6a and 1c. Altogether, the solution to our puzzle is 1c, 2f, 3, 4f, 5a, 6a, 7b, 8, 9a. Discussion Since formula satisfiability is hard, solving such jigsaw puzzles will in general require some form of backtrack. But in the above example the simple strategy "always try to replace single elements (or short strings) by longer strings" worked successfully. For a frightening example, see Taking on a killer Tso ... . Prev Up Next
190526	https://www3.nd.edu/~jhauenst/preprints/chSVDhomotopy.pdf	A singular value homotopy for ﬁnding critical parameter values J.B. Collins Jonathan D. Hauenstein May 18, 2020 Abstract Various applications in science and engineering depend upon solving nonlinear systems of equations which depend upon parameters. Locally in the parameter space, the qualitative behavior of the solutions remains the same except at critical parameter values. This ar-ticle develops a singular value homotopy that aims to compute critical parameters values. Several examples are presented including computing critical parameter values for nonlinear boundary value problems, turning points for a steady-state system connected to learning and memory, and computing the maximum Gaussian curvature of a surface. 1 Introduction Systems of equations arising in science and engineering typically depend upon parameters such as temperature, pressure, concentration, and length. The solutions to these equations vary with the parameters with the implicit function theorem, e.g., see [22, Thm. A.2.8], showing that the qualitative behavior of the solutions remains the same except at critical parameter values. Therefore, computing critical parameter values is important for creating a complete understanding of the solutions over the parameter space. Two main applications that beneﬁt from computing critical parameter values are steady-state bifurcations and diﬀerential equations with a multiplicity of solutions. Bifurcation diagrams demonstrate the steady-state solutions of diﬀerential equations as they depend on one or more parameters of the equations. While most points on a bifurcation diagram can be found through typical computations, the critical parameter values correspond to limit points or turning points which are more diﬃcult to compute as this is where the number of solutions change. Most of the work in determining bifurcation diagrams is in ﬁnding these critical parameter values . Many nonlinear diﬀerential equations can be shown to have multiple solutions for a particular range of parameter values. For instance, the two-point boundary value problem y′′(x) = −λ(1 + y2), 0 < x < 1 y(0) = 0 (1) y(1) = 0 Department of Mathematics, University of Mary Washington (jcollin2@umw.edu) Department of Applied and Computational Mathematics and Statistics, University of Notre Dame (hauen-stein@nd.edu, www.nd.edu/∼jhauenst). This author was partially supported by NSF grant CCF-1812746 and ONR grant N00014-16-1-2722. 1 is known to have a critical parameter value λ∗> 0 such that there are two solutions for 0 < λ < λ∗and no solutions for λ > λ∗[15, 17]. While the existence of such a critical pa-rameter value λ∗is known, there is typically no analytical method for computing the value. Therefore, it is left to numerical methods to approximate λ∗via discretization. Since the computation of critical parameter values is used in a variety of contexts, there are numerous approaches such as [6, 7, 12, 20]. In this work, we develop a homotopy continuation algorithm (see [5, 22] for a general overview) based on the singular value decomposition which aims to drive the minimum singular value of the Jacobian matrix to zero corresponding to a crit-ical parameter value. This idea can be easily modiﬁed when the Jacobian matrix is symmetric, the minimum singular value is degenerate, i.e., has two or more linearly independent singular vectors, or to create a degenerate singular value. The structure of the paper is as follows. Section 2 provides a short introduction to homotopy continuation and discretization. Section 3 describes the singular value homotopy and some modiﬁcations. Section 4 presents several applications of the singular value homotopy applied to diﬀerential equations. 2 Homotopy continuation and discretization The following provides some background information on homotopy continuation and the dis-cretization of diﬀerential equations. 2.1 Homotopy continuation and path tracking The essential idea of homotopy continuation, e.g., see [5, 22] for more details, is to track along a solution path over a segment which, without loss of generality, we may assume is the in-terval [0, 1]. In particular, suppose that H(y; t) = 0 is a system of n analytic equations with variables y = (y1, . . . , yn) and parameter t ∈[0, 1]. Let JyH(y; t) and JtH(y; t) be the Jacobian matrix and vector with respect to y and t, respectively. Suppose that one is given a point z such that H(z; 1) = 0 and JyH(z; 1) is invertible, i.e., z is a nonsingular solution of H(•; 1) = 0. Then, the implicit function theorem shows that there is an analytic solution path y(t) such that y(1) = z and H(y(t); t) ≡0 for t in an open neighborhood of 1. In particular, this open neighborhood can be extended from 1 towards 0 until there exists t∗< 1 such that either lim t→(t∗)+ ∥y(t)∥= ∞ or det JyH(y(t∗); t∗) = 0. (2) If there is no t∗∈(0, 1] such that (2) occurs, then the path y(t) is said to be track-able [14, Defn. 4.5]. Therefore, one can use predictor-corrector path tracking methods as il-lustrated in Figure 1 to compute y(t) for t ∈(0, 1] as well as the endpoint y(0) = lim t→0+ y(t). Since H(y(t); t) ≡0, the predictor part arises from applying standard numerical solvers for ordinary diﬀerential equations, e.g., Euler or Runge-Kutta, to d dtH(y(t); t) ≡0 = ⇒ ˙ y(t) = −JyH(y(t); t)−1 · JtH(y(t); t). (3) Given a ﬁxed value of t and a numerical approximation of y(t), the corrector part arises from applying standard local numerical solvers, e.g., Newton’s method, to H(•; t) = 0. 2 Predictor Corrector 1 t y Figure 1: Illustration of predictor-corrector path tracking for path y(t). Example 1. As an illustrative example, consider H(y; t) = y2 −t with z = 1. Clearly, the analytic solution path y(t) with y(1) = 1 and H(y(t); t) ≡0 is y(t) = √ t. Since the critical parameter value is t∗= 0, this solution path is trackable with ˙ y(t) = 1 2y(t). For singular endpoints, i.e., y(0) such that JyH(y(0); 0) is singular, endgame procedures (see [22, Chap. 10] for a general overview) can be used to produce accurate approximations of y(0). 2.2 Discretizing diﬀerential equations For parameterized systems of diﬀerential equations, critical parameter values are of great in-terest for fully describing their bifurcation diagrams or ﬁnding where the number of solutions changes [10, 17, 18, 21]. One way to construct systems of equations from diﬀerential equations which can be used to numerically approximate critical parameter values is by using the ﬁnite diﬀerence method [2, 9, 16]. The problems presented in Section 4 utilize ﬁnite diﬀerence discretization to construct sys-tems of equations. To illustrate, consider the boundary value problem (1). Using n + 1 dis-cretization nodes on the interval [0, 1], the spatial step size is h = (n + 1)−1 and the spatial nodes are xj = j · h for j = 0, 1, . . . , n + 1. Then, yj denotes an approximation of y(xj) at each grid point for j = 1, . . . , n with y0 = yn+1 = 0 since y(0) = y(1) = 0. Using a central diﬀerence approximation for y′′(xj), namely y′′(xj) ≈yj+1 −2yj + yj−1 h2 , one obtains a system of equations in variables y = (y1, . . . , yn) and parameter λ: y0 −2y1 + y2 + h2λ(1 + y2 1) = 0 y1 −2y2 + y3 + h2λ(1 + y2 2) = 0 (4) . . . yn−1 −2yn + yn+1 + h2λ(1 + y2 n) = 0 3 with y0 = yn+1 = 0. Therefore, (4) yields a system of n polynomial equations F(y; λ) = 0 in n variables y = (y1, . . . , yn) and parameter λ. 3 Singular value homotopy The basic setup is as follows. First, one has a system of n analytic equations F(y; λ) = 0 in n variables y = (y1, . . . , yn) and parameter λ. Second, one is given a parameter value λ and a nonsingular solution y to F(•; λ) = 0. With this setup, the goal is to track along the solution path through y to ﬁnd a critical parameter value λ∗which is a parameter value such that F(•; λ∗) = 0 has a singular solution, i.e., a solution with a singular Jacobian matrix JyF. A common technique used in a variety of problems, e.g., [3, 6, 7, 11, 12, 20], is to incremen-tally move, i.e., “sweep,” along the solution path looking for critical parameter values. This is inherently sensitive to how one numerically steps along the solution path and the numerical criterion for identifying critical parameter values. Therefore, an alternative is to use a homo-topy that forces the determinant of the n × n Jacobian matrix JyF to be zero. Due to potential numerical computational issues with evaluating and diﬀerentiating det JyF when n is large, we instead propose a homotopy-based approach based on the minimum singular value. Although this adds more variables and equations, the highly-structured nature of the equations permits large-scale computations as demonstrated on examples in Section 4. Since the applications in Section 4 depend on tracking real solution paths and computing real solutions, we formulate everything in terms of real computations. The singular value de-composition of a matrix A ∈Rn×n is A = U · Σ · V T where U, V ∈Rn×n are orthogonal, i.e., U T · U = V T · V = I, and Σ =    σ1 ... σn   ∈Rn×n such that σ1 ≥· · · ≥σn ≥0. The columns of U and V are called the left and right singular vectors, respectively, and the nonnegative numbers σ1, . . . , σn are called the singular values. The Eckart-Young theorem shows that σn is equal to the 2-norm distance between A and the set of singular n × n matrices. Moreover, a correlating set of left and right singular vectors and singular value (ui, vi, σi) satisﬁes uT i · ui = vT i · vi = 1, A · vi = σi · ui, and AT · ui = σi · vi. (5) If σi ̸= 0, then uT i · ui = 1 is redundant. Removing it from (5) results in 2n + 1 equations in the 2n + 1 variables (ui, vi, σi). A singular value σi is called nondegenerate if there is unique (up to sign) left and right singular vectors ui and vi. In Section 3.1, we consider the case when the minimum singular value σn is nondegenerate with Section 3.3 considering the degenerate case. Section 3.4 considers the case of making a singular value degenerate. A simpliﬁcation for the symmetric case, i.e., when AT = A, is provided in Section 3.2. 4 3.1 Construction of singular value homotopy With this setup, the following homotopy aims to drive the minimum singular value to zero: H(y, u, v, λ; t) =     F(y; λ) JyF(y; λ) · v −σ(t) · u JyF(y; λ)T · u −σ(t) · v vT · v −1    = 0. (6) The start point (y, u, v, λ) at t = 1 and σ(t) are constructed as follows. Since y and λ are given such that JyF(y; λ) is full rank, let σn be minimum singular value of JyF(y; λ) with corresponding left and right singular vectors u and v, respectively. Finally, σ(t) = t · σn. As with iterative methods such as Newton’s method, the starting solution (y, λ) will impact the trackability of the corresponding solution path to (6). In particular, there is no guarantee that the corresponding solution path will be trackable (as deﬁned in Section 2.1) as shown in Ex. 3. However, if the path is trackable and the endpoint is ﬁnite, then the solution path ends at a point yielding a critical parameter value. This is summarized in the following. Theorem 1. With the setup above, if the solution path for H in (6) starting at (y, u, v, λ) is trackable for t ∈(0, 1] and the endpoint (y∗, u∗, v∗, λ∗) is ﬁnite, then λ∗is a critical parameter value such that y∗is a singular solution of F(•; λ∗) = 0. Proof. Note that trackable solution paths to a homotopy deﬁned by real equations that start at a real point remain real over t ∈(0, 1]. Hence, the result follows since H ensures F(y∗; λ∗) = 0 and 0 is a singular value of JyF(y∗; λ∗). The next two examples are an illustrative example followed by a system where the solution path is not trackable. Example 2. Consider the polynomial system F(y; λ) = y2 1 + y2 2 −1 y1 + y2 −λ = 0 which describes the intersection of the unit circle and the family of lines with slope m = −1. The following illustrates using (6) with two diﬀerent starting points to compute critical parameter values in which the line is tangent to the circle as shown in Figure 2. (a) (b) Figure 2: Computing tangent lines starting with (a) λ = 1.3 and (b) λ = −1.3. 5 First, consider λ = 1.3. Rounding to four decimal places, we take y = (0.9284, 0.3716). Then, the minimum singular value of JyF(y; λ) is σ2 = 0.4629 with left and right singular values u = (0.5661, −0.8243)T and v = (0.4901, −0.8717)T , respectively. Applying (6) produces a trackable path which ends with the critical parameter value λ∗= √ 2 which corresponds with a line that is tangent to the circle as shown in Figure 2(a). Similarly, consider λ = −1.3 with y = (−0.9284, −0.3716). Then, the minimum singular value of JyF(y; λ) is σ2 = 0.4629 with left and right singular values u = (0.5661, 0.8243)T and v = (−0.4901, 0.8717)T , respectively. Applying (6) produces a trackable path which ends with the critical parameter value λ∗= − √ 2 which corresponds with a line that is tangent to the circle as shown in Figure 2(b). Example 3. Consider the system F(y; λ) = 3y1 −1 (1 + λ2)y2 −1 = 0 with JyF(y; λ) = 3 0 0 1 + λ2 . Clearly, for λ ∈R, F(y; λ) = 0 has no singular solutions. Thus, every path of a singular value homotopy H in (6) starting at a real solution can not be trackable. For example, with λ = 1 and y = (1/3, 1/2), one has u = v = (0, 1)T and σ2 = 2. The corresponding solution path of H in (6) is not trackable at t = 1/2. This corresponds with λ = 0 which is the minimum of 1 + λ2. 3.2 Symmetric problems The homotopy H in (6) depends on 3n + 1 variables. One way to reduce this is to exploit symmetry of the problem. A common structure that arises from discretizations such as the one in (4) is a symmetric Jacobian matrix. When the Jacobian matrix JyF(y; λ) is symmetric, the left and right singular vectors agree up to sign. Hence, one can identify the proper sign choice δ ∈{+1, −1} based on the start point (y, λ) and remove n variables and equations yielding H(y, v, λ; t) =   F(y; λ) JyF(y; λ) · v −δ · σ(t) · v vT · v −1  = 0. (7) Theorem 1 can naturally be extended to (7). 3.3 Degenerate minimum singular value The construction of the singular value homotopy in (6) is applied when the minimum singular value at the start point and along the solution path is nondegenerate. The following considers the degenerate case. Suppose that σn is the minimum singular value of JyF(y; λ) which we aim to drive to zero. Let r ≥1 be such that σn−r+1 = · · · = σn, i.e., the minimum singular value σn is repeated r times so that one needs to consider r left and right singular vectors. One obvious generalization of (6) is to replace vectors u and v by n × r matrices U and V yielding H(y, U, V, λ; t) =       F(y; λ) JyF(y; λ) · V −σ(t) · U JyF(y; λ)T · U −σ(t) · V U T · U −I V T · V −I       = 0. (8) 6 However, the corresponding starting values of U and V associated with y and λ need not be isolated as shown in the following. Example 4. Consider the system F(y; λ) = λ(y1 −1) λ(y2 −1) = 0 with JyF(y; λ) = λ 0 0 λ . For y = (1, 1) and t = λ = σ1 = σ2 = 1, i.e., r = 2, the constraints from (8) on U and V simplify to U = V and U T ·U = I which deﬁnes the one-dimensional family of 2×2 orthogonal matrices. Since path tracking using (3) needs the Jacobian matrix with respect to the variables to be nonsingular, this implies that the solution is isolated. Therefore, one needs to remove the extra degrees of freedom as illustrated in Ex. 4. Theoretically, this is accomplished by working in a corresponding Grassmannian which is a space whose points correspond with linear spaces. To avoid ununnecessary technical details, the following works on an aﬃne patch of the Grassman-nian . The key observation is that, for any invertible r × r matrix K, the following sets of equations are equivalent: JyF(y; λ) · V = σ(t) · U JyF(y; λ)T · U = σ(t) · V and JyF(y; λ) · V · K = σ(t) · U · K JyF(y; λ)T · U · K = σ(t) · V · K. One approach to remove this freedom is to write the n×r matrix U in a particular way. Since U has rank r, one can always pick K to be the inverse of an r × r submatrix of U. Therefore, we can select a permutation matrix P ∈Rn×n and write U(Λ) = P · I Λ (9) where Λ is an (n −r) × r matrix of variables. The disadvantage of this representation is that one loses the orthogonality of both U(Λ) and corresponding V . Therefore, (8) simpliﬁes to H(y, Λ, V, λ; t) =   F(y; λ) JyF(y; λ) · V −σ(t) · U(Λ) JyF(y; λ)T · U(Λ) −σ(t) · V  = 0 (10) where U(Λ) as in (9) and σ(t) = t · σn as before. Although U(Λ) and V are not orthogonal, one still maintains the relationship U(Λ)T · U(Λ) = V T · V. (11) One can easily translate between the two representations: (1) orthogonal U and V typically computed by singular value decomposition algorithms and (2) Λ and V satisfying (11). Suppose that (Uo, Vo) and (Λp, Vp) are two such representations, respectively. Given (Uo, Vo), one can compute a permutation matrix Q ∈Rn×n such that the ﬁrst r rows of Q · Uo form an invertible r × r matrix, say K. Let M ∈R(n−r)×r such that Q · Uo = K M . Then, P = QT , Λp = M · K−1 and Vp = Vo · K−1 so that Uo = U(Λp) · K and Vo = Vp · K. 7 Conversely, given (Λp, Vp), compute a QR decomposition Vp = Q · R where Q ∈Rn×r orthogonal and R ∈Rr×r upper triangular and invertible. Then, both Uo = U(Λd) · R−1 and Vo = Vd · R−1 = Q are orthogonal. Theorem 1 can naturally be extended to (10). Moreover, note that (10) has n+2nr equations in n + 2nr + 1 −r2 variables. Since this is overdetermined for r > 1, provides adaptive techniques for constructing a well-constrained homotopy. Example 5. Reconsider the system from Ex. 4. Using the orthogonal approach from (8), one has U = V is any 2×2 orthogonal matrix. However, for (10), one has n = r = 2 so that one can take P = I and there is no Λ. Hence, one has U(Λ) = I and V = I at the start point. Running the homotopy (10) yields a trackable path that ends at a critical parameter value λ∗= 0. 3.4 Two equal singular values The previous approaches were focused on driving the minimum singular value to zero. However, this method can be easily modiﬁed to instead drive two consecutive singular values towards each other using the following homotopy: H(y, u, v, σ, µ, ν, s, λ; t) =             F(y; λ) JyF(y; λ) · v −σ · u JyF(y; λ)T · u −σ · v vT · v −1 JyF(y; λ) · ν −s · µ JyF(y; λ)T · µ −s · ν νT · ν −1 σ −s −t · (σ −s)             = 0. (12) In particular, the second and third blocks of equations enforce that (u, v, σ) and (µ, ν, s) are both correlating collections of left and right singular vectors and singular value for the ma-trix JyF(y; λ). The last equation aims to drive singular values σ and s together where σ and s are the corresponding starting values. A natural extension of Theorem 1 can be applied to (12). 4 Numerical results The following presents four applications of singular value homotopies for diﬀerential equations. The ﬁrst two are boundary value problems. The third is a biological model for signal transduction and gene network used to study long-term memory. The last is a fully nonlinear, second order elliptic partial diﬀerential equation which computes a surface of constant Gaussian curvature. For each of these problems, Bertini was used to perform the path tracking. For the bound-ary value problems, Matlab was used to determine determine the start point by computing a corresponding singular value decomposition. The start point for the biological model was computed in Maple. The Gaussian curvature problem was implemented in Python. The start point for the corresponding discretized system was calculated using Newton’s method in Python which was also used to generate the input ﬁles for Bertini. 4.1 Boundary value problem As mentioned in the Introduction, the two-point boundary value problem (1) has a critical parameter value λ∗> 0 such that (1) has two solutions for 0 < λ < λ∗which merge at λ∗and 8 2.5 3 3.5 4 4.5 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y(0.25) Solution 1 Solution 2 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 x 0.5 1 1.5 2 2.5 y(x) = 4 = 4 = 4.755 (a) (b) Figure 3: (a) Plot of y(0.25) for the two solutions to (1) as a function of λ. (b) The two solutions y(x) to (1) plotted for selected values of λ which converge to each other as λ →λ∗≈4.755. has no solutions for λ > λ∗as illustrated in Figure 3(a) which plots y(0.25) as a function of λ. For selected values of n, we applied a (symmetric) singular value homotopy to the discretization in (4) to approximate λ∗. To generate a start point, one can, for example, use shooting methods or Newton’s method at λ = 0.1 which was chosen to ensure that λ < λ∗. Note that suggests λ∗= 4. Table 1 shows the results of computing λ∗for various values of n showing that the theoretical value of λ∗is approximately 4.755 and Figure 3(b) shows how the two solutions y(x) to (1) converge to each other as λ →λ∗≈4.755. n λ∗ 10 4.73438 50 4.75415 100 4.75476 200 4.75491 400 4.75495 Table 1: Value of λ∗for (1) using increasingly reﬁned discretizations. 4.2 Another boundary value problem Consider computing radially symmetric solutions u(x, y) on the annulus Ω= {(x, y) \| 1 < x2 + y2 < 2} satisfying the following boundary value problem: ∆u = −λ 1 + u + u2 2 + u3 6 in Ω (13) u(x, y) = x2 + y2 on ∂Ω. Similar to (1), there exists critical parameter value λ∗> 0 such that there are two solutions for 0 < λ < λ∗which merge at λ∗and no solutions for λ > λ∗. Using a discretization with a central diﬀerence scheme and a singular value homotopy, Table 2 shows that the theoretical value of λ∗is approximately 2.292. 9 n λ∗ 10 2.28757 50 2.29225 100 2.29240 200 2.29243 Table 2: Value of λ∗for (13) using increasingly reﬁned discretizations. 4.3 Biological example The next application is to compute bifurcations for the steady-state behavior of a a biologi-cal model connected to learning and memory . This model simulates the levels of various proteins that are responsible for long-term facilitation after exposure to serotonin and is math-ematically represented by a system of 15 ordinary diﬀerential equations with 40 parameters. In particular, studies steady-state behavior under diﬀerent model parameter values. They utilized singularity theory which generates bifurcation diagrams as one particular parameter, called the bifurcation parameter λ, is varied. The bifurcation parameter was chosen to be the level of serotonin as this is what is most often varied in experiments. Another branch of solutions were found in . Although the original system depended upon rational functions, one can, at steady-state, clear denominators to produce the following parameterized system of 15 polynomial equations in state variables y1, . . . , y15, bifurcation parameter λ, and other model parameters k1, . . . , k39: 3.6λ −(y1 −0.06)(λ + 14) = 0 k1 + k2y4y2 1 −k7y2y3 −k3y2(y15 −k4) −k5y2 = 0 k1 + k2y4y2 1 −k7y2y3 + k3y2(y15 −k4) −k5y3 = 0 k7y2y3 −k2y4y2 1 −k3y4(y15 −k4) −k5y4 = 0 k9y7(y12 + k29)(y5 + k13) −k12y5 −k14y5(y5 + k13) = 0 k10(y12 + k29)(y5 −y6)(y5 + k13) −k12y6(y5 −y6 + k11) = 0 −k14y6(y5 + k13)(y5 −y6 + k11) = 0 k15 −λk16y7 −k18y7(y7 + k17) = 0 −λk19y8 + k22(k25 −y8) = 0 −k20(k25 −y8)y9(k26 −y9 −y10 + k28) + k23(k26 −y9 −y10)(y9 + k28) = 0 k20(k25 −y8)(y10 + k28) −k23y10(k26 −y9 −y10 + k28) = 0 −k21y10y11(k27 −y11 −y12 + k28) + k24(k27 −y11 −y12)(y11 + k28) = 0 k21y10(k27 −y11 −y12)(y12 + k28) −k24y12(k27 −y11 −y12 + k28) = 0 k30y3(1 −y13) −k31k32y13 = 0 k33(y12 + k29)(1 −y14) −k34k32y14 = 0 k35y2 13y2 14 + k36(y2 13 + k2 37)(y2 14 + k2 38) −k39y15(y2 13 + k2 37)(y2 14 + k2 38) = 0 All model parameters except k35 are ﬁxed to be the constants given in with k35 ﬁxed at diﬀerent values to obtain qualitatively diﬀerent diagrams. In particular, for k35 = 0.015, 0.020, and 0.030, we utilized a singular value homotopy to determine a critical parameter value λ∗ corresponding with a turning point of the bifurcation diagram for three values of k35 from . 10 While only provides drawings without giving explicit values, the values presented in Table 3 appear to be in correspondence with their drawings. k35 λ∗ 0.015 0.116 0.020 0.105 0.030 0.0859 Table 3: Critical level of serotonin λ∗at the turning points of the bifurcation diagrams for selected values of k35 from . 4.4 Maximum Gaussian curvature The ﬁnal application considers K-surfaces which are surfaces with constant Gaussian curva-ture K > 0. For a domain Ω⊂R2 and a function u(x, y), consider the surface S ⊂R3 given by the graph of u over Ω, namely S = {(x, y, u(x, y)) \| (x, y) ∈Ω}. (14) The Gaussian curvature on S is given by K = det(D2u) (1 + ∥∇u∥2)2 where D2u = uxxuyy −u2 xy and ∥∇u∥2 = ∇u · ∇u. Therefore, the surfaces S represented in (14) which have a constant Gaussian curvature K > 0 arise from functions u(x, y) on Ωwhich solve the following partial diﬀerential equation: det(D2u) = K · (1 + ∥∇u∥2)2 in Ω (15) u(x, y) = g(x, y) on ∂Ω for some boundary function g(x, y). Hence, given a domain Ω, boundary function g(x, y), and curvature K > 0, one obtains a K-surface by solving (15). Figures 4 and 5 show solutions to (a) (b) (c) Figure 4: K-surfaces obtained by solving (15) on Ω= [−0.57, 0.57]2 with boundary condition g1(x, y) = p 1 −x2 −y2 and curvature (a) K = 1, (b) K = 1.5, and (c) K = 2. 11 (a) (b) (c) Figure 5: K-surfaces obtained by solving (15) on Ω= [−0.57, 0.57]2 with boundary condition g2(x, y) = 1 −x2 −y2 and curvature (a) K = 1, (b) K = 1.5, and (c) K = 2.2. this problem for Ω= [−0.57, 0.57]2 on two diﬀerent sets of boundary conditions for various values of K. Moreover, if (15) has a solution for some K0 > 0, then it has a solution for all K ∈(0, K0] . Due to geometric constraints, there is a maximum curvature Kmax > 0 for which a well-deﬁned solution to (15) exists for K ∈(0, Kmax] and no solution for K > Kmax. The value of Kmax can be computed explicitly for simple cases, but there is no analytic method for com-puting Kmax in general. In , a type of “sweeping” method was used to approximate Kmax by incrementing the value of K on a discretization of (15) until some termination criteria suggesting that no solution could exist. However, it is unclear what termination criteria was used. Since the value of Kmax is a critical parameter value, we utilize a singular value homotopy applied to a discretization to approximate it. Assuming that the domain is rectangular, say Ω= [ax, bx] × [ay, by], we following a setup similar to Section 2.2. In particular, using n + 1 of grid points in each direction, the mesh size in the x and y directions, respectively, are hx = bx −ax n + 1 and hy = by −ay n + 1 . With grid points xj = ax + j · hx and yj = ay + j · hy for j = 0, . . . , n + 1, let ui,j be an approximation of u(xi, yj). We utilized the following standard discretized approximations: uxx(xi, yj) ≈ui+1,j −2ui,j + ui−1,j h2 x , uyy(xi, yj) ≈ui,j+1 −2ui,j + ui,j−1 h2 y , uxy(xi, yj) ≈ui+1,j+1 −ui−1,j+1 −ui+1,j−1 + ui−1,j−1 4hxhy , ux(xi, yj) ≈ui+1,j −ui−1,j 2hx , uy(xi, yj) ≈ui,j+1 −ui,j−1 2hy . 12 To compare with results in , we consider two problems from on the square domain Ω= [−.57, .57]2 with boundary conditions g1(x, y) = p 1 −x2 −y2 and g2(x, y) = 1 −x2 −y2. (16) For each boundary condition, we started with the parameter value K = 1.8 and computed a starting solution obtained using Newton’s method applied to the discretized system. Table 4 shows the computed values for Kmax for selected values of n which are similar to the values reported in , namely 2.10 and 2.24, respectively. g1(x, y) n Kmax 16 2.12 24 2.085 32 2.07 g2(x, y) n Kmax 16 2.24 24 2.21 32 2.20 Table 4: Computed values of maximum Gaussian curvature for domain Ω= [−.57, .57]2 with boundary condition g1(x, y) and g2(x, y) from (16) for selected values of n. 5 Conclusion The singular value homotopy is an approach for determining critical parameter values by forcing the minimum singular value to zero. These critical parameter values are of great interest in many applications as they can determine where the qualitative behavior, such as the number of solutions, can change. Classical approaches for computing critical parameter values use a “sweeping” approach along the parameter space. For instance, the maximum Gaussian curvature Kmax can be found by ﬁnd a starting solution for some K < Kmax and then slowly increasing the parameter value K. The disadvantage of this approach is one has to decide how to increase the parameter and when one reaches the critical parameter value. In , K was increased by a ﬁxed value of ∆K = 0.02. Using a singular value homotopy, one can utilize homotopy continuation path tracking techniques (e.g., see [5, Chaps. 2-3]) to adaptively determine the step size for predictor-corrector path tracking and endgames to accurately compute the endpoint corresponding with the critical parameter value. A variety of problems arising from diﬀerential equations are used to demonstrate some of the many applications of singular value homotopies. References E. L. Allgower, D. J. Bates, A. J. Sommese, and C. W. Wampler. Solution of polynomial systems derived from diﬀerential equations. Computing, 76(1-2):1–10, 2006. U. M. Ascher and L. R. Petzold. Computer methods for ordinary diﬀerential equations and diﬀerential-algebraic equations, volume 61. Siam, 1998. F. E. Baginski and N. Whitaker. Numerical solutions of boundary value problems for K-surfaces in R3. Numerical Methods for Partial Diﬀerential Equations: An International Journal, 12(4):525–546, 1996. 13 D. J. Bates, J. D. Hauenstein, C. Peterson, and A. J. Sommese. Numerical decomposi-tion of the rank-deﬁciency set of a matrix of multivariate polynomials. In L. Robbiano and J. Abbott, editors, Approximate Commutative Algebra, pages 55–77. Springer Vienna, Vienna, 2010. D. J. Bates, J. D. Hauenstein, A. J. Sommese, and C. W. Wampler. Numerically solving polynomial systems with Bertini, volume 25 of Software, Environments, and Tools. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2013. A. Dhooge, W. Govaerts, and Y. A. Kuznetsov. Matcont: A matlab package for numerical bifurcation analysis of odes. SIGSAM Bull., 38(1):21–22, 2004. E. Doedel. AUTO: a program for the automatic bifurcation analysis of autonomous systems. Congr. Numer., 30:265–284, 1981. C. Eckart and G. Young. The approximation of one matrix by another of lower rank. Psychometrika, 1:211–218, 1936. W. Gautschi. Numerical analysis. Springer Science & Business Media, 1997. J. R. Graef, L. Kong, and H. Wang. Existence, multiplicity, and dependence on a parameter for a periodic boundary value problem. J. Diﬀerential Equations, 245(5):1185–1197, 2008. W. Hao, J. D. Hauenstein, B. Hu, and A. J. Sommese. A three-dimensional steady-state tumor system. Applied Mathematics and Computation, 218(6):2661–2669, 2011. H. A. Harrington, D. Mehta, H. M. Byrne, and J. D. Hauenstein. Decomposing the pa-rameter space of biological networks via a numerical discriminant approach. In J. Gerhard and I. Kotsireas, editors, Maple in Mathematics Education and Research, pages 114–131, Cham, 2020. Springer International Publishing. J. D. Hauenstein and M. H. Regan. Adaptive strategies for solving parameterized systems using homotopy continuation. Applied Mathematics and Computation, 332:19–34, 2018. J. D. Hauenstein, A. J. Sommese, and C. W. Wampler. Regeneration homotopies for solving systems of polynomials. Math. Comp., 80(273):345–377, 2011. T. Laetsch. On the number of solutions of boundary value problems with convex nonlin-earities. J. Math. Anal. Appl., 35:389–404, 1971. R. J. LeVeque. Finite diﬀerence methods for ordinary and partial diﬀerential equations: steady-state and time-dependent problems, volume 98. Siam, 2007. S. S. Lin. Positive radial solutions and nonradial bifurcation for semilinear elliptic equations in annular domains. J. Diﬀerential Equations, 86(2):367–391, 1990. N. S. Papageorgiou and G. Smyrlis. A bifurcation-type theorem for singular nonlinear elliptic equations. Methods Appl. Anal., 22(2):147–170, 2015. D. B. Pettigrew, P. Smolen, D. A. Baxter, and J. H. Byrne. Dynamic properties of regulatory motifs associated with induction of three temporal domains of memory in aplysia. Journal of computational neuroscience, 18(2):163–181, 2005. 14 K. Piret and J. Verschelde. Sweeping algebraic curves for singular solutions. J. Comput. Appl. Math., 234(4):1228–1237, 2010. J. Shi and M. Yao. On a singular nonlinear semilinear elliptic problem. Proc. Roy. Soc. Edinburgh Sect. A, 128(6):1389–1401, 1998. A. J. Sommese and C. W. Wampler, II. The numerical solution of systems of polynomials arising in engineering and science. World Scientiﬁc Publishing Co. Pte. Ltd., Hackensack, NJ, 2005. H. Song, P. Smolen, E. Av-Ron, D. A. Baxter, and J. H. Byrne. Bifurcation and singularity analysis of a molecular network for the induction of long-term memory. Biophysical journal, 90(7):2309–2325, 2006. S. H. Strogatz. Nonlinear dynamics and chaos. Westview Press, Boulder, CO, second edition, 2015. With applications to physics, biology, chemistry, and engineering. 15
190527	https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/0576eaa972df1b7d8c170a9a59c8d241_MIT18_03SCF11_s35_1text.pdf	General Linear ODE Systems and Independent Solutions We have studied the homogeneous system of ODE’s with constant co efﬁcients, x0 = A x , (1) where A is an n × n matrix of constants (n = 2, 3). We described how to calculate the eigenvalues and corresponding eigenvectors for the matrix A, and how to use them to ﬁnd n independent solutions to the system (1). With this concrete experience in solving low-order systems with con stant coefﬁcients, what can be said when the coefﬁcients are functions of the independent variable t? We can still write the linear system in the ma trix form (1), but now the matrix entries will be functions of t: x0 = a(t)x + b(t)y , x 0 = a(t) b(t) . x , (2) y0 = c(t)x + d(t)y y c(t) d(t) y or in more abridged notation, valid for n × n linear homogeneous systems, x0 = A(t) x . (3) Note how the matrix becomes a function of t — we call it a matrix-valued function of t, since to each value of t the function rule assigns a matrix: t0 A(t0) = a(t0) b(t0) → c(t0) d(t0) In the rest of this chapter we will often not write the variable t explicitly, but it is always understood that the matrix entries are functions of t. We will sometimes use n = 2 or 3 in the statements and examples in order to simplify the exposition, but the deﬁnitions, results, and the argu ments which prove them are essentially the same for higher values of n. Deﬁnition 1 Solutions x1(t), . . . , xn(t) to (3) are called linearly dependent if there are constants ci, not all of which are 0, such that c1x1(t) + . . . + cnxn(t) = 0, for all t. (4) If there is no such relation, i.e., if c1x1(t) + . . . + cnxn(t) = 0 for all t all ci = 0, (5) ⇒ the solutions are called linearly independent, or simply independent. General Linear ODE Systems and Independent Solutions OCW 18.03SC The phrase for all t is often in practice omitted, as being un derstood. This can lead to ambiguity. To avoid it, we will use the symbol ≡ 0 for identically 0, meaning zero for all t; the symbol 6≡ 0 means not identically 0, i.e., there is some t-value for which it is not zero. For example, (4) would be written c1x1(t) + . . . + cnxn(t) ≡ 0 . Theorem 1 If x1, . . . , xn is a linearly independent set of solutions to the n × n system x0 = A(t)x, then the general solution to the system is x = c1x1 + . . . + cnxn. (6) Such a linearly independent set is called a fundamental set of solutions. This theorem is the reason for expending so much effort to ﬁnd two independent solutions, when n = 2 and A is a constant matrix. In this chapter, the matrix A is not con stant; nevertheless, (6) is still true. Proof. There are two things to prove: (a) All vector functions of the form (6) really are solutions to x0 = A x. This is the superposition principle for solutions of the system; it’s true because the system is linear. The matrix notation makes it really easy to prove. We have (c1x1 + . . . + cnxn)0 = c1x1 0 + . . . + cnx0 n = c1 A x1 + . . . + cnA xn, since xi 0 = A xi ; = A (c1x1 + . . . + cnxn), by the distributive law. (b) All solutions to the system are of the form (6). This is harder to prove and will be the main result of the next note. 2 MIT OpenCourseWare 18.03SC Differential Equations Fall 2011 For information about citing these materials or our Terms of Use, visit:
190528	https://www.kristakingmath.com/blog/work-done-on-elastic-springs-with-hookes-law	Work done on elastic springs, and Hooke's law — Krista King Math \| Online math help Work done on elastic springs, and Hooke's law The work required to stretch or compress a spring To calculate the work done when we stretch or compress an elastic spring, we’ll use the formula W=∫abF(x) dx where W is the work done, F(x) is the force equation, and [a,b] is the distance over which the spring is stretched or compressed. Every spring has its own spring constant k. This spring constant is part of Hooke’s Law, which states that F(x)=kx where F(x) is the force required to stretch or compress the spring, k is the spring constant, and x is the difference between the natural length and the stretched or compressed length. Since k is unique to each spring, we’ll need to calculate it prior to determining work, unless it’s given in the problem. Keep in mind that we’ll want to find work in terms of Joules J, which is the same as Newton-meters N-m. Video example of work done on elastic springs Take the course Want to learn more about Calculus 2? I have a step-by-step course for that. :) Learn More When you know natural length and the force required to stretch the spring Example A spring has a natural length of 30 cm. A 50 N force is required to stretch and hold the spring at a length of 40 cm. How much work is done to stretch the spring from 42 cm to 48 cm? How much work is done to compress the spring from 30 cm to 25 cm? We’ll use Hooke’s Law to find F(x), but first we need to find k. Since we know that a 50 N force is required to stretch and hold the spring at a length of 40 cm, from its natural length of 30 cm, we’ll set F(x)=50 and x=0.10 m, which is the difference between the natural length and the stretched length, converted from cm to m. Remember that we’ll be finding work in terms of Newtons and meters, which is why we converted 10 cm to 0.10 m. 50=0.10k k=500 With k, we can develop a generic equation for our spring using Hooke’s Law. F(x)=kx F(x)=500x Work done to stretch the spring To calculate the work required to stretch the spring from 42 cm to 48 cm, we pretend that the spring at its natural length of 30 cm ends at the origin, which means that stretching it to 42 cm means we’ve stretched it to 12, because 42−30=12. Stretching it to 48 cm means we’ve stretched it from the origin to 18, because 48−30=18. Therefore, the work equation would be W=∫abF(x) dx W=∫1218500x dx But we need to convert the units from cm to m, so the interval becomes 0.12 m to 0.18 m. W=∫0.120.18500x dx W=250x2∣∣∣0.120.18 W=250(0.18)2−250(0.12)2 W=4.5 The work done to stretch a spring with natural length 30 cm and spring constant k=500 from 42 cm to 48 cm is 4.5 J. Work done to compress the spring To calculate the work required to compress the spring from 30 cm to 25 cm, we pretend that the spring ends at the origin, which means that compressing it to 25 cm means we’ve compressed it to −5, because 25−30=−5. Therefore, the work equation would be W=∫abF(x) dx W=∫0−5500x dx But we need to convert the units from cm to m, so the interval becomes 0 m to −0.05 m. W=∫0−0.05500x dx W=250x2∣∣∣0−0.05 W=250(−0.05)2−250(0)2 W=0.625 The work done to compress a spring with natural length 30 cm and spring constant k=500 from 30 cm to 25 cm is 0.625 J. Get access to the complete Calculus 2 course Get started Learn mathKrista Kingmath, learn online, online course, online math, work, work required, work done, elastic springs, work done on elastic springs, hooke's law, stretch, compress, stretch a spring, compress a spring, calculus 2, calculus ii, calc 2, calc ii, applications of integrals, integral applications, integrals, integrate, integration Facebook0TwitterLinkedIn0RedditTumblrPinterest00 Likes
190529	https://artofproblemsolving.com/wiki/index.php/Proportion?srsltid=AfmBOop5I4MxAUCl9M8MTPkdX-W_r4AZovbLzIGq3s8mcd2AYbrRI2zN	Art of Problem Solving Proportion - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Proportion Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Proportion Two numbers are said to be in proportion to each other if some numeric relationship exists between them. There are several types of proportions, each defined by a separate class of function. Contents 1 Direct Proportion 2 Inverse Proportion 3 Exponential Proportion 4 Problems 4.1 Introductory 4.2 Intermediate 4.3 Olympiad 5 See Also Direct Proportion Direct proportion is a proportion in which one number is a multiple of the other. Direct proportion between two numbers and can be expressed as: where is some real number. The graph of a direct proportion is always linear. Often, this will be written as . Inverse Proportion Inverse proportion is a proportion in which as one number's absolute value increases, the other's decreases in a directly proportional amount. It can be expressed as: Exponential Proportion A proportion in which one number is equal to a constant raised to the power of the other, or the logarithm of the other, is called an exponential proportion. It can be expressed as either: or for some real number , where is not zero or one. Problems Introductory Suppose is either or in the following system: Find the possible values of . (Source) Intermediate is directly proportional to the sum of the squares of and and inversely proportional to and the square of . If when and , find when and , what is ? (Source) (Thanks to Bicameral of the AoPS forum for this one) Olympiad See Also Ratio Fraction Retrieved from " Categories: Algebra Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
190530	https://math.mit.edu/~goemans/18310S15/chernoff-notes.pdf	18.310 lecture notes February 21, 2015 Chernoﬀbounds, and some applications Lecturer: Michel Goemans 1 Preliminaries Before we venture into Chernoﬀbound, let us recall Chebyshev’s inequality which gives a simple bound on the probability that a random variable deviates from its expected value by a certain amount. Theorem 1 (Chebyshev’s Inequality). Let X : S →R be a random variable with expectation E(X) and variance Var(X). Then, for any a ∈R: P(\|X −E(X)\| ≥a) ≤Var(X) a2 . We gave a proof from ﬁrst principles, but we can also derive it easily from Markov’s inequality which only applies to non-negative random variables and gives us a bound depending on the expectation of the random variable. Theorem 2 (Markov’s Inequality). Let X : S →R be a non-negative random variable. Then, for any a > 0, P(X ≥a) ≤E(X) a . Proof. Let A denote the event {X ≥a}. Then: E(X) = X s∈S p(s)X(s) = X x∈A p(s)X(s) + X s∈¯ A p(s)X(s). As X is non-negative, we have P s∈¬A p(s)X(s) ≥0. Hence: E(X) ≥ X s∈A p(s)X(s) ≥a X s∈A p(s) = a · P(A). Chebyshev’s inequality requires the variance of the random variable but can be derived from Markov’s inequality. Proof. (of Chebyshev’s inequality.) Apply Markov’s Inequality to the non-negative random variable (X − E(X))2. Notice that E (X −E(X))2 = Var(X). Even though Markov’s and Chebyshev’s Inequality only use information about the expectation and the variance of the random variable under consideration, they are essentially tight for a general random variable. Exercise. Verify this by constructing non-trivial (i.e. non-constant) random variables for which Theorem 2 and Theorem 1 are tight, i.e. hold with equality. Chernoﬀ-1 2 Deviation of a sum on independent random variables As we are not able to improve Markov’s Inequality and Chebyshev’s Inequality in general, it is worth to consider whether we can say something stronger for a more restricted, yet interesting, class of random variables. This idea brings us to consider the case of a random variable that is the sum of a number of independent random variables. This scenario is particularly important and ubiquitous in statistical applications. Examples of such random variables are the number of heads in a sequence of coin tosses, or the average support obtained by a political candidate in a poll. Can Markov’s and Chebyshev’s Inequality be improved for this particular kind of random variable? Before confronting this question, let us check what Chebyshev’s Inequality (the stronger of the two) gives us for a sum of independent random variables. Theorem 3. Let X1, X2, . . . , Xn be independent random variables with E(Xi) = µi and Var(Xi) = σ2 i . Then, for any a > 0: P(\| n X i=1 Xi − n X i=1 µi\| ≥a) ≤ Pn i=1 σ2 i a2 Proof. This follows from Chebyshev’s Inequality applied to Pn i=1 Xi and the fact that Var(Pn i=1 Xi) = Pn i=1 Var(Xi) for independent variables. In particular, for identically distributed random variables with expectation µ and variance σ2, we obtain P Pn i=1 Xi n −µ ≥ϵ ≤σ2 nϵ2 for any ϵ > 0. We have dervied this when discussing the Weak Law of Large Numbers. Can this result be improved or is it tight? At a ﬁrst glance, you may suspect that this is tight, as we have made use of all our assumptions. In particular, we exploited the independence of the variables {Xi} to get Var(Pn i=1 Xi) = Pn i=1 Var(Xi). Notice, however, that this last step actually only uses the pairwise independence of the variables {Xi}, i.e. the fact that, for all couples i ̸= j ∈[n] and all x, y ∈R: P(Xi = x ∧Xj = y) = P(Xi = x) · P(Xj = y). (1) Indeed, it is possible to show that Theorem 3 is tight when all the variables {Xi} are just guaranteed to be pairwise independent. Hard Exercise Let X1, . . . , Xd be independent random variables that take value 1 or −1, each with probability 1/2. For each S ⊆[d], deﬁne the random variable YS = Q i∈S Xi. i) Show that the variables {YS} are pairwise independent. ii) Let Z = P S⊆D YS. Show that Chebyshev’s Inequality is asymptotically tight for Z. We are now ready to tackle the case of a sum of independent random variables. Recall that we are now using the following strong version of independence (also known as joint or mutual independence), which guarantees the same property of Equation 1 for any subset S ⊆[n] of random variables: ∀S ⊆[n], P( ^ i∈S Xi = xi) = Y i∈S P(Xi = xi). In this case, the proof of Theorem 3 is too weak as it does not rely on the joint independence. In the next section, we will see that we can indeed obtain stronger bounds under this stronger assumpiton. These bounds are known as Chernoﬀbounds, after Herman Chernoﬀ, Emeritus Professor of Applied Mathematics here at MIT! Chernoﬀ-2 3 ChernoﬀBound There are many diﬀerent forms of Chernoﬀbounds, each tuned to slightly diﬀerent assumptions. We will start with the statement of the bound for the simple case of a sum of independent Bernoulli trials, i.e. the case in which each random variable only takes the values 0 or 1. For example, this corresponds to the case of tossing unfair coins, each with its own probability of heads, and counting the total number of heads. Theorem 4 (ChernoﬀBounds). Let X = Pn i=1 Xi, where Xi = 1 with probability pi and Xi = 0 with probability 1 −pi, and all Xi are independent. Let µ = E(X) = Pn i=1 pi. Then (i) Upper Tail: P(X ≥(1 + δ)µ) ≤e−δ2 2+δ µ for all δ > 0; (ii) Lower Tail: P(X ≤(1 −δ)µ) ≤e−µδ2/2 for all 0 < δ < 1; Notice that the lower and upper tail take slightly diﬀerent forms. Curiously, this is necessary and boils down to the use of diﬀerent approximation of the logarithmic function. There exist more general versions of this bound, where this asymmetry is not present, but they are more complicated, as the involve the entropy of the distribution at the exponent. For δ ∈(0, 1), we can combine the lower and upper tails in Theorem 4 to obtain the following simple and useful bound: Corollary 5. With X and X1, . . . , Xn as before, and µ = E(X), P(\|X −µ\| ≥δµ) ≤2e−µδ2/3 for all 0 < δ < 1. Example application: coin tossing Suppose we have a fair coin. Repeatedly toss the coin, and let Sn be the number of heads from the ﬁrst n tosses. Then the weak law of large numbers tells us that P(\|Sn/n −1/2\| ≥ϵ) →0 as n →∞. But what can we say about this probability for some ﬁxed n? If we go back to the proof of the weak law that we gave in terms of Chebyshev’s inequality, we ﬁnd that it tells us that P(\|Sn/n −1/2\| ≥ϵ) ≤ 1 4nϵ2 . So for example, P(\|Sn/n −1/2\| ≥1/4) ≤4 n. But we can apply Chernoﬀinstead of Chebyshev; what do we get then? From Corollary 5, using E(Sn) = n/2, P(\|Sn −n/2\| ≥δ(n/2)) ≤2e−nδ2/6. Taking δ = 1/2 we obtain P(\|Sn/n −1/2\| ≥1/4) ≤2e−n/24. This is a massive improvement over the Chebyshev bound! Let’s try this now with a much smaller δ: let δ = p 6 ln n/n. Then we obtain P(\|Sn/n −1/2\| ≥1 2 p 6 ln n/n) ≤2e−ln n = 2 1 n. If instead we take δ just twice as large, δ = 2 p 6 ln n/n, P(\|Sn/n −1/2\| ≥ p 6 ln n/n) ≤2e−4 ln n = 2 1 n4 . Chernoﬀ-3 3.1 Proof idea and moment generating function For completeness, we give a proof of Theorem 4. Let X be any random variable, and a ∈R. We will make use of the same idea which we used to prove Chebyshev’s inequality from Markov’s inequality. For any s > 0, P(X ≥a) = P(esX ≥esa) ≤E(esX) esa by Markov’s inequality. (2) (Recall that to obtain Chebyshev, we squared both sides in the ﬁrst step, here we exponentiate.) So we have some upper bound on P(X > a) in terms of E(esX). Similarly, for any s > 0, we have P(X ≤a) = P(e−sX ≥e−sa) ≤E(e−sX) e−sa The key player in this reasoning is the moment generating function MX of the random variable X, which is a function from R to R deﬁned by MX(s) = E esX . The reason for the name is related to the Taylor expansion of esX; assuming it converges, we have MX(s) = E 1 + sX + 1 2s2X2 + 1 3!s3X3 + · · · = ∞ X i=0 1 i!siE(Xi). The terms E(Xi) are called “moments” and encode important information about the distribution; notice that the ﬁrst moment (i = 1) is just the expectation, and the second moment is closely related to the variance. So the moment generating function encodes information of all of these moments in some way. Moment generating functions behave wonderfully with respect to addition of independent random vari-ables: Lemma 1. If X = Pn i=1 Xi where X1, X2, . . . , Xn are independent random variables, then MX(s) = n Y i=1 MXi(s). Proof. MX(s) = E(esX) = E es Pn i=1 Xi = E n Y i=1 esXi ! = n Y i=1 E(esXi) by independence = n Y i=1 MXi(s). This lemma allows us to prove a Chernoﬀbound by bounding the moment generating function of each Xi individually. Chernoﬀ-4 3.2 Proof of Theorem 4 Before proceeding to prove the theorem, we compute the form of the moment generating function for a single Bernoulli trial. Our goal is to then combine this expression with Lemma 1 in the proof of Theorem 4. Lemma 2. Let Y be a random variable that takes value 1 with probability p and value 0 with probability 1 −p. Then, for all s ∈R: MY (s) = E(esY ) ≤ep(es−1). Proof. We have: MY (s) = E(esY ) = p · es + (1 −p) · 1 by deﬁnition of expectation = 1 + p(es −1) ≤ep(es−1) using 1 + y ≤ey with y = p(es −1). We are now ready to prove Theorem 4 by combining Lemma 1 and 2. Proof of Theorem 4. Applying Lemma 1 and Lemma 2, we obtain MX(s) ≤ n Y i=1 epi(es−1) = e(es−1) Pn i=1 pi ≤e(es−1)µ, (3) using that Pn i=1 pi = E(X) = µ. For the proof of the upper tail, we can now apply the strategy described in Equation 2, with a = (1+δ)µ and s = ln(1 + δ). P(X ≥(1 + δ)µ) ≤e−s(1+δ)µe(es−1)µ) = eδ (1 + δ)1+δ µ . Our choice of s is motivated as follows: we are trying to make our upper bound for the tail probability to be as small as possible. To do this, we can minimize our expression for the upper bound as a function of s. Taking the derivative of the exponent shows that this minimum is achieved exactly at s = log(1 + δ). Taking the natural logarithm of the right-hand side yields µ(δ −(1 + δ) ln(1 + δ)). Using the following inequality for x > 0(left as an exercise): ln(1 + x) ≥ x 1 + x/2, we obtain µ(δ −(1 + δ) ln(1 + δ)) ≤−δ2 2 + δ µ. Hence, we have the desired bound for the upper tail: P(X ≥(1 + δ)µ) ≤ eδ (1 + δ)1+δ µ ≤e−δ2 2+δ µ. Chernoﬀ-5 The proof of the lower tail is entirely analogous. It proceeds by taking s = ln(1 −δ) and applies the following inequality for the logarithm of (1 −δ) in the range 0 < δ < 1 : ln(1 −δ) ≥−δ + δ2 2 . Details are left as an exercise. 4 Other versions of ChernoﬀBound Chernoﬀbound can be applied to more general settings than that of Bernoulli variables. In particular, the following version of the bound applies to bounded random variables, regardless of their distribution! Theorem 6. Let X1, X2, . . . , Xn be random variables such that a ≤Xi ≤b for all i. Let X = Pn i=1 Xi and set µ = E(X). Then, for all δ > 0 : (i) Upper Tail: P(X ≥(1 + δ)µ) ≤e − 2δ2µ2 n(b−a)2 ; (ii) Lower Tail: P(X ≤(1 −δ)µ) ≤e − δ2µ2 n(b−a)2 . Chernoﬀ-6
190531	https://topos.institute/blog/2023-01-10-recursive-types/	Recursive Types via Domain Theory – Topos Institute Home Work Team Events Blog Get Involved Table of contents 1 Introduction 2 Directed-Complete Partial Orders 3 Diverging Computations and Partial Functions 4 Domains as Retracts of a Universal Space 5 Recursive Types 6 Domains via the Untyped λ\lambda λ-Calculus 7 Conclusion Recursive Types via Domain Theory programming languages type theory lambda calculus logic Author Harrison Grodin Published 2023-01-10 Abstract Functional programming languages often include recursive types, allowing programmers to define types satisfying a given isomorphism. Such types cannot all be interpreted as sets, but they can be understood using domain theory. We consolidate and present foundational techniques in domain theory to understand eager programming languages with recursive types from a categorical perspective. We define a category of domains with finite coproducts, a symmetric monoidal closed structure for eager products and functions, and least solutions of recursive domain equations, whose objects are retracts of a universal space and whose morphisms are continuous with respect to a topology of approximations. 1 Introduction When writing code, it is common to use recursive types to represent data structures. As a simple example, one might want a type representing the natural numbers. The type of natural numbers can be characterized as the least type such that N≅1+N N \cong 1 + N N≅1+N; we will refer to constraints of the form X≅F(X)X \cong F(X)X≅F(X) as domain equations. In ML-style pseudocode, a least (i.e., inductive) solution to this type equation may be defined as follows: datatype N = Fold of 1 + N val unfold = fn (Fold x) => x Here, Fold : 1 + N -> N is the name of the constructor. Via pattern matching, we define its inverse, unfold : N -> 1 + N. Sometimes, types are informally thought of as sets; this type N N N is thought of as the set N\mathbb{N}N of natural numbers. For a more unusual example, non-total programming languages (such as Standard ML, OCaml, and Haskell) allow the following pseudocode: datatype C = Fold of C -> 2 val unfold = fn (Fold x) => x This defines a type C C C that is isomorphic to the “function space” from C C C to 2 2 2! In this case, set-oriented reasoning breaks down: by Cantor’s Theorem, there is no set C C C isomorphic to its power set, 2 C 2^C 2 C. However, types of this form are very powerful, underpinning unbounded recursion. For example, using C C C, we may write a diverging (i.e., “infinitely looping”) program by applying an element c:C c : C c:C to itself (up to the given isomorphism): ``` ( unroll : C -> 2 ) val unroll = fn (c : C) => (unfold c) c ( val diverge : 2 ) val diverge = unroll (Fold unroll) ``` This example begins to demonstrate the power granted by recursive types.1 In order to give a categorical account of languages with such unusual type isomorphisms, we must find a suitable category to work in. The category of sets has too many morphisms between each pair of objects, which prevents us from defining types like C C C; thus, we will work in a subcategory of domains that imposes further topological restrictions. In particular, this subcategory will be capable of expressing that some programs may be partially defined (i.e., diverge sometimes). To model a reasonable eager functional programming language, we will ensure that this subcategory has finite coproducts, a symmetric monoidal closed structure for eager products and functions, and least fixed points of domain equations. We will not give a formal interpretation of a programming language into the category, but we will treat its objects as types and its morphisms as programs in the standard way. 2 Directed-Complete Partial Orders We begin by defining a subcategory of P o s e t\mathbf{Poset}Poset, called D C P O\mathbf{DCPO}DCPO, whose objects will ultimately correspond to types. Objects of D C P O\mathbf{DCPO}DCPO will be posets with particular sensible least upper bounds. Definition Let X=(U X,≤X)X = (U X, \le_X)X=(U X,≤X) be a poset. Say X X X is directed iff every finite subset of U X U X U X has an upper bound with respect to the ≤X\le_X≤X ordering. Corollary Every directed set X X X is nonempty, since the empty set is finite and its upper bound must therefore exist in X X X. Definition A directed-complete partial order (DCPO) is a poset D=(U D,≤D)D = (U D, \le_D)D=(U D,≤D) such that all directed subsets X⊆U D X \subseteq U D X⊆U D have a least upper bound, ⨆D X\bigsqcup_D X⨆DX, with respect to the ≤D\le_D≤D ordering. We will understand each type as a DCPO D D D, where the elements of a type are simply the elements of U D UD U D. For x,y∈U D x, y \in UD x,y∈U D, we understand x≤D y x \le_D y x≤Dy as “x x x is at most as defined as y y y”. For example, a diverging computation (i.e., “infinite loop”) will be less defined than a result value. Example Every discrete poset (A,=)(A, =)(A,=) is a DCPO, the discrete DCPO. The only directed subsets of A A A are singleton sets of the form {a}{{a}}{a} for a∈A a \in A a∈A, which have least upper bound a a a. For readability, we will abbreviate DCPO (A,=)(A, =)(A,=) as A A A. Discrete DCPOs will correspond to positive types. For example, 2={0,1}2 = {{0, 1}}2={0,1} will correspond to the Boolean type, the type type with two values. Example Consider the following poset, T\mathbb{T}T: Since every directed subset has a least upper bound, T\mathbb{T}T is a DCPO. Note, however, that not every subset has a least upper bound: the non-directed subset {0,1}{{0, 1}}{0,1} has no upper bound at all. T\mathbb{T}T will be the type of computations of a Boolean value, which can be in {0,1}{{0, 1}}{0,1} or can be a computation ∙\bullet∙ that diverges. Divergence is less defined than any Boolean value, but both Boolean values are on equal footing when it comes to definedness. Well-behaved functions between DCPOs, known as Scott-continuous functions, should preserve the DCPO structure. Definition Let D D D and E E E be DCPOs. Then, a Scott-continuous function f:D→E f : D \to E f:D→E is a (monotone) function D→E D \to E D→E that preserves directed suprema; i.e., f(⨆D X)=⨆E f(X)f(\bigsqcup_D X) = \bigsqcup_E f(X)f(⨆DX)=⨆Ef(X) for all directed X⊆U D X \subseteq U D X⊆U D. Remark Continuity will be used here to approximate computability. In particular, if f:D→E f : D \to E f:D→E is continuous, we may think about f f f as a computable function from type D D D to type E E E. We may now define the category D C P O\mathbf{DCPO}DCPO of DCPOs in the obvious way. Definition D C P O\mathbf{DCPO}DCPO is the category with DCPOs as objects and Scott-continuous functions as morphisms. Remark Each DCPO D D D has an associated topology (the Scott topology), built from continuous maps from D D D into the DCPO of truth values {⊥≤⊤}{{\bot \le \top}}{⊥≤⊤}, for which the given notion of Scott continuity is the same as topological continuity. Therefore, D C P O\mathbf{DCPO}DCPO is a full subcategory of T o p\mathbf{Top}Top, the category of topological spaces. The category D C P O\mathbf{DCPO}DCPO has plenty of structure, which will let us understand sum, eager product, and function types. Theorem D C P O\mathbf{DCPO}DCPO is bicartesian closed. Products and coproducts are inherited from P o s e t\mathbf{Poset}Poset, and the exponential D→E D \to E D→E is the DCPO consisting of all Scott-continuous functions from D D D to E E E, ordered pointwise. (Abramsky and Jung 1994) 3 Diverging Computations and Partial Functions As discussed in the introduction, in a programming language with recursive types, a computation may diverge (i.e., “infinitely loop”). While a value of Boolean type must be either 0 0 0 or 1 1 1, a computation of a Boolean may be a value 0 0 0 or 1 1 1, or it may diverge. Thus, for each DCPO D D D of values, we will define a corresponding DCPO D⊥{D}_\bot D⊥ of potentially-diverging computations. Definition Let D D D be a DCPO. The lifting of D D D, written D⊥{D}\bot D⊥, is a DCPO consisting of D D D with a fresh least element adjoined. Concretely: U(D⊥)≜{⊥}+D x≤D⊥y≜(x=⊥)∨(x≤D y) \begin{aligned} U({D}\bot) &\triangleq{{\bot}} + D \ x \le_{{D}_\bot} y &\triangleq(x = \bot) \lor (x \le_D y) \end{aligned} U(D⊥)x≤D⊥y≜{⊥}+D≜(x=⊥)∨(x≤Dy) Remark The DCPO T\mathbb{T}T is simply 2⊥{2}_\bot 2⊥. In an eager functional programming language, a function evaluates its definition and performs any suspended behaviors when applied to an argument. For example, values of the thunk type unit -> t are computations of type t, which may diverge when the function is applied to the unique value of type unit (typically written ()). If t corresponds to a DCPO D D D, we will see that unit -> t corresponds to D⊥{D}\bot D⊥. We may understand this phenomenon by viewing partiality as an effect, embodied by the _lift monad. Definition The lift monad(−)⊥:D C P O→D C P O{(-)}\bot : \mathbf{DCPO}\to \mathbf{DCPO}(−)⊥:DCPO→DCPO is the functor that sends a DCPO D D D to D⊥{D}\bot D⊥. Remark The lift monad is similar to the common maybe monad, (−)+1(-) + 1(−)+1, in that it also adjoins a fresh new element. However, ⊥\bot⊥ is less than all existing elements, whereas the element adjoined by (−)+1(-) + 1(−)+1 is incomparable to all existing elements. This expresses the subtle difference between the recoverable failure effect and partiality. For example, it is trivial to define a function of type D+1→2 D + 1 \to 2 D+1→2 that computes whether its input “failed”. However, it is impossible to write a function D⊥→2{D}_\bot \to 2 D⊥→2 that computes whether its input “failed” (was ⊥\bot⊥); such a function would not even be monotone. Informally, this corresponds to the undecidability of the Halting problem, which says that it is impossible to write a program that decides whether its input computation terminates or diverges. Now, we may work in the Kleisli category of this monad, following the typical development by Moggi (1991) of categorical models of effectful eager functional programming languages. Definition The Kleisli category of the lift monad, D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥, has the same objects as D C P O\mathbf{DCPO}DCPO. However, a morphism D→E D \to E D→E in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is a morphism D→E⊥D \to {E}_\bot D→E⊥ in D C P O\mathbf{DCPO}DCPO, a partial function continuous on a Scott-closed subset of D D D into E E E. Remark The category D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is equivalent to the subcategory of D C P O\mathbf{DCPO}DCPO where an object is a pointed DCPO (i.e., a DCPO with a least element) and morphisms are required to be _strict (i.e., least element preserving). This alternative presentation is common in the literature (Plotkin 1978), but we find the Kleisli category presentation more elegant: it aligns with the operational semantics of an eager functional programming language (where divergence is not legal to substitute as an input) and the usual semantics for monadic effects (Moggi 1991). Explanation of the significance behind this perspective is explained further by Levy (2003). Theorem Since D C P O\mathbf{DCPO}DCPO has coproducts, so too does the Kleisli category D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥. Although D C P O\mathbf{DCPO}DCPO has products, the object D×E D \times E D×E from D C P O\mathbf{DCPO}DCPO is not a categorical product in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥: the presence of the partiality effect breaks the existence condition of the universal property for products.2 Although D C P O\mathbf{DCPO}DCPO products are not categorical products in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥, they will still be useful for understanding eager product types. Theorem The categorical product in D C P O\mathbf{DCPO}DCPO is a symmetric monoidal product in D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥, written (⊗,I)(\otimes, I)(⊗,I). D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is also closed with respect to this monoidal structure. Let the internal hom D⊸E D \multimap E D⊸E be defined as D⇒D C P O E⊥D \Rightarrow\mathbf{DCPO}{E}_\bot D⇒DCPOE⊥. Then, (−)⊗D(-) \otimes D(−)⊗D is left adjoint to D⊸(−)D \multimap(-)D⊸(−). We will understand eager product and function types as ⊗\otimes⊗ and ⊸\multimap⊸, respectively. 4 Domains as Retracts of a Universal Space We may now define a category of domains in which we can solve domain equations, which will be a subcategory of D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥. Informally, a domain will be an DCPO whose elements can be encoded as sequences of bits. Definition Let U\mathbb{U}U be the DCPO N⊸2\mathbb{N}\multimap 2 N⊸2, partial functions from N\mathbb{N}N to 2 2 2.3 In other words, the elements of U\mathbb{U}U are the infinite streams of elements of 2⊥{2}_\bot 2⊥. We will call U\mathbb{U}U the universal domain. We will sometimes write elements of U\mathbb{U}U as infinite streams s s s of the form (x,s′)({x}, {s'})(x,s′), where x∈2⊥x \in {2}\bot x∈2⊥ and s′s's′ is another infinite stream, in place of a function i↦{x i=0 s′(i−1)i>0. i \mapsto \begin{cases} x & i = 0 \ s'(i - 1) & i > 0. \end{cases} i↦{x s′(i−1)i=0 i>0. Additionally, to disambiguate between the bottom element available from the Kleisli category and the bottom element of 2⊥{2}\bot 2⊥, we will refer to the latter as ∙\bullet∙ (as in T\mathbb{T}T). A domain will be a DCPO that can be encoded in the universal space U\mathbb{U}U. We formalize this encoding property using the categorical notion of retract. Definition Let A A A and B B B be objects in a category. A A A is a retract of B B B if there exist maps i A:A→B i_{A} : A \to B i A:A→B and r A:B→A r_{A} : B \to A r A:B→A such that i A;r A=id A i_{A};r_{A} = \text{id}_{A}i A;r A=id A, where ;;; denotes forward composition. Call i A i_{A}i A the encoder (or section) and r A r_{A}r A the decoder (or retraction). With this definition in hand, we may define “domain” in this setting. Definition A DCPO D D D is a domain iff it is a retract of U\mathbb{U}U in D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥. In other words, a domain is a DCPO that can be faithfully encoded into U\mathbb{U}U. Definition Define D o m\mathbf{Dom}Dom to be the full subcategory of D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥ consisting of domains. This category D o m\mathbf{Dom}Dom is what we were originally looking for: recursive types will make sense within D o m\mathbf{Dom}Dom. Let us now consider some examples of domains. Example The discrete DCPO 2 2 2 of Boolean values is a domain, with encoder and decoder defined by: i 2≜x↦(x,∙‾)r 2≜(x,s)↦x \begin{aligned} i_{2} &\triangleq x \mapsto ({x}, {\overline{\bullet}}) \ r_{2} &\triangleq({x}, {s}) \mapsto x \end{aligned} i 2r 2≜x↦(x,∙)≜(x,s)↦x where ∙‾\overline{\bullet}∙ is the infinite stream of ∙∈2⊥\bullet\in {2}\bot∙∈2⊥. Clearly, i 2;r 2=id 2 i{2};r_{2} = \text{id}_{2}i 2;r 2=id 2. In other words, a Boolean value is encoded as a single bit, the first in the infinite stream (followed by ∙\bullet∙ values). When working with coproducts, we write [−,−][-, -][−,−] for copairing, ι 1\iota_1 ι 1 for the left injection, and ι 2\iota_2 ι 2 for the right injection. Example If D D D and E E E are domains, then the coproduct D+E D + E D+E in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is a domain. Its encoder and decoder defined by: i D+E≜[d↦(0,i D d),e↦(1,i E e)]r D+E≜(x,s)↦{ι 1(r D s)if x=0 ι 2(r E s)if x=1⊥if x=∙ \begin{aligned} i{D + E} &\triangleq[d \mapsto ({0}, {i_{D}d}), e \mapsto ({1}, {i_{E}e})] \ r_{D + E} &\triangleq({x}, {s}) \mapsto \ \begin{cases} \iota_1(r_{D}s) &\text{if } x = 0 \ \iota_2(r_{E}s) &\text{if } x = 1 \ \bot &\text{if } x = \bullet \end{cases} \end{aligned} i D+Er D+E≜[d↦(0,i Dd),e↦(1,i Ee)]≜(x,s)↦⎩⎨⎧ι 1(r Ds)ι 2(r Es)⊥if x=0 if x=1 if x=∙ Informally, this encodes an element of D+E D + E D+E by using the first bit as a “tag bit” and encoding the data in the remaining bits. The nullary coproduct 0 0 0 is a domain, as well, with i 0≜¡i_{0} \triangleq¡i 0≜¡ and r 0≜s↦⊥r_{0} \triangleq s \mapsto \bot r 0≜s↦⊥. Example If D D D and E E E are domains, then the monoidal (“eager”) product D⊗E D \otimes E D⊗E in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is a domain. Its encoder and decoder defined by: i D⊗E≜⟨d,e⟩↦(h−1;[i D d,i E e])r D⊗E≜s↦⟨r D(ι 1;h;s),r E(ι 2;h;s)⟩ \begin{aligned} i{D \otimes E} &\triangleq\langle d, e \rangle \mapsto (h^{-1};[i_{D}d, i_{E}e]) \ r_{D \otimes E} &\triangleq s \mapsto \langle r_{D}(\iota_1;h;s), r_{E}(\iota_2;h;s) \rangle \end{aligned} i D⊗Er D⊗E≜⟨d,e⟩↦(h−1;[i Dd,i Ee])≜s↦⟨r D(ι 1;h;s),r E(ι 2;h;s)⟩ where h:N+N→N h : \mathbb{N}+ \mathbb{N}\to \mathbb{N}h:N+N→N is a standard isomorphism (e.g., interleaving via evens and odds). Informally, this encodes an element of D⊗E D \otimes E D⊗E by encoding its components and interleaving their bit representations. The nullary monoidal product I={⋆}I = {{\star}}I={⋆} is a domain, as well, with i I≜⋆↦∙‾i_{I} \triangleq\star \mapsto \overline{\bullet}i I≜⋆↦∙ and r I≜s↦⋆r_{I} \triangleq s \mapsto \star r I≜s↦⋆. Example If D D D and E E E are domains, then the internal hom D⊸E≜D⇒D C P O E⊥D \multimap E \triangleq D \Rightarrow_\mathbf{DCPO}{E}\bot D⊸E≜D⇒DCPOE⊥ in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ is a domain. We elide the details of the construction here, although they are described in full by Plotkin (1978). Informally, an element of D⊸E D \multimap E D⊸E is encoded as its “graph”, expressing all of its input-output pairs as a bit stream. Details about Scott’s model, a similar construction, are given in a recent blog post by my co-mentee. 5 Recursive Types We will now see that the category D o m\mathbf{Dom}Dom has least fixed point solutions to domain equations. Definition Say encoding/decoding pair (i:D→E,r:E→D)(i_{} : D \to E,\ r_{} : E \to D)(i:D→E,r:E→D) is an embedding-projection pair if r;i≤id E r_{};i_{} \le \text{id}{E}r;i≤id E. Define D o m◃{\mathbf{Dom}\triangleleft}Dom◃ to be the wide subcategory of D o m\mathbf{Dom}Dom where a morphism D→D o m◃E D \to_{\mathbf{Dom}_\triangleleft}E D→Dom◃E is an embedding-projection pair, written D◃E{D} \triangleleft {E}D◃E. Theorem In D o m◃{\mathbf{Dom}_\triangleleft}Dom◃, every ω\omega ω-chain has a colimit. (Smyth and Plotkin 1982) In other words, every diagram of the following shape has a colimit: Expressed in D o m\mathbf{Dom}Dom: Lemma D o m◃{\mathbf{Dom}_\triangleleft}Dom◃ inherits coproducts and monoidal products from D o m\mathbf{Dom}Dom. Corollary Every endofunctor F:D o m◃→D o m◃F : {\mathbf{Dom}\triangleleft}\to {\mathbf{Dom}\triangleleft}F:Dom◃→Dom◃ has a least fixed point object μ F≅F(μ F)\mu{F} \cong F(\mu{F})μ F≅F(μ F), defined as the colimit of the following ω\omega ω-chain: D n≜F n(0)u n≜F n(¡F D)\begin{aligned} D_n &\triangleq F^n(0) \ u_n &\triangleq F^n(¡_{FD}) \end{aligned}D nu n≜F n(0)≜F n(¡F D) Using this construction, we can build inductive types in D o m\mathbf{Dom}Dom. Example Let F(X)=I+X F(X) = I + X F(X)=I+X, where I={⋆}I = {{\star}}I={⋆} is the monoidal unit (eager nullary product). Then, the following diagram has a colimit μ F\mu{F}μ F: Here, μ F≅N\mu{F} \cong \mathbb{N}μ F≅N, the discrete domain of natural numbers. Example Let F(X)=I+(X⊗X)F(X) = I + (X \otimes X)F(X)=I+(X⊗X). Then, the following diagram has a colimit μ F\mu{F}μ F: Here, μ F\mu{F}μ F is the discrete domain of binary trees. In these examples, F F F is built from coproducts and (monoidal) products. Our original goal was to define a type C C C isomorphic to C⊸2 C \multimap 2 C⊸2; however, this places C C C in a contravariant position, seemingly violating functoriality of F F F. However, in this category, we can understand internal homs as covariant in both positions. Definition Let ⊸:D o m op×D o m→D o m\multimap: {\mathbf{Dom}}^\text{op} \times \mathbf{Dom}\to \mathbf{Dom}⊸:Dom op×Dom→Dom be the internal hom bifunctor. We may define the covariant bifunctor ⊸◃:D o m◃×D o m◃→D o m◃\multimap_\triangleleft: {\mathbf{Dom}\triangleleft}\times {\mathbf{Dom}\triangleleft}\to {\mathbf{Dom}\triangleleft}⊸◃:Dom◃×Dom◃→Dom◃ as follows (Smyth and Plotkin 1982): D⊸◃E≜D⊸E(i D,r D)⊸◃(i E,r D)≜(r D;(−);i E,i D;(−);r E) \begin{aligned} D \multimap\triangleleft E &\triangleq D \multimap E \ (i_{D}, r_{D}) \multimap_\triangleleft(i_{E}, r_{D}) &\triangleq(r_{D};(-);i_{E},\ i_{D};(-);r_{E}) \end{aligned} D⊸◃E(i D,r D)⊸◃(i E,r D)≜D⊸E≜(r D;(−);i E,i D;(−);r E) Here, we have: i D:D→D o m D′r D:D′→D o m D i E:E→D o m E′r E:E′→D o m E r D;(−);i E:(D⊸E)→D o m(D′⊸E′)i D;(−);r E:(D′⊸E′)→D o m(D⊸E)\begin{aligned} i_{D} &: D \to_\mathbf{Dom}D' \ r_{D} &: D' \to_\mathbf{Dom}D \ i_{E} &: E \to_\mathbf{Dom}E' \ r_{E} &: E' \to_\mathbf{Dom}E \ r_{D};(-);i_{E} &: (D \multimap E) \to_\mathbf{Dom}(D' \multimap E') \ i_{D};(-);r_{E} &: (D' \multimap E') \to_\mathbf{Dom}(D \multimap E) \end{aligned} i Dr Di Er Er D;(−);i Ei D;(−);r E:D→DomD′:D′→DomD:E→DomE′:E′→DomE:(D⊸E)→Dom(D′⊸E′):(D′⊸E′)→Dom(D⊸E) Now, we may finally find the type C C C. Example Let F(X)=X⊸◃2 F(X) = X \multimap_\triangleleft 2 F(X)=X⊸◃2. Then, the following diagram has a colimit μ F\mu{F}μ F: Here, μ F\mu{F}μ F is the type C C C. 6 Domains via the Untyped λ\lambda λ-Calculus Domains can alternatively be characterized as subsets of the universal domain, U\mathbb{U}U. Theorem Let r:U→U r_{} : \mathbb{U}\to \mathbb{U}r:U→U in D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ be an idempotent map, with r;r=r r{}; r_{} = r_{}r;r=r. Define FP(r)\text{FP}({r_{}})FP(r) to be the set of fixed points of r r_{}r: FP(r)≜{x∈U∣x=r(x)}\text{FP}({r_{}}) \triangleq{{x \in \mathbb{U}\mid x = r(x)}}FP(r)≜{x∈U∣x=r(x)} In other words, FP(r)\text{FP}({r_{}})FP(r) is the equalizer of id U\text{id}{\mathbb{U}}id U and r r{}r in D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥. Then, FP(r)\text{FP}({r_{}})FP(r) is is a retract of U\mathbb{U}U, where the encoder is the inclusion i:FP(U)↪U i_{} : \text{FP}({\mathbb{U}}) \hookrightarrow \mathbb{U}i:FP(U)↪U and the decoder is r r_{}r. Example Define: ff≜(0,∙‾)tt≜(1,∙‾)if((x,s);s 1;s 0)≜{s 1 x=1 s 0 x=0⊥x=∙\begin{aligned} \textsf{ff}&\triangleq({0}, {\overline{\bullet}}) \ \textsf{tt}&\triangleq({1}, {\overline{\bullet}}) \ \textsf{if}({({x}, {s})};\ {s_1};\ {s_0}) &\triangleq \begin{cases} s_1 & x = 1 \ s_0 & x = 0 \ \bot & x = \bullet \end{cases} \end{aligned}ff tt if((x,s);s 1;s 0)≜(0,∙)≜(1,∙)≜⎩⎨⎧s 1s 0⊥x=1 x=0 x=∙ and let r 2(s)≜if(s;tt;ff)r_{2}(s) \triangleq\textsf{if}({s};\ {\textsf{tt}};\ {\textsf{ff}})r 2(s)≜if(s;tt;ff) Then, we have FP(r 2)≅2\text{FP}({r_{2}}) \cong 2 FP(r 2)≅2 in D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥. Other domains are defined similarly, “traversing” the data structure such that the fixed points are precisely the desired elements. In fact, we can use the untyped λ\lambda λ-calculus to define such maps r r r. The untyped λ\lambda λ-calculus can be interpreted into a category with a reflexive object. Definition In a category, a reflexive object is an object D D D such that the internal hom D⊸D D \multimap D D⊸D is a retract of D D D. (Scott 1980) Theorem The universal domain, U\mathbb{U}U, is a reflexive object. This fact is straightforward to observe. Every object is a retract of itself, so U\mathbb{U}U is a domain. Since internal homs of domains are domains, U⊸U\mathbb{U}\multimap\mathbb{U}U⊸U is a domain, too. By definition, domains are retracts of U\mathbb{U}U, so U⊸U\mathbb{U}\multimap\mathbb{U}U⊸U is a retract of U\mathbb{U}U. Theorem Terms of the untyped λ\lambda λ-calculus can be interpreted as elements of a reflexive object D D D, such as D≜U D \triangleq\mathbb{U}D≜U. We define ⟦−⟧ε\llbracket {-} \rrbracket_\varepsilonε as the interpretation of an untyped λ\lambda λ-term with free variables in the domain of ε\varepsilon ε, which maps from variables to D D D: ⟦x⟧ε≜ε(x)⟦λ(x.e)⟧ε≜i(u x↦⟦e⟧ε,x↦u x)⟦e 1(e 2)⟧ε≜r(⟦e 1⟧ε)(⟦e 2⟧ε) \begin{aligned} \llbracket {x} \rrbracket_\varepsilon &\triangleq\varepsilon(x) \ \llbracket {\lambda({x}.\ {e})} \rrbracket_\varepsilon &\triangleq{i_{}}(u_x \mapsto \llbracket {e} \rrbracket_{\varepsilon, x \mapsto u_x}) \ \llbracket {e_1(e_2)} \rrbracket_\varepsilon &\triangleq{r_{}}(\llbracket {e_1} \rrbracket_\varepsilon)(\llbracket {e_2} \rrbracket_\varepsilon) \end{aligned} εεε≜ε(x)≜i(u x↦ε,x↦u x)≜r(ε)(ε) The interpretations respect the α\alpha α- and β\beta β-equivalence principles and function extensionality. Going forward, we will omit ⟦−⟧ε\llbracket {-} \rrbracket_\varepsilonε when defining maps using closed λ\lambda λ-terms. For example, we define the domain 2 2 2 as FP(λ(s.if(s;tt;ff)))\text{FP}({\lambda({s}.\ {\textsf{if}({s};\ {\textsf{tt}};\ {\textsf{ff}})})})FP(λ(s.if(s;tt;ff))). More strikingly, we may use the strict Y\textsf{Y}Y fixed-point combinator to solve recursive domain equations. Definition Define the strict Y\textsf{Y}Y combinator as follows: Y:((U⊸U)⊸(U⊸U))→(U⊸U)Y≜λ(f.∇f(∇f))\begin{aligned} \textsf{Y}&: ((\mathbb{U}\multimap\mathbb{U}) \multimap(\mathbb{U}\multimap\mathbb{U})) \to (\mathbb{U}\multimap\mathbb{U}) \ \textsf{Y}&\triangleq\lambda({f}.\ {\nabla_f(\nabla_f)}) \end{aligned}Y Y:((U⊸U)⊸(U⊸U))→(U⊸U)≜λ(f.∇f(∇f)) where ∇f≜λ(x.f(λ(y.(x(x))(y))))\nabla_f \triangleq\lambda({x}.\ {f(\lambda({y}.\ {(x(x))(y)}))})∇f≜λ(x.f(λ(y.(x(x))(y)))). Example We may solve recursive domain equations using the Y\textsf{Y}Y combinator. To solve for domain N≅I+N\mathbb{N}\cong I + \mathbb{N}N≅I+N, define: r N≜Y(λ(r N.r I+N))r_{\mathbb{N}} \triangleq\textsf{Y}(\lambda({r_{\mathbb{N}}}.\ {r_{I + \mathbb{N}}}))r N≜Y(λ(r N.r I+N)) where r I+N r_{I + \mathbb{N}}r I+N is the retraction for sums. Note that on the right side of the definition, r N r_{\mathbb{N}}r N is a bound variable name, to be used within r I+N r_{I + \mathbb{N}}r I+N. Then, N\mathbb{N}N may be defined as FP(r N)\text{FP}({r_{\mathbb{N}}})FP(r N). This technique works for recursive domain equations involving internal hom, as well. To solve for domain C≅C⊸2 C \cong C \multimap 2 C≅C⊸2, define: r C≜Y(λ(r C.r C⊸2))r_{C} \triangleq\textsf{Y}(\lambda({r_{C}}.\ {r_{C \multimap 2}}))r C≜Y(λ(r C.r C⊸2)) where r C⊸2 r_{C \multimap 2}r C⊸2 is the retraction for internal homs. Then, C C C may be defined as FP(r C)\text{FP}({r_{C}})FP(r C). 7 Conclusion In this blog post, we presented a well-known technique from domain theory for solving domain equations in a particular subcategory of P o s e t\mathbf{Poset}Poset. This subcategory, D o m\mathbf{Dom}Dom, has coproducts, is symmetric closed monoidal, and has least fixed points of domain equations. Of course, this is only the very foundations of domain theory; there are many other connections to be explored. For example, the approach we considered here can be extended to support polymorphism via partial equivalence relations. Additionally, there are deep connections to computability theory discussed in the given references. This post represents some of the many things I learned at the Topos Institute as a Summer Research Associate. I owe a great deal of gratitude to Dana Scott, who spent countless hours throughout the summer (and beyond!) sharing with me his vast experience in and passion for domain theory and programming languages. Additionally, I would like to thank David Spivak, Brendan Fong, and everyone at the Topos Institute for the support, guidance, and insightful conversations. References Abramsky, Samson, and Achim Jung. 1994. “Domain Theory.” In Handbook of Logic in Computer Science, 3:167. Gunter, C. A., and Dana Scott. 1990. “Semantic Domains.” In Formal Models and Semantics, 633–74. Elsevier. Harper, Robert. 2016. Practical Foundations for Programming Languages. Cambridge University Press. Levy, Paul Blain. 2003. Call-By-Push-Value. Dordrecht: Springer Netherlands. Moggi, Eugenio. 1991. “Notions of Computation and Monads.”Information and Computation 93 (1): 55–92. Plotkin, Gordon. 1978. “T ω\mathbb{T}^\omega T ω as a Universal Domain.”Journal of Computer and System Sciences 17 (2): 209–36. Scott, Dana. 1976. “Data Types as Lattices.”SIAM Journal on Computing 5 (3): 522–87. ———. 1980. “Relating Theories of the λ-Calculus.”To H.B. Curry: Essays on Combinatory Logic, Lambda Calculus and Formalism. Smyth, Michael, and Gordon Plotkin. 1982. “The Category-Theoretic Solution of Recursive Domain Equations.”SIAM Journal on Computing 11 (4): 23. Footnotes For a more detailed account of recursive types, including program-level recursion via such types, see Harper (2016), Chapter 20.↩︎ In fact, D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥_does have categorical products, defined as D×D C P O⊥E≜D⊥×D C P O E⊥D \times_{\mathbf{DCPO}\bot}E \triangleq{D}\bot \times_\mathbf{DCPO}{E}\bot D×DCPO⊥E≜D⊥×DCPOE⊥. This is the “lazy product”, sometimes written &\&& (e.g., in substructural type theory). However, D C P O⊥{\mathbf{DCPO}\bot}DCPO⊥ does not have exponential objects. (Smyth and Plotkin 1982)↩︎ In Plotkin (1978), this object is written T ω\mathbb{T}^\omega T ω. While it is understood in the category of pointed DCPOs with strict maps (and thus is not the equivalent object to U\mathbb{U}U here), an object is a retract of T ω\mathbb{T}^\omega T ω in this category iff it is a retract of U\mathbb{U}U in D C P O⊥{\mathbf{DCPO}_\bot}DCPO⊥.↩︎ Leaving a comment will set a cookie in your browser. For more information, see our cookies policy. Legal status and annual reportsCookies policy
190532	https://www.turingtalks.ai/p/cosine-similarity-explained-the-math-behind-llms	Cosine Similarity Explained: The Math Behind LLMs Search Home Videos Author Login Home Posts Cosine Similarity Explained: The Math Behind LLMs Cosine Similarity Explained: The Math Behind LLMs Cosine similarity measures the angle between vectors to compare meaning in text. This simple math powers LLMs, enabling search, recommendation systems, and semantic understanding. Sep 10, 2025 When you talk to a large language model (LLM), it feels like the model understands meaning. But under the hood, the system relies on numbers, vectors, and math to find relationships between words and sentences. One of the most important tools that makes this possible is cosine similarity. If you want to know how an LLM can judge that two sentences mean almost the same thing, cosine similarity is the key. This article explains cosine similarity in plain language, shows the math behind it, and connects it to the way modern language models work. By the end, you will see why this simple idea of measuring angles between vectors powers search, chatbots, and many other AI systems. What Is Cosine Similarity Imagine you have two sentences. To a computer, they are not words but vectors, a long lists of numbers that capture meaning. Cosine similarity measures how close these two vectors are, not by their length, but by the angle between them. Cosine Similarity Think of two arrows starting from the same point. If they point in the same direction, the angle between them is zero, and cosine similarity is one. If they point in opposite directions, the angle is 180 degrees, and cosine similarity is negative one. If they are at a right angle, the cosine similarity is zero. So, cosine similarity tells us whether two vectors are pointing in the same general direction. In language tasks, this means it tells us whether two pieces of text carry a similar meaning. The Math Behind Cosine Similarity To understand cosine similarity, we need to look at a bit of math. The cosine of an angle in geometry is the ratio between the dot product of two vectors and the product of their magnitudes. Written as a formula, cosine similarity looks like this: cosine_similarity(A, B) = (A · B) / (\|\|A\|\| \|\|B\|\|) Here: A · B is the dot product of vectors A and B. \|\|A\|\| is the magnitude (length) of vector A. \|\|B\|\| is the magnitude of vector B. The dot product multiplies corresponding numbers in the two vectors and adds them up. The magnitude of a vector is like finding the length of an arrow, using the Pythagorean theorem. This formula always gives a value between -1 and 1. A value close to 1 means the vectors are pointing in nearly the same direction. A value close to 0 means they are unrelated. A value close to -1 means they are opposite. A Simple Example Let’s see a short example using Python. Suppose you want to check how similar two short texts are. We can use scikit-learn to turn them into vectors and then compute cosine similarity. ``` Import the TfidfVectorizer to convert text into numerical vectors using TF-IDF from sklearn.feature_extraction.text import TfidfVectorizer Import cosine_similarity to measure similarity between two vectors from sklearn.metrics.pairwise import cosine_similarity Two example sentences to compare texts = [ "I love machine learning", "I love deep learning" ] Convert the text data into TF-IDF vectors TF-IDF (Term Frequency - Inverse Document Frequency) gives more importance to unique words vectorizer = TfidfVectorizer().fit_transform(texts) Convert the sparse matrix into a dense array for easier handling vectors = vectorizer.toarray() Compute cosine similarity between the first and second sentence Cosine similarity measures how close the vectors are (1 means identical, 0 means no similarity) similarity = cosine_similarity([vectors], [vectors]) Print the similarity score print("Cosine similarity:", similarity) ``` The code starts by importing two important tools. TfidfVectorizer is responsible for turning text into numbers, while cosine_similarity measures how similar two sets of numbers are. Together, they let us compare text in a way a computer can understand. Next, we define the sentences we want to compare. In this example, we use “I love machine learning” and “I love deep learning.” These two sentences share some words such as “I,” “love,” and “learning,” while differing in one word: “machine” versus “deep.” This makes them good examples to test, because they are clearly related but not exactly the same. The vectorizer then builds a vocabulary from all the unique words across the two sentences. For these inputs, the vocabulary becomes ["deep", "learning", "love", "machine"]. This means the program now has a list of all the words it will track when building the numerical representation of the sentences. Each sentence is then converted into a vector, which is simply a list of numbers. These numbers are not just raw word counts. Instead, they are weighted using TF-IDF, which stands for Term Frequency–Inverse Document Frequency. TF-IDF gives more importance to words that matter in a sentence and less importance to very common words. In simplified form, the first sentence becomes something like [0. 0.50154891 0.50154891 0.70490949], while the second becomes [0.70490949 0.50154891 0.50154891 0. ]. The numbers may look small, but what matters is their relative values. The .toarray() method then converts these vectors into standard Python arrays. This makes them easier to handle, since the TF-IDF output is stored in a special sparse format by default. Once the sentences are represented as vectors, cosine similarity is applied. This step checks the angle between the two vectors. If the vectors point in exactly the same direction, their similarity score will be one. If they are unrelated, the score will be close to zero. If they point in opposite directions, the score will be negative. In this case, because the two sentences share most of their words, the vectors point in a similar direction, so the cosine similarity falls somewhere around 0.5 to 0.7. In simple terms, this code shows how a computer can compare two sentences by turning them into vectors of numbers and then checking how close those vectors are. By using cosine similarity, the program can judge not just whether the sentences share words, but also how strongly they overlap in meaning. Cosine Similarity in Embeddings In practice, LLMs like GPT or BERT do not use simple word counts. Instead, they use embeddings. An embedding is a dense vector that captures meaning. Each word, phrase, or sentence is turned into a set of numbers that place it in a high-dimensional space. In this space, words with similar meaning are close together. For example, the embeddings for “king” and “queen” are closer than the embeddings for “king” and “table.” Cosine similarity is the tool that allows us to measure how close two embeddings are. When you search for “dog,” the system can look for embeddings that point in a similar direction. That way, it finds results about “puppy,” “canine,” or “pet” even if those exact words are not in your query. How LLMs Use Cosine Similarity Large language models use cosine similarity in many ways. When you ask a question, the model encodes your input into a vector. It then compares this vector with stored knowledge or with candidate answers using cosine similarity. For semantic search, cosine similarity helps rank documents. A system can embed all documents into vectors, then embed your query and compute similarity scores. The documents with the highest scores are the most relevant. In clustering, cosine similarity helps group sentences that have related meaning. In recommendation systems, it helps match users to items by comparing their preference vectors. Even when generating answers, LLMs rely on vector similarity to decide which words or phrases best follow in context. Cosine similarity gives the model a simple but powerful way to measure closeness of meaning. Limits of Cosine Similarity While cosine similarity is powerful, it has limits. It depends heavily on the quality of embeddings. If embeddings fail to capture meaning well, similarity scores may not reflect real-world closeness. Also, cosine similarity only measures direction. Sometimes, magnitude contains useful information too. For example, a sentence embedding might have a length that reflects confidence. By ignoring it, cosine similarity may lose part of the picture. Still, despite these limits, cosine similarity remains one of the most widely used methods in natural language processing. Why It Matters for LLMs Cosine similarity is not just a math trick. It is a bridge between human language and machine understanding. It allows a model to treat meaning as geometry, turning questions and answers into points in space. Without cosine similarity, embeddings would be less useful, and tasks like semantic search, clustering, and ranking would be harder. By reducing the problem to measuring angles, we make meaning measurable and usable. Every time you search on Google, chat with an AI, or use a recommendation engine, cosine similarity is at work behind the scenes. Conclusion Cosine similarity explains how LLMs judge the closeness of meaning between words, sentences, or even whole documents. It works by comparing the angle between vectors, not their length, which makes it ideal for text. With embeddings, cosine similarity becomes the foundation of semantic search, clustering, recommendations, and many other tasks in natural language processing. The next time an AI gives you an answer that feels “close enough,” remember that a simple mathematical idea, measuring the angle between two arrows, is doing much of the heavy lifting. Keep Reading Jul 16, 2025 • 6 min read How to Build a Simple Sentiment Analyzer Using Hugging Face Transformer In this article, we will look at writing a sentiment analyzer using Hugging Face Transformer, a powerful tool in the world of NLP. Manish Shivanandhan Aug 20, 2025 • 6 min read Bag of Words: The Foundation of Language Models Bag of Words explained—how it works, why it matters, and how it laid the foundation for today’s large language models. Manish Shivanandhan Aug 13, 2025 • 8 min read How AI Agents Remember Things: The Role of Vector Stores in LLM Memory Learn how vector databases like FAISS and Pinecone give AI agents a sense of memory. Manish Shivanandhan View more TuringTalks Learn to build and deploy LLMs to production. Subscribe © 2025 TuringTalks. 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190533	https://www.geogebra.org/m/NWv6tz3p	RELACIONES EQUIVALENTES A LA PENDIENTE DE UNA RECTA OBJETIVO OBSERVA TIPOS DE PENDIENTES EXPLICACIÓN SOBRE LA PENDIENTE EN UN PLANO CARTESIANO Graficas_Lineas Graficas_Lineas.pdf CONOCE RAZON DE CAMBIO (FUNCIÓN LINEAL)Las funciones lineales sirven para describir una gran cantidad de fenómenos en los que se relacionan dos magnitudes que varían proporcionalmente. Es decir, un cambio en la variable independiente induce un cambio proporcional en la variable dependiente . Una razón de cambio es una medida de la variación de la variable dependiente comparada con la variación que se produce en la variable independiente. En una función lineal la razón de cambio permanece constante: Ya sabes que en una ecuación lineal el valor de la pendiente (m) se obtiene con la ecuación de la recta: Leer más: APRENDE EJERCITA TUS HABILIDADES CONOCE Nuevos recursos Descubrir recursos Descubre temas Descarga aquí nuestras aplicaciones: © 2025 GeoGebra®
190534	https://www.chegg.com/homework-help/questions-and-answers/-z0-boson-rest-mass-energy-91187-gev-produced-collisions-positrons-electrons-e-e-z0-much-k-q6079263	Solved a. Z0 boson has a rest mass energy of 91.187 GeV. \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Physics Physics questions and answers a. Z0 boson has a rest mass energy of 91.187 GeV. It is produced in collisions of positrons and electrons. e- + e+ --> Z0 How much kinetic energy must the positrons and electrons in symmetric colliding beams have to produce the Z0 boson? The rest mass energy for electrons and positrons is 0.511 MeV. Creating Z0 boson via positron beam striking Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: a. Z0 boson has a rest mass energy of 91.187 GeV. It is produced in collisions of positrons and electrons. e- + e+ --> Z0 How much kinetic energy must the positrons and electrons in symmetric colliding beams have to produce the Z0 boson? The rest mass energy for electrons and positrons is 0.511 MeV. Creating Z0 boson via positron beam striking a. Z 0 boson has a rest mass energy of 91.187 GeV. It is produced in collisions of positrons and electrons. e- + e+ --> Z 0 How much kinetic energy must the positrons and electrons in symmetric colliding beams have to produce the Z 0 boson? The rest mass energy for electrons and positrons is 0.511 MeV. Creating Z 0 boson via positron beam striking electron target b. Instead, suppose a beam of positrons strike electrons that are at rest. How much energy must the positrons have to produce the Z 0 boson? c. Compare and comment on your answers to part a and b. Here’s the best way to solve it.Solution Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! To find the kinetic energy required to produce a Z0 boson in symmetric colliding beams, consider the conservation of energy and the fact that the total energy in the center of mass frame is twice the kinetic energy of one particle plus their rest mass energies, which should equal the rest mass energy of the Z0 boson. a)Photon-nucleus pair production can only occur if the photons have an energy exceeding twice the rest energy (mec2) of an electron (0.511 MeV rest energy doubled to1.022 MeV) b) There are different processes how an electron-po… View the full answer Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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190535	https://pleclair.ua.edu/PH253/Homework/Fall_2013/HW7/HW7_18Nov13_SOLN.pdf	PH 253 / LeClair Fall 2013 Problem Set 7 Solutions 1. The speciﬁc heat at constant volume per electron is deﬁned as cV = d⟨E⟩ dT (1) where ⟨E⟩is the average energy of electrons. A classical electron would have an average energy of 3 2kBT. Presuming the electrons follow a Fermi-Dirac distribution, estimate the speciﬁc heat for a collection of free electrons. Presume low temperatures, i.e., kBT ≪EF . Hint: electrons that have been thermally excited above the Fermi energy behave basically as free electrons. What fraction of electrons are excited above EF , roughly? Solution: Roughly speaking, only electrons within a few kBT of the Fermi energy EF will con-tribute to the speciﬁc heat, because only those electrons will have available states at a nearby energy. How many kBT should we take? We could characterize the ’width’ of the distribution f(E) by taking d f/dE (a function peaked at EF ) and ﬁnding its full width at half maximum, which is about 3.5kBT. About half of that width (the upper half) ought to be electrons available. (Or you could have just said “about three kBT” and it amounts to the same thing.)The fraction of electrons contributing then would be fraction = 3.5kBT 2EF (2) The average energy of these electrons would then be the classical result 3 2kBT times the fraction of electrons actually available we just found: ⟨E⟩= 3 2kBT 3.5kBT 2EF = 21 8 k2 BT 2 EF (3) The speciﬁc heat is then approximately cV = d⟨E⟩ dT ≈21 4 k2 BT EF = 5.25k2 BT EF (4) We ﬁnd cV linear in T, which agrees with experiments at low temperature. The exact result using the full Fermi-Dirac distribution is π2 2 k2 BT EF ≈4.9k2 BT EF , oﬀby about 7%. Of course, we could just have easily decided to use 2 or 3 kBT as our fraction, so the agreement is a bit spurious, but the facts that cV ∝T and that we have the right order of magnitude still hold true. 2. Following section 10.6 in your textbook (ﬁnd one), you can ﬁnd the total energy of a photon gas U as well as the total number of photons in the gas N.i Recall that a photon gas was our model for blackbody radiation! (a) What is the total speciﬁc heat CV for the photon gas, noting CV = ∂U ∂T (5) (b) How does the average energy per photon (U/N) vary with temperature? (c) Given this average energy per photon, estimate the speciﬁc heat per photon using the expression given in problem 1. Solution: The internal energy of a photon gas of volume V , using the link in the footnotes, is U = 8π5k4 B 15c3h3 V T 4 (6) You’d ﬁnd this by integrating the Planck formula for the energy density over all frequencies, by the way. The total speciﬁc heat (or heat capacity) is then CV = ∂U ∂T = 32π5k4 B 15c3h3 V T 3 (7) The same link gives us the expected number of photons in the gas: N = 16πk3 Bζ(3) c3h3 V T 3 (8) where ζ(n) is the Riemann zeta function, and ζ(3) ≈1.202. The average energy per photon then goes as U N = π4kBT 30ζ(3) (9) The speciﬁc heat per photon is then readily found. cV = ∂(U/N) ∂T = πkB 240ζ(3) (10) The heat capacity per photon is a constant, independent of temperature. 3. Use the free electron theory to determine the Fermi energy and density of states vs. energy for a two-dimensional metal. Take N as the average number of electrons per unit area. Solution: In two dimensions, we can write the energy as a function of ⃗ k in terms of the x and y iFor N, use the expression for dN in Eq. 10.38 and integrate it over all energies. Or look here: wikipedia.org/wiki/Photon_gas. Always read the footnotes. components of the k vector: E(⃗ k) = ℏ2 2m k2 x + k2 y (11) What we need to ﬁnd is the density of allowed k values. Given a crystal of side L, our boundary conditions dictate that the k components must come in integer multiples of 2π/L. If we made a plot with axes kx and ky, there would be allowed states at every integer multiple of 2π/L along each axis, making a square lattice of such points. The area each allowed state takes up is then dA=(2π/L)2. That’s how much “k-space” each allowed state takes up. What about the total number of states and their area? We have as many states as it takes to ﬁll all of them up to the Fermi wave vector kF , so all the allowed states lie within a circle of radius kF . Their total area is thus A=πk2 F . The number of states is then the total area of all states divided by the area of a single state, times 2 to account for the fact that each state can have one spin up and one spin down electron: N = 2 A dA = 2 πk2 F (2π/L)2 = k2 F L2 2π (12) Inverting, we can ﬁnd what kF must be given N electrons in a square of side L in two dimensions: \|⃗ kF \| = kF = √ 2πN L (13) The Fermi energy is then readily found from the energy-wavevector relationship EF = ℏ2k2 F 2m = πNℏ2 mL2 (14) How about the density of states? From the preceding equation, at an energy E we know what the energy is as a function of N by virtue of knowing E(k) and k(N). Equation 13 tells us that for a given N, k2 =2πN/L2, thus E(N) = ℏ2 2m (k(N))2 = ℏ2 2m 2πN L2 = πℏ2N mL2 (15) The density of states per unit volume just depends on dE/dN and the volume V =L2: g(E) = 1 V dN dE = 1 L2 1 dE/dN = 1 L2 mL2 πℏ2 = m πℏ2 (16) In 2D, the density of states is a constant, independent of energy. 4. (a) Obtain an expression for the Fermi energy at T = 0 K for an electron gas in a three dimensional metal in terms of the total number of electrons, the volume, and fundamental constants. (b) At T = 0 K, what is the average speed, in terms of the Fermi energy, of a three-dimensional electron gas in a metal?ii Solution: The Fermi energy at T =0 is given by EF T=0 = ℏ2 2m 3π2N V !2/3 (17) where N is the number of electrons in a volume V . A derivation may be found in your textbook, or here: This is the energy of the highest-energy electrons in a metal at T =0. If all this energy is available as kinetic energy, Ef = 1 2mv2 = ⇒ v = s 2EF m (18) 5. Show that the average kinetic energy of an electron in a three-dimensional electron gas at 0 K is Eav = 3 5EF .iii Solution: The Fermi energy, the energy of the highest occupied quantum state in a system of fermions at absolute zero temperature, is a function of the number of fermions as found in the last problem. If there are N particles in a volume V , then the highest occupied state has an energy EF (T = 0) = ℏ2 2m 3π2N V !2/3 (19) Thus, as more and more particles are added, it takes more and more energy to add the last particle. The total energy of the system, if there are enough particles to consider the distribution of states to be quasi-continuous, is given by Etot = N Z 0 EF (N′) dN′ (20) iiOpen your textbook to 10.7. iiiOpen your textbook to 10.7. The average energy is simply the total energy divided by the number of particles: Eavg = 1 N N Z 0 EF (N′) dN′ = 1 N N Z 0 ℏ2 2m 3π2N′ V !2/3 dN′ = 1 N ℏ2 2m 3π2 V !2/3 3 5N5/3 = 3 5EF (21) 6. Now that you’ve followed the derivation for the three dimensional case from the textbook, repeat the previous two problems for a two dimensional electron gas, using your results from problem 3. Solution: We already found the Fermi energy in problem 3, conveniently enough. EF = ℏ2k2 F 2m = πNℏ2 mL2 (22) The average speed at the Fermi energy is the same as in three dimensions (in terms of EF anyway): v = s 2EF m = s 2πNℏ2 m2L2 = √ 2πN ℏ mL (23) If the density of states is constant, the average energy is easy - it must be half the maximum value, 1 2EF . As a sanity check, we can proceed as in the previous problem: Eavg = 1 N N Z 0 EF (N′) dN′ = 1 N N Z 0 πN′ℏ2 mL2 dN′ = 1 N πN2ℏ2 2mL2 = πNℏ2 2mL2 = 1 2EF (24)
190536	https://www.weather.gov/media/epz/wxcalc/tempConvert.pdf	Temperature Conversion To convert between Fahrenheit (°F) and degrees Celsius (°C): = 𝑇𝑐 5 9 × 𝑇𝑓 −32 ( ) + 32 𝑇𝑓= 9 5 ( ) × 𝑇𝐶 Where: is temperature in Celsius 𝑇𝑐 is temperature in Fahrenheit 𝑇𝑓 To convert between degrees Fahrenheit (°F) and Kelvin (K): = + 32 𝑇𝑓 9 5 × 𝑇𝑘 −273. 15 ( ) ( ) 𝑇𝐾= 5 9 × 𝑇𝑓 − 32 ( ) ( ) + 273. 15 Where: is temperature in Fahrenheit 𝑇𝑓 is temperature in Kelvin 𝑇𝐾 To convert between degrees Fahrenheit (°F) and Rankine (R): 𝑇𝑓 = 𝑇𝑅 − 459. 67 𝑇𝑅 = 𝑇𝑓 + 459. 67 Where: is temperature in Fahrenheit 𝑇𝑓 is temperature Rankine 𝑇𝑅 To convert between degrees Celsius (°C) and Kelvin (K): 𝑇𝑐 = 𝑇𝐾 − 273. 15 𝑇𝐾 = 𝑇𝑐 + 273. 15 Where: is temperature in Celsius 𝑇𝑐 is temperature in Kelvin 𝑇𝐾 To convert between degrees (°C) and Rankine (R): 𝑇𝐶= 5 9 × 𝑇𝑅 − 459. 67 ( ) − 32 ( ) + 459.67 𝑇𝑅 = 9 5 × 𝑇𝑐 + 32 ( ) Where: is temperature in Celsius 𝑇𝑐 is temperature in Rankine 𝑇𝑅 To convert between degrees Kelvin (K) and Rankine (R): 𝑇𝐾 = 5 9 × 𝑇𝑅 − 459. 67 ( ) − 32 ( ) ( ) + 273. 15 𝑇𝑅 = 9 5 × 𝑇𝐾 − 273. 15 ( ) + 32 ( ) + 459. 67 Where: is temperature in Kelvin 𝑇𝐾 is temperature in Rankine 𝑇𝑅
190537	https://plato.stanford.edu/archives/win2015/entries/goedel-incompleteness/sup1.html	Stanford Encyclopedia of Philosophy Archive Winter 2015 Edition Browse Table of Contents New in this Archive Chronological Archives About Editorial Information About the SEP Editorial Board How to Cite the SEP Special Characters Contact Support SEP Support the SEP PDFs for SEP Friends Make a Donation SEPIA for Libraries Entry Navigation Back to Entry Entry Contents Entry Bibliography Academic Tools Friends PDF Preview Author and Citation Info Back to Top Supplement to Gödel's Incompleteness Theorems Gödel Numbering A key method in the usual proofs of the first incompleteness theorem is the arithmetization of the formal language, or Gödel numbering: certain natural numbers are assigned to terms, formulas, and proofs of the formal theory F. There are different ways of doing this; one standard approach is sketched here (for a rather different method of coding, see, e.g., Boolos & Jeffrey 1989). An essential requirement is that the method is effective, that is, a purely mechanical routine. The method proceeds in two steps: 1. Symbol numbers To begin with, to each primitive symbol s of the language of the formalized system F at stake, a natural number #(s), called the symbol number of s, is attached. It does not matter how and in which order this is done—that is arbitrary—but once it is done, it is obviously kept fixed. For example, in the case of the standard language of arithmetic, with symbols 0, ′, +, ×, =, (, ), ¬, →, ∀ (for simplicity, let us assume that ¬, → and ∀ are the only primitive logical symbols, and that ∧, ∨, ↔ and ∃ are defined with the help of them), one might proceed, e.g., as follows: \| \| \| \| --- \| #(‘0’) = 1 \| #(‘=’) = 5 \| #(‘¬’) = 9 \| \| #(‘ ′ ’) = 2 \| #(‘(’ ) = 6 \| #(‘∀’) = 10 \| \| #(‘+’) = 3 \| #(‘)’) = 7 \| #(‘xi’) = 11 + i \| \| #(‘×’) = 4 \| #(‘→’) = 8 \| 2. Coding sequences Some way of coding finite sequences of numbers by single numbers is also fixed. There are indefinitely many possible ways to achieve that; one common approach (used also originally by Gödel himself) is based on the products of the powers of the prime numbers. Recall that a prime number is a natural number which is greater than 1 and can be divided only by 1 and the number itself (all the other numbers greater than 1 are “composite”). There are infinitely many prime numbers; the beginning of the sequence is 2, 3, 5, 7, 11, 13, 17, … The fundamental theorem of arithmetic (or the unique-prime-factorization theorem) states that any natural number greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers. Let then p1 be the first prime number, p2 the second prime number, and so forth. In general, pn is the n-th prime number, and p1, p2, …, pn is the sequence of the first n primes in the increasing order. Given an arbitrary finite sequence of positive numbers (0 would cause complications) with length k + 1, (n0, n1, …, nk), it can be uniquely coded as a product of powers of the prime numbers p1, p2, …, pk+1 as follows: c = 2n0 × 3n1 × 5n2 ×… × pnkk+1, For example, the sequence ⟨ 3, 1, 2 ⟩ is coded as 23 × 31 × 52, that is, 8 × 3 × 25, which equals 576. Combining the two steps: Given these two methods, it is possible to code an arbitrary expression of the language by a single number: first, replace each symbol s by it symbol number #(s). This way a sequence of symbols becomes a sequence of numbers. Second, using the above powers of primes coding, associate to this sequence of numbers a unique single number as its code—its “Gödel number”. For example, take the simple expression ‘0 = 0’. As it was stipulated above that the symbol numbers of the symbols ‘0’ and ‘=’ are 1 and 5, respectively, the corresponding sequence of the symbol numbers is: ⟨ 1, 5, 1 ⟩. The code (that is, the Gödel number) of ‘0 = 0’ then is: 21 × 35 × 51 = 2 × 243 × 5 = 2430. In general, the Gödel number of a formula (sentence, derivation) A is denoted by ⌈A⌉. For example, ⌈0 = 0⌉ refers, under the coding we have fixed, to 2430. In this way we can assign Gödel numbers to formulas, sequences of formulas (once a method for distinguishing when one formula ends and another begins has been adopted), and most notably, proofs, or derivations. We can also go to the other direction: It is an essential part of the method that if a code number is given (many numbers simply do not code anything, but it can be decided which ones do), it is also possible to decode it in a unique way, that is, to reconstruct the unique original expression (or derivation) that it encodes. For example, let the Gödel number that has been given be 18. The prime factorization of it, which is unique (by the fundamental theorem of arithmetic; see above), can then be determined. It is: 21 × 32 Focusing on the powers, we see that this represents the sequence ⟨ 1, 2 ⟩, and recalling the relevant symbol numbers: 1 = #(‘0’) 2 = #(‘ ′ ’) it can be seen that the encoded sequence of symbols is ⟨ 0, ′ ⟩, that is, the expression that is coded by 18 is ‘ 0′ ’. 3. Defining syntactical properties and operations It is then possible to develop the key notions of exact syntax in the arithmetized form, mimicking their ordinary definitions, though the rigorous definitions tend to get a bit complicated. That is, it is possible to define, in an arithmetical language, properties such as: Const(x) : x is (a Gödel number of) a constant. Var(x) : x is (a Gödel number of) a variable. Term(x) : x is (a Gödel number of) a term. Form(x) : x is (a Gödel number of) a formula. The following operations on (the Gödel numbers of) formulas can also be easily defined in the language of arithmetic: neg(x) : the arithmetical function that sends the Gödel number of a formula to the Gödel number of its negation: > neg(⌈A⌉) = (⌈¬A⌉); impl(x, y) : the function which maps the Gödel numbers of a pair of formulas to the Gödel number of the implication of the formulas: > impl(⌈A⌉, ⌈B⌉) = ⌈A → B⌉; Given that y (z) is (a Gödel number of) a formula, the following syntactic relations can be also defined in arithmetic: Free(x, y) : x is a free variable in y. FreeFor(x, y, z) : x is free for y in z. The following operation on Gödel numbers plays a particularly important role: subst(x, y) = z : the operation which maps the pair with the Gödel number of a formula with one free variable and the Gödel number of a numeral to the Gödel number of the closed formula which results from the original formula when the given numeral is substituted for the free variable: > subst(⌈A(x)⌉, ⌈n⌉) = ⌈A(n)⌉ Once the background logic, its axioms and rules of inference have been fixed, the following can also be defined: LogAx(x) : x is the Gödel number of a logical axiom. Rules of inference also get expressed in the arithmetized form: For example, there is an arithmetical formula M(x, y, z) which is true exactly when one has an application of a standard rule of inference “Modus Ponens” at hand; i.e., for some formulas A and B, x = ⌈A⌉, y = ⌈A → B⌉ and z = ⌈B⌉. When the specific formal system F at stake has been fixed, the following properties and relations can also be defined: AxiomF(x) : x is the Gödel number of a non-logical axiom of F. PrfF(x, y) : x is the Gödel number of a derivation (in F) of the formula with the Gödel number y. ProvF(x) : ∃xPrfF(x, y), i.e., the formula (with the Gödel number) x is provable (derivable) in F. All the above properties and relations, except the last one, provability, are decidable and can be, not only defined but, strongly represented in any sufficiently strong F. Copyright © 2015 by Panu Raatikainen <panu.raatikainen@helsinki.fi> This is a file in the archives of the Stanford Encyclopedia of Philosophy. Please note that some links may no longer be functional. Browse Table of Contents New in this Archive Chronological Archives About Editorial Information About the SEP Editorial Board How to Cite the SEP Special Characters Contact Support SEP Support the SEP PDFs for SEP Friends Make a Donation SEPIA for Libraries Mirror Sites View this site from another server: USA (Main Site) CSLI, Stanford University Australia SETIS, University of Sydney Netherlands ILLC, University of Amsterdam The Stanford Encyclopedia of Philosophy is copyright © 2014 by The Metaphysics Research Lab, Center for the Study of Language and Information (CSLI), Stanford University Library of Congress Catalog Data: ISSN 1095-5054 [an error occurred while processing the directive]
190538	https://researcher.life/blog/article/what-is-inductive-reasoning-definition-types-examples/	Skip to content Educational resources and simple solutions for your research journey Home » R Discovery » What is Inductive Reasoning? Definition, Types and Examples Getting Published R Discovery Research tips What is Inductive Reasoning? Definition, Types and Examples Dr. Renu Bisht Inductive reasoning is a type of reasoning method that analyzes specific evidence or observations to form general inferences. In inductive reasoning, a bottom-up approach is used, in which specific instances are considered to reach a broader conclusion—often to predict future outcomes or infer general rules or principles. Many often confuse inductive reasoning with deductive reasoning, as both are logical reasoning methods but with different approaches and outcomes. In deductive reasoning, conclusions are drawn from a set of premises or assumptions by moving from general assumptions to more specific conclusions. Hence, deductive reasoning is a top-down approach in which one moves from more general assumptions to specific conclusions, usually to prove a fact or hypothesis. In this article, we will discuss inductive and deductive reasoning, evaluate the benefits and limitations of inductive and deductive approaches, and provide examples of the two logical reasoning methods. Table of Contents Toggle What is inductive reasoning? Inductive reasoning is a logical reasoning method with widespread usage in day-to-day decision-making, statistics, research, and probability analysis. It makes use of more specific observations or instances to derive general principles or conclusions. Because of these features, inductive reasoning is a valuable tool to form hypotheses for research and experimental studies. It is interesting to note that, without really thinking about it, each of us uses inductive reasoning in our lives to reach specific conclusions. Before, we discuss how we use the inductive reasoning approach, it is wise to first understand the basic elements of an inductive reasoning approach. The inductive reasoning approach can be broken down into a series of three progressive statements, leading from a more specific observation to a general assumption. Here are some inductive reasoning examples: \| \| \| \| \| --- --- \| \| Stages \| Specific observation \| Pattern recognition \| General conclusion \| \| Example 1 \| The sun rises in the east each morning \| The sun continues to rise in the east each morning \| The sun will always rise in the east each morning \| \| Example 2 \| Eating spicy food makes you sweat with a hot sensation in your mouth \| Every time you eat spicy food it makes you sweat and causes a hot sensation in your mouth \| Spicy food triggers sweating and a hot sensation in your mouth \| \| Example 3 \| Watering plants makes them grow and become taller \| A positive correlation is found between watering plants and their growth patterns \| It can be concluded that consistent watering leads to taller plants and improved flower production \| Inductive reasoning in research Now, we have seen how inductive reasoning is used in day-to-day life to make assumptions and reach conclusions. Next, we will explore how to use inductive reasoning in research. Inductive reasoning is widely used in exploratory (qualitative) research where the goal is to understand a phenomenon and form a hypothesis for further studies. However, inductive reasoning can also be used in quantitative research in the early stages of hypothesis formation. The inductive reasoning in research is usually done in three parts: data (observations) collection, pattern recognition, and general conclusions based on the patterns. Let’s have a look at an example of inductive reasoning in research: Research Question: Does regular exercise improve mental health? Part I: Observation and data collection: To address the question, you would give surveys to people who workout to collect their responses on whether regular exercise helps them improve their mental health. Part II: Pattern recognition: Next, you will try to identify any patterns and correlations between exercising and mental health by analyzing the data. For example, you identify a consistent pattern showing a positive correlation between regular exercise and mental health indicators such as reduced stress levels, improved mood and sleep, and enhanced cognitive function. Part III: General conclusion: Based on the research findings and observed patterns, you form a general conclusion that regular exercise is associated with improved mental health outcomes. However, the conclusion does not imply a causal relationship; it simply suggests a correlation between exercise and mental health. Tip: It is important to be open-minded while applying inductive reasoning in research, as preconceived notions or biases can influence the interpretation of data and introduce research biases such as confirmation bias. You should allow the patterns to emerge naturally from the data to avoid any biases while also being open to alternate explanations as new data become available in iterative stages. Inductive vs deductive reasoning Inductive reasoning and deductive reasoning are both logical forms of reasoning approaches that are used to make conclusions—the difference is that they operate differently. The question of inductive vs deductive reasoning has puzzled many as they are often confused with one another. Here is a rule of thumb to differentiate between the two: for inductive reasoning, the reasoning follows a bottoms-up approach, meaning the logic flows from a specific to a more general conclusion. In the case of deductive reasoning, the reasoning uses the top-down approach in which the logic flows from general to more specific observation. See the diagram below to easily remember the difference between deductive and inductive research methods. Tip: Inductive reasoning is used to make an educated guess about the outcome by using experiences and proven observations. Deductive reasoning uses theories and general observations to reach a specific conclusion. The idea is to prove a fact. Let’s look at inductive and deductive reasoning in detail for better understanding: Inductive reasoning Definition: Inductive reasoning involves drawing general conclusions from specific observations or examples. It moves from particular to general. Process: The process of inductive reasoning starts with specific instances or observations and then the stage of pattern/trend identification among them. From these specific observations, a general hypothesis or conclusion is then formed. Example: I get sad when the weather is gloomy. Many people feel sad when there is no sun and the weather is gloomy. People usually feel depressed when the weather is gloomy and no sunshine. Strengths: Inductive reasoning is useful for generating hypotheses, exploring new phenomena, and identifying patterns in data. Weaknesses: Inductive reasoning does not guarantee certainty; conclusions drawn from inductive reasoning are probabilistic rather than definite. Deductive reasoning Definition: Deductive reasoning involves deriving specific conclusions from general principles or premises. It moves from the general to the particular (you would have noticed deductive reasoning skills portrayed by Sherlock Holmes to solve a mystery). Process: In deductive reasoning, the reasoning makes use of general principles or known facts to apply logical rules and derive specific conclusions that necessarily follow from those premises. Example: All calico cats are female. My (fostered) kitten Luna is a calico. Hence, Luna must be a female. Strengths: Deductive reasoning ensures certainty; if the premises are true and the reasoning is valid, then the conclusion must also be true. It is often used in mathematics, logic, and formal systems. Weaknesses: Deductive reasoning relies strongly on the notion that the premise is true; if the premises are incorrect, the conclusion will also be incorrect. It may not be suitable for exploring new phenomena or generating hypotheses. Inductive vs abductive reasoning Inductive vs abductive reasoning are other similar terms people often confuse. Like inductive reasoning, abductive reasoning is also a form of analyzing premises or observations to predict outcomes. One of the major differences between inductive and abductive reasoning is that the latter uses incomplete information to reach the conclusion. Abductive reasoning allows one to conclude in lieu of complete information, providing freedom. However, this feature makes it unreliable as it might lead to several wrong conclusions before reaching the true answer. Abductive reasoning is commonly used in the medical field especially for diagnosis purposes in the absence of more concrete information such as lab tests. For example: Doctors often use abductive reasoning to diagnose their patients based on their knowledge and logical guesswork considering the patient’s symptoms. In such scenarios, patients often present some symptoms, which might or might not reflect the exact problem. Benefits of inductive reasoning Inductive reasoning is a valuable reasoning tool to form hypotheses for exploratory and other kinds of research. The major benefits of inductive reasoning are listed below: Flexibility: Inductive reasoning workflow begins with a specific observation leading to a more general conclusion, thus, providing ranges of probabilities and possibilities to work with. Generate new hypothesis: As inductive reasoning involves observing specific instances and patterns it can be a valuable tool to generate new hypotheses, sometimes an entirely novel hypothesis that has not been explored earlier. This allows researchers to innovate solutions to problems that advance the field further. Applicability to diverse fields: Inductive reasoning has applications in a wide range of disciplines such as science, social sciences, humanities, and everyday decision-making. It allows researchers and practitioners in various fields to derive insights and make predictions based on observed patterns. Encourages exploration: Inductive reasoning encourages exploration, as analyzing patterns and existing information provides new hypotheses, which open up avenues for future research and opportunities for newer discoveries. Limitations of inductive reasoning Now, that we have learned about the benefits of inductive reasoning, let’s also talk about the limitations of inductive reasoning which you should take into consideration before applying inductive reasoning in your research. Inaccurate hypothesis generation: If limited information is available to form a hypothesis, it may lead to the generation of a wrong theory, which further results in incorrect interpretation as you generalize the findings. Findings are subject to change: Inductive reasoning-based hypothesis generation runs a risk of becoming obsolete if contradicting new data becomes available. Care must be taken while forming a hypothesis with limited information or instances. Uncertainty: Inductive reasoning cannot guarantee the truth of its conclusions. The conclusions can be wrong even if all the premises are true. As you move from specific to general principles, there is always a possibility that new observations could contradict those principles. Overgeneralizations: This approach can lead to overgeneralizations where broad conclusions are drawn from limited observations, leading to stereotypes and incorrect assumptions about groups of people, objects, or phenomena. Key takeaways The key takeaways of inductive reasoning are listed below for your easy reference: Inductive reasoning is a logical bottom-up reasoning method that uses specific observations to make generalized assumptions. For example: Specific observation: Noticing that the supermarket near my house in Milan closes at 7 pm every day. Pattern recognition: All the supermarkets in my locality close at 7 pm. Generalized assumption: All the supermarkets in my city close at 7 pm. Inductive and deductive reasoning are both logical reasoning methods, but vice–versa approaches to addressing the phenomenon. Deductive reasoning is a top-down approach in which reasoning follows from general observations to more specific inferences. For example: General observation: If it rains, the streets will be wet. Pattern recognition: It is raining now. Specific observation: The streets were wet as it rained. Since inductive reasoning is exploratory while deductive reasoning is confirmatory in nature, the former is usually used for hypothesis generation while the latter is used for validating facts. However, both can be used in combination within a single research study. Frequently asked questions What is inductive reasoning? Inductive reasoning is a method of reasoning where conclusions are drawn from specific observations or patterns to form general principles or hypotheses. What are types of inductive reasoning? There are different types of inductive reasoning, including: Inductive generalization: Generalizing from specific cases or observations to form assumptions about the population. Statistical generalization: Making predictions/generalization about the population by applying statistical analysis on the observed data. It is more specific than inductive generalization as it deals with numerical data. Causal relationships: Inferring causal relationships based on correlations observed in data. It is used to make a cause-and-effect relationship between two different observations to reach a general conclusion Analogical reasoning: This type of inductive reasoning is applied to draw conclusions about a certain element based on its similarities with another. If a particular conclusion holds true for one element, it should be true for the other. What is the process of inductive reasoning in research? The process of inductive reasoning in research typically involves: Observing specific instances or phenomena. Identifying patterns or regularities in the observed data. Formulating a general hypothesis or principle based on these patterns. Testing the hypothesis through further observation, experimentation, or data collection. Refining or revising the hypothesis based on new evidence or findings. Iterating the process to continually refine and develop a deeper understanding of the phenomenon under study. Can inductive reasoning be combined with deductive reasoning in research? Yes, inductive reasoning and deductive reasoning can be combined in research to form a comprehensive approach when conducting a study. Researchers often use inductive reasoning to generate hypotheses or theories based on observed patterns or data. Then, they use deductive reasoning to test these hypotheses by making specific predictions and designing experiments or studies to gather evidence. After collecting data through deductive methods, researchers may analyze it inductively to identify further patterns or refine their hypotheses. This iterative process allows for the integration of both inductive and deductive reasoning in the research endeavor, leading to a more robust understanding of the phenomena under investigation. What are some common misconceptions about inductive reasoning in research? Some misconceptions about inductive reasoning in research are mentioned below. Inductive reasoning guarantees conclusions: Despite generating hypotheses from observed patterns, inductive reasoning doesn’t always lead to definite conclusions; hypotheses are subject to revision. Inductive reasoning lacks scientific validity: Despite its association with deductive reasoning, inductive reasoning is a fundamental part of the scientific method, generating testable hypotheses. Inductive reasoning is entirely subjective: Inductive reasoning can exhibit subjectivity due to the interpretation involved, potentially leading to confirmation bias.The usage of systematic methods can reduce subjectivity in drawing conclusions from data. Inductive conclusions are always valid: The reliability of inductive conclusions can be influenced by factors like sample size and data representativeness. Inductive and deductive reasoning are opposites: Although different, inductive and deductive reasoning often complement each other in research, forming a comprehensive approach. How is inductive reasoning used in qualitative research? Inductive reasoning in qualitative research involves: Formulating research questions based on specific observations Collecting descriptive data Analyzing data to recognize patterns or themes Developing theories grounded in the data Iteratively refining understanding through reflexivity and iterations We hope this article has provided you with the information you need about inductive and deductive reasoning to allow you to apply these effectively in your research. R Discovery is a literature search and research reading platform that accelerates your research discovery journey by keeping you updated on the latest, most relevant scholarly content. With 250M+ research articles sourced from trusted aggregators like CrossRef, Unpaywall, PubMed, PubMed Central, Open Alex and top publishing houses like Springer Nature, JAMA, IOP, Taylor & Francis, NEJM, BMJ, Karger, SAGE, Emerald Publishing and more, R Discovery puts a world of research at your fingertips. 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190539	https://www.wordhippo.com/what-is/another-word-for/candor.html	What is another word for candor? \| Candor Synonyms - WordHippo Thesaurus What's another word for What's the opposite of Meaning of the word Words that rhyme with Sentences with the word Translate to Find Words Use for blank tiles (max 2)Use for blank spacesAdvanced Word Finder Find the of Pronounce the word in Find Names Synonyms Antonyms Definitions Rhymes Sentences Translations Find Words Word Forms Pronunciations ☀ \| \| Appearance \| --- \| \| ✓ \| Use device theme \| \| ✓ \| Dark theme \| \| ✓ \| Light theme \| What is another word for candor? ==================================== Need synonyms for candor? Here's a list of similar words from our thesaurus that you can use instead. ================================================================================================================= Contexts ▼▲ The state of being sincere and frank or blunt in speech The state or fact of being fair or impartial The quality or state of being bright or illuminated The state or fact of being honest and upright, not capable of being corrupted Openness or familiarity in speech or behavior … more ▼▲ Noun ▲ The state of being sincere and frank or blunt in speech frankness honesty openness forthrightness bluntness straightforwardness outspokenness candidness directness unreservedness unreserve openheartedness probity unguardedness truthfulness plumpness sincerity plainness plainspokenness artlessness guilelessness simplicity parrhesia unequivocalness complete honesty candourUK ingenuousness genuineness truth integrity naturalness veracity trustworthiness lack of deceit plain speaking innocence authenticity reliability unaffectedness seriousness absence of reserve open-heartedness freeness verity bluffness candidity accuracy faithfulness laying it on the line correctness exactitude naivety simpleness naivete unworldliness naiveness greenness viridity simplemindedness unsophistication naiveté inexperience unpretentiousness spontaneity lack of sophistication honourUK honorUS impartiality wholeheartedness earnestness justice goodwill singleness heart sincereness bona fides good faith gullibility callowness credulity credulousness immaturity trustfulness ignorance childlikeness lack of experience rawness plain-spokenness fairness unsophisticatedness over-trustfulness blind faith lack of suspicion transparency straightness full disclosure unsuspiciousness trustingness childishness purity humbleness gullibleness unassumingness trust puerility jejunosity jejuneness amateurishness naïveté naiveté newness juvenileness childlike simplicity lack of guile inexpertness freshness total revelation absolute honesty whole truth complete truthfulness full report utmost frankness complete disclosure more ❯ “Her many contributions to the press are marked with vigor and originality, with a blunt candor that assures one of her sincerity.” Noun ▲ The state or fact of being fair or impartial impartiality equity objectivity detachment uprightness fairness objectiveness neutrality evenhandedness nonalignment fair-mindedness open-mindedness even-handedness lack of bias lack of prejudice justice disinterestedness disinterest equitableness honesty righteousness rightness dispassion nonpartisanship integrity rightfulness dispassionateness goodness decency consideration justness honourUK reasonableness honorUS right legitimacy moderation propriety impartialness truth tolerance veracity decorum civility charity suitability courtesy exactitude rationality humanity seemliness due charitableness duty square deal fair shake give and take good faith neutralism candourUK balance equitability independence equality unbiasedness indifference egalitarianism distance unselfishness absence of prejudice absence of bias unprejudicedness uninvolvement autonomy noninterference cool coldness noninterventionism selflessness candidness noninvolvement considerateness evenness without favoritism rectitude equal opportunity honourableness scrupulousness conscientiousness properness sensibleness parity non-partisanship fair treatment fair dealing fair play lack of discrimination lack of bigotry non-alignment non-intervention non-interventionism non-involvement non-combativeness non-participation more ❯ “I have very great respect for the learning and candor of the judge who made that decision, but after careful consideration, I cannot yield to the soundness of his reasoning.” Noun ▲ The quality or state of being bright or illuminated brilliance brightness radiance dazzle luminosity effulgence lustreUK refulgence illumination lambency luminousness splendourUK splendorUS lustrousness lightness brilliancy lusterUS luminance shine glow sparkle glitter gleam flash resplendence incandescence light blaze vividness luminescence shimmer glare intensity fluorescence coruscation sheen glaze glint flare iridescence patina opalescence afterglow candescence translucence beam gloss lucidity flame flashing glittering richness photometric brightness fire reflection phosphorescence shining reflectivity visibility reflectiveness clarity lucency irradiance radiation beams rays candourUK glimmer burnish polish twinkle fulguration scintillation glister shininess finish ray glossiness lighting glisten smoothness glowing strength silkiness varnish flicker glimmering gleaming lumination lights sunlight irradiation brightening warmth vibrancy sleekness slickness veneer gilt blink heat spark glitz glassiness burst beam of light ray of light blossom burning twink shaft of light display firelight lamplight fulgor electric light flash of light starlight moonlight gaslight magnitude dazzling light strong light blinding light vividity wax illumining streak lighting up liveliness appearance colorfulness seriousness darkness extremity intenseness deepness keenness severity sun daylight sunbeams depth sunshine natural light light of day more ❯ “One side of the planet is constantly being exposed to the candor of the sun, while the other sits in cold darkness.” Noun ▲ The state or fact of being honest and upright, not capable of being corrupted upstandingness integrity honourUK morality uprightness honorUS virtue principle probity rectitude goodness trustworthiness fairness equity incorruptibility scrupulousness justness faithfulness loyalty conscientiousness fidelity reputability veracity trustiness honorableness evenhandedness impeccability sincerity right straightness frankness soundness straightforwardness genuineness bluntness confidence responsibility openness outspokenness plainness self-respect honesty righteousness decency truthfulness verity correctness rightness nobility truth actuality right-mindedness high-mindedness credibility impartiality respectability authenticity trueness moral fibreUK ethics moral fiberUS reliability virtuousness veraciousness veridicality real McCoy reality accuracy fact character honourableness principles purity morals truism verisimilitude noble-mindedness incorruption forthrightness good character honestness sportsmanship gospel precision exactness word exactitude straight stuff honest-to-god truth like it is candourUK dependability worthiness justice dignity ethicalness propriety blamelessness nobleness irreproachableness excellence superiority magnanimity greatness graciousness pureness honourability lack of corruption ethicality good gallantry moral virtue worth generosity gentry illustriousness sublimity exaltation ennoblement elevation glorification merit meritoriousness faith lawfulness godliness saintliness innocence nobility of spirit nobility of soul fair play high principles sense esteem reverence veneration regard respect dutifulness dedication devotion constancy sublime royalty stateliness loftiness peerage magnificence majesty grandeur kindness love charity fortitude ideal ethic moral correctness incorruptness proper conduct moral integrity legitimacy reasonableness rightfulness guiltlessness self-discipline moral excellence good conscience moral rectitude moral courage fair shake what is right noblesse sinlessness irreproachability bravery spotlessness chastity wholesomeness quality gentility chasteness manners gentleness scruples conduct eminence selflessness inculpability stainlessness good behaviorUS good behaviourUK good habits innocency blue blood high society faultlessness value legality equitableness more ❯ Noun ▲ Openness or familiarity in speech or behavior freedom naturalness familiarity openness forwardness impudence informality casualness spontaneity ease candidness candourUK cheek closeness forthrightness genuineness intimacy overfamiliarity poise presumption composure directness fellowship frankness ingenuousness abandon coolness insouciance lack of ceremony lack of inhibition lack of reserve nonchalance abandonment free expression indulgence lack of restraint laissez faire looseness readiness relaxed manner unconstraint boldness brazenness disrespect impertinence laxity relaxedness liberty unceremoniousness friendliness simplicity comfortableness sociability freshness acquaintanceship absence of reserve unpretentiousness lack of pretension lack of affectation modesty unassumingness unostentatiousness unselfconsciousness humbleness unfussiness unaffectedness straightforwardness unstudiedness naturality lack of sophistication more ❯ Find more words! Use for blank tiles (max 2)Advanced SearchAdvanced Search Use for blank spacesAdvanced Search Advanced Word Finder Related Words and Phrases ------------------------- candid candour candidness candidly candidity candidest candider candors candours See Also -------- What is the opposite of candor? Sentences with the word candor Words that rhyme with candor What is the plural of candor? What is the adverb for candor? What is the adjective for candor? What is the noun for candor? What is another word for candour? Translations for candor Use our Synonym Finder Nearby Words candors candour candours candy candy apple candy apples can-doer can-do attitude can-do can do candling candlewax 6-letter Words Starting With c ca can cand candoFind Synonyms go Word Tools Finders & Helpers Apps More Synonyms Synonyms Antonyms Rhymes Sentences Nouns VerbsAdjectives Adverbs Plural Singular Past Tense Present TenseWord Unscrambler Words With Friends Scrabble Crossword / CodewordiOS / Apple AndroidHome About Us Terms of Use Privacy Statement Don't Sell Personal DataABCDEFGHIJKLM NOPQRSTUVWXYZ Antonyms ABCDEFGHIJKLM NOPQRSTUVWXYZ Copyright WordHippo © 2025
190540	https://www.hmhco.com/blog/teaching-x-and-y-axis-graph-on-coordinate-grids?srsltid=AfmBOoqz41RuRsy2NmGvg_4LVnL2GyUI_1rJDj7Gha27pay24fPknvcq	Teaching x- and y-Axis Graphing on Coordinate Grids \| HMH Curriculum Literacy Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path Math Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment Science Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 Social Studies Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable Professional Development For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools More AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA Shop Support Platform Login From setup to support, HMH will ensure your back-to-school season runs smoothly. 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Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA HMH Careers Exploring a career at HMH? Do work that matters. Learn about our culture, benefits, and available job opportunities. Accessibility Explore HMH’s approach to designing affirming and accessible curriculum materials and learning tools for students and teachers. Shop Support 0 Log in Back to Shaped Math Teaching x- and y-Axis Graphing on Coordinate Grids Richard Blankman October 22, 2021 8 Min Read Coordinate graphing can sound very daunting for students in Grades 4–9, but it's actually just a visual method for showing relationships between numbers. The relationships are shown on a coordinate grid. A coordinate grid has two perpendicular lines, or axes(pronounced AX-eez), labeled just like number lines. The horizontal axis is usually called the x-axis. The vertical axis is usually called the y-axis. The point where the x- and y-axis intersect is called the origin. Drawing a Coordinate Graph The numbers on a coordinate grid are used to locate points. Each point can be identified by an ordered pair of numbers; that is, a number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate. Ordered pairs are written in parentheses (x-coordinate, y-coordinate). The origin is located at (0,0). Note that coordinates are often written with no space after the comma. The location of (2,5) is shown on the coordinate grid below. The x-coordinate is 2. The y-coordinate is 5. To locate (2,5), move 2 units to the right on the x-axis and 5 units up on the y-axis. The order in which you write x- and y-coordinates in an ordered pair is very important. The x-coordinate always comes first, followed by the y-coordinate. As you can see in the coordinate grid below, the ordered pairs (3,4) and (4,3) are two different points! Describing a Linear Relationship The function table below shows the x- and y-coordinates for five ordered pairs. You can describe the relationship between the x- and y-coordinates for each of these ordered pairs with this rule: the x-coordinate plus two equals the y-coordinate. You can also describe this relationship with the algebraic equation x + 2 = y. x-coordinatex + 2 = yy-coordinateordered pair 00 + 2 = 22(0,2) 11 + 2 = 33(1,3) 22 + 2 = 44(2,4) 33 + 2 = 55(3,5) 44 + 2 = 66(4,6) To graph the equation x + 2 = y, each ordered pair is located on a coordinate grid, then the points are connected. In fact, the graph forms a straight line. The arrows indicate that the line goes on in both directions. For students who are ready to take it to the next level, consider explaining that the graph for any equation that can be written as ax + by = c, where a, b, and c are numbers, forms a straight line. Notice how x + 2 = y can also be written as ax + by = c, where a = 1, b = –1, and c = –2. Introducing the Concept Finding and Graphing Points for Linear Relationships Your students may have encountered ordered pairs last year, but it's a good idea to start by reviewing how to locate a point on a grid from an ordered pair. A day spent plotting coordinates that fall in a straight line will be a day well spent. Key Standard:Graph points on the coordinate plane. (5.G.A.1) Materials: Poster paper or a way to display a coordinate grid publicly for the class; straightedge Preparation: Draw a large coordinate grid that the entire class can see. Label the x- and y-axes from 0 through 10. Prerequisite Skills and Concepts: Students should know about ordered pairs and locating points on a grid. Write these ordered pairs where all students can see them: (6,4); (7,5); (8,6); and (9,7). Point to the ordered pair (6,4). Ask:What rule describes the relationship between the numbers in this ordered pair? Although many rules work for this pair in isolation, elicit from students this rule: the first number minus two equals the second number. Ask:Does the same rule apply to the other ordered pairs? Students should notice that each ordered pair follows this rule. You can help them by using the rule to write each ordered pair as an equation: 6 – 2 = 4, 7 – 2 = 5, 8 – 2 = 6, 9 – 2 = 7. Say:Let's locate these ordered pairs on a grid. Ask:How would you locate the point for (6,4) on the grid? Students should say to "start at 0, move 6 units to the right, then 4 units up." Mark this point on the grid for the class to see. Have students verbalize how to locate the point for each of the other ordered pairs. Then mark each point on the grid. Emphasize the importance of moving right for the first number in the ordered pair and up for the second number. Ask:What figure do you think will be formed by connecting the points on the grid? Students should see that a line will be formed. Use a straightedge to connect the points. Provide students with other examples of ordered pairs that follow a rule. Have students identify the rule and explain how to graph the points. One example could read, "Rule: The first number plus three equals the second number; ordered pairs: (2,5); (3,6); (4,7); and (5,8)." Developing the Concept Finding and Graphing Points for Linear Relationships At this level, students will begin to see the relationship between equations and straight-line graphs on a coordinate grid. Key Standard:Interpret an equation as a linear function, whose graph is a straight line. (8.F.A.3) Materials:Poster paper or a way to display a coordinate grid publicly for the class; straightedge; one copy of a coordinate grid, a straightedge, and lined paper for each student Preparation:Draw a coordinate grid where all students can see it. Label the x- and y-axes from 0 through 10. Ensure all students have a copy of the grid. Prerequisite Skills and Concepts: Students should know about ordered pairs and locating points on a grid. They should also be able to recognize and interpret an equation. Write the equation x + 5 = y publicly for the class to see. Ask:How could you say this equation in words? Students should say that the equation means "a number plus five equals another number," or a comparable statement. Draw a table with four columns and five rows. Have students draw their own table. Label the first column x, the second column x + 5, and the third column y. Leave the fourth column blank for now. Write "1" in the first column below x. Ask:What happens to the equation if we replace x with 1? Elicit from students the equation 1 + 5 = 6. Write "1 + 5" in the second column below "x + 5." Then write "6" in the third column below y. Continue to replace x with 2, 3, then 4. Have students complete the first three columns of their tables on their own. Then ask for a volunteer to complete the table publicly for the class. Say:Let's write ordered pairs using the values of x and y . Label the fourth column of your table "Ordered Pairs." Remind students that when they locate points on a grid, they first move right on the x-axis, then up on the y-axis. Therefore, the first number in an ordered pair is a value for x, and the second number is a value for y. These numbers are called the x- and y-coordinates. Ask:What is the first number we used for x? (1) What is the first number we calculated for y? (6) So, what is the first ordered pair? (1,6) Have students complete their tables. When they are finished, record the ordered pairs in the table publicly for the class. Say:Now we're going to graph the equation x + 5 = y on a grid. (Point to the grid you made.) This grid is called a coordinate grid. Let's take a closer look at the different parts of the grid. Point to the horizontal line on the grid. Say:This line is called the x-axis. Point to the vertical line on the grid. Ask:What do you think this line is called? Students should make the connection to the y-axis. Say:Now, let's locate the ordered pairs on the grid. Who can find (1,6)? Have a volunteer describe the location of the ordered pair. Mark the location on the coordinate grid for all students to see. Then have students locate the rest of the ordered pairs on their own grids. Say:Let's connect all of the points. What figure did we make? Have students use a straight edge to connect the points. Show students how extending both ends of the line slightly, and drawing arrows, shows that the line goes on in both directions. Students should identify the figure as a straight line. Have students repeat this activity with the equation x – 2 = y. Use the numbers 5, 6, 7, 8, and 9 for x. Wrap-Up and Assessment Hints These skills will need lots of practice. Reinforce the need for students to work carefully so their graph is accurate. When you assess students' progress, keep the number of exercises small enough that they have time to complete each step without rushing. This blog post, originally published in 2020, has been updated for 2021. Read Next Teaching Quadrant Numbers on a Graph With this math lesson, you can teach students about coordinates, integers, coordinate planes, and functions on a graph. Richard Blankman September 1, 2021 Read More Free Math Activities for Middle School Try these free middle school math activities that cover budgeting, taxes, social media engagement, coding, and many more topics. Shaped Staff January 10, 2025 Read More Healthy Habits for a Growth Mindset in Math Here's a helpful list of healthy habits to enable a growth mindset in math, plus a free downloadable poster for your classroom. 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190541	http://hyperphysics.phy-astr.gsu.edu/hbase/Bohr.html	\| \| \| \| --- \| Quantized Energy StatesThe electrons in free atoms can only be found in certain discrete energy states. These sharp energy states are associated with the orbits or shells of electrons in an atom, e.g., a hydrogen atom. One of the implications of these quantized energy states is that only certain photon energies are allowed when electrons jump down from higher levels to lower levels, producing the hydrogen spectrum. The Bohr model successfully predicted the energies for the hydrogen atom, but had significant failures that were corrected by solving the Schrodinger equation for the hydrogen atom. \| \| \| Scaled energy levels \| \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| --- --- \| \| Angular Momentum QuantizationIn the Bohr model, the wavelength associated with the electron is given by the DeBroglie relationship and the standing wave condition that circumference = whole number of wavelengths. In the hydrogenic case, the number n is the principal quantum number. These can be combined to get an expression for the angular momentum of the electron in orbit. (Note that this assumes a circular orbit, a generally unwarranted assumption.) \| \| \| --- \| \| \| Use in Bohr orbit \| Thus L is not only conserved, but constrained to discrete values by the quantum number n. This quantization of angular momentum is a crucial result and can be used in determining the Bohr orbit radii and Bohr energies. \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| \| \| --- --- --- \| \| Classical Electron Orbit In the Bohr theory, this classical result was combined with the quantization of angular momentum to get an expression for quantized energy levels. \| \| \| \| \| --- --- \| \| Bohr orbit \| Centripetal force \| Newton's 2nd law \| Electric potential \| \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| \| --- --- \| \| Bohr OrbitCombining the energy of the classical electron orbit with the quantization of angular momentum, the Bohr approach yields expressions for the electron orbit radii and energies: Substitution for r gives the Bohr energies and radii: Although the Bohr model of the atom was shown to have many failures, the expression for the hydrogen electron energies is amazingly accurate. The Schrodinger equation for the hydrogen atom actually gave the same energies, so the Bohr model was a helpful step along the way to developing a quantum mechanical model for hydrogen. \| \| \| --- \| \| Energy level structure \| Associated electron wavelengths \| \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| --- \| Wavelengths for Different States \| \| \| Bohr model of the atom \| \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \| \| \| \| \| --- \| Failures of the Bohr Model While the Bohr model was a major step toward understanding the quantum theory of the atom, it is not in fact a correct description of the nature of electron orbits. Some of the shortcomings of the model are: 1. It fails to provide any understanding of why certain spectral lines are brighter than others. There is no mechanism for the calculation of transition probabilities. 2. The Bohr model treats the electron as if it were a miniature planet, with definite radius and momentum. This is in direct violation of the uncertainty principle which dictates that position and momentum cannot be simultaneously determined. The Bohr model gives us a basic conceptual model of electron orbits and energies. The precise details of spectra and charge distribution must be left to quantum mechanical calculations, as with the Schrodinger equation. \| \| \| Bohr model of the atom \| \| Index Bohr model concepts \| \| \| \| \| --- \| \| HyperPhysics Quantum Physics \| R Nave \| \| Go Back \|
190542	https://www.youtube.com/watch?v=kI5QxBGQBfM	Openstax Precalculus Ch 5.2 Unit Circle and Trig functions Kwai Chan 4770 subscribers 2 likes Description 295 views Posted: 9 Apr 2021 Transcript: okay precalculus student this video is for chapter 5.2 unicircle definition of six trig functions um so in the last video we talked about how we can define the six trig function using an angle and then any point on the terminal side um the the definition is actually the ultimate definition of how you can do periodic function but what happened is you've uh another definition would be actually fix the point that you want to choose instead of picking any point on the terminal side we want to intersect it with a unit circle so i'll have the unit circle get intersect with that terminal side and you would have one point right so that point is x y it is still right it's still good to use that point because it's on the terminal side but the good thing is r is equal to one so we don't we don't have to calculate r any time like every time what you need to do is just look at that point that point is uh um i always have r equal to 1 and it is simplest because simpler here because sine t would you go to y over r used to be just y over r now you become y over one so you just y cosine t would be x over r now is just x so basically the point of intersection with the unit circle of your terminal side is actually cosine t sine t cosine t is x so your x y point on the terminal side this point is directly sine t of sine t and cosine t you don't have to do our calculation so that's so so you can say that this particular definition is not anything new it's just restricting the r to be one and if that's the case tangent and cotangent has no change the way you define it is still y over x for tangent and cotangent is still x over y these pairs are still reciprocal but for cosecant is 1 over y secant is 1 over x of course we want to make sure that x and y is not zero otherwise it's called undefined because this function is actually defined at it that way okay um the easiest way is to actually look at correct angles that means angles that are in the x and y axis so if i am to find cosine zero so i would have you know in the unit circle the four points that we know on the unit circle is one zero zero one negative one zero and 0 negative 1. so 0 would be here 0 degrees here so the x coordinate is cosine 0. the y coordinate is sine 0. so so i would say okay 1 would be here i choose the point of intersection of the unit circle is one comma zero the first is cosine so that's equal to one sine three pi over two three pi over two is right here right so the point on the unit circle is zero zero comma negative one sine is the y coordinate so that is equal to negative one so just read it off you don't have to go through r and everything so that's why it's easier now how do i so that would be very nice how do i feel other points so that i can i can actually find more sine and cosine of an angle it turns out that for angles special angles of 30 degree 45 degree and 60 degree and the corresponding one on the other coordinate i can actually find those points okay so let's do um let's do the first quadrant 30 60 and 30 degrees 45 degree and 60 degree and you have to recall if i have a 45 let's do 45 degree first if i have a 45 degree angle um i can make a right triangle and from the last video you remember that if it's 45 45 it doesn't look like 45 but uh i would have a a a root two right so anytime you have a short side or the two sides are a because it's 45 is isosceles so this two should be equal so this is 40 this is like this is five this is five then the hypotenuse is 5 over 2. if this is 7 this is 7 then it's 7 root 2. so what i want to do is put my right triangle here so that this is a 1. right so how do i change it so the the hypotenuse is one is if i multiply a by if i get a equal to one over root two right one over root two if a is equal to one over root two so the hypotenuse become one a will be one over root two and one over root two rationalize it that's actually equal to root two over two root two over two so if i put a triangle over here i would imagine a point intersect here as root 2 over 2 root 2 over 2 why because my x a length is root 2 over 2 this is also root 2 over 2 so the point x y is root 2 over 2. and so that would be our cosine cosine 45 and sine 45 right so this right triangle is very useful now that would just be one one point what about these what about this 135 degree can i use the same thing so what i'll do is we'll put to flip this triangle or flip the triangle over so that um it the 45 degree is here how do i know because 135 degree so i'll say this is 45 degree which is pi over 2 right and this would no pi over 4 and so this would be 180 degree minus pi over 4 or 180 degree minus 45 so that would be 135 degree or 3 pi over 4 right pi pi minus pi over plus 3 pi over 4 and we would have the same root 2 over 2 root 2 of 2 except on the y axis this is negative y x axis on the x axis this is negative this is negative one negative two so my two coordinate is negative root two over two the y coordinate is still positive is root two over two so i would know the angle is actually 135 right is 135 is sine sine 135 is root 2 for 2 cosine 135 is negative with 2 over 2. and then there's two more that i can do here you still remember the two angles of 45 degree 180 at 45 degree is 2 to 5 and pi at pi over 4 remember this is pi over 4 it's 5 pi over 4 so again it is still root 2 over 2 root 2 over 2. so this coordinate would be negative root 2 over 2 and negative root 2 over 2 y because this is the the x-axis is negative the y-axis is negative similarly this angle is pi over 4 i would have 360 or 2 pi minus 45 or minus pi over 4 it becomes 15 degree and um what 7 pi over 4 and i would have the same i would put the triangle over here because this is this is pi over 4 here the whole angle is actually think of it like this right so but i only care about the coordinate the coordinate would be positive x root 2 over 2 negative y because it's y is negative negative root 2 for 2. so i have a lot of x y and you can say you can actually use this to read off sine 3 15 oh sine root 15. cosine for 15 or cosine 7 pi over 4 sine 7 pi over 4 so this is very useful you can use this two triangle uh insert on it you can find those coordinates because the radius is the hypotenuse is equal to one so that's how we construct the unit circle and we would be able to get how many uh sine and cosine four more sine and cosine in additional to the current angle let's do the 360 and and the 30 and 60. so if i have a 30 degree angle let's figure out the first the first quadrant first so this is 30 degree and we know that for 3016 we have a uh with 3a and 2a that is the template right and this is 60. um so if i let a be one-half let a be one-half a equal to one half a the last one we have a is equal to one over root two now i have a equal to one half all of a sudden the hypotenuse is one because we want to put it in the in here to get like a thirty degree like that right so this would be one and this would be one half a is one half this would be with three over two so do i know the coordinate the coordinate will be horizontally is root three over 2 vertically it's one half right because i can actually imagine myself put my triangle over here like that so i have 2 2 coordinate and at the same time i can get four uh three more by thinking of so this would be in terms of degree is 30 degree and pi over six pi over six is 30 degree what is this degree this what is this angle this angle represent from from the x-axis all the way but i have the 30 degree over here so it's 150 and what is 150 equivalent to is 5 pi over 6 because it's just pi or 180 minus pi minus pi over 6 is 5 pi over 6. so it's that's the and then can i write the coordinate the coordinate would be the same as what you have here except the x coordinate is negative so i have negative with 3 over 2 comma one half and this would be both a negative right negative x axis negative y coordinate but the same size one half with three over two this is one half this is one half this is the shorter side right so i would have um the x coordinate with three over two negative with three over two and negative one half so basically you're copying the same coordinate except for the sign so so pi over 6 we call it is like the reference angle this would be what from 360 degree or 2 pi minus 30 so this is 330 or what um 2 pi minus pi over 6 is we talked about this in the last video is 11 pi over 6 and the coordinate is the same as the x coordinate let's go the same with 3 over 2 but the y is negative negative one half so i have sine 30 d a cosine 30 degrees sine 30 degree cosine 150 sine 150 oh what is what angle is this this is 180 plus 30 is 210 pi plus pi over 6 is 7 pi over 6 so i know sine 7 pi over 6 is negative one half and the same for the 60 degree so 60 degrees should be here following the same idea i'll go a little faster um so what what kind of angle do i have like this this is 60 degree right so this this was 30 degree and so my my triangle would be like like this right so where the 30 degree is here and we have a the vertical is square with 3 a the hypotenuse is 2a again i change it into 1 right change it into 1 by plugging in x 0 with 3 so i would have um change it into vertical like kind of like longer like this but with 3 over 2 as a vertical the horizontal is one half so my point is with b flip over one half with 3 over 2 right and this is 60 which is pi over 3. and then i would just copy the same thing so this would be 180 i would put actually the triangle over here 180 minus 60 is 120 which is also pi minus pi over 3 is 2 pi over 3 and the 2 8 2 coordinate is the same as here except the x is negative negative one half with 3 over 2 right and then the bottom two would be 180 at 60 is 240 and pi at pi over 3 is 4 pi over 3. the coordinate is the same as here example positive and negative uh both x and y and a negative negative one half negative with three over two right x is shorter is negative one half and what about this one is 300 minus 60 so it's uh 360 minus 60 is 300 and in terms of the radian is 2 pi minus pi over 3 is 5 pi over 3. the two coordinate is exactly the same as here except the x coordinate is now positive so i have one half with three over two so all the all of these are generated from the first chord from the first quadrant and um you need to remember those six pairs of uh x and y coordinate and you can generate the other sine and cosine so this is um we'll combine it in into a unit circle uh i have a blank unit circle you can use it to
190543	https://www.youtube.com/watch?v=kNRcb3KX4U4	A Polynomial Functional System of Equations SyberMath 155000 subscribers 733 likes Description 20321 views Posted: 28 Jul 2021 Join this channel to get access to perks:→ My merch → Follow me → Subscribe → Suggest → If you need to post a picture of your solution or idea: Integral of Powers of sec(x): ChallengingMathProblems #FunctionalEquations EXPLORE 😎: Solving a Quick and Easy Functional Equation: Solving a Polynomial System in Two Ways: A Trigonometry Challenge: PLAYLISTS 🎵 : Number Theory Problems: Challenging Math Problems: Trigonometry Problems: Diophantine Equations and Systems: Calculus: 57 comments Transcript: hello everyone in this video we're going to be solving a polynomial system we have p of p of x equals 4x plus 7 and p of q of x equals negative 6x plus 5. px and qx are polynomials so this could be considered a functional system as well with functions being some special type so if you consider the result from here we get p of p of x so a polynomial composed with itself gives us 4x plus 7. how is that possible so there's there are some different possible scenarios so let's go ahead and take a look at them but first of all here's one thing i want you to consider with polynomials one of the most important things to talk about is the degree the degree of the polynomial gives us a lot of information and a lot of clues for problem solving in this case we are composing a function or polynomial with itself so what is that supposed to mean suppose you had p of x equals x squared plus one i'm just testing it doesn't have to be the same p of x if you don't like this change it into i don't know t of x or something like that or a of x how about that let's do it so people don't get confused they're gonna say how did you change the p of x okay let's suppose we have a polynomial a of x equals x squared plus 1 and i want to compose a with itself so it's kind of like a of a of x how do you evaluate that well just replace t x with a of x which is x squared plus one and by using the definition of a of x this gives you x squared plus one squared plus one and then from here you get x to the fourth power plus two x squared plus 1 plus 1 is going to give you 2. so now notice that when i compose a with itself i got a quartic polynomial a was quadratic then the degree just doubled if i take a cubic and compose it with itself i'm going to be getting something six power so it's basically going to uh well i shouldn't say double so if you had a cubic polynomial let's let's think about another scenario like suppose b of x is equal to x cubed then b of b of x would be b of x cubed and that would equal x cubed quantity cubed and that will be x to the ninth power so the power basically is the square of the degree so now here we have a linear result what is that supposed to mean the degree of this polynomial is 1 and the square root of 1 is 1. that means that p of x needs to be linear all right so that's the first assumption we have to make otherwise we're going to have to go through infinitely many possibilities and obviously none of them is going to work except for the linear case so p of x needs to be linear let's go ahead and solve the first equation first this is not something you can do elimination or substitution with yet let's go ahead and find out p of x first so if p of x is linear can i write it as a x plus p right where obviously a is not equal to zero so on and so forth now if you compose p with itself which is p of p of x we're supposed to get 4x plus 7. now replace p with ax plus b and you're supposed to get 4x plus 7. now apply the definition of p to a x plus b so that's going to be a times the quantity a x plus b plus b so that's and you can keep doing this sometimes you're going to have a problem like f of f of f of dot dot n times that's basically what it means okay and of course in these cases hopefully there's some type of pattern otherwise it's going to take it's going to be crazy all right so let's go ahead and distribute a squared x and then i get a b plus b let's go ahead and factor out b so that's going to give us a plus one in parentheses this is going to make a solution easier so now we have the following and what is that supposed to mean if two polynomials are equal for all values of x that means the coefficients of x are equal and the constant terms are equal so from here we get two results a squared equals four that means a is equal to two or a is equal to negative two obviously there's no reason why we would pick one over the other so we're going to consider both if a is equal to 2 i can just go ahead and substitute that here while this is supposed to equal 7 and if a is 2 then i'm getting something like 3 b equals 7 and this implies that b is equal to 7 3. from the second one i get a nicer expression if a is equal to negative 2 negative 2 plus 1 is negative 1 so i get negative b equals 7 which implies b equals negative 7. so this a value should be taken with that b value and the second one should be taken with the second value so this gives us two options okay two cases let's take a look at both cases case number one case number one is where a is to and b is 7 thirds and that gives us p of x equals 2 x plus 7 thirds now you can obviously test this like plug it in for example if i had to do the composition of this with itself you'll notice that it works you get 4x plus 14 thirds plus 7 thirds which is 21 thirds and that is equal to 7. and obviously the other one is going to work as well so you can test it you don't have to but it just shows you that you got it right so now p of x can be this one and what about the q of x right well p is composed with q and you know the composition is from right to left so p of q of x let's go ahead and write that down as an equation p of q of x is given to us as negative six x plus five right so we're supposed to find q of x from here where when p of x is equal to two x plus seven thirds so let's go ahead and apply the definition of p to q of x that gives us two times q of x plus seven thirds is equal to negative six x plus five obviously there's more than one way to do it you could also assume that hey q of x must be a linear polynomial as well and suppose we call it mx plus n substitute solve for m and n and so on so but i just want to do it this way because it's kind of cooler anyways so you can go ahead and solve for q of x isolated subtract seven thirds 15 minus 7 is 8. so it's going to be negative 6x plus 8 thirds and then you can just divide everything by 2 here to find q of x and that is going to be negative 3x plus 4 thirds so for this p of x value we get this q of x value that gives us a solution as an ordered pair and we're going to write these at the end so don't worry about keeping track of them right now let's go ahead and take a look at the second case now remember the second case involves a being equal to negative 2. so if a is equal to negative 2 then this gives us what it should give us a b value and it did right it was b equals negative 7 great so b equals negative 7 in this case so that gives us p of x as negative 2x minus seven okay great now i wanna use my q of x equation again one more time p of q of x is given to be uh is given to be equal to what is that equal to okay p of q of x is given as negative six x plus five okay negative six x plus five so now i'm going to replace the x here with q of x and on the right hand side as well so it's gonna give me negative two times q of x minus seven is equal to negative 6x plus 5. again this solution avoids the assumption that q of x equals mx plus n but it's pretty much the same thing so add 7 to both sides and then divide both sides by negative 2 and you get q of x equals 3x minus 6. so if p of x is equal to negative 2x minus 7 then q of x is equal to 3x minus 6. let's go ahead and write down the solution or solutions as ordered pairs so the solutions are going to look like this with px being the first one if px is 2x plus 7 thirds then q of x is going to be negative 3x plus 4 thirds and if p of x is equal to negative 2x minus 7 then q of x is going to be 3x minus 6. and let's not forget to close the braces because i believe last time i forgot to brace uh close the braces anyways this brings us to the end of this video well thank you for watching i hope you enjoyed it please let me know don't forget to comment like and subscribe i'll see you now i'll see you in another video tomorrow until then be safe take care and bye bye
190544	https://www.masterorganicchemistry.com/2018/01/19/hybridization-and-bond-strengths/	On Hybrid Orbitals And Bond Strengths – Master Organic Chemistry Skip to content Study Guides Reactions and Mechanisms Quizzes Free Stuff MOC Membership Tutoring Search Type and Press “enter” to Search Searching... Tutoring Login Home / Orbital Hybridization And Bond Strengths Bonding, Structure, and Resonance By James Ashenhurst Orbital Hybridization And Bond Strengths Previous How To Determine Hybridization: A Shortcut Next Sigma bonds come in six varieties: Pi bonds come in one Last updated: December 29th, 2024 \| How Orbital Hybridization Affects Bond Strengths Understanding the concept of hybrid orbitals (see previous post) lets you make accurate predictions about trends in bond strengths. In this post we’ll give several examples of how to do this. Table of Contents A Quick Quiz (On Bond Dissociation Energy) Key Principle On Orbital Hybridization And Bond Strengths: The Greater The s-character, The Stronger The Bond Electrons In s-Orbitals Are Closer To the Nucleus Than Electrons In The Corresponding p-Orbitals Summary: All Else Being Equal, The Greater The s-Character, The Stronger The Bond Bonus Section: An Answer To A Question A Few People Might Be Asking Homolytic versus Heterolytic Bond Cleavage A Test For You One Last Thing: This Concept Also Explains Why CH 3+ Is More Stable Than NH 3+ Notes 1. A Quick Quiz Let’s start with a quick quiz. What’s the strongest C–H bond, below? (In other words, which C–H bond has the highest bond-dissociation energy?) The answer is C. In a), the carbon is sp 3 hybridized and the bond dissociation energy is 105 kcal/mol In b), the carbon is sp 2 hybridized and the bond dissociation energy is 110 kcal/mol In c), the carbon is sp hybridized and the bond dissociation energy is 126 kcal/mol 2. Key Principle On Orbital Hybridization And Bond Strengths: The Greater The s-character, The Stronger The Bond Note that the trend for bond strengths, above, is sp > sp 2> sp 3 In other words, the more s-character on carbon, the stronger the bond. Let’s try this again. What about these three C–C bonds? Hopefully you ranked them A > B > C . All else being equal: an sp-sp 3 bond is stronger than an sp 2-sp 3 bond,which in turn is stronger than an sp 3-sp 3 bond. Why? 3. Electrons In s-Orbitals Are Closer To the Nucleus Than Electrons In The Corresponding p-Orbitals On average: electrons in s orbitals are closer to the nucleus (and therefore feel a greater effective positive charge) than electrons in the corresponding p orbitals. Electrons in an sp orbital (50% s-character) will therefore experience a greater attractive force to the nucleus than electrons in an sp 2 (33% s-character) or sp 3 (25% s-character) orbital. “Bond dissociation energy” (BDE) mentioned above, is a measure of the energy required for homolytic cleavage of a bond (homo = same;lysis= breaking). [See post: Bond Dissociation Energies = Homolytic Cleavage] That is, it measures the energy required for breakage of the bond such that each atom ends up with the same number of electrons, such as the reactions below: In the homolytic cleavage reactions above, a single electron is removed from the C–H molecular orbital and placed on hydrogen, which dissociates. The other electron remains on carbon. The bond dissociation energy for the sp–H bond is higher than that for sp 3–H because more force (energy) is required to remove an electron from the more tightly-held sp–H molecular orbital and place it exclusively on the hydrogen atom. So electrons in an orbital with more s character will be closer to the nucleus and feel a stronger electrostatic force. This helps to explain the higher bond energy. Recognizing the type of bonds in molecules is a key skill.(see also:C-C sigma bonds come in six varieties, C-C pi bonds come in one). For most purposes, this ends the key lesson of the post. 4. Summary: All Else Being Equal, The Greater The s-Character, The Stronger The Bond But… if you find yourself trying to reconcile an apparent contradiction that comes from another part of the course, read on. 5. Bonus Section: An Answer To A Question A Few People Might Be Asking Alkynes (pK a 25) are stronger acids than typical alkanes (pK a> 50). For example, alkynes are readily deprotonated by strong bases such as NaNH 2, whereas alkanes are not: Why? Because the C–H bond in an alkyne has more s-character, and the resulting lone pair on carbon is held more tightly to the nucleus, rendering the conjugate base more stable. Oh dear. There seems to be an angry mob approaching. WAIT!You just said that alkyne C–H bonds are stronger than alkane C–H bonds because they have more s-character and now you are saying that they are easier to break because they have more s-character. Shouldn’t that make alkynesless acidic because the C–H bonds have more s-character? Put away the pitchforks! There is a perfectly logical explanation for this! 6. Homolytic versus Heterolytic Bond Cleavage What’s the source of the confusion here? Let’s follow the electrons. Ultimately the resolution to this dilemma is recognizing the difference between homolytic cleavage (which is what bond dissociation energy measures) and heterolytic cleavage (which is what occurs in an acid-base reaction, the loss of H+ ). Let’s look at these two processes. In homolytic cleavage of a C–H bond, an electron is completely removed from the vicinity of the carbon and placed on hydrogen. Because of the greater s-character of the bond, it is more difficult to remove an electron from an sp-hybridized carbon than from an sp 3 hybridized carbon. As we said above, that’s why the bond-dissociation energies for alkyne C–H bonds are higher than for alkane C–H bonds. In an acid-base reaction, the C–H bond also breaks, but it breaks in such a way that the pair of electrons stays on the carbon atom. Since the bond breaks in a way that leads to an uneven distribution of electrons, it is called heterolytic bond cleavage (hetero = different, lysis = breaking). Let’s say that again: in an acid base reaction, the pair of electrons stays on carbon, resulting in a negatively-charged carbon atom (a “carbanion”). Acidity, as measured by pK a, is a measure of the equilibrium between an acid and its conjugate base. The more that the equilibrium favours the conjugate base, the lower the pK a and the stronger the acid. In other words: Any factor which makes the conjugate base more stable, increases acidity. In the conjugate base of an alkyne, the lone pair is held in an sp- orbital with 50% s-character. In the conjugate base of an alkane, the lone pair is held in an sp 3 orbital with 25% s-character. What lone pair will be more stable? The lone pair held more tightly to the nucleus – that is, the sp-orbital. That’s why alkynes are more acidic than alkanes: the conjugate base is more stable. Hopefully this makes it clearer that stronger C–H bonds and greater C–H acidity are two sides of the same phenomenon. [You might find it helpful to think of sp orbitals as having a greater effective electronegativity than sp 2 orbitals, which in turn have a greater electronegativity than sp 3 orbitals. So the reason for the greater acidity of alkynes relative to alkenes is not really so different than the greater acidity of H–F relative to H 2 O. ] 7. A Test For You At the risk of droning on, there is a third side of this phenomenon to consider. Homolytic cleavage can only happen one way. But there are two ways to draw heterolytic cleavage of a C–H bond, and you might have noticed that I only showed one. There’s a second (albeit unlikely) way heterolytic cleavage can happen. The pair of electrons in the C–H bond could migrate instead to hydrogen, not carbon, resulting in a hydride anion and a carbocation. [Again, very unlikely, but bear with me here]. Here’s a test I have for you. Based on everything we’ve gone through so far, which of the two products below (A or B) would be more stable? Think it through. To form A, we have to remove a pair of electrons from an sp-hybridized carbon. To form B, we have to remove a pair of electrons from an sp 3-hybridized carbon. What’s more favourable? Reaction to give A would require removing two electrons from an sp-hybridized carbon. Reaction to give B would require removing two electrons from an sp 3 hybridized carbon. The reaction to give B should be much easier, because the electrons are less tightly held. And indeed, sp 3-hybridized carbocations have been observed and even isolated. In contrast, sp-hybridized carbocations such as A are extremely unstable and have never been observed directly. [Note 1] This observation is also consistent with all the observations above! 8. One Last Thing: This Concept Also Explains Why CH 3+ Is More Stable Than NH 3+ As I said above, it can be helpful to think of sp-hybridized carbons as having a greater effective electronegativity than sp 2-hybridized carbons, which in turn have greater electronegativity than sp 3 hybridized carbons. We used this to rationalize why alkynes are stronger acids than alkanes, similarly to why H-F is a stronger acid than H 2 O. This can also be flipped around, just as in our carbocation example above. The greater the electronegativity of the atom, the more unstable electron-deficient species become. This helps to explain why, for example, H 3 C+ (with six valence electrons) is much more stable than H 3 N+ (also with six valence electrons) which in turn is far more stable than oxygen or fluorine with six valence electrons; increasing electronegativity has the same effect on the stability of electron-deficient species as increasing s-character. Notes Related Articles Hybrid Orbitals and Hybridization How To Determine Hybridization: A Shortcut How Do We Know Methane (CH4) Is Tetrahedral? 7 Factors That Stabilize Positive Charge in Organic Chemistry 7 Factors that stabilize negative charge in organic chemistry 3 Factors That Stabilize Carbocations Bond Dissociation Energies = Homolytic Cleavage The Marriage May Be Bad, But the Divorce Still Costs Money Note 1: Here’s a notable attempt to make sp-hybridized carbocations. The only application of helium in organic chemistry that I have ever seen. 00 General Chemistry Review Lewis Structures Ionic and Covalent Bonding Chemical Kinetics Chemical Equilibria Valence Electrons of the First Row Elements How Concepts Build Up In Org 1 ("The Pyramid") 01 Bonding, Structure, and Resonance How Do We Know Methane (CH4) Is Tetrahedral? Hybrid Orbitals and Hybridization How To Determine Hybridization: A Shortcut Orbital Hybridization And Bond Strengths Sigma bonds come in six varieties: Pi bonds come in one A Key Skill: How to Calculate Formal Charge The Four Intermolecular Forces and How They Affect Boiling Points 3 Trends That Affect Boiling Points How To Use Electronegativity To Determine Electron Density (and why NOT to trust formal charge) Introduction to Resonance How To Use Curved Arrows To Interchange Resonance Forms Evaluating Resonance Forms (1) - The Rule of Least Charges How To Find The Best Resonance Structure By Applying Electronegativity Evaluating Resonance Structures With Negative Charges Evaluating Resonance Structures With Positive Charge Exploring Resonance: Pi-Donation Exploring Resonance: Pi-acceptors In Summary: Evaluating Resonance Structures Drawing Resonance Structures: 3 Common Mistakes To Avoid How to apply electronegativity and resonance to understand reactivity Bond Hybridization Practice Structure and Bonding Practice Quizzes Resonance Structures Practice 02 Acid Base Reactions Introduction to Acid-Base Reactions Acid Base Reactions In Organic Chemistry The Stronger The Acid, The Weaker The Conjugate Base Walkthrough of Acid-Base Reactions (3) - Acidity Trends Five Key Factors That Influence Acidity Acid-Base Reactions: Introducing Ka and pKa How to Use a pKa Table The pKa Table Is Your Friend A Handy Rule of Thumb for Acid-Base Reactions Acid Base Reactions Are Fast pKa Values Span 60 Orders Of Magnitude How Protonation and Deprotonation Affect Reactivity Acid Base Practice Problems 03 Alkanes and Nomenclature Meet the (Most Important) Functional Groups Condensed Formulas: Deciphering What the Brackets Mean Hidden Hydrogens, Hidden Lone Pairs, Hidden Counterions Don't Be Futyl, Learn The Butyls Primary, Secondary, Tertiary, Quaternary In Organic Chemistry Branching, and Its Affect On Melting and Boiling Points The Many, Many Ways of Drawing Butane Wedge And Dash Convention For Tetrahedral Carbon Common Mistakes in Organic Chemistry: Pentavalent Carbon Table of Functional Group Priorities for Nomenclature Summary Sheet - Alkane Nomenclature Organic Chemistry IUPAC Nomenclature Demystified With A Simple Puzzle Piece Approach Boiling Point Quizzes Organic Chemistry Nomenclature Quizzes 04 Conformations and Cycloalkanes Staggered vs Eclipsed Conformations of Ethane Conformational Isomers of Propane Newman Projection of Butane (and Gauche Conformation) Introduction to Cycloalkanes Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes Calculation of Ring Strain In Cycloalkanes Cycloalkanes - Ring Strain In Cyclopropane And Cyclobutane Cyclohexane Conformations Cyclohexane Chair Conformation: An Aerial Tour How To Draw The Cyclohexane Chair Conformation The Cyclohexane Chair Flip The Cyclohexane Chair Flip - Energy Diagram Substituted Cyclohexanes - Axial vs Equatorial Ranking The Bulkiness Of Substituents On Cyclohexanes: "A-Values" Cyclohexane Chair Conformation Stability: Which One Is Lower Energy? Fused Rings - Cis-Decalin and Trans-Decalin Naming Bicyclic Compounds - Fused, Bridged, and Spiro Bredt's Rule (And Summary of Cycloalkanes) Newman Projection Practice Cycloalkanes Practice Problems 05 A Primer On Organic Reactions The Most Important Question To Ask When Learning a New Reaction Curved Arrows (for reactions) Nucleophiles and Electrophiles The Three Classes of Nucleophiles Nucleophilicity vs. Basicity What Makes A Good Nucleophile? What Makes A Good Leaving Group? 3 Factors That Stabilize Carbocations Equilibrium and Energy Relationships 7 Factors that stabilize negative charge in organic chemistry 7 Factors That Stabilize Positive Charge in Organic Chemistry What's a Transition State? Hammond's Postulate Learning Organic Chemistry Reactions: A Checklist (PDF) Introduction to Oxidative Cleavage Reactions 06 Free Radical Reactions Free Radical Reactions 3 Factors That Stabilize Free Radicals Bond Strengths And Radical Stability Free Radical Initiation: Why Is "Light" Or "Heat" Required? Initiation, Propagation, Termination Monochlorination Products Of Propane, Pentane, And Other Alkanes Selectivity In Free Radical Reactions Selectivity in Free Radical Reactions: Bromination vs. Chlorination Halogenation At Tiffany's Allylic Bromination Bonus Topic: Allylic Rearrangements In Summary: Free Radicals Synthesis (2) - Reactions of Alkanes Free Radicals Practice Quizzes 07 Stereochemistry and Chirality Types of Isomers: Constitutional Isomers, Stereoisomers, Enantiomers, and Diastereomers How To Draw The Enantiomer Of A Chiral Molecule How To Draw A Bond Rotation Introduction to Assigning (R) and (S): The Cahn-Ingold-Prelog Rules Assigning Cahn-Ingold-Prelog (CIP) Priorities (2) - The Method of Dots Enantiomers vs Diastereomers vs The Same? Two Methods For Solving Problems Assigning R/S To Newman Projections (And Converting Newman To Line Diagrams) How To Determine R and S Configurations On A Fischer Projection The Meso Trap Optical Rotation, Optical Activity, and Specific Rotation Optical Purity and Enantiomeric Excess What's a Racemic Mixture? Chiral Allenes And Chiral Axes Stereochemistry Practice Problems and Quizzes 08 Substitution Reactions Nucleophilic Substitution Reactions - Introduction Two Types of Nucleophilic Substitution Reactions The SN2 Mechanism Why the SN2 Reaction Is Powerful The SN1 Mechanism The Conjugate Acid Is A Better Leaving Group Comparing the SN1 and SN2 Reactions Polar Protic? Polar Aprotic? Nonpolar? All About Solvents Steric Hindrance is Like a Fat Goalie Common Blind Spot: Intramolecular Reactions Substitution Practice - SN1 Substitution Practice - SN2 09 Elimination Reactions Elimination Reactions (1): Introduction And The Key Pattern Elimination Reactions (2): The Zaitsev Rule Elimination Reactions Are Favored By Heat Two Elimination Reaction Patterns The E1 Reaction The E2 Mechanism E1 vs E2: Comparing the E1 and E2 Reactions Antiperiplanar Relationships: The E2 Reaction and Cyclohexane Rings Bulky Bases in Elimination Reactions Comparing the E1 vs SN1 Reactions Elimination (E1) Reactions With Rearrangements E1cB - Elimination (Unimolecular) Conjugate Base Elimination (E1) Practice Problems And Solutions Elimination (E2) Practice Problems and Solutions 10 Rearrangements Introduction to Rearrangement Reactions Rearrangement Reactions (1) - Hydride Shifts Carbocation Rearrangement Reactions (2) - Alkyl Shifts Pinacol Rearrangement The SN1, E1, and Alkene Addition Reactions All Pass Through A Carbocation Intermediate 11 SN1/SN2/E1/E2 Decision Identifying Where Substitution and Elimination Reactions Happen Deciding SN1/SN2/E1/E2 (1) - The Substrate Deciding SN1/SN2/E1/E2 (2) - The Nucleophile/Base SN1 vs E1 and SN2 vs E2 : The Temperature Deciding SN1/SN2/E1/E2 - The Solvent Wrapup: The Key Factors For Determining SN1/SN2/E1/E2 Alkyl Halide Reaction Map And Summary SN1 SN2 E1 E2 Practice Problems 12 Alkene Reactions E and Z Notation For Alkenes (+ Cis/Trans) Alkene Stability Alkene Addition Reactions: "Regioselectivity" and "Stereoselectivity" (Syn/Anti) Stereoselective and Stereospecific Reactions Hydrohalogenation of Alkenes and Markovnikov's Rule Hydration of Alkenes With Aqueous Acid Rearrangements in Alkene Addition Reactions Halogenation of Alkenes and Halohydrin Formation Oxymercuration Demercuration of Alkenes Hydroboration Oxidation of Alkenes m-CPBA (meta-chloroperoxybenzoic acid) OsO4 (Osmium Tetroxide) for Dihydroxylation of Alkenes Palladium on Carbon (Pd/C) for Catalytic Hydrogenation of Alkenes Cyclopropanation of Alkenes A Fourth Alkene Addition Pattern - Free Radical Addition Alkene Reactions: Ozonolysis Summary: Three Key Families Of Alkene Reaction Mechanisms Synthesis (4) - Alkene Reaction Map, Including Alkyl Halide Reactions Alkene Reactions Practice Problems 13 Alkyne Reactions Acetylides from Alkynes, And Substitution Reactions of Acetylides Partial Reduction of Alkynes With Lindlar's Catalyst Partial Reduction of Alkynes With Na/NH3 To Obtain Trans Alkenes Alkyne Hydroboration With "R2BH" Hydration and Oxymercuration of Alkynes Hydrohalogenation of Alkynes Alkyne Halogenation: Bromination and Chlorination of Alkynes Oxidation of Alkynes With O3 and KMnO4 Alkenes To Alkynes Via Halogenation And Elimination Reactions Alkynes Are A Blank Canvas Synthesis (5) - Reactions of Alkynes Alkyne Reactions Practice Problems With Answers 14 Alcohols, Epoxides and Ethers Alcohols - Nomenclature and Properties Alcohols Can Act As Acids Or Bases (And Why It Matters) Alcohols - Acidity and Basicity The Williamson Ether Synthesis Ethers From Alkenes, Tertiary Alkyl Halides and Alkoxymercuration Alcohols To Ethers via Acid Catalysis Cleavage Of Ethers With Acid Epoxides - The Outlier Of The Ether Family Opening of Epoxides With Acid Epoxide Ring Opening With Base Making Alkyl Halides From Alcohols Tosylates And Mesylates PBr3 and SOCl2 Elimination Reactions of Alcohols Elimination of Alcohols To Alkenes With POCl3 Alcohol Oxidation: "Strong" and "Weak" Oxidants Demystifying The Mechanisms of Alcohol Oxidations Protecting Groups For Alcohols Thiols And Thioethers Calculating the oxidation state of a carbon Oxidation and Reduction in Organic Chemistry Oxidation Ladders SOCl2 Mechanism For Alcohols To Alkyl Halides: SN2 versus SNi Alcohol Reactions Roadmap (PDF) Alcohol Reaction Practice Problems Epoxide Reaction Quizzes Oxidation and Reduction Practice Quizzes 15 Organometallics What's An Organometallic? Formation of Grignard and Organolithium Reagents Organometallics Are Strong Bases Reactions of Grignard Reagents Protecting Groups In Grignard Reactions Synthesis Problems Involving Grignard Reagents Grignard Reactions And Synthesis (2) Organocuprates (Gilman Reagents): How They're Made Gilman Reagents (Organocuprates): What They're Used For The Heck, Suzuki, and Olefin Metathesis Reactions (And Why They Don't Belong In Most Introductory Organic Chemistry Courses) Reaction Map: Reactions of Organometallics Grignard Practice Problems 16 Spectroscopy Degrees of Unsaturation (or IHD, Index of Hydrogen Deficiency) Conjugation And Color (+ How Bleach Works) Introduction To UV-Vis Spectroscopy UV-Vis Spectroscopy: Absorbance of Carbonyls UV-Vis Spectroscopy: Practice Questions Bond Vibrations, Infrared Spectroscopy, and the "Ball and Spring" Model Infrared Spectroscopy: A Quick Primer On Interpreting Spectra IR Spectroscopy: 4 Practice Problems 1H NMR: How Many Signals? Homotopic, Enantiotopic, Diastereotopic Diastereotopic Protons in 1H NMR Spectroscopy: Examples 13-C NMR - How Many Signals Liquid Gold: Pheromones In Doe Urine Natural Product Isolation (1) - Extraction Natural Product Isolation (2) - Purification Techniques, An Overview Structure Determination Case Study: Deer Tarsal Gland Pheromone 17 Dienes and MO Theory What To Expect In Organic Chemistry 2 Are these molecules conjugated? Conjugation And Resonance In Organic Chemistry Bonding And Antibonding Pi Orbitals Molecular Orbitals of The Allyl Cation, Allyl Radical, and Allyl Anion Pi Molecular Orbitals of Butadiene Reactions of Dienes: 1,2 and 1,4 Addition Thermodynamic and Kinetic Products More On 1,2 and 1,4 Additions To Dienes s-cis and s-trans The Diels-Alder Reaction Cyclic Dienes and Dienophiles in the Diels-Alder Reaction Stereochemistry of the Diels-Alder Reaction Exo vs Endo Products In The Diels Alder: How To Tell Them Apart HOMO and LUMO In the Diels Alder Reaction Why Are Endo vs Exo Products Favored in the Diels-Alder Reaction? Diels-Alder Reaction: Kinetic and Thermodynamic Control The Retro Diels-Alder Reaction The Intramolecular Diels Alder Reaction Regiochemistry In The Diels-Alder Reaction The Cope and Claisen Rearrangements Electrocyclic Reactions Electrocyclic Ring Opening And Closure (2) - Six (or Eight) Pi Electrons Diels Alder Practice Problems Molecular Orbital Theory Practice 18 Aromaticity Introduction To Aromaticity Rules For Aromaticity Huckel's Rule: What Does 4n+2 Mean? Aromatic, Non-Aromatic, or Antiaromatic? Some Practice Problems Antiaromatic Compounds and Antiaromaticity The Pi Molecular Orbitals of Benzene The Pi Molecular Orbitals of Cyclobutadiene Frost Circles Aromaticity Practice Quizzes 19 Reactions of Aromatic Molecules Electrophilic Aromatic Substitution: Introduction Activating and Deactivating Groups In Electrophilic Aromatic Substitution Electrophilic Aromatic Substitution - The Mechanism Ortho-, Para- and Meta- Directors in Electrophilic Aromatic Substitution Understanding Ortho, Para, and Meta Directors Why are halogens ortho- para- directors? Disubstituted Benzenes: The Strongest Electron-Donor "Wins" Electrophilic Aromatic Substitutions (1) - Halogenation of Benzene Electrophilic Aromatic Substitutions (2) - Nitration and Sulfonation EAS Reactions (3) - Friedel-Crafts Acylation and Friedel-Crafts Alkylation Intramolecular Friedel-Crafts Reactions Nucleophilic Aromatic Substitution (NAS) Nucleophilic Aromatic Substitution (2) - The Benzyne Mechanism Reactions on the "Benzylic" Carbon: Bromination And Oxidation The Wolff-Kishner, Clemmensen, And Other Carbonyl Reductions More Reactions on the Aromatic Sidechain: Reduction of Nitro Groups and the Baeyer Villiger Aromatic Synthesis (1) - "Order Of Operations" Synthesis of Benzene Derivatives (2) - Polarity Reversal Aromatic Synthesis (3) - Sulfonyl Blocking Groups Birch Reduction Synthesis (7): Reaction Map of Benzene and Related Aromatic Compounds Aromatic Reactions and Synthesis Practice Electrophilic Aromatic Substitution Practice Problems 20 Aldehydes and Ketones What's The Alpha Carbon In Carbonyl Compounds? Nucleophilic Addition To Carbonyls Aldehydes and Ketones: 14 Reactions With The Same Mechanism Sodium Borohydride (NaBH4) Reduction of Aldehydes and Ketones Grignard Reagents For Addition To Aldehydes and Ketones Wittig Reaction Hydrates, Hemiacetals, and Acetals Imines - Properties, Formation, Reactions, and Mechanisms All About Enamines Breaking Down Carbonyl Reaction Mechanisms: Reactions of Anionic Nucleophiles (Part 2) Aldehydes Ketones Reaction Practice 21 Carboxylic Acid Derivatives Nucleophilic Acyl Substitution (With Negatively Charged Nucleophiles) Addition-Elimination Mechanisms With Neutral Nucleophiles (Including Acid Catalysis) Basic Hydrolysis of Esters - Saponification Transesterification Proton Transfer Fischer Esterification - Carboxylic Acid to Ester Under Acidic Conditions Lithium Aluminum Hydride (LiAlH4) For Reduction of Carboxylic Acid Derivatives LiAlH[Ot-Bu]3 For The Reduction of Acid Halides To Aldehydes Di-isobutyl Aluminum Hydride (DIBAL) For The Partial Reduction of Esters and Nitriles Amide Hydrolysis Thionyl Chloride (SOCl2) And Conversion of Carboxylic Acids to Acid Halides Diazomethane (CH2N2) Carbonyl Chemistry: Learn Six Mechanisms For the Price Of One Making Music With Mechanisms (PADPED) Carboxylic Acid Derivatives Practice Questions 22 Enols and Enolates Keto-Enol Tautomerism Enolates - Formation, Stability, and Simple Reactions Kinetic Versus Thermodynamic Enolates Aldol Addition and Condensation Reactions Reactions of Enols - Acid-Catalyzed Aldol, Halogenation, and Mannich Reactions Claisen Condensation and Dieckmann Condensation Decarboxylation The Malonic Ester and Acetoacetic Ester Synthesis The Michael Addition Reaction and Conjugate Addition The Robinson Annulation Haloform Reaction The Hell–Volhard–Zelinsky Reaction Enols and Enolates Practice Quizzes 23 Amines The Amide Functional Group: Properties, Synthesis, and Nomenclature Basicity of Amines And pKaH 5 Key Basicity Trends of Amines The Mesomeric Effect And Aromatic Amines Nucleophilicity of Amines Alkylation of Amines (Sucks!) Reductive Amination The Gabriel Synthesis Some Reactions of Azides The Hofmann Elimination The Hofmann and Curtius Rearrangements The Cope Elimination Protecting Groups for Amines - Carbamates The Strecker Synthesis of Amino Acids Introduction to Peptide Synthesis Reactions of Diazonium Salts: Sandmeyer and Related Reactions Amine Practice Questions 24 Carbohydrates D and L Notation For Sugars Pyranoses and Furanoses: Ring-Chain Tautomerism In Sugars What is Mutarotation? Reducing Sugars The Big Damn Post Of Carbohydrate-Related Chemistry Definitions The Haworth Projection Converting a Fischer Projection To A Haworth (And Vice Versa) Reactions of Sugars: Glycosylation and Protection The Ruff Degradation and Kiliani-Fischer Synthesis Isoelectric Points of Amino Acids (and How To Calculate Them) Carbohydrates Practice Amino Acid Quizzes 25 Fun and Miscellaneous A Gallery of Some Interesting Molecules From Nature Screw Organic Chemistry, I'm Just Going To Write About Cats On Cats, Part 1: Conformations and Configurations On Cats, Part 2: Cat Line Diagrams On Cats, Part 4: Enantiocats On Cats, Part 6: Stereocenters Organic Chemistry Is Shit The Organic Chemistry Behind "The Pill" Maybe they should call them, "Formal Wins" ? Why Do Organic Chemists Use Kilocalories? The Principle of Least Effort Organic Chemistry GIFS - Resonance Forms Reproducibility In Organic Chemistry What Holds The Nucleus Together? How Reactions Are Like Music Organic Chemistry and the New MCAT 26 Organic Chemistry Tips and Tricks Common Mistakes: Formal Charges Can Mislead Partial Charges Give Clues About Electron Flow Draw The Ugly Version First Organic Chemistry Study Tips: Learn the Trends The 8 Types of Arrows In Organic Chemistry, Explained Top 10 Skills To Master Before An Organic Chemistry 2 Final Common Mistakes with Carbonyls: Carboxylic Acids... Are Acids! Planning Organic Synthesis With "Reaction Maps" Alkene Addition Pattern #1: The "Carbocation Pathway" Alkene Addition Pattern #2: The "Three-Membered Ring" Pathway Alkene Addition Pattern #3: The "Concerted" Pathway Number Your Carbons! The 4 Major Classes of Reactions in Org 1 How (and why) electrons flow Grossman's Rule Three Exam Tips A 3-Step Method For Thinking Through Synthesis Problems Putting It Together Putting Diels-Alder Products in Perspective The Ups and Downs of Cyclohexanes The Most Annoying Exceptions in Org 1 (Part 1) The Most Annoying Exceptions in Org 1 (Part 2) The Marriage May Be Bad, But the Divorce Still Costs Money 9 Nomenclature Conventions To Know Nucleophile attacks Electrophile 27 Case Studies of Successful O-Chem Students Success Stories: How Corina Got The The "Hard" Professor - And Got An A+ Anyway How Helena Aced Organic Chemistry From a "Drop" To B+ in Org 2 – How A Hard Working Student Turned It Around How Serge Aced Organic Chemistry Success Stories: How Zach Aced Organic Chemistry 1 Success Stories: How Kari Went From C– to B+ How Esther Bounced Back From a "C" To Get A's In Organic Chemistry 1 And 2 How Tyrell Got The Highest Grade In Her Organic Chemistry Course This Is Why Students Use Flashcards Success Stories: How Stu Aced Organic Chemistry How John Pulled Up His Organic Chemistry Exam Grades Success Stories: How Nathan Aced Organic Chemistry (Without It Taking Over His Life) How Chris Aced Org 1 and Org 2 Interview: How Jay Got an A+ In Organic Chemistry How to Do Well in Organic Chemistry: One Student's Advice "America's Top TA" Shares His Secrets For Teaching O-Chem "Organic Chemistry Is Like..." - A Few Metaphors How To Do Well In Organic Chemistry: Advice From A Tutor Guest post: "I went from being afraid of tests to actually looking forward to them". Comments Comment section 16 thoughts on “Orbital Hybridization And Bond Strengths” Deeptisays: May 29, 2024 at 12:02 pm Thank you. I would also like to say this is a great website.. I really liked the explanations and the dash of humour in between :) Reply Deeptisays: September 5, 2023 at 3:26 am How did the sp3 hybridization of C in CH3Cl change to sp2 hybridization in CH3+ obtained after heterolytic cleavage? CH3Cl –> CH3+ + Cl- How is there suddenly an unhybridised p orbital present on C of CH3+ (perpendicular to the molecular plane) and the C-H bond angles are 120°. Did C-H bonds break when sp3 hybridization changed to sp2? Reply 1. James Ashenhurstsays: September 5, 2023 at 1:42 pm No C-H bonds broke. The C-Cl bond broke. The unhybridized p orbital used to be the sp3-hybridized C-Cl bond. Imagine bonds as being tiny springs that attach atoms together – the C-Cl bond is always vibrating back and forth. Then imagine a vibration where the C-Cl bond becomes so long that the Cl- is able to break free, leaving with a pair of electrons. An empty orbital will remain, and the three filled C-H sp3 bonding pairs will adopt a geometry where they are slightly further apart (120°) Reply Ayush sharmasays: February 24, 2023 at 10:13 am is hybridisation limited to carbon only? Reply James Ashenhurstsays: February 24, 2023 at 11:39 am No – nitrogen, oxygen, sulfur, other atoms exhibit orbital hybridization Reply Duffsays: May 4, 2021 at 8:55 pm In part 7 you asked which product was most stable but your discussion dealt with the reaction and every point you made was sensible. The sp3 orbital in the (B) reaction would lend itself to a more favorable reaction in comparison to the sp orbital in the (A) reaction. My problem is, and maybe I’m overthinking this from a lot of experience with phenyl ethynyl endcaps and fluorinated cyclic olefins, wouldn’t the electron rich triple bond tend to lend electrons across the molecule to de-localize the carbocation thus making the methylacetylene product more stable. Victor made a good point about how messing with triple bonds can lead to quick reactions/polymerizations. I know the answer wouldn’t be as troublesome with just acetylene but in my addled mind the methyl group muddies up the result. Or I’m throwing arrows around willy-nilly like I did decades ago in school. Also, this site is amazing. Reply nigatu amsalusays: April 27, 2021 at 7:24 am thank you it is good post can I use bond dissociation energy and bond energy interachegeablly ? Reply 1. James Ashenhurstsays: May 3, 2021 at 2:02 pm Yes, that is correct. Reply zazaisays: March 30, 2020 at 12:24 am which sigma bond is strongest A – s_s overlap B – s-p overlap C – p-p overlap Reply 7. Georgesays: March 4, 2020 at 2:32 pm Jim, I tried posting a comment twice. I carefully copy and paste and read it in this box just to check it is OK. But each time some of my comment gets deleted after it is sent and it reads horrible. Ugh! Why?!? Third time lucky (probably not) Jim, First, thank you for your excellent posts (and figures!) which clarify so many points. My question concerns the second part of your post. I have no trouble understanding that in an acid/base reaction an electron pair would prefer to sit in an orbital with more s character rather than one with more p. But I guess my question is when you say think of alkane/alkyne acid base properties as similar to electronegativity across a row, the alternative might be to think of going up a column, where smaller, more electronegative atoms are worse acids (HF<< HCl < HBr < HI). So wouldn't a C-H bond with more s character also stabilize the bond better? Consistent with this C-H bonds do decrease slightly in length from alkanes to alkynes. Now obviously the experimental numbers support your way of thinking… but is there any chemical intuition you would use to explain why one should immediately think of this as going across a row rather than up a column? The best I can come up with is, well the changes in size here are very modest as you move from alkanes to alkynes and more in line with going across a row, i.e. from carbon to fluorine covalent radii go: 84 pm to 57 pm, a difference of only 50%, while going up a column, iodine to fluorine size goes from 139 to 57 pm, a 144% difference (and the shell number changes). Is there any other intuition you might offer of why electronegativity defeats the column size argument here? Sorry for the length… and again… thank you for your wonderful site. Regards, George Reply 8. Sarveshwar Pandeysays: December 2, 2019 at 12:32 pm Thank you sir for the above explanation. But don’t we compare the bond strength based on overlap extent of orbitals. As my book remarks, the relative strength of bond resulting from a p-p overlap is greater than s-s overlap due to greater extent of overlapping in p-p orbital. And according to this logic can we say that a sp2-s overlap will be greater than a sp-s overlap thus resulting in higher bond energy of C-H bond in alkenes as compared to that in alkynes ( believing that an orbital with greater p character has a better overlap). Can u plz clear this confusion… Reply 1. James Ashenhurstsays: December 2, 2019 at 12:39 pm I am using, “bond dissociation energies” interchangeably with “bond strengths” here. Comparing bond dissociation energies leads to the conclusion that bond strength increases as p-orbital character decreases. What book are you using that states p-p orbital overlap as being more favorable than s-s ? If you look in the carbon series (alkane alkene alkyne) the progression is clear. Be aware that in O-O and N-N it may appear that pi-bonds are stronger but this is an artefact from the relief of lone pair repulsion. Reply Kristynsays: February 8, 2018 at 11:21 am Hello, as far as the s-character goes, should sp = 50% s-character, sp^2 = 33.3% s-character and sp^3 = 25% s-character? You said sp has the greatest s character so the other hybridizations should be lower in s-character. Reply Jamessays: February 8, 2018 at 12:39 pm Fixed the typo. Thanks!!!! Reply Victorsays: January 19, 2018 at 5:53 pm Nice post, James, quite as usual. Spent a couple of years of my life putting EWG’s (like F, boranes, and CF3) on the terminal acetylenes to make them more reactive. Half of the time, the product would polymerize or do something else before I could snap-freeze it with liq. nitrogen. On a side note, you’re missing the Note 2. So, I had to note that your note 2 is, in fact, note 1. Pun intended ;) Reply 1. Jamessays: January 23, 2018 at 12:57 pm Thank you – note 1 was expunged, since I managed to cover it in the text. Reply Leave a Reply Your email address will not be published.Required fields are marked Comment Name Email Website [x] Save my name, email, and website in this browser for the next time I comment. [x] Notify me via e-mail if anyone answers my comment. Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. MOC About Mission Privacy Terms Contact Categories Study and Exam Tips Case Studies of Successful O-Chem Students Videos Site Reaction Guide Login Report a typo or mistake Share Your O-Chem Story For Educators Products & Services Online Tutoring Reagent Guide Org 1 Summary Sheets Org 2 Summary Sheets Reaction Guide Subscription Spectroscopy Pack Master Organic Chemistry LLC, 1831 12th Avenue South, #171, Nashville TN, USA 37203 © Copyright 2025, Master Organic Chemistry Organic Chemistry Is Awesome
190545	https://www.askiitians.com/iit-jee-s-and-p-block-elements/group-15-elements/	Group 15 Elements - Study Material for IIT JEE \| askIITians help@askiitians.com 1800-150-456-789 Live Courses Resources Exam Info Forum Notes Upcoming Examination Login Book a free demo of live class Group 15 Elements Table of Content Introduction to Group 15 Elements The Study of Group 15 Elements Trends of Some Atomic Properties Atomic Radii Ionization Enthalpy Electro-Negativity Physical Properties Trends in Melting and Boiling Points Allotropy Oxidation States Anomalous Properties Chemical Reactivity Introduction to Group 15 Elements The p-block elements are put to the right-hand side of the periodic table in groups from 13 to 18. In the iotas of p-block elements, the separating electron enters the valence p subshell. In this manner, in these elements, the n p subshell is step by step filled. The general valence shell electronic setup of group fifteen elements is ns 2, np 1-6. The electronic design of helium is 1s 2. Disregarding the way that helium does not have p orbitals, it is a p-block element since it takes after that of other p-block elements of the eighteenth group concerning their physical and compound properties. P-block elements are generally non-metals, while the remaining are metalloidsand metals. Fig. 1: Elements of group 15 with their atomic number, electronic configuration, group number and period number. The Study ofGroup 15 Elements Occurrence: Group 15 elements include nitrogen, phosphorus, arsenic, antimonyand bismuth. Nitrogen is the real constituent of the world's air, and records for 78% of it by volume. It is the primary member of this group and happens in a free state as a diatomic gas, N2. Minerals of Nitrogen: Indian saltpeter and Chile saltpeter. It is additionally found as the fundamental constituent of proteins, nucleic acids, amino acids, and catalysts. The following element in the group is phosphorus. It is the eleventh most copious element in the crust of the earth. In the consolidated state, it occurs as minerals as phosphates. Example: Fluoroapatite, Chlorapatite, and Hydroxyapatite are a few. Phosphorus is a fundamental constituent of animal and plant matter. Phosphate groups are constituents of nucleic acids, that is, DNA and RNA. Around 60% of bones and teethare made out of phosphates. Phosphoproteins are available in egg yolk, milk, and bone marrow. The rest of the elements of the group, that is, arsenic, antimony, and bismuth, mostly happen as sulfides. Example: Stibnite, Arsenopyrite, and bismuth glance. Fig. 2: Group 15 elements The separating electron in the elements of group 15 enters the n p subshell. Thus, the valence shell electronic arrangement for these elements becomes ns2 np3. There are five electrons in the valence shell of these elements. Because of the precisely half-filled electronic arrangement of the 'n p' subshell, the elements of this group are genuinely steady and stable. Example: Dinitrogen is an inactive gas under ordinary conditions. Fig. 3: Electronic configuration of group 15 elements Trends of Some Atomic Properties Atomic Radii Moving down the group, the ionic radii, and atomic radii increases because of the expansion of another main energy level in each progressive element. Ionization Enthalpy These elements demonstrate higher values of ionization enthalpy when contrasted with group 14 elements. This is because of their higher atomic charge, smaller nuclear radii, and stable half-filled electronic setups. As we move down thegroup, the ionization enthalpy values diminish. This is because of the progressive increment in the nuclear size. Electro-Negativity Electro negativity is the inclination of a particle to pull in a shared pair of electrons more towards itself. The electro negativity diminishes bit by bit on moving down the group because of the increase in atomic radius. Physical Properties Physical properties incorporate physical state, boiling and melting points, metallic character, allotropy, and density. Nitrogen is a diatomic gas, while the rest of the elements are solids in nature. Moving down a group, metallic character increments and the ionization enthalpy of the elements diminish with an increase in their nuclear size. Trends in Melting and Boiling Points The melting point increments from nitrogen to arsenic because of the continuous increment in nuclear size. The low melting point of nitrogen is because of its discrete diatomic particles. Then again, the high melting point of arsenic is ascribed to its goliath layered structure in which the layers are firmly pressed. In spite of the fact that the nuclear size increments from arsenic to antimony, there is a decrease in their melting points. Despite the fact that antimony has a layered structure, it has a low melting point than arsenic on account of the generally free pressing of particles. Furthermore, the melting point of bismuth is not as much as antimony because of the packing of atoms loosely by metallic holding. Then again, the boiling point step by step increments from nitrogen to bismuth. The density of these elements increases from nitrogen to bismuth. Allotropy All group fifteen elements, aside from bismuth, indicate allotropy. Nitrogen is found in two allotropic structures, that is, alpha nitrogenand beta nitrogen. Phosphorus exists in numerous allotropic structures. Of these, the two critical allotropic structures are red phosphorusand white phosphorus. Arsenic exists in three essential allotropic structures - black, grey, and yellow. Antimony additionally has three essential allotropic structures, to be specific, yellow, metallicand explosive. Oxidation States Every one of the elements of group 15 has 5 electrons in their outermost circle. They require just 3 electrons to finish their octet setup. The octet can be accomplished either by picking up 3 electrons or by sharing 3 electrons by a method for covalent bonds. Accordingly, the basic negative oxidation state of these elements is - 3. Moving down the group, the propensity to display - 3 oxidation state diminishes. This is because of the expansion in nuclear size and metallic character. Group 15 elements additionally indicate positive oxidation states of +3 and +5 by developing covalent bonds. Because of the inert pair effect, the +5 oxidation state stability diminishes down the group, while that of +3 oxidation state increments. Nitrogen has just s-and p-orbitals, yet no d-orbitals in its valence shell. Therefore, nitrogen can demonstrate a most extreme covalency of 4. A covalency of four is reached by the sharing of its lone pair of electron to another iota or particle. Phosphorus and the rest of the elements can display a covalency of five and a most extreme covalency, additionally called extended covalency of six. This is conceivable as a result of the nearness of empty d-orbitals in the valence shell. Every one of the compounds of group 15 elements, which display a +5 oxidation state, are covalent. Fig. 4: In case of +3 oxidation state, both ionic and covalent compounds are formed. Fig. 5: In group 15 elements, the covalent character decreases from nitrogen to bismuth. Nitrogen, in light of its small size, high electro negativity and solid propensity to shape p pi – p pi numerous bonds, it shows different oxidation states from - 3 to +5. Compound#### Oxidation State NH 3 Ammonia-3 N 2 H 4 Hydrazine-2 NH 2 OH Hydroxylamine-1 N 2 Dinitrogen 0 N 2 O Nitrous oxide 1 NO Nitric oxide 2 N 2 O 3 Nitrogen trioxide 3 N 2 O 4 Nitrogen tetroxide 4 N 2 O 5 Nitrogen pentoxide 5 If there should arise an occurrence of nitrogen, all oxidation states from +1 to +4 in acidic solution tend to disproportionate. In phosphorus, almost all transitional oxidation states in alkali and acid disproportionate into +5 and - 3. Fig. 5: Oxidation states of Nitrogen The general properties of group fifteen elements are tabulated below: Anomalous Properties Generally, the principal element of any group varies from rest of the elements regarding their properties. Nitrogen varies from other elements in its properties. The exceptional properties of nitrogen are credited to its small nuclear size, high ionization enthalpy or high electro negativity, the non–availability of d-orbitals and the possibility to shape various bonds. No one but nitrogen can shape nitride particles by picking up electrons because of its high electro negativity and small size. Nitrogen, being smaller in size, can successfully shape p pi – p pi bonds with different molecules of different elements with a small size and high electro-negativity, for example, oxygen and carbon. Alternate elements don't shape p pi - p pi bonds in view of their moderately bigger size. Synonymously the p orbitals of these elements are bigger in size. Dinitrogen is a diatomic particle with a triple bond between the two molecules. Fig. 6: Triple covalent bond between two atoms of nitrogen The bond enthalpy of a triple bond is extremely high around 941.4 KJ/mole. Dinitrogen is stable under conventional conditions. For a similar reason, it shapes a few compounds. They are: Compound#### Formulae Sodium nitrate NaNO 3 Potassium nitrite KNO 2 Sodium azide (azo)NaN 3 Nitrous oxide N 2 O Nitric acid (oxo acid)HNO 3 Potassium cyanide KCN The different elements in the group, for example, phosphorus, arsenic and antimony, exist as tetra atomic particles. In every one of these particles, just single bonds are available between the iotas or atoms. Fig. 7: Three tetraatomic molecules of group 15 But bismuth in its elemental state shapes metallic bonds. The catenation inclination is less for nitrogen when contrasted with alternate elements of the group. This is on the grounds that there are higher inter- electronic repulsions amongst the lone pair of electrons present on the nitrogen atoms. The high inter- electronic aversions in dinitrogen are credited to its small bond length or little size of nitrogen particles. Nitrogen does not shape d pi – p pi bonds because of the missing d orbitals. Phosphorus can frame d pi – p pi bonds. Example: Triethyl phosphate and Phosphorus Oxo chloride. Due to the accessibility of empty d orbitals in the rest of the elements of group 15 they frames compounds, for example, triphenyl arsine and triethyl phosphine, shape d pi – d pi bonds with transition elements. Chemical Reactivity Reactivity towards Hydrogen: Every one of the elements of group 15 forms EH3 type hydrides.Here E can be any element of group 15 such as nitrogen, phosphorus, arsenic, antimony or bismuth. 2E + 3H 2 → 2EH 3 Hydrogen Hydride N2 + 3H 2 → 2NH 3 Nitrogen Ammonia P 4 + 6H 2 →4PH 3 PhosphorusPhosphine AS 4+ 6H 2 →4ASH 3 Arsenic Arsine 4Sb + 6H 2 →4SbH 3 Antimony Stibine 4Sb + 6H 2 →4SbH3 Bismuth Bismuthine Stability: The inertness of hydrides decreases from ammonia to bismuth. This is on account of the fact that the central atom E increases in size down the group. With this increase in the central atom's size, the E – H bond gets to be distinctly weaker. With the expansion in size, the covering between the orbitals of E and hydrogen will not be powerful. This is the reason for the decrease in the inertness of hydrides moving down the group. Fig. 8: The trend of hydrides stability and bond dissociation enthalpy for the elements of group 15 Reducing Character: Hydrides formed from group 15 elements are very strong reducing agents. There is an increase in the reducing character of hydrides from ammonia to bismuth because of a reduction in the quality and strength of the E – H bond down the group. Bismuth is the strongest reducing agent among every one of the hydrides of group 15 elements. Basic Nature: The hydrides of these elements are basic in nature. They go about as Lewis bases because of the accessibility of a lone pair of electrons present on the central atom. With the increasing size of the central atom, there is a decrease in the basic character as we move down the group. Boiling Point: The boiling point of hydrides decreases from ammonia to phosphine and afterward increments from phosphine to bismuth. A similar pattern is watched for their melting points. Reactivity towards Oxygen: Two types of oxides are formed in group 15 elements.They are E 2 O 3 and E 2 O 5. pπ-pπ bonding tendency with oxygen is very high in Nitrogen. This is the reason why nitrogen forms a variety of oxides.Nitrogen forms 5 stable oxides. They are: Oxidation StateOxides NPAsSbBi +1 N 2 O---- Nitrous oxide +2 NO---- Nitric oxide +3 N 2 O 3 P 4 O 6 As 4 O 6 Sb 4 O 6 Bi 2 O 3 Dinitrogen trioxide +4 N 2 O 4---- Dinitrogen tetroxide +5 N 2 O 5 P 4 O 10 As 4 O 10 Sb 4 O 10⊗ Dinitrogen pentoxide Electronic Configuration of Bi: [Xe]4f 14 5d 10 6S 2 6P 3 ↓ Inert Pair Due to the inert pair effect Bismuth is not able to form oxides in +5 oxidation.Oxides formed from elements when in higher oxidation state are more acidic than that of the lower oxidation state. The acidic strength of oxides of nitrogen increases from N 2 O 3 to N 2 O 5. As we move down the group, the acidic character diminishes. As such, the basic character of oxides increases on moving down the group. Reactivity towards halogens: On reaction with halogens all the elements of group 15 form trihalides and pentahalides with the general formula EX 3and EX 5. 2E + 2X 2 → 2EX 3 Halogen Trihalide 2E + 5X 2 → 2EX 5 Halogen Pentahalide Example:NF 3, PF 3, AsF 3, SbF 3 and BiF 3 are trihalides. All the trihalides of these elements, beside those of nitrogen, are consistent. Phosphorus, arsenic and antimony shape pentahalides in light of the closeness of empty d orbitals in their valence shells. Nitrogen does not shape pentahalides as a result of the nonappearance of a d orbital in its valence shell. Pentahalides are more covalent than the relating trihalides. Moreover, the covalent character of halides decreases from nitrogen to bismuth. Each element of group 15 react with metals to frame their binary compounds demonstrating - 3 oxidation state with the general equation,M3E2. Here, M remains for metals while E remains for an element of group 15. Example: calcium phosphide, calcium nitride, and so forth. 3M + 2E → M 3 E 2 Binary compound 3Ca + N 2 → Ca 3 N 2 Calcium nitride 6Ca + P 2 → 2Ca 3 P 2 Calcium phosphide 6Zn + 4Sb → 2Zn 3 Sb 2 Zinc antimonide 6Mg + 4Bi → 2Mg 3 Bi 2 Magnesium bismuthide Watch this Video for more reference More Readings Group 15 Elements Complete Your Profile Phone Number +91 Name Email Select grade of the Student 6 th 7 th 8 th 9 th 10 th 11 th 12 th Submit View courses by askIITians ### Design classes One-on-One in your own way with Top IITians/Medical Professionals Click Here Know More ### Complete Self Study Package designed by Industry Leading Experts Click Here Know More ### Live 1-1 coding classes to unleash the Creator in your Child Click Here Know More ### a Complete All-in-One Study package Fully Loaded inside a Tablet! Click Here Know More Book a Live classLive coursesAsk question We help you live your dreams. 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190546	https://quantum.ch.ntu.edu.tw/online_courses/PCIIQC2009/PDF/Lecture8.pdf	Lecture 8 Properties of Quantum Harmonic Oscillator Study Goal of This Lecture • Energy level and vibrational states • Expectation values 8.1 Energy Levels and Wavefunctions We have ”solved” the quantum harmonic oscillator model using the operator method. Again, the mathematics is not diﬃcult but the ”logic” needs some eﬀort to get used to it. Think it through. Now we are ready to examine the rules. The quantum harmonic oscillator model yields. Hamiltonian: ˆ H = ˆ p2 2m + 1 2mω2ˆ x2, ω = s k µ, (8.1) k is force constant which related to bond energy and the µ is reduced mass for di-atomics. µ = m1m2 m1+m2 Schr¨ odinger Equation: ˆ Hψn = Enψn. (8.2) 1 Energy: En = (n + 1 2)ℏω, n = 0, 1, 2, 3, · · · . (8.3) Wavefunctions: ψ0(x) = (µω πℏ) 1 4 · e−µω 2ℏx2, ψn(x) = h 1 √ 2( rµω ℏx − s ℏ µω d dx) in · ψ0(x). (8.4) Energy levels and stationary wave functions: Figure 8.1: Wavefunctions of a quantum harmonic oscillator. Figure’s author: Al-lenMcC. The features of harmonic oscillator: 1. ψ0(x) is non-degenerate, all levels are non-degenerate. 2. Energy levels are equally spaced. The vibrational quanta = ℏω and n is the number of vibrational energy in the oscillator. The zero point energy = 1 2ℏω. 3. Wave function can be constructed by considering number of nodes. 4. At classical turning points: kinetic energy = 0, V (xt) = En ⇒1 2µω2x2 t = (n + 1 2)ℏω, (8.5) 2 x(n) t = s 2ℏ µω(n + 1 2). (8.6) Still, \|ψn(x > xn t )\|2 > 0, there is pobability of ﬁnding the particle in the classical forbidden region ⇒quantum tunneling. 5. Vibrational spectrum (IR/Raman) from transition between nearest vibrational levels. ∴∆E = ℏω = hν, ν = ω 2π = 1 2π s k µ, note ν ∝ q k µ, for stronger bond: ν ↑and for larger mass: ν ↓. This related to isotope eﬀect. (See Exercise) 8.2 Expectation Values In addition to thest properties of energy levels and wave functions, we can also quantitative calculate any expectation values of observables using the results we have so far. Let’s recall the properties of ladder operators: ˆ a = rmω 2ℏ(ˆ x + i ˆ p mω), lowing operator, ˆ a† = rmω 2ℏ(ˆ x −i ˆ p mω), raising operator, ˆ x = r ℏ 2mω(ˆ a† + ˆ a), ˆ p = i r ℏmω 2 (ˆ a† −ˆ a), [ˆ a, ˆ a†] = 1, ˆ a†ψn = √ n + 1ψn+1, ˆ aψn = √nψn−1. These results allow us to rewrite any observable in terms of ˆ a and ˆ a†, then we can calculate expectation values easily! Let’s calculate the position and momentum expectation values for the vibrational states ﬁrst. Before we start, note that {ψn} form a orthonormal set ∴ Z ψ∗ nˆ aψndτ = Z ψ∗ n( √ n + 1)ψn+1dτ = √ n + 1 Z ψ∗ nψn+1dτ = 0. (8.7) 3 Similarly, ∴ Z ψ∗ nˆ a†ψndτ = 0. (8.8) Expectation value of position The expectation value of position of wavefunction with n quantum number. ⟨ˆ x⟩n = Z ψ∗ n(x)ˆ xψm(x)dx = Z ψ∗ n(x) r ℏ 2mω(ˆ a† + ˆ a)ψn(x)dx = r ℏ 2mω Z ψ∗ n(x) · [ √ n + 1ψn+1(x) + √nψn−1(x) ]dx = 0. (8.9) Notice that only the term which the num-ber of ˆ a and ˆ a† match will contribute.(or say-ing nonzero) ⟨ˆ x2⟩= Z ψ∗ n(x) ℏ 2mω(ˆ a† + ˆ a2)ψn(x)dx = ℏ 2mω Z ψ∗ n(x){(ˆ a† + ˆ a)[ √ n + 1ψn+1(x) + √nψn−1(x)]}dx = ℏ 2mω Z ψ∗ n(x){ p (n + 1)(n + 2)ψn+2(x) + nψn(x)+ (n + 1)ψn(x) + p n(n −1)ψn−2(x)}dx = ℏ 2µω(2n + 1) = 1 µω2 ℏω(n + 1 2) = En µω2 . (8.10) With these result, we obtain the potential energy: ⟨ˆ V ⟩n = ⟨1 2µω2ˆ x2⟩n = 1 2µω2⟨ˆ x2⟩n = 1 2En. (8.11) We ﬁnd that it is half the total energy →make sense! Try to explain why it is one half the total energy in your own language! The expectation value of momentum Since the potential energy is known, the kinetic energy can be determined easily: ⟨ˆ T⟩n = En −⟨ˆ V ⟩n = 1 2En. (8.12) So the expectation value of momentum square: ∴⟨ˆ p2 2µ⟩n = ⟨ˆ T⟩n, ⟨ˆ p2⟩n = 2µ · 1 2En = µℏω(n + 1 2). (8.13) The above fomula can be checked using the same trick of ladder operator. Ex-pectation value of momentum can be calculated too and one will ﬁnd it is is zero (⟨ˆ p⟩n = 0). 4 Uncertainty of harmonic oscillator Recall that for any phycial observable ˆ A, the uncertainty is (∆ˆ A)2 = ⟨( ˆ A −⟨A⟩)2⟩= ⟨ˆ A2⟩−⟨A⟩2, (8.14) so for the uncertainty of position and momentum: ∆ˆ x = p ⟨ˆ x2⟩−⟨x⟩2 = s ℏ 2µω, (8.15) ∆ˆ p = p ⟨ˆ p2⟩−⟨p⟩2 = r µℏω 2 . (8.16) Multiply them, we obtain ∆ˆ x∆ˆ p = ℏ 2. (8.17) This is the minimal possible value allowed by the Heisenberg uncertainty principle. This is a special property of the ground state of the harmonic osillator model. It is also called a ”minimal uncetainty wavefunction”, ”coherent state”, · · · . We will show what’s special about it when we discuss time-evolution of it. The key for calculating the expectation value of quantum harmonic oscillator is to use ˆ a and ˆ a†. 8.2.1 Evaluate the Expectation Value of Superposition State The above calculation is not restricted to eigenstate. Since ˆ Hψn = Enψn, {ψn} form a complete orthonormal set, any wavefunction can be written as superposition of {ψn}: Ψ(x) = X n Cnψn, where Cn = Z ψ∗ nΨ(x)dx. (8.18) Then any ˆ A(x, p) can be evaluated: ⟨ˆ A⟩= Z Ψ∗(x) · ˆ A(ˆ x, ˆ p) · Ψ(x)dx. (8.19) Let’s consider an example, for: Ψ = 1 √ 2(ψ0 + ψ1), (8.20) then the energy will be ⟨E⟩= Z Ψ∗ˆ HΨdτ = 1 2 Z (ψ∗ 0 + ψ∗ 1) ˆ H(ψ0 + ψ1)dτ. (8.21) 5 Since ˆ H(ψ0 + ψ1) = ˆ Hψ0 + ˆ Hψ1 = E0ψ0 + E1ψ1 ⟨E⟩= Z Ψ∗ˆ HΨdτ = 1 2 Z (ψ∗ 0 + ψ∗ 1)(E0ψ0 + E1ψ1)dτ = 1 2[ E0 Z ψ∗ 0ψ0dτ + E1 Z ψ∗ 0ψ1dτ + E0 Z ψ∗ 1ψ0dτ + E1 Z ψ∗ 1ψ1dτ ] = 1 2(E0 + E1) ←averaged energy! (8.22) and for the position: ⟨x⟩= 1 2 Z (ψ∗ 0 + ψ∗ 1) r ℏ 2mω(ˆ a† + ˆ a)(ψ0 + ψ1)dτ = 1 2 r ℏ 2mω Z (ψ∗ 0 + ψ∗ 1)(ψ1 + √ 2ψ2 + ψ0)dτ = 1 2 r ℏ 2mω ̸= 0. (8.23) Figure 8.2: Ψ = 1 √ 2(ψ0 + ψ1). This state can exist but it is not a stationary state. 6
190547	https://www.igb-berlin.de/en/news/watch-out-birds-how-fish-evade-their-predators	Watch out – birds! How fish evade their predators \| IGB Skip to main content Suche Englisch EN Deutsch DE Secondary Menu Intranet Contact Jobs Structure Bluesky Youtube Mastodon LinkedIn Menü Research Programme areas Departments Research groups Projects Publications Selected publications All publications Infrastructures Data resources Networks Newsroom Topics Angling Adaptation and evolution Aquaculture and aquaponics Biodiversity Freshwater ecosystems Use and management Multiple stressors and pollutants Environmental change Behavioural ecology and swarm intelligence Water and matter cycles News Downloads Events Videos Newsletter Career IGB as employer Working conditions Human resources development (HRS) Equal opportunities Jobs Guest programme Doctoral education Apprenticeships Studies, internships and FÖJ People About us Contact Media representatives Politics and authorities Associations, societies, initiatives, foundations, think tanks Economy Schools Museums and artists Visitors Profile Library Structure Locations History of IGB Deutsch DE Intranet Contact Jobs Structure Breadcrumb Home Newsroom News 4 August 2021 short news Angelina Tittmann Watch out – birds! How fish evade their predators Many fish species are hunted by birds. Thus, it is essential for fish to recognise when they are in danger from avian predators. This is not trivial: not all birds prey on fish and even those fish-eating ones are not always in hunting mode, yet may fly over the water with no intention of attacking. Juliane Lukas, Jens Krause and David Bierbach from IGB have investigated whether fish are able to differentiate between cues that suggest a real bird attack and others that may indicate non-threatening events such as bird overflights. It turned out that fish responded with diving even when encountering minor disturbances. In the case of a bird’s predation attempt, however, they dove deeper, faster and longer. To decide whether and how to respond to disturbances, they used a combination of visual and acoustic cues. Sulphur mollies (Poecilia sulphuraria) react to fish-eating birds, such as the green heron (Butorides virescens), with fast dives. To avoid wasting energy, the fish have to distinguish between real attacks and non-threatening overflights. \| Photo: Juliane Lukas The ability of animals to effectively avoid predators is crucial for their survival. Prey must therefore be able to accurately interpret cues of potential danger, and distinguish them from disturbances that pose no threat. The international research team led by IGB researcher David Bierbach has investigated the antipredator behaviour of fish – the sulphur molly (Poecilia sulphuraria) – to avoid bird predation in a field environment and under lab conditions. In their natural habitat – sulphidic springs in southern Mexico – sulphur mollies are exposed to high rates of bird predation and evolved strategies to adapt their shoaling behaviour to this threat. Escape response in energy-saving mode Piscivorous birds must first enter the water before they can catch their prey. As such, they provide both visual and acoustic cues upon impact with the water surface or when diving into water. By contrast, overflying birds only produce visual cues. In a laboratory experiment, the researchers exposed groups of sulphur mollies to three different cue modalities: a visual cue, an acoustic cue and both cues combined. As the combined cue mimicked a real bird attack, it was predicted that fish should respond strongest to this treatment. And indeed, the research team found that sulphur mollies, which spend much of their time at the water surface, respond in a fine-tuned way to visual and acoustic cues: “Interestingly, fish responded with diving in 96 per cent of all cases, regardless of stimulus type. However, depth, duration and velocity of the dive were dependent on whether fish were responding to a single or the combined cue. For fish, this seems to be an optimal strategy: They are highly responsive to all cues, but do not waste too much energy when it turns out that there was no bird attack,” explained IGB researcher Juliane Lukas, lead author of the study. Multistage antipredator responses are a common strategy among prey In other words, the fish respond in two stages: the “all-or-nothing” response pattern of unselective dive initiation is followed by fine-tuning of certain dive parameters to the observed cues. Such multistage antipredator responses have also been described in other animals. “By performing successive behaviours, animals can update information about the predator or gather new information, optimising their behaviour at each stage – which reduces their risk of being eaten and minimises the efforts if it turns out to be less dangerous after all,” concluded Juliane Lukas. Read the study > Contact person David Bierbach -------------- Guest Scientist Research group Mechanisms and Functions of Group-Living Behavioural Biology Related news 26 September 2025 short news Open data: A roadmap that facilitates data reuse ------------------------------------------------ In an article in Nature Microbiology, the authors – including researchers from IGB – discuss improvements for the exchange of microbiome data between institutions. 15 September 2025 short news Recommendation for measurement method European Water Framework Directive ------------------------------------------------------------------------ An international research team recommends a genetic approach for the obligatory measurements under the European Water Framework Directive (WFD). 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190548	https://www.physicsforums.com/threads/find-integer-solutions-to-k-dfrac-ab-2-1-a-2b-1.1036773/	Find Integer Solutions to $k=\dfrac{ab^2-1}{a^2b+1}$ • Physics Forums Insights Blog-- Browse All Articles --Physics ArticlesPhysics TutorialsPhysics GuidesPhysics FAQMath ArticlesMath TutorialsMath GuidesMath FAQEducation ArticlesEducation GuidesBio/Chem ArticlesTechnology GuidesComputer Science Tutorials ForumsGeneral MathCalculusDifferential EquationsTopology and AnalysisLinear and Abstract AlgebraDifferential GeometrySet Theory, Logic, Probability, StatisticsMATLAB, Maple, Mathematica, LaTeX Trending Log inRegister What's new General Math Calculus Differential Equations Topology and Analysis Linear and Abstract Algebra Differential Geometry Set Theory, Logic, Probability, Statistics MATLAB, Maple, Mathematica, LaTeX Menu Log in Register Navigation More options Style variation SystemLightDark Contact us Close Menu Forums Mathematics General Math MHB Find Integer Solutions to $k=\dfrac{ab^2-1}{a^2b+1}$ Thread starter Albert1 Start date Feb 14, 2017 TagsInteger Feb 14, 2017 1 Albert1 1,221 0 $a,b\in N$ $k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$ find pair(s) of $(a,b)$ Discover more Online tutoring services Physics Forums Subscription Insights Blog Access Physics Articles Science-themed apparel math Math puzzle books Science forum sciences Mathematics news on Phys.org ChatGPT appears to improvise when put through ancient Greek math puzzle Systematic fraud uncovered in mathematics publications The science of sacrifice: How altruism and evolution can work in tandem Feb 17, 2017 2 Albert1 1,221 0 Albert said: $a,b\in N$ $k=\dfrac {ab^2-1}{a^2b+1}\,\, \,also \,\,\in N$ find pair(s) of $(a,b)$ $hint:$ Spoiler $if \,\,a=1\,\, then \,\,b=?$ $if \,\,a>1\,\, then \,\,no \,\,solution.\,\, why ?$ Thread 'Non-orthogonal bases' Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual... View full post » G By geordief Sep 18, 2025 Replies: 11 Thread 'Fermat's Last Theorem' Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation has no solutions with positive integers if It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et... 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190549	https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/	What Planck Length Is and It’s Common Misconceptions /29 Comments/in Physics Articles/by Alex Klotz Estimated Read Time: 7 minute(s) Common Topics: planck, length, quantum, constant, gravity The Planck length is an extremely small distance constructed from physical constants. Many misconceptions generally overstate its physical significance, stating that it’s the inherent pixel size of the universe. The Planck length does have physical significance, and I’ll discuss what it is and what it isn’t. Table of Contents What is the Planck Length? Planck units are defined based on physical constants rather than human-scale phenomena. So while the second is originally one-86400th of a day, the Planck time is based on the speed of light, Newton’s gravitational constant, and Planck’s (reduced) constant, which is twice the angular momentum of an electron. Hypothetically, if we met a group of aliens and wanted to discuss weights and measures, we could use Planck units and they’d know what we are talking about. There is a push towards making our human units based on physical constants, like defining the meter in terms of the speed of light, but at this time the kilogram is still the mass of a brick in France. “Natural” units still have a bit of choice regarding their definitions. Convention has chosen Planck’s reduced constant over Planck’s regular constant (they differ by a factor of 2##\pi##), and chosen to use the Coulomb constant instead of the dielectric constant or the fundamental charge for electromagnetic units. The latter provides a great example showing that Planck units are not inherently fundamental quantities: the Planck charge is roughly 11.7 times the actual fundamental charge of the universe. So what is the Planck length? It is defined as: This is how far light can go in a unit of Planck time because the speed of light is the “Planck speed.” In SI units, this is on the order of 10-35 meters. By comparison, one of the smallest lengths that have been “measured” is the upper bound on the electron’s radius (if an electron has a radius, what can we certainly say it is smaller than?) is 10-22 meters, about ten-trillion Planck lengths. It is really small. And so far, it is just a unit. The meter is a useful unit for measuring length, but there’s nothing inherently special about it. The Planck length is not useful for measuring any length, but is there anything special about it? How is it relevant to physics? The Planck length is the length scale at which quantum gravity becomes relevant. It is roughly the distance things have to be before you start to consider “Hmm I wonder if there’s a chance this whole system randomly forms a black hole.” I did not understand this until I convinced myself with the following derivation, which was the main inspiration for this article. Consider the energy (E) between two charges (let’s say they’re electrons) at some distance r. Doesn’t matter if they’re attracting or repelling right now. Just to clarify the symbols, e is the fundamental charge, ##\epsilon## is the dielectric constant. Now let’s change the units around, using the definition of the fine structure constant ##\alpha##, which is roughly 1/137. This lets us swap out the electromagnetic constants e and ##\epsilon## with the more “general” constants ##\hbar## and c. The Coulomb energy now looks like this: This is where the hand-waving will begin. If a given volume at rest has a certain amount of energy within, it will have a rest mass m=E/c##^2##. From Newtonian gravity, we can calculate the gravitational energy associated with our charges. We are neglecting the rest masses of the charges, but those are much smaller than the interaction energy. The question now is: at what distance is the electrostatic energy equal to the gravitational energy it causes? So we solve for r… and we find that the radius at which the gravitation of the interaction energy is as important as the interaction energy itself is roughly the Planck length (divided by the 11.7, the square root of 137, but we’ll hand-wave that away for now). This is where it is important: if things are interacting at distances close to the Planck length, you will have to take quantum gravity into account. One of the only physical systems where quantum gravity is relevant is the black hole. When calculating the entropy of a black hole, Hawking and Bekenstein found that it was equal to the number of Planck areas (Planck lengths squared) that can fit in the cross-sectional area of a Schwartzschild black hole (or a quarter of its total surface area), in units of the Boltzmann constant. The Hawking temperature of a black hole is one of the only equations where ##\hbar##, c, and G all appear, making it a quantum relativistic gravitational equation. However, the mass of a black hole can be continuous so the number of Planck areas on its surface need not be an integer. How is it not relevant to physics? There is a misconception that the universe is fundamentally divided into Planck-sized pixels, that nothing can be smaller than the Planck length, and that things move through space by progressing one Planck length every Planck time. Judging by the ultimate source, a cursory search of Reddit questions, the misconception is fairly common. There is nothing in established physics that says this is the case, nothing in general relativity or quantum mechanics pointing to it. I have an idea as to where the misconception might arise, that I can’t back up but I will state anyway. I think that when people learn that the energy states of electrons in an atom are quantized and that Planck’s constant is involved, a leap is made toward the pixel fallacy. I remember in my early teens reading about the Planck time in National Geographic, and hearing about Planck’s constant in high school physics or chemistry, and thinking they were the same. As I mentioned earlier, just because units are “natural” it doesn’t mean they are “fundamental,” due to the choice of constants used to define the units. The simplest reason that Planck-pixels don’t make up the universe is special relativity and the idea that all inertial reference frames are equally valid. If there is a rest frame in which the matrix of these Planck-pixels is isotropic, in other frames they would be length contracted in one direction, and moving diagonally concerning his matrix might impart angle-dependence on how you experience the universe. If an electromagnetic wave with the wavelength of one Planck length was propagating through space, its wavelength could be made even smaller by transforming to a reference frame in which the wavelength is even smaller, so the idea of rest-frame equivalence and a minimal length is inconsistent with one another. To add to people’s confusion, a lot of the Wikipedia articles on the Planck length were corrupted by one person trying to promote his papers by posting them on Wikipedia, making nonsensical claims with “proof” that a Planck-wavelength photon will collapse into a black hole (again, Lorentz symmetry explains why this doesn’t make sense). There is a surreal and amusing dialogue trying to get to the bottom of this, that you can still read in the discussion section of the Planck length Wikipedia page. There was an analysis recently of gamma-ray arrival times from a burst in a distant galaxy. The author considered what effect a discretization of space might have on the travel speed of photons of differing energy (it would no longer necessarily be constant), and found that to explain the observations the length scale of the discretization must be at least 525 smaller than the Planck-length. I’m not too sure how seriously people in the field take this paper. How might it be relevant to physics? Lorentz symmetry explains why Planck-pixels don’t make sense within current physics, however, current physics is incomplete, especially about quantum gravity. Going beyond established physics, is there more of a roll for the Planck length? I’m a bit out of my element talking about this, so I’ll be brief. The closest beyond-standard theory to the Planck-pixel idea is Loop Quantum Gravity and the concept of quantum foam. At least that is what I thought before John Baez corrected me. One of the features of Loop Quantum Gravity is that for something to have a surface area or a volume, it must have at least a certain quantum value of surface area or volume, but will not necessarily have integer values of that quantum, and the quantum is not exactly the square or cube of the Planck length, although it is of that order. Another potential model of quantum gravity is string theory, based on the dynamics of really small strings. To have these dynamics explain gravity, they are of order Planck length, but not specifically the Planck length. The first iteration of string theory was theorized to explain nuclear physics rather than gravity, and the length scale of the strings was much much larger. So to summarize, the Planck length is an important order of magnitude when quantum gravity is being discussed, but it is not the fundamental pixel size of the universe. Thanks to John Baez and Nima Lashkari for answering some questions about quantum gravity. Click For Forum Comments Alex Klotz Ph.D. McGill University, 2015 Assistant Professor, California State University, Long Beach My research is at the interface of biological physics and soft condensed matter. I am interested in using tools provided from biology to answer questions about the physics of soft materials. In the past I have investigated how DNA partitions itself into small spaces and how knots in DNA molecules move and untie. Moving forward, I will be investigating the physics of non-covalent chemical bonds using “DNA chainmail” and exploring non-equilibrium thermodynamics and fluid mechanics using protein gels. More Related Articles Blockworld and its Foundational Implications: Delayed Choice and No Counterfactual Definiteness Does the Block Universe of Physics Mean Time is an Illusion? Blockworld and its Foundational Implications: General Relativity and the Big Bang A Principle Explanation of the “Mysteries” of Modern Physics Struggles With the Continuum: Is Spacetime Really a Continuum? Why Is Quantum Mechanics So Difficult? 135 240 Alex Klotz Alex Klotz2015-09-09 14:00:532025-01-26 09:16:48What Planck Length Is and It’s Common Misconceptions You might also like Tools to Enrich our Quantum Mechanics Interpretations Discourse Either the Sun Is Getting Smaller or Gravity Is Getting Weaker The Unreasonable Effectiveness of the Popescu-Rohrlich Correlations Struggles with the Continuum: General Relativity Understanding the General Relativity View of Gravity on Earth Does Gravity Gravitate? 29 replies Benwade says: “I do understand the argument that the Planck length is not fundamental cause there is quite some choice left when it comes to defining such a length. What I don’t understand is how you can take arguments from the continuous paradigm (which is theories in terms of differential equations on real numbers) and argue about the invalidity of ideas from the discrete paradigm (universe being pixelated, things moving at the speed of light one unit at a time, …). From my point of view this chain of argument is invalid, exactly because the continuous paradigm breaks down around the scale when spacetime supposedly becomes discrete. As for myself I’m taking serious the idea, that all our established physical theories (including GR and QM) are effective theories in the sense, that they don’t express anything fundamental about the ultimate nature of reality, but instead are approximations to the inner workings of reality in the discrete paradigm. Any thoughts?”Hi, I am a complete physics idiot, but I read your posting. Are you saying that the equations the author of this article uses break down/or do not apply in this situation? I would be interested in hearing more about this. Thank you. Log in to Reply 2. Nugatory says: ”As for myself I’m taking serious the idea, that all our established physical theories (including GR and QM) are effective theories in the sense, that they don’t express anything fundamental about the ultimate nature of reality, but instead are approximations to the inner workings of reality in the discrete paradigm. Any thoughts?” Could be… There’s no way of disproving the possibility. But absent a candidate theory based on this discrete paradigm, there’s also nothing to discuss under the Physics Forums rules. This thread is closed. As always, PM me or another mentor if you have more to add and want it reopened. Log in to Reply 3. Ken G says: If there is anything that the history of physics has shown us, it is that we don’t shoot with high percentage when we try to anticipate the behavior in fundamentally new regimes. So I think what we really need are experiments that are capable of looking for evidence of discreteness. Until we have that, any theory will be pretty much guessing, in my opinion. But I do agree that all theories should be regarded as effective theories until demonstrated otherwise, with attention to the fact that they are impossible to demonstrate otherwise! Log in to Reply 4. kalimaa says: I do understand the argument that the Planck length is not fundamental cause there is quite some choice left when it comes to defining such a length. What I don’t understand is how you can take arguments from the continuous paradigm (which is theories in terms of differential equations on real numbers) and argue about the invalidity of ideas from the discrete paradigm (universe being pixelated, things moving at the speed of light one unit at a time, …). From my point of view this chain of argument is invalid, exactly because the continuous paradigm breaks down around the scale when spacetime supposedly becomes discrete. As for myself I’m taking serious the idea, that all our established physical theories (including GR and QM) are effective theories in the sense, that they don’t express anything fundamental about the ultimate nature of reality, but instead are approximations to the inner workings of reality in the discrete paradigm. Any thoughts? Log in to Reply 5. Benwade says: “I would probably go the other way… Obviously if your theory implies that something is turning into a black hole according to one observer, but is not turning into a black hole according to another observer, then your theory has been essentially discounted by reductio ab adsurdam. I believe the problem is with the premise than an object’s mass increases as it approaches the speed of light. An object’s MOMENTUM increases as [tex]p = frac{m}{sqrt{1 – (frac v c)^2}}v[/tex]; I feel that has been pretty well reasoned out. But the claim that an object’s actual mass has increased (and hence it’s capacitiy to pull other objects toward it by gravity) is NOT well supported by any reasoning I’m familiar with. I’m pretty sure I’ve seen this point made explicitly in some texts, but at 43, I’m well into my fifth decade of memory failure.” I (a complete physics idiot) actually posted a question that made the assumption that objects gained mass as they approached the speed of light. I was soon set right. Thank you for your explication, hand-wavey or not, of the Planck length, because I was a victim of the (erroneous) Planck-length = pixel size fiction as well. Log in to Reply 6. mfb says: A “classical” 4D planck volume of one planck length in spatial directions and one planck time in time direction would be crossed by light diagonally, as light moves by one planck length per planck time. A transformed planck volume with a shorter distance but a longer time loses this property. Log in to Reply 7. Ken G says: “I’m not a fan of this theory, but there is an idea that spacetime is divided into pre-existing irregular grains of 1 Planck volume. This is called spacetime “glass” quantization, as opposed to “crystal” quantization should the grains be regular. The glassy properties of the quantization help it escape the usual problems with Lorentz invariance.”Thank you for that insight. I would indeed think that if one wishes to regard spacetime as in some sense “coarse-grained” at the Planck scale, one must use a version of coarse-graining that is Lorentz invariant, meaning that the grains are defined by their volume but not their shape. This is hardly unprecedented– the same thing is done to “coarse grain” phase space for statistical mechanical calculations, since there is no need to use a cubic tiling of “equal lengths” of distance and momentum when deciding how to count states. I don’t mean to be unresponsive to the comment[quote=mfb] To make it worse, if you transform pixels, the relation between (dilated) Planck time and (contracted in one dimension) distance does not hold any more.[/quote]I simply didn’t understand it. It was my impression that volumes in spacetime would be Lorentz invariant, but perhaps there is something I am missing. Log in to Reply 8. haael says: “On the topic of the “Planck pixel,” perhaps this overall idea is being rejected too sweepingly. Presumably, the “pixels” would be in 4-D spacetime, not 3-D space, and volumes in 4-D spacetime are invariant, are they not? So I would imagine that if someone wanted to formulate a theory that said spacetime itself was parceled into “Planck pixels”, they would play the usual game that in different reference frames, meaning along different world lines, the “pixels” would distort, but they’d still tile the spacetime in the same way. Yes that means objects don’t “move one Planck length every Planck time”, but that’s obvious– any such object would be perceived as moving at the speed of light. Instead, a “Planck pixel” idea could say that spacetime is discretely tiled, in the sense that world lines cannot be defined with finer precision than that– similar to the way quantum mechanics “tiles” phase space in statistical mechanics. Also, if we think of the “Planck pixels” as being in spacetime, their 1-D version also takes on some kind of meaning. If we choose c=1, it is often said that all objects seem to “move through” spacetime at a rate of 1 unit of spacetime displacement per unit of coordinate time. In that sense, an object could appear to move one Planck length each Planck time, and not seem to move at the speed of light, if the “Planck length” was interpreted broadly as also existing in the time dimension. It seems to me that could all be formulated in an invariant way, though its usefulness and/or ramifications I could not say. Most likely it would be some kind of “ultraviolet cutoff” to doing path integrals in spacetime, or some such thing.”I’m not a fan of this theory, but there is an idea that spacetime is divided into pre-existing irregular grains of 1 Planck volume. This is called spacetime “glass” quantization, as opposed to “crystal” quantization should the grains be regular. The glassy properties of the quantization help it escape the usual problems with Lorentz invariance. Log in to Reply 9. OCR says: “To see how the calculation works, go here:[URL] [RIGHT]Last edited by a moderator: Yesterday at 1:24 PM[/RIGHT]” “[URL] Fixed that for you…[COLOR=black]..[/COLOR] :oldsmile:” Aww, gee… thanks for the help…[COLOR=black].:oldeyes:..[/COLOR]“Last edited by a moderator: Yesterday at 1:24 PM”[COLOR=black]”..”…[/COLOR] Log in to Reply 10. BiGyElLoWhAt says: “Hint: compare the user name with the url.[SIZE=2]Sorry, could not resist.[/SIZE]”Hahahaha! Observation OP! Log in to Reply mfb says: “That’s not how I interpreted that link. It seems to me what the author is saying […]”Hint: compare the user name with the url.[size=2]Sorry, could not resist.[/size] Log in to Reply 12. BiGyElLoWhAt says: ” ..it takes approximately enough energy to create a black hole whose Schwarzschild radius is… the Planck length!… ” That’s not how I interpreted that link. It seems to me what the author is saying is that if you try to measure a black hole of the plank scale within the accuracy of a radius, then there is enough uncertainty in the momentum that there [i]could exist[/i] another black hole due to the corresponding energy uncertainty of the system (differing by a factor of v/2, classically) Log in to Reply 13. klotza says: “Nice post! Another way to think about the Planck length is that if you try to measure the position of an object to within in accuracy of the Planck length, it takes approximately enough energy to create a black hole whose Schwarzschild radius is… the Planck length! So, one can argue that it’s impossible to measure distances shorter than this – though the argument is a bit hand-wavy. To see how the calculation works, go here: [URL] Hand-wavy is the name of the game here! Thanks for the link, and for the advice. Log in to Reply 14. OCR says: “To see how the calculation works, go here:” [URL] Fixed that for you…[COLOR=black]..[/COLOR] :oldsmile: BTW, I’ve been there many, many times…[COLOR=black]…:oldwink:…[/COLOR] Log in to Reply 15. BiGyElLoWhAt says: “Eisberg?”Indeed it is. Log in to Reply 16. JDoolin says: “[USER=268035]@JDoolin[/USER]: That neutrino would need an incredible energy. Neglecting factors of 2, we have ##m_nu m_P = 3 eV cdot E_nu## where the lightest neutrino mass is probably of the order of 1 meV.” I’m not sure if I’m doing this right, but I just googled “energy of a neutrino collision” and found mention of an apparent 5000-10,000 TeV neutrino. at [URL] So with a bit of estimation, assuming (1) the rest mass energy of a neutrino is about equal to 1 meV, (2) oncoming blueshift is approximately equal to the lorentz contraction factor here. (3) [itex]gamma approx frac{10 times 10^{12} }{1times 10^{-3}}=10^{16}[/itex] Yes, if we started with visible light, at around [itex]10^{-7}[/itex] meters, it would be blueshifted to a wavelength around [itex]10^{-23}[/itex] meters; a trillion times longer than the Planck Length. To do what I imagined and have a neutrino observer see my ordinary light-bulb-photon have a wavelength at the Planck Length, it would have to be a Yotta-eV neutrino. So yes, as you say, “an incredible energy” Log in to Reply 17. BiGyElLoWhAt says: Eisberg? Log in to Reply 18. BiGyElLoWhAt says: I’m not going to argue within the last 30 years. I thinknthe book was ’56 or there abouts. Fundamentals of modern physics, and it was by a german author, I’ll try to dig it up here sometime soon. Log in to Reply 19. mfb says: “Mass increasing is definitely included in some texts, so you’re not losing that memory just yet! My first text that I read on SR had a thought experiment with 2 bouncing balls and 2 observers, and used it to demonstrate relativistic mass.”Try to find any publication of the last 30 years using that concept. “and volumes in 4-D spacetime are invariant, are they not?”You would still get different pixels in each frame. To make it worse, if you transform pixels, the relation between (dilated) Planck time and (contracted in one dimension) distance does not hold any more. Log in to Reply 20. Ken G says: On the topic of the “Planck pixel,” perhaps this overall idea is being rejected too sweepingly. Presumably, the “pixels” would be in 4-D spacetime, not 3-D space, and volumes in 4-D spacetime are invariant, are they not? So I would imagine that if someone wanted to formulate a theory that said spacetime itself was parceled into “Planck pixels”, they would play the usual game that in different reference frames, meaning along different world lines, the “pixels” would distort, but they’d still tile the spacetime in the same way. Yes that means objects don’t “move one Planck length every Planck time”, but that’s obvious– any such object would be perceived as moving at the speed of light. Instead, a “Planck pixel” idea could say that spacetime is discretely tiled, in the sense that world lines cannot be defined with finer precision than that– similar to the way quantum mechanics “tiles” phase space in statistical mechanics. Also, if we think of the “Planck pixels” as being in spacetime, their 1-D version also takes on some kind of meaning. If we choose c=1, it is often said that all objects seem to “move through” spacetime at a rate of 1 unit of spacetime displacement per unit of coordinate time. In that sense, an object could appear to move one Planck length each Planck time, and not seem to move at the speed of light, if the “Planck length” was interpreted broadly as also existing in the time dimension. It seems to me that could all be formulated in an invariant way, though its usefulness and/or ramifications I could not say. Most likely it would be some kind of “ultraviolet cutoff” to doing path integrals in spacetime, or some such thing. Log in to Reply 21. BiGyElLoWhAt says: Mass increasing is definitely included in some texts, so you’re not losing that memory just yet! My first text that I read on SR had a thought experiment with 2 bouncing balls and 2 observers, and used it to demonstrate relativistic mass. Log in to Reply 22. mfb says: The use of relativistic mass is purely historic (and in bad popular science). General relativity predicts that objects can collapse under certain conditions, usually described as sufficient energy density in their rest-frame. GR does not predict the collapse of something just because it moves at high speed, independent of the reference frame chosen to describe the system. [USER=268035]@JDoolin[/USER]: That neutrino would need an incredible energy. Neglecting factors of 2, we have ##m_nu m_P = 3 eV cdot E_nu## where the lightest neutrino mass is probably of the order of 1 meV. Log in to Reply 23. JDoolin says: “I can’t remember what it’s called, even enough to search it via google, but there is actually a solution to this problem. The example provided on the wiki page that I remember used larger masses, as opposed to photons. Basically it says that as you approach the speed of light and pass a large mass, it can’t turn into a black hole due to your reference frame. I really wish I could remember what it was called. If I remember correctly (I very well could not), it has to do something with the geodesics of spacetime warping under the energy tensor from the relative speed of you and the mass you’re observing. Now, this doesn’t necessarily apply when we’re talking photons. Darn my memory, and I’m only 23! I guess it’s all downhill from here =/” I would probably go the other way… Obviously if your theory implies that something is turning into a black hole according to one observer, but is not turning into a black hole according to another observer, then your theory has been essentially discounted by reductio ab adsurdam. I believe the problem is with the premise than an object’s mass increases as it approaches the speed of light. An object’s MOMENTUM increases as [tex]p = frac{m}{sqrt{1 – (frac v c)^2}}v[/tex]; I feel that has been pretty well reasoned out. But the claim that an object’s actual mass has increased (and hence it’s capacitiy to pull other objects toward it by gravity) is NOT well supported by any reasoning I’m familiar with. I’m pretty sure I’ve seen this point made explicitly in some texts, but at 43, I’m well into my fifth decade of memory failure. Log in to Reply 24. JDoolin says: Have you considered the idea of extremely high blueshift reference frames? I have a common ordinary lightbulb producing wavelengths of light between 400 to 700 nanometers. However, from the point-of-view of a passing neutrino; with it’s velocity negligibly below the speed of light, that same light bulb could be producing light with wavelengths less than the Planck Length. So, perhaps the light from my lightbulb is producing a black hole in some frames of reference, but producing ordinary visible light in other frames of reference? I’m highlighting the issue with a rather extreme case–the observer on the neutrino. Some people may argue that neutrino observers are not valid, because they have no ears, no eyes, and no souls, and that their reference frame doesn’t exist. But consider if we took a light of wavelength JUST OVER the planck length, and had one observer fly away from it, while another flew toward it. The observer flying toward it would find that the wavelength of the photon was smaller than the Schwarzschild radius of the photon’s energy. But the observer flying away would find that the wavelength of the same photon was larger than the Schwarzschild radius of the photon’s energy. Well, I guess my point is that radiant energy– E = hf = hc/lambda, is simply not the same as mass energy E=mc^2. The mass has its own reference frame independent of everything else in the universe–mass is an intrinsic property. Also, being a black hole, or NOT being a black hole is an intrinsic feature of matter. The light only has a reference frame in reference to its source and its observer, and frequency and wavelength of light are extrinsic features–observer dependent… Relatively moving observers are going to measure different wavelengths of the same light, so if this idea is accurate, they would also disagree on whether the light spontaneously collapsed into a black hole. Log in to Reply 25. jtbell says: [quote]There is a misconception that the universe is fundamentally divided into Planck-sized pixels, that nothing can be smaller than the Planck length, that things move through space by progressing one Planck length every Planck time. Judging by the ultimate source, [URL=’ cursory search of reddit questions[/URL], the misconception is fairly common.[/quote] This misconception turns up a lot here on PF, too: [URL] I’m glad to have a good article now to point people to, when it comes up again. Thanks! :biggrin: Log in to Reply 26. John Baez says: Nice post! Another way to think about the Planck length is that if you try to measure the position of an object to within in accuracy of the Planck length, it takes approximately enough energy to create a black hole whose Schwarzschild radius is… the Planck length! So, one can argue that it's impossible to measure distances shorter than this – though the argument is a bit hand-wavy. To see how the calculation works, go here: Log in to Reply 27. Tyler Witte says: I can't remember what it's called, even enough to search it via google, but there is actually a solution to this problem. The example provided on the wiki page that I remember used larger masses, as opposed to photons. Basically it says that as you approach the speed of light and pass a large mass, it can't turn into a black hole due to your reference frame. I really wish I could remember what it was called. If I remember correctly (I very well could not), it has to do something with the geodesics of spacetime warping under the energy tensor from the relative speed of you and the mass you're observing. Now, this doesn't necessarily apply when we're talking photons. Darn my memory, and I'm only 23! I guess it's all downhill from here =/ Log in to Reply 28. JDoolin says: Ah, upon rereading the article, I see that you really pretty much hit on my issues in my last post. Log in to Reply 29. Greg Bernhardt says: Nice work! Log in to Reply Leave a Reply Want to join the discussion?Feel free to contribute! Leave a Reply Cancel reply You must be logged in to post a comment. Struggles With The Continuum: Quantum Mechanics of Charged ParticlesWhy Does C Have a Particular Value, and Can It Change? Scroll to top
190550	https://stats.stackexchange.com/questions/600339/ranking-genes-within-gsea	p value - Ranking genes within GSEA - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Ranking genes within GSEA Ask Question Asked 2 years, 9 months ago Modified1 month ago Viewed 916 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I know you can rank genes by p-/q-value, or by log2 fold change. But can you combine the two methods somehow? I have read about combining fold change and p-value here but what is one advantage and one disadvantage of using this ranking method? p-value ranking bioinformatics combining-p-values Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Oct 17, 2023 at 17:08 EdM 111k 11 11 gold badges 120 120 silver badges 353 353 bronze badges asked Dec 29, 2022 at 13:10 katarakatara 111 1 1 bronze badge Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. I know you can rank genes by p-/q-value, or by log2 fold change. But can you combine the two methods somehow? As you may have noted, p-values tell you nothing about effect size (although they are correlated with it, of course). Conversely, raw effect size (fold-change in the case of gene expression) doesn't tell you anything about the uncertainty around its estimate. For this reason, I prefer to work with shrunk log-fold-changes which effectively combine p-value and fold change. This not only for ranking, but also for most other purposes. Besides, (shrunk) fold-changes have a clear biological interpretation, while p-values only tell you how much evidence you have against the hypothesis of no change. Since p-values get smaller and smaller with larger samples, you never get to a "true", biologically meaningful p-value. In fact, the astronomically small p-values you sometimes see (p << 1e9) are effectively pure numbers with no biological relevance. P-values for genes that do not change will not converge to zero with increasing sample size, but I would contend that such genes are unusual since genes are part of a network and ultimately most of them must change by a tiny amount. Having said that, @Gordon Smyth who is active on this forum and knows quite a lot about this topic, is more skeptical about shrinkage in RNAseq data - see here Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Aug 12 at 18:02 darioberdariober 5,913 19 19 silver badges 25 25 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Consider the classic paper explaining Gene Set Expression Analysis (GSEA) (Reference 1) . Given two classes of samples (representing 2 phenotypes, 2 experimental treatments, etc.), GSEA first places all genes into a ranked numeric list L, based on differential expression between the two classes, and then determines whether a pre-defined set of genes S tends to be found at the extremes of that list. The gene sets are defined based on prior biological knowledge, e.g., published information about biochemical pathways or coexpression in previous experiments. The goal of GSEA is to determine whether members of a gene set S tend to occur toward the top (or bottom) of the list L, in which case the gene set is correlated with the phenotypic class distinction. That's different from other "over-representation" approaches that use only a subset of genes in the list L, selected as highly or poorly expressed based on some cutoff criterion. That cutoff criterion is typically based on some estimate of "statistical significance" like a false discovery rate. GSEA is thus less concerned with issues of "statistical significance" of differential expression of individual genes than those other approaches. Given the list L, the process continues: We calculate an enrichment score (ES) that reflects the degree to which a set S is overrepresented at the extremes (top or bottom) of the entire ranked list L. The score is calculated by walking down the list L, increasing a running-sum statistic when we encounter a gene in S and decreasing it when we encounter genes not in S. The magnitude of the increment depends on the correlation of the gene with the phenotype. The enrichment score is the maximum deviation from zero encountered in the random walk; it corresponds to a weighted Kolmogorov–Smirnov-like statistic. (Emphasis added) The emphasized sentence is a major distinction from an earlier version of GSEA, in which equal weights were used at each step. In the original implementation, the running-sum statistic used equal weights at every step, which yielded high scores for sets clustered near the middle of the ranked list (Fig. 2 and Table 1). These sets do not represent biologically relevant correlation with the phenotype. We addressed this issue by weighting the steps according to each gene’s correlation with a phenotype. For the weights: Genes are ranked based on the correlation between their expression and the class distinction by using any suitable metric. So the question is: what is a suitable metric for ranking the gene list L and weighting the GSEA running sum? The paper linked by the OP (Reference 2) seems focused on identifying "differentially expressed" genes, which is not required for GSEA. Thus that paper's emphasis on the area under the receiver operating characteristic curve (AUC) as a metric for evaluating different ranking methods seems a bit misplaced, if your primary interest is in GSEA with its ranked list of all genes. A moderated t-statistic on log-transformed expression values (e.g., the empirical Bayes or "treat" methods provided by the Bioconductor limma package) provides a useful way to rank all genes for GSEA. That tends to minimize over-ranking of genes based on their individual t-statistics and thus minimizes the "small fold change, small variance" problem noted in Reference 2. If you are worried about the "large fold change, large variance" problem noted in that reference, you could consider just using log-fold changes as the metric. If those are problems for just a few genes in a gene set S, your choice of ranking metric might not matter much at all. That's a strength of GSEA with its emphasis on the entire set of gene-expression values. Ref. 1. Subramanian et al., PNAS 102: 15545–15550, 2005. Ref. 2. Xiao et al., Bioinformatics 30: 801–807, 2014 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 17, 2023 at 16:59 EdMEdM 111k 11 11 gold badges 120 120 silver badges 353 353 bronze badges Add a comment\| Your Answer Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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190551	https://www.xometry.com/resources/materials/strain/	Solutions Additive Manufacturing 3D Printing Service Plastic 3D Printing Service Fused Deposition Modeling HP Multi Jet Fusion Selective Laser Sintering Production 3D Printing Service Carbon DLS Metal 3D Printing Service Direct Metal Laser Sintering Metal Binder Jetting Vapor Smoothing 3D Prints CNC Machining CNC Machining CNC Milling CNC Turning Wire EDM Machining Medical CNC CNC Routing Sheet and Tube Fabrication Sheet Metal Fabrication Sheet Cutting Laser Cutting Waterjet Cutting Laser Tube Cutting Tube Bending Plastic Part Production Plastic Injection Molding Quick-Turn Molding Prototype Molding Bridge Molding Production Molding Overmolding Insert Molding Compression Molding Urethane and Silicone Casting Plastic Extrusion Injection Molded Surface Finishes Custom Plastic Fabrication Micro Molding Metal Part Production Die Casting Metal Stamping Metal Extrusion Value-Added Solutions Assembly Services Weldments Rapid Prototyping High-Volume Production Precision Grinding Surface Grinding Powder Coating Industries Industries We Serve Aerospace and Defense Automotive Consumer Products Design Agencies Education Electronics and Semiconductors Energy Hardware Startups Industrial Lighting Medical and Dental Robotics Supply Chain and Purchasing Manufacturing Resources Learn All Technical Guides Design Guides eBooks Library 3D Printing Articles Injection Molding Articles Machining Articles Sheet Cutting Articles Xometry Production Guide Tools CAD Add-ins Materials Finishes Manufacturing Standards Standard Sheet Thicknesses Standard Tube Sizes Standard Threads Standard Inserts Explore More FAQs Case Studies Tariff Resources ITAR and Certifications Supplier Community Company Release Notes Contact Us For Support Call: +1-800-983-1959 Email: support@xometry.com Help Center Enterprise Xometry Enterprise Discover Xometry Teamspace Meet An Account Rep eProcurement Integrations Bulk Upload for Production Quotes Onboard Xometry As Your Vendor How Xometry Works Xometry's Platform How to Use the Xometry Instant Quoting Engine® Test Drive Xometry Tool Library and Tool Detail Pages From Our FAQ Part Revisions & Same-Suppliers for Repeat Orders Xometry's Privacy and Security About Xometry Xometry’s Quality Assurance Xometry’s Supplier Network Xometry's Machine Learning Xometry's Manufacturing Community Become a Supplier ResourcesMaterialsStrain: Definition, Importance, Calculation, Types, and Examples Strain: Definition, Importance, Calculation, Types, and Examples Written by Dean McClements Updated by Phillip Keane 18 min read Published May 18, 2023 Updated August 5, 2025 Strain is a key concept in engineering and physics that describes how a material deforms in response to applied stress. The ratio of an object's change in length or shape to its original length or shape is known as strain. In many disciplines, including structural engineering, mechanical engineering, and materials science, an understanding of strain is essential because it determines how a material will behave under load, and enables the design of structures that can withstand the expected service stresses. When calculating strain, the amount of material deformation is measured. This deformation is usually expressed as a percentage change in the material's length, area, or volume. Engineers identify several types of strain, based on the kind of stress that produces the particular kind of deformation: longitudinal, lateral, shear, and volumetric strain. Many structures, including bridges, vehicles, and aircraft, experience strain as a result of applied stress in the real world. This article will define strain, its types, and their significance, provide calculation techniques for each type, and give examples of practical applications of strain analysis. What Is Strain in Mechanics? In mechanics, strain is a dimensionless measure of the deformation that a material experiences in response to an external force or load. It is described as the ratio of the change in the material's length or shape to that of the original. Stress, or the force applied per unit area, is related to strain by the material's stiffness or elasticity. Epsilon (ε), the symbol for strain, can be written as a percentage or a fractional value. A fundamental idea in mechanics, strain is important in the planning and evaluation of mechanical systems and structures. What Is the Other Term for Strain? The other word for strain is "deformation", which describes how a material changes in size or shape as a result of the application of a load or force from the outside. In engineering and physics, deformation is commonly used interchangeably with strain and can take many different forms, including stretching, compression, and bending. How Does Strain in Mechanics Work? In mechanics, strain is a concept that describes how a material deforms or changes shape as a result of an external force or stress. Strain is measured using strain gauges, which are tools for detecting and quantifying material deformation. The surface of the material being tested is covered with a grid of fine wire or foil. The strain gauge's wire or foil deforms along with the material being measured, resulting in changes in the gauge's electrical resistance. The degree of strain can be calculated by measuring these variations in resistance. What Is the Importance of Strain in Materials? Strain in materials is important because it explains how materials deform under stress. Designing structures that can withstand expected loads without failing requires a thorough understanding of a material's strain behavior. Strain is used to evaluate a material’s ductility and allows engineers to calculate how much energy a material or structure can absorb when being deformed. What Is the Use of Strain in Mechanics? Engineers use strain as a key parameter to forecast how materials will behave when subjected to loads. Knowledge of the details of strain behavior is essential for the design and analysis of mechanical parts and structures. The main application of strain in mechanics is to comprehend how materials respond to stress. Any time a force is applied to a substance, deformation or a change in shape results. This deformation can be measured in terms of strain. To ensure that structures can withstand the loads they are intended to carry, engineers use strain measurements to assess the strength and stiffness of materials. Mechanical parts like gears, bearings, and shafts are also designed using stress and strain concepts. Engineers can make sure that these parts are sturdy enough to withstand the stresses they will experience during operation by measuring the strain imposed on them by particular levels of stress. How is Strain Measured in Mechanics? Strain gauges and extensometers are the most common methods used to measure both elastic and plastic deformation. Extensometers directly measure a material's change in length, as opposed to strain gauges, which convert mechanical deformation into an electrical signal. Optical methods, such as digital image correlation, can also be used to measure strain by analyzing images of the material under stress. What Are the Units of Strain? The quantity of strain has no dimensions, so it has no units. However, strain is usually expressed as a fractional or percentage change in length. Length changes involved in strain calculations may be measured in meters, millimeters, or micrometers, but strain itself remains unitless. Why Does the Strain in Materials Need To Be Measured? When designing structures that can withstand expected loads, it's important to understand how materials behave mechanically under stress. Structures deform under stress, whether elastically (temporarily) or plastically (permanently). Measurements of strain can also reveal details of material characteristics, such as elasticity, plasticity, and fracture toughness, which are essential for choosing the right materials for engineering applications. What Are the Different Causes of Strain in Materials? There are many causes of strain in materials, including: Mechanical Stress: Mechanical stress is the most common cause of strain in materials. When a force is applied to a material, it undergoes deformation, or strain, which can be either elastic or plastic. Electromagnetic Fields: Certain materials, such as magnetostrictive or piezoelectric materials, can experience strain when exposed to electromagnetic fields due to internal interactions between magnetic or electric fields and their crystal structures. Radiation: When metals are exposed to radiation, they can undergo a process known as radiation hardening. The material may become more brittle but also stronger as a result of this process. As a result, it is less likely to deform under stress, which can result in brittle fractures. The radiation damages the metal's atoms by interacting with them, leading to flaws and dislocations in the lattice structure. This is comparable to other hardening processes like precipitation hardening and work hardening. Fatigue: Fatigue is the gradual weakening of a material due to repeated loading and unloading cycles. This can cause strain in the material, leading to failure over time. Creep: Creep is the slow deformation of a material over time due to a constant load or stress. This can cause strain in the material, leading to failure over time. Manufacturing Processes: Manufacturing processes such as casting, forging, and welding can cause strain on materials due to the high temperatures and stresses involved in these processes. What Are the Different Effects of Strain on Materials? Here are a few typical outcomes of material strain: Elastic Deformation: When a material is subjected to a small amount of strain, it may exhibit elastic deformation. It will return to its original shape when the stress is removed. The amount of strain is proportional to the stress applied, according to Hooke's Law. Plastic Deformation: If the strain applied to a material exceeds a certain threshold, it may undergo plastic deformation. It will remain deformed even when the stress is removed. The material’s capacity to reorganize its atomic configuration without rupturing the interatomic bonds holding it together determines the degree of plastic deformation. Fracture: When a material is subjected to a large amount of strain, it may fracture or break. The amount of dislocation pinning, available slip systems, and atomic bonding in the crystal lattice all affect the amount of strain that can be applied before fracture. Thermal Expansion: Materials may exhibit expansion or contraction in response to changes in temperature. This is known as thermal expansion and can result in significant strains in some materials. Work Hardening: Some materials may become harder and stronger after undergoing plastic deformation. This is known as work hardening or strain hardening and can occur in metals, polymers, and other materials. What Is the Formula for Strain? Strain is a unit of measurement for how much a material deforms in response to stress. Strain is calculated as follows: ε = ΔL / L₀ Where: ε represents strain ΔL represents the change in length of the material L₀ represents the original length of the material Strain is equal to the change in length divided by the original length. This formula is commonly used to calculate strain in materials, such as metals, polymers, and composites. How Is Strain Calculated? The steps in calculating strain (ε) are listed below: Measure the initial length (L0) of the object or material for which you want to calculate the strain. Measure the final length (Lf) of the object or material after it has been deformed. Subtract the initial length from the final length to get the change (the delta) in length: ΔL = Lf - L0. Divide the change in length by the initial length: (ΔL) / L0. Multiply the result by 100 to get the strain as a percentage: ((ΔL) / L0) x 100. Here's an example: Suppose you have a steel wire with an initial length of 10 cm. After stretching it, you measure its length to be 13 cm. L0 = 10 cm (initial length) Lf = 13 cm (final length) Lf - L0 = ΔL = 3 cm (change in length) (ΔL) / L0 = 0.3 (strain as a decimal fraction) 0.3 x 100 = 30% (strain as a percentage) So the strain (ε) in this case is 30%. What Are the Different Types of Strain That Can Occur in Materials? The various types of strains that can develop in materials are as follows: 1. Normal Strain Normal strain describes how a solid material responds to a force that is applied perpendicular to its cross-sectional area. It is measured as the ratio of the change in length to the original length of the material. The formula for normal strain is: ε = ΔL / L₀ Understanding normal strain is crucial for designing and testing materials in various industries. It can help ensure the safety and reliability of structures and products. 2. Tensile Strain Tensile strain is the amount of deformation that a material undergoes when it is pulled or stretched. It is measured as the ratio of the change in length of the material to its original length. The formula for tensile strain is: ε = ΔL / L₀ The measurement of tensile strain is important in materials science and engineering, as it helps to determine a material's ductility and elasticity. 3. Shear Strain Shear strain is a type of strain that occurs when the particles of a material slide past each other in a parallel direction, causing a change in the shape of the material. It is measured as the change in angle between two originally perpendicular lines in the material. The formula for shear strain is given by: γ = tan θ Where: γ is the shear strain θ is the shear angle Shear strain is used in the design of various structures and mechanical systems. 4. Thermal Strain Thermal strain refers to the deformation or change in the size of a material due to temperature changes. Thermal expansion can occur under two different circumstances: non-constrained and constrained. When materials are not constrained, they can move freely in all directions, but when they are, certain directions are restricted, which can cause stress, deformation, or failure. When designing materials and structures, engineers must take into account both conditions to make sure they can withstand the stresses and strains brought on by temperature changes. Thermal strain can be measured using strain gauges or interferometry techniques. The formula for calculating thermal strain is: ε = αΔT Where: α is the coefficient of thermal expansion ΔT is the temperature change Understanding thermal strain is critical in designing and building structures that can withstand temperature variations. Accurate measurement and calculation of thermal strain can help prevent material failure and extend the lifespan of structures. 5. Compressive Strain Compressive strain is the deformation that takes place in a material when it is subjected to a contracting force. Under compression, the space between the atoms of a material is reduced, causing the material to become compacted. It is measured by comparing the original and deformed lengths of a test specimen or structure under compression. The formula for compressive strain is: ε = ΔL / L₀ Compressive strain is important in materials science as it helps determine a material's compressive strength and its ability to resist deformation. 6. Residual Strain The term "residual strain" describes the internal stress that remains in a material after it has been deformed or subjected to external forces. It can be measured using techniques such as X-ray diffraction or neutron diffraction. The formula for calculating residual strain involves measuring the change in lattice spacing due to deformation. An accurate measure of residual strain can help an engineer understand the residual stress state in a structure, providing needed information for avoiding potential failures. It can also provide valuable information for improving manufacturing processes and designing structures with enhanced performance. 7. Volumetric Strain Volumetric strain is the term for the volume change that occurs in a material as a result of uniformly applied external forces (hydrostatic strain). It is determined by dividing the volume change by the starting volume. The formula for calculating volumetric strain is: ε = ΔV/V Where: ΔV is the change in volume V is the volume Volumetric strain is important in various fields such as engineering, geology, and materials science. It aids in predicting how materials will react to stress and in designing structures appropriately. “Designing for load is the easy part. Designing for deformation is where it gets complex. Strain dictates whether interfaces misalign, seals fail, or fasteners loosen under cyclic loading. If you're not thinking about strain compatibility between materials and assemblies, you're not thinking long term.” Phillip Keane, Design Engineer & Prototyping Specialist, Note from the Editor What Are Examples of Materials Used in 3D Printing, and How Do They Handle Strain? Engineering employs a variety of materials, each of which has particular properties and strain behavior. Here are some examples of material types commonly used in 3D printing, and examples of how they handle strain: Steel: Steel is a common engineering material renowned for its ductility and strength. Steel is the perfect material for high-stress applications like construction, bridges, and heavy machinery because it is ductile enough to accommodate some plastic deformation before final failure. Aluminum: Aluminum is frequently used to take advantage of its light weight and corrosion resistance. Aluminum has a high strength-to-weight ratio and adequate ductility for the kinds of construction, automotive, and aerospace applications that demand consideration of the weight of the structure. Concrete: Concrete is a widely used construction material due to its strength and durability. Although it has a low tensile strength, concrete can handle high compressive loads. Reinforcing materials, such as steel, are added to concrete to increase its tensile strength. Plastics: Plastics are suitable for a variety of applications because they are lightweight and simple to manufacture. They can withstand mild amounts of strain and have good ductility. Plastics are not suited for high-stress applications due to their low strength in comparison to other materials. Ceramics: Ceramics are strong and tough materials that work well in applications where high temperatures and wear resistance are necessary. Ceramics are susceptible to failure under stress because they are brittle and have low ductility. What Are Examples of Polymer Materials Used in 3D Printing, and How Do They Handle Strain? Here are five polymer materials commonly used in 3D printing and their strain-handling characteristics: PLA (Polylactic Acid): 3D printing often uses PLA, a biodegradable and environmentally friendly material. Its moderate tensile strength and relatively high brittleness make it easily breakable when put under stress. ABS (Acrylonitrile Butadiene Styrene): A common material for 3D printing, ABS is a solid and durable plastic. It can withstand moderate strain without breaking because of its relatively high tensile strength. Nylon: A popular material for 3D printing that is both durable and flexible is nylon. It has the highest tensile strength of the polymers mentioned here. When nylon is subjected to strain, it exhibits high elasticity and toughness, meaning that it can stretch without breaking and can absorb significant amounts of energy before failing. PETG (Polyethylene Terephthalate Glycol): PETG is a sturdy and durable material that is widely utilized in 3D printing. It has a moderate tensile strength and can withstand moderate levels of strain without breaking. TPU (Thermoplastic Polyurethane): TPU is a versatile and elastic material that can be utilized in 3D printing. Although it has a low tensile strength, its elastic properties allow it to withstand large amounts of strain without cracking. For more information, see our 3D printer guide. How Does Strain Relate to Stress? Strain is the result of the material's deformation or changes in shape as a result of forces that were applied to it. The elastic modulus of the material describes the relationship between stress and strain. Strain is a measure of deformation, whereas stress is a measure of force per unit area. What Comes First, Stress or Strain? When a material sample is subjected to a force, stress comes first, then strain follows. Stress is defined as the force per unit area applied to an object, whereas strain is the resultant deformation of the object. Stress is the cause, and strain is the effect of stress on the material. What Is the Stress-Strain Curve? A stress-strain curve illustrates how a material responds to an applied force. Figure 1 below is an example of a tensile stress-strain curve: Stress strain curve It is made up of a region where the material is elastic and returns to its original shape, a region where it is plastic and deforms permanently, and a region where it necks and experiences significant deformation and potential fracture. When designing and engineering materials for different applications, it is a crucial tool for engineers and materials scientists. What Is Strain in Hooke's Law? According to Hooke's law, strain is the degree of elastic material deformation that results from the application of a force. In more detail, strain is defined as the ratio of a material’s changed length to its initial length. A material may change in size or shape when subjected to a force, and the amount of strain it undergoes is determined by the strength of the applied force and the characteristics of the material. The modulus of elasticity, also known as Young’s modulus, is a material characteristic that describes how much a material will deform under a specific amount of stress. It describes the relationship between the amount of strain and the amount of stress applied. Mathematically, strain is represented as: ε = ΔL / L₀ What Are the Benefits of Strain Analysis in Mechanics? Among the advantages of strain analysis in mechanics are the following: Ability to predict material behavior under load (evaluation of material properties such as Young's modulus and Poisson's ratio) Design of safe and efficient structures Analysis of deformation and failure mechanisms Optimization of manufacturing processes for specific materials and applications What Are the Limitations of Strain Analysis in Mechanics? The following are some of the limitations of strain analysis in mechanics: External elements like vibration, electromagnetic interference, and other environmental elements can have an impact on strain measurements. This can make it challenging to measure strain accurately in some circumstances. Although strain is an important idea in mechanics, it only represents a small portion of the whole. Other elements, including stress, are crucial in describing how materials behave when they are under load. Strain analysis has limited applicability for nonlinear material behavior, especially when materials do not follow linear-elastic assumptions. Temperature is an additional factor that can (a) cause strain in materials, independent of externally applied forces; or (b) cause the relationship between stress and strain to change with temperature, and become more difficult to predict in some cases. Is Tensile Stress the Same as Tensile Strain? No, tensile stress and strain are not the same. Tensile stress is the amount of force per unit area applied in a direction normal to the cross-section of interest, while tensile strain is the material's elastic or plastic deformation response to the application of a tensile force. However, the two quantities, tensile stress and tensile strain, are related to each other by the elastic modulus of the material. Young’s Modulus, calculated as stress divided by strain in the elastic regime, characterizes a material’s stiffness. Is Shear Stress the Same as Shear Strain? No, shear strain and shear stress are not the same. Shear strain is the shape change caused by an applied shear force, whereas shear stress is the force per unit area applied parallel to a cross-sectional surface of interest. Summary This article presented strain, explained what it is, and discussed how to calculate it. To learn more about strain, contact a Xometry representative. Xometry provides a wide range of manufacturing capabilities and other value-added services for all of your prototyping and production needs. Visit our website to learn more or to request a free, no-obligation quote. Disclaimer The content appearing on this webpage is for informational purposes only. Xometry makes no representation or warranty of any kind, be it expressed or implied, as to the accuracy, completeness, or validity of the information. Any performance parameters, geometric tolerances, specific design features, quality and types of materials, or processes should not be inferred to represent what will be delivered by third-party suppliers or manufacturers through Xometry’s network. Buyers seeking quotes for parts are responsible for defining the specific requirements for those parts. Please refer to our terms and conditions for more information. Dean McClements Dean McClements is a B.Eng Honors graduate in Mechanical Engineering with over two decades of experience in the manufacturing industry. His professional journey includes significant roles at leading companies such as Caterpillar, Autodesk, Collins Aerospace, and Hyster-Yale, where he developed a deep understanding of engineering processes and innovations. Read more articles by Dean McClements
190552	https://www.cuemath.com/ncert-solutions/ncert-solutions-class-9-maths-chapter-2-polynomials/	NCERT Solutions Class 9 Maths Chapter 2 Polynomials NCERT solutions for class 9 maths Chapter 2 Polynomials are all about the basics of polynomials like the different types of polynomials, finding roots, or solutions to a polynomial equation. Polynomials are algebraic expressions having one variable or more. These NCERT solutions class 9 maths Chapter 2 also explain the remainder theorem and factor theory of polynomials in detail, the algebraic identities, and polynomials of various degrees. Class 9 Maths NCERT Solutions Chapter 2 polynomials illustrate the difference between linear, quadratic, and cubic polynomials. Important theorems mentioned are the Remainder theorem and the Factor theorem, which help identify the factors of a polynomial. Students can access the solutions from the pdf links given below and also find some of these in the exercises given below. NCERT Solutions for Class 9 Maths Chapter 2 PDF The exercises related to identifying the type of polynomial, finding the roots or solution of a polynomial equation, and finding factors of the polynomial are available for free pdf download using the four links provided below: ☛ Download Class 9 Maths NCERT Solutions Chapter 2 Polynomials NCERT Class 9 Maths Chapter 2 Download PDF NCERT Solutions for Class 9 Maths Chapter 2 Polynomials These fundamental properties and theorems of polynomials form the building blocks for higher mathematics. Thus, it is very important to master the fundamentals by solving many different example exercises using the links provided above. These NCERT Solution exercises will help understand the properties of polynomials better, as well as how to utilize them. Chapter-wise detailed analysis of NCERT Solutions Class 9 Maths Chapter 2 Polynomials is given below. ☛ Download Class 9 Maths Chapter 2 NCERT Book Topics Covered: The topics that are covered under the chapter on polynomials include an explanation of polynomials as a special set of algebraic equations, different types of polynomials, solutions of polynomial equations, factor theorem, and remainder theorem. Also, these class 9 maths NCERT solutions Chapter 2 define the algebraic identities, which help in factorizing the algebraic equations. Total Questions: Class 9 Maths Chapter 2 Polynomials consists of a total of 45 questions, of which 31 are easy, 9 are moderate, and 5 are long answer type questions. Cuemath is one of the world's leading math learning platforms that offers LIVE 1-to-1 online math classes for grades K-12. Our mission is to transform the way children learn math, to help them excel in school and competitive exams. Our expert tutors conduct 2 or more live classes per week, at a pace that matches the child's learning needs. List of Formulas in NCERT Solutions Class 9 Maths Chapter 2 NCERT solutions class 9 maths Chapter 2 covers lots of important concepts crucial for understanding higher grade maths. By learning to factorize a polynomial expression, one can find the roots of the polynomial equation. This is a relatively simple process that can greatly improve an individual's understanding of polynomial equations. Some important algebraic identities or formulas which help in factorization and are covered in NCERT solutions for class 9 maths chapter 2 are given below. Important Questions for Class 9 Maths NCERT Solutions Chapter 2 CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.1 Question 1Question 2Question 3Question 4Question 5 CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.2 Question 1Question 2Question 3Question 4 CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.3 Question 1Question 2Question 3 CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.4 Question 1Question 2Question 3Question 4Question 5 CBSE Important Questions for Class 9 Maths Chapter 2 Exercise 2.5 Question 1Question 2Question 3Question 4Question 5Question 6Question 7Question 8Question 9Question 10Question 11Question 12Question 13Question 14Question 15Question 16 Video Solutions for Class 9 Maths NCERT Chapter 2 NCERT Video Solutions for Class 9 Maths Chapter 2 Video Solutions for Class 9 Maths Exercise 2.1 Chapter 2 Exercise 2.1 Question 1 \| Chapter 2 Exercise 2.1 Question 4 Chapter 2 Exercise 2.1 Question 2 \| Chapter 2 Exercise 2.1 Question 5 Chapter 2 Exercise 2.1 Question 3 \| Video Solutions for Class 9 Maths Exercise 2.2 Chapter 2 Exercise 2.2 Question 1 \| Chapter 2 Exercise 2.2 Question 3 Chapter 2 Exercise 2.2 Question 2 \| Chapter 2 Exercise 2.2 Question 4 Video Solutions for Class 9 Maths Exercise 2.3 Chapter 2 Exercise 2.3 Question 1 \| Chapter 2 Exercise 2.3 Question 3 Chapter 2 Exercise 2.3 Question 2 \| Video Solutions for Class 9 Maths Exercise 2.4 Chapter 2 Exercise 2.4 Question 1 \| Chapter 2 Exercise 2.4 Question 4 Chapter 2 Exercise 2.4 Question 2 \| Chapter 2 Exercise 2.4 Question 5 Chapter 2 Exercise 2.4 Question 3 \| Video Solutions for Class 9 Maths Exercise 2.5 Chapter 2 Exercise 2.5 Question 1 \| Chapter 2 Exercise 2.5 Question 9 Chapter 2 Exercise 2.5 Question 2 \| Chapter 2 Exercise 2.5 Question 10 Chapter 2 Exercise 2.5 Question 3 \| Chapter 2 Exercise 2.5 Question 11 Chapter 2 Exercise 2.5 Question 4 \| Chapter 2 Exercise 2.5 Question 12 Chapter 2 Exercise 2.5 Question 5 \| Chapter 2 Exercise 2.5 Question 13 Chapter 2 Exercise 2.5 Question 6 \| Chapter 2 Exercise 2.5 Question 14 Chapter 2 Exercise 2.5 Question 7 \| Chapter 2 Exercise 2.5 Question 15 Chapter 2 Exercise 2.5 Question 8 \| FAQs on NCERT Solutions Class 9 Maths Chapter 2 How CBSE Students can utilize NCERT Solutions Class 9 Maths Chapter 2 effectively? Algebra forms the basis of higher mathematical studies. Hence, students should focus on the important terms defined in this chapter, like the degree of a polynomial, the difference between constant and variable, to get a clear understanding of the polynomials. This will help them to make their base strong to appear for their board exams and face any kind of difficult questions. Why are Class 9 Maths NCERT Solutions Chapter 2 Important? The NCERT Solutions Class 9 Maths Chapter 2 includes a detailed explanation of the remainder and the factor theorem, which hold an important place in algebra. Also, the crucial algebraic identities are discussed in an elaborate manner with plenty of questions to solve for the students. A list of all key equations and concepts is available at the end of the chapter. This is a significant benefit because students can use this list whenever required instead of figuring it out from between the lengthy chapter text. Overall, these solutions cover all of the major concepts, approaches, and formulas, making them of utmost importance for class 9 math students. How Many Questions are there in NCERT Solutions Class 9 Maths Chapter 2 Polynomials? Overall the NCERT Solutions Class 9 Maths Chapter 2 has 98 questions that can be categorized as easy, medium, and difficult ones. Roughly 70 questions are straightforward and easy to solve, 20 questions are of medium difficulty level while 8 would require some thinking as they are long-form questions. What are the Important Topics Covered in NCERT Solutions Class 9 Maths Chapter 2? The important topics that are covered under the NCERT Solutions Class 9 Maths Chapter 2 include the basic understanding of polynomials, the components of algebraic expressions, and their definitions. The chapter focuses on the types of polynomials and how to solve them, with special emphasis on factor and remainder theorem and the algebraic entities. What are the Important Formulas in NCERT Solutions Class 9 Maths Chapter 2? Since the NCERT Solutions Class 9 Maths Chapter 2 covers the polynomials from their basic structure, several definitions of important terms have been explained with their formulas, like the factor and the remainder theorem. But the most important formula would be the algebraic identities as they help in factorization itself. For example, (a + b)2 = a2 + 2ab + b2 Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths Polynomials? NCERT Solutions Class 9 Maths Polynomials encompass a variety of questions that explore all the algebraic concepts related to polynomials. Hence, it would be good if the students make use of this resource and start practicing by solving the examples first, which will help them in getting an idea of what steps are to be followed when questions related to polynomials are solved. NCERT Solutions Class 9 Maths Chapters Chapter 1 Number Systems \| Chapter 3 Coordinate Geometry Chapter 4 Linear Equation in Two Variables \| Chapter 5 Introduction to Euclids Geometry Chapter 6 Lines and Angles \| Chapter 7 Triangles Chapter 8 Quadrilaterals \| Chapter 9 Areas Of Parallelograms and Triangles Chapter 10 Circles \| Chapter 11 Constructions Chapter 12 Herons Formula \| Chapter 13 Surface Areas and Volumes Chapter 14 Statistics \| Chapter 15 Probability
190553	https://fluka-forum.web.cern.ch/t/microscopic-cross-section/2340	Microscopic cross section Dear Cerutti: Yes, this method can get the average macroscopic cross section。 However, I want get the microscopic cross section（p,n）depending on the energy. How I can get this cross section? Expect your reply! Kind regards, Xiaohe Wang Well, the relationship between macroscopic cross section Sigma (cm^-1) and microscopic cross section sigma (cm^2) is a basic concept introduced in physics books: sigma = A Sigma / (rho N_A), where A is the molar mass (g/mol), N_A is the Avogadro constant (mol^-1), and rho is the material density (g/cm^3). From the inelastic scattering length lambda = 1/Sigma printed in the output file, you can calculate the reaction cross section of beam particles, which is not the (p,n) cross section. This is just a fraction of the reaction cross section, which one can estimate from the yield of the (p,n) products (e.g. Cu-64 in Ni-64, as given by RESNUCLEi) divided by the number of beam proton reactions (provided that the contribution of reactions by secondary particles and beam protons of lower energy is negligible). Dear Ceruttif: Sorry , I am still not understand. From the output file, I can get the value of the Inelastic Scattering Length, shown in the picture. QQ图片202110180858001236×707 94.7 KB However, this is an average value, without energy information. This can not give the reaction cross section with respect to the energy. Meanwhile, according to your advice, I am still not understand how to get the (p,n) cross section? Is there a way to give the(p,n) cross sections directly? Kind regards, Xiaohe Wang However, this is an average value, without energy information. This can not give the reaction cross section with respect to the energy. As you can read in the header you highlighted, this is the inelastic scattering length at beam energy. So, the energy information is there. Is there a way to give the(p,n) cross sections directly? No. I understand. Thank you verty much! Powered by Discourse, best viewed with JavaScript enabled
190554	https://openstax.org/books/calculus-volume-1/pages/5-6-integrals-involving-exponential-and-logarithmic-functions	Learning Objectives 5.6.1 Integrate functions involving exponential functions. 5.6.2 Integrate functions involving logarithmic functions. Exponential and logarithmic functions are used to model population growth, cell growth, and financial growth, as well as depreciation, radioactive decay, and resource consumption, to name only a few applications. In this section, we explore integration involving exponential and logarithmic functions. The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, y=e x,y=e x, is its own derivative and its own integral. Rule: Integrals of Exponential Functions Exponential functions can be integrated using the following formulas. ∫e x d x=e x+C∫a x d x=a x ln a+C∫e x d x=e x+C∫a x d x=a x ln a+C (5.21) Example 5.37 Finding an Antiderivative of an Exponential Function Find the antiderivative of the exponential function _e_−_x_. Checkpoint 5.31 Find the antiderivative of the function using substitution: x 2 e−2 x 3.x 2 e−2 x 3. A common mistake when dealing with exponential expressions is treating the exponent on _e_ the same way we treat exponents in polynomial expressions. We cannot use the power rule for the exponent on _e_. This can be especially confusing when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we should always double-check to make sure we’re using the right rules for the functions we’re integrating. Example 5.38 Square Root of an Exponential Function Find the antiderivative of the exponential function e x 1+e x.e x 1+e x. Checkpoint 5.32 Find the antiderivative of e x(3 e x−2)2.e x(3 e x−2)2. Example 5.39 Using Substitution with an Exponential Function Use substitution to evaluate the indefinite integral ∫3 x 2 e 2 x 3 d x.∫3 x 2 e 2 x 3 d x. Checkpoint 5.33 Evaluate the indefinite integral ∫2 x 3 e x 4 d x.∫2 x 3 e x 4 d x. As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number _e_ is often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let’s look at an example in which integration of an exponential function solves a common business application. A price–demand function tells us the relationship between the quantity of a product demanded and the price of the product. In general, price decreases as quantity demanded increases. The marginal price–demand function is the derivative of the price–demand function and it tells us how fast the price changes at a given level of production. These functions are used in business to determine the price–elasticity of demand, and to help companies determine whether changing production levels would be profitable. Example 5.40 Finding a Price–Demand Equation Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is 50 tubes per week at $2.35 per tube, given that the marginal price—demand function, p′(x),p′(x), for _x_ number of tubes per week, is given as p'(x)=−0.015 e−0.01 x.p'(x)=−0.015 e−0.01 x. If the supermarket chain sells 100 tubes per week, what price should it set? Example 5.41 Evaluating a Definite Integral Involving an Exponential Function Evaluate the definite integral ∫1 2 e 1−x d x.∫1 2 e 1−x d x. Checkpoint 5.34 Evaluate ∫0 2 e 2 x d x.∫0 2 e 2 x d x. Example 5.42 Growth of Bacteria in a Culture Suppose the rate of growth of bacteria in a Petri dish is given by q(t)=3 t,q(t)=3 t, where _t_ is given in hours and q(t)q(t) is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function Q(t)Q(t) that gives the number of bacteria in the Petri dish at any time _t_. How many bacteria are in the dish after 2 hours? Checkpoint 5.35 From Example 5.42, suppose the bacteria grow at a rate of q(t)=2 t.q(t)=2 t. Assume the culture still starts with 10,000 bacteria. Find Q(t).Q(t). How many bacteria are in the dish after 3 hours? Example 5.43 Fruit Fly Population Growth Suppose a population of fruit flies increases at a rate of g(t)=2 e 0.02 t,g(t)=2 e 0.02 t, in flies per day. If the initial population of fruit flies is 100 flies, how many flies are in the population after 10 days? Checkpoint 5.36 Suppose the rate of growth of the fly population is given by g(t)=e 0.01 t,g(t)=e 0.01 t, and the initial fly population is 100 flies. How many flies are in the population after 15 days? Example 5.44 Evaluating a Definite Integral Using Substitution Evaluate the definite integral using substitution: ∫1 2 e 1/x x 2 d x.∫1 2 e 1/x x 2 d x. Checkpoint 5.37 Evaluate the definite integral using substitution: ∫1 2 1 x 3 e 4 x−2 d x.∫1 2 1 x 3 e 4 x−2 d x. Integrals Involving Logarithmic Functions Integrating functions of the form f(x)=x−1 f(x)=x−1 result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as f(x)=ln x f(x)=ln x and f(x)=log a x,f(x)=log a x, are also included in the rule. Rule: Integration Formulas Involving Logarithmic Functions The following formulas can be used to evaluate integrals involving logarithmic functions. ∫x−1 d x=ln\|x\|+C∫ln x d x=x ln x−x+C=x(ln x−1)+C∫log a x d x=x ln a(ln x−1)+C∫x−1 d x=ln\|x\|+C∫ln x d x=x ln x−x+C=x(ln x−1)+C∫log a x d x=x ln a(ln x−1)+C (5.22) Example 5.45 Finding an Antiderivative Involving ln x ln x Find the antiderivative of the function 3 x−10.3 x−10. Checkpoint 5.38 Find the antiderivative of 1 x+2.1 x+2. Example 5.46 Finding an Antiderivative of a Rational Function Find the antiderivative of 2 x 3+3 x x 4+3 x 2.2 x 3+3 x x 4+3 x 2. Example 5.47 Finding an Antiderivative of a Logarithmic Function Find the antiderivative of the log function log 2 x.log 2 x. Checkpoint 5.39 Find the antiderivative of log 3 x.log 3 x. Example 5.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration. Example 5.48 Evaluating a Definite Integral Find the definite integral of ∫0 π/2 sin x 1+cos x d x.∫0 π/2 sin x 1+cos x d x. Section 5.6 Exercises In the following exercises, compute each indefinite integral. 320. ∫e 2 x d x∫e 2 x d x 321. ∫e−3 x d x∫e−3 x d x 322. ∫2 x d x∫2 x d x 323. ∫3−x d x∫3−x d x 324. ∫1 2 x d x∫1 2 x d x 326. ∫1 x 2 d x∫1 x 2 d x In the following exercises, find each indefinite integral by using appropriate substitutions. 328. ∫ln x x d x∫ln x x d x 329. ∫d x x(ln x)2∫d x x(ln x)2 330. ∫d x x ln x(x>1)∫d x x ln x(x>1) 331. ∫d x x ln x ln(ln x)∫d x x ln x ln(ln x) 332. ∫tan θ d θ∫tan θ d θ 333. ∫cos x−x sin x x cos x d x∫cos x−x sin x x cos x d x 334. ∫ln(sin x)tan x d x∫ln(sin x)tan x d x 335. ∫ln(cos x)tan x d x∫ln(cos x)tan x d x 336. ∫x e−x 2 d x∫x e−x 2 d x 337. ∫x 2 e−x 3 d x∫x 2 e−x 3 d x 338. ∫e sin x cos x d x∫e sin x cos x d x 339. ∫e tan x sec 2 x d x∫e tan x sec 2 x d x 340. ∫e ln x d x x∫e ln x d x x 341. ∫e ln(1−t)1−t d t∫e ln(1−t)1−t d t In the following exercises, verify by differentiation that ∫ln x d x=x(ln x−1)+C,∫ln x d x=x(ln x−1)+C, then use appropriate changes of variables to compute the integral. 342. ∫x ln x d x∫x ln x d x(H i n t:∫x ln x d x=1 2∫x ln(x 2)d x)(H i n t:∫x ln x d x=1 2∫x ln(x 2)d x) 343. ∫x 2 ln(x 2)d x∫x 2 ln(x 2)d x 344. ∫ln x x 2 d x∫ln x x 2 d x(H i n t:Set u=1 x.)(H i n t:Set u=1 x.) 345. ∫ln x x d x∫ln x x d x(H i n t:Set u=x.)(H i n t:Set u=x.) 346. Write an integral to express the area under the graph of y=1 t y=1 t from t=1 t=1 to _e x_ and evaluate the integral. 347. Write an integral to express the area under the graph of y=e t y=e t between t=0 t=0 and t=ln x,t=ln x, and evaluate the integral. In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms. 348. ∫tan(2 x)d x∫tan(2 x)d x 349. ∫sin(3 x)−cos(3 x)sin(3 x)+cos(3 x)d x∫sin(3 x)−cos(3 x)sin(3 x)+cos(3 x)d x 350. ∫x sin(x 2)cos(x 2)d x∫x sin(x 2)cos(x 2)d x 351. ∫x csc(x 2)d x∫x csc(x 2)d x 352. ∫ln(cos x)tan x d x∫ln(cos x)tan x d x 353. ∫ln(csc x)cot x d x∫ln(csc x)cot x d x 354. ∫e x−e−x e x+e−x d x∫e x−e−x e x+e−x d x In the following exercises, evaluate the definite integral. 355. ∫1 2 1+2 x+x 2 3 x+3 x 2+x 3 d x∫1 2 1+2 x+x 2 3 x+3 x 2+x 3 d x 356. ∫0 π/4 tan x d x∫0 π/4 tan x d x 357. ∫0 π/3 sin x−cos x sin x+cos x d x∫0 π/3 sin x−cos x sin x+cos x d x 358. ∫π/6 π/2 csc x d x∫π/6 π/2 csc x d x 359. ∫π/4 π/3 cot x d x∫π/4 π/3 cot x d x In the following exercises, integrate using the indicated substitution. 360. ∫x x−100 d x;u=x−100∫x x−100 d x;u=x−100 361. ∫y−1 y+1 d y;u=y+1∫y−1 y+1 d y;u=y+1 362. ∫1−x 2 3 x−x 3 d x;u=3 x−x 3∫1−x 2 3 x−x 3 d x;u=3 x−x 3 363. ∫sin x+cos x sin x−cos x d x;u=sin x−cos x∫sin x+cos x sin x−cos x d x;u=sin x−cos x 364. ∫e 2 x 1−e 2 x d x;u=e 2 x∫e 2 x 1−e 2 x d x;u=e 2 x 365. ∫ln(x)1−(ln x)2 x d x;u=ln x∫ln(x)1−(ln x)2 x d x;u=ln x In the following exercises, does the right-endpoint approximation overestimate or underestimate the exact area? Calculate the right endpoint estimate _R_ 50 and solve for the exact area. 366. [T]y=e x y=e x over [0,1][0,1] 367. [T]y=e−x y=e−x over [0,1][0,1] 368. [T]y=ln(x)y=ln(x) over [1,2][1,2] 369. [T]y=x+1 x 2+2 x+6 y=x+1 x 2+2 x+6 over [0,1][0,1] 370. [T]y=2 x y=2 x over [−1,0][−1,0] 371. [T]y=−2−x y=−2−x over [0,1][0,1] In the following exercises, f(x)≥0 f(x)≥0 for a≤x≤b.a≤x≤b. Find the area under the graph of f(x)f(x) between the given values _a_ and _b_ by integrating. 372. f(x)=log 10(x)x;a=10,b=100 f(x)=log 10(x)x;a=10,b=100 373. f(x)=log 2(x)x;a=32,b=64 f(x)=log 2(x)x;a=32,b=64 374. f(x)=2−x;a=1,b=2 f(x)=2−x;a=1,b=2 375. f(x)=2−x;a=3,b=4 f(x)=2−x;a=3,b=4 376. Find the area under the graph of the function f(x)=x e−x 2 f(x)=x e−x 2 between x=0 x=0 and x=5.x=5. 377. Compute the integral of f(x)=x e−x 2 f(x)=x e−x 2 and find the smallest value of _N_ such that the area under the graph f(x)=x e−x 2 f(x)=x e−x 2 between x=N x=N and x=N+1 x=N+1 is, at most, 0.01. 378. Find the limit, as _N_ tends to infinity, of the area under the graph of f(x)=x e−x 2 f(x)=x e−x 2 between x=0 x=0 and x=N.x=N. 379. Show that ∫a b d t t=∫1/b 1/a d t t∫a b d t t=∫1/b 1/a d t t when 00 f(x)>0 for all _x_ and that _f_ and _g_ are differentiable. Use the identity f g=e g ln f f g=e g ln f and the chain rule to find the derivative of f g.f g. 381. Use the previous exercise to find the antiderivative of h(x)=x x(1+ln x)h(x)=x x(1+ln x) and evaluate ∫2 3 x x(1+ln x)d x.∫2 3 x x(1+ln x)d x. 382. Show that if c>0,c>0, then the integral of 1/x 1/x from _ac_ to _bc_(00.ln x y=ln x+ln y for x,y>0. 385. Use the identity ln x=∫1 x d t t ln x=∫1 x d t t to show that ln(x)ln(x) is an increasing function of _x_ on [0,∞),[0,∞), and use the previous exercises to show that the range of ln(x)ln(x) is (−∞,∞).(−∞,∞). Without any further assumptions, conclude that ln(x)ln(x) has an inverse function defined on (−∞,∞).(−∞,∞). 386. Pretend, for the moment, that we do not know that e x e x is the inverse function of ln(x),ln(x), but keep in mind that ln(x)ln(x) has an inverse function defined on (−∞,∞).(−∞,∞). Call it _E_. Use the identity ln x y=ln x+ln y ln x y=ln x+ln y to deduce that E(a+b)=E(a)E(b)E(a+b)=E(a)E(b) for any real numbers _a_, _b_. 387. Pretend, for the moment, that we do not know that e x e x is the inverse function of ln x,ln x, but keep in mind that ln x ln x has an inverse function defined on (−∞,∞).(−∞,∞). Call it _E_. Show that E'(t)=E(t).E'(t)=E(t). 388. The sine integral, defined as S(x)=∫0 x sin t t d t S(x)=∫0 x sin t t d t is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large _x_. Show that for k≥1,\|S(2 π k)−S(2 π(k+1))\|≤1 k(2 k+1)π.k≥1,\|S(2 π k)−S(2 π(k+1))\|≤1 k(2 k+1)π.(H i n t:sin(t+π)=−sin t)(H i n t:sin(t+π)=−sin t) 389. [T] The normal distribution in probability is given by p(x)=1 σ 2 π e−(x−μ)2/2 σ 2,p(x)=1 σ 2 π e−(x−μ)2/2 σ 2, where _σ_ is the standard deviation and _μ_ is the average. The _standard normal distribution_ in probability, p s,p s, corresponds to μ=0 and σ=1.μ=0 and σ=1. Compute the right endpoint estimates R 10 and R 100 R 10 and R 100 of ∫−1 1 1 2 π e−x 2 2 d x.∫−1 1 1 2 π e−x 2 2 d x. 390. [T] Compute the right endpoint estimates R 50 and R 100 R 50 and R 100 of ∫−3 5 1 2 2 π e−(x−1)2/8.∫−3 5 1 2 2 π e−(x−1)2/8.
190555	https://www.vocabulary.com/dictionary/tenable	Compete in the Vocabulary Bowl: 192 countries, 50 states, countless prizes. Register now! SKIP TO CONTENT Random Word tenable Add to list /ˈtɛnəbəl/ IPA guide Other forms: tenably If your teacher says your position on the benefits of abolishing homework is not tenable, she means it is not based in fact. To be tenable is to be evidence-based and well-founded. Tenable comes from the Latin root tenir which means "to hold," as in "hold together." If your plan is tenable, it will probably hold together when you execute it, or hold up to scrutiny. It's good to note that the opposite of tenable is untenable and not intenable, though throughout history both have been used. Definitions of tenable adjective based on sound reasoning or evidence synonyms: well-founded reasonable, sensible showing reason or sound judgment Cite this entry Style: MLA MLA APA Chicago "Tenable." Vocabulary.com Dictionary, Vocabulary.com, Accessed 03 Sep. 2025. Copy citation Try a free round of VocabTrainer, vocabulary learning that adapts to you! Think you know tenable? Answer a question to start your personalized learning plan. ASSESSMENT : 100 POINTS In addition to this cause of error, there have crept into the discussion several misstatements of fact that have been commonly adopted, and it is evident that in order to reach any reliable conclusion now, it will be necessary to examine the facts critically and ascertain what are ________. Source: The Word Hoosier; John Finley Indiana Historical Society Publications, Volume IV, Number 2 effeminate tenable taciturn virulent Not sure? Get a hint: 50/50 Definition Be a Vocabulary Bowl champion Do you have what it takes? Register now Examples from books and articles All sources All sources Fiction Arts / culture News Business Sports Science / Med Technology loading examples... We were finding no common ground, and I was becoming convinced that the marriage was no longer tenable. Long Walk to Freedom by Nelson Mandela As soon as you start to think about where meat comes from, putting it in your mouth no longer feels tenable. The GuardianJan 6, 2018 As for the thriller component, it has less to do with any actual action than with the unraveling of political delusions, step by step, until Nixon realizes that hanging onto power simply isn't tenable. Seattle TimesFeb 22, 2012 That one phone call transmuted what had been a difficult but tenable distance into a tick-tock torture. The GuardianJul 16, 2020 < prev \| next > DISCLAIMER: These example sentences appear in various news sources and books to reflect the usage of the word ‘tenable'. Views expressed in the examples do not represent the opinion of Vocabulary.com or its editors. Send us feedback Word Family tenabletenablyuntenabletenabilitytenableness the "tenable" family Vocabulary lists containing tenable The Vocabulary.com Top 1000 The top 1,000 vocabulary words have been carefully chosen to represent difficult but common words that appear in everyday academic and business writing. These words are also the most likely to appear on the SAT, ACT, GRE, and ToEFL. To create this list, we started with the words that give our users the most trouble and then ranked them by how frequently they appear in our corpus of billions of words from edited sources. If you only have time to study one list of words, this is the list. 100 SAT Words Beginning with "T" Find lists of SAT words organized by every letter of the alphabet here: A, B, C, D, E, F, G, H, I, J, K & L, M, N, O, P, Q, R, S, T, U, V, and W, X, Y & Z. This Week In Culture: October 5–11, 2019 Vampires at Downton Abbey? Cybernetic implants that unlock your house? David Hasselhoff is still relevant somewhere? We've gathered the shiniest vocabulary words from this week's sports, tech, and entertainment news and arranged them for your appreciation and edification. MORE VOCABULARY LISTS 2 million people are mastering new words. Master a word Sign up now (it’s free!) Whether you’re a teacher or a learner, Vocabulary.com can put you or your class on the path to systematic vocabulary improvement. Get started
190556	https://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem	Jump to content Proofs of Fermat's little theorem Español Italiano Magyar Edit links From Wikipedia, the free encyclopedia This article collects together a variety of proofs of Fermat's little theorem, which states that for every prime number p and every integer a (see modular arithmetic). Simplifications [edit] Some of the proofs of Fermat's little theorem given below depend on two simplifications. The first is that we may assume that a is in the range 0 ≤ a ≤ p − 1. This is a simple consequence of the laws of modular arithmetic; we are simply saying that we may first reduce a modulo p. This is consistent with reducing modulo p, as one can check. Secondly, it suffices to prove that for a in the range 1 ≤ a ≤ p − 1. Indeed, if the previous assertion holds for such a, multiplying both sides by a yields the original form of the theorem, On the other hand, if a = 0, the theorem holds trivially. Combinatorial proofs [edit] Proof by counting necklaces [edit] This is perhaps the simplest known proof, requiring the least mathematical background. It is an attractive example of a combinatorial proof (a proof that involves counting a collection of objects in two different ways). The proof given here is an adaptation of Golomb's proof. To keep things simple, let us assume that a is a positive integer. Consider all the possible strings of p symbols, using an alphabet with a different symbols. The total number of such strings is ap since there are a possibilities for each of p positions (see rule of product). For example, if p = 5 and a = 2, then we can use an alphabet with two symbols (say A and B), and there are 25 = 32 strings of length 5: : AAAAA, AAAAB, AAABA, AAABB, AABAA, AABAB, AABBA, AABBB, : ABAAA, ABAAB, ABABA, ABABB, ABBAA, ABBAB, ABBBA, ABBBB, : BAAAA, BAAAB, BAABA, BAABB, BABAA, BABAB, BABBA, BABBB, : BBAAA, BBAAB, BBABA, BBABB, BBBAA, BBBAB, BBBBA, BBBBB. We will argue below that if we remove the strings consisting of a single symbol from the list (in our example, AAAAA and BBBBB), the remaining ap − a strings can be arranged into groups, each group containing exactly p strings. It follows that ap − a is divisible by p. Necklaces [edit] Let us think of each such string as representing a necklace. That is, we connect the two ends of the string together and regard two strings as the same necklace if we can rotate one string to obtain the second string; in this case we will say that the two strings are friends. In our example, the following strings are all friends: : AAAAB, AAABA, AABAA, ABAAA, BAAAA. In full, each line of the following list corresponds to a single necklace, and the entire list comprises all 32 strings. : AAAAB, AAABA, AABAA, ABAAA, BAAAA, : AAABB, AABBA, ABBAA, BBAAA, BAAAB, : AABAB, ABABA, BABAA, ABAAB, BAABA, : AABBB, ABBBA, BBBAA, BBAAB, BAABB, : ABABB, BABBA, ABBAB, BBABA, BABAB, : ABBBB, BBBBA, BBBAB, BBABB, BABBB, : AAAAA, : BBBBB. Notice that in the above list, each necklace with more than one symbol is represented by 5 different strings, and the number of necklaces represented by just one string is 2, i.e. is the number of distinct symbols. Thus the list shows very clearly why 32 − 2 is divisible by 5. One can use the following rule to work out how many friends a given string S has: : If S is built up of several copies of the string T, and T cannot itself be broken down further into repeating strings, then the number of friends of S (including S itself) is equal to the length of T. For example, suppose we start with the string S = ABBABBABBABB, which is built up of several copies of the shorter string T = ABB. If we rotate it one symbol at a time, we obtain the following 3 strings: : ABBABBABBABB, : BBABBABBABBA, : BABBABBABBAB. There aren't any others because ABB is exactly 3 symbols long and cannot be broken down into further repeating strings. Completing the proof [edit] Using the above rule, we can complete the proof of Fermat's little theorem quite easily, as follows. Our starting pool of a p strings may be split into two categories: Some strings contain p identical symbols. There are exactly a of these, one for each symbol in the alphabet. (In our running example, these are the strings AAAAA and BBBBB.) The rest of the strings use at least two distinct symbols from the alphabet. If we can break up a given string S into repeating copies of some string T, the length of T must divide the length of S. But since the length of S is the prime p, the only possible length for T is also p. Therefore, the above rule tells us that S has exactly p friends (including S itself). The second category contains a p − a strings, and they may be arranged into groups of p strings, one group for each necklace. Therefore, a p − a must be divisible by p, as promised. Proof using dynamical systems [edit] This proof uses some basic concepts from dynamical systems. We start by considering a family of functions Tn(x), where n ≥ 2 is an integer, mapping the interval [0, 1] to itself by the formula where {y} denotes the fractional part of y. For example, the function T3(x) is illustrated below: A number x0 is said to be a fixed point of a function f(x) if f(x0) = x0; in other words, if f leaves x0 fixed. The fixed points of a function can be easily found graphically: they are simply the x coordinates of the points where the graph of f(x) intersects the graph of the line y = x. For example, the fixed points of the function T3(x) are 0, 1/2, and 1; they are marked by black circles on the following diagram: We will require the following two lemmas. Lemma 1. For any n ≥ 2, the function Tn(x) has exactly n fixed points. Proof. There are 3 fixed points in the illustration above, and the same sort of geometrical argument applies for any n ≥ 2. Lemma 2. For any positive integers n and m, and any 0 ≤ x ≤ 1, In other words, Tmn(x) is the composition of Tn(x) and Tm(x). Proof. The proof of this lemma is not difficult, but we need to be slightly careful with the endpoint x = 1. For this point the lemma is clearly true, since So let us assume that 0 ≤ x < 1. In this case, so Tm(Tn(x)) is given by Therefore, what we really need to show is that To do this we observe that {nx} = nx − k, where k is the integer part of nx; then since mk is an integer. Now let us properly begin the proof of Fermat's little theorem, by studying the function Tap(x). We will assume that a ≥ 2. From Lemma 1, we know that it has ap fixed points. By Lemma 2 we know that so any fixed point of Ta(x) is automatically a fixed point of Tap(x). We are interested in the fixed points of Tap(x) that are not fixed points of Ta(x). Let us call the set of such points S. There are ap − a points in S, because by Lemma 1 again, Ta(x) has exactly a fixed points. The following diagram illustrates the situation for a = 3 and p = 2. The black circles are the points of S, of which there are 32 − 3 = 6. The main idea of the proof is now to split the set S up into its orbits under Ta. What this means is that we pick a point x0 in S, and repeatedly apply Ta(x) to it, to obtain the sequence of points This sequence is called the orbit of x0 under Ta. By Lemma 2, this sequence can be rewritten as Since we are assuming that x0 is a fixed point of Ta p(x), after p steps we hit Tap(x0) = x0, and from that point onwards the sequence repeats itself. However, the sequence cannot begin repeating itself any earlier than that. If it did, the length of the repeating section would have to be a divisor of p, so it would have to be 1 (since p is prime). But this contradicts our assumption that x0 is not a fixed point of Ta. In other words, the orbit contains exactly p distinct points. This holds for every orbit of S. Therefore, the set S, which contains ap − a points, can be broken up into orbits, each containing p points, so ap − a is divisible by p. (This proof is essentially the same as the necklace-counting proof given above, simply viewed through a different lens: one may think of the interval [0, 1] as given by sequences of digits in base a (our distinction between 0 and 1 corresponding to the familiar distinction between representing integers as ending in ".0000..." and ".9999..."). Tan amounts to shifting such a sequence by n many digits. The fixed points of this will be sequences that are cyclic with period dividing n. In particular, the fixed points of Tap can be thought of as the necklaces of length p, with Tan corresponding to rotation of such necklaces by n spots. This proof could also be presented without distinguishing between 0 and 1, simply using the half-open interval [0, 1); then Tn would only have n − 1 fixed points, but Tap − Ta would still work out to ap − a, as needed.) Multinomial proofs [edit] Proofs using the binomial theorem [edit] Proof 1 [edit] This proof, due to Euler, uses induction to prove the theorem for all integers a ≥ 0. The base step, that 0p ≡ 0 (mod p), is trivial. Next, we must show that if the theorem is true for a = k, then it is also true for a = k + 1. For this inductive step, we need the following lemma. Lemma. For any integers x and y and for any prime p, (x + y)p ≡ xp + yp (mod p). The lemma is a case of the freshman's dream. Leaving the proof for later on, we proceed with the induction. Proof. Assume kp ≡ k (mod p), and consider (k+1)p. By the lemma we have Using the induction hypothesis, we have that kp ≡ k (mod p); and, trivially, 1p = 1. Thus which is the statement of the theorem for a = k+1. ∎ In order to prove the lemma, we must introduce the binomial theorem, which states that for any positive integer n, where the coefficients are the binomial coefficients, described in terms of the factorial function, n! = 1×2×3×⋯×n. Proof of Lemma. We consider the binomial coefficient when the exponent is a prime p: The binomial coefficients are all integers. The numerator contains a factor p by the definition of factorial. When 0 < i < p, neither of the terms in the denominator includes a factor of p (relying on the primality of p), leaving the coefficient itself to possess a prime factor of p from the numerator, implying that Modulo p, this eliminates all but the first and last terms of the sum on the right-hand side of the binomial theorem for prime p. ∎ The primality of p is essential to the lemma; otherwise, we have examples like which is not divisible by 4. Proof 2 [edit] Using the Lemma, we have: : . Proof using the multinomial expansion [edit] The proof, which was first discovered by Leibniz (who did not publish it) and later rediscovered by Euler, is a very simple application of the multinomial theorem, which states where and the summation is taken over all sequences of nonnegative integer indices k1, k2, ..., km such the sum of all ki is n. Thus if we express a as a sum of 1s (ones), we obtain Clearly, if p is prime, and if kj is not equal to p for any j, we have and if kj is equal to p for some j then Since there are exactly a elements such that kj = p for some j, the theorem follows. (This proof is essentially a coarser-grained variant of the necklace-counting proof given earlier; the multinomial coefficients count the number of ways a string can be permuted into arbitrary anagrams, while the necklace argument counts the number of ways a string can be rotated into cyclic anagrams. That is to say, that the nontrivial multinomial coefficients here are divisible by p can be seen as a consequence of the fact that each nontrivial necklace of length p can be unwrapped into a string in p many ways. This multinomial expansion is also, of course, what essentially underlies the binomial theorem-based proof above) Proof using power product expansions [edit] An additive-combinatorial proof based on formal power product expansions was given by Giedrius Alkauskas. This proof uses neither the Euclidean algorithm nor the binomial theorem, but rather it employs formal power series with rational coefficients. Proof as a particular case of Euler's theorem [edit] This proof, discovered by James Ivory and rediscovered by Dirichlet, requires some background in modular arithmetic. Let us assume that a is positive and not divisible by p. The idea is that if we write down the sequence of numbers \| \| \| \| --- \| \| \| A \| and reduce each one modulo p, the resulting sequence turns out to be a rearrangement of \| \| \| \| --- \| \| \| B \| Therefore, if we multiply together the numbers in each sequence, the results must be identical modulo p: Collecting together the a terms yields Finally, we may “cancel out” the numbers 1, 2, ..., p − 1 from both sides of this equation, obtaining There are two steps in the above proof that we need to justify: Why the elements of the sequence (A), reduced modulo p, are a rearrangement of (B), and Why it is valid to “cancel” in the setting of modular arithmetic. We will prove these things below; let us first see an example of this proof in action. An example [edit] If a = 3 and p = 7, then the sequence in question is reducing modulo 7 gives which is just a rearrangement of Multiplying them together gives that is, Canceling out 1 × 2 × 3 × 4 × 5 × 6 yields which is Fermat's little theorem for the case a = 3 and p = 7. The cancellation law [edit] Let us first explain why it is valid, in certain situations, to “cancel”. The exact statement is as follows. If u, x, and y are integers, and u is not divisible by a prime number p, and if \| \| \| \| --- \| \| \| C \| then we may “cancel” u to obtain \| \| \| \| --- \| \| \| D \| Our use of this cancellation law in the above proof of Fermat's little theorem was valid because the numbers 1, 2, ..., p − 1 are certainly not divisible by p (indeed they are smaller than p). We can prove the cancellation law easily using Euclid's lemma, which generally states that if a prime p divides a product ab (where a and b are integers), then p must divide a or b. Indeed, the assertion (C) simply means that p divides ux − uy = u(x − y). Since p is a prime which does not divide u, Euclid's lemma tells us that it must divide x − y instead; that is, (D) holds. Note that the conditions under which the cancellation law holds are quite strict, and this explains why Fermat's little theorem demands that p is a prime. For example, 2×2 ≡ 2×5 (mod 6), but it is not true that 2 ≡ 5 (mod 6). However, the following generalization of the cancellation law holds: if u, x, y, and z are integers, if u and z are relatively prime, and if then we may “cancel” u to obtain This follows from a generalization of Euclid's lemma. The rearrangement property [edit] Finally, we must explain why the sequence when reduced modulo p, becomes a rearrangement of the sequence To start with, none of the terms a, 2a, ..., (p − 1)a can be congruent to zero modulo p, since if k is one of the numbers 1, 2, ..., p − 1, then k is relatively prime with p, and so is a, so Euclid's lemma tells us that ka shares no factor with p. Therefore, at least we know that the numbers a, 2a, ..., (p − 1)a, when reduced modulo p, must be found among the numbers 1, 2, 3, ..., p − 1. Furthermore, the numbers a, 2a, ..., (p − 1)a must all be distinct after reducing them modulo p, because if where k and m are one of 1, 2, ..., p − 1, then the cancellation law tells us that Since both k and m are between 1 and p − 1, they must be equal. Therefore, the terms a, 2a, ..., (p − 1)a when reduced modulo p must be distinct. To summarise: when we reduce the p − 1 numbers a, 2a, ..., (p − 1)a modulo p, we obtain distinct members of the sequence 1, 2, ..., p − 1. Since there are exactly p − 1 of these, the only possibility is that the former are a rearrangement of the latter. Applications to Euler's theorem [edit] This method can also be used to prove Euler's theorem, with a slight alteration in that the numbers from 1 to p − 1 are substituted by the numbers less than and coprime with some number m (not necessarily prime). Both the rearrangement property and the cancellation law (under the generalized form mentioned above) are still satisfied and can be utilized. For example, if m = 10, then the numbers less than m and coprime with m are 1, 3, 7, and 9. Thus we have: Proof as a corollary of Euler's criterion [edit] Main article: Euler's criterion § Alternative proof Proofs using group theory [edit] Standard proof [edit] This proof requires the most basic elements of group theory. The idea is to recognise that the set G = {1, 2, ..., p − 1}, with the operation of multiplication (taken modulo p), forms a group. The only group axiom that requires some effort to verify is that each element of G is invertible. Taking this on faith for the moment, let us assume that a is in the range 1 ≤ a ≤ p − 1, that is, a is an element of G. Let k be the order of a, that is, k is the smallest positive integer such that ak ≡ 1 (mod p). Then the numbers 1, a, a2, ..., ak −1 reduced modulo p form a subgroup of G whose order is k and therefore, by Lagrange's theorem, k divides the order of G, which is p − 1. So p − 1 = km for some positive integer m and then To prove that every element b of G is invertible, we may proceed as follows. First, b is coprime to p. Thus Bézout's identity assures us that there are integers x and y such that bx + py = 1. Reading this equality modulo p, we see that x is an inverse for b, since bx ≡ 1 (mod p). Therefore, every element of G is invertible. So, as remarked earlier, G is a group. For example, when p = 11, the inverses of each element are given as follows: : \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| a \| 1 \| 2 \| 3 \| 4 \| 5 \| 6 \| 7 \| 8 \| 9 \| 10 \| \| a −1 \| 1 \| 6 \| 4 \| 3 \| 9 \| 2 \| 8 \| 7 \| 5 \| 10 \| Euler's proof [edit] If we take the previous proof and, instead of using Lagrange's theorem, we try to prove it in this specific situation, then we get Euler's third proof, which is the one that he found more natural. Let A be the set whose elements are the numbers 1, a, a2, ..., ak − 1 reduced modulo p. If A = G, then k = p − 1 and therefore k divides p −1. Otherwise, there is some b1 ∈ GA. Let A1 be the set whose elements are the numbers b1, ab1, a2b1, ..., ak − 1b1 reduced modulo p. Then A1 has k distinct elements because otherwise there would be two distinct numbers m, n ∈ {0, 1, ..., k − 1} such that amb1 ≡ anb1 (mod p), which is impossible, since it would follow that am ≡ an (mod p). On the other hand, no element of A1 can be an element of A, because otherwise there would be numbers m, n ∈ {0, 1, ..., k − 1} such that amb1 ≡ an (mod p), and then b1 ≡ anak − m ≡ an + k − m (mod p), which is impossible, since b1 ∉ A. So, the set A∪A1 has 2k elements. If it turns out to be equal to G, then 2k = p −1 and therefore k divides p −1. Otherwise, there is some b2 ∈ G(A∪A1) and we can start all over again, defining A2 as the set whose elements are the numbers b2, ab2, a2b2, ..., ak − 1b2 reduced modulo p. Since G is finite, this process must stop at some point and this proves that k divides p − 1. For instance, if a = 5 and p = 13, then, since 52 = 25 ≡ 12 (mod 13), 53 = 125 ≡ 8 (mod 13), 54 = 625 ≡ 1 (mod 13), we have k = 4 and A = {1, 5, 8, 12}. Clearly, A ≠ G = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Let b1 be an element of GA; for instance, take b1 = 2. Then, since 2×1 = 2, 2×5 = 10, 2×8 = 16 ≡ 3 (mod 13), 2×12 = 24 ≡ 11 (mod 13), we have A1 = {2, 3, 10, 11}. Clearly, A∪A1 ≠ G. Let b2 be an element of G(A∪A1); for instance, take b2 = 4. Then, since 4×1 = 4, 4×5 = 20 ≡ 7 (mod 13), 4×8 = 32 ≡ 6 (mod 13), 4×12 = 48 ≡ 9 (mod 13), we have A2 = {4, 6, 7, 9}. And now G = A∪A1∪A2. Note that the sets A, A1, and so on are in fact the cosets of A in G. Notes [edit] ^ Golomb, Solomon W. (1956), "Combinatorial proof of Fermat's "Little" Theorem" (PDF), American Mathematical Monthly, 63 (10): 718, doi:10.2307/2309563, JSTOR 2309563 ^ Iga, Kevin (2003), "A Dynamical Systems Proof of Fermat's Little Theorem", Mathematics Magazine, 76 (1): 48–51, doi:10.2307/3219132, JSTOR 3219132 ^ Jump up to: a b c Dickson, Leonard Eugene (2005) , "Fermat's and Wilson's theorems, generalizations, and converses; symmetric functions of 1, 2, ..., p − 1 modulo p", History of the Theory of Numbers, vol. I, Dover, ISBN 978-0-486-44232-7, Zbl 1214.11001 ^ Vacca, Giovanni (1894), "Intorno alla prima dimostrazione di un teorema di Fermat", Bibliotheca Mathematica, 2nd series (in Italian), 8 (2): 46–48 ^ Alkauskas, Giedrius (2009), "A Curious Proof of Fermat's Little Theorem", American Mathematical Monthly, 116 (4): 362–364, arXiv:0801.0805, doi:10.4169/193009709x470236, JSTOR 40391097 ^ Hardy, G. H.; Wright, E. M. (2008), "Fermat's Theorem and its Consequences", An Introduction to the Theory of Numbers (6th ed.), Oxford University Press, ISBN 978-0-19-921986-5 ^ Ivory, James (1806), "Demonstration of a theorem respecting prime numbers", The Mathematical Repository, New Series, 1 (II): 6–8 ^ Lejeune Dirichlet, Peter Gustav (1828), "Démonstrations nouvelles de quelques théorèmes relatifs aux nombres", Journal für die reine und angewandte Mathematik (in French), 3: 390–393 ^ Weil, André; Rosenlicht, Maxwell (1979), "§ VIII", Number Theory for beginners, Springer-Verlag, doi:10.1007/978-1-4612-9957-8, ISBN 978-0-387-90381-1, Zbl 0405.10001 ^ Weil, André (2007) , "§ III.VI", Number theory: An approach through history; from Hammurapi to Legendre, Birkhäuser, ISBN 978-0-8176-4565-6, Zbl 1149.01013 ^ Euler, Leonhard (1761), "Theoremata circa residua ex divisione potestatum relicta" (PDF), Novi Commentarii Academiae Scientiarum Petropolitanae (in Latin), 7: 49–82 Retrieved from " Categories: Modular arithmetic Number theory Article proofs Hidden categories: CS1 Italian-language sources (it) CS1 French-language sources (fr) CS1 Latin-language sources (la) Articles with short description Short description with empty Wikidata description
190557	https://www.ramauniversity.ac.in/online-study-material/fet/me/btech/vsemester/machinedesign-i/lecture-30.pdf	Lecture Machine Design Department of Mechanical Engineering 146 Lecture Machine Design • GERBER EQUATION • The Soderberg line and Goodman line illustrated in • Fig. 5.39 are straight lines. The theories using such • straight lines for predicting fatigue failure are called • ‘linear’ theories. There are some theories that use • parabolic or elliptical curves instead of straight lines. • These theories are called ‘non-linear’ theories. One • of the most popular non-linear theories is the Gerber • theory that is based on parabolic curve. The Gerber • curve is shown in Fig. 5.52. The equation for the • Gerber curve is as followsTheories based on the • Soderberg line or the • Goodman line, as failure criteria are conservative • theories. This results in increased dimensions of the • component. The Gerber curve takes the mean path • through failure points. It is therefore more accurate • in predicting fatigue failure. Department of Mechanical Engineering 147 Lecture Machine Design • Example 5 A machine component is subjected • to fl uctuating stress that varies from 40 to 100 • N/mm2. The corrected endurance limit stress for • the machine component is 270 N/mm2. The ultimate • tensile strength and yield strength of the material • are 600 and 450 N/mm2 • respectively. Find the factor • of safety using(i) Gerber theory(ii) Soderberg line • (iii) Goodman lineAlso, fi nd the factor of • safety against staticfailure. Department of Mechanical Engineering 148 Step V Factor of safety against static failure The factor of safety against static failure is given by, Lecture Machine Design • FATIGUE DESIGN UNDER COMBINED STRESSES • In practice, the problems are more complicated because the component may be subjected to two-dimensional stresses, or to combined bending and torsional moments. In case of two-dimensional stresses, each of the two stresses may have two components— mean and alternating. Similarly, the bending moment as well as torsional moment may have two components— mean and alternating. Such problems involving combination of stresses are solved by the distortion energy theory of failure. Department of Mechanical Engineering 149 Lecture Machine Design • Example A machine component is subjected to two-dimensional stresses. The tensile stress in the X direction varies from 40 to 100 N/mm2 while the tensile stress in the Y direction varies from 10 to 80 N/mm2. The frequency of variation of these stresses is equal. The corrected endurance limit of the component is 270 N/mm2. The ultimate tensile strength of the material of the component is 660 N/mm2. Determine the factor of safety used by the designer. Department of Mechanical Engineering 150
190558	https://www.verywellhealth.com/h2-blockers-vs-ppi-8605888	H2 Blockers vs PPIs: Which Is Best for Acid Reflux? A pharmacists explains the difference Acid reflux (gastroesophageal reflux) occurs when stomach contents enter your esophagus, causing symptoms like heartburn, regurgitation, and chest pain. Medications are often necessary to manage discomfort and prevent complications of acid reflux, such as gastroesophageal reflux disease (GERD), gastrointestinal (GI) bleeding, ulcers, and more. Histamine-2 receptor antagonists (H2 blockers) and proton pump inhibitors (PPIs) are the two main classes of medications used for acid reflux relief. These medicines can be prescribed by your healthcare provider or purchased over the counter (OTC). Read on for more details. Bloomberg / Contributor / Getty Images H2 Blockers vs PPIs: How They Work Both H2 blockers and PPIs decrease stomach acid but work differently. H2 blockers reduce the production of stomach acid through a histamine receptor blockade. Medications in this class include the following: In contrast to H2 blockers, PPIs directly block the proton pump responsible for producing stomach acid. For best results, take PPIs 30 to 60 minutes before your meal. Medications in this class include the following: Which Is More Effective? PPIs are the most effective medicines for acid reflux across the board. They relieve heartburn in up to 77% of people with esophagitis (inflammation of the esophagus) and 60% without esophagitis. They are even more effective in healing the esophagus and work in about 80% of esophagitis cases. H2 blockers are less effective at treating reflux symptoms than PPIs. H2 blockers relieve heartburn in about 50% of people who take them for up to 12 weeks. They heal esophagitis in about 40% of people who take them. A Word From Verywell When choosing between a PPI or an H2 blocker, the decision depends on your symptoms. If you experience two or fewer GERD episodes per week and have not been diagnosed with esophagitis, H2 blockers should be your first choice. On the other hand, if you have severe or frequent GERD symptoms, erosive esophagitis, or Barrett's esophagus, then a PPI is the way to go. How Long Do They Take to Work? H2 blockers start working faster than PPIs, typically within one to three hours. In contrast, it may take up to four days to see the effects of PPIs. The effects of PPIs last longer than those of H2 blockers. PPIs block stomach acid 15 to 21 hours a day versus about eight hours for H2 blockers. Indications and Usage PPIs are recommended for most cases of acid reflux. They are broadly used for conditions like the following: However, there are some situations in which H2 blockers are typically recommended instead. Some of these include the following: Side Effects and Safety Taking prescription or OTC H2 blockers or PPIs may cause side effects and drug interactions. Some precautions for certain groups are also important to note. Side Effects The side effects of these two drug classes are similar, primarily affecting the gastrointestinal tract. Common side effects of H2 blockers include the following: More serious side effects of these medicines include the following: Tolerance (tachyphylaxis) to H2 blockers can develop within as little as three days. This means the medicine doesn't work as well after the body gets used to it, and higher doses may be necessary to get the same results. PPIs can cause the following side effects: Long-term use of PPIs is associated with some significant risks, which include the following: Deciding how long to use PPIs can be a tricky situation. Drug manufacturers recommend taking PPIs for less than eight weeks to reduce the risk of side effects. However, many people who use PPIs for GERD will stay on them indefinitely (lifelong) because stopping treatment often causes reflux symptoms to return. Be sure to discuss the risks and benefits of prolonged PPI therapy with your healthcare provider for optimal safety. Precautions Keep in mind these precautions for the following groups: Interactions H2 blockers and PPIs can cause some significant drug interactions. Here are a few important ones to keep in mind: Recall Information In 2020, the H2 blocker Zantac (ranitidine) was pulled from the market due to the presence of N‐nitrosodimethylamine (NDMA). NDMA is thought to be a carcinogen (cancer-causing agent). The U.S. Food and Drug Administration (FDA) determined that levels of NDMA found in ranitidine could increase if stored at high temperatures. Keep in mind that the product Zantac 360 that's currently available contains famotidine, not ranitidine. How to Choose The choice of which medicine to take for acid reflux may come down to factors such as the following: Consult your healthcare professional or pharmacist for personalized evaluation and treatment recommendations. Lifestyle Modifications and Alternative Treatments Lifestyle modifications, alone or in conjunction with medications, can help control symptoms of acid reflux. Some recommended strategies include the following: Additionally, supplements that may help with acid reflux symptoms include the following: More research is necessary before these can be routinely recommended. Remember that the FDA doesn't approve supplements before products are marketed like they do for prescription medications. This means they don't have to be proven safe or effective. Summary PPIs and H2 blockers are the most commonly taken medications to treat acid reflux and its complications, including GERD and esophagitis. In many cases, PPIs are recommended first because they are more effective than H2 blockers at reducing stomach acid and symptoms like heartburn. Though the side effects of these two classes are similar, the prolonged use of PPIs is associated with severe adverse effects like fractures, infections, and even dementia. If you are experiencing acid reflux, discuss treatment options with your healthcare provider to help you achieve your health goals. Clarrett DM, Hachem C. Gastroesophageal reflux disease (GERD). Mo Med. 2018;115(3):214-218. National Institute of Diabetes and Digestive and Kidney Diseases. Histamine type-2 receptor antagonists (H2 blockers). In: LiverTox: Clinical and Research Information on Drug-Induced Liver Injury. Bethesda (MD): January 25, 2018. Strand DS, Kim D, Peura DA. 25 years of proton pump inhibitors: a comprehensive review. Gut Liver. 2017;11(1):27-37. doi:10.5009/gnl15502 Liang SW, Wong MW, Yi CH, et al. Current advances in the diagnosis and management of gastroesophageal reflux disease. Tzu Chi Med J. 2022;34(4):402-408. Published 2022 May 16. doi:10.4103/tcmj.tcmj_323_21 Vordenberg SE. Nonprescription heartburn medications for adults. JAMA. 2023;330(15):1499-1499. doi:10.1001/jama.2023.16280 Shim YK, Kim N. The effect of H2 receptor antagonist in acid inhibition and its clinical efficacy. Korean J Gastroenterol. 2017;70(1):4-12. doi:10.4166/kjg.2017.70.1.4 Shin JM, Sachs G. Pharmacology of proton pump inhibitors. Curr Gastroenterol Rep. 2008;10(6):528-534. doi:10.1007/s11894-008-0098-4 Altuwaijri M. Evidence-based treatment recommendations for gastroesophageal reflux disease during pregnancy: a review. Medicine (Baltimore). 2022;101(35):e30487. doi:10.1097/MD.0000000000030487 Leung AK, Hon KL. Gastroesophageal reflux in children: an updated review. Drugs Context. 2019;8:212591. Published 2019 Jun 17. doi:10.7573/dic.212591 American College of Gastroenterology. Acid reflux/GERD (gastroesophageal reflux disease). U.S. Food and Drug Administration. Prilosec (omeprazole) prescribing information. Lehault WB, Hughes DM. Review of the long-term effects of proton pump inhibitors. Fed Pract. 2017;34(2):19-23. Northuis CA, Bell EJ, Lutsey PL, et al. Cumulative use of proton pump inhibitors and risk of dementia: the atherosclerosis risk in communities study. Neurology. 2023;101(18):e1771-e1778. doi:10.1212/WNL.0000000000207747 Koyyada A. Long-term use of proton pump inhibitors as a risk factor for various adverse manifestations. Therapie. 2021;76(1):13-21. doi:10.1016/j.therap.2020.06.019 Wagner JA, Dinh JC, Lightdale JR, et al. Is this the end for ranitidine? NDMA presence continues to confound. Clin Transl Sci. 2021;14(4):1197-1200. doi:10.1111/cts.12995 Schulz RM, Ahuja NK, Slavin JL. Effectiveness of nutritional ingredients on upper gastrointestinal conditions and symptoms: a narrative review. Nutrients. 2022;14(3):672. Published 2022 Feb 5. doi:10.3390/nu14030672 By Megan Nunn, PharmD Nunn is a community pharmacist in Tennessee with 12 years of experience in medication counseling and immunization. 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190559	https://fluidmech.ninja/fluidmech.pdf	Fluid Dynamics for Engineers Olivier Cleynen A one-semester course for students in the Chemical and Energy Engineering program at the University Otto von Guericke of Magdeburg 2015-2020 last edited October 27, 2022 Contents Contents 2 About this course (syllabus) 6 1 Basic fow quantities 9 1.1 Concept of a fuid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2 Fluid dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3 Important concepts in mechanics . . . . . . . . . . . . . . . . . . . . . . . . 12 1.4 Properties of fuids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.5 Forces on fuids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 1.6 Basic fow quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.7 Four balance equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.8 Classifcation of fuid fows . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.9 Limits of fuid dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.10 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 1.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2 Analysis of existing fows with one dimension 33 2.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.2 One-dimensional fow problems . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3 Balance of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.4 Balance of momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.5 Balance of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.6 The Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.7 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3 Analysis of existing fows with three dimensions 51 3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.2 The Reynolds transport theorem . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.3 Balance of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.4 Balance of momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.5 Balance of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.6 Balance of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.7 Limits of integral analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.8 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 4 Efects of pressure 73 4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.2 Pressure forces on walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.3 Pressure felds in fuids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.4 Special case: pressure in static fuids . . . . . . . . . . . . . . . . . . . . . . . 79 4.5 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 4.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 5 Efects of shear 91 5.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3 5.2 Shear forces on walls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.3 Shear felds in fuids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 5.4 Resistance to shear: viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 5.5 Special case: shear in simple laminar fows . . . . . . . . . . . . . . . . . . . 101 5.6 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 6 Prediction of fuid fows 111 6.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 6.2 Organizing calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 6.3 Equations for all fows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.4 Equations for incompressible fow . . . . . . . . . . . . . . . . . . . . . . . . 123 6.5 CFD: the Navier-Stokes equations in practice . . . . . . . . . . . . . . . . . . 127 6.6 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 6.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 7 Pipe fows 135 7.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.2 Frictionless fow in pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 7.3 Parameters to quantify losses in pipes . . . . . . . . . . . . . . . . . . . . . . 137 7.4 Laminar fow in pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 7.5 Turbulent fow in pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 7.6 Engineer’s guide to pipe fows . . . . . . . . . . . . . . . . . . . . . . . . . . 146 7.7 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 7.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 8 Engineering models 161 8.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 8.2 Comparing infuences: the weighted momentum balance . . . . . . . . . . . 161 8.3 Making models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 8.4 Comparing results: coefcients . . . . . . . . . . . . . . . . . . . . . . . . . . 169 8.5 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 8.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 9 Dealing with turbulence 179 9.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 9.2 Recognizing turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 9.3 The efects of turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 9.4 Quantifying turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 9.5 Computing turbulent fow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 9.6 Commented bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 9.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 10 Flow near walls 199 10.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 10.2 The concept of boundary layer . . . . . . . . . . . . . . . . . . . . . . . . . . 199 10.3 Laminar boundary layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 10.4 Boundary layer transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 10.5 Turbulent boundary layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 10.6 Flow separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 10.7 Solved problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 10.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 11 Large- and small-scale fows 221 11.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 11.2 Flow at large scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 11.3 Plotting velocity with functions . . . . . . . . . . . . . . . . . . . . . . . . . 223 11.4 Flow at very small scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 11.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 Appendix 245 A1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 A2 Vector operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 A3 Field operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 A4 Derivations of the Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . 253 A5 Flow parameters as force ratios . . . . . . . . . . . . . . . . . . . . . . . . . . 256 A6 Details of the winter 2020-2021 fnal examination (updated February 2021) . 259 A7 Example of previous examinations . . . . . . . . . . . . . . . . . . . . . . . . 261 A8 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 About this course (syllabus) Fluid dynamics for engineers by Olivier Cleynen, PhD University of Magdeburg, 2015-2020 Welcome! From 2015 to 2020, these notes were the core of the Fluid Dynamics course of the Chemical and Energy Engineering program at the University of Magdeburg, Germany. Objectives Starting with little or no experience with fuid mechanics, after taking this course: • you should have a good understanding of what can, and cannot, be calculated with fuid mechanics in engineering: how we approach problems depending on how much information is available. • you should be able to solve several real-world engineering fuid mechanics prob-lems with confdence: calculating forces within fuids and on objects, predicting fow in pipes, near walls, at small and large scales. My objective is to enable you to get there with the minimum amount of your time and energy (but not minimum power!). If all goes well, at the end of the semester, you should be well-prepared to begin a course in Computational Fluid Dynamics, where the knowledge and skills you acquire here can be used to solve applied problems in great detail. About these notes and the author My name is Olivier Cleynen. I had the pleasure and priviledge of directing this course from 2015 to 2020 while I was a PhD student at the University Otto von Guericke of Magdeburg. In 2021, my colleagues at the fuid dynamics laboratory took over. I am no longer involved in teaching the course, but am delighted that these course materials live on. If you are a student enrolled in this course, you should defnitely stop reading now and check with your course instructor so you understand how the course will go (where to get help, what chapters are examinable, and so on). I am not at all up-to-date with that information, and unfortunately cannot help at all! I obtained my Master’s in 2006, then went on to found and work for a non-proft organization, and then became a university teacher in France. I arrived in Magdeburg in 2015, and currently live here with my partner and her twelve year-old child. I completed my PhD in 2022, and I now am a math and physics teacher in a local high school. I am delighted that the present course notes at keep being useful to students around the world. Contact My email is olivier cleynen.fr. I am available to answer queries about re-using the course materials. Unfortunately, I cannot provide assistance with the content itself (e.g. how to solve the end-of-chapter problems) or answer any questions about the course. If you would like such assistance, please contact my colleagues at the fuid dynamics laboratory of the University of Magdeburg. Copyright, remixing, and authors This document is mainly authored by Olivier Cleynen. Substantial contributions have been made by colleagues Germán Santa-Maria , Jochen König, and Arjun Neyyathala. Numerous improvements have been contributed by students over the years. Many fgures from authors not asso ciated with this course are included; the author, license, and a link to the source are indicated every time. A few fgures still remain which are extracted from cited, fully-copyrighted works, as indicated. Most portrait illustrations are authored by Oksana Grivina; they are fully-copyrighted and used under a commercial license in this project. Other portrait illustrations are authored by Agustin Dede Prasetyo and Olivier Cleynen under a CC-by license. The text of this document is licensed under a Creative Commons CC-by-nc (attribution, non-commercial) license. This means you are free to make copies of it, but not for com-mercial purposes. This license also does not permit uploading this document to content platforms such as Facebook, Academia.edu or Scribd. The Latex sources of this document can be accessed from the git repository accessed from the course homepage. A number of YouTube videos, shot during the 2020 German lockdown, are linked along the text. If you use this document in other works, please cite it as “Olivier Cleynen. Fluid dynamics for engineers. Under CC-by-nc license. 2020. url: Conclusion Welcome to this course! (I recorded this YouTube video in April 2020) I hope you have a great semester! Fluid mechanics is one of the most exciting disciplines out there. Now, let us begin! Olivier Cleynen April 2020 (updated October 2022) 8 Fluid Dynamics Chapter 1 – Basic fow quantities last edited April 18, 2021 by Olivier Cleynen — 1.1 Concept of a fuid 9 1.2 Fluid dynamics 9 1.2.1 Solution of a fow 9 1.2.2 Modeling of fuids 10 1.2.3 Theory, numerics, and experiment 11 1.3 Important concepts in mechanics 12 1.3.1 Position, velocity, acceleration 12 1.3.2 Forces and moments 12 1.3.3 Energy 13 1.4 Properties of fuids 14 1.4.1 Density 14 1.4.2 Phase 14 1.4.3 Temperature 15 1.4.4 Perfect gas model 15 1.4.5 Speed of sound 16 1.4.6 Viscosity 17 1.5 Forces on fuids 18 1.5.1 Gravity 18 1.5.2 Pressure 18 1.5.3 Shear 18 1.6 Basic fow quantities 19 1.7 Four balance equations 20 1.8 Classifcation of fuid fows 21 1.9 Limits of fuid dynamics 23 1.10 Solved problems 24 1.11 Problems 27 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 1.1 Concept of a fuid We call fuidw a type of matter which is continuously deformable, and which spontaneously tends to adapt its shape to its container by occupying all of the space made available to it. 1.2 Fluid dynamics 1.2.1 Solution of a fow Fluid dynamics (or generally, fuid mechanics) is the study of the movement of fuids. The most common type of problem in this discipline is the search for a complete description of the fuid fow around or through a solid object. This problem is solved when the entire set of velocities of fuid particles has been described. This set of velocities, which is often a function of time, 9 can be described either as a set of discrete values (“pixelized” data) or as a mathematical function; it is called the fow solution. If the solution is known, the shear and pressure eforts generated on the surface of the object can be calculated. Other quantities, such as the force and moments applying on the object, or the fuid’s energy gains or losses, can also be calculated. 1.2.2 Modeling of fuids Like all matter, fuids are made of discrete, solid molecules. However, in fuid mechanics, we work at the macroscopic scale: at that scale, matter can be treated like a continuum, in which all physical properties of interest can be continuously diferentiated. Video: why not to calculate the movement of molecules by Olivier Cleynen (CC-by) There are about 2 ⋅1022 molecules in the air within an “empty” 1-liter bottle at ambient temperature and pressure. Even when the air within the bottle is completely still, these molecules are constantly colliding with each other and with the bottle walls; on average, their speed is equal to the speed of sound: approximately 1 000 km/h. Despite the complexity of individual molecule movements, even the most turbulent fows can be accurately described and solved by considering the velocities of groups of several millions of molecules collectively, which we name fuid particles.w By doing so, we never fnd out the velocity of individual molecules: instead, those are averaged in space and time and result in much simpler and smoother trajectories, which are those we can observe with macroscopic instruments such as video cameras and pressure probes. XKCD #2283: how (not) to pic-ture big numbers by Randall Munroe (CC-by-nc) Our selection of an appropriate fuid particle size (in efect defning the lower boundary of the macroscopic scale), is illustrated in fg. 1.1. We choose to reduce our volume of study to the smallest possible size before the efect of individual molecules becomes meaningful. Adopting this point of view, which is named the continuum abstraction, is not a trivial decision, because the physical laws which determine the behavior of molecules are very diferent from those which determine the behavior of elements of fuid. For example, in fuid mechanics we never consider any inter-element attraction or repulsion forces; while new forces “appear” due to pressure or shear efects that do not exist at a molecular level. Figure 1.1: Measurement of the average value of a property (here, velocity 푉; but it could be pressure, or temperature) inside a given volume. As the volume shrinks towards zero, the fuid can no longer be treated as a continuum; and property measurements will oscillate wildly. Figure CC-by-sa Olivier Cleynen 10 A direct beneft of the continuum abstraction is that the mathematical com-plexity of our problems is greatly simplifed. Finding the solution for the bottle of “still air” mentioned above, for example, requires only a single equation (eq. 4/15 p. 80) instead of a system of 2 ⋅1022 equations with 2 ⋅1022 unknowns (all leading up to ⃗ 푉average,푥,푦,푧,푡= ⃗ 0!). Another consequence is that we cannot treat a fuid as if it were a mere set of marbles with no interaction which would hit objects as they move by. Instead we must think of a fuid –even a low-density fuid such as atmospheric air– as an infnitely-fexible medium able to fll in almost instantly all of the space made available to it. 1.2.3 Theory, numerics, and experiment SMBC #2010-08-29: cooperation between theoretical and experi-mental scientists by Zach Weinersmith Today, fuid dynamicists are typically specializing in any one of three sub-disciplines: Analytical fuid mechanics which is the main focus of these lectures and which consists in predicting fuid fows mathematically. As we shall see, it is only able to provide (exact) solutions for very simple fows. In fuid mechanics, theory nevertheless allows us to understand the mechanisms of complex fuid phenomena, describe scale efects, and predict forces associated with given fuid fows; Numerical fuid mechanics also called Computational Fluid Dynamicsw or cfd, which consists in solving problems using very large numbers of discrete values. Initiated as a research topic in the 1970s, cfd is now omnipresent in the industry; it allows for excellent visualization and parametric studies of very complex fows. Nevertheless, computational solutions obtained within practical time frames are inherently approxi-mate: they need to be challenged using analysis, and calibrated using experimental measurements; Experimental fuid mechanics which consists in reproducing phenom-ena of interest within laboratory conditions and observing them using experimental techniques. A very mature branch (it frst provided useful results at the end of the 19th century), it is unfortunately associated with high human, equipment and fnancial costs. Experimental mea-surements are indispensable for the validation of computational simu-lations; meanwhile, the design of meaningful experiments necessitates a good understanding of scale efects. Our study of analytical fuid mechanics should therefore be a useful tool to approach the other two sub-disciplines of fuid mechanics. 11 1.3 Important concepts in mechanics Mechanics in general deals with the study of forces and motion of bodies. A few concepts relevant for us are recalled here. 1.3.1 Position, velocity, acceleration The description of the movement of bodies (without reference to the causes and efects of that movement) is called kinematics.w The position in space of an object can be fully expressed using three com-ponents (one for each dimension) This is usually done with a vector (here written ⃗ 푥). If the object moves, then this vector varies with time. The velocity ⃗ 푉of the object is the rate of change in time of its position: ⃗ 푉≡d⃗ 푥 d푡 (1/1) The length 푉of the velocity vector ⃗ 푉is measured in m s−1: 푉≡\|\| ⃗ 푉\|\| (1/2) In this document, 푉always is a positive number. Its formal name is speed, but in practice the term velocity is used to designate either the vector or its length, according to context. In order to express the velocity vector completely, three distinct values (each having positive or negative values in m s−1) must be expressed. In this document, this is done with diferent notations. In Cartesian coordinates, we have: ⃗ 푉= ⎛ ⎜ ⎜ ⎝ 푉푥 푉푦 푉푧 ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ 푢 푣 푤 ⎞ ⎟ ⎟ ⎠ In cylindrical coordinates, we write: ⃗ 푉= ⎛ ⎜ ⎜ ⎝ 푢푟 푢휃 푢푧 ⎞ ⎟ ⎟ ⎠ The acceleration ⃗ 푎of the object is the rate of change in time of its velocity: ⃗ 푎≡d ⃗ 푉 d푡 (1/3) Acceleration is especially important in mechanics because it can be deduced from Newton’s second law (see eq. 1/25 p. 20 below) if the forces applying on the object are known. Acceleration can then be integrated with respect to time to obtain velocity, which can be integrated with respect to time to obtain position. Like velocity, acceleration has three components (each measured in m s−2). It shows at which rate each component of velocity is changing. It may not always point in the same direction as velocity (fg. 1.2). 12 Figure 1.2: A body (black dot) is following a trajectory, plotted in blue. The accel-eration vectors are plotted in red for two cases: on top, when the magnitude of its velocity remains constant, and on the bottom, when this magnitude changes continuously. Figure CC-0 Olivier Cleynen 1.3.2 Forces and moments A force expresses an efort exerted on a body. Force has a direction as well as a magnitude, and so it is expressed with a vector (typically noted ⃗ 퐹). The three main types of forces relevant to fuid mechanics are those due to pressure, those due to shear, and those due to gravity (the force due to gravity is called weight). When a force exerts at a position ⃗ 푟away from a reference point, it exerts a torsion (or “twisting”) efort named moment.w Like a force, a moment has a direction as well as a magnitude, and so is best expressed with a vector. This vector ⃗ 푀is expressed as the cross product of the arm ⃗ 푟and the force ⃗ 퐹: ⃗ 푀≡⃗ 푟∧⃗ 퐹 (1/4) See Appendix A2.2 p. 248 for a short briefng about the cross product of vectors. 1.3.3 Energy Energy, measured in joules (J), is in most general terms the ability of a body to set other bodies in motion. It can be accumulated or spent by bodies in a large number of diferent ways. The most relevant forms of energy in fuid mechanics are: Kinetic energyw noted 퐸k, accumulated as motion: 퐸k ≡1 2푚푉2 (1/5) 13 Workw noted 푊, which is energy spent on displacing an object over a distance 푙with a force 퐹: 푊≡⃗ 퐹⋅⃗ 푙 (1/6) where 푊is the work (J); ⃗ 퐹is the force (vector with magnitude in N); and ⃗ 푙is the movement distance (vector with magnitude in m). See Appendix A2.1 p. 247 for a short briefng about the dot product of vectors. Internal energyw noted 퐼stored as heat within the body itself. As long as no phase changes occurs, the internal energy 퐼of fuids is roughly proportional to their absolute temperature 푇. 1.4 Properties of fuids Beyond velocity, which is the primary unknown for us in fuid mechanics, there are a few other important fuid properties. 1.4.1 Density The density 휌(Greek letter rho) is the amount of mass per unit volume: 휌≡푚  (1/7) where 휌is the density (kg m−3); 푚is the considered mass (kg); and is the considered volume (m3). Two orders of magnitude that are useful to remember: at ambient atmospheric conditions, air has a density of approximately 휌air = 1,2 kg m−3; that of water is almost a thousand times greater, at 휌water = 1 000 kg m−3. Advice from an expert Never confuse pressure 푝, which is how hard a fuid pushes on walls, with density 휌, which is how much of the fuid there is per unit volume. Not many fuid dynamicists will confess to it, but most have, as beginner students, mixed up the two symbols once in a moment of weakness. Make sure you write those two letters very clearly, so you distinguish them easily even when stressed or distracted. 1.4.2 Phase Fluids can be broadly classifed into phases,w which are loosely-defned sets of physical behaviors. Typically one distinguishes liquids which are fuids with large densities on which surface tension efects play an important role, from gases or vapors which have low densities and no surface tension efects. Phase changes are often brutal (but under specifc conditions can be blurred or smeared-out); they usually involve large energy transfers. The presence of 14 multiple phases in a fow is an added layer of complexity in the description of fuid phenomena. 1.4.3 Temperature Temperaturew is a scalar property measured in Kelvins (an absolute scale). It represents a body’s potential for receiving or providing heat and is defned, in thermodynamics, based on the transformation of heat and work. We convert from Kelvins to degrees Celsius by subtracting 273,15 units: 푇(°C) = 푇(K) −273,15 (1/8) Although we can “feel” temperature in daily life, it must be noted that the human body is a very poor thermometer in practice. This is because we constantly produce heat, and we infer temperature by the power our body loses or gains as heat. This power not only depends on our own body temperature (hot water “feels” hotter when we are cold), but also on the heat capacity of the fuid (cold water “feels” colder than air at the same temperature) and on the amount of convection taking place (ambient air “feels” colder on a windy day). In spite of these impressions, the fact is that the heat capacity of fuids is extremely high (푐air ≈1 kJ kg−1 K, 푐water ≈4 kJ kg−1 K). Unless very high velocities are attained, the temperature changes associated with fuid fow are much too small to be measurable in practice. Advice from an expert Never make guesses about the dynamics of a fow using your own perception of temperature. The human body constantly rejects or absorbs heat, and makes for a very bad thermometer. Wind may just feel cold because the air movement increases heat transfer of your skin, and not because of changes in air pressure or in density. If needed, use a real thermometer! 1.4.4 Perfect gas model When a gas has relatively simple molecules, moderate temperature and low pressure, several of its properties can be related easily to one another with the perfect gas model.w Their absolute temperature 푇is then modeled as a function of their pressure 푝with a single, approximately constant parameter 푅specifc ≡푝/휌푇: 푝 휌= 푅specifc푇 (1/9) where 푅specifc depends on the state and nature of the gas (J K−1 kg−1); and 푝is the pressure (Pa, see §1.5.2 further down). Note that 푅specifc here is a specifc gas constant (whose value depends on the gas); chemists often instead use a universal defnition of 푅in J mol−1 K−1. This type of model (relating temperature to pressure and density) is called an equation of state. When 푅remains approximately constant, the fuid is 15 said to behave as a perfect gas. The properties of air can satisfactorily be predicted using this model. Other models existw which predict the properties of gases over larger property ranges, at the cost of increased mathematical complexity. Many fuids, especially liquids, do not follow this equation and their tem-perature must be determined in another way, most often with the help of laboratory measurements. 1.4.5 Speed of sound An important property of fuids is the speed at which pressure changes can travel within the fuid (these pressure changes may for example be caused by the movement of an object). This speed is equal to the average speed of molecules within the fuid, and it is called the speed of sound,w noted 푐. In fuid dynamics, we often quantify how fast the fuid is fowing relative to the speed of sound. For this, we defne the Mach numberw noted [Ma] as the ratio of the local fuid speed 푉to the local speed of sound 푐: [Ma] ≡푉 푐 (1/10) Since both 푉and 푐can be functions of space in a given fow, [Ma] may not be uniform (e.g. the Mach number around an aircraft in fight is diferent at the nose and above its wings). Nevertheless, a single value is typically chosen to identify “the” representative Mach number of any given fow. It is observed that providing no heat or work transfer occurs, when fuids fow at [Ma] ≤0,3, their density 휌stays constant. Density variations in practice can be safely neglected below [Ma] = 0,6. When the density is uniform, the fow is said to be incompressible. Above these Mach numbers, it is observed that when subjected to pressure variations, fuids exert work upon themselves, which translates into measurable density and temperature changes: these are called compressibility efects, and we will not study them in this course. In most fows, the density of liquids is almost invariant – so that water fows are generally entirely incompressible. When the fuid is air (and generally within a perfect gas), it can be shown that 푐depends only on the absolute temperature: 푐= √ 훾푅푇 (1/11) in all cases for a perfect gas, where 푐is the local speed of sound (m s−1); 훾is a gas property, approx. constant (dimensionless); and 푇is the local temperature (K). Advice from an expert When you are showering and decrease the temperature at the knob, it takes some time for the water to feel colder: this is because it travels at speed 푉in the pipe to the showerhead. But when you decrease the fow rate at the knob, the response is instantaneous: this is because the decrease in pressure travels at speed 푐in the pipe (5 000 km h−1). The dynamics of fuids become funky 16 when the speed of the fuid and the speed of sound are comparable: this is why the Mach number is so important to us. 1.4.6 Viscosity Video: how to make sense of vis-cosity by Olivier Cleynen (CC-by) We have said above that a fuid element can deform continuously under pressure and shear eforts: it will never “snap” or break apart. However this deformation is not “for free”: it will require force and energy inputs which are not reversible (they are not reversed if the motion is reversed). Resistance to straining in a fuid is measured with a property named viscosity.w In informal terms, viscosity is the “stickiness” of fuids: for example, honey and sugar syrups are more viscous than water. More formally, viscosity is quantifed as follows. Imagine a small brick-shaped element of fuid, which is deformed (strained) horizontally, as shown in fg. 1.3. The continuous straining of the brick requires a force 퐹per unit area 퐴, called shear stress 휏(see also §1.5.3 further down). We expect that the shear increases with both the “stickiness” of the fuid —the viscosity— and with the speed Δ푣at which the brick is strained. Conversely, we expect that the shear will decrease when the element height Δ푦is increased. We defne the viscosity 휇as the ratio between the required shear stress, 휏= 퐹/퐴, and the rate at which the brick is strained, Δ푣/Δ푦: 휇≡ 휏 ( Δ푣 Δ푦) (1/12) where 휇 is the viscosity (N s m−2 or Pa s); 휏 is the horizontal shear in fg. 1.3 (Pa); Δ푢is the velocity diference between top and bottom planes in fg. 1.3 (m s−1); and Δ푦is the height diference between top and bottom planes in fg. 1.3 (m). The dimension of viscosity is in (force per area) divided by (velocity per distance), and so this turns out as Pascal −seconds in si units. We will come back to shear and viscosity in chapter 5 (Efects of shear). Figure 1.3: A brick-shaped element of fuid is strained, by applying a velocity diference Δ푣between the top and bottom surface. The horizontal force 퐹required on the top edge, divided by the area 퐴, is the shear stress 휏(see also §1.5.3 p. 18 further down). The higher the required shear stress for a given strain rate, the more viscous the fuid is. Figure CC-0 Pm.schroeder/Olivier Cleynen 17 1.5 Forces on fuids Fluids are subjected to, and subject their surroundings and themselves to forces. Identifying and quantifying those forces allows us to determine how they will fow. Three types of forces are relevant in fuid dynamics: gravity, pressure and shear. 1.5.1 Gravity Gravity (here, the attraction efort exerted on fuids at a distance by the Earth) is expressed with a vector ⃗ 푔pointing downwards. Within the Earth’s atmosphere, the magnitude 푔of this vector varies extremely slightly with altitude, and it may be considered constant at 푔= 9,81 m s−2. The weight force exerted on an object of mass 푚is then simply quantifed as ⃗ 퐹weight = 푚⃗ 푔 (1/13) 1.5.2 Pressure The concept of pressurew can be approached with the following conceptual experiment: if a fat solid surface is placed in a fuid at zero relative velocity, the pressure 푝will be the ratio of the perpendicular force 퐹⟂to the surface area 퐴: 푝≡퐹⟂ 퐴 (1/14) where 푝 is the pressure (N m−2 or Pascals, 1 Pa ≡1 N m−2); 퐹⟂is the component of force perpendicular to the surface (N); and 퐴is the surface area (m2). Although in si units pressure is measured in Pascals (1 Pa ≡1 N m−2), in prac-tice it is often measured in bars (1 bar ≡1 ⋅105 Pa). Ambient atmospheric pres-sure at normal altitude varies with the weather and is approximately 1 bar. The pressure distribution within a fuid is related to their velocity distribution according to relations that we will study later. We shall then be looking for a pressure feld 푝(푥,푦,푧,푡), a function of space an time. The pressure 푝exerting on an element of fuid can be thought of as the time- and space-average of the perpendicular component of impact force of its molecules on its neighbors. It is strictly a macroscopic property, that is, it cannot be defned at a microscopic level (there is no such thing as the “pressure of a molecule”). In subsonic fows, it is a scalar property, meaning that for a given particle, it is the same in all directions. Pressure efects are explored in further detail in chapter 4 (Efects of pressure). 1.5.3 Shear In the same thought experiment as above, the shearw, noted 휏(greek letter “tau”), expresses the eforts of a force parallel to a surface of interest: 휏≡퐹∥ 퐴 (1/15) where 휏is the shear (N m−2 or Pa); 퐹∥is the component of force parallel to the surface (N); and 퐴is the surface area (m2). 18 Contrary to pressure, shear is not a scalar: it can (and often does) take diferent values in diferent directions: on the fat plate above, it would have two components and could be represented by a vector ⃗ 휏= (휏푥, 휏푦). We will explore shear in further detail in chapter 5 (Efects of shear). 1.6 Basic fow quantities Video: how to deal with the ⟂ symbol when calculating mass fow by Olivier Cleynen (CC-by) A few fuid-fow related quantities can be quantifed easily and are worth listing here. Mass foww is noted ̇ 푚and represents the amount of mass fowing through a chosen surface per unit time. When the velocity across the surface is uniform, it can be quantifed as: ̇ 푚= 휌푉⟂퐴 (1/16) where ̇ 푚 is the mass fow (kg s−1); 휌 is the fuid density (kg m−3); 퐴 is the area of the considered surface (m2); and 푉⟂is the component of velocity perpendicular to the surface (m s−1). Instead of 푉⟂퐴, we might pick 푉퐴⟂: the velocity and the area of a surface perpendicular to that velocity: ̇ 푚= 휌푉퐴⟂ (1/17) where 푉 is the fow speed (m s−1); and 퐴⟂is the area of a surface perpendicular to the fow velocity (m2). Volume foww is noted ̇ and represents the volume of the fuid fowing through a chosen surface per unit time. Much like mass fow, when the velocity is uniform, it is quantifed as: ̇ = 푉⟂퐴= 푉퐴⟂= ̇ 푚 휌 (1/18) where ̇ is the volume fow (m3 s−1). Power to cross a surface is a time rate of energy transfer. The power ̇ 푃pressure necessary to force a mass fow of fuid through a chosen surface at a pressure 푝is: ̇ 푃pressure = ⃗ 퐹pressure ⋅⃗ 푉fuid (1/19) = ̇ 푝 (1/20) = ̇ 푚 휌푝 (1/21) where ̇ 푃pressure is the power required to cross the surface (W); and 푝is the mean pressure at the surface (Pa). If a fuid passes across a volume, the net power ̇ 푃pressure, net required to both enter and leave the volume may be expressed as ̇ 푃pressure, net = ̇ 푚 휌Δ푝 (1/22) where Δ푝is the pressure diference between outlet and inlet. 19 Power to increase temperature is also a time rate of energy transfer. If the temperature of a fuid changes when it fows through a volume, as long as no phase change occurs, the associated power ̇ 푃temperature is: ̇ 푃temperature = ̇ 푚푐fuid Δ푇 (1/23) where ̇ 푃temperature is the power to increase temperature (W); Δ푇 is the temperature change occurring in the fuid (K); and 푐fuid is the specifc heat capacity of the fuid (J K−1 kg−1). The heat capacityw 푐fuid of fuids varies strongly according to the amount of work that they are performing. When no work is performed in a steady fow, the heat capacity is termed 푐푝. In fuids such as liquid water and air, this capacity is almost independent of temperature. As we will see in chapter 2 (Analysis of existing fows with one dimension), fuid fow involves many forms of energy changes. We will learn to combine and compare them progressively. 1.7 Four balance equations Most problems in fuid mechanics are solved by applying basic physical prin-ciples. We write out those principles in the form of four balance equations. 1. Mass balance: The total amount of matter at hand in a given phenomenon must remain constant (since in fuid mechanics, we do not usually consider nuclear reactions). This statement can be expressed as: 푚system = cst d푚system d푡 = 0 (1/24) 2. Balance of linear momentum: The momentum balance equation is a formulation of Newton’s second law,w which states that the net force ⃗ 퐹net ≡Σ ⃗ 퐹applying to any given system is equal to its mass 푚times its acceleration. In fuid mechanics, the most useful formulation of this physical law uses the change in time of the system’s linear momentum,w the quantity 푚⃗ 푉: ⃗ 퐹net = d d푡(푚⃗ 푉) (1/25) 3. Balance of angular momentum: This is a diferent form of Newton’s second law, useful in situations where rotation about an axis, or moments (“twisting” eforts) are ap-plied about an axis. It states that the net moment ⃗ 푀net, X ≡Σ ⃗ 푀X applied on a system about a point X is equal to the time change of its angular momentumw about this same point, the quantity ⃗ 푟∧푚⃗ 푉: ⃗ 푀net, X = d d푡(⃗ 푟∧푚⃗ 푉) (1/26) 4. Balance of energy: This equation, also known as the “frst principle of thermodynamics”,w 20 states that the total amount of energy within an isolated system must remain constant: d퐸isolated system d푡 = 0 (1/27) In special cases, further equations are used to describe other phenomena af-fecting the fuid fow (e.g. chemical reactions, or interaction between phases). In most cases however, the four equations above are the only important equa-tions written in fuid mechanics. We usually apply those balance statements to our problem in either one of two ways: • We may have information about a fow which already exists, and want to calculate how fuid properties change as it fows through the area of interest, and what the related forces are. In that case, we write the equations in an integral form: we will do this in chapters 2 and 3 (Analysis of existing fows). • We may instead wish to predict how the fuid is going to fow through our zone of interest. In order to do this, we need to calculate fow properties in an extensive manner, aiming to obtain vector felds for the velocity and pressure everywhere, at all times. To this efect, we write the equations in a diferential form: we will do this in chapter 6 (Prediction of fuid fows). 1.8 Classifcation of fuid fows As we will see progressively, it is extremely difcult to obtain general so-lutions for fuid fow. Thus, whenever possible or reasonable, simplifying hypothesis are made about the behavior of any particular fow, that allow us to proceed with the analysis and obtain a reasonable, if inexact, specifc solution. It is therefore a habit of fuid dynamicists to classify fuid fows in various categories, which are not necessarily incompatible. When approach-ing a given problem, we typically look out for the following characteristics: Time dependence Flows which do not vary with time are called steady. Steadiness is dependent on the chosen point of view: for example, the air fow around an aircraft in cruise fight is seen to be steady from within the aircraft, but is obviously highly unsteady from the point of view of the air particles directly in the path of the airliner. Steadiness is also dependent on the selection of a suitable time frame: time-variations in a fow may be negligible if the time window used in the study is short enough. Mathematically, fows are steady when either the partial time deriva-tive 휕/휕푡or the total time derivative D/D푡—concepts we will study in chapter 6 (Prediction of fuid fows) — of properties is zero: this consid-erably eases the search for solutions. It is important not to confuse steadiness (the felds of properties remain the same as time passes) with uniformity (the properties are the same everywhere in space). To fnd out whether or not a fow is steady, compare instantaneous representations (photos, measurement readings, vector feld represen-tations, etc.) taken at diferent times. If they are all identical, the fow is steady. There does not exist an analytical method, however, that allows us to predict whether a fow will be steady. 21 Compressibility When the density of the fuid is uniform, the fow is said to be incompressible. The term is treacherous, because it refers to density, not pressure (incompressible fows almost always feature non-uniform pressure felds). To fnd out whether a gas fow is compressible, compute the Mach number (eq. 1/10 p. 16). Below 0,3 the fow is always incompressible. Compressibility efects can be reasonably neglected below [Ma] = 0,6. Unless very specifc phenomena such as phase changes or extreme speeds occur, the fow of liquids is always incompressible. Advice from an expert The word “incompressible” is a really mean false friend: it has nothing to do with pressure 푝, and all with density 휌. Most fows with moderate speeds and powers are incompressible: the density does not change. Pressure will always change in space or even in time if there is a fuid fow, so there is no word for “pressure remains the same” in fuid mechanics. Temperature distribution In a fuid fow, temperature changes can occur due to three phenomena: • Heat transfer from solid bodies; • Changes in pressure and density due to work being performed on the fuid (by moving solid bodies, or by the fuid itself); • Heat created through internal friction within the fuid. To be certain that a fuid fow will have uniform temperature (i.e. whether it is isothermal), therefore, we must take three steps: 1. Quantify the heat transfer from external bodies. If that is zero, the fow is at least adiabatic, meaning there is no heat transfer; 2. Find out whether the fow is incompressible. As long as it is, the density changes are negligibly small and no temperature changes will occur due to compression or expansion of the fuid; 3. Quantify the mechanical energy lost every second by the fuid as it fows through the domain of interest. This power can be either transmitted to a moving part (e.g. a turbine), or dissipated internally through friction, as heat. Because the heat capacity of fuids is generally very high, temper-ature changes due to internal friction are usually negligibly small. This is assessed with an example in exercise 1.7 p. 29. Turbulence One last characteristic that we systematically attempt to iden-tify in fuid fows is turbulencew (or its opposite, laminarity). While laminar fows are generally very smooth and steady, turbulent fows feature multiple, chaotic velocity feld variations in time and space. We shall frst approach the concept of turbulence in chapter 7 (Pipe fows), and study it more formally in chapter 9 (Dealing with turbulence). In the meantime, we can predict whether a fow will become turbulent 22 by using a non-dimensional parameter named the Reynolds number, noted [Re]: [Re] ≡휌푉퐿 휇 (1/28) To fnd out whether a fow will become turbulent, quantify [Re] by using the fuid properties 휌and 휇, a representative fuid speed 푉, and a length 퐿which is representative of the fow domain (for example, the length or width of an obstacle in the fow). If the result is on the order of 104 or more, the fow is very likely to become turbulent over the length 퐿. By contrast, with [Re] of the order of 102 or less, the fow is very likely to remain laminar over this length. This crude quantifcation, of course, deserves more explanation — we will be coming back to the Reynolds number in chapters 7 and following. Advice from an expert The Reynolds number is the best mea-sure of how complex a fow is — how intricate and chaotic the movement of the fuid will be. When they want to compare themselves to their peers (and maybe try to impress them), fuid dynamicists won’t ask about fow speed or mass fow, but instead just ask: “what’s your Reynolds number?” 1.9 Limits of fuid dynamics Fluid dynamics is a complex discipline. It is easy to observe that fows as ordinary as sea waves crashing on a reef, water fowing down a river stream, or air blown into one’s hands, display tremendous geometrical complexity. Even after choosing to describe only the movement of macroscopic fuid particles instead of individual molecules (and thereby avoiding studying thousands of billions of individual movements), we still need to describe a three-dimensional feld of properties (pressure, temperature, etc.), one of which, velocity, is itself three-dimensional. Thus, even before we begin describing the exact problem and a procedure to obtain its solution, we know that the mere description of a solution can have tremendous complexity. Additionally, we have to admit that in practice much progress can be made in the feld of fuid dynamics. For example, our weather forecasts have almost no value beyond one week, and aircraft manufacturers with budgets measured in billions of dollars still make extensive use of wind tunnel models – this despite our staggering continuous rate of progress in computing technology, and many decades of eforts dedicated to analytical fuid dynamics. In our present study of fuid dynamics, therefore, we shall proceed modestly, and will always take care to show the limits of our analysis. 23 1.10 Solved problems Force due to pressure on a plate Figure 1.4: Pressure distribution on a plate: on the left, exerted by water; on the right, exerted by air. Figure CC-0 Olivier Cleynen A large, 3×3 m square aquarium window (fg. 1.4) is subjected to pressure from fuids on each side. On the right side, the atmosphere exerts uniform pressure 푝air as: 푝air = 푝atmosphere (1/29) On the other side, water from the aquarium exerts non-uniform pressure 푝water expressed in Pascals as: 푝water = 1,2 ⋅105 + 9,81 ⋅103 × 푥 (1/30) What is the force resulting from fuid pressure on each side of the plate? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Note: Unfortunately Olivier made an error in this video: the end computation is incorrect! The correct result is 1,2124 MN. The method, numbers, equations etc. are all ok — only the fnal result is afected. Many thanks to the students who double-checked and reported the problem! 24 Power and moment resulting from a force Figure 1.5: A boat seen from above, trav-eling at velocity ⃗ 푉. The wind exerts a force ⃗ 퐹on the boat, a little forward of its center of gravity. Figure CC-0 Olivier Cleynen A sailboat travels at velocity ⃗ 푉 with 푉=1,5 m s−1 (see fg. 1.5). The relative wind comes from the back and from the left; it acts on the sail and results in an aerody-namic force ⃗ 퐹. The magnitude is 퐹=13 kN; the force acts at an angle 휃=30°, 2 m ahead of the center of gravity. What is the power contributed by the wind, and what is the moment exerted by the aerodynamic force about the center of gravity? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 25 26 Problem sheet 1: Basic fow quantities last edited September 3, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 1.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 1.2 Compressibility efects An aircraft is fying in air with density 0,9 kg m−3 and temperature −5 °C. Above which fight speed would you expect the air fow over the wings to become compressible? 1.3 Pressure-induced force A fat, 2 m-by-2 m panel is used as the wall of a swimming pool (fg. 1.6). On the left side, the pressure is uniform at 1 bar. Figure 1.6: Pressure distribution on a fat panel that is part of the wall of a swimming pool Figure CC-0 Olivier Cleynen 1.3.1. What is the pressure force (i.e. the force resulting from the pressure) exerted on the left side of the plate? On the right side of the plate, the water exerts a pressure which is not uniform: it increases with depth. The relation, expressed in pascals, is: 푝water = 1,3 ⋅105 −9,81 ⋅103 × 푧 (1/31) 27 1.3.2. What is the pressure force exerted on the right side of the plate? [Hint: we will explore the required expression in chapter 4 (Efects of pressure) as eq. 4/3 p. 74] 1.4 Shear-induced force A fuid fows over a 3 m by 3 m fat horizontal plate, in the 푥-direction as shown in fg. 1.7. Because of this fow, the plate is subjected to uniform shear 휏푧푥= 1,65 Pa. Figure 1.7: Shear force exerting on a plate Figure CC-0 Olivier Cleynen 1.4.1. What is the shear force applying on the plate? 1.4.2. What would be the shear force if the shear was not uniform, but instead was a function of 푥expressed (in pascals) as 휏푧푥= 1,65 −0,01 × 푥2? [Hint: we will explore the required expression in chapter 5 (Efects of shear) as eq. 5/3 p. 92] 1.5 Speed of sound White P1.87 Isaac Newton measured the speed of sound by timing the interval between observing smoke produced by a cannon blast and the hearing of the detonation. The cannon is shot 8,4 km away from Newton. What is the air temperature if the measured interval is 24,2 s? What is the temperature if the interval is 25,1 s? 1.6 Wind on a truck A truck moves with constant speed ⃗ 푉on a road, with 푉= 50 km h−1. It experiences strong relative wind coming from the back and from the right: this results in an aerodynamic force ⃗ 퐹with 퐹= 5 kN at an angle 휃= 20°, as shown in fg. 1.8. Figure 1.8: Top view of a truck traveling at velocity ⃗ 푉and subject to a aerodynamic force ⃗ 퐹 Figure CC-0 Olivier Cleynen 28 1.6.1. What is the power given by the wind to the truck? The force ⃗ 퐹is applying at a distance 0,8 m behind the center of gravity of the truck. 1.6.2. What are the magnitude and the direction of the moment exerted by the aerody-namic force ⃗ 퐹about the center of gravity? 1.7 Go-faster exhaust pipe The engine exhaust gases of a student’s hot-rod car are fowing quasi-steadily in a cylindrical outlet pipe, whose outlet is slanted at an angle 휃= 25° to improve the good looks of the car and provide the opportunity for an exercise. Figure 1.9: Exhaust gas pipe of a car. The outlet cross-section is at an angle 휃relative to the axis of the pipe. Figure CC-0 Olivier Cleynen Photo cropped, mirrored and edited from an original CC-by-sa by kazandrew2 The outlet velocity is measured at 15 m s−1, and the exhaust gas density is 1,1 kg m−3. The slanted outlet section area 퐴is 420 cm2. 1.7.1. What is the mass fow ̇ 푚through the pipe? 1.7.2. What is the volume fow ̇ of exhaust gases? Because of the shear within the exhaust gases, the fow through the pipe induces a pressure loss of 21 Pa – we will learn to quantify this in chapter 7 (Pipe fows). In these conditions, the specifc heat capacity of the exhaust gases is 푐푝gases = 1 100 J kg−1 K−1. 1.7.3. What is the power required to carry the exhaust gases through the pipe? 1.7.4. What is the gas temperature increase due to the shear in the fow? 29 1.8 Acceleration of a particle Inside a complex, turbulent water fow, we are studying the trajectory of a cubic fuid particle of width 0,1 mm. The particle is accelerating at a rate of 2,5 m s−2. 1.8.1. What is the net force applying to the particle? 1.8.2. In practice, which types of forces could cause it to accelerate? 1.9 Flow classifcations 1.9.1. Can an incompressible fow also be unsteady? 1.9.2. Can a very viscous fuid fow in a turbulent manner? 1.9.3. [more difcult] Can a compressible fow also be isothermal? 1.9.4. Give an example of an isothermal fow, of an unsteady fow, of a compressible fow, and of an incompressible fow. 30 Answers 1.2 If you adopt [Ma] = 0,6 as an upper limit, you will obtain 푉max = 709 km h−1 (eqs. 1/10 & 1/11 p. 16). Note that propellers, fan blades etc. will meet compressiblity efects far sooner. 1.3 1) 퐹left = 400 kN (eq. 1/14 p. 18); 2) 퐹right = 480 kN (eq. 4/3 p. 74). 1.4 1) 퐹1 = 14,85 N (eq. 1/15 p. 18); 2) 퐹2 = 14,58 N (eq. 5/3 p. 92). 1.5 26,7 °C & 5,6 °C. 1.6 1) ̇ 푊= ⃗ 퐹aero ⋅⃗ 푉truck = 65,3 kW. See Appendix A2.1 p. 247 for a short briefng about the dot product of vectors; 2) 푀= \|\|⃗ 푟∧⃗ 퐹aero\|\| = 1 368 N m, ⃗ 푀= ⎛ ⎜ ⎜ ⎝ 0 0 −1 368 ⎞ ⎟ ⎟ ⎠ (points vertically upwards). See Appendix A2.2 p. 248 for a short briefng about the cross product of vectors. 1.7 1) ̇ 푚= 0,2929 kg s−1 (eq. 1/16 p. 19); 2) ̇ = 266,2 L s−1 (eq. 1/18 p. 19); 3) ̇ 푊= 5,59 W (eq. 1/20 p. 19); 4) Δ푇= +0,0174 K (eq. 1/23 p. 20), an illustration of remarks made in §1.8 p. 22 regarding temperature distribution. 1.8 1) 퐹net = 2,5 ⋅10−9 N (eq 1/25 p. 20), such are the orders of magnitude involved in cfd calculations! 2) Only three kinds: forces due to pressure, shear, and gravity. 1.9 1) yes, 2) yes if [Re] is high enough, 3) yes (in very specifc cases such as high pressure changes combined with high heat transfer or high irreversibility, therefore generally no), 4) open the cap of a water bottle and turn it upside down: you have an isothermal, unsteady, incompressible fow. An example of compressible fow could be the expansion in a jet engine nozzle. 31 32 Fluid Dynamics Chapter 2 – Analysis of existing fows with one dimension last edited May 22, 2020 by Olivier Cleynen — 2.1 Motivation 33 2.2 One-dimensional fow problems 33 2.3 Balance of mass 35 2.3.1 Mass balance equation 35 2.3.2 Problems with the mass balance equation 36 2.4 Balance of momentum 37 2.5 Balance of energy 39 2.6 The Bernoulli equation 41 2.6.1 Theory 41 2.6.2 Reality 42 2.7 Solved problems 44 2.8 Problems 45 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 2.1 Motivation In this chapter, we learn to analyze fuid fows for which a lot of information is already available. We want, when confronted to a simple fow (for example, fow entering and leaving a machine), to be able to answer three questions: • What is the mass fow in each inlet and outlet? • What is the force required to move the fow? • What energy transfers are required for this movement? 2.2 One-dimensional fow problems Abstruse Goose #237: how many dimensions are enough for you? by an anonymous artist (CC-by-nc) The method we develop here is called integral analysis, because it involves calculating the overall (integral) efect of the fuid fowing through a consid-ered volume. In this chapter, we consider one-dimensional fows (at least in a loose defnition); we will consider more advanced cases in chapter 3 (Analysis of existing fows with three dimensions). For now, we are interested in fows where four conditions are met: 1. There is a clearly identifed inlet and outlet; 2. At inlet and outlet, the fuid properties are uniform, so that they can be evaluated in bulk (e.g. the inlet has only one velocity, one temperature etc.); 3. There are no signifcant changes in fow direction; 33 4. A lot of information is available about the fuid properties at inlet. Providing that those conditions are met, we can answer the question: what is the net efect of the fuid fow through the considered volume? In order to write useful equations, we need to begin with rigorous defnitions, with the help of fgure 2.1: • We call control volume a certain volume we are interested in. Fluid fows through the control volume. In this chapter, the control volume does not change with time. The fuid fow does not change in time, either (i.e. the fow is steady). The fuid enters and leaves the control volume at a clearly-identifable inlet and outlet. • At a certain instant, the mass of fuid that is inside the control volume is called the system. The system is traveling. At a later point in time, it has moved and deformed. The three equations that we write in this chapter state that basic physical laws apply to the system. They are balance equations (see §1.7 p. 20). Each time, we will express what is happening to the system, as a function of the fuid properties at the inlet and outlet of the control volume. This will allow us to answer three questions: • What is the mass fow entering and leaving the control volume? • What is the force required to move the fow through the control vol-ume? • What energy transfer is required to move the fow through the control volume? Figure 2.1: A control volume within a fow. The system is the amount of mass included within the control volume at a given time. Because mass enters and leaves the control volume, the system is being moved and deformed (bottom). Figure CC-0 Olivier Cleynen 34 2.3 Balance of mass 2.3.1 Mass balance equation How much mass fow is coming in and out of the control volume? We answer this question by writing a mass balance equation. It compares the rate of change of the system’s mass (which by defnition is zero, see eq. 1/24 p. 20), to the fow of mass through the borders of the control volume: d푚sys d푡 = 0 = d d푡푚CV + ̇ 푚net (2/1) the rate of change of the fuid’s mass as it transits = 0 = the rate of change of mass inside the considered volume + the net mass fow at the borders of the considered volume Since we are here interested only in steady fows, d푚CV/ d푡= 0. Furthermore, we have clearly-identifed inlets and outlets allowing us to re-express the net mass fow ̇ 푚net. Equation 2/1 becomes: 0 = Σ ̇ 푚incoming + Σ ̇ 푚outgoing (2/2) time rate of creation or destruction of mass = the sum of incoming mass fows (negative terms) + the sum of outgoing mass fows (positive terms) For steady fow through a fxed considered volume. The sign convention is counter-intuitive: mass fows are negative inwards and positive outwards. For example, in a case where there were two inlets and two outlets, we could write: 0 = ̇ 푚in 1 + ̇ 푚in 2 + ̇ 푚out 1 + ̇ 푚out 2 0 = −\| ̇ 푚\|in 1 −\| ̇ 푚\|in 2 + \| ̇ 푚\|out 1 + \| ̇ 푚\|out 2 We can substitute ̇ 푚= 휌푉⟂퐴(eq. 1/16 p. 19) into the equations above, obtain-ing: 0 = Σ [휌푉⟂퐴]incoming + Σ [휌푉⟂퐴]outgoing (2/3) For steady fow through a fxed considered volume. Looking again at an example case where there were two inlets and two outlets, this equation 2/3 would become: 0 = 휌in 1푉⟂in 1퐴in 1 + 휌in 2푉⟂in 2퐴in 2 + 휌out 1푉⟂out 1퐴out 1 + 휌out 2푉⟂out 2퐴out 2 0 = (휌푉⟂퐴)in 1 + (휌푉⟂퐴)in 2 + (휌푉⟂퐴)out 1 + (휌푉⟂퐴)out 2 0 = −(휌\|푉⟂\|퐴)in 1 − (휌\|푉⟂\|퐴)in 2 + (휌\|푉⟂\|퐴)out 1 + (휌\|푉⟂\|퐴)out 2 For steady fow through a fxed considered volume with two inlets and two outlets. In a simple case where there is only one inlet and one outlet, this last equation can be rewritten as (휌\|푉⟂\|퐴)1 = (휌\|푉⟂\|퐴)2 (2/4) 35 For steady fow through a fxed considered volume with one inlet and one outlet. 2.3.2 Problems with the mass balance equation Video: Dangers associated with the mass balance equation by Olivier Cleynen (CC-by) The equation 2/4 above is interesting, but also treacherous. The best way to mis-use this equation is to draw the conclusion that “if 퐴decreases, then 푉 must increase”. This is only true some of the time, and here are two reasons why: 1. The density 휌may change between inlet and outlet. In low-speed fows without heat transfer, 휌does not vary signifcantly (see §1.8 p. 1.8). But in compressible fows (for example when combustion is involved, or when compressed air is expanded), 휌may vary together with 퐴 and 푉. Typically, in supersonic fows (where [Ma]>1, and the fuid moves faster than the speed of sound), increases in 퐴lead to increases in 푉, because of a decrease in 휌. 2. There is no causal relationship in equation 2/4. In an incompressible fow, it may well be that reducing 퐴2 leads to an increase in 푉2, but nothing guarantees that the product of the two remains constant. In other words, reducing 퐴2 may both increase 푉2 and decrease ̇ 푚. In-creases in velocity are not “for free”: they require force be applied and energy be spent. The mass balance equation cannot account for those phenomena. Advice from an expert Remember the title of the chapter: the tools here are for analyzing existing fows: those for which we can, if needed, gather more information by making mea-surements. If you fnd yourself predicting velocity in a machine you design with just a mass balance equation, then you might quickly fnd yourself making unrealistic assumptions. Immediately check what force and power are required to generate this velocity. For this, you need a momentum balance equation, and an energy balance equation. 36 2.4 Balance of momentum What force is applied to the fuid for it to travel through the control volume? We answer this question by writing a momentum balance equation. It compares the rate of change of the system’s momentum (which by defnition is the net force applying to it, see eq. 1/25 p. 20), to the fow of momentum through the borders of the control volume: d(푚⃗ 푉sys) d푡 = ⃗ 퐹net = d d푡(푚⃗ 푉)CV + ( ̇ 푚⃗ 푉)net (2/5) the rate of change of the fuid’s momentum as it transits = ⃗ 퐹net = the rate of change of momentum within the considered volume + the net fow of momentum through the boundaries of the considered volume Since we are here interested only in steady fows, and we have clearly-identifed inlets and outlets, this becomes: ⃗ 퐹net = Σ ( ̇ 푚⃗ 푉)incoming + Σ ( ̇ 푚⃗ 푉)outgoing (2/6) the vector sum of forces on the fuid = the sum of incoming momentum fows (with negative ̇ 푚terms) + the sum of outgoing momentum fows (with positive ̇ 푚terms) The same convention as above is applied for the sign of the mass fow ̇ 푚. For example, in a case where there were one inlet and one outlet, we would write: ⃗ 퐹net = ( ̇ 푚⃗ 푉)in + ( ̇ 푚⃗ 푉)out ⃗ 퐹net = −(\| ̇ 푚\| ⃗ 푉)in + (\| ̇ 푚\| ⃗ 푉)out For steady fow through a fxed considered volume with one inlet and one outlet. As before, we can substitute ̇ 푚= 휌푉⟂퐴(eq. 1/16 p. 19) into the equations above, obtaining: ⃗ 퐹net on fuid = Σ [휌푉⟂퐴⃗ 푉]incoming + Σ [휌푉⟂퐴⃗ 푉]outgoing (2/7) For steady fow through a fxed considered volume, where 푉⟂is negative inwards, positive outwards. In the example case where there is one inlet and one outlet, we would write: ⃗ 퐹net = (휌푉⟂퐴⃗ 푉)in + (휌푉⟂퐴⃗ 푉)out ⃗ 퐹net = −(휌\|푉⟂\|퐴⃗ 푉)in + (휌\|푉⟂\|퐴⃗ 푉)out To make clear a few things, let us focus on the simple case where a considered volume is traversed by a steady fow with mass fow ̇ 푚, with one inlet (point 1) and one outlet (point 2). The net force ⃗ 퐹net applying on the fuid is ⃗ 퐹net = \| ̇ 푚\| ( ⃗ 푉2 −⃗ 푉1) (2/8) 37 Video: The net force thing in the momentum balance equation by Olivier Cleynen (CC-by) Three remarks can be made about this equation. First, we need to be aware that this is not one, but three equations, one for each dimension. In order to express ⃗ 퐹net, we need to calculate its three components: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 퐹net푥 = \| ̇ 푚\| (푉2푥−푉1푥) 퐹net푦 = \| ̇ 푚\| (푉2푦−푉1푦) 퐹net푧 = \| ̇ 푚\| (푉2푧−푉1푧) (2/9) Second, there are two reasons why we could calculate a non-zero net force on the fuid in equation 2/8, as illustrated in fg. 2.2. 1. Even if ⃗ 푉2 is aligned and in the same direction as ⃗ 푉1, they can be of diferent magnitude. A force is required to accelerate or decelerate the fuid (more precisely, an acceleration or deceleration of the fuid is equivalent to a force); 2. Even if ⃗ 푉2 has the same magnitude as ⃗ 푉1, they can have diferent directions. A force is required to change the direction in which a fow is fowing (or more precisely, a change of direction is equivalent to a force). We will explore these phenomena in greater detail in chapter 3 (Analysis of ex-isting fows with three dimensions). In this current chapter, we are mostly interested in one-dimensional fows, and it will sufce for us to solve equation 2/8 in one suitable direction only, for example, 퐹net푥= \| ̇ 푚\| (푉2푥−푉1푥) (2/10) The fnal remark is that the equation does not describe a cause-efect re-lationship. The net force does not cause the change in velocity any more than the change in velocity causes the net force: they are both equivalent and simultaneous. Similarly, we have no way to know what ⃗ 퐹net is made of. The exact mechanism which adds up to a net force (pressure change, shear applied through a static wall, the movement of a turbine, etc.) is “hidden” in the control volume, and unknown to us. In order to fnd out what happens in the control volume, we need a diferent type of analysis, which we will approach in chapter 6 (Prediction of fuid fows). Figure 2.2: Two reasons can explain why a net force ⃗ 퐹net appears in eq. 2/8. On the left, ⃗ 푉2 and ⃗ 푉1 are aligned, but have diferent lengths. On the right ⃗ 푉2 and ⃗ 푉1 have the same length, but diferent directions. We will look at the second case in chapter 3 (Analysis of existing fows with three dimensions). Figure CC-0 Olivier Cleynen 38 2.5 Balance of energy What power is applied to the fuid for it to travel through the control volume? We answer this question by writing an energy balance equation. It compares the rate of change of the system’s energy, to the fow of energy through the borders of the control volume. For this we prefer to express the energy 퐸as the specifc energy 푒(in J kg−1) multiplied by the mass 푚(kg). The time rate change of 푚푒is measured in watts (1 W ≡1 J s−1). The energy balance equation is then: d푚푒sys d푡 = Σ ( ̇ 푄+ ̇ 푊) = d d푡(푚푒)CV + ( ̇ 푚푒)net (2/11) the rate of change of the fuid’s energy as it transits = the sum of powers as heat and work = the rate of change of energy within the considered volume + the net fow of energy through the boundaries of the considered volume where ̇ 푄is the power transferred as heat (W) ̇ 푊is the power transferred as work (W) Let us examine the terms of this equation. The sum of powers as heat and work can be broken down as three compo-nents: • the net power transferred as heat ̇ 푄net (positive inwards); • the net power transferred as work with moving solid surfaces ̇ 푊surfaces, net (for example, a moving piston, turbine blade, or rotating shaft, positive inwards); • and the net power transferred to and from the fuid by the fuid itself, in order to enter and leave the considered volume. This power is called power to cross a surface (see §1.6 p. 1.6); for each inlet or outlet we have ̇ 푊pressure = −̇ 푚(푝/휌). We can thus write: Σ ( ̇ 푄+ ̇ 푊) = ̇ 푄net + ̇ 푊shaft, net + ̇ 푊pressure (2/12) = ̇ 푄net + ̇ 푊shaft, net −( ̇ 푚푝 휌)net (2/13) Turning now to the specifc energy 푒, we break it down into three components (see also §1.3.3 p. 13): • the specifc internal energy 푖, which represents the energy per unit mass contained as stored heat within the fuid itself. In thermodynam-ics, this is often noted 푢, but in fuid mechanics we reserve this symbol to note the 푥-component of velocity. In a perfect gas, 푖is simply pro-portional to absolute temperature (푖= 푐푣푇), but for other fuids such as water, it cannot be easily measured, and precomputed tables relating 푖 to other properties must be used; • the specifc kinetic energy 푒푘, 푒푘≡1 2푉2 (2/14) 39 • the specifc potential energy 푒푝, related to gravity 푔(m s−2) and altitude 푧(m) as: 푒푝≡푔푧 (2/15) We thus write out specifc energy 푒(in J kg−1) as: 푒≡푖+ 푒푘+ 푒푝 (2/16) Now, we focus on steady fows (for which energy in the control volume does not change with time), and we can come back to eq. 2/11 to rewrite it as: ̇ 푄net + ̇ 푊shaft, net + ̇ 푊presure = ( ̇ 푚푒)net ̇ 푄net + ̇ 푊shaft, net = ( ̇ 푚푒)net + ( ̇ 푚푝 휌)net ̇ 푄net + ̇ 푊shaft, net = [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]net (2/17) Rewriting this into one general, usable form, we obtain: ̇ 푄net + ̇ 푊shaft, net = Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]in +Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]out (2/18) For steady fow through a fxed considered volume, where ̇ 푚= 휌푉⟂퐴is negative inwards, positive outwards. This equation 2/18 is known in thermodynamics as the steady fow energy equation (in thermodynamics, it is usually expressed with the help of the concept of enthalpy ℎ≡푖+ 푝/휌, which we do not use here). As usual, let us focus on a case where there is only one inlet and one outlet. We obtain: ̇ 푄net + ̇ 푊shaft, net = −[\| ̇ 푚\| (푖+ 푝 휌+ 1 2푉2 + 푔푧)]in + [\| ̇ 푚\| (푖+ 푝 휌+ 1 2푉2 + 푔푧)]out ̇ 푄net + ̇ 푊shaft, net = \| ̇ 푚\| [Δ푖+ Δ푝 휌+ Δ ( 1 2푉2) + Δ(푔푧)] (2/19) This equation is very useful in principle, but not so much in practice, for two reasons: 1. It contains a lot of terms. There are fve fuid properties at inlet and outlet which afect energy, and it is difcult to predict which one will be afected by a heat or work transfer. For example, consider a simple water pump with known powers ̇ 푄in and ̇ 푊shaft, in. An efcient pump will generate large increases in 푝(or 푉and 푧), while an inefcient pump will generate large increases in 푖and 1/휌. The energy balance equation, in this form, tells us nothing about how energy input to the control volume is redistributed. 40 Advice from an expert Again, remember the title of the chapter. To cal-culate the value of any one property in equa-tion 2/18, you need to input the value of the eleven other ones. It is tempting to take shortcuts while doing so (“oh, the pressure is probably the same”), with disastrous consequences. There is no solu-tion to this. If you are attempting to predict fuid fow, and are missing information, better stop without a result than take hazardous attempts at using equation 2/18. 2. The terms have disproportionate values in practice. The heat capacity of ordinary fuids is very large, and so 푖is usually hundreds of times larger than the four terms in the brackets of equation 2/18. In water for example, an increase of temperature of 0,1 °C (with the term Δ푖) requires the same energy as increasing its velocity from 30 km h−1 to 110 km h−1 (with the term Δ푒푘). This is not an issue in thermodynamics, where heat, work and temperature are the most important parameters. But in fuid mechanics, velocity is of great interest, and the energy balance is not always useful to predict its changes. Advice from an expert In fuid dynamics, fuid movement usually in-volves relatively small amounts of energy. You can convince yourself of this by dropping milk into a bowl of water: minuscule amounts of potential energy as Δ(푔푧) are converted into an incredibly complex distribution of veloci-ties, before slowly dissipating into internal energy as Δ푖. While in principle, we could calculate pressure drops or velocity changes by measuring temperature diferences (and thus Δ푖), in practice this only works when very high powers are involved, such as in a compressor or in a rocket engine nozzle. For ordinary fow (say, air fow around a car, or water fow in a pipe), the temperature changes are much too small to be measured. See exercise 2.3 p. 46 for an example of this. 2.6 The Bernoulli equation 2.6.1 Theory The Bernoulli equation is the energy equation applied to specifc cases. To derive the Bernoulli equation, we will start from equation 2/18 and add fve constraints: 1. Steady fow. (We had already implemented this restriction, when we set d(푚푒)CV d푡 from eq. 2/11 to zero in order to obtain eq. 2/18.) 41 2. Incompressible fow. Thus, 휌stays constant; 3. No heat or work transfer. Thus, both ̇ 푄net and ̇ 푊shaft, net are zero; 4. No friction. Thus, the fuid internal energy 푖cannot increase; 5. One-dimensional fow. Thus, our considered volume has only one inlet (labeled 1) and one outlet (labeled 2): all fuid particles move together with the same transit time, and the overall trajectory is already known. With these fve restrictions, equation 2/18 simply becomes: 0 + 0 = [ ̇ 푚(푖cst. + 푝 휌cst. + 1 2푉2 + 푔푧)]1 +[ ̇ 푚(푖cst. + 푝 휌cst. + 1 2푉2 + 푔푧)]2 Dividing by \| ̇ 푚\| and canceling 푖cst., as follows, 0 = −(푖cst. + 푝 휌cst. + 1 2푉2 + 푔푧)1 + (푖cst. + 푝 휌cst. + 1 2푉2 + 푔푧)2 0 = −( 푝 휌cst. + 1 2푉2 + 푔푧)1 + ( 푝 휌cst. + 1 2푉2 + 푔푧)2 and multiplying by the (constant and uniform) density 휌, we obtain the Bernoulli equation, with all terms having dimensions of pressure: (푝+ 1 2휌푉2 + 휌푔푧)1 = (푝+ 1 2휌푉2 + 휌푔푧)2 (2/20) This equation describes the properties of a fuid particle in a steady, incom-pressible, friction-less fow with no energy transfer. 2.6.2 Reality Video: The Bernoulli equation will kill you by Olivier Cleynen (CC-by) Let us insist on the incredibly frustrating restrictions brought by the fve conditions above: 1. Steady fow. This constrains us to continuous fows with no transition efects, which is a reasonable limit; 2. Incompressible fow. We cannot use this equation to describe fow in compressors, turbines, difusers, nozzles, nor in fows where 푀> 0,6. 3. No heat or work transfer. We cannot use this equation in a machine (e.g. in pumps, turbines, combustion chambers, coolers). 4. No friction. This is a tragic restriction! We cannot use this equation to describe a turbulent or viscous fow, e.g. near a wall or in a wake. 42 5. One-dimensional fow. This equation is only valid if we know precisely the trajectory of the fuid whose properties are being calculated. Of course, we can overcome those shortcomings by adding one extra (nega-tive) term called “Δ푝loss” to eq. 2/20, which lumps together all of the efects unaccounted for. In this way, we obtain the Bernoulli equation with losses: (푝+ 1 2휌푉2 + 휌푔푧)1 = (푝+ 1 2휌푉2 + 휌푔푧)2 + Δ푝loss (2/21) There are indeed cases where the pressure losses due to the imperfection of the fow are well-understood, and can be easily quantifed. This is true of fow in pipes, for example (we study those in chapter 7). In those cases, eq. 2/21 is extremely useful. Nevertheless, this approach is also easily misused. In a fuid fow where several of the restrictions above do not hold —and many such fows can be found in everyday life as well as engineering applications— equation 2/21 will betray its users. Convince yourself that any wrong equation can be made correct by adding an unknown “bucket” term at the end: for example 2 + 3 = −18 + Δ푝loss. Advice from an expert In case you are not sure whether the Bernoulli equa-tion applies, start from an energy balance equation. Crossing out the terms that do not apply will force you to question their importance (e.g. is heat transfer really negligible? etc.). If you do not come to a conclu-sive end, do not remove terms that are inconvenient. The unfortunate reality is that in fuid mechanics, the energy balance equation contains many terms, with disproportionate values, and using it alone is not enough to solve most practical problems. Advice from an expert Among the fve restrictions listed, the last is the most severe, and the most often forgotten: the Bernoulli equation does not allow us to predict the trajectory of fuid particles. Just like all of the other equations in this chapter, it requires a control volume with a known inlet and a known outlet. If you fnd yourself drawing out fow streamlines and interpreting the result with the Bernoulli equation, you are running astray. The tools you need to do this correctly are waiting for us in chapter 6 (Prediction of fuid fows). 43 2.7 Solved problems Flow in a nozzle Water is fowing through a nozzle, where the diameter decreases gently from 2 m2 to 1 m2. The fow is so smooth that energy dissipation due to wall friction is negligible. The water enters the nozzle with a uniform velocity of 3 m s−1. What is the mass fow? What is the outlet velocity? And what is the pressure change across the pipe? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Flow through a valve Water is fowing through a straight pipe with constant diameter. The mass fow entering the pipe is 2 kg s−1, and it enters the pipe with a uniform velocity of 2 m s−1. In the middle of the pipe length, a valve is installed, which causes the pressure drop: Δ푝valve = −3,5 kPa. What is the outlet velocity? What is the net force on the fuid as it transits? What is the power dissipated as friction? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 44 Problem sheet 2: Analysis of existing fows with one dimension last edited April 15, 2021 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Balance of mass in a fxed control volume with steady fow: 0 = Σ [휌푉⟂퐴]incoming + Σ [휌푉⟂퐴]outgoing (2/3) where 푉⟂is negative inwards, positive outwards. Balance of momentum in a fxed control volume with steady fow: ⃗ 퐹net on fuid = Σ [휌푉⟂퐴⃗ 푉]incoming + Σ [휌푉⟂퐴⃗ 푉]outgoing (2/7) where 푉⟂is negative inwards, positive outwards. Balance of energy in a fxed control volume with steady fow: ̇ 푄net + ̇ 푊shaft, net = Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]in +Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]out (2/18) where ̇ 푚is negative inwards, positive outwards. 2.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 2.2 Pipe expansion without losses Water fows from left to right in a pipe, as shown in fg. 2.3. On the left, the diameter is 8 cm, the water arrives with a uniform velocity of 1,5 m s−1. The diameter increases gently until it reaches 16 cm; the expansion is smooth, so that losses (specifcally, energy losses due to wall friction and fow separation) are negligible. 2.2.1. What are the mass and volume fows at inlet and outlet? 2.2.2. What is the average velocity of the water at the right end of the expansion? 45 Figure 2.3: A simple pipe expansion, with water fowing from left to right. CC-0 Olivier Cleynen 2.2.3. What is the pressure change in the water across the expansion? The volume fow of water in the pipe is now doubled. 2.2.4. What is the new pressure change? The water is drained from the pipe, and instead, air with density 1,225 kg m−3 is fowed in the pipe, incoming with a uniform velocity of 1,5 m s−1. 2.2.5. What is the new pressure change? 2.3 Pipe fow with losses Water fows in a long pipe which has constant diameter; a valve is installed in the middle of the pipe length (fg. 2.4). Water arrives the pipe with a uniform velocity of 1,5 m s−1 and the pipe diameter is 250 mm. The pipe itself and the valve, together, induce a pressure loss which can be quantifed using the dimensionless loss coefcient 퐾valve (we later will later encounter it as eq. 7/6 p. 137). With this tool, the pressure loss Δ푝valve is related to the mean incoming speed 푉incoming as: 퐾valve ≡ \|Δ푝valve\| 1 2휌푉2 incoming = 2,6 (2/22) 2.3.1. What is the outlet velocity of the water? (note: this is a classical “trick” question! :-) 2.3.2. What is the drop in water pressure across the valve? 2.3.3. What is the power dissipated into internal energy? 2.3.4. If the heat losses of the pipe and valve are negligible, what is the temperature increase of the water? Figure 2.4: A simple, straight pipe, featuring a partially-open valve in the center CC-0 Olivier Cleynen 46 2.4 Combustor from a jet engine A jet engine is equipped with several combustors (sometimes also called combustion chambers). We are interested in fuid fow through one such combustor, shown in fg. 2.5. Air from the compressor enters the combustor, is mixed with fuel, and combustion occurs, which greatly increases the temperature and specifc volume of the mix, before it is run through the turbine. The conditions at inlet are as follows: • Air mass fow: 0,5 kg s−1; • Air properties: 25 bar, 1 050 °C, 12 m s−1 • Fuel mass fow: 5 g s−1. At the outlet, the hot gases have pressure 24,5 bar and temperature 1 550 °C, and exit with a speed of 50 m s−1. We assume that the incoming air and outgoing gas have the same thermodynamic properties: 푐v = 718 J kg−1 K−1, 푅air = 287 J kg−1 K−1. 2.4.1. What are the mass fows at inlet and outlet? 2.4.2. What are the volume fows at inlet and at outlet? 2.4.3. What is the power provided to the fow as heat? 2.4.4. What is the net force exerted on the gas as it travels through the combustor? Figure 2.5: A combustor in a sectioned jet engine (here, a Turboméca Adour). Air enters from the left, out of the compressor (whose blades are painted blue). It leaves the combustor on the right side, into the turbine. In the combustor, high-temperature, steady combustion takes place. Photo CC-by-sa Olivier Cleynen 47 2.5 Water jet on a truck A water nozzle shoots water towards the back of a small stationary van. It has a 3 cm2 cross-sectional area, and the water speed at the nozzle outlet is 푉jet = 20 m s−1. As the horizontal water jet hits the back of the van, it is split in two symmetrical vertical fows (fg. 2.6). The two opposite vertical jets have same mass fow and same velocity (푉2 = 푉3 = 20 m s−1). 2.5.1. What is the net force exerted on the water by the truck? 2.5.2. What is the net force exerted on the truck by the water? Now, the truck moves longitudinally in the same direction as the water jet, with a speed 푉truck = 15 m s−1. (This is a crude conceptual setup, which allows us to approach conceptually the case where water acts on the blades of a turbine.) 2.5.3. What is the new force exerted by the water on the truck? 2.5.4. What is the mechanical power transmitted to the truck? 2.5.5. How would the power be modifed if the volume fow was kept constant, but the diameter of the nozzle was reduced? (briefy justify your answer, e.g. in 30 words or less) Figure 2.6: A water jet fowing out of a nozzle (left), and impacting the vertical back surface of a small electric truck, on the right. Figure CC-0 Olivier Cleynen 48 2.6 High-speed gas fow Scientists build a very high-speed wind tunnel. For this, they build a large compressed air tank. Air escapes from the tank into a pipe which decreasing cross-section, as shown in fg. 2.7. The pipe diameter reaches a minimum (at the tunnel throat), and then it expands again, before discharging into the atmosphere. Figure 2.7: A converging-diverging nozzle. Air fows from the left tank to the right outlet, with a contraction in the middle. Figure CC-0 Olivier Cleynen For simplicity, we assume that heat losses through the tunnel walls are negligible, and that the fuid has uniformly-distributed velocity in cross-sections of the pipe. In the tank (point 1), the air is stationary, with pressure 7,8 bar and temperature 246,6 °C. At the throat (point 2), the pressure and temperature have dropped to 4,2 bar and 160 °C. The velocity has reached 417,2 m s−1. The throat cross-section is 0,01 m2. 2.6.1. What is the mass fow through the tunnel? 2.6.2. What is the kinetic energy per unit mass of the air at the throat? Downstream of the throat, the pressure keeps dropping. By the time it reaches a point 3, the air has seen its pressure and temperature drop to 1,38 bar and 43 °C. 2.6.3. What is the fuid velocity at point 3? (if you need to convince yourself that 퐴3 > 퐴1, you may also calculate the cross-section area) 2.6.4. What is the net force exerted on the fuid between the points 2 and 3? 2.6.5. What is the kinetic energy per unit mass of the air at point 3? Once it has passed point 3, the air undergoes complex loss-inducing evolutions (including going through a shock wave, where its properties change very suddenly), before it discharges into the atmosphere (point 4) with pressure 1 bar and temperature 165 °C. 2.6.6. What is the fuid velocity at outlet? 2.6.7. What is the outlet cross-section area? 2.6.8. What is the net force exerted on the fuid between section 3 and the outlet? 2.6.9. What is the kinetic energy per unit mass of the air at the outlet? 49 Answers 2.2 p. 45 2.2.1 At both inlet and outlet, ̇ 푚= 7,53 kg s−1 and ̇ = 7,53 L s−1 2.2.2 푉2 = 0,375 m s−1 2.2.3 Δ푝1→2 = +1 054 Pa 2.2.4 Δ푝3→4 = +4 218 Pa 2.2.5 Δ푝5→6 = +1,29 Pa 2.3 p. 46 2.3.1 푉2 = 푉1 = 1,5 m s−1 by application of the mass balance equation; although a mis-application of the energy equation would suggest otherwise 2.3.2 With eq. 2/22, Δ푝valve = −2 925 Pa 2.3.3 ̇ 푚Δ푖= +215,37 W 2.3.4 With eq. 2/18, Δ푇= +0,7 mK (very small!) 2.4 p. 47 2.4.1 \| ̇ 푚1\| = 0,005 + 0,5 kg s−1 and \| ̇ 푚2\| = 0,505 kg s−1 2.4.2 ̇ 1 = 0,0759 m3 s−1 & ̇ 2 = 0,1078 m3 s−1 (there is no volume balance equa-tion!) 2.4.3 ̇ 푄= +261 kW (using 푉2 = 50 m s−1) 2.4.4 퐹net = +19,25 N (in fow-wise direction) 2.5 p. 48 2.5.1 퐹net on water = −120 N 2.5.2 ⃗ 퐹water/truck = −⃗ 퐹net on water 2.5.3 퐹net on water = −7,5 N 2.5.4 ̇ 푊truck = 112,5 W 2.6 p. 49 2.6.1 With eq. 2/18, 푉2 = 417,2 m s−1, and so ̇ 푚2 = ̇ 푚= 14,1 kg s−1 2.6.2 푒푘2 = 87,03 kJ kg−1 2.6.3 With eq. 2/18, 푉3 = 638,71 m s−1 (you may then calculate 휌3 and obtain 퐴3 > 퐴2 even though 푉3 > 푉2, a classical feature of supersonic fows) 2.6.4 퐹net2→3 = +3,137 kN 2.6.5 푒푘3 = 204,61 kJ kg−1 2.6.6 푉4 = 405 m s−1 2.6.7 휌4 = 0,7952 kg s−1 and so 퐴4 = 0,0438 m2 2.6.8 퐹net3→4 = −3,309 kN (so, against the fow direction) 2.6.9 푒푘4 = 82,01 kJ kg−1 50 Fluid Dynamics Chapter 3 – Analysis of existing fows with three dimensions last edited May 22, 2020 by Olivier Cleynen — 3.1 Motivation 51 3.2 The Reynolds transport theorem 51 3.2.1 Control volume 51 3.2.2 Rate of change of an additive property 52 3.3 Balance of mass 54 3.4 Balance of momentum 56 3.5 Balance of angular momentum 57 3.6 Balance of energy 59 3.7 Limits of integral analysis 59 3.8 Solved problems 60 3.9 Problems 63 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 3.1 Motivation In this chapter, we use the same tools that we developed in chapter 2, but we improve them so we can apply them to more complex cases. Specifcally, we would like to answer the following questions: 1. What are the mass fows and forces involved when a fow has non-uniform velocity? 2. What are the forces and moments involved when a fow changes direc-tion? 3.2 The Reynolds transport theorem 3.2.1 Control volume Let us begin, this time, by building a control volume in any arbitrary fow: we are no longer limited to one-inlet, one-oulet steady-fow situations. Instead, we will write equations that work inside any generic velocity feld ⃗ 푉= (푢, 푣, 푤) which is a function of space and time: ⃗ 푉= 푓(푥, 푦, 푧, 푡). Within this fow, we draw an arbitrary volume named control volume (CV) which is free to move and change shape (fg. 3.1). We are going to measure the properties of the fuid at the borders of this volume, which we call the control surface (CS), in order to compute the net efect of the fow through the volume. At a given time, the control volume contains a certain amount of mass which we call the system (sys). Thus the system is a fxed amount of mass transiting 51 Figure 3.1: A control volume within an arbitrary fow (compare with fgure 2.1 p. 34). The system is the amount of mass included within the control volume at a given time. At a later time (bottom), it may have left the control volume, and its shape and properties may have changed. The control volume may also change shape with time, although this is not represented here. Figure CC-0 Olivier Cleynen through the control volume at the exact time when we write the equation, and its properties (volume, pressure, velocity etc.) may change in the process. All along the chapter, we are focusing on the question: based on measured fuid properties at some chosen area in space and time (the properties at the control surface), how can we quantify what is happening to the system (the mass inside the control volume)? 3.2.2 Rate of change of an additive property In order to proceed with our calculations, we need to make our accounting methodology even more robust. We start with a “dummy” fuid property 퐵, which we will later replace with physical variables of interest. Let us therefore consider an arbitrary additive property 퐵of the fuid. By the term additive property, we mean that the total amount of property is divided if the fuid is divided. For instance, this works for properties such as mass, volume, energy, entropy, but not pressure or temperature. The specifc (i.e. per unit mass) value of 퐵is designated 푏≡퐵/푚. We now want to compute the variation of a system’s property 퐵based on measurements made at the borders of the control volume. We will achieve this with an equation containing three terms: • The time variation of the quantity 퐵within the system is measured with the term d퐵sys/ d푡. We will use this term to represent, for example, the rate of change of the fuid’s energy or momentum as it travels through a jet engine. 52 • Within the control volume, the enclosed quantity 퐵CV can vary by accumulation (for example, mass may be increasing in an air tank fed with compressed air): we measure this with the term d퐵CV/ d푡. • Finally, a mass fux may be fowing through the boundaries of the control volume, carrying with it some amount of 퐵every second: we write that net fow out of the system as ̇ 퐵net ≡̇ 퐵out −̇ 퐵in. We can now link these three terms with the simple equation: d퐵sys d푡 = d퐵CV d푡 + ̇ 퐵net (3/1) the rate of change of the fuid’s 퐵 as it transits = the rate of change of 퐵inside considered volume + the net fow of 퐵 at the borders of the considered volume Since 퐵may not be uniformly distributed within the control volume, we like to express the term 퐵CV as the integral of the volume density 퐵/with respect to volume : 퐵CV = ∭ CV 퐵 d= ∭ CV 휌푏d (3/2) d퐵CV d푡 = d d푡∭ CV 휌푏d (3/3) where CV is the control volume, and  is volume (m3). The second term of eq. 3/1, ̇ 퐵net, can be evaluated by quantifying, for each area element d퐴of the control volume’s surface, the surface fow rate 휌푏푉⟂ of property 퐵that fows through it, as shown in fg. 3.2. The integral over the entire control volume surface CS of this term is: ̇ 퐵net = ∬ CS 휌푏푉⟂d퐴= ∬ CS 휌푏( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/4) where CS is the control surface (enclosing the control volume CV), ⃗ 푛 is a unit vector on each surface element d퐴pointing outwards, ⃗ 푉rel is the local velocity of fuid relative to the control surface, and 푉⟂≡⃗ 푉rel ⋅⃗ 푛is the local cross-surface speed (positive outwards, neg. inwards) Figure 3.2: Part of the system may be fowing through an arbitrary piece of the control surface with area d퐴. The ⃗ 푛vector defnes the orientation of d퐴surface, and by convention is always pointed outwards. The amount of 퐵fowing through this small area per unit time is d퐴is 휌푉⟂d퐴푏 Figure CC-0 Olivier Cleynen 53 By inserting equations 3/3 and 3/4 into equation 3/1, we obtain: d퐵sys d푡 = d d푡∭ CV 휌푏d+ ∬ CS 휌푏( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/5) Equation 3/5 is named the Reynolds transport theoremw; it stands now as a general, abstract accounting tool, but as we soon replace 퐵by meaningful variables, it will prove extremely useful, allowing us to quantify the net efect of the fow of a system through a volume for which border properties are known. In the following sections we are going to use this equation to write out four key balance equations (see §1.7 p. 20): • balance of mass; • balance of linear momentum; • balance of angular momentum; • balance of energy. Advice from an expert Take some time to observe the (sometimes curious!) terms of equation 3/5, and learn the associated vocab-ulary. Just like an accountant in a company would like to have a very clear method for counting how money is spent and earned, fuid dynamicists need very good tools to describe what’s coming in and out of their fows. When you’ll be hurling three-dimensional vector operations at swirling fows down this chapter, you’ll be glad you learned about control surfaces and sign conventions earlier on. 3.3 Balance of mass What is the balance of mass for fuid fowing through any arbitrary volume? We answer this question by writing out a mass balance equation in the template provided by the Reynolds transport theorem (eq. 3/5). We now state that the placeholder variable 퐵is mass 푚. It follows that d퐵/ d푡becomes d푚sys/ d푡, which by defnition is zero (see eq. 1/24 p. 20). Also, 푏≡퐵/푚= 푚/푚= 1 and now the Reynolds transport theorem becomes: d푚sys d푡 = 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/6) the rate of change of the fuid’s mass as it transits = 0 = the rate of change of mass inside considered volume + the net mass fow at the borders of the considered volume 54 Video: with sufcient skills (and lots of practice!), it is possible for a musician to produce an un-interrupted stream of air into an instrument while still con-tinuing to breathe, a technique called circular breathing.w Can you identify the diferent terms of eq. 3/8 as they apply to the saxophonist’s mouth? by David Hernando Vitores (CC-by-sa) This equation 3/6 is often called continuity equation. It allows us to compare the incoming and outgoing mass fows through the borders of the control volume. Sometimes, the control volume has well-defned inlets and outlets through which the volume fow 휌( ⃗ 푉rel ⋅⃗ 푛) is uniform: this is illustrated in fgure 3.3. In that case equation 3/6 reduces to forms that we have already identifed in the previous chapter (see §2.3 p. 35): 0 = d d푡∭ CV 휌d+ ∑ out {휌푉⟂퐴} + ∑ in {휌푉⟂퐴} (3/7) = d d푡∭ CV 휌d+ ∑ out {휌\|푉⟂\|퐴} −∑ in {휌\|푉⟂\|퐴} 0 = d d푡푚CV + ∑ out {\| ̇ 푚\|} −∑ in {\| ̇ 푚\|} (3/8) Figure 3.3: A control volume for which the system’s properties are uniform at each inlet and outlet. Here eq. 3/6 translates as 0 = d d푡∭CV 휌d+ 휌3\|푉⟂3\|퐴3 + 휌2\|푉⟂2\|퐴2 − 휌1\|푉⟂1\|퐴1. Figure CC-0 Olivier Cleynen As before, in equation 3/7, the term 휌푉⟂퐴at each inlet or outlet corresponds to the local mass fow ± ̇ 푚(negative inwards, positive outwards) through the boundary. Advice from an expert If you are thinking that equation 3/6 is just a very fancy way to write the “good” mass balance equation we al-ready wrote in chapter 2 (eq. 2/3 p. 35), you are not completely wrong. There are two useful improvements, however: First, having ∬휌푉⟂d퐴instead of just 휌푉⟂퐴allows us to handle cases where the incoming/outgoing velocities are not uniform. Outside of fuid mechanics textbooks, very few fows have a nice smooth uniform outlet! Second, there is an unsteady term d푚CV/ d푡. It is not used to solve prob-lems in this course, but one day when you are confronted to a case where your inlet and outlet mass fows are not equal, it will save your day! 55 3.4 Balance of momentum Abstruse Goose #338: Newton’s laws of motion almost didn’t happen by an anonymous artist (CC-by-nc) What force is applied to the fuid for it to travel through the control volume? We answer this question by writing out a mass balance equation in the template provided by the Reynolds transport theorem (eq. 3/5). We now state that the placeholder variable 퐵is momentum 푚⃗ 푉. It follows that d퐵/ d푡becomes d푚⃗ 푉sys)/ d푡, which by defnition is the net force ⃗ 퐹net applying to the system (see eq. 1/25 p. 20). Also, 푏≡퐵/푚= 푚⃗ 푉/푚= ⃗ 푉and now the Reynolds transport theorem becomes: d(푚⃗ 푉sys) d푡 = ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d + ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/9) the vector sum of forces on the fuid = the rate of change of momentum within the considered volume + the net fow of momentum through the boundaries of the considered volume Video: Making sense of the 3D linear momentum balance equa-tion by Olivier Cleynen (CC-by) Sometimes, the control volume has well-defned inlets and outlets through which the fow 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) is uniform: this is illustrated again in fgure 3.4. In that case equation 3/9 reduces to forms that we have already identifed in the previous chapter (see §2.4 p. 37): ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∑ out { (휌\|푉⟂\|퐴) ⃗ 푉 } −∑ in { (휌\|푉⟂\|퐴) ⃗ 푉 } (3/10) = d d푡(푚⃗ 푉)CV + ∑ out { \| ̇ 푚\| ⃗ 푉 } −∑ in { \| ̇ 푚\| ⃗ 푉 } (3/11) Figure 3.4: The same control volume as in fg. 3.3. Here, since the system’s properties are uniform at each inlet and outlet, eq. 3/9 translates as ⃗ 퐹net = d d푡∭CV 휌⃗ 푉d+ 휌3\|푉⟂3\|퐴3 ⃗ 푉3 + 휌2\|푉⟂2\|퐴2 ⃗ 푉2 −휌1\|푉⟂1\|퐴1 ⃗ 푉1. Figure CC-0 Olivier Cleynen 56 Video: as a person walks, the de-fection of the air passing around their body can be used to sustain the fight of a paper airplane (a walkalong gliderw). Can you fg-ure out the momentum fow en-tering and leaving a control vol-ume surrounding the glider, and the resulting net force? by Y:sciencetoymaker (styl) To make clear a few things, let us focus on the simple case where a con-sidered volume has only one inlet (point 1) and one outlet (point 2). From equation 3/9, the net force ⃗ 퐹net applying on the fuid is: ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬휌2\|푉⟂2\| ⃗ 푉2 d퐴2 −∬휌1\|푉⟂1\| ⃗ 푉1 d퐴1 (3/12) In this equation 3/12, what could cause ⃗ 퐹net to be non-zero? • The frst term, which we could informally write as d(푚⃗ 푉)CV/ d푡, could be non-zero. This happens when the momentum inside the control volume changes. This may occur if the distribution of velocities ⃗ 푉 within the control volume is changing, such as when the fuid in a tank sloshes back and forth against the walls. • The sum of the last two terms, which we could informally write as \| ̇ 푚\|2 ⃗ 푉2 −\| ̇ 푚\|1 ⃗ 푉1, could also be non-zero. This happens when the fux of momentum entering the control volume is diferent from the one leaving it: – It may be because the mass fow ̇ 푚is diferent at inlet and outlet, even if the two velocity distributions are the same; – It may be because the velocities have diferent length, and the fow is speeding up or slowing down; – It may be because the velocities are aligned diferently, and the fow is changing directions; – It may be because the velocities are non-uniform and distributed diferently, even if their average (and thus the mass fows) are the same. As you can see, a lot of diferent things may be happening at once! We will study (separately) the most relevant of those efects in the problem section of this chapter. 3.5 Balance of angular momentum Video: rocket landing gone wrong. Can you compute the moment exerted by the top thruster around the base of the rocket as it (unsuccessfully) at-tempts to compensate for the collapsed landing leg? by Y:SciNews (styl) What moment (“twisting efort”) is applied to the fuid for it to travel through the control volume? We answer this question by writing an angular momen-tum balance (see eq. 1/26 p. 20) in the template provided by the Reynolds transport theorem (eq. 3/5). We position ourselves at a point 푋, about which we measure all moments. All positions are measured with a position vector ⃗ 푟X푚. We now state that the placeholder variable 퐵is angular momentum ⃗ 푟X푚∧푚⃗ 푉. It follows that d퐵/ d푡becomes d⃗ 푟X푚∧푚⃗ 푉sys/ d푡, which by defnition is the net moment ⃗ 푀net applying to the system (see again eq. 1/26 p. 20). Also, 푏≡퐵/푚= ⃗ 푟X푚∧⃗ 푉and now the Reynolds transport theorem becomes: 57 d(⃗ 푟X푚∧푚⃗ 푉)sys d푡 = ⃗ 푀net,X = d d푡∭ CV ⃗ 푟X푚∧휌⃗ 푉d+∬ CS ⃗ 푟X푚∧휌( ⃗ 푉rel ⋅⃗ 푛) ⃗ 푉d퐴 (3/13) the vector sum of moments on the fuid = the rate of change of the angular momentum within the considered volume + the net fow of angular momentum through the boundaries of the considered volume in which ⃗ 푟X푚is a vector giving the position of any mass 푚relative to point 푋. Video: Making sense of the an-gular momentum balance equa-tion by Olivier Cleynen (CC-by) Sometimes, the control volume has well-defned inlets and outlets through which the fow ⃗ 푟X푚∧휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) is uniform: this is illustrated in fgure 3.5. In that case equation 3/13 reduces to a more readable form: ⃗ 푀net,X = d d푡∭ CV ⃗ 푟X푚∧휌⃗ 푉d+ ∑ out { ⃗ 푟X푚∧\| ̇ 푚\| ⃗ 푉 } −∑ in { ⃗ 푟X푚∧\| ̇ 푚\| ⃗ 푉 } (3/14) Figure 3.5: A control volume for which the properties of the system are uniform at each inlet or outlet. Here the moment about point X in the bottom right is ⃗ 푀net,X = d d푡∭CV ⃗ 푟X푚∧휌⃗ 푉d+ ⃗ 푟2 ∧\| ̇ 푚2\| ⃗ 푉2 −⃗ 푟1 ∧\| ̇ 푚1\| ⃗ 푉1. Figure CC-0 Olivier Cleynen The same remarks we made for linear momentum earlier apply here: there are many possible phenomena which may equate to a moment on the fuid. We will explore this equation rather shyly in the problem section of the chapter. Advice from an expert The angular momentum balance equation is useful in cases where we attempt to balance a machine using fuid fows. Moments can be added and subtracted as vectors, just like forces. It’s an extra layer of ab-straction to learn (just about every fuid dynamicist has sufered at frst with the clockwise-is-positive convention). You’ll be glad you learned it when you attempt to prevent the helicopter from spinning un-controllably — or when you ask your cfd software to calculate a moment, and it spits out a vector! 58 3.6 Balance of energy What power is applied to the fuid for it to travel through the control volume? We can answer this question by writing an energy balance in the template provided by the Reynolds transport theorem (eq. 3/5). The derivation of the equation we look for is the same as the derivation of equation 2/18 p. 40 in the last chapter. We may therefore directly jump to the result: d퐸sys d푡 = ̇ 푄net + ̇ 푊shaft, net = d d푡∭ CV 휌푒푓d+ ∬ CS 휌푒푓( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/15) with the term 푒푓carrying all of the energy terms relevant for us in fuid mechanics: 푒푓≡푖+ 푝 휌+ 1 2푉2 + 푔푧 (3/16) It can be seen here that this equation only difers from equation 2/18 in the last chapter in the two following ways: • There is an unsteady term (frst term on the right-hand side), account-ing for the accumulation or depletion of energy within the control volume; • The energy fows through the inlet and outlet are expressed as integrals, allowing us to account for non-uniform distributions. Save for those two diferences, the equation is not any diferent — and, it must be admitted, not much more useful in practice. In this course, we will not be using this equation to solve problems. 3.7 Limits of integral analysis Integral analysis is an incredibly useful tool in fuid dynamics: in any given problem, it allows us to rapidly describe and calculate the main fuid phenom-ena at hand. The net force exerted on the fuid as it is defected downwards by a helicopter, for example, can be calculated using just a loosely-drawn control volume and a single vector equation. As we progress through the end-of-chapter problems, however, the limits of this method slowly become apparent. There are two of them: • First, we are confned to calculating the net efect of fuid fow. The net force, for example, encompasses the integral efect of all forces —due to pressure, shear, and gravity— applied on the fuid as it transits through the control volume. Integral analysis gives us absolutely no way of distinguishing between those sub-components. In order to do that (for example, to calculate which part of a pump’s mechanical power is lost to internal viscous efects), we would need to look within the control volume. 59 • Second, all four of our equations in this chapter only work in one direction. The value d퐵sys/ d푡of any fnite integral cannot be used to fnd which function 휌푏푉⟂d퐴was integrated over the control surface to obtain it. For example, there are an infnite number of velocity profles which will result in a net force of +12 N. Knowing the net value of an integral, we cannot deduce the conditions which lead to it. In practice, this is a major limitation on the use of integral analysis, because it confnes us to working with large swaths of experimental data gathered at the borders of our control volumes. From the wake below the helicopter, we deduce the net force; but the net force tells us nothing about the shape of the wake. Clearly, in order to overcome these limitations, we are going to need to open up the control volume, and look at the details of the fow within — perhaps by dividing it into a myriad of sub-control volumes. This is what we set ourselves to in chapter 6 (Prediction of fuid fows), with a thundering and formidable methodology we shall call derivative analysis. 3.8 Solved problems Flow through pipe bend A chemical is fowing through a pipe with 90° bend, with a mass fow of 200 kg s−1. Its incoming velocity (uniform) is 2 m s−1, and its outgoing velocity is 3 m s−1. What is the net force exerting on the fuid? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 60 Outlet with a non-uniform velocity Water is fowing through a straight rectangular pipe. At inlet, its velocity is uniform, with 푉1 = 20 m s−1. At the outlet, the velocity is the same on average; however, it is not uniformly distributed. We have 푉2 = 19 + 푦 (m s−1), with 푦the vertical coordinate (in meters). The width of the pipe in the 푧-direction is Δ푧= 1 m. What is the net force exerting on the fuid? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Force and power on a moving turbine blade In a jet engine, a row of blades is moving with great speed (푉blade = 442,4 m s−1). Each blade receives 1,7 kg s−1 of high-speed, high-temperature, high-pressure gas. Through both the movement and the shape of the blade, this gas is deviated and their properties change. For simplicity, we assume the properties at inlet and outlet are uniform. The incoming and outgoing velocities are described below two times: once (left) from the point of view of the stationary engine, and once (right) from the point of view of the moving blades. What is the net force exerting on each blade? What is the power trans-mitted to the blade? 61 See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Mass fows in a boundary layer A fat plate is positioned parallel to a steady, uniform air fow (푢1 = 푈= 25 m s−1). At the trailing edge of the plate, the velocity 푢2 is measured. It is a function of height 푦, with thickness 훿= 2 cm, following the distribution: 푢2 = 푈( 푦 훿) 1 6 (3/17) The plate has length 퐿1 = 50 cm (in 푥-direction, along the fow) and width 퐿2 = 80 cm (in 푧-direction, across the fow). The density of air is 1,225 kg m−3. The layer has thickness 훿at outlet. What is the inlet height of a control volume that has the mass fow of this layer? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Shear force in a boundary layer In the fat plate problem from above, what is the net force exerted on the air by the plate? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 62 Problem sheet 3: Analysis of existing fows with three dimensions last edited May 15, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Mass balance through an arbitrary volume: 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/6) Momentum balance through an arbitrary volume: ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 (3/9) Angular momentum balance through an arbitrary volume: ⃗ 푀net,X = d d푡∭ CV ⃗ 푟X푚∧휌⃗ 푉d+ ∬ CS ⃗ 푟X푚∧휌( ⃗ 푉rel ⋅⃗ 푛) ⃗ 푉d퐴 (3/13) 3.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. (the quiz for chapter 3 will open from May 10 to May 20) 3.2 Pipe bend CC-0 Olivier Cleynen A pipe with diameter 30 mm has a bend with angle 휃= 130°, as shown in fg. 3.6. Wa-ter enters and leaves the pipe with the same speed 푉1 = 푉2 = 1,5 m s−1. The velocity distribution at both inlet and outlet is uniform. 3.2.1. What is the mass fow traveling through the pipe? 3.2.2. What is the force exerted by the pipe bend on the water? 3.2.3. Represent the force vector qualitatively (i.e. without numerical data). 3.2.4. What would be the new force if all of the speeds were doubled? 63 Figure 3.6: A pipe bend, through which water is fowing. We assume that the velocity distribution at inlet and outlet is identical. Figure CC-0 Olivier Cleynen 3.3 Exhaust gas defector A defector is used behind a stationary aircraft during ground testing of a jet engine (fg. 3.7). Figure 3.7: A mobile exhaust gas defector, used to defect hot jet engine exhaust gases upwards during ground tests. Figure CC-0 Olivier Cleynen The defector is fed with a horizontal air jet with a quasi-uniform velocity profle; the speed is 푉jet = 600 km h−1, temperature 400 °C and the pressure is atmospheric. As the exhaust gases travel through the pipe, their heat losses are negligible. Gases are rejected with a 40° angle relative to the horizontal. The inlet diameter is 1 m and the horizontal outlet surface is 6 m2. 3.3.1. What is the force exerted on the ground by the defection of the exhaust gases? 3.3.2. Describe qualitatively (i.e. without numerical data) a modifcation to the defec-tor that would reduce the horizontal component of force. 3.3.3. What would the force be if the defector traveled rearwards (positive 푥-direction) with a velocity of 10 m s−1? 3.4 Pelton water turbine White P3.56 A water turbine is modeled as the following system: a water jet exiting a stationary nozzle hits a blade which is mounted on a rotor (fg. 3.8). In the ideal case, viscous efects can be neglected, and the water jet is defected entirely with a 180° angle. The nozzle has a cross-section diameter of 5 cm and produces a water jet with a speed 푉jet = 15 m s−1. The rotor diameter is 2 m and the blade height is negligibly small. We frst study the case in which the rotor is prevented from rotating, so that the blade is stationary (푉blade = 0 m s−1). 64 Figure 3.8: Schematic drawing of a water turbine blade. This type of turbine is called Pelton turbine. Figure CC-0 Olivier Cleynen 3.4.1. What is the force exerted by the water on the blade? 3.4.2. What is the moment exerted by the blade around the rotor axis? 3.4.3. What is the power transmitted to the rotor? We now let the rotor rotate freely. Friction losses are negligible, and it accelerates until it reaches maximum velocity. 3.4.4. What is the rotor rotation speed? 3.4.5. What is the power transmitted to the rotor? The rotor is now coupled to an electrical generator. 3.4.6. Show that the maximum power that can be transmitted to the generator occurs for 푉blade = 1 3푉water. 3.4.7. What is the maximum power that can be transmitted to the generator? 3.4.8. How would the above result change if viscous efects were taken into account? (briefy justify your answer, e.g. in 30 words or less) 3.5 Snow plow derived from Gerhart & Gross Ex5.9 A road-based snow plow (fg. 3.9) is clearing up the snow on a fat surface. We wish to quantify the power required for its operation. The snow plow is advancing at 25 km h−1; its blade has a frontal-view width of 4 m. The snow on the ground is 30 cm deep and has density 300 kg m−3. The snow is pushed along the blade and is rejected horizontally with a 30° angle to the left of the plow. Its density has then risen to 450 kg m−3. The cross-section area 퐴outlet of the outfowing snow in the 푥-푦plane is 1,1 m2. 3.5.1. What is the force exerted on the blade by the defection of the snow? (Indicate its magnitude and coordinates) 3.5.2. What is the power required for the operation of the snow plow? 3.5.3. If the plow velocity was increased by 10 %, what would be the increase in power? 65 Figure 3.9: Outline schematic of a blade snow plow. Figure CC-0 Olivier Cleynen 3.6 Inlet of a pipe Based on White Water is circulated inside a cylindrical pipe with diameter 1 m (fg. 3.10). At the entrance of the pipe, the speed is uniform: 푢1 = 푈av. = 5 m s−1. Shear applies on the fuid from the walls, where the velocity is zero. This strains the fuid particles, and changes the velocity distribution. At the outlet of the pipe, the velocity profle is no longer uniform. It can be modeled as a function of the radius with the relationship: 푢2(푟) = 푈center (1 −푟 푅) 1 7 (3/18) What is the center velocity 푈center at the outlet? 66 Figure 3.10: Velocity profles at the inlet and outlet of a circular pipe. Figure CC-0 Olivier Cleynen 3.7 Drag on a cylindrical profle In order to measure the drag on a cylindrical profle, a cylindrical tube is positioned perpendicular to the air fow in a wind tunnel (fg. 3.11), and the longitudinal component of velocity is measured across the tunnel section. Figure 3.11: A cylinder profle set up in a wind tunnel, with the air fowing from left to right. Figure CC-0 Olivier Cleynen Upstream of the cylinder, the air fow velocity is uniform (푢1 = 푈= 30 m s−1). Downstream of the cylinder, the speed is measured across a 2 m height interval. Hori-zontal speed measurements are gathered and modeled with the following relationship: 푢2(푦) = 29 + 푦2 (3/19) The width of the cylinder (perpendicular to the fow) is 2 m. The Mach number is very low, and the air density remains constant at 휌= 1,23 kg m−3; pressure is uniform all along the measurement feld. 3.7.1. What is the drag force applying on the cylinder? 3.7.2. How would this value change if the fow in the cylinder wake was turbulent, and the function 푢2(푦) above only modeled time-averaged values of the horizontal velocity? (briefy justify your answer, e.g. in 30 words or less) 67 3.8 Drag on a fat plate We wish to measure the drag applying on a thin plate positioned parallel to an air stream. In order to achieve this, measurements of the horizontal velocity 푢are made around the plate (fg. 3.12). Figure 3.12: Side view of a plate positioned parallel to the fow. Figure CC-0 Olivier Cleynen At the leading edge of the plate, the horizontal velocity of the air is uniform: 푢1 = 푈= 10 m s−1. At the trailing edge of the plate, we observe that a thin layer of air has been slowed down by the efect of shear. This layer, called boundary layer, has a thickness of 훿= 1 cm. The horizontal velocity profle can be modeled with the relation: 푢2(푦) = 푈( 푦 훿) 1 7 (3/20) The width of the plate (perpendicular to the fow) is 30 cm and it has negligible thickness. The fow is incompressible (휌= 1,23 kg m−3) and the pressure is uniform. 3.8.1. What is the drag force applying on the plate? 3.8.2. What is the power required to compensate the drag? 3.8.3. Under which form is the kinetic energy lost by the fow carried away? Can this new form of energy be measured? (briefy justify your answer, e.g. in 30 words or less) 68 3.9 Drag measurements in a wind tunnel A group of students proceeds with speed measurements in a wind tunnel. The objective is to measure the drag applying on a wing profle positioned across the tunnel test section (fg. 3.13). Figure 3.13: Wing profle positioned across a wind tunnel. The horizontal velocity distributions upstream and downstream of the profle are also shown. Figure CC-0 Olivier Cleynen Upstream of the profle, the air fow velocity is uniform (푢1 = 푈= 50 m s−1). Downstream of the profle, horizontal velocity measurements are made every 5 cm across the fow; the following results are obtained: vertical position (cm) horizontal speed 푢2 (m s−1) 0 50 5 50 10 49 15 48 20 45 25 41 30 39 35 40 40 43 45 47 50 48 55 50 60 50 The width of the profle (perpendicular to the fow) is 50 cm. The airfow is incompressible (휌= 1,23 kg m−3) and the pressure is uniform across the measurement surface. 3.9.1. What is the drag applying on the profle? 3.9.2. How would the above calculation change if vertical speed measurements were also taken into account? 69 3.10 Moment on gas defector non-examniable We revisit the exhaust gas defector of exercise 3.3 p. 64. Figure 3.14 below shows the defector viewed from the side. The midpoint of the inlet is 2 m above and 5 m behind the wheel labeled “A”, while the center axis of the outlet passes 1,72 m away from it, as represented in fg 3.14 Figure 3.14: Side view of the mobile exhaust gas defector which was shown in fg. 3.7 p. 64 Figure CC-0 Olivier Cleynen What is the moment generated by the gas fow about the axis of the wheel labeled “A”? 3.11 Helicopter tail moment non-examinable In a helicopter, the role of the tail is to counter exactly the moment exerted by the main rotor about the main rotor axis. This is usually done using a tail rotor which is rotating around a horizontal axis. A helicopter, shown in fgure 3.15, is designed to use a tail without a rotor, so as to reduce risks of accidents when landing and taking of. The tail is a long hollow cylindrical tube with two inlets and one outlet. To simplify calculations, we consider that pressure is atmospheric at every inlet and outlet. • inlet A has a cross-section area of 0,2 m2. It contributes hot exhaust gases of density 0,8 kg m−3 and velocity 12 m s−1, aligned with the (푥) axis of the tail; • inlet B contributes 25 kg s−1 of atmospheric air incoming at an angle 훼= 130° relative to the axis of the tail, with a velocity of 3 m s−1. The mix of exhaust gases and atmospheric air is rejected at the tip of the tail (outlet C) with a fxed velocity of 45 m s−1. The angle 휃at which gases are rejected is controlled by the fight computer. 3.11.1. What is the rejection angle 휃required so that the tail generates a moment of +6 kN m around the main rotor (푦) axis? 70 Figure 3.15: Top-view of a helicopter using a rotor-less tail. Figure derived from a fgure CC-by-sa by Commons User:FOX 52 3.11.2. Propose and quantify a modifcation to the tail geometry or operating conditions that would allow the tail to produce no thrust (that is to say, zero force in the 푥-axis), while still generating the same moment. Remark: this system is commercialized by MD Helicopters as the notar. The use of exhaust gases was abandoned, however, a clever use of air circulation around the tail pipe axis contributes to the generated moment; this efect is explored in chapter 11 (Large- and small-scale fows) (§11.3.4 p. 228). 71 Answers 3.2 1) ̇ 푚= 1,0603 kg s−1 2) ⃗ 퐹net = (−0,5681; −1,2184) N 4) The force is quadrupled. 3.3 1) ⃗ 퐹net = (−9,532; +1,479) kN : \|\| ⃗ 퐹net\|\| = 9,646 kN (force on ground is opposite: ⃗ 퐹fuid on pipe = −⃗ 퐹pipe on fuid); 3) \|\| ⃗ 퐹net 2\|\| = 8,525 kN. 3.4 1) 퐹net = (−883,6; 0) N; 2) 푀net 푋= \|퐹net\|푅= 883,6 N m; 3) ̇ 푊rotor = 0 W; 4) 휔= 143,2 rpm (퐹net = 0 N); 5) ̇ 푊rotor = 0 W again; 6) ̇ 푊rotor, max = 1,963 kW @ 푉blade, optimal = 1 3푉water jet. 3.5 1) ̇ 푚= 2 500 kg s−1; 퐹net 푥= +10,07 kN, 퐹net 푧= −12,63 kN (force on blade is opposite); 2) ̇ 푊= ⃗ 퐹net ⋅⃗ 푉plow = 퐹net푥\|푉1\| = 69,94 kW 3) ̇ 푊2 = 1,13 ̇ 푊(+33 %) 3.6 푉center = 1,2245푈 3.7 퐹net푥= 휌퐿∫푆2 (푢2 2 −푈푢2) d푦= −95,78 N. 3.8 퐹net푥= 휌퐿∫ 훿 0 (푢2 (푦) −푈푢(푦)) d푦= −7,175 ⋅10−2 N : ̇ 푊drag = 푈\|퐹net푥\| = 0,718 W. 3.9 퐹net푥≈휌퐿Σ푦[(푢2 2 −푈푢2) δ푦] = −64,8 N. 3.10 Re-use ̇ 푚= 67,76 kg s−1, 푉1 = 166,7 m s−1, 푉2 = 33,94 m s−1 from ex. 3.3. With 푅2⟂푉2 = 1,717 m, plug in numbers in eq. 3/14 p. 58: 푀net = +18,64 kN m in 푧-direction. 3.11 1) Work eq. 3/13 down to scalar equation (in 푦-direction), solve for 휃: 휃= 123,1°. 2) There are multiple solutions which allow both moment and force equations to be solved at the same time. 푟C can be shortened, the fow in C can be split into forward and rearward components, or tilted downwards etc. Reductions in ̇ 푚B or 푉C are also possible, but quantifying them requires solving both equations at once. 72 Fluid Dynamics Chapter 4 – Efects of pressure last edited September 3, 2020 by Olivier Cleynen — 4.1 Motivation 73 4.2 Pressure forces on walls 73 4.2.1 Magnitude of the pressure force 73 4.2.2 Position of the pressure force 74 4.3 Pressure felds in fuids 75 4.3.1 The direction of pressure 75 4.3.2 Pressure on an infnitesimal volume 76 4.4 Special case: pressure in static fuids 79 4.4.1 Forces in static fuids 79 4.4.2 Pressure and depth 80 4.4.3 Buoyancy 82 4.5 Solved problems 83 4.6 Problems 85 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 4.1 Motivation In fuid mechanics, only three types of forces apply to fuid particles: forces due to gravity, pressure, and shear. This chapter focuses on pressure (we will address shear in chapter 5), and should allow us to answer two questions: • How is the efect of pressure described and quantifed? • What are the pressure forces generated on walls by static fuids? 4.2 Pressure forces on walls 4.2.1 Magnitude of the pressure force Video: when pressure-induced forces in static fuids matter: 24 hours of heavy tonnage transit through the Mirafores locks in Panama. Can you quantify the force applying on a single lock door? by Y:Pancho507 (styl) What is the force with which a fuid pushes against a wall? When the pressure 푝exerted is uniform and the wall is fat, the resulting force 퐹is easily calculated: 퐹pressure = 푝uniform 푆fat wall (4/1) When the fuid pressure 푝is not uniform (for example, as depicted on the right side of the wall in fgure 4.1), the situation is more complex: the force must be obtained by integration. The surface is split in infnitesimal portions of area d푆, and the corresponding forces are summed up as: 퐹pressure = ∫ 푆 d퐹pressure = ∫ 푆 푝d푆 (4/2) for a fat surface. 73 Figure 4.1: Pressure distribution on a fat plate. We already studied this situation in chapter 1, problem 1.3 p. 27. Figure CC-0 Olivier Cleynen where the 푆-integral denotes an integration over the entire surface. What is required to calculate the scalar 퐹in eq. 4/2 is an expression of 푝as a function of 푆. In a static fuid, this expression will often be easy to fnd, as we see later on. Typically, in two dimensions 푥and 푦we re-write d푆as d푆= d푥d푦and we may then proceed with the calculation starting from 퐹pressure = ∬푝(푥,푦) d푥d푦 (4/3) The above equations work only for a fat surface. When we consider a two-or three-dimensional object immersed in a fuid with non-uniform pressure, the integration must be carried out with vectors. ⃗ 퐹pressure = ∫ 푆 d ⃗ 퐹= ∫ 푆 푝⃗ 푛d푆 (4/4) where the 푆-integral denotes an integration over the entire surface; and ⃗ 푛is a unit vector describing, on each infnitesimal surface element d푆, the direction normal to the surface. 4.2.2 Position of the pressure force We are often interested not only in the magnitude of the pressure force, but also its position. This position can be evaluated by calculating the magnitude of the moment generated by the pressure forces about any chosen point X. This moment ⃗ 푀X, using notation shown in fg. 4.2, is expressed as: ⃗ 푀X = ∫ 푆 d ⃗ 푀X = ∫ 푆 ⃗ 푟X퐹∧d ⃗ 퐹= ∫ 푆 ⃗ 푟X퐹∧푝⃗ 푛d푆 (4/5) where ⃗ 푟X퐹is a vector expressing the position of each infnitesimal surface relative to point X. Much like eq. 4/4 above, this eq. 4/5 is easily implemented in a software algorithm but not very approachable on paper. In this course however, we want to study the simple case where the surface is fat, and where the reference point X is in the same plane as the surface. Equation 4/5 is then a 74 Figure 4.2: Moment generated about an arbitrary point X by the pressure exerted on an arbitrary surface (left: perspective view; right: side view). The vector ⃗ 푛is a convention unit vector everywhere perpendicular to the infnitesimal surface d푆 considered. Figure CC-0 Olivier Cleynen great deal simpler, and we can calculate the magnitude 푀X as: 푀X = ∫ 푆 d푀X = ∫ 푆 푟X퐹d퐹= ∫ 푆 푟X퐹푝d푆 (4/6) for a fat surface, with X in the plane of the surface. Once both 퐹pressure and 푀X pressure have been quantifed, the distance 푅X퐹be-tween point X and the application point of the net pressure force is easily computed: 푅X퐹= 푀X pressure 퐹pressure (4/7) 4.3 Pressure felds in fuids We approached the concept of pressure in chapter 1 (Basic fow quantities) with the notion that it represented force perpendicular to a given fat surface (eq. 1/14), for example a fat plate of area 퐴: 푝≡퐹⟂ 퐴 (4/8) To appreciate the concept of pressure in fuid mechanics, we need to go beyond this equation. 4.3.1 The direction of pressure An important concept is that in continuum mechanics, the fat surface is imaginary. More precisely, a fuid is able to exert pressure not only on solid 75 surfaces, but also upon and within itself. In this context, we need to rework eq. 4/8 so that now pressure is defned as perpendicular force per area on an infnitesimally small surface of fuid: 푝≡lim 퐴→0 퐹⟂ 퐴 (4/9) Video: Two weird things about pressure in fuid dynamics by Olivier Cleynen (CC-by) Equation 4/9 may appear unsettling at frst sight, because as 퐴tends to zero, 퐹⟂also tends to zero; nevertheless, in any continuous medium, the ratio of these two terms tends to a single non-zero value: the local pressure. This brings us to the second particularity of pressure in fuids: the pressure on either side of the infnitesimal fat surface is the same regardless of its orientation. In other words, pressure has no direction: there is only one (scalar) value for pressure at any one point in space. Thus, in a fuid, pressure applies not merely on the solid surfaces of its container, but also everywhere within itself. We need to think of pressure as a scalar property feld 푝(푥,푦,푧,푡). 4.3.2 Pressure on an infnitesimal volume While pressure has no direction, it may not have the same value everywhere in a fuid, and so the gradient (the rate of change with respect to distance) of pressure may not be null. For example, in a static water pool, pressure is uniform in the two horizontal directions, but it increases along with depth. Instead of a fat plate, let us now consider an infnitesimally small cube within the fuid (fg. 4.3). Because the cube is placed in a scalar feld, the pressure exerting on each of its six faces may be diferent. The net efect of pressure will therefore have three components: one for each pair of opposing faces. What are those three components? In the 푥-direction, the pressure on faces 1 and 4 act upon a surface of area d푦d푧: 퐹net, pressure,푥= d푦d푧[푝1 −푝4] Figure 4.3: The pressure on each face of an infnitesimal volume may have a diferent value. The net efect of pressure will depend on how the pressure varies in space. These changes are labeled d푝\|푖in each of the 푖= 푥, 푦, 푧directions. fgure CC-0 Olivier Cleynen 76 We express 푝4 −푝1 as the derivative of pressure in the 푥-direction (휕푝/휕푥), multiplied by the distance d푥which separates points 1 and 4, obtaining: 퐹net, pressure,푥= d푦d푧[−휕푝 휕푥d푥] = d−휕푝 휕푥 (4/10) where d≡d푥d푦d푧is the volume of the infnitesimal cube. Now generalizing eq. 4/10 for the other two directions, we can write: 퐹net, pressure,푥= d−휕푝 휕푥 퐹net, pressure,푦= d−휕푝 휕푦 퐹net, pressure,푧= d−휕푝 휕푧 Video: How many dimensions does it take to describe the efect of pressure? by Olivier Cleynen (CC-by) This is tedious to write, but we recognize a pattern. And indeed, we introduce the concept of gradient, a mathematical operator, defned as so (see also Appendix A3 p. 250): ⃗ ∇≡⃗ 푖휕 휕푥+ ⃗ 푗휕 휕푦+ ⃗ 푘휕 휕푧 (4/11) With this cool new tool, we elegantly re-write the group of equations above: ⃗ 퐹net, pressure = −d⃗ ∇푝 (4/12) Finally, we obtain: 1 d ⃗ 퐹net, pressure = −⃗ ∇푝 (4/13) This last equation reads “the pressure force per unit volume is the opposite of the pressure gradient”. It shows us that in any fuid and any situation, the force due to pressure points the opposite way of the pressure gradient. Thus, if a particle of any kind is “dropped” into a fuid fow, we can quantify in which direction, and with which magnitude, pressure (a scalar feld) is going to “push” it. This is given by equation 4/13, which quantifes this efect as a vector feld (see fgures 4.4 & 4.5). Advice from an expert Ponder for a moment what the dimensions of the physical properties in this equation 4/13 are. If you were to record on a usb stick the values for pressure in a given fuid domain, you would have to store one value (in Pa) for each point in space (푥, 푦, 푧), once for each time point (푡). If you were to calculate the efect of pressure, you would calculate minus the gradient of pressure. To record this information, you would need to store three values (the three components, in 푥, 푦, and 푧, each in Pa m−1) for each point in space, once for each time point. This feld of vectors 77 would indicate the amount of force per unit volume with which pressure is pushing the fuid. Having a good grasp of the tools we are using here is important, because things will soon get more complicated when shear is added to the equa-tion (in chapter 5), ultimately leading to the all-powerful Navier-Stokes equation in chapter 6. Figure 4.4: Water fow in a two-dimensional water tank, visualized with a computa-tional fuid dynamics (cfd) software package. The fow is from left to right: water enters with a velocity of 10 m s−1 in a 10 m-high tunnel, and fows around a “bump” at the bottom. On the top, the magnitude of velocity is represented (background color), with white lines indicating fow direction. On the bottom, pressure is displayed. The values for pressure have been arbitrarily adjusted for visual purposes so that the minimum pressure in the fow is zero Pascal. Figure CC-by by Arjun Neyyathala 78 Figure 4.5: The negative of the pressure gradient feld in the two-dimensional fow from fgure 4.4. All the vectors are represented with the same length, but their magnitude is coded as color. The arrows indicate the local force per unit volume with which pressure is acting on the fuid. Figure CC-by by Arjun Neyyathala 4.4 Special case: pressure in static fuids 4.4.1 Forces in static fuids Fluid statics is the study of fuids at rest, i.e. those whose velocity feld ⃗ 푉is everywhere null and constant: { ⃗ 푉 = ⃗ 0 휕⃗ 푉 휕푡 = ⃗ 0 (4/14) We choose to study this type of problem now, because it makes for a concep-tually and mathematically simple case with which we can practice calculating fuid-induced forces. What are the forces applying on an arbitrary particle in a static fuid? • The force due to pressure is related to the pressure gradient: we just quantifed this with equation 4/13 above. • The force due to shear is zero. We will indeed see in chapter 5 (Efects of shear) that shear eforts can be expressed as a function of viscosity and velocity. All ordinary fuids are unable to exert shear when they are static. • The force due to gravity is easy to quantify: it is the mass 푚of the fuid particle multiplied by the gravity vector ⃗ 푔. In a moving fuid, the sum of these forces would add up to the mass of the particle times its acceleration. But in a static fuid, the velocity is zero and never changes. We can thus write: ⃗ 퐹net, pressure + ⃗ 퐹shear + ⃗ 퐹gravity = ⃗ 0 −d⃗ ∇푝+ ⃗ 0 + 푚⃗ 푔 = ⃗ 0 −⃗ ∇푝 + ⃗ 0 + 휌⃗ 푔 = ⃗ 0 79 We can fnally rewrite this as: ⃗ ∇푝= 휌⃗ 푔 (4/15) in a static fuid. This is a very useful equation, which states that in a static fuid, the only parameter afecting pressure is gravity. More precisely, the fuid density times the gravity vector is equal to the change in space of the pressure. We will see in chapter 6 (Prediction of fuid fows) that equation 4/15 is the specifc case for a much larger general and powerful equation, the Navier-Stokes equation. But more on that later! Advice from an expert What this equation 4/15 is really saying is that in a static fuid, “pressure changes only with altitude”. This sounds trivial, but consider the consequences. For example, when you swim under an anchored boat, you can’t “feel” the presence of the boat: only your own depth matters. Or this: the pressure at the inlet of the turbines of a gigawatt-class hydraulic dam power station depends on the depth of the reservoir, but not at all on its overall size. So much consequence for such a small equation! 4.4.2 Pressure and depth It is now easy to quantify pressure everywhere inside a static fuid. Very often in studies of static fuids, the 푧-axis is oriented vertically, positive downwards. With this convention, there is no need for a vector equation to quantify pressure, and equation 4/15 becomes: d푝 d푧= 휌푔 (4/16) in a static fuid, when 푧is oriented positive downwards. The frst consequence we draw from equation 4/16 is that in a static fuid (e.g. in a glass of water, in a swimming pool, in a calm atmosphere), pressure depends solely on height. Within a static fuid, at a certain altitude, we will measure the same pressure regardless of the surroundings (fg. 4.6). How is pressure distributed within static liquid water bodies? The density of liquid water is approximately constant: 휌water = 1 000 kg m−3. In a water reservoir, equation 4/16 becomes: ( d푝 d푧)water = 휌water 푔 ( d푝 d푧)water = 1 000 × 9,81 = 9,81 ⋅103 Pa m−1 = 9,81 ⋅10−2 bar m−1 (4/17) 80 Figure 4.6: Pressure at a given depth (or height) in a static fuid does not depend on the environment. Here, as long as the fuid remains static, 푝A = 푝B = 푝C = 푝D. Figure CC-0 Olivier Cleynen Therefore, in static water, pressure increases by approximately 0,1 bar/m as depth increases. For example, at a depth of 3 m, the pressure will be approximately 1,3 bar (which is the atmospheric pressure plus Δ푧× d푝/ d푧). In the atmosphere, the situation is more complex, because the density 휌air of atmospheric air is not uniform. If we model atmospheric air as a perfect gas, once again orienting 푧vertically downwards, we can express the pressure gradient as: ( d푝 d푧)atm. = 휌air 푔= 푝1 푇 푔 푅 (4/18) This time, the variation of pressure with respect to distance depends on pressure itself (and it is proportional to it). A quick numerical investigation for ambient temperature and pressure (1 bar, 15 °C) yields: ( d푝 d푧)atm. ambient = 1 ⋅105 × 1 288,15 × 9, 81 287 = 11,86 Pa m−1 = 1,186 ⋅10−4 bar m−1 (4/19) This rate (approximately 0,1 mbar/m) is almost a thousand times smaller than that of water (fg. 4.7). Since the rate of pressure change depends on pressure, it also varies with altitude, and the calculation of pressure diferences in the atmosphere is a little more complicated than for water. XKCD #2153: efects of high alti-tude by Randall Munroe (CC-by-nc) If we focus on a moderate height change, it may be reasonable to consider that temperature 푇, the gravitational acceleration 푔and the gas constant 푅are uniform. In this (admittedly restrictive) case, equation 4/18 can be integrated as so: d푝 d푧= 푔 푅푇cst. 푝 ∫ 2 1 1 푝d푝= 푔 푅푇cst. ∫ 2 1 d푧 ln 푝2 푝1 = 푔 푅푇cst. Δ푧 푝2 푝1 = exp [ 푔Δ푧 푅푇cst.] (4/20) 81 Figure 4.7: Variation of pressure as a function of altitude for water and air at the surface of a water reservoir. The gradient of pressure with respect to altitude is almost a thousand times larger in water than in air. Figure CC-0 Olivier Cleynen This is only a simplifed model: in reality, air temperature varies signifcantly within the atmosphere (at moderate altitudes the change with altitude is approximately −6 K km−1). Adapting equation 4/20 for a uniform temperature gradient (instead of uniform temperature) is the subject of problem 4.8 p. 88. Advice from an expert Here, we see how the “푝=휌푔ℎ” equation that many of us learned in our frst class of fuid mechanics is dangerous: if 휌is not uniform, everything falls apart. To be safe, when you study a static fuid, always start from equa-tion 4/16. As you integrate 푝with respect to 푧, you will be forced to consider how 휌falls into the picture. In wa-ter, 휌is uniform (just a number). But in the atmosphere, air is “squished” by gravity, and much denser near the ground. In practice, the atmosphere also features signifcant lateral pressure gradients (which are strongly related to the wind) and its internal fuid mechanics are complex and fascinating. Equation 4/20 is a useful and convenient model, but refnements must be made if precise results are to be obtained. 4.4.3 Buoyancy Video: playing around with an air pump and a vacuum chamber by Y:Roobert33 (styl) Any solid body immersed within a fuid is subjected to pressure on its walls. When the pressure is not uniform (for example because the fuid is subjected to gravity, although this may not be the only cause), then the net force due to fuid pressure on the body walls will be non-zero. When the fuid is purely static, this net pressure force is called buoyancy. Since in this case, the only cause for the pressure gradient is gravity, the net pressure force is oriented upwards. The buoyancy force is completely independent from (and may or may not compensate) the object’s weight. Since it comes from equation 4/15 that the variation of pressure within a fuid is caused solely by the fuid’s weight, we can see that the force exerted on an immersed body is equal to the weight of the fuid it replaces (that is to say, the weight of the fuid that would occupy its own volume were it not 82 there). This relationship is sometimes named Archimedes’ principle. The force which results from the static pressure gradient applies to all immersed bodies: a submarine in an ocean, an object in a pressurized container, and of course, the reader of this document as presently immersed in the earth’s atmosphere. Figure 4.8: Immersion in a static fuid results in forces that depend on the body’s vol-ume. They can evidenced by the removal of the fuid (for example in a depressurized semi-spherical vessel). Figure CC-0 Olivier Cleynen 4.5 Solved problems Pressure at the bottom of a lake A lake has the dimensions shown in the fgure above. What is the pressure at the bottom of the lake? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 83 Pressure force on a wall The side wall of a water tank has the dimensions shown in the fgure above. What is the force exerted due to the pressure of the water on the wall? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Position of pressure force on a wall In the problem above, at what height above the ground does the force due to pressure apply? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 84 Problem sheet 4: Efects of pressure last edited April 25, 2021 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 4.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 4.2 Pressure in a static fuid A small water container whose geometry is described in fg. 4.9 is flled with water. What is the pressure at the bottom of the container? Figure 4.9: A small water container. Figure CC-0 Olivier Cleynen 4.3 Pressure measurement with a U-tube A tube is connected to a pressurized air vessel as shown in fg. 4.10. The U-tube is flled with water. What is the pressure 푝int. in the vessel? Figure 4.10: Working principle of a simple liquid tube manometer. The outlet is at atmospheric pressure 푝atm.. Figure CC-0 Olivier Cleynen What would be the height diference shown for the same internal pressure if mercury (휌mercury = 13 600 kg m−3) was used instead of water? 85 4.4 Straight water tank door A water tank has a window on one of its straight walls, as shown in fgure 4.11. The window is 3 m high, 4 m wide, and is positioned 0,4 m above the bottom of the tank. Figure 4.11: An aquarium tank with a window installed on one of its walls Figure CC-0 Olivier Cleynen 4.4.1. What is the magnitude of the force applying on the window due to the pressure exerted by the water? 4.4.2. At which height does this force apply? 4.5 Access door on a water channel wall An open water channel used in a laboratory is flled with stationary water (fg. 4.12). An observation window is installed on one of the walls of the channel, to enable observation and measurements. The window is hinged on its bottom face. The hinge stands 1,5 m below the water surface. The window has a length of 0,9 m and a width of 2 m. The walls of the channel are inclined with an angle 휃= 60° relative to horizontal. Figure 4.12: A door installed on the wall of a water channel. The water in the canal is perfectly still. Figure CC-0 Olivier Cleynen 4.5.1. Represent graphically the pressure of the water and atmosphere on each side of the window. 4.5.2. What is the magnitude of the moment exerted by the pressure of the water about the axis of the window hinge? 4.5.3. If the same door was positioned at the same depth, but the angle 휃was decreased, would the moment be modifed? (briefy justify your answer, e.g. in 30 words or less) 86 4.6 Pressure force on a cylinder An idealized fow over a cylinder is depicted in fgure 4.13. This is a very primitive fow solution, obtained using a model (the potential fow model, which we mention in chapter 11) which cannot account for viscous efects or fow separation. Nevertheless, it provides a good frst “ideal fow” situation to compute surface pressure forces in fuid fows. Figure 4.13: Idealized fow around a cylinder, as predicted by potential fow theory. The stream-lines are represented only in a plane crossing the center of the cylinder, but they are identical all along the 푧direction. Figure CC-by-sa by Commons User:Kraaiennest & Olivier Cleynen The cylinder has diameter 10 cm, and it spans 50 cm across the fow (in the 푧direction). We start by considering the case where there is no fow: the velocity is everywhere 푉= 0 m s−1 and the air pressure is everywhere 푝∞= 1 bar. We would like to calculate the forces applying represented in fgure 4.14, both caused by the air pressure. Figure 4.14: Left: vertical force on the top half of the cylinder. Right: horizontal force on “nose” of the cylinder, on an area spanning 20° around the leading edge. CC-0 Olivier Cleynen 4.6.1. What is the magnitude of ⃗ 퐹top? 4.6.2. What is the magnitude of ⃗ 퐹upstream? 87 (a couple of hints to help with the algebra: ∫sin 푥d푥= −cos 푥+ 푘and ∫sin3 푥d푥= 1 3 cos3 푥−cos 푥+ 푘) We now consider the case there there is fuid fow: air with density 휌atm. = 1,225 kg m−3 is coming in at 푉∞= 50 km h−1. In this case, the pressure 푝푠on the surface of the cylinder is no longer uniform (see also problem 11.3 p. 236). It is expressed as a function of the coordinate 휃as: 푝푠= 푝∞+ 1 2휌(푉2 ∞−4푉2 ∞sin2 휃) (4/21) 4.6.3. What is the new magnitude of ⃗ 퐹top? 4.6.4. What is the new magnitude of ⃗ 퐹upstream? (a couple of hints to help with the algebra: ∫cos 푥d푥= sin 푥+ 푘and ∫cos 푥sin2 푥d푥= 1 3 sin3 푥+ 푘) 4.7 Buoyancy of a barge A barge of very simple geometry is moored in a water reservoir (fg. 4.15). Figure 4.15: Basic layout of a barge foating in water. Figure CC-0 Olivier Cleynen 4.7.1. Sketch the distribution of pressure on each of the immersed walls of the barge (left and right sides, rear , bottom and slanted front). 4.7.2. What is the magnitude of the force resulting from pressure eforts on each of these walls? 4.7.3. What is the weight of the barge? 4.8 Atmospheric pressure distribution non-examinable The integration we carried out in with equation 4/20 p. 81 to model the pressure distribu-tion in the atmosphere was based on the hypothesis that the temperature was uniform and constant (푇= 푇cst.). In practice, this may not always be the case. 4.8.1. If the atmospheric temperature decreases with altitude at a constant rate (e.g. of −7 K km−1), how can the pressure distribution be expressed analytically? 88 A successful fuid dynamics lecturer purchases an apartment at the top of the Burj Khalifa tower (800 m above the ground). Inside the tower, the temperature is controlled everywhere at 18,5 °C. Outside, the ground temperature is 30 °C and it decreases linearly with altitude (gradient: −7 K km−1). A door is opened at the bottom of the tower, so that at zero altitude the air pressure (1 bar) is identical inside and outside of the tower. For the purpose of the exercise, we pretend the tower is entirely hermetic (meaning air is prevented from fowing in or out of its windows). 4.8.2. What is the pressure diference between each side of the windows in the apart-ment at the top of the tower? 89 Answers 4.2 푝A = 푝atm. + 0,039 bar ≈1,039 bar. 4.3 1) 푝inside = 푝atm. + 0,0157 bar ≈1,0157 bar; 2) Δ푧2 = 1,1765 cm. Is having both of those results right enough to call yourself a U-tube star? 4.4 1) 퐹net = 휌푔퐿(푍max퐿max −1 2퐿2 max) = 176,6 kN; 2) 푀net, bottom hinge = 휌푔퐿( 1 2푍max퐿2 max −1 3퐿3 max) = 176,6 kN m so the force exerts at 푅= 푀net/퐹net = 1 m above the bottom hinge. 4.5 2) 푀net = 7,79 kN m; 3) observe the equation used to calculate 푀net to answer this question. If needed, ask for help in class! 4.6 1) 퐹top = ∫ 휃=휋 휃=0 푅퐿푝푠sin 휃d휃= −5 000 N (downwards); 2) 퐹front = +868 N (in downstream direction); 3) 퐹top = −4 990 N downwards (a lift force of 10 N); 4) 퐹front = +869 N (1 N of drag!). 4.7 2) 퐹rear = 0,2453 MN, 퐹side = 2,1714 MN, 퐹bottom = 13,734 MN, 퐹front = 0,3468 MN; 3) 퐹buoyancy = 13,979 MN (1 425 t). 4.8 1) 푝2 푝1 = (1 + 푘푧2 푇1 ) 푔 푘푅 2) Δ푝= +247,4 Pa inwards. 90 Fluid Dynamics Chapter 5 – Efects of shear last edited September 19, 2020 by Olivier Cleynen — 5.1 Motivation 91 5.2 Shear forces on walls 91 5.2.1 Magnitude of the shear force 91 5.2.2 Direction and position of the shear force 92 5.3 Shear felds in fuids 92 5.3.1 The direction of shear 93 5.3.2 Shear on an infnitesimal volume 93 5.4 Resistance to shear: viscosity 97 5.4.1 Viscosity 97 5.4.2 Kinematic viscosity 98 5.4.3 Turbulent viscosity 99 5.4.4 Non-Newtonian fuids 100 5.4.5 The no-slip condition 100 5.5 Special case: shear in simple laminar fows 101 5.6 Solved problems 103 5.7 Problems 105 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 5.1 Motivation In fuid mechanics, only three types of forces apply to fuid particles: forces due to gravity, pressure, and shear. This chapter focuses on shear, and should allow us to answer two questions: • How is the efect of shear described and quantifed? • What are the shear forces generated on walls by simple fows? 5.2 Shear forces on walls 5.2.1 Magnitude of the shear force What is the force which which a fuid shears (i.e. “rubs”) against a wall? When the shear 휏exerted is uniform and the wall is fat, the resulting force 퐹 in the direction 푖is easily calculated: 퐹shear, direction 푖= 휏uniform, direction 푖푆fat wall (5/1) When the shear 휏exerted by the fuid is not uniform (for example, because more friction is occurring on some parts of the surface than on others), the situation is more complex: the force must be obtained by integration. The 91 surface is split in infnitesimal portions of area d푆, and the corresponding forces are summed up as: 퐹shear, direction 푖= ∫ 푆 d퐹shear, direction 푖 (5/2) 퐹shear, direction 푖= ∬ 푆 휏direction 푖d푆 (5/3) for a fat surface, where the 푆-integral denotes an integration over the entire surface. What is required to calculate the scalar 퐹in eq. 5/3 is an expression of 휏as a function of 푆. In a simple laminar fow, this expression will often be relatively easy to fnd, as we see later on. 5.2.2 Direction and position of the shear force The above equations work only for a fat surface, and in a chosen direction 푖. When we consider a two- or three-dimensional object immersed in a fuid with non-uniform shear, the integration must be carried out with vectors. We will not attempt this in this course, but the expression is worth writing out in order to understand how computational fuid dynamics (cfd) software will proceed with the calculation. In a general case, the shear on any infnitesimal surface d푆needs to be expressed as a vector ⃗ 휏푛, where 푛is the direction perpendicular to the surface. The net force due to shear on the surface is then: ⃗ 퐹shear = ∫ 푆 ⃗ 휏푛d푆 (5/4) Much like equation 4/4 in the previous chapter, eq. 5/4 is not too hard to implement as a software algorithm to obtain numerically, for example, the force resulting from shear due to fuid fow around a body such as the body of a car. Its computation by hand, however, is far too tedious for us to even attempt. The position of the shear force is obtained with two moment vector equations, in a manner similar to that described in §4.2.2 p. 74 with pressure. This is outside of the scope of this course. 5.3 Shear felds in fuids We approached the concept of shear in chapter 1 (Basic fow quantities) with the notion that it represented force parallel to a given fat surface (eq. 1/15), for example a fat plate of area 퐴: 휏≡퐹∥ 퐴 (5/5) Like we did with pressure, to appreciate the concept of shear in fuid mechan-ics, we need to go beyond this equation. 92 5.3.1 The direction of shear Video: cloud movements in a time-lapse video on an interest-ing day are evidence of a highly-strained atmosphere: pilots and meteorologists refer to this as wind shear. by Y:StormsFishingNMore (styl) Already from the defnition in eq. 5/5 we can appreciate that “parallel to a fat plate” can mean a multitude of diferent directions, and so that we need more than one dimension to represent shear. Furthermore, much in the same way as we did for pressure, we do away with the fat plate and accept that shear is a feld, i.e. it is an efort applying not only upon solid objects but also upon and within fuids themselves. We replace eq. 5/5 with a more general defnition: ⃗ 휏≡lim 퐴→0 ⃗ 퐹∥ 퐴 (5/6) Contrary to pressure, shear is not a scalar, i.e. it can (and often does) take diferent values in diferent directions. At a given point in space we represent it as a vector ⃗ 휏= (휏푥, 휏푦, 휏푦), and in a fuid, there is a shear vector feld: ⃗ 휏(푥,푦,푧,푡) ≡ ⎛ ⎜ ⎜ ⎝ 휏푥 휏푦 휏푧 ⎞ ⎟ ⎟ ⎠(푥,푦,푧,푡) (5/7) 5.3.2 Shear on an infnitesimal volume We the Robots #20071126: using the right terminology is impor-tant (or not?). by Chris Harding Describing the changes in space of the shear vector feld requires another mathematical dimension (called order). Instead of a fat plate, let us consider an infnitesimally small cube within the fuid, as shown in fgure 5.1. Because the cube is immersed inside a vector feld, it may have a diferent the shear vector exerting on each of its six faces. In order to express the eforts on any given face, we express a component of shear with two subscripts: • The frst subscript indicates the direction normal to the surface in which we are interested; • The second subscript indicates the direction in which the shear is applying. Figure 5.1: Shear eforts on a cubic fuid particle (with only the eforts on the visible faces 1 to 3 represented). The shear tensor ⃗ 휏푖푗has six members of three components each. Figure CC-0 Olivier Cleynen 93 Video: the net efect of shear by Olivier Cleynen (CC-by) For example, ⃗ 휏푥푦represents the shear in the 푦-direction on a surface perpen-dicular to the 푥-direction. On this face, the shear vector would be: ⃗ 휏푥푗= ⃗ 휏푥푥+ ⃗ 휏푥푦+ ⃗ 휏푥푧 (5/8) = 휏푥푥⃗ 푖+ 휏푥푦⃗ 푗+ 휏푥푧⃗ 푘 (5/9) where the subscript 푥푗indicates all of the directions (푗= 푥, 푦, 푧) on a face perpendic-ular to the 푥-direction. Video: what is 휏푥푥, perpendicu-lar shear? by Olivier Cleynen (CC-by) In eq. 5/8, a surprising term appears: 휏푥푥. It is the shear efort perpendicular to the surface of interest. How is this possible? The answer is that the faces of the infnitesimal cube studied here are not solid. They are permeable, and the local velocity may include have a component through the face of the cube (in fact, this must happen for any fow to occur at all). Therefore, there is no reason for the shear efort, which is three-dimensional, to be aligned along each fat surface. As the fuid travels across any face, it can be sheared and strained in any arbitrary direction, and therefore, shear can and most often does have a component (휏푖푖) perpendicular to an arbitrary surface inside a fuid. Now, the net shear efect on the cube will have eighteen components: one tree-dimensional vector for each of the six faces. Each of those components may take a diferent value. The net shear could perhaps be represented as en entity —a tensor— containing six vectors ⃗ 휏1, ⃗ 휏2, ⃗ 휏3... ⃗ 휏6. By convention, however, shear is notated using only three vector components: one for each pair of faces. Shear eforts on a volume are thus represented with a tensor feld ⃗ 휏푖푗: ⃗ 휏푖푗≡ ⎛ ⎜ ⎜ ⎝ ⃗ 휏푥푗 ⃗ 휏푦푗 ⃗ 휏푧푗 ⎞ ⎟ ⎟ ⎠ ≡ ⎛ ⎜ ⎜ ⎝ ⃗ 휏푥푗{1,4} ⃗ 휏푦푗{2,5} ⃗ 휏푧푗{3,6} ⎞ ⎟ ⎟ ⎠ ⃗ 휏푖푗≡ ⎛ ⎜ ⎜ ⎝ 휏푥푥 휏푥푦 휏푥푧 휏푦푥 휏푦푦 휏푦푧 휏푧푥 휏푧푦 휏푧푧 ⎞ ⎟ ⎟ ⎠ (5/10) In this last equation 5/10, each of the nine components of the tensor acts as the container for two contributions: one for each of the two faces perpendicular to the direction expressed in its frst subscript. So much for the shear efort on an element of fuid. What about the net force due to shear on the fuid element? Not every element counts: part of the shear will accelerate (change the velocity vector) the particle, while part of it will merely strain (deform) the particle. Quantifying this force thus requires making a careful selection within the eighteen components of ⃗ 휏푖푗. We may start with the 푥-direction, which consists of the sum of the component of shear in the 푥-direction on each of the six cube faces: ⃗ 퐹shear 푥= 푆3⃗ 휏푧푥3 −푆6⃗ 휏푧푥6 +푆2⃗ 휏푦푥2 −푆5⃗ 휏푦푥5 +푆1⃗ 휏푥푥1 −푆4⃗ 휏푥푥4 (5/11) 94 Given that 푆3 = 푆6 = d푥d푦, that 푆2 = 푆5 = d푥d푧and that 푆1 = 푆4 = d푧d푦, this is re-written as: ⃗ 퐹shear 푥= d푥d푦(⃗ 휏푧푥3 −⃗ 휏푧푥6) + d푥d푧(⃗ 휏푦푥2 −⃗ 휏푦푥5) + d푧d푦(⃗ 휏푥푥1 −⃗ 휏푥푥4) (5/12) In the same way we did with pressure in chapter 4 (§4.4.2 p. 80), we express each pair of values as a derivative with respect to space multiplied by an infnitesimal distance: ⃗ 퐹shear 푥= d푥d푦(d푧휕⃗ 휏푧푥 휕푧) + d푥d푧(d푦휕⃗ 휏푦푥 휕푦) + d푧d푦(d푥휕⃗ 휏푥푥 휕푥) = d( 휕⃗ 휏푧푥 휕푧+ 휕⃗ 휏푦푥 휕푦+ 휕⃗ 휏푥푥 휕푥) (5/13) Video: the divergent of shear by Olivier Cleynen (CC-by) We can see with this equation 5/13 that shear in the 푥-direction has three contributors (one for each pair of cube faces). Each of the contributors is a derivative in space of a shear component which points in the 푥-direction. Now, we introduce the operator divergent (see also Appendix A3 p. 250), written ⃗ ∇⋅: ⃗ ∇⋅≡휕 휕푥 ⃗ 푖⋅+ 휕 휕푦 ⃗ 푗⋅+ 휕 휕푧 ⃗ 푘⋅ (5/14) ⃗ ∇⋅⃗ 퐴≡휕퐴푥 휕푥 + 휕퐴푦 휕푦 + 휕퐴푧 휕푧 (5/15) ⃗ ∇⋅⃗ 퐴푖푗≡ ⎛ ⎜ ⎜ ⎜ ⎝ 휕퐴푥푥 휕푥+ 휕퐴푦푥 휕푦+ 휕퐴푧푥 휕푧 휕퐴푥푦 휕푥+ 휕퐴푦푦 휕푦+ 휕퐴푧푦 휕푧 휕퐴푥푧 휕푥+ 휕퐴푦푧 휕푦+ 휕퐴푧푧 휕푧 ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ ⃗ ∇⋅⃗ 퐴푖푥 ⃗ ∇⋅⃗ 퐴푖푦 ⃗ ∇⋅⃗ 퐴푖푧 ⎞ ⎟ ⎟ ⎟ ⎠ (5/16) With this new tool, we can go back to equation 5/13 to see that the net shear force in the 푥-direction is equal to the particle volume times the divergent of the shear in the 푥-direction: ⃗ 퐹shear 푥= d⃗ ∇⋅⃗ 휏푖푥 (5/17) So much for the 푥-direction. The 푦- and 푧-direction are taken care of in the same fashion, so that we can gather up our puzzle pieces and express the force per unit volume due to shear as the divergent of the shear tensor: ⃗ 퐹shear = ⎛ ⎜ ⎜ ⎝ 퐹shear 푥 퐹shear 푦 퐹shear 푧 ⎞ ⎟ ⎟ ⎠ = d ⎛ ⎜ ⎜ ⎝ \|⃗ ∇⋅⃗ 휏푖푥\| \|⃗ ∇⋅⃗ 휏푖푦\| \|⃗ ∇⋅⃗ 휏푖푧\| ⎞ ⎟ ⎟ ⎠ = d⃗ ∇⋅⃗ 휏푖푗 (5/18) 1 d ⃗ 퐹net, shear = ⃗ ∇⋅⃗ 휏푖푗 (5/19) An example of a divergent of shear vector feld is shown in fgures 5.2 and 5.3. This equation 5/19 is more than we really need to go through the problems in this chapter, but we will come back to it when we will want to calculate 95 Figure 5.2: The divergent of shear in the fow feld of the computed fow described in fgure 4.4 p. 78. Vectors with magnitude lower than 1 Pa m−1 are not represented. The arrows indicate the local force per unit volume with which shear is acting on the fuid. Figure CC-by by Arjun Neyyathala Figure 5.3: The magnitude of the divergent of shear in the fow feld of the computed fow described in fgure 4.4 p. 78 (the magnitude of the vector feld represented in fgure 5.2 above). In the top image, the color scale is saturated at 1 ⋅105 Pa m−1, while on the bottom image, it reaches a value 20 times higher (2 ⋅106 Pa m−1), showing the very high local values attained very close to the wall. Figures CC-by by Arjun Neyyathala the dynamics of fuid particles in chapter 6 (Prediction of fuid fows), where the divergent of shear be a building block of the glorious Navier-Stokes equations. For now, it is enough to sum up our fndings as follows: • Shear at a point in space has three components — it is a vector feld; • The efect of shear on a volume of fuid has eighteen components – it is a second-order tensor feld; 96 • The net force due to shear on a volume of fuid, expressed using the divergent of the shear tensor, has three components — it is a vector feld. Advice from an expert Again, take a moment to consider the dimen-sions involved here. If you were to store in-formation about shear in a fow in a fle on a usb key, how many columns would you need? Shear is a vector feld, which means it has three components (each in Pa) at every point in space and time: that’s three values to store for each combination of 푥,푦,푧and 푡. If you were to calculate the efect of shear, then you would calculate its divergent, which is also a three-dimensional vector feld. To record this information, you would, likewise, have to store three values (one for each of \|⃗ ∇⋅⃗ 휏푖푥\|, \|⃗ ∇⋅⃗ 휏푖푦\|, and \|⃗ ∇⋅⃗ 휏푖푧\|, all in Pa m−1), for each combination of 푥,푦,푧 and 푡. Just like the negative of the gradient of pressure, the divergent of shear represents a force per unit volume, showing in which direction shear is “pushing” the fuid particles as they fow. 5.4 Resistance to shear: viscosity 5.4.1 Viscosity In chapter 1, we saw already that viscosity 휇is a fuid property that quantifes its resistance to shear (see §1.4.6 p. 17). More precisely, we quantifed 휇as the ratio of shear stress to strain rate with equation 1/12, reproduced here: 휇≡ 휏 ( Δ푣 Δ푦) (5/20) Now, we generalize this equation: we shrink down the “brick” of fuid from chapter 1 down to an infnitesimal volume of fuid inside an arbitrary fow (fg. 5.4). The strain rate is now 휕푉푗/휕푖, which is the rate of change in the 푖-direction of the velocity in the 푗-direction. Viscosity is the ratio between shear in the shear in the 푗-direction and this strain rate: 휇≡\|\|⃗ 휏푖푗\|\| ( 휕푉푗 휕푖) (5/21) If we turn this equation around, we fnd that we can express the local shear by diferentiating the local velocity with respect to distance: \|\|⃗ 휏푖푗\|\| = 휇휕푉푗 휕푖 (5/22) in which the subscript 푖is an arbitrary direction (푥, 푦or 푧) and 푗is the direction following it in order (e.g. 푗= 푧when 푖= 푦); and where 휇is the viscosity (or “dynamic viscosity”) (Pa s). 97 Figure 5.4: Any velocity gradient 휕푉푦/휕푥= 휕푣/휕푥in the fow results in a shear force 퐹∥in the direction 푦. The ratio between the shear and the velocity gradient is called viscosity. Figure CC-0 Olivier Cleynen Viscosity 휇is measured in Pa s, which is the same as N s m−2 or kg m−1 s−1. It has historically been measured in poise (1 poise ≡0,1 Pa s). Video: viscosity for the engineer by Olivier Cleynen (CC-by) The values of viscosity vary very strongly from one fuid to another: for example, honey is roughly ten thousand times more viscous than water, which is roughly a hundred times more viscous than ambient air. The viscosities of four relevant fuids are quantifed in fgure 5.6 (where they can easily be quantifed) and fgure 5.5 (where the relative values are better observed). −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 0 5×10−3 10−2 1.5×10−2 Viscosity 휇of liquids in Pa s Water Crude Oil Air CO2 Figure 5.5: The viscosity of four fuids (crude oil, water, air, and C02) as a function of temperature, plotted on a linear scale. This makes clearly visible the diference in the order of magnitudes of the viscosities of the four fuids, but air and CO2 are indistinguishably close to the zero axis. A more useful version of this fgure is shown as fgure 5.6. Figure CC-by by Arjun Neyyathala & Olivier Cleynen 98 −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 5.6: The viscosity of four fuids (crude oil, water, air, and C02) as a function of temperature. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. Figure CC-by by Arjun Neyyathala & Olivier Cleynen 5.4.2 Kinematic viscosity Sometimes, the concept of kinematic viscosity is used. Kinematic viscosity is written 휈: the Greek letter nu, an unfortunate choice because it is easy to mis-read as the 푦-component of velocity, 푣≡푉푦. Kinematic viscosity 휈is defned as 휈≡휇 휌 (5/23) where 휈is measured in m2 s−1. Kinematic viscosity is formulated that way because it is the part of the Reynolds number (see eq. 1/28 p. 23) that depends on the fuid properties. We do not make use of it in this course. 5.4.3 Turbulent viscosity In computational fuid dynamics (cfd) simulations, use is made of turbulent viscosity 휇T (in Pa s just like viscosity). This is because the bulk efect of turbulence is to increase dissipation, in a way that is similar to viscous dissipation. Turbulent viscosity is therefore a property of the fow, with values that vary very strongly with space and time within the fow (unlike viscosity which is usually only a property of the fuid). We will encounter it in chapter 9 (Dealing with turbulence). 99 5.4.4 Non-Newtonian fuids Viscosity is said to be a property of the fuid, and not of the fow, because for most fuids, it has a value that varies only with temperature. Those fuids are formally called Newtonian fuids. Most fuids of interest in engineering fuid mechanics (air, water, exhaust gases, pure gases) are Newtonian fuids. Some fuids, however, feature a strain rate that depends on the value of shear stress: one could say that they have “variable viscosity”, depending on how they are strained. Those fuids are called non-Newtonian fuids. The study of the viscosity characteristics of such fuids is called rheologyw. Some fuids, like oil-based paint, jelly-based fuids, or tomato juice, are shear-thinning: they become progressively less viscous when the strain rate is increased. Some fuids, like blood, are shear-thickening: they become progressively thicker when the strain rate is increased. A few representative viscosity characteristics are displayed in fgure 5.7. Advice from an expert “Non-Newtonian” does not mean “sticky”; it merely means that the fuid is “variably sticky”. Some, like plant resin, resist being strained too quickly, while others, like paint, resist being strained too slowly. It is the behavior which matters, and not the actual value of their viscosities (which may be generally high or low). Mix some starch and water in a bowl, and pour some paint in another: time spent playing with those two fuids in your kitchen totally counts as serious fuid mechanics study time. Figure 5.7: Various possible viscosity characteristics of fuids. Those for which the slope of the curve (휇) varies with 휕푉푗/휕푖(so, those which do not feature straight lines on this diagram) are called non-Newtonian. Figure CC-0 Olivier Cleynen 100 5.4.5 The no-slip condition We observe that whenever we measure the velocity of a fuid fow along a solid wall, the speed tends to zero as we approach the wall surface. In other words, the fuid adheres to the surface regardless of the overall faraway fow velocity. This phenomenon, called the no-slip condition, is extremely important in fuid dynamics. One consequence of this is that fuid fows near walls are dominated by viscous efects (internal friction) due to the large strain rates there. Advice from an expert The no-slip condition is actually pretty freaky: no mat-ter how fast you go, and no matter how smooth the surface, the fuid velocity on the surface of a solid ob-ject will always be zero! Engineers like to have polished, low-roughness surfaces on machines like aircraft, be-cause this reduces the thickness of the shear layer, and how turbulent it becomes, thus reducing shear — we will study this in chapter 10 (Flow near walls). But there is no escaping the fact that fuid particles next to the wall will always feature zero relative velocity. 5.5 Special case: shear in simple laminar fows In any ordinary fuid fow, the velocity feld is complex, and it is difcult to express shear and its net efect on particles. Since this requires expressing three values at each point in (three-dimensional) space and time, this would require complex mathematics or large amounts of discrete data. In simple cases, however, it is possible to express and calculate shear relatively easily. This is especially true in simple, steady, laminar (smooth) fows —typically fows for which the Reynolds number (eq.1/28 p. 23) is low. In those cases, we can guess a reasonably realistic velocity distribution, and then derive an expression for the distribution of shear from it. One such classical example is the Couette fow, where fuid is imprisoned between a static fat surface and another fat surface moving parallel to it, as illustrated in fgure 5.8. In this case, the bottom wall and top wall velocities, as well as the spacing 퐻, are known. Figure 5.8: A simple fow. The bottom wall is stationary, while the top wall slides from left to right. In between the walls, fuid is strained uniformly. Figure CC-by-sa Commons User:Kulmalukko 101 A reasonable guess for the velocity distribution in steady laminar regime is: { 푉푥 = 푉bottom wall + 푘푦 푉푦 = 0 By applying boundary conditions (푉bottom wall = 0 and 푉푥@ 푦=퐻= 푉top wall) we can re-write this as: { 푉푥 = 0 + 푉top wall 퐻 푦 푉푦 = 0 And now that the velocity feld is known, the shear everywhere in the fuid can be computed. The shear in the 푥-direction is proportional to the derivative in the 푦-direction of the velocity in the 푥-direction: 휏푦푥= 휇d d푦(0 + 푉top wall 퐻 푦) = 휇푉top wall 퐻 Thus, we see here that the shear applied in the fuid is the same everywhere (it is independent of 푦and 푥). A few slightly more complex cases are waiting for us in the problem sheet; but to handle more realistic shear distributions, what is needed is a software able to compute the behavior of fuids. The basic but formidable equations to be solved for this are the topic of the upcoming chapter 6 (Prediction of fuid fows). Advice from an expert Note that this method only works because we have guessed the velocity feld before we calcu-lated shear. We can only do this when fows are trivially simple. What if we don’t know what the velocity is to begin with? Things will be much harder once we have to eval-uate shear based on a calculated velocity feld, so that ⃗ 휏and ⃗ 푉depend on one another. Not to say, but it looks like the math in chapter 6 will be of its kadoova. 102 5.6 Solved problems Shear force on a fat plate A fat plate with dimensions 200 mm×400 mm is moved 1 mm above the ground at 0,5 m s−1, in a fuid with viscosity 1 ⋅10−5 Pa s. What is the shear force applying on the plate? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Cylindrical viscometer A moment of 0,9 N m is required to turn the inner cylinder of a viscometer (a device designed to measure viscosity) is turned at 150 rpm. The inner cylinder has diameter 20 cm, the height of the liquid is 80 cm, and the spacing between the two cylinders (greatly exaggerated on the drawing) is 2 mm. What is the viscosity of the fuid? 103 See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Note: Unfortunately Olivier made an error in this video: the height 퐻1 is forgotten in the last few lines. The fnal numerical result is nevertheless correct (휇= 2,27 ⋅10−2 Pa s). Many thanks to the students who double-checked and reported the problem! 104 Problem sheet 5: Efects of shear last edited June 5, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Shear force on a fat solid surface: 퐹shear, direction 푖= ∬ 푆 휏direction 푖d푆 (5/3) Shear in the direction 푗, on a plane perpendicular to direction 푖: \|\|⃗ 휏푖푗\|\| = 휇휕푉푗 휕푖 (5/22) −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 5.9: The viscosity of four fuids (crude oil, water, air, and C02) as a function of temperature. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. This is a reproduction of fgure 5.6 p. 99. Figure CC-by by Arjun Neyyathala & Olivier Cleynen 105 5.1 Quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 5.2 Flow in between two plates Munson & al. Ex1.5 A fuid is forced to fow between two stationary plates (fg. 5.10). We observe that the fow is laminar (smooth and fully steady), with a univorm velocity profle 푢= 푓(푦) which is linked to the average fuid velocity 푉average by the relationship: 푢= 3 2푉average [1 −( 푦 퐻) 2 ] (5/24) where 푦is measured from the middle of the gap; and 퐻is half of the gap length. Figure 5.10: Velocity distribution for laminar fow in between two plates, also known as Couette fow. Figure CC-0 Olivier Cleynen The fuid is water at 40 °C, the average velocity is 0,6 m s−1 and the two plates are 1,5 mm apart. 5.2.1. What is the shear efort 휏푦푥plate generated on the lower plate? 5.2.2. What is the shear efort 휏푦푥in the middle plane of the fow? 106 5.3 Friction on a plate A plate the size of an A4 sheet of paper (210 mm × 297 mm) is moved horizontally at constant speed above a large fat surface (fg. 5.11). We assume that the velocity profle of the fuid betweeen the plate and the fat surface is entirely uniform, smooth, and steady. Figure 5.11: A plate moved horizontally across a fat surface. Figure CC-0 Olivier Cleynen 5.3.1. Express the force 퐹푦푧due to shear on the plate as a function of its velocity 푈plate, the gap height 퐻, and the properties of the fuid. 5.3.2. The plate speed is 푈plate = 1 m s−1 and the gap height is 퐻= 5 mm. What is the shear force 퐹푦푧when the fuid is air at 40 °C, and when the fuid is crude oil at the same temperature? 5.3.3. If a very long and thin plate with the same surface area was used instead of the A4-shaped plate, would the shear force be diferent? (briefy justify your answer, e.g. in 30 words or less) 5.4 Viscometer Çengel & al. 2-78 An instrument designed to measure the viscosity of fuids (named viscometer) is made of two coaxial cylinders (fg. 5.12). The inner cylinder is immersed in a liquid, and it rotates within the stationary outer cylinder. The two cylinders are 75 cm tall. The inner cylinder diameter is 15 cm and the spacing is 1 mm. When the inner cylinder is rotated at 300 rpm, a friction-generated moment of 0,8 N m is measured. 5.4.1. If the fow in between the cylinders corresponds to the simplest possible fow case (steady, uniform, fully-laminar), what is the viscosity of the fuid? 5.4.2. Would a non-Newtonian fuid induce a higher moment? (briefy justify your answer, e.g. in 30 words or less) [Note: in practice, when the inner cylinder is turned at high speed, the fow displays mesmerizing patterns called Taylor—Couette vortices, the description of which is much more complex!] 107 Figure 5.12: Sketch of a cylinder viscometer. The width of the gap has been greatly exaggerated for clarity. Figure CC-0 Olivier Cleynen 5.5 Boundary layer White P1.56 A laminar fuid fow occurs along a wall (fg. 5.13). Close to the wall (푦< 훿), we observe that viscous efects dominate the mechanics of the fow. This zone is designated boundary layer. The speed 푢(푦) can then be modeled with the relation: 푢= 푈sin ( 휋푦 2훿) (5/25) in which 푈is the fow speed far away from the wall. Figure 5.13: Velocity profle across the boundary layer. Figure CC-0 Olivier Cleynen The fuid is CO2 at 20 °C; measurements yield 푈= 10,8 m s−1 and 훿= 3 cm. 5.5.1. What is the shear efort 휏푦푥on the wall? 5.5.2. At which height 푦1 above the surface will the shear efort be half of this value? 5.5.3. What would be the wall shear if the CO2 was replaced with water at the same temperature? 108 5.6 Clutch Çengel & al. 2-74 Two aligned metal shafts are linked by a clutch, which is made of two disks very close one to another, rotating in the same direction at similar (but not identical) speeds. The disk diameters are both 30 cm and the gap between them is 2 mm; they are submerged in crude oil with temperature 80 °C. Figure 5.14: Sketch of the two disks constituting the clutch. The gap width has been exaggerated for clarity. Figure CC-0 Olivier Cleynen The power shaft rotates at 1 450 rpm, while the powered shaft rotates at 1 398 rpm. We consider the simplest possible fow case (steady, laminar) in between the two disks. 5.6.1. What is the moment imparted by one disk to the other? 5.6.2. How would the moment change if the radius of each disk was doubled? 5.6.3. What is the transmitted power and the clutch efciency? 5.6.4. Briefy (e.g. in 30 words or less) propose one reason why in practice the fow in between the two disks may be diferent from the simplest-case fow used in this exercise. 109 Answers 5.2 1) 휏푦푥\|푦=−퐻= 1,44 N m−2; 2) 휏푦푥\|푦=0 = 0 N m−2. 5.3 1) 퐹푦푧= 퐿1퐿2휇푈 퐻= 2,38 ⋅10−4 N for air, and 6,55 ⋅10−2 N for oil. 5.4 휇= 푀퐿Δ푅/ (2휋휔푅3 1퐻) = 1,281 ⋅10−2 N s m−2. 5.5 1) 휏푦푥,wall CO2 = 8,369 ⋅10−3 N m−2; 2) 푦1 = 2 3훿= 2 cm; 3) and 휏푦푥,wall water = 0,565 N m−2 5.6 1) 푀= 휋 2 휇휔 ℎ푅4 = 7,145 ⋅10−3 N m; 3) ̇ 푊2 = 휔2푀= 1,05 W; 휂clutch = ̇ 푊2 ̇ 푊1 = 96,4 % (adequate for this very low-power, low-relative speed, laminar-fow case). 110 Fluid Dynamics Chapter 6 – Prediction of fuid fows last edited September 24, 2021 by Olivier Cleynen — 6.1 Motivation 111 6.2 Organizing calculations 112 6.2.1 Problem description 112 6.2.2 The total time derivative 112 6.3 Equations for all fows 116 6.3.1 Balance of mass 116 6.3.2 Balance of linear momentum 118 6.3.3 Balance of energy 120 6.3.4 Other terms and equations 121 6.3.5 Interlude: where does this leave us? 122 6.4 Equations for incompressible fow 123 6.4.1 Balance of mass 123 6.4.2 Balance of linear momentum 124 6.4.3 The Bernoulli equation (again) 127 6.5 CFD: the Navier-Stokes equations in practice 127 6.6 Solved problems 128 6.7 Problems 131 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 6.1 Motivation In this chapter we assign ourselves the daunting task of predicting the movement of fuids completely. We wish to express formally, and calculate, the dynamics of fuids —the velocity feld as a function of time— in any arbitrary situation. For this, we develop a methodology named derivative analysis. Let us start with the unfortunate truth: not only are the methods developed here are incredibly complex, but they are also very inefective to solve fuid fow problems with a pen and paper. Despite this, this chapter is extraordi-narily important, for two reasons: • derivative analysis allows us to formally describe and relate the key parameters that regulate fuid fow, and so, it is the key to developing an understanding of any fuid phenomenon, even when solutions cannot be derived; • it is the backbone for computational fuid dynamics (cfd) in which fow solutions are obtained using numerical procedures, in every problem of interest in research and industry today. 111 6.2 Organizing calculations 6.2.1 Problem description From now on, we wish to describe the velocity and pressure felds of a fuid with the highest possible resolution. For this, we aim to predict and describe the trajectory of fuid particles (recall §1.2.2 p. 10) as they travel. Newton’s second law (recall eq. 1/25 p. 20) allows us to quantify how the velocity vector of a particle varies with time. If we know all of the forces to which one particle is subjected, we can obtain a streak of velocity vectors ⃗ 푉particle = (푢, 푣, 푤) = 푓(푥, 푦, 푧, 푡) as the particle moves through our area of interest. This is a description of the velocity of one particle; we obtain a function of time which depends on where and when the particle started its travel: ⃗ 푉particle = 푓(푥0, 푦0, 푧0, 푡0, 푡). This is the process used in solid mechanics when we wish to describe the movement of one object, for example, a satellite in orbit. This kind of description, however, is poorly suited to the description of fuid fow, for three reasons: • Firstly, in order to describe a given fuid fow (e.g. air fow around the side mirror of a car), we would need a large number of initial points (푥0, 푦0, 푧0, 푡0), and we would then obtain as many trajectories ⃗ 푉particle. It then becomes very difcult to study and describe a problem that is local in space (e.g. the wake immediately behind the car mirror), because this requires fnding out where the particles of interest originated, and accounting for the trajectories of each of them. • Secondly, the concept of a “fuid particle” is not well-suited to the draw-ing of trajectories. Indeed, not only can particles strain indefnitely, but they can also difuse into the surrounding particles, “blurring” and blending themselves one into another. • Finally, the velocity of a given particle is very strongly afected by the properties (velocity, pressure) of the surrounding particles. We have to resolve simultaneously the movement equations of all of the particles. A space-based description of properties —one in which we describe properties at a chosen fxed point of coordinates 푥point, 푦point, 푧point, 푡point— is much more useful than a particle-based description which depends on departure points 푥0, 푦0, 푧0, 푡0. It is easier to determine the acceleration of a particle together with that of its current neighbors, than together with that of its initial (former) neighbors. What we are looking for, therefore, is a description of the velocity felds that is expressed in terms of a fxed observation point ⃗ 푉point = (푢, 푣, 푤) = 푓(푥point, 푦point, 푧point, 푡), through which particles of many diferent origins may be passing. This is termed a Eulerian fow description, as opposed to the particle-based Lagrangian description.w Grouping all of the point velocities in our fow study zone, we will obtain a velocity feld ⃗ 푉point that is a function of time. 6.2.2 The total time derivative Let us imagine a canal in which water is fowing at constant and uniform speed 푢= 푈canal (fg. 6.1). The temperature 푇water of the water is constant 112 (in time), but not uniform (in space). We measure this temperature with a stationary probe, reading 푇probe = 푇water on the instrument. Even though the temperature 푇water is constant, when reading the value measured at the probe, temperature will be changing with time: { 휕푇water 휕푡 \| \|particle = 0 휕푇water 휕푡 \| \|probe = −휕푇water 휕푥 푢water Figure 6.1: A one-dimensional water fow, for example in a canal. The water has a non-uniform temperature, which, even if it is constant in time, translates in a temperature rate change in time at the probe. Figure CC-0 Olivier Cleynen Advice from an expert The total time derivative does not describe some kind of funky physics phenomenon, like general relativity. It is very straightforward. It is important to realize that properties are not changed by the reference frame. In the canal example above, the temperature measured at the probe (푇probe) is always the temperature of the water (푇water), regardless of the fow conditions. By contrast, it is the change in time of the properties which is diferent. This is because as time passes, diferent particles keep hitting the probe, and they already have difering temperatures. Nothing spooky here! Advice from an expert Do not let yourself be fooled by the idea that this is just an abstract curiosity. Using the total time derivative is what enables modern computational fuid dynamics (cfd) to work at all. In those simulations, cells in a fxed, stationary grid are attributed coordinates. The fow through those cells is computed without ever following 113 the particles (this would rapidly become messy, as particles mix and tangle up). Cool kids do not ever “fow” the fuid fow. Instead, they oscillate vectors, like beautiful three-dimensional wheat felds gently oscillating with the wind. Video: fguring out the total time derivative by Olivier Cleynen (CC-by) Let us now study the case where the water temperature, in addition to being non-uniform, is also decreasing everywhere because the canal is cooling down. A particle’s temperature will then be changing at a non-zero rate d푇/ d푡, which also afects the reading at the probe: { 휕푇water 휕푡 \| \|particle ≠ 0 휕푇water 휕푡 \| \|probe = 휕푇water 휕푡 \| \|particle −휕푇water 휕푥푢water Re-arranging this last equation, we obtain the time change of the particle’s temperature, expressed from the reference frame of the probe: 휕푇water 휕푡 \| \|particle = 휕푇water 휕푡 \| \|probe + 휕푇water 휕푥푢water (6/1) We must keep in mind that all those derivatives can themselves be functions of time and space; in equation 6/1, it is their value at the position of the probe and at the time of measurement which is taken into account. Video: half-century-old, but timeless didactic exploration of the concept of Lagrangian and Eulerian derivatives, with ac-companying notes by Lumley by the National Committee for Fluid Mechanics Films (ncfmf, 1969) (styl) This line of thought can be generalized for three dimensions and for any property 퐴of the fuid (including vector properties). The property 퐴of one individual particle can vary as it is moving, so that it has a distribution 퐴= 푓(푥, 푦, 푧, 푡) within the fuid. The time rate change of 퐴expressed at a point fxed in space is named the total time derivative or simply total derivative1of 퐴and written D퐴/D푡: D D푡≡휕 휕푡+ 푢휕 휕푥+ 푣휕 휕푦+ 푤휕 휕푧 (6/2) D퐴 D푡= 휕퐴 휕푡+ 푢휕퐴 휕푥+ 푣휕퐴 휕푦+ 푤휕퐴 휕푧 (6/3) D ⃗ 퐴 D푡= 휕⃗ 퐴 휕푡+ 푢휕⃗ 퐴 휕푥+ 푣휕⃗ 퐴 휕푦+ 푤휕⃗ 퐴 휕푧 (6/4) = ⎛ ⎜ ⎜ ⎜ ⎝ 휕퐴푥 휕푡+ 푢휕퐴푥 휕푥+ 푣휕퐴푥 휕푦+ 푤휕퐴푥 휕푧 휕퐴푦 휕푡+ 푢휕퐴푦 휕푥+ 푣휕퐴푦 휕푦+ 푤휕퐴푦 휕푧 휕퐴푧 휕푡+ 푢휕퐴푧 휕푥+ 푣휕퐴푧 휕푦+ 푤휕퐴푧 휕푧 ⎞ ⎟ ⎟ ⎟ ⎠ (6/5) In equations 6/3 and 6/4, it is possible simplify the notation of the last three terms. For this, we have to recall two ingredients: the coordinates of the velocity vector ⃗ 푉(by defnition), and the components of the operator gradient ⃗ ∇(previously introduced in eq. 4/11 p. 77), writing them out as so: ⃗ 푉≡⃗ 푖푢+ ⃗ 푗푣+ ⃗ 푘푤 ⃗ ∇≡⃗ 푖휕 휕푥+ ⃗ 푗휕 휕푦+ ⃗ 푘휕 휕푧 1Unfortunately this term has many denominations across the literature, including advective, convective, hydrodynamic, Lagrangian, particle, substantial, substantive, or Stokes derivative. In this document, the term total derivative is used. 114 We can now defne the advective operator,w written ( ⃗ 푉⋅⃗ ∇) (see also Ap-pendix A3.3 p. 251): ⃗ 푉⋅⃗ ∇≡푢휕 휕푥+ 푣휕 휕푦+ 푤휕 휕푧 (6/6) We can now rewrite eqs. 6/2 and 6/3 in a more concise way: D D푡≡휕 휕푡+ ( ⃗ 푉⋅⃗ ∇) (6/7) D퐴 D푡≡휕퐴 휕푡+ ( ⃗ 푉⋅⃗ ∇)퐴 (6/8) D ⃗ 퐴 D푡≡휕⃗ 퐴 휕푡+ ( ⃗ 푉⋅⃗ ∇) ⃗ 퐴 (6/9) Advice from an expert To make sure you master the concept of total deriva-tive, try thinking of examples where only one member of equation 6/8 is zero. A fow that is steady from the point of view of the particle (D/D푡= 0) may be unsteady in the reference frame of the laboratory (휕/휕푡≠0). This happens for example where a fuid with non-uniform temperature fows at constant velocity. Conversely, a fow that is unsteady from the point of view of the particle may be steady in the reference frame of the laboratory. For example, in a jet engine nozzle, the air particles accelerate sharply (D/D푡≠0), however the fow is steady from the point of view of the jet engine (휕/휕푡= 0). The total time derivative is the tool that we were looking for. From now on, we can study fuid fows from a stationary reference frame, instead of in the reference frame of a moving particle. When we do so, all properties remain the same, but all the time derivatives d/ d푡are replaced with total derivatives D/ D푡. This allows us to compute the change in time of a property locally, without the need to track the movement of particles along our feld of study. All computational fuid dynamics (cfd) simulations work in this manner. The frst and most important such property we are interested in is accel-eration. Instead of solving for the acceleration of each of many particles (d ⃗ 푉particle/ d푡), we will instead calculate focus on calculating the acceleration feld D푉/ D푡: D ⃗ 푉 D푡= 휕⃗ 푉 휕푡+ ( ⃗ 푉⋅⃗ ∇) ⃗ 푉 (6/10) An illustration of an acceleration feld is shown further down in fgure 6.5 p. 126. 115 6.3 Equations for all fows 6.3.1 Balance of mass How can we write a balance of mass equation for an entire complex, unsteady, three-dimensional fuid fow? Let’s begin by considering a fuid particle of volume d, at a given instant in time (fg. 6.2). We can reproduce our analysis from chapter 3 (Analysis of existing fows with three dimensions) by quantifying the mass fows passing through an infnitesimal volume. In the present case, the control volume is stationary and the particle (our system) is fowing through it. We start with eq. 3/6 p. 54: d푚particle d푡 = 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (6/11) XKCD #1524: changes in time by Randall Munroe (CC-by-nc) The frst of these two integrals can be rewritten using the Leibniz integral rule: d d푡∭ CV 휌d= ∭ CV 휕휌 휕푡d+ ∬ CS 휌푉푆d퐴 = ∭ CV 휕휌 휕푡d (6/12) where 푉푆is the speed of the control volume wall; and where the term∬CS 휌푉푆d퐴is simply zero because we chose a fxed control volume, such as a fxed computation grid.1 Now we turn to the second term of equation 6/11, ∬CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴, which represents the mass fow ̇ 푚net fowing through the control volume. Figure 6.2: Conservation of mass within a fuid particle. In the 푥-direction, a mass fow ̇ 푚1 = ∬휌1푢1 d푧d푦is fowing in, and a mass fow ̇ 푚2 = ∬휌2푢2 d푧d푦is fowing out. These two fows may not be equal, since mass may also fow in the 푦- and 푧-directions. Figure CC-0 Olivier Cleynen 1In a cfd calculation in which the grid is deforming, this term ∬CS 휌푉푆d퐴will have to be re-introduced in the continuity equation. 116 In the direction 푥, the mass fow ̇ 푚net 푥fowing through our control volume can be expressed as: ̇ 푚net 푥= ∬ CS −휌1\|푢1\| d푧d푦+ ∬ CS 휌2\|푢2\| d푧d푦 = ∬ CS ∫ 휕 휕푥(휌푢) d푥d푧d푦 = ∭ CV 휕 휕푥(휌푢) d (6/13) The same applies for directions 푦and 푧, so that we can write: ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴= ̇ 푚net = ̇ 푚net 푥+ ̇ 푚net 푦+ ̇ 푚net 푧 = ∭ CV [ 휕 휕푥(휌푢) + 휕 휕푦(휌푣) + 휕 휕푧(휌푤)] d = ∭ CV ⃗ ∇⋅(휌⃗ 푉) d (6/14) (note that here we have again used the operator divergent, which we frst used in chapter 5 p. 95 — see also Appendix A3 p. 250) Now, with these two equations 6/12 and 6/14, we can come back to equa-tion 6/11, which becomes: d푚particle d푡 = 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 0 = ∭ CV 휕휌 휕푡d+ ∭ CV ⃗ ∇⋅(휌⃗ 푉) d (6/15) Since we are only concerned with a very small volume d, we drop the integrals, obtaining: 휕휌 휕푡+ ⃗ ∇⋅(휌⃗ 푉) = 0 (6/16) for all fows, with all fuids. This equation 6/16 is named continuity equationw and is of crucial importance in fuid mechanics. It sums up two terms: • The frst term on the left, 휕휌/휕푡, is the local time-change of density. For example, when a gas contracts as it cools down, its density increases, and the term becomes positive. In an incompressible fow, it is always zero. • The second term is the divergent of density times velocity, ⃗ ∇⋅휌⃗ 푉= 휕휌푢/휕푥+ 휕휌푣/휕푦+ 휕휌푤/휕푧. It sums up the changes in space of the mass fuxes 휌푉푖. For example, if a particle in a static fuid is heated up suddenly, it will expand and the divergent of 휌⃗ 푉will have positive value. 117 6.3.2 Balance of linear momentum What is the force feld applying to the fuid everywhere in space and time, and how does that afect its velocity feld? To answer this question, we write out a momentum balance equation. We start by writing Newton’s second law (eq. 1/25 p. 20) as it applies to a fuid particle of mass 푚particle, as shown in fg. 6.3. Fundamentally, the forces on a fuid particle are of only three kinds, namely weight, pressure, and shear:1 푚particle d ⃗ 푉 d푡= ⃗ 퐹weight + ⃗ 퐹net, pressure + ⃗ 퐹net, shear (6/17) We now write this equation from the point of view of a stationary cube of infnitesimal volume d, which is traversed by a fuid particle. We measure the time-change of velocity from the reference frame of the cube, making good use of the total time derivative tool we developed earlier p. 115: 푚particle D ⃗ 푉 D푡= ⃗ 퐹weight + ⃗ 퐹net, pressure + ⃗ 퐹net, shear 푚particle d D ⃗ 푉 D푡= 1 d ⃗ 퐹weight + 1 d ⃗ 퐹net, pressure + 1 d ⃗ 퐹net, shear 휌D ⃗ 푉 D푡= 1 d ⃗ 퐹weight + 1 d ⃗ 퐹net, pressure + 1 d ⃗ 퐹net, shear (6/18) Now, we rewrite the forces term on the right, and the hard work we did in the previous chapters is paying of. The force due to gravity is of course the weight. We have: 1 d ⃗ 퐹weight = 1 d푚⃗ 푔= 휌⃗ 푔 (6/19) Figure 6.3: In our study of fuid mechanics, we consider only forces due to gravity, shear, or pressure. Figure CC-0 Olivier Cleynen 1In some special applications, additional forces may also apply, see §6.3.4 p. 121 further down. 118 The force due to pressure was dealt with in chapter 4 (Efects of pressure). Back then, we expressed it with the help of the gradient of pressure in equation 4/13 p. 77, which we repeat here: 1 d ⃗ 퐹net, pressure = −⃗ ∇푝 (6/20) And fnally, we had dealt with the shear force in chapter 5 (Efects of shear). With somewhat efort, we had expressed it as a function of the divergent of shear in equation 5/19 p. 95, which we repeat here: 1 d ⃗ 퐹net, shear = ⃗ ∇⋅⃗ 휏푖푗 (6/21) Now, we can put together all of our fndings back into equation 6/18, we obtain the Cauchy equation:w 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ ⃗ ∇⋅⃗ 휏푖푗 (6/22) for all fows, with all fuids. The Cauchy equation is a formulation of Newton’s second law applied to a fuid particle. It expresses the time change of the velocity feld measured from a fxed reference frame (the acceleration feld D ⃗ 푉/D푡) as a the sum of the contributions of gravity, pressure and shear efects. This is quite a breakthrough. Within the chaos of an arbitrary fow, in which fuid particles are shoved, pressurized, squeezed, and distorted, we know precisely what we need to look for in order to quantify the time-change of velocity: gravity, the gradient of pressure, and the divergent of shear. Nevertheless, while it is an excellent start, this equation isn’t detailed enough for us. In our search for the velocity feld ⃗ 푉, the changes in time and space of the shear tensor ⃗ 휏푖푗and pressure 푝are unknowns. Ideally, those two terms should be expressed solely as a function of the fow’s other properties. Obtaining such an expression is what Claude-Louis Navier and Gabriel Stokes set themselves to in the 19th century: we follow their footsteps in the next paragraphs. The Navier-Stokes equation is the Cauchy equation (eq. 6/22) applied to Newtonian fuids. In Newtonian fuids, which we encountered in chapter 5 (§5.4.4 p. 100), shear eforts are simply proportional to the rate of strain; thus, the shear component of eq. 6/22 can be re-expressed usefully. We had seen with eq. 5/22 p. 97 that the norm \|\|⃗ 휏푖푗\|\| of shear component in direction 푗along a surface perpendicular to 푖depended on the viscosity and the velocity: \|\|⃗ 휏푖푗\|\| = 휇휕푢푗 휕푖 This is a one-dimensional (scalar) equation. Unfortunately, it does not trans-late easily into three dimensions. The required vector algebra far exceeds our level for this course, and we are interested only in the result (the derivation of this equation in Cartesian coordinates is covered in Anderson and Ver-steeg & Malalasekra , and the vector form can be found in Batchelor ). We obtain the heavy-handed result, in the form of a (three-dimensional) 119 vector feld: ⃗ ∇⋅⃗ 휏푖푗= 휇⃗ ∇2 ⃗ 푉+ 1 3휇⃗ ∇(⃗ ∇⋅⃗ 푉) (6/23) The details of the notation (which includes the Laplacian operator ⃗ ∇2) do not interest us at the moment; we will explore them later on. Adding this relationship between shear and the velocity feld into the last term of equation 6/22, we obtain the Navier-Stokes equation for compressible fow:w 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉+ 1 3휇⃗ ∇(⃗ ∇⋅⃗ 푉) (6/24) for all fows of a Newtonian fuid. This three-dimensional vector equation sets the conditions that are to be followed by the velocity feld ⃗ 푉, in all possible fows of a Newtonian fuid. We will use a simplifed version of this equation in section 6.4.2 below. 6.3.3 Balance of energy This topic is well covered in Anderson and Versteeg & Malalasekra How much energy is expended or received by the fuid particles as they travel through a complex, arbitrary fow? We answer this question with an energy balance equation. Once again, we start from the analysis of transfers on an infnitesimal control volume. We are going to relate three energy terms in the following form, naming them 퐴, 퐵and 퐶for clarity: the rate of change of energy inside the fuid element = the net fux of heat into the element + the rate of work done on the element due to body and surface forces 퐴 = 퐵 + 퐶 (6/25) Let us frst evaluate term 퐶. The rate of work done on the particle is the dot product of its velocity ⃗ 푉and the net force ⃗ 퐹net applying to it: 퐶= ⃗ 푉⋅( ⃗ 퐹weight d + ⃗ 퐹net, pressure d + ⃗ 퐹net, shear d ) d We replace the content of the parentheses with the right part of the Navier-Stokes equation above, obtaining the scalar term: 퐶= ⃗ 푉⋅[휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉+ 1 3휇⃗ ∇(⃗ ∇⋅⃗ 푉)] d (6/26) We now turn to term 퐵, the net fux of heat into the element. We attribute this fux to two contributions (i.e. 퐵= ̇ 푄radiation + ̇ 푄conduction). The frst contributor is the heat transfer ̇ 푄radiation from the emission or absorption of radiation, which we shyly express as: ̇ 푄radiation = 휌̇ 푞radiation d (6/27) in which ̇ 푞radiation is the local power per unit mass (in W kg−1) transfered to the element, to be determined from the boundary conditions and fow temperature distribution. 120 The second contributor the term 퐵is the named ̇ 푄conduction, attributed to thermal conduction through the faces of the element. In the 푥-direction, thermal conduction through the faces of the element causes a net fow of heat ̇ 푄conduction,푥expressed as a function of the power per area 푞(in W m−2) through each of the two faces perpendicular to 푥: ̇ 푄conduction,푥= [푞푥−(푞푥+ 휕푞푥 휕푥d푥)] d푦d푧 = −휕푞푥 휕푥d푥d푦d푧 (6/28) = −휕푞푥 휕푥d (6/29) In turn, the fuxes 푞푖can be expressed as a function of the local temperature gradients according to the Fourier law, 푞푖= −휅휕푇 휕푖 (6/30) where 휅is the conductivity of the fuid (W m−1 K−1). so that now we may write the heat transfer due to conduction as ̇ 푄conduction = ̇ 푄conduction,푥+ ̇ 푄conduction,푦+ ̇ 푄conduction,푧 = −[ 휕푞푥 휕푥+ 휕푞푦 휕푦+ 휕푞푧 휕푧] d (6/31) = 휅[ 휕2푇 (휕푥)2 + 휕2푇 (휕푦)2 + 휕2푇 (휕푧)2] d (6/32) Abstruse Goose #275: how scien-tists see the world by an anonymous artist (CC-by-nc) Finally, term 퐴, the rate of change of energy inside the fuid element, can be expressed as a function of the specifc internal energy 푖specifc kinetic energy 푒푘and : 퐴= 휌D D푡(푖+ 1 2푉2) d (6/33) We are therefore able quantify the change of energy of fuid particle with a scalar feld equation as follows: 휌D D푡(푖+ 1 2푉2) = 휌̇ 푞radiation + 휅[ 휕2푇 (휕푥)2 + 휕2푇 (휕푦)2 + 휕2푇 (휕푧)2] + ⃗ 푉⋅[휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉+ 1 3휇⃗ ∇(⃗ ∇⋅⃗ 푉)] (6/34) 6.3.4 Other terms and equations The more is happening in a given fow, and the more equations and equation terms are needed to describe it. Depending on the applications, new forces may become dominant and might be added to gravity, pressure and shear in our equations. For example: • In a description of an ocean fow, a tidal force may be added; 121 • In a description of a large atmospheric fow, a Coriolis force may be added; • In a description of sap fow in a tree or plant, forces related to surface tension may be added. Additionally, modeling some more advanced fuid phenomena requires alto-gether new equations, for example: • Equations to describe chemical reactions, such as combustion; • Equations to distinguish and model the interaction of several fuids (e.g. in fow featuring drops, bubbles, or non-uniform mixes). 6.3.5 Interlude: where does this leave us? So far, we have written fve equations: • One equation for balance of mass (eq. 6/16 p. 117); • Three equations for balance of momentum (the three components of the vector equation 6/24 p. 120); • One equation for balance of energy (eq. 6/34 p. 121). In order to solve an arbitrary fow involving transfers of both momentum and energy —for example, the fow of air in a room when an electric fan heater is turned on— we need to solve all fve equations simultaneously. It is hard to hide that the mathematical complexity of the problem is simply intractable. No method is available to fnd a solution of this problem “by hand”, and the computational costs to fnd numerical solutions with computers are enormous. This is the reason why in the section below, we narrow down our scope down to more restrictive conditions: incompressible fows in which the energy does not signifcantly vary. We will write again a balance of mass and a balance of momentum, and take more time to explore the shape and behavior of the resulting four equations. Advice from an expert Fluid dynamicists are not easy to observe in the wild, but if you get the chance, you will likely fnd them busy trying to not solve equations! At the start of every problem, the efective fuid dynamicist will try to fgure which terms s/he can neglect, and reduce the number of equations to solve. This is because in practice, both experiments and computations are prohibitively expensive, and answers need to be ob-tained in reasonable amounts of time. In fact, we will soon see how we can “weigh” the relative importance of equation terms. In chapter 8 (Engineering models) and following, we’ll be removing terms from eq. 6/24 until only the signifcant ones remain. 122 6.4 Equations for incompressible fow For the rest of this document, we focus on fuid fows that are incompressible (§1.8 p. 21): those for which 휌is uniform and constant. 6.4.1 Balance of mass To write a balance of mass for incompressible fow, we begin where we left of with the general mass balance (eq. 6/16 p. 117), which we re-write here: 휕휌 휕푡+ ⃗ ∇⋅(휌⃗ 푉) = 0 In this equation, we see that if 휌is uniform and constant, the frst term will vanish. Once this happens, 휌can be simply dropped from the second term. This leaves us with the (much simpler) mass balance equation for incompressible fow, also called incompressible continuity equation: ⃗ ∇⋅⃗ 푉= 0 (6/35) for any incompressible fow. This equation states that “the divergent of velocity is zero” and it is a scalar equation. In three Cartesian coordinates, it can be re-expressed as: 휕푢 휕푥+ 휕푣 휕푦+ 휕푤 휕푧= 0 (6/36) In spite of its elegance, equation 6/35 is not very talkative: it gives us no particular information about the orientation of ⃗ 푉or about its change in time. How should we calculate 푢, 푣or 푤if we have only one information about the sum of their derivatives in space? In practice, the equation is insufcient to solve the majority of problems in fuid mechanics, and it is used as a kinematic constraint to solutions used to evaluate their physicality or the quality of the approximations made to obtain them. Advice from an expert As strange as it sounds, the continuity equation is a constant source of frustration for fuid dynamicists. If you already have the three felds 푢, 푣or 푤, you can plug them into equation 6/36, and hope it adds up to zero. But if you don’t (for example, if you are missing 푤), how do you get to the missing element? Only one partial derivative is specifed for each of the elements, leaving a lot of unknown dependencies (e.g. the functions relating 푤to 푥, 푦and 푡). A great deal of completely nonsensical fuid fows will abide by the continuity equation. They cannot be realized, because they would require non-physical changes in fuid momentum. How do we determine what is possible? We need a momentum balance equation. 123 6.4.2 Balance of linear momentum Video: worst one ever? The incompressible Navier-Stokes equation by Olivier Cleynen (CC-by) Here, we start from the balance of momentum equation which we obtained for general fow as the Cauchy equation, eq. 6/22 p. 119: 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ ⃗ ∇⋅⃗ 휏푖푗 Clearly, the “hard” term in this equation is the last one: the divergent of shear ⃗ ∇⋅⃗ 휏푖푗, which has three components: ⃗ ∇⋅⃗ 휏푖푥, ⃗ ∇⋅⃗ 휏푖푦and ⃗ ∇⋅⃗ 휏푖푧. Let us frst focus on the 푥-direction, using the diagram in fgure 6.4. In the 푥-direction, the net efect of shear on the six faces is a vector expressed as the divergent ⃗ ∇⋅⃗ 휏푖푥, which, in incompressible fow, can be expressed with with the help of viscosity 휇as: ⃗ ∇⋅⃗ 휏푖푥= 휕⃗ 휏푥푥 휕푥+ 휕⃗ 휏푦푥 휕푦+ 휕⃗ 휏푧푥 휕푧 = 휕(휇휕푢 휕푥⃗ 푖) 휕푥 + 휕(휇휕푢 휕푦⃗ 푖) 휕푦 + 휕(휇휕푢 휕푧⃗ 푖) 휕푧 = 휇휕( 휕푢 휕푥) 휕푥 ⃗ 푖+ 휇 휕( 휕푢 휕푦) 휕푦 ⃗ 푖+ 휇휕( 휕푢 휕푧) 휕푧 ⃗ 푖 = 휇( 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2) ⃗ 푖 (6/37) To tidy up this tedious equation structure, we introduce the Laplacian oper-ator (see also Appendix A3 p. 250) to represent the spatial variation of the spatial variation of an object: ⃗ ∇2 ≡⃗ ∇⋅⃗ ∇ (6/38) ⃗ ∇2퐴≡⃗ ∇⋅⃗ ∇퐴 (6/39) ⃗ ∇2 ⃗ 퐴≡ ⎛ ⎜ ⎜ ⎝ ⃗ ∇2퐴푥 ⃗ ∇2퐴푦 ⃗ ∇2퐴푧 ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ ⃗ ∇⋅⃗ ∇퐴푥 ⃗ ∇⋅⃗ ∇퐴푦 ⃗ ∇⋅⃗ ∇퐴푧 ⎞ ⎟ ⎟ ⎠ (6/40) Figure 6.4: Shear eforts on a cubic fuid particle (already represented in fg. 5.1 p. 93). Figure CC-0 Olivier Cleynen 124 And now, we can re-write eq. 6/37 more elegantly and generalize to three dimensions: ⃗ ∇⋅⃗ 휏푖푥= 휇⃗ ∇2푢⃗ 푖= 휇⃗ ∇2 ⃗ 푢 ⃗ ∇⋅⃗ 휏푖푦= 휇⃗ ∇2푣⃗ 푗= 휇⃗ ∇2 ⃗ 푣 ⃗ ∇⋅⃗ 휏푖푧= 휇⃗ ∇2푤⃗ 푘= 휇⃗ ∇2 ⃗ 푤 The three last equations are grouped together simply as ⃗ ∇⋅⃗ 휏푖푗= 휇⃗ ∇2 ⃗ 푉 (6/41) With this new expression, we can come back to the Cauchy equation (eq. 6/22 p. 119), in which we can replace the shear term with eq. 6/41. This produces the beautiful Navier-Stokes equation for incompressible fow: 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (6/42) for all incompressible fows of a Newtonian fuid. XKCD #435: scientifc felds sorted by purity by Randall Munroe (CC-by-nc) This simplifed but still formidable equation describes the property felds of all incompressible fows of Newtonian fuids. It expresses the acceleration feld (left-hand side) as the sum of three contributions (right-hand side): those of gravity, gradient of pressure, and divergent of shear. The solutions we look for in equation 6/42 are the velocity (vector) feld ⃗ 푉= (푢, 푣, 푤) = 푓1(푥, 푦, 푧, 푡) and the pressure feld 푝= 푓2(푥, 푦, 푧, 푡), given a set of constraints to represent the problem at hand. Though it is without doubt charming, equation 6/42 should be remembered for what it is really: a three-dimensional system of coupled equations. In Cartesian coordinates this complexity is more apparent: 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦+ 푤휕푢 휕푧] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2] (6/43) 휌[ 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦+ 푤휕푣 휕푧] = 휌푔푦−휕푝 휕푦+ 휇[ 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2 + 휕2푣 (휕푧)2] (6/44) 휌[ 휕푤 휕푡+ 푢휕푤 휕푥+ 푣휕푤 휕푦+ 푤휕푤 휕푧] = 휌푔푧−휕푝 휕푧+ 휇[ 휕2푤 (휕푥)2 + 휕2푤 (휕푦)2 + 휕2푤 (휕푧)2] (6/45) Advice from an expert The incompressible Navier-Stokes equation is the crux of modern, practical fuid dynamics. Don’t be one of those students who just casu-ally dismiss it, saying “computers will solve it”. To understand why real-world fuid dynamicists work so hard rigging and confguring their sim-ulations and experiments, and then spend so much time waiting for their computational simulations to complete, you must have minimal experience playing with the mathematics. Take 125 the vector equation 6/42, and practice expanding it into eqs. 6/43-6/45. It’s a necessary step, like the frst time jumping in the deep end of the swimming pool! Today indeed, 150 years after it was frst written, no general expression has been found for velocity or pressure felds that would solve this vector equation in the general case. Nevertheless, in this course we will use it directly: • to understand and quantify the importance of key fuid fow parameters, in chapter 8 (Engineering models); • to fnd analytical solutions to fows in a few selected cases, in the other remaining chapters. After this course, the reader might also engage into computational fuid dynamics (cfd) a discipline entirely architectured around this equation, and to which it purposes to fnd solutions as felds of discrete values. Figure 6.5: The velocity (top) and acceleration (bottom) felds in the fow feld of the computed fow described in fgure 4.4 p. 78. The velocity feld is the result of a computational fuid dynamics simulation (steady, Reynolds-Averaged: a modifed version of eq. 6/42). The acceleration feld is obtained based on that solution, with equation 6/10, but could also have been obtained with the right side of equation 6/42. Figure CC-by by Arjun Neyyathala 126 6.4.3 The Bernoulli equation (again) Video: Bernoulli wants to help Navier and Stokes by Olivier Cleynen (CC-by) We had made clear in chapter 2 (Analysis of existing fows with one dimension) that the Bernoulli equation was very limited in scope, and that it was always safer to approach a problem from an energy equation instead (§2.6 p. 41). As a reminder of this fact, and as an illustration of the bridges that can be built between integral and derivative analysis, it can be instructive to derive the Bernoulli equation directly from the Navier-Stokes equation. This derivation is not difcult to follow; it is covered in Appendix A4.3 p. 253. 6.5 CFD: the Navier-Stokes equations in practice This topic is well covered in Versteeg & Malalasekera Video: the very basics of CFD by Olivier Cleynen (CC-by) In our analysis of fuid fow from a derivative perspective, our fve physical principles from §1.7 have been condensed into three balance equations (often loosely referred together to as the Navier-Stokes equations). Out of these, the frst two, for conservation of mass (6/35) and linear momentum (6/42) in incompressible fows, are often enough to characterize most free fows, and should in principle be enough to fnd the primary unknown, which is the velocity feld ⃗ 푉: 0 = ⃗ ∇⋅⃗ 푉 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 We know of many individual analytical solutions to this mathematical prob-lem: they apply to simple cases, and we shall describe several such fows in the upcoming chapters. However, we do not have one general solution: one that would encompass all of them. For example, in solid mechanics we have long understood that all pure free fall movements can be described with the solution 푥= 푥0 + 푢0푡and 푦= 푦0 + 푣0푡+ 1 2푔푡2, regardless of the particularities of each fall. In fuid mechanics, even though our analysis was carried out in the same manner, we have yet to fnd such one general solution — or even to prove that one exists at all. It is therefore tempting to attack the above pair of equations from the nu-merical side, with a computer algorithm. If one discretizes space and time in small increments 훿푥, 훿푦, 훿푧and 훿푡, we could re-express the 푥-component of eq. 6/42 as: 휌[ 훿푢\|푡 훿푡+ 푢훿푢\|푥 훿푥+ 푣훿푢\|푦 훿푦+ 푤훿푢\|푧 훿푧] = 휌푔푥−훿푝\|푥 훿푥+ 휇 ⎡ ⎢ ⎢ ⎣ 훿훿푢\|푥 훿푥 \| \| \|푥 훿푥 + 훿훿푢\|푦 훿푦 \| \| \|푦 훿푦 + 훿훿푢\|푧 훿푧 \| \| \|푧 훿푧 ⎤ ⎥ ⎥ ⎦ (6/46) If we start with a known (perhaps guessed) initial feld for velocity and pressure, this equation 6/46 allows us to isolate and solve for 훿푢\|푡, and therefore predict what the 푢velocity feld would look like after a time increment 훿푡. The same can be done in the 푦- and 푧-directions. Repeating the process, we then proceed to the next time step and so on, marching in time, obtaining at every new time step the value of 푢, 푣and 푤at each position within our computation grid. The discretization schemes and solver algorithms used in practice in con-temporary software are of course more subtle than those described here; 127 nevertheless, this single-page brief does lay out the fundamental working principle of computational fuid dynamics (cfd) today. Diving into the intricacies of cfd is beyond the scope of our study. Never-theless, the remarks in this last section should hopefully hint at the fact that an understanding of the mathematical nature of the diferential conservation equations is of great practical importance in fuid dynamics. It is for that reason that the problems in this chapter are dedicated to playing with the mathematics of our two main equations. 6.6 Solved problems Acceleration feld A fow has the velocity feld ⃗ 푉= (2 + 3푥)⃗ 푖+ (4 −3푦)⃗ 푗(in si units). What is the acceleration feld? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Playing with the continuity equation An incompressible fow has the velocity feld defned as follows: 푢= 2푥2 −푦2 + 푧2 푣= 3푥푦+ 3푦푧+ 푧 푤= ? How must 푤be to satisfy the mass balance equation? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Note: Unfortunately Olivier made an error in this video: at 4:00 the term 2푥2 is incorrectly derived into 2푥. The fnal result should have −7푥푧instead of −5푥푧 as the frst term. Many thanks to the students who double-checked and reported the problem! 128 Playing with the Navier-Stokes equation An incompressible fow with no gravity has the velocity feld defned as follows: ⃗ 푉= (푎푥+ 푏)⃗ 푖+ (−푎푦+ 푐푥)⃗ 푗 Does a function exist to describe the pressure feld, and if so, what is it? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 129 130 Problem sheet 6: Prediction of fuid fows last edited June 12, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Continuity equation for incompressible fow: ⃗ ∇⋅⃗ 푉= 0 (6/35) Navier-Stokes equation for incompressible fow: 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (6/42) 6.1 Quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 6.2 Revision questions For the continuity equation (eq. 6/35), and then for the incompressible Navier-Stokes equation (eq. 6/42), 6.2.1. Write out the equation in its fully-developed form in three Cartesian coordinates; 6.2.2. State in which fow conditions the equation applies. Also, in order to revise the notion of total (or substantial) derivative: 6.2.3. Describe a situation in which the total time derivative D/D푡of a property is non-zero, even though the fow is entirely steady (휕/휕푡= 0). 6.2.4. Describe a situation in which the the fow is unsteady, although some property of the fuid, when measured from the point of view of a fuid particle, is not changing with time. 6.3 Acceleration feld Çengel & al. E4-3 A fow is described with the velocity feld ⃗ 푉= (0,5 + 0,8푥)⃗ 푖+ (1,5 −0,8푦)⃗ 푗(in si units, in the laboratory frame of reference). What is the acceleration of a particle positioned at (2; 2; 2) at 푡= 3 s ? 131 6.4 Volumetric dilatation rate der. Munson & al. 6.4 A fow is described by the following feld (in si units): 푢= 푥3 + 푦2 + 푧 푣= 푥푦+ 푦푧+ 푧3 푤= −4푥2푧−푧2 + 4 What is the volumetric dilatation rate feld (the divergent of the velocity feld)? What is the value of this rate at {2;2;2}? 6.5 Incompressibility Çengel & al. 9-28 Does the vector feld ⃗ 푉= (1,6 + 1,8푥)⃗ 푖+ (1,5 −1,8푦)⃗ 푗satisfy the continuity equation for two-dimensional incompressible fow? 6.6 Missing components Munson & al. E6.2 + Çengel & al. 9-4 Two fows are described by the following felds: 푢1 = 푥2 + 푦2 + 푧2 푣1 = 푥푦+ 푦푧+ 푧 푤1 = ? 푢2 = 푎푥2 + 푏푦2 + 푐푧2 푣2 = ? 푤2 = 푎푥푧+ 푏푦푧2 What must 푤1 and 푣2 be so that these fows be incompressible? 6.7 Another acceleration feld White E4.1 Given the velocity feld ⃗ 푉= (3푡)⃗ 푖+ (푥푧)⃗ 푗+ (푡푦2) ⃗ 푘(si units), what is the acceleration feld, and what is the value measured at {2;4;6} and 푡= 5 s? 6.8 Vortex Çengel & al. 9.27 A vortex is modeled with the following two-dimensional fow: 푢= 퐶 푦 푥2 + 푦2 푣= −퐶 푥 푥2 + 푦2 Verify that this feld satisfes the continuity equation for incompressible fow. 132 6.9 Pressure felds Çengel & al. E9-13, White 4.32 & 4.34 We consider the four (separate and independent) incompressible fows below: ⃗ 푉1 = (푎푥+ 푏)⃗ 푖+ (−푎푦+ 푐푥)⃗ 푗 ⃗ 푉2 = (2푦)⃗ 푖+ (8푥)⃗ 푗 ⃗ 푉3 = (푎푥+ 푏푡)⃗ 푖+ (푐푥2 + 푒푦)⃗ 푗 ⃗ 푉4 = 푈0 (1 + 푥 퐿) ⃗ 푖−푈0 푦 퐿 ⃗ 푗 The infuence of gravity is neglected on the frst three felds. Does a function exist to describe the pressure feld of each of these fows, and if so, what is it? 133 Answers 6.2 1) Continuity: eq. 6/36 p. 123. Navier-Stokes: see eqs. 6/43, 6/44 and 6/45 p. 125; 2) Read §6.3.1 p. 116 for continuity, and §6/22 p. 119 for Navier-Stokes; 3) and 4) see §6.2.2 p. 112. 6.3 D ⃗ 푉 D푡= (0,4 + 0,64푥)⃗ 푖+ (−1,2 + 0,64푦)⃗ 푗. At the probe it takes the value 1,68⃗ 푖+ 0,08⃗ 푗 (length 1,682 m s−2). 6.4 ⃗ ∇⋅⃗ 푉= −푥2 + 푥−푧; thus at the probe it takes the value (⃗ ∇⋅⃗ 푉)probe = −4 s−1. 6.5 Apply equation 6/36 p. 123 to ⃗ 푉: the answer is yes. 6.6 1) Applying equation 6/36: 푤1 = −3푥푧−1 2푧2 + 푓(푥,푦,푡); 2) idem, 푣2 = −3푎푥푦−푏푧푦2 + 푓(푥,푧,푡). 6.7 D ⃗ 푉 D푡= (3)⃗ 푖+(3푧+푦2푥)푡⃗ 푗+(푦2+2푥푦푧푡) ⃗ 푘. At the probe it takes the value 3⃗ 푖+250⃗ 푗+496 ⃗ 푘. 6.8 Apply equation 6/36 to ⃗ 푉to verify incompressibility. 6.9 Note: the constant (initial) value 푝푖푛푖is sometimes implicitly written in the un-known functions 푓. 1) 푝= −휌[푎푏푥+ 1 2푎2푥2 + 푏푐푦+ 1 2푎2푦2] + 푝푖푛푖+ 푓(푡); 2) 푝= −휌(8푥2 + 8푦2) + 푝푖푛푖+ 푓(푡); 3) 휕 휕푥( 휕푝 휕푦) ≠ 휕 휕푦( 휕푝 휕푥), thus we cannot describe the pressure with a mathematical function; 4) 푝= −휌[ 푈2 0 퐿(푥+ 푥2 2퐿+ 푦2 2퐿) −푔푥푥−푔푦푦] + 푝푖푛푖+ 푓(푡). 134 Fluid Dynamics Chapter 7 – Pipe fows last edited September 3, 2020 by Olivier Cleynen — 7.1 Motivation 135 7.2 Frictionless fow in pipes 135 7.3 Parameters to quantify losses in pipes 137 7.4 Laminar fow in pipes 137 7.4.1 Laminar fow between plates 137 7.4.2 Laminar fow in pipes 139 7.5 Turbulent fow in pipes 142 7.5.1 When is a pipe fow turbulent? 142 7.5.2 Characteristics of turbulent fow 143 7.5.3 Velocity profle in turbulent pipe fow 144 7.5.4 Pressure losses in turbulent pipe fow 144 7.6 Engineer’s guide to pipe fows 146 7.6.1 Summary so far 146 7.6.2 Choosing laminar or turbulent fow 146 7.6.3 Pressure losses in laminar fow 147 7.6.4 Pressure losses in turbulent fow 147 7.6.5 Calculating pumping and turbining power 148 7.7 Solved problems 148 7.8 Problems 151 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 7.1 Motivation In this chapter we focus on fuid fow in pipes. This topic allows us to explore several important phenomena with only very modest mathematical complexity. In particular, we are trying to answer two questions: 1. What does it take to describe fuid fow in ducts? 2. How can we quantify pressure changes in pipes and the power neces-sary to overcome them? 7.2 Frictionless fow in pipes We begin with the simplest possible ducted fow case: a purely hypothetical fully-inviscid, incompressible, steady fuid fow in a one-dimensional pipe. Since there are no shear forces, the velocity profle across the duct remains uniform (fat) all along the fow, as shown in fgure 7.1. If the cross-sectional area 퐴is changed, then the principle of mass conserva-tion (eqs. 1/24, 3/6) is enough to allow us to compute the change in velocity 푢= 푉: 휌푉1퐴1 = 휌푉2퐴2 (7/1) in steady pipe fow. 135 Figure 7.1: Inviscid fuid fow in a one-dimensional duct. In this purely hypothetical case, the velocity distribution is uniform across a cross-section of the duct. The average velocity and pressure change with cross-section area, but the total pressure 푝0 = 푝+ 1 2휌푉2 av. remains constant. Figure CC-0 Olivier Cleynen Abstruse Goose #517: Daniel Bernoulli was part of a big familyw by an anonymous artist (CC-by-nc) This information, in turn allows us to compute the pressure change between two sections of diferent areas by using the principle of energy conservation. We notice that the fow is so simple that the fve conditions associated with the use of the Bernoulli equation (see §2.6 p. 41) are fulflled: the fow is steady, incompressible, one-dimensional, has known trajectory, and does not feature friction or energy transfer. A simple application of eq. 2/20 p. 42 between any two points 1 and 2 gives us: 푝2 −푝1 = −1 2휌[푉2 2 −푉2 1 ] −휌푔(푧2 −푧1) (7/2) in steady, incompressible, inviscid pipe fow without heat or work transfer. Thus, in this kind of simple fow, pressure increases everywhere the velocity decreases, and vice-versa. Another way of writing this equation is by stating that at constant altitude, the total or dynamic pressure 푝total ≡푝0 ≡푝+ 1 2휌푉2 remains constant: 푝0 = cst. (7/3) at constant altitude, in laminar inviscid straight pipe fow. Inviscid fows are nice, but real fows are more interesting. Real fows feature losses due to viscosity, and with viscous efects, one key assumption of the Bernoulli equation breaks down. An additional term will appear in the Bernoulli equation, as we have seen in chapter 2 with equation 2/21 p. 43. What does this extra term Δ푝loss depend on, and how can we quantify it? This is what the rest of the chapter is about. 136 7.3 Parameters to quantify losses in pipes Hydraulics is the oldest branch of fuid dynamics, and much of the notation used to describe pressure losses predates modern applications. The most widely-used parameters for quantifying losses due to friction in a duct are the following: The pressure loss is the most intuitive way of quantifying the net efect of friction in pipes. Engineers and physicists usually quantify Δ푝loss as a negative number (i.e. Δ푝loss ≡푝2 −푝1 with 푝2 < 푝1). However, in hydraulics, the historical precedent is to quantify pressure loss with a positive number. To make clear this convention, we will always refer to pressure loss as \|Δ푝loss\|. The elevation loss which we note \|Δ푙\| (in the literature, often noted Δℎ), is defned as \|Δ푙\| ≡\|Δ푝loss\| 휌푔 (7/4) It represents the hydrostatic height loss (with a positive number) caused by the fuid fow in the duct, and is measured in meters. The reference density 휌in this defnition is taken as the density of the fuid, the density of water, or the density of mercury, depending on cases. We do not use this defnition in this course. The Darcy friction factor noted 푓is defned as 푓≡\|Δ푝loss\| 퐿 퐷 1 2휌푉2 av. (7/5) where 푉av. is the average fow velocity in the pipe. In pipe fows of relevance to the engineer, 푓has values between 5 ⋅10−5 and 5 ⋅10−2. Those values can be calculated or read from experimental data, as explained further down. The loss coefcient noted 퐾퐿is defned as 퐾퐿≡\|Δ푝loss\| 1 2휌푉2 av. (7/6) Components found in pipe networks, such as bends, flter screens, valves, or junction screens, all result in losses that remain roughly proportional to the square of the average velocity. Typically, 퐾퐿values range between 0,3 (smooth bend) and 2 (partially-closed valve). 7.4 Laminar fow in pipes 7.4.1 Laminar fow between plates Before we study fuid fow in a circular pipe, let us begin with a simpler case: fow between two parallel plates. This is a good place to start, because we can work with Cartesian coordinates, and focus on two dimensions only. 137 Let us frst try a qualitative description of the fow, as displayed in fgure 7.2. Because of the no-slip condition at the walls, the velocity distribution within any cross-fow section cannot be uniform. Shear occurs, which translates into a pressure decrease along the fow. The faster the fow, and the higher the gradient of velocity. Thus, shear within the fow, and the resulting pressure loss, both increase when the cross-sectional area is decreased. Figure 7.2: Viscous laminar fuid fow in a one-dimensional pipe. This time, the no-slip condition at the wall creates a viscosity gradient across the duct cross-section. This in turn translates into pressure loss. Sudden duct geometry changes such as represented here would also disturb the fow further, but the efect was neglected here. Figure CC-0 Olivier Cleynen How can we now describe quantitatively the velocity profle and the pressure loss? We need to clearly sketch the fow we are interested in, which we do in fgure 5.10. We also need a powerful, extensive mathematical tool to describe the fow: we turn to the Navier-Stokes equation which we derived in the previous chapter as eq. 6/42 p. 125: 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (7/7) Figure 7.3: Two-dimensional laminar fow between two plates, also called Poiseuille fow. We already studied this fow case in fg. 5.10 p. 106; this time, we wish to derive an expression for the velocity distribution. Figure CC-0 Olivier Cleynen 138 Since we are applying this tool to the simple case of fully-developed, two-dimensional incompressible fuid fow between two parallel plates (fg. 7.3), we need only two Cartesian coordinates, so that the vector equation trans-lates to: 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2] (7/8) 휌[ 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦] = 휌푔푦−휕푝 휕푦+ 휇[ 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2] (7/9) In this particular fow, we have restricted ourselves to a fully-steady (d/ d푡= 0), horizontal (푔= 푔푦), one-directional fow (푣= 0). When the fow is fully de-veloped, 휕푢/휕푥= 0 and 휕2푢/(휕푥)2 = 0, and the system above shrinks down to: 0 = −휕푝 휕푥+ 휇[ 휕2푢 (휕푦)2] (7/10) 0 = 휌푔−휕푝 휕푦 (7/11) We only have to integrate equation 7/10 twice with respect to 푦to come to the velocity profle across two plates separated by a height 2퐻: 푢= 1 2휇( 휕푝 휕푥) (푦2 −퐻2) (7/12) Now, the longitudinal pressure gradient 휕푝/휕푥can be evaluated by working out the volume fow rate ̇ for any given width 푍with one further integration of equation 7/12: ̇  푍= 2 푍∫ 퐻 0 푢푍d푦= −2퐻3 3휇( 휕푝 휕푥) 휕푝 휕푥= −3 2 휇 푍퐻3 ̇  (7/13) In this section, the overall process is more important than the result: by starting with the Navier-Stokes equations, and adding known constraints that describe the fow of interest, we can predict analytically all of the characteristics of a laminar fow. 7.4.2 Laminar fow in pipes We now turn to studying fow in cylindrical pipes, which are widely used; frst considering laminar fow, and then expanding to turbulent fow. The process is identical to above, only applied to cylindrical instead of Cartesian coordinates. We focus on the fully-developed laminar fow of a fuid in a cylindrical pipe without gravity (fg. 7.4). For this fow, we wish to work out the velocity profle and calculate the pressure loss related to the fow. 139 Figure 7.4: A cylindrical coordinate system to study laminar fow in a cylindrical duct. Figure CC-0 Olivier Cleynen We once again start from the Navier-Stokes vector equation, choosing this time to develop it using cylindrical coordinates: 휌[ 휕푣푟 휕푡+ 푣푟 휕푣푟 휕푟+ 푣휃 푟 휕푣푟 휕휃−푣2 휃 푟+ 푣푧 휕푣푟 휕푧] = 휌푔푟−휕푝 휕푟+ 휇[ 1 푟 휕 휕푟(푟휕푣푟 휕푟) −푣푟 푟2 + 1 푟2 휕2푣푟 (휕휃)2 −2 푟2 휕푣휃 휕휃+ 휕2푣푟 (휕푧)2] (7/14) 휌[ 휕푣휃 휕푡+ 푣푟 휕푣휃 휕푟+ 푣휃 푟 휕푣휃 휕휃+ 푣푟푣휃 푟 + 푣푧 휕푣휃 휕푧] = 휌푔휃−1 푟 휕푝 휕휃+ 휇[ 1 푟 휕 휕푟(푟휕푣휃 휕푟) −푣휃 푟2 + 1 푟2 휕2푣휃 (휕휃)2 + 2 푟2 휕푣푟 휕휃+ 휕2푣휃 (휕푧)2] (7/15) 휌[ 휕푣푧 휕푡+ 푣푟 휕푣푧 휕푟+ 푣휃 푟 휕푣푟 휕휃+ 푣푧 휕푣푧 휕푧] = 휌푔푧−휕푝 휕푧+ 휇[ 1 푟 휕 휕푟(푟휕푣푧 휕푟) + 1 푟2 휕2푣푧 (휕휃)2 + 휕2푣푧 (휕푧)2] (7/16) XKCD #1230: polar coordinates by Randall Munroe (CC-by-nc) This mathematical arsenal does not frighten us, for the simplicity of the fow we are studying allows us to bring in numerous simplifcations. First, we have 푔= 0. Second, we have 푣푟= 0 and 푣휃= 0 everywhere. Thus, by continuity, 휕푣푧/휕푧= 0. Furthermore, since our fow is symmetrical, 푣푧is independent from 휃. With these two conditions, the above system shrinks down to: 0 = 0 (7/17) 0 = 0 (7/18) 0 = −휕푝 휕푧+ 휇[ 1 푟 휕 휕푟(푟휕푣푧 휕푟)] (7/19) Now, with equation 7/19, we work towards obtaining an expression for 푣푧by integrating twice our expression for 휕푣푧/휕푟: 휕 휕푟(푟휕푣푧 휕푟) = 푟 휇 휕푝 휕푧 (푟휕푣푧 휕푟) = 푟2 2 휇( 휕푝 휕푧) + 푘1 푣푧= 푟2 4 휇( 휕푝 휕푧) + 푘1 ln 푟+ 푘2 (7/20) 140 We have to use boundary conditions so as to unburden ourselves from integration constants 푘1 and 푘2. By setting 푣푧@푟=0 as fnite, we deduce that 푘1 = 0 (because ln(0) →−∞). By setting 푣푧@푟=푅= 0 (no-slip condition), we obtain 푘2 = −푅2 4 휇 휕푝 휕푧. This simplifes eq. (7/20) and brings us to our objective, an extensive expres-sion for the velocity profle across a pipe of radius 푅when the fow is laminar: 푣푧= 푢(푟) = −1 4휇( 휕푝 휕푧) (푅2 −푟2) (7/21) This equation is parabolic (fg. 7.5). It tells us that in a pipe of given length 퐿 and radius 푅, a given velocity profle will be achieved which is a function only of the ratio Δ푝/휇. Figure 7.5: The velocity profle across a cylindrical pipe featuring laminar viscous fow. Figure CC-0 Olivier Cleynen We can also express the pressure gradient in the pipe as a function of the volume fow rate. This is done through integration of velocity with respect to density and cross-section area. We obtain: ̇ 푚= −휋휌퐷4 128 휇( 휕푝 휕푧) Δ푝loss = −128 휋 휇퐿̇ 푚 휌퐷4 (7/22) This equation is interesting in several respects. For a given pipe length 퐿and pressure drop Δ푝loss, the volume fow ̇ increases with the power 4 of the diameter 퐷. In other words, the volume fow is multiplied by 16 every time the diameter is doubled. We also notice that the pipe wall roughness does not appear in equation 7/22. In a laminar fow, increasing the pipe roughness has no efect on the velocity distribution in the pipe. Out of curiosity, we may translate the result in equation 7/22 into a friction factor equation, with the help of defnition 7/5, obtaining 푓laminar cylinder fow = 32푉av.휇퐿 퐿 퐷 1 2휌푉2 av.퐷2 = 64 휇 휌푉av. 퐷= 64 [Re]퐷 (7/23) in which we inserted the average velocity 푉av. = ̇  휋푅2 = −Δ푝퐷2 32휇퐿. 141 7.5 Turbulent fow in pipes 7.5.1 When is a pipe fow turbulent? Video: very basic visualization of laminar and turbulent fow regimes in a transparent pipe by Engineering Fundamentals (styl) It has long been observed that pipe fow can have diferent regimes. In some conditions, the fow is unable to remain laminar (one-directional, fully-steady); it becomes turbulent. Although the fow is steady when it is averaged over short time period (e.g. a few seconds), it is subject to constant, small-scale, chaotic and spontaneous velocity feld changes in all directions. In 1883, Osborne Reynolds published the results of a meticulous investigation into the conditions in which the fow is able, or not, to remain laminar (fgs. 7.6 and 7.7). He showed that they could be predicted using a single non-dimensional parameter, later named Reynolds number, which, as we have seen already with eq. 1/28 p. 23, is expressed as: [Re] ≡휌푉퐿 휇 (7/24) In the case of pipe fow, the representative length 퐿is conventionally set to the pipe diameter 퐷and the velocity to the cross-section average velocity: [Re]퐷≡휌푉av. 퐷 휇 (7/25) where 푉av. is the average velocity in the pipe (m s−1), and 퐷is the pipe diameter (m). The occurrence of turbulence is very well documented. The following values are widely accepted: • Pipe fow is laminar for [Re]퐷≲2 300 ; Figure 7.6: Illustration published by Reynolds in 1883 showing the installation he set up to investigate the onset of turbulence. Water fows from a transparent rectangular tank down into a transparent drain pipe, to the right of the picture. Colored die is injected at the center of the pipe inlet, allowing for the visualization of the fow regime. Image by Osborne Reynolds (1883, public domain) 142 Figure 7.7: Illustration published by Reynolds in 1883 showing two diferent fow regimes observed in the installation from fg. 7.6. Image by Osborne Reynolds (1883, public domain) • Pipe fow is turbulent for [Re]퐷≳4 000. The signifcance of the Reynolds number extends far beyond pipe fow; we shall explore this in chapter 8 (Engineering models). 7.5.2 Characteristics of turbulent fow This topic is well covered in Tennekes & Lumley Video: turbulence for those who don’t have time to read chapter 9 by Y:Veritasium (styl) Turbulence is a complex topic which is still not fully described analytically today. Although it may display steadiness when time-averaged, a turbulent fow is highly three-dimensional, unsteady, and chaotic in the sense that the description of its velocity feld is carried out with statistical, instead of analytic, methods. In the scope of our study of fuid dynamics, the most important characteristics associated with turbulence are the following: • A strong increase in mass and energy transfer within the fow. Slow and rapid fuid particles have much more interaction (especially momentum transfer) than within laminar fow; • A strong increase in losses due to friction (typically by a factor 2). The increase in momentum exchange within the fow creates strong dissipa-tion through viscous efects, and thus transfer (as heat) of macroscopic forms of energy (kinetic and pressure energy) into microscopic forms (internal energy, translating as temperature); 143 • Internal fow movements appear to be chaotic (though not merely random, as would be white noise), and we do not have mathematical tools to describe them analytically. Consequently, solving a turbulent fow requires taking account of fow in all three dimensions even for one-directional fow! We will come back to this topic in chapter 9 (Dealing with turbulence). 7.5.3 Velocity profle in turbulent pipe fow Video: highly-turbulent fow ex-iting the food discharge ducts of the Tarbela dam in northeastern Pakistan by Y:Beauty Of Pakistan (styl) In order to deal with the vastly-increased complexity of turbulent fow in pipes, we split each velocity component 푣푖in two parts, a time-averaged component 푣푖and an instantaneous fuctuation 푣′ 푖: 푣푟= 푣푟+ 푣′ 푟 푣휃= 푣휃+ 푣′ 휃 푣푧= 푣푧+ 푣′ 푧 In our case, 푣푟and 푣휃are both zero, but the fuctuations 푣′ 푟and 푣′ 휃are not, and will cause 푣푧to difer from the laminar fow case. The extent of turbulence is often measured with the concept of turbulence intensity 퐼: 퐼≡[푣′2 푖] 1 2 푣푖 (7/26) Regrettably, we have not found a general analytical solution to turbulent pipe fow — note that if we did, it would likely exhibit complexity in proportion to that of such fows. A widely-accepted average velocity profle constructed from experimental observations is: 푢(푟) = 푣푧= 푣푧max (1 −푟 푅) 1 7 (7/27) While it closely and neatly matches experimental observations, this model is nowhere as potent as an analytical one and must be seen only as an approximation. For example, it does not allow us to predict internal energy dissipation rates (because it describes only time-averaged velocity), or even wall shear stress (because it yields (휕푢/휕푟)푟=푅= ∞, which is not physical). The following points summarize the most important characteristics of turbu-lent velocity profles: • They continuously fuctuate in time and we have no means to predict them extensively; • They are much “fatter” than laminar profles (fg. 7.8); • They depend on the wall roughness; • They result in shear and dissipation rates that are markedly higher than laminar profles. 144 Figure 7.8: Velocity profles for laminar (A), and turbulent (B and C) fows in a cylindrical pipe. B represents the time-averaged velocity distribution, while C shows several arbitrary instantaneous distributions (blurred) as well as their average in time. Turbulent fow in a pipe also features velocities in the radial and angular directions, which are not shown here. Figure CC-0 Olivier Cleynen 7.5.4 Pressure losses in turbulent pipe fow Losses caused by turbulent fow depend on the wall roughness 휖and on the diameter-based Reynolds number [Re]퐷. For lack of an analytical solution, we are not able to predict the value of the friction factor 푓anymore. Several empirical models can be built to obtain 푓, the most important of which is known as the Colebrook equation expressed as: 1 √ 푓 = −2 log ( 1 3,7 휖 퐷+ 2,51 [Re]퐷 √ 푓) (7/28) The structure of this equation makes it inconvenient to solve for 푓. To circumvent this difculty, equation 7/28 can be solved graphically on the Moody diagram, fg. 7.9. This classic document allows us to obtain numerical values for 푓(and thus predict the pressure losses Δ푝loss) if we know the diameter-based Reynolds number [Re]퐷and the relative roughness 휖/퐷. 145 Figure 7.9: A Moody diagram, which presents values for 푓measured experimentally, as a function of the diameter-based Reynolds number [Re]퐷, for diferent values of the relative roughness 휖/퐷. This fgure is reproduced with a larger scale as fgure 7.11 p. 152. Diagram CC-by-sa S Beck and R Collins, University of Shefeld 7.6 Engineer’s guide to pipe fows 7.6.1 Summary so far The world of pipe fows is a good playground for us to experiment with practical fuid mechanics. The main lessons learned in this chapter are as follows: • From an engineering point of view, pipe fows are one-dimensional. We push a volume fow of fuid in on one side and expect to receive it on the other, perhaps with diferent properties. The integral “cost” of doing so is quantifed with a single value, Δ푝loss. • When the fow is laminar, we can solve for the fow in the pipe. For this, we write the Navier-Stokes equations and apply known boundary conditions. We obtain the velocity everywhere in the pipe, which gives us a wealth of information about the fow, including the Δ푝loss. • When the fow is turbulent, this analysis method does not work any-more. Even though the conditions are in principle simple, the fow is mesmerizingly complex. We resort to building models based on time-averaged measurement data. Those models work for a wide number of pipes and conditions because they relate parameters which are non-dimensionalized. We obtain the desired Δ푝loss with a single diagram reading and a simple algebraic manipulation. These broad trends apply to many other sub-areas of fuid mechanics. 7.6.2 Choosing laminar or turbulent fow When designing a pipe system, an engineer may have the opportunity to create laminar or turbulent fow. Inserting an expression for mass fow 146 ̇ 푚= 휌푉av.(휋/4)퐷2 into the defnition of the Reynolds number, we obtain, for pipe fow: [Re] = ̇ 푚 퐷 4 휋휇 (7/29) This equation is telling: for a given fuid (휇) and a given mass fow ( ̇ 푚), the only way to ensure that the fow is laminar (with low Reynolds number) is to increase the diameter 퐷. Doing so also increases installation costs; therefore, there is a balance to strike between initial installation costs (increasing with diameter) and operating costs due to pressure losses (which decrease as the diameter increases). In practice, except for cases where very small mass fows and velocities are involved (e.g. medical fuid fows), most piping installations feature turbulent fow. 7.6.3 Pressure losses in laminar fow The pressure losses in laminar fow are summed up with equation 7/22 p. 141, which we repeat here with the most important terms positioned frst: Δ푝loss = −퐿̇ 푚 퐷4 128휇 휋휌 (7/30) for laminar pipe fow. Again, it is visible here that in laminar pipe fow, losses per unit pipe length increase linearly with mass fow ̇ 푚, and with the power −4 (!) of the diame-ter 퐷. 7.6.4 Pressure losses in turbulent fow The picture for losses in turbulent fow is harder to draw. Re-arranging the defnition 7/5 p. 137 to include the mass fow, we obtain: Δ푝loss = −퐿̇ 푚2 퐷5 푓 8 휋2휌 (7/31) for turbulent pipe fow. The factor 푓in this equation generally varies according to the Reynolds number and to the roughness of the pipe, as described in the Moody diagram. At very turbulent regimes (in the upper right area of the diagram), 푓becomes independent of [Re]퐷and proportional to the relative roughness, so that we can write: Δ푝loss = −퐿̇ 푚2 퐷5 (푐1 + 푐2 휖 퐷) 8 휋2휌 (7/32) for turbulent pipe fow in very turbulent regimes, where 푐1 and 푐2 are approximately constant. Based on this equation, we can see that pressure losses in highly-turbulent pipe fow increase approximately with the square of mass fow ̇ 푚and the power −5 of the diameter 퐷. In between this regime and the laminar regime, a variety of intermediary states are quantifed using the Moody diagram. 147 7.6.5 Calculating pumping and turbining power The tools above are all we need to calculate, given the geometry of an installation, how much pumping or turbining power is involved in moving a given mass fow of liquid though a pipe system. The steps are as follows: 1. Calculate the hydrostatic pressure drop across the device. This is done by imagining that there is no fow, and calculating the static pressure which would then exert on each side of the device, using equation 4/15 p. 80. 2. Calculate the pressure losses due to friction. This is done by calculating the Reynolds number, using the Moody diagram to read the corresponding friction factor 푓, and calculating the corresponding Δ푝loss using the defnition 7/5. Care must be taken with the sign of Δ푝loss, which is always negative by defnition, but very often expressed as a positive number in the literature. Pressure losses induced by bends, junctions and obstacles are likewise calculated using their 퐾퐿values and the defnition 7/6. 3. Summing up the pressure diferences. The complete pressure diference across the device is Δ푝device = −Δ푝losses+ Δ푝hydrostatic. Pressure losses due to friction are always negative, and hydrostatic pressure diferences may have either sign. The power is recovered using equation 1/22 p. 19, ̇ 푃device = Δ푝device ̇ 푚/휌. If this power is negative, the liquid is losing energy, and the device is acting as a turbine. If the power is positive, the liquid is gaining energy, and the device is acting as a pump. 7.7 Solved problems Hydrostatic pressure on a turbine A turbine is installed as shown above. What is the hydrostatic pressure diference available to the turbine? 148 See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Pressure loss in a pipe In the piping installation from the previous example, water at 20 °C is fowing with a volume fow of 800 L s−1. The pipe has roughness 휖= 0,25 mm and a diameter 퐷= 1,1 m. The bends each induce a loss coefcient 퐾퐿= 0,75. What is the pressure drop due to friction losses in the pipe? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Turbining power In the piping installation from the previous examples, what is the power made available to the turbine? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Pressure distribution in a pipe In the piping installation from the previous examples, what is the pres-sure distribution along the pipe? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 149 150 Problem sheet 7: Pipe fows last edited June 30, 2021 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 In cylindrical pipe fow, we assume the fow is always laminar for [Re]퐷≲2 300, and always turbulent for [Re]퐷≳4 000. The Darcy friction factor 푓is defned as: 푓≡\|Δ푝loss\| 퐿 퐷 1 2휌푉2 av. (7/5) The loss coefcient 퐾퐿is defned as: 퐾퐿≡\|Δ푝loss\| 1 2휌푉2 av. (7/6) Viscosities of various fuids are given in fg. 7.10. Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in fg. 7.11 p. 152. −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 7.10: The viscosity of four fuids (crude oil, water, air, and C02) as a function of temperature. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. Figure reproduced from fgure 5.6 p. 99; CC-by by Arjun Neyyathala & Olivier Cleynen 151 Figure 7.11: A Moody diagram, which presents values for 푓measured experimentally, as a function of the diameter-based Reynolds number [Re]퐷, for diferent relative roughness values. Diagram CC-by-sa S Beck and R Collins, University of Shefeld 152 7.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 7.2 Revision questions non-examinable The Moody diagram (fg. 7.11 p. 152) is simple to use, yet it takes practice to understand it fully... here are three questions to guide your exploration. They can perhaps be answered as you work through the other examples. 7.2.1. Why is there no zero on the diagram? 7.2.2. Why are the curves sloped downwards — should friction losses not instead increase with increasing Reynolds number? 7.2.3. Why can the pressure losses Δ푝losses be calculated given the volume fow ̇ , but not the other way around? 7.3 Air fow in a small pipe Munson & al. E8.5 A machine designed to assemble micro-components uses an air jet. This air is driven through a 10 cm-long cylindrical pipe with a 4 mm diameter, roughness 0,0025 mm, at an average speed of 50 m s−1. The inlet air pressure and temperature are 1,2 bar and 60 °C; the viscosity of air is quantifed in fg. 7.10 p. 151. 7.3.1. What is the pressure loss caused by the fow through the pipe? 7.3.2. What is the maximum average fow speed for which the fow would remain laminar? 7.3.3. What would this speed be if the pipe diameter was 4 cm instead of 4 mm? 7.4 Water piping A long pipe is installed to carry water from one large reservoir to another (fg. 7.16). The total length of the pipe is 10 km, its diameter is 0,5 m, and its roughness is 휖= 0,5 mm. It must climb over a hill, so that the altitude changes along with distance. The pump must be powerful enough to push 1 m3 s−1 of water at 20 °C. Figure 7.10 p. 151 quantifes the viscosity of various fuids, and fg. 7.11 p. 152 quantifes losses in cylindrical pipes. 7.4.1. Will the fow in the water pipe be turbulent? 7.4.2. What is the pressure loss caused by the fow through the pipe? 7.4.3. What is the pumping power required to meet the design requirements? 7.4.4. What would be the power required for the same volume fow if the pipe diameter was doubled? 153 Figure 7.12: Layout of the water pipe. For clarity, the vertical scale is greatly exaggerated. The diameter of the pipe is also exaggerated. Figure CC-0 Olivier Cleynen 7.5 Design of a water piping system You start your career as a junior engineer in a company that designs piping and pumping systems. Your frst assignment is to choose the dimensions of a system which should carry 3 m3 h−1 of water (10−3 Pa s) across a horizontal distance of 1 km. Fresh from reading through chapter 7, you design the system to feature laminar fow only. You begin with a revision of the relevant theory, starting, of course, from the Navier-Stokes equations and a simple diagram (fg. 7.13), obtaining the velocity distribution for laminar fow in a circular pipe: 푣푧= 푢(푟) = −1 4휇( 휕푝 휕푧) (푅2 −푟2) (7/21) Figure 7.13: A cylindrical coordinate system to study laminar fow in a horizontal cylindrical duct. Figure CC-0 Olivier Cleynen 7.5.1. What is the minimum pipe diameter for which the fow will remain laminar? 7.5.2. Starting from equation 7/21, show that the pressure loss per unit length in the pipe with laminar fow is expressed as a function of the volume fow ̇ and the diameter 퐷as: Δ푝loss = −128 휋 휇퐿̇ 푚 휌퐷4 (7/22) 7.5.3. With the diameter chosen above, what is the pressure loss in the pipe? 7.5.4. What is the pumping power required? With those results in hand, you turn to your colleagues — but with a smile, they suggest you try a design with turbulent fow instead. 154 7.5.5. What would be the pressure loss if an 8 cm-diameter plastic pipe was used? 7.5.6. What would then be the pumping power required? 7.5.7. What is one advantage of using a pipe with smaller diameter? (briefy justify your answer, e.g. in 30 words or less) 7.6 Major oil pipeline Strong from your experience working through problem 7.5, you join the team in charge of designing one very large oil pipeline system (fg. 7.14). Your design must safely carry 700 thousand barrels of oil (110 000 m3) per day along a length of 1 200 km. The crude oil has density 900 kg m−3 and its viscosity is quantifed in fg. 7.10 p. 151. The average temperature of the oil during the transit is 60 °C. The landscape is fat for most of the journey, with a 200 km-wide mountain range in the middle that reaches 1 400 m altitude. Your team selects a cylindrical, smooth steel duct with 1,22 m diameter, average roughness 휖= 0,15 mm. Because the pipeline passes through ecologically fragile areas, as well as a seismically-active region, you decide to never exceed 200 psi (13,8 bar) of gauge pressure in the pipeline. To prevent oil cavitation (a change of state with destructive consequences), you decide to never reach below 0,8 bar of absolute pressure in the pipeline. 7.6.1. How much time does the average oil particle need to travel across the line? 7.6.2. How much pumping power is required in total? 7.6.3. How far apart should the pumping stations be laid out in the fat sections of the pipeline? 7.6.4. How far apart should the pumping stations be laid out in the ascending section of the pipeline? 7.6.5. Propose a pumping station arrangement, and calculate the power required for each pump. Figure 7.14: The Trans-Alaska Pipeline System, which inspired this problem. It was built in the 1970s, at tremendous fnancial, political and social cost. Photo CC-by-sa by Luca Galuzzi – www.galuzzi.it 155 Before you start building the pipeline, the operator would like to know how the system would perform at half-capacity (i.e. with half the volume fow). 7.6.6. If none of the other input data changes, what is the new pumping power? 7.6.7. Propose one reason why in practice, the pumping power may be higher than you just calculated (briefy justify your answer, e.g. in 30 words or less). 7.7 Pump with pipe expansion A pump is used to carry a volume fow of 200 L s−1 from one large water reservoir to another (fg. 7.15). The altitude of the water surface in both reservoirs is the same. Figure 7.15: Layout of the water pipe. For clarity, the diameter of the pipe and the vertical scale are exaggerated. Figure CC-0 Olivier Cleynen The pipe connecting the reservoirs is made of concrete (휖= 0,25 mm); it has a diameter of 50 cm on the frst half, and 100 cm on the second half. In the middle, the conical expansion element induces a loss coefcient of 0,8. At the outlet (at point D), the pressure is approximately equal to the corresponding hydrostatic pressure in the outlet tank. The inlet is 14 m below the surface. The total pipe length is 400 m; the altitude change between inlet and outlet is 12 m. 7.7.1. Represent qualitatively (that is to say, showing the main trends, but without displaying accurate values) the water pressure as a function of pipe distance, when the pump is turned of. 7.7.2. On the same graph, represent qualitatively the water pressure when the pump is switched on. 7.7.3. What is the water pressure at points A, B, C and D? 7.8 Piping and power of a water turbine from 2018-07 fnal examination A water turbine is installed between two reservoirs in order to extract power from the fow of water. The water is guided to the turbine through a pipe, as shown in fgure 7.16. The pipe is made of reinforced concrete, with a total length 퐿= 0,8 km, a diameter 퐷= 푑= 1,2 m, and an interior surface roughness 휖= 6 mm. The pipe has altitude variations along its length, as indicated in fgure 7.16. The turbine is designed to handle 5 000 L s−1 of water at 20 °C. 156 Figure 7.16: Layout of the water pipe. For clarity, the vertical scale and the diameter of the pipe are greatly exaggerated. 7.8.1. On a diagram, represent qualitatively (i.e. without numerical data) the pressure distribution along the length of the pipe. 7.8.2. What is the pressure drop due to friction losses generated by the water fow in the pipe? 7.8.3. What is the power developed by the turbine? 7.8.4. What would be the new power developed by the turbine if the volume fow was divided by two? 7.9 Politically incorrect fuid mechanics non-examinable In spite of the advice of their instructor, a group of students attempts to apply fuid mechanics to incommendable activities. Their objective is to construct a drinking straw piping system that can mix a drink of vodka and tonic water in the correct proportions (fg. 7.17). They use a “Strawz” kit of connected drinking straws, two bottles, and a glass full of ice and liquid water to cool the mix. For simplicity, the following information is assumed about the setup: Figure 7.17: Conceptual sketch of a student experiment. Figure CC-0 Olivier Cleynen 157 Percentage of alcohol by weight Viscosity in centipoise 0 1,005 10 1,538 20 2,183 30 2,71 40 2,91 50 2,87 60 2,67 70 2,37 80 2,008 90 1,61 100 1,2 Table 7.1: Viscosity of a mix of ethanol and water at 20 °C. data from Bingham, Trans. Chem. Soc, 1902, 81, 179. • Vodka is modeled as 40 % pure alcohol (ethanol) with 60 % water by volume; • Ethanol density is 0,8 kg m−3; • Tonic water is modeled as pure water; • The viscosity of water and alcohol mixes is described in table 7.1 (use the nearest relevant value); • The pipe bends induce a loss coefcient factor 퐾퐿bend = 0,5 each; • The pipe T-junction induces a loss coefcient factor 퐾퐿= 0,3 in the line direction and 1 in the branching fow; • The pipe has inner diameter 퐷= 3 mm and roughness 휂= 0,0025 mm. The students wish to obtain the correct mix: one quarter vodka, three quarters tonic water. For given levels of liquid in the bottles, is there a straw pipe network confguration that will yield the correct mix, and if so, what is it? 158 Answers 7.3 1) Calculating inlet density with the perfect gas model, [Re]퐷= 14 263 (turbulent), a Moody diagram read gives 푓≈0,029, so Δ푝friction = −1 292,6 Pa = −0,0129 bar. 7.4 \|Δ푝alt.\| = 휌푔(26 −8 + 5 −7) = 1,57 bar and \|Δ푝friction\| = 51,87 bar : ̇ 푊pump = 5,345 MW. 7.5 1) 퐷min = 0,46 m 2) Follow the process used p. 139 to go from eq. 7/10 to eq. 7/13: the only diference is the use of cylindrical (instead of rectangular) coordinates. 3) \|Δ푝\|loss = 0,75 Pa 4) ̇ 푊= 0,63 mW 5) \|Δ푝\|loss 2 = 4,82 kPa (8 000 times more) 6) ̇ 푊2 = 4,2 W 7.6 1) Approximately 12 days 18 hours 2) ̇ 푊total = 10,02 MW 3) Δ퐿fat terrain = 213 km 4) Δ퐿ascending terrain = 11,96 km 7) Hint: in normal operation, what happens with the 10 MW of power — in which form is this energy converted? 7.7 3) Δ푝푓A→B = −3 735 Pa, Δ푝B→C = +71 Pa (the sum of losses due to friction and gains due to decrease in kinetic energy), Δ푝푓C→D = −117 Pa. Add hydrostatic pressure changes, working backwards from D to A, to obtain pressure in all four points. 7.8 2) Δ푝friction losses = −2,0196 bar 3) ̇ 푊turbine = −1,0503 MW 4) The power will decrease by 14 % 7.9 The author cannot remember which exercise you are referring to. 159 160 Fluid Dynamics Chapter 8 – Engineering models last edited April 3, 2021 by Olivier Cleynen — 8.1 Motivation 161 8.2 Comparing infuences: the weighted momentum balance161 8.2.1 Principle 161 8.2.2 The non-dimensional Navier-Stokes equation 162 8.2.3 The fow parameters of Navier-Stokes 164 8.2.4 Flow parameters obtained as force ratios 166 8.2.5 The Reynolds number in practice 166 8.3 Making models 167 8.4 Comparing results: coefcients 169 8.4.1 Principle 169 8.4.2 Force coefcients 169 8.4.3 Power coefcient, and other coefcients 170 8.4.4 Non-dimensionalizing all the problems 171 8.5 Solved problems 172 8.6 Problems 173 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 8.1 Motivation In this chapter, we develop tools to analyze scale efects in fuid mechanics. This study should allow us to answer two questions: • How can we adequately reproduce a fow in an enlarged or reduced version? • How do forces and powers change when the fow is scaled? 8.2 Comparing infuences: the weighted momentum balance 8.2.1 Principle Video: when their movement is flmed and then viewed sped-up, fog banks can appear to fow like water. What time-lapse rate is required to achieve physical sim-ilarity? by Simon Christen (stvl) In many practical engineering situations which involve fuid fow, investi-gating the fow in the real application is impossible or impractical. In those cases it is common practice to build scaled-up or scaled-down versions of the fow in a laboratory. This brings up the question: how should the model fow properties be adapted to represent the original fow? For example, if the model is half as small as the original, should the velocity be halved? Or perhaps doubled? The answer to this problem is as follows: two fows are dynamically similar (i.e. representative of one another) when their fow parameters are the same. In order to understand what this means, we will need to look back at the Navier-Stokes equation (the momentum balance equation we derived in chapter 6), re-writing it in a non-dimensional form. Onwards! 161 Advice from an expert Making models is useful when you have a large machine, say, a jet airliner or a submarine, and you want to study diferent confgurations without building a new full-scale machine each time. But there are other applications: imagine trying to study the fow of molten metal in a furnace, the fow of blood in a beating heart, or the fow of air around a mosquito’s wings. Making models and understanding scale efects will allow you to investi-gate fuid fow in all kind of inaccessible locations. That makes the math worth the efort! 8.2.2 The non-dimensional Navier-Stokes equation Here, we want to obtain an expression for the Navier-Stokes equation for incompressible fow which allows for an easy comparison of its constituents. We start with the original equation, which we derived in chapter 6 as equa-tion 6/42 p. 125: 휌휕⃗ 푉 휕푡+ 휌( ⃗ 푉⋅⃗ ∇) ⃗ 푉= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (8/1) What we would now like to do is separate the geometry from the scalars in this equation. The principle is to express each vector ⃗ 퐴as the multiple of its length 퐴and a non-dimensional vector ⃗ 퐴∗, which has the same direction as ⃗ 퐴but only unit length. In order to achieve this, we introduce a series of non-dimensional physical terms, starting with non-dimensional time 푡∗, defned as time 푡multiplied by the frequency 푓(in Hertz) at which the fow repeats itself: 푡∗≡푓푡 (8/2) A fow with a very high frequency is highly unsteady, and the changes in time of the velocity feld will be relatively important. On the other hand, the lower the frequency, and the closer the fow is to being steady. In all cases, as we observe the fow, non-dimensional time 푡∗progresses from 0 to 1, after which the solution is repeated. We then introduce non-dimensional velocity ⃗ 푉∗, a unit vector feld equal to the velocity vector feld divided by its own (scalar feld) length 푉: ⃗ 푉∗≡ ⃗ 푉 푉 (8/3) Pressure 푝is non-dimensionalized diferently, since in fuid mechanics, it is the pressure changes across a feld, not their absolute value, that infuence the velocity feld. For example, in eq. 8/1 ⃗ ∇푝can be replaced by ⃗ ∇(푝−푝∞) (in which 푝∞can be any constant faraway pressure). Now, the pressure feld 푝−푝∞is non-dimensionalized by dividing it by a reference pressure diference 162 푝0 −푝∞, obtaining: 푝∗≡푝−푝∞ 푝0 −푝∞ (8/4) If 푝0 is taken to be the maximum pressure in the studied feld, then 푝∗is a scalar feld whose values can only vary between 0 and 1. Non-dimensional gravity ⃗ 푔∗is simply a unit vector equal to the gravity vector ⃗ 푔divided by its own length 푔: ⃗ 푔∗≡⃗ 푔 푔 (8/5) And fnally, we defne a new operator, the non-dimensional del ⃗ ∇∗, ⃗ ∇∗≡퐿⃗ ∇ (8/6) which ensures that vector felds obtained with a non-dimensional gradient, and the scalar felds obtained with a non-dimensional divergent, are “scaled” by a reference length 퐿. These new terms allow us to replace the constituents of equation 8/1 each by a non-dimensional “unit” term multiplied by a scalar term representing its length or value: 푡= 푡∗ 푓 ⃗ 푉= 푉⃗ 푉∗ 푝−푝∞= 푝∗(푝0 −푝∞) ⃗ 푔= 푔⃗ 푔∗ ⃗ ∇= 1 퐿 ⃗ ∇∗ Now, inserting these in equation 8/1, and re-arranging, we obtain: 휌휕 휕1 푓푡∗(푉⃗ 푉∗) + 휌(푉⃗ 푉∗⋅1 퐿 ⃗ ∇∗) (푉⃗ 푉∗) = 휌푔⃗ 푔∗−1 퐿 ⃗ ∇∗[푝∗(푝0 −푝∞) + 푝∞] + 휇 ⃗ ∇∗2 퐿2 (푉⃗ 푉∗) 휌푉푓휕⃗ 푉∗ 휕푡∗+ 휌푉푉1 퐿( ⃗ 푉∗⋅⃗ ∇∗) ⃗ 푉∗= 휌푔⃗ 푔∗−1 퐿 ⃗ ∇∗[푝∗(푝0 −푝∞)] + 휇푉1 퐿2 ⃗ ∇∗2 ⃗ 푉∗ [휌푉푓] 휕⃗ 푉∗ 휕푡∗+ [ 휌푉2 퐿] ( ⃗ 푉∗⋅⃗ ∇∗) ⃗ 푉∗= [휌푔] ⃗ 푔∗−[ 푝0 −푝∞ 퐿 ] ⃗ ∇∗푝∗+ [ 휇푉 퐿2 ] ⃗ ∇∗2 ⃗ 푉∗ (8/7) Video: Figuring out the non-di-mensional Navier-Stokes equa-tion by Olivier Cleynen (CC-by) In equation 8/7, the terms in brackets each appear in front of non-dimen-sional (unit) vectors. These bracketed terms all have the same dimension, i.e. kg m−2 s−2. Multiplying each by 퐿 휌푉2 (of dimension m2 s2 kg−1), we obtain a purely non-dimensional equation: [ 푓퐿 푉] 휕⃗ 푉∗ 휕푡∗+ ( ⃗ 푉∗⋅⃗ ∇∗) ⃗ 푉∗= [ 푔퐿 푉2] ⃗ 푔∗−[ 푝0 −푝∞ 휌푉2 ] ⃗ ∇∗푝∗+ [ 휇 휌푉퐿] ⃗ ∇∗2 ⃗ 푉∗ (8/8) 163 Equation 8/8 does not bring any information on top of the original incom-pressible Navier-Stokes equation (eq. 8/1). Instead, it merely separates it into two distinct kinds of components. The frst, in square brackets, are a scalar felds (purely numbers), which indicate the magnitude of the acceleration feld. The others are unit vector felds (felds of oscillating vectors, all of length 1, and noted with stars), which represent the geometry (direction) of the acceleration feld. In this form, we can more easily observe and quantify the weight of the diferent terms relative to one another. At this point, it is time to introduce the notion of fow parameters. 8.2.3 The fow parameters of Navier-Stokes From here on, in equation 8/8, we convene to call the terms written in brackets fow parameters, and label them with the following defnitions: [St] ≡푓퐿 푉 (8/9) [Eu] ≡푝0 −푝∞ 휌푉2 (8/10) [Fr] ≡ 푉 √ 푔퐿 (8/11) [Re] ≡휌푉퐿 휇 (8/12) This allows us to re-write eq. 8/8 as: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= [ 1 Fr2] ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ [ 1 Re] ⃗ ∇∗2 ⃗ 푉∗ And fnally, we arrive at the simple, elegant and formidable non-dimensional incompressible Navier-Stokes equation expressed with fow parameters: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (8/13) XKCD #762: comparing things by Randall Munroe (CC-by-nc) Equation 8/13 is an incredibly useful tool in the study of fuid mechanics, for two reasons: 1. It allows us to quantify the relative weight of the diferent terms in a given fow. In this way, it serves as a compass, helping us determine which terms can safely be neglected in our attempts to fnd a particular fow solution, merely by quantifying the four parameters [St], [Eu], [Fr], and [Re]. 2. It allows us to obtain dynamic similarity between two fows at diferent scales. In order that a fuid fow be representative of another fow (for example in order to simulate airfow around an aircraft with a model in a wind tunnel), it must generate the same vector feld ⃗ 푉∗. In order to do this, we must generate an incoming fow with the same four parameters [St], [Eu], [Fr], and [Re]. 164 Let us therefore take the time to explore the signifcation of these four parameters. The Strouhal number [St] ≡푓퐿/푉(eq. 8/9) quantifes the infuence of un-steadiness efects over the acceleration feld. It does this by evaluating the importance of the representative frequency 푓at which the fow pattern repeats itself. Very high frequencies denote highly unsteady fows. When the frequency is very low, [St] is very small; the fow is then quasi-steady and it can be solved at a given moment in time as if it was entirely steady. The Euler number [Eu] ≡푝0 −푝∞/휌푉2 (eq. 8/10) quantifes the infuence of the pressure gradient over the acceleration feld. It does this by comparing the largest relative pressure 푝0−푝∞to the fow of momentum in the fow feld. The greater [Eu] is, and the more the changes in the velocity feld ⃗ 푉are likely to be caused by pressure gradients rather than viscosity, convection or unsteadiness. The Froude number [Fr] ≡푉/ √ 푔퐿(eq. 8/11) quantifes the relative impor-tance of gravity efects. In practice, gravity efects only play an impor-tant role in free surface fows, such as waves on the surface of a water reservoir. In most other cases, gravity contributes only to a hydrostatic efect, which has little infuence over the velocity feld. The Reynolds number [Re] ≡휌푉퐿/휇(eq. 8/12, also eqs. 1/28 p. 23 & 7/24 p. 142) quantifes the infuence of viscosity over the acceleration feld. It does this by comparing the magnitude of inertial efects (휌푉퐿) to viscous efects (휇). When [Re] is very large, viscosity plays a negligible role and the velocity feld is mostly dictated by the inertia of the fuid. We return to the signifcance of the Reynolds number in §8.2.5 below. The Mach number [Ma] ≡푉/푐(eq. 1/10 p. 16) compares the fow speed 푉with that of the molecules within the fuid particles (the speed of sound 푐). [Ma] does not appear in equation 8/13, because we already decided to restrict ourselves to non-compressible fows. If we hadn’t, we would be re-expressing the pressure term in that equation as a function of [Ma], quantifying the efect of changes in density. These fve fow parameters should be thought of scalar felds within the studied fow domain: there is one distinct Reynolds number, one Mach number etc. for each point in space and time. Nevertheless, when describing fuid fows, the custom is to choose for each parameter a single representative value for the whole fow. For example, when describing pipe fow, it is customary to quantify a representative Reynolds number [Re]퐷based on the average fow velocity 푉av. and the pipe diameter 퐷(as we have seen with eq. 7/25 p. 142), while the representative Reynolds number for fow over an aircraft wing is often based on the free-stream velocity 푉∞and the wing chord length. Similarly, the fight Mach number [Ma]∞displayed on an aircraft cockpit instrument is computed using the relative free-stream air speed 푉∞and the free-stream speed of sound 푐∞, rather than particular values measured closer to the aircraft. 165 8.2.4 Flow parameters obtained as force ratios Instead of the mathematical approach covered above, the concept of fow parameter can be approached by comparing forces in fuid fows. This method is described for reference in Appendix A5 p. 256. 8.2.5 The Reynolds number in practice Video: half-century-old, but timeless didactic exploration of how the dynamics of fuids change with the Reynolds num-ber, with accompanying notes by G. I. Taylor by the National Committee for Fluid Mechanics Films (ncfmf, 1967) (styl) Among the fve non-dimensional parameters described above, the Reynolds number [Re] is by far the most relevant in the study of most fuid fows, and it deserves a few additional remarks. As we have seen, the Reynolds number is a measure of how little efect the viscosity has on the time-change of the velocity vector feld: • With low [Re], the viscosity 휇plays an overwhelmingly large role, and the velocity of fuid particles is largely determined by that of their own neighbors; • With high [Re], the momentum 휌푉of the fuid particles plays a more important role than the viscosity 휇, and the inertia of fuid particles afects their trajectory much more than the velocity of their neighbors. In turn, this gives the Reynolds number a new role in characterization of fows: it can be thought of as the likeliness of the fow being turbulent over the length 퐿. Indeed, from a kinematic point of view, viscous efects are highly stabilizing: they tend to harmonize the velocity feld and smooth out disturbances. On the contrary, when these efects are overwhelmed by inertial efects, velocity non-uniformities have much less opportunity to dissipate, and at the slightest disturbance the fow will become and remain turbulent (fg. 8.1). This is the reason why the quantifcation of a representative Reynolds number is often the frst step engineers and scientists take when studying a fuid fow. 166 Figure 8.1: A viscous opaque fuid is dropped into a clearer receiving static fuid with identical viscosity. The image shows four diferent experiments photographed after the same amount of time has elapsed. The viscosity is decreased from left to right, yielding Reynolds numbers of 0,05, 10, 200 and 3 000 respectively. As described in eq. A/36 p. 256, a low Reynolds number indicates that viscous efects dominate the acceleration feld. As the Reynolds number increases, the nature of the velocity feld changes until it becomes clearly turbulent. By the National Committee for Fluid Mechanics Films (ncfmf, 1967), with accompanying notes by Taylor Image ©1961-1969 Educational Development Center, Inc., reproduced under Fair Use doctrine A screen capture from flm Low Reynolds Number Flow at 8.3 Making models We now know that when we create a scaled-down or scaled-up version of one fow (as shown for example in fg. 8.2), we attempt to keep all fow parameters identical. In practice, this is extremely difcult to do, as we will see while going through this chapter’s problem sheet, especially if ordinary fuids (air or water) are to be used. The practice in science and engineering is usually to focus on one or two key parameters, while ignoring others. If the budget allows for it, several models may be built, each focusing on one parameter (e.g. one model for compressibility efects, one model for viscous efects). Figure 8.2: In order for the fow around a wind tunnel model to be representative of the fow around the real-size aircraft (here, a 48 m-wide Lockheed C-141 Starlifter), dynamic similarity must be obtained. The value of all fow parameters must be kept identical. This is not always feasible in practice. Wind tunnel photo by NASA (public domain) Full-size aircraft photo CC-by by Peter Long 167 In the design and construction of models, practical constraints must be balanced against the need to reproduce the dynamics of fuids accurately. They include: • Cost of production. The volume of a model typically increases with its length cubed, i.e. doubling its length multiplies its volume by a factor 8; • Precision of manufacturing. Usually, the smaller the model, and the smaller the geometrical accuracy that can be reached; • Ease of instrumenting. Carrying out measurements over extremely large or extremely small models may be challenging; • Ease of optical access. Optical fow measurement devices are usually preferred because they do not obstruct the fow, but when the fow is internal to a model, they require transparent, fat walls to work correctly; • Performance of large-scale laboratory equipment. Wind tunnels and water channels are expensive installations. Often, the characteristics of models is adapted to the equipment’s availability and performance, rather than the other way around. Finally, it must be noted that in this chapter, we focus on reproducing fow dynamics only. Other kinds of forces applying on models, and other physical phenomena, are not taken into account. For example, an airplane model in a tunnel may not be able to fy, and a boat model may not be able to foat, because they are made of diferent materials than the original objects, and because weight and lift forces do not scale together (this is also the reason why large birds such as condors or swans do not look like, and cannot fy as slowly as mosquitoes and bugs! — these ideas are beautifully and smartly explored by Hendrik Tennekes [8, 23]). Similarly, heat transfer or chemical reaction rates, may be completely of in the model, unless the physical laws that govern them are also taken into account. Advice from an expert When your small wind tunnel already cost half a million euros to build, is run by fans with hundreds of kilowatts of power, and your new laser-Doppler velocity mea-surement system cost the lab 200 k€, it is clear that the available laboratory equipment will dictate the model size and the experimental fow conditions, not the other way around. A well-designed experiment will always take advantage of the best per-formance of the available equipment. This is why fuid dynamicists like to measure the performance of their wind tunnel not with size or speed, but with a tunnel Reynolds number — the maximum Reynolds number you can subject a model to in the test section. 168 8.4 Comparing results: coefcients 8.4.1 Principle The general rule that we follow when we compare measurements from experiments at diferent scale is that in two physically similar experiments, the force and power coefcients are the same. In other words, for scientists and engineers, the hard work is making a model where the fow coefcients are the same (and thus the original fow and the model are physically similar). Once this is done, the comparison of results is easy: the force and power coefcients are the same on the model and in the original fow. 8.4.2 Force coefcients Video: Comparing fuid-induced forces: using force coefcients by Olivier Cleynen (CC-by) In order to compare forces, we want to have a sense of how fow-induced forces scale when fuid fows are scaled. A look back on chapter 2 (Analysis of existing fows with one dimension), and in particular eq. 2/7 p. 37, shows us that for a steady fow through a given control volume, the net force induced on the fuid is expressed by: ⃗ 퐹net on fuid = Σnet [휌푉⟂퐴⃗ 푉] (8/14) This equation tells us that the norm of the net force vector, 퐹net, is directly related to a term with dimensions of 휌\|푉⟂\|퐴푉. In our selection of a scale by which to measure 퐹net, it is therefore sensible to include a term proportional to the the density 휌, a term proportional to the area 퐴, and a term proportional to the square of velocity 푉. This “scale of fuid-induced force” is conventionally measured using the force coefcient 퐶퐹: 퐶퐹≡ 퐹 1 2휌푆푉2 (8/15) where 퐹is the considered fuid-induced force (N); 휌is a reference fuid density (kg m−3); 푆is a reference surface area (m2); and 푉is a reference velocity (m s−1). In efect, the force coefcient relates the magnitude of the force exerted by the fuid on an object (퐹) to the rate of fow of momentum towards the object (휌푆푉2 = ̇ 푚푉). It is worth making a few remarks about this equation. First, it is important to realize that eq. 8/15 is a defnition: while the choice of terms is guided by physical principles, it is not a physical law in itself, and there are no reasons to expect 퐶퐹(which has no dimension, and thus no unit) to reach any particular value in any given case. The choice of terms is also worth commenting: 169 • 퐹can be any fuid-induced force; generally we are interested in quan-tifying either the drag 퐹D (the force component parallel to the free-stream velocity) or the lift 퐹L (force component perpendicular to the free-stream velocity); • the reference area 푆is chosen arbitrarily. It matters only that this area grow and shrink in proportion to the studied fow case. In the automobile industry, it is customary to choose the vehicle frontal area as a reference, while in the aeronautical industry the top-view wing area is customarily used. The square of a convenient reference length 퐿 can also be chosen instead; • the choice of reference velocity 푉and density 휌is also arbitrary, since both these properties may vary in time and space within the studied fuid fow case. It matters merely that the chosen reference values are representative of the case; typically the free-stream (faraway) condi-tions 푉∞and 휌∞are used; • the term 1/2 in the denominator is a purely arbitrary and conventional value. Force coefcients are meaningful criteria to compare and relate what is going on in the wind tunnel and on the full-size object: in each case, we scale the measured force according to the relevant local fow conditions. Advice from an expert Just like when selecting the length 퐿in the Reynolds number, there is no absolute physical law that would indicate which reference 푆or 푉should be taken when calculating a force coefcient. The only important thing is that the defnition should remain the same across all scaled fows. However, a large body of professional customs and cultures usually exists in each sub-area of fuid dynamics. Make sure you consult the available literature and feld conventions, so your results ft among them. Calculating the drag coefcient of a car using the car side-view area is not wrong, but it would go a long way to annoy colleagues and competitors! Video: practical application of scale efects: translating mea-surements made on a 60 % size formula one car wind tunnel model into their “real size” race car values, as in problem 8.5 by the Sauber F1 team (styl) For example, a fow case around a car A may be studied using a model B. If dynamic similarity is maintained (and that is by no means an easy task!), then the fow dynamics will be identical. The drag force 퐹D B measured on the model can then be compared to the force 퐹D A on the real car, using coefcients: since 퐶퐹D B = 퐶퐹D A we have: 퐹D A = 퐶퐹D A 1 2휌퐴푆A푉2 A = 퐶퐹D B 1 2휌퐴푆A푉2 A = 퐹D B 1 2휌퐵푆B푉2 B 1 2휌퐴푆A푉2 A 퐹D A = ( 휌퐴 휌퐵 푆A 푆B 푉2 A 푉2 B ) 퐹D B 8.4.3 Power coefcient, and other coefcients 170 Video: Understanding scale ef-fects will save you a lot of money and embarrassment: the caution-ary tale of Howard Hughes by Olivier Cleynen (CC-by) During our investigation of the fow using a small-scale model, we may be interested in measuring not only forces, but also other quantities such as power — this would be an important parameter, for example, when studying a pump, an aircraft engine, or a wind turbine. We again go back to chapter 2 (Analysis of existing fows with one dimension), and in particular eq. 2/18 p. 40. This helps us recall that power gained or lost by a fuid fowing steadily through a control volume could be expressed as: ̇ 푄net + ̇ 푊shaft, net = Σnet [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)] (8/16) Thus, the power gained or lost by the fuid is directly related to the magnitude of the scalar Σnet 휌\|푉⟂\|퐴푉2. A meaningful ”scale” for the power of a machine can therefore be the amount of energy in the fuid that is made available to it every second, a quantity that grows proportionally to 휌퐴푉3. This “scale of fuid fow-related power” is conventionally measured using the power coefcient 퐶P: 퐶P ≡ ̇ 푊 1 2휌푆푉3 (8/17) where ̇ 푊is the power added to or subtracted from the fuid (W). 8.4.4 Non-dimensionalizing all the problems XKCD #687: abusing dimen-sional analysis by Randall Munroe (CC-by-nc) In this chapter, we have focused on scaling forces and powers from one fow to the other. There are of course other parameters of interest, and for each, it is common practice to defne a coefcient. We have already used the pressure loss coefcient 퐾퐿≡\|Δ푝\|/1 2휌푉2 av. in chapter 7 (eq. 7/6 p. 137), and we shall soon use the shear coefcient 푐푓in the forthcoming chapter (eq. 10/6 p. 203). There are many other examples; in fact, in fuid mechanics, the non-dimen-sionalization of problems is a very important work methodology. This helps us extrapolate from one or two experiments up to entire families of fows. Advice from an expert Now you should begin to see that there is more to di-mensional analysis than just making models. Being good at comparing the infuences of various factors on a fuid fow phenomenon means that you can gen-eralize results. Pressure loss in a pipe, for example, is quantifed with a single, suitably-designed param-eter 푓, irrespective of say the fow speed or fuid density. Take a look back on the Moody diagram page 152: it has non-dimensional terms on all three sides! This is the reason why we only need one diagram, instead of a catalog of diagrams (e.g. one for each fow speed and diameter). And so, you should not be surprised when, after you show them the results of your frst experimental fuid fow measurement campaign, your colleagues in the fuid dynamics laboratory ask you: “can you non-dimensionalize that diagram?” 171 8.5 Solved problems Scale model of a car If we want to reproduce turbulent efects in the airfow around a car using a 1/20 model, how fast should the fow around the model be? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Wind tunnel speed We reproduce the airfow around a full-size car moving in air at 25 °C. We use a 1/20 model in a wind tunnel. This tunnel is limited to 20 m s−1, but the air is cooled (푇= 0 °C). What is the real-car speed that is being simulated in the wind tunnel? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Wind tunnel measurement In the setup from the previous example, how should the force measure-ments on the car be translated to values corresponding to the real car? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 172 Problem sheet 8: Engineering models last edited September 19, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 The non-dimensional incompressible Navier-Stokes equation: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (8/13) in which [St] ≡푓퐿 푉, [Eu] ≡푝0−푝∞ 휌푉2 , [Fr] ≡ 푉 √푔퐿, and [Re] ≡휌푉퐿 휇 . The force coefcient 퐶퐹and power coefcient 퐶P are defned as: 퐶퐹≡ 퐹 1 2휌푆푉2 퐶P ≡ ̇ 푊 1 2휌푆푉3 (8/17) The speed of sound 푐in air is modeled as: 푐= √ 훾푅푇 (1/11) Figure 8.3 quantifes the viscosity of various fuids as a function of temperature. 173 −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 8.3: The viscosity of four fuids (crude oil, water, air, and CO2) as a function of temperature. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. Figure reproduced from fgure 5.6 p. 99; CC-by by Arjun Neyyathala & Olivier Cleynen 8.1 Reading quiz Once you are done with reading the content of this chapter, you can go take the associated quiz at In the winter semester, quizzes are not graded. 8.2 Scaling a golf ball CC-0 Olivier Cleynen 8.2.1. Write the defnition of the Reynolds number and indicatie the si units for each term. 8.2.2. What is the consequence on the velocity feld of having a low Reynolds number? (Briefy justify your answer, e.g. in 30 words or less) 8.2.3. Give one example of a high-Reynolds number fow, and one of a low-Reynolds number fow. 174 The standard golf ball has a diameter of 42,67 mm and a mass of 45,93 g (fgure 8.4). A typical maximum velocity for such balls is 200 km h−1. Figure 8.4: A golf ball Photo CC-by by Bradley P. Johnson A student wishes to investigate the fow over a golf ball using a model in a wind tunnel. Using a 3D printer, s/he prints an enlarged model with a diameter of 50 cm. 8.2.4. If the atmospheric conditions are identical, what fow velocity needs to be generated during the experiment in order to reproduce the fow patterns around the real ball? 8.2.5. Would the Mach number for the real ball then be reproduced? 8.2.6. If the temperature on the golf course is −5 °C, and the temperature in the wind tunnel is 31 °C (with identical pressure, 1 bar), what should the new wind tunnel velocity be? 8.2.7. If the enlarged model was made out of the same materials as the real ball, how heavy would it be? 8.3 Fluid mechanics of a giant airliner non-examinable, CC-0 Olivier Cleynen An Airbus A380 airliner has an 80 m-wingspan and is designed to cruise at [Ma] = 0,85 where the air temperature and pressure are −40 °C and 0,25 bar. A group of students wishes to study the fow feld around the aircraft, with a wind tunnel whose test section has a diameter of 1 m (in which, obviously, the model has to ft!). Together, they brainstorm various possibilities. They assume that the viscosity of the air is always 1,5 ⋅10−5 Pa s 8.3.1. If the temperature inside the tunnel is maintained at 푇tunnel = 20 °C, propose a combination of wind tunnel velocity and pressure which would enable the team to adequately model the efects of viscosity. 8.3.2. Which fow conditions would be required in a water tunnel with the same dimensions? Realizing the conditions calculated in the questions above cannot be realized in practice, the group of students tries instead to study the efect of compressibility (while accepting that viscous efects may not be adequately modeled). 8.3.3. If the maximum velocity attainable in the wind tunnel is 80 m s−1, which tunnel air temperature is required for compressibility efects to be modeled? 175 8.4 Scale efects on a dragonfy A dragonfy (sketched in fg. 8.5) has a 10 cm wingspan, a mass of 80 mg, and cruises at 4 m s−1, beating its four wings 20 times per second. Figure 8.5: Plan view of a dragonfy Figure by Olivier Cleynen, A. Plank, Drury, Dru, Westwood, J. O. (CC-0) You are tasked with the investigation of the fight performance of the dragonfy, and have access to a wind tunnel with a test section of diameter 1 m. 8.4.1. Which model size and fow velocity would you use for this experiment? 8.4.2. How many wing beats per second would then be required on the model? 8.4.3. What would be the lift developed by the model during the experiment? 8.4.4. How much mechanical power would the model require, compared to the real dragonfy? 8.5 Formula One testing You are leading the Aerodynamics team in a successful Formula One racing team. Your team races a car that is 5,1 m long, 1,8 m wide, has a 610 kW power plant, and a mass of 750 kg (fgure 8.6). The car can reach a top speed of 310 km h−1; at that speed, the drag force is 7,1 kN. In order to test diferent aerodynamic confgurations for the car, your team invests in a 50-million-euro wind tunnel, in which you will run tests on a model. You currently have the choice between two model sizes: a 60 % model and a (smaller) 50 % model. 8.5.1. What would be the frontal area of each model, in proportion to the frontal area of the real car? 8.5.2. What would be the volume of each model, compared to the volume of the real car? 8.5.3. How much less weight would the 50 % model have than the 60 % model? 176 Figure 8.6: A 2017 Formula One racing car, driven by Pascal Wehrlein Photo CC-by by Jake Archibald From here on, your team decides to use the 50 % car model. 8.5.4. If the ambient atmospheric conditions cannot be changed, which fow speed in the wind tunnel is required, so that the air fow around the real car is reproduced around the model? You are preparing for a race where the air temperature will be 30 °C. The Formula One regulations forbid you from running your wind tunnel faster than 50 m s−1. 8.5.5. Your team considers modifying the wind tunnel air temperature, in order to compensate for the limit in the air speed. If the temperature in the tunnel can be controlled between −10 °C and 40 °C, but the pressure remains atmospheric (1 bar), what is the maximum race-track speed that can be simulated in the wind tunnel? 8.5.6. In that case, by which factor should the model drag force measurements be multiplied in order to correspond to the real car? 8.5.7. The wind tunnel has a 2 m by 2 m square test cross-section. What is the power required to bring stationary atmospheric air outside (푇atm. = 15 °C, 푉atm. = 0 m s−1) to the desired speed and temperature in the test section? NB: this exercise is inspired by an informative and entertaining video by the Sauber F1 team about their wind tunnel testing, which you are encouraged to watch at 177 Answers 8.2 1) See equation 8/12 and subsequent comments; 2) See §8.2.5, in practice this can be summarized in two or three sentences; 3) High-Re: air fow around an airliner in cruise; Low-Re: air fow around a dust particle falling to the ground. 4) Match [Re]: 푉B = 4,74 m s−1 5) No, but it is very low anyway (no compressibility efects to be reproduced) 6) Increase the velocity by 25 % 7) 푚C = 73,9 kg 8.3 1) With a half-aircraft model of half-width 퐿2/2 = 80 cm, at identical speed and ambient temperature, we would need an air pressure 푝2 ≈20 bar! 2) In water, the density cannot be reasonably controlled, and we need a velocity 푉3 = 418 m s−1! 3) In air, at 푉4 = 80 m s−1, Mach number can be reproduced at 푇4 = −251 °C (although the Reynolds number is of). Wind tunnels used to investigate compressible fow around aircraft have very powerful coolers. 8.4 1) With e.g. a model of span 60 cm, match the Reynolds number: 푉2 = 0,67 m s−1; 2) Match the Strouhal number: 푓2 = 0,56 Hz. Mach, Froude and Euler numbers will have no efect here; 3) and 4) are left as a surprise! 8.5 3) The third model would have 42 % less mass than the second; 4) Maintaining [Re] requires 푉3 = 620 km h−1 (fast!); 5) Now the fastest track speed that can be studied is 31,7 m s−1; 6) Multiply force measurements by 1,39 to scale up to reality; 7) The required work as power is 331 kW and the power to change temperature is −6,654 MW (better keep the cooled air in a closed circuit than feed the tunnel with outside air!) 178 Fluid Dynamics Chapter 9 – Dealing with turbulence last edited August 13, 2019 by Olivier Cleynen — 9.1 Motivation 179 9.2 Recognizing turbulence 180 9.2.1 A brief defnition 180 9.2.2 Chaos, not randomness 180 9.2.3 Growth and decay 181 9.2.4 A cascade of vortices 182 9.2.5 Not turbulence 182 9.3 The efects of turbulence 183 9.3.1 Dissipation (losses) 183 9.3.2 Main fow patterns 183 9.3.3 Mixing 184 9.4 Quantifying turbulence 184 9.4.1 Average and fuctuation 184 9.4.2 Turbulence intensity 185 9.4.3 The size of eddies 185 9.4.4 Turbulent kinetic energy and dissipation rate 186 9.4.5 Turbulence anisotropy and inhomogeneity 188 9.5 Computing turbulent fow 189 9.5.1 Basic premise 189 9.5.2 Accounting for turbulence 189 9.6 Commented bibliography 190 9.7 Problems 193 9.1 Motivation Most fows of interest to engineers and scientists are turbulent. Fluid fow in industrial and domestic piping, in engines and in turbomachinery, is turbulent. Flow close to solid surfaces and in the wake of objects is turbulent at all but the slowest speeds. Blood fow in our largest veins and arteries, and air fow in our nostrils and tracheae, are turbulent. River fows, ocean currents, and all but the calmest winds are turbulent. Turbulence may be ubiquitous, but it remains an incredibly complex phe-nomenon, and describing it accurately requires either extraordinarily power-ful numerical computations, or advanced mathematics. Neither of those is available in this course. We are going to treat turbulence not as a topic of research, but instead merely as an occurrence that, as engineers, we need to account for. We will try to answer the following two questions: In a fow, how to quantify and measure the degree of turbulence? And how to predict its degree and its rate of decay? In answering those, we will keep in mind that at our level, turbulence char-acterization is not an exact science: we expect that our models may be of by a factor of 10 or so. An imprecise or inexact understanding is better than none at all, and, at the very least, we are developing some familiarity with turbulence that will be useful in further studies, especially when exploring computational fuid dynamics or experimental fuid dynamics. 179 Figure 9.1: Plankton blooming in the Atlantic ocean reveals the complexity of the fow passing over the coast of Argentina. The scale of the image is so that the height covers approximately 500 km in this take. Image by Jacques Descloitres, MODIS Rapid Response Team NASA GSFC (public domain) 9.2 Recognizing turbulence This topic is well covered in Tennekes & Lumley 9.2.1 A brief defnition We defne turbulence as fuid motion in multiple scales which is chaotic and dissipative. 9.2.2 Chaos, not randomness This topic is well covered in Leschziner Turbulence is made possible by kinematic instability in the fow; this means that small disturbances, such as non-uniformities in the velocity distribution, are not signifcantly damped. The main dampening factor in fuids is viscos-ity 휇, and so the main parameter which determines how stable a given fow is the Reynolds number (eq. 1/28): [Re] ≡휌푉퐿 휇 (9/1) Typically, when [Re] exceeds 1 000, the fow is very likely to be or become turbulent (fgure 9.2). We have seen in chapter 8 (Engineering models) that this is because the magnitude of the Laplacian of the velocity feld is 1 000 times smaller than the magnitude of its advective. (Dampening factors other than viscosity sometimes also exist, such as density gradients or interaction with soft solid surfaces: in those cases, other non-dimensional parameters are used). This instability is what gives turbulence its chaotic characteristic. If a rider-less bicycle is rolled forward and left to itself, it will continue rolling and eventually fall to the side. Which side, left or right, depends on the initial 180 conditions: even a minute modifcation to the start position is likely to infuence the result. The fall is deterministic, and can be calculated very precisely, but with a very strong dependence on the initial conditions. Figure 9.2: A column of hot air from burning incense rising through cold air is an unstable situation. After a certain length, the fow breaks down into chaotic patterns. The occurrence is predicted across all fuids, plume diameters, and velocities: every time, the Reynolds number is the determining parameter. Photo CC-by by Rafa Espada Video: simulation of two misci-ble fuids of diferent densities layered one on top of the other (color representing density). The “perfect” uniform initial situa-tion is unstable and leads to chaotic (hard to predict) pat-terns whose details will depend strongly on minute changes in the initial conditions. (A 2d dns simulation performed with Mi-croHH) by Chiel van Heerwaarden (CC-by) Turbulent motion in fuids has the same properties. Fluid motion follows laws which are fully deterministic (the Navier-Stokes equation, eq. 6/42), but the exact patterns in situations where the Reynolds number is high cannot be predicted because, much like the for the bicycle above, they depend very minutely on the initial confguration. Thus, in two identical turbulent fow experiments, the details of the fow will be diferent. In this sense, turbulent fow is chaotic (depending extremely sensitively on initial conditions) but not random: it remains predictable, governed by well-known deterministic laws in which chance does not play a role. The efective engineer will determine 1) what general characteristics of turbulence do remain identical in both fows, and 2) how they afect the main, global fow characteristics. 9.2.3 Growth and decay Turbulence occurs because of shear (sometimes also pressure) applied non-uniformly in a fuid. Typical sources are a sudden turn at the downstream side of an obstacle, and shear alongside solid walls. Turbulence is only sustained if shear is continually applied to the fow (for example by the walls of a long pipe). If no sustaining source is provided, turbulence decays and eventually dies. This is a critical property: turbulence is dissipative. Large-scale motion created by an initial obstacle continually 181 breaks down into ever smaller-scale motion. Ultimately, the motion becomes so small and slow that the energy is dissipated by viscosity into molecular-scale motion. In this sense, turbulence is not magic: it occurs in circumstances that are well understood and documented. Turbulence takes energy out of the main fow (this is usually measurable as a streamwise pressure drop) and ultimately transforms it into heat. 9.2.4 A cascade of vortices This topic is well covered in Davidson It is useful to think of turbulence in terms of vortices (also named eddies): parcels of rotating fuid. In a frst approach, we can consider a turbulent fuid fow as made of two components: one main, mostly steady component (the time-average), and an additional secondary component, made of a complex chaotic tangle of vortices of multiple sizes. Turbulence begins when large vortices are created, whose size is approxi-mately that of the largest obstacle in the fow path. When those vortices are stretched, they speed up (by virtue of conservation of angular momentum). When they are compressed, they buckle and twist, deforming into complex shapes and interacting with other vortices; this leads to their breakup into smaller vortices. A turbulent fow thus consists of a cascade of structures of decreasing size. Energy is always passed down from larger into smaller structures. Smaller vortices have lower diameter and feature lower velocities: they are then more strongly afected by viscosity. Ultimately, the smallest vortices disappear, their energy dissipated down into heat. 9.2.5 Not turbulence Not all unsteady fows are turbulent. Well-known patterns such as a von Kármán vortex street (fgure 9.3) or a series of wave clouds, for example, are not turbulent. Those oscillations occur at a single recognizable frequency and scale, and once the phenomena has begun, their evolution is easily predictable. Most surface waves on a body of water (e.g. waves in open sea, fgure 9.3) are not turbulence: although they may be partly chaotic, they dissipate very little energy (unless they crash on a shore) and propagate over very large distances. Molecular motion is, at the macroscopic scale, completely random, and will appear in measurements as Gaussian white noise with no distinguishable range of frequencies, and no dissipation phenomenon: it is not turbulence, either. Although more formal defnitions of turbulence exist (see e.g. [13, 12, 31]), we are satisfed with focusing on just three requisites: a fow is turbulent if the motion contains many scales, is chaotic, and is dissipative. 182 Figure 9.3: Not turbulence: surface waves on the ocean (top) are complex, but not fully-chaotic, and they feature very little dissipation. Well-known oscillatory patterns such as the von Kármán vortex street feature one dominant frequency and one dominant vortex size: they are not turbulent either. Sea wave photo CC-by-sa by Tiago Fioreze (cropped) Wake photo CC-by-sa by Jürgen Wagner 9.3 The efects of turbulence This topic is well covered in Tennekes & Lumley 9.3.1 Dissipation (losses) Turbulence extracts energy from the main fow and dissipates it at a molecular level. This energy is never recovered. Turbulence therefore causes additional losses in the fow, that would not be present if it were to remain laminar. Typically, friction losses are increased by a factor 2 to 10. In bulk, turbulence has an efect similar to an increase in viscosity. A word of warning must follow this sentence: the degree of turbulence in a given fow is rarely uniform, and so this efect can rarely be quantifed as a single value for the entire fow; instead, it must be computed locally. The increase in temperature caused by turbulent dissipation is not usually measurable, due to the high heat capacity of fuids. It usually does not signifcantly alter their density. Turbulence therefore acts as a mechanical energy dissipator for the fow. In some special cases, turbulence leads to a reduction in losses. We shall study one such occurrence in chapter 10 (Flow near walls), where we sometimes intentionally create turbulent boundary layers. This is not important at the moment. 9.3.2 Main fow patterns In any given fow, the laminar and turbulent regimes result in markedly diferent fow patterns. Turbulent fows always lead to wider shear zones and very unsteady patterns, which, when averaged over time, have more uniform velocity profles. 183 When the fow is clearly turbulent, increases in the Reynolds number do not lead to main fow pattern changes anymore. For example, jet plumes at [Re]푑= 105 and 107 have nearly identical spread, length, and time-averaged velocity distribution. Some properties of turbulence still change with [Re], as we will see below. 9.3.3 Mixing Turbulence tremendously increases mixing. The large range of the scales of motion (i.e. the many diferent sizes of vortices which occur simultaneously) increases the contact surface between two mixing fuids, for example. This makes turbulence a desired property in many chemical reactors, or in cases where pollutants have to be dissipated (e.g. for exhaust gases). The same features of turbulence greatly enhance heat transfer compared to laminar fow. Most heat exchangers for which space is important, such as radiators, feature turbulent fow. The increased mixing also afects exchange of momentum, with increase in the interaction between slower and faster fuid particles. This tends to widen areas of interaction between fuids of diferent velocities, such as plumes, exhaust stacks and shear layers. 9.4 Quantifying turbulence 9.4.1 Average and fuctuation For the purpose of quantifying turbulence, we distinguish, in a given fow, be-tween the average velocity and the “turbulent part” of velocity, as illustrated in fgure 9.4. We thus decompose the velocity feld into two components: one is the average fow (푢, 푣, 푤), and the other the instantaneous fuctuation fow (푢′, 푣′, 푤′): 푢푖≡푢푖+ 푢′ 푖 (9/2) 푢′ 푖≡0 (9/3) Figure 9.4: An example of the separation between instantaneous and average values, here for temperature. The instantaneous temperature 푇is decomposed as the sum of the time-averaged temperature 푇(blue curve) and the fuctuation 푇′, whose average 푇′ is zero (red curve). Figure CC-0 Olivier Cleynen 184 9.4.2 Turbulence intensity This topic is well covered in De Nevers We defne turbulence intensity 퐼as the average of the root-mean-square of the velocity fuctuations: 퐼≡1 푉[ 1 3 [(푢′2) + (푣′2) + (푤′2)]] 1 2 (9/4) Turbulence intensity is a dimensionless local property representing the “strength” or “violence” of turbulence, usually expressed as a percentage. At each point in space, the root-mean-square of each of the three compo-nents of the velocity fuctuation is compared to the magnitude of the velocity. Typically, a low-turbulence channel fow would feature 퐼= 3 %, while a highly-turbulent channel fow could display for example 퐼= 80 %. 9.4.3 The size of eddies This topic is well covered in Leschziner When observing turbulence from the engineer’s point of view, it is useful to quantify two characteristic lengths: the size of the largest turbulent structures, and the size of the smallest structures. In that view, the cascade of turbulent eddies is contained between two limits: the turbulence-producing (large) scale, and the viscous dissipation (small) scale. The largest structures have a length whose order of magnitude is desig-nated as 퐿max ≡Λ. It is observed that Λ is typically 50 % of the largest scale of the fow (somewhat more in unconfned fows, and somewhat less in very confned fows). For example, wind fow around a building will feature eddies whose maximum size is half the size of the building. Those eddies also feature a characteristic velocity 푢Λ (typically, the maximum measurable velocity in the eddy) and a characteristic time scale 푡Λ (typically, a duration sufcient to describe their movement fully), linked by a straightforward relation: 푣Λ = Λ 푡Λ (9/5) It follows that one may quantify a large-scale eddy Reynolds number based on those quantities: [Re]Λ ≡휌푢ΛΛ 휇 (9/6) The smallest structures have a length whose order of magnitude is desig-nated as 퐿min ≡휂. Likewise, they feature a characteristic velocity 푢휂and a characteristic time scale 푡휂. In the 1940s, Andrey Kolmogorov and his team developed a model to relate both the large and the small scales in the simplest occurrences of turbulence. In particular, Kolmogorov postulated that the characteristic Reynolds number of the smallest vortices (based on their characteristic length 휂and speed 푢휂) is approximately equal to 1: [Re]휂≡휌푢휂휂 휇 ≈1 (9/7) 185 Figure 9.5: Surface-relative vorticity in the Atlantic ocean. Blue color indicates clockwise rotation, and red color anticlockwise rotation. Rotating structures of many diferent sizes can be observed. In homogeneous isentropic turbulence (and in this fow case by approximation), the size of the largest and smallest vortices are related to one another through the Reynolds number. doi:10.1038/s41467-018-02983-w CC-by by Z. Su, J. Wang, P. Klein, A. F. Thompson & D. Menemenlis Based on this postulate, when the turbulence has been given time and space enough to develop fully, is homogeneous, and isotropic (has identical proper-ties in all three directions) —these are important restrictions—, Kolmogorov and his peers showed using dimensional analysis that 퐿min 퐿max = 휂 Λ = [Re]−3/4 Λ (9/8) 푢휂 푢Λ = [Re]−1/4 Λ (9/9) 푡휂 푡Λ = [Re]−1/2 Λ (9/10) These three equations are a very important result. They show that as the Reynolds number of a fow increases, the size and time scale of the smallest structures in the fow decreases. The higher the Reynolds number, the more complex and more minute the details of the fow become. 9.4.4 Turbulent kinetic energy and dissipation rate This topic is well covered in De Nevers We defne turbulent kinetic energy 푘as: 푘≡1 2 ((푢′2) + (푣′2) + (푤′2)) (9/11) 186 Turbulent kinetic energy, measured in J kg−1, represents the amount of energy per unit mass contained in the chaotic (turbulent) component of the fuid fow velocity. To convince oneself that the parameters are related, one may insert 퐼(eq. 9/4) into eq. 9/11 to show that 푘= 3/2 푉 2퐼2. We defne the turbulent dissipation rate 휖as the rate at which turbulent kinetic energy is dissipating to heat. When no turbulence is produced, so that turbulence is simply left to decay, then 휖is the time rate change of 푘: 휖= −휕푘 휕푡 (9/12) when no new turbulence is produced. The dissipation rate is measured in W kg−1 and represents the local amount of turbulent kinetic energy that is currently being converted to heat through viscosity. Through dimensional analysis, Kolmogorov and his peers showed that in ho-mogeneous, fully-developed and isotropic turbulence, the size, characteristic velocity, and characteristic time scale of the smallest eddies could be related to the dissipation rate with the relationships: 휂= ( 휇3 휌3 1 휖) 1 4 (9/13) 푢휂= ( 휇 휌 1 휖) 1 4 (9/14) 푡휂= ( 휇 휌 1 휖) 1 2 (9/15) In other words, for completely-developed, homogeneous isotropic turbulence, if the characteristic size and speed of the largest turbulent structures are known, then the size of the smallest structures and the dissipation power can be known. In homogeneous, fully-developed, isotropic turbulence, it is interesting to observe how the turbulent kinetic energy and dissipation rate are distributed across the scales of the vortices. One can defne the kinetic energy density 퐸as the kinetic energy of all the eddies in an eddy-size increment d푙in a volume of interest, divided by that increment: 퐸≡휕푘 휕푙 In this way, the kinetic energy 푘is recovered as the integral of 퐸with respect to d푙: 푘= ∫ Λ 휂 퐸d푙 (9/16) In the same way, we may defne the dissipation density 퐷as 휕휖/휕푙, so that the dissipation rate would be recovered as: 휖= ∫ Λ 휂 퐷d푙 (9/17) 187 These two integral equations are only useful to understand the meaning of fgure 9.6, where the distribution of 푘and 휖across the scales of eddies is plotted. Figure 9.6: Distribution of turbulent kinetic energy (left) and of turbulent dissipation rate (right) in fully-developed homogeneous isotropic turbulence. The top diagrams are in linear scale, while the bottom diagrams are in logarithmic scale. In those diagrams, the horizontal axis displays 1/푙, so that the small-scale eddies are on the right side, and large-scale eddies are on the left side. Those energy and dissipation distributions are for the simplest occurrences of turbulence; their features (in particular, the curves’ slopes and the ratios between Λ and 휖) are used as reference cases in the study of more complex cases. Figure CC-by-sa by Olivier Cleynen In this fgure 9.6, one may see that most of the energy is contained in the large eddies, while most of the dissipation occurs in the small eddies. 9.4.5 Turbulence anisotropy and inhomogeneity The relationships between largest and smallest scales derived by Kolmogorov only hold true for very “straightforward” turbulence. In most fows of interest, turbulence is being created and dissipated in more complex ways. In such cases, which are outside the scope of this course, important properties to quantify are: Inhomogeneity (uneven distribution in space). The properties quantifed above are observed in diferent places, often revealing that production and dissipation of turbulence occur in diferent areas; Anisotropy (dependence on direction). The properties quantifed above are observed in diferent directions. Anisotropy is most often quantifed by evaluating correlations between velocity fuctuations, and the skewness (asymmetry) of their distributions. 188 Development. Sometimes, not enough time has passed for the full spectrum of scales to appear, because the energy has not cascaded down to the smallest scales yet. In most of those cases, the basic models studied above serve as reference cases, against which comparisons can be made. 9.5 Computing turbulent fow 9.5.1 Basic premise We have seen in §6.5 that in principle, the fow of fuids can be computed for any given fow by solving for the change in time of the unknowns 푢, 푣 and 푤in the Navier-Stokes equation. Such a formulation is called a Direct Numerical Simulation (dns); it allows solving for all fows and will very well describe turbulent fows. Video: dns simulation of air fow over an airfoil at relatively low speeds ([Re] = 4 ⋅105, so 푉≈ 50 km h−1). Because the com-plete details of the fow are solved, 35 million cpu-hours were needed for this calculation in 2015. On an ordinary desktop computer, this would take 500 years to complete. by Linné FLOW Centre (styl) Unfortunately, the reality is that the computational cost of dns is enormous, and precludes us from solving most fows of interest. Turbulence exacerbates the problem. With eqs. 9/8 & 9/10 we can see that as the Reynolds number increases, the spatial and temporal discretization of the computation must increase, too. Every decrease in the size of the grid cell and in the length of the time step increases the total number of equations to be solved by the algorithm. Halving each of 훿푥, 훿푦, 훿푧and 훿푡multiplies the total number of equations by 16, so that soon enough the designer of the simulation will wish to know what maximum (coarsest) grid size is appropriate or tolerable. Furthermore, in many practical cases, we may not even be interested in an exhaustive description of the velocity feld, and just wish to obtain a general, coarse description of the fuid fow. 9.5.2 Accounting for turbulence This topic is well covered in Versteeg et al. , Leschziner , and Wilkes The decomposition of the fow into two components 푢and 푢′ which we performed in eq. 9/2 earlier is useful when we wish to simulate the fow numerically (performing computational fuid dynamics or cfd). Let us now say that 푢is the component of velocity that is captured by the simulation, while 푢′ is the component which is too small, or occurring too quickly, for the simulation to capture. Inserting the defnition 9/2 into the 푥-component of the Navier-Stokes equation for incompressible fow, we obtain: 휌[ 휕(푢+ 푢′) 휕푡 + (푢+ 푢′)휕(푢+ 푢′) 휕푥 + (푣+ 푣′)휕(푢+ 푢′) 휕푦 + (푤+ 푤′)휕(푢+ 푢′) 휕푧 ] = 휌푔푥−휕(푝+ 푝′) 휕푥 + 휇[ 휕2(푢+ 푢′) (휕푥)2 + 휕2(푢+ 푢′) (휕푦)2 + 휕2(푢+ 푢′) (휕푧)2 ] Taking the average of this equation —thus expressing the dynamics of the fow as we calculate them with a fnite, coarse grid— yields, after some intimidating but easily conquerable algebra: 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦+ 푤휕푢 휕푧] + 휌[푢′휕푢′ 휕푥+ 푣′휕푢′ 휕푦+ 푤′휕푢′ 휕푧] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2] (9/18) 189 Equation 9/18 is the 푥-component of the Reynolds-averaged Navier-Stokes equation (rans). It shows that when one observes the fow in terms of the sum of an average and an instantaneous component, the dynamics cannot be expressed solely according to the average component. Comparing eqs. 9/18 and 6/43 we fnd that an additional term has appeared on the left side. This term, called the Reynolds stress, is often re-written as 휌휕푢′ 푖푢′ 푗/휕푗. In turbulent fow, it is not zero, because the instantaneous fuctuations of velocity (e.g. 푢′ and 푣′) are strongly correlated: they are each zero on average, but their multiples are not. The diference between eqs. 9/18 and 6/43 can perhaps be expressed difer-ently: the time-average of a turbulent fow cannot be calculated by solving for the time-average velocities. Or, more bluntly: the average of the solution can-not be obtained with only the average of the fow. This is a tremendous burden in computational fuid dynamics, where limits on the available computational power prevent us in practice from solving for these fuctuations. In the overwhelming majority of computations, the Reynolds stress has to be approximated in bulk with schemes named turbulence models. That is, a local value for 휌휕푢′ 푖푢′ 푗/휕푗is estimated everywhere, depending on the average values (푢푖). The most well-known method for doing this is the 푘-epsilon turbulence model, which involves solving partly arbitrary transport equations for both 푘and 휖. The delights, shortcomings and mysteries of that method and more are left for the reader to discover in a good hands-on course on the youngest and most promising area of this discipline, cfd. 9.6 Commented bibliography Most books on turbulence spill into mathematical intricacies which are irrel-evant to the engineer; it is regrettable that some of the very best documenta-tion on turbulence is in the user manuals of cfd software and experimental measurement devices. It is also regrettable that few books provide actual applied problems to be solved quantitatively. A commented (and necessarily subjective) bibliography for learning about turbulence is proposed here: • General understanding (encyclopedic knowledge) The introductions of Mathieu & Scott , of Tennekes & Lumley , and of Davidson will provide excellent information. • Reference works Unfortunately, no book truly aimed at engineers is known to the au-thor. The following books provide in-depth insight over the physics of turbulence: – Tennekes & Lumley : despite its age, an outstanding book, in particular for its frst chapter. – Leschziner : despite its focus on cfd, the book serves as a great step-by-step exploration of turbulence. – Davidson (reference book). Exquisitely referenced. – Mathieu & Scott (reference book). An appropriate amount of comment is provided around mathematical expressions. – Pope (reference book). Terser than its counterparts above. 190 • Works containing useful problems De Nevers (a single chapter within) is a very applied, down-to-earth and useful treatment; Tennekes & Lumley also contains good examples in the frst chapter. • Dealing with turbulence in cfd The best coverage is probably in Leschziner . Some useful (passing) information can be found in Tu et al. and Versteeg et al. . • Books with parts useful for specifc purposes include: – Bernard & Wallace , for theory about experiments; – Libby , for its introduction to applied statistics; – Sagaut , for its coverage of the notion of spectra, and its excellent illustrations; – Cebeci , for its rigorous introduction into basic metrics, and excellent fundamental data and diagrams. 191 192 Problem sheet 9: Dealing with turbulence last edited August 13, 2019 by Olivier Cleynen — These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Turbulence intensity 퐼: 퐼≡1 푉[ 1 3 [(푢′2) + (푣′2) + (푤′2)]] 1 2 (9/4) Turbulent kinetic energy 푘: 푘≡1 2 ((푢′2) + (푣′2) + (푤′2)) (9/11) In homogeneous, isotropic, fully-developed turbulence, the following relationships apply between the largest-scale and smallest-scale eddies: 퐿min 퐿max = 휂 Λ = [Re]−3/4 Λ (9/8) 푢휂 푢Λ = [Re]−1/4 Λ (9/9) 푡휂 푡Λ = [Re]−1/2 Λ (9/10) 휂= ( 휇3 휌3 1 휖) 1 4 (9/13) 푢휂= ( 휇 휌 1 휖) 1 4 (9/14) 푡휂= ( 휇 휌 1 휖) 1 2 (9/15) 193 9.1 Hypothetical fow From De Nevers Ex 18.1 We imagine a turbulent fow described at some point with the equations (in m s−1) 푢= 10 + sin 푡 (9/19) 푣= 0 (9/20) 푤= 0 (9/21) (No real turbulent fow can be described by equations this simple — but this is a nice frst basis for practice) What are the values of 푢, 푢′, 푢′, 퐼푥, 퐼, and 푘? Hint: ∫sin2 푥d푥= 1 2 (푥+ sin 2푥 2 ) + 푏 9.2 Turbulent channel fow From De Nevers Ex 18.2 A wind tunnel carries air through a channel which is 1 m wide and 0,24 m high. The average velocity is 0,82 m s−1. The pressure drop caused by both friction on the walls and turbulent dissipation is measured at −0,0286 Pa m−1. 9.2.1. What is the non-turbulent kinetic energy per unit mass of the fow? 9.2.2. At what average rate does this kinetic energy degrade into heat? 9.2.3. If there was no heat transfer, what would be the rate of temperature increase of the air? Measurements are carried out to measure the turbulent intensity through the channel. Those are displayed in fgure 9.7. Figure 9.7: Measurements of turbulent intensity in 푥and 푦directions in a rectangular channel 1 m wide and 0,24 m wide, in which the centerline velocity 푣is 1 m s−1. Here 푢′ and 푣′ are written 푣푥and 푣푦respectively. Figure extracted from De Nevers , with source data from Reichardt 1938, Naturwissenschaften 26:407 9.2.4. What is the value of 푘at a point 2 cm from the wall? 194 9.3 Cumulus cloud From Tennekes & Lumley P1.1 A cumulus cloud (one of those “fufy” summer clouds, fgure 9.8) has roughly the size of a sphere of diameter 퐷= 50 m. To a good approximation, it features isotropic, homogeneous, fully-developed turbulence. The largest-scale air currents in the cloud reach a maximum velocity 푉= 3 m s−1. Figure 9.8: Summer clouds (Cumulus humilis) form when hot moist air convected from the ground is cooled down when it rises. Photo CC-by-sa by en:Wikipedia User:Dwindrim 9.3.1. What is approximately the size of the smallest eddies in the cloud? 9.3.2. What is approximately the dissipation power, per unit mass of air and for the entire cloud? 9.3.3. What will those three values become once the cloud has grown to a diameter of 퐷2 = 100 m? 9.4 Reactor tank A tank used to store chemical reactants has roughly the size of a cube of side length 퐿= 2 m (fgure 9.9). The tank is flled with a water-like liquid and vigorously stirred with a large agitator propeller for a prolonged amount of time. The propeller induces a maximum fuid velocity of 1,5 m s−1. 9.4.1. What is approximately the size of the smallest eddies in the tank? 9.4.2. What is approximately the specifc dissipation power? A full-scale simulation (dns) of the fow was carried out, which required 500 hours of computing time on a supercomputer. Now the same simulation is to be carried out again, for the same fow, but with a fuid whose viscosity is half that of water. 9.4.3. What do you expect the new computation time to be? 195 Figure 9.9: Schematic drawing of a cubic tank stirred with an agitator propeller Figure CC-0 Olivier Cleynen 196 Answers 9.1 p. 194 9.1.1 푢= 10 m s−1 9.1.2 푢′ = sin(푡) 9.1.3 푢′ = 0 m s−1 (as always) 9.1.4 퐼푥= 7,07 % 9.1.5 퐼= 4,08 % 9.1.6 푘= 0,25 J kg−1 9.2 p. 194 9.2.1 ̇ 푒푚main = 0,336 J kg−1 9.2.2 휖= 0,0191 W kg−1 9.2.3 ̇ 푇= 0,02 mK s−1 9.2.4 푘= 5,76 mJ kg−1 9.3 p. 195 9.3.1 Since [Re]Λ ≡6,1 ⋅103, 휂≈0,2 mm 9.3.2 휖≈1,147 W kg−1 & ̇ 푊휖≈92 kW 9.3.3 휂2 = 0,6휂1, 휖2 = 8휖1, ̇ 푊휖2 = 64 ̇ 푊휖1 9.4 p. 195 9.4.1 휂≈0,023 mm (width of human hair) 9.4.2 휖≈3,4 W kg−1 9.4.3 Increase resolution in all three directions according to 휂, and the time resolution according to 푡휂: the computation time increases by a factor 6,7. 197 198 Fluid Dynamics Chapter 10 – Flow near walls last edited September 19, 2020 by Olivier Cleynen — 10.1 Motivation 199 10.2 The concept of boundary layer 199 10.2.1 Rationale 199 10.2.2 Why do we study the boundary layer? 201 10.2.3 Characterization of the boundary layer 201 10.3 Laminar boundary layers 203 10.3.1 Governing equations 203 10.3.2 Blasius’ solution 204 10.4 Boundary layer transition 205 10.5 Turbulent boundary layers 206 10.6 Flow separation 207 10.7 Solved problems 210 10.8 Problems 213 These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . 10.1 Motivation In this chapter, we focus on fuid fow close to solid walls. In these regions, viscous efects dominate the dynamics of fuids. This study should allow us to answer two questions: • How can we quantify shear-induced friction on solid walls? • How can we describe and predict fow separation? 10.2 The concept of boundary layer 10.2.1 Rationale At the very beginning of the 20th century, Ludwig Prandtl observed that for most ordinary fuid fows, viscous efects played almost no role outside of a very small layer of fuid along solid surfaces. In this area, shear between the zero-velocity solid wall and the outer fow dominates the fow structure. He named this zone the boundary layer.w We indeed observe that around any solid object within a fuid fow, there exists a thin zone which is signifcantly slowed down because of the object’s presence. This deceleration can be visualized by measuring the velocity profle (fg. 10.1). The boundary layer is a concept, a thin invisible layer whose upper limit (termed 훿, as we will see in §10.2.3) is defned as the distance where the fuid velocity in direction parallel to the wall is 99 % of the outer (undisturbed) fow velocity. 199 Figure 10.1: A typical velocity profle in a boundary layer. Only the horizontal component of velocity (푉푥= 푢) is represented. Figure CC-by Olivier Cleynen Figure 10.2: The thickness of the boundary layer depends strongly on the faraway incoming fow velocity 푈∞. Figure CC-by-sa Olivier Cleynen & Commons User:F l a n k e r Upon investigation, we observe that the boundary layer thickness depends strongly on the main fow characteristics. In particular, it decreases when speed increases or when viscosity is decreased (fg. 10.2). As we travel downstream along a boundary layer, we observe experimentally that the fow regime is always laminar at frst. Then, at some distance downstream which we name transition point, the boundary layer becomes turbulent. The fow-wise position of the transition point depends on the fow properties and is somewhat predictable. Further downstream, the boundary layer becomes fully turbulent. It has larger thickness than in the laminar regime, and grows at a faster rate. Like all turbulent fows, it then features strong energy dissipation and its analytical description becomes much more difcult. The fow within the boundary layer, and the main external fow (outside of it) afect one another, but may be very diferent in nature. A laminar boundary layer may exist within a turbulent main fow; while turbulent boundary layers are commonplace in laminar fows. 200 10.2.2 Why do we study the boundary layer? Video: the basics of the bound-ary layer by Olivier Cleynen (CC-by) Expending our energy on solving such a minuscule area of the fow may seem counter-productive, yet three great stakes are at play here: • First, a good description allows us to avoid having to solve the Navier-Stokes equations in the whole fow. Indeed, outside of the boundary layer and of the wake areas, viscous efects can be safely neglected. Fluid fow can then be described with 휌D ⃗ 푉 D푡= −⃗ ∇푝(the Euler equation, which we will introduce as eq. 11/3 in the next chapter) with acceptably low inaccuracy. As we shall soon see, this allows us to fnd many interesting solutions, all within reach of human comprehension and easily obtained with computers. Unfortunately, they cannot account for shear on walls. Thus, solving the fow within the boundary layer, whether analytically or experimentally, allows us to solve the rest of the fow in a simplifed manner (fg. 10.3). Figure 10.3: Fluid fow around a wing profle. When analyzing the fow, whether analytically or within a computational fuid dynamics (cfd) simulation, the fow domain is frequently split into three distinct areas. In the boundary layer (B), fuid fow is dominated by viscosity. Outside of the boundary layer (A), viscous efects are very small, and the fow can be approximately solved using the Euler equation. Lastly, in the turbulent wake (C), characterization of the fow is very difcult and typically necessitates experimental investigations. Figure CC-by-sa Olivier Cleynen • Secondly, the boundary layer is the key to quantifying friction. A good resolution of the boundary layer allows us to precisely quantify the shear forces generated by a fuid on an object. • Finally, a good understanding of the mechanisms at hand within the boundary layer allows us to predict fow separation, which is the di-vergence of streamlines relative to the object. Control of the boundary layer is key to ensuring that a fow will follow a desired trajectory! 10.2.3 Characterization of the boundary layer Three diferent parameters are typically used to quantify how thick a bound-ary layer is at any given position. 201 The frst is the thickness 훿, 훿≡푦\|푢=0,99푈 (10/1) which is, as we have seen above, equal to the distance away from the wall where the speed 푢is 99 % of 푈. The second is the displacement thickness 훿∗, which corresponds to the vertical distance by which streamlines outside of the boundary layer have been displaced. This vertical “shifting” of the fow occurs because the inner fuid is slowed down near the wall, creating some blockage for the outer fow, which then proceeds to avoid it partially by deviating outwards (fg. 10.4). Integral analysis performed on a control volume enclosing the boundary layer (as for example in exercise 3.8 from chapter 3) allows us to quantify the displacement thickness as: 훿∗≡∫ ∞ 0 (1 −푢 푈) d푦 (10/2) In practice, this integral can be calculated on a fnite interval (instead of using the ∞limit), as long as the upper limit exceeds the boundary layer thickness. The third and last parameter is the momentum thickness 훿∗∗(sometimes written 휃) which is equal to the thickness of a corresponding layer of fuid at velocity 푈which would possess the same amount of momentum as the boundary layer. The momentum thickness can be thought of as the thickness of the fuid that would need to be entirely stopped (for example by pumping it outside of the main fow) in order to generate the same drag as the local boundary layer. A review of the experience we gathered in chapter 3 while solving prob-lems 3.7 to 3.9 p. 67 allows us to to quantify the momentum thickness as: 훿∗∗≡∫ 훿 0 푢 푈(1 −푢 푈) d푦 (10/3) with the same remark regarding the upper limit. The momentum thickness, in particular when compared to the displacement thickness, is an important parameter used in prediction models for boundary layer separation. Once these three thicknesses have been quantifed, we are generally looking for a quantifcation of the shear term 휏wall. Since we are working with the Figure 10.4: Thickness 훿and displacement thickness 훿∗of a boundary layer. It is important to understand that the boundary layer is not a closed zone: streamlines (drawn blue) penetrate it and the vertical velocity 푣, although very small compared to 푢, is not zero. Figure CC-by-sa Olivier Cleynen & Commons User:F l a n k e r 202 hypothesis that the fuid is Newtonian, we merely have to know 푢(푦) to quantify shear, according equation 5/22 which we wrote way back p. 97: 휏wall 푦푥= 휇휕푢 휕푦 (10/4) In a boundary layer, the shear 휏wall will decrease with longitudinal distance 푥, because the velocity gradient above it also decreases. Consequently, 휏wall will become a function of 푥, so that the entire shear force will be obtained by integration (reusing eq. 5/3 p. 92): 퐹shear푦푥= ∫ 푆 휏wall 푦푥d푥d푧 (10/5) As we have seen in the previous chapter, fuid dynamicists like to quantify phenomena with non-dimensional parameters. The wall shear exerted by the boundary layer is typically non-dimensionalized with the shear coefcient 푐푓, 푐푓(푥) ≡휏wall 1 2휌푈2 (10/6) where 푈is the outer-layer (free-stream) velocity. The shear coefcient, just like the shear, remains a function of the fow-wise distance 푥. 10.3 Laminar boundary layers 10.3.1 Governing equations What is happening inside a laminar, steady boundary layer? We begin by writing out the Navier-Stokes for incompressible isothermal fow in two Cartesian coordinates (eqs. 6/42 p. 125): 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2] (10/7) 휌[ 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦] = 휌푔푦−휕푝 휕푦+ 휇[ 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2] (10/8) Building from these two equations, we are going to add three simplifcations, which are hypotheses based on experimental observation of fuid fow in boundary layers: 1. Gravity plays a negligible role; 2. The component of velocity perpendicular to the wall (in our conven-tion, 푣) is very small (푣≪푢). Thus, its stream-wise spatial variations can also be neglected: 휕푣/휕푥≈0 and 휕2푣/(휕푥)2 ≈0. The same goes for the derivatives in the 푦-direction: 휕푣 휕푦≈0 and 휕2푣/(휕푦)2 ≈0. 3. The component of velocity parallel to the wall (in our convention, 푢) varies much more strongly in the 푦-direction than in the 푥-direction: 휕2푢/(휕푥)2 ≪휕2푢/(휕푦)2. 203 With all of these simplifcations, equation 10/8 shrinks down to 휕푝 휕푦≈0 (10/9) which tells us that pressure is a function of 푥only (휕푝/휕푥= d푝/ d푥). We now turn to equation 10/7, frst to obtain an expression for pressure by applying it outside of the boundary layer where 푢= 푈: d푝 d푥= −휌푈d푈 d푥 (10/10) and secondly to obtain an expression for the velocity profle: 푢휕푢 휕푥+ 푣휕푢 휕푦= −1 휌 휕푝 휕푥+ 휇 휌 휕2푢 (휕푦)2 = 푈d푈 d푥+ 휇 휌 휕2푢 (휕푦)2 (10/11) Thus, the velocity feld ⃗ 푉= (푢; 푣) = 푓(푥, 푦) in a steady laminar boundary layer is driven by the two following equations: a balance of momentum, and a balance of mass: 푢휕푢 휕푥+ 푣휕푢 휕푦= 푈d푈 d푥+ 휇 휌 휕2푢 (휕푦)2 (10/12) 휕푢 휕푥+ 휕푣 휕푦= 0 (10/13) The main unknown in this system is the longitudinal speed profle across the layer, 푢(푥,푦). Unfortunately, over a century after it has been written, we still have not found an analytical solution to it. 10.3.2 Blasius’ solution Heinrich Blasius undertook a PhD thesis under the guidance of Prandtl, in which he focused on the characterization of laminar boundary layers. As part of his work, he showed that the geometry of the velocity profle (i.e. the velocity distribution) within such a layer is always the same, and that regardless of the fow velocity or the position, 푢can be simply expressed as a function of non-dimensionalized distance away from the wall termed 휂: 휂≡푦 √ 휌푈 휇푥 (10/14) Blasius was able to show that 푢is a function such that 푢/푈= 푓′ (휂), with 푓′′′ + 1 2푓푓′′ = 0. Unfortunately, no known analytical solution to this equation is known. However, it has now long been possible to obtain numerical values for 푓′ at selected positions 휂. Those are plotted in fg. 10.5. Based on this work, it can be shown that for a laminar boundary layer fowing along a smooth wall, the four parameters of interest for the engineer are 204 Figure 10.5: The velocity profle obtained by Blasius (an exact solution to the Navier-Stokes equations simplifed with laminar boundary-layer hypothesis). Figure CC-0 Olivier Cleynen simply functions of the distance-based Reynolds number [Re]푥: 훿 푥= 4,91 √ [Re]푥 (10/15) 훿∗ 푥= 1,72 √ [Re]푥 (10/16) 훿∗∗ 푥= 0,664 √ [Re]푥 (10/17) 푐푓(푥) = 0,664 √ [Re]푥 (10/18) 10.4 Boundary layer transition After it has traveled a certain length along the wall, the boundary layer becomes very unstable and it transits rapidly from a laminar to a turbulent regime (fg. 10.6). We have already described the characteristics of turbulence in broadly in chapter 7 (Pipe fows) and more extensively in chapter 9 (Dealing with turbulence); they apply to turbulence within the boundary layer. It is worth reminding ourselves that the boundary layer may be turbulent in a globally laminar fow (e.g. around an aircraft in fight, the boundary layer is turbulent, but the main fow is laminar). Here, we refer to the regime of the boundary layer only, not the outer fow. We observe that the distance 푥transition at which the boundary layer changes regime is reduced when the velocity is increased, or when the viscosity is decreased. In practice this distance depends on the distance-based Reynolds number [Re]푥≡휌푈푥/휇. The most commonly accepted prediction for the transition position is: [Re]푥transition ≈5 ⋅105 (10/19) 205 Figure 10.6: Transition of a boundary layer from laminar to turbulent regime. Figure CC-by Olivier Cleynen Transition can be generated earlier if the surface roughness is increased, or if obstacles (e.g. turbulators, vortex generators, trip wires) are positioned within the boundary layer. Conversely, a very smooth surface and a very steady, uniform incoming fow will result in delayed transition. 10.5 Turbulent boundary layers Video: visualizing the turbu-lent boundary layer on fat body moved in a stationary tank by Lee, Kwon, Hutchins, and Monty (styl) The extensive description of turbulent fows remains an unsolved problem. As we have seen in chapter 9 (Dealing with turbulence), by contrast with laminar counterparts, turbulent fows result in • increased mass, energy and momentum exchange; • increased losses to friction; • apparently chaotic internal movements. Instead of resolving the entire time-dependent fow in the boundary layer, we satisfy ourselves with describing the average component of the longitudinal speed, 푢. A widely-accepted velocity model is: 푢 푈≈( 푦 훿) 1 7 (10/20) for turbulent boundary layer fow over a smooth surface. Video: a look inside a turbu-lent boundary layer, with a com-pletely resolved (dns) fuid fow simulation by Linné Flow Centre & Serc KTH [26, 27] (styl) This profle has a much fatter geometry near the wall than its laminar counterpart (fg. 10.7). In the same way that we have worked with the laminar boundary layer profles, we can derive models for our characteristics of interest from this velocity profle: 훿 푥≈0,16 [Re] 1 7 푥 (10/21) 훿∗ 푥≈0,02 [Re] 1 7 푥 (10/22) 훿∗∗ 푥≈0,016 [Re] 1 7 푥 (10/23) 푐푓(푥) ≈0,027 [Re] 1 7 푥 (10/24) 206 Figure 10.7: Comparison of laminar and turbulent boundary layer profles, both scaled to the same size. It is important to remember that in practice turbulent boundary layers are much thicker than laminar ones, and that 푢is only a time-average velocity. Figure CC-0 Olivier Cleynen 10.6 Flow separation Video: fow separation: what it is, how to predict and avoid it by Olivier Cleynen (CC-by) Under certain conditions, fuid fow separates from the wall. The boundary layer then disintegrates and we observe the appearance of a turbulent wake near the wall. Separation is often an undesirable phenomenon in fuid mechanics: it may be thought of as the point where we fail to impart a desired trajectory to the fuid. When the main fow speed 푈along the wall is varied, we observe that the geometry of the boundary layer changes. The greater the longitudinal speed gradient (d푈/ d푥> 0), and the fatter the profle becomes. Conversely, when the longitudinal speed gradient is negative, the boundary layer velocity profle straightens up. When it becomes perfectly vertical at the wall, it is such that streamlines separate from the wall: this is called separationw (fg. 10.8). The occurrence of separation can be predicted if we have a robust model for the velocity profle inside the boundary layer. For this, we go back to fundamentals, stating that at the separation point, the shear efort on the surface must be zero: 휏wall at separation = 0 = 휇( 휕푢 휕푦)@푦=0 (10/25) At the wall surface (푢= 0 and 푣= 0), equation 10/12 p. 204 becomes: 휇( 휕2푢 (휕푦)2)@푦=0 = d푝 d푥= −휌푈d푈 d푥 (10/26) Thus, as we progressively increase the term d푝/ d푥, the term 휕2푢/(휕푦)2 reaches higher (positive) values on the wall surface. Nevertheless, we know 207 Figure 10.8: Separation of the boundary layer. The main fow is from left to right, and fowing towards a region of increasing pressure. For clarity, the 푦-scale is greatly exaggerated. Figure CC-by Olivier Cleynen that it must take a negative value at the exterior boundary of the boundary layer. Therefore, it must change sign somewhere in the boundary. This point where 휕2푢/(휕푦)2 changes sign is called infexion point. The existence of the infection point within the boundary layer tells us that at the wall (푦= 0) the term 휕푢/휕푦tends towards ever smaller values. Given enough distance 푥, it will reach zero value, and the boundary layer will separate (fg. 10.9). Therefore, the longitudinal pressure gradient, which in practice determines the longitudinal velocity gradient, is the key factor in the analytical prediction of separation. We shall remember two crucial points regarding the separation of boundary layers: 1. Separation occurs in the presence of a positive pressure gradient, which is sometimes named adverse pressure gradient. Separation points along a wall (e.g. a car bodywork, an aircraft wing, rooftops, mountains) are always situated in regions where pressure increases (positive d푝/ d푥or negative d푈/ d푥). If pressure remains constant, or if it decreases, then the boundary layer cannot separate. Figure 10.9: The infection point within a boundary layer about to separate. Figure CC-by Olivier Cleynen 208 2. Laminar boundary layers are much more sensitive to separation than turbulent boundary layers (fg. 10.10). A widely-used technique to reduce or delay the occurrence of separa-tion is to make boundary layers turbulent, using low-height artifcial obstacles positioned in the fow. By doing so, we increase shear-based friction (which increases with turbulence) as a trade-of for better resistance to stall. Figure 10.10: The efect of decreasing Reynolds number on fow attachment over an airfoil at constant angle of attack, with the transition point highlighted. Laminar boundary layers are much more prone to separation than turbulent boundary layers. Figure CC-by-sa Olivier Cleynen, based on Barlow & Pope 1999 Video: a football with a rough edge on one side will see tur-bulent boundary layer on that side and laminar boundary layer on the other. When the lami-nar layer separates before the turbulent one, the main fow around the ball becomes asym-metric and deviates the ball side-ways, a phenomenon that can be exploited to perform impressive tricks. by freekikerz (styl) Predicting in practice the position of a separation point is difcult, because an intimate knowledge of the boundary layer profle and of the (external-fow-generated) pressure feld are required — and as the fow separates, these are no longer independent. Resorting to experimental measurements, in this case, is often a wise idea! 209 10.7 Solved problems Advertising board on a car A successful fuid dynamics professor advertises for their course using a board above their car. They drive at 10 m s−1; the board is 3 m long and 1,5 m high. Drawing CC-by-sa by Imad Kharkouk Will the boundary layer on the board become turbulent? How thick will it become? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Shear force on a board (laminar part) In the example above, what is the shear force on the laminar layer part of the board? See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) Note: Unfortunately Olivier made an error in this video: the fnal expression is correct, but improperly calculated. The correct result is 0,107 N. Many thanks to the students who double-checked and reported the problem! Shear force on a board (turbulent part) In the example above, what is the shear force on the turbulent layer part of the board? And what would be the power lost to friction on the entire board? 210 See this solution worked out step by step on YouTube (CC-by Olivier Cleynen) 211 212 Problem sheet 10: Flow near walls last edited July 9, 2020 by Olivier Cleynen — Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 In boundary layer fow, we assume that transition occurs at [Re]푥≳5 ⋅105. The wall shear coefcient 푐푓, a function of distance 푥, is defned based on the free-stream fow velocity 푈: 푐푓(푥) ≡휏wall 1 2휌푈2 (10/6) Exact solutions to the laminar boundary layer along a smooth surface yield: 훿 푥= 4,91 √ [Re]푥 훿∗ 푥= 1,72 √ [Re]푥 (10/16) 훿∗∗ 푥= 0,664 √ [Re]푥 푐푓(푥) = 0,664 √ [Re]푥 (10/18) Solutions to the turbulent boundary layer along a smooth surface yield the following time-averaged characteristics: 훿 푥≈0,16 [Re] 1 7 푥 훿∗ 푥≈0,02 [Re] 1 7 푥 (10/22) 훿∗∗ 푥≈0,016 [Re] 1 7 푥 푐푓(푥) ≈0,027 [Re] 1 7 푥 (10/24) Figure 10.11 quantifes the viscosity of various fuids as a function of temperature. 213 −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 10.11: The viscosity of four fuids (crude oil, water, air, and CO2) as a function of tempera-ture. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. Figure reproduced from fgure 5.6 p. 99; CC-by by Arjun Neyyathala & Olivier Cleynen 10.1 Water and air fow White E7.2 A fat plate of length 0,3 m is placed parallel to a uniform fow with speed 0,3 m s−1. How thick can the boundary layer become: 10.1.1. if the fuid is air at 1 bar and 20 °C? 10.1.2. if the fuid is water at 20 °C? 10.2 Boundary layer sketches CC-0 Olivier Cleynen A thin and long horizontal plate is moved horizontally through a stationary fuid. 10.2.1. Sketch the velocity profle of the fuid: • at the leading edge; • at a point where the boundary layer is laminar; 214 • and at a point further downstream where the boundary layer is turbulent. 10.2.2. Draw a few streamlines, indicate the boundary layer thickness 훿, and the displacement thickness 훿∗. 10.2.3. Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime can be triggered. 10.2.4. Explain shortly (e.g. in 30 words or less) how the transition to turbulent regime could instead be delayed. 10.3 Shear force due to boundary layer White E7.3 A thin and smooth plate of dimensions 0,5 × 3 m is placed with a zero angle of attack in a fow incoming at 1,25 m s−1, as shown in fg. 10.12. Figure 10.12: A thin plate positioned parallel to an incoming uniform fow. Two confgurations are studied in this exercise. Figure CC-0 Olivier Cleynen 10.3.1. What is the shear force exerted on the top surface of the plate for each of the two confgurations shown in fg. 10.12, when the fuid is air of viscosity 휇air = 1,5 ⋅10−5 Pa s? 10.3.2. What are the shear forces when the fuid is water of viscosity 휇water = 1 ⋅10−3 Pa s? 10.3.3. [difcult question] How would these shear eforts evolve if the plate was tilted with an angle of 20° relative to the fow? 10.4 Wright Flyer I CC-0 Olivier Cleynen The Wright Flyer I, the frst airplane capable of sustained controlled fight (1903), was a biplane with a 12 m wingspan (fg. 10.13). It had two wings of chord length 1,98 m stacked one on top of the other. The wing profle was extremely thin and it could only fy at very low angles of attack. Its fight speed was approximately 40 km h−1. 10.4.1. If the fow over the wings can be treated as if they were fat plates, what is the power necessary to compensate the shear exerted by the airfow on the wings during fight? 10.4.2. Which other forms of drag would also be found on the aircraft? (give a brief answer, e.g. in 30 words or less) 215 Figure 10.13: The Wright Flyer I, frst modern airplane in history. Built with meticulous care and impeccable engineering methodology by two bicycle makers, it made history in December 1903. Photo by Orville Wright, 1908 (public domain) 10.5 Power lost to shear on an airliner fuselageCC-0 Olivier Cleynen An Airbus A340-600 (fg. 10.14) is cruising at [Ma] = 0,82 at an altitude of 10 000 m (where the air has viscosity 1,457 ⋅10−5 N s m−2, temperature 220 K, density 0,4 kg m−3). Figure 10.14: The Airbus A340-600, a large airliner frst fown in 2001. Photo CC-by-sa by Iberia Airlines (retouched) The cylindrical part of the fuselage has diameter 5,6 m and length 65 m. 10.5.1. What is approximately the maximum boundary layer thickness around the fuselage? 10.5.2. What is approximately the average shear applying on the fuselage skin? 10.5.3. Estimate the power dissipated to friction on the cylindrical part of the fuselage. 10.5.4. In practice, in which circumstances could fow separation occur on the fuselage skin? (give a brief answer, e.g. in 30 words or less) 10.6 Laminar wing profle non-examinable. Based on a diagram from Bertin et al. 2010 The characteristics of a so-called “laminar” wing profle are compared in fgs. 10.15 to 10.17 with those of an ordinary profle. On the graph representing the pressure coefcient 퐶푝≡푝−푝∞ 1 2휌푉2, identify the curve corre-sponding to each profle. What advantages and disadvantages does the laminar wing profle have, and how can they be explained? In which applications will it be most useful? 216 Figure 10.15: Comparison of the thickness distribution of two uncambered wing profles: an ordinary medium-speed naca 0009 profle, and a “laminar” naca 66-009 profle. Figure © Bertin & Cummings 2010 Figure 10.16: Static pressure distribution (represented as a the local non-dimensional pressure coefcient 퐶푝≡푝−푝∞ 1 2휌푉2 ) as a function of distance 푥(non-dimensionalized with the chord 푐) over the surface of the two airfoils shown in fg. 10.15. Figure © Bertin & Cummings 2010 217 Figure 10.17: Values of the section drag coefcient 퐶푑≡ 푑 1 2푐휌푉2 as a function of the section lift coefcient 퐶푙≡ 푙 1 2푐휌푉2 for both airfoils presented in fg. 10.15. Figure © Bertin & Cummings 2010 , based on data by Abott & Von Doenhof 1949 10.7 Separation mechanism non-examinable, CC-0 Olivier Cleynen Sketch the velocity profle of a laminar or turbulent boundary layer shortly upstream of, and at a separation point. The two equations below describe fow in laminar boundary layer: 푢휕푢 휕푥+ 푣휕푢 휕푦= 푈d푈 d푥+ 휇 휌 휕2푢 (휕푦)2 (10/12) 휕푢 휕푥+ 휕푣 휕푦= 0 (10/13) Identify these two equations, list the conditions in which they apply, and explain shortly (e.g. in 30 words or less) why a boundary layer cannot separate when a favorable pressure gradient is applied along the wall. 218 Answers 10.1 1) At trailing edge [Re]푥= 5 348 thus the layer is laminar everywhere. 훿will grow from 0 to 2,01 cm (eq. 10/15 p. 205); 2) For water: 훿trailing edge = 4,91 mm. 10.2 1) See fg. 10.6 p. 206. At the leading-edge the velocity is uniform. Note that the 푦-direction is greatly exaggerated, and that the outer velocity 푈is identical for both regimes; 2) See fg. 10.4 p. 202. Note that streamlines penetrate the boundary layer; 3) and 4) See §10.4 p. 205. 10.3 푥transition, air = 4,898 m and 푥transition, water = 0,4 m. In a laminar boundary layer, insert-ing equation 10/18 into equation 10/6 into equation 10/5 yields 퐹휏= 0,664퐿푈1,5√휌휇[ √푥] 푥transition 0 . In a turbulent boundary layer, we use equation 10/24 instead and get 퐹휏= 0,01575퐿휌 6 7푈 13 7 휇 1 7 [푥 6 7] 푥trailing edge 푥transition . These expressions allow the calculation of the forces below, for the top surface of the plate: 1) (air) First case: 퐹= 3,445 ⋅10−3 N; second case 퐹= 8,438 ⋅10−3 N (who would have thought eh?); 2) (water) First case: 퐹= 3,7849 N; second case 퐹= 2,7156 N. 10.4 Using the expressions developed in exercise 10.3, ̇ 푊friction ≈255 W. 10.5 1) 푥transition = 7,47 cm (the laminar part is negligible). With the equations developed in exercise 7.3, we get 퐹= 24,979 kN and ̇ 푊= 6,09 MW. Quite a jump from the Wright Flyer I! 2) When the longitudinal pressure gradient is zero, the boundary layer cannot separate. Thus separation from the fuselage skin can only happen if the fuselage is fown at an angle relative to the fight direction (e.g. during a low-speed maneuver). 219 220 Fluid Dynamics Chapter 11 – Large- and small-scale fows last edited June 26, 2019 by Olivier Cleynen — 11.1 Motivation 221 11.2 Flow at large scales 221 11.2.1 Problem statement 221 11.2.2 Investigation of inviscid fows 222 11.3 Plotting velocity with functions 223 11.3.1 Kinematics that ft conservation laws 223 11.3.2 Strengths and weaknesses of potential fow 224 11.3.3 Superposition: the lifting cylinder 225 11.3.4 Circulating cylinder 227 11.3.5 Modeling lift with circulation 229 11.4 Flow at very small scales 232 11.5 Problems 235 11.1 Motivation This exploratory chapter is not a critical component of fuid dynamics; in-stead, it is meant as a brief overview of two extreme cases: fows for which viscous efects are negligible, and fows for which they are dominant. This exploration should allow us to answer two questions: • How can we model large-scale fows? • How can we model small-scale fows? 11.2 Flow at large scales 11.2.1 Problem statement In this section, we are interested in fow at very large scales: those for which a representative length 퐿is very large. In particular, when 퐿is large enough, the infuence of viscosity is reduced. Formally, this corresponds to the case where the Reynolds number [Re] ≡휌푉퐿/휇is very large. To examine the mechanics of such a fow, we turn to our beloved non-dimensional Navier-Stokes equation for incompressible fow derived as eq. 8/13 p. 164, [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (11/1) We saw in chapter 8 (Engineering models) that we could compare the relative weight of terms: when the Reynolds number [Re] is very large, the last term 221 becomes negligible relative to the other four. Thus, our governing equation can be reduced as follows: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗≈ 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗ (11/2) Now, converting eq. 11/2 back to dimensional terms, the governing momen-tum equation for large-scale fow becomes: 휌D ⃗ 푉 D푡= 휌⃗ 푔−⃗ ∇푝 (11/3) We see that with the starting proposition that [Re] was large, we have re-moved altogether the viscous (last) term from the Navier-Stokes equation. Flows governed by this equation are called inviscid fows. Equation 11/3 is named the Euler equation; it stipulates that the acceleration feld is driven only by gravity and by the pressure feld. 11.2.2 Investigation of inviscid fows From what we studied in chapter 8 (Engineering models), we recognize immedi-ately that fows governed by eq. 11/3 are troublesome: the absence of viscous efects facilitates the occurrence of turbulence and makes for much more chaotic behaviors. Although the removal of shear from the Navier-Stokes equation simplifes the governing equation, the solutions to this new equation become even harder to fnd and describe. What can be done in the other two branches of fuid mechanics? • Large-scale fows are difcult to investigate experimentally. As we have seen in chapter 8 (Engineering models), scaling down a fow (e.g. so it may ft inside a laboratory) while maintaining constant [Re] requires increasing velocity by a corresponding factor. • Large-scale fows are also difcult to investigate numerically. At high [Re], the occurrence of turbulence makes for either an exponential increase in computing power (we saw in chapter 9 that direct numerical simulation computing power increases with [Re]3,5), or for increased reliance on hard-to-calibrate turbulence models (in Reynolds-averaged simulations). All three branches of fuid mechanics, therefore, struggle with large-scale fows, because of turbulence. In spite of this, large-scale fows are undeniably important. In chapter 10 we were able to understand and describe fuid fow very close to walls. Now we wish to be able to to the same for large structures, for example, in order to describe the broad patterns of fuid fow (in particular, pressure distribution) in the wake of an aircraft, around a wind turbine, or within a hurricane. It is clear that we have no hope of accounting easily for turbulence, but we can at least describe the main features of such fows by restricting ourselves to laminar cases. In the following sections, we will model such laminar solutions directly, based on intuition and observation, and make sure that they match the condition described by eq. 11/3 above. 222 11.3 Plotting velocity with functions 11.3.1 Kinematics that ft conservation laws For simple fow structures, the velocity feld can be simply described based on observation and intuition. This is done for example in exercise 11.6, where we reconstruct the fow feld within and around a tornado using very simple, almost primitive, kinematics. Without so much as a small increase in complexity, this approach becomes untenable. In the last exercises of chapter 6 (ex. 6.8 & 6.9 p. 133) we have seen that it is easy to propose a velocity feld that does not satisfy (is not a solution of) either the mass balance equation or the momentum balance equation. For example, if one considers two of the tornado fows mentioned above together, the velocity feld cannot be described easily anymore. One approach has been developed in the 17th century to overcome this problem. It consists in fnding a family of fows, all steady, that always satisfy the conservation equations. Those fows can then be added to one another to produce new fows which satisfy the balance equations. Such fows are called potential fows. Two conditions need to be fulflled for this approach to work: 1. The velocity feld must always be describable with a function; that is, there must correspond a single value of 푢, of 푣and of 푤at each of the coordinates 푥푖, 푦푖, or 푧푖in space. This means in practice that we cannot account for fows which “curl up” on themselves, occasionally recirculating back on their path. 2. The velocity feld must conserve mass. In incompressible fow, this is achieved if the continuity equation ⃗ ∇⋅⃗ 푉= 0 (eq. 6/35 p. 123) is respected. With potential fow, these two conditions are addressed as follows: 1. We restrict ourselves to irrotational fows, those in which the curl of velocity (see Appendix A3 p. 250) is always null: ⃗ ∇× ⃗ 푉= ⃗ 0 (11/4) by defnition, for an irrotational fow. It can be shown that fows are irrotational when there exists a scalar function 휙(pronounced “phi” and named potential function) of which the gradient is the velocity vector feld: ⃗ ∇휙≡⃗ 푉 (11/5) In the case of two-dimensional fow, this translates as: 휕휙 휕푥≡푢 (11/6) 휕휙 휕푦≡푣 (11/7) When plotted out, lines of constant 휙(named equipotential lines) are always perpendicular to the streamlines of the fow. 223 2. The continuity equation is satisfed by referring to stream functions. It can be shown that the divergent of velocity is null when there exists a vector feld function ⃗ 휓(pronounced “psi” and named stream function) of which the curl is the velocity vector feld: ⃗ ∇× ⃗ 휓= ⃗ 푉 (11/8) In the case of two-dimensional fow, 휓is a scalar feld and eq. 11/8 translates as: 휕휓 휕푦≡푢 (11/9) −휕휓 휕푥≡푣 (11/10) When plotted out, lines of of constant 휓value are streamlines – in other words, as they travel along, fuid particles follow paths of constant 휓 value. In summary, we have shifted the problem from looking for 푢and 푣, to looking for 휓and 휙. The existence of such functions ensures that fows can be added and subtracted from one another yet will always result in mass-conserving, mathematically-describable fows. If such two functions are known, then the velocity components can be obtained (recovered) easily either in Cartesian coordinates, 푢= 휕휙 휕푥= 휕휓 휕푦 (11/11) 푣= 휕휙 휕푦= −휕휓 휕푥 (11/12) or angular coordinates: 푣푟= 휕휙 휕푟= 1 푟 휕휓 휕휃 (11/13) 푣휃= 1 푟 휕휙 휕휃= −휕휓 휕푟 (11/14) 11.3.2 Strengths and weaknesses of potential fow The potential fow methodology allows us to fnd solutions to the Euler equation: fows in which the Reynolds number is high enough that viscosity has no signifcant role anymore. Potential fows are the simplest solutions that we are able to come up with. They are: • strictly steady; • inviscid; • incompressible; • devoid of energy transfers; • two-dimensional (in the scope of this course at least). This is quite convenient for the academician, who recognizes immediately four of the fve criteria which we set forth in chapter 2 (Analysis of existing 224 fows with one dimension) for using the Bernoulli equation. Along a streamline, the ffth condition is met, and Euler’s equation reduces to eq. 2/20 (p. 42), reproduced here: 푝1 휌+ 1 2푉2 1 + 푔푧1 = 푝2 휌+ 1 2푉2 2 + 푔푧2 = cst. (11/15) along a streamline in a steady incompressible inviscid fow. and so it follows that if the solution to a potential fow is known, the pressure is known everywhere, and the forces due to pressure can be calculated with relative ease. Nevertheless, from a science and engineering point of view, potential fows have only limited value, because they are entirely unable to account for turbulence, which we have seen is an integral feature of high-[Re] fows. We should therefore use them only with great caution. Potential fows help us model large-scale structures with very little computational cost, but this comes with strong limitations. 11.3.3 Superposition: the lifting cylinder It is possible to describe a handful of basic potential fows called elementary fows as fundamental ingredients that can be added to one another to create more complex and interesting fows. Without going into much detail, the most relevant elementary fows are: • Uniform longitudinal fow, 휙= 푉푟cos 휃 (11/16) 휓= 푉푟sin 휃 (11/17) • Sources and sinks (fg. 11.1) which are associated with the appearance of a (positive or negative) volume fow rate ̇ from a single point in the fow: 휙= ̇  2휋ln 푟 (11/18) 휓= ̇  2휋휃 (11/19) Figure 11.1: Concept of a source inside a two-dimensional potential fow. A sink would display exactly opposed velocities. Figure CC-by-sa Commons User:Nicoguaro 225 • Irrotational vortices, rotational patterns which impart a rotational velocity 푣휃= 푓(푟, 휃) on the fow, in addition to which there may exist a radial component 푣푟: 휙= Γ 2휋휃 (11/20) 휓= −Γ 2휋ln 푟 (11/21) in which Γ (termed circulation) is a constant proportional to the strength of the vortex. • Doublets, which consist in a source and a sink of equal volume fow rate positioned extremely close one to another (fg. 11.2): 휙= 퐾cos 휃 푟 (11/22) 휓= −퐾sin 휃 푟 (11/23) in which 퐾is a constant proportional to the source/sink volume fow rate ̇ . It was found in the 17th Century that combining a doublet with uniform fow resulted in fow patterns that imiated “perfect” fow around a cylinder: a fow where the fuid fows smoothly and steadily everywhere (fg. 11.3). The stream function of that fow is: 휓= 푈∞sin 휃(푟−푅2 푟) (11/24) This stream function allows us to describe the velocity everywhere: 푣푟= 1 푟 휕휓 휕휃= 푈∞cos 휃(1 −푅2 푟2 ) (11/25) 푣휃= −휕휓 휕푟= −푈∞sin 휃(1 + 푅2 푟2 ) (11/26) We can even calculate the lift and drag applying on the cylinder surface. Indeed, along the cylinder wall, 푟= 푅and 푣푟\|푟=푅= 0 (11/27) 푣휃\|푟=푅= −2푈∞sin 휃 (11/28) Since the Bernoulli equation can be applied along any streamline in this (steady, constant-energy, inviscid, incompressible) fow, we can express the Figure 11.2: Left: a simple uniform steady fow; Right: a doublet, the result of a source and a sink brought very close one to another Figure CC-0 Olivier Cleynen 226 Figure 11.3: The addition of a doublet and a uniform fow produces streamlines for an (idealized) fow around a cylinder. Figure CC-by-sa by Commons User:Kraaiennest pressure 푝s on the cylinder surface as a function of 휃: 푝∞+ 1 2휌푈2 ∞= 푝s + 1 2휌푣2 휃 푝s (휃) = 푝∞+ 1 2휌(푈2 ∞−푣2 휃) (11/29) Now, a relatively simple integration gives us the net forces exerted by the fuid on the cylinder per unit width 퐿, in each of the two directions 푥and 푦: 퐹net,푥 퐿 = −∫ 2휋 0 푝s cos 휃푅d휃= 0 (11/30) 퐹net,푦 퐿 = −∫ 2휋 0 푝s sin 휃푅d휃= 0 (11/31) The results are interesting, and at the time they were obtained by their author, Jean le Rond D’Alembert, were devastating: both lift and drag are zero. This inability to reproduce the well-known phenomena of drag is often called the d’Alembert paradox. To fnd out why the solution is not realistic, we can plot the resulting surce pressure distribution graphically, and compared to experimental measure-ments: this is done in fg. 11.4. Good agreement is obtained on the leading edge of the cylinder; but as the pressure gradient becomes unfavorable, in practice the boundary layer separates –a phenomenon that cannot be de-scribed with inviscid fow— and a low-pressure area forms on the downstream side of the cylinder. 227 Figure 11.4: Pressure distribution (relative to the far-fow pressure) on the surface of a cylinder, with fow from left to right. On the left is the potential fow case, purely symmetrical. On the right (in blue) is a measurement made at a high Reynolds number. Boundary layer separation occurs on the second half of the cylinder, which prevents the recovery of leading-edge pressure values, and increases drag. Figure CC-by-sa Commons User:BoH & Olivier Cleynen 11.3.4 Circulating cylinder Video: watch the Brazilian foot-ball team show their French counterparts how circulation (in-duced through friction by ball ro-tation) is associated to dynamic lift on a circular body by TF1, 1997 (styl) An extremely interesting “hack” can be implemented with the potential cylinder fow above if an irrotational vortex of stream function 휓= −Γ 2휋ln 푟 (eq. 11/21) is added to it. The overall fow feld becomes: 휓= 푈∞sin 휃(푟−푅2 푟) −Γ 2휋ln 푟 (11/32) With this function, several key characteristics of the fow feld can be obtained. The frst is the velocity feld at the cylinder surface: 푣푟\|푟=푅= 0 (11/33) 푣휃\|푟=푅= −2푈∞sin 휃+ Γ 2휋푅 (11/34) and we immediately notice that the velocity distribution is no longer sym-metrical with respect to the horizontal axis (fg. 11.5): the fuid is defected, and so there will be a net force on the cylinder. This time, the net pressure forces on the cylinder have changed: 퐹net,푥 퐿 = −∫ 2휋 0 푝s cos 휃푅d휃= 0 (11/35) 퐹net,푦 퐿 = −∫ 2휋 0 푝s sin 휃푅d휃= −휌푈∞Γ (11/36) We thus fnd out that the drag is once again zero —as for any potential fow— but that lift occurs which is proportional to the free-stream velocity 푈 and to the circulation Γ. In practice, such a fow can be generated by spinning a cylindrical object in a uniform fow. A lateral force is then obtained, which can be used as a propulsive or sustaining force. Several boats and even an aircraft have been used in practice to demonstrate this principle. Naturally, fow separation from the cylinder profle and the high shear eforts generated on the surface 228 Figure 11.5: The addition of an irrotational vortex on top of the cylinder fow described in fg. 11.3 distorts the fow feld and it becomes asymmetrical: a lift force is developed, which depends directly on the circulation Γ. Figure © White 2008 cause real fows to difer from the ideal case described here, and it turns out that rotating cylinders are a horribly uneconomical and unpractical way of generating lift. 11.3.5 Modeling lift with circulation Video: acting on swimming pool water with a round plate sheds a half-circular vortex that is ex-tremely stable and can be inter-acted with quite easily. Such sta-ble laminar structures are excel-lent candidates for analysis us-ing potential fow. by Physics Girl (Dianna Cowern) (styl) Fluid fow around cylinders may have little appeal for the modern student of fuid mechanics, but the methodology above has been taken much further. With further mathematical manipulation called conformal mapping, potential fow can be used to described fow around geometrical shapes such as airfoils (fg. 11.6). Because the fow around such streamlined bodies usually does not feature boundary layer separation, the predicted fow felds everywhere except in the close vicinity of the solid surface are accurately predicted. There again, it is observed that regardless of the constructed geometry, no lift can be modeled unless circulation is also added within the fow. The amount of circulation needed so that results may correspond to experimental obser-vations is found by increasing it progressively until the the rear stagnation point reaches the rear trailing edge of the airfoil, a condition known as the Kutta condition. Regardless of the amount of circulation added, potential fow remains entirely reversible, both in a kinematic and a thermodynamic sense, thus, care must be taken in the problem setup to make sure the model is realistic (fg. 11.7). It is then observed in general that any dynamic lift generation can be modeled as the superposition of a free-stream fow and a circulation efect (fg. 11.8). With such a tool, potential fow becomes an extremely useful tool, mathemat-ically and computationally inexpensive, in order to model and understand the cause and efect of dynamic lift in fuid mechanics. In particular, it has been paramount in the description of aerodynamic lift distribution over aircraft wing surfaces (fg. 11.9), with a concept called the Lifting-line theory. 229 Figure 11.6: Potential fow around an airfoil without (top) and with (bottom) circu-lation. Much like potential fow around a cylinder, potential fow around an airfoil can only result in a net vertical force if an irrotational vortex (with circulation Γ) is added on top of the fow. Only one value for Γ will generate a realistic fow, with the rear stagnation point coinciding with the trailing edge, a occurrence named Kutta condition. Figure CC-0 Olivier Cleynen Figure 11.7: Potential fow allows all velocities to be inverted without any change in the fow geometry. Here the fow around an airfoil is reversed, displaying unphysical behavior. Figure CC-0 Olivier Cleynen Figure 11.8: A numerical model of fow around an airfoil. In the left fgure, the velocity vectors are represented relative to a stationary background. In the right fgure, the velocity of the free-stream fow has been subtracted from each vector, bringing the circulation phenomenon into evidence. Figures 1 & 2 CC-by-sa by en:Wikipedia User:Crowsnest 230 Figure 11.9: From top to bottom, the lift distribution over the wings of a glider is modeled with increasingly complex (and accurate) lift and circulation distributions along the span. The Lifting-line theory is a method associating each element of lift with a certain amount of circulation. The efect of each span-wise change of circulation is then mapped onto the fow feld as a trailing vortex. Figures 1, 2 & 3 CC-by-sa Olivier Cleynen 231 11.4 Flow at very small scales At the complete opposite of the spectrum, we fnd fow at very small scales: fows around bacteria, dust particles, and inside very small ducts. In those fows the representative length 퐿is extremely small, which makes for small values of the Reynolds number. Such fows are termed creeping or Stokes fows. What are their main characteristics? Looking back once again at the non-dimensional Navier-Stokes equation for incompressible fow derived as eq. 8/13 p. 164, [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (11/37) we see that creeping fow will occur when the Reynolds number is much smaller than 1. The relative weight of the term (1/[Re]) ⃗ ∇∗2 ⃗ 푉∗then becomes overwhelming. In addition to cases where [Re] ≪1, we focus our interest on fows for which: • gravitational efects have negligible infuence over the velocity feld; • the characteristic frequency is extremely low (quasi-steady fow). With these characteristics, the terms associated with the [St] (Strouhal) and [Fr] (Froude) numbers become very small with respect to the other terms, and our non-dimensionalized Navier-Stokes equation (eq. 11/37) is approximately reduced to: ⃗ 0 ≈−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (11/38) We can now come back to dimensionalized equations, concluding that for a fuid fow dominated by viscosity, the pressure and velocity felds are linked together by the approximate relation: ⃗ ∇푝= 휇⃗ ∇2 ⃗ 푉 (11/39) In this type of fow, the pressure feld is entirely dictated by the Laplacian of velocity, and the fuid density has no importance. Micro-organisms, for which the representative length 퐿is very small, spend their lives in such fows (fg. 11.10). At the human scale, we can visualize the efects of these fows by moving an object slowly in highly-viscous fuids (e.g. a spoon in honey), or by swimming in a pool flled with plastic balls. The inertial efects are almost inexistent, drag is extremely important, and the object geometry has comparatively small infuence. In 1851, George Gabriel Stokes worked through equation 11/39 for fow around a sphere, and obtained an analytical solution for the fow feld. This allowed him to show that the drag 퐹D sphere applying on a sphere of diameter 퐷in creeping fow (fg. 11.11) is: 퐹D sphere = 3휋휇푈∞퐷 (11/40) 232 Figure 11.10: Micro-organisms carry themselves through fuids at extremely low Reynolds numbers, since their scale 퐿is very small. For them, viscosity efects dominate inertial efects. Photo by Yutaka Tsutsumi, M.D., Fujita Health University School of Medicine Inserting this equation 11/40 into the defnition of the drag coefcient 퐶퐹D ≡ 퐹D/1 2휌푆frontal푈2 ∞(from eq. 8/15 p. 169) then yields: 퐶퐹D = 퐹D sphere 1 2휌푈2 ∞ 휋 4 퐷2 = 24휇 휌푈∞퐷= 24 [Re]퐷 (11/41) These equations are specifc to fow around spheres, but the trends they describe apply well to most bodies evolving in highly-viscous fows, such as dust or liquid particles traveling through the atmosphere. Drag is only proportional to the speed (as opposed to low-viscosity fows in which it grows with velocity squared), and it does not depend on fuid density. Figure 11.11: Flow at very low Reynolds numbers around a sphere. In this regime, the drag force is proportional to the velocity. Figure CC-by-sa by Olivier Cleynen & Commons User:Kraaiennest 233 234 Problem sheet 11: Large- and small-scale fows last edited June 26, 2019 by Olivier Cleynen — These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 °C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 In a highly-viscous (creeping) steady fow, the drag 퐹D exerted on a spherical body of diameter 퐷at by fow at velocity 푈∞is quantifed as: 퐹D sphere = 3휋휇푈∞퐷 (11/40) 11.1 Volcanic ash from the Eyjafjallajökull Çengel & al. E10.2 In 2010, a volcano with a complicated name and unpredictable mood decided to ground the entire European airline industry for fve days. We consider a microscopic ash particle released at very high altitude (−50 °C, 0,55 bar, 1,474 ⋅10−5 N s m−2). We model it as a sphere with 50 µm diameter. The density of volcanic ash is 1 240 kg m−3. 11.1.1. What is the terminal velocity of the particle? 11.1.2. Will this terminal velocity increase or decrease as the particle progresses towards the ground? (briefy justify your answer, e.g. in 30 words or less) 11.2 Water drop Çengel & al. 10-21 A rainy day provides yet another opportunity for exploring fuid dynamics (fg. 11.12). A water drop with diameter 42,4 µm is falling through air at 25 °C and 1 bar. 11.2.1. Which terminal velocity will it reach? 11.2.2. Which velocity will it reach once its diameter will have doubled? 235 Figure 11.12: A sketched diagram showing the geometry of water drops of various sizes in free fall. When their diameter is lower than 2 mm, water drops are approximately spherical (B). As they grow beyond this size, their shape changes and they eventually break-up (C-E). They never display the “classical” shape displayed in A, which is caused only by surface tension efects when they drip from solid surfaces. Figure CC-by-sa by Ryan Wilson 11.3 Idealized fow over a hangar roof based on White P8.54 Certain fows in which both compressibility and viscosity efects are negligible can be described using the potential fow assumption (the hypothesis that the fow is everywhere irrotational). If we compute the two-dimensional laminar steady fuid fow around a cylinder profle, we obtain the velocities in polar coordinates as: 푣푟= 푉∞cos 휃(1 −푅2 푟2 ) (11/25) 푣휃= −푉∞sin 휃(1 + 푅2 푟2 ) (11/26) where the origin (푟= 0) is at the center of the cylinder profle; 휃 is measured relative to the free-stream velocity vector; 푉∞is the incoming free-stream velocity; and 푅 is the (fxed) cylinder radius. In this exercise, we study the air fow over a hangar roof with this model. We use the equations above to describe the air velocity everywhere, pretending the as the wind blows about a large semi-cylindrical solid structure — an idealized description of an otherwise complex fow. Wind with a nearly-uniform velocity 푈∞= 100 km h−1 is blowing across a 50 m-long hangar with a semi-cylindrical geometry, as shown in fg. 11.13. The radius of the hangar is 푅= 20 m. 236 Figure 11.13: A semi-cylindrical hangar roof. Wind with uniform velocity 푈fows perpendicular to the cylinder axis. Figure CC-0 Olivier Cleynen 11.3.1. Starting from eqs. 11/25 and 11/26, show that the pressure 푝푠on the surface on the roof is distributed as: 푝푠= 푝∞+ 1 2휌(푉2 ∞−4푉2 ∞sin2 휃) (11/42) 11.3.2. The pressure inside the hangar is set to 푝∞. What is the total lift force on the hangar? (see also problem 4.6 p. 87) (a couple of hints to help with the algebra: ∫sin 푥d푥= −cos 푥+푘and ∫sin3 푥d푥= 1 3 cos3 푥−cos 푥+ 푘). 11.3.3. At which position on the roof is the 푝푠= 푝∞? 11.3.4. Describe briefy (e.g. in 30 words or less) two reasons why the results above would not correspond to reality. 11.4 Cabling of the Wright Flyer derived from Munson & al. 9.106 The Wright Flyer I, the frst powered and controlled aircraft in history, was subjected to multiple types of drag. We have already studied viscous friction on its thin wings in exercise 7.4. The data in fgure 11.14 provides the opportunity to quantify drag due to pressure. A network of metal cables with diameter 1,27 mm criss-crossed the aircraft in order to provide structural rigidity. The cables were positioned perpendicularly to the air fow, which came at 40 km h−1. The total cable length was approximately 60 m. What was the drag generated by the cables? 237 Figure 11.14: Experimental measurements of the drag coefcient applying to a cylinder and to a sphere as a function of the diameter-based Reynolds number [Re]퐷, shown together with schematic depictions of the fow around the cylinder. By convention, the drag coefcient 퐶D ≡퐶퐹D ≡ 퐹D 1 2휌푆푈2 ∞ (eq. 8/15 p. 169) compares the drag force 퐹D with the frontal area 푆. Both fgures © from Munson & al. 238 11.5 Ping pong ball Munson & al. E9.16 A series of experiments is conducted in a wind tunnel on a large cast iron ball with a smooth surface; the results are shown in fg. 11.15. These measurement data are used to predict the behavior of a ping pong ball. Table tennis regulations constrain the mass of the ball to 2,7 g and its diameter to 40 mm. 11.5.1. Is it possible for a ball thrown at a speed of 50 km h−1 to have a perfectly horizontal trajectory? 11.5.2. If so, what would be its deceleration? 11.5.3. How would the drag and lift applying on the ball evolve if the air viscosity was progressively decreased to zero? Figure 11.15: Experimental measurements of the lift and drag coefcients applying on a rotating sphere in an steady uniform fow. Figure © from Munson & al. 239 11.6 Flow feld of a tornado Çengel & al. E9-5, E9-14 & E10-3 In this problem, we attempt to model a very large-scale fow: that of a tornado (fg. 11.16). We begin by pretending the tornado is one perfectly straight, stationary structure. We divide the fow into two regions: a core cylinder that rotates almost like a solid body, and an outer region where fow spins in an irrotational matter. This model is called the Rankine vortex (displayed in fg. 11.17) and is used widely as a simple, frst approximation to model fows as large as a hurricane and as small as turbulence-induced vortices. Figure 11.16: Photo of an approaching tornado in Manitoba, Canada Photo CC-by-sa by Commons User:Grhu Figure 11.17: Modeled angular velocity in a vortex, according to the Rankine vortex model Figure CC-by-sa by en:Wikipedia User:Justin1569 We are frst interested in the outer region of the tornado fow feld. We model the fow as being steady, two-dimensional (neglecting any movement in the vertical, 푧-direction), and having a rotational velocity 푣휃such that: 푣휃= Γ 2휋푟 (11/43) in which Γ is the circulation (measured in s−1) and remains constant and uniform. 11.6.1. The mass balance equation for incompressible fow (eq. 6/35 p. 123) is developed in cylindrical coordinates as follows: 1 푟 휕푟푣푟 휕푟 + 1 푟 휕푣휃 휕휃+ 휕푣푧 휕푧= 0 (11/44) According to this mass balance equation, what form must the radial velocity 푣푟 take? 240 Among all the possibilities for 푣푟, we choose the simplest form, so that from now on, we model radial velocity as: 푣푟= 0 (11/45) 11.6.2. The momentum balance equation for incompressible fow (eq. 6/42 p. 125) is developed in cylindrical coordinates are as follows: 휌[ 휕푣푟 휕푡+ 푣푟 휕푣푟 휕푟+ 푣휃 푟 휕푣푟 휕휃−푣2 휃 푟+ 푣푧 휕푣푟 휕푧] = 휌푔푟−휕푝 휕푟+ 휇[ 1 푟 휕 휕푟(푟휕푣푟 휕푟) −푣푟 푟2 + 1 푟2 휕2푣푟 (휕휃)2 −2 푟2 휕푣휃 휕휃+ 휕2푣푟 (휕푧)2] (11/46) 휌[ 휕푣휃 휕푡+ 푣푟 휕푣휃 휕푟+ 푣휃 푟 휕푣휃 휕휃+ 푣푟푣휃 푟 + 푣푧 휕푣휃 휕푧] = 휌푔휃−1 푟 휕푝 휕휃+ 휇[ 1 푟 휕 휕푟(푟휕푣휃 휕푟) −푣휃 푟2 + 1 푟2 휕2푣휃 (휕휃)2 + 2 푟2 휕푣푟 휕휃+ 휕2푣휃 (휕푧)2] (11/47) 휌[ 휕푣푧 휕푡+ 푣푟 휕푣푧 휕푟+ 푣휃 푟 휕푣푟 휕휃+ 푣푧 휕푣푧 휕푧] = 휌푔푧−휕푝 휕푧+ 휇[ 1 푟 휕 휕푟(푟휕푣푧 휕푟) + 1 푟2 휕2푣푧 (휕휃)2 + 휕2푣푧 (휕푧)2] (11/48) Starting from those equations, show that the pressure distribution in the outer region of the tornado can be expressed as: 푝= 푝∞−1 2휌Γ2 1 푟2 (11/49) where 푝∞is the atmospheric pressure far away from the tornado. We now turn to the core of the tornado, which we model as if it were a rotating solid (a vortex core). 11.6.3. What is the radial velocity 푣휃distribution? 11.6.4. What is the pressure feld within the rotational core of the tornado? (hint: you may start directly from an energy balance equation, eq. 2/18 p. 40, without having to use the Navier-Stokes equations above). 11.6.5. Make a simple, qualitative sketch (i.e. without numerical data) of the pressure as a function of radius throughout the entire tornado fow feld. It is fnally time to calibrate and exploit our model. We estimate the tornado diameter to be 50 m and the maximum wind velocity to be 180 km h−1. 11.6.6. According to the model, what is the lowest pressure attained by the air? 11.6.7. According to the model, at what distance from the core are winds lower than 50 km h−1? (curious students may play with the above model by adding a non-zero radial velocity, and look up the phenomenon of vortex stretching) 241 11.7 Lift on a symmetrical object non-examinable Briefy explain (e.g. with answers 30 words or less) how lift can be generated on a sphere or a cylinder, • with diferential control boundary layer control; • with the efect of rotation. Draw a few streamlines in a two-dimensional sketch of the phenomenon. 11.8 Air fow over a wing profle From Munson & al. 9.109 The characteristics of a thin, fat-bottomed airfoil are examined by a group of students in a wind tunnel. The frst investigations focus on the boundary layer, and the research group evaluate the boundary layer thickness and make sure that it is fully attached. Once this is done, the group proceeds with speed measurements all around the airfoil. Measurements of the longitudinal speed 푢just above the boundary layer on the top surface are tabulated below: 푥/푐(%) 푦/푐(%) 푢/푈 0 0 0 2,5 3,72 0,971 5 5,3 1,232 7,5 6,48 1,273 10 7,43 1,271 20 9,92 1,276 30 11,14 1,295 40 10,49 1,307 50 10,45 1,308 60 9,11 1,195 70 6,46 1,065 80 3,62 0,945 90 1,26 0,856 100 0 0,807 On the bottom surface, the speed is measured as being constant (푢= 푈) to within experimental error. What is the lift coefcient of the airfoil? 242 Answers 11.1 1) At terminal velocity, the weight of the sphere equals the drag. This allows us to obtain 푈= 푔휌sphere 퐷2 18휇= 0,1146 m s−1: unbearably slow when you are stuck in an airport! With 푈, check that the Reynolds number indeed corresponds to creeping fow: [Re]퐷= 0,334. 11.2 Same as previous exercise: 푈1 = 4,578 ⋅10−2 m s−1 and 푈2 = 0,183 m s−1, with Reynolds numbers of 0,113 and 0,906 respectively (thus creeping fow hypothesis valid). 11.3 1) Integrate the vertical component of force due to pressure: 퐹L roof = 1,575 MN. 11.4 A simple reading of fg. 11.14 gives 퐹D = 6,9 N, ̇ 푊= 76 W. 11.5 Yes — a reading of fg. 11.13 gives 휔= 83 rev/s. 243 244 Appendix last edited February 12, 2021 A1 Notation 246 A2 Vector operations 247 A2.1 Vector dot product 247 A2.2 Vector cross product 247 A3 Field operators 250 A3.1 Gradient 250 A3.2 Divergent 250 A3.3 Advective 251 A3.4 Laplacian 251 A3.5 Curl 252 A4 Derivations of the Bernoulli equation 253 A4.1 The Bernoulli equation from the energy equation 253 A4.2 The Bernoulli equation from the integral momentum equation 253 A4.3 The Bernoulli equation from the Navier-Stokes equation 253 A5 Flow parameters as force ratios 256 A5.1 Acceleration vs. viscous forces: the Reynolds number 256 A5.2 Acceleration vs. gravity force: the Froude number 257 A5.3 Acceleration vs. elastic forces: the Mach number 257 A5.4 Other force ratios 258 A6 Details of the winter 2020-2021 fnal examination (updated Febru-ary 2021) 259 A7 Example of previous examinations 261 A8 References 292 A1 Notation ≡ By defnition. The ≡symbol sets the defnition of the term on its left (which does not depend on previous equations). ̇ (dot above symbol) Time rate: ̇ ≡ d d푡. For example, ̇ 푄is the rate of heat (in watts) representing a heat quantity 푄(in joules) every second. (bar above symbol) Time-average: 퐴≡avg(퐴) = avg(퐴+퐴′). The prime symbol indicates the instantaneous fuctuation around the average. Δ Indicates a net diference between two values: (Δ푋)A→B ≡푋B −푋A. Can be negative. italics Physical properties (e.g. mass 푚, temperature 푇). straight subscripts Points in space or in time (temperature 푇A at point A). Subscripts “cst” indicate a constant property, “in” indicates “incoming” and “out” is “outgoing”. Subscript “av.” indicates “average”. operators Diferential d, partial diferential 휕, fnite diferential 훿, total (alt.: substantial) derivative D/D푡(def. eq. 6/7 p. 115), exponential exp 푥≡푒푥, natural logarithm ln 푥≡log푒푥. vectors Vectors are written with an arrow. Velocity is ⃗ 푉≡(푢, 푣, 푤), alter-natively written 푢푖≡(푢, 푣, 푤). The norm of a vector ⃗ 퐴(positive or negative) is \| ⃗ 퐴\|, its length (always positive) is \|\| ⃗ 퐴\|\|. vector calculus Dot product ⃗ 퐴⋅⃗ 퐵(see §A2.1 p. 247); Cross product ⃗ 퐴∧⃗ 퐵(see §A2.2 p. 248); Gradient ⃗ ∇퐴(def. eq. 4/11 p. 77, see also §A3.1 p. 250); Divergent ⃗ ∇⋅⃗ 퐴(def. eq. 5/14 p. 95, see also §A3.2 p. 250); Laplacian ⃗ ∇2 ⃗ 퐴(def. eq. 6/38 p. 124, see also §A3.4 p. 251); Curl ⃗ ∇× ⃗ 퐴(def. eq. A/32 p. 252, see also §A3.5 p. 252). units Units are typed in roman (normal) font and colored gray (1 kg). In sentences units are fully-spelled and conjugated (one hundred watts). The liter is noted L to increase readability (1 L ≡10−3 m3). Units in equations are those from système international (si) unless otherwise indicated. numbers The decimal separator is a comma, the decimal exponent is pre-ceded by a dot, integers are written in groups of three (1,234 ⋅103 = 1 234). Numbers are rounded up as late as possible and never in series. Leading and trailing zeros are never indicated. 246 A2 Vector operations For a step-by-step revision of those notions and many more, written in a progressive, nonjudgmental way, with plenty of worked-out exercises, you can try John Bird’s Higher Engineering Mathematics . A2.1 Vector dot product The dot productw of two vectors is a number defned as: ⃗ 푎⋅⃗ 푏= \|⃗ 푎\| \|⃗ 푏\| cos 휃 (A/1) where 휃is the angle separating the two vectors ⃗ 푎and ⃗ 푏. In this document, the dot product is always written with a median dot (⃗ 푎⋅⃗ 푏), but in other literature, it is sometimes written with the × symbol. Take care not to confuse it with the vector cross product (see §A2.2 p. 248). It can be shown that the dot product of two vectors ⃗ 푎{푥푎, 푦푎, 푧푎} and ⃗ 푏{푥푏, 푦푏, 푧푏} can be quantifed as: ⃗ 푎⋅⃗ 푏= 푥푎푥푏+ 푦푎푦푏+ 푧푎푧푏 (A/2) The dot product of two vectors is the same regardless of the order in which they are multiplied: ⃗ 푎⋅⃗ 푏= ⃗ 푏⋅⃗ 푎 (A/3) Its is easily shown using eq. (A/1) that: (− → 푎⋅− → −푏) = −(⃗ 푎⋅⃗ 푏) (A/4) Figure A.1: Two vectors ⃗ 푎and ⃗ 푏. 247 A2.2 Vector cross product The cross productw of two vectors is a vector written as: ⃗ 푎∧⃗ 푏= ⃗ 푐 (A/5) The vector ⃗ 푐is so that: its length is equal to 푐= 푎푏sin 휃 (A/6) its direction is perpendicular to ⃗ 푎and ⃗ 푏; its orientation is so that if ⃗ 푏is positioned at the end of ⃗ 푎, then ⃗ 푐points away from a point from which the rotation generated ⃗ 푏is in the clockwise direction. Describing ⃗ 푐requires a third dimension, even if ⃗ 푎et ⃗ 푏have only two dimen-sions. In this document, the cross product is written with a wedge symbol (⃗ 푎∧⃗ 푏) but in the literature, it is often written with the symbol ×. Make sure you do not confuse it with the dot product (§A2.2 p. 248). It can be shown that the cross product ⃗ 푐of two vectors ⃗ 푎{푥푎, 푦푎, 푧푎} et ⃗ 푏{푥푏, 푦푏, 푧푏} is: ⃗ 푐= \| \| \| \| \| \| \| ⃗ 푖 ⃗ 푗 ⃗ 푘 푥푎 푦푎 푧푎 푥푏 푦푏 푧푏 \| \| \| \| \| \| \| (A/7) So that one obtains: ⃗ 푐= \| \| \| \| 푦푎 푧푎 푦푏 푧푏 \| \| \| \| ⃗ 푖− \| \| \| \| 푥푎 푧푎 푥푏 푧푏 \| \| \| \| ⃗ 푗+ \| \| \| \| 푥푎 푦푎 푥푏 푦푏 \| \| \| \| ⃗ 푘 (A/8) = (푦푎푧푏−푦푏푧푎)⃗ 푖−(푥푎푧푏−푥푏푧푎)⃗ 푗+ (푥푎푦푏−푥푏푦푎) ⃗ 푘 (A/9) The two vectors ⃗ 푎∧⃗ 푏and ⃗ 푏∧⃗ 푎are pointing away one from the other (fg. A.3): ⃗ 푏∧⃗ 푎= −(⃗ 푎∧⃗ 푏) (A/10) Figure A.2: Two vectors ⃗ 푎and ⃗ 푏. The vector product ⃗ 푎∧⃗ 푏has length the product of the lengths 푏⟂et 푎. In the case shown here, the vector ⃗ 푐= ⃗ 푎∧⃗ 푏is going into through the document plane, going away from the reader. 248 Figure A.3: The vectors ⃗ 푎∧⃗ 푏and ⃗ 푏∧⃗ 푎have the same length but are pointing directions opposite one from the other (the frst away from the reader, and the other towards the reader). If any vector changes direction, the cross product also changes direction (fg. A.4): ⃗ 푎∧− → −푏= −(⃗ 푎∧⃗ 푏) (A/11) Figure A.4: The vector ⃗ 푎∧⃗ −푏is pointing away from the vector ⃗ 푎∧⃗ 푏. 249 A3 Field operators Four operators which apply on vector or scalar felds are important in fuid mechanics: gradient, divergent, Laplacian and curl. A3.1 Gradient The mathematical operator gradientw (frst introduced as eq. 4/11 p. 77) is written ⃗ ∇. It applies on a scalar feld and produces a vector feld. It is defned as: ⃗ ∇≡⃗ 푖휕 휕푥+ ⃗ 푗휕 휕푦+ ⃗ 푘휕 휕푧 (A/12) ⃗ ∇퐴≡휕퐴 휕푥 ⃗ 푖+ 휕퐴 휕푦 ⃗ 푗+ 휕퐴 휕푧 ⃗ 푘= ⎛ ⎜ ⎜ ⎝ 휕퐴 휕푥 휕퐴 휕푦 휕퐴 휕푧 ⎞ ⎟ ⎟ ⎠ (A/13) For example, the gradient of a pressure feld is the vector feld ⃗ ∇푝: ⃗ ∇푝≡휕푝 휕푥 ⃗ 푖+ 휕푝 휕푦 ⃗ 푗+ 휕푝 휕푧 ⃗ 푘= ⎛ ⎜ ⎜ ⎝ 휕푝 휕푥 휕푝 휕푦 휕푝 휕푧 ⎞ ⎟ ⎟ ⎠ (A/14) A3.2 Divergent The mathematical operator divergentw (frst introduced as eq. 5/14 p. 95) is written ⃗ ∇⋅and is defned as: ⃗ ∇⋅≡휕 휕푥 ⃗ 푖⋅+ 휕 휕푦 ⃗ 푗⋅+ 휕 휕푧 ⃗ 푘⋅ (A/15) When applied on a vector feld, it produces a scalar feld: ⃗ ∇⋅⃗ 퐴≡휕 휕푥 ⃗ 푖⋅⃗ 퐴+ 휕 휕푦 ⃗ 푗⋅⃗ 퐴+ 휕 휕푧 ⃗ 푘⋅⃗ 퐴 (A/16) = 휕퐴푥 휕푥 + 휕퐴푦 휕푦 + 휕퐴푧 휕푧 (A/17) When applied on a 2nd order tensor feld, it produces a vector feld: ⃗ ∇⋅⃗ 퐴푖푗≡ ⎛ ⎜ ⎜ ⎜ ⎝ 휕퐴푥푥 휕푥+ 휕퐴푦푥 휕푦+ 휕퐴푧푥 휕푧 휕퐴푥푦 휕푥+ 휕퐴푦푦 휕푦+ 휕퐴푧푦 휕푧 휕퐴푥푧 휕푥+ 휕퐴푦푧 휕푦+ 휕퐴푧푧 휕푧 ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ ⃗ ∇⋅⃗ 퐴푖푥 ⃗ ∇⋅⃗ 퐴푖푦 ⃗ ∇⋅⃗ 퐴푖푧 ⎞ ⎟ ⎟ ⎟ ⎠ (A/18) For example, the divergent of a velocity feld is the scalar feld ⃗ ∇⋅⃗ 푉: ⃗ ∇⋅⃗ 푉≡휕푢 휕푥+ 휕푣 휕푦+ 휕푤 휕푧 (A/19) 250 A3.3 Advective The advective operator,w ( ⃗ 푉⋅⃗ ∇) is defned as follows: ⃗ 푉⋅⃗ ∇≡푢휕 휕푥+ 푣휕 휕푦+ 푤휕 휕푧 (A/20) Do not confuse the advective operator with the divergent of velocity, ⃗ ∇⋅⃗ 푉 (see Appendix A3.2 above, including eq. A/19), which is a scalar feld. The advective operator can be applied to a scalar feld 퐴: ( ⃗ 푉⋅⃗ ∇)퐴= 푢휕퐴 휕푥+ 푣휕퐴 휕푦+ 푤휕퐴 휕푧 (A/21) It can also be applied to a vector feld ⃗ 퐴: ( ⃗ 푉⋅⃗ ∇) ⃗ 퐴= 푢휕⃗ 퐴 휕푥+ 푣휕⃗ 퐴 휕푦+ 푤휕⃗ 퐴 휕푧 (A/22) = ⎛ ⎜ ⎜ ⎜ ⎝ 푢휕퐴푥 휕푥+ 푣휕퐴푥 휕푦+ 푤휕퐴푥 휕푧 푢휕퐴푦 휕푥+ 푣휕퐴푦 휕푦+ 푤휕퐴푦 휕푧 푢휕퐴푧 휕푥+ 푣휕퐴푧 휕푦+ 푤휕퐴푧 휕푧 ⎞ ⎟ ⎟ ⎟ ⎠ (A/23) A3.4 Laplacian The mathematical operator Laplacianww (frst introduced as eq. 6/38 p. 124) is written ⃗ ∇2 and defned as: ⃗ ∇2 ≡⃗ ∇⋅⃗ ∇ (A/24) When applied to a scalar feld, it is equal to the divergent of the gradient of the feld, and produces a scalar feld: ⃗ ∇2퐴≡⃗ ∇⋅⃗ ∇퐴 (A/25) = 휕2퐴 (휕푥)2 + 휕2퐴 (휕푦)2 + 휕2퐴 (휕푧)2 (A/26) When applied to a vector feld, the general expression uses the curl operator (we never use this expression in this course), and produces a vector feld: ⃗ ∇2 ⃗ 퐴≡(⃗ ∇⋅⃗ ∇) ⃗ 퐴−⃗ ∇× (⃗ ∇× ⃗ 퐴) (A/27) In Cartesian coordinates, this simplifes as: ⃗ ∇2 ⃗ 퐴= ≡ ⎛ ⎜ ⎜ ⎝ ⃗ ∇2퐴푥 ⃗ ∇2퐴푦 ⃗ ∇2퐴푧 ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎝ ⃗ ∇⋅⃗ ∇퐴푥 ⃗ ∇⋅⃗ ∇퐴푦 ⃗ ∇⋅⃗ ∇퐴푧 ⎞ ⎟ ⎟ ⎠ (A/28) = ⎛ ⎜ ⎜ ⎜ ⎝ 휕2퐴푥 (휕푥)2 + 휕2퐴푥 (휕푦)2 + 휕2퐴푥 (휕푧)2 휕2퐴푦 (휕푥)2 + 휕2퐴푦 (휕푦)2 + 휕2퐴푦 (휕푧)2 휕2퐴푧 (휕푥)2 + 휕2퐴푧 (휕푦)2 + 휕2퐴푧 (휕푧)2 ⎞ ⎟ ⎟ ⎟ ⎠ (A/29) 251 For example, the Laplacian of a velocity feld is the vector feld ⃗ ∇2 ⃗ 푉: ⃗ ∇2 ⃗ 푉≡ ⎛ ⎜ ⎜ ⎝ ⃗ ∇2푢 ⃗ ∇2푣 ⃗ ∇2푤 ⎞ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2 + 휕2푣 (휕푧)2 휕2푤 (휕푥)2 + 휕2푤 (휕푦)2 + 휕2푤 (휕푧)2 ⎞ ⎟ ⎟ ⎟ ⎠ (A/30) A3.5 Curl The mathematical operator curlw (sometimes named rotational) is written ⃗ ∇× . It applies to a vector feld and produces a vector feld. It is defned as: ⃗ ∇× ≡ \| \| \| \| \| \| \| ⃗ 푖 ⃗ 푗 ⃗ 푘 휕 휕푥 휕 휕푦 휕 휕푧 \| \| \| \| \| \| \| (A/31) ⃗ ∇× ⃗ 퐴≡ \| \| \| \| \| \| \| ⃗ 푖 ⃗ 푗 ⃗ 푘 휕 휕푥 휕 휕푦 휕 휕푧 퐴푥 퐴푦 퐴푧 \| \| \| \| \| \| \| = ( 휕퐴푧 휕푦−휕퐴푦 휕푧) ⃗ 푖+ (−휕퐴푧 휕푥+ 휕퐴푥 휕푧) ⃗ 푗+ ( 휕퐴푦 휕푥−휕퐴푥 휕푦) ⃗ 푘 (A/32) For example, the curl of velocity is the vector feld ⃗ ∇× ⃗ 푉: ⃗ ∇× ⃗ 푉= \| \| \| \| \| \| \| ⃗ 푖 ⃗ 푗 ⃗ 푘 휕 휕푥 휕 휕푦 휕 휕푧 푢 푣 푤 \| \| \| \| \| \| \| = ( 휕푤 휕푦−휕푣 휕푧) ⃗ 푖+ (−휕푤 휕푥+ 휕푢 휕푧) ⃗ 푗+ ( 휕푣 휕푥−휕푢 휕푦) ⃗ 푘 (A/33) 252 A4 Derivations of the Bernoulli equation A4.1 The Bernoulli equation from the energy equation This is covered in section 2.6 p. 41. A4.2 The Bernoulli equation from the integral momen-tum equation We begin with the integral linear momentum equation (eq. 3/9 p. 56): ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 When considering a fxed, infnitely short control volume along a known streamline 푠of the fow, this equation becomes: d ⃗ 퐹pressure + d ⃗ 퐹shear + d ⃗ 퐹gravity = d d푡∭ CV 휌⃗ 푉d+ 휌푉퐴d ⃗ 푉 along a streamline, where the velocity ⃗ 푉is aligned (by defnition) with the streamline. Now, adding the restrictions of steady fow ( d/ d푡= 0) and no friction ( d ⃗ 퐹shear = ⃗ 0), we already obtain: d ⃗ 퐹pressure + d ⃗ 퐹gravity = 휌푉퐴d ⃗ 푉 The projection of the net force due to gravity d ⃗ 퐹gravity on the streamline segment d푠has norm d ⃗ 퐹gravity ⋅d⃗ 푠= −푔휌퐴d푧, while the net force due to pressure is aligned with the streamline and has norm d퐹pressure,푠= −퐴d푝. Along this streamline, we thus have the following scalar equation, which we integrate from points 1 to 2: −퐴d푝−휌푔퐴d푧= 휌푉퐴d푉 −1 휌d푝−푔d푧= 푉d푉 −∫ 2 1 1 휌d푝−∫ 2 1 푔d푧= ∫ 2 1 푉d푉 The last obstacle is removed when we consider fows without heat or work transfer, where, therefore, the density 휌is constant. In this way, we arrive to equation. 2/20 p. 42 again: 푝1 휌+ 1 2푉2 1 + 푔푧1 = 푝2 휌+ 1 2푉2 2 + 푔푧2 (푝+ 1 2휌푉2 + 휌푔푧)1 = (푝+ 1 2휌푉2 + 휌푔푧)2 A4.3 The Bernoulli equation from the Navier-Stokes equation We start by following a particle along its path in an arbitrary fow, as dis-played in fg. A.5. The particle path is known (condition 5 in §2.6 p. 41), but its speed 푉is not. 253 Figure A.5: Diferent pathlines in an arbitrary fow. We follow one particle as it travels from point 1 to point 2. An infnitesimal path segment is named d⃗ 푠. Figure CC-0 Olivier Cleynen We are now going to project every component of the Navier-Stokes equation (eq. 6/42 p. 125) onto an infnitesimal portion of trajectory d⃗ 푠. Once all terms have been projected, the Navier-Stokes equation becomes a scalar equation: 휌휕⃗ 푉 휕푡+ 휌( ⃗ 푉⋅⃗ ∇) ⃗ 푉= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 휌휕⃗ 푉 휕푡⋅d⃗ 푠+ 휌( ⃗ 푉⋅⃗ ∇) ⃗ 푉⋅d⃗ 푠= 휌⃗ 푔⋅d⃗ 푠−⃗ ∇푝⋅d⃗ 푠+ 휇⃗ ∇2 ⃗ 푉⋅d⃗ 푠 Because the velocity vector ⃗ 푉of the particle, by defnition, is always aligned with the path, its projection is always equal to its norm: ⃗ 푉⋅d⃗ 푠= 푉d푠. Also, the downward gravity 푔and the upward altitude 푧have opposite signs, so that ⃗ 푔⋅d⃗ 푠= −푔d푧; we thus obtain: 휌휕푉 휕푡d푠+ 휌d푉 d푠푉d푠= −휌푔d푧−d푝 d푠d푠+ 휇⃗ ∇2 ⃗ 푉⋅d⃗ 푠 When we restrict ourselves to steady fow (condition 1 in §2.6), the frst left-hand term vanishes. Neglecting losses to friction (condition 4) alleviates us from the last right-hand term, and we obtain: 휌d푉 d푠푉d푠= −휌푔d푧−d푝 d푠d푠 휌푉d푉= −휌푔d푧−d푝 This equation can then be integrated from point 1 to point 2 along the pathline: 휌∫ 2 1 푉d푉= −∫ 2 1 휌푔d푧−∫ 2 1 d푝 When no work or heat transfer occurs (condition 3) and the fow remains incompressible (condition 2), the density 휌remains constant, so that we indeed have returned to eq. 2/20 p. 42: Δ ( 1 2푉2) + 푔Δ푧+ 1 휌Δ푝= 0 (A/34) (푝+ 1 2휌푉2 + 휌푔푧)1 = (푝+ 1 2휌푉2 + 휌푔푧)2 (A/35) 254 Thus, we can see that if we follow a particle along its path, in a steady, incompressible, frictionless fow with no heat or work transfer, its change in kinetic energy is due only to the result of gravity and pressure, in accordance with the Navier-Stokes equation. 255 A5 Flow parameters as force ratios This topic is well covered in Massey Instead of the mathematical approach covered in §8.2.2 p. 162, the concept of fow parameter can be approached by comparing forces in fuid fows. Fundamentally, understanding the movement of fuids requires applying Newton’s second law of motion: the sum of forces which act upon a fuid particle is equal to its mass times its acceleration. We have done this in an aggregated manner with integral analysis (in chapter 3, eq. 3/9 p. 56), and then in a precise and all-encompassing way with diferential analysis (in chapter 6, eq. 6/42 p. 125). With the latter method, we obtain complex mathematics suitable for numerical implementation, but it remains difcult to obtain rapidly a quantitative measure for what is happening in any given fow. In order to obtain this, an engineer or scientist can use force ratios. This involves comparing the magnitude of a type of force (pressure, viscous, gravity) either with another type of force, or with the mass-times-acceleration which a fuid particle is subjected to as it travels. We are not interested in the absolute value of the resulting ratios, but rather, in having a measure of the parameters that infuence them, and being able to compare them across experiments. A5.1 Acceleration vs. viscous forces: the Reynolds number The net sum of forces acting on a particle is equal to its mass times its acceleration. If a representative length for the particle is 퐿, the particle mass grows proportionally to the product of its density 휌and its volume 퐿3. Meanwhile, its acceleration relates how much its velocity 푉will change over a time interval Δ푡: it may be expressed as a ratio Δ푉/Δ푡. In turn, the time interval Δ푡may be expressed as the representative length 퐿divided by the velocity 푉, so that the acceleration may be represented as proportional to the ratio 푉Δ푉/퐿. Thus we obtain: \|net force\| = \|mass × acceleration\| ∼휌퐿3푉Δ푉 퐿 \| ⃗ 퐹net\| ∼휌퐿2푉Δ푉 We now observe the viscous force acting on a particle: it is proportional to the shear efort and a representative acting surface 퐿2. The shear can be modeled as proportional to the viscosity 휇and the rate of strain, which will grow proportionally to Δ푉/퐿. We thus obtain a crude measure for the magnitude of the shear force: \|viscous force\| ∼휇Δ푉 퐿퐿2 \| ⃗ 퐹viscous\| ∼휇Δ푉퐿 The magnitude of the viscous force can now be compared to the net force: \|net force\| \|viscous force\| ∼휌퐿2푉Δ푉 휇Δ푉퐿 = 휌푉퐿 휇 = [Re] (A/36) 256 and we recognize the ratio as the Reynolds number (8/12 p. 164). We thus see that the Reynolds number can be interpreted as the inverse of the infuence of viscosity. The larger [Re] is, and the smaller the infuence of the viscous forces will be on the trajectory of fuid particles. A5.2 Acceleration vs. gravity force: the Froude number The weight of a fuid particle is equal to its mass, which grows with 휌퐿3, multiplied by gravity 푔: \|weight force\| = \| ⃗ 퐹W\| ∼휌퐿3푔 The magnitude of this force can now be compared to the net force: \|net force\| \|weight force\| ∼휌퐿2푉2 휌퐿3푔= 푉2 퐿푔= [Fr]2 (A/37) and here we recognize the square of the Froude number (8/11 p. 164). We thus see that the Froude number can be interpreted as the inverse of the infuence of weight on the fow. The larger [Fr] is, and the smaller the infuence of gravity will be on the trajectory of fuid particles. A5.3 Acceleration vs. elastic forces: the Mach number In some fows called compressible fows the fuid can perform work on itself, and and the fuid particles then store and retrieve energy in the form of changes in their own volume. In such cases, fuid particles are subject to an elastic force in addition to the other forces. We can model the pressure resulting from this force as proportional to the bulk modulus of elasticity 퐾 of the fuid (formally defned as 퐾≡휌휕푝/휕휌); the elastic force can therefore be modeled as proportional to 퐾퐿2: \|elasticity force\| = \| ⃗ 퐹elastic\| ∼퐾퐿2 The magnitude of this force can now be compared to the net force: \|net force\| \|elasticity force\| ∼휌퐿2푉2 퐾퐿2 = 휌푉2 퐾 This ratio is known as the Cauchy number; it is not immediately useful because the value of 퐾in a given fuid varies considerably not only according to temperature, but also according to the type of compression undergone by the fuid: for example, it grows strongly during brutal compressions. During isentropic compressions and expansions (isentropic meaning that the process is fully reversible, i.e. without losses to friction, and adiabatic, i.e. without heat transfer), it can be shown that the bulk modulus of elasticity is proportional to the square of the speed of sound 푐: 퐾\|reversible = 푐2휌 (A/38) The Cauchy number calibrated for isentropic evolutions is then \|net force\| \|elasticity force\|reversible ∼휌푉2 퐾 = 푉2 푐2 = [Ma]2 (A/39) 257 and here we recognize the square of the Mach number (1/10 p. 16). We thus see that the Mach number can be interpreted as the infuence of elasticity on the fow. The larger [Ma] is, and the smaller the infuence of elastic forces will be on the trajectory of fuid particles. A5.4 Other force ratios The same method can be applied to reach the defnitions for the Strouhal and Euler numbers given in §8.2 p. 161. Other numbers can also be used which relate forces that we have ignored in our study of fuid mechanics. For exam-ple, the relative importance of surface tension forces or of electromagnetic forces are quantifed using similarly-constructed fow parameters. In some applications featuring rotative motion, such as fows in centrifugal pumps or planetary-scale atmospheric weather, it may be convenient to ap-ply Newton’s second law in a rotating reference frame. This results in the appearance of new reference-frame forces, such as the Coriolis or centrifugal forces; their infuence can then be studied using additional fow parameters. In none of those cases can fow parameters give enough information to predict solutions. They do, however, provide quantitative data to indicate which forces are relevant in which places: this not only helps us understand the mechanisms at work, but also distinguish the negligible from the infuential, a key characteristic of efcient scientifc and engineering work. 258 A6 Details of the winter 2020-2021 fnal examination (updated February 2021) Video: The companion video to this exam briefng (2020) by Olivier Cleynen (CC-by) The fnal examination for this course in the winter semester 2020-2021 is an open-book, take-home exam which will take place on February 18, 2021 from 14:00 to 16:00. The most important information is as follows: • This is a take-home exam: you may consult your notes, books, use software and resources ofine or online. • You must take this exam alone: you cannot interact with anyone online or ofine during the exam. • The exam lasts 90 minutes. Additionally, 30 minutes are provided for the the scanning and upload of your answer. • You will receive the assignment per email shortly before the start. You must scan your answer with your student card on each page, and send it as a single-PDF fle to fuidmech@ovgu.de using your academic email before the end of the exam. • In the winter semester, this online examination grade is the only grade you receive in this course. You can download the template of the exam at Video: How to survive the exam-ination (2020) by Olivier Cleynen (CC-by) The complete list of examinable problems (unchanged from the 2020 exam) is as follows: • 2.2 2.3 2.4 2.5 2.6 • 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 • 4.4 4.5 4.61 4.7 • 5.2 5.3 5.4 5.5 5.6 • 6.2 6.3 6.4 6.5 6.6 6.7 6.9 • 7.3 7.4 7.5 7.6 7.7 7.8 • 8.2 8.4 8.5 • Chapter 9 is not examinable this semester • 10.2 10.3 10.4 10.5 10.6 • Chapter 11 is not examinable this semester Three problems will be given, all mandatory. The frst problem (10 pts is exercise 6.2 p.131. The two other problems (45 pts each) are extracted from the list above, and modifed slightly. Typically, the input data is changed, as well as the problem geometry. The method for solving the problems remains the same. A formula sheet is provided. It is the sum of the preambles of every problem sheet in the lecture notes. It includes the Moody diagram and the viscosity 1In exercise 4.6, only the calculation of the vertical force 퐹top is examinable. 259 diagram used in the problem sheets. You should defnitely have a calculator with you, to facilitate calculations. The criteria for grading your answers are: • You must show your work in all answers; • Answers to questions starting with “show that” should be particularly well-developed and continuous; • Illegible or ambiguous answers are always discarded. Examinations from previous years, and their full solution, are available on the course website ( Since the course content has changed over time, you might fnd a few diferences: • Problems involving calculating compressible air fow using tables are no longer examinable; • A problem involving a ball fountain (“Kugel fountain”) is no longer examinable; • Viscosity values were read in a diferent diagram, and may not match values read in the 2020 viscosity diagram. You are welcome (and in fact encouraged!) to ask me questions of all sorts about the exam. You may contact me as described in the introduction, page 7. I wish you to have productive and joyful revisions! Olivier February 2021 (updated February 12, after cancellations following covid-19 restrictions) 260 A7 Example of previous examinations The following pages present the fnal examinations for this course in 2020 and 2021, and their full solutions. 261 Fluid Mechanics examination — July 11, 2019 Fluid Mechanics for Master Students Solve problem 1, plus three other problems among problems 2 to 6. Duration: 2 h – Use of calculator is authorized; documents are not authorized. Except otherwise indicated, we assume that: Fluids are Newtonian The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 ◦C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Balance of mass in a considered volume with steady fow: 0 = Σ [휌푉⟂퐴]incoming + Σ [휌푉⟂퐴]outgoing (1) where 푉⟂is negative inwards, positive outwards. Balance of momentum in a considered volume with steady fow: ⃗ 퐹net on fuid = Σ [휌푉⟂퐴⃗ 푉]incoming + Σ [휌푉⟂퐴⃗ 푉]outgoing (2) where 푉⟂is negative inwards, positive outwards. Balance of energy in a considered volume with steady fow: ̇ 푄net + ̇ 푊shaft, net = Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]in +Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]out (3) where ̇ 푚is negative inwards, positive outwards. Mass balance through an arbitrary volume: 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (4) Momentum balance through an arbitrary volume: ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 (5) Angular momentum balance through an arbitrary volume: ⃗ 푀net,X = d d푡∭ CV ⃗ 푟X푚∧휌⃗ 푉d+ ∬ CS ⃗ 푟X푚∧휌( ⃗ 푉rel ⋅⃗ 푛) ⃗ 푉d퐴 (6) 1 Continuity equation for incompressible fow: ⃗ ∇⋅⃗ 푉= 0 (7) Navier-Stokes equation for incompressible fow: 휌D ⃗ 푉 D푡 = 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (8) In a highly-viscous (creeping) steady fow, the drag 퐹퐷exerted on a spherical body of diameter 퐷at by fow at velocity 푉∞is quantifed as: 퐹퐷sphere = 3휋휇푉∞퐷 (9) In cylindrical pipe fow, we assume the fow is always laminar for [Re]퐷≲2 300, and always turbulent for [Re]퐷≳4 000. The Darcy friction factor 푓is defned as: 푓≡ \|Δ푝loss\| 퐿 퐷 1 2휌푉2 av. (10) The loss coefcient 퐾퐿is defned as: 퐾퐿≡\|Δ푝loss\| 1 2휌푉2 av. (11) Viscosities of various fuids are given in fg. 1. Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in fg. 2 p.4. Non-dimensional incompressible Navier-Stokes equation: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (12) in which [St] ≡ 푓퐿 푉, [Eu] ≡ 푝0−푝∞ 휌푉2 , [Fr] ≡ 푉 √푔퐿and [Re] ≡ 휌푉퐿 휇 . The force coefcient 퐶퐹and power coefcient 퐶P are defned as: 퐶퐹≡ 퐹 1 2휌푆푉2 퐶P ≡ ̇ 푊 1 2휌푆푉3 (13) The speed of sound 푐in air is modeled as: 푐= √ 훾푅푇 (14) 2 In boundary layer fow, we assume that transition occurs at [Re]푥≳5 ⋅105. The wall shear coefcient 푐푓, a function of distance 푥, is defned based on the free-stream fow velocity 푈: 푐푓(푥) ≡ 휏wall 1 2휌푈2 (15) Exact solutions to the laminar boundary layer along a smooth surface yield: 훿 푥= 4,91 √ [Re]푥 훿∗ 푥 = 1,72 √ [Re]푥 (16) 훿∗∗ 푥 = 0,664 √ [Re]푥 푐푓(푥) = 0,664 √ [Re]푥 (17) Solutions to the turbulent boundary layer along a smooth surface yield the following time-averaged characteristics: 훿 푥≈ 0,16 [Re] 1 7 푥 훿∗ 푥 ≈ 0,02 [Re] 1 7 푥 (18) 훿∗∗ 푥 ≈0,016 [Re] 1 7 푥 푐푓(푥) ≈0,027 [Re] 1 7 푥 (19) In a highly-viscous (creeping) steady fow, the drag 퐹D exerted on a spherical body of diameter 퐷at by fow at velocity 푈∞is quantifed as: 퐹D sphere = 3휋휇푈∞퐷 (20) Figure 1 – Viscosity of various fuids at a pressure of 1 bar (in practice viscosity is almost independent of pressure). Figure © White 2008 3 Figure 2 – A Moody diagram, which presents values for 푓measured experimentally, as a function of the diameter-based Reynolds number [Re]퐷, for diferent relative roughness values. Diagram CC-by-sa S Beck and R Collins, University of Shefeld 4 Solve problem 1, and three other problems among problems 2 to 6. The following marking guidelines will be used: • Answers to questions starting with “show that” should be fully-developed and continuous; • In all other questions, the correct result with the correct unit is enough to obtain full points; • Illegible or ambiguous answers are always discarded. 5 1 Governing equation 1.1. [5 pts] Write out equation (8), the Navier-Stokes equation for incompressible fow, in its fully-developed form in three Cartesian coordinates. 1.2. [5 pts] Write out equation (7), the continuity equation for incompressible fow, in its fully-developed form in three Cartesian coordinates. 2 Observation window in a water tank A water tank used in a laboratory is flled with stationary water (fg. 3). A window is installed on one of the walls of the canal, to enable observation. The window is hinged on its top face. The window has a height of 1,5 m and a width of 3,5 m. The walls of the tank are inclined with an angle 휃= 70° relative to horizontal. Figure 3 – A door installed on the wall of a water tank. 2.1. [15 pts] What is the magnitude of the net force applying on the tank window? 2.2. [10 pts] At what distance away from the hinge does this force apply? Water is added to the tank, so that the water level increases. 2.3. [5 pts] How will the distance calculated above change as water is added? (briefy justify your answer, e.g. in 30 words or less) 6 3 Piping leading to a turbine A pipe leads water from one reservoir to a turbine, which discharges into another reservoir, as shown in fgure 4. Figure 4 – Layout of the water pipe. For clarity, in this fgure, the vertical scale is greatly exaggerated. In the vertical scale, the diameter of the pipe is also greatly exaggerated. The pipe is made of coarse concrete (roughness 0,25 mm) and carries 800 L s−1 of water at 20 ◦C. It has a diameter of 1,1 m and features four elbow bends with sharp angles, each inducing a loss coefcient 퐾퐿of 0,75. 3.1. [10 pts] Represent qualitatively (i.e. without numerical data) the pressure distribu-tion along the length of the pipe, both when the turbine is shut down (without any fow), and when it is operating. 3.2. [15 pts] What is the hydraulic power available to the turbine? The outlet tank on the right is very large, so that its water level does not vary. The source water tank on the left, however, sees its height decrease as the water is emptied through the turbine. Ultimately, as the water level decreases, the water stops fowing entirely. 3.3. [5 pts] When the water stops fowing, what will be the height of the water level in the source tank on the left? 7 4 Boundary layer on a fat plate A thin and smooth plate with width 푊= 0,6 m and length 퐿= 2 m is placed with a zero angle of attack in atmospheric air fow incoming at 21 m s−1, as shown in fgure 5. We would like to study the shear exerted by the fow over the top surface of the plate. Figure 5 – A thin plate positioned parallel to an incoming uniform fow. Figure CC-0 o.c. 4.1. [5 pts] At what distance 푥tr. along the plate, approximately, will the boundary layer transit and become turbulent? 4.2. [10 pts] Starting from equation (21), which quantifes the friction factor 푐푓(see defnition 15) in a laminar boundary layer, 푐푓(푥) = 0,664 √ [Re]푥 (21) show that the shear force 퐹휏laminar exerted in the laminar section of the boundary layer is: 퐹휏laminar = 0,664√휌휇푈 3 2푊푥 1 2 tr. (22) 4.3. [5 pts] What is the shear force exerted on the top surface of the plate by the laminar section of the boundary layer? 4.4. [5 pts] What is the shear force exerted on the top surface of the plate by the turbulent section of the boundary layer? 4.5. [5 pts] Would the boundary layer become thicker if the velocity was increased? (briefy justify your answer, e.g. in 30 words or less). 8 5 Velocity measurements in a tunnel A group of students proceeds with speed measurements in a water tunnel. The objective is to measure the drag applying on a an object with constant cross-section, positioned across the tunnel test section (fg. 6). Figure 6 – An object with constant cross-section positioned across a water tunnel. The object spans completely across the tunnel (in the 푧-direction). The horizontal velocity distributions upstream and downstream of the profle are also shown. Figure CC-0 o.c. Upstream of the object, the water fow velocity is uniform (푢1 = 푈= 3,2 m s−1). Downstream of the object, horizontal velocity measurements are made every 5 cm across the fow; the following results are obtained: vertical position 푦(cm) horizontal speed 푢2 (m s−1) 0 3,2 5 3,2 10 3,15 15 3,14 20 3,03 25 2,92 30 2,81 35 2,87 40 2,89 45 2,97 50 3,19 55 3,2 60 3,2 9 The width of the profle (perpendicular to the fow, in the 푧-direction) is 70 cm. The water has uniform temperature and density (20 ◦C, 999 kg m−3) and the pressure is uniform across the measurement surface. 5.1. [20 pts] What is the drag force applying on the profle? 5.2. [10 pts] If water was replaced with a fuid with higher viscosity, how would you expect the drag force to change? (briefy justify your answer, e.g. in 30 words or less) 6 Lift and drag on a rotating football A group of fuid dynamicists investigates the air fow around a football. In particular, they are interested in the forces applying on the ball when it has been kicked and is fying through the air. The football has diameter 22 cm, a weight of 430 g; it is traveling at 70 km h−1. In order to observe the fow, they install a steel sphere in a wind tunnel (fgure 7). The sphere has a diameter of 1,1 m. Drag force measurements are carried out in the tunnel. Figure 7 – A steel sphere positioned in a wind tunnel. The sphere is maintained stationary, while the air travels with speed 푉tunnel. Force measurements are carried out on the ball. Figure CC-0 o.c. 10 6.1. [5 pts] What is the wind tunnel speed required, so that the fow around the real football is reproduced around the sphere in the tunnel? 6.2. [5 pts] With the speed calculated above, by which factor should the drag force measured in the wind tunnel be multiplied, in order to obtain the drag force on the real football? The fuid dynamicists now investigate the efect of spin on the ball. When the football is rotated along a horizontal axis during travel, a lift force exerts laterally on the ball, curving its trajectory. This is represented, from above, in fgure 8. In order to quantify this efect, the wind tunnel sphere is rotated in the wind tunnel, and measurements are carried out; the results are plotted in fgure 9. Figure 8 – Trajectory of a rotating football in free fight, as seen from above. A lift force exerts towards the left, and deviates the trajectory towards the left. Figure CC-0 o.c. 11 Figure 9 – Experimental measurements of the lift and drag coefcients applying on a rotating sphere in an steady uniform fow. Figure © from Munson & al. 2013 6.3. [10 pts] How many rotations per second are required in order to generate a lift force of 3,1 N on the real football when it travels? 6.4. [5 pts] What is then the corresponding drag force ? 6.5. [5 pts] Propose and quantify one possibility for the football player to double the lift force applying on the ball. 12 Solution: Fluid Mechanics examination — July 11, 2019 Fluid Mechanics for Master Students fuidmech.ninja 1 Governing equation 1.1 N-S equation question 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦+ 푤휕푢 휕푧] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2] (1) 휌[ 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦+ 푤휕푣 휕푧] = 휌푔푦−휕푝 휕푦+ 휇[ 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2 + 휕2푣 (휕푧)2] (2) 휌[ 휕푤 휕푡+ 푢휕푤 휕푥+ 푣휕푤 휕푦+ 푤휕푤 휕푧] = 휌푔푧−휕푝 휕푧+ 휇[ 휕2푤 (휕푥)2 + 휕2푤 (휕푦)2 + 휕2푤 (휕푧)2] (3) 1.2 Continuity question 휕푢 휕푥+ 휕푣 휕푦+ 휕푤 휕푧 = 0 (4) 1 2 Observation window in a water tank 2.1 Net force We defne coordinates 푟and 푧, and the length 퐿1, as shown in the fgure below. The force on a small section of door with length d푟and width 푊is d퐹. On the complete door, the force applying due to the net pressure 푝net of water and air is: 퐹net = ∫ 푟=푅max 푟=0 d퐹 (5) = ∫ 푟=푅max 푟=0 푝net d푆 (6) = ∫ 푟=푅max 푟=0 푝net푊d푟 (7) = ∫ 푟=푅max 푟=0 휌푔푧푊d푟 (8) A coordinate transform is needed to solve the integral, expressing 푧as a function of 푅. This is obtained by geometry: 푧= 푍min + 푟sin 휃 (9) 2 Inserting eq. 9 into eq. 8, we continue with: 퐹net = ∫ 푟=푅max 푟=0 휌푔(푍min + 푟sin 휃) 푊d푟 (10) = 휌푔푊∫ 푟=푅max 푟=0 (푍min + 푟sin 휃) d푟 (11) = 휌푔푊∫ 푟=푅max 푟=0 (퐿1 sin 휃+ 푟sin 휃) d푟 (12) = 휌푔푊sin 휃[퐿1푟+ 1 2푟2] 푟=푅max 푟=0 (13) = 휌푔푊sin 휃(퐿1푅max + 0,5푅max 2) (14) = 103 × 9,81 × 3,5 × sin(70°) (0,8 × 1,5 + 0,5 × 1,52) (15) = 75 014 N (16) 퐹net = 75,01 kN (17) 2.2 Distance from hinge We frst calculate the moment exerting about the hinge due to the net pressure of air and water, using the same notation as above: 푀net = ∫ 푟=푅max 푟=0 푟d퐹 (18) = ∫ 푟=푅max 푟=0 푟휌푔푧푊d푟 (19) = ∫ 푟=푅max 푟=0 푟휌푔(푍min + 푟sin 휃) 푊d푟 (20) = 휌푔푊∫ 푟=푅max 푟=0 푟(퐿1 sin 휃+ 푟sin 휃) d푟 (21) = 휌푔푊sin 휃[ 1 2퐿1푟2 + 1 3푟3] 푟=푅max 푟=0 (22) = 103 × 9,81 × 3,5 × sin(70°) (0,5 × 0,8 × 1,52 + 1 3 × 1,53) (23) = 65 335 N m (24) 푀net = 65,34 kN m (25) The distance away from the hinge 푅퐹is obtained by dividing the moment by the force: 푅퐹= 푀net 퐹net (26) = 75 014 65 335 (27) 푅퐹= 0,87 m (28) 3 2.3 Change in distance Increasing water height translates (only) in an increase of the value of 퐿1 in the equations above. Both 푀net and 퐹net increase together with 퐿1 (eqs. 14 & 22), so it is not immediately apparent how 푅퐹changes in eq. 26. Several possibilities can be used to fnd the answer: • Substituting 푥퐿1 instead of 퐿1 and comparing the two radiuses, one can write: 푅퐹2 < 푅퐹1 (29) 1 2푥퐿1푅2 max + 1 3푅3 max 푥퐿1푅max + 1 2푅2 max < 1 2퐿1푅2 max + 1 3푅3 max 퐿1푅max + 1 2푅2 max (30) 1 < 푥 (31) (it is even possible to show, using this equation 30, that 푅퐹tends towards 푅max/2 as 퐿1 increases); • It is possible to observe graphically that the center of application of the force moves closer to the hinge when the net pressure distribution is changed due to the increase in 퐿1; • It is also possible to calculate manually one or several new values for 푅퐹; All those methods will provide some evidence that the distance 푅퐹will in fact decrease when 퐿1 is increased. 4 3 Piping leading to a turbine 3.1 Pressure distribution 3.2 Turbine power We want to calculate three pressure drops: • Pressure drop due to wall friction losses along the pipe, Δ푝푓: The average velocity in the pipe is 푉av. = ̇  푆 (32) = ̇  휋퐷2 4 (33) = 0,8 휋× 1,12 4 (34) 푉av. = 0,842 m s−1 (35) 5 The Reynolds number is [Re]퐷= 휌푉av.퐷 휇 (36) = 103 × 0,842 × 1,1 10−5 (37) [Re]퐷= 9,251 ⋅105 (38) The relative roughness is 휖 퐷= 0,25 ⋅10−3 1,1 (39) 휖 퐷= 2,27 ⋅10−4 (40) With those values, the Moody diagram reads: 푓= 0,0158 (41) Finally, the wall friction losses along the pipe are calculated as: Δ푝푓= −푓1 2휌푉2 av. 퐿 퐷 (42) = −0,01581 21030,84224 ⋅103 1,1 (43) Δ푝푓= −2,03 ⋅104 Pa (44) • Pressure drop due to losses in the four bends, Δ푝bends: Δ푝bends = −4 × 퐾퐿 1 2휌푉2 av. (45) = −4 × 0,75 × 1 2 × 103 × 0,8422 (46) Δ푝bends = −1,06 ⋅103 Pa (47) • Pressure drop due to hydrostatic pressure change across the turbine, Δ푝ℎ: Δ푝ℎ= 휌푔(Δ푧) (48) = 103 × 9,81 [4 −(25 + 51)] (49) Δ푝ℎ= −7,06 ⋅105 Pa (50) 6 Finally, the turbine hydraulic power is obtained as: ̇ 푊turbine = ̇ (Δ푝turbine) (51) = ̇ (Δ푝ℎ−Δ푝푓−Δ푝bends) (52) = 0,8 × [−7,06 ⋅105 −(−2,03 ⋅104) −(−1,06 ⋅10−3)] (53) = −5,479 ⋅105 W (54) ̇ 푊turbine = −547,9 kW (55) 3.3 Residual water height The water will fow until air is entrained (“sucked”) into the pipe inlet. At this point, the residual water height in the left tank will be 8 m. 7 4 Boundary layer on a fat plate 4.1 Transition point The transition point occurs at [Re]푥≈5 ⋅105. Solving for 푥tr., we have: [Re]푥tr. = 휌푈푥tr. 휇 (56) 푥tr. = [Re]푥tr.휇 휌푈 (57) = 5 ⋅105 × 1,5 ⋅10−5 1,225 × 21 (58) 푥tr. = 0,29 m (59) 4.2 Shear in laminar section We start with the given equation and implement the defnition (15) of the formula sheet, as well as the defnition of the distance-based Reynolds number [Re]푥: 푐푓(푥) = 0,664 √ [Re]푥 (60) 휏wall, laminar 1 2휌푈2 = 0,664 √ 휌푈푥 휇 (61) 휏wall, laminar = 0,664 ( 휌푈푥 휇 ) −1 2 1 2휌푈2 (62) 휏wall, laminar = 0,332√휌휇푈1,5푥−1 2 (63) The shear force is the integral of the shear with respect to area: 퐹shear, laminar = ∫휏wall, laminar d푆 (64) We split the total area covered by the laminar boundary layer in strips of width 푊and length d푥, with 푥ranging from 0 (leading edge) to 푥tr. (where the laminar part of the 8 boundary ends), obtaining: 퐹shear, laminar = ∫ 푥=푥tr. 푥=0 휏wall푊d푥 (65) = ∫ 푥=푥tr. 푥=0 0,332√휌휇푈1,5푥−1 2푊d푥 (66) = 0,332√휌휇푈1,5푊∫ 푥tr. 0 푥−0,5 d푥 (67) = 0,332√휌휇푈1,5푊[ 1 −0,5 + 1푥−0,5+1 ] 푥tr. 0 (68) = 0,332√휌휇푈1,5푊2 푥0,5 tr. (69) 퐹휏laminar = 0,664√휌휇푈 3 2푊푥 1 2 tr. (70) 4.3 Value of shear in the laminar section We simply insert values into eq. 70: 퐹휏laminar = 0,664 √ 1,225 × 1,5 ⋅10−5 21 3 2 × 0,6 × 0,292 1 2 (71) 퐹휏laminar = 0,0887 N (72) 4.4 Shear in turbulent section The process is the same in the turbulent part of the layer, with 휏wall, turbulent derived from equation (19) in the formula sheet: 휏wall, turbulent = 0,0271 2휌푈2 ( 휌푈 휇) −1 7 푥−1 7 (73) This is integrated with respect to area, with 푥ranging from 푥tr. (where the turbulent section begins) to 푥max (the trailing edge of the plate): 퐹shear, turbulent = ∫ 푥=푥max 푥=푥tr. 휏wall푊d푥 (74) = 0,0135 휌 6 7푈 13 7 휇 1 7푊∫ 푥max 푥tr. 푥−1 7 d푥 (75) = 0,01575 휌 6 7푈 13 7 휇 1 7푊[푥 6 7] 푥max 푥tr. (76) = 0,01575 × 1,225 6 7 × 21 13 7 × (1,5 ⋅10−5) 1 7 × 0,6 × (2 6 7 −0,292 6 7) (77) 퐹shear, turbulent = 0,961 N (78) 9 4.5 Thickness Models for the boundary layer thickness are given in the formula sheet as equations 16 and 18 for 훿(one for the laminar section, the other for the turbulent section). In both, the Reynolds number [Re]푥appears in the denominator (the lower part of the fraction). As 푈is increased, [Re]푥will increase too, and consequently, the thickness of the boundary layer will decrease. 10 5 Velocity measurements in a tunnel 5.1 Drag force We build a control volume around the object: We use a mass balance equation to quantify the height ℎ1 of the inlet: 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (79) = −∬휌\|푉in\| d퐴+ ∬휌\|푉out\| d퐴 (80) = −휌퐿∫ ℎ1 0 푈d푦+ 휌퐿∫ ℎ2 0 푢2(푦) d푦 (81) = −휌퐿푈ℎ1 + 휌퐿∫ ℎ2 0 푢2(푦) d푦 (82) ℎ1 = 1 푈∫ ℎ2 0 푢2(푦) d푦 (83) The drag force is quantifed using a momentum balance equation, which reduces to a scalar equation in the 푥-direction: ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 (84) −퐹net = −∬휌\|푉in\|2 d퐴+ ∬휌\|푉out\|2 d퐴 (85) = −휌퐿ℎ1푈2 1 + 휌퐿∫ ℎ2 0 푢2 2(푦) d푦 (86) 11 We insert the expression for ℎ1 obtained above in this last expression, continuing as: −퐹net = −휌퐿1 푈∫ ℎ2 0 푢2(푦) d푦+ 휌퐿∫ ℎ2 0 푢2 2(푦) d푦 (87) = 휌퐿∫ ℎ2 0 (푢2 2(푦) −푈1푢2(푦)) d푦 (88) Instead of a function 푢2 = 푓(푦), we have discrete values. The integral is therefore approximated as: −퐹net = 휌퐿∑ 푦(푢2 2 −푈1푢2) δ푦 (89) = 휌퐿∑ 푦 [푢2(푢2 −푈1)] δ푦 (90) −퐹net = 휌퐿∑ 푦 [3,2 (3,2 −3,2) +3,2 (3,2 −3,2) +3,15 (3,15 −3,2) +3,14 (3,14 −3,2) +3,03 (3,03 −3,2) +2,92 (2,92 −3,2) +2,81 (2,81 −3,2) +2,87 (2,87 −3,2) +2,89 (2,89 −3,2) +2,97 (2,97 −3,2) +3,19 (3,19 −3,2) +3,2 (3,2 −3,2) +3,2 (3,2 −3,2) ] δ푦 (91) = 999 × 0,7 × (−5,3325) × 0,05 (92) = −186,45 N (93) 퐹net = 186,45 N (94) 퐹net is the net force exerted on the fuid by the object. It is positive in the 푥-direction. The drag force is the force exerted on the object on the fuid, and so is pointing in the opposite direction (fow-wise direction): 퐹drag = −186,45 N. 12 5.2 Dependence on viscosity Viscosity does not appear in equation (90) above. Nevertheless, an increase in viscosity will translate into higher shear, and so it is likely that the object will afect a larger amount of fuid around itself. This will result in a larger velocity defcit (reduced values of 푢2 in the tabled measurement values). The expression for 퐹drag will not change, but its value will increase. 13 6 Lift and drag on a rotating football 6.1 Required wind tunnel speed The two fows will have identical behavior if the Reynolds numbers are equal. With 1 denoting the real football, and 2 denoting the wind tunnel sphere, we have: [Re]1 = [Re]2 (95) 휌푉1퐷1 휇 = 휌푉2퐷2 휇 (96) 푉2 = 푉1 퐷1 퐷2 (97) = 70 3,6 0,22 1,1 (98) = 3,89 m s−1 (99) 푉2 = 14 km h−1 (100) 6.2 Ratio of forces Since the two fows are dynamically similar, the force coefcients are the same. Consid-ering the lift coefcients, 퐶퐿1 = 퐶퐿2 (101) 퐿1 1 2휌푆1푉2 1 = 퐿2 1 2휌푆2푉2 1 (102) 퐿1 퐷2 1푉2 1 = 퐿2 퐷2 2푉2 1 (103) 퐿1 퐿2 = 퐷2 1 퐷2 2 푉2 1 푉2 2 (104) = 푉2 2 푉2 1 푉2 1 푉2 2 (105) 퐿1 퐿2 = 1 (106) So, the forces will be identical on both the wind tunnel model and the real football. 14 6.3 Rotation speed The desired lift force is 퐿1 = 3,1 N. This corresponds to a lift coefcient of: 퐶퐿1 = 퐿1 1 2휌푆1푉2 1 (107) = 퐿1 1 2휌휋퐷2 1 4 푉2 1 (108) = 3,1 0,5 × 1,225 × 휋× 0,222 4 × 19,442 (109) 퐶퐿1 = 0,352 (110) Inputting this value in fgure 9, one corresponding value of 휔퐷/2푈is 1,51. This allows us to obtain a value for 휔(other higher values also work): 휔1퐷1 2푈1 = 1,51 (111) 휔1 = 1,51 × 2푈1 퐷1 (112) = 1,51 × 2 × 19,44 0,22 (113) = 266 rad s−1 (114) 휔1 = 42,5 rotations/s (115) 6.4 Drag force The chosen value of 휔퐷/2푈corresponds to a drag coefcient reading of 0,56 in fgure 9. Inputting this in the defnition for the drag coefcient, we can solve for the drag 퐹퐷: 0,56 = 퐶퐷1 = 퐹퐷1 1 2휌푆1푉2 1 (116) 퐹퐷1 = 0,56 × 1 2휌휋퐷2 1 4 푉2 1 (117) = 0,56 × 0,5 × 1,225 × 휋0,222 4 × 19,442 (118) 퐹퐷1 = 4,93 N (119) 6.5 Doubling of lift force The force is “easily” doubled by multiplying the speed 푉1 by a factor √2: 푉3 = √2푉1. We obtain the same lift coefcient (퐶퐿3 = 퐶퐿1 = 0,352). The rotation speed has to be adapted according to the expression 112: we obtain 휔3 = 60,1 rotations/s. 15 Fluid dynamics examination — September 21, 2020 Fluid Dynamics for Engineers by Olivier Cleynen Solve problem 1, plus three other problems among problems 2 to 6. Duration: 2 h – Use of calculator is authorized; documents are not authorized. Except otherwise indicated, assume that: The atmosphere has 푝atm. = 1 bar; 휌atm. = 1,225 kg m−3; 푇atm. = 11,3 ◦C; 휇atm. = 1,5 ⋅10−5 Pa s Air behaves as a perfect gas: 푅air=287 J kg−1 K−1; 훾air=1,4; 푐푝air=1 005 J kg−1 K−1; 푐푣air=718 J kg−1 K−1 Liquid water is incompressible: 휌water = 1 000 kg m−3, 푐푝water = 4 180 J kg−1 K−1 Balance of mass in a fxed control volume with steady fow: 0 = Σ [휌푉⟂퐴]incoming + Σ [휌푉⟂퐴]outgoing (1) where 푉⟂is negative inwards, positive outwards. Balance of momentum in a fxed control volume with steady fow: ⃗ 퐹net on fuid = Σ [휌푉⟂퐴⃗ 푉]incoming + Σ [휌푉⟂퐴⃗ 푉]outgoing (2) where 푉⟂is negative inwards, positive outwards. Balance of energy in a fxed control volume with steady fow: ̇ 푄net + ̇ 푊shaft, net = Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]in +Σ [ ̇ 푚(푖+ 푝 휌+ 1 2푉2 + 푔푧)]out (3) where ̇ 푚is negative inwards, positive outwards. Mass balance through an arbitrary volume: 0 = d d푡∭ CV 휌d+ ∬ CS 휌( ⃗ 푉rel ⋅⃗ 푛) d퐴 (4) Momentum balance through an arbitrary volume: ⃗ 퐹net = d d푡∭ CV 휌⃗ 푉d+ ∬ CS 휌⃗ 푉( ⃗ 푉rel ⋅⃗ 푛) d퐴 (5) Angular momentum balance through an arbitrary volume: ⃗ 푀net,X = d d푡∭ CV ⃗ 푟X푚∧휌⃗ 푉d+ ∬ CS ⃗ 푟X푚∧휌( ⃗ 푉rel ⋅⃗ 푛) ⃗ 푉d퐴 (6) 1 Shear force on a fat solid surface: 퐹shear, direction 푖= ∬ 푆 휏direction 푖d푆 (7) Shear in the direction 푗, on a plane perpendicular to direction 푖: \|\|⃗ 휏푖푗\|\| = 휇휕푉푗 휕푖 (8) Continuity equation for incompressible fow: ⃗ ∇⋅⃗ 푉= 0 (9) Navier-Stokes equation for incompressible fow: 휌D ⃗ 푉 D푡 = 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (10) In a highly-viscous (creeping) steady fow, the drag 퐹퐷exerted on a spherical body of diameter 퐷at by fow at velocity 푉∞is quantifed as: 퐹퐷sphere = 3휋휇푉∞퐷 (11) In cylindrical pipe fow, we assume the fow is always laminar for [Re]퐷≲2 300, and always turbulent for [Re]퐷≳4 000. The Darcy friction factor 푓is defned as: 푓≡ \|Δ푝loss\| 퐿 퐷 1 2휌푉2 av. (12) The loss coefcient 퐾퐿is defned as: 퐾퐿≡\|Δ푝loss\| 1 2휌푉2 av. (13) Viscosities of various fuids are given in fg. 1 p. 4. Pressure losses in cylindrical pipes can be calculated with the help of the Moody diagram presented in fg. 2 p.5. The non-dimensional incompressible Navier-Stokes equation: [St] 휕⃗ 푉∗ 휕푡∗+ ⃗ 푉∗⋅⃗ ∇∗⃗ 푉∗= 1 [Fr]2 ⃗ 푔∗−[Eu] ⃗ ∇∗푝∗+ 1 [Re] ⃗ ∇∗2 ⃗ 푉∗ (14) in which [St] ≡ 푓퐿 푉, [Eu] ≡ 푝0−푝∞ 휌푉2 , [Fr] ≡ 푉 √푔퐿and [Re] ≡ 휌푉퐿 휇 . 2 The force coefcient 퐶퐹and power coefcient 퐶P are defned as: 퐶퐹≡ 퐹 1 2휌푆푉2 퐶P ≡ ̇ 푊 1 2휌푆푉3 (15) The speed of sound 푐in air is modeled as: 푐= √ 훾푅푇 (16) In boundary layer fow, we assume that transition occurs at [Re]푥≈5 ⋅105. The wall shear coefcient 푐푓, a function of distance 푥, is defned using the free-stream fow velocity 푈: 푐푓(푥) ≡ 휏wall 1 2휌푈2 (17) Exact solutions to the laminar boundary layer along a smooth surface yield: 훿 푥= 4,91 √ [Re]푥 훿∗ 푥 = 1,72 √ [Re]푥 (18) 훿∗∗ 푥 = 0,664 √ [Re]푥 푐푓(푥) = 0,664 √ [Re]푥 (19) Solutions to the turbulent boundary layer along a smooth surface yield the following time-averaged characteristics: 훿 푥≈ 0,16 [Re] 1 7 푥 훿∗ 푥 ≈ 0,02 [Re] 1 7 푥 (20) 훿∗∗ 푥 ≈0,016 [Re] 1 7 푥 푐푓(푥) ≈0,027 [Re] 1 7 푥 (21) 3 −20 0 20 40 60 80 100 120 Temperature 푇in degree Celsius (◦C) 10−4 10−3 10−2 2×10−4 3×10−4 4×10−4 5×10−4 6×10−4 7×10−4 8×10−4 9×10−4 2×10−3 3×10−3 4×10−3 5×10−3 6×10−3 7×10−3 8×10−3 9×10−3 2×10−2 Viscosity 휇of liquids in Pa s ⟵Water ⟵Crude Oil 10−5 1.2×10−5 1.4×10−5 1.6×10−5 1.8×10−5 2×10−5 2.2×10−5 2.4×10−5 Viscosity 휇of gases in Pa s Air ⟶ CO2 ⟶ Figure 1 – The viscosity of four fuids (crude oil, water, air, and C02) as a function of temperature. The scale for liquids is logarithmic and displayed on the left; the scale for gases is linear and displayed on the right. Figure CC-by by Arjun Neyyathala & Olivier Cleynen 4 Figure 2 – A Moody diagram, which presents values for 푓measured experimentally, as a function of the diameter-based Reynolds number [Re]퐷, for diferent relative roughness values. Diagram CC-by-sa S Beck and R Collins, University of Shefeld 5 Solve problem 1, and three other problems among problems 2 to 6. The following marking guidelines will be used: • Answers to questions starting with “show that” should be fully-developed and continuous; • In all other questions, the correct result with the correct unit is enough to obtain full points; • Illegible or ambiguous answers are always discarded. 6 1 Navier-Stokes equation 1.1. [5 pts] Write out equation (10), the Navier-Stokes equation for incompressible fow, in its fully-developed form in three Cartesian coordinates. 1.2. [5 pts] In which fow conditions does this equation apply? 2 Pressure forces on the panels of a barge A large barge is being built with the dimensions shown in fgure 3. Once completed, it will be foated in a lake. The bottom panel of the barge will then sit horizontally, 2 m below the surface of the water. Figure 3 – Basic layout of a barge 2.1. [15 pts] What is the magnitude of the force resulting from pressure eforts on each of the panels labeled A, B and C? 2.2. [5 pts] How would the force on panel C change if the angle 휃was increased? (briefy justify your answer, e.g. in 30 words or less) 2.3. [10 pts] What is the weight of the barge? 7 3 Governing equations We consider a two-dimensional fuid fow described with the following velocity feld, described in Cartesian coordinates 푥and 푦in m s−1: ⃗ 푉= (2퐴푥−퐶)⃗ 푖+ (−2퐴푦−퐵푥)⃗ 푗 where 퐴, 퐵, and 퐶are all constants. 3.1. [5 pts] Show that this fow satisfes the continuity equation for incompressible fow (equation 9). 3.2. [10 pts] What is the acceleration feld corresponding to this fow? 3.3. [5 pts] What is the value of acceleration at a point of coordinates (3; 3)? 3.4. [10 pts] Does a function exist to describe the pressure feld of this fow, and if so, what is it? 8 4 Pipe installation with a pump-turbine Engineers in a chemical company would like to pump 50 L s−1 of water at 20 ◦C from reservoir A to reservoir B in the installation drawn below in fgure 4. The network is built with a cylindrical pipe made out of concrete (surface roughness 휖= 0,25 mm), with a diameter of 12 cm. The Y-junction induces a loss coefcient of 0,2. Figure 4 – Basic layout of the piping network The installation lengths are as follows: 퐿1 = 15 m 퐿5 = 16 m 퐿2 = 12 m 퐿6 = 6 m 퐿3 = 8 m 퐿7 = 5 m 퐿4 = 14 m 4.1. [10 pts] What is the hydraulic power that the pump must provide, in order to deliver the required volume fow? 4.2. [10 pts] On a diagram, represent qualitatively (i.e. without numerical data) the pressure distribution along the length of the pipe, indicating both the top and the bottom paths followed by the water. 4.3. [10 pts] If the water was to be transferred back from reservoir B to reservoir A, the pump/turbine device would be operated as a turbine. In that case, what would be the hydraulic power available to the turbine? 9 5 Friction on a fat plate A laboratory is developing a special oil with Newtonian fuid characteristics. An experi-ment is set up to measure its viscosity. For this, a thin layer of oil is poured at the bottom of a tank with rectangular walls. A fat plate is moved perfectly horizontally, sitting fush with the surface of the oil. A force measurement is carried out. The plate has width 푊= 40 cm and length 퐿= 60 cm. It is moved at 15 cm s−1 at the surface of the oil, which is at a distance 퐷= 4 mm above the bottom of the tank. Figure 5 – Flat plate moved horizontally at the surface of a layer of oil, at the bottom of a tank The fow below the plate is smooth, steady, and laminar, so that the velocity distribution in the oil between the plate and the bottom of the tank is entirely uniform. 5.1. [10 pts] What is the relationship between the viscosity of the oil and the drag force due to shear on the bottom side of the plate? The magnitude of the drag force resulting from the shear exerted by the oil is measured as 퐹drag = 0,47 N. 5.2. [5 pts] What is the viscosity of the oil? 10 On the top surface of the plate, in the air, there is enough space for a boundary layer to develop. The scientists in the laboratory would like to check that the drag force generated by this boundary layer does not infuence the measurement. The air in the room has properties 1 bar and 20 ◦C. 5.3. [10 pts] Starting with equation 19 for the friction coefcient in a laminar boundary layer: 푐푓(푥) = 0,664 √ [Re]푥 (19) show that the relationship between the shear force on the top side of the plate and the properties of the air is: 퐹top = 0,664푊( 푝air휇air퐿 푅푇air ) 1 2 푉 3 2 plate (22) 5.4. [5 pts] What is the magnitude of the force exerted by the air on the top surface of the plate? 11 6 Pickup truck with snow plow A pickup truck is equipped with a snow plow blade. It travels steadily at 30 km h−1 in 20 cm of snow with density 휌snow ground = 400 kg m−3, with the blade angled at 훼= 60°. The snow is defected along the blade, compacted by the movement, and it exits with density 휌snow compacted = 600 kg m−3 at velocity ⃗ 푉snow/truck relative to the truck. The cross-sectional area of the rejected snow is 퐴outlet = 0,55 m2, measured in a plane perpendicular to ⃗ 푉snow/truck. Figure 6 – A pickup truck using a large blade to clear snow Pickup drawing CC-0 by en:Wikipedia User:Wikideas1; diagram CC-0 Olivier Cleynen 6.1. [15 pts] What is the force exerted on the blade by the snow? 6.2. [5 pts] What is the power required for the truck to plow the snow? 6.3. [10 pts] How would the power change if the angle 훼was reduced? (briefy justify your answer, e.g. in 30 words or less) 12 Solution: Fluid Dynamics examination — Sept. 21, 2020 Fluid Dynamics for Engineers by Olivier Cleynen 1 Governing equation 1.1 N-S equation question 휌[ 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦+ 푤휕푢 휕푧] = 휌푔푥−휕푝 휕푥+ 휇[ 휕2푢 (휕푥)2 + 휕2푢 (휕푦)2 + 휕2푢 (휕푧)2] (1) 휌[ 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦+ 푤휕푣 휕푧] = 휌푔푦−휕푝 휕푦+ 휇[ 휕2푣 (휕푥)2 + 휕2푣 (휕푦)2 + 휕2푣 (휕푧)2] (2) 휌[ 휕푤 휕푡+ 푢휕푤 휕푥+ 푣휕푤 휕푦+ 푤휕푤 휕푧] = 휌푔푧−휕푝 휕푧+ 휇[ 휕2푤 (휕푥)2 + 휕2푤 (휕푦)2 + 휕2푤 (휕푧)2] (3) 1.2 Conditions for this equation This equation applies to all incompressible fows of a Newtonian fuid. 1 2 Pressure forces on the panels of a barge 2.1 Magnitude of the forces We defne coordinates 푟, 푟퐶, and 푧, as well as the lengths 퐿퐴, 퐿퐵and 퐿퐶, as shown in the fgure below. On each wall, the net pressure applying is that due to water and air from both sides, 푝net = 푝water + 푝atm. −푝atm. (4) = 휌푔푧 (5) Wall A The force on a small horizontal strip of panel with height d푟and width 퐿A is d퐹A. On the complete door, the force applying due to the net pressure 푝net of water and air is: 퐹netA = ∫ 푟=푅max 푟=0 d퐹A (6) = ∫ 푟=푅max 푟=0 푝net d푆 (7) = ∫ 푟=푅max 푟=0 푝net퐿A d푟 (8) = ∫ 푟=푅max 푟=0 휌푔푧퐿A d푟 (9) 2 A coordinate transform is needed to solve the integral, expressing depth 푧as a function of 푟. This is obtained by geometry: 푧= 푍max −푟 (10) Inserting eq. 10 into eq. 9, we continue with: 퐹net A = ∫ 푟=푅max 푟=0 휌푔(푍max −푟) 퐿A d푟 (11) = 휌푔퐿A ∫ 푟=푅max 푟=0 (푍max −푟) d푟 (12) = 휌푔퐿A [푍max푟−1 2푟2] 푟=푅max 푟=0 (13) = 휌푔퐿A (푍max푅max −0,5푅max 2) (14) = 103 × 9,81 × 20 × (2 × 2 −0,5 × 22) (15) = 3,924 ⋅105 N (16) 퐹net A = 392,4 kN (17) Wall B 퐹net B is found with a similar calculation. This time, 퐿B is a function of 푟, which can be found by trigonometry: 퐿B = 푟 tan 휃 (18) Picking up eq. 9 applied to panel B, and inserting eqs. 10 and 18, we get: 퐹net B = ∫ 푟=푅max 푟=0 휌푔(푍max −푟) 푟 tan 휃d푟 (19) = 휌푔 1 tan 휃∫ 푟=푅max 푟=0 (푍max푟−푟2) d푟 (20) = 휌푔 1 tan 휃[ 1 2푍max푟2 −1 3푟3] 푟=푅max 푟=0 (21) = 휌푔 1 tan 휃( 1 2푍max푅2 max −1 3푅max 3) (22) = 103 × 9,81 × 1 tan 30° × (0,5 × 2 × 22 −23 3 ) (23) = 2,266 ⋅104 N (24) 퐹net B = 22,66 kN (25) 3 Wall C 퐹net C is found with a similar calculation. This time, the relationship between 푟C and 푧is a little more complicated, and found by trigonometry: 푧= 푍max −푟C sin 휃 (26) Picking up eq. 9 applied to panel C and inserting eqs. 10 and 26, we get: 퐹netC = ∫ 푟C=푅Cmax 푟C=0 휌푔(푍max −푟C sin 휃) 퐿C d푟C (27) = 휌푔퐿C ∫ 푟C=푅Cmax 푟C=0 (푍max −푟C sin 휃) d푟C (28) = 휌푔퐿C [푍max푟C −1 2 sin 휃푟2 C] 푟C=푅Cmax 푟C=0 (29) = 휌푔퐿C (푍max푅Cmax −1 2 sin 휃푅2 Cmax) (30) = 휌푔퐿C (푍max 푍max sin 휃−1 2 sin 휃푍2 max (sin 휃)2) (31) = 휌푔퐿C 1 sin 휃(푍2 max −1 2푍2 max) (32) = 휌푔퐿C 1 sin 휃 1 2푍2 max (33) = 103 × 9,81 × 6 × 1 sin 30° × 0,5 × 22 (34) = 2,3544 ⋅105 N (35) 퐹net C = 235,4 kN (36) 2.2 Change in force with angle If the angle 휃was increased, the term 1/ sin 휃in eq. 33 would decrease, and so the calculated force would decrease. However, the barge would also sink in more deeply (because the immersed volume, which is responsible for buoyancy, would have to remain constant). This would result in an increase in 푍2 max, likely smaller than the decrease caused by the term 1/ sin 휃. 2.3 Weight of the barge The weight of the barge is equal to the net force due to pressure on all panels. This can be calculated either one of two ways: • Using force components. The weight is equal to the force on the bottom panel, 퐹D = 휌푔푍max퐿C퐿A, plus the vertical component of the force on panel C, 퐹Cvertical = 4 퐹netC cos 휃. The sum is: 퐹weight = 퐹D + 퐹Cvertical (37) = 휌푔푍max퐿C퐿A + 퐹netC cos 휃 (38) = 2,558 ⋅106 N (39) 퐹weight = 2,558 MN (40) • Using the immersed volume and multiplying it by the density of water and gravity to obtain the buoyancy: 퐹weight = immersed휌푔 (41) = 휌푔(퐿A + 1 2퐿Bmax) 퐿C푍max (42) = 휌푔(퐿A + 1 2 푍max tan 휃) 퐿C푍max (43) = 103 × 9,81 × (20 + 1 2 2 tan 30°) × 6 × 2 (44) = 2,558 ⋅106 N (45) 퐹weight = 2,558 MN (46) The mass of the barge is approximately 260 tons. 5 3 Governing equations 3.1 Continuity equation The incompressible continuity equation (eq. 9 in the formula sheet) translates in two dimensions as: 휕푢 휕푥+ 휕푣 휕푦= 0 (47) In this case, we have 휕푢 휕푥+ 휕푣 휕푦= 휕(2퐴푥−퐶) 휕푥 + 휕(−2퐴푦−퐵푥) 휕푦 (48) = 2퐴−2퐴 (49) = 0 s−1 (50) So the incompressible continuity equation is satisfed, and the fow is incompressible. 3.2 Acceleration feld The acceleration feld ⃗ 푎(in m s−2) is: ⃗ 푎= D ⃗ 푉 D푡 (51) = 휕⃗ 푉 휕푡+ ( ⃗ 푉⋅⃗ ∇) ⃗ 푉 (52) The 푥-component of ⃗ 푎is 푎푥= 휕푢 휕푡+ 푢휕푢 휕푥+ 푣휕푢 휕푦 (53) = 휕(2퐴푥−퐶) 휕푡 + (2퐴푥−퐶)휕(2퐴푥−퐶) 휕푥 + (−2퐴푦−퐵푥)휕(2퐴푥−퐶) 휕푦 (54) = 0 + 2퐴(2퐴푥−퐶) + 0 (55) 푎푥= 4퐴2푥−2퐴퐶 (56) 6 The 푦-component of ⃗ 푎is 푎푦= 휕푣 휕푡+ 푢휕푣 휕푥+ 푣휕푣 휕푦 (57) = 휕(−2퐴푦−퐵푥) 휕푡 + (2퐴푥−퐶)휕(−2퐴푦−퐵푥) 휕푥 + (−2퐴푦−퐵푥)휕(−2퐴푦−퐵푥) 휕푦 (58) = 0 −퐵(2퐴푥−퐶) −2퐴(−2퐴푦−퐵푥) (59) = −2퐴퐵푥+ 퐵퐶+ 4퐴2푦+ 2퐴퐵푥 (60) 푎푦= 4퐴2푦+ 퐵퐶 (61) So, the acceleration feld is, in m s−2: ⃗ 푎= ( 4퐴2푥−2퐴퐶 4퐴2푦+ 퐵퐶) (62) 3.3 Acceleration at one point At point with 푥= 3 and 푦= 3, we have ⃗ 푎point = ( 12퐴2 −2퐴퐶 12퐴2 + 퐵퐶) m s−2 (63) 3.4 Function for pressure We use the incompressible Navier-Stokes equation (eq. 10 of the equation sheet) to express the acceleration feld as a function of the gravity, pressure and shear terms: 휌D ⃗ 푉 D푡 = 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (64) 휌⃗ 푎= 휌⃗ 푔−⃗ ∇푝+ 휇⃗ ∇2 ⃗ 푉 (65) Splitting that last equation into two components, we get, for the 푥-component: 휌푎푥= 휌푔푥−휕푝 휕푥+ 휇휕2푢 (휕푥)2 (66) 휌(4퐴2푥−2퐴퐶) = 휌푔푥−휕푝 휕푥+ 휇휕2(2퐴푥−퐶) (휕푥)2 (67) 휌(4퐴2푥−2퐴퐶) = 휌푔푥−휕푝 휕푥+ 0 (68) 휕푝 휕푥= 휌(−4퐴2푥+ 2퐴퐶+ 푔푥) (69) 7 Likewise in the 푦-direction, we have 휌푎푦= 휌푔푦−휕푝 휕푦+ 휇휕2푣 (휕푦)2 (70) 휌(4퐴2푦+ 퐵퐶) = 휌푔푦−휕푝 휕푦+ 휇휕2(−2퐴푦−퐵푥) (휕푦)2 (71) 휌(4퐴2푦+ 퐵퐶) = 휌푔푦−휕푝 휕푦+ 0 (72) 휕푝 휕푦= 휌(−4퐴2푦−퐵퐶+ 푔푦) (73) We have: 휕( 휕푝 휕푥) 휕푦 = 0 = 휕( 휕푝 휕푦) 휕푥 (74) which guarantees that there exists a function to describe 푝. We proceed to integrate eq. 69 with respect to 푦: 푝= 휌[−2퐴2푥2 + (2퐴퐶+ 푔푥) 푥] + 푓(푦,푡) (75) with 푓a function of 푦and 푡only. To obtain function 푓, we derivate this equation 75 with respect to 푦, and compare with eq. 73: 휕푝 휕푦= d (휌[−2퐴2푥2 + (2퐴퐶+ 푔푥) 푥]) d푦 + 푓′ (푦,푡) = 휌(−4퐴2푦−퐵퐶+ 푔푦) (76) 0 + 푓′ (푦,푡) = 휌(−4퐴2푦−퐵퐶+ 푔푦) (77) We can now integrate this equation 77 with respect to 푦to fnd the function 푓: 푓(푦,푡) = 휌[−2퐴2푦2 + (−퐵퐶+ 푔푦) 푦] + 푖(푡) + 푝0 (78) with 푖 a function of 푡only, and 푝0 an arbitrary integration constant. Inserting this equation 78 into eq. 75, we fnally get the function for pressure (in Pa) corresponding to the given velocity feld: 푝= 휌[−2퐴2푥2 + (2퐴퐶+ 푔푥) 푥] + 휌[−2퐴2푦2 + (−퐵퐶+ 푔푦) 푦] + 푖(푡) + 푝0 (79) 푝= 휌[−2퐴2푥2 −2퐴2푦2 + (2퐴퐶+ 푔푥) 푥+ (−퐵퐶+ 푔푦) 푦] + 푖(푡) + 푝0 (80) All functions 푖of time, and any initial value 푝0, may be inserted in equation 80, and it will still satisfy the incompressible Navier-Stokes equation. 8 4 Pipe installation with a pump-turbine 4.1 Hydraulic power of pump We want to calculate four pressure diferences: • Pressure drop due to wall friction losses along the horizontal pipe, Δ푝푓1: The average velocity in the pipe is 푉av.1 = ̇  푆 (81) = ̇  휋퐷2 4 (82) = 0,05 휋× 0,122 4 (83) 푉av.1 = 4,42 m s−1 (84) The Reynolds number is [Re]퐷1 = 휌푉av.1퐷 휇 (85) = 103 × 4,42 × 0,12 10−3 (86) [Re]퐷1 = 5,305 ⋅105 (87) The relative roughness is 휖 퐷= 0,25 ⋅10−3 0,12 (88) 휖 퐷= 2,08 ⋅10−3 (89) With those values, the Moody diagram reads: 푓1 = 0,0241 (90) Finally, the wall friction losses along the pipe are calculated as: Δ푝푓1 = −푓1 1 2휌푉2 av. 퐿1 퐷 (91) = −0,0241 × 0,5 × 103 × 4,422 × 15 0,12 (92) Δ푝푓1 = −2,944 ⋅104 Pa (93) = −0,29 bar (94) 9 • Pressure drop due to wall friction losses in each of the two branches, Δ푝푓2: The average velocity in the pipe is 푉av.2 = 푉av.1 2 (95) 푉av.2 = 2,21 m s−1 (96) The Reynolds number is [Re]퐷2 = [Re]퐷1 2 (97) [Re]퐷2 = 2,653 ⋅105 (98) With the same relative roughness, the Moody diagram reads: 푓2 = 0,025 (99) Finally, the wall friction losses along the pipe are calculated as: Δ푝푓2 = −푓2 1 2휌푉2 av.2 퐿2 퐷 (100) = −0,025 × 0,5 × 103 × 2,212 × 12 0,12 (101) Δ푝푓2 = −6,108 ⋅103 Pa (102) = −0,06 bar (103) • Pressure drop due to losses in the junction, Δ푝junction: Δ푝junction = 퐾퐿 1 2휌푉2 av.1 (104) = 0,2 × 1 2 × 103 × 4,4212 (105) Δ푝junction = −1,954 ⋅103 Pa (106) = −0,02 bar (107) • Pressure drop due to hydrostatic pressure change across the pump, Δ푝ℎ: Δ푝ℎ= 휌푔(퐿5 + 퐿6 −퐿3) (108) = 103 × 9,81 × (16 + 6 −8) (109) Δ푝ℎ= +1,373 ⋅105 Pa (110) = +1,37 bar (111) 10 Finally, the pump hydraulic power is obtained as: ̇ 푊pump = ̇ (Δ푝pump) (112) = ̇ (Δ푝ℎ−Δ푝푓1 −Δ푝푓2 −Δ푝junction) (113) = 0,05 × [+1,373 ⋅105 −(−2,944 ⋅104) −(−6,108 ⋅103) −(−1,954 ⋅103)] (114) = 0,05 × [+1,373 + 0,2944 + 0,0611 + 0,01954] × 105 (115) = +8,742 ⋅103 W (116) ̇ 푊pump = +8,74 kW (117) 4.2 Pressure distribution 4.3 Hydraulic power of turbine If the fow runs from B to A, the magnitude of the pressure diferences remains the same. The hydrostatic pressure diference changes sign, but the pressure losses due to friction do not: ̇ 푊turbine = ̇ (Δ푝turbine) (118) = ̇ (Δ푝ℎturbine −Δ푝푓1 −Δ푝푓2 −Δ푝junction) (119) = ̇ (−Δ푝ℎpump −Δ푝푓1 −Δ푝푓2 −Δ푝junction) (120) = 0,05 × [−1,373 + 0,2944 + 0,0611 + 0,01954] × 105 (121) = −4,992 ⋅103 W (122) ̇ 푊turbine = −4,99 kW (123) So, pumping the fow from A to B costs 8,7 kW of power, but turbining the fow back down from B to A only generates 5 kW of power. 11 5 Friction on a fat plate The fow can be represented as shown below: 5.1 Shear force on bottom side In order to calculate shear on the bottom face, we express the horizontal fuid velocity below the plate (relative to the plate) in a very simple velocity feld: 푉푥= 푢= 0 + 푘푦 (124) The constant 푘is found using boundary conditions: at a distance 퐷away from the plate, the velocity of the fuid relative to the plate is 푉plate: 푢\|@ y = D = 푉plate = 0 + 푘퐷 (125) 푘= 푉plate 퐷 (126) So the velocity feld becomes: 푢= 1 퐷푉plate푦 (127) 12 Now, the shear on the bottom plate is found through integration, starting with equation 7 from the formula sheet: 퐹bottom = ∫ d퐹= ∬ 푆 휏plate d푆 (128) = ∬휏plate d푆 (129) = ∫ 푥=퐿 푥=0 휏plate푊d푥 (130) Expressing the shear 휏plate as a function of the velocity feld in a Newtonian fuid, we continue to get: 퐹bottom = ∫ 푥=퐿 푥=0 휇oil 휕푢 휕푦 \| \| \| \|@푦=0 푊d푥 (131) = 휇oil푊∫ 푥=퐿 푥=0 휕( 1 퐷푉plate푦) 휕푦 \| \| \| \| \|@푦=0 d푥 (132) = 휇oil푊∫ 푥=퐿 푥=0 ( 1 퐷푉plate)@푦=0 d푥 (133) 퐹bottom = 휇oil푊1 퐷푉plate퐿 (134) Rearranging this equation to solve for 휇, we get: 휇oil = 퐹bottom 퐷 푊푉plate퐿 (135) 5.2 Viscosity of oil The viscosity of the oil is obtained by plugging in values in eq. 135: 휇oil = 0,47 × 4 ⋅10−3 0,4 × 0,15 × 0,6 (136) 휇oil = 5,22 ⋅10−2 Pa s (137) 5.3 Shear force on top side A boundary layer develops on the top side, in air. That boundary layer would transit at [Re]푥≈5 ⋅105. Solving for 푥tr. with 푈(the faraway velocity) being equal to 푉plate, and 13 reading the value 휇air = 1,85 ⋅10−5 Pa s in fgure 1, we have: [Re]푥tr. = 휌air푈푥tr. 휇air (138) 푥tr. = [Re]푥tr.휇air 휌air푈 (139) = 푝air[Re]푥tr.휇air 푅푇air푈 (140) = 1 ⋅105 × 5 ⋅105 × 1,85 ⋅10−5 287 × (273,15 + 20) × 0,15 (141) 푥tr. = 73,3 m (142) Since 푥tr. > 퐿, the boundary layer never transits and remains completely laminar. To fnd an expression for 휏, we start with the given equation and implement the def-nition 15 of the formula sheet, as well as the defnition of the distance-based Reynolds number [Re]푥: 푐푓(푥) = 0,664 √ [Re]푥 (143) 휏wall, laminar 1 2휌air푈2 = 0,664 √ 휌air푈푥 휇air (144) 휏wall, laminar = 0,664 ( 휌air푈푥 휇air ) −1 2 1 2휌air푈2 (145) 휏wall, laminar = 0,332√휌air휇air 푈1,5푥−1 2 (146) The shear force on the top surface is the integral of the shear with respect to area: 퐹top = ∫휏wall d푆 (147) We split the total area covered by the laminar boundary layer in strips of width 푊and length d푥, with 푥ranging from 0 (leading edge) to 퐿(trailing edge), obtaining: 퐹top = ∫ 푥=퐿 푥=0 휏wall푊d푥 (148) = ∫ 푥=퐿 푥=0 0,332√휌air휇air 푈1,5푥−1 2푊d푥 (149) = 0,332√휌air휇air 푈1,5푊∫ 퐿 0 푥−0,5 d푥 (150) = 0,332√휌air휇air 푈1,5푊[ 1 −0,5 + 1푥−0,5+1 ] 퐿 0 (151) = 0,332√휌air휇air 푈1,5푊2 퐿0,5 (152) 14 = 0,664√휌air휇air 푈 3 2푊퐿 1 2 (153) = 0,664푊(휌air휇air퐿) 1 2 푈 3 2 (154) = 0,664푊( 푝air휇air퐿 푅푇air ) 1 2 푈 3 2 (155) 퐹top = 0,664푊( 푝air휇air퐿 푅푇air ) 1 2 푉 3 2 plate (156) 5.4 Force exerted by the air We simply insert values into eq. 156: 퐹top = 0,664 × 0,4 × ( 1 ⋅105 × 1,85 ⋅10−5 × 0,6 287 × (273,15 + 20) ) 0,5 × 0,151,5 (157) 퐹top = 5,6 ⋅10−5 N (158) This force is is very much smaller than the force generated by the oil, so the air’s infuence in the viscosity measurement experiment is indeed very small. 15 6 Pickup truck with snow plow We defne a control volume moving together with the truck, as shown below. The snow enters with velocity ⃗ 푉1 and leaves with velocity ⃗ 푉2 = ⃗ 푉out. 6.1 Force on blade We want expressions for the mass fow ̇ 푚, as well as for the two vectors ⃗ 푉1 and ⃗ 푉2. The mass fow is found using the inlet conditions: ̇ 푚= 휌1푉1퐴⟂1 (159) = 휌1푉1푊sweepℎsnow (160) = 400 × 30 3,6 × 3,5 × 0,2 (161) ̇ 푚= 2,333 ⋅103 kg s−1 (162) The outlet velocity can then be obtained: ̇ 푚= 휌2푉2퐴⟂2 (163) ̇ 푚= 휌2푉2퐴out (164) 푉2 = ̇ 푚 휌2퐴out (165) = 2,333 ⋅103 600 × 0,55 (166) 푉2 = 7,07 m s−1 (167) 16 The velocity vector ⃗ 푉2 is then expressed according to its two components: 푉2푥= +푉2 sin 훼 (168) = 7,07 × sin(60°) (169) 푉2푥= +6,12 m s−1 (170) 푉2푧= +푉2 cos 훼 (171) = 7,07 × cos(60°) (172) 푉2푧= +3,54 m s−1 (173) So, we have all the necessary information about the fow: ̇ 푚= 2,333 ⋅103 kg s−1 (174) ⃗ 푉1 = ⎛ ⎜ ⎜ ⎜ ⎝ 0 0 +8,33 ⎞ ⎟ ⎟ ⎟ ⎠ m s−1 (175) ⃗ 푉2 = ⎛ ⎜ ⎜ ⎜ ⎝ +6,12 0 +3,54 ⎞ ⎟ ⎟ ⎟ ⎠ m s−1 (176) Now, applying eq. 2 from the equation sheet, we can express the net force ⃗ 퐹net on the snow: ⃗ 퐹net = Σ [휌푉⟂퐴⃗ 푉]incoming + Σ [휌푉⟂퐴⃗ 푉]outgoing (177) = ̇ 푚( ⃗ 푉2 −⃗ 푉1) (178) = ̇ 푚 ⎛ ⎜ ⎜ ⎜ ⎝ 푉2푥−푉1푥 푉2푦−푉1푦 푉2푧−푉1푧 ⎞ ⎟ ⎟ ⎟ ⎠ (179) = 2,333 ⋅103 ⎛ ⎜ ⎜ ⎜ ⎝ (+6,12) −0 0 −0 (+3,54) −(+8,33) ⎞ ⎟ ⎟ ⎟ ⎠ (180) 17 = ⎛ ⎜ ⎜ ⎜ ⎝ +1,429 ⋅104 0 −1,12 ⋅104 ⎞ ⎟ ⎟ ⎟ ⎠ N (181) ⃗ 퐹net = ⎛ ⎜ ⎜ ⎜ ⎝ +14,3 0 −11,2 ⎞ ⎟ ⎟ ⎟ ⎠ kN (182) The force exerted on the blade by the snow is exactly the opposite, i.e. ⃗ 퐹snow on blade = −⃗ 퐹net = ⎛ ⎜ ⎜ ⎜ ⎝ −14,3 0 +11,2 ⎞ ⎟ ⎟ ⎟ ⎠ kN (183) 6.2 Power required of truck The power ̇ 푊truck provided by the truck to compensate for the force is ̇ 푊truck = ⃗ 퐹net ⋅⃗ 푉truck on ground (184) = −푉truck퐹net 푧 (185) = −8,333 × (−1,12 ⋅104) (186) = 9,329 ⋅104 W (187) ̇ 푊truck = 93,3 kW (188) (approximately 120 horsepower) 6.3 Power change with angle The power is quantifed using equation 185, in which we include eq. 179, eq. 171, and then eq. 160: ̇ 푊truck = −푉truck퐹net 푧 (189) = −푉truck ̇ 푚(푉2푧−푉1푧) (190) = −푉truck ̇ 푚(푉2 cos 훼−푉1푧) (191) = −푉truck ̇ 푚( ̇ 푚 휌2퐴out cos 훼−푉truck) (192) 18 = −푉truck ̇ 푚( 휌1푉1푊sweepℎsnow 휌2퐴out cos 훼−푉truck) (193) = −푉truck휌푉truckℎsnow푊sweep ( 휌1푉truck푊sweepℎsnow 휌2퐴out cos 훼−푉truck) (194) = −푉3 truckℎsnow푊sweep ( 휌1푊sweepℎsnow 휌2퐴out cos 훼−1) (195) In equation 195, decreasing 훼would: • increase the term cos 훼; • decrease the efective width 푊sweep (unless a new, larger blade is used). Both factors tend to decrease the magnitude of ̇ 푊truck. Therefore, the power would decrease, likely faster than 훼. 19 A8 References These notes are based on textbooks by White , Çengel & al., Munson & al., and de Nevers . Ira Herbert Abbott and Albert Edward Von Doenhof. Theory of wing sections, including a summary of airfoil data. Courier, 1959. George Keith Batchelor. An introduction to fuid dynamics. Cambridge University Press, 1967. isbn: 9780521663960. Geofrey Ingram Taylor. “Film notes for low-Reynolds-number fows”. In: National Committee for Fluid Mechanics Films. Illustrated experi-ments in fuid mechanics: the NCFMF book of flm notes. MIT Press, 1967. isbn: 9780262640121. url: John Leask Lumley. “Film notes for Eulerian and Lagrangian descrip-tions in fuid mechanics”. In: National Committee for Fluid Mechanics Films. Illustrated experiments in fuid mechanics: the NCFMF book of flm notes. MIT Press, 1969. isbn: 9780262640121. url: edu/hml/ncfmf/01ELDFM.pdf. Hendrik Tennekes and John Leask Lumley. A frst course in turbulence. MIT press, 1972. isbn: 978-0262200196. Bernard Stanford Massey. Mechanics of fuids. 5th ed. Van Nostrand Reinhold, 1983. isbn: 0442305524. Phillip M. Gerhart and Richard J. Gross. Fundamentals of fuid dynamics. 1st ed. Addison-Wesley, 1985. isbn: 0201114100. Hendrik Tennekes. The simple science of fight. From insects to jumbo jets. 1st ed. MIT Press, 1992. isbn: 0262201054. John David Anderson. Computational fuid dynamics. McGraw-Hill, 1995. isbn: 0071132104. Paul A. Libby. An introduction to turbulence. CRC Press, 1996. isbn: 9781560321002. Jewel B. Barlow, William H. Jr. Rae, and Alan Pope. Low-speed wind tunnel testing. 3rd ed. Wiley & Sons, 1999. isbn: 0471557749. Jean Mathieu and Julian Scott. An introduction to turbulent fow. Cam-bridge University Press, 2000. isbn: 0521775388. Stephen B. Pope. Turbulent fows. Cambridge University Press, 2000. isbn: 9780521177849. Peter S. Bernard and James M. Wallace. Turbulent fow: analysis, mea-surement, and prediction. John Wiley & Sons, 2002. isbn: 9780471332190. John Bird. Higher engineering mathematics. 4th ed. Newnes, 2004. isbn: 9780750662666. Tuncer Cebeci. Analysis of turbulent fows. 2nd ed. Elsevier, 2004. isbn: 9780080443508. Noel de Nevers. Fluid mechanics for chemical engineers. 3rd ed. McGraw-Hill, 2004. isbn: 9781259002380. Pierre Sagaut. Large eddy simulation for incompressible fows: an in-troduction. Ed. by Charles Meneveau. 3rd ed. Springer, 2006. isbn: 9783540263449. 292 James O. Wilkes. Fluid mechanics for chemical engineers. with microfu-idics and CFD. 2nd ed. with contributions by S. G. Birmigham, B. J. Kirby, COMSOL (FEMLAB), and C-Y. Cheng. Prentice Hall, 2006. isbn: 9780131482128. Henk Kaarle Versteeg and Weeratunge Malalasekera. An introduction to computational fuid dynamics: the fnite volume method. 2nd ed. Pearson Education, 2007. isbn: 9780131274983. Ascher Shapiro and the Massachusetts Institute of Technology. Na-tional Committee for Fluid Mechanics Films. 2008. url: edu/hml/ncfmf.html (visited on 09/23/2017). Frank M. White. Fluid mechanics. 7th ed. McGraw-Hill, 2008. isbn: 9780071311212. Hendrik Tennekes. The simple science of fight. From insects to jumbo jets. 2nd ed. MIT Press, 2009. isbn: 9780262513135. J.J Bertin and R.M. Cummings. Aerodynamics for engineers. 5th ed. Pearson/Prentice Hall, 2010. isbn: 978-0132272681. Yunus A. Çengel and John M. Cimbala. Fluid mechanics. Fundamentals and applications. 2nd ed. McGraw-Hill, 2010. isbn: 9780070700345. Philipp Schlatter, Mattias Chevalier, Miloš Ilak, and Dan S. Henning-son. The structure of a turbulent boundary layer studied by numerical simulation. 2010. arXiv: 1010.4000 [physics.flu-dyn]. Philipp Schlatter and Ramis Örlü. “Assessment of direct numerical simulation data of turbulent boundary layers”. In: Journal of Fluid Mechanics 659 (2010), p. 116. doi: 10.1017/S0022112010003113. J. H. Lee, Y. S. Kwon, N. Hutchins, and J. P. Monty. Spatially devel-oping turbulent boundary layer on a fat plate. 2012. arXiv: 1210.3881 [physics.flu-dyn]. Bruce R. Munson, Theodore H. Okiishi, Wade W. Huebsch, and Alric P. Rothmayer. Fluid mechanics. 7th ed. Wiley, 2013. isbn: 9781118318676. Jiyuan Tu, Guan Heng Yeoh, and Chaoqun Liu. Computational fuid dynamics: a practical approach. 2nd ed. Butterworth-Heinemann, 2013. isbn: 9789382291787. Peter Alan Davidson. Turbulence: an introduction for scientists and engineers. 2nd ed. Oxford University Press, 2015. isbn: 9780198779469. Michael Leschziner. Statistical turbulence modelling for fuid dynam-ics demystifed. An introductory text for graduate engineering students. World Scientifc, 2015. isbn: 9781783266630. Zhan Su, Jinbo Wang, Patrice Klein, Andrew F Thompson, and Dimitris Menemenlis. “Ocean submesoscales as a key component of the global heat budget”. In: Nature communications 9.1 (2018), p. 775. doi: 10. 1038/s41467-018-02983-w. Olivier Cleynen. Git repository: Fluid dynamics. Under CC-by-nc license. 2020. url: https : / / framagit . org / olivier / fluidmech (visited on 04/20/2020). Olivier Cleynen, Germán Santa-Maria, Mathias Magdowski, and Do-minique Thévenin. “Peer-graded individualized student homework in a single-instructor undergraduate engineering course”. In: Research in Learning Technology 28 (2020). doi: 10.25304/rlt.v28.2339. 293
190560	https://intro2r.com/correlation.html	6.2 Correlation \| An Introduction to R An Introduction R Preface 0.1 The aim of this book 0.2 Who is this book for? 0.3 Why an open book? 0.4 Who are we? 0.5 How to use this book 0.6 Book website 0.7 Some R pointers 0.8 Thanks 0.9 License 1 Getting started with R and RStudio 1.1 Installing R 1.1.1 Windows users 1.1.2 Mac users 1.1.3 Linux users 1.1.4 Testing R 1.2 Installing RStudio 1.2.1 Windows and Mac users 1.2.2 Linux users 1.2.3 Testing RStudio 1.3 RStudio orientation 1.3.1 Console 1.3.2 Environment/History/Connections 1.3.3 Files/Plots/Packages/Help/Viewer 1.4 Alternatives to RStudio 1.4.1 Advanced text editors 1.4.2 Integrated development environments 1.5 R packages 1.5.1 CRAN packages 1.5.2 Bioconductor packages 1.5.3 GitHub packages 1.5.4 Using packages 1.6 Projects in RStudio 1.7 Working directories 1.8 Directory structure 1.9 File names 1.10 Project documentation 1.11 R style guide 1.12 Backing up projects 1.13 Citing R 1.14 Exercise 1 2 Some R basics 2.1 Getting started 2.2 Objects in R 2.2.1 Creating objects 2.2.2 Naming objects 2.3 Using functions in R 2.4 Working with vectors 2.4.1 Extracting elements 2.4.2 Replacing elements 2.4.3 Ordering elements 2.4.4 Vectorisation 2.4.5 Missing data 2.5 Getting help 2.5.1 R help 2.5.2 Other sources of help 2.6 Saving stuff in R 2.7 Exercise 2 3 Data in R 3.1 Data types 3.2 Data structures 3.2.1 Scalars and vectors 3.2.2 Matrices and arrays 3.2.3 Lists 3.2.4 Data frames 3.3 Importing data 3.3.1 Saving files to import 3.3.2 Import functions 3.3.3 Common import frustrations 3.3.4 Other import options 3.4 Wrangling data frames 3.4.1 Positional indexes 3.4.2 Logical indexes 3.4.3 Ordering data frames 3.4.4 Adding columns and rows 3.4.5 Merging data frames 3.4.6 Reshaping data frames 3.5 Summarising data frames 3.6 Exporting data 3.6.1 Export functions 3.6.2 Other export functions 3.7 Exercise 3 4 Graphics with base R 4.1 Getting started 4.2 Simple base R plots 4.2.1 Scatterplots 4.2.2 Histograms 4.2.3 Box and violin plots 4.2.4 Dot charts 4.2.5 Pairs plots 4.2.6 Coplots 4.2.7 Lattice plots 4.3 Customising plots 4.3.1 Customising with arguments 4.3.2 Building plots 4.4 Multiple graphs 4.5 Exporting plots 4.6 Exercise 4 5 Graphics with ggplot 5.1 Beginning at the end 5.2 The start of the end 5.2.1 The purest of ggplots 5.2.2 Wrapping grids 5.2.3 Plotting multiple ggplots 5.2.4 Make it your own 5.2.5 Setting the theme 5.2.6 Prettification 5.3 Tips and tricks 5.3.1 Statistics layer 5.3.2 Axis limits and zooms 5.3.3 Layering layers 5.3.4 Continuous colours 5.3.5 Size of points 5.3.6 Moving the legend 5.3.7 Hiding the legend 5.3.8 Hiding part of the legend 5.3.9 Writing on a figure 5.3.10 Axes tick marks and tick labels 5.3.11 More advanced patchwork 5.4 A ggplot bestiary 5.4.1 Density plot 5.4.2 Histogram 5.4.3 Frequency polygons 5.4.4 Boxplot 5.4.5 Violin plots 5.4.6 Barchart 5.4.7 Quantile lines 5.4.8 Heatmap 5.4.9 Hex map 5.4.10 Contour map 5.4.11 Cleveland dotplot 5.4.12 Pairs plot 5.5 Exercise 5 6 Simple Statistics in R 6.1 One and two sample tests 6.2 Correlation 6.3 Simple linear modelling 6.4 Other modelling approaches 6.5 Exercise 6 7 Programming in R 7.1 Looking behind the curtain 7.2 Functions in R 7.3 Conditional statements 7.4 Combining logical operators 7.5 Loops 7.5.1 For loop 7.5.2 While loop 7.5.3 When to use a loop? 7.5.4 If not loops, then what? 7.6 Exercise 7 8 Reproducible reports with R markdown 8.1 What is R markdown? 8.2 Why use R markdown? 8.3 Get started with R markdown 8.4 Create an R markdown document 8.5 R markdown anatomy 8.5.1 YAML header 8.5.2 Formatted text 8.5.3 Code chunks 8.5.4 Adding figures 8.5.5 Adding tables 8.5.6 Inline R code 8.6 Some tips and tricks 8.7 Further Information 9 Version control with Git and GitHub 9.1 What is version control? 9.2 Why use version control? 9.3 What is Git and GitHub? 9.4 Getting started 9.4.1 Install Git 9.4.2 Configure Git 9.4.3 Configure RStudio 9.4.4 Register a GitHub account 9.5 Setting up a project in RStudio 9.5.1 Option 1 - GitHub first 9.5.2 Option 2 - RStudio first 9.6 Using Git 9.6.1 Tracking changes 9.6.2 Commit history 9.6.3 Reverting changes 9.6.4 Collaborate with Git 9.6.5 Git tips 9.7 Further resources Appendix A Installing R Markdown A.1 MS Windows A.2 Mac OSX Published with bookdown A A Serif Sans White Sepia Night An Introduction to R 6.2 Correlation In R, the Pearson’s product-moment correlation coefficient between two continuous variables can be estimated using the cor() function. Using the trees data set again, we can determine the correlation coefficient of the association between tree Height and Volume. ``` data(trees) str(trees) 'data.frame': 31 obs. of 3 variables: $ Girth : num 8.3 8.6 8.8 10.5 10.7 10.8 11 11 11.1 11.2 ... $ Height: num 70 65 63 72 81 83 66 75 80 75 ... $ Volume: num 10.3 10.3 10.2 16.4 18.8 19.7 15.6 18.2 22.6 19.9 ... cor(trees$Height, trees$Volume) 0.5982497 ``` or we can produce a matrix of correlation coefficients for all variables in a data frame ``` cor(trees) Girth Height Volume Girth 1.0000000 0.5192801 0.9671194 Height 0.5192801 1.0000000 0.5982497 Volume 0.9671194 0.5982497 1.0000000 ``` Note that the correlation coefficients are identical in each half of the matrix. Also, be aware that, although a matrix of coefficients can be useful, a little commonsense should be used when using cor() on data frames with numerous variables. It is not good practice to trawl through these types of matrices in the hope of finding large coefficients without having an a priori reason for doing so and remember the correlation coefficient assumes that associations are linear. If you have missing values in the variables you are trying to correlate, cor() will return an error message (as will many functions in R). You will either have to remove these observations (be very careful if you do this) or tell R what to do when an observation is missing. A useful argument you can use with the cor() function is use = "complete.obs". ``` cor(trees, use = "complete.obs") Girth Height Volume Girth 1.0000000 0.5192801 0.9671194 Height 0.5192801 1.0000000 0.5982497 Volume 0.9671194 0.5982497 1.0000000 ``` The function cor() will return the correlation coefficient of two variables, but gives no indication whether the coefficient is significantly different from zero. To do this you need to use the function cor.test(). ``` cor.test(trees$Height, trees$Volume) Pearson's product-moment correlation data: trees$Height and trees$Volume t = 4.0205, df = 29, p-value = 0.0003784 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.3095235 0.7859756 sample estimates: cor 0.5982497 ``` Two non-parametric equivalents to Pearson correlation are available within the cor.test() function; Spearman’s rank and Kendall’s tau coefficient. To use either of these simply include the argument method = "spearman" or method = "kendall" depending on the test you wish to use. For example ``` cor.test(trees$Height, trees$Volume, method = "spearman") Warning in cor.test.default(trees$Height, trees$Volume, method = "spearman"): Cannot compute exact p-value with ties Spearman's rank correlation rho data: trees$Height and trees$Volume S = 2089.6, p-value = 0.0006484 alternative hypothesis: true rho is not equal to 0 sample estimates: rho 0.5787101 ```
190561	https://www.greenemath.com/Algebra1/42/SolvingQuadraticEquationsbyFactoringLesson.html	GreeneMath.com Master Math & Ace Your Exam! Solving Quadratic Equations by Factoring Lesson Lesson Objectives Demonstrate an understanding of how to Factor Trinomials Learn how to write a Quadratic Equation in Standard Form Learn about the Zero-Product Property Learn how to solve a Quadratic Equation using Factoring How to Solve Quadratic Equations Using Factoring Over the course of the last few lessons, we have learned to factor quadratic expressions. A quadratic expression contains a squared variable and no term with a higher degree. We will expand on this knowledge and learn how to solve a quadratic equation using factoring. A quadratic equation is an equation that contains a squared variable and no other term with a higher degree. Generally, we think about a quadratic equation in standard form: ax2 + bx + c = 0 a ≠ 0 (since we must have a variable squared) a, b, and c are any real numbers (a can't be zero) Some examples of a quadratic equation are: 4x2 + 7x - 15 = 0 5x2 + 18x + 9 = 0 Zero-Product Property Up to this point, we have not attempted to solve an equation in which the exponent on a variable was not 1. For these types of problems, obtaining a solution can be a bit more work than what we have seen so far. When a quadratic equation is in standard form and the left side can be factored, we can solve the quadratic equation using factoring. This works based on the zero-product property (also known as the zero-factor property). The zero product property tells us if the product of two numbers is zero, then at least one of them must be zero: xy = 0 x could be 0, y could be a non-zero number y could be 0, x could be a non-zero number x and y could both be zero We can apply this to more advanced examples. Suppose we saw the following: (x - 2)(x + 3) = 0 In this case, we have a quantity (x - 2) multiplied by another quantity (x + 3). The result of this multiplication is zero. This means we can use our zero-product property. To do this, we set each factor equal to zero and solve: (x - 2) = 0 (x + 3) = 0 x - 2 = 0 x = 2 x + 3 = 0 x = -3 Essentially, x could be 2 or x could be -3. In either scenario, the equation would be true: Let's check x = 2: (x - 2)(x + 3) = 0 (2 - 2)(2 + 3) = 0 0(5) = 0 0 = 0 Let's check x = -3: (-3 - 2)(-3 + 3) = 0 (-5)(0) = 0 0 = 0 Solving a Quadratic Equation using Factoring Place the quadratic equation in standard form Factor the left side Use the zero-product property and set each factor with a variable equal to zero Check the result Let's look at a few examples. Example 1: Solve each quadratic equation using factoring. x2 + 3x = 18 Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation: x2 + 3x - 18 = 0 Step 2) Factor the left side: x2 + 3x - 18 » (x + 6)(x - 3) (x + 6)(x - 3) = 0 Step 3) Use the zero-product property and set each factor with a variable equal to zero: x + 6 = 0 x - 3 = 0 x = -6 x = 3 We can say that x = -6, 3 This means x can be -6 or x can be 3. Either will work as a solution. Step 4) Check the result: Plug in a -6 for x: (-6)2 + 3(-6) - 18 = 0 36 - 18 - 18 = 0 36 - 36 = 0 0 = 0 Plug in a 3 for x: (3)2 + 3(3) - 18 = 0 9 + 9 - 18 = 0 18 - 18 = 0 0 = 0 Example 2: Solve each quadratic equation using factoring. 3x2 - 5 = -14x Step 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation: 3x2 + 14x - 5 = 0 Step 2) Factor the left side: 3x2 + 14x - 5 » (3x - 1)(x + 5) (3x - 1)(x + 5) = 0 Step 3) Use the zero-product property and set each factor with a variable equal to zero: 3x - 1 = 0 x + 5 = 0 x = 1/3 x = -5 We can say that x = -5, 1/3 This means x can be -5 or x can be 1/3. Either will work as a solution. Step 4) Check the result: Plug in a -5 for x: 3(-5)2 + 14(-5) - 5 = 0 3(25) - 70 - 5 = 0 75 - 75 = 0 0 = 0 Plug in a 1/3 for x: 3(1/3)2 + 14(1/3) - 5 = 0 3(1/9) + (14/3) - 5 = 0 1/3 + 14/3 - 5 = 0 15/3 - 5 = 0 5 - 5 = 0 0 = 0 Skills Check: Example #1 Solve each equation. $$x^{2}+ 2x - 8=0$$ Please choose the best answer. A $$x=5, -1$$ B $$x=-4, 2$$ C $$x=4, -2$$ D $$x=5, -5$$ E $$x=\frac{1}{2}, -3$$ Example #2 Solve each equation. $$x^{2}- 5x=6$$ Please choose the best answer. A $$x=-1, 6$$ B $$x=-4, -7$$ C $$x=1, 4$$ D $$x=4, 2$$ E $$x=-\frac{2}{5}, 6$$ Example #3 Solve each equation. $$3x^{2}+ 26x + 17=1$$ Please choose the best answer. A $$x=-\frac{1}{3}, 4$$ B $$x=\frac{2}{3}, 8$$ C $$x=-\frac{2}{3}, -8$$ D $$x=-\frac{6}{5}, -2$$ E $$x=-4, \frac{1}{3}$$ Congrats, Your Score is 100% Better Luck Next Time, Your Score is % Try again? Ready for more?Watch the Step by Step Video Lesson \| Take the Practice Test
190562	https://www.cnblogs.com/Roni-i/p/8904097.html	求解点关于直线的距离、垂足、对称点公式 - Roni_i - 博客园会员众包新闻博问闪存赞助商 HarmonyOS Chat2DB 所有博客当前博客我的博客我的园子账号设置会员中心简洁模式 ...退出登录注册登录 Roni coding, design, style and best practices 博客园首页联系管理随笔 - 1039 文章 - 0 评论 - 25 阅读 - 40万求解点关于直线的距离、垂足、对称点公式下面通过两种直线方程的形式，求解点关于直线的距离、垂足、对称点公式。问题描述1：已知点的坐标（x0，y0），直线的方程为Ax+By+C = 0；求点到直线上的距离d、点在直线上的垂足（x, y）、点关于直线的对称点(x’, y’)。解决方法：（1）距离： d = ( Ax0 + By0 + C ) / sqrt ( AA + BB ); 这个“距离”有符号，表示点在直线的上方或者下方，取绝对值表示欧式距离。（2）垂足：求解两个方程：（a） Ax + By + C = 0; （b） (y - y0) / (x - x0) = B / A; 解得，x = ( BBx0 - ABy0 - AC ) / ( AA + BB ); y = ( -ABx0 + AAy0 - BC ) / ( AA + BB ); （3）对称点：方法一：求解两个方程：(a)、A( x’+x0 ) / 2 + B( y‘+y0 ) / 2 + C = 0; (b)、(y’ - y0) / (x‘ - x0) = B / A; 方法二：把问题转化为求解已知点关于垂足的对称点：首先，求出垂足；则x’ = 2x - x0; y‘ = 2y - y0; 解得，x’ = ( (BB - AA)x0 - 2ABy0 - 2AC ) / ( AA + BB ); y‘ = ( -2ABx0 + (AA - BB) y0 - 2BC ) / ( AA+BB ); 方法三：首先，求一系数k，k = - 2 (Ax0 + By0 + C) / (AA+BB); 则，x' = x0 + k A; y' = y0 + k B; 此证明详见资源：问题描述2：已知点的坐标（x0，y0），直线上的两点（x1，y1）、（x2，y2）；求点到直线上的距离d、点在直线上的垂足（x, y）、点关于直线的对称点(x’, y‘)。解决方法：方法一：把直线化两点式为一般式，则一般式中的A = y2 -y1;B = x1 - x2;C = x2y1 - x1y2; 带入上面的公式，即可求出相应的距离、垂足、对称点。方法二：（a）距离：首先，求出垂足的坐标；则d = sqrt((x - x0) (x - x0) + (y - y0) (y - y0)); （b）垂足：首先，求一系数 k：设直线的起点和终点分别为A（x1， y1）、B（x2， y2），直线外一点为C（x0， y0），垂足为D；并设k = \|AD\| / \|AB。则，k AB = AD = AC+CD，又 AB CD= 0；所以，k ABAB=ACAB，故 k =ACAB/ （ABAB）。带入坐标，即得， k = ( (x0- x1) (x2 - x1) + (y0 - y1) (y2 - y1) ) / ((x2 - x1) (x2 - x1) +(y2 - y1) (y2 - y1)) ; 则 x = x1 + k(x2 - x1); y = y1 + k(y2 - y1); （c）对称点：同问题描述1中的方法。分类: 计算几何-几何公式, AAA模版, 计算几何好文要顶关注我收藏该文微信分享 Roni_i 粉丝 - 116关注 - 24 +加关注 0 0 « 上一篇： Educational Codeforces Round 40 (Rated for Div. 2) » 下一篇：任意选若干个不相邻的数得到的和最大【dp】 posted @ 2018-04-21 22:40Roni_i 阅读(1973) 评论(0)收藏举报刷新页面返回顶部登录后才能查看或发表评论，立即登录或者逛逛博客园首页【推荐】100%开源！大型工业跨平台软件C++源码提供，建模，组态！【推荐】HarmonyOS 专区 —— 闯入鸿蒙：浪漫、理想与「草台班子」【推荐】2025 HarmonyOS 鸿蒙创新赛正式启动，百万大奖等你挑战【推荐】天翼云爆款云主机2核2G限时秒杀，28.8元/年起！立即抢购编辑推荐： · 无意中在应用层瞥见了一个微内核的操作系统调度器 · Runtime Async - 步入高性能异步时代 · 基于 IOCP 的协程调度器——零基础深入浅出 C++20 协程 · 有有点意思！Java8后最有用新特性排行榜 · Ribbon LoadBalancer: 开源的客户端式负载均衡框架 HarmonyOS专区： · 闯入鸿蒙：浪漫、理想与「草台班子」 · 3天赚2万！开发者的梦想也可以掷地有声！ · Flutter 适配 HarmonyOS 5 开发知识地图 · HEIF：更高质量、更小体积，开启 HarmonyOS 图像新体验 · CodeGenie 的 AI 辅助调优让你问题定位效率大幅提升公告 -人生需要一点唐吉诃德，需要一点阿甘精神。人无所痴，难有所成。《网易云音乐》 -阅读是拓展人生空间的一个主要途径，每本书都是一扇门，通向我们没有涉足的新世界，在阅读中，我们经历了我们本来不可能在有生之年经历的事情，游历了那些我们的肉身永远无法到达的地方，愿大家在阅读中建立美好的生活！——刘慈欣世之奇伟、瑰怪、非常之观、常在于险远，而人之所罕至焉，故非有志者不能至也。尽吾志也而不能至者,可以无悔矣,其孰能讥之乎？力尚足以入，火尚足以明。盖余所至，比好游者尚不能十一，然视其左右，来而记之者已少。学者不可以不深思而慎取之也。 [游褒禅山记-王安石] 快速链接： CodeForces洛谷日报底特律新闻模板排版区域赛题 zj赛题单调队列题单 wiki Camp ⭐ 14杭电-苏州微软->日本谷歌-王凯晨 14电科-头条/b站cf-qsc 14浙江财经-京东->苏州微软-周甄陶 14长理-face++-綦文博 18哈工大阮行止 16北大-AI-黄哲威 OI-iwtwiioi ACdreamer 上海大学-谷歌-匡斌英雄哪里出来 ⭐ 15林科大-中南研-yj 15烟台大-华东师研-千千 16华交-考研-Angel 16南理-头条-hpc 16南华010 16台州hrb 16南大葫芦爷 16 Wisdom+.+ 17南大Fancy 17长理Dillonh 北化maggch 九月旧约忘乎所以 Hua-dong ACjiumeng MingSD 16读研九戒青大键盘上的青春伟大的蚊子 tokitsukaze OI-北大-Candy 电科B站 ⭐ CSU15-zjh-浙大读研 15xp-sxf-长沙蜜獾 15lyg-锐捷 16lzj 16hz-OI讲师 16czq-sxf实习 16ly 16yb-金山云 17lfw 昵称： Roni_i 园龄： 8年3个月粉丝： 116 关注： 24 +加关注搜索积分与排名积分 - 369741 排名 - 2422 随笔分类 (1935) AAA计算机基础-JavaWeb(1) AAA计算机基础-操作系统(10) AAA计算机基础-计算机网络(5) AAA计算机基础-数据库(6) AAA模版(129) ACM-训练计划(61) Java-多线程/高并发(5) Java-基础知识(17) Java-面试笔试(84) Java-实例代码(30) OJ-51nod(38) OJ-BZOJ(2) OJ-codeforces(142) OJ-CSU(9) OJ-FOJ(9) OJ-HDU(128) OJ-hihoCoder(6) OJ-Newcoder(56) OJ-PAT(20) OJ-POJ(36) OJ-UVA刷书计划(10) OJ-计蒜客(5) OJ-洛谷(55) OJ-其他学校(48) STL(23) STL-map(10) STL-next_permutation(4) STL-priority_queue(7) STL-queue/双端队列(2) STL-set(5) STL-stack(7) 常用技巧-打表(5) 常用技巧-预处理(10) 常用技巧-折半枚举（双向搜索）(3) 动态规划(69) 动态规划-背包问题-01背包(22) 动态规划-背包问题-多重背包(2) 动态规划-背包问题-分组背包(1) 动态规划-背包问题-完全背包(4) 动态规划-概率/期望DP(2) 动态规划-区间DP(2) 动态规划-树形DP(5) 动态规划-数位DP(4) 动态规划-线性DP(32) 动态规划-状压DP(4) 动态规划-最长公共子序列(3) 动态规划-最长上升子序列(10) 基本算法-递归和分治(4) 基本算法-递推(17) 基本算法-二分(36) 基本算法-高精度计算(8) 基本算法-构造(8) 基本算法-快速幂(8) 基本算法-枚举(54) 基本算法-模拟(110) 基本算法-排序(21) 基本算法-贪心(45) 基本算法-位运算(8) 计算几何(4) 计算几何-叉积和点积(2) 计算几何-多边形(2) 计算几何-几何公式(5) 计算几何-凸包(1) 数据结构(18) 数据结构-LCA（最近公共祖先）(1) 数据结构-RMQ(1) 数据结构-并查集(15) 数据结构-单调队列/单调栈(4) 数据结构-前后缀维护&差分标记(26) 数据结构-树状数组(1) 数据结构-一维线段树&二维线段树(6) 数据库设计(1) 数论(52) 数论-博弈论(4) 数论-进制问题(11) 数论-矩阵快速幂(2) 数论-欧拉函数&降幂(4) 数论-数学概率&期望(2) 数论-素数与整除问题(21) 数论-同余模&模线性方程(3) 数论-拓展欧几里得(5) 搜索(29) 搜索-BFS(32) 搜索-DFS(44) 搜索-记忆化搜索(2) 搜索-双向搜索(2) 图论(42) 图论-二分图(2) 图论-割边割点(1) 图论-强连通分量(2) 图论-拓扑排序(14) 图论-网络流之最大流(2) 图论-最短路&次短路&k短路(15) 图论-最小&次小&k小生成树(8) 字符串(40) 字符串-KMP(10) 字符串-manacher（马拉车）(1) 字符串-后缀数组(1) 字符串-字典树/01Trie(2) 字符串-字符串Hash(16) 组合数学-莫比乌斯反演(2) 组合数学-母函数/生成函数(1) 组合数学-排列组合&Lucas定理(7) 组合数学-线性递推/打表(2) 更多随笔档案 (1033) 2023年3月(3) 2022年2月(1) 2020年10月(1) 2020年7月(1) 2020年6月(3) 2020年5月(4) 2020年4月(4) 2020年3月(1) 2020年2月(1) 2020年1月(5) 2019年12月(7) 2019年11月(1) 2019年10月(2) 2019年9月(3) 2019年8月(4) 2019年7月(10) 2019年6月(7) 2019年5月(24) 2019年4月(15) 2019年3月(37) 2019年2月(42) 2019年1月(23) 2018年12月(14) 2018年11月(21) 2018年10月(7) 2018年9月(1) 2018年8月(60) 2018年7月(65) 2018年6月(47) 2018年5月(66) 2018年4月(81) 2018年3月(60) 2018年2月(13) 2018年1月(20) 2017年12月(92) 2017年11月(79) 2017年10月(44) 2017年9月(32) 2017年8月(38) 2017年7月(94) 更多阅读排行榜 1. 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190563	https://www.youtube.com/watch?v=_nAgDNz6ETQ	Brahmagupta's formula and the Quadruple Quad Formula (I) \| Rational Geometry Math Foundations 125 Insights into Mathematics 129000 subscribers 157 likes Description 9842 views Posted: 18 Jan 2015 In this video we introduce Brahmagupta's celebrated formula for the area of a cyclic quadrilateral in terms of the four sides. This is an obvious extension of Heron's formula. We are interested in finding a rational variant of it, that will be independent of a prior theory of real numbers',square roots' and `lengths'. For motivation we look at the situation of four affine 1-points. Is there an analog of the Triple Quad Formula? Yes there is: it is the more mysterious and complicated Quadruple Quad Formula. To get it, we will examine some important manipulations for a pair of quadratic equations which are of independent interest. This lecture has some more serious algebra in it: a great place to practice your manipulation and organizational skills. If you are teaching college mathematics, please consider doing yourself and your students a favour: teach them some of the material of this lecture carefully and explicitly! Video Content: 00:00 Intro to Brahmagupta's formula 3:40 Brahmagupta's cyclic quadrilateral 8:21 Analog of Archimedes' formula 10:09 Some natural questions 14:37 Pairs of quadratic equations 18:09 Proof of BCQ formula 22:59 Archimedes' function 27:34 Two quad triples 33:47 Quadruple Quad Formula 36:19 Proof of QQF Screenshot PDFs for my videos are available at the website These give you a concise overview of the contents of the lectures for various Playlists: great for review, study and summary. My research papers can be found at my Research Gate page, at My blog is at where I will discuss lots of foundational issues, along with other things. Online courses will be developed at openlearning.com. The first one, already underway is Algebraic Calculus One at Please join us for an exciting new approach to one of mathematics' most important subjects! If you would like to support these new initiatives for mathematics education and research, please consider becoming a Patron of this Channel at Your support would be much appreciated. Here are the Insights into Mathematics Playlists: 1 comments Transcript:
190564	https://sentence.yourdictionary.com/syntactic	Make Our Dictionary Yours Sign up for our weekly newsletters and get: By signing in, you agree to our Terms and Conditions and Privacy Policy. Success! We'll see you in your inbox soon. Syntactic Sentence Examples The goal of algebraic semantics is to capture the semantics of behavior by a set of axioms with purely syntactic properties. Nonetheless, a simplification of lexical and possibly syntactic coverage can be expected in less extreme cases. Tone sandhi in Shanghai is a morpho-phonological process to produce prosodic words, while compounding is a syntactic means to make lexical words. Syntactic sequents are defined (in our case) in terms of tableaux. Again the predominance of the 6H polytype is emphasized by its frequency and obvious syntactic coalescence to the long period polytypes. The syntax with its syntactic constituent leads to the construction of the language component and therefore helps to convey the meaning. These words mostly fulfill syntactic functions; they have elaborate lemmas but lean lexical concepts. It reports two studies which showed evidence for syntactic priming in comprehension. Semantically, they are just syntactic sugar for a normal function definition. It is only syntactic irregularity, then, that requires a special poetic license. On the syntactic level, Heston (1976) studies the comparative typology of Middle Iranian languages, viz. The child with SLI often has difficulties learning language "incidentally," (picking up the meaning of a new word from context or generalizing a new syntactic form). The child with SLI also often has difficulties learning language incidentally, that is, in picking up the meaning of a new word from context or generalizing a new syntactic form. Browse other sentences examples The word usage examples above have been gathered from various sources to reflect current and historical usage. They do not represent the opinions of YourDictionary.com. Related Articles Many people believe that grammar and syntax are the same concepts. While they both dictate the construction of a sentence, it is more accurate to say that syntax is a part of grammar. Read on to learn the key elements of both grammar and syntax, and how to tell the difference between the two. Along with diction, syntax is a fundamental part of written language. Diction defines the style of written or spoken language, representing the choices a speaker or writer makes within the rules of grammar and syntax. Syntax states the rules for using words, phrases, clauses and punctuation, specifically to form sentences. Words near syntactic in the Dictionary
190565	https://emedicine.medscape.com/article/216089-overview	Tools & Reference>Infectious Diseases Diphyllobothriasis Updated: Jun 10, 2021 Author: Shireen R Chacko, MBBS; Chief Editor: Pranatharthi Haran Chandrasekar, MBBS, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Diphyllobothriasis Sections Diphyllobothriasis Overview Practice Essentials Background Pathophysiology Etiology Epidemiology Prognosis Patient Education Show All Presentation History Physical Examination Complications Show All DDx Workup Laboratory Studies Other Studies Show All Treatment Approach Considerations Pharmacologic Therapy Consultations Long-Term Monitoring Complications Prevention Show All Medication Medication Summary Anthelmintics Show All Media Gallery;) Tables) References;) Overview Practice Essentials Diphyllobothriasis is a parasitic infection caused by cestodes of the genus, Diphyllobothrium, also known as “broad tapeworms” or “fish tapeworms”. There are numerous species of Diphyllobothrium, the commonest being Diphyllobothrium latum, with others including Diphyllobothrium nihonkaiense, Diphyllobothrium dendriticum, and Adenocephalus pacificus,(syn. Diphyllobothrium pacificum), being less common. [2, 3, 4] Humans are the definitive host, infected by ingestion of raw fish containing the larval forms of the cestode that evolve into the adult worm finally residing in the intestines. Although the manifestations may not be life-threatening, patients typically present with symptoms of diarrhea, abdominal discomfort, vitamin B12 deficiency, or rarely, intestinal obstruction. Through globalization, human migration, and increased consumption of raw fish, the disease that was traditionally limited to the circumpolar areas of the world, has re-emerged with increased prevalence in other regions, stimulating new interest and research in its prevention. Egg of Diphyllobothrium latum with arrow pointing to operculum. View Media Gallery) Signs and symptoms Most infections with diphyllobothriasis are asymptomatic. In symptomatic persons, the following are the most common symptoms: Diarrhea Abdominal pain or discomfort Constipation Fatigue Passage of proglottids Headache Allergic reactions Prolonged infection can lead to low vitamin B12 levels in 40% of infected individuals, as the parasite leads to dissociation of the vitamin B12-intrinsic factor complex in the intestinal lumen and subsequently absorbs most of the B12. Many patients with diphyllobothrasis have no signs of illness. Rare physical findings (likely related to vitamin B12 deficiency) that may be noted include the following: Pallor Glossitis Disturbances of movement and coordination, loss of vibratory sense and proprioception Rarely, infection with this parasite can lead to obstruction of lumens, resulting in intestinal obstruction, cholangitis, appendicitis and cholecystitis. [1, 5] See Presentation for more detail. Diagnosis Laboratory studies that may be helpful for diphyllobothriasis include the following: Microscopic stool examination for ova and parasites (the usual basis for the diagnosis) Complete blood count (CBC) demonstrating anemia Mean cell volume demonstrating macrocytosis Peripheral smear demonstrating macrocytosis (unless also iron deficient) and hypersegmented polymorphonuclear neutrophils Vitamin B12 level of less than 150 pmol/L and total serum homocysteine level of more than 13 µmol/L or methylmalonic acid level of more than 0.4 µmol/L in the absence of renal failure and folate and vitamin B-6 deficiencies. Vitamin B12 deficiency is reported more frequently in D latum infections than with other species. [6, 7] Other studies that may be considered are as follows: Diagnostic imaging modalities - Not typically required, except as clinically indicated by other aspects of the presentation (eg, obstruction); abdominal ultrasonography, a real-time, non-invasive test might detect the hyperechoic, ribbon-like structure freely floating in the intestinal lumen. Endoscopic evaluation, particularly capsule endoscopy - Usually performed to evaluate unexplained anemia, with an unexpected discovery of worms. [9, 10, 11, 12] See Workup for more detail. Management Most patients with diphyllobothriasis, unless they have severe symptoms, can be safely treated as outpatients. Treatment of the infection is pharmacologic, involving one of the following agents: Praziquantel (drug of choice) 5-10 mg/kg orally in single dose; taken with liquids during a meal Niclosamide (acceptable alternative, but not available in the United States) If the first course of treatment fails, a second identical course of therapy may be administered. Other aspects of treatment include the following: Surgical treatment is not required unless otherwise indicated (eg, intestinal obstruction). Vitamin supplementation may be required in severe cases of vitamin B12 deficiency. As a rule, no activity limitations or restrictions are necessary. See Treatment and Medication for more detail. Next: Background Diphyllobothriasis is defined as human intestinal infection with the cestode D latum, D nihonkaiense, or other broad tapeworm species. It is acquired by ingestion of inadequately cooked or unfrozen freshwater, anadromous, or marine fish containing larvae called plerocercoids, either in fish muscle or on serosal surfaces. Most individuals with diphyllobothriasis have minimal or minor gastrointestinal symptoms or present with passage of worm segments (ie, proglottids) in stool. Adult D latum is the largest human parasite and can grow to a length of greater than 20 meters and live for decades. It is capable of causing vitamin B12 deficiency through dissociation of the vitamin from intrinsic factor and consumption of the vitamin. This is rare today, but low vitamin B12 levels have been reported in up to 40% of patients with D latum infestation and 5% of patients with A pacificus infestation. Clinical anemia and neurological disease is much rarer (< 2%). Evidence of human A pacificus infestation and D latum infestation dates to at least 6,000 years ago in Peru and Germany, respectively, and D nihonkaiense to 1,000 years ago in Japan, with the first clinical description in Switzerland in 1592. Diphyllobothriasis’s relationship to fish was first noted by Sporing in 1747. The disease appears to have been particularly widespread in Baltic and Alpine freshwater areas up through the early 20th century. In the early 1970s, an estimated 9 million people were infested with D latum worldwide, with 5 million in Europe, 4 million in Asia, and the remainder in the Americas. The disease is notable for its association with temperate climate and rarity in the tropics.With improvement in human sewerage treatment, prevalence has significantly decreased in developed areas; however, some zoonotic infection persists, especially with species other than D latum, owing to nonhuman definitive hosts, including bears, wolves, marine birds, and sea lions. With the increased popularity of raw seafood as part of ethnic foods, wider and faster food distribution networks, and, possibly, fish aquaculture, diphyllobothriasis seems to be extending its traditional geography and resurging, particularly with species other than D latum. [3, 15] Previous Next: Pathophysiology Fish tapeworms have a complex life cycle, with humans, other fish-eating mammals, and birds as their definitive hosts. Diphyllobothrium life cycle. This illustration depicts the life cycle of different species of Diphyllobothrium parasitic cestodes, the causal agents of the disease diphyllobothriasis. Courtesy of the Centers for Disease Control and Prevention (CDC) ( View Media Gallery) Adults are long ribbonlike creatures originally identified based on scolex and egg morphology; increasingly, cytochrome oxidase (COX1) molecular testing has clarified nosology and epidemiology. Except for D latum, and possibly D nihonkaiense, which appears specifically adaptive to humans as the preferential host and are by far the most common species involved in human infection, most other species rely on non-human definitive hosts, and humans are incidentally infected. A full-grown Diphyllobothrium worm is the longest human tapeworm; it can range from 1-15 m in length and 1-2 cm in width. It consists of up to 3000-4000 proglottids. The scolex has 2 sucking grooves, also called bothria, that attach to small-intestinal mucosa. Proglottids are typically wider than they are long and contain hermaphroditic sexual parts. Humans and other animals may be infested with multiple worms simultaneously. In the gravid state, the worms have a distinctive rosettelike uterus in the center. Each adult worm sheds around one million operculated eggs every day. To complete their maturation, the eggs must reach water to be eaten by one of 40 species of crustacean copepods and cyclops, within which the procercoid matures. Copepods that contain the procercoid are then eaten by small fish that function as the second intermediate host. Here, the procercoid matures into the plerocercoid in fish muscle or, in the case of D dendriticumand A pacificus, a serosal body surface such as the peritoneum or an organ such as the liver. [2, 4] Fish surveys of infected lakes can be performed, monitoring risk via inspection of fish muscle or organs for encysted plerocercoids. A study of fish caught in Lake Como, Italy, demonstrated infection rates of D latumin pike (Esox lucius) and perch (Perca fluviatilis) of 84% and 25%, respectively.This prevalence was attributed to the popularity of a local dish and the breakdown of sewerage effectiveness. Progressively larger fish that are paratenic hosts (hosts in which the larvae do not mature further) consume the infected fish. In this manner, the plerocercoid are passed on until, finally, the fish is consumed by a human, the definitive host. Over the subsequent 2-4 weeks, the plerocercoid larva attaches to the host intestine and matures into an adult that can live for up to 20 years. Table 1. Prevalence of Diphyllobothriasis Infections (Open Table in a new window)) \| \| \| \| \| \| --- --- \| Species \| Geography \| Fish \| Definitive Host \| Human Prevalence \| \| D latum \| Finland, Baltic States, Danube Delta, Karelia, Manitoba, Great Lakes \| Perch, pike, burbot \| Humans, dogs, wolves \| Millions \| \| D nihonkaiense \| North Pacific Coast, Japan, Korea, East Russia \| Salmon (cherry, chum, pink, sockeye) \| Brown and black bears, wolves, dogs, foxes, mink \| Thousands (March-June peak) \| \| D dendriticum \| Lake Baikal region, Subarctic North America, Russia, Rocky Mountains, Patagonia \| Whitefish, salmonids \| Gulls, bears, otter \| Thousands (600/year mostly) \| \| A pacificus \| Peru, Chile, Ecuador, Sakhalin Island \| Marine fish (bonito, mackerel, lorna drum, croaker) \| Sea lions, seals \| Thousands \| \| D balanopterae \| Japan, Spain, Norway, Korea \| Anchovy, sardines, skipjack tuna \| Whales \| Hundreds (June-July peak) \| Plerocercoid larvae infect humans who have ingested infected and inadequately cooked or unfrozen fish. Because of the requirement for intermediate hosts, direct human-to-human transmission does not occur; therefore, no isolation measures are required. Most humans with diphyllobothriasis (all forms) are asymptomatic, and the most common presentation is passage of worm parts, which can be alarming. Many patients retrospectively report diarrhea or may present with symptoms of obstruction or even appendicitis. Both vitamin B12 deficiency and iron deficiency anemia have been attributed to infestation with D latum, but not as often with other species. Because of this, infection might be a concern for pregnant women. Previous Next: Etiology Diphyllobothriasis is caused by ingestion of raw, undercooked, or unfrozen infected fish and subsequent intestinal infection. The main causative organisms are D latum and D nihonkaiense, but other Diphyllobothrium and similar species have also been reported as infecting agents, albeit much less frequently. Examples include the following : Diphyllobothrium dendriticum Adenocephalus pacificus Diphyllobothrium balanopterae Diphyllobothrium alascense Diphyllobothrium cameroni Diphyllobothrium cordatum Diphyllobothrium dalliae Diphyllobothrium elegans Diphyllobothrium lanceolatum Diphyllobothrium orcini Diphyllobothrium scoticum Diphyllobothrium stemmacephalum Diphyllobothrium ursi (possible synonym of D dendritum) Ligula intestinalis Schistocephalus solidus Previous Next: Epidemiology Diphyllobothriasis has been traditionally considered an endemic disease of specific locations where consumption of uncooked fish is common, such as Finland, Scandinavia, alpine Europe, North American lakes (D latum), northern Japan (D nihonkaiense), and Peru (A pacificus). With improvements in sanitation and understanding of disease pathogenesis, the prevalence of infection in endemic areas appears to have fallen dramatically during the 20th century. Although Japan has mandated case reporting since 2012, little is known about the current prevalence and incidence of various fish tapeworm infections except for case reports and occasional surveys. Increasing travel, migration, and popularity of ethnic foods have contributed to a broadening of interest and, possibly, exposure to raw fish delicacies, particularly in Asia, but also elsewhere owing to rapid food transport and fish aquaculture. Examples of foods responsible for infection include raw and salted or marinated fish filets in Scandinavia, Jewish gefilte fish,Italian carpaccio,French tartare maison(raw salmon) and poisson du lac façon nordique, Japanese sushi (raw fish on rice cakes or balls) and sashimi(sliced raw fish), and ceviche(lightly marinated fish) in Latin America. The quality and type of fish, as well as other animal products in dishes, varies by locale and cost, with the highest risk apparently associated with rural restaurants and street vendors. Types of fish and risk may vary by season. The peak risk for D nihonkaienseis associated with the spring catch of cherry and immature chum salmon in Northern Japan. The peak risk for D balanopterae in Northern Japan is associated with the harvest of anchovies from June to August. Concerns have been raised regarding salmon and other fish aquaculture, particularly in Chile, because of growing salmon smolt in lakes endemic for D latumand potential contamination and escape of infected fish grown in oceanic pens. Such infection might account for infections with D latumin Brazil and other tropical areas such as Taiwan and South India. Similarly, infection with A pacificusin Spain might be associated with heavy importation of chilled fish from Chile, Ecuador, and Peru. [2, 27] Although salmonids are not the preferred intermediate host for D latum in the Northern hemisphere, recent reports of D latum plerocercoid larvae adapting to salmonids are especially alarming in South America and may have important epidemiologic consequences as the consumption of salmonids increases. Placing fish on ice for transport is inadequate to kill plerocercoids and therefore, adequate freezing of fish and parasitologic screening before exportation may be warranted to prevent the spread of disease. In addition to human migration and importation of disease, other factors might account for the increased incidence and range of disease. These include, in the case of D dentriticum, migrating birds such as gulls around Lake Baikal and climate change affecting intermediate hosts. In the absence of effective surveillance and reporting, these diseases will primarily be discovered incidentally. Fish surveys show variable positivity, with D latum plerocercoid larvae in 84% of pike in Lake Como in 2013-2014 and 51% of Tokyo chum salmon in 2001 to 20% of various species of Peruvian marine fish infested with A pacificusin 2014. Workers in the fish industry may be at some risk for diphyllobothriasis, particularly for serosal or organ-encysted worms such as D dendriticumand A pacificus, from gutting and either contamination or habits of consumption of roe and fish organs while working. [2, 4] Diphyllobothriasis has no reported age predilection or sexual predilection. In addition, it has no known racial predilection, except as would be expected based on geographic and cultural factors. Previous Next: Prognosis Diphyllobothriasis carries an excellent prognosis. D latum is not invasive, and mortality due to diphyllobothriasis is rare. Patients are often frightened and emotionally upset at the passage of worm parts for a long period of time unless treated. Single-dose therapy is usually effective, although some treatment failures have been reported, and repeat treatment is occasionally needed. Occasionally, infestation can lead to severe megaloblastic anemia or intestinal obstruction. Although it is well described, macrocytic anemia and neurological disease is extremely rare.Gastrointestinal (GI) obstruction is also rare but may occur, especially when numerous worms are present. Previous Next: Patient Education Because reinfection is possible, patients should be advised to modify their dietary habits to minimize the potential for reexposure. Proper food preparation and hygiene should be encouraged, particularly during travel within endemic areas. Preventive public education has been generally inadequate, and the food and restaurant industry has generally been slow to adopt recommendations for freezing of fish. Simple icing of fish is insufficient to prevent infection. The US Centers for Disease Control and Prevention (CDC) recommends cooking fish to an internal temperature of 63°C (145°F) or higher, freezing fish to -4°F (-20°C) for 7 days or -31°F (-35°C) or less until solid, and storing at either -31°F (-65°C) or below for 15 hours or -4°F (-20°C) for 24 hours. These precautions kill the plerocercoid larvae. The fish must never be sampled before it is properly prepared. Previous Clinical Presentation References Scholz T, Garcia HH, Kuchta R, Wicht B. Update on the human broad tapeworm (genus diphyllobothrium), including clinical relevance. Clin Microbiol Rev. 2009 Jan. 22 (1):146-60, Table of Contents. [QxMD MEDLINE Link]. Kuchta R, Serrano-Martínez ME, Scholz T. Pacific Broad Tapeworm Adenocephalus pacificus as a Causative Agent of Globally Reemerging Diphyllobothriosis. Emerg Infect Dis. 2015 Oct. 21 (10):1697-703. [QxMD MEDLINE Link]. Scholz T, Kuchta. 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The Things They Carry: Diphyllobothriasis at Sea, a Case Report. Mil Med. 2020 Mar 2. 185 (3-4):e510-e512. [QxMD MEDLINE Link]. Björkenheim B. Optic neuropathy caused by vitamin-B12 deficiency in carriers of the fish tapeworm, Diphyllobothrium latum. Lancet. 1966 Mar 26. 1 (7439):688-90. [QxMD MEDLINE Link]. Craig N. Fish tapeworm and sushi. Can Fam Physician. 2012 Jun. 58 (6):654-8. [QxMD MEDLINE Link]. Media Gallery Life cycle of diphyllobothrium. Egg of Diphyllobothrium latum with arrow pointing to operculum. Proglottids of Diphyllobothrium latum. Diphyllobothrium life cycle. This illustration depicts the life cycle of different species of Diphyllobothrium parasitic cestodes, the causal agents of the disease diphyllobothriasis. Courtesy of the Centers for Disease Control and Prevention (CDC) ( of 4 Tables Table 1. Prevalence of Diphyllobothriasis Infections;) Table 1. Prevalence of Diphyllobothriasis Infections \| \| \| \| \| \| --- --- \| Species \| Geography \| Fish \| Definitive Host \| Human Prevalence \| \| D latum \| Finland, Baltic States, Danube Delta, Karelia, Manitoba, Great Lakes \| Perch, pike, burbot \| Humans, dogs, wolves \| Millions \| \| D nihonkaiense \| North Pacific Coast, Japan, Korea, East Russia \| Salmon (cherry, chum, pink, sockeye) \| Brown and black bears, wolves, dogs, foxes, mink \| Thousands (March-June peak) \| \| D dendriticum \| Lake Baikal region, Subarctic North America, Russia, Rocky Mountains, Patagonia \| Whitefish, salmonids \| Gulls, bears, otter \| Thousands (600/year mostly) \| \| A pacificus \| Peru, Chile, Ecuador, Sakhalin Island \| Marine fish (bonito, mackerel, lorna drum, croaker) \| Sea lions, seals \| Thousands \| \| D balanopterae \| Japan, Spain, Norway, Korea \| Anchovy, sardines, skipjack tuna \| Whales \| Hundreds (June-July peak) \| Back to List Contributor Information and Disclosures Author Shireen R Chacko, MBBS Resident Physician, Department of Internal Medicine, Albert Einstein Medical Center Disclosure: Nothing to disclose. Coauthor(s) Antonette B Climaco, MD Attending Physician, Division of Infectious Diseases, Department of Medicine, Einstein Healthcare Network Antonette B Climaco, MD is a member of the following medical societies: American Academy of HIV Medicine, HIV Medicine Association, Infectious Diseases Society of America, Philippine Medical Association Disclosure: Nothing to disclose. Chief Editor Pranatharthi Haran Chandrasekar, MBBS, MD Professor, Chief of Infectious Disease, Department of Internal Medicine, Wayne State University School of Medicine Pranatharthi Haran Chandrasekar, MBBS, MD is a member of the following medical societies: American College of Physicians, American Society for Microbiology, International Immunocompromised Host Society, Infectious Diseases Society of America Disclosure: Nothing to disclose. Additional Contributors David R Haburchak, MD, FACP Key Faculty, Wellstar Kennestone Internal Medicine Residency Program, Wellstar Health System David R Haburchak, MD, FACP is a member of the following medical societies: American College of Physicians, American Osler Society, American Society for Microbiology, Christian Medical and Dental Associations, Infectious Diseases Society of America Disclosure: Nothing to disclose. Derek Ryan Linklater, MD Assistant Professor of Military and Emergency Medicine, F Edward Hebert School Of Medicine; Assistant Clinical Professor of Emergency Medicine, Texas A&M Health Science Center College of Medicine; Clinical Instructor of Pediatrics, Baylor College of Medicine Derek Ryan Linklater, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Emergency Medicine, Society for Academic Emergency Medicine, American College of Emergency Physicians Disclosure: Nothing to disclose. Coburn H Allen, MD Assistant Professor of Pediatrics, Dell Children's Hospital Coburn H Allen, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Pediatrics, Infectious Diseases Society of America, Pediatric Infectious Diseases Society Disclosure: Nothing to disclose. Acknowledgements Thomas M Kerkering, MD Chief of Infectious Diseases, Virginia Tech Carilion School of Medicine Thomas M Kerkering, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American Public Health Association, American Society for Microbiology, American Society of Tropical Medicine and Hygiene, Infectious Diseases Society of America, Medical Society of Virginia, and Wilderness Medical Society Disclosure: Nothing to disclose. Alia Rai, MD Adolescent Medicine Fellow, Department of Pediatrics, West Virginia University Alia Rai, MD is a member of the following medical societies: American Academy of Pediatrics, American Medical Association, and Society for Adolescent Medicine Disclosure: Nothing to disclose. Russell W Steele, MD Head, Division of Pediatric Infectious Diseases, Ochsner Children's Health Center; Clinical Professor, Department of Pediatrics, Tulane University School of Medicine Russell W Steele, MD is a member of the following medical societies: American Academy of Pediatrics, American Association of Immunologists, American Pediatric Society, American Society for Microbiology, Infectious Diseases Society of America, Louisiana State Medical Society, Pediatric Infectious Diseases Society, Society for Pediatric Research, and Southern Medical Association Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Medscape Salary Employment Mark R Wallace, MD, FACP, FIDSA Clinical Professor of Medicine, Florida State University College of Medicine; Head of Infectious Disease Fellowship Program, Orlando Regional Medical Center Mark R Wallace, MD, FACP, FIDSA is a member of the following medical societies: American College of Physicians, American Medical Association, American Society of Tropical Medicine and Hygiene, and Infectious Diseases Society of America Disclosure: Nothing to disclose. Martin Weisse, MD Program Director, Associate Professor, Department of Pediatrics, West Virginia University Martin Weisse, MD is a member of the following medical societies: Ambulatory Pediatric Association, American Academy of Pediatrics, and Pediatric Infectious Diseases Society Disclosure: Nothing to disclose. Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Nothing to disclose. encoded search term (Diphyllobothriasis) and Diphyllobothriasis What to Read Next on Medscape
190566	https://blog.truegeometry.com/calculators/A_Diesel_engine_has_an_indicated_thermal_efficiency_of_35_The_brake_power_developed_by_the_engine_is.html	Thermo-Economic Characterization of Diesel Fuel \| True Geometry’s Blog Home Authors Calculators Designs3d Engineering Posts Tutorials Thermo-Economic Characterization of Diesel Fuel 11 Oct 2024 Tags: MechanicalEngineeringThermodynamicsThermodynamicsCalorificValueofFuel Popularity: ⭐⭐⭐ Calorific Value Calculation for Diesel Engine This calculator determines the calorific value of the fuel used in a diesel engine based on its indicated thermal efficiency, brake power, and fuel consumption rate. Explanation Calculation Example: The calorific value (CV) of a fuel is the amount of heat energy released when a unit mass of the fuel is completely burned. In this context, we are using the brake power (BP) and indicated thermal efficiency (η_ith) to determine the heat input to the engine. The fuel consumption rate (m_dot) provides the mass of fuel consumed per unit time. By dividing the heat input by the fuel consumption rate, we obtain the calorific value of the fuel. Related Questions Q: What is the difference between indicated thermal efficiency and brake thermal efficiency? A: Indicated thermal efficiency refers to the efficiency of the engine based on the work done by the piston within the cylinder, while brake thermal efficiency considers the power delivered to the crankshaft after accounting for mechanical losses. Brake thermal efficiency is typically lower than indicated thermal efficiency. Q: How does the calorific value of fuel affect engine performance? A: The calorific value of fuel directly impacts engine performance. A higher calorific value means more energy is released per unit mass of fuel, potentially leading to higher power output and efficiency. However, it is important to consider other factors like combustion characteristics and emissions when selecting a fuel. Variables \| Symbol \| Name \| Unit \| --- \| η_ith \| Indicated Thermal Efficiency \| % \| \| BP \| Brake Power \| kW \| \| m_dot \| Fuel Consumption Rate \| kg/h \| Calculation Expression Calorific Value: The calorific value of the fuel is calculated using the formula: CV = (BP / (η_ith / 100)) / (m_dot / 3600) (BP / (η_ith / 100)) / (m_dot / 3600) Calculator Indicated Thermal Efficiency (%): Brake Power (kW): Fuel Consumption Rate (kg/h): Solve Calculated values Considering these as variable values: η_ith=35.0, m_dot=0.2, BP=15.0, the calculated value(s) are given in table below \| Derived Variable \| Value \| --- \| \| Calorific Value \| 5400000.0/(m_Dotη_ith) \| Similar Calculators Calcination Thermodynamics Calculations in Thermodynamics Thermodynamic Energies Thermodynamic Systems Analysis Thermodynamic Equations and Calculations Thermodynamic Calculations in Physics Thermodynamic Energy Calculations Quantitative Thermodynamic Analysis Quantitative Thermodynamics Analysis Thermodynamic Process Analysis Explore Diesel engine performance analysis Thermodynamics of combustion Fuel properties and characteristics Calculator Apps Calorific Value Calculation for Diesel Engine AI supported calculatorn Gear Design in 3D & Learning True Geometry's Blog True Geometry's Blog truegeometryopc@gmail.com TrueGeometry TrueGeometryOPC We do reverse engineering on popularly known 3D file formats. We apply state of the art machine learning algorithms to do so.
190567	https://math.stackexchange.com/questions/4216888/derivative-chain-rule-for-x-as-exponent	algebra precalculus - Derivative Chain Rule for $x$ as Exponent - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Derivative Chain Rule for x x as Exponent Ask Question Asked 4 years, 1 month ago Modified4 years, 1 month ago Viewed 3k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Let's say I have a simple problem like this: d d x[2 x]d d x[2 x] I would apply the exponent rule and work it like this: d d x[2 x]d d x[2 x] =x⋅2 x−1=x⋅2 x−1 I don't know how to simplify that further so I will leave it as is. If anyone knows, just comment. But the real answer turns out to be ln(2)⋅2 x ln⁡(2)⋅2 x. How is this worked out and how can I apply this rule to complex problems like: d d x[(2 x+4)x+1]d d x[(2 x+4)x+1] I want to know the following: How to use this ln(x)ln⁡(x) rule? When to use this rule? Why you can't use the standard exponent rule? algebra-precalculus derivatives logarithms exponentiation chain-rule Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Aug 4, 2021 at 22:44 amWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges asked Aug 4, 2021 at 19:34 DragonflyRoboticsDragonflyRobotics 131 1 1 silver badge 6 6 bronze badges 3 1 You are making a common (and easy to make) mistake. If consider the BASE of x k x k to be our variable and the exponent k k to be a CONSTANT, then, yes, the derivative if k x k−1 k x k−1. But if it is the base that is CONSTANT and it is our EXPONENT that is x x then the derivative of b x b x is NOT x b x−1 x b x−1. b x b x and x k x k are ENTIRELY different functions with entirely different behaviors and that just will not work.fleablood –fleablood 2021-08-04 22:50:19 +00:00 Commented Aug 4, 2021 at 22:50 The reason d x k d x=k x k−1 d x k d x=k x k−1 is because (x+h)k−x k h=[x k+h x k−1+(k 2)h 2 x k−2........+k x h k−1+h k]−x k h=x k−1+(k 2)h x k−2........+k x h k−2+h k−1→x k−1(x+h)k−x k h=[x k+h x k−1+(k 2)h 2 x k−2........+k x h k−1+h k]−x k h=x k−1+(k 2)h x k−2........+k x h k−2+h k−1→x k−1. But d b x d x=b x+h−b x h=b x b h−b x h=b x⋅(b h−1 h)→b x⋅ln b d b x d x=b x+h−b x h=b x b h−b x h=b x⋅(b h−1 h)→b x⋅ln⁡b. Notice it is an entirely different type of calculation.fleablood –fleablood 2021-08-04 22:56:11 +00:00 Commented Aug 4, 2021 at 22:56 BTW x b x b is consider a POWER and you use the POWER rule. b x b x is considered and EXPONENT and you use the EXPONENT rule. The POWER rule is d x k d x=k x k−1 d x k d x=k x k−1. The EXPONENT rule is d b x d x=b x⋅ln b d b x d x=b x⋅ln⁡b.fleablood –fleablood 2021-08-04 23:01:07 +00:00 Commented Aug 4, 2021 at 23:01 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. I think that you are confusing the rules. The rule you are misquoting is d d x x a=a x a−1 d d x x a=a x a−1 (Note that x x is not in the exponent). To calculate the derivative of a x a x we will use the special property of e e. More precisely, we have: d d x a x=d d x e x ln a=e x ln a(d d x x ln a)=e x ln a ln a=a x ln a d d x a x=d d x e x ln⁡a=e x ln⁡a(d d x x ln⁡a)=e x ln⁡a ln⁡a=a x ln⁡a So for the more complex example, we have: d d x[(2 x+4)x+1]=d d x e(x+1)ln(2 x+4)=(d d x((x+1)ln(2 x+4)))e(x+1)ln(2 x+4)=(ln(2 x+4)+x+1 2 x+4)e(x+1)ln(2 x+4)=(ln(2 x+4)+1 2 x+1 x+2)(2 x+4)x+1 d d x[(2 x+4)x+1]=d d x e(x+1)ln⁡(2 x+4)=(d d x((x+1)ln⁡(2 x+4)))e(x+1)ln⁡(2 x+4)=(ln⁡(2 x+4)+x+1 2 x+4)e(x+1)ln⁡(2 x+4)=(ln⁡(2 x+4)+1 2 x+1 x+2)(2 x+4)x+1 If you want a more general rule we can write: d d x f(x)g(x)=d d x e g(x)ln(f(x))=(g′(x)ln(f(x))+g(x)1 f(x))e g(x)ln(f(x))=(g′(x)ln(f(x))+g(x)1 f(x))f(x)g(x)d d x f(x)g(x)=d d x e g(x)ln⁡(f(x))=(g′(x)ln⁡(f(x))+g(x)1 f(x))e g(x)ln⁡(f(x))=(g′(x)ln⁡(f(x))+g(x)1 f(x))f(x)g(x) Hope that helps. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 4, 2021 at 23:43 Graham Kemp 133k 7 7 gold badges 55 55 silver badges 128 128 bronze badges answered Aug 4, 2021 at 19:44 Maths WizzardMaths Wizzard 2,418 1 1 gold badge 12 12 silver badges 26 26 bronze badges 3 Great answer but can you give me a more complex answer where this could be applied. A good example would be the second example in my question.DragonflyRobotics –DragonflyRobotics 2021-08-04 19:49:26 +00:00 Commented Aug 4, 2021 at 19:49 I will add it now.Maths Wizzard –Maths Wizzard 2021-08-04 20:00:44 +00:00 Commented Aug 4, 2021 at 20:00 1 @DragonflyRobotics I have added your complex example as well as a proof for a more general rule.Maths Wizzard –Maths Wizzard 2021-08-04 20:15:46 +00:00 Commented Aug 4, 2021 at 20:15 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The "exponent rule" is for d d x x a d d x x a not d d x a x d d x a x To handle a x a x, note a x=e x ln(a)a x=e x ln⁡(a) and use the "chain rule" and d d x e x=e x d d x e x=e x so d d x a x=d d x e x ln(a)=ln(a)e x ln(a)=ln(a)a x d d x a x=d d x e x ln⁡(a)=ln⁡(a)e x ln⁡(a)=ln⁡(a)a x Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2021 at 19:46 HenryHenry 171k 10 10 gold badges 139 139 silver badges 297 297 bronze badges 1 Great answer but can you give me a more complex answer where this could be applied. A good example would be the second example in my question.DragonflyRobotics –DragonflyRobotics 2021-08-04 19:49:28 +00:00 Commented Aug 4, 2021 at 19:49 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. The derivative of 2 x 2 x is ln(2)2 x ln⁡(2)2 x, because 2 x 2 x is an exponential function. Now when dealing with an exponential function of the form y=a x y=a x, you can write a=e l n(a)a=e l n(a), so we can write a x=e x ln(a)a x=e x ln⁡(a). Then using the chain rule you get a x ln(a)a x ln⁡(a). For your example: d d x(2 x+4)x+1 d d x(2 x+4)x+1 And from the exponent rule we know e(x+1)ln(2 x+4)=(2 x+4)x+1 e(x+1)ln⁡(2 x+4)=(2 x+4)x+1 Then apply the chain rule e(x+1)ln(2 x+4)d d x(x+1)ln(2 x+4)e(x+1)ln⁡(2 x+4)d d x(x+1)ln⁡(2 x+4) Using the product rule, this gives e(x+1)ln(2 x+4)(ln(2 x+4)+x+1 x+2)e(x+1)ln⁡(2 x+4)(ln⁡(2 x+4)+x+1 x+2) And we saw that e(x+1)ln(2 x+4)=(2 x+4)x+1 e(x+1)ln⁡(2 x+4)=(2 x+4)x+1 So this ends up in (ln(2 x+4)+x+1 x+2)(2 x+4)x+1(ln⁡(2 x+4)+x+1 x+2)(2 x+4)x+1 You often, with more complex functions, need to use several rules. Just learn all the rules by heart and the rest is just practice, practice, practice, ... Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 4, 2021 at 20:18 answered Aug 4, 2021 at 19:48 Steven01123581321Steven01123581321 1,457 1 1 gold badge 10 10 silver badges 18 18 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The rules for x a x a and a x a x only work when a a is NOT a function of x x. If a a is a function of x x, then we exponentiate, like for example in the case of (2 x+4)x+1(2 x+4)x+1: We rewrite it as e ln((2 x+4)x+1)=e(x+1)ln(2 x+4)e ln⁡((2 x+4)x+1)=e(x+1)ln⁡(2 x+4). To differentiate this, we use the rule for the exponential, so the derivative is e(x+1)ln(2 x+4)((x+1)ln(2 x+4))′e(x+1)ln⁡(2 x+4)((x+1)ln⁡(2 x+4))′, which, by the product rule for the derivative, becomes e(x+1)ln(2 x+4)((x+1)′ln(2 x+4)+(x+1)ln(2 x+4)′)=e(x+1)ln(2 x+4)(ln(2 x+4)+(x+1)⋅2 2 x+4)=(e ln(2 x+4))x+1(ln(2 x+4)+(x+1)⋅2 2 x+4)=(2 x+4)x+1(ln(2 x+4)+x+1 x+2 e(x+1)ln⁡(2 x+4)((x+1)′ln⁡(2 x+4)+(x+1)ln⁡(2 x+4)′)=e(x+1)ln⁡(2 x+4)(ln⁡(2 x+4)+(x+1)⋅2 2 x+4)=(e ln⁡(2 x+4))x+1(ln⁡(2 x+4)+(x+1)⋅2 2 x+4)=(2 x+4)x+1(ln⁡(2 x+4)+x+1 x+2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2021 at 20:21 WolfuryoWolfuryo 587 5 5 silver badges 14 14 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. You seem to be confusing two different rules: if f(x)=x a f(x)=x a, then f′(x)=a x a−1 f′(x)=a x a−1; however, if f(x)=a x f(x)=a x, then f′(x)=a x⋅ln(a)f′(x)=a x⋅ln⁡(a). This is because if f(x)=e x f(x)=e x, then by definition, f′(x)=e x f′(x)=e x, and if f(x)=a x f(x)=a x, then we can rewrite f(x)f(x) as e x ln(a)e x ln⁡(a) and use the chain rule and product rule to obtain f′(x)=e x ln(a)⋅ln(a)=a x⋅ln(a)f′(x)=e x ln⁡(a)⋅ln⁡(a)=a x⋅ln⁡(a). You can't apply the standard exponent rule because the function x↦2 x x↦2 x has a constant base and a variable exponent, not a variable base and constant exponent. Every time you have a function of the form x↦a x x↦a x, you should use the "ln(x)ln⁡(x) rule". If f(x)=g(x)h(x)f(x)=g(x)h(x), then things become even trickier: now, both the base and the exponent are variable. To get around this, you should rewrite f(x)f(x) as e h(x)⋅ln(g(x))e h(x)⋅ln⁡(g(x)). In the rewritten form, only the exponent is variable, and so we can use the "ln(x)ln⁡(x) rule", along with the chain and product rule. Let y=e h(x)⋅ln(g(x))y=e h(x)⋅ln⁡(g(x)) and u=h(x)⋅ln(g(x))u=h(x)⋅ln⁡(g(x)). Then, d y d x=d y d u⋅d u d x=e u⋅d d x(h(x)⋅ln(g(x)))=e h(x)⋅ln(g(x))⋅(h′(x)ln(g(x))+h(x)h′(x)g(x))=g(x)h(x)⋅(h′(x)ln(g(x))+h(x)h′(x)g(x)).d y d x=d y d u⋅d u d x=e u⋅d d x(h(x)⋅ln⁡(g(x)))=e h(x)⋅ln⁡(g(x))⋅(h′(x)ln⁡(g(x))+h(x)h′(x)g(x))=g(x)h(x)⋅(h′(x)ln⁡(g(x))+h(x)h′(x)g(x)). In the case g(x)=2 x+4 g(x)=2 x+4 and h(x)=x+1 h(x)=x+1, this becomes d y d x=(2 x+4)x+1⋅(ln(2 x+4)+x+1 2 x+4)d y d x=(2 x+4)x+1⋅(ln⁡(2 x+4)+x+1 2 x+4) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 4, 2021 at 20:50 JoeJoe 23.6k 4 4 gold badges 58 58 silver badges 104 104 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus derivatives logarithms exponentiation chain-rule See similar questions with these tags. 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190568	https://theses.hal.science/tel-04208334/file/2023UPSLD004.pdf	HAL Id: tel-04208334 Submitted on 15 Sep 2023 HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. Constructive and asymptotic estimates for the solutions of some linear and nonlinear PDEs Claudia Fonte Sánchez To cite this version: Claudia Fonte Sánchez. Constructive and asymptotic estimates for the solutions of some linear and nonlinear PDEs. Analysis of PDEs [math.AP]. Université Paris sciences et lettres, 2023. English. NNT : 2023UPSLD004. tel-04208334 Préparée à Université Paris Dauphine-PSL Constructive and asymptotic estimates for the solutions of some linear and nonlinear PDEs Soutenue par Claudia Fonte Sánchez Le 18/04/2023 École doctorale no543 Sciences de la Décision, des Organisations, de la Société et de l’Echange Spécialité Mathématiques appliquées Composition du jury : Delphine SALORT Professeur, Sorbonne Université Président Vincent BANSAYE Professeur, Ecole polytechnique Rapporteur José A. CAÑIZO Professeur, Universidad de Granada Rapporteur Clément MOUHOT University of Cambridge, Centre for Mathematical Sciences Examinateur Benoît PERTHAME Professeur, Sorbonne-Université Examinateur Tony LELIEVRE Professeur, Ecole des Ponts ParisTech Examinateur Emeric BOUIN Professor, Université Paris Dauphine-PSLment Examinateur Stephane MISCHLER Professor, Université Paris Dauphine-PSL Directeur de thèse Contents 1 Introduction 5 1.1 Dynamical systems and mean-ﬁeld limit . . . . . . . . . . . . . . . . . . . . . . . 5 1.1.1 Two dynamical systems on neuroscience . . . . . . . . . . . . . . . . . . . 7 1.2 Partial diﬀerential equations models . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2.1 Function spaces and weak solutions . . . . . . . . . . . . . . . . . . . . . . 10 1.2.2 Diﬀusion models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.3 Kinetic Fokker-Planck model . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.2.4 Selection Mutation model . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3 Long time behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.3.1 The eigentriplet problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.3.2 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.3.3 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 1.4 Thesis plan and contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2 Existence of solution of the eigentriplet problem 21 2.1 Existence through a stationary problem approach . . . . . . . . . . . . . . . . . . 21 2.1.1 The Banach lattice framework . . . . . . . . . . . . . . . . . . . . . . . . 21 2.1.2 Existence part of the Krein-Rutman theorem . . . . . . . . . . . . . . . . 24 2.1.3 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.2 Existence through a dynamical approach . . . . . . . . . . . . . . . . . . . . . . . 38 2.2.1 About dissipativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.2.2 Existence in the dissipative case . . . . . . . . . . . . . . . . . . . . . . . 40 2.2.3 About weak dissipativity . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.2.4 First existence result in the weakly dissipative case . . . . . . . . . . . . . 46 2.2.5 Second existence result in the weakly dissipative case . . . . . . . . . . . . 48 3 Irreducibility and the geometry of the ﬁrst eigenvalue 51 3.1 More about positivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.2 Irreducibility and strong maximum principle . . . . . . . . . . . . . . . . . . . . . 53 3.3 The geometry of the ﬁrst eigenvalue problem . . . . . . . . . . . . . . . . . . . . 56 3.4 Mean ergodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 4 The geometry of the boundary point spectrum 65 4.1 Complexiﬁcation and the sign operator . . . . . . . . . . . . . . . . . . . . . . . . 65 4.2 On the point boundary spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.3 Stronger positivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.4 On the triviality of the boundary spectrum . . . . . . . . . . . . . . . . . . . . . 73 4.5 Ergodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 4.6 A word about spectral analysis argument . . . . . . . . . . . . . . . . . . . . . . 79 1 2 CONTENTS 5 Quantitative stability 83 5.1 About quantiﬁed positivity conditions . . . . . . . . . . . . . . . . . . . . . . . . 83 5.2 Asymptotic stability under Doblin condition . . . . . . . . . . . . . . . . . . . . . 84 5.3 Asymptotic stability under Harris condition . . . . . . . . . . . . . . . . . . . . . 85 5.4 The weak dissipativity case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6 Application to Diﬀusion models 91 6.1 Diﬀusion with rough coeﬃcients in a bounded domain . . . . . . . . . . . . . . . 91 6.2 Diﬀusion in Rd with strong potential conﬁnement . . . . . . . . . . . . . . . . . . 105 6.3 Diﬀusion in Rd with weak potential conﬁnement . . . . . . . . . . . . . . . . . . 108 6.4 Diﬀusion in Rd with drift conﬁnement . . . . . . . . . . . . . . . . . . . . . . . . 111 7 The kinetic Fokker-Planck equation 117 7.1 The trace problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 7.2 Well-posedness problem with inﬂow term at the boundary . . . . . . . . . . . . . 120 7.3 Well-posedness problem with reﬂection condition . . . . . . . . . . . . . . . . . . 124 7.4 The ﬁrst eigenvalue with reﬂection at the boundary . . . . . . . . . . . . . . . . . 128 8 A mutation-selection model 139 9 Time elapse model with memory 147 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 9.1.1 The age- and leaky memory-structured model . . . . . . . . . . . . . . . . 147 9.1.2 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 9.1.3 Assumptions and main results . . . . . . . . . . . . . . . . . . . . . . . . . 151 9.1.4 Discussion of the methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 9.1.5 Plan of the chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 9.2 Well-posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 9.3 Exponential stability in the non-interacting case . . . . . . . . . . . . . . . . . . 160 9.4 Stationary solutions for arbitrary connectivity strength . . . . . . . . . . . . . . . 163 9.4.1 Doeblin-Harris method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 9.4.2 Alternative proof for the existence using Schauder’s ﬁxed-point theorem . 166 9.4.3 Formula in the case of short-term synaptic depression . . . . . . . . . . . 171 9.5 Exponential stability in the weak connectivity regime . . . . . . . . . . . . . . . . 172 10 Concentration inequality 177 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 10.1.1 The FitzHugh-Nagumo model . . . . . . . . . . . . . . . . . . . . . . . . . 178 10.1.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 10.1.3 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 10.1.4 Plan of the Chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 10.2 Brief summary of classical results . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 10.3 Existence and uniqueness proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 10.4 Bernstein’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 10.5 FitzHugh-Nagumo parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 10.6 Nonparametric estimation of µt . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 10.7 Testing of the interaction between agents . . . . . . . . . . . . . . . . . . . . . . 200 10.8 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 10.8.1 Proof of lemma 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 Remerciements Tout d’abord, je voudrais remercier mon directeur de thèse, Stéphane Mischler, qui n’est pas seulement un scientiﬁque impressionnant, mais aussi une personne merveilleuse. J’admire vraiment ses capacités d’analyse ainsi que l’étendue et la profondeur de ses connaissances, que je trouve particulièrement inspirantes. En même temps, je lui suis profondément reconnaissant pour sa sollicitude et sa gentillesse. Je remercie également les rapporteurs, José A. Cañizo et Vincent Bansaye, ainsi que les autres membres du jury, Delphine Salort, Emeric Bouin, Benot Perthame, Clément Mouhot et Tony Lelievre, d’avoir pris le temps de lire et d’évaluer ce manuscrit. J’adresse également mes remerciements à Valentin, avec qui j’ai fait l’expérience du "travail en temps covid" - nous ne nous sommes pas encore rencontrés -, à Pierre Gabriel, pour tous ses conseils et ses discussions mathématiques animées, et surtout à Marc Hoﬀmann, qui m’a conseillé et guidé en tant que second directeur de thèse. Dans une autre dimension, je remercie tous ceux qui m’ont soutenu logistiquement et administrativement tout au long de ces trois ans et demi. Tout d’abord, le FSMP et le programme de recherche et d’innovation Horizon 2020 de l’Union européenne dans le cadre de la convention de subvention Marie Skodowska-Curie nř 754362 pour le ﬁnancement de cette thèse, et plus particulièrement Ariela, pour toute son aide. Je tiens également à remercier César, Isabelle, Marie et Béatrice Baeza pour tout le travail administratif qu’ils eﬀectuent au quotidien. Je remercie également mes collègues de bureau : Amirali, pour son accueil chaleureux et son séminaire de bureau ; Adrien H. pour toute l’aide et les conseils sur la bureaucratie française dans et hors du cadre universitaire ; Quan, pour sa compagnie tout au long de ces deux années, presque sans absences, et bien que nous ayons moins coincidé, Adrien S. et plus récemment Lucas. Merci aussi à tous les autres membres du lab et aux autres doctorants. Merci à vous tous pour les déjeuners, ces agréables pauses de midi, le séminaire, l’école d’hiver, les pique-niques, les après-midi de jeux de société,... les parties de dominos ici et là. J’aimerais remercier tout particulièrement l’équipe italienne, qui fait vivre le laboratoire grâce à son charisme particulier, et en particulier Giovanni, toujours un rival aux dominos et un grand allié de Crew. Je remercie également Changqing, que j’ai rencontré en premier et avec qui j’ai cheminé depuis le master. Je voudrais également remercier les doctorants de Jussieu, en particulier Chourouk, Elena et Giorgia. Je remercie également mes nouveaux amis, Alberto, Agustin et Cata - il y a déjà un an. Extiendo mis agradecimientos en español ahora para la gran familia cubana. Para el polo de Toulouse, incluidos aquellos que ya no están en Toulouse, muchas gracias por su cálida acogida en mi visita hace ya unos años, y, sobre todo, por los activos grupos de comunicación, donde se solucionan tantos problemas y se comparten tan buenas ideas y material; en especial un agradecimiento a Josue. Al polo de Compiegne, a la Madrina, a Pedro y las niñas, por ser la familia lejos de casa, los padrinos afectuosos, los oídos comprensivos y la opinión esclarecedora. Finalmente, al polo de Paris, incluidos los ausentes, por el recibimiento, el acompañamiento en esos primeros días en que todo es nuevo, en los días que siguen, los paseos, los café, las quedadas, la amistad: gracias a todos, en especial a Nani, Mariano, Ale y Frank. Para la muchachada, Jesus, Agua, Capi, Monra, Chamonix (el chaparrón, el profesor, 3 4 CONTENTS Patrañas), Cossio, Ana, Suney y Paula, un grupo de amigos para reunir una y otra vez a través de los años. À Loria, le premier ami que je me suis fait à Paris et qui m’a laissé une marque indélébile. À Kathi, qui est devenue si proche en si peu de temps. A Suney, quien ha tirado de mi en los momentos más tensos. A Nora y Yanni, conﬁdentes, consejeros y amigos. A Milena, que siempre ha estado ahí para mi. A mi familia, a mis abuelos, que realmente leyeron fragmentos de demostraciones y evacuaron dudas de análisis durante el doctorado, a Paula, a mi lado en las buenas y las malas, a mamá, con quien tanto he debatido en cada incertidumbre desde que decidí seguir sus pasos y a papá, quien me enseñó a pensar desde pequeña. Al resto de mi familia que ha estado siempre presente desde la distancia para acompañarme. A todos, muchas gracias. Chapter 1 Introduction This thesis is devoted to the study of asymptotic estimates for partial diﬀerential equations (PDE). These estimates can be in two directions: 1. The study of the limit when the number of individuals n included in a dynamic system becomes large. 2. The study of the asymptotic behavior when the time t becomes large. In the ﬁrst direction we go from the individual behavior of diﬀerent agents at a microscopic level to the description of the collective behavior at a mesoscopic level through the probability density of the distribution of a population in certain states. In the second direction, we start in this work from the partial derivative model that describes the mesoscopic behavior and we study the existence of stationary or quasi-stationary solutions and their stability. In this chapter, we introduce the diﬀerent elements to be considered for our work, as well as the main results that will be presented in the following chapters. 1.1 Dynamical systems and mean-ﬁeld limit Dynamical system is the study of the particularity of time dependent evolving systems. The modern theory of it originated at the end of the 19th century with fundamental questions concerning the stability of the solar system. The questions raised in this context led to the emergence of a wide theoretical ﬁeld with applications in physics, biology, astronomy, sociology and other areas. We are interested in the evolution over time t of N variables X(N) = (X1, . . . , XN) according to the rates of change dX1 dt , . . . , dXN dt . Thus we obtain n diﬀerential equations dXi dt = Fi(X(N)). (1.1.1) State changes can be deterministic or random. When Fi referred to deterministic functions, it leads to a system of ordinary diﬀerential equations (ODE) as in the foundation work include on Birkhoﬀ’s 1927 seminal book . On the other hand, the inclusion of random elements such as Gaussian noise or some stochastic jump process, leads us to consider stochastic diﬀerential equations (SDE) . It is this second case in which we are interested here. 5 6 CHAPTER 1. INTRODUCTION We talk then of N random variables Xi evolving in E ⊂Rd (frequently E = Rd or E = Rd + := {x ∈Rd : xi > 0}) and we deﬁne P(E) the space of probability measures on E and P(P(E)) the space or probability measures on P(E). To go further, we consider that each individual is indistinguishable and follows a “typical behavior”. This is naturally described by exchangeable random variables (i.e. such that the probability laws are invariant under coordinates permutations) which are driven by a model equation with identical structure and parameters. This conditions are usually referred to as the “mean-ﬁeld approximation”, we recommend F. Golse’s lecture . We deﬁne the empirical measure µN ∈P(E) of mean ﬁeld approximation X(N) by µN = 1 N N X j=1 δXj. We say that the system (1.1.1) has a mean-ﬁeld limit if there exist µ ∈P(E) such that µN converge to µ. The equation satisﬁed by µ is called then the mean-ﬁeld limit equation. The literature on mean-ﬁeld approximation and mean-ﬁeld limits is vast, see and references therein. One of the motivation of these works being the possibility of going from one measurement scale to another: from the microscopic level in which the trajectory of each individual can be separated to a mesoscopic level in which the behavior of a population is studied instead; this is from the trajectories of the evolution of diﬀerent individuals to the density of the distribution of these in the states of interest. The applications of this type of result are varied, among them, • The possibility of understanding how a change at the individual level aﬀects the behavior of a population. • Make estimates related to individual behavior from measurements on a larger scale (sometimes easier to obtain in practice). • Model simpliﬁcation: it is usually easier to carry out a qualitative study of the properties of a PDE with low dimensionality, than with a system of many particles. On the other hand, we remark that the mean-ﬁeld approximation assumptions do not imply the independence of the random variables Xi, however, under certain assumptions, asymptotic independence is obtained. This property is known as chaos propagation and is closed related to the mean-ﬁeld limit. Indeed, let GN ∈P(E) be the law of X(n), the mean-ﬁeld approximation of the system, GN j the jth-marginal of GN deﬁned by GN j = Z EN−j GNdxj+1 . . . dxN then we say that there is chaos propagation if ∀j ≥1, GN j →µ⊗j weakly in P(Ej) as N →∞. It is know that the convergence of the empirical measure to the mean ﬁeld limit is equivalent to the chaos propagation property above, see and reference therein. This brings us to another important application for statistics: • The possibility of transferring properties of systems of independent variables to non-independent systems. For example, the law of large numbers or the central limit theorem. Below we present two examples of dynamic systems describing the behavior of a neuronal population and discuss about their mean-ﬁeld approximations. 1.1. DYNAMICAL SYSTEMS AND MEAN-FIELD LIMIT 7 1.1.1 Two dynamical systems on neuroscience We present here two models, the FitzHugh-Nagumo (FhN) model and the time elapse model (TEM). Before describing each of then in details, we would like to comment on the common and divergent points between them. In the ﬁrst of these models (FhN) the activity of the neurons is modeled through the evolution of the voltage of each neuron, while in the second the variable of interest is the time elapsed between two spikes. To understand the biological interest of both of them, we would like to emphasize that information between one neuron and another is transmitted through discharges occurring at the synapse sites. Moreover, neuroscientists have come to the conclusion that most of the information can be encoded as a function of the time between spikes. When the variable being modeled is voltage, discharge times can be obtained by supposing that, if the voltage value exceeds a certain threshold, a discharge occurs. On the other hand, directly modeling the time between spikes is a relevant simpliﬁcation. In both cases when describing the behavior of each neuron, three elements are determinant i The behavior of a neuron when no discharge occurs. ii The interaction between neurons. iii Random noise (which can come from errors in transmission or small random external signals). The FhN model is a simpliﬁcation of the 4-dimensional Huxley-Hodgkin model (HH) that describes the physiological process of a discharge. In the HH model the one of the 4 variables correspond to the voltage and the other three to model the calcium, potassium and sodium channels involved in the discharge. We can say that the FhN model tries to make a more precise description of the way the spike is generated. Thus, in the absence of any other element, the model presents a description of the behavior of an isolated neuron that can (and has been) of interest on its own. In the TEM model the spike generation process is not modeled. About (ii) in the FhN model, electrical synaptic connections between neurons are considered, whereby one neuron has an inhibitory eﬀect on another if its voltage is lower, and excitatory if the voltage is higher. We emphasize that this term inﬂuences during the whole time, the neurons are constantly interacting, which is less realistic since the neurons interact only during the discharge. In the TEM model the behavior of the neuron is described through the ﬁring rate at which the neurons spike, which is a function of all the spikes that have occurred up to the current time. In this sense, TEM oﬀers a better description of the interaction, easily adaptable to the case of a population of inhibitory or excitatory neurons. In Raad et. al. work , they present a version that includes both inhibitory and excitatory populations. It also allows modeling eﬀects such as delayed detection. In the FhN model the randomness is introduced by the addition of a Gaussian noise modulated by a constant sigma, while in TEM it comes from considering a Hawkes process. Consequently, the ﬁrst model is continuous in time while the second one has jumps every time a spike occurs. From the biological point of view, we should note that when a spike occurs the time between spikes is reset (to 0), which means that at the level of a neuron, the classical TEM model loses the memory of its previous activity, which makes it diﬃcult to model phenomena such as adaptability and short term plasticity, which is the advantage of the version of the model presented in Chapter 7. FitzHugh-Nagumo model The FitzHugh-Nagumo (FhN) model was ﬁrst introduced in the works of FitzHugh and Nagumo as a simpliﬁcation of the Huxley-Hodgkin model (HH) that describes the evolution 8 CHAPTER 1. INTRODUCTION of the membrane potential of a neuron . The dynamics is based on two variables, a variable x which corresponds to the membrane potential and a recovery variable y, which satisfy the equations ˙ x = F(x, y), ˙ y = G(x, y), for F(x, y) = x−x3/3−y+I and G(x, y) = 1/c(x+a−by) with a, I, c > 0, b ∈R. On this model, I denotes the total membrane current and is a stimulus applied to the neuron, c determines the strength of damping while a and b govern two important characteristics of the oscillating solution, namely spike rate and spike duration . With only these elements the system shows the most important properties of the 4-dimensional HH model such as refractoriness, insensitivity to further immediate stimulation after one discharge, and excitability, the ability to generate a large, rapid change of membrane voltage in response to a very small stimulus. Numerous works have been aimed at studying the ODE model and their properties, we reference the book of Rocsoreanu et al. and the references therein. More recently, specialists have been interested in the passage of the behavior of a neuron to neural networks, see for example Mischler et al. , Lucon and Poquet and Baladron et al. . When neurons interact through electrical synapses, it has been proposed that the evolution of N neurons satisﬁes ( dXi t = (F(Xi t, Y i t ) −PN j=1 Jij(Xi t −Xj t ))dt + σdBi t dY i t = G(Xi t, Y i t )dt , (1.1.2) where coeﬃcients Jij > 0 represent the eﬀect of the interconnection between the neurons, and the term Bi t refers, as usual, to independent Brownian motions. We consider that the interactions are symmetric and identical for every pair of neurons in the network, which in particular implies that all neurons are connected. The strength of the interaction is measured by the parameter Jij = J that we re-parametrize as a function of a new parameter λ such that J = λ N , where N is the number of neurons in the network. In this form, we obtain the mean ﬁeld approximation, ( dXi t = (F(Xi t, Y i t ) −λ N PN j=1(Xi t −Xj t ))dt + σdBi t dY i t = G(Xi t, Y i t )dt . (1.1.3) Note that this equation can be rewrite as    dXi t = F(Xi t, Y i t ) −λ(Xi t − R R xµN t (dx)) dt + σdBi t dY i t = G(Xi t, Y i t )dt, which gives us an intuition about the form of the mean ﬁeld limit equation ( dXt = F(Xt, Yt) −λ(Xt −E[Xt]) dt + σdBt dYt = G(Xt, Yt)dt. (1.1.4) As a consequence the law µt of (Xt, Yt) satisﬁed the mean-ﬁeld limit equation, ( ∂tµ(x, y) = −∇((F, G)µ(x, y)) −∇(µ(x, y) R (x −¯ x)µ(d¯ x, d¯ y)) + 1 2σ2∂xxµ(x, y) µt=0 = µ0. (1.1.5) Indeed, in Baladron et. al. work [15, theo. 4], the authors prove that the process (Xi,N, Y i,N) the solution Eq. (1.1.3) for the neuron i converge in law to the process (X, Y ), the solution of the mean-ﬁeld limit equation (1.1.4) when N →∞for any i ∈Z+. There they prove the mean-ﬁeld limit convergence by a coupling argument. In chapter 10, we present a diﬀerent approach to arrive at this result, that allows us to obtain in addition a control of the deviation of the probability measures through a concentration inequality in the spirit of Bernstein inequality. 1.2. PARTIAL DIFFERENTIAL EQUATIONS MODELS 9 Time elapse model The second example we present here is the time elapse model (TEM) also knows as “age dependent Hawkes process for neurons” . In eﬀect, it is a Hawkes process described by the system of equations          Zi t = R t 0 R ∞ 0 1{z≤f(Ai s,Xs)}πi(ds, dz) Ai t = Ai 0 + t − R t 0 Ai sZi(ds) Xt = 1 N Pn j=1 R t 0 h(t −s, Aj s)Zj(ds), with (A0, M0) ∼u0 and where π is a Poisson random measure on R+ × R+ with intensity dsdz (see ). Here Zi is a point process indicating the moment of the spike of the neuron i, Ai t is the elapse time since the previous spikes of the neuron i at time t, and Xt gathers information on all network activity. We note that Xt is a function of all the spike until time t modulated by the kernel h that allows to consider some delay (e. g. exponential delay on t). In Chevallier’s work [58, theo. 4.1], the author proves the mean-ﬁeld convergence of the law of the stochastic process Ai toward ρt, the solution of the mean-ﬁeld limit equation            ∂tρt(a) + ∂aρt(a) = −f(a, εxt)ρt(a), ρt(0) = R ∞ 0 f(a, εxt)ρt(a)da, xt = R t 0 R ∞ 0 h(t −s)f(a, εxs)ρs(a)dads, ρ0(a) = u0(a), (1.1.6) also by a coupling argument. The transport equation (1.1.6) has been widely used to study the properties of the behavior of neuronal population, in particular the convergence toward a stationary state when the interaction between neuron (here modulated by the inﬂuence of xt in the ﬁring rate f) is weak and the generation of oscillatory behavior when it is not. We can see for example the works of Pakdaman et. al. for the study of this model and , , for versions of it. In Chapter 9, we present a diﬀerent version of this model that deals with the lack of memory of the previous spike of a particular neuron in this model, to allow the description of eﬀects of real neurons such that adaptability and short time plasticity. Then we make use of the mean ﬁeld limit equation to study the long time behavior under weak interactions. 1.2 Partial diﬀerential equations models As mentioned in the previous section, we refer to models at the mesoscopic level, when they describe the behavior of a given population by studying the probability density of ﬁnding an individual from that population in a given state. We have seen that such models can arrive from the mean ﬁeld limit of a particle system when the number of individuals becomes large. In general, these models assume that we have a homogeneous population satisfying certain common rules and allow us to study the global behavior. In the previous section we saw two ﬁrst examples of these models. In the ﬁrst case, from the FhN model with Brownian noise in only one of the variables, a kinetic Fokker-Planck type model is obtained for the probability density. In the second of a model based on a Poisson process, we arrive at a transport equation. Here are three more models that we will introduce in the following sections and to which we will dedicate chapters 6, 7 and 8 respectively. • Diﬀusion models (which generally comes up considering a particle system with Gaussian noise in all variables i.e. considering a strictly positive sigma matrix aﬀecting all states of particle i). 10 CHAPTER 1. INTRODUCTION • A general version of the Kinetic Fokker-Planck model, considering in addition the case of a bounded domain with diﬀerent boundary conditions. • Mutation-Selection model, which is used in evolutionary theory. 1.2.1 Function spaces and weak solutions In order to carry out a rigorous deﬁnition of our model (to ensure the existence and uniqueness of the PDE solution), we frequently work with the so-called “weak solutions” in a certain function space, even measures (distributions) spaces. Let us introduce these spaces and concepts ﬁrst. In general we will work in Banach lattice X, that is a Banach space (X, ∥·∥) endowed with a compatible order relation ≥, which satisﬁes either X = Y ′ or X′ = Y for another dual Banach lattice Y . On the applications, the Banach lattices X will be one the following examples (or small variation of it) : • X := C0(E), the space of continuous functions which tend to 0 at inﬁnity (when E is not a compact set) endowed with the uniform norm ∥·∥∞; • X := Lp(E) = Lp(E, E , µ), the Lebesgue space of functions associated to the Borel σ-algebra E , a positive σ-ﬁnite measure µ and an exponent p ∈[1, ∞]; in some cases, for a weight function m we will alternatively use the Lp m := {f : fm ∈Lp} weighted space with the norm ∥f∥Lp m= ∥fm∥Lp; • X := M1(E) = (C0(E))′, the space of Radon measures deﬁned as the dual space of C0(E). In all the above examples, E denotes a σ-compact metric space, and we write E = ∪ER, with ER ⊂ER+1, ER compact (generaly E ⊂Rd). We always consider the test function space X := C∞ c (E) of the inﬁnity diﬀerentiable functions with compact support in E and occasional, the function space X := H1(E) = {f ∈L2(E) : ∂f ∈L2(E)}, where ∂f is considered in the distribution sense i. e. g = ∂f if Z E gϕ = − Z E f∂ϕ; ∀ϕ ∈C∞ c (E). For a bounded domain Ω⊂Rd we could also consider H1 0(Ω) = {f ∈H1 : f\|∂Ω= 0}. Let L : D(L) ⊂X →X be a linear operator, we say say that f is a weak solution of the equation ∂tf = Lf (1.2.7) if for all test function ϕ ∈C∞ c (E) we have ⟨fT , ϕT ⟩= ⟨f0, ϕ0⟩+ Z t 0 ⟨ft, L∗ϕt⟩dt, where ⟨·, ·⟩denote the dual application and L∗the dual operator of L (note that in any case the test function space is contained in the dual of X). In what follows we present each model by deﬁning the operator L and in every case we consider the model represented by the evolution equation (1.2.7). 1.2.2 Diﬀusion models A general elliptic operator in divergence form is given by Lf := ∂i(aij∂jf) + bi∂if + ∂i(βif) + cf, f ∈H1 0(Ω), (1.2.8) 1.2. PARTIAL DIFFERENTIAL EQUATIONS MODELS 11 where Ω⊂Rd is a bounded domain (i.e. an open and connected set) or Ω= Rd, and for simplicity we always assume d ≥3. We also always assume at least a boundedness and ellipticity condition on the (aij) matrix, namely aij ∈L∞(Ω), ∃ν > 0, ∀ξ ∈Rd, aijξiξj ≥ν\|ξ\|2, (1.2.9) and some conditions on the coeﬃcients bi, βj and c which will be described below depending on each particular case. We present four situations that we will consider in Chapter 6. Diﬀusion with rough coeﬃcients in a bounded domain In the ﬁrst case, we consider the general elliptic operator in divergence form (1.2.8) in the case of a bounded and smooth enough domain Ω⊂Rd with general elliptic condition on aij as formulated above. We further assume bi, βj ∈Lr(Ω), c ∈Lr/2(Ω), r > d, (1.2.10) as well as c + divβ ≤c0 and c −divb ≤c0, (1.2.11) for some c0 ∈R. Diﬀusion in Rd with strong potential conﬁnement In the second case, we deal with an unbounded domain Ω= Rd which brings us new diﬃculties. For a solution to be well deﬁned, it must be ensured that the growth of the solutions in the region outside compact sets is controlled. One way to do this is to add a conﬁnement condition. We consider then the elliptic operator in the form Lf := ∆f + b · ∇f + cf, f ∈H1(Rd), (1.2.12) with b ∈L∞ loc(Rd), c ∈L2 loc(Rd) and a conﬁnement condition that we impose through the properties of the potential function c, which is roughly speaking c →−∞as \|x\|→∞. More precisely, we assume σi+ ∈Ld/2, meas{σi ≥K} < ∞, ∀K < 0, (1.2.13) with either σ1 := c + \|b\|2/κ for some constant κ ∈(0, 4) or either σ2 := c + divb/2. For example, when we assume that c ∼−\|x\|γ and b ∼x\|x\|β−1 as \|x\|→∞, the condition (1.2.13) for σ1 is reached when γ > max(0, 2β) or when γ = 2β > 0 and some conditions on the constants involved in the behavior of the coeﬃcients. In that context, the condition (1.2.13) for σ2 is more general since it is reached when γ > max(0, β −1) or when γ = β −1 > 0 and some conditions on the constants involved in the behavior of the coeﬃcients. Diﬀusion in Rd with weak potential conﬁnement In the third case, we return to the previous context of the elliptic operator (1.2.12) but now with a weaker conﬁnement condition: we assume that c converges to a constant. With no loss of generality, we may assume c →0. More precisely, we consider the elliptic operator Lf := ∆f + b · ∇f + rcf, (1.2.14) now with c ∈C0(Rd), b ∈C0(Rd) and r ∈R+ a parameter that we will make move. 12 CHAPTER 1. INTRODUCTION Diﬀusion in Rd with drift conﬁnement Finally, we consider the elliptic operator Lf := ∆f + b · ∇f + cf, with a drift conﬁnement as it is the case for the Fokker-Planck operator. More precisely, and for the sake of simplicity, we assume here b = ∇U, U(x) = 1 γ ⟨x⟩γ, γ > 0. (1.2.15) When γ = 2 and c = x, that operator corresponds to the classical harmonic Fokker-Planck operator which is known to be related to the standard Poincaré inequality and to the standard log-Sobolev inequality, see [11, 12, 233] or more recently [14, 138] and the references therein. In the present situation, we impose that the contribution of c has lower inﬂuence at the inﬁnity that the drift term b and we assume c ∈L∞ loc(Rd), ∃C0, R0 > 0, ∀x ∈Bc R0, \|c(x)\|= o(\|x\|2(γ−1)). (1.2.16) We further assume that c ≥divb when γ ∈(0, 1]. (1.2.17) The action of the drift term will be revealed through the choice of a convenient “conﬁning space". More precisely, for a weight function m : Rd →[1, ∞), we will work in a weighted Lebesgue space. 1.2.3 Kinetic Fokker-Planck model We present now a general version of the kinetic Fokker-Planck evolution equation, associated to the operator Lf := −v · ∇xf + ∆vf + b · ∇vf + cf, (1.2.18) on functions f : O →R, where O := Ω× Rd, Ω⊂Rd is a domain, b : O →Rd is a given vector ﬁeld and c : O →R is a given function. It can be interpreted as a simple model of particles dynamic which move and interacts with a thermal bath. Collisions are typically modelized by a Fokker-Planck operator ∆vf + divv(vf) (when b = v and c = d) which takes into account a thermal bath of (Gaussian) whitenoise. We will consider the case when Ωis a bounded domain and the equation is complemented with a boundary condition. More precisely, we assume the classical balance between the values of the trace γf of f on the outgoing and incoming velocities subsets of the boundary (γ−f)(x, v) = Rx(γ +f(x, .))(v) on Σ−, (1.2.19) where in this context we deﬁne Σx ± := {v ∈Rd; ± v · νx > 0} the sets of outgoing (Σx +) and incoming (Σx −) velocities at point x ∈∂Ω, next the sets Σ± = {(x, v) ∈Σ; ±νx · v > 0} = {(x, v); x ∈∂Ω, v ∈Σx ±}, and ﬁnally the outgoing and incoming trace functions γ±f := 1Σ± γf. Here and in the sequel, νx denotes the unit normal outward vector ﬁeld deﬁned on the boundary set ∂Ω. We similarly deﬁne the grazing velocity set Σ0 = {(x, v) ∈Σ; νx · v = 0}. 1.2. PARTIAL DIFFERENTIAL EQUATIONS MODELS 13 The reﬂection operator Rx is local in position, but can be local or nonlocal in the velocity variable, so that it writes (Rxg)(v) := Z Σx + r(x, v, v∗)g(v∗)v∗· νx dv∗, for a reﬂection kernel r : ∂Ω× Rd × Rd →R. Some classical general assumptions on r are r ≥0, R∗ x1 = 1, RxM = M , (1.2.20) for some positive function M = M (v), see for instance [52, 53, 54]. The second (normalisation) condition corresponds to the fact that all the particles reaching the outgoing boundary are put back on the incoming boundary (no mass is lost) while the third (reciprocity) condition means (when M is a Gaussian function) that the wall is in a local equilibrium state and is not inﬂuenced by the incoming particles. The normalization condition implies the local mass conservation Z Σx − Rxg\|ν · v\|dv = Z Σx + gν · vdv, (1.2.21) while the three assumptions (1.2.20) together also imply Z Σx − (Rxg)2M −1\|ν · v\|dv ≤ Z Σx + g2M −1 ν · vdv, (1.2.22) In the sequel, we will rather consider the possibly position dependent Maxwell boundary condition operator Rxg = α(x)Dxg + β(x)Γxg, (1.2.23) where the accommodation coeﬃcients α, β : ∂Ω→[0, 1] satisfy α(x) + β(x) =: ζ(x) ≤1, Γx is the specular reﬂection operator Γx(g(x, ·))(v) = g(x, Vxv), Vxv = v −2ν(x)(ν(x) · v), (1.2.24) and Dx is the diﬀusive operator Dx(g(x, ·))(v) = cM M (v)e g(x), e g(x) = Z Σx + g(x, w) ν(x) · w dw. (1.2.25) Here the constant cM := (2π)1/2 is such that cM f M = 1 and M stands for the standard Maxwellian M (v) := (2π)−d/2 exp(−\|v\|2/2), (1.2.26) or, more generally, M = M (\|v\|) ≥0 is such that D∗ x1 = 1, DxM = M , ⟨v⟩ϑM ∈L1(Rd), (1.2.27) with ϑ ≥1 (that last condition is necessary in order that the second relation above makes sense). The boundary condition (1.2.23) corresponds to the pure specular reﬂection boundary condition when β ≡1 and it corresponds to the pure diﬀusive boundary condition when α ≡1. When ζ ≡1, the Maxwell boundary condition operator (1.2.23) satisﬁes (1.2.20). On the contrary, when ζ ̸≡1, the L2 estimate (1.2.22) holds but not anymore the mass conservation (1.2.21). However, the following L1 estimate Z Σx − \|Rxg\|\|ν · v\|dv ≤ζ∗ Z Σx + \|g\| ν · vdv (1.2.28) holds, with 0 ≤sup ζ ≤ζ∗≤1. Finally, the case ζ ≡0 corresonds to the zero inﬂow problem. 14 CHAPTER 1. INTRODUCTION 1.2.4 Selection Mutation model The last example that we present here is the mutation-selection evolution equation associated to the mutation-selection operator Lf := J ∗f −W(x)f (1.2.29) deﬁned on functions f : Rd →R, f = f(x), where J is a the mutation kernel, ∗stands for the convolution operator and W is a conﬁning potential. This equation appears in the modeling of genetic variability in evolutionary biology. In this context, ft(x) represents the density of a population, at time t, of phenotypic trait x on the multi-dimensional phenotypic trait space Rd. Here, the rate of change in f per generation is considered given by the following two components: 1. Mutation. Considering µ be the mutation rate per gene per generation, the eﬀect of mutation on the present of a phenotypic trait x is usually modeling as −µf(x)+µ(J ∗f)(x), where the ﬁrst term gives the rate of loss of alleles having an average eﬀect x by mutation to other alleles having a diﬀerent eﬀect and the second term gives the rate of gain from the mutation to alleles with average eﬀect x from other alleles. Note that as we allow J to have a singular part, all this component can be include on the term J ∗f. 2. Selection The function W(x) represents the ﬁtness of the phenotype x and models the individual reproductive success, considering 0 as the optimum phenotypic value. This model has been widely used in the literature; we refer, for example, to the works of M. Kimura , R. Lande , W.H. Fleming and R. Bürger as examples of biological applications. 1.3 Long time behavior We now focus on the second asymptotic estimate that we address in this thesis, which will be the subject of most of the following chapters: the study of long time behavior for the evolution equations in the general form (1.2.7). Roughly speaking, we are interested in understanding under what conditions there exists a proﬁle f∞such that every solution ft of Eq. (1.2.7) converges to f∞when t tends to inﬁnity. When the equation (1.2.7) is mass conservative, this is ⟨ft⟩:= Z E ft = Z E f0, there is a wide range of literature that establish that under certain conditions, there is an unique positive stationary normalized solution f∞(with ∥f∞∥X= 1) such that we have the convergence ∥ft −⟨f0⟩f∞∥X≤Θ(t)∥f0 −f∞∥X; ∀f0 ∈X, with a rate Θ(t). If the problem is not mass conservative, we could not expect that ft converge to a stationary solution anymore (as the mass growth or decrease depending on t). Yet, we ﬁrst study the existence of a solution to the eigentriplet problem of ﬁnding (λ1, f1, φ1) ∈R × X × Y such that Lf1 = λ1f1, f1 ≥0, ∥f1∥= 1, (1.3.30) L∗φ1 = λ1φ1, φ1 ≥0, ∥φ1∥= 1. (1.3.31) we can easily check that the quantity ⟨e−λ1tft, φ1⟩is constant. Next, under a positive answer, one might expect that, granted the existence of such a triplet, a set of hypotheses could be arranged to obtain that ∥e−λ1tft −⟨f, φ1⟩f1∥≤Θ(t)∥f0 −⟨f0, φ1⟩f1∥, ∀t ≥0, ∀f ∈X. (1.3.32) 1.3. LONG TIME BEHAVIOR 15 Such is the aim of the theoretical results presented in this thesis from chapter 2 to 5. More in details, in this chapters, we revisit the Krein-Rutman theory for semigroups of positive operators in a Banach lattice framework and we provide some very general, eﬃcient and handy results with constructive estimates about 1. the existence of a solution for the eigentriplet problem (1.3.30)-(1.3.31); 2. the geometry of the ﬁrst eigenvalue; 3. the asymptotic stability of the ﬁrst eigenvector with possible constructive rate of conver-gence. This abstract theory is motivated and illustrated in chapters 6, 7 and 8 by the application of the general theory to the three models presented in the previous section. The present work is motivated by new problems and ideas presented in the course on the Krein-Rutman theorem by P.-L. Lions at Collège de France and by the recent contributions by Bansaye et al and by Cañizo and Mischler developing Harris techniques. Bringing and developing these ideas and techniques together with the more classical spectral analysis approach developed or synthesized in previous contributions by Krein and Rutman , by Arendt et al , by Mischler and Scher , by Bátkai et al and many others, we are then able to signiﬁcantly generalize and improve the Krein-Rutman theory for positive semigroups. In what follows in this section, we present the results of the ﬁrst four chapters and compare them with those existing in the literature. 1.3.1 The eigentriplet problem We next consider a positive semigroup S = SL on X, and we denote by L its generator, by D(L) ⊂X the domain of L, by ρ(L) ⊂C the resolvent set of L and by Σ(L) = C\ρ(L) the spectrum of L. We also denote by S∗and L∗the corresponding semigroup and generator on the dual space Y . We refer to the end of Section 2.1.1, for deﬁnition and related notations. We also remark that under the well-posedness of equation (1.2.7), the operator L generate a semigroup associated with the solution of such equation, see for example the text book . We now present some minimal condition for the existence and uniqueness of the triplet, in the spirit of what is established in the course of Lions . First we ask, (H1) ∃κ1 ∈R such that λ −L is invertible and (λ −L)−1 : X+ →X+ for any λ ≥κ1. We then set I := {κ ∈R; λ −L is invertible, (λ −L)−1 ≥0 for any λ ≥κ}, (1.3.33) which is a non empty and non upper bounded interval due to (H1). We ﬁnally set λ1 := inf I ∈[−∞, κ1]. (1.3.34) (H2)Second, we ask ∃κ0 ∈R such that inf I ≥κ0. Under conditions (H1) and (H2), there hold λ1 ∈[κ0, κ2] (1.3.35) and ∃λn ↘λ1, ∃ˆ fn ∈D(L) ∩X+, εn := λn ˆ fn −L ˆ fn ≥0, ∥ˆ fn∥= 1, ∥εn∥→0. (1.3.36) Then we demand (H3) for any sequence ( ˆ fn) of X such that (2.1.20) holds, there exist f1 ∈X+{0} and a subsequence ( ˆ fn′) such that ˆ fn′ ⇀f1 for the weak convergence or the weak ∗convergence. 16 CHAPTER 1. INTRODUCTION Under these condition for the primal problem, asking (H3) also for the dual, we can prove that the triplet exists. This is show in Theorem 2.1.20 in Chapter 2, where a prove is presented following the ideas of Lions’ course. The conditions (H1)-(H2)-(H3) are really weak, but somehow leaves room for the question of how to check it in applications. A possible more concrete conditions are (H1∗) L is the generator of a positive semigroup S = SL; (H2∗) ∃κ0 ∈R, ∃ψ0 ∈Y+{0} such that L∗ψ0 ≥κ0ψ0; (H3∗) there exist κB < κ0 and a splitting L = A + B such that A ∈B(X), B −α is invertible and RB(α) := (α −B)−1 is uniformly bounded for any α ≥κB and there is N ≥1 such that (RB(α)A)N : C →C is positive and uniformly bounded in α ≥κ0 (1.3.37) (RB∗(α)A∗)N : C∗→C∗is positive and uniformly bounded in α ≥κ0 (1.3.38) for C and C∗are relatively sequentially compact for the corresponding weak topology in X and its dual Y and 0 / ∈C, where the closure is taken in the sense of the weak topology in both case. Then we have the following result Theorem 1.3.1. Under (H1∗)-(H2∗)-(H3∗) the ﬁrst eigentriplet problem (1.3.30)-(1.3.31) admits a solution (λ1, f1, φ1) ∈R × X × Y . This result generalize the one of German school, see Büguer , Greiner et. al. , and Webb , that assume that C have strong compactness and include the case of weak compactness in M1 addressed by the probabilistic group [167, 168, 169, 170, 119, 50]. We also study the problem without spectral gap through a dynamic approach (see Theorem 2.2.4 in Chapter 2), which we believe is a novelty with respect to the literature. For a more detailed discussion on the subject, please refer to the section 2.1.3 of chapter 2. 1.3.2 Geometry A second issue is about the geometry of the spectrum and the principal eigenvalue problem. One way to characterize λ1, and that is related to the way we will solve the ﬁrst eigentriplet problem, is λ1 := sup{ℜeλ; λ ∈Σ(L)} = inf{κ ∈R; κ′ ∈ρ(L), ∀κ′ ≥κ}. The boundary spectrum is then deﬁned by Σ+(L) := Σ(L) ∩∆λ1 = {λ ∈Σ(L); ℜeλ = λ1}, where ∆α is the open half plan ∆α := {z ∈C; ℜez > a}. We are next interested by describing Σ+(L) and to ﬁnd conditions on L so that at least we are in one of the following cases 1. λ1 is algebraically and geometrically simple, 2. (better) Σ+(L) is trivial, namely Σ+(L) ∩ΣP (L) = {λ1}, (1.3.39) where ΣP is the point spectrum set, 3. (even better) we have ∃κ < λ1; Σ(L) ∩∆κ = {λ1}. (1.3.40) 1.4. THESIS PLAN AND CONTRIBUTIONS 17 In the second situation (1.3.40), we say that there is a spectral gap: a band separates the spectral value λ1 to the remainder of the spectrum, while there is no spectral gap when (1.3.39) holds but (1.3.40) does not. In Chapter 3 we will see that (1) is a consequence of the Strong Maximum Principle, see Theorem 3.3.11 while (2) is related with the “reverse Kato’s inequality condition”, see deﬁnition 4.3.10 and Theorem 4.4.14. The case (3) on the existence of spectral gap is addressed in Chapter 5 through some generalized Doeblin-Harris condition. 1.3.3 Stability Three types of results are obtained closely linked to the geometry of the ﬁrst eigenvalue. In the ﬁrst case in which we have that λ1 is algebraically and geometrically simple, we will see that under some weak compactness assumption we arrive to the conclusion (CE1) for any f ∈X, there holds 1 T Z T 0 e Stfdt →⟨f, φ1⟩f1, as T →∞. (1.3.41) See Theorem 3.4.19. If in addition we have the condition that assure that Σ+(L) is trivial, then we arrive to (CE2) for any f ∈X, there holds e Stf →⟨f, φ1⟩f1, as T →∞, (1.3.42) in a sense to be speciﬁed. See the Subsection 4.5 in Chapter 3. Finally, when we can assure the existence of spectral gap, then under some condition to be speciﬁed later, we have (CE3) there exist some constants κ < 0 and C ≥1 such that for any f ∈X, there holds ∥e S(t)f −⟨f, φ1⟩f1∥≤Θ(t)∥f −⟨f, φ1⟩f1∥, ∀t ≥0, ∀f ∈X, (1.3.43) with the decay rate function Θ(t) := C eκt. At the end of Chapter 3 we give some ﬁrst result in this direction without quantitative estimation of κ, then we dedicate Chapter 5 to constructive ones. The results in Chapter 3 are general well know. However, we rewrite them from diﬀerent sources, placing them in the same framework and, in some cases, we provide new proofs that simplify the existing one. On the other hand, we consider the results in the Chapter 5 to be new. 1.4 Thesis plan and contributions As mentioned above, from chapters 2 to 5 are dedicated to the theoretical development of the study of long time behavior. In succession, we deal with the existence of a solution to the eigenvalue triplet problem, then with the geometry and asymptotic behavior, without and with constructive quantiﬁcation of convergence. After that, we apply these results to the three examples introduced in section 1.2 in chapters 6, 7 and 8 respectively. Contribution of chapter 6 Chapter 6 addressed the study of the diﬀusion models presented in section 1.2.2. For each of the introduced cases we prove the existence of the eigentriplet. In addition, we go through all the program until conclusion (CE3) with constructive estimates in the ﬁrst case. In the case of the 18 CHAPTER 1. INTRODUCTION second, third and fourth model, we prove all the condition until reaching conclusion (CE2), of asymptotic convergence without spectral gap. Further to this, we consider that the method proposed in Chapter 5 , which we exempliﬁed in the ﬁrst case, is adaptable to obtain estimates of constructive exponential convergence also in these cases. We believe that these results are new and are an improvement over those previously reported in the literature. Contribution of chapter 7 Chapter 7 is devoted to the Kinetic Fokker-Planck model, presented in section 1.2.3. In this case we will complete the whole program up to (CE3) without constructive rate. This result generalizes [152, Theorem 2.18] and [116, Theorem 6.8] for the zero inﬂow condition and [1, Theorems 1.6 & 1.7] for the torus case. Contribution of chapter 8 We continue in Chapter 8 with the case of the Selection mutation model. Here we also complete the whole program up to (CE3) with constructive exponential rate. As far as we know, this result improves on the existing ones in the literature by weakening the conditions required of the kernel J. In particular, it generalizes the recent result obtained by Gabriel in , extending it to non-symmetric kernels. Chapters 9 and 10 can be read independently from the previous 7 chapters, since they correspond to two separate works. Even if chapter 8 still deals with long time behavior. Contribution of chapter 9 In chapter 9, we present a modiﬁed version of the TEM model introduced in section 1.1.1, which incorporates a memory variable. This new addition is signiﬁcant from a biological point of view because it allows the model to describe phenomena such as adaptability and short term plasticity, as explained in Section 9.1.2. Next, we study the long time behavior of this model and prove that the conclusion (CE3) with constructive exponential rate is valid in the case of weak interaction between neurons. Two aspects should be commented in the context of the above results. On one hand, this model is mass-conservative, so the classical Doeblin-Harris theory as developed in and applies. This corresponds to the case where (λ1, φ1) = (0, 1). The proof will ﬁrst establish that there is a stationary solution f1 = f∞and then an asymptotic result. On the other hand, we should emphasize that TEM model is nonlinear. To deal with this, we will divide the problem into two parts: ﬁrst we will study a linearization of the problem, so we can applied the strategy described above and, in a second stage, we will use a perturbation argument to arrive at the conclusion for the nonlinear problem, when the nonlinear part is small. Contribution of chapter 10 Finally, in Chapter 10, we study the particle system on the form (1.1.2) for general locally Lipschitz terms F and G and establish a Bernstein inequality-type concentration result that we use in two applications: First, to create an estimator of the parameters of the FhN model introduced in section 1.1.1; and then to extend the results of Della Maestra and Hoﬀmann in to the non-globally Lipschitz coeﬃcient case in position-velocity models; we create a nonparametric estimator of the measure µt. Finally, we conclude the chapter by performing an exploratory study of the feasibility of constructing a hypothesis test to check the existence of interactions between neurons through numerical simulations. A more extensive explanation of the results of this part can be found in the introduction of chapter 10. 1.4. THESIS PLAN AND CONTRIBUTIONS 19 All chapters from 2 to 8 are the result of a work in collaboration with Stephane Mischler and Pierre Gabriel. 20 CHAPTER 1. INTRODUCTION Chapter 2 Existence of solution of the eigentriplet problem In this chapter, we address the existence of a solution to the ﬁrst eigentriplet problem, as deﬁned in the introduction. To do so, we take two approaches. In a ﬁrst section we consider a stationary approach in which we look at the operator L and its resolvent RL, i.e. we look directly for a solution of the problem L −λ1 = 0. In a second section, we look at it from a dynamical point of view: we deal with the semigroup SL(t) and the associated evolution problem proving the existence of the eigentriplet problem through a ﬁxed point theorem. 2.1 Existence through a stationary problem approach We start by presenting the basic material about the Banach lattice framework and conclude with a comparison with several previous works. 2.1.1 The Banach lattice framework We start recalling the Banach lattice framework by stating (most of the time without proof) some well-known facts that one can ﬁnd in reference monographs as [38, Chapitre II: Espaces de Riesz] or [222, 7, 16, 20]. Banach lattice. A real Banach lattice is a real Banach space (X, ∥·∥) endowed with a partial order denoted by ≥(or ≤) such that the following holds: (1) The set X+ := {f ∈X; f ≥0} is a nonempty convex cone (compatibility of the order with the vector space structure). (2) For any f ∈X, there exist some unique positive part f+ ∈X+ and negative part f−∈X+ such that f = f+ −f−which are minimal: f = g −h, g, h ≥0 imply g ≥f+ and h ≥f− (generation and properness of the positive cone). We set \|f\|:= f+ + f−∈X+ the absolute value of f ∈X. (3) For any f, g ∈X, \|f\|≤\|g\| implies ∥f∥≤∥g∥(compatibility of norm and order structures). Under these assumptions, one can show that - The convex cone X+ is closed, pointed X+ ∩(−X+) = {0} and generating X = X+ −X+. - The lattice operations f 7→f+, f 7→f−and f 7→\|f\| are continuous (1-Lipschitz). - The order intervals {h ∈X; g ≤h ≤f} are closed and bounded for any given f, g ∈X, f ≥g. It is worth emphasizing that one commonly deﬁnes the supremum and inﬁmum operations by f ∨g := g + (f −g)+ ≥f, g, f ∧g := g −(g −f)+ ≤f, g, 21 22 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM for any f, g ∈X, and these operations can be used as an alternative way for deﬁning a Banach lattice (the lattice structure referees indeed to these supremum and inﬁmum operations). We may note the following elementary formulas f+ ∧f−= 0, ∥\|f\|∥= ∥f∥, ∀f ∈X. (2.1.1) We write f ⊥g when \|f\|∧\|g\|= 0 or equivalently when \|f\|+\|g\|= \|f\|∨\|g\|. In that case, we have \|f\|+\|g\|= \|f + g\|. Dual Banach lattice. On the dual space X′, we may naturally associate a dual order ≥ (or ≤) by writing for ϕ ∈X′ ϕ ≥0 (or ϕ ∈X′ +) iﬀ ∀f ∈X+ ⟨ϕ, f⟩≥0. For ϕ ∈X′, there exist some unique ϕ± ∈X′ + such that ϕ = ϕ+ −ϕ−which also satisfy (and are deﬁned by) ∀f ∈X+, ⟨ϕ±, f⟩= sup 0≤g≤f ⟨±ϕ, g⟩. One can show that the above conditions (1), (2) and (3) of a Banach lattice are fulﬁlled, and thus X′ = (X′, ∥·∥X′, ≥) is a Banach lattice. We observe that for any f ∈X+ there exists f∗∈X′ + such that ⟨f∗, f⟩= ∥f∥2= ∥f∗∥2 X′, (2.1.2) as a classical corollary of the Hahn-Banach dominated extension theorem. Moreover, for any f ∈X, f ≥0 iﬀ ⟨ϕ, f⟩≥0, ∀ϕ ∈X′ +, (2.1.3) as an immediate application of the Hahn-Banach separation theorem. In other words, the restriction to X of the dual order in X′′ associated to the order deﬁned (by duality) on X′ is nothing but the initial order, in particular the positive cone X′ + is weakly ∗closed. The functional framework : The duality bracket. We consider X, Y such that X = Y ′ with Y separable or Y = X′. We emphasize on the facts that for f ∈X : f ∈X+ iﬀ⟨f, ϕ⟩≥0, ∀ϕ ∈Y+, (2.1.4) for ϕ ∈Y : ϕ ∈Y+ iﬀ⟨f, ϕ⟩≥0, ∀f ∈X+, (2.1.5) which are immediate consequences of (2.1.3) and of the deﬁnition of the dual order. Examples. For the space C0(E), the order is deﬁned by f ≥0 iﬀf(x) ≥0 for any x ∈E. For a space Lp(E, E , µ), 1 ≤p ≤∞, the order is deﬁned by f ≥0 iﬀf(x) ≥0 for µ-a.e. x ∈E. For the space M1(E), the order is deﬁned by f ≥0 iﬀin the Hahn decomposition f = f+ −f−, there holds f−= 0, or equivalently, by duality: f ≥0 iﬀ⟨f, ϕ⟩≥0 for any ϕ ∈C0(E), ϕ ≥0. Because conﬁnement will play a major role in our analysis, we will use some weighted version of the above space associated to a weight (continuous or Borel measurable) function m : E →(0, ∞) that we introduce now. We recall that E always denotes a σ-compact metric space, and we write E = ∪ER, with ER ⊂ER+1, ER compact. In that context, we write xn →∞if for any R ≥1 there exists nR such that xn / ∈ER for any n ≥nR. • We denote by Cm,0(E) the space Cm,0(E) := {ϕ ∈C(E); \|ϕ(x)\|/m(x) →0 as x →∞} endowed with the norm ∥ϕ∥Cm,0:= ∥ϕ/m∥C0. • We denote by M1 m(E) := (Cm,0(E))′ the associated space of Radon measures. 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 23 • We denote by Lp m(E) = Lp m(E, E , µ) the space Lp m(E) := {f ∈L1 loc(E); ∥f∥Lp m:= ∥fm∥Lp< ∞}. It is worth emphasizing that Lp m(E, E , µ) = Lp(E, E , mpµ) when p ∈[1, ∞). Positive operator. We denote by B(X) the set of linear and bounded operators on X. We also denote by K (X) the subspace of compact operators. We say that a bounded operator A ∈B(X) is positive, we write A ≥0, if Af ∈X+, ∀f ∈X+. We will also sometimes abuse notations by writing A ∈B(X+) for meaning that A ≥0. For a positive operator A ∈B(X), we have \|Af\|≤A\|f\|, ∀f ∈X, and ∥A∥= sup 0≤f∈BX ∥Af∥, (2.1.6) where BX is the unit closed ball. More generally, we have (Af) ∨(Ag) ≤A(f ∨g), ∀f, g ∈X. (2.1.7) For A ∈B(X) and Y in duality with X, we uniquely deﬁne A∗∈B(Y ) by ⟨Af, φ⟩= ⟨f, A∗φ⟩, ∀f ∈X, φ ∈Y. For A ∈B(X) and A∗∈B(Y ), there holds A ≥0 iﬀ A∗≥0. (2.1.8) Let us present the elementary and classical but instructive proof of the direct impication, the reciprocal sense being similar. We assume thus A ≥0. We take ϕ ∈Y+ and we deﬁne ψ := A∗ϕ. We then take f ∈X+ and we deﬁne g := Af, so that g ≥0 by assumption. We compute ⟨ψ, f⟩= ⟨A∗ϕ, f⟩= ⟨ϕ, Af⟩= ⟨ϕ, g⟩≥0. Since f ∈X+ is arbitrary, we get ψ ∈Y+, and thus A∗≥0. Semigroup, generator and spectrum. In this work, a semigroup S = S(t) = (St) on X will always denote a semigroup of linear and bounded operators on a Banach lattice X which trajectories are - either strongly continuous, namely, the mapping t 7→Stf is continuous for the norm of X for any ﬁxed f ∈X; - either weakly ∗continuous, namely X = Y ′ for some separable Banach lattice Y such that ∀f ∈X, ∀φ ∈Y , t 7→⟨Stf, φ⟩X,Y is continuous and ∀t ≥0, ∀φ ∈Y , f 7→⟨Stf, φ⟩X,Y is continuous. That is in particular the case when there exists a strongly continuous semigroup P on Y such that St = P ∗ t for any t ≥0. For a semigroup S, we denote by L its generator and D(L) the associated domain, and thus we sometimes write S = SL. We also denote the iterated domain deﬁned recursively by D(Lk) := {f ∈D(Lk−1), Lf ∈D(Lk−1)} for any k ≥2 and D(L∞) := T k≥1 D(Lk). We recall that D(L) is dense in X and the graph of L is closed in X × X. We deﬁne the growth bound ω = ω(S) := lim sup t→∞ 1 t log∥S(t)∥∈R ∪{−∞}, (2.1.9) so that ∀ω′ > ω, ∃M ≥1, ∥S(t)∥B(X)≤M eω′t, ∀t ≥0, (2.1.10) 24 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM and ω is the inﬁmum of ω′ ∈R such that (2.1.10) holds. We say that S is a semigroup of contractions when S satisﬁes (2.1.10) with M = 1 and ω′ = 0. The resolvent set ρ(L) is the set of z ∈C such that if z −L : D(L) →X is bijective and its inverse belongs to B(X). We deﬁne the resolvent operator by R(z) = RL(z) := (z −L)−1, ∀z ∈ρ(L) (2.1.11) and the spectrum by Σ(L) := C\ρ(L). Denoting the half complex plane of abscissa α ∈R by ∆α := {z ∈C; ℜe(z) > α}, (2.1.12) we have ρ(L) ⊃∆ω and for any z ∈∆ω there holds R(z) = Z ∞ 0 S(t)e−zt dt. (2.1.13) Positive semigroup. We say that a semigroup (St) on a Banach lattice X is positive if St ≥0, ∀t ≥0. Lemma 2.1.1. For a semigroup S on a Banach lattice X, there is equivalence between (a) S is positive; (b) the associate resolvent operator R is positive: R(κ) ≥0 for some (and thus any) κ > ω. It is immediate from Hille’s identity (2.1.13) that (a) implies (b). The reciprocal implication comes from the relation S(t) = limn→∞[n/tR(n/t)]n at the foundation of the Hille-Yosida theory, see for instance [203, Theorem I.8.3]. 2.1.2 Existence part of the Krein-Rutman theorem From now-on in this section, we consider a Banach lattice X and an operator L with dense domain and closed graph. Our goal is mainly to prove the existence part for the primal problem in the KR theorem, namely ∃λ1 ∈R, ∃f1 ∈X+{0}, Lf1 = λ1f1. (2.1.14) We will also discuss the existence part for the dual problem at the end of the section. We ﬁrst assume (H1) ∃κ1 ∈R such that λ −L is invertible and (λ −L)−1 : X+ →X+ for any λ ≥κ1. We then set I := {κ ∈R; λ −L is invertible, (λ −L)−1 ≥0 for any λ ≥κ}, (2.1.15) which is a non empty and non upper bounded interval due to (H1). We ﬁnally set λ1 := inf I ∈[−∞, κ1]. (2.1.16) For the sake of completeness, we recall now some well known (which probably goes back to [111, 112]) and general facts about I and λ1 when L is the generator of a positive semigroup. We also refer to [82, Section 1.b, Chapter VI] or [20, Chapter 12] and the references therein for more details. 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 25 Lemma 2.1.2. When L is the generator of a positive semigroup S = SL, then (i) (H1) automatically holds with any κ1 > ω(S), so that λ1 ≤ω(S); (ii) Σ(L) ∩∆λ1 = ∅and the representation formula (2.1.13) holds true for any z ∈∆λ1; (iii) it may happen that λ1 = −∞. The important property (b) is probably due to . Proof of Lemma 2.1.2. The claim (i) is an immediate consequence of the representation formula (2.1.13) for any κ1 > ω(S) and the positivity of S(t) for any t ≥0 (that is nothing but Lemma 2.1.1). We prove (ii). Take λ > λ1. From the classical identity S(t)e−λt −I = (L −λ) Z t 0 S(s)e−λs ds, ∀t ≥0, and the positivity property of S, we have 0 ≤V (t) := Z t 0 S(s)e−λs ds = R(λ) −R(λ)S(t)e−λt ≤R(λ), for any t ≥0. From that estimate, we get ∥V (t)∥≤∥R(λ)∥. For any z ∈∆λ, an integration by part yields Z t 0 e−zsS(s) ds = e−(z−λ)tV (t) + (z −λ) Z t 0 e−(z−λ)sV (s) ds. The estimate on V makes possible to pass to the limit t →∞in the above identity, and we deduce U(z) := Z ∞ 0 e−zsS(s) ds = (z −λ) Z ∞ 0 e−(z−λ)sV (s) ds ∈B(X). In that situation, one classically knows that z ∈ρ(L) and (z −L)−1 = U(z). We have thus established Σ(L) ∩∆λ = ∅and we conclude the proof of (ii) by observing that (2.1.13) is then nothing but the above formula. (iii) On Lp(0, 1), 1 ≤p < ∞, the translation semigroup deﬁned for a > 0 by S(t)f(x) := f(x + at)1x+at≤1, ∀t ≥0, x ∈(0, 1), is strongly continuous and positive. Since S(t) ≡0 for any t ≥1/a, we have ω(S) = −∞, and thus λ1 = −∞because of (i). For further discussion, we give some probably classical results about the condition (H1) and some equivalent deﬁnitions of the set I. Lemma 2.1.3. The operator L satisﬁes (H1) if and only if the operator L∗satisﬁes (H1). Furthermore, under condition (H1) for L (or L∗), we have I = Ii, ∀i = 2, 3, 4, (2.1.17) with I2 := {κ ∈R; λ −L is invertible for any λ ≥κ}, I3 := {κ ∈R; λ −L∗is invertible, (λ −L∗)−1 ≥0 for any λ ≥κ}, I4 := {κ ∈R; λ −L∗is invertible for any λ ≥κ}. 26 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM Proof of Lemma 2.1.3. The equivalence of condition (H1) for the operators L and L∗is an immediate consequence of the identity ρ(L) = ρ(L∗) (see for instance [136, Theorem III.6.22]) and the fact that (λ −L)−1 : X+ →X+ iﬀ(λ −L∗)−1 : Y+ →Y+ as recalled in (2.1.8). As a consequence, we have I = I3 and I2 = I4. We obviously have I2 ⊂I and let us show the reverse inclusion. We denote R = RL. On the one hand, for any z0 ∈ρ(L) and any z ∈C, \|z −z0\|< ∥R(z0)∥−1, we know that R(z) = R(z0) ∞ X k=0 (z0 −z)kR(z0)k, (2.1.18) which gives a proof of the fact that resolvent set ρ(L) is open and that R is an holomorphic function on ρ(L). Formula (2.1.18) also ensures that for λ0, λ ∈R, the condition R(λ0) ≥0 implies that R(λ) ≥0 provided that λ0 −λ > 0 is small enough and thus R(λ) ≥0 for any λ in the not upper bounded connected component of the set ρ(L) ∩R thanks to a continuation argument. In particular, I is an open set and I = I2. We next assume (H2) ∃κ0 ∈R such that inf I ≥κ0. We point out several conditions under which (H2) is satisﬁed. Lemma 2.1.4. Condition (H2) holds under one of the four following conditions (i) ∃κ0 ∈R, ∃ψ0 ∈Y+{0} such that L∗ψ0 ≥κ0ψ0, which means ∀f ∈D(L) ∩X+, ⟨ψ0, (κ0 −L)f⟩≤0; (ii) ∃κ0 ∈R, ∃f0 ∈X+{0} such that Lf0 ≥κ0f0, which means ∀ψ ∈D(L∗) ∩Y+, ⟨(κ0 −L∗)ψ, f0⟩≤0; (iii) L∗is the generator of a positive semigroup S∗= (S∗ t ) and ∃κ0 ∈R, ∃ψ0 ∈Y+{0}, ∃T > 0 such that S∗ T ψ0 ≥eκ0T ψ0; (iv) L is the generator of a positive semigroup S = (St) and ∃κ0 ∈R, ∃f0 ∈X+{0}, ∃T > 0 such that ST f0 ≥eκ0T f0. Proof of Lemma 2.1.4. In the three cases, we claim that κ0 / ∈I and thus inf I ≥κ0. We argue by contradiction, assuming λ1 < κ0, so that κ0 ∈I = Ii for any i = 2, 3, 4. We assume (i). For any g ∈X+, we deﬁne f := (κ0 −L)−1g ∈X+ and we compute 0 ≤⟨ψ0, g⟩= ⟨ψ0, (κ0 −L)f⟩≤0. That implies ⟨ψ0, g⟩= 0 for any g ≥0, so that ψ0 = 0 and a contradiction. We assume (ii). For any ξ ∈Y+, we deﬁne ψ := (κ0 −L∗)−1ξ ∈Y+ and we compute 0 ≤⟨ξ, f0⟩= ⟨(κ0 −L∗)ψ, f0⟩≤0. That implies ⟨ξ, f0⟩= 0 for any ξ ≥0, so that f0 = 0 and a contradiction. We assume ﬁrst that (iii) holds for any T > 0. For any f ∈D(L) ∩X+{0}, we compute ⟨ψ0, (κ0 −L)f⟩= −d dt⟨ψ0, e−κ0tStf⟩≤0, 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 27 which is precisely (i). We assume now that (iii) holds. If κ0 ∈I, for any g ∈X+, we may deﬁne f = (κ0 −L)−1g ∈X+ ∩D(L) and from condition (iii), we have 0 ≤⟨e−nκ0T SnT f −f, ψ0⟩= D (L −κ0) Z nT 0 e−κ0tStf dt, ψ0 E , for any n ∈N. From the very deﬁnition of f, we also have (L −κ0) Z nT 0 e−κ0tStf dt = Z nT 0 e−κ0tSt(L −κ0)f dt = − Z nT 0 e−κ0tStg dt ≤0. The two pieces of information together imply D Z nT 0 e−κ0tStg dt, ψ0 E = 0. Passing to the limit n →∞thanks to Lemma 2.1.2-(ii) and using (2.1.11)-(2.1.13), we obtain 0 = D Z ∞ 0 e−κ0tStg dt, ψ0 E = ⟨f, ψ0⟩= ⟨g, (κ0 −L∗)−1ψ0⟩. That implies (κ0 −L∗)−1ψ0 = 0 since g is arbitrary, what is not possible since ψ0 ̸= 0. The proof of (H2) under assumption (iv) is similar and thus skiped. Remark 2.1.5. In practice, we may build f0 or ψ0 through an explicit computation or use a barier fonction and strong maximum principle techniques. We refer to Lemma 3.2.10 for a possible general result in that direction. Lemma 2.1.6. Under conditions (H1) and (H2), there hold λ1 ∈[κ0, κ1] (2.1.19) and ∃λn ↘λ1, ∃ˆ fn ∈D(L) ∩X+, εn := λn ˆ fn −L ˆ fn ≥0, ∥ˆ fn∥= 1, ∥εn∥→0. (2.1.20) Proof of Lemma 2.1.6. We obviously have λ1 ≤κ1 from assumption (H1) and λ1 ≥κ0 by assumption (H2), so that (2.1.19) is proved. Consider now a sequence (λn)n≥2 such that λn ↘λ1 as n →∞. We eventually have ∥R(λn)∥→∞as n →∞. On the contrary, we would have ∥R(λn′)∥≤M for some subsequence λn′ ↘λ1 and some constant M > 0. Because of (2.1.18) this implies that (λn′ −ε, λn′) ⊂I for any n′ and some ε > 0, and this is in contradiction with the deﬁnition of λ1. The blow up ∥R(λn)∥→∞means that ∃fn ∈D(L), ∃gn ∈X, R(λn)gn = fn, ∥fn∥→∞, ∥gn∥≤1. By splitting gn = g+ n −g− n , we get fn = R(λn)g+ n −R(λn)g− n with ∥g± n ∥≤1 and (∥R(λn)g+ n ∥→∞or ∥R(λn)g− n ∥→∞). Changing notations, we have thus ∃fn ≥0, ∃gn ≥0, R(λn)gn = fn, ∥fn∥→∞, ∥gn∥≤1. We get (2.1.20) by deﬁning ˆ fn := fn/∥fn∥and εn := gn/∥fn∥. 28 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM We learn a very similar proof in , from which our own proof is adapted. The same type of arguments can also be found in [20, proof of Theorem 12.15]. We ﬁnally assume that (H3) for any sequence ( ˆ fn) of X such that (2.1.20) holds, there exist f1 ∈X+{0} and a subsequence ( ˆ fn′) such that ˆ fn′ ⇀f1 for the weak convergence or the weak ∗convergence. We discuss several situations in which assumption (H3) holds. We start with a very classical framework due to Voigt , which is however quite restrictive since it is based on a strong compactness property assumed at the level of the associated semigroup of operators. Lemma 2.1.7. We assume that L generates a positive semigroup S, that (H2) holds for a constant κ0 ∈R and that there exists T > 0 such that the splitting ST = VT + KT , (2.1.21) holds with ∥VT ∥B(X)≤eκT , κ < κ0, and KT ∈K (X). Then condition (H3) holds for the primal and the dual problems. Proof of Lemma 2.1.7. The condition (H1) holds beacuse of Lemma 2.1.2-(i). Let us then consider three sequences (λn), ( ˆ fn) and (εn) satisfying (2.1.20). Integrating along the rescaled ﬂow, this yields e−λnT ST ˆ fn −ˆ fn = Z T 0 e−λntSt(L −λn) ˆ fndt = − Z T 0 e−λntStεn dt =: ˜ εn, which also reads V ˆ fn + K ˆ fn −eλnT ˆ fn = eλnT ˜ εn. Since eλnT ≥eκ0T > eκT , the operator eλnT −VT is invertible with inverse Q(λn) := (eλnT −VT )−1 uniformly bounded and converging in B(X) to Q(λ1) = (eλ1T −VT )−1. We thus have ˆ fn = wn + vn, wn := Q(λn)KT ˆ fn, vn := −Q(λn)eλnT ˜ εn, with ∥vn∥X→0 and (wn) relatively compact in X, since there exist a subsequence ( ˆ fnk) and g ∈X such that KT ˆ fnk →g and next wnk −Q(λ1)g = (Q(λnk) −Q(λ1))KT ˆ fnk + Q(λ1)(KT ˆ fnk −g) →0. We deduce that ˆ fnk →f1 strongly in X. Because of the positivity and normalized properties of ˆ fn, we get f1 ∈X+, ∥f1∥X= 1, and we conclude that (H3) holds for the primal problem Observing that the dual semigroup S∗satisﬁes S∗ T = V ∗ T + K∗ T with ∥V ∗ T ∥B(Y )≤eκT and K∗ T ∈K (Y ), the same proof implies that (H3) holds for the dual problem. In the six next lemmas, we will assume that (H1)-(H2) holds associated to some constants κi ∈R, κ0 < κ1, and we always make the following splitting structure hypothesis (HS1) there exists a splitting L = A + B such that B −α is invertible for any α ≥κ0 and V(α) := N−1 X i=0 (RB(α)A)iRB(α), W(α) := (RB(α)A)N, (2.1.22) are bounded in B(X) uniformly with respect to α ≥κ0 and for some N ≥1, where we recall that RB(α) := (α −B)−1. We ﬁrst present a result also based on a strong compactness property which is assumed to hold however at the level of the resolvent operator. We will be able to use that result in most of the applications. 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 29 Lemma 2.1.8. (1) We assume (H1)-(H2)-(HS1) and there exists N ≥1 such that W(α) is strongly compact locally uniformly on α ≥κ0, (2.1.23) in the sense that if αn →α, αn ≥κ0, and (gn) is a bounded sequence in X, then there exist f ∈X and a subsequence (gnk) such that W(αnk)gnk →f strongly in X. Then condition (H3) holds. (2) We assume (H1)-(H2) and (HS1) where RB(α) is bounded uniformly in α ≥κ0, A ∈B(X) and W(α) ∈K (X) for any ﬁxed α ≥κ0 and some N ≥1. Then condition (H3) holds both for the primal and the dual problems. Remark 2.1.9. (1) The property (2.1.23) holds if we assume W(α) : X →X1 is bounded uniformly in α ≥κ0 and X1 ⊂X with strong compact embedding. (2) The property (2.1.23) holds if we assume (H1)-(H2)-(HS1) together with the facts that RB(α) and RB(α)A are bounded uniformly in α ≥κ0 and W(α) ∈K (X) for any ﬁxed α ≥κ0. Consider indeed αn →α, αn ≥κ0, and (gn) a bounded sequence in X. On the one hand, there exist f ∈X and a subsequence (gnk) such that W(α)gnk →f strongly in X, because W(α) ∈ K (X). On the other hand, using the resolvent identity RB(λ) −RB(µ) = (µ −λ)RB(λ)RB(µ), we have W(α) −W(αn) = (αn −α) N X j=1 (RB(α)A)N−jRB(α)(RB(αn)A)j →0, so that W(αnk)gnk →f strongly in X, and (2.1.23) holds true. Proof of Lemma 2.1.8. We ﬁrst assume (1). Taking advantage of the splitting structure (HS1), we write equation (2.1.20) as (λn −B) ˆ fn = A ˆ fn + εn, or equivalently ˆ fn = RB(λn)A ˆ fn + RB(λn)εn. Iterating that last identity and using the notations (2.1.22), we get ˆ fn = wn + vn, wn := W(λn) ˆ fn, vn := V(λn)εn. (2.1.24) We observe that (wn) is strongly relativelly compact from (2.1.23) and ∥ˆ fn∥X= 1, so that there exist a subsequence (wnk) and f1 ∈X such that wnk →f1 strongly in X. Since vn →0 strongly in X, we deduce that ˆ fnk →f1 strongly in X. We conclude that condition (H3) holds as in the proof of Lemma 2.1.7. We next assume (2). As observed in Remark 2.1.9-(2), the property (2.1.23) holds and thus also the condition (H3) for the primal problem. We claim that the same locally uniform strong compactness property (2.1.23) holds for the dual problem at order N + 1 and thus condition (H3) holds for the dual problem. We may indeed use Remark 2.1.9-(2) since then RB∗(α) and A∗RB∗(α) are bounded uniformly in α ≥κ0 and (A∗RB∗(α))N+1 = A∗W(α)∗RB∗(α) ∈K (Y ), ∀α ≥κ0, as a product of two bounded operator with a compact operator. As we see now, strong compactness is not really necessary. Lemma 2.1.10. We assume (H1)-(H2)-(HS1) and there exists N ≥1 such that W(α) : X →X1 ⊂X is positive and uniformly bounded in α ≥κ0 30 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM and, denoting X0 := X, we assume that for any R1 ≥R0 > 0 the set C = CR0,R1 := {g ∈X+; ∥g∥X0≥R0, ∥g∥X1≤R1} (2.1.25) is relatively sequentially compact for the weak topology σ(X, Y ) and 0 / ∈C, where the closure is taken in the sense of the weak topology σ(X, Y ). Then condition (H3) holds. Remark 2.1.11. When X1 ⊂X0 with strongly compact embedding the above set C clearly satisﬁes the required conditions. In particular, Lemma 2.1.10 generalizes the result stated in Remark 2.1.9-(1). Proof of Lemma 2.1.10. We go back to the proof of Lemma 2.1.8 and we start with (2.1.24). We recall that ∥ˆ fn∥X0= 1 and ∥vn∥X0→0 from (2.1.20) and that wn ≥0 because W(λn) is a postive operator. We also observe that ∥wn∥X1≤CW∥ˆ fn∥X0= CW and ∥wn∥X0≥1 −∥vn∥X0≥1/2 for any n ≥n∗, with n∗≥1 large enough, so that wn ∈C := C1/2,CW for any n ≥n∗. By the compactness properties of C, there exist a subsequence (wnk) and f1 ∈X+{0} such that wnk ⇀f1 weakly σ(X, Y ). Since vn →0 strongly in X, we deduce that ˆ fnk ⇀f1 weakly σ(X, Y ) and that ends the proof of (H3). We present a typical concrete application of the preceding result. Lemma 2.1.12. We assume X = Lp(E, E , µ), p ∈[1, ∞), (H1)-(H2)-(HS1) with A ≥0, RB(α) ≥0 for α ≥κ0, and there exists N ≥1 such that W(α) : X →X1 is uniformly bounded in α ≥κ0, (2.1.26) for a subspace X1 ⊂X such that {gp; g ≥0, ∥g∥X1≤R1} is a weakly compact subset of L1(E) for any R1 > 0. Then condition (H3) holds. Remark 2.1.13. (1) A typical example in the above statement is X1 := Lq ∩Lp m for some exponent q > p and some weight function m : E →[1, ∞) such that m(x) →∞as x →∞. (2) The same result holds under the condition that if (un) is a nonnegative and bounded sequence in Lp then the nonnegative sequence wn := W(λn)un is such that wp n is weakly compact in L1. Proof of Lemma 2.1.12. For 0 < R0 < R1, we deﬁne C by (2.1.25) with X0 := Lp. From the weak compacteness property made on X1, we observe that α(R) := sup g∈C ∥g1Ec R∥Lp→0, as R →∞, and β(M) := sup g∈C ∥g1g≥M∥Lp→0, as M →∞. For g ∈C, we may then write R0 ≤∥g∥Lp≤∥g ∧M1ER∥Lp+∥g1Ec R∥Lp+∥g1g≥M∥Lp and thus M1−1/p∥g1ER∥1/p L1 ≥∥g ∧M1ER∥Lp≥R0 −α(R) −β(M) ≥R0/2, 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 31 for some R, M > 0 large enough. On the one hand, from the reﬂexivity of Lp or the Dunford-Pettis theorem, the set C is relatively sequentially compact for the weak topology σ(Lp, Lp′). On the other hand, because 1ER ∈Lp′ the last estimate implies that any element g∗∈C, where the closure is taken in the sense of the weak topology σ(Lp, Lp′), satisﬁes ⟨g∗, 1ER⟩= ∥g∗1ER∥L1≥Mp−1(R0/2)p > 0, and in particular 0 / ∈C. We deduce that (H3) holds as a consequence of Lemma 2.1.10. We present a second kind of result where some weak compacteness is required. Lemma 2.1.14. We assume (H1)-(H2)-(HS1) and there exists N ≥1 such that W(α) : X0 →X ⊂X0 is uniformly bounded in α ≥κ0 (2.1.27) and, denoting X1 := X, the set C deﬁned by (2.1.25) satisﬁes the same properties as the ones stated in Lemma 2.1.10. Then condition (H3) holds. Remark 2.1.15. If we replace the norm ∥·∥X0 by a seminorm ∥f∥X0:= ⟨\|f\|, ϕ0⟩, ϕ0 ∈Y+, and we deﬁne C accordingly by (2.1.25), and if we assume that X = Y ′ with Y separable, then C satisﬁes the same compactness properties as required in the statement of Lemma 2.1.10. If we further assume that (2.1.27) holds where X0 is endowed with the above seminorm, we may repeat the proof below in order to obtain that (H3) holds in that situation (see also Lemma 2.1.18 and its proof for a slightly generalized situation). Proof of Lemma 2.1.14. We start here again with (2.1.24). We have 1 = ∥ˆ fn∥X1≤CW∥ˆ fn∥X0+∥vn∥X1, and thus ∥ˆ fn∥X0≥C−1 W (1 −∥vn∥X1) ≥(2CW)−1 for any n ≥n∗, with n∗≥1 large enough, so that ˆ fn ∈C := C(2CW)−1,1, for n ≥n∗. By the compactness properties of C, there exist a subsequence ( ˆ fnk) and f1 ∈X+{0} such that ˆ fnk ⇀f1 weakly σ(X, Y ). We present a strong variant of Lemma 2.1.12 which is also a concrete consequence of Lemma 2.1.10 and Lemma 2.1.14. Corollary 2.1.16. We assume (H1)-(H2)-(HS1) in X = Lp0 m0, 1 ≤p0 < ∞, together with the facts that RB(α) is positive and bounded in B(Lp0 m0) and B(Lp1 m1) uniformly in α ≥κ0, 0 ≤A ∈B(Lp0 m0) and (RB(α)A)N is bounded in B(Lp0 m0, Lp1 m1) uniformly in α ≥κ0 for some N ≥1, with p1 > p0 and m1 such that m0/m1 ∈Lϑ, 1/ϑ := 1/p0 −1/p1. Then condition (H3) holds for both the primal and the dual problems. Proof of Corollary 2.1.16. On the one hand, we have RB(α) + · · · + (RB(α)A)N−1RB(α) is bounded in B(X) uniformly in α ≥κ0, W(α) := (RB(α)A)N is bounded in B(X, X1) uniformly in α ≥κ0, with X1 := Lp1 m1 ⊂X and thus {(gm0)p0; g ≥0, ∥g∥X1≤R1} is a weakly compact subset of L1(E) for any R1 > 0. Condition (H3) holds for the direct problem thanks to Lemma 2.1.12. On the other hand, we set Y := X′ = Lq0 ν0, q0 := p′ 0, ν0 := m−1 0 , and we ﬁrst observe that RB∗(α) + · · · + (RB∗(α)A∗)N−1RB∗(α) is bounded in B(Y ) uniformly in α ≥κ0. 32 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM We next observe that (A∗RB∗(α))N+1 = A∗W(α)∗RB∗(α) is bounded in B(Y0, Y ) uniformly in α ≥κ0, with Y0 := Lq1 ν1, q1 := p′ 1, ν1 := m−1 1 . Because {(gν1)q1; g ≥0, ∥g∥Y ≤R1} is a weakly compact subset of L1(E) for any R1 > 0, we have from the proof of Lemma 2.1.12 that the set C deﬁned by (2.1.25) for the norms of Y0 and Y1 := Y satisﬁes the weak compacteness property required in the statement of Lemma 2.1.10. We may thus apply Lemma 2.1.14 and we deduce that condition (H3) holds for the dual problem. Another concrete consequence of Lemma 2.1.10 and Lemma 2.1.14 is the following. Lemma 2.1.17. We assume X = M1 mi(E) for a continuous weight function mi on E, i = 0 or i = 1, (H1)-(H2)-(HS1) and there exists N ≥1 such that (RB(α)A)N : M1 m0(E) → M1 m1(E) uniformly in α ≥κ0 for another continuous weight function m1−i on E such that m1(x)/m0(x) →∞as x →∞. We additionally assume that A ≥0 and RB(α) ≥0 for α ≥κ0 when i = 0. Then condition (H3) holds. Proof of Lemma 2.1.17. We deﬁne Xi := M1 mi(E) and we consider the set C deﬁned by (2.1.25) which is clearly compact for the weak ∗σ(M1 m1, Cm1,0) topology. When X = M1 m0, the result follows from Lemma 2.1.10 while when X = M1 m1, the result is a consequence of Lemma 2.1.14. We may slightly improve the preceding results by considering a more abstract framework and a somehow more general boundedness condition. Lemma 2.1.18. We assume X = Y ′, Y separable, (H1)-(H2)-(HS1) and there exist N ≥1, γ ∈[0, 1) and ϕ ∈Y+ \ {0} such that for any α ≥κ0, there holds ∥W(α)f∥X≤γ∥f∥X+⟨f, ϕ⟩X,Y , (2.1.28) for all f ∈X+, or there holds ∥W(α)f∥X≤γ∥f∥X+⟨W(α)f, ϕ⟩X,Y , (2.1.29) for all f ∈X+. Then condition (H3) holds true, and the limit f1 satisﬁes ⟨f1, ϕ⟩X,Y ≥1−γ > 0. The case X = M1 m1(E) in Lemma 2.1.17 corresponds here to the ﬁrst situation where (2.1.28) holds with X := M1 m1(E), γ := 0, Y := Cm0,0(E) and ϕ := m0/m1. Proof of Lemma 2.1.18. Starting with (2.1.24) and using (2.1.28), we have ∥ˆ fn∥X ≤ ∥W(λn) ˆ fn∥X+∥V(λn)εn∥X ≤ γ∥ˆ fn∥X+⟨ˆ fn, ϕ⟩X,Y + ∥vn∥X, so that ⟨ˆ fn, ϕ⟩X,Y ≥1 −γ −∥vn∥X. By compactness, there are f1 ≥0 and a subsequence ( ˆ fn′) such that ˆ fn′ ⇀f1 weak ∗σ(X, Y ). Passing to the limit as n′ →∞in the above estimate, we ﬁnd ⟨f1, ϕ⟩X,Y = lim n′→∞⟨ˆ fn′, ϕ⟩X,Y ≥1 −γ, (2.1.30) and in particular f1 ̸= 0. Under the assumption (2.1.29), modifying slightly the previous argument, we have ∥ˆ fn∥X≤γ∥ˆ fn∥X+⟨wn, ϕ⟩X,Y + ∥vn∥X, 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 33 which, together with ⟨ˆ fn, ϕ⟩X,Y = ⟨wn, ϕ⟩X,Y + ⟨vn, ϕ⟩X,Y , implies ⟨ˆ fn, ϕ⟩X,Y ≥1 −γ −∥vn∥X+⟨vn, ϕ⟩X,Y . By compactness again, there are f1 ≥0 and a subsequence ( ˆ fn′) such that ˆ fn′ ⇀f1 weak ∗σ(X, Y ), and passing to the limit n′ →∞in the above estimate, we conclude again to (2.1.30). Let us comment on Lemma 2.1.18 and in particular the condition (2.1.29). In the case when X = L∞(E, E , µ) = (L1(E, E , µ))′, we can relate condition (2.1.29) to the assumption that there exist f0 ∈X+ and ψ0 ∈Y+ \ {0} such that ∥SL(t)f0∥X≤⟨SL(t)f0, ψ0⟩, ∀t ≥0. (2.1.31) This last condition is reminiscent from conditions that appear in probabilistic inspired methods for the ergodicity of semigroups, see [55, Assumption (A2)] and [18, Assumption (A4)], which essentially build on [71, Condition Z]. Assume indeed (2.1.31), let η > κ1 −κ0 > 0 and consider the trivial decomposition L = A + B = η + (L −η). Then set κB := κ1 −η < κ0, so that for any α > κB, B −α = L −(η + α) is invertible since η + α > η + κB = κ1. We thus have for any α > κB W(α) := η(α −B)−1 = η Z ∞ 0 e−(η+α)tSL(t) dt and (2.1.31) then ensures that ∥W(α)f0∥X≤⟨W(α)f0, ψ0⟩. We recover (2.1.29) with γ = 0 and the diﬀerence that f0 is ﬁxed here. As a Corollary of Lemma 2.1.17 or Lemma 2.1.18 and anticipating on the material of part 2.2, we present now a situation very classical in stochastic processes theory. Corollary 2.1.19. We consider a positive semigroup S = SL deﬁned on a Radon space X = M1 ψ(E) for some weight functions ψ on E, in particular (H1) holds. We also assume that (H2) holds for some κ0 ∈R. We ﬁnally assume the Lyapunov condition L∗ψ ≤κBψ + Mχ, (2.1.32) with κB < κ0, M ≥0 and χ ∈Cψ,0(E), 0 ≤χ ≤ψ. Then condition (H3) holds true. Let us emphasize that we may assume some regularity on ψ by considering ψ ∈D(L∗) so that (2.1.32) makes sense in X or just understand (2.1.32) in the weak sense: ⟨Lf, ψ⟩≤κB⟨f, ψ⟩+ M⟨f, χ⟩, ∀f ∈D(L) ∩X+. Proof of Corollary 2.1.19. We introduce the splitting L = A + B where A is the multiplicator (and thus bounded) operator A := Mχ/ψ. As a bounded perturbation of L, the operator B is the generator of a semigroup SB. Deﬁning e St := SL(t)e−Mt ≥0 and Ac := M(1 −χ/ψ) ≥0, we have the Duhamel formula SB = e S + e SAc ∗SB and iterating inﬁnitly many times, we deduce the Dyson-Philips formula SB = ∞ X k=0 ( e SAc)(∗k) ∗e S. 34 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM That implies that SB ≥0 as a combination of positive operators. Alternatively, from the very deﬁnition of B, we have κ −B ≤(M + κ) −L for any κ ∈R. Choosing κ > max(ω(SL), ω(SB)) and using the direct implication in Lemma 2.1.1, we have RB(κ) ≥RL(M + κ) ≥0. Using the reciprocal implication in Lemma 2.1.1, we obtain again that SB ≥0. Now, for 0 ≤f0 ∈D(B) and setting ft := SB(t)f0, we may compute d dt⟨ft, ψ⟩= ⟨Bft, ψ⟩≤κB⟨ft, ψ⟩, so that ∥SB(t)f0∥M1 ψ≤eκBt∥f0∥M1 ψ. Using (2.1.13) we immediately and classically deduce ∥RB(α)∥B(M1 ψ)≤ 1 α −κB , ∀α > κB, so that RB(α) is bounded in B(M1 ψ) and RB(α)A is bounded in B(M1 χ, M 1 ψ) uniformly for α ≥κ0. We apply Lemma 2.1.17 or Lemma 2.1.18 (with N = 1, γ = 0 and ϕ = M α−κB χ) in order to conclude. In the proof of Corollary 2.1.19, we may alternatively use the trivial splitting L = ˜ A + ˜ B = r + (L −r) for arbirary r > 0, and reformulate the Lyapunov condition (α −˜ B∗)ψ ≥(α + r −κB)ψ −Mχ, for any α ≥κ0. Observing that f W(α) := ˜ AR ˜ B(α) = r(α −˜ B)−1, we deduce f W∗(α)ψ ≤ r r + α −κB ψ + M r + α −κB f W∗(α)χ. We equivalently have ∥f W(α)f∥M1 ψ≤γ∥f∥M1 ψ+⟨W(α)f, ˜ χ⟩, uniformly for any α ≥κ0, with γ := r r+κ0−κB < 1 and ϕ := M r+κ0−κB χ, which is nothing but condition (2.1.29). We ﬁnally come to the existence of a solution to the ﬁrst eigenvalue problem and the ﬁrst eigentriplet problem. Theorem 2.1.20. Under conditions (H1)-(H2)-(H3), the ﬁrst eigenvalue problem (2.1.14) has a solution (λ1, f1) with λ1 satisfying (2.1.19). When furthermore (H3) holds for the dual problem, then the ﬁrst eigentriplet problem (1.3.30)-(1.3.31) admits a solution (λ1, f1, φ1) ∈R × X × Y . Theorem 2.1.20 generalizes some known versions of the existence part of the Krein-Rutman Theorem where either L is assumed additionally to be the generator of a semigroup or to have strongly power compact resolvent or even where some additional conditions are made on the positive cone X+. As mentioned in the introduction, some possible references for these previous results are Krein-Rutman , Greiner in [109, Cor 1.2] and in [7, C-IV Thm 2.1] and Webb [244, Prop. 2.5], see also [43, Thm. 2], [177, Thm. 5.3], , [18, Thm 2.1], the textbook [20, Theorem 12.15] and the references therein. Proof of Theorem 2.1.20. We ﬁrst assume (H1)-(H2)-(H3). Because of Lemma 2.1.6, there exists a sequence ( ˆ fn) of X such that (2.1.20) holds, and in particular ⟨λn ˆ fn, φ⟩−⟨ˆ fn, L∗φ⟩= ⟨εn, φ⟩, (2.1.33) for any φ ∈D(L∗). Because of condition (H3), we may pass to the limit n′ →∞in equation (2.1.33) and we deduce that (λ1, f1) satisﬁes (2.1.14) and (2.1.19). 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 35 We now additionally assume that (H3) holds for the dual problem. As recalled during the proof of Lemma 2.1.3 and by deﬁnion of λ1, we have (λ1, +∞) ⊂ρ(L) = ρ(L∗) and λ1 ∈Σ(L) = Σ(L∗). Taking λn ↘λ1, we argue as in the proof of Lemma 2.1.6 and we get ∃b φn ≥0, λn b φn −L∗b φn →0, ∥b φn∥= 1. Thanks to (H3) for the dual problem, we deduce that there exist a subsequence (b φnk) and φ1 ∈X′, ∥φ1∥= 1 such that b φnk →φ1. We thus conclude that φ1 is a solution to the dual problem (1.3.31) (for the same eigenvalue λ1). Let us conclude this section by some remarks. Remark 2.1.21. (1) - As seen above, the condition (H1)-(H2) for the primal and the dual problems are equivalent, and thus one only has to check (H1)-(H2)-(H3) for the primal problem and (H3) for the dual problem in order to solve the ﬁrst eigentriplet problem. It is worth emphasizing that condition (H3) on the dual problem is not a consequence of the condition (H3) on the primal problem. However, as presented in Lemma 2.1.7, Lemma 2.1.8 and Corollary 2.1.16, there exist several natural situations where both conditions (H3) for the primal and the dual problems hold together. (2) - Alternatively, one may also assume (H1)-(H2)-(H3) for the dual problem, and then use a more classical ﬁxed point theorem for proving the existence of a steady state for the rescaled semigroup by using for instance the MarkovKakutani ﬁxed point theorem as in [138, Theorem 5.1], the Tychonov ﬁxed point theorem as in or [83, Theorem 1.2] or a Birkhoﬀ-Von Neumann type Theorem as in [50, Theorem 6.1]. For these last techniques, we also refer to Section 2.2, where such a dynamical approach is adapted to the present context. One may also use the usual Doblin-Harris theory, see for instance [119, 50] and the references therein. 2.1.3 Discussion We discuss now the existence results presented in the preceding section. For further references, let us ﬁrst recall that when X is a Hilbert space and L is self-adjoint, the ﬁrst eigenvalue may be simply obtained thanks to the variational problem λ1 = inf f∈X+{0} ⟨Lf, f⟩ ∥f∥2 . (2.1.34) We now explain how Theorem 2.1.20 is a generalization of the classical Krein-Rutman theorem stated in Theorem ??. We thus consider a Banach lattice X such that X++ := intX+ ̸= ∅ and an operator L such that, for κ1 ∈R and any κ > κ1, R := (κ −L)−1 : X →X is compact and R : X+{0} →X++, in particular (H1) holds true. As a ﬁrst step, we recall the following very classical technical lemma of the KR theory. Lemma 2.1.22. Assume X++ := int X+ ̸= ∅. For g ∈X+ and f ∈X++, there exists C ≥0 such that g ≤Cf. Proof of Lemma 2.1.22. We argue by contradiction. Otherwise, for any n ≥1, we would have f −g/n ∈Xc + ⊂Xc ++ and that last set is closed. Passing to the limit, we get f ∈Xc ++, which is in contradiction with the assumption f ∈X++. For a given g0 ∈X+{0}, we set f0 := Rg0 ∈X++. From Lemma 2.1.22, there exists C0 ≥0 such that (κ −L)f0 = g0 ≤C0f0. That implies that (ii) holds with κ0 := κ −C0, and thus (H2) also holds. One may then deﬁne µ1 := κ −λ1, with λ1 := inf{λ ∈R; (λ′ −L)−1 ∈B(X), ∀λ′ ∈[λ, κ]} ≥κ0. 36 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM We recall that because of Lemma 2.1.6 (or its proof), there exist (λn), ( ˆ fn) and (εn) such that (2.1.20) holds, namely λn ↘λ1, ˆ fn ≥0, εn := λn ˆ fn −L ˆ fn ≥0, ∥ˆ fn∥= 1, ∥εn∥→0. We may rewrite the equation as ˆ fn = R[εn + (κ −λn) ˆ fn], so that ( ˆ fn) belongs to a compact set of X because of the compactness assumption made on R, so that (H3) holds true. Because of Theorem 2.1.20, we deduce that there exists f1 ∈X+ such that ∥f1∥= 1 and Lf1 = λ1. That implies f1 = µ1Rf1, and thus that the existence part of Theorem ?? is a consequence of Theorem 2.1.20 for an operator R which is the positive resolvent of an operator L. We would like to emphasize on the fact that our deﬁnition of the ﬁrst eigenvalue by (2.1.15)-(2.1.16) bears some strong similarity with the deﬁnition of the ﬁrst eigenvalue for elliptic operators in non divergence form as presented in . Indeed, if λ ∈I, then ∃f ∈X+{0}, Lf ≤λf. Assuming now that X is a space of functions (deﬁned on a set E) and that f(x) > 0 for any x ∈E, we deduce that λ ≥sup E Lf f , and thus λ1 is characterized by λ1 = inf f>0 sup E Lf f , which is nothing but [23, (1.13)] (with a change of sign because of a diﬀerent sign convention). We thus see that our formulation is a generalization at a more abstract level and for operators with positive inverse of that classical min-max approach for elliptic operators. Some more or less classical references on that subject are [78, 77], , , and . In particular in , two generalized principal eigenvalues λ′ 1 := sup{κ0 ∈R; ∃g0 ∈C0 Lg0 ≤κ0g0} and λ′′ 1 := inf{κ1 ∈R; ∃g1 ∈C1 Lg1 ≥κ1g1} are deﬁned for appropriate cones Ci ⊂X+{0} for problems with lack of compactness. The links between the three quantities λ1, λ′ 1 and λ′′ 1 are discussed as well as the possible not existence of a related generalized principal eigenfunction f1. The non existence of associated generalized principal eigenfunction should not be a surprise since it is the case when one considers L = ∆in X = L2(Rd) where Lg1 = L∗g1 = λ′′ 1ψ with 0 < ψ = 1 / ∈X = X′ and λ′′ 1 = 0 but no associated generalized principal eigenfunction exists in X. We also refer to where some examples of such a situation are discussed. For its own interest and further discussions, we ﬁnally state and prove a slightly less general variant of Theorem ??. Theorem 2.1.23. Consider a Banach lattice with positive cone X+ and a linear and bounded operator R : X →X such that (i) R : X+ →X+; 2.1. EXISTENCE THROUGH A STATIONARY PROBLEM APPROACH 37 (ii) ∃g2 ∈X+{0}, ∃C2 > 0 such that Rg2 ≤C2g2. We deﬁne K2 := {g ∈X+; ∃a > 0, g ≤ag2}, and next A(g) := inf{a > 0; g ≤ag2}, if g ∈K2, as well as J := {µ ≥0; ∃h ∈K2, h ≥µRh + g2}. We further assume (iii) µ1 := sup J < +∞. (iv) Any upper bounded and increasing sequences (gn) of X is convergent in the weak sense σ(X, Y ). More precisely, if gn ≤gn+1 ≤¯ g ∈X for any n ≥1, there exists g ∈X, g ≤¯ g, such that gn ⇀g. (v) Any sequence (gn) of normalized almost ﬁrst eigenvectors is relatively compact. More precisely, for any sequence (gn) of K2 such that A(gn) = 1, gn = µnRgn + εn, µn ↗µ1 and εn →0, there exists g ∈K2 and a subsequence (gnk) such that gnk →g and A(g) = 1. Then there exists f1 ∈X2 such that f1 = µ1Rf1 and A(f1) = 1. Proof of Theorem 2.1.23. We split the proof into three steps. Step 1. We ﬁrst establish that for any µ ∈J , there exists ˜ g = ˜ gµ ∈K2 such that ˜ g = µR˜ g + g2. (2.1.35) We set ˜ g0 = 0, ˜ g1 = g2, and we deﬁne (˜ gn) recursively by ˜ gn+1 = µR˜ gn + g2, for any n ≥1. We claim that 0 ≤˜ gn ≤˜ gn+1 ≤h, for any n ≥0, where h enters in the deﬁnition of µ ∈J . That is obviously true at order n = 0. Assuming that last inequality is proved at order n −1 for n ≥1, we compute ˜ gn+1 = µR˜ gn + g2 ≥µR˜ gn−1 + g2 = ˜ gn and ˜ gn+1 = µR˜ gn + g2 ≤µRh + g2 ≤h, which proves the same inequality at order n, and thus for any n ≥0. Using the convergence property (iv) of upper bounded increasing sequences in X, we deduce that there exists ˜ g ∈X2 such that ˜ gn →˜ g and thus (2.1.35) holds. Step 2. We obviously have 0 ∈J and J is an interval because if (µ, h) satisﬁes the condition µ ∈J then so do (µ′, h) for any µ′ ∈[0, µ]. We ﬁnally claim that J is open. Take indeed µ ∈J and ˜ g ∈K2 such that (2.1.35) holds, what is possible due to Step 1. By deﬁnition, there would exist A > 0 such that ˜ g ≤Ag2. Choosing 0 < ε ≤1/(2AC2) and M ≥2, we compute (M˜ g) −(µ + ε)R(M˜ g) = Mg2 −MεR˜ g ≥ Mg2 −MεARg2 ≥M(1 −εAC2)g2 ≥g2, so that µ + ε ∈J . Step 3. We ﬁrst establish by contradiction that A(˜ gµ) ↗∞when µ ↗µ1. If it was not the case, there exists A ∈(0, ∞) and a sequence (µn) such that A(˜ gµn) ≤A as µn ↗µ1. Choosing 0 < ε ≤1/(2AC2) and M ≥2 as in Step 2, the same computation gives (M˜ gµn) −(µ + ε)R(M˜ gµn) ≥g2, 38 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM so that µn + ε ∈J . That means that µn + ε ≤µ1, and a contradiction with the fact µn ↗µ1. We next consider µn ↗µ1 and we deﬁne An := A(˜ gµn), ˆ gn := ˜ gµn An , εn := g2 An , ˆ gn = µnT ˆ gn + εn. We observe that εn →0 and A(ˆ gn) = 1. Because of the compactness assumption (v), we deduce that there exists f1 ∈K2 and a subsequence (ˆ gnk) such that ˆ gnk →f1 and A(f1) = 1. We conclude by passing to the limit in the above almost ﬁrst eigenvalue equations. We may compare Theorem 2.1.23 with the results presented in the previous section. When L satisﬁes condition (H1), we may set R := RL(κ1) so that R ∈B(X) and R satisﬁes (i). In that case, Theorem 2.1.23 claims the existence of f1 ∈K2 such that Lf1 = λ1f1, with λ1 := κ1 −µ1. The condition (ii) on R translates as Lg2 ≤(κ1 −1/C2)g2 which may be seen as an equivalent of condition (H1) (when working in the space X2 := K2 −K2 with norm ∥g∥2:= A2(\|g\|) and L generates a semigroup S. The hypothesis (iii) is nothing but (H2) and the hypothesis (iv) is very natural: it holds in the space Lp(E) and M1(E) without additional condition on R and it holds in a space of continuous functions when some additional uniform continuity assumption is made on the range of R. Assumption (v) has to be compared with condition (H3). It is worth emphasizing that when X ⊂Lp(E) and g2 > 0 a.e., we simply have A(g) = ∥g/g2∥L∞for any g ∈X+. As a conclusion, although Theorem 2.1.20 and Theorem 2.1.23 bear some similarity none seems to be a consequence of the other. We believe that Theorem 2.1.20 is more ﬂexible since it does not impose to work with the normalization associated to the seminorm g 7→A(\|g\|) of L∞-type. It is also worth emphasizing that it is shown in that Theorem ?? is also a particular case of Theorem 2.1.23 by essentially exploiting the fundamental Lemma 2.1.22. 2.2 Existence through a dynamical approach In this part, we develop a dynamical approach for proving the existence part of the Krein-Rutman Theorem. We thus always consider a positive semigroup S = SL on a Banach lattice X. We recover Theorem 2.1.20 under slightly reinforced assumptions. Above all, we are able to extend the existence part of the Krein-Rutman Theorem to a more general framework, namely to the case when L only enjoys a suitable weakly dissipative condition. 2.2.1 About dissipativity Let us start by recalling some classical deﬁnitions and results. We say that an operator L deﬁned in a Banach space X is dissipative if there is some number κ ∈R such that ∀f ∈D(L), ∃f∗∈Jf, ℜe⟨f∗, Lf⟩≤κ∥f∥2, where we deﬁne the associated dual set Jf ⊂X′ of f by Jf := {ϕ ∈X′; ⟨ϕ, f⟩= ∥f∥= ∥ϕ∥X′}. (2.2.1) In that situation and in order to be more precise, we should say that L −κ is dissipative. It is worth emphasizing that Jf ̸= ∅thanks to the corollary (2.1.2) of the Hahn-Banach dominated extension theorem. We say that an operator L is hypodissipative in a Banach space X if there exist an equivalent norm \|\|\|·\|\|\| in X and a number κ ∈R such that ∀f ∈D(L), ∃f∗∈Jf,\|\|\|·\|\|\|, ℜe⟨ ⟨f∗, Lf⟩ ⟩≤κ\|\|\|f\|\|\|2, (2.2.2) where Jf,\|\|\|·\|\|\| := {ϕ ∈X′; ⟨ ⟨ϕ, f⟩ ⟩= \|\|\|f\|\|\|2= \|\|\|ϕ\|\|\|2 X′}. (2.2.3) 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 39 The only diﬀerence between the two deﬁnitions (2.2.1) and (2.2.3) comes from the norms in which the normalization is performed. When L is the generator of a semigroup SL, one can show that the growth bound ω = ω(SL) deﬁned in (2.1.9) also satisﬁes ω = inf{κ ∈R, (2.2.2) holds for some equivalent norm \|\|\|·\|\|\|}, and SL is a semigroup of contraction when L is dissipative with κ = 0. At least formally, denoting ft := S(t)f, for f ∈D(L), we deduce from (2.2.2) that 1 2 d dt\|\|\|ft\|\|\|2= ℜe⟨ ⟨(ft)∗, Lft⟩ ⟩≤κ\|\|\|ft\|\|\|2, and together with the Grönwall lemma, we deduce \|\|\|S(t)f\|\|\|≤eκt\|\|\|f\|\|\|, ∀t ≥0, which is nothing but (2.1.10). That last estimate is actually equivalent to the hypodissipativity estimate (2.2.2). Quite similarly, when ∃ψ ∈Y+{0}, ∃κ ∈R, ±L∗ψ ≤κψ, (2.2.4) we may compute ± d dt⟨ft, ψ⟩= ±⟨Lft, ψ⟩= ±⟨ft, L∗ψ⟩≤κ⟨ft, ⟩, and together with the Grönwall lemma, we get ±⟨Stf, ψ⟩≤±e±κt⟨f, ψ⟩, ∀t ≥0. (2.2.5) Two important more accurate versions of the previous ones are presented now. They will be on the main importance in the sequel. On the one hand, we may assume that L satisﬁes a Lyapunov type condition, namely there exists ψi ∈Y+ and κ ∈R such that L∗ψ2 ≤κψ2 + ψ0, (2.2.6) with ψ2 > 0 and ψ0/ψ2 →0 at inﬁnity. For ft = SL(t)f, f ∈D(L) ∩X+, a similar computation as above gives d dt⟨ft, ψ2⟩= ⟨ft, L∗ψ2⟩≤κ⟨ft, ψ2⟩+ ⟨ft, ψ0⟩. Denoting [f]i := ⟨\|f\|, ψi⟩and using the Grönwall lemma, we classically deduce [S(t)f]2 ≤eκt[f]2 + Z t 0 eκ(t−s)[S(s)f]0 ds, ∀t ≥0. (2.2.7) The Lyapunov condition (2.2.6) is particularly relevant and useful in a Radon measures space framework X = M1 ψ2(E) for some weight function ψ2 on E. On the other hand, we may generalize the above Lyapunov condition by assuming the structure condition (HS2) there exist a splitting L = A + B and κB ∈R such that A is B-bounded, that means ∃C ≥0, ∀f ∈X, ∥Af∥≤C(∥f∥+∥Bf∥), the operator B generates a semigroup SB and ∥(SBA)(∗ℓ) ∗SB(t)∥B(X)= O(eαt), ∀t > 0, (2.2.8) 40 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM for any ℓ≥0 and α > κB. Here and below, for two functions U : R+ →B(X0, X1) and V : R+ →B(X1, X2), we deﬁne the convolution function (V ∗U)(t) := Z t 0 V (t −s)U(s) ds, when the integral is well-deﬁned. For U : R+ →B(X), we also recursively deﬁne U(∗0) = I and U(∗(ℓ+1)) = U(∗ℓ) ∗U. Using this convolution notation, the Duhamel formula writes SL = SB + SBA ∗SL, and iterating this formula, for any N ≥1, we get the following iterated Duhamel formula SL = SB + · · · + (SBA)∗(N−1) ∗SB + (SBA)(∗N) ∗SL. (2.2.9) When SL is well deﬁned in another space X0 ⊃X and the last iterated convolution term enjoys the regularity property ∥(SBA)(∗N)(t)∥B(X0,X)= O(eαt) for all t > 0 and α > κB, we deduce from the above iterated Duhamel formula, the estimate ∥S(t)f∥≤C0eαt∥f∥+C1 Z t 0 eα(t−s)∥S(s)f∥0 ds, ∀t ≥0, α > κB, ∀f ∈X, (2.2.10) for some constants Ci ≥1 and where ∥·∥0 stands for the norm in X0. We may observe that the estimate (2.2.7) in the case of a Lyapunov condition is a particular case of (2.2.10) corresponding to the norms ∥·∥= [·]2 and ∥·∥0= [·]0. More speciﬁcally, in a Radon measures space framework, the splitting condition (HS2) is obtained by introducing the bounded operator Af := fψ0 and the generator B := L −A. Because of (2.2.6), we have B∗ψ2 ≤κψ2, and arguing as for establishing (2.2.7), we have [SB(t)f]2 ≤eκt[f]2 for any t ≥0 and f ∈X. That last growth condition is equivalent to assuming that B −κ is dissipative for the norm [·]2, so that we have established that L enjoys the splitting condition (HS2). 2.2.2 Existence in the dissipative case In this section, we give an existence result for a positive semigroup SL on a Banach lattice X satisfying a kind of regularity/compactness assumption in the spirit of the structure condition (HS2) discussed above. Theorem 2.2.1. On a Banach lattice X = Y ′, with Y separable Banach lattice, consider a positive semigroup S = SL satisfying the growth bound (2.1.10), and set κ1 := ω′ + log M for some ω′ > ω(SL). We assume (1) ∃ψ0 ∈Y+{0}, ∃κ0 ∈R such that [S(t)f]0 ≥eκ0t[f]0 for any t ≥0 and f ∈X+, where we denote [f]0 := ⟨\|f\|, ψ0⟩; (2) there exist κ, C0, C1 ∈R with κ < κ0, C0 ≥1 and C1 ≥0, such that ∥S(t)f∥≤C0eκt∥f∥+C1 Z t 0 eκ(t−s)[S(s)f]0 ds, ∀t ≥0, ∀f ∈X. (2.2.11) Then there exist λ1 ∈[κ0, κ1] and f1 ∈X+{0} such that Lf1 = λ1f1. Remark 2.2.2. (1) Assumption (2) in the statement of Theorem 2.2.1 holds when there exist V, W such that S = V + W ∗S, W ≥0, (2.2.12) and there exist κ, CV , CW ∈R, κ < κ0, CV ≥1, CW > 0 such that ∥V (t)∥B(X)≤CV eκt, ∥W(t)∥B(X0,X)≤CW eκt, (2.2.13) 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 41 (2) Under the structural condition (HS2) together with some regularization eﬀect on the semigroup of the type ∥(SBA)(∗N)(t)∥B(X0,X)= O(eκt), ∀t > 0, κ ∈(κB, κ0), we recover the above condition (2.2.12)-(2.2.13) with V := SB + · · · + (SBA)∗(N−1) ∗SB, W := (SBA)(∗N) (2.2.14) because of the iterated Duhamel formula (2.2.9). In that case, the representation formula (2.1.13) holds true for any z > λ1 from Lemma 2.1.2-(ii) and we easilly compute RL(z) = V(z) + W(z)RL(z), ∀z > λ1, with V(z) := Z ∞ 0 e−λtV (t)dt, W(z) := Z ∞ 0 e−λtW(t)dt, ∀z > κ. We observe that W satisﬁes (2.1.27) in Lemma 2.1.14 if W satisﬁes (2.2.13) and the set C deﬁned by (2.1.25) satisﬁes the same compactness properties as required in the statement of Lemma 2.1.10. We may thus apply Lemma 2.1.14 (see also Remark 2.1.15) and deduce that (H3) holds for the primal problem. We ﬁnally obtain the same conclusion as in Theorem 2.2.1 thanks to Theorem 2.1.20. (3) Under the same structural condition (HS2) as above, but assuming now that ∥W(t)∥B(X,X1)= O(eκt), ∀t > 0, κ ∈(κB, κ0), with W := (SBA)(∗N) and X1 ⊂X with strongly compact embedding, we observe that S does not necessary satisﬁes the assumptions of Theorem 2.2.1, but it rather satisﬁes the assumptions of Lemma 2.1.7 with KT := (W ∗S)(T) and T > 0 large enough. In that situation, we also obtain the same conclusion as in Theorem 2.2.1 thanks to Lemma 2.1.7 and Theorem 2.1.20. Proof of Theorem 2.2.1. We split the proof into two steps. Step 1. We deﬁne the set C := {f ∈X+, [f]0 = 1, ∥f∥≤R}, for a convenient constant R > 0 to be ﬁxed later. For any ﬁxed t > 0, we next deﬁne the nonlinear weakly σ(X, Y ) continus mapping Φt : C →X, f 7→ Stf [Stf]0 . Thanks to assumption (1), we may observe that it is well deﬁned because [Stf]0 ≥eκ0t[f]0 = eκ0t > 0. (2.2.15) For any f ∈C, we thus immediately have Φtf ≥0 and [Φtf]0 = 1. On the other hand, from assumption (1) again and the semigroup property, we have [S(t)f]0 ≥eκ0(t−s)[S(s)f]0. (2.2.16) For f ∈C and t ≥0, we next compute ∥Φtf∥ ≤ C0e−αt∥f∥+C1 Z t 0 e−α(t−s) ds ≤ C0e−αtR + C1 α , 42 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM where we have set α := κ0 −κB > 0. Fixing T0 such that C0e−αT0 ≤1/2 and next R ≥2C1/α, we have thus ΦT0 : C →C. Thanks to the Tykonov ﬁxed point Theorem, there exists fT0 ∈C such that ΦT0fT0 = fT0. In other words, we have established the existence of fT0 ∈X such that fT0 ≥0, [fT0]0 = 1, ST0fT0 = eλ1T0fT0, (2.2.17) with λ1 := (1/T0) log[ST0fT0]0 ∈[κ0, κ1]. Step 2. Rewriting equation (2.2.17) as 0 = e−λ1T0ST0fT0 −fT0 = (L −λ1) Z T0 0 e−λ1tStfT0dt and deﬁning f1 := Z T0 0 e−λ1tStfT0dt, we get that f1 ∈X+{0} and satisﬁes Lf1 = λ1f1. We present now a second proof based on a large times dynamical argument which is classical in the mean ergodicity theory of Von Neumann and Birkhoﬀintroduced in [194, 30] and which will be adaped in the weak dissipativity case in Section 2.2.5 below. Alternative Step 2. We deﬁne e St := Ste−λ1t, so that fT0 becomes a periodic state for e St from (2.2.17), namely e StfT0 = e St−kT0fT0, k := [t/T0], ∀t > 0. Using (2.2.15) and the above relation, we have [ e StfT0]0 = [ e St−kT0fT0]0 ≥ e(κ0−λ1)(t−kT0)[fT0]0 ≥e(κ0−λ1)T0 =: r∗> 0, for any t ≥0. On the other hand, thanks to the growth bound (2.1.10), we have ∥e StfT0∥ = ∥e St−kT0fT0∥ ≤ Me(κ−λ1)(t−kT0)∥fT0∥≤Me(κ−λ1)T0R =: R∗< ∞, for any t ≥0. We ﬁnally deﬁne uT := 1 T Z T 0 e StfT0 dt. From the previous estimates, both sequences ( e StfT0) and (uT ) are bounded in K := {f ∈X; f ≥0, [f]0 ≥r∗, ∥f∥≤R∗}. By compactness, there exists a subsequence (uTk) and f1 ∈K such that uTk ⇀f1 in a weak sense as k →∞. For any ﬁxed t > 0, we observe that e Stf1 −f1 = lim k→∞ n 1 Tk Z Tk 0 e St e SsfT0ds −1 Tk Z Tk 0 e SsfT0ds o = lim k→∞ n 1 Tk Z Tk+t Tk e SsfT0ds −1 Tk Z t 0 e SsfT0ds o = 0, where we have used that ( e SsfT0) is uniformly bounded in the last line. As a consequence, f1 is a stationary state for the rescaled semigroup e St, and thus an eigenfunction associated to the eigenvalue λ1 for the operator L. 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 43 2.2.3 About weak dissipativity In this section, we recall some deﬁnitions and results about the weak dissipativity. We say that the generator B of a semigroup SB is weakly dissipative in a Banach space Xi if there exist a second Banach space Xi−1 ⊃Xi and some numbers κ ∈R and σ > 0 such that ∀f ∈D(B\|Xi), ∃f∗∈Jf,Xi, ⟨f∗, Bf⟩≤κ∥f∥2 Xi−σ∥f∥2 Xi−1, where we deﬁne the associated dual set Jf,Xi ⊂X′ i of f (for the norm ∥·∥Xi) by Jf,Xi := {ϕ ∈X′ i; ⟨ϕ, f⟩= ∥f∥2 Xi= ∥ϕ∥2 X′ i}. (2.2.18) By translation, we may assume that κ = 0, an hypothesis we will always make in the sequel of this section. We will furthermore assume the splitting structure L = A + B with A B-bounded and B weakly dissipative. More precisely, we assume that there exists one more Banach lattice X0 ⊃X1 ⊃X2 := X, with norm denoted by ∥·∥k:= ∥·∥Xk, such that B generates a semigroup and is weakly dissipative in each Xk: for any k = 1, 2 ∀f ∈D(B\|Xk), ∃f∗∈Jf,Xk, ⟨f∗, Bf⟩X′ k,Xk ≤−σ∥f∥2 k−1. (2.2.19) This classically implies (or we can take the next inequality as a deﬁnition of the weak dissipativity condition) that d dt∥SB(t)f∥k+σ∥SB(t)f∥k−1≤0, ∀t ≥0, ∀f ∈Xk, ∀k = 1, 2. (2.2.20) We assume that Xk is dense into Xk−1 for k = 1, 2 and that X1 is an interpolated space between X0 and X2 in the sense that there exists a continuous and strictly decreasing function η : (0, 1] →[0, ∞), η(ε) →∞when ε →0, η(1) = 0, such that ∥f∥1≤ε∥f∥2+η(ε)∥f∥0, ∀ε ∈(0, 1], ∀f ∈X2. (2.2.21) From (2.2.20) with k = 2, we deduce ∥SB(t)f∥2≤∥f∥2, ∀t ≥0, ∀f ∈X2. (2.2.22) Next, for k = 1, gathering the weak dissipativity condition (2.2.20), the interpolation condition (2.2.21) and the non expansion property (2.2.22) in the space X2, we get d dt∥SB(t)f∥1+ σ η(ε) ∥SB(t)f∥1 ≤ σε η(ε) ∥SB(t)f∥2 ≤ σε η(ε) ∥f∥2, for any t ≥0, ε ∈(0, 1) and f ∈X2. We deduce d dt ∥SB(t)f∥1e σ η(ε) t ≤σε η(ε)e σ η(ε) t∥f∥2, and thanks to the Grönwall lemma, we obtain ∥SB(t)f∥1≤Θ(t)∥f∥2, (2.2.23) for any t ≥0 and f ∈X2, with Θ(t) := inf ε∈(0,1)(e− σ η(ε) t + ε) →0 as t →+∞. (2.2.24) 44 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM On the other hand, using the representation formula RB(z)f = Z ∞ 0 e−ztSB(t)f dt, ∀z ∈∆0, ∀f ∈X2, together with (2.2.20), we get σ∥RB(z)f∥1≤ Z ∞ 0 σ∥SB(t)f∥1 dt ≤∥f∥2, for any z ∈∆0 and f ∈X2. We next assume that Θ(t)−1∥ASB(t)f∥1+ Z ∞ 0 ∥ASB(t)f∥1dt ≲∥f∥1, (2.2.25) that there exist α > 1, N ≥1, C ≥1 such that sup x+iy∈∆0 ∥AR1+ε1 B (x + iy) . . . AR1+εN B (x + iy)f∥2≤ C ⟨y⟩α ∥f∥2, (2.2.26) for any ε ∈{0, 1}N, ε1 + · · · + εN ≤1, and that sup z∈∆0 ∥(RB(z)A)Nf∥X1≤∥f∥1, (2.2.27) with X1 compactly imbedded in X1. The necessity to add (εi) in (2.2.26) is probably purely technical and not restrictive for applications. In examples, we can take N = 2N′, when sup x+iy∈∆0 ∥(ARB)N′(x + iy)f∥3≤ C ⟨y⟩α ∥f∥2, (2.2.28) for some convenient space X3 such that A : X1 →X3 and supz∈∆0∥RB(z)∥B(X3,X2)< ∞. At the level of the semigroup, (2.2.28) is typically a consequence of ∥(ASB)(∗N′′)(t)∥B(X2,Xζ 3 )∈L1(R+), with ζ > 0, where Xζ 3 := {f ∈X3, Lζf ∈X3} stands for the (possibly fractional) domain for the operator deﬁned in X3. However, (2.2.26) is a bit more general than that last estimate. We refer to [177, 179, 178, 176] for precise deﬁnition, examples and discussion. For further references, we observe that (2.2.23) and (2.2.25) together imply 1 T Z T 0 ∥(SB ∗ASB)(t)f∥1 dt ≤ 1 T Z T 0 Z t 0 ∥SB(t −s)ASB(s)f∥1 dsdt ≤ 1 T Z T 0 Z T 0 ∥SB(u)∥B(X2,X1)∥ASB(s)f∥1 duds ≲ 1 T Z T 0 Θ(u) du ∥f∥2. Arguing in a similar way for any ℓ≥1, we establish 1 T Z T 0 ∥(SB ∗(ASB)(∗ℓ))(t)f∥1 dt ≲1 T Z T 0 Θ du ∥f∥2→0 as T →∞. (2.2.29) For synthesizing and for further references, let us now bring out some possible general framework for semigroup enjoying weak dissipativity. We introduce the following structure condition on a semigroup SL and its generator L by assuming 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 45 (HS3) there exist a splitting L = A + B, some Banach lattices X2 ⊂X1, an integer N ≥1 and some decaying functions Θi : R+ →R+ with Θ1(t) →0 as t →∞, Θ2 ∈L1(R+) such that A is positive, B generates a positive semigroup SB and the following estimates hold ∥(SBA)(∗ℓ) ∗SB∥B(X2,X1)= O(Θ1), ∀ℓ∈{0, . . . , N −1}, (2.2.30) ∥(SBA)(∗N)∥B(X1,X2)= O(Θ2). (2.2.31) We now particularize our discussion to a Radon measures framework. We assume that there exist some weight functions ψi on E, ψ0 ≲ψ1 ≤ψ2, with ψ2(x)/ψ1(x) →∞as x →∞so that M1 ψ2 ⊂⊂M1 ψ1 (compact imbedding for the weak convergence), a function χ ∈Cc(E), 0 ≤χ ≤1, and a constant M ≥0 such that (i) L∗ψ1 ≤−ψ0 + Mχ; (ii) L∗ψ2 ≤Mχ; (iii) ψ1 ≤εψ2 + η(ε)ψ0 for any ε > 0, for a function η : (0, 1] →(0, ∞) such that η(1) = 0, η(ε) →∞when ε →0, and t 7→Θ(t) := inf ε∈(0,1)(e− t η(ε) + ε) ∈L1(0, ∞). (2.2.32) It is worth emphasizing that from the very deﬁnition, we have automatically that Θ is positive and decreasing, Θ(0) = 1 and Θ(t) →0 as t →∞. Arguing similarly as we did during the proof of Corollary 2.1.19 and the end of Section 2.2.1, we introduce the splitting A := Mχ, B := L −A, and we establish that SB is a positive semigroup on X = M1 ψ2(E). More precisely, for 0 ≤f0 ∈ D(B) in the domain of SB and denoting ft := SB(t)f0, we may compute d dt Z ft ψ2 ≤ Z ft B∗ψ2 ≤0 and similarly d dt Z ft ψ1 ≤ Z ft B∗ψ1 ≤− Z ft ψ0. Integrating both diﬀerential inequalities, we deduce SB ∈L∞ t (B(M1 ψi)), i = 1, 2 and Z ∞ 0 ∥SB(t)f0∥M1 ψ0dt ≤∥f0∥M1 ψ1, ∀f0 ∈M1 ψ1. We may make a slight (but important) improvement of the previous estimate by proceeding similarly as we did for proving (2.2.23). Using the same notations as in the above computation, we indeed have d dt Z ft ψ1 + 1 η(ε) Z ft ψ1 ≤ ε η(ε) Z ft ψ2 ≤ ε η(ε) Z f0 ψ2, where we have used (i) and (iii) in the ﬁrst inequality and the previous L∞ t (B(M1 ψ2)) bound in the second inequality. Integrating in time, we deduce ∥SB(t)f∥M1 ψ1≤Θ(t)∥f∥M1 ψ2, ∀t > 0. Taking Xi := M1 ψi and N = 1, we see that L then satisﬁes (HS3) with Θi = Θ. 46 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM 2.2.4 First existence result in the weakly dissipative case We ﬁrst come back to the proof of Theorem 2.1.20 and explain what goes wrong when we try to adapt it to a weak dissipativity context. More precisely, we assume that SL is a positive semigroup (so that (H1) holds) satisfying L∗ψ0 ≥0 for some ψ0 ∈X′{0} (so that (H2) holds) and the splitting structure (HS3) for some bounded operator A and some weakly dissipative operator B, in the sense that (2.2.19) holds. In such a situation, we may deﬁne λ1 := inf{λ ∈R; RL(κ) ∈B(X), ∀κ ≥λ} ≥0, and there exist sequences (λn) of R and ( ˆ fn) of X+ such that λn ↘λ1 ≥0, ∥ˆ fn∥= 1, εn := λn ˆ fn −L ˆ fn →0 in X, thanks to Lemma 2.1.6. In the simplest situation, we may further assume that RB(κ) : X1 →X0 is uniformly bounded in κ ≥λ1 and A : X0 →X1 with X = X1 ⊂X0. The issue is that even in that case, we may write ˆ fn = RB(λn)A ˆ fn + RB(λn)εn, but it is not clear how to conclude that ( ˆ fn) belongs to a compact set in X because it is not clear that RB(λn)εn →0 in X. The next result aim precisely to establish that last convergence under suitable quite strong (although natural and true in some examples) assumptions on the operator L. The proof is adapted from [138, Section 6.3] and mixes some dynamical argument together with the stationary approach developed in Section 2.1.2. Theorem 2.2.3. Consider a positive semigroup SL in a Banach lattice X = X2 ⊂X1 ⊂X0 such that its generator L satisﬁes (1) there exists ψ0 ∈D(L∗), ψ0 ≥0, ψ0 ̸= 0, such that L∗ψ0 ≥0. (2) L = A + B with A and B satisfying (2.2.23), (2.2.25), (2.2.26) and (2.2.27); Then, there exist λ1 ≥0 and f1 ∈X1 such that ∥f1∥X1= 1, f1 ≥0, Lf1 = λ1f1. (2.2.33) Proof of Theorem 2.2.3. We split the proof into four steps. Step 1. We know from Lemma 2.1.2 and Lemma 2.1.4-(i) that (H1) and (H2) hold. We may then deﬁne λ1 ≥0 with the help of (2.1.16). If λ1 > 0, we see that V(α) deﬁned in (2.1.22) is bounded in B(X) uniformly on α ≥κ0 := λ1/2 because of (2.2.23) and (2.2.25), and that W(α) also deﬁned in (2.1.22) satisﬁes (2.1.23) because of (2.2.25) and Remark 2.1.9-(1). Using Lemma 2.1.8, we get that (H3) holds, and we conclude thanks to Theorem 2.1.20 in that case. In the sequel, we always assume λ1 = 0. Step 2. Let us ﬁx f0 ∈D(L) such that f0 ≥0 and C0 := ⟨f0, ψ0⟩> 0, which exists by deﬁnition of ψ0. Denoting f(t) := SL(t)f0, we have d dt⟨f(t), ψ0⟩= ⟨Lf(t), ψ0⟩= ⟨f(t), L∗ψ0⟩≥0, which in turns implies ⟨f(t), ψ0⟩≥C0, ∀t ≥0. Step 3. We claim that ∥RL(0)∥B(X2,X1)= +∞. That in particular implies ∥RL(0)∥B(X)= +∞ and thus 0 ∈Σ(L). We assume by contradiction that K2,1 := ∥RL(0)∥B(X2,X1)< +∞. First, because SL is positive, we have \|RL(z)f\|≤ Z ∞ 0 e−tℜezSL(t)\|f\| dt = \|RL(ℜez)\|f\|, 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 47 from which we deduce ∥RL(z)∥B(X2,X1)≤∥RL(ℜez)∥B(X2,X1), ∀z ∈∆0. As a consequence, we have sup y∈R ∥RL(iy)∥B(X2,X1)≤K2,1. (2.2.34) We write the representation formulas (taken from [177, (2.21)]) SL(t)f = T0(t) + lim M→∞T1,M(t) with T0(t) := N−1 X ℓ=0 SB ∗(ASB)(∗ℓ)(t)f and T1,M(t) := i 2π Z a+iM a−iM ezt RL(z) (ARB(z))Nf dz, for any f ∈D(L), t ≥0 and a > 0. On the one hand, from (2.2.29), we have 1 T Z T 0 T0(t) dt →0 in X1, as T →∞. (2.2.35) On the other hand, we estimate the contribution of the Cesàro mean of T1,M. Integrating by part, we have T1,M(t) = 1 t i 2π Z a+iM a−iM ezt d dz [RL(z) (ARB(z))N]f dz, with d dz [RL(z) (ARB(z))N] = X ε∈NN+1, \|ε\|=1 RL(z)1+ε0AR1+ε1 B (z) . . . AR1+εN B (z). Together with condition (2.2.26) and estimate (2.2.34), we get d dz [RL(z) (ARB(z))N]f 1 ≤(K2,1 + K2 2,1)N sup ε∈NN, \|ε\|≤1 ∥AR1+ε1 B (z) . . . AR1+εN B (z)f∥2 ≤C1 ⟨y⟩α ∥f∥2, uniformly for any z = x + iy ∈∆0, for some constant C1 > 0. We deduce ∥lim M→∞T1,M(t)∥1≤1 t 1 2π Z R C1 ⟨y⟩α dy∥f∥2→0, (2.2.36) as t →∞. Gathering (2.2.35) and (2.2.36), we conclude in particular that 1 T Z T 0 SL(t)f0 dt →0 in X1, as T →∞, which is in contradiction with the estimate of Step 2. Step 4. Conclusion. Taking advantage of the convenient blow up of RL(λ) as λ ↘0 established in the previous Step 2, we may now argue similarly as in the proof of Theorem 2.1.20. From Step 2, there exists a sequence (λn) such that λn →0 and ∥RL(λn)∥B(X2,X1)→∞. 48 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM That means that there exist (fn) and (gn) such that ∥fn∥X1→∞, ∥gn∥X2= 1, fn = RL(λn) gn, or equivalently that there exist ( ˆ fn) and (εn) (by deﬁning ˆ fn := fn±/∥fn±∥X1, εn := gn±/∥fn±∥X1) satisfying ∥ˆ fn∥X1= 1, ˆ fn ≥0, ∥εn∥X2→0, εn = (λn −L) ˆ fn. (2.2.37) As in the proof of Lemma 2.1.8, we deduce that (2.1.24) holds, that is ˆ fn = N−1 X ℓ=0 (RB(λn)A)ℓRB(λn)εn + (RB(λn)A)N ˆ fn. (2.2.38) Using the uniform boundedness (RB(λn)A)ℓRB(λn) ∈B(X2, X1), (RB(λn)A)N ∈B(X1, X1), X1 ⊂⊂X1, we deduce that ( ˆ fn) belongs to a compact set of X1, or in other words, that there exist a subsequence of ( ˆ fn) (not relabeled) and f1 ∈X1 such that ˆ fn →f1 in X1. We may pass to the limit in (2.2.37), and we get (2.2.33). 2.2.5 Second existence result in the weakly dissipative case Using a pure dynamical approach adapted from the second proof of Theorem 2.2.1 and from [50, Theorem 6.1], we establish a second existence result which is less demanding in terms of conditions on the semigroup SL. Theorem 2.2.4. Consider a positive semigroup S = SL on a Banach lattice X = Y ′ for a separable Banach lattice Y . We assume (i) there exists ψ0 ∈Y+{0} such that [Stf]0 ≥[f]0 for any f ∈X+ and f 7→[f]0 := ⟨\|f\|, ψ0⟩ is a norm on X. We then denotes X the vector space X endowed with this norm [·]0; (ii) there exist v ∈L∞(R+; B(X)) and 0 ≤w ∈L1(R+; B(X, X)) such that S = V + W ∗S, (2.2.39) and we set M := sup t≥0 ∥v(t)∥B(X)< ∞, Θ(t) := ∥w(t)∥B(X,X)∈L1(R+). (2.2.40) Then there exists a pair (λ1, f1) ∈R+ × X+{0} such that Lf1 = λ1f1. Remark 2.2.5. (1) When SL satisﬁes (HS3) then (2.2.39) holds with V := N−1 X ℓ=0 SB ∗(ASB)(∗ℓ), W := (SBA)(∗N). (2.2.41) (2) By deﬁnition of the norm [·]0 of X, we see that X is a weighted L1 space or a weighted Radon measures space. In many applications, when both X and X are Radon measures spaces, one can choose N = 1. On the other hand, when X is for instance a (possibly weighted) Lp space, one must take N ≥2 in most of the applications. In condition (ii), the ﬁrst bound is not really demanding and almost automatic in view of the estimates exhibited in Section 2.2.3. The second bound is a kind of regularity estimate reminiscent of the enlarging and shrinkage technique developed in [188, 113, 174]. 2.2. EXISTENCE THROUGH A DYNAMICAL APPROACH 49 Proof of Theorem 2.2.4. We split the proof into three steps. Step 1. We deﬁne R := max(2∥Θ∥L1, ∥g0∥), for some g0 ∈X+ such that [g0]0 = 1, and next the nonempty convex and compact (in the sense of X) set C := {f ∈X+; [f]0 = 1, ∥f∥≤R}, as well as the increasing function λ(t) := inf f∈C[S(t)f]0, ∀t ≥0. We have the alternative • (1) sup λ > 2M, • (2) sup λ ≤2M. Step 2. We assume that the ﬁrst term (1) of the alternative holds true, or in other words, there exists T0 > 0 such that ∀f ∈C, [ST0f]0 ≥2M. (2.2.42) We deﬁne as before ΦT0f := ST0f [ST0f]0 , ∀f ∈C. By construction, for any f ∈C, we have ΦT0f ≥0 and [ΦT0f]0 = 1. On the other hand, using the splitting structure (2.2.39) and the estimates (2.2.40), we have ∥S(t)f∥≤M∥f∥+ Z t 0 Θ(t −s)[S(s)f]0 ds. From hypothesis (i) and the semigroup property, we also have [Stf]0 ≥[Ssf]0, ∀t ≥s ≥0. The two above estimates together imply ∥ΦT0f∥ ≤ M∥f∥ [ST0f]0 + Z T0 0 Θ(T0 −s) [Ssf]0 [ST0f]0 ds ≤ 1 2∥f∥+∥Θ∥L1≤R, for any f ∈C. We have thus proved ΦT0 : C →C. Thanks to the Tykonov ﬁxed point Theorem, there exists fT0 ∈C such that ΦT0fT0 = fT0. In other words, we have built a pair of “almost eigenvalue and eigenfunction” fT0 ≥0, [fT0]0 = 1, ST0fT0 = eλ1T0fT0, with eλ1T0 = [ST0f]0 and thus λ1 ∈[0, κ1]. We conclude to the existence of f1 ∈C such that Lf1 = λ1f1 really similarly as in Step 2 of the Second proof of Theorem 2.2.1. Step 3. We assume that the second term (2) of the alternative holds true. In that case, for any n ≥1, there exists fn ∈C such that [S(n)fn]0 ≤2M. By compactness, there exists f0 ∈C and a subsequence (fnk) such that fnk ⇀f0 ∈C and ∀t ≥0, ∀k (nk ≥t), [S(t)fnk]0 ≤[S(nk)fnk]0 ≤2M, so that ∀t ≥0, [S(t)f0]0 ≤2M. (2.2.43) 50 CHAPTER 2. EXISTENCE OF SOLUTION OF THE EIGENTRIPLET PROBLEM Using this particular initial datum, we argue similarly as in [50, proof of Theorem 6.1], and we conclude to the existence of a stationary state. More precisely, we come back to the splitting structure (2.2.39) of the semigroup S and we introduce the associated Cesàro means UT := 1 T Z T 0 S(t) dt, VT := 1 T Z T 0 v(t) dt, KT := 1 T Z T 0 (w ∗S)(t) dt, (2.2.44) for any T > 0. We obviously have ∥VT ∥B(X)≤1 T Z T 0 ∥v(t)∥B(X) dt ≤M. On the other hand, we have Z T 0 (w ∗S)(t) dt = Z T 0 Z T s w(t −s)dtS(s) ds ≤ Z T 0 w(τ)dτ Z T 0 S(s) ds, thanks to the Fubini theorem and the positivity of the two operators involved in this integral formula. We deduce ∥KT f0∥ ≤ Z T 0 w(τ)dτ 1 T Z T 0 S(s) f0 ds ≤ Z ∞ 0 ∥w(τ)∥B(X,X)dτ h 1 T Z T 0 S(s)f0 ds i 0 = ∥Θ∥L1 [UT f0]0, thanks to assumption (ii), so that KT f0 is uniformly bounded in X thanks to (2.2.43) and the elementary estimate [UT f0]0 ≤[ST f0]0. We then deduce that UT = VT + KT satisﬁes ∥UT f0∥≤M∥f0∥+2M∥Θ∥L1 and 1 ≤[ST f0]0 ≤2M, for any T > 0. By compactness, there exists Tk →+∞and f1 ∈X+ such that UTkf ⇀f1 weakly in X. Thanks to the second inequality, we have [f1]0 ≥1. We then argue thanks to the usual mean ergodic theorem trick. For any ﬁxed s > 0, we observe that S(s)f1 −f1 = lim k→∞ n 1 Tk Z Tk 0 S(s)S(t)f0dt −1 Tk Z Tk 0 S(t)f0 dt o = lim k→∞ n 1 Tk Z Tk+s Tk S(t)f0dt −1 Tk Z s 0 S(t)f0 dt o weakly in X. By the lower semicontinuous property of the norm [·]0, we deduce [S(s)f1 −f1]0 ≤ lim inf k→∞ n 1 Tk Z Tk+s Tk [S(t)f0]0dt + 1 Tk Z s 0 [S(t)f0]0 dt o = 0, so that f1 is a stationary solution, and thus f1 is an eigenfunction associated to the eigenvalue λ1 = 0. Chapter 3 Irreducibility and the geometry of the ﬁrst eigenvalue In this chapter, we are concerned with the geometric part of the Krein-Rutman theorem for an unbounded operator L on a Banach lattice X. We assume that the conclusions of the existence part are achieved, namely (C1) the ﬁrst primal and dual eigenvalue problem has a solution (λ1, f1, φ1): there exist λ1 ∈R, f1 ∈X+ ∩D(L), φ1 ∈Y+ ∩D(L∗) such that ∥f1∥= 1, Lf1 = λ1f1, ∥φ1∥= 1, L∗φ1 = λ1φ1. (3.0.1) By construction, we also have Σ(L) ⊂{z ∈C, ℜe(z) ≤λ1}. Assuming that S is positive as for the existence part and an additional strong maximum principle property, we analyze the ﬁrst eigenvalue problem. 3.1 More about positivity For further references, we introduce several notions which are strongly related to the positivity property for semigroups. The signum operator sign. In a real Banach lattice X, we say that sign f ∈B(X, X′′) is a signum operator for f ∈X, if it satisﬁes the following properties (sign f) f = \|f\|, (sign f) g ≤\|g\|, ∀g ∈X. In the sequel, we will always assume that such an operator exists. We refer to [7, Sections C.I & C.II] for a general introduction to the topic. In practice, we will only need a weak formulation of the sign operator (see below) which may be deﬁned only in some subspace X ⊂X. We always additionally assume that the signum operator satisﬁes (sign (−f)) (−g) = (sign f)g, ∀g ∈X, (sign f) g = g, ∀g ∈X, if f ∈X+, We also deﬁne sign+f := 1 2(I + signf). • When X is a space of functions, the sign operator sign f associated to f ∈X corresponds to the multiplication by the function sign f := 1f>0 −1f<0. When X := Lp(E), we obviously see that sign f ∈B(Lp(E)) for any f ∈Lp(E). On the other hand, when X := C0(E), we only have 51 52CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE sign f ∈B(C0(E); M∞(E)), where M∞(E) denotes the space of uniformly bounded measurable functions, so that M∞(E) ⊂(C0(E))′′. In a space of bounded measures X = M1(E), we may deﬁne the sign operator by means of the Radon-Nikodym theorem. For a given f ∈M1(E), using Hahn decomposition, there exists indeed a measurable function α : E →{−1, 1} such that f = α\|f\|, and we then deﬁne (sign f)g = αg for any g ∈M1(E). • When X is σ-order complete, in the sense that any increasing and upper bounded sequence has a supremum (a common least upper bound), the operator sign exists and is more regular, namely signf ∈B(X) for any f ∈X, see and also [7, Section C.I.8]. We recover in particular that sign f ∈B(Lp(E)) for any f ∈Lp(E). Weak principle maximum and Kato’s inequality. We introduce now two deﬁnitions formulated on an operator L which are almost equivalent to the positivity property of the semigroup S when L is the generator of S. • We say that the operator L satisﬁes the weak maximum principle when κ ∈R, f ∈D(L) and (κ −L)f ≥0 imply f ≥0; (3.1.2) • We say that the operator L satisﬁes Kato’s inequality when ∀f ∈D(L), L\|f\|≥(sign f)Lf. (3.1.3) Since \|f\| does not necessarily belong to D(L), the correct way to understand Kato’s inequality is ∀f ∈D(L), ∀ψ ∈D(L∗) ∩Y+, ⟨\|f\|, L∗ψ⟩≥⟨(sign f)Lf, ψ⟩. (3.1.4) We immediately see from the deﬁnitions that (3.1.3) is equivalent to assuming ∀f ∈D(L), Lf+ ≥(sign+ f)Lf. (3.1.5) Remark 3.1.1. We complement Lemma 2.1.1, by claiming that for a semigroup S = SL on a Banach lattice X, there is equivalence between the fact that S is positive and κ −L satisﬁes the weak maximum principle for any κ > ω(L), what is straightforward using that these properties are equivalent to the fact that RL(κ) ≥0 for any κ > ω(L). These properties also imply that Kato’s inequality holds true, see [190, 5], [8, Proposition 1.1], [6, Remark 3.10] and the textbook [7, Theorems C.II.2.4, C.II.2.6 and Remark C-II.3.12]. We end this chapter by introducing notions of strict positivity which strengthen the previous deﬁned positivity condition. Strict order. We may deﬁne a ﬁrst strict order > (or <) on X by writing for f ∈X f > 0 if f ∈X+{0} and similarly a strict order > (or <) on X′ by writing for ψ ∈X′ ψ > 0 if ψ ∈X′ +{0}. We may next deﬁne a second (and stronger) strict order ≫(or ≪) on X by writing for f ∈X f ≫0 or f ∈X++ iﬀ ∀ψ ∈X′ +{0}, ⟨ψ, f⟩> 0, and similarly a strict order ≫(or ≪) on X′ by writing for ψ ∈X′ ψ ≫0 or ψ ∈X′ ++ iﬀ ∀g ∈X+{0}, ⟨ψ, g⟩> 0. On the two Banach lattices X and Y , we have thus three positivity notions with ≫(associated to X++ and Y++) stronger than > (associated to X+{0} and Y+{0}) which itself is stronger than ≥(associated to X+ and Y+). Let us comment on the notion of strict positivity. 3.2. IRREDUCIBILITY AND STRONG MAXIMUM PRINCIPLE 53 Examples 3.1.2. In the space C0(E), the strict order is deﬁned by f ≫0 iﬀf(x) > 0 for any x ∈E. In a space Lp(E, E , µ), 1 ≤p ≤∞, the strict order is deﬁned by f ≫0 iﬀf(x) > 0 for µ-a.e. x ∈E. In the space M1(E), the strict order is deﬁned by duality by f ≫0 iﬀ⟨f, ϕ⟩> 0 for any ϕ ∈C0(E), ϕ ≥0, ϕ ̸≡0. Remark 3.1.3. In a Banach lattice X such that intX+ ̸= ∅, the common deﬁnition of the strict order is X++ := intX+. In particular, in the case when E is compact and X = C0(E) = C(E), we have intX+ ̸= ∅and the deﬁnition of X++ introduced in Examples 3.1.2 coincides with intX+. In all the other examples considered, we have int(X+) = ∅, and thus our deﬁnition of the strict order does not coincide with the one deﬁned through the set int(X+). Remark 3.1.4. Another notion of strict order can be deﬁned through the notions of ideals and quasi-interior points as brieﬂy explained now, see or [20, Chapter 10] and the references therein for details. Deﬁning the segment [g1, g2] and the set If for g1, g2 ∈X and f ∈X+{0} by [g1, g2] := {g ∈X; g1 ≤g ≤g2}, If := [ k≥0 [−kf, kf], one shows that If is an ideal in the sense that g ∈If implies \|g\|∈If and 0 ≤g ≤f implies g ∈If. We say that f is an order unit if If = X. When intX+ ̸= ∅, we ﬁnd that f is an order unit iﬀf ∈intX+ from Lemma 2.1.22, so that we recover the notion of strict positivity deﬁned above. On the other hand, we say that f is a quasi-interior point if ¯ If = X. When X = Lp(E, E , µ), 1 ≤p < ∞, µ is a σ-ﬁnite diﬀuse (or atomless) measure, one shows that f is a quasi-interior point iﬀf > 0 a.e., see [20, Examples 10.16], so that we also recover the notion of strict positivity deﬁned above by deﬁning f ∈X++ iﬀ¯ If = X. We ﬁnally point out the following result. For a semigroup S = SL in a Banach lattice, under the mild assumption that there exists a strictly positive subeigenvector for the dual problem, namely ∃φ ∈X′ ++, ∃b ∈R, L∗φ ≤b φ, Kato’s inequality implies that S is positive, see [8, Theorem 1.6]. 3.2 Irreducibility and strong maximum principle We present some material involving the strict positivity. In the sequel, for both spaces we will always assume X++ ̸= ∅, Y++ ̸= ∅and f+ ∈X++ implies f ∈X++, φ+ ∈Y++ implies φ ∈Y++. (3.2.6) Lemma 3.2.5. The property (3.2.6) holds true when X is a reﬂexive space, X = Lp, p ∈[1, ∞], X = C0 or X = M1. Proof of Lemma 3.2.5. Assume ﬁrst that X is reﬂexive and ﬁx f ∈X such that f+ ∈X++. There exists 0 ≤ϕ∗∈X′, ∥ϕ∗∥= 1 such that ∥f−∥= ⟨ϕ∗, f−⟩= sup 0≤ψ≤ϕ∗⟨ψ, −f⟩, where we have used the corollary (2.1.2) of the Hahn-Banach dominated extension theorem in the ﬁrst equality and the deﬁnition f−as an element of X′′ in the second equality. There next exists 0 ≤ψ∗≤ϕ∗, such that ∥f−∥= ⟨ψ∗, −f⟩= ⟨ψ∗, f−−f+⟩, 54CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE where we have used that BX′ is compact for the weakly ∗topology σ(X′, X) in the ﬁrst equality. We deduce 0 ≤⟨ϕ∗−ψ∗, f−⟩= −⟨ψ∗, f+⟩≤0, with ⟨ψ∗, f+⟩< 0 if ψ∗̸= 0. That implies ψ∗= 0 and f−= 0. We next assume X = Lp(E, E , µ) and we take f ∈X such that f+ ∈X++. From the deﬁnition of X++ made explicit in Examples 3.1.2, we have f+(x) = max(f(x), 0) > 0 a.e., so that f(x) > 0 a.e. and ﬁnally f−(x) = 0 a.e.. We last assume X = M1(E, E ) and we take again f ∈X such that f+ ∈X++. Writing f = α\|f\|, where α : E →{−1, 1} is the measurable function related to Hahn decomposition as introduced in Section 3.1, the condition f+ = α+\|f\|> 0 means that α+(x) > 0 for \|f\|-a.e. x ∈E. We deduce again α(x) > 0 for \|f\|-a.e. x ∈E and thus f−= α−\|f\|= 0. For an operator A ∈B(X), we have yet formalized a positivity condition in section 2.1.1, by (P1) A ≥0 if A : X+ →X+. Other possible deﬁnition of positivity may be (P2) A : X+{0} →X+{0}; (P3) A : X++ →X++. We now deﬁne a stronger notion of positivity, named as strong positivity condition, by (P4) A > 0 if A : X+{0} →X++. We list without proof some elementary properties about these diﬀerent notions and also refer to Section 5.1 for further discussion. We have (P2) ⇒(P1), (P3) ⇒(P1) as well as (P4) ⇒((P3), (P2)). We also have A : X+ →X+ iﬀA∗: Y+ →Y+; A : X++ →X++ iﬀA∗: Y++ →Y++; A : X+{0} →X++ iﬀA∗: Y+{0} →Y++. We say that λ −L satisﬁes the strong maximum principle if f ∈X+ ∩D(L), (λ −L)f ≥0 imply f ≫0 or f = 0. (3.2.7) We say that a positive semigroup S is irreducible if ∀f ∈X+{0}, ∀φ ∈Y+{0}, ∃τ > 0 ⟨Sτf, φ⟩> 0. (3.2.8) A semigroup S is classically said to be irreducible and aperiodic if the above positivity condition holds for all suﬃciently large times, namely ∀f ∈X+{0}, ∀φ ∈Y+{0}, ∃T > 0, ∀τ ≥T ⟨Sτf, φ⟩> 0. (3.2.9) Other notions of strong positivity for the semigroup S are ∃T > 0, ST : X+{0} →X++, (3.2.10) ∃T > 0, Z T 0 S(t)dt : X+{0} →X++. (3.2.11) We summarize some possible implications between the previous positivity notions. Lemma 3.2.6. For a positive semigroup S, the following hold: (1) The pointwise strong positivity condition (3.2.10) implies the condition (3.2.11); (2) The integral strong positivity condition (3.2.11) implies the irreducibility condition (3.2.8), but the reverse implication is false. Similarly, the irreducibility and aperiodicity condition (3.2.9) implies the irreducibility condition (3.2.8), but the reverse implication is false; (3) The irreducibility condition (3.2.8) is equivalent to the fact that RL(λ) : X+{0} →X++, for any λ > λ1 as well as to the fact that λ −L satisﬁes the strong maximum principle (3.2.7) for any λ > λ1. 3.2. IRREDUCIBILITY AND STRONG MAXIMUM PRINCIPLE 55 The result is very classic, at least for strongly positive semigroup, see e.g. [7, Deﬁnition C.3.1] or [20, Proposition 14.10]. For the sake of completeness, we however present a short proof. Proof of Lemma 3.2.6. We prove (1). We assume (3.2.10) and we ﬁx g ∈X+{0}, φ ∈Y+{0}, so that ⟨S(T)g, φ⟩> 0. Observing that the function t 7→⟨S(t)g, φ⟩is continuous, there exists ε > 0 such that ⟨S(t)g, φ⟩> 0 for any t ∈[T −ε, T], so that DZ T 0 S(t)dtg, φ E = Z T 0 ⟨S(t)g, φ⟩dt > 0. Because φ ∈Y+{0} may be chosen arbitrary, we deduce (3.2.11). We prove (2). We assume now (3.2.11) and we ﬁx g ∈X+{0}, φ ∈Y+{0}, so that Z T 0 ⟨S(t)g, φ⟩dt = DZ T 0 S(t)dtg, φ E > 0, by assumption. We get (3.2.8) by observing that the function t 7→⟨S(t)g, φ⟩must be positive somewhere on [0, T]. For the reverse implication we refer to [27, 93], where is studied an example of growth-fragmentation operator associated to mitosis satisfying the irreducibility condition (3.2.8) but not the integral strong positivity condition (3.2.11) nor the irreducibility and aperiodicity condition (3.2.9), see also Section ??. We prove (3). We ﬁnally assume (3.2.8). From the above continuity argument, for any g ∈X+{0}, φ ∈Y+{0} there exist τ > ε > 0 such that ⟨S(t)g, φ⟩> 0 for any t ∈[τ −ε, τ + ε]. As a consequence and thanks to the representation formula (2.1.13) for some ﬁxed λ > λ1 which holds thanks to Lemma 2.1.2-(ii), we have ⟨RL(λ)g, φ⟩= DZ ∞ 0 e−λtS(t)dtg, φ E > 0. Because φ ∈Y+{0} is arbitrary, we have established that RL(λ)g ∈X++ for any g ∈X+{0}. In other words, when λ > λ1 and f ∈X+ ∩D(L) satisfy g := (λ −L)f ≥0, we deduce that f = RL(λ)g ∈X++, what is the strong maximum principle. This one is obviously equivalent to the strong positivity property RL(λ) : X+{0} →X++. On the other way round, writing the above identity as Z ∞ 0 e−λt⟨S(t)g, φ⟩dt = ⟨RL(λ)g, φ⟩, we see that the strong maximum principle implies that the RHS term is positive for any g ∈X+{0}, φ ∈Y+{0}. As a consequence, the LHS term is positive and there exists τ > 0 such that ⟨S(τ)g, φ⟩> 0, which is nothing but the irreducibility condition (3.2.8). We present two other elementary results about the strong maximum principle. Lemma 3.2.7. Consider L satisfying (H1) and λ ∈R. Then the following assertions are equivalent (1) λ −L satisﬁes the strong maximum principle for any f ∈D(L) ∩X+; (2) λ −L satisﬁes the strong maximum principle for any f ∈D(Lk) ∩X+ for some k ≥1; (3) λ −L∗satisﬁes the strong maximum principle for any φ ∈D(L∗) ∩Y+; (4) λ −L satisﬁes the strong maximum principle for any φ ∈D((L∗)ℓ) ∩Y+ for some ℓ≥1. Proof of Lemma 3.2.7. Assume that λ−L satisﬁes the strong maximum principle for some λ ∈R and k ≥1 and consider φ ∈D(L∗) ∩Y+{0} such that (λ −L∗)φ ≥0. For any κ > max(λ, λ1) and g ∈D(Lk−1) ∩X+{0}, thanks to (H1) and the strong maximum principle, there exists f ∈D(Lk) ∩X++ such that (κ −L)f = g. As a consequence, we have ⟨φ, g⟩ = ⟨φ, (κ −L)f⟩ = ⟨(κ −L∗)φ, f⟩≥(κ −λ) ⟨φ, f⟩> 0. 56CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE Since g ∈D(Lk−1) ∩X+ is arbitrary and D(Lk−1) ∩X+ is dense in X+, we deduce that φ ≫0. We have proved that λ −L∗satisﬁes the strong maximum principle. The other implications can be proved similarly. Remark 3.2.8. We may replace the condition (1) by assuming that λ −L satisﬁes the strong maximum principle for f ∈C ∩X+ for a subspace C ⊂D(L) such that (λ −L)−1 ∈B(C) and C is dense in X. The strong maximum principle can be seen as a consequence of the weak maximum principle together with the existence of a family of strictly positive barrier functions. We give now a typical result which can be applied (or modiﬁed in order to be applied) in many situations. Lemma 3.2.9. We assume that (i) the operator λ −L satisﬁes the weak maximum principle; (ii) there exists a subset G ⊂X++ ∩D(L) such that ∀f ∈D(L) ∩X+{0}, ∃g ∈G such that (g −f)+ ∈D(L) and (L −λ)g ≥0. Then λ −L satisﬁes the strong maximum principle. Proof of Lemma 3.2.9. We consider f ∈D(L) ∩X+{0} such that (λ −L)f ≥0 and choose g ∈G such that h := (g −f)+ ∈D(L). We remark that from Kato’s inequality (L −λ)h ≥sign+(g −f)(L −λ)(g −f) ≥0. As a consequence of the weak maximum principle, we have h ≤0. That implies h = 0, so that g −f ≤0 and ﬁnally f ≫0. The above barrier functions technique is also useful for obtaining the condition (H2) (possibly in a constructive way). Lemma 3.2.10. For an operator L, we assume that (i) the condition (H1) holds with a constant κ1 ∈R; (ii) the hypothesis (ii) in Lemma 3.2.9 holds; (iii) there exists h0 ∈X+{0} such that for any g ∈G there exists ε > 0 such that g ≥εh0. Then the property (H2) holds true. Proof of Lemma 3.2.10. Thanks to assumption (i), we may deﬁne f0 ∈D(L) ∩X+{0} as the solution to the equation (κ1 −L)f0 = h0. From the proof of Lemma 3.2.9 and condition (iii), there exists g ∈G and next ε > 0 such that f0 ≥g ≥εh0. Coming back to the equation, we have Lf0 = κ1f0 −h0 ≥(κ1 −ε−1)f0, so that condition (H2) holds true with κ0 := κ1 −ε−1 thanks to Lemma 2.1.4-(ii). 3.3 The geometry of the ﬁrst eigenvalue problem We come back on and state a result about the geometry of the ﬁrst eigenvalue. On the one hand, we assume that X is a Banach lattice such that (X1) the signum operator is well deﬁne in X, X++ ̸= ∅and the property (3.2.6) holds true. On the other hand, we consider an operator L on X which satisﬁes the conclusion (C1) about the existence of a solution (λ1, f1, φ1) to the ﬁrst eigentriplet problem. We also assume (H1′) the weak maximum principle λ > λ1, f ∈D(L), (λ −L)f ≥0 imply f ≥0 (3.3.12) 3.3. THE GEOMETRY OF THE FIRST EIGENVALUE PROBLEM 57 and its Kato’s inequalities counterpart (signf)Lf ≤L\|f\|, (sign+f)Lf ≤Lf+, (3.3.13) as well as (H4) the strong maximum principle λ ≥λ1, f ∈X+ ∩D(L), (λ −L)f ≥0 imply f ≫0 or f = 0. (3.3.14) We may then state our main result in this section, where we recall that N(A) denotes the null space associated to the operator A. Theorem 3.3.11. We assume that X is a Banach lattice satisfying (X1). We consider an unbounded operator L on X which satisﬁes the conclusion (C1) about the existence of a solution (λ1, f1, φ1) to the ﬁrst eigentriplet eigenvalue problem, the weak maximum principles and Kato’s inequality (H1′), as well as the strong maximum principle (H4). Then the following hold i) f1 ≫0, φ1 ≫0 and λ1 is the unique eigenvalue associated to a positive eigenvector. ii) λ1 is algebraically simple: N((L −λ1)k) = Span(f1), ∀k ≥1, N((L∗−λ1)k) = Span(φ1), ∀k ≥1, in particular f1 (resp. φ1) if the unique positive and normalized eigenvector of L (resp. L∗) associated to λ1. Finally, the projection on the ﬁrst eigenspace (associated to λ1) writes Π f := ⟨f, φ1⟩f1. Remark 3.3.12. (1) It is worth emphasizing again that (3.3.12) is true when L is the generator of a positive semigroup, and that (3.3.14) is true when SL enjoys additional strong positivity (or irreducibility) condition as formulated in (3.2.8), (3.2.9), (3.2.10) or (3.2.11). As a consequence, the conclusion of Theorem 3.3.11 holds true when L is the generator of a positive semigroup which satisﬁes the hypotheses of the existence part of the Krein-Rutman Theorem 2.1.20 and one of the additional above strict positivity conditions. (2) Theorem 3.3.11 has to be compared with the seminal Krein and Rutman Theorem ?? (), to the many results gathered in [7, Part C-III] (see in particular [7, Proposition C.3.5], [7, Theorem C.3.8] and the original paper ) and to the more recent contributions [177, Theorem 5.3], [20, Theorem 14.15] and [138, Theorem 5.1]. Probably many of the conclusions of Theorem 3.3.11 are very similar (or even included) in the material of [7, Part C-III]. However, our assumptions slightly diﬀerent since we do no make explicit reference to a positive semigroup but rather refer to the weak and strong maximum principle. (3) Our proof is quite direct and elementary and uses similar arguments as those used during the proof of [177, Theorem 4.3] and [138, Theorem 5.1]. We learnt this kind of technique in the (less abstract and general) proof of the uniqueness part of [204, Lemma 2.1]. (4) From ii), we deduce that L decomposes acording to X = X0 ⊕X1 with X1 := Span f1 and X0 := (Span φ1)⊥= {f ∈X; ⟨f, φ1⟩= 0} in the sense of [136, § III.5.6]. More precisely, X = X0 ⊕X1 is a topological direct sum, L : X0 ∩D(L) →X0 and L : X1 →X1. The proof of Theorem 3.3.11 is split into the following Lemma 3.3.13, Lemma 3.3.15, Lemma 3.3.16 and Lemma 3.3.18. Lemma 3.3.13. Under assumptions (C1) and (H4), the solution (λ1, f1, φ1) to the ﬁrst eigentriplet problem satisﬁes f1 ≫0 and φ1 ≫0. (3.3.15) 58CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE Proof of Lemma 3.3.13. By deﬁnition of f1, we have f1 ∈X+{0}, (λ1 −L)f1 ≥0, and thus f1 ≫0 from (H4). Thanks to Lemma 3.2.7, the strong maximum principle holds for λ1 −L∗, and the same proof leads to the same conclusion, namely φ1 ≫0. Remark 3.3.14. It is worth emphasizing that the same conclusion holds when we only assume that f1 ∈X+ and φ1 ∈Y+ satisfy ∥f1∥= 1, Lf1 = λ1f1, ∥φ1∥= 1, L∗φ1 = λ∗ 1φ1, (3.3.16) what we see by just repeating the argument. In that case, we deduce that λ∗ 1 = λ1 by writing λ1⟨f1, φ1⟩= ⟨Lf1, φ1⟩= ⟨f1, L∗φ1⟩= λ∗ 1⟨f1, φ1⟩, and observing that ⟨f1, φ1⟩̸= 0. Lemma 3.3.15. Under assumptions (C1) and (H4), λ1 is the unique eigenvalue associated to a positive eigenvector for L (resp. for L∗). Proof of Lemma 3.3.15. Consider λ ∈C and f ∈X+{0} such that Lf = λf. We compute 0 = ⟨(λ −L)f, φ1⟩= ⟨f, (λ −L∗)φ1⟩= (λ −λ1)⟨f, φ1⟩, and thus λ = λ1 since ⟨f, φ1⟩> 0. The same proof applies to the dual problem. Lemma 3.3.16. Under assumptions (C1), (H1′) and (H4), we have N(L −λ1) = Span(f1) (resp. N(L∗−λ1) = Span(φ1)). In particular, f1 (resp. φ1) is unique (because of the positivity and normalization condition). Proof of Lemma 3.3.16. Consider a eigenfunction f ∈X{0} associated to the eigenvalue λ1. First, we observe from Kato’s inequality that λ1\|f\|= λ1sign(f)f = sign(f)Lf ≤L\|f\|. That inequality is in fact an equality, otherwise we would have λ1⟨\|f\|, φ1⟩̸= ⟨L\|f\|, φ1⟩= ⟨\|f\|, L∗φ1⟩= λ1⟨\|f\|, φ1⟩, and a contradiction. As a consequence, \|f\| is a solution to the eigenvalue problem λ1\|f\|= L\|f\|, so that λ1f± = Lf±, by writing f± = (\|f\|±f)/2. The strong maximum principle assumption implies f± ≫0 or f± = 0, and thus f+ ≫0 or f−≫0 thanks to Lemma 3.3.13. Without loss of generality we may assume f+ ≫0. From (3.2.6), we then deduce f ≫0. We introduce the normalized eigenfunctions ˜ f := rf and ˜ f1 = r1f1 with ⟨˜ f, φ1⟩= ⟨˜ f1, φ1⟩= 1. (3.3.17) Now, thanks to Kato’s inequality again, we write λ1( ˜ f −˜ f1)+ = sign+( ˜ f −˜ f1)L( ˜ f −˜ f1) ≤L( ˜ f −˜ f1)+, and for the same reason as above that last inequality is in fact an inequality. The strong maximum principle implies that either ( ˜ f −˜ f1)+ = 0, or in other words ˜ f ≤˜ f1, either ( ˜ f −˜ f1)+ > 0 or in other words ˜ f > ˜ f1. Because of the identity (3.3.17) and the fact that φ1 ∈X′ +{0}, the second case in the above alternative is not possible. Repeating the same argument with ( ˜ f1 −˜ f)+ we get that ˜ f1 ≤˜ f and we conclude with ˜ f = ˜ f1. The same proof applies to the dual problem. 3.4. MEAN ERGODICITY 59 Remark 3.3.17. Under the same hypothesizes as in Lemma 3.3.16, we have ψ ∈span(φ1) if ψ ∈Y+ satisﬁes L∗ψ ≥λ1ψ and g ∈span(f1) if g ∈X+ satisﬁes Lg ≥λ1g. In the second case, we indeed cannot have L∗g −λ1g ∈X+{0}, since this would implies ⟨Lg −λ1g, φ1⟩> 0, and this would be in contradiction with the fact that ⟨Lg −λ1g, φ1⟩= ⟨g, L∗φ1 −λ1φ1⟩= 0. We thus must have Lg −λ1g = 0 and we conclude thanks to Lemma 3.3.16. The same proof applies to the dual problem. Lemma 3.3.18. Under assumptions (C1), (H1′) and (H4), λ1 is algebraically simple. Proof of Lemma 3.3.18. We use an induction argument. We have already proved that N((L −λ1)k) = Span(f1) for k = 1. Assume then the result proved for any ℓ, 1 ≤ℓ≤k, and consider f ∈N((L −λ1)k+1). That means that (L −λ1)f ∈N((L −λ1)k), and thus (L −λ1)f = rf1, with r ∈R, thanks to the induction hypothesis. If r = 0, then f ∈N(L −λ1) = Span(f1). Otherwise, r ̸= 0, and then λ1⟨f, φ1⟩= ⟨f, L∗φ1⟩= ⟨Lf, φ1⟩= ⟨λ1f + rf1, φ1⟩, which in turn implies r⟨f1, φ1⟩= 0 and a contradiction. That concludes the proof. 3.4 Mean ergodicity We deduce from the above analysis a ﬁrst classical and general but rough information about the long-time behaviour of the trajectories associated to a semigroup. More precisely, assuming the existence and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1) for the generator L of a semigroup S and introducing the rescaled semigroup e St := e−λ1t S(t), we wish to establish the following mean ergodic property (CE1) for any f ∈X, there holds 1 T Z T 0 e Stfdt →⟨f, φ1⟩f1, as T →∞, (3.4.18) in a sense to be speciﬁed. Theorem 3.4.19. Consider a positive semigroup S on a Banach lattice X and assume that its generator L satisﬁes the conclusions of Theorem 3.3.11 about the existence and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1). With the above notations, we assume furthermore that (1) ( e St) is bounded; (2) BX is weakly compact for a topology which makes f 7→⟨f, φ1⟩continuous. Then, the above mean ergodic property (CE1) holds for the topology introduced in (2). The result is not new, see for instance [82, Corollary V.4.6]. We present however the very short proof for completeness and further discussion. Proof of Theorem 3.4.19. Fix f ∈X and deﬁne uT := 1 T Z T 0 e Stf dt. From (1), we have ∥uT ∥≤1 T Z T 0 ∥e Stf∥dt ≤M ∥f∥, ∀T > 0. 60CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE We also compute ⟨uT , φ1⟩= 1 T Z T 0 ⟨e Stf, φ1⟩dt = ⟨f, φ1⟩, ∀T > 0. Thanks to assumption (2), we deduce that there exists f∗∈X and a sequence (Tk) such that uTk →f∗ and ⟨f∗, φ1⟩= ⟨f, φ1⟩. Because ( e Stf) is bounded, we may use the usual ergodicity trick as in the second proof of Theorem 2.2.1 and for any t > 0, we have e Stf∗−f∗= lim k→∞ 1 Tk nZ Tk+t Tk e Ssfds − Z t 0 e Ssfds o = 0. We have established (L −λ1)f∗= 0, so that f∗∈Span(f1) and more precisely f∗= ⟨f, φ1⟩f1. By uniqueness of the limit, it is the whole family (uT ) which converges to f∗. We present a variant of the previous result in which we see that in a very general framework (including all the applications we present in the second part of this work) the above hypothesizes (1) and (2) are not needed (or more precisely are automatically satisﬁed). Theorem 3.4.20. (1) Consider a Banach lattice X ⊂L1 loc(E, E , µ) and a positive semigroup S on X such that its generator L satisﬁes the conclusions of Theorem 3.3.11 about the existence, positivity and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1), then the mean ergodic convergence (CE1) holds for the weak topology of L1 φ1. (2) If we additionally assume that X k := {f ∈X; Ljf ∈X, ∀j ≤k} ⊂L1 loc, (3.4.19) with strong compact embedding for some k ≥1, where we denote by X the space X endowed with the norm L1 φ1, then the mean ergodic convergence (CE1) holds for the strong topology of L1 φ1 for any f belonging to the closure of D(Lk) (for the strong topology of X). Proof of Theorem 3.4.20. Step 1. We ﬁrst recall a very classical result about conservative semigroups. Denoting e St := e−λ1t S(t), we observe that this rescaled semigroup satisﬁes (i) e St ≥0; (ii) e Stf1 = f1 for any t ≥0; (iii) ⟨e Stg, φ1⟩= ⟨g, φ1⟩for any g ∈X and t ≥0. We denote [f]1 := ⟨\|f\|, φ1⟩which is a norm on X (we use here that φ1 ≫0) and e St is obviously a contraction for this one since for any f ∈X there holds \| e Stf\|= \| e Stf+ −e Stf−\|≤e Stf+ + e Stf−= e St\|f\| using (i) in the inequality, and next [ e Stf]1 = ⟨\| e Stf\|, φ1⟩≤⟨e St\|f\|, φ1⟩= [f]1, (3.4.20) using (iii) in the last equality. Denoting by X the completion of X for this norm (so that we main identify X to a closed subspace of L1 φ1), we may extend e St to X by uniform continuity and this extension still satisﬁes the properties (i)-(ii)-(iii) on X. Consider now f ∈X such that H(f/f1)f1 ∈X for some convex function H : R →R, where we use here that X ⊂L1 loc, and thus in particular f1 > 0 a.e. on E, in order to give a sense to the term H(f/f1)f1. From (ii), we have ℓ[( e Stf)/f1]f1 = e St[ℓ(f/f1)f1], 3.4. MEAN ERGODICITY 61 for any real aﬃne function ℓ. Next from (i) and (2.1.7), we have H[( e Stf)/f1]f1 ≤e St[H(f/f1)f1], because of H = supℓ≤H ℓand the supremum can be taken on a numerable set of aﬃne functions. Thanks to (iii), we conclude that ⟨H[( e Stf)/f1]f1, φ1⟩≤⟨H[f/f1]f1, φ1⟩, ∀t ≥0. (3.4.21) Step 2. We normalize ⟨f1, φ1⟩= 1. For f ∈X ⊂L1 φ1 so that fφ1 = (f/f1)f1φ1 ∈L1, the de la Vallée Poussin theorem tells us that there exists an even and convex function H : R →R+ such that H(s)/s →+∞as s →∞and H(f/f1)f1φ1 ∈L1. Using the notations of the proof of Theorem 3.4.19, the Jensen inequality and the above estimate (3.4.21), we deduce Z E H(uT /f1)f1φ1dµ ≤1 T Z T 0 Z E H( e Stf)/f1)f1φ1dµdt ≤ Z E H(f/f1)f1φ1dµ, for any T > 0. Now, for any A ∈E and T, K > 0, we have Z E uT 1Aφ1dµ = Z E uT f1 1 \|uT \| f1 >K1Af1φ1dµ + Z E uT f1 1 \|uT \| f1 ≤K1Af1φ1dµ ≤ K H(K) Z E H(uT /f1)f1φ1dµ + K Z E 1Af1φ1dµ ≤ K H(K) Z E H(f/f1)f1φ1dµ + K Z E 1Af1φ1dµ, from what we immediately deduce that (uT ) belongs to a weak compact set of L1 φ1. We conclude that (3.4.18) holds for the weak convergence in L1 φ1 as in the proof of Theorem 3.4.19. Step 3. We now additionally assume that (3.4.19) holds with strong compact embedding for some k ≥1. Taking f ∈D(Lk), we compute ⟨\|Lj( e Stf)\|, φ1⟩= ⟨\| e St(Ljf)\|, φ1⟩≤⟨\|Ljf\|, φ1⟩, for any j ≤k and any t ≥0, and thus the same bound holds for (uT ). From (3.4.19), we deduce that up to the extraction of a subsequence, (uT ) converges a.e. on E. Together with the weak convergence in L1 φ1 yet established, we classically deduce that the whole familly (uT ) converges for the strong toplogy in L1 φ1. We conclude that the same holds for any f ∈X by taking advantage of the fact that D(Lk) is dense in X for the strong topology of X and thus for the strong topology of X. Remark 3.4.21. (1) A similar conclusion holds as in Theorem 3.4.20 when we assume X ⊂M1 loc, D(Lk) ⊂L1 loc and D(L∗k) ⊂L1 loc for some k ≥1 instead of X ⊂L1 loc. For f ∈D(Lk) ⊂L1 loc, we may indeed repeat the proof of Theorem 3.4.20 and we obtain the same conclusion. We next deﬁne X as the closure of D(Lk) for the norm [·]1. We conclude that (3.4.18) holds weakly in L1 φ1 for any f ∈X by a density argument. (2) The proof of Theorem 3.4.20 is based on so-called General Relative Entropy (GRE) techniques as developped for instance in , and . These ones are known to be usefull for some classes of PDEs and for stochastic semigroups in order to establish uniform in time estimates and longtime convergence results. The main interest of the two previous results is that they do not ask any new information on the semigroup but they are just based on the eigentriplet stationary problem. The shortcoming is that they are formulated in terms of the norm [·]1 instead of the norm of X. We present a second variant of Theorem 3.4.19 which is well adapted to the splitting framework developed in Sections 2.1 and 2.2 and is precisely formulated in a weak or strong topology of a space X0 ⊃X. 62CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE Theorem 3.4.22. Consider a positive semigroup S = SL such that L satisﬁes the conclusions of Theorem 3.3.11 about the existence and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1). Assume furthermore that S satisﬁes the splitting structure introduced in (HS2) in section 2.2.2 or (HS3) in Section 2.2.2, or more precisely, there exist two families of operators (V (t)) and (W(t)) such that S = V + W ∗S, a real number κ ≤λ1 and some decaying functions Θi : R+ →R+ with Θ1(t) →0 as t →∞, Θ2 ∈L1(R+) such that the following estimates hold ∥V (t)e−κt∥B(X)= O(1), ∥V (t)e−κt∥B(X,X0)= O(Θ1), (3.4.22) ∥W(t)e−κt∥B(X0,X1)= O(Θ2), (3.4.23) with X1 ⊂X0 ⊂X0, where X0 is the space X endowed with the norm [g]1 := ⟨\|g\|, φ1⟩. (1) Assuming furthermore that X1 ⊂X0 with compact embedding for the weak or the strong topology in X0 and this topology makes f 7→⟨f, φ1⟩continuous. Then the conclusion of Theorem 3.4.19 holds true for the above strong or weak topology. (2) Assuming furthermore that X ⊂L1 loc and that the space X k deﬁned by (3.4.19) is strongly compact embedded in L1 loc for some k ≥1. Then the conclusion of Theorem 3.4.20-(2) holds true for the strong topology of X0. Proof of Theorem 3.4.22. We deﬁne e V (t) := V (t)e−λ1t, f W(t) := W(t)e−λ1t, so that e S = e V + f W ∗e S, and M := sup t≥0 ∥e V (t)∥B(X)< ∞, ∥e V ∥B(X,X0)≲e Θ1 ∈C0(R+), (3.4.24) e Θ2(t) := ∥f W(t)∥B(X0,X1)∈L1(R+). Step 1. We furthermore assume (1) and that the weak topology of X0 makes f 7→⟨f, φ1⟩ continuous. We denote by T the weak or the strong toplogy X0 (depending of the assumption made on the embedding X1 ⊂X0). For f0 ∈X, we split f(t) := e Stf0 = v(t) + k(t), v(t) := e V (t)f0, k(t) := (f W ∗e S)(t)f0, and we observe that ∥v(t)∥X0→0 as t →∞from the second estimate in (3.4.24). On the other hand, we have sup t≥0 ∥k(t)∥X1≤∥f W∥L1sup t≥0 ∥e Stf0∥X0≤∥f W∥L1∥f0∥X0, from (3.4.20). In particular, k(t) belongs to a compact set of T , so that (f(t))t≥0 also belongs to a compact set for the same topology T . The same argument used on the Cesàro function (uT ) deﬁned during the proof of Theorem 3.4.19 implies that there exist f∗∈X and a sequence (Tk) such that uTk →f∗in the sense of T and ⟨f∗, φ1⟩= ⟨f, φ1⟩, the last identity following from the assumption that f 7→⟨f, φ1⟩continuous for T . We may then conclude as in the proof of Theorem 3.4.19. 3.4. MEAN ERGODICITY 63 Step 2. We furthermore assume (2), and by linearity we may assume f0 ∈X, ⟨f0, φ1⟩= 0. We recall that (3.4.18) holds for the strong toplogy of L1 φ1 from Therorem 3.4.20 and that ∥v(t)∥X0→0 as t →∞from Step 1. Arguing as in Step 3 of the proof of Theorem 2.2.4, we have K(T) := 1 T Z T 0 (W ∗e S)(t) dt = 1 T Z T 0 W(s) Z T−s 0 e S(u) duds = Z T 0 W(s)T −s T e U(T −s) ds, where e UT := UT e−λ1T , Ut is deﬁned by (2.2.44), so that uT = e UT f0 and [uT ]1 →0 as T →∞ from Theorem 3.4.20. As a consequence, we have ∥K(T)f0∥X1 ≤ Z T/2 0 Θ2(s)[ e U(T −s)]0 ds + Z T T/2 Θ(s)[ e U(T −s)]0 ds ≤ ∥e Θ2∥L1 sup t≥T/2 [ e U(t)]0 + Z ∞ T/2 e Θ2(s)ds sup t≥0 [ e U(t)]0 →0, as T →∞. All together, we have established that ∥uT ∥X0→0 as T →∞. 64CHAPTER 3. IRREDUCIBILITY AND THE GEOMETRY OF THE FIRST EIGENVALUE Chapter 4 The geometry of the boundary point spectrum We summarize the results established up to now by assuming that the main conclusions in the previous sections are achieved, namely (C2) the ﬁrst eigentriplet problem (3.0.1) has a unique solution (λ1, f1, φ1), and fur-thermore, f1 ≫0 and φ1 ≫0. In that situation, we usually normalize the eigenvectors by ⟨f1, φ1⟩= 1 and either ∥f1∥= 1 or ∥φ1∥= 1. In this chapter, we aim to describe one step further the geometry of the spectrum and more precisely to get some some information on the boundary point spectrum Σ+ P (L) := ΣP (L) ∩∆λ1. That will be possible by introducing ﬁrst a suitable and usual complexiﬁcation framework and next by assuming a stronger positivity property on L or on the associated semigroup. Here and for further references below, we recall that we deﬁne the sets Σd(L) ⊂ΣP (L) ⊂Σ(L), where the point spectrum set ΣP (L) is the set of eigenvalues, namely λ ∈ΣP (L) if N(L−λ) ̸= {0}, and the discret spectrum set Σd(L) is the set of eigenvalues which are isolate and have ﬁnite algebraic multiplicity. 4.1 Complexiﬁcation and the sign operator We present some materials, most of them being very classical, about the sign operator in a complex Banach lattice and we refer to [222, 7] for more details. Complexiﬁcation. The complexiﬁcation space XC associated to a real Banach lattice X is deﬁned by XC := X + iX so that f ∈XC if f = g + ih, g, h ∈X. In general, we just write X without mentioning the ﬁeld, although when we need to specify it we write XC or XR. We extend on XC the order deﬁned on XR by writing f = g + ih ≥0 if g ≥0 and h = 0. The conjugate ¯ f of a complex vector f = g + ih is classically deﬁned by ¯ f = g −ih. We then deﬁne the modulus \|f\|:= sup θ∈[0,2π] (g cos θ + h sin θ), (4.1.1) 65 66 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM which indeed exists for such a family of vectors. One checks the usual modulus properties: \|f\|≥0, \|f\|= 0 iﬀf = 0, \|λf\|= \|λ\| \|f\|, \|f + g\|≤\|f\|+\|g\|, for any f, g ∈X and λ ∈C. We ﬁnally deﬁne the norm on XC by ∥f∥:= ∥\|g + ih\|∥XR, and we observe that XC has then a complex Banach lattice structure. We extend the deﬁnition of A ∈B(XR) to XC by setting A(g + ih) = Ag + iAh, ∀g + ih ∈XC. The operator sign. We may classically extend the sign operator deﬁned in Section 3.1 to the present complex Banach lattice framework. Instead of dealing with the most general case, we will use some regularity assumption on the Banach lattice X which is suitable for our purpose and that we present below. Similarly as in Remark 3.1.4, for f ∈X, we deﬁne Xf := [ n {g ∈X; \|g\|≤n\|f\|}, and next, similarly as in Theorem 2.1.23, we deﬁne Af[g] := inf{C > 0; \|g\|≤C\|f\|}, ∀g ∈Xf. We summarize the regularity assumption on the Banach lattice X we need by assuming : (X2) For any f ∈X such that \|f\|∈X++, there exists a sign operator sign f ∈B(X), with the following properties sign f ◦sign ¯ f = I, (sign f)f = \|f\|, (4.1.2) (sign f) g = (sign (uf)) (ug), \|(sign f) g\|≤\|g\|, ∀g ∈X, ∀u ∈S1. (4.1.3) Furthermore, X = Xf, the closure of Xf, and for all f ∈X and g ∈Xf ( g ∈XR and \|g\|≤C\|f\| ) ⇔Af[g −ir\|f\|] ≤ p C2 + r2, ∀r ∈R. (4.1.4) For a space of functions, the sign operator is deﬁned as the multiplication by (abusing notations) sign f := ¯ f/\|f\|, ∀f ∈X, \|f\|∈X++. (4.1.5) Lemma 4.1.1. With (4.1.5), the property (X2) holds when X = Lp(E, E , µ) or X = C0(E). Proof of Lemma 4.1.1. For f ∈X, \|f\|∈X++, we just indicate the proof of Xf = X, the other algebraic properties being clear from the deﬁnition (4.1.5). When f ∈Lp such that \|f\|> 0 µ-a.e. and 0 ≤g ∈Lp, we set gn := g ∧(n\|f\|). We have 0 ≤gn ≤g and gn →g strongly Lp if p < ∞ and weakly ∗L∞if p = ∞. The general case g ∈Lp is dealt in the usual way by introducing positive and negative parts and next real and imaginary part. That concludes the proof of Xf = Lp. The proof of Xf = C0(E) is similar. A sign operator satisfying (X2) can actually be built by using Kakutani’s theorem in general Banach lattices whenever \|f\| is a quasi-interior point, see for instance [20, Chapter 14.3]. In a space of measures X = M1(E), there is no quasi-interior point and no sign operator can be deﬁned in that way. Yet, we can associate to f such that \|f\|≫0 a sign operator by means of the Radon-Nikodym theorem. Denoting α : E →S1 the measurable function such that f = α\|f\|, the multiplication by ¯ α/\|α\| deﬁnes a sign operator sign f ∈B(X) which satisﬁes the ﬁrst properties in (X2). However, it does not satisfy X = Xf, since there is no quasi-interior point in M1(E). Measure spaces thus not enjoy (X2). 4.2. ON THE POINT BOUNDARY SPECTRUM 67 Lemma 4.1.2. Assume (X2) and f ∈X++. Consider a linear operator Q : Xf →Xf such that Qf = f and Af(Qg) ≤Af(g) for any g ∈Xf. Then Q ≥0. Proof of Lemma 4.1.2. Take 0 ≤g ∈Xf such that g ≤2Cf, C > 0, and observe that −Cf ≤g −Cf ≤Cf. For any r ∈R, we compute Af[(Qg) −Cf −irf] = Af[Q(g −Cf −irf)] ≤ Af[g −Cf −irf] ≤ p C2 + r2, by using the non expansion property of Q and the claim (4.1.4). Using again (4.1.4), we deduce −Cf ≤(Qg) −Cf ≤Cf and the conclusion. We generalize Kato’s inequality (3.1.3) to the present complex framework by saying that an operator L on X satisﬁes (the complex) Kato’s inequality if ∀f ∈D(L), ℜe(signf)Lf ≤L\|f\|, (4.1.6) possibly in a dual sense as in (3.1.4). As for the real Kato’s inequality, when L is the generator of a semigroup, Kato’s inequality (4.1.6) is a consequence of the positivity of the semigroup, and we refer to Remark 3.1.1 for references about this claim. 4.2 On the subgroup and discrete structure of the point bound-ary spectrum In this section, we establish that the point boundary spectrum enjoys a subgroup structure under the same kind of hypothesis as considered in the previous sections. Lemma 4.2.3. Under assumptions (X2), (C1), (H1′) and (H4), for any λ ∈Σ+ P (L){0} the associated normalized eigenfunction f satisﬁes \|f\|= f1. Proof of Lemma 4.2.3. By deﬁnition Lf = λf and f ∈D(L). By linearity of the operator sign and thanks to (4.1.2) and Kato’s inequality (4.1.6), we have λ1\|f\|= ℜe[λ(signf)f] = ℜe(signf)(λf) = ℜe(signf)Lf ≤L\|f\|. By the duality argument introduced during the proof of Lemma 3.3.16, we must have λ1\|f\|= L\|f\| and the conclusion. Theorem 4.2.4. Assume (X2) and (C2), and denote e L = L −λ1. Under assumptions (H1′) and (H4), the set S := ΣP ( e L) ∩iR is an additive subgroup and dimN( e L −iα)k = 1 for any iα ∈S and k ≥1. Theorem 4.2.4 is similar to [7, C–III, Cor. 2.12] and [20, Prop. 14.15]. Our proof is very similar to the proof of [20, Proposition 14.15]. However, it is more direct and avoid the use of the C(K) algebra and Kakutani Theorem (see also [165, Theorem 2.1.3]). Proof of Theorem 4.2.4. We split the proof into three steps. Step 1. We consider f associated to an eigenvalue iα ∈ΣP ( e L){0}, and we deﬁne T(t) := (sign f)e−iαt e S(t)(sign ¯ f). 68 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Observing that e S(t)f = eiαtf, we have T(t)\|f\|= (sign f)e−iαt e S(t)f = (sign f)f = \|f\|= e S(t)\|f\|. On the other hand we have \|T(t)g\|≤\| e S(t)(sign ¯ f)g\|≤e S(t)\|g\|, ∀g ∈X, which, by positivity of e S(t), yields \|T(t)g\|≤Af(g) e S(t)\|f\|= Af(g)\|f\|, ∀g ∈Xf. Because \|f\|= f1 ≫0 from Lemma 4.2.3 and (C2), we can apply Lemma 4.1.2 to \|f\| and Q := T(t). We deduce that T(t) ≥0 on X\|f\| = Xf, and then on X = Xf. As a consequence, 0 ≤T(t)g = \|T(t)g\|≤e S(t)g for any g ≥0. In other words, we have 0 ≤e S(t) −T(t) and then 0 ≤e S(t)∗−T(t)∗. We must have e S(t)∗−T(t)∗= 0. Otherwise, there would exist ψ ∈X′ +{0} such that ( e S(t)∗−T(t)∗)ψ ∈X′ +{0}, and we ﬁnd a contradiction by computing 0 < ⟨( e S(t)∗−T(t)∗)ψ, f1⟩= ⟨ψ, ( e S(t) −T(t))f1⟩= 0. We have thus established that e S(t) = T(t). Step 2. Consider α, β ∈R and f, g ∈X{0} such that e Lf = iαf and e Lg = iβg, and suppose ﬁrst that (sign ¯ f) : D(L) →D(L). From Step 1 and the fact that (sign ¯ f) ◦sign f = I, for any h ∈D(L), we may compute e Lh = lim t→0 1 t ( e S(t)h −h) = (sign f) lim t→0 1 t (e−iαt e S(t)(sign ¯ f)h −(sign ¯ f)h) = (sign f)( e L −iα)(sign ¯ f)h, or in other words e L −iα = (sign ¯ f) e L(sign f). We have similarly e L −iβ = (sign ¯ g) e L(sign g). Both equations together imply e L −i(α + β) = (sign ¯ f)(sign ¯ g) e L(sign g)(sign f). Deﬁning h := (sign ¯ f)(sign ¯ g)f1, so that (sign g)(sign f)h = f1, we get e Lh = i(α + β)h, and ﬁnally i(α + β) ∈S, so that the additive subgroup structure is established. When the condition (sign ¯ f) : D(L) →D(L) is not granted, we modify the above argument by using a resolvent approach. For some λ > 0, we compute (λ −e L)−1 = Z ∞ 0 e−λt e S(t) dt = (sign f) Z ∞ 0 e−(λ+iα)t e S(t) dt (sign ¯ f) = (sign f)(λ + iα −e L)−1(sign ¯ f). Repeating the argument, we obtain (λ + i(α + β) −e L)−1 = (sign ¯ f)(sign ¯ g)(λ −e L)−1(sign g)(sign f). Applying that last identity to the vector h = (sign ¯ f)(sign ¯ g)f1 and using that (λ −e L)−1f1 = λ−1f1, we deduce (λ + i(α + β) −e L)−1h = λ−1h. In other words, we have again e Lh = i(α + β)h, and we conclude as above. 4.2. ON THE POINT BOUNDARY SPECTRUM 69 Step 3. From the fact that (sign f) is an invertible operator and the equation ( e L −iα)k = (sign f)−1( e L)k(sign f), we see from Theorem 3.3.11-(ii) that N( e L −iα)k = (sign f)−1N( e L)k = (sign f)−1Spanf1 for any k ≥1, so that its dimension is one. Making an additional splitting structure hypothesis as yet introduced in Section 2.1.2, we may signiﬁcantly improve the conclusion. We ﬁrst recall a classical result on the spectrum of an operator which holds under some (power) compactness assumption on the resolvent. Theorem 4.2.5. We assume that L satisﬁes the splitting structure (HS1) with (ARB(z))N ∈ K (X) for some N ≥1 and any z ∈∆κ0, where we recall that K (X) denote the space of compact operators. Then Σ(L) ∩∆κ0 ⊂Σd(L). Theorem 4.2.5 is a consequence of [242, Corrollary 1.1]. For the sake of completeness, we present the idea of the proof. We also refer to [177, proof of Theorem 3.1] for a possible elementary proof. A sketch of the proof of Theorem 4.2.5. Iterating the formula RL = RB + RBARL, we deduce K(z)RL(z) = V(z) with K := I −(ARB)N, V := RB + · · · + RB(ARB)N−1. Because K is holomorphic on ∆κ0, it is a compact perturbation of the identity and K(z) →I when ℜez →∞, one may use the theory of degenerate-meromorphic functions of Ribarič and Vidav , and conclude that K(z) is invertible outside of a discrete set D of ∆κ0. That implies that Σ(L) ∩∆κ0 = D is a discrete set of ∆κ0. On the other hand, thanks to the Fredholm alternative , one deduces that the eigenspace associated to each spectral value λ ∈D is non zero and ﬁnite dimensional, so that λ ∈Σd(L). We end this section by a result which gives a more accurate description of the geometry of the boundary spectrum, and is a variant of the classical results [7, C–III, Thm. 3.12], [82, Thm. VI.1.12], [20, Thm. 14.17]. Theorem 4.2.6. Assume (X2) for both the spaces X and Y , (C2), (H1′), (H4) and additionally that the splitting structure (HS1) holds with (ARB(z))N ∈K (X) for some N ≥1 and any z ∈∆κ0. Then the set ΣP ( e L) ∩iR is a discrete additive subgroup of iR and any of its elements is an algebraically simple eigenvalue. More precisely, - either ΣP ( e L) ∩iR = {0} and the projection on the ﬁrst eigenspace (associated to λ1) writes Π f := ⟨f, φ1⟩f1; - or ΣP ( e L) ∩iR = iαZ for some α > 0 and there exists a sequence (gk, ψk)k∈Z such that Lgk = (λ1 + ikα)gk, L∗ψk = (λ1 + ikα)ψk, and ⟨gk, ψℓ⟩= δkℓ. Proof of Theorem 4.2.6. Combining Theorem 4.2.4 and Theorem 4.2.5, we immediately get that the subgroup S := ΣP ( e L) ∩iR satisﬁes S ⊂Σd(L), it is thus discrete and made of algebraically simple eigenvalues. The ﬁrst case ΣP ( e L) ∩iR = {0} falls yet in the conclusions of Theorem 3.3.11. In the second case, where the boundary spectrum is not trivial, the existence of a projection on the boundary eigenspace Span(gk)k∈Z is ensured by the Jacobs–de Leeuw–Glicksberg theorem provided that L is the generator of a relatively compact semigroup, see for instance [20, 70 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Theorem A.39 and Proposition A.40] and the references therein. We also refer to [136, paragraphs III.6.4 and III.6.5] for very classical results on the projector on the direct sum of eigenspaces associated to eigenvalues belonging to a subset of the spectrum. We can even give an explicit expression of this projection in terms of (gk) and (ψk) under the form of a Fejér type sum, see Theorem 4.5.19. 4.3 Stronger positivity In order to go one step further and establish the triviality of the boundary point spectrum, we need to reinforce the positivity of the semigroup or its generator. One possible condition will be the following. The reverse strong positivity condition For A ≥0, we recall that from (2.1.6), we have \|Af\|≤A\|f\|, ∀f ∈X, (4.3.7) and we observe that the above inequality is an equality when Af = uA\|f\| for some u ∈S1. We focus now on the case of equality in (4.3.7). Deﬁnition 4.3.7. We say that A satisﬁes the “reverse strong positivity condition" for a subclass of vectors C ⊂X if for any f ∈C \|Af\|= A\|f\| implies ∃u ∈S1, Af = uA\|f\|. (4.3.8) We observe that A ≫0 implies the strict positivity for non-signed vectors in XR. Lemma 4.3.8. Consider an operator A ≫0 and assume X is reﬂexive. For f ∈XR such that ±f / ∈X+, there holds \|Af\|≪A\|f\|. Proof of Lemma 4.3.8. We consider f ∈XR such that f± ̸= 0. We claim that \|Af\|≪A\|f\|. We ﬁrst recall that Af+ = Af + Af−≥Af, so that Af+ ≥(Af)+, and similarly Af−≥(Af)−. Consider the case (Af)+ > 0. We argue similarly as in the proof of Lemma 3.2.5. By deﬁnition, for any φ ≫0, we have 0 < ⟨(Af)+, φ⟩ = sup 0≤ψ≤φ ⟨Af, ψ⟩= ⟨Af, ψ∗⟩= ⟨f, A∗ψ∗⟩ < ⟨f+, A∗ψ∗⟩= ⟨Af+, ψ∗⟩≤⟨Af+, φ⟩, where the strict inequality comes from the fact that ψ∗> 0 because ⟨(Af)+, φ⟩> 0, so that A∗ψ∗≫0 and ﬁnally ⟨f−, A∗ψ∗⟩> 0 because f−> 0. We have thus (Af)+ < Af+. Similarly, we establish (Af)−< Af−when (Af)−> 0. As a conclusion, in the three cases Af = 0, (Af)+ ̸= 0 and (Af)−̸= 0, we have \|Af\|= (Af)+ + (Af)−≪Af+ + Af−= A\|f\|, which is the desired strict inequality. We believe that a similar result also holds true for complex vectors in a general Banach lattice framework. We do not try to prove such a statement but we rather establish the corresponding complex version for at least one of our standard examples of concrete Banach spaces in which the deﬁnition of the absolute value \|f\| of a vector f ∈X is more tractable. We start with the example of continuous functions. 4.3. STRONGER POSITIVITY 71 Lemma 4.3.9. Consider an operator A ≫0 on X ⊂Cb(E) endowed with the uniform norm, for some locally and σ-compact metric space E. For f ∈X such that \|f\|≫0, we have \|Af\|= A\|f\| implies ∃u ∈S1, Af = uA\|f\|. Proof of Lemma 4.3.9. We assume by contradiction that f = g + ih ∈X, A\|f\|≡\|Af\| on E and ∀v ∈S1, \|f\|> ℜe(fv) (in the sense ≥0 and ̸≡0). Pick up x ∈E and denote by v = eiθ the element of S1 such that \|(Af)(x)\|= v(Af)(x). We thus have A\|f\|(x) = \|(Af)(x)\|= ℜe(v(Af)(x)) = cos θ(Ag)(x) −sin θ(Ah)(x). On the other hand, because of A ≫0 and A is linear, we have A\|f\|(x) > (Aℜe(vf))(x) = (A(cos θg −sin θh))(x) = cos θ(Ag)(x) −sin θ(Ah)(x), and a contradiction. We have established that there exists v ∈S1 such that \|f\|≡ℜe(fv). Now, we have q (ℜe(fv))2 + (ℑm(fv))2 = \|fv\|= \|f\|= ℜe(fv), which in turn implies ℑm(fv) = 0, since ℜe(fv) ≫0. That is here that we use the assumption \|f\|≫0 and not only f ∈X+{0}. We conclude that \|f\|= fv and thus that f = u\|f\|, with u := v−1 ∈S1. The reverse Kato’s inequality condition We recall that it has been stated in section 3.1 that the generator L of a positive semigroup S(t) satisﬁes Kato’s inequality which in a complex framework writes ∀f ∈X, ℜe(signf)Lf ≤L\|f\|. (4.3.9) We observe that if f = u\|f\| for some u ∈S1, we have ℜe(sign f)Lf = sign(u−1f)L(u−1f) = L\|f\|, which is the case of equality in Kato’s inequality. Deﬁnition 4.3.10. We say that L satisﬁes a “reverse Kato’s inequality condition” for a class of vectors C ⊂D(L) if for any f ∈C the case of equality in Kato’s inequality L\|f\|= ℜe(signf)Lf implies ∃u ∈C, f = u\|f\|. In some situation, we may prove directly that the “reverse Kato’s inequality condition” holds by reasoning at the level of the operator L, see for instance [138, Proof of Theorem 5.1]. That is also a consequence of the strong positivity of the semigroup as we see below. Lemma 4.3.11. Consider a semigroup S and its generator L. On the set C of vectors f ∈X{0} such that ∃λ ∈C, Lf = λf, L\|f\|= (ℜeλ)\|f\|, (4.3.10) there is equivalence between: (i) S(t) satisﬁes the “reverse strong positivity condition" for some (and thus any) t > 0; (ii) L satisﬁes the “reverse Kato’s inequality condition”. 72 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Remark 4.3.12. When X ⊂L1 loc, the “reverse Kato’s inequality condition” (ii) implies the “reverse strong positivity condition" (i) on the class C of vectors such that f ∈D(L), 0 ≪\|f\|∈ D(L). Assume indeed that L satisﬁes (ii) and consider f ∈C such that St\|f\|= \|Stf\| for any t ≥0. By diﬀerenciating, we get (sign f)Lf = L\|f\|. (4.3.11) From the “reverse Kato’s inequality condition”, we deduce that f = u\|f\| for some u ∈S1, so that (i) holds. Proof of Lemma 4.3.11. In what follows, we ﬁx f ∈X{0} such that (4.3.10) holds, and we compute ℜe(sign f)Lf = ℜe(sign f)(λf) = (ℜeλ)\|f\|= L\|f\|. (4.3.12) For any t > 0, we also have Stf = eλtf, St\|f\|= eℜeλt\|f\|, and thus \|Stf\|= St\|f\|. (4.3.13) Assuming the “reverse Kato’s inequality condition”, we deduce from (4.3.12) that f = u\|f\| for some u ∈S1, thus Stf = uSt\|f\| for some u ∈S1, which is the conclusion of the “reverse strong positivity condition" when (4.3.13) holds. On the other way round, assuming the “reverse strong positivity condition" for some T > 0, we deduce from (4.3.13) for T > 0 that there exists v ∈S1 such that eλT f = ST f = vST \|f\|= veℜeλT \|f\|. That implies that f = u\|f\| with u = ve−i(ℑmλ)T , which is nothing but the conclusion of the “reverse Kato’s inequality condition” when (4.3.12) holds. We summarize the material developed above in the following main result of the section. Theorem 4.3.13. Assume that S is a positive semigroup on X with X = Lp(E, E , µ), X = M1(E, E ) or X = C0(E), for some locally and σ-compact metric space E and denote by (Ek) a sequence of increasing compact sets such that E = lim Ek. We furthermore assume that for any k ≥1 there exists T > 0 such that ST is strictly positive on Ek, in the sense that ∀f ∈X+{0}, f\|Ek ̸≡0, ∀φ ∈X′ +{0}, supp φ ⊂Ek, ⟨ST f, φ⟩> 0. (4.3.14) Then L satisﬁes the “reverse Kato’s inequality condition” on the set C of eigenvectors introduced in Lemma 4.3.11. Proof of Theorem 4.3.13. Let us consider f ∈X{0} such that (4.3.10) holds, so that St\|f\|= e(ℜeλ)t\|f\| for any t ≥0. On the one hand, we may ﬁx k ≥1 such that \|f\|̸≡0 on Ek. Then for any ℓ≥k, there exists Tℓ> 0 such that (4.3.14) holds, so that e(ℜeλ)Tℓ⟨\|f\|, φ⟩= ⟨STℓ\|f\|, φ⟩> 0, for any φ ∈X′ +{0}, supp φ ⊂Eℓ. That implies \|f\|> 0 on Eℓfor any ℓ≥k, and thus \|f\|≫0. Next, as in the proof of Lemma 4.3.11, we observe that \|STℓf\|= STℓ\|f\|, ∀ℓ≥k. Repeating the proof of Lemma 4.3.9, Lemma ?? and Lemma ??, we deduce that there exists uℓ∈S1 such that STℓf = uℓSTℓ\|f\| on Eℓ, or equivalently there exists vℓ∈S1 such that f = vℓ\|f\| on Eℓ, with vℓ:= uℓe−i(ℑmλ)Tℓ. Because Eℓ⊃E1, we have established that f = v1\|f\| on E which is the conclusion of the “reverse Kato’s inequality condition” when (4.3.10) holds. 4.4. ON THE TRIVIALITY OF THE BOUNDARY SPECTRUM 73 4.4 On the triviality of the boundary spectrum As in section 3.3, we still assume the existence (C1) of a solution (λ1, f1, φ1) ∈R × X+ × Y+ to the ﬁrst eigenvalue problem (3.0.1), that the condition (H1′) holds, so that L enjoys the weak maximum principle (3.3.12) and its complex Kato’s inequalities counterpart (4.3.9) as well as the strong maximum principle (H4). We introduce the additional assumption: (H5) the“reverse Kato’s inequality condition” (as deﬁned in Deﬁnition 4.3.10) holds true for the class C deﬁned in Lemma 4.3.11: for f ∈X{0} such that ∃λ ∈C, Lf = λf, L\|f\|= (ℜeλ)\|f\|= ℜe(signf)Lf, (4.4.15) we have ∃u ∈C, f = u\|f\|. On the other hand, we do not need the structure assumption (X2). We are then able to make a more accurate analyse of the geometry of the spectrum. Theorem 4.4.14. Consider an unbounded operator L on a Banach lattice X which satisﬁes (C1), (H1′), (H4) and (H5). Then λ1 is the unique eigenvalue with largest real part: Σ+ P (L) = {λ1}. Remark 4.4.15. (1) It is worth emphasizing again that (4.3.9) is true when L is the generator of a positive semigroup and that (4.4.15) is true when SL(T) satisﬁes the “reverse strong positivity condition" for some T > 0 as a consequence of Lemma 4.3.11, see also Theorem 4.3.13. (2) During the proof we use similar arguments as in [138, Theorem 5.1]. (3) When instead of (H4) and (H5) we only assume that L is the generator of a positive semigroup, we ﬁnd Σ(L) ∩∆λ1 = ∅as established in Lemma 2.1.2-(ii). Proof of Theorem 4.4.14. Consider an eigenvalue λ ∈C with normalized eigenvector f ∈X{0}, so that ∥f∥= 1, Lf = λf. Thanks to the complex Kato’s inequality, we have (ℜeλ)\|f\|= ℜe sign(f)(λf) = ℜe sign(f)(Lf) ≤L\|f\|. We consider two cases: When the above inequality is not an equality, we have (ℜeλ)⟨\|f\|, φ1⟩< ⟨L\|f\|, φ1⟩= ⟨\|f\|, L∗φ1⟩= λ1⟨\|f\|, φ1⟩, and thus ℜeλ < λ1. When on the contrary the above inequality is an equality, then \|f\| is a positive eigenvector associated to the eigenvalue ℜeλ. Lemma 3.3.15 implies ℜeλ = λ1 and Lemma 3.3.16 together with (2.1.1) imply \|f\|= f1. In other words, f satisﬁes (4.4.15) and thus f ∈Span(f1) from assumption (H5), in particular λ = λ1. An alternative result is the following, where we make the alternative assumption: (H5′) a positive and irreducible semigroup S is aperiodic if it enjoys (3.2.9), namely ∀f ∈X+{0}, ∀φ ∈Y+{0}, ∃T > 0, ∀τ ≥T ⟨Sτf, φ⟩> 0. Theorem 4.4.16. On a Banach lattice X, consider a positive semigroup S which satisﬁes the irreducibility and aperiodicity condition (H5′), and such that its generator L satisﬁes (C2). Then λ1 is the unique eigenvalue with largest real part. 74 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Proof of of Theorem 4.4.16. We introduce the notations e St := Ste−λ1t and e L := L−λ1. Assume that f = g +ih ∈X, g, h ∈XR, is an eigenfunction associated to the eigenvalue λ = λ1 +iα ∈C, α > 0, so that e L(g + ih) = iα(g + ih) = 2πi t0 (g + ih), for some t0 > 0. On the one hand, because α ̸= 0, we must have g ̸= 0 and h ̸= 0, and because of α⟨g, φ1⟩= ⟨e Lh, φ1⟩= ⟨h, e L∗φ1⟩= 0, and φ1 ≫0, we have g+ ̸= 0 and g−̸= 0. As a consequence and because of (3.2.6), there exists ψ ∈X+{0} such that ⟨g+, ψ⟩= 0. On the other hand, we compute e St0(g + ih) = eiαt0(g + ih) = g + ih, from what we deduce e St0g = g, because St is real. On the other hand, because St is positive, we have g+ = ( e St0g)+ ≤e St0g+, and next ⟨φ1, g+⟩≤⟨φ1, e St0g+⟩= ⟨e S∗ t0φ1, g+⟩= ⟨φ1, g+, ⟩, so that the inequality is an equality (remind that φ1 ≫0), and thus e St0g+ = g+. We conclude that ⟨e Skt0g+, ψ⟩= ⟨g+, ψ⟩= 0, ∀k ≥0, what is in contradiction with (H5′). We have established that Σ+(L) = {λ1}. We end this section with a third situation where the triviality of the boundary spectrum can be shown as a immediate consequence of Theorem 4.2.6. Theorem 4.4.17. We assume (X2) and there exists M > 0 large enough such that λ −L is invertible in B(X) for any λ ∈C, ℜeλ = λ1, \|λ\|≥M. Then λ1 is the unique eigenvalue with largest real part. When furthermore the same holds for any λ ∈C, ℜeλ ≥λ1 −ε, \|λ\|≥M and W ∈K(X) then L enjoys a spectral gap. We summarize the main results established in this section as follows. (C3) the ﬁrst eigentriplet problem (3.0.1) has a solution (λ1, f1, φ1), furthermore this one is unique, f1 ≫0, φ1 ≫0, λ1 is algebraically simples (for both L and L∗) and Σ+ P (L) = {λ1}. 4.5 Ergodicity Thanks to the above analyze, we are able to formulate some convergence results on the trajectories associated to a semigroup. More precisely, assuming the existence and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1) for the generator L of a semigroup S and still denoting the rescaled semigroup e St := e−λ1t S(t), we wish in particular to establish the following ergodic property (CE2) for any f ∈X, there holds e Stf →⟨f, φ1⟩f1, as T →∞, (4.5.16) in a sense to be speciﬁed. We start with a simple result which take advantage of some dissipativity property of the semigroup formulated by a "reverse positivity condition". We next present some more involved results which use directly the spectral information. It is worth emphasizing that our results in this section do not use any spectral gap property what contrasts with the results we will present in the next section. 4.5. ERGODICITY 75 Theorem 4.5.18. Consider a positive semigroup S on a Banach lattice X such that its generator L enjoys the conclusions (C2) of existence, uniqueness and strict positivity of the ﬁrst eigentriplet (λ1, f1, φ1) and let us set e St := e−λ1tSt. We denote X0 the space X endowed with the norm [·], with [f] := ⟨\|f\|, φ1⟩. Assume furthermore that (1) for any f ∈X, the trajectory ( e Stf)t≥0 belongs to a compact set of a normed space X, with X ⊂X0; (2) (St) satisﬁes the reverse positivity condition for semigroups \|Stf\|= St\|f\|, ∀t > 0, implies ∃T > 0, ∃uT ∈S1, ST f = uT ST \|f\|. (4.5.17) Then, the ergodicity property (CE2) holds in the sense of the norm of X. Let us comment on hypotheses made in the statement of Theorem 4.5.18. Hypothesis (1) can be obtained as a consequence of a Lyapunov (or growth) condition reminiscent of the structure condition (HS3) introduced in Section 2.2.3 and an irreducibility condition. Typically, we assume ﬁrst ∥e S(t)f∥≤M∥f∥+K[ e S(t)f]0, with [g]0 := ⟨\|g\|, ψ0⟩, ψ0 ∈Y+{0}, what can be established under the very general condition (ii) of Theorem 2.2.4. Next we need to be able to prove that ψ0 ≤rφ1 for some r > 0. In concrete situations, we may take ψ0 with compact support and then the above inequality is a consequence of the standard strong maximum principle. We deduce ∥e S(t)f∥ ≤ M∥f∥+Kr⟨\| e S(t)f\|, φ1⟩ ≤ M∥f∥+Kr⟨e S(t)\|f\|, φ1⟩ = M∥f∥+Kr⟨\|f\|, φ1⟩, so that ( e St) is bounded. The hypothesis (1) is in fact a bit more demanding, but also quite natural. Assume that SL enjoys the splitting structure introduced in section 2.2.1 and section 2.2.3, so that e S = V + K, (4.5.18) with V := e SB + · · · + ( e SBA)∗(N−1) ∗¯ SB, K := ( e SBA)(∗N) ∗e S, e SB(t) := e−λ1tSB(t). In some applications, we typically have ∥V (t)f0∥≤Θ(t)∥f0∥, ∥( e SBA)(∗N)∥B(X0,X1)≤Θ with Θ ∈L1(R+) ∩C0(R+), X1 ⊂X0 compact. In that situation, we deduce (1). Proof of Theorem 4.5.18. We ﬁx f ∈X and without loss of generality, we may assume that ⟨f, φ1⟩= 0. We observe that ⟨\| e Stf\|, φ1⟩= ⟨\| e St−s e Ssf\|, φ1⟩≤⟨e St−s\| e Ssf\|, φ1⟩= ⟨\| e Ssf\|, φ1⟩, (4.5.19) for any t ≥s. We deduce that ( e St) is a dynamical system with compact trajectoires in X and H(f) := ⟨\|f\|, φ1⟩is a Lyapunov functional. As a consequence, from the La Salle invariance principle, we have inf g∈ωH⟨\| e Stf −g\|, φ1⟩→0 as t →∞, (4.5.20) with ωH := {g ∈X; ⟨g, φ1⟩= 0, ∀t ∈R, H( e Stg) = inf s>0 H( e Ssf)}. (4.5.21) 76 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM We characterize ωH. Picking up g ∈ωH, we observe that ⟨\| e Stg\|, φ1⟩= ⟨\|g\|, φ1⟩= ⟨\|g\|, e S∗ t φ1⟩= ⟨e St\|g\|, φ1⟩, ∀t ≥0, so that ⟨e St\|g\|−\| e Stg\|, φ1⟩= 0, ∀t ≥0. In particular, using that \| e Stg\|≤e St\|g\|, we have e St\|g\|= \| e Stg\|, ∀t ≥0. (4.5.22) Because of the reverse positivity condition for semigroups (4.5.17), there exist T > 0 and uT ∈S1 such that e Stg = uT e St\|g\|. As a consequence, by deﬁnition of the set ωH, we have 0 = ⟨g, φ1⟩= ⟨e Stg, φ1⟩= uT ⟨e St\|g\|, φ1⟩= uT ⟨\|g\|, φ1⟩. Because uT ̸= 0, we conclude that g = 0. In other words, we have established that ωH = {0} and together with (4.5.20), we obtain (4.5.16). In the next variant, we formulate the aperiodicity condition in term of inverse Kato’s inequality stated at the level of the generator instead of the above reverse positivity condition stated at the level of the semigroup. We present now a variant of the previous result which holds under the sole assumption (C2). Theorem 4.5.19. Consider a strongly continuous semigroup S on a Banach lattice X such that its generator L enjoys the conclusions (C2) on the existence and uniqueness of the ﬁrst eigentriplet problem (λ1, f1, φ1). We set e St := e−λ1tSt and we assume that the trajectories ( e Stf)t≥0 are relatively compact for all f ∈X. Then we have the following convergence results. 1. If the (point) boundary spectrum is trivial (in the sense of conclusion (C3)), then the ergodicity property (CE2) holds in the sense of the norm of X. 2. If the boundary spectrum is discrete, non-trivial, and made of algebraically simple eigen-values, then for all f ∈X the projection Πf = lim n→∞ 1 n n X ℓ=0 ℓ X k=−ℓ ⟨f, ψk⟩gk is well deﬁned and ∥e Stf −e StΠf∥→0 as t →+∞. The previous result can be compared for instance to [20, Theorems 14.19], although our hypothesises are slightly more general. Our proof is also more direct than in and it additionally provides an explicit expression of the projection on the boundary eigenspace Span(gk)k∈Z. The proof of this result relies on the theory of almost periodic functions. The concept of almost periodicity dates back to H. Bohr and there is a large literature on the subject. We refer for instance to the book of Corduneanu for a comprehensive introduction. There are several equivalent deﬁnitions of almost periodic functions and we will use the following one. A function f ∈Cb(R, X), i.e. a bounded continuous function from R to X, is said to be almost periodic if the set {f(· + τ), τ ∈R} is relatively compact in Cb(R, X). The set of almost periodic functions is a sub-algebra of Cb(R, X), and also the closure of the space of periodic functions in Cb(R, X). 4.5. ERGODICITY 77 Proof of Theorem 4.5.19. Step 1. Let f ∈X. Since the trajectory ( e Stf)t≥0 is relatively compact, we infer from [120, Theorem 8] (with U(τ, t) = St and thus no periodicity condition on U) the existence of an almost periodic eternal solution g of the rescaled semigroup e S, i.e. a function g : R →X such that g(t + τ) = e Sτg(t) for all t ∈R and τ ≥0, such that lim t→+∞∥e Stf −g(t)∥= 0. The end of the proof consists in characterizing the function g in the two cases (1) and (2). For λ ∈R, we deﬁne the Bohr transformation of the almost-periodic function g by cλ(g) = lim T→+∞ 1 T Z T 0 e−iλtg(t) dt, which is known to exists, see [61, Theorem 3.4], since e−iλtg(t) is also almost periodic. Since e−iλtg(t) is besides an eternal solution of the semigroup e−iλt e St with inﬁnitesimal generator Lλ = L −λ1 −iλ, we have that Lλ Z T 0 e−iλtg(t) dt = g(T) −g(0). Dividing by T the above expession, and passing to the limit T →+∞and using that Lλ is a closed operator, we get Lλcλ(g) = 0. In other words, we have established Lcλ(g) = (λ1 + iλ)cλ(g) and λ1 + iλ is an eigenvalue of L if cλ(g) ̸= 0. Step 2. We deduce that if the boundary spectrum is trivial, then necessarily cλ(g) = 0 for all λ ̸= 0. By the uniqueness theorem, see for instance [61, Theorem 4.7], we get that g is constant. Due to the conservation law ⟨e Stf, φ1⟩= ⟨f, φ1⟩and the simplicity of the eigenvalue 0, we get that g = ⟨f, φ1⟩f1 and the result of case (1) is proved. Step 3. In the case (2), the boundary spectrum is discrete and given by {λ1 + iαk, k ∈Z} for some α > 0. As a consequence, any λ such that cλ(g) ̸= 0 is necessarily of the form λ = αk for some k ∈Z. By the uniqueness theorem, g is then a α-periodic function which is given, by virtue of Fejér’s theorem, by g(t) = lim n→∞ 1 n n X ℓ=0 ℓ X k=−ℓ cαk(g)eiαkt. Consider (gk, ψk) two positive direct and dual eigenvectors of L associated to the eigenvalue iαk such that ⟨gk, ψk⟩= 1. From the conservation laws ⟨e Stf, ψk⟩= ⟨f, ψk⟩eiαkt and the algebraic simplicity of the eigenvalues iαk, we get that cαk(g) = ⟨f, φk⟩gk, and the result is proved. We will many time apply the following variant and immediate consequence of Theorem 4.5.19. Theorem 4.5.20. Consider a positive semigroup S on a Banach lattice X such that its generator L enjoys the conclusions (C2) of existence, uniqueness and strict positivity of the ﬁrst eigentriplet (λ1, f1, φ1) and let us set e St := e−λ1tSt. We assume that X ⊂L1 loc and that for some k ≥1 the space X k deﬁned in (3.4.19) satisﬁes X k ⊂L1 loc with strongly compact embedding. We also assume that L satisﬁes the reverse Kato’s inequality, in the sense that the conditions f ∈D(L∞), 0 ≪\|f\|∈D(L), L\|f\|= ℜe(signf)Lf, (4.5.23) implies f = u\|f\| for some u ∈S1. Then, for any f ∈X, there holds e Stf →f1⟨f, φ1⟩strongly in L1 φ1 as t →∞. (4.5.24) 78 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Remark 4.5.21. Theorem 4.5.20 enjoys some strong similarities with the main consequences of the General Relative Entropy technique developped in , see in particular [171, Theorem 3.2], [171, Theorem 4.3] and [171, Theorem 5.2]. In particular, the aperiodicity condition (4.5.23) may be compared with the fact that the ﬁrst eigenvector f1 is the unique (normalized and nonnegative) vector f ∈X with vanishing dissipation of entropy D(f) = 0 as deﬁned in or more generally that spanf1 is the unique invariant space on which the functional D vanishes. The present formulation is more abstract and probably more general. The drawback is the condition X k ⊂L1 loc with strongly compact embedding which can be avoided in , by using some weak compacteness argument and the lower semicontinuity property of D. That is explained by the fact that our proof uses rather the La Salle invariance principle (similarly as in the proof of [83, Theorem 3.2]) instead of a entropy dissipation argument. Although Theorem 4.5.19 deals with strong compacteness assumption, it also allows handling weak compacteness conditions, as attested by the following variant. Theorem 4.5.22. Consider a positive semigroup S on a Banach lattice X such that its generator L satisﬁes the conclusions (C3) about the existence, positivity and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1) and the triviality of the boundary spectrum. We set e St := e−λ1tSt. We assume furthermore that (1) either X = Y ′, Y separable, and the trajectory ( e Stf)t≥0 is bounded for any f ∈X. We denote by T the weak ∗σ(Y ′, Y ) topology; (2) or X ⊂L1 loc(E, E , µ) and we denote by T the weak topology of L1 φ1. In both case, the ergodicity property (CE2) holds in the sense of the weak convergence in T . Remark 4.5.23. When instead of assumption (C3) in the statement of Theorem 4.5.22, we assume (C2) and Σ+ P (L) is discrete and made of algebraically simple eigenvalues, then the conclusion (2) of Theorem 4.5.19 holds in the sense of the weak convergence in T . Proof of Theorem 4.5.22. We split the proof into two steps. Step 1. We ﬁrst assume (1). Since Y is separable, we can ﬁnd a sequence (ϕn)n≥1 ⊂Y which satisﬁes ∥ϕn∥= 1 and span(ϕn) is dense in Y . Then we can deﬁne on X the norm ∥·∥∗by setting ∥f∥∗= ∞ X n=1 2−n\|⟨f, ϕn⟩\|. (4.5.25) On bounded subsets of X the topology of this norm is the same as the weak- topology, it is worth emphasizing fn ⇀f ∗σ(Y ′, Y ) ⇔(sup∥fn∥< ∞ and ∥fn −f∥∗→0). Since by assumption the trajectory ( e Stf) is bounded, it is weakly- relatively compact, and so relatively compact in (X, ∥·∥∗). It is also clear that the semigroup S is continuous for the weak norm ∥·∥∗. The normed space (X, ∥·∥∗) is not a Banach space, but the proof of Theorem 4.5.19 actually only requires, for applying [120, Theorem 8], that the closed balls of X are complete metric spaces, which is the case for the distance induced by ∥·∥∗. Applying Theorem 4.5.19-(1) then yields the result. Step 2. We next assume (2). We consider f ∈X and repeating the proof of Step 2 in Theorem 3.4.20, we get that (Stf)t≥0 belongs to a weak compact set of L1 φ1. We deﬁne the norm ∥·∥∗by (4.5.25) for a sequence (ϕn)n≥1 ⊂Cc(E) which satisﬁes ∥ϕn∥L∞= 1 and span(ϕn) is dense in Cc(E). The trajectory ( e Stf) is then relatively compact in (X, ∥·∥∗) and the semigroup S is continuous for the weak norm ∥·∥∗. We conclude as in the previous step. 4.6. A WORD ABOUT SPECTRAL ANALYSIS ARGUMENT 79 4.6 A word about spectral analysis argument The aim of this section is to recall some more or less classical results which makes possible to slightly improve the conclusions of the results presented in the previous section by additionally assume some spectral gap at the level of the operator or the semigroup. More precisely, we are interested by some accurate versions of a partial, but principal spectral mapping theorem which asserts that Σ(etL) ∩∆eκt = etΣ(L)∩∆κ, ∀t ≥0, (4.6.26) for some κ < λ1, and even more precisely, we wish to establish the following geometric (or exponential) asymptotic stability (CE3) there exist some constants κ < 0 and C ≥1 such that for any f ∈X, there holds ∥e S(t)f −⟨f, φ1⟩f1∥≤Θ(t)∥f −⟨f, φ1⟩f1∥, ∀t ≥0, ∀f ∈X, (4.6.27) with the decay rate function Θ(t) := C eκt. In order to discuss the several results we present, we recall the spliting framework S = V + W ∗S, ∥V (t)∥B(X)+∥W(t)∥B(X)≲eκt, (4.6.28) for the same κ ∈R as above. We start by recalling the spectral mapping theorem for the point spectrum, and its proof, which is instructive. Lemma 4.6.24 (Spectral mapping theorem for point spectrum). For a semigroup (St)t≥0 with inﬁnitesimal generator L we have ΣP (St) \ {0} = etΣP (L), ∀t ≥0. Proof of Lemma 4.6.24. The inclusion etΣP (L) ⊂ΣP (St) \ {0} is clear. Now let ξ ∈ΣP (St) \ {0}, that is ξ ∈C \ {0} such that Stf = ξf for some f ∈XC \ {0}, and let λ ∈C such that ξ = eλt and φ ∈X′ such that ⟨φ, f⟩̸= 0. For any k ∈Z we have ξ = eλt+2ikπ and so 0 = e−(λ+ 2ikπ t )tStf −f = L −λ −2ikπ t Z t 0 e−(λ+ 2ikπ t )sSsf ds. If the last integral is non-zero for some k ∈Z, we deduce that λ + 2ikπ t is an eigenvalue of L and the result is proved. Assume by contradiction that R t 0 e−(λ+ 2ikπ t )sSsf ds = 0 for all k ∈Z. This means that the continuous t-periodic complex-valued function s 7→e−λs⟨φ, Ssf⟩has all its Fourier coeﬃcients equal to zero, which is not possible since this function is not equally zero (its value at s = 0 is not zero). We next present a very classical result about the exponential stability of f1 which is based on the quasi-compact semigroup framework of Voigt (see also [7, B-IV-2] and [82, Section V.3]) and which is a more accurate version of Lemma 2.1.7 and Theorem 4.2.6. Theorem 4.6.25. Let (St)t≥0 be a positive irreducible semigroup on a Banach lattice X satisfying the hypothesizes of Lemma 2.1.7, namely: (H2) holds for a constant κ0 ∈R and there exists T > 0 such that the splitting ST = VT + KT , (4.6.29) holds with ∥VT ∥B(X)≤eκT , κ < κ0, and KT ∈K (X). Then there exists a unique solution (λ1, f1, φ1) to the eigentriplet and the exponential stability (4.6.27) holds (without constructive estimate). 80 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Remark 4.6.26. In the splitting framework (4.6.28) the critical hypothseis KT ∈K (X) may be obtained by assuming (and proved the applications) that ∥W(t)∥B(X,X1)≲eκt, ∀t ≥0, X1 ⊂X compact. In fact, in many applications, we are also able to estrablish X1 ⊂D(Lβ), for some β > 0, without too much more work. Theorem 4.6.25 is in fact nothing but [20, Theorems 14.18] (see also [244, Section 2], [82, Theorem V.3.7] or [7, C–IV, Thm. 2.1 & Rk. 2.2]). We give however a short proof of Theorem 4.6.25 since it is simpler and more direct than the ones we usual ﬁnd in the literature and in particular does not refer to subtil results about the spectrum and its essential part. Proof of Theorem 4.6.25. First step. From Lemma 2.1.7, we already know that (H1), (H2) and (H3) hold. Together with the irreducibility which is nothing but (H4) from Lemma 3.2.6, we may apply Theorem 3.3.11 and conclude to the existence, uniqueness and strict positivity result about the eigentriplet solution (λ1, f1, φ1). Second step. We claim that Σ(L) ∩{z ∈C, ℜe(z) ≥κ0} is also made of a ﬁnite number of isolated eigenvalues with ﬁnite geometric multiplicity. We indeed set β0 := eκ0T . Since for any λ ∈Bc β0 := {z ∈C, \|z\|≥β0} the operator λ −VT is invertible, we easily see that λ ∈Bc β0 is in the spectrum of ST if and only if 0 is in the spectrum of I −(λ −VT )−1KT , or in the spectrum of I −KT (λ −VT )−1. Indeed, solving (λ −S)f = g is equivalent to, on the one hand, (I −(λ −VT )−1KT )f = (λ −VT )−1g, and in the other hand, (I −KT (λ −VT )−1)(λ −VT )f = g. So if λ ∈Σ(ST ) ∩Bc β0 then 1 ∈Σ((λ −VT )−1KT ). Since (λ −VT )−1KT is a compact operator, the classical Fredholm alternative (see for instance [40, Theorem 6.6]) asserts that its spectrum is made of eigenvalues with ﬁnite geometric multiplicity, and then so does for Σ(ST ) ∩Bc β0. We can also prove, by adapting the proof of [40, Lemma VI.2], that these eigenvalues are isolated, and thus Σ(ST ) ∩Bc β0 is made of a ﬁnite number of isolated eigenvalues with ﬁnite geometric multiplicity. Since eTΣ(L) ⊂Σ(ST ), we deduce that Σ(L) ∩{z ∈C, ℜe(z) ≥κ0} is also made of a ﬁnite number of isolated eigenvalues with ﬁnite geometric multiplicity. Third step. We prove the existence of a spectral gap and we conclude. Since Σ(L) ∩{z ∈C, ℜe(z) ≥κ0} is ﬁnite, λ1 is simple, and the boudary spectrum of L is a group, we deduce the existence of ε > 0 such that Σ(L) ∩{z ∈C, ℜe(z) ≥λ1 −ε} = {λ1}. The spectral mapping theorem in Lemma 4.6.24 then ensures that Σ(ST ) ∩{z ∈C, \|z\|≥e(λ1−ε)T } = {eλ1T } and that eλ1T is simple with eigenspace spanned by f1. The restriction S⊥ T of ST to the invariant subspace X⊥:= {f ∈X, ⟨φ1, f⟩= 0} thus has a spectral radius smaller than e(λ1−ε)T . The spectral radius formula (see [221, Thm 10.13] for instance) then ensures that lim n→∞∥S⊥ nT ∥1/n= r(S⊥ T ) ≤e(λ1−ε)T . This guarantees, for any η ∈(0, ε), the existence of a constant Cη > 0 such that for all f ∈X⊥ and all t ≥0 ∥e−λ1tStf∥≤Cηe−ηt∥f∥, and the proof is complete. Let us now present a variant of another classical result known as the Gearhart-Prüss Theorem in [98, 210], see also the contributions of Herbst and Greiner [7, A-III.7] as well as the more constructive proof [82, Theorem V.1.11] based on techniques developed in or related to [247, 31]. 4.6. A WORD ABOUT SPECTRAL ANALYSIS ARGUMENT 81 Theorem 4.6.27. Consider a positive semigroup S on a Banach lattice X such that its generator L satisﬁes the conclusions (C2) about the existence, positivity and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1). We assume furthermore that X is an Hilbert space and that there exist κ < λ1 and R > 0 such that (i) supz∈∆κ\BR∥RL(z)∥B(X)< ∞; (ii) Σ(L) ∩∆κ ⊂Σd(L) ∩BR. Then the exponential stability (4.6.27) holds (without constructive estimate). Proof of Theorem 4.6.27. The spectral information (C2) and (ii) together imply (C3) (because of Theorem 4.2.4) and that there exists κ∗∈(κ, λ1), such that Σ(L)∩∆κ∗= {λ1}. The operator L on X0 := (vect{f1})⊥thus satisﬁes supz∈∆κ∗∥RL(z)∥B(X0)< ∞, and we conclude thanks to [82, Theorem V.1.11]. The lack of constructivity here only comes from the fact that our assumptions do not provide any information on the spectral gap λ1 −κ > 0. Remark 4.6.28. Except of the Hilbert space framework, the assumptions made in Theorem 4.6.27 are slightly weaker than those of Theorem 4.6.25, and are indeed established during the proof of Theorem 4.6.25: such an information at the level of the resolvent is a bit easier to establish than a similar estimate at the level of the semigroup. In the splitting framework (4.6.28) and its resolvent counterpart (2.1.22), we typically only have to show sup κ≤ℜez≤κ1 ∥V(z)∥B(X)< ∞, lim r→∞ sup κ≤ℜez≤κ1, \|ℑmz\|≥r ∥W(z)∥B(X)= 0, (4.6.30) for some κ < λ1, and W(z) ∈K (X) for any z ∈∆κ. That last claim is classical (see for instance ) and we only brieﬂy sketch the proof. On the one hand, from the ﬁrst and the last estimates, we deduce that Σ(L) ∩∆κ ⊂Σd(L) thanks to Theorem 4.2.5. As in the proof of Theorem 4.2.5 and with the usual notations, we also have (I −W(z))RL(z) = V(z), ∀z ∈∆κ, where I −W(z) is invertible and ∥(I −W(z))−1∥B(X)≤2 for any z ∈C such that κ ≤ℜez ≤κ1, \|ℑmz\|≥R and R is large enough. We immediately deduce that the condition (i) in Theorem 4.6.27 holds. We end this section by a more recent result which is similar to the Gearhart-Prüss Theorem but is not restricted to an Hilbert space. Theorem 4.6.29. Consider a positive semigroup S on a Banach lattice X such that its generator L satisﬁes the conclusions (C2) about the existence, positivity and uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1). We further assume that L = A + B with 0 ≤A ∈B(X), SB ≥0 and the associated operators V and W deﬁned by (2.2.14) satisfy (4.6.28) for some κ < λ1 and that the resolvent counterpart W deﬁned by (2.1.22) satisﬁes (4.6.30) and more precisely sup κ≤ℜez≤κ1 ∥⟨z⟩αW(z)∥B(X)< ∞, with α > 1. Then the exponential stability (4.6.27) holds (without constructive estimate). The proof of Theorem 4.6.29 is a mere adaptation of [177, Theorem 3.1] (see also ) and it is thus skipped. The needed estimates are a bit stronger than those of Remark 4.6.28, but in the applications, they are not really more demanding. They also holds at the level of the resolvent instead of what is assumed in the statement of Theorem 4.6.25. We conclude by emphasizing again on the fact that all the above results are not constructive. We propose in the next part an alternative approach which is constructive. 82 CHAPTER 4. THE GEOMETRY OF THE BOUNDARY POINT SPECTRUM Chapter 5 Quantitative stability In this chapter we establish some quantitative stability results in the spirit of the Doblin, Harris, Meyn-Tweedie theory for Markov semigroup. 5.1 About quantiﬁed positivity conditions We brieﬂy discuss some positivity conditions related to the strong maximum principle and barriers techniques. The issue is about how quantify the strong maximum principle f ∈X+{0}, (κ1 −L)f ≥0 imply f > 0 or f ≫0 or the related strong positivity of the associated semigroup. A possible way can be achieved with the help of a barrier functions familly G ⊂X+ and a second weaker (semi)norm [·] used for normalization. Let us then introduce the two conditions ∀R > 0, ∃gi ∈G, ∀f ∈X+, [f] = 1, ∥f∥≤R, (5.1.1) we have (i) ST f ≥g1 (for some T > 0) or (ii) f ≥g2 if (κ1 −L)f ≥0. Point (ii) is a quantiﬁed version of the strong maximum principle when G ⊂X++ and it is always a consequence of the positivity condition (i). Assume indeed that (i) holds (for some T > 0) and that f satisﬁes the requirements (5.1.1) and (κ1 −L)f ≥0. We then write d dt(e(L−κ1)tf) = e(L−κ1)t(L −κ1)f ≤0, so that f ≥e(L−κ1)T f = e−κ1T ST f ≥e−κ1T g2 =: g1, with g2 given by condition (i). The recipocal implication is not clear, see however Lemma 3.2.6-(3). Let us now make a list of possible quantiﬁed positivity conditions for a linear (and continuous) operator A : X →X: (P1′) ∃g0 ∈X+{0}, ∃ψ0 ∈X+{0}, ∀f ∈X+, Af ≥g0⟨f, ψ0⟩(Harris condition); (P2′) ∃g0 ∈X+{0}, ∃ψ0 ∈X′ ++, ∀f ∈X+, Af ≥g0⟨f, ψ0⟩(Doblin condition); (P3′) ∃g0 ∈X++, ∃ψ0 ∈X′ +{0}, ∀f ∈X+, Af ≥g0⟨f, ψ0⟩(dual Doblin condition); (P4′) ∃g0 ∈X++, ∃ψ0 ∈X′ ++, ∀f ∈X+, Af ≥g0⟨f, ψ0⟩(strong Doblin condition). We summarize some elementary relations between these conditions and those listed in Section 3.2. 83 84 CHAPTER 5. QUANTITATIVE STABILITY Lemma 5.1.1. We have (P2′) ⇒(P2) ⇒(P1), (P3′) ⇒(P3) ⇒(P1), (P4′) ⇒((P4), (P3′), (P2′)) as well as (P4) ⇒((P3), (P2)). We also have: A satisﬁes (P2′) iﬀA∗satisﬁes (P3′); A satisﬁes (P3′) iﬀA∗satisﬁes (P2′); A satisﬁes (P4′) iﬀA∗satisﬁes (P4′). We ﬁnally have: A satisﬁes (P2′) implies ∃g0 ∈X+{0}, ∃κ > 0, Ag0 ≥κg0 Proof of Lemma 5.1.1. We assume Af ≥g0⟨f, ψ0⟩for any f ∈X+ and some g0 ∈X+, ψ0 ∈X′ +. For any φ ∈X′{0} and f ∈X+, we have ⟨A∗φ, f⟩= ⟨φ, Af⟩≥⟨φ, g0⟨f, ψ0⟩⟩, which implies A∗φ ≥ψ0⟨φ, g0⟩. We thus deduce that A satisﬁes (P2′) (resp. (P3′), (P4)) implies that A∗satisﬁes (P3′) (resp. (P2′), (P4)). We conclude this introductory section by emphasizing on the fact (as already mentioned above) that SL satisﬁes (Pi′) implies RL(λ) satisﬁes (Pi′) for any λ ≥λ1 and i = 1, . . . , 4. 5.2 Asymptotic stability under Doblin condition We start with a simple situation. We assume the Doblin condition, namely ∃T > 0, ∃ψ0 ≫0, ∃g0 > 0, ∀f ≥0, ST f ≥g0⟨ψ0, f⟩, (5.2.2) together with the companion positivity condition ∃r0 > 0, ⟨φ1, g0⟩≥r0, (5.2.3) as well as the strong additional boundedness assumption ∃R0 > 0, φ1 ≤R0ψ0. (5.2.4) When ψ0 := 1 ∈X′ ⊂L∞, the condition in (5.2.4) is automatically satisﬁed with R0 := ∥φ1∥= 1. In the case when S∗ L enjoys a splitting structure similar to (4.5.18) and more precisely ∥e S∗(t)φ∥≤Θ(t)∥φ∥+ Z t 0 Θ(t −s)[ e S∗(s)φ]g0 ds, with Θ ∈L1(R+) ∩C0(R+), we deduce that 1 = ∥φ1∥= ∥e S∗(t)φ1∥≤Θ(t) + Z t 0 Θ(t −s)[φ1]g0 ds, ∀t > 0. Passing to the limit t →∞, we get (5.2.3) with r0 := ∥Θ∥−1 L1 . Another condition can be formulated as follows. We assume that D(L∞) ⊂L1 loc and the splitting property L = A + B with A ∈B(X), RB(λ) ∈B(X) ∩B(X+) for any λ ≥κ, with κ < κ0 ≤λ1, and the additionnal regularity condition (RB∗(λ)A∗)N : L1 g0 →L∞ ψ−1 0 , ∀λ > κ. (5.2.5) The dual eigenvector φ1 satisﬁes (λ −B∗)φ1 = A∗φ1, λ1 > κ, and then φ1 = (RB∗(λ1)A∗)Nφ1, so that (5.2.3)-(5.2.4) holds with the normalization condition r0 := 1 and R0 := ∥(RB∗(λ)A∗)N∥B(L1 g0,L∞ ψ−1 0 ). We are then able to formulate a ﬁrst quantiﬁed stability result. 5.3. ASYMPTOTIC STABILITY UNDER HARRIS CONDITION 85 Theorem 5.2.2. Consider a semigroup S on a Banach lattice X such that its generator L enjoys the conclusion (C1) on the existence of the ﬁrst eigentriplet (λ1, f1, φ1). We assume furthermore the Doblin condition (5.2.2)–(5.2.4)-(5.2.3). Thenthe exponential stability (4.6.27) in the norm [·]ψ0 holds true, with constructive constants. The proof closely follows the usual contraction argument in the Doblin result, see for instance , [92, Theorem 11] or [50, Theorem 2.1]. We do not explicitly assume the irreducibility of the semigroup, but the Doblin condition (5.2.2)–(5.2.4)-(5.2.3) is in many aspects a strong positivity condition. In particular, our result implies the uniqueness of the ﬁrst eigentriplet (λ1, f1, φ1) and the triviality of the boundary spectrum. Proof of Theorem 5.2.2. The two conditions (5.2.2) and (5.2.4) together imply the modiﬁed Doblin condition ∃T > 0, ∃g1 > 0, ∀f ≥0, ST f ≥g1⟨φ1, f⟩, with g1 := g0/R0. Take f such that ⟨φ1, f⟩= 0, so that ⟨φ1, f±⟩= r := ⟨φ1, \|f\|⟩/2 ≥0 and thus ST f± ≥g1⟨φ1, f±⟩= rg1. We write \|ST f\|≤\|ST f+ −rg1\|+\|ST f−−rg1\|= ST \|f\|−2rg1. We deduce ⟨φ1, \|ST f\|⟩≤⟨S∗ T φ1, \|f\|⟩−2r⟨φ1, g1⟩= eλ1T −⟨φ1, g1⟩ ⟨φ1, \|f\|⟩. In other words, setting e St := e−λ1tSt, we have [ e ST f]φ1 ≤γ[f]φ1, with γ < 1 which depends explicitly of r0, R0, T and the estimates on λ1. We then classically deduce the exponential convergence in the [·]φ1 norm. Now, the dual condition associated to the Doblin hypothesis (5.2.2) is ∀ψ ∈X′ +, S∗ T ψ ≥ψ0⟨ψ, g0⟩. In particular, the ﬁrst dual eigenvector φ1 satisﬁes φ1 = e−λ1T S∗ T φ1 ≥e−λ1T ψ0⟨φ1, g0⟩= e−λ1T r0ψ0. (5.2.6) Together with condition in (5.2.4), we see that [·]φ1 and [·]ψ0 are equivalent norm, and we immediately obtain the exponential convergence in the [·]ψ0 norm (with constructive constants). 5.3 Asymptotic stability under Harris condition The Doblin condition (5.2.2)–(5.2.4)-(5.2.3) is too much demanding for many applications. In this section, we make the following somehow more general Harris type condition    ∃T > 0, ∀R > 0, ∃gR > 0 such that ∀f ≥0, ∥f∥≤R[f]φ1 there holds ST f ≥gR[f]φ1, (5.3.7) that we complement with the following Lyapunov condition ∥e Sf∥≤γL∥f∥+K[f]φ1, (5.3.8) with e S = ST e−λ1T , γL ∈(0, 1), K ≥0. We repleace the positivity condition (5.2.3) by ∃rR > 0, ⟨φ1, gR⟩≥rR. (5.3.9) As we have seen several times, condition (5.3.8) is some kind of regularity hypothesis which is natural under a splitting structure on the semigroup SL. 86 CHAPTER 5. QUANTITATIVE STABILITY Theorem 5.3.3. Consider a semigroup S on a Banach lattice X such that its generator L enjoys the conclusions (C1) on the existence of the ﬁrst eigentriplet (λ1, f1, φ1). We assume furthermore the Harris condition (5.3.7) together with the Lyapunov condition (5.3.8) and the the positivity condition (5.3.9). Then the exponential stability (4.6.27) in the norm of X holds true, with constructive constants. Of course, in order that Theorem 5.3.3 really gives a constructive convergence result, we have to establish (5.3.7), (5.3.8) and (5.3.9) in a constructive way. Proof of Theorem 5.3.3. On the one hand, we have [ e Sf]φ1 ≤⟨e S\|f\|, φ1⟩= ⟨\|f\|, e S∗φ1⟩= [f]φ1. (5.3.10) On the other hand, we wish to establish the coupling property [ e Sf]φ1 ≤γH[f]φ1 if ∥f∥≤A[f]φ1 and ⟨f, φ1⟩= 0, (5.3.11) for any A > K/(1 −γL) and some γH ∈(0, 1). In order to do so we ﬁx A > K/(1 −γL) and we set R := 2A. We next consider f ∈X, such that ⟨f, φ1⟩= 0 and ∥f∥≤A[f]φ1, so that ∥f±∥≤∥f∥≤A[f]φ1 = R[f±]φ1. Using the Harris condition (5.3.7), we deduce e Sf± ≥ϑgR, ϑ := 1 2e−λ1T [f]φ1. Similarly as in the proof of f of Theorem 5.2.2, we next compute \| e Sf\|≤\| e Sf+ −ϑgR\|+\| e Sf−−ϑgR\|≤e S\|f\|−2ϑgR and then [ e Sf]φ1 ≤ ⟨e S\|f\|−2ϑgR, φ1⟩ = ⟨\|f\|, e S∗φ1⟩−2ϑ⟨gR, φ1⟩ = (1 −e−λ1T ⟨gR, φ1⟩)[f]φ1, which in turn implies (5.3.11) with γH := 1 −e−κ0T rR. Now, the two estimates (5.3.10) and (5.3.11) together give [Sf]φ1 ≤γH[f]φ1 + 1 −γH A ∥f∥. (5.3.12) From (5.3.12) and the Lyapunov condition (5.3.8), we deduce that Un+1 = MUn with Un := ∥Snf∥ [Snf]φ1 ! and M := γL K 1−γH A γH ! . The eigenvalues of M are µ± := 1 2(T ± p T 2 −4D), with T := trM = γL + γH, D := detM = γLγH −(1 −γH)K A . 5.4. THE WEAK DISSIPATIVITY CASE 87 We observe that γLγH > D > γLγH −(1 −γH)(1 −γL) = T −1, so that (γH −γL)2 = T 2 −4γLγH < T 2 −4D < T 2 −4(T −1) = (T −2)2 and ﬁnally α := max(\|µ+\|, \|µ−\|) < max(γH, γL, \|T −1\|, 1) = 1. We conclude that ∥Mn∥≲αn, from what we immediately conclude. 5.4 The weak dissipativity case In this section, we consider a weak dissipative semigroup (St) as considered in Section 2.2.3 and in a sense we make precise now. We consider four Banach lattices X3 ⊂X2 ⊂X1 ⊂X0 = X. We ﬁrst make the same kind of Harris type condition as in the previous section, namely Hypothesis (H) (Doblin-Harris) condition (5.3.7) holds for the same time T > 0 and for both norms ∥·∥= ∥·∥X0 and ∥·∥= ∥·∥X2 as well as the companion positivity condition (5.3.9) holds; Instead of the strong Lyapunov condition (5.3.8), we assume Hypothesis (L) (weak Lyapunov) there exist a constant K ≥0 such that ∥e Sf∥1+∥e Sf∥0 ≤ ∥f∥1+K[f]φ1, ∀f ∈X1, ∥e Sf∥3+∥e Sf∥2 ≤ ∥f∥3+K[f]φ1, ∀f ∈X3, with e S = ST e−λ1T ; Hypothesis (I) (interpolation) there exists an increasing function ξ : R+ →R+, λ 7→ξλ, such that λ∥f∥1≤∥f∥0+ξλ∥f∥3, ∀λ > 0, ξλ/λ →0 as λ →0. Theorem 5.4.4. Consider a semigroup S on a Banach lattice X such that its generator L enjoys the conclusions (C1) on the existence of the ﬁrst eigentriplet (λ1, f1, φ1). We assume furthermore the three above conditions of weak conﬁnement (L), Doblin-Harris strong irreducibility (H) and interpolation (I). Then, there exist some constructive decay rate functions Θ and e Θ such that ∥Snf∥X1≲Θ(n)∥f∥X3, ∀n ≥1, (5.4.13) and ∥Snf∥≲e Θ(n)∥f∥X3, ∀n ≥1, (5.4.14) for any f ∈X3, ⟨f, φ1⟩= 0. More precisely, the decay rate functions Θ and e Θ are deﬁned by Θ(t) := inf λ Θζλ(t), e Θ(t) := t−1Θ([t/2]), (5.4.15) for a constructive constant ζ ∈(0, 1), the inﬁmum being taken over all the decreasing function λ : R+ →R+, t 7→λt, and Θλ is deﬁned by Θ(t) :≃inf λ>0(e−λt + ξλ λ ). (5.4.16) The proof follows closely the proof of [50, Thm 4.8]. We start with the following key argument of non expansive mapping result on a well chosen norm. 88 CHAPTER 5. QUANTITATIVE STABILITY Proposition 5.4.5. Consider a positive semigroup (St) which satisﬁes both above conditions of weak conﬁnement (L) and Doblin-Harris strong irreducibility (H). There exist some equivalent norms \|\|\|·\|\|\|1 to ∥·∥1 and \|\|\|·\|\|\|3 to ∥·∥3 such that e St is a non expansive mapping for the two new norms \|\|\|·\|\|\|1 and \|\|\|·\|\|\|3. More precisely, there exists α > 0 such that \|\|\| e Sf\|\|\|1+α∥e Sf∥0 ≤ \|\|\|f\|\|\|1, ∀f ∈X1, ⟨f, φ1⟩= 0, (5.4.17) \|\|\| e Sf\|\|\|3+α∥e Sf∥2 ≤ \|\|\|f\|\|\|3, ∀f ∈X3, ⟨f, φ1⟩= 0. (5.4.18) Proof of Proposition 5.4.5. We deﬁne \|\|\|f\|\|\|1:= [f]φ1 + δ∥f∥0+β∥f∥1, (5.4.19) with β > δ > 0 conveniently chosen. We take β := (1 −γH)/K, δ := (1 −γH)/A. We deﬁne \|\|\|·\|\|\|3 in the same way. In what follows, we then only establish (5.4.17), the proof of (5.4.18) being exactly the same. We ﬁx f ∈X1, ⟨f, φ1⟩= 0, and we recall [ e Sf]φ1 ≤[f]φ1. (5.4.20) We also recall that from (5.3.11), for any A > 0, there exists γH = γH(A) ∈(0, 1) such that the following coupling property holds [ e Sf]φ1 ≤γH[f]φ1 if ∥f∥0≤A[f]φ1. (5.4.21) We ﬁx A > K and we observe that the following alternative holds ∥f∥0≤A[f]φ1 (5.4.22) or ∥f∥0> A[f]φ1. (5.4.23) Case 1. Under condition (5.4.22), we use (5.4.21) and the ﬁrst estimate in (L), and we deduce \|\|\| e Sf\|\|\|1 = [ e Sf]φ1 + δ∥e Sf∥0+β∥e Sf∥1 ≤ γH[f]φ1 + β∥f∥1+βK[f]φ1 −(β −δ)∥e Sf∥0. From our choice of β > 0 we have γH + βK = 1, and we conclude that (5.4.17) holds with α := β −δ > 0. Case 2. Under condition (5.4.23), the ﬁrst Lyapunov condition in (L) implies ∥e Sf∥1+∥e Sf∥0≤∥f∥1+K A ∥f∥0. Together with the non expansivity estimate (5.4.20), we get [ e Sf]φ1 + β∥e Sf∥1+β∥e Sf∥0≤[f]φ1 + β∥f∥1+δ∥f∥0, and we conclude to (5.4.17) again. The subgeometric convergence result is a straightforward consequence of Proposition 5.4.5 and an interpolation argument. Proposition 5.4.6. Assume that S satisﬁes the hypotheses of Theorem 5.4.4. Then (5.4.13) and (5.4.14) hold true with the same decay rate functions Θ and e Θ given by (5.4.15) (up to a modiﬁcation of the constant ζ). 5.4. THE WEAK DISSIPATIVITY CASE 89 Proof of Proposition 5.4.6. We recall that we have already proven (5.4.17) and (5.4.18). From (5.4.17) and the interpolation condition (I), we deduce \|\|\| e Sf\|\|\|1+λα∥e Sf∥1≤\|\|\|f\|\|\|1+ξλα∥e Sf∥3. We observe next that from the very deﬁnition of the \|\|\|·\|\|\|1 norm \|\|\| e Sf\|\|\|1+α λ∥e Sf∥1≥Zλ\|\|\| e Sf\|\|\|1, Zλ = 1 + κλ ∈(1, 2], for some κ > 0 and any λ ∈(0, λ0), λ0 > 0, and that from the very deﬁnition of the \|\|\|·\|\|\|3 norm αξλ∥e Sf∥3≤Bξλ\|\|\| e Sf\|\|\|3, for some B > 0. The three above estimates together imply Zλ\|\|\| e Sf\|\|\|1≤\|\|\|f\|\|\|1+Bξλ\|\|\| e Sf\|\|\|3. Using the second estimate (5.4.18) and repeating the same proof, we have Zλn+1\|\|\| e Sn+1f\|\|\|1≤\|\|\| e Snf\|\|\|1+Bξλn+1\|\|\|f\|\|\|3, for any n ≥0 and for any λn+1 > 0. The discrete Grönwall lemma implies \|\|\| e Snf\|\|\|1≤An\|\|\|f\|\|\|1+ n X k=1 Ak,nξλkB\|\|\|f\|\|\|3, ∀n ≥0, (5.4.24) where we have deﬁned An := n Y k=1 ak, Ak,n = An/Ak = n Y i=k+1 ai, ai := Z−1 λi . Observing that Ak,n ≲e−κPn i=k λi ≲eκ(Λ(k)−Λ(n)), with Λ(t) := Z t 0 λs ds and λs := λi if s ∈(i −1, i], we immediately conclude that the ﬁrst estimate (5.4.13) holds true. We come back to the ﬁrst inequality in (5.4.17) that we iterate and sum up in order to obtain \|\|\| e Snf\|\|\|1+α n X k=[n/2]+1 ∥e Skf∥0≤\|\|\| e S[n/2]f\|\|\|1, for any n ≥1. Together with the non expansion inequality [ e Snf]φ1 ≤[ e Skf]φ1 ≲∥e Skf∥0, ∀n ≥k, and the ﬁrst estimate (5.4.13), we deduce (n −[n/2] −1)α[ e Snf]φ1 ≲Θ([n/2])\|\|\|f\|\|\|3, which is nothing but (5.4.14). 90 CHAPTER 5. QUANTITATIVE STABILITY Chapter 6 Application to Diﬀusion models In this part, we consider a general elliptic operator in divergence form Lf := ∂i(aij∂jf) + bi∂if + ∂i(βif) + cf, f ∈H1 0(Ω), (6.0.1) where Ω⊂Rd is a bounded domain (i.e. an open and connected set) or Ω= Rd, and for simplicity we always assume d ≥3. We also always assume at least a boundedness and ellipticity condition on the (aij) matrix, namely aij ∈L∞(Ω), ∃ν > 0, ∀ξ ∈Rd, aijξiξj ≥ν\|ξ\|2, (6.0.2) and some conditions on the coeﬃcients bi, βj and c which will be described below. We aim to establish the existence of (λ1, f1, φ1) solution to the ﬁrst eigentriplet problem λ1 ∈R, 0 < f1 ∈H1 0, Lf1 = λ1f1, 0 < φ1 ∈H1 0, L∗φ1 = λ1φ1, (6.0.3) and the existence of some (constructive) reate function Θ such that the rescaled semigroup e S associated to the generator e L = L −λ1 satisﬁes ∥e S(t)f −⟨f, φ1⟩f1∥H0≤Θ(t)∥f −⟨f, φ1⟩f1∥H, (6.0.4) for any t ≥0 and any f ∈H, with H ⊂H0 ⊂L2. 6.1 Diﬀusion with rough coeﬃcients in a bounded domain In this section, we consider the general elliptic operator in divergence form (1.2.8) in the case of a bounded and smooth enough domain Ω⊂Rd with general elliptic condition on aij as formulated above. We further assume bi, βj ∈Lr(Ω), c ∈Lr/2(Ω), r > d, (6.1.5) as well as c + divβ ≤c0 and c −divb ≤c0, (6.1.6) for some c0 ∈R. In that situation, the ﬁrst eigentriplet problem (6.0.3) in a slightly less general framework (all the coeﬃcients belong to L∞) is considered in . We explain with all details the existence proof by following more or less the arguments presented in and next we present a proof of the geometric part and the stability part by taking advantage of the abstract material developed in the previous sections. It is worth emphasizing that our proof of the uniqueness of the ﬁrst eigenfunction signiﬁcantly diﬀers from the one presented in which is based on a dissipativity argument, probably related to the reverse Kato’s inequality condition. The framework considered 91 92 CHAPTER 6. APPLICATION TO DIFFUSION MODELS here is the usual generalized solutions or weak solutions framework which goes back at least to Stampacchia [226, 225], but it is reminiscent of previous contributions by Friedrichs [91, 90], Gårding , De Giorgi , Nash , Morrey , Moser [186, 187, 185], Ladyzhenskaya, Solonnikov, Ural’ceva [148, 146], Oleinik, Kruzhkov and many others. Lot of the functional arguments are picked up from the book of Gilbarg and Trudinger, and more speciﬁcally from [104, Chapter 8], and also in recent notes by Kavian and Vasseur . It is worth emphasizing that the present analysis does not apply directly to elliptic operator in non divergence form, although this framework is considered in . We expect that all the results developed below can be generalized to a non divergence form framework, for example the one developed in , but we do not follow this line of research in the present work. The proof of (6.0.3) and (6.0.4) are straightforward consequences of the abstract results developed in the previous sections once we have been able to check that the corresponding hypotheses are fulﬁlled. In the sequel, we will then show how these hypotheses are met in the present context. Condition (H1). We recall that a weak (or variational) solution to the elliptic equation Lf = g ∈H−1(Ω), f ∈H1 0(Ω), is a function f ∈H1 0(Ω) such that DL(f, w) = ⟨g, w⟩, ∀w ∈H1 0(Ω), (6.1.7) where the (negative) Dirichlet form DL is deﬁned by DL(f, w) := − Z Ω (aij∂jf + βif)∂iw + Z Ω (bi∂ifw + cfw), for any f, w ∈H1 0(Ω). Most of the time, we will simply write ⟨Lf, w⟩= ⟨g, w⟩, ∀w ∈H1 0(Ω), (6.1.8) instead of (6.1.7). For the reader convenience, we repeat here some estimates picked up in . For λ ∈R and f ∈H1 0(Ω), we start with ⟨(λ −L)f, f⟩ = Z Ω aij∂if∂jf + Z Ω (βi −bi)∂iff + Z Ω (λ −c)f2 ≥ ∥f√c−∥2 L2+ν∥∇f∥2 L2−∥\|β −b\|f∥L2∥∇f∥L2−∥√c+f∥2 L2+λ∥f∥2 L2 ≥ ∥f√c−∥2 L2+ν 2∥∇f∥2 L2−1 2ν ∥\|β −b\|f∥2 L2−∥√c+f∥2 L2+λ∥f∥2 L2, using the Cauchy-Schwarz inequality and the Young inequality, and next ⟨(λ −L)f, f⟩ ≥ ∥f√c−∥2 L2+ν 4∥∇f∥2 L2+(λ −M 2ν −M1/2)∥f∥2 L2 +ν 4CΩ∥f∥2 L2∗−1 2ν ∥\|β −b\|1\|β−b\|≥Mf∥2 L2−∥√c+1c+≥Mf∥2 L2 ≥ ∥f√c−∥2 L2+ν 4∥∇f∥2 L2+(λ −M 2ν −M1/2)∥f∥2 L2 +(ν 4CΩ−1 2ν ∥\|β −b\|1\|β−b\|≥M∥2 Ld−∥c+1c+≥M∥Ld/2)∥f∥2 L2∗, using the Sobolev inequality (with associated constant CΩ) and the Holder inequality. Choosing M > 0 large enough in such a way that the last term is positive, and next κ1 > 0 large enough, we deduce for instance that ⟨(λ −L)f, f⟩≥∥f√c−∥2 L2+ν 4∥∇f∥2 L2+∥f∥2 L2, ∀λ ≥κ1. (6.1.9) 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 93 Thanks to the Lax-Milgram theorem and the above coercivity estimate, we deduce that λ −L is invertible, and more precisely the mapping (λ −L)−1 : H−1 →H1 0(Ω) is well deﬁned. We also claim that λ −L enjoys a weak principle maximum, and more precisely f ∈H1 0(Ω), (λ −L)f ≥0 imply f ≥0. (6.1.10) Indeed, for such a function f ∈H1 0(Ω), we take w = f−∈H1 0(Ω), as a test function, and elementary Sobolev space calculus together with the previous estimate yields 0 ≤ ⟨(λ −L)f, f−⟩= −⟨(λ −L)f−, f−⟩ ≤ −∥f− √c−∥2 L2−ν 4∥∇f−∥2 L2−∥f−∥2 L2≤0, so that f−= 0 and f ≥0. We thus deduce (λ −L)−1 : L2 + →L2 +, and from J.-L. Lions theory on parabolic equation (see for instance [155, Chapter 3]), we next deduce that L is the generator in L2 of a positive semigroup SL, so that (H1) holds. It is worth emphasizing at this point that the semigroup S built thanks to Lions’ theory is deﬁned by S(t)f0 = f for any f0 ∈L2, where f ∈E := C([0, ∞); L2) ∩L2 loc([0, ∞); H1 0) ∩H1 loc([0, ∞); H−1) is the unique (variational) solution to the equation (f(T), g(T))L2 −(f0, g(0))L2 = Z T 0 {⟨∂tg, f⟩H−1,H1 0 + DL(f, g)}ds, (6.1.11) for any T > 0 and g ∈E. Choosing g = f in the above equation, we classically compute 1 2∥f(t)∥2 L2−1 2∥f0∥2 L2− Z t 0 DL(f, f)ds = 0, ∀t > 0, which together with (6.1.9) imply 1 t Z t 0 ν 4∥∇f∥2 L2ds ≤−(f(t) −f0 t , f(t) + f0 2 )L2 + κ1 t Z t 0 ∥f∥2 L2ds, ∀t > 0. When f0 ∈D(L), the RHS is bounded and there thus exists a sequence tn →0 such that ∥∇f(tn)∥L2 is bounded. That implies f0 ∈H1 0(Ω) and thus D(L) ⊂H1 0(Ω). Similarly, we may consider the dual Dirichlet form D∗(f, g) := DL(g, f) and build an associated positive semigroup S∗thanks to Lions’ theory discribed above. More precisely S∗(t)g0 = g for any t ≥0 and g0 ∈L2, where g ∈E is the unique (variational) solution to the equation (g(t), f(t))L2 −(g0, f(0))L2 = Z t 0 {⟨∂tf, g⟩H−1,H1 0 + D∗(g, f)}ds, for any t > 0 and f ∈E. Now, we ﬁx T > 0, gT ∈L2 and we set g(t) := S∗(T −t)gT , so that g is a solution to the backward evolution equation −∂tg = L∗g, g(T) = gT , with L∗g := ∂j(aij∂ig) −∂i(big) −βi∂ig + cg. The variational formulation of this last problem is (gT , f(T))L2 −(g(0), f(0))L2 = Z T 0 {⟨∂tf, g⟩H−1,H1 0 −D∗(g, f)}ds, (6.1.12) for any f ∈E. Summing up (6.1.11) and (6.1.12) with f(t) := S(t)f0 for f0 ∈L2 and g(t) := S∗(T −t)gT for gT ∈L2, we deduce (S(T)f0, gT )L2 = (S∗(T)gT , f0)L2. 94 CHAPTER 6. APPLICATION TO DIFFUSION MODELS In other words, we have established that S∗= (SL)∗and thus that L∗is the generator of the semigroup S∗. Condition (H2). Let us consider a ball BR, R > 0, such that B4R ⊂Ωand next the solution f0 ∈H1 0(Ω), (κ1 −L)f0 = 1BR, (6.1.13) which exists from the above discussion. We next recall some classical results. On the one hand, from [226, Section 3 & Section 4] or [104, Theorem 8.15] (see also the original papers [68, 192, 186]), the following global L∞De Gorgi-Nash-Moser type estimate ∥f+∥L∞(Ω)≲∥f+∥L2(Ω)+∥g∥Lr/2(Ω) (6.1.14) holds for any subsolution f ∈H1 0(Ω), (λ −L)f ≤g ∈Lr/2(Ω). The local estimate variant [104, Theorem 8.18] (or weak Harnack inequality) ∥f∥Lp(B2R)≲inf BR f + ∥g∥Lr/2(Ω), ∀p ∈[1, 2∗/2), (6.1.15) also holds for a nonnegative supersolution f ∈H1(Ω), f ≥0 on B4R ⊂Ω, (λ −L)f ≥g ∈Lr/2(Ω), from what one deduces that a strong maximum principle [104, Theorem 8.19] holds, and more precisely, for any f ∈H1 0(Ω), we have Lf ≤0 in Ω, f ≥0 in Ω imply f ≡0 or f > 0 a.e. in Ω. (6.1.16) When indeed f ̸≡0, we may choose B4R ⊂Ωsuch that ∥f∥L1(B2R)> 0 and thus infBR f > 0 from (6.1.15) (with g = 0). We conclude that f is positive thanks to a connexity argument. Although the statements in [104, Chapter 8] are written assuming L∞bound on the coeﬃcients b, β, c, it is observed at the end of [104, section 8.5] that the theory may be extended under less restrictive conditions on the coeﬃcients as those assumed here. This theory is developed in although the above strong maximum principle is not explicitly written. On the other hand, the following Hölder regularity estimate [226, Théorème 7.1] and [104, Theorem 8.29] (see also the original papers [68, 192, 186]) of De Gorgi-Nash-Moser type ∥f∥Cα(Ω)≤C ∥(λ −L)f∥L∞(Ω) (6.1.17) holds true for some α = α(aij) ∈(0, 1) and C > 0. These last two pieces of information together and the fact that f0 ̸≡0 imply that there exists a constant θ > 0 such that f0 ≥θ1BR, and thus Lf0 ≥(κ1 −θ−1)f0. That is condition (i) in Lemma 2.1.4, so that condition (H2) holds thanks to Lemma 2.1.4. Presented in that way, the above estimate is not really constructive, but the constant θ := infBR(κ1 −L)−11BR can also be considered as a geometric quantity associated to geometric properties of the operator and the domain. First constructive argument for (H2). In the case when L is self-adjoint, that corresponds to the case aij = aji and bi + βi = 0, we classically know (that has been recalled in Section 2.1.3, see (2.1.34)) that λ1 = inf f∈X+{0} ⟨Lf, f⟩ ∥f∥2 = inf f∈H1 0, ∥f∥L2=1 Z O {a∇f · ∇f + cf2}, 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 95 from what and the Sobolev imbedding, we get λ1 ≥ inf f∈H1 0, ∥f∥L2=1{(νCΩ−∥c−1c−≥M∥Ld/2)∥f∥2 L2∗−M} ≥−M, by choosing M large enough. That gives an explicit lower bound on λ1. Second constructive argument for (H2). We give another constructive argument without assuming any self-adjointness property. We rather assume (∂ibi −c)+ ∈M1(Ω), bi + βi −∂jaij ∈M1(Ω). (6.1.18) We ﬁx h0 ∈C2 c (Ω) such that c01Bρ ≤h0 ≤c01B3ρ/2 with B8ρ ⊂Ωand ∥h0∥L2= 1. We next deﬁne f0 as the (positive) solution to f0 ∈H1 0(Ω), (κ1 −L)f0 = h0, (6.1.19) so that f0 ∈Cα(Ω) from (6.1.14) and (6.1.17), and similarly e f0 ∈H1 0(B2ρ), (κ1 −L) e f0 = h0, (6.1.20) so that e f0 ∈Cα(B2ρ) from (6.1.14) and (6.1.17). We observe that 0 ≤e f0 ≤f0 thanks to the weak maximum principle. We then compute 1 = ∥h0∥2 L2= Z B2ρ h0(κ1 −L) e f0 = Z B2ρ e f0(κ1 −L∗)h0 ≤∥e f0∥L∞∥(κ1 −L∗)h0∥M1, where the last term is ﬁnite because of the additional hypothesis (6.1.18). We conclude to a ﬁrst constructive lower bound ∥e f0∥L∞(B2ρ)≥c1 > 0. Because of the Holder continuity, we also have ∥e f0∥L1(B2ρ)≥c2 with constructive constante c2 = c2(c1, α, d) > 0. Thanks to (6.1.15) (with g = 0), we obtain f0 ≥ 1B2ρ inf B2ρ f0 ≥1B2ρCwH∥f0∥L1(B2ρ) ≥ 1B2ρCwH∥e f0∥L1(B2ρ)≥CwHc2c−1 0 h0. Because all the inequalities are constructive and proceeding as above, we deduce that condition (ii) in Lemma 2.1.4 holds and thus also (H2) with constructive constant κ0 := κ1 −C−1 wHc−1 2 c0. Finally, because of (κ1 −L)f0 = 0 on Ω\B2ρ, we may apply the Harnack inequality [104, Corollary 8.21], and we classically deduce there exist constructive constants C > 0 and Cϱ > 0 for any ϱ > 0 such that Cϱ1ωϱ ≤f0 ≤C, (6.1.21) with ωϱ := {x ∈Ω; δ(x) > ϱ} and δ(x) := d(x, ∂Ω) is the distance to the boundary function. We can also get a constructive argument for (H2) by asking that condition (i) in Lemma 2.1.4 holds. We may for instance verify that the dual counterpart of the above constructive argument holds when (c + ∂iβi)−∈M1 and bi + βi + ∂jaji ∈M1. More precisely, we establish in a similar way as above that the solution to the problem ψ0 ∈H1 0(Ω), (κ1 −L∗)ψ0 = h0, (6.1.22) satisﬁes κ0ψ0 ≤L∗ψ0 ≤κ1ψ0, (6.1.23) for some constructive constants κ0 ≤κ1. Similarly as above again, there exist constructive constants C > 0 and Cϱ > 0 for any ϱ > 0 such that Cϱ1ωϱ ≤ψ0 ≤C. (6.1.24) 96 CHAPTER 6. APPLICATION TO DIFFUSION MODELS Third constructive argument for (H2). We write Lf = aij∂2 ijf + ˜ bi∂if + ˜ cf, (6.1.25) with ˜ bi := bi + ∂jaji + βi and ˜ c := c + ∂iβi. We further assume ˜ bi, ˜ c ∈L∞. In that case, we may also obtain an explicit lower bound on λ1 by proceeding in the following way. We deﬁne f0(x) := χ(\|x\|) with χ ∈C1 c (R+) ∩W 2,∞(R+), 1[0,1/3] ≤χ ≤1[0,1], χ′ ≤0 on [0, 1], χ(s) := n2(1 −s)2/2 on [ιn, 1], ιn := 1 −1/(2n), for some n ≥1 to be chosen. As a consequence, χ′′ = n2 on [ιn, 1], \|χ′\|≤n on [ιn, 1] and χ ≥1/2 on [0, ιn]. Denoting s := \|x\|, we compute Lf0 = aij{χ′′(s)ˆ xiˆ xj + χ′(s)δij −ˆ xiˆ xj s } + ˜ b(x) · ˆ xχ′(s) + ˜ c(x)χ(s). For n large enough, we get Lf0 ≥ n2ν −n2A −nB −C ≥0 on B1\Bιn, Lf0 ≥ −A{∥χ′′∥L∞+∥χ′(s)/s∥L∞} −B∥χ′∥L∞−C ≥κ0χ on Bιn, with A := ∥a∥L∞(B1), B := ∥˜ b∥L∞(B1), C := ∥˜ c∥L∞(B1) and κ0 ∈R−. As a conclusion, we have again established condition (ii) in Lemma 2.1.4, so that condition (H2) holds. Fourth constructive argument for (H2). We present a last situation when we are able to prove a quantitative version of condition (H2). We assume that a ∈C0(¯ Ω), divβ ∈Lr/2, as well as ˜ bi ∈Lr and ˜ c ∈Lr/2 in the deﬁnition of (6.1.25). We deﬁne h0 and f0 as in the second constructive argument for (H2), so that (6.1.18) holds. Choosing p ∈(1, 2) deﬁned by 1/p := 1/r + 1/2 > 2/r + 1/2∗, we observe that ∥κ1f0 −˜ bi∂if0 −˜ cf0 −h0∥Lp ≲ κ1∥f0∥L2+∥˜ bi∥Lr∥∂if0∥L2+∥˜ c∥Lr/2∥f0∥L2∗+∥h0∥L2 ≲ ∥h0∥L2, from equation (6.1.19) and the coercivity estimate (6.1.9). From the Calderon-Zygmond regularity theory or [104, Theorem 9.14], we also know that ∥f0∥W 2,p(Ω)≲∥aij∂2 ijf0∥Lp(Ω). (6.1.26) Writing aij∂2 ijf0 = κ1f0 −˜ bi∂if0 −˜ cf0 −h0 and using the two above estimates, we deduce ∥f0∥W 2,1(Ω)≲∥h0∥L2(Ω). (6.1.27) On the other hand, from (6.1.8) and the Poincaré inequality, we have 1 = ∥h0∥2 L2= ⟨(κ1 −L)f0, h0⟩≲∥∇f0∥L2∥∇h0∥L2. Together with the estimate (6.1.27) and the Gagliardo-Niremberg inequality ∥∇f∥L2≲∥D2f∥1/2 L1 ∥f∥1/2 L∞, we obtain a lower bound ∥f0∥L∞≥C0 > 0. We then conclude as in the second constructive argument for (H2). Condition (H3). Because of Rellich-Kondrachov theorem on the compact embedding H1 0 ⊂L2, the mapping (λ −L)−1 : L2 →L2 is compact for any λ ≥κ1. As a consequence, introducing the splitting L = A + B with A := κ1 −κB, κB ∈R arbitrary, the operator RB(λ) = (λ + κ1 −κB −L)−1 is bounded unformly on λ ≥κB and it is compact for any λ ≥κB. We deduce from Lemma 2.1.8-(2) that (H3) holds for both the primal and the dual problems. 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 97 We may thus apply Theorem 2.1.20 and deduce the existence of a solution (λ1, f1, φ1) to the ﬁrst eigentriplet problem λ1 ∈R, 0 ≤f1 ∈H1 0, Lf1 = λ1f1, 0 ≤φ1 ∈H1 0, L∗φ1 = λ1φ1, (6.1.28) where both equations must be understood in the variational sense as a consquence of the discussion at the end of the proof of condition (H1). Condition (H4). The strong maximum principle holds as already mentioned in the paragraph dedicated to condition (H2). As a consequence and thanks to Theorem 3.3.11, we know that the ﬁrst eigentriplet problem (6.0.3) has a unique solution (λ1, f1, φ1) which satisﬁes f1 > 0, φ1 > 0, N(L −λ1)k = Span(f1) and N(L∗−λ1)k = Span(φ1) for any k ≥1. Condition (H5). Consider f ∈D(L∞) such that 0 < \|f\|∈D(L∞) and L\|f\|= ℜe(signf)Lf, so that multiplying both term of the equation by \|f\| and integrating, we have ℜe⟨Lf, ¯ f⟩= ⟨L\|f\|, \|f\|⟩. We next compute ℜe⟨Lf, ¯ f⟩= − Z Ω akjℜe(∂jf∂k ¯ f) + Z Ω (bk −βk)ℜe( ¯ f∂kf) + Z Ω c\|f\|2, and ⟨L\|f\|, \|f\|⟩= − Z Ω akj∂j\|f\|∂k\|f\|+ Z Ω (bk −βk)ℜe( ¯ f∂kf) + Z Ω c\|f\|2, where in the last equality, we have used that ∂k\|f\|= 1 \|f\|ℜe( ¯ f∂kf). From the three above equations, we deduce Z Ω akj[∂j\|f\|∂k\|f\|−ℜe(∂jf∂k ¯ f)] = 0. Introducing the real and complex part decomposition f = u + iv, and similarly as in [138, Proof of Theorem 5.1], we next compute ∂j\|f\|∂k\|f\|−ℜe(∂jf∂k ¯ f) = 1 \|f\|2 [uv(∂ku∂jv + ∂kv∂ju) −u2∂jv∂kv −v2∂ju∂ku] = 1 \|f\|2 (u∂jv −v∂ju)(u∂kv −v∂ku), so that from the ellipticity condition on a, we have u∂kv−v∂ku = 0 a.e. on Ω. On the other hand, from De Girogi-Nash-Moser regularity estimates (6.1.14) and (6.1.17), f has Hölder regularity. In particular both functions u and v are continuous. Because \|f\|̸≡0, one of the two function is not identically vanishing, say for instance v ̸≡0. There exists some points x0 ∈Ωsuch that v(x0) ̸= 0, say for instance v(x0) > 0. Denoting by ω the connected component of the set {x ∈Ω; v(x) > 0} containing x0, we have ∇(u/v) = 0 on ω. Hence u = α v on ω for some α ∈R, which implies that there exists σ ∈S1 such that f = σ\|f\| on ω. If ω ̸= Ω, we would have \|f\|= 0 on ∂ω ∩Ω̸= ∅, which would be a contradiction with the fact that \|f\|> 0. We conclude that ω = Ωand thus that f = σ\|f\|, which is nothing but the reverse Kato’s inequality condition (H5). At this stage, we may use Theorem 4.4.14, in order to get the conclusion (C3) on the triviality of the boundary punctual spetrum. In order to go one step further and establish the asymptotic stability of f1, we may use the two following approaches which are consequences respectively of Lemma 6.1.1 and Lemma 6.1.2. 98 CHAPTER 6. APPLICATION TO DIFFUSION MODELS Lemma 6.1.1. For any R > 0, the set K := {f ∈D(L); [f] ≤R, [Lf] ≤R} is strongly compact in L1 loc(Ω), where [g] := ∥g∥L1 φ1. Proof of Lemma 6.1.1. Consider f ∈K so that f ∈H1 0(Ω) and ∂i(aij∂jf) + bi∂if + ∂i(βif) + cf = g ∈L2(Ω). From the renormalization theory of elliptic equations and the GRE trick (see for instance and the references therein) for any renormalizing function H ∈C2(R), there holds H′′(u)f1φ1a∇u · ∇u = div(aφ1∇(H(u)f1)) −div(f1H(u)a∇φ1) + div((b + β)H(u)f1φ1) + gH′(u)f1φ1, with u := f/f1. Considering H ∈W 2,∞the even (and convex) function such that H(0) = 0 and H′′ := 1[n,n+1], so that in particular \|H′(s)\|≤1, and integrating the previous equation, we deduce ν Z \|u\|∈[n,n+1] \|∇u\|2f1φ1 ≤ Z \|g\|f1φ1 ≤∥f1∥L∞R. We proced along the line of the proof of [32, Theorem 1]. For a ﬁxed ω ⊂⊂Ω, we deﬁne Bn := {x ∈ω; \|u(x)\|∈[n, n + 1]}. Using that f1 > 0 and φ1 > 0, there exists a constructive constant Cω,R > 0 such that Z Bn \|∇u\|2≤C2 ω, ∀n ≥0. From the Cauchy-Schwarz inequality, we have Z Bn \|∇u\|≤Cωmeas(Bn)1/2, ∀n ≥0. (6.1.29) On the other hand, denoting by 1∗:= d/(d −1) the Sobolev exponent, we have Z Bn \|∇u\|≤Cω,R n−1∗Z Bn \|u\|1∗1/2. Summing up and using the Cauchy-Schwarz inequality again, we have X n≥1 Z Bn \|∇u\| ≤ Cω,R X n≥1 n−1∗1/2X n≥1 Z Bn \|u\|1∗1/2 ≤ Cω,R X n≥1 n−1∗1/2∥u∥1∗/2 L1∗. Together with (6.1.29) for n = 0, we deduce ∥∇u∥L1(ω)≤C′ ω,R(1 + ∥∇u∥1∗/2 L1(ω)). Because 1∗/2 ≤3/4 < 1 (recall that d ≥3), we can kill the last term, and we obtain the estimate ∥∇(f/f1)∥L1(ω)≤C′′, ∀f ∈K, for some constant C′′ := C′′ ω,R > 0. We classically conclude thanks to the Rellich-Kondrachov theorem. From the above lemma and Theorem 4.5.20, we deduce that e S(t)f →⟨f, φ1⟩f1 in the L1 φ1 norm sense as t →∞for any f ∈L2(Ω). The alternative approach is based on the following result. 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 99 Lemma 6.1.2. Setting κ := κ0 −1, there exist A, α, R > 0 such that (i) supz∈∆κ⟨y⟩α∥RB(z)∥B(L2;H1 0)+ supz∈∆κ\BR∥RL(z)∥B(L2;H1 0)< ∞, (ii) Σ(L) ∩∆κ ⊂Σd(L) ∩BR, where B := L −A and z = x + iy, x, y ∈R. Proof of Lemma 6.1.2. Let us consider an a priori solution to the stationary problem f ∈H1 0, z = x + iy ∈∆κ, (L + z)f = g ∈L2. This one satisﬁes − Z (a∇f + βf) · ∇¯ f + Z b · ∇f ¯ f + (c + z)\|f\|2 = Z g ¯ f . Using the elliptic condition, the Cauchy-Schwarz inequality and triangular inequalities, we get Z g ¯ f ≥ Z a∇f∇¯ f + ((c + x)+ + iy)\|f\|2 − Z b · ∇f ¯ f −βf · ∇¯ f + (c + x)−\|f\|2 ≥ ν 2∥∇f∥2 L2+(\|y\| 2 −x−)∥f∥2 L2−∥(\|b\|+\|β\|)f∥L2∥∇f∥L2−∥√c−f∥2 L2. Using next similar arguments and those introduced in the paragraph dedicated to condition (H1) and with similar deﬁnition for the constant M := M(b, β, c) > 0, we deduce Z g ¯ f ≥ (\|y\| 2 −x−−M)∥f∥2 L2+ν 4∥∇f∥2 L2. Deﬁning the sectorial set S := {z = x + iy ∈C; \|y\|> 2x−+ M}, we have established the a priori estimates ∥f∥L2 ≤ (\|y\| 2 −x−−M)−1/2∥g∥L2, ∥∇f∥L2 ≤ 2ν−1/2(\|y\| 2 −x−−M)−1/4∥g∥L2, for any z ∈S. We classically and immediately deduce that ρ(L) ⊃S and the resolvent estimate ∥RL(z)∥B(L2,H1 0)≲(\|y\| 2 −x−−M)−1/2 + (\|y\| 2 −x−−M)−1/4 for any z ∈S, and in particular the estimate (i) holds true. On the other hand, because L has compact resolvent as established just above or during the proof of (H3) and using the Fredholm alternative, we have Σ(L) = Σd(L) and Σ(L) ∩∆κ is ﬁnite for any κ ∈R, what is nothing but the property (ii). From the above lemma and Theorem 4.6.27 or Theorem 4.6.29, we deduce that e S(t)f → ⟨f, φ1⟩f1 in the L2 norm sense as t →∞for any f ∈L2(Ω) with exponential rate. We may summarize our analysis in the following result. Theorem 6.1.3. Consider the elliptic operator (1.2.8) in a bounded domain and assume that the coeﬃcients satisfy (6.0.2), (6.1.5), (6.1.6) and (6.1.18). Then the conclusions (C3) holds as well as (CE2) in L1 φ1 norm and (CE3) in L2 with non constructive rate. 100 CHAPTER 6. APPLICATION TO DIFFUSION MODELS It is however worth emphasizing again that the above approach is deﬁnitively not constructive. We propose now an alternative approach which is constructive. Quantitative estimate of stability. Using the Doblin-Harris type approach presented in Section 5, we are able to establish a rate of convergence to the principal dynamic, at least in a regular framework. We thus make some regularity assumptions on Ωand additional regularity assumptions on the coeﬃcients. - For the domain, we asssume that that there exist constant rΩ> 0 such that for any x ∈Ω there is y ∈Ωsuch that x ∈B(y, rΩ) ⊂Ω, in particular, for any x ∈∂Ωthere is y ∈Ωsuch that x ∈∂B(y, rΩ), B(y, rΩ) ⊂Ω. We also assume that Ωis C1,1. - For the coeﬃcients, we assume aij ∈C(¯ Ω), ˜ bi, ˜ c ∈L∞(Ω), where ˜ bi and ˜ c are deﬁned in (6.1.25). Theorem 6.1.4. Consider the elliptic operator (1.2.8) in a bounded domain and assume that the assumptions of Theorem 6.1.3 hold togetrher with the additional regularity assumptions on the coeﬃcients and the boundary. Then the conclusion (CE3) holds with constructive exponential rate. The proof of Theorem 6.1.4 follows from Theorem 5.3.3. We split the proof into several steps. - Step 1. Regularity estimates. Thanks to De Giorgi-Nash-Moser regularity technique for parabolic equations developed for instance in (in Russian), [237, Theorem 1.3, Theorem 2.2] as well as more recently in [137, Lemma 2.7] and [114, Theorem 1.1], there exists α = α(aij) ∈(0, 1) and for any T1 > T0 > 0 and any ϱ ∈(0, 1), there exist constructive constants Ci = Ci(∥f∥L∞ t L2 x, T, τ, r) such that any solution f ∈L∞(0, ∞; L2(Ω)) to the parabolic equation ∂tf = Lf satisﬁes ∥f∥L∞([T0,T1]×Ω)≤C1, ∥f∥Cα([T0,T1]×ωϱ)≤C2, (6.1.30) with ωr := {x ∈Ω; d(x, ∂Ω) > r}. More precisely, in order to establish the second estimate in (6.1.30) with constructive constant, one may observe that the proof of [114, Proposition 2.4] may be repeated in order to get that solutions to the parabolic equation considered in the present framework fall into De Giorgi classes as deﬁned in [114, Deﬁnition 2.3], and thus [114, Theorem 1.1] applies. On the other hand, in this context and because of the regularity assumptions, we may establish a more accurate regularity estimate. More precisely, by gathering the Sobolev inequality and the Calderon-Zygmond estimate (6.1.26), we obtain the classical constructive regularity estimate ∥u∥C0,1(Ω)≲∥u∥W 2,d+1(Ω)≲∥(κ1 −L)u∥Ld+1(Ω), (6.1.31) see for instance Theorem 7.10, Theorem 7.25 and Lemma 9.17 in . Iterating the same kind of arguments, we get ∥u∥C0,1(Ω)≤C∥(κ1 −L)ku∥L2(Ω), (6.1.32) with constructive constants C and k. - Step 2. Harnack estimate. We claim that for any 0 ≤t0 < T and ϱ > 0, there exist a constant CH > 0 such that, for any f0 ∈L2, the associated solution f := SLf0 satisﬁes sup ωϱ ft0 ≤CH inf ωϱ fT . (6.1.33) The proof mainly follows form Aronson-Serrin (see also [185, 128, 130, 143, 129, 236, 237, 144] for similar results). First, we know from [10, Theorem 3] that max Q∗(ρ) f ≤C min Q(ρ) f, (6.1.34) 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 101 for any ρ > 0, t > 0 such that Q∗(3ρ) ⊂(0, ∞) × Ω, where Q(ρ) := [t −ρ2, t] × C(ρ), Q∗(ρ) := [t −8ρ2, t −7ρ2] × C(ρ) and C(ρ) is a cube with length ρ. To avoid technical issues we assume that wϱ is convex. In other case, the geometrical condition given above implies that there is N ∈Z+ such that any two points x, y ∈Ωcan be connected by a polygonal path of at most N segments, and we can argue as follows for any segment. We deﬁne D := supa,b∈Ωd(a, b) the diameter of Ωand we choose r′ < ϱ/7 such that 7(⌊D 2r′ ⌋+ 1)(r′)2 < T −t0. For any x, y ∈ωϱ, we also deﬁne Nc = ⌊\|x−y\| r′ ⌋. Since ωϱ is convex, r′ < ϱ/7, we have that the familly of cubes {C(xi, 2r′)}i=0,Nc of center xi and length 2r′ for xi = x + (x−y)i Nc satisfy that C(xi, 6r′) ⊂Ωand C(xi, 2r′) ∩C(xi+1, 2r′) ̸= ∅for any i = 0, . . . , Nc. As a consequence, we can apply Aronson-Serrin estimate (6.1.34) for each cube to obtain max C(xi,2r′) fti ≤C2r′ min C(xi,2r′) fti+1, with ti = t0 + 7i(2r′)2. Taking yi ∈C(xi, 2r′) ∩C(xi+1, 2r′), we deduce max C(xi,2r′) fti ≤C2r′ min C(xi,2r′) fti+1 ≤C2r′fti+1(yi) ≤C2r′ max C(xi+1,2r′) fti+1 ≤C2 2r′ min C(xi+1,2r′) fti+2. By induction, we obtain ft0(x) ≤ max C(x1,2r′) ft1 ≤CNc 2r′ min C(xNc,2r′) ftNc ≤CNc 2r′ftNc(y), with tNc = t0 + 7Ncr ′2 ≤T. Note that in any case the constant C2r′ is the same since it only depends on the length 2r′ and the coeﬃcient of the equation. We have thus established (6.1.33) with CH := C ⌊D 2r′ ⌋+1 2r′ . On the other hand, we state an improved version of the already mentioned stationary Harnack inequality. Because of the interior ball condition the Hopf Lemma (see for instance the proof of [104, Lemma 3.4]) claims that for any ϱ ∈(0, rΩ/2] there exists a constructive constant α > 0 such that if u ∈W 2,p(Ω), p > d, is such that u ≥1ωϱ, (κ1 −L∗)u ≥0, then u satisﬁes u ≥χ(x) := e−α(2ϱ−δ(x))2 −e−α(2ϱ)2 on ωc ϱ. (6.1.35) Let us give two applications of the above sharp regularity and positivity estimates. First, recalling (6.1.24) and using (6.1.31) and (6.1.35), we deduce that there exist two constructive constants ci ∈(0, ∞) such that c0δ ≤ψ0 ≤c1δ on Ω. (6.1.36) Consider now f1 ∈H1 0(Ω) the positive ﬁrst eigenfunction with normalization ∥f1∥L2= 1. Using the estimate of regularity (6.1.32) on the iterated equation (κ1 −L)kf1 = (κ1 −λ1)kf1, we have ∥f1∥L∞(Ω)≤∥f1∥C0,1(Ω)≤C1, for some constructive constant C1 ∈(0, ∞). Next using the elementary inequality 1 = Z Ω f2 1 ≤∥f1∥L∞∥f1∥L1≤C1∥f1∥L1, 102 CHAPTER 6. APPLICATION TO DIFFUSION MODELS we deduce \|Ω\|sup ωϱ f1 ≥ Z ωϱ f1 = ∥f1∥L1− Z ωc ϱ f1 ≥ 1/C1 −C1\|ωc ϱ\|≥1/(2C1), by choosing ϱ ∈(0, rΩ/2) small enough. Then, from the Harnack inequality [104, Corollary 8.21], we deduce inf ωϱ f1 ≥CH sup ωϱ f1 ≥CH(2C1\|Ω\|)−1. Finally, from the above Hopf lemma and the above Lipschitz continuity, we have established c0δ ≤f1 ≤c1δ on Ω, (6.1.37) for two constructive constants ci ∈(0, ∞). The same arguments on the normalized and positive ﬁrst dual eigenfunction φ1 lead to the samùe estimùate c0δ ≤φ1 ≤c1δ on Ω. (6.1.38) In particular, for any such ϱ ∈(0, rΩ/2), we have ⟨φ1, 1ωϱ⟩≥rϱ, (6.1.39) with constructive constant rϱ, what is nothing but condition (5.3.9) in the Harris theorem that we will use below. - Step 3. Splitting of L. We introduce the splitting L = A + B, with Af = M 1ωϱf, M ≥0 large enough and ϱ > 0 small enough that we ﬁx just below. Using (6.1.9), we observe that (Bf, f)L2 = (Lf, f)L2 −M ∥f∥L2+M Z ωc ϱ f2 ≤ −ν 4∥∇f∥2 L2+(κ1 −M )∥f∥2 L2+M \|ωc ϱ\|4/d∥f∥2 L2∗≤κ0∥f∥2 L2, by choosing ﬁrst M ≥κ1 −κ0 and next ϱ > 0 small enough in order to be able to throw away the last term using the negative ﬁrst term and the Sobolev inequality. We deduce SB(t) : L2 →L2 with bound O(eκ0t). (6.1.40) On the other hand, denoting ft := SL(t)f for f ∈L2(Ω) and recalling that ψ0 deﬁned by (6.1.22) satisﬁes (6.1.23), we have d dt Z \|ft\|ψ0 ≤ Z L\|ft\|ψ0 ≤ Z \|ft\|L∗ψ0 ≤κ1 Z \|ft\|ψ0, so that Z \|ft\|ψ0 ≤eκ1t Z \|f0\|ψ0. (6.1.41) Arguing in the same way for SB and using (6.1.36), we have established SL(t), SB(t) : L1 δ →L1 δ with bound O(eκ1t). (6.1.42) For a solution to the evolution equation ∂tf = Cf, C = L or C = B, we also classically compute d dt Z f2ψ0 = 2 Z (Cf)fψ0 = −2 Z (∇f · a∇f)ψ0 + Z f2C∗ψ0. 6.1. DIFFUSION WITH ROUGH COEFFICIENTS IN A BOUNDED DOMAIN 103 Thanks to (6.1.23) again, we have d dt Z f2ψ0 ≤−2ν Z \|∇f\|2ψ0 + κ1 Z f2ψ0, (6.1.43) from what we deduce SL(t), SB(t) : L2(δ) →L2(δ) with bound O(eκ1t/2). In the sequel, we will need the following version of Nash inequality. Lemma 6.1.5 (weighted Nash inequality). There exists a constructive constant CN such that ∥f∥L2(δ)≤CN∥∇f∥ d+1 d+2 L2(δ)∥f∥ 1 d+2 L1 δ , ∀f ∈H1(δ). (6.1.44) Proof of Lemma 6.1.5. For ε > 0, we deﬁne fε(x) := 1 δε(x) Z B(x,ε) f(y) δ(y)dy, δε(x) = δ(B(x, ε)) := Z B(x,ε) δ(y)dy, and B(x, ε) := {y ∈Ω; \|x −y\|< ε}. It is worth emphasizing that εd+1 ≲δε(x) ≲εd, ∀ε > 0. (6.1.45) For f ∈H1(δ), we compute ∥f −fε∥2 L2(δ) = Z Ω 1 δε(x) Z B(x,ε) (f(y) −f(x)) δ(y)dy 2δ(x)dx ≤ Z Ω Z Ω 1\|y−x\|≤ε\|f(y) −f(x)\|2 δ(y) δε(x)δ(x)dxdy ≤ ε2 Z 1/2 0 Z Ω Z Ω \|∇f((1 −t)x + ty)\|2 δ(y) δε(x)δ(x)dxdydt +ε2 Z 1 1/2 Z Ω Z Ω \|∇f((1 −t)x + ty)\|2 δ(y) δε(x)δ(x)dxdydt ≲ ε2 Z 1/2 0 Z Ω Z Ω \|∇f(z)\|2 δ(y) εd+1 dy2δ(z) dz (1 −t)d dt +ε2 Z 1 1/2 Z Ω Z Ω \|∇f(z)\|2 2δ(z) εd+1 δ(x)dz td dtdx where for the last inequality we have used the ﬁrst inequality in (6.1.45), the fact that δ(x) ≤2δ(z) when 0 < t < 1/2 and the fact that δ(y) ≤2δ(z) when 1/2 < t < 1. Using the second inequality in (6.1.45), we straightforwardly obtain ∥f −fε∥2 L2(δ)≤C1 ε ∥∇f∥2 L2(δ), ∀ε > 0, for a constant C1 > 0. On the other hand, we also observe that ∥fε∥L∞≤C2 εd+1 ∥f∥L1 δ. Writing now f2 = f(f −fε) + ffε and using the above two estimates, we deduce ∥f∥2 L2 δ ≤ ∥f∥L2 δ ∥f −fε∥L2 δ+∥f∥L1 δ ∥fε∥L∞. ≤ ∥f∥L2 δ C1 ε1/2∥∇f∥L2 δ+C2 ε−d−1 ∥f∥2 L1 δ ≤ 1 2∥f∥2 L2 δ+C1 2 ε ∥∇f∥2 L2 δ+C2 ε−d−1 ∥f∥2 L1 δ, 104 CHAPTER 6. APPLICATION TO DIFFUSION MODELS and we obtain the weighted Nash inequality (6.1.44) by choosing ε := (∥f∥2 L1 δ/∥∇f∥2 L2 δ)1/(d+2). Deﬁning u := Z \|ft\|ψ0dxe−2κt, v := Z f2 t ψ0dxe−2κt, with κ := κ1+, coming back to (6.1.43) and using (6.1.36), the Nash inequality (6.1.44) and the estimate (6.1.41), we get v′(t) ≤ −2νc0 Z \|∇ft\|2δe−2κt ≤ −2νc0C −2d + 2 d + 1 N ∥ft∥2 L2(δ)e−2κtd + 2 d + 1 ∥ft∥2 L1(δ)e−2κt 1 d + 1 ≤ −C v(t)1+α u(0)2α , with C := 2νC −2d + 2 d + 1 N c 1+2d + 2 d + 1 0 c − 2 d + 1 1 and α := 1/(d + 1). Integrating in time, we deduce v(t) ≤ α1/α C1/α u(0)2 t1/α , ∀t > 0. We have thus established that there exist constructive constants K > 0 and κ ≥0 such that ∥SC(t)f∥L2(ψ0)≤K eκt t(d+1)/2 ∥f∥L1(ψ0), ∀f ∈L1(ψ0). (6.1.46) From thar last result, the estimates (6.1.36) and the properties of A, we deduce that for N ≥1 large enough (SBA)(∗N) : L1(δ) →L2(δ) with bound O(eκt). (6.1.47) We refer to [113, Proposition 3.9], [175, Proposition 2.5] and [138, Lemma 2.4] for details. - Step 4. Lyapunov condition. We may next write e SL = V + W ∗e SL, with V := e SB + · · · + ( e SBA)(∗(N−1)), V := ( e SBA)(∗N). On the one hand, using that A : L2 →L2 is bounded and (6.1.40), we deduce that V : L2 →L2, with bound O(eκt), for any κ ∈(κ0 −κ1, 0). On the other hand, using that A : L2 δ →L2 is bounded as well as (6.1.42) for SL, (6.1.47), (6.1.38), (6.1.36) and (6.1.40), we deduce that W ∗e SL : L1 φ1 →L2, with bound O(eκ′t), for any κ′ > κ1 −κ0. We may thus ﬁx t = T large enough such that the following Lyapunov inequality holds ∥e SL(T)f∥L2≤1 2∥f∥L2+MT ∥f∥L1 φ1, (6.1.48) which is nothing but (5.3.8) in the hypothesis of the Harris theorem. 6.2. DIFFUSION IN RD WITH STRONG POTENTIAL CONFINEMENT 105 - Step 5. Harris condition Let R > 0 and consider 0 ≤f0 ∈L2 such that ∥f0∥2≤R⟨f0, ψ0⟩. We set e ft := e−λ1tSL(t)f0. From the ﬁrst inequality in (6.1.23), we have d dt⟨e ft, ψ0⟩= ⟨e ft, (L∗−λ1)ψ0⟩≥−(λ1 −κ0)⟨e ft, ψ0⟩, and then, thanks to Gronwall lemma again, we obtain, ⟨e ft, ψ0⟩≥e−(λ1−κ0)t⟨f0, ψ0⟩. This estimate, together with the previous step, shows that Z ωϱ e ft0(x)ψ0dx = Z Ω e ft0(x)ψ0dx − Z ωc ϱ e ft0(x)ψ0dx ≥e−(λ1−κ0)t0⟨f0, ψ0⟩−∥e ft0∥2∥ψ0∥∞\|ωc ϱ\|1/2 ≥e−(κ1−κ0)t0⟨f0, ψ0⟩−e(κ1−κ0)t0∥f0∥2∥ψ0∥∞\|ωc ϱ\|1/2 ≥ e−(κ1−κ0)t0 −Re(κ1−κ0)t0∥ψ0∥∞\|ωc ϱ\|1/2 ⟨f0, ψ0⟩. Choosing ϱ > 0 small enough, we get Z ωϱ e ft0(x)ψ0dx ≥γ⟨f0, ψ0⟩, γ := 1 2e−(λ1−κ0)t0. As a consequence, there is xf t0 ∈ωϱ such that e ft0(xf t0) ≥ 1 \|ωϱ\| Z ωϱ e ft0(x)dx ≥ 1 \|Ω\|c1ϱ Z ωϱ e ft0(x)ψ0dx ≥ γ \|Ω\|c1ϱ⟨f0, ψ0⟩. On the other hand, from the Harnack inequality (6.1.33) established in Step 2, we know that for any T > t0 exits CH such that e ft0(xf t0) ≤sup ωϱ e ft0 ≤CH inf ωϱ e fT . The two last estimates together with (6.1.38) and (6.1.36) imply the Harris type estimate e fT = e S(T)f0 ≥gR⟨f0, φ1⟩, (6.1.49) with gR := c0γ CH\|Ω\|c2 1ϱ1wϱ, which is nothing but (5.3.7) in Harris theorem. - Step 6. Conclusion. Because of the constructive estimates (6.1.39), (6.1.48) and (6.1.49), we may apply the Harris type Theorem 5.3.3, and we conclude to the exponential stability (4.6.27) in the norm of L2(Ω) with constructive constants. 6.2 Diﬀusion in Rd with strong potential conﬁnement We consider in this section the elliptic operator Lf := ∆f + b · ∇f + cf, f ∈H1(Rd), (6.2.50) with b ∈L∞ loc(Rd), c ∈L2 loc(Rd) and a conﬁnement condition that we impose through the properties of the potential function c, which is roughly speaking c →−∞as \|x\|→∞. More precisely, we assume σi+ ∈Ld/2, meas{σi ≥K} < ∞, ∀K < 0, (6.2.51) 106 CHAPTER 6. APPLICATION TO DIFFUSION MODELS with either σ1 := c + \|b\|2/κ for some constant κ ∈(0, 4) or either σ2 := c + divb/2. When we assume that c ∼−\|x\|γ and b ∼x\|x\|β−1 as \|x\|→∞, the condition (6.2.51) for σ1 is reached when γ > max(0, 2β) or γ = 2β > 0 and some conditions on the constants involved in the behavior of the coeﬃcients. In that context, the condition (6.2.51) for σ2 is more general since it is reached when γ > max(0, β −1) or γ = β −1 > 0 and some conditions on the constants involved in the behavior of the coeﬃcients. A similar framework is considered in and for the reader convenance we just brieﬂy check that it falls in the framework developed before by slightly modifying the arguments presented in the previous section. The integrability conditions on b and c may be probably weaken. For the sake of clarity we do not follow this line of research but rather focus on the new arguments which are necessary in order to deal with the unbounded domain Ω= Rd. Condition (H1). The deﬁnition of the operator is still made through the formula (6.1.8). Under assumption (6.2.51) on σ1, denoting θ1 := 1 −κ/4 and proceeding exactly as in the previous section during the proof of (6.1.9), for any f ∈H1(Rd) and λ ∈R, we have ⟨(λ −L)f, f⟩ = Z Rd\|∇f\|2+ Z Rd f b · ∇f + Z Rd(λ −c)f2 ≥ θ1 Z Rd\|∇f\|2+ Z Rd(λ −σ1)f2, by using successively the Cauchy-Schwarz inequality and the Young inequality. On the other hand, under assumption (6.2.51) on σ2, denoting θ2 := 1, for any f ∈H1(Rd) and λ ∈R, we write ⟨(λ −L)f, f⟩ = θ2 Z Rd\|∇f\|2+ Z Rd(λ −σ2)f2, by performing one integration by part in the previous equation. In both cases, for and any M > 0, proceeding again as in the previous section during the proof of (6.1.9), and denoting from now on σ = σi, θ = θi we have ⟨(λ −L)f, f⟩ ≥ θ 2∥∇f∥2 L2+∥√σ−f∥2 L2+(λ −M)∥f∥2 L2+(θCS 2 −∥σ1σ≥M∥Ld/2)∥f∥2 L2∗, by using the Sobolev inequality (with associated constant CS) and the Holder inequality. Taking M > 0 large enough, and next κ1 > 0 large enough, we ﬁnally obtain ⟨(λ −L)f, f⟩≥θ 2∥∇f∥2 L2+∥√σ−f∥2 L2+∥f∥2 L2, ∀λ ≥κ1. (6.2.52) With the same arguments as in the previous section, we conclude that L is the generator in L2 of a positive semigroup SL, so that (H1) holds. Condition (H2). We may for instabce use the third constructive argument presented in section 6.1 and we establish ∃f0 ∈H1 0{0}, f0 ≥0, ∃κ0 ∈R, Lf0 ≥κ0f0. That is condition (ii) in Lemma 2.1.4, so that condition (H2) holds. Condition (H3). We introduce again the splitting L = A+B with A := κ1 −κ0 +1, so that from (6.2.52), the operator λ −B = (λ −κ0 + 1) + (κ1 −L) is invertible for any λ ≥κB := κ0 −1. We claim that the operator (λ −B)−1 is compact for any λ ≥κB. For that purpose, let us consider a sequence (fn) such that (λ −B)fn is bounded in L2 and we have to prove that (fn) is 6.2. DIFFUSION IN RD WITH STRONG POTENTIAL CONFINEMENT 107 relatively strongly compact. When condition (6.2.51) holds and because of the estimate (6.2.52) and the very deﬁnition of B, we have θ 2∥∇fn∥2 L2+∥√σ−fn∥2 L2+∥fn∥2 L2≤C, (6.2.53) for some constant C ∈R+. Because of the Rellich-Kondrachov theorem, we just have to show that lim R→∞sup n Z Bc R f2 n = 0. But that last convergence may be established using the assumption (6.2.51) in the following way. We write Z Bc R f2 n = Z Bc R∩{σ≥K} f2 n + Z Bc R∩{σ<K} f2 n ≤ ∥fn∥ d−2 d L2∗[meas(Bc R ∩{σ ≥K})] 2 d + 1 \|K\| Z σ−f2 n, for any K < 0, by using the Holder inequality. Using next the Sobolev inequality, the estimate (6.2.53) and the assumption (6.2.51), we deduce lim sup R→∞ Z Bc R f2 n ≲ lim sup R→∞ inf K<0 n [meas(Bc R ∩{σ ≥K})] 2 d + 1 \|K\| o = 0, and the claim is proved. As a consequence, we may apply Lemma 2.1.8-(2) and we deduce that (H3) holds for both the primal and the dual problems. Condition (H4). As in [138, Proposition 5.4], we establish the strong maximum principle by exhibiting a barrer function and using Lemma 3.2.9. An alternative argument should be to adapt the proof based on the Harnack inequality as presented in the previous section. Let us then consider f ∈D(Lk) ∩X+{0} such that (λ −L)f ≥0 with k large enough (k > d/2 must be suitable) and λ ≥λ1 large enough but ﬁxed (λ ≥κ1 is suitable). Using a very classically bootstrap argument based on iterated application of the Calderon-Zygmond elliptic regularity theorem and the Morrey estimate, we have f ∈C(Rd). By assumption, there thus exist x0 ∈Rd, and two constants τ, r > 0 such that f ≥τ on B(x0, r) and we take choose x0 = 0 in order to simplify the notations. We next ﬁx R > r and we observe that the function g(x) := τ ∗(g0(\|x\|) −g0(R)), g0(s) := exp(σr2/2 −σs2/2), satisﬁes (τ ∗)−1(λ −L)g = (λ −c)(g0 −g0(R)) + (dσ −σb · x −σ2r2) g0 ≤ [2(\|λ\|+∥c∥L∞(BR)) + σ(d + ∥b · x∥L∞(BR)) −σ2r2] g0 ≤0 on O := B(0, R)\B(0, r) for σ > 0 large enough. We next ﬁx τ ∗such that g = τ on ∂B(x0, r). We also observe that from (6.2.52), λ −L is coercive on O, in the sense that ∀h ∈H1 0(O) ((λ −L)h, h)L2(O) ≥∥h∥L2(O). In particular, λ −L satisﬁes the weak maximum principle as explained in the proof of (6.1.10). Arguing as in the proof of Lemma 3.2.9, we deduce that f ≥g > 0 on O, what we also see directly by observing that h := (g −f)+ ∈H1 0(O), (λ −L)h ≤0 and using the weak maximum principle implies h ≤0, thus h ≡0 and ﬁnally f ≥g. Because R > r can be choosen arbitrarily large, we conclude with f > 0 on Rd. Condition (H5). The reverse Kato’s inequality condition is proved by using local arguments, so that it holds for the same reasons as in the previous section. Similarly, because the argument are local, the conclusion of Lemma 6.1.1 holds here. As a consequence, using Theorem 2.1.20, Theorem 3.3.11, Theorem 4.4.14 and Theorem 4.5.20, we may summarize our analysis in the following result. 108 CHAPTER 6. APPLICATION TO DIFFUSION MODELS Theorem 6.2.6. Consider the elliptic operator (6.2.50) in the whole space and assume that the coeﬃcients satisfy (6.2.51). Then the conclusions (C3) holds as well as (CE2) in L1 φ1. We do not present an exponential constructive estimate, which we believe is possible to prove, but would require signiﬁcantly more development. 6.3 Diﬀusion in Rd with weak potential conﬁnement We consider in this section the same elliptic operator (6.2.50) with now a weak conﬁnement condition assuming that c converges to a constant. With no loss of generality, we may assume c →0. More precisely, we consider the elliptic operator Lf := ∆f + b · ∇f + rcf, (6.3.54) with c ∈C0(Rd), b ∈C0(Rd) and r ∈R+ a parameter. When not necessary in the discusion we will take r = 1. The associated ﬁrst eigenvalue problem in such a situation has been studied in [156, 8th and 9th courses] to which we refer for more details. We deﬁne λ1 = λ1(r) := inf{κ ∈R; (λ −L)−1 well deﬁned and positive for any λ ≥κ}. Proceeding exactly as in the proof of (H1) in the preceding section, we see that the operator λ −L is invertible for any λ > ∥c+∥L∞, and then its inverse is positive. Because the proof of (H2) in the preceding section also applies here, we deduce that the inﬁmum λ1 of the set I of real resolvent values is well deﬁned with λ1 ∈(κ0, κ1), for some constructive constants κi ∈R. We split now the discussion into two cases. Case 1. We start considering the case b = 0. In that case, L is self-adjoint so that λ1 is also characterized by λ1 = sup ∥f∥L2=1 E(f), with E(f) := (Lf, f) = r Z cf2 − Z \|∇f\|2. We make the following elementary observations : • We claim that λ1 ≥0. Taking fn(x) := n−d/2u(x/n) for some function u ∈H1(Rd), ∥u∥L2= 1, we compute −E(fn) = Z \|∇fn\|2− Z BR rcf2 n − Z Bc R rcf2 n ≤ 1 n2 Z \|∇u\|2+∥rc∥L∞(BR) Z BR/n u2 + ∥rc∥L∞(Bc R), for any R > 0, so that −λ1 ≤lim sup(−E(fn)) ≤0. • We claim that λ1 = 0 when c ≤0. In that case, we have E(f) ≤0 for any f ∈H1(Rd), and we deduce the reverse inequality λ1 ≤0. In particular, as a function λ1 = λ1(r) of r ≥0, we have λ1(0) = 0. We also claim that λ1(r) →∞as r →∞when c+ ̸≡0. We may indeed ﬁx f ∈H1(Rd), ∥f∥L2= 1, supp f ⊂supp c+, and we compute E(f) = r Z Rd c+f2 − Z \|∇f\|2→∞, as r →∞. • We ﬁnally observe that λ1 : R+ →R+ is convex since it is deﬁned as the supremum of linear functions r →E(f) for any ﬁxed f ∈H1(Rd). As a consequence, we have the following alternative: 6.3. DIFFUSION IN RD WITH WEAK POTENTIAL CONFINEMENT 109 - λ1 ≡0; - ∃r0 ∈[0, ∞) such that λ1(r) = 0 for r ≤r0 and λ1(r) > 0 for r > r0. Concerning the value of r0, it may happen that r0 > 0, and that is the case when c ∈Ld/2 because of the Sobolev inequality, or that r0 = 0, and that is the case for instance when c ≥0, c(x) = \|x\|−m for x ∈Bc R, m ∈(0, 2), R > 0. To prove that last claim, we may take the same sequence (fn) as above, and we compute E(fn) ≥ Z Bc R r\|x\|−mf2 n − Z \|∇fn\|2dx = r nm Z Bc R/n \|x\|−mu2 −1 n2 Z \|∇u\|2dx > 0, for n large enough (whatever is the value of r > 0). About condition (H3). It is established in that when λ1 = 0, the condition (H3) is not satisﬁed and there does not exist a ﬁrst eigenfunction f1 ∈L2(Rd) to the operator L deﬁned by (6.3.54). We refer to [156, 8th course] for a proof of that result. On the other hand, we claim that the condition (H3) is satisﬁed when λ1 > 0. Consider indeed three sequences (λn) of R, (fn) of H1(Rd) and (εn) of L2(Rd) such that (λn −L)fn = εn, εn, fn ≥0, ∥fn∥L2= 1, for any n ≥1, λn →λ1 and εn →0 in L2 as n →∞. We then have λn −E(fn) = ⟨(λn −L)fn, fn⟩= ⟨εn, fn⟩→0, as n →∞. By deﬁnition of E and boundedness of c, we see that (fn) is bounded in H1. As a consequence, up to the extraction of a subsequence, we have fn →f1 ≥0 in L2 loc and thus next (λ1 −L)f1 = 0 in the variational sense and Z cf2 n → Z cf2 1 , ∥∇f1∥L2≤lim inf∥∇fn∥L2, where we have used the dominated convergence theorem of Lebesgue and the fact that c →0 at inﬁnity in order to get the ﬁrst convergence. We ﬁnally deduce E(f1) ≥lim sup E(fn) = λ1 > 0, so that f1 ̸≡0, and the condition (H3) is veriﬁed. As a conclusion, for a self-adjoint operator, condition (H3) is automatically fulﬁlled by its adjoint and the conditions (H4) and (H5) have been proved in a general situation, including the present framework. The same conclusions of existence and uniqueness of a ﬁrst eigentriplet solution (λ1, f1, φ1) as in section 6.2 hold true when λ1 > 0. Case 2. We consider the general case b ∈C0(Rd). • We claim that λ1 ≥0. Adapting the second constructive argument in the proof of (H2) in Section 6.1, we consider χ ∈C1 c (R+) ∩W 2,∞(R+) such that 1[0,1/2] ≤χ ≤1[0,1], χ′ ≤0 on [0, 1], χ(s) := (1 −s)2/2 on [η, 1] with η ∈(1/2, 1) large enough in such a way that χ′′(s) + (d −1)χ′(s)/s ≥1/2, ∀s ∈(η, 1), (6.3.55) and deﬁne f0(x) := χ(\|x −x0\|/n) for \|x0\| large enough to be chosen later. We have supp f0 ⊂ Bn(x0) for any n ≥1 and we compute Lf0(x) = 1 n2 {χ′′(r/n) + d −1 r/n χ′(r/n)} + 1 nb(x) · ˆ y χ′(r/n) + c(x)χ(r/n) where y = x −x0 and r = \|y\|. On Bnη(x0), we have Lf0(x) ≥−∥χ′′∥∞ n2 −d −1 n2 χ′(r) r ∞−∥χ′∥∞ n sup Bn(x0) \|b\|−∥χ∥∞sup Bn(x0) \|c\|. 110 CHAPTER 6. APPLICATION TO DIFFUSION MODELS On Bn(x0) \ Bηn(x0), thanks to (6.3.55), we have Lf0(x) ≥ 1 2n2 −∥χ′∥∞ n sup Bn(x0) \|b\|−∥χ∥∞sup Bn(x0) \|c\|. Let now ﬁx ϵ > 0 and choose ﬁrst n large enough so that −∥χ′′∥∞ n2 −d −1 n2 χ′(r) r ∞≥−ϵ 2 inf (0,η) χ. Then, using that b, c ∈C0(Rd), we can take \|x0\| large enough so that −∥χ′∥∞ n sup Bn(x0) \|b\|−∥χ∥∞sup Bn(x0) \|c\|≥−ϵ 2 inf (0,η) χ and ∥χ′∥∞ n sup Bn(x0) \|b\|+ sup Bn(x0) \|c\|≤ 1 2n2 . Gathering the above inequalities, we obtain Lf0 ≥−ϵf0, and the condition (H2) is veriﬁed with κ0 = −ϵ. Because ϵ > 0 can be choose arbitrarily small, we conclude with λ1 ≥0. • We claim that λ1 = 0 when σ2 ≤0. Indeed, we have already seen that ⟨Lf, f⟩= − Z Rd\|∇f\|2+ Z Rd σ2f2, from which we deduce that d dt∥Stf∥2= 2⟨Lf, f⟩≤0. This ensures that (H1) is veriﬁed with κ1 = 0 and so λ1 ≤0. • We claim that λ1 > 0 when c+ ̸≡0 and r > 0 is large enough. For simplifying notations and up to translation and dilatation, we may reduce to the case c ≥c01B(0,1) with c0 > 0. Adapting the second constructive argument in the proof of (H2) in Section 6.1, we consider χ ∈C1 c (R+) ∩W 2,∞(R+), 1[0,1/2] ≤χ ≤1[0,1], suppχ = [0, 1], χ′′(1) = 1, χ′ ≤0 on [0, 1] and we set f0(x) := χ(\|x\|). We compute Lf0 = χ′′(\|x\|) + χ′(\|x\|)((d −1)/\|x\|+b · ˆ x) + rc(x)χ(\|x\|). On the one hand, we ﬁx η ∈(1/2, 1), 1 −η small enough, in such a way that ∥χ′∥L∞(η,1)(2(d −1) + ∥b∥L∞) ≤1/4, 1/2 ≤∥χ′′∥L∞(η,1), and thus Lf0 ≥1 4 ≥1 4f0 on B(0, η)c. On the other hand, we ﬁx r > 0, large enough, in such a way that ∥χ′′∥L∞+∥χ′∥L∞(2(d −1) + ∥b∥L∞) ≤κ(r) := 1 2rc0 inf [0,η) χ, and thus Lf0 ≥κ(r) ≥κ(r)f0 on B(0, η). 6.4. DIFFUSION IN RD WITH DRIFT CONFINEMENT 111 As a conclusion, we have established that condition (ii) in the statement Lemma 2.1.4 holds with κ0 := min(1/4, κ(r)), and that ends the constructive proof of condition (H2) by using Lemma 2.1.4. That implies in particular the claim since then λ1 ≥κ0 > 0. • We ﬁnally claim again that (H3) holds when λ1 > 0. To see that, we consider again three sequences (λn) of R, (fn) of H1(Rd) and (εn) of L2(Rd) such that (λn −L)fn = εn, εn, fn ≥0, ∥fn∥L2= 1, for any n ≥1, λn ↘λ1 and εn →0 in L2 as n →∞. As a consequence, we have λn + Z \|∇fn\|2− Z fnb · ∇fn − Z cf2 n = ((λn −L)fn, fn) = ⟨εn, fn⟩→0, as n →∞. Using the boundedness of c, b and λn, we see that (fn) is bounded in H1. As a consequence, up to the extraction of a subsequence, we have fn →f1 ≥0 in L2 loc. We assume by contradiction that f1 ≡0. We deduce that Z cf2 n →0, Z fnb · ∇fn →0, where we have used the dominated convergence theorem of Lebesgue and the fact that b, c →0 at inﬁnity. We thus obtain 0 < λ1 ≤λn + Z \|∇fn\|2= Z fnb · ∇fn + Z cf2 n + ⟨εn, fn⟩→0, and our contradiction. So that f1 ̸≡0, and the condition (H3) is veriﬁed. For the dual problem, from the above analysis, we know that there exist two sequences (φn) of H1(Rd), (εn) of L2(Rd) such that (λn −L∗)φn = εn, εn, φn ≥0 and ∥φn∥L2= 1, for any n ≥1, and εn →0 in L2 as n →∞. But we face the same situation as previously, since again λn + Z \|∇φn\|2− Z φnb · ∇φn − Z cφ2 n = ((λn −L∗)φn, φn) = (εn, φn) →0, and thus the same conclusion, namely φn →φ1, with φ1 ∈H1(Rd), φ1 ≥0, φ1 ̸≡0. Conclusion. The conditions (H4) and (H5) have been proved in a general situation, including the present framework. The same conclusions as in section 6.2 hold true when r > 0 is large enough (and thus λ1 > 0). 6.4 Diﬀusion in Rd with drift conﬁnement We now consider the elliptic operator Lf := ∆f + b · ∇f + cf, with a drift conﬁnement as it is the case for the Fokker-Planck operator. More precisely, and for the sake of simplicity, we assume here b = ∇U, U(x) = 1 γ ⟨x⟩γ, γ > 0. (6.4.56) When γ = 2 and c = x, that operator corresponds to the classical harmonic Fokker-Planck operator which is known to be related to the standard Poincaré inequality and to the standard log-Sobolev inequality, see [11, 12, 233] or more recently [14, 138] and the references therein. When c = divb, the operator L is on divergence form and L∗1 = 0, so that (0, 1) ∈R × L∞(Rd) is a solution to the dual ﬁrst eigenvalue problem. Existence of stationary solution f1 (which is also the ﬁrst eigenfunction) and its stability have been widely studied. We refer for instance to [234, 217, 94, 13] as well as to [138, 174, 113] which techniques will be adapted here. 112 CHAPTER 6. APPLICATION TO DIFFUSION MODELS In the present situation, we impose that the contribution of c has lower inﬂuence at the inﬁnity that the drift term b and we assume c ∈L∞ loc(Rd), ∃C0, R0 > 0, ∀x ∈Bc R0, \|c(x)\|= o(\|x\|2(γ−1)). (6.4.57) We further assume that c ≥divb when γ ∈(0, 1]. (6.4.58) The action of the drift term will be revealed through the choice of a convenient “conﬁning space". More precisely, for a weight function m : Rd →[1, ∞), we will work in a weighted Lebesgue space. Our analysis is based on the following elementary computation which can be readily adapted from [138, Lemma 3.8], [174, Lemma 3.8] and [113, Lemma 2.1]. Lemma 6.4.7. For any p ∈[1, ∞), any weight function m and any smooth, rapidly decaying function f, we have Z (Lf) f\|f\|p−2mp = −(p −1) Z \|∇f\|2\|f\|p−2mp + Z \|f\|pmpϕ1, (6.4.59) with ϕ1 := (p −1) \|∇m\|2 m2 + ∆m m + c −1 pdiv b ! −b · ∇m m (6.4.60) as well as Z (Lf) f\|f\|p−2mp = −(p −1) Z \|∇(fm)\|2\|fm\|p−2+ Z \|f\|pmpϕ2, (6.4.61) with ϕ2 := 2(1 −1 p)\|∇m\|2 m2 + (2 p −1)∆m m + c −1 pdiv b ! −b · ∇m m . (6.4.62) In order to simplify the discussion, we restrict ourself to the exponent p = 2 and to the exponential weight function m = ea⟨x⟩s, s ∈(0, γ], a > 0. We thus work in the Banach lattice X := L2 m. We observe that ∇m m = sax⟨x⟩s−2 ∼sa\|x\|s−1, ∆m m = sad⟨x⟩s−2 + s(s −2)a\|x\|2⟨x⟩s−4 + (sa)2\|x\|2⟨x⟩2s−4 ∼(sa)2\|x\|2s−2, divb = d⟨x⟩γ−2 + (γ −2)\|x\|2⟨x⟩γ−4 ∼(d + γ −2)\|x\|γ−2, b · ∇m m = sax⟨x⟩s−2 · x⟨x⟩γ−2 ∼sa\|x\|s+γ−2, so that the contribution of (c −divb/2) is always negligible at inﬁnity, and we get ϕi ∼2(sa)2\|x\|2s−2−sa\|x\|s+γ−2. (6.4.63) We denote a′ := sa > 0 if s ∈(0, γ), a′ := aγ −2(aγ)2 > 0 if s = γ and a ∈(0, 1/( √ 2γ)). We then face to three cases : (i) γ > 1 : taking s ∈((2 −γ)+, γ), we have ϕi ∼−a′\|x\|s+γ−2→−∞with s + γ −2 > 0; (ii) γ = 1 : taking s = γ, a < 1/( √ 2γ), we have ϕi →−a′; (iii) γ ∈(0, 1) : taking s = γ, a < 1/( √ 2γ), we have ϕi ∼−a′\|x\|2γ−2→0 with 2γ −2 < 0. 6.4. DIFFUSION IN RD WITH DRIFT CONFINEMENT 113 Condition (H1). In any of the above cases, we have from (6.4.59) ((λ −L)f, f) = Z \|∇f\|2m2 + Z (λ −ϕ1)f2m2, for λ ∈R, with inf(λ −ϕ1) > 0 for λ ≥κ1 and κ1 > 0 large enough. We deduce that λ −L is coercive for λ ≥κ1. With the same arguments as in section 6.1, we conclude that L is the generator in L2 m of a positive semigroup SL, so that (H1) holds. Condition (H2). When γ > 1, the same arguments as in Section 6.2 imply that condition (H2) holds for some κ0 ∈R. When γ ∈(0, 1], we have L∗1 = c −divb ≥0 from (6.4.58) and (H2) holds with κ0 = 0. Conditions (H4) and (H5). The strong maximum principle holds here because for instance we may apply the same barrer function argument as presented in Section 6.2. The reverse Kato’s inequality condition is proved by using local arguments, so that it holds for the same reasons as in the previous section. Condition (H3). We deﬁne the multiplication operator A and the elliptic operator B by A := MχR, B := L −A, for M, R > 0 and χR(x) := χ(x/R) with χ ∈D(Rd), 1B1 ≤χ ≤1B2. We ﬁx κB < κ0 in case (i), κB := −a′/4 in case (ii) and κB := 0 in case (iii), and we set a′′ := a′/2. Choosing M, R > 0 large enough, from Lemma 6.4.7 and the discussion which follows, we deduce that ((B −α)f, f) ≤− Z \|∇f\|2m2 −a′′ Z f2(1B1 + 1Bc 1\|x\|s+γ−2)m2, (6.4.64) for any α ≥κB and any nice function f. We classically deduce that α −B is coercive and thus invertible. We discuss the three diﬀerent cases. - In the ﬁrst case γ > 1, so that s + γ −2 > 0, we see that the operator RB(α) is compact from Rellich-Kondrachov theorem, so that also W(α) := RB(α)A for any α ≥κB. We may thus apply Lemma 2.1.8-(2) and we deduce that (H3) holds for both the primal and the dual problems. - In the case γ = 1, so that 2γ −2 ≤0, the operator RB(α) is not compact anymore. However, for any sequence (fn) which is bounded in L2 m, we deﬁne the sequence (gn) by gn := Afn, and (gn) is bounded in L2 ˜ m with ˜ m := e˜ a⟨x⟩γ, ˜ a ∈(a, 1/√2γ). Using the dissipativity estimate (6.4.64) in L2 ˜ m, we see that B −α is dissipative in L2 ˜ m for any α ≥κB, and more precisely the sequence (hn) deﬁned by hn := RB(α)gn satisﬁes Z \|∇hn\|2m2 + ˜ a′′ Z h2 n(1B1 + 1Bc 1\|x\|2γ−2) ˜ m2 ≤ Z g2 n ˜ m2. Using that \|x\|2γ−2 ˜ m2/m2 →∞as \|x\|→∞, that implies that (hn) is relatively compact in L2 m. More precisely, the above estimates show that W(α) := RB(α)A : L2 m →H1 m ∩L2 m♯with m♯:= m1/2 ˜ m1/2 and in particular we have established that W(α) := RB(α)A is a compact operator in L2 m uniformly on α ≥κB because of the Rellich-Kondrachov theorem and the fact that m = o(m♯). Since RB(α) is bounded in B(L2 m) uniformly for any α > κB, the operator L satisﬁes the splitting structure (HS1) and, applying Lemma 2.1.8-(2), we deduce that (H3) holds for both the primal and the dual problems. At this stage, when γ ≥1, we obtain a solution (λ1, f1, φ1) to the ﬁrst eigentriplet problem (6.0.3) by using Theorem 2.1.20. Condition (HS3). In the case γ ∈(0, 1), the same as in the case γ = 1 holds except that RB(α) is not uniformly bound in B(L2 m) for α ≥κB because we are in the critical case κB = κ0. We do not know how to adapt the stationary approach in that situation and we thus aim to use a dynamical approach through the use of Theorem 2.2.4 with the above splitting L = A + B and N := [d/4] + 1. We set X = X1 := L2 m and X0 := L1. The proof of condition (HS3) is an immediate consequence of the following estimate. 114 CHAPTER 6. APPLICATION TO DIFFUSION MODELS Proposition 6.4.8. We deﬁne Θζ(t) := e−ζtγ/(2−γ). For N := [d/4] + 1, there hold (i) SB ∈L∞ t (B(X1)); (ii) SBAΘ−1 ζ ∈L∞ t (B(Xi)) for i = 0, 1 and any ζ ∈(0, ζ∗); (iii) (SBA)(∗N)Θ−1 ζ ∈L∞ t (B(X0, X1)) for any ζ ∈(0, ζ∗/2). The proof of Proposition 6.4.8 is similar to the proofs of [138, Lemma 2.1], [138, Lemma 2.2], [138, Lemma 2.3] and [138, Lemma 2.4]. For the sake of completeness we however present the main lines of the proof. We start with a technical result that we will use during the proof of Proposition 6.4.8. Lemma 6.4.9. Consider two Banach spaces X0, X1 and a function u : R+ →B(X0) + B(X1) which satisﬁes (a) uΘ−1 ∈L∞(0, ∞; B(X0) ∩B(X1)); (b) u℘∈L∞(0, ∞; B(X0, X1)); for any exponentially decaying function Θ = Θζ = e−ζtς, ∀ζ ∈(0, ζ∗), and for the power function ℘:= t−α, with ζ∗> 0, ς ∈(0, 1] and α ≥0 ﬁxed. Then (c) there exists N such that u(∗N) e Θ ∈L∞(0, ∞; B(X0, X1)), for any e Θ = Θ˜ ζ, ˜ ζ ∈(0, ζ∗/2). Proof of Lemma 6.4.9. A similar argument is developed in [113, Lemma 2.17], [174, Lemma 2.4], [175, Proposition 2.5] and [138, Lemma 2.4]. Step 1. Consider two functions v and w which satisfy the estimate (a). For X = X0 or X = X1, we compute ∥v ∗w(t)∥X→X ≤ Z t 0 ∥v(t −s)w(s)∥X→X ds ≤ Z t 0 Cv X Θ(t −s) Cw X Θ(s) ds ≤Cv X Cw X t Θ(t), with obvious notation and where we have used that Θ(t −s) Θ(s) ≤Θ(t) for any 0 < s < t. Since for any ζ′ ∈(0, ζ), there exists a constant C such that tΘζ(t) ≤C Θζ′(t) for any t ≥0, we see that the function v ∗w satisﬁes the same estimate (a) for any Θ = Θζ, ζ ∈(0, ζ∗). Step 2. Consider two functions v and w which satisfy the estimates (a) and (b) with α ≥1. We compute ∥v ∗w(t)∥X0→X1 ≤ Z t/2 0 ∥v(t −s)w(s)∥X0→X1 ds + Z t t/2 ∥v(t −s)w(s)∥X0→X1 ds ≤ Z t/2 0 Cv 01(t −s)−α Cw 0 Θ(s) ds + Z t t/2 Cv 1Θ(t −s) Cw 01s−α ds = [Cv 1 Cw 01 + Cv 01 Cw 0 ] Θ(0) t−α+1 Z 1/2 0 (1 −τ)−α dτ, with obvious notation and we have used that Θ is a decaying function. As a consequence, the function v ∗w satisﬁes estimate (b) with an exponent α −1 instead of α. 6.4. DIFFUSION IN RD WITH DRIFT CONFINEMENT 115 Step 3. Consider two functions v and w which satisfy the estimates (a) and (b) with α ∈[0, 1). We compute ∥v ∗w(t)∥X0→X1 ≤ Z t/2 0 ∥v(t −s)w(s)∥X0→X1 ds + Z t t/2 ∥v(t −s)w(s)∥X0→X1 ds ≤ Z t/2 0 Cv 1Θ(t −s) Cw 01s−α ds + Z t t/2 Cv 01(t −s)−α Cw 0 Θ(s) ds ≤ Cv 1 Cw 01 Θ(t/2) Z t/2 0 s−α ds + Cv 01 Cw 0 Θ(t/2) Z t t/2 (t −s)−α ds = [Cv 1 Cw 01 + Cv 01 Cw 0 ] Θ(t/2) t1−α 1 −α, with the same obvious notation and we have used again that Θ is a decaying function. Step 4. Iterating n := [α] times steps 1 and 2, we get that u(∗n) still satisﬁes estimate (a) and satisﬁes the estimate (b) for the exponent α −[α] ∈(0, 1). We then conclude that (c) holds with N := n + 1 and any ˜ ζ ∈(0, ζ∗/2) by using the third step. Proof of Proposition 6.4.8. We classically establish that B generates a positive semigroup SB in both spaces Xi and we thus concentrate on the announced estimates. On the one hand, proceeding as for the proof of (6.4.64), we have Z (Bf)(sign f)m ≤−a′′ Z \|f\|(1B1 + 1Bc 1\|x\|s+γ−2)m, (6.4.65) for any nice function f and any weight function m = ma, with ma(x) := ea⟨x⟩γ, a ∈(a1, a2), 0 < a1 < a2 < 1/( √ 2γ), where we deﬁne a′′ := aγ/2 −(aγ)2. That exactly means that B is weakly dissipative in L1 m as deﬁned in (2.2.19). From the discussion in Section 2.2.3, we deduce that SB is a semigroup of contractions and satisﬁes the associated decay estimate (2.2.23), (2.2.24), and more precisely ∥SB(t)f∥L1 ma≤∥f∥L1 ma, ∥SB(t)f∥L1 ma≤Θζ(t)∥f∥L1 ma′ , (6.4.66) for any a, a′ ∈(a1, a2), a < a′, ζ ∈(0, ζ∗), ζ∗:= (a′ −a)(2−2γ)/(2−γ)(a′γ(1 −a′γ))γ/(2−γ). We refer to [138, Lemma 2.1] for details. Using that A : L1 →L1 m is bounded, that establishes (ii) in X0. Similarly, starting from (6.4.61) and proceeding as in the proof of (6.4.64), we get (Bf, f)L2 m ≤− Z \|∇(fm)\|2−a′′ Z f2(1B1 + 1Bc 1\|x\|s+γ−2)m2, (6.4.67) for any nice function f and any weight function m = ma as above. Throwing away the ﬁrst term at the RHS and arguing as we did in L1 m, we obtain that SB satisﬁes ∥SB(t)f∥L2 ma≤∥f∥L2 ma, ∥SB(t)f∥L2 ma≤Θζ(t)∥f∥L2 ma′ , (6.4.68) for any a, a′ ∈(a1, a2). Using that A : L2 ma →L2 m′ a is bounded, that establishes (i) and (ii) in X1. On the other hand, throwing away the second term at the RHS in (6.4.67), for any trajectory ft = SB(t)f0, f0 in the domain of B in L2 m, we have 1 2 d dt Z Rd f2 t m2dx ≤− Z Rd\|∇(ftm)\|2dx. Using Nash’s inequality which for some constant CN ∈(0, ∞) stipulates that Z Rd g2dx ≤CN Z Rd\|∇g\|2dx d d+2 Z Rd\|g\|dx 4 d+2 , ∀g, 116 CHAPTER 6. APPLICATION TO DIFFUSION MODELS with g := ftm and the ﬁrst estimate in (6.4.66), we deduce F ′(t) ≤−2 C′ N F(t)−4/d G(t)1+ 2 d ≤−2 C′ N F(t)−4/d G(0)1+ 2 d , (6.4.69) with C′ N := C−1−2/d N and where for brevity of notations we have set F(t) := ∥ft∥2 L2(m), G(t) := ∥ft∥L1(m). Integrate the diﬀerential inequality (6.4.69), we ﬁnd ∥SB(t)f0∥2 L2 m≲t−d/4 ∥f0∥L1 m, ∀t > 0, and using that A : L1 →L1 m, we next obtain SB(t)Atd/4 ∈L∞(0, ∞; B(X0, X1)). (6.4.70) Setting with u(t) := SB(t)A, we see that u satisﬁes (a) in Lemma 6.4.9 thanks to (ii) in X0 and X1. Furthermore, u satisﬁes (b) in Lemma 6.4.9 thanks to (6.4.70). Using Lemma 6.4.9, we conclude that condition (iii) holds. We come back to the proof of (HS3). Gathering (i) and (ii) in X1 in Proposition 6.4.8, we get that (SBA)(∗ℓ) ∗SB ∈L∞ t (B(X1)) for any ℓ∈{0, ·, N −1}, N := [d/4] + 1. Using that Θ ∈L1(0, ∞) and (iii) in Proposition 6.4.8, we deduce that (SBA)(∗N) ∈L1(0, ∞; B(X0, X1)). We may now handle the existence part of the ﬁrst eigenvalue problem. On the one hand, recalling (H2), we have L∗ψ0 ≥0 with ψ0 = 1 so that the condition (i) in Theorem 2.2.4 holds. On the other hand, the condition (ii) in Theorem 2.2.4 is an immediate consequence of (HS3) as emphasized in Remark 2.2.5-(1). As a conclusion, the hypotheses of Theorem 2.2.4 are thus met, and we deduce that there exists (λ1, f1) ∈R+ × L2 m+ solution to the ﬁrst eigenvalue problem. Because the strong maximum principle (H4) holds, we have f1 > 0 on Rd. In order to prove the existence of a ﬁrst positive eigenfunction for the dual problem, we argue in the following way. We start observing that we have the alternative: λ1 = 0 or λ1 > 0. - In the ﬁrst case, we may argue as in Remark 3.3.17. We indeed have in the same time L∗ψ0 ≥0 and ⟨L∗ψ0, f1⟩= ⟨ψ0, Lf1⟩= 0, so that L∗ψ0 = 0 because f1 > 0. The function φ1 := ψ0 is thus a solution to the ﬁrst dual eigenvalue problem. - In the second case λ1 > 0, we may argue as in the case γ = 1 above. On the one hand, the operator RB(α) is uniformly bounded in L2 m for any α ≥κB := λ1/2 > 0 and on the other hand the operator W(α) := RB(α)A : L2 m →H1 m ∩L2 m♯is uniformly bounded for any α ≥κB with m = o(m♯), so that H1 m ∩L2 m♯⊂L2 m is compact. We may thus apply Theorem 2.1.20 and we conclude to the existence of a solution (λ′ 1, f′ 1, φ′ 1) to the eigentriplet problem. The conditions (H4) and (H5) being true in a general situation as well as the conclusions of Lemma 6.1.1, as an intermediate conclusion, we have established under the general condition γ > 0 in (6.4.56) that yet the same conclusions as in section 6.2 hold true. Chapter 7 The kinetic Fokker-Planck equation In this chapter, we consider the kinetic Fokker-Planck evolution equation associated to the operator Lf := −v · ∇xf + ∆vf + b · ∇vf + cf, (7.0.1) on functions f : O →R, where O := Ω× Rd, Ω⊂Rd is a domain, b : O →Rd is a given vector ﬁeld and c : O →R is a given function. As we said in the introduction, we will consider the case when Ωis a bounded domain and the equation is complemented with a boundary condition. For ease of reading, we will reintroduce the key notation and elements of this chapter. We assume the classical balance between the values of the trace γf of f on the outgoing and incoming velocities subsets of the boundary (γ−f)(x, v) = Rx(γ +f(x, .))(v) on Σ−, (7.0.2) where in this context we deﬁne Σx ± := {v ∈R3; ± v · νx > 0} the sets of outgoing (Σx +) and incoming (Σx −) velocities at point x ∈∂Ω, next the sets Σ± = {(x, v) ∈Σ; ±νx · v > 0} = {(x, v); x ∈∂Ω, v ∈Σx ±}, and ﬁnally the outgoing and incoming trace functions γ±f := 1Σ± γf. Here and in the sequel, νx denotes the unit normal outward vector ﬁeld deﬁned on the boundary set ∂Ω. We similarly deﬁne the grazing velocity set Σ0 = {(x, v) ∈Σ; νx · v = 0}. The reﬂection operator Rx is local in position, but can be local or nonlocal in the velocity variable, so that it writes (Rxg)(v) := Z Σx + r(x, v, v∗)g(v∗)v∗· νx dv∗, for a reﬂection kernel r : ∂Ω× Rd × Rd →R. Some classical general assumptions on r are r ≥0, R∗ x1 = 1, RxM = M , (7.0.3) for some positive function M = M (v), see for instance [52, 53, 54]. The second (normalisation) condition corresponds to the fact that all the particles reaching the outgoing boundary are put back on the incoming boundary (no mass is lost) while the third (reciprocity) condition means (when M is a Gaussian function) that the wall is in a local equilibrium state and is not inﬂuenced by the incoming particles. The normalization condition implies the local mass conservation Z Σx − Rxg\|ν · v\|dv = Z Σx + gν · vdv, (7.0.4) 117 118 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION while the three assumptions (7.0.3) on r together also imply Z Σx − (Rxg)2M −1\|ν · v\|dv ≤ Z Σx − (Rx(g2/M ))(RxM )M −1\|ν · v\|dv = Z Σx + g2 ∗M −1 ∗ (R∗1)ν · v∗dv∗, where we have used the Cauchy-Schwarz inequality (and the fact that r ≥0) in the ﬁrst line and the reciprocity condition in the second line. As a consequence, we have Z Σx − (Rxg)2M −1\|ν · v\|dv ≤ Z Σx + g2M −1 ν · vdv, (7.0.5) where we have used the normalization condition in that last step. In the sequel, we will rather consider the possibly position dependent Maxwell boundary condition operator Rxg = α(x)Dxg + β(x)Γxg, (7.0.6) where the accommodation coeﬃcients α, β : ∂Ω→[0, 1] satisfy α(x) + β(x) =: ζ(x) ≤1, Γx is the specular reﬂection operator Γx(g(x, ·))(v) = g(x, Vxv), Vxv = v −2ν(x)(ν(x) · v), (7.0.7) and Dx is the diﬀusive operator Dx(g(x, ·))(v) = cM M (v)e g(x), e g(x) = Z Σx + g(x, w) ν(x) · w dw. (7.0.8) Here the constant cM := (2π)1/2 is such that cM f M = 1 and M stands for the standard Maxwellian M (v) := (2π)−d/2 exp(−\|v\|2/2), (7.0.9) or, more generally, M = M (\|v\|) ≥0 is such that D∗ x1 = 1, DxM = M , ⟨v⟩ϑM ∈L1(Rd), (7.0.10) with ϑ ≥1 (that last condition is necessary in order that the second relation above makes sense). The boundary condition (7.0.6) corresponds to the pure specular reﬂection boundary condition when β ≡1 and it corresponds to the pure diﬀusive boundary condition when α ≡1. When ζ ≡1, the Maxwell boundary condition operator (7.0.6) satisﬁes (7.0.3). On the contrary, when ζ ̸≡1, the L2 estimate (7.0.5) holds but not anymore the mass conservation (7.0.4). However, the following L1 estimate Z Σx − \|Rxg\|\|ν · v\|dv ≤ζ∗ Z Σx + \|g\| ν · vdv (7.0.11) holds, with 0 ≤sup ζ ≤ζ∗≤1. Finally, the case ζ ≡0 corresonds to the zero inﬂow problem. Let us ﬁnally mention that the regularity needed on the domain Ωmay be formulated in the following way: we assume that Ωis locally on one side of ∂Ωand there exists a function δ = δΩ∈W 2,∞(Rd) such that for all x in an interior neighborhood of ∂Ωone has δ(x) = dist(x, ∂Ω) and the vector ﬁeld ν deﬁned on Rd by x 7→ν(x) = νx := −∇xδ(x) coincides with the previously deﬁned unit outward normal vector ﬁeld on ∂Ωand satisﬁes ∥ν∥L∞= 1. We also assume that the Lebesgue measure on ∂Ωis well deﬁned and it is denoted by dσ x. 7.1. THE TRACE PROBLEM 119 7.1 The trace problem We consider in this section the trace problem for a solution g = g(x, v) to the stationary Vlasov-Fokker-Planck equation Mg := v · ∇xg −b · ∇vg −∆vg = G in O, (7.1.12) for a given a vector ﬁeld b = b(x, v), a source term G = G(x, v) and for a solution g = g(t, x, v) to the evolution Vlasov-Fokker-Planck equation ∂tg + v · ∇xg −b · ∇vg −∆vg = G in (0, T) × O, (7.1.13) for a given a vector ﬁeld b = b(t, x, v), a source term G = G(t, x, v). The second problem has been considered ﬁrst in and next in [172, Section 4], where a strong (renormalized) trace function is proved to exist. In the sequel, we recall these results and slightly extending them by considering a possible L2H−1 source term. We introduce some notations. We denote dξ := \|ν(x) · v\|dvdσ x and dξ2 := (ν(x) · ˆ v)2dvdσ x the measures on the boundary set Σ. We denote by B1 the class of renormalized functions β ∈W 2,∞ loc (R) such that β′′ has a compact support, by B2 the class of functions β ∈W 2,∞ loc (R) such that β′′ ∈L∞(R) and by D0( ¯ O) the space of test functions ϕ ∈D( ¯ O) such that ϕ = 0 on Σ0. We ﬁnally deﬁne the dual operator M∗ϕ := −v · ∇xϕ + divv(bϕ) −∆vϕ. Theorem 7.1.1. We consider g, b ∈L2 loc,xH1 loc,v, G ∈L2 loc,xH−1 loc,v and we assume that g is a solution to the stationary Vlasov-Fokker-Planck equation (7.1.12). Then there exists γg ∈ L2 loc(Σ, dξ2) such that the following Green renormalized formula ZZ O (β(g) M∗ϕ −β′′(g) \|∇v g\|2ϕ) dvdx + ⟨G, β′(g)ϕ⟩= (7.1.14) = ZZ Σ β(γ g) ϕ ν(x) · v dvdσ x holds for any renormalized function β ∈B1 and any test functions ϕ ∈D( ¯ O), as well as for any renormalized function β ∈B2 and any test functions ϕ ∈D0( ¯ O). It is worth emphasizing that β′(g)ϕ ∈L2 xH1 v so that the duality product ⟨G, β′(g)ϕ⟩is well deﬁned. If furthermore γ∓g ∈L2 loc(Σ; dξ) then γ±g ∈L2 loc(Σ; dξ) and (7.1.14) holds for any renor-malized function β ∈B2 and any test functions ϕ ∈D( ¯ O). Proof of Theorem 7.1.1. We only allude the proof which uses very similar arguments as those presented in [75, 172]. In the one hand, considering the molliﬁer (ρε)ε>0 deﬁned by ρε(z) = 1 εd ρ(z/ε), 0 ≤ρ ∈D(Rd), supp ρ ⊂B1, Z RN ρ(z) dz = 1, with z := (x, y), we get that gε is smooth and satisﬁes gε →g in L2 loc,xH1 loc,v, Mgε = Gε →G in L2 loc,xH−1 loc,v, which is nothing but a variant of [76, Lemma II.1]. The function gε being smooth, for any β ∈C2 such that β′ ∈C1 b , we may diﬀerentiate β(gε) and we get Mβ(gε) + β′′(gε)\|∇vgε\|2= β′(gε)Gε in O, with. We may thus pass to the limit as ε →0 and we obtain (7.1.14). We also write without proof a stability result that we will use several times in the sequel. 120 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Proposition 7.1.2. Let us consider three sequences (gk), (bk) of L2 loc,xH1 loc,v and (Gk) of L2 loc,xH−1 loc,v such that v · ∇xgk −bk · ∇vgk −∆vgk = Gk in D′(O) for any k ≥1 and three functions g, b ∈L2 loc,xH1 loc,v and G ∈L2 loc,xH−1 loc,v such that gk →g strongly in L2 loc,xH1 loc,v, bk ⇀b weakly in L2 loc( ¯ O) and Gk →G strongly in L2 loc,xH−1 loc,v. Then g satisﬁes (7.1.12) and, up to the extaction of a subsequence, γgk →γg a.e. on Σ\Σ0. (2) If gk ⇀g weakly in L1 loc( ¯ O) then g satisﬁes (7.1.14) and, up to the extaction of a subsequence, γgk r ⇀γg on Σ\Σ0. Where gn ⇀r g denote the renormalized convergence to be deﬁned as follows: a sequence (gn) of L(E) converges in the renormalized sense to g if for any δ ∈(0, 1] (or for any δ ∈D ⊂(0, 1], D := {δm} for a sequence δm ↘0) there exists ¯ βδ ∈L∞(E) such that βδ(gn) ⇀¯ βδ ∗σ(L∞, L1) as n →∞and β1(¯ βδ) →β1(g) L1 loc( ¯ O) as δ →0. We may observe that gn ⇀g weakly L1(E) implies gn r ⇀g up to the extaction of a susbequence and that gn →g a.e. in E implies gn r ⇀g. We refer to for more material about renormalized convergence. 7.2 Well-posedness problem with inﬂow term at the boundary We consider the kinetic Fokker-Planck operator L deﬁned in (7.0.1) and we start revisiting the well posedness problem (λ −L)f = F in O, γ−f = g on Σ−, (7.2.15) for given data F : O →R and g : Σ−→R. For a given weight function m : Rd →[1, ∞), we deﬁne the measure dξm := m2\|ν(x)·v\| dvdσ x on the boundary Σ. We next deﬁne L2H1 m = L2H1 m(O) the space associated to the Hilbert norm deﬁned by ∥f∥2 L2H1 m:= ∥f∥2 L2 m+∥∇vf∥2 L2 m, and we assume that m satisﬁes the Poincaré type inequality ∥f ∇m m ∥L2 m≲∥f∥L2H1 m, ∀f ∈L2H1 m. (7.2.16) Such a Poincaré inequality is classically known to be true when m := M −ϑ, M is the Maxwellian (7.0.9) and ϑ > 0. We also deﬁne L2H−1 m := {F = g + divvG; g, Gi ∈L2 m(O)}, so that when m = 1 the space L2H−1 m is nothing but the space of continuous and linear mappings on L2H1. For F ∈L2H−1 m and f ∈L2H1 m, we may thus write ⟨F, fm2⟩≤∥F∥L2H−1 m ∥f∥L2H1 m. We ﬁnally deﬁne in this context W2 := {f ∈L2H1 m; ˆ v · ∇xf ∈L2H−1 m }, and W2,Σ := {g ∈W2; γg ∈L2(Σ; dξm)}, with W2,Σ ̸= W2 in general. 7.2. WELL-POSEDNESS PROBLEM WITH INFLOW TERM AT THE BOUNDARY 121 Theorem 7.2.3. Let us ﬁx a vector ﬁeld b ∈H1 loc( ¯ O), a function c ∈L∞(O), a weight function m : Rd →[1, ∞) and let us assume that b/⟨v⟩∈L∞(O), that (7.2.16) holds and that λ∗:= ess sup ϖ < ∞, ϖ := c + ∆m2 2m2 −1 2divb −b · ∇m m . (7.2.17) For any F ∈L2H−1 m , g ∈L2(Σ−; dξm) and λ > λ∗, there exists a unique solution f ∈W2,Σ to the Dirichlet problem (7.2.15). We have furthermore f ≥0 if F ≥0 and g ≥0. A similar result is proved in [70, Appendix A] in the case Ω= Rd. Also observe that (7.2.17) holds with m := M −1/2 when M is the standard Maxwellian (7.0.9) and b(v) = ϑv, with ϑ > 1/2. Proof of Theorem 7.2.3. We split the proof into ﬁve steps. Step 1. A priori estimates. Multiplying the ﬁrst equation in (7.2.15) by fm2, performing several integrations by part in the velocity variable and using the Green formula, we have Z O (λ −ϖ)f2m2 + 1 2 Z Σ (γf)2m2ν · v + Z O \|∇vf\|2m2 = ⟨F, fm2⟩. Fixing λ > λ∗, using the Young inequality ∥F∥L2H−1 m ∥f∥L2H1 m≤ 1 2(λ −λ∗) + 1 2 ∥F∥2 L2H−1 m +λ −λ∗ 2 ∥f∥2 L2 m+1 2∥∇vf∥2 L2 m and the boundary condition on the incoming set Σ−in (7.2.15), we deduce (λ−λ∗) Z O f2m2 + Z Σ+ (γ+f)2dξm + Z O \|∇vf\|2m2 ≤1 + λ −λ∗ λ −λ∗ ∥F∥2 L2H−1 m + Z Σ− g2dξm. (7.2.18) • Because of the ﬁrst equation in (7.2.15) and the above estimate, we ﬁnd ˆ v · ∇xf = 1 ⟨v⟩(F −λf + ∆vf + b · ∇vf + cf) ∈L2H−1 m , (7.2.19) so that f ∈W2. • Multiplying the ﬁrst equation in (7.2.15) by fψ, ψ := ν(x) · ˜ vm2 where here and below we use the notations ˆ v := v/⟨v⟩, ˜ v := v/⟨v⟩2, ⟨v⟩2 := 1 + \|v\|2, and using the Green formula and one integration by part in the velocity variable, we get 1 2 Z Σ (γf)2(ν · ˆ v)2m2 = 1 2 Z O f2ˆ v · Dxνxˆ v m2 − Z O \|∇vf\|2ψ + Z O f∇vf(bψ −∇vψ) + Z O f2ψ(c −λ) + ⟨F, fψ⟩. Observing that \|⟨F, fψ⟩\|≤∥F∥L2H−1 m ∥fν(x) · ˜ v∥L2H1 m≲∥F∥L2H−1 m ∥f∥L2H1 m and ∥f∇vψ∥L2≲∥f∥2 L2H1 m, recalling that b/⟨v⟩∈L∞(O) and using the Cauchy-Schwarz inequality, we deduce ∥γf∥2 L2(Σ;dξ2 m)≤C(1 + \|λ\|)∥f∥2 L2H1 m+C∥F∥L2H−1 m ∥f∥L2H1 m, (7.2.20) for some constant C = C(b, c, m, ν), with dξ2 m := (ν · ˆ v)2m2dvdσ x. 122 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION • For latter reference, we establish an estimate about the behaviour of the solution near the boundary. We now introduce the following Lions-Perthame type weight function ψ := 2δ(x)1/2ν(x) · ˜ v, (7.2.21) and we observe that ψ = 0 on Σ, ⟨v⟩ψ ∈L∞(O), ∇vψ ∈L∞(O) and v · ∇xψ = 1 δ(x)1/2 (ˆ v · ν(x))2 + 2δ(x)1/2ˆ v · Dxν(x)ˆ v. Multiplying the ﬁrst equation in (7.2.15) by fψ, we have 1 2 Z O v · ∇xf2ψ − Z O f b ⟨v⟩· ∇vf⟨v⟩ψ + Z O ∇v(fψ) · ∇vf + Z O (λ −c)f2ψ = ⟨F, fψ⟩. Using Cauchy-Schwarz and Young inequalities, we deduce Z O f2 (ˆ v · ν(x))2 δ(x)1/2 dvdx ≤C(1 + \|λ\|)(∥f∥2 L2H1+∥F∥2 L2H−1), (7.2.22) for some constant C = C(b, c, n). • We ﬁnally state a somehow classical regularity estimate when F ∈L2 m(O). Taking advantage of the fact that F ∈L2 m and f ∈L2H1 m and localizing the problem by introducing the function g := fχε ∈L2 xH1 v(Rd × Rd), χε ∈C2 c (O), 1Oε ≤χε ≤1, Oε := {(x, v) ∈O; δ(x) > ε, \|v\|≤1/ε}, we have v · ∇xg −∆vg + ⟨v⟩2g = G in D′(Rd × Rd), with G := (F −λf −cf −b · ∇vf)χε −2∇vf · ∇vχε + ⟨v⟩2fχε ∈L2(Rd × Rd). From the quantitative Hormander’s hypoellipticity estimate of Hérau & Pravda-Starov [122, Proposition 2.1], we then have ∥D2/3 x g∥L2+∥D2 vg∥L2≲∥G∥L2+∥g∥L2. Coming back to the function f and using the previous estimates, we deduce ∥D2/3 x f∥L2(Oε)+∥D2 vf∥L2(Oε)≤C(∥Lf∥L2(O)+∥f∥L2(O)), (7.2.23) for a constant C = C(λ, ε) > 0. Step 2. Existence. We assume g = 0. A possible way for proving the existence is to use Lions’ variant of the Lax-Milgram theorem [154, Chap III, §1] as in [19, 70] and as we proceed now. Deﬁning the bilinear form E : L2H1 m(O) × C1 c (O ∪Σ−) →R, by E(f, ϕ) = Z O (λ −L)fϕm2 := Z O (λf −b · ∇vf −cf)ϕm2 + ∇vf · ∇v(ϕm2) −f(v · ∇xϕ)m2, for any f ∈L2H1 m(O) and ϕ ∈C1 c (O), we observe that this one is coercive, namely E(ϕ, ϕ) = Z O (λ −ϖ)ϕ2m2 + Z O \|∇vϕ\|2m2 + 1 2 Z Σ− (γ−ϕ)2dξm ≥ κ∥ϕ∥2 L2H1 m, for any ϕ ∈C1 c (O ∪Σ−), with κ := min(λ −λ∗, 1) > 0. From the above metionned Lions’ theorem, for any F ∈L2H−1 m , there exists f ∈L2H1 m such that E(f, ϕ) = ⟨F, ϕm2⟩, ∀ϕ ∈C1 c (O ∪Σ−). (7.2.24) 7.2. WELL-POSEDNESS PROBLEM WITH INFLOW TERM AT THE BOUNDARY 123 In particular, f satisﬁes the ﬁrst equation in (7.2.15) in the distributional sense D′(O), and thus from (7.2.19), we deduce that f ∈W2. Thanks to the trace Theorem 7.1.1 and the estimate (7.2.20), the function f admits a trace γf ∈L2(Σ; dξ2 m). Using the Green formula (7.1.14) with β = id ∈B1, we have ZZ O (f(L∗−λ)ϕ + Fϕ) dvdx = ZZ Σ γf ϕ ν(x) · v dvdσ x, (7.2.25) for any ϕ ∈D0( ¯ O). Particularizing to ϕ ∈D(O ∪Σ−) and comparing with (7.2.24), we deduce that γ−f = 0. Step 3. Existence. The general case g ∈L2(Σ−; dξm). When g ∈C2 c (Σ−), there exists a function h ∈C2 c (O∪Σ−) such that h\|Σ−= g and we consider the source term G := F+(L−λ)h ∈ L2H−1 m as well as the problem (λ −L)g = G in O, γ−g = 0 on Σ−. From Step 2, there exists a solution g ∈W2,Σ to this problem and we set f := g + h. In such a way that f ∈W2,Σ and satisﬁes Z O f(λ −L∗)ϕ = Z O g(λ −L∗)ϕ + Z O h(λ −L∗)ϕ = Z O Gϕ + Z O (λ −L)hϕ − Z Σ h\|Σϕ ν · v, and thus Z O f(λ −L∗)ϕ = Z O Fϕ − Z Σ− gϕ ν · v, (7.2.26) for any ϕ ∈C2 c (O ∪Σ−). Together with (7.2.25), we get that γ−f = g on Σ−. In order to deal with the general case g ∈L2(Σ−; dξm), we introduce a sequence (gn) of C2 c (Σ−) such that gn →g in L2(Σ−, dξm) and we next consider the associated sequence of solutions (fn) of W2,Σ just built above. Using the estimates exhibited in Step 1, we get that (fn) is a Cauchy sequence in W2, so that it converges to a limit f ∈W2,Σ. We may pass to the limit in (7.2.26) written for the sequence (fn) and deduce that the same equation holds at the limit for f. Step 4. Uniqueness. Consider two weak solutions fi ∈W2 to the equation (7.2.15) in the sense that E(fi, ϕ) = ⟨F, ϕm2⟩, ∀ϕ ∈C1 c (O ∪Σ−). In particular, the diﬀerence f := f2 −f1 ∈W2 satisﬁes E(f, ϕ) = 0, ∀ϕ ∈C1 c (O ∪Σ−), and from the above discussion γ−f = 0 ∈L2(Σ−; dξm). Thanks to the trace Theorem 7.1.1, we deduce that γf ∈L2 loc(Σ; dξm) and we may choose β(s) = s2 in the Green formula (7.1.14): we get Z O f2{v · ∇xϕ −divv(bϕ) + ∆vϕ + 2f(c −λ)ϕ} −2\|∇vf\|2ϕ = Z Σ+ (γf)2ν · vϕ, for any test function ϕ ∈C2 c ( ¯ O). Choosing ϕ = m2χϱ, with χϱ(v) := χ(v/ϱ), χ ∈C2 c (Rd), 1B1 ≤χ ≤1B2, we deduce Z O f2m2{(λ −ϖ)χϱ + 1 2b · ∇χϱ −∇m m · ∇χϱ −∆χϱ} ≤0. Because f ∈L2H1 m, we may pass to the limit ϱ →∞thanks to the dominated convergence theorem and we obtain Z O f2m2(λ −ϖ) ≤0, 124 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION and thus f = 0. Step 5. Positivity. We assume now that F ≥0 and g ≥0. We proceed similarly as in the previous step by considering β(s) = s2 −, ϕ = m2χM. Letting M →∞, we deduce Z O f2 −m2(λ −ϖ) ≤0, and thus f−= 0. Summing up, gathering the above estimates (7.2.18), (7.2.19), (7.2.20), (7.2.22), (7.2.23), we see that there exists a constant C > 0 such that any function f ∈D(L) satisﬁes ∥f∥L2H1 m+∥ˆ v · ∇xf∥L2H−1 m +∥f ˆ v · ν δ1/4 ∥L2 (7.2.27) + ∥γf∥L2(Σ;dξ2 m)+∥γ+f∥L2(Σ;dξm)≤C(∥f∥L2+∥Lf∥L2) and for any ε > 0 there exists a constant Cε such that any function f ∈D(L) satisﬁes ∥D2/3 x f∥L2(Oε)+∥D2 vf∥L2(Oε)≤Cε(∥f∥L2+∥Lf∥L2). 7.3 Well-posedness problem with reﬂection condition at the boundary We consider now the well posedness problem associated to the stationary equation (λ −L)f = F in O, γ−f = Rγ+f on Σ−, (7.3.28) for a given datum F : O →R, where the kinetic Fokker-Planck operator L is still deﬁned by (7.0.1) and the reﬂexion operator R is described in (7.0.6), (7.0.7), (7.0.8). Theorem 7.3.4. Let us ﬁx a vector ﬁeld b ∈H1 loc( ¯ O), a function c ∈L∞(O) which satisfy the assumptions of Theorem 7.2.3 with a given weight function m : Rd →[1, ∞) for the pure specular reﬂection case α ≡0 and with the weight function m := M −1/2 when α ̸≡0, where M is the Gaussian function (7.0.9) or a more general equilibrium satsifying (7.0.10). In that last case, we furthermore assume one of the two following conditions (i) 1 −ζ + α2/2 ≥δ∗> 0, and we observe that L2(Σ; dξm) ⊂L1(Σ; dξ), (ii) ⟨v⟩2M ∈L1, and we observe that L2(Σ; dξ2 m) ⊂L1(Σ; dξ), where we recall that we have deﬁned dξm := m2\|ν(x)·v\|dvdσ x and dξ2 m := m2(ν(x)· ˆ v)2dvdσ x. For any F ∈L2H−1 m and λ > λ∗, there exists at least one solution f ∈W2 to the Dirichlet problem (7.3.28). Assuming furthermore that λ∗∗:= ess sup (c −divb) < ∞, (7.3.29) and λ > λ∗∗, the solution f is unique and f ≥0 if F ≥0. It is worth emphasizing that the assumptions of Theorem 7.3.4 hold when b = v and m := M −1/2. We also emphasize on the fact that the additional assumptions (i) or (ii) are made in order to prove the uniqueness of the solution during the proof. Proof of Theorem 7.3.4. We split the proof into four steps. 7.3. WELL-POSEDNESS PROBLEM WITH REFLECTION CONDITION 125 Step 1. A priori estimates. We multiply the ﬁrst equation in (7.3.28) by fm2. As in Step 1 of the proof of Theorem 7.2.3, we get Z O (λ −ϖ)f2m2 + 1 2 Z Σ (γf)2m2ν · v + Z O \|∇vf\|2m2 = ⟨F, fm2⟩. Using for instance [28, Lemma 3.1], we have Z Σ (γf)2m2ν · v ≥ Z Σ+ [(1 −ζ)(γ+f)2 + α(D⊥γ+f)2]dξm =: Eζ,α(γ+f) ≥0, (7.3.30) with D⊥g := g −Dg. Using that the contribution of the boundary is nonnegative in the ﬁrst estimate, we ﬁrst deduce (λ −λ∗)∥f∥2 L2 m+∥∇f∥2 L2 m≤∥F∥L2H−1 m ∥f∥L2H1 m, for λ > λ∗, so that min(λ −λ∗, 1)∥f∥L2H1 m≤∥F∥L2H−1 m . From the three above estimates together, for λ > λ∗, we obtain Z O (λ −ϖ)+f2m2 + Z O \|∇vf\|2m2 + 1 2Eζ,α(γ+f) ≤ 1 min(λ −λ∗, 1)∥F∥2 L2H−1 m . (7.3.31) There is no diﬃculty for also getting the pieces of information (7.2.19), (7.2.20), (7.2.22) and (7.2.23), so that in particular f ∈W2. It is worth emphasizing here that when ⟨v⟩2M ∈L1, we have L2(dξ2 m) ⊂L1(Σ; dξ) by using the Cauchy-Schwarz and (7.2.20), so that in particular the boundary condition is well deﬁned. Let us show now how the last conclusion also holds under condition (i) in the statement of the Theorem. We then assume ϑ = 1 in (7.0.10) and we show how to establish an additionnal a priori estimate. We indeed know from (7.2.20) that Z Σ− (αD(γ+f))2(ν · ˆ v)2m2dvdσ x ≤ Z Σ (γf)2(ν · ˆ v)2m2dvdσ x ≤Cλ∥F∥2 L2H−1 m , and similarly as in or [172, proof of Lemma 2.2] that 1 = Z Σx − \|ν(x) · v\|M dv = C Z Σx − (ν(x) · ˆ v)2M dv, ∀x ∈∂Ω, for some constant C ∈(0, ∞), so that Z Σ− (αD(γ+f))2dξm = C Z Σ− (αD(γ+f))2(ν · ˆ v)2m2 ≤CCλ∥F∥2 L2H−1 m . (7.3.32) Summing up (7.3.31) and (7.3.32), and using that (γ+f)2 ≤2(D⊥γ+f)2 + 2(Dγ+f)2, we deduce that Z Σ+ 1 −ζ + α2/22dξm ≤Cλ∥F∥2 L2H−1 m . (7.3.33) Deﬁning f ∈W2,R := {g ∈W2; γ−g = Rγ+g}, we see that W2,R = W2,Σ if 1 −ζ + α2/2 ≥δ∗> 0, but it is worth emphasizing that we may have W2,R ̸= W2,Σ in the general case. 126 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Step 2. Existence when F ≥0. With the help of Theorem 7.2.3, we deﬁne f0 = 0 and, recursively for any n ≥1, we deﬁne fn ∈W2,Σ as the solution of (λ −L)fn = F in O, γ−fn = Rγ+fn−1 on Σ−. (7.3.34) It is worth emphasizing here that γ+fn−1 ∈L2(Σ+; dξm) implies R(γ+fn−1) ∈L2(Σ−; dξm) because of (7.0.5). We also notice that fn ≥0 because F ≥0. By linearity (λ −L)(fn+1 −fn) = 0 in O, γ−(fn+1 −fn) = Rγ+(fn −fn−1) on Σ−, and we thus show recursevily that fn+1 −fn ≥0. In other words, (fn) is an increasing sequence and thus also is (γfn). From (7.3.30), we have Z Σ (γfn)2dξm = Z Σ+ (γ+fn)2dξm − Z Σ− (Rγ−fn−1)2dξm ≥ Z Σ+ (γ+fn)2dξm − Z Σ− (Rγ−fn)2dξm ≥Eζ,α(γ+fn), so that the estimate (7.3.31) holds true for fn (instead of f) uniformly in n ≥1. From the monotonous convergence theorem, there exists f ∈L2H1 m satisfying (7.3.31), (7.3.33), (7.2.20) and such that fn ↗f a.e. Thanks to Proposition 7.1.2, we have γfn ↗γf a.e. on Σ, from what we deduce that Rγ+fn →Rγ+f in L2(Σ−; dξ2 m) thanks to the monotonous convergence theorem. As a consequence, we may pass to the limit in the weak formulation of (7.3.34), and we get that f is a solution of (7.3.28). We may also pass to the liminf in the estimate (7.3.31) written for fn, and we thus deduce that the same estimate holds for f. Step 3. Existence when F ∈L2H−1 m . When F ∈L2 m, we may introduce the splitting F = F+ −F−, just use the previous step for F± and conclude by linearity of the equation. When F / ∈L2 m, we proceed similarly as in and in the following way. We ﬁrst assume ζ ≤ζ∗∈[0, 1) and we consider the mapping Ψ : W2,Σ →W2,Σ, g 7→f = Ψ(g), where f is the solution to the stationary problem    (λ −L)f = F in O γ−f = Rγ+g on Σ−. (7.3.35) The space W2,Σ is endowed with the norm ∥·∥W2,Σ deﬁned by ∥g∥2 W2,Σ= ∥g∥2 L2 m+∥∇vg∥2 L2 m+∥γ+g∥2 L2 m(dξ1). From (7.2.18) and the estimate ∥Rg∥L2(Σ−;dξm)≤ζ∗∥g∥L2(Σ+;dξm) what we obtain by repeating the proof of (7.0.5), we deduce 1 Cλ ∥f∥2 L2H1 m+∥γ+f∥2 L2(Σ+;dξm) ≤ Cλ∥F∥2 L2H−1 m +∥Rγ+g∥L2(Σ−;dξm) ≤ Cλ∥F∥2 L2H−1 m +ζ∗∥γ+g∥L2(Σ+;dξm), for some cosntant Cλ > 0. By linearity of (7.3.35), we deduce that for two functions g1, g2 ∈W2,Σ, and denoting fi := Ψ(gi), we have 1 Cλ ∥f2 −f1∥2 L2H1 m+∥γ+f2 −γ+f1∥2 L2(Σ+;dξm)≤ζ∗∥γ+g2 −γ+g1 ∥2 L2(Σ+;dξm), so that Ψ is a contraction in W2,Σ. By the Banach ﬁxed point theorem, we deduce that there exists a solution f ∈W2,Σ to the equation (7.3.28) in that case. Finally, in order to deal with the case ζ∗= 1, we consider a sequence (ζ∗ n) of [0, 1) such that ζ∗ n ↗1 and the associated sequence 7.3. WELL-POSEDNESS PROBLEM WITH REFLECTION CONDITION 127 (fn) of solutions in W2,Σ associated to the equation (7.3.28) with the modiﬁed reﬂection kernel Rng := ζ∗ nRg. From (7.3.31) and (7.2.20), that sequence satisﬁes ∥fn∥2 L2H1 m+∥γfn∥2 L2(Σ;dξ2 m)+E1,α(γ+fn) ≤Cλ∥F∥2 L2H−1 m . When α ̸≡0, the above estimate or (7.3.33) also implies that (γ+fn) belongs to a weakly compact set of L1(Σ+; dξ). As a consequence, there exist f ∈W2 and ¯ γ± two functions deﬁned on Σ± such that, up to the extraction of a subsequence, fn ⇀f L2H1 m, γ±fn ⇀¯ γ± L2(Σ±; dξ2 m), γ+fn ⇀¯ γ+ L1(Σ+; dξ), Rγ+fn ⇀R¯ γ+ L1(Σ−; dξ), where we have used (7.0.11) for the last convergence. Using Proposition 7.1.2, we may thus pass to the limit in the equation (7.3.28) satisﬁed by fn with modiﬁed reﬂection kernel Rn and we get that f is a solution of (7.3.28). In the pure specular reﬂection case α ≡0, only the ﬁrst line of convergences holds, but that it is enough in order to pass to the limit in the equations (we refer to [173, 172] for similar arguments). Step 4. Other properties. We further assume λ > λ∗∗. We proceed similarly as in . Consider two weak solutions fi ∈W2 to the equation (7.3.28). In particular, the diﬀerence f := f2 −f1 ∈W2 satisﬁes (λ −L)f = 0 in O, γ−f = Rγ+f on Σ−. Using the Green renormalized formula (7.1.14), we have 0 = Z O β′(f)(λ −c)fϕ + β′′(f)\|∇f\|2ϕ + β(f)(divv(bϕ) −v · ∇xϕ −∆vϕ) + Z Σ β(γf)ν · vϕ. for any β ∈C2(R), β′ ∈C1 b (R) and any test function ϕ ∈C2 c ( ¯ O). We choose ϕ = ϕ(v) ≥0, β ≥0 and β′′ ≥0, so that 0 ≥ Z O β′(f)(λ −c)fϕ + β(f)(divv(bϕ) −∆vϕ) + Z Σ β(γf)ν · vϕ. By an approximation argument, we may now take β(s) = \|s\|, and we get 0 ≥ Z O \|f\|{(λ −c)ϕ + (divv(bϕ) −∆vϕ)} + Z Σ \|γf\|ν · vϕ. We observe that in any cases we have f ∈L2 m(O) ⊂L1(O) and γf ∈L1(Σ; dξ). By an approximation argument, we may now take ϕ = 1 and using the L1 estimate (7.0.11) on R (with ζ∗= 1), we get 0 ≥ Z Σ− \|Rγ+f\|\|ν · v\|− Z Σ+ \|γ+f\|\|ν · v\| ≥ Z O \|f\|{λ −c + divvb} ≥(λ −λ∗∗) Z O \|f\|. We deduce that f = 0. The proof of the positivity property follows the same arguments but choosing β(s) = s−. For latter reference, we state the counterpart of the preceding result for the kinetic Fokker-Planck evolution equation. 128 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Theorem 7.3.5. Let us make the same assumptions as in Theorem 7.3.4. For any f0 ∈L2 m, there exists a unique solution f ∈C([0, T); L2 m) ∩L2(0, T; H1 m) for any T > 0 to the kinetic Fokker-Planck evolution equation    ∂tf = Lf in (0, ∞) × O γ−f = Rγ+f on (0, ∞) × Σ−, (7.3.36) with L deﬁned in (7.0.1) and R deﬁned in (7.0.6). The proof of Theorem 7.3.5 is skipped since it is a mere adaptation of the proof of Theo-rem 7.3.4. We refer to [248, Cor. 2 7, Lem. 2.8 and Cor. 2.8] where similar well-posedness results are established (see also for the existence part). 7.4 The ﬁrst eigenvalue problem in a domain with reﬂection at the boundary We consider now the ﬁrst eigenvalue problem for the kinetic Fokker-Planck operator (7.0.1) in a domain with reﬂection at the boundary, namely    λf + v · ∇xf −∆vf −b · ∇vf −cf = 0 in O γ−f = Rγ+f on Σ−, (7.4.37) and the associated dual problem. In this section, we assume that b and c satisfy the assumptions of Theorem 7.2.3 with the weight function m := M −1/2 when α ̸≡0 and for a given weight function m : Rd →[1, ∞) when α ≡0 and R is given by (7.0.6). We additionnally assume that lim inf \|(x,v)\|→∞ϖ(x, v) = −∞, (7.4.38) where we recall that ϖ is deﬁned in (7.2.17). When M is the Gaussian function, we ﬁnd ϖ = c + \|v\|2+d 2 −1 2divb −b · v, so that (7.4.38) holds when b is typically a bounded perturbation of the vector ﬁeld b0(v) = ϑ0v, ϑ0 > 1/2, and more precisely divvb ∈L∞(O) and inf x∈Ωlim inf \|v\|→∞(b · v⟨v⟩−2) ≥ϑ0 > 1/2. The above condition is quite technical but can be seen as a compatibility condition between the thermalization due to the boundary and to the Fokker-Planck collisional operator. We are then able to work in the functional space X := L2 m(O). Theorem 7.4.6. Under the above conditions, the ﬁrst eigentriplet problem associated to (7.0.1) has a unique solution (λ1, f1, φ1) ∈R × X × X′ with f1 > 0 and φ1 > 0. The proof of Theorem 7.4.6 follows from Theorem 2.1.20, Theorem 3.3.11 and Theorem 4.4.14 as a consequence of conditions (H1)–(H5). We prove now that each of these conditions is satisﬁed. Theorem 7.4.6 generalizes [152, Theorem 2.12] where the same problem is tackled for the zero inﬂow condition (α = β = 0) with b = v and c = 0 by using the classical Krein-Rutman theorem in the space X = Cb( ¯ O). We also refer to [116, Theorem 6.8] for a variant and somehow generalisation of . 7.4. THE FIRST EIGENVALUE WITH REFLECTION AT THE BOUNDARY 129 Condition (H1). From Theorem 7.3.4, the operator L satisﬁes (H1) with κ1 := max(λ∗, λ∗∗), with λ∗deﬁned by (7.2.17) and λ∗∗deﬁned by (7.3.29). For later reference, let us state more precisely, the available estimates for f. On the one hand, repeating the proof of Step 1 in the proof of Theorem 7.3.4, we establish that for any λ > κ1 and F ∈L2 m, the solution f ∈W2 to the Dirichlet problem (7.3.28) satisﬁes Z O (λ −ϖ)+f2m2 + Z O \|∇vf\|2m2 + 1 2Eζ,α(γ+f) ≤ 1 λ −λ∗∥F∥2 L2 m. (7.4.39) On the other hand, adapting the proof of (7.2.22), we straightforwardly obtain Z O f2 (ˆ v · ν(x))2 δ(x)1/2 dvdx ≤C Z F2m2, (7.4.40) for some constant C = C(b, c, ν, λ). For εx, εv, ϱ > 0, let us now deﬁne U := {(x, v) ∈O; d(x, ∂Ω) > εx, \|v\|< ϱ}, (7.4.41) and compute Z Uc f2m2 ≤ Z f2m21\|v\|≥ϱ + Z f2m21Ax + Z f2m21B, with Ax := {v ∈Bϱ, (ˆ v · ν(x))2 ≤ε2 v}, B := {(x, v); \|v\|≤ϱ, (ˆ v · n)2 ≥ε2 v, d(x, ∂Ω) ≤εx}. For the second term, we have Z f2m21Ax ≤ Z \|Ax\|2/r′∥f(x, ·)∥2 Lr vdx ≲ (ϱd−1εv)2/r′∥f∥2 L2H1 m, where we have used the Holder inequality with r ∈(1, 2∗/2) in the ﬁrst line and the Sobolev inequality in the second line. For the third term, we have Z f2m21B ≤m2(ϱ)ε1/2 x ε2 v Z O f2 n (ˆ v · ν(x))2 δ(x)1/2 . Gathering these last estimates with (7.4.39) and (7.4.40), we have established that the solution f to equation (7.3.28) furthermore satisﬁes Z Uc f2m2 ≤C( 1 ⟨ϱ⟩2 + ϱd−1εv + m2(ϱ)ε1/2 x ε2 v ) Z F2m2, (7.4.42) for a constant C = C(b, c, Ω, λ) and for any εx, εv, ϱ > 0. The strong maximum principle. Let us now consider a function 0 ≤f ∈W2{0} which satisﬁes the Dirichlet problem (7.3.28) associated to λ > κ1 and a source term 0 ≤F ∈L2 m ∩L∞. In order to simplify the discussion, we assume that the normalization ∥f∥L2 m= 1 holds. For proving the strong maximum principle, we brieﬂy explain how we may adapt the arguments we have presented for the diﬀusive equation in Chapter 6 by taking in particular advantage of the above established estimates, the regularity results established in [106, 115] and some spreading positivity results we learnt in [241, Corollary A.20]. We proceed in three steps. 130 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Step 1. On the one hand, from (7.4.42), we may choose conveniently ϱ−1, εv, εx > 0 small enough in such a way that Z Uc f2m2 ≤1 2∥f∥2 L2 m, where U is deﬁned by (7.4.41). Because of the normalization condition, we have Z U f2m2 = ∥f∥2 L2 m (7.4.43) and consequently f(x0, v0)2 ≥δ2 0 := ∥f∥2 L2 m(2∥1U∥2 L2 m)−1 for at least one point (x0, v0) ∈U. Step 2. On the other hand, let us recall some integrability and regularity results established in for a solution g to the kinetic Fokker-Planck evolution equation ∂tg + v · ∇xg = ∆vg + B · ∇vg + s in V, or a sub-solution ∂tg + v · ∇xg ≤∆vg + B · ∇vg + s in V, for some bounded set V ⊂(0, T) × O, s ∈L2(V) and B ∈L∞(V). For that purpose, given some (t∗, x∗, v∗), we deﬁne Qr := {(t, x, v); t ∈(t∗−r2, t∗], \|x −x∗−(t −t∗)v∗\|< r3, \|v −v∗\|< r}. We claim then that there exist 2 < p < q < ∞, α ∈(0, 1) and for any 0 < r1 < r0 there exists C such that ∥g∥Lp(Qr1)≤C (∥g∥L2(Qr0)+∥s∥L2(Qr0)) (7.4.44) for any nonnegative subsolution g on Qr0 from [106, Theorem 6], ∥g∥L∞(Qr1)≤C (∥g∥L2(Qr0)+∥s∥Lq(Qr0)) (7.4.45) for any nonnegative subsolution g on Qr0 from [106, Theorem 12] and ∥g∥Cα(Qr1)≤C (∥g∥L2(Qr0)+∥s∥L∞(Qr0)) (7.4.46) for any solution g on Qr0 from [106, Theorem 3]. As a consequence of (7.4.44) and a classical covering argument, for any bounded set U ⊂¯ U ⊂O, there exist C0 = C0(U) and C1 = C1(U, λ) such that ∥f∥Lp(U)≤C0 (∥f∥L2(O)+∥F + cf −λf∥L2(O)) ≤C1(∥f∥L2(O)+∥F∥L2(O)). Observing that for ϱ = p/2 > 1, we have v · ∇xfϱ −∆vfϱ −b · ∇vfϱ + ϱfϱ−1(λf −cf −F) = −4(ϱ −1) ϱ \|∇(fϱ/2)\|2≤0, so that fϱ is a weak sub-solution to the kinetic Fokker-Planck equation, we may repeat the argument and obtain in that way that f ∈Lpk(U) for any k ≥1, with pk := ϱk2. Now, choosing k such that pk ≥q and using (7.4.45) (as well as again a classical covering argument), we get ∥f∥L∞(U)≲∥f∥L2(O)+∥F + cf −λf∥Lq(O)≲∥f∥L2(O)+∥F∥Lq(O). Using ﬁnally (7.4.46), we deduce that there exists a constant C = C(U, λ) such that ∥f∥Cα(U)≲∥f∥L2(O)+∥F∥L∞(O). 7.4. THE FIRST EIGENVALUE WITH REFLECTION AT THE BOUNDARY 131 Together with the conclusion of the ﬁrst step, we deduce that there exists a constructive constante r0 > 0 such that f ≥δ01B((x0,v0),r0). Step 3. From [241, Corollary A.20], we deduce that for any bounded set U ⊂¯ U ⊂O, there exists a constructive constant δ = δ(δ0, r0, U) > 0 such that f(x, v) ≥δ for any (x, v) ∈U, where it is worth emphasizing that the hypothesis b, c ∈C(O) made in [241, Corollary A.20] is not really necessary and can be replaced by b, c ∈L∞(U). Because U may be choosen arbitrary, we have established that f > 0 on O and the strong maximum principle. Condition (H2). For a given function 0 ≤h0 ∈C2 c (O) normalized by ∥h0∥L2 m= 1, we deﬁne f0 ∈D(L) as the solution to (κ1 −L)f0 = h0 in O, γ−f0 = Rγ+f0 on Σ−. Taking advantage of the fact that h0 has compact support, we compute 1 = Z O h2 0m2 = Z O (κ1 −L)f0 h0m2 = Z O f0(κ1 −L∗)(h0m2) ≤C1∥f0∥L2 m, with C1 := ∥m−1(κ1 −L∗)(h0m2)∥L2. On the other hand, from (7.2.27), we have ∥f0∥L2H1 m+∥f0 ˆ v · ν δ1/4 ∥L2≤C2, (7.4.47) for a constant C2 only depending on ∥h0∥L2 m, κ1 and the constant C which appears in (7.2.27). Arguing as in (7.4.43), we deduce that Z U f2 0 m2 ≥(2C1)−1, supp h0 ⊂U, (7.4.48) with U = Uϱ deﬁned in (7.4.41) and ϱ > 0 small enough (chosen constructively from C2 and C1). From the above constructive strong maximum principle, we deduce that f0 ≥ε1U ≥1/C0h0 for some ε, C0 > 0. We conclude as in the Second constructive argument for (H2) in Section 6.1. Coming back indeed to the equation, we have Lf0 = κ1f0 −h0 ≥κ1f0 −∥h0∥L∞1U ≥(κ1 −∥h0∥L∞C0)f0, so that (H2) holds with κ0 := κ1 −∥h0∥L∞C0 from Lemma 2.1.4-(ii). Condition (H3). Let us ﬁx κ < κ0 arbitrary. We deﬁne Bf := Lf −nχR(v)f for any f ∈W2,R, with χR ∈D(Rd) such that 1BR ≤χR ≤1B2R and for some given n, R ≥0 to be speciﬁed below. We observe that, at least formally, Z fm2(B −κ)f = Z O (ϖ −κ −nχR)f2m2 −1 2 Z Σ (γf)2m2ν · v − Z O \|∇vf\|2m2. Thanks to (7.4.38), there exists a constant R > 0 such that sup v∈Rd\BR ϖ ≤κ. Choosing n := sup ϖ+ −κ, we deduce that ϖ −κ −nχR ≤0. On the other hand, because of (7.3.30), the contribution of the boundary term in the above identity is non positive. We thus deduce that (B −κ) is dissipative in L2 m. We now establish that the associated operator B has compact resolvent. For F ∈L2 m, we consider f ∈L2 m the solution to −Bf = F in O, γ−f = Rγ+f on Σ−, (7.4.49) 132 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION which existence follows from Theorem 7.3.4. From the above discussion (with κ = −1) and the same arguments as in Step 1 of the proof of Theorem 7.2.3, we have Z f2⟨ϖ⟩−m2 + 2 Z \|∇vf\|2m2 ≤ Z F2m2. (7.4.50) Together with the regularity estimate (7.2.23) and the compact imbedding H2/3(U) ⊂L2(U), we conclude that B has compact resolvent. The operator A on L2 m deﬁned by Af := nχR(v)f being bounbded, we may apply Lemma 2.1.8-(2) and we deduce that (H3) holds for both the primal and the dual problems. Condition (H4) is nothing but the yet established strong maximum principle. A variant of condition (H5). Consider (f, λ) a pair of eigenfunction and eigenvalue such that λ ∈ΣP+(L). Arguing similarly as in the proof of condition (H5) in Section 6.1, we know that e Lf = iϑf, ϑ ∈R, e L\|f\|= 0 and introducing the real and complex part decomposition f = g + ih, we have Z O 1 \|f\|2 \|g∇vh −h∇vg\|2= 0, and ﬁnally g∇vh −h∇vg = 0 a.e. on O. Because of the regularity estimate presented during the above proof of the strong maximum principle, the functions f has Hölder regularity, and thus g and h are continuous on O. Because \|f\|̸≡0, we may claim that there exists a point (x0, v0) ∈O such that h(x0, v0) > 0 for instance. Denoting by ω the connected component of {(x, v) ∈O; h(x, v) > 0} containing (x0, v0), we have ∇(g/h) = 0 on ω, and thus g = α(x)h on ω for some continuous function α : Ω→R. Coming back to the eigenvalue equation that we may write in the following system form e Lg = −ϑh, e Lh = ϑg, we compute −ϑh = e L(αh) = α e Lh −hv · ∇xα = αϑg −hv · ∇xα on ω, so that −ϑ = α2ϑ −v · ∇xα on ω. We deduce that α is a constant on ω and ﬁnally ϑ = 0. We have thus established that λ = λ1. At this stage, we may use Theorem 2.1.20, Theorem 3.3.11 and Theorem 4.4.14, in order to get the conclusions (C1), (C2) and (C3) about the existence and uniqueness of the eigentriplet (λ1, f1, φ1) which satisﬁes f1 > 0, φ1 > 0, λ1 is algebraically simple and on the triviality of the boundary punctual spetrum. We brieﬂy explain how we may deduce the stability of f1 by adapting some arguments developped in and already mentioned. On the one hand, we know from [171, Lemma 1.1] that any solution f to the rescaled evolution equation (7.3.36) with L replaced by e L = L −λ1 satisﬁes ∂t(H(X)f1φ1) + divx(vH(X)f1φ1) −divv(φ2 1∇v(H(X)f1/φ1)) = −H′′(X)f1φ1\|∇vX\|2, for any convex function H : R →R and with X := f/f1. After integration, we get d dt Z O H(X)f1φ1 + Z Σ ν · vH(X)f1φ1 = − Z O H′′(X)f1φ1\|∇vX\|2, (7.4.51) 7.4. THE FIRST EIGENVALUE WITH REFLECTION AT THE BOUNDARY 133 When H(s) := \|s\|, the boundary term is Z Σ \|γf\|γφ1ν · v = Z Σ+ \|γ+f\|R∗γ−φ1ν · v − Z Σ− \|Rγ+f\|γ−φ1\|ν · v\| ≥ Z Σ+ \|γ+f\|R∗γ−φ1\|ν · v\|− Z Σ− R\|γ+f\|γ−φ1 \|ν · v\|= 0, from what we deduce the non expansive property Z O \|ft1\|φ1 ≤ Z O \|ft0\|φ1, ∀t1 ≥t0 ≥0. (7.4.52) On the other hand, from the Cauchy-Schwarz inequality, we have (Rγ+f)2 ≤(Rγ+f1)R(γ+f2/γ+f1), so that Z Σ− (Rγ+f)2 Rγ+f1 γ−φ1\|ν · v\|≤ Z Σ− R(γ+f2/γ+f1)γ−φ1\|ν · v\| and ﬁnally Z Σ (γf)2(γf1)−1γφ1 ν · v ≤0. When H(s) = s2, the equation (7.4.51) and the last inequality imply dt dt Z O f1φ1(f/f1)2 + 2 Z O f1φ1\|∇v(f/f1)\|2≤0. (7.4.53) We next recall a classical compactness result. Lemma 7.4.7. Let (gn) be a sequence of functions such that (gn) is bounded in L∞(0, T; L2 xv,loc) ∩L2(0, T; L2 x,locH1 v,loc) and ∂tgn + v · ∇xg −∆vgn = Gn bounded in L2 loc, then (gn) belongs to a strong compact set of L2 loc. Proof of Lemma 7.4.7. We just sketch it. Because ∂tgn + v · ∇xg = ∆vgn + Gn bounded in L2 txH−1 v , the usual averaging lemma in [107, 74] implies that (gn ∗ρ) belongs to a strong compact set of L2 loc, for any ρ ∈D(Rd). On the other hand, introducing a molliﬁers sequence (ρε) and writing then gn = (gn −gn ∗ρε) + gn ∗ρε, we see that the ﬁrst term is small uniformly in n as ε →0 and the second term is relativelly compact thanks to the ﬁrst step, from what we immediately conclude. Now, for 0 ≤f0 ∈L1 φ1, we introduce the sequence f0,k := (f0 ∧k)1Uk ∈L2(f−1 1 φ1) ∩L2, with Uk := {(x, v) ∈O; δ(x) > 1/k, \|v\|≤k}, and the associated solution fk ∈L∞(0, T; L2) ∩ L2(0, ∞; L2 xH1 v). Because of (7.4.53), for any increasing sequence (tn) which converges to ∞and for any function ϕm ∈D(O), 1Um ≤ϕm ≤1, the rescaled and truncated function gn(t) := fk(t+tn)f−1 1 e−λ1(t+tn)ϕm meet the hypothesis of Lemma 7.4.7, from what we classically deduce that the sequence of ˜ fn(t) := fk(t + tn)f−1 1 e−λ1(t+tn) is relatively strongly compact in L2 loc. Repeating the proof of Theorem 3.4.20 and Theorem 4.5.20 (see also [171, Thm. 3.2]), we deduce that ˜ fn(t) →⟨f0,k, φ1⟩f1 as t →∞. Together with the above non expansive property (7.4.53), we deduce that ft →⟨f0, φ1⟩f1 in L1 φ1 as t →∞. We summarize our convergence result in the following theorem. 134 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Theorem 7.4.8. For any f0 ∈L2, the holds ft →⟨f0, φ1⟩f1 in L1 φ1 as t →∞. Theorem 7.4.8 generalizes [152, Theorem 2.18] and [116, Theorem 6.8] for the zero inﬂow condition and [1, Theorems 1.6 & 1.7] for the torus case. It is worth emphasizing that in these papers the longtime convergence is established with exponantial rate (with constructive estimate in ). In the proof is based on a representation formula for the associated semigroup S which is proved to have a kernel pt ∈(L1 ∩L∞∩C∞)(O) for any t > 0 (see [152, Thm. 2.4 & 2.6] as well as [214, 124, 151]). One then classically deduces that St ∈K (X) for any t > 0 and X = Lp, p ∈[1, ∞], or X = C0 (see [152, Thm. 2.18]), and next one may apply Theorem 4.6.25. We follow now a similar approach as in [152, 116]. We start with a series of technical results. Here, we make the additional assumption ϖ♯(x, v) := sup 1≤p≤∞ wp(x, v) ≤κ2 < ∞, (7.4.54) with ϖp := (2 −p) p ∆m m + 2 p′ \|∇m\|2 m2 + c −1 p div(bmp) mp . Lemma 7.4.9. For any ﬁxed κ < κ0 there exists ϱx > 0, ϱv > 0 and κ2 ∈R such that deﬁning Af := nξϱv(v)ζϱx(x)f with n ≥0, ξρv ∈D(Rd), 1\|v\|≤ρv ≤ξρv ≤1\|v\|≤2ρv, ζρx ∈D(Ω), 1δ(x)≥ρx/2 ≤ζϱx ≤1δ(x)≥ϱx, and next B := L −A, there hold ∥SB(t)∥B(L2 m) ≲ eκt, ∀t ≥0, (7.4.55) ∥SB(t)∥B(Lp mp) ≲ eκ2t, ∀t ≥0, ∀p ∈(2, ∞], (7.4.56) with mp := M 1−1/p. Proof of Lemma 7.4.9. We ﬁrst recall from Step 1 of the proof of Theorem 7.2.3 and (7.3.30) that (Lf, f)L2 m = − Z \|∇f\|2m2 −1 2 Z Σ (γf)2m2ν · v + Z f2ϖm2 ≤ − Z \|∇f\|2m2 + Z f2ϖm2 and, with ψ deﬁned in (7.2.21), (−Lf, f)L2 ψ = −1 2 Z f2(v · ∇xψ) − Z f b ⟨v⟩· ∇vf⟨v⟩ψ + Z ∇v(fψ) · ∇vf − Z cf2ψ ≤ − Z f2 (ˆ v · ν(x))2 δ(x)1/2 dvdx + C Z (f2 + \|∇f\|2). Deﬁning then ˜ m := m −βψ, with β > 0 small enough, and summing up the two previous estimates, we get (Lf, f)L2 ˜ m ≤ −β Z f2 (ˆ v · ν(x))2 δ(x)1/2 −1 2 Z \|∇f\|2m2 + Z f2(ϖm2 + 1). Similarly as in (7.4.41), we deﬁne U := {(x, v) ∈O; δ(x) > ϱx, \|v\|< ϱv}, and we observe that Uc ⊂A ∪B ∪C, 7.4. THE FIRST EIGENVALUE WITH REFLECTION AT THE BOUNDARY 135 with A := {v ∈Bϱv, \|ˆ v · ν(x)\|≤εv}, B := {v ∈Bϱv, \|ˆ v · n\|≥εv, δ(x) ≤ϱx}, for some εx > 0, and C := Bc ρv. We next repeat the proof of (7.4.42), and we get Z Uc f2m2 ≲ (ϱd−1 v εv)2/r′ Z \|∇vf\|2+m(ϱv)2 ϱ1/2 x ε2 v Z f2 (ˆ v · ν(x))2 δ(x)1/2 + 1 ϖ−(ρv) Z f2ϖ−m2. Observing that Z f2(ϖm2 + 1) ≤κ Z f2 ˜ m2 + Cκ Z U f2m2 + Cκ Z Uc f2m2 with Cκ := sup(ϖ + 2 −κ)+ < ∞, and A ≥Cκ1U for n := Cκ, altogether, we conclude with (Bf, f)L2 ˜ m ≤κ∥f∥L2 ˜ m. We then classically deduce that (7.4.55) holds. Similarly as for the ﬁrst estimate and in the proof of [174, Lem. 3.8], for any smooth, rapidly decaying and positive function f, we have Z (Lf)fp−1mp = − Z Σ (mγf)p p ν · v −(p −1) Z \|∇(mf)\|2 (mf)p−2dx + Z fpϖpmp. From Darozès-Guiraud (or Jensen) inequality, we know that the ﬁrst (boundary) term is nonpositive (see or [172, Rem. 6.4]) and we then classically conclude to (7.4.56). Lemma 7.4.10. There exists a ﬁnite family 2 = p0 < p1 < · · · < pk < ∞and α ∈(0, 1) such that such that both C = B, L, for any T > τ > 0 and V ⊂⊂O, Z T τ ∥ASC(t)f0∥L pj m dt ≤Cpj pj−1∥f0∥L pj−1 m , j = 1 . . . , k, (7.4.57) sup t∈[τ,T] ∥ASB(t)f0∥L∞≤C∞ pk∥f0∥Lpk, (7.4.58) sup t∈[τ,T] ∥SB(t)f0∥Cα(V)≤Cα ∞∥f0∥L∞. (7.4.59) Proof of Lemma 7.4.10. For 0 ≤f0 ∈L2 m, let us denote f := SBf0 which thus satisﬁes the PDE ∂tf −Bf = s := cf in D′((0, T) × O). Let us ﬁx two open sets Ui such that [τ, T] × suppξ × suppζ ⊂U0 ⊂⊂U1 ⊂⊂(0, T) × O. From [106, Thm 6] and a covering lemma, there exists a constant ¯ C0 > 0 and p1 > 2 such that ∥f∥Lp1(U0)≤¯ C0(∥f∥L2(U1)+∥s∥L2(U1)). The estimate (7.4.57) for j = 1 then follows from Theorem 7.3.5 (and the classical underlying energy estimate). On the other hand, [106, Thm 12] similarly implies that there exists a constant ¯ Ck > 0 and pk ∈(p1, ∞) such that ∥f∥L∞(U0)≤¯ Ck(∥f∥L2(U1)+∥s∥Lpk(U1)), and interpolating with the previous estimate, we get ∥f∥Lpj (U0)≤¯ Cj−1(∥f∥L2(U1)+∥s∥Lpj−1(U1)), ∀j, 2 ≤j ≤k −1. The growth bound (7.4.56) and the two last estimates imply (7.4.58) and (7.4.57) for any 2 ≤j ≤k −1. Finally, [106, Thm 3] similarly implies that there exists a constant ¯ Ck+1 > 0 and α ∈(0, 1) such that ∥f∥Cα(U0)≤¯ Ck+1(∥f∥L2(U1)+∥s∥L∞(U1)), from what we deduce (7.4.59) in the same way. 136 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Theorem 7.4.11. The conclusion (CE3) holds in L2 m with exponential rate but non constructive rate. Proof of Theorem 7.4.11. We introduce the splitting SL = V + Kc 1 + Kc 2 + K, where we set K := ΥSB ∗τ Kk+2, Kc 1 := W ∗SL −SB ∗τ Kk+2, Kc 2 := (1 −Υ)SB ∗τ Kk+2. Here we deﬁne V and W with the usual notations (2.2.41) where we take N = k + 2. Next for two functions (of operators) a and b and 0 < τ < T, we deﬁne (a ∗τ b)(t) := Z t−τ τ a(t −s)b(s) ds if t ∈[τ, T −τ], (a ∗τ b)(t) := 0 if t ∈[τ, T −τ]c, and then recursively K0 = Id, K1 := ASL, Kℓ:= (ASB) ∗τ Kℓ−1 for ℓ≥2. We ﬁnally deﬁne Υg := χεg with χε ∈D(O), 1U2ε ≤χε ≤1Uε, with Uε := {\|v\|≤1/ε, δ(x) > ε}. The sequel of the proof is split into two steps. Step 1. On the one hand, we compute ∥(SB ∗τ Kk+2)f0∥Lp1 m ≤ ∥SB∥L∞(B(Lp1 m )) Z T−τ τ ∥ Z t−τ τ ASB(t −s)Kk+1(s)dsf0∥Lp1 m dt ≤ CT Z T τ Z T τ ∥ASB(t)Kk+1(s)f0∥Lp1 m dtds ≤ CT Cp1 2 Z T τ ∥Kk+1(s)f0∥L2 mds, and thus ∥(SB ∗τ Kk+2)f0∥Lp1 m ≤CT ∥f0∥L2 m, (7.4.60) where we have used (7.4.56) in the ﬁrst line, the Fubini theorem in the second line, (7.4.57) with j = 1 in the third line and several times (7.4.55) in the last line. From the very deﬁnition of A and SB as well as (7.4.55), we may ﬁx κB ∈(κ, κ0) and next T > 0 such that ∥V (T)∥B(L2 m)≤1 3eκBT . We may next use (7.4.55) and ﬁx τ > 0 small enough such that ∥Kc 1∥B(L2 m)≤τCT ≤1 3eκBT . Last, because of (7.4.60), we may ﬁx ε > 0 small enough, in such a way that ∥Kc 2f0∥L2 m≤η(ε)∥(SB ∗τ Kk+2)f0∥Lp1 m ≤1 3eκBT ∥f0∥L2 m. The three last estimates together, we have established ∥V (T) + Kc 1 + Kc 1∥B(L2 m)≤eκBT . (7.4.61) Step 2. Performing the same kind of computatiuons as for proving (7.4.60), we get Z T 0 ∥(ASB ∗τ Kj−1)(s)f0∥L pj m ds ≤ Z T 0 Z T−τ τ ∥ASB(t)Kj−1(s)f0∥L pj m dtds ≤ Cpj pj−1 Z T 0 ∥Kj−1(s)f0∥L pj−1 m ds, 7.4. THE FIRST EIGENVALUE WITH REFLECTION AT THE BOUNDARY 137 for j = 1, . . . , k + 1, pk+1 := ∞. Iterating, and using (7.4.57) with j = 1, we get Z T 0 ∥Kk+1(s)f0∥L∞ m ds ≲∥f0∥L2 m. Similarly, we may write sup [τ,T] ∥Kk+2f0∥L∞ m ≤ sup t∈[τ,T] Z t τ ∥ASB(t −s)Kk+1(s)f0∥L∞ m ds ≤ sup t∈[τ,T] ∥ASB(s)∥B(L∞ m ) Z T τ ∥Kk+1(s)f0∥L∞ m ds and ∥Kf0∥Cα(O) ≤ Z T−τ τ ∥SB(t −s)Kk+2(s)f0∥Cα(Uε) ds ≤ Cα ∞T sup [τ,T] ∥Kk+2f0∥L∞ m . The three last estimates together and the compast suport condition suppχε ⊂⊂O imply ∥Kf0∥Cα∩L2 m≲∥f0∥L2 m, ∀f0 ∈L2 m, from what we deduce that K ∈K (L2 m). We may apply Theorem 4.6.25 in order to conclude. 138 CHAPTER 7. THE KINETIC FOKKER-PLANCK EQUATION Chapter 8 A mutation-selection model In this chapter, we consider the mutation-selection evolution equation associated to the mutation-selection operator Lf := J ∗f −W(x)f (8.0.1) deﬁned on functions f : Rd →R, f = f(x), where J is a the mutation kernel, ∗stands for the convolution operator and W is a conﬁning potential. This equation appears in the modeling of genetic variability in evolutionary biology. As we say in the introduction, ft(x) represents the density of a population, at time t, of phenotypic trait x on the multi-dimensional phenotypic trait space Rd. the kernel J is related to mutations and the potential W to the selection process. Diﬀerent selection potentials have been considered. The case of linear ﬁtness W(x) = −x is completely studied by the works of M. Alfaro and R. Carles 2 and R. Burger and M.E. Gil whereas the quadratic case, that is W(x) := x2 (say N = 1), was considered in M. Kimura and analyzed by R. Bürger for Equation (1.1) and by M. Alfaro and R. Carles . We also can ﬁnd more realistic cases for which ﬁtness functions are deﬁned by a linear combination of two components (e.g. birth and death rates), each maximized by diﬀerent optimal values of the underlying trait, a typical case being W(x) := x4 −x2 with is studied in Alfaro and M. Veruete . The study of the eigenvalue problem associated with this model has been of interest to the scientiﬁc community for some time. In particular the works of R. Bürger and I.M. Bomze and also O. Bonnefon, J. Coville, and G. Legendre for bounded domains, show that the lack of regularizing eﬀect of the integral operator compared to the Laplace operator can arrive at a concentration phenomenon which leads to the formation of singular measure for the ﬁrst eigenfunction. A ﬁrst fundamental question is to identify sharp conditions on J and W that ensure the existence of a principal eigenfunction. Some of those conditions have been presented in . In this chapter, we present a relaxation of these conditions, mainly by removing the J symmetry assumption, see remark 12.2 (3) below. Under these relaxed assumptions, we present a quantiﬁcation of the spectral gap, and its stability result that generalizes the one shown in . We assume that the mutation kernel J is a positive ﬁnite measure of Rd which is lower bounded on a neighborhood of the origin, or in other words 0 ≤J ∈M1(Rd), J ≥J∗1Br, (8.0.2) for some constants J∗, r > 0. We also assume that the selection potential W : Rd →R is continuous and satisﬁes W(x) > W(0) = 0, ∀x ̸= 0, W(x) →+∞as \|x\|→∞. (8.0.3) 139 140 CHAPTER 8. A MUTATION-SELECTION MODEL We ﬁnally assume the following compatibility condition between mutation and selection: there exist β > 0 and a bounded Borel set A ⊂Rd such that a := ess inf x∈Aβ Z x−Aβ J(dz) W(x −z) > 1, (8.0.4) J = J1 + J2, J1 ∈C1 c (Rd), κ∗:= ∥J2∥1:= Z RddJ2 < κ0 := (a −1)β, (8.0.5) where we use the notation Aβ = A ∩{W ≥β}. In the sequel, we work in the Banach lattice X := L1(Rd). Theorem 8.0.1. Under the above assumptions, we can arrive to the following conclusions, 1. The ﬁrst eigentriplet problem (1.3.30)-(1.3.31) admits a unique solution (λ1, f1, φ1) ∈R × X+×X′ +, with λ1 ≥κ0, 0 < f1 ∈L1 ⟨W⟩(Rd)∩L∞ ⟨W⟩(Rd) and 0 < φ1 ∈L1 ⟨W⟩(Rd)∩L∞ ⟨W⟩(Rd). 2. Moreover, L generates a semigroup SL on X = L1(Rd) and for any f0 ∈X, there holds ∥e−λ1tSL(t)f0 −⟨φ1, f0⟩f1∥L1≤Ce−αt∥f0 −⟨φ1, f0⟩f1∥L1, (8.0.6) for any t ≥0 and for some constructive constants C, α > 0. Let us comment on the above result which is our main result in this section. Remark 8.0.2. (1) Assumption (8.0.4) is satisﬁes when W is small enough in a neighborhood of the origin. It is for instance satisﬁed if W −1 / ∈L1(B1). That is in particular the case in dimension d = 1 when W is Lipschitz, because of the condition W(0) = 0. (2) Assume J(x) = ε−dρ(ε−1x) with ρ ∈C1 c (Rd) ∩P(Rd) and ρ ≥ρ∗1B1, ρ∗> 0, so that J = J1 and J2 = 0, and W = W(\|x\|). We may observe that for β > 0 and ε > 0 small enough inf β≤W(x)<2β Z β≤W(y)<2β J(x −y) W(y) dy =: a ≥ρ∗ 2β meas{Rd + ∩B1} > 1, so that (8.0.4) holds with A := {W(x) < 2β}. (3) Assumption (8.0.4) is similar to [153, Condition (2.3)], see also [3, Assumption 2.6] and the comparaison with [3, Assumption 2.4], as well as [43, Condition (3.7)-(3.8)] and [46, p. 250, Note added in proof.]. On the other hand, the conditions on J are relaxed since J may have singular part, and more importantly, it is not assumed to be symmetric as in [3, Assumption 2.2]. (4) Optimal conditions linking J and W for the existence of a spectral gap are still unknown. In the recent paper , using variational methods in a L2 framework, the authors obtain a quantiﬁed spectral gap and the associated exponential stability when the mutation kernel J is additionally assumed to be symmetric. Up to our knowledge, Theorem 8.0.1 is the very ﬁrst result providing a quantiﬁed spectral gap for a non-symmetric mutation kernel J. (5) Condition (8.0.4) can be compared to the condition ¯ a := ess sup x∈Rd Z Rd J(x −y) W(y) dy < 1, under which no ﬁrst eigenfunction may exist in X. First, we claim that λ1 ≥0. Indeed, considering ϵ > 0 and fϵ = 1Bϵ, we have Lfϵ ≥−( inf Bϵ W)fϵ, 141 so that the condition (H2) holds for κ0 = −infBϵ W for any ϵ > 0. Since W is continuous and W(0) = 0, we deduce that λ1 ≥0 by passing to the limit ϵ →0. Assume now by contradiction that there exists f1 ∈X+ \ {0} such that λ1f1(x) = Lf1(x) = Z Rd J(x −y)f1(y) dy −W(x)f1(x) (8.0.7) and deﬁne, for any ϵ > 0, the function ϕϵ(x) = 1 ϵ+W(x) ∈L∞(Rd). Testing (8.0.7) against ϕϵ we get for any ϵ ∈(0, 1) 0 ≤λ1⟨f1, ϕ1⟩≤λ1⟨f1, ϕϵ⟩= ZZ J(x −y) ϵ + W(x)f1(y) dx dy − Z W(x) ϵ + W(x)f1(x) dx ≤¯ a Z f1 − Z W(x) ϵ + W(x)f1(x) dx and passing to the limit ϵ →0 we obtain the contradiction 0 ≤λ1⟨f1, ϕ1⟩≤(¯ a −1) R f1 < 0. However, there always exists a principal eigenvector f1 in M1(Rd). The proof of Theorem 8.0.1 follows from Theorem 2.1.20, Theorem 3.3.11 and Theorem 4.4.14 as a consequence of conditions (H1)–(H5) that we establish now. Setting D(L) := L1 ⟨W⟩(Rd), we observe that L is an unbounded closed operator with dense domain D(L). Condition (H1) and (H1′). We deﬁne the semigroup SW (t)f(x) := e−W(x)tf(x), ∀f ∈Lp, p ∈[1, ∞], which is clearly a positive semigroup of contractions. We next deﬁne SL as a bounded perturbation of SW . It is also positive and it satisﬁes the growth estimate ∥SL(t)∥B(Lp)≤e∥J∥1t, where we recall that ∥J∥1 stands for the L1 norm or the total variation norm of J. We deduce that (H1) holds true with κ1 := 1 = ∥J∥1 thanks to Lemma 2.1.2-(i). Multiplying Lf by sign f, for f ∈D(L), we immediately get Kato’s inequality (sign f)Lf = (sign f)J ∗f −W\|f\|≤J ∗\|f\|−W\|f\|= L\|f\|. Condition (H2). Let us deﬁne f0 := 1 W(x)1Aβ, where Aβ is introduced in condition (8.0.4). We compute Lf0 = J ∗ 1Aβ 1 W −1Aβ ≥ J ∗ 1Aβ 1 W −1 1Aβ ≥ ess inf x∈Aβ h J ∗(1Aβ 1 W ) i −1 1Aβ = (a −1)1Aβ ≥(a −1) β W 1Aβ = κ0f0, where in the second equality we have used the very deﬁnition of a in assumption (8.0.4). We conclude that (H2) holds thanks to Lemma 2.1.4-(ii). Condition (H3). We introduce the splitting L = A + B, Af := J1 ∗f, Bf := J2 ∗f −W(x)f. (8.0.8) Arguing as in the proof of condition (H1), we see that B is the generator a positive semigroup in Lp(Rd), 1 ≤p ≤∞, with growth bound ω(SB) ≤κ∗and thus (α −B) is invertible for any α ≥κ0 > κ∗, with ∥(α −B)−1∥B(Lp)≤ 1 α −κ∗ . (8.0.9) 142 CHAPTER 8. A MUTATION-SELECTION MODEL Next, observing that (W + α)h = (α −B)h + J2 ∗h, for any h ∈D(L) and α ≥κ0, we deduce that (W + α)(α −B)−1g = g + J2 ∗((α −B)−1g), (8.0.10) for any g ∈X and α ≥κ0. Together with (8.0.9), we deduce ∥(α −B)−1g∥Lp W ≤∥g∥Lp+∥J2 ∗((α −B)−1g)∥Lp≤ α α −κ∗ ∥g∥Lp, (8.0.11) for any g ∈Lp and α ≥κ0. Deﬁning W(α) := (α −B)−1A, we ﬁnally deduce from (8.0.10) the identity W(α)f = 1 W + αAf + 1 W + αJ2 ∗((α −B)−1Af), for any f ∈X and α ≥κ0. We may then compute ∥W(α)f∥L∞≤1 α∥Af∥L∞+ 1 α∥J2∥1∥(α −B)−1Af∥L∞, and together with (8.0.9) for p = ∞, we deduce ∥W(α)f∥L∞≤∥J1∥∞ 1 α −κ∗ ∥f∥L1, (8.0.12) for any f ∈X and α ≥κ0. Starting from the same identity, we prove in a similar way ∥W(α)f∥L∞ W ≤∥J1∥∞ α α −κ∗ ∥f∥L1, (8.0.13) for any f ∈X and α ≥κ0. As a conclusion and gathering (8.0.9), (8.0.11), (8.0.12) and (8.0.13), we have established that W(α) : L1 →L1 ⟨W⟩∩L∞ ⟨W⟩, (8.0.14) with uniform bound for any α ≥κ0. Observing that L1 ⟨W⟩∩L∞ ⟨W⟩⊂L1 is weakly compact and using Lemma 2.1.12 with p = 1, we deduce that (H3) holds. We can actually strengthen the compactness by noticing that A : L1 →L1 W ∩W 1,1 is bounded because of assumption (8.0.5). This ensures that A : L1 →L1 is compact, from what we deduce that W(α) : L1 →L1 is strongly compact for all α ≥κ0. We may thus apply Lemma 2.1.8-(2) to infer that condition (H3) holds for both the primal and the dual problems. Condition (H4). Assume that λ ≥λ1 and f ∈D(L) = L1 ⟨W⟩satisfy ∥f∥L1= 1, f ≥0, (λ −L)f ≥0. (8.0.15) Denoting WR := infBc R W, we compute Z BR f ≥ Z Rd f − 1 WR Z Bc R fW ≥1 − 1 WR ∥f∥L1 ⟨W ⟩≥1/2, for R > 0 large enough by taking advantage of the fact that W(x) tend to inﬁnity when \|x\|→∞. In particular, there exists xf 0 ∈BR such that Z Br/2(xf 0) f ≥δ := 1 2 r 2R d > 0, where we recall that r is deﬁned in (8.0.2). We deduce that (J ∗f)(x) ≥J∗ Z Br/2(xf 0) f(y)dy1Br/2(xf 0)(x) ≥J∗δ1Br/2(xf 0)(x). 143 Using the equation (8.0.15), we obtain f(x) ≥(J ∗f)(x) W(x) + λ ≥ J∗δ W[R] + λ1Br/2(xf 0)(x), for W[R] = supBR W(x). With that last information and (8.0.2) again, we have now J ∗f ≥J∗ 2d J∗δ W[R] + λ 1Br(xf 0), and, iterating the argument, we deduce f ≥ Jm ∗ 2(m−1)d(W[R] + λ)m−1 δ1Bmr/2(xf 0) ≥¯ γ1BR, with ¯ γ = ¯ γ(R) > 0 for m = m(R) large enough. Choosing R an integer, we have proved that f ≥h0 := ¯ γ(R)1BR + X n≥R ¯ γ(n + 1)1Bn+1\Bn > 0. (8.0.16) That means that the (H4) holds, with constructive lower bound. Condition (H5). Let us consider f ∈L1 ⟨W⟩{0} and λ ∈C such that (4.4.15) holds, in particular L\|f\|= (ℜeλ)\|f\| and L\|f\|= ℜe(signf)Lf. (8.0.17) The ﬁrst equality means that ℜeλ is an eigenvalue associated to a positive eigenfunction, and Lemma 3.3.15 then enforces ℜeλ = λ1. Lemma 3.3.16 subsequently ensures that \|f\|∈ (Span f1)+ \ {0}, and in particular \|f\|≫0. Throwing away the term W\|f\| in each side of the second identity in (8.0.17), we have ℜe ¯ f \|f\| (J ∗f) = J ∗\|f\|. Integrating this equation, we get Z R2d J(x −y)ℜe h \|f(y)\|− ¯ f(x) \|f(x)\|f(y) i dy = 0. From the positivity condition (8.0.2) on J, we deduce \|f(y)\|− ¯ f(x) \|f(x)\|f(y) = ℜe h \|f(y)\|− ¯ f(x) \|f(x)\|f(y) i = 0, ∀x, y ∈Rd, \|x −y\|< r, and thus ¯ f(x)/\|f(x)\|= ¯ u for any x ∈Rd for a constant u ∈C. That ends the proof of the reverse Kato’s inequality (H5). Proof of theorem 8.0.1 part (1). We may use Theorem 2.1.20 in order to establish the existence of a solution (λ1, f1, φ1) ∈(0, +∞) × L1 × L∞to the ﬁrst eigentriplet problem (1.3.30)-(1.3.31). From Theorem 3.3.11 and Theorem 4.4.14, this solution is unique, f1 ≫0, φ1 ≫0, λ1 is algebraically simple for both L and L∗and it is the unique eigenvalue in Σ+(L). Due to (8.0.14), we actually have f1 ∈L1 ⟨W⟩∩L∞ ⟨W⟩. Observing that L∗is of the same type as L, L∗φ = ˇ J ∗φ −Wφ, ˇ J(x) := J(−x), and considering the dual problem as a primal problem in L1, Theorem 2.1.20 also provides the existence of λ∗ 1 > 0 and 0 ≪φ∗ 1 ∈L1 ⟨W⟩∩L∞ ⟨W⟩such that L∗φ∗ 1 = λ∗ 1φ∗ 1. 144 CHAPTER 8. A MUTATION-SELECTION MODEL Because of the f1 and φ1 are strictly positive, we have in fact λ∗ 1 = λ1 and the simplicity of λ1 then yields that Span φ∗ 1 = Span φ1. This ensures that φ1 ∈L1 ⟨W⟩∩L∞ ⟨W⟩and also that φ1 enjoys the explicit lower bound (8.0.16). Besides, we can prove ∥φ1∥L∞ ⟨W ⟩≤∥J1∥L1 λ1 λ1 −κ∗ ∥φ1∥L∞≤∥J1∥L1 κ1 κ0 −κ∗ ∥φ1∥L∞ by arguing similarly as for (8.0.13). In order to prove Theorem 8.0.1 part (2) with constructive constants we use a Doeblin-Harris type argument Lemma 8.0.3 (Lyapunov Condition). Under the above assumptions, there are γL ∈(0, 1), T > 0 and K > 0 such that ∥e ST f∥L1≤γL∥f∥L1+K∥f∥φ1. Proof. Writing ft = S(t)f, we have d dt Z Rd\|ft\| ≤ Z Rd J ∗\|ft\|− Z Rd(W + λ1)\|ft\| ≤∥J∥1 Z Rd\|ft\|− Z Rd(W + λ1)\|ft\| ≤ Z Rd\BR (∥J∥1−W −λ1)\|ft\|+∥J∥1−λ1 αR Z BR \|ft\|φ1 ≤− Z Rd\BR \|ft\|+∥J∥1−λ1 αR Z Rd\|ft\|φ1, for some αR the bound by below of φ1 in BR. Since Z Rd\|ft\|φ1 ≤ Z Rd S\|f0\|φ1 = Z Rd\|f0\|φ1, we get ∥S(t)f∥≤e−t∥f∥+ 1 αR (1 −e−t)∥f∥φ1, by Grönwall’s lemma. Lemma 8.0.4 (Harris’ condition). Under the assumption above, there exists ψ0 ∈X′ ++, g0 ∈X+ and T > 0 such that ST f ≥g0⟨f, ψ0⟩, ∀f ∈X+. (8.0.18) Proof. Step 1. proof of (8.0.18). From Duhamel’s formula (2.2.9) we have SL = SB + · · · + (SBA)∗(N−1) ∗SB + (SBA)(∗N) ∗SL. We note that (SBA ∗SB)f(x) = Z t 0 SB(t −s)ASB(s)fds = Z t 0 [A(feW(x)s)]e−W(x)(t−s)ds. For any R > r, x ∈BR, it is satisﬁed that A(feWs)(x) = Z Rd J(x −y)f(y)e−W(y)sdy ≥J∗e−W[2R]s Z Br(x) f(y)dy with W[R] deﬁned as in the proof of (8.0.16). Then we get (SBA ∗SB)f(x) ≥1BR(x)J∗te−W[2R]t Z Br(x) f(y)dy. 145 Subsequently, we obtain that SBA ∗(SBA ∗SB)f(x) ≥1BR−r(x) Z t 0 J∗se−W[2R]tA 1BR(x) Z Br(x) f(y)dy ! ds, with A 1BR(x) Z Br(x) f(y)dy ! = Z Rd J(x −y)1BR(y) Z Br(y) f(z)dzdy ≥J∗ Z Br(x) Z Br(y) f(z)dzdy. We claim that for all a ≥r, Z Br(x) Z Ba(y) f(z)dzdy ≥\|Br/4\| Z Ba+r/2(x) f(z)dz. Indeed, we deduce Z Br(x) Z Ba(y) f(z)dzdy = Z Br(x) Z Rd 1Ba(y)(z)f(z)dzdy = Z Rd f(z) Z Br(x) 1Ba(z)(y)dy dz and, since for all z ∈Ba+r/2(x), B r 4 z −x \|z −x\| 3r 4 + x ! ⊂Br(x) ∩Ba(z), we have Z Br(x) 1Ba(z)(y)dy ≥\|Br/4\|1Ba+r/2(x)(z), and consequently, Z Br(x) Z Ba(y) f(z)dzdy ≥\|Br/4\| Z Ba+r/2(x) f(z)dz. We have obtained that SBA ∗(SBA ∗SB)f(x) ≥1BR−r(x)J2 ∗t2/2e−W[2R]t Z Br+r/2(x) f(y)dy. Iterating the same argument we arrive to (SBA)(∗n)SBf(x) ≥1BR−nr(x)Jn ∗ tn n!e−W[2R]t Z Br+(n−1)r/2(x) f(y)dy. In consequence, for R = (n + 1)r, we get (SBA)(∗n)SBf(x) ≥1Br(x)Jn ∗ tn n!e−W[2(n+1)r]t Z B(n−1)r/2(0) f(y)dy. Coming back to the Duhamel formula (2.2.9), we deduce SLf(x) ≥g0(x) ∞ X n=2 (J∗t)n n! e−W[2(n+1)r]t Z B(n−1)r/2 f(y)dy, from where (8.0.18) follows with ψ0 := ∞ X n=2 (J∗t)n n! e−W[2(n+1)r]t1B(n−1)r/2 146 CHAPTER 8. A MUTATION-SELECTION MODEL Proof of Theorem 8.0.1 part (2). Let us consider A > 0 and f ∈X+ such that ∥f∥≤A[f]φ1. For any integer n ≥1, we have [f]φ1 = Z Bn fφ1 + Z Bc n fφ1 ≤αn⟨f, ψ0⟩+ βn∥f∥ ≤ αn⟨f, ψ0⟩+ βnA[f]φ1, with αn = ∥φ1∥L∞ ⟨W ⟩/infBn ψ0 and βn = ∥φ1∥L∞ ⟨W ⟩/infBc n W. Choosing nA such that βnAA ≤1/2, we deduce the constructive estimate [f]φ1 ≤2αnA⟨f, ψ0⟩, and thus that (5.3.7) holds with gA := (2αnA)−1g0. Because of the constructive lower bound (8.0.16) on φ1, we have ⟨φ1, gR⟩≥(2αnA)−1⟨h0, g0⟩=: rA, which provides (5.3.9) in a quantiﬁed way. The two above estimates and the Lyapunov condition established in Lemma 8.0.3 ensure that we may apply the Harris-Doblin Theorem 5.3.3 and thus conclude to (8.0.6) with constructive rate. Chapter 9 Time elapse model with memory This chapter is the result of a collaboration with Valentin Schmutz. 9.1 Introduction Multidimensional mean-ﬁeld models in theoretical neuroscience are challenging to analyse [216, 240, 21, 189] but their study is a necessary step towards understanding how multiple timescales present at the single-neuron level [208, 230] aﬀect the dynamics of large networks of neurons. One-dimensional mean-ﬁeld equations for populations of spiking neurons with deterministic drift with stochastic jumps have been a subject of mathematical studies since the works of Pakdaman, Perthame and Salort [199, 202, 198], providing rigorous foundations to earlier works in theoretical neuroscience [245, 101, 100, 99]. These population equations correspond to the mean-ﬁeld limit of large networks of interacting neurons [69, 88, 59]. However, they are derived from spiking neuron models that are of the ‘renewal’ type (with the exception of ), which means that, while they capture the eﬀect of neuronal refractoriness, they neglect slower neuronal timescales, like those of spike frequency adaptation and short-term synaptic plasticity. To take into account slow neuronal timescales, state-of-the-art phenomenological spiking neuron models are multidimensional [140, 230] or kernel-based [235, 205, 207] (and see [102, Ch. 6.4]). In the following, we consider a class of neuron models that characterize neuronal refractoriness by an ‘age’ variable (the time elapsed since last spike) and eﬀects of spike frequency adaptation or short-term synaptic plasticity by a ‘leaky memory’ variables. For this class of neuron models, the mean-ﬁeld limit is characterized by a multidimensional transport equation with a nonlocal boundary condition . In this work, we study the long time behavior of the solutions to the equation proposed in , in the two-dimensional case. 9.1.1 The age- and leaky memory-structured model The population model we consider describes the evolution of a density ρt over the state-space (a, m) ∈R+ × R∗ +, where a and m are the ‘age’ and ‘leaky memory’ variables of the neuron, and ρt(a, m) represents the density of neurons in state (a, m) at time t. The nonlinear evolution problem for the density ρt, for the initial datum u0, writes ∂tρt + ∇· (bρt) = −f(a, m, εxt)ρt, (9.1.1a) ρt(0, m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), εxt)ρt(a, γ−1(m))da, (9.1.1b) xt = Z t 0 Z ∞ 0 Z ∞ 0 h(t −s, a, m)f(a, m, εxs)ρs(a, m)dadmds, (9.1.1c) ρ0 = u0. (9.1.1d) 147 148 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY The dynamics of the model can be decomposed in three elements: (i) the behavior of neurons between spikes, (ii) the spike-triggered jumps and (iii) the interaction between neurons, which we discuss in turn. (i) Between spikes, neurons are transported along the vector ﬁeld b(a, m) = (1, −λm), with λ > 0 (∇· denotes the divergence operator over the state-space). (ii) Neurons spike at a rate f(a, m, εxt), where f : R+ × R∗ + × R →R+ is the ‘ﬁring rate function’ corresponding to the stochastic intensity of the spike generation process and ε ∈R is the connection strength. When a neuron spikes, its age a is reset to 0 and its leaky memory variable m jumps to γ(m), where γ : R+ →R∗ + is the ‘jump mapping’ and is assumed to be a strictly increasing C1−diﬀeomorphism. As a consequence, the border condition (9.1.1b) has a simple interpretation: the density of neurons in state (0, m) at time t is equal to the marginal density of those neurons that have their leaky memory variable in state γ−1(m) and spike at time t. The indicator function 1m>γ(0) reﬂects the fact that m is always strictly positive and the term (γ−1)′(m) is necessary to guarantee the conservation of the total mass of neurons. Indeed, formally, ∂t Z ρt = Z 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), εxt)ρt(a, γ−1(m))dadm − Z f(a, m, εxt)ρt = 0, by a change of variable. (iii) Neurons interact through the ‘total postsynaptic potential’ xt, which integrates the past spiking activity of the population, ﬁltered by the ‘interaction function’ h : R+ × R+ × R∗ + →R. xt, weighted by the connection strength ε ∈R, inﬂuences the ﬁring rate f. If we write N(t) for the mean ﬁring rate N(t) := Z ∞ 0 Z ∞ 0 f(a, m, εxt)ρt(a, m)dadm, and if we take h independent of a and m, then xt takes the form xt = Z t 0 h(t −s)N(s)ds, where now h is simply a delay kernel, as in [101, 100, 99, 199]. In our formulation, h in Eq. (9.1.1c) allows to model more general interactions. For example, in Sec. 9.1.2, we show that by choosing h(t, a, m) = ˆ h(t)(1 −m), we can include the eﬀects of a classical short-term synaptic plasticity model . 9.1.2 Motivation The model (9.1.1) extends the time elapsed neuron network model (see also [100, 99]) by the addition of a leaky memory variable which can accumulate over spikes (as opposed to the age variable which is reset to 0 at each spike) and hence introduces a slow timescale in the population dynamics. Such a slow timescale is typically used to account for some form of fatigue mechanism, which can act on the spiking activity (spike frequency adaptation) or on synaptic transmission (short-term synaptic depression). Slow fatigue at the single neuron level can lead to nontrivial emergent behaviors at the population level, like population bursts [243, 103, 97] (see Fig. 1), which have not been observed in the age- or voltage-structured models of and (but see ). Even though some population equations have been successfully used in the computational neuroscience literature to study emergent behaviors in networks of neurons with fatigue, these population equations were obtained at the cost of a timescale separation approximation [103, 97] or a ‘mixing’ assumption [193, 224], making them inexact. In contrast, the model (9.1.1) is the exact mean-ﬁeld limit for spiking neuron models with spike-frequency adaptation or short-term synaptic depression, as we discuss now. 9.1. INTRODUCTION 149 Spike frequency adaptation The recent spike history of a neuron can modulate its ﬁring rate f, leading to spike frequency adaptation . If h is independent of a and m and if γ(m) = m + ˆ Γ, for a ﬁxed ˆ Γ>0, (9.1.1) becomes ∂tρt + ∇· (bρt) = −f(a, m, εxt)ρt, (9.1.2a) ρt(0, m) = 1m>ˆ Γ Z ∞ 0 f(a, m −ˆ Γ, εxt)ρt(a, m −ˆ Γ)da, (9.1.2b) xt = Z t 0 h(t −s) Z ∞ 0 Z ∞ 0 f(a, m, εxs)ρs(a, m)dadmds, (9.1.2c) ρ0 = u0. (9.1.2d) If η : R+ →R is a bounded function such that lima→+∞η(a) = 0 (η is the ‘refractory kernel’ [102, Sec. 9.3]), we can deﬁne f more explicitly: f(a, m, εxt) := ˆ f(η(a) −m + εxt), (9.1.2e) where ˆ f : R →R+ is typically a non-decreasing function. Since m makes jumps of size ˆ Γ > 0 at each spike and decays exponentially at rate λ between spikes, m accumulates over spikes, which decreases the ﬁring rate f (Eq. (9.1.2e)), leading to spike frequency adaptation . More speciﬁcally, Eq. (9.1.2) is a population equation for adaptive SRM0 (Spike Response Model) neurons [132, 102]. Populations of spiking neurons with spike frequency adaptation exhibit self-sustained popu-lation bursts when the connectivity strength is suﬃciently strong [243, 103, 97]. In Fig. 1, we show simulations of (9.1.2) for two diﬀerent connectivity strengths ε. For large ε, we observe self-sustained bursts, whereas for small ε, we observe relaxation to a stationary state. For comparison, in the Appendix, we show similar simulations for the time elapsed neuron network model , where, as expected, we only observe self-sustained oscillations or relaxation to a stationary state. Short-term synaptic depression The recent spike history of a presynaptic neuron can modulate the synaptic transmission, leading to short-term synaptic plasticity . We will consider here the case of depressive synapses and use the model of (with a change of variable for convenience). In this case, the state-space is (a, m) ∈R+×]0, 1[. Taking f independent of m, and choosing h and γ of the the form h(t, a, m) := ˆ h(t)(1 −m) and γ(m) := 1 −υ + υm for a ﬁxed υ ∈]0, 1[, (9.1.1) becomes ∂tρt + ∇· (bρt) = −f(a, m, εxt)ρt, (9.1.3a) ρt(0, m) = 1m>γ(0) 1 υ Z ∞ 0 f(a, εxt)ρt(a, γ−1(m))da, (9.1.3b) xt = Z t 0 ˆ h(t −s) Z 1 0 Z ∞ 0 (1 −m)f(a, εxs)ρs(a, m)dadmds, (9.1.3c) ρ0 = u0. (9.1.3d) Note that the term 1 υ on the RHS of Eq. (9.1.3b) simply comes from the fact that \|(γ−1)′(m)\|= 1 υ, for all m ∈]0, 1[. Here, at each spike, m makes strictly positive jumps which size tends to 0 as m tends to 1 (since γ(1) = 1) and decays exponentially at rate λ between spikes. If m is close to 1, synaptic transmission is weak because of the factor (1 −m) in Eq. (9.1.3c). As observed in , the stationary state of populations of neurons with short-term synaptic plasticity can be described by a simple formula, which we prove in Sec. 9.4.3. 150 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY 0 5 0 5 10 15 0 5 0 5 10 15 Figure 9.1: Depending on the connectivity strength ε, a population of adaptive SRM0 can exhibit self-sustained bursts (ε ≫0) or relaxation to a stationary state (small ε). We show simulations of a network of 5·105 adaptive SRM0 neurons, approximating the mean-ﬁeld limit Eq. (9.1.2), with identical parameters (except for ε) and identical initial conditions. The raster plots below the plots for the time-evolution of the total postsynaptic potential xt represent the spikes of 100 randomly selected neurons. 9.1. INTRODUCTION 151 9.1.3 Assumptions and main results The main result of this work is the exponential stability of (9.1.1) in the weak connectivity regime (Theorem 3) – or, more explicitly, there exists ε∗∗> 0 such that (9.1.1) is exponentially stable for all connectivity strength ε ∈] −ε∗∗, +ε∗∗[. Before proving the exponential stability, we ﬁrst establish the well-posedness of (9.1.1) in the appropriate function space (Theorem 1) and show that stationary solutions exist and are unique for suﬃciently weak connectivity (Theorem 2). Here, we study the weak solutions to (9.1.1) for an initial datum in L1 + := L1(R+ × R∗ +, R+) and write L1 +(R∗ +) := L1(R∗ +, R+). Deﬁnition (Solutions). (ρ, x) ∈C(R+, L1 +) × C(R+) is a solution to (9.1.1), for the initial datum u0 ∈L1 +, if xt = Z t 0 Z ∞ 0 Z ∞ 0 h(t −s, a, m)f(a, m, εxs)ρs(a, m)dadmds, ∀t ≥0, (9.1.4a) and if for all ϕ ∈C∞ c (R+ × R+ × R∗ +), 0 = Z ∞ 0 Z ∞ 0 u0(a, m)ϕ(0, a, m)dadm + Z ∞ 0 Z ∞ 0 Z ∞ 0 ρt(a, m) n [∂t +∂a −λm∂m]ϕ+(ϕ(t, 0, γ(m))−ϕ(t, a, m))f(a, m, εxt) o dadmdt. (9.1.4b) To prove the well-posedness of (9.1.1), we need some simple assumptions of the ﬁring rate function f and the interaction function h: Assumptions 1. f is bounded and Lf-Lipschitz, i.e. \|f(a, m, x) −f(a∗, m∗, x∗)\|≤Lf(\|a −a∗\|+\|m −m∗\|+\|x −x∗\|), and h is bounded and continuous. Since we want to apply Harris’ theorem, the well-posedness in L1 (which is treated in ) is not enough and we need the well-posedness in a weighted L1 space (where the weight satisﬁes a Lyapunov condition ) with a global-in-time estimate in the weighted L1 norm. Using the weight function w : R+ × R+ →[1, ∞), (a, m) 7→1 + m, we deﬁne the function space L1 +(w) := g ∈L1(R+ × R∗ +, R+) ∥g∥L1(w) := Z ∞ 0 Z ∞ 0 g(a, m)w(a, m)dadm < ∞ . To obtain a global-in-time estimate in the L1 +(w) norm, we further need that the jump sizes of γ are bounded: Assumptions 2. There exists a bounded function Γ : R∗ + →R∗ + such that for all m ∈R∗ +, γ(m) = m + Γ(m). Theorem 1 (Well-posedness). Grant Assumption 1. For any initial datum u0 ∈L1 +, there exists a unique weak solution (ρ, x) to (9.1.1). This solution satisﬁes I (L1-stability) ∥ρt∥L1 =∥u0∥L1, ∀t > 0, 152 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY II (Global bound in L1 +(w)) if, in addition, Assumption 2 holds and u0 ∈L1 +(w), then ∀t > 0, ∥ρt∥L1(w) ≤∥u0∥L1(w) e−αt + b α(1 −e−αt), (9.1.5) for some constants α > 0 and b ∈R. In contrast to , the well-posedness proof presented here does not involve any probabilistic argument. The proof consists of two consecutive applications of Banach’s ﬁxed-point theorem, where a ﬁrst ﬁxed-point gives the unique solution to a linearized version of (9.1.1) which is then used in a second ﬁxed-point treating the nonlinearity of (9.1.1). The second step towards the exponential stability proof is the study of the existence and uniqueness of the stationary solutions to (9.1.1). For this step, we require: Assumptions 3. nothing I There exists ∆abs > 0 and σ > 0 such that f(a, m, x) ≥σ, ∀(a, m, x) ∈[∆abs, +∞[×R∗ + × R. II There exists Cγ ∈]0, 1] such that Cγ ≤γ′ ≤1. III ¯ h(a, m) = R ∞ 0 h(t, a, m)dt is bounded. The ﬁrst point of Assumption 3 sets a lower bound on the ﬁring rate function f for any a ≥∆abs and hence allows for an absolute refractory period ∆abs > 0, i.e. a period of time following a spike during which f = 0 (which is an important neurodynamical feature [102, Sec. 1.1]). This assumption is also used in . In the second point of Assumption 3, the lower bound 0 < Cγ ≤γ′ guarantees that γ is strictly increasing, which reﬂects the idea that m is a ‘leaky memory’ variable of the past neuronal activity. On the other hand, the upper bound γ′ ≤1, which can be rewritten in terms of the jump size function Γ as Γ′ ≤0, prevents the variable m from growing too fast and allows for a potential saturation of the memory, as in the example with short-term synaptic plasticity (9.1.3). The third point of Assumption 3 reﬂects the fact that a single spike has a ﬁnite impact on the neuron that receives it. We emphasize that the two examples shown above, spike frequency adaptation (9.1.2) and short-term synaptic depression (9.1.3), satisfy Assumption 3. Theorem 2 (Stationary solutions). Grant Assumptions 1 – 3. I There exists a stationary solution to (9.1.1). II There exists ε∗> 0 such that for all ε ∈] −ε∗, +ε∗[, the stationary solution to (9.1.1) is unique. Over the course of this work, we obtained the existence of the stationary solution by two diﬀerent approaches. The ﬁrst approach is based on the Doeblin-Harris method and is similar to that of . First, we show that when xt is ﬁxed and time-invariant in (9.1.1) (neurons are non-interacting), the system satisﬁes a Harris condition – this constitutes a key result of this work –, and we can use Harris’ theorem to get the stationary solution. Then, we use the Lipschitz continuity the stationary solutions with respect to the ﬁxed x to prove the existence of a stationary solution for arbitrary connectivity strengths ε. Finally, for ε small enough, we also get the uniqueness of the stationary solution, by Banach’s ﬁxed-point theorem. The second approach relies on the fact that the stationary solutions solve an integral equation, for which we can show that a solution exists by Schauder’s ﬁxed-point theorem. In the process, 9.1. INTRODUCTION 153 we get several estimates on the stationary solutions, namely that they are continuous, bounded, and exponentially decaying in m. However, this approach does not give uniqueness. As mentioned above, the application of Harris’ theorem requires us to consider solutions in the weighted space L1(w). However, in the case where the state-space of the leaky memory variable m is bounded, the situation is simpler: we can use Doeblin’s theorem in L1. The following assumption guarantees that m stays in a bounded state-space: Assumptions 4. There exists G > 0 such that for all m ∈R∗ +, γ(m) < G. Note that this assumption is satisﬁed in the example with short-term synaptic plastic-ity (9.1.3), with G = 1. Finally, to study the exponential stability of (9.1.1), we need an exponential decay on h: Assumptions 5. There exists h, Ch > 0 such that h(t, a, m) ≤Che−ht, ∀(t, a, m). By a perturbation argument similar to that of , we obtain our main result: Theorem 3 (Exponential stability in the weak connectivity regime). Grant Assumptions 1 – 3 and 5. For any W > 0, there exists ε∗∗ W > 0 such that for ε ∈] −ε∗∗ W , +ε∗∗ W [, there exists C ≥1 and cW > 0 such that for all initial data u0 ∈L1 +(w) with ∥u0∥L1 = 1 and ∥u0∥L1(w) ≤W, the solution (ρ, x) to (9.1.1) satisﬁes ∥ρt −ρ∞∥L1(w) + \|xt −x∞\|≤Ce−cW t ∥u0 −ρ∞∥L1(w) + 1 , ∀t ≥0, (9.1.6) where (ρ∞, x∞) is the unique stationary solution given by Theorem 2 (ii). If, in addition, we grant Assumption 4, then there exists ε∗∗> 0 such that for all ε ∈ ] −ε∗∗, +ε∗∗[, there exists C′ ≥1 and c > 0 such that for all initial data u0 ∈L1 + with ∥u0∥L1 = 1, ∥ρt −ρ∞∥L1 + \|xt −x∞\|≤C′e−ct ∥u0 −ρ∞∥L1 + 1 , ∀t ≥0. (9.1.7) From the neuronal modeling point of view, this result is not surprising: when the connection strength is weak enough, neurons do not synchronize and the population activity converges to a stationary state. This was already proved for simpler one-dimensional models (see below) and the addition of a leaky memory variable carrying the eﬀect of spike frequency adaptation or short-term synaptic plasticity does not change this behavior. 9.1.4 Discussion of the methods The asymptotic stability of the age-structured model of in the weak connectivity regime has been studied using entropy methods (assuming that f is a step-function) [199, 202], spectral analysis of semigroups in Banach spaces [182, 181] or Doeblin’s theorem . For the treatment of the strong connectivity regime, we refer to [199, 181]. On the closely related voltage-structured model , also proved asymptotic stability in the weak connectivity regime using Laplace transform techniques. For this model, the nonlinear stability of the stationary solutions has been recently studied in (see also ) and can identify Hopf bifurcations . Doeblin’s theorem has also been used in in the case of the ‘threshold crossing’ neuronal population equation of . Note that closely related methods have been used by probabilists to study the ergodicity of single neuron models [127, 80]. Our approach combines strategies from and , even though uses Doeblin’s instead of Harris’ theorem. On the one hand, our proof is based on the application of Harris’ theorem for the linear problem, which simpliﬁes the proof of . On the other hand, we use an argument from in to deal with delay eﬀects, which are not considered in . Note that our model is two-dimensional (by the addition of the leaky memory variable), whereas the aforementioned works only considered one-dimensional models. 154 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY 9.1.5 Plan of the chapter The proof of Theorem 1 (Well-posedness) is presented in Section 9.2. In Section 9.3, we prove the exponential stability of (9.1.1) in the non-interacting case ε = 0 using Harris’ or Doeblin’s theorem. The proof of Theorem 2 (Stationary solutions) is presented in Section 9.4 which is divided in three parts: in the ﬁrst part, we present a proof which uses the exponential stability of the non-interacting case; in the second part, we present an alternative proof for the existence of stationary solutions which does not involve the Doeblin-Harris method; and in the last part, we present a proof for the formula of in the case of short-term synaptic plasticity (9.1.3). Finally, Section 9.5 is dedicated to the proof of Theorem 3 (Exponential stability in the weak connectivity regime). 9.2 Well-posedness This section is dedicated to the proof of Theorem 1, which we decompose is several lemmas. First, we verify the a priori L1-stability of the solutions to (9.1.1), a technical result we use later in the proof. Then, we introduce a linearized version of (9.1.1) and show that it is well-posed by an application of Banach’s ﬁxed-point theorem. Another Banach’s ﬁxed-point argument is used to treat the nonlinearity of (9.1.1) and concludes the proof of the well-posedness in L1. Finally, we prove the global bound in L1 +(w) (point (ii) of Theorem 1), which we will use to apply Harris’ theorem in the next sections. Lemma 1 (A priori L1-stability). Grant Assumption 1. If (ρ, x) is a weak solution to (9.1.1) for the initial datum u0 ∈L1 +, then ∥ρt∥L1 =∥u0∥L1 , ∀t > 0. Proof. By a standard cut-oﬀin time argument, we have that for all T > 0 and for all ϕ ∈ C∞ c (R+ × R+ × R∗ +), Z ∞ 0 Z ∞ 0 ρT (a, m)ϕ(T, a, m)dadm − Z ∞ 0 Z ∞ 0 u0(a, m)ϕ(0, a, m)dadm = Z T 0 Z ∞ 0 Z ∞ 0 ρt(a, m) n [∂t + ∂a −λm∂m]ϕ + (ϕ(t, 0, γ(m)) −ϕ(t, a, m))f(a, m, εxt) o dadmdt. Let χ be a function in C∞ c (R+ × R∗ +, R+) such that χ(a, m) = 1, for all a2 + m2 ≤1. For all n ∈N∗, we write e ϕn ∈C∞(R+ ×R+ ×R∗ +) the classical solution to the transport equation ∂t e ϕn(t, a, m) + ∂a e ϕn(t, a, m) −λm∂m e ϕn(t, a, m) = 0, (9.2.8a) e ϕn(0, a, m) = χ(a/n, m/n). (9.2.8b) Because of the ﬁnite speed of propagation of the transport equation, for all n, there exists a function ϕn ∈C∞ c (R+ × R+ × R∗ +) such that ϕn(t, a, m) = e ϕn(t, a, m), for all (t, a, m) ∈ [0, T] × R+ × R∗. Hence, for all n ∈N∗, Z ∞ 0 Z ∞ 0 ϕn(T, a, m)ρT (a, m)dadm − Z ∞ 0 Z ∞ 0 ϕn(0, a, m)u0(a, m)dadm = Z T 0 Z ∞ 0 Z ∞ 0 ∂tϕn+∂aϕn−λm∂mϕn+ ϕn(t, 0, γ(m)) −ϕn(t, a, m) f(a, m, εxt) ρt(a, m)dadmdt. 9.2. WELL-POSEDNESS 155 As ϕn is a solution to Eq. (9.2.8a) on time [0, T], we get Z ∞ 0 Z ∞ 0 ϕn(T, a, m)ρT (a, m)dadm − Z ∞ 0 Z ∞ 0 ϕn(0, a, m)u0(a, m)dadm = Z T 0 Z ∞ 0 Z ∞ 0 ϕn(t, 0, γ(m)) −ϕn(t, a, m) f(a, m, εxt) ρt(a, m)dadmdt. For all (t, a, m) ∈[0, T] × R+ × R∗ +, ϕn(t, a, m) − − − → n→∞1, since the initial datum tends to 1 as n →∞(Eq. (9.2.8b)) and by ﬁnite speed of propagation. Thus, by dominated convergence, we get Z ∞ 0 Z ∞ 0 ρT (a, m)dadm − Z ∞ 0 Z ∞ 0 u0(a, m)dadm = 0. (9.2.9) Since ρ is nonnegative, this concludes the proof. Lemma 1 will allow us to prove the well-posedness of (9.1.1) by the means of ﬁxed-point arguments. Let us ﬁrst introduce a linearized version of Eq. (9.1.1): for all x ∈C(R+), we consider the linear evolution problem ∂tρt + ∇· (bρt) = −f(a, m, εxt)ρt, (9.2.10a) ρt(0, m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), εxt)ρt(a, γ−1(m))da, (9.2.10b) ρ0 = u0. (9.2.10c) We can see Eq. (9.2.10) as the Kolmogorov forward equation of a time-dependent Makrov process. Indeed, we can rewrite Eqs. (9.2.10a) and (9.2.10b) as ∂ρt = Ltρt (9.2.11) where, for all suitable test function φ : R+ × R∗ + →R, L∗ t φ(a, m) = b(a, m) · ∇φ(a, m) + [φ(0, γ(m)) −φ(a, m)]f(a, m, εxt). (9.2.12) L∗ t is the time-dependent generator of a piecewise deterministic Markov process with degenerate jumps. The linearized equation (9.2.10) will play a special role in the following sections and it therefore deserves its own proposition: Proposition 1 (Well-posedness of the linearized equation (9.2.10)). Grant Assumption 1. For any initial datum u0 ∈L1 + and any x ∈C(R+), there exists a unique weak solution ρx ∈C(R+, L1 +) to Eq. (9.2.10). Furthermore, ρx satisﬁes I For all t > 0 and for all m ∈R∗ +, ρx t (0, m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), εx)ρx t (a, γ−1(m))da, ρx t (a, m) =    u0(a −t, eλtm) exp λt − R t 0 f(a −t + s, eλ(t−s)m, εx)ds if a ≥t, ρx t−a(0, eλam) exp λa − R t t−a f(a −t + s, eλ(t−s)m, εx)ds if 0 < a < t. II For all t > 0 and for all φ ∈C∞ c (R+ × R∗ +), ⟨ρx t , φ⟩= ⟨u0, φ⟩+ Z t 0 ⟨ρx t , L∗ xφ⟩ds. (9.2.14) 156 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY Proof. Fix x ∈C(R+). For all p ∈C(R+, L1 +(R∗ +)) and u0 ∈L1 +, we know, from the standard theory of transport equations, that there is a unique weak solution to ∂tρt + ∇· (bρt) = −f(a, m, εxt)ρt, ρt(0, m) = pt(m), ρ0 = u0, which we denote ρx,p and is given by the representation formula, ρx,p t (a, m) :=    u0(a −t, eλtm) exp λt − R t 0 f(a −t + s, eλ(t−s)m, εxs)ds if a ≥t, pt−a(eλam) exp λa − R t t−a f(a −t + s, eλ(t−s)m, εxs)ds if 0 < a < t. ρx,p is in C(R+, L1), since ∀t ∈R+, ρx,p t L1 ≤∥u0∥L1 + Z t 0 ∥ps∥L1 ds. We have 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), x)ρx,p t (a, γ−1(m))da (t,m)∈R+×R∗ + ∈C(R+, L1 +(R∗ +)) since ∀t ∈R+, Z ∞ γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), x)ρx,p t (a, γ−1(m))dadm ≤∥f∥∞ ρx,p t L1 ≤∥f∥∞ ∥u0∥L1 + Z t 0 ∥ps∥L1 ds ! . Hence, we can deﬁne, for any T > 0, the operator Φx T : C([0, T], L1 +(R∗ +)) →C([0, T], L1 +(R∗ +)) p 7→ 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), x)ρx,p t (a, γ−1(m))da (t,m)∈[0,T]×R∗ + . For any p, q ∈C([0, T], L1 +(R∗ +)), Φx T (p) −Φx T (q) C([0,T],L1) ≤∥f∥∞sup t∈[0,T] ρx,p t −ρx,q t L1 ≤∥f∥∞ Z T 0 ∥ps −qs∥L1 ds ≤T∥f∥∞∥p −q∥C([0,T],L1) . Therefore, if 0 < T <∥f∥−1 ∞, Φx T is a contraction. By Banach’s ﬁxed-point theorem, there exists a unique ρx ∈C([0, T], L1 +) solving Eq. (9.2.10). Since the choice of the contracting T does not depend on the initial datum, we can iterate the above argument on successive time intervals of length T and conclude that there exists a unique ρx ∈C(R+, L1 +) solving Eq. (9.2.10) for which the formula (i) is satisﬁed. Then, (ii) follows from a standard cut-oﬀin time argument. Now, we can prove the existence and uniqueness of a solution to the nonlinear problem (9.1.1) by the means of a second application of Banach’s ﬁxed-point theorem. 9.2. WELL-POSEDNESS 157 Proof of the well-posedness of (9.1.1) in L1. For any x ∈C(R+), we take the ρx given by Propo-sition 1. We have   Z t 0 Z R+×R∗ + h(t −s)f(εxs)ρx s dadmds   t∈R+ ∈C(R+) since ∀t ∈R+, Z t 0 Z R+×R∗ + h(t −s)f(εxs)ρx s dadmds ≤∥h∥∞∥f∥∞ Z t 0 ∥ρx s∥L1 ds. Hence, for any T > 0, we can deﬁne the operator ΨT : C([0, T]) →C([0, T]) x 7→   Z t 0 Z R+×R∗ + h(t −s)f(εxs)ρx s dadmds   t∈[0,T] . For any x, y ∈C([0, T]), we have ΨT (x) −ΨT (y) C([0,T]) ≤T∥h∥∞sup t∈[0,T] Z R+×R∗ + \|f(εxt)ρx t −f(εyt)ρy t \| dadm ≤T∥h∥∞sup t∈[0,T] εLf\|xt −yt\|∥ρx t ∥L1 +∥f∥∞ ρx t −ρy t L1 . By Grönwall’s lemma, ∥ρx t ∥L1 ≤∥u0∥L1 exp( ∥f∥∞t), since ∀t ∈[0, T], ∥ρx t ∥L1 ≤∥u0∥L1 +∥f∥∞ Z t 0 ∥ρx s∥L1 ds. On the other hand, we have, for all t ∈[0, T], ρx t −ρy t L1 ≤ Z t 0 Z ∞ 0 ρx s(0, m) exp − Z t s f(u −s, e−λ(u−s)γ(m), εxu)du ! −ρy s(0, m) exp − Z t s f(u −s, e−λ(u−s)γ(m), εyu)du ! dmds ≤∥f∥∞ Z t 0 ∥ρx s −ρy s∥L1 ds + tε∥f∥∞Lf∥x −y∥C([0,T]) Z t 0 ∥ρx s∥L1 ds. Hence, by Grönwall’s lemma, for all t ∈[0, T], ρx t −ρy t L1 ≤εLf∥u0∥L1 exp( ∥f∥∞t) −1 2 ∥f∥∞ ∥x −y∥C([0,T]) . Gathering the bounds, we get ΨT (x) −ΨT (y) C([0,T]) ≤Tε∥h∥∞Lf∥u0∥L1 exp( ∥f∥∞T) 1 + exp( ∥f∥∞T) ∥x −y∥C([0,T]) . For T small enough, ΨT is a contraction and, by Banach’s ﬁxed-point theorem, has a unique ﬁxed-point. Thus, there exists a unique solution (ρ, x) ∈C([0, T], L1 +). Since, by Lemma 1, ∥ρT ∥L1 = ∥u0∥L1, we can iterate this argument on successive time intervals of length T and conclude that there exists a unique solution in C(R+, L1 +). 158 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY To conclude the proof of Theorem 1, it remains to show the estimate Eq. (9.1.5). Under Assumption 2, the weight function w : R+ × R+ →[1, ∞), (a, m) 7→1 + m satisﬁes w(a, m) →∞when m →∞and the Lyapunov condition on m: ∃α > 0, b ≥0 such that L∗ t w ≤−αw + b. (9.2.16) Indeed, for all (t, a, m) ∈R+ × R+ × R∗ +, L∗ t w(a, m) = −λm + Γ(m)f(a, m, εxt) ≤−λw(a, m) + λ +∥Γ∥∞∥f∥∞. Importantly, the constants α and b do not depend on x. Lemma 2 (Global bound in L1 +(w)). Grant Assumptions 1 and 2. If the initial datum u0 is in L1 +(w), then ρt ∈L1 +(w) for all t ≥0. Moreover, ∀t > 0, ∥ρt∥L1(w) ≤∥u0∥L1(w) e−αt + b α(1 −e−αt), (9.2.17) where the constants α and b are taken from the Lyapunov condition (9.2.16). Proof. We divide the proof in two steps: ﬁrst, we prove that the solution is stable in L1 +(w) with a weaker and time dependent bound; then, we use this ﬁrst bound to apply the dominated convergence theorem and obtain Eq. (9.2.17) by Grönwall’s lemma. Step 1. Fix any T > 0. Let χ ∈C∞ c (R+, R+) be a non-increasing function such that χ(x) = 1 if 0 ≤x ≤1 and χ(x) = 0 if x > 2. For all n ∈N∗, let us write ϕk(a)χn(m) := χ(a/k)χ(m/n). We also consider gM(w) a smooth approximation of w ∧M, such that ∥g′∥∞≤1 and M1w≥M ≤g(w) ≤M. For all n, k and M, gM(w)χnϕk ∈C∞ c (R+ × R+, R+). Hence, by Proposition 1 (ii), ∀n ∈N∗, ⟨ρT , gM(w)χnϕk⟩= ⟨u0, gM(w)χnϕk⟩+ Z T 0 ⟨ρt, L∗ ˜ x(gM(w)χnϕk)⟩dt, where L∗ ˜ x(gM(w)χnϕk) = ∂a(gM(w)χnϕk) −λm∂m(gM(w)χnϕk) + gM(w(γ(m)))χn(γ(m))ϕk(0) −gM(w)χnϕk f = gM(w)χn 1 kχ′(a/k) −λmgM(w)ϕk 1 nχ′(m/n) −λmg′ M(w)χnϕk + gM(w(γ(m)))χn(γ(m))ϕk(0) −gM(w)χnϕk f. From the L1-stability and the fact that both gM(w)∂mχn and gM(w)χn are bounded and have compact support, we can go to the limit in k using the dominated convergence theorem: ⟨ρT , gM(w)χn⟩= ⟨u0, gM(w)χn⟩ + Z T 0 ρt, −λmgM(w) 1 nχ′(m/n) −λmg′ M(w)χn dt + Z T 0 ρt, (gM(w(γ(m)))χn(γ(m)) −gM(w)χn)f dt, (9.2.18) On the other hand, from the properties of χ and gM, we have λmgM(w) 1 nχ′(m/n) ≤λgM(w)2n n ∥χ′∥∞≤2λM∥χ′∥∞ 9.2. WELL-POSEDNESS 159 and \|λmg′ M(w)χn\|≤λgM(w) ≤λM, whence ⟨ρT , gM(w)χn⟩≤⟨u0, gM(w)χn⟩ + Z T 0 ⟨ρt, λgM(w)∥χ′∥∞+λgM(w)⟩dt + Z T 0 ⟨ρt, (gM(w(γ(m)))χn(γ(m)) −gM(w)χn)f⟩dt and we can take the limit in n using the dominated convergence theorem: ⟨ρT , gM(w)⟩≤⟨u0, gM(w)⟩ + Z T 0 ⟨ρt, λgM(w)∥χ′∥∞+λgM(w)⟩dt + Z T 0 ⟨ρt, (gM(w(γ(m))) −gM(w))f⟩dt. From the properties of γ, we get w(0, γ(m)) ≤w(0, m +∥Γ∥∞) ≤(1 +∥Γ∥∞)w(a, m) and gM(w(0, γ(m))) ≤(1 +∥Γ∥∞)gM(w(m)). This, together with the fact that f is bounded, shows that there exists a constant C, which does not depend on M, such that ⟨ρT , gM(w)⟩≤⟨u0, gM(w)⟩+ C Z T 0 ⟨ρt, gM(w)⟩dt, and we can apply Grönwall’s lemma to obtain ⟨ρT , gM(w)⟩≤⟨u0, gM(w)⟩eCt. Then, it follows from Fatou lemma that ρT ∈L1 +(w). Step 2. To improve the previous estimate, we come back to (9.2.18) and use dominated convergence in n and M (Step 1. guarantees domination) to show ⟨ρT , w⟩= ⟨u0, w⟩+ Z T 0 ⟨ρt, L∗ ˜ xw⟩dt. By the Lyapunov condition (9.2.16), ∥ρT ∥L1(w) ≤∥u0∥L1(w) −α Z T 0 ∥ρt∥L1(w) dt + Tb. and by Grönwall’s lemma, we have, for all t ∈[0, T], ∥ρT ∥L1(w) ≤∥u0∥L1(w) e−αt + b α(1 −e−αt). Since T can be chosen arbitrarily large, this achieves the proof. Remark. Following the same steps as in the proof above, we can show that the bound Eq. (9.2.17) also holds for the linearized equation (9.2.10) and does not depend on x nor the constants α and b. 160 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY 9.3 Exponential stability in the non-interacting case If x ∈C(R+) in the linearized equation (9.2.10) is time-invariant, i.e. x ≡˜ x for some ˜ x ∈R, then Eq. (9.2.10) can be seen as the dynamics of a non-interacting population of neurons. In this section, we prove the exponential stability in the non-interacting case using Harris’ or Doeblin’s theorem. This is the key result of this work and will allow us to prove the existence and uniqueness of the stationary solution to (9.1.1) (Section 9.4) and the exponential convergence to it (Section 9.5). For ˜ x ∈R, u0 ∈L1, we denote ρ˜ x t the unique solution to Eq. (9.2.10) for the initial datum u0 and x ≡˜ x, given by Proposition 1. We write, using the semigroup notation, S˜ x t u0 := ρ˜ x t , ∀t ≥0. (9.3.19) To show that the Eq. (9.3.19) is exponentially stable we will use Harris’ theorem in the general case or Doeblin’s theorem if Assumption 4 is granted. In both cases, the main technical diﬃculty is to verify the Doeblin minoration condition (Lemma 3) as the jumps of the process described by Eq. (9.2.12) are degenerate and the model is two-dimensional. Lemma 3 (Doeblin minoration condition). Grant Assumptions 1 and 3. Fix any x ∈R. For all R > 0, there exists T > 0 and a positive non-zero measure ν such that ∀u0 ∈L1 +, S˜ x T u0 ≥ν Z R+×]0,R] u0 dadm. (9.3.20) Proof. We proceed in two steps. First (Step 1), we choose a time T > 0 and a rectangle [0, ¯ a]×[m, m] ⊂R+×R∗ + (with nonzero Lebesgue measure) and show that the density S˜ x T u0 ∈L1 has a lower bound on [0, ¯ a] × [m, m] which depends on a Lebesgue integral in R2 + involving u0. Then (Step 2), we perform a change of variable to express this lower bound in terms of R R+×]0,R] u0 dadm. The proof only relies on the expression of S˜ x t u0 given by the method of characteristics (see Proposition 1) and this allows treating a typically probabilistic question – the Doeblin minoration condition – from a transport point of view. This is possible because S˜ x t is the stochastic (mass-conservative) semigroup of a piecewise deterministic Markov process. The constants ∆abs, σ and Cγ are taken from Assumption 3. Step 1: Fix R > 0. Since γ(e−λ∆absγ(0)) > γ(0) and γ(e−λtγ(e−λ∆absR)) →γ(0) as t →∞, there exists ¯ a > 0 and T > ¯ a + ∆abs such that m =: γ(e−λ(T−¯ a−∆abs)γ(e−λ∆absR)) < e−λ¯ aγ(e−λ∆absγ(0)) =: m. (9.3.21) Eq. (9.3.21) has the following heuristic interpretation: if we see S˜ x t as the stochastic semigroup of the piecewise deterministic Markov process deﬁned by the generator Eq. (9.2.12), for any initial point (a0, m0) ∈R+×]0, R] and any landing point (a, m) ∈[0, ¯ a] × [m, m] at time T, there is a ‘possible’ trajectory going from (a0, m0) to (a, m), with exactly two jumps (spikes). Since the trajectories of the process are determined by the jump times, we will exploit the fact that these ‘possible’ trajectories correspond to jump times with strictly positive probability density. Below, we take a transport point of view on this probabilistic argument. 9.3. EXPONENTIAL STABILITY IN THE NON-INTERACTING CASE 161 For all (a, m) ∈[0, ¯ a] × [m, m], (S˜ x T u0)(a, m) ≥1{a<T}(S˜ x T−au0)(0, eλam) exp λa − Z T T−a f(a −T + s, eλ(T−s)m, ˜ x)ds ! ≥1{a<T}e− ∥f∥∞T eλa(S˜ x T−au0)(0, eλam) ≥1{a<T}e− ∥f∥∞T σeλa (γ−1)′(eλam) Z ∞ ∆abs (S˜ x T−au0)(a′, γ−1(eλam))da′ = 1{a 0. We can therefore deﬁne a∗ a,m := 1 λ log γ(0) −log γ−1(eλam) . Note that a∗ a,m satisﬁes γ−1(eλa∗ a,mγ−1(eλam)) = 0. In other words, a∗ a,m is the minimal time between the last and second last jumps for a trajectory landing at (a, m) at time T. We can easily verify that, by our choice of {T, ¯ a, m, m}, ∆abs ≤a∗ a,m < T −a −∆abs. This guarantees that it is possible to make two jumps in [0, T] and land at (a, m) at time T while respecting the absolute refractoriness of the neuron (i.e. there needs to be a time interval ≥∆abs between jumps). This allows us to go further back in time to the second last jump: For all a′ ∈[a∗ a,m, T −a −∆abs], (⋆) ≥1{a′ 0: For all a′ ∈[a∗ a,m, T −a], (ψT a,m)′(a′) = λeλ(T−a−a′) (γ−1)′(eλa′γ−1(eλam))eλa′γ−1(eλam) −γ−1(eλa′γ−1(eλam)) . (9.3.23) As Γ > 0 and γ′ ≤1 (Assumption 3), we have (ψT a,m)′(a′) > λeλ(T−a−a′) (γ−1)′(eλa′γ−1(eλam))eλa′γ−1(eλam) −eλa′γ−1(eλam) = λeλ(T−a)γ−1(eλam) (γ−1)′(eλa′γ−1(eλam)) \| {z } ≥1 −1 ≥0. Therefore, ψT a,m is a strictly increasing C1-diﬀeomorphism from [a∗ a,m, T−a−∆abs] to [ψT a,m(a∗ a,m), ψT a,m(T− a −∆abs)]. We can now rewrite Eq. (9.3.22): (S˜ x T u0)(a, m) ≥e(λ−3 ∥f∥∞)T σ2 Z T−a−∆abs a∗ a,m Z ∞ 0 u0(a0, ψT a,m(a′))da0da′ = e(λ−3 ∥f∥∞)T σ2 Z ψT a,m(T−a−∆abs) ψT a,m(a∗ a,m) Z ∞ 0 u0(a0, m0) ((ψT a,m)−1)′(m0) da0dm0. Going back to Eq. (9.3.23), and using the fact that there exists Cγ such that Cγ ≤γ′ ≤1 (Assumption 3), we have, for all a′ ∈[a∗ a,m, T −a −∆abs], (ψT a,m)′(a′) ≤λeλ(T−a−a′)C−1 γ eλa′γ−1(eλam) ≤λeλT C−1 γ m. Hence, (S˜ x T u0)(a, m) ≥e−3 ∥f∥∞T σ2Cγ λm Z ψT a,m(T−a−∆abs) ψT a,m(a∗ a,m) Z ∞ 0 u0(a0, m0)da0dm0. In addition, by our choice of {T, ¯ a, m, m}, we have ψT a,m(a∗ a,m) = 0, ψT a,m(T −a −∆abs) = eλ∆absγ−1(eλ(T−a−∆abs)γ−1(eλam)) > R. Therefore, (S˜ x T u0)(a, m) ≥e−3 ∥f∥∞T σ2Cγ λm Z R 0 Z ∞ 0 u0(a0, m0)da0dm0. Since we have supposed that (a, m) ∈[0, ¯ a] × [m, m], this concludes the proof. With the Lyapunov condition (9.2.16) and the Doeblin minoration condition (9.3.20), we can apply a version of Harris’ theorem: 9.4. STATIONARY SOLUTIONS FOR ARBITRARY CONNECTIVITY STRENGTH 163 Theorem 4 (Harris). Grant Assumptions 1 – 3. For all ˜ x ∈R, there exists a unique ρ˜ x ∞∈L1 +(w) with ρ˜ x ∞ L1 = 1 such that S˜ x t ρ˜ x ∞= ρ˜ x ∞, for all t ≥0, and there exists K ≥1 and a > 0 such that for all initial data u0 ∈L1 +(w) with ∥u0∥L1 = 1, S˜ x t u0 −ρ˜ x ∞ L1(w) ≤Ke−at u0 −ρ˜ x ∞ L1(w) , ∀t ≥0. (9.3.24) Furthermore, by Lemma 2, we have that ρ˜ x ∞ L1(w) ≤b α, where the constants α and b are taken from the Lyapunov condition (9.2.16). Proof. This is a classic result which proof can be found in the work of Hairer and Mattingly . If, in addition, Assumption 4 holds, we can simply apply Doeblin’s theorem: Theorem 5 (Doeblin). Grant Assumptions 1, 3 and 4. For all ˜ x ∈R, there exists a unique ρ˜ x ∞∈L1 + with ρ˜ x ∞ L1 = 1 such that S˜ x t ρ˜ x ∞= ρ˜ x ∞, for all t ≥0, and there exists K ≥1 and a > 0 such that for all initial data u0 ∈L1 + with ∥u0∥L1 = 1, S˜ x t u0 −ρ˜ x ∞ L1 ≤Ke−at u0 −ρ˜ x ∞ L1 , ∀t ≥0. (9.3.25) Proof. See, for example, Theorem 2.3 in . We say that ρ˜ x ∞is the invariant probability measure of the semigroup (S˜ x t )t∈R+. Note that both theorems imply the next corollary. Corollary 1. Grant Assumptions 1 – 3. For all ˜ x ∈R, there exists a unique ρ˜ x ∞∈L1 +(w) with ρ˜ x ∞ L1 = 1 solving ∂aρ˜ x ∞(a, m) −λ∂m(mρ˜ x ∞(a, m)) = −f(a, m, ˜ x)ρ˜ x ∞(a, m), (9.3.26a) ρ˜ x ∞(0, m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), ˜ x)ρ˜ x ∞(a, γ−1(m))da, (9.3.26b) in the weak sense. Furthermore, we have that ρ˜ x ∞∈C(R+, L1 +(R∗ +)) ∩L∞(R+, L1 +(R∗ +)). 9.4 Stationary solutions for arbitrary connectivity strength In this section, we study the stationary solutions to (9.1.1), namely the solution to ∂aρ∞(a, m) −λ∂m(mρ∞(a, m)) = −f(a, m, εx∞)ρ∞(a, m), (9.4.27a) ρ∞(0, m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), εx∞)ρ∞(a, γ−1(m))da, (9.4.27b) x∞= Z ∞ 0 Z ∞ 0 ¯ h(a, m)f(a, m, εx∞)ρ∞(a, m)dadm. (9.4.27c) Deﬁnition. (ρ∞, x∞) ∈L1 +(w)∩C(R+, L1 +(R∗ +))∩L∞(R+, L1 +(R∗ +))×R+ is a stationary solution to (9.1.1) if ∥ρ∞∥L1 = 1 and if it solves Eq. (9.4.27) in the weak sense. 164 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY 9.4.1 Existence and uniqueness using the Doeblin-Harris method. We present two Lipschitz continuity results, which will allow us to prove the existence (and the uniqueness when ε is small) of stationary solutions. The following lemma plays the same role as Theorem 4.5 in : Lemma 4 (Lipschitz continuity at ﬁnite T). Grant Assumptions 1 – 3. For all initial data u0 ∈L1 +(w) and for all T > 0, there exists a constant CT, ∥u0∥L1(w) > 0 such that ∀f x1, f x2 ∈R, S e x1 T u0 −S e x2 T u0 L1(w) ≤CT, ∥u0∥L1(w)\|f x1 −f x2\|. (9.4.28) Proof. For all t > 0, S e x1 t u0 −S e x2 t u0 L1(w) = Z ∞ 0 Z ∞ t u0(a −t, eλtm) exp λt − Z t 0 f(a −t + s, eλ(t−s)m, f x1)ds ! −u0(a −t, eλtm) exp λt − Z t 0 f(a −t + s, eλ(t−s)m, f x2)ds ! w(a, m)dadm + Z ∞ 0 Z t 0 ρ e x1 t−a(0, eλam) exp λa − Z t t−a f(a −t + s, eλ(t−s)m, f x1)ds ! −ρ e x2 t−a(0, eλam) exp λa − Z t t−a f(a −t + s, eλ(t−s)m, f x2)ds ! w(a, m)dadm =: Q1 + Q2. Q1 = Z ∞ 0 Z ∞ 0 u0(a, m) exp − Z t 0 f(a + s, e−λsm, f x1)ds ! −exp − Z t 0 f(a + s, e−λsm, f x2)ds ! w(a + t, e−λtm)dadm ≤ Z ∞ 0 Z ∞ 0 u0(a, m) Z t 0 f(a + s, e−λsm, f x1) −f(a + s, e−λsm, f x2) ds ! w(a + t, e−λtm)dadm ≤tLf\|f x1 −f x2\| Z ∞ 0 Z ∞ 0 u0(a, m)w(a + t, e−λtm)dadm ≤tLf∥u0∥L1(w) \|f x1 −f x2\|, where in the last inequality we used w(a + t, e−λtm) ≤w(a, m), ∀a ≥0, m ≥0. (9.4.29) Q2 = Z ∞ 0 Z t 0 ρ e x1 t−a(0, m) exp − Z t t−a f(a −t + s, eλ(t−s−a)m, f x1)ds ! −ρ e x2 t−a(0, m) exp − Z t t−a f(a −t + s, eλ(t−s−a)m, f x2)ds ! w(a, e−λam)dadm. 9.4. STATIONARY SOLUTIONS FOR ARBITRARY CONNECTIVITY STRENGTH 165 By changes of variables, Q2 = Z ∞ 0 Z t 0 ρ e x1 s (0, m) exp − Z t−s 0 f(u, e−λum, f x1)du ! −ρ e x2 s (0, m) exp − Z t−s 0 f(u, e−λum, f x2)du ! w(t −s, e−λ(t−s)m)dsdm ≤ Z ∞ 0 Z t 0 ρ e x1 s (0, m) exp − Z t−s 0 f(u, e−λum, f x1)du ! −exp − Z t−s 0 f(u, e−λum, f x2)du ! w(t −s, e−λ(t−s)m)dsdm + Z ∞ 0 Z t 0 ρ e x1 s (0, m) −ρ e x2 s (0, m) w(t −s, e−λ(t−s)m)dsdm =: Q2,1 + Q2,2 Q2,1 ≤t∥f∥∞Lf\|f x1 −f x2\| Z t 0 Z ∞ 0 Z ∞ 0 (γ−1)′(m) ρ e x1 s (a, γ−1(m))w(t, m)dadmds ≤t∥f∥∞Lf\|f x1 −f x2\| Z t 0 Z ∞ 0 Z ∞ 0 ρ e x1 s (a, m)w(t, m +∥Γ∥∞)dadmds ≤t(1 +∥Γ∥∞)∥f∥∞Lf\|f x1 −f x2\| Z t 0 ρ e x1 s L1(w) ds, where in the last inequality we used w(t, m +∥Γ∥∞) = 1 + m +∥Γ∥∞≤(1 +∥Γ∥∞)w(a, m), ∀a ≥0, m ≥0. (9.4.30) By Lemma 2, Q2,1 ≤t2(1 +∥Γ∥∞)∥f∥∞Lf ∥u0∥L1(w) + b α ! \|f x1 −f x2\|. Q2,2 ≤∥f∥∞ Z t 0 Z ∞ 0 Z ∞ 0 (γ−1)′(m) ρ e x1 s (a, γ−1(m)) −ρ e x2 s (a, γ−1(m)) w(t, m)dadmds ≤∥f∥∞ Z t 0 Z ∞ 0 Z ∞ 0 ρ e x1 s (a, m) −ρ e x2 s (a, m) w(t, m +∥Γ∥∞)dadmds ≤(1 +∥Γ∥∞)∥f∥∞ Z t 0 S e x1 s u0 −S e x2 s u0 L1(w) ds, where again, in the last inequality, we used Eq. (9.4.30). Fix T > 0. Gathering the bounds for Q1, Q2,1 and Q2,2 we see that there exists constants C > 0 and C′ T, ∥u0∥L1(w) > 0 such that, for all t ∈[0, T], S e x1 t u0 −S e x2 t u0 L1(w) ≤C Z t 0 S e x1 s u0 −S e x2 s u0 L1(w) ds + tC′ T, ∥u0∥L1(w)\|f x1 −f x2\|. By Grönwall’s lemma, for all t ∈[0, T], S e x1 t u0 −S e x2 t u0 L1(w) ≤ C′ T, ∥u0∥L1(w)\|f x1 −f x2\| C exp(Ct) −1 . (9.4.31) Since Eq. (9.4.31) holds for all t ∈[0, T], this achieves the proof. 166 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY Lemma 5 (Lipschitz continuity at T = ∞). Grant Assumptions 1 – 3. Writing ρ˜ x ∞∈L1 +(w) the invariant probability measure given by Theorem 4 for any ˜ x ∈R, the function Υ : R+ →R+, Υ(x) = Z ∞ 0 Z ∞ 0 ¯ h(a, m)f(a, m, εx)ρεx ∞(a, m)dadm is Lipschitz and there exists C > 0 such that ∀x1, x2 ∈R+, \|Υ(x1) −Υ(x2)\|≤\|ε\|C\|x1 −x2\|. Proof. Since f is Lipschitz in x, we have, for any x1, x2 ∈R+, \|Υ(x1) −Υ(x2)\| ≤ ¯ h ∞ n ∥f∥∞∥ρεx1 ∞−ρεx2 ∞∥L1 + Lf\|ε\|\|x1 −x2\| o ≤ ¯ h ∞ n ∥f∥∞∥ρεx1 ∞−ρεx2 ∞∥L1(w) + Lf\|ε\|\|x1 −x2\| o , from where we only need to bound the ﬁrst term on the right hand side. We can use Theorem 4 and Lemma 4: for any T ∈R+, ∥ρεx1 ∞−ρεx2 ∞∥L1(w) = ∥Sεx1 T ρεx1 ∞−Sεx1 T ρεx2 ∞+ Sεx1 T ρεx2 ∞−Sεx2 T ρεx2 ∞∥L1(w) ≤Ke−aT ∥ρεx1 ∞−ρεx2 ∞∥L1(w)+CT \|ε\|\|x1 −x2\|, where K and a are the exponential stability constants of Theorem 4. Choosing T such that Ke−aT = 1/2, we get ∥ρεx1 ∞−ρεx2 ∞∥L1(w) ≤2CT \|ε\|\|x1 −x2\|. Gathering the bounds concludes the proof. Theorem 6 (Stationary solutions). Grant Assumptions 1 – 3. We have I There exists a stationary solution to (9.1.1). II There exists ε∗> 0 such that for all ε ∈] −ε∗, +ε∗[, the stationary solution to (9.1.1) is unique. Proof. For all ˜ x ∈R, let us write ρ˜ x ∞∈L1 +(w) the unique invariant measure given by Theorem 4 and let us also take the function Υ from Lemma 5. By Corollary 1, (ρ∞, x∞) ∈L1 +(w) ∩ C(R+, L1 +(R∗ +))∩L∞(R+, L1 +(R∗ +))×R+ is a weak solution to Eq. (9.4.27) if and only if ρ∞= ρεx∞ ∞ and x∞is a ﬁxed-point of Υ. Hence, the study of the existence and the uniqueness of stationary solutions is reduced to the study of the existence and the uniqueness of the ﬁxed-point of Υ. Since for all x ∈R+,∥ρεx ∞∥L1 = 1, we have that for all x ∈R+, Υ(x) ≤ ¯ h ∞∥f∥∞. Therefore, the set [0, ¯ h ∞∥f∥∞] (which is compact and convex) is stable by Υ. Then, the continuity of Υ guarantees the existence of a ﬁxed-point, which proves (i). To obtain (ii), we observe that the Lipschitz constant of Υ is \|ε\|C: if we take \|ε\|< ε∗:= C−1, Υ is a contraction and we can apply Banach’s ﬁxed-point theorem to conclude. 9.4.2 Alternative proof for the existence using Schauder’s ﬁxed-point theorem We include here an alternative proof for the existence of a stationary solution, which is interesting for two reasons: on the one hand, it does not rely on the Harris-Doeblin method, and on the other hand, it provides some estimates on the stationary solutions. For any (˜ u, ˜ x) ∈L1 +(]γ(0), +∞[) × R, consider the transport equation ∂aϱ(a, m) −λ∂m(mϱ(a, m)) = −f(a, m, ˜ x)ϱ(a, m), ϱ(0, m) = ˜ u(m). 9.4. STATIONARY SOLUTIONS FOR ARBITRARY CONNECTIVITY STRENGTH 167 It has a unique weak solution ρ˜ u,˜ x ∞∈C(R+, L1 +(R∗ +)) ∩L∞(R+, L1 +(R∗ +)) given by the method of characteristics: for all (a, m) ∈R+ × R∗ +, ρ˜ u,˜ x ∞(a, m) = ˜ u(eλam) exp λa − Z a 0 f(s, eλ(a−s)m, ˜ x)ds . (9.4.32) We can now deﬁne the operator Φ := (Φ1, Φ2) on L1 +(]γ(0), +∞[) × R where, for all (˜ u, ˜ x) ∈L1 +(]γ(0), +∞[) × R, Φ1(˜ u, ˜ x)(m) := 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), ˜ x)ρ(˜ u,˜ x) ∞ (a, γ−1(m))da, (9.4.33a) Φ2(˜ u, ˜ x) := Z ∞ 0 Z ∞ 0 ¯ h(a, m)f(a, m, ˜ x)ρ˜ u,˜ x ∞(a, m)dadm. (9.4.33b) (ρ∞, x∞) is a stationary solution if and only if it is a ﬁxed-point of Φ. Whence, we get the a priori estimates: Lemma 6. Grant Assumptions 1 and 3. There exists θ ∈]0, 1[ such that for all (˜ u, ˜ x) ∈ L1 +(]γ(0), +∞[) × R, I Φ1(˜ u, ˜ x) L1 =∥˜ u∥L1. II For all m ∈R∗ +, \|Φ1(˜ u, ˜ x)(m)\|≤1m>γ(0) ∥f∥∞ λγ−1(m)∥˜ u∥L1. III Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm ≤max Z ∞ 0 ˜ u(m)mdm, γ(0) 1 −θ∥˜ u∥L1 ! . IV For all β ∈]0, min(f) λ [, Z ∞ γ(0) Φ1(˜ u, ˜ x)(m) γ−1(m)β dm ≤∥f∥∞ λγ(0)β min(f) λ −β ! ∥˜ u∥L1 . V Φ2(˜ u, ˜ x) ≤ ¯ h ∞∥˜ u∥L1. Proof. (i) By changes of variables on m, Φ1(˜ u, ˜ x) L1 = Z ∞ 0 Z ∞ 0 f(a, m, ˜ x)˜ u(eλam) exp λa − Z a 0 f(s, eλ(a−s)m, ˜ x)ds dadm = Z ∞ 0 ˜ u(m) Z ∞ 0 f(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds da \| {z } =1 (by Assumption 3 (i)) dm. (ii) \|Φ1(˜ u, ˜ x)(m)\| ≤1m>γ(0)∥f∥∞ Z ∞ 0 ˜ u(eλaγ−1(m)) exp(λa)da = 1m>γ(0) ∥f∥∞ λγ−1(m) Z ∞ 0 ˜ u(eλaγ−1(m))γ−1(m)λ exp(λa)da = 1m>γ(0) ∥f∥∞ λγ−1(m) Z ∞ γ−1(m) ˜ u(y)dy \| {z } ≤ ∥˜ u∥L1 , where for the last equality we used the change of variable y = eλaγ−1(m). 168 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY (iii) Performing the same change of variable as for (i) and using the fact that γ(m) ≤γ(0)+m, ∀m ∈R+ (since γ′ ≤1), we have Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm = Z ∞ 0 ˜ u(m) Z ∞ 0 γ(e−λam)f(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds dadm ≤ Z ∞ 0 ˜ u(m)m Z ∞ 0 e−λaf(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds da \| {z } =:ϑ(m) dm + γ(0)∥˜ u∥L1 . There exists θ ∈]0, 1[ such that for all m ∈R∗ +, ϑ(m) < 1: Fix ϵ > 0. ϑ(m) ≤ Z ϵ 0 f(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds da + Z ∞ ϵ e−λϵf(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds da = 1 −(1 −e−λϵ) Z ∞ ϵ f(a, e−λam, ˜ x) exp − Z a 0 f(s, e−λsm, ˜ x)ds da = 1 −(1 −e−λϵ) exp − Z ϵ 0 f(s, e−λsm, ˜ x)ds ≤1 −(1 −e−λϵ) exp(−∥f∥∞ϵ) =: θ < 1. Whence, Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm ≤θ Z ∞ 0 ˜ u(m)mdm + γ(0)∥˜ u∥L1 . To see that Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm ≤max Z ∞ 0 ˜ u(m)mdm, γ(0) 1 −θ∥˜ u∥L1 ! , we can distinguish three cases: if R ∞ 0 ˜ u(m)mdm = ∞, the inequality is trivial; if γ(0) 1−θ ∥˜ u∥L1 ≤ R ∞ 0 ˜ u(m)mdm < +∞, then Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm ≤ Z ∞ 0 ˜ u(m)mdm −(1 −θ) Z ∞ 0 ˜ u(m)mdm + γ(0)∥˜ u∥L1 ≤ Z ∞ 0 ˜ u(m)mdm; and ﬁnally if R ∞ 0 ˜ u(m)mdm < γ(0) 1−θ ∥˜ u∥L1, then Z ∞ 0 Φ1(˜ u, ˜ x)(m)mdm ≤θ γ(0) 1 −θ∥˜ u∥L1 + γ(0)∥˜ u∥L1 = γ(0) 1 −θ∥˜ u∥L1 . (iv) Z ∞ γ(0) Φ1(˜ u, ˜ x)(m) γ−1(m)β dm = Z ∞ 0 Z ∞ 0 1 mf(a, m, ˜ x)˜ u(eλam) exp λa − Z a 0 f(s, eλ(a−s)m, ˜ x) dadm ≤∥f∥∞ Z ∞ 0 Z ∞ 0 1 mβ ˜ u(eλam) exp λa −min(f)a dadm, making the change of variable y = eλam: =∥f∥∞ Z ∞ 0 Z ∞ m 1 λm1+β ˜ u(y) exp −min(f) 1 λ ln y m ! dydm = ∥f∥∞ λ Z ∞ 0 Z ∞ m mmin(f)/λ−1−β˜ u(y)y−min(f)/λdydm, 9.4. STATIONARY SOLUTIONS FOR ARBITRARY CONNECTIVITY STRENGTH 169 using Fubini’s theorem and the fact that min(f)/λ −β > 0: = ∥f∥∞ λ Z ∞ 0 ˜ u(y)y−min(f)/λ Z y 0 mmin(f)/λ−1−βdm \| {z } = ymin(f)/λ−β min(f)/λ−β dy = ∥f∥∞ λ min(f) λ −β ! Z ∞ 0 ˜ u(y)y−βdy. Finally, it is easy to check that R ∞ 0 ˜ u(y)y−βdy ≤γ(0)−β∥˜ u∥L1. (v) Use Eq. (9.4.33b) and see the proof of (i). By these estimates, we see that there exists β, C1, C2, C3, C4 > 0 such that the set C × B ∈ L1(]γ(0), +∞[) × R, where C := u ∈L1 +(]γ(0), +∞[) ∥u∥L1 ≤1; u ≤ C1 γ−1(·) a.e.; Z ∞ 0 u(m)mdm ≤C2; Z ∞ γ(0) u(m) γ−1(m)β dm ≤C3 and B := [−C4, +C4], is stable by the operator Φ. In order to apply Schauder’s ﬁxed-point theorem, we will need Lemma 7. Grant Assumptions 1 and 3. C is convex, closed and compact for the weak topology σ(L1, L∞). Proof. It is easy to verify that C is convex. Since C is convex, if suﬃces to show that it is strongly closed to show that it is weakly closed. Let un be a sequence of elements of C which converge strongly to u ∈L1(]γ(0), +∞[). By the strong convergence, ∥u∥L1 ≤1. We can extract a subsequence unk such that unk converges to u a.e. Taking the pointwise limit, we have that u ≤ C1 γ−1(·) a.e. Furthermore, by Fatou’s lemma, Z ∞ γ(0) u(m)mdm ≤lim inf k→+∞ Z ∞ γ(0) unk(m)mdm ≤C2 and Z ∞ γ(0) u(m) γ−1(m)β dm ≤lim inf k→+∞ Z ∞ γ(0) unk(m) γ−1(m)β dm ≤C3. Hence, C is strongly closed. To show that C is weakly compact, we will show that a. supu∈C ∥u∥L1 < ∞, b. ∀ϵ > 0, ∃R > 0 such that R ∞ R u(m)dm < ϵ for all u ∈C , c. C is equi-integrale, i.e. ∀ϵ > 0, ∃δ > 0 such that for all Borel set A ⊂R+ with \|A\|≤δ and for all u ∈C , R A u(m)dm ≤ϵ, and use Dunford-Pettis theorem. (a.) is clearly veriﬁed. (b.) is also veriﬁed since for all R > 0, R ∞ R u(m)dm ≤1 R R ∞ 0 u(m)mdm ≤C2 R . To show (c.), let us ﬁrst observe that for all δ1 > 0, Z γ(0)+δ1 γ(0) u(m)dm ≤γ−1(γ(0) + δ1)β Z ∞ γ(0) u(m) γ−1(m)β dm ≤γ−1(γ(0) + δ1)βC3. 170 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY For any ϵ > 0, let us choose δ1 > 0 such that γ−1(γ(0) + δ1)βC3 ≤ϵ 2. Then, for all Borel set A ⊂R+ with \|A\|≤δ, Z A u(m)dm ≤ Z γ(0)+δ1 γ(0) u(m)dm + Z A[0,γ(0)+δ1] u(m)dm ≤ϵ 2 + δ C1 γ−1(γ(0) + δ1). Hence, we can choose δ = min δ1, ϵγ−1(γ(0)+δ1) 2C1 and (c.) is veriﬁed. By the Dunford-Pettis theorem, C is weakly relatively compact. Finally, since C is weakly closed, C is weakly compact. We can now give an alternative proof of the existence of stationary solutions to (9.1.1) for arbitrary connectivity strength ε: Proof of Theorem 2 (i). We verify that the operator Φ is weakly continuous: For any sequence (un, xn) →(u, x) in C × R and for any ϕ ∈L∞(R+), Z (Φ1(un, xn) −Φ1(u, x))ϕ(m)dm ≤Qn 1 + Qn 2 + Qn 3, where Qn 1 := Z ∞ 0 Z ∞ 0 (un(eλam) −u(eλam))ϕ(γ(m))eλaf(a, m, x)e−R a 0 f(τ,eλ(a−τ)m,x)dτdadm , Qn 2 :=∥ϕ∥∞ Z ∞ 0 Z ∞ 0 un(eλam)eλa\|f(a, m, x) −f(a, m, xn)\|e−R a 0 f(τ,eλ(a−τ)m,x)dτdadm, Qn 3 :=∥ϕ∥∞ Z ∞ 0 Z ∞ 0 un(eλam)eλaf(a, m, xn) e−R a 0 f(τ,eλ(a−τ)m,x)dτ −e−R a 0 f(τ,eλ(a−τ)m,xn)dτ dadm. Making the change of variable ydy = eλamdm in Q1 we get Qn 1 = Z ∞ 0 (un(y) −u(y)) Z ∞ 0 ϕ(γ(ye−λa))f(a, ye−λa, x)e−R a 0 f(τ,e−λτy,x)dadm . Since un converges to u in σ(L1, L∞) and Z ∞ 0 ϕ(γ(ye−λa))f(a, ye−λa, x)e−R a 0 f(τ,e−λτy,x)dτda ≤∥ϕ∥∞ Z ∞ 0 f(a, ye−λa, x)e−R a 0 f(τ,e−λτy,x)dτda =∥ϕ∥∞, Qn 1 converges to 0. On the other hand, since f is bounded and Lipschitz, Qn 2, Qn 3 ≤∥un∥L1C\|xn− x\|≤C\|xn−x\|. Whence, Φ1 is a continuous operator with respect to the weak topology σ(L1, L∞). The continuity of Φ2 is shown analogously, taking ϕ = h (h is a bounded). Since C is stable by Φ, convex and weakly compact (Lemma 7), we can apply Schauder’s ﬁxed-point theorem to obtain the existence of a ﬁxed-point, which gives the existence of a stationary solution. Corollary 2. Grant Assumptions 1 and 3. If f is of class Ck, then u(m) is a function of class Ck for all m > γ(0). Consequently, the stationary solutions of (9.1.1) are of class Ck. Proof. If (u, ˜ x) is a ﬁxed-point of Φ, then u(m) = 1m>γ(0) (γ−1)′(m) Z ∞ 0 f(a, γ−1(m), ˜ x)u(eλaγ−1(m)) exp λa − Z a 0 f(s, eλ(a−s)γ−1(m), ˜ x)ds da. (9.4.34) 9.4. STATIONARY SOLUTIONS FOR ARBITRARY CONNECTIVITY STRENGTH 171 Making the change of variable y = eλaγ−1(m) in a, as in the estimate (ii) of Lemma 6, we obtain u(m) = 1m>γ(0) (γ−1)′(m) λγ−1(m) Z ∞ γ−1(m) f(g(y, m), y, ˜ x)u(y) exp − Z g(y,m) 0 f(s, esy, ˜ x)ds ! dy, (9.4.35) where g(y, m) = ln y λ(γ−1(m)). We conclude with a bootstrap argument: if u is L1, then the right hand side of Eq. (9.4.35) is a continuous function of m, meaning that u is continuous. But if u is continuous, then the right hand side is of class C1, etc. Corollary 3. Grant Assumptions 1 and 3. There exists a constant C > 0, such that the stationary solution ρ∞satisﬁes, ρ∞(a, m) ≤Ce−σ(a−∆abs) m . (9.4.36) Proof. From the previous theorem it follows that there is C such that u(m) ≤C/m, which, together with (9.4.32), implies ρ∞(a, m) ≤C e−R a 0 f(s,eλ(a−s)γ−1(m),˜ x)ds m . (9.4.37) The estimate follows from Assumption 3 (i). 9.4.3 Formula in the case of short-term synaptic depression In general, there is no explicit formula for the invariant probability measure solving Eq. (9.3.26). However, in the case of short-term synaptic depression Eq. (9.1.3), we can derive an explicit expression for the total postsynaptic potential X(˜ x) := Z ∞ 0 ˆ h(t) Z 1 0 Z ∞ 0 (1 −m)f(a, ˜ x)ρ˜ x ∞(a, m)dadmdt, (9.4.38) for any ˜ x ∈R. This fact has been reported in the theoretical neuroscience literature ; we provide here a rigorous and analytic justiﬁcation for it. For all ˜ x ∈R, let us introduce the quantities I ˜ x := Z ∞ 0 af(a, ˜ x) exp − Z a 0 f(s, ˜ x)ds da = Z ∞ 0 exp − Z a 0 f(s, ˜ x)ds da, P ˜ x(λ) := Z ∞ 0 e−λaf(a, ˜ x) exp − Z a 0 f(s, ˜ x)ds da. I ˜ x can be interpreted as the mean inter-spike interval of a neuron receiving a constant input ˜ x. P ˜ x(λ) can be seen as the Laplace transform of the inter-spike interval distribution of that neuron, evaluated in λ. Proposition 2. Grant Assumptions 1 and 3. For all ˜ x ∈R, X(˜ x) = Z ∞ 0 ˆ h(t)dt 1 I ˜ x ( 1 −P ˜ x(λ) 1 −υP ˜ x(λ) ) . Proof. Using the method of characteristics (i.e. combining Eqs. (9.3.26b) and (9.4.32)), we have 1 = Z 1 0 Z ∞ 0 ρ˜ x ∞(a, m)dadm = Z 1 0 Z ∞ 0 1eλam<1ρ˜ x ∞(0, eλam) exp λa − Z a 0 f(s, ˜ x)ds dadm = Z 1 0 Z ∞ 0 ρ˜ x ∞(0, m) exp − Z a 0 f(s, ˜ x)ds dadm = I ˜ x Z 1 0 ρ˜ x ∞(0, m)dm. 172 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY Whence, Z 1 0 Z ∞ 0 f(a, ˜ x)ρ˜ x ∞(a, m)dadm = Z 1 0 ρ˜ x ∞(0, m)dm = 1 I ˜ x . On the other hand, Z 1 0 Z ∞ 0 mf(a, ˜ x)ρ˜ x ∞(a, m)dadm = Z 1 0 Z ∞ 0 1eλam<1mf(a, ˜ x)ρ˜ x ∞(0, eλam) exp λa − Z a 0 f(s, ˜ x)ds dadm = Z 1 0 Z ∞ 0 e−λamf(a, ˜ x)ρ˜ x ∞(0, m) exp − Z a 0 f(s, ˜ x)ds dadm = P ˜ x(λ) Z 1 0 mρ˜ x ∞(0, m)dm and Z 1 0 mρ˜ x ∞(0, m)dm = Z 1 0 m1m>1−υ 1 υ Z ∞ 0 f(a, ˜ x)ρ˜ x ∞ a, 1 −1 −m υ dadm = Z 1 0 (1 −υ + υm) Z ∞ 0 f(a, ˜ x)ρ˜ x ∞(a, m)dadm = 1 −υ I ˜ x + υP ˜ x(λ) Z 1 0 mρ˜ x ∞(0, m)dm. Whence, Z 1 0 mρ˜ x ∞(0, m)dm = 1 −υ I ˜ x(1 −υP ˜ x(λ)) and Z 1 0 Z ∞ 0 mf(a, ˜ x)ρ˜ x ∞(a, m)dadm = P ˜ x(λ)(1 −υ) I ˜ x(1 −υP ˜ x(λ)). Finally, we have X(˜ x) = Z ∞ 0 ˆ h(t)dt (Z 1 0 Z ∞ 0 f(a, ˜ x)ρ˜ x ∞(a, m)dadm − Z 1 0 Z ∞ 0 mf(a, ˜ x)ρ˜ x ∞(a, m)dadm ) = Z ∞ 0 ˆ h(t)dt 1 I ˜ x ( 1 −P ˜ x(λ)(1 −υ) (1 −υP ˜ x(λ)) ) = Z ∞ 0 ˆ h(t)dt 1 I ˜ x ( 1 −P ˜ x(λ) 1 −υP ˜ x(λ) ) . 9.5 Exponential stability in the weak connectivity regime To study the long time behavior (9.1.1) in the weak connectivity regime, we perturb the non-interacting case (9.3.19), taking ˜ x = εx∞, where x∞is given by the unique stationary solution to (9.1.1) when ε ∈] −ε∗, +ε∗[ (ε∗is taken from Theorem 2 (ii)). In this section, we keep the small ε ﬁxed and we work under Assumptions 1 – 3 and 5. We roughly follow the same line of argument as [181, Sec. 5]. For convenience, we ﬁrst rewrite (9.1.1) in a more formal and compact form: ∂tρt = −∂aρt + λ∂m(mρt) −f(εxt)ρt + δa 0(γ ◦Π)∗ f(εxt)ρt , (9.5.39a) xt = Z t 0 Z h(t −s)f(εxs)ρs dadm ds, (9.5.39b) ρ0 = u0, (9.5.39c) 9.5. EXPONENTIAL STABILITY IN THE WEAK CONNECTIVITY REGIME 173 where δa 0 indicates that (singular) mass enters in a = 01, Π : (a, m) 7→m is the projection on m and ∗denotes the pushforward measure. To write Eq. (9.5.39) as an evolution equation, we introduce an auxiliary transport equation on R+ × R+ × R∗ + ∂tζt = −∂sζt + δs 0f(εxt)ρt, ζ0 = 0, which solution is given by the method of characteristics: ζt(s) = 1s≤tf(εxt−s)ρt−s, ∀(t, s) ∈R∗ + × R+. Using the auxiliary equation, Eq. (9.5.39) is equivalent to ∂t(ρt, ζt) = −∂aρt + λ∂m(mρt) −f(εxt)ρt + δa 0(γ ◦Π)∗ f(εxt)ρt , −∂sζt + δs 0f(εxt)ρt , (9.5.40a) (ρ0, ζ0) = (u0, 0) , (9.5.40b) where xt := R ∞ 0 R h(s)ζt(s) dadm ds. By Theorem 2, for all ε ∈] −ε∗, +ε∗[, there exists a unique stationary solution (ρ∞, x∞) and we have −∂aρ∞+ λ∂m(mρ∞) −f(εx∞)ρ∞+ δa 0(γ ◦Π)∗ f(εx∞)ρ∞ = 0. (9.5.41) Now, we write Eq. (9.5.40) as the sum of a linear equation and a perturbation: ∂t(ρt, ζt) = Λ(ρt, ζt) + (Z(1) t , Z(2) t ), (9.5.42a) (ρ0, ζ0) = (u0, 0) , (9.5.42b) where Λ(ρt, ζt) := −∂aρt + λ∂m(mρt) −f(εx∞)ρt + δa 0(γ ◦Π)∗ f(εx∞)ρt , −∂sζt + δs 0f(εx∞)ρt , Z(1) t := [f(εx∞) −f(εxt)]ρt + δa 0(γ ◦Π)∗([f(εxt) −f(εx∞)]ρt), Z(2) t := δs 0[f(εxt) −f(εx∞)]ρt. Let us put ζ∞(s) := f(εx∞)ρ∞, ∀s ∈R+. Then, using Eq. (9.5.41), by the linearity of the operator Λ and writing ¯ ρt := ρt −ρ∞and ¯ ζt := ζt −ζ∞, we get ∂t(¯ ρt, ¯ ζt) = Λ(¯ ρt, ¯ ζt) + (Z(1) t , Z(2) t ), (9.5.43a) (¯ ρ0, ¯ ζ0) = (u0 −ρ∞, −ζ∞) . (9.5.43b) Writing (SΛ t )t∈R+ the semigroup associated with the operator Λ, we have, by Duhamel’s formula, (¯ ρt, ¯ ζt) = SΛ t (¯ ρ0, ¯ ζ0) + Z t 0 SΛ t−s(Z(1) s , Z(2) s )ds, ∀t ≥0. (9.5.44) Let us deﬁne the weighted space L1 +(µ) := ζ ∈L1(R+ × R+ × R∗ +, R+) Z ∞ 0 ζ(s) L1∥h∥∞e−hsds < ∞ . 1δa 0 should not be confused with the Dirac distribution δ0=a. Using δ0=a, by integration by parts of weak solutions, Eq. (9.5.39a) should write ∂tρt = −∂aρt + λ∂m(mρt) −f(εxt)ρt + δ0=a n (γ ◦Π)∗ f(εxt)ρt −ρt(0, ·) o . 174 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY Note that, for all t ≥0, \|xt −x∞\| = Z ∞ 0 Z h(s)ζt(s) dadm ds − Z ∞ 0 Z h(s)ζ∞(s) dadm ds ≤ Z ∞ 0 ∥h∥∞e−hs ζt(s) −ζ∞(s) L1 ds = ¯ ζt L1(µ) . Also, we have, for all t ≥0, Z(1) t L1 ≤\|ε\|2Lf∥ρt∥L1 \|xt −x∞\|≤\|ε\|2Lf ¯ ζt L1(µ) , (9.5.45a) Z(1) t L1(w) ≤\|ε\|2Lf∥ρt∥L1(w) \|xt −x∞\|≤\|ε\|2Lf ∥u0∥L1(w)+ b α ¯ ζt L1(µ) , (9.5.45b) Z(2) t L1(µ) ≤\|ε\| ∥h∥∞Lf∥ρt∥L1 \|xt −x∞\|≤\|ε\| ∥h∥∞Lf ¯ ζt L1(µ) , (9.5.45c) where we have used Theorem 1 (ii) in the ﬁrst line and Lemma 1 in the second. Lemma 8. Grant Assumptions 1 – 3 and 5 and take (¯ ρ0, ¯ ζ0) as in Eq. (9.5.43). There exists K1 ≥1 and a1 > 0 such that, for all initial data u0 ∈L1 +(w) with ∥u0∥L1 = 1, SΛ t (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) ≤K1e−a1t (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) , ∀t ≥0. (9.5.46) If in addition, we grant Assumption 4, then there exists K2 ≥1 and a2 > 0 such that, for all initial data u0 ∈L1 + with ∥u0∥L1 = 1, SΛ t (¯ ρ0, ¯ ζ0) L1×L1(µ) ≤K2e−a2t (¯ ρ0, ¯ ζ0) L1×L1(µ) , ∀t ≥0. (9.5.47) Proof. We write (SΛ t (¯ ρ0, ¯ ζ0)(1), SΛ t (¯ ρ0, ¯ ζ0)(2)) := SΛ t (¯ ρ0, ¯ ζ0) the ﬁrst and second component of SΛ t (¯ ρ0, ¯ ζ0). By Theorem 4, there exists K ≥0 and a > 0 such that, SΛ t (¯ ρ0, ¯ ζ0)(1) L1(w) ≤Ke−at∥¯ ρ0∥L1(w) , ∀t ≥0. Then, SΛ t (¯ ρ0, ¯ ζ0)(2) L1(µ) = Z t 0 f(εx∞)SΛ t−s(¯ ρ0, ¯ ζ0)(1) L1 Che−hsds + Z ∞ t ¯ ζ0(s) L1 Che−hsds ≤Ch ( ∥f∥∞K Z t 0 e−a(t−s)e−hsds∥¯ ρ0∥L1(w) + e−ht ¯ ζ0 L1(µ) ) . Gathering the bounds on the two components and observing that the function t 7→ R t 0 e−a(t−s)e−hsds decays exponentially, we conclude that there exists K1 ≥1 and a1 > 0 such that Eq. (9.5.46) holds. For Eq. (9.5.47), we use Theorem 5 and follow the same argument. We can now prove our main result: 9.5. EXPONENTIAL STABILITY IN THE WEAK CONNECTIVITY REGIME 175 Proof of Theorem 3. By Duhamel’s formula (9.5.44), Eq. (9.5.46) in Lemma 8 and the bounds Eqs. (9.5.45), for all t ≥0, (¯ ρt, ¯ ζt) L1(w)×L1(µ) ≤ SΛ t (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) + Z t 0 SΛ t−s(Z(1) s , Z(2) s ) L1(w)×L1(µ) ds ≤K1e−a1t (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) + K1 Z t 0 e−a1(t−s) (Z(1) s , Z(2) s ) L1(w)×L1(µ) ds ≤K1e−a1t (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) + \|ε\| ˜ CW Z t 0 e−a1(t−s) (¯ ρs, ¯ ζs) L1(w)×L1(µ) ds =: Q(t), where ˜ CK is a constant depending on W. We have, for all t ≥0, d dtQ(t) = −a1Q(t) + \|ε\| ˜ CW (¯ ρt, ¯ ζt) L1(w)×L1(µ) ≤ −a1 + \|ε\| ˜ CW Q(t). Whence, by Grönwall’s lemma, ∀t ≥0, Q(t) ≤K1 (¯ ρ0, ¯ ζ0) L1(w)×L1(µ) exp −a1 + \|ε\| ˜ CW t . For all t ≥0, we have ∥ρt −ρ∞∥L1(w) + \|xt −x∞\|≤ (¯ ρt, ¯ ζt) L1(w)×L1(µ) ≤Q(t) and ¯ ζ0 L1(µ) ≤ Z ∞ 0 f(εx∞)ρ∞ L1 Che−hsds ≤∥f∥∞Ch h . Whence, choosing ε∗∗ W := a1 ˜ CW ∧ε∗, we easily see that there exists C ≥1 and cW > 0 such that Eq. (9.1.6) holds. For Eq. (9.1.7), we use Eq. (9.5.47) instead of Eq. (9.5.46) and follow the same argument. Appendix Here, we compare simulations of Eq. (9.1.2) with simulations of the time elapsed neuron network model . If, the ﬁring rate function f does not depend on m and if we put f(a, εxt) := ˆ f(η(a) + εxt), (9.5.48) then Eq. (9.1.2) reduces to the time elapsed neuron network model ∂tρt(a) + ∂aρt(a) = −f(a, εxt)ρt(a), (9.5.49a) ρt(0) = Z ∞ 0 f(a, εxt)ρt(a)da, (9.5.49b) xt = Z t 0 h(t −s) Z ∞ 0 Z ∞ 0 f(a, εxs)ρs(a)dads, (9.5.49c) ρ0(a) = u0(a). (9.5.49d) Eq. (9.5.49) is the population equation for non-adaptive SRM0 neurons (or age-dependent nonlinear Hawkes processes) . As reported previously, Eq. (9.5.49) exhibits self-sustained oscillations for large ε or relaxation to a stationary state for small ε (see Fig. 2). 176 CHAPTER 9. TIME ELAPSE MODEL WITH MEMORY 0 5 0 5 10 15 0 5 0 5 10 15 Figure 9.2: Same as Fig. 1 but for the time elasped neuron network model Eq. (9.5.49). Simulations of a network of 5 · 105 non-adaptive SRM0 neurons, approximating Eq. (9.5.49), with identical parameters (except for ε) and identical initial conditions. Neuronal parameters are the same is in Fig. 1, expect that f is replaced by Eq. (9.5.48). The ε have also been adapted. Chapter 10 Concentration inequality for dynamical systems and applications This chapter is the result of a work in progress in collaboration with Sylvain Delattre and Marc Hoﬀmann. 10.1 Introduction We are interested in studying the stochastic system of N interacting agents Zt = (Z1 t , . . . , ZN t ), t ∈[0, T], (10.1.1) where each agent Zi t = (Xi t, Y i t ) is described by two states Xi t, Y i t evolving in the Euclidean space Rd and solves        dXi t = F(Xi t, Y i t )dt + 1 N PN j=1 H(Xi t −Xj t , Y i t −Y j t ) + σdBi t dY i t = G(Xi t, Y i t )dt L(X0, Y0) = µ⊗N 0 , (10.1.2) where Bi are independent Rd-valued Brownian motions, a constant σ > 0, and the coeﬃcient F, G and H satisfy regularity and the growth conditions to be speciﬁed below. This type of model was introduced back in the 1960s , originally in work in statistical physics to describe the interaction of particles in plasma physics, the state (X, Y) corresponding to the position and velocity of electrons or ions. This remains one of the main current applications and continues to be of interest to the mathematical community as illustrated by the works of Guillin et al., Bresch et al. and Boley et al. . Beyond that, the 2010s brought an expansion of the ﬁeld of application, spreading its use to collective animal behavior and population dynamics Bolley et al. , Mogilner et al. ; opinion dynamics Chazelle et al. , ﬁnance Fouque and Sun and neuroscience Baladron et al. . The latter is the main motivation for this work, in particular, the FitzHugh-Nagumo model that we describe below in details and has already been presented in the work of Mischler et al. , Lucon and Poquet , and Quininao and Touboul. The word agents, in this case, corresponds to neurons interacting in the same network interacting through electrical synapses. The coeﬃcients F and G describe the part of the behavior of each agent that depends only on its own state, we call it "single agent behavior", while H describes the interaction between the agents and the last term introduces some randomness, whose intensity is modulated by the σ parameter. The objective of this work goes in three related directions. First of all, we are interested in studying the behavior of such a system when the number of agents N is large. More in details, 177 178 CHAPTER 10. CONCENTRATION INEQUALITY we prove that the empirical measure of the system µN t = 1 N N X j=1 δXj (10.1.3) is close to µt the solution of the associated Fokker Planck equation ∂tµ(x, y) = −∇((F, G)µ(x, y)) −∇(µ(x, y) Z H(x −¯ x, y −¯ y)µ(d¯ x, d¯ y)) + 1/2σ2∂xxµ(x, y), (10.1.4) In order to do so, we analyze the mean-ﬁeld limit of the system and establish a concentration inequality. Extensive methodology has been developed to address these problems, see for example Tanaka and Hitsuda , Sznitman , , Fernandez et al. , Bolley , Massart among myriad references. Our result follows in spirit that of Della Maestra and Hoﬀmann comparable to the Bernstein deviation inequality for independent data as it appears in Boucheron et al. . The model (10.1.2) considered here is similar to the one studied by Bolley et al. in . In that work, the authors prove the existence of a limit equation, and then use classical coupling techniques to establish a mean ﬁeld limit. Here we present a diﬀerent strategy to arrive at the limit equation and we obtain a new concentration inequality that relates the empirical measure of the system and the resulting law of this limit equation in a stronger result. Our perspective follows the strategy proposed by the recent work of Lacker and Della Maestra and Hoﬀmann , which we extend here to accommodate to continuous coeﬃcients not necessarily bounded (as in Lacker) or globally Lipschitz (as in Della Maestra), and we deal with the degeneracy that arises when considering a two-state model like the one described above. As a ﬁrst application of the concentration result, we estimate the parameters of the FitzHugh-Nagumo model. As a second application, we take advantage of it to arrive at a nonparametric estimate of µt, following the arguments of Della Maestra and Hoﬀmann and extending them to our conditions. Finally, on the three and last stage of this chapter, we move towards the construction of a statistical test to answer the question of the existence or not of interactions between the agents. For diﬀerent models in the ﬁeld of neuroscience, it is widely accepted that the behavior of the voltage and the time between discharges of a population of neurons diﬀers signiﬁcantly when the connectivity between their is low or high. In chapter 1, we see that when the interaction is weak, the distribution of the state of the neurons, the elapse time between discharges, converges to a steady state, while when the connectivity increases the distribution depends on time and some patterns can appear, describing, for example, oscillations. The same behavior is observed in the time elapse model proposed by Pakdaman et al., , in the integrate and ﬁre model, see for example Cormier et al. , and for the FhN model, Mischler et al. . This leads us to the question, how connected are neurons in practice? when one sees the appearance of patterns in the measurements, is this because the population of neurons is receiving diﬀerent stimulation every short time, so that although the connectivity is weak, it never manages to reach the steady state or is it a stable pattern without external intervention caused by the connectivity between neurons? Our interest is to create a test that, based on the experience in a short window of time, can tell us if the interaction between the agents is weak or strong. In that sense, we use our ﬁrst result to anticipate a suitable candidate of an estimator and a rejection zone and we test it in diﬀerent simulations, testing its validity at least numerically. 10.1.1 The FitzHugh-Nagumo model The main motivation for this work is the Fitzhugh-Nagumo model (FhN) for populations of neurons. The FhN model was ﬁrst introduced in the works of FitzHugh and Nagumo 10.1. INTRODUCTION 179 as a simpliﬁcation of the Huxley-Hodgkin model (HH) that describes the evolution of the membrane potential of a neuron . The dynamics is based on two variables, a variable x which corresponds to the membrane potential and a recovery variable y, which satisfy the equations ˙ x = F(x, y), ˙ y = G(x, y), for F(x, y) = x−x3/3−y+I and G(x, y) = 1/c(x+a−by) with a, I, c > 0, b ∈R. On this model, I denotes the total membrane current and is a stimulus applied to the neuron, c determines the strength of damping while a and b govern two important characteristics of the oscillating solution, namely spike rate and spike duration . With only these elements the system shows the most important properties of the 4-dimensional HH model such as refractoriness, insensitivity to further immediate stimulation after one discharge, and excitability, the ability to generate a large, rapid change of membrane voltage in response to a very small stimulus. Numerous works have been aimed at studying the ODE model and their properties, we reference the book of Rocsoreanu et al. and the references therein. More recently, specialists have been interested in the passage of the behavior of a neuron to neural networks, see for example Mischler et al. , Lucon and Poquet and Baladron et al.. When neurons interact through electrical synapses, it has been proposed that the evolution of N neurons satisﬁes ( dXi t = (F(Xi t, Y i t ) −PN j=1 Jij(Xi t −Xj t ))dt + σdBi t dY i t = G(Xi t, Y i t )dt (10.1.5) where coeﬃcients Jij > 0 represent the eﬀect of the interconnection between the neurons, and the term Bi t refers, as usual, to independent Brownian motions. We consider that the interactions are symmetric and identical for every pair of neurons in the network, which in particular implies that all neurons are connected. The strength of the interaction is measured by the parameter Jij = J that we re-parametrize as a function of a new parameter λ such that J = λ N , where N is the number of neurons in the network. In section 10.5 we propose a method to estimate the parameters θ = (I, a, b, c, λ, σ2) based on the observation of the moments of the activity of a neuronal population. The estimation of the parameters of the FhN model has been approached through diﬀerent methods, see Che et al. for the least squares method, Rudi et al. for Neural Networks and Jensen and Ditlevsen for Markov chain Monte Carlo approach. Yet in most cases the estimations target a selection of the parameters of the FnH equation for a single neuron from measurements of the voltage of neurons whose activity is assumed to be independent. Here we present an alternative method that includes the interaction between neurons. 10.1.2 Notation Let us introduce some notation and deﬁnitions. We set an integer d ≥1 and a time horizon T > 0. We denote by C the space of the continuous paths from [0, T] to R2d. For the rest of this work, we use the notation Z = (Zt)t∈[0,T] = (Xt, Yt)t∈[0,T] for the canonical process in C. The existence of a solution for any of the stochastic equation that appear in this work, can be understood equivalently as the existence of a probability such that the canonical process is a solution of the equation. We work in the probability space P(C), the space of all the probabilities over C, endowed with the Wasserstein 1-metric W1(µ, ν) = inf m∈Γ(µ,ν) Z C×C \|z1 −z2\|m(dz1, dz2) = sup \|ϕ\|Lip≤1 Z ϕd(µ −ν), (10.1.6) 180 CHAPTER 10. CONCENTRATION INEQUALITY where Γ(µt, νt) denote the set of probability measures on C × C with marginals µ and ν. Also, we will use the notation CX to refer to the space of continuous functions of the continuous path from [0, T] to Rd and P(CX) for the corresponding probability space. Finally we denote CN the space of the continuous path from [0, T] to (R2d)N and P(CN) the associated probabilities space. We call P the probability such that Wt = σ−1Xt is a standard Brownian motion and consider (Ft)t≥0 the natural ﬁltration of Wt. 10.1.3 Main results Let µN t be the empirical measure as deﬁned in (10.1.3), we write the previous system as      dXi t = (F(t, Xi t, Y i t ) + R H(Xi t −x, Y i t −y)µN t (dx, dy))dt + σdBi t dY i t = G(t, Xi t, Y i t )dt L(X0, Y0) = µ⊗N 0 . (10.1.7) We are interested in the relation when N is going to inﬁnity between µN t and µt, the unique weak solution of ( ∂tµ(x, y) = −∇((F, G)µ(x, y)) −∇(µ(x, y) R H(x −¯ x, y −¯ y)µ(d¯ x, d¯ y)) + 1 2σ2∂xxµ(x, y) µt=0 = µ0 , (10.1.8) which correspond to the law of the stochastic process that solves      dXt = (F(Xt, Y t) + R H(Xt −x, Y t −y)µt(dx, dy))dt + σdBt dY t = G(Xt, Y t)dt L(X0, Y 0) = µ0, (10.1.9) for Bt an Rd-value Brownian motion and µt = L(Xt, Y t) i.e. the kinetic Mckean-Vlasov equation. We begin by proving the well-posedness of equation (10.1.9). Next, we prove a Bernstein inequality which establishes the concentration of the empirical measure close to the measure µt. The limit equation In fact, our ﬁrst result deals with a more general framework, we prove the existence and uniqueness of the solution to the equation      dXt = b1(t, Xt, Yt, µt)dt + σdBt dYt = b2(t, Xt, Yt)dt L(X0, Y0) = µ0, (10.1.10) where µt denote the law of Xt, under the following set of assumptions Assumptions 6. For some k4 > 0 and all p ≥1 the initial condition µ0 satisﬁes Z R2d\|x\|2pµ0(dz) ≤k4p! . Assumptions 7. (A) b1 is locally Lipschitz and can be written as b1(t, x, y, µ) = Z Rd ˜ b1(t, x, y, u, v)µ(du, dv) for some measurable ˜ b1(t, x, y, u, v) : [0, T] × Rd × Rd × P(Rd) →Rd which is uniformly Lipschitz on the variables (u, v). More precisely, there exists k1 > 0 such that, for all (t, x, y), \|˜ b1(t, x, y, u1, v1) −˜ b1(t, x, y, u2, v2)\|≤k1(\|u1 −u2\|+\|v1 −v2\|). 10.1. INTRODUCTION 181 (B) b2(t, x, y) is a Lipschitz continuous function, this is, there exist k2 > 0 such that \|b2(t1, x1, y1) −b2(t2, x2, y2)\|≤k2(\|t1 −t2\|+\|x1 −x2\|+\|y1 −y2\|). (C) There is a constant k3 > 0 that such that x˜ b1(t, x, y, u, v) ≤k3(1 + \|x\|2+\|y\|2+\|u\|2+\|v\|2), and yb2(t, x, y) ≤k3(1 + \|x\|2+\|y\|2). Let us note that equation (10.1.9) is contained in the framework of equation (10.1.10) with the correspondence ˜ b1(t, x, y, u, v) = F(t, x, y) + H(x −u, y −v), b2 = G. Then, Assumption 7(B) requires a locally Lipschitz condition for the transport coeﬃcient F and globally Lipschitz condition for the interaction coeﬃcient H. In addition, Assumption 7(C) asks a growth condition to be satisﬁed on all the coeﬃcients. Under such assumptions we are able to prove the following theorem, Theorem 7. Grant Assumptions 6 and 7, the Eq. (10.1.10) has a unique strong solution. Bernstein inequality In a similar spirit that in Eq. (10.1.10) we consider a more compact formulation of Eq. (10.1.2), given by      dXi t = b1(t, Xi t, Y i t , µN t )dt + σ(t, Xi t)dBi t dY i t = b2(t, Xi t, Y i t )dt L(Xi 0, Y i 0) = µ⊗N 0 , i = 1, . . . , N, (10.1.11) for initial variable Zi = (Xi, Y i) which satisﬁes E\|Zi 0\|p≤kp 0. The existence and uniqueness of a solution for this system under Assumptions 6 and 7 is a classical result, see for example and . Equivalently, there is a probability, that we denote PN, such that Z = (Z1, . . . , ZN) the canonical process in CN is a solution of this equation. Let ρ(dt) denote a probability measure in [0, T]; deﬁne, νN(dt, dz) = µN t (dz) ⊗ρ(dt), ν(dt, dz) = µt(dz) ⊗ρ(dt). We argue as in in order to proof the next result. Theorem 8. Grant Assumptions 6 and 7, there exist α1, α2, R > 0 such that PN Z [0,T]×R2d φ(t, z)(νN(dt, dz) −ν(dt, dz)) ≥γ ! ≤α1 exp  −α2 Nγ2 \|φ\|2 L2(ν)+CRγ   (10.1.12) for all γ ≥0 and every φ : [0, T]×R2d →R such that there is k > 0 which satisﬁes \|φ(t, z)\|≤C\|z\|k for some C > 0. Remark. Theorem 8 extend [73, theo. 18] to accommodate locally Lipschitz coeﬃcient and the position-velocity scheme of (10.4.41). We also remark that the conclusion is slightly stronger as we allow the function φ to be unbounded (with polynomial growth). 182 CHAPTER 10. CONCENTRATION INEQUALITY FithHugh-Nagumo model As we mention above, we apply the previous result to estimete the parameters of the FhN model described by equation (10.1.5). For simplicity, we re-parameterize the model in order to obtain linear dependence on the parameters, we make ¯ c = 1 c, ¯ a = 1 ca and ¯ b = 1 cb, for which G take the form G(x, y) = ¯ cx + ¯ a −¯ by. The goal of this section is to ﬁnd an estimator of the parameter vector θ = (I, ¯ a,¯ b, ¯ c, λ, σ2) considering as known data the evolution of the network up to a time T ﬁltered by a smooth function in the sense: we know the elements of the set K = {⟨µN t , ϕ⟩: ϕ ∈C∞(R2d), ∀t ∈[0, T]} where ⟨µ, ϕ⟩= R R2d ϕµ(dz). We introduce the notation Oi t = Z R2 xi + yiµt(dx, dy); mi,j t = Z R2 xiyjµt(dx, dy), (10.1.13) for the momentum of µt, b ON and d mi,jN by integrating with respect to the empirical measure instead of µt, we deﬁne the matrix M such that each row is given by Mi = (i Z T 0 mi−1,0 s , i Z T 0 m1,i−1 s , i Z m0,i−1, −i Z T 0 m0,i, −i Z T 0 (mi,0 + m1,0mi−1,0), i(i −1) 2 Z T 0 mi−2,i), the independent term Λ = (Λ1, . . . , Λ6), with Λi = Oi 0 + i R T 0 (mi,0 −mi+2,0 s 3 + mi−1,1 s ) and the corresponding c MN, b ΛN, replacing µt by µN t . The vector b θN = c M−1 N ( b ON −b ΛN)1∥b M−1 N ∥F>δN . is an estimator of the parameter vector θ. Indeed, the following theorem is satisﬁed. Theorem 9. Under Assumption 6, let θ be the parameter vector corresponding to the FitzHugh-Nagumo model deﬁned by equation (10.1.5) and b θN deﬁned as above, it is satisﬁed that, PN(∥b θN −θ∥∞≥γ) ≤c1e−c2Nγ1/2 + c′ 1e−c′ 2N + c ′′ 1e−c ′′ 2 Nγ. (10.1.14) Nonparametric oracle estimation of µt As a consequence of Theorem 8 and [141, Theo. 2.1] we can follow step by step the proof of the oracle estimate [73, Theo. 7] to obtain the same oracle inequality that is obtained in such result. For the sake of completeness we include this theorem, slightly adapted when needed. Let K1 : (0, T) →R and K2 : Rd →R be two bounded and compactly supported kernel functions such that Z T 0 K1(t)dt = Z R2d K2(z)dz = 1. For h > 0 we denote, K(h) 1 (t) = h−1K1(h−1t); K(h) 2 (z) = h−dK2(h−1z). We construct a family of estimators of µt0(z0) depending on h by setting ˆ µN h (t0, z0) = Z R2d K(h) 2 (z0 −z)µN t0(dx) (10.1.15) 10.1. INTRODUCTION 183 We ﬁx (t0, x0) ∈(0, T] × Rd and a discrete set HN 1 ⊂ h N−1/d(log N)2/d, 1 i , of admissible bandwidths such that Card HN 1 ≲N. The algorithm, based on Lepski’s principle, requires the family of estimators b µN h (t0, x0) , h ∈HN 1 deﬁned in 10.1.15 and selects an appropriate bandwidth b hN from data µN t0(dx). Writing {x}+ = max(x, 0), deﬁne AN h = max h′≤h,h′∈HN 1 b µN h (t0, x0) −b µN h′ (t0, x0) 2 − VN h + VN h′ + , where VN h = ϖ1\|K\|2 2(log N)N−1h−d, ϖ1 > 0 (10.1.16) Let b hN ∈argminh∈HN 1 AN h + VN h . (10.1.17) The data driven Goldenshluger-Lepski estimator of µt0 (x0) deﬁned by b µN GL (t0, x0) = b µN b hN (t0, x0) is speciﬁed by K2 and ϖ1. We deﬁne BN h (µ) (t0, x0) = sup h′≤h,h′∈HN 1 Z Rd Kh′ (x0 −x) µt0(x)dx −µt0 (x0) . (10.1.18) Theorem 10 (Oracle estimate). Grant Assumptions 6 and 7. Let (t0, x0) ∈(0, T] × R2d, the following oracle inequality holds true: EPN b µN GL (t0, x0) −µt0 (x0) 2 ≲min h∈HN 1 BN h (µ) (t0, x0)2 + VN h , for large enough N, up to a constant depending on (t0, x0) , \|K\|∞and b, provided b µN GL (t0, x0) is calibrated with ϖ1 ≥16κ−1 2 κ3, where κ2 is speciﬁed in Theorem 18 and κ3 is a (local) upper bound of µt0, see Lemma 23 below. 10.1.4 Plan of the Chapter The sections of this chapter are organized as follows: in Section 10.2 we recall some classical result that will be useful for the next section. In continuation, in Section 10.3 we will deal with proving Theorem 7 in the general context of formulation (10.1.10) under Assumptions 6 and 7. Then in section 10.4 we will prove the Bernstein’s inequality stated in theorem 8. Section 10.5 is dedicated to the proof of Theorem 9 and Section 10.6 of this of Theorem 10. Finally in section 10.7 we explore numerically the possibility of testing the interaction bases on the previous section. 184 CHAPTER 10. CONCENTRATION INEQUALITY 10.2 Brief summary of classical results Two classical results are essential for the development of this chapter, Girsanov’s theorem and Bernstein’s inequality. For a smoother reading, we brieﬂy introduce both results in this section. Let (Ω, F, (Ft)t≥0, P) be a ﬁltered probability space, W a centered d-dimensional Brownian motion with respect to the ﬁltration F and Lt an adapted mesurable process satisfying P Z T 0 \|Lt\|2< ∞ ! = 1; ∀T ≥0. (10.2.19) We introduce the process E(Lt) = exp(Lt −1 2⟨L⟩t), (10.2.20) where ⟨L⟩t denote the quadratic variation of L. This process is well deﬁned under condition (10.2.19), see for example [135, Sec. 3.5]. The Girsanov theorem can be stated as follows Theorem 11 (Girsanov (1960)). Assume that E(Lt) deﬁned above is a martingale, then the process deﬁned by ˜ W (i) t = W (i) t − Z t 0 L(i) s ds; 1 ≤i ≤d, 0 ≤t < ∞ (10.2.21) is a Brownian motion under the probability Q given by Q\|Ft= E(Lt)P\|Ft. Proof. A proof of such a result can be found in Karatzas’ book [ section 3.5] as well as in Le Gall’s book [ section 5.5]. In practice, we would like to start from a convenient process L such that the theorem provides us with a probability such that the canonical process solve our stochastic equation. That leaves us with the need to prove that E(Lt) is a martingale for our choice of L. We add here a suﬃcient condition for that. Let L be a continuous local martingale such that L0 = 0 then E(Lt) is also a local martingale, hence a supermartingale, and is a martingale if and only if E[E(Lt)] = 1; ∀t ≥0. (10.2.22) One condition to assure this is the following, Lemma 9 (Novikov Condition). Let L be a continuous martingale such that L0 = 0 and E[exp 1 2⟨L⟩t] < 0, ∀t ≥0. then E[E(Lt)] = 1, and thus E(Lt) is a martingale. Proof. The proof of this lemma can be found in Karatzas’s book [ prop 5.12]. The second result that is important to us is the Bernstein’s inequality. We conclude this section by stating the result as presented in Boucheron et al. . Theorem 12 (Bernstein’s inequality). Let X1, . . . , Xn be independent real-valued random vari-ables. Assume that there exist positive numbers v and c such that Pn i=1 E[X2 i ] ≤v and n X i=1 E[(Xi)q +] ≤q! 2 vcq−2, ∀q ≥3, then for all t ≥0, P   n X i=1 (Xi −EXi) ≥t  ≤e− t2 2(v+ct) . (10.2.23) 10.3. EXISTENCE AND UNIQUENESS PROOF 185 For one proof of this result we reference [ theorem 2.10 and corollary 2.11]. We ﬁnish this section by recalling a classical deﬁnition of a sub-Gaussian random variable, see for example . Deﬁnition 10.2.5. A real-valued random variable Z such that E[Z] = 0 is λ2 sub-Gaussian if one of the following conditions is satisﬁed, each statement implying the next: (i) Laplace transform condition E[exp(zZ)] ≤exp 1 2λ2z2 for every z ∈R. (ii) Moment condition E h Z2pi ≤p! 4λ2p for every integer p ≥1. (iii) Orlicz condition E " exp 1 8λ2 Z2 # ≤2. (iv) Laplace transform condition (bis) E[exp(zZ)] ≤exp 24 2 λ2z2 for every z ∈R. We will also use the following additive property of sub-Gaussian random variables: if the random variables Zi are independent and λ2 i sub-Gaussian, then ρ (Z1 + Z2) is \|ρ\|2 λ2 1 + λ2 2 subGaussian for every ρ ∈R. 10.3 Existence and uniqueness proof This section is dedicated to the proof of theorem 7. It is based on an argument derived from Girsanov’s theorem in a similar way to how it is done in Lacker . However, as we mentioned before, our work extends to equations with unbounded coeﬃcients, locally Lipschitz. We emphasize that, for us, the proof of the existence of a solution is understood as ﬁnding a probability such that the canonical process is the solution of equation (10.1.10). The proof is divided into two main stages. In a ﬁrst step, we start from a ﬁxed measure ¯ µ, which belongs to a suitable closed subset Ξ of P(C) with respect to the metric W1 deﬁned in (10.1.6). We consider the equation      dXt = b1(t, Xt, Yt, ¯ µt)dt + σdBi t dYt = b2(t, Xt, Yt)dt L(X0, Y0) = µ0 , (10.3.24) for this ﬁxed ¯ µ. According to the classical theory of stochastic equations, which can be reviewed for example in Protter’s textbook , a solution of such equation exists up to a stopping time T with respect to the ﬁltration (Ft)t≥0 generated by the Brownian Bt. We prove that once Assumptions 6 and 7 are satisﬁed, the probability of the event {T = ∞} is 1, and, furthermore, the probability P ¯ µ that solves the equation (10.3.24) has a representation from the probability P, for which σ−1Xt is a Brownian motion. In a second stage, the problem is posed as the existence of a ﬁxed point of the operator Φ(µ) that associates the corresponding solution of the equation (10.3.24) to each µ ∈Ξ. The existence of the ﬁxed point is proved thanks to Banach’s theorem using the representation found in the ﬁrst stage. 186 CHAPTER 10. CONCENTRATION INEQUALITY Preparation of the proof Before we begin the proof of the theorem, let us make a series of observations, some of them written as lemmas. One ﬁrst remark is that, given Xt : [0, T] →Rd a continuous function, since b2 is a globally Lipschitz function, the equation dYt = b2(t, Xt, Yt)dt (10.3.25) has a unique solution associated to the initial data Y0 = y0 ∈Rd, which does not explode in ﬁnite time. In other words, the random variable Yt can be written as a function of the path of X until the time t; we use the notation Yt(X0≤s≤t) to make it clear when necessary. Consequently, the equations (10.3.24) and (10.1.10) can be written just in terms of X. Moreover we have the following lemma. Lemma 10. Let ∥X·∥L∞[0,T]≤R, Yt deﬁned from X0≤s≤t by Eq. (10.3.25); then there is a constant CR depending on T and the Lipschitz constant k2 given by Assumption 7-(2), such that ∥Y·∥L∞[0,T]≤CR(T). Proof. By deﬁnition \|Yt\|≤\|Y0\|+ Z t 0 \|b2(s, Xs, Ys)\|ds ≤\|Y0\|+ Z t 0 (k2(s + 2∥X∥∞+\|Ys\|+\|Y0\|) + \|b2(0, X0, Y0)\|)ds we conclude by Gronwall’s lemma. In what follows, we will use the notation µZ = µ(dx, dy) to emphasize that that we are referring to the joint law of X and Y while the notations µX and µY stand for the marginal distribution with respect to X and Y respectively. In term of probability, the previous key observation shows that for the solution µZ of Eq. (10.1.9) we can ﬁnd a probability µX ∈P(CX) such that Z C g(z)µZ(dz) = Z C g(x, y)δy=y(x)(dy)µX(dx) = Z CX g(x, y(x))µX(dx), where y(x) is the solution of Eq. (10.3.25) corresponding to x = (xt)0≤t≤T ∈CX. To refer to the relation between µZ and µX we will use the notation µZ = µX ⊗δY (X). Next, we present an important consequence of Assumptions 6 and 7-(3) concerning the second moment of the solution of Eq. (10.3.24). Lemma 11. There exist a function C(t) : [0, T] →R+, depending only on k3, σ, µ0, such that if Z Rd×Rd\|z\|2¯ µt(dz) ≤C(t), for all 0 ≤t ≤T, then the canonical process Zt has bounded second moment under P ¯ µ. More precisely, let µ = LP ¯ µ(Zt) be the law of the solution of Eq. (10.3.24) then Z Rd×Rd\|z\|2µt(dz) ≤C(t). Proof. Let µ = LP ¯ µ(Zt), by Ito’s formula, µ satisﬁes the equation ∂tµ = −∇(bµ) + 1 2σ2∂xxµ, (10.3.26) 10.3. EXISTENCE AND UNIQUENESS PROOF 187 from where we obtain, after integrating by part, that d dt Z Rd×Rd\|z\|2µt(dz) = Z Rd×Rd 2zb(t, z, ¯ µt)µt(dz) + 1/2 Z Rd×Rd 2σ2µt(dz), = Z Rd×Rd 2x Z Rd×Rd b1(t, z, w)¯ µt(dw)µt(dz) + 2k3 + σ2, ≤ 2k3 Z Rd×Rd Z Rd×Rd(\|z\|2+\|w\|2) ¯ µt(dw)µt(dz) + σ2, = 2k3 Z Rd×Rd\|z\|2µt(dz) + 2k3C(t) + 2k3 + σ2, where we have made use of Assumptions 7-(1),(3) We get a problem of the form u′(t) ≤C1(u(t) + C(t) + 1), (10.3.27) with u(t) = R Rd×Rd\|z\|2µt(dz), C1 = 2k3 + σ2. We would like to ﬁnd C(t) such that the equation (10.3.27) implies that u(t) ≤C(t). This is satisﬁed if C(t) = (M + 1 2)e2Mt −1 2, with M = max{C1, u(0)}. We consider Ξ ⊂P(C) the space of all the probabily measures such that R Rd×Rd\|z\|2µt(dz) ≤ C(t) for all t ∈[0, T], Ξ = {µ ∈P(C) : Z Rd×Rd\|z\|2µt(dz) ≤C(t), ∀t ∈[0, T]}. This is a closed subspace of P(C). Indeed, let µn ∈Ξ be converging to µ for the distance W1, then for all t ∈[0, T], µn t converge weakly to µt, this is Z ϕµn t (dz) → Z ϕµt(dz) for all test function ϕ ∈Cc(R2d). Taking ϕR = χ(z/R) ≥0 with χ(z) = 1 for all \|z\|≤1, χ(z) = 0 for all \|z\|> 2 we get Z mϕRµn t (dz) → Z mϕRµt(dz); z ∈R2d which implies that R mϕRµ(dz) ≤C1, and then, applying Fatou lemma, we obtain Z mµt(dz) ≤lim inf Z mϕRµt(dz) ≤C(t). One of the consequence of the previous lemma is the following observation. Corollary 4. For all µ ∈Ξ, it is satisﬁed that zb(t, z, µ) ≤k3(1 + \|z\|2+C(t)) ≤k3(1 + C(T) + \|z\|2). (10.3.28) Lemma 12. Grant Assumptions 6 and 7, for all T > 0, there is a constant C2(k3) > 0 such that for all p ≥2, we have E\| ¯ Zt\|p≤σ2(p/2)! Cp/2 2 (10.3.29) for all t ∈[0, T], ¯ Zt = ( ¯ Xt, ¯ Yt) solution of 10.3.24. The proof is classical, yet we present here a version adapted from the one of Mao [161, Chap. 2 Theo 4.1] to our condition. 188 CHAPTER 10. CONCENTRATION INEQUALITY Proof. Applying Ito’s Formula (1 + \|Zt\|2) p 2 = (1 + \|Z0\|2) p 2 + p Z t 0 (1 + \|Zt\|2) p−2 2 zT b(s, z, ¯ µ) +σ2 p 2 Z t 0 (1 + \|Zs\|2) p−2 2 + σ2 p(p −2) 2 Z t 0 (1 + \|Zs\|2) p−4 2 \|Zs\|2 +p Z t 0 (1 + \|Zs\|2) p−2 2 ZT s σdBs ≤ 2 p−2 2 (1 + \|Z0\|p) + p Z t 0 (1 + \|Zs\|2) p−2 2 (ZT b(s, Zs, ¯ µ) + σ2 p −1 2 ) +p Z t 0 (1 + \|Zs\|2) p−2 2 ZT s σdBs From Young’s inequality we have that p−1 2 (1 + \|Zs\|2) p−2 2 ≤(1 + \|Zs\|2) p 2 + 2 (p/2) p 2 p , plugging this and the result in corollary 4 to the previous estimate, we obtain (1 + \|Zt\|2) p 2 ≤ 2 p−2 2 (1 + \|Z0\|p) + p Z t 0 (1 + \|Zs\|2) p−2 2 k3(1 + ∥C∥+\|Zs\|2)) + Z t 0 (p(1 + \|Zs\|2) p 2 + 2σ2(p/2) p 2 ) + p Z t 0 (1 + \|Zs\|2) p−2 2 ZT s σdBs From where, it is some constant α such that, (1 + \|Zt\|2) p 2 ≤ 2 p−2 2 (1 + \|Z0\|p) + pα Z t 0 (1 + \|Zs\|2) p 2 (10.3.30) +2σ2(p/2) p 2 T + p Z t 0 (1 + \|Zs\|2) p−2 2 ZT s σdBs (10.3.31) Let us deﬁne for n ≥0 the stopping time τn = T ∧ inf t∈[0,T]{\|Z(t)\|≥n}. clearly, from the non-explosion of the solution, τn →T a.s., moreover from (10.3.30) and the properties of Ito’s integral, we get E(1 + \|Zt∧τn\|2) p 2 ≤ 2 p−2 2 (1 + E\|Z0\|p) + pαE Z t∧τn 0 (1 + \|Zs\|2) p 2 ds + 2σ2(p/2) p 2 t ≤ 2 p−2 2 (1 + E\|Z0\|p) + pαE Z t 0 (1 + \|Zs∧τn\|2) p 2 ds + 2σ2(p/2) p 2 t. Applying Gronwall’s lemma we obtain E(1 + \|Zt∧τn\|2) p 2 ≤2 p−2 2 epαt(1 + E\|Z0\|p) + (p/2) p 2 σ2epαt ≤2 p−2 2 (1 + E\|Z0\|p)epαt + (p/2)! σ2ep(αt+1/2). Letting n →∞yields E(1 + \|Zt\|2) p 2 ≤σ2(p/2)! Cp/2 2 . for some constant C2 which come from E\|Z0\|p≤C(p 2)!. In consequence, we have the next corollary. Corollary 5. Let the initial condition Z0 be such that E\|Z0\|p≤kp 0 for some constant k0. Zt satisﬁes sup [0,T] E[e 1 2C2 \|Zt\|2 ] ≤1 + σ2. 10.3. EXISTENCE AND UNIQUENESS PROOF 189 Proof. From lemma 12, for p ≥1, we have sup [0,T] E[e 1 2C2 \|Zt\|2 ] = 1 + X p≥1 2−p p! Cp 2 E\|Zt\|2p\|≤1 + σ2 X p≥1 2−p = 1 + σ2. (10.3.32) From the last corollary and proposition 6.3 of , it follows that there is a constant k5 > 0 that only depends on C2 such that for any measures µ and ν in Ξ, it is satisﬁed that W1(µt, νt) ≤k5 q Ht(µ\|ν); ∀t ∈[0, T]. where µt = µ·∧t, νt = ν·∧t ∈P(C) and Ht(µ\|ν) = H(µt\|νt) denote the relative entropy H(µ\|ν) = Z C dµ dν log dµ dν dν. Proof of theorem 7 Proof. As we mention above, we divide the proof in two steps: ﬁrst, we ﬁx ¯ µ ∈Ξ and prove the existence and uniqueness of a solution of (10.3.24) for this ﬁxed ¯ µ. Then we prove through a ﬁxed point argument that there exists µ such that it is the law of the component Xt of the corresponding solution of (10.3.24) for this µ. Step 1. Let ¯ µ ∈Ξ be ﬁxed, we use a Girsanov’s argument to prove that we have a solution P ¯ µ ⊗δY (X0≤s≤T ) with P ¯ µ that satisﬁes P ¯ µ\|Ft= MtP\|Ft for Mt = Et Z · 0 σ−1b1(s, Xs, Ys, ¯ µ) · dWs , (10.3.33) where we deﬁne Et(M) = exp (Mt −1 2⟨M⟩t). We recall that P is the measure such that Wt = σ−1Xt is a Brownian motion, with Xt the canonical process in CX, and (Ft)t≥0 the natural ﬁltration of Wt. Grant Assumptions 6 and 7, there is a probability such that Z is a solution of (10.1.10) until a time of explosion T. Since Yt is unequivocally determined by X0≤s≤t, we can write this probability as P ¯ µ ⊗δY (X0≤s≤t), with P ¯ µ ∈P(CX). We deﬁne, for R > 0, τR = inf{t ≥0, \|Xt\|≥R}. By lemma 10, there is CR such that \|Yt\|≤CR in [0, τR], which implies that τ = supR>0 τR ≤T. We note that for any R, τR is a stopping time with respect to the ﬁltration (Ft)0≤t≤T . Let us deﬁne fR(x) = ( x if \|x\|≤R Rx/\|x\| if \|x\|> R and b1R(t, Xt, Yt, ¯ µ) = b1(t, fR(X), Yt(fR(X)), ¯ µ). The process MR t = Et Z · 0 σ−1b1R(s, Xs, Ys(X·∧t), ¯ µ) · dWs . (10.3.34) is a martingale under P. Indeed, by the deﬁnition of b1R, MR t satisﬁes Novikov criterion, lemma 9, i.e. E " exp 1 2 Z T 0 \|σ−1b1R(t, Xt, Yt(X·∧t), ¯ µ)\|2dt # < ∞ (10.3.35) 190 CHAPTER 10. CONCENTRATION INEQUALITY for all 0 ≤T < ∞. It follows then that MR t∧τR is also a martingale and, as they coincide in [0, τR], Mt∧τR is also a martingale. Girsanov theorem shows then that there is the probability measure PR PR\|Ft∧τR= Mt∧τRP on (C, FT ) such that W µ t∧τR = Wt∧τR − R t∧τR 0 σ−1b1(s, Xs, Ys, ¯ µ)ds is a standard Brownian motion. By uniqueness of the solution in [0, T) we get P ¯ µ\|Ft∧τR= PR\|Ft∧τR= Mt∧τRP. For all h ∈C2(R2) we deﬁne, Lth(x, y) := 1 2σ∂xxh(x, y) + b(t, x, y, µ) · ∇h(x, y). The key part of the argument is to ﬁnd f such that • Ltf ≤Cf, • f(ZτR) ≥g(R) for some g such that g(R) →∞when R →∞. Then we can proceed in a similar way as it is done in to conclude that limR→∞P ¯ µ(τR > t) = 1 for all t ≥0. Let f(z) = \|z\|2+1 2 ; by corollary 4, we have Ltf ≤Cf for C = 2(∥σ2∥∞+k3). Applying Ito’s formula to e−Ctf(x, y), we get for 0 ≤s ≤t e−Ctf(Xt, Yt) = e−Csf(Xs, Ys) + Z t s e−CuLufdu −C Z t s e−Cufdu + Z t s ∇fσdW µ ≤ e−Csf(Xs, Ys) + Z t s ∇fσdW µ Evaluating in s ∧τR, t ∧τR and taking conditional expectation with respect to the σ-ﬁeld Fs, we obtain EP ¯ ⪵[e−Ct∧τRf(Xt, Yt)\|Fs] ≤e−Cs∧τRf(Xs, Ys) since the ﬁrst term is Fs-measurable and the second is a centred Gaussian with respect to P ¯ µ. From the fact that e−Ct∧τRf(x, y) is a supermartingale, we infer EP ¯ ⪵[e−Ct∧τRf(Xt, Yt)] ≤EP ¯ ⪵[f(X0, Y0)]. Since f(ZτR) ≥g(R) = R2/2 in [τR < +∞], we get P ¯ µ(τR ≤t) = eCt/g(R)EP ¯ µ[1τR≤te−Ctg(R)] ≤eCt/g(R)EP ¯ µ[1τR≤te−Ct∧τRf(Zt∧τR)] ≤ eCt/g(R)EP ¯ µ[e−Ct∧τRf(Zt∧τR)] ≤ eCt/g(R)f(Z0) which goes to 0 when R goes to inﬁnity. From lim P ¯ µ(τR > t) = 1 we conclude, Z MtdP ≥ Z 1τR>tMt∧τRdP = P ¯ µ(τR > t) →1 which implies that Mt is a martingale. We can then apply Girsanov’s theorem directly to Mt, to conclude that P ¯ µ\|Ft= MtP and W ¯ µ t = Wt − R t 0 σ−1b1(s, Xs, Ys, ¯ µ)ds is a Brownian Motion. Step 2. We deﬁne Φ(µ) = P µ ⊗δY (X). Let A be a subset of the σ-ﬁeld generated by Xt, CY = C([0, T]; Rd), we have Z A dΦ(ν)t dΦ(µ)t (x, y(x))dP µ(x) = Z A⊗CY dΦ(ν)t dΦ(µ)t (x, y)dΦ(ν)t = Φ(µ)t(A ⊗CY ) = P µ(A). 10.4. BERNSTEIN’S INEQUALITY 191 It follows, dP ν t dP µ t (x) = dΦ(ν)t dΦ(µ)t (x, y(x)). In consequence, we have that Ht(Φ(µ)\|Φ(ν)) = − Z C log dΦ(ν)t dΦ(µ)t dΦ(µ)t = − Z CX log dΦ(ν)t dΦ(µ)t (x·∧t, y(x·∧t))P µ(dx), = −EP µ " log dP ν t dP µ t (X·∧t) # , = −EP µ  log EP µ " dP ν t dP µ t \|X·∧t # . Since Xt and Wt generate the same ﬁltration we get Ht(Φ(µ)\|Φ(ν)) = −EP µ  log EP µ " dP ν t dP µ t \|F·∧t #  = 1 2EPµ "Z t 0 \|σ−1b1(s, Z, ν) −σ−1b1(s, Z, µ)\|2ds # where we use the result from the previous step for the second equality. On the other hand, we have that if b1 satisﬁes Assumption 7-(1), then \|b1(s, z, ν) −b1(s, z, µ)\|= \| Z Rd ˜ b1(t, z, u)d(µ −ν)\|≤k1 sup \|ϕ\|Lip≤1 Z ϕd(µ −ν) = k1W1(µ, ν). Gathering everything, we get, W1(Φ(µ), Φ(ν)) ≤k1k5 q Ht(Φ(µ)\|Φ(ν)) ≤k1k5 q 1/2\|σ\|2tW1(µ, ν). We use Banach’s ﬁxed point theorem to conclude. 10.4 Bernstein’s Inequality This section is dedicated to the proof of theorem 2. This is based on a change of measures argument which allows us to use a Bernstein inequality. Here is a short sketch of the proof to make it easier to read. 1. First we rewrite equation (10.1.12) in terms of Bernstein inequality. Let us deﬁne AN = (Z [0,T]×R2d φ(t, z)(νN(dt, dz) −ν(dt, dz)) ≥γ ) (10.4.36) =    N X i=1 Z [0,T] φ(t, Zi)ρ(dt) − Z [0,T]×Rd φ(t, z)ρ(dt)µ(dz) ! ≥Nγ    (10.4.37) =    N X i=1 βi ≥Nγ   , (10.4.38) for βi = R [0,T] φ(t, Zi)ρ(dt) − R [0,T]×R2d φ(t, z)ρ(dt)µ(dz). 192 CHAPTER 10. CONCENTRATION INEQUALITY We use one generalization of Bernstein’s Inequality which can be found in Boucheron et al. . This establishes that if the βi are independent (in our case this is equivalent to says that the Zi are independent), centred (E(βi) = 0), with uniformly bounded second moment, and satisfy that for all integers p ≥0 E\|βi\|p≤p! Rp−2V (βi) (10.4.39) for some constant R, V denoting the variance, then P( N X i=1 βi ≥Nγ) ≤2 exp − (Nγ)2 2(PN i=1 E[β2 i ] + Rγ) ! what is the result. The diﬃculty here is that the variable Zi, solutions of (10.1.11), are not independent. In order to solve this, the strategy is based on two more steps: 2. We consider a new probability measure ¯ PN for which the variables Zi are independent, E¯ PN[βi] = 0 and the random variables βi satisfy the moment estimates, so that the inequality is valid for this new probability. 3. We prove that PN(A) ≤C¯ PN(A)δ, (10.4.40) for some C, δ > 0, in order to obtain the desired inequality for the original probability measure. Deﬁnition of the probability measure ¯ P N Let µ be the law of the unique solution of Eq. (10.1.10). Deﬁne ¯ PN = µ⊗N t as the unique probability measure for which the canonical process solve      dXi t = b1(t, Xi t, Y i t , µt)dt + σ(t, Xi t)d ¯ Bi t dY i t = b2(t, Xi t, Y i t )dt L(Xi 0, Y i 0) = µ⊗N 0 , (10.4.41) where ¯ Bi t = Z t 0 σ−1(s, Zi s)(dZi s −b1(s, Zi s, µs)ds), 1 ≤i ≤N, (10.4.42) are independent Brownian motions. That is, the variables Zi = (Xi, Y i) correspond to N independent samples of the solution of equation 10.1.10, whose well-posedness is given by theorem 7. Since L(Zi) = µt, is clear from its deﬁnition that E¯ PN [βi] = 0. Next lemma conclude with the second step establish above. Lemma 13. Under the probability ¯ PN, the second moment of βi is bounded by a constant V and all the remaining moments satisfy the estimate (10.4.39) for a proper constant R. Proof. Let ξ = R [0,T]×R2d φ(t, z)ρ(dt)µ(dz), p a positive integer, then E¯ PN \|βi\|p= Z R2d Z [0,T] φ(t, z)ρ(dt) −ξ p µ(dz). Since \|φ(t, z)\|≤C\|z\|k and \|ξ\|p≤ R Cp\|z\|p+kµ(dz), using the inequality \|a + b\|p≤2p−1(\|a\|p+\|b\|p), we get E¯ PN \|βi\|p≤2pCp Z R2d\|z\|p+kµ(dz). 10.4. BERNSTEIN’S INEQUALITY 193 From where, considering the estimates of the moment of Zt given by lemma 12, we obtain, E¯ PN \|βi\|2 ≤ 4 Z R2d\|z\|2+kµ(dz) ≤ C222+k(1 + k2+k 0 )eα(2+k)T = V E¯ PN \|βi\|p ≤ 2p+k(1 + E\|Z0\|p+k)eα(p+k)T ≤ V (2eαT k0)p−2 = V (CR)p−2, for R = 2eαT k0. Relation between the probability measures: proof of inequality (10.4.40) Let us deﬁne ¯ MN t = N X i=1 Z t 0 ((σ−1b1)(s, Xs, Ys, µN s ) −(σ−1b1)(s, Xs, Ys, µs))d ¯ Bi s, which is a local martingale with respect to ¯ PN. Moreover, as one consequence of the next proposition, Norikov’s criterion, Et( ¯ MN) = exp ( ¯ MN t −1 2⟨¯ MN⟩t) is a martingale under ¯ PN. Then, applying Girsanov’s theorem, since the solution of (10.1.11) is unique, we get dPN d¯ PN = Et( ¯ MN). Proposition 3. For every τ > 0 there exist δ0 > 0 such that sup N≥1 sup [0,T] E[exp(τ(⟨¯ MN⟩t+δ −⟨¯ MN⟩t))] ≤¯ C, for every 0 ≤δ ≤δ0 and some constant ¯ C. Proof. Let EN s (x, y) = b1(s, x, y, µN s ) −b1(s, x, y, µs), thanks to Assumption 7-(2), \|EN s (Xs, Ys)\|2≤ Z \| ¯ b1(s, Xs, Ys, u)(µN s −µs)(du)\|2. From where τ(⟨¯ MN⟩t+δ −⟨¯ MN⟩t) = N X i=1 Z t+δ t ((σ−1b1)(s, Xi s, Y i s , µN s ) −(σ−1b1)(s, Xi s, Y i s , µs))ds ≤ τ\|σ−1\|∞ N X i=1 Z t+δ t \|EN s (Xi s, Y i s )\|2 ≤ κ Z t+δ t N X i=1 \| Z ¯ b1(s, Xi s, Y i s , u)(µN s −µs)(du)\|2 with κ = τ\|σ−1\|∞. By Jensen’s inequality together with the exchangeability of Zt we have E[exp(τ(⟨¯ MN⟩t+δ −⟨¯ MN⟩t))] ≤ 1 δ Z t+δ t E[exp(κδ N X i=1 Z \| ¯ b1(s, Xi s, Y i s , u)(µN s −µs)(du)\|2)] ≤ sup [0,T] EN ¯ P [exp(κδN Z \| ¯ b1(s, X1 s , Y 1 s , u)(µN s −µs)(du)\|2)] ≤ sup [0,T] EN ¯ P [exp(κδ 1 N N X j=1 \|A1,j\|2)] ≤ sup [0,T] 1 2  EN ¯ P [exp(κδ 1 N \|A1,1\|2)] + EN ¯ P [exp(κδ 1 N N X j=2 \|A1,j\|2)]  , 194 CHAPTER 10. CONCENTRATION INEQUALITY where, Ai,j s = ¯ b1(s, Xi s, Y i s , Xj s) − Z Rd ¯ b1(s, Xi s, Y i s , u)µs(du). As a consequence of lemma 12 and the Assumption 7(1) on b1 we have, E\| ¯ Ai,j t \|2p≤σ2p! Cp 2, (10.4.43) which is the moment condition on the deﬁnition of a sub-Gaussian random variable. In consequences, as the variables A1,j s for j = 2, . . . , N are independent under ¯ ¶N, PN j=2 A1,j z is a (N −1)c′ sub-Gaussian for some constant C′ depending on C2 and by the moment condition again E[\| N X j=2 A1,j s (X, Y )\|2p] ≤p! 4p(N −1)pC′. Hence, EN ¯ P [exp(κδ 1 N N X j=2 \|A1,j\|2)] = 1 X p≥1 (κδ)p p! 1 Np p! 4p(n −1)pC′p < ∞, and since the ﬁrst term, EN ¯ P [exp(κδ 1 N \|A1,1\|2)], is clearly bounded as a consequence of (10.4.43) the result follows. Let now A ∈FT , since ¯ PN and PN coincide in F0, we have PN(A) = EPN [PN(A\|F0)] = E¯ PN [PN(A\|F0)]. The next result is the key to arrive at the relation 10.4.40, Lemma 14. For any division 0 = t0 < · · · < tK ≤T and A ∈FT , it is satisﬁed that E¯ PN [P(A\|F0)] = E¯ PN [P(A\|FtK)]1/4K K Y j=1 E¯ PN [exp(2(⟨¯ MN · ⟩tj −⟨¯ MN · ⟩tj−1))]j/4. Indeed, from this result, and proposition 3 with τ = 2, tj = jT/K and K large enough so tj −tj−1 ≤δ0, we can conclude that PN(A) ≤ E¯ PN [P(A\|FtK)]1/4K K Y j=1 E¯ PN [exp(2(⟨¯ MN · ⟩tj −⟨¯ MN · ⟩tj−1))]j/4 ≤ ¯ PN(A)1/4K sup N≥1 sup t∈[0,T−δ0] E¯ PN [exp(2(⟨¯ MN · ⟩t+δ0 −⟨¯ MN · ⟩t))] K(K+1)/8 ≤ ¯ CK(K+1)/8¯ PN(A)1/4K. The proof of lemma 14 can be found in Lacker and Della Maestra . For completeness, we include a version of this in the appendix. Conclusion, proof of theorem 8 We now have all the ingredients to complete the proof of Theorem 8. Let AN be deﬁned as in 10.4.36, we have from step 3 that, PN(AN) ≤¯ CK(K+1)/8¯ PN(AN)1/4K. Moreover, from step 2, the random variables βi satisfy the conditions to apply Bernstein Inequality under ¯ PN, from where, PN(AN) ≤¯ CK(K+1)/8  2 exp − (Nγ)2 2(PN i=1 E[β2 i ] + Rγ) !  1/4K , which conclude the result for α1 = ¯ CK(K+1)/821/4K and α2 = 1 2 1 4K . 10.5. FITZHUGH-NAGUMO PARAMETERS 195 10.5 Application to the parameter estimation of FitzHugh-Nagumo model Here, we present an application of the concentration inequality (10.1.12) to the estimation of the parameters of the FitzHugh-Nagumo model presented in the introduction. Given µt the solution of equation (10.1.5), let us compute the moments of order i. We deﬁne Oi t and mi,j t as in (10.1.13). Since µt is a solution of the equation, it is satisﬁed that Oi T = Oi 0 + Z T 0 Z R2{i(x −x3/3 −y + I)xi−1 + i(¯ cx + ¯ a −¯ by)yi−1 + 1 2σ2i(i −1)xi−2}µt(dx, dy) = Oi 0 + i Z T 0 (mi,0 −mi+2,0 3 −mi−1,1 + Imi−1.0) + ¯ c Z T 0 m1,i−1 + ¯ a Z t 0 m0,i−1 −¯ b Z T 0 m0,i −λ Z T 0 (mi,0 + m1,0mi−1.0) + σ2 Z T 0 i(i −1) 2 mi−2,0. Let us consider the six ﬁrst moment Oi T with i = 1, 6, we obtain a linear system of six equation, which we write with the matrix formulation O = Mθ + Λ, with O = (O1 T , . . . , O6 T )T , the rows of M given by Mi,... = (i Z T 0 mi−1,0 s , i Z T 0 m1,i−1 s , i Z m0,i−1, −i Z T 0 m0,i, −i Z T 0 (mi,0 + m1,0mi−1,0), i(i −1) 2 Z T 0 mi−2,i) and the independent term Λ = (Λ1, . . . , Λ6), with Λi = Oi 0 + i R T 0 (mi,0 −mi+2,0 s 3 + mi−1,1 s ). Note that all moments are well deﬁned, according to lemma 12. In consequence, if O, M and Λ are known, we can estimate the parameter vector as θ = M−1(O −Λ). Let µN t the empirical measure, b ON, c MN and b ΛN the associated approximations from replacing µ for the empirical measure, then, we obtain an estimator for the parameter vector given by b θN = c M−1 N ( b ON −b ΛN)1∥b M−1 N ∥>δN . Proof of theorem 3 Proof. Since ∥b θN −θ∥∞ = ∥M−1(O −Λ) −c M−1 N ( b ON −b ΛN)∥∞ ≤ ∥(M−1 −c M−1 N )( b ON −b ΛN)∥∞+∥M−1( b ON −O)∥∞+∥M−1(b ΛN −Λ)∥∞, it is satisﬁed that, P(∥b θN −θ∥∞≥γ) ≤ P(∥(M−1 −c M−1 N )( b ON −b ΛN)∥∞≥γ 3) +P(∥M−1( b ON −O)∥∞≥γ 3) + P(∥M−1(b ΛN −Λ)∥∞≥γ 3) = I + II + III. We now proceed to bound each of these terms. For the last two, let us observe that ∥b ON −O∥∞= max i=1,6 ⟨µN T −µT , xi + yi⟩; (10.5.44) ∥b ΛN −Λ∥∞= max i=1,6 i Z T 0 ⟨µN t −µt, xi −xi+2 3 + xi−1y⟩dt. (10.5.45) 196 CHAPTER 10. CONCENTRATION INEQUALITY Then we get the estimate by applying the Bernstein Inequality shown in theorem 8. Indeed, taking φ(x, y) = xi + yi, ρ = δT in theorem 8 we obtain P ⟨µN T −µT , xi + yi⟩≥ γ 3∥M−1∥∞ ! ≤α1 exp ( −α2 N(γ/(3∥M−1∥∞))2 C1 + RCγ/(3∥M−1∥∞) ) , where C1 is an upper bound for all moments of µ less than 12. Note that \|φ\|≤C\|z\|i for some constant C (uniformly for all i). Analogously, for φ(x, y) = xi −xi+2 3 + xi−1y, ρ the Lebesgue’s measure in [0, T], C′ such that \|φ\|≤C′\|z\|2 we get P i Z T 0 ⟨µN t −µt, xi −xi+2 3 + xi−1y⟩dt ≥ γ 3∥M−1∥∞ ! ≤α1 exp ( −α2 N(γ/(18∥M−1∥∞))2 C′ 1 + RC′γ/(18∥M−1∥∞) ) , for C′ 1 upper bound of 16th-moment of µ. I ≤P(∥b ON −O∥∞≥ γ 3∥M−1∥∞ ) ≤6α1 exp ( −α2 N(γ/(3∥M−1∥∞))2 C1 + RCγ/(3∥M−1∥∞) ) , (10.5.46) II ≤P(∥b ΛN −Λ∥∞≥ γ 3∥M−1∥∞ ) ≤6α1 exp ( −α2 N(γ/(18∥M−1∥∞))2 C′ 1 + RC′γ/(18∥M−1∥∞) ) , (10.5.47) It only remains then to bound I. Let us denote A = {∥(M−1 −c M−1 N )( b ON −b ΛN)∥∞≥γ 3}, for all L > 0, we can write A as A = A ∩{∥b ON −b ΛN∥≥L} [ {∥b ON −b ΛN∥< L} ∩{∥(M−1 −c M−1 N )∥≥γ 3L} . This implies that P(A) ≤ P({∥b ON −b ΛN∥≥L}) + P({∥(M−1 −c M−1 N )∥≥γ 3L}) ≤ P({∥b ON∥≥L}) + P({∥b ΛN∥≥L}) + P({∥(M−1 −c M−1 N )∥≥γ 3L}) Let’s set L = γ1/2 + ∥O∥+∥Λ∥then P(A) ≤P({∥b ON −O∥≥γ1/2}) + P({∥b ΛN −Λ∥≥γ1/2}) + P({∥(M−1 −c M−1 N )∥≥γ1/2 3 }) The ﬁrst two terms can be bounded using Bernstein’s inequality as above. To proceed with the third one, we must ﬁrst perform the following algebraic trick. Let us denote ξ = M −c MN, then c M−1 N −M−1 = (M −ξ)−1 −M−1 = (1 −M−1ξ)−1M−1 −M−1. There are two possibilities, or ∥M−1ξ∥F < 1 2 or ∥M−1ξ∥F ≥1 2. In the ﬁrst case, we have that (1 −M−1ξ) = P j≥0(M−1ξ)j, from where c M−1 N −M−1 = ( X j≥0 (M−1ξ)j)M−1 −M−1 = ( X j≥1 (M−1ξ)j)M−1, ∥c M−1 N −M−1∥F ≤ ∥ξ∥∥M−1∥2X j≥0 ∥ξM−1∥j= ∥ξ∥ ∥M−1∥2 1 −∥ξM−1∥≤2∥ξ∥∥M−1∥2. In conclusion, P(∥(M−1 −c M−1 N )∥≥γ1/2 3 ) ≤P(∥ξ∥F ≥γ1/2∥M∥2 F 6 ) + P(∥ξ∥F ≥1/2∥M∥F ) 10.6. NONPARAMETRIC ESTIMATION OF µT 197 since ∥ξ∥F ≤6 max1≤ij≤6\|mij\|, applying again Bernstein inequality for each coeﬃcient of the matrix, we obtain, P(∥(M−1 −c M−1 N )∥≥γ1/2 3 ) ≤ α1 exp ( −α2 Nγ(∥M∥F /6)2 C′′ 1 + RC′′γ1/2(∥M∥F /6) ) +α1 exp ( −α2 N(1/12∥M∥F )2 C′′ 2 + RC′′1/12∥M∥F ) Gathering all together, we conclude PN(∥b θN −θ∥∞≥γ) ≤α1′ exp ( −α′ 2 Nγ 1 + R ˆ Cγ1/2 ) + α1 exp ( −α2 N(1/12∥M∥F )2 C′′ 2 + RC′′1/12∥M∥F ) 10.6 Nonparametric estimation of µt Let’s start by stating the following lemma, which is a consequence of [141, Theo. 2.1] and take the role of [73, lemm. 23]. Lemma 15. Grant Assumptions 6, 7 and (t0, z0) ∈(0, T] × R2d. Let r > 0 and [r1, r2] ⊂(0, T), there exist κ3, κ4 depending on z0, r1, r2, r and b such that 0 < κ5 ≤ inf t∈[r1,r2],\|z−z0\|≤r µt(z) ≤ sup t∈[r1,r2],\|z−z0\|≤r µt(z) ≤κ6. In turn, for a compactly supported kernel K, this implies the existence of r = r(K) such that the estimate Kh (z0 −·) 2 L2µt0 = Z R2d h−4dK h−1z 2 µt0 (z0 −z) dz ≤κ3(r)h−2d\|K\|2 2 (10.6.48) holds true. Lemma 16. In the setting of Theorem 10. if K is a bounded and compactly supported kernel and h ∈HN 1 , we have EPN b µN h (t0, z0) −µt0 (z0) 2 ≲BN h (µ) (t0, z0)2 + VN h , up to a constant that depends (continuously) on (t0, z0) , \|K\|∞and b, and where BN h (µ) (t0, z0) is deﬁned in (10.1.18) and VN h in (10.1.16). Proof. Write b µN h (t0, z0) −µt0 (z0) = I + II, with I = Z Rd Kh (z0 −z) µt0(z)dz −µt0 (z0) and II = Z Rd Kh (z0 −z) µN t0(dz) −µt0(z)dz . We have I2 ≤BN h (µ) (t0, z0)2 for the squared bias term. For the variance term, using successively Theorem 8 and the estimate (10.6.48), we have 198 CHAPTER 10. CONCENTRATION INEQUALITY EPN h II2i = Z ∞ 0 PN \|II\|≥u1/2 du ≤2κ1 Z ∞ 0 exp   − κ2Nu Kh (z0 −·) 2 L2µt0 + Kh (z0 −·) ∞u1/2   du ≤2κ1 Z ∞ 0 exp − κ2Nh2du κ3\|K\|2 2+\|K\|∞u1/2 ! du ≲ Nh2d−1 1 + Nh2d−1 ≲VN h where we used (48) and the fact that maxh∈HN 1 Nh2d−1 ≲1. Completion of proof of Theorem 10 Recall that b hN denotes the data-driven bandwidth deﬁned in 10.1.17. Step 1: For h ∈HN 1 , we successively have EPN b µN GL (t0, z0) −µt0 (z0) 2 ≲EPN b µN GL (t0, z0) −b µN h (t0, z0) 2 + EPN b µN h (t0, z0) −µt0 (z0) 2 ≲EPN " b µN b hN (t0, z0) −b µN h (t0, z0) 2 −VN h −VN b hN + + VN h + VN b hN # + EPN b µN h (t0, z0) −µt0 (z0) 2 ≲EPN  AN max b hN,h + VN h + VN b hN  + EPN b µN h (t0, z0) −µt0 (z0) 2 ≲EPN h AN h i + VN h + EPN h AN b hN + VN b hN i + EPN b µN h (t0, z0) −µt0 (z0) 2 ≲EPN h AN h i + VN h + BN h (µ) (t0, z0)2 , where we applied Lemma 16 in order to obtain the last line. Step 2: We ﬁrst estimate AN h . Write µh (t0, z0) for R R2d Kh (z0 −z) µt0(z)dz. For h, h′ ∈HN 1 with h′ ≤h, since b µN h (t0, z0) −b µN h′ (t0, z0) 2 ≤4 b µN h (t0, z0) −µh (t0, z0) 2 + 4 µh (t0, z0) −µt0 (z0) 2 + 4 µh′ (t0, z0) −µt0 (z0) 2 + 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 we have b µN h (t0, z0) −b µN h′ (t0, z0) 2 −VN h −VN h′ ≤8BN h (µ) (t0, z0)2 + 4 b µN h (t0, z0) −µh (t0, z0) 2 −VN h + 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 −VN h′ using h′ ≤h in order to bound b µN h′(t, a) −µh′ (t0, z0) 2 by the bias at scale h. Taking maximum 10.6. NONPARAMETRIC ESTIMATION OF µT 199 over h′ ≤h, we obtain max h′≤h b µN h (t0, z0) −b µN h′ (t0, z0) 2 −VN h −VN h′ + ≤8BN h (µ) (t0, z0)2 + 4 b µN h (t0, z0) −µh (t0, z0) 2 −VN h + + max h′≤h 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 −VN h′ + (10.6.49) Step 3: We estimate the expectation of the ﬁrst stochastic term in the right-hand side of (10.6.49). We reﬁne the computation of the term II in the proof of Lemma 16. By Theorem 8 and using estimates Z ∞ ν exp (−ur) du ≤2r−1ν1−r exp (−νr) , ν, r > 0, ν ≥(2/r)1/r, and Z ∞ 0 exp − aup b + cup/2 ! du ≤Cp max a b −1/p , a c −2/p! , a, b, c, p > 0, with Cp = 2 R ∞ 0 exp −1 2(min(√u, u))p du, stemming from the rough bound exp − aup b + cup/2 ! ≤exp −aup 2b ! + exp −aup/2 2c ! , u > 0. we have EPN " 4 b µN h (t0, z0) −µh (t0, z0) 2 −VN h + # = Z ∞ 0 PN 4 b µN h (t0, z0) −µh (t0, z0) 2 −VN h ≥u du = Z ∞ 0 PN b µN h (t0, z0) −µh (t0, z0) ≥1 2 VN h + u 1/2 du ≤2κ1 Z ∞ VN h exp − κ2Nh2d 1 4u κ3\|K\|2 2+\|K\|∞1 2u1/2 ! du ≲ Z ∞ VN h exp −κ2Nh2du 8κ3\|K\|2 2 ! du + Z ∞ VN h exp −κ2Nh2du1/2 4\|K\|∞ ! du ≲ Nh2d−1 exp −κ2Nh2d WN h 8κ3\|K\|2 2 ! + Nh2d−2 Nh2d VN h 1/2 exp    − κ2Nh2d VN h 1/2 4\|K\|∞     ≲ Nh2d−1 N−ϖ1κ2/(8κ3) + Nh2d−3/2 (log N)1/2 exp  −κ2\|K\|2ϖ1/2 1 4\|K\|∞ (log N)5/2  , ≲N−2 as soon as ϖ1 ≥16κ−1 2 κ3, thanks to maxh∈HN 1 Nh2d−1 ≲1, and using minh∈HN 1 h ≥ N−1(log N)21/d to show that the second term is negligible in front of N−2. Step 4: For the second stochastic term, we have the rough estimate EPN " max h′≤h 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 −VN h′ + # 200 CHAPTER 10. CONCENTRATION INEQUALITY ≤ X h′≤h EPN " 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 −VN h′ + # ≲Card HN 1 N−2 ≲N−1 where we used Step 3 to bound each term EPN " 4 b µN h′ (t0, z0) −µh′ (t0, z0) 2 −VN h′ + # inde-pendently of h together with Card HN 1 ≲N. In conclusion, we have through Steps 2-4 that EPN h AN h i ≲N−1 + BN h (µ) (t0, z0)2. Therefore, from Step 1 , we conclude EPN b µN GL (t0, z0) −µt0 (z0) 2 ≲BN h (µ) (t0, z0)2 + VN h + N−1 for any h ∈HN 1 . Since N−1 ≲VN h always, the proof of Theorem 10 is complete. 10.7 Testing of the interaction between agents the last part of this chapter if exploratory only, and is an joint ongoing work with S. Delattre and M. Hoﬀmann. We propose the construction of a statistical test to decide whether we have existence of connectivity between agents in a dynamic system such as the one presented in model (10.1.2), assuming that the coeﬃcients F, G and H are unknown. To understand the diﬃculties that this objective presents and to introduce our strategy, let us begin with an important observation related to propagation of chaos for mean ﬁeld problems. One might think that a possible strategy for testing connectivity is to test the independence between agents, since it is intuitive to assume that if such interaction exists, the state of an agent will depend on the state of those connected to it and vice versa. However, the phenomenon of propagation of chaos indicates that when the number of agents increases, it is impossible to distinguish between a system of N interrelated particles and that of N independent particles, each of them following the law of limit distribution. More precisely, let us call P N, the joint distribution interrelated according to the system of equations (10.1.2) and P ⊗N the distribution of N independent particles each of them following the law (10.1.9), in the article of Della Maestra and Hoﬀmann, it has been proved that, for not large coeﬃcients or short time windows, lim sup N→∞ ∥P N −P ⊗N∥TV < 1 (10.7.50) where ∥·∥TV denote the total variation distance. In particular, this implies that there exist no test of the null hypothesis H0 : P N = P ⊗N against H1 : P N ̸= P ⊗N such that the ﬁrst error and the second error converge both asymptotically to zero. More precisely, for all test T and rejection zone Ω, the ﬁrst error α = P(T ∈Ω\|H0) = P ⊗N(Ω) and second error β = P(T / ∈Ω\|H1) = P N(Ωc) satisfy P ⊗N(Ω) + P N(Ωc) ≥1 −∥P N −P ⊗N∥TV > 0, for all N large enough. Consequently, another type of strategy has to be considered. We suggest to take advantage of the mean ﬁeld convergence and redeﬁne our problem in a new approach, related to the closeness of the system to equilibrium. Indeed, we would like to use the result on theorem 8 an transfer properties from µt, the limit measure given by the solution of equation (10.1.9), to the empirical measure µN t . More in details, if we take a look at equations (10.1.9) and (10.4.41), we notice that under the hypothesis of no interactions H0 : H = 0, the coeﬃcient b does not depend on time, while in the case 10.7. TESTING OF THE INTERACTION BETWEEN AGENTS 201 of interactions, b depends on t through µt. This motive the construction of a test that is interesting on the hypothesis "b no depending on t", against the alternative of the dependence. This approach, however, presents a new diﬃculty: in case the system is in an equilibrium state, µt = µ∞, the null hypothesis is again indistinguishable from the alternative. We conjecture that in the case where the system is in equilibrium it is not possible to create a test that distinguishes whether there are interactions, but we lack of a proof at the present time. We nevertheless can propose a strategy based on a hierarchical test, where we ﬁrst test that whether the system is at equilibrium or not, and then, based on a rejection of this hypothesis, we can check whether or not there are interactions, by testing whether the drift coeﬃcient depends on time or not. In the following we perform some exploratory work based on numerical simulations to underline the idea of such a test construction. After which we present a ﬁrst theoretical result in a simpliﬁed model. It has been proved in Mischler for the FhN model presented in the introduction [citation], that if the interaction between neurons is weak or null, the system converges to an equilibrium state, while when the interaction grows diﬀerent patterns may appear such as synchronous behavior of neurons. In Figure 10.7, we present the typical behavior of the voltage of a neuronal population (result of one simulation), when there is not interaction. We show three cases, in top we observe the trajectories and a density plot of a population of 1000 neurons without interaction and with an initial distribution close to equilibrium (that we have obtained by letting the system evolve for a long time; in the middle row, we show the same plots for a similar population, also without considering interaction between neurons but with an uniform initial distribution in [−1, 1]; then in the last row, we see a case when we start far from equilibrium all neurons start in the same state Z0 = (2, 2)-. We see how when we star from equilibrium the distribution is maintained over time, when in the other two cases, the population is moving towards the steady state. In the other hand, in Figure 10.7 we show, ﬁrst the typical evolution in long time without interaction and with increasing interaction (J = 1 and J = 10). We see that, when the interactions are weaker, we still have an equilibrium state while when the interactions are stronger a time-dependent pattern appears. We will explore our test idea on several simulations of this cases. Equilibrium test To start with it, we pick two bounded and compactly supported kernel functions K1 : (0, T) →R and K2 : Rd →R such that Z T 0 K1(t)dt = Z R2d K2(z)dz = 1. In our simulation we will consider the Epanechnikov kernel, K(u) = 3 4(1 −u2); supp K = {\|u\|≤1}. for t and each component of z. For h > 0 we denote, K(h) 1 (t) = h−1K1(h−1t); K(h) 2 (z) = h−dK2(h−1z). We construct a family of estimators of µt0(z0) depending on h by setting ˆ µN h (t0, z0) = Z R2d K(h) 2 (z0 −z)µN t0(dx) (10.7.51) 202 CHAPTER 10. CONCENTRATION INEQUALITY Figure 10.1: We show the voltage of three typical simulations of the model without interaction with three diﬀerent initial conditions. In the top row, from the steady state; in the middle row, from a uniform distribution in [−1, 1]2 and ﬁnally in the last row, when all neurons initialize in the same state (corresponding to a Dirac distribution in (2, 2)). Figure 10.2: We show the voltage of three typical simulations evolving in a long time (in [0, 100]) without interaction, with weak interaction and with strong interaction. 10.7. TESTING OF THE INTERACTION BETWEEN AGENTS 203 Figure 10.3: Box plot showing the distribution of T N 1 test values starting at equilibrium or not. The leftmost plot corresponding to the case when we start at the equilibrium, in the middle, when we start from the uniform distribution, and the last one when the system is far from the equilibrium (corresponding to the third case presented). how is suggested in section 10.6. The performance beneﬁts of such an estimator are shown by the oracle inequality given in Lemma 16. We regularise it on t by taking the convolution with the kernel K1, ˜ µN h (t0, z0) = Z T 0 Z R2d(K1 ⊗K2)¯ h(t0 −t, z0 −z)µN t0(dz)dt, (10.7.52) where for ¯ h = (h1, h2), hi > 0, we deﬁne (K1 ⊗K2)¯ h(t, z) = K(h1) 1 (t)K(h2) 2 (z). Under the hypothesis of being on the equilibrium, µt = µ∞, it is expected that ∂tµt = 0. We rely our construction on this argument by considering that the equilibrium test will be a function of T1 = R \|∂tµ(z)dz\|2 which can be estimated by T N 1 (t) = Z R2d\|∂t˜ µN h (t, z)\|2dz. (10.7.53) In Figure 3, we can compare the values of T N 1 , over 20 simulations of the three cases without interaction. From the box plot we can see that there is signiﬁcant statistical diﬀerent of the indicator T N 1 between the case when the system is close to the equilibrium with respect to the case in where is not from, increasing as we move further away from equilibrium. We see this behavior again when we include interactions (Figure 4) although in this case with more variability. In conclusion, these explorations suggest the validity of such an indicator as the basis for the creation of a test for the equilibrium hypothesis. Interaction test We continue now considering that we are not close to equilibrium and exploring the possibility of testing the connectivity between the agents. Let (t0, z0) ∈(0, T) × R2d we denote πN(dt, dz) = N−1 N X i=1 Zi(dt)δZi(dz), 204 CHAPTER 10. CONCENTRATION INEQUALITY Figure 10.4: Box plot showing the distribution of T N 1 test values starting at equilibrium in the case without interaction or when there is not equilibrium (cases with interaction). The leftmost plot corresponding to the case when we start at the equilibrium with J = 0, and the middle and right, to increase the interaction in correspondence with the models presented in Figure 2. the Rd-value random measure deﬁned by Z [0,T]×R2d φ(t, z)πN(dt, dz) = 1 N N X i=1 Z T 0 φ(t, Zi t)dZi t for a test function φ(t, z) : [0, T] × R2d →R. We recall that Zi = (Xi, Yi) and rewriting equation 10.1.7 in a more synthetic form we have dZi = b(Zi t, µN t )dt + ¯ σd ¯ Bi t where ¯ σ is a diagonal 2d-matrix with σ in the diagonal elements corresponding to X and 0 in the diagonal elements corresponding to Y , and ¯ Bi t being a 2ddimensional Brownian motion. We have that, at least formally, 1 N N X i=1 Z T 0 φ(t, Zi t)dZi t = 1 N N X i=1 Z T 0 φ(t, z)b(z, µN t )µN t dt + 1 N N X i=1 Z T 0 ¯ σφ(t, Zi t)d ¯ Bi t with E  1 N N X i=1 Z T 0 ¯ σφ(t, Zi t)d ¯ Bi t  = 0, since φ is bounded. This suggests the convergence in expected value of πN(t, z) to π(t, z) = b(z, µt(z))µt(z). Such kind of result has been proven in Della Maestra and Hoﬀmann in [73, Teorem 9] for elliptic case (non-degenerated matrix ¯ σ) and Lipschitz coeﬃcient b, as it is the case of the linear model (??). More in details, they propose for b the estimator ˆ bN ¯ h (t0, x0)w = ˆ πN ¯ h (t0, x0) ˆ µN ¯ h (t0, x0) ∨w, where πN ¯ h (t0, x0) = Z [0,T]×Rd(K1 ⊗K2)¯ h(t0 −t, x0 −x)πN(dt, dx). 10.8. APPENDIX 205 Figure 10.5: We compare the values of T N 2 when there is and there is not interaction. We see signiﬁcant diﬀerence between the simulation with large interaction and without interaction but this diﬀerence is less clear when interaction are small. Following this approach, we have ∂tb(z, µt(z)) = ∂t πt(z) µt(z) ! = ∂tπt(z)µt(z) −πt(z)∂tµt(z) µ2 t (z) , and the hypothesis b is not depending on t coincide with T2(t, z) = ∂tπt(z)µt(z) −πt(z)∂tµt(z) = 0, that could be approximate by ˆ T N 2 (t, z) = ∂tˆ πN t (z)ˆ µN t (z) −ˆ πN t (z)∂tˆ µN t (z). In Figure 5 we explore the viability of such approach on simulation of the FhN model. We realize several simulations for each of the case, with and without interaction and for both models, and we compute the value of the indicator ˆ T N 2 for each of them. We see on the boxplot graphics that we have signiﬁcant diﬀerence between the simulation with large interaction and without interaction but this diﬀerence is less clear when interaction are small. This could be explained by the fact that when interactions are weak the system behaves similarly to when there are no interactions. This suggests that this approach may be successful in detecting when we are in the presence of strong interactions but perhaps less eﬀective in detecting weak interactions. 10.8 Appendix 10.8.1 Proof of lemma 14 We now turn to the proof of lemma 5. The proof follows the arguments of Della Maestra and Lacker . 206 CHAPTER 10. CONCENTRATION INEQUALITY Proof. Let it be the partition 0 = t0 < · · · < tK ≤T, for j −1 we deduce, E¯ PN [PN(AN\|Ftj−1)] = E¯ PN [EPN [PN(AN\|Ftj)\|Ftj−1]] = E¯ PN [E¯ PN [ Etj( ¯ MN) Etj−1( ¯ MN)PN(AN\|Ftj)\|Ftj−1]] = E¯ PN [ Etj( ¯ MN) Etj−1( ¯ MN)PN(AN\|Ftj)], where the second equality comes from applying the lemma 3.5.3 of . Following its deﬁnition we can write that Etj( ¯ MN) Etj−1( ¯ MN) = Etj(2( ¯ MN · −¯ MN tj−1))1/2(exp (⟨¯ MN⟩tj −⟨¯ MN⟩tj−1))1/2. As we have proven that E( ¯ MN · ) is a martingale under ¯ PN, then, for t ≥tj−1 we have that Etj(2( ¯ MN · −¯ MN tj−1)) is a martingale and E¯ PN [Etj(2( ¯ MN · −¯ MN tj−1))1/2] = 1. 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Trudinger. “Pointwise estimates and quasilinear parabolic equations”. In: Comm. Pure Appl. Math. 21 (1968), pp. 205–226. issn: 0010-3640. doi: 10.1002/cpa.3160210302. url: Misha Tsodyks, Klaus Pawelzik, and Henry Markram. “Neural networks with dynamic synapses”. In: Neural Comput. 10.4 (1998), pp. 821–835. Alexis Vasseur. “The De Giorgi method for elliptic and parabolic equations and some appli-cations”. Sebastian Vellmer and Benjamin Lindner. “Theory of spike-train power spectra for multidimensional integrate-and-ﬁre neurons”. In: Phys. Rev. Res. 1.2 (2019), p. 023024. Cédric Villani. “Hypocoercivity”. In: Mem. Amer. Math. Soc. 202.950 (2009), pp. iv+141. issn: 0065-9266. doi: 10.1090/S0065-9266-09-00567-5. url: proxy.bu.dauphine.fr/10.1090/S0065-9266-09-00567-5. Jürgen Voigt. “A perturbation theorem for the essential spectral radius of strongly continuous semigroups”. In: Monatsh. Math. 90.2 (1980), pp. 153–161. issn: 0026-9255. doi: 10.1007/BF01303264. url: 1007/BF01303264. 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BIBLIOGRAPHY 223 Yuzhe Zhu. “Velocity averaging and Hölder regularity for kinetic Fokker-Planck equations with general transport operators and rough coeﬃcients”. In: SIAM J. Math. Anal. 53.3 (2021), pp. 2746–2775. issn: 0036-1410. doi: 10.1137/20M1372147. url: Robert S Zucker and Wade G Regehr. “Short-term synaptic plasticity”. In: Annu. Rev. Physiol. 64.1 (2002), pp. 355–405. MOTS CLÉS PDE, neurosciences, comportement asymptotique, comportement à long terme, inégalité de Bernstein, mod-èle de diffusion, cinétique de Fokker-Planck RÉSUMÉ Cette thèse est consacrée à l'étude des estimations asymptotiques pour les équations aux dérivées partielles (EDP) dans deux directions : l'analyse du comportement à long terme des solutions des EDP et l'étude de la limite lorsque le nombre d'individus inclus dans un système dynamique devient grand. Dans une première partie de la thèse, nous abordons la première de ces estimations. Pour cela, nous étudions successivement le problème des éléments propres en étendant le théorème de Krein-Rutman, puis la géométrie de la première valeur propre et enfin, ses implications pour le comportement asymptotique en temps long. En particulier, nous étendons la théorie de Doeblin-Harris pour les semigroupes non-conservateurs, en obtenant des résultats de couverture exponentielle avec un taux constructif. Nous appliquons ces résultats à quatre exemples : des modèles de diffusion, un modèle cinétique de Fokker-Planck, un modèle de sélection-mutation et un modèle de temps écoulé pour une population neuronale. Dans une seconde partie, nous étudions l'existence d'un résultat de concentration de type inégalité de Bernstein pour les modèles position-vitesse que nous appliquons ensuite à trois problèmes : à l'estimation des paramètres du modèle de FitzHugh-Nagumo, à l'estimation non paramétrique de coefficients de modèles généraux et, enfin, un travail exploratoire pour la construction d'un test statistique de la connectivité entre particules dans un système dynamique. ABSTRACT This thesis is devoted to the study of asymptotic estimates for partial differential equations (PDEs) in two directions, the analysis of the long time behavior of PDEs solutions and the study of the limit when the number of individuals included in a dynamical system becomes large. In a first part of the thesis, we address the first of these estimates. For this, we study successively the eigenelement problem by extending the Krein-Rutman theorem, then the geometry of the first eigenvalue and finally, its implications for the asymptotic behavior in long time. In particular, we extend the Doeblin-Harris theory for nonconservative semigroups, obtaining exponential covergence results with constructive rate. We apply these results to four examples: diffusion models, a kinetic Fokker-Planck model, a selection-mutation model and an elapsed time model for neuronal population. In a second part, we study the existence of a concentration result of the Bernstein inequality type for position-velocity models which we then apply to three problems: to the estimation of the parameters of the FitzHugh-Nagumo model, to the nonparametric estimation of general model coefficients and, finally, an exploratory work for the construction of a statistical test of the connectivity between particles in a dynamical system. KEYWORDS PDE, neuroscience, asymptotic behavior, Long-term behavior, diffusion model, Bernstein inequality Fokker-Planck kinetics
190569	https://openstax.org/books/college-physics-2e/pages/31-introduction-to-radioactivity-and-nuclear-physics	Skip to ContentGo to accessibility pageKeyboard shortcuts menu College Physics 2e Introduction to Radioactivity and Nuclear Physics College Physics 2eIntroduction to Radioactivity and Nuclear Physics Search for key terms or text. Figure 31.1 The synchrotron source produces electromagnetic radiation, as evident from the visible glow. (credit: United States Department of Energy, via Wikimedia Commons) Chapter Outline 31.1 Nuclear Radioactivity 31.2 Radiation Detection and Detectors 31.3 Substructure of the Nucleus 31.4 Nuclear Decay and Conservation Laws 31.5 Half-Life and Activity 31.6 Binding Energy 31.7 Tunneling There is an ongoing quest to find substructures of matter. At one time, it was thought that atoms would be the ultimate substructure, but just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a tiny nucleus. The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, such as why the core of the earth remains molten and how the sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus revealed fundamental and previously unknown particles, forces, and conservation laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this chapter, the fundamentals of nuclear radioactivity and the nucleus are explored. The following two chapters explore the more important applications of nuclear physics in the field of medicine. We will also explore the basics of what we know about quarks and other substructures smaller than nuclei. PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Peter Urone, Roger Hinrichs Publisher/website: OpenStax Book title: College Physics 2e Publication date: Jul 13, 2022 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
190570	https://blog.csdn.net/beyondqinghua/article/details/105171466	博客下载学习社区 GitCode InsCodeAI 会议 AI 搜索最新推荐文章于 2024-09-20 16:12:03 发布转载于 2020-03-28 23:40:31 发布 · 1.1w 阅读 · 5 · 24 数学专栏收录该内容 11 篇文章华先生考虑生产工艺的(局部)层面，如何选取工艺参数和工艺过程，以快速提高产品质量。由于线性规划和其它一些方法要求繁杂的计算(当时没有计算机)，国民教育水平不高，线性规划的方案无法难以进行大规模的推广。华先生总结经验，不断思考，要寻求一些易于被人接受、应用面广的数学方法，提出了优选法，可以用最少的试验次数来找出最优点。这种方法非常简单，对当时中国既缺乏数学人才，又缺乏计算机的企事业单位提高效率起到了巨大的作用。我们假定1公斤面粉，放碱的重量范围为0～10克之间，精准度到0.1克。当然碱放得太多太少都不行。我们还假定用不同碱量做出来的馒头的口味是可以量化度量的：黄金分割有一个特别好的性质，如1:0.618=1.618:1，(1-0.618)/0.618=0.618，这样一来，0到6.18克的黄金分割点正好是10-6.18= 3.82克的位置，这就使得这前后两次找到的黄金分割点，6.18和3.82中间出现了中间点，恰好是5.0克，这两个点距离0-1的中间点0.5距离相同，也就是说5.0成了两次黄金分割点的对称点：黄金分割法通常适用于不预先限定试验次数的，单峰的函数，如这样的：比较两次试验，如果(2)好于(3)，则剪掉(3)左边的纸条，留下0.618的长度，然后再对折，找到下一个试验点，直至找到最佳点。这样的方法需要的试验次数不多，5次试验就可以将范围缩小到原来的0.618^5=0.09，6次可以将范围缩小到原来的0.618^6=0.056，呈指数收敛效果。黄金分割法与二分法适应场景差异单峰函数找极值适用黄金分割法，单调函数找零点适合二分法。单峰函数找极值的时候，二分法是不能够一次实验就把可行区域一分为二，我们不能断言极值一定在右边半个区域。我们需要四个点，极值一定在第二个点到第四个点之间，但不能确定是第三个点左侧还是右侧。使用黄金分割法的时候，第二个点和第三个点构成区间的黄金分割，当确定是在第二个点到第四个点的时候，第三个点刚好是新区间的黄金分割，可以在下一次实验中复用，这样就大大减少了实验次数。确定要放弃本次机会？福利倒计时 : : 立减 ¥ 普通VIP年卡可用立即使用 beyondqinghua 关注关注 5 点赞踩 24 收藏觉得还不错? 一键收藏 1 评论分享复制链接分享到 QQ 分享到新浪微博扫一扫举报举报专栏目录线性与非线性规划：黄金分割方法 qq_45397810的博客 08-11 910 本系列涉及线性与非线性规划中的几种规划算法 1.本节介绍黄金分割（0.618）方法与进退法 2.参考：陈宝林-最优化理论与算法 3.采用python编程实现，已测试，代码可行。 1 条评论您还未登录，请先登录后发表或查看评论数学发展简史 9-13 战国时期的中国,筹算成为当时的主要计算方法;出现《庄子》、《考工记》记载中的极限概念、分数运算法、特殊角度概念及对策论的例证。公元前230年,古希腊的埃拉托色尼提出素数概念,并发明了寻找素数的筛法。公元前三至前二世纪,古希腊的阿波罗尼发表了八本《圆锥曲线学》,这是最早关于椭圆、抛物线和双曲线的论著。黄金分割:美学与数学的交汇 9-12 到19世纪黄金分割这一名称才逐渐通行。黄金分割数有许多有趣的性质,人类对它的实际应用也很广泛:最著名的例子是优选学中的黄金分割法或0.618法,是由美国数学家基弗于1953年首先提出的,70年代由华罗庚提倡在中国推广。编辑本段几何作法已知线段AB,按照如下方法作图: ... 数学归纳法（华罗庚） 08-28 同时，华罗庚先生在书中提出，仅仅理解归纳法的运用是不够的，更重要的是要理解归纳法背后的基本原理和思想。文章中还提到了数学归纳法的基本步骤：首先验证命题在基础情况下成立（通常是n=1），然后假设在n=k时... 【最优化】一维最优化方法：华罗庚黄金分割优选法 TuTuTuhong的博客 09-20 910 本文简单介绍了黄金分割法，但是对于为什么一定是黄金分割点还有待学习。华罗庚数学学校数学课本高3年级.pdf资源 9-16 《华罗庚文集:应用数学卷1》分卷Ⅰ、卷Ⅱ两卷,卷Ⅰ主要内容包括近似分析中的数论方法和应用统计中的数论方法,卷Ⅱ主要内容包括计划经济大范围最优化数学理论、关于经济优化平衡的数学理论、数学普及之初简介、统筹方法平话及补充、优选法平话及补充、优选学等。从卷Ⅰ、卷Ⅱ可以看出华罗庚在中国发展应用数学... 5看电视(2).ppt资源 9-15 最后,提到了著名数学家华罗庚和他的“优选法”,这种方法强调通过最优选择和时间规划提高工作效率,其应用领域... 格式:ppt 资源大小:4.5MB 页数:18 敏捷之道:从理论到实践浏览:3 ### 深入了解客户:产品成功的关键要素在产品开发过程中,准确识别目标客户至关重要。下面我们将深... 资源评论 #完美解决问题 ... 优选法中的黄金分割法川流的专栏 01-08 7318 2. 1 　黄金分割法的基本思想黄金分割法是优化方法中的经典算法,以算法简单、效果显著而著称,是许多优化算法的基础.但它只适用于一维区间[a,b] 上的凸函数.其基本思想是:依照“去坏留好”原则、对称原则以及等比收缩原则来逐步缩小搜索范围.以具体的单参数变量优选来说,根据工程经验选取搜索区间为[a,b],并在该变量区间内评价函数 Q (x) 存在单值极点.在[a,b] 中取试验点 x 优选法案例分析 weixin_30339969的博客 03-26 2610 无标题文档 0.618法，又叫黄金分割法，是优选法的一种。它在试验时，把试点安排在黄金分割点上来寻找最佳点。而生产生活中，我们常常取其近似值0.618，因此得名。0.618法是最常用的单因素单峰目标函数优选法之一。历史 1953年，美国数学家基弗提出了0.618法。20世纪60、70年代，中国数学家华罗庚先生对其作了简化和补充，并在全中国范围内推广，取得了令人满意的结... 对于单峰函数(有唯一极值的函数),黄金分割法比二分法能用更少的搜索次 ... 9-21 华罗庚在《优选法》中阐述了在一定条件下使用golden ratio黄金分割法搜索单峰函数的最值,比二分法搜索更快,可用更少的步数找到最值。当目标函数可导时,可使用梯度下降类的方法搜索极值点,但当目标函数不可导时,只能使用这些非梯度类的搜索方法。由下图中推导可知,使用黄金分割向下(即向自变量的减小方向)搜索时,该自... 优选法 weixin_30624825的博客 07-19 583 优选法 Tags：搜索作业部落评论地址三分法没什么好讲的，分析一波时间复杂度（网上很多是错的）二分： n∗(12)k=1,k=log121n=log2n 三分：[n(\frac{2}{3})^k=1,k=log_{\frac{2}{3}}{\frac{1}{n}}=log_{\frac{3}{2}}n=... PID控制优选法（黄金分割法）具体试验方法 Drop_maples的博客 08-24 1933 B = 起点 + [终点-起点] 0.618 A = 起点 + 终点 - B点波动很小，有效。（4到5个点基本确定了，5到6个点不错哦~）新点 = 起点+终点-保留点优选法介绍实验经验说明，当实验4~5个点后，基本上已找到最优点所在范围，在这个范围内，各个点的实验结果都差不多了，因此也就没有必要为找出最优点而去做更多的实验了。只要... 优选法基础知识PPT学习教案.pptx 10-02 3. 选择优化方法：对于单调的目标函数，可以使用平分法或黄金分割法。平分法通过不断将试验范围平均划分，逐步缩小范围；黄金分割法则利用0.618比例原理来选取试验点，以此提高效率。平分法的操作流程是每次在... 华罗庚统筹方法.pdf 09-26 作者提供了三种不同的方法来泡茶，每种方法都有其优缺，通过对这些方法的分析，作者指出第一种方法是最优的，因为它可以缩短工时。在华罗庚统筹方法中，任务之间的相互关系非常重要。通过绘制箭头图，可以清晰地... 华罗庚著作《数学归纳法》PDF扫描版介绍数学归纳法是数学中的一种证明技巧，它用于证明与自然数相关的命题。这种方法特别适用于证明一些从有限步骤推广到无限步骤的数学问题。...华罗庚的《数学归纳法》一书，无疑为学习这一方法提供了宝贵的资源和指导。《华罗庚文集.应用数学卷I》作者:华罗庚著;杨德庄主编出版年: 2010年 05-26 《华罗庚文集:应用数学卷1》分卷Ⅰ、卷Ⅱ两卷，卷Ⅰ主要内容包括近似分析中的数论方法和应用统计中的数论方法，卷Ⅱ主要内容包括计划经济大范围最优化数学理论、关于经济优化平衡的数学理论、数学普及之初简介、统筹... 黄金分割法知识总结 markconca的博客 11-25 1万+ 黄金分割法黄金分割法也称为中外比，指把一条线段分割为两部分，使其中一部分与全长之比等于另一部分与这部分之比。其比值是一个无理数，取其前三位数字的近似值是0.618，所以也称为0.618法。数学·黄金分割法其比值是一个十分有趣的数字，我们以0.618来近似，通过简单的计算就可以发现： 1/0.618=1.618 500/309=1.618 (1-0.618)/0.618... 黄金分割法求解局部最小值——python实现热门推荐 oldsoliderneverdie的博客 11-21 1万+ 黄金分割法搜索局部最小值的原理是基于单峰函数的特性。 1、单峰函数定义：设f是定义在闭区间[a,b]上的一元函数，x是f在[a,b]上的极小点，并且对于任意的x1,x2属于[a,b],x1&amp;amp;amp;amp;lt;x2; 若当x2&amp;amp;amp;amp;lt;=x时，f(x1)&amp;amp;amp;amp;gt;f(x2), 当x&amp;amp;amp;amp;lt;=x1时，f 华罗庚统筹法与计算机专业,华罗庚的优选法与统筹法 weixin_29525745的博客 06-29 1445 优化问题是人们经常要遇到的问题，例如，我们出门旅行就要考虑选择怎样的路线和交通工具，才能使旅行所需费用最少或者所花的时间最短；又如著名的邮递员送信最短路线问题。在经济建设、工农业生产、交通运输、军事国防等各行各业都会面临优化的问题。华罗庚爷爷是我国著名的数学家，他就是在研究泡茶、烙饼等问题中，最早在我国提出了数学中的“优选法”理论——宣传推广“合理安排”的思想和方法，为人们节约资源，提高效率作出了... MATLAB黄金分割法标注最优点,matlab黄金分割法课程论文--.doc weixin_36364022的博客 03-22 884 matlab黄金分割法课程论文--.doc 中南林业科技大学本科课程论文学院：理学院专业年级： 14 级信息与计算科学 2 班学生姓名：邱文林学号： 20144349 课程： MATLAB 程序设计教程设计题目：基于 MATLAB 的黄金分割法与抛物线插值法指导教师：龚志伟 2016 年 4 月中文摘要为了求解最优化模型的最优解，可使用基于 MATLAB算法编... 【最优化】黄金分割法 Zimu的博客 11-23 3852 【黄金分割法】【0.618】一维搜索是最优化结构里的重要且基础的模块，学习一位搜索对研究最优性条件、收敛性的证明、算法速度的设计有着极其重要的作用。精确一维搜索通常有两种实现方式：试探法和函数逼近法。其中试探法常见的有黄金分割法（0.618法）和斐波那契法（Fibonacci法）函数逼近法常见的有牛顿法、割线法、抛物线法、插值法等。本文只浅谈试探法众多黄金分割法。基于MATLAB开发的图形用户界面数字信号处理仿真平台_信号生成与调制解调滤波器设计与频谱分析时频域变换与波形可视化音频与图像信号处理通信系统仿真与算法验证_为高校教学提供.zip 09-27 基于MATLAB开发的图形用户界面数字信号处理仿真平台_信号生成与调制解调滤波器设计与频谱分析时频域变换与波形可视化音频与图像信号处理通信系统仿真与算法验证_为高校教学提供.zip 数据结构教程(第6版)-课本所有源程\第2章最新发布 09-27 数据结构教程(第6版)-课本所有源程\第2章关于我们招贤纳士商务合作寻求报道 400-660-0108 kefu@csdn.net 在线客服工作时间 8:30-22:00 公安备案号11010502030143 京ICP备19004658号京网文〔2020〕1039-165号经营性网站备案信息北京互联网违法和不良信息举报中心家长监护网络110报警服务中国互联网举报中心 Chrome商店下载账号管理规范版权与免责声明版权申诉出版物许可证营业执照 ©1999-2025北京创新乐知网络技术有限公司登录后您可以享受以下权益：免费复制代码和博主大V互动下载海量资源发动态/写文章/加入社区 × 评论 1 被折叠的条评论为什么被折叠? 到【灌水乐园】发言查看更多评论添加红包发出的红包实付元钱包余额 0 1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。 2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。余额充值确定取消举报包含不实信息涉及个人隐私请选择具体原因（必选）侮辱谩骂诽谤请选择具体原因（必选）搬家样式博文样式点击体验 DeepSeekR1满血版专业的中文 IT 技术社区，与千万技术人共成长客服返回顶部
190571	https://www.sciencedirect.com/topics/medicine-and-dentistry/mononeuropathy-multiplex	Skip to Main content My account Sign in Mononeuropathy Multiplex In subject area:Medicine and Dentistry Mononeuritis multiplex is defined as an asymmetric sensory and motor peripheral neuropathy that involves two or more isolated peripheral nerves. AI generated definition based on: Surgical Clinics of North America, 2014 How useful is this definition? Add to Mendeley Also in subject areas: Nursing and Health Professions Pharmacology, Toxicology and Pharmaceutical Science Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Mononeuropathy Multiplex 2021, Decision-Making in Adult NeurologyNoah Levinson, Raymond S. Price Mononeuropathy multiplex (sometimes referred to as mononeuritis multiplex) is defined as lesions affecting two or more nerves that cannot be explained by a single root or plexus injury. The involvement of multiple nerves is usually in an asymmetric, non–length-dependent pattern and develops subacutely. If this process continues with additional individual nerve involvement, it can progress to appear clinically and electrophysiologically identical to a more common polyneuropathy. An extensive evaluation for the etiology of mononeuropathy multiplex should be performed, as treatable autoimmune and inflammatory causes are frequently identified. A. : Nerve conduction studies and electromyography (NCS/EMG) should be performed in all patients with suspected mononeuropathy multiplex to confirm the pattern of nerve injury and to characterize the specific nerve pathology (e.g., demyelinating vs. axonal). B. : In a patient with multiple mononeuropathies at common sites of compression (e.g., ulnar neuropathy at the elbow) and a family history of polyneuropathy, hereditary neuropathy with liability to pressure palsies (HNPP) and familial amyloidosis should be considered. HNPP is caused by a deletion of the PMP22 gene and would be suggested by moderate diffuse slowing of the motor conduction velocities that do not meet criteria for demyelination. Familial amyloidosis, which is most commonly caused by a transthyretin mutation, would be suspected in patients with autonomic dysfunction or systemic disease (e.g., cardiac abnormalities). C. : Neurolymphomatosis is the direct invasion of nerve roots or peripheral nerves by lymphoma cells. Depending on the extent of involvement, this can present as an individual mononeuropathy, radiculopathy, or mononeuropathy multiplex. In bone marrow transplant patients, graft-versus-host disease occurs when the transplanted immune cells attack organs of the bone marrow recipient. The skin, gastrointestinal tract, and liver are the most commonly affected organs, but rarely peripheral nerves can also be injured in this disease. Diagnosis of these conditions requires nerve biopsy. D. : Leprosy is rare in the United States, but should be considered as a cause of axonal mononeuritis multiplex in endemic areas in Mexico, Central America, South America, and Africa. In tuberculoid leprosy, nerve injury usually occurs around skin lesions. In lepromatous leprosy, nerve injury can be much more diffuse. E. : Lead toxicity is a very rare cause of neuropathy in adults. It has an unusual pattern that is classically an isolated motor neuropathy with a predilection for extensor muscles of the arms and legs. Gasoline and certain adhesives contain n-hexane, a chemical that is toxic to peripheral nerves. Chronic sniffing, or huffing, of these substances classically causes a motor-predominant axonal multifocal neuropathy. F. : Primary vasculitides frequently cause neuropathy in addition to other end-organ injury, and neuropathy often occurs early in the course of systemic vasculitis. Subacute onset and rapid progression of symptoms, severe pain, and the presence of concurrent systemic complaints suggest vasculitis. Peripheral nervous system involvement most commonly occurs in small and medium-vessel primary vasculitides, which include polyarteritis nodosa (PAN), granulomatosis with polyangiitis (formerly Wegner's granulomatosis), microscopic polyangiitis, and eosinophilic granulomatosis with polyangiitis (formerly Churg-Strauss syndrome). Treatment is typically high-dose steroids and cyclophosphamide. A secondary vasculitic neuropathy can be caused by cryoglobulinemia and numerous autoimmune diseases, most commonly lupus, rheumatoid arthritis, Sjögren syndrome, and Behçet disease. Treatment is targeted to the underlying autoimmune condition. Patients with peripheral nerve injury from either primary or secondary vasculitides will likely also need physical therapy, bracing, and neuropathic pain management for the nerve injury. G. : Nonsystemic vasculitic neuropathy is an isolated vasculitis of the peripheral nerves without other end-organ involvement. These patients must be carefully monitored because up to 10% of these patients will eventually develop systemic vasculitis. Nerve and muscle biopsy is necessary for diagnosis. Biopsy of both muscle and nerve is recommended to increase the diagnostic yield of the biopsy. Treatment is similar to systemic vasculitis with high-dose oral steroids and cyclophosphamide. H. : Amyloidosis should be suspected in patients with autonomic dysfunction or systemic disease such as cardiac abnormalities or renal failure. Serum protein electrophoresis with immunofixation may show elevated serum free light chains. Amyloid deposition may be seen on a fat pad or nerve and muscle biopsy. I. : Isolated demyelination of the motor neurons (multifocal motor neuropathy, MMN) may clinically mimic anterior horn cell disease. MMN causes slowly progressive asymmetric weakness, typically beginning in the upper extremities. Anti-GM1 bodies are seen in some but not all patients. J. : Acquired demyelinating neuropathies can rarely cause mononeuropathy multiplex. Guillain-Barré syndrome is classically symmetric, but a small subset of patients may have asymmetric features. When chronic inflammatory demyelinating polyneuropathy presents as mononeuropathy multiplex, this variant is known as multifocal acquired demyelinating sensory and motor polyneuropathy (MADSAM). The key to distinguishing these disorders from vasculitis is evidence of multifocal demyelinating features on the NCS. View chapterExplore book Read full chapter URL: Book2021, Decision-Making in Adult NeurologyNoah Levinson, Raymond S. Price Review article General Orthopedics 2024, Physician Assistant ClinicsAmy Dix PhD, PA-C, Stephanie Kubiak PhD, OTR/L Mononeuropathy Multiplex Mononeuropathy multiplex or multifocal nerve injury is a result of vascular, immunologic, or infectious processes.4 This type of neuropathy exhibits a combination of motor, sensory, and/or autonomic deficits that affect individual adjacent nerves versus a diffuse and/or symmetric nerve pattern (eg, polyneuropathy). Mononeuropathy multiplex cause includes vasculopathy (eg, arteritis), infiltration process (eg, leprosy, sarcoidosis), radiation damage, or immunologic disorders (eg, brachial plexopathy).4 View article Read full article URL: Journal2024, Physician Assistant ClinicsAmy Dix PhD, PA-C, Stephanie Kubiak PhD, OTR/L Review article Neuromuscular Complications of Systemic Conditions 2008, Physical Medicine and Rehabilitation Clinics of North AmericaJessica Robinson-Papp MD, David M. Simpson MD Mononeuropathy multiplex Mononeuropathy multiplex (MM), an uncommon complication of HIV infection, is defined as dysfunction of two or more peripheral nerves. As the individual nerves involved are variable, presenting complaints differ between patients. Common features include focal, asymmetric, sensory and motor deficits consistent with the distribution of one or more peripheral nerves. The incidence of MM has a bimodal distribution, with peaks in early and advanced HIV infection. In relatively immunocompetent patients, MM typically presents with dysfunction attributable to a limited number of cranial or peripheral nerves. The deficits typically resolve spontaneously over several months . Although the initial presentation of MM in patients with advanced immunosuppression may be similar, the process rapidly spreads to involve multiple nerves, and the prognosis for recovery is much poorer . The phenotypic dichotomy between early and late MM likely reflects differing pathogenic mechanisms. An autoimmune mechanism is postulated for early MM, whereas late MM may be related to CMV infection . In both forms axonal degeneration and inflammation are seen in nerve biopsy specimens. This pattern is distinctive from MM in the general population, which is more often vasculitic. NCS and EMG are the most helpful diagnostic studies in confirming the diagnosis of MM and assessing its severity. NCS reveals decreased motor and sensory evoked response amplitudes in specific nerves, with relative sparing of conduction velocities. Normal NCS in other nerves confirms the multifocality of the disorder and helps to distinguish it from polyneuropathy. EMG shows denervation in muscles supplied by affected nerves. Cerebrospinal fluid examination is helpful to document a positive CMV polymerase chain reaction in late MM; other changes are nonspecific. Nerve biopsy may be considered to rule out a vasculitic etiology. Early MM does not necessarily require treatment; however, corticosteroids, IVIg, or plasmapheresis may hasten recovery. Numerous investigators recommend that late MM should be treated with empiric antiviral therapy, such as ganciclovir, foscarnet, and cidofovir, for presumed CMV infection, regardless of whether the organism can be demonstrated in cerebrospinal fluid, (preferably with PCR assay) or nerve biopsy. View article Read full article URL: Journal2008, Physical Medicine and Rehabilitation Clinics of North AmericaJessica Robinson-Papp MD, David M. Simpson MD Chapter Mononeuropathy Multiplex 2014, Encyclopedia of the Neurological Sciences (Second Edition)M.P. Collins Abstract Mononeuropathy multiplex refers to an anatomical pattern of neuropathy in which the pathological process affects individual nerves rather than many nerves diffusely, contrasting with a general polyneuropathy. With disease progression, contiguous nerves become affected, producing an overlapping or confluent mononeuropathy multiplex. In longstanding disease, the neuropathy mimics a generalized polyneuropathy but for the presence of asymmetries. Pathogenic mechanisms underlying multiple mononeuropathies include ischemia, inflammatory demyelination, other autoimmune processes, peripheral nervous system (PNS) infections, entrapments, other compressive nerve lesions, trauma, burns, PNS tumors, neoplastic infiltration, deposition of toxic proteins, and intraneural hemorrhages. Multiple mononeuropathy is often attributed to vasculitis, but many nonvasculitic etiologies exist. View chapterExplore book Read full chapter URL: Reference work2014, Encyclopedia of the Neurological Sciences (Second Edition)M.P. Collins Review article Tickborne Borrelia Infections 2015, Clinics in Laboratory MedicineJohn J. Halperin MD Mononeuropathy Some patients develop a typical mononeuropathy or mononeuropathy multiplex: focal involvement of one or several individual peripheral nerves. Occasionally, this can involve the brachial or lumbosacral plexus, the complex structures in which the nerves to the limbs form from the originating spinal nerve roots. View article Read full article URL: Journal2015, Clinics in Laboratory MedicineJohn J. Halperin MD Chapter Connective Tissue Diseases, Vasculitis, and the Nervous System 2014, Aminoff's Neurology and General Medicine (Fifth Edition)Richard B. Rosenbaum Mononeuropathy Multiplex Vasculitis is a common cause of mononeuritis multiplex. About one-fourth of patients with mononeuropathy multiplex have nonsystemic vasculitic neuropathy and another one-fourth have some form of systemic vasculitis.48 Vasculitis of the nerve can also be secondary to other systemic diseases, such as rheumatoid arthritis, infections (especially in association with hepatitis B, hepatitis C, and human immunodeficiency virus), sarcoidosis,95 or paraneoplastic syndromes (especially in association with T-cell lymphomas).54 Diabetic amyotrophy, diabetic radiculopathy, and other manifestations of multifocal asymmetric diabetic neuropathy are often associated with vasculitis in the affected nerves, a pattern that is unusual in diabetics whose neuropathy is limited to the more common distal symmetric pattern.96 Mononeuritis multiplex is unusual in patients with SLE or Sjögren syndrome. Other causes of asymmetric multiple neuropathies include Lewis–Sumner syndrome, multifocal motor neuropathy with conduction block, neuralgic amyotrophy, compression neuropathies superimposed on more symmetric neuropathies especially in patients with hereditary liability to pressure palsies, amyloidosis, neoplastic nerve infiltration, Lyme disease, porphyria, and leprosy. View chapterExplore book Read full chapter URL: Book2014, Aminoff's Neurology and General Medicine (Fifth Edition)Richard B. Rosenbaum Review article Neuromuscular diseases associated with Human Immunodeficiency Virus infection 2018, Journal of the Neurological SciencesDevin E. Prior, ... Jeffrey A. Cohen 6Mononeuropathy multiplex Mononeuropathy multiplex also presents at all stages of HIV infection due to varying etiologies. In patients undergoing ART, mononeuropathy multiplex is often secondary to immunological factors sometimes involved in seroconversion and also may be secondary to vasculitis thought to be caused by immune complexes from HIV or from concomitant hepatitis B or C infections depositing in the perineurial vessels causing damage . In underserved areas and in developing countries where patients do not have access to ART, mononeuropathy multiplex may result from opportunistic infections such as CMV, Herpes Simplex virus, and VZV. These viral pathogens may infect the nerves themselves or the vascular supply in several nerves resulting in the clinical syndrome of multiple mononeuropathies . Patients present with painful asymmetric polyneuropathy affecting multiple nerves in a stepwise fashion. The neuropathy progresses slowly involving sensory, motor, and autonomic functions . Diagnosis is based on clinical presentation and electrophysiological studies. Electrophysiological studies show multifocal distribution of neuropathy, often axonal. Nerve biopsy may be diagnostic showing epineurial and endoneurial perivascular inflammatory infiltration and at times necrotizing vasculitis. Biopsy is not required for diagnosis but may be helpful for guiding further treatment. Hepatitis B and C serologies should be checked as a part of work up . It is important to assess for CMV as an etiology with cerebrospinal fluid PCR. However, CMV presence in cerebrospinal fluid may not be sensitive for the diagnosis of mononeuropathy multiplex, as the viral infection may be restricted to PNS. If patients are CMV positive they should be treated with ganciclovir . In patients without evidence of CMV, neuropathy may be immune-mediated due to HIV seroconversion and involves milder symptoms that resolve without therapy over several months. In patients who do not recover after several months, immunomodulatory therapy including IVIG, plasmapheresis, and corticosteroids may be used particularly if there is evidence of vasculitis . View article Read full article URL: Journal2018, Journal of the Neurological SciencesDevin E. Prior, ... Jeffrey A. Cohen Review article Clinical approach to neuro-rheumatology 2021, Journal of the Neurological SciencesSarah Goglin, Tracey A. Cho 1.1.7Mononeuritis multiplex Mononeuritis multiplex, or multiple mononeuropathy, is characterized by pain and sensory loss, with or without weakness, in multiple individual peripheral nerves. Onset is typically acute to subacute, and nerves may be injured simultaneously in sequence. Patients often report a deep aching pain without clear localization, followed by more typical burning paresthesias along with negative symptoms. Mononeuritis multiplex is characteristic of vasculitic neuropathies, especially in the setting of systemic vasculitis. Clues include constitutional symptoms such as fatigue, weight loss, and night sweats, or involvement of other organs such as sinuses, lungs, and kidneys. Nerve conduction studies and electromyography can confirm axonal injury to multiple individual nerves in a patchy and non-length dependent pattern, but later in the course some patients accrue sequential nerve injuries that may mimic a more typical generalized polyneuropathy both clinically and on neurophysiological testing . The presence of antineutrophil cytoplasmic autoantibodies in the setting of classic mononeuritis multiplex may obviate the need for biopsy, but in cases of uncertainty biopsy of cutaneous sensory nerve and surrounding muscle may be necessary to secure the diagnosis. In rare cases, biopsy shows vasculitic neuropathy but no systemic involvement or antibodies is detected (non-systemic vasculitic neuropathy). View article Read full article URL: Journal2021, Journal of the Neurological SciencesSarah Goglin, Tracey A. Cho Review article Neurologic Manifestations of Rheumatic Diseases 2017, Rheumatic Disease Clinics of North AmericaMary Margaretten MD, MAS Mononeuritis Multiplex Mononeuritis multiplex encompasses the vasculitic neuropathies, and, although rare in primary Sjögren syndrome, when promptly diagnosed and treated, mononeuritis multiplex may be the peripheral nerve manifestation that exhibits the most dramatic and durable response to therapy.27 Because the longest nerves in the body are affected first, foot drop is the most common manifestation of mononeuritis multiplex. Patients are often unaware of dorsiflexion weakness; patients with tibialis anterior weakness must raise their knee on the affected side high to avoid tripping over the weakened foot, leading to a “foot-slapping” gait. Nerve infarction eventually leads to muscle wasting, which, in many cases, is permanent. There is decreased survival in patients with primary Sjögren syndrome with peripheral neuropathy, especially in those with mononeuritis multiplex and sensorimotor polyneuropathy, in comparison to patients with primary Sjögren syndrome without peripheral neuropathy.28 Nerve and muscle biopsy may be required to confirm the diagnosis of vasculitic neuropathy unless vasculitis is confirmed by biopsy of other involved organs. Biopsy of affected nerves often shows axonal degeneration and a perivascular inflammatory infiltrate, suggesting an underlying vasculitis cause in many cases. Patients with primary Sjögren syndrome mononeuritis multiplex should be evaluated for underlying cryoglobulinemia. View article Read full article URL: Journal2017, Rheumatic Disease Clinics of North AmericaMary Margaretten MD, MAS Review article Clinical Electromyography 2012, Neurologic ClinicsMark A. Ross MD Multiple Mononeuropathies The pattern of multiple mononeuropathies, also known as mononeuropathy multiplex, represents an important pattern of peripheral nerve disease whereby the neuropathic manifestations occur in the territory of individual nerves, often with a stepwise progression, rather than simultaneously involving distal nerves symmetrically. The underlying disease mechanism typically involves nerve ischemia, nerve infiltration, or multifocal nerve inflammation. Disorders, which are known to commonly manifest with a multiple mononeuropathy pattern, are listed in Box 4. Box 4 Neuropathies causing a pattern of multiple mononeuropathies : Axon loss : Vasculitis : Amyloidosis : Lymphoma : Diabetes : Sarcoidosis : Leprosy : Demyelination : CIDP : GBS : Multifocal motor neuropathy : Hereditary liability to pressure palsies The EDX approach to the evaluation of patients with suspected multiple mononeuropathies involves testing multiple nerves and muscles to demonstrate the multifocal and asymmetric pattern of neuropathy. The abnormal findings on NCS and needle EMG can be localized to an individual nerve territory while a neighboring nerve traveling in the same limb is spared. However, in severe or chronic cases, the stepwise involvement of multiple, bilateral lower extremity nerves over time may result in a clinical picture that is similar to a distal symmetric polyneuropathy. Side-to-side comparison of NCS and needle EMG in the lower extremity may identify subtle differences in the degree of abnormalities within nerves (eg, peroneal more severe than tibial), helping to raise the suspicion of multiple mononeuropathies. Most cases of multiple mononeuropathies produce findings of axonal loss on EDX testing. Exceptions include hereditary neuropathy with liability to pressure palsies (HNPP) or multifocal acquired demyelinating sensory and motor neuropathy (MADSAM). View article Read full article URL: Journal2012, Neurologic ClinicsMark A. Ross MD Related terms: Central Nervous System Peripheral Nervous System Granulomatosis with Polyangiitis Polyneuropathy Polyarteritis nodosa Neuropathy Sjoegren Syndrome Mononeuropathy Microscopic Polyangiitis Peripheral Nerve View all Topics
190572	https://mrbartonmaths.com/topics/number-skills/operations-with-negative-numbers/richtasks.html	Operations with Negative Numbers: Rich Tasks You cannot beat a good rich task! For me, a rich task is one that both stimulates and challenges students of all ages and abilities. Here is a selection of some of my favourites. Contents Median Rich Tasks and Purposeful Practise keyboard_arrow_up Back to Top There are no words to express how much I love Don Steward and his amazing resources on his Median Blog. They allow for purposeful practise, to enable students to gain fluency in key skills. But they also have more than a sprinkling of richness, that leads students merrily along the path towards hypotheses and generalisations. Amazing. NRICH keyboard_arrow_up Back to Top NRICH is simply one of the best websites for rich maths problems in the whole wide world. Other Rich Tasks keyboard_arrow_up Back to Top A selection of some of my favourite rich tasks, including – I must confess – some of my own. For my complete collection, including my thoughts on what makes a good rich task, please visit the Rich Tasks page.
190573	https://stackoverflow.com/questions/45633852/program-to-check-whether-point-lies-on-x-axis-y-axis-or-origin	c - program to check whether point lies on x-axis , y-axis or origin - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more program to check whether point lies on x-axis , y-axis or origin Ask Question Asked 8 years, 1 month ago Modified5 years, 4 months ago Viewed 3k times This question shows research effort; it is useful and clear -1 Save this question. Show activity on this post. I have just started learning c programming. For the question, I wrote the following code. Can you please help me find the error? I am not getting the desired result and the statement in the last else gets executed always. ```c include include void dummy(float a) { float b=a; //perform some floating access dummy (&b); //calling a floating point function } void main() { double x,y; clrscr(); scanf("%lf %lf",x,y); if(x==0 && y!=0) { printf("The point lies on the y-axis."); } else if(y==0 && x!=0 ) { printf("The point lies on the x-axis."); } else if(x==0 && y==0) { printf("The point is the origin"); } else { printf("The point lies neither on the x nor the y axis "); } getch(); } ``` c Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Aug 11, 2017 at 15:04 Martin Verjans 4,836 1 1 gold badge 26 26 silver badges 49 49 bronze badges asked Aug 11, 2017 at 11:14 joey langjoey lang 187 2 2 silver badges 9 9 bronze badges 8 2 Please edit your question to include input that causes the faulty behavior, as well as the actual and expected output.Some programmer dude –Some programmer dude 2017-08-11 11:16:23 +00:00 Commented Aug 11, 2017 at 11:16 hi. whatever be the 2 numbers i enter ,the statement"the point lies neither on the x nor the y axis" gets executed joey lang –joey lang 2017-08-11 11:18:46 +00:00 Commented Aug 11, 2017 at 11:18 tht's what must happen when you don;t check the return value of scanf.Sourav Ghosh –Sourav Ghosh 2017-08-11 11:19:42 +00:00 Commented Aug 11, 2017 at 11:19 return value of scanf?joey lang –joey lang 2017-08-11 11:20:45 +00:00 Commented Aug 11, 2017 at 11:20 2 Never compare floating point values directly for (un)equality: like it's said here: stackoverflow.com/a/4858547/8051589. This may could be the problem.Andre Kampling –Andre Kampling 2017-08-11 11:22:47 +00:00 Commented Aug 11, 2017 at 11:22 \|Show 3 more comments 4 Answers 4 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. while reading values from keyboard with scanf you need to add & infront of variable. instead c scanf("%lf %lf",x,y); Use c scanf("%lf %lf",&x,&y); Update you dont have to check every time for both y and x instead if(x==0 && y!=0) use only one, if(x==0) or if(y==0) Try: ```c void main() { double x,y; clrscr(); scanf("%lf %lf",&x,&y); if(x==0 && y==0) { printf("points lies on origin."); } else if(y==0) { printf("points lies on y-axis."); } else if(x==0) { printf("points lies on x-axis"); } else { printf("The point lies neither on the x nor the y axis "); } getch(); } ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Aug 11, 2017 at 11:40 answered Aug 11, 2017 at 11:28 user6448640 user6448640 1 Comment Add a comment user6448640 user6448640Over a year ago check my answer now, you need to check only x==0 or y==0 at a time 2017-08-11T11:41:30.663Z+00:00 2 Reply Copy link This answer is useful 1 Save this answer. Show activity on this post. For to check if is equal use macro or function like ```c define FEQUAL(x,y,err) (fabs((x) - (y)) < (err)) ``` Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Aug 11, 2017 at 11:38 0_____0_____ 70.9k 4 4 gold badges 41 41 silver badges 90 90 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. Undefined reference to clrscr() function. So, you can try removing clrscr() function from your code. ```c include include void dummy(float a) { float b=a; //perform some floating access dummy (&b); //calling a floating point function } void main() { double x,y; scanf("%lf %lf",&x,&y); if(x==0 && y!=0) { printf("The point lies on the y-axis."); } else if(y==0 && x!=0 ) { printf("The point lies on the x-axis."); } else if(x==0 && y==0) { printf("The point is the origin"); } else { printf("The point lies neither on the x nor the y axis "); } getch(); } ``` It will work. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered Oct 25, 2019 at 6:20 JagadishJagadish 45 6 6 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. A simple solution may be as ```c include int main() { int x,y; printf("Enter the point "); scanf("%d,%d",&x,&y); if (y==0) { if (x==0) { printf("\nPonit lies on the origin\n"); } else printf("\nPoint lies on X-axis\n"); } else if (x==0) { if (y==0) { printf("\nPoint lies on the origin\n"); } else printf("\nPoint lies on the Y-axis\n"); } else printf("\nPoint lies in X-Y coordinate\n"); return 0; } ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited May 17, 2020 at 10:41 Piotr Labunski 1,656 4 4 gold badges 21 21 silver badges 27 27 bronze badges answered May 17, 2020 at 9:38 Ashfi RahmanAshfi Rahman 43 2 2 gold badges 3 3 silver badges 9 9 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. 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190574	https://www.hellenic.org.au/post/follow-the-money-trail-a-history-of-coins	top of page Search Follow the Money Trail: A History of Coins Hellenic Museum Nov 5, 2021 3 min read Updated: Nov 9, 2021 Coins and their inscriptions are one of the main archaeological sources by which we reconstruct the culture, society and history of the ancient world. The images and their inscriptions provide a wealth of knowledge, which allows us to trace chronologies of kings, hierarchies of gods and the ideology of ancient people. If you look at the coins and notes in your wallet (if you still carry one), you will notice that they often depict the profile or bust of a ruler or important person on the obverse, while the reverse bears the image of an important civic symbol – this might be a building or an animal. This tradition began with the ancient Greeks. Xenophanes, writing in the sixth century BCE, claimed that coinage was an Anatolian invention, originating in ancient Lydia. Research of archaeological finds shows Lydian coins are the earliest dateable, additionally Lydia had ready access to a natural source of electrum, a gold and silver alloy, from which coins were first minted. Despite this, the way in which coinage spread through the Greek world and how readily it was accepted made coinage a truly Greek phenomenon. Coins are eloquent testimony to the autonomy of the city-states. Almost from their earliest appearance, Greek coins used a combination of figures, symbols, and inscriptions to emphasise the independence and individuality of the poleis, depicting their patron deities and heroes, products, even visual puns on the cities' names. For example: Delphi ~ Dolphin Leontini ~ Lion Rhodos ~ Rose Selinus ~ Parsley-rock celery Tauromention ~ Bull Melos ~ Apple Given the expense of electrum and gold, even small denominations were worth a considerable amount. For coins to be practical in a commercial context, fractional coinage was introduced during the archaic period, the smallest of which was one-sixteenth of an obol weighing a tiny 0.44mg. How coins as small as this were handled in everyday life is a mystery. As a result, it was not long before a token coinage made from a base metal, usually bronze, became readily available and these were significantly easier to handle. There is some suggestion that during this period, coins were first designed to be used by weight and that these tiny coins were a way of bringing the scale into balance. At the other end of the denominational scale were the gold coins minted by Phillip II of Macedon. These coins took on a role of international currency which was promulgated under Alexander the Great. These types of coins were widely circulated and used as payments between states. Additionally, many more were hoarded by cities like Athens and Rome against potential disasters or wars. Greek coins clearly reflect the changes in artistic styles over time. Archaic art slowly transformed into the more refined classical style and then into the later Hellenistic period. Ancient Greek coins were all handmade using metals including electrum, gold, silver and copper alloys. As early as the Archaic period, the hand of individual die engravers can be seen, but it was during the classical period city-states began to employ the most skilled engravers, providing more detail into each die, and thus into each coin. Ancient Greek coins have made a significant contribution to art history. They not only represent the aesthetic ideals of the time but they provide miniature representations of numerous sculptural and architectural marvels, both sacred and secular. Until coins were discovered, we only had verbal descriptions. The remarkable details in portraiture hold a wealth of information about the artistic styles of the time, in addition to being tools for the expression of art and the communication of religious devotion and civic pride. Due to their uniform value, coins were also a tool with which a social hierarchy of wealth could be definitively calculated. This was invaluable in a society that valued the moral authority of the patrician classes as a way of determining the worth of each group. Coins were also used for political representation, albeit in a more subtle fashion. Coins were used to signify the autonomy of a particular polis or the legitimacy of a ruler. As a historical resource, coins cannot compare with literary sources in revealing the depth and intricacy of political machinations, although they often serve to support the official rhetoric. In some cases, a political alliance between cities will be promoted through the minting of shared coins with symbolism that alluded to their cause. When portraits were introduced on coins, with recognisable individual characteristics, they signified that this person was representative of the state and claimed a right to leadership. The Hellenic Museum is proudly supported by the Bank of Sydney. We are delighted to continue our partnership with aninstitution that supports both the arts and the Greek community in Australia. Written by Sarah Craig, CEO and Head of Curation Recent Posts See All Behind the 'Mobile Museum': how our aged care outreach initiative came to life Dining and Reclining The Babylonian Engine – Why this Akkadian AI translator is a groundbreaking tool for scholars bottom of page
190575	https://www.weizmann.ac.il/math/klartag/sites/math.klartag/files/uploads/log_concave.pdf	Geometry of log-concave functions and measures B. Klartag, V.D. Milman School of Mathematical Sciences Tel Aviv University Tel Aviv 69978, Israel Abstract We present a view of log concave measures, which enables one to build an isomorphic theory for high dimensional log-concave measures, analogous to the corresponding theory for convex bodies. Concepts such as duality and the Minkowski sum are described for log-concave functions. In this context, we interpret the Brunn-Minkowski and the Blaschke-Santal´ o inequalities and prove the two corresponding reverse inequalities. We also prove an analog of Milman’s quotient of subspace theorem, and present a functional version of the Urysohn inequality. 1 Introduction A measure µ on Rn is log-concave if for any measurable A, B ⊂Rn and 0 < λ < 1, µ (λA + (1 −λ)B) ≥µ(A)λµ(B)1−λ whenever λA + (1 −λ)B is measurable, where A + B = {a + b; a ∈ A, b ∈B} and λA = {λa; a ∈A}. Such measures naturally appear in convex geometry, since the Brunn-Minkowski inequality states that uniform measures on convex sets are log-concave measures (including the Lebesgue measure on Rn). The Brunn-Minkowski inequality also implies that lower-dimensional marginals of uniform measures on con-vex bodies are log-concave. In fact, marginals of uniform measures on convex bodies are essentially the only source for log-concave measures, as these marginals form a dense subset in the class of log-concave mea-sures. A function f : Rn →[0, ∞) is log-concave if log f is concave. As was shown in [Bo2], a measure µ on Rn whose support is not contained in any aﬃne hyperplane is a log-concave measure if and only if it is absolutely continuous with respect to the Lebesgue measure, and its density is a log-concave function. 1 As log-concave measures retain some features of uniform measures on convex bodies, many results on uniform measures on convex bod-ies may be generalized to log-concave measures (two samples among many are [Bo] and [BN]). However, it has recently become clear to the authors that such a generalization may shed new light on uniform measures on convex bodies, and may help clarify the diﬃcult open problems regarding such measures. Such an approach is demonstrated in [K1], and has led there to some progress regarding the slicing prob-lem. Therefore we believe that a systematic study of the geometry of log-concave measures is essential in order to understand uniform measures on convex bodies. In this paper we present some steps in this direction, and we recover most of the isomorphic results for convex bodies in the context of log-concave functions. When trying to generalize the geometry of convex bodies to log-concave measures, the ﬁrst problem we encounter is that of duality. For a convex body K ⊂Rn which is centrally-symmetric (i.e. K = −K), its polar is deﬁned by K◦= {x ∈Rn; ∀y ∈K, ⟨x, y⟩≤ 1}. The polar body is a fundamental tool in convex geometry. We show that a suitable variation of the Legendre transform may constitute a proper replacement in the context of functions. Given a function f : Rn →R, its Legendre transform is deﬁned by Lf(x) = sup y∈Rn [⟨x, y⟩−f(y)] . The function L(f) is convex. If f is convex as well as continuous, then L(L(f)) = f. The Legendre transform is a classical operation, which was used, for example, in the derivation of Hamilton equations in classical mechanics (e.g. [Ar]). Since the most natural domain for the Legendre transform is convex functions, we deﬁne the dual of a log-concave function f : Rn →[0, ∞) by f ◦= e−L(−log f). This deﬁnition is closely related to the duality of convex bodies. Let ∥·∥, ∥·∥∗be the norms that K, K◦are their unit balls, correspondingly. Then the dual functions to 1K, e−∥x∥, e−1 2 ∥x∥2 are exactly the functions e∥x∥∗, 1K◦, e−1 2 ∥x∥2 ∗respectively. Let us demonstrate the usefulness of this deﬁnition with the Blaschke-Santal´ o inequality and its converse. These inequalities state that there exists a numerical constant c > 0 such that for any centrally-symmetric convex body K ⊂Rn, c < V ol(K) V ol(D) V ol(K◦) V ol(D) 1 n ≤1 (1) where D is the standard Euclidean unit ball in Rn. The right-most inequality is due to Santal´ o (see e.g. [MeP] for a clear presentation), 2 and the left-most one was proved by Bourgain and Milman [BM]. Log-concave functions satisfy corresponding inequalities, which are func-tional analogs of Santal´ o and reverse-Santal´ o inequalities: Theorem 1.1 There exist universal constants c, C > 0 such that for any dimension n and for any f : Rn →[0, ∞), an even log-concave function with 0 < R f < ∞, we have c < Z Rn f Z Rn f ◦ 1 n ≤C. The optimal value of the constant C from Theorem 1.1 is known to be 2π (see [Ba1, AKM]). Regarding the equality case in the right-most inequality in Theorem 1.1; In the case of convex bodies, it is known that V ol(K) V ol(D) V ol(K◦) V ol(D) = 1 if and only if K is an ellipsoid. In the functional version of the Santal´ o inequality, the role of ellipsoids is replaced by gaussian functions (functions of the form ce−⟨Ax,x⟩for a positive-deﬁnite matrix A and a positive c > 0). Note that the standard gaussian e−\|x\|2 2 is the only function which is dual to itself. As is proved in [Ba1] (the equality case appears in [AKM]), Theorem 1.2 Let f : Rn →[0, ∞) be an even function such that 0 < R f < ∞. Then, Z Rn f Z Rn f ◦≤(2π)n where equality holds exactly for gaussians. An operation related to Legendre transform is the Asplund sum. In a completely analogous way to the deﬁnition of the Asplund sum, we deﬁne the Asplund product of two functions f, g : Rn →[0, ∞) as f ⋆g(x) = sup x1+x2=x f(x1)f(x2). The Asplund product preserves log-concavity. Also, (f⋆g)◦= f ◦g◦, i.e. the dual to the Asplund product is simply the usual product of the dual functions. The Asplund product of log-concave functions is analogous to the Minkowski sum of convex bodies. Indeed, 1A ⋆1B = 1A+B for any A, B ⊂Rn. A central inequality connected with the Minkowski sum of two bodies A, B ⊂Rn and a parameter 0 ≤λ ≤1 is the Brunn-Minkowski inequality: V ol(λA + (1 −λ)B) ≥V ol(A)λV ol(B)1−λ. 3 The Brunn-Minkowski inequality is also known in the following equiv-alent formulation, V ol(A + B) 1 n ≥V ol(A) 1 n + V ol(B) 1 n for any A, B ⊂Rn. Deﬁne λ · f = f λ x λ . Note that f ⋆f = 2 · f for a log-concave f, and that (λ · f)◦= (f ◦)λ. The function λ · f is the analog of a λ-homothety of a convex body. The functional analog of the Brunn-Minkowski inequality is the Prekopa-Leindler inequality (e.g. [P]). In our terminology, it states that given f, g : Rn →[0, ∞) and 0 ≤λ ≤1, Z [λ · f] ⋆[(1 −λ) · g] ≥ Z f λ Z g 1−λ . (2) Therefore, the Prekopa-Leindler inequality constitutes a complete ana-log to the Brunn-Minkowski inequality for bodies, where the Minkowski sum of bodies is replaced by an Asplund product of functions (see also [AKM]). Here we prove the analog of the inverse Brunn-Minkowski in-equality (see [M2] or the book [P]), as follows. We denote (f ◦T)(x) = f(Tx). Theorem 1.3 Let f, g : Rn →[0, ∞) be even log-concave functions with f(0) = g(0) = 1. Then there exist Tf, Tg ∈SL(n) such that ˜ f = f ◦Tf and ˜ g = g ◦Tg satisfy Z ˜ f ⋆˜ g 1 n < C "Z ˜ f 1 n + Z ˜ g 1 n # where C > 0 is a universal constant, Tf depends solely on f, and Tg depends solely on g. Further inequalities which are the analogs to the quotient of sub-space theorem and related results are formulated and proved in Section 2. In Section 2 we also prove Theorem 1.1 and Theorem 1.3. The main tool in the proofs of these isomorphic results is a method of attaching a convex body to any log-concave function, which is due to Ball [Ba2]. In Section 3 a functional analog of the mean width is introduced. We show that an analog of the Urysohn inequality holds in this set-ting. Throughout this paper, the letters c, C, c1, c′ etc. denote positive universal constants whose value is not necessarily the same in various appearances. A ≈B means that cA < B < CA for some universal constants c and C. A convex body is a convex set with a non-empty interior in Rn. 4 2 Convex bodies Let f : Rn →R be an even log-concave function. For x ∈Rn deﬁne ∥x∥f = Z ∞ 0 f(rx)rn−1dr −1 n . By [Ba2], ∥· ∥f is a norm on Rn. Denote its unit ball by Kf, the convex body that is associated with f. Then Kf is convex, centrally-symmetric, and V ol(Kf) = 1 n Z Sn−1 Z ∞ 0 f(rx)rn−1drdx = 1 n Z Rn f where Sn−1 = ∂D. Next, we shall elaborate on some connections between the body Kf and the log-concave function f. We start with a one dimensional lemma, in the spirit of the Laplace method. Recall (e.g. [R]) that if g : R →[0, ∞] is convex, then its left and right derivatives, denoted here as gL and gR, exist whenever g is ﬁnite. The function ϕ(t) = g(t) −n log t is convex in (0, ∞), and if g ̸≡Const then ϕ(t) t→0,∞ − →∞. By strict convexity, there exists a unique critical point t0 of ϕ such that g(t) −n log t is non-increasing for t < t0 and non-decreasing for t > t0. Note that it is possible that g(t) = ∞for t ≥t0, however, lim t→t− 0 [g(t) −n log t] = inf t∈R [g(t) −n log t] . Lemma 2.1 Let g : [0, ∞) →[0, ∞] be a non-decreasing convex func-tion such that g(0) = 0 and g ̸≡0. Denote M = supt>0 e−g(t)tn, and let t0 be the corresponding (unique) critical point. Then, M t0 n + 1 ≤ Z ∞ 0 e−g(t)tndt < cM t0 √n g(t0) ≤n, g(2t0) ≥n and g(ln) ≥(l −1)n for any l > 1. In addition, Z ∞ 5t0 e−g(t)tndt < e−2n Z ∞ 0 e−g(t)tndt. Proof: The left-most inequality is straightforward: since g(t) is non-decreasing, Z ∞ 0 e−g(t)tndt ≥e −limt→t− 0 g(t) Z t0 0 tndt = M t0 n + 1. 5 To prove the right-most inequality, recall that t0 is a critical point of the convex function ϕ(t) = g(t) −n log t. Hence ϕL(t0) ≤0 ≤ϕR(t0). We conclude that gL(t0) ≤ n t0 ≤gR(t0) and g(t0) + n t0 (t −t0) is a supporting line to g at t0. Since g is convex, g(t) ≥g(t0) + n t0 (t −t0) for every t, and Z ∞ 0 tne−g(t)dt < en−g(t0) Z ∞ 0 tne−nt t0 dt = en−g(t0) t0 n n+1 Z ∞ 0 tne−tdt = e−g(t0)tn 0 enn! nn t0 n ≈M t0 √n. Additionally, for t < t0, we have gR(t) ≤n t0 , and hence g(t0) ≤g(0) + R t0 0 n t0 = n. Also, g(2t0) ≥g(t0) + n t0 (2t0 −t0) ≥n. The estimate for g(lt0) follows the same argument. The last assertion follows from Z ∞ 5t0 e−g(t)tndt ≤en−g(t0) Z ∞ 5t0 tne−tn t0 dt << e−2nMt0. □ A convex function is diﬀerentiable almost everywhere (e.g. [Sch]). Yet, we still need a notion of a gradient for the relatively few non-smooth points. For a convex function g we deﬁne its gradient in a non-smooth point x (see e.g. [Sch]) as ∇g(x) = {y ∈Rn; g(z) ≥ g(x0) + ⟨y, z −x⟩}. For an even log-concave function f deﬁne ¯ Kf = {x ∈Rn; ⟨∇(−logf)(x), x⟩≤n −1} where for a non-smooth point x, the condition ⟨∇(−logf)(x), x⟩≤n−1 should be understood as ∃y ∈∇(−log f)(x), ⟨y, x⟩≤n−1. Deﬁne also ¯ ¯ Kf = {x ∈Rn; f(x) > e−n}. Then ¯ ¯ Kf is clearly convex, but ¯ Kf is not necessarily convex. Never-theless, we show that Kf, ¯ Kf and ¯ ¯ Kf are close to each other. The radial function of a convex body K in direction θ is r(K, θ) = sup{r > 0; rθ ∈K}. Lemma 2.2 Assume that f : Rn →[0, ∞) is an even log-concave function with f(0) = 1. Then, Kf ⊂¯ Kf ⊂¯ ¯ Kf ⊂cKf for some universal constant c > 0. 6 Proof: Fix θ ∈Sn−1 and let g(r) = −log f(rθ). If g ≡0 then r(Kf, θ) = r( ¯ Kf, θ) = r( ¯ ¯ Kf, θ) = ∞. Otherwise, denote M = supt>0 e−g(t)tn−1, and let t0 be the corresponding critical point. By Lemma 2.1, r(Kf, θ) ≈(Mt0)1/n = e−g(t0)tn 0 1/n ≈t0 and actually, r(Kf, θ) < t0. On the other hand, since gL(t0)t0 ≤ n −1 ≤gR(t0)t0, we have r( ¯ Kf, θ) = t0, and since t0 ≤g−1(n) ≤2t0, then also t0 ≤r( ¯ ¯ Kf, θ) ≤2t0. □ Corollary 2.3 Let f : Rn →[0, ∞) be an even log-concave function with f(0) = 1. Let E ⊂Rn be a λn-dimensional subspace, for some 0 < λ < 1. Then c1λKf ∩E ⊂Kf\|E ⊂c2Kf ∩E where f\|E is the restriction of f to the subspace E and c1, c2 > 0 are universal constants. Proof: By the log-concavity of f, ¯ ¯ Kf ∩E = {x ∈E; f(x) > e−n} ⊂c λ{x ∈E; f(x) > e−λn} = c λ ¯ ¯ Kf\|E. According to Lemma 2.2, Kf\|E ∩E ⊂c ¯ ¯ Kf\|E ⊂c ¯ ¯ Kf ∩E ⊂c′ λ ¯ ¯ Kf\|E ⊂c′′ λ Kf\|E and since c ¯ ¯ Kf ⊂Kf ⊂c′ ¯ ¯ Kf, the corollary follows. □ By a polar integration of the last inequality in Lemma 2.1, we obtain the following: Corollary 2.4 Let f : Rn →[0, ∞) be an even log-concave function with f(0) = 1 and a ﬁnite integral. Then Z cKf f ≥ 1 −e−2n Z Rn f. Next, we should exhibit a connection between Kf ◦and K◦ f. Proposition 2.5 Assume that f : Rn →R is even and log-concave. Then, c1nK◦ f ⊂Kf ◦⊂c2nK◦ f where c1, c2 > 0 are universal constants. 7 Proof: Since (cf)◦= 1 cf ◦and Kcf = c1/nKf, multiplying f by a scalar if needed, we may assume that f(0) = 1. Denote g = −log f. Assume ﬁrst that g is smooth and strictly convex. A crucial simple observation is that ∇L(g)(x) = (∇g)−1(x) (e.g. [R]). Hence, (∇g) ¯ Kf = {(∇g)x; ⟨x, (∇g)x⟩≤n−1} = {x; ⟨(∇g)−1x, x⟩≤n−1} = ¯ Kf ◦. Let x0 ∈∂¯ Kf. Then, ⟨x0, ∇g(x0)⟩= n −1. Denote by ∥y∥= supx∈¯ Kf ⟨x, y⟩, the norm that ¯ Kf ◦is its unit ball. Then, x0 ∈∂¯ Kf ⇒ ∥∇g(x0)∥≥n −1 and hence (n −1) ¯ Kf ◦⊂(∇g) ¯ Kf = ¯ Kf ◦. By Lemma 2.2, cnK◦ f ⊂(n −1) ¯ Kf ◦⊂¯ Kf ◦⊂c′Kf ◦. Regarding the opposite inclusion, since g is convex, for any y ∈Rn, g(y) ≥g(x0) + ⟨∇g(x0), y −x0⟩= ⟨∇g(x0), y⟩+ g(x0) + 1 −n. If furthermore y ∈¯ ¯ Kf then g(y) ≤n and ⟨∇g(x0), y⟩≤g(y) + n −1 −g(x0) < 2n −1. Hence ∇g(x0) ∈2n ¯ ¯ Kf ◦ and ¯ Kf ◦⊂2n ¯ ¯ Kf ◦ . An application of Lemma 2.2 concludes the proof under the assumption that g is smooth and strictly convex. For an arbitrary function, an approximation argu-ment is needed. For instance, we may deﬁne fε = f ⋆e−ε\|x\|2 e−ε\|x\|2. Then fε is smooth and strictly log-concave for any ε > 0. If ε is small enough, the bodies Kfε, Kf ◦ ε are close to Kf, Kf ◦, and the proposition follows. □ Proof of Theorem 1.1: By Proposition 2.5, Z Rn f Z Rn f ◦ 1 n = n2V ol(Kf)V ol(Kf ◦) 1 n ≈ V ol(Kf)V ol(cnK◦ f) 1 n and by Santal´ o’s inequality and its converse (e.g. [MeP, BM]), Z f Z f ◦ 1 n ≈nV ol(D) 2 n ≈c. □ The reverse Brunn-Minkowski inequality for convex bodies (see [M2]) is the following theorem: 8 Theorem 2.6 Let K, P ⊂Rn be centrally-symmetric convex bodies. Then there exist invertible linear transformations TK, TP of determi-nant one, such that ˜ K = TK(K), ˜ P = TP (P) satisfy V ol( ˜ K + ˜ P) 1 n < C h V ol( ˜ K) 1 n + V ol( ˜ P) 1 n i where C > 0 is a numerical constant, TK depends solely on K and TP depends solely on P. Lemma 2.7 Let f, g be even log-concave functions with f(0) = g(0) = 1. Then, c1Kf⋆g ⊂Kf + Kg ⊂c2Kf⋆g where c1, c2 > 0 are numerical constants. Proof: x ∈ ¯ ¯ Kf⋆g implies that there exists x1 + x2 = x with f(x1)g(x2) ≥e−n. Since the functions are not larger than one, nec-essarily x1 ∈¯ ¯ Kf and x2 ∈¯ ¯ Kg, hence ¯ ¯ Kf⋆g ⊂¯ ¯ Kf + ¯ ¯ Kg. Combining this with Lemma 2.2 we conclude the left-most inclusion. The other inclusion follows from the fact that for any x ∈¯ ¯ Kf + ¯ ¯ Kg we have that (f ⋆g)(x) ≥e−2n. From Lemma 2.1 it follows that ¯ ¯ Kf + ¯ ¯ Kg ⊂3 ¯ ¯ Kf⋆g. □ Proof of Theorem 1.3: Note that for any T a linear transforma-tion, Kf◦T = T −1(Kf). By Theorem 1.3 we may choose Tf, Tg, linear transformations of determinant one, such that K ˜ f and K˜ g satisfy V ol K ˜ f + K˜ g 1 n < C V ol K ˜ f 1 n + V ol (K˜ g) 1 n . According to Lemma 2.7, Z ˜ f ⋆˜ g 1 n = n 1 n V ol K ˜ f⋆˜ g 1 n ≈V ol K ˜ f + K˜ g 1 n < C V ol K ˜ f 1 n + V ol (K˜ g) 1 n ≤C "Z ˜ f 1 n + Z ˜ g 1 n # . □ Given two functions f, g : Rn →[0, ∞) with ﬁnite mass, we say that f ≺α g if there exists a set A ⊂Rn such that for any x ∈A, f 1 n (x) ≤eg 1 n x α and R A f > 1 −e−2n R f. We say that f ∼α g if f ≺α g and g ≺α f. If f ∼α g for α being a numerical constant, we say that f and g are “roughly-isomorphic”. 9 Lemma 2.8 Let f, g : Rn →[0, ∞) be even log-concave functions with f(0) = g(0) = 1 and ﬁnite, positive integrals. Then for any α > 1, Kf ⊂c1αKg = ⇒ f ≺α g = ⇒ Kf ⊂c2αKg where c1, c2 > 0 are universal constants. Proof: Assume that Kf ⊂cαKg. By Corollary 2.4, Z Rn\c′ ¯ ¯ Kf f < e−2n Z Rn f. Denote A = c′ ¯ ¯ Kf. If c, c′ > 0 are chosen properly, for any x ∈A we have that x α ∈¯ ¯ Kg, and hence g x α 1 n ≥1 e ≥1 ef(x) 1 n . Therefore f ≺α g. Regarding the other statement, assume that f ≺α g and let A be the corresponding witness set. If x ∈1 α h ¯ ¯ Kf ∩A i then g(x) 1 n ≥ 1 ef(αx) 1 n ≥ 1 e2 , and by Lemma 2.1 we get that x ∈2 ¯ ¯ Kg. Since ¯ ¯ Kg is a convex set, we conclude that conv ¯ ¯ Kf ∩A ⊂2α ¯ ¯ Kg. It remains to show that c′ ¯ ¯ Kf ⊂conv ¯ ¯ Kf ∩A . This would follow if we prove that V ol ¯ ¯ Kf ∩A > 1 −e−n 2 V ol ¯ ¯ Kf (e.g. Lemma 2.2 in [K2]). Finally, note that R A f > 1 −e−2n R f and that f(x) > e−n for any x ∈¯ ¯ Kf. We conclude that e−nV ol ¯ ¯ Kf \ A < Z Rn\A f < e−2n Z Rn f and hence V ol ¯ ¯ Kf \ A < e−n Z f = ne−nV ol(Kf) < e−n 2 V ol( ¯ ¯ Kf). □ Lemma 2.8 implies that if Kf = Kg, then f ∼c g for some universal c > 0. In particular, if Kf is a Euclidean ball, then f is roughly-isomorphic to a gaussian. We may now formulate more analogs of isomorphic results from the asymptotic theory of convex bodies. It is known (see [M3]) that given a centrally-symmetric convex body K ⊂ Rn, there exists ˜ K, a linear image of K, and two rotations U1, U2 ∈ O(n) such that if we deﬁne T = U1( ˜ K) + ˜ K and P = U2(T) ∩T, then c1D ⊂P ⊂c2D for some universal c1, c2 > 0. The functional analog is presented below. 10 Proposition 2.9 Let f : Rn →[0, ∞) be an even log-concave function with f(0) = 1. Then there exists ˜ f(= f ◦Tf) a linear image of f and two rotations U1, U2 ∈O(n) such that if g = ˜ f ◦U1 ⋆˜ f and h = (g ◦U2) · g, then h ∼C G where G(x) = e−\|x\|2 2 is the standard gaussian, and C > 0 is a numerical constant. Proof: By Lemma 2.7, Kf⋆g is close to Kf + Kg. It is equally easy to realize that Kfg is close to Kf ∩Kg in the same sense. Using the corresponding result for convex bodies, we may conclude that c1D ⊂ Kh ⊂c2D for some universal constants c1, c2 > 0. The proposition follows by Lemma 2.8. □. Milman’s quotient of subspace theorem [M1] is the following state-ment. Theorem 2.10 Let K ⊂Rn be a convex centrally-symmetric body. Then there exist subspaces E ⊂F ⊂Rn with dim(E) > n/2 and an ellipsoid E ⊂E such that c1E ⊂ProjE(K ∩F) ⊂c2E where c1, c2 > 0 are universal constants. Let E ⊂Rn be a subspace. Since (f\|E)◦(x) = supy∈E⊥f ◦(x + y), we naturally deﬁne ProjE(f) = supy∈E⊥f(x + y). Assume that dim(E) = ⌊n 2 ⌋. Note that according to Corollary 2.3, c1Kf\|E ⊂Kf ∩ E ⊂c2Kf\|E for some universal constants c1, c2 > 0. We can now formulate the functional analog of the Quotient of subspace theorem. The proof is omitted, as it follows from Theorem 2.10 in a similar way to the previous proofs. Proposition 2.11 Let f : Rn →[0, ∞) be an even log-concave func-tion with 0 < R f < ∞. Then there exist two subspaces E ⊂F ⊂Rn such that dim(E) > n 2 and ProjE(f\|F ) ∼c G where G is some gaussian measure, and c > 0 is a numerical constant. Remark. There is nothing special about the dimension 1 2n. For any 0 < λ < 1 one may ﬁnd subspaces E ⊂F ⊂Rn such that dim(E) > λn and the conclusion of Proposition 2.11 holds with a constant c(λ) that depends solely on λ. This follows from a corresponding sharpening of Theorem 2.10 (see [M1]). 11 3 Urysohn inequality Let K, T ⊂Rn be two convex, centrally-symmetric bodies. A classical theorem due to Minkowski (e.g. [Sch]) states that V ol(K + λT) is a polynomial in λ. In particular, V ol(K + εT) = V ol(K) + εnV (T, 1; K, n −1) + O(ε2) where V (T, 1; K, n−1) is the corresponding mixed volume (e.g. [Sch]). Let us deﬁne an analogous quantity for the log-concave case. If f, H : Rn →[0, ∞) are even log-concave functions of ﬁnite positive masses, we deﬁne VH(f) = lim ε→0+ R H ⋆[ε · f] − R H ε . This limit always exists, because by the Prekopa-Leindler inequality R H ⋆[ε · f] is a log-concave function of ε. Denote G(x) = e−\|x\|2 2 , the standard gaussian, and consider the case of VG(f), which may be viewed as the “mean width” of a log-concave function, up to some normalization. We denote M ∗(f) = 2VG(f) n R G = VG(f) n 2 (2π) n 2 . For a centrally-symmetric convex body K ⊂Rn, the usual deﬁnition is M ∗(K) = R Sn−1 supy∈K⟨x, y⟩dσ(x) where σ is the unique rotation invariant probability measure on Sn−1. In the case f = 1K, a straight-forward calculation yields VG(1K) = lim ε→0+ R Rn e−d(x,εK)2 2 − R Rn e−\|x\|2 2 ε = (2π) n−1 2 nκn κn−1 M ∗(K) where d(x, K) = infy∈K \|x −y\| and κm is the volume of the unit ball in Rm. We conclude that M ∗(K) = cnM ∗(1K) for some normalization constant cn ≈√n. Next we present an analytic formula for the mean width of a smooth log-concave function. Lemma 3.1 Let f : Rn →[0, ∞) be a log-concave function, strictly log-concave on its support, with continuous second derivatives, such that 0 < R f < ∞, and that satisﬁes Hess(L(−log f))(x) ≤Kexp(K\|x\|)Id in the sense of positive matrices for some K = K(f). Then, M ∗(f) = 2 n(2π)n/2 Z Rn e−\|y\|2 2 △(−log f ◦) + \|y\|2Dr log f ◦(y) \|y\| dy. 12 Proof: Denote g = −log f. Then g is strictly convex and smooth on its support, hence (e.g. [R]) g(x) = ⟨x, ∇g(x)⟩−Lg(∇g(x)) and Z Rn f(x)dx = Z Rn e−⟨x,∇g(x)⟩+Lg(∇g(x))dx. Substituting y = (∇g)x and recalling that ∇Lg = (∇g)−1 we get Z Rn f(x)dx = Z Rn e−⟨y,(∇Lg)y⟩+Lg(y)det(Hess(Lg))dy. Denote the radial derivative by Dr (i.e. Dr(g)(x) = ⟨∇g(x), x \|x\|⟩). Since Dr Lg(y) \|y\| = ⟨∇Lg(y), y \|y\| ⟩\|y\|−Lg(y) \|y\|2 , we obtain Z Rn f(x)dx = Z Rn e−(\|y\|2Dr Lg(y) \|y\| )det(Hess(Lg))dy. (3) This is true for any smooth and strictly log-concave function f. The function G ⋆[ε · f] is also smooth and strictly log-concave, and by (3), Z G ⋆[ε · f] = Z Rn e − \|y\|2Dr 1 2 \|y\|2+εLg(y) \|y\| det(Id + εHess(Lg))dy. (4) We would like to ﬁnd an expression for M ∗(f). Since det(Id+εHess(Lg)) = 1+εTr(Hess(Lg))+O(ε2) = 1+ε△Lg+O(ε2), e − \|y\|2Dr 1 2 \|y\|2+εLg(y) \|y\| = e−\|y\|2 2 1 −ε\|y\|2Dr Lg(y) \|y\| + O(ε2) , the integrand in (4) is e−\|y\|2 2 + εe−\|y\|2 2 △Lg −\|y\|2Dr Lg(y) \|y\| + O(ε2). Using our assumption on the growth of Hess(Lg), by the dominated convergence theorem, VG(f) = Z Rn e−\|y\|2 2 △Lg −\|y\|2Dr Lg(y) \|y\| dy. □ From lemma 3.1 it follows that M ∗(G) = 1. It also follows that M ∗(f ⋆g) = M ∗(f) + M ∗(g) and M ∗(λ · f) = λM ∗(f) for λ > 0 and f that satisﬁes the requirements of Lemma 3.1. However, glancing at Lemma 3.1, it is not obvious why M ∗(f) should be positive. This follows from our next proposition, which is the functional analog of Urysohn inequality. 13 Proposition 3.2 Let f : Rn →[0, ∞) be any even log-concave func-tion such that R f = R G. Then M ∗(f) ≥M ∗(G). Proof: By the Prekopa-Leindler inequality, Z G⋆[ε·f] ≥ Z 1 1 −ε · G 1−ε Z f ε = Z e−1−ε 2 \|x\|2dx 1−ε Z f ε and computing the gaussian integral we obtain Z G ⋆[ε · f] ≥ 2π 1 −ε n(1−ε) 2 Z f ε . Since R f = (2π) n 2 we conclude that VG(f) ≥lim ε→0+ 2π 1−ε n(1−ε) 2 (2π) nε 2 −(2π) n 2 ε = (2π) n 2 n 2 and the proposition is proved. □ Remarks. 1. Note that if f(0) = 1 and R f = R G, then M ∗(f) > cM ∗(1Kf ) for some universal constant c > 0. Indeed, if M ∗(1Kf ) ≈M ∗(G) = 1 this follows from Proposition 3.2. Otherwise M ∗(1Kf ) > C, and since f ≥e−n1 ¯ ¯ Kf we conclude that M ∗(f) ≥M ∗ e−n1 ¯ ¯ Kf = M ∗(1 ¯ ¯ Kf ) −2 > cM ∗(1Kf ). 2. Formally, the results in this paper are formulated and proved for even functions. Yet, the evenness is never essentially used, and in fact the results hold for an arbitrary log-concave function, provided that the origin is suitably chosen. If f : Rn →[0, ∞) is log-concave and has a ﬁnite, positive mass, then it must be a bounded function, and its supremum is attained at some point. All of our results hold, with the same proofs, for log-concave functions that reach their maximum at the origin (note that the dual function also reaches its maximum at the origin). 14 References [Ar] Arnold, V. I., Mathematical methods of classical mechanics. Translated from the Russian by K. Vogtmann and A. Weinstein. Second edition. Graduate Texts in Mathematics, 60. Springer-Verlag, New York, 1989. [AKM] Artstein, S., Klartag B., V.D. Milman, The Santal´ o point of a function, and a functional form of Santal´ o inequality, preprint. [Ba1] Ball, K., PhD dissertation, Cambridge. [Ba2] Ball, K. Logarithmically concave functions and sections of con-vex sets in Rn. Studia Math. 88, no. 1 (1988) 69–84. [BN] Bobkov, S.G.; Nazarov, F.L. On convex bodies and log-concave probability measures with unconditional basis. Milman, V. D. (ed.) et al., Geom. Aspects of Funct. Anal. Israel seminar 2001-2002. Springer, Lect. Notes Math. 1807 (2003) 53-69. [Bo] Borell, C., Convex measures on locally convex spaces. Ark. Mat. 12 (1974) 239–252. [Bo2] Borell, C., Convex set functions in d-space. Period. Math. Hun-gar. 6, no. 2 (1975) 111–136. [BM] Bourgain, J.; Milman, V. D. New volume ratio properties for convex symmetric bodies in Rn. Invent. Math. 88, no. 2 (1987) 319–340. [K1] B. Klartag, An isomorphic version of the slicing problem, to ap-pear in J. Funct. Anal. [K2] B. Klartag, A geometric inequality and a low M estimate. Proc. Amer. Math. Soc. 132, no. 9 (2004) 2619–2628. [MeP] Meyer, Mathieu; Pajor, Alain, On Santal´s inequality. Geo-metric aspects of functional analysis (1987–88), Lecture Notes in Math., 1376, Springer, Berlin, (1989) 261–263. [M1] Milman, V.D. Almost Euclidean quotient spaces of subspaces of a ﬁnite-dimensional normed space. Proc. Amer. Math. Soc. 94, no. 3 (1985) 445–449. [M2] V.D. Milman, In´ egalit´ e de Brunn-Minkowski inverse et applica-tions ` a le th´ eorie locale des espaces norm´ es. C.R. Acad. Sci. Paris, Ser. I 302 (1986) 25–28. [M3] Milman, V. D. Isomorphic symmetrizations and geometric in-equalities. Geometric aspects of functional analysis (1986/87), Lecture Notes in Math., 1317, Springer, Berlin (1988) 107–131. [P] Pisier, Gilles, The volume of convex bodies and Banach space geometry. Cambridge Tracts in Mathematics, 94. Cambridge Uni-versity Press, Cambridge, 1989. 15 [R] Rockafellar, R. Tyrrell Convex analysis. Princeton Mathematical Series, No. 28 Princeton University Press, Princeton, N.J. 1970. [Sch] Schneider, Rolf Convex bodies: the Brunn-Minkowski theory. Encyclopedia of Mathematics and its Applications, 44. Cambridge University Press, Cambridge, 1993. 16
190576	https://math.libretexts.org/Bookshelves/Geometry/Euclidean_Plane_and_its_Relatives_(Petrunin)/14%3A_Affine_Geometry/14.01%3A_Affine_transformations	Skip to main content 14.1: Affine transformations Last updated : Sep 5, 2021 Save as PDF 14: Affine Geometry 14.2: Constructions Page ID : 23668 Anton Petrunin Pennsylvannia State University ( \newcommand{\kernel}{\mathrm{null}\,}) Affine geometry studies the so-called incidence structure of the Euclidean plane. The incidence structure sees only which points lie on which lines and nothing else; it does not directly see distances, angle measures, and many other things. A bijection from the Euclidean plane to itself is called affine transformation if it maps lines to lines; that is, the image of any line is a line. So we can say that affine geometry studies the properties of the Euclidean plane preserved under affine transformations. Exercise 14.1.1 Show that an affine transformation of the Euclidean plane sends any pair of parallel lines to a pair of parallel lines. Hint : Assume the two distinct lines ℓ and m are mapped to the intersecting lines ℓ′ and m′. Suppose that P′ denotes their point of intersection. Let P be the inverse image of P′. By the definition of affine map, it has to lie on both ℓ and m; that is, ℓ and m are intersecting. Hence the result. The observation below follows since the lines are defined using the metric only. Observation 14.1.1 Any motion of the Euclidean plane is an affine transformation. The following exercise provides more general examples of affine transformations. Exercise 14.1.2 The following maps of a coordinate plane to itself are affine transformations: (a) Shear map defined by (x,y)↦(x+k⋅y,y) for a constant k. (b) Scaling defined by (x,y)↦(a⋅x,a⋅y) for a constant a≠0. (c) x-scaling and y-scaling defined respectively by (x,y)↦(a⋅x,y), and (x,y)↦(x,a⋅y) for a constant a≠0. (d) A transformation defined by (x,y)↦(a⋅x+b⋅y+r,c⋅x+d⋅y+s) for constants a,b,c,d,r,z such that the matrix acbd is invertible. Hint : In each case check that the map is a bijection and apply Exercise 7.6.3 From the fundamental theorem of affine geometry (Theorem 14.3.1), it will follow that any affine transformation can be written in the form (d). Recall that points are collinear if they lie on one line. Exercise 14.1.3 Suppose P↦P′ is a bijection of the Euclidean plane that maps collinear triples of points to collinear triples. Show that P↦P′ maps noncollinear triples to noncollinear. Conclude that P↦P′ is an affine transformation. Hint : Choose a line (AB). Assume that X′∈(A′B′) for some X∉(AB). Since P↦P′ maps colliner points to collinear, the three lines (AB),(AX), and (BX) are mapped to (A′B′). Further, any line that connects a pair of points on these three lines is also mapped to (A′B′). Use it to show that the whole plane is mapped to (A′B′). The latter contradicts that the map is a bijection. By assumption, if X∈(AB), then X′∈(A′B′). Form above the converse holds as well. Use it to prove the second statement. 14: Affine Geometry 14.2: Constructions
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Your next lesson will play in 10 seconds 0:01 What are Aromatic Hydrocarbons 3:46 Examples 6:04 Uses 8:28 Lesson Summary QuizCourseView Video OnlySaveTimeline 122K views Aromatic Hydrocarbon Lessons ##### Acenaphthylene: Uses & Aromaticity ##### Trinitrotoluene Properties, Uses & Preparation ##### Phenyl Formula, Structure & Applications ##### 2,4-Dinitrophenylhydrazine \| Overview, Structure & Test What Are Aromatic Hydrocarbons? ------------------------------- The general aromatic hydrocarbons definition state that such compounds are organic molecules that possess aromatic characteristics and consist of no other atoms different from hydrogen and carbons. Coal and petroleum are the primary sources for aromatic hydrocarbons; these compounds are liberated when coal is burnt. Upon liberation, they disperse in the atmosphere and float as small particles in air. Many of these compounds are classified as toxic, which is why exposure to these compounds must be limited if not avoided. Properties of Aromatic Hydrocarbons Aromatic hydrocarbons are unique in having conjugated double bonds throughout their rings. The conjugated system in these compounds greatly enhances their stability through resonance, which is the delocalization of π electrons by distributing the electrons in the p orbitals over the adjacent atoms. The shared characteristics between all aromatic hydrocarbons are the following: They are made entirely of carbon and hydrogen atoms; they do not contain any other atom nor functional groups because they are hydrocarbons. The carbon atoms are linked together in a closed loop, forming a cyclic structure. All aromatic hydrocarbons are cyclic compounds. The atoms are bonding together through the mutual sharing of electrons, through covalent bonds. They all contain alternating single and double bonds within the ring structure. They are classified as unsaturated hydrocarbons; the presence of a pi bonds prevents the carbon atoms from bonding with the maximum number of hydrogen atoms. Most aromatic hydrocarbons have a pleasant aromatic scent. Their enjoyable aroma is how these compounds earned their name, aromatic. Most of the aromatic hydrocarbons are highly toxic and carcinogenic; they mutate the cells and cause cancer. Aromatic hydrocarbons are extremely flammable; they form flammable mixtures with air and ignite spontaneously in the presence of a heat source. These compounds are reactive towards electrophilic aromatic substitution reactions. These compounds are very stable and tend to be unreactive, which is why they are used as organic solvents. Aromatic Hydrocarbons Examples ------------------------------ The renowned aromatic hydrocarbons examples are benzene, toluene and naphthalene. The following subsections presents a list of aromatic compounds. Benzene Benzene is the simplest aromatic hydrocarbon; many other aromatic hydrocarbons are benzene-based, they contain a benzene ring. The molecular formula for benzene is C 6 H 6. The structure of benzene is presented in figure 1. Notice that the structure of the benzene molecule resembles a hexagon. The double bonds are not fixed at the same atoms; they are delocalized constantly from one atom to another through resonance. Figure 2 shows the two possible resonance structures for benzene. Benzene is primarily found in crude oil and petroleum products. This compound is used as a gasoline additive and as an organic solvent. The following list summarizes a few of benzene's many uses: It is used in the polymers industry in producing some resins and plastics. It is used as a gasoline additive to decrease the chances of premature ignition. Benzene used to be used in coffee due to its nice aroma long ago. But that stopped when scientist learned that benzene is a carcinogen. Figure 1: Benzene is the simplest aromatic hydrocarbon Figure 2: Benzene has two resonance structures Toluene Toluene, alternatively known as methylbenzene, is another example of an aromatic hydrocarbon. Figure 3 shows that toluene is basically a benzene ring with a methyl group −C H 3 attached to one of the carbons in the ring. The chemical formula for toluene is C 7 H 8. Like benzene, the π bonds in toluene delocalize constantly as well. Toluene is also present naturally in crude oil. This compound has similar uses as benzene; it is used as a nonpolar solvent, as an additive in gasoline and in producing plastics. Toluene is also used in the production of synthetic aromas, adhesives, and explosives. Figure 3 Toluene is an aromatic hydrocarbon that is made of a benzene ring and a methyl group Naphthalene Naphthalene is another common example of an aromatic hydrocarbon. Its chemical formula is C 10 H 8. Naphthalene structure is basically a double benzene ring, as shown in figure 4. Naphthalene is used primarily in eliminating bad odors in toilets, killing insects and as a precursor of various organic chemicals. Naphthalene is the main ingredient in mothballs, which are used in preventing the molding of clothing and to kill insects and moths. Figure 4: Naphthalene is made of two benzene rings Naturally Occurring Aromatic Compounds Petroleum and crude oil are not the sole source for aromatic hydrocarbons; there are other biological sources where aromatic compounds are present in great quantities. One such example is the DNA; DNA is composed of nucleotides bases. These nucleotides are aromatic compounds, as shown in figure 5. They are not considered as hydrocarbons because they contain functional groups and atoms different from carbons and hydrogens. Some aromatic compounds also occur naturally in plants as chlorophyll, which is the compound that helps absorb energy from light and make the photosynthesis process possible. Chlorophyll molecules give plants their green pigment. It can be confirmed that aromatic compounds are generally extremely stable. This explains how they occur naturally in the DNA and in chlorophyll. Figure 5: Some of the nucleotide bases in DNA Figure 6: Chlorophyll is an aromatic compound that is present in plants Vanillin is an aromatic compound that is used in vanilla extract. It consists of a phenol group, which is a benzene attached to a hydroxyl group -OH. In general, aromatic compounds are primarily used in the polymers and in the chemical synthesis industries. They are used as organic solvents because of their stability and their moderately unreactive nature. They are used in synthesizing many chemicals, especially benzene-based compounds, through electrophilic aromatic substitution reactions. Which are reactions that involve substituting a weakly electronegative atom (most of the cases hydrogen) with an electrophile, which is a molecule that has a tendency to accept electrons. Aromatic hydrocarbons, like toluene and benzene, are used in the manufacturing of plastics, dyes, medicines and pesticides. Lesson Summary -------------- Aromatic hydrocarbons are organic cyclic molecules that are made of carbon and hydrogen atoms. They are unsaturated hydrocarbons because they contain π bonds in their ring structure. These compounds are characterized by having conjugated double bonds. The electrons are constantly delocalized through resonance, which means that they are moving between the p orbitals of the parent carbon atoms. The electron delocalization makes these compounds extremely stable, which explains how they occur naturally in DNA as nucleotide bases and in chlorophyll. Aromatic hydrocarbons are classified as hydrocarbons because they are made of no other elements but carbon and hydrogen. The atoms in aromatic compounds are bonding together through covalent bonds. These compounds earned their title 'aromatic' because they are characterized by their distinct pleasant aromas. Generally, aromatic hydrocarbons are mainly found in crude oil and petroleum. They are extremely flammable and toxic compounds. Benzene is the simplest aromatic hydrocarbon; its structure resembles a hexagon with alternating double and single bonds. Toluene is the second simplest aromatic hydrocarbon. Toluene is basically a benzene ring combined with a methyl group. The following are examples of aromatic compounds: naphthalene, phenol, vanillin, purine, adenine, and guanine. Aromatic compounds are primarily used as organic solvents due to their stability. They are also used in various other applications; they are used to synthesize other chemicals by undergoing electrophilic aromatic substitution. They are also used in making plastics, medicines, dyes, and pesticides. Video Transcript What Are Aromatic Hydrocarbons? Once upon a time, smell was one of a chemist's best tools. Some chemists even went so far as to taste their chemicals! Some people think that the Swedish chemist Carl Scheele died because he happened to taste an especially poisonous chemical while working alone in his lab. Robert Bunsen, best known for his development of the Bunsen burner, was partial to smelling the noxious arsenic compounds he worked with. Just the smell of these compounds would make his hands and feet tingle, and at least once the smell turned his tongue black! It was during this era that chemists started using the term aromatic to describe certain carbon-based compounds with distinct odors. Thanks to the odors of compounds, like toluene and benzene (which apparently smell sweet), certain compounds made primarily of carbon and hydrogen are now called aromatic hydrocarbons. The definition of aromatic hydrocarbon is quite specific. Generally speaking, aromatic hydrocarbons are especially stable, unsaturated cyclic compounds made primarily of hydrogen and carbon atoms. Let's take a look at what it means when a compound is especially stable. I bet you already have a good idea. Check out these three hydrocarbon structures. Each structure contains six carbons. Which one do you think is more stable (meaning harder to break): A, B or C? Why do you think so? If you said C is more stable, well done! What made you think so? Was it the presence of double bonds? Or that it is a ringed compound? It turns out that both factor into the stability of aromatic hydrocarbons. Just to review, we refer to ringed structures sometimes made of more than one ring as cyclic compounds. When a hydrocarbon contains one or more double bonds, and therefore, does not contain the maximum number of hydrogens, it is known to be unsaturated. The unsaturated quality of aromatic hydrocarbons is special. The double bonds in these compounds are not fixed in one place; the electrons making up the double bonds are delocalized and can move from parent atom to parent atom. This delocalization happens in a rather organized fashion, so that double bonds are always alternating with single bonds. The phenomenon of delocalization or having more than one acceptable bond structure is called resonance. It lends to an average bond length in between that of a single bond and double bond. The cyclic nature of aromatic hydrocarbons allows these bonds to click and clack back and forth between atoms that are all in the same plane. I say atoms instead of carbons because sometimes the ring structure may include a nitrogen, oxygen or sulfur. Before we move on, let's look at delocalization in a tiny bit more detail. The electron orbitals involved in double bonds are called p orbitals. They extend at 90-degree angles from each atom in the ringed structure. The p orbitals in aromatic compounds overlap, creating a cloud that allows the electrons to move back and forth. Aromatic hydrocarbons are known to be hard to break in chemical reactions. Unlike their cousins the aliphatic hydrocarbons, aromatics don't readily undergo addition reactions in which an element is added to the structure. Examples Not that it's a popularity contest, but benzene is probably the most well-known aromatic hydrocarbon. Benzene has the formula C 6 H 6. It is a six-carbon ring; each carbon is bonded to two other carbons and one hydrogen. There are three double bonds in benzene - but these bonds are not fixed in place! The electrons in the bonds are delocalized and constantly are moving between the p orbitals of neighboring carbon atoms. The two different bond structures of benzene - together known as resonance structures - are often shown together with a double-headed arrow between them. This double-headed arrow lets us know that both of these structures are favorable and possible. Far more often, though, we see benzene depicted as a hexagon with a circle in the middle. This circle lets us know that the double bonds are moving between positions. Notice that there are no hydrogens drawn into this structure. It is assumed that they are there. Toluene is another example of an aromatic hydrocarbon. It's just like benzene, but with a methyl (CH 3) group attached in place of one of the carbons. The double bonds on the toluene ring are constantly moving, so toluene is represented with a circle in the middle. Naphthalene is an example of an aromatic compound made of two benzene rings fused together. Anthracene is made of three benzene rings fused together. Furan is an example of an aromatic hydrocarbon that contains an element other than carbon and hydrogen. Oxygen makes up part of this five-sided ring. If you've ever seen a structure like the one on the left before, then you've been in biology! Where do you recognize this from? What about the one on the right? Maybe DNA? DNA is made of nucleotide bases that are in turn made up of aromatic hydrocarbons. Purines (Guanine and Adenine) are made of two rings, one is hexagonal, one is a pentagon. Notice that there are four nitrogens that are part of the rings. Pyrimidines (Cytosine and Thymine) are made of one hexagonal ring that contains two nitrogen atoms. Uses Aromatic hydrocarbons are everywhere, literally. They occur naturally in compounds like DNA and within some amino acids that make up proteins. Chlorophyll, a pigment used by plants to absorb light, contains aromatic groups, and so do the heme groups that help our blood cells carry oxygen. The spicy compounds in hot peppers, ginger and black pepper are all aromatic compounds. Not so spicy (but delicious) compounds like vanillin (artificial vanilla) and nutmeg are aromatic compounds that contain a phenol group. A phenol is a benzene ring with an -OH group attached in place of a hydrogen. Many aromatic compounds are used as solvents to remove or thin out oil- or grease-based compounds. Toluene, for example, is an ingredient in paint thinners. Benzene is a gasoline additive that reduces knocking in engines. Benzene and toluene are widely used to make other chemicals including dyes and plastic products. Plastic products made from benzene range from PVC pipes to the thin plastic used in packaging. That Styrofoam that's keeping your to-go food warm? It's made of aromatic compounds. Aromatic compounds are abundant in medicines. Prozac, an antidepressant, is an aromatic compound. Phenol and its derivatives have quite an interesting history. Mentioned previously, a phenol consists of a benzene ring with an -OH group attached in place of a hydrogen. Phenol was originally used by Joseph Lister to make antiseptic pastes in the mid to late 1800s. Though the paste worked to an extent, patients had to endure the toxicity of phenol as a side effect. Their skin cracked, turned white and became numb. Today, much safer derivatives of phenol are used in medicine for their antimicrobial properties. The first plastics - Bakelite - were created from phenol by a man named Leo Baekeland in the early 1900s. Interestingly, Bakelite was first synthesized as an alternative to ivory to meet demands of billiards players. The fantastic properties of Bakelite launched it into many other uses, from buttons to furniture. Many modern plastics are still made from phenol derivatives. Lesson Summary Aromatic Hydrocarbons are especially stable, unsaturated cyclic compounds made primarily of hydrogen and carbon atoms. Cyclic Compounds are ringed structures sometimes made of more than one ring. Unsaturated hydrocarbons contain one or more double bonds, and therefore, do not contain the maximum number of hydrogens. Resonance occurs when a compound has more than one acceptable bond structure. Delocalization occurs when electrons can move between the p orbitals of parent atoms. Aromatic hydrocarbons are especially stable because they are both cyclic and unsaturated. The electrons that create the double bonds are delocalized and can move between parent atoms. Examples of aromatic hydrocarbons include benzene, toluene, purines and pyrimidines. Aromatic hydrocarbons occur naturally in DNA and in chlorophyll. They also are used by humans to make plastic, medicine, dyes and paint thinner. Learning Outcomes After this lesson, you should have the ability to: Define aromatic hydrocarbons Describe the properties of aromatic hydrocarbons and explain what makes them especially stable Recall what resonance and delocalization refer to Identify natural aromatic hydrocarbons and manufactured goods made from them Register to view this lesson Are you a student or a teacher? I am a student I am a teacher Unlock your education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now. Become a member Already a member? Log in Go back Resources created by teachers for teachers Over 30,000 video lessons & teaching resources—all in one place. Video lessons Quizzes and worksheets Classroom integration Lesson plans I would definitely recommend Study.com to my colleagues. It’s like a teacher waved a magic wand and did the work for me. I feel like it’s a lifeline. Jennifer B. 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Try it now Science 102: Principles of Physical Science 19 chapters 151 lessons 13 flashcard sets Chapter 1 Force, Mass & Newton's Laws of Motion Applied Force \| Definition, Types & Examples 7:02 min Net Force: Definition and Calculations 6:16 min Newton's First Law of Motion \| Overview & Examples 8:25 min Newton's Second Law of Motion \| Overview & Examples 8:04 min Acceleration \| Definition, Formula & Examples 6:56 min Newton's Third Law of Motion \| Definition, Application & Examples 4:24 min Newton's Law of Gravitation \| Overview, Calculation & Examples 6:35 min Uniform Circular Motion \| Formula & Examples 7:00 min Mass to Weight Conversion \| Calculation & Examples 5:44 min Chapter 2 Thermodynamics & Energy Work \| Definition, Formula & Examples 4:38 min Kinetic vs Potential Energy \| Definition, Differences & Examples 5:59 min Power in Physics \| Definition, Units & Formula 5:24 min Linear Momentum \| Definition, Formula & Examples 4:58 min Mechanical Equivalent of Heat 6:09 min First Law of Thermodynamics \| Definition, Formula & Examples 7:42 min Second Law of Thermodynamics: Entropy and Systems 6:21 min Chapter 3 Principles of Thermodynamics Heat Transfer \| Definition, Diagram & Types 8:09 min Specific Heat Capacity \| Definition, Formula & Calculation 7:30 min Heat Measurement of Calorimeter \| Unit & Substances 8:49 min Phase Changes and Heating Curves 5:38 min Heat Transfer & Phase Changes 4:52 min Enthalpy: Energy Transfer in Physical and Chemical Processes 8:50 min Enthalpy of Reaction \| Formula, Equations & Examples 8:44 min Entropy Change Overview, Factors & Prediction 7:29 min Enthalpy, Entropy & Free Energy \| Overview & Relationship 8:33 min Chapter 4 Sound Waves & Optics Vibrations & Waves \| Overview & Examples 6:26 min Wave Parameters \| Amplitude, Frequency & Period 6:54 min Speed of Sound Formula, Equation & Examples 8:25 min Pitch and Volume in Sound Waves 7:36 min Constructive & Destructive Interference \| Overview, Differences & Examples 8:46 min Electromagnetic Waves \| Definition, Composition & Types 6:25 min Reflection: Angle of Incidence and Curved Surfaces 6:14 min Mirrors: Difference Between Plane & Spherical 4:20 min Thin Lens Equation \| Formula, Problems & Examples 6:16 min Refraction & Dispersion: Definition, Snell's Law & Index of Refraction 9:08 min Diffraction \| Light, Sound & Wavelength 6:33 min Chapter 5 Basics of Electric Power Electric Power Definition, Formula & Examples 6:41 min Electric Circuit \| Definition, Components & Types 9:38 min Electric Current \| Definition, Types & Examples 7:59 min Ohm's Law \| Relationship Between Voltage, Current & Resistance 7:17 min Conductors & Insulators \| Definition, Characteristics & Examples 6:38 min Coulomb's Law \| Overview, Formula & Equation 8:04 min Electric Potential: Charge Collections and Volt Unit 4:38 min Chapter 6 Magnetic Forces & Fields Magnetism Definition, Causes & Examples 6:09 min Magnetic Field \| Overview & Ferromagnetic Materials 6:47 min Faraday's Law \| Formula, Applications & Examples 8:12 min Technological Applications of Electromagnetism 4:05 min Chapter 7 Experimental Chemistry in the Laboratory Scientific Thinking Definition, Method & Examples 6:04 min Scientific Method \| Definition, Steps & Examples 8:43 min Unit Conversion and Dimensional Analysis 10:29 min Chemistry Lab Equipment \| Names, Types & Uses 8:44 min Metric System \| Measurements, Conversion & Units 8:47 min Significant Figures and Scientific Notation \| Rules & Examples 10:12 min Chapter 8 Properties of Matter in Chemistry Physical & Chemical Properties \| Overview, Difference & Examples 7:22 min States of Matter \| Characteristics, Comparison & Examples 5:15 min Chemical Reaction Overview, Types & Examples 5:10 min Endothermic vs. Exothermic Reactions \| Process & Examples 5:11 min Chapter 9 Compounds & Concentration Molarity vs Molality \| Comparison, Formula & Examples 6:03 min Naming Ionic Compounds \| Binary, Transition Metals & Polyatomic 8:11 min Ionic Compound \| Formation, Properties & Examples 5:34 min Chapter 10 Basics of Chemical Bonding Octet Rule & Lewis Structure \| Definition & the Periodic Table 4:19 min Ions \| Formation, Types & Charge 4:45 min Lewis Structures \| Overview, Structural Formula & Examples 5:09 min Lewis Dot Structures: Polyatomic Ions 9:36 min Lewis Dot Structures \| Drawing, Resonance & Examples 7:01 min Ionic Character \| Bond Polarity, Electronegativity & Trend 4:56 min Chemical Bond \| Definition, Types & Formation 5:08 min VSEPR Theory \| Chart & Model 9:09 min Dipole & Dipole Moment \| What is Molecular Polarity? 8:31 min Intermolecular Forces \| Types, Effects & Examples 6:12 min London Dispersion Forces: Causes, Importance & Examples 4:31 min Determining Orbital Hybridization \| Overview & Examples 5:31 min Molecular Orbital Theory \| Concept & Diagrams 4:35 min Chapter 11 Properties of Gases & Gas Laws Kinetic Molecular Theory of Gases \| Properties & Examples 6:49 min Dalton's Law of Partial Pressure \| Formula & Examples 8:39 min Graham's Law of Diffusion vs. Effusion \| Formula & Differences 6:57 min Avogadro's Law & Molar Volume \| Overview, Formula & Units 9:09 min Boyle's Law \| Overview, Equation & Examples 6:48 min Charles' Law \| Formula, Units & Application 8:13 min Gay-Lussac's Gas Law \| Equation and Examples 6:42 min Ideal Gas Law & Constant \| Formula & Examples 8:03 min Pressure, Temperature & Volume of a Gas \| Formula & Calculation 3:42 min Ideal vs. Real Gas Laws \| Differences, Formula & Assumptions 7:39 min Van der Waals Equation \| Definition & Examples 6:48 min Chapter 12 Kinetics in Chemistry Rate of Reaction \| Definition, Factors & Formula 8:44 min Rate Constant & Rate Law \| Definition, Differences & Examples 6:35 min Rate of a Chemical Reaction: Effect of Temperature 4:00 min Activation Energy and Catalysts \| Definition, Relation & Examples 5:29 min Reaction Mechanisms and The Rate Determining Step 4:34 min Collision Theory \| Definition, Examples & Significance 6:07 min Chemical Kinetics, Reaction Rate Constant & Equilibrium Constant 9:34 min Chapter 13 The Periodic Table Periodic Table Groups vs. Periods \| Properties & History 7:06 min Main Group Elements & Transition Metals \| Definition & Properties 6:31 min Valence Electrons \| Definition, Role & Examples 9:34 min Atomic & Ionic Radius Trend \| Definition, Differences & Chart 6:48 min Ionization Energy \| Definition, Trends & Factors 7:44 min Electronegativity \| Definition, Periodic Trend & Example 6:17 min Diagonal Relationship, Boiling Point & Metallic Character Trend 5:47 min Chapter 14 Atoms & Atomic Theory Atom \| Definition, Structure & Examples 4:09 min Atomic & Mass Number \| Overview & Difference 9:15 min Early Atomic Theory \| History, Scientists & Models 6:35 min Atomic Mass vs Atomic Number vs Atomic Weight \| Meaning & Example 7:29 min Moles to Atoms \| Formula, Calculation & Avogadro's Number 9:15 min Electron Configuration \| Overview, Levels & Patterns 10:40 min Quantum Numbers on the Periodic Table \| Definition & Overview 10:05 min Using Atoms and Ions to Determine Molecular Formulas 9:49 min Bohr Atomic Model \| Overview & Examples 8:41 min Chapter 15 Understanding Stoichiometry Molarity & Stoichiometry \| Definition, Formula & Calculation 11:07 min Excess & Limiting Reactants \| Formula, Steps & Example 7:12 min Mole Ratio \| Definition, Formula & Examples 10:13 min Percent Composition \| Formula & Calculation 11:04 min Percent Yield Formula \| How to Calculate Yield 7:07 min Balancing Chemical Equations \| Steps & Examples 6:07 min Chapter 16 Radioactivity Nuclear Reaction Definition, Types & Equations 5:30 min Fusion, Fission, Carbon Dating, Tracers & Imaging: Applications of Nuclear Chemistry 6:41 min Mass Defect Definition, Formula & Examples 7:44 min Half-life & Radioactive Decay \| Equation, Calculations & Graphs 6:33 min Balancing Nuclear Equations \| Steps & Examples 7:37 min Radioactive Decay \| Definition, Occurrence & Types 8:26 min Chapter 17 Acid-Base Chemical Reactions The Arrhenius of Acids & Bases \| Lists, Concept & Properties 7:53 min The Bronsted-Lowry and Lewis Definition of Acids and Bases 6:15 min Acid-Base Neutralization Reactions \| Overview & Examples 5:43 min Autoionization & Dissociation Constant of Water 6:41 min The pH Scale \| Definition, Equation & Examples 9:49 min Strong & Weak Acids & Bases \| Definition & Examples 7:45 min Oxidation Numbers \| Rules, Equations & Examples 6:50 min Oxidizing vs. Reducing Agents \| Overview & Identification 6:38 min Electrochemical Cells and Electrochemistry 4:22 min Anode vs. Cathode in Electrochemical Cells \| Reaction & Notation 9:45 min Combustion Reaction \| Definition, Formula & Reactants 9:19 min Chapter 18 Chemical Equilibrium Solubility Product Constant (Ksp) Overview & Formula 10:06 min Titrations of Weak & Strong Acids & Bases 12:39 min Dynamic & Chemical Equilibrium \| Definition & Examples 6:31 min Le Chatelier's Principle \| Definition, Equation & Examples 6:04 min Common Ion Effect \| Definition, Examples & Exceptions 6:43 min Strong Acid-Base Titration \| Overview & Examples 12:27 min Chapter 19 Studying for Science 102 Explaining Properties & Relationships in Physics & Chemistry: Practice Problems Measuring & Classifying in Physics & Chemistry: Practice Problems Force, Mass & Newton's Laws of Motion Flashcards Thermodynamics & Energy Flashcards Sound Waves & Optics Flashcards Electric Power & Magnetic Forces Flashcards Experimental Chemistry in the Laboratory Flashcards Properties of Matter in Chemistry Flashcards Basics of Chemical Bonding Flashcards Kinetics & Stoichiometry in Chemistry Flashcards Understanding the Periodic Table Flashcards Atoms & Atomic Theory Flashcards Radioactivity & Solar Radiation Flashcards Acid-Base Chemical Reactions Flashcards Chemical Equilibrium & Concentration Flashcards More Aromatic Hydrocarbon Lessons Acenaphthylene: Uses & Aromaticity Trinitrotoluene Properties, Uses & Preparation Phenyl Formula, Structure & Applications 2,4-Dinitrophenylhydrazine \| Overview, Structure & Test Acetanilide Structure, Formula & Properties Acetanilide Structure, Preparation & Hazards Alizarin: Structure, Properties & PreparationAnthracene Hazards, Preparation & PropertiesAnthracene: Lewis Structure, Formula & ResonanceAromatic Hydrocarbons: Properties, Uses & Examples Benzene Derivatives \| List, Formulas & RisksFluorene \| Overview, Polarity & StructureNaming Aromatic Compounds \| Systems & Examples Naphthalene \| Formula, Structure & Uses Naphthalene: Uses, Hazards & Safety Naphthol Structure, Melting Point & Solubility Nitrobenzene Structure, Production & Uses Pyridine Formula, Structure & Uses TNT Overview, Chemical Structure & Formula Aromatic Hydrocarbon Courses Test Prep Alberta Education Diploma - Chemistry 30: Exam Prep & Study Guide View course Chemistry 9th Grade Physical Science Textbook View course Chemistry BITSAT Exam - Chemistry: Study Guide & Test Prep View course Chemistry ISC Chemistry: Study Guide & Syllabus View course Chemistry Organic Chemistry I View course FAQ What is meant by aromatic hydrocarbons? Aromatic hydrocarbons are organic molecules that are made of carbon and hydrogen atoms only. The following are the main characteristics of aromatic hydrocarbons: They are cyclic compounds. They contain conjugated double bonds. The pi electrons constantly delocalize. They are extremely stable because of the electron delocalization. They are highly flammable and extremely toxic compounds. Why are they called aromatic hydrocarbons? Aromatic hydrocarbons are named this way for two reasons, which are the following: They are named hydrocarbons because they do not contain any other atom different from carbon and hydrogen. Aromatic compounds have a very pleasant aromatic scent. They earned their title 'aromatic' because their nice aroma. What is the most common aromatic hydrocarbon? Aromatic hydrocarbons are conjugated cyclic compounds made entirely of carbon and hydrogen atoms. The most common and basic aromatic hydrocarbon is the benzene molecule; it is composed of six carbon atoms bonding in a closed loop. Many aromatic hydrocarbons are benzene-based, meaning they contain a benzene ring. All Hydrocarbon Topics Aliphatic Compound Alkane Alkene Alkynes Cycloalkanes Create an account to start this course today Used by over 30 million students worldwide Create an account Explore our library of over 88,000 lessons Search Browse Browse by subject College Courses Business English Foreign Language History Humanities Math Science Social Science See All College Courses High School Courses AP Common Core GED High School See All High School Courses Other Courses College & Career Guidance Courses College Placement Exams Entrance Exams General Test Prep K-8 Courses Skills Courses Teacher Certification Exams See All Other Courses Study.com is an online platform offering affordable courses and study materials for K-12, college, and professional development. 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190578	https://www.teacherspayteachers.com/Product/Simplifying-Algebraic-Expressions-Algebra-1-Math-Worksheets-3599053	Simplifying Algebraic Expressions Algebra 1 Math Worksheets Description Simplifying Algebraic Expressions by Combining Like Terms Algebra 1 Simplifying algebra and combining like terms Collecting like terms including positive and negative numbers. Grade 6 math Grade 7 Math For example: 3p x 2t and 3a +2b -a -5b 4 worksheets with a total of 57 questions. ✎ Contents: This resource includes: ✦ Objectives: To practice: - ✽ Directions: Print out a set of worksheets for each student. ✦ Notes: All answers are included, so you can easily mark students work. ♥ You may also like: Algebra Review Combining Like Terms Go Fish Earth Day Algebra Solving Equations Coloring Activity Algebra Substitution Game Algebra Substitution Worksheets Combining Like Terms Coloring Activity Distributive Property Simplifying Expressions Word problems involving formulas Factoring polynomials GCF Equivalent Expressions ☞ Visit Math Shop to see more of my resources. Simplifying Algebraic Expressions Algebra 1 Math Worksheets Save even more with bundles Reviews Questions & Answers Standards
190579	https://www.khanacademy.org/math/cc-fourth-grade-math/imp-fractions-2/imp-adding-and-subtracting-fractions-with-like-denominators/e/adding_fractions_with_common_denominators	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
190580	https://pocketdentistry.com/anatomy-and-root-canal-morphology/	Anatomy and Root Canal Morphology (1) Canberra, ACT, Australia The majority of endodontic failures can often be attributed to the inability of localising and treating all the canals of the root canal system. Root canal systems are commonly complex with the teeth often having lateral ramifications, extra roots or additional canals. Molar and premolar teeth can present with the highest incidence of aberrant morphology. A thorough knowledge of expected anatomy and variations from the norm are essential when undertaking root canal therapy to ensure success. Clinical Relevance The clinical impact of missed anatomy may result in failure and the necessity to carry out costly root canal re-treatment. Prevention of missed anatomy begins with a thorough knowledge of common tooth morphology. The clinician must be aware of the complexities of the root canal system and anatomical variations of the norm that may be encountered according to tooth type. The human dental pulp manifests multiple configurations and shapes which can vary from one individual to another. Careful interpretation of preoperative angled radiographs, correct access extension and a detailed exploration of the interior of the tooth using magnification and illumination are key steps for achieving clinical and radiographic success. 13.1 Overview of Root Canal Anatomy In 1925 Hess and Zurcher first published a study showing us how the teeth had complicated root canal systems rather than the simplified canals that had been previously described . For the first time we were able to see the intricate complexities of the internal root canal anatomy using vulcanised Indian rubber following dissolution of the surrounding tooth using 50 % hydrochloric acid. Even roots with a single canal will have lateral (an accessory canal located in the coronal or middle third of the root, usually extending horizontally from the main canal) and accessory canals leaving the main canal and communicating with the external surface of the root . A thorough understanding of the complexities of the root canal system is essential for understanding the principles and problems of shaping and cleaning and for performing successful microsurgical procedures . In order to grasp the varying anatomy that can occur with the teeth, a number of classification systems have been proposed. Weine’s classification of root canal morphology was based on the number of canal orifices, the number of canals and the number of foramina in each tooth (see Fig. 13.1). A more comprehensive classification was provided by Vertucci, resulting in eight categories based on 2,400 extracted teeth that were rendered transparent and treated with a dye injection technique (see Fig. 13.2). These classification systems provide us with a further insight into the complexities associated within the root canal system and help visualise the internal anatomy that we may become accustomed to encountering when carrying out preparation procedures. Fig. 13.1 Schematic diagrams representing Weine’s classification of root canal configurations. Type I – single canal from pulp chamber to apex. Type II – two canals leaving the chamber and merging to form a single canal short of the apex. Type III – two separate and distinct canals from chamber to apex. Type IV – one canal leaving the chamber and dividing into two separate and distinct canals Fig. 13.2 Vertucci classification. Type I – a single canal with one foramen. Type II – two canals that join in the apical third. Type III – one canal that divides into two that subsequently reunites and exits as one. Type IV – two separate canals all the way to the apex. Type V – one canal that divides just short of the apex. Type VI – two canals that unite in the root and divide again at the apex. Type VII – one canal that divides, reunites and finally exits through two apical foramina. Type VIII – three separate canals in one root Maxillary central and lateral incisors generally have one canal [3, 5]. When more than one canal is present, the possible configurations include two canals joining into one apical foramen (Vertucci type II) , two separate canals in one root (Vertucci type IV) , two canals in two separated roots or two or more canals associated with developmental abnormalities such as germination, fusion and dens invaginatus [7, 8]. Maxillary incisors with more than one root canal are rare and adequate access opening and observations of intra-operative radiographs should help in these difficult cases. The majority of maxillary canine teeth have one canal in one root . Morphological variations do exist but are rare where more than one canal is present . The most common canal configuration for the maxillary first premolar is two separate canals in one root (Vertucci type IV) with a frequency of about 60–65 %. One canal that extends from the pulp chamber and dividing in the mid-root into two canals (Vertucci type V) can occur in 6–7 % of cases. In about 8–9 % of cases, the maxillary first premolar can have one canal and in 16–18 % two canals joining into one. On rare occasions three canals (mesio-buccal, disto-buccal and palatal) can occur in 2–6 % with a root disposition similar to the first permanent maxillary molar tooth [5, 10–12]. Maxillary second premolars usually have one canal in one root (Vertucci type I) in 38–48 % of cases, two canals joining in one root (Vertucci type II) in 20–22 % of cases and one canal separating into two canals that rejoin in the apical third in 5–10 % of cases (Vertucci type III). Two canals in two roots can occur in the region of 10–20 % of cases and one canal that splits and exits as two (Vertucci type V) in 6–9 % of cases. Rarely two canals that rejoin and split again (Vertucci type VI) occur in 2–5 % of cases. Three separate canals occur in the frequency of 1–2 %. Due to the aberrant canal morphology in these teeth, clinicians should pay close attention to both pre- and intra-operative radiographic examinations and clinical pulpal floor anatomy [5, 10–12]. A wide range of variation exists for the maxillary first permanent molar in the literature. It is generally accepted that this tooth has three roots (95 %) and four canals. Four percent of teeth have only two roots. The broad bucco-lingual dimension of the mesio-buccal root and associated concavities on the mesial and distal surface is consistent with the majority of mesio-buccal roots having two canals. The incidence of root fusion of two or three roots is approximately 5.2 %. Conical and C-shaped root morphology is very rare (<1 %). The incidence of two canals in the mesio-buccal root is 60 % and one canal to be about 40 %. The incidence of more than one disto-buccal or palatal root is rare (<1 %) [13–21]. The majority of maxillary second permanent molars have three roots with four root canals (90 %). Six percent may present with two roots and 3 % with a single root. Vertucci described the mesio-buccal root of having one canal in 71 % of cases (type I), two canals joining in 17 % (type II) of cases and two separate canals in 12 % of cases (type IV). The incidence of C-shaped canals is very rare in this subgroup of teeth [5, 15, 22]. The morphology of mandibular central and lateral incisors is not very dissimilar with the majority of teeth having two canals (25–40 %) [5, 23, 24]. The mandibular permanent canine tooth often has a single root (98 %) with a single canal and one apical foramen (92 %). Five percent of cases have two canals, which are confluent in the apical third forming one canal. The incidence of two canals and two separate roots is rare (<3 %) [5, 25]. A literature review of the root canal morphology of the mandibular first permanent premolar revealed that these teeth were one (98 %), two (1.8 %) and rarely three or four rooted (<0.1 %). Internal canal morphology revealed the majority of canals were single (76 % type I). Studies and case reports have also confirmed variation of canal morphology where division of the canal occurs in the middle or apical third into buccal and lingual branches (type II and IV). Rarely three canals can be present and occasionally the presence of C-shaped canals [5, 26–30]. The majority of mandibular second permanent premolars have a single root canal system (97.5 % type I) and two canals in the remainder (2.5 % type IV). The frequency of three of more canals is very rare (<0.4 % type VIII) [5, 29–31]. The majority of mandibular first permanent molars are often found to be two rooted with various canal configurations existing. Often two separate canals occur in 59 % of teeth (type IV), two canals conjoining apically in 28 % of teeth (type II) and a single canal system in 12 % of teeth. Occasionally three canals may be present in 1 % of teeth (type VIII). The frequency of the middle mesial canal varies between 1 and 7 % with variations of either conjoining in the middle or apical third or rarely with three separate exits apically. Distal roots often have one single canal in 70 % of teeth (type I), two canals that join into one in 15 % of teeth (type II), two separate canals in 5 % of teeth (type IV), one canal that divides into two in 8 % of teeth and one canal that divides into two then joins again in the apical third, forming a single canal in 2 % of teeth (type III). Overall the distal canal often presents with two or more canals in 30 % of cases. A major variant of root morphology known as radix entomolaris can occur rarely in 5 % of Caucasians but more often in populations with mongoloid traits (5–30 %) [5, 32–37]. Mandibular second permanent molars usually have two roots (mesial and distal). In the mesial root a type I canal configuration may be present (27 %), two canals which co-join apically (38 % type II) and two separate canals (38 % type IV). In the distal root, variations of type I (92 %), type II (3 %) and type IV (5 %) can be found. The incidence of a C-shaped canal system in this particular tooth group varies between 2 and 30 % and found often with a higher prevalence in the Chinese population [5, 38–42]. The root morphology and canal anatomy of maxillary and mandibular third molars is highly variable with often fusion involving one or more of the roots . To ensure that all internal anatomy is correctly identified and treated, good preoperative radiographs, correct access cavity preparation, enhanced lighting, use of the dental operating microscope and controlled use of ultrasonic instruments are necessary tools that must be adopted. The clinical impact of missed anatomy is likely a failure in the future, resulting in either costly re-treatment or worse still extraction of the tooth. On the other hand, localisation and treatment of all canal anatomy will typically lead to a favourable outcome with complete clinical and radiographic healing. Cone-beam technology has been present for the last 30 years, but recent convergence and innovations have lead to the viable application of cone-beam volumetric tomography (CBVT) or cone-beam computed tomography (CBCT) in the dental office. Together these advantages have allowed CBVT to become a valuable tool in the modern endodontic practice, allowing us to truly appreciate the difficulties of treating these three-dimensional spaces that are more complex than we could have imagined [44–46] (see Chap. 12). 13.2 Maxillary Central Incisor Teeth The root of the maxillary central incisor is often bulkier than the lateral with cross-sectional anatomy varying from triangular, circular or oval in shape. The canal is often tapering towards the apex in mature teeth with minimal curvatures in the apical portion (Fig. 13.3). Fig. 13.3 Overview of maxillary permanent central incisor teeth The access cavity is initiated by penetrating the bur occlusal to the cingulum, avoiding the incisal edge. Once penetration to the root canal is achieved, the access cavity must be refined in a mesio-distal direction to remove the entire roof associated with the pulp horns. The access cavity achieves a roughly triangular shape with this preparation, which mirrors the anatomy of the pulp chamber (see Fig. 13.4). Fig. 13.4 Clinical photographs showing step-by-step access preparation in an anterior maxillary central incisor. Note (a) preoperative view (dotted line representing intended triangular access incorporating pulp horns), (b) initial bur penetration, (c) root canal penetrated, (d) widening of access in a mesial and distal direction to ensure pulp horns are incorporated, (e) use of ultrasonics to remove overhanging dentine and to ensure pulp horns are free of any tissue remnants and (f) final access preparation completed Lateral canals may be present in the middle or apical third and the occurrence of a second canal is very rare. Due attention must be given to canals with very wide-open apices (blunderbuss teeth) and also narrow calcified canals (calcific metamorphosis) typically encountered following a traumatic incident. A sharp bend or curvature in the root would indicate previous trauma or bony interference during root formation (dilacerations), which may also affect management. 13.3 Maxillary Lateral Incisor Teeth The lateral incisor is often smaller in comparison to its neighbouring central counterpart. The root is often slender with an apical curvature generally in a disto-palatal direction. The presence of extra roots or occurrence of developmental grooves is more likely in this tooth group. Tooth invaginations (dens in dente) are also a common finding, which complicates endodontic management. Early recognition is essential and referral is probably best considered (Fig. 13.5). Fig. 13.5 Overview of maxillary permanent lateral incisor teeth In this tooth, the access cavity is created in the same way as in the central incisor. The pulp horns are often closely situated or singular in this tooth, resulting in a final shape that is more likely to be ovoid as opposed to triangular. Care must be taken when negotiating the apical curvature in this tooth, particularly with larger file sizes, which can result in canal transportation or ledging if not correctly identified (see Fig. 13.6). Fig. 13.6 Clinical radiographs demonstrating management of maxillary lateral incisor. Note the apical curvature that cannot be appreciated due to the distal and palatal curvature. (a) Preoperative view with large peri-radicular radiolucent lesion and (b) completed endodontic treatment of tooth 22. Note distal curvature of apical third maintained. Significant reduction of peri-radicular lesion achieved following chemo-mechanical debridement and interim intra-canal dressing of calcium hydroxide over a 3–month period, (c) 6–month follow-up and (d) 18–month follow-up showing continued healing 13.4 Maxillary Canine Teeth These teeth can be one of the longest encountered with roots that are irregularly tapered and wide in a labio-palatal direction. The canal is typically a type I configuration with a root curvature apically often in a distal direction although can occur in any direction (Fig. 13.7). Fig. 13.7 Overview of maxillary permanent canine teeth The access cavity begins about halfway up the crown on the palatal side (see Fig. 13.8). With an ovoid pulp chamber and a single horn, the access cavity is oval in outline and shape. The root canal is quite straight in the coronal and middle third and long enough to require the use of long 30-mm instruments. Fig. 13.8 Clinical photographs demonstrating access cavity preparation in a maxillary canine tooth. Note (a) preoperative view and (b) completed access preparation showing ovoid outline Canines with a normal crown and root but with two canals are rare and difficult to identify because the canals may be superimposed. An adequate pulp chamber opening and careful observation of intra-operative radiographs may help in these difficult cases. 13.5 Maxillary Premolar Teeth The maxillary first premolar has variable morphology but is generally considered to have two roots and two canals (see Fig. 13.9). The second premolar although variable at times generally presents with a single-rooted single canal system (see Fig. 13.10). Fig. 13.9 Overview of maxillary permanent first premolar teeth Fig. 13.10 Overview of maxillary permanent second premolar teeth Careful radiographic assessment is essential in order to ascertain the likely number of roots and canal morphology before any operative intervention. Peri-apical radiographs give a two-dimensional image of a three-dimensional root canal system, so a minimum of two radiographs should be taken to reveal external and internal features of the tooth. A parallel radiograph should be taken with either a mesial or distal horizontal tube shift. The second angled radiograph may help visualise superimposed roots, displace the zygomatic process of the maxillary bone and increase the chances of observing additional root canals present. The intra-oral radiograph should be carefully inspected for any sudden changes in the radiographic density of the root canal space, which may indicate bifurcation or trifurcation of the roots. Any sudden narrowing or disappearance may indicate branching of the root canal system often encountered in these teeth. The same principles can be applied during peri-operative treatment whereby any asymmetry of the file placed in the canal can also indicate the presence of a second root system. The pulp chamber of the upper first and second premolar is oriented more towards the bucco-lingual direction since in the great majority of cases, it has two canals beneath the respective cusps (see Figs. 13.11 and 13.12). The point of entry of the bur is the middle of the central sulcus drilling parallel to the long axis of the tooth. Fig. 13.11 Preoperative clinical photograph showing (a) occlusal anatomy of an upper maxillary premolar tooth and (b) completed ovoid access preparation extended in a bucco-palatal direction Fig. 13.12 Clinical radiographs and photographs demonstrating endodontic treatment of a maxillary first premolar tooth with two canals. Note (a) preoperative radiograph demonstrating a two-rooted tooth, (b) MAF, (c) cone fit and (d) final obturation showing Vertucci IV canal configuration The floor of the pulp chamber should be carefully examined looking at the position and symmetry of the canals. Asymmetry is often a good indication of additional anatomy that may be present. Upper premolars with three canals require a modified access cavity with a mesio-distal extension in the buccal portion of the traditional cavity. This modification, resulting in a T-shaped access, permits good access to both buccal canals. The presence of three canals is higher in the first premolar compared to the second but nevertheless rare. The anatomical roots are similar to the first molar whereby an MB, a DB and a P root may be present. Treatment of maxillary premolars requires extra attention due to their extreme variability that can be encountered. Caution is advised due to the higher risk of failing to treat all internal anatomy, and a dental operating microscope or other forms of magnification are a necessary tool when dealing with these teeth. 13.6 Maxillary Molar Teeth The literature shows a wide variation in anatomy of the maxillary molars although the average tooth often has three roots and four canals (see Figs. 13.13 and 13.14). The palatal root is often the longest with a curvature in the apical third towards the buccal. The disto-buccal root, which is shorter in comparison, may curve towards the mesial or distal in the apical third. The mesio-buccal root shows the greatest variation with a root that is broad in the bucco-palatal plane and narrowing in the mesio-distal place. This ribbon-shaped root will generally exit the crown mesially but can commonly abruptly curve towards the distal. Curvatures are often encountered in the apical third with possibilities of convergence and joining of additional mesio-palatal canals if present (see Fig. 13.15). Fig. 13.13 Overview of maxillary permanent first molar teeth Fig. 13.14 Clinical radiographs demonstrating root canal treatment carried out in a permanent maxillary first molar with three roots and four canals. Note (a) preoperative view, (b) IAF, (c) mid-fill and (d and e) final obturation using warm vertical compaction. (f) MB2 canal was located following ultrasonic troughing in this area (arrowhead) Fig. 13.15 Clinical photographs demonstrating cleared maxillary first permanent molar teeth showing complex canal anatomy. (a) Maxillary first permanent molar showing the MB root. Note Vertucci classification VI (2–1–2) with a lateral canal from the MB1 canal. (b) Maxillary first permanent molar with two palatal canals. Note Vertucci classification II (2–1) with interconnecting fins and communication between both canals Access preparation will be mesial to the external oblique ridge, allowing it to be maintained during preparation whereby enhancing the structural strength of the tooth. The orifice of the palatal canal is often the most prominent beneath the mesio-palatal cusp. The mesio-buccal canal often lies directly beneath the mesio-buccal cusp tip and the disto-buccal canal can often be found 2–3 mm distal to this canal and slightly towards the palatal. Careful inspection of the floor of the pulp chamber may reveal additional anatomy. Recommended clinical approaches when treating maxillary molars include: 1. Take two diagnostic radiographs, one parallel and one either with a mesial or distal tube horizontal tube shift, to correctly assess and identify external and internal anatomy. 2. Careful removal of the pulp chamber roof is carried out using a non-end cutting to avoid damage to the pulpal floor. Dark developmental lines on the floor of the pulp chamber should be observed like a ‘road map’ to identify possible root canals present. 3. Careful observation of the pulp chamber floor may also indicate additional root canal anatomy using sodium hypochlorite which may effervesce. 4. Use of the dental operating microscope or dental loupes with adequate illumination is essential to correctly identify internal anatomy and avoid iatrogenic damage to the floor of the pulp chamber. 5. Access cavity preparation does not need to be extended beyond the marginal ridge (see Fig. 13.16). Loss of marginal ridge greatly reduces the strength of the tooth. Fig. 13.16 Clinical photographs demonstrating access cavity preparation and refinement during the treatment of a maxillary permanent first molar. Note (a) calcified floor of pulp chamber with only DB canal located, (b) ultrasonic troughing carried out, (c) DB and P canal orifices located, (d) MB1 canal orifice located, (e) MB2 canal orifice located and (f) final four canals cleaned and shaped. Access cavity is not extended beyond marginal ridge (red dash line) 6. Refinement of the access cavity will be required in the mesio-palatal direction using a rhomboid access. The MB2 canal is often palatal and often mesial in a line drawn between the MB1 and the palatal canal (see Fig. 13.16). 7. Use of methylene blue dye stains can be useful to highlight the pulp chamber anatomy. 8. Trough and search with a low-speed bur or ultrasonic tip beginning from the MB1 orifice. Be careful not to exceed a depth of 2–3 mm. Only gold members can continue reading. Log In or Register to continue Share this: Click to share on Twitter (Opens in new window) Click to share on Facebook (Opens in new window) ## Stay updated, free dental videos. Join our Telegram channel Tags: Endodontic Diagnosis Pathology and Treatment Planning, ## VIDEdental - Online dental courses Get VIDEdental app for watching clinical videos Get VIDEdental app for watching clinical videos Công ty Cổ phần Nguyên Phát Hà Nội Số 10 ngõ 243a đường Xun Khanh, Phường Xun Khanh, Thị Xã Sơn Ty, Hà Nội
190581	http://ltcconline.net/greenl/courses/115/functionGraphLimit/cont.htm	Continuity Continuity Continuity If the limit exists at x = c, the function has some nice properties. However, even if the limit exists, there may still be a hole. We define a function to be continuous at x = c is the limit exists and the function agrees with the limit at x = c. More formally Definition of a Continuous Function A function is continuous at x = c if the the limit exists there and A function is called continuous if it is continuous for all real numbers. In other words, if the graph has no holes asymptotes, or ,breaks then the function is continuous. If you can draw the function without lifting your pencil then it is continuous. Below are some examples of continuous functions. Continuous Functions 1) Polynomials 2) sin and cos 3) Rational Functions where the denominator is nonzero 4) Sums, Differences, and Products of continuous functions 5) Quotients of continuous functions 6) Compositions of continuous functions Examples: The following are continuous: A) y = x 2 + 3x - 4 B) y = x sin x C) 1 y = 1+ x 2 Exercises:Determine whether the following are continuous. If they are not continuous, at which points are they discontinuous? Hold the mouse on the yellow rectangle for the answer. A) x - 1 y = x + 1 B) x y = x 2 + 3x - 4 C) y ={2x + 3 for x < 1 3x - 2 for x > 1 D) y ={x 2 for x < 2 5x - 6 for x > 2 E) For what value of k is the function continuous? y ={3x 2 - 5 for x < 1 5x - k for x > 1 Applications: Suppose that your account initially has $10,00 in it. The account pays 5% annual interest compounded monthly. Sketch the graph of your balance as a function of time. Is it continuous? Sketch the graph of the population of the earth as a function of time. Is this a continuous function. Why is it reasonable to represent this graph as a continuous function? Sketch the graph of your telephone costs for using the phone. Is this a continuous function? Sketch the graph of your blood pressure as a function of time while on a bicycle ride. Is this a continuous function? Come up with a continuous function and a discontinuous function that occurs in the real world. Back to the calculus home page Back to the math department home page e-mail Questions and Suggestions
190582	https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/8.1/primary/lesson/systems-of-two-equations-and-two-unknowns-pcalc/	Systems of Two Equations and Two Unknowns \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Start Practice 8.1 Systems of Two Equations and Two Unknowns FlexBooks 2.0> CK-12 Precalculus Concepts 2.0> Systems of Two Equations and Two Unknowns Written by:CK-12 \|Mark Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Lesson Review Asked on Flexi Related Content Lesson [Figure 1] Which cell service plan is cheaper? The cost of two cell phone plans can be written as a system of equations based on the number of minutes used and the base monthly rate. As a consumer, it would be useful to know when the two plans cost the same and when one plan is cheaper. Plan A costs $40 per month plus $0.10 for each minute of talk time. Plan B costs $25 per month plus $0.50 for each minute of talk time. Plan B has a lower starting cost, but since it costs more per minute, it may not be the right plan for someone who likes to spend a lot of time on the phone. When do the two plans cost the same amount? Solving Systems of Equations with Two Unknowns There are many ways to solve a system that you have learned in the past including substitution and graphical intersection. Here you will focus on solving using elimination because the knowledge and skills used will transfer directly into using matrices. When solving a system, the first thing to do is to count the number of variables that are missing and the number of equations. The number of variables needs to be the same or fewer than the number of equations. Two equations and two variables can be solved, but one equation with two variables cannot. Here's the procedure for solving a system using the elimination method: Step 1: Write both equations with two variables in standard form,A x+B y=C.This form helps to align the variables. Step 2: Determine which variable you want to eliminate. Step 3: Scale each equation as necessary by multiplying through by constants. Step 4: Add the equations together. This should reduce both the number of equations and the number of variables leaving one equation and one variable. Step 5: Solve and substitute to determine the value of the second variable. Here is a system of two equations and two variables in standard form:5 x+12 y=72 and 3 x−2 y=18. Notice that there is an x column and a y column on the left hand side and a constant column on the right hand side when you rewrite the equations as shown. Also notice that if you add the system as written no variable will be eliminated. Equation 1: 5 x+12 y=72 Equation 2: 3 x−2 y=18 Strategically choose to eliminate y by scaling the second equation by 6 so that the coefficient of y will match at 12 and -12. 5 x+12 y=72 18 x−12 y=108 Add the two equations: 23 x=180 x=180 23 The value for x could be substituted into either of the original equations and the result could be solved for y;however, since the value is a fraction it will be easier to repeat the elimination process in order to solve for x. This time you will take the first two equations and eliminate x by making the coefficients of x to be 15 and -15. Scale the first equation by a factor of 3 and scale the second equation by a factor of -5. Equation 1:15 x+36 y=216 Equation 2:−15 x+10 y=−90 Adding the two equations: 0 x+46 y=126 y=126 46=63 23 The point (180 23,63 23)is where these two lines intersect. Examples Example 1 Earlier, you were asked about two phone plans. Plan A costs $40 per month plus $0.10 for each minute of talk time. Plan B costs $25 per month plus $0.50 for each minute of talk time. If you want to find out when the two plans cost the same, you can represent each plan with an equation and solve the system of equations. Let y represent cost and x represent number of minutes. y=0.10 x+40 y=0.50 x+25 First you put these equations in standard form. x−10 y=−400 x−2 y=−50 Then you scale the second equation by -1 and add the equations together and solve for y. −8 y=−350 y=43.75 To solve for x, you can scale the second equation by -5, add the equations together and solve for x. −4 x=−150 x=37.5 The equivalent costs of plan A and plan B will occur at 37.5 minutes of talk time with a cost of $43.75. Example 2 Solve the following system of equations: 6 x−7 y=8 15 x−14 y=21 Scaling the first equation by -2 will allow the y term to be eliminated when the equations are summed. −12 x+14 y=−16 15 x−14 y=21 The sum is: 3 x=5 x=5 3 You can substitute x into the first equation to solve for y. 6⋅5 3−7 y=8 10−7 y=8−7 y=−2 y=2 7 The point (5 3,2 7)is where these two lines intersect. Example 3 Solve the following system using elimination: 5 x−y=22−2 x+7 y=19 Start by scaling the first equation by 7 and notice that the y coefficient will immediately be eliminated when the equations are summed. 35 x−7 y=154−2 x+7 y=19 Add, solve for x=173 33. Instead of substituting, practice eliminating x by scaling the first equation by 2 and the second equation by 5. 10 x−2 y=44−10 x+35 y=95 Add, solve for y. Final Answer:(173 33,139 33) Example 4 Solve the following system of equations: 5⋅1 x+2⋅1 y=11 1 x+1 y=4 The strategy of elimination still applies. You can eliminate the 1 y term if the second equation is scaled by a factor of -2. 5⋅1 x+2⋅1 y=11−2(1 x)−2(1 y)=−2⋅(4) Add the equations together and solve for x. 3⋅1 x+0⋅1 y=3 3⋅1 x=3 1 x=1 x=1 Substitute into the second equation and solve for y. 1(1)+1 y=4 1+1 y=4 1 y=3 y=1 3 The point (−1,1 3)is the point of intersection between these two curves. Example 5 Solve the following system using elimination: 11⋅1 x−5⋅1 y=−38 9⋅1 x+2⋅1 y=−25 To eliminate 1 y, scale the first equation by 2 and the second equation by 5. To eliminate 1 x, scale the first equation by -9 and the second equation by 11. Final Answer:(−1 3,1) Summary To solve a system of equations using the elimination method Step 1: Write both equations with two variables in standard form, A x+B y=C. Step 2: Determine which variable you want to eliminate. Step 3: Scale each equation as necessary by multiplying through by constants. Step 4: Add the equations together, leaving one equation and one variable. Step 5: Solve and substitute to determine the value of the second variable. Review Solve each system of equations using the elimination method. x+y=−4;−x+2 y=13 3 2 x−1 2 y=1 2;−4 x+2 y=4 6 x+15 y=1;2 x−y=19 x−2 y 3=−2 3;5 x−2 y=10 −9 x−24 y=−243;1 2 x+y=21 2 5 x+28 3 y=176 3;y+x=10 2 x−3 y=50;7 x+8 y=−10 2 x+3 y=1;2 y=−3 x+14 2 x+3 5 y=3;3 2 x−y=−5 5 x=9−2 y;3 y=2 x−3 How do you know if a system of equations has no solution? If a system of equations has no solution, what does this imply about the relationship of the curves on the graph? Give an example of a system of two equations with two unknowns with an infinite number of solutions. Explain how you know the system has an infinite number of solutions. Solve: 12⋅1 x−18⋅1 y=4 8⋅1 x+9⋅1 y=5 Solve: 14⋅1 x−5⋅1 y=−3 7⋅1 x+2⋅1 y=3 Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. Add Note CancelSave Discuss with Flexi NOTES / HIGHLIGHTS Please Sign In to create your own Highlights / Notes Add Note Edit Note Remove Highlight Image Attributions Asked by Students Here are the top questions that students are asking Flexi for this concept: Can Omari's strategy work for solving the given system of equations 4x+2y=62 and -8x-y=59 by multiplying the second equation by 2 and then subtracting the first equation? Explain your thinking. Omari's strategy is to multiply the second equation by 2 and then subtract the first equation from the result. Let's see if this will work. First, multiply the second equation by 2: −16 x−2 y=118 Then, subtract the first equation from the result: (−16 x−2 y)−(4 x+2 y)=118−62 Simplify the left side: −20 x=56 So, yes, Omari's strategy will work because it eliminates the y variable and allows us to solve for x. How do you solve equations with the elimination method? The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. Steps for Solving Simultaneous Equations by Elimination Method: Step 1: Choose a variable to eliminate. Step 2: Find the LCM of the coefficients of that variable. Step 3: Multiply both sides of the equations to make the coefficient of the variable you want to eliminate the LCM. Make sure these terms have opposite signs, so they are additive inverses of each other. Step 4: Add the equations and solve for the remaining variable if necessary. Step 5: Once you solve the equations for one of the variables, substitute the solution into either equation to determine the value of the other variable. Step 6: The solution to the system is the intersection point of the two equations and it represents the coordinates of the ordered pair. Click here to learn more about solving a pair of linear equations by elimination method! Two whole numbers have a sum of 30 and a difference of 6. Let x be the larger number and y be the smaller number. Which system of equations can be used to find the numbers? The system of equations that can be used to find the numbers is: x+y=30 x−y=6 Explain the "criss cross" method. Thecriss cross method, also known as the cross multiplication method, is a technique used to solve proportions or find the common denominator of fractions. Here's a step-by-step explanation: Write the two fractions or ratios in the form a b=c d. Cross multiply, which means you multiply the numerator of the first fraction by the denominator of the second fraction, and the denominator of the first fraction by the numerator of the second fraction. Set these two products equal to each other: a⋅d=b⋅c. Solve for the unknown variable. Example: For a proportion like 2 3=x 6, you would cross multiply to get: 2⋅6=3⋅x Then solve for x to get x=4. Consider the following pair of linear equations. 2x + ay = 5 , 4x + 6y = 10. If this pair has a unique solution, what can we say about a? For a pair of linear equations to have a unique solution, the ratios of the coefficients of x and y in the two equations should not be equal. In other words, if we have two equations a x+b y=c and d x+e y=f, they will have a unique solution if and only if a d≠b e. In this case, the coefficients of x in the two equations are 2 and 4, so their ratio is 2 4=0.5. The coefficients of y in the two equations are a and 6, so their ratio is a 6. For the system to have a unique solution, we must have 2 4≠a 6, or equivalently, 0.5≠a 6. Solving this inequality for a gives us: 0.5≠a 6 3≠a So, for the system to have a unique solution, a must not be equal to 3. The value of a must not be 3 for the system to have a unique solution. Overview To solve a system of equations using the elimination method Step 1: Write both equations with two variables in standard form, A x+B y=C. Step 2: Determine which variable you want to eliminate. Step 3: Scale each equation as necessary by multiplying through by constants. Step 4: Add the equations together, leaving one equation and one variable. Step 5: Solve and substitute to determine the value of the second variable. Vocabulary system of equations rate intersection Equation elimination method Standard form Variable substitute constant Coefficient fraction order factor Term Test Your Knowledge Question 1 Double-scoop frozen yogurt cones cost $4, and frozen fruit cakes are $11. If you spend $273 on 35 items, how many of each did you buy? a 5 frozen yogurt cones and 30 frozen fruit cakes b 20 frozen yogurt cones and 15 frozen fruit cakes c 10 frozen yogurt cones and 25 frozen fruit cakes d 16 frozen yogurt cones and 19 frozen fruit cakes Check It Let x represent the number of frozen yogurt cones. Let y represent the number of frozen fruit cakes. From the question: “Double-scoop frozen yogurt cones cost $4, and frozen fruit cakes are $11. You spend $273 on 35 items.” x+y=35…(1)4 x+11 y=273…(2) Solve the equation (1) for x, x+y=35 x+y−y=35−y x=35−y…(3) Substitute the equation (3) into the equation (2), 4 x+11 y=273 4(35−y)+11 y=273 140−4 y+11 y=273 140+7 y=273 140−140+7 y=273−140 7 y=133 7 y 7=133 7 y=19 Substitute the value of y into the equation (3), x=35−y x=35−19 x=16 You bought 16 frozen yogurt cones and 19 frozen fruit cakes. FlexCard™ Question 2 At a local gym you go to, you can either pay $90 for a monthly pass, then pay $4 for each time you decide to go to the gym, or pay $13 for each time you go to the gym. Which of the following choices below is a system of equations representing this situation? a y=90 x−4 x and y=13 x b y=4 x+13 and y=90 x c y=4 x and y=90+13 x d y=90+4 x and y=13 x Check It Let x represent the number of visits. Let y represent the total cost. Given that, At a local gym, you go to, you can either pay $90 for a monthly pass, then pay $4 each time you decide to go to the gym. Total cost=Monthly pass+Cost for per visit×Number of visits y=90+4×x y=90+4 x Pay $13 each time you go to the gym. Total cost=Cost for per visit×Number of visits y=13×x y=13 x The equations for the given situation are y=90+4 x and y=13 x. FlexCard™ Asked by Students Ask your question Here are the top questions that students are asking Flexi for this concept: Can Omari's strategy work for solving the given system of equations 4x+2y=62 and -8x-y=59 by multiplying the second equation by 2 and then subtracting the first equation? Explain your thinking. Omari's strategy is to multiply the second equation by 2 and then subtract the first equation from the result. Let's see if this will work. First, multiply the second equation by 2: −16 x−2 y=118 Then, subtract the first equation from the result: (−16 x−2 y)−(4 x+2 y)=118−62 Simplify the left side: −20 x=56 So, yes, Omari's strategy will work because it eliminates the y variable and allows us to solve for x. How do you solve equations with the elimination method? The elimination method for solving a system of two equations involves combining the two equations in order to produce one equation in one variable. Steps for Solving Simultaneous Equations by Elimination Method: Step 1: Choose a variable to eliminate. Step 2: Find the LCM of the coefficients of that variable. Step 3: Multiply both sides of the equations to make the coefficient of the variable you want to eliminate the LCM. Make sure these terms have opposite signs, so they are additive inverses of each other. Step 4: Add the equations and solve for the remaining variable if necessary. Step 5: Once you solve the equations for one of the variables, substitute the solution into either equation to determine the value of the other variable. Step 6: The solution to the system is the intersection point of the two equations and it represents the coordinates of the ordered pair. Click here to learn more about solving a pair of linear equations by elimination method! Two whole numbers have a sum of 30 and a difference of 6. Let x be the larger number and y be the smaller number. Which system of equations can be used to find the numbers? The system of equations that can be used to find the numbers is: x+y=30 x−y=6 Explain the "criss cross" method. Thecriss cross method, also known as the cross multiplication method, is a technique used to solve proportions or find the common denominator of fractions. Here's a step-by-step explanation: Write the two fractions or ratios in the form a b=c d. Cross multiply, which means you multiply the numerator of the first fraction by the denominator of the second fraction, and the denominator of the first fraction by the numerator of the second fraction. Set these two products equal to each other: a⋅d=b⋅c. Solve for the unknown variable. Example: For a proportion like 2 3=x 6, you would cross multiply to get: 2⋅6=3⋅x Then solve for x to get x=4. Consider the following pair of linear equations. 2x + ay = 5 , 4x + 6y = 10. If this pair has a unique solution, what can we say about a? For a pair of linear equations to have a unique solution, the ratios of the coefficients of x and y in the two equations should not be equal. In other words, if we have two equations a x+b y=c and d x+e y=f, they will have a unique solution if and only if a d≠b e. In this case, the coefficients of x in the two equations are 2 and 4, so their ratio is 2 4=0.5. The coefficients of y in the two equations are a and 6, so their ratio is a 6. For the system to have a unique solution, we must have 2 4≠a 6, or equivalently, 0.5≠a 6. Solving this inequality for a gives us: 0.5≠a 6 3≠a So, for the system to have a unique solution, a must not be equal to 3. The value of a must not be 3 for the system to have a unique solution. 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190583	https://people.highline.edu/iglozman/classes/physnotes/efield.htm	Coulomb force We find in nature that a positive charge repels another positive charge, a negative charge repels another negative charge, and a positive charge is attracted to a negative charge. The magnitude of this attractive or repulsive force is given by Coulomb's law. The inverse-square nature of this force is similar to that for the law of gravity. Unlike the case for gravity, however, the electric force can be either attractive or repulsive. Sample problem: Compare the electrostatic force between an electron and proton to gravitational force between them when the two particles are separated by 0.5x10-10 m. Solution: This example shows that the gravitational forces between charged microscopic objects are much greater than the gravitational forces between them. That's why gravitational interaction is usually neglected in these problems. Sample problem: Honeybees actively foraging in the field have been found to be electrically charged, due largely to air resistance as they fly. The charge carried by a bee is thought to play a significant role in pollination—the bee can detach grains of pollen from a distance, like a charged comb attracting bits of paper. If a bee carries a charge of 93 pC and the force required to detach pollen from an avocado stigma is 4.0x10-10 N, estimate the maximum distance at which the electrostatic force between a bee and a grain of pollen is sufficient to detach the pollen. Solution: To simplify the calculation, we approximate each charge as a point charge and assume that the pollen has a charge opposite in sign and equal in magnitude to that of the bee. Electric field Different charge distributions produce different electric fields. The fields for the most common geometries--point charges, plates, spheres, and cylinders--are shown here. Sample problem: In fair weather, the atmospheric field of the Earth near the surface is about 100 N/C; it points vertically downward. What will be the acceleration of a grain of dust of mass 10-18 kg carrying a single electron charge? Solution: It would appear that the electrostatic forces generated by charge separation between the ground and the clouds is big enough to overcome the force of gravity and keep the dust particles suspended in air. Sample problem: Two identical point charges are positioned at two corners of a square of side r. What is the direction of the electric field at an adjacent corner? Solution: This a vector geometry problem best approached by figuring out the components of the resultant and using those to figure out the angle. Sample problem: Estimate the width of a lightning column, which has a linear charge density of about –1x10-3 C/m. Assume the electric field at which ionization begins to occur is 3x106 N/C. Assume the electric field at which ionization begins to occur is 3x106 N/C. Solution: Here we use the expression for electric field due to an infinite line as a approximation to the lightning bolt. Return to class notes TOC. Page last modified: 05/07/2021 22:14:05
190584	https://www.slideserve.com/kayla/inclusion-exclusion-principle	PPT - Inclusion-Exclusion Principle PowerPoint Presentation, free download - ID:5373823 Browse Recent Presentations Recent Articles Content Topics Updated Contents Featured Contents PowerPoint Templates Create Presentation Article Survey Quiz Lead-form E-Book Presentation Creator Pro Upload Download Presentation Discover more Presentation presentation Mathematics mathematics Slide templates Buy interactive quiz software Presentation software Presentation creator Download 1 / 24 Download Download Presentation >> Discover more mathematics Mathematics Presentation presentation Quiz tools Visual content tools Purchase quiz builder platforms Find marketing tool subscriptions Discover more Mathematics presentation Presentation mathematics presentations Inclusion-Exclusion Principle Oct 10, 2014 • 2.45k likes • 6.08k Views Inclusion-Exclusion Principle. Lecture 14: Oct 28. Sum Rule. If sets A and B are disjoint, then \| A  B \| = \| A \| + \| B \|. A. B. What if A and B are not disjoint ?. Inclusion-Exclusion (2 sets). For two arbitrary sets A and B. A. B. Inclusion-Exclusion (2 sets). Share Presentation Embed Code Link Download Presentation kayla + Follow Download Presentation Inclusion-Exclusion Principle An Image/Link below is provided (as is) to download presentationDownload Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.Content is provided to you AS IS for your information and personal use only. Download presentation by click this link.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.During download, if you can't get a presentation, the file might be deleted by the publisher. Discover more Presentation mathematics presentation Mathematics Presentation downloads Find survey creation software Document sharing tools Find online presentation courses Buy presentation templates online Content marketing software E N D Presentation Transcript Inclusion-Exclusion Principle Lecture 14: Oct 28 Sum Rule If sets A and B are disjoint, then \|AB\| = \|A\| + \|B\| A B • What if A and B are not disjoint? Inclusion-Exclusion (2 sets) For two arbitrary sets A and B A B Inclusion-Exclusion (2 sets) Let S be the set of integers from 1 through 1000 that are multiples of 3 or multiples of 5. Let A be the set of integers from 1 to 1000 that are multiples of 3. Let B be the set of integers from 1 to 1000 that are multiples of 5. It is clear that S is the union of A and B, but notice that A and B are not disjoint. A B \|A\| = 1000/3 = 333 \|B\| = 1000/5 = 200 A Å B is the set of integers that are multiples of 15, and so \|A Å B\| = 1000/15 = 66 So, by the inclusion-exclusion principle, we have \|S\| = \|A\| + \|BA Å B\| = 467. Inclusion-Exclusion (3 sets) \|A [B [C\| = \|A\| + \|B\| + \|C\| – \|AÅB\| – \|AÅC\| – \|BÅC\| + \|AÅBÅC\| A B C Inclusion-Exclusion (3 sets) \|A\| 30 know Java 18 know C++ 26 know C# 9 know both Java and C++ 16 know both Java and C# 8 know both C++ and C# 47 know at least one language. From a total of 50 students: \|B\| \|C\| \|AÅB\| How many know none? How many know all? \|AÅC\| \|BÅC\| \|AÅBÅC\| \|A[B[C\| \|A[B[C\| = \|A\| + \|B\| + \|C\|– \|AÅB\| – \|AÅC\| – \|BÅC\| + \|AÅBÅC\| 47 = 30 + 18 + 26 – 9 – 16 – 8 + \|AÅBÅC\| \|AÅBÅC\| = 6 Inclusion-Exclusion (4 sets) \|A [ B [ C [ D\| = \|A\| + \|B\| + \|C\| + \|D\| – \|A Å B\| – \|A Å C\| – \|A Å D\| – \|B Å C\| – \|B Å D\| – \|C Å D\| + \|A Å B Å C\| + \|A Å B Å D\| + \|A Å C Å D\| + \|B Å C Å D\| – \|A Å B Å C Å D \| A B C D Inclusion-Exclusion (n sets) What is the inclusion-exclusion formula for the union of n sets? Inclusion-Exclusion (n sets) sum of sizes of all single sets – sum of sizes of all 2-set intersections + sum of sizes of all 3-set intersections – sum of sizes of all 4-set intersections … + (–1)n+1 × sum of sizes of intersections of all n sets Inclusion-Exclusion (n sets) \|A1[ A2[ A3[ … [ An\| sum of sizes of all single sets – sum of sizes of all 2-set intersections + sum of sizes of all 3-set intersections – sum of sizes of all 4-set intersections … + (–1)n+1 × sum of sizes of intersections of all n sets We want to show that every element is counted exactly once. Consider an element which belongs to exactly k sets, say A1, A2, A3, …, Ak. In the formula, such an element is counted the following number of times: Therefore each element is counted exactly once, and thus the formula is correct Inclusion-Exclusion (n sets) Plug in x=1 and y=-1 in the above binomial theorem, we have Christmas Party In a Christmas party, everyone brings his/her present. There are n people and so there are totally n presents. Suppose the host collects and shuffles all the presents. Now everyone picks a random present. What is the probability that no one picks his/her own present? Let the n presents be {1, 2, 3, …, n}, where the present i is owned by person i. Now a random ordering of the presents means a permutation of {1, 2, 3, …, n}. e.g. (3,2,1) means the person 1 picks present 3, person 2 picks present 2, etc. And the question whether someone picks his/her own present becomes whether there is a number i which is in position i of the permutation. Fixed Points in a Permutation Given a random permutation of {1, 2, 3, …, n}, what is the probability that a permutation has no fixed point (i.e number i is not in position i for all i)? e.g. {2, 3, 1, 5, 6, 4} has no fixed point, {3, 4, 7, 5, 2, 6, 1} has a fixed point, {5, 4, 3, 2, 1} has a fixed point. You may wonder why we are suddenly asking a probability question. Actually, this is equivalent to the following counting question: What is the number of permutations of {1,2,3,…,n} with no fixed point? Fixed Points in a Permutation What is the number of permutations of {1,2,3,…,n} with no fixed point? For this question, it is more convenient to count the complement. Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let A1 be the set of permutations in which the number 1 is in position 1. … Let Aj be the set of permutations in which the number j is in position j. … Let An be the set of permutations in which the number n is in position n. S = A1[ A2[ … [ An Note that Ai and Aj are not disjoint, and so we need inclusion-exclusion. Fixed Points in a Permutation Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let Aj be the set of permutations in which the number j is in position j. S = A1[ A2[ … [ An How large is \|Aj\|? Once we fixed j, we can have any permutation on the remaining n-1 elements. Therefore, \|Aj\| = (n-1)! How large is \|AiÅ Aj\|? Once we fixed i and j, we can have any permutation on the remaining n-2 elements. Therefore, \|AiÅ Aj\| = (n-2)! Fixed Points in a Permutation Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let Aj be the set of permutations in which the number j is in position j. S = A1[ A2[ … [ An How large is the intersection of k sets? In the intersection of k sets, there are k positions being fixed. Then we can have any permutation on the remaining n-k elements. Therefore, \|the intersection of k sets\| = (n-k)! Fixed Points in a Permutation Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let Aj be the set of permutations in which the number j is in position j. S = A1[ A2[ … [ An \|the intersection of k sets\| = (n-k)! \|S\| = \|A1[ A2[ … [ An\| \|A1[ A2[ A3[ … [ An\| sum of sizes of all single sets – sum of sizes of all 2-set intersections + sum of sizes of all 3-set intersections – sum of sizes of all 4-set intersections … + (–1)n+1 × sum of sizes of intersections of n sets … Fixed Points in a Permutation Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let Aj be the set of permutations in which the number j is in position j. S = A1[ A2[ … [ An \|the intersection of k sets\| = (n-k)! \|S\| = \|A1[ A2[ … [ An\| \|S\| = \|A1[ A2[ … [ An\| = n! – n!/2! + n!/3! +… (-1)i+1 n!/i! + … + (-1)n+1 … Fixed Points in a Permutation Let S be the set of permutations of {1,2,3…,n} with some fixed point(s). Let Aj be the set of permutations in which the number j is in position j. S = A1[ A2[ … [ An \|S\| = n! – n!/2! + n!/3! +… (-1)i+1 n!/i! + … + (-1)n+1 The number of permutations with no fixed points = n! – \|S\| = n! – n! + n!/2! – n!/3! +… (-1)i n!/i! + … + (-1)n = n! (1 – 1/1! + 1/2! – 1/3! + … + (-1)i 1/i! … + (-1)n 1/n!) -> n!/e (where e is the constant 2.71828…) Euler Function Given a number n, how many numbers from 1 to n are relatively prime to n? When n is a prime number, then every number from 1 to n-1 is relatively prime to n, and so When n is a prime power, then p, 2p, 3p, 4p, …, n are not relatively prime to n, there are n/p = pc-1 of them, and other numbers are relatively prime to n. Therefore, Euler Function Given a number n, how many numbers from 1 to n are relatively prime to n? Suppose Then p, 2p, 3p, 4p, …, n are not relatively prime to n, there are n/p of them. Also, q, 2q, 3q, 4q, …, n are not relatively prime to n, and there are n/q of them. Other numbers are relatively prime to n. Therefore, The numbers pq, 2pq, 3pq, …, n are subtracted twice, and there are n/pq of them. So the correct answer is Euler Function Given a number n, how many numbers from 1 to n are relatively prime to n? Let Let S be the set of numbers from 1 to n that are not relatively prime to n. Let Ai be the set of numbers that are a multiple of pi. S = A1[ A2[ … [ An For the intersection of k sets, say A1, A2, A3,…, Ak then every number in A1Å A2Å … Å Ak is a multiple of p1p2…pk then \|A1Å A2Å … Å Ak\| = n/p1p2…pk Euler Function Given a number n, how many numbers from 1 to n are relatively prime to n? Let Let S be the set of numbers from 1 to n that are not relatively prime to n. Let Ai be the set of numbers that are a multiple of pi. S = A1[ A2[ … [ An \|A1Å A2Å … Å Ak\| = n/p1p2…pk \|A1[A2[A3\| = \|A1\| + \|A2\| + \|A3\| – \|A1 ÅA2\| – \|A1ÅA3\| – \|A2ÅA3\| + \|A1ÅA2ÅA3\| When r=3 (only 3 distinct factors), \|A1[ A2[ A3\| = n/p1 + n/p2 + n/p3 - n/p1p2 – n/p1p3 – n/p2p3 + n/p1p2p3 = n(1-p1)(1-p2)(1-p3) Euler Function Given a number n, how many numbers from 1 to n are relatively prime to n? Let Let S be the set of numbers from 1 to n that are not relatively prime to n. Let Ai be the set of numbers that are a multiple of pi. S = A1[ A2[ … [ An \|A1Å A2Å … Å Ak\| = n/p1p2…pk \|A1[ A2[ A3[ … [ An\| \|S\| = \|A1[ A2[ … [ An\| sum of sizes of all single sets – sum of sizes of all 2-set intersections + sum of sizes of all 3-set intersections – sum of sizes of all 4-set intersections … + (–1)n+1 × sum of sizes of intersections of n sets calculations… = n(1-p1)(1-p2)…(1-pn) Related More by User #### INCLUSION/EXCLUSION CRITERIA INCLUSION/EXCLUSION CRITERIA. 1. 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Dzhevid Mahmud CITIZEN PARTICIPATION UNIVERSITY 2014. First ever excluded were?. Adam and Eve It brings to self-identification . What?. Oxford dictionary 952 views • 34 slides #### G/T: Inclusion, not Exclusion G/T: Inclusion, not Exclusion. by Janet Newton. HB5 Section 46. SCHOOL DISTRICT EVALUATION OF PERFORMANCE IN COMMUNITY AND STUDENT ENGAGEMENT; COMPLIANCE. 304 views • 13 slides #### Inclusion-Exclusion Formula Inclusion-Exclusion Formula. S k. Let A 1 , A 2 , …, A n be n sets in a universe U of N elements. Let: S 1 = \|A 1 \| + \|A 2 \| + …+ \|A n \| S 2 = \|A 1 A 2 \| + \|A 1 A 3 \| + …+ \|A n-1 A n \|, i.e., the size of all A i A j for i  j. 493 views • 18 slides #### Pauli Exclusion Principle Pauli Exclusion Principle. by Robert Nemiroff Michigan Tech. Physics X: About This Course. Officially "Extraordinary Concepts in Physics" Being taught for credit at Michigan Tech Light on math, heavy on concepts Anyone anywhere is welcome No textbook required 3.11k views • 15 slides #### Chapter 6 The principle of inclusion and exclusion Chapter 6 The principle of inclusion and exclusion. 6.1 排容原則 (the principle of inclusion and exclusion) 6.2 亂序 (derangements) 6.3 城堡多項式 (the rook polynomial). 6.1 排容原則. Determine the number of positive integers n , 1  n  30, that are divisible by 2, 3, or 5. Ans: 15+10+6-5-3-2+1=22. 661 views • 13 slides #### Section 3.3 Principle of Inclusion and Exclusion Section 3.3 Principle of Inclusion and Exclusion. ( 2 more counting principles ) Principle of Inclusion and Exclusion Given finite sets A 1 , A 2 , …A n , n 2, then \|A 1 A 2 …  A n \| =. + (-1) n+1 \| A 1 A 2  …  A n \|. A-B, A B, B-A are mutually disjoint sets. 353 views • 15 slides #### Principle of Inclusion-Exclusion Principle of Inclusion-Exclusion. Principle of Inclusion and Exclusion. 644 views • 46 slides #### Binomial Coefficients, Inclusion-exclusion principle Binomial Coefficients, Inclusion-exclusion principle. A. B. Lecture 13: Oct 31. C. D. Plan. Binomial coefficients, combinatorial proof Inclusion-exclusion principle. Binomial Theorem. We can compute the coefficients c i by counting arguments. e.g. 530 views • 26 slides #### Inclusion-Exclusion Formula Inclusion-Exclusion Formula. S k. Let A 1 , A 2 , …, A n be n sets in a universe U of N elements. Let: S 1 = \|A 1 \| + \|A 2 \| + …+ \|A n \| S 2 = \|A 1 A 2 \| + \|A 1 A 3 \| + …+ \|A n-1 A n \|, i.e., the size of all A i A j for i  j. 437 views • 18 slides Loading... More Related English Français About Privacy DMCA Blog Contact © 2025 SlideServe. All rights reserved
190585	https://www.universalclass.com/articles/writing/how-to-teach-prepositions-of-place-to-esl-students.htm	Browse Courses My Classes Using realia makes a lesson more interesting and exciting for the students, but the real benefit is that the use of physical objects can really help students to learn and remember vocabulary. A class on the vocabulary of fruits and vegetables is totally enhanced by real-life examples of each word you are teaching that day. Handing out a potato and getting the students to say "potato" while they pass it around really reinforces the word in their brain. Always take every chance to use real objects in the class whenever practical. Once in the Presentation stage on the first day of a one-on-one class, I took the realia of the day and played catch with the student, who was a young beginner. I lightly tossed each object to her, saying its name, and had her repeat it when she tossed it back. She was right in front of me and, of course, I tossed each object very gently. I tossed her a small book and said "a book," and she tossed it back as she said "a book." For other objects that weren't so appropriate for tossing, I handed them to her. "A glass," I said as I gave it to her, and she repeated it when she handed it back. Besides helping the student learn the vocabulary, it was fun and an ice-breaker for the youngster. If your solo student is not so young or otherwise might be adverse to playing catch, handing the object back and forth works just as well. Using Realia in the Deserted Island Exercise Deserted Island is a classic English teaching class that is fun and really encourages student participation. It can be adjusted to practice different grammar forms, and realia can really help grab student interest while they learn new words. Because it departs from some of the strict structure of PPP methodology, it can be used to change up the routine of your classroom lessons. First, you draw a picture on the board of a sinking ship with stick-figure people on it. As you draw it, say what it is. For example, "You are on a ship. The ship is sinking." You can also draw an arrow showing the ship is going down. Then, next to it, draw an island. On the island draw a river (a source of water), coconut trees on the beach (a possible food source) and a forest (a possible food source but also a possible source of danger). Again, say the words as you draw them. Now ask the students what they need on the island. This can also be done in a level-appropriate way. For students that already have been taught conditionals, you could say "If this is you, what would you need?" For lower-level students, you could way, "This is you. What do you need?" Students will start to name things they want on the island. If they don't immediately start, help them with your realia. Some things you can bring for this class: a box of matches, a lighter, a Swiss Army knife, a small radio, a mobile phone, a can of food, an airplane bottle of liquor, various tools, etc. Elicit the word as you show the each object. Soon the students will be naming their own. Students will often say a gun or rifle (for hunting in the forest or for protection). Some will say a "satellite phone." Usually someone will say "coffee" and "cigarettes." Sometimes a student will say "a friend" but it's best to limit the list to objects. Write all the words on the board, and draw a picture for the words that you don't have realia for. This list can be as long as you like, which might break the rule of thumb concerning the ideal number of vocabulary words per classroom lesson, but it's okay in this case especially since the students will be offering up most of the vocabulary. However, limit the length of time for this part of the class. It should be no more than one-third of the class. So you now have a list like this: a box of matches a knife a Swiss Army knife a gun a rifle bullets canned food (etc.) After this part of the class, I sometimes amend the words to make the exercise go smoother during the next part. For example, I change "canned food" to "a box of canned food." You will see why in a moment. Now, next to the list of words write "Why?" and elicit reasons to have each object. For example: What? Why? a box of matches to make a fire a knife to cut food a gun to hunt / to protect yourself a rifle " a box of canned food to eat (etc.) This part can be adjusted according to the class level and your target language. For example, under "Why?" the list could say "for protection" instead of "to protect." At this point, you can do the standard vocabulary drill we have learned in previous lessons. Have the students write the words on the board in their notebooks while you go around and check for misspellings, then have them close their books and verbally test them on the words, etc. Now announce that because the ship in sinking, the students can only bring five of the objects from the list with them. Tell the students to make a list in their notebooks of the five objects they choose. If you want the students to practice modals, you can have them write five sentences such as: If I was alone on a deserted island, I would need a box of matches to make a fire. With a box of matches, I could make a fire. I should have a gun so I can hunt. For a class on comparisons, you could make them write sentences like this: A box of matches is more useful than a gun. For a more basic class, sentences like this: I need a box of matches to make a fire. Next, put the class into groups. For a larger class, you can make two or three groups. For a small class, two or three groups of two or three. Tell the class that each group must discuss and agree on a new list of the five objects they want on the island. Make sure their discussion is in English ("English, please") but otherwise let the students talk freely in their groups. This can be the end of the class, with discussion within groups being a de factor Production phase of the class. Or you can leave time for a final part of the class, where you compare the group results. In this case, give the students enough time to finish their group discussions by saying "you have five minutes" (or ten for a longer class). Then have someone from each group read their list out loud while you write each list on the board. Point out the differences and similarities. You can adjust this part according to the level of your class. If your class has already studied comparison, you can lead the open discussion by asking about the differences in the list by saying "Why is a gun more important than a phone?" or "Why is having a lighter better than having a radio?" At the end, you can have the class vote in order to choose between the differences, and then cross out the losing objects, until there are only five objects left on the list. Realia also helps in understanding concepts, as you will see below in the lesson plan on prepositions of place. A Lesson Plan on Prepositions of Place Classroom Objectives: -For students to learn and be able to use the prepositions of place in, on, under, next to, behind, in front of and between -Vocabulary: dog, cat, box, house, ball and bird. Presentation First teach the vocabulary with the techniques we have covered in previous lessons. Then take out two empty boxes and some object. In this case, we will use a ball, but it could be a mobile phone, a book, or anything you choose. A new technique for eliciting Then place the ball in one of the boxes. Say "the ball is mmm the box." The "mmm" is a noise you make to indicate that there is a missing word in the sentence. This is another elicitation technique which often comes in handy in the classroom. You can also write on the board "the ball is _________ the box" and point to each word as you read it aloud, but say "mmm" for the blank. This will help your student to understand what you mean by the "mmm." So you say "the ball is mmm the box" and look to the class for a response. If they get it and someone knows the answer, someone will offer "in" (or maybe an incorrect answer, in which case we use the techniques discussed in previous lessons, such as avoiding the word "no" etc.) Instead of this eliciting that way, you can simply write on the board "Where is the ball?" and then ask the question to the class. If they don't get it or you get no correct answers, say "the ball is IN the box." Go to the board and draw a simple sketch of a square with a circle in the box, and write "in" next to the drawing. Have the students repeat "in" and "in the box." Repeat for the words on, under, next to, behind, in front of and between (using two boxes with the ball in the middle for between.) Using the techniques we have covered, have the students write the words in their notebooks, then test them on the vocabulary. You do this by erasing the words next to the drawings and asking them to identify the right word that goes with each drawing. Then you can use your realia and act out each scenario, quickly and wordlessly, for example by putting the ball in the box and eliciting a response. Some will call out "in the box" while others just "in." Both are fine. Practice This first exercise is a transition from Presentation to Practice. Practice 1 You can use an open pairs to demonstrate this task. (Later you will have the class perform in closed pairs for the Presentation.) In this case, choose one student with an open hand, and then indicate "Where is the ball?" to the first student, getting the student to repeat. Then choose another student to reply based on the arrangement of the objects, for example "it is in the box." Get the first student to repeat the question and the second student to repeat the answer, several times. Then change the location of the box and ball, have the first student repeat the question, and the second student to reply with the new answer, "it's on the box" (for example.) Then change the arrangement of the box and ball, choose two new students, indicate the question, have the first student ask, and the second answer based on the new arrangement. As soon as it's clear the students understand the task, quickly move through the class, choosing pairs of students, until each student has asked questions and given answers. Practice 2 The second Practice exercise can be a fill-in-the-blank exercise. A fun one is to hand out a quizzes with drawings of a dog, a cat, a bird and a house in the various positions (in the house, on the house etc.) The first part can be a simple fill-in-the-blank quiz for choosing the correct preposition. For example, Fill in the blank with on, under, next to, behind, in front of or between 1. The dog is ____________ the house. 2. The cat is ____________ the houses. (Etc.) As you are handing out the quiz, announce "four minutes!" (or whatever is appropriate for the length of the quiz.) During this time you walk around the class and encourage students while also encouraging self-correction of mistakes, as we have learned in previous lessons. You could even do this without a handout by numbering images on the board and writing the example sentence also on the board. Have the students write the answers in their notebooks. For example, 1. IN 2. BETWEEN (etc.) Finish this practice quickly by asking the whole class for the correct answers. You can simply say, "Number one?" and let the students say the answer out loud. If anyone says the wrong answer, encourage a self-correction. Practice 3 The third exercise can be a full, if short, writing exercise. Using images similar to the one in the previous exercise (or the same ones already on the board), have the students write full sentence to describe each image. Again, announce a time limit and afterwards get students to read their answers, on line at a time, quickly going through the whole class. Production Now you can put the class into closed pairs. Have the students use their book (or notebook) and their pens as their own realia. Model the activity by putting a pen on a notebook and asking "where is the pen?" After they answer, put the book between two pens, and ask "where is the book?" After they answer, have the students do the activity in their pairs. As always in the Production, go through the class and encourage the activity, in this case making sure they are taking turns asking and answering as well as using all the prepositions. In addition or instead of this exercise, you can do any of the following: Further activities for the Production -Drawing pictures: In their pairs, each student draws a picture suggesting a preposition of place, and asks the other student "where is the...?" while taking turns. To present this activity, you could do it a couple times on the board. Encourage fun by drawing odd pictures. Where is the bird? The bird is in the car. Where is the man? The man is on the big bird, etc. An alternative to this exercise is to have the students draw themselves in various places. Then the person who draws says "Where am I?" and the other student looks at the drawing and answers. Another idea is to have students draw their rooms. Then the students can ask and answer questions such as "What is on the table? Where is the lamp?" -Describing pictures: This is similar to the above exercise, but instead you hand out pictures that show common objects in a common setting. They could be photos of an open refrigerator, fruits and vegetables in a supermarket, objects on shelves. Again the exercise is the have one student ask "where is the..?" while the other answers, taking turns. How to Write a Short Report (Over Email) Basic Punctuation Usage Everyone Should Know How to Write a Successful Proposal through Email How to Teach English Language Conditionals to ESL Students What Makes Good Business Writing? 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190586	https://www.ck12.org/flexi/geometry/counting-techniques/an-ice-cream-shop-offers-5-flavors-of-ice-cream-and-2-types-of-toppings-how-many-different-ice-cream-combinations-can-you-have-with-one-flavor-and-one-topping/	Flexi answers - An ice cream shop offers 5 flavors of ice cream and 2 types of toppings. How many different ice cream combinations can you have with one flavor and one topping? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects Geometry Counting Techniques Question An ice cream shop offers 5 flavors of ice cream and 2 types of toppings. How many different ice cream combinations can you have with one flavor and one topping? Flexi Says: The number of combinations can be found by multiplying the number of choices for each category. In this case, there are 5 choices for ice cream flavors and 2 choices for toppings. So, the total number of combinations is: 5×2 10 different ice cream combinations can be made with one flavor and one topping. Analogy / Example Try Asking: At a music festival, there are eight bands scheduled to play, numbered 1 through 8. If band 2 is performing first and band 8 last, then how many ways can their appearances be scheduled?A student is selecting books from a library shelf. There are 6 fiction books and 4 non-fiction books. If the student needs to choose 2 books, how many ways can it be done such that both of the selected books are either fiction or non-fiction?A camp counselor and four campers are to be seated along a picnic bench. In how many ways can this be done if the counselor must be seated in the fourth seat and a camper who has a tendency to engage in food fights must sit to the counselor's immediate left? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy
190587	https://math.stackexchange.com/questions/4301522/complex-numbers-in-system-of-linear-equations	Skip to main content Complex numbers in System of Linear Equations Ask Question Asked Modified 3 years, 9 months ago Viewed 565 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Good evening, Summing up the problem I have is the following: Is it possible to have complex numbers in a System of Linear Equations? It is hard for me to visualize this problem as well, since if we have a System of Linear Equations with, for example, three unknowns and three equations, it is possible to find the relationship between those three equations of lines. Visualizing this is doable as well. But, trying to do this with complex numbers is confusing me. I do think it is possible but an approach is not coming to mind. Kind regards, Tema. linear-algebra Share CC BY-SA 4.0 Follow this question to receive notifications edited Nov 9, 2021 at 18:41 MachineLearner 4,09711 gold badge88 silver badges1515 bronze badges asked Nov 9, 2021 at 18:33 TemaTema 1322 bronze badges 2 If you have three real, linear equations of three variables, then you don't have lines, you have planes. – Arthur Commented Nov 9, 2021 at 19:00 Oh, I apologize for the mis-use of that term! But what would your explanation be if you added complex numbers to those planes, then? And how is that to visualize? – Tema Commented Nov 9, 2021 at 19:03 Add a comment \| 2 Answers 2 Reset to default This answer is useful 0 Save this answer. Show activity on this post. Yes, you can have sets of complex linear equations. For instance, {(5+3i)x+(2−i)y=1(−7+4i)x+(8+9i)y=4−3i Yes, it's difficult to visualise. This one in particular requires you to see in your mind's eye two planes in four-dimensional space, and find their intersession point. And it doesn't get easier for higher number of unknowns. However, the algebra doesn't care. You solve this more or less exactly the same way you would solve a set of two real equations in two variables. Get rid of one variable through either substitution, or adding some suitable multiple of one equation to the other equation, then solve for the other variable. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Nov 9, 2021 at 19:05 ArthurArthur 205k1414 gold badges185185 silver badges331331 bronze badges 2 Thanks a lot, Arthur ^^ What do you mean exactly with higher dimensions though? If we have even more unknowns? – Tema Commented Nov 9, 2021 at 19:15 @Tema Yes. Just like each new variable increases the geometric dimension of the problem by 1 in the real case, it increases the geometric dimension by 2 in the complex case. – Arthur Commented Nov 9, 2021 at 19:39 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Indeed, this is possible but you should not expect to get a nice visualisation. First of all, all the arithmetic operations that you need to solve linear equations over the reals also work for complex numbers. So if you know the arithmetical rules for complex numbers then you can solve such a system of linear equations exactly in the same way as for real numbers. But why is it harder to visualize this compared to a system over real numbers? Assume that you have a system of linear equations over the complex number with 3 variables and 3 equations. Then every variable can have a real and imaginary part now. Thus you can translate this into an equivalent system with 6 variables and 6 equations now. Let me give an example. Imagine you have the following system of linear equations over the complex numbers: z+iu=1,z−u=i We can transform this into an equivalent system over the reals by writing z=αz+iβz and u=αu+iβu. Plugging this into the equations yields: αz+iβz+i(αu+iβu)=1,αz+iβz−αu+iβu=i Now we can group this a bit differently: αz+−βu+i(βz+αu)=1,αz−αu+i(βz+βu)=i Now each of these equations is actually two equations (because we have a constraint on both the imaginary and the real part in each equation). So we can write this equivalently as: αz+−βu=1,i(βz+αu)=0,αz−αu=0,i(βz+βu)=i As a last step we observe that we can divide the equations for the imaginary parts by i in order to get a system of linear equations over the reals: αz+−βu=1,βz+αu=0,αz−αu=0,βz+βu=1 This last system of equations (over the reals) is equivalent to what we started from, but it has now four variables and four constraints whereas our initial system had two variables and two constraints. Note that you now can visualize this system of real equations (imagine 4 lines in the plane). But it is not clear (at least to me) how this would translate to a visualization of the behavior of the complex variables z and u. Share CC BY-SA 4.0 Follow this answer to receive notifications answered Nov 9, 2021 at 19:04 sebastiansebastian 2,19911 gold badge88 silver badges2121 bronze badges 1 Thank you for this insightful answer, araomis! – Tema Commented Nov 9, 2021 at 19:18 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra See similar questions with these tags. 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190588	https://www.opm.gov/forms/pdf_fill/sf182.pdf	Authorization, Agreement, and Certification of Training A. Agency code, agency sub-element and submitting office number B. Request Status Section A – Trainee Information 1. Applicant's Name (Last, First, Middle Initial) 2. Social Security Number (###-##-####) (Agency Use Only) 3. Date of Birth (Enter Date as yyyy-mm-dd) (Agency Use Only) 4. Home Address (Optional) (Number, Street, City, State, ZIP Code) 5. Home Telephone (Optional) (Include Area Code) 6. Position Level (See page 4 for additional instructions) 7. Organization Mailing Address (Branch-Division/Office/Bureau/Agency) 8. Office Telephone (Include Area Code and Extension) 9. Work Email Address 10. Position Title 11. Does applicant need special accommodation? If yes, please describe the requirements below 12. Type of Appointment 13. Education Level (See page 4 for additional instructions) 14. Pay Plan 15. Series 16. Grade 17. Step Section B – Training Course Data 1a. Name and Mailing Address of Training Vendor (No., Street, City, State, ZIP Code) 1.b Location of Training Site (if different form 1a) 1c. Vendor Telephone Number 1d. Vendor Email Address 1e. Vendor website 1f. Vendor Point-of-Contact (POC) 2a. Course Title 2b. Course Number Code 3. Training Start Date (Enter Date as yyyy-mm-dd) 4. Training End Date (Enter Date as yyyy-mm-dd) 5. Training Duty Hours 6. Training Non-Duty Hours 7. Training Purpose Type (See page 6 for additional instructions) 8. Training Type Code (See page 6 for additional instructions) 9. Training Sub Type Code (See page 6 for additional instructions) 10. Training Delivery Type Code (See page 8 for additional instructions) 11. Training Designation Type Code 12. Training Credit 13. Training Credit Type Code 14. Training Accreditation Indicator 15. Continued Service Agreement Required Indicator (Agency Use Only) (See page 8 for additional instructions) 16. Continued Service Agreement Expiration Date (Enter date as yyyy-mm-dd) 17. Training Source Type Code (See page 8 for additional instructions) 18. Individual or Group Training 19. Student/ Membership ID 20. Skill Learning Objective 21. Agency Use Only (For use by agency as needed) Section C – Costs and Billing Information 1. Direct Costs and Appropriation/Fund Chargeable 2. Indirect Costs and Appropriation/Fund Chargeable Item Amount Appropriation Fund Item Amount Appropriation Fund a. Tuition and Fees $ a. Travel $ b. Books & Material Costs $ b. Per Diem $ c. Total $ c. Total $ 3. Total Training Non-Government Contribution Cost 6. Billing Instructions (Furnish invoice to) 4. Document/Purchasing Order/Requisition Number 5. 8-Digit Station Symbol (Example: 12-34-5678) U.S. Office of Personnel Management Page 1 NSN 7540-01-008-3901 Standard Form 182 Revised March 2020 All previous editions not usable. Select one Select one Select one Select one Select one Select one Select one Select one Select one Select one Select one Select one Select one U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 2 Section D – Approvals Complete the appropriate number of approvals your agency requires (e.g. first, second, and/or third level approval) before submission of this form to the Agency Training Office. 1a. Immediate Supervisor/First-line Supervisor (Name and Title) 1b. Telephone Number (Include Area Code and Extension) 1c. Email Address 1d. Signature 1e. Date (Enter Date as yyyy-mm-dd) 2a. Second-line Supervisor (Name and Title) 2b. Telephone Number (Include Area Code and Extension) 2c. Email Address 2d. Signature 2e. Date (Enter Date as yyyy-mm-dd) 3a Training Officer (Name and Title) 3b. Telephone Number (Include Area Code and Extension) 3c. Email Address 3d. Signature 3e. Date (Enter Date as yyyy-mm-dd) Section E – Approvals/Concurrence To be completed by the nominating Agency Official authorized to approve or disapprove training requests. 1a. Authorizing Official (Name and Title) 1b. Telephone Number (Include Area Code and Extension) 1c. Email Address 1d. Signature 1e. Date Section F – Certification of Training Completion and Evaluation 1a. Authorizing Official (Name and Title) 1b. Telephone Number (Include Area Code and Extension) 1c. Email Address 1d. Signature 1e. Date Training Facility: Bills should be sent to office indicated in item C6. Please refer to number given in item C4 to assure prompt payment. Agency Certifying Officials are certifying the employee has completed the requirements for the training and an evaluation has been completed. The requirement to evaluate training is found in 5 CFR 410.202. The agency head shall evaluate training to determine how well it meets short and long-range program needs of the agency and the individual. The needs should be aligned with the strategic plan to strengthen and develop the performance and behavior of the individual whose positive results will impact the performance of the agency. SIGN SIGN SIGN SIGN SIGN Print Form Clear Form U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 3 Privacy Act Statement Authority ─ This information is being collected under the authority of 5 U.S.C. § 4115, a provision of The Government Employees Training Act. Purposes and Uses ─ The primary purpose of the information collected is to document the approval and completion of employee training within an agency. This completed training information must be electronically provided to the Governmentwide system where it will become a part of the permanent employment record and subject to all published routine uses of that system of records. Information collected may also be provided to other agencies and to Congress upon request. Effects and Nondisclosure ─ Providing the personal information requested, in paper form, is voluntary and at the agency’s discretion. However, failure to provide this information electronically to the Governmentwide system may result in errors in processing and documenting the training you have completed. Information Regarding Disclosure of your Social Security Number (SSN) Under Public Law 93-579, Section 7(b) ─ Solicitation of SSNs by the U.S. Office of Personnel Management (OPM) is authorized under provisions of the Executive Order 9397, dated November 22, 1943. However, agencies must also safeguard such Personally Identifiable Information (PII) when providing completed training information to the Governmentwide system. Employee SSNs will be used primarily to give proper recognition for completed training and to accumulate Governmentwide statistical data and information. U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 4 Agency Training Electronic Reporting Instructions General Instructions: 1. You must complete all questions in sections A-E on the training application. In addition, your financial institution must complete Section F, Certification of Training Completion and Evaluation section. 2. Electronic Requirements - An agency should only submit data for completed training events for which all mandatory data elements have been recorded. Additional Instructions for Section A - Trainee Information: 6. Position Level - Select whether the employee's position level is one of the following: 6a. Non-supervisory - Anyone who does not have supervisory/team leader responsibilities. 6b. Supervisory - First-line supervisors who do not supervise other supervisors; typically, those who are responsible for an employee's performance appraisal or approval of their leave. 6c. Manager - Those in management positions who typically supervise one or more supervisors. 6d. Executive - Members of the Senior Executive Service (SES) or equivalent. 13. Education Level - Use the employee educational level codes listed below. Code Short Description Long Description (if applicable) 1 No formal education or some elementary school--did not complete Elementary school means grades 1 through 8, or equivalent, not completed. 2 Elementary school completed--no high school Grade 8 or equivalent completed. 3 Some high school--did not graduate High school means grades 9 through 12, or equivalent. 4 High school graduate or certificate equivalency N/A 5 Terminal occupational program--did not complete Program extending beyond grade 12, usually no more than three years; designed to prepare students for immediate employment in an occupation or cluster of occupations; not designed as the equivalent of the first two or three years of a baccalaureate degree program. Includes cooperative training or apprenticeship consisting of formal classroom instruction coupled with on-the-job training. 6 Terminal occupational program--certificate of completion, diploma or equivalent See code 5 above for definition of terminal occupational program. Two levels are recognized: (1) The technical and/or semi-professional level preparing technicians or semi-professional personnel in engineering and non-engineering fields; and (2) the craftsman/clerical level training artisans, skilled operators, and clerical workers. 7 Some college--less than one year Less than 30 semester hours completed. 8 One year of college 0-59 semester hours or 45-89 quarter hours completed. 9 Two years of college 60-89 semester hours or 90-134 quarter hours completed. 10 Associate Degree 2-year college degree program completed. 11 Three years of college 90-119 semester hours or 135-179 quarter hours completed. 12 Four years of college 120 or more semester hours or 180 or more quarter hours completed--no baccalaureate (Bachelor's) degree. 13 Bachelor’s Degree Requires completion of at least four, but no more than five, years of academic work; includes a Bachelor's degree conferred in a cooperative business, industry, or Government to allow student to combine actual work experience with college studies. U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 5 Code Short Description Long Description (if applicable) 14 Post-Bachelor's Some academic work beyond (at a higher level than) the Bachelor's degree but no additional higher degree. 15 First professional Signifies the completion of academic requirements for selected professions that are based on programs requiring at least two academic years of previous college work for entrance and a total of at least six academic years of college work for completion, e.g., Dentistry (D.D.S. or D.M.D.), Law (LL. B. or J.D.), Medicine (M.D.), Theology (B.D.), Veterinary Medicine (D.V.M.), Chiropody or Podiatry (D.S.C. or D.P.), Optometry (O.D.), and Osteopathy (D.O.). 16 Post-first professional Some academic work beyond (at a higher level than) the first professional degree but no additional higher degree. 17 Master's degree For liberal arts and sciences customarily granted upon successful completion of one (sometimes two) academic years beyond the Bachelor's degree. In professional fields, an advanced degree beyond the first professional but below the Ph.D., e.g., the LL.M.; M.S. in surgery following the M.D.; M.S.D., Master of Science in Dentistry; M.S.W., Master of Social Work, and MA, Master of Arts. 18 Post-Master's Some academic work beyond (at a higher level than) the Master's degree but no additional higher degree. 19 Sixth-year degree Includes such degrees as Advanced Certificate in Education, Advanced Master of Education, Advanced Graduate Certificate, Advanced Specialist in Education Certificate, Certificate of Advanced Graduate Study, Certificate of Advanced Study, Advanced Degree in Education, Specialist in Education, Licentiate in Philosophy, Specialist in Guidance and Counseling, Specialist in Art, Specialist in Science, Specialist in School Administration, Specialist in School Psychology, and Licentiate in Sacred Theology. 20 Post-sixth year Some academic work beyond (at a higher level than) the sixth-year degree but no additional higher degree. 21 Doctorate degree Includes such degrees as Doctor of Education, Doctor of Juridical Science, Doctor of Public Health, and the Ph.D. (or equivalent) in any field. Does not include a Doctor's degree that is a first professional degree, per code 15. 22 Post-Doctorate Work beyond the Doctorate. U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 6 Additional Instructions for Section B - Training Course Data: 7. Training Purpose Type - Select and insert the purpose for taking this course or program using the appropriate training purpose type code from the list below. Code Short Description Long Description (if applicable) 01 Program/Mission Training to provide the knowledge, skills and abilities needed as a result of agency mission, policies, or procedures. 02 New Work Assignment Training to acquire the knowledge, skills and abilities needed as a result of assignment to new duties and responsibilities when such training is not part of a planned, career development program (e.g., training provided to a staffing specialist who has been newly assigned to a position involving classification duties). 03 Improve/Maintain Present Performance Training to provide the knowledge, skills and abilities needed to improve or maintain proficiency in present job. 04 Future Staffing Needs Training to provide the knowledge, skills, and abilities needed to meet future staffing needs (e.g., to implement succession planning). 05 Develop Unavailable Skills Training to acquire the knowledge, skills and abilities needed for fields of work for which the labor market cannot produce a sufficient number of trained candidates (e.g., air traffic controllers or Information Technology [IT] professionals). 06 Retention Training/education used to address staffing issue of retaining an employee (e.g., academic degree training). 9. Training Sub-Type Code - There are Sub-Type Categories for each of the three (3) different Training Type Codes. Select one (1) Sub-Type Category code that applies to the training type code you selected. Training Type Code Training Sub Type Code Competency 01 - Training Program Area 01 - Legal Education or training in the concepts, principles, and theories, or techniques of law. Technical 02 - Medical and Health Education or training in the concepts, principles, and theories, or techniques of medicine or health sciences. Technical 03 - Scientific Education or training in the concepts, principles, and theories, or techniques of disciplines such as physical, biological, natural, and social sciences; education; mathematics or statistics. Technical 04 - Engineering or Architecture Education or training in the concepts, principles, and theories, or techniques of disciplines such as architecture and engineering. Technical 05 - Human Resources Education or training in the concepts, principles, and theories of such fields as Human Resources Management and/or Strategic Human Capital Management. Human Resources 06 - Budget/Finance Business Administration Education or training in the concepts, principles, and theories of business administration, accounting and finance. Financial 07 - Planning and Analysis Education or training in the concepts, principles, and theories of systems analysis; policy, program or management analysis; or planning, including strategic planning. Technical 08 - Information Technology Education and training in the concepts and application of all facets of Information Technology. Technical U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 7 Training Type Code Training Sub Type Code Competency 01 - Training Program Area (continued) 09 - Project Management Education and training in the concepts, principles, and theories necessary to plan, develop, and modify the management of products, services, or systems with regards to scope, resources, and time. Technical 10 - Acquisition Education or training in the concepts, principles, and theories or techniques related to acquiring and procuring goods and services for the government. Financial 11 - Logistic Specialty Training for professional skills of a specialized nature in the methods and techniques of such fields as supply, procurement, transportation, or air traffic control. Technical 12 - Security Training of a specialized nature in the methods and techniques of investigation in cybersecurity, physical security, personal security, and police science. Technical 13 - Administration (Non-supervisory administrative & program/project support) Training in skills used for administrative tasks associated with business/office management and program/project support. Human Resources 14 - Trade and Craft Training in the knowledge, skills, and abilities needed to perform and accomplish required tasks in trade and craft occupations (i.e. Carpentry, Plumbing, Electrical, etc.). Technical 15 - Foreign Affairs Training for professional skills of a specialized nature in the methods and techniques of such fields as foreign languages, foreign culture, intelligence (foreign affairs/intelligence), diplomacy, or strategic studies. Technical 02 - Developmental Training Area Description: Formal developmental/training programs. 20 - Supervisory Program/First-Line Supervisors Development/training program which provides education or training in supervisory principles and techniques in such subjects as personnel policies and practices (including equal employment opportunity, merit promotion, and labor relations); human behavior and motivation; diversity and inclusion; communication processes in supervision, work planning, scheduling, and review; and performance evaluation for first-line supervisors. Leadership 21 - Management Program Development/training program which provides mid-management level education or training in the concepts, principles, and theories of such subject matters as public policy formulation and implementation, management principles and practices, quantitative approaches to management, or management planning organizing and controlling. (Supervisors of supervisors; GS-14/15 supervisors; GS-14/15 direct reports to SES). Leadership 22 - Leadership/Manager/Pre-supervisory Development Formal developmental program that provides mid-level and leadership training and development opportunities for supervisors and non-supervisors; which includes coursework that addresses skill areas such as Leadership/Management and Communication (e.g., written, oral and interpersonal). Leadership 23 - SES Candidate Development OPM-approved program to prepare potential SES members. Leadership 24 - Executive Development Continuing development for leaders above the GS-15 level. Leadership 25 - Mentoring Program Formal stand-alone program with established goals and measured outcomes. Open to all who qualify; protégées and mentors paired to facilitate compatibility, training and support provided, and company benefits directly. Employee Development 26 - Coaching Program Formal stand-alone program which provides ongoing partnership with an employee and coach that helps employee produce desired results in professional life. Employee Development U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 8 Training Type Code Training Sub Type Code Competency 03 - Basic Training Area Description: Fundamental and/or required training programs. 30 - Employee Orientation Training of a general nature to provide an understanding of the organization and missions of the Federal Government, employing agency or activity, or a broad overview and understanding of matters of public policy. Employee Development 31 - Adult Basic Education Education or training to provide basic completeness in such subjects as remedial reading, grammar, arithmetic, lip reading or Braille. Employee Development 32 - Federally Mandated Training Mandatory training for all employees Governmentwide. This includes training mandated by federal statute or regulation; such as in the areas of computer security awareness (5 CFR 930.301-305), ethics (5 CFR 2638.703 and 704), or executives, managers, and supervisors (5 CFR Part 412). Federally Mandatory 33 - Work-life Training to promote work-life (e.g., health and wellness training, employee retirement/benefits training, etc.). Human Resources 34 - Professional Skills Training on non-technical skills related to employee performance such as communication, conflict resolution, emotional intelligence, organization, time management, etc. Fundamental Skills 35 - Agency Required Training Agency-specific training required by the agency and provided to Federal employees in order to achieve the goals and objectives of the Agency as needed. For example: agency training based on Inspector General's Audit; agency training aimed at improving individual's needs based on Performance Improvement Plan (PIP); agency training based on signing agreement between Union and Management. Note: Unlike Federally mandated training, this type of training is not mandated by Federal statute for all Federal employees. Agency Required 10. Training Delivery Type Code Code Short Description Long Description (if applicable) 1 Traditional Classroom (no technology) Individual or multiple person led, face-to-face training. 2 On the Job Formal methods/activities planned and structured to promote learning by doing; e.g., detail assignments/programs. 3 Technology Based Methods mainly using technology, which may include tutorials embedded in software, CD ROM products, Web-based courses, and interactive media. 4 Conference/Workshop An organized learning event which has an announced educational or instructional purpose; more than half the time is scheduled for a planned, organized exchange of information between presenters and audience which meets the definition of training in 5 U.S.C. 4110; content of the conference/retreat is germane to improving individual and/or organizational performance; and developmental benefits will be derived through the employee's attendance. 5 Blended Training that requires two or more methods of delivery that must be completed in order to satisfy the educational requirements. 6 Correspondence Self-study course material: Training provided via the assignment of non-interactive methods such as a book, document, regulation, or manual. U.S. Office of Personnel Management Standard Form 182 Revised March 2020 All previous editions not usable. Page 9 11. Training Designation Type Code - Select and insert the appropriate training credit designation type code: Code Short Description 01 Undergraduate Credit 02 Graduate Credit 03 Continuing Education Credit 04 Post Graduate Credit 05 N/A 13. Training Credit Type Code - Select and insert the appropriate training credit designation type code only if “training credit” is greater than zero: Code Short Description 1 Semester Hours 2 Quarter Hours 3 Continuing Education Unit 4 Professional Development Unit 5 Continuing Professional Education 6 Continuous Learning Points 7 Other 15. Continued Service Agreement Required Indicator - (Agency response.) An agency may require an employee to enter into a continued service agreement (CSA) with the agency prior to authorization for training. According to 5 CFR 410.309, the head of the agency determines the requirements for CSA’s. Typically, when an employee signs a CSA, an employee agrees to remain in the agency’s service for a specified period of time after receipt of training. Indicate whether, or not, the employee signed an agreement to remain in the service of the agency for a specified amount of time as a condition of participating in the authorized training. Select Yes (Y), No (N) or non-applicable (N/A). 17. Training Source Type Code: Code Short Description Long Description (if applicable) 1 Government Internal Training provided by a Federal department, agency, or independent establishment for its own employees. 2 Government External Training provided by an interagency training activity, or a Federal department, agency, or independent establishment other than the one which currently employs the trainee. 3 Non-government Sources include commercial or industrial concern, educational institutions, professional societies or associations, or consultants or individuals who are not Government employees, (but are contracted to develop and/or provide training course or program.) 4 Government: State/Local Training provided by a state, county, or municipal Government. Education provided by State-operated or other public educational institutions is reported as non-Government. 5 Foreign Governments and Organizations Training provided by non-United States entities which may or may not be outside the United States.
190589	http://mathcentral.uregina.ca/QandQ/topics/logic%20puzzle	Quandaries & Queries at Math Central Math Central - mathcentral.uregina.ca Quandaries & Queries Q & Q Topic: logic puzzlelist of topics start over 3 items are filed under this topic. Page 1/1 Two logic problems2007-12-07 From Grace: 1. You are an archaeologist that has just unearthed a long-sought triplet of ancient treasure chests. One chest is plated with silver, one with gold, and one with bronze. According to legend, one of the three chests is filled with great treasure, whereas the other two chests house a man-eating python that can rip your head off. Faced with a dilemma, you then notice that there are inscriptions on the chests: Silver Chest - Treasure is in this Chest. Gold Chest - Treasure is not in this Chest. Bronze Chest - Treasure is not in the Gold Chest . You know that at least one of the inscriptions is true, and at least one of the inscriptions is false. Which chest do you open? (a) Silver (b) Gold (c) Bronze Answered by Penny Nom. 8 squares from 12 sticks2002-10-08 From A student: If you have 12 sticks the same size, how do you make them into 8 squares? Answered by Claude Tardif. The community garden2002-09-08 From Michele: Bryce, Duanne, Juliana and Sonya share a 40ft by 48 ft rectangular portion of the community garden in the neighborhood. With this space each has their own rectangular garden plot. -Bryce is the only one with a square plot. its are is 1/2 the area of Duanne's -The area of Bryce's plot is 2/3 the area of Julianna's plot. Their plots have one side in common. -Sonya's garden has an area twice that of Julianna's garden. What are the dimensions of each person's garden plot? Answered by Leeanne Boehm. Page 1/1 Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. about math central :: site map :: links :: notre site français
190590	https://www.chemteam.info/Stoichiometry/Limiting-Reagent.html	Stoichiometry Limiting Reagent Examples \| \| \| --- \| \| Limiting Reagent Problems #1-10 \| Problems solved using dimensional analysis only \| \| Limiting Reagent Problems #11-20 \| Stoichiometry Menu \| First comment before starting: Just a bit below, I'm going to tell you (several times) how to determine the limiting reagent in a chemistry problem. I certainly hope it is something you pay attention to and remember. Figuring out which substance is the limiting reagent is an area that many students struggle with. You will see the word "excess" used in this section and in the problems. It is used several different ways: (a) Compound A reacts with an excess of compound B. In this case, mentally set compound B aside for the moment. Since it is "in excess," this means there is more than enough of it. The other compound will run out first. (b) 20 grams of A and 20 grams of B react. Which is in excess? What we will do below is find out which substance runs out first (called the limiting reagent). Obviously (I hope), the other compound is seen to be in excess. (c) After 20 gm. of A and 20 gm. of B react, how much of the excess compound remains? To answer this problem, a subtraction will be involved. This is a part of many limiting reagent problems and it causes difficult with students. Expect it to be on your test. Second comment before starting: What is the Limiting Reagent? It is simply the substance in a chemical reaction that runs out first. It seems to be a simple concept, but it does cause people problems. Let's try a simple non-chemical example. Reactant A is a test tube. I have 20 of them. Reactant B is a stopper. I have 30 of them. Product C is a stoppered test tube. The reaction is: A + B ---> C test tube plus stopper gives stoppered test tube. So now we let them "react." The first stopper goes in, the second goes in and so on. Step by step we use up stoppers and test tubes (the amounts go down) and make stoppered test tubes (the amount goes up). Suddenly, we run out of one of the "reactants." Which one? We run out of test tubes first. Seems obvious, doesn't it? We had 20 test tubes, but we had 30 stoppers. So when the test tubes are used up, we have 10 stoppers sitting there unused. And we also have 20 test tubes with stoppers firmly inserted. So, which "reactant" is limiting and which is in excess? The test tubes are limiting (they ran out first) and the stoppers are in excess (we have some left over when the limiting reagent ran out). There are two techniques for determine the limiting reagent in chemical problems. The first technique is discussed as part of the solution to the first example. Make sure you take a close look at it. The second technique will make its first appearance in Example #6. Example #1: Here's a nice limiting reagent problem we will use for discussion. Consider the reaction: 2Al + 3I2 ------> 2AlI3 Determine the limiting reagent and the theoretical yield of the product if one starts with: (a) 1.20 mol Al and 2.40 mol iodine. (b) 1.20 g Al and 2.40 g iodine (c) How many grams of Al are left over in part b? Solution for part (a): We already have moles as the unit, so we use those numbers directly. 1) Here is how to find out the limiting reagent: take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent. 2) Let's say that again: to find the limiting reagent, take the moles of each substance and divide it by its coefficient in the balanced equation. The substance that has the smallest answer is the limiting reagent. You're going to need that technique, so remember it. By the way, did you notice that I bolded the technique to find the limiting reagent? I did this so as to emphasize its importance to you when learning how to do limiting reagent problems. 3) Resuming with the problem solution: Aluminum ---> 1.20 / 2 = 0.60 Iodine ---> 2.40 / 3 = 0.80 4) The lowest number indicates the limiting reagent. Aluminum will run out first in part (a) of the question. Why? 1.20/2 means there are 0.60 "groupings" of 2 and 2.40/3 means there are 0.80 "groupings" of 3. If they ran out at the same time, we'd need one "grouping" of each. Since there is less of the "grouping of 2," it will run out first. If you're not sure what I just said, that's OK. The technique works, so remember it and use it. 5) The second part of the question "theoretical yield" depends on finding out the limiting reagent. Once we do that, it becomes a stoichiometric calculation. Al and AlI3 stand in a one-to-one molar relationship, so 1.20 mol of Al produces 1.20 mol of AlI3. Notice that the amount of I2 does not play a role, since it is in excess. Solution for part (b): 1) Since we have grams, we must first convert to moles. The we solve just as we did in part (a) just above. For the mole calculation: Aluminum ---> 1.20 g / 26.98 g mol¯1 = 0.04477 mol Iodine ---> 2.4 g / 253.8 g mol¯1 = 0.009456 mol 2) To determine the limiting reagent: Aluminum ---> 0.04477 / 2 = 0.02238 Iodine ---> 0.009456 / 3 = 0.003152 The lower number is iodine, so we have identified the limiting reagent. 3) Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum. I2 and AlI3 stand in a three-to-two molar relationship, so 0.009456 mol of I2 produces 0.006304 mol of AlI3. Again, notice that the amount of Al does not play a role, since it is in excess. From here figure out the grams of AlI3 and you have your answer. Solution for part (c): Since we have moles, we calculate directly and then convert to grams. Al and I2 stand in a two-to-three molar relationship, so 0.009456 mol of I2 uses 0.006304 mol of Al. Convert this aluminum amount to grams and subtract it from 1.20 g and that's the answer. Just above was some discussion on a way to determine the limiting reagent in a chemistry problem. This particular thing (determine the limiting reagent) is a real stumbling block for students. Be aware! Example #2: 15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. Here is the balanced equation for the reaction: Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S (a) Which is the limiting reagent? (b) What is the maximum mass of H2S which can be formed from these reagents? (c) How much excess reagent remains after the reaction is complete? Some comments first: The key to this problem is the limiting reagent, part (a). Once you know that, part (b) becomes "How much H2S can be made from the limiting reagent?" Part (c) becomes two connected questions: first, "How much Al2S3 is used up when reacting with the limiting reagent?" then second, "What is 15.00 minus the amount in the first part?" Make sure you note that second part. The calculation o be performed gives you the answer to "How much reacted?" but the question is "How much remained?" Lots of students forget to do the second part (the 15 minus part) and so get graded down. Note: I'm carrying a guard digit or two through the calculations. The final answers will appear with the proper number of significant figures. Solution for limiting reagent, part (a): 1) Determine the moles of Al2S3 and H2O Aluminum sulfide ---> 15.00 g ÷ 150.158 g/mol = 0.099895 mol Water ---> 10.00 g ÷ 18.015 g/mol = 0.555093 mol 2) Divide each mole amount by equation coefficient: Aluminum sulfide ---> 0.099895 mol ÷ 1 mol = 0.099895 Water --> 0.555093 mol ÷ 6 mol = 0.0925155 3) The water is the lesser amount; it is the limiting reagent. Solution for mass of H2S formed, part (b) Now that we know the limiting reagent is water, this problem becomes "How much H2S is produced from 10.00 g of H2O and excess aluminum sulfide?" 1) Determine moles of 10.00 g of H2O Water ---> 10.00 g ÷ 18.015 g/mol = 0.555093 mol 2) Use molar ratios to determine moles of H2S produced from above amount of water. (a) the H2O/H2S molar ratio is 6/3, a 2/1 ratio. (b) water is associated with the two. This means the H2S amount is one-half the water value = 0.2775465 mol. 3) Convert moles of H2S to grams. (0.2775465 mol) (34.0809 g/mol) = 9.459 g Solution for excess reagent remaining, part (c) We will use the amount of water to calculate how much Al2S3 reacts, then subtract that amount from 15.00 g. 1) Determine moles of 10.00 g of H2O Water ---> 10.00 g ÷ 18.015 g/mol = 0.555093 mol 2) Use molar ratios to determine moles of Al2S3 that reacts with the above amount of water. (a) the Al2S3/H2O ratio is 1/6 (b) water is associated with the 6. This means the Al2S3 amount is one-sixth the water value = 0.09251447 mol 3) Convert moles of Al2S3 to grams. (0.09251447 mol) (150.158 g/mol) = 13.891943 g 4) However, we are not done. We were asked for the amount remaining and the answer just above is the amount which was used up, so the final step is: 15.00 g − 13.891943 g = 1.108 g Example #3: If there is 35.0 grams of C6H10 and 45.0 grams of O2, how many grams of the excess reagent will remain after the reaction ceases? 2C6H10 + 17O2 ---> 12CO2 + 10H2O Solution: 1) Convert each substance to moles: C6H10 ---> 35.0 g / 82.145 g/mol = 0.426 mol O2 ---> 45.0 g / 31.998 g/mol = 1.406 mol 2) Determine the limiting reagent: C6H10 ---> 0.426 mol / 2 = 0.213 O2 ---> 1.406 mol / 17 = 0.083 O2 is the limiting reagent. Comment: the units don't matter in this step. What we are looking for is the smallest number after carrying out the divisions. The value of 0.083 is the important thing. Not if it has a unit attached to it or not. 3) Determine how many moles of the excess reagent is used up when the limiting reagent is fully consumed: the mole ratio we desire is 2/17 (C6H10 to O2) \| \| \| \| --- \| 2 \| \| x \| \| –––– \| = \| ––––––– \| \| 17 \| \| 1.406 mol \| x = 0.1654 mol of C6H10 consumed 4) Determine grams of C6H10 remaining: 0.426 mol − 0.1654 mol = 0.2606 mol of C6H10 remaining (0.2606 mol) (82.145 g/mol) = 21.4 g remaining (to three sig figs) Example #4: (a) What mass of Al2O3 can be produced from the reaction of 10.0 g of Al and 19.0 g of O3? (b) How much of the excess reagent remains unreacted? Solution to a: 1) Write balanced chemical equation: 2Al + O3 ---> Al2O3 2) Convert grams to moles: Al ---> 10.0 g / 26.982 g/mol = 0.37062 mol O3 ---> 19.0 g / 47.997 g/mol = 0.39586 mol 3) Determine limiting reagent: Al ---> 0.37062 / 2 = 0.18531 O3 ---> 0.39586 / 1 = 0.39586 Al is the limiting reagent 4) Determine moles of product formed: Al to Al2O3 molar ratio is 2 to 1. \| \| \| \| --- \| 2 \| \| 0.37062 mol \| \| –––– \| = \| –––––––––– \| \| 1 \| \| x \| x = 0.18531 mol 5) Determine grams of product: (0.18531 mol) (101.961 g/mol) = 18.8944 g To three sig figs, 18.9 g Solution to b: 1) Determine moles of ozone that reacted: Al to O3 molar ratio is 2 to 1 \| \| \| \| --- \| 2 \| \| 0.37062 mol \| \| –––– \| = \| –––––––––– \| \| 1 \| \| x \| x = 0.18531 mol 2) Determine moles of ozone remaining: 0.39586 mol − 0.18531 mol = 0.21055 mol 3) Determine grams of ozone remaining: (0.21055 mol) (47.997 g/mol) = 10.1 g (to three sig figs) Example #5: Based on the balanced equation: C4H8 + 6O2 ---> 4CO2 + 4H2O Calculate the number of excess reagent units remaining when 28 C4H8 molecules and 228 O2 molecules react? Solution: Remember, numbers of molecules are just like moles, so treating the 28 and 228 as moles is perfectly acceptable. This is because I could divide the 28 and the 228 by Avogadro's Number to obtain the moles. Those mole amounts could be used in the calculation below and the final answer could then be multiplied by Avogadro's Number to obtain the answer of 60. 1) Determine the limiting reagent: butane ---> 28 / 1 = 28 oxygen ---> 228 / 6 = 38 Butane is the limiting reagent. 2) Determine how much oxygen reacts with 28 C4H8 molecules: the butane to oxygen molar ratio is 1:6 28 x 6 = 168 oxygen molecules react 3) Determine excess oxygen: 228 − 168 = 60 Here's aother way to consider this: The 38 above means that there are 38 "groupings" of six oxygen molecules. 38 − 28 = 10 oxygen "groupings" remain after the butane is used up 10 x 6 = 60 Example #6: Determine the maximum mass of TiCl4 that can be obtained from 35.0 g of TiO2, 45.0 g Cl2 and 11.0 g of C. (See comment below problem.) 3TiO2 + 4C + 6Cl2 ---> 3TiCl4 + 2CO2 + 2CO Solution: 1) Assume each reactant is the limiting reagent. Determine the moles of product produced by each assumption: Note: the first factor in each case converts grams of each reactant to moles. The second factor uses a molar ratio from the chemical equation to convert from moles of the reactant to moles of product. There is no need to convert to grams because all three calculations yield moles of the same compound (the TiCl4). \| \| \| \| \| \| --- --- \| \| 1 mole Cl2 \| \| 3 mole TiCl4 \| \| \| 45.0 g Cl2 x \| ––––––––––– \| x \| ––––––––––– \| = 0.31732 mol TiCl4 \| \| \| 70.9064 g Cl2 \| \| 6 mol Cl2 \| \| \| \| \| \| \| \| --- --- \| \| 1 mole C \| \| 3 mole TiCl4 \| \| \| 11.0 g C x \| ––––––––––– \| x \| ––––––––––– \| = 0.68688 mol TiCl4 \| \| \| 12.01078 g C \| \| 4 mol C \| \| \| \| \| \| \| \| --- --- \| \| 1 mole TiO2 \| \| 3 mole TiCl4 \| \| \| 35.0 g TiO2 x \| ––––––––––– \| x \| ––––––––––– \| = 0.438235 mol TiCl4 \| \| \| 79.8658 g TiO2 \| \| 3 mol TiO2 \| \| Cl2 makes the least amount of TiCl4, so Cl2 is the limiting reactant. 2) The mass of TiCl4 produced is: (0.31732 mol TiCl4) (189.679 g TiCl4/mol) = 60.2 g TiCl4 (to three sig figs) Note that the "divide moles by coefficient" was not used to determine the limiting reagent. Instead, a full calculation was done and the least amount of product identified the limiting reagent. Here is what the "divide moles by coefficient" set up looks like: Cl2 ---> 0.63464 / 6 = 0.10577 <--- there's our limiting reagent C ---> 0.915844 / 4 = 0.228961 TiO2 ---> 0.438235 / 3 = 0.14608 Example #7: Determine the starting mass of each reactant if 46.3 of K3PO4 is produced and 92.8 of H3PO4 remains unreacted. 3KOH(aq) + H3PO4(aq) ---> K3PO4(aq) + 3H2O(ℓ) Solution: 1) The fact that some phosphoric acid remains tells us it is the excess reagent. Let us determine the amount of KOH (the limiting reagent) required to produce the 46.3 g of K3PO4. 46.3 g / 212.264 g/mol = 0.2181246 mol of K3PO4 Three moles of KOH are required to produce one mole of K3PO4 (3) (0.2181246 mol) = 0.6543738 mol of KOH required (0.6543738 mol) (56.1049 g/mol) = 36.7 g (to thee sig figs) 2) Determine the starting mass of H3PO4 0.2181246 mol of K3PO4 requires 0.2181246 mol of H3PO4 based on the 1:1 molar ratio from the balanced equation. (0.2181246 mol) (97.9937 g/mol) = 21.4 g (to three sig figs) 21.4 + 92.8 = 114.2 g Example #8: Determine the limiting reagent of this reaction: Na2B4O7 + H2SO4 + 5H2O ---> 4H3BO3 + Na2SO4 There are 5.00 g of each reactant. Solution: 1) Convert everything into moles, by dividing each 5.00 g by their respective molar masses: Na2B4O7 ---> 0.02485 mol H2SO4 ---> 0.05097 mol H2O ---> 0.2775 mol 2) Note that there are three reactants. How is the limiting reagent determined when there are three reactants? Answer: determine the limiting reagent between the first two: Na2B4O7 ---> 0.02485 / 1 = 0.02485 H2SO4 ---> 0.05097 / 1 = 0.05097 Na2B4O7 is the limiting reagent when compared to H2SO4 3) Now, compare the "winner" to the third reagent: Na2B4O7 ---> 0.02485 / 1 = 0.02485 H2O ---> 0.2775 / 5 = 0.0555 Na2B4O7 is the limiting reagent between itself and H2O. Na2B4O7 is the overall limiting reagent in this problem. Example #9: How much O2 could be produced from 2.45 g of KO2 and 4.44 g of CO2? 4KO2 + 2CO2 ---> 2K2CO3 + 3O2 Solution: I will do a solution assuming KO2 is the limiting reagent, then I will do a solution assuming CO2 is the limiting reagent. The reactant that produces the lesser amount of oxygen is the limiting reagent and that lesser amount will be the answer to the question. 1) Solution using KO2: 2.45 g / 71.096 g/mol = 0.03446045 mol \| \| \| \| --- \| 4 \| \| 0.03446045 mol \| \| ––– \| = \| –––––––––––––– \| \| 3 \| \| x \| x = 0.02584534 mol (0.02584534 mol) (31.998 g/mol) = 0.827 g of O2 2) Solution using CO2: 4.44 g / 44.009 g/mol = 0.10088845 mol \| \| \| \| --- \| 2 \| \| 0.10088845 mol \| \| ––– \| = \| –––––––––––––– \| \| 3 \| \| x \| x = 0.151332 mol (0.151332 mol) (31.998 g/mol) = 4.84 g of O2 3) 0.827 g is the answer. Note that I could have calculated the mole amounts, used the "divide moles by coefficient" to determine the limiting reagent, and then done just one complete calculation. Example #10: (a) What mass of hydrogen peroxide should result when 1.45 g of barium peroxide is treated with 25.5 mL of hydrochloric acid solution containing 0.0277 g of HCl per mL? (b) How much of the excess reactant is left? BaO2(s) + 2HCl(aq) ---> H2O2(aq) + BaCl2(aq) Solution: Calculate the amount of product using each reactant. The reactant that produces the lesser of the two amounts will tell you the limiting reactant. This solution will use dimensional analysis (also called the unit-factor, or unit-label, method) for the proposed solution. 1) First, determine the mass of HCl that reacts: (0.0277 g/mL) (25.5 mL) = 0.70635 g 2) The barium peroxide solution: \| \| \| \| \| \| --- --- \| (1.45 g BaO2) \| (1 mol BaO2 / 169.3 g BaO2) \| (1 mol H2O2 / 1 mol BaO2) \| (34.0 g H2O2 / 1 mol H2O2) \| = 0.291 g H2O2 \| \| \| ↑ convert grams to moles ↑ \| ↑ molar ratio ↑ from equation \| ↑ convert moles to grams ↑ \| \| 3) The hydrochloric acid solution: (0.70635 g) (1 mol HCl / 36.46 g HCl) (1 mol H2O2 / 2 mol HCl) (34.0 g H2O2 / 1 mol H2O2) = 0.332 g H2O2 4) Since 0.291 g is less than 0.332 g, the BaO2 is the limiting reactant. 5) The other method to determine the limiting reagent is to divide the moles of each reactant by their respective coefficient in the balanced equation: BaO2 ---> 1.45 g / 169.3 g/mol = 0.008565 mol HCl ---> 0.70635 g / 36.46 g/mol = 0.01937 mol 0.008565 / 1 = 0.008565 0.01937 / 2 = 0.009685 BaO2 (the 0.008565) is the lesser amount, so it is the limiting reagent. 6) To solve part (b), we observe that 0.008565 mol of BaO2 was used. Using a 1:2 molar ratio, we can determine the amount of HCl that was used: \| \| \| \| --- \| 1 \| \| 0.008565 mol \| \| ––– \| = \| ––––––––––– \| \| 2 \| \| x \| x = 0.01713 mol of HCl used up in the reaction 7) Next, we subtract the amount used up from the total amount that was present: 0.01937 mol − 0.01713 mol = 0.00224 mol of HCl remains after reaction stops 8) Convert moles to grams: (0.00224 mol) (36.46 g/mol) = 0.0817 g (to three sig figs) Bonus Example: Consider the following reaction at 1.10 atm and 19.0 °C: 2NaCl(s) + 2NH3(g) + CO2(g) + H2O(ℓ) ---> 2NH4Cl(aq) + Na2CO3(s) 0.218 mol of sodium chloride, 2.55 L of ammonia, 2.00 L of carbon dioxide, and an unlimited amount of water react to form aqueous ammonium chloride and solid sodium bicarbonate. How many moles of ammonium chloride are formed in the reaction? Comment: this question was asked and answered on a now-defunct "answers" website and the one answer given (besides mine) totally missed the point of the question. The answerer focused on the non-realistic nature of the above chemical equation. However, the point of the question is to determine the limiting reagent and the non-realistic nature of the chemical equation is completely beside the point. Solution: 1) Use PV = nRT to determine moles of ammonia and carbon dioxide: ammonia: (1.10 atm) (2.55 L) = (n) (0.08206 L atm / mol K) (292 K) n = 0.11706 mol carbon dioxide: (1.10 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (292 K) n = 0.091814 mol 2) Determine the limiting reagent: 0.218 / 2 = 0.109 0.11706 / 2 = 0.05853 0.091814 / 1 = 0.091814 Ammonia is the limiting reagent. 3) Now, the problem becomes this: 0.11706 moles of ammonia produces how many moles of ammonium chloride? The molar ratio between ammonia and ammonium chloride is 1:1. 0.11706 moles of ammonia produces 0.117 moles of ammonium chloride (rounded off to three significant figures). And we are done. \| \| \| --- \| \| Limiting Reagent Problems #1-10 \| Problems solved using dimensional analysis only \| \| Limiting Reagent Problems #11-20 \| Stoichiometry Menu \|
190591	https://askfilo.com/user-question-answers-algebra-2/solve-the-formula-for-35373639373935	Question asked by Filo student Solve the formula V=LWH for H . Views: 5,376 students Updated on: Oct 21, 2023 Text SolutionText solutionverified iconVerified Step 1. Definition: In the given formula V=LWH, V represents the volume of the object which is found by multiplying its length, width, and height. Here, we need to solve for the height of the object, represented by H. Step 2. Simplification: To solve for H, we need to isolate it on one side of the equation. So, we first divide both sides of the formula by LW. LWV=LWLWH Then, simplify the right-hand side of the equation by canceling out LW from the numerator and denominator on the right-hand side of the equation. LWV=H Thus, we have the value of H expressed in terms of V, L, and W. Step 3. Solution: The solution for H is: H=LWV Therefore, we can find the height of the object by dividing its volume by the product of its length and width. Students who ask this question also asked Views: 5,513 Topic: Algebra 2 View solution Views: 5,338 Topic: Algebra 2 View solution Views: 5,628 Topic: Algebra 2 View solution Views: 5,660 Topic: Algebra 2 View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Solve the formula V=LWH for H . \| \| Updated On \| Oct 21, 2023 \| \| Topic \| All Topics \| \| Subject \| Algebra 2 \| \| Class \| High School \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
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190593	https://math.stackexchange.com/questions/3865771/how-many-balls-to-draw-to-ensure-balls-of-same-color	pigeonhole principle - How many balls to draw to ensure balls of same color? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How many balls to draw to ensure balls of same color? [closed] Ask Question Asked 4 years, 11 months ago Modified4 years, 11 months ago Viewed 987 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Closed. This question does not meet Mathematics Stack Exchange guidelines. It is not currently accepting answers. Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. Closed 4 years ago. Improve this question We have a bag which contains 5 5 red balls, 8 8 blue balls, 10 10 white balls, 12 12 green balls and 7 7 yellow balls. How many balls should we pick to ensure a) at least 4 balls have the same color? b) at least 6 balls have the same color? c) at least 7 balls have the same color? d) at least 9 balls have the same color? I think I solved the a) part. If we pick a ball every time till we reach 4 4 balls with the same color it takes us 15+1 15+1 balls. But I don't understand the other parts. Can you guys help me? It will mean A LOT to me. Thanks pigeonhole-principle Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Oct 14, 2020 at 19:12 cosmo5 10.8k 2 2 gold badges 11 11 silver badges 36 36 bronze badges asked Oct 14, 2020 at 17:31 HosseinHossein 63 5 5 bronze badges 2 Part (b) works exactly like part (a). Part (c) and (d), note that there are only so many red and yellow balls available. Think of the "worst case scenario" where you have drawn as many balls as possible of each color while still avoiding the "at least X balls of the same color" condition.JMoravitz –JMoravitz 2020-10-14 17:36:10 +00:00 Commented Oct 14, 2020 at 17:36 While the "worst case scenario" for drawing at least four of some color is to have drawn three of every color ball, the "worst case scenario" for drawing at least 9 balls of each color does not involve drawing 8 red balls since that is impossible but rather to have drawn all five possible red balls, and then the eight blue balls, eight of the white balls, etc...JMoravitz –JMoravitz 2020-10-14 17:37:45 +00:00 Commented Oct 14, 2020 at 17:37 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. We look at the worst possible case, and draw one more ball. Worst possible case was drawing 3 3 balls of each of five colors first, before reaching to fourth ball of some color. Hence 3⋅5+1 3⋅5+1 Worst case includes drawing all 5 of red balls, then five each of other colors. Hence 5⋅5+1 5⋅5+1 …… Similarly, can you proceed? Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Oct 14, 2020 at 17:39 cosmo5cosmo5 10.8k 2 2 gold badges 11 11 silver badges 36 36 bronze badges 11 if we omit 5 red balls because its impossible to have 6 red balls the we start picking colors one by one till we reach 6 balls with same color will be 24+1 balls which is 25 right?Hossein –Hossein 2020-10-14 18:01:20 +00:00 Commented Oct 14, 2020 at 18:01 But we are picking out balls without knowing the color. So red balls can be picked; they must be taken into account.cosmo5 –cosmo5 2020-10-14 18:03:28 +00:00 Commented Oct 14, 2020 at 18:03 For b), as we have only four other colors, 5+5⋅4+1=26 5+5⋅4+1=26.cosmo5 –cosmo5 2020-10-14 18:13:30 +00:00 Commented Oct 14, 2020 at 18:13 We pick one less, then add one in the end. That final ball counts as 6th ball.cosmo5 –cosmo5 2020-10-14 18:15:20 +00:00 Commented Oct 14, 2020 at 18:15 so for d it will be 5(red)+7(blue)+8(yellow)+82+1=37 Hossein –Hossein 2020-10-14 18:21:51 +00:00 Commented Oct 14, 2020 at 18:21 \|Show 6 more comments Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions pigeonhole-principle See similar questions with these tags. 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190594	https://enthu.com/blog/chemistry/chlorine-electron-configuration?srsltid=AfmBOorvkReHJMtbzG-UsQ4ZEZRA2jb292VJi0sYeQ7mLHx-cnx2NDEA	Chlorine Electron Configuration About Our Charter Courses Live Learning Co-horts Enthu App Membership LightDark Search chemistry Chlorine Electron Configuration Learn the electron configuration of Chlorine and how its electrons are arranged in different shells. This simple guide explains Chlorine’s atomic structure, making it easier for students and chemistry learners to understand. ByPK Mar 1, 2023 The electron configuration of chlorine is: 1s22s22p63s23p5. In this article, we will study chlorine electron configuration and see how electrons are arranged in different shells and subshells in Chlorine. Chlorine 1. Electron Configuration of Chlorine 2. Distribution of Electrons 3. Valence Electrons in Chlorine 4. Occurrence of Chlorine Key Takeaways Conclusion FAQs 1. What is the electronic configuration of Chlorine? 2. How many shells does Chlorine have? 3. How many valence electrons does Chlorine have? 4. Why is the electronic configuration of Chlorine important? 5. What is the significance of the p orbital in the electronic configuration of Chlorine? 6. How does the electronic configuration of Chlorine relate to its position in the periodic table? Chlorine Chlorine (Cl) is a chemical element with the atomic number 17 and the symbol Cl. It is a non-metal in group 17 of the periodic table, also known as the halogens. Chlorine is a highly reactive element and exists in nature primarily as the diatomic molecule Cl2. Chlorine is a greenish-yellow gas at room temperature and pressure. It has a pungent odor and is highly toxic. Chlorine is used in a variety of industrial applications, such as in the production of bleach, solvents, and plastics. It is also commonly used as a disinfectant in water treatment and swimming pools. In its natural form, chlorine is not found as a free element on Earth. It is primarily found in the form of chloride compounds in salt water and in the Earth's crust. Chlorine is a necessary nutrient for most plants and animals, including humans, in small quantities. However, in larger concentrations, chlorine can be harmful and even deadly. 1. Electron Configuration of Chlorine The electron configuration of chlorine is: 1s22s22p63s23p5 Image source: terpconnect Terpconnect This means that chlorine has 17 electrons. The first two electrons are in the 1s orbital, the next two are in the 2s orbital, the next six are in the 2p orbital, the next two are in the 3s orbital, and the last five are in the 3p orbital. The outermost shell, which is the 3p shell, has five electrons, and these are thevalence electrons of chlorine. 2. Distribution of Electrons Chlorine (Cl) has an atomic number of 17, which means it contains 17 electrons distributed across different energy levels or shells. The first shell (K) holds 2 electrons, the second shell (L) holds 8 electrons, and the third shell (M) holds 7 electrons. This distribution can be expressed as K(2), L(8), M(7) or simply 2, 8, 7. The presence of 7 electrons in the outermost shell makes chlorine a highly reactive nonmetal, as it readily gains one electron to achieve a stable octet configuration. 3. Valence Electrons in Chlorine To determine the number of valence electrons in chlorine, we need to look at the outermost shell, which is the third shell (designated as "M"). The M shell has a total of 7 electrons, so chlorine has 7 valence electrons in its outermost shell. Valence Electrons Valence Electrons enthu.com Valence Electrons Herefore, the valency of the chlorine atom becomes 1. Valence electrons are the electrons in the outermost shell of an atom that are involved in chemical bonding. In the case of chlorine, the 7 valence electrons make it highly reactive and able to form a variety of chemical bonds with other elements. 4. Occurrence of Chlorine Chlorine (Cl) is a relatively abundant element on Earth and is the second lightest element in the halogen group (after fluorine). Chlorine does not occur naturally as a free element in nature, but it is commonly found in the form of chloride compounds in various minerals and in seawater The most common chloride compound is sodium chloride (NaCl), or table salt, which is found in large deposits in underground salt mines and in the seawater of saltwater bodies. Other chloride minerals, such as sylvite (KCl), carnallite (KMgC 3·6H 2 O), and halite (NaCl), are also found in the Earth's crust. Chlorine is also produced commercially by the electrolysis of brine (salty water). This is a major source of chlorine for industrial purposes, such as the production of bleach, solvents, and plastics. Chlorine is also used in the production of chlorofluorocarbons (CFCs), which were once used as refrigerants, but are now known to be harmful to the environment. Small amounts of chlorine are found in the Earth's atmosphere, primarily as a result of the reaction of sea spray with atmospheric ozone. Chlorine is also an essential element for life and is present in many biological compounds, such as chlorophyll, which gives plants their green color. KeyTakeaways The chemical symbol for the chlorine atom is Cl. The electronic configuration of chlorine is 1s 2 2s 2 2p 6 3s 2 3p 5 The distribution of electrons in the shell of the chlorine atom is 2, 8, 7. The valence electrons in chlorine are 7. Conclusion In conclusion, crafting engaging and informative blog content is essential for connecting with your audience and establishing your brand's presence online. By focusing on quality writing, understanding your target audience, and utilizing effective SEO practices, you can create posts that not only provide value but also encourage interaction and sharing. Remember to stay consistent with your posting schedule and continuously analyze the performance of your content to refine your strategies. With these tips in mind, you can elevate your blogging journey and achieve your desired outcomes. Happy blogging! FAQs 1. What is the electronic configuration of Chlorine? The electronic configuration of Chlorine is 1s 2 2s 2 2p 6 3s 2 3p 5. 2. How many shells does Chlorine have? Chlorine has three electron shells, which are designated as K, L, and M shells. 3. How many valence electrons does Chlorine have? Chlorine has 7 valence electrons in its outermost shell (the M shell). 4. Why is the electronic configuration of Chlorine important? The electronic configuration of Chlorine determines its chemical properties and behavior, such as its reactivity and ability to form chemical bonds. 5. What is the significance of the p orbital in the electronic configuration of Chlorine? The p orbital in the electronic configuration of Chlorine refers to the set of three p orbitals in the third shell (designated as 3p). These orbitals allow Chlorine to form covalent bonds with other elements and participate in chemical reactions. 6. How does the electronic configuration of Chlorine relate to its position in the periodic table? The electronic configuration of Chlorine follows the pattern of the periodic table, which arranges elements in order of increasing atomic number and places elements with similar chemical properties in the same group. Chlorine is a member of group 17 (also known as the halogens), which all have 7 valence electrons in their outermost shell. 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190595	https://www.youtube.com/watch?v=m2JS3Ab4CRA	DEFORMACIÓN POR CARGA AXIAL \| MECÁNICA DE MATERIALES \| BEER \| EJERCICIO 2.28 dcahue-ingeniería 24600 subscribers 153 likes Description 8068 views Posted: 5 Oct 2021 📕 📘 TEMA: DEFORMACIÓN POR CARGA AXIAL. 🔴 Ejercicio resuelto 2.28. PROBLEMA La longitud del alambre de acero de 2 mm de diámetro CD ha sido ajustada de forma que, si no se aplica ninguna carga, existe una distancia de 1.5 mm entre el extremo B de la viga rígida ACB y un punto de contacto E. Si se sabe que E = 200 GPa, determine el sitio sobre la viga donde debe colocarse un bloque de 20 kg para provocar un contacto entre B y E. SOLUCIÓN ⏱️📍 00:00 DATOS del problema. ⏱️📍 01:45 Relación de DEFORMACIONES. ⏱️📍 05:30 Obtención de DEFORMACIÓN C. ⏱️📍 07:20 Obtención de FUERZA CD. ⏱️📍 12:15 Obtención de DEFORMACIÓN TOTAL. ⏱️📍 12:35 Diagrama de CUERPO LIBRE. ⏱️📍 15:50 Obtención de DISTANCIA Y POSICIÓN. ✅ SUSCRÍBETE 💬 BLOG: 🔴 YOUTUBE: 👥 Facebook: 📸 Instagram: 📕📗 📘 REFERENCIAS:📕📗 📘 -Beer, F. P., Johnston, E. R., DeWolf, J. T., & Mazurek, D. F. (2009). -Mecánica De Materiales. -McGrawHill, Ed. -5th ed. Ingeniería mecánica, mecatrónica y civil. Resistencia de materiales #Mecania de materiales 2 comments Transcript: la longitud del alambre de acero de 2 milímetros de diámetro cede ha sido ajustada de tal forma que si no se aplica ninguna carga existe una distancia de 1.5 milímetros entre el extremo b de la viga rígida y el punto de contacto sí se sabe que el módulo de las ti cidades de 200 llega pascal es determine en sitios sobre la viga donde se debe colocar el bloque de 20 kilogramos para provocar un contacto entre b y e como pueden apreciar ingenieros nosotros tenemos una viga la cual está siendo sostenida por este alambre sin embargo como ya sabemos conforme nosotros vamos acercando a la orilla este bloque ve o este bloque mejor dicho conforme lo vamos acercando ve el momento va aumentando porque porque va aumentando el brazo de palanca y entonces que trae como consecuencia que la carga en este alambre vaya aumentando de tal manera de que va a llegar un momento en el que el bloque ya esté muy cercano a la orilla y entonces la deformación de este alambre va a permitir que esté en contacto esta parte interna o esta parte inferior de la viga con el suelo la pregunta es qué distancia debe de tener este bloque para que haya ese contacto para dar inicio con este problema primero necesitamos determinar las deformaciones por lo tanto vamos a hacer un diagrama de deformaciones para ver más o menos cuál es el comportamiento de estas deformaciones ingenieros entonces presten mucha atención vamos a suponer inglés que yo voy a dibujar una deformación exagerada de la viga esta deformación exagerada la voy a poner así miren cómo pueden ver a es un apoyo los apoyos decimos por definición que son totalmente indeformables entonces observen el fenómeno de deformación va a ser el siguiente en el momento en el que el bloque b se mueve hacia la orilla pues lo que va a pasar es que esta viga va a tender a bajar pues que creen en que es a esta deformación que va a ir hacia abajo por cierto la vamos a poner en esta dirección a esta deformación nosotros le vamos a llamar delta ahora qué pasa que como pueden apreciar a esta altura porque vamos a poner el eje neutro de la viga para que vaya quedando más claro todo esto el eje neutro de la viga es este que se muestra a continuación aquí está entonces ingenieros todas las dimensiones hacia abajo del eje neutro son nuestras deformaciones por lo tanto aquí nosotros vamos a tener una longitud qué longitud es ésta ésta va a ser o le vamos a llamar deformación ence y como ya se imaginan esta deformación en ce es lo que está bajando el cable una vez que el bloque está haciendo contacto con el suelo como pueden ver ingenieros nosotros hemos formado un triángulo de hecho hemos formado triángulos congruentes entre sí estos triángulos los voy a pasar en otra ilustración para que nos permitan identificar de mejor manera las deformaciones que se están llevando a cabo en el elemento observen como de este triángulo que conocemos pues en realidad las deformaciones todavía no las conocemos bueno tenemos este desplazamiento n en realidad ahorita le vamos a dar un valor y aquí nosotros tenemos el desplazamiento en c afortunadamente ingenieros vean como el desplazamiento debe si lo conocemos porque está limitado bueno el desplazamiento debe es el mismo que el desplazamiento porque porque está limitado por el suelo nosotros sabemos que este desplazamiento es de 1.5 milímetros este desplazamiento en c la realidad es que yo no lo conozco que si conocemos de manera adicional de este triángulo las dimensiones esto es muy importante estas dimensiones ingenieros claro que las conocemos esta es de 0.08 metros sin embargo mucha atención para trabajar con unidades que sean consistentes vamos a pasar estos puntos 0 8 metros a milímetros eso quiere decir link es que vamos a tener 80 milímetros aquí y esta longitud de este triángulo o más bien de lo que queda del triángulo va a tener una longitud de 320 milímetros ahora bien nosotros afortunadamente conocemos las dos longitudes de cada uno de los triángulos y conocemos otro cada uno de sus catetos si nosotros aplicamos las semejanzas de triángulos que obtendríamos mucha atención vamos a decir que al desplazamiento en c es el mismo cateto vean el desplazamiento en c es el cateto opuesto del triángulo pequeño y este desplazamiento le corresponde este cateto por ser su análogo o su homólogo entonces que tenemos pues vamos a tener que es 1.5 si observan yo ya no voy a poner del tae pues estoy poniendo solamente su valor y sé que está en milímetros continuando ahora vamos a decir que entonces como la parte de arriba habla del triángulo pequeñito pues vamos a decir que la base de este triángulo pequeñito que le vamos a poner 80 milímetros es la misma que la base o más bien le corresponde la base del triángulo grande la base del triángulo grande como pueden apreciar es de 400 milímetros no voy a poner unidades ahí está vean como en la parte de arriba son los datos del triángulo pequeño en la parte de abajo son los datos del triángulo grande gracias a esta semejanza de triángulos yo puedo determinar de manera muy simple ese desplazamiento en ce y como me quedaría ese desplazamiento pues vamos a tener que multiplicar 80 por 1.5 y después lo vamos a dividir entre 400 el desplazamiento me da un valor de 0.3 milímetros hasta aquí yo ya obtuve el desplazamiento en ce esto es algo bueno ahora que otra cosa podemos hacer vamos a ver nosotros sabemos que el desplazamiento de hecho vamos a decir que el desplazamiento en c lo podemos encontrar con la carga que alguien se de hecho esa carga que hay a quién se yo la voy a identificar como cd porque si observan este desplazamiento corresponde a el alargamiento del cable y este cable obviamente está sometido a una fuerza que yo le estoy llamando cede esto lo vamos a multiplicar por la longitud cede y lo vamos a dividir inglés entre el área cede y el módulo de elasticidad cede qué datos tenemos tenemos la longitud el área la podemos calcular y tenemos el módulo de elasticidad eso quiere decir que de esta ecuación yo sin problema puedo despejar la fuerza en cedé que es la fuerza que me está provocando esta deformación o esta este desplazamiento de punto tres milímetros entonces ingenieros vamos a calcular la que tenemos vamos a decir que la fuerza cede es igual a que esta fuerza se dé es igual al desplazamiento es muy importante observar que las unidades sean consistentes el desplazamiento es de 0.3 milímetros el área el área lo pasamos para el otro lado pasa multiplicando y vamos a tener que el área es de pi de hecho aquí vamos a poner un corchete esto es corchete quedaría como pi por radio en este caso a mí me están dando el diámetro de un milímetro eso quiere decir que el radio es de un milímetro vamos a tener que speed por radio al cuadrado después el módulo de elasticidad también lo pasamos para arriba este módulo de elasticidad sabemos que es de 200 giga pascales pero mucha atención con lo siguiente cuando nosotros decimos que hay 200 giga pascal es ingenieros la realidad de las cosas es que estamos diciendo que hay 200 por 10 a la 9 newton entre metro cuadrado así es nosotros estamos trabajando en milímetros por lo tanto no podemos poner estos 200 por 10 a la 9 porque entonces eso quiere decir que estaríamos trabajando con metros es perentorio que nosotros hagamos la conversión de de newton sobre metro cuadrado a newton entre milímetro cuadrado y para hacer con esta conversión es muy simple solamente debemos de convertir esta o más bien debemos de dividir esta cantidad entre mil al cuadrado lo cual me da como resultado doscientos mil y vamos a tener que son newton entre milímetro al cuadrado vamos a poner esa cantidad tenemos que son 200.000 newtons / milímetros al cuadrado después ingenieros esto lo vamos a dividir entre que pues muy bien como estamos despejando la fuerza lo único que nos hace falta pasar hacia el otro lado es la longitud como pueden apreciar la longitud está en metros pues nosotros la vamos a pasar dividiendo en milímetros es decir vamos a tener 250 milímetros de longitud nunca está además el hecho de revisar que todas las unidades que nosotros tenemos realmente sean consistentes si es así entonces vamos a hacer la operación ingenieros el resultado de esta fuerza que tenemos como sede va a ser d 700 53 punto 98 las unidades van a ser newtons como pueden ver estos milímetros cuadrados con estos milímetros cuadrados se van estos metros estos milímetros con estos milímetros también se van y la única cantidad que me queda es el newton entonces ingenieros hasta aquí yo ya tengo la carga que está soportando el alambre cede sin embargo todavía no he encontrado o no he determinado la distancia para que esta caja me produzca esta deflexión de tal manera que al producir esta deflexión sabemos que esta deflexión ocurre cuando la viga está en contacto con el suelo para ello qué vamos a hacer necesitamos hacer un nuevo diagrama de cuerpo libre pero ahora no lo vamos a hacer con deformaciones lo vamos a hacer con cargas y este diagrama quedaría más o menos de la siguiente forma aquí nosotros podemos apreciar a nuestra vida las cargas vamos a ver en el punto a nosotros vamos a tener dos cargas porque como pueden apreciar nosotros tenemos un perno los pernos me generan dos reacciones una en equis y una ingle esto lo pueden ver en cualquier libro de estática vienen algunas tablas para saber qué tipo de apoyos existen y cuántas reacciones me genera un apoyo un apoyo de perno ingenieros me va a generar dos reacciones a esta le voy a llamar reacción y en ésta le voy a llamar reacción a equis ahora cómo quedaría el tema de esta fuerza este alambre me está produciendo una fuerza hacia arriba la cual va a tener un valor que ya conocemos a esta fuerza nosotros le llamamos fuerza cede atención como sabemos que esta fuerza cede va hacia arriba esto es muy simple ingenieros si yo la pongo hacia abajo se va con todo y bloque exacto porque porque el peso del bloque va hacia abajo y entonces si esta fuerza fuera hacia abajo se va con todo y bloque y no estaríamos hablando de un elemento estático porque el equilibrio se está perdiendo estaríamos hablando de un problema dinámico por eso es que aquí es importante saber identificar en qué dirección va por qué porque esto ya no es un incógnita ingenieros desde el momento en el que nosotros encontramos su valor tenemos que exponer de manera precisa en qué dirección va ahora vamos a poner también la carga que va a generar el bloque esa carga que va a generar el bloque la vamos a poner como efe de fuerza de bloque y listo las distancias muy bien aquí si va a ser necesario poner las distancias porque si las vamos a necesitar sabemos que la distancia de aquí hasta donde termina la viga que es más o menos aquí es de cuánto aquí es de 80 milímetros y esta distancia es de 320 milímetros ahora bien también tenemos una distancia más que es la distancia que hay de la parte superior la voy a poner en la parte superior que va desde el lado derecho de la viga hasta donde iría el bloque y esta distancia sería mi distancia x así como no lo está ilustrando el problema sin embargo esta es la distancia que yo requiero encontrar así que bien como vamos a encontrar esa distancia como saben nosotros ya tenemos nuestro diagrama de cuerpo libre y algo que puedo hacer es simple voy a hacer una sumatoria de momentos y esta sumatoria de momentos la voy a hacer en qué punto en el punto a porque la vamos a hacer en el punto a ingenieros porque si ustedes observan al yo hacer la sumatoria de momentos en el punto a en ese que en ese momento yo estoy eliminando o estoy dejando no estoy eliminando estoy dejando fuera de la ecuación estas dos cargas o esas dos incógnitas voy a decir que los momentos que vayan en esta dirección van a ser positivos entonces veamos esta fuerza se ve como va hacia arriba provocaría que mi elemento girará en esta dirección es decir en sentido contrario a mi convención pues vamos a poner que la fuerza se ve es de menos por el signo negativo por el giro 753 punto 98 y esto nos va a quedar como newtons después ingenieros esto lo vamos a multiplicar por el brazo de palanca que es de 80 milímetros ahora qué pasa ahora nosotros necesitamos encontrar este valor de x la fuerza del bloque va hacia abajo claramente va hacia abajo la fuerza del bloque eso quiere decir que me va a dar un giro positivo vamos a poner que es más la fuerza del bloque pues sabemos que el bloque pesa 20 kilogramos pues vamos a poner los 20 kilogramos del bloque pero esto es masa de hecho corrijo no pesa 20 kilogramos tiene una masa de 20 kilogramos si queremos el peso lo vamos a multiplicar por la aceleración de la gravedad entonces redondeando vamos a tener que es 9.81 metros sobre segundo al cuadrado hasta aquí yo tengo fuerza solamente me falta el brazo de palanca ingenieros y presten atención porque aquí está un punto medular en este problema cuál sería el brazo de palanca pues bien en mi brazo de palanca sería el siguiente resulta que toda la viga toda la viga tiene 400 milímetros de longitud pero yo estoy buscando un brazo de palanca que llegue hasta aquí hasta la carga entonces cómo quedaría esta longitud pues muy simple vamos a tener que esa longitud es de 400 observen que es toda esta distancia menos así es ingreso 400 es toda esta distancia - x que es este segmento que nosotros estamos aquí vean que es lo que me queda me queda este brazo de palanca entonces pues nosotros vamos a poner que son nuestro brazo de palanca quedaría como 400 - x y todo esto ingeniero sería igual a cero porque nuestro elemento se encuentra en equilibrio ahora como pueden apreciar inglés pues nosotros vamos a pasar esto hacia el otro lado y que tendríamos vamos a tener los 753 punto 98 x 80 este signo negativo pasaría como positivo y entonces vamos a tener que es 60.000 318 punto 4 esto me queda en unidades de newton por milímetro pero no las voy a poner todavía esto que sería igual como esto lo pasamos al otro lado esto se queda con los mismos signos y que vamos a tener pues primero vamos a tener que son 20 x 9.81 que es el la fuerza atención nos vamos a poner a multiplicar primero por 400 eso quiere decir que vamos a tener 7000 no es cierto 78 mil 480 es la multiplicación de estos dos términos por el primer término de este binomio después vamos a poner el signo negativo y sabemos que 20 por 9.81 me da una un valor de 196 punto 2 ese valor lo vamos a multiplicar ahora por la equis eso quiere decir que vamos a tener 190 y 6.2 multiplicando a la equis y listo interest tenemos una ecuación de las baratas una ecuación lineal con una incógnita que vamos a tener entonces como resultado pues bien como pueden apreciar nosotros tendríamos que 6000 a más bien a 60 mil 318 le restaría mos porque esto pasa hacia el otro lado restando le restaría mos los 78 1480 y esto que nos da como resultado ingenieros esto nos va a dar como resultado un valor negativo de cuánto de menos 18 mil 162 y esto lo vamos a dividir entre que como estamos despejando x entre menos 190 y 6.2 eso quiere decir ingenieros que la distancia que el problema me está solicitando es de 92 punto 56 milímetros esta distancia que nosotros acabamos de obtener es prácticamente recuérdenlo del final de la viga a la aplicación de la fuerza del bloque sin embargo si yo deseo conocer la distancia a partir de a pues sería muy simple a ese valor o más bien a 400 le vamos a restar ese valor que quiere decir que este brazo de palanca que nosotros buscábamos inicialmente va a tener un valor ingenieros de 300 7.43 entonces observen ahora zinc es como pueden ver el brazo de palanca más bien la ubicación del bloque para que este elemento haga contacto con el suelo tiene que ser a 307 punto 43 milímetros como comentario final es muy importante que ustedes noten como el diagrama de deformaciones fue una parte fundamental para la solución de este problema siempre debemos de poner hacia dónde van las deformaciones y cómo es que se están relacionando entre sí con esto ingenieros damos por concluido todo lo que el problema nos había solicitado por lo tanto te invito a suscribirte a este canal y ver cada una de las listas de reproducción que aparecen en el seguramente van a ser de gran utilidad en tu estancia como estudiante de ingeniería así que regalarnos un like comparte los vídeos y te deseamos mucho éxito que la fuerza te acompañe
190596	https://en.wikipedia.org/wiki/Closed_system	Jump to content Closed system العربية Български Català Čeština Deutsch Eesti Ελληνικά Español Euskara فارسی Français 한국어 हिन्दी Hrvatski Bahasa Indonesia Italiano ქართული Latviešu Magyar മലയാളം Nederlands Norsk bokmål Polski Português Romnă Русский Simple English Slovenčina Slovenščina Српски / srpski Svenska Türkçe Українська 粵語中文 Edit links From Wikipedia, the free encyclopedia Type of system This article is about the physical sciences concept. For the computing concept, see Closed source software. A closed system is a natural physical system that does not allow transfer of matter in or out of the system, although – in the contexts of physics, chemistry, engineering, etc. – the transfer of energy (e.g. as work or heat) is allowed. Physics [edit] In classical mechanics [edit] In nonrelativistic classical mechanics, a closed system is a physical system that does not exchange any matter with its surroundings, and is not subject to any net force whose source is external to the system. A closed system in classical mechanics would be equivalent to an isolated system in thermodynamics. Closed systems are often used to limit the factors that can affect the results of a specific problem or experiment. In thermodynamics [edit] Main article: Thermodynamic system In thermodynamics, a closed system can exchange energy (as heat or work) but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter. (This scheme of definition of terms is not uniformly used, though it is convenient for some purposes. In particular, some writers use 'closed system' where 'isolated system' is used here.) For a simple system, with only one type of particle (atom or molecule), a closed system amounts to a constant number of particles. However, for systems which are undergoing a chemical reaction, there may be all sorts of molecules being generated and destroyed by the reaction process. In this case, the fact that the system is closed is expressed by stating that the total number of each elemental atom is conserved, no matter what kind of molecule it may be a part of. Mathematically: where is the number of j-type molecules, is the number of atoms of element in molecule and is the total number of atoms of element in the system, which remains constant, since the system is closed. There will be one such equation for each different element in the system. In thermodynamics, a closed system is important for solving complicated thermodynamic problems. It allows the elimination of some external factors that could alter the results of the experiment or problem thus simplifying it. A closed system can also be used in situations where thermodynamic equilibrium is required to simplify the situation. In quantum physics [edit] Further information: Quantum field theory This equation, called Schrödinger's equation, describes the behavior of an isolated or closed quantum system, that is, by definition, a system which does not interchange information (i.e. energy and/or matter) with another system. So if an isolated system is in some pure state \|ψ(t) ∈ H at time t, where H denotes the Hilbert space of the system, the time evolution of this state (between two consecutive measurements). where i is the imaginary unit, ħ is the Planck constant divided by 2π, the symbol ⁠∂/∂t⁠ indicates a partial derivative with respect to time t, Ψ (the Greek letter psi) is the wave function of the quantum system, and Ĥ is the Hamiltonian operator (which characterizes the total energy of any given wave function and takes different forms depending on the situation). In chemistry [edit] In chemistry, a closed system is where no reactants or products can escape, only heat can be exchanged freely (e.g. an ice cooler). A closed system can be used when conducting chemical experiments where temperature is not a factor (i.e. reaching thermal equilibrium). In engineering [edit] In an engineering context, a closed system is a bound system, i.e. defined, in which every input is known and every resultant is known (or can be known) within a specific time. See also [edit] Glossary of systems theory Dynamical system Isolated system Open system (systems theory) Sense and Respond Thermodynamic system References [edit] ^ Rana, N.C.; P.S. Joag (1991). Classical Mechanics. McGraw-Hill Education (India) Pvt Limited. p. 78. ISBN 978-0-07-460315-4. ^ Landau, L.D.; E.M. Lifshitz (1976). Mechanics (third ed.). Butterworth-Heinemann. p. 8. ISBN 978-0-7506-2896-9. ^ Prigogine, I., Defay, R. (1950/1954). Chemical Thermodynamics, Longmans, Green & Co, London, p. 66. ^ Tisza, L. (1966). Generalized Thermodynamics, M.I.T Press, Cambridge MA, pp. 112–113. ^ Guggenheim, E.A. (1949/1967). Thermodynamics. An Advanced Treatment for Chemists and Physicists, (1st edition 1949) 5th edition 1967, North-Holland, Amsterdam, p. 14. ^ Münster, A. (1970). Classical Thermodynamics, translated by E.S. Halberstadt, Wiley–Interscience, London, pp. 6–7. ^ Haase, R. (1971). Survey of Fundamental Laws, chapter 1 of Thermodynamics, pages 1–97 of volume 1, ed. W. Jost, of Physical Chemistry. An Advanced Treatise, ed. H. Eyring, D. Henderson, W. Jost, Academic Press, New York, lcn 73–117081, p. 3. ^ Tschoegl, N.W. (2000). Fundamentals of Equilibrium and Steady-State Thermodynamics, Elsevier, Amsterdam, ISBN 0-444-50426-5, p. 5. ^ Silbey, R.J., Alberty, R.A., Bawendi, M.G. (1955/2005). Physical Chemistry, fourth edition, Wiley, Hoboken NJ, p. 4. ^ Callen, H.B. (1960/1985). Thermodynamics and an Introduction to Thermostatistics, (1st edition 1960) 2nd edition 1985, Wiley, New York, ISBN 0-471-86256-8, p. 17. ^ ter Haar, D., Wergeland, H. (1966). Elements of Thermodynamics, Addison-Wesley Publishing, Reading MA, p. 43. ^ Rivas, Ángel; Huelga, Susana F. (October 2011). Open Quantum Systems. Berlin Heidelberg: Springer-Verlag. ISBN 978-3-642-23354-8. Retrieved from " Categories: Cybernetics Systems theory Thermodynamic systems Hidden categories: Articles with short description Short description is different from Wikidata
190597	https://pubmed.ncbi.nlm.nih.gov/9819691/	Costs of low-temperature plasma sterilization compared with other sterilization methods - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Costs of low-temperature plasma sterilization compared with other sterilization methods S Adler1,M Scherrer,F D Daschner Affiliations Expand Affiliation 1 Institute for Environmental Medicine, University Hospital, Freiburg, Germany. PMID: 9819691 DOI: 10.1016/s0195-6701(98)90091-3 Item in Clipboard Comparative Study Costs of low-temperature plasma sterilization compared with other sterilization methods S Adler et al. J Hosp Infect.1998 Oct. Show details Display options Display options Format J Hosp Infect Actions Search in PubMed Search in NLM Catalog Add to Search . 1998 Oct;40(2):125-34. doi: 10.1016/s0195-6701(98)90091-3. Authors S Adler1,M Scherrer,F D Daschner Affiliation 1 Institute for Environmental Medicine, University Hospital, Freiburg, Germany. PMID: 9819691 DOI: 10.1016/s0195-6701(98)90091-3 Item in Clipboard Full text links Cite Display options Display options Format Abstract Plasma sterilization is a new technique for decontaminating thermolabile products without the severe drawbacks associated with gas sterilization methods (residues, environment compatibility). The actual costs, per sterilization unit, of three sterilization techniques--plasma, ethylene oxide and formaldehyde--were compared. As plasma sterilization is an alternative to steam sterilization for sterilizing thermostable but easily corroding products or electronic instruments, costs for steam sterilization were calculated and compared as well. If one considers only the cost of the sterilization procedure itself, without taking into account the time-saving element of plasma sterilization, then ethylene oxide sterilization proves to be the most expensive procedure, followed by plasma sterilization; sterilization with formaldehyde was the least expensive. Inclusion of the time required to sterilize an instruments, however, altered the relative costs of the three methods. Because plasma sterilization takes less time to perform than either ethylene oxide or formaldehyde sterilization, fewer instruments need be procured. In order to measure and compare the time-saving advantage that plasma sterilization affords, five groups of instruments were assembled and the total cost of sterilizing an instrument of each group was calculated. The five groups included (1) disposable, (2) electronic, (3) endoscopic, (4) sharp and (5) standard instruments. In all cases, ethylene oxide sterilization was the most expensive method. Formaldehyde sterilization was, in four out of five cases, more expensive than plasma sterilization. Steam sterilization proved to be the cheapest method of sterilizing a laparoscopic set, even when costs due to damage inflicted on the optical instruments were calculated. In the case of a vitrectome, however, plasma sterilization costs were nearly the same as steam sterilization when the lower rate of damage by plasma sterilization was taken into account. PubMed Disclaimer Comment in Costs of low temperature plasma sterilization.Trispel S.Trispel S.J Hosp Infect. 1999 Jul;42(3):247-8. doi: 10.1053/jhin.1999.0600.J Hosp Infect. 1999.PMID: 10439999 No abstract available. Similar articles A comparative study of ethylene oxide gas, hydrogen peroxide gas plasma, and low-temperature steam formaldehyde sterilization.Kanemitsu K, Imasaka T, Ishikawa S, Kunishima H, Harigae H, Ueno K, Takemura H, Hirayama Y, Kaku M.Kanemitsu K, et al.Infect Control Hosp Epidemiol. 2005 May;26(5):486-9. doi: 10.1086/502572.Infect Control Hosp Epidemiol. 2005.PMID: 15954488 An economic analysis of the benefits of sterilizing medical instruments in low-temperature systems instead of steam.McCreanor V, Graves N.McCreanor V, et al.Am J Infect Control. 2017 Jul 1;45(7):756-760. doi: 10.1016/j.ajic.2017.02.026. Epub 2017 Mar 29.Am J Infect Control. 2017.PMID: 28365144 [Comparison between steam and gas sterilized optics. Costs, availability, maintenance].Conrad C, Pem M.Conrad C, et al.Chirurg. 1998 Feb;69(2):212-4. doi: 10.1007/s001040050400.Chirurg. 1998.PMID: 9551273 German. Sterilization of endoscopic instruments.Sabnis RB, Bhattu A, Vijaykumar M.Sabnis RB, et al.Curr Opin Urol. 2014 Mar;24(2):195-202. doi: 10.1097/MOU.0000000000000034.Curr Opin Urol. 2014.PMID: 24451088 Review. Choosing a low-temperature sterilization technology.[No authors listed][No authors listed]Health Devices. 1999 Nov;28(11):430-55.Health Devices. 1999.PMID: 10609148 Review. See all similar articles Cited by Current Trends in the Management of Port-Site Infections: A Case Series and a Review of Published Work.Chaudhary R, Tak V, Dutt A, Rodha MS, Meena SP, Badkur M, Kobade SP, Lodha M, Sharma N, Nag VL, Puranik A.Chaudhary R, et al.Cureus. 2023 Jun 25;15(6):e40936. doi: 10.7759/cureus.40936. eCollection 2023 Jun.Cureus. 2023.PMID: 37496535 Free PMC article. Selective Apoptotic Effect of Plasma Activated Liquids on Human Cancer Cell Lines.Sersenová D, Machala Z, Repiská V, Gbelcová H.Sersenová D, et al.Molecules. 2021 Jul 13;26(14):4254. doi: 10.3390/molecules26144254.Molecules. 2021.PMID: 34299530 Free PMC article. Effect of low-pressure cold plasma on surface microflora of meat and quality attributes.Ulbin-Figlewicz N, Brychcy E, Jarmoluk A.Ulbin-Figlewicz N, et al.J Food Sci Technol. 2015 Feb;52(2):1228-32. doi: 10.1007/s13197-013-1108-6. Epub 2013 Jul 24.J Food Sci Technol. 2015.PMID: 25694745 Free PMC article. Tailoring the composition of biocopolyester blends for dimensionally accurate extrusion-based printing, annealing and steam sterilization.Burkhardt F, Schmidt VD, Wesemann C, Schirmeister CG, Rothlauf S, Pieralli S, Brandenburg LS, Kleinvogel L, Vach K, Spies BC.Burkhardt F, et al.Sci Rep. 2022 Nov 25;12(1):20341. doi: 10.1038/s41598-022-24991-z.Sci Rep. 2022.PMID: 36434090 Free PMC article. Reprocessing and reuse of urological armamentarium: How correct are we!Raval KV, Chaudhari R, Khant SR, Joglekar O, Patel D.Raval KV, et al.Urol Ann. 2017 Apr-Jun;9(2):117-124. doi: 10.4103/UA.UA_167_16.Urol Ann. 2017.PMID: 28479760 Free PMC article.Review. 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190598	https://mitocw.ups.edu.ec/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-13-implicit-differentiation	Session 13: Implicit Differentiation \| Part B: Implicit Differentiation and Inverse Functions \| 1. 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Differentiation Part A: Definition and Basic Rules Part B: Implicit Differentiation and Inverse Functions Exam 1 2. Applications of Differentiation Part A: Approximation and Curve Sketching Part B: Optimization, Related Rates and Newton's Method Part C: Mean Value Theorem, Antiderivatives and Differential Equa Exam 2 3. The Definite Integral and its Applications Part A: Definition of the Definite Integral and First Fundamental Part B: Second Fundamental Theorem, Areas, Volumes Part C: Average Value, Probability and Numerical Integration Exam 3 4. Techniques of Integration Part A: Trigonometric Powers, Trigonometric Substitution and Com Part B: Partial Fractions, Integration by Parts, Arc Length, and Part C: Parametric Equations and Polar Coordinates Exam 4 5. Exploring the Infinite Part A: L'Hospital's Rule and Improper Integrals Part B: Taylor Series Final Exam Download Course Materials « Previous \| Next » Overview "Implicit Differentiation" is a way of applying the chain rule. 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190599	https://www.youtube.com/watch?v=0sYC5bRYFkI	Hydrostatic Pressure Distribution and Resultant Hydrostatic Force - Part 2 HydroMohsen 41700 subscribers 25 likes Description 1527 views Posted: 10 Feb 2023 Hydrostatic pressure distribution refers to the distribution of pressure within a fluid that is in a state of static equilibrium, meaning the fluid is not flowing and is not undergoing any changes in shape. In this state, the pressure at a given point in the fluid is determined by the weight of the fluid above it and the gravitational force acting on that fluid. The pressure in a fluid decreases with increasing height above the surface, and it increases with increasing depth below the surface. This pressure distribution is a result of the fluid's weight, which creates a force that acts in a downward direction, and the buoyant force that acts in an upward direction, which opposes the weight of the fluid and the object immersed in it. In a closed container, the pressure at any point within the fluid is equal in all directions, and this pressure is known as the hydrostatic pressure. The hydrostatic pressure distribution is an important concept in fields such as fluid mechanics, oceanography, and civil engineering, where it is used to analyze the behavior of fluid systems, determine the stability of structures, and design containers to withstand external pressures. Hydrostatic force is the force exerted by a fluid in static equilibrium on an object that is submerged or partially submerged in it. The hydrostatic force is proportional to the fluid's density, the volume of the fluid that is in contact with the object, and the pressure difference between the top and bottom of the object. Transcript: in this short video I'm going to talk about an example of hydrostatic pressure distribution and how to calculate the magnitude and the location of the resultant Force as a result of hydrostatic pressure distribution so we have a tank of water and then here there is a circular plate we want to calculate what is the resultant Force as a result of the hydro static pressure Distribution on this circular plate so the first thing that I want to do I want to sketch a better picture of our my problem so this is a circular plate and let me actually show you this is a circular plate like this right but when you I want to take a look a side look at this problem from this direction so it would be if the surface of water is right over here and I'm looking at um the side of this gate the side of this gate is going to be I'm going to turn it so it would be something like this so what you see although this is a circular plate when you look at it from the side because it has a specific thickness you see it as a rectangular area right so this would be the circular gate when I look at it from the side okay now I need to um apply the pressures on top of this and that would be actually pretty easy so because the pressure over here the magnitude is going to be this and the magnitude here is going to be higher so this is essentially your hydrostatic pressure distribution okay and the C centroid is going to be right in the middle of the circle right in the middle of the circle and the center of pressure is going to be a little bit lower than that so what I want to find is the magnitude and the location of this resultant Force right over here this is vertically submerged underwater remember from the previous video whenever we have a vertically submerged plate underwater H bar is equal to Y Bar and HP is equal to y p because it's vertically suppressed okay so let's go back to the main equations so when we want to calculate the magnitude of resultant force that would be gamma in this case we have water so gamma of water times HR times area of this plate over here this plate is a circular plate so area of a circular circular plate is going to be pi r to the power 2 okay I'm going to put the numbers right over here again GMA of water 9,810 har bar H bar is the depth from the surface all the way to centroid and the centroid is right over here so H bar we do have that value to be 3 m and then this value is H subp that we'll find later okay so H bar is 3 m times area diameter is 2 m so radius is 1 met 1 to the^ 2 so if you calculate this value it will give you Newton or we can write this as 92.5 konon okay so this is the magnitude of the force that is acting on this circular plate now the next thing that I want to find is the location of that and the location would be H subp okay so H subp if you don't know why I instead of writing Y Bar I wrote H bar you should watch the previous video that we talked about the theory of this okay so we do have the value of H bar we do have the value of cross-sectional area the only thing that we don't have is I bar or uh moment of inertia for a circular plate so for a circular plate if you take a look at your book it will give you the value of pi r to^ 4 / 4 take this and put it in this equation and calculate the depth to cent off pressure here here so when you calculate this it will give you the value of as we expected the center of pressure is going to be a little bit lower than the centroid of the shape so HP as we expected is going to be a little bit larger than three so in this problem we calculated the value of HP Center of pressure we also calculate the magnitude of the force that is acting on this um surface let's take a look at another example all right so in this example we have a reservoir which is filled with water and there is a gase that is connected to the end of this pipe over here and this gate has an elliptical shape and the pipe is a circular pipe with the diameter of four meters so we know that the center or centroid of this is going to be right in the middle of it by the way the elliptical shaped gate is going to look like this so if I want to show it to you it would be something like this the centroid is going to be right in the middle and then we call this distance little B and then this distance is going to be little a so if I want to write to you what would be the moment of inertia for this elliptical shaped gate it's going to be Pi / 4 a to ^ 3 B you can find this in your textbook as well okay so we do have this uh what we want to find out is the magnitude and the location of the resultant force on this elliptical shaped gate so I know that the resultant Force this is the centroid is going to be acting on the center of pressure so I'm going to find the magnitude and also the center of pressure over here find that as well let's start with resultant Force so F fr the equation for it is gamma of w because our fluid is water H bar time area Okay so let's first of all talk about H bar H bar is the depth from the surface of the fluid in our case water all the way to uh the centroid so I need to this says 8 m obviously I have that I just need to find this if I find this then h plus this would be H bar and that's pretty easy because this is a circular pipe so and because the centroid of this elliptical shape is right in the middle of it this is going to be four divided two or 2 met right so in in my case over here H bar is going to be 8 + 2 = 10 all right the next thing to find is area what area the area of the gate this again from your textbook the area of this elliptical shaped um gate is going to be Pi a B okay that's easy to find so a is half of its length and I know that its length is 5 Metter so half of that is going to be 2.5 and B is going to be equal to the diameter of this pipe which is four so B is 4 / 2 because it's half of it so would be < 2.5 2 or 5 Pi this is the area of the gate okay so area is equal to 5 pi now if I want to write this and the resultant Force eventually will be equal to all right so we have the magnitude of the force that is acting on this the next thing that I want to find is um the location okay so the location right now take a look at this gate this gate is not vertical right so when the gate is not vertical it's slanted I need to find YP and YP is equal to Y Bar plus I bar / Y Bar time a area okay so now I need to find the value of Yar so I know that um Yar is has a relationship with h so essentially H is equal to Y Bar Time s of theta and Theta is this angle or this angle right so of s OFA in this example would be okay this would be four divided by the hypotenuse which is five so four divided by five um this is s of theta so in order to find Y Bar I can write it like this so the H bar was 10 okay great so this is y bar now I'm going to put the Y Bar in this equation and I'll also have the value of I bar right over here so it's gonna be like this okay so this term over here if you calculate this term it gives you the value of 0.125 12.5 plus .125 YP essentially would be 12.625 M and this would be the value of YP that um we calculated now if you want to calculate the value of HP and HP would be this depth from the surface right over here this is HP so HP would be equal to YP times s of theta that we have over here I will let you to calculate this okay so in this video we talked about two examples to understand how we can apply the hydrostatic resultant force and how to find the location that the hydrostatic resultant force is acting on

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