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Relativity, Gravitation and Cosmology: A Basic Introduction Einstein's general theory of relativity is introduced in this advanced undergraduate and beginning graduate level textbook. Topics include special relativity, in the formalism of Minkowski's four-dimensional space-time, the principle of equivalence, Riemannian geometry and tensor analysis, Einstein field equation, as well as many modern cosmological subjects, from primordial inflation and cosmic microwave anisotropy to the dark energy that propels an accelerating universe. The author presents the subject with an emphasis on physical examples and simple applications without the full tensor apparatus. The reader first learns how to describe curved spacetime. At this mathematically more accessible level, the reader can already study the many interesting phenomena such as gravitational lensing, precession of Mercury's perihelion, black holes, and cosmology. The full tensor formulation is presented later, when the Einstein equation is solved for a few symmetric cases. Many modern topics in cosmology are discussed in this book: from inflation, cosmic microwave anisotropy to the "dark energy" that propels an accelerating universe. Mathematical accessibility, together with the various pedagogical devices (e.g., worked-out solutions of chapter-end problems), make it practical for interested readers to use the book to study general relativity and cosmology on their own. Download link: Buy Premium To Support Me & Get Resumable Support & Fastest Speed!
Mean Curvature Flow and Isoperimetric Inequalities Geometric flows have many applications in physics and geometry. The mean curvature flow occurs in the description of the interface evolution in certain physical models. This is related to the property that such a flow is the gradient flow of the area functional and therefore appears naturally in problems where a surface energy is minimized. The mean curvature flow also has many geometric applications, in analogy with the Ricci flow of metrics on abstract riemannian manifolds. One can use this flow as a tool to obtain classification results for surfaces satisfying certain curvature conditions, as well as to construct minimal surfaces. Geometric flows, obtained from solutions of geometric parabolic equations, can be considered as an alternative tool to prove isoperimetric inequalities. On the other hand, isoperimetric inequalities can help in treating several aspects of convergence of these flows. Isoperimetric inequalities have many applications in other fields of geometry, like hyperbolic manifolds. Buy Premium To Support Me & Get Resumable Support & Max Speed
Differential Geometry and Topology, Discrete and Computational Geometry The aim of this volume is to give an introduction and overview to differential topology, differential geometry and computational geometry with an emphasis on some interconnections between these three domains of mathematics.
The chapters give the background required to begin research in these fields or at their interfaces. They introduce new research domains and both old and new conjectures in these different subjects show some interaction between other sciences close to mathematics. Topics discussed are; the basis of differential topology and combinatorial topology, the link between differential geometry and topology, Riemanian geometry (Levi-Civita connextion, curvature tensor, geodesic, completeness and curvature tensor), characteristic classes (to associate every fibre bundle with isomorphic fiber bundles), the link between differential geometry and the geometry of non smooth objects, computational geometry and concrete applications such as structural geology and graphism. IOS Press is an international science, technical and medical publisher of high-quality books for academics, scientists, and professionals in all fields. Buy Premium To Support Me & Get Resumable Support & Max Speed
Modelling Electroanalytical Experiments by the Integral Equation Method This comprehensive presentation of the integral equation method as applied to electro-analytical experiments is suitable for electrochemists, mathematicians and industrial chemists. The discussion focuses on how integral equations can be derived for various kinds of electroanalytical models. The book begins with models independent of spatial coordinates, goes on to address models in one dimensional space geometry and ends with models dependent on two spatial coordinates. Bieniasz considers both semi-infinite and finite spatial domains as well as ways to deal with diffusion, convection, homogeneous reactions, adsorbed reactants and ohmic drops. Bieniasz also discusses mathematical characteristics of the integral equations in the wider context of integral equations known in mathematics. Part of the book is devoted to the solution methodology for the integral equations. As analytical solutions are rarely possible, attention is paid mostly to numerical methods and relevant software. This book includes examples taken from the literature and a thorough literature overview with emphasis on crucial aspects of the integral equation methodology. DOWNLOAD Register an account and upgrade to premium to get unlimited download speed and resume capability
Entropy, Compactness and the Approximation of Operators Entropy quantities are connected with the 'degree of compactness' of compact or precompact spaces, and so are appropriate tools for investigating linear and compact operators between Banach spaces. The main intention of this Tract is to study the relations between compactness and other analytical properties, e.g. approximability and eigenvalue sequences, of such operators. The authors present many generalized results, some of which have not appeared in the literature before. In the final chapter, the authors demonstrate that, to a certain extent, the geometry of Banach spaces can also be developed on the basis of operator theory. All mathematicians working in functional analysis and operator theory will welcome this work as a reference or for advanced graduate courses. Download link: Buy Premium To Support Me & Get Resumable Support & Fastest Speed!
D-Modules, Perverse Sheaves, and Representation Theory D-modules continues to be an active area of stimulating research in such mathematical areas as algebraic, analysis, differential equations, and representation theory. Key to D-modules, Perverse Sheaves, and Representation Theory is the authors' essential algebraic-analytic approach to the theory, which connects D-modules to representation theory and other areas of mathematics.
To further aid the reader, and to make the work as self-contained as possible, appendices are provided as background for the theory of derived categories and algebraic varieties. The book is intended to serve graduate students in a classroom setting and as self-study for researchers in algebraic geometry, representation theory. Buy Premium To Support Me & Get Resumable Support & Max Speed
Barycentric Calculus in Euclidean and Hyperbolic Geometry: A Comparative Introduction The word barycentric is derived from the Greek word barys (heavy), and refers to center of gravity. Barycentric calculus is a method of treating geometry by considering a point as the center of gravity of certain other points to which weights are ascribed. Hence, in particular, barycentric calculus provides excellent insight into triangle centers. This unique book on barycentric calculus in Euclidean and hyperbolic geometry provides an introduction to the fascinating and beautiful subject of novel triangle centers in hyperbolic geometry along with analogies they share with familiar triangle centers in Euclidean geometry. As such, the book uncovers magnificent unifying notions that Euclidean and hyperbolic triangle centers share.
In his earlier books the author adopted Cartesian coordinates, trigonometry and vector algebra for use in hyperbolic geometry that is fully analogous to the common use of Cartesian coordinates, trigonometry and vector algebra in Euclidean geometry. As a result, powerful tools that are commonly available in Euclidean geometry became available in hyperbolic geometry as well, enabling one to explore hyperbolic geometry in novel ways. In particular, this new book establishes hyperbolic barycentric coordinates that are used to determine various hyperbolic triangle centers just as Euclidean barycentric coordinates are commonly used to determine various Euclidean triangle centers.
The hunt for Euclidean triangle centers is an old tradition in Euclidean geometry, resulting in a repertoire of more than three thousand triangle centers that are known by their barycentric coordinate representations. The aim of this book is to initiate a fully analogous hunt for hyperbolic triangle centers that will broaden the repertoire of hyperbolic triangle centers provided here. Buy Premium To Support Me & Get Resumable Support & Max Speed
Who Gave you the Epsilon?: & Other Tales of Mathematical History This book picks up the history of mathematics from where Sherlock Holmes in Babylon left it. The forty articles of Who Gave You the Epsilon? continue the story of the development of mathematics into the nineteenth and twentieth centuries.
The articles have all been published in the Mathematical Association of America journals and are in many cases written by distinguished mathematicians such as G. H. Hardy and B. van der Waerden. The articles are arranged thematically to show the development of analysis, geometry, algebra and number theory through this period of time. Each chapter is preceded by a foreword, giving the historical background and setting and the scene, and is followed by an afterword, reporting on advances in our historical knowledge and understanding since the articles first appeared. This book is ideal for anyone wanting to explore the history of mathematics. Download: Buy Premium To Support Me & Get Resumable Support & Max Speed Uploaded.net:
Valuation Theory in Interaction For more than a century, valuation theory has had its classical roots in algebraic number theory, algebraic geometry and the theory of ordered fields and groups. In recent decades it has seen an amazing expansion into many other areas. Moreover, having been dormant for a while in algebraic geometry, it has now been reintroduced as a tool to attack the open problem of resolution of singularities in positive characteristic and to analyze the structure of singularities. Driven by this topic, and by its many new applications in other areas, the research in valuation theory itself has also been intensified, with a particular emphasis on the deep open problems in positive characteristic. The multifaceted development of valuation theory has been monitored by two International Conferences and Workshops: the first in 1999 in Saskatoon, Canada, and the second in 2011 in Segovia and El Escorial in Spain. This book grew out of the second conference and presents high quality papers on recent research together with survey papers that illustrate the state of the art in several areas and applications of valuation theory. This book is addressed to researchers and graduate students who work in valuation theory or the areas where it is applied, as well as a general mathematical audience interested in the expansion and usefulness of the valuation theoretical approach, which has been called the "most analytic" form of algebraic reasoning. For young mathematicians who want to enter these areas of research, it provides a valuable source of up-to-date information. A publication of the European Mathematical Society (EMS). Distributed within the Americas by the American Mathematical Society. Buy Premium To Support Me & Get Resumable Support & Max Speed
Eighth Grade Common Core Math: 16 Days to Mastering by John D. Forlini English | December 9, 2014 | ISBN: 1505452384 | 146 Pages | EPUB/MOBI/AZW3/PDF (Converted) | 4 MB This book is a Common Core course for students taking 8th grade math. It can be used prior or during the 8th grade to help students excel. It teaches students the material that will be covered in the classrooms that follow Common Core curriculum.
Eighth Grade Common Core Math: 16 Days to Mastering Besides including a complete Common Core curriculum, there are just the right amount of problems and solutions provided to orient the student for this course. This book is divided into 16 distinct days for learning. A student and teacher can develop their own pace. At the conclusion of the 16th day there is a chapter titled "Beyond Common Core Math (Prep for 9th grade Algebra"). The book is designed for self-teaching by a student or-for a teacher or parent working with the student. The ideal time to read this material would be the summer before taking the actual 8th grade course. However, it can be used any time during the 8th grade. In addition, there is a method revealed in the book that will truly show students how to get the most out of their study time and classroom learning experience towards obtaining an A. It's like getting a 10 minute head start in a 5 k race. This 5 k race is the basis for the rest of a student's life. As mentioned, it also includes a final chapter that is titled "Beyond Common Core". While Common Core does a good job of covering basics, and including introduction to geometry and statistics, it is weak in introducing algebra. Since algebra precedes geometry in high school, this section will introduce the student to algebra with emphasis on word problems. This book is also valuable for preparing for the SAT and ACT tests. It is a solid refresher for these. Download: uploaded_net: | 677.169 | 1 |
All course information is on Moodle, but since some students could not register and therefore do not have access,
I am posting course materials here for the time being.
First handout. Class 1 Class 2
Suggested exercises
These exercises range from straightforward computations to fairly nontrivial theorems that will eventually be covered in class. It is strongly recommended that you think about all exercises and work out some of the computations.
It is strongly recommended not to read the solutions to the exercises prior to thinking about them hard enough to solve some of them. If you find yourself reading most of the solutions and solving very few exercises, then you are not preparing for the tests and the exams. If you are having difficulties with the material, contact me or go to Math Lab, Ross S525. Once the semester is under way teaching assistants will be available in this room during the working hours).
Some of the exercises are starred (like this: 1*). These are theoretical exercises
that may sometimes sound as if they have little to do with the material taught in class, but they do. It is up to you to discover the connections. (Technically speaking, students who expect to merely pass the course need not lose their sleep over these exercises. However, doing so will turn beautiful mathematics into a chore and rob you of most of the benefits of taking this course.)
Suggested exercises
1.1, Solutions and elementary operations
4-8, 12, 15, 17, 18
1.2, Gaussian elimination
1,2, 5-11, 12*, 13-16
1.3 Homogeneous equations
1-3, 5,6,7*, 8-10?
2.1 Matrix addition, scalar multiplication, and transposition
1-5, 8, 9*, 10-21
2.2 Equations, matrices, and transformations
1-7, 9,10,12,13*,14,16-18,20,21
2.3 Matrix multiplication
1-4,6-21,23,24,27,28*,29*,30*,31*,32-34
2.4 Matrix inverses
1,2,3ab,4ac,5ab,8-12,14,17-21,24-28,30-34,36,37,39,41
2.5 Elementary matrices
1-4,6,8,9,12-14,16,17ab,18,21,22
2.6 Linear transformations
1,3-7,9,11-18,20-23
2.7 LU-factorization
1-4,5*,6*,9
Suggested exercises from sections 3.1-3.4, 4.1-4.4, 5.1-5.6 will be posted soon.
Home | 677.169 | 1 |
Maths GCSE
GCSE Mathematics
Pupils in Years 10 and 11 are assigned to sets according to their performance and progress. Pupils have three hour long Mathematics lessons per week and two pieces of homework, each of which should be expected to represent around 40 minutes' work.
Pupils in Year 10 in the academic year 2015-16 are following the Edexcel GCSE Mathematics (2015) syllabus and will all sit the Higher GCSE at the end of Year 11 in June 2017.
Pupils in Year 11 Sets 3, 4 and 5 in the academic year 2015-16 are following the Edexcel GCSE Mathematics A (2010) syllabus and will sit the Higher GCSE in June 2016.
See below for suggestions for becoming an independent learner of GCSE Mathematics.
Year 11 Additional Mathematics
Pupils in Year 11 Sets 1 and 2 in the academic year 2015-16 have already taken GCSE Mathematics in June 2015. In Year 11 they will consolidate their mathematical knowledge and skills from GCSE as well as meeting some key topics from the A Level syllabus such as calculus and the binomial expansion. In June 2016 they will sit either the OCR Free Standing Mathematics Qualification (FSMQ) in Additional Mathematics (Set 1) or the AQA Level 2 Certificate in Further Mathematics (Set 2).
The recommendations below for becoming an independent learner of GCSE Mathematics are equally relevant to pupils studying for either of the further qualifications. | 677.169 | 1 |
Ratings and reviews
Most relevant reviews
Really like this book. I've been struggling to help my 13 year old with maths homework. This book explains the topics in an easy to understand way and with lots of examples and a few have a go questions
This is very useful book for KS3. Explained in detail,even if as a parent you don't know how to figure out any sum, now you can do it by the help of this book. Clearly written with examples on all topics Maths,Algabra,geometry,Shapes.It covers everything for all levels of pupils. | 677.169 | 1 |
Proof and Proving in Mathematics Education
ISBN 9789401780667
ISBN-10
9401780668
Binding
Paperback
Number of Pages
488 Pages
Language
(English)
Subject
Teaching of a specific subject
Proof and Proving in Mathematics Education argues that deductive reasoning and proof should be integral parts of any mathematics curriculum. It identifies the resources teachers needed to enact such instruction, and facilitates the design of teacher education and development programs that provide them. | 677.169 | 1 |
This ebook is available for the following devices:
iPad
Windows
Mac
Sony Reader
Cool-er Reader
Nook
Kobo Reader
iRiver Story
This book deals with the twistor treatment of certain linear and non-linear partial differential equations. The description in terms of twistors involves algebraic and differential geometry, algebraic topology and results in a new perspective on the properties of space and time. The authors firstly develop the mathematical background, then go on to discuss Yang-Mills fields and gravitational fields in classical language, and in the final part a number of field-theoretic problems are solved. Issued here for the first time in paperback, this self-contained volume should be of use to graduate mathematicians and physicists and research workers in theoretical physics, relativity, and cosmology.
In the press
"... skillfully written. It will serve as a relatively accessible introduction to twistor theory for many readers who have not studied the subject before. Others will find it useful as a refresher and as a source of many valuable insights." Nature | 677.169 | 1 |
Description of FastCalc 1.9
Sponsored Links
FastCalc is an intelligent Calculator that is appreciated for its large keys and the constant display of its memories.
To select a side symbol, drag your finger toward it from the center of the key.
To gradually erase an entered expression, drag your finger slowly to the left on the display. To erase only ONE character, do a small swipe to the left.
To clear the entire expression, drag your finger quickly horizontally. The result of the last expression erased is stored in ANS.
Touch the result to store it in one of the memories. | 677.169 | 1 |
Differential Calculus
Calculus is a mathematical tool used to analyse
quantities that keeps _ .
Example:
The slope of a line y = mx + c is _ .
Velocity of an object is the _ of change
of _ of the object with respect to time.
The _ is a measure of the ra
7.1 Recurrence Relation
1
CPT112
Recurrence Relation RR
RR for sequence cfw_an, is the expression that
defines an by using one or more previous terms.
E.g. a0, a1, , an-1, for integer n, n n0, where
n0 is a non-negative integer.
A sequence is a solution f
CPT114
Classical Logic GKH
UNIVERSITI SAINS MALAYSIA
CPT114: Logic and Applications
Classical Logic
Tutorial 5
A. Fill in the blanks.
1. In a categorical syllogism, the three terms each occur in exactly two of the constituent
propositions.
2. The premises
CPT114
Classical Logic GKH
UNIVERSITI SAINS MALAYSIA
School of Computer Sciences
CPT114: Logic and Applications
Categorical Proposition
Tutorial 4
A. Identify the subject and predicate terms in, and name the form of, each of the following
propositions.
1.
1) What are the responsibilities of the file manage r?
The File Manager has a complex job. Its in charge of the systems physical components, its
information resources.
1. Keep track of where each file is stored.
2. Use a policy that will determine where a
CST 232 Tutorial 6 Questions (Concurrent Processes)
1. a. Explain the 3 multiprocessing configurations in common use.
b. For each configuration, give a real life example where that particular configuration
is best suited for addressing the problem
2. Give
CST 232 Tutorial Blended Learning Student Guide
How tutorials are conducted for CST232
1. Tutorial Questions will be posted as a separate Resource file in Moodle.
2. There would normally be 4 questions for each tut
CST 232 Operating Systems
Tutorial 1 (week 3 21-9-2015 to 25-9-2015)
1. Explain the impact of the evolution of computer hardware in the 1960s and
1970s and the accompanying evolution of operating systems software.
2. Explain the fundamental differences be
CST 232 Tutorial 9 Questions (File Management)
1. CD-R disks are write-once disks. However, it is possible to update the files stored in a
CD-R disks. Explain how the file system keeps track of these updated files, and what
happens if there are many small
CST 232 Operating Systems
Tutorial 3 (week 5 5-10-2015 to 9-10-2015)
1. a. Compare and contrast internal fragmentation and external fragmentation.
Explain the circumstances where one might be preferred over the other.
b. Describe how the function of the P
CST 232 Tutorial 8 Questions (Device Management)
1. For each of the following technologies, rank them (1:best-5:worst) based on the
given metrics, and explain what are the main factors that affects the ranking:
Technology / Rank
Sequential Random Random
R | 677.169 | 1 |
Personalized Help at Your PaceSometimes class lectures can move too fast and homework can be bomework. Our tutors will help you master Algebra 2 formulas and homework problems at a pace that works for you. Your tutor will work one-to-one with you and can even help you find Algebra 2 worksheets and practice problems for extra practice. He is enjoying math and understanding also. We have tried other online math services but nothing compares to this.
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Each akgebra shows how to solve a textbook problem, apgebra step at a time help for algebra 2 homework we can help with. Worksheets Need to practice a new type of problem. We havetons of problems in the Worksheets section.
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The Language of Algebra. The Basics ofAlgebra. How can I help you. Enter a problem. (. Algebra 1: Homework Practice Workbook. The. | 677.169 | 1 |
Physical Mathematics
ISBN
9781107005211
Publisher
Cambridge University Press
Author(s)
Kevin Cahill
Publication Date
14 Mar 2013
Overview
Unique in its clarity, examples and range, Physical Mathematics explains as simply as possible the mathematics that graduate students and professional physicists need in their courses and research. The author illustrates the mathematics with numerous physical examples drawn from contemporary research. In addition to basic subjects such as linear algebra, Fourier analysis, complex variables, differential equations and Bessel functions, this textbook covers topics such as the singular-value decomposition, Lie algebras, the tensors and forms of general relativity, the central limit theorem and Kolmogorov test of statistics, the Monte Carlo methods of experimental and theoretical physics, the renormalization group of condensed-matter physics and the functional derivatives and Feynman path integrals of quantum field theory. | 677.169 | 1 |
MATH2221 (alternatively MATH2121) is a compulsory course for Mathematics majors.
If you are currently enrolled in MATH2221, you can log into UNSW Moodle for this course.
Course Overview
A differential equation arises naturally whenever one seeks a mathematical description of a phenomenon that changes continuously in time or space. As a consequence, solutions of differential equations exhibit a great variety of features and behaviours, and not surprisingly a great variety of techniques are used in this field of mathematics. In this course, we cover the classical foundations of the subject, restricting our attention to analytical methods of solution and concentrating on linear differential equations. As such, MATH2221 will prepare you for a range of later-year courses offered by the School of Mathematics and Statistics that treat topics in the theory and applications of differential equations. The methods we cover are also used in many courses taught by other schools. | 677.169 | 1 |
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
How neural networks work - a glimpse into math for beginners
a practical crash course in the mathematical concept used in neural networks for beginnersWhat is machine learning / ai ? How to lean machine learning in practice?
machine learning / ai (artificial intelligence) is the hottest topic in this century - for good reasons. Some people conceive it the "steam engine" of our century and one thing is certain: It will drastically change the world.
Neural Networks (often referred to as deep learning) are particular interesting. But there are several questions to answer.
One of those is the math part involved in the construction of these neural networks. This is exactly were we want to start here.
This course is designed to give beginners a practical introduction to the mathematical concept. In this course we will use linear regression as an example to understand the math. The idea is completely transferable to neural networks. The change in coefficient works exactly the same.
If you are a complete beginner and want to get the main idea of neural networks first then I suggest to start with my other course
"A crash course in neural networks for beginners"
first.
Do you want to take your chance get a glimpse into the mathematical concept and expose yourself to this interesting topic which will change the world forever? Then join me and other students to dive deeper into neural networks right now. What are you waiting for?
See you in the first lecture
Who is the target audience?
beginners in neural networks who want to dig a little bit into math
beginners who want to learn the mathematical background applied in neural networks | 677.169 | 1 |
Tough Test Questions? Missed Lectures? Not Enough Time? Fortunately, there's Schaum's. This all-in-one-package includes more than 65065 fully solved problems Concise explanations of all geometry concepts Support for all major textbooks for geometry courses Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time--and get your best test scores! 400,000 sold! This review of standard college courses in geometry has been updated to reflect the latest course scope and sequences. The new edition includes an added chapter on Solid Geometry and a chapter on Transformation, plus expanded explanations of particularly difficult topics, as well as many new worked-out and supplementary problems.
Tough Test Questions? Missed Lectures? Not Enough Time? Fortunately18 fully solved problems to reinforce knowledge Concise explanations of all trigonometry concepts Updates that reflect the latest course scope and sequences, with coverage of periodic functions and curve graphing. Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time--and get your best test scores! Schaum's Outlines-- 830 fully solved problems with complete solutions Clear, concise explanations of all course concepts Coverage of biochemical signaling, genetic engineering, the human genome project, and new recombinant DNA techniques and sequencing b>Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time-and get your best test scores! Schaum's Outlines--See how to solve precalculus problems with this enhanced ebook that features 30 videos of professors working through solutions! Confusing textbooks? Missed lectures? Not enough time? Fortunately for you, there's Schaum's Outlines. More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Each Outline presents all the essential course information in an easy-to-follow, topic-by-topic format. And there are plenty of problems for you to practice on, with more than 700 precalculus problems with fully worked solutions so you can check your work, or get help when you need it. Plus, this new enhanced edition features video solutions of professors showing exactly how to solve problems. If you want top grades and a thorough understanding of precalculus, this powerful study tool is the best tutor you can have!
Principles and problems of plane geometry with coordinate geometry has been writing in one form or another for most of life. You can find so many inspiration from Principles and problems of plane geometry with coordinate geometry also informative, and entertaining. Click DOWNLOAD or Read Online button to get full Principles and problems of plane geometry with coordinate geometry book for free.
Tough Test Questions? Missed Lectures? Not Enough Time? Textbook too pricey? Fortunately, there's Schaum's. This all-in-one-package includes more than 650 Geometry, Sixth Edition features: • Updated content to matches the latest curriculum • Over 650 problems, solved step by step • An accessible format for quick and easy review • Clear explanations for all geometry concepts • Access to revised Schaums.com website and new app with access to 25 problem-solving videos, and moreYou donÕt need to be a math genius to do well in algebra! This plain-English guide to intermediate algebra explains what algebra is all about, in language thatÕs easy to understand. Then it shows you how to solve every kind of problem that youÕll find on your tests, step-by-step! You get 885 completely worked problems, with each step in their solutions. More than a thousand additional problems let you test your skills, then check the answers. So comprehensive that it can be used alone as a complete independent study course, this solved-problem study guide is the perfect review for college algebra. It even includes instructions on using calculators, as well as effective study skill suggestions!Schaum s Outline of Theory and Problems of Basic Mathematics has been writing in one form or another for most of life. You can find so many inspiration from Schaum s Outline of Theory and Problems of Basic Mathematics also informative, and entertaining. Click DOWNLOAD or Read Online button to get full Schaum s Outline of Theory and Problems of Basic Mathematics book for free.
This lucid introduction for undergraduates and graduates proves fundamental for pactitioners of theoretical physics and certain areas of engineering, like aerodynamics and fluid mechanics, and exteremely valuable for mathematicians. This study guide teaches all the basics and efective problem-solving skills too. | 677.169 | 1 |
Terrific Triangles is a math acts drill-and-practice program that can present addition/subtraction and multiplication/division facts. There are 2 playing modes. In all cases the triangle presents two parts of a fact family and the player must determine the missing third part. The sum and product always appear at the top of the triangle. The addends and factors always go in the two lower corners. Student mode allows individual facts practice...
Mathematics
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dhrystone Benchmark 1.0
dhrystone Benchmark is a general-performance benchmark test originally developed by Reinhold Weicker in 1984. dhrystone benchmark is used to measure and compare the performance of different computers or, the efficiency of the code generated for...
60 KB
Mathematics
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Boolean Calculator 2.0
Boolean Calculator is a simple utility for technical informatics on university. You can entera formular like "a and b and (not c)" and it shows you a table where youcan see on which boolean combination of a, b and c the term is true.
Mathematics
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RPN Calculator X 1.9.6
RPN Calculator is a simple but powerful RPN calculator for MacOS with a scalable interface via selectable function palettes. This calculator is one of the only mac calculator programs that handles complex arithmetic.
Mathematics
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RPN Calculator 1.9.2
RPN Calculator is a simple but powerful RPN calculator for MacOS with a scalable interface via selectable function palettes. This calculator is one of the only mac calculator programs that handles complex arithmetic.
Mathematics
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Protractor 2.4.1
Protractor is a software tool with which to measure angles. It has a semitransparent view that mimics real protractors. The program can number the angle along gradations and draw gradations clockwise and counterclockwise. Using this tool with a...
749 KB
Mathematics
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Triangle 0.4
Triangle is a triangle solver. If you know three attributes of a triangle (sides or angles) it finds the rest unless you specified three angles. It also draws and prints the found triangle(s).
Mathematics
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Matlab 7.4
Matlab integrates mathematical computing, visualization, and a powerful language to provide a flexible environment for technical computing. The open architecture makes it easy to use Matlab and its companion products to explore data, create...
12.2 MB
Mathematics
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Spoke Length Calculator 1.0.1
Spokcalc calculates spoke lengths for wire spoked wheels. Spokcalc includes a database of over 400 hubs and 550 rims. The Spokcalc database is also user-modifiable and -expandable so you can add more rims and hubs.
116 KB
Mathematics
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Metes and Bounds 2.7.2
Metes and Bounds turns data into a plot map.It can calculate square feet, acreage and hectares.Enter measurements in rods, chains, meters, yards, inches, fathoms, hands, spans or furlongs. Angles can be given down to minutes and seconds. Can... | 677.169 | 1 |
3. Extract, interpret, evaluate, communicate, and apply quantitative information and methods to solve problems, evaluate claims, and support decisions in their academic, professional and private lives. (Quantitative Literacy)
Not Addressed
4. Appreciate cultural diversity and constructively address issues that arise out of cultural differences in the workplace and community. (Cultural Awareness)
Minimally
5. Recognize the consequences of human activity upon our social and natural world. (Community and Environmental Responsibility)
Outcome Assessment Strategies
At least one project plus some combination of the following:
Class participation
Group projects
Presentations
Portfolios
Research papers
Homework assignments
Written paper
Quizzes
Exams
Other assessments of the instructors choosing
Course Activities and Design
This course will be delivered through a combination of lecture and student activities including group and individual problem solving during class. Emphasis is to be given to applications from outside the mathematics classroom. Applications will come from the broadest possible range of disciplines.
Course Content (Themes, Concepts, Issues and Skills)
Integration
Considering the question of area – using limits
The definite integral
Fundamental Theorem of Calculus
Anti-derivatives and indefinite integrals
Techniques of integration:
substitution
integration by parts
trigonometric integrals/trig substitution/partial fractions
Numerical Integration and approximation
Improper Integrals
Applications using Integration Directly
Area under and between functions
Volumes:
Rotation about the x-axis
Rotation about the y-axis
Slicing
Arc-length and Surface Area
Mean Value Theorem for Integrals
Work
Force of Water Behind a Dam
Centroids and Center of Mass
Statistics
Applications of Integration in Business
Other Applications of Integration
Applications of Integration in solving Basic Differential Equations
What are differential equations?
Differential Equations and Assumptions about Growth
Slope Fields
Solutions
Separable Differential Equations
Continuous Growth Model
Logistics Model
Phase Diagrams
Predator-Prey Model
Department Notes
Answers to all application problems will be given in complete sentences with correct units. The grade will include at least one project | 677.169 | 1 |
Showing 1 to 30 of 103
CS1231
Tutorial 2Terence Sim
11 Aug 2016
I can do it!
CS1231 Discrete Structures
^
Message of the Day
I can do it!
I can do it!
I can do it!
Dont feel like this
But like this
http:/
Think
CS1231
Tutorial 3CS1231
Tutorial 1
Informal Introduction to Methods of Proof
The purpose of this tutorial is to learn to read and write proofs in the format prescribed. It is recommended that you explore the available resources in the library and on the Internet to answer
CHAPTER 4 INDUCTION
SECTION 4.1 MATHEMATICAL INDUCTION
Mathematical induction is used to prove statements that asserts that P (n) is true for
all n Z+ where P (n) is a propositional function. It is an extremely important proof
technique.
PRINCIPLE OF MATH
CHAPTER 5: COUNTING
SECTION 5.1 BASICS
PRODUCT RULE
Suppose an operation can be broken down into a sequence of 2 steps. If the rst step can
be done in r ways and the second step can be done in s ways (regardless of how the rst
step was done), then the ent
CHAPTER 7
SECTION 7.1
DEFINITION:
A
GRAPH
G = (V, E) consists of:
V , a nonempty nite set of
E, a set of
VERTICES
AND
EDGES.
Each edge is associated with either one or two vertices called its
with one endpoint is called a LOOP.
ENDPOINTS.
An edge
Someti
CHAPTER 3 THE INTEGERS
SECTION 3.1 DIVISIBILITY
DEFINITION:
Let n, d Z with d = 0. We say that d DIVIDES n if n = dk for some k Z or equivalently,
n/d Z.
Other ways of saying include: n is DIVISIBLE by d, or n is a
FACTOR of n, or d is a DIVISOR of n.
MUL
CHAPTER 8 TREES
SECTION 8.1
DEFINITION:
A
TREE
is a connected (undirected) graph with no cycles.
REMARK
A tree cannot contain loops or multiple edges since they are cycles.
EXAMPLE
G1 , G2 are trees. G3 is not a tree as it contains a cycle. G4 is not a tr
Unit 3
Algebra and Number Sense:
Proportions
and Percents
Develop an understanding
of and apply proportionality.
CHAPTER 6
Ratios and Proportions
Use ratio and
proportionality to solve problems,
including those with tables and
graphs.
CHAPTER 7
Applying P
A Brief Introduction to Fitting Models and Parameter Estimation
Least Squares Fitting
Suppose we have N pairs of observations, (x1 , y1 ), . . . (xN , yN ) and we expect x and y to be
related according to some function, g, which itself involves one or mor
Wireless Connection Guide
See over for wired instructions
Connecting to the Internet 2016/17
There are two ways to connect to the Internet in your room: the first is using eduroam
wireless and the second is by connecting to the ResNet socket on your wall.
Congestion Control and QoS
CS158a
Chris Pollett
Apr 11, 2007.
Outline
Congestion Control Algorithms
Quality of Service
Congestion Prevention Policies
We briefly point out some policy issues in
each layer which can affect congestion.
The choice is like
Recent Documents
Cardinality
Read Rosen, 2.1, 2.4; Epp (2nd ed.), 7.6
1
Finite and Infinite Sets
Definitions.
A set S is called finite if S is the empty set or there is a 1-to-1
correspondence from S to cfw_1, 2, . . . , n where n is a positive
integer. In the first case,
CS1231
Tutorial 7
Graph Theory
This tutorial must be prepared in writing. Make two copies of your answers: one for you and
one for the tutor. Bring your lecture notes to the tutorial. The answers to some of these questions
require the use of the theoretic
Combinations
Read Rosen, 5.3, 5.5
1
r-Combinations
Definition. An r-combination of a set of n elements is an unordered selection (subset) of r elements taken from the set of
n elements. The number of r-combinations (subsets of size r)
of a set of n elemen
CS1231
Tutorial 8
Trees
This tutorial must be prepared in writing. Make two copies of your answers: one for you and
one for the tutor. Bring your lecture notes to the tutorial. The answers to some of these questions
require the use of the theoretical resu
Chapter 7
Relations, Equivalence Relations and Orders
A friend to all is a friend to none.
Corpus Aristotelicum, by Aristotle
7.1
Relations
Ordered pairs give us the opportunity to associate mathematical objects. Sets of ordered
pairs are collections of s | 677.169 | 1 |
Basic Linear Algebra is a text for first year students, working from concrete examples towards abstract theorems, via tutorial-type exercises. The book explains the algebra of matrices with applications to analytic geometry, systems of linear equations, difference equations, and complex numbers.
This is a short, readable introduction to basic linear algebra, as usually encountered in a first course. The development of the subject is integrated with a large number of worked examples that illustrate the ideas and methods. | 677.169 | 1 |
Our Guides
H2 Mathematics Handy Guide: Statistics
Nothing less, nothing more. That's what this book is written for.
H2 Mathematics Handy Guide: Statistics is written in a simple manner, based on the needs of a statistics student. This pocket guidebook is packed with a number of tips and tricks, cherry-picked to help students maximise their precious time during examinations. From permutations and combinations to hypothesis testing, we have included key concepts, definitions and almost every formula you will ever need. | 677.169 | 1 |
Ideal (solved examples) companion to your calculus textbook
If you are studying calculus, the books from Schaum will certainly help you to develop your problem solving ability.
Not what I expected
This is a good book if you are not looking for true advanced calculus material.This is just calculus--with some tougher topics, but is not a true advanced calculus book (which is much more proof based).
A Different Take on Advanced Calculus
After finishing the first 3 semesters of what is basically applied Calculus, majors in pure mathematics part company with their fellow students and move on to what is called "advanced calculus" or "functional analysis". This jump to the next level of mathematics, in which basically students have to be taught calculus all over again from a rigorous proof-based method, is a rude awakening to many. Most textbooks used in this situation, such as those by Rudin, assume that a talented and enthusiastic instructor will be acting as tour guide for the student, and will fill in the gaps. Unfortunately, this is not always the case. This Schaum's outline works well for such a situation. It not only covers what you would expect to see in undergraduate analysis, it mixes proofs with solved numerical problemsand applications so that the student does not get bogged down or bored as can be the case with many of the texts on the subject that have nothing but proofs as exercises. In that way, it is unconventional in its treatment of the subject. Also, it provides more explanation and tutorial than you will find in the average Schaum's outline, and is pretty good at illustrating the correct way to perform a proof. If you can afford it, an excellent companion text to this Schaum's outline is Abbott's "Understanding Analysis", which can be used for self teaching of analysis because it is so detailed and clear.
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Too many errors
i cannot believe the amount of incorrect math in this book. and that no one bothered to check it before they published it
no so good, don't buy this book!!
I am studying calculus currently in my ap class. I bought this book. And I found out there are many misprints and errors there.
Should be called "1 Step to a 1"
Avoid this review book.I am a teacher of Calculus, and I am going to teach AP Calculus next year.Unlike most students who need to review, I already know the material 100%.I was SHOCKED by the amount of errors and misprints that are in this book.If you are a student who doesn't know Calculus that well, you are going to think that YOU are the one making errors, but it is the book that is wrong.I contacted the publisher, and they have yet to give me a decent response.I feel this book should be pulled from the shelves.They should title it "1 Step to a 1", and the one step is to buy this book.
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Stewart Solution Manual for Calculus
This is NOT the solution manual for the second half of Stewart's Calculus, 5th Edition (Thomson Publishing).It's part 2 for an earlier (obsolete?) calculus book.If you need the solution's manual for Edition 5, order Dan Clegg's Student Solutions Manual for James Stewart's (Multivariable) Calculus (Thomson Publishing, 2003).
Not too bad...
I thought the text wasn't too bad, but considering the material until about chapter 17 was quite easy, that is really not saying much. The book didn't give appropriate examples of all the different special cases of line and surface integrals, which made 17.3-17.7 quite difficult; I had to learn surface integrals elsewhere. This book wasn't horrible, but I wouldn't recommend it.
could be better but does the job.
the chapter on vector calculus was confusing and the examples were ridiculous in that they didn't fully teach the concepts. if you want a real calculus book, try Anton. half of our class used Anton's Calculus to learn the material since the chapters are in the same order pretty much.
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Great side reader for a calculus course
I wish this book had been around when I was taking calculus a few decades ago.It is extremenly well written and explains all the reasons why mathematicians had to introduce all the concepts and definitions you encounter in a calculus course.Reading this book on the side will tell you exactly why you're doing what you're doing, and where you are going.All students of calculus will benefit from this book.
A Great Book For Math Fans
If you like math, I guarantee you'll like this book! The author starts out with some very nice infinite sums from 17th and 18th century mathematics (Newton, Leibniz, the Bernoullis, and Euler's Gamma function). He continues into the 19th century with Riemann and Lebesque integrals, Weierstrass' pathological functions, Cantor's set theory, and winds up with Baire's category theorem.
Calculus and the Masters
I haven't had a math class in 30 years.I wish I had this book when I was in college.The author mentions studying the art masters if you're an art student, so why not study the math masters if you're a math student?Sometimes the reading gets tedious during one sitting, but after a break it's fascinating once again.Be sure that a few calculus memories are floating in the back of your head before you start, and you'll be well-rewarded.
... Read more
What were they thinking
I am a student and do to me being more of a visual learner I end up teaching myself most courses from the book.I am usually very succesful in this manner except when a come across a book as poorly pute together as this one.In each section the authors will have about two pages to explain a concept in which they fill with mostly erroneous information.They then will have 3 or 4 pages of problems in which they didnt explain how to do at all.I am taking a statistics class alongside this and am having absolutely no problems figuring out using the book exclusively.Maybie the internet material will help a little thougheather way the book should be able to stand alone without the aid of a solutions manuel which as far as im concerned is just another ploy to make a textbook into a package that ultimately costs more.
... Read more
Ancillaries for Calculus (6th edition)
If you already own or plan on purchasing Swokowski's "Calculus" (6th edition) or his "Calculus of a Single Variable" (2nd edition), then you will probably want to obtain some of the following ancillaries:
Amazing book!!
This book was adopted in my college, but it is out of print in Brazil. Thus, the depth Math solved replace it by Anton's book. 1280 pages pure and applied calculus + answers and appendix
This book is so comprehensive you can use it for any engineering and general purposes. You must also get it`s study guides and instructor`s book for an efficient study... Their study guide 1 ISBN: 0-534-936-261, 2 ISBN: 0-534-936-27X, Instructor`s manual 1 ISBN: 0-534-936-30X and 2 ISBN: 0-534-936-318.
... Read more
A wonderful calculus book, must have if you love good book
I first saw this book as a MIT textbook.Read it briefly and absolutely loved it.I learnt Calculus many years ago and hope I had this book then.The book is so clear written and easy to understand. I am buying this book for my son now.I am sure it will be a great help for anyone reads this.
Strang Calculus Book
This book has excellent explanations, and is well organized.Usefull as a reference, and to teach yourself calculus.I used it all through college as a supplement (and sometimes replacement) for the assigned text.
An excellent text
If you intend to teach yourself calculus, or if you are looking for a text for review, this one would be an excellent choice.The topics are well explained and well motivated by applications.The book covers a wide array of topics and each of them is clearly developed.I would choose this as the text for a class if I were to teach it.I certainly recommend it for those learning outside the structure of a classroom.
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One reviewer said :"by carefully developing only what is essential." which is best thing to say about this book
So far Im at chapter 2 (just finished it). So Im going to update this once im done with the book.
Consice
This is essentially preparation for differential geometry/topology.It is definitely for math students (I would say familiarity with calc 3 and a good intro to analysis course are essential prerequisites).It is very consice, and so not easy to read (few or no examples or worked out exercises!).Though only about 130 pages, it covers something like a semester and a half (an estimate... I took a year long course which used it) of material.Nonetheless, it is a great little book.The exercises are excellent.
A Ph.D. in Electrical Engineering
I haven't read the book yet, but I browsed the contents and it seems to be at the same level of other graduate textbooks in applied and theoretical math such as those that deal with Wavelets, Hilbert Spaces and Functional Analysis. It seems very strange to me to read the despective comments from pure mathematicians about the "low" level of mathematics that engineers have. Engineering and Mathematics share some roots but they are different professions. One of the greatest mathematicians in the whole history of mankind, Paul Dirac, was an E.E. and a Physicist. If it wasn't for engineering, mathematics would be in the same planet with poetry. Some respect please.
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Written with the reader in mind.
I think this is one of the most understandable books in signal processing that I've ever come across.I get the feeling that Brigham had been frustrated by technical texts that were poorly written, and decided he wasn't going to commit the same sin.Plenty of carefully planned illustrations designed to help the reader start from a known place, and move step-by-step to an understanding of something new.Not just a bunch of faceless equations.I think this would be an excellent college text.
Very good, but be careful, though
The book I have is ISBN 0-13-307496. It was published in 1974.
Outstanding classic
I am mainly a neurobiologist by training, but transform theory is very useful in visual neurobiology and visual psychophysics, and I've looked at and read dozens of books on various aspects of signal processing and transform theory. Much of this is surprisingly applicable to the brain science area, as the revolutionary work of David Marr and other scientists showed a quarter of a century ago.
I recommend this text for self-study
This collection of problems is really quite good. First, the problem is spelled out entirely, so you don't need to refer to the text. It is self-contained, for the most part. Each of the solutions is worked out with generous detail and the problems are chosen to accentuate certain basic results that are easy to take for granted unless you see the difficulties that arise in their absence. I recommend this text for self-study as a running start to Polya and Szego's more difficult problem book.
Excellent companion to Lang's Analysis book
This book is not just a "solutions manual."There are detailed proofs for all of the problems in the book.This book is essential if you want to truly learn real analysis.
Finaly an outstanding problems and solutions guide
The only way to study mathematics is by solving a large number of problems: this book permits the eager student to challenge his acquired knowledge. A must to all the students wanting to really masterundergraduate analysis.
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Poor presentation
I used this text in high school for AP Calculus AB.
One of the best introductory calculus texts ever
If you want to learn calculus from first-rate experts and also first-rate explainers, you will love this text. The authors have gone to great lengths to motivate and elucidate the ideas behind calculus, and they do so in a brilliant and very readable fashion. They take the reader by the hand and explain, step by step, what you need to know. Countless applications from science, engineering, and economics let you get an idea of how calculus is really used out there in the field.
Not bad, but needs better explanations
This book was good for understanding the basics of all the concepts, but if you wanted to go more in depth, it was not a very good learning resource. The explanations were not very clear and detailed, and there were also not enough examples.Often, I felt that it would have been much more helpful to have detailed explanations of the more difficult problems.
... Read more
Calculus boot camp
Each chapter gives a brief over view of the topic ( limits, continuity, etc etc) then immediately runs you through the paces with a dozen or so problems over the subject. The problems are wide ranging and better than what most texts will give you. Each problem is then followed with a step by step guide to solvingWITH EXPLANATIONS! (whatta concept)They say repetition is good for learning and this gives it to you. If you don't know the subjects well and you use this book, you will know them. If you know calculus and just need to be stronger in some areas, this can do that too. If you need a refresher this works in that capacity as well.
Forget this, if you want the best and most comprehensive TRY
THE KING OF ALL MATHEMATICAL PROBLEMS!!!!!!!!!
Practice is key
One thing students of calculus need to do is practice, practice, practice problems. A book with lots of problems and clearly written, detailed solutions is part of the required program. Having used this book myself as a student, I completely recommend to my own students. As always, try to work out the problems yourself before peeking in on the solutions.
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great book
i found the first three chapters of this book very clear and well written. i'd strongly recommend it for someone looking to learn about analysis on the real line.
Good book for reading and as a graduate student
Easy to read. My university is using this book to get the graduate students ready for the real analysis qualifying exam. So go ahead and buy this book if you're planning to work on a PhD in mathematics. If you're not planning to work on a PhD in math, this is still a good book to read if you enjoy studying about the real line.
Suffers from all the flaws of a 1st edition
This book has a lot of problems. Several sections are poorly written/edited. Several important named theorems are not clearly labeled. Also some of the proofs contain typos or errors. The chapter on differentiation is particularly lacking. The chapter is poorly organized and presented. There is also a glaring TeX error in the chapter.
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good book...
My knowledge of this book originates from a course in measure and integration theory a couple of years ago. Back then I honestly didn't like the book very much. It is very terse, sometimes tending towards incomprehensible. The exercises were great though - hard. Exercises should be hard - that's the best way to learn mathematics.
A compact and good book
I like this book! Of course the book demand some work of the reader, and it is not the easiest book to read, but its easier to get an overview of an subject, when the material is compact as in this book!
Frustrating
It seems that the higher up you go in the Mathematics curriculum, the poorer the books you meet.In my honest opinion, a book should help you learn and understand the material as quickly as possible.Otherwise, you might as well be given a list of definitions, stuck in a closed padded room and asked to come up with all the theorems by yourself.Unfortunately, there are too many graduate textbooks out there written by individuals who seem to have no desire to make the ideas they are trying to present as clear as possible.There's no educational philosophy.This book falls under that category.For example, this book is almost completely devoid of any examples.I don't know about you, but from example, is how I learn.I could go through this book much faster, if there were some decent examples.You can tell me a thousand times what a sigma algebra is, but if you don't give me some decent, worked-out examples which might tell me why tell me why I should learn it (other than because I'll fail the course), I'm going to forget the definition after 5 minutes.Secondly, it would help if there were more pictures.A picture is worth a thousand words.Third, some of the definitions are not worded as well as they should be.last night I spent ten minutes trying to figure out whether the definition for x-section Ex = {y in Y : (x,y) in E} meant that "for all x," or just "for some x?"It turned out it meant "for fixed x."But nowhere was that little tidbit of information written.Ten minutes may not sound like much, but if you have to read 10 pages before you get to pleasure of spending 10 hours with the homework problems, that translates into a lot of time you could spend doing other things if only this book were presented in a manner which would enable you to learn the material more efficiently.I give it two stars primarily because some of the homework problems aren't too bad.If you have a choice, have a look at Kolmogorov and Fomins book on Real analysis.It's not perfect, but the material in it is organized better.(It's not as DENSE)Plus it's a Dover book, and therefore much cheaper.
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Not great, but not too crappy
I used this book for calc 2 and am currently using it for calc 3 (I have a test tomorrow....AHHHHHH).I don't really like it.It doesn't explain a lot of stuff in a way easy to understand.Also there aren't good examples for the problems, so you get to homework and you have no idea how to actually do it because there isn't an example of it.But it's not the worst book ever.If you think hard enough you can usually follow what the text is attempting to explain to you.The Student Solutions Manual is helpful, but can lead to more confusion when it's wrong or it solves something in some obscure totally weird way.
If All Mathematics Texts Could Be This Good
I have two copies, because the first became worn out.
The Best! there is
I was fortunate to have the co-author of this book as a professor at the University of Houston. The book is a wonderful combination of: 1) clear and concise presentation of theory; 2) examples and applications relevant to science, mathematics and business; 3) analysis, in fact most results are derived and there are problems which enable a motivated student to begin building his/her derivation and mathematical reasoning skills; 4) diagrams. The book is also comprehensive. It "constructs" calculus from the theory of limits, provides the foundation of calculus in the form of various theorems, including but not limited to the Mean Value Thoerem, the Intermediate Value Theorem, Rolle's theorem, L'Hospital's theorem, Riemann Integral, Fundamental Theorem of Calculus. Also provided are the resulting powerful derivative, integration and function analysis tools. The book includes calulus of one variable as well as many variables in the Vector Calculus section of the book. Finally, it ends with an introduction to Ordinary Differential Equations. It takes three semesters: Calculus I, II and III to get through this book at the University of Houston. So, although the price seems high, there is a tremendous amount here. I am now a graduate student in physics taking an Analysis course with the great author of this book (G. Etgen) this summer. This is not a course I need, so this gives you an idea of how stimulating Dr. Etgen's presentation of calculus is. You must master calculus before you can be competent in applied physics (or other science field), advanced finance, ... If you study each page of this book and work many of the problems in each section, you will be well on your way to being accomplished in a career in applied or theoretical mathematics. Also, the student solutions manual to the book is excellent and recommended.
Read | 677.169 | 1 |
ISC Syllabus 2018, 2017, 2016 Class XII – Mathematics
Important Talent Search Exams
ISC Syllabus for year 2018, 2017, 2016 is given below. Students of Class XII can use this syllabus for their annual examination preparation.
Board: Indian School Certificate Class: XII Subject: Mathematics
ISC Mathematics Syllabus
Aims:
To enable candidates to acquire knowledge and to develop an understanding of the terms, concepts, symbols, definitions, principles, processes and formulae of Mathematics at the Senior Secondary stage.
To develop the ability to apply the knowledge and understanding of Mathematics to unfamiliar situations or to new problems.
To develop skills of –
(a) computation.
(b) reading tables, charts, graphs, etc.
To develop an appreciation of the role of Mathematics in day-to-day life.
To develop an interest in Mathematics.
To develop a scientific attitude through the study of Mathematics.
As regards to the standard of algebraic manipulation, students should be taught:
(i) To check every step before proceeding to the next particularly where minus signs are involved.
(ii) To attack simplification piecemeal rather than en block.
(iii) To observe and act on any special features of algebraic form that may be obviously present.
The syllabus is divided into three sections A, B and C –
Section A is compulsory for all candidates. Candidates will have a choice of attempting questions from
EITHER Section B OR Section C.
There will be one paper of three hours duration of 100 marks.
Section A (80 marks) Candidates will be required to attempt all questions. Internal choice will be provided in three questions of four marks each and two questions of six marks each.
Section B / C (20 marks) Marks): Candidates will be required to attempt all questions EITHER from Section B OR Section C. Internal choice will be provided in two questions of four marks each. | 677.169 | 1 |
Unit 1: Modeling withThis unit from the Mathematics Vision Project has activities which could be used to highlight certain features of functions. One needs to be careful about prior knowledge when using these activities as their units are sequenced differently than ours | 677.169 | 1 |
The School Review Volume 30or who intends to enter it possess a knowledge of and a skill in the fundamental operations of arithmetic and the ability to apply this knowledge and skill in practical business affairs. In order to acquire skill in the fundamental operations and to be able to apply this skill in actual business life, it is necessary that the skill be acquired in working out real problems such as will be encountered in business surroundings. The aim of the author of a recent text in business arithmetic1 has "been.... to meet the educational needs of all those who wish to enter the business 1 Caul Holltday and Sophia Camenisch, English Grammar Drills. Chicago: Laird & Lee, 1922. Pp. 149. 'frank M. Rich, Morning Readings. Boston: Richard G. Badger, 1920. Pp-355-t' oo. 'helen J. Kigcen, Practical Business Arithmetic. New York: Macmillan Co., 1922. Pp. xi+404. world and those who have already entered it without sufficient knowledge of the fundamental operations in arithmetic or who lack the power to apply their knowledge to the solution of present day business problems" (p. v). As to topics treated, the book is not very different from many of our textbooks on business arithmetic. The chief differences between this book and others on the same subject are the method by which the author connects each topic with practical business affairs and the consistency in the practical application of the work. This method of treatment may be illustrated from the chapter on "Units of Measure and Their Application." In this chapter, the methods of measuring dress goods, carpeting, papering, flooring, boards or lumber, time, heat, humidity, water, gas, electricity, wood, and coal are fully explained. In connection with each of the methods, exercises and...show more | 677.169 | 1 |
The California Mathematics Standards - PowerPoint PPT Presentation
The California Mathematics Standards. The California Standards come in two flavors. General Standards Green dot standards. In California the green dot standards are the focus of most instruction since these topics comprise 85% of the state exams in grades 2 – 7 | 677.169 | 1 |
The master theme of the Eleventh Grade will be to encourage students, through the curriculum, to take stock of the higher cognitive and affective skills which they have developed, as they more deftly use these skills in the preparatory stages of not only discovering, but embarking upon, the vocation which God has pre-ordained for them. Students will come to understand more fully just how their abilities and skills can be brought to bear in focusing on the challenging task of applying principles, concepts, and ideals toward the ultimate discovery of God's plan for each of them. Moreover, as they prepare themselves for their vocations, they will work toward becoming productive members of their communities, equipped with all the tools and information to allow them to defend their Faith and apply traditional Catholic principles and values, no matter where their vocation takes them.
Theology: This year the students will continue the Quest forHappiness Series, with an emphasis on the salvatory role of the Church, guidedby the Holy Ghost. This third bookof this series, The Ark and the DoveText, reveals the love which the Holy Ghost has for us. Students will studythe Third Person of the Holy Trinity, the Holy Ghost, as well as the nature,origin, structure and history of the Church. Also discussed will be thesacrament of Penance, and the fifth, sixth, and ninth commandments. The Ark& the DoveAnswer Keyis a convenient aid for parents or in-home tutors to checkover student responses. Also available are The Rosary Novena Bookletand The Way of the Cross Booklet,according to the method of St. Alphonsus Liguori.
Advanced Mathematics I: The course is centered around the first half of Saxon'sAdvanced Math Text, with the second half being covered in the 12th Grade. This course should only be attempted by those students who have successfully completed Saxon Algebra II. General topics covered include: advanced algebra, geometry, trigonometry, discrete mathematics, and mathematical analysis, all interwoven to form a fully integrated text. Specific topics treated in the text include: permutations and combinations, trigonometric identities, inverse trigonometric functions, conic sections, graphs of sinusoids, rectangular and polar representation of complex numbers, theorems, matrices, and determinants. A rigorous treatment of Euclidean geometry is presented, as well. Word problems are developed throughout the problems sets and become progressively more elaborate. The graphing calculator is used to graph functions and perform data analysis. Conceptually-oriented problems prepare students for college entrance exams. The student edition contains answers to odd-numbered problems. The 3 book set includes: Saxon Advanced Math Text, Saxon Advanced Math Answer Key, and Saxon Advanced Math Test Booklet. Saxon Advanced MathSolutions Manual is also available, and strongly recommended.
Public Speaking: As Grammar/Composition is required each year, all students must take this 11th Grade Grammar course. Through this course students will develop their abilities in verbal persuasion. Ultimately, they will be able to present oral arguments for their Faith, in support of apologetics. They will soon be competing with thousands of others for jobs and advancement, along with meeting new friends. To develop the ability of expressing themselves well, they will study the textbook, Basic Speech Experiences Text, along with a very special book on political communication, Slightly to the Right, which will give students unique knowledge on how to discuss controversial subjects with "liberals and fuzzy minded friends." As in all of the OLVS courses, students will learn to spell hundreds of new words and their meanings, including Catholic terms, all of which will be found in the Traditional Catholic Speller 11 booklet.
Literature: In the first quarter, the students will read, Great Expectations, by Charles Dickens. A terrifying encounter with an escaped convict on the wild Kent marshes; a summons to meet the bitter, decaying Miss Havisham and her beautiful, cold-hearted ward, Estella; the sudden generosity of a mysterious benefactor – these form a series of events that changes the orphaned Pip's life forever, and he eagerly abandons his humble origins to begin a new life as a gentleman. Dickens's haunting late novel depicts Pip's education and development through adversity as he discovers the true nature of his 'great expectations.' Next, the student will read, Silas Marner by George Eliot (pen name for Mary Ann (Marian) Evans. George Eliot combined a formidable intelligence with imaginative sympathy and acute poser of observation, and became one of the greatest and most influential of English novelists. In Silas Marner, "she combines humour, rich symbolism and pointed social criticism to create an unsentimental but affectionate portrait of rural life." During the last semester, the students will read The Portable Dante. As a philosopher, Dante wedded classical methods of inquiry to a Christian faith. As an autobiographer, he looked unsparingly at his own failures to depict universal moral struggles. As a visionary, he dared draw maps of Hell, with Purgatory and Paradise, and populate all three realms with recognizable human beings. In all of these things, Dante Alighieri paved the way for modern literature, while creating verse and prose that remain unparalleled for formal elegance, intellectual depth, and emotional grandeur. The Portable Dante captures the scope and fire of Dante's genius as thoroughly as any single volume can. It contains complete verse translations of Dante's masterwork, The Divine Comedy, plus a bibliography, notes, and an introduction by the eminent scholar and translator Mark Musa.
History: West vs. Communism: This year's course is entitled "The Free World vs. Communism". Merging on both a history and a civics course, this year's social studies is a survey of the antagonism between East and West during most of the 20th Century. Notwithstanding the fall of the Berlin Wall in 1989, and the 1991 implosion of the former Soviet Union, we must still be aware that hundreds of millions of people are still enslaved by Communism in Asia, Africa, Europe and Latin America. This course will impart to students important information about not only how their government was formed, how it operates, and how they can actively participate in it, but, of equal significance, students will learn what evil forces have threatened (and to a certain extent, continue to threaten) the existence of their God-given freedoms. That knowledge will fortify them for the intelligent battle they must wage no matter where they live. The Catholic Church, and many patriots, have tried valiantly to warn us of the devilish threat from socialism and communism which has permeated the whole world. We will learn of that threat by studying The Naked Communist, by W. Cleon Skousen, a former F.B.I. agent, and the anti-Communist classic, Masters of Deceit, by J. Edgar Hoover, the former Director of the F.B.I. Other books used: Pope Pius XI's Encyclical on Atheistic Communism, With God in Russia, 1917 Red Banners White Mantle.
Chemistry: This year's science course will be centered on Dr. Jay Wile's Chemistry Text. As a Christian scientist whose area of expertise is nuclear chemistry, Dr. Wile introduces the students to general concepts of measurements and units, as well as reviews of energy, heat, and temperature. Then, in his informal and affable writing style, by which Dr. Wile gives the reader the impression that he is sitting across the table, tutoring the students one-on-one, he delves into atoms, molecules, matter classification and changes, stoichiometry, atomic and molecular structure, and numerous other fascinating topics. A rudimentary laboratory equipment set is available from the publisher at a modest price, or equipment can be bought separately for even more economy. Labs are thoughtfully designed, and often use nothing more than household materials or items that can readily be found at a pharmacy. Chemistry Solutions and Tests Book contains: (1) answers to the Study Guide found at the end of each module in the text; (2) tests for each module; and (3) answers to each of the modular tests. | 677.169 | 1 |
Introduction to Finite Elements And the Big Picture
Introduction on Introduction to Finite Elements And the Big Picture :
Application of physical principles, such as mass balance, energy conservation, and equilibrium, naturally leads many engineering analysis situations into differential equations. The exact solutions for differential equations are not applicable to many practical problems because either their governing differential equations does not have an exact solution or they involve complex geometries.
The finite element method is one of the numerical methods for obtaining approximate solution of ordinary and partial differential equations. It is especially powerful when dealing with boundary conditions defined over complex geometries that are common in practical applications.
Concepts and Formulas of Introduction to Finite Elements And the Big Picture:
Complete solution procedure:
The approximate solution of any problem governed by a differential equation can be obtained by using the element equations by following the usual finite element steps:
Development of element equations
Discretization of solution domain into a finite element mesh
Assembly of element equations
Introduction of boundary conditions
Solution for nodal unknowns
Computation of solution and related quantities over each element
Advantages of using Finite Element Method:
The subdivision of a whole domain into simpler parts has several advantages:
Accurate representation of complex geometry; while most analytical solutions are restricted to simplified geometries.
Inclusion of dissimilar material properties; while most analytical methods are applicable to homogeneous materials. | 677.169 | 1 |
College Algebra
Fifth Edition
James Stewart
Lothar Redlin
Saleem Watson
1
Equations and
Inequalities
Introduction
In Section 1.3, we saw that, if the
discriminant of a quadratic equation is
negative, the equation has no real solution.
For example, the eq
College Algebra Advice
Showing 1 to 3 of 8
This course covers selected mathematical topics in an effort to acquaint students with examples of mathematical reasoning. Upon successful completion of the course, students should be able to: model applied problems symbolically and perform manipulations on the symbols within an appropriate mathematical or logical formal system; distinguish between a rigorous proof and a conjecture; author an elementary proof; apply formal rules or algorithms to solve numeric, symbolic, graphical and/or applied problems
Hours per week:
6-8 hours
Advice for students:
the book, and the mind set to succeed because it takes more than just a smile and a wave this is not high school. Paying attention in class = less time spent on homework = more fun. If you focus in class, you might end up making new friends much more easily. But don't always lock yourself in your room to study, take breaks and relax a little.
Course Term:Fall 2017
Professor:RAFFERTY, KATIE
Course Required?Yes
Course Tags:Great Intro to the SubjectGo to Office HoursAlways Do the Reading
Jun 30, 2017
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
Mr. Chackman is more knowledgeable about the subject than you will originally think and he show and explain different methods in order for you to understand problems quicker and more efficiently. He does goes off tangent a little bit, but overall he is a very good teacher to have and learn from.
Course highlights:
His methods would help out the class a lot and it is less time consuming once you've got it down easily. On exams, he does throw in some hints here and there to the class so we are not stuck on a problem for a long time.
Hours per week:
3-5 hours
Advice for students:
For this class, it's best to go to class everytime you have it. Each class, he does three to four chapters alone so just missing one day can set you behind immensely.
Course Term:Fall 2017
Professor:GEORGECHACKMAN
Course Required?Yes
Course Tags:Lots of WritingAlways Do the ReadingGreat Discussions
Jun 05, 2017
| Would highly recommend.
Pretty easy, overall.
Course Overview:
I love this math class it was very simple but for me it was a necessity because I took so many math classes in high school and finished with them early i needed a recap of simple math.
Course highlights:
Knowing that I knew a lot more than i expected was a highlight for me. I was able to help others and just simply get better and faster with my math.
Hours per week:
3-5 hours
Advice for students:
study and use the tutoring on campus, even though I didn't need them I was able to help others and it is a big help. | 677.169 | 1 |
Level:
Instructors:
Course Features
Course Description
This course provides an elementary introduction to number theory with no algebraic prerequisites. Topics include primes, congruences, quadratic reciprocity, diophantine equations, irrational numbers, continued fractions and elliptic curves. | 677.169 | 1 |
NCERT Exemplar Problems class 10 Maths
NCERT Exemplar Problems class 10 Maths in PDF form to free download. Each question is solved and verified properly by the subject experts subject teachers. If you still find any mistake, please inform us we will try to rectify the problems. NCERT Solutions and exemplar problems both collectively provides a good practice in Maths and also in other subjects.
NCERT Exemplar ProblemsClass 10 Maths
Download Exemplar Problems here
Most of the chapter's solutions of class x maths NCERT Exemplar are already uploaded. The remaining chapters will be uploaded in the current academic session 2017 – 2018. We also update the solutions already uploaded, if required. We will upload the chapter in sequence like 1, 2, 3 and so on, if some one desire to upload any particular chapter, the priority will be given to that chapter.
Why NCERT Exemplar Problems are important?
NCERT Exemplar problems contains easy, average and difficult all the three categories of questions. These question are very useful in order to prepare for board examination conducted by CBSE. In every question papers of CBSE board, there are so many questions picked from exemplar books. There are so many questions which are totally application based.
NCERT books solutions are essential for CBSE exams but exemplar books provide a supplementary material for the preparation of examination. There are MCQ questions which are easy as well as tricky. Short answers questions (2 marks or 3 marks) and long answers questions (4 marks questions). These books also follow the latest cbse/ncert syllabus.
For class 10, there will be a board examination including full syllabus from the session 2017-18 onward. So, it become more important for class tenth for there exams. | 677.169 | 1 |
Mathematics
MATH 94 INTRODUCTION TO ALGEBRA (5)
Four hours lecture and one hour activity per week
A review of fundamental concepts of arithmetic, geometry and elementary algebra. Students who earn Credit in this course and in MATH 095 satisfy the Entry Level Mathematics (ELM) requirement. This course is offered Credit/No Credit only. Credit will not apply toward the baccalaureate degree but will apply as 5 units of University Credit.
A review of concepts of geometry and intermediate algebra with applications. Students who earn Credit in this course satisfy the ELM requirement. This course is offered Credit/No Credit only. Credit will not apply toward the baccalaureate degree but will apply as 5 units of University Credit.
Prerequisite: A passing score on the Entry Level Mathematics Examination
Topics include: number systems and their algebraic properties; systems of equations and inequalities; basic analytic geometry of lines and conic sections; elementary functions including polynomial, rational, exponential, and logarithmic, with emphasis on trigonometric functions, fundamental theorem of algebra and theory of equations; polar equations and curves.
Prerequisite: A passing score on the Entry Level Mathematics examination or MATH 95
Presents the diversity of mathematics and the spirit, in which it is employed in various situations, including different problem solving strategies, inductive- deductive reasoning, paradoxes, puzzles and mathematical modeling. The contributions of various cultures and influences of other disciplines to mathematical thinking are studied.
Prerequisites: A passing score on the Entry Level Mathematics Examination
This course introduces mathematics to the analysis of games. The principles of game theory including graphs, logic, algebra, geometry and probability are connected to game design, computer graphics and game strategies in various contexts. Applicable algorithms and techniques are demonstrated through appropriate computer gaming examples.
Prerequisite: A passing score on the Calculus Placement Examination or MATH 101 or MATH 105
An integrated course in analytic geometry and calculus in the context of business and economics applications. Functions, limits, derivatives, integrals and mathematical modeling are used in problem solving in decision making context.
Prerequisite: A passing score on the Entry Level Mathematics Exam (ELM) or MATH 105 or equivalent
Critical reasoning using a quantitative and statistical problem-solving approach to solve real-world problems. Uses probability and statistics to describe and analyze biological data collected from laboratory or field experiments. Course will cover descriptions of sample data, probability and empirical data distributions, sampling techniques, estimation and hypothesis testing, ANOVA, and correlation and regression analysis. Students will use standard statistical software to analyze real world and simulated data.
Prerequisite: A passing score on the Entry Level Mathematics Examination or Math 95
Current issues of modern math curriculum including abstract thinking and problem solving approaches to teaching. Content covers systems of numeration, nature of numbers and fundamental operations, relations and functions, properties of integers, rational and real numbers, and mathematical modeling. Problem solving strategies and geometric interpretations are stressed. Designed for students intending to teach in K-8. This course is not open to students who have credit for Calculus.
Introduction to modern deductive logic. Critical thinking and abstract approaches to common language. Includes abstract sets and number sets, relations, prepositional logic, common language cases, and theory of quantification.
Current issues of modern math curriculum including abstract thinking and problem solving approaches to teaching. Content covers systems of geometry and geometric interpretation of real numbers, geometric constructions, mathematical modeling, basic probability and statistics. Problem solving strategies are stressed. Designed for students intending to become elementary school teachers.
Current issues of modern secondary school math curriculum including abstract thinking, technology use and problem solving approaches to teaching. Content is geometry based, but selected topics from algebra, precalculus, and calculus will be discussed. designed for students intending to teach. service learning project required.
Introduction to modern statistical methods used in business and economic analysis. Topics include: sampling, probability, various distributions, correlation and regression, statistical inferences, hypothesis testing, problem solving and the consequences to underlying economical systems.
Prerequisite: A passing score on the Entry Level Mathematics examination, or MATH 95
The course is specially designed for students interested in fine arts, with the emphasis on understanding geometric patterns and concepts by self-explorations. The course creates a vast reservoir of art-related examples and hands-on experiences, and will give an innovative mathematical background for future artistic endeavors of students.
Study of breakthrough mathematical ideas and their creators, including historical and scientific context. Important concepts of current mathematics are studied: inception, development, difficulties, significance and various viewpoints will be presented. Lecture-discussion. At least one significant writing assignment is required.
The course addresses the issue of analyzing the pattern content within an image. Pattern recognition consists of image segmentation, feature extraction and classification. The principles and concepts underpinning pattern recognition, and the evolution, utility and limitations of various techniques (including neural networks) will be studied. Programming exercises will be used to implement examples and applications of pattern recognition processes, and their performance on a variety of diverse synthetic and real images will be studied
Prerequisite: Admission to the Computer Science or Mathematics Graduate Program
Mathematical research methods in education. Current issues of college level curriculum including systems of geometry, algebra, precalculus, calculus, probability and statistics, linear algebra, differential equations, and discrete mathematics. | 677.169 | 1 |
Computational Geometry
Computational
geometry is the field of theoretical computer science devoted
to the design, analysis, and implementation of algorithms and data
structures to solve geometric problems. These problems arise in
several different areas, including computer graphics, robotics,
databases, data mining, parallel computing, statistics, and pure math.
Their solutions combine traditional algorithmic techniques with
beautiful results from combinatorics, geometry, and other mathematical
areas. Computational geometry was the breeding ground for several
important techniques that later spread to the broader algorithms
community - for example, dynamic data structures, randomized
algorithms, and external memory computing - and continues to be one of
the most active, interesting, and applicable areas of algorithms
research today.
The exact course topics will depend on the interests of the
participants, but here is a sample of possibilities. Please let me
know if there's something specific you want to talk about! | 677.169 | 1 |
Help in algebra 1
Help in algebra 1
Hereare a few of the ways you can learn here.Lessons Explore one of our dozens of lessons on keyalgebra topics like Equations,Simplifying and Factoring. Checkout the entire list of lessons. Over the years, algebrahelp.com has helpedstudents solve over 15 million equations. See allthe problems we can help with. Worksheets Need to practice a new type of problem. We havetons of problems in the Worksheets section. We can grade youranswers automatically, or you can compare against the answer key.
Browse the list of worksheets to getstarted.LessonsBasAlgebra 1 is the second math course in high school and will guide you through among other things expressions, systems of equations, functions, real numbers, inequalities, exponents, polynomials, radical help in algebra 1 rational expressions.This Algebra 1 math course help in algebra 1 divided into 12 chapters algebfa each chapter is divided into several lessons. Then you found the right place to get help.
We have more than forty free, text-based algebra lessons listed on the left. If not, try the site search at the top of every page. She was falling behind. Since subscribing to your program she has been 100% self-sufficient. After my son bombed the first quarter in Algebra 1, I decided to sign up. I wish I knew about you the last 2 years when my son was in Middle School. Your site is so user friendly and interactive. Everything is right there. Createdby our FREE tutors.Solvers with work shown, write algebra lessons,help ln solve your homework help in algebra 1. Interactive solvers foralgebra word problems.
Ask questions aalgebra our question board.Created hslp the people. Each sectionhas solvers (calculators), lessons, and a place where you cansubmit your problem to our free mathtutors. Most sections have archives withhundreds of problems solved by the tutors | 677.169 | 1 |
Erica makes mistakes; lots of mistakes. You have her homework on every topic covered in the first year of her A level mathematics course where she consistently makes mistakes. Your job, or more accurately, the students in your classes' job is to correct Erica's errors and explain where she's gone wrong so that she doesn't make the same mistakes again. These have gone down well in my classes and really encourage discussion about the mathematics and should embed a deeper understanding.
Erica is struggling with the binomial expansion; can you (and your class) help Erica correct the answers she has arrived at and explain to her what she has done incorrectly. This is designed to create discussion and help students become familiar with new style questions.
There are two codebreakers, each furnished with a terrible joke. The first deals with positive powers only, the second deals with fractional and negative powers. Designed for a quick homework or plenary.
Structured worksheet to help pupils take key notes and examples for the topic of Algebra and Functions (Polynomial Long Division, Remainder and Factor Theorem and Binomial Expansion). Designed for the new A Level Maths (2017+), Edexcel Specification.
Got fed up of my classroom looking the same and bored of the same rubbish maths posters, so decided to create my own. (First sheet for preview only)
I have created another four posters based on my excellent resources to help reinforce what the pupils are being taught. This little bundle includes five posters that have been DESIGNED TO BE A3 in size, but will come out A4 or A2 if that is your preference.
1. Linear Graphs 2. Sequences 3. Solving Equations 4. Binomial Expansion
NOTE: Feel free to browse my shop for more excellent free and premium resources and as always please rate and feedback, thanks.
Preview
Files included (4)Including all the Power point presentations for the Topic 1 - Algebra based on the IB Mathematics SL Syllabus. The sub topics are:
Patterns and Sequences
Arithmetic sequences
Geometric sequences
Sigma notation
Arithmetic series
Geometric series
Convergent series and sums to infinity
Applications of geometric and arithmetic patterns
Exponents
Solving exponential equations
Properties of logarithms
Laws of logarithms
Change of base
Exponential and logarithmic equations
Pascal triangle
Binomial expansion
This is an extremely useful resource to help prepare for teaching the new maths A Level. It can be used to help structure the teaching of all pure topics.
With a site license, teachers can distribute the document freely to students within their school either electronically or in print.
This document makes a great reference tool for students to use throughout the course.
Approximately 45 pages. | 677.169 | 1 |
4
4 What is Calculus? Calculus is a branch of mathematics that deals with limits and the differentiation and integration of functions of one or more variables
5
5 Real Definition A calculus is just a bunch of rules for manipulating symbols. People can give meaning to those symbols, but that's not part of the calculus. Differential calculus is a bunch of rules for manipulating symbols. There is an interpretation of those symbols corresponds with physics, slopes, etc. | 677.169 | 1 |
unique text provides a geometric approach to group theory and linear algebra, bringing to light the interesting ways in which these subjects interact. Requiring few prerequisites beyond understanding the notion of a proof,
This is the second in a series of three volumes dealing with important topics in algebra. Volume 2 is an introduction to linear algebra (including linear algebra over rings), Galois theory, representation theory, and the theory of group extensions. The section on linear algebra (chapters 1–5) does not require any background material from Algebra 1, except an understanding of set theory | 677.169 | 1 |
This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles. This course is designed to help prepare students to enroll for a first semester course in single variable calculus.
Calculus is about the very large, the very small, and how things change. The surprise is that something seemingly so abstract ends up explaining the real world. Calculus plays a starring role in the biological, physical, and social sciences. By focusing outside of the classroom, we will see examples of calculus appearing in daily life.
Learn basic engineering mathematics and how to apply basic mathematics to solve engineering problems. The goal of this mathematics course is to provide high school students and college freshmen an introduction to basic mathematics and especially show how mathematics is applied to solve fundamental engineering problems.
This course provides a brisk, entertaining treatment of differential and integral calculus, with an emphasis on conceptual understanding and applications to the engineering, physical, and social sciences | 677.169 | 1 |
Searching For Algebra Homework Assistance – Useful Directions
Algebra is not supposed to be one of those subjects that you fear. As a matter of fact, there is so much more that you can do in algebra that you might never have thought possible. This is what makes the difference between students who are looking for and hungry for success, and the rest of the student population. One of the biggest challenges perhaps that most students tend to go through is the fact that when they have algebra homework, they do not know what to do.
The thing about assignments like these is that they are basically a continuation of the work that you have been doing in class. Because of this reason, it is important for you to give it some thought, and make sure that you reflect on the things that you have been doing in class, and then from there you can easily get through this task.
If you have been having challenges from time to time with your algebra homework, the following are some useful directions that might spare you the blushes when you send in your work for marking in the morning:
Get as much done in school
Carry your class notes home
Refer to some study text
Consult your classmates
Get as much done in school
The more work you can do while you are in school, the easier it will be for you to complete the task as fast as possible. The good thing about starting this task in school is that there are lots of support networks and services that you can rely on while you are still in class, so make sure you think about this and you will be much safer.
Carry your class notes home
Do not leave your notes in school. In fact, carry them with you at home and you will manage to have something that you can refer to. These notes will usually help you out a great deal because they are direct from the teacher.
Refer to some study text
Use the text books that your teachers usually recommend. These books can help you find out so much more. Other than that, they will also make your work easy since they have good examples you can use for reference.
Consult your classmates
Talk to your classmates and get them to share some of the options that they use when they need help with algebra. This will save you a lot of time too. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
This book provides thorough coverage of the main topics of abstract algebra while offering nearly 100 pages of applications. A repetition and examples first approach introduces learners to mathematical rigor and abstraction while teaching them the basic notions and results of modern algebra. Chapter topics include group theory, direct products and Abelian groups, rings and fields, geometric constructions, historical notes, symmetries, and coding theory. For future teachers of algebra and geometry at the high school level.
Synopsis
For a one-semester course covering groups and rings or a two-semester course in Abstract Algebra.This text provides thorough coverage of the main topics of abstract algebra while offering nearly 100 pages of applications. A repetition and examples first approach introduces students to mathematical rigor and abstraction while teaching them the basic notions and results of modern algebra.
Group Actions and Cayley's Theorem. Stabilizers and Orbits in a Group Action. Burnside's Theorem and Applications. Conjugacy Classes and the Class Equation. Conjugacy in Sn and Simplicity of A5. The Sylow Theorems. Applications of the Sylow Theorems. 5. Composition Series. | 677.169 | 1 |
focuses on traditional algebra topics that have found greatest application in science and engineering as well as in mathematics. This undergraduate course focuses on traditional algebra topics that have found greatest application in science and engineering as well as in mathematics individ 160 to explain the parallax angle, comprising an interactive task using flash which explains the parallax in visual terms. A number of online calculations using arc seconds and degrees are included. There is also a definition of the Parsec and how it can be calculatedA mini-lecture about Networking. This short lecture will introduce you to the concept and practice of networking and look at why it is useful to you as soon to be, or recent graduate. We will then look at ways that you can develop and build your network and networking skills in order to become more successful in your personal and professional life | 677.169 | 1 |
already exists as an alternate
of this question.
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Well, you have to buy materials, which means you have to deal with money, so you have to be able to do arithmetic.
If you're making lots of clothes, you'll want to be able to find ways to lay patterns out efficiently - that will require geometry, probably algebra, and maybe even calculus.
If you want to set up a manufacturing process, you would probably want to be familiar with control theory, and perhaps calculus.
If you're really trying to be creative, topology might be useful in helping you think up new ways to get clothes onto and off of bodies. (For example, pants normally have three 'holes' in them - one for the waist, and one for each leg. But diapers have _no_ holes in them - you wrap them around the baby and fasten them. It might be interesting to take a few minutes to think about why they are designed that way.)
If you want to design cloth patterns, it would be useful to study tessellation's, and perhaps abstract algebra.
If you are thinking of becoming a designer of clothes then you need math to measure garments, see proportions on the model and if you are running the business you will need ma…th to do company bookkeeping and your taxes.You need to know math to take measurements and take stock. You also need it to do taxes and know how much of what you need to order.also you need it to add tax and minus the discountsplus if you are a customer and u need to pay for something u have to beable to count how much u owe before u get up to the counter so u wont hold up the line
Math is used in designing because when you cut the fabric and make measurements. It is also used when you cut the fabrics and certain parts of it have to be certain siz…es
Ratios when it comes to gears, a lot of small units of measuring for cylinder boring, percentage problems, weight ratios, displacement or cubic measuring of cylinders, limits …when it comes to car weight. Physics would describe the type of math for a mechanical engineer which goes hand in hand with math.
Depends on the angle you are taking. If you are the "designer" - designing the clothes that your models will be wearing, well, you have to know inches and such, to be a…ble to size up your clothes properly. Now, if you are the "model" - than I'm not too sure ;) Sure, you probably don't need to know calculus, but basic arithmetic, algebra would be good to know at least.
When cutting out patterns to make clothing,the measurments of every cut need to be exact. Just one inch could ruin the entire piece. Also,when displaying clothing, the clothi…ng needs to fit a model perfectly. It needs to hug in all the right places. A model has 3 basic body measurments. Idealy 32-24-32 first number her breast measurments..second number her waist ..third her hips. You need to make sure the clothing fits exactly how you want it on her... but hey dont say you cant design because your not sure how to do the math. fashion is a way of expressing yourself.
Fashion designers us math in many ways... they use math to measure how much material they need and how long the material they are using needs to be. They also need to make… different sized clothing and measure the size of the clothing they are making so it fits the average person....for example if they need to measure a dress they need to measure the chest area and the waist area and about how big around and how long the material needs to be. | 677.169 | 1 |
Dr. Hanley's Science Graphs Enter first and second order nonlinear ordinary differential in analytical form. The program will generate a numerical solution to the equations and graph multiple curves on a single graph. Can also be used for algebra and calculus formulas. Includes illustrations and free demo.
Calculi For Mobile Processes Books. The Picalculus - A Theory of Mobile Processes (by Davide Sangiorgi and DavidWalker), June 2001. A pi-course introducing the -calculus, given at SICS.
Pitt Distance Learning Program Internet calculus. UNIVERSITY OF PITTSBURGH. College in High School.This is the home Other calculus Links at Pitt. calculus Home Page The
QuickMath Automatic Math Solutions QuickMath allows students to get instant solutions to all kinds of math problems,from algebra and equation solving right through to calculus and matrices.
Technology Based Problems calculus and differential calculus problems are presented. The possible solution or solutions are given.Category Science Math calculusWelcome to the Complex, Technology Based Problems in calculus HomePage. What we're all about We offer complex, technologybased
An Invitation To Mathematics A free tutorial that explains difficult algebra, trigonometry and calculus concepts to beginning middle/high school students in a simplified way that they can understand and use.
Algebra And Calculus Sketches For calculus, gives the tangent line problem and its solution.
Learning Calculus LEARNING calculus. Prepared by Susan Hermiller, Melanie Martin, EricYork. Welcome to the wonderful world of calculus. In order
History Of Calculus Guide To History of calculus. Topic essays and biographies keyed to the chapters and content of the 10th edition of Thomas's calculus. | 677.169 | 1 |
Spectrum Algebra includes lessons, perfect for students in grades 6—8, that strengthen math skills by focusing on factors and fractions, equalities and inequalities, functions, graphing, proportion, interest, and more. Spectrum Middle School Math helps st | 677.169 | 1 |
Algebra Explained c. 1 Order of Operations
This application does a fantastic job of explaining elementary math concepts, through video with tons of examples and application. Not only do you get a LOT of visual instruction from Jamie, but she does it in a very fun way!
The app gives a lot of opportunities to practice and tests your knowledge before moving you to the next section! I have personally never seen algebra so well communicated and simplified. I have to admit, from having reviewed this app fully, including watching all the videos and doing the practice problems, that I have whole new view of algebra and just how simple the subject is, when properly communicated. I actually took the time to write to the developers and let them know how impressed I am with their apps.
This is the first in a series of Algebra Explained apps. To get a complete understanding, one should take the time to go through all of them!
Compatible with iPhone, iPod touch, and iPad.
Features
Video instruction for each lesson
Teaches: an introduction to exponents, order of operations, evaluating algebraic expressions, Euler's number "e"
Review of fractions
10 practice problems per lesson
Calculator
Quizzes and tests
Electronic scratch and graph paper
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Chapter 5
There are some references in chapters 4 and 5 to material from chapter 3 (some linear algebra),
but these references are not absolutely essential and can be skimmed over, so chapter 3 can safely
be dropped, while still covering chapters 4 and 5. The notes are done for two types of courses.
Either at 4 hours a week for a semester (Math 286 at UIUC):
Introduction,
chapter 1 (plus the two IODE labs),
chapter 2,
chapter 3,
chapter 4,
chapter 5.
Or a shorter version (Math 285 at UIUC) of the course at 3 hours a week for a semester:
Introduction,
chapter 1 (plus the two IODE labs),
chapter 2,
chapter 4.
For the shorter version some additional material should be covered. IODE need not be used for
either version. If IODE is not used, some additional material should be covered instead.
There is a short introductory chapter on Laplace transform (chapter 6 that could be used as
additional material. The length of the Laplace chapter is about the same as the Sturm-Liouville
chapter (chapter 5). While Laplace transform is not normally covered at UIUC 285/286, I think
it is essential that any notes for Differential equations at least mention Laplace and/or Fourier
transforms.
0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 7
0.2 Introduction to differential equations
Note: more than 1 lecture, §1.1 in EP
0.2.1 Differential equations
The laws of physics are generally written down as differential equations. Therefore, all of science
and engineering use differential equations to some degree. Understanding differential equations is
essential to understanding almost anything you will study in your science and engineering classes.
You can think of mathematics as the language of science, and differential equations are one of
the most important parts of this language as far as science and engineering are concerned. As
an analogy, suppose that all your classes from now on were given in Swahili. Then it would be
important to first learn Swahili, otherwise you will have a very tough time getting a good grade in
your other classes.
You have already seen many differential equations without perhaps knowing about it. And
you have even solved simple differential equations when you were taking calculus. Let us see an
example you may not have seen.
dx
dt
+ x = 2 cos t. (1)
Here x is the dependent variable and t is the independent variable. Equation (1) is a basic example
of a differential equation. In fact it is an example of a first order differential equation, since it
involves only the first derivative of the dependent variable. This equation arises from Newton's
law of cooling where the ambient temperature oscillates with time.
0.2.2 Solutions of differential equations
Solving the differential equation means finding x in terms of t. That is, we want to find a function
of t which we will call x such that when we plug x, t, and
dx
dt
into (1) the equation holds. It is the
same idea as it would be for a normal (algebraic) equation of just x and t. In this case we claim
that
x = x(t) = cos t + sin t
is a solution. How do we check? Just plug it back in! First you need to compute
dx
dt
. We find that
dx
dt
= −sin t + cos t. Now let us compute the left hand side of (1)
dx
dt
+ x = (−sin t + cos t) + (cos t + sin t) = 2 cos t.
Yay! We got precisely the right hand side. There is more! We claim x = cos t + sin t + e
−t
is also a
solution. Let us try,
dx
dt
= −sin t + cos t − e
−t
.
8 INTRODUCTION
Again plugging into the left hand side of (1)
dx
dt
+ x = (−sin t + cos t − e
−t
) + (cos t + sin t + e
−t
) = 2 cos t.
And it works yet again!
So there can be many different solutions. In fact, for this equation all solutions can be written
in the form
x = cos t + sin t + Ce
−t
for some constant C. See Figure 1 for the graph of a few of these solutions. We will see how we
can find these solutions a few lectures from now.
It turns out that solving differential equations
0 1 2 3 4 5
0 1 2 3 4 5
-1
0
1
2
3
-1
0
1
2
3
Figure 1: Few solutions of
dx
dt
+
y
2
= cos t.
can be quite hard. There is no general method
that solves any given differential equation. We
will generally focus on how to get exact formu-
las for solutions of differential equations, but we
will also spend a little bit of time on getting ap-
proximate solutions.
For most of the course we will look at ordi-
nary differential equations or ODEs, by which
we mean that there is only one independent vari-
able and derivatives are only with respect to this
one variable. If there are several independent
variables, we will get partial differential equa-
tions or PDEs. We will briefly see these near the
end of the course.
Even for ODEs, which are very well under-
stood, it is not a simple question of turning a crank to get answers. It is important to know when
it is easy to find solutions and how to do this. Even if you leave much of the actual calculations
to computers in real life, you need to understand what they are doing. For example, it is often
necessary to simplify or transform your equations into something that a computer can actually
understand and solve. You may need to make certain assumptions and changes in your model to
achieve this.
To be a successful engineer or scientist, you will be required to solve problems in your job
which you have never seen before. It is important to learn problem solving techniques, so that
you may apply those techniques to new problems. A common mistake is to expect to learn some
prescription for solving all the problems you will encounter in your later career. This course is no
exception to this.
0.2.3 Differential equations in practice
So how do we use differential equations in science and engineering? You have some real
0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 9
world problem that you want to understand. You make some simplifying assumptions and create
a mathematical model. That is, you translate your real world situation into a set of differential
equations. Then you apply mathematics to get some sort of mathematical solution. There is still
something left to do. You have to interpret the results. You have to figure out what the mathematical
solution says about the real world problem you started with.
Learning how to formulate the mathematical
solve
Mathematical
Real world problem
interpret
Mathematical
solution model
abstract
model and how to interpret the results is essen-
tially what your physics and engineering classes
do. In this course we will mostly focus on the
mathematical analysis. Sometimes we will work
with simple real world examples so that we have
some intuition and motivation about what we are
doing.
Let us look at an example of this process. One of the most basic differential equations is the
standard exponential growth model. Let P denote the population of some bacteria on a petri dish.
Let us suppose that there is enough food and enough space. Then the rate of growth of bacteria
will be proportional to population. I.e. a large population growth quicker. Let t denote time (say in
seconds). Hence our model will be
dP
dt
= kP
for some positive constant k > 0.
Example 0.2.1: Suppose there are 100 bacteria at time 0 and 200 bacteria at time 10s. How many
bacteria will there be in 1 minute from time 0 (in 60 seconds)?
First we have to solve the equation. We claim that a solution is given by
P(t) = Ce
kt
,
where C is a constant. Let us try,
dP
dt
= Cke
kt
= kP.
And it really is a solution.
OK, so what now? We do not know C and we do not know k. Well we know something. We
know that P(0) = 100 and we also know that P(10) = 200. Let us plug these in and see what
happens.
100 = P(0) = Ce
k0
= C,
200 = P(10) = 100 e
k10
.
Therefore, 2 = e
10k
or
ln 2
10
= k ≈ 0.069. So we know that
P(t) = 100 e
(ln 2)t/10
≈ 100 e
0.069t
.
10 INTRODUCTION
At one minute, t = 60, the population is P(60) = 6400. See Figure 2.
OK, let us talk about the interpretation of the results. Does this mean that there must be exactly
6400 bacteria on the plate at 60s? No! We have made assumptions that might not be true. But if
our assumptions are reasonable, then there will be about 6400 bacteria. Also note that P in real life
is a discrete quantity, not any real number, but our model has no problem saying that for example
at 61 seconds, P(61) ≈ 6859.35.
Normally, the k in P
·
= kP will be known,
0 10 20 30 40 50 60
0 10 20 30 40 50 60
0
1000
2000
3000
4000
5000
6000
0
1000
2000
3000
4000
5000
6000
Figure 2: Bacteria growth in the first 60 sec-
onds.
and you will want to solve the equation for dif-
ferent initial conditions. What does that mean?
Suppose k = 1 for simplicity. So we want to
solve
dP
dt
= P subject to P(0) = 1000 (the ini-
tial condition). Then the solution turns out to be
(exercise)
P(t) = 1000 e
t
.
We will call P(t) = Ce
t
the general solution,
as every solution of the equation can be writ-
ten in this form for some constant C. Then you
will need an initial condition to find out what C
is to find the particular solution we are looking
for. Generally when we say particular solution
we just mean some solution.
Let us get to what we will call the 4 fundamental equations. These appear very often and it
is useful to just memorize what their solutions are. These solutions are reasonably easy to guess
by recalling properties of exponentials, sines, and cosines. They are also simple to check, which
is something that you should always do. There is no need to wonder if you have remembered the
solution correctly.
First such equation is,
dy
dx
= ky,
for some constant k > 0. Here y is the dependent and x the independent variable. The general
solution for this equation is
y(x) = Ce
kx
.
We have already seen that this is a solution above with different variable names.
Next,
dy
dx
= −ky,
for some constant k > 0. The general solution for this equation is
y(x) = Ce
−kx
.
0.2. INTRODUCTION TO DIFFERENTIAL EQUATIONS 11
Exercise 0.2.1: Check that the y given is really a solution to the equation.
Next, take the second order differential equation
d
2
y
dx
2
= −k
2
y,
for some constant k > 0. The general solution for this equation is
y(x) = C
1
cos(kx) + C
2
sin(kx).
Note that because we have a second order differential equation we have two constants in our general
solution.
Exercise 0.2.2: Check that the y given is really a solution to the equation.
And finally, take the second order differential equation
d
2
y
dx
2
= k
2
y,
for some constant k > 0. The general solution for this equation is
y(x) = C
1
e
kx
+ C
2
e
−kx
,
or
y(x) = D
1
cosh(kx) + D
2
sinh(kx).
For those that do not know, cosh and sinh are defined by
cosh x =
e
x
+ e
−x
2
,
sinh x =
e
x
− e
−x
2
.
These functions are sometimes easier to work with than exponentials. They have some nice familiar
properties such as cosh 0 = 1, sinh 0 = 0, and
d
dx
cosh x = sinh x (no that is not a typo) and
d
dx
sinh x = cosh x.
Exercise 0.2.3: Check that both forms of the y given are really solutions to the equation.
An interesting note about cosh: The graph of cosh is the exact shape a hanging chain will make
and it is called a catenary. Contrary to popular belief this is not a parabola. If you invert the graph
of cosh it is also the ideal arch for supporting its own weight. For example, the gateway arch in
Saint Louis is an inverted graph of cosh (if it were just a parabola it might fall down). This formula
is actually inscribed inside the arch:
y = −127.7 ft cosh(x/127.7 ft) + 757.7 ft.
12 INTRODUCTION
0.2.4 Exercises
Exercise 0.2.4: Show that x = e
4t
is a solution to x
···
− 12x
··
+ 48x
·
− 64x = 0.
Exercise 0.2.5: Show that x = e
t
is not a solution to x
···
− 12x
··
+ 48x
·
− 64x = 0.
Exercise 0.2.6: Is y = sin t a solution to
dy
dt
2
= 1 − y
2
? Justify.
Exercise 0.2.7: Let y
··
+ 2y
·
− 8y = 0. Now try a solution y = e
rx
. Is this solution for some r? If
so, find all such r.
Exercise 0.2.8: Verify that x = Ce
−2t
is a solution to x
·
= −2x. Find C to solve the initial condition
x(0) = 100.
Exercise 0.2.9: Verify that x = C
1
e
−t
+ C
2
e
2t
is a solution to x
··
− x
·
− 2x = 0. Find C
1
and C
2
to
solve the initial condition x(0) = 10.
Exercise 0.2.10: Using properties of derivatives of functions that you know try to find a solution
to (x
·
)
2
+ x
2
= 4.
Chapter 1
First order ODEs
1.1 Integrals as solutions
Note: 1 lecture, §1.2 in EP
A first order ODE is an equation of the form
dy
dx
= f (x, y),
or just
y
·
= f (x, y).
In general, there is no simple formula or procedure one can follow to find solutions. In the next
few lectures we will look at special cases where solutions are not difficult to obtain. In this section,
let us assume that f is a function of x alone, that is, the equation is
y
·
= f (x). (1.1)
We could just integrate (antidifferentiate) both sides here with respect to x.
y
·
(x) dx =
f (x) dx + C,
that is
y(x) =
f (x) dx + C.
This y(x) is actually the general solution. So to solve (1.1) find some antiderivative of f (x) and
then you add an arbitrary constant to get the general solution.
Now is a good time to discuss a point about calculus notation and terminology. Calculus text-
books muddy the waters by talking about integral as primarily the so-called indefinite integral. The
13
14 CHAPTER 1. FIRST ORDER ODES
indefinite integral is really the antiderivative (in fact the whole one parameter family of antideriva-
tives). There really exists only one integral and that is the definite integral. The only reason for
the indefinite integral notation is that you can always write an antiderivative as a (definite) integral.
That is, by fundamental theorem of calculus you can always write
f (x) dx + C as
x
x
0
f (t) dt + C.
Hence, the terminology integrate when you may really mean antidifferentiate. Integration is just
one way to compute the antiderivative (and it is a way that always works, see the following exam-
ple). Integration is defined as the area under the graph, it only happens to also compute antideriva-
tives. For sake of consistency, we will keep using the indefinite integral notation when we want an
antiderivative, and you should always think of the definite integral.
Example 1.1.1: Find the general solution of y
·
= 3x
2
.
We see that the general solution must be y = x
3
+ C. Let us check: y
·
= 3x
2
. We have gotten
precisely our equation back.
Normally, we also have an initial condition such as y(x
0
) = y
0
for some two numbers x
0
and y
0
(x
0
is usually 0, but not always). We can write the solution as a definite integral in a nice way.
Suppose our problem is y
·
= f (x), y(x
0
) = y
0
. Then the solution is
y(x) =
x
0
x
0
f (x) dx + y
0
= y
0
. And it is!
Do note that the definite integral and indefinite integral (antidifferentiation) are completely
different beasts. The definite integral always evaluates to a number. Therefore, (1.2) is a formula
you can plug into the calculator or a computer and it will be happy to calculate specific values for
you. You will easily be able to plot the solution and work with it just like with any other function.
It is not so crucial to find a closed form for the antiderivative.
Example 1.1.2: Solve
y
·
= e
−x
2
, y(0) = 1.
By the preceeding discussion, the solution must be
y(x) =
x
0
e
−s
2
ds + 1.
Here is a good way to make fun of your friends taking second semester calculus. Tell them to find
the closed form solution. Ha ha ha (bad math joke). It is not possible (in closed form). There is
absolutely nothing wrong with writing the solution as a definite integral. This particular integral is
in fact very important in statistics.
1.1. INTEGRALS AS SOLUTIONS 15
We can also solve equations of the form
y
·
= f (y)
using this method. Let us write it in Leibniz notation
dy
dx
= f (y)
Now use the inverse function theorem to switch roles of x and y.
dx
dy
=
1
f (y)
What we are doing seems like algebra with dx and dy. It is tempting to just do algebra with dx
and dy as if they were numbers. And in this case it does work. Be careful, however, as this sort of
hand-waving calculation can lead to trouble, especially when more than one independent variable
is involved. Now we can just integrate
x(y) =
1
f (y)
dy + C
Next, we try to solve for y.
Example 1.1.3: We guessed y
·
= ky has solution Ce
kx
. We can actually do it now. First note that
y = 0 is a solution. Henceforth, assume y 0. We write
dx
dy
=
1
ky
.
Now integrate and get
x(y) = x =
1
k
ln|ky| + C
·
.
we solve for y
ke
kC
·
e
kx
= |y|.
If we replace ke
kC
with an arbitrary constant C we can get rid of the absolute value bars. In this
way we also incorporate the solution y = 0, and we get the same general solution as we guessed
before, y = Ce
kx
.
Example 1.1.4: Find the general solution of y
·
= y
2
.
First note that y = 0 is a solution. We can now assume that y 0. Write
dx
dy
=
1
y
2
16 CHAPTER 1. FIRST ORDER ODES
Now integrate to get
x =
−1
y
+ C.
Solve for y =
1
C−x
. So the general solution is
y =
1
C − x
or y = 0.
Note the singularities of the solution. If for example C = 1, then the solution blows up as we
approach x = 1. It is hard to tell from just looking at the equation itself how the solution is going
to behave sometimes. The equation y
·
= y
2
is very nice and defined everywhere, but the solution
is only defined on some interval (−∞, C) or (C, ∞).
Classical problems leading to differential equations solvable by integration are problems deal-
ing with velocity, acceleration and distance. You have surely seen these problems before in your
calculus class.
Example 1.1.5: Suppose a car drives at a speed e
t/2
meters per second, where t is time in seconds.
How far did the car get in 2 seconds? How far in 10 seconds.
Let x denote the distance the car travelled. The equation is
x
·
= e
t/2
.
We can just integrate this equation to get that
x(t) = 2e
t/2
+ C.
Note that we still need to figure out C. But we know that when t = 0 then x = 0, that is: x(0) = 0
so
0 = x(0) = 2e
0/2
+ C = 2 + C.
So C = −2 and hence
x(t) = e
t/2
− 2.
Now we just plug in to get that at 2 seconds (and 10), the car has travelled
x(2) = 2e
2/2
− 2 ≈ 3.44 meters, x(10) = 2e
10/2
− 2 ≈ 294 meters.
Example 1.1.6: Suppose that the car accelerates at the rate t
2
m/s
2
. At time t = 0 the car is at the
1 meter mark and is travelling at 10 m/s. Where is the car at time t = 10.
Well this is actually a second order problem. If x is the distance travelled, then x
·
is the velocity,
and x
··
is the acceleration. The equation with initial conditions is
x
··
= t
2
, x(0) = 1, x
·
(0) = 10.
Well, what if we call x
·
= v and then we have the problem
v
·
= t
2
, v(0) = 10.
Once we solve for v, we can then integrate and find x.
Exercise 1.1.1: Solve for v and then solve for x.
1.1. INTEGRALS AS SOLUTIONS 17
1.1.1 Exercises
Exercise 1.1.2: Solve
dy
dx
= x
2
+ x for y(1) = 3.
Exercise 1.1.3: Solve
dy
dx
= sin 5x for y(0) = 2.
Exercise 1.1.4: Solve
dy
dx
=
1
x
2
−1
for y(0) = 0.
Exercise 1.1.5: Solve y
·
= y
3
for y(0) = 1.
Exercise 1.1.6: Solve y
·
= (y − 1)(y + 1) for y(0) = 3.
Exercise 1.1.7: Solve
dy
dx
=
1
y
2
+1
for y(0) = 0.
Exercise 1.1.8: Solve y
··
= sin x for y(0) = 0.
18 CHAPTER 1. FIRST ORDER ODES
1.2 Slope fields
Note: 1 lecture, §1.3 in EP
At this point it may be good to first try the Lab I and/or Project I from the IODE website:
As we said, the general first order equation we are studying looks like
y
·
= f (x, y).
In general we cannot really just solve these kinds of equations explicitly. It would be good if we
could at least figure out the shape and behavior of the solutions or even find approximate solutions
for any equation.
1.2.1 Slope fields
As you have seen in IODE Lab I (if you did it), this means that at each point in the (x, y)-plane
we get a slope. We can plot the slope at lots of points as a short line with this given slope. See
Figure 1.11: Slope field of y
·
= xy2: Slope field of y
·
= xy with a graph
of solutions satisfying y(0) = 0.2, y(0) = 0, and
y(0) = −0.2.
We call this the slope field of the equation. Then if we are given a specific initial condition
y(x
0
) = y
0
, we can really just look at the location (x
0
, y
0
) and follow the slopes. See Figure 1.2.
By looking at the slope field we can find out a lot about the behavior of solutions. For example,
in Figure 1.2 we can see what the solutions do when the initial conditions are y(0) > 0, y(0) = 0
1.2. SLOPE FIELDS 19
and y(0) < 0. Note that a small change in the initial condition causes quite different behavior. On
the other hand, plotting a few solutions of the of the equation y
·
= −y, we see that no matter where
we start, all solutions tend to zero as x tends to infinity. See Figure 1.33: Slope field of y
·
= −y with a graph of a few solutions.
1.2.2 Existence and uniqueness
We wish to ask two fundamental questions about the problem
y
·
= f (x, y), y(x
0
) = y
0
.
(i) Does a solution exist?
(ii) Is the solution unique (if it exists)?
What do you think is the answer? The answer seems to be yes to both does it not? Well, pretty
much. But there are cases when the answer to either question can be no.
Since generally the equations come from real life situation, then it seems logical that a solution
exists. It also has to be unique if we believe our universe is deterministic. If the solution does not
exist, or if it does is not unique, we have probably not devised the correct model. Hence, it is good
to know when things go wrong and why.
Example 1.2.1: Attempt to solve:
y
·
=
1
x
, y(0) = 0.
Integrate to find the general solution y = ln |x| +C Note that the solution does not exist at x = 0.
See Figure 1.4 on the next page.
20 CHAPTER 1. FIRST ORDER4: Slope field of y
·
=
1
x
5: Slope field of y
·
= 2
|y|, y(0) = 0.
Note that y = x
2
is a solution and y = 0 is a solution (but note x
2
is a solution only for x > 0).
See Figure 1.5.
It is actually hard to tell from the slope field that the solution will not be unique. Is there any
hope? Of course there is. It turns out that the following theorem is true. It is known as Picard's
theorem
∗
.
Theorem 1.2.1 (Picard's theorem on existence and uniqueness). If f (x, y) is continuous (as a
function of two variables) and
∂f
∂y
exists and is continuous near some (x
0
, y
0
), then a solution to
y
·
= f (x, y), y(x
0
) = y
0
,
exists (at least for some small interval of x's) and is unique.
Note that y
·
=
1
x
, y(0) = 0 and y
·
= 2
|y|, y(0) = 0 do not satisfy the theorem. But we ought to
be careful about this existence business. It is quite possible that the solution only exists for a short
while.
Example 1.2.3:
y
·
= y
2
, y(0) = A,
for some constant A.
∗
Named after the French mathematician Charles Émile Picard (1856 – 1941)
1.2. SLOPE FIELDS 21
We know how to solve this equation. First assume that A 0, so y is not equal to zero at least
for some x near 0. So x
·
=
1
y
2
, so x =
−1
y
+ C, so y =
1
C−x
. If y(0) = A, then C =
1
A
so
y =
1
1
A
− x
.
Now if A = 0, then y = 0 is a solution.
For example, when A = 1 the solution "blows up" at x = 1. Hence, the solution does not exist
for all x even if the equation is nice everywhere. y
·
= y
2
certainly looks nice.
For the most of this course we will be interested in equations where existence and uniqueness
holds, and in fact will hold "globally" unlike for the y
·
= y
2
.
1.2.3 Exercises
Exercise 1.2.1: Sketch direction field for y
·
= e
x−y
. How do the solutions behave as x grows? Can
you guess a particular solution by looking at the direction field?
Exercise 1.2.2: Sketch direction field for y
·
= x
2
.
Exercise 1.2.3: Sketch direction field for y
·
= y
2
.
Exercise 1.2.4: Is it possible to solve the equation y
·
=
xy
cos x
for y(0) = 1? Justify.
22 CHAPTER 1. FIRST ORDER ODES
1.3 Separable equations
Note: 1 lecture, §1.4 in EP
When the equation is of the form y
·
= f (x), we can just integrate: y =
f (x) dx + C. Unfor-
tunately this method no longer works for the general form of the equation y
·
= f (x, y). Integrating
both sides yields
y =
f (x, y) dx + C.
Notice dependence on y in the integral.
1.3.1 Separable equations
On the other hand, what if the equation is separable, that is, if it looked like
y
·
= f (x)g(y),
for some functions f (x) and g(y). Let us write the equation in Leibniz notation
dy
dx
= f (x)g(y).
Then we rewrite the equation as
dy
g(y)
= f (x) dx.
Now both sides look like something we can integrate. We obtain
dy
g(y)
=
f (x) dx + C.
If we can explicitly solve this integral we can maybe solve for y.
Example 1.3.1: Take the equation
y
·
= xy
First note that y = 0 is a solution, so assume y 0 from now on. Write the equation as
dy
dx
= xy,
then
dy
y
=
x dx + C.
We compute the antiderivatives to get
ln |y| =
x
2
2
+ C.
1.3. SEPARABLE EQUATIONS 23
Or
|y| = e
x
2
2
+C
= e
x
2
2
e
C
= De
x
2
2
,
where D > 0 is some constant. Because y = 0 is a solution and because of the absolute value we
actually can write:
y = De
x
2
2
,
for any number D (including zero or negative).
We check:
y
·
= Dxe
x
2
2
= x(De
x
2
2
) = xy.
Yay!
We should be a little bit more careful about the method. Because we were integrating in two
different variables, that does not sound right. We seemed to be doing a different operation to each
side. Let us see work out this method more rigorously.
dy
dx
= f (x)g(y)
We rewrite the equation as follows. Note that y = y(x) is a function of x and so is
dy
dx
!
1
g(y)
dy
dx
= f (x)
We integrate both sides with respect to x.
1
g(y)
dy
dx
dx =
f (x) dx + C.
We can use the change of variables formula.
1
g(y)
dy =
f (x) dx + C.
And we are done.
1.3.2 Implicit solutions
It is clear that we might sometimes get stuck even if we can do the integration. For example, take
the separable equation
y
·
=
xy
y
2
+ 1
.
We separate variables
y
2
+ 1
y
dy =
¸
y +
1
y
dy = x dx.
24 CHAPTER 1. FIRST ORDER ODES
Now we integrate to get
y
2
2
+ ln |y| =
x
2
2
+ C.
Or maybe the easier looking expression:
y
2
+ 2 ln |y| = x
2
+ C.
It is not easy to find the solution explicitly as it is hard to solve for y. We will, therefore, call this
solution an implicit solution. It is easy to check that implicit solutions still satisfy the differential
equation. In this case, we differentiate to get
y
·
¸
2y +
2
y
= 2x.
It is simple to see that the differential equation holds. If you want to compute values for y you
might have to be tricky. For example, you can graph x as a function of y, and then flip your paper.
Computers are also good at some of these tricks, but you have to be careful.
We note above that the equation also has a solution y = 0. In this case, it turns out that the
general solution is y
2
+2 ln |y| = x
2
+C together with y = 0. These outlying solutions such as y = 0
are sometimes called singular solutions.
1.3.3 Examples
Example 1.3.2: Solve x
2
y
·
= 1 − x
2
+ y
2
− x
2
y
2
, y(1) = 0.
First factor the right hand side to obtain
x
2
y
·
= (1 − x
2
)(1 + y
2
).
Now we separate variables, integrate and solve for y
y
·
1 + y
2
=
1 − x
2
x
2
y
·
1 + y
2
=
1
x
2
− 1
arctan(y) =
−1
x
− x + C
y = tan
¸
−1
x
− x + C
.
1.3. SEPARABLE EQUATIONS 25
Example 1.3.3: Suppose Bob made a cup of coffee, and the water was boiling (100 degrees Cel-
sius) at time t = 0. Suppose Bob likes to drink his coffee at 70 degrees. Let the Ambient (room)
temperature be 26 degrees. Furthermore, suppose Bob measured the temperature of the coffee at 1
minute (t = 60) and found that it dropped to 95 degrees. When should Bob start drinking?
Let T be the temperature of coffee, let A be the ambient (room) temperature. Then for some k
the temperature of coffee is:
dT
dt
= k(A − T).
For our setup A = 26, T(0) = 100, T(1) = 95. We separate variables and integrate (C and D will
denote arbitrary constants)
1
A − T
dT
dt
= k,
ln A − T = −kt + C,
A − T = De
−kt
,
T = A − De
−kt
.
That is T = 26 − De
−kt
. We plug in the first condition 100 = T(0) = 26 − D and hence D = −74.
Now we have T = 26 + 74e
−kt
. We plug in 95 = T(1) = 26 + 74e
−k
. Solving for k we get
k = −ln(95 − 26)/74 ≈ 0.07. Now to solve for which t gives me 70 degrees. That is we solve
70 = 26 + 74e
−0.07t
to get t = −
ln(70−26)/74
0.07
≈ 7.43 minutes. So Bob can begin to drink the coffee at
about 7 and a half minutes from the time Bob made it. Probably about the amount of time it took
us to calculate how long it would take.
Example 1.3.4: Solve y
·
=
−xy
2
3
.
First note that y = 0 is a solution (a singular solution). So assume that y 0 and write
−3
y
2
y
·
= x,
1
y
3
=
x
2
2
+ C,
y =
1
(
x
2
2
+ C)
1/3
.
1.3.4 Exercises
Exercise 1.3.1: Solve y
·
=
x
y
.
Exercise 1.3.2: Solve y
·
= x
2
y.
Exercise 1.3.3: Solve
dx
dt
= (x
2
− 1) t, for x(0) = 0.
26 CHAPTER 1. FIRST ORDER ODES
Exercise 1.3.4: Solve
dx
dt
= x sin(t), for x(0) = 1.
Exercise 1.3.5: Solve
dy
dx
= xy + x + y + 1. Hint: Factor the right hand side.
Exercise 1.3.6: Find an implicit solution to xy
·
= y + 2x
2
y, where y(1) = 1.
Exercise 1.3.7: Solve x
dy
dx
− y = 2x
2
y, for y(0) = 10.
1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 27
1.4 Linear equations and the integrating factor
Note: more than 1 lecture, §1.5 in EP
One of the most important types of equations we will learn how to solve are so-called linear
equations. In fact the majority of this course will focus on linear equations. In this lecture we will
focus on the first order linear equation. That is a first order equation is linear if we can put it into
the following form:
y
·
+ p(x)y = f (x). (1.3)
The word "linear" here means linear in y. The dependence on x can be more complicated.
Solutions of linear equations have nice properties. For example, the solution exists wherever
p(x) and f (x) are defined, and has the same regularity (read: it is just as nice). But most importantly
for us right now, there is a method for solving linear first order equations.
What we will do is to multiply both sides of (1.3) by some function r(x) such that
r(x)y
·
+ r(x)p(x)y =
d
dx
,
r(x)y
¸
.
We can then integrate both sides of
d
dx
,
r(x)y
¸
= r(x) f (x).
Note that the right hand side does not depend on y and the left hand side is written as a derivative
of a function. We can then solve for y. The function r(x) is called the integrating factor and the
method is called the integrating factor method.
So we are looking for a function r(x) such that if we differentiate it, we get the same function
back multiplied by p(x). That seems like a job for the exponential function!
r(x) = e
p(x)dx
Let us do the calculation.
y
·
+ p(x)y = f (x),
e
p(x)dx
y
·
+ e
p(x)dx
p(x)y = e
p(x)dx
f (x),
d
dx
,
e
p(x)dx
y
¸
= e
p(x)dx
f (x),
e
p(x)dx
y =
e
p(x)dx
f (x) dx + C,
y = e
−
p(x)dx
¸
e
p(x)dx
f (x) dx + C
.
Of course, to get a closed form formula for y we need to be able to find a closed form formula
for the two integrals.
28 CHAPTER 1. FIRST ORDER ODES
Example 1.4.1: Solve
y
·
+ 2xy = e
x−x
2
y(0) = −1.
First note that p(x) = 2x and f (x) = e
x−x
2
. The integrating factor is r(x) = e
p(x)dx
. You can always
add a constant of integration, but those constants will not matter in the end.
Exercise 1.4.1: Try it! Add a constant of integration to the integral in the integrating factor and
show that the solution you get in the end is the same as what we got above.
An advice: Do not try to remember the formula itself, that is way too hard. It is easier to
remember the process and repeat it.
Since we cannot always evaluate the integrals in closed form, it is useful to know how to write
the solution in definite integral form. A definite integral is something that you can plug into a
computer or a calculator. Suppose we are given
y
·
+ p(x)y = f (x) y(x
0
) = y
0
.
Look at the solution and write the integrals as definite integrals.
y(x) = e
−
x
x
0
p(s) ds
¸
x
x
0
e
t
x
0
p(s) ds
f (t) dt + y
0
. (1.4)
You should be careful to properly use dummy variables here. If you now plug that into a computer
of a calculator, it will be happy to give you numerical answers.
Exercise 1.4.2: Check that y(x
0
) = y
0
in formula (1.4).
1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 29
Exercise 1.4.3: Write the solution of the following problem as a definite integral, but try to simplify
as far as you can. You will not be able to find the solution in closed form.
y
·
+ y = e
x
2
−x
y(0) = 10.
Example 1.4.2: The following is a simple application of linear equations and this type of a prob-
lem is used often in real life. For example, linear equations are used in figuring out the concentra-
tion of chemicals in bodies of water.
A 100 liter tank contains 10 kilograms of salt dissolved in 60 liters of water. Solution of water
and salt (brine) with concentration of 0.1 kg / liter is flowing in at the rate of 5 liters a minute. The
solution in the tank is well stirred and flows out at a rate of 3 liters a minute. How much salt is in
the tank when the tank is full?
Let us come up with the equation. Let x denote the kg of salt in the tank, let t denote the time
in minutes. Then for a small change ∆t in time, the change in x (denoted ∆x) is approximately
∆x ≈ (rate in × concentration in)∆t − (rate out × concentration out)∆t
Taking the limit ∆t → 0 we see that
dx
dt
= (rate in × concentration in) − (rate out × concentration out)
We have
rate in = 5
concentration in = 0.1
rate out = 3
concentration out =
x
volume
=
x
60 + (5 − 3)t
Our equation is, therefore,
dx
dt
= (5 × 0.1) −
0.5(60 + 2t)
3/2
dt + C(60 + 2t)
−3/2
x = 0.5(60 + 2t)
−3/2
2
5
(60 + 2t)
5/2
+ C(60 + 2t)
−3/2
x =
60 + 2t
5
+ C(60 + 2t)
−3/2
Now to figure out C. We know that at t = 0, x = 10. So
10 = x(0) =
60
5
+ C(60)
−3/2
= 12 + C(60)
−3/2
or
C = −2(60
3/2
) ≈ −929.5
We are interested in x when the tank is full. So we note that the tank is full when 60+2t = 100,
or when t = 20. So
x(20) =
60 + 40
5
+ C(60 + 40)
−3/2
≈ 20 − 929.5(100)
−3/2
≈ 19.07
The concentration at the end is approximately 0.19 kg/liter and we started with
1
6
or 0.167
kg/liter.
1.4.1 Exercises
In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot be
found. If you can find a closed form solution, you should give that.
Exercise 1.4.4: Solve y
·
+ xy = x.
Exercise 1.4.5: Solve y
·
+ 6y = e
x
.
Exercise 1.4.6: Solve y
·
+ 3x
2
y = sin(x) e
−x
3
, with y(0) = 1.
Exercise 1.4.7: Solve y
·
+ cos(x)y = cos(x).
Exercise 1.4.8: Solve
1
x
2
+1
y
·
+ xy = 3, with y(0) = 0.
1.4. LINEAR EQUATIONS AND THE INTEGRATING FACTOR 31
Exercise 1.4.9: Suppose there are two lakes. The output of one is flowing to the other. The in
and out flow from each lake is 500 liters per hour. The first lake contains 100 thousand liters of
water and the second lake contains 200 thousand liters of water. A truck with 500 kg of toxic
substance crashes into the first lake. Assume that the water is being continually mixed perfectly by
the stream. a) Find the concentration of toxic substance as a function of time (in seconds) in both
lakes. b) When will the concentration in the first lake be below 0.01 kg per liter. c) When will the
concentration in the second lake be maximal.
Exercise 1.4.10: Newton's law of cooling states that
dx
dt
= −k(x − A) where x is the temperature,
t is time, A is the ambient temperature, and k > 0 is a constant. Suppose that A = A
0
cos ωt for
some constants A
0
and ω. That is the ambient temperature oscillates (for example night and day
temperatures). a) Find the general solution. b) In the long term, will the initial conditions make
much of a difference? Why or why not.
32 CHAPTER 1. FIRST ORDER ODES
1.5 Substitution
Note: 1 lecture, §1.6 in EP
Just like when solving integrals, one method is to try to change variables to end up with a
simpler equation that can be solved.
1.5.1 Substitution
The equation
y
·
= (x − y + 1)
2
.
is neither separable nor linear. What can we do? How about trying to change variables, so that in
the new variables the equation is simpler. We will use another variable v, which we will treat as a
function of x. Let us try
v = x − y + 1.
Now we need to figure out y
·
in terms of v
·
, v and x. We differentiate (in x) to obtain v
·
= 1 − y
·
.
So y
·
= 1 − v
·
. We plug this into the equation to get
1 − v
·
= v
2
.
In other words, v
·
= 1 − v
2
. Such an equation we know how to solve.
1
1 − v
2
dv = dx.
So
1
2
ln
.
Note that the integral in this expression is not possible to find in closed form. But again, as we said
before, it is perfectly fine solution to have a definite integral in our solution. Now unsubstitute
y
−4
= e
4x
x
4
¸
4
x
1
e
−4s
s
4
ds + 1
,
y =
e
−x
x
4
x
1
e
−4s
s
4
ds + 1
1/4
.
1.5.3 Homogeneous equations
Another type of equations we can solve are the so-called homogeneous equations. Suppose that
we can write the differential equation as
y
·
= F
x
2
+ y
2
.
Exercise 1.5.5: Solve y
·
= (x + y − 1)
2
.
Exercise 1.5.6: Solve y
·
=
x+y
2
y
√
y
2
+1
, with y(0) = 1.
36 CHAPTER 1. FIRST ORDER ODES
1.6 Autonomous equations
Note: 1 lecture, §2.2 in EP
Let us consider problems of the form
dx
dt
= f (x),
where the derivative of solutions depends only on x (the dependent variable). These types of
equations are called autonomous equations. If we think of t as time, the naming comes from the
fact that the equation is independent of time.
Let us come back to the cooling coffee problem. Newton's law of cooling says that
dx
dt
= −k(x − A),
where x is the temperature, t is time, k is some constant and A is the ambient temperature. See
Figure 1.6 for an example.
Note the solution x = A (in the example A = 5). We call these types of solutions equilibrium
solutions. The points on the x axis where f (x) = 0 are called critical points. The point x = A is
a critical point. In fact, each critical point corresponds to an equilibrium solution. Note also, by
looking at the graph, that the solution x = A is "stable" in that small perturbations in x do not lead
to substantially different solutions as t grows. If we change the initial condition a little bit, then as
t → ∞ we get x → A. We call such critical points stable. In this simple example it turns out that
all solutions in fact go to A as t → ∞. If a critical point is not stable we would say it is unstable.
0 5 10 15 20
0 5 10 15 20
-10
-5
0
5
10
-10
-5
0
5
10
Figure 1.6: Slope field and some solutions of
x
·
= −0.3(x − 5).
0 5 10 15 20
0 5 10 15 20
-5
0
5
10
-5
0
5
10
Figure 1.7: Slope field and some solutions of
x
·
= −0.1x(5 − x).
1.6. AUTONOMOUS EQUATIONS 37
Let us consider the logistic equation
dx
dt
= kx(M − x),
for some positive k and M. This equation is commonly used to model population if you know the
limiting population M, that is the maximum sustainable population. This scenario leads to less
catastrophic predictions on world population. Note that in the real world there is no such thing as
negative population, but we will still consider negative x for the purposes of the math.
See Figure 1.7 on the facing page for an example. Note two critical points, x = 0 and x = 5.
The critical point at x = 5 is stable. On the other hand the critical point at x = 0 is unstable.
It is not really necessary to find the exact solutions to talk about the long term behavior of the
solutions. For example, from the above we can easily see that
lim
t→∞
x(t) =
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
5 if x(0) > 0,
0 if x(0) = 0,
DNE or − ∞ if x(0) < 0.
Where DNE means "does not exist." From just looking at the slope field we cannot quite decide
what happens if x(0) < 0. It could be that the solution does not exist t all the way to ∞. Think
of the equation y
·
= y
2
, we have seen that it only exists for some finite period of time. Same can
happen here. In our example equation above it will actually turn out that the solution does not exist
for all time, but to see that we would have to solve the equation. In any case, the solution does go
to −∞, but it may get there rather quickly.
Many times are interested only in the long term behavior of the solution and hence we would
just be doing way too much work if we tried to solve the equation exactly. It is easier to just
look at the phase diagram or phase portrait, which is a simple way to visualize the behavior of
autonomous equations. In this case there is one dependent variable x. So draw the x axis, mark
all the critical points and then draw arrows in between. Mark positive with up and negative with
down.
y = 0
y = 5
Armed with the phase diagram, it is easy to approximately sketch how the solutions are going
to look.
38 CHAPTER 1. FIRST ORDER ODES
Exercise 1.6.1: Try sketching a few solutions. Check with the graph above if you are getting the
same answers.
Once we draw the phase diagram, we can easily classify critical points as stable or unstable.
unstable stable
Since any mathematical model we cook up will only be an approximation to the real world,
unstable points are generally bad news.
Let us think about the logistic equation with harvesting. Logistic equations are commonly used
for modelling population. Suppose an alien race really likes to eat humans. They keep a planet
with humans on it and harvest the humans at a rate of h million humans per year. Suppose x is the
number of humans in millions on the planet and t is time in years. Let M be the limiting population
when no harvesting is done. k > 0 is some constant depending on how fast humans multiply. Our
equation becomes
dx
dt
= kx(M − x) − h.
Multiply out and solve for critical points
dx
dt
= −kx
2
+ kMx − h.
Critical points A and B are
A =
kM +
(kM)
2
− 4hk
2k
B =
kM −
(kM)
2
− 4hk
2k
.
Exercise 1.6.2: Draw the phase diagram for different possibilities. Note that these possibilities
are A > B, or A = B, or A and B both complex (i.e. no real solutions).
It turns out that when h = 1, then A and B are distinct and positive. The graph we will
get is given in Figure 1.8 on the next page. As long as the population stays above B which is
approximately 1.55 million, then the population will not die out. If ever the population drops
below B, humans will die out, and the fast food restaurant serving them will go out of business.
When h = 1.6, then A = B. There is only one critical point which is unstable. When the
population is above 1.6 million it will tend towards this number. If it ever drops below 1.6 million,
humans will die out on the planet. This scenario is not one that we (as the human fast food
proprietor) want to be in. A small perturbation of the equilibrium state and we are out of business.
There is no room for error. See Figure 1.9 on the facing page
Finally if we are harvesting at 2 million humans per year, the population will always plummet
towards zero, no matter how well stocked the planet starts. See Figure 1.10 on the next page.
1.6. AUTONOMOUS EQUATIONS 39
0 5 10 15 20
0 5 10 15 20
0
2
5
8
10
0
2
5
8
10
Figure 1.8: Slope field and some solutions of
x
·
= −0.1x(8 − x) − 1.
0 5 10 15 20
0 5 10 15 20
0
2
5
8
10
0
2
5
8
10
Figure 1.9: Slope field and some solutions of
x
·
= −0.1x(8 − x) − 1.6.
0 5 10 15 20
0 5 10 15 20
0
2
5
8
10
0
2
5
8
10
Figure 1.10: Slope field and some solutions of x
·
= −0.1x(8 − x) − 2.
1.6.1 Exercises
Exercise 1.6.3: Let x
·
= x
2
. a) Draw the phase diagram, find the critical points and mark them
stable or unstable. b) Sketch typical solutions of the equation. c) Find lim
t→∞
x(t) for the solution
with the initial condition x(0) = −1.
Exercise 1.6.4: Let x
·
= sin x. a) Draw the phase diagram for −4π ≤ x ≤ 4π. On this interval
mark the critical points stable or unstable. b) Sketch typical solutions of the equation. c) Find
lim
t→∞
x(t) for the solution with the initial condition x(0) = 1.
Exercise 1.6.5: Suppose f (x) is positive for 0 < x < 1 and negative otherwise. a) Draw the phase
40 CHAPTER 1. FIRST ORDER ODES
diagram for x
·
= f (x), find the critical points and mark them stable or unstable. b) Sketch typical
solutions of the equation. c) Find lim
t→∞
x(t) for the solution with the initial condition x(0) = 0.5.
Exercise 1.6.6: Start with the logistic equation
dx
dt
= kx(M − x). Suppose that we modify our
harvesting. That is we will only harvest only an amount proportional to current population, that
we harvest hx for some h > 0. a) Construct the differential equation. b) Show that if kM > h, then
the equation is still logistic. c) What happens when kM < h?
1.7. NUMERICAL METHODS: EULER'S METHOD 41
1.7 Numerical methods: Euler's method
Note: 1 lecture, §2.4 in EP
At this point it may be good to first try the Lab II and/or Project II from the IODE website:
The first thing to note is that, as we said before, it is generally very hard if not impossible to
get a nice formula for the solution of the problem
y
·
= f (x, y) y(x
0
) = y
0
.
What if we want to find out the value of the solution at some particular x. Or perhaps we even
want to produce a graph of the solution to inspect the behavior.
Euler's method
‡
: We take x
0
and compute the slope k = f (x
0
, y
0
). The slope is the change in
y per unit change in x. We follow the line for an interval of length h. Hence if y = y
0
at x
0
, then
we will say that y
1
(the approximate value of y at x
1
= x
0
+ h) will be y
1
= y
0
+ hk. Rinse repeat!
That is, compute x
2
and y
2
using x
1
and y
1
. For an example of the first two steps of the method see
Figure 1.11.11: First two steps of Euler's method with h = 1 for the equation y
·
=
y
2
3
with initial
conditions y(0) = 1.
More abstractly we compute
x
i+1
= x
i
+ h, y
i+1
= y
i
+ h f (x
i
, y
i
).
By connecting the dots we get an approximate graph of the solution. Do note that this is not exactly
the solution. See Figure 1.12 on the next page for the plot of the real solution.
‡
Named after the Swiss mathematician Leonhard Paul Euler (1707 – 1783). Do note the correct pronunciation of
the name sounds more like "oiler."
42 CHAPTER 1. FIRST ORDER ODES12: Two steps of Euler's method (step size 1) and the exact solution for the equation
y
·
=
y
2
3
with initial conditions y(0) = 1.
Let us see what happens with the equation y
·
=
y
2
3
, y(0) = 1. Let us try to approximate y(2)
using Euler's method. In Figures 1.11 and 1.12 we have essentially graphically approximated
y(2) with step size 1. With step size 1 we have y(2) ≈ 1.926. The real answer is 3. So we are
approximately 1.074 off. Let us halve the step size. If you do the computation you will find that
y(2) ≈ 2.209, so error of about 0.791. Table 1.1 on the facing page gives the values computed for
various parameters.
Exercise 1.7.1: Solve this equation exactly and show that y(2) = 3.
The difference between the actual solution and the approximate solution we will call the error.
We will usually talk about just the size of the error and we do not care much about its sign. The
main point is, that we usually do not know the real solution, so we only have a vague understanding
of the error. If we knew the error exactly ... what is the point of doing the approximation.
We notice that except for the first few times, every time we halved the interval the error approx-
imately halved. This halving of the error is a general feature of Euler's method as it is a first order
method. In the IODE Project II you are asked to implement a second order method. A second
order method reduces the error to approximately one quarter every time you halve the interval.
Note that to get the error to be within 0.1 of the answer we had to already do 64 steps. To
get it to within 0.01 we would have to halve another 3 or four times, meaning doing 512 to 1024
steps. That is quite a bit to do by hand. The improved Euler method should quarter the error every
time you halve the interval, so you would have to approximately do half as many "halvings" to
get the same error. This reduction can be a big deal. With 10 halvings (starting at h = 1) you
have 1024 steps, whereas with 5 halvings you only have to do 32 steps, assuming that the error
was comparable to start with. A computer may not care between this difference for a problem this
simple, but suppose each step would take a second to compute (the function may be substantially
1.7. NUMERICAL METHODS: EULER'S METHOD 43
h Approximate y(2) Error
Error
Previous error
1 1.92592592593 1.074074074070
0.5 2.20861152999 0.791388470013 0.736809954840
0.25 2.47249414666 0.527505853335 0.666557415634
0.125 2.68033658758 0.319663412423 0.605990266083
0.0625 2.82040079550 0.179599204497 0.561838476090
0.03125 2.90412106479 0.095878935207 0.533849442573
0.015625 2.95035498158 0.049645018422 0.517788587396
0.0078125 2.97472419486 0.025275805142 0.509130743538
Table 1.1: Euler's method approximation of y(2) where of y
·
=
y
2
3
, y(0) = 1.
more difficult to compute than y
2
/3). Then the difference is 32 seconds versus about 17 minutes.
Note: We are not being altogether fair, a second order method would probably double the time to
do each step. Even so, it is 1 minute versus 17 minutes. Next, suppose that you have to repeat such
a calculation for different parameters a thousand times. You get the idea.
Note that we do not know the error! How do you know what is the right step size? Essentially
you keep halving the interval and if you are lucky you can estimate the error from a few of these
calculations and the assumption that the error goes down by a factor of one half each time (if you
are using standard Euler).
Exercise 1.7.2: In the table above, suppose you do not know the error. Take the approximate
values of the function in the last two lines, assume that the error goes down by a factor of 2. Can
you estimate the error in the last time from this? Does it agree with the table? Now do it for the
first two rows. Does this agree with the table?
Let talk a little bit more about this example y
·
=
y
2
3
, y(0) = 1. Suppose that instead of y(2)
we wish to find y(3). Results of this effort are listed in Table 1.2 on the next page for successive
halvings of h. What is going on here? Well, you should solve the equation exactly and you will
notice that the solution does not exist at x = 3. In fact the solution blows up.
Another case when things can go bad is if the solution oscillates wildly near some point. Such
an example is given in IODE Project II. In this case, the solution may exist at all points, but even
a better approximation method than Euler would need an insanely small step size to compute the
solution with reasonable precision. And computers might not be able to handle such a small step
size anyway.
In real applications you would not use a simple method such as Euler's. The simplest method
that would probably be used in a real application is the standard Runge-Kutta method (we will not
describe it here). That is a fourth order method, that means that if you halve the interval, the error
generally goes down by a factor of 16.
44 CHAPTER 1. FIRST ORDER ODES
h Approximate y(3)
1 3.16232281664
0.5 4.54328915766
0.25 6.86078752222
0.125 10.8032064113
0.0625 17.5989264104
0.03125 29.4600446195
0.015625 50.4012144477
0.0078125 87.7576927770
Table 1.2: Attempts to use Euler's to approximate y(3) where of y
·
=
y
2
3
, y(0) = 1.
Choosing the right method to use and the right step size can be very tricky. There are several
competing factors to consider.
• Computational time: Each step takes computer time. Even if the function f is simple to
compute, you do it many times over. Large step size means faster computation, but perhaps
not the right precision.
• Roundoff errors: Computers only compute with a certain number of significant digits. Errors
introduced by rounding numbers off during your computations become noticeable when the
step size becomes too small relative to the quantities you are working with. So reducing step
size may in fact make errors worse.
• Stability: Certain equations may be numerically unstable. Small errors lead to large errors
down the line. Or in the worst case the numerical computations might be giving you bogus
numbers that look like a correct answer. Just because the numbers have stabilized after
successive halving, does not mean that you must have the right answer. Or what may happen
is that the numbers may never stabilize no matter how many times you halve the interval.
You have seen just the beginnings of the challenges that appear in real applications. There is
ongoing active research by engineers and mathematicians on how to do numerical approximation
in the best way. For example, the general purpose method used for the ODE solver in Matlab and
Octave (as of this writing) is a method that appeared only in the literature only in the 1980s.
1.7.1 Exercises
Exercise 1.7.3: Consider
dx
dt
= (2t − x)
2
, x(0) = 2. Use Euler's method with step size h = 0.5 to
approximate x(1).
1.7. NUMERICAL METHODS: EULER'S METHOD 45
Exercise 1.7.4: Consider
dx
dt
= t − x, x(0) = 1. a) Use Euler's method with step sizes h = 1,
1
2
,
1
4
,
1
8
to approximate x(1). b) Solve the equation exactly. c) Describe what happens to the errors for each
h you used. That is, find the factor by which the error changed each time you halved the interval.
46 CHAPTER 1. FIRST ORDER ODES
Chapter 2
Higher order linear ODEs
2.1 Second order linear ODEs
Note: less than 1 lecture, first part of §3.1 in EP
Let us consider the general second order linear differential equation
A(x)y
··
+ B(x)y
·
+ C(x)y = F(x).
We usually divide through by A to get
y
··
+ p(x)y
·
+ q(x)y = f (x), (2.1)
where p = B/A, q = C/A, and f = F/A. The word linear means that the equation contains no
powers nor functions of y, y
·
, and y
··
.
In the special case when f (x) = 0 we have a homogeneous equation
y
··
+ p(x)y
·
+ q(x)y = 0. (2.2)
We have already seen some second order linear homogeneous equations.
y
··
+ ky = 0 Two solutions are: y
1
= cos kx, y
2
= sin kx.
y
··
− ky = 0 Two solutions are: y
1
= e
kx
, y
2
= e
−kx
.
If we know two solutions two a linear homogeneous equation, we know a lot more of them.
Theorem 2.1.1 (Superposition). Suppose y
1
and y
2
are two solutions of the homogeneous equation
(2.2). Then
y(x) = C
1
y
1
(x) + C
2
y
2
(x),
also solves (2.2) for arbitrary constants C
1
and C
2
.
47
48 CHAPTER 2. HIGHER ORDER LINEAR ODES
That is, we can add together solutions and multiply by constants to obtain new different solu-
tions. We will prove this theorem because the proof is very enlightening and illustrates how linear
equations work.
Proof: Let y = C
1
y
1
+ C
2
y
2
. Then
y
··
+ py
·
+ qy = (C
1
y
1
+ C
2
y
2
)
··
+ p(C
1
y
1
+ C
2
y
2
)
·
+ q(C
1
y
1
+ C
2
y
2
)
= C
1
y
··
1
+ C
2
y
··
2
+ C
1
py
·
1
+ C
2
py
·
2
+ C
1
qy
1
+ C
2
qy
2
= C
1
(y
··
1
+ py
·
1
+ qy
1
) + C
2
(y
··
2
+ py
·
2
+ qy
2
)
= C
1
0 + C
2
0 = 0
The proof becomes even simpler to state if we use the operator notation. An operator is an
object that eats functions and spits out functions (kind of like what a function is, but a function eats
numbers and spits out numbers). Define the operator L by
Ly = y
··
+ py
·
+ qy.
L being linear means that L(C
1
y
1
+ C
2
y
2
) = C
1
Ly
1
+ C
2
Ly
2
. Hence the proof simply becomes
Ly = L(C
1
y
1
+ C
2
y
2
) = C
1
Ly
1
+ C
2
Ly
2
= C
1
0 + C
2
0 = 0.
Two other solutions to the second equation y
··
− ky = 0 are y
1
= cosh kx and y
2
= sinh kx.
Let us remind ourselves of the definition, cosh x =
e
x
+e
−x
2
and sinh x =
e
x
−e
−x
2
. Therefore, these are
solutions by superposition as they are linear combinations of the two exponential solutions.
As sinh and cosh are sometimes more convenient to use than the exponential, let us review
some of their properties.
cosh 0 = 1 sinh 0 = 0
d
dx
cosh x = sinh x
d
dx
sinh x = cosh x
cosh
2
x − sinh
2
x = 1
Exercise 2.1.1: Derive these properties from the definitions of sinh and cosh in terms of exponen-
tials.
Linear equations have nice and simple answers to the existence and uniqueness question.
Theorem 2.1.2 (Existence and uniqueness). Suppose p, q, f are continuous functions and a, b
0
, b
1
are constants. The equation
y
··
+ p(x)y
·
+ q(x)y = f (x),
has exactly one solution y(x) satisfying the initial conditions
y(a) = b
0
y
·
(a) = b
1
.
2.1. SECOND ORDER LINEAR ODES 49
For example, the equation y
··
+ y = 0 with y(0) = b
0
and y
·
(0) = b
1
has the solution
y(x) = b
0
cos x + b
1
sin x.
Or the equation y
··
− y = 0 with y(0) = b
0
and y
·
(0) = b
1
has the solution
y(x) = b
0
cosh x + b
1
sinh x.
Here note that using cosh and sinh allows us to solve for the initial conditions much more easily
than if we have used the exponentials.
Note that the initial condition for a second order ODE consists of two equations. So if we have
two arbitrary constants we should be able to solve for the constants and find a solution satisfying
the initial conditions.
Question: Suppose we find two different solutions y
1
and y
2
to the homogeneous equation (2.2).
Can every solution be written (using superposition) in the form y = C
1
y
1
+ C
2
y
2
?
Answer is affirmative! Provided that y
1
and y
2
are different enough in the following sense. We
will say y
1
and y
2
are linearly independent if one is not a constant multiple of the other. If you find
two linearly independent solutions, then every other solution is written in the form
y = C
1
y
1
+ C
2
y
2
.
In this case y = C
1
y
1
+ C
2
y
2
is the general solution.
For example, we found the solutions y
1
= sin x and y
2
= cos x for the equation y
··
+ y = 0. It
is obvious that sin and cos are not multiples of each other. If sin x = Acos x for some constant A,
we let x = 0 and this would imply A = 0 = sin x, which is preposterous. So y
1
and y
2
are linearly
independent. Hence
y = C
1
cos x + C
2
sin x
is the general solution to y
··
+ y = 0.
2.1.1 Exercises
Exercise 2.1.2: Show that y = e
x
and y = e
2x
are linearly independent.
Exercise 2.1.3: Take y
··
+ 5y = 10x + 5. Can you find guess a solution?
Exercise 2.1.4: Prove the superposition principle for nonhomogeneous equations. Suppose that
y
1
is a solution to Ly
1
= f (x) and y
2
is a solution to Ly
2
= g(x) (same operator L). Show that y
solves Ly = f (x) + g(x).
Exercise 2.1.5: For the equation x
2
y
··
− xy
·
= 0, find two solutions, show that they are linearly
independent and find the general solution. Hint: Try y = x
r
.
50 CHAPTER 2. HIGHER ORDER LINEAR ODES
Note that equations of the form ax
2
y
··
+ bxy
·
+ cy = 0 are called Euler's equations or Cauchy-
Euler equations. They are solved by trying y = x
r
and solving for r (we can assume that x ≥ 0 for
simplicity).
Exercise 2.1.6: Suppose that (b − a)
2
− 4ac > 0. a) Find a formula for the general solution
of ax
2
y
··
+ bxy
·
+ cy = 0. Hint: Try y = x
r
and find a formula for r. b) What happens when
(b − a)
2
− 4ac = 0 or (b − a)
2
− 4ac < 0?
We will revisit the case when (b − a)
2
− 4ac < 0 later.
Exercise 2.1.7: Suppose that (b − a)
2
− 4ac = 0. Find a formula for the general solution of
ax
2
y
··
+ bxy
·
+ cy = 0. Hint: Try y = x
r
ln x for the second solution.
If you have one solution to a second order linear homogeneous equation you can find another
one. This is the reduction of order method.
Exercise 2.1.8: Suppose y
1
is a solution to y
··
+ p(x)y
·
+ q(x)y = 0. Show that
y
2
(x) = y
1
(x)
e
p(x) dx
(y
1
(x))
2
dx
is also a solution.
Let us solve some famous equations.
Exercise 2.1.9 (Chebychev's equation of order 1): Take (1 − x
2
)y
··
− xy
·
+ y = 0. a) Show that
y = x is a solution. b) Use reduction of order to find a second linearly independent solution. c)
Write down the general solution.
Exercise 2.1.10 (Hermite's equation of order 2): Take y
··
−2xy
·
+4y = 0. a) Show that y = 1−2x
2
is a solution. b) Use reduction of order to find a second linearly independent solution. c) Write
down the general solution.
2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 51
2.2 Constant coefficient second order linear ODEs
Note: more than 1 lecture, second part of §3.1 in EP
Suppose we have the problem
y
··
− 6y
·
+ 8y = 0, y(0) = −2, y
·
(0) = 6.
This is a second order linear homogeneous equation with constant coefficients. Constant coeffi-
cients means that the functions in front of y
··
, y
·
, and y are constants, not depending on x.
Think about a function that you know that stays essentially the same when you differentiate it,
so that we can take the function and its derivatives, add these together, and end up with zero.
Let us try a solution y = e
rx
. Then y
·
= re
rx
and y
··
= r
2
e
rx
. Plug in to get
y
··
− 6y
·
+ 8y = 0,
r
2
e
rx
− 6re
rx
+ 8e
rx
= 0,
r
2
− 6r + 8 = 0 (divide through by e
rx
),
(r − 2)(r − 4) = 0.
So if r = 2 or r = 4, then e
rx
is a solution. So let y
1
= e
2x
and y
2
= e
4x
.
Exercise 2.2.1: Check that y
1
and y
2
are solutions.
The functions e
2x
and e
4x
are linearly independent. If they were not we could write e
4x
= Ce
2x
,
which would imply that e
2x
= C which is clearly not possible. Hence, we can write the general
solution as
y = C
1
e
2x
+ C
2
e
4x
.
We need to solve for C
1
and C
2
. To apply the initial conditions we first find y
·
= 2C
1
e
2x
+ 4C
2
e
4x
.
We plug in x = 0 and solve.
−2 = y(0) = C
1
+ C
2
,
6 = y
·
(0) = 2C
1
+ 4C
2
.
Either apply some matrix algebra, or just solve these by high school algebra. For example, divide
the second equation by 2 to obtain 3 = C
1
+ 2C
2
, and subtract the two equations to get 5 = C
2
.
Then C
1
= −7 as −2 = C
1
+ 5. Hence, the solution we are looking for is
y = −7e
2x
+ 5e
4x
.
Let us generalize this example into a method. Suppose that we have an equation
ay
··
+ by
·
+ cy = 0, (2.3)
52 CHAPTER 2. HIGHER ORDER LINEAR ODES
where a, b, c are constants. Try the solution y = e
rx
to obtain
ar
2
e
rx
+ bre
rx
+ ce
rx
= 0
ar
2
+ br + c = 0.
The equation ar
2
+ br + c = 0 is called the characteristic equation of the ODE. Solve for the r by
using the quadratic formula.
r
1
, r
2
=
−b ±
√
b
2
− 4ac
2a
.
Therefore, we have e
r
1
x
and e
r
2
x
as solutions. There is still a difficulty if r
1
= r
2
, but it is not hard
to overcome.
Theorem 2.2.1. Suppose that r
1
and r
2
are the roots of the characteristic equation.
(i) If r
1
and r
2
are distinct and real (b
2
− 4ac > 0), then (2.3) has the general solution
y = C
1
e
r
1
x
+ C
2
e
r
2
x
.
(ii) If r
1
= r
2
(b
2
− 4ac = 0), then (2.3) has the general solution
y = (C
1
+ C
2
x) e
r
1
x
.
For another example of the first case, note the equation y
··
− k
2
y = 0. Here the characteristic
equation is r
2
−k
2
= 0 or (r −k)(r +k) = 0 and hence e
−kx
and e
kx
are the two linearly independent
solutions.
Example 2.2.1: Find the general solution of
y
··
− 8y
·
+ 16y = 0.
The characteristic equation is r
2
−8r +16 = (r −4)
2
= 0. Hence a double root r
1
= r
2
= 4. The
general solution is, therefore,
y = (C
1
+ C
2
x) e
4x
= C
1
e
4x
+ C
2
xe
4x
.
Exercise 2.2.2: Check that e
4x
and xe
4x
are linearly independent.
That e
4x
solves the equation is clear. If xe
4x
solves the equation then we know we are done. Let
us compute y
·
= e
4x
+ 4xe
4x
and y
··
= 8e
4x
+ 16xe
4x
. Plug in
y
··
− 8y
·
+ 16y = 8e
4x
+ 16xe
4x
− 8(e
4x
+ 4xe
4x
) + 16xe
4x
= 0.
We should note that in practice, doubled root rarely happens. If you pick your coefficients truly
randomly you are very unlikely to get a doubled root.
Let us give a short "proof" for why the solution xe
rx
works when the root is doubled. Since
this case is really a limiting case of when cases the two roots are distinct and very close. Note that
e
r
2
x
−e
r
1
x
r
2
−r
1
is a solution when the roots are distinct. When r
1
goes to r
2
in the limit this is like taking
derivative of e
rx
using r as a variable. This limit is xe
rx
, and hence this is also a solution in the
doubled root case.
2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 53
2.2.1 Complex numbers and Euler's formula
It may happen that a polynomial has some complex roots. For example, the equation r
2
+ 1 = 0
has no real roots, but it does have two complex roots. Here we review some properties of complex
numbers.
Complex numbers may seem a strange concept especially because of the terminology. There is
nothing imaginary or really complicated about complex numbers. A complex number is really just
a pair of real numbers, (a, b). We can think of a complex number as a point in the plane. We add
complex numbers in the straightforward way. We define a multiplication by
(a, b) × (c, d)
def
= (ac − bd, ad + bc).
It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further,
and most importantly (0, 1) × (0, 1) = (−1, 0).
Generally we just write (a, b) as a + ib, and we treat i as if it were an unknown. You can just
do arithmetic with complex numbers just as you would do with polynomials. The property we just
mentioned becomes i
2
= −1. So whenever you see i
2
you can replace it by −1. Also, for example
i and −i are roots of r
2
+ 1 = 0.
Note that engineers often use the letter j instead of i for the square root of −1. We will use the
mathematicians convention and use i.
Exercise 2.2.3: Make sure you understand (that you can justify) the following identities:
• i
2
= −1, i
3
= −i, i
4
= 1,
•
1
i
= −i,
• (3 − 7i)(−2 − 9i) = = −69 − 13i,
• (3 − 2i)(3 + 2i) = 3
2
− (2i)
2
= 3
2
+ 2
2
= 13,
•
1
3−2i
=
1
3−2i
3+2i
3+2i
=
3+2i
13
=
3
13
+
2
13
i.
We can also define the exponential e
a+ib
of a complex number. We can do this by just writing
down the Taylor series and plugging in the complex number. Because most properties of the
exponential can be proved by looking at the Taylor series, we note that many properties still hold
for the complex exponential. For example, e
x+y
= e
x
e
y
. This means that e
a+ib
= e
a
e
ib
and hence if
we can compute e
ib
easily, we can compute e
a+ib
. Here we will use the so-called Euler's formula.
Theorem 2.2.2 (Euler's formula).
e
iθ
= cos θ + i sin θ and e
−iθ
= cos θ − i sin θ.
54 CHAPTER 2. HIGHER ORDER LINEAR ODES
Exercise 2.2.4: Using Euler's formula, check the identities:
cos θ =
e
iθ
+ e
−iθ
2
and sin θ =
e
iθ
− e
−iθ
2i
.
Exercise 2.2.5: Double angle identities: Start with e
i(2θ)
=
e
iθ
2
. Use Euler on each side and
deduce:
cos 2θ = cos
2
θ − sin
2
θ and sin 2θ = 2 sin θ cos θ.
We also will need some notation. For a complex number a + ib we call a the real part and b
the imaginary part of the number. In notation this is
Re(a + bi) = a and Im(a + bi) = b.
2.2.2 Complex roots
So now suppose that the equation ay
··
+ by
·
+ cy = 0 has a characteristic equation ar
2
+ br + c = 0
which has complex roots. That is, by quadratic formula the roots are
−b±
√
b
2
−4ac
2a
. These are complex
if b
2
− 4ac < 0. In this case we can see that the roots are
r
1
, r
2
=
−b
2a
± i
√
b
2
− 4ac
2a
.
As you can see, you will always get a pair of roots of the form α±iβ. In this case we can still write
the solution as
y = C
1
e
(α+iβ)x
+ C
2
e
(α−iβ)x
.
However, the exponential is now complex valued. We would need to choose C
1
and C
2
to be
complex numbers to obtain a real valued solution (which is what we are after). While there is
nothing particularly wrong with this, it can make calculations harder and it would be nice to find
two real valued solutions.
Here we can use Euler's formula. First let
y
1
= e
(α+iβ)x
and y
2
= e
(α−iβ)x
.
Then note that
y
1
= e
αx
cos βx + ie
αx
sin βx,
y
2
= e
αx
cos βx − ie
αx
sin βx.
We note that linear combinations of solutions are also solutions. Hence
y
3
=
y
1
+ y
2
2
= e
αx
cos βx,
y
4
=
y
1
− y
2
2i
= e
αx
sin βx,
are also solutions. And furthermore they are real valued. It is not hard to see that they are linearly
independent (not multiples of each other). Therefore, we have the following theorem.
2.2. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES 55
Theorem 2.2.3. Take the equation
ay
··
+ by
·
+ cy = 0.
If the characteristic equation has the roots α ± iβ, then the general solution is
y = C
1
e
αx
cos βx + C
2
e
αx
sin βx.
Example 2.2.2: Find the general solution of y
··
+ k
2
y = 0, for a constant k > 0.
The characteristic equation is r
2
+ k
2
= 0. Therefore, the roots are r = ±ik and by the theorem
we have the general solution
y = C
1
cos kx + C
2
sin kx.
Example 2.2.3: Find the solution of y
··
− 6y
·
+ 13y = 0, y(0) = 0, y
·
(0) = 10.
The characteristic equation is r
2
−6r+13 = 0. By completing the square we get (r−3)
2
+2
2
= 0
and hence the roots are r = 3 ± 2i. By the theorem we have the general solution
y = C
1
e
3x
cos 2x + C
2
e
3x
sin 2x.
To find the solution satisfying the initial conditions, we first plug in zero to get
0 = y(0) = C
1
e
0
cos 0 + C
2
e
0
sin 0 = C
1
.
Hence C
1
= 0 and hence y = C
2
e
3x
sin 2x. We differentiate
y
·
= 3C
2
e
3x
sin 2x + 2C
2
e
3x
cos 2x.
We again plug in the initial condition and obtain 10 = y
·
(0) = 2C
2
, or C
2
= 5. Hence the solution
we are seeking is
y = 5e
3x
sin 2x.
2.2.3 Exercises
Exercise 2.2.6: Find the general solution of 2y
··
+ 2y
·
− 4y = 0.
Exercise 2.2.7: Find the general solution of y
··
+ 9y
·
− 10y = 0.
Exercise 2.2.8: Solve y
··
− 8y
·
+ 16y = 0 for y(0) = 2, y
·
(0) = 0.
Exercise 2.2.9: Solve y
··
+ 9y
·
= 0 for y(0) = 1, y
·
(0) = 1.
Exercise 2.2.10: Find the general solution of 2y
··
+ 50y = 0.
Exercise 2.2.11: Find the general solution of y
··
+ 6y
·
+ 13y = 0.
56 CHAPTER 2. HIGHER ORDER LINEAR ODES
Exercise 2.2.12: Find the general solution of y
··
= 0 using the methods of this section.
Exercise 2.2.13: The method of this section applies to equations of other orders than two. We will
see higher orders later. Try to solve the first order equation 2y
·
+ 3y = 0 using the methods of this
section.
Exercise 2.2.14: Let us revisit Euler's equations of Exercise 2.1.6 on page 50. Suppose now that
(b − a)
2
− 4ac < 0. Find a formula for the general solution of ax
2
y
··
+ bxy
·
+ cy = 0. Hint: Note
that x
r
= e
r ln x
.
2.3. HIGHER ORDER LINEAR ODES 57
2.3 Higher order linear ODEs
Note: 2 lectures, §3.2 and §3.3 in EP
In general, most equations that appear in applications tend to be second order. Higher order
equations do appear from time to time, but it is a general assumption of modern physics that the
world is "second order."
The basic results about linear ODEs of higher order are essentially exactly the same as for
second order equations with 2 replaced by n. The important new concept here is the concept of
linear independence. This concept is used in many other areas of mathematics and even other
places in this course, and it is useful to understand this in detail.
For constant coefficient ODEs, the methods are slightly harder, but we will not dwell on these.
You can always use the methods for systems of linear equations we will learn later in the course to
solve higher order constant coefficient equations.
So let us start with a general homogeneous linear equation
y
(n)
+ p
n−1
(x)y
(n−1)
+ + p
1
(x)y
·
+ p
0
(x)y = 0. (2.4)
Theorem 2.3.1 (Superposition). Suppose y
1
, y
2
, . . . , y
n
are solutions of the homogeneous equation
(2.4). Then
y(x) = C
1
y
1
(x) + C
2
y
2
(x) + + C
n
y
n
(x),
also solves (2.4) for arbitrary constants C
1
, . . . , C
n
.
We also have the existence and uniqueness theorem for nonhomogeneous linear equations.
Theorem 2.3.2 (Existence and uniqueness). Suppose p
0
through p
n−1
, and f are continuous func-
tions and a, b
0
, b
1
, . . . , b
n−1
are constants. The equation
y
(n)
+ p
n−1
(x)y
(n−1)
+ + p
1
(x)y
·
+ p
0
(x)y = f (x), .
has exactly one solution y(x) satisfying the initial conditions
y(a) = b
0
, y
·
(a) = b
1
, . . . , y
(n−1
)(a) = b
n−1
.
2.3.1 Linear independence
When we had two functions y
1
and y
2
we said they were linearly independent if one was not the
multiple of the other. Same idea holds for n functions. In this case it is easier to state as follows.
The functions y
1
, y
2
, . . . , y
n
are linearly independent if
c
1
y
1
+ c
2
y
2
+ + c
n
y
n
= 0,
has only the trivial solution c
1
= c
2
= = c
n
= 0. If we can write the equation with a nonzero
constant, say c
1
0, then we can solve for y
1
as a linear combination of the others. If the functions
are not linearly independent, way say they are linearly dependent.
58 CHAPTER 2. HIGHER ORDER LINEAR ODES
Example 2.3.1: Show e
x
, e
2x
, e
3x
are linearly independent.
Let us give several ways to do this. Most textbooks (including [EP] and [F]) introduce Wron-
skians, but that is really not necessary here.
Let us write down
c
1
e
x
+ c
2
e
2x
+ c
3
e
3x
= 0.
Use rules of exponentials and write z = e
x
. Then we have
c
1
z + c
2
z
2
+ c
3
z
3
= 0.
The left hand side is is a third degree polynomial in z. It can either be identically zero or have at
most 3 zeros. Therefore, it is identically zero and c
1
= c
2
= c
3
= 0 and the functions are linearly
independent.
Let us try another way. Write
c
1
e
x
+ c
2
e
2x
+ c
3
e
3x
= 0.
This equation has to hold for all x. What we could do is divide through by e
3x
to get
c
1
e
−2x
+ c
2
e
−x
+ c
3
= 0.
This is true for all x, therefore, let x → ∞. After taking the limit we see that c
3
= 0. Hence our
equation becomes
c
1
e
x
+ c
2
e
2x
= 0.
Rinse, repeat!
How about yet another way. Write
c
1
e
x
+ c
2
e
2x
+ c
3
e
3x
= 0.
We could evaluate at several different x to get equations for c
1
, c
2
and c
3
. That might be a lot of
computation. We can also take derivatives of both sides and then evaluate. Let us first divide by e
x
for simplicity.
c
1
+ c
2
e
x
+ c
3
e
2x
= 0.
Set x = 0 to get the equation c
1
+ c
2
+ c
3
= 0. Now differentiate both sides
c
2
e
x
+ 2c
3
e
2x
= 0,
and set x = 0 to get c
2
+2c
3
= 0. Finally divide by e
x
again and differentiate to get 4c
3
e
2x
= 0. It is
clear that c
3
is zero. Then c
2
must be zero as c
2
= −2c
3
and c
1
must be zero because c
1
+c
2
+c
3
= 0.
There is no one good way to do it. All of these methods are perfectly valid.
Example 2.3.2: On the other hand, the functions e
x
, e
−x
, and cosh x are linearly dependent. Simply
apply definition of the hyperbolic cosine:
cosh x =
e
x
+ e
−x
2
.
2.3. HIGHER ORDER LINEAR ODES 59
2.3.2 Constant coefficient higher order ODEs
When we have a higher order constant coefficient homogeneous linear equation. The song and
dance is exactly the same as it was for second order. We just need to find more solutions. If the
equation is n
th
order we need to find n linearly independent solutions. It is best seen by example.
Example 2.3.3: Find the general solution to
y
···
− 3y
··
− y
·
+ 3y = 0. (2.5)
Try: y = e
rx
. We plug in and get
r
3
e
rx
− 3r
2
e
rx
− re
rx
+ 3e
rx
= 0.
We divide out by e
rx
. Then
r
3
− 3r
2
− r + 3 = 0.
The trick now is to find the roots. There is a formula for degree 3 and 4 equations but it is very
complicated. There is no formula for higher degree polynomials. That does not mean that the roots
do not exist. There are always n roots for an n
th
degree polynomial. They might be repeated and
they might be complex. Computers are pretty good at finding roots approximately for reasonable
size polynomials.
Best place to start is to plot the polynomial and check where it is zero. Or you can try plugging
in. Sometimes it is a good idea to just start plugging in numbers r = −2, −1, 0, 1, −1, 2, . . . and
see if you get a hit. There are some signs that you might have missed a root. For example, if you
plug in −2 into our polynomial you get −15. If you plug in 0 you get 3. That means there is a root
between −2 and 0 because the sign changed.
A good strategy at first is to look for roots −1, 1, or 0, these are easy to see. When check our
polynomial we note that r
1
= −1 and r
2
= 1 are roots. The last root is then reasonably easy to
find. We note that the constant term in a polynomial is the multiple of the negations of all the roots
because r
3
− 3r
2
− r + 3 = (r − r
1
)(r − r
2
)(r − r
3
). In our case we see that
3 = (−r
1
)(−r
2
)(−r
3
) = (1)(−1)(−r
3
) = r
3
.
You should check that r
3
= 3 is a root. Hence we know that e
−x
, e
x
and e
3x
are solutions to (2.5).
They are linearly independent as can easily be checked, and there is 3 of them, which happens to
be exactly the number we need. Hence the general solution is
y = C
1
e
−x
+ C
2
e
x
+ C
3
e
3x
.
Suppose we were given some initial conditions y(0) = 1, y
·
(0) = 2, and y
··
(0) = 3. This leads
to
1 = y(0) = C
1
+ C
2
+ C
3
,
2 = y
·
(0) = −C
1
+ C
2
+ 3C
3
,
3 = y
··
(0) = C
1
+ C
2
+ 9C
3
.
60 CHAPTER 2. HIGHER ORDER LINEAR ODES
It is possible to find the solution by high school algebra, but it would be a pain. The only sensible
way to solve a system of equations such as this is to use matrix algebra, see § 3.2. For now we note
that the solution is C
1
= −
1
4
, C
2
= 1 and C
3
=
1
4
. With this the specific solution is
y =
−1
4
e
−x
+ e
x
+
1
4
e
3x
.
Next, suppose that we have real roots, but they are repeated. Let us say we have a root r
repeated k times. In this case, in the spirit of the second order solution we note the solutions
e
rx
, xe
rx
, x
2
e
rx
, . . . , x
k−1
e
rx
.
We take a linear combination of these solutions to find the general solution.
Example 2.3.4: Solve
y
(4)
− 3y
···
+ 3y
··
− y
·
= 0.
We note that the characteristic equation is
r
4
− 3r
3
+ 3r
2
− r = 0.
By inspection we note that r
4
− 3r
3
+ 3r
2
− r = r(r − 1)
3
. Hence the roots given with multiplicity
are r = 0, 1, 1, 1. Thus the general solution is
y = (c
0
+ c
1
x + c
2
x
2
) e
x
....
terms coming from r = 1
+ c
4
....
from r = 0
.
Similarly to the second order case we can handle complex roots and we really only need to talk
about how to handle repeated complex roots. Complex roots always come in pairs r = α ± iβ. The
corresponding solution is
(c
0
+ c
1
x + + c
k−1
x
k
) e
αx
cos βx + (d
0
+ d
1
x + + d
k−1
x
k
) e
αx
sin βx.
where c
0
, . . . , c
k−1
, d
0
, . . . , d
k−1
are arbitrary constants.
Example 2.3.5: Solve
y
(4)
− 4y
···
+ 8y
··
− 8y
·
+ 4y = 0.
The characteristic equation is
r
4
− 4r
3
+ 8r
2
− 8r + 4 = 0,
(r
2
− 2 + 2)
2
= 0,
(r − 1)
2
+ 2
2
= 0.
Hence the roots are 1 ± i with multiplicity 2. Hence the general solution is
y = (c
0
+ c
1
x) e
x
cos x + (d
0
+ d
1
x) e
x
sin x.
The way we solved the characteristic equation above is really by guessing or by inspection. It is
not so easy in general. You could also have asked a computer or an advanced calculator for the
roots.
2.3. HIGHER ORDER LINEAR ODES 61
2.3.3 Exercises
Exercise 2.3.1: Find the general solution for y
···
− y
··
+ y
·
− y = 0.
Exercise 2.3.2: Find the general solution for y
(4)
− 5y
···
+ 6y
··
= 0.
Exercise 2.3.3: Find the general solution for y
···
+ 2y
··
+ 2y
·
= 0.
Exercise 2.3.4: Suppose that the characteristic equation for an equation is (r −1)
2
(r −2)
2
= 0. a)
Find such an equation. b) Find its general solution.
Exercise 2.3.5: Suppose that a fourth order equation has the following solution. y = 2e
4x
x cos x.
a) Find such an equation. b) Find the initial conditions which the given solution satisfies.
Exercise 2.3.6: Find the general solution for the equation of Exercise 2.3.5.
Exercise 2.3.7: Let f (x) = e
x
−cos x, g(x) = e
x
+cos x, and h(x) = cos x. Are f (x), g(x), and h(x)
linearly independent? If so, show it, if not, find the linear combination that works.
Exercise 2.3.8: Let f (x) = 0, g(x) = cos x, and h(x) = sin x. Are f (x), g(x), and h(x) linearly
independent? If so, show it, if not, find the linear combination that works.
Exercise 2.3.9: Are x, x
2
, and x
4
linearly independent? If so, show it, if not, find the linear
combination that works.
Exercise 2.3.10: Are e
x
, xe
x
, and x
2
e
x
linearly independent? If so, show it, if not, find the linear
combination that works.
62 CHAPTER 2. HIGHER ORDER LINEAR ODES
2.4 Mechanical vibrations
Note: 2 lectures, §3.4 in EP
We want to look at some applications of linear second order constant coefficient equations.
2.4.1 Some examples
Our first example is a mass on a spring. Suppose we have a
damping c
m
k
F(t)
mass m > 0 (in kilograms for instance) connected by a spring
with spring constant k > 0 (in Newtons per meter perhaps) to a
fixed wall. Furthermore, there is some external force F(t) acting
on the mass. Finally, there is some friction in the system and this
is measured by a constant c ≥ 0.
Let x be the displacement of the mass (x = 0 is the rest position). With x growing to the right
(away from the wall). The force exerted by the spring is proportional to the compression of the
spring by Hooke's law. Therefore, it is kx in the negative direction. Similarly the amount of force
exerted by friction is proportional to the velocity of the mass. By Newton's second law we know
that force equals mass times acceleration and hence
mx
··
+ cx
·
+ kx = F(t).
This is a linear second order constant coefficient ODE. We set up some terminology about this
equation. We say the motion is
(i) forced, if F 0 (F not identically zero),
(ii) unforced or free, if F ≡ 0,
(iii) damped, if c > 0, and
(iv) undamped, if c = 0.
This system is appears in lots of applications even if it does not at first seems like it. Many
real world scenarios can be simplified to a mass on a spring. For example, a bungee jump setup is
essentially a spring and mass system (you are the mass). It would be good if someone did the math
before you jump off right? Let us just give 2 other examples.
Here is an example for electrical engineers. Suppose that you have the
E
L
C
R
pictured RLC circuit. There is a resistor with a resistance of R ohms, an
inductor with an inductance of L henries, and a capacitor with a capacitance
of C farads. There is also an electric source (such as a battery) giving a
voltage of E(t) volts at time t (measured in seconds). Let Q(t) be the charge
in columbs on the capacitor and I(t) be the current in the circuit. The relation between the two is
2.4. MECHANICAL VIBRATIONS 63
Q
·
= I. Furthermore, by elementary principles we have that LI
·
+RI +Q/C = E. If we differentiate
we get
LI
··
(t) + RI
·
(t) +
1
C
I(t) = E
·
(t).
This is an nonhomogeneous second order constant coefficient linear equation. Further, as L, R,
and C are all positive, this system behaves just like the mass and spring system. The position of
the mass is replaced by the current. Mass is replaced by the inductance, damping is replaced by
resistance and the spring constant is replaced by one over the capacitance. The change in voltage
becomes the forcing function. Hence for constant voltage this is an unforced motion.
Our next example is going to behave like a mass and spring system only approximately. Sup-
pose we have a mass m on a pendulum of length L. We wish to find an equation for the angle θ(t).
Let g be the force of gravity. Elementary physics mandates that the equation is of the form
θ
··
+
g
L
sin θ = 0.
This equation can be derived using Newton's second law, where force
θ
L
equals mass times acceleration. Note that acceleration is Lθ
··
and mass
is m. This has to be equal to the tangential component of the force given
by the gravity. This is mg sin θ in the opposite direction. The m curiously
cancels from the equation.
Now we make our approximation. For small θ we have that approxi-
mately sin θ ≈ θ. This can be seen by looking at the graph. In Figure 2.1
we can see that for approximately −0.5 < θ < 0.5 (in radians) the graphs of sin θ and θ are almost
the same.
-1.0 -0.5 0.0 0.5 1.0
-1.0 -0.5 0.0 0.5 11: The graphs of sin θ and θ (in radians).
64 CHAPTER 2. HIGHER ORDER LINEAR ODES
Therefore, when the swings are small, θ is always small and we can model the behavior by the
simpler linear equation
θ
··
+
g
L
θ = 0.
Note that the errors that we get from the approximation build up so over a very long time, the
behavior might change more substantially. Also we will see that in a mass spring system, the
amplitude is independent of the period, this is not true for a pendulum. But for reasonably short
periods of time and small swings (for example if the length of the pendulum is very large), the
behavior is reasonably close.
In real world problems it is very often necessary to make these types of simplifications. There-
fore, it is good to understand both the mathematics and the physics of the situation to see if the
simplification is valid in the context of the questions we are trying to answer.
2.4.2 Free undamped motion
In this section we will only consider free or unforced motion, as we cannot yet solve nonhomo-
geneous equations. First let us start with undamped motion and hence c = 0, so we have the
equation
mx
··
+ kx = 0.
If we divide out by m and let ω
0
be a number such that ω
2
0
=
k
m
we can write the equation as
x
··
+ ω
2
0
x = 0.
The general solution to this equation is
x(t) = Acos ω
0
t + Bsin ω
0
t.
First we notice that by a trigonometric identity we have that for two other constants C and γ we
have
Acos ω
0
t + Bsin ω
0
t = C cos(ω
0
t − γ).
It is not hard to compute that C =
√
A
2
+ B
2
and tan γ =
B
A
. Therefore, we can write x(t) =
C cos(ω
0
t − γ), and let C and γ be our arbitrary constants.
Exercise 2.4.1: Justify this identity and verify the equations for C and γ.
While it is generally easier to use the first form with A and B to find these constants given the
initial conditions, the second form is much more natural. The constants C and γ have very nice
interpretation. If we look at the form of the solution
x(t) = C cos(ω
0
t − γ)
2.4. MECHANICAL VIBRATIONS 65
We can see that the amplitude is C, ω
0
is the (angular) frequency, and γ is the so-called phase
shift. It just shifts the graph left or right. We call ω
0
is called the natural (angular) frequency. The
motion is usually called simple harmonic motion.
A note about the word angular before the frequency. ω
0
is given in radians per unit time, not
in cycles per unit time as is the usual measure of frequency. But because we know one cycle is 2π,
the usual frequency is given by
ω
0
2π
. It is simply a matter of where we put the constant 2π, and that
is a matter of taste.
The period of the motion is one over the frequency (in cycles per unit time) and hence
2π
ω
0
. That
is the amount of time it takes to complete one full oscillation.
Example 2.4.1: Suppose that m = 2kg and k = 8N/m. Suppose the whole setup is on a truck
which was travelling at 1m/s and suddenly crashes and hence stops. The mass was rigged 0.5
meters forward from the rest position, and gets loose in the crash and starts oscillating. What is
the frequency of the resulting oscillation and what is the amplitude. The units are the mks units
(meters-kilograms-seconds).
Well the setup means that the mass was at half a meter in the positive direction during the crash
and relative to the wall the spring is mounted to, the mass was moving forward (in the positive
direction) at 1m/s. This gives us the initial conditions.
So the equation with initial conditions is
2x
··
+ 8x = 0, x(0) = 0.5, x
·
(0) = 1.
We can directly compute ω
0
=
k
m
=
√
4 = 2. Hence the angular frequency is 2. The usual
frequency in Hertz (cycles per second) is
2
2π
=
1
π
≈ 0.318
The general solution is
x(t) = Acos 2t + Bsin 2t.
Letting x(0) = 0 means A = 0.5. Then x
·
(t) = −0.5 sin 2t + Bcos 2t. Letting x
·
(0) = 1 we get
B = 1. Therefore, the amplitude is C =
√
A
2
+ B
2
=
√
1.25 ≈ 1.118. The solution is
x(t) = 0.5 cos 2t + sin 2t.
A plot is shown in Figure 2.2 on the following page.
For the free undamped motion, if the solution is of the form
x(t) = Acos ω
0
t + Bsin ω
0
t,
this corresponds to the initial conditions x(0) = A and x
·
(0) = B.
This makes it much easier to figure out A and B, rather than the amplitude and phase shift. In the
example, we have already found C. Let us compute the phase shift. We know that tan γ = B/A = 2.
We take the arctangent of 2 and get approximately 1.107. Unfortunately if you remember, we still
66 CHAPTER 2. HIGHER ORDER LINEAR ODES2: Simple undamped oscillation.
need to check if this γ is in the right quadrant. Since both B and A are positive, then γ should be in
the first quadrant, and 1.107 radians really is in the first quadrant.
Note: Many calculators and computer software do not only have the atan function for arctan-
gent, but also what is sometimes called atan2. This function takes two arguments, B and A and
returns a γ in the correct quadrant for you.
2.4.3 Free damped motion
Let us now focus on damped motion. Let us rewrite the equation
mx
··
+ cx
·
+ kx = 0,
as
x
··
+ 2px
·
+ ω
2
0
x = 0,
where
ω
0
=
p
2
− ω
2
0
.
The form of the solution depends on whether we get complex or real roots and this depends on the
sign of
p
2
− ω
2
0
=
c
2m
2
−
k
m
=
c
2
− 4km
4m
2
.
2.4. MECHANICAL VIBRATIONS 67
The sign of p
2
− ω
2
0
is the same as the sign of c
2
− 4km.
Overdamping
When c
2
−4km > 0, we say the system is overdamped. In this case, there are two distinct real roots
r
1
and r
2
. Notice that both are negative, as
.
It is not hard to see that this satisfies the initial conditions.
Critical damping
When c
2
− 4km = 0, we say the system is critically damped. In this case, there is one root of
multiplicity 2 and this root is −p. Therefore, our solution is
x(t) = C
1
e
−pt
+ C
2
te
−pt
.
The behavior of a critically damped system is very similar to an overdamped system. After all a
critically damped system is in some sense a limit of overdamped systems. Since these equations
are really only an approximation to the real world, in reality we are never critically damped, it
is only a place you can reach in theory. You are always a little bit underdamped or a little bit
overdamped. It is better not to dwell on critical damping.
68 CHAPTER 2. HIGHER ORDER LINEAR ODES
Underdamping
When c
2
−4km < 0, we say the system is un-
0 5 10 15 20 25 30
0 5 10 15 20 25 30
-1.0
-0.5
0.0
0.5
1.0
-1.0
-0.5
0.0
0.5
1.0
Figure 2.4: Underdamped motion with the en-
velope curves shown.
derdamped. In this case, the roots are complex.
r = −p ±
p
2
− ω
2
0
= −p ±
√
−1
ω
2
0
− p
2
= −p ± iω
1
,
where ω
1
=
ω
2
0
− p
2
. Our solution is
x(t) = e
−pt
(Acos ω
1
t + Bsin ω
1
t) ,
Or
x(t) = Ce
−pt
cos(ω
1
t − γ).
An example plot is given in Figure 2.4. Note that
we still have that x(t) → 0 as t → ∞.
In the figure we also show the envelope curves Ce
−pt
and −Ce
−pt
. The solution is the oscillating
plot between the two curves. The envelope curves give the maximum amplitude of the oscillation
at any given point in time. For example if you are bungee jumping, you are really interested in
computing the envelope curve so that you do not hit the concrete with your head.
The phase shift γ just shifts the graph left or right but within the envelope curves (the envelope
curves do not change of course if γ changes).
Finally note that the angular pseudo-frequency (we do not call it a frequency since the solution
is not really a periodic function) ω
1
becomes lower when the damping c (and hence p) becomes
larger. This makes sense since if we keep changing c at some point the solution should start looking
like the solution for critical damping or overdamping which do not oscillate at all. When we change
the damping just a little bit, we do not expect the behavior to change dramatically.
On the other hand when c becomes smaller, ω
1
approaches ω
0
(it is always smaller) and the
solution looks more and more like the steady periodic motion of the undamped case. The envelope
curves become flatter and flatter as p goes to 0.
2.4.4 Exercises
Exercise 2.4.2: Consider a mass and spring system with a mass m = 2, spring constant k = 3,
and damping constant c = 1. a) Set up and find the general solution of the system. b) Is the system
underdamped, overdamped or critically damped? c) If the system is not critically damped, find a c
which makes the system critically damped.
Exercise 2.4.3: Do Exercise 2.4.2 for m = 3, k = 12, and c = 12.
2.4. MECHANICAL VIBRATIONS 69
Exercise 2.4.4: Using the mks units (meters-kilograms-seconds) Suppose you have a spring of
with spring constant 4N/m. You want to use it to weight items. Assume no friction. Suppose you
you place the mass on the spring and put it in motion. a) You count and find that the frequency is
0.8 Hz (cycles per second) what is the mass. b) Find a formula for the mass m given the frequency
ω in Hz.
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not
know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your
setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you
measured 0.8 Hz, for the 2 kg weight you measured 0.39 Hz. a) Find k (spring constant) and c
(damping constant). b) Find a formula for the mass in terms of the frequency in Hz. c) For an
unknown mass you measured 0.2 Hz, what is the weight?
70 CHAPTER 2. HIGHER ORDER LINEAR ODES
2.5 Nonhomogeneous equations
Note: 2 lectures, §3.5 in EP
2.5.1 Solving nonhomogeneous equations
You have seen how to solve the linear constant coefficient homogeneous equations. Now suppose
that we drop the requirement of homogeneity. This usually corresponds to some outside input to
the system we are trying to model, like the forcing function for the mechanical vibrations of last
section. That is, we have an equation such as
y
··
+ 5y
·
+ 6y = 2x + 1. (2.6)
Note that we still say this equation is constant coefficient equation. We only require constants in
front of the y
··
, y
·
, and y.
We will generally write Ly = 2x + 1 instead when the operator is not important. The way we
solve (2.6) is as follows. We find the general solution y
c
to the associated homogeneous equation
y
··
+ 5y
·
+ 6y = 0. (2.7)
We also find a single particular solution y
p
to (2.6) in some way and then we know that
y = y
c
+ y
p
is the general solution to (2.6). We call y
c
the complementary solution.
Note that y
p
can be any solution. Suppose you find a different particular solution ˜ y
p
. Then write
the difference w = y
p
− ˜ y
p
. Then plug w into the left hand side of the equation and get
w
··
+ 5w
·
+ 6w = (y
··
p
+ 5y
·
p
+ 6y
p
) − (˜ y
··
p
+ 5˜ y
·
p
+ 6˜ y
p
) = (2x + 1) − (2x + 1) = 0.
In other words, using the operator notation the calculation becomes simpler. Note that L is a linear
operator and so we could just write.
Lw = L(y
p
− ˜ y
p
) = Ly
p
− L˜ y
p
= (2x + 1) − (2x + 1) = 0.
So w = y
p
− ˜ y
p
is a solution to (2.7). So any two solutions of (2.6) differ by a solution to the
homogeneous equation (2.7). The solution y = y
c
+y
p
includes all solutions to (2.6), since y
c
is the
general solution to the homogeneous equation.
Moral of the story is that you can find the particular solution in any old way, and you might
find a different one by a different method (or by guessing) and still get the right general solution
to the whole problem even if it looks different and the constants you will have to choose given the
initial conditions will be different.
2.5. NONHOMOGENEOUS EQUATIONS 71
2.5.2 Undetermined coefficients
So the trick is to somehow in a smart way guess a solution to (2.6). Note that 2x+1 is a polynomial,
and the left hand side of the equation will be a polynomial if we let y be a polynomial of the same
degree. So we will try
y = Ax + B.
we plug in
y
··
+ 5y
·
+ 6y = (Ax + B)
··
+ 5(Ax + B)
·
+ 6(Ax + B) = 0 + 5A + 6Ax + 6B = 6Ax + (5A + 6B).
So 6Ax + (5A + 6B) = 2x + 1. So A =
1
3
and B =
−1
9
. That means that y
p
=
1
3
x −
1
9
=
3x−1
9
. Solving
the complementary problem we get (Exercise!)
y
c
= C
1
e
−2x
+ C
2
e
−3x
.
Hence the general solution to (2.6) is
y = C
1
e
−2x
+ C
2
e
−3x
+
3x − 1
9
.
Now suppose we are further given some initial conditions y(0) = 0 and y
·
(0) =
1
3
. First find
y
·
= −2C
1
e
−2x
− 3C
2
e
−3x
+
1
3
Then
0 = y(0) = C
1
+ C
2
−
1
9
1
3
= y
·
(0) = −2C
1
− 3C
2
+
1
3
We solve to get C
1
=
1
3
and C
2
=
−2
9
. Hence our solution is
y(x) =
1
3
e
−2x
−
2
9
e
−3x
+
3x − 1
9
=
3e
−2x
− 2e
−3x
+ 3x − 1
9
.
Exercise 2.5.1: Check that y really solves the equation.
Note: A common mistake is to solve for constants using the initial conditions with y
c
and only
adding the particular solution y
p
after that. That will not work. You need to first compute y = y
c
+y
p
and only then solve for the constants using the initial conditions.
Similarly a right hand side consisting of exponentials or sines and cosines can be handled. For
example:
y
··
+ 2y
·
+ 2y = cos 2x
Let us just find y
p
in this case. We notice that we may have to also guess sin 2x since derivatives
of cosine are sines. So we guess
y = Acos 2x + Bsin 2x.
72 CHAPTER 2. HIGHER ORDER LINEAR ODES
Plug in to the equation and we get
−4Acos 2x − 4Bsin 2x − 4Asin 2x + 4Bcos 2x + 2Acos 2x + 2Bsin 2x = cos 2x.
Since the left hand side must equal to right hand side we group terms and we get that −4A + 4B +
2A = 1 and −4B−4A+2B = 0. So −2A+4B = 1 and 2A+B = 0 and hence A =
−1
10
and B =
1
5
. So
y
p
=
−cos 2x + 2 sin 2x
10
.
And in a similar way if the right hand side contains exponentials we guess exponentials. For
example, if the equation is (where L is a linear constant coefficient operator)
Ly = e
3x
we will guess y = Ae
3x
. We note also that using the multiplication rule for differentiation gives us
a way to combine these guesses. Really if you can guess a form for y such that Ly has all the terms
needed to for the right hand side, that is a good place to start. For example for:
Ly = (1 + 3x
2
) e
−x
cos πx
we will guess
y = (A + Bx + Cx
2
) e
−x
cos πx + (D + Ex + Fx
2
) e
−x
sin πx.
We will plug in and then hopefully get equations that we can solve for A, B, C, D, E, F. As you can
see this can make for a very long and tedious calculation very quickly. C'est la vie!
There is one hiccup in all this. It could be that our guess actually solves the associated homo-
geneous equation. That is, suppose we have
y
··
− 9y = e
3x
.
We would love to guess y = Ae
3x
, but if we plug this into the left hand side of the equation we get
y
··
− 9y = 9Ae
3x
− 9Ae
3x
= 0 e
3x
.
There is no way we can choose A to make the left hand side be e
3x
. The trick in this case is to
multiply our guess by x until we get rid of duplication with the complementary solution. That is
first we compute y
c
(solution to Ly = 0)
y
c
= C
1
e
−3x
+ C
2
e
3x
and we note that the e
3x
term is a duplicate with our desired guess. We modify our guess to
y = Axe
3x
and notice there is no duplication. Now we can go forward and try it. Note that
y
·
= Ae
3x
+ 3Axe
3x
and y
··
= 4Ae
3x
+ 9Axe
3x
. So
y
··
− 9y = 4Ae
3x
+ 9Axe
3x
− 9Axe
3x
= 4Ae
3x
2.5. NONHOMOGENEOUS EQUATIONS 73
Then we note that this is supposed to be e
3x
and hence we find that 4A = 1 and so A =
1
4
. Thus we
can now write the general solution as
y = y
c
+ y
p
= C
1
e
−3x
+ C
2
e
3x
+
1
4
xe
3x
.
Now what about the case when multiplying by x does not get rid of duplication. For example,
y
··
− 6y
·
+ 9 = e
3x
.
Note that y
c
= C
1
e
3x
+C
2
xe
3x
. So guessing y = Axe
3x
would not get us anywhere. In this case you
want to guess y = Ax
2
e
3x
. Basically, you want to multiply your guess by x until all duplication is
gone. But no more! Multiplying too many times will also make the process not work.
Finally what if the right hand side is several terms, such as
Ly = e
2x
+ cos x.
In this case find u that solves Lu = e
2x
and v that solves Lv = cos x (do each terms separately).
Then we note that if y = u + v, then Ly = e
2x
+ cos x. This is because L is linear and this is just
superposition again. We have Ly = L(u + v) = Lu + Lv = e
2x
+ cos x.
See Edwards and Penney [EP] for more detailed and complete information on undetermined
coefficients.
2.5.3 Variation of parameters
It turns out that undetermined coefficients will work for many basic problems that crop up. It does
not work all the time. Really it only works when the right hand side of the equation Ly = f (x) has
only finitely many linearly independent derivatives, so that you can write a guess that consists of
them all. But some equations are a bit tougher. Consider
y
··
+ y = tan x.
Note that each new derivative of tan x looks completely different and cannot be written as a linear
combination of the previous derivatives. We get sec
2
x, 2 sec
2
x tan x, etc. . .
This equation calls for a different method. We present the method of variation of parameters
which will handle all the cases Ly = f (x) provided you can solve certain integrals. For simplicity
we will restrict ourselves to second order equations, but the method will work for higher order
equations just as well (but the computations will be more tedious).
Let us try to solve the example.
Ly = y
··
+ y = tan x.
74 CHAPTER 2. HIGHER ORDER LINEAR ODES
First we find the complementary solution Ly = 0. This is reasonably simple we get y
c
= C
1
y
1
+C
2
y
2
where y
1
= cos x and y
2
= sin x. Now to try to find a solution to the nonhomogeneous equation we
will try
y
p
= y = u
1
y
1
+ u
2
y
2
,
where u
1
and u
2
are functions and not constants. We are trying to satisfy Ly = tan x. That gives us
one condition on the functions u
1
and u
2
. First compute (note the product rule!)
y
·
= (u
·
1
y
1
+ u
·
2
y
2
) + (u
1
y
·
1
+ u
2
y
·
2
).
Since we can still impose at our will to simplify computations (we have two unknown functions,
so we are allowed two conditions), we impose that (u
·
1
y
1
+ u
·
2
y
2
) = 0. This makes computing the
second derivative easier.
y
·
= u
1
y
·
1
+ u
2
y
·
2
,
y
··
= (u
·
1
y
·
1
+ u
·
2
y
·
2
) + (u
1
y
··
1
+ u
2
y
··
2
).
Now since y
1
and y
2
are solutions to y
··
+ y = 0, we know that y
··
1
= −y
1
and y
··
2
= −y
2
. (Note: If
the equation was instead y
··
+ ay
·
+ by = 0 we would have y
··
i
= −ay
·
i
− by
i
.)
So
y
··
= (u
·
1
y
·
1
+ u
·
2
y
·
2
) − (u
1
y
1
+ u
2
y
2
).
Now note that
y
··
= (u
·
1
y
·
1
+ u
·
2
y
·
2
) − y,
and hence
y
··
+ y = Ly = u
·
1
y
·
1
+ u
·
2
y
·
2
.
For y to satisfy Ly = f (x) we must have f (x) = u
·
1
y
·
1
+ u
·
2
y
·
2
.
So what we need to solve are the two equations (conditions) we imposed on u
1
and u
2
u
·
1
y
1
+ u
·
2
y
2
= 0,
u
·
1
y
·
1
+ u
·
2
y
·
2
= f (x).
We can now solve for u
·
1
and u
·
2
in terms of f (x), y
1
and y
2
. You will always get these formulas for
any Ly = f (x). There is a general formula for the solution you can just plug into, but it is better to
just repeat what we do below. In our case the two equations become
u
·
1
cos x + u
·
2
sin x = 0,
−u
·
1
sin x + u
·
2
cos x = tan x.
Hence
u
·
1
cos x sin x + u
·
2
sin
2
x = 0,
−u
·
1
sin x cos x + u
·
2
cos
2
x = tan x cos x = sin x.
2.5. NONHOMOGENEOUS EQUATIONS 75
And thus
u
·
2
(sin
2
x + cos
2
x) = sin x,
u
·
2
= sin x,
u
·
1
=
−sin
2
x
cos
x = −tan x sin x.
Now we need to integrate u
·
1
and u
·
2
to get u
1
and u
2
.
u
1
=
.
2.5.4 Exercises
Exercise 2.5.2: Find a particular solution of y
··
− y
·
− 6y = e
2x
.
Exercise 2.5.3: Find a particular solution of y
··
− 4y
·
+ 4y = e
2x
.
Exercise 2.5.4: Solve the initial value problem y
··
+ 9y = cos 3x + sin 3x for y(0) = 2, y
·
(0) = 1.
Exercise 2.5.5: Setup the form of the particular solution but do not solve for the coefficients for
y
(4)
− 2y
···
+ y
··
= e
x
.
Exercise 2.5.6: Setup the form of the particular solution but do not solve for the coefficients for
y
(4)
− 2y
···
+ y
··
= e
x
+ x + sin x.
Exercise 2.5.7: a) Using variation of parameters find a particular solution of y
··
−2y
·
+y = e
x
. b)
Find a particular solution using undetermined coefficients. c) Are the two solutions you found the
same? What is going on?
Exercise 2.5.8: Find a particular solution of y
··
− 2y
·
+ y = sin x
2
. It is OK to leave the answer as
a definite integral.
76 CHAPTER 2. HIGHER ORDER LINEAR ODES
2.6 Forced oscillations and resonance
Note: 2 lectures, §3.6 in EP
Before reading the lecture, it may be good to first try Project III from the IODE website:
Let us return back to the mass on a spring example. We will
damping c
m
k
F(t)
now consider the case of forced oscillations. That is, we will con-
sider the equation
mx
··
+ cx
·
+ kx = F(t)
for some nonzero F(t). In the mass on a spring example, the setup is again, m is mass, c is friction,
k is the spring constant and F(t) is an external force acting on the mass.
Usually what we are interested in is some periodic forcing, such as noncentered rotating parts,
or perhaps even loud sounds or other sources of periodic force. Once we will learn about Fourier
series we will see that we will essentially cover every type of periodic function by considering
F(t) = F
0
cos ωt (or sin instead of cosine, the calculations will be essentially the same).
2.6.1 Undamped forced motion and resonance
First let us consider undamped (c = 0) motion as this is simpler. We have the equation
mx
··
+ kx = F
0
cos ωt.
This has the complementary solution (solution to the associated homogeneous equation)
x
c
= C
1
cos ω
0
t + C
2
sin ω
0
t,
where ω
0
=
k
m
. ω
0
is said to be the natural frequency (angular). It is essentially the frequency at
which the system "wants to oscillate" without external interference.
Let us suppose that ω
0
ω. Now try the solution x
p
= Acos ωt and solve for A. Note that we
need not have sine in our trial solution as on the left hand side we will only get cosines anyway. If
you include a sine it is fine; you will find that its coefficient will be zero (I cannot find a rhyme).
So we solve as in the method of undetermined coefficients with the guess above and we find
that
x
p
=
F
0
m(ω
2
0
− ω
2
)
cos ωt.
We leave it as an exercise to do the algebra required here.
The general solution is
x = C
1
cos ω
0
t + C
2
sin ω
0
t +
F
0
m(ω
2
0
− ω
2
)
cos ωt.
2.6. FORCED OSCILLATIONS AND RESONANCE 77
or written another way
x = C cos(ω
0
t − γ) +
F
0
m(ω
2
0
− ω
2
)
cos ωt.
Hence it is a superposition of two cosine waves at different frequencies.
Example 2.6.1: Suppose
0.5x
··
+ 8x = 10 cos πt
and let us suppose that x(0) = 0 and x
·
(0) = 0.
Well let us compute. First we read off the parameters: ω = π, ω
0
=
.
Notice that x is now a high frequency wave mod-
ulated by a low frequency wave.
Now suppose that ω
0
= ω. Obviously in this
case we cannot try the solution Acos ωt and use
undetermined coefficients. In this case we see
that cos ωt solves the homogeneous equation. Therefore, we need to try x
p
= At cos ωt +Bt sin ωt.
This time we need the sin term since two derivatives of t cos ωt do contain sines. We write the
equation
x
··
+ ω
2
x =
F
0
m
cos ωt
Then plugging into the left hand side we get
2Bωcos ωt − 2Aωsin ωt =
F
0
m
cos ωt
78 CHAPTER 2. HIGHER ORDER LINEAR ODES
Hence A = 0 and B =
F
0
2mω
. Our particular solution is
F
0
2mω
t sin ωt and our general solution is
x = C
1
cos ωt + C
2
sin ωt +
F
0
2mω
t sin ωt.
The important term is the last one (the par-6: Graph of
1
π
t sin πt.
ticular solution we found). We can see that this
term grows without bound as t → ∞. In fact
it oscillates between
F
0
t
2mω
and
−F
0
t
2mω
. The first two
terms only oscillate between ±
C
2
1
+ C
2
2
, which
becomes smaller and smaller in proportion to the
oscillations of the last term as t gets larger. In
Figure 2.6 we see the graph with C
1
= C
2
= 0,
F
0
= 2, m = 1, ω = π.
By forcing the system in just the right fre-
quency we produce very wild oscillations. This
kind of behavior is called resonance or some-
times pure resonance and is sometimes desired.
For example, remember when as a kid you could
start swinging by just moving back and forth on
the swing seat in the correct "frequency"? You were trying to achieve resonance. The force of each
one of your moves was small but after a while it produced large swings.
On the other hand resonance can be destructive. After an earthquake some buildings are col-
lapsed and others may be relatively undamaged. This is due to different buildings having different
resonance frequencies. So figuring out the resonance frequency can be very important.
A common (but wrong) example of destructive force of resonance is the Tacoma Narrows
bridge failure. It turns out, there was an altogether different phenomenon at play there
∗
.
2.6.2 Damped forced motion and practical resonance
Of course in real life things are not as simple as they were above. There is of course some damping.
That is our equation becomes
mx
··
+ cx
·
+ kx = F
0
cos ωt, (2.8)
for some c > 0. We have solved the homogeneous problem before. We let
p =
c
2m
ω
0
=
p
2
− ω
2
0
. The form of the general solution of the associated homogeneous equation depends
on the sign of p
2
− ω
2
0
, or equivalently on the sign of c
2
− 4km, as we have seen before. That is
x
c
=
ω
2
0
− p
2
. In any case, we can see that x
c
(t) → 0 as t → ∞. Furthermore, there can
be no conflicts when trying to solve for the undetermined coefficients by trying x
p
= Acos ωt +
Bsin ωt. Let us plug in and solve for A and B. We get (the tedious details are left to reader)
(2ωp)
2
+ (ω
2
0
− ω
2
)
2
cos(ωt − γ).
If ω = ω
0
we see that A = 0, B = C =
F
0
2mωp
and γ = π/2.
The exact formula is not as important as the idea. You should not memorize the above formula,
you should remember the ideas involved. Even if you change the right hand side a little bit you
will get a different formula with different behavior. So there is no point in memorizing this specific
formula. You can always recompute it later or look it up if you really need it.
For reasons we will explain in a moment we will call x
c
the transient solution and denote it by
x
tr
and we will call the x
p
we found above the steady periodic solution and denote it by x
sp
. The
general solution to our problem is
x = x
c
+ x
p
= x
tr
+ x
sp
.
We note that x
c
= x
tr
goes to zero as t → ∞as7: Solutions with different initial con-
ditions for parameters k = 1, m = 1, F
0
= 1,
c = 0.7, and ω = 1.1.
all the terms involve an exponential with a nega-
tive exponent. Hence for large t, the effect of x
tr
is negligible and we will essentially only see x
sp
.
Notice that x
sp
involves no arbitrary constants,
and the initial conditions will only affect x
tr
. This
means that the effect of the initial conditions will
be negligible after some period of time. Hence
the name transient. Because of this behavior, we
might as well focus on the steady periodic solu-
tion and ignore the transient solution. See Fig-
ure 2.7 for a graph of different initial conditions.
Notice that the speed at which x
tr
goes to zero
depends on p (and hence c). The bigger p is (the
bigger c is), the "faster" x
tr
becomes negligible.
So the smaller the damping, the longer the "tran-
sient region." This agrees with the observation
that when c = 0, the initial conditions affect the behavior for all time (i.e. an infinite "transient
region").
Let us describe what do we mean by resonance when damping is present. Since there were no
conflicts when solving with undetermined coefficient, there is no term that goes to infinity. What
we will look at however is the maximum value of the amplitude of the steady periodic solution. Let
C be the amplitude of x
sp
. If we plot C as a function of ω (with all other parameters fixed) we can
find its maximum. This maximum is said to be practical resonance (we call the ω that achieves
this maximum the practical resonance frequency). A sample plot for three different values of c
2.6. FORCED OSCILLATIONS AND RESONANCE 81
is given in Figure 2.8. As you can see the practical resonance amplitude grows as damping gets
smaller, and any practical resonance can disappear when damping is large.
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
0.5
1.0
1.5
2.0
2.5
0.0
0.5
1.0
1.5
2.0
2.5
Figure 2.8: Graph of C(ω) showing practical resonance with parameters k = 1, m = 1, F
0
= 1.
The top line is with c = 0.4, the middle line with c = 0.8, and the bottom line with c = 1.6.
To find the maximum it turns out we need to find the derivative C
·
(ω). This is easily computed
to be
C
·
(ω) =
−4ω(2p
2
+ ω
2
− ω
2
0
)F
0
m
(2ωp)
2
+ (ω
2
0
− ω
2
)
2
3/2
.
This is zero either when ω = 0 or when 2p
2
+ ω
2
− ω
2
0
= 0. In other words when
ω =
ω
2
0
− 2p
2
or 0
It can be shown that if ω
2
0
− 2p
2
is positive then
ω
2
0
− 2p
2
is the practical resonance frequency
(that is the point where C(ω) is maximal, note that in this case C
·
(ω) > 0 for small ω). If ω = 0 is
the maximum, then essentially there is no practical resonance since we assume that ω > 0 in our
system. In this case the amplitude gets larger as the forcing frequency gets smaller.
If practical resonance occurs, the frequency is smaller than ω
0
. As damping c (and hence p)
becomes smaller, the closer the practical resonance frequency comes to ω
0
. So when damping
is very small, ω
0
is a good estimate of the resonance frequency. This behavior agrees with the
observation that when c = 0, ω
0
is the resonance frequency.
The behavior will be more complicated if the forcing function is not an exact cosine wave, but
for example a square wave. It will be good to come back to this section once you have learned
about the Fourier series.
82 CHAPTER 2. HIGHER ORDER LINEAR ODES
2.6.3 Exercises
Exercise 2.6.1: Derive a formula for x
sp
if the equation is mx
··
+ cx
·
+ kx = F
0
sin ωt. Assume
c > 0.
Exercise 2.6.2: Derive a formula for x
sp
if the equation is mx
··
+cx
·
+kx = F
0
cos ωt +F
1
cos 3ωt.
Assume c > 0.
Exercise 2.6.3: Take mx
··
+ cx
·
+ kx = F
0
cos ωt. Fix m > 0 and k > 0. Now think of the function
C(ω). For what values of c (solve in terms of m, k, and F
0
will there be no practical resonance (for
what values of c is there no maximum of C(ω) for ω > 0).
Exercise 2.6.4: Take mx
··
+ cx
·
+ kx = F
0
cos ωt. Fix c > 0 and k > 0. Now think of the function
C(ω). For what values of m (solve in terms of c, k, and F
0
will there be no practical resonance (for
what values of m is there no maximum of C(ω) for ω > 0).
Exercise 2.6.5: Suppose a water tower in an earthquake acts as a mass-spring system. Assume
that the container on top is full and the water does not move around. The container then acts as a
mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that
the container with water has a mass of m =10,000 kg. It takes a force of 1000 newtons to displace
the container 1 meter. For simplicity assume no friction.
Suppose that an earthquake induces an external force F(t) = mAω
2
cos ωt.
a) What is the natural frequency of the water tower.
b) If ω is not the natural frequency, find a formula for the amplitude of the resulting oscillations
of the water container.
c) Suppose A = 1 and an earthquake with frequency 0.5 cycles per second comes. What is the
amplitude of the oscillations. Suppose that if the water tower moves more than 1.5 meter, the tower
collapses. Will the tower collapse?
Chapter 3
Systems of ODEs
3.1 Introduction to systems of ODEs
Note: 1 lecture, §4.1 in EP
Often we do not have just one dependent variable and one equation. And as we will see, we
may end up with systems of several equations and several dependent variables even if we start with
a single equation.
If we have several dependent variables, suppose y
1
, y
2
, . . . , y
n
we can have a differential equa-
tion involving all of them and their derivatives. For example, y
··
1
= f (y
·
1
, y
·
2
, y
1
, y
2
, x). Usually,
when we have two dependent variables we would have two equations such as
y
··
1
= f
1
(y
·
1
, y
·
2
, y
1
, y
2
, x),
y
··
2
= f
2
(y
·
1
, y
·
2
, y
1
, y
2
, x),
for some functions f
1
and f
2
. We call the above a system of differential equations. More precisely,
it is a second order system. Sometimes a system is easy to solve by solving for one variable and
then for the second variable.
Example 3.1.1: Take the first order system
y
·
1
= y
1
,
y
·
2
= y
1
− y
2
,
with initial conditions of the form y
1
(0) = 1, y
2
(0) = 2.
We note that y
1
= C
1
e
x
is the general solution of the first equation. We can then plug this y
1
into the second equation and get the equation y
·
2
= C
1
e
x
− y
2
, which is a linear first order equation
that is easily solved for y
2
. By the method of integrating factor we get
e
x
y
2
=
C
1
2
e
2x
+ C
2
,
83
84 CHAPTER 3. SYSTEMS OF ODES
or y
2
=
C
1
2
e
x
+ C
2
e
−x
. The general solution to the system is, therefore,
y
1
= C
1
e
x
,
y
2
=
C
1
2
e
x
+ C
2
e
−x
.
We can now solve for C
1
and C
2
given the initial conditions. We substitute x = 0 and find that
C
1
= 1 and C
2
=
3
2
.
Generally, we will not be so lucky to be able to solve like in the first example, and we will have
to solve for all variables at once.
As an example application, let us think of mass and spring sys-
k
m
2
m
2
tems again. Suppose we have one spring with constant k but two
masses m
1
and m
2
. We can think of the masses as carts, and we
will suppose that they ride along with no friction. Let x
1
be the
displacement of the first cart and x
2
be the displacement of the second cart. That is, we put the two
carts somewhere with no tension on the spring, and we mark the position of the first and second
cart and call those the zero position. That is, x
1
= 0 is a different position on the floor than the
position corresponding to x
2
= 0. The force exerted by the spring on the first cart is k(x
2
−x
1
), since
x
2
− x
1
is how far the string is stretched (or compressed) from the rest position. The force exerted
on the second cart is the opposite, thus the same thing with a negative sign. Using Newton's second
law, we note that force equals mass times acceleration.
m
1
x
··
1
= k(x
2
− x
1
),
m
2
x
··
2
= −k(x
2
− x
1
).
In this system we cannot solve for the x
1
variable separately. That we must solve for both x
1
and x
2
at once is intuitively obvious, since where the first cart goes depends exactly on where the
second cart goes and vice versa.
Before we talk about how to handle systems, let us note that in some sense we need only
consider first order systems. Take an n
th
order differential equation
y
(n)
= F(y
(n−1)
, . . . . , y
·
, y, x).
Define new variables u
1
, . . . , u
n
and write the system
u
·
1
= u
2
u
·
2
= u
3
.
.
.
u
·
n−1
= u
n
u
·
n
= F(u
n
, u
n−1
, . . . , u
2
, u
1
, x).
3.1. INTRODUCTION TO SYSTEMS OF ODES 85
Now try to solve this system for u
1
, u
2
, . . . , u
n
. Once you have solved for the u's, you can discard
u
2
through u
n
and let y = u
1
. We note that this y solves the original equation.
A similar process can be done for a system of higher order differential equations. For example,
a system of k differential equations in k unknowns, all of order n, can be transformed into a first
order system of n × k equations and n × k unknowns.
Example 3.1.2: Sometimes we can use this idea in reverse as well. Let us take the system
x
·
= 2y − x, y
·
= x,
where the independent variable is t. We wish to solve for the initial conditions x(0) = 1, y(0) = 0.
We first notice that if we differentiate the first equation once we get y
··
= x
·
and now we know
what x
·
is in terms of x and y.
y
··
= x
·
= 2y − x = 2y − y
·
.
So we now have an equation y
··
+ y
·
− 2y = 0. We know how to solve this equation and we find
that y = C
1
e
−2t
+ C
2
e
t
. Once we have y we can plug in to get x.
x = y
·
= −2C
1
e
−2t
+ C
2
e
t
.
We solve for the initial conditions 1 = x(0) = −2C
1
+C
2
and 0 = y(0) = C
1
+C
2
. Hence, C
1
= −C
2
and 1 = 3C
2
. So C
1
=
−1
3
and C
2
=
1
3
. Our solution is:
x =
2e
−2t
+ e
t
3
, y =
−e
−2t
+ e
t
3
.
Exercise 3.1.1: Plug in and check that this really is the solution.
It is useful to go back and forth between systems and higher order equations for other reasons.
For example, the ODE approximation methods are generally only given as solutions for first order
systems. It is not very hard to adapt the code for the Euler method for a first order equation to first
order systems. We essentially just treat the dependent variable not as a number but as a vector. In
many mathematical computer languages there is almost no distinction in syntax.
In fact, this is what IODE was doing when you had it solve a second order equation numerically
in the IODE Project III if you have done that project.
The above example was what we will call a linear first order system, as none of the dependent
variables appear in any functions or with any higher powers than one. It is also autonomous as the
equations do not depend on the independent variable t.
For autonomous systems we can easily draw the so-called direction field or vector field. That
is, a plot similar to a slope field, but instead of giving a slope at each point, we give a direction (and
a magnitude). The previous example x
·
= 2y − x, y
·
= x says that at the point (x, y) the direction
in which we should travel to satisfy the equations should be the direction of the vector (2y − x, x)
with the speed equal to the magnitude of this vector. So we draw the vector (2y − x, x) based at the
86 CHAPTER 3. SYSTEMS OF ODES
point (x, y) and we do this for many points on the xy-plane. We may want to scale down the size
of our vectors to fit many of them on the same direction field. See Figure 3.1.
We can now draw a path of the solution in the plane. That is, suppose the solution is given by
x = f (t), y = g(t), then we can pick an interval of t (say 0 ≤ t ≤ 2 for our example) and plot all
the points ( f (t), g(t)) for t in the selected range. The resulting picture is usually called the phase
portrait (or phase plane portrait). The particular curve obtained we call the trajectory or solution
curve. An example plot is given in Figure 3.2. In this figure the line starts at (1, 0) and travels
along the vector field for a distance of 2 units of t. Since we solved this system precisely we can
compute x(2) and y(2). We get that x(2) ≈ 2.475 and y(2) ≈ 2.457. This point corresponds to the
top right end of the plotted solution curve in the figure.
-1 0 1 2 3
-1 0 1 2 3
-1
0
1
2
3
-1
0
1
2
3
Figure 3.1: The direction field for x
·
= 2y − x,
y
·
= x.
-1 0 1 2 3
-1 0 1 2 3
-1
0
1
2
3
-1
0
1
2
3
Figure 3.2: The direction field for x
·
= 2y − x,
y
·
= x with the trajectory of the solution start-
ing at (1, 0) for 0 ≤ t ≤ 2.
Notice the similarity to the diagrams we drew for autonomous systems in one dimension. But
now note how much more complicated things become if we allow just one more dimension.
Also note that we can draw phase portraits and trajectories in the xy-plane even if the system is
not autonomous. In this case however we cannot draw the direction field, since the field changes
as t changes. For each t we would get a different direction field.
3.1.1 Exercises
Exercise 3.1.2: Find the general solution of x
·
1
= x
2
− x
1
+ t, x
·
2
= x
2
.
Exercise 3.1.3: Find the general solution of x
·
1
= 3x
1
− x
2
+ e
t
, x
·
2
= x
1
.
Exercise 3.1.4: Write ay
··
+ by
·
+ cy = f (x) as a first order system of ODEs.
Exercise 3.1.5: Write x
··
+y
2
y
·
−x
3
= sin(t), y
··
+(x
·
+y
·
)
2
−x = 0 as a first order system of ODEs.
3.2. MATRICES AND LINEAR SYSTEMS 87
3.2 Matrices and linear systems
Note: 1 and a half lectures, first part of §5.1 in EP
3.2.1 Matrices and vectors
Before we can start talking about linear systems of ODEs, we will need to talk about matrices, so
let us review these briefly. A matrix is an m × n array of numbers (m rows and n columns). For
example, we denote a 3 × 5 matrix as follows
A =
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
a
11
a
12
a
13
a
14
a
15
a
21
a
22
a
23
a
24
a
25
a
31
a
32
a
33
a
34
a
35
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
.
By a vector we will usually mean a column vector which is an n × 1 matrix. If we mean a row
vector we will explicitly say so (a row vector is a 1 × n matrix). We will usually denote matrices
by upper case letters and vectors by lower case letters with an arrow such as x or
b. By
0 we will
mean the vector of all zeros.
It is easy to define some operations on matrices. Note that we will want 1 ×1 matrices to really
act like numbers, so our operations will have to be compatible with this viewpoint.
First, we can multiply by a scalar (a number). This means just multiplying each entry by the
same number. For example,
2
,
1 2 3
4 5 6
¸
=
,
2 4 6
8 10 12
¸
.
Matrix addition is also easy. We add matrices element by element. For example,
,
1 2 3
4 5 6
¸
+
,
1 1 −1
0 2 4
¸
=
,
2 3 2
4 7 10
¸
.
If the sizes do not match, then addition is not defined.
If we denote by 0 the matrix of with all zero entries, by c, d some scalars, and by A, B, C some
matrices, we have the following familiar rules.
A + 0 = A = 0 + A,
A + B = B + A,
(A + B) + C = A + (B + C),
c(A + B) = cA + cB,
(c + d)A = cA + dA.
88 CHAPTER 3. SYSTEMS OF ODES
Another operation which is useful for matrices is the so-called transpose. This operation just
swaps rows and columns of a matrix. The transpose of A is denoted by A
T
. Example:
,
1 2 3
4 5 6
¸
T
=
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
1 4
2 5
3 6
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
3.2.2 Matrix multiplication
Next let us define matrix multiplication. First we define the so-called dot product (or inner product)
of two vectors. Usually this will be a row vector multiplied with a column vector of the same size.
For the dot product we multiply each pair of entries from the first and the second vector and we
sum these products. The result is a single number. For example,
,
a
1
a
2
a
3
¸
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
b
1
b
2
b
3
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
= a
1
b
1
+ a
2
b
2
+ a
3
b
3
.
And similarly for larger (or smaller) vectors.
Armed with the dot product we can define the product of matrices. First let us denote by
row
i
(A) the i
th
row of A and by column
j
(A) the j
th
column of A. Now for an m×n matrix A and an
n × p matrix B we can define the product AB. We let AB be an m × p matrix whose i j
th
entry is
row
i
(A) column
j
(B).
Note that the sizes must match. Example:
,
1 2 3
4 5 6
¸
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
1 0 −1
1 1 1
1 0 0
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
=
=
,
1 1 + 2 1 + 3 1 1 0 + 2 1 + 3 0 1 (−1) + 2 1 + 3 0
4 1 + 5 1 + 6 1 4 0 + 5 1 + 6 0 4 (−1) + 5 1 + 6 0
¸
=
,
6 2 1
15 5 1
¸
For multiplication we will want an analogue of a 1. Here we use the so-called identity matrix.
The identity matrix is a square matrix with 1s on the main diagonal and zeros everywhere else. It
is usually denoted by I. For each size we have a different matrix and so sometimes we may denote
the size as a subscript. For example, the I
3
would be the 3 × 3 identity matrix
I = I
3
=
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
1 0 0
0 1 0
0 0 1
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
.
3.2. MATRICES AND LINEAR SYSTEMS 89
We have the following rules for matrix multiplication. Suppose that A, B, C are matrices of the
correct sizes so that the following make sense. c is some scalar (number).
A(BC) = (AB)C,
A(B + C) = AB + AC,
(B + C)A = BA + CA,
c(AB) = (cA)B = A(cB),
IA = A = AI.
A few warnings are in order however.
(i) AB BA in general (it may be true by fluke sometimes). That is, matrices do not commute.
(ii) AB = AC does not necessarily imply B = C even if A is not 0.
(iii) AB = 0 does not necessarily mean that A = 0 or B = 0.
For the last two items to hold we would need to essentially "divide" by a matrix. This is where
matrix inverse comes in. Suppose that A is an n×n matrix and that there exists another n×n matrix
B such that
AB = I = BA.
Then we call B the inverse of A and we denote B by A
−1
. If the inverse of A exists, then we call A
invertible. If A is not invertible we say A is singular or just say it is not invertible.
If A is invertible, then AB = AC does imply that B = C (the inverse is unique). We just
multiply both sides by A
−1
to get A
−1
AB = A
−1
AC or IB = IC or B = C. It is also not hard to see
that (A
−1
)
−1
= A.
3.2.3 The determinant
We can now talk about determinants of square matrices. We define the determinant of a 1 × 1
matrix as the value of its own entry. For a 2 × 2 matrix we define
det
¸,
a b
c d
¸
= ad − bc.
Before trying to compute determinant for larger matrices, let us first note the meaning of the
determinant. Consider an n × n matrix as a mapping of R
n
to R
n
. For example, a 2 × 2 matrix A
is a mapping of the plane where x gets sent to Ax. Then the determinant of A is then the factor
by which the volume of objects gets changed. For example, if we take the unit square (square of
sides 1) in the plane, then A takes the square to a parallelogram of area |det(A)|. The sign of det(A)
denotes changing of orientation (if the axes got flipped). For example,
A =
,
1 1
−1 1
¸
.
90 CHAPTER 3. SYSTEMS OF ODES
Then det(A) = 1 + 1 = 2. Now let us see where the square with vertices (0, 0), (1, 0), (0, 1) and
(1, 1) gets sent. Obviously (0, 0) gets sent to (0, 0). Now
,
1 1
−1 1
¸ ,
1
0
¸
=
,
1
−1
¸
,
,
1 1
−1 1
¸ ,
0
1
¸
=
,
1
1
¸
,
,
1 1
−1 1
¸ ,
1
1
¸
=
,
2
0
¸
.
So it turns out that the image of the square is another square. This one has a side of length
√
2 and
is therefore of area 2.
If you think back to high school geometry, you may have seen a formula for computing the
area of a parallelogram with vertices (0, 0), (a, c), (b, d) and (a + b, c + d). And it is precisely
det
¸,
a b
c d
¸
.
The vertical lines here mean absolute value. The matrix
,
a b
c d
¸
carries the unit square to the given
parallelogram.
Now we can define the determinant for larger matrices. We define A
i j
as the matrix A with the
i
th
row and the j
th
column deleted. To compute the determinant of a matrix, pick one row, say the
i
th
row and compute.
det(A) =
n
¸
j=1
(−1)
i+j
a
i j
det(A
i j
).
For example, for the first row we get
det(A) = a
11
det(A
11
) − a
12
det(A
12
) + a
13
det(A
13
) −
¸
¸
¸
¸
¸
¸
+a
1n
det(A
1n
) if n is odd,
−a
1n
det(A
1n
) if n even.
We alternately add and subtract the determinants of the submatrices A
i j
for a fixed i and all j. For
example, for a 3×3 matrix, picking the first row, we would get det(A) = a
11
det(A
11
)−a
12
det(A
12
)+
a
13
det(A
13
). For example,
det
b is the vector
,
2
5
10
,
. The solution can be also computed with the
inverse,
x = A
−1
Ax = A
−1
b.
One last note to make about linear systems of equations is that it is possible that the solution is
not unique (or that no solution exists). It is easy to tell if a solution does not exist. If during the
row reduction you come up with a row where all the entries except the last one are zero (the last
entry in a row corresponds to the right hand side of the equation) the system is inconsistent and
has no solution. For example if for a system of 3 equations and 3 unknowns you find a row such
as [ 0 0 0 1 ] in the augmented matrix, you know the system is inconsistent.
You generally try to use row operations until the following conditions are satisfied. The first
nonzero entry in each row is called the leading entry.
(i) There is only one leading entry in each column.
(ii) All the entries above and below a leading entry are zero.
(iii) All leading entries are 1.
Such a matrix is said to be in reduced row echelon form. The variables corresponding to columns
with no leading entries are said to be free variables. Free variables mean that we can pick those
variables to be anything we want and then solve for the rest of the unknowns.
Example 3.2.1: The following augmented matrix is in reduced row echelon form.
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
1 2 0 3
0 0 1 1
0 0 0 0
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
If the variables are named x
1
, x
2
, and x
3
, then x
2
is the free variable and x
1
= 3 − 2x
2
and x
3
= 1.
On the other hand if during the row reduction process you come up with the matrix
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
1 2 13 3
0 0 1 1
0 0 0 3
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
,
there is no need to go further. The last row corresponds to the equation 0x
1
+ 0x
2
+ 0x
3
= 3 which
is preposterous. Hence, no solution exists.
94 CHAPTER 3. SYSTEMS OF ODES
3.2.5 Computing the inverse
If the coefficient matrix is square and there exists a unique solution x to Ax =
b for any
b, then A
is invertible. In fact by multiplying both sides by A
−1
you can see that x = A
−1
b. So it is useful to
compute the inverse, if you want to solve the equation for many different right hand sides
f be a linear system of ODEs. Suppose x
p
is one particular solution.
Then every solution can be written as
x = x
c
+ x
p
,
where x
c
is a solution to the associated homogeneous equation (x
·
= Px).
So the procedure will be exactly the same. We find a particular solution to the nonhomogeneous
equation, then we find the general solution to the associated homogeneous equation and we add
the two.
Alright, suppose you have found the general solution x
·
= Px +
f . Now you are given an
initial condition of the form x(t
0
) =
b for some constant vector
b. Now suppose that X(t) is
3.3. LINEAR SYSTEMS OF ODES 97
the fundamental matrix solution of the associated homogeneous equation (i.e. columns of X are
solutions). The general solution is written as
x(t) = X(t)c + x
p
(t).
Then we are seeking a vector c such that
f (t).
Exercise 3.3.2: a) Verify that the system x
·
=
,
1 3
3 1
¸
x has the two solutions
,
1
1
¸
e
4t
and
,
1
−1
¸
e
−2t
. b)
Write down the general solution. c) Write down the general solution in the form x
1
=?, x
2
=? (i.e.
write down a formula for each element of the solution).
Exercise 3.3.3: Verify that
,
1
1
¸
e
t
and
,
1
−1
¸
e
t
are linearly independent. Hint: Just plug in t = 0.
Exercise 3.3.4: Verify that
,
1
1
0
,
e
t
and
,
1
−1
1
,
e
t
and
,
1
−1
1
,
e
2t
are linearly independent. Hint: You must
be a bit more tricky than in the previous exercise.
Exercise 3.3.5: Verify that
,
t
t
2
¸
and
,
t
3
t
4
¸
are linearly independent.
3.4. EIGENVALUE METHOD 99
3.4 Eigenvalue method
Note: 2 lectures, §5.2 in EP
In this section we will learn how to solve linear homogeneous constant coefficient systems of
ODEs by the eigenvalue method. Suppose you have a linear constant coefficient homogeneous
system
x
·
= Px.
Now suppose we try to adapt the method for single constant coefficient equations by trying the
function e
λt
. However, x is a vector. So we try ve
λt
, where v is an arbitrary constant vector. We
plug into the equation to get
λve
λt
= Pve
λt
.
We divide by e
λt
and notice that we are looking for a λ and v that satisfy the equation
λv = Pv.
To solve this equation we need a little bit more linear algebra which we review now.
3.4.1 Eigenvalues and eigenvectors of a matrix
Let A be a square constant matrix. Suppose there is a scalar λ and a nonzero vector v such that
Av = λv.
We then call λ an eigenvalue of A and v is called the corresponding eigenvector.
Example 3.4.1: The matrix
,
2 1
0 1
¸
has an eigenvalue of λ = 2 with the corresponding eigenvector
,
1
0
¸
because
,
2 1
0 1
¸ ,
1
0
¸
=
,
2
0
¸
= 2
,
1
0
¸
.
If we rewrite the equation for an eigenvalue as
(A − λI)v =
0.
We notice that this has a nonzero solution v only if A − λI is not invertible. Were it invertible, we
could write (A−λI)
−1
(A−λI)v = (A−λI)
−1
0 which implies v =
0. Therefore, A has the eigenvalue
λ if and only if λ solves the equation
det(A − λI) = 0.
Note that this means that we will be able to find an eigenvalue without finding the correspond-
ing eigenvector. The eigenvector will have to be found later, once λ is known.
100 CHAPTER 3. SYSTEMS OF ODES
Example 3.4.2: Find all eigenvalues of
,
2 1 1
1 2 0
0 0 2
,
.
We write
det
0.
It is obvious that the equations iv
1
+ v
2
= 0 and −v
1
+ iv
2
= 0 are multiples of each other. So we
only need to consider one of them. After picking v
2
= 1, for example, we have the eigenvector
v =
,
i
1
¸
. In similar fashion we find that
,
−i
1
¸
is an eigenvector corresponding to the eigenvalue 1+i.
We could write the solution as
x = c
1
,
i
1
¸
e
(1−i)t
+ c
2
,
−i
1
¸
e
(1+i)t
=
,
c
1
ie
(1−i)t
− c
2
ie
(1+i)t
c
1
e
(1−i)t
+ c
2
e
(1+i)t
1
¸
.
But then we would need to look for complex values c
1
and c
2
to solve any initial conditions. And
even then it is perhaps not completely clear that we get a real solution. We could use Euler's
formula here and do the whole song and dance we did before, but we will do something a bit
smarter first.
We claim that we did not have to look for the second eigenvector (nor for the second eigen-
value). All complex eigenvalues come in pairs (because the matrix P is real).
First a small side note. The real part of a complex number z can be computed as
z+¯ z
2
, where the
bar above z means a + ib = a − ib. This operation is called the complex conjugate. Note that for
a real number a, ¯ a = a. Similarly we can bar whole vectors or matrices. If a matrix P is real then
P = P. We note that Px = Px = Px. Or
(P − λI)v = (P −
¯
λI)v.
So if v is an eigenvector corresponding to eigenvalue a +ib, then v is an eigenvector corresponding
to eigenvalue a − ib.
Now suppose that a + ib is a complex eigenvalue of P, v the corresponding eigenvector and
hence
x
1
= ve
(a+ib)t
3.4. EIGENVALUE METHOD 103
is a solution (complex valued) of x
·
= Px. Then note that e
a+ib
= e
a−ib
and hence
x
2
= x
1
= ve
(a−ib)t
is also a solution. Now take the function
x
3
= Re x
1
= Reve
(a+ib)t
=
x
1
+ x
1
2
=
x
1
+ x
2
2
.
Is also a solution. And it is real valued! Similarly as Imz =
z−¯ z
2i
is the imaginary part we find that
x
4
= Imx
1
=
x
1
− x
2
2i
.
is also a real valued solution. It turns out that x
3
and x
4
are linearly independent.
Returning to our problem, we take
x
1
=
,
i
1
¸
e
(1−i)t
=
,
i
1
¸
e
t
cos t + ie
t
sin t
=
,
ie
t
cos t − e
t
sin t
e
t
cos t + ie
t
sin t
¸
It is easy to see that
Re x
1
=
,
−e
t
sin t
e
t
cos t
¸
,
Imx
1
=
,
e
t
cos t
e
t
sin t
¸
,
are the solutions we seek.
Exercise 3.4.4: Check that these really are solutions.
The general solution is
x = c
1
,
−e
t
sin t
e
t
cos t
¸
+ c
2
,
e
t
cos t
e
t
sin t
¸
=
,
−c
1
e
t
sin t + c
2
e
t
cos t
c
1
e
t
cos t + c
2
e
t
sin t
¸
.
This solution is real valued for real c
1
and c
2
. Now we can solve for any initial conditions that we
have.
The process is this. When you have complex eigenvalues, you notice that they always come in
pairs. You take one λ = a + ib from the pair, you find the corresponding eigenvector v. You note
that Reve
(a+ib)t
and Imve
(a+ib)t
are also solutions to the equation, are real valued and are linearly
independent. You go on to the next eigenvalue which is either a real eigenvalue or another complex
eigenvalue pair. Hence, you will end up with n linearly independent solutions if you had n distinct
eigenvalues (real or complex).
You can now find a real valued general solution to any homogeneous system where the matrix
has distinct eigenvalues. When you have repeated eigenvalues, matters get a bit more complicated
and we will look at that situation in §3.7.
104 CHAPTER 3. SYSTEMS OF ODES
3.4.4 Exercises
Exercise 3.4.5: Let A be an 3 ×3 matrix with an eigenvalue of 3 and a corresponding eigenvector
v =
,
1
−1
3
,
. Find Av.
Exercise 3.4.6: a) Find the general solution of x
·
1
= 2x
1
, x
·
2
= 3x
2
using the eigenvalue method
(first write the system in the form x
·
= Ax). b) Solve the system by solving each equation separately
and verify you get the same general solution.
Exercise 3.4.7: Find the general solution of x
·
1
= 3x
1
+ x
2
, x
·
2
= 2x
1
+ 4x
2
using the eigenvalue
method.
Exercise 3.4.8: Find the general solution of x
·
1
= x
1
− 2x
2
, x
·
2
= 2x
1
+ x
2
using the eigenvalue
method. Do not use complex exponentials in your solution.
Exercise 3.4.9: a) Compute eigenvalues and eigenvectors of A =
,
9 −2 −6
−8 3 6
10 −2 −6
,
. b) Find the general
solution of x
·
= Ax.
Exercise 3.4.10: Compute eigenvalues and eigenvectors of
,
−2 −1 −1
3 2 1
−3 −1 0
,
.
3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 105
3.5 Two dimensional systems and their vector fields
Note: 1 lecture, should really be in EP §5.2, but is in EP §6.2
Let us take a moment to talk about homogeneous systems in the plane. We want to think about
how the vector fields look and how this depends on the eigenvalues. So we have a 2 × 2 matrix P
and the system
,
x
y
¸
·
= P
,
x
y
¸
. (3.2)
We will be able to visually tell how the vector field looks once we find the eigenvalues and eigen-
vectors of the matrix.
Case 1. Suppose that the eigenvalues are real3: Eigenvectors of P.
and positive. Find the two eigenvectors and plot
them in the plane. For example, take the matrix
,
1 1
0 2
¸
. The eigenvalues are 1 and 2 and the corre-
sponding eigenvectors are
,
1
0
¸
and
,
1
1
¸
. See Fig-
ure 3.3.
Now suppose that x and y are on the line de-
termined by an eigenvector v for an eigenvalue λ.
That is,
,
x
y
¸
= av for some scalar a. Then
,
x
y
¸
·
= P
,
x
y
¸
= P(av) = a(Pv) = aλv.
The derivative is a multiple of v and hence points
along the line determined by v. As λ > 0, the
derivative points in the direction of v when a is
positive and in the opposite direction when a is
negative. Let us draw arrows on the lines to indicate the directions. See Figure 3.4 on the following
page.
We fill in the rest of the arrows and we also draw a few solutions. See Figure 3.5 on the next
page. You will notice that the picture looks like a source with arrows coming out from the origin.
Hence we call this type of picture a source or sometimes an unstable node.
Case 2. Suppose both eigenvalues were negative. For example, take the negation of the matrix
in case 1,
,
−1 −1
0 −2
¸
. The eigenvalues are −1 and −2 and the corresponding eigenvectors are the same,
,
1
0
¸
and
,
1
1
¸
. The calculation and the picture are almost the same. The only difference is that the
eigenvalues are negative and hence all arrows are reversed. We get the picture in Figure 3.6 on the
following page. We call this kind of picture a sink or sometimes a stable node.
Case 3. Suppose one eigenvalue is positive and one is negative. For example the matrix
,
1 1
0 −2
¸
.
The eigenvalues are 1 and −2 and the corresponding eigenvectors are the same,
,
1
0
¸
and
,
1
−3
¸
. We
106 CHAPTER 3. SYSTEMS OF4: Eigenvectors of P with directions5: Example source6: Example sink7: Example saddle vector field with
eigenvectors and solutions.
reverse the arrows on one line (corresponding to the negative eigenvalue) and we obtain the picture
in Figure 3.7. We call this picture a saddle point.
The next three cases we will assume the eigenvalues are complex. In this case the eigenvectors
are also complex and we cannot just plot them on the plane.
Case 4. Suppose the eigenvalues are purely imaginary. That is, suppose the eigenvalues are
±ib. For example, let P =
,
0 1
−4 0
¸
. The eigenvalues turn out to be ±2i and the eigenvectors are
,
1
2i
¸
and
,
1
−2i
¸
. We take the eigenvalue 2i and its eigenvector
,
1
2i
¸
and note that the real an imaginary
3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 107
parts of ve
i2t
are
Re
,
1
2i
¸
e
i2t
=
,
cos 2t
−2 sin 2t
¸
Im
,
1
2i
¸
e
i2t
=
,
sin 2t
2 cos 2t
¸
.
Note that which combination of them we take just depends on the initial conditions. So we might
as well just take the real part. If you notice this is a parametric equation for an ellipse. Same with
the imaginary part and in fact any linear combination of them. It is not difficult to see that this is
what happens in general when the eigenvalues are purely imaginary. So when the eigenvalues are
purely imaginary, you get ellipses for your solutions. This type of picture is sometimes called a
center. See Figure 3.88: Example center vector field9: Example spiral source vector field.
Case 5. Now the complex eigenvalues have positive real part. That is, suppose the eigenvalues
are a ± ib for some a > 0. For example, let P =
,
1 1
−4 1
¸
. The eigenvalues turn out to be 1 ± 2i and
the eigenvectors are
,
1
2i
¸
and
,
1
−2i
¸
. We take 1 + 2i and its eigenvector
,
1
2i
¸
and find the real and
imaginary of ve
(1+2i)t
are
Re
,
1
2i
¸
e
(1+2i)t
= e
t
,
cos 2t
−2 sin 2t
¸
Im
,
1
2i
¸
e
(1+2i)t
= e
t
,
sin 2t
2 cos 2t
¸
.
Now note the e
t
in front of the solutions. This means that the solutions grow in magnitude while
spinning around the origin. Hence we get a spiral source. See Figure 3.9.
108 CHAPTER 3. SYSTEMS OF ODES
Case 6. Finally suppose the complex eigenvalues have negative real part. That is, suppose the
eigenvalues are −a ± ib for some a > 0. For example, let P =
,
−1 −1
4 −1
¸
. The eigenvalues turn out
to be −1 ± 2i and the eigenvectors are
,
1
−2i
¸
and
,
1
2i
¸
. We take −1 − 2i and its eigenvector
,
1
2i
¸
and
find the real and imaginary of ve
(1+2i)t
are
Re
,
1
2i
¸
e
(−1−2i)t
= e
−t
,
cos 2t
2 sin 2t
¸
,
Im
,
1
2i
¸
e
(−1−2i)t
= e
−t
,
−sin 2t
2 cos 2t
¸
.
Now note the e
−t
in front of the solutions. This means that the solutions shrink in magnitude while
spinning around the origin. Hence we get a spiral sink. See Figure 3.1010: Example spiral sink vector field.
We summarize the behavior of linear homogeneous two dimensional systems in Table 3.1.
Eigenvalues Behavior
real and both positive source / unstable node
real and both negative sink / stable node
real and opposite signs saddle
purely imaginary center point / ellipses
complex with positive real part spiral source
complex with negative real part spiral sink
Table 3.1: Summary of behavior of linear homogeneous two dimensional systems.
3.5. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS 109
3.5.1 Exercises
Exercise 3.5.1: Take the equation mx
··
+cx
·
+kx = 0, with m > 0, c ≥ 0, k > 0 for the mass-spring
system. a) Convert this to a system of first order equations. b) Classify for what m, c, k do you
get which behavior. c) Can you explain from physical intuition why you do not get all the different
kinds of behavior here?
Exercise 3.5.2: Can you find what happens in the case when P =
,
1 1
0 1
¸
. In this case the eigenvalue
is repeated and there is only one eigenvector. What picture does this look like?
Exercise 3.5.3: Can you find what happens in the case when P =
,
1 1
1 1
¸
. Does this look like any of
the pictures we have drawn?
110 CHAPTER 3. SYSTEMS OF ODES
3.6 Second order systems and applications
Note: more than 2 lectures, §5.3 in EP
3.6.1 Undamped mass spring systems
While we did say that we will usually only look at first order systems, it is sometimes more con-
venient to study the system in the way it arises naturally. For example, suppose we have 3 masses
connected by springs between two walls. We could pick any higher number, and the math would
be essentially the same, but for simplicity we pick 3 right now. And let us assume no friction,
that is, the system is undamped. The masses are m
1
, m
2
, and m
3
and the spring constants are k
1
,
k
2
, k
3
, and k
4
. Let x
1
be the displacement from rest position of the first mass and, x
2
and x
3
the
displacement of the second and third mass. We will make, as usual, positive values go right (as x
1
grows, mass 1 is moving right). See Figure 3.11.
k
1
m
1
k
2
m
2
k
3
m
3
k
4
Figure 3.11: System of masses and springs.
This simple system turns up in unexpected places. Note for example that our world really
consists of small particles of matter interacting together. When we try this system with many more
masses, this is a good approximation to how an elastic material will behave. In fact by somehow
taking a limit of the number of masses going to infinity we obtain the continuous one dimensional
wave equation. But we digress.
Let us set up the equations for the three mass system. By Hooke's law we have that the force
acting on the mass equals the spring compression times the spring constant. By Newton's second
law we again have that force is mass times acceleration. So if we sum the forces acting on each
mass and put the right sign in front of each depending on the direction in which it is acting, we end
up with the system.
m
1
x
··
1
= −k
1
x
1
+ k
2
(x
2
− x
1
) = −(k
1
+ k
2
)x
1
+ k
2
x
2
,
m
2
x
··
2
= −k
2
(x
2
− x
1
) + k
3
(x
3
− x
2
) = k
2
x
1
− (k
2
+ k
3
)x
2
+ k
3
x
3
,
m
3
x
··
3
= −k
3
(x
3
− x
2
) − k
4
x
3
= k
3
x
2
− (k
3
+ k
4
)x
3
.
We define the matrices
M =
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
m
1
0 0
0 m
2
0
0 0 m
3
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
and K =
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
−(k
1
+ k
2
) k
2
0
k
2
−(k
2
+ k
3
) k
3
0 k
3
−(k
3
+ k
4
)
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
.
3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 111
We write the equation simply as
Mx
··
= Kx.
At this point we could introduce 3 new variables and write out a system of 6 equations. We claim
this simple setup is easier to handle as a second order system. We will call x the displacement
vector, M the mass matrix, and K the stiffness matrix.
Exercise 3.6.1: Do this setup for 4 masses (find the matrix M and K). Do it for 5 masses. Can
you find a prescription to do it for n masses?
As before we will want to "divide by M." In this case this means computing the inverse of M.
All the masses are nonzero and it is easy to compute the inverse, as the matrix is diagonal.
M
−1
=
,
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
1
m
1
0 0
0
1
m
2
0
0 0
1
m
3
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
.
This fact follows readily by how we multiply diagonal matrices. You should verify that MM
−1
=
M
−1
M = I as an exercise.
We let A = M
−1
K and we look at the system x
··
= M
−1
Kx, or
x
··
= Ax.
Many real world systems can be modeled by this equation. For simplicity we will keep the given
masses-and-springs setup in mind. We try a solution of the form
x = ve
αt
.
We note that for this guess, x
··
= α
2
ve
αt
. We plug into the equation and get
α
2
ve
αt
= Ave
αt
.
We can divide by e
αt
to get that α
2
v = Av. Hence if α
2
is an eigenvalue of A and v is the corre-
sponding eigenvector, we have found a solution.
In our example, and in many others, it turns out that A has negative real eigenvalues (and
possibly a zero eigenvalue). So we will study only this case here. When an eigenvalue λ is negative,
it means that α
2
= λ is negative. Hence there is some real number ω such that −ω
2
= λ. Then
α = ±iω. The solution we guessed was
x = v(cos ωt + i sin ωt).
By again taking real and imaginary parts (note that v is real), we again find that v cos ωt and
v sin ωt are linearly independent solutions.
If an eigenvalue was zero, it turns out that v and vt are solutions if v is the corresponding
eigenvector.
112 CHAPTER 3. SYSTEMS OF ODES
Exercise 3.6.2: Show that if A has a zero eigenvalue and v is the corresponding eigenvector, then
x = v(a + bt) is a solution of x
··
= Ax for arbitrary constants a and b.
Theorem 3.6.1. Let A be an n×n with n distinct real negative eigenvalues we denote by −ω
2
1
, −ω
2
2
,
. . . , −ω
2
n
, and corresponding eigenvectors v
1
, v
2
, . . . , v
n
. Then
x(t) =
n
¸
i=1
v
i
(a
i
cos ω
i
t + b
i
sin ω
i
t),
is the general solution of
x
··
= Ax,
for some arbitrary constants a
i
and b
i
. If A has a zero eigenvalue and all other eigenvalues are
distinct and negative, that is ω
1
= 0, then the general solution becomes
x(t) = v
1
(a
1
+ b
1
t) +
n
¸
i=2
v
i
(a
i
cos ω
i
t + b
i
sin ω
i
t).
Now note that we can use this solution and the setup from the introduction of this section even
when some of the masses and springs are missing. Simply when there are say 2 masses and only 2
springs, take only the equations for the two masses and set all the spring constants that are missing
to zero.
3.6.2 Examples
Example 3.6.1: Suppose we have the system in Figure 3.12, with m
1
= 2, m
2
= 1, k
1
= 4, and
k
2
= 2.
k
1
m
1
k
2
m
2
Figure 3.12: System of masses and springs.
The equations we write down are
,
2 0
0 1
¸
x
··
=
,
−(4 + 2) 2
2 −2
¸
x.
or
x
··
=
,
−3 1
2 −2
¸
x.
3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 113
We find the eigenvalues of A to be λ = −1, −4 (exercise). Now we find the eigenvectors to be
,
1
2
¸
and
,
1
−1
¸
respectively (exercise).
We check the theorem and note that ω
1
= 1 and ω
2
= 2. Hence the general solution is
x =
,
1
2
¸
(a
1
cos t + b
1
sin t) +
,
1
−1
¸
(a
2
cos 2t + b
2
sin 2t) .
The two terms in the solution represent the two so-called natural or normal modes of oscilla-
tion. And the two (angular) frequencies are the natural frequencies. The two modes are plotted in
Figure 3.13.
0.0 2.5 5.0 7.5 10.0
0.0 2.5 5.0 7.5 10.0
-2
-1
0
1
2
-2
-1
0
1
2 3.13: The two modes of the mass spring system. In the left plot the masses are moving in
unison and the right plot are masses moving in the opposite direction.
Let us write the solution as
x =
,
1
2
¸
c
1
cos(t − α
2
) +
,
1
−1
¸
c
2
cos(2t − α
1
).
The first term,
x
1
=
,
1
2
¸
c
1
cos(t − α
1
) =
,
c
1
cos(t − α
1
)
2c
1
cos(t − α
1
)
¸
,
corresponds to the mode where the masses move synchronously in the same direction.
On the other hand the second term,
x
1
=
,
1
−1
¸
c
2
cos(2t − α
2
) =
,
c
2
cos(2t − α
2
)
−c
2
cos(2t − α
2
)
¸
,
corresponds to the mode where the masses move synchronously but in opposite directions.
The general solution is a combination of the two modes. That is, the initial conditions determine
the amplitude and phase shift of each mode.
114 CHAPTER 3. SYSTEMS OF ODES
Example 3.6.2: Let us do another example. In this example we have two toy rail cars. Car 1 of
mass 2 kg is travelling at 3 m/s towards the second rail car of mass 1 kg. There is a bumper on the
second rail car which engages one the cars hit (it connects to two cars) and does not let go. The
bumper acts like a spring of spring constant k = 2 N/m. The second Car is 10 meters from a wall.
See Figure 3.14.
m
1
k
m
2
10 meters
Figure 3.14: The crash of two rail cars.
We want to ask several question. At what time after the cars link does impact with the wall
happen? What is the speed of car 2 when it hits the wall?
OK, let us first set the system up. Let us assume that time t = 0 is the time when the two cars
link up. Let x
1
be the displacement of the first car from the position at t = 0, and let x
2
be the
displacement of the second car from its original location. Then the time when x
2
(t) = 10 is exactly
the time when impact with wall occurs. For this t, x
·
2
(t) is the speed at impact. This system acts
just like the system of the previous example but without k
1
. Hence the equation is
,
2 0
0 1
¸
x
··
=
,
−2 2
2 −2
¸
x.
or
x
··
=
,
−1 1
2 −2
¸
x.
We compute the eigenvalues of A. It is not hard to see that the eigenvalues are 0 and −3
(exercise). Furthermore, the eigenvectors are
,
1
1
¸
and
,
1
−2
¸
respectively (exercise). We note that
ω
2
=
√
3 and we use the second part of the theorem to find our general solution to be
x =
,
1
1
¸
(a
1
+ b
1
t) +
,
1
−2
¸
0 = x(0) =
,
a
1
+ a
2
a
1
− 2a
2
¸
.
3.6. SECOND ORDER SYSTEMS AND APPLICATIONS 115
It is not hard to see that this implies that a
1
= a
2
= 0. We plug a
1
and a
2
and differentiate to get
x
·
(t) =
,
b
1
+
√
3 b
2
cos
√
3 t
b
1
− 2
√
3 b
2
cos
√
3 t
¸
.
So
,
3
0
¸
= x
·
(0) =
,
b
1
+
√
3 b
2
b
1
− 2
√
3 b
2
¸
.
It is not hard to solve these two equations to find b
1
= 2 and b
2
=
1
√
3
. Hence the position of our
cars is (until the impact with the wall)
x =
,
¸
¸
¸
¸
¸
¸
2t +
1
√
3
sin
√
3 t
2t −
2
√
3
sin
√
3 t
¸
¸
¸
¸
¸
¸
¸
.
Note how the presence of the zero eigenvalue resulted in a term containing t. This means that the
carts will be travelling in the positive direction as time grows, which is what we expect.
What we are really interested in is the second expression, the one for x
2
. We have x
2
(t) =
2t −
2
√
3
sin
√
3 t. See Figure 3.15 for the plot of x
2
versus time.
0 1 2 3 4 5 6
0 1 2 3 4 5 6
0.0
2.5
5.0
7.5
10.0
12.5
0.0
2.5
5.0
7.5
10.0
12.5
Figure 3.15: Position of the second car in time (ignoring the wall).
Just from the graph we can see that time of impact will be a little more than 5 seconds from
time zero. For this you have to solve the equation 10 = x
2
(t) = 2t −
2
√
3
sin
√
3 t. Using a computer
(or even a graphing calculator) we find that t
impact
≈ 5.22 seconds.
As for the speed we note that x
·
2
= 2 − 2 cos
√
3 t. At time of impact (5.22 seconds from t = 0)
we get that x
·
2
(t
impact
) ≈ 3.85.
The maximum speed is the maximum of 2 − 2 cos
√
3 t which is 4. We are travelling at almost
the maximum speed when we hit the wall.
116 CHAPTER 3. SYSTEMS OF ODES
Now suppose that Bob is a tiny person sitting on car 2. Bob has a Martini in his hand and would
like to not spill it. Let us suppose Bob would not spill his martini when the first car links up with
car 2, but if car 2 hits the wall at any speed greater than zero, Bob will spill his drink. Suppose
Bob can move the car 2 a few meters back and forth from the wall (he cannot go all the way to the
wall, nor can he get out of the way of the first car). Is there a "safe" distance for him to be in? A
distance such that the impact with the wall is at zero speed?
Actually, the answer is yes. From looking at Figure 3.15 on the preceding page, we note the
"plateau" between t = 3 and t = 4. There is a point where the speed is zero. We just need to
solve x
·
2
(t) = 0. This is when cos
√
3 t = 1 or in other words when t =
2π
√
3
,
4π
√
3
, etc. . . If we plug in
x
2
2π
√
3
=
4π
√
3
≈ 7.26. So a "safe" distance is about 7 and a quarter meters from the wall.
Alternatively Bob could move away from the wall towards the incoming car 2 where another
safe distance is
8π
√
3
≈ 14.51 and so on, using all the different t such that x
·
2
(t) = 0. Of course t = 0
is always a solution here, corresponding to x
2
= 0, but that means standing right at the wall.
3.6.3 Forced oscillations
Finally we move to forced oscillations. Suppose that now our system is
x
··
= Ax +
F cos ωt. (3.3)
That is, we are adding periodic forcing to the system in the direction of the vector
F.
Just like before this system just requires us to find one particular solution x
p
, add it to the
general solution of the associated homogeneous system x
c
and we will have the general solution to
(3.3). Let us suppose that ω is not one of the natural frequencies of x
··
= Ax, then we can guess
x
p
= c cos ωt,
where c is an unknown constant vector. Note that we do not need to use sine since there are only
second derivatives. We solve for c to find x
p
. This is really just the method of undetermined
coefficients for systems. Let us differentiate x
p
twice to get
x
p
··
= −ω
2
c cos ωt.
Now plug into the equation
−ω
2
c cos ωt = Ac cos ωt +
d cos ωt.
That is assuming that all eigenvalues of the coefficient matrix are distinct. Note that the amplitude
of this solution grows without bound as t grows.
118 CHAPTER 3. SYSTEMS OF ODES
3.6.4 Exercises
Exercise 3.6.3: Find a particular solution to
x
··
=
,
−3 1
2 −2
¸
x +
,
0
2
¸
cos 2t.
Exercise 3.6.4: Let us take the example in Figure 3.12 on page 112 with the same parameters as
before: m
1
= 2, k
1
= 4, and k
2
= 2, except for m
2
which is unknown. Suppose that there is a force
cos 5t acting on the first mass. Find an m
1
such that there exists a particular solution where the
first mass does not move.
Note: This idea is called dynamic damping. In practice there will be a small amount of damp-
ing and so any transient solution will disappear and after long enough time, the first mass will
always come to a stop.
Exercise 3.6.5: Let us take the example 3.6.2 on page 114, but that at time of impact, cart 2 is
moving to the left at the speed of 3m/s. a) Find the behavior of the system after linkup. b) Will the
second car hit the wall, or will it be moving away from the wall as time goes on. c) at what speed
would the first car have to be travelling for the system to essentially stay in place after linkup.
Exercise 3.6.6: Let us take the example in Figure 3.12 on page 112 with parameters m
1
= m
2
= 1,
k
1
= k
2
= 1. Does there exist a set of initial conditions for which the first cart moves but the second
cart does not? If so find those conditions, if not argue why not.
3.7. MULTIPLE EIGENVALUES 119
3.7 Multiple eigenvalues
Note: 1–2 lectures, §5.4 in EP
It may very well happen that a matrix has some "repeated" eigenvalues. That is, the character-
istic equation det(A − λI) = 0 may have repeated roots. As we have said before, this is actually
unlikely to happen for a random matrix. If you take a small perturbation of A (you change the
entries of A slightly) you will get a matrix with distinct eigenvalues. As any system you will want
to solve in practice is an approximation to reality anyway, it is not indispensable to know how to
solve these corner cases. But it may happen on occasion that it is easier or desirable to solve such
a system directly.
3.7.1 Geometric multiplicity
Take the diagonal matrix
A =
,
3 0
0 3
¸
.
A has an eigenvalue 3 of multiplicity 2. We usually call the multiplicity of the eigenvalue in the
characteristic equation the algebraic multiplicity. In this case, there exist 2 linearly independent
eigenvectors,
,
1
0
¸
and
,
0
1
¸
. This means that the so-called geometric multiplicity of this eigenvalue
is 2.
In all the theorems where we required a matrix to have n distinct eigenvalues, we only really
needed to have n linearly independent eigenvectors. For example, let x
·
= Ax has the general
solution
x = c
1
,
1
0
¸
e
3t
+ c
2
,
0
1
¸
e
3t
.
Let us restate the theorem about real eigenvalues. In the following theorem we will repeat eigen-
values according to (algebraic) multiplicity. So for A above we would say that it has eigenvalues 3
and 3.
Theorem 3.7.1. Take x
·
= Px. If P is n × n and has n real eigenvalues (not necessarily distinct),
λ
1
, . . . , λ
n
, and if there are n linearly independent corresponding eigenvectors v
1
, . . . , v
n
, and the
general solution to the ODE can be written as.
x = c
1
v
1
e
λ
1
t
+ c
2
v
2
e
λ
2
t
+ + c
n
v
n
e
λ
n
t
.
The geometric multiplicity of an eigenvalue of algebraic multiplicity n is equal to the number of
linearly independent eigenvectors we can find. It is not hard to see that the geometric multiplicity is
always less than or equal to the algebraic multiplicity. Above we, therefore, handled the case when
these two numbers are equal. If the geometric multiplicity is equal to the algebraic multiplicity we
say the eigenvalue is complete.
120 CHAPTER 3. SYSTEMS OF ODES
The hypothesis of the theorem could, therefore, be stated as saying that if all the eigenvalues
of P are complete then there are n linearly independent eigenvectors and thus we have the given
general solution.
Note that if the geometric multiplicity of an eigenvalue is 2 or greater, then the set of linearly
independent eigenvectors is not unique up to multiples as it was before. For example, for the
diagonal matrix A above we could also pick eigenvectors
,
1
1
¸
and
,
1
−1
¸
, or in fact any pair of two
linearly independent vectors.
3.7.2 Defective eigenvalues
If an n × n matrix has less than n linearly independent eigenvectors, it is said to be deficient.
Then there is at least one eigenvalue with algebraic multiplicity that is higher than the geometric
multiplicity. We call this eigenvalue defective and the difference between the two multiplicities we
call the defect.
Example 3.7.1: The matrix
,
3 1
0 3
¸
has an eigenvalue 3 of algebraic multiplicity 2. Let us try to compute the eigenvectors.
,
0 1
0 0
¸ ,
v
1
v
2
¸
=
0.
We must have that v
2
= 0. Hence any eigenvector is of the form
,
v
1
0
¸
. Any two such vectors are
linearly dependent, and hence the geometric multiplicity of the eigenvalue is 1. Therefore, the
defect is 1, and we can no longer apply the eigenvalue method directly to a system of ODEs with
such a coefficient matrix.
The key observation we will use here is that if λ is an eigenvalue of A of algebraic multiplicity
m, then we will be able to find m linearly independent vectors solving the equation (A−λI)
m
v =
0, and (A − 3I)v
2
= v
1
.
If these two equations are satisfied, then x
2
is a solution. We know the first of these equations is
satisfied because v
1
is an eigenvector. If we plug the second equation into the first we find that
(A − 3I)(A − 3I)v
2
=
0, or (A − 3I)
2
v
2
=
0.
If we can, therefore, find a v
2
which solves (A − 3I)
2
v
2
=
0, and such that (A − 3I)v
2
= v
1
, we are
done. This is just a bunch of linear equations to solve and we are by now very good at that.
We notice that in this simple case (A−3I)
2
is just the zero matrix (exercise). Hence, any vector
v
2
solves (A − 3I)
2
v
2
=
e
λt
.
This machinery can also be generalized to larger matrices and higher defects. We will not go
over, but let us just state the ideas. Suppose that A has a multiplicity m eigenvalue λ. We find
vectors such that
(A − λI)
k
v =
0, but (A − λI)
k−1
v
0.
Such vectors are called generalized eigenvectors. For every eigenvector v
1
we find a chain of
generalized eigenvectors v
2
through v
k
such that:
(A − λI)v
1
=
e
λt
.
We proceed to find chains until we form m linearly independent solutions (m is the multiplicity).
You may need to find several chains for every eigenvalue.
3.7.3 Exercises
Exercise 3.7.2: Let A =
,
5 −3
3 −1
¸
. Solve x
·
= Ax.
Exercise 3.7.3: Let A =
,
5 −4 4
0 3 0
−2 4 −14: Let A =
,
2 1 0
0 2 0
0 0 2
,
. a) What are the eigenvalues? b) What is/are the defect(s) of the
eigenvalue(s)? c) Solve x
·
= Ax in two different ways and verify you get the same answer.
3.7. MULTIPLE EIGENVALUES 123
Exercise 3.7.5: Let A =
,
0 1 2
−1 −2 −2
−4 4 76: Let A =
,
0 4 −2
−1 −4 1
0 0 −27: Let A =
,
2 1 −1
−1 0 2
−1 −2 48: Suppose that A is a 2 × 2 matrix with a repeated eigenvalue λ. Suppose that there
are two linearly independent eigenvectors. Show that the matrix is diagonal, in particular A = λI.
124 CHAPTER 3. SYSTEMS OF ODES
3.8 Matrix exponentials
Note: 2 lectures, §5.5 in EP
3.8.1 Definition
In this section we present a different way of finding the fundamental matrix solution of a system.
Suppose that we have the constant coefficient equation
x
·
= Px,
as usual. Now suppose that this was one equation (P is a number or a 1 × 1 matrix). Then the
solution to this would be
x = e
Pt
.
It turns out the same computation works for matrices when we define e
Pt
properly. First let us write
down the Taylor series for e
at
for some number a.
e
at
= 1 + at +
(at)
2
2
+
(at)
3
6
+
(at)
4
24
+ =
∞
¸
k=0
(at)
k
k!
.
Recall k! = 1 2 3 k, and 0! = 1. Now if we differentiate this series
a + a
2
t +
a
3
t
2
2
+
a
4
t
3
6
+ = a
¸
1 + at +
(at)
2
2
+
(at)
3
6
+
= ae
at
.
Maybe we can write try the same trick here. Suppose that for an n × n matrix A we define the
matrix exponential as
e
A
def
= I + A +
1
2
A
2
+
1
6
A
3
+ +
1
k!
A
k
+
Let us not worry about convergence. The series really does always converge. We usually write Pt
as tP by convention when P is a matrix. With this small change and by the exact same calculation
as above we have that
d
dt
e
tP
= Pe
tP
.
Now P and hence e
tP
is an n × n matrix. What we are looking for is a vector. We note that in the
1 × 1 case we would at this point multiply by an arbitrary constant to get the general solution. In
the matrix case we multiply by a column vector c.
Theorem 3.8.1. Let P be an n × n matrix. Then the general solution to x
·
= Px is
x = e
tP
c,
where c is an arbitrary constant vector. In fact x(0) = c.
3.8. MATRIX EXPONENTIALS 125
Let us check.
d
dt
x =
d
dt
e
tP
c
= Pe
tP
c = Px.
Hence e
tP
is the fundamental matrix solution of the homogeneous system. If we find a way
to compute the matrix exponential, we will have another method of solving constant coefficient
homogeneous systems. It also makes it easy to solve for initial conditions. To solve x
·
= Ax,
x(0) =
b, we take the solution
x = e
tA
b.
This equation follows because e
0A
= I, so x(0) = e
0A
b =
b.
We mention a drawback of matrix exponentials. In general e
A+B
e
A
e
B
. The trouble is that
matrices do not commute, that is, in general AB BA. If you try to prove e
A+B
e
A
e
B
using the
Taylor series, you will see why the lack of commutativity becomes a problem. However, it is still
true that if AB = BA, that is, if A and B commute, then e
A+B
= e
A
e
B
. We will find this fact useful.
Let us restate this as a theorem to make a point.
Theorem 3.8.2. If AB = BA then e
A+B
= e
A
e
B
. Otherwise e
A+B
e
A
e
B
in general.
3.8.2 Simple cases
In some instances it may work to just plug into the series definition. Suppose the matrix is diagonal.
For example, D =
,
a 0
0 b
¸
. Then
D
k
=
,
a
k
0
0 b
k
¸
,
and
e
D
= I + D +
1
2
D
2
+
1
6
D
3
+ =
,
1 0
0 1
¸
+
,
a 0
0 b
¸
+
1
2
,
a
2
0
0 b
2
¸
+
1
6
,
a
3
0
0 b
3
¸
+ =
,
e
a
0
0 e
b
¸
.
So by this rationale we have that
e
I
=
,
e 0
0 e
¸
and e
aI
=
,
e
a
0
0 e
a
¸
.
This makes exponentials of certain other matrices easy to compute. Notice for example that
the matrix A =
,
5 −3
3 −1
¸
can be written as 2I + B where B =
,
3 −3
3 −3
¸
. Notice that 2I and B commute,
and that B
2
=
,
0 0
0 0
¸
. So B
k
= 0 for all k ≥ 2. Therefore, e
B
= I + B. Suppose we actually
want to compute e
tA
. 2tI and tB still commute (exercise: check this) and e
tB
= I + tB, since
(tB)
2
= t
2
B
2
= 0. We write
e
tA
= e
2tI+tB
= e
2tI
e
tB
=
,
e
2t
0
0 e
2t
¸
(I + tB) =
=
,
e
2t
0
0 e
2t
¸ ,
1 + 3t −3t
3t 1 − 3t
¸
=
,
(1 + 3t) e
2t
−3te
2t
3te
2t
(1 − 3t) e
2t
¸
.
126 CHAPTER 3. SYSTEMS OF ODES
So we have found the fundamental matrix solution for the system x
·
= Ax. Note that this matrix
has a repeated eigenvalue with a defect; there is only one eigenvector for the eigenvalue 2. So we
have found a perhaps easier way to handle this case. In fact, if a matrix A is 2 × 2 and has an
eigenvalue λ of multiplicity 2, then either it is diagonal, or A = λI + B where B
2
= 0. This is a
good exercise.
Exercise 3.8.1: Suppose that A is 2×2 and λ is the only eigenvalue. Then show that (A−λI)
2
= 0.
Then we can write A = λI + B, where B
2
= 0. Hint: First write down what does it mean for the
eigenvalue to be of multiplicity 2. You will get an equation for the entries. Now compute the square
of B.
Matrices B such that B
k
= 0 for some k are called nilpotent. Computation of the matrix
exponential for nilpotent matrices is easy by just writing down the first k terms of the Taylor series.
3.8.3 General matrices
In general, the exponential is not as easy to compute as above. We cannot usually write any matrix
as a sum of commuting matrices where the exponential is simple for each one. But fear not, it
is still not too difficult provided we can find enough eigenvectors. First we need the following
interesting result about matrix exponentials. For any two square matrices A and B, we have
e
BAB
−1
= Be
A
B
−1
.
This can be seen by writing down the Taylor series. First note that
(BAB
−1
)
2
= BAB
−1
BAB
−1
= BAIAB
−1
= BA
2
B
−1
.
And hence by the same reasoning (BAB
−1
)
k
= BA
k
B
−1
. So now write down the Taylor series for
e
BAB
−1
e
BAB
−1
= I + BAB
−1
+
1
2
(BAB
−1
)
2
+
1
6
(BAB
−1
)
3
+
= BB
−1
+ BAB
−1
+
1
2
BA
2
B
−1
+
1
6
BA
3
B
−1
+
= B
b is
,
4
2
¸
. Then the particular solution
we are looking for is
,
x
y
¸
=
,
e
3t
+e
−t
2
e
3t
−e
−t
2
e
3t
−e
−t
2
e
3t
+e
−t
2
¸ ,
4
2
¸
=
,
2e
3t
+ 2e
−t
+ e
3t
− e
−t
2e
3t
− 2e
−t
+ e
3t
+ e
−t
¸
=
,
3e
3t
+ e
−t
3e
3t
− e
−t
¸
.
3.8.4 Fundamental matrix solutions
We note that if you can compute the fundamental matrix solution in a different way, you can use
this to find the matrix exponential e
tA
. The fundamental matrix solution of a system of ODEs is
not unique. The exponential is the fundamental matrix solution with the property that for t = 0
we get the identity matrix. So we must find the right fundamental matrix solution. Let X be any
fundamental matrix solution to x
·
= Ax. Then we claim
e
tA
= X(t) [X(0)]
−1
.
Obviously if we plug t = 0 into X(t) [X(0)]
−1
we get the identity. It is not hard to see that we can
multiply a fundamental matrix solution on the right by any constant invertible matrix and we still
get a fundamental matrix solution. All we are doing is changing what the arbitrary constants are in
the general solution x(t) = X(t)c.
3.8.5 Approximations
If you think about it, the computation of any fundamental matrix solution X using the eigenvalue
method is just as difficult as computation of e
tA
. So perhaps we did not gain much by this new tool.
However, the Taylor series expansion actually gives us a very easy way to approximate solutions,
which the eigenvalue method did not.
3.8. MATRIX EXPONENTIALS 129
The simplest thing we can do is to just compute the series up to a certain number of terms. There
are better ways to approximate the exponential
∗
. In many cases however, few terms of the Taylor
series give a reasonable approximation for the exponential and may suffice for the application. For
example, let us compute the first 4 terms of the series for the matrix A =
,
1 2
2 1
¸
.
e
tA
≈ I + tA +
t
2
2
A
2
+
t
3
6
A
3
= I + t
,
1 2
2 1
¸
+ t
2
,
5
2
2
2
5
2
¸
+ t
3
,
13
6
7
3
7
3
13
6
¸
=
=
,
1 + t +
5
2
t
2
+
13
6
t
3
2 t + 2 t
2
+
7
3
t
3
2 t + 2 t
2
+
7
3
t
3
1 + t +
5
2
t
2
+
13
6
t
3
¸
.
Just like the Taylor series approximation for the scalar version, the approximation will be better
for small t and worse for larger t. For larger t, you will generally have to compute more terms.
Let us see how we stack up against the real solution with t = 0.1. The approximate solution is
approximately (rounded to 8 decimal places)
e
0.1 A
≈ I + 0.1 A +
0.1
2
2
A
2
+
0.1
3
6
A
3
=
,
1.12716667 0.22233333
0.22233333 1.12716667
¸
.
And plugging t = 0.1 into the real solution (rounded to 8 decimal places) we get
e
0.1 A
=
,
1.12734811 0.22251069
0.22251069 1.12734811
¸
.
This is not bad at all. Although if you take the same approximation for t = 1 you get (using the
Taylor series)
,
6.66666667 6.33333333
6.33333333 6.66666667
¸
,
while the real value is (again rounded to 8 decimal places)
,
10.22670818 9.85882874
9.85882874 10.22670818
¸
.
So the approximation is not very good once we get up to t = 1. To get a good approximation at
t = 1 (say up to 2 decimal places) you would need to go up to the 11
th
power (exercise).
3.8.6 Exercises
Exercise 3.8.2: Find a fundamental matrix solution for the system x
·
= 3x + y, y
·
= x + 3y.
Exercise 3.8.3: Find e
At
for the matrix A =
,
2 3
0 2
¸
.
∗
C. Moler and C.F. Van Loan, Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years
Later, SIAM Review 45 (1), 2003, 3–49
130 CHAPTER 3. SYSTEMS OF ODES
Exercise 3.8.4: Find a fundamental matrix solution for the system x
·
1
= 7x
1
+ 4x
2
+ 12x
3
, x
·
2
=
x
1
+ 2x
2
+ x
3
, x
·
3
= −3x
1
− 2x
2
− 5x
3
. Then find the solution that satisfies x =
,
0
1
−2
,
.
Exercise 3.8.5: Compute the matrix exponential e
A
for A =
,
1 2
0 1
¸
.
Exercise 3.8.6: Suppose AB = BA (matrices commute). Show that e
A+B
= e
A
e
B
.
Exercise 3.8.7: Use exercise 3.8.6 to show that (e
A
)
−1
= e
−A
. In particular this means that e
A
is
invertible even if A is not.
Exercise 3.8.8: Suppose A is a matrix with eigenvalues −1, 1, and corresponding eigenvectors
,
1
1
¸
,
,
0
1
¸
. a) Find matrix A with these properties. b) Find the fundamental matrix solution to x
·
= Ax.
c) Solve the system in with initial conditions x(0) =
,
2
3
¸
.
Exercise 3.8.9: Suppose that A is an n × n matrix with a repeated eigenvalue λ of multiplicity n.
Suppose that there are n linearly independent eigenvectors. Show that the matrix is diagonal, in
particular A = λI. Hint: Use diagonalization and the fact that the identity matrix commutes with
every other matrix.
3.9. NONHOMOGENEOUS SYSTEMS 131
3.9 Nonhomogeneous systems
Note: 3 lectures (may have to skip a little), somewhat different from §5.6 in EP
3.9.1 First order constant coefficient
Integrating factor
Let us first focus on the nonhomogeneous first order equation
x
·
(t) = Ax(t) +
f (t),
where A is a constant matrix. The first method we will look at is the integrating factor method. For
simplicity we rewrite the equation as
x
·
(t) + Px(t) =
f (t),
where P = −A. We multiply both sides of the equation by e
tP
(being mindful that we are dealing
with matrices which may not commute) to obtain
e
tP
x
·
(t) + e
tP
Px(t) = e
tP
f (t) dt + e
−tP
c.
132 CHAPTER 3. SYSTEMS OF ODES
Perhaps it is better understood as a definite integral. In this case it will be easy to also solve for
the initial conditions as well. Suppose we have the equation with initial conditions
x
·
(t) + Px(t) =
f (t), x(0) =
b.
The solution can then be written as
x(t) = e
−tP
t
0
e
sP
f (s) ds + e
−tP
b. (3.5)
Again, the integration means that each component of the vector e
sP
f (s) is integrated separately. It
is not hard to see that (3.5) really does satisfy the initial condition x(0) =
P(t) dt
,
because matrices generally do not commute.
Eigenvector decomposition
For the next method, we note that the eigenvectors of a matrix give the directions in which the
matrix acts like a scalar. If we solve our system along these directions these solutions would be
simpler as we can treat the matrix as a scalar. We can put those solutions together to get the general
solution.
Take the equation
x
·
(t) = Ax(t) +
e
−λ
k
t
g
k
(t) dt + C
k
e
λ
k
t
.
Note that if you are looking for just any particular solution, you could set C
k
to be zero. If we leave
these constants in, we will get the general solution. Write x(t) = v
1
ξ
1
(t) +v
2
ξ
2
(t) + +v
n
ξ
n
(t),
and we are done.
Again, as always, it is perhaps better to write these integrals as definite integrals. Suppose that
we have an initial condition x(0) =
=
,
e
4t
−e
−2t
3
+
3−12t
16
e
−2t
+e
4t
+2e
t
3
+
4t−5
16
¸
.
That is, x
1
=
e
4t
−e
−2t
3
+
3−12t
16
and x
2
=
e
−2t
+e
4t
+2e
t
3
+
4t−5
16
.
Exercise 3.9.1: Check that x
1
and x
2
solve the problem. Check both that they satisfy the differential
equation and that they satisfy the initial conditions.
Undetermined coefficients
The method of undetermined coefficients also still works. The only difference here is that we
will have to take unknown vectors rather than just numbers. Same caveats apply to undetermined
coefficients for systems as they do for single equations. This method does not always work, fur-
thermore if the right hand side is complicated, you will have lots of variables to solve for. In this
case you can think of each element of an unknown vector as an unknown number. So in system of
3 equations if you have say 4 unknown vectors (this would not be uncommon), then you already
have 12 unknowns that you need to solve for. The method can turn into a lot of tedious work. As
this method is essentially the same as it is for single equations, let us just do an example.
Example 3.9.3: Let A =
,
−1 0
−2 1
¸
. Find a particular solution of x
·
= Ax +
f where
f (t) =
,
e
t
t
¸
.
Note that we can solve this system in an easier way (can you see how), but for the purposes of
the example, let us use the eigenvalue method plus undetermined coefficients.
The eigenvalues of A are −1 and 1 and the corresponding eigenvectors are
,
1
1
¸
and
,
0
1
¸
respec-
tively. Hence our complementary solution is
x
c
= α
1
,
1
1
¸
e
−t
+ α
2
,
0
1
¸
e
t
,
for some arbitrary constants α
1
and α
2
.
We would want to guess a particular solution of
x = ae
t
+
bt +c.
However, something of the form ae
t
appears in the complementary solution. Because we do not
yet know the vector if the a is a multiple of
,
0
1
¸
we do not know if a conflict arises. It may very
3.9. NONHOMOGENEOUS SYSTEMS 137
well not, but to be safe we should also try
bte
t
. Here we find the crux of the difference for systems.
You want to try both ae
t
and
bte
t
really was needed.
138 CHAPTER 3. SYSTEMS OF ODES
Exercise 3.9.2: Check that x
1
and x
2
solve the problem. Also try setting a
2
= 1 and again check
these solutions. What is the difference between the two solutions we can obtain in this way?
As you can see, other than the handling of conflicts, undetermined coefficients works exactly
the same as it did for single equations. However, the computations can get out of hand pretty
quickly for systems. The equation we had done was very simple.
3.9.2 First order variable coefficient
Just as for a single equation, there is the method of variation of parameters. In fact for constant
coefficient systems, this is essentially the same thing as the integrating factor method we discussed
earlier. However this method will work for any linear system, even if it is not constant coefficient,
provided you have somehow solved the associated homogeneous problem.
Suppose we have the equation
x
·
= A(t) x +
f (t). (3.9)
Further, suppose that you have solved the associated homogeneous equation x
·
= A(t) x and found
the fundamental matrix solution X(t). The general solution to the associated homogeneous equa-
tion is X(t)c for a constant vector c. Just like for variation of parameters for single equation we try
the solution to the nonhomogeneous equation of the form
x
p
= X(t) u(t),
where u(t) is a vector valued function instead of a constant. Now substitute into (3.9) to obtain
x
p
·
(t) = X
·
(t) u(t) + X(t) u
·
(t) = A(t) X(t) u(t) +
F
0
cos ωt, then you can try a solution of the
form
x
p
= c cos ωt,
140 CHAPTER 3. SYSTEMS OF ODES
and you do not need to introduce sines.
If the
F is a sum of cosines, you note that we still have the superposition principle, so if
F(t) =
F
0
cos ω
0
t +
F
1
cos ω
1
t, you could try a cos ω
0
t for the problem x
··
= Ax +
F
0
cos ω
0
t, and
you would try
b cos ω
1
t for the problem x
··
= Ax +
F
0
cos ω
1
t. Then sum the solutions.
However, if there is duplication with the complementary solution, or the equation is of the form
x
··
= Ax
·
+ Bx +
F(t), then you need to do the same thing as you do for first order systems.
Actually you will never go wrong with putting in more terms than needed into your guess. You
will just find that the extra coefficients will turn out to be zero. But it is useful to save some time
and effort.
Eigenvector decomposition
If we have the system
x
··
= Ax +
=
,
1
20
−3
10
¸
cos 3t.
This matches what we got previously in § 3.6.
3.9.4 Exercises
Exercise 3.9.4: Find a particular solution to x
·
= x+2y+2t, y
·
= 3x+2y−4. a) Using integrating
factor method, b) using eigenvector decomposition, c) using undetermined coefficients.
Exercise 3.9.5: Find the general solution to x
·
= 4x + y − 1, y
·
= x + 4y − e
t
. a) Using integrating
factor method, b) using eigenvector decomposition, c) using undetermined coefficients.
Exercise 3.9.6: Find the general solution to x
··
1
= −6x
1
+ 3x
2
+ cos t, x
··
2
= 2x
1
− 7x
2
+ 3 cos t. a)
using eigenvector decomposition, b) using undetermined coefficients.
142 CHAPTER 3. SYSTEMS OF ODES
Exercise 3.9.7: Find the general solution to x
··
1
= −6x
1
+ 3x
2
+ cos 2t, x
··
2
= 2x
1
− 7x
2
+ 3 cos 2t.
a) using eigenvector decomposition, b) using undetermined coefficients.
Exercise 3.9.8: Take the equation
x
·
=
,
1
t
−1
1
1
t
¸
x +
,
t
2
−t
¸
.
a) Check that
x
c
= c
1
,
t sin t
−t cos t
¸
+ c
2
,
t cos t
t sin t
¸
is the complementary solution. b) Use variation of parameters to find a particular solution.
Chapter 4
Fourier series and PDEs
4.1 Boundary value problems
Note: 2 lectures, similar to §3.8 in EP
4.1.1 Boundary value problems
Before we tackle the Fourier series, we need to study the so-called boundary value problems (or
endpoint problems). For example, suppose we have
x
··
+ λx = 0, x(a) = 0, x(b) = 0,
for some constant λ, where x(t) is defined for t in the interval [a, b]. Unlike before when we
specified the value of the solution and its derivative at a single point, we now specify the value of
the solution at two different points. Note that x = 0 is a solution to this equation, so existence of
solutions is not an issue here. Uniqueness is another issue. The general solution to x
··
+ λx = 0
will have two arbitrary constants present, so it is natural to think that requiring two conditions will
guarantee a unique solution.
Example 4.1.1: However take λ = 1, a = 0, b = π. That is,
x
··
+ x = 0, x(0) = 0, x(π) = 0.
Then x = sin t is another solution satisfying both boundary conditions. In fact, write down the
general solution of the differential equation, which is x = Acos t + Bsin t. The condition x(0) = 0
forces A = 0. But letting x(π) = 0 does not give us any more information as x = Bsin t already
satisfies both conditions. Hence there are infinitely many solutions x = Bsin t for an arbitrary
constant B.
143
144 CHAPTER 4. FOURIER SERIES AND PDES
Example 4.1.2: On the other hand, change to λ = 2.
x
··
+ 2x = 0, x(0) = 0, x(π) = 0.
Then the general solution is x = Acos
√
2 t + Bsin
√
2 t. Letting x(0) = 0 still forces A = 0. But
now letting 0 = x(π) = Bsin
√
2 π. sin
√
2 π 0 and hence B = 0. So x = 0 is the unique solution
to this problem.
So what is going on? We will be interested in classifying which constants λ imply a nonzero
solution, and we will be interested in finding those solutions. This problem is an analogue of
finding eigenvalues and eigenvectors of matrices.
4.1.2 Eigenvalue problems
In general we will consider more equations, but we will postpone this until chapter 5. For the basic
Fourier series theory we will need only the following three cases.
x
··
+ λx = 0, x(a) = 0, x(b) = 0, (4.1)
x
··
+ λx = 0, x
·
(a) = 0, x
·
(b) = 0, (4.2)
and
x
··
+ λx = 0, x(a) = x(b), x
·
(a) = x
·
(b), (4.3)
A number λ will be considered an eigenvalue of (4.1) (resp. (4.2) or (4.3)) if and only if there
exists a nonzero solution to (4.1) (resp. (4.2) or (4.3)) given that specific λ. The nonzero solution
we found will be said to be the corresponding eigenfunction.
Note the similarity to eigenvalues and eigenvectors of matrices. The similarity is not just
coincidental. If we think of the equations as differential operators, then we are doing the same
exact thing. For example, let L = −
d
2
dt
2
, then we are looking for eigenfunctions f satisfying certain
endpoint conditions that solve (L − λ) f = 0. A lot of the formalism from linear algebra can still
apply here, though we will not pursue this line of reasoning too far.
Example 4.1.3: Let us find the eigenvalues and eigenfunctions of
x
··
+ λx = 0, x(0) = 0, x(π) = 0.
We will have to handle the cases λ > 0, λ = 0, λ < 0 separately.
First suppose that λ > 0, then the general solution to x
··
+ λx = 0 is
x = Acos
√
λ t + Bsin
√
λ t.
The condition x(0) = 0 implies immediately A = 0. Next
0 = x(π) = Bsin
√
λ π.
4.1. BOUNDARY VALUE PROBLEMS 145
If B is zero then x is not a nonzero solution. So to get a nonzero solution we must have that
sin
√
λ π = 0. Hence,
√
λ π must be an integer multiple of π, or
√
λ = k for a positive integer k.
Hence the positive eigenvalues are k
2
for all integers k ≥ 1. The corresponding eigenfunctions can
be taken as x = sin kt. Just like for eigenvectors, we get all the multiples of an eigenfunction, so
we only need to pick one.
Now suppose that λ = 0. In this case the equation is x
··
= 0 and the general solution is
x = At + B. x(0) = 0 implies that B = 0 and then x(π) = 0 implies that A = 0. This means that
λ = 0 is not an eigenvalue.
Finally, let λ < 0. In this case we have the general solution
x = Acosh
√
−λ t + Bsinh
√
−λ t.
Letting x(0) = 0 implies that A = 0 (recall cosh 0 = 1 and sinh 0 = 0). So our solution must be
x = Bsinh
√
−λ t and satisfy x(π) = 0. This is only possible if B is zero. Why? Because sinh ξ
is only zero for ξ = 0, you should plot sinh to see this. Also we can just look at the definition
0 = sinh t =
e
t
−e
−t
2
. Hence e
t
= e
−t
which implies t = −t and that is only true if t = 0. So there are
no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
λ
k
= k
2
with an eigenfunction x
k
= sin kt for all integers k ≥ 1.
Example 4.1.4: Let us also compute the eigenvalues and eigenfunctions of
x
··
+ λx = 0, x
·
(0) = 0, x
·
(π) = 0.
Again we will have to handle the cases λ > 0, λ = 0, λ < 0 separately.
First suppose that λ > 0, then the general solution to x
··
+λx = 0 is x = Acos
√
λ t +Bsin
√
λ t.
So
x
·
= −Asin
√
λ t + Bcos
√
λ t
The condition x
·
(0) = 0 implies immediately B = 0. Next
0 = x
·
(π) = −Asin
√
λ π.
Again A should not be zero, and sin
√
λ π is only zero if
√
λ = k for a positive integer k. Hence the
positive eigenvalues are again k
2
for all integers k ≥ 1. And the corresponding eigenfunctions can
be taken as x = cos kt.
Now suppose that λ = 0. In this case the equation is x
··
= 0 and the general solution is
x = At + B so x
·
= A. x
·
(0) = 0 implies that A = 0. Obviously setting x
·
(π) = 0 does not get
us anything new. This means that B could be anything (let us take it to be 1). So λ = 0 is an
eigenvalue and x = 1 is the corresponding eigenfunction.
Finally, let λ < 0. In this case we have the general solution x = Acosh
√
−λ t + Bsinh
√
−λ t
and hence
x
·
= Asinh
√
−λ t + Bcosh
√
−λ t.
146 CHAPTER 4. FOURIER SERIES AND PDES
We have already seen (with roles of A and B switched) that for this to be zero at t = 0 and t = π it
implies that A = B = 0. Hence there are no negative eigenvalues.
In summary, the eigenvalues and corresponding eigenfunctions are
λ
k
= k
2
with an eigenfunction x
k
= sin kt for all integers k ≥ 1,
and there is another eigenvalue
λ
0
= 0 with an eigenfunction x
0
= 1.
We could also do this for a little bit more complicated boundary value problem. This problem
is be the one that leads to the general Fourier series.
Example 4.1.5: Let us compute the eigenvalues and eigenfunctions of
x
··
+ λx = 0, x(−π) = x(π), x
·
(−π) = x
·
(π).
You should notice that we have not specified the values or the derivatives at the endpoints, but
rather that they are the same at the beginning and at the end of the interval.
Let us skip λ < 0. The computations are the same and again we find that there are no negative
eigenvalues.
For λ = 0, the general solution is x = At + B. The condition x(−π) = x(π) implies that A = 0
(Aπ + B = −Aπ + B implies A = 0). The second condition x
·
(−π) = x
·
(π) says nothing about B and
hence λ = 0 is an eigenvalue with a corresponding eigenfunction x = 1.
For λ > 0 we get that x = Acos
√
λ t + Bsin
√
λ t. Now
Acos −
√
λ π + Bsin −
√
λ π = Acos
√
λ π + Bsin
√
λ π.
We remember that cos −θ = cos θ and sin −θ = −sin θ. Therefore,
Acos
√
λ π − Bsin
√
λ π = Acos
√
λ π + Bsin
√
λ π.
and hence either B = 0 or sin
√
λ π = 0. Similarly (exercise) if we differentiate x and plug in the
second condition we find that A = 0 or sin
√
λ π = 0. Therefore, unless we want A and B to both
be zero (which we do not) we must have sin
√
λ π = 0. Therefore,
√
λ is an integer and hence the
eigenvalues are yet again λ = k
2
for an integer k ≥ 1. In this case however, x = Acos kt +Bsin kt is
an eigenfunction for any A and any B. So we have two linearly independent eigenfunctions sin kt
and cos kt. Remember that for a matrix we could also have had two eigenvectors corresponding to
an eigenvalue if the eigenvalue was repeated.
In summary, the eigenvalues and corresponding eigenfunctions are
λ
k
= k
2
with the eigenfunctions cos kt and sin kt for all integers k ≥ 1,
λ
0
= 0 with an eigenfunction x
0
= 1.
4.1. BOUNDARY VALUE PROBLEMS 147
4.1.3 Orthogonality of eigenfunctions
Something that will be very useful in the next section is the orthogonality property of the eigen-
functions. This is an analogue of the following fact about eigenvectors of a matrix. A matrix is
called symmetric if A = A
T
. Eigenvectors for two distinct eigenvalues of a symmetric matrix are
orthogonal. That symmetry is required. We will not prove this fact here. The differential operators
we are dealing with act much like a symmetric matrix. We, therefore, get the following theorem.
Theorem 4.1.1. Suppose that x
1
(t) and x
2
(t) are two eigenfunctions of the problem (4.1), (4.2) or
(4.3) for two different eigenvalues λ
1
and λ
2
. Then they are orthogonal in the sense that
b
a
x
1
(t)x
2
(t) dt = 0.
Note that the terminology comes from the fact that the integral is a type of inner product. We
will expand on this in the next section. The theorem has a very short, elegant, and illuminating
proof so let us give it here. First note that we have the following two equations.
x
··
1
+ λ
1
x
1
= 0 and x
··
2
+ λ
2
x
2
= 0.
Multiply the first by x
2
and the second by x
1
and subtract to get
(λ
1
− λ
2
)x
1
x
2
= x
··
2
x
1
− x
2
x
··
1
.
Now integrate both sides of the equation.
(λ
1
− λ
2
)
π
−π
(cos mt)(sin nt) dt = 0.
4.1.4 Fredholm alternative
We now touch on a very useful theorem in the theory of differential equations. The theorem holds
in a more general setting than we are going to state it, but for our purposes the following statement
is sufficient. We will give a slightly more general version in chapter 5.
Theorem 4.1.2 (Fredholm alternative
∗
). Suppose p and q are continuous on [a, b]. Then either
x
··
+ λx = 0, x(a) = 0, x(b) = 0 (4.4)
has a nonzero solution, or
x
··
+ λx = f (t), x(a) = 0, x(b) = 0 (4.5)
has a unique solution for every continuous function f .
The theorem is also true for the other types of boundary conditions we considered. The theorem
means that if λ is not an eigenvalue, the nonhomogeneous equation (4.5) has a unique solution for
every right hand side. On the other hand if λ is an eigenvalue, then (4.5) need not have a solution
for every f , and furthermore, even if it happens to have a solution, the solution is not unique.
We also want to reinforce the idea here that linear differential operators have much in common
with matrices. So it is no surprise that there is a finite dimensional version of Fredholm alternative
for matrices as well. Let A be an n × n matrix. The Fredholm alternative then states that either
(A − λI)x =
0 has a nontrivial solution, or (A − λI)x =
b has a solution for every
b.
A lot of intuition from linear algebra can be applied for linear differential operators, but one
must be careful of course. For example, one obvious difference we have already seen is that
in general a differential operator will have infinitely many eigenvalues, while a matrix has only
finitely many.
∗
Named after the Swedish mathematicain Erik Ivar Fredholm (1866 – 1927).
4.1. BOUNDARY VALUE PROBLEMS 149
4.1.5 Application
Let us consider a physical application of an endpoint problem. Suppose we have a tightly stretched
quickly spinning elastic string or rope of uniform linear density ρ. Let us put this problem into the
xy-plane. The x axis represents the position on the string. The string rotates at angular velocity
ω, so we will assume that the whole xy-plane rotates at angular velocity ω along. We will assume
that the string stays in this xy-plane and y will measure its deflection from the equilibrium position,
y = 0, on the x axis. Hence, we will find a graph which gives the shape of the string. We will
idealize the string to have no volume to just be a mathematical curve. If we take a small segment
and we look at the tension at the endpoints, we see that this force is tangential and we will assume
that the magnitude is the same at both end points. Hence the magnitude is constant everywhere and
we will call its magnitude T. If we assume that the deflection is small then we can use Newton's
second law to get an equation
Ty
··
+ ρω
2
y = 0.
Let L be the length of the string and the string is fixed at the beginning and end points. Hence,
y(0) = 0 and y(L) = 0. See Figure 4.1.
L x
y
y
0
Figure 4.1: Whirling string.
We rewrite the equation as y
··
+
ρω
2
T
y = 0. The setup is similar to Example 4.1.3 on page 144,
except for the interval length being L instead of π. We are looking for eigenvalues of y
··
+ λy =
0, y(0) = 0, y(L) = 0 where λ =
ρω
2
T
. As before there are no nonpositive eigenvalues. With λ > 0,
the general solution to the equation is y = Acos
√
λ x + Bsin
√
λ x. The condition y(0) = 0 implies
that A = 0 as before. The condition y(L) = 0 implies that sin
√
λ L = 0 and hence
√
λ L = kπ for
some integer k > 0, so
ρω
2
T
= λ =
k
2
π
2
L
2
.
What does this say about the shape of the string? It says that for all parameters ρ, ω, T not
satisfying the above equation, the string is in the equilibrium position, y = 0. When
ρω
2
T
=
k
2
π
2
L
2
,
then the string will "pop out" some distance B at the midpoint. We cannot compute B with the
information we have.
Let us assume that ρ and T are fixed and we are changing ω. For most values of ω the string
is in the equilibrium state. When the angular velocity ω hits a value ω =
kπ
√
T
L
√
ρ
, then the string will
150 CHAPTER 4. FOURIER SERIES AND PDES
pop out and will have the shape of a sin wave crossing the x axis k times. When ω changes again,
the string returns to the equilibrium position. You can see that the higher the angular velocity the
more times it crosses the x axis when it is popped out.
4.1.6 Exercises
Hint for the following exercises: Note that cos
√
λ (t − a) and sin
√
λ (t − a) are also solutions of
the homogeneous equation.
Exercise 4.1.2: Compute all eigenvalues and eigenfunctions of x
··
+ λx = 0, x(a) = 0, x(b) = 0.
Exercise 4.1.3: Compute all eigenvalues and eigenfunctions of x
··
+ λx = 0, x
·
(a) = 0, x
·
(b) = 0.
Exercise 4.1.4: Compute all eigenvalues and eigenfunctions of x
··
+ λx = 0, x
·
(a) = 0, x(b) = 0.
Exercise 4.1.5: Compute all eigenvalues and eigenfunctions of x
··
+ λx = 0, x(a) = x(b), x
·
(a) =
x
·
(b).
Exercise 4.1.6: We have skipped the case of λ < 0 for the boundary value problem x
··
+ λx =
0, x(−π) = x(π), x
·
(−π) = x
·
(π). So finish the calculation and show that there are no negative
eigenvalues.
4.2. THE TRIGONOMETRIC SERIES 151
4.2 The trigonometric series
Note: 2 lectures, §9.1 in EP
4.2.1 Periodic functions and motivation
As motivation for studying Fourier series, suppose we have the problem
x
··
+ ω
2
0
x = f (t), (4.6)
for some periodic function f (t). We have already solved
x
··
+ ω
2
0
x = F
0
cos ωt. (4.7)
One way to solve (4.6) is to decompose f (t) as a sum of of cosines (and sines) and then solve many
problems of the form (4.7). We then use the principle of superposition, to sum up all the solutions
we got to get a solution to (4.6).
Before we proceed, let us talk a little bit more in detail about periodic functions. A function
is said to be periodic with period P if f (t) = f (t + P) for all t. For brevity we will say f (t) is P-
periodic. Note that a P-periodic function is also 2P-periodic, 3P-periodic and so on. For example,
cos t and sin t are 2π-periodic. So are cos kt and sin kt for all integers k. The constant functions are
an extreme example. They are periodic for any period (exercise).
Normally we will start with a function f (t) defined on some interval [−L, L] and we will want
to extend periodically to make it a 2L-periodic function. We do this extension by defining a new
function F(t) such that for t in [−L, L], F(t) = f (t). For t in [L, 3L], we define F(t) = f (t −2L), for
t in [−3L, −L], F(t) = f (t + 2L), and so on.
Example 4.2.1: Defined f (t) = 1−t
2
on [−1, 1]. Now extend periodically to a 2-periodic function.
See Figure 4.2 on the following page.
You should be careful to distinguish between f (t) and its extension. A common mistake is to
assume that a formula for f (t) holds for its extension. It can be confusing when the formula for
f (t) is periodic, but with perhaps a different period.
Exercise 4.2.1: Define f (t) = cos t on [−π/2, π/2]. Now take the π-periodic extension and sketch
its graph. How does it compare to the graph of cos t.
4.2.2 Inner product and eigenvector decomposition
Suppose we have a symmetric matrix, that is A
T
= A. We have said before that the eigenvectors of
A are then orthogonal. Here the word orthogonal means that if v and ware two distinct eigenvectors
of A, then (v, w) = 0. In this case the inner product (v, w) is the dot product, which can be computed
as v
T
w.
152 CHAPTER 4. FOURIER SERIES AND PDES
-3 -2 -1 0 1 2 3
-3 -2 -1 0 1 2 3
-0.5
0.0
0.5
1.0
1.5
-0.5
0.0
0.5
1.0
1.5
Figure 4.2: Periodic extension of the function 1 − t
2
.
To decompose a vector v in terms of mutually orthogonal vectors w
1
and w
2
we write
v = a
1
w
1
+ a
2
w
2
.
Let us find the formula for a
1
and a
2
. First let us compute
(v, w
1
) = (a
1
w
1
+ a
2
w
2
, w
1
) = a
1
( w
1
, w
1
) + a
2
( w
2
, w
1
) = a
1
( w
1
, w
1
).
Therefore,
a
1
=
(v, w
1
)
( w
1
, w
1
)
.
Similarly
a
2
=
(v, w
2
)
( w
2
, w
2
)
.
You probably remember this formula from vector calculus.
Example 4.2.2: Write v =
,
2
3
¸
as a linear combination of w
1
=
,
1
−1
¸
and w
2
=
,
1
1
¸
.
First note that w
1
and w
2
are orthogonal as ( w
1
, w
2
) = 1(1) + (−1)1 = 0. Then
a
1
=
(v, w
1
)
( w
1
, w
1
)
=
2(1) + 3(−1)
1(1) + (−1)(−1)
=
−1
2
.
a
2
=
(v, w
2
)
( w
2
, w
2
)
=
2 + 3
1 + 1
=
5
2
.
Hence
,
2
3
¸
=
−1
2
,
1
−1
¸
+
5
2
,
1
1
¸
.
4.2. THE TRIGONOMETRIC SERIES 153
4.2.3 The trigonometric series
Now instead of decomposing a vector in terms of the eigenvectors of a matrix, we will decompose
a function in terms of eigenfunctions of a certain eigenvalue problem. In particular, the eigenvalue
problem we will use for the Fourier series is the following
x
··
+ λx = 0, x(−π) = x(π), x
·
(−π) = x
·
(π).
We have previously computed that the eigenfunctions are 1, cos kt, sin kt. That is, we will want to
find a representation of a 2π-periodic function f (t) as
f (t) =
a
0
2
+
∞
¸
n=1
a
n
cos nt + b
n
sin nt.
This series is called the Fourier series
†
or trigonometric series for f (t). Note that here we have
used the eigenfunction
1
2
instead of 1. This is for convenience. We could also think of 1 = cos 0t,
so that we only need to look at cos kt and sin kt.
Just like for matrices we will want to find a projection of f (t) onto the subspace generated by
the eigenfunctions. So we will want to define an inner product of functions. For example, to find
a
n
we want to compute ( f (t) , cos nt ). We define the inner product as
( f (t) , g(t) )
def
=
1
π
π
−π
t dt = 0.
4.2. THE TRIGONOMETRIC SERIES 155
We will often use the result from calculus that the integral of an odd function over a symmetric
interval is zero. Recall that an odd function is a function ϕ(t) such that ϕ(−t) = −ϕ(t). For example
the function t, the function sin t, or more to the point the function t cos mt are all odd.
a
m
=
1
π
π
−π
t cos mt dt = 0.
Let us move to b
m
. Another useful fact from calculus is that the integral of an even function over a
symmetric interval is twice the integral of the same function over half the interval. Recall an even
function is a function ϕ(t) such that ϕ(−t) = ϕ(t). For example t sin mt is even.
b
m
=
1
π
¸
¸
¸
¸
¸
¸
0 if t = −π or t = π,
t otherwise.
and extend periodically, then the series equals the extended f (t) everywhere, including the dis-
continuities. We will generally not worry about changing the function at several (finitely many)
points.
We will say more about convergence in the next section. Let us however mention briefly an
effect of the discontinuity. Let us zoom in near the discontinuity in the square wave. Further, let
us plot the first 100 harmonics, see Figure 4.7. You will notice that while the series is a very good
approximation away from the discontinuities, the error (the overshoot) near the discontinuity at
t = π does not seem to be getting any smaller. This behavior is known as the Gibbs phenomenon.
The region where the error is large gets smaller and smaller, however, the more terms in the series
you take.
1.75 2.00 2.25 2.50 2.75 3.00 3.25
1.75 2.00 2.25 2.50 2.75 3.00 3.25
2.75
3.00
3.25
3.50
2.75
3.00
3.25
3.50
Figure 4.7: Gibbs phenomenon in action.
We can think of a periodic function as a "signal" being a superposition of many signals of pure
frequency. That is, we could think of say the square wave as a tone of certain frequency. It will be
in fact a superposition of many different pure tones of frequency which are multiples of the base
frequency. On the other hand a simple sine wave is only the pure tone. The simplest way to make
4.2. THE TRIGONOMETRIC SERIES 159
sound using a computer is the square wave, and the sound will be a very different from nice pure
tones. If you have played video games from the 1980s or so you have heard what square waves
sound like.
4.2.4 Exercises
Exercise 4.2.3: Suppose f (t) is defined on [−π, π] as sin 5t + cos 3t. Extend periodically and
compute the Fourier series of f (t).
Exercise 4.2.4: Suppose f (t) is defined on [−π, π] as |t|. Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.5: Suppose f (t) is defined on [−π, π] as |t|
3
. Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.6: Suppose f (t) is defined on [−π, π] as
f (t) =
¸
¸
¸
¸
¸
¸
−1 if −π < t ≤ 0,
1 if 0 < t ≤ π.
Extend periodically and compute the Fourier series of f (t).
Exercise 4.2.7: Suppose f (t) is defined on [−π, π] as t
3
. Extend periodically and compute the
Fourier series of f (t).
Exercise 4.2.8: Suppose f (t) is defined on [−π, π] as t
2
. Extend periodically and compute the
Fourier series of f (t).
There is another form of the Fourier series using complex exponentials that is sometimes easier
to work with.
Exercise 4.2.9: Let
f (t) =
a
0
2
+
∞
¸
n=1
a
n
cos nt + b
n
sin nt.
Use Euler's formula e
iθ
= cos θ + i sin θ, show that there exist complex numbers c
m
such that
f (t) =
∞
¸
m=−∞
c
m
e
imt
.
Note that the sum now ranges over all the integers including negative ones. Do not worry about
convergence in this calculation. Hint: It may be better to start from the complex exponential form
and write the series as
c
0
+
∞
¸
m=1
c
m
e
imt
+ c
−m
e
−imt
.
160 CHAPTER 4. FOURIER SERIES AND PDES
4.3 More on the Fourier series
Note: 2 lectures, §9.2 – §9.3 in EP
Before reading the lecture, it may be good to first try Project IV (Fourier series) from the
IODE website: After reading the lecture it may be good
to continue with Project V (Fourier series again).
4.3.1 2L-periodic functions
We have computed the Fourier series for a 2π-periodic function, but what about functions of dif-
ferent periods. Well, fear not, the computation is a simple case of change of variables. We can just
rescale the independent axis. Suppose that you have the 2L-periodic function f (t) (L is called the
half period). Let s =
π
L
t, then the function
g(s) = f
L
π
s
is 2π-periodic. We want to also rescale all our sines and cosines. We will want to writeIf we change variables to s we see that
g(s) =
a
0
2
+
∞
¸
n=1
a
n
cos ns + b
n
sin ns.
So we can compute a
n
and b
n
as before. After we write down the integrals we change variables
back to t.
a
0
=
1
π
π
−π
g(s) ds =
1
L
L
−L
f (t) dt,
a
n
=
1
π
π
−π
g(s) cos ns ds =
1
L
L
−L
f (t) cos
nπ
L
t dt,
b
n
=
1
π
π
−π
g(s) sin ns ds =
1
L
L
−L
f (t) sin
nπ
L
t dt.
The two most common half periods that show up in examples are π and 1 because of the sim-
plicity. We should stress that we have done no new mathematics, we have only changed variables.
If you understand the Fourier series for 2π-periodic functions, you understand it for 2L-periodic
functions. All that we are doing is moving some constants around, but all the mathematics is the
same.
4.3. MORE ON THE FOURIER SERIES 161
Example 4.3.1: Let
f (t) = |t| for −1 < t < 1,
extended periodically. The plot of the periodic extension is given in Figure 4.8. Compute the
Fourier series of f (t).8: Periodic extension of the function f (t).
We will write f (t) =
a
0
2
+
¸
∞
n=1
a
n
cos nπt + b
n
sin nπt. For n ≥ 1 we note that |t| cos nπt is even
and hence
a
n
=
1
−1
|t| dt = 1.
Note: You should be able to find this integral by thinking about the integral as the area under the
graph without doing any computation at all. Finally we can find b
n
. Here, we notice that |t| sin nπt
is odd and, therefore,
b
n
=
1
−1
f (t) sin nπt dt = 0.
162 CHAPTER 4. FOURIER SERIES AND PDES
Hence, the series is
f (t) =
1
2
+
¸
n=1
n odd
−4
n
2
π
2
cos nπt.
Let us explicitly write down the first few terms of the series up to the 3
rd
harmonic.
f (t) ≈
1
2
−
4
π
2
cos πt −
4
9π
2
cos 3πt −
The plot of these few terms and also a plot up to the 20
th
harmonic is given in Figure 4.9. You
should notice how close the graph is to the real function. You should also notice that there is no
"Gibbs phenomenon" present as there are no discontinuities.9: Fourier series of f (t) up to the 3
rd
harmonic (left graph) and up to the 20
th
harmonic
(right graph).
4.3.2 Convergence
We will need the one sided limits of functions. We will use the following notation
f (c−) = lim
t↑c
f (t), and f (c+) = lim
t↓c
f (t).
If you are unfamiliar with this notation, lim
t↑c
f (t) means we are taking a limit of as t approaches c
from below (i.e. t < c) and lim
t↓c
f (t) means we are taking a limit of as t approaches c from above
(i.e. t > c). For example, for the square wave function
f (t) =
¸
¸
¸
¸
¸
¸
0 if −π < t ≤ 0,
π if 0 < t ≤ π,
(4.8)
4.3. MORE ON THE FOURIER SERIES 163
we have f (0−) = 0 and f (0+) = π.
Let f (t) be a function defined on an interval [a, b]. Suppose that we find finitely many points
a = t
0
, t
1
, t
2
, . . . , t
k
= b in the interval, such that f (t) is continuous on the intervals (t
0
, t
1
), (t
1
, t
2
),
. . . , (t
k−1
, t
k
). Also suppose that f (t
k
−) and f (t
k
+) exists for each of these points. Then we say f (t)
is piecewise continuous.
If moreover, f (t) is differentiable at all but finitely many points, and f
·
(t) is piecewise contin-
uous, then f (t) is said to be piecewise smooth.
Example 4.3.2: The square wave function (4.8) is piecewise smooth on [−π, π] or any other inter-
val. In such a case we just say that the function is just piecewise smooth.
Example 4.3.3: The function f (t) = |t| is piecewise smooth.
Example 4.3.4: The function f (t) =
1
t
is not piecewise smooth on [−1, 1] (or any other interval
containing zero). In fact, it is not even piecewise continuous.
Example 4.3.5: The function f (t) =
3
√
t is not piecewise smooth on [−1, 1] (or any other interval
containing zero). f (t) is continuous, but the derivative of f (t) is unbounded near zero and hence
not piecewise continuous.
Piecewise smooth functions have an easy answer on the convergence of the Fourier series.
Theorem 4.3.1. Suppose f (t) is a 2L-periodic piecewise smooth function. Let
a
0
2
+
∞
¸
n=1
a
n
cos
nπ
L
t + b
n
sin
nπ
L
t
be the Fourier series for f (t). Then the series converges for all t. If f (t) is continuous near t, thenOtherwise
f (t−) + f (t+)
2
=
a
0
2
+
∞
¸
n=1
a
n
cos
nπ
L
t + b
n
sin
nπ
L
t.
If we happen to have that f (t) =
f (t−)+f (t+)
2
at all the discontinuities, the Fourier series converges
to f (t) everywhere. We can always just redefine f (t) by changing the value at each discontinuity
appropriately. Then we can write an equals sign between f (t) and the series without any worry.
We mentioned this fact briefly at the end last section.
Note that the theorem does not say how fast the series converges. Think back the discussion of
the Gibbs phenomenon in last section. The closer you get to the discontinuity, the more terms you
need to take to get an accurate approximation to the function.
164 CHAPTER 4. FOURIER SERIES AND PDES
4.3.3 Differentiation and integration of Fourier series
Not only does Fourier series converge nicely, but it is easy to differentiate and integrate the series.
We can do this just by differentiating or integrating term by term.
Theorem 4.3.2 continuous function and the derivative f
·
(t) is piecewise smooth. Then the
derivative can be obtained by differentiating term by term.
f
·
(t) =
∞
¸
n=1
−a
n
nπ
L
sin
nπ
L
t +
b
n
nπ
L
cos
nπ
L
t.
It is important that the function is continuous. It can have corners, but no jumps. Otherwise the
differentiated series will fail to converge. For an exercise, take the series obtained for the square
wave and try to differentiate the series. Similarly, we can also integrate a Fourier series.
Theorem 4.3.3 function. Then the antiderivative is obtained by antidifferentiating term by
term and so
F(t) =
a
0
t
2
+ C +
∞
¸
n=1
a
n
L
nπ
sin
nπ
L
t +
−b
n
L
nπ
cos
nπ
L
t.
where F
·
(t) = f (t) and C is an arbitrary constant.
Note that the series for F(t) is no longer a Fourier series as it contains the
a
0
t
2
term. The
antiderivative of a periodic function need no longer be periodic and so we should not expect a
Fourier series.
4.3.4 Rates of convergence and smoothness
Let us do an example of a periodic function with one derivative everywhere.
Example 4.3.6: Take the function
f (t) =
¸
¸
¸
¸
¸
¸
(1 − t) t if 0 < t < 1,
(t + 1) t if −1 < t < 0,
and extend to a 2-periodic function. The plot is given in Figure 4.10 on the facing page.
Note that this function has a derivative everywhere, but it does not have two derivatives at all
the integers.
4.3. MORE ON THE FOURIER SERIES 165
-2 -1 0 1 2
-2 -1 0 1 2
-0.50
-0.25
0.00
0.25
0.50
-0.50
-0.25
0.00
0.25
0.50
Figure 4.10: Smooth 2-periodic function.
Exercise 4.3.1: Compute f
··
(0+) and f
··
(0−).
Let us compute the Fourier series coefficients. The actual computation involves several inte-
gration by parts and is left to student.
a
0
=
1
−1
f (t) dt =
0
−1
(t + 1) t dt +
1
0
(1 − t) t dt = 0,
a
n
=
1
−1
f (t) cos nπt dt =
0
−1
(t + 1) t cos nπt dt +
1
0
(1 − t) t cos nπt dt = 0
b
n
=
1
−1
f (t) sin nπt dt =
0
−1
(t + 1) t sin nπt dt +
1
0
(1 − t) t sin nπt dt
=
4(1 − (−1)
n
)
π
3
n
3
=
¸
¸
¸
¸
¸
¸
8
π
3
n
3
if n is odd,
0 if n is even.
This series converges very fast. If you plot up to the third harmonic, that is the function
8
π
3
sin πt +
8
27π
3
sin 3πt,
it is almost indistinguishable from the plot of f (t) in Figure 4.10. In fact, the coefficient
8
27π
3
is
already just 0.0096 (approximately). The reason for this behavior is the n
3
term in the denominator.
The coefficients b
n
in this case go to zero as fast as
1
n
3
goes to zero.
It is a general fact that if you have one derivative, the Fourier coefficients will go to zero ap-
proximately like
1
n
3
. If you have only a continuous function, then the Fourier coefficients will go
to zero as
1
n
2
, and if you have discontinuities then the Fourier coefficients will go to zero approx-
imately as
1
n
. Therefore, we can tell a lot about the smoothness of a function by looking at its
Fourier coefficients.
166 CHAPTER 4. FOURIER SERIES AND PDES
To justify this behavior take for example the function defined by the Fourier series
f (t) =
∞
¸
n=1
1
n
3
sin nt.
When we differentiate term by term we notice
f
·
(t) =
∞
¸
n=1
1
n
2
cos nt.
Therefore, the coefficients now go down like
1
n
2
, which we said means that we have a continuous
function. That is, the derivative of f
·
(t) may be defined at most points, but at least at some points
it is not defined. If we differentiate again we find that f
··
(t) really is not defined at some points as
we get a piecewise differentiable function
f
··
(t) =
∞
¸
n=1
−1
n
sin nt.
This function is similar to the sawtooth. If we tried to differentiate again we would obtain
∞
¸
n=1
−cos nt,
which does not converge!
Exercise 4.3.2: Use a computer to plot f (t), f
·
(t) and f
··
(t). That is, plot say the first 5 harmonics
of the functions. At what points does f
··
(t) have the discontinuities.
4.3.5 Exercises
Exercise 4.3.3: Let
f (t) =
¸
¸
¸
¸
¸
¸
−t
10
if −10 < t < 0,
t
10
if 0 ≤ t < 10,
extended periodically (period is 20). a) Compute the Fourier series for f (t). b) Write out the series
explicitly up to the 3
rd
harmonic.
Exercise 4.3.6: Let f (t) =
¸
∞
n=1
1
n
3
cos nt. Is f (t) continuous and differentiable everywhere? Find
the derivative (if it exists) or justify if it does not exist.
Exercise 4.3.7: Let f (t) =
¸
∞
n=1
(−1)
n
n
sin nt. Is f (t) differentiable everywhere? Find the derivative
(if it exists) or justify if it does not exist.
168 CHAPTER 4. FOURIER SERIES AND PDES
4.4 Sine and cosine series
Note: 2 lectures, §9.3 in EP
4.4.1 Odd and even periodic functions
You may have noticed by now that an odd function has no cosine terms in the Fourier series and
an even function has no sine terms in the Fourier series. This observation is not a coincidence. Let
us look at even and odd periodic function in more detail.
Recall a function f (t) is odd if f (−t) = −f (t). A function f (t) is even if f (−t) = f (t). For
example, cos nt is even and sin nt is odd. Similarly the function t
k
is even if k is even and odd when
k is odd.
Exercise 4.4.1: Take two functions f (t) and g(t) and define their product h(t) = f (t)g(t). a) Sup-
pose both are odd, is h(t) odd or even? b) Suppose one is even and one is odd, is h(t) odd or even?
c) Suppose both are even, is h(t) odd or even.
If f (t) is odd and g(t) we cannot in general say anything about the sum f (t) + g(t). In fact, the
Fourier series of a function is really a sum of an odd (the sine terms) and an even (the cosine terms)
function.
In this section we are of course interested in odd and even periodic functions. We have previ-
ously defined the 2L-periodic extension of a function defined on the interval [−L, L]. Sometimes
we are only interested in the function in the range [0, L] and it would be convenient to have an odd
(resp. even) function. If the function is odd, all the sine (resp. cosine) terms will disappear. What
we can do is take the odd (resp. even) extension of the function to [−L, L] and then we can extend
periodically to a 2L-periodic function.
Take a function f (t) defined on [0, L]. On (−L, L] define the functions
F
odd
(t)
def
=
L
−L
f (t) cos
nπ
L
t dt = 0.
That is, there are no cosine terms in a Fourier series of an odd function. The integral is zero because
f (t) cos nπL t is an odd function (product of an odd and an even function is odd) and the integral
of an odd function over a symmetric interval is always zero. Furthermore, the integral of an even
function over a symmetric interval [−L, L] is twice the integral of the function over the interval
[0, L]. The function f (t) sin
nπ
L
t is the product of two odd functions and hence even.
b
n
=
1
L
L
−L
f (t) sin
nπ
L
t dt =
2
L
L
0
f (t) sin
nπ
L
t dt.
We can now write the Fourier series of f (t) as
∞
¸
n=1
b
n
sin
nπ
L
t.
Similarly, if f (t) is an even 2L-periodic function. For the same exact reasons as above, we find
that b
n
= 0 and
a
n
=
2
L
L
0
f (t) dt.
170 CHAPTER 4. FOURIER SERIES AND PDES
The Fourier series is then
a
0
2
∞
¸
n=1
a
n
cos
nπ
L
t.
An interesting consequence is that the coefficients of the Fourier series of an odd (or even)
function can be computed by just integrating over the half interval. Therefore, we can compute the
odd (or even) extension of a function as a Fourier series by computing certain integrals over the
interval where the original function is defined.
Theorem 4.4.1. Let f (t) be a piecewise smooth function defined on [0, L]. Then the odd extension
of f (t) has the Fourier series
F
odd
(t) =
∞
¸
n=1
b
n
sin
nπ
L
t,
where
b
n
=
2
L
L
0
f (t) cos
nπ
L
t dt.
The series
¸
∞
n=1
b
n
sin
nπ
L
t is called the sine series of f (t) and the series
a
0
2
+
¸
∞
n=1
a
n
cos
nπ
L
t
is called the cosine series of f (t). It is often the case that we do not actually care what happens
outside of [0, L]. In this case, we can pick whichever series fits our problem better.
It is not necessary to start with the full Fourier series to obtain the sine and cosine series. The
sine series is really the eigenfunction expansion of f (t) using the eigenfunctions of the eigenvalue
problem x
··
+ λx = 0, x(0) = 0, x(L) = L. The cosine series is the eigenfunction expansion of f (t)
using the eigenfunctions of the eigenvalue problem x
··
+ λx = 0, x
·
(0) = 0, x
·
(L) = L. We could
have, therefore, have gotten the same formulas by defining the inner product
( f (t), g(t)) =
L
0
f (t)g(t) dt,
and following the procedure of § 4.2. This point of view is useful because many times we use
a specific series because our underlying question will lead to a certain eigenvalue problem. In
fact, if the eigenvalue value problem is not one of the three we covered so far, you can still do
an eigenfunction expansion, generalizing the results of this chapter. We will deal with such a
generalization in chapter 5.
4.4. SINE AND COSINE SERIES 171
Example 4.4.2: Find the Fourier series of the even periodic extension of the function f (t) = t
2
for
0 ≤ t ≤ π.
We will write
f (t) =
a
0
2
+
∞
¸
n=1
a
n
cos nt,
where
a
0
=
2
π
π
0
t
2
dt =
2π
2
3
,
and
a
n
=
2
π
π
0
t
2
cos nt dt =
2
π
,
t
2
1
n
sin nt
¸
π
0
−
4
nπ
π
0
t sin nt dt
=
4
n
2
π
,
t cos nt
¸
π
0
+
4
n
2
π
π
0
cos nt dt =
4(−1)
n
n
2
.
Note that we have detected the "continuity" of the extension since the coefficients decay as
1
n
2
. That
is, the even extension of t
2
has no jump discontinuities. Although it will have corners, since the
derivative (which will be on odd function and a sine series) will have a series whose coefficients
decay only as
1
n
so it will have jumps.
Explicitly, the first few terms of the series are
π
2
3
− 4 cos t + cos 2t −
4
9
cos 3t +
Exercise 4.4.3: a) Compute the derivative of the even extension of f (t) above and verify it has
jump discontinuities. Use the actual definition of f (t), not its cosine series! b) Why is it that the
derivative of the even extension of f (t) is the odd extension of f
·
(t).
4.4.3 Application
We have said that Fourier series ties in to the boundary value problems we studied earlier. Let us
see this connection in more detail.
Suppose we have the boundary value problem for 0 < t < L,
x
··
(t) + λ x(t) = f (t),
for the Dirichlet boundary conditions x(0) = 0, x(L) = 0. By using the Fredholm alternative
(Theorem 4.1.2 on page 148) we note that as long as λ is not an eigenvalue of the underlying
homogeneous problem, there will exist a unique solution. Note that the eigenfunctions of this
eigenvalue problem were the functions sin
nπ
L
t. Therefore, to find the solution, we first find f (t) in
terms of the Fourier sine series. We write x as a sine series as well with unknown coefficients. We
substitute into the equation and solve for the Fourier coefficients of x.
If on the other hand we have the Neumann boundary conditions x
·
(0) = 0, x
·
(L) = 0. We do
the same procedure using the cosine series. These methods are best seen by examples.
172 CHAPTER 4. FOURIER SERIES AND PDES
Example 4.4.3: Take the boundary value problem for 0 < t < 1,
x
··
(t) + 2x(t) = f (t),
where f (t) = t on 0 < t < 1. We want to look for a solution x satisfying the Dirichlet conditions
x(0) = 0, x(1) = 0. We write f (t) as a sine series
f (t) =
∞
¸
n=1
c
n
sin nπt,
where
c
n
= 2
k
m
. So any solution to mx
··
(t)+kx(t) = F(t) will be of the form Acos ω
0
t +Bsin ω
0
t +
x
sp
, where x
sp
is the particular steady periodic solution. The steady periodic solution will always
have the same period as F(t).
In the spirit of the last section and the idea of undetermined coefficients we will first write
F(t) =
c
0
2
+
∞
¸
n=1
c
n
cos
nπ
L
t + d
n
sin
nπ
L
t.
Then we write
x(t) =
a
0
2
+
∞
¸
n=1
a
n
cos
nπ
L
t + b
n
sin
nπ
L
t,
and we plug in x into the differential equation and solve for a
n
and b
n
in terms of c
n
and d
n
. This is
perhaps best seen by example.
Example 4.5.1: Suppose that k = 2, and m = 1. The units are the mks units (meters-kilograms-
seconds) again. There is a jetpack strapped to the mass, which fires with a force of 1 Newtons for
1 second and then is off for 1 second. We want to find the steady periodic solution.
176 CHAPTER 4. FOURIER SERIES AND PDES
The equation is, therefore,
x
··
+ 2x = F(t),
where F(t) is the step function
F(t) =
+
∞
¸
n=1
nN
a
n
cos
nπ
L
t + b
n
sin
nπ
L
t.
In other words, we multiply the offending term by t. From then on, we proceed as before.
Of course, the solution will not be a Fourier series (it will not even be periodic) since it contains
these terms multiplied by t. Further, the terms t
a
N
cos
Nπ
L
t + b
N
sin
Nπ
L
t
will eventually dominate
and lead to wild oscillations. As before, this behavior is called pure resonance or just resonance.
Note that there now may be infinitely many resonance frequencies to hit. That is, as we change
the frequency of F (we change L), different terms from the Fourier series of F may interfere
with the complementary solution and will cause resonance. However, we should note that since
everything is an approximation and in particular c is never actually zero but something very close
to zero, only the first few resonance frequencies will matter.
Example 4.5.2: Find the steady periodic solution to the equation
2x
··
+ 18π
2
x = F(t),
where
F(t) =
¸
¸
¸
¸
¸
¸
1 if 0 < t < 1,
−1 if − 1 < t < 0,
extended periodically. We note that
F(t) =
∞
¸
n=1
n odd
4
πn
sin nπt.
4.5. APPLICATIONS OF FOURIER SERIES 179
Exercise 4.5.1: Compute the Fourier series of F to verify.
The solution must look like
x(t) = c
1
cos 3πt + c
2
sin 3πt + x
p
(t)
for some particular solution x
p
.
We note that if we just tried a Fourier series with sin nπt as usual, we would get duplication
when n = 3. Therefore, we pull out that term and multiply by t. And we have add a cosine term to
get everything right. That is, we must try
x
p
(t) = a
3
t cos 3πt + b
3
t sin 3πt +
∞
¸
n=1
n odd
n3
b
n
sin nπt.
Let us compute the second derivative.
x
··
p
(t) = −6a
3
π sin 3πt − 9π
2
a
3
t cos 3πt + 6b
3
π cos 3πt − 9π
2
b
3
t sin 3πt+
+
∞
¸
n=1
n odd
n3
(−n
2
π
2
b
n
) sin nπt.
We now plug into the differential equation.
2x
··
p
+ 18π
2
x = − 12a
3
π sin 3πt − 18π
2
a
3
t cos 3πt + 12b
3
π cos 3πt − 18π
2
b
3
t sin 3πt+
+ 18π
2
a
3
t cos 3πt + 18π
2
b
3
t sin 3πt+
+
∞
¸
n=1
n odd
n3
(−2n
2
π
2
b
n
+ 18π
2
b
n
) sin nπt.
If we simplify we obtain
2x
··
p
+ 18π
2
x = −12a
3
π sin 3πt + 12b
3
π cos 3πt +
∞
¸
n=1
n odd
n3
(−2n
2
π
2
b
n
+ 18π
2
b
n
) sin nπt.
This series has to equal to the series for F(t). We equate the coefficients and solve for a
3
and b
n
.
a
3
=
4/(3π)
−12π
=
−1
9π
2
b
3
= 0
b
n
=
4
nπ(18π
2
− 2n
2
π
2
)
=
2
π
3
n(9 − n
2
)
for n odd and n 3.
180 CHAPTER 4. FOURIER SERIES AND PDES
That is,
x
p
(t) =
−1
9π
2
t cos 3πt +
∞
¸
n=1
n odd
n3
2
π
3
n(9 − n
2
)
sin nπt.
When c > 0, you will not have to worry about pure resonance. That is, there will never be
any conflicts and you do not need to multiply any terms by t. There is a corresponding concept of
practical resonance and it is very similar to the ideas we already explored in chapter 2. We will not
go into details here.
4.5.3 Exercises
Exercise 4.5.2: Let F(t) =
1
2
+
¸
∞
n=1
1
n
2
cos nπt. Find the steady periodic solution to x
··
+2x = F(t).
Express your solution as a Fourier series.
Exercise 4.5.3: Let F(t) =
¸
∞
n=1
1
n
3
sin nπt. Find the steady periodic solution to x
··
+ x
·
+ x = F(t).
Express your solution as a Fourier series.
Exercise 4.5.4: Let F(t) =
¸
∞
n=1
1
n
2
cos nπt. Find the steady periodic solution to x
··
+ 4x = F(t).
Express your solution as a Fourier series.
Exercise 4.5.5: Let F(t) = t for −1 < t < 1 and extended periodically. Find the steady periodic
solution to x
··
+ x = F(t). Express your solution as a Fourier series.
Exercise 4.5.6: Let F(t) = t for −1 < t < 1 and extended periodically. Find the steady periodic
solution to x
··
+ π
2
x = F(t). Express your solution as a Fourier series.
4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 181
4.6 PDEs, separation of variables, and the heat equation
Note: 2 lectures, §9.5 in EP
Let us recall that a partial differential equation or PDE is an equation containing the partial
derivatives with respect to several independent variables. Solving PDEs will be our main applica-
tion of Fourier series.
A PDE is said to be linear if the dependent variable and its derivatives appear only to the first
power and in no functions. We will only talk about linear PDEs here. Together with a PDE, we
usually have specified some boundary conditions, where the value of the solution or its derivatives
is specified along the boundary of a region, and/or some initial conditions where the value of the
solution or its derivatives is specified for some initial time. Sometimes such conditions are mixed
together and we will refer to them simply as side conditions.
We will study three partial differential equations, each one representing a more general class of
equations. First, we will study the heat equation, which is an example of a parabolic PDE. Next,
we will study the wave equation, which is an example of a hyperbolic PDE. Finally, we will study
the Laplace equation, which is an example of an elliptic PDE. Each of our examples will illustrate
behaviour that is typical for the whole class.
4.6.1 Heat on an insulated wire
Let us first study the heat equation. Suppose that we have a wire (or a thin metal rod) that is
insulated except at the endpoints. Let x denote the position along the wire and let t denote time.
See Figure 4.13.
0
L x
insulation
temperature u
Figure 4.13: Insulated wire.
Now let u(x, t) denote the temperature at point x at time t. It turns out that the equation govern-
ing the this system is the so-called one-dimensional heat equation:
∂u
∂t
= k
∂
2
u
∂x
2
,
for some k > 0. That is, the change in heat at a specific point is proportional to the second derivative
of the heat along the wire. This makes sense. You would expect that if the heat distribution had a
maximum (was concave down), then heat would flow away from the maximum. And vice versa.
182 CHAPTER 4. FOURIER SERIES AND PDES
We will generally use a more convenient notation for partial derivatives. We will write u
t
instead of
∂u
∂t
and we will write u
xx
instead of
∂
2
u
∂x
2
. With this notation the equation becomes
u
t
= ku
xx
.
For the heat equation, we must also have some boundary conditions. We assume that the wire
is of length L and the ends are either exposed and touching some body of constant heat, or the ends
are insulated. If the ends of the wire are for example kept at temperature 0, then we must have the
conditions
u(0, t) = 0 and u(L, t) = 0.
If on the other hand the ends are also insulated we get the conditions
u
x
(0, t) = 0 and u
x
(L, t) = 0.
In other words, heat is not flowing in nor out of the wire at the ends. Note that we always have two
conditions along the x axis as there are two derivatives in the x direction. These side conditions
are called homogeneous.
Furthermore, we will suppose we know the initial temperature distribution.
u(x, 0) = f (x),
for some known function f (x). This initial condition is not a homogeneous side condition.
4.6.2 Separation of variables
First we must note the principle of superposition still applies. The heat equation is still called
linear, since u and its derivatives do not appear to any powers or in any functions. If u
1
and u
2
are
solutions and c
1
, c
2
are constants, then u = c
1
u
1
+ c
2
u
2
is still a solution.
Exercise 4.6.1: Verify the principle of superposition for the heat equation.
Superposition also preserves some of the side conditions. In particular, if u
1
and u
2
are solutions
that satisfy u(0, t) = 0 and u(L, t) = 0, and c
1
, c
2
are constants, then u = c
1
u
1
+ c
2
u
2
is still a
solution that satisfies u(0, t) = 0 and u(L, t) = 0. Similarly for the side conditions u
x
(0, t) = 0 and
u
x
(L, t) = 0. In general, superposition preserves all homogeneous side conditions.
The method of separation of variables is to try to find solutions that are sums or products of
functions of one variable. For example, for the heat equation, we try to find solutions of the form
u(x, t) = X(x)T(t).
That the desired solution we are looking for is of this form is too much to hope for. However, what
is perfectly reasonable to ask is to find enough "building-block" solutions u(x, t) = X(x)T(t) using
4.6. PDES, SEPARATION OF VARIABLES, AND THE HEAT EQUATION 183
this procedure so that the desired solution to the PDE is somehow constructed from these building
blocks by the use of superposition.
Let us try to solve the heat equation
u
t
= ku
xx
with u(0, t) = 0 and u(L, t) = 0 and u(x, 0) = f (x).
Let us guess u(x, t) = X(x)T(t). We plug into the heat equation to obtain
X(x)T
·
(t) = kX
··
(x)T(t).
We rewrite as
T
·
(t)
kT(t)
=
X
··
(x)
X(x)
.
As this equation is supposed to hold for all x and t. But the left hand side does not depend on t and
the right hand side does not depend on x. Therefore, each side must be a constant. Let us call this
constant −λ (the minus sign is for convenience later). Thus, we have two equations
T
·
(t)
kT(t)
= −λ =
X
··
(x)
X(x)
.
Or in other words
X
··
(x) + λX(x) = 0,
T
·
(t) + λkT(t) = 0.
The boundary condition u(0, t) = 0 implies X(0)T(t) = 0. We are looking for a nontrivial solution
and so we can assume that T(t) is not identically zero. Hence X(0) = 0. Similarly, u(L, t) = 0
implies X(L) = 0. We are looking for nontrivial solutions X of the eigenvalue problem X
··
+λX = 0,
X(0) = 0, X(L) = 0. We have previously found that the only eigenvalues are λ
n
=
n
2
π
2
L
2
, for integers
n ≥ 1, where eigenfunctions are sin
nπ
L
x. Hence, let us pick the solutions
X
n
(x) = sin
nπ
L
x.
The corresponding T
n
must satisfy the equation
T
·
n
(t) +
n
2
π
2
L
2
kT
n
(t) = 0.
By the method of integrating factor, the solution of this problem is easily seen to be
T
n
(t) = e
−n
2
π
2
L
2
kt
.
It will be useful to note that T
n
(0) = 1. Our building-block solutions are
u
n
(x, t) = X
n
(x)T
n
(t) =
C(ξ) dξ + B(η). The solution
must then be of the following form for some functions A(ξ) and B(η).
y = A(ξ) + B(η) = A(x − at) + B(x + at).
4.8.2 The formula
We know what any solution must look like, but we need to solve for the given side conditions. We
will just give the formula and see that it works. First let F(x) denote the odd extension of f (x), and
let G(x) denote the odd extension of g(x). Now define
A(x) =
1
2
F(x) −
1
2a
x
0
G(s) ds is an even function of x because G(x) is odd (to see this fact, do the substitution
§
We can just as well integrate with ξ first, if we wish.
200 CHAPTER 4. FOURIER SERIES AND PDES
s = −v). So
y(0, t) =
1
2
F(−at) −
1
2a
at
0
G(s) ds = 0.
And voilà, it works.
Example 4.8.1: What the d'Alembert solution says is that the solution is a superposition of two
functions (waves) moving in the opposite direction at "speed" a. To get an idea of how it works,
let us do an example. Suppose that we have the simpler setup
y
tt
= y
xx
,
y(0, t) = y(1, t) = 0,
y(x, 0) = f (x),
y
t
(x, 0) = 0.
Here f (x) is an impulse of height 1 centered at x = 0.5:
f (x) =
0 if x < −1,
x + 1 if −1 ≤ x < 0,
−x + 1 if 0 ≤ x < 1,
0 if x > 1.
Solve using the d'Alembert solution. That is, write down a piecewise definition for the solution.
Then sketch the solution for t = 0, t =
1
2
, t = 1, and t = 2.
204 CHAPTER 4. FOURIER SERIES AND PDES
4.9 Steady state temperature, Laplacian, and Dirichlet prob-
lems
Note: 1 lecture, §9.7 in EP
Suppose we have an insulated wire, a plate, or a 3-dimensional object. We apply certain fixed
temperatures on the ends of the wire, the edges of the plate or on all sides of the 3-dimensional
object. We wish to find out what is the steady state temperature distribution. That is, we wish to
know what will be the temperature after long enough period of time.
We are really looking for a solution to the heat equation that is not dependent on time. Let us
first do this in one space variable. We are looking for a function u that satisfies
u
t
= ku
xx
,
but such that u
t
= 0 for all x and t. Hence, we are looking for a function of x alone that satisfies
u
xx
= 0. It is easy to solve this equation by integration and we see that u = Ax + B for some
constants A and B.
Suppose we have an insulated wire, and we apply constant temperature T
1
at one end (say
where x = 0) and T
2
and the other end (at x = L where L is the length of the wire). Then our steady
state solution is
u(x) =
T
2
− T
1
L
x + T
1
.
This solution agrees with our common sense intuition with how the heat should be distributed in
the wire. So in one dimension, the steady state solutions are basically just straight lines.
Things are more complicated in two or more space dimensions. Let us restrict to two space
dimensions for simplicity. The heat equation in two variables is
u
t
= k(u
xx
+ u
yy
), (4.18)
or more commonly written as u
t
= k∆u or u
t
= k∇
2
u. Here the ∆ and ∇
2
symbols mean
∂
2
∂x
2
+
∂
2
∂y
2
.
We will use ∆ from now on. The reason for that notation is that you can define ∆ to be the right
thing for any number of space dimensions and then the heat equation is always u
t
= k∆u. The ∆ is
called the Laplacian.
OK, now that we have notation out of the way, let us see what does an equation for the steady
state solution look like. We are looking for a solution to (4.18) which does not depend on t. Hence
we are looking for a function u(x, y) such that
∆u = u
xx
+ u
yy
= 0.
This equation is called the Laplace equation
]
. Solutions to the Laplace equation are called har-
monic functions and have many nice properties and applications far beyond the steady state heat
problem.
]
Named after the French mathematician Pierre-Simon, marquis de Laplace (1749 – 1827).
4.9. STEADY STATE TEMPERATURE 205
Harmonic functions in two variables are no longer just linear (plane graphs). For example, you
can check that the functions x
2
− y
2
and xy are harmonic. However, if you remember your multi-
variable calculus we note that if u
xx
is positive, u is concave up in the x direction, then u
yy
must be
negative and u must be concave down in the y direction. Therefore, a harmonic function can never
have any "hilltop" or "valley" on the graph. This observation is consistent with our intuitive idea
of steady state heat distribution.
Commonly the Laplace equation is part of a so-called Dirichlet problem
|
. That is, we have
some region in the xy-plane and we specify certain values along the boundaries of the region. We
then try to find a solution u defined on this region such that u agrees with the values we specified
on the boundary.
For simplicity, we will consider a rectangular region. Also for simplicity we will specify bound-
ary values to be zero at 3 of the four edges and only specify an arbitrary function at one edge. As
we still have the principle of superposition, you can use this simpler solution to derive the gen-
eral solution for arbitrary boundary values by solving 4 different problems, one for each edge, and
adding those solutions together. This setup is left as an exercise.
We wish to solve the following problem. Let h and w be the height and width of our rectangle,
with one corner at the origin and lying in the first quadrant.
∆u = 0, (4.19)
u(0, y) = 0 for 0 < y < h, (4.20)
u(x, h) = 0 for 0 < x < w, (4.21)
u(w, y) = 0 for 0 < y < h, (4.22)
u(x, 0) = f (x) for 0 < x < w. (4.23)
(0, 0)
(0, h)
u = 0
u = 0
u = f (x) (w, 0)
u = 0
(w, h)
The method we will apply is separation of variables. Again, we will come up with enough
building-block solutions satisfying all the homogeneous boundary conditions (all conditions except
(4.23)). We notice that superposition still works for all the equation and all the homogeneous
conditions. Therefore, we can use the Fourier series for f (x) to solve the problem as before.
We try u(x, y) = X(x)Y(y). We plug into the equation to get
X
··
Y + XY
··
= 0.
We put the Xs on one side and the Ys on the other to get
−
X
··
X
=
Y
··
Y
.
|
Named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805 – 1859).
206 CHAPTER 4. FOURIER SERIES AND PDES
The left hand side only depends on x and the right hand side only depends on y. Therefore, there
is some constant λ such that λ =
−X
··
X
=
Y
··
Y
. And we get two equations
X
··
+ λX = 0,
Y
··
− λY = 0.
Furthermore, the homogeneous boundary conditions imply that X(0) = X(w) = 0 and Y(h) = 0.
Taking the equation for X we have already seen that we have a nontrivial solution if and only if
λ = λ
n
=
n
2
π
2
w
2
and the solution is a multiple of
X
n
(x) = sin
nπ
w
x.
For these given λ
n
, the general solution for Y (one for each n) is
Y
n
(y) = A
n
cosh
nπ
w
y + B
n
sinh
nπ
w
y. (4.24)
We only have one condition on Y
n
and hence we can pick one of A
n
or B
n
constants to be whatever
is convenient. It will be useful to have Y
n
(0) = 1, so we let A
n
= 1. Setting Y
n
(h) = 0 and solving
for B
n
we get that
B
n
=
−cosh
nπh
w
sinh
nπh
w
.
After we plug the A
n
and B
n
we into (4.24) and simplify, we find
Y
n
(y) =
sinh
nπ(h−y)
w
sinh
nπh
w
.
We define u
n
(x, y) = X
n
(x)Y
n
(y). And note that u
n
satisfies (4.19)–(4.22).
Observe that
u
n
(x, 0) = X
n
(x)Y
n
(0) = sin
nπ
n
x.
Suppose that
f (x) =
∞
¸
n=1
b
n
sin
nπx
w
.
Then we get a solution of (4.19)–(4.23) of the following form.
u(x, y) =
∞
¸
n=1
b
n
u
n
(x, y) =
∞
¸
n=1
b
n
sin
nπ
w
x
¸
¸
¸
¸
¸
¸
sinh
nπ(h−y)
w
sinh
nπh
w
¸
¸
¸
¸
¸
¸
.
As u
n
satisfies (4.19)–(4.22) and any linear combination (finite or infinite) of u
n
must also satisfy
(4.19)–(4.22), we see that u must satisfy (4.19)–(4.22). By plugging in y = 0 it is easy to see that
u satisfies (4.23) as well.
4.9. STEADY STATE TEMPERATURE 207
Example 4.9.1: Suppose that we take w = h = π and we let f (x) = π. We compute the sine series
for the function π (we will get the square wave). We find that for 0 < x < π we have
f (x) =
∞
¸
n=1
n odd
4
n
sin nx.
Therefore the solution u(x, y), see Figure 4.22, to the corresponding Dirichlet problem is given as
u(x, y) =
∞
¸
n=1
n odd
4
n
(sin nx)
¸
sinh n(π − y)
sinh nπ
λ
n
x.
0 2 4 6
0 2 4 6
-4
-2
0
2
4
-4
-2
0
2
4
Figure 5.1: Plot of
1
x
and tan x.
5.1.2 Orthogonality
We have seen the notion of orthogonality before. For example, we have shown that sin nx are
orthogonal for distinct n on [0, π]. For general Sturm-Liouville problems we will need a more
general setup. Let r(x) be a weight function (any function, though generally we will assume it is
positive) on [a, b]. Then two functions f (x), g(x) are said to be orthogonal with respect to the
weight function r(x) when
b
a
f (x) g(x) r(x) dx,
and then say f and g are orthogonal whenever ( f , g) = 0. The results and concepts are again
analogous to finite dimensional linear algebra.
The idea of the given inner product is that those x where r(x) is greater have more weight.
Nontrivial (nonconstant) r(x) arise naturally, for example from a change of variables. Hence, you
could think of a change of variables such that dξ = r(x) dx.
We have the following orthogonality property of eigenfunctions of a regular Sturm-Liouville
problem.
Theorem 5.1.2. Suppose we have a regular Sturm-Liouville problem
d
dx
¸
p(x)
dy
dx
b
a
y
j
(x) y
k
(x) r(x) dx = 0,
that is, y
j
and y
k
are orthogonal with respect to the weight function r.
Proof is very similar to the analogous theorem from § 4.1. It can also be found in many books
including, for example, Edwards and Penney [EP].
5.1.3 Fredholm alternative
We also have the Fredholm alternative theorem we talked about before for all regular Sturm-
Liouville problems. We state it here for completeness.
Theorem 5.1.3 (Fredholm alternative). Suppose that we have a regular Sturm-Liouville problem.
Then either
d
dx
¸
p(x)
dy
dx
− q(x)y + λr(x)y = f (x),
α
1
y(a) − α
2
y
·
(a) = 0,
β
1
y(b) + β
2
y
·
(b) = 0,
has a unique solution for any f (x) continuous on [a, b].
This theorem is used in much the same way as we did before in § 4.4. It is used when solving
more general nonhomogeneous boundary value problems. Actually the theorem does not help us
solve the problem, but it tells us when does a solution exist and is unique, so that we know when
to spend time looking for a solution. To solve the problem we decompose f (x) and y(x) in terms
of the eigenfunctions of the homogeneous problem, and then solve for the coefficients of the series
for y(x).
5.1.4 Eigenfunction series
What we want to do with the eigenfunctions once we have them is to compute the eigenfunction
decomposition of an arbitrary function f (x). That is, we wish to write
f (x) =
∞
¸
n=1
c
n
y
n
(x), (5.3)
where y
n
(x) the eigenfunctions. We wish to find out if we can represent any function f (x) in
this way, and if so, we wish to calculate c
n
(and of course we would want to know if the sum
converges). OK, so imagine we could write f (x) as above. We will assume convergence and the
ability to integrate term by term. Because of orthogonality we have
( f , y
m
) =
2
r(x) dx
. (5.4)
5.1. STURM-LIOUVILLE PROBLEMS 217
Note that y
m
are known up to a constant multiple, so we could have picked a scalar multiple
of an eigenfunction such that (y
m
, y
m
) = 1 (if we had an arbitrary eigenfunction ˜ y
m
, divide it
by
.
The point is that X
n
T
n
is a solution that satisfies all the homogeneous conditions (that is, all condi-
tions except the initial position). And since and T
n
(0) = 1, we have
y(x, 0) =
∞
¸
n=1
b
n
X
n
(x)T
n
(0) =
∞
¸
n=1
b
n
X
n
(x) =
∞
¸
n=1
b
n
sin nπx = f (x).
So y(x, t) solves (5.5).
Note that the natural (circular) frequency of the system is n
2
π
2
a
2
. These frequencies are all
integer multiples of the fundamental frequency π
2
a
2
, so we will get a nice musical note. The exact
frequencies and their amplitude are what we call the timbre of the note.
The timbre of a beam is different than for a vibrating string where we will get "more" of the
smaller frequencies since we will get all integer multiples, 1, 2, 3, 4, 5, . . . For a steel beam we will
get only the square multiples 1, 4, 9, 16, 25, . . . That is why when you hit a steel beam you hear a
very pure sound. The sound of a xylophone or vibraphone is, therefore, very different from a guitar
or piano.
5.2. APPLICATION OF EIGENFUNCTION SERIES 221
Example 5.2.1: Let us assume that f (x) =
x(x−1)
10
. On 0 < x < 1 we have (you know how to do this
by now)
f (x) =
∞
¸
n=1
n odd
4
5π
3
n
3
sin nπx.
Hence, the solution to (5.5) with the given initial position f (x) is
y(x, t) =
∞
¸
n=1
n odd
4
5π
3
n
3
(sin nπx)
cos n
2
π
2
a
2
t
.
5.2.1 Exercises
Exercise 5.2.2: Suppose you have a beam of length 5 with free ends. Let y be the transverse
deviation of the beam at position x on the beam (0 < x < 5). You know that the constants are such
that this satisfies the equation y
tt
+4y
xxxx
= 0. Suppose you know that the initial shape of the beam
is the graph of x(5 − x), and the initial velocity is uniformly equal to 2 (same for each x) in the
positive y direction. Set up the equation together with the boundary and initial conditions. Just set
up, do not solve.
Exercise 5.2.3: Suppose you have a beam of length 5 with one end free and one end fixed (the
fixed end is at x = 5). Let u be the longitudinal deviation of the beam at position x on the beam
(0 < x < 5). You know that the constants are such that this satisfies the equation u
tt
= 4u
xx
.
Suppose you know that the initial displacement of the beam is
x−5
50
, and the initial velocity is
−(x−5)
100
in the positive u direction. Set up the equation together with the boundary and initial conditions.
Just set up, do not solve.
Exercise 5.2.4: Suppose the beam is L units long, everything else kept the same as in (5.5). What
is the equation and the series solution.
Exercise 5.2.5: Suppose you have
a
4
y
xxxx
+ y
tt
= 0 (0 < x < 1, t > 0),
y(0, t) = y
xx
(0, t) = 0,
y(1, t) = y
xx
(1, t) = 0,
y(x, 0) = f (x), y
t
(x, 0) = g(x).
That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did
for the wave equation.
222 CHAPTER 5. EIGENVALUE PROBLEMS
5.3 Steady periodic solutions
Note: 1–2 lectures, §10.3 in EP
5.3.1 Forced vibrating string.
Suppose that we have a guitar string of length L. We have studied the wave equation problem in
this case, where x was the position on the string, t was time and y was the displacement of the
string. See Figure 5.3.
L x
y
y
0
Figure 5.3: Vibrating string.
The problem is governed by the equations
y
tt
= a
2
y
xx
,
y(0, t) = 0, y(L, t) = 0,
y(x, 0) = f (x), y
t
(x, 0) = g(x).
(5.6)
We saw previously that the solution is of the form
y =
∞
¸
n=1
A
n
cos
nπa
L
t + B
n
sin
nπa
L
t
sin
nπ
L
x
where A
n
and B
n
were determined by the initial conditions. The natural frequencies of the system
are the (circular) frequencies
nπa
L
for integers n ≥ 1.
But these are free vibrations. What if there is an external force acting on the string. Let us
assume say air vibrations (noise), for example a second string. Or perhaps a jet engine. For
simplicity, assume nice pure sound and assume the force is uniform at every position on the string.
Let us say F(t) = F
0
cos ωt as force per unit mass. Then our wave equation becomes (remember
acceleration is force times mass)
y
tt
= a
2
y
xx
+ F
0
cos ωt. (5.7)
with the same boundary conditions of course.
5.3. STEADY PERIODIC SOLUTIONS 223
We will want to find the solution here that satisfies the above equation and
y(0, t) = 0, y(L, t) = 0, y(x, 0) = 0, y
t
(x, 0) = 0. (5.8)
That is, the string is initially at rest. First we find a particular solution y
p
of (5.7) that satisfies
y(0, t) = y(L, t) = 0. We define the functions f and g as
f (x) = −y
p
(x, 0), g(x) = −
∂y
p
∂t
(x, 0).
We then find solution y
c
of (5.6). If we add the two solutions, we find that y = y
c
+ y
p
solves (5.7)
with the initial conditions.
Exercise 5.3.1: Check that y = y
c
+ y
p
solves (5.7) and the side conditions (5.8).
So the big issue here is to find the particular solution y
p
. We look at the equation and we make
an educated guess
y
p
(x, t) = X(x) cos ωt.
We plug in to get
−ω
2
X cos ωt = a
2
X
··
cos ωt + F
0
cos ωt
or −ω
2
X = a
2
X
··
+ F
0
after cancelling the cosine. We know how to find a general solution to
this equation (it is an nonhomogeneous constant coefficient equation) and we get that the general
solution is
X(x) = Acos
ω
a
x + Bsin
ω
a
x −
F
0
ω
2
.
The endpoint conditions imply that X(0) = X(L) = 0, so
0 = X(0) = A −
F
0
ω
2
or A =
F
0
ω
2
and
0 = X(L) =
F
0
ω
2
cos
ωL
a
+ Bsin
ωL
a
−
F
0
ω
2
.
Assuming that sin
ωL
a
is not zero we can solve for B to get
B =
−F
0
¸
¸
¸
¸
¸
cos ωt.
224 CHAPTER 5. EIGENVALUE PROBLEMS
Exercise 5.3.2: Check that y
p
works.
Now we get to the point that we skipped. Suppose that sin
ωL
a
= 0. What this means is that ω
is equal to one of the natural frequencies of the system, i.e. a multiple of
πa
L
. We notice that if ω is
not equal to a multipe of the base frequency, but is very close, then the coefficient B in (5.9) seems
to become very large. But let us not jump to conclusions just yet. When ω =
nπa
L
for n even, then
cos
ωL
a
= 1 and hence we really get that B = 0. So resonance occurs only when both cos
ωL
a
= −1
and sin
ωL
a
= 0. That is when ω =
nπa
L
for odd n.
We could again solve for the resonance solution if we wanted to, but it is, in the right sense, the
limit of the solutions as ω gets close to a resonance frequency. In real life, pure resonance never
occurs anyway.
The above calculation explains why a string will begin to vibrate if the identical string is
plucked close by. In the absence of friction this vibration would get louder and louder as time
goes on. On the other hand, you are unlikely to get large vibration if the forcing frequency is not
close to a resonance frequency even if you have a jet engine running close to the string. That is,
the amplitude will not keep increasing unless you tune to just the right frequency.
Similar resonance phenomena occur when you break a wine glass using human voice (yes this
is possible, but not easy
†
) if you happen to hit just the right frequency. Remember a glass has much
purer sound, i.e. it is more like a vibraphone, so there are far fewer resonance frequencies to hit.
When the forcing function is more complicated, you decompose it in terms of the Fourier series
and apply the above result. You may also need to solve the above problem if the forcing function
is a sine rather than a cosine, but if you think about it, the solution is almost the same.
Example 5.3.1: Let us do the computation for specific values. Suppose F
0
= 1 and ω = 1 and
L = 1 and a = 1. Then
y
p
(x, t) =
¸
cos x −
cos 1 − 1
sin 1
sin x − 1
cos t.
It is not hard to compute specific values for an odd extension of a function and hence (5.10) is
a wonderful solution to the problem. For example it is very easy to have a computer do it, unlike
series solutions. A plot is given in Figure 5.4.
5.3.2 Underground temperature oscillations
Let u(x, t) be the temperature at a certain location at depth x underground at time t. See Figure 5.5
on the following page.
226 CHAPTER 5. EIGENVALUE PROBLEMS
depth x
Figure 5.5: Underground temperature.
The temperature u satisfies the heat equation u
t
= ku
xx
, where k is the diffusivity of the soil.
We know the temperature at the surface u(0, t) from weather records. Let us assume for simplicity
that
u(0, t) = T
0
+ A
0
cos ωt.
For some base temperature T
0
, then t = 0 is midsummer (could put negative sign above to make it
midwinter). A
0
is picked properly to make this the typical variation for the year. That is, the hottest
temperature is T
0
+ A
0
and the coldest is T
0
− A
0
. For simplicity, we will assume that T
0
= 0. ω is
picked depending on the units of t, such that when t = 1year then ωt = 2π.
It seems reasonable that the temperature at depth x will also oscillate with the same frequency.
And this in fact will be the steady periodic solution, independent of the initial conditions. So we
are looking for a solution of the form
u(x, t) = V(x) cos ωt + W(x) sin ωt.
for the problem
u
t
= ku
xx
, u(0, t) = A
0
cos ωt. (5.11)
We will employ the complex exponential here to make calculations simpler. Suppose we have
a complex valued function
h(x, t) = X(x) e
iωt
.
We will look for an h such that Re h = u. To find an h, whose real part satisfies (5.11), we look for
an h such that
h
t
= kh
xx
, h(0, t) = A
0
e
iωt
. (5.12)
Exercise 5.3.3: Suppose h satisfies (5.12). Use Euler's formula for the complex exponential to
check that u = Re h satisfies (5.11).
Substitute h into (5.12).
iωXe
iωt
= kX
··
e
iωt
Hence,
kX
··
− iωX = 0,
5.3. STEADY PERIODIC SOLUTIONS 227
or
X
··
− α
2
X = 0,
where α = ±
ω
2k
= π
we find the depth in centimeters where the seasons are reversed. That is, we get the depth at which
summer is the coldest and winter is the warmest. We get approximately 700 centimeters which is
approximately 23 feet below ground.
But be careful. The temperature swings decay rapidly as you dig deeper. The amplitude of the
temperature swings is A
0
e
−
√
ω
2k
x
. This decays very quickly as x grows. Let us again take typical
parameters as above. We also will assume that our surface temperature temperature swing is ±15
◦
Celsius, that is, A
0
= 15. Then the maximum temperature variation at 700 centimeters is only
±0.66
◦
Celsius.
You need not dig very deep to get an effective "refrigerator." I.e. Why wines are kept in a cellar;
you need consistent temperature. The temperature differential could also be used to for energy. A
home could be heated or cooled by taking advantage of the above fact. Even without the earth core
you could heat a home in the winter and cool it in the summer. There is also the earth core, so
temperature presumably gets higher the deeper you dig. We did not take that into account above.
228 CHAPTER 5. EIGENVALUE PROBLEMS
5.3.3 Exercises
Exercise 5.3.5: Suppose that the forcing function for the vibrating string is F
0
sin ωt. Derive the
particular solution y
p
.
Exercise 5.3.6: Take the forced vibrating string. Suppose that L = 1, a = 1. Suppose that the
forcing function is the quare wave which is 1 on the interval 0 < x < 1 and −1 on the interval
−1 < x < 0. Find the particular solution. Hint: you may want to use result of Exercise 5.3.5.
Exercise 5.3.7: The units are cgs (centimeters, grams, seconds). For k = 0.005, ω = 1.991×10
−7
,
A
0
= 20. Find the depth at which the temperature variation is half (±10 degrees) of what it is on
the surface.
Exercise 5.3.8: Derive the solution for underground temperature oscillation without assuming
that T
0
= 0.
Chapter 6
The Laplace transform
6.1 The Laplace transform
Note: 2 lectures, §10.1 in EP
6.1.1 The transform
In this chapter we will discuss the Laplace transform
∗
. The Laplace transform turns out to be a very
efficient method to solve certain ODE problems. In particular, the transform can take a differential
equation and turn it into an algebraic equation. If the algebraic equation can be solved, applying
the inverse transform gives us our desired solution. The Laplace transform is also useful in the
analysis of certain systems such as electrical circuits, NMR spectroscopy, signal processing and
others. Finally, understanding the Laplace transform will also help with understanding the related
Fourier transform, which, however, requires more understanding of complex numbers. We will not
cover the Fourier transform.
The Laplace transform also gives a lot of insight into the nature of the equations we are dealing
with. It can be seen as converting between the time and the frequency domain. For example, take
the standard equation
mx
··
(t) + cx
·
(t) + kx(t) = f (t).
We can think of t as time and f (t) as incoming signal. The Laplace transform will convert the
equation from a differential equation in time to an algebraic (no derivatives) equation, where the
new independent variable s is the frequency.
We can think of the Laplace transform as a black box. It eats functions and spits out functions
in a new variable. We write !¦ f (t)¦ = F(s). It is common to write lower case letters for functions
in the time domain and upper case letters for functions in the frequency domain. We will use the
∗
Just like the Laplace equation and the Laplacian, also named after Pierre-Simon, marquis de Laplace (1749 –
1827).
229
230 CHAPTER 6. THE LAPLACE TRANSFORM
same letter to denote that one function is the Laplace transform of the other, for example F(s) is
the Laplace transform of f (t). Let us define the transform.
!¦ f (t)¦ = F(s)
def
=
∞
0
e
−st
f (t) dt.
We note that we are only considering t ≥ 0 in the transform. Of course, if we think of t as time there
is no problem, we are generally interested in finding out what will happen in the future (Laplace
transform is one place where it is safe to ignore the past). Let us compute the simplest transforms.
Example 6.1.1: Suppose f (t) = 1, then
!¦1¦ =
¸
¸
¸
¸
¸
¸
0 if t < 0,
1 if t ≥ 0.
†
The function is named after Oliver Heaviside (1850–1925). Only by coincidence is the function "heavy" on "one
side."
6.1. THE LAPLACE TRANSFORM 231
Let us find the Laplace transform of u(t − a), where a ≥ 0 is some constant. That is, the function
which is 0 for t < a and 1 for t ≥ a.
!¦u(t − a)¦ =
∞
0
e
−st
f (t) dt = C!¦ f (t)¦.
So we can "pull out" a constant out of the transform. Similarly we have linearity. Since linearity
is very important we state it as a theorem.
Theorem 6.1.1 (Linearity of Laplace transform). Suppose that A, B, and C are constants, then
!¦Af (t) + Bg(t)¦ = A!¦ f (t)¦ + B!¦g(t)¦,
and in particular
!¦Cf (t)¦ = C!¦ f (t)¦.
232 CHAPTER 6. THE LAPLACE TRANSFORM
Exercise 6.1.2: Verify the theorem. That is, show that !¦Af (t) + Bg(t)¦ = A!¦ f (t)¦ + B!¦g(t)¦.
These rules together with Table 6.1 on the previous page make it easy to already find the
Laplace transform of a whole lot of functions already. It is a common mistake to think that Laplace
transform of a product is the product of the transforms. But in general
!¦ f (t)g(t)¦ !¦ f (t)¦!¦g(t)¦.
It must also be noted that not all functions have Laplace transform. For example, the function
1
t
does not have a Laplace transform as the integral diverges. Similarly tan t or e
t
2
do not have
Laplace transforms.
6.1.2 Existence and uniqueness
Let us consider in more detail when does the Laplace transformexist. First let us consider functions
of exponential order. f (t) is of exponential order as t goes to infinity if
| f (t)| ≤ Me
ct
,
for some constants M and c, for sufficiently large t (say for all t > t
0
for some t
0
). The simplest
way to check this condition is to try and compute
lim
t→∞
f (t)
e
ct
.
If the limit exists and is finite (usually zero), then f (t) is of exponential order.
Exercise 6.1.3: Use L'Hopital's rule from calculus to show that a polynomial is of exponential
order. Hint: Note that a sum of two exponential order functions is also of exponential order. Then
show that t
n
is of exponential order for any n.
For an exponential order function we have existence and uniqueness of the Laplace transform.
Theorem 6.1.2 (Existence). Let f (t) be continuous and of exponential order for a certain constant
c. Then F(s) = !¦ f (t)¦ is defined for all s > c.
You may have existence of the transform for other functions, that are not of exponential order,
but that will not relevant to us. Before dealing with uniqueness, let us also note that for exponential
order functions you also obtain that their Laplace transform decays at infinity:
lim
s→∞
F(s) = 0.
Theorem 6.1.3 (Uniqueness). Let f (t) and g(t) be continuous and of exponential order. Suppose
that there exists a constant C, such that F(s) = G(s) for all s > C. Then f (t) = g(t) for all t ≥ 0.
6.1. THE LAPLACE TRANSFORM 233
Both theorems hold for piecewise continuous functions as well. Recall that piecewise contin-
uous means that the function is continuous except perhaps at a discrete set of points where it has
jump discontinuities like the Heaviside function. Uniqueness however does not "see" values at the
discontinuities. So you can only conclude that f (t) = g(t) outside of discontinuities. For example,
the unit step function is sometimes defined using u(0) =
1
2
. This new step function, however, we
defined has the exact same Laplace transform as the one we defined earlier where u(0) = 1.
6.1.3 The inverse transform
As we said, the Laplace transform will allow us to convert a differential equation into an algebraic
equation which we can solve. Once we do solve the algebraic equation in the frequency domain we
will want to get back to the time domain, as that is what we are really interested in. We, therefore,
need to also be able to get back. If we have a function F(s), to be able to find f (t) such that
!¦ f (t)¦ = F(s), we need to first know if such a function is unique. It turns out we are in luck by
Theorem 6.1.3. So we can without fear make the following definition.
If F(s) = !¦ f (t)¦ for some function f (t). We define the inverse Laplace transform as
!
−1
¦F(s)¦
def
= f (t).
There is an integral formula for the inverse, but it is not as simple as the transform itself (requires
complex numbers). The best way to compute the inverse is to use the Table 6.1 on page 231.
Example 6.1.5: Take F(s) =
1
s+1
. Find the inverse Laplace transform.
We look at the table and we find
!
−1
= 1 + sin t.
A useful property is the so-called shifting property or the first shifting property
!¦e
−at
f (t)¦ = F(s + a),
where F(s) is the Laplace transform of f (t).
Exercise 6.1.4: Derive this property from the definition.
The shifting property can be used when the denominator is a more complicated quadratic that
may come up in the method of partial fractions. You always want to write such quadratics as
(s + a)
2
+ b by completing the square and then using the shifting property.
Example 6.1.7: Find !
−1
¸
1
s
2
+4s+8
¸
.
First we complete the square to make the denominator (s + 2)
2
+ 4. Next we find
!
−1
1
s
2
+ 4
=
1
4
sin 2t.
Putting it all together with the shifting property we find
!
−1
1
s
2
+ 4s + 8
= !
−1
1
(s + 2)
2
+ 4
=
1
4
e
−2t
sin 2t.
In general, we will want to be able to apply the Laplace transform to rational functions, that is
functions of the form
F(s)
G(s)
where F(s) and G(s) are polynomials. Since normally (for functions that we are considering) the
Laplace transform goes to zero as s → ∞, it is not hard to see that the degree of F(s) will always
be smaller than that of G(s). Such rational functions are called proper rational functions and we
will always be able to apply the method of partial fractions. Of course this means we will need to
be able to factor the denominator into linear and quadratic terms, which involves finding the roots
of the denominator.
6.1.4 Exercises
Exercise 6.1.5: Find the Laplace transform of 3 + t
5
+ sin πt.
Exercise 6.1.6: Find the Laplace transform of a + bt + ct
2
for some constants a, b, and c.
6.1. THE LAPLACE TRANSFORM 235
Exercise 6.1.7: Find the Laplace transform of Acos ωt + Bsin ωt.
Exercise 6.1.8: Find the Laplace transform of cos
2
ωt.
Exercise 6.1.9: Find the inverse Laplace transform of
4
s
2
−9
.
Exercise 6.1.10: Find the inverse Laplace transform of
2s
s
2
−1
.
Exercise 6.1.11: Find the inverse Laplace transform of
1
(s−1)
2
(s+1)
.
236 CHAPTER 6. THE LAPLACE TRANSFORM
6.2 Transforms of derivatives and ODEs
Note: 2 lectures, §7.2 –7.3 in EP
6.2.1 Transforms of derivatives
Let us see how the Laplace transform is used for differential equations. First let us try to find
the Laplace transform of a function that is a derivative. That is, suppose g(t) is a continuous
differentiable function of exponential order.
!¦g
·
(t)¦ =
∞
0
e
−st
g
·
(t) dt =
,
e
−st
g(t)
¸
∞
t=0
−
∞
0
(−s) e
−st
g(t) dt = −g(0) + s!¦g(t)¦.
We can keep doing this procedure for higher derivatives. The results are listed in Table 6.2. The
procedure also works for piecewise smooth functions, that is functions which are piecewise con-
tinuous with a piecewise continuous derivative. The fact that the function is of exponential order
is used to show that the limits appearing above exist. We will not worry much about this fact.
f (t) !¦ f (t)¦ = F(s)
g
·
(t) sG(s) − g(0)
g
··
(t) s
2
G(s) − sg(0) − g
·
(0)
g
···
(t) s
3
G(s) − s
2
g(0) − sg
·
(0) − g
··
(0)
Table 6.2: Laplace transforms of derivatives (G(s) = !¦g(t)¦ as usual).
Exercise 6.2.1: Verify Table 6.2.
6.2.2 Solving ODEs with the Laplace transform
If you notice, the Laplace transform turns differentiation essentially into multiplication by s. Let
us see how to apply this to differential equations.
Example 6.2.1: Take the equation
x
··
(t) + x(t) = cos 2t, x(0) = 0, x
·
(0) = 1.
We will take the Laplace transform of both sides. By X(s) we will, as usual, denote the Laplace
transform of x(t).
!¦x
··
(t) + x(t)¦ = !¦cos 2t¦,
s
2
X(s) − sx(0) − x
·
(0) + X(s) =
s
s
2
+ 4
.
6.2. TRANSFORMS OF DERIVATIVES AND ODES 237
We can plug in the initial conditions now (this will make computations more streamlined) to obtain
s
2
X(s) − 1 + X(s) =
s
s
2
+ 4
.
We now solve for X(s),
X(s) =
s
(s
2
+ 1)(s
2
+ 4)
+
1
s
2
+ 1
.
We use partial fractions (exercise) to write
X(s) =
1
3
s
s
2
+ 1
−
1
3
s
s
2
+ 4
+
1
s
2
+ 1
.
Now take the inverse Laplace transform to obtain
x(t) =
1
3
cos t −
1
3
cos 2t + sin t.
The procedure is as follows. You take an ordinary differential equation in the time variable
t. You apply the Laplace transform to transform the equation into an algebraic (non differential)
equation in the frequency domain. All the x(t), x
·
(t), x
··
(t), and so on, will be converted to X(s),
sX(s) − x(0), s
2
X(s) − sx(0) − x
·
(0), and so on. If the differential equation we started with was
constant coefficient linear equation, it is generally pretty easy to solve for X(s) and we will obtain
some expression for X(s). Then taking the inverse transform if possible, we find x(t).
It should be noted that since not every function has a Laplace transform, not every equation can
be solved in this manner.
6.2.3 Using the Heaviside function
Before we move on to more general functions than those we could solve before, we want to con-
sider the Heaviside function. See Figure 6.1 on the following page for the graph.
u(t) =
¸
¸
¸
¸
¸
¸
0 if t < 0,
1 if t ≥ 1.
This function is useful for putting together functions, or cutting functions off. Most commonly
it is used as u(t − a) for some constant a. This just shifts the graph to the right by a. That is, it is a
function which is zero when t < a and 1 when t ≥ a. Suppose for example that f (t) is a "signal"
and you started receiving the signal sin t at time t. The function f (t) should then be defined as
f (t) =
.
Hence it is useful to know how the Heaviside function interacts with the Laplace transform.
We have already seen that
!¦u(t − a)¦ =
e
−as
s
.
This can be generalized into a shifting property or second shifting property.
!¦ f (t − a)u(t − a)¦ = e
−as
!¦ f (t)¦. (6.1)
Example 6.2.2: Suppose that the forcing function is not periodic. For example, suppose that we
had a mass spring system
x
··
(t) + x(t) = f (t), x(0) = 0, x
·
(0) = 0,
where f (t) = 1 if 1 ≤ t < 3 and zero otherwise. We could imagine a mass and spring system where
a rocket was fired for 2 seconds starting at t = 1. Or perhaps an RLC circuit, where the voltage was
6.2. TRANSFORMS OF DERIVATIVES AND ODES 239
being raised at a constant rate for 2 seconds starting at t = 1 and then held steady again starting at
t = 3.
We can write f (t) = u(t − 1) − u(t − 3). We transform the equation and we plug in the initial
conditions as before to obtain
s
2
X(s) + X(s) =
e
−s
s
−
e
−3s
s
.
We solve for X(s) to obtain
X(s) =
e
−s
s(s
2
+ 1)
−
e
−3s
s(s
2
+ 1)
.
We leave it as an exercise to the reader to show that
!
−1
u(t − 2).
The plot of this solution is given in Figure 6.2 on the next page.
6.2.4 Transforms of integrals
A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is,
equations in which integrals rather than derivatives of functions appear. The basic property, which
can be proven by applying the definition and again doing integration by parts, is the following.
!
t
0
f (τ) dτ
=
1
s
F(s).
It is sometimes useful for computing the inverse transform to write
∞
∞
f (τ)g(t−τ) dτ.
This definition agrees with (6.2) if you define f (t) and g(t) to be zero for t < 0. When discussing the Laplace transform
the definition we gave is sufficient. Convolution does occur in many other applications, however, where you may have
to use the more general definition with infinities.
6.3. CONVOLUTION 243
The convolution has many properties that make it behave like a product. Let c be a constant
and f , g, and h be functions then
f ∗ g = g ∗ f ,
(c f ) ∗ g = f ∗ (cg) = c( f ∗ g),
( f ∗ g) ∗ h = f ∗ (g ∗ h).
The most interesting property for us, and the main result of this section is the following theorem.
Theorem 6.3.1. Let f (t) and g(t) be of exponential type, then
!¦( f ∗ g)(t)¦ = !
t
0
f (τ)g(t − τ) dτ
= !¦ f (t)¦!¦g(t)¦.
In other words, the Laplace transform of a convolution is the product of the Laplace transforms.
The simplest way to use this result is in reverse.
Example 6.3.3: Suppose we have the function of s defined by
1
(s + 1)s
2
=
1
s + 1
1
s
2
.
We recognize the two entries of Table 6.2. That is
!
−1
1
s + 1
= e
−t
and !
−1
1
s
2
= t.
Therefore,
!
−1
1
s + 1
1
s
2
=
t
0
τe
t−τ
dτ = 2e
t
− t
2
− 2t − 2.
Where the calculation of the integral of course involved an integration by parts.
6.3.2 Solving ODEs
The next example will demonstrate the full power of the convolution and Laplace transform. We
will be able to give a solution to the forced oscillation problem for any forcing function as a definite
integral.
Example 6.3.4: Find the solution to
x
··
+ ω
2
0
x = f (t), x(0) = 0, x
·
(0) = 0,
for an arbitrary function f (t).
244 CHAPTER 6. THE LAPLACE TRANSFORM
We first apply the Laplace transform to the equation. Denote the transform of x(t) by X(s) and
the transform of f (t) by F(s) as usual.
s
2
X(s) + ω
2
0
X(s) = F(s),
or in other words
X(s) = F(s)
1
s
2
+ ω
2
0
.
We know
!
−1
=
1
2ω
0
t sin ω
0
t.
Note the t in front of the sine. This solution will, therefore, grow without bound as t gets large,
meaning we get resonance.
Using convolution you can also find a solution as a definite integral for arbitrary forcing func-
tion f (t) for any constant coefficient equation. A definite integral is usually enough for most
practical purposes. It is usually not hard to numerically evaluate a definite integral.
6.3.3 Volterra integral equation
One of the most common integral equations is the Volterra integral equation
§
:
x(t) = f (t) +
t
0
g(t − τ)x(τ) dτ,
§
Named for the Italian mathematician Vito Volterra (1860 – 1940).
6.3. CONVOLUTION 245
where f (t) and g(t) are known functions and x(t) is an unknown. To solve this equation we apply
the Laplace transform to get
X(s) = F(s) + G(s)X(s)
where X(s), F(s), and G(s) are the Laplace transforms of x(t), f (t), and g(t) respectively. We find
X(s) =
F(s)
1 − G(s)
if we can find the inverse Laplace transform now we obtain the result.
Example 6.3.5: Solve
x(t) = e
−t
+
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These are class notes from teaching Math 286, differential equations at the University of Illinois at Urbana-Champaign in fall 2008 and spring 2009. These originated from my lecture notes. There is usually a little more "padding" material than I can cover in the time alloted. There are still not enough exercises throughout. Some of the exercises in the notes are things I do explicitly in class depending on time, or let the students work out in class themselves. The book used for the class is Edwards and Penney, Differential Equations and Boundary Value Problems [EP], fourth edition, from now on referenced just as EP. The structure of the notes, therefore, reflects the structure of this book, at least as far as the chapters that are covered in the course. Many examples and applications are taken more or less from this book, though they also appear in many other sources, of course. Other books I have used as sources of information and inspiration are E.L. Ince's classic (and inexpensive) Ordinary Differential Equations [I], and also my undergraduate textbooks, Stanley Farlow's Differential Equations and Their Applications [F], which is now available from Dover, and Berg and McGregor's Elementary Partial Differential Equations [BM]. See the Further Reading section at the end of these notes. I taught the course with the IODE software ( IODE is a free software package that is used either with Matlab (properietary) or Octave (free software). Projects and labs from the IODE website are referenced throughout the notes. They need not be used for this course, but I think it is better to use them. The graphs in the notes were made with the Genius software (see I have used Genius in class to show essentially these and similar graphs. I would like to acknowledge Rick Laugesen. I have used his handwritten class notes on the first go through the course. My organization of these present notes, and the choice of the exact material covered, is heavily influenced by his class notes. Many examples and computations are taken from his notes. The organization of these notes to some degree requires that they be done in order. Hence, later chapters can be dropped. The dependence of the material covered is roughly given in the the following diagram: 5
The notes are done for two types of courses. chapter 4. IODE need not be used for either version. If IODE is not used. While Laplace transform is not normally covered at UIUC 285/286.
. chapter 2. chapter 3. chapter 4. Either at 4 hours a week for a semester (Math 286 at UIUC): Introduction. chapter 1 (plus the two IODE labs). some additional material should be covered instead. The length of the Laplace chapter is about the same as the Sturm-Liouville chapter (chapter 5). chapter 5.6
INTRODUCTION
Introduction Chapter 1 Chapter O 2
Chapter 6
o ooo ooo o wooo
OOO OOO OOO O'
Chapter 4
oo oo wo o
Chapter 3
Chapter 5 There are some references in chapters 4 and 5 to material from chapter 3 (some linear algebra). Or a shorter version (Math 285 at UIUC) of the course at 3 hours a week for a semester: Introduction. There is a short introductory chapter on Laplace transform (chapter 6 that could be used as additional material. For the shorter version some additional material should be covered. but these references are not absolutely essential and can be skimmed over. I think it is essential that any notes for Differential equations at least mention Laplace and/or Fourier transforms. chapter 1 (plus the two IODE labs). so chapter 3 can safely be dropped. chapter 2. while still covering chapters 4 and 5.
2. Let us try. How do we check? Just plug it back in! First you need to compute dx = − sin t + cos t.2 Solutions of differential equations
Solving the differential equation means finding x in terms of t. we want to find a function of t which we will call x such that when we plug x.1 in EP
0.1 Differential equations
The laws of physics are generally written down as differential equations. INTRODUCTION TO DIFFERENTIAL EQUATIONS
7
0. dt
We find that
. You have already seen many differential equations without perhaps knowing about it. Understanding differential equations is essential to understanding almost anything you will study in your science and engineering classes. In this case we claim that x = x(t) = cos t + sin t is a solution. Then it would be important to first learn Swahili. And you have even solved simple differential equations when you were taking calculus. It is the dt same idea as it would be for a normal (algebraic) equation of just x and t.0. In fact it is an example of a first order differential equation. dt Yay! We got precisely the right hand side. Equation (1) is a basic example of a differential equation. dx = − sin t + cos t − e−t . otherwise you will have a very tough time getting a good grade in your other classes. (1) dt Here x is the dependent variable and t is the independent variable. dt
dx .2
Introduction to differential equations
Note: more than 1 lecture. Now let us compute the left hand side of (1) dt dx + x = (− sin t + cos t) + (cos t + sin t) = 2 cos t. and differential equations are one of the most important parts of this language as far as science and engineering are concerned. Therefore. t.2.2. §1. Let us see an example you may not have seen. since it involves only the first derivative of the dependent variable. all of science and engineering use differential equations to some degree. and dx into (1) the equation holds. This equation arises from Newton's law of cooling where the ambient temperature oscillates with time. suppose that all your classes from now on were given in Swahili.
0. That is. You can think of mathematics as the language of science. dx + x = 2 cos t. As an analogy. There is more! We claim x = cos t + sin t + e−t is also a solution.
it is not a simple question of turning a crank to get answers. you need to understand what they are doing. In fact. This course is no exception to this. dt And it works yet again! So there can be many different solutions.3 Differential equations in practice
So how do we use differential equations in science and engineering? You have some real
. See Figure 1 for the graph of a few of these solutions. we will get partial differential equations or PDEs. We will see how we can find these solutions a few lectures from now. There is no general method that solves any given differential equation. it is often necessary to simplify or transform your equations into something that a computer can actually understand and solve. You may need to make certain assumptions and changes in your model to achieve this. so that you may apply those techniques to new problems. For example. It is important to know when it is easy to find solutions and how to do this. We will generally focus on how to get exact formulas for solutions of differential equations. Even for ODEs. It is important to learn problem solving techniques. For most of the course we will look at ordinary differential equations or ODEs. which are very well understood. Even if you leave much of the actual calculations to computers in real life. A common mistake is to expect to learn some prescription for solving all the problems you will encounter in your later career. but we will also spend a little bit of time on getting approximate solutions. you will be required to solve problems in your job which you have never seen before.2. It turns out that solving differential equations can be quite hard. If there are several independent variables. by which we mean that there is only one independent variable and derivatives are only with respect to this one variable. dt end of the course. We will briefly see these near the y Figure 1: Few solutions of dx + 2 = cos t.
0 1 2 3 4 5 3 3 2 2 1 1 0 0 -1 -1 0 1 2 3 4 5
0. To be a successful engineer or scientist.8
INTRODUCTION
Again plugging into the left hand side of (1) dx + x = (− sin t + cos t − e−t ) + (cos t + sin t + e−t ) = 2 cos t. for this equation all solutions can be written in the form x = cos t + sin t + Ce−t for some constant C.
Therefore. Let us plug these in and see what happens.2. Sometimes we will work solve Mathematical Mathematical with simple real world examples so that we have model solution some intuition and motivation about what we are doing. One of the most basic differential equations is the standard exponential growth model. You have to interpret the results. Let us suppose that there is enough food and enough space. There is still something left to do. Let us look at an example of this process. 100 = P(0) = Cek0 = C.2.069. That is. Let us try. So we know that P(t) = 100 e(ln 2)t/10 ≈ 100 e0.069t . Let P denote the population of some bacteria on a petri dish. where C is a constant. you translate your real world situation into a set of differential equations. We claim that a solution is given by P(t) = Cekt .1: Suppose there are 100 bacteria at time 0 and 200 bacteria at time 10s. How many bacteria will there be in 1 minute from time 0 (in 60 seconds)? First we have to solve the equation. so what now? We do not know C and we do not know k. Let t denote time (say in seconds). In this course we will mostly focus on the mathematical analysis. dP = Ckekt = kP. Example 0. Then you apply mathematics to get some sort of mathematical solution. a large population growth quicker.
.0. 2 = e10k or
ln 2 10
= k ≈ 0. You have to figure out what the mathematical solution says about the real world problem you started with. Then the rate of growth of bacteria will be proportional to population. OK. Hence our model will be dP = kP dt for some positive constant k > 0. I. dt And it really is a solution.e. 200 = P(10) = 100 ek10 . INTRODUCTION TO DIFFERENTIAL EQUATIONS
9
world problem that you want to understand. You make some simplifying assumptions and create a mathematical model. Well we know something. Learning how to formulate the mathematical Real world problem model and how to interpret the results is essentially what your physics and engineering classes abstract interpret do. We know that P(0) = 100 and we also know that P(10) = 200.
not any real number. t = 60.
0 10 20 30 40 50 60
6000
6000
5000
5000
4000
4000
3000
3000
We will call P(t) = Cet the general solution. The general solution for this equation is y(x) = Cekx . dy = −ky. Also note that P in real life is a discrete quantity. Generally when we say particular solution onds. First such equation is. They are also simple to check. What does that mean? Suppose k = 1 for simplicity. and you will want to solve the equation for different initial conditions. Normally.35. Then the solution turns out to be (exercise) P(t) = 1000 et . But if our assumptions are reasonable. We have already seen that this is a solution above with different variable names. The general solution for this equation is y(x) = Ce−kx . the population is P(60) = 6400.
. then there will be about 6400 bacteria. P(61) ≈ 6859. Here y is the dependent and x the independent variable. we just mean some solution. and cosines. There is no need to wonder if you have remembered the solution correctly. dx for some constant k > 0. OK. sines.10
INTRODUCTION
At one minute. These solutions are reasonably easy to guess by recalling properties of exponentials.
1000 1000 0 0 0 10 20 30 40 50 60
2000
2000
Let us get to what we will call the 4 fundamental equations. but our model has no problem saying that for example at 61 seconds. let us talk about the interpretation of the results. the k in P = kP will be known. Next. which is something that you should always do. Then you will need an initial condition to find out what C is to find the particular solution we are looking Figure 2: Bacteria growth in the first 60 secfor. See Figure 2. These appear very often and it is useful to just memorize what their solutions are. dx for some constant k > 0. Does this mean that there must be exactly 6400 bacteria on the plate at 60s? No! We have made assumptions that might not be true. dy = ky. as every solution of the equation can be written in this form for some constant C. So we want to solve dP = P subject to P(0) = 1000 (the inidt tial condition).
0. Contrary to popular belief this is not a parabola.2.2.
11
Note that because we have a second order differential equation we have two constants in our general solution. dx2 for some constant k > 0. This formula is actually inscribed inside the arch: y = −127. An interesting note about cosh: The graph of cosh is the exact shape a hanging chain will make and it is called a catenary.7 ft) + 757. For those that do not know. take the second order differential equation d2 y = −k2 y.7 ft · cosh(x/127. The general solution for this equation is y(x) = C1 cos(kx) + C2 sin(kx). Next. If you invert the graph of cosh it is also the ideal arch for supporting its own weight. Exercise 0. dx Exercise 0.2. 2 e x − e−x .2. the gateway arch in Saint Louis is an inverted graph of cosh (if it were just a parabola it might fall down). For example. The general solution for this equation is y(x) = C1 ekx + C2 e−kx . INTRODUCTION TO DIFFERENTIAL EQUATIONS Exercise 0.2: Check that the y given is really a solution to the equation. or y(x) = D1 cosh(kx) + D2 sinh(kx). cosh and sinh are defined by cosh x = e x + e−x . They have some nice familiar d properties such as cosh 0 = 1.3: Check that both forms of the y given are really solutions to the equation. sinh 0 = 0. sinh x = 2
These functions are sometimes easier to work with than exponentials. take the second order differential equation d2 y = k2 y. And finally. dx2 for some constant k > 0. and dx cosh x = sinh x (no that is not a typo) and d sinh x = cosh x.
.1: Check that the y given is really a solution to the equation.7 ft.
Chapter 1 First order ODEs
1. f (x) dx + C. let us assume that f is a function of x alone. dx or just y = f (x. there is no simple formula or procedure one can follow to find solutions. In the next few lectures we will look at special cases where solutions are not difficult to obtain. y). We could just integrate (antidifferentiate) both sides here with respect to x.2 in EP A first order ODE is an equation of the form dy = f (x. y). In general.1) find some antiderivative of f (x) and then you add an arbitrary constant to get the general solution. y (x) dx = that is y(x) = f (x) dx + C. that is. Calculus textbooks muddy the waters by talking about integral as primarily the so-called indefinite integral. The 13
. §1. (1. the equation is y = f (x). Now is a good time to discuss a point about calculus notation and terminology.1 Integrals as solutions
Note: 1 lecture.1)
This y(x) is actually the general solution. So to solve (1. In this section.
We see that the general solution must be y = x3 + C. Normally. And it is! 0 Do note that the definite integral and indefinite integral (antidifferentiation) are completely different beasts. this is a solution. we also have an initial condition such as y(x0 ) = y0 for some two numbers x0 and y0 (x0 is usually 0. by fundamental theorem of calculus you can always write f (x) dx + C as
x
f (t) dt + C. Example 1.2: Solve y = e−x .
By the preceeding discussion. the terminology integrate when you may really mean antidifferentiate.2) is a formula you can plug into the calculator or a computer and it will be happy to calculate specific values for you. It is not possible (in closed form). Tell them to find the closed form solution. This particular integral is in fact very important in statistics. The only reason for the indefinite integral notation is that you can always write an antiderivative as a (definite) integral. it only happens to also compute antiderivatives. That is.1.14
CHAPTER 1. For sake of consistency. Example 1. see the following example).1. You will easily be able to plot the solution and work with it just like with any other function. the solution must be y(x) =
0
e−s ds + 1.1: Find the general solution of y = 3x2 . There is absolutely nothing wrong with writing the solution as a definite integral. Integration is defined as the area under the graph. Integration is just one way to compute the antiderivative (and it is a way that always works. Ha ha ha (bad math joke). Let us check: y = 3x2 . (1. FIRST ORDER ODES
indefinite integral is really the antiderivative (in fact the whole one parameter family of antiderivatives). There really exists only one integral and that is the definite integral.
x
2
y(0) = 1.
x0
Hence. y(x0 ) = y0 . The definite integral always evaluates to a number. we will keep using the indefinite integral notation when we want an antiderivative.2)
Let us check! y = f (x) (by fundamental theorem of calculus) and by Jupiter. It is not so crucial to find a closed form for the antiderivative. Then the solution is
x
y(x) =
x0
f (s) ds + y0 . Is x0 it the one satisfying the initial condition? Well. We can write the solution as a definite integral in a nice way. y(x0 ) = x f (x) dx + y0 = y0 .
2
Here is a good way to make fun of your friends taking second semester calculus. but not always). and you should always think of the definite integral. Suppose our problem is y = f (x).
(1. Therefore.
. We have gotten precisely our equation back.
Let us write it in Leibniz notation dy = f (y) dx Now use the inverse function theorem to switch roles of x and y. We write dx 1 = . Example 1.1. dy ky Now integrate and get x(y) = x = we solve for y kekC ekx = |y|. y = Cekx . as this sort of hand-waving calculation can lead to trouble.4: Find the general solution of y = y2 . k 1 dy + C f (y)
. Example 1. First note that y = 0 is a solution.3: We guessed y = ky has solution Cekx . Now we can just integrate x(y) = Next. dx 1 = dy f (y)
15
What we are doing seems like algebra with dx and dy. First note that y = 0 is a solution. we try to solve for y. It is tempting to just do algebra with dx and dy as if they were numbers.1. Be careful. If we replace kekC with an arbitrary constant C we can get rid of the absolute value bars. INTEGRALS AS SOLUTIONS We can also solve equations of the form y = f (y) using this method. and we get the same general solution as we guessed before. In this way we also incorporate the solution y = 0. And in this case it does work. Henceforth. especially when more than one independent variable is involved. We can actually do it now.1.1. however. Write 1 ln|ky| + C . We can now assume that y 1 dx = 2 dy y 0. assume y 0.
1.
. then the solution blows up as we approach x = 1.1: Solve for v and then solve for x. y
So the general solution is y=
1 or y = 0. Well.1. x(10) = 2e10/2 − 2 ≈ 294 meters.
Now we just plug in to get that at 2 seconds (and 10). Classical problems leading to differential equations solvable by integration are problems dealing with velocity. the car has travelled x(2) = 2e2/2 − 2 ≈ 3. You have surely seen these problems before in your calculus class. Exercise 1. If x is the distance travelled. If for example C = 1. and x is the acceleration. where t is time in seconds. At time t = 0 the car is at the 1 meter mark and is travelling at 10 m/s. x (0) = 10. then x is the velocity. It is hard to tell from just looking at the equation itself how the solution is going to behave sometimes. v = t2 . FIRST ORDER ODES
−1 + C. C−x
CHAPTER 1. Where is the car at time t = 10. we can then integrate and find x.44 meters. C−x Note the singularities of the solution. Note that we still need to figure out C. So C = −2 and hence x(t) = et/2 − 2. what if we call x = v and then we have the problem v(0) = 10. Well this is actually a second order problem. Let x denote the distance the car travelled. Once we solve for v. But we know that when t = 0 then x = 0. but the solution is only defined on some interval (−∞. that is: x(0) = 0 so 0 = x(0) = 2e0/2 + C = 2 + C.5: Suppose a car drives at a speed et/2 meters per second. C) or (C. The equation is x = et/2 . The equation with initial conditions is x = t2 . How far did the car get in 2 seconds? How far in 10 seconds.6: Suppose that the car accelerates at the rate t2 m/s2 . x(0) = 1. acceleration and distance. The equation y = y2 is very nice and defined everywhere.16 Now integrate to get x= Solve for y =
1 . Example 1. We can just integrate this equation to get that x(t) = 2et/2 + C. Example 1. ∞).1.
Then if we are given a specific initial condition y(x0 ) = y0 . y).3 in EP At this point it may be good to first try the Lab I and/or Project I from the IODE website: For example.edu/iode/.uiuc. y(0) = 0
.2. See Figure 1.1. See Figure 1. we can really just look at the location (x0 .2. It would be good if we could at least figure out the shape and behavior of the solutions or even find approximate solutions for any equation. the general first order equation we are studying looks like y = f (x.2
Slope fields
Note: 1 lecture. y)-plane we get a slope. and y(0) = −0.1 Slope fields
As you have seen in IODE Lab I (if you did it). y0 ) and follow the slopes. We can plot the slope at lots of points as a short line with this given slope.2: Slope field of y = xy with a graph of solutions satisfying y(0) = 0. in Figure 1.2.math. As we said. By looking at the slope field we can find out a lot about the behavior of solutions. this means that at each point in the (x.18
CHAPTER 1.2. y(0) = 0.
We call this the slope field of the equation.1: Slope field of y = xy. In general we cannot really just solve these kinds of equations explicitly. FIRST ORDER ODES
1.
-3 3 -2 -1 0 1 2 3 3 3 -3 -2 -1 0 1 2 3 3
2
2
2
2
1
1
1
1
0
0
0
0
-1
-1
-1
-1
-2
-2
-2
-2
-3 -3 -2 -1 0 1 2 3
-3
-3 -3 -2 -1 0 1 2 3
-3
Figure 1.
1.
Figure 1. §1.2 we can see what the solutions do when the initial conditions are y(0) > 0.
1. or if it does is not unique. It also has to be unique if we believe our universe is deterministic. it is good to know when things go wrong and why. we see that no matter where we start.1: Attempt to solve: 1 y = . Note that a small change in the initial condition causes quite different behavior.3. (i) Does a solution exist? (ii) Is the solution unique (if it exists)? What do you think is the answer? The answer seems to be yes to both does it not? Well. y(x0 ) = y0 . plotting a few solutions of the of the equation y = −y.
Integrate to find the general solution y = ln |x| +C Note that the solution does not exist at x = 0.
1.
. If the solution does not exist. See Figure 1. all solutions tend to zero as x tends to infinity. But there are cases when the answer to either question can be no.2.3: Slope field of y = −y with a graph of a few solutions.2. SLOPE FIELDS
19
and y(0) < 0. Hence.2. pretty much.4 on the next page. we have probably not devised the correct model. Since generally the equations come from real life situation. See Figure 1.Figure 1. then it seems logical that a solution exists. On the other hand. y).2 Existence and uniqueness
We wish to ask two fundamental questions about the problem y = f (x. x y(0) = 0. Example 1.
2.2.2.1.1: Sketch direction field for y = e x−y . Exercise 1. For the most of this course we will be interested in equations where existence and uniqueness holds.2. Exercise 1. So x = y12 .3: Sketch direction field for y = y2 . when A = 1 the solution "blows up" at x = 1. the solution does not exist for all x even if the equation is nice everywhere. so y is not equal to zero at least 1 1 for some x near 0. SLOPE FIELDS
21
We know how to solve this equation.
1.2.
.2.3 Exercises
Exercise 1. How do the solutions behave as x grows? Can you guess a particular solution by looking at the direction field? Exercise 1. then C = A so y y=
1 A
1 . so x = −1 + C. so y = C−x . If y(0) = A. then y = 0 is a solution. −x
Now if A = 0. Hence. y = y2 certainly looks nice. First assume that A 0.4: Is it possible to solve the equation y =
xy cos x
for y(0) = 1? Justify.2: Sketch direction field for y = x2 . and in fact will hold "globally" unlike for the y = y2 . For example.
g(y) Now both sides look like something we can integrate.
.4 in EP When the equation is of the form y = f (x). We obtain dy = g(y) f (x) dx + C. 2 0 from now on. if it looked like y = f (x)g(y). Unfortunately this method no longer works for the general form of the equation y = f (x. §1.3
Separable equations
Note: 1 lecture. we can just integrate: y = f (x) dx + C. what if the equation is separable.22
CHAPTER 1. that is.3. FIRST ORDER ODES
1. Write the equation as x dx + C. f (x. so assume y then dy = y We compute the antiderivatives to get ln |y| = x2 + C.1: Take the equation y = xy First note that y = 0 is a solution. Integrating both sides yields y= Notice dependence on y in the integral.3.1 Separable equations
On the other hand. y). for some functions f (x) and g(y).
dy dx
= xy. dx Then we rewrite the equation as dy = f (x) dx.
If we can explicitly solve this integral we can maybe solve for y. Let us write the equation in Leibniz notation dy = f (x)g(y). y) dx + C.
1. Example 1.
3.1. 1 dy dx = g(y) dx We can use the change of variables formula. f (x) dx + C.3. that does not sound right. We seemed to be doing a different operation to each side. Because y = 0 is a solution and because of the absolute value we actually can write: x2 y = De 2 . f (x) dx + C. Note that y = y(x) is a function of x and so is 1 dy = f (x) g(y) dx We integrate both sides with respect to x. For example.2 Implicit solutions
It is clear that we might sometimes get stuck even if we can do the integration. take the separable equation xy y = 2 .
dy ! dx
1.
where D > 0 is some constant. SEPARABLE EQUATIONS Or
x2 x2 x2
23
|y| = e 2 +C = e 2 eC = De 2 . y y
. Because we were integrating in two different variables. Let us see work out this method more rigorously. We check: x2 x2 y = Dxe 2 = x(De 2 ) = xy. Yay! We should be a little bit more careful about the method. dy = f (x)g(y) dx We rewrite the equation as follows. for any number D (including zero or negative). y +1 We separate variables y2 + 1 1 dy = y + dy = x dx. 1 dy = g(y) And we are done.
FIRST ORDER ODES
It is not easy to find the solution explicitly as it is hard to solve for y. We note above that the equation also has a solution y = 0. you can graph x as a function of y. y(1) = 0. For example.3. Now we separate variables. y = tan x
.. If you want to compute values for y you might have to be tricky.
1. 2 2 Or maybe the easier looking expression: y2 + 2 ln |y| = x2 + C.. Computers are also good at some of these tricks. therefore. but you have to be careful. We will.3 Examples
Example 1. These outlying solutions such as y = 0 are sometimes called singular solutions. It is easy to check that implicit solutions still satisfy the differential equation. y
It is simple to see that the differential equation holds.). and then flip your paper. etc. therefore. 0 = tan(−2 + C) to get C = 2 (or 2 + π. The solution we are seeking is. call this solution an implicit solution. it turns out that the general solution is y2 + 2 ln |y| = x2 + C together with y = 0. In this case. First factor the right hand side to obtain x2 y = (1 − x2 )(1 + y2 ).3. In this case.2: Solve x2 y = 1 − x2 + y2 − x2 y2 .
CHAPTER 1. we differentiate to get y 2y + 2 = 2x. −1 −x+2 . integrate and solve for y 1 − x2 y = 1 + y2 x2 y 1 = 2 −1 2 1+y x −1 arctan(y) = − x+C x −1 y = tan − x+C x Now solve for the initial condition.24 Now we integrate to get y2 x2 + ln |y| = + C.
the solution exists wherever p(x) and f (x) are defined.4. The function r(x) is called the integrating factor and the method is called the integrating factor method. That seems like a job for the exponential function! r(x) = e Let us do the calculation.4
Linear equations and the integrating factor
Note: more than 1 lecture. f (x). That is a first order equation is linear if we can put it into the following form: y + p(x)y = f (x). to get a closed form formula for y we need to be able to find a closed form formula for the two integrals. For example.
Of course.
.1. We can then solve for y. there is a method for solving linear first order equations. In this lecture we will focus on the first order linear equation. e
p(x)dx
y= y = e−
e
p(x)dx
p(x)dx
f (x) dx + C . But most importantly for us right now. Solutions of linear equations have nice properties. What we will do is to multiply both sides of (1. LINEAR EQUATIONS AND THE INTEGRATING FACTOR
27
1. f (x) dx + C.3) by some function r(x) such that r(x)y + r(x)p(x)y = We can then integrate both sides of d r(x)y = r(x) f (x). e
p(x)dx p(x)dx
d r(x)y . we get the same function back multiplied by p(x). In fact the majority of this course will focus on linear equations. §1. dx Note that the right hand side does not depend on y and the left hand side is written as a derivative of a function. dx
y + e p(x)dx p(x)y = e d e p(x)dx y = e dx e
p(x)dx
p(x)dx p(x)dx
f (x). The dependence on x can be more complicated. y + p(x)y = f (x). So we are looking for a function r(x) such that if we differentiate it. (1.5 in EP One of the most important types of equations we will learn how to solve are so-called linear equations.3) The word "linear" here means linear in y. and has the same regularity (read: it is just as nice).
2 2 2 2 2 2 2
= e x .
p(x)dx
2 2
. An advice: Do not try to remember the formula itself. Suppose we are given y + p(x)y = f (x) y(x0 ) = y0 . FIRST ORDER ODES
y(0) = −1. that is way too hard.4. but those constants will not matter in the end.4. we solve for the initial condition −1 = y(0) = 1 + C.
p(x) dx
First note that p(x) = 2x and f (x) = e x−x . The integrating factor is r(x) = e multiply both sides of the equation by r(x) to get e x y + 2xe x y = e x−x e x . Note that we do not care which antiderivative we take when computing e add a constant of integration.28 Example 1.4)
You should be careful to properly use dummy variables here.4.
(1. y(x) = e
−
x x0
p(s) ds
x x0
t
e x0
p(s) ds
f (t) dt + y0 . Exercise 1. Since we cannot always evaluate the integrals in closed form. We
2
Next.1: Solve y + 2xy = e x−x
2 2
CHAPTER 1. It is easier to remember the process and repeat it. You can always
Exercise 1.
Look at the solution and write the integrals as definite integrals. If you now plug that into a computer of a calculator. A definite integral is something that you can plug into a computer or a calculator. The solution is y = e x−x − 2e x . y = e x−x + Ce x .4). so C = −2. d x2 e y = ex . dx We integrate e x y = e x + C.
.1: Try it! Add a constant of integration to the integral in the integrating factor and show that the solution you get in the end is the same as what we got above.2: Check that y(x0 ) = y0 in formula (1. it is useful to know how to write the solution in definite integral form. it will be happy to give you numerical answers.
The integrating factor is r(t) = exp 3 3 dt = exp ln(60 + 2t) = (60 + 2t)3/2 60 + 2t 2 x x = volume 60 + (5 − 3)t
. Let x denote the kg of salt in the tank. Solution of water and salt (brine) with concentration of 0. therefore. but try to simplify as far as you can. the change in x (denoted ∆x) is approximately ∆x ≈ (rate in × concentration in)∆t − (rate out × concentration out)∆t Taking the limit ∆t → 0 we see that dx = (rate in × concentration in) − (rate out × concentration out) dt We have rate in = 5 concentration in = 0.1 kg / liter is flowing in at the rate of 5 liters a minute.1 rate out = 3 concentration out = Our equation is. You will not be able to find the solution in closed form. Then for a small change ∆t in time.5 dt 60 + 2t Let us solve.1) − 3 dt 60 + 2t Or in the form (1. let t denote the time in minutes. A 100 liter tank contains 10 kilograms of salt dissolved in 60 liters of water.
Example 1. LINEAR EQUATIONS AND THE INTEGRATING FACTOR
29
Exercise 1.2: The following is a simple application of linear equations and this type of a problem is used often in real life.4.4.3: Write the solution of the following problem as a definite integral.4. How much salt is in the tank when the tank is full? Let us come up with the equation. y + y = e x −x
2
y(0) = 10. linear equations are used in figuring out the concentration of chemicals in bodies of water. x dx = (5 × 0. For example.1.3) dx 3 + x = 0. The solution in the tank is well stirred and flows out at a rate of 3 liters a minute.
LINEAR EQUATIONS AND THE INTEGRATING FACTOR
31
Exercise 1. The output of one is flowing to the other. dt t is time.9: Suppose there are two lakes.4. A is the ambient temperature. c) When will the concentration in the second lake be maximal. The first lake contains 100 thousand liters of water and the second lake contains 200 thousand liters of water. Suppose that A = A0 cos ω t for some constants A0 and ω.4. a) Find the general solution. b) When will the concentration in the first lake be below 0. Assume that the water is being continually mixed perfectly by the stream. and k > 0 is a constant. A truck with 500 kg of toxic substance crashes into the first lake. will the initial conditions make much of a difference? Why or why not. That is the ambient temperature oscillates (for example night and day temperatures). Exercise 1. The in and out flow from each lake is 500 liters per hour.01 kg per liter.
.4.1.10: Newton's law of cooling states that dx = −k(x − A) where x is the temperature. a) Find the concentration of toxic substance as a function of time (in seconds) in both lakes. b) In the long term.
There are some general things to look for. a change of coordinates v = y1−n transforms the Bernoulli equation into a linear equation.5. 4
There are several things called Bernoulli equations. −xy5 v + y(x + 1) + xy5 = 0.1: Solve xy + y(x + 1) + xy5 = 0.1. Example 1. If a substitution does not work (it does not make the equation any simpler). Substitution in differential equations is applied in much the same way that it is applied in calculus. This equation looks a lot like a linear equation except for the yn . These particular equations are named for Jacob Bernoulli (1654 – 1705). For example. this is just one of them. When you see yy y2 y (cos y)y (sin y)y y ey Try substituting y2 y3 sin y cos y ey
Usually you try to substitute in the "most complicated" part of the equation with the hopes of simplifying it.
First we note this is Bernoulli (p(x) = (x + 1)/x and q(x) = −1). v = y1−5 = y−4 . y + p(x)y = q(x)yn . try a different one. The above table is just a rule of thumb. Several different substitutions might work. the so-called Bernoulli equations† . We substitute v = −4y−5 y .
= y . Otherwise.5. SUBSTITUTION
33
Note that D = 0 gives y = x + 2. So xy + y(x + 1) + xy5 = 0. The Bernoullis were a prominent Swiss family of mathematicians. In other words. 4 −x v + v(x + 1) + x = 0.
1. If n = 0 or n = 1 then the equation is linear and we can solve it. You guess. Note that n need not be an integer.
−y5 v 4
y(1) = 1. 4 −x v + y−4 (x + 1) + x = 0. You might have to modify your guesses. We summarize a few of these in a table.2 Bernoulli equations
There are some forms of equations where there is a general rule for substitution which always works. but no value of D gives the solution y = x.5.
†
.
§2. We call these types of solutions equilibrium solutions. We call such critical points stable. t is time. then as t → ∞ we get x → A. by looking at the graph. that the solution x = A is "stable" in that small perturbations in x do not lead to substantially different solutions as t grows.7: Slope field and some solutions of x = −0.6 for an example.
Figure 1. If a critical point is not stable we would say it is unstable.
0 10 5 10 15 20 10
0 10
5
10
15
20 10
5
5
5
5
0
0
0
-5 -5
0
-10 0 5 10 15 20
-10
-5 0 5 10 15 20
-5
Figure 1. the naming comes from the fact that the equation is independent of time. dt where x is the temperature.3(x − 5). The points on the x axis where f (x) = 0 are called critical points.6
Autonomous equations
Note: 1 lecture. Note also. These types of equations are called autonomous equations. In this simple example it turns out that all solutions in fact go to A as t → ∞. Newton's law of cooling says that dx = −k(x − A).2 in EP Let us consider problems of the form dx = f (x). If we change the initial condition a little bit. In fact.1x(5 − x). Note the solution x = A (in the example A = 5). each critical point corresponds to an equilibrium solution.6: Slope field and some solutions of x = −0. If we think of t as time. FIRST ORDER ODES
1. k is some constant and A is the ambient temperature. dt where the derivative of solutions depends only on x (the dependent variable). Let us come back to the cooling coffee problem.36
CHAPTER 1. The point x = A is a critical point. See Figure 1.
.
It is easier to just look at the phase diagram or phase portrait. Note two critical points. It is not really necessary to find the exact solutions to talk about the long term behavior of the solutions. but it may get there rather quickly. This equation is commonly used to model population if you know the limiting population M.1.
. we have seen that it only exists for some finite period of time. x = 0 and x = 5. Think of the equation y = y2 . On the other hand the critical point at x = 0 is unstable." From just looking at the slope field we cannot quite decide what happens if x(0) < 0. For example. In this case there is one dependent variable x. AUTONOMOUS EQUATIONS Let us consider the logistic equation dx = kx(M − x). The critical point at x = 5 is stable.7 on the facing page for an example. but we will still consider negative x for the purposes of the math.6. In our example equation above it will actually turn out that the solution does not exist for all time. lim x(t) = 0 if x(0) = 0. mark all the critical points and then draw arrows in between. dt
37
for some positive k and M. In any case. So draw the x axis. Where DNE means "does not exist. Note that in the real world there is no such thing as negative population. it is easy to approximately sketch how the solutions are going to look.
y=5 y=0
Armed with the phase diagram. Same can happen here. but to see that we would have to solve the equation. Mark positive with up and negative with down. Many times are interested only in the long term behavior of the solution and hence we would just be doing way too much work if we tried to solve the equation exactly. See Figure 1. which is a simple way to visualize the behavior of autonomous equations. from the above we can easily see that 5 if x(0) > 0. This scenario leads to less catastrophic predictions on world population. It could be that the solution does not exist t all the way to ∞. the solution does go to −∞. t→∞ DNE or − ∞ if x(0) < 0. that is the maximum sustainable population.
and the fast food restaurant serving them will go out of business. FIRST ORDER ODES
Exercise 1. no matter how well stocked the planet starts.9 on the facing page Finally if we are harvesting at 2 million humans per year. See Figure 1.6. If ever the population drops below B. Suppose x is the number of humans in millions on the planet and t is time in years.6. then A and B are distinct and positive. or A = B. no real solutions). A= It turns out that when h = 1. See Figure 1. unstable points are generally bad news.6 million.e. Let us think about the logistic equation with harvesting. or A and B both complex (i. the population will always plummet towards zero.
unstable
stable
Since any mathematical model we cook up will only be an approximation to the real world. then the population will not die out. Our equation becomes dx = kx(M − x) − h.1: Try sketching a few solutions. dt Critical points A and B are kM − (kM)2 − 4hk (kM)2 − 4hk B= .6 million it will tend towards this number. Note that these possibilities are A > B. Logistic equations are commonly used for modelling population. They keep a planet with humans on it and harvest the humans at a rate of h million humans per year. A small perturbation of the equilibrium state and we are out of business. If it ever drops below 1. humans will die out. Let M be the limiting population when no harvesting is done. This scenario is not one that we (as the human fast food proprietor) want to be in.6. then A = B. When the population is above 1. As long as the population stays above B which is approximately 1. There is only one critical point which is unstable. Check with the graph above if you are getting the same answers. The graph we will get is given in Figure 1. we can easily classify critical points as stable or unstable. dt Multiply out and solve for critical points dx = −kx2 + kMx − h.10 on the next page. humans will die out on the planet.38
CHAPTER 1. kM +
. Once we draw the phase diagram. When h = 1. There is no room for error.8 on the next page.2: Draw the phase diagram for different possibilities. 2k 2k Exercise 1.55 million. Suppose an alien race really likes to eat humans. k > 0 is some constant depending on how fast humans multiply.
That is we will only harvest only an amount proportional to current population. Suppose that we modify our dt harvesting. find the critical points and mark them stable or unstable.6.6: Start with the logistic equation dx = kx(M − x). Exercise 1.40
CHAPTER 1. b) Sketch typical solutions of the equation.5. b) Show that if kM > h. c) Find limt→∞ x(t) for the solution with the initial condition x(0) = 0. then the equation is still logistic. a) Construct the differential equation. FIRST ORDER ODES
diagram for x = f (x). that we harvest hx for some h > 0. c) What happens when kM < h?
.
Rinse repeat! That is. y0 ). y) y(x0 ) = y0 .4 in EP At this point it may be good to first try the Lab II and/or Project II from the IODE website: it is generally very hard if not impossible to get a nice formula for the solution of the problem y = f (x.0
2.edu/iode/. Do note that this is not exactly the solution.7.0 -1 0 1 2 3
0.5
0.5
2.0 3.0
2. NUMERICAL METHODS: EULER'S METHOD
41
1.
-1 3.5
2. Or perhaps we even want to produce a graph of the solution to inspect the behavior.5
1. The slope is the change in y per unit change in x.11.0
2. then we will say that y1 (the approximate value of y at x1 = x0 + h) will be y1 = y0 + hk. More abstractly we compute xi+1 = xi + h.5
0.0 -1 0 1 2 3 3.0
1.math. yi+1 = yi + h f (xi .12 on the next page for the plot of the real solution.
Named after the Swiss mathematician Leonhard Paul Euler (1707 – 1783).0
1.5
1.5
0. We follow the line for an interval of length h.uiuc. See Figure 1.5
0. compute x2 and y2 using x1 and y1 .5
2. For an example of the first two steps of the method see Figure 1.1.11: First two steps of Euler's method with h = 1 for the equation y = conditions y(0) = 1. as we said before.5
1. The first thing to note is that. Hence if y = y0 at x0 .0
0.0
1. Do note the correct pronunciation of the name sounds more like "oiler.0
0.0 0 1 2 3 3.0
1."
‡
.0
2.
What if we want to find out the value of the solution at some particular x.5
2. §2. Euler's method‡ : We take x0 and compute the slope k = f (x0 .7
Numerical methods: Euler's method
Note: 1 lecture. yi ).0 -1 0 1 2 3
0.5
1.
y2 3
with initial
By connecting the dots we get an approximate graph of the solution.0
Figure 1.
so we only have a vague understanding of the error. Exercise 1. The main point is. so error of about 0. The difference between the actual solution and the approximate solution we will call the error. Let us try to approximate y(2) using Euler's method. If we knew the error exactly . Table 1. assuming that the error was comparable to start with. With step size 1 we have y(2) ≈ 1. Let us halve the step size.0
Figure 1. The real answer is 3.1 on the facing page gives the values computed for various parameters. We will usually talk about just the size of the error and we do not care much about its sign.12 we have essentially graphically approximated y(2) with step size 1.11 and 1. This halving of the error is a general feature of Euler's method as it is a first order method.5
1. The improved Euler method should quarter the error every time you halve the interval. That is quite a bit to do by hand. This reduction can be a big deal.5
0..42
-1 3.209.0 -1 0 1 2 3
0. To get it to within 0. Note that to get the error to be within 0..12: Two steps of Euler's method (step size 1) and the exact solution for the equation 2 y = y3 with initial conditions y(0) = 1. meaning doing 512 to 1024 steps. We notice that except for the first few times. every time we halved the interval the error approximately halved. that we usually do not know the real solution.926. whereas with 5 halvings you only have to do 32 steps.5
2. but suppose each step would take a second to compute (the function may be substantially
2
.5
0. FIRST ORDER ODES
3 3.
Let us see what happens with the equation y = y3 . so you would have to approximately do half as many "halvings" to get the same error.5
1.0
1.5
2.0
2. y(0) = 1. So we are approximately 1.791. A computer may not care between this difference for a problem this simple.074 off. With 10 halvings (starting at h = 1) you have 1024 steps.1 of the answer we had to already do 64 steps. A second order method reduces the error to approximately one quarter every time you halve the interval.0
2.1: Solve this equation exactly and show that y(2) = 3. what is the point of doing the approximation. If you do the computation you will find that y(2) ≈ 2. In Figures 1.0
1. In the IODE Project II you are asked to implement a second order method.7.0 0 1 2
CHAPTER 1.01 we would have to halve another 3 or four times.0
0.
Then the difference is 32 seconds versus about 17 minutes.509130743538
y2 .0078125 2. Next. assume that the error goes down by a factor of 2.1: Euler's method approximation of y(2) where of y =
y(0) = 1. you should solve the equation exactly and you will notice that the solution does not exist at x = 3.791388470013 0. suppose you do not know the error.049645018422 0. Can you estimate the error in the last time from this? Does it agree with the table? Now do it for the first two rows.179599204497 0. the error generally goes down by a factor of 16.7. suppose that you have to repeat such a calculation for different parameters a thousand times.97472419486 Error 1. Results of this effort are listed in Table 1. it is 1 minute versus 17 minutes.5 2.2: In the table above.95035498158 0.82040079550 0.25 2.0625 2.7.533849442573 0.125 2. a second order method would probably double the time to do each step.92592592593 0. but even a better approximation method than Euler would need an insanely small step size to compute the solution with reasonable precision. You get the idea. And computers might not be able to handle such a small step size anyway. Another case when things can go bad is if the solution oscillates wildly near some point.095878935207 0. In real applications you would not use a simple method such as Euler's. Take the approximate values of the function in the last two lines. Note that we do not know the error! How do you know what is the right step size? Essentially you keep halving the interval and if you are lucky you can estimate the error from a few of these calculations and the assumption that the error goes down by a factor of one half each time (if you are using standard Euler). That is a fourth order method.2 on the next page for successive halvings of h.68033658758 0.47249414666 0.561838476090 0. Even so. 3
Table 1. NUMERICAL METHODS: EULER'S METHOD h Approximate y(2) 1 1. Exercise 1.03125 2.
more difficult to compute than y2 /3).90412106479 0.527505853335 0. Suppose that instead of y(2) we wish to find y(3). In fact the solution blows up. In this case. that means that if you halve the interval. The simplest method that would probably be used in a real application is the standard Runge-Kutta method (we will not describe it here).015625 2. Does this agree with the table? Let talk a little bit more about this example y = y3 . Such an example is given in IODE Project II. What is going on here? Well.025275805142
Error Previous error
43
0.20861152999 0.
2
.319663412423 0.605990266083 0. y(0) = 1.074074074070 0. Note: We are not being altogether fair.517788587396 0.736809954840 0. the solution may exist at all points.666557415634 0.1.
4600446195 0.
Choosing the right method to use and the right step size can be very tricky. There are several competing factors to consider.86078752222 0. does not mean that you must have the right answer.44
CHAPTER 1.2: Attempts to use Euler's to approximate y(3) where of y =
y2 . Use Euler's method with step size h = 0. There is ongoing active research by engineers and mathematicians on how to do numerical approximation in the best way. Errors introduced by rounding numbers off during your computations become noticeable when the step size becomes too small relative to the quantities you are working with. So reducing step size may in fact make errors worse.3: Consider = (2t − x)2 . FIRST ORDER ODES h Approximate y(3) 1 3.4012144477 0.7.5989264104 0. Small errors lead to large errors down the line. x(0) = 2. 3
y(0) = 1.54328915766 0. Even if the function f is simple to compute.03125 29. Just because the numbers have stabilized after successive halving.015625 50. the general purpose method used for the ODE solver in Matlab and Octave (as of this writing) is a method that appeared only in the literature only in the 1980s. You have seen just the beginnings of the challenges that appear in real applications.8032064113 0.5 4.
1.0625 17. but perhaps not the right precision. For example.1 Exercises
dx Exercise 1.7. Or what may happen is that the numbers may never stabilize no matter how many times you halve the interval. • Computational time: Each step takes computer time.0078125 87.5 to dt approximate x(1).16232281664 0. Large step size means faster computation. • Roundoff errors: Computers only compute with a certain number of significant digits.25 6. Or in the worst case the numerical computations might be giving you bogus numbers that look like a correct answer.
.7576927770 Table 1. • Stability: Certain equations may be numerically unstable.125 10. you do it many times over.
1. Exercise 2. So y1 and y2 are linearly independent. Hence y = C1 cos x + C2 sin x is the general solution to y + y = 0. It is obvious that sin and cos are not multiples of each other.
2.1.1. Hint: Try y = xr . For example.2: Show that y = e x and y = e2x are linearly independent. Or the equation y − y = 0 with y(0) = b0 and y (0) = b1 has the solution y(x) = b0 cosh x + b1 sinh x.
49
Here note that using cosh and sinh allows us to solve for the initial conditions much more easily than if we have used the exponentials. we found the solutions y1 = sin x and y2 = cos x for the equation y + y = 0. Show that y solves Ly = f (x) + g(x). In this case y = C1 y1 + C2 y2 is the general solution.1 Exercises
Exercise 2. If sin x = A cos x for some constant A.5: For the equation x2 y − xy = 0. the equation y + y = 0 with y(0) = b0 and y (0) = b1 has the solution y(x) = b0 cos x + b1 sin x.1. then every other solution is written in the form y = C1 y1 + C2 y2 . we let x = 0 and this would imply A = 0 = sin x. Exercise 2.1.
. Question: Suppose we find two different solutions y1 and y2 to the homogeneous equation (2. Note that the initial condition for a second order ODE consists of two equations. Suppose that y1 is a solution to Ly1 = f (x) and y2 is a solution to Ly2 = g(x) (same operator L). If you find two linearly independent solutions.2. which is preposterous. show that they are linearly independent and find the general solution.1.2).3: Take y + 5y = 10x + 5. So if we have two arbitrary constants we should be able to solve for the constants and find a solution satisfying the initial conditions. Can you find guess a solution? Exercise 2.4: Prove the superposition principle for nonhomogeneous equations. find two solutions. Can every solution be written (using superposition) in the form y = C1 y1 + C2 y2 ? Answer is affirmative! Provided that y1 and y2 are different enough in the following sense. SECOND ORDER LINEAR ODES For example. We will say y1 and y2 are linearly independent if one is not a constant multiple of the other.
6: Suppose that (b − a)2 − 4ac > 0. If you have one solution to a second order linear homogeneous equation you can find another one. c) Write down the general solution. c) Write down the general solution. Exercise 2. Exercise 2. b) What happens when (b − a)2 − 4ac = 0 or (b − a)2 − 4ac < 0? We will revisit the case when (b − a)2 − 4ac < 0 later. They are solved by trying y = xr and solving for r (we can assume that x ≥ 0 for simplicity). b) Use reduction of order to find a second linearly independent solution. e p(x) dx dx (y1 (x))2
. Hint: Try y = xr and find a formula for r. Exercise 2. Hint: Try y = xr ln x for the second solution. HIGHER ORDER LINEAR ODES
Note that equations of the form ax2 y + bxy + cy = 0 are called Euler's equations or CauchyEuler equations. a) Show that y = x is a solution. b) Use reduction of order to find a second linearly independent solution.9 (Chebychev's equation of order 1): Take (1 − x2 )y − xy + y = 0. Show that y2 (x) = y1 (x) is also a solution.1. Exercise 2. Exercise 2. a) Find a formula for the general solution of ax2 y + bxy + cy = 0.1. a) Show that y = 1 − 2x2 is a solution. Let us solve some famous equations. This is the reduction of order method.8: Suppose y1 is a solution to y + p(x)y + q(x)y = 0.1.1. Find a formula for the general solution of ax2 y + bxy + cy = 0.7: Suppose that (b − a)2 − 4ac = 0.50
CHAPTER 2.1.10 (Hermite's equation of order 2): Take y − 2xy + 4y = 0.
If they were not we could write e4x = Ce2x . To apply the initial conditions we first find y = 2C1 e2x + 4C2 e4x . Constant coefficients means that the functions in front of y . Then C1 = −7 as −2 = C1 + 5. y . r2 erx − 6rerx + 8erx = 0. Plug in to get y − 6y + 8y = 0.2. which would imply that e2x = C which is clearly not possible. −2 = y(0) = C1 + C2 . add these together.1: Check that y1 and y2 are solutions. y (0) = 6.3)
. We need to solve for C1 and C2 . the solution we are looking for is y = −7e2x + 5e4x . Exercise 2. Either apply some matrix algebra. 6 = y (0) = 2C1 + 4C2 . and subtract the two equations to get 5 = C2 .2. y(0) = −2. second part of §3. and end up with zero. we can write the general solution as y = C1 e2x + C2 e4x . not depending on x. Think about a function that you know that stays essentially the same when you differentiate it. and y are constants. So let y1 = e2x and y2 = e4x . The functions e2x and e4x are linearly independent.
So if r = 2 or r = 4. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES
51
2.2. Suppose that we have an equation ay + by + cy = 0. Then y = rerx and y = r2 erx . Hence. We plug in x = 0 and solve. For example. then erx is a solution. r2 − 6r + 8 = 0 (r − 2)(r − 4) = 0. so that we can take the function and its derivatives.
This is a second order linear homogeneous equation with constant coefficients. (2.1 in EP Suppose we have the problem y − 6y + 8y = 0. or just solve these by high school algebra. (divide through by erx ). divide the second equation by 2 to obtain 3 = C1 + 2C2 .2
Constant coefficient second order linear ODEs
Note: more than 1 lecture. Let us try a solution y = erx . Hence. Let us generalize this example into a method.
Let us compute y = e4x + 4xe4x and y = 8e4x + 16xe4x .3) has the general solution y = C1 er1 x + C2 er2 x . The characteristic equation is r2 − 8r + 16 = (r − 4)2 = 0. When r1 goes to r2 in the limit this is like taking r2 −r1 derivative of erx using r as a variable. This limit is xerx . Let us give a short "proof" for why the solution xerx works when the root is doubled. HIGHER ORDER LINEAR ODES
where a. note the equation y − k2 y = 0. Note that er2 x −er1 x is a solution when the roots are distinct. Suppose that r1 and r2 are the roots of the characteristic equation.2. The equation ar2 + br + c = 0 is called the characteristic equation of the ODE. (ii) If r1 = r2 (b2 − 4ac = 0). Hence a double root r1 = r2 = 4. b.1. c are constants. √ −b ± b2 − 4ac r1 . Example 2. doubled root rarely happens. There is still a difficulty if r1 = r2 . and hence this is also a solution in the doubled root case. then (2. Exercise 2. We should note that in practice.2: Check that e4x and xe4x are linearly independent. we have er1 x and er2 x as solutions. (i) If r1 and r2 are distinct and real (b2 − 4ac > 0). The general solution is. Here the characteristic equation is r2 − k2 = 0 or (r − k)(r + k) = 0 and hence e−kx and ekx are the two linearly independent solutions. Since this case is really a limiting case of when cases the two roots are distinct and very close.2.3) has the general solution y = (C1 + C2 x) er1 x . r2 = . then (2. but it is not hard to overcome.52
CHAPTER 2. Plug in y − 8y + 16y = 8e4x + 16xe4x − 8(e4x + 4xe4x ) + 16xe4x = 0.2. That e4x solves the equation is clear. If xe4x solves the equation then we know we are done. Theorem 2.1: Find the general solution of y − 8y + 16y = 0. y = (C1 + C2 x) e4x = C1 e4x + C2 xe4x .
. therefore. For another example of the first case. 2a Therefore. Solve for the r by using the quadratic formula. Try the solution y = erx to obtain ar2 erx + brerx + cerx = 0 ar2 + br + c = 0. If you pick your coefficients truly randomly you are very unlikely to get a doubled root.
13
We can also define the exponential ea+ib of a complex number. A complex number is really just a pair of real numbers. i3 = −i. b). e x+y = e x ey . b) as a + ib. We define a multiplication by (a. •
1 3−2i
=
1 3+2i 3−2i 3+2i
=
3+2i 13
=
3 13
+
2 i.
. Further. You can just do arithmetic with complex numbers just as you would do with polynomials. b) × (c. This means that ea+ib = ea eib and hence if we can compute eib easily. 1) × (0. Theorem 2. 0). we can compute ea+ib .3: Make sure you understand (that you can justify) the following identities: • i2 = −1. We will use the mathematicians convention and use i. Complex numbers may seem a strange concept especially because of the terminology. we note that many properties still hold for the complex exponential.2. For example. for example i and −i are roots of r2 + 1 = 0. 1) = (−1. ad + bc). We can think of a complex number as a point in the plane. (a. and we treat i as if it were an unknown. eiθ = cos θ + i sin θ and e−iθ = cos θ − i sin θ.2. all the standard properties of arithmetic hold. For example. Generally we just write (a.2. So whenever you see i2 you can replace it by −1. but it does have two complex roots. There is nothing imaginary or really complicated about complex numbers. Note that engineers often use the letter j instead of i for the square root of −1. Here we review some properties of complex numbers.2. Exercise 2. Here we will use the so-called Euler's formula.1 Complex numbers and Euler's formula
It may happen that a polynomial has some complex roots. Because most properties of the exponential can be proved by looking at the Taylor series. Also. CONSTANT COEFFICIENT SECOND ORDER LINEAR ODES
53
2. i4 = 1.2. i
def
• (3 − 7i)(−2 − 9i) = · · · = −69 − 13i.2 (Euler's formula). The property we just mentioned becomes i2 = −1. • (3 − 2i)(3 + 2i) = 32 − (2i)2 = 32 + 22 = 13. d) = (ac − bd. We can do this by just writing down the Taylor series and plugging in the complex number. It turns out that with this multiplication rule. and most importantly (0. • 1 = −i. the equation r2 + 1 = 0 has no real roots. We add complex numbers in the straightforward way.
check the identities: cos θ = eiθ + e−iθ 2 and sin θ = eiθ − e−iθ . That is. r2 = ±i . and y2 = e(α−iβ)x . We note that linear combinations of solutions are also solutions. However.
2. you will always get a pair of roots of the form α ± iβ. In this case we can still write the solution as y = C1 e(α+iβ)x + C2 e(α−iβ)x .2.
. First let y1 = e(α+iβ)x Then note that y1 = eαx cos βx + ieαx sin βx.4: Using Euler's formula. We would need to choose C1 and C2 to be complex numbers to obtain a real valued solution (which is what we are after).54
CHAPTER 2.2. In this case we can see that the roots are √ −b b2 − 4ac r1 . Use Euler on each side and deduce: cos 2θ = cos2 θ − sin2 θ and sin 2θ = 2 sin θ cos θ. HIGHER ORDER LINEAR ODES
Exercise 2. the exponential is now complex valued. These are complex if b2 − 4ac < 0. 2a 2a As you can see. For a complex number a + ib we call a the real part and b the imaginary part of the number. 2i
2
Exercise 2.5: Double angle identities: Start with ei(2θ) = eiθ . In notation this is Re(a + bi) = a and Im(a + bi) = b. We also will need some notation. While there is nothing particularly wrong with this. we have the following theorem. Therefore.2. by quadratic formula the roots are −b± 2a −4ac .2 Complex roots
So now suppose that the equation ay + by + cy = 0 has a characteristic equation ar2 + br + c = 0 √ b2 which has complex roots. y3 = 2 y1 − y2 y4 = = eαx sin βx. y2 = eαx cos βx − ieαx sin βx. it can make calculations harder and it would be nice to find two real valued solutions. Hence y1 + y2 = eαx cos βx. And furthermore they are real valued. Here we can use Euler's formula. 2i are also solutions. It is not hard to see that they are linearly independent (not multiples of each other).
12: Find the general solution of y = 0 using the methods of this section.56
CHAPTER 2. Suppose now that (b − a)2 − 4ac < 0. We will see higher orders later. HIGHER ORDER LINEAR ODES
Exercise 2.13: The method of this section applies to equations of other orders than two. Exercise 2. Exercise 2.
.2. Find a formula for the general solution of ax2 y + bxy + cy = 0.6 on page 50. Try to solve the first order equation 2y + 3y = 0 using the methods of this section.14: Let us revisit Euler's equations of Exercise 2. Hint: Note that xr = er ln x .2.2.1.
. y2 . . has only the trivial solution c1 = c2 = · · · = cn = 0. . and f are continuous functions and a. The important new concept here is the concept of linear independence. If we can write the equation with a nonzero constant.3 in EP In general. .4) for arbitrary constants C1 . In this case it is easier to state as follows.3
Higher order linear ODEs
Note: 2 lectures. the methods are slightly harder. yn are linearly independent if c1 y1 + c2 y2 + · · · + cn yn = 0. . Suppose y1 .
.. Theorem 2.
2. . and it is useful to understand this in detail. The functions y1 .2. Higher order equations do appear from time to time. . (2. So let us start with a general homogeneous linear equation y(n) + pn−1 (x)y(n−1) + · · · + p1 (x)y + p0 (x)y = 0. yn are solutions of the homogeneous equation (2.." The basic results about linear ODEs of higher order are essentially exactly the same as for second order equations with 2 replaced by n. . .2 and §3. b0 . Then y(x) = C1 y1 (x) + C2 y2 (x) + · · · + Cn yn (x). This concept is used in many other areas of mathematics and even other places in this course. §3. HIGHER ORDER LINEAR ODES
57
2.1 Linear independence
When we had two functions y1 and y2 we said they were linearly independent if one was not the multiple of the other. but we will not dwell on these.3. . has exactly one solution y(x) satisfying the initial conditions y(a) = b0 . bn−1 are constants. . For constant coefficient ODEs. y(n−1 )(a) = bn−1 . We also have the existence and uniqueness theorem for nonhomogeneous linear equations. but it is a general assumption of modern physics that the world is "second order.. You can always use the methods for systems of linear equations we will learn later in the course to solve higher order constant coefficient equations.4). b1 . way say they are linearly dependent.2 (Existence and uniqueness). .1 (Superposition). Same idea holds for n functions. The equation y(n) + pn−1 (x)y(n−1) + · · · + p1 (x)y + p0 (x)y = f (x).4)
Theorem 2. .3. . Cn . most equations that appear in applications tend to be second order. also solves (2.3. say c1 0. y (a) = b1 . then we can solve for y1 as a linear combination of the others. . . Suppose p0 through pn−1 . If the functions are not linearly independent. . y2 .3. .
Now differentiate both sides c2 e x + 2c3 e2x = 0. and cosh x are linearly dependent. Hence our equation becomes c1 e x + c2 e2x = 0. Set x = 0 to get the equation c1 + c2 + c3 = 0. therefore. That might be a lot of computation.3. Let us first divide by e x for simplicity. Finally divide by e x again and differentiate to get 4c3 e2x = 0. it is identically zero and c1 = c2 = c3 = 0 and the functions are linearly independent.2: On the other hand. Use rules of exponentials and write z = e x . Most textbooks (including [EP] and [F]) introduce Wronskians. Rinse. Then we have c1 z + c2 z2 + c3 z3 = 0. Let us write down c1 e x + c2 e2x + c3 e3x = 0. c2 and c3 . Therefore. c1 + c2 e x + c3 e2x = 0. It is clear that c3 is zero. e2x .58
CHAPTER 2. After taking the limit we see that c3 = 0. Let us give several ways to do this. There is no one good way to do it. Let us try another way. HIGHER ORDER LINEAR ODES
Example 2. What we could do is divide through by e3x to get c1 e−2x + c2 e−x + c3 = 0.1: Show e x . This is true for all x. let x → ∞. We can also take derivatives of both sides and then evaluate. This equation has to hold for all x. Example 2. All of these methods are perfectly valid. but that is really not necessary here. Then c2 must be zero as c2 = −2c3 and c1 must be zero because c1 +c2 +c3 = 0. e3x are linearly independent. Simply apply definition of the hyperbolic cosine: cosh x = e x + e−x . 2
. The left hand side is is a third degree polynomial in z. the functions e x . Write c1 e x + c2 e2x + c3 e3x = 0. e−x . and set x = 0 to get c2 + 2c3 = 0.3. repeat! How about yet another way. We could evaluate at several different x to get equations for c1 . Write c1 e x + c2 e2x + c3 e3x = 0. It can either be identically zero or have at most 3 zeros.
1. For example.3: Find the general solution to y − 3y − y + 3y = 0.3. these are easy to see. We just need to find more solutions. 1. Then r3 − 3r2 − r + 3 = 0. The last root is then reasonably easy to find. 2 = y (0) = −C1 + C2 + 3C3 . if you plug in −2 into our polynomial you get −15.5). A good strategy at first is to look for roots −1. There is no formula for higher degree polynomials. They are linearly independent as can easily be checked. When check our polynomial we note that r1 = −1 and r2 = 1 are roots. Computers are pretty good at finding roots approximately for reasonable size polynomials. The song and dance is exactly the same as it was for second order. or 0. The trick now is to find the roots. y (0) = 2. Hence the general solution is y = C1 e−x + C2 e x + C3 e3x . There are always n roots for an nth degree polynomial. That means there is a root between −2 and 0 because the sign changed. Or you can try plugging in. Best place to start is to plot the polynomial and check where it is zero. and see if you get a hit.3.2. You should check that r3 = 3 is a root.2 Constant coefficient higher order ODEs
When we have a higher order constant coefficient homogeneous linear equation.3. That does not mean that the roots do not exist. Example 2. There are some signs that you might have missed a root. Sometimes it is a good idea to just start plugging in numbers r = −2. . If the equation is nth order we need to find n linearly independent solutions. . −1. . e x and e3x are solutions to (2. −1. 2. HIGHER ORDER LINEAR ODES
59
2. and y (0) = 3. It is best seen by example. There is a formula for degree 3 and 4 equations but it is very complicated. Try: y = erx . In our case we see that 3 = (−r1 )(−r2 )(−r3 ) = (1)(−1)(−r3 ) = r3 . Suppose we were given some initial conditions y(0) = 1. We note that the constant term in a polynomial is the multiple of the negations of all the roots because r3 − 3r2 − r + 3 = (r − r1 )(r − r2 )(r − r3 ). and there is 3 of them. We plug in and get r3 erx − 3r2 erx − rerx + 3erx = 0.5)
. We divide out by erx . They might be repeated and they might be complex. This leads to 1 = y(0) = C1 + C2 + C3 . 0. Hence we know that e−x . 3 = y (0) = C1 + C2 + 9C3 . If you plug in 0 you get 3. which happens to be exactly the number we need. (2.
It would be good if someone did the math before you jump off right? Let us just give 2 other examples. Let x be the displacement of the mass (x = 0 is the rest position). it is kx in the negative direction. if c > 0. Suppose we have a mass m > 0 (in kilograms for instance) connected by a spring with spring constant k > 0 (in Newtons per meter perhaps) to a fixed wall. there is some friction in the system and this
is measured by a constant c ≥ 0. Similarly the amount of force exerted by friction is proportional to the velocity of the mass.62
CHAPTER 2. There is also an electric source (such as a battery) giving a voltage of E(t) volts at time t (measured in seconds). For example. There is a resistor with a resistance of R ohms. and (iv) undamped. This system is appears in lots of applications even if it does not at first seems like it. The relation between the two is
.4. (iii) damped. Let Q(t) be the charge in columbs on the capacitor and I(t) be the current in the circuit. Finally. a bungee jump setup is essentially a spring and mass system (you are the mass). if F 0 (F not identically zero). and a capacitor with a capacitance R of C farads. Suppose that you have the pictured RLC circuit. if c = 0. By Newton's second law we know that force equals mass times acceleration and hence mx + cx + kx = F(t). The force exerted by the spring is proportional to the compression of the spring by Hooke's law. Here is an example for electrical engineers.4
Mechanical vibrations
Note: 2 lectures. With x growing to the right (away from the wall). if F ≡ 0. §3. We say the motion is (i) forced. there is some external force F(t) acting on the mass. (ii) unforced or free. Many real world scenarios can be simplified to a mass on a spring. HIGHER ORDER LINEAR ODES
2. We set up some terminology about this equation.1 Some examples
k m damping c F(t) Our first example is a mass on a spring. Therefore.
2. This is a linear second order constant coefficient ODE. Furthermore. an C E L inductor with an inductance of L henries.4 in EP We want to look at some applications of linear second order constant coefficient equations.
This can be seen by looking at the graph. Mass is replaced by the inductance. Furthermore. This is mg sin θ in the opposite direction. C This is an nonhomogeneous second order constant coefficient linear equation. R.
-1. The change in voltage becomes the forcing function.5 0.5 (in radians) the graphs of sin θ and θ are almost the same. Hence for constant voltage this is an unforced motion.5
-0. The m curiously θ cancels from the equation. Further.1: The graphs of sin θ and θ (in radians). as L. The position of the mass is replaced by the current.2. This has to be equal to the tangential component of the force given L by the gravity.5 1. Note that acceleration is Lθ and mass is m.5 0. MECHANICAL VIBRATIONS
63
Q = I. damping is replaced by resistance and the spring constant is replaced by one over the capacitance.0
0.0 0.5
0. In Figure 2. If we differentiate we get 1 LI (t) + RI (t) + I(t) = E (t). For small θ we have that approximately sin θ ≈ θ. Elementary physics mandates that the equation is of the form θ + g sin θ = 0. We wish to find an equation for the angle θ(t). by elementary principles we have that LI +RI + Q/C = E.0
-1.5
-1.
.0 -0.5 1.0 -1. Now we make our approximation. and C are all positive.4.5 < θ < 0. where force equals mass times acceleration.5
0. Suppose we have a mass m on a pendulum of length L.0 1.0
Figure 2.0 -0.0 0. Our next example is going to behave like a mass and spring system only approximately.0
-0. L
This equation can be derived using Newton's second law.0
0. this system behaves just like the mass and spring system. Let g be the force of gravity.0 1.1 we can see that for approximately −0.
First let us start with undamped motion and hence c = 0. Therefore. when the swings are small. so we have the equation mx + kx = 0. Therefore. But for reasonably short periods of time and small swings (for example if the length of the pendulum is very large). this is not true for a pendulum.2 Free undamped motion
In this section we will only consider free or unforced motion. the behavior is reasonably close. HIGHER ORDER LINEAR ODES
Therefore. √ B It is not hard to compute that C = A2 + B2 and tan γ = A .1: Justify this identity and verify the equations for C and γ. Also we will see that in a mass spring system. the amplitude is independent of the period.4. If we look at the form of the solution x(t) = C cos(ω0 t − γ)
k m
we can write the equation as
. 0 The general solution to this equation is x(t) = A cos ω0 t + B sin ω0 t. θ is always small and we can model the behavior by the simpler linear equation g θ + θ = 0. First we notice that by a trigonometric identity we have that for two other constants C and γ we have A cos ω0 t + B sin ω0 t = C cos(ω0 t − γ). as we cannot yet solve nonhomogeneous equations. and let C and γ be our arbitrary constants. L Note that the errors that we get from the approximation build up so over a very long time. The constants C and γ have very nice interpretation.
2. the second form is much more natural. In real world problems it is very often necessary to make these types of simplifications.4. Exercise 2. While it is generally easier to use the first form with A and B to find these constants given the initial conditions.64
CHAPTER 2. the behavior might change more substantially. we can write x(t) = C cos(ω0 t − γ). If we divide out by m and let ω0 be a number such that ω2 = 0 x + ω2 x = 0. it is good to understand both the mathematics and the physics of the situation to see if the simplification is valid in the context of the questions we are trying to answer.
We know that tan γ = B/A = 2.
=
√ 4 = 2. 2π The period of the motion is one over the frequency (in cycles per unit time) and hence ω0 . Example 2. Then x (t) = −0. the mass was moving forward (in the positive direction) at 1m/s.4. Therefore. For the free undamped motion. This makes it much easier to figure out A and B. In the example. That is the amount of time it takes to complete one full oscillation. This gives us the initial conditions.5 sin 2t + B cos 2t. not in cycles per unit time as is the usual measure of frequency.318 The general solution is x(t) = A cos 2t + B sin 2t. But because we know one cycle is 2π. A note about the word angular before the frequency. Letting x (0) = 1 we get √ √ B = 1.5 cos 2t + sin 2t.107. ω0 is given in radians per unit time. We call ω0 is called the natural (angular) frequency. and that 2π is a matter of taste. A plot is shown in Figure 2.2 on the following page. MECHANICAL VIBRATIONS
65
We can see that the amplitude is C. Unfortunately if you remember. It is simply a matter of where we put the constant 2π.4. The units are the mks units (meters-kilograms-seconds). Let us compute the phase shift.5. we still
. and gets loose in the crash and starts oscillating.
x (0) = 1. So the equation with initial conditions is 2x + 8x = 0. Hence the angular frequency is 2.5. The solution is x(t) = 0. Suppose the whole setup is on a truck which was travelling at 1m/s and suddenly crashes and hence stops. the usual frequency is given by ω0 . We can directly compute ω0 =
k m
x(0) = 0. this corresponds to the initial conditions x(0) = A and x (0) = B. The motion is usually called simple harmonic motion. The usual
1 2 frequency in Hertz (cycles per second) is 2π = π ≈ 0. rather than the amplitude and phase shift. Well the setup means that the mass was at half a meter in the positive direction during the crash and relative to the wall the spring is mounted to.1: Suppose that m = 2kg and k = 8N/m. It just shifts the graph left or right.5 meters forward from the rest position. and γ is the so-called phase shift. the amplitude is C = A2 + B2 = 1.2.118. we have already found C. if the solution is of the form x(t) = A cos ω0 t + B sin ω0 t. The mass was rigged 0.
Letting x(0) = 0 means A = 0. ω0 is the (angular) frequency. We take the arctangent of 2 and get approximately 1.25 ≈ 1. What is the frequency of the resulting oscillation and what is the amplitude.
0
2. Since both B and A are positive.0
0.0 2. as x + 2px + ω2 x = 0. 0 k .66
0.2: Simple undamped oscillation.
need to check if this γ is in the right quadrant. 2m
The form of the solution depends on whether we get complex or real roots and this depends on the sign of c 2 k c2 − 4km p2 − ω2 = − = .4.107 radians really is in the first quadrant.5
-0. Note: Many calculators and computer software do not only have the atan function for arctangent.0
7. This function takes two arguments. and 1. 0 where ω0 = The characteristic equation is r2 + 2pr + ω2 = 0.0
0. Let us rewrite the equation mx + cx + kx = 0. m p= c .3 Free damped motion
Let us now focus on damped motion.0
0. then γ should be in the first quadrant. HIGHER ORDER LINEAR ODES
5. 0 Using the quadratic formula we get that the roots are r = −p ± p2 − ω2 .5
CHAPTER 2.5
-1.0
Figure 2.0
-1. but also what is sometimes called atan2.0 7.5
10.5 10.0
1.0
1.5
5. B and A and returns a γ in the correct quadrant for you.5
0. 0 2m m 4m2
.0
-0.
2.5
0.
r2 are negative. You are always a little bit underdamped or a little bit overdamped. Then x0 x(t) = r1 er2 t − r2 er1 t .0
1. we try to solve 0 = C1 er1 t + C2 er2 t . Example 2. there is one root of multiplicity 2 and this root is −p. To see this fact. After all a critically damped system is in some sense a limit of overdamped systems. Hence the solution is x(t) = C1 er1 t + C2 er2 t .2.3.3: Overdamped motion for several different initial conditions. r1 − r2 It is not hard to see that this satisfies the initial conditions. as always negative. Do note that no oscillation happens. and hence −C1 = e(r2 −r1 )t . Critical damping When c2 − 4km = 0. p2 − ω2 is always less than p so −p ± 0 p2 − ω2 is 0
0 1.4. C2 This has at most one solution t ≥ 0.
. our solution is x(t) = C1 e−pt + C2 te−pt . Notice that both are negative. 0 Overdamping
67
When c2 − 4km > 0. For a few sample plots for different initial conditions see Figure 2. Since these equations are really only an approximation to the real world. it is only a place you can reach in theory.5
0.0
0
25
50
75
100
Figure 2.4. in reality we are never critically damped.5
0. That is x(0) = x0 and x (0) = 0. The behavior of a critically damped system is very similar to an overdamped system. there are two distinct real roots r1 and r2 .5
1. It is better not to dwell on critical damping. x(t) → 0 as t → ∞. In fact the graph will cross the x axis at most once. In this case. Note that since r1 .0
0. In this case. This means that the mass will just tend towards the rest position as time goes to infinity.5
25
50
75
100 1. we say the system is overdamped.2: Suppose the mass is released from from rest. So C1 er1 t = −C2 er2 t . MECHANICAL VIBRATIONS The sign of p2 − ω2 is the same as the sign of c2 − 4km.0
0. Therefore. we say the system is critically damped.
The envelope curves become flatter and flatter as p goes to 0.
2.4. In the figure we also show the envelope curves Ce−pt and −Ce−pt . we say the system is underdamped.4: Underdamped motion with the envelope curves shown.
. overdamped or critically damped? c) If the system is not critically damped.5 -0.3: Do Exercise 2. and damping constant c = 1.5
where ω1 =
ω2 − p2 . you are really interested in computing the envelope curve so that you do not hit the concrete with your head. b) Is the system underdamped. When we change the damping just a little bit. In this case.4.68 Underdamping
CHAPTER 2. and c = 12. The solution is the oscillating plot between the two curves.
An example plot is given in Figure 2.4. Or x(t) = Ce
−pt
cos(ω1 t − γ). On the other hand when c becomes smaller.2: Consider a mass and spring system with a mass m = 2.2 for m = 3. Exercise 2. Our solution is 0
-1. a) Set up and find the general solution of the system.0
= −p ± iω1 . This makes sense since if we keep changing c at some point the solution should start looking like the solution for critical damping or overdamping which do not oscillate at all.4. find a c which makes the system critically damped.
-0.0 0 5 10 15 20 25 30 -1.0
5
10
15
20
25
30 1. For example if you are bungee jumping. Finally note that the angular pseudo-frequency (we do not call it a frequency since the solution is not really a periodic function) ω1 becomes lower when the damping c (and hence p) becomes larger. r = −p ± = −p ± p2 − ω2 0 √ −1 ω2 − p2 0
0 1.4 Exercises
Exercise 2. the roots are complex.0
0. The envelope curves give the maximum amplitude of the oscillation at any given point in time. k = 12. ω1 approaches ω0 (it is always smaller) and the solution looks more and more like the steady periodic motion of the undamped case. The phase shift γ just shifts the graph left or right but within the envelope curves (the envelope curves do not change of course if γ changes). spring constant k = 3. we do not expect the behavior to change dramatically.5
0.0
0.
Figure 2.4. Note that we still have that x(t) → 0 as t → ∞.5
0. HIGHER ORDER LINEAR ODES
When c2 − 4km < 0.0
x(t) = e−pt (A cos ω1 t + B sin ω1 t) .
4. Assume no friction. MECHANICAL VIBRATIONS
69
Exercise 2. suppose you do not know the spring constant.8 Hz (cycles per second) what is the mass.4: Using the mks units (meters-kilograms-seconds) Suppose you have a spring of with spring constant 4N/m. a) Find k (spring constant) and c (damping constant). a) You count and find that the frequency is 0.4.39 Hz. Exercise 2. what is the weight?
. You put each in motion on your spring and measure the frequency. c) For an unknown mass you measured 0.5: Suppose we add possible friction to Exercise 2.8 Hz.2 Hz. For the 1 kg weight you measured 0. but you have two reference weights 1 kg and 2 kg to calibrate your setup. b) Find a formula for the mass in terms of the frequency in Hz. Further.4. Suppose you you place the mass on the spring and put it in motion.2. You want to use it to weight items. for the 2 kg weight you measured 0. b) Find a formula for the mass m given the frequency ω in Hz.4.4.
The solution y = yc + y p includes all solutions to (2. Lw = L(y p − y p ) = Ly p − L˜ p = (2x + 1) − (2x + 1) = 0.6). like the forcing function for the mechanical vibrations of last section. Note that L is a linear operator and so we could just write.70
CHAPTER 2. We will generally write Ly = 2x + 1 instead when the operator is not important.5 in EP
2.6)
Note that we still say this equation is constant coefficient equation.6) in some way and then we know that y = yc + y p is the general solution to (2.5
Nonhomogeneous equations
Note: 2 lectures.1 Solving nonhomogeneous equations
You have seen how to solve the linear constant coefficient homogeneous equations. we have an equation such as y + 5y + 6y = 2x + 1. We find the general solution yc to the associated homogeneous equation y + 5y + 6y = 0.6) differ by a solution to the ˜ homogeneous equation (2. Suppose you find a different particular solution y p . using the operator notation the calculation becomes simpler. We call yc the complementary solution. (2. and you might find a different one by a different method (or by guessing) and still get the right general solution to the whole problem even if it looks different and the constants you will have to choose given the initial conditions will be different. since yc is the general solution to the homogeneous equation. ˜ y So w = y p − y p is a solution to (2. So any two solutions of (2. (2. This usually corresponds to some outside input to the system we are trying to model. Note that y p can be any solution.7)
. We only require constants in front of the y . Then plug w into the left hand side of the equation and get ˜ w + 5w + 6w = (y p + 5y p + 6y p ) − (˜ p + 5˜ p + 6˜ p ) = (2x + 1) − (2x + 1) = 0. y y y In other words.6). §3. Then write ˜ the difference w = y p − y p . HIGHER ORDER LINEAR ODES
2.7). The way we solve (2. and y.7). Moral of the story is that you can find the particular solution in any old way.5. Now suppose that we drop the requirement of homogeneity. We also find a single particular solution y p to (2. y .6) is as follows. That is.
We note also that using the multiplication rule for differentiation gives us a way to combine these guesses. Now we can go forward and try it. That is. So −2A + 4B = 1 and 2A + B = 0 and hence A = −1 and B = 1 . We will plug in and then hopefully get equations that we can solve for A. It could be that our guess actually solves the associated homogeneous equation. As you can see this can make for a very long and tedious calculation very quickly. C'est la vie! There is one hiccup in all this. that is a good place to start. 10
And in a similar way if the right hand side contains exponentials we guess exponentials. Really if you can guess a form for y such that Ly has all the terms needed to for the right hand side. Note that y = Ae3x + 3Axe3x and y = 4Ae3x + 9Axe3x . if the equation is (where L is a linear constant coefficient operator) Ly = e3x we will guess y = Ae3x . We modify our guess to y = Axe3x and notice there is no duplication. For example for: Ly = (1 + 3x2 ) e−x cos πx we will guess y = (A + Bx + Cx2 ) e−x cos πx + (D + Ex + F x2 ) e−x sin πx. B. The trick in this case is to multiply our guess by x until we get rid of duplication with the complementary solution. F. Since the left hand side must equal to right hand side we group terms and we get that −4A + 4B + 2A = 1 and −4B − 4A + 2B = 0. We would love to guess y = Ae3x . suppose we have y − 9y = e3x . For example.72 Plug in to the equation and we get
CHAPTER 2. So y − 9y = 4Ae3x + 9Axe3x − 9Axe3x = 4Ae3x
. That is first we compute yc (solution to Ly = 0) yc = C1 e−3x + C2 e3x and we note that the e3x term is a duplicate with our desired guess. but if we plug this into the left hand side of the equation we get y − 9y = 9Ae3x − 9Ae3x = 0 e3x .
There is no way we can choose A to make the left hand side be e3x . C. So 10 5 yp = − cos 2x + 2 sin 2x . D. E. HIGHER ORDER LINEAR ODES
−4A cos 2x − 4B sin 2x − 4A sin 2x + 4B cos 2x + 2A cos 2x + 2B sin 2x = cos 2x.
5. . 4
Now what about the case when multiplying by x does not get rid of duplication. But no more! Multiplying too many times will also make the process not work. 2 sec2 x tan x. In this case find u that solves Lu = e2x and v that solves Lv = cos x (do each terms separately). We present the method of variation of parameters which will handle all the cases Ly = f (x) provided you can solve certain integrals. So guessing y = Axe3x would not get us anywhere. We have Ly = L(u + v) = Lu + Lv = e2x + cos x. Ly = y + y = tan x. such as Ly = e2x + cos x. Note that yc = C1 e3x + C2 xe3x . Let us try to solve the example. Note that each new derivative of tan x looks completely different and cannot be written as a linear combination of the previous derivatives. For simplicity we will restrict ourselves to second order equations. Consider y + y = tan x. etc. you want to multiply your guess by x until all duplication is gone.
2.2.3 Variation of parameters
It turns out that undetermined coefficients will work for many basic problems that crop up. but the method will work for higher order equations just as well (but the computations will be more tedious). Really it only works when the right hand side of the equation Ly = f (x) has only finitely many linearly independent derivatives. so that you can write a guess that consists of them all. In this case you want to guess y = Ax2 e3x . . Thus we can now write the general solution as
y = yc + y p = C1 e−3x + C2 e3x +
1 3x xe . Basically. This equation calls for a different method.
. Then we note that if y = u + v. Finally what if the right hand side is several terms. See Edwards and Penney [EP] for more detailed and complete information on undetermined coefficients. We get sec2 x. then Ly = e2x + cos x.5. It does not work all the time. This is because L is linear and this is just superposition again. For example. y − 6y + 9 = e3x . NONHOMOGENEOUS EQUATIONS
73
1 Then we note that this is supposed to be e3x and hence we find that 4A = 1 and so A = 4 . But some equations are a bit tougher.
We have the equation mx + kx = F0 cos ω t. Now try the solution x p = A cos ω t and solve for A.6 in EP Before reading the lecture.
k where ω0 = m .edu/iode/. xp = 2 m(ω0 − ω2 )
We leave it as an exercise to do the algebra required here. Let us suppose that ω0 ω. This has the complementary solution (solution to the associated homogeneous equation) xc = C1 cos ω0 t + C2 sin ω0 t. you will find that its coefficient will be zero (I cannot find a rhyme). k m damping c F(t) Let us return back to the mass on a spring example. Once we will learn about Fourier series we will see that we will essentially cover every type of periodic function by considering F(t) = F0 cos ω t (or sin instead of cosine. The general solution is x = C1 cos ω0 t + C2 sin ω0 t + F0 cos ω t. it may be good to first try Project III from the IODE website: §3. So we solve as in the method of undetermined coefficients with the guess above and we find that F0 cos ω t.uiuc.76
CHAPTER 2. such as noncentered rotating parts. the setup is again. ω0 is said to be the natural frequency (angular). m is mass. In the mass on a spring example. the calculations will be essentially the same). Note that we need not have sine in our trial solution as on the left hand side we will only get cosines anyway. m(ω2 − ω2 ) 0
.
2.math. That is. Usually what we are interested in is some periodic forcing. c is friction. we will consider the equation mx + cx + kx = F(t)
for some nonzero F(t).6
Forced oscillations and resonance
Note: 2 lectures. HIGHER ORDER LINEAR ODES
2. It is essentially the frequency at which the system "wants to oscillate" without external interference. k is the spring constant and F(t) is an external force acting on the mass.6. We will now consider the case of forced oscillations. If you include a sine it is fine. or perhaps even loud sounds or other sources of periodic force.1 Undamped forced motion and resonance
First let us consider undamped (c = 0) motion as this is simpler.
6 we see the graph with C1 = C2 = 0.0 0. Tacoma Narrows Bridge Failure. 59(2). (2. For example. 2mω
CHAPTER 2. F0 = 2.8) for some c > 0. In Figure 2. remember when as a kid you could start swinging by just moving back and forth on the swing seat in the correct "frequency"? You were trying to achieve resonance. Scanlan.5 -2.ketchum. HIGHER ORDER LINEAR ODES Our particular solution is
F0 2mω
t sin ω t and our general solution is F0 t sin ω t. 2mω
5 10 15 20
x = C1 cos ω t + C2 sin ω t + The important term is the last one (the particular solution we found). A common (but wrong) example of destructive force of resonance is the Tacoma Narrows bridge failure.5 0. 1991. and Undergraduate Physics Textbooks.0 0 5 10 15 20
2. 1 Figure 2. By forcing the system in just the right frequency we produce very wild oscillations. American Journal of Physics. there was an altogether different phenomenon at play there∗ . It turns out. ω = π. There is of course some damping. 118–124.org/billah/Billah-Scanlan. which becomes smaller and smaller in proportion to the oscillations of the last term as t gets larger. The force of each one of your moves was small but after a while it produced large swings. On the other hand resonance can be destructive.pdf
.5 -5. Resonance. The first two 2mω
0
5. We let p=
∗
c 2m
ω0 =
k .0 -5.0 -2.6: Graph of π t sin πt.
2. m = 1. That is our equation becomes mx + cx + kx = F0 cos ω t. So figuring out the resonance frequency can be very important. In fact F0 t it oscillates between 2mω and −F0 t . We can see that this term grows without bound as t → ∞. After an earthquake some buildings are collapsed and others may be relatively undamaged. We have solved the homogeneous problem before.6.5 2. Billah and R. This is due to different buildings having different resonance frequencies.2 Damped forced motion and practical resonance
Of course in real life things are not as simple as they were above.78 Hence A = 0 and B =
F0 . m
K. This kind of behavior is called resonance or sometimes pure resonance and is sometimes desired.0
2 2 terms only oscillate between ± C1 + C2 .0
5.
Even if you change the right hand side a little bit you will get a different formula with different behavior. F0 = 1. This maximum is said to be practical resonance (we call the ω that achieves this maximum the practical resonance frequency). we might as well focus on the steady periodic solution and ignore the transient solution. If we plot C as a function of ω (with all other parameters fixed) we can find its maximum. Hence the name transient. m = 1.1.depends on p (and hence c). So the smaller the damping. Because of this behavior. The exact formula is not as important as the idea.
m (2ωp)2 + (ω2 − ω2 )2 0
F0 If ω = ω0 we see that A = 0. and the initial conditions will only affect xtr .0 5. What we will look at however is the maximum value of the amplitude of the steady periodic solution.5 -2.e.
For reasons we will explain in a moment we will call xc the transient solution and denote it by xtr and we will call the x p we found above the steady periodic solution and denote it by x sp . Let C be the amplitude of x sp .
0 5 10 15 20 5. Hence for large t. So there is no point in memorizing this specific formula.0 2. the longer the "tranc = 0. there is no term that goes to infinity.0 -5. and ω = 1. We note that xc = xtr goes to zero as t → ∞ as all the terms involve an exponential with a negative exponent. This means that the effect of the initial conditions will be negligible after some period of time. A sample plot for three different values of c
.0 0. See Figure 2. Since there were no conflicts when solving with undetermined coefficient. The general solution to our problem is x = xc + x p = xtr + x sp . Notice that x sp involves no arbitrary constants.7: Solutions with different initial con.7 for a graph of different initial conditions. Notice that the speed at which xtr goes to zero Figure 2. you should remember the ideas involved." This agrees with the observation that when c = 0. HIGHER ORDER LINEAR ODES
cos(ω t − γ). B = C = 2mωp and γ = π/2. the effect of xtr is negligible and we will essentially only see x sp . the "faster" xtr becomes negligible.0 -2.0 0 5 10 15 20
Let us describe what do we mean by resonance when damping is present.5 2. You can always recompute it later or look it up if you really need it.7. bigger c is). the initial conditions affect the behavior for all time (i.80 Hence we have xp = F0
CHAPTER 2. an infinite "transient region"). The bigger p is (the ditions for parameters k = 1.5 -5.5 0. sient region. You should not memorize the above formula.
This is easily computed to be −4ω(2p2 + ω2 − ω2 )F0 0 C (ω) = .5 3.0 0. As damping c (and hence p) becomes smaller. The behavior will be more complicated if the forcing function is not an exact cosine wave.4. The top line is with c = 0.8.0 0.5
2.0 0. In other words when 0 ω= ω2 − 2p2 or 0 0
It can be shown that if ω2 − 2p2 is positive then ω2 − 2p2 is the practical resonance frequency 0 0 (that is the point where C(ω) is maximal.0 1. If ω = 0 is the maximum.
.2. As you can see the practical resonance amplitude grows as damping gets smaller.5 1.6.5
2. ω0 is the resonance frequency.0
0. 2 + (ω2 − ω2 )2 3/2 m (2ωp) 0 This is zero either when ω = 0 or when 2p2 + ω2 − ω2 = 0.5
1.8. In this case the amplitude gets larger as the forcing frequency gets smaller. ω0 is a good estimate of the resonance frequency. the frequency is smaller than ω0 .5
0.5 3.0 1.
To find the maximum it turns out we need to find the derivative C (ω).8: Graph of C(ω) showing practical resonance with parameters k = 1.
0. then essentially there is no practical resonance since we assume that ω > 0 in our system. This behavior agrees with the observation that when c = 0.0
2.0
0. m = 1. but for example a square wave. the middle line with c = 0. and any practical resonance can disappear when damping is large.0 2.5 1.5 2. and the bottom line with c = 1. note that in this case C (ω) > 0 for small ω).6.0
Figure 2. FORCED OSCILLATIONS AND RESONANCE
81
is given in Figure 2.5
1.0
1.0 2.0
2. the closer the practical resonance frequency comes to ω0 . F0 = 1. If practical resonance occurs.5
0.0
1. So when damping is very small. It will be good to come back to this section once you have learned about the Fourier series.5 2.
6.5: Suppose a water tower in an earthquake acts as a mass-spring system.1: Derive a formula for x sp if the equation is mx + cx + kx = F0 sin ω t. k.6. and F0 will there be no practical resonance (for what values of c is there no maximum of C(ω) for ω > 0).6.5 meter. Will the tower collapse?
. Assume that the container on top is full and the water does not move around. where the induced vibrations are horizontal.4: Take mx + cx + kx = F0 cos ω t.2: Derive a formula for x sp if the equation is mx +cx +kx = F0 cos ω t+ F1 cos 3ω t. Exercise 2.3 Exercises
Exercise 2. It takes a force of 1000 newtons to displace the container 1 meter. Exercise 2. Assume c > 0. Now think of the function C(ω). HIGHER ORDER LINEAR ODES
2. The container then acts as a mass and the support acts as the spring.5 cycles per second comes. What is the amplitude of the oscillations. Now think of the function C(ω). b) If ω is not the natural frequency. For what values of m (solve in terms of c.000 kg. a) What is the natural frequency of the water tower. For what values of c (solve in terms of m. c) Suppose A = 1 and an earthquake with frequency 0. Assume c > 0.3: Take mx + cx + kx = F0 cos ω t. find a formula for the amplitude of the resulting oscillations of the water container.6.6. Suppose that if the water tower moves more than 1. For simplicity assume no friction. k.82
CHAPTER 2. Suppose that an earthquake induces an external force F(t) = mAω2 cos ω t. Fix c > 0 and k > 0. Exercise 2. the tower collapses.6. Fix m > 0 and k > 0. Exercise 2. and F0 will there be no practical resonance (for what values of m is there no maximum of C(ω) for ω > 0). Suppose that the container with water has a mass of m =10.
y2 .Chapter 3 Systems of ODEs
3. for some functions f1 and f2 . . For example. we may end up with systems of several equations and several dependent variables even if we start with a single equation. Sometimes a system is easy to solve by solving for one variable and then for the second variable. §4. Example 3. y2 . y2 . with initial conditions of the form y1 (0) = 1. y2 . y1 .1 in EP Often we do not have just one dependent variable and one equation. y2 . y1 . x). y2 = f2 (y1 . which is a linear first order equation that is easily solved for y2 . By the method of integrating factor we get e x y2 = C1 2x e + C2 . suppose y1 . it is a second order system. . y1 = f (y1 . y2 = y1 − y2 . x). 2 83
. yn we can have a differential equation involving all of them and their derivatives. Usually.1. We note that y1 = C1 e x is the general solution of the first equation. If we have several dependent variables.1 Introduction to systems of ODEs
Note: 1 lecture.1: Take the first order system y1 = y1 . x). y2 (0) = 2. . . And as we will see. y2 . y1 . We can then plug this y1 into the second equation and get the equation y2 = C1 e x − y2 . y2 . More precisely. We call the above a system of differential equations. when we have two dependent variables we would have two equations such as y1 = f1 (y1 .
That is. and we will suppose that they ride along with no friction. . Suppose we have one spring with constant k but two m2 m2 masses m1 and m2 . We can think of the masses as carts. Let x1 be the displacement of the first cart and x2 be the displacement of the second cart. y1 = C1 e x . and we mark the position of the first and second cart and call those the zero position. . Define new variables u1 . . Take an nth order differential equation y(n) = F(y(n−1) . and we will have to solve for all variables at once. As an example application. x1 = 0 is a different position on the floor than the position corresponding to x2 = 0. . let us think of mass and spring systems again. thus the same thing with a negative sign. The general solution to the system is. we will not be so lucky to be able to solve like in the first example. . SYSTEMS OF ODES + C2 e−x . y. k m1 x1 = k(x2 − x1 ). Using Newton's second law. y . un and write the system u1 = u2 u2 = u3 . m2 x2 = −k(x2 − x1 ). . The force exerted by the spring on the first cart is k(x2 − x1 ). let us note that in some sense we need only consider first order systems. since x2 − x1 is how far the string is stretched (or compressed) from the rest position. .84 or y2 =
C1 x e 2
CHAPTER 3. . x). we note that force equals mass times acceleration. y2 = C1 x e + C2 e−x . un−1 = un un = F(un . Before we talk about how to handle systems. . we put the two carts somewhere with no tension on the spring. therefore. That is. . That we must solve for both x1 and x2 at once is intuitively obvious. . We substitute x = 0 and find that 3 C1 = 1 and C2 = 2 . un−1 . . . u2 . . x). . Generally. In this system we cannot solve for the x1 variable separately. u1 . The force exerted on the second cart is the opposite. 2
We can now solve for C1 and C2 given the initial conditions. since where the first cart goes depends exactly on where the second cart goes and vice versa.
.
In fact.1. . For autonomous systems we can easily draw the so-called direction field or vector field. this is what IODE was doing when you had it solve a second order equation numerically in the IODE Project III if you have done that project. So we now have an equation y + y − 2y = 0. y) the direction in which we should travel to satisfy the equations should be the direction of the vector (2y − x.3. We essentially just treat the dependent variable not as a number but as a vector. We know how to solve this equation and we find that y = C1 e−2t + C2 et . For example.2: Sometimes we can use this idea in reverse as well. you can discard u2 through un and let y = u1 . Hence. Once we have y we can plug in to get x. Example 3. can be transformed into a first order system of n × k equations and n × k unknowns. That is.1: Plug in and check that this really is the solution. y = x says that at the point (x.
where the independent variable is t. we give a direction (and a magnitude). 3
Exercise 3. 3 y= −e−2t + et . but instead of giving a slope at each point. So we draw the vector (2y − x. It is also autonomous as the equations do not depend on the independent variable t. C1 = −C2 and 1 = 3C2 . So C1 = −1 and C2 = 1 . a system of k differential equations in k unknowns. In many mathematical computer languages there is almost no distinction in syntax. For example. y = x. y = x = 2y − x = 2y − y . x) based at the
. . x) with the speed equal to the magnitude of this vector.1. We note that this y solves the original equation. a plot similar to a slope field. u2 .1. It is useful to go back and forth between systems and higher order equations for other reasons. as none of the dependent variables appear in any functions or with any higher powers than one. y(0) = 0. un . Let us take the system x = 2y − x. the ODE approximation methods are generally only given as solutions for first order systems. A similar process can be done for a system of higher order differential equations. It is not very hard to adapt the code for the Euler method for a first order equation to first order systems. We wish to solve for the initial conditions x(0) = 1. . . Our solution is: 3 3 x= 2e−2t + et . all of order n. We solve for the initial conditions 1 = x(0) = −2C1 + C2 and 0 = y(0) = C1 + C2 . x = y = −2C1 e−2t + C2 et . The above example was what we will call a linear first order system. We first notice that if we differentiate the first equation once we get y = x and now we know what x is in terms of x and y. The previous example x = 2y − x. INTRODUCTION TO SYSTEMS OF ODES
85
Now try to solve this system for u1 . Once you have solved for the u's.
457.1.
Figure 3.1.1. The resulting picture is usually called the phase portrait (or phase plane portrait).
Notice the similarity to the diagrams we drew for autonomous systems in one dimension. We can now draw a path of the solution in the plane. Since we solved this system precisely we can compute x(2) and y(2).1. We get that x(2) ≈ 2. y + (x + y )2 − x = 0 as a first order system of ODEs. 0) for 0 ≤ t ≤ 2. x2 = x2 . g(t)) for t in the selected range.86
CHAPTER 3.1 Exercises
Exercise 3. See Figure 3. But now note how much more complicated things become if we allow just one more dimension. then we can pick an interval of t (say 0 ≤ t ≤ 2 for our example) and plot all the points ( f (t). In this case however we cannot draw the direction field. Exercise 3.1: The direction field for x = 2y − x. For each t we would get a different direction field.4: Write ay + by + cy = f (x) as a first order system of ODEs. 0) and travels along the vector field for a distance of 2 units of t. y) and we do this for many points on the xy-plane. The particular curve obtained we call the trajectory or solution curve.2: The direction field for x = 2y − x.1. In this figure the line starts at (1.2.1. That is. suppose the solution is given by x = f (t).
. y = x with the trajectory of the solution starting at (1.3: Find the general solution of x1 = 3x1 − x2 + et .5: Write x + y2 y − x3 = sin(t). Exercise 3. This point corresponds to the top right end of the plotted solution curve in the figure. We may want to scale down the size of our vectors to fit many of them on the same direction field. y = x. SYSTEMS OF ODES
point (x. y = g(t). Also note that we can draw phase portraits and trajectories in the xy-plane even if the system is not autonomous.2: Find the general solution of x1 = x2 − x1 + t.
-1 3 0 1 2 3 3 3 -1 0 1 2 3 3
2
2
2
2
1
1
1
1
0
0
0
0
-1 -1 0 1 2 3
-1
-1 -1 0 1 2 3
-1
Figure 3. Exercise 3. x2 = x1 . An example plot is given in Figure 3.475 and y(2) ≈ 2.
3. since the field changes as t changes.
A matrix is an m × n array of numbers (m rows and n columns).2.2
Matrices and linear systems
Note: 1 and a half lectures. so our operations will have to be compatible with this viewpoint.1 in EP
3. (A + B) + C = A + (B + C). we denote a 3 × 5 matrix as follows a11 a12 a13 a14 a15 A = a21 a22 a23 a24 a25 . 4 5 6 0 2 4 4 7 10 If the sizes do not match. First. For example. d some scalars. We add matrices element by element.2. and by A. we will need to talk about matrices. By 0 we will mean the vector of all zeros. B.1 Matrices and vectors
Before we can start talking about linear systems of ODEs. We will usually denote matrices by upper case letters and vectors by lower case letters with an arrow such as x or b. 1 2 3 1 1 −1 2 3 2 + = . For example. a31 a32 a33 a34 a35 By a vector we will usually mean a column vector which is an n × 1 matrix. first part of §5. A + 0 = A = 0 + A. Note that we will want 1 × 1 matrices to really act like numbers.
. so let us review these briefly. we have the following familiar rules. MATRICES AND LINEAR SYSTEMS
87
3. 4 5 6 8 10 12 Matrix addition is also easy. C some matrices. It is easy to define some operations on matrices. If we mean a row vector we will explicitly say so (a row vector is a 1 × n matrix). (c + d)A = cA + dA. then addition is not defined.3. A + B = B + A. by c. 1 2 3 2 4 6 2 = . For example. If we denote by 0 the matrix of with all zero entries. we can multiply by a scalar (a number). c(A + B) = cA + cB. This means just multiplying each entry by the same number.
a 2 × 2 matrix A is a mapping of the plane where x gets sent to Ax. −1 1
.3. For the last two items to hold we would need to essentially "divide" by a matrix. if we take the unit square (square of sides 1) in the plane. Then the determinant of A is then the factor by which the volume of objects gets changed. IA = A = AI.2. then AB = AC does imply that B = C (the inverse is unique). then A takes the square to a parallelogram of area |det(A)|. For example. A= 1 1 . That is. A(B + C) = AB + AC. If A is not invertible we say A is singular or just say it is not invertible. c is some scalar (number). Then we call B the inverse of A and we denote B by A−1 .
(ii) AB = AC does not necessarily imply B = C even if A is not 0. If A is invertible. matrices do not commute. It is also not hard to see that (A−1 )−1 = A. C are matrices of the correct sizes so that the following make sense. (B + C)A = BA + CA. We just multiply both sides by A−1 to get A−1 AB = A−1 AC or IB = IC or B = C. (iii) AB = 0 does not necessarily mean that A = 0 or B = 0.
Before trying to compute determinant for larger matrices. A few warnings are in order however. c(AB) = (cA)B = A(cB). MATRICES AND LINEAR SYSTEMS
89
We have the following rules for matrix multiplication.
3.2. We define the determinant of a 1 × 1 matrix as the value of its own entry. For example. If the inverse of A exists. Suppose that A. let us first note the meaning of the determinant. then we call A invertible. Suppose that A is an n × n matrix and that there exists another n × n matrix B such that AB = I = BA. This is where matrix inverse comes in. For example. A(BC) = (AB)C. B. The sign of det(A) denotes changing of orientation (if the axes got flipped). For a 2 × 2 matrix we define det a b c d = ad − bc. Consider an n × n matrix as a mapping of Rn to Rn . (i) AB BA in general (it may be true by fluke sometimes).3 The determinant
We can now talk about determinants of square matrices.
for the first row we get +a det(A ) if n is odd. Obviously (0. 1n 1n We alternately add and subtract the determinants of the submatrices Ai j for a fixed i and all j. (1. 1) gets sent. c).
For example.
n
det(A) =
j=1
(−1)i+ j ai j det(Ai j ). It is also possible to compute the determinant by expanding along columns (picking a column instead of a row above).
carries the unit square to the given
Now we can define the determinant for larger matrices. c + d). If you think back to high school geometry. 1 2 3 5 6 4 6 4 5 − 2 · det + 3 · det det 4 5 6 = 1 · det 8 9 7 9 7 8 7 8 9 = 1(5 · 9 − 6 · 8) − 2(4 · 9 − 6 · 7) + 3(4 · 8 − 5 · 7) = 0. 1) and (1. 0). (b. 1n 1n det(A) = a11 det(A11 ) − a12 det(A12 ) + a13 det(A13 ) − · · · −a det(A ) if n even. 0). −1 1 1 1 1 1 1 2 = . SYSTEMS OF ODES
Then det(A) = 1 + 1 = 2. The numbers (−1)i+ j det(Ai j ) are called cofactors of the matrix and this way of computing the determinant is called the cofactor expansion. d) and (a + b.
a b c d
The vertical lines here mean absolute value. We define Ai j as the matrix A with the ith row and the jth column deleted. 0). we would get det(A) = a11 det(A11 )−a12 det(A12 )+ a13 det(A13 ). And it is precisely det a b c d . Note that a common notation for the determinant is a pair of vertical lines. say the ith row and compute. To compute the determinant of a matrix. −1 1 1 0
√ So it turns out that the image of the square is another square. c d c d
.90
CHAPTER 3. Now let us see where the square with vertices (0. For example. 0). (a. Now 1 1 1 1 . pick one row. For example. This one has a side of length 2 and is therefore of area 2. picking the first row. 0) gets sent to (0. you may have seen a formula for computing the area of a parallelogram with vertices (0. (0. for a 3×3 matrix. = −1 −1 1 0 1 1 0 1 = . a b a b = det . The matrix parallelogram.
� 1 4 1 x3 10 To solve the system we put the coefficient matrix (the matrix on the left hand side of the equation) together with the vector on the right and side and get the so-called augmented matrix 2 2 2 2 1 1 3 5 . we have a formula for the inverse of a 2 × 2 matrix a b c d
−1
0.4 Solving linear systems
One application of matrices we will need is to solve systems of linear equations. An n × n matrix A is invertible if and only if det(A) In fact. and we could add a multiple of one equation to another equation. x1 + x2 + 3x3 = 5. MATRICES AND LINEAR SYSTEMS
91
I personally find this notation confusing since vertical lines for me usually mean a positive quantity. The formula only works if the determinant is nonzero.3. Theorem 3.2. One of the most important properties of determinants (in the context of this course) is the following theorem. (i) Swap two rows. So I will not ever use this notation in these notes. Suppose that we have the following system of linear equations 2x1 + 2x2 + 2x3 = 2. ad − bc −c a
Notice the determinant of the matrix in the denominator of the fraction. 1 4 1 10 We then apply the following three elementary operations.2. we could multiply any of the equations by a nonzero number. Note that the system can be written as 2 2 2 x1 2 1 1 3 x2 = 5 . This may be best shown by example. Without changing the solution. otherwise we are dividing by zero.2.
=
1 d −b . It is easier to write this as a matrix equation. while determinants can be negative.1.
3. we note that we could do swap equations in this system. x1 + 4x2 + x3 = 10. It turns out these operations always suffice to find a solution.
.
Example 3. and x3 . One last note to make about linear systems of equations is that it is possible that the solution is not unique (or that no solution exists). (i) There is only one leading entry in each column. The variables corresponding to columns with no leading entries are said to be free variables. The last row corresponds to the equation 0x1 + 0x2 + 0x3 = 3 which is preposterous. (iii) All leading entries are 1. 0 0 0 3 there is no need to go further.3.2. The solution can be also computed with the
x = A−1 Ax = A−1 b. (ii) All the entries above and below a leading entry are zero. For example if for a system of 3 equations and 3 unknowns you find a row such as [ 0 0 0 1 ] in the augmented matrix. Such a matrix is said to be in reduced row echelon form.
2 2 2 1 1 3 1 4 1
93
2 5 10
and b is the vector
. 1 2 0 3 0 0 1 1 0 0 0 0 If the variables are named x1 . MATRICES AND LINEAR SYSTEMS where A is the matrix inverse. Free variables mean that we can pick those variables to be anything we want and then solve for the rest of the unknowns. no solution exists.
.2.1: The following augmented matrix is in reduced row echelon form. The first nonzero entry in each row is called the leading entry. You generally try to use row operations until the following conditions are satisfied. you know the system is inconsistent. On the other hand if during the row reduction process you come up with the matrix 1 2 13 3 0 0 1 1 . Hence. If during the row reduction you come up with a row where all the entries except the last one are zero (the last entry in a row corresponds to the right hand side of the equation) the system is inconsistent and has no solution. x2 . It is easy to tell if a solution does not exist. then x2 is the free variable and x1 = 3 − 2x2 and x3 = 1.
. let us touch on how to do it. .2.2: Solve
1 2 3 4
x=
5 6
by using matrix inverse.
1 2 3
x=
2 0 0
. then the inverse is the matrix with the columns xk for k = 1. then the matrix will be of the form [ I A−1 ] if and only if A is invertible.
. Finding the inverse of A is actually just solving a bunch of linear equations. to find the inverse we can write a larger n × 2n augmented matrix [ A I ].8: Solve Exercise 3.5: Compute inverse of Exercise 3. .
2 0 4 1
x=
.2. if you want to solve the equation for many different right hand sides b.2.
x=
0 3 2 3
.
9 −2 −6 −8 3 6 10 −2 −6 1 4 6 8 2 0 0 0 3 5 7 10 1 0 0 1
Exercise 3. SYSTEMS OF ODES
3.6: For which h is Infinitely many. Exercise 3. The 2 × 2 inverse is basically given by a formula. n (exercise: why?). then A is invertible.94
CHAPTER 3. So it is useful to compute the inverse.2.
not invertible? Is there only one such h? Are there several?
h 1 1 0 h 0 1 1 h
not invertible? Find all such h. Hint: expand along the proper row or column
.3: Compute determinant of Exercise 3.2. In fact by multiplying both sides by A−1 you can see that x = A−1 b.4: Compute determinant of to make the calculations simpler.2.7: For which h is Exercise 3.2.10: Solve
9 −2 −6 −8 3 6 10 −2 −6 5 3 7 8 4 4 6 3 3 3 3 0 2 2 3 2 3 3 3 4 4 1 2 3 4 5 6 7 8 h 1 2 3 1 1 1 0 1 0
.
3. If you can solve Axk = ek where ek is the vector with all zeros except a 1 at the kth position.2. If you do row reduction and put the matrix in reduced row echelon form.2. Exercise 3. . so you can just read off the inverse A−1 . While we will not have too much occasion to compute inverses for larger matrices than 2 × 2 by hand. where I is the identity.5 Computing the inverse
If the coefficient matrix is square and there exists a unique solution x to Ax = b for any b.2. Therefore.6 Exercises
Exercise 3.9: Solve Exercise 3. but it is not hard to also compute inverses of larger matrices. .2.
Now suppose that X(t) is
. X(t) is called the fundamental matrix. If furthermore this is a system of n equations (P is n × n). That is. −1 e
We will mostly concentrate on equations that are not just linear. . . xn are n solutions of the equation. The linear combination c1 x1 + c2 x2 + · · · + cn xn could always be written as X(t) c. Now you are given an initial condition of the form x(t0 ) = b for some constant vector b. cn . . then x = c1 x1 + c2 x2 + · · · + cn xn . then we say the system is homogeneous. So the procedure will be exactly the same.2. . xn . . Theorem 3. SYSTEMS OF ODES
et t2 x+ t . xn are linearly independent if and only if c1 x1 + c2 x2 + · · · + cn xn = 0 has only the solution c1 = c2 = · · · = cn = 0. just like for single homogeneous equations.1 (Superposition). . . or fundamental matrix solution. . For homogeneous linear systems we still have the principle of superposition. Suppose that x1 . suppose you have found the general solution x = Px + f . . . We apply the same technique as we did before. then we find the general solution to the associated homogeneous equation and we add the two. xn are linearly independent. Suppose x p is one particular solution. . Then every solution can be written as (3. the matrix P will be a constant and not depend on t.1). Let x = Px be a linear homogeneous system of ODEs. . We find a particular solution to the nonhomogeneous equation. When f = 0 (the zero vector). . . . .3. and c is the column vector with entries c1 . and x1 . but are in fact constant coefficient equations. To solve nonhomogeneous first order linear systems. Then every solution can be written as x = xc + x p . Let x = Px+ f be a linear system of ODEs. Alright.3.96 can be written as x = 2t
1 t
CHAPTER 3. where xc is a solution to the associated homogeneous equation (x = Px). (3. Linear independence for vector valued functions is essentially the same as for normal functions. . . x1 . . . where X(t) is the matrix with columns x1 .1)
is also a solution. Theorem 3.
x2 =? (i. c) Write down the general solution in the form x1 =?.3. SYSTEMS OF ODES
3.1 Exercises
Exercise 3. b) 3 1 1 Write down the general solution. Hint: You must
and
t3 t4
are linearly independent.4: Verify that 1 et and −1 et and 0 1 be a bit more tricky than in the previous exercise.1: Write the system x1 = 2x1 − 3tx2 + sin t.
Exercise 3.e.3.
1 Exercise 3. x2 = et x1 + 3x2 + cos t as in the form x = P(t)x + f (t).3. Exercise 3.
.5: Verify that
t t2
e2t are linearly independent. write down a formula for each element of the solution).
1 −1 1
Exercise 3.3.3: Verify that
1 1 1
et and
1 −1 1
et are linearly independent.98
CHAPTER 3.3.3. Hint: Just plug in t = 0.2: a) Verify that the system x = 1 3 x has the two solutions 1 e4t and −1 e−2t .
We then call λ an eigenvalue of A and v is called the corresponding eigenvector. Therefore.
3. EIGENVALUE METHOD
99
3. Were it invertible. §5. Now suppose we try to adapt the method for single constant coefficient equations by trying the function eλt . Suppose you have a linear constant coefficient homogeneous system x = Px. x is a vector. where v is an arbitrary constant vector. Suppose there is a scalar λ and a nonzero vector v such that Av = λv. once λ is known.4. A has the eigenvalue λ if and only if λ solves the equation det(A − λI) = 0. 0 1 0 0 0
If we rewrite the equation for an eigenvalue as (A − λI)v = 0. Example 3. we could write (A − λI)−1 (A − λI)v = (A − λI)−1 0 which implies v = 0.
. The eigenvector will have to be found later.4
Eigenvalue method
Note: 2 lectures. Note that this means that we will be able to find an eigenvalue without finding the corresponding eigenvector. So we try veλt .1: The matrix 1 because 0
2 1 0 1
has an eigenvalue of λ = 2 with the corresponding eigenvector 2 1 1 2 1 = =2 .2 in EP In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. However. To solve this equation we need a little bit more linear algebra which we review now. We notice that this has a nonzero solution v only if A − λI is not invertible.4.4.3. We plug into the equation to get λveλt = Pveλt .1 Eigenvalues and eigenvectors of a matrix
Let A be a square constant matrix. We divide by eλt and notice that we are looking for a λ and v that satisfy the equation λv = Pv.
So if v is an eigenvector corresponding to eigenvalue a + ib. Now suppose that a + ib is a complex eigenvalue of P. We note that Px = P x = Px.4. So we only need to consider one of them.102
CHAPTER 3. −1 i It is obvious that the equations iv1 + v2 = 0 and −v1 + iv2 = 0 are multiples of each other. We claim that we did not have to look for the second eigenvector (nor for the second eigenvalue). but we will do something a bit smarter first. Similarly we can bar whole vectors or matrices. 1 We could write the solution as x = c1 i (1−i)t −i (1+i)t c ie(1−i)t − c2 ie(1+i)t e + c2 e = 1 (1−i)t .3 Complex eigenvalues
A matrix might very well have complex eigenvalues even if all the entries are real. suppose that we have the system 1 1 x = x. z First a small side note.
From this we note that λ = 1 ± i. After picking v2 = 1. The corresponding eigenvectors will also be complex (P − (1 − i)λ)v = 0 i 1 v = 0. All complex eigenvalues come in pairs (because the matrix P is real). This operation is called the complex conjugate. v the corresponding eigenvector and hence x1 = ve(a+ib)t
. for example. where the 2 bar above z means a + ib = a − ib. We could use Euler's formula here and do the whole song and dance we did before. −1 1 Let us compute the eigenvalues of the matrix P = det(P − λI) = det 1−λ 1 −1 1 − λ
1 1 −1 1
.
= (1 − λ)2 + 1 = λ2 − 2λ + 2 = 0. For example. we have the eigenvector i v = 1 . 1 1 c1 e + c2 e(1+i)t 1
But then we would need to look for complex values c1 and c2 to solve any initial conditions. And even then it is perhaps not completely clear that we get a real solution. then v is an eigenvector corresponding to eigenvalue a − ib. If a matrix P is real then ¯ P = P. Or ¯ (P − λI)v = (P − λI)v. The real part of a complex number z can be computed as z+¯ . Note that for a real number a. a = a. SYSTEMS OF ODES
3. In similar fashion we find that −i is an eigenvector corresponding to the eigenvalue 1 + i.
And it is real valued! Similarly as Im z = x4 = Im x1 = x1 − x2 . are real valued and are linearly independent. you will end up with n linearly independent solutions if you had n distinct eigenvalues (real or complex). You go on to the next eigenvalue which is either a real eigenvalue or another complex eigenvalue pair. You can now find a real valued general solution to any homogeneous system where the matrix has distinct eigenvalues. Exercise 3. Then note that ea+ib = ea−ib and hence x2 = x1 = ve(a−ib)t is also a solution. You take one λ = a + ib from the pair. EIGENVALUE METHOD is a solution (complex valued) of x = Px.3. The process is this.4.
. you notice that they always come in pairs. 2 2
z−¯ z 2i
103
Is also a solution. When you have repeated eigenvalues.4: Check that these really are solutions. you find the corresponding eigenvector v. 2i
is the imaginary part we find that
is also a real valued solution. matters get a bit more complicated and we will look at that situation in §3. When you have complex eigenvalues. et cos t et cos t . et sin t i (1−i)t i t iet cos t − et sin t e = e cos t + iet sin t = t 1 1 e cos t + iet sin t
This solution is real valued for real c1 and c2 .4. t e cos t e sin t c1 et cos t + c2 et sin t −et sin t . Returning to our problem. Hence. The general solution is x = c1 −et sin t et cos t −c1 et sin t + c2 et cos t + c2 t = . Now we can solve for any initial conditions that we have. It turns out that x3 and x4 are linearly independent.7. You note that Re ve(a+ib)t and Im ve(a+ib)t are also solutions to the equation. Now take the function x3 = Re x1 = Re ve(a+ib)t = x1 + x1 x1 + x2 = . we take x1 = It is easy to see that Re x1 = Im x1 = are the solutions we seek.
Now suppose that x and y are on the line determined by an eigenvector v for an eigenvalue λ. So we have a 2 × 2 matrix P and the system x x . Case 1. For example. x That is.3: Eigenvectors of P. See Figure 3. along the line determined by v. We 0
. As λ > 0. We get the picture in Figure 3. 0 1 The eigenvalues are 1 and −2 and the corresponding eigenvectors are the same.5
Two dimensional systems and their vector fields
Note: 1 lecture. The eigenvalues are 1 and 2 and the corre0 2 sponding eigenvectors are 1 and 1 . We call this kind of picture a sink or sometimes a stable node. Suppose one eigenvalue is positive and one is negative. See Figure 3. Suppose that the eigenvalues are real and positive. Then x x = P(av) = a(Pv) = aλv. The only difference is that the 0 1 eigenvalues are negative and hence all arrows are reversed. TWO DIMENSIONAL SYSTEMS AND THEIR VECTOR FIELDS
105
3. take the matrix 1 1 .6 on the following page.2) =P y y We will be able to visually tell how the vector field looks once we find the eigenvalues and eigenvectors of the matrix. Case 2. 0 −2 1 and 1 . Let us draw arrows on the lines to indicate the directions. We want to think about how the vector fields look and how this depends on the eigenvalues. You will notice that the picture looks like a source with arrows coming out from the origin.
1 Case 3. Suppose both eigenvalues were negative. y = av for some scalar a.3. (3. For example the matrix 1 −2 . the derivative points in the direction of v when a is positive and in the opposite direction when a is negative. The eigenvalues are −1 and −2 and the corresponding eigenvectors are the same. Hence we call this type of picture a source or sometimes an unstable node. The calculation and the picture are almost the same. 1 and −3 . See Fig0 1 ure 3. take the negation of the matrix in case 1. −1 −1 . For example.3. but is in EP §6.4 on the following page. We fill in the rest of the arrows and we also draw a few solutions. should really be in EP §5.5.5 on the next page. Find the two eigenvectors and plot them in the plane.2 Let us take a moment to talk about homogeneous systems in the plane. =P y yThe derivative is a multiple of v and hence points Figure 3.2.
Note that which combination of them we take just depends on the initial conditions. So we might as well just take the real part. If you notice this is a parametric equation for an ellipse. Same with the imaginary part and in fact any linear combination of them. It is not difficult to see that this is what happens in general when the eigenvalues are purely imaginary. So when the eigenvalues are purely imaginary, you get ellipses for your solutions. This type of picture is sometimes called a center. See Figure 3.8.
-3 3 -2 -1 0 1 2 3 3 3 -3 -2 -1 0 1 2 3 3
3.5.1 Exercises
Exercise 3.5.1: Take the equation mx + cx + kx = 0, with m > 0, c ≥ 0, k > 0 for the mass-spring system. a) Convert this to a system of first order equations. b) Classify for what m, c, k do you get which behavior. c) Can you explain from physical intuition why you do not get all the different kinds of behavior here? Exercise 3.5.2: Can you find what happens in the case when P = 1 1 . In this case the eigenvalue 0 1 is repeated and there is only one eigenvector. What picture does this look like? Exercise 3.5.3: Can you find what happens in the case when P = the pictures we have drawn?
1 1 1 1
. Does this look like any of
110
CHAPTER 3. SYSTEMS OF ODES
3.6
Second order systems and applications
Note: more than 2 lectures, §5.3 in EP
3.6.1 Undamped mass spring systems
While we did say that we will usually only look at first order systems, it is sometimes more convenient to study the system in the way it arises naturally. For example, suppose we have 3 masses connected by springs between two walls. We could pick any higher number, and the math would be essentially the same, but for simplicity we pick 3 right now. And let us assume no friction, that is, the system is undamped. The masses are m1 , m2 , and m3 and the spring constants are k1 , k2 , k3 , and k4 . Let x1 be the displacement from rest position of the first mass and, x2 and x3 the displacement of the second and third mass. We will make, as usual, positive values go right (as x1 grows, mass 1 is moving right). See Figure 3.11. k1 m1 k2 m2 k3 m3 k4
Figure 3.11: System of masses and springs. This simple system turns up in unexpected places. Note for example that our world really consists of small particles of matter interacting together. When we try this system with many more masses, this is a good approximation to how an elastic material will behave. In fact by somehow taking a limit of the number of masses going to infinity we obtain the continuous one dimensional wave equation. But we digress. Let us set up the equations for the three mass system. By Hooke's law we have that the force acting on the mass equals the spring compression times the spring constant. By Newton's second law we again have that force is mass times acceleration. So if we sum the forces acting on each mass and put the right sign in front of each depending on the direction in which it is acting, we end up with the system. m1 x1 = −k1 x1 + k2 (x2 − x1 ) m2 x2 = −k2 (x2 − x1 ) + k3 (x3 − x2 ) m3 x3 = −k3 (x3 − x2 ) − k4 x3 We define the matrices m1 0 0 M = 0 m2 0 0 0 m3 = −(k1 + k2 )x1 + k2 x2 , = k2 x1 − (k2 + k3 )x2 + k3 x3 , = k3 x2 − (k3 + k4 )x3 .
3.6. SECOND ORDER SYSTEMS AND APPLICATIONS We write the equation simply as Mx = K x.
111
At this point we could introduce 3 new variables and write out a system of 6 equations. We claim this simple setup is easier to handle as a second order system. We will call x the displacement vector, M the mass matrix, and K the stiffness matrix. Exercise 3.6.1: Do this setup for 4 masses (find the matrix M and K). Do it for 5 masses. Can you find a prescription to do it for n masses? As before we will want to "divide by M." In this case this means computing the inverse of M. All the masses are nonzero and it is easy to compute the inverse, as the matrix is diagonal. 1 m1 0 0 0 1 0 . −1 M = m2 1 0 0 m3 This fact follows readily by how we multiply diagonal matrices. You should verify that MM −1 = M −1 M = I as an exercise. We let A = M −1 K and we look at the system x = M −1 K x, or x = Ax. Many real world systems can be modeled by this equation. For simplicity we will keep the given masses-and-springs setup in mind. We try a solution of the form x = veαt . We note that for this guess, x = α2 veαt . We plug into the equation and get α2 veαt = Aveαt . We can divide by eαt to get that α2 v = Av. Hence if α2 is an eigenvalue of A and v is the corresponding eigenvector, we have found a solution. In our example, and in many others, it turns out that A has negative real eigenvalues (and possibly a zero eigenvalue). So we will study only this case here. When an eigenvalue λ is negative, it means that α2 = λ is negative. Hence there is some real number ω such that −ω2 = λ. Then α = ±iω. The solution we guessed was x = v(cos ω t + i sin ω t). By again taking real and imaginary parts (note that v is real), we again find that v cos ω t and v sin ω t are linearly independent solutions. If an eigenvalue was zero, it turns out that v and vt are solutions if v is the corresponding eigenvector.
is the general solution of x = Ax, for some arbitrary constants ai and bi . If A has a zero eigenvalue and all other eigenvalues are distinct and negative, that is ω1 = 0, then the general solution becomes
n
x(t) = v1 (a1 + b1 t) +
i=2
vi (ai cos ωi t + bi sin ωi t).
Now note that we can use this solution and the setup from the introduction of this section even when some of the masses and springs are missing. Simply when there are say 2 masses and only 2 springs, take only the equations for the two masses and set all the spring constants that are missing to zero.
For this t. k
m1
m2 10 meters
Figure 3. the eigenvectors are 1 and −2 respectively (exercise).14. First the cars start at position 0 so x1 (0) = 0 and x2 (0) = 0. The bumper acts like a spring of spring constant k = 2 N/m. At what time after the cars link does impact with the wall happen? What is the speed of car 2 when it hits the wall? OK. Car 1 of mass 2 kg is travelling at 3 m/s towards the second rail car of mass 1 kg. Furthermore. Hence the equation is 2 0 −2 2 x = x. In this example we have two toy rail cars. We note that 1 √ ω2 = 3 and we use the second part of the theorem to find our general solution to be x= √ √ 1 1 (a1 + b1 t) + a2 cos 3 t + b2 sin 3 t = 1 −2
√ √ 3 a1 + b1 t + a2 cos √3 t + b2 sin √ t = a1 + b1 t − 2a2 cos 3 t − 2b2 sin 3 t
We now apply the initial conditions. It is not hard to see that the eigenvalues are 0 and −3 1 (exercise). a1 − 2a2
. so x2 (0) = 0. Let us assume that time t = 0 is the time when the two cars link up. This system acts just like the system of the previous example but without k1 . The first car is travelling at 3 m/s. so x1 (0) = 3 and the second car starts at rest.2: Let us do another example. The first conditions says a + a2 0 = x(0) = 1 . and let x2 be the displacement of the second car from its original location. Let x1 be the displacement of the first car from the position at t = 0. We want to ask several question. See Figure 3. 2 −2
We compute the eigenvalues of A. 0 1 2 −2 or x = −1 1 x. let us first set the system up. There is a bumper on the second rail car which engages one the cars hit (it connects to two cars) and does not let go. Then the time when x2 (t) = 10 is exactly the time when impact with wall occurs. The second Car is 10 meters from a wall.14: The crash of two rail cars. SYSTEMS OF ODES
Example 3. x2 (t) is the speed at impact.114
CHAPTER 3.6.
15: Position of the second car in time (ignoring the wall).22 seconds.0 0 1 2 3 4 5 6
0.0
2.
. We plug a1 and a2 and differentiate to get √ √ b1 + √ b2 cos √ t 3 3 x (t) = .5
2.6.3.22 seconds from t = 0) we get that x2 (timpact ) ≈ 3. What we are really interested in is the second expression.0
Figure 3.
0 1 2 3 4 5 6 12. This means that the carts will be travelling in the positive direction as time grows.5 12.5
0.5
5. 0 b1 − 2 3 b2
1 √ .0
5. SECOND ORDER SYSTEMS AND APPLICATIONS
115
It is not hard to see that this implies that a1 = a2 = 0. 2 3
Hence the position of our
Note how the presence of the zero eigenvalue resulted in a term containing t. the one for x2 . which is what we expect.0
7. At time of impact (5. Just from the graph we can see that time of impact will be a little more than 5 seconds from √ 2 time zero. √ As for the speed we note that x2 = 2 − 2 cos 3 t. We have x2 (t) = √ 2 2t − √3 sin 3 t. We are travelling at almost the maximum speed when we hit the wall. 3
It is not hard to solve these two equations to find b1 = 2 and b2 = cars is (until the impact with the wall) √ 1 2t + √ sin 3 t 3 x= 2t − √ sin √3 t .5
10.0
10. See Figure 3.85. b1 − 2 3 b2 cos 3 t So √ b1 + √ b2 3 3 = x (0) = .5
7.15 for the plot of x2 versus time. √ The maximum speed is the maximum of 2 − 2 cos 3 t which is 4. Using a computer (or even a graphing calculator) we find that timpact ≈ 5. For this you have to solve the equation 10 = x2 (t) = 2t − √3 sin 3 t.
116
CHAPTER 3. SYSTEMS OF ODES
Now suppose that Bob is a tiny person sitting on car 2. Bob has a Martini in his hand and would like to not spill it. Let us suppose Bob would not spill his martini when the first car links up with car 2, but if car 2 hits the wall at any speed greater than zero, Bob will spill his drink. Suppose Bob can move the car 2 a few meters back and forth from the wall (he cannot go all the way to the wall, nor can he get out of the way of the first car). Is there a "safe" distance for him to be in? A distance such that the impact with the wall is at zero speed? Actually, the answer is yes. From looking at Figure 3.15 on the preceding page, we note the "plateau" between t = 3 and t =√ There is a point where the speed is zero. We just need to 4. 2π 4π solve x2 (t) = 0. This is when cos 3 t = 1 or in other words when t = √3 , √3 , etc. . . If we plug in
4π = √3 ≈ 7.26. So a "safe" distance is about 7 and a quarter meters from the wall. Alternatively Bob could move away from the wall towards the incoming car 2 where another 8π safe distance is √3 ≈ 14.51 and so on, using all the different t such that x2 (t) = 0. Of course t = 0 is always a solution here, corresponding to x2 = 0, but that means standing right at the wall.
That is, we are adding periodic forcing to the system in the direction of the vector F. Just like before this system just requires us to find one particular solution x p , add it to the general solution of the associated homogeneous system xc and we will have the general solution to (3.3). Let us suppose that ω is not one of the natural frequencies of x = Ax, then we can guess x p = c cos ω t, where c is an unknown constant vector. Note that we do not need to use sine since there are only second derivatives. We solve for c to find x p . This is really just the method of undetermined coefficients for systems. Let us differentiate x p twice to get x p = −ω2 c cos ω t. Now plug into the equation −ω2 c cos ω t = Ac cos ω t + F cos ω t We can cancel the cosine and rearrange to obtain (A + ω2 I)c = −F. So c = (A + ω2 I)−1 (−F).
3.6. SECOND ORDER SYSTEMS AND APPLICATIONS
117
Of course this means that (A + ω2 I) = (A − (−ω2 )I) is invertible. That matrix is invertible if and only if −ω2 is not an eigenvalue of A. That is true if and only if ω is not a natural frequency of the system. Example 3.6.3: Let us take the example in Figure 3.12 on page 112 with the same parameters as before: m1 = 2, m2 = 1, k1 = 4, and k2 = 2. Now suppose that there is a force 2 cos 3t acting on the second cart. The equation is 0 −3 1 cos 3t. x+ x = 2 2 −2 We have solved the associated homogeneous equation before and found the complementary solution to be 1 1 (a cos t + b1 sin t) + (a cos 2t + b2 sin 2t) . xc = 2 1 −1 2 We note that the natural frequencies were 1 and 2. Hence 3 is not a natural frequency, we can try c cos 3t. We can invert (A + 32 I) −3 1 + 32 I 2 −2 Hence, c = (A + ω2 I)−1 (−F) =
7 40 −1 20 −1 40 3 20 −1
The constants a1 , a2 , b1 , and b2 must then be solved for given any initial conditions. If ω is a natural frequency of the system resonance occurs because you will have to try a particular solution of the form x p = c t sin ω t + d cos ω t. That is assuming that all eigenvalues of the coefficient matrix are distinct. Note that the amplitude of this solution grows without bound as t grows.
Exercise 3.6.4: Let us take the example in Figure 3.12 on page 112 with the same parameters as before: m1 = 2, k1 = 4, and k2 = 2, except for m2 which is unknown. Suppose that there is a force cos 5t acting on the first mass. Find an m1 such that there exists a particular solution where the first mass does not move. Note: This idea is called dynamic damping. In practice there will be a small amount of damping and so any transient solution will disappear and after long enough time, the first mass will always come to a stop. Exercise 3.6.5: Let us take the example 3.6.2 on page 114, but that at time of impact, cart 2 is moving to the left at the speed of 3m/s. a) Find the behavior of the system after linkup. b) Will the second car hit the wall, or will it be moving away from the wall as time goes on. c) at what speed would the first car have to be travelling for the system to essentially stay in place after linkup. Exercise 3.6.6: Let us take the example in Figure 3.12 on page 112 with parameters m1 = m2 = 1, k1 = k2 = 1. Does there exist a set of initial conditions for which the first cart moves but the second cart does not? If so find those conditions, if not argue why not.
3.7. MULTIPLE EIGENVALUES
119
3.7
Multiple eigenvalues
Note: 1–2 lectures, §5.4 in EP It may very well happen that a matrix has some "repeated" eigenvalues. That is, the characteristic equation det(A − λI) = 0 may have repeated roots. As we have said before, this is actually unlikely to happen for a random matrix. If you take a small perturbation of A (you change the entries of A slightly) you will get a matrix with distinct eigenvalues. As any system you will want to solve in practice is an approximation to reality anyway, it is not indispensable to know how to solve these corner cases. But it may happen on occasion that it is easier or desirable to solve such a system directly.
3.7.1 Geometric multiplicity
Take the diagonal matrix A= 3 0 . 0 3
A has an eigenvalue 3 of multiplicity 2. We usually call the multiplicity of the eigenvalue in the characteristic equation the algebraic multiplicity. In this case, there exist 2 linearly independent eigenvectors, 1 and 0 . This means that the so-called geometric multiplicity of this eigenvalue 0 1 is 2. In all the theorems where we required a matrix to have n distinct eigenvalues, we only really needed to have n linearly independent eigenvectors. For example, let x = Ax has the general solution 1 3t 0 3t x = c1 e + c2 e . 0 1 Let us restate the theorem about real eigenvalues. In the following theorem we will repeat eigenvalues according to (algebraic) multiplicity. So for A above we would say that it has eigenvalues 3 and 3. Theorem 3.7.1. Take x = Px. If P is n × n and has n real eigenvalues (not necessarily distinct), λ1 , . . . , λn , and if there are n linearly independent corresponding eigenvectors v1 , . . . , vn , and the general solution to the ODE can be written as. x = c1 v1 eλ1 t + c2 v2 eλ2 t + · · · + cn vn eλn t . The geometric multiplicity of an eigenvalue of algebraic multiplicity n is equal to the number of linearly independent eigenvectors we can find. It is not hard to see that the geometric multiplicity is always less than or equal to the algebraic multiplicity. Above we, therefore, handled the case when these two numbers are equal. If the geometric multiplicity is equal to the algebraic multiplicity we say the eigenvalue is complete.
120
CHAPTER 3. SYSTEMS OF ODES
The hypothesis of the theorem could, therefore, be stated as saying that if all the eigenvalues of P are complete then there are n linearly independent eigenvectors and thus we have the given general solution. Note that if the geometric multiplicity of an eigenvalue is 2 or greater, then the set of linearly independent eigenvectors is not unique up to multiples as it was before. For example, for the 1 diagonal matrix A above we could also pick eigenvectors 1 and −1 , or in fact any pair of two 1 linearly independent vectors.
3.7.2 Defective eigenvalues
If an n × n matrix has less than n linearly independent eigenvectors, it is said to be deficient. Then there is at least one eigenvalue with algebraic multiplicity that is higher than the geometric multiplicity. We call this eigenvalue defective and the difference between the two multiplicities we call the defect. Example 3.7.1: The matrix 3 1 0 3 has an eigenvalue 3 of algebraic multiplicity 2. Let us try to compute the eigenvectors. 0 1 v1 = 0. 0 0 v2 We must have that v2 = 0. Hence any eigenvector is of the form v01 . Any two such vectors are linearly dependent, and hence the geometric multiplicity of the eigenvalue is 1. Therefore, the defect is 1, and we can no longer apply the eigenvalue method directly to a system of ODEs with such a coefficient matrix. The key observation we will use here is that if λ is an eigenvalue of A of algebraic multiplicity m, then we will be able to find m linearly independent vectors solving the equation (A − λI)m v = 0. We will call these the generalized eigenvectors. Let us continue with the example A = 3 1 and the equation x = Ax. We have an eigenvalue 0 3 λ = 3 of (algebraic) multiplicity 2 and defect 1. We have found one eigenvector v1 = 1 . We have 0 the solution x1 = v1 e3t . In this case, let us try (in the spirit of repeated roots of the characteristic equation for a single equation) another solution of the form x2 = (v2 + v1 t) e3t .
So our general solution to x = Ax is 1 x = c1 1 3t 0 1 c e3t + c2 te3t e + c2 + t e3t = 1 .1: Solve x = 3 1 x by first solving for x2 and then for x1 independently.
If we can. If we plug the second equation into the first we find that (A − 3I)(A − 3I)v2 = 0. Now check 0 3 that you got the same solution as we did above. we are done. or (A − 3I)2 v2 = 0. Hence we can take v2 = 0 .7. and (A − 3I)v2 = v1 . therefore. and such that (A − 3I)v2 = v1 . 0 1 0 c2 e3t
Let us check that we really do have the solution. any vector v2 solves (A − 3I)2 v2 = 0. First for λ of multiplicity 2.7. Now find a vector v2 such that Find v2 such that (A − 3I)2 v2 = 0. We notice that in this simple case (A − 3I)2 is just the zero matrix (exercise). This means that (A − 3I)v1 = 0. the equation for x2 does not depend on x1 . Let us describe the general algorithm. good.
. find a v2 which solves (A − 3I)2 v2 = 0. Now x2 = 3c2 e3t = 3x2 . defect 1. (A − 3I)v2 = v1 . Note that the system x = Ax has a simpler solution since A is a triangular matrix.
If these two equations are satisfied. Write 0 1 a 1 = . Exercise 3. First find an eigenvector v1 of λ. In particular. then x2 is a solution. First x1 = c1 3e3t + c2 e3t + 3c2 te3t = 3x1 + x2 .
121
By looking at the coefficients of e3t and te3t we see 3v2 + v1 = Av2 and 3v1 = Av1 . We know the first of these equations is satisfied because v1 is an eigenvector. Hence.3. This is just a bunch of linear equations to solve and we are by now very good at that. and Ax2 = A(v2 + v1 t) e3t = Av2 e3t + Av1 te3t . MULTIPLE EIGENVALUES We differentiate to get x2 = v1 e3t + 3(v2 + v1 t) e3t = (3v2 + v1 ) e3t + 3v1 te3t . 0 0 b 0 By inspection we see that letting a = 0 (a could be anything in fact) and b = 1 does the job. good. So we just have to make sure that (A − 3I)v2 = v1 . x2 must equal Ax2 .
Now suppose that this was one equation (P is a number or a 1 × 1 matrix).8. Then the general solution to x = Px is x = etP c. Suppose that we have the constant coefficient equation x = Px.8
Matrix exponentials
Note: 2 lectures. where c is an arbitrary constant vector.8.1 Definition
In this section we present a different way of finding the fundamental matrix solution of a system. What we are looking for is a vector. We usually write Pt as tP by convention when P is a matrix. Let P be an n × n matrix. It turns out the same computation works for matrices when we define ePt properly. k!
Recall k! = 1 · 2 · 3 · · · k. In fact x(0) = c.124
CHAPTER 3. First let us write down the Taylor series for eat for some number a. Now if we differentiate this series (at)2 (at)3 a3 t2 a4 t3 + + · · · = a 1 + at + + + · · · = aeat . §5. Then the solution to this would be x = ePt . a+a t+ 2 6 2 6
2
Maybe we can write try the same trick here. We note that in the 1 × 1 case we would at this point multiply by an arbitrary constant to get the general solution. With this small change and by the exact same calculation as above we have that d tP e = PetP . eat = 1 + at + (at)2 (at)3 (at)4 + + + ··· = 2 6 24
∞
k=0
(at)k . dt Now P and hence etP is an n × n matrix. and 0! = 1. Suppose that for an n × n matrix A we define the matrix exponential as 1 1 1 def eA = I + A + A2 + A3 + · · · + Ak + · · · 2 6 k! Let us not worry about convergence.
. SYSTEMS OF ODES
3. The series really does always converge. Theorem 3.1. In the matrix case we multiply by a column vector c. as usual.5 in EP
3.
8.126
CHAPTER 3. there is only one eigenvector for the eigenvalue 2. where B2 = 0. Then we can write A = λI + B. Matrices B such that Bk = 0 for some k are called nilpotent. then either it is diagonal. In fact. the exponential is not as easy to compute as above. if a matrix A is 2 × 2 and has an eigenvalue λ of multiplicity 2. If we can do that. This is a good exercise. it is still not too difficult provided we can find enough eigenvectors. And hence by the same reasoning (BAB−1 )k = BAk B−1 . or A = λI + B where B2 = 0. Then show that (A − λI)2 = 0. You will get an equation for the entries.
−1
. Now compute the square of B. Note that this matrix has a repeated eigenvalue with a defect. Computation of the matrix exponential for nilpotent matrices is easy by just writing down the first k terms of the Taylor series.3 General matrices
In general. But fear not. This can be seen by writing down the Taylor series. For any two square matrices A and B.1: Suppose that A is 2 × 2 and λ is the only eigenvalue.
3. So now write down the Taylor series for −1 eBAB 1 1 −1 eBAB = I + BAB−1 + (BAB−1 )2 + (BAB−1 )3 + · · · 2 6 1 2 −1 1 3 −1 = BB−1 + BAB−1 + BA B + BA B + · · · 2 6 1 2 1 3 = B I + A + A + A + · · · B−1 2 6 A −1 = Be B . where D is diagonal. Exercise 3. This procedure is called diagonalization. you can see that the computation of the exponential becomes easy. Now we will write a general matrix A as EDE −1 . we have eBAB = BeA B−1 . We cannot usually write any matrix as a sum of commuting matrices where the exponential is simple for each one. First note that (BAB−1 )2 = BAB−1 BAB−1 = BAIAB−1 = BA2 B−1 . Hint: First write down what does it mean for the eigenvalue to be of multiplicity 2. Adding t into the mix we see that etA = EetD E −1 .8. SYSTEMS OF ODES
So we have found the fundamental matrix solution for the system x = Ax. So we have found a perhaps easier way to handle this case. First we need the following interesting result about matrix exponentials.
So we must find the right fundamental matrix solution. Then the particular solution 2 we are looking for is x = y
e3t +e−t 2 e3t −e−t 2 e3t −e−t 2 e3t +e−t 2
4 2e3t + 2e−t + e3t − e−t 3e3t + e−t = = . Then we claim etA = X(t) [X(0)]−1 .4 Fundamental matrix solutions
We note that if you can compute the fundamental matrix solution in a different way. However. We first compute (exercise) that the eigenvalues are 3 and 2 1 1 −1 and the corresponding eigenvectors are 1 and −1 .8. So perhaps we did not gain much by this new tool. 2 2e3t − 2e−t + e3t + e−t 3e3t − e−t
3. The fundamental matrix solution of a system of ODEs is not unique. All we are doing is changing what the arbitrary constants are in the general solution x(t) = X(t)c. Hence we write 1 etA = 1 1 e3t 0 1 1 1 −1 0 e−t 1 −1 1 1 e3t 0 −1 −1 −1 = 1 −1 0 e−t 2 −1 1 −1 e3t e−t −1 −1 = 2 e3t −e−t −1 1 = −1 −e3t − e−t −e3t + e−t = 2 −e3t + e−t −e3t − e−t
e3t +e−t 2 e3t −e−t 2 e3t −e−t 2 e3t +e−t 2 −1
. SYSTEMS OF ODES
Then compute the particular solution for the initial conditions x(0) = 4 and y(0) = 2.128
CHAPTER 3.
. you can use this to find the matrix exponential etA . Let X be any fundamental matrix solution to x = Ax. which the eigenvalue method did not. The exponential is the fundamental matrix solution with the property that for t = 0 we get the identity matrix.8. It is not hard to see that we can multiply a fundamental matrix solution on the right by any constant invertible matrix and we still get a fundamental matrix solution. Hence. Obviously if we plug t = 0 into X(t) [X(0)]−1 we get the identity. by the property that e0A = I we find that the particular solution we are looking for is etA b where b is 4 .
3. the Taylor series expansion actually gives us a very easy way to approximate solutions. the computation of any fundamental matrix solution X using the eigenvalue method is just as difficult as computation of etA .
The initial conditions are x(0) = 4 and y(0) = 2. Let A be the coefficient matrix 1 2 .5 Approximations
If you think about it.
1. Nineteen Dubious Ways to Compute the Exponential of a Matrix. 3–49
.8. Exercise 3. 7 3 5 2 2t + 2t + 3 t 1 + t + 2 t + 13 t3 6
Just like the Taylor series approximation for the scalar version.F.1 A = 1.3: Find eAt for the matrix A =
∗
2 3 0 2
.22670818 9.33333333 6.13 3 1.8.12716667 0.22233333 1.
3. For larger t.1 A + A + A = .8. the approximation will be better for small t and worse for larger t.
C.22251069 .12734811 0. 2 1
5 1 2 t2 2 t3 3 2 2 2 3 e ≈ I + tA + A + A = I + t +t 5 +t 2 6 2 1 2 2 tA 13 6 7 3 7 3 13 6 5 2 t 2 2
=
=
1+t+ + 13 t3 2 t + 2 t2 + 7 t3 6 3 .33333333 .85882874 . y = x + 3y.12734811
This is not bad at all.6 Exercises
Exercise 3. 2003.12 2 0. There are better ways to approximate the exponential∗ . Moler and C.2: Find a fundamental matrix solution for the system x = 3x + y. MATRIX EXPONENTIALS
129
The simplest thing we can do is to just compute the series up to a certain number of terms. 6.22251069 1. The approximate solution is approximately (rounded to 8 decimal places) e
0.1 into the real solution (rounded to 8 decimal places) we get e0.66666667 6. Let us see how we stack up against the real solution with t = 0. 9. Although if you take the same approximation for t = 1 you get (using the Taylor series) 6. Van Loan. SIAM Review 45 (1).12716667 2 6
And plugging t = 0. few terms of the Taylor series give a reasonable approximation for the exponential and may suffice for the application. In many cases however.22233333 ≈ I + 0.8.85882874 10. To get a good approximation at t = 1 (say up to 2 decimal places) you would need to go up to the 11th power (exercise).1 A
0. let us compute the first 4 terms of the series for the matrix A = 1 2 . 0. you will generally have to compute more terms. 0. For example.3.22670818 So the approximation is not very good once we get up to t = 1. Twenty-Five Years Later.66666667 while the real value is (again rounded to 8 decimal places) 10.
1 1
for some arbitrary constants α1 and α2 . let us just do an example. something of the form aet appears in the complementary solution. The only difference here is that we will have to take unknown vectors rather than just numbers. As ξ1 (0) = ξ2 = e4t As ξ2 (0) =
1 16 1 4
CHAPTER 3. We would want to guess a particular solution of x = aet + bt + c. Similarly 4
e−4t (et + t) dt + C2 e4t = −
−1 16
et t 1 − − + C2 e4t . −2 1 Note that we can solve this system in an easier way (can you see how). Undetermined coefficients The method of undetermined coefficients also still works.3: Let A = −1 0 . x1 =
e4t −e−2t 3
et − e−2t 1 − 2t 1 + + 1 3 4
3−12t 16
e4t − et 4t + 1 = − 3 16
4t−5 . Example 3. The method can turn into a lot of tedious work. furthermore if the right hand side is complicated.1: Check that x1 and x2 solve the problem. 3 4 16
we have that
=
−1 3
−
1 16
1 + C2 and hence C2 = 3 . you will have lots of variables to solve for.9. SYSTEMS OF ODES then
1 4
=
1 3
1 + 1 + C1 and hence C1 = − 3 . Check both that they satisfy the differential equation and that they satisfy the initial conditions. Because we do not yet know the vector if the a is a multiple of 0 we do not know if a conflict arises. The solution is
1 x(t) = −1 That is. 16
e4t −e−2t + 3−12t 3 16 −2t +e4t +2et e + 4t−5 3 16
. This method does not always work.136 C1 is the constant of integration. The eigenvalues of A are −1 and 1 and the corresponding eigenvectors are 1 and 0 respec1 1 tively. As this method is essentially the same as it is for single equations.
+
and x2 =
e−2t +e4t +2et 3
+
Exercise 3. So in system of 3 equations if you have say 4 unknown vectors (this would not be uncommon). then you already have 12 unknowns that you need to solve for. but for the purposes of the example. It may very 1
. Hence our complementary solution is
t
xc = α1
1 −t 0 t e + α2 e. In this case you can think of each element of an unknown vector as an unknown number. However.9. Find a particular solution of x = Ax + f where f (t) = et . Same caveats apply to undetermined coefficients for systems as they do for single equations. let us use the eigenvalue method plus undetermined coefficients.
suppose that you have solved the associated homogeneous equation x = A(t) x and found the fundamental matrix solution X(t).9) to obtain x p (t) = X (t) u(t) + X(t) u (t) = A(t) X(t) u(t) + f (t). then u (t) = [X(t)]−1 f (t). The equation we had done was very simple. then [X(t)]−1 = e−tA and hence we get a solution x p = etA e−tA f (t) dt which is precisely what we got using the integrating factor method. Now integrate to obtain u and we have the particular solution x p = X(t) u(t). this is essentially the same thing as the integrating factor method we discussed earlier. Hence X(t) u (t) = f (t). But X is the fundamental matrix solution to the homogeneous problem so X (t) = A(t)X(t). the computations can get out of hand pretty quickly for systems. Also try setting a2 = 1 and again check these solutions. The general solution to the associated homogeneous equation is X(t)c for a constant vector c. Just like for variation of parameters for single equation we try the solution to the nonhomogeneous equation of the form x p = X(t) u(t). If we compute [X(t)]−1 . undetermined coefficients works exactly the same as it did for single equations. even if it is not constant coefficient. However this method will work for any linear system. In fact for constant coefficient systems.
Note that if A is constant and you let X(t) = etA . there is the method of variation of parameters. Let us write this as a formula x p = X(t) [X(t)]−1 f (t) dt. However.
3.9) Further. other than the handling of conflicts.2: Check that x1 and x2 solve the problem. where u(t) is a vector valued function instead of a constant.9.2 First order variable coefficient
Just as for a single equation. Example 3.9. Suppose we have the equation x = A(t) x + f (t). (3. and thus X (t) u(t) + X(t) u (t) = X (t) u(t) + f (t).4: Find a particular solution to x = t2 1 t −1 t 2 (t + 1).9. SYSTEMS OF ODES
Exercise 3. x+ 1 t 1 +1 (3. Now substitute into (3.10)
.138
CHAPTER 3. What is the difference between the two solutions we can obtain in this way? As you can see. provided you have somehow solved the associated homogeneous problem.
That is. Hence there are infinitely many solutions x = B sin t for an arbitrary constant B. where x(t) is defined for t in the interval [a. b]. which is x = A cos t + B sin t. similar to §3.1. For example. The condition x(0) = 0 forces A = 0. write down the general solution of the differential equation. we need to study the so-called boundary value problems (or endpoint problems).1 Boundary value problems
Note: 2 lectures. In fact.1 Boundary value problems
Before we tackle the Fourier series. we now specify the value of the solution at two different points.Chapter 4 Fourier series and PDEs
4. b = π. x(π) = 0.1. Unlike before when we specified the value of the solution and its derivative at a single point. so existence of solutions is not an issue here.8 in EP
4. x(0) = 0. The general solution to x + λx = 0 will have two arbitrary constants present. But letting x(π) = 0 does not give us any more information as x = B sin t already satisfies both conditions. so it is natural to think that requiring two conditions will guarantee a unique solution.1: However take λ = 1. x(b) = 0. Example 4. x + x = 0. x(a) = 0. Note that x = 0 is a solution to this equation. a = 0.
for some constant λ. Uniqueness is another issue. 143
. suppose we have x + λx = 0.
Then x = sin t is another solution satisfying both boundary conditions.
let L = − dt2 .2)
We will have to handle the cases λ > 0. This problem is an analogue of finding eigenvalues and eigenvectors of matrices. x(π) = 0.2: On the other hand. So x = 0 is the unique solution to this problem. x (a) = x (b).1.1.1) (resp. λ = 0.1) (4.
4.3) A number λ will be considered an eigenvalue of (4. (4. then the general solution to x + λx = 0 is √ √ x = A cos λ t + B sin λ t. but we will postpone this until chapter 5. sin 2 π 0 and hence B = 0. λ < 0 separately. x(0) = 0. x + λx = 0. So what is going on? We will be interested in classifying which constants λ imply a nonzero solution. The condition x(0) = 0 implies immediately A = 0. The similarity is not just coincidental. Example 4. For the basic Fourier series theory we will need only the following three cases.2 Eigenvalue problems
In general we will consider more equations.3)) if and only if there exists a nonzero solution to (4. and we will be interested in finding those solutions.2) or (4. x (b) = 0. x(a) = x(b). (4. and x + λx = 0. x(a) = 0. change to λ = 2.3)) given that specific λ. x (a) = 0. then we are doing the same d2 exact thing.1) (resp. If we think of the equations as differential operators. x(0) = 0. The nonzero solution we found will be said to be the corresponding eigenfunction. though we will not pursue this line of reasoning too far. But √ now letting 0 = x(π) = B sin 2 π. A lot of the formalism from linear algebra can still apply here.144
CHAPTER 4. FOURIER SERIES AND PDES
Example 4. Next √ 0 = x(π) = B sin λ π.1.
. Note the similarity to eigenvalues and eigenvectors of matrices. x(b) = 0. (4. Letting x(0) = 0 still forces A = 0. x + 2x = 0. x + λx = 0. x(π) = 0. First suppose that λ > 0. (4.3: Let us find the eigenvalues and eigenfunctions of x + λx = 0.2) or (4. √ √ Then the general solution is x = A cos √ 2 t + B sin 2 t. For example. then we are looking for eigenfunctions f satisfying certain endpoint conditions that solve (L − λ) f = 0.
This means that B could be anything (let us take it to be 1). So to get a nonzero solution we must have that √ √ sin λ π = 0. Finally. √ √ First suppose that λ > 0. you should plot sinh to see this. so we only need to pick one. x (0) = 0. Now suppose that λ = 0. Now suppose that λ = 0. In this case the equation is x = 0 and the general solution is x = At + B.1. x (0) = 0 implies that A = 0. Hence the positive eigenvalues are k2 for all integers k ≥ 1. or λ = k for a positive integer k. Letting x(0)√= 0 implies that A = 0 (recall cosh 0 = 1 and sinh 0 = 0). Hence et = e−t which implies t = −t and that is only true if t = 0. So x = −A sin √ √ λ t + B cos λ t
The condition x (0) = 0 implies immediately B = 0. then the general solution to x + λx = 0 is x = A cos λ t + B sin λ t.
Again we will have to handle the cases λ > 0. let λ < 0. √ √ Again A should not be zero. Just like for eigenvectors.4: Let us also compute the eigenvalues and eigenfunctions of x + λx = 0. Hence. we get all the multiples of an eigenfunction. So our solution must be x = B sinh −λ t and satisfy x(π) = 0. λ π must be an integer multiple of π.
Example 4. This means that λ = 0 is not an eigenvalue. This is only possible if B is zero. x (π) = 0.
. The corresponding eigenfunctions can be taken as x = sin kt. In this case we have the general solution x = A cosh −λ t + B sinh −λ t and hence √ √ x = A sinh −λ t + B cosh −λ t. Hence the positive eigenvalues are again k2 for all integers k ≥ 1. λ = 0. Next √ 0 = x (π) = −A sin λ π. So there are 2 no negative eigenvalues. In this case we have the general solution √ √ x = A cosh −λ t + B sinh −λ t. BOUNDARY VALUE PROBLEMS
145
If B√is zero then x is not a nonzero solution. And the corresponding eigenfunctions can be taken as x = cos kt. In this case the equation is x = 0 and the general solution is x = At + B so x = A. λ < 0 separately. Why? Because sinh ξ is only zero for ξ = 0. Obviously setting x (π) = 0 does not get us anything new.1.4. x(0) = 0 implies that B = 0 and then x(π) = 0 implies that A = 0. and sin λ π is only zero if λ = k for a positive integer k. So λ = 0 is an eigenvalue and x = 1 is the corresponding eigenfunction. √ √ Finally. Also we can just look at the definition t −t 0 = sinh t = e −e . In summary. let λ < 0. the eigenvalues and corresponding eigenfunctions are λk = k2 with an eigenfunction xk = sin kt for all integers k ≥ 1.
but rather that they are the same at the beginning and at the end of the interval. Hence there are no negative eigenvalues. √ √ √ √ A cos λ π − B sin λ π = A cos λ π + B sin λ π. the eigenvalues and corresponding eigenfunctions are λk = k2 with an eigenfunction xk = sin kt for all integers k ≥ 1. Similarly (exercise) if we differentiate x and plug in the √ second condition we find that A = 0 or sin √ π = 0. the eigenvalues and corresponding eigenfunctions are λk = k2 λ0 = 0 with the eigenfunctions with an eigenfunction cos kt and x0 = 1.146
CHAPTER 4. In summary. Therefore. In this case however. the general solution is x = At + B. The computations are the same and again we find that there are no negative eigenvalues. So we have two linearly independent eigenfunctions sin kt and cos kt. Let us skip λ < 0. FOURIER SERIES AND PDES
We have already seen (with roles of A and B switched) that for this to be zero at t = 0 and t = π it implies that A = B = 0. x(−π) = x(π). Remember that for a matrix we could also have had two eigenvectors corresponding to an eigenvalue if the eigenvalue was repeated.5: Let us compute the eigenvalues and eigenfunctions of x + λx = 0. Therefore. sin kt for all integers k ≥ 1.1. unless we want A and B to both λ √ be zero (which we do not) we must have sin λ π = 0.
and there is another eigenvalue λ0 = 0 with an eigenfunction x0 = 1. The second condition x (−π) = x (π) says nothing about B and hence λ = 0 is an eigenvalue with a √ corresponding eigenfunction x = 1. x (−π) = x (π). √ For λ > 0 we get that x = A cos λ t + B sin λ t. √ and hence either B = 0 or sin λ π = 0. Therefore. The condition x(−π) = x(π) implies that A = 0 (Aπ + B = −Aπ + B implies A = 0). This problem is be the one that leads to the general Fourier series.
You should notice that we have not specified the values or the derivatives at the endpoints.
We could also do this for a little bit more complicated boundary value problem. x = A cos kt + B sin kt is an eigenfunction for any A and any B. Example 4. Now √ √ √ √ A cos − λ π + B sin − λ π = A cos λ π + B sin λ π.
. We remember that cos −θ = cos θ and sin −θ = − sin θ. In summary. λ is an integer and hence the eigenvalues are yet again λ = k2 for an integer k ≥ 1. For λ = 0.
x(π) = 0. First note that we have the following two equations. Then they are orthogonal in the sense that
b
x1 (t)x2 (t) dt = 0. We will not prove this fact here. For example. Theorem 4.1. The theorem has a very short.3).1.
Multiply the first by x2 and the second by x1 and subtract to get (λ1 − λ2 )x1 x2 = x2 x1 − x2 x1 .1 (easy): Finish the theorem (check the last equality in the proof) for the cases (4.1. if we consider (4.4. and illuminating proof so let us give it here.
The last equality holds because of the boundary conditions.
a
Note that the terminology comes from the fact that the integral is a type of inner product. x1 + λ1 x1 = 0 and x2 + λ2 x2 = 0.
. As λ1 λ2 . therefore.1. (4.1. A matrix is called symmetric if A = AT .
0
when m
n. get the following theorem. This is an analogue of the following fact about eigenvectors of a matrix.1).
b b
(λ1 − λ2 )
a
x1 x2 dt =
a b
x2 x1 − x2 x1 dt d x x1 − x2 x1 dt dt 2
b t=a
=
a
= x2 x1 − x2 x1
= 0. Now integrate both sides of the equation. Hence we have the integral
π
(sin mt)(sin nt) dt = 0.3) for two different eigenvalues λ1 and λ2 .1) we have x1 (a) = x1 (b) = x2 (a) = x2 (b) = 0 and so x2 x1 − x2 x1 is zero at both a and b. That symmetry is required. x(0) = 0. the theorem follows. Eigenvectors for two distinct eigenvalues of a symmetric matrix are orthogonal.2) or (4. elegant.2) and (4. Exercise 4. We. We have seen previously that sin nt was an eigenfunction for the problem x +λx = 0. Suppose that x1 (t) and x2 (t) are two eigenfunctions of the problem (4. The differential operators we are dealing with act much like a symmetric matrix. BOUNDARY VALUE PROBLEMS
147
4.3 Orthogonality of eigenfunctions
Something that will be very useful in the next section is the orthogonality property of the eigenfunctions. We will expand on this in the next section.
A lot of intuition from linear algebra can be applied for linear differential operators.
−π
4.
(cos mt)(cos nt) dt = 0. The Fredholm alternative then states that either (A − λI)x = 0 has a nontrivial solution. On the other hand if λ is an eigenvalue.2 (Fredholm alternative∗ ). even if it happens to have a solution.5) has a unique solution for every right hand side. And finally we also get
π
when m
n. So it is no surprise that there is a finite dimensional version of Fredholm alternative for matrices as well. We will give a slightly more general version in chapter 5. Theorem 4.1. x(b) = 0 (4.
−π π
when m when m
n. We also want to reinforce the idea here that linear differential operators have much in common with matrices.4)
has a unique solution for every continuous function f .5) x(a) = 0. Then either x + λx = 0. then (4. n.5) need not have a solution for every f .
. x(a) = 0. but one must be careful of course.148 Similarly
0 π
CHAPTER 4.
−π
and
π
(cos mt)(sin nt) dt = 0.1. while a matrix has only finitely many. b]. x(b) = 0 (4. Let A be an n × n matrix. one obvious difference we have already seen is that in general a differential operator will have infinitely many eigenvalues. For example. or x + λx = f (t). The theorem holds in a more general setting than we are going to state it. Suppose p and q are continuous on [a.
∗
Named after the Swedish mathematicain Erik Ivar Fredholm (1866 – 1927). the nonhomogeneous equation (4. the solution is not unique. has a nonzero solution. but for our purposes the following statement is sufficient. The theorem means that if λ is not an eigenvalue. and furthermore. FOURIER SERIES AND PDES
(cos mt)(cos nt) dt = 0. The theorem is also true for the other types of boundary conditions we considered.4 Fredholm alternative
We now touch on a very useful theorem in the theory of differential equations. or (A − λI)x = b has a solution for every b.
(sin mt)(sin nt) dt = 0.
we will find a graph which gives the shape of the string.1.1.3 on page 144. With λ > 0. Suppose we have a tightly stretched quickly spinning elastic string or rope of uniform linear density ρ.5 Application
Let us consider a physical application of an endpoint problem. T not 2 2 2 satisfying the above equation. y = 0. As before there are no nonpositive eigenvalues. the string is in the equilibrium position. T except for the interval length being L instead of π. then the string will L
2
L
x
. ω. BOUNDARY VALUE PROBLEMS
149
4. on the x axis. Let us put this problem into the xy-plane. For most√values of ω the string T is in the equilibrium state. When the angular velocity ω hits a value ω = kπ √ρ . Hence the magnitude is constant everywhere and we will call its magnitude T . If we assume that the deflection is small then we can use Newton's second law to get an equation T y + ρω2 y = 0.1: Whirling string. The condition y(0) = 0 implies λ √ that A = 0 as before. so ρω2 k2 π2 =λ= 2 . When ρω = kLπ . 2 T then the string will "pop out" some distance B at the midpoint. Hence.4. The x axis represents the position on the string. y = 0.1.1. y(L) = 0 where λ = ρω . T L What does this say about the shape of the string? It says that for all parameters ρ. we see that this force is tangential and we will assume that the magnitude is the same at both end points. Let L be the length of the string and the string is fixed at the beginning and end points. so we will assume that the whole xy-plane rotates at angular velocity ω along. We will assume that the string stays in this xy-plane and y will measure its deflection from the equilibrium position. We cannot compute B with the information we have. See Figure 4. y y 0 Figure 4. We will idealize the string to have no volume to just be a mathematical curve. Let us assume that ρ and T are fixed and we are changing ω. The condition y(L) = 0 implies that sin λ L = 0 and hence λ L = kπ for some integer k > 0. y(0) = 0 and y(L) = 0. If we take a small segment and we look at the tension at the endpoints. T √ √ the general solution to the equation is y = A cos λ x + B sin √ x. We are looking for eigenvalues of y + λy = 2 0. The setup is similar to Example 4. y(0) = 0. We rewrite the equation as y + ρω y = 0. Hence. The string rotates at angular velocity ω.
You can see that the higher the angular velocity the more times it crosses the x axis when it is popped out.
.1.1. When ω changes again. x(−π) = x(π). x (−π) = x (π).150
CHAPTER 4. x(a) = x(b). x (b) = 0.2: Compute all eigenvalues and eigenfunctions of x + λx = 0. x(b) = 0.6 Exercises
Hint for the following exercises: Note that cos the homogeneous equation.1. x (a) = 0. FOURIER SERIES AND PDES
pop out and will have the shape of a sin wave crossing the x axis k times. x (a) = 0. √ √ λ (t − a) and sin λ (t − a) are also solutions of
Exercise 4. x(a) = 0.1.6: We have skipped the case of λ < 0 for the boundary value problem x + λx = 0. Exercise 4.1. Exercise 4.4: Compute all eigenvalues and eigenfunctions of x + λx = 0. x (a) = x (b).3: Compute all eigenvalues and eigenfunctions of x + λx = 0. So finish the calculation and show that there are no negative eigenvalues. x(b) = 0.1. the string returns to the equilibrium position.5: Compute all eigenvalues and eigenfunctions of x + λx = 0. Exercise 4.
4. Exercise 4.
6)
One way to solve (4. let us talk a little bit more in detail about periodic functions. 3L]. 1]. for t in [−3L.2. F(t) = f (t + 2L). w is the dot product.1: Defined f (t) = 1−t2 on [−1.2 Inner product and eigenvector decomposition
Suppose we have a symmetric matrix. 0 for some periodic function f (t). but with perhaps a different period. It can be confusing when the formula for f (t) is periodic.2.6) is to decompose f (t) as a sum of of cosines (and sines) and then solve many problems of the form (4. For t in [L.1: Define f (t) = cos t on [−π/2. to sum up all the solutions we got to get a solution to (4. which can be computed as vT w.1 Periodic functions and motivation
As motivation for studying Fourier series. For example.2. You should be careful to distinguish between f (t) and its extension. A common mistake is to assume that a formula for f (t) holds for its extension. and so on. How does it compare to the graph of cos t. Example 4. −L].2 on the following page. Now take the π-periodic extension and sketch its graph. L] and we will want to extend periodically to make it a 2L-periodic function. Exercise 4. We then use the principle of superposition. So are cos kt and sin kt for all integers k.7) (4. The constant functions are an extreme example. A function is said to be periodic with period P if f (t) = f (t + P) for all t. We do this extension by defining a new function F(t) such that for t in [−L. We have said before that the eigenvectors of A are then orthogonal.4.
4. For brevity we will say f (t) is Pperiodic. cos t and sin t are 2π-periodic.6). Normally we will start with a function f (t) defined on some interval [−L. we define F(t) = f (t − 2L). In this case the inner product v. §9. suppose we have the problem x + ω2 x = f (t). Before we proceed. Now extend periodically to a 2-periodic function. that is AT = A. 0 (4. Here the word orthogonal means that if v and w are two distinct eigenvectors of A. w = 0.7). THE TRIGONOMETRIC SERIES
151
4.1 in EP
4.
. Note that a P-periodic function is also 2P-periodic. L]. 3P-periodic and so on.2. See Figure 4. then v. F(t) = f (t).2. π/2].2
The trigonometric series
Note: 2 lectures. They are periodic for any period (exercise). We have already solved x + ω2 x = F0 cos ω t.
sin nt = 1. cos nt . 1 = 2. or more simply 1 π f (t) dt.
n=1
This series is called the Fourier series† or trigonometric series for f (t). So we will want to define an inner product of functions.4. to find an we want to compute f (t) . This is for convenience. For example. x (−π) = x (π). sin nt 1 bn = = sin nt . We define the inner product as f (t) .
We have previously computed that the eigenfunctions are 1. Just like for matrices we will want to find a projection of f (t) onto the subspace generated by the eigenfunctions. cos nt = an = cos nt . so that we only need to look at cos kt and sin kt.
−π
With this definition of the inner product.3 The trigonometric series
Now instead of decomposing a vector in terms of the eigenvectors of a matrix.2.
.
−π
Compare these expressions with the finite dimensional example. sin nt = 0 sin mt . The coefficients are given by 1 f (t) . x(−π) = x(π). We could also think of 1 = cos 0t. a0 = π −π
†
Named after the French mathematician Jean Baptiste Joseph Fourier (1768 – 1830).
−π π
f (t) sin nt dt. sin nt π
π
f (t) cos nt dt. and sin kt are orthogonal in the sense that cos mt . cos kt.2. for all m and n. cos nt = 0 for m n. THE TRIGONOMETRIC SERIES
153
4. The formula above also works for n = 0. That is. we will decompose a function in terms of eigenfunctions of a certain eigenvalue problem. cos nt = 0 sin mt . we will want to find a representation of a 2π-periodic function f (t) as a0 f (t) = + 2
∞
an cos nt + bn sin nt. Note that here we have 1 used the eigenfunction 2 instead of 1. For the constant we get that 1 . we have seen in the previous section that the eigenfunctions cos kt (this includes the constant eigenfunction). g(t) =
def
1 π
π
f (t)g(t) dt. cos nt π f (t) .
By elementary calculus we have that cos nt . sin kt. for m n. the eigenvalue problem we will use for the Fourier series is the following x + λx = 0. cos nt = 1 (except for n = 0) and sin nt . In particular.
2.
−π
Let us move to bm . or more to the point the function t cos mt are all odd. is f (t) =
n=1
2 (−1)n+1 sin nt.4.4 on the following page. Recall an even function is a function ϕ(t) such that ϕ(−t) = ϕ(t). bm = = = = = We have used the fact that 1 π t sin mt dt π −π 2 π t sin mt dt π 0 2 −t cos mt π 1 π + cos mt dt π m m 0 t=0 2 −π cos mπ +0 π m −2 cos mπ 2 (−1)m+1 = . THE TRIGONOMETRIC SERIES
155
We will often use the result from calculus that the integral of an odd function over a symmetric interval is zero. Another useful fact from calculus is that the integral of an even function over a symmetric interval is twice the integral of the same function over half the interval.4: Take the function 0 if −π < t ≤ 0.
∞
The series. therefore. cos mπ = (−1) = −1 if m odd. For example the function t. f (t) = 2 sin t − sin 2 t + 2 sin 3 t + · · · 3
The plot of these first three terms of the series. n
Let us write out the first 3 harmonics of the series for f (t). the function sin t. For example t sin mt is even. m m
m
1 if m even. Extend f (t) periodically and write it as a Fourier series. along with a plot of the first 20 terms is given in Figure 4.
. Example 4.2. This function or its variants appear often in applications and the function is called the square wave. Recall that an odd function is a function ϕ(t) such that ϕ(−t) = −ϕ(t). f (t) = π if 0 < t ≤ π. am = 1 π
π
t cos mt dt = 0.
f (t) = t otherwise.
1.158
CHAPTER 4.50 2. including the discontinuities. see Figure 4. The simplest way to make
. the more terms in the series you take. then the series equals the extended f (t) everywhere.50
3.75
2. however. That is. n
If we redefine f (t) on [−π.75
3. π] as 0 if t = −π or t = π. then
∞
n=1
2 (−1)n+1 sin nt = 0. π and all the other discontinuities of f (t).25
3.00 3.7.7: Gibbs phenomenon in action.25
3. It is not hard to see that when t is an integer multiple of π (which includes all the discontinuities). let us plot the first 100 harmonics.00
3.25
2.75 3. Let us however mention briefly an effect of the discontinuity. It will be in fact a superposition of many different pure tones of frequency which are multiples of the base frequency. we do not get an equality for t = −π. This behavior is known as the Gibbs phenomenon. We will generally not worry about changing the function at several (finitely many) points.00 2. FOURIER SERIES AND PDES
is only an equality for t where the sawtooth is continuous. We can think of a periodic function as a "signal" being a superposition of many signals of pure frequency. the error (the overshoot) near the discontinuity at t = π does not seem to be getting any smaller.00
2. Further.75 2. On the other hand a simple sine wave is only the pure tone.00
3. We will say more about convergence in the next section.50
3. we could think of say the square wave as a tone of certain frequency. Let us zoom in near the discontinuity in the square wave.25
Figure 4.75
1.50
2. and extend periodically. That is.75
2.25
3.00
2. The region where the error is large gets smaller and smaller. You will notice that while the series is a very good approximation away from the discontinuities.25 2.
3: Suppose f (t) is defined on [−π. Hint: It may be better to start from the complex exponential form and write the series as ∞ c0 +
m=1
cm eimt + c−m e−imt . show that there exist complex numbers cm such that
∞
f (t) =
m=−∞
cm eimt . There is another form of the Fourier series using complex exponentials that is sometimes easier to work with.2.2. Extend periodically and compute the Fourier series of f (t). Extend periodically and compute the Fourier series of f (t). Exercise 4.4: Suppose f (t) is defined on [−π.
4. π] as t3 . π] as |t|3 . Extend periodically and compute the Fourier series of f (t).
. Exercise 4. Exercise 4.
Note that the sum now ranges over all the integers including negative ones.2. π] as sin 5t + cos 3t.2.2.2.7: Suppose f (t) is defined on [−π.6: Suppose f (t) is defined on [−π. Extend periodically and compute the Fourier series of f (t).2. π] as t2 .9: Let a0 f (t) = + 2
∞
an cos nt + bn sin nt.4 Exercises
Exercise 4. Exercise 4.4. Do not worry about convergence in this calculation. Exercise 4. π] as |t|.2. π] as −1 if −π < t ≤ 0.5: Suppose f (t) is defined on [−π.
n=1
Use Euler's formula eiθ = cos θ + i sin θ.2. f (t) = 1 if 0 < t ≤ π. Extend periodically and compute the Fourier series of f (t). Extend periodically and compute the Fourier series of f (t).8: Suppose f (t) is defined on [−π. Exercise 4. THE TRIGONOMETRIC SERIES
159
sound using a computer is the square wave. If you have played video games from the 1980s or so you have heard what square waves sound like. and the sound will be a very different from nice pure tones.
L L
If we change variables to s we see that a0 g(s) = + 2
∞
an cos ns + bn sin ns. π −π L −L 1 π 1 L nπ an = g(s) cos ns ds = f (t) cos t dt. the computation is a simple case of change of variables. Let s = L t. then the function g(s) = f L s π
is 2π-periodic. We can just rescale the independent axis. you understand it for 2L-periodic functions. After we write down the integrals we change variables back to t. We will want to write f (t) = a0 + 2
∞
an cos
n=1
nπ nπ t + bn sin t. FOURIER SERIES AND PDES
4. g(s) sin ns ds = f (t) sin π −π L −L L The two most common half periods that show up in examples are π and 1 because of the simplicity.3. 1 L 1 π a0 = g(s) ds = f (t) dt. but what about functions of different periods. but all the mathematics is the same.uiuc. it may be good to first try Project IV (Fourier series) from the IODE website: Suppose that you have the 2L-periodic function f (t) (L is called the π half period). All that we are doing is moving some constants around.3 in EP Before reading the lecture. π −π L −L L 1 π 1 L nπ bn = t dt.2 – §9.
4. we have only changed variables. We should stress that we have done no new mathematics. If you understand the Fourier series for 2π-periodic functions.160
CHAPTER 4.edu/iode/.math. Well. §9. fear not.
n=1
So we can compute an and bn as before. After reading the lecture it may be good to continue with Project V (Fourier series again).1 2L-periodic functions
We have computed the Fourier series for a 2π-periodic function.3
More on the Fourier series
Note: 2 lectures.
. We want to also rescale all our sines and cosines.
Note: You should be able to find this integral by thinking about the integral as the area under the graph without doing any computation at all. Finally we can find bn . Here, we notice that |t| sin nπt is odd and, therefore,
1
The plot of these few terms and also a plot up to the 20th harmonic is given in Figure 4.9. You should notice how close the graph is to the real function. You should also notice that there is no "Gibbs phenomenon" present as there are no discontinuities.
-2 -1 0 1 2 -2 -1 0 1 2
1.00
1.00
1.00
1.00
0.75
0.75
0.75
0.75
0.50
0.50
0.50
0.50
0.25
0.25
0.25
0.25
0.00
0.00
0.00
0.00
-2
-1
0
1
2
-2
-1
0
1
2
Figure 4.9: Fourier series of f (t) up to the 3rd harmonic (left graph) and up to the 20th harmonic (right graph).
4.3.2 Convergence
We will need the one sided limits of functions. We will use the following notation f (c−) = lim f (t),
t↑c
we have f (0−) = 0 and f (0+) = π. Let f (t) be a function defined on an interval [a, b]. Suppose that we find finitely many points a = t0 , t1 , t2 , . . . , tk = b in the interval, such that f (t) is continuous on the intervals (t0 , t1 ), (t1 , t2 ), . . . , (tk−1 , tk ). Also suppose that f (tk −) and f (tk +) exists for each of these points. Then we say f (t) is piecewise continuous. If moreover, f (t) is differentiable at all but finitely many points, and f (t) is piecewise continuous, then f (t) is said to be piecewise smooth. Example 4.3.2: The square wave function (4.8) is piecewise smooth on [−π, π] or any other interval. In such a case we just say that the function is just piecewise smooth. Example 4.3.3: The function f (t) = |t| is piecewise smooth. Example 4.3.4: The function f (t) = 1 is not piecewise smooth on [−1, 1] (or any other interval t containing zero). In fact, it is not even piecewise continuous. √ Example 4.3.5: The function f (t) = 3 t is not piecewise smooth on [−1, 1] (or any other interval containing zero). f (t) is continuous, but the derivative of f (t) is unbounded near zero and hence not piecewise continuous. Piecewise smooth functions have an easy answer on the convergence of the Fourier series. Theorem 4.3.1. Suppose f (t) is a 2L-periodic piecewise smooth function. Let a0 + 2
∞
an cos
n=1
nπ nπ t + bn sin t L L
be the Fourier series for f (t). Then the series converges for all t. If f (t) is continuous near t, then f (t) = Otherwise a0 + 2
∞
an cos
n=1
nπ nπ t + bn sin t. L L nπ nπ t + bn sin t. L L
f (t−) + f (t+) a0 = + 2 2
∞
an cos
n=1
If we happen to have that f (t) = f (t−)+ f (t+) at all the discontinuities, the Fourier series converges 2 to f (t) everywhere. We can always just redefine f (t) by changing the value at each discontinuity appropriately. Then we can write an equals sign between f (t) and the series without any worry. We mentioned this fact briefly at the end last section. Note that the theorem does not say how fast the series converges. Think back the discussion of the Gibbs phenomenon in last section. The closer you get to the discontinuity, the more terms you need to take to get an accurate approximation to the function.
164
CHAPTER 4. FOURIER SERIES AND PDES
4.3.3 Differentiation and integration of Fourier series
Not only does Fourier series converge nicely, but it is easy to differentiate and integrate the series. We can do this just by differentiating or integrating term by term. Theorem 4.3.2. Suppose a0 f (t) = + 2
∞
an cos
n=1
nπ nπ t + bn sin t, L L
is a piecewise smooth continuous function and the derivative f (t) is piecewise smooth. Then the derivative can be obtained by differentiating term by term.
∞
f (t) =
n=1
−an nπ nπ bn nπ nπ sin t+ cos t. L L L L
It is important that the function is continuous. It can have corners, but no jumps. Otherwise the differentiated series will fail to converge. For an exercise, take the series obtained for the square wave and try to differentiate the series. Similarly, we can also integrate a Fourier series. Theorem 4.3.3. Suppose f (t) = a0 + 2
∞
an cos
n=1
nπ nπ t + bn sin t, L L
is a piecewise smooth function. Then the antiderivative is obtained by antidifferentiating term by term and so ∞ a0 t nπ −bn L nπ an L F(t) = +C + sin t+ cos t. 2 nπ L nπ L n=1 where F (t) = f (t) and C is an arbitrary constant.
0 Note that the series for F(t) is no longer a Fourier series as it contains the a2 t term. The antiderivative of a periodic function need no longer be periodic and so we should not expect a Fourier series.
and extend to a 2-periodic function. The plot is given in Figure 4.10 on the facing page. Note that this function has a derivative everywhere, but it does not have two derivatives at all the integers.
4.3. MORE ON THE FOURIER SERIES
-2 0.50 -1 0 1 2 0.50
165
0.25
0.25
0.00
0.00
-0.25
-0.25
-0.50 -2 -1 0 1 2
-0.50
Figure 4.10: Smooth 2-periodic function.
Exercise 4.3.1: Compute f (0+) and f (0−). Let us compute the Fourier series coefficients. The actual computation involves several integration by parts and is left to student.
1 0 1
a0 =
−1 1
f (t) dt =
−1
(t + 1) t dt +
0 0
(1 − t) t dt = 0,
1
an =
−1 1
f (t) cos nπt dt =
−1 0
(t + 1) t cos nπt dt +
0 1
(1 − t) t cos nπt dt = 0 (1 − t) t sin nπt dt
0
bn =
f (t) sin nπt dt = 4(1 − (−1)n ) π38n3 = = 0 3 n3 π
−1
(t + 1) t sin nπt dt +
−1
if n is odd, if n is even.
This series converges very fast. If you plot up to the third harmonic, that is the function 8 8 sin πt + sin 3πt, 3 π 27π3
8 it is almost indistinguishable from the plot of f (t) in Figure 4.10. In fact, the coefficient 27π3 is already just 0.0096 (approximately). The reason for this behavior is the n3 term in the denominator. The coefficients bn in this case go to zero as fast as n13 goes to zero.
It is a general fact that if you have one derivative, the Fourier coefficients will go to zero approximately like n13 . If you have only a continuous function, then the Fourier coefficients will go to zero as n12 , and if you have discontinuities then the Fourier coefficients will go to zero approximately as 1 . Therefore, we can tell a lot about the smoothness of a function by looking at its n Fourier coefficients.
4. a) Compute the Fourier series for f (t). but at least at some points it is not defined.3.
n=1
which does not converge! Exercise 4. f (t) = t if 0 ≤ t < 1.
. f (t) and f (t).5 Exercises
Exercise 4. n3
When we differentiate term by term we notice
∞
f (t) =
n=1
1 cos nt. the derivative of f (t) may be defined at most points. the coefficients now go down like n12 .3. At what points does f (t) have the discontinuities. Exercise 4.3: Let 0 if −1 < t < 0. plot say the first 5 harmonics of the functions.3. if 0 ≤ t < 1.2: Use a computer to plot f (t).
extended periodically. n
This function is similar to the sawtooth. FOURIER SERIES AND PDES
To justify this behavior take for example the function defined by the Fourier series
∞
f (t) =
n=1
1 sin nt. n2
Therefore.4: Let −t f (t) = 2 t if −1 < t < 0. If we differentiate again we find that f (t) really is not defined at some points as we get a piecewise differentiable function
∞
f (t) =
n=1
−1 sin nt. That is. If we tried to differentiate again we would obtain
∞
− cos nt. b) Write out the series explicitly up to the 3rd harmonic. which we said means that we have a continuous function.
extended periodically. That is.3.166
CHAPTER 4. b) Write out the series explicitly up to the 3rd harmonic. a) Compute the Fourier series for f (t).
cos nt is even and sin nt is odd. In this section we are of course interested in odd and even periodic functions. is h(t) odd or even. L] and it would be convenient to have an odd (resp. Similarly the function tk is even if k is even and odd when k is odd.4. A function f (t) is even if f (−t) = f (t). Recall a function f (t) is odd if f (−t) = − f (t).4. We have previously defined the 2L-periodic extension of a function defined on the interval [−L. Figure 4. even) extension of the function to [−L.1: Take two functions f (t) and g(t) and define their product h(t) = f (t)g(t). Take a function f (t) defined on [0. §9. If f (t) is odd and g(t) we cannot in general say anything about the sum f (t) + g(t). cosine) terms will disappear. This observation is not a coincidence. and Feven (t) is called the even periodic extension of f .2: Check that Fodd (t) is odd and that Feven (t) is even. In fact. is h(t) odd or even? c) Suppose both are even. def f (t) Fodd (t) = − f (−t) if −L < t < 0. is h(t) odd or even? b) Suppose one is even and one is odd. For example. What we can do is take the odd (resp. L] define the functions if 0 ≤ t ≤ L.168
CHAPTER 4. L]. if 0 ≤ t ≤ L.1: Take the function f (t) = t(1 − t) defined on [0. all the sine (resp.4. L]. a) Suppose both are odd.11 on the facing page shows the plots of the odd and even extensions of f (t). If the function is odd.4. Exercise 4. FOURIER SERIES AND PDES
4.1 Odd and even periodic functions
You may have noticed by now that an odd function has no cosine terms in the Fourier series and an even function has no sine terms in the Fourier series. the Fourier series of a function is really a sum of an odd (the sine terms) and an even (the cosine terms) function. Then Fodd (t) is called the odd periodic extension of f (t). L] and then we can extend periodically to a 2L-periodic function. 1]. Let us look at even and odd periodic function in more detail. On (−L. Example 4.
. And extend Fodd (t) and Feven (t) to be 2L-periodic. even) function. Sometimes we are only interested in the function in the range [0. def f (t) Feven (t) = f (−t) if −L < t < 0.3 in EP
4. Exercise 4.4
Sine and cosine series
Note: 2 lectures.
L
f (t) cos
0
nπ t dt. It is not necessary to start with the full Fourier series to obtain the sine and cosine series.
. L]. generalizing the results of this chapter. L
The series ∞ bn sin nπ t is called the sine series of f (t) and the series a20 + ∞ an cos nπ t n=1 n=1 L L is called the cosine series of f (t).1. we can pick whichever series fits our problem better. x(L) = L. Then the odd extension of f (t) has the Fourier series
∞
Fodd (t) =
n=1
bn sin
nπ t.170 The Fourier series is then a0 2
∞
CHAPTER 4. Let f (t) be a piecewise smooth function defined on [0. g(t) =
0
f (t)g(t) dt. The cosine series is the eigenfunction expansion of f (t) using the eigenfunctions of the eigenvalue problem x + λx = 0. Theorem 4. We could have. FOURIER SERIES AND PDES
an cos
n=1
nπ t. The sine series is really the eigenfunction expansion of f (t) using the eigenfunctions of the eigenvalue problem x + λx = 0. L]. x(0) = 0. x (L) = L.2. therefore. It is often the case that we do not actually care what happens outside of [0. x (0) = 0. This point of view is useful because many times we use a specific series because our underlying question will lead to a certain eigenvalue problem. In fact.4. if the eigenvalue value problem is not one of the three we covered so far. We will deal with such a generalization in chapter 5. L
An interesting consequence is that the coefficients of the Fourier series of an odd (or even) function can be computed by just integrating over the half interval. In this case. you can still do an eigenfunction expansion. L
The even extension of f (t) has the Fourier series a0 Feven (t) = + 2 where an = 2 L
L ∞
an cos
n=1
nπ t. Therefore.
and following the procedure of § 4. have gotten the same formulas by defining the inner product
L
f (t). we can compute the odd (or even) extension of a function as a Fourier series by computing certain integrals over the interval where the original function is defined. L
where 2 bn = L
L
f (t) sin
0
nπ t dt.
By using the Fredholm alternative (Theorem 4. Use the actual definition of f (t). we first find f (t) in L terms of the Fourier sine series. + 2 n=1 where a0 = and
π 2 21 4 2 π 2 t sin nt − t cos nt dt = t sin nt dt an = π 0 π n nπ 0 0 π π 4 4 4(−1)n = 2 t cos nt + 2 cos nt dt = . That is. the first few terms of the series are π2 4 − 4 cos t + cos 2t − cos 3t + · · · 3 9 Exercise 4.
4. SINE AND COSINE SERIES
171
Example 4. Note that the eigenfunctions of this eigenvalue problem were the functions sin nπ t.4. Suppose we have the boundary value problem for 0 < t < L. We write x as a sine series as well with unknown coefficients. for the Dirichlet boundary conditions x(0) = 0.2 on page 148) we note that as long as λ is not an eigenvalue of the underlying homogeneous problem.4.4. If on the other hand we have the Neumann boundary conditions x (0) = 0. x (t) + λ x(t) = f (t).4. We substitute into the equation and solve for the Fourier coefficients of x.4. the even extension of t2 has no jump discontinuities. there will exist a unique solution. We do the same procedure using the cosine series.3 Application
We have said that Fourier series ties in to the boundary value problems we studied earlier. 0 nπ nπ 0 n2 π
2 π
π
t2 dt =
0
2π2 .1. n Explicitly.
. Although it will have corners. Let us see this connection in more detail. not its cosine series! b) Why is it that the derivative of the even extension of f (t) is the odd extension of f (t). since the derivative (which will be on odd function and a sine series) will have a series whose coefficients decay only as 1 so it will have jumps. x(L) = 0. We will write ∞ a0 f (t) = an cos nt. 3
Note that we have detected the "continuity" of the extension since the coefficients decay as n12 . x (L) = 0. to find the solution.3: a) Compute the derivative of the even extension of f (t) above and verify it has jump discontinuities. Therefore.2: Find the Fourier series of the even periodic extension of the function f (t) = t2 for 0 ≤ t ≤ π. These methods are best seen by examples.
The units are the mks units (meters-kilogramsseconds) again. where x sp is the particular steady periodic solution. with damping c. We have already seen this problem in chapter 2 with a simple F(t).5. Now suppose that the forcing function F(t) is 2L-periodic for some L > 0.1: Suppose that k = 2. This is perhaps best seen by example. k m damping c F(t)
(4. L L
an cos
n=1
and we plug in x into the differential equation and solve for an and bn in terms of cn and dn .5. we are mostly interested in the part of x p which does not decay. So any solution to mx (t)+kx(t) = F(t) will be of the form A cos ω0 t+ B sin ω0 t+ x sp . The equation mx + kx = 0 has the general solution x(t) = A cos ω0 t + B sin ω0 t. The difference in what we will do now is that we consider an arbitrary forcing function F(t).5. We want to find the steady periodic solution. APPLICATIONS OF FOURIER SERIES
175
4. We call this x p the steady periodic solution as before. L L nπ nπ t + bn sin t. Since the complementary solution xc will decay as time goes on. The steady periodic solution will always have the same period as F(t). which fires with a force of 1 Newtons for 1 second and then is off for 1 second. The problem with c > 0 is very similar.9) we will call x p . Example 4.
k where ω0 = m . There is a jetpack strapped to the mass. In the spirit of the last section and the idea of undetermined coefficients we will first write
F(t) = Then we write x(t) =
c0 + 2 a0 + 2
∞
cn cos
n=1 ∞
nπ nπ t + dn sin t. We have a mass spring system as before. and a force F(t) applied to the mass. where we have a mass m on a spring with spring constant k. For simplicity. and a particular solution of (4.4.
.1 Periodically forced oscillation
Let us return to the forced oscillations. §9. and m = 1.9)
We know that the general solution will consist of xc which solves the associated homogeneous equation mx + cx + kx = 0.5
Applications of Fourier series
Note: 2 lectures.4 in EP
4. The equation that governs this particular setup is mx (t) + cx (t) + kx(t) = F(t). let us suppose that c = 0.
they will disappear when we plug into the left hand side and we will get a contradictory equation (such as 0 = 1). we cannot use those terms in the guess. 2 n=1
n odd
We plug into the differential equation and obtain
∞ ∞
x + 2x =
n=1 n odd
−bn n2 π2 sin nπt + a0 + 2
n=1 n odd ∞
bn sin nπt
= a0 +
n=1 n odd
bn (2 − n2 π2 ) sin nπt
∞
1 = F(t) = + 2 So a0 =
1 2
n=1 n odd
2 sin nπt.12 on the following page for the plot of this solution.
4. See Figure 4.5. Similarly bn = 0 for n even. When we expand F(t) and find that some of its terms coincide with the complementary solution to mx + kx = 0. Hence we try ∞ a0 x(t) = + bn sin nπt. That is. suppose xc = A cos ω0 t + B sin ω0 t. take the equation mx (t) + kx(t) = F(t). πn
and bn =
2 . Again. πn(2 − n2 π2 )
We know this is the steady periodic solution as it contains no terms of the complementary solution and is periodic with the same period as F(t) itself. Just like before.
. πn(2 − n2 π2 ) The steady periodic solution has the Fourier series 1 x sp (t) = + 4
∞
n=1 n odd
2 sin nπt.4.2 Resonance
Just like when the forcing function was a simple cosine.5. resonance could still happen. Let us assume c = 0 and we will discuss only pure resonance. APPLICATIONS OF FOURIER SERIES
177
We notice that once we plug into the differential equation x + 2x = F(t) it is clear that an = 0 for n ≥ 1 as there are no corresponding terms in the series for F(t).
Of course. However.0
Figure 4. this behavior is called pure resonance or just resonance.4
0. We note that F(t) =
n=1 n odd
4 sin nπt. the terms t aN cos Nπ t + bN sin Nπ t will eventually dominate L L and lead to wild oscillations.0 0. we multiply the offending term by t. In this case we have to modify our guess and try
∞
a0 Nπ Nπ x(t) = + t aN cos t + bN sin t + 2 L L
an cos
n=1 n N
nπ nπ t + bn sin t. FOURIER SERIES AND PDES
7.
∞
extended periodically.5 5. as we change the frequency of F (we change L). Example 4.3
0.2
0.5. the solution will not be a Fourier series (it will not even be periodic) since it contains these terms multiplied by t.178
0.12: Plot of the steady periodic solution x sp of Example 4. πn
.3
0.5
0. That is. we proceed as before.2: Find the steady periodic solution to the equation 2x + 18π2 x = F(t). Further. As before.1
0. Note that there now may be infinitely many resonance frequencies to hit.5 10.0 0.0
CHAPTER 4.5 2.5.0 0.1
0. where 1 if 0 < t < 1.5
0. From then on.0 7.0 2.4
0. we should note that since everything is an approximation and in particular c is never actually zero but something very close to zero. different terms from the Fourier series of F may interfere with the complementary solution and will cause resonance.
where ω0 =
Nπ L
for some positive integer N. L L
In other words. F(t) = −1 if − 1 < t < 0.2
0.5 5. only the first few resonance frequencies will matter.0 10.1.
which is an example of a hyperbolic PDE. and/or some initial conditions where the value of the solution or its derivatives is specified for some initial time. You would expect that if the heat distribution had a maximum (was concave down). t) denote the temperature at point x at time t. we usually have specified some boundary conditions. SEPARATION OF VARIABLES. we will study the wave equation. §9. Suppose that we have a wire (or a thin metal rod) that is insulated except at the endpoints. And vice versa. which is an example of a parabolic PDE. Let x denote the position along the wire and let t denote time.1 Heat on an insulated wire
Let us first study the heat equation. which is an example of an elliptic PDE.6. then heat would flow away from the maximum. Sometimes such conditions are mixed together and we will refer to them simply as side conditions. AND THE HEAT EQUATION
181
4. we will study the heat equation.13: Insulated wire.
L
x
Now let u(x. We will study three partial differential equations.6
PDEs. Together with a PDE. It turns out that the equation governing the this system is the so-called one-dimensional heat equation: ∂u ∂2 u = k 2. First. ∂t ∂x for some k > 0. We will only talk about linear PDEs here.5 in EP Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. Each of our examples will illustrate behaviour that is typical for the whole class.13. where the value of the solution or its derivatives is specified along the boundary of a region. each one representing a more general class of equations. A PDE is said to be linear if the dependent variable and its derivatives appear only to the first power and in no functions. Solving PDEs will be our main application of Fourier series. That is. See Figure 4. PDES.
. temperature u
0 insulation Figure 4.4. and the heat equation
Note: 2 lectures. we will study the Laplace equation. Finally. separation of variables.6. Next. the change in heat at a specific point is proportional to the second derivative of the heat along the wire. This makes sense.
4.
0) = f (x). For the heat equation. Superposition also preserves some of the side conditions. However. We will write ut 2 instead of ∂u and we will write u xx instead of ∂ u . If the ends of the wire are for example kept at temperature 0. In general. t) = 0 and u x (L.
4. t) = 0. t) = X(x)T (t) using
. u(x. t) = 0. c2 are constants. Note that we always have two conditions along the x axis as there are two derivatives in the x direction. Similarly for the side conditions u x (0. c2 are constants. This initial condition is not a homogeneous side condition. In particular. t) = 0 and u(L. Furthermore. t) = 0 and u x (L. If on the other hand the ends are also insulated we get the conditions u x (0. for the heat equation. we will suppose we know the initial temperature distribution. Exercise 4. since u and its derivatives do not appear to any powers or in any functions.
In other words. t) = 0. if u1 and u2 are solutions that satisfy u(0. then u = c1 u1 + c2 u2 is still a solution. These side conditions are called homogeneous.2 Separation of variables
First we must note the principle of superposition still applies. then we must have the conditions u(0. then u = c1 u1 + c2 u2 is still a solution that satisfies u(0. heat is not flowing in nor out of the wire at the ends. That the desired solution we are looking for is of this form is too much to hope for. t) = 0 and u(L. and c1 . t) = 0 and u(L. we must also have some boundary conditions.6. With this notation the equation becomes ∂t ∂x2 ut = ku xx . The heat equation is still called linear. The method of separation of variables is to try to find solutions that are sums or products of functions of one variable.1: Verify the principle of superposition for the heat equation. For example.182
CHAPTER 4. We assume that the wire is of length L and the ends are either exposed and touching some body of constant heat. superposition preserves all homogeneous side conditions. If u1 and u2 are solutions and c1 . we try to find solutions of the form u(x. FOURIER SERIES AND PDES
We will generally use a more convenient notation for partial derivatives.6. or the ends are insulated. t) = 0. t) = 0. t) = X(x)T (t). what is perfectly reasonable to ask is to find enough "building-block" solutions u(x. for some known function f (x).
It will be useful to note that T n (0) = 1. The boundary condition u(0. let us pick the solutions L Xn (x) = sin The corresponding T n must satisfy the equation n2 π2 T n (t) + 2 kT n (t) = 0. 0) = f (x). t) = Xn (x)T n (t) = sin
−n2 π2 nπ x e L2 kt . SEPARATION OF VARIABLES. the solution of this problem is easily seen to be T n (t) = e
−n2 π2 kt L2
We rewrite as
nπ x. t) = 0 implies X(0)T (t) = 0. Thus. L
. T (t) X (x) = . We have previously found that the only eigenvalues are λn = nLπ . kT (t) X(x) As this equation is supposed to hold for all x and t. t) = 0 implies X(L) = 0.
Let us guess u(x. We plug into the heat equation to obtain X(x)T (t) = kX (x)T (t). t) = X(x)T (t). u(L. where eigenfunctions are sin nπ x. AND THE HEAT EQUATION
183
this procedure so that the desired solution to the PDE is somehow constructed from these building blocks by the use of superposition. Hence. X(L) = 0. L By the method of integrating factor. Our building-block solutions are un (x. Let us call this constant −λ (the minus sign is for convenience later). But the left hand side does not depend on t and the right hand side does not depend on x. We are looking for nontrivial solutions X of the eigenvalue problem X +λX = 0. Hence X(0) = 0. 2 2 X(0) = 0.6. Therefore. L
. Let us try to solve the heat equation ut = ku xx with u(0. T (t) + λkT (t) = 0. each side must be a constant. we have two equations X (x) T (t) = −λ = . kT (t) X(x) Or in other words X (x) + λX(x) = 0. t) = 0 and u(x. for integers 2 n ≥ 1. t) = 0 and u(L. Similarly. PDES. We are looking for a nontrivial solution and so we can assume that T (t) is not identically zero.4.
5/2 = 6. in Figure 4.003 t .003 t .5
10.5
5. Getting back to the question of when is the maximum temperature one half of the initial maximum temperature. π That is.16: Temperature at the midpoint of the wire (the bottom curve).5
7.0
10. and the approximation of this temperature by using only the first term in the series (top curve).5 we get
∞
CHAPTER 4. the temperature at the midpoint of the wire at time t. FOURIER SERIES AND PDES
u(0.5. we solve 400 2 6.5.5.5
2.
0 25 50 75 100 12. the terms of the series are insignificant compared to the first term. The figure also plots the approximation by the first term. when is the temperature at the midpoint 12.5. It would be hard to tell the difference after t = 5 or so between the first term of the series representation of u(x. Let us plot the function u(0.0
7.16. If you are interested in behavior for large enough t. t) =
n=1 n odd
400 2 2 (sin nπ 0.003
.25 = 3 e−π 0.5 12. 3 n3 π
For n = 3 and higher (remember we are taking only odd n). Therefore. −π2 0.5) e−n π 0. t) and the real solution. We notice from the graph that if we use the approximation by the first term we will be close enough. only the first one or two terms may be necessary. π The approximation gets better and better as t gets larger as the other terms decay much faster. The first term in the series is already a very good approximation of the function and hence 400 2 u(0.003 t . This behavior is a general feature of solving the heat equation.25 π 400 t= ≈ 24. That is.0
2. t).5
0
25
50
75
100
Figure 4. 3 ln 6.186 If we plug in x = 0.25.0
5. t) ≈ 3 e−π 0.
6 in EP Suppose we have a string such as on a guitar of length L. w(x.
Note that we always have two conditions along the x axis as there are two derivatives in the x direction. It will be easier to solve two separate problems and add their solutions. L x
The equation that governs this setup is the so-called one-dimensional wave equation: ytt = a2 y xx . w(0. 0) = g(x)
for 0 < x < L. 0) = 0 wt (x. t) = 0. t) = 0 and y(L. The two problems we will solve are wtt = a2 w xx .18: Vibrating string.
(4. y(x. There are also two derivatives along the t direction and hence we will need two further conditions here. ONE DIMENSIONAL WAVE EQUATION
191
4. §9. There is one change however. As the equation is again linear. t) = w(L. Suppose we only consider vibrations in one direction. We will assume that the ends of the string are fixed and hence we get y(0. And again we will use separation of variables to find enough building-block solutions to get the overall solution.10)
.4. 0) = g(x). t) = 0. y y 0 Figure 4.18. for some a > 0. for 0 < x < L.
for some known functions f (x) and g(x). let t denote time and let y denote the displacement of the string from the rest position. We will need to know the initial position and the initial velocity of the string.7
One dimensional wave equation
Note: 1 lecture. superposition works just as it did for the heat equation. 0) = f (x) and yt (x. See Figure 4.7. That is let x denote the position along the string.
That is.7. t) for the shape of the string at time t. Find the solution y(x. 0) = 0
for 0 < x < 1.7. for 0 < x < 1. the string was moving at some velocity at impact (t = 0). 0) = f (x) yt (x. say yt (x. Suppose that the length of the string is 1 and a = 1. 0) = 0.
.
Suppose that 0 < k < 2πa. Exercise 4.4.7. 0) = −1. However.6: Suppose that a stringed musical instrument falls on the floor.7 (challenging): Suppose that you have a vibrating string and that there is air resistance proportional to the velocity. t) = 0. you have ytt = a2 y xx − kyt . y(0. y(x. ONE DIMENSIONAL WAVE EQUATION
197
Exercise 4. Any coefficients in the series should be expressed as integrals of f (x). t) = y(1. Derive a series solution to the problem. When the musical instrument hits the ground the string was in rest position and hence y(x.
¶
Named after the French mathematician Pierre-Simon. Then our steady state solution is T2 − T1 u(x) = x + T1. Hence we are looking for a function u(x. the edges of the plate or on all sides of the 3-dimensional object. the steady state solutions are basically just straight lines. Here the ∆ and 2 symbols mean ∂x2 + ∂y2 .18)
2
∂ ∂ or more commonly written as ut = k∆u or ut = k 2 u. That is. The reason for that notation is that you can define ∆ to be the right thing for any number of space dimensions and then the heat equation is always ut = k∆u. now that we have notation out of the way. or a 3-dimensional object. Let us first do this in one space variable. We wish to find out what is the steady state temperature distribution. L This solution agrees with our common sense intuition with how the heat should be distributed in the wire.7 in EP Suppose we have an insulated wire. and Dirichlet problems
Note: 1 lecture. Hence. and we apply constant temperature T 1 at one end (say where x = 0) and T 2 and the other end (at x = L where L is the length of the wire).
2
(4. let us see what does an equation for the steady state solution look like. So in one dimension. Things are more complicated in two or more space dimensions. FOURIER SERIES AND PDES
4. We are looking for a function u that satisfies ut = ku xx . Suppose we have an insulated wire. Let us restrict to two space dimensions for simplicity. This equation is called the Laplace equation¶ . We are looking for a solution to (4. We are really looking for a solution to the heat equation that is not dependent on time. we are looking for a function of x alone that satisfies u xx = 0. but such that ut = 0 for all x and t. We will use ∆ from now on. Laplacian.204
CHAPTER 4.
. The heat equation in two variables is ut = k(u xx + uyy ).9
Steady state temperature. marquis de Laplace (1749 – 1827). a plate. we wish to know what will be the temperature after long enough period of time. Solutions to the Laplace equation are called harmonic functions and have many nice properties and applications far beyond the steady state heat problem. The ∆ is called the Laplacian.18) which does not depend on t. We apply certain fixed temperatures on the ends of the wire. OK. y) such that
∆u = u xx + uyy = 0. It is easy to solve this equation by integration and we see that u = Ax + B for some constants A and B. §9.
For simplicity. We put the Xs on one side and the Ys on the other to get − X Y = .
. We try u(x. we will consider a rectangular region. Also for simplicity we will specify boundary values to be zero at 3 of the four edges and only specify an arbitrary function at one edge. (4. a harmonic function can never have any "hilltop" or "valley" on the graph. Therefore. h)
u=0
u=0
(0. u is concave up in the x direction.4. For example. we will come up with enough building-block solutions satisfying all the homogeneous boundary conditions (all conditions except (4. y) = 0 for 0 < y < h. u(x. We then try to find a solution u defined on this region such that u agrees with the values we specified on the boundary. Let h and w be the height and width of our rectangle.22) (4. We wish to solve the following problem. This observation is consistent with our intuitive idea of steady state heat distribution. y) = X(x)Y(y). 0) = f (x) for 0 < x < w. Again. you can use this simpler solution to derive the general solution for arbitrary boundary values by solving 4 different problems. then uyy must be negative and u must be concave down in the y direction. u(x. X Y
Named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805 – 1859). That is. STEADY STATE TEMPERATURE
205
Harmonic functions in two variables are no longer just linear (plane graphs). We plug into the equation to get X Y + XY = 0.21) (4.23) u=0 (w. one for each edge. Therefore.9. u(w. This setup is left as an exercise. 0)
The method we will apply is separation of variables. Commonly the Laplace equation is part of a so-called Dirichlet problem .20) (4. 0)
u = f (x)
(w. However. we have some region in the xy-plane and we specify certain values along the boundaries of the region. with one corner at the origin and lying in the first quadrant. and adding those solutions together. if you remember your multivariable calculus we note that if u xx is positive. We notice that superposition still works for all the equation and all the homogeneous conditions. y) = 0 for 0 < y < h. (0. we can use the Fourier series for f (x) to solve the problem as before. h) ∆u = 0. As we still have the principle of superposition.19) (4. u(0. you can check that the functions x2 − y2 and xy are harmonic.23)). h) = 0 for 0 < x < w.
we see that u must satisfy (4. Y − λY = 0. It will be useful to have Yn (0) = 1. Furthermore. y) = Xn (x)Yn (y).19)–(4. nπh sinh w
As un satisfies (4. f (x) = bn sin w n=1 Then we get a solution of (4. Observe that nπ un (x.19)–(4. Setting Yn (h) = 0 and solving for Bn we get that − cosh nπh w . y) = bn sin x w n=1 n=1
∞ ∞
sinh nπ(h−y) w . we find Yn (y) = sinh nπ(h−y) w sinh nπh w .24) and simplify.23) of the following form.
We define un (x. n Suppose that ∞ nπx .19)–(4.19)–(4.22). 0) = Xn (x)Yn (0) = sin x.19)–(4. Therefore. And note that un satisfies (4. nπ u(x.
.22). w
For these given λn . y) = bn un (x. the general solution for Y (one for each n) is Yn (y) = An cosh nπ nπ y + Bn sinh y. And we get two equations X X + λX = 0.22) and any linear combination (finite or infinite) of un must also satisfy (4. Bn = nπh sinh w After we plug the An and Bn we into (4.23) as well. w w (4.24)
We only have one condition on Yn and hence we can pick one of An or Bn constants to be whatever is convenient. so we let An = 1. the homogeneous boundary conditions imply that X(0) = X(w) = 0 and Y(h) = 0. Taking the equation for X we have already seen that we have a nontrivial solution if and only if 2 2 λ = λn = nwπ and the solution is a multiple of 2 Xn (x) = sin nπ x. By plugging in y = 0 it is easy to see that u satisfies (4. FOURIER SERIES AND PDES
The left hand side only depends on x and the right hand side only depends on y.22). there is some constant λ such that λ = −X = YY .206
CHAPTER 4.
Hint: Use superposition. It turns out that this soap film is precisely the graph of the solution to the Laplace equation. u(x. y) = u(1.9. 0) = u(x. Take a wire and bend it in just the right way so that it corresponds to the graph of the temperature above the boundary of your region. 0) = 0. each using one piece of nonhomogeneous data.9.9. Then dip the wire in soapy water and let it form a soapy film streched between the edges of the wire. 0) = sin x. Solve the problem u xx + uyy = 0. π) = π. Harmonic functions come up frequently in problems when we are trying to minimize area of some surface or minimize energy in some system. u(0. Hint: Guess. u(0.
Hint: Try a solution of the form u(x. u(x.4: Let R be the region described by 0 < x < π and 0 < y < π.9. u(0. then we solve four problems.3: Let R be the region described by 0 < x < 1 and 0 < y < 1. u(x. Exercise 4.9. u(x.1: Let R be the region described by 0 < x < π and 0 < y < π.9.
u(x. y) = 0. y) = y. y) = y.
. y) = 0.1 Exercises
Exercise 4. for some constant C.2: Let R be the region described by 0 < x < 1 and 0 < y < 1. u(x.208
CHAPTER 4. u(π. y) = 0. 1) = 0. u(x. FOURIER SERIES AND PDES
This scenario corresponds to the steady state temperature on a square plate of width π with 3 sides held at 0 degrees and one side held at π degrees. Solve the problem ∆u = 0. π) = 0.4 to solve ∆u = 0.
Exercise 4. 0) = sin x. y) = X(x) + Y(y) (different separation of variables).5: Use the solution of Exercise 4. then check your intuition.9. Exercise 4. y) = C. π) = π. u(x. Then we use the principle of superposition to add up all four solutions to have a solution to the original problem. y) = y. y) = 0. y) = y. Solve the problem u xx + uyy = 0. 0) = sin πx − sin 2πx. If we have arbitrary initial data on all sides. u(1. u(0. u(x.
Exercise 4. Solve ∆u = 0. 1) = u(0. u(π. There is another way to visualize the solutions.
4. u(π.
For example. EIGENVALUE PROBLEMS √ h √ = tan λ.1. Easiest method is to plot the functions h/x and tan x and √ for which x they intersect. π].
0 4
2
4
6 4
2
2
0
0
-2
-2
-4 0 2 4 6
-4
Figure 5. For general Sturm-Liouville problems we will need a more general setup. .
a
.2 Orthogonality
We have seen the notion of orthogonality before.43. then B = √λ ) is yn (x) = cos h λn x + √ sin λn λn x. For example.214 or
CHAPTER 5. by λ2 the second intersection. and λ2 ≈ 3. g(x) are said to be orthogonal with respect to the weight function r(x) when
b
f (x) g(x) r(x) dx = 0. though using a computer or a graphing calculator will probably be far more convenient nowdays.1. Let r(x) be a weight function (any function. b]. . etc. So denote by λ1 the first intersection. There are tables available. we get that λ1 ≈ 0. we have shown that sin nx are orthogonal for distinct n on [0. Then two functions f (x). when h = 1. There will be an infinite number of see √ intersections. λ
Now use a computer to find λn .86.
5. A plot for h = 1 is given in Figure 5. h The appropriate eigenfunction (let A = 1 for convenience.1: Plot of
1 x
and tan x. though generally we will assume it is positive) on [a.
The results and concepts are again analogous to finite dimensional linear algebra. Nontrivial (nonconstant) r(x) arise naturally. dx dx α1 y(a) − α2 y (a) = 0.
a
that is.1. you could think of a change of variables such that dξ = r(x) dx.
and then say f and g are orthogonal whenever f. Then either dy d p(x) − q(x)y + λr(x)y = 0. Theorem 5. Edwards and Penney [EP]. We have the following orthogonality property of eigenfunctions of a regular Sturm-Liouville problem. β1 y(b) + β2 y (b) = 0. y j and yk are orthogonal with respect to the weight function r. Suppose that we have a regular Sturm-Liouville problem. for example from a change of variables. β1 y(b) + β2 y (b) = 0.1. Theorem 5. Hence.5. we define the inner product as f. Then
b
y j (x) yk (x) r(x) dx = 0. dx dx α1 y(a) − α2 y (a) = 0. Proof is very similar to the analogous theorem from § 4.1. The idea of the given inner product is that those x where r(x) is greater have more weight.2.1. It can also be found in many books including. STURM-LIOUVILLE PROBLEMS In this setting. for example.3 Fredholm alternative
We also have the Fredholm alternative theorem we talked about before for all regular SturmLiouville problems.1.
. We state it here for completeness.3 (Fredholm alternative). Let y j and yk be two distinct eigenfunctions for two distinct eigenvalues λ j and λk .
5. g =
def a b
215
f (x) g(x) r(x) dx. Suppose we have a regular Sturm-Liouville problem d dy p(x) − q(x)y + λr(x)y = 0. g = 0.
we wish to write
∞
f (x) =
n=1
cn yn (x).
b
f.4. EIGENVALUE PROBLEMS
d dy p(x) − q(x)y + λr(x)y = f (x).
(5. and then solve for the coefficients of the series for y(x).216 has a nonzero solution.
5. we wish to calculate cn (and of course we would want to know if the sum converges). This theorem is used in much the same way as we did before in § 4. It is used when solving more general nonhomogeneous boundary value problems. β1 y(b) + β2 y (b) = 0.4)
.3)
where yn (x) the eigenfunctions. but it tells us when does a solution exist and is unique. b]. ym =
a ∞
f (x) ym (x) r(x) dx
b
=
n=1
cn
a b
yn (x) ym (x) r(x) dx ym (x) ym (x) r(x) dx = cm ym .1. We will assume convergence and the ability to integrate term by term. has a unique solution for any f (x) continuous on [a. ym
= cm
a
Hence. We wish to find out if we can represent any function f (x) in this way.
(5. OK. Because of orthogonality we have
b
f. To solve the problem we decompose f (x) and y(x) in terms of the eigenfunctions of the homogeneous problem. ym cm = = ym . so that we know when to spend time looking for a solution. and if so.4 Eigenfunction series
What we want to do with the eigenfunctions once we have them is to compute the eigenfunction decomposition of an arbitrary function f (x). so imagine we could write f (x) as above. That is. ym
a a
f (x) ym (x) r(x) dx
b
ym (x) r(x) dx
2
. or
CHAPTER 5. Actually the theorem does not help us solve the problem. dx dx α1 y(a) − α2 y (a) = 0.
yt (x. X aT We note that we want T + λa4 T = 0. t) measure the displacement of the point x on the beam at time t. APPLICATION OF EIGENFUNCTION SERIES
219
5.
(5.2 in EP The eigenfunction series can arise even from higher order equations.2. t) = y xx (0. Suppose we have an elastic beam (say made of steel). Suppose the beam is displaced by some function f (x) at time t = 0 and then let go (initial velocity is 0).
The equation that governs this setup is a4 ∂4 y ∂2 y + = 0. §10. That is. Similarly λ = 0 will not occur. Then y satisfies: a4 y xxxx + ytt = 0 (0 < x < 1. assume the beam lies along the x axis and let y(x. t) = X(x)T (t) and plug in to get a4 X (4) T + XT = 0 or as usual X (4) −T = 4 = λ.5. 0) = f (x). 0) = 0. t) = 0. Again we try y(x. t) = y xx (1. ∂x4 ∂t2
for some constant a (a4 = EI/ρ in EP).2: Transversal vibrations of a beam. Let us assume that λ > 0. t > 0). We can argue that we expect vibration and not exponential growth nor decay in the t direction (there is no friction in our model for instance). See Figure 5. y(1.2 y
y x Figure 5. y(x. We will study the transversal vibrations of the beam. Suppose the beam is of length 1 simply supported (hinged) at the ends.2
Application of eigenfunction series
Note: 1 lecture. y(0. t) = 0.5)
.
. 1.5) is
∞ ∞
y(x. These frequencies are all integer multiples of the fundamental frequency π2 a2 . . So our solutions are T n (t) = cos n2 π2 a2 t. 5. The sound of a xylophone or vibraphone is. 0) =
n=1
bn Xn (x)T n (0) =
n=1
bn Xn (x) =
n=1
bn sin nπx = f (x). so we will get a nice musical note. and 0 = X (1) = Aω2 (eω − e−ω ) − Cω2 sin ω. So the eigenvalues are λn = n4 π4 and the eigenfunctions are sin nπx. The general solution is X(x) = Aeωx + Be−ωx + C sin ωx + D cos ωx.2. For a steel beam we will get only the square multiples 1.220
CHAPTER 5.1: Try to justify λ > 0 just from the equations. . Also ω must be an integer multiple of π hence ω = nπ and n ≥ 1 (as ω > 0). 3. so that we do not need to write the fourth root all the time.5).
The point is that Xn T n is a solution that satisfies all the homogeneous conditions (that is. Since the eigenfunctions are just sines again. Write ω4 = λ. all conditions except the initial position). . This means that C sin ω = 0 and A(eω − e−ω ) = A2 sinh ω = 0. We can take C = 1. 16. . That is why when you hit a steel beam you hear a very pure sound. t) solves (5. The general solution is T (t) = A sin n2 π2 a2 t + B cos n2 π2 a2 t. 9. EIGENVALUE PROBLEMS
Exercise 5. very different from a guitar or piano. or B = −A. If ω > 0. 4. Now T + n4 π4 a4 T = 0.
So y(x. But T (0) = 0 and hence we must have A = 0 and we can take B = 1 to make T (0) = 1 for convenience. we can decompose the function f (x) on 0 < x < 1 using the sine series as Now we note that on 0 < x < 1 we have (you know how to do this by now)
∞
f (x) =
n=1
bn sin nπx. Hence. And since and T n (0) = 1. So we have X(x) = Aeωx − Ae−ωx + C sin ωx. 4. . then sinh ω 0 and so A = 0. Now 0 = X(1) = A(eω − e−ω ) + C sin ω. The timbre of a beam is different than for a vibrating string where we will get "more" of the smaller frequencies since we will get all integer multiples. therefore. we have
∞ ∞ ∞
y(x. D = 0 and A + B = 0. Now 0 = X(0) = A + B + D. t) =
n=1
bn Xn (x)T n (t) =
n=1
bn (sin nπx) cos n2 π2 a2 t . 0 = X (0) = ω2 (A + B − D).
Then the solution to (5. This means that C 0 else we do not have an eigenvalue. 25. The exact frequencies and their amplitude are what we call the timbre of the note. Note that the natural (circular) frequency of the system is n2 π2 a2 . 2. For X we get the equation X (4) − ω4 X = 0. .
Exercise 5.3: Suppose you have a beam of length 5 with one end free and one end fixed (the fixed end is at x = 5). t) = y xx (0. t) = 0.2.4: Suppose the beam is L units long. do not solve.5. Let u be the longitudinal deviation of the beam at position x on the beam (0 < x < 5).2. t) = 0. t) =
n=1 n odd
4 (sin nπx) cos n2 π2 a2 t . the solution to (5.1: Let us assume that f (x) = by now) f (x) =
n=1 n odd x(x−1) . That is. t > 0).
.2. Suppose you know that the initial shape of the beam is the graph of x(5 − x). 5π3 n3
5. Find a series solution. y(0. everything else kept the same as in (5. You know that the constants are such that this satisfies the equation utt = 4u xx . Hint: Use the same idea as we did for the wave equation. y(1. y(x. Let y be the transverse deviation of the beam at position x on the beam (0 < x < 5). t) = y xx (1. 0) = g(x).2.5). 10 ∞
221
On 0 < x < 1 we have (you know how to do this
4 sin nπx. Set up the equation together with the boundary and initial conditions.2. and the initial velocity is uniformly equal to 2 (same for each x) in the positive y direction. do not solve. and the initial velocity is −(x−5) 50 100 in the positive u direction. You know that the constants are such that this satisfies the equation ytt + 4y xxxx = 0.1 Exercises
Exercise 5.5: Suppose you have a4 y xxxx + ytt = 0 (0 < x < 1. 5π3 n3
Hence. Exercise 5. What is the equation and the series solution. Exercise 5. 0) = f (x). Suppose you know that the initial displacement of the beam is x−5 .5) with the given initial position f (x) is
∞
y(x. Just set up.2. you have also an initial velocity.2. Just set up. yt (x.2: Suppose you have a beam of length 5 with free ends. APPLICATION OF EIGENFUNCTION SERIES Example 5. Set up the equation together with the boundary and initial conditions.
We saw previously that the solution is of the form
∞
(5. See Figure 5.3. assume nice pure sound and assume the force is uniform at every position on the string.1 Forced vibrating string. L x
The problem is governed by the equations ytt = a2 y xx . §10. What if there is an external force acting on the string. t) = 0. Then our wave equation becomes (remember acceleration is force times mass) ytt = a2 y xx + F0 cos ω t.222
CHAPTER 5. (5. y y 0 Figure 5. L But these are free vibrations. For simplicity. EIGENVALUE PROBLEMS
5. y(L.3. Let us say F(t) = F0 cos ω t as force per unit mass. t was time and y was the displacement of the string.6)
y=
n=1
An cos
nπa nπ nπa t + Bn sin t sin x L L L
where An and Bn were determined by the initial conditions. yt (x.
Suppose that we have a guitar string of length L.7)
. Let us assume say air vibrations (noise).3
Steady periodic solutions
Note: 1–2 lectures. We have studied the wave equation problem in this case. 0) = g(x). y(x.3: Vibrating string. 0) = f (x). t) = 0. The natural frequencies of the system are the (circular) frequencies nπa for integers n ≥ 1. for example a second string. where x was the position on the string.3 in EP
5. y(0. with the same boundary conditions of course. Or perhaps a jet engine.
Suppose F0 = 1 and ω = 1 and L = 1 and a = 1. then the coefficient B in (5. What this means is that ω a is equal to one of the natural frequencies of the system.9) seems to become very large. sin 1 Then plug in t = 0 to get f (x) = −y p (x. episode 31. 0) = − cos x + B sin x + 1 and after differentiating in t we see that g(x) = − ∂tp (x. 0) = − cos x + B sin x + 1 yt (x. On the other hand. EIGENVALUE PROBLEMS
Now we get to the point that we skipped.3. originally aired may 18th 2005. Discovery Channel. That is when ω = nπa for odd n. t) = 0 y(x.1: Let us do the computation for specific values. That is. 0) = 0
†
cos 1 − 1 sin x − 1 cos t sin 1
∂y
Mythbusters. so there are far fewer resonance frequencies to hit. i.
. The above calculation explains why a string will begin to vibrate if the identical string is plucked close by. but is very close.3. Remember a glass has much purer sound. We notice that if ω is L not equal to a multipe of the base frequency. You may also need to solve the above problem if the forcing function is a sine rather than a cosine. Then y p (x. Similar resonance phenomena occur when you break a wine glass using human voice (yes this is possible. t) = cos x − Call B = cos 1−1 for simplicity. then L cos ωL = 1 and hence we really get that B = 0. pure resonance never occurs anyway. but it is.224 Exercise 5. the solution is almost the same. Suppose that sin ωL = 0. a multiple of πa . In the absence of friction this vibration would get louder and louder as time goes on. in the right sense. but if you think about it. When ω = nπa for n even.
CHAPTER 5. In real life. a L We could again solve for the resonance solution if we wanted to. Hence to find yc we need to solve the problem ytt = y xx y(0. the limit of the solutions as ω gets close to a resonance frequency. the amplitude will not keep increasing unless you tune to just the right frequency. Example 5. But let us not jump to conclusions just yet. So resonance occurs only when both cos ωL = −1 a a and sin ωL = 0.e. 0) = 0. you are unlikely to get large vibration if the forcing frequency is not close to a resonance frequency even if you have a jet engine running close to the string.2: Check that y p works. but not easy† ) if you happen to hit just the right frequency. t) = 0 y(1.e. When the forcing function is more complicated. it is more like a vibraphone. you decompose it in terms of the Fourier series and apply the above result. i.
Let us assume for simplicity that u(0. the hottest temperature is T 0 + A0 and the coldest is T 0 − A0 . (5. (5.11). iωXeiω t = kX eiω t Hence. And this in fact will be the steady periodic solution. It seems reasonable that the temperature at depth x will also oscillate with the same frequency. Substitute h into (5. We know the temperature at the surface u(0. then t = 0 is midsummer (could put negative sign above to make it midwinter). kX − iωX = 0. independent of the initial conditions.11) We will employ the complex exponential here to make calculations simpler.12). where k is the diffusivity of the soil.3: Suppose h satisfies (5.12) Exercise 5.226
CHAPTER 5. ω is picked depending on the units of t. we look for an h such that ht = kh xx . We will look for an h such that Re h = u. t) = A0 cos ω t. That is. t) = A0 eiω t . for the problem ut = ku xx . For simplicity.11). u(0. whose real part satisfies (5. t) from weather records. EIGENVALUE PROBLEMS
depth x
Figure 5. we will assume that T 0 = 0. t) = T 0 + A0 cos ω t.3.
. Use Euler's formula for the complex exponential to check that u = Re h satisfies (5. t) = V(x) cos ω t + W(x) sin ω t. t) = X(x) eiω t . A0 is picked properly to make this the typical variation for the year. h(0.
The temperature u satisfies the heat equation ut = ku xx .12). For some base temperature T 0 . Suppose we have a complex valued function h(x. So we are looking for a solution of the form u(x. such that when t = 1year then ωt = 2π. To find an h.5: Underground temperature.
The amplitude of the √ω temperature swings is A0 e− 2k x .
ω . t) should be bounded (we are not worrying about the earth core!).3. X(0) = A0 since h(0.e.66◦ Celsius. √ω Exercise 5.3. We did not take that into account above. Why wines are kept in a cellar. while √ω −(1+i) 2k x e will be bounded as x → ∞. so temperature presumably gets higher the deeper you dig. We also will assume that our surface temperature temperature swing is ±15◦ Celsius. A0 = 15. That is. Then the maximum temperature variation at 700 centimeters is only ±0. so we apply Euler's formula to get √ω ω ω h(x. You need not dig very deep to get an effective "refrigerator. A home could be heated or cooled by taking advantage of the above fact. seconds) we have k = 0." I. Let us again take typical parameters as above. If you use Euler's formula to expand the complex exponentials. t) = A0 e−(1+i) 2k x eiω t = A0 e−(1+i) 2k x+iω t = A0 e− 2k x ei(ω t− 2k x) . we get the depth at which summer is the coldest and winter is the warmest. t) = A0 e− √ω
2k
x
cos ω t −
ω x . We get approximately 700 centimeters which is approximately 23 feet below ground. 2π ω ω = seconds2π a year = 31. We will need to get the real part of h. But be careful.341 ≈ 1. t) = Re h(x.557. grams. The temperature differential could also be used to for energy.
. For example in cgs units (centimeters. t) = A0 eiω t .99 × 10−7 . Then if we compute where the phase shift x 2k = π in we find the depth in centimeters where the seasons are reversed. Even without the earth core you could heat a home in the winter and cool it in the summer. X − α2 X = 0. Thus A = A0 . The temperature swings decay rapidly as you dig deeper. There is also the earth core. At depth x the phase is delayed by x 2k .5. you will note that the second term will be unbounded (if B 0). Hence B = 0. 2k
Yay! ω Notice the phase is different at different depths. 2k 2k Then finally u(x. that is.005 (typical value for soil). Note that ± i = ± 1+i so you could simplify to α = ±(1 + i) k 2 general solution is √ω √ω X(x) = Ae−(1+i) 2k x + Be(1+i) 2k x .4: Use Euler's formula to show that e(1+i) 2k x will be unbounded as x → ∞. This decays very quickly as x grows. STEADY PERIODIC SOLUTIONS or √ √ where α = ± iω . This means that √ω √ω √ω √ω h(x. you need consistent temperature. 2k
227
Hence the
We assume that an X(x) that solves the problem must be bounded as x → ∞ since u(x. while the first term is always bounded. Furthermore. t) = A0 e− 2k x cos ω t − x + i sin ω t − x .
ω = 1. A0 = 20. Exercise 5.3. Exercise 5.7: The units are cgs (centimeters.3.228
CHAPTER 5.5. Find the depth at which the temperature variation is half (±10 degrees) of what it is on the surface. EIGENVALUE PROBLEMS
5.3. Suppose that the forcing function is the quare wave which is 1 on the interval 0 < x < 1 and −1 on the interval −1 < x < 0. Suppose that L = 1. Derive the particular solution y p . seconds).3. a = 1.3.6: Take the forced vibrating string. grams. Exercise 5. For k = 0.8: Derive the solution for underground temperature oscillation without assuming that T 0 = 0.3.
. Find the particular solution.991 × 10−7 .005.5: Suppose that the forcing function for the vibrating string is F0 sin ωt.3 Exercises
Exercise 5. Hint: you may want to use result of Exercise 5.
We will use the
Just like the Laplace equation and the Laplacian. It can be seen as converting between the time and the frequency domain. In particular. take the standard equation mx (t) + cx (t) + kx(t) = f (t).
∗
229
. The Laplace transform turns out to be a very efficient method to solve certain ODE problems. understanding the Laplace transform will also help with understanding the related Fourier transform. If the algebraic equation can be solved. the transform can take a differential equation and turn it into an algebraic equation. NMR spectroscopy. however. also named after Pierre-Simon. where the new independent variable s is the frequency. The Laplace transform also gives a lot of insight into the nature of the equations we are dealing with. which. We can think of t as time and f (t) as incoming signal. Finally. The Laplace transform will convert the equation from a differential equation in time to an algebraic (no derivatives) equation. requires more understanding of complex numbers. It is common to write lower case letters for functions in the time domain and upper case letters for functions in the frequency domain.1 The transform
In this chapter we will discuss the Laplace transform∗ . We can think of the Laplace transform as a black box.1 in EP
6. marquis de Laplace (1749 – 1827). For example. We write L{ f (t)} = F(s). It eats functions and spits out functions in a new variable. signal processing and others. §10. The Laplace transform is also useful in the analysis of certain systems such as electrical circuits. applying the inverse transform gives us our desired solution.Chapter 6 The Laplace transform
6. We will not cover the Fourier transform.1.1 The Laplace transform
Note: 2 lectures.
This function is generally given as 0 if t < 0. if we think of t as time there is no problem. Let us define the transform.1: Suppose f (t) = 1. L{ f (t)} = F(s) =
def 0 ∞
e−st f (t) dt. for example F(s) is the Laplace transform of f (t). then
∞
L{1} =
0
e−st dt =
e−st −s
∞ t=0
1 = . the limit only exists if s > 0. which is sometimes called the Heaviside function† . s
Of course.
We note that we are only considering t ≥ 0 in the transform.230
CHAPTER 6. Let us compute the simplest transforms. So L{e−at } is only defined for s + a > 0.1. the limit only exists if s > 0. we are generally interested in finding out what will happen in the future (Laplace transform is one place where it is safe to ignore the past). Example 6. Example 6.4: A common function is the unit step function. Only by coincidence is the function "heavy" on "one side."
†
.
The function is named after Oliver Heaviside (1850–1925). again.1. Of course. So L{1} is only defined for s > 0.3: Suppose f (t) = t.1.
e−st dt
0
Example 6. s+a
Of course. then using integration by parts
∞
L{t} =
0
e−st t dt
∞ ∞
1 −te−st + = s t=0 s ∞ 1 e−st =0+ s −s t=0 1 = 2. u(t) = 1 if t ≥ 0. Example 6.1. s Of course. THE LAPLACE TRANSFORM
same letter to denote that one function is the Laplace transform of the other.2: Suppose f (t) = e−at . the limit only exists if s + a > 0. then
∞ ∞
L{e } =
−at 0
e e
−st −at
dt =
0
e
−(s+a)t
e−(s+a)t dt = −(s + a)
∞
=
t=0
1 .
But in general L{ f (t)g(t)}
1 t
L{ f (t)}L{g(t)}. Similarly tan t or et do not have Laplace transforms. let us also note that for exponential order functions you also obtain that their Laplace transform decays at infinity:
s→∞
lim F(s) = 0.2 Existence and uniqueness
Let us consider in more detail when does the Laplace transform exist. For example. Let f (t) and g(t) be continuous and of exponential order. Before dealing with uniqueness.232
CHAPTER 6. That is. Hint: Note that a sum of two exponential order functions is also of exponential order. ect
t→∞
If the limit exists and is finite (usually zero). For an exponential order function we have existence and uniqueness of the Laplace transform. It is a common mistake to think that Laplace transform of a product is the product of the transforms.1 on the previous page make it easy to already find the Laplace transform of a whole lot of functions already. Then show that tn is of exponential order for any n.3: Use L'Hopital's rule from calculus to show that a polynomial is of exponential order. Suppose that there exists a constant C. such that F(s) = G(s) for all s > C.2: Verify the theorem.1. the function 2 does not have a Laplace transform as the integral diverges. for some constants M and c. THE LAPLACE TRANSFORM
Exercise 6. but that will not relevant to us.3 (Uniqueness).
It must also be noted that not all functions have Laplace transform. that are not of exponential order.1.1. Theorem 6. Then f (t) = g(t) for all t ≥ 0.
6. The simplest way to check this condition is to try and compute lim f (t) . Let f (t) be continuous and of exponential order for a certain constant c. f (t) is of exponential order as t goes to infinity if | f (t)| ≤ Mect . show that L{A f (t) + Bg(t)} = AL{ f (t)} + BL{g(t)}. Exercise 6.1.
. First let us consider functions of exponential order. then f (t) is of exponential order.2 (Existence).1.
Theorem 6. for sufficiently large t (say for all t > t0 for some t0 ). You may have existence of the transform for other functions. Then F(s) = L{ f (t)} is defined for all s > c. These rules together with Table 6.
Once we do solve the algebraic equation in the frequency domain we will want to get back to the time domain.
6. Recall that piecewise continuous means that the function is continuous except perhaps at a discrete set of points where it has jump discontinuities like the Heaviside function. we need to first know if such a function is unique.1. So we can without fear make the following definition. This new step function. s3 + s s s +1
. therefore. If we have a function F(s).1. In other words. We look at the table and we find def
L−1
1 = e−t . We. A + B = 1.1. F(s) = s2 + s + 1 1 1 = + 2 . For example. Let us demonstrate how linearity is used by the following example. That is. It turns out we are in luck by Theorem 6.1. There is an integral formula for the inverse. as that is what we are really interested in.1.3. A = 1. We define the inverse Laplace transform as L−1 {F(s)} = f (t). however. need to also be able to get back. THE LAPLACE TRANSFORM
233
Both theorems hold for piecewise continuous functions as well. but it is not as simple as the transform itself (requires complex numbers). The best way to compute the inverse is to use the Table 6. the inverse Laplace transform is also linear. Uniqueness however does not "see" values at the discontinuities. Example 6. s+1
We note that because the Laplace transform is linear. 3 +s First we use the method of partial fractions to write F in a form where we can use Table 6. to be able to find f (t) such that L{ f (t)} = F(s). Find the inverse Laplace transform. C = 1. Therefore. we defined has the exact same Laplace transform as the one we defined earlier where u(0) = 1. So you can only conclude that f (t) = g(t) outside of discontinuities. Find the inverse Laplace transform. We factor the denominator as s(s2 + 1) and write
2
s2 + s + 1 A Bs + C = + 2 . s3 + s s s +1 Hence A(s2 − 1) + s(Bs + C) = s2 + s + 1.5: Take F(s) = s+1 .1 on page 231. If F(s) = L{ f (t)} for some function f (t).6. 1 the unit step function is sometimes defined using u(0) = 2 .3 The inverse transform
As we said.6: Take F(s) = s s+s+1 . the Laplace transform will allow us to convert a differential equation into an algebraic equation which we can solve.
1 Example 6. We can of course also just pull out constants.1 on page 231. L−1 {AF(s) + BG(s)} = AL−1 {F(s)} + BL−1 {G(s)}.
The shifting property can be used when the denominator is a more complicated quadratic that may come up in the method of partial fractions.
6. where F(s) is the Laplace transform of f (t).1. Such rational functions are called proper rational functions and we will always be able to apply the method of partial fractions. that is functions of the form F(s) G(s) where F(s) and G(s) are polynomials. Since normally (for functions that we are considering) the Laplace transform goes to zero as s → ∞. 2+4 + 4s + 8 (s + 2) 4
In general. and c. 3+ s s s s +1
A useful property is the so-called shifting property or the first shifting property L{e−at f (t)} = F(s + a).1. which involves finding the roots of the denominator.4: Derive this property from the definition. +4 4
Putting it all together with the shifting property we find L−1 s2 1 1 1 = L−1 = e−2t sin 2t.5: Find the Laplace transform of 3 + t5 + sin πt. it is not hard to see that the degree of F(s) will always be smaller than that of G(s).1.1. First we complete the square to make the denominator (s + 2)2 + 4. Next we find
L−1
s2
1 1 = sin 2t. You always want to write such quadratics as (s + a)2 + b by completing the square and then using the shifting property.4 Exercises
Exercise 6. we will want to be able to apply the Laplace transform to rational functions.1.7: Find L−1 s2 +4s+8 . THE LAPLACE TRANSFORM
By linearity of Laplace transform (and thus of its inverse) we get that L
−1
s2 + s + 1 1 1 = L−1 + L−1 2 = 1 + sin t. b. Exercise 6. Of course this means we will need to be able to factor the denominator into linear and quadratic terms.6: Find the Laplace transform of a + bt + ct2 for some constants a.234
CHAPTER 6. Exercise 6.
1 Example 6.
.
s s2 X(s) − sx(0) − x (0) + X(s) = 2 .2. the Laplace transform turns differentiation essentially into multiplication by s. f (t) g (t) g (t) g (t) L{ f (t)} = F(s) sG(s) − g(0) s2G(s) − sg(0) − g (0) s3G(s) − s2 g(0) − sg (0) − g (0)
Table 6. The results are listed in Table 6. L{x (t) + x(t)} = L{cos 2t}. that is functions which are piecewise continuous with a piecewise continuous derivative.2. s +4
.
We can keep doing this procedure for higher derivatives. By X(s) we will. §7. Example 6. We will not worry much about this fact.2
Transforms of derivatives and ODEs
Note: 2 lectures. as usual. First let us try to find the Laplace transform of a function that is a derivative.1: Take the equation x (t) + x(t) = cos 2t.
6.1: Verify Table 6.
∞
L {g (t)} =
0
e−st g (t) dt = e−st g(t)
∞ t=0
∞
−
0
(−s) e−st g(t) dt = −g(0) + sL{g(t)}.2.3 in EP
6.2. The procedure also works for piecewise smooth functions. That is.236
CHAPTER 6. x (0) = 1.2 –7. suppose g(t) is a continuous differentiable function of exponential order.1 Transforms of derivatives
Let us see how the Laplace transform is used for differential equations. denote the Laplace transform of x(t). Let us see how to apply this to differential equations.2 Solving ODEs with the Laplace transform
If you notice.2: Laplace transforms of derivatives (G(s) = L{g(t)} as usual). x(0) = 0.
We will take the Laplace transform of both sides. The fact that the function is of exponential order is used to show that the limits appearing above exist. THE LAPLACE TRANSFORM
6. Exercise 6.2.2.
2+1 2+4 3 s 3 s s +1
Now take the inverse Laplace transform to obtain x(t) = 1 1 cos t − cos 2t + sin t.6.
. If the differential equation we started with was constant coefficient linear equation. it is a function which is zero when t < a and 1 when t ≥ a. The function f (t) should then be defined as 0 if t < π. x (t). All the x(t). Then taking the inverse transform if possible.2. will be converted to X(s). This just shifts the graph to the right by a. we want to consider the Heaviside function. Suppose for example that f (t) is a "signal" and you started receiving the signal sin t at time t. It should be noted that since not every function has a Laplace transform. s2 X(s) − sx(0) − x (0). 2 + 4) + 1)(s s +1 s2 s .2. TRANSFORMS OF DERIVATIVES AND ODES
237
We can plug in the initial conditions now (this will make computations more streamlined) to obtain s2 X(s) − 1 + X(s) = We now solve for X(s). and so on. x (t). f (t) = sin t if t ≥ π.
6. it is generally pretty easy to solve for X(s) and we will obtain some expression for X(s). or cutting functions off. X(s) = (s2 1 s + 2 . we find x(t). 3 3
The procedure is as follows. sX(s) − x(0). Most commonly it is used as u(t − a) for some constant a. That is. and so on. This function is useful for putting together functions. You apply the Laplace transform to transform the equation into an algebraic (non differential) equation in the frequency domain.3 Using the Heaviside function
Before we move on to more general functions than those we could solve before. u(t) = 1 if t ≥ 1. not every equation can be solved in this manner. +4
We use partial fractions (exercise) to write X(s) = 1 s 1 s 1 − + 2 . 0 if t < 0. You take an ordinary differential equation in the time variable t. See Figure 6.1 on the following page for the graph.
00
-1. If you want the function t on when t is in [0.5 0.00
0.5 1. L{u(t − a)} = s This can be generalized into a shifting property or second shifting property.75
0.0
Figure 6. We could imagine a mass and spring system where a rocket was fired for 2 seconds starting at t = 1. Similarly the step function which is 1 on the interval [1. For example. We have already seen that e−as . f (t) can be written as f (t) = u(t − π) sin t.5
1. x (0) = 0. 1] and the function −t + 2 when t is in [1.
where f (t) = 1 if 1 ≤ t < 3 and zero otherwise.1: Plot of the Heaviside (unit step) function u(t). x(0) = 0. you can use the expression t u(t) − u(t − 1) + (−t + 2) u(t − 1) − u(t − 2) . (6.25
0.0 -0. Hence it is useful to know how the Heaviside function interacts with the Laplace transform.1)
Example 6.50
0.5
0. Or perhaps an RLC circuit.00
1.
Using the Heaviside function. suppose that we had a mass spring system x (t) + x(t) = f (t). where the voltage was
.0
-0.2. L{ f (t − a)u(t − a)} = e−as L{ f (t)}.75
0. The Heaviside function is useful to define functions defined piecewise.00
0.2: Suppose that the forcing function is not periodic.0
0.25
0.0
1. THE LAPLACE TRANSFORM
0. 2] and zero otherwise.50
0.0
CHAPTER 6. 2) and zero everywhere else can be written as u(t − 1) − u(t − 2).238
-1.
2)
So the convolution of two functions of t is another function of t. 2
t
. you may have seen it defined as ( f ∗g)(t) = ∞ f (τ)g(t−τ) dτ.2 in EP
6.242
CHAPTER 6.5 lectures.
Now we use the identity cos θ sin ψ = Hence. ( f ∗ g)(t) =
For those that have seen convolution defined before.1 The convolution
We have said that the Laplace transformation of a product is not the product of the transforms.2: Take f (t) = sin ωt and g(t) = cos ωt for t ≥ 0. We did assume that f and g are zero (or just not defined) for negative t. however. Then
t
( f ∗ g)(t) =
0
eτ (t − τ) dτ = et − t − 1.3.
‡ ∞
1 sin(θ + ψ) − sin(θ − ψ) . Then
t
( f ∗ g)(t) =
0
sin ωτ cos ω0 (t − τ) dτ. All hope is not lost however. Define the convolution‡ of f (t) and g(t) as ( f ∗ g)(t) =
def 0 t
f (τ)g(t − τ) dτ. There exists a very important type of a product which works. Take two functions f (t) and g(t) defined for t ≥ 0. THE LAPLACE TRANSFORM
6. When discussing the Laplace transform the definition we gave is sufficient. §7.1: Take f (t) = et and g(t) = t for t ≥ 0. Convolution does occur in many other applications.2) if you define f (t) and g(t) to be zero for t < 0.3. 1 sin(ω0 t) − sin(ω0 t − 2ω0 τ) dτ 0 2 t 1 1 = τ sin ω0 t + cos(2ω0 τ − ω0 t) 2 4ω0 τ=0 1 = t sin ω0 t.3
Convolution
Note: 1 or 1. Example 6.
Where we of course did one integration by parts. where you may have to use the more general definition with infinities. 2 Of course the formula only holds for t ≥ 0.3. Example 6. This definition agrees with (6.
(6.
6.3. Example 6. (s + 1)s2 s + 1 s2 We recognize the two entries of Table 6.3.
In other words.3. the Laplace transform of a convolution is the product of the Laplace transforms.
6.
Where the calculation of the integral of course involved an integration by parts. Example 6.3.2. ( f ∗ g) ∗ h = f ∗ (g ∗ h). That is L−1 Therefore.3: Suppose we have the function of s defined by 1 1 1 = . and h be functions then f ∗ g = g ∗ f.3.2 Solving ODEs
The next example will demonstrate the full power of the convolution and Laplace transform. L−1 1 1 = s + 1 s2
0
1 = e−t s+1
t
and
L−1
1 = t. The simplest way to use this result is in reverse. The most interesting property for us. and the main result of this section is the following theorem. then
t
L {( f ∗ g)(t)} = L
0
f (τ)g(t − τ) dτ = L{ f (t)}L{g(t)}. s2
τet−τ dτ = 2et − t2 − 2t − 2. x(0) = 0. We will be able to give a solution to the forced oscillation problem for any forcing function as a definite integral.
.1. Let f (t) and g(t) be of exponential type. Theorem 6. (c f ) ∗ g = f ∗ (cg) = c( f ∗ g). x (0) = 0. Let c be a constant and f .4: Find the solution to x + ω2 x = f (t). 0 for an arbitrary function f (t). g. CONVOLUTION
243
The convolution has many properties that make it behave like a product. | 677.169 | 1 |
Program Guide: Mathematics Learning Centre
The Mathematics Learning Centre is a
special unit within the University of Sydney, set up to assist students whose
background has not adequately prepared them for first year university
mathematics.
Mature-age students, those who have not studied mathematics for some years,
those who do not have the assumed knowledge for their chosen mathematics
course, or those who were educated interstate or overseas, are all welcome to
drop in at the MLC on a regular basis throughout their first year, for
individual consultations, supplementary tutorials and self-study materials. | 677.169 | 1 |
Right Triangles and Intro to Polynomials
This week the students finished up the unit on right triangles and radicals and started reviewing for their test on Monday, April 25. Please remind your student to study over this weekend and review the trigonometric and special right triangle ratios. Next week, we are going start our unit on polynomials. We will be discussing what makes an expression a polynomial and use the basic operations (addition, subtraction, multiplication, and division) with polynomials. | 677.169 | 1 |
Revision Notes for Class 11 Mathematics Mathematicians seek out patterns and use them to formulate new conjectures.This subject resolve the truth or falsity of conjectures by mathematical proof.when mathematical reasoning can provide insight or predictions about nature through the use of abstraction and logic so that if you take more details then our websites may help for you every time so you may visit our websites here. | 677.169 | 1 |
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Overview
Exploring Numerical Methods: An Introduction to Scientific Computing stresses insight and hands-on experience as it provides students in mathematics, engineering, computer science, and the physical sciences a well-crafted introduction to traditional numerical analysis topics. The text includes an exploration section with problems that deal with the practical issues of software evaluation, selection, modification, and solutions to not-entirely-open-ended problems. Designed to accommodate the needs of beginner and more experienced students, the text encourages active learning, and uses MATLAB to implement the methods and algorithms that it discusses. Students are provided with a strong experiential foundation for future study. | 677.169 | 1 |
Providing a thorough groundingExam SAM's Praxis Core Math Study Guide with Mathematics Workbook and Practice Tests - Academic Skills for Educators (5732) book helps you learn the skills, methods, and formulas that you need to answer all of the types of questions on the Praxis Core Math Test. The book contains 168 Praxis Core Math practice problems with answers and step-by-step explanations and solutions. To see a free sample of this study guide, please click on the "Look Inside" icon at the top of the book cover image at the left. The book is new and up-to-date for the Praxis Core Math Test. The format of the Praxis Core exam is different than the now outdated Praxis I test (also called the Praxis 1 or PPST). So, you can ace your Praxis Core test with our new and up-to-date practice materials. Exam SAM's unique study system gives you in-depth focus on just the math part of the Praxis Core, letting you perfect the skills in the areas of math that students find the most troublesome. Practice Test 1 is in workbook format with exam tips, formulas, explanations, and solutions after each question. You can refer back to the formulas and explanations in the workbook section of the study guide as you complete the remaining practice tests in the book. The practice tests are in the same format and cover the same skill areas as the actual exam, so each practice test has: 17 number and quantity questions 17 algebra and function questions 11 geometry questions 11 statistics and probability questions You may also be interested in our other publication: Praxis Core Reading & Writing Practice Tests: Study Guide for Preparation for Academic Skills for Educators 5712 & 5722 Please visit Exam SAM at Praxis Mathematics: Content Knowledge (5161) Test Prep with Online Practice Tests 3rd Edition - Completely Aligned withCirrus Test Prep's Praxis Core Academic Skills for Educators (5712, 5722, 5732) Study Guide: Test Prep and Practice Test Questions for the Praxis Core Reading, Math and Writing Exams will provide you with a detailed overview of the Praxis Core Academic Skills for Educators, so you know exactly what to expect on test day. We'll take you through all the concepts covered on the test and give you the opportunity to test your knowledge with Praxis Core Academic Skills for Educators practice questions. Even if it's been a while since you last took a major test, don't worry; we'll make sure you're more than ready Cirrus Test Prep's Praxis Core Academic Skills for Educators (5712, 5722, 5732) Study Guide: Test Prep and Practice Test Questions for the Praxis Core Reading, Math and Writing Exams includes: A comprehensive REVIEW of: READING Reading Skills WRITING Language and Research Skills Writing the Essay MATHEMATICS Numbers and Operations Algebra Geometry Statistics and Probability ...as well as a FULL Praxis Core Academic Skills for Educators practice test | 677.169 | 1 |
R in a Nutshell: A Desktop Quick Reference (Softcover)
"Why learn R? Because it's rapidly becoming the standard for developing statistical software. R in a Nutshell provides a quick and practical way to learn this increasingly popular open source language and environment." — From the back cover
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GOOD: The book is in good condition, showing some signs of use and wear (namely, bent corners); however, the pages are clean, and the RepKover™ durable and flexible lay-flat binding is intact.
Description
From the back cover: "You'll not only learn how to program in R, but also how to find the right user-contributed R packages for statistical modeling, visualization, and bioinformatics.
""Understand the basics of the language, including the nature of R objects.
Learn how to write R functions and build your own packages.
Work with data through visualization, statistical analysis, and other methods.
Explore the wealth of packages contributed by the R community.
Become familiar with the lattice graphics package for high-level data visualization.
Products related to "R in a Nutshell: A Desktop Quick Reference (Softcover)"Mathematics plays an important role in many scientific and engineering disciplines. This book deals with the numerical solution of differential equations, a very important branch of mathematicsStatistics and computing share many close relationships. Computing now permeates every aspect of statistics, from pure description to the development of statistical theory. At the same time, the computational methods used in statistical work span much of computer science volume develops the classical theory of the Lebesgue integral and some of its applications. The integral is initially presented in the context However, non-profit consignors have the option to reveal their names in their dedicated pages of the catalog. | 677.169 | 1 |
Introduction to Lebesgue integration theory; measure, σ-algebra, σ-finite measures. Different notion of convergences; product spaces, signed measure, Radon-Nikodym derivative, Fubini and Riesz theorems; Weierstrass approximation theorem. Solid foundation in the Lebesgue integration theory, basic techniques in analysis. It also enhances student's ability to make their own notes.At the end of the course students are expected to understand the difference between "naive" and rigorous modern analysis. Should have a glimpse into the topics of functional analysis as well. They must know and recall the main results, proofs, definition.
Learning Outcomes:
By the end of the course, students areexperts on the topic of the course, and how to use these methods to solve specific problems. In addition, they develop some special expertise in the topics covered, which they can use efficiently in other mathematical fields, and in applications, as well. They also learn how the topic of the course is interconnected to various other fields in mathematics, and in science, in general. At the end of the course students are expected to understand the difference between "naive" and rigorous modern analysis. They should have a glimpse into the topics of functional analysis as well. They must know and recall the main results, proofs, definition. | 677.169 | 1 |
3 Resources Tips from Someone With Experience
Available Types of Calculators and Their Primary Uses Nowadays, the size of the calculator has become very small and this has made it easier for users to fit these devices in smaller storage spaces. Although they are smaller in size, modern day calculators perform many different functions, including those that were previously a reserve for computers. These calculators also have better programming capabilities which make them more powerful. Calculators have made it easier to calculate swiftly and accurately due to the improvement in the technology used in their manufacture. This post features some of the calculators and their functions. The first type is the scientific calculator which offers other complex functions like powers, cosines, and logarithms other than the normal arithmetic functions. These calculators perform these functions using series expansion. The ones who benefit the most from using them are mathematics, science and engineering students. Other functions this type of calculator performs is combinatorial calculations, possible linear regression and standard deviations. When choosing a calculator, make sure to have an idea of all the functions it performs before you settle for the right one. Most of the calculators in the market can do application of logic operators and conversions between varied numbers. These features are useful to those working with computers and logic design. Scientific calculators seem to be more complicated to use because of their layout as compared to simple arithmetic calculators since one button in a scientific calculator is designed to do more than one task as is the norm. The Essentials of Products – Breaking Down the Basics
Advanced scientific calculators display two-lines. This is an attempt by manufacturers to make them easier to operate. The response is on one line while the algebraic style expression is on the other. This makes re-calculation of the answer and editing of the expression possible. The correct syntax should be considered. Some users however, have a preference for the more conventional style of the calculator. These advanced scientific calculators give the user the ability to program them and calculate fractions. With limited language and display capability, programming is not easily done on calculators. What I Can Teach You About Products
The kind of calculators that permit you to show an expression in graphical form are known as graphing scientific calculators. It's mostly useful in education as students can swiftly show a transcendental function, this aids them in comprehension. One can also see the effects of changing values in the expression. This is what makes graphic calculators even more popular for educational purposes. Quite a large number of these calculators contain a menu system. This is in addition to more features which had more complexity to the key layout of the calculator. Since graphic calculators have a somewhat big display it is easier for the user to interact with the programs. | 677.169 | 1 |
Description:
About this title:
Part of a new generation of textbooks for in-service and pre-service teachers at the junior-senior level, this text teaches in three main ways: it extends students' breadth of knowledge beyond, but related to, the topics covered in elementary and middle-grade curriculums; it increases prospective teachers' depth of mathematical understanding by providing problems rich in exploration and mathematical communication; and it models the most current ways of teaching mathematics.
Many Section Openers begin with a motivating lesson that introduces a new topic in an understandable, real-world context.
Exercise Sets at the end of every section provide more traditional practice and are labeled either Proof Exercises or Writing Exercises.
Exploratory Exercises at the end of every section lead students to investigate topics outside the framework presented in the section. The final exercise in each section is a writing exercise.
Book Description Book Condition: Good. Writing on one or more of the edges. There is handwriting, stickers or numbers inside the front cover. Cover has some rubbing and edge wear. Access codes, CDs, and other accessories may not be included. All items ship Mon-Fri. Bookseller Inventory # 2Y68LJ003Y33
Book Description Houghton Mifflin Company. Book Condition: Acceptable. Book contains slight water damage, but it does not interfere with the text.Binding is worn. Used books cannot guarantee unused access codes or working CD's! Ships fast. Ships fast! Ships from USA!. Bookseller Inventory # HSU-F-076-448B 183081 | 677.169 | 1 |
Showing 1 to 11 of 11
Applications of College Algebra
Author:
Joselyn Cendejas
*MAT250 Project 1
Grand Canyon University, Phoenix, AZ, USA
Introduction. By doing this project student will be able to calculate and learn how to use
excel. The two parts of this project was learne
Deliverable 02 Worksheet
Instructions: The following worksheet is shown to you by a student who is asking for
help. Your job is to help the student walk through the problems by showing the student
how to solve each problem in detail. You are expected to e
Deliverable 03 Worksheet
1. <b> Discuss the importance of constructing confidence intervals for the
population mean by answering these questions.
o
What are confidence intervals?
o
What is a point estimate?
o
What is the best point estimate for the popula
MATH 250 Advice
Showing 1 to 3 of 4
it is something that is needed because you need to be able to do calculations especially if your going to be a manager because you have to be able to know budgets and how to stay within the allotted budget.
Course highlights:
The highlight was that I finished the class with a decent grade. It also challenges you.
Hours per week:
9-11 hours
Advice for students:
This course will build patience. You must be diligent in completing assignments because it is easy fall behind.
Course Term:Spring 2013
Professor:John Nicholson
Course Required?Yes
Course Tags:Math-heavyMany Small Assignments
Jan 23, 2017
| No strong feelings either way.
Not too easy. Not too difficult.
Course Overview:
This was a required course that I had to take, and it is a bit hard, and there is a lot of studying.
Course highlights:
from this course I learned many new mathematical formulas. I also learned how to use Microsoft excel as a calculator, and it is very helpful, and it is used for completing homework.
Hours per week:
6-8 hours
Advice for students:
Make sure that you understand the different formulas, when to use which formula. Also learn how to use excel, it will be your best friend. Make sure that you also understand the different symbols and when to use them. Do not be scared to ask questions, or bother the instructor, that is what they are there for. | 677.169 | 1 |
Program in Mathematics Education
Pages
Mathematics is the easiest subject to learn with practice. Different mathematicians in the history came and designed different techniques to solve polynomials. The general form of the equation of degree "2" is, "ax^2+bx+c=0" with the condition that "a" cannot be equal to zero. This equation is also called quadratic equation because of its degree, which is equal to "2". In this article, we will discuss three methods to solve the polynomials of degree "2". These methods include completing square method, factorization and quadratic formula. The easiest of the three methods is using quadratic formula.
The first method of solving polynomials of degree "2" is "completing square method". Before proceeding towards the solution, you should make sure that the leading coefficient of the equation is "1". If it is not "1", then you should divide each term of the equation with the leading coefficient.
School level math progresses from arithmetic to algebra, which is comparatively tougher and takes longer to understand. Algebra is a different ball game from what students are used to and it takes some time and effort to understand and do well in this subject. Pre - algebra introduces some of the basic concepts and algebra 1 builds on that. Students who do well in algebra 1 and 2 find college level algebra courses easier than their peers. There are a lot of different areas of math which are covered in algebra class and students are more likely than not to find something that interests them.
Many students often find themselves clueless about algebra and struggle to get passing grades. This could be because students are unable to follow the lessons in class. Not everyone understands algebra the same way and in the rush to finish the curriculum for the year, some students lag behind.
Math tutoring is one of the most frequently requested services for children looking to succeed in school. There are now many different private teaching services around the country that target specific areas of education. Mathematics is one of the subjects that many kids have difficulty with. Many children are shy to actually say that they have issues with this subject. It is always a good idea for the parents to be supportive and understanding of the child's problems in school.
Services
Remedial math tutoring is one of the services that private tutors often offer their clients. This is necessary if the child has difficulty understanding lesson plans in school. This usually happens when the teacher explains the topic and reinforces it with the kids, but some kids still need more explanation. For the most part, many teachers are willing to help, but sometimes they run out of time, or the students themselves may be embarrassed to come out and ask for help in front of the entire class.
Statistics is a discipline that includes the collection, organization, analysis and interpretation of data. Statistical methods are used to design and analyze experiments and surveys. The theory and methods of statistics are applied to a wide variety of fields because data is involved in almost all areas of human endeavors. So, the application of statistical principles helps us in understanding more about the world around us.
Why Study Statistics?
Large complex data is ubiquitous in today's world and to deal with data the knowledge of statistics is very important. | 677.169 | 1 |
SPECIAL PRICE ONLY TODAY!! 95% off original prices. Buy whilst you can before the price goes up.This Bundle Contains 16 of the best Algebra Resources. Ideal for whole lessons or revision. Very well planned so ideal for either mode.
Solving Quadratic Equations (Factorising, Completing Square and Formula),
Great for whole lesson or quick fire revision as questions contain in depth answers.
Solving Simultaneous Equations (Elimination, Substitution, Linear and Quadratics) comes with Questions and Solutions. Bonus Hexagon Puzzle for Indices.
It also now includes Inequalities.
Great for Revision or Whole Lessons.
Read on for full details.
Quadratic Equations:
Written by an Outstanding practitioner from an Outstanding School.
This resource contains:
1) A 23 slide power point containing explanation, questions and in depth answers/solutions on: great for you and your students to check answers and use as a guide to support and differentiate.
There is 3 hours worth of lesson resources.
Simultaneous Equations:
Written by an Outstanding practitioner from an Outstanding School.
This resource contains:
1) A 17 slide power point containing explanation, questions and in depth answers/solutions on:2) 3 Exercises/questions on the below 3 points: The exercises are on the power point and on a word document for you to choose how to use them.3) Full in depth answers to the above 3 exercises, this is fantastic tool for support and extension. The answers are on a word document, great for you and your students to check answers and use as a guide to support and differentiate.
4) 3 Exam Questions with Mark Scheme.
There is minimum 2 hours worth of lesson resources.
Kindly review after purchasing. Thanks.
All Quadratics covered, written by an Outstanding practitioner from an Outstanding School.
Great for whole lesson or quick fire revision. As it is very detailed it can be used for revision lessons or whole lessons. Can be printed and used to provide revision material to students.
This resource contains:
1) A 23 slide power point containing explanation, questions and in depth answers/solutions on:
a) Drawing quadratic graphs.
b) Factorising and solving quadratic Equations.
c) Completing the square
d) Solving quadratic equation after completing the square.
e) Solving quadratic equationgreat for you and your students to check answers and use as a guide to support and differentiate.
There is 3 hours worth of lesson resources.
As it is in depth can be used for revision or whole lessons.
Kindly review after purchasing. Thanks. | 677.169 | 1 |
8th Grade Algebra Basic Pre-Requisites Pre/Post Mini-Assessments
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This mini assessment was designed to be used before instruction on multi-step equations in the 8th grade. Ideally, students can take the pre-assessment, be grouped by need over the next day or two reviewing the following concepts, and then take the post assessment to determine if they are ready for the 8th grade equation standards. Concepts include:
Integers
Combining Like Terms
Distributive Property
One-Step Equations
Two-Step Equations | 677.169 | 1 |
Like Physics, calculus has taken a bad rap for being difficult in the initial learning, and this bad reputation has kept many from getting into the twin disciplines, even more engineering technology students from delaying the learning benefits for the second and much later year. We will show right now how easy differential equations are to get into {large part of the applications of calculus to technical programming}, especially with the ease of accessibility of MATLAB to your finger tips. Something must definitely be said about the foundation of mathematics in the history and evolution over the last 50 years of numerical analysis for data analysis. You will read more about this in later chapters, especially the one on "The Digital Atomic Revolution", but it can be simplified by considering math as passing from an emphasis on differential equations {which of course like all motion and physical phenomenon itself are analogous} to the matrix of linear algebra; of course with mathematical techniques like Taylor series and the LaPlace transform to convert from an analog to a digital format, or from differential to linear format. That is where MATLAB in technical computer applications, much like Microsoft in total computer applications, has ridden the back of the Digital Atomic Revolution: MATLAB has provided the tools in a single concise language {many statements called commands like "diff" to differentiate and "FF" for the fast fourier transform of the LaPlace equation} to easily provide calculations in analog, digital, and both. Then you throw in the simulation part of MATLAB with "Simulink", the built in simulator of MATLAB, and you have a language and software, albeit quite expensive except for large companies except for the student version used primarily in this book, and MATLAB can do just about anything for you, including your calculus and linear algebra homework, except cook your meals. Even as Microsoft developed into the more popular software than MAC while not being at all any more functional, so other software today like FORTRAN, Mathematica, or BASIC depending on the technical application can be as fucntional as MATLAB, we just have in MATLAB and Microsoft the more popular means of communicating with our technical, social and work environment. Let us begin right now, to take some of the mystery of differential equations with introduction to a very simple one of y' = f(x) = sin(pi*x), where y' is an easy and standard way to write dy/dt. What this equation means then is the first derivative of y, which is a function of x, that is f(x), which in turn is the speific function of x which is sin(pi*x).
NOTE: Get use right at this beginning chapter, that all equations will be written in the language of MATLAB; that is * for multiplication, and pi for pie. And what a way to start feeling at ease with differential equations in terms of what you know so well for fundamental mathematics, the most common functions of sine and cosine. These by the way, written in the language of MATLAB, are among the many built-in functions of MATLAB.
If you back to the basic definition of the derivative, y' = dy/dt, the basic concept of which comes to mind as the rate of change, the rate of change of y, or delta y, as compared to the change in time, delta time. What is different about the calculus of derivations is that the delta change is extremely small; in fact, a small change that approach infinitesimal. {It is about the smallest change that you can conceive of.} When you operate on a function to get it out of its derivative back into the original function, like to integrate y' = -sin(pi*x) to get y {notice we are passing in integration from y' to y}, or y = -1/pi * cos(pi*x) + c. Even as it use to be the custom to take semesters of "analytical geometry and calculus", since so many of the concepts of the calculus of differential equations comes from geometrical visualization and foundations. For example in the MATLAB plot above {edited over in the figure windown to show slopes and places where y is equal to zero}, you can visualize y' as the slope at different x points on the y = f(x) = sin(x) curve. Left to right, and without numbers, you can see that between 0 and 1/2 pi the slope {delta y/delta x, or so much rise over run} is max; then at pi, zero slope; then through the sequence of minus max, zero slope, and max postive. And intuitively you can see that these slope values in turn plot a cosine function. You can also think of differentiation and integration as the opposites: diff takes the rate of change of the curve, and int is the anti derivative. MATLAB makes it easy for you to switch between functions like y and y' with the "int" and "diff" commands. As we spell out the process of gong back to y from y' with the proper symbology of differential equations, we get:
MATLAB makes it so easy to pass from y' to y by use of the "int" command with the following code {check it out in the command window}.
You will find many threads that run consistently through the book which will not be in the chapter titles as at the highest level this is a book of Technical Applications; however you will see them at the Section Heading Level, like 1-6 above: numerical methods, Excel, Mathematica, MINITAB, optimization, sytems ID, linear theory, and advanced engineering math, even data analysis itself and systems integration. You might call some of those threads of book specialization, the "continuity in the book" like: optimization, simulation of a transport aircraft and F-16 in MATLAB, data analysis, the programming of advanced engineering math, systems integration {closely related to the process}, parameter estimation and modeling. These threads of speciality, along with the engineering and teaching experience of the author, help to make this book unqiquely what it is. Many more specialized, and much more difficult, books have been written on the specialized and other threads; but it is the unique selection with applications that makes TECHNICAL APPLICATIONS OF COMPUTERS. {Therefore each thread, especially the specialized threads, will be introduced as early as possible in the chapters of the book; and then hung onto and progressively applied.} And yes, along the way I will share with you some of the intuitive excitement of 50 years of working in technology, teaching and engineering. It takes one who has experienced, and still remembers often, the excitements of learning physics, different equations, computers, electronics, and sophisticated systems to boot, to have those memories; and it is hoped that some of that emotion, with the technical content, is communicated in this book. One might call it true learning or scholarship as contrasted to a priority for gaining a piece of paper to earn money, and might have its roots in thinking like of Ralph Waldo Emerson in "The American Scholar", perhaps you would like to call it "The American Technical Scholar". This helps to make the subject of TECHNICAL APPLICATIONS OF COMPUTERS more interesting; and you will enjoy simultaneously technically working with your computer while you make graphs and plots like the MATLAB logo.
1-7: Sharing of Learning Theory. Another consistent thread throughout this book will be sharing on learning theory. Finally, it is apparent that the "you are picking on me" attitude of the boomer generation just did not work, as the cover-up in education had been blown, even bigger than the cover-ups in financial institutions and government related to the economy, American now recognizing that over one/third of high school students do not graduate. The proof in is in the pudding, and how can you claim to teach and the system is working when the results are so miserable. Poor education of teachers in colleges and university, and the practicing of poor teaching in high schools and other levels of education, has sacrificed generations of student drop- outs on the altar of "you are picking on us". As one successful teacher in California recently said, "The education system is broken. Teachers can't teach, and students can't learn." There are several obvious reasons for this, but more of the responsibility should fall on the education departments of universities across the nation who have so poorly trained teachers. It is a well known fact among college students and alumnus that the easies route to go in college is to major in eduction, and likewise that when a student has taken approximately 3 education courses for then on it is a repeat. Of course, the real problem, a weakness that has been known for years by these money-making education organizations like Sylvan, is that of the four levels of learning, teachers normally take the path of least resistance and use only the lower two levels. Of the four levels of learning--rote memory, example, trial and error, and problem solving-- memory work and example learning are the easiest, require less time, and take less preparation. An extension of this problem is in continuing education. The average knowledge and skill of the average college degree now has a half-life of approximately 8 years. What that means is simple: in 8 years after graduation over half of what you have learned is outdated. Increasing the average worker, especially in the more challenging skills of Aerospace, must find practical ways of continue learning in what is generally called adult education; and yet at the same time, he must avoid the same systems of education that have made teachers what they are not today! And the solution is obviously not in more and higher pieces of paper. In this book, all four levels of learning will be encouraged and promoted, especially the highest levels of trial and error, also of the creative learning of problem solving. Every 10 years or so America spouts a new cure-it-all for schools, and they never work because they come from educators of the education departments and not from science and math departments. Everyone knows when you go to college, next to Basketweaving II and Greenhouse IV, education courses from the education departments are the easiest, also that once you have taken 3 or 4 of them, you then just repeat subject matter. The ideal combination for teachers in STEM would be 3 courses in the technical discipline for every one course from the education department. {Even more potential for excellence would be teachers required to have equal employment time in the technical industries for each year of teaching experience, but the solution every 10 years follow the path of least resistance, not the path of challenge and excellence.} STEM, the newest approach to fix the American education system may offer something, we should give it a chance with one foot solidly in one of the technical disciplines, and you will find on the google plus Technical Corner for this website, an invitation for STEM to show its wares. However, although there are definitely a few good people and teachers in both education and education departments, just like in civil service, the system itself is the big problem that is always seeking the lowest possible level of operation. NEVER forget that graduates of education departments by and large of the American education system thinks if you can teach you can teach anything, the technical subject matter is not that important, so that increasingly {and the trend continues as in many decades of the past where there is a failure to modify the university and college education departments, concentrating rather on the unfortunate school teachers of high school and below that have been trained by these universities in the first place} in superficial recommendations for excellence like New Math, CORE, and STEM. Can we learn something from them, perhaps, but let us not put all our eggs in one basket as learning is a many faceted, interesting, and challenging experience. It goes with growth from youth to adulthood and with the excitements of continuing education years. Since you teachers are most often good at the lowest levels of learning, memory work and example learning, and practice the same, in this TAC book, you are offered an example of learning by "Systems Integration", such applying not only to the included technical subjects, but also learning.
1-8 Ups and Downs of UAV Testing by John Del Frate of NASA.
You will enjoy for perspective on UAV testing the very personable report of Del Frate in newspaper/magazine format in February 2008, the "Abstract" of which is: If you could see the road ahead, you might just pass up a fantastic opportunity because you're blinded by the potential pitfalls. In my case, I was testing the project management waters at the NASA Dryden Flight Research Center after ten years of being a research engineer. I was an eager (but ignorant) rookie project manager (PM) and I was willing to engage in just about any project without knowing what it would entail. The assignment I accepted was to help NASA's Environment Research Aircraft and Sensor Technology (ERAST) Project, a partnership with a fledgling Uninhabited Aerial Vehicle (UAV) industry, to tackle stratospheric flight. I remember one of our industrial partners querying me about whether or not I understood what I was getting into. Like one of those bobble-head toys that have become quite popular, I nodded. But in reality, I didn't have a clue. His response was, "Hang on, it's going to be a wild ride." He was right. In retrospect, if I had clearly understood the ten years of pitfalls that were coming, I might not have "hung on." Now I can look back and say that I would not trade the experience for anything. The lows included the destruction of a number of UAVs on my watch. Later someone told me that we should not be surprised if we lost one UAV for every ten flights. We wrote many chapters in the book on what can go wrong with UAVs-and we are still writing. As you can imagine, each mishap was accompanied by an investigation. What an education!
Jerry V. McMichael May 8, 2010 Portales, New Mexico
1-9: Drilling with FreeMat, MATLAB, SciLab, and Octave Basics.
For some these drills to follow, and at the end of many chapters, may be a review, and for others they may be a first time; however without the memorization, example of knowledge and skill in MATLAB and other program languages and algorithms, there can be no trial and error and creativity. 1. Some Basic Built In Data Analysis Functions in MATLAB. {This also applies to the other free and open software utilized in this course, that is primarily SciLab, FreeMat, and Octave.} MATLAB, among many other things, is made for basic data analysis with built-in functions like mean, median, std {standard deviation}, sum, trapz {trapezoidal integration}, and others. You will begin to think, rightly so, that MATLAB is almost magic, but there is also a built-in function, not so much for data analysis but will allow us to quickly set up a matrix of data for analysis, which is v = magic(3), where the 3 in parenthesis stands for a 3x3 matrix of 3 columns and 3 rows. Perhaps you know by now that MATLAB stands for MATrix LABoratory, and as such you will find much of the math, like in the basic and built-in functions of data analysis and also for the magic function are columns oriented. For example if you used the plot function on a magic function, it will automatically plot a line for each column as shown below.
All the figures for this chapter with the chapter itself is available free as a PDF. You can download it by clicking here, and then either save it to your computer or view. Remember that you must have Adobe Reader on your computer, also free from adobe at . | 677.169 | 1 |
Computing in the Classroom
Below we present two problems of the kind students learn
to solve in our sophomore mathematics courses. Our philosophy is to
use problems from the sciences whenever possible
use the computer to solve problems that would otherwise
be out of reach
use the computer to think about problems in new ways, e.g.,
via numerical experimentation and visualization.
Computing now a standard part of the principal sophomore courses,
Calculus II, Engineering Mathematics, and is used selected freshman
courses (some sections of Calculus), Math 108. It is used extensively
in upper division and graduate courses where appropriate,
e.g., the CES program.
The first is taken from the Calculus II course (Mathematics 221), the
second from Mathematics 217 (Introduction to Computation). It is also
treated in Mathematics 108 (Introduction to Maple).
This presentation also shows how the World Wide Web can be used
in education. The syllabus, notes, info on office hours, etc. for
Mathematics 216 and 217
are available on-line. They can be used by students using any computer
attached to the campus network, or by a computer with a dial-in connection.
Students can ask questions and get answers by electronic mail.
The Physics
The temperature distribution of a thin conducting
plate insulated on both sides is determined by the temperature
along the edges. The physics of the plate is governed
by the mean value principle:
The temperature at a given point is the same as the
average temperature of nearby points.
This expresses a kind of balance between inflows and outflows
of heat: a point at a temperature lower than the local average
temperature will warm up; a point at a temperature higher than
the local average will cool down.
The Mathematics
One way to express the mean value principle
in mathematical form is to divide the plate into a grid of small
square cells, as in the figure on the right. Then
The temperature of a cell is one-quarter the sum of
the temperatures of the four nearest neighbors.
For our example this gives the temperature of the middle cell
as
T = (1 + 1 + 5 + 5)/4 = 3
When there is just one interior cell, as in the
three-by-three grid above,
simple arithmetic is enough to solve the problem
using the mean value principle.
When the grid is larger, say, four-by-four, as in the
figure below, the mean value principle yields a system
of four equations in four unknowns.
In general there is one equation for each interior cell.
For example, in the figure below we have an eight-by-eight
grid, hence a six-by-six grid of interior cells. Consequently
the mean-value principle yields a system of thirty-six
equations in thirty-six unknowns.
The Computation
On the one hand an eight-by-eight grid is too small in the sense that it
does not give a grid fine enough to determine the interior
temperatures with accuracy. On the other hand it is too large
in the sense that thirty-six equations is already too large to
be solved by hand. However, with a computer we can solve large
systems and so resolve the dilemma. The first task is to find
an efficient way of writing the equations.
If T[i,j] denotes the temperature of the cell in the i-th
row and the j-th column, then we have
T[i,j] = (1/4)( T[i-1,j] + T[i+1,j] + T[i,j-1] + T[i,j+1] )
Because the notation is so systematic, we can use a
computer program
both to write the equations and to solve them using,
for example, the method of elimination discovered by
Gauss (1777-1855).
The figure below illustrates what happens when the plate is divided
into a 22-by-22 grid with temperature T = 0 along the top and left and
temperature T = 1 along the bottom and right. To determing the
interior temperatures we solve a system of 400 equations in 400
unknowns. Since it is difficult to interpret a table of 400 numbers
we represent the result graphically, coding the numerical temperature
by a color between red and blue, where red is the hottest temperature
and blue is the coldest. We can also use the computer to trace the
isotherms --- the lines along which temperature is constant.
Consider a community of insects whose population
at time t = n is x_n. A simple model for the dynamics of this population
assumes that the population at time t = n+1 is completely determined
by the population at time t = n. In other words, x_{n+1} is a function
of x_n, a fact we can write as
x_{n+1} = f(x_n)
This choice for the next-state function f is
f(x) = kx
For example, if k = 2, then
x_n = 2^nx_0
This kind of function models bacterial populations
as long as the bacteria are not too crowded. Growth
is exponential, and quite rapid. For example, if
the unit of time is four hours, then the sequence of populations
for the first day is
Such growth, illustrated in the figure below, cannot continue for long, so a more realistic model is needed.
We will examine the model introduced by the Princeton biologist Robert May
in his study of insect populations. The idea is to use quadratic functions
f(x) = kx(1-x)
Unfortunately, there is no simple formula for the population sequence { x_n },
so we must use other tools to analyze its behavior. One approach
is to generate the sequence using a simple computer program, then
graph the results. When the initial population is 0.1 units and k = 2,
we get this:
.1, .18, .2952, .4161, .4859, .4996, .4999, .5000, .5000, ...
It is clear what happens: the population quickly approaches
an "equilibrium population" of 0.5 units. The
equilibrium value is in fact a solution of the quadratic
equation
x = f(x)
It is a fixed point of the function. A graph of population
versus time looks like this:
This is how many populations behave: there is rapid approach to
equilibrium. For example, if we start above equilibrium, we have
something like the graph below.
This is pretty tame. However, as k increases, i.e., as the
fecundity of the population increases, thing become more interesting.
Here is what we find when k = 3.2:
The population rapidly settles into a cycle of period 2.
Let's see what happens when the k = 3.5:
The eventual dynamic is cyclic, but with double the previous
one: each four years the pattern repeats itself. It is worth
exploring the behavior for other values of k. Here is what
we find for k = 4:
The behavior is chaotic. This means several things.
First, small changes in initial conditions lead to large changes
in the population level for a fixed time in the "distant"
future. These changes do not follow a simple pattern and so are
"unpredictable". Second, there is no discernible periodicity.
This is rather natural, since as k gets closer to 4, the period gets
larger and larger, i.e., tends to infinity.
May's model shows that very simple, deterministic systems
can generate extremely complicated behavior, behavior which is
sometimes so sensitive to initial conditions that it cannot
be predicted from measured data in any realistic way. | 677.169 | 1 |
MATH AP Calc Questions & Answers
MATH AP Calc Flashcards
MATH AP Calc Advice
MATH AP Calc Advice
Showing 1 to 3 of 8
This class furthers your understanding of mathematics and opens the doors of calculus to you!
Course highlights:
The highlights were learning the fundamentals of calculus, derivatives and integrals, and doing a final math project at the end.
Hours per week:
3-5 hours
Advice for students:
If you like math at all, take calculus! It is very interesting and you get to build and expand on what you already know about math. It also will help you a lot in science classes.
Course Term:Fall 2017
Professor:Esbensen
Course Tags:Background Knowledge ExpectedGreat Intro to the SubjectMany Small Assignments
Feb 06, 2017
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
I would recommend it, especially if you love math. Calculus is like solving puzzles but in the form of rigorous mathematics.
Course highlights:
The highlights of the course were learning chain rule and reverse chain rule. It is difficult at first, but when you learn it, it is so satisfying.
Hours per week:
3-5 hours
Advice for students:
Pay attention in class and ask questions. It's the best way to learn.
Course Term:Winter 2015
Professor:Esbensen
Course Tags:Math-heavyMany Small AssignmentsCompetitive Classmates
Jan 30, 2017
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
With Mrs. Dorothy Fleck as a teacher you definitely won't have problems learning in my opinion. Although, when you are coming into the course itself [keeping in mind that Calculus, whether AB Calculus (moves at a slower pace) or BC Calculus (moves at a faster pace), is a year long course split up into two sections. You must take Differential Calculus the first semester in the fall and then you go to either Calculus AB or Calculus BC.]
--------------------What You Need To Do To Prepare For AP Calculus
• Before I even start, please note that the amount of hours you may spend on material when it comes to Calculus all depends of what your strengths and weaknesses are. Whether or not you grasp the information and how to do it quickly or not. Nevertheless, without further ado how to start begins right after this upcoming period.
• From the beginning you need to really have a good Pre-calculus Honors background. Now you can take the route of doing Pre-calculus CP but it will be much more difficult on you if you decide to do calculus. I know this because I started in Pre-calculus CP and you don't learn the information needed to prepare you for Calculus (most math teachers know this). However, most advisors or counselors will not know this because they assume that Pre-calculus, regardless of whether its CP or Honors, prepares you for Calculus (not their fault).
• Even so if your are looking to join, there are some key things one needs to know from the start:
1. KNOWLEDGE on the UNIT CIRCLE including how to count is RADIANS as well as degrees.
2. The "basic" parent functions and their characteristics (type in Google 'Harold's Cheat Sheet' & make sure to know each parent function & its characteristics on the list except for the arc- & hyperbolic).
3. Piecewise functions
4. The domain & range of those functions.
5. How to sketch those functions without a graphing calculator.
6. Whether a function is odd, even, neither.
7. Function Inverses, Opposites, & Reciprocals
8. Function compositions
9. Properties of Logarithms
10. Factoring
11. Period, Amplitude, Phase Shift, & Vertical Shift
12. Making lines that are *just* parallel & perpendicular to a line OR making lines that are parallel or perpendicular to a line at a point.
13. Parametrization (only needed if you plan to do Calculus AB)
--------------------Overall Advice & What I Wish I Knew
• Other than these things, make sure you actually do the homework and study otherwise you will become lost very easily. Even if you know or think you know it well practice it, you might just find an area you're a little bit rusty in. I think everyone would agree that you would rather be able to ask how to do the problem before the quiz or test rather than wondering how to do it during the quiz or test.
• The list you see above were things I had little to no knowledge on mostly but was faced with at the start of Calculus. Mainly, I did not know them because Pre-calculus CP did not prepare me for all the things. So try if you can to get in an Honors Pre-calculus class, but if you can't then you could always use Google or any other means to look up the information listed above about what you need to know to be prepared for the start of Calculus if you are serious about it. Maybe even so much as ask your Pre-calculus teacher what you can do to prepare, or even the Calculus teacher. | 677.169 | 1 |
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In this Gr 10 Mathematics revision show we look at Algebraic Expressions & Exponents. In this lesson we revise: Algebraic Expressions: Products & Surds, Algebraic Expressions: Simplifying Fractions as well as Exponents. | 677.169 | 1 |
Modeling and Simulation in Science and Mathematics Education: Macintosh/Windows Version(Hardback)
Synopsis
The role of simulation modeling in understanding dynamic processes is now extending beyond research and university curricula to pre-college education. Computer modeling offers the promise of transforming teaching of many subjects, notably science, and is enhanced by the adoption of new standards for science education, the increasing access to scientific information through computer networks, and the availability of powerful, user-friendly modeling software. This book and its accompanying software will bring the tools and excitement of modeling to pre-college teachers, to researchers involved in curriculum development, and to software developers interested in the pre-college | 677.169 | 1 |
only text that takes basic math concepts and applies them specifically to HVAC! This unique text covers the entire range of mathematical problems and subjects encountered by HVAC technicians in real-world situations. With practice problems, a review unit, and three review tests, students can easily assimilate the material presented and visualize its use in the field. A glossary defines terms specific to HVAC, while conversion charts present critical field information at a glance. This book works well as a math text all by itself, or in conjunction with a general math text. An instructor's guide includes two achievement review tests with answer keys, as well as answer keys to problems in the book.
Meet the Authors
Russell B. DeVore,
Russell B. DeVore [retired] was a Shift Technical Advisor and Simulator Instructor at the training center of a nuclear power plant for the Pennsylvania Power and Light Company. He is former Chair of the Arts & Sciences Division at Trident Technical College, and a former physics teacher at Bloomsburg University. Currently, he teaches on campus at a community college and online at a four-year state college. Dr. DeVore has worked in the energy and education fields for many years, contributing both academic and practical expertise to this book. He is also the author of PRACTICAL PROBLEMS IN MATHEMATICS FOR HEATING AND COOLING TECHNICIANS1111541392 | 677.169 | 1 |
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Algebra online
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Awarding Body
Length of Course
AS: 1 year
A2: 1 year (to make a full A Level)
Teaching and Learning Style
Direct teaching, investigation and consolidation of new topics.
The Subject
At A Level, Maths students learn to recognise the key elements in real life problems and to develop solutions in an organised and efficient way. Employers value these skills, knowing they can be used in many diverse situations. Core Maths is the study of the rules and structures that make maths consistent and powerful. You will develop techniques and ideas in the use of these rules. Applied Maths is all about solving real life problems, such as making informed decisions, analysing statistical data and the mechanics of moving or stationary bodies.
Assessment
By end of module examination with related coursework in Year 13.
Unit Descriptions
AS Level
Unit 1: Statistics Broad range of statistical themes many from GCSE Maths and only slightly extended | 677.169 | 1 |
This course runs on a Tuesday (Kilwinning) and Thursday (Ayr/Kilmarnock) between 6pm - 9pm for 32 weeks.
Entry Requirements
It is recommended that students have a pass at National 4 or Intermediate 1 A or Standard Grade General or students without any formal qualifications, but with life experience and a demonstrable interest in the subject area will also be considered for entry.
What's Involved
This unit seeks to extend the mathematical skills learned at General level of Standard Grade, National 4 or Intermediate 1, including:
percentage calculations
volumes of solids
linear relationships
algebraic operations
properties of the circle trigonometry and solution of simultaneous equations and further aspects of statistics
This is an entry qualification for many courses including Primary Educations and involves three internal unit assessments and an external examination. This course is excellent preparation to study Maths at Higher level. It is fast moving and requires some home study.
Cost
£210 | 677.169 | 1 |
Quizzes Functions and Limits
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This is 3 quizzes for Calculus.
One quiz is a review of functions from precalculus or algebra II (domain, composition, inverses, and zeros). It is 5 multiple choice questions (A - E as the AP test would look.)
Another quiz is also a review. There are 2 review multiple choice questions (A - E as on an AP test) of domain and solving a quadratic. There is a piecewise function and 4 questions of plugging in values. There is 1 velocity question from a graph. And one writing an equation from a word problem.
Another quiz is all on limits. There are 3 graphs with 6 questions on each graph, 1 determine a limit using a chart, and 6 questions determining limits however they decide. | 677.169 | 1 |
This semester course is a study of problem solving, analysis, and interpretation of Algebra and Geometry concepts within real world applications so that students can gain a deeper understanding and appreciation of mathematics. Topics of study will include patterning and number theory within bar codes, zip codes, and banking; similarity and fractals; exponential growth and decay in business, population, and finance; and statistics within politics, marketing, and sports. | 677.169 | 1 |
Linear Algebra with Applications: Alternate Edition, Eighth Edition
The more abstract material on vector spaces starts later, in Chapter 4, with the introduction of the vector space Rn. This leads directly into general vector spaces and linear transformations. This alternate edition is especially appropriate for students preparing to apply linear equations and matrices in their own fields.
Clear, concise, and comprehensive--the Alternate Eighth Edition continues to educate and enlighten students, leading to a mastery of the mathematics and an understanding of how to apply it. Features & Benefits
New and Key Features of the Alternate Eighth Edition:
- Updated and revised throughout with new section material and exercises included in every chapter.
- Provides students with a flexible blend of theory, important numerical techniques and interesting relevant applications.
- Includes discussions of the role of linear algebra in many areas such as the operation of the Google search engine and the global structure of the worldwide air transportation network.
- A MATLAB manual that ties into the regular course material is included as an appendix. These ideas can be implemented on any matrix algebra software package. A graphing calculator manual is also included.
- An Instructor Complete Solutions Manual containing worked solutions to all exercises is also available. | 677.169 | 1 |
"College Algebra" sequence is designed to provide scholars an excellent grounding in pre-calculus themes in a straight forward demeanour. The sequence emphasizes computational talents, principles, and challenge fixing instead of conception. Explore/Discuss packing containers built-in all through each one textual content inspire scholars to imagine severely approximately mathematical options. All labored examples are by means of Matched difficulties that toughen the techniques being taught. New to those variants, expertise Connections illustrate how recommendations that have been formerly defined in an algebraic context can also be solved utilizing a graphing calculator. scholars are consistently proven the underlying algebraic tools first so they don't develop into calculator-dependent. moreover, every one textual content within the sequence comprises an abundance of routines - together with a variety of calculator-based and reasoning and writing workouts - and a large choice of real-world functions illustrating how math comes in handy.
Speakout is a brand new common English path that is helping grownup newbies achieve self belief in all ability parts utilizing real fabrics from the BBC. With its wide selection of help fabric, it meets the varied wishes of inexperienced persons in quite a few educating occasions and is helping to bridge the space among the study room and the true international.
The exciting chilly warfare masterwork through the Nobel Prize winner, released in complete for the 1st time Moscow, Christmas Eve, 1949. The Soviet mystery police intercept a choice made to the yankee embassy via a Russian diplomat who gives you to convey secrets and techniques in regards to the nascent Soviet Atomic Bomb application. On that very same day, a super mathematician is locked away within a Moscow criminal that homes the country's brightest minds.
Simplify and express answers using positive exponents. 83. (a3րnb3րm)1ր3 84. (anր2bnր3)1րn 85. (xmր4ynր3)Ϫ12 86. (amր3bnր2)Ϫ6 87. If possible, find a real value of x such that: (A) (x2)1ր2 x (B) (x2)1ր2 ϭ x (C) (x3)1ր3 x 88. If possible, find a real value of x such that: (A) (x2)1ր2 Ϫx (B) (x2)1ր2 ϭ Ϫx (C) (x3)1ր3 ϭ Ϫx 89. If n is even and b is negative, then b1րn is not real. If m is odd, n is even, and b is negative, is (bm)1րn real? 90. If we assume that m is odd and n is even, is it possible that one of (b1րn)m and (bm)1րn is real and the other is not? | 677.169 | 1 |
A comprehensive reference to category theory for students and researchers in mathematics, computer science, logic, cognitive science, linguistics, and philosophy. Useful for self-study and as a course text, the book includes all basic definitions and theorems (with full proofs), as well as numerous examples and exercises. more...
Just as athletes stretch their muscles before every game and musicians play scales to keep their technique in tune, mathematical thinkers and problem solvers can benefit from daily warm-up exercises. Jessica Shumway has developed a series of routines designed to help young students internalize and deepen their facility with numbers. The daily... more...
Introduction to Algebra and Trigonometry provides a complete and self-contained presentation of the fundamentals of algebra and trigonometry. This book describes an axiomatic development of the foundations of algebra, defining complex numbers that are used to find the roots of any quadratic equation. Advanced concepts involving complex numbers are... more...... more...
This book is a systematic treatment of barrelled spaces, and of structures in which barrelledness conditions are significant. It is a fairly self-contained study of the structural theory of those spaces, concentrating on the basic phenomena in the theory, and presenting a variety of functional-analytic techniques. Beginning with some basic and important... more...
A broad range of topics is covered here, including commutative monoid rings, the Jacobson radical of semigroup rings, blocks of modular group algebras, nilpotency index of the radical of group algebras, the isomorphism problem for group rings, inverse semigroup algebras and the Picard group of an abelian group ring. The survey lectures provide an... more... | 677.169 | 1 |
Thursday, May 20, 2010
The most obvious benefit of this particular book is its size. At fewer than five hundred pages it is a comparatively light book. Despite its compact nature, it does include some clear and helpful illustrations, and features some interesting applications However, the layout is far from appealing, and at first glance can be confusing. The answers to the random exercises can be difficult to locate, and the solutions are often too brief to be of any use. In general the proofs are difficult to follow because they are in compact notation, and like many books of this type, it makes little provision for students struggling to remember previously mentioned results and definitions. In some cases a theorem is stated on page X and then the proof of this theorem is given on page X+2 without restating the theorem. This means the reader is constantly flicking between pages X and X+2. The inclusion of repeated theorems would not hamper the confident mathematician, but would be invaluable to a less self assured student. There seems to be little attempt made to inspire the reader, and its approach is largely clinical and less personal.
This book has an attractive layout, with its theorems and definitions clearly and repetitively presented in highlighted boxes. It is easy to follow and could certainly prove useful to a weak student. However, many results are stated without providing proof, leaving the student to produce the proof as part of the exercise. Underpinning this type of mathematics with proof is certainly a fundamental part of the education process, but in my experience many students find this an intimidating step. I believe it is vital that students have proofs demonstrated repeatedly, until they become confident enough to formulate their own. This book could undoubtedly benefit from the inclusion of more examples. In addition, there is a noticeable lack of illustrations, giving the book a very dense feel. Generally speaking, while this book might provide a valuable handbook for some students, I don't believe its style engages the attention of the reader and it doesn't attempt to provide a thorough explanation of linear algebra.
This is a polished, well written text and features a wide variety of interesting contemporary applications towards the end of the book Its definitions and theorems are highlighted and boxed, and it makes excellent use of illustrations. However, after their initial introduction, the definitions and results are not reprinted. This makes the book awkward to use and the reader is often forced to search back through the book to find a particular formula or theorem. It is my opinion that the average student is unlikely to absorb this information at their first attempt, and I think it is important to repeat important points until the student is completely familiar with them. This book also fails to summarise each chapter. My experience of students is that they find a modular approach to mathematics easier to absorb, and I think it is imperative that at the conclusion of each section, new information is condensed and consolidated before the next module commences. Its explanations are somewhat brief in places, and it uses a compact notation. A confident mathematician would find this method of presentation clear and concise, while the nervous undergraduate finds this kind of brevity difficult to interpret. | 677.169 | 1 |
Funtions Foldable Flip Book
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In this foldable, students will learn how to identify a function on a graph, a table and mapping diagram. It includes a vocabulary section (relation, function, domain, range).
This foldable is made up of two pages. When you print or copy this foldable you must copy page 3 -4 (front and back) and page 5-6 (front to back). Fold both papers on the line that states "fold here". Page 3-4 is the outside flaps and 5-6 is the inside flap. Staple on the dotted line. Please see pictures of the example from my student's notes (zoom in to see the notes clearly). | 677.169 | 1 |
Explore Infinite, Knowledge and more!
Explore related topics
The students read a situation that leads them through a "treasure hunt" in order to save their hometown. The students must have previous knowledge.
Power point Special Types of Linear Systems with GUIDED NOTES
The following lesson is crafted around "The Special Types of Linear Systems!" * Students will find this lesson to be the "kingpin" of all Linear System lessons, as it includes quick and efficient ways for one to recognize a Linear System with ONE solution, NO solution, and INFINITELY MANY solutions (please note the cut pieces of lumber on the floor of the carpenter's shop denoting one solution, infinitely many solutions, and no solution). * Students will surely find themselves…
Algebra that Functions Linear Equations Open-Ended Real World Application Math Project (Previously Titled End of the Year Activity: Math Project Linear Equations Real World Application) This open-ended project is a great way for your | 677.169 | 1 |
Recognition of certain events in the history of Integral Calculus – Matemáticas 11°
The teacher guides the student in watching an animation that describes an example of Integral Calculus. Questions for the student: From what height is Felipe taking photos? What is the distance between where Felipe is and San Antonio? What is the inclination of the mountain, with respect to San Antonio? | 677.169 | 1 |
Introductory Algebra For College Students 6th Edition - hjutr.herokuapp.com
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Cambridge Checkpoint Maths Student's Book 2 by Pimentel, Ric 1444143972143973
EAN:
9781444143973
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Description
This widely-used and highly-respected Student's Book, for Cambridge Secondary 1 Maths, is fully matched to the Curriculum Framework, Cambridge Checkpoint Tests and the Cambridge Progression Tests. It includes sections on calculations and mental strategies that provide accessible guidance through these difficult topics. There are also chapters that focus on ICT, investigations and problem-solving, helping your students to apply Maths to real-life situations.
Ric Pimentel and Terry Wall have extensive teaching experience and have been Heads of Department. In a thirty year teaching career Terry Wall has taught in Turkey and the USA. He ran training courses for teachers in England and abroad including courses for the University of Cambridge International Examinations at the International Conference for Teachers held at Robinson College in Cambridge. Ric Pimentel was a teacher trainer specialising in IT applications for three years, and is currently teaching Mathematics in Cambridge. In addition to the first editions of Checkpoint Maths, other titles by these authors also endorsed by the University of Cambridge International Examinations are IGCSE Mathematics (Extended Syllabus) and IGCSE Core Mathematics | 677.169 | 1 |
Outcomes
All aspects of Working Mathematically, as described in this syllabus document, are integral to the outcomes of the Mathematics Advanced Stage 6 course, in particular outcomes MA11-8, MA11-9, MA12-9 and MA12-10.
develop the ability to use mathematical concepts and skills and apply complex techniques to the modelling and solution of problems in algebra and functions, measurement, financial mathematics, calculus, data and statistics and probability
Year 11 outcomes A student:
Year 12 outcomes A student:
MA11-2 uses the concepts of functions and relations to model, analyse and solve practical problems
MA12-4 applies the concepts and techniques of arithmetic and geometric sequences and series in the solution of problems
MA11-3 uses the concepts and techniques of trigonometry in the solution of equations and problems involving geometric shapes
MA12-5 applies the concepts and techniques of periodic functions in the solution of problems involving trigonometric graphs
MA11-4 uses the concepts and techniques of periodic functions in the solutions of trigonometric equations or proof of trigonometric identities
MA11-5 interprets the meaning of the derivative, determines the derivative of functions and applies these to solve simple practical problems
MA12-6 applies appropriate differentiation methods to solve problems
MA11-6 manipulates and solves expressions using the logarithmic and indicial laws, and uses logarithms and exponential functions to solve practical problems
MA12-7 applies the concepts and techniques of indefinite and definite integrals in the solution of problems
MA11-7 uses concepts and techniques from statistics and probability to present and interpret data and solve problems in a variety of contexts, including the use of probability distributions
MA12-8 solves problems using appropriate statistical processes
Objective
Students:
develop the ability to use advanced mathematical models and techniques, aided by appropriate technology, to organise information, investigate, model and solve problems and interpret a variety of practical situations
Year 11 outcomes A student:
Year 12 outcomes A student:
MA11-8 uses appropriate technology to investigate, organise, model and interpret information in a range of contexts
MA12-9 chooses and uses appropriate technology effectively in a range of contexts, models and applies critical thinking to recognise appropriate times for such use
Objective
Students:
develop the ability to communicate and interpret mathematics logically and concisely in a variety of forms
Year 11 outcomes A student:
Year 12 outcomes A student:
MA11-9 provides reasoning to support conclusions which are appropriate to the context
MA12-10 constructs arguments to prove and justify results and provides reasoning to support conclusions which are appropriate to the context | 677.169 | 1 |
Mathematics (MT)
MT-123 College Algebra (3 credits)
The student develops conceptual understanding of algebra as a language to model real-world problems together with algebraic skills to solve those problems. She develops competance in algebra skills related to solving equations and graphing in the Cartesian plane. She studies elementary functions and their graphs (including polynomial, rational, exponential, and logarithmic functions). She develops her analytic and problem-solving abilities while working to formulate and solve problems, applying skills to solve standard and novel mathematical problems.
The student learns to use the right triangle and unit circle definitions of trigonometric functions, together with their graphs, to reason about the behavior of the functions and solve applied problems. She develops her analytic and problem-solving abilities using trigonometric functions to model realistic periodic phenomena.
Offered Spring Term & Summers only. The student builds on her previous algebra knowledge (solving equations, elementary functions and their graphs) to develop deeper knowledge of mathematical functions and to use them to crate quantitative models of phenomena in science, business, and everyday life. Emphasis is placed on the use of technology tools to understand, use, and apply the function concept. Problem-solving and analytical abilities are developed throughout the work of the course. This course prepares the science or mathematics major for calculus. For the elementary education student pursuing a mathematics support, this course helps her to integrate her algebra knowledge and serves as the bridge to further mathematics courses.
Course Offered Fall Term only. The student studies functions and their rates of change in the context of applied problems, using the ideas and techniques of differential calculus. Topics include derivatives of elementary functions (polynomial, exponential, rational, logarithmic, trigonometric) and their compositions in a variety of representations (graphical, numeric, and symbolic); limits; differential equations as mathematical models for changing phenomena; and antidifferentiation. The student develops her problem-solving, analytic, and communication skills by working both independently and collaboratively to understand, formulate, and solve problems from a variety of disciplines such as physics, chemistry, biology, social science, and management. Computers and calculators are used as tools to computation, communication, and exploration of mathematical ideas.
Prerequisite(s):MT-148 completed or MP Level 3 or higher. Sign up for a Calculus lab (MT-152L), Take MT-152LPrerequisite(s): For Educational Licensure & LTM Students only. This is a hybrid course with significant independent work. Some online work is expected. See Moodle for preliminary assignment.
MT-244 Fundumental Concepts/Mathematics 2 (3 credits)Prerequisite(s):MT-243A completed. Limited to licensure or LTM students only. This course is a hybrid; significant independent work online is expected.
MT-253 Calculus 2 (4 credits)
Offered Spring Term only. The student extends her knowledge of calculus by exploring the ideas and techniques of integral calculus. Topics include differential equations as mathematical models of changing phenomena, the definite integral and its standard applications, techniques of antidifferentiation, Taylor polynomial approximations, improper integrals, and representations of functions by infinite series. The student builds knowledge and skill using technology tools to solve problems.
Offered Fall Term only. The student studies the calculus of multivariate functions with emphasis on functions of two independent variables and their three-dimensional graphs. Further topics include parametric equations; conic sections; polar, cylindrical, and spherical coordinate systems; the calculus of vectors and vector-valued functions; multiple integrals; and line integrals. She continues to develop her analytic and problem-solving abilities, working purposefully on generalization skills, algorithm and formula development, and understanding and applying theorems. In individual and group work, she solved applied problems that arise from the areas of physics, chemistry, biology, management, and mathematics itself.
The student studies the mathematics of matrix algebra; the structure and operations of vector spaces, including use of determinants, eigenvalues, and eigenvectors; and linear transformations. She learns the basic concepts and computational procedures associated with these structures, including the use of computer and calculator technology. Linear algebra is applied to problems in areas including linear programming, graph theory, theory of games, least squares regression, linear economic models, traffic flow, and scheduling.
The student engages in the systematic collection, presentation, and characterization of statistical information for the purpose of decision making. She develops the mathematical skills and knowledge necessary for problem solving in statistical contexts. Both descriptive and inferential statistics are studied. Knowledge of the mathematics of probability support conceptual understanding of statistical methods. Data analysis, graphical representation, correlation, regression, and reliability and validity issues are considered. Technology tools are used.
This introductory course in programming introduces the student to an object-oriented program design paradigm. With Java a student can create World Wide Web and stand-alone applications. She develops a number of projects that lead to an independent final project. This course assumes no previous experience with programming.
This course teaches the programming language Python. It was created to solve real world problems, but, at the same time, be easy to learn and use. Python is designed to process large amounts of data where critical understanding of the latest research can be found, whether it be in Chemistry, Biology, or Business Analytics. It is considered the best programming language choice for computational chemistry, an essential part of the tool kit for biologists of all types, and as the language with the flexibility and speed needed for business analytics to make sense of big data and answer business' most important questions. Prerequisite: QL-156
This introductory course in programming introduces the student to an object-oriented program design paradigm. With Java a student can create World Wide Web and stand-alone applications. She develops a number of projects that lead to an independent final project. This course assumes no previous experience with programming.
Prerequisite(s): CM 156Q completed
MT-297 Independent Study (1 credit)
Under the approval and direction of a faculty member, independent study is available to students.
MT-340 History of Mathematics (2 credits)
Offered Spring Term only. The student studies the history of mathematics as it is embedded in the development of world cultures. Contemporary mathematical events and trends are placed in historical context. The student spends significant time analyzing problem-solving methods and using them to solve problems from given times and cultures.
In this course, the student works with Euclidean geometry in two dimensions. She uses visualization, spatial reasoning, and geometric modeling to solve problems. Technology tools are employed to explore ideas and generate conjectures, leading to mathematical proofs.
Course Offered Fall Term alternate years only. The student learns to identify abstract algebraic structures such as groups, rings, and fields and to use their defining axioms. She explores the properties of these systems and examines others, applying the properties. Foundational work involves sets, mappings, and relations. The student gains experience reading and applying theorems, examining proofs for understanding, and constructing her own proofs.
Offered Spring Term only in alternate years. The student learns about differential equations as descriptions of changing phenomena. She studies solutions from several perspectives, surveying basic analytic methods for solving differential equations, learning to use graphical and qualitative approaches to analyze behavior of solutions, and using the computer to obtain numerical solutions. She works to interpret mathematical results in realistic contexts.
Under the approval and direction of a faculty member, independent study is available to students.
MT-399 Formal Introduction to Advanced Work (0 credits)
Course Offered Fall Term only. TheOffered Fall Term in alternate years. The student learns about the structure and scope of mathematical axiom systems in the context of modern geometries. She expands her analytic-thinking and problem-solving abilities as she reads, understands, and writes mathematical theorems and their proofs.
Course Offered Fall Term only. The mathematics education student develops an in-depth understanding in the area of algebraic thinking, with emphasis given to proportional reasoning. Other middle-school mathematics topics, such as geometry in two and three dimensions and probability and statistics, may be studied. The historical development of the elementary and middle school curriculum is examined, with emphasis on the teaching, learning, and assessment processes highlighted in local, state, and national standards documents.
Students will prepare a portfolio that demonstrates that they have met the outcomes for the mathematics major. Students will meet with the mathematics faculty to discuss their portfolio.
Prerequisite(s): Taken by Mathematic Majors in their second to last semester.
MT-460 Introduction to Real Analysis (3 credits)
Offered Spring Term in alternate years. In this course, the student studies functions of real variables from an advanced viewpoint. She examines the concepts of sequence, limit, continuity, and derivative in a mathematically rigorous setting. She gains experience in mathematical thinking and writing, developing an appreciation of the nature and role of mathematical proof. | 677.169 | 1 |
Description
Appropriate for one- or two-semester Advanced EngineeringMathematics courses in departments of Mathematics and Engineering. This clear, pedagogically rich book develops a strong understanding of the mathematical principles and practices that today's engineers and scientists need to know. Equally as effective as either a textbook or reference manual, it approaches mathematical concepts from an engineering perspective, making physical applications more vivid and substantial | 677.169 | 1 |
For courses covering general topics in math course, often called liberal arts math, contemporary math, or survey of math. Everyday math, everyday language. The Tenth Edition of A Survey of Mathematics with Applications continues Also available with MyMathLab My NEW! This edition's MyMathLab course provides additional tools to help with understanding and preparedness115767 / 9780134115764 * A Survey of Mathematics with Applications plus MyMathLab Student Access Card -- Access Code Card Package Package consists of: 0134112105 / 9780134112107 * A Survey of Mathematics with Applications 0321431308 / 9780321431301 * MyMathLab -- Glue-in Access Card 0321654064 / 9780321654069 * MyMathLab Inside Star Sticker
In a Liberal Arts Math course, a common question students ask is, "Why do I have to know this?" A Survey of Mathematics with Applications continues to be a best-seller because it shows students how we use mathematics in our daily lives and why this is important. The Ninth Edition further emphasizes this with the addition of new "Why This Is Important" sections throughout the text. Real-life and up-to-date examples motivate the topics throughout, and a wide range of exercises help students to develop their problem-solving and critical thinking skills. Angel, Abbott, and Runde present the material in a way that is clear and accessible to non-math majors. The text includes a wide variety of math topics, with contents that are flexible for use in any one- or two-semester Liberal Arts Math course.
Economics impacts every sector of the marketplace. A solid understanding of basic economic principles will give you better insight into the market, industry, and world around you--equipping you with the tools to make more informed decisions as both a professional and a consumer. Noted for its clarity, sensibility, and focused approach, Tucker's SURVEY OF ECONOMICS, Sixth Edition, emphasizes the basics of economics, presenting concepts in the context of real-world situations--and helping you apply them to your own environment. Dr. Tucker emphasizes application and student learning, making the study of economics accessible, interesting, and--with its new four-color format--eye-catching. An award-winning instructor, Dr. Tucker draws from his own experience to implement unique teaching tools and methodologies that drive the learning process and help you sharpen your critical analysis skills--which will benefit you in any career path. His book is renowned for its lively presentation, active learning environment, highly motivational pedagogy, unparalleled visual support, and thorough self-check reviews. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
SURVEYING: PRINCIPLES & APPLICATIONS, 9/e is the clearest, easiest to understand, and most useful introduction to surveying as it is practiced today. It brings together expert coverage of surveying principles, remote sensing and other new advances in technological instrumentation, and modern applications for everything from mapping to engineering. Designed for maximum simplicity, it also covers sophisticated topics typically discussed in advanced surveying courses. This edition has been reorganized and streamlined to align tightly with current surveying practice, and to teach more rapidly and efficiently. It adds broader and more valuable coverage of aerial, space and ground imaging, GIS, land surveying, and other key topics. An extensive set of appendices makes it a useful reference for students entering the workplace.
Survey Sampling Theory and Applications offers a comprehensive overview of survey sampling, including the basics of sampling theory and practice, as well as research-based topics and examples of emerging trends. The text is useful for basic and advanced survey sampling courses. Many other books available for graduate students do not contain material on recent developments in the area of survey sampling. The book covers a wide spectrum of topics on the subject, including repetitive sampling over two occasions with varying probabilities, ranked set sampling, Fays method for balanced repeated replications, mirror-match bootstrap, and controlled sampling procedures. Many topics discussed here are not available in other text books. In each section, theories are illustrated with numerical examples. At the end of each chapter theoretical as well as numerical exercises are given which can help graduate students. Covers a wide spectrum of topics on survey sampling and statistics Serves as an ideal text for graduate students and researchers in survey sampling theory and applications Contains material on recent developments in survey sampling not covered in other books Illustrates theories using numerical examples and exercises 0321935446 / 9780321935441 Mathematics with Applications In the Management, Natural, and Social Sciences Plus NEW MyMathLab with Pearson eText1931076 / 9780321931078 Mathematics with Applications In the Management, Natural and Social Sciences
Susanna Epp's DISCRETE MATHEMATICS WITH APPLICATIONS, FOURTH EDITION provides a clear introduction to discrete mathematics. Renowned for her lucid, accessible prose, Epp explains complex, abstract concepts Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
For Principles of Management courses. The Practical Tools of Management Presented Through In-depth Practice Fundamentals of Management is the most engaging and up-to-date introduction to management resource on the market today. Covering the essential concepts of management, it provides a solid foundation for understanding the key issues and offers a strong, practical focus, including the latest research on what works for managers and what doesn't. The Tenth Edition has been updated with the latest coverage on hot topics such as sustainability, holacracy, the sharing economy, gamification, data analytics/big data, BYOD (bring your own device), and wearable technology. Engaging and fun videos and exercises motivate readers and give them the practice they need to become successful managers. Also Available with MyManagementLabTMManagementLab does not come packaged with this content. If you would like to purchase both the physical text and MyManagementLab search for: 0134303172 / 9780134303178 Fundamentals of Management: Essential Concepts and Applications Plus MyManagementLab with Pearson eText -- Access Card Package Package consists of: 0134237471 / 9780134237473 Fundamentals of Management: Essential Concepts and Applications 0134240693 / 9780134240695 MyManagementLab with Pearson eText -- Access Card -- for Fundamentals of Management: Essential Concepts and Applications | 677.169 | 1 |
Overview Features a unique guided approach, with an introduction to explain each skill in student-friendly language, a step-by-step process that demonstrates how to work an example, and computation and word problems for achieving mastery. The Solutions Key gives students the correct answer as well as an explanation so they understand the problem's solution | 677.169 | 1 |
Algebra B Math: Complex Numbers and the Quadratic Formula
In this topic, students are introduced to the division of complex numbers. The student also learns how to solve all quadratic equations. This series of four lessons provides the student with the tools to solve any quadratic equation.
Dividing Complex Numbers – The Complex Conjugate
This lesson introduces the student to dividing complex numbers. The complex conjugate is explained. The student learns how to use the complex conjugate to divide complex numbers.
The Discriminant of the Quadratic Formula
This lesson introduces the student to the discriminat of the quadratic formula. The student learns the equation for the discriminat. The student also learns the role the discriminat plays in determining the number of solutions and whether the solutions will be real or complex.
The Quadratic Formula with Complex Solutions
In this lesson, the student learns how to use the quadratic formula to solve quadratic equations where the solutions are complex. The student learns they can solve all quadratic equations using the quadratic formula, whether the solutions are real or complex.
Solving General Quadratic Equations
This lesson teaches the student how to solve quadratic equations that are not already written in standard form. From this lesson, the student can solve any quadratic equation. | 677.169 | 1 |
What is a function?
How do you define function? Do you teach relation first before teaching function? Does knowing about relation a pre-requisite to function understanding?
The concept of function "was born as a result of a long search after a mathematical model for physical phenomena involving variable quantities" (Sfard, 1991, p. 14). In 1755, Euler (1707-1783) elaborated on this conception of function as a dependence relation. He proposed that, "a quantity should be called a function only if it depends on another quantity in such a way that if the latter is changed the former undergoes change itself" (p. 15). Seventy-five years later, Dirichlet (1805-1859) introduced the notion of function as an arbitrary correspondence between real numbers. About a hundred years later in 1932, with the rise of abstract algebra, the Bourbaki generalised Dirichlet's definition. Thus, function came to be defined as a correspondence between two sets (Kieran, 1992). This formal set-theoretic definition is very different from its original definition. Function is no longer associated with numbers only and the notion of dependence between two varying quantities is now only implied (Markovits, Eylon, & Bruckheimer, 1986). The Direchlet-Bourbaki definition allows function to be conceived as a mathematical object, which is the weakness of the early definition. However, the set-theoretic definition is too abstract for an initial introduction to students and is inconsistent with their experiences in the real world (Freudenthal, 1973; Leinhardt, Zaslavsky, & Stein, 1990; Sfard, 1992).
Textbooks, which often define function as a set of ordered pairs usually start the discussion with relation and introduce function as a special kind of relation. But relation is more abstract than function. Thus the supposed pedagogical value of having to learn relation first before one understands function is, in the opinion of Thorpe (1989), wrong. Freudenthal (1973) also expressed strongly that "to introduce function, relations can be dismissed" (p. 392). Thorpe went on to say that the use of the set-theoretic definition which defines function as a set of ordered pairs "was certainly one of the errors of the sixties and it is time that it were laid to rest" (p. 13). Amen to that.
8 Responses to What is a function?
You never mention the notion that a function maps a given input value to a single output value. Whereas any "correspondence between two sets" will be a relation, not all are functions. A relation can map an input value to multiple outputs. I perceive this as the critical distinction between functions and relations, one that is necessary to establish before the notions of continuity and differentiability can be tackled without (additional) potential confusion. I have not researched the history as you have, so I would be interested in learning when this distinction came to be accepted.
Maybe I should have. I assumed the readers already know that. What I was trying to point out in my post are the two notions of function: dependence vs correspondence and that the former is easier to make sense of for students initially hence we should not start our teaching with the notion of correspondence.
It was not mentioned explicitly when function was formally defined function with that condition that for every x, there's only and only one y in the materials I considered. I think right at the very start that condition is already there because there lies the power of this mathematical model. It would not be a very useful model for predicting if not for that. My theory is that function came before relation in the consciousness of mathematicians. Relation, I suppose was born during the time when they want to define function formally and logically. Definitions of concepts should always be part of a bigger set. Hence we have "function is a relation …." Indeed it's more mathematical sounding definition if you say that.
Another potentially interesting angle to the "function vs relation" concept would be to take a statistical approach: statistics is full of correlations, but proving cause and effect is a completely different matter. Correlations without cause & effect are comparable to relations. Correlations WITH cause & effect are functions providing that any differing output values for a given input are measurement errors (i.e. correlations that are not perfect are due entirely to measurement errors).
I guess the notion of a function producing one output for a given input seems more intuitive to me than the notion of a relation… then again, I had been programming computers for several years before functions were introduced in math class – so my thinking was already warped into seeing the world around me as a series of functions… | 677.169 | 1 |
This course aims to convey the richness, diversity, connectedness,
depth and pleasure of mathematics. The title is taken from the classic
book by Hilbert and Cohn-Vossen, ``Geometry and the Imagination'. Geometry is taken in a broad sense, as used by mathematicians, to
include such fields as topology and differential geometry as well as
more classical geometry. Imagination, an essential part of
mathematics, means not only the facility which is imaginative, but also
the facility which calls to mind and manipulates mental images. One aim
of the course is to develop the imagination.
While the mathematical content of the course will be high, we will try
to make it as independent of prior background as possible. Calculus,
for example, is not a prerequisite.
We will emphasize the process of thinking about mathematics.
Assignments will involve thinking and writing, not just grinding
through formulas. There will be a strong emphasis on projects and
discussions rather than lectures. All students are expected to get
involved in discussions, within class and without. A Geometry Room
on the fifth floor will be reserved for students in the course. The
room will accrete mathematical models, materials for building models,
references related to geometry, questions, responses and (most
important) people. There will be computer workstations in or near the
geometry room. You are encouraged to spend your afternoons on the
fifth floor.
The spirit of mathematics is not captured by spending 3 hours solving
20 look-alike homework problems. Mathematics is thinking, comparing,
analyzing, inventing, and understanding. The main point is not
quantity or speed-the main point is quality of thought. The goal is
to reach a
more complete and a better understanding. We will use materials such
as mirrors, Polydrons, scissors and tissue paper not because we
think this is easier than solving algebraic equations and differential
equations, but because we think that this is the way to bring thinking
and reasoning to the course.
We are very enthusiastic about this course, and we have many plans to
facilitate your taking charge and learning. While you won't need a
heavy formal background for the course, you do need a commitment of
time and energy. | 677.169 | 1 |
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
calculus for physics, vector calculus part 1
What's a vector, what's the difference between a vector and a scalar ? an electricity and magnetism course, or who have never had an advanced course on vector calculus. Anyone can use it just needs a HighSchool Diploma level, this is a free course with 34 minutes videos.
Before any electricity and magnetism course we need mathematical tools like :
Vector calculus
Coordinate system
Relation between cartesienne and cylindrical coordinates
Differential calculus
Vector operators
In this course we present all this concept for beginners
Our method is detailed the calculation from A to Z .
We start with defintion of a vector and the difference with a vector and scalar, after that we see a sample example, we will see representation of a vector, the magnitude of a vector, addition of vectors and finally vector multiplication | 677.169 | 1 |
College Business Math Textbooks
Culinary Math Worksheets
Download On Cooking A Textbook of Culinary Fundamentals 5th Edition.
Download and Read Culinary Artistry. culinary calculations simplified math for culinary. culinary-calculations-simplified-math-for-culinary-professionals.pdf.Dazzling Decimals and Percentages: Games and Activities that Make Math Easy and Fun by Lynette Long.
Culinary Math Conversion Chart Recipe
Professional Cooking 7th Edition
Colette Peters Cakes to Dream On
Food and Culinary Arts (Field Guides to. questions and interviews with professionals in the field to.CURRENTLY SOLD OUT Culinary Calculations: Simplified Math for Culinary Professionals by Terri Jones (2003, Paperback). | 677.169 | 1 |
This study guide is a must have reference when teaching and reviewing for the Texas STARR Algebra I End of Course Exam or any Algebra I EOC. I created this product after working with average students and special populations. I address the most frequent areas of confusion and difficulty. I added, changed then edited again to give you a most effective tool for students. Print it as a wall poster for easy reference or print it in black & white then have students use their own colored pencils to connect and remember! Your students will use it and love it! Also, use it as a quick reference when teaching Algebra II. Try it! | 677.169 | 1 |
Comparing Functions Activity
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This activity forces students to look at functions in different modes including a graph, equation, and a table. The students then have to compare the functions in their different states and analyze the information presented in all forms. | 677.169 | 1 |
Complete Mathematics is written by experienced Cambridge examiners to give students complete preparation for the revised Cambridge Secondary 1 curriculum. Methodical, logical and closely matching the order of the new framework this series offers rigorous and thorough preparation for the Cambridge Checkpoint test, as well as a flying start for the IGCSE. Internationally focused and relevant, all questions have been carefully developed to build knowledge and really stretch and challenge students. Complete Mathematics is the new name for Oxford International Maths.
Deborah is an experienced teacher, principal examiner and maths consultant with a passion for active learning in mathematics. She has a great deal of experience working with gifted and talented students using innovative methods to stretch those with high ability in the subject. Deborah has experience as an author, writing activities for GCSE and A level and has written Content and Language Integrated Learning resources for ESOL students. She also conducts endorsement reviews for maths books for CIE.
Review:
These are the best Maths resources for this level that I have ever taught from. We love the homework books which offer extra practice to consolidate core content. * Carolyn Hatton, Head of Mathematics, King Richard III College, Spain * Our middle school teachers have been very happy with the format and are looking forward to using this resource next academic year. Some of the things they like about it are that it does have more practice problems than our present resource and that all the topics explored at each level are in a single book which makes it easier for students to manage. * Phil Bennett, Academic Leader Mathematics, International School of Luxembourg * This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. BTE9780199137053 | 677.169 | 1 |
This textbook is designed to college sophomores and juniors the basics of linear algebra and the techniques of formal mathematics. There are no prerequisites other than ordinary algebra. The text has two goals: to teach the fundamental concepts and…
Essential Physics 1, is an intensive introduction to classical and special relativity, Newtonian dynamics and gravitation, Einsteinian dynamics and gravitation, and wave motion. Mathematical methods are discussed, as needed; they include: elements of…
This book is a survey of abstract algebra with emphasis on linear algebra. It is intended for students in mathematics, computer science, and the physical sciences. The first three or four chapters can stand alone as a one semester course in abstract… | 677.169 | 1 |
Archive for the 'Math' Category
Yes, it is true that you can raise your math grade without actually learning another formula, proof, or mastering your multiplication table (even though that would be a good idea!) Follow these 10 tips and approach math homework in an organized and systematic method, and we guarantee your grades will rise.
These little things can make a huge difference at the end of a semester.
Clearly label homework at the top of your homework with section, pages, problems, and date – this will help with organization in your folder as well.
Be NEAT! In algebra, go for the "V-shape." If you are drawing graphs, shapes, or number lines, then use a ruler.
NEVER use a pen. Always use a pencil.
Give yourself space to work– you can recycle your homework at the end of the year
Show all of your steps, and model your answers after the book.
Use graph paper when required.
Always, box or circle your answer.
Listen to what your teacher wants (they are the ones grading it) and showing all your work, circling your answers, and putting it in the correct final form (fraction, decimal, etc.) is critical
Finally, check your answer. Do this yourself, or look for help from the book – worked out problems and answers in the back.
Got it wrong? Look back at the examples and follow the same steps. Still no help? Call a friend, ask your teacher, or go on-line for other examples
A few words on calculators…
Most teachers will let you use calculators for your homework, and on exams. This doesn't mean that you have to use a calculator for EVERYTHING! Use a calculator only for large, complex calculations. Do not use a calculator for algebra, ever, even if you can program it to do it. If you do multiplication and division on paper it will help you down the road. Be Careful! Always ask, "Does that make sense?" when you get an answer. It is easy to make mistakes when using a calculator. Don't blindly trust the result. | 677.169 | 1 |
Our sources of geometric thought and writing go back to
ancient civilizations, most notably the Greek and the Chinese. As an introduction
to this vast field, we recommend the book by Hilbert and Cohn-Vossen [1],
which illustrates the variety of geometric thinking within mathematics.
We group our recommendations into six subareas.
Elementary Geometry. This
field studies simple geometric figures and their relationships. We mention
Coxeter and Greitzer [2] as an excellent source. We also
recommend the book by Pedoe [3], which contains a wealth
of material on circles and spheres.
Discrete Geometry. This
area in geometry is dominated by the Hungarian school of thought, which
includes work on packings and coverings as studied by Laszlo Fejes-Toth
and combinatorial extremum problems as popularized by Paul Erdos. We recommend
the text by Pach and Agarwal [4], which discusses a broad
range of problems and results in the area.
Convex Geometry. Convex
sets have been studied from a variety of angles. A big topic in this field
is convex polytopes, and we recommend the relatively recent text by Ziegler
[5]. Other topics include cross-sections and projections
of convex bodies and their measurement, and we recommend the book by Schneider
[6] as an introduction to that literature.
Stochastic Geometry. The
combinatorial viewpoint in discrete geometry is related to, but not the
same as the probabilistic focus we find in stochastic geometry. A fairly
recent book on the topic is Klain and Rota [7]. A related
area is the study of shape spaces understood as the set of deformations
of a collection of measurements. We mention the text by Kendall [8]
as an introduction to this school of thought.
Differential Geometry. This
is probably the largest subarea in geometry. At the more elementary end
of the spectrum it includes vector calculus, and at the other end it connects
to Riemannian geometry and other advanced topics. We recommend the text
by O'Neill [9] for a general introduction, and the book
by Morgan [10] for an introduction to Riemannian geometry.
Singularity Theory. This
is another advanced topic in differential geometry and concerns itself
with the classification of metamorphisms that occur in the generic deformation
of smooth shapes. A delightful text in this field is the book by Bruce
and Giblin [11], which studies curves and surfaces.
Catastrophe theory is a synonymous name for the field and also the title
of an introductory booklet by Arnol'd [12]. | 677.169 | 1 |
Summary
Algebra is the study of operations (such as addition,
multiplication, composition) on sets of objects (such as numbers,
polynomials, matrices, permutations). In addition to studying
specific operations on specific sets, we also abstract properties
that such operations commonly satisfy and the implications of
these properties, thereby unifying the study of a wide variety of
mathematical objects. In addition to being a beautiful subfield of
mathematics, algebra has numerous applications in science and
engineering. It is extremely useful for studying symmetries of
physical objects, and for encoding data and computations to
provide properties such as error-correction and privacy.
In this course, we will cover:
The basics of abstract algebra (groups, rings, fields)
Algorithmic aspects of algebra: which algebraic problems have
(efficient) algorithms, and which do not.
The
topics above might be too much to cover in the course, but we will
try to get at least a sample of all the above.
Prerequisites
The formal prerequisite for the course is
(Applied) Math 21ab or equivalent, but general "mathematical
maturity" is more important than the specific material in these
courses. At times,
we will assume familiarity with basic linear algebra as covered in
Math 21b, but students who have instead taken a prior proof-based
course on a different topic (such as AM 107, Math 101, CS 121, or
CS 124) should also be adequately prepared.
Grading
Weekly problem
sets: 50%(lowest
score dropped)
Two in-class
quizzes: 10% each
Final exam: 25%
Class
participation: 5%
Your class participation grade is based on participation in
lecture, but can also be boosted by participation in section
and/or coming to office hours or section with questions or
comments that show genuine interest in the material (i.e. is not
just aimed to help you answer questions on the problem set or
exam). Do not be afraid of asking "stupid" questions!
Problem
Sets
& Collaboration Policy
The course will have weekly problem sets, due by 11:59PM sharp electronically
via Canvas. You are allowed 6 late days for the semester, of
which at most 2 can be used on any individual problem set. (1
late day = 24 hours exactly). For any exceptions to these
rules, I require a note from your senior tutor.
Students are encouraged to discuss the course material and the
homework problems with each other in small groups (2-3
people). Discussion of homework problems may include
brainstorming and verbally walking through possible solutions, but
should not include one person telling the others how to solve the
problem. In addition, each person must write up their
solutions independently, and these write-ups should not be checked
against each other or passed around.
While
working on your problem sets, you may not refer to existing
solutions, whether from other students, past offerings of this
course, materials available on the internet, or elsewhere.
All sources of ideas, including the names of any collaborators,
must be listed on your homework paper.
Sections
There will be weekly sections, which will be
used to clarify difficult points from lecture, review background
material, go over previous homework solutions, and sometimes
provide interesting supplementary material.
Readings
The handouts for the course will be made
available at the course website. These will be required
reading. The recommended text is:
It has been ordered at the Coop, and placed on reserve in the
libraries. This book will include most of the topics covered in the
course, and a lot more too, but it will not cover all the topics in
the course. In particular it does not cover some of the applications
and the algorithmic discussions. So it is important that you also
attend lecture.
Related Courses at Harvard
Math 122
&123: Algebra I & II. A full-year course in
abstract algebra. Because it is longer and assumes more
background, Math 122-123 covers a number of topics that we
cannot fit in AM 106, such as group representations, modules,
and Galois theory. (Students
taking AM 206 can study some of these topics for their
additional assignments.) On the other hand, it usually
does not include the algorithmic aspects and applications of
algebra that we will cover in AM 106.
Math 152:
Discrete Mathematics. A seminar-style course covering a
variety of related topics in abstract algebra and discrete
mathematics. My guess is that it has significant but not
complete overlap with the "general theory" we cover, but that it
has not much overlap with the applications and algorithmic
issues we cover.
Computer
Science theory (x2x) courses. A number of the applications
and algorithmic aspects of algebra that we will cover (and
others) also appear scattered in various theoretical computer
science courses. In AM 106, we will have time to do a more
systematic treatment, in which these applications are integrated
with the general study of abstract algebra (rather than taking
algebraic facts on faith, or doing a quick "crash course"). | 677.169 | 1 |
Calculus--Basic Integral Answer Hunt
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Are you tired of students just sitting around doing book work or worksheets? Then my Answer Hunt activities are for you. Students must work with partners to "hunt" for answers to various problems around the room.
This activity has basic integrals. Using no more than the integral power rule and trig integrals, students will work to find the integrals of various functions.
Students start with a poster and complete the integral on the provided sheet; they then search for the poster that matches that answer. They then do the integral there. You get the idea now, right? | 677.169 | 1 |
Description
Clear prose, tight organization, and a wealth of examples and computational techniques make Basic Matrix Algebra with Algorithms and Applications an outstanding introduction to linear algebra. The author designed this treatment specifically for freshman majors in mathematical subjects and upper-level students in natural resources, the social sciences, business, or any discipline that eventually requires an understanding of linear models. With extreme pedagogical clarity that avoids abstraction wherever possible, the author emphasizes minimal polynomials and their computation using a Krylov algorithm. The presentation is highly visual and relies heavily on work with a graphing calculator to allow readers to focus on concepts and techniques rather than on tedious arithmetic. Supporting materials, including test preparation Maple worksheets, are available for download from the Internet. This unassuming but insightful and remarkably original treatment is organized into bite-sized, clearly stated objectives. It goes well beyond the LACSG recommendations for a first course while still implementing their philosophy and core material.
Classroom tested with great success, it prepares readers well for the more advanced studies their fields ultimately will require.show more
Review quote
"In brief, I think the book is wonderful! It's pedagogically excellent The examples are very thoughtfully chosen, anticipating possible misunderstandings and illuminating both the main idea and some subtleties The emphasis on recursion is unusual and valuable It's rich in topics not often, if at all, treated effectively in texts at this level Again, I think this book is terrific" -Harriet Pollatsek, Professor of Mathematics and Julia and Sarah Ann Adams Professor of Science, Mount Holyoke College, South Hadley, Massachusetts, USAshow more | 677.169 | 1 |
Editorial Reviews
From the Back Cover
The present book is intended as a text in basic mathematics. As such, it can have multiple use: for a one-year course in the high schools during the third or fourth year (if possible the third, so that calculus can be taken during the fourth year); for complementary reference in earlier high school grades (elementary algebra and geometry are covered); for a one-semester course at the college level, to review or to get a firm foundation in the basic mathematics to go ahead in calculus, linear algebra, or other topics.
Top customer reviews
I love the way this book is written! It really circumvents a lot of the dry technical material that you see in a lot of math textbooks but still gives you the most necessary information. The exercises are great for practicing the skills covered. This book is good to have on reference as well as to use for a prep for higher level math like calculus.
If you are going back to college get this book it covers almost everything you need to know minus some trig. and things like summations. Also great if you want to get an idea of what proofs are and a feel for more rigorous math. Almost every subject is covered more succinctly and clearly that it will be the mast majority of professors and textbooks and cheaper too.
This text is awesome. It has wonderful explanations of mathematics ranging from simple addition/subtraction properties to working with polar format. Great introductory text for mathematics. Lang is clear and consistent.
Serge Lang has written one of the better books out there for learning high-school level math well. The explanations are clear and the problems are graded from easy to hard well. It covers everything you would need to know to start higher-level math studies. Some of the problems are really hard and require a good deal of thought to solve. I would recommend this book to anyone, but especially future or current math majors. The reason that this book gets 4 stars is because several sections have few answers in the back of the book.
Very good as a reference book for all those subjects I studied long ago. But unlike some other books I haven't found this one to be "fun." It's not the one I'd recommend if you're looking for something you can read cover to cover. It could certainly be done, there's just other books I've found to be better suited for working your way through like that (Stillwell's Numbers and Geometry or Hardy's Course of Pure Mathematics maybe).
This one is a pleasure to dip into when you've got a specific topic you need to review (i.e. trig or something) but if you're looking to buy one book to inspire you and take your mathematical knowledge and thinking to new depths, I'd go with something else. | 677.169 | 1 |
provides students with the elementary mathematical tools that are needed to study economics. Students will be introduced to basic mathematical concepts and manipulations. Examples of applying these techniques to a variety of economical and management problems are given throughout the course. | 677.169 | 1 |
Examining the pioneering ideas, works, and applications that have made math the language of science, Mathematics and the Laws of Nature looks at the many ways in which so-called ''pure'' math has been used in the applied sciences. For example, the volume explores how mathematical theories contributed to the development of Kepler's laws of planetary... more...
The Kenneth May Lectures have never before been published in book form Important contributions to the history of mathematics by well-known historians of science Should appeal to a wide audience due to its subject area and accessibility more...
Writings by early mathematicians feature language and notations that are quite different from what we're familiar with today. Sourcebooks on the history of mathematics provide some guidance, but what has been lacking is a guide tailored to the needs of readers approaching these writings for the first time. How to Read Historical Mathematics fills... more...
The updated new edition of the classic and comprehensive guide to the history of mathematics For more than forty years, A History of Mathematics has been the reference of choice for those looking to learn about the fascinating history of humankind's relationship with numbers, shapes, and patterns. This revised edition features up-to-date... more...
This book provides students of mathematics with the minimum amount of knowledge in logic and set theory needed for a profitable continuation of their studies. There is a chapter on statement calculus, followed by eight chapters on set theory. more... | 677.169 | 1 |
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