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Lesson content has been expanded and includes more video instruction for each lesson. In addition to the video examples each lesson includes interactive practice problems, challenge questions, and worksheets.
Description
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Alison has three diverse interests: mathematics, German, and music. They may seem unrelated on the surface, but she managed to tie them together during her time as a WLC student and continues to do so as a TA at Purdue University...
Mathematics
Mathematics extends beyond mere numerical computation. Students learn to work with abstract concepts and relationships, to recognize patterns, to analyze complexity, and to communicate their ideas professionally.
As an application tool, mathematics uncovers the structures and relationships that help us understand
the nature and problems of the world around us. Mathematics expresses its concepts in symbols and
graphic representations within a framework of logical inference that serves as the universal language
for communicating technical knowledge. A major in mathematics is one of the finest courses of study
for developing ability in analytical thinking. Combined with an appropriate concentration of application
courses, a major in mathematics prepares the student for a wide range of career and educational
opportunities.
As an academic discipline, mathematics shapes and sharpens the rational capabilities of the mind. It is
a mode of thinking that empowers the individual to absorb new ideas, adapt to rapid change, cope with
ambiguity, recognize patterns, solve unconventional problems, analyze complexity, work with abstract
concepts and relationships, detect bias, assess risks, and suggest alternatives. Increasingly these are the
capacities of mind that are needed for responsible citizenship and successful careers.
What Makes the Program Distinctive?
WLC's mathematics program is led by professors who have been recognized for excellence in teaching and
who provide students a solid foundation in mathematics. Students will learn from experts in the mathematics
discipline, as well as study from textbooks written by WLC mathematics faculty. In addition, students have
an opportunity to study cryptography, a practical course of study for those interested in computer security.
Cryptography is not commonly found in the undergraduate mathematics curriculum.
Interdisciplinary Opportunities
The mathematics program at WLC can serve as a stand-alone major or be combined with an additional major or minor for interdisciplinary study. In the past, mathematics majors have included study in computer science, business, and education to their overall study at Wisconsin Lutheran College. Students studying mathematics and secondary education have traditionally scored high within the mathematics portion of the PRAXIS standardized test. A recent mathematics minor and psychology major is enrolled in the doctoral psychometric methods program at the University of Minnesota, one of the renowned programs in the nation.
BEYOND THE CLASSROOM
Competitions
Mathematics majors at WLC have repeatedly performed well in various regional, state, and national competitions. Students recently earned high marks in the Face Off! mathematics competition held at St. Norbert College.
Campus Opportunities
WLC math majors have an opportunity to participate in Math Club, a student organization designed to provide further information about mathematical careers, graduate school studies, and offer social activities for those interested in math. The mathematics department also employs several student tutors, an opportunity that improves credentials for graduate school and employment applications.
GRADUATE SCHOOL AND CAREERS
Graduate School
WLC's mathematics major boasts an unusually high percentage of students electing to continue studies at the doctoral level. In addition, the major successfully prepares students for acceptance to some of the finest universities in the nation. In recent years, WLC's mathematics majors have been offered scholarships and teaching assistantships for doctoral programs at a number of universities, including:
Georgia Institute of Technology
Iowa State University
Oklahoma State University
Purdue University
The Ohio State University
University of Nebraska
University of North Carolina-Chapel Hill
University of Washington-Seattle
University of Wisconsin-Madison
Washington University in St. Louis
Students also may pursue graduate study in pure or applied mathematics, operations research, statistics, cryptography, business, computing, and economics as well as many areas in the natural, life, and social sciences.
Careers
According to JobsRated.com, mathematicians and statisticians rank among the highest for levels of job satisfaction. In addition, the National Association of Colleges and Employers ranks mathematics majors as having the second highest median starting salary among bachelor's degree employees, second only to computer science. WLC mathematics majors currently work as university professors and U.S. Defense Department employees, as well as serve in positions within such industries as banking and finance, computer programming, software development, and engineering. Graduates currently work at corporations in the Milwaukee area, including U.S. Bank and Northwestern Mutual. | 677.169 | 1 |
At a Glance - Logarithmic Functions
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Math Courses
TAP TO EXPAND +
For an official and updated listing of courses, please refer to the current academic catalog.
MAT 101-College Mathematics (3)
Logical structure of the decimal system. Designed to acquaint the student with meaning, development, and communication of number ideas and the logical structure of number systems; the how and why of the basic algorithms of arithmetic. Fundamental concepts of elementary algebra and informal geometry. Prerequisite: Math SAT scores 200-400.
MAT 110-College Algebra (3)
The study of topics from Algebra including basic concepts, equations and inequalities of the first and second degree, functions and graphs, linear and quadratic functions, higher degree polynomial and rational functions, exponential and logarithmic functions, systems of equations and inequalities, sequences and series. Prerequisite: Math SAT 400-525.
MAT 211-Math for the Elementary School Teacher (3)
The study of topics from the National Council of Teachers of Mathematics for grades K-6 will be covered. Such topics will include math as problem solving, math as communication, estimation, numbers and numeration, probability and statistics. Education majors only. Prerequisite: MAT 101 or equivalent.
An introductory course in the theory and applications of differential calculus including concept of limits, introduction to the derivative, techniques of differentiation, integration and integration methods, the fundamental theorem of the calculus, and the study of exponential and logarithmic functions. Prerequisite: MAT 110 or MAT 231 or instructor's permission.
MAT 237-Calculus II (3)
A course in advanced techniques in the application of calculus to a better understanding of God's world. Topics include calculating the areas bounded by curves, volumes of solids of revolution, arc lengths, and surface areas of various functions, trigonometric integrals, and L'Hopital's Rule. Prerequisite: MAT 236.
MAT 238-Calculus III (3)
The study of infinite series, parametric curves and vectors in the plane, vectors, curves, and surfaces in space, partial differentiation, and multiple integration. Prerequisite: MAT 237.
Informal and formal development of propositional calculus, predicate calculus, and predicate calculus with equality. The study of the completeness theorem and some consequences. Prerequisite: MAT 345.
MAT 458-Operations Management and Linear Programming (3)
(For course description see BUS 458)
MAT 470-Mathematics Seminar (3)
In-depth study of some topic or topics of current interest to Mathematics faculty and advanced students. Students will be expected to do independent research and to present their findings in a small-group setting. Instructor's permission required.
MAT 490-Internship (1-3)
The practical application of mathematical knowledge in an applied setting will be studied. The location and nature of the internship for the Learning Contract must be approved by the Department Head and the Dean. | 677.169 | 1 |
Find an AlamedaNot being ready, students can't study calculus successfully. Therefore, in Precalculus, students will be introduced to the important and basic mathematical concepts inquired before in algebra with deeper and higher details. They comprise, but not limited in, inequalities, equations, absolute values, and graphs of lines and circles | 677.169 | 1 |
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year, thousands of students in the USA declare mathematics as their major. Many are extremely intelligent and hardworking. However, even the best will encounter challenges, because upper-level mathematics involves not only independent study and learning from lectures, but also a fundamental shift from calculation to proof.This shift is demanding but it need not be mysterious — research has revealed many insights into the mathematical thinking required, and this book translates these into practical advice for a student audience. It covers every aspect of studying as a mathematics major, from tackling abstract intellectual challenges to interacting with professors and making good use of study time. Part 1 discusses the nature of upper-level mathematics, and explains how students can adapt and extend theirexisting skills in order to develop good understanding. Part 2 covers study skills as these relate to mathematics, and suggests practical approaches to learning effectively while enjoying undergraduate life.As the first mathematics-specific study guide, this friendly, practical text is essential reading for any mathematics major. less | 677.169 | 1 |
Pre-Algebra Mathematics Support Curriculum Unit 1: Wind Power Summary The first unit of the summer mathematics support program asks students to design a wind turbine ...
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Do you find it hard to solve certain mathematical exercises? Maths not your favourite? Then Algebra NET will help you, as it lets you do algebraic reductions, find factorials, do simplifications and solve equations.
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QUIZZES:
You should expect a 20 minute quiz each Friday (20 points).
I do not give makeup quizzes-if you
have a valid excuse for missing a quiz
you should check with me.
HONORS:
The honors part of the course (Math 297)
will use these notes:
The Calculus of Polynomials.
We will usually cover honors material on Tuesdays.
GENERAL ADVICE:
The key ideas of calculus are not the main barrier in the course--they
can be explained rather intuitively.
In my experience,
students have trouble in this course either
(1) because of a weak background in algebra or
(2) because they fall behind.
I can help with your algebra,
but you are responsible for keeping up,
by attending every class and doing all of the homework problems.
The WEB site
Understanding Mathematics: a study guide
has a good discussion about studying and learning mathematics.
There is also a list of
resources
on the WEB.
CALCULUS I (4)
A first course in calculus.
PRQ:
A level placement on the Math Placement Exam
or MATH 155 (with at least a C)
TEXT:Calculus,
5th Ed.,
by James Stewart, McMaster University
COURSE OBJECTIVES:
You are expected to acquire not only
computational facility with the topics introduced,
but also a basic understanding of the concepts and theory of calculus.
Mathematics is a truly universal language;
we want you to become more fluent in reading, writing,
and using this precise language 229H.
Introductory lecture
The study of calculus is divided into two parts:
(1) derivatives and (2) integrals.
The derivative of a function measures its rate of growth.
The integral of a function (over a given interval)
involves finding an average value over the interval.
The original motivation for studying derivatives came from physics,
at a time when people were trying to understand things in motion.
In this context, the derivative measures velocity,
which we can think of as an instantaneous rate of change.
The acceleration of an object measures
how fast its velocity is changing,
and so this can also be measured by a derivative
(the second derivative).
When working with the graph of a function,
the derivative can be very useful.
For a straight line, the derivative is just the slope of the line,
and a positive slope corresponds to a graph that moves up
(from left to right),
while a negative slope corresponds to a graph that moves down.
For more general functions,
just knowing the sign of the derivative is useful,
since the graph goes up when the derivative is positive,
and down when it is negative.
This allows to find maximum and minimum values for a function.
The indefinite integral is the opposite of the derivative.
It allows us to solve the following problem:
given a formula that tells the rate of growth at any point in time,
try to reconstruct the original function.
The definite integral is an averaging process.
For example, the something is moving at a constant rate,
the distance it travels is just the rate multiplied by the time it takes.
But what if the rate is variable?
That is where the definite integral can be used.
As another example,
in physics the work done by a force is defined to be
the magnitude of the force times the distance through which it acts
(provided the force is constant).
But even in stretching a spring the force is variable,
and this simple formula can't be used.
The definite integral of the force
finds the average force times the distance,
and that is why physicists usually define work in terms of an integral.
Finally, the integral allows us to find the area under a curve
because it can find an average height,
multiplied by the width of the interval.
Understanding the product rule
To take the derivative of the sum of two functions,
we use the rule that the derivative
of a sum is the sum of the derivatives.
Unfortunately, the rule for taking the derivative of a product of two functions
is more complicated.
A simple example for which we already know the answer
gives a good illustration of the difficulties.
Suppose that f(x) = x2 and g(x) = x3.
Then the individual derivatives are f '(x) = 2x and g '(x) = 3x2.
We know that the derivative of the product
f(x)g(x) = x2x3 = x5
should be 5x4,
but it is difficult to see how to combine f '(x) and g '(x) to get this answer.
It turns out that we also have to use f(x) and g(x) in the formula.
One approach to understanding the formula for differentiating products
comes from working with tangent lines.
Remember that the equation of the line through the point (a,b)
with slope m is
y = m(x-a) + b.
At the point (a,f(a)) on the curve y = f(x),
we have the y-coordinate b = f(a),
and the slope of the tangent line at this point
is given by m = f '(a).
This allows us to calculate
the equation of the line tangent to the curve y = f(x) at x=a as
y = f '(a)(x-a) + f(a).
In the same way, the line tangent to the curve y = g(x) at x=a is
y = g '(a)(x-a) + g(a).
To get the formula for the tangent line to y = f(x)g(x) at x=a,
we could try multiplying the two formulas.
We would get the following quadratic expression.
f '(a)g '(a)(x-a)2 + ( f '(a)g(a) + f(a)g '(a) )(x-a) + f(a)g(a)
Because the tangent line has to be linear,
this doesn't solve the problem,
unless we can change it in some way.
The tangent line is intended to be the line
that gives the best approximation to the curve,
near the point (a,f(a)).
When x is very close to a, the factor (x-a) is very small.
In comparison, the factor (x-a)2 is even smaller;
in fact, it is very much smaller.
For example, if x-a = .1, then (x-a)2 = .01;
if x-a = .01, then (x-a)2 = .0001;
if x-a = .001, then (x-a)2 = .000001.
Because the tangent line is just an approximation to the curve,
it turns out that we can ignore the quadratic term.
Notice that the part we had to ignore contains the term f '(a)g '(a),
which would have given us a "nice" formula.
Conclusion:
the tangent line to y = f(x)g(x) at x=a is
y = ( f '(a)g(a) + f(a)g '(a) )(x-a) + f(a)g(a).
The general formula is best given using x instead of a.
The Product Rule
The product rule for taking the derivative of the product of two functions
f(x) and g(x) can be stated this way:
d/dx ( f(x)g(x) ) = f '(x)g(x) + f(x)g '(x).
Besides remembering the product rule in the "visual" form given above,
it is good to remember it in words, as an algorithm.
The derivative of the product of two functions is equal to
the derivative of the first times the second,
plus the first times the derivative of the second.
Example 1.
Going back to the first example, we can now compute
d/dx ( x2 x3 )
= (2x)(x3) + (x2)(3x2)
= 2x4 + 3x4
= 5x4.
Our formula gives the expected answer,
but has to include f(x) and g(x) along with f '(x) and g '(x).
Example 2.
It is interesting to see the formula for the derivative
of the product of three functions.
It clearly shows the pattern in which we add up terms
where we differentiate one function at a time.
Understanding the chain rule
The rule for taking the derivative of the composition of two functions
is also complicated.
Again, a simple example for which we already know the answer
gives a good illustration of the difficulties.
Suppose that f(y) = y3 and u(x) = x4.
The formula for the composite function is
f(u(x)) = (x4)3.
The individual derivatives are f '(y) = 3y2 and
u '(x) = 4x3.
We know that the derivative of the composite function
f(u(x)) = x12
should be 12x11.
In this case we just need to substitute y=u(x) into f '(y)
and multiply by u '(x).
We can show this by looking at the tangent lines.
The line tangent to the curve y = u(x) at x=a is
y = u '(a)(x-a) + u(a).
At the point (b,f(b)) on the curve z = f(y),
the slope of the tangent line at this point is given by f '(b).
Thus the equation of the line tangent to the curve z = f(y) at y=b as
z = f '(b)(y-b) + f(b).
To compute the composite function f(u(x)),
we substitute the formula for u(x) into the formula for f(y).
If x=a, then we evaluate f(y) at y=u(a).
This means that in the equation of the tangent line,
we must take b=u(a).
The equation becomes
z = f '(u(a))(y-u(a)) + f(u(a)).
To get the formula for the tangent line to z = f(u(x)) at x=a,
we can substitute the formula
y = u '(a)(x-a) + u(a)
into the formula
z = f '(u(a))(y-u(a)) + f(u(a)).
This gives us
z = f '(u(a))(u '(a)(x-a) + u(a) -u(a)) + f(u(a))
or
z = f '(u(a))(u '(a)(x-a)) + f(u(a))
The derivative is the slope f '(u(a))(u '(a).
The general formula is best given using x instead of a.
The Chain Rule
The chain rule for taking the derivative of the composite of two functions
f(y) and u(x) can be stated this way:
d/dx ( f(u(x)) ) = f '(u(x)) u '(x).
Besides remembering the product rule in the "visual" form given above,
it is good to remember it in words, as an algorithm.
The derivative of the composite of two functions is found by
taking the derivative of the first function
and substituting in the second,
the multiplying by the derivative of the second.
Example. 1
Going back to the first example, we can now compute
d/dx ( (x4)3 )
= 3(x4)2(4x3)
= (3x8)(4x3)
= 12x11.
Our formula gives the expected answer.
It is important to remember to multiply the term
f '(u(x)) by u '(x).
Example. 2
To give another point of view,
we can check the chain rule in a case
where we can actually apply the product rule.
If f(u(x)) = u(x)3,
we can write | 677.169 | 1 |
College Algebra and Trigonometry / Edition 2
Overview use, and each section can be covered in one class. Clearly marked subsections make it easy to omit more basic topics when necessary. The material is carefully organized and paced, offering thoughtful explanations through a combination of examples and theory. Contains an excellent review of basic algebra, with coverage of equations and inequalities, graphs and functions, complex numbers and more. This edition contains more exercises requiring the use of a calculator, new and numerous examples, and end-of-section exercises that provide a good test of the student's progress.
Customer Reviews
Most Helpful Customer Reviews
College Algebra and Trigonometry 4 out of 5based on
0 ratings.
1 reviews.
Guest
More than 1 year ago
This book is your basic text covering all College Algebra and Pre-calculus topics. It covers the entire gammet of topics from basic algebra on up inluding trig, matrices, vectors, probabilty, etc. It's a good reference book for the basics, with lots of examples and tons of problems (with answers included). It is easy to understand and problems explanations are clear. I would recommend it as a good book for the youngsters (or oldsters) just learning the topic. It doesn't assume you have a vast background in algebra. | 677.169 | 1 |
Is Algebra too Easy?
There are math-phobes everywhere that are just shaking their heads, but bear with me. I just started tutoring a geometry student, and one of the first questions I asked him was how he did in Algebra. He said it was a piece of cake and now he's hitting a wall in geometry. Why? Because sometimes Algebra can be too easy.
You see it all the time - people skip to the end of the problem and they don't know how they got the right answer. Typical math teaching follows a pattern - introduce the concept, explain the concept, show a simple example and then get students to solve.
Head First books don't do that. And one of the big reasons that they don't is because when you follow that pattern, the simple example part can be a motivation killer. The problem is that it's really easy to SEE the answer and not understand it.
For example:
X + 2 = 5
X is 3
No steps, no explanations, but if you have experience with your fingers many students know it's 3. Now, if you're just starting Algebra, how does it help for you to just know the answer is 3? You haven't learned any Algebra at all. You used inverse operations and you maintained equality, but do you know that or what it's called or how to do that again?
It's a shame, because using simple examples is really a great way to help people who are scared of math to get easily into the material, but it can derail them in the long run. You end depriving people of a true learning experience because they can skip the process, and the process is what Algebra is all about. The numerical answer they get without understanding or context isn't really an answer at all.
Here's what Head First has done. If you look at Head First Algebra, in Chapter 1, we do have a couple of pretty intuitive equations to start, but the rest of the book - lots of decimals and fractions. You'll see a lot of x = 3.445. Why? Because you can't just look at 160 + 410g = 1330 and figure out what g represents. You have to work through the equation, using the rules, to figure out the solution. The end result? You have to learn the steps because you can't get the answer without them. No skipping- ha!
Here's another reason we stayed away from simple, whole numbers when we developed the book. The problems in Head First Algebra deal with things like car payments, road trip costs, and projectile motion. These situations can't generally be summed up in whole numbers, so we didn't. The reality of dealing with the world forces some messy situations, and we've embraced it.
Working with Head First Algebra gets you two advantages - you learn to work with real world numbers and real world situations, and doing that means that you have to learn the Algebra. Learning the skills and the concepts? That's Head First.
3 Comments
The world is made up of two kinds of people: those that find algebra intuitive (not me) and those that find geometry intuitive (me). Never fails. Alas, being algebra intuitive is a greater asset for higher maths topology, analysis and the like (as typically taught). Einstein said he world out the Special Theory geometrically, but eventually had to produce the algebra.
When I was in 6th grade I did Algebra Topics, which is computer program on the computer which teaches you how to do algebra. I didn't like how they taught, because they just don't explain it clearly. But now that I am in 7th grade and actually have a real person teaching me it is really easy! I hope I don't hit a wall once I get into Geometry! To be truthful, I want to skip Algebra and just go straight into Geometry, unfortunately it is already way into the 2nd semester and I cannot change my classes anymore.
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Details about Matrix Algebra:
Matrix algebra is one of the most important areas of mathematics for data analysis and for statistical theory. This much-needed work presents the relevant aspects of the theory of matrix algebra for applications in statistics. It moves on to consider the various types of matrices encountered in statistics, such as projection matrices and positive definite matrices, and describes the special properties of those matrices. Finally, it covers numerical linear algebra, beginning with a discussion of the basics of numerical computations, and following up with accurate and efficient algorithms for factoring matrices, solving linear systems of equations, and extracting eigenvalues and eigenvectors.
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Rent Matrix Algebra 1st edition today, or search our site for other textbooks by James E. Gentle. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Springer.
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Popular Textbooks
Details about Hutchison's Basic Mathematical Skills with Geometry:
Basic Mathematical Skills with Geometry, 8/e by Baratto/Bergman is part of the latest offerings in the successful Hutchison Series in Mathematics. The eigth edition continues the hallmark approach of encouraging the learning of mathematics by focusing it
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Rent Hutchison's Basic Mathematical Skills with Geometry 8th edition today, or search our site for other textbooks by Stefan Baratto. Every textbook comes with a 21-day "Any Reason" guarantee. Published by McGraw-Hill Higher Education. | 677.169 | 1 |
easy-to-understand trigonometry text makes learning trigonometry an engaging, simple process. The book contains many examples that parallel most problems in the problem sets. There are many application problems that show how the concepts can be applied to the world around you, and review problems in every problem set after Chapter 1, which make review part of your daily schedule. If you have been away from mathematics for awhile, study skills listed at the beginning of the first six chapters give you a path to success in the course. Finally, the authors have included some historical notes in case you are interested in the story behind the mathematics you are learning. This text will leave you with a well-rounded understanding of the subject and help you feel better prepared for future mathematics courses. | 677.169 | 1 |
MATH Documents
Showing 1 to 30 of 2,267
Name . Perm .
Second Test Math. 122A, summer session A, 2016, Akemann
You may use your calculator, but you must not store notes on it. All action (points, sets, and whatever)
takes place in the complex plane C.
[8] 1a. State the Cauchy-Riemann differentia
Math. 122A, summer 2016, Chuck Akemann, First test
The complex numbers will be denoted by C. U will denote an open region in C.
1. [14 ] Complete the following definitions.
a. If is a positive number, the neighborhood of a point z0 C is
b. A function f :
MATH 4BDifferential Equations, Spring 2016
Midterm 1Midterm 2Final Exam Study Guide
GENERAL INFORMATION AND FINAL EXAM RULES
The exam will have a duration of 3 hours. No extra time will be given. Failing to
submit your solutions within 3 hours will result in your exam not(i) Name the feature marked P.
(1mk)
(ii) Describe how the feature P is formed.
(6mks)
(c) (i) Name two warm ocean currents along the coast of Africa.
(ii) Explain how ocean currents influence the climate of the surrounding.
(d) (i) Name two types of move
(i) Deflation hollow
- is a feature of wind erosion by deflation wind carries away the loosely held soil particles from a particular point in
a desert
- wind deflation taking place for along time lead to formation of a hollow which is enlarged by further
2.
(i)
Which movement of the earth is represented by the diagram?
(1mk)
(ii)
Give two effects of the movement represented by the diagram
(2mks)
The diagram below represents the structure of the earth. Use it to answer question
(a) Name
(i)
The parts marke
High rates charged on local tourists.
Perimeter fencing of game parks and reserves.
Establishing a strong anti-poaching police unit.
Employment adequate game wardens.
Enacting legislation against poaching.
Providing mass education on the value of wild ga
Math 6B: PDE Quiz Solution
May 28, 2016
1. Vibration of a stretched string with fixed ends: The ends of a stretched string of length L = 1 are fixed
at x = 0 and x = 1. The string is set to vibrate from rest by releasing it from an initial triangular
shap
Math 6B: Series Quiz Solutions
April 21, 2016
Determine if the following series converge or diverge. State what test you used to come to your conclusion.
X
arctan n
1.
n1.2
n=1
Proof. Since arctan n
2,
we have
The series
arctan n
2
.
n1.2
n1.2
X
n=1
2
n1
HW 6
Due Friday July 15th.
1. The function f (t) = cos2 (t) can be regarded as either periodic or 2
periodic. Choose one of these periods and find the fourier series of f (t).
Hint: Simplify with trig identities first.
t,
f or
<t<
2. The function f(t) =
Math 6B: Sequences Quiz Solutions
April 21, 2016
Determine if the following sequences converge or diverge. If the sequence converges, find its limit. Be sure
to explain what theorems you used to reach your conclusion.
1. an =
3 + 5n2
n2 + n
Solution. an i
Midterm practice:
1. Use Stokes theorem. Either parameterize the surface given (in polar coordinates works Best), find its
normal, and integrate, or choose a second surface which is three flat faces in the xy, the yz, and the xz
planes having the same bou
MATH Flashcards
Showing 1 to 1 of 1
-when AC = I or BA = I.
-Let A be an nxn matrix then A is invertible if there is another nxn matrix A^-1 such that AA^-1 = Identity and A^-1A = Identity
-A has a pivot in every row and column
-reduced echelon form of A is Identity
-linear dependence = not invertible.definition | 677.169 | 1 |
This book explores the many disciplinary and theoretical links between language, linguistics, and mathematics. It examines trends in linguistics, such as structuralism, conceptual metaphor theory, and other relevant theories,to show that language and mathematics have a similar structure, but differential functions, even though one without the other would not exist.
From Limits and Continuity to Integration, Areas and Volumes, there is a lot of subject matter to know if you want to succeed on your AP Calculus AB/BC exams. That's why we've selected these 500 AP-style questions and answers that cover all topics found on these exams. The targeted questions will prepare you for what you'll see on test day, help you study more effectively, and use your review time wisely to achieve your best score.
This book is an ideal text for an advanced course in the theory of complex functions. The author successfully enables the reader to experience function theory personally and to participate in the work of the creative mathematician. Unlike the first volume, it contains numerous glimpses of the function theory of several complex variables emphasizing how autonomous this discipline has become. Covered are Weierstrass' product theorem, Mittag-Leffler's theorem, the Riemann mapping theorem, and Runge's theorems on approximations of analytic functions.
What can a physicist gain by studying mathematics? By gathering together everything a physicist needs to know about mathematics in one comprehensive and accessible guide, this is the question Mathematics for Physics and Physicists successfully takes on. Mathematics for Physics and Physicists is the resource today's physicists must have to strengthen their math skills and to gain otherwise unattainable insights into their fields of study
This book provides an overview of current research on a variety of topics related to both large-scale and classroom assessment. First, the purposes, traditions and principles of assessment are considered, with particular attention to those common to all levels of assessment and those more connected with either classroom or large-scale assessment. Assessment design based on sound assessment principles is discussed, differentiating between large-scale and classroom assessment, but also examining how the design principles overlap.
Geometry can be a daunting subject. That's why our new High School Unlocked series focuses on giving you a wide range of key techniques to help you tackle subjects like Geometry. If one method doesn't "click" for you, you can use an alternative approach to understand the concept or problem, instead of painfully trying the same thing over and over without success. Trust us—unlocking geometric secrets doesn't have to hurt! | 677.169 | 1 |
Higher Algebra with Visual Basic
Here's something for the math teachers. And the people who are teaching computer science but who would really rather be teaching math. Or perhaps the people who just want to learn how to do some really mathematical things in a computer program.
Brian Beckman is a physicist and computer scientist who works on programming language development. In a new video on Channel 9 he mixes an education in advanced algebra (vectors and matrices) with some advanced programming concepts (generics and operator overloading).
It's all done with Visual Basic .NET which Brian is a huge proponent of using. I think you'll find the interview and demo shows that there is an awful lot of power and flexibility in Visual Basic. But perhaps better still the demo ties math and programming together very well.
Brian has written more details and shows sample code here at the VB Team blog. Its a good companion piece to the interview. This is not beginner stuff BTW. Not in math or in programming. But it is interesting. | 677.169 | 1 |
Algebraic geometry
It can be seen as the study of solution sets of systems of polynomials.
When there is more than one variable, geometric considerations enter and are important to understand the phenomenon.
One can say that the subject starts where equation solving leaves off, and it becomes at least as important to understand the totality of solutions of a system of equations as to find some solution; this does lead into some of the deepest waters in the whole of mathematics, both conceptually and in terms of technique.
June 11, 2009 — Hirzebruch's problem at the interface of topology and algebraic geometry has occupied mathematicians for more than 50 years. A professor of mathematics at the Ludwig-Maximilians-Universitaet in ... read more
Sep. 16, 2013 — Researchers have found high school students in the United States achieve higher scores on a standardized mathematics test if they study from a curriculum known as integrated ... read more
Aug. 3, 2009 — As mathematics continues to become an increasingly important component in undergraduate biology programs, a more comprehensive understanding of the use of algebraic models is needed by the next ... read more
Jan. 18, 2016 — Over centuries, humans have tried to discover a Theory of Everything. Possible candidates for this cachet, such as String Theory and Grand Unified Theory, require higher dimensions or ... read more
Aug. 16, 2006 — Studying complex systems, such as the movement of robots on a factory floor, the motion of air over a wing, or the effectiveness of a security network, can present huge challenges. Mathematician ... read more
June 20, 2011 — A new study finds important differences in math curricula across US states and school districts. The findings suggest that many students across the country are placed at a disadvantage by less ... read more
Mar. 4, 2009 — Greek mathematics is considered one of the great intellectual achievements of antiquity. It has been decisive to the academic and cultural development of Western civilization. The three Roman authors | 677.169 | 1 |
books.google.com -,... Darstellende Geometrie Ein Grundriss Für Vorlesungen an Technischen Hochschulen Und Zum Selbststuduim | 677.169 | 1 |
I don't think you will find that a big problem. None of the main ideas of the book require any mathematical knowledge.
Some of the examples are mathematical in nature. For instance, there is about twelve pages in chapter 6 that concerns using an infinite list object to represent a sequence of increasingly accurate approximations to the solution of a certain financial problem. People who have studied calculus will immediately understand the ideas here; people who haven't might understand them anyway, and even if you don't you can always skip those examples; most of them have nothing in particular to do with math.
My own tendency is to write a lot of math stuff, because I find it very interesting, but while I was writing HOP I tried really hard to get rid of the mathematics, because I knew that a lot of people don | 677.169 | 1 |
Details about Mathematics for Economics and Finance:
Mathematics has become indispensable in the modelling of economics, finance, business and management. Without expecting any particular background of the reader, this book covers the following mathematical topics, with frequent reference to applications in economics and finance: functions, graphs and equations, recurrences (difference equations), differentiation, exponentials and logarithms, optimisation, partial differentiation, optimisation in several variables, vectors and matrices, linear equations, Lagrange multipliers, integration, first-order and second-order differential equations. The stress is on the relation of maths to economics, and this is illustrated with copious examples and exercises to foster depth of understanding. Each chapter has three parts: the main text, a section of further worked examples and a summary of the chapter together with a selection of problems for the reader to attempt. For students of economics, mathematics, or both, this book provides an introduction to mathematical methods in economics and finance that will be welcomed for its clarity and breadth.
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Rent Mathematics for Economics and Finance 1st edition today, or search our site for other textbooks by Anthony Martin. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Cambridge University Press.
Need help ASAP? We have you covered with 24/7 instant online tutoring. Connect with one of our Finance tutors now. | 677.169 | 1 |
For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.
ISBN ISBN
981-238-291-7 981-238-292-5 (pbk)
Printed in Singapore.
PREFACE
The objective of this book is to provide a mathematical text at the third year level and beyond, appropriate for students of engineering and sciences. It is a book of applicable mathematics. We have avoided the approach of listing only the techniques followed by a few examples, without explaining why the techniques work. Thus we have provided not only the know how but also the know why. Equally it is not written as a book of pure mathematics with a list of theorems followed by their proofs. Our emphasis is to help students develop an understanding of mathematics and its applications. We have refrained from using cliches like "it is obvious" and "it can be shown", which might be true only to a mature mathematician. In general, we have been generous in writing down all the steps in solving the example problems. Contrary to the opinion of the publisher of S. Hawking's book, A Short History of Time, we believe that, for students, every additional equation in the worked examples will double the readership. Many engineering schools offer little mathematics beyond the second year level. This is not a desirable situation as junior and senior year courses have to be watered-down accordingly. For graduate work, many students are handicapped by a lack of preparation in mathematics. Practicing engineers reading the technical literature, are more likely to get stuck because of a lack of mathematical skills. Language is seldom a problem. Further self-study of mathematics is easier said than done. It demands not only a good book but also an enormous amount of self-discipline. The present book is an appropriate one for self-study. We hope to have provided enough motivation, however we cannot provide the discipline! The advent of computers does not imply that engineers need less mathematics. On the contrary, it requires more maturity in mathematics. Mathematical modelling can be more sophisticated and the degree of realism can be improved by using computers. That is to say, engineers benefit greatly from more advanced mathematical training. As Von Karman said: "There is nothing more practical than a good theory". The black box approach to numerical simulation, in our opinion, should be avoided. Manipulating sophisticated software, written by others, may give the illusion of doing advanced work, but does not necessarily develop one's creativity in solving real problems. A careful analysis of the problem should precede any numerical simulation and this demands mathematical dexterity. The book contains ten chapters. In Chapter one, we review freshman and sophomore calculus and ordinary differential equations. Chapter two deals with series solutions of differential equations. The concept of orthogonal sets of functions, Bessel functions, Legendre polynomials, and the Sturm Liouville problem are introduced in this chapter. Chapter three covers complex variables: analytic functions, conformal mapping, and integration by the method of residues. Chapter four is devoted to vector and tensor calculus. Topics covered include the divergence and Stokes' theorem, covariant and contravariant components, covariant differentiation, isotropic and objective tensors. Chapters five and six consider partial differential equations, namely Laplace, wave, diffusion and Schrodinger equations. Various analytical methods, such as separation of variables, integral transforms, Green's functions, and similarity solutions are discussed. The next two chapters are devoted to numerical methods. Chapter seven describes methods of solving algebraic and ordinary differential equations. Numerical integration and interpolation are also included in this chapter. Chapter eight deals with
id
ADVANCED MATHEMATICS
numerical solutions of partial differential equations: both finite difference and finite element techniques are introduced. Chapter nine considers calculus of variations. The Euler-Lagrange equations are derived and the transversality and subsidiary conditions are discussed. Finally, Chapter ten, which is entitled Special Topics, briefly discusses phase space, Hamiltonian mechanics, probability theory, statistical thermodynamics and Brownian motion. Each chapter contains several solved problems clarifying the introduced concepts. Some of the examples are taken from the recent literature and serve to illustrate the applications in various fields of engineering and science. At the end of each chapter, there are assignment problems labeled a or b. The ones labeled b are the more difficult ones. There is more material in this book than can be covered in a one semester course. An example of a typical undergraduate course could cover Chapter two, parts of Chapters four, five and six, and Chapter seven. A list of references is provided at the end of the book. The book is a product of close collaboration between two mathematicians and an engineer. The engineer has been helpful in pinpointing the problems engineering students encounter in books written by mathematicians. We are indebted to many of our former professors, colleagues, and students who indirectly contributed to this work. Drs. K. Morrison and D. Rodrigue helped with the programming associated with Chapters seven and eight. Ms. S. Boily deserves our warmest thanks for expertly typing the bulk of the manuscript several times. We very much appreciate the help and contribution of Drs. D. Cartin , Q. Ye and their staff at World Scientific. New Orleans December 2002 C.F. Chan Man Fong D. De Kee P.N. Kaloni
References Appendices Appendix I - The equation of continuity in several coordinate systems Appendix II - The equation of motion in several coordinate systems Appendix III - The equation of energy in terms of the transport properties in several coordinate systems Appendix IV - The equation of continuity of species A in several coordinate systems Author Index Subject Index
861 867 867 868 871 873 875 877
CHAPTER 1
REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
1.1 FUNCTIONS OF ONE REAL VARIABLE
The search for functional relationships between variables is one of the aims of science. For simplicity, we shall start by considering two real variables. These variables can be quantified, that is to say, to each of the two variables we can associate a set of real numbers. The rule which assigns to each real number of one set a number of the other set is called a function. It is customary to denote a function by f. Thus if the rule is to square, we write
y
= f(x) = x 2
(l.l-la,b)
The variable y is known as the dependent variable and x is the independent variable. It is important not to confuse the function (rule) f with the value f(x) of that function at a point x. A function does not always need to be expressed as an algebraic expression as in Equation (1.1-1). For example, the price of a litre of gas is a function of the geographical location of the gas station. It is not obvious that we can express this function as an algebraic expression. But we can draw up a table listing the geographical positions of all gas stations and the price charged at each gas station. Each gas station can be numbered and thus to each number of this set there exists another number in the set of prices charged at the corresponding gas station. Thus, the definition of a function as given above is general enough to include most of the functional relationships between two variables encountered in science and engineering. The function f might not be applicable (defined) over all real numbers. The set of numbers for which f is applicable is called the domain of f. Thus if f is extracting the square root of a real number, f is not applicable to negative numbers. The domain of f in this case is the set of non-negative numbers. The range of f is the set of values that f can acquire over its domain. Figure 1.1-1 illustrates the concept of domain and range. The function f is said to be even if f (-x) = f (x) and odd if f(-x) = -f(x). Thus, f(x) = x 2 is even since f(-x) = (-x) 2 = x 2 = f(x) while the function f(x) = x3 is odd because f(-x) = (-x) 3 = - x 3 = -f(x) (l.l-3a,b,c) (l.l-2a,b,c)
2
ADVANCF.n
MATHEMATICS
A function is periodic and of period T if f(x + T) = f(x) (1.1-4)
y ;,
RANGE
OF f
I
\
i \
I
y= f (X )
>w
"I"""
I L _
DOMAIN OF f
FIGURE 1.1-1
•
j ^
X
Domain and range of a function
An example of a periodic function is sinx and its period is 2K. A function f is continuous at the point x 0 if f(x) tends to the same limit as x tends to x 0 from both sides of x 0 and the limit is f(x 0 ). This is expressed as lim
X—^XQ_J_
f(x) = f(x 0 ) = lim
X—^XQ_
f(x)
(l.l-5a,b) lim
x—>xo_
The notation lim
x—>x 0+
means approaching x 0 from the right side of x 0 (or from above) and
the limit as x 0 is approached from the left (or from below). An alternative equivalent definition of continuity of f (x) at x = x 0 is, given e > 0, there exists a number 8 (which can be a function of 8) such that whenever I x - xol < 8, then | f ( x ) - f ( x o ) | <£ This is illustrated in Figure 1.1-2. If the function f (x) is continuous in a closed interval [a, b], it is continuous at every point x in the interval a < x < b. (1.1-6)
REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
3
^o+€
y^
ijl
*o-a! i !««*> ^ X
FIGURE 1. . Likewise higher dx 2 derivatives can be defined and the n th derivative is written either as f ^ or dx n Geometrically.
. For example. one might want to know. waiting at the corner of "walk and don't walk". x 0 ] is defined as Af Ax
=
f (xQ + Ax) . The derivative of f with respect to x is also denoted as 4*-. is defined as f(
} =
*L Ax->0 Ax
lim
=
f(x o + Ax)-f(x o ) Ax-*O AX
lim
(i
22ah)
We have assumed that the limit in Equations (1.2 DERIVATIVES
Continuity of a function
We might be interested not only in the values of a function at various points x but also at its rate of change. The average rate of change of f (x) in an interval [x0 + Ax. which is the derivative of f at x 0 .2-2a. not only the position of a car but also its speed before crossing the road.1-2 1. dx d2f The second derivative of f is the derivative of f and is denoted by f" or . f'(x 0 ) is the tangent to the curve f(x) at the point x = x 0 .b) exists and f is thus differentiable at x = x 0 .f (x0) Ax
( i 2 i )
The rate of change of f at x 0 .
Taylor's Theorem
If f(x) is continuous and differentiable f (x) = f (x0) + (x .. the remainder term for each point x in the interval will generally be different.x 0 ) f'(x 0 ) + (X " 2XQ) where R n is the remainder term. But if the lim —— exists x->x0 x->x0 x->x0 g(x) x->xo g'( x )
lim ^
x^x 0 g(x)
= lim I M
x^x0
(1.
= ..2-9) can be If lim ——-1.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
5
L'Hopital's Rule
lim f(x) = 0 and lim g(x) = 0.is indeterminate.2-9)
g'( x )
f'(x) f(n) (x) — exists for some value of n. the lim ~~. There are various ways of expressing the remainder term R n . R n is the result of a summation of the remaining terms. and represents the error made by truncating the series at the n * term.does not exist but lim x->x0 g(n) ^ x^x 0 g (x) replaced by
to M. ta V®. Other indeterminate forms. x-»x o g(n)( x )
(1.2-11)
The simplest one is probably
\n+^
°-
(n+1)!
f ( n + 1 ) [*o
+ 6(*-xo)]
0-2-12)
where 0 < 6 < 1. must first be reduced to one of the indeterminate forms discussed here before applying the rule.2. + (X " J ° ) n f(n) (x 0 ) + R n
(1. such as the difference of two quantities tending to infinity. Lagrange's expression which may be written as
(
R
f "(X 0 ) + . the lim -y-^r is indeterminate. The
.
x-^x o g(x. Equation (1. Note that we are expanding about a point x 0 which belongs to an interval (x 0 .10)
ffx) The same rule applies if the lim f (x) = °° and lim g(x) = °o. x). The
x-»x 0
x -^ x o x~*xo
S( x )
rule holds for x —> °° or x —> -°°. Therefore.
2-1
Extremum of a function f
.2-11).x 0 .
y=f(x)
/\
j |^
Af\^l
h h
j j I
h h
x
FIGURE 1.. if the function f has an extremum at xQ Af = f ( x o + h ) . If Af is positive. In Equation (1.f " ( x 0 ) + . where h = x .2-11). Figure 1. associated with the evaluation of a function.2-14)
y<
MAX. From Equation (1. f has a minimum at x 0 and if Af is negative. and if x 0 is the origin.2-13)
(1. determines the value of 0 in Equation (1. f has a maximum at x 0 .f ( x o ) must have the same sign irrespective of the sign of h. The Taylor series expansion is widely used as a method of approximating a function by a polynomial.6
ADVANCED MATHEMATICS
maximum truncation error.2-12).2-1 defines such extrema.. we are dealing with Maclaurin series.
h2
(1. we have expanded f(x) about the point xQ. we see that Equation (1. Thus.2-13) can be written as Af = hf(x o ) + ^ . Maximum and Minimum We might need to know the extreme (maximum or minimum) values of a function and this can be obtained by finding the derivatives of the function. at different values of x within the considered interval.
c)
From Equation (1. we have f'(x) = 3x 2 . Thus.. We need to consider higher derivatives until we obtain a f (n) (x 0 ) which is non-zero. f(x) does not have an extreme value at x = x 0 .3
INTEGRALS
An integral can be considered to be an antiderivative. Since the derivative of a constant is zero.2-1. From the criteria given earlier we deduce that f does not have an extreme value at the origin.b). we see that f(O) = f"(O) = O (1.2-17) (1.
1. and if f ^ ( x 0 ) exists and is non-zero. as can be seen by drawing the curve given by f (x) = x 3 .2-19a.2-18 a. f has a maximum at x = x o and if f (n) (x0) > 0.2-15)
But if f "( x o) = 0> we cannot deduce that f has an extreme value at xQ. the derivative of ^-x 3 is x 2 .b. we write
. f (x) has an extreme value at x 0 if n is even.b)
Thus we need to consider higher derivatives and the next one f"'(0) happens to be non-zero. and an integral of x 2 is ^. f has a minimum at x = x Q .2-14) we deduce that the condition for f to have an extremum at xQ is f'(x o )=O The conditions for f to have a maximum or minimum at x = x 0 are f"(x o )>O. b). The origin is neither a maximum nor a minimum. Example 1. If f( n )(x 0 )<0. the general criteria for extreme values are if f(x) is defined in [a. Find the extreme values of f (x) = x3. it is a point of inflection. if they exist. an integral of f(x) is F(x). The integral we have defined is known as an indefinite integral which is usually denoted by the symbol I . = f ^~^ (x 0 ) = 0.. For example. if we know that the derivative of F (x) is f(x) [=F'(x)]. If n is odd.2-16) (1. Thus. F (x) is arbitrary to the extent of an arbitrary constant.x 3 . Note that we have used the article an. fhasaminimum f hasamaximum (1. but f'(x 0 ) = f"(x 0 ) = . Thus.2-18a.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
_J_
From Equation (1. On differentiating. f"(x o )<O. f"(x) = 6x. f'"(x) = 6 (1.b] and x 0 is an interior point of (a.
For the area to be definite.3-4) that
(1.3-2)
Equation (1.3-3a. There are tables of integrals.3-2) defines a definite integral.b)
Thus to evaluate a definite integral analytically.3-1) to a definite integral if x = b. Below we list some of the general methods of integration. Integration by Parts If f and g are functions of x
£-(fg) = fg + fg'
It follows from Equation (1.3-1. x = a and x = b. x = b
A=
/a
f(x)dx
(1. Carrying
Q.b)
where a is an arbitrary constant of integration. we usually write
F(b)-F(a) = Ja
f(x)dx = [F(x)J a
(1. In this case. the limits x = a and x = b are given. where the indefinite integrals of standard functions are given.b) define a function of x in terms of a dummy variable t.3-5)
Example
1. we first need to find an indefinite integral. if A is the area bounded by the curve y = f(x).8
ADVANCED MATHEMATICS
F(x)= I f(x)dx= I f(t)dt
(1.3-la.3-la. Thus.X
out the integration. identifying from Equation (1. Integrate I e ax sin bx dx.3-4)
|f|dx
= [fg]-|f gdx
(1. Equations (1.3-5) e ax as f(x) and sinbx as -§-. The integral may be interpreted as the area enclosed by the curve y = f(x) and the x-axis. We can convert the indefinite integral in Equation (1. we have
. such as. the x-axis and the ordinates x = a.
We integrate by parts. we need to fix the ordinates.
The dependent variable u is said to be a function of the independent variables x and y if to every pair of values of (x. x n is obvious.5 DERIVATIVES (1. we often encounter one variable which depends on several other independent variables. there exists a 8. the volume of gas depends on the temperature and the pressure. y0) if given e > 0.4-1)
The domain of f is the set of values of (x. x2. y) over which f is applicable.y) (1. Thus J dx dx x x df=]im
ox AX->0
f(x
+
Ax.y) Ay
(1
5 2 )
8f The computation of ^— is the same as in the case of one independent variable. Similarly to compute jr-. y) one can assign a value of u.. their partial derivatives with respect to x and y may exist. ^— is defined as dy 9f oy
= lim
f(x.4-3)
Since u is a function of two variables x and y.y
+
Ay-40
Ay)-f(x. y o ) | <£ 1.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
13. Here. In this case we write u = f(x. In science and engineering. The range of f is the set of values that u may have over the domain of f.
1.4
FUNCTIONS OF SEVERAL VARIABLES
So far. such that whenever V(x-xo)2 + (y-yo)2 < 5 | f ( x . we consider x to be a constant. y is treated as a constant. Since fx and f are functions of x and y. we may calculate the rate of change of u with respect to x. the extension to n variables Xj. . holding y fixed. y ) . For example. For simplicity. we shall consider functions of two independent variables x and y. we have considered functions of one variable only.y)
Ax
(1
5 1 )
Similarly.4-2) (1. f or u Y . Other notations are: ^—.y)-f(x. y) is continuous at (x 0 .. The function f (x. This is the partial derivative of u with respect to x and is denoted by ^r—. .f ( x 0 . In most cases. They are defined as
.
5. In an xy-coordinate system.5-6a.b) (1. where the vector n is parallel to the line along which we wish to determine the rate of change. whereas for f the order of differentiation is reversed. We can also define and compute the rate of change of f along any arbitrary line in the xy-plane.b)
We note that f means taking the partial derivative of f with respect to x first and then with respect to y. Such a rate of change is known as a directional derivative and is denoted by 5—. In practice.(x.5-8a. If the partial derivatives are continuous. y0) along a line that makes an angle 0 with the x-axis is defined as
<*L = Urn dn p_>o L
f ( X ° + P C0S 6 ' Yo + P Sl" 6 )
~f ( X ° ' y o ) J
(1-5-10)
P
.14
ADVANCED MATHEMATICS
j-(£U
dx \ dx /
l i m f.5-9)
Likewise higher partial derivatives can be defined and computed.(^Ax.y)
AX->0 Ax
(1. the first order partial derivatives fx and f may be regarded as the rate of change of f along the x and y-axis respectively. If f is continuous then the order is not important.4)
The second-order partial derivatives are denoted as
|_(|L] = ^Uf
dx \dx )
5x2 xx
(l-5-5a. One may wonder if the order of differentiation is important.y)-f. this is generally the case and fxy = fyx (1. Thus 5— at a point (x0.=fx y
(L5"7a'b)
|_f|l)=^l=f
(i.b)
f f|f)=-^-=f
dy \ dx / dydx ^x
d7(d7) = axay. then the order of differentiation is not important.5-3)
a [*)
dy \ dx /
= lm
A y->0
y^±Mzifcy>
Ay
(1.
6-2).6-1)
[V-nb]-nRT = 0
(1. We now consider an implicit function written as f(x. But as can be seen from
Equation (1. y and u. In theory.c) (1. y and u are not all independent. we can obtain ^-
and other partial derivatives. There are examples where it is more convenient to express u implicitly as a function of x and y. in thermodynamics an equation of state which could be given as T = f (P. V is the volume and T is the temperature. We are free to choose any one of them as the dependent variable. y = V.6-2) partially oT with respect to P and then deduce 5= . c. The two-parameter Redlich and Kwong (1949) equation is expressed as P+ — T 1 / 2 V(V + nb) where a and b are two parameters. d) can be obtained by substituting Equations (1.5-24) into Equation (1. Since f = 0.6-2) it is not easy to solve for T.6 IMPLICIT FUNCTIONS
So far we have considered u as an explicit function of x and y [u = f (x. Then
rfT
(1.6-2) and express T as a function of P and V. u=T (1. we can solve for T in Equation (1.b. implicitly as f(P. It is simpler to differentiate the expression in Equation (1.6-2)
by partial differentiation.5-15). We shall show how this is done by reverting to the variables x. the resulting function will be even more complicated than Equation (1. df = 0 and from Equation (1.5-23) and thus f is expressed explicitly as a function of r and 0 and the partial differentiation can be carried out.u) = 0 In terms of the thermodynamic example.6-4a. and finding ^ r
or
(say) will be time consuming. In many cases. we can think of x = P. For example.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
_£9
Equations (1. R is the gas constant and n the number of moles.VT) = 0 where P is the pressure.y. it implies solving a cubic equation. we obtain
.5-27a. the substitution can be very complicated.6-3)
The variables x. Even if we solve for T. y)]. V) is usually written. b. 1.
Using the results in Section 1.5.8-3)
Note that if we were integrating at another point on the same curve. I may be treated as a function of three variables x.8-2) Ja If the limits of integration are not fixed but are functions of x.8-7a. identifying y with t and v (x) with x.c.8-6a. v(x) and u(x). via Equation (1.y)dy = f[x. we have
|1 = | -
I
Ju (x)
f(x. Equation (1.3-1) yields dv f
x
^ p
=|)a
f(t)dt = |^[F(x)-F(a)] =F'(x) = f(x)
(l. v(x)]
(l.8-5) yu(x) Note that u(x) and v(x) are not fixed! To evaluate the partial derivative —-. the limits of integration would read u(xi) and u(x2) which is equivalent to integrating from a to b. dL is given by dx di_ dx
=
a i _ + a ] [ d v + ai_du dx dv dx du dx
(1.b.U(X)
(1.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
21
Differentiating I with respect to x results in
(1. we fix the variables x and u(x). 5— and v~.d)
Therefore. y)dy . we integrate with respect to y from a point on a curve given by y = u (x) to a point on another curve given by y = v (x)
/•v(x)
I(x)=| f(x. we have.8-4)
31 dl 91 From the definitions of the partial derivatives 5—.8-2)
/•v(x)
(1.b)
Similarly
.
The highest order derivative occurring in the differential equation determines the order of the differential equation.D. economics. engineering. Otherwise it is non-linear. The degree of a differential equation is determined by the power to which the highest derivative is raised. There are instances when it is dx not possible to evaluate I explicitly and one has to use Leibnitz's rule.D.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
25
dL dx
=
5xl_3x^ 2 2
(1. it is an ordinary differential equation (O.
yf
y = x2
x
FIGURE 1.8-1 1.E. it is a partial differential equation (P. A differential equation is linear only if the dependent variable and its derivatives occur to the first degree.E. and other fields of applied science are expressed as differential equations.E. Many laws and relations in science. If only one independent variable is involved.8-10)
ORDINARY DIFFERENTIAL EQUATIONS (O.
.DEFINITIONS
A differential equation is an equation involving one dependent variable and its derivatives with respect to one or more independent variables.) and if more than one independent variable is involved.8-12C)
Both methods result in the same expression for dL ? as they should.9
Integration described by Equation (1.) .D.).
Examples of first-order differential equations are separable equations. standardized. For each of these types. Starting at Section 1. 1. we can easily verify that f (x) = e ax is a solution of the equation y' = ay. (y is a function of x only) of order one (y' is the highest order derivative) and of degree one (y' is raised to the power one) and is linear.11-1. a straightforward integration will yield a result. f'(x) = a e a x = y' and the right side (ay) is of course af (x) = ae ax . Solve y dx .10.E.26
ADVANCED
MATHEMATICS
For example.D. linear differential equations.19. there exists a known. we look at the modeling problem.11 SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
For the case where M is a function of x only and N is a function of y only. as follows
I M(x)dx = .10-2)
Equations of this type occur in problems dealing with orthogonal trajectories. Indeed. of degree three and is non-linear. 1. and chemical reactions.^ = 0
* y
(1. A function f(x) is a solution of a given differential equation on some interval.11-1)
(1.I N(y)dy
Example 1. There are several types of ordinary differential equations.11-2)
(1. leading to the solution of several of the types of ordinary differential equations encountered in practice.10 FIRST-ORDER DIFFERENTIAL EQUATIONS The standard form of a first-order differential equation is as follows M(x. y)
(1. Dividing both sides of the equation by x 2 y results in the appropriate form ^ . procedure to arrive at a solution. y' = 5y is an O.x 2 dy = 0. In Section 1.10-1)
or
y'=-^H
N(x. homogeneous linear equations. we summarize the approach. y) dy = 0 (1. (y " ) 3 + y = x is of order two.11-3)
. etc. decay. if f(x) is defined and differentiable on that interval and if the equation becomes an identity when y and y ^ are replaced by f(x) and f^(x) respectively. For example. exact differential equations. y) dx + N(x. growth.
\ that is (1.11-11) (1.12 HOMOGENEOUS FIRST-ORDER DIFFERENTIAL EQUATIONS (1.3 xy + Y' is an example of a homogeneous polynomial of degree two. y) in Equation (1.11-7)
y = ce"1/x
Example 1.
. In a constant volume batch reactor.11-10) (1. then the substitution y = ux or x = vy will generate a separable first-order differential equation.^ =-kf(cA) Solve Equation (1.11-6)
(1. the rate of disappearance of reactant A can be given by . x 2 .10-1) are homogeneous polynomials of the same degree. This can be written as iny-inc = . y) and N(x.1 is not a homogeneous polynomial. dcA .REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
27
Integration yields iny = i n c .11-2.11-4)
in£
or
=-\
(1.^ =-kdt i n cA =-kt + i n c in^=-kt c A = ce~ kt 1. x + y .11-8)
If M(x.11-8) for the case where f (cA) = c A .11-9) (1.11-12) (1.11-5) (1.Y where c is the constant of integration.
12-19. we manage to do away with the constants Cj and c 2 .REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
22
^-=l-2y_y! x x x or x2-2xy-y2 = c
(1.12-16) yields d(Y+p) d Y _ a1(X + a ) + b 1 ( Y + p ) + c 1 d(X + a) = dX = a 2 (X + a) + b2(Y + p) + c 2
.12-21a.12-14)
(1. We are left with the homogeneous equation
.12-15)
NOTE: An equation such as
dy
a1x + b 1 y + c 1 (1.b)
dX = a2X + b 2 Y + a 2 a + b 2 p + c 2
(1.18) with solutions x = a and y = p.12-17.18)
we then obtain a situation where the directions of the lines are preserved (aj andfyremain unchanged) and the coefficients Cj vanish.Y ajX+bjY+aja + bjP + Cj
(1. we perform the following change of variables
JX=x-a {Y=y-p
Substitutions in Equation (1. Since the coordinates of this point of intersection are (a.12-22)
where the underlined terms add up to zero by virtue of system (1.12-17. If we translate the origin of the coordinate system to their point of intersection. b j . Think of the numerator and the denominator as representing two intersecting straight lines. c 1? a^ b 2 and c 2 are constants can be reduced to a homogeneous equation if through a change of variables. P). p) of the system
{
ajX +bjy +Cj =0
a 2 x+b 2 y + c2 = 0
(1.12-16)
dx ~ a 2 x + b 2 y + c 2
where a}.20)
(1. that is to the solution (a.
2 x y .
£ .12-27)
A problem arises if the determinant (ajb 2 .p respectively. The final solution is thus given by x 2 .12-24)
dY = ——— X — Y using the substitution Y = u X.12-23)
determining the solution (a and (3) of system (1.30
ADVANCED
MATHEMATICS
dY
a
l
X +
b
l
Y
d X ~ a 2 X + b2Y The solution of such a reducible equation is thus obtained by i) ii) solving equation (1. the coefficients of x and y in the linear equations are multiples of one another. A.18).y 2 .^4
dx i) x+ y. iii) We now substitute X and Y in the solution of part (i) by (x .12-23) by x . that is. Introducing a variable z = ajx + bjy yields a relation of the form — — = f and f is a function of z only.
(1. That is to say.bja 2 ) = 0.12-2. This requires the determinant
al bl
*0
a2 b2
and
iii)
replacing X and Y in the solution of (1.7 = constant (1.
.
ii)
Next we determine the values a and (3 by solving the system
[x-y = 3
|
x + y = 1
(1.26)
We obtain a = 2 and P = .2) and (y + 1).a and y . This leads to Equation We first solve -rrr (1. the two lines are parallel and do not intersect.12-23) [a first-order homogeneous equation]. Solve
Example 1.12-17. a separable first-order equation. + Y
d.l .1
QA.12-15).6 x + 2y = c .12-25.
9)
The following test allows one to determine if the equation Mdx + Ndy = 0 (1. Equation (1.13-4)
An equation such as x 2 dy + 2 x d x = 0 (1.12)
So M and N are partial derivatives of the same function F. of at least order two.= .13-6) (1.13-3)
or
y = f
(1. assuming that F and its partial derivatives.10-1) could be represented by dF (x.13-8. That is.2 i n ex x2 (1.13-5)
can be solved by inspection. Furthermore. after multiplication by an appropriate "integrating" factor. This can be written as
f
Therefore
dx+^dy=0
(U3_1O)
M=||
and N = J £
(1. are continuous in the region of interest. multiplication by e^ results in the following total differential equation d(x2ey) = 0 thus and x 2 ey = c y = i n .13-11. We suspect that the equation is exact.13-2)
(1.10-1)
is a total (or exact) differential equation.° .22
ADVANCED
MATHEMATICS
d(xy) = O Integration yields xy = constant
(1.13-7) (1. y) = 0.13-13)
. In this particular case. we note that
dxdy ~ dydx That is to say
d F _ 3 F
2
2
(1.
Then 5 — = 0 .3 x2y + 4 x cos y + y2 .J f(y) = y 2 .10-1) becomes IMdx + INdy = 0 which is exact if j . Therefore. and Equation (1.13-25)
If Equation (1. the integrating factor I can be determined and Equation (1.i n y + c Hence. (i) I is a function of x only.13-29) becomes 1 dl _ dM/dy-dN/dx I dx N ^
(1.13-28)
dy or I \ dx
dy
dy j ~ dy
dx
(1.(IM) = ^ That is
T3M A/r ai T
(1.13-27)
3 N XT ai
dx dx (1. we can try to make it exact by multiplying with an integrating factor I.10-1) is not exact. Equation (1. the solution is x3 .3d
ADVANCED
MATHEMATICS
f'(y) = 2 y .
.13-30) is also a function of x only.13-29)
In genera] it is not easy to find I but there are special cases when there is a standard procedure to obtain I.13-24)
(1.13-26) is exact and can be solved.13-30)
The assumption that I is a function of x only implies that the right side of Equation (1.13-23) (1.Jin y = constant •
(1.13-26)
(IN)
(1.
ii)
two equal real roots (the case where D = 0).16-10). Since we are dealing with a second order equation.16-11)
iii)
two complex conjugate roots (the case where D 2 < 0) y h is now given by yh = Cj e(a+ib>x + c 2 e(a-ib>x (1. We propose y h to be of the form e a x . In this case. In this case. (Xj and 0C2. we have that a 2 + Aot + B = 0 (1. y h is given by y h = c1e0ClX + c 2 xe aiX (1. using the Euler formula to yield
.16-12)
where (Xj
2
= -y ± i
z
= a ± ib
This complex solution can be transformed into a real one.16-10)
The second order characteristic equation could have
2 2
i)
two real and distinct roots (the case where D = A . Substitution of this function and its derivatives into the homogeneous equation yields e ax (oc 2 + A a + B) = 0 Since e a x * 0 .16-9) (1.16-8) (1. we also have two constants of integration and the solution y h is given by yh =
Cleaix
+ c 2 e a2X
(1.4B > 0).42
ADVANCED MATHEMATICS
Solving the homogeneous equation y"+Ay'+By = 0 will generate y h . y h is given by equation (1.16-7)
This is known as the characteristic or auxiliary equation. This equation has two solutions.
3 y ' + 2y = x 2 + 1 The characteristic equation a2 .16-21)
.16-22) (1.16-23) (1.3 b + 2c = x 2 + l Equating the coefficients of x 2 yields 2a = 1 and a= y (1. This solution technique is referred to as the method of undetermined coefficients.3 ( 2 a x + b) + 2(ax 2 + bx + c) = x 2 + 1 2a . Solve (1.3b + 2ax2 + 2bx + 2c = x 2 + l 2ax2 + x ( 2 b .16-16)
y " .
(1. Example 1. we propose for y the quadratic polynomial ax 2 + bx + c.16-17)
+ c2e2x
(1.16-20) (1.16-18)
Since the right side of the equation is of order two.16-14. This can be done by proposing a solution.6 a ) + 2 a .16-1.16-24) (1. Substitution of the proposed form and of its derivatives into the equation to be solved followed by equating the coefficients of the like terms allows one to determine the values of the introduced constants as illustrated next. b and c. the right side of the equation.b) (1. and proceed via substitution in the given equation to determine the values of the constants a.3 a + 2 = 0 has the following roots: a 1 = 1 and a 2 = 2 so that yh =
Clex
(1. based on the form of Q(x). y p = ax 2 + bx + c y p = 2ax + b yp'= 2a Substitution yields 2 a .16-19) (1.16-25a.16-13) (1. 15)
Next we have to determine the particular solution y .REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
43
y h = e M (c 3 cos bx + c 4 sin bx) where c 3 and c 4 are constants and real.6ax .
A mass balance on A leads to the following differential equation which we wish to solve
-J9 ^AB
CA
dc A
2
I / v \ CA I i{C .16-43) in standard form as follows
. then y is y p = x(a o e ( l x ) If xe qx is also part of the homogeneous solution.431 U .i o is)
We can write Equation (1. The concentration of A is given at the inlet (z = 0) of the reactor by c A and at the outlet (z = L) by c ^ . Levenspiel (1972) describes a flow problem with diffusion.16-37)
If cos qx (or/and sin qx) is not a solution of the homogeneous equation.o \ z ' dz A ~
(] 16. The average velocity in the tubular reactor is (v z ). The particular integral will be of the form y p = aoe«lx if e^* is not a term in y h .16-42) (1.16-39)
We can also determine the particular solution for a combination of the cases examined in (a to c) by intelligent guess work. Example 1. We need to find the number of undetermined coefficients that can be determined to fit the differential equation. 18-1).16-2. If e1 x occurs in y h . involving a firstorder chemical reaction.14-3.16-41) (1.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
41
In this case an appropriate y is y p = a 0 cosqx + b 0 s i n q x (1.16-38) (1. The particular solution can also be obtained by the method of variation of parameters (see Examples 1. then y p = x 2 (a 0 e1x) (1. If cos qx (or/and sin qx) is present in y h then we need to try y p = x (a 0 cos qx + b 0 sin qx) (c) Q(x) = e ^ .16-40) (1.k c A . The reaction is characterized by the rate expression — ^ = .
16-47) (1.16-44)
JJ^
We now propose a solution of the form cA = e a z .st dt
(1. The Laplace transform is linear. substitution leads to the characteristic (or auxiliary) equation a2^ a--^=0 (1.16-45)
The result in terms of the concentration cA is given by cA = c 1 e a i z + c 2 e a 2 z (1. The Laplace transform L[f(t)] of a function f(t) is defined as
L[f(t)] = F(s) = I
f(t)e. That is to say
.16-48)
This leads to the following final result /cAL-cA eaiMl cAI-cA eaiL ^ A° eaiz+ ^ Ao ea2z ea2L-eaiL ea2L-eaiL
c A
-
cA A°
(1.46
ADVANCED MATHEMATICS
^_KideA_^_
dz ^ ^ az
C A = 0
(1.17
SOLUTIONS BY LAPLACE TRANSFORM
Linear differential equations can sometimes be reduced to algebraic equations. which are easier to solve.17-1)
Jo
where the integral is assumed to converge. A way of achieving this is by performing a so called Laplace transform. As before.
2
= .^ -
1 ±A
/ I + ^ ® A B
The constants Cj and c 2 are evaluated from the conditions cA = CAQ
CA = CAL
at z = 0 at z = L
(1.16-46)
where a.16-49)
1.
except at t = x.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
57
1. t) is the Green's function for this problem. Once a Green's function has been obtained for a given operator and a set of boundary conditions. Thus the Green's function is equivalent to the integrating factor in the first-order equation. as a differential equation in t. G is continuous at t = x. That is G = G t= x_ t= x +
(1. subject to the conditions y = 0 at x = xQ and x = Xj can be written as
y(x) =
P G(x. This requires that during the process of constructing G.16-4). x and t. The Green's function is a function of two variables. In Equation (1.t)f(t)dt
(1.18-2)
G satisfies the boundary conditions. The solution of Equation (1. The Green's function for any x has to satisfy conditions such as i) G(x. we regard the differential equation which G has to satisfy. The solution of a non-homogeneous boundary value problem can be obtained if the Green's function G is known for the homogeneous equation. The boundary conditions will be applied at t = x0 and t = Xj . Our purpose in introducing the Green's function now is to show the generality of the method.18-1). then the solution to the boundary value problem can be written as an integral.18-1)
where G (x.18 SOLUTIONS USING GREEN'S FUNCTIONS The Green's function method is a powerful method to solve boundary value problems and can be used not only for ordinary differential equations but also for partial differential equations and integral equations.16-1) with Q(x) replaced by f(x). t) is a solution of the homogeneous Equation (1. That is to say d_G_ + A dG_ + BG dt 2 ii) iii)
dt =
o
(1. we integrate with respect to t so as to obtain a solution y at point x.18-3)
. That is G = 0 at t = xQ and t = x 1 .
t) = G(t. x) Example 1.1.18-2). Having obtained the solution to Equation (1.18-9) (1. We try a solution of the form y p = a (x) cos x + b (x) sin x where we replace the constants Cj and c 2 by functions of x.18-6) for any f (x). We can then show that the Green's function satisfies conditions (i to v). We then calculate yp = -a(x) sin x + b(x) cos x + a'(x) cos x + b'(x) sinx (1.18-4)
More generally.18-7) (1. dG
dt
t=x +
dG
dt
=1
t=x_
(1.58
ADVANCED MATHEMATICS
iv)
The first derivative is discontinuous at t = x.18-6). The solution to the differential equation is given.18-5)
(1. in general. by Equation (1.18-1.18-6)
The particular solution y p can be obtained via the method of variation of parameters. Having dx 2 obtained the Green's function allows us to write down the solution to Equation (1. The solution to the homogeneous equation is y h = Cj cos x + c 2 sin x (1. v) G is symmetric. it will be possible to deduce the Green's function for the operator d2 L = \. G(x.18-8)
. as follows. Solve the equation d2v H r + Y = f(x) dx 2 subject to y (0) = y (1) = 0 and identify the Green's function. this difference equals the reciprocal of the coefficient of the highest derivative in Equation (1.16-6).
c o s x sin ( x .18-2 and in Problem 37b.18-1. b) we can deduce that G satisfies conditions (iv. t) should be chosen from the right side column in Table 1.18-25a. b).
. t) is the Green's function and is given by /sirU^inCx^) sin(l) G(x. The Green's function can of course also be constructed from the condition given by Equations (1. b)
From Equations (1.18-25a. via Equations (1.62
ADVANCED MATHEMATICS
= (Xf(t)[sintsin(x-l)dtldt Jo L sin(l) J
+
fl ) x
rsinxsin(t-l)ldt [ sm(l) J
^
y(x) = I
f(t) G(x. is given by Equation (1.l )
t=x _"
=
n is o&»\
(L18'26a)
sinTI)
sinx cos(x-1) sm(l)
t=x+ and G' t= x +
.18-24f) where the appropriate function G(x.l ) sin(l)
Y < t < 1 Q
^
x
(1. which is independent of f (t). That is to say. t) = sinx s i n ( t .G1
= sinx cos ( x . the solution to a given problem is to be written as in Equation (1.18-1) where G(x.1 ) .18-25a.18-25a. As mentioned earlier. The Green's functions for several frequently used operators and boundary conditions are given in the following table.18-2 to 5). b). we deduce that r1
G
G. v) given earlier.
_ cosx s i n ( x . t).1 ) t=x_ sin(l)
=
{
^
The solution to the problem therefore is given by Equation (1. t) dt
(1. The boundaries in Table 1. The boundary conditions will be satisfied automatically. In particular. This exercise is addressed in Example 1.18-6) can now be determined for any f (x). the solution of Equation (1.18-1)
where G (x.18-1 are all chosen to be x = 0 and x = 1.
18-1.b.x . the force acts at one point only. t) is given by G(x. c3 = . Thus f(x) will be zero every where except at the point x = £.18-31)
G (x.18-35) (1. That is to say.18-38) t<x t>x (1. = c3(x) + c4(x) t .18-36a. we can consider f (x) to be an impulsive force. t<x t>x (1.18-37a) (1.18-32a) (1. t) is given by G(x.18-1) as /•I y(x)= .18-33a) (1. The displacement y at any point x is then given by Equation (1.d) (1. = x(t-l).l ) .18-34) (1.t) = t ( x .18-32b)
Applying the boundary conditions on t yields Cj = 0 c3 + c4 = 0 The continuity condition implies xc 2 = c 3 + xc 4 The jump discontinuity can be expressed as c4-c2 = 1 From Equations (1. which is given in Table 1. t) = C!(x) + c 2 (x)t . say at x = £.c. We write f(x) as
.t) ^ dt (1. The result are Cj=O.l ) .18-33 to 35) we can solve for c1 toc 4 .18-33b)
Thus G (x.18-37b)
Jo
As an example. c2 = ( x . Mathematically f(x) can be represented by the Dirac delta function 8.| G(x.64
ADVANCED MATHEMATICS
Since G has to satisfy
ft d2G -^-=°
(1. c4 = x (1. where it is not defined.
.
5.17. a Reynolds number. one can compare flow situations in pipes of small to very large diameters in terms of a dimensionless variable £. The main properties of the Dirac delta function are given in Section 1. For example. Draw a sketch of the problem.
Jo
(. it is useful to proceed according to the following steps. Formulate a model in mathematical terms. etc. That is to say.U e x p . is to consider the Dirac delta function as a functional or distribution. 2. G represents the displacement at x due to a point force applied at "t. This may reduce the number of variables and it allows for the identification of controlling variables such as.f ' G ( * .18-41b) provides a physical interpretation of the Green's function. 3.^ d . we obtain
y(x) = ." T 5 ( .19 MODELLING OF PHYSICAL SYSTEMS
In order to generate the equation(s) representing the physical situation of interest.
An alternative approach. 1.8-4^)
(U8-4.b)
-5%i2
Equation (1. Combining Equations (1.18-38. = £ which will always vary from 0 to 1. introduced by Schwartz (1957).REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
65. for example.18-40)
e.18-39)
The delta function is not an ordinary function but it can be regarded as a limit of a sequence of functions. given by Equations (1. determine the equation(s) as well as the boundary and/or initial condition(s).5 )
(1. Indicate information such as pressures.
. R Determine the limiting forms (asymptotic solutions). temperatures.17-14c). It also allows one to easily compare situations which are dimensionally quite different. An example of such a sequence is lim .
4. Non-dimensionalize the equation(s). Make sure you understand the physical process(es) involved.
f(x) = 5 ( x . 1. 39).( M 2
e->0 VTCE L
v
= 5(x)
(1. Both approaches produce equivalent results..
19-1 illustrates the problem which is to be modelled subject to the following assumptions steady-state flow. the fluid is incompressible: p = constant.19-1
Flow in a pipe with associated velocity profile v z (r) and shear stress profile x r z
.66
ADVANCED
MATHEMATICS
6.
<
i
FIGURE 1. there are no end effects. Figure 1.
Solve the equation(s). We can imagine thin concentric cylindrical sheets of liquid sliding past each other. the piece of pipe of length L which we are considering is located far from either end of the pipe which is very long. 7. that is. laminar flow: Re = — ^ — — — < 2100 and vz is a function of r only. we consider the flow of a Newtonian fluid in a pipe. Verify your solution(s)! Do they make sense physically? Do all terms have the same dimensions?
As a first example.
r. The forces acting on the system (cylindrical shell of thickness Ar and length L) are: (i) the pressure force in the z-direction. and (ii) by viscous (molecular) transport. c) z-momentum in by viscous transport at r. Since the flow is laminar. and (ii) the z-component of the gravity force.19-1) is zero because of the steady-state assumption (no accumulation!).REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
67
Cylindrical coordinates are used since the geometry of the problem is cylindrical. That is z-momentum in T rz (27crL)| r by viscous transport at r
L o
. The differential equation representing this physical situation is obtained by applying a momentum (or force) balance on a shell of thickness Ar. n. This makes it mathematically easier (less calculations) to apply the boundary conditions. That is z-momentum in (v z 27crAr)pv z | z = by convection at z = 0 b) Similarly we write z-momentum out (v z 27trAr)pv z | z = by convection at z = L Note that the quantities 2. The individual contributions making up the force balance are a) z-momentum in by convection at z = 0. with pvz. Analyzing the problem. v z is not going to change with z so that the net contribution to the force balance of the zmomentum associated with convection is zero. Ar and p are not changing. This contribution is obtained by multiplying the shear stress x rz (force per area) by the appropriate area (2 TtrL). one has to realize that z momentum is transported according to two mechanisms: (i) by convection due to bulk flow. This contribution is obtained by multiplying the volumetric flowrate associated with the appropriate area (27trAr). The balance is given as follows / rate of z \_l rate of z \ + [net sum of forces] _ Q vmomentumin/ Imomentum out/ I in z-direction / Q 19-1) "
The right side of Equation (1.
That is pressure force at z = 0 f) Similarly we write pressure force at z = L g) P L (27rrAr) PQ (27trAr)
The z-contribution to the force balance due to the gravity acting on the system (shell) is given by the weight of the fluid in the shell.19-2) has to be a function of r and we will have a net contribution from the z-momentum by viscous transport to the force balance.
. That is. e).19-2)
dv vz is a function of r. We established earlier that Trz is a function of r.Jg^)r
Ar L
+ rpgKz!!L) = 0 ' ^ L '
(1.19-1.19. The Newtonian viscosity |I is constant and therefore the left side of Equation (1.h L . f) and g) into the force balance yields t r z (27crL)| r -x r z (27 t rL)| r + A r +(P 0 -P L )27crAr + 27crpg(h 0 -h L )Ar = 0 We can divide Equation (1.19-1) by the appropriate area (27trAr). d). e) The contribution from the pressure force is obtained by multiplying the pressure (force per area) at z = 0 (Po in Figure 1. the difference in height of the positions at z = 0 and z = L in Figure 1.19-3)
[^-^UA.68
ADVANCED MATHEMATICS
d)
Similarly we write z-momentum in
T r z (27irL)| r + A r
by viscous transport at r + Ar Note that since we are dealing with a Newtonian fluid.h L ) Ar. the shear stress t r z is given by dv^ ^ dr
XfZ
(1. the volume (2rcrArL) multiplied by pg and by cos a (z-contribution) gravity force (27trArL) pg cos a
Note that L cos a = h 0 . Substitution of the terms in c).4)
Note that we did not divide by r. The gravity force can therefore be written as: 2nr pg (h 0 .19-3) by 2ftLAr to obtain (1. otherwise —-^ would be zero.
\z = § Px^)
(1. We cannot accept an infinitely large force (per area). vz = 0 at r = R.19-16) allows us to evaluate the integration constant C 2 . we look at the physical situation for r = 0. Therefore Q = 0 and Trz is a linear function of r. That is.19-13)
dv The velocity profile is obtained by replacing the left side of Equation (1.19)
H^MR2
Substitution into Equation (1.70
_ _
ADVANCED
MATHEMATICS
To evaluate Cx.19-1). (1960). Applying this to Equation (1. v z is related to the square of r (a parabola as in Figure 1. The relation for v z is usually written as (1.17 >
(ii9-i8)
(L19. The minus sign is as a result of following the convention by Bird et al.19-14) This equation is solved as follows
\dv*" " F i ^ ) Irdr
^-(^MT^
( 1 1 9 -15)
(U9-I6)
The physics of the situation allows us to assume that the velocity is zero at the wall.19-13) by -fi — £ . We now have a differential equation relating vz to r given by (1.19-20)
.19-16) yields
Vi =
|?ci) R 'Jti| r >
That is.
° = -(^)f+c^
So
" 19 .
as it should.19-23)
Note also that the limits of integration have to be adjusted according to the newly introduced variable £. The lower limit remains unchanged. That is. Indeed. Equation (1. as follows
e2n /-R
I vrdrdG <vz) = *>-» I r dr d6 Jo Jo
(1. as it should.19-20. The integral can then be written as R
I (l-^ 2 )^d^ = l Jo
The constant R2 necessary to non-dimensionalize r dr is obtained from the denominator. we consider the flow of a non-Newtonian fluid in a pipe. goes to one. at r = 0. As a second example.19-21)
The denominator yields n R2. We will assume the fluid to obey the power-law. 21) yields
=
\ 4. Combining Equations (1.19-2.b) Example 1. That is to say
\z = . = £•. the new variable £.
(1. )
h 1 lR>]
TCR2
(1 19 _ 22)
We introduce a dimensionless variable £.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
77
Note that the maximum velocity occurs at the center of the tube.^ |
n
a-19-25)
.0. Note that this integral is now independent of R and is valid for pipes of any radius.19-22) reduces to (1. Integrating this product yields a flow rate. The numerator is the product of a point velocity (vz) and an infinitesimal area (r dr d0).19-24a. The average velocity (vz) is obtained by dividing the flow rate by the cross-sectional area. as the upper limit goes to R.
19-26) /dv \ and the shear rate — — is given by
(1. Equation (1. one should realize that Equation (1.
vz = i^f^f11
1/n
| -r1/n dr
r /n+
(1.12-1.19-28a)
p _p \ _ l l\ (-fen TTT +C2
< 1 1 9 "28b)
n / The constant C2 is obtained.19-29) reduces to Equation (1. it is valid for any fluid. In fact.19-13) now becomes (1.19-30b)
. We now proceed by computing the velocity distribution.19-30a)
1 +3 n
2\iL
J
(1. using the same (no slip) boundary condition as in Example 1.19-21). and the velocity profile is
(1.19-29) Note that for n = 1. the power-law fluid reduces to the Newtonian fluid.
(1. The average velocity is obtained by substituting Equation (1. Equation (1.19-13) is still valid.19-1. following the procedure given in Example 1. since no assumption as to the type of fluid has to be made to derive the shear stress distribution.19-20).72
ADVANCED
MATHEMATICS
At this point. That is to say.19-29) into Equation (1.19-27) Integration yields the velocity profile.
. note that the dimensions of ji depend on n and jx has the same dimension in both examples when n = 1.19-30c) yields
i = 2a±jJi_(i.19-2. In this example. 0<JL< 1
0
1 l-(if
=
\ R
R
2 ['-(it) 2
We note that for n = 1.) 1/n+1 ] n + 1 [ lR) J (vj
(L19-31) v '
Different values of n yield different profiles. However. For n = 0. The asymptotic solutions (linear profiles) are obtained for n = 0 and n = <x>.
Vz
(l.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIA!. For intermediate values of n. as n tends to zero and to infinity. the slope is zero and for n = °°. we obtain the Newtonian result (parabolic profile). To achieve this.19-1 and are shown in Figure 1. we continued to use parameter \JL as in Example 1. In practice. TABLE 1. These profiles are illustrated in Figure 1.19-29) by Equation (1. we plot j-^-r versus —.19-2.19-1.
\ v z)
R
Dividing Equation (1. EQUATIONS
73
We are now in a position to compare the velocity profiles in dimensionless form and to evaluate
y
asymptotic behavior.19-1 Velocity profiles n
r -. the slope must lie between 0 and 3. n is usually between 0 and 1 and the observed velocity profiles are indeed more blunt than the Newtonian one. we obtain a maximum slope of 3. The ones of interest are tabulated in Table 1.
The volumetric flow rate Q of a non-Newtonian fluid in a circular tube of radius R is given by
Q(R. 15 a.xR) = ^ [ R Y r z t r 2 z d t r z
TR JO
where t R and t r z are the shear stress at the wall and at any point in the tube respectively.REVIEW OF CALCULUS
AND ORDINARY
DIFFERENTIAL
EQUATIONS
7J_
a2u
= c 3t 2
2a2u
. The pressure P. can be written as
. c is a constant.and -^ . 16b. (3. " 3T . which is a function of the shear stress. By differentiating with respect to i R . 3x 2
is transformed to
35 an
14a. y ^ is the shear rate. show that the viscosity r\ (y R ). show that the shear rate at the wall YR is given by
y
=
JM3Q
+
TtR3 k
dQ-l
dtJ
Also. which is defined as the ratio of the shear stress to the shear rate. the volume V and the temperature T are given by
|p + -0LJ(V-p)
=
RT
where a. dy Compute —^ if dx y 2 .9V Compute ^p. This is the Van der Waals' equation.sin xy + x 2 = 4 . and R are constants.
P-33 Motion of a manometer fluid 35a.
. Consider separately the cases when t > a and t < a. a shear of magnitude Yo w a s suddenly imposed. it was deduced that the constitutive equation of a Maxwell fluid can be written as
v . Using the properties of the Heaviside and Dirac functions. deduce the shear stress Tyx at time t. In Problem 27b. at time t = a.82
ADVANCED MATHEMATICS
P I
FIGURE l.f f<=-(t-t')A»v<t'>d'1
A Maxwell fluid has been at rest and.
a > 0 .
Construct the Green's function.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
£2
wL I t )
1
•Wo It)
V
FIGURE l. y = e~l + e a . That is to say (i) (ii) determine the solution to the homogeneous equation for the cases t < x and t >x.
37b. Answer: t > a. by applying the conditions given by Equations (1.18-2 to 5).
(iii)
.
Mixing process at constant volumetric flow rate
Solve the differential equation
£ + y = «(. (i) (ii) by the Laplace transform method.-a)
with y(0)= 1.18-6). y = e.P-34
36b.1
Consider the cases t > a and t < a separately. use the jump discontinuity condition of Equation (1.18-4) to determine the remaining equation needed to solve for the constants that were generated in step (i). needed to solve Equation (1. use the boundary and continuity conditions to determine the relations which have to be satisfied by the constants. by finding the integrating factor.t t < a.
Show that this can be done in the following way. As the air penetration in the liquid film depends on the film thickness. we wish to establish a relation between the film thickness 8 and the velocity V. up to a wall C as illustrated.
In order to accelerate the aeration of a Newtonian fluid. The belt has a width W and a velocity V. (i) Perform a momentum (or force) balance over a length L to obtain the following shear stress distribution txz = pgxcosp List all assumptions made.P-41
Aeration of a fluid
. (ii) Combine Newton's law with this equation and show the velocity profile to be
Vz =
_v+pgcosP(g2_x2)
(iii)
Calculate the flow rate
r f8
Q = v z dxdy
Jo Jo
and deduce from the physics of the situation that
62 =
_3MV_ p g cos P
FIGURE l. The belt transports a laminar film of liquid.P-41.REVIEW OF CALCULUS AND ORDINARY DIFFERENTIAL EQUATIONS
§A
41b. a continuous belt has been introduced in the fluid reservoir. as shown in Figure l.
the sensible heat required to cool the droplet from To to TM is negligible compared to the latent heat of fusion. The constant k is the thermal conductivity. note that the heat flux qr is linearly related to the temperature gradient ^J.
In the design of spray dryers.86
ADVANCED MATHEMATICS
42b. one is interested in knowing the time required to solidify the liquid droplets.
(ii) (iii)
Perform an energy balance on the solidified spherical portion and show that the temperature profile is given by T-To
T —T
ls l0 =
R'f1-!-1
1 1
R^-R-1
In performing this calculation. show that the time required to solidify the droplet is given by
tf =
f
\6 k
/lRJi + l|[p R A f i f
S/^To-Tj
where AHf is the latent heat of solidification per unit mass.P-42. as follows dr
4r
dr
This is Fourier's law. Estimate the time tf required to solidify a droplet of radius R if (i) the droplet is initially at the melt temperature T o and the surrounding air is at T^ as shown in Figure 1 . the heat transfer coefficient h at the solid gas interface is constant.
. Show that the heat loss to the surrounding air is given by
9
47tR2h(Tn-Tj
Given that the heat loss at the liquid-solid interface (Rf) is -pAH f 47tRf —j-£.
The form of this series depends on the nature of the point xQ.'s.D. If the coefficients of the differential equation are not constants.E.1-1) can be written in standard form as y " + p 1 y 1 + p2y =b where pj = aj/a 2 .E.1-2a) is homogeneous. Since the derivatives of exponential functions are also exponential functions.16-8. Higher order O.1-2a) can be written as y = yh + yP ( 2 -i-3)
where y^ is the homogeneous solution and y p is the particular integral. The point x 0 is an ordinary point if both pj and p 2 are analytic at the point xQ.x 0 ). If b is zero. we consider second order O. Equation (2.18-1 and Section 2.E.c. The solution of Equation (2. We recall that a
. The particular integral (or particular solution) is usually obtained by the method of variation of parameters which is described in Example 1.'s).d) (2. we have seen that second order linear differential equations with constant coefficients admit a solution in the form of an exponential function. p 2 = a Q /a 2 . we develop methods of solving O. 9)].1-2a) (2.D. the differential equation reduces to an algebraic equation [see Equations (1. The solution of the homogeneous equation in the neighborhood of a point x 0 is assumed to be given by a power series in (x .1-2b.E. then the solution is not of an exponential form.CHAPTER 2
SERIES SOLUTIONS AND SPECIAL FUNCTIONS
2.D. we reviewed several of the standard techniques used to solve ordinary differential equations (O. we have only the homogeneous solution.E. As in Chapter 1.1-1)
We recall that if b is zero.'s with variable coefficients. b = a/a2 (2.D. In particular. Equation (2.D. can be written as a 2 (x)y" + a 1 (x)y' + a o (x)y =a(x) where a2 (x) * 0 and ' denotes differentiation with respect to x. In this chapter. A second order O.5.'s can be solved by the same method.1 DEFINITIONS
In Chapter 1.
iii)
The standard form of this equation is given by (2.b.1-5b) (2. i) This equation is in standard form and P! = x . P2. ii) The standard form of this equation is given by
""•TV+M^T)^0 p-=^T^2-3^).1-5a) (2. xQ is a singular point. i) ii) iii) y"+xy' + (x2-4)y =0 (x-l)y" + x y ' + i y =0 x 2 ( x .1-4a)
(2. c). From Equations (2. «"••»
(21-7a)
<2-'-™.cd)
Pl is not analytic at x = 1 and P2 is not analytic at x = 0 and at x = 1.1-2b. Analyze the following equations for ordinary and singular points. The singular points are therefore at x = 0 and at x = 1 and all other points are ordinary points. then x 0 is a singular point. That is to say.1-1.P2) is not analytic at x 0 . and b and by determining the presence or absence of singular points.1-5c)
We proceed by evaluating the functions p^.2 ) 2 y " + 2 ( x . f(x) can be represented by the series
oo
f(x) = X a n (x-x 0 ) n
n=0
(2. Example 2. it can be seen that if a2(x 0 ) is zero and either a^ (x 0 ) or a o ( x o ) is non-zero.1-6a. and b = 0 (2.1-4b)
an = f ( n ) ( x 0 ) / n !
If one or the other (or both) coefficients (pi. P2 = x 2 .2Q
ADVANCED
MATHEMATICS
function f(x) is analytic at x 0 if its Taylor series about x 0 exists.c)
These functions are analytic everywhere and all points are ordinary points.4 .2 ) y ' + (x + l ) y = 0 (2.1-8a)
x 2 (x-2) 2
x 2 (x-2r
.
92
ADVANCED
MATHEMATICS
2. The third series diverges for all values of x (* 0). It is usual to state that R is zero for series that converge only at x = 0 and R is infinite for series that converge for all values of x. No conclusion can be drawn if L = 1. There are three possibilities: (a) the series converges for all values of x.. and (c) the series diverges for all non-zero values of x.2-lc)
The first series represents exp (x) and is convergent for all values of x. The series converges at a point x if the
lim
m->0°
2L an x "
n=0
the sum of the series is the value of this limit. We illustrate this situation by the following examples. If a series converges for |xj < R and diverges for I x I > R.
(2.
(2. we can write the series as ^
m n=0
e x i s t s and
an x n .
(2.2-lb)
(c)
£
n=0
n ! x n = l + x + 2 ! x 2 + 3 ! x 3 + .. In many cases.1-4a) is a power series in (x ... Thus all power series have a radius of convergence. If L > 1.Not all power series converge for all non-zero values of x.. A series may or may not converge at its radius of convergence. Every power series
converges at x = 0 and its sum is a0. (b) the series converges for some values of x and diverges for other values of x.
00
Applying this test to the series ^
n=0
an x n .2-2)
is less than one (L < 1).
00
(a)
]£
n=o
00
x "/( n ! )
= 1+x + x 2 /2!+x 3 /3! + . By translating the
oo
origin to x 0 .2-1 a)
(b)
2L x "
n=0
00
=
l + x + x 2 + x 3 + .2
POWER SERIES
The series on the right side of Equation (2. we have (x * 0)
. The second series is a geometric series and its sum is 1 /(I .. R is the radius of convergence.x 0 ).x) and is convergent if |x| < 1 and diverges if | x| > 1. We recall that the
00
series X
n=0
un
is convergent if the ratio Hfl±i _
u n
lim
n —> 00
L
(2. the radius of convergence R can be determined by the ratio test. the series is divergent.
that is to say
.2| < 1 and is divergent if |x | > 1 or |x . the series
oo oo oo
becomes ^T 1/(1 + n) and since ^
n=0 n=l
1/n is divergent. For
oo
x = 1. we translate the origin and write
x* = x .2-8)
As in a).
oo
If the series ^
n=0
a n x n converges for | x | < R (R •£• 0) and its sum is denoted by f (x). At both of these
oo
values of x. if u n > u n + 1 > 0
n=0
and lim u
n—>~
is zero. the series is convergent.b)
n=0
The series becomes £
R = lim 3 + 2n _ j n —>~ l + 2n
The series is convergent if |x | < 1 or |x .l .
oo
the series
1/n .2
oo
(2. which states that for an alternating series j ^ ( . If x = . the series ^ ( .1 .2-9) (x*) 2n /(l +2n). the series is convergent
if x = .2| > 1. the series is convergent if | x| < 1 and divergent if | x| > 1. We apply
n=0
oo
Leibnitz's test. In the present example.l ) n u n . the series behaves as ^
n=0
1/n and is divergent.
We make use of the following properties of power series.
(2.l . it follows that ^
n=0
OO
1/(1 + n) is also
divergent.94
ADVANCED
MATHEMATICS
The series is convergent if |x| < 1 and is divergent if |x| > 1. To x = 1 corresponds x = 1 or x = 3.l ) n / ( l + n ) is an alternating series. For x = 1. b) R = Urn
n —> oo
f±Jl
1 + n
= 1
(2. In this example x* is raised to an even power and there is no need to separately consider the cases x = 1 and x = .2-10a. c) In this case. we
becomes
jT 1 / ( I + n )
n=0
and by comparison with the convergent series
^
n=l
deduce that the series is also convergent if |x| = 1.
d)
where p l n = pf°(O) and p 2 n = p f (0)
. If we are required to find the solution at points near infinity. Without loss of generality. we can translate the origin and write x* = x .3-2c.
2. The differentiated series have the same radius of convergence R and converge to the corresponding derivatives of f (x). The functions pj and P2 are analytic at x 0 (= 0) and their Taylor series are
oo
(2. as given by Equation (2.3-lb)
Pi 00 = X Pln X "
n=0
oo
<2-3"2a)
P200 = Z
n=0
P2nx"
(2.E.SERIFS SOLUTIONS AND SPECIAL FUNCTIONS
oo
91
f(x) = X
n=0
anx"
(2.3-2b) (2. the two series ^
n=0
a n x n and
^
n=0
b n x n can be added and multiplied in the same way as polynomials. If x 0 is non-zero. we can set x 0 to be zero.
If two power series converge to the same sum throughout their interval of convergence. The series can also be integrated term by term and the resulting series represents the integral of f (x).x 0 In the new variable x .3-la)
(2.D.
oo oo
If ^
n=0
OO
b n x n is another power series with radius of convergence R. x = x 0 corresponds to x = 0 . their coefficients are equal.3
ORDINARY POINTS
Consider the standard form of the second order O.2-11)
the series can be differentiated term by term as many times as is required. We seek a solution in the neighborhood of XQ.1-2a). we change the independent variable from x to Xj and write X! = 1/x To points near x at infinity correspond to points near Xj at the origin. Further discussions on power series are given in Chapter 3.
.3-6)
y'
1+x2
2—y 1+x2
=
o
(2.3-6) is written as y"+
2x
(2.96
ADVANCED MATHEMATICS
The homogeneous form of Equation (2. = 0 where kj are known expressions in terms of p^:.3-4a to c) into Equation (2. P2k ...2
n=2
Substituting Equations (2. ) is zero. In standard form.3-7)
.. Equation (2.3-1. and c^ . 1...3-4a)
where c n (n = 0.3-5) is true for all x and this implies that each of the kj (i = 0. The recurrence equation (kj = O) allows us to determine c n in terms of pjj and P2j . Example 2.3-3) results in an equation of the form k 0 + kj x + k 2 x 2 + .. Differentiating term by term yields
oo
/ = X ^n^"1
n= l
oo
(2-3"4b)
(2.3-5)
(2..1-2a) can be written as
y"+ X Pm x " yf+ X P 2 « 4
\n=0 / Vn=O /
=
°
(2-3"3)
The solution y also has a Taylor series near the origin and y can be expressed as
y = X cnxn
n=0
(2. Obtain the power series solution of (l+x2)y"+2xy'-2y = 0 in the neighborhood of the origin. Equation (2.3-4c)
y" = X n(n-l)c n x n . 1. + k n xn + . The next example illustrates the method of obtaining a series solution.. .) are constants.
.. On expanding p^ (x) and p 2 (x) in powers of x. b) are convergent for | x| < 1 and pj (x) and p 2 (x) are analytic in the interval | x| < 1.3-1.. it can be seen that yj can be written as yx = 1 + x (x ..x 3 /3 + x 5 /5 . In Example 2.D.3-15a. In Example 2. we obtain Pj (x) = 2x (1 .2 (1 . ) p 2 (x) = .x 4 /3 + x 6 /5 . the series is an alternating series and by Leibnitz's test. ) (2...98
ADVANCED MATHEMATICS
where c 0 andcj are arbitrary constants. It must be pointed out that it does not follow that there is no analytic solution that goes beyond the critical point of the differential equation. the solution y 2 (= x) is valid for all values of x.x 2 + x 4 .. There is no loss of generality in setting c 0 and Cj to be equal to 1. The general solution is given by the linear combination of yj and y 2 .x 2 + x 4 . We observe that the series solution about an ordinary point is valid in an interval that extends at least up to the nearest singular point of the differential equation.6-5). The general solution of a second order equation involves two arbitrary constants (A and B) and they are determined by the initial conditions [y (0) and y'(0)] or by the boundary conditions.3-12a.3-15b)
We note that the series in Equations (2.3-14) (2. From Equation (2. The solution yi is in the form of an infinite series and is valid as long as the series is convergent. The general solution y is expressed as y = Ay1+By2 where A and B are arbitrary constants.x 6 + . At |x| = 1.x 6 + .3-1.b)
The infinite series is convergent for |x| < 1. . The
.E and has two linearly independent solutions yj and y 2 . it is convergent..3-1 and can be generalized to all power series solutions about an ordinary point. The fundamental solutions y^ and y 2 can be written as yj = 1 + x 2 . • Note that Equation (2. The solution yj is valid for | x | < l . y 2 (= x) has only one term and is a valid solution for all values of x. The point ( x = l ) is a singular point (see Example 3. This observation is not restricted to Example 2.3-15a) (2.3-11)..3-6) is a second order O.3-13) y2 = x (2..) (2.
b)
This means that if Equation (2. x r and ^r.4-16a. f (r0) = 0
(2.4-20a.4-21)
where c 0 is not zero and r is any real or complex number.(x r ) = j .4-la.[exp (r i n x)] = ( i n x) exp (r i n x) = xrinx The other linearly independent solution is y2 = x i n x (2. b). by changing the independent variable x to t [Equation (2.4-17c)
Alternatively. To differentiate with respect to r.4-15)
(2.
. That is to say j .(x r ) are solutions of the or differential equation.6-11)] Yi = el = x y2 = t e l = x i n x • Note that the exponent r is not necessarily an integer. we note that if f (r) has a double root at r 0 .4-18) (2.b) (2.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
]QJ
Yl = x To obtain the other solution. then f(r 0 ) = 0. we write x r as exp (r i n x). This suggests that if we retain all the terms on the right side of Equation (2.4-10b) can be written as *-JL _ 2 ^ + y = 0 dr dt The solutions are [see Equation (1.4-19)
(2.4-6) has a double root. we should try a solution of the form
CO
(2.4-20c.d)
y = x r X cnxn
n=0
(2.4-17b) (2. Equation (2.4-17a) (2.4-7)].
4-22a)
yi = X
n=0
cnx"+r2
(2. xpj(x) = ( x + l ) / 2 . Obtain a power series solution to the following equation 2xy" + (x + 1) y' + 3y = 0 in the neighborhood of the origin.25 )
On differentiating. On comparing powers of x. the two roots are coincident.b) (2.102
ADVANCED MATHEMATICS
The series is differentiated term by term and substituted in the differential equation. p 2 (x) = 3/2x x 2 p 2 (x) = 3x/2 (2. The other equations are the recurrence formulae and are used to determine the coefficients c n [see Equations (2. In the examples that follow. we consider the three possible cases: the roots of the indicial equation are distinct and do not differ by an integer. we deduce that x = 0 is a regular singular point.4-6)] and is the indicial equation. We seek a solution of the form
oo
y = Z c n* n + r
n=0
( 2 . In this example p : (x) = ( x + l ) / 2 x .4-23)
From Equations (2.4-2. This method of solving a differential equation is called the method of Frobenius. yj and y2 are not linearly independent and we have to modify our method.4-24c. The two linearly independent solutions are
oo
yi = S
n=0
Cnx"+Fl
(2.4-24a to d).3-9atoh)]. The equation associated with the lowest power of x is a quadratic equation in r [see Equation (2. and the two roots differ by an integer.4-24a.4-lb) or the two roots differ by an integer.4-22D)
If the two roots coincide (see Example 2. we obtain
oo
y1 = X
n=0
(n + r)cnxn+T-1
(2. In general.d) (2.4 . the indicial equation yields two distinct values of r which are denoted rj and r2. Example 2. we obtain a set of algebraic equations.4-26a)
.
The general solution of Equation (2.4-44) is the linear combination of Jo and y2 and is y = AJ 0 + By 2 where A and B are constants. Note the presence of Jtn x in y 2 . This implies that y2 is singular at the origin. If the physics of the problem require y to be finite at the origin, we require B to be zero. Example 2.4-4. Solve the equation x 2 (x2 - 1) y" - (x2 + 1) x y' + (x2 + 1) y = 0 in the neighborhood of the origin. In this example, pi and p 2 are given respectively by (2.4-68) (2.4-67)
p , - - - ^ - .
x(x2-l)
P2 = f ^
x2(x2-l)
(2.4-69a,b)
The singular points are at x = 0, x = 1, and x = —1. We further note that x = 0 is a regular singular point and we assume a solution of the form
oo
To determine y 2 , the solution corresponding to r = - 1 , we substitute this value of r in Equation (2.4-72h) and we observe that the numerator ( r - 1 ) 2 is non-zero and the denominator [(2 + r) 2 - 1] is zero. This implies that c 2 is infinity and we must seek a solution of the form
00
Without loss of generality, we can set c 0 = b Q = 1 and the two linearly independent solutions y \ and y 2 can be written as
yi
= x,
y 2 = x i n x + 1/(2x)
(2.4-80a,b)
The general solution of Equation (2.4-68) is y = Ayi+By2 where A and B are arbitrary constants. Note that y 2 is singular at the origin and if the general solution y is finite at the origin, B has to be zero. Example 2.4-5. Find a series solution to the equation x 2 y" + x y ' + ( x 2 - l / 4 ) y = 0 valid near the origin. The origin is a regular singular point [xpj(x) = 1, x 2 p 2 (x) = (x2 - 1/4)] and we seek a solution in the form of Equation (2.4-70). Proceeding as in the previous example, we obtain
oo
The general solution of Equation (2.4-82) is y = Ay!+By2 where A and B are arbitrary constants. Observe that y 2 is singular at the origin and y^ is finite at the origin. For solutions which remain finite at the origin, we require B to be zero. Example 2.4-6. Gupta and Douglas (1967) considered a steady state diffusion problem, associated with a first order irreversible reaction involving isobutylene and spherical cation exchange resin particles. Equation (1) of their paper can be written as (see Equation A.IV-3) (2.4-96)
Ul
ADVANCED MATHEMATICS
(2.4-97a) |_r where R A = - k c A dr v dr / (2.4-97b)
Assuming c A to be a function of r only (symmetry!), Equation (2.4-97a) reduces to an O.D.E. given by ^AB r \
2
We now solve Equations (2.4-98a-c) for c A via the method of Frobenius. Note that the origin (r = 0) is a regular singular point. We seek a solution of the form
oo
C A=
X
n=0
Cnr"+P
(2-4-99)
In Equation (2.4-99), the exponent involves p, so as to avoid confusion with the radial variable r. Differentiating Equation (2.4-99) and substituting the resulting expressions in Equation (2.4-98a) yields
oo
(s - 1) (s) In this case, we have two arbitrary constants, c 0 and C}. This implies an even as well as an odd solution. To compute the even solution, we replace s by 2m and Equation (2.4-107) becomes (2.4-108)
116
ADVANCED MATHEMATICS
As before, we obtain Kmc c2m=^/ For the case where s is odd, we write (2-4-109)
^
" ( 2 m ) ( 2 ^ 1)
(2-4-U0)
In terms of c^, we obtain Kmc
C*-=(2^7TJ! <2 - 4 - ul)
Another solution to Equation (2.4-98a) is
' A , " c. 2 £ ^ - + 0 , 5 ) ^ ^
m=0 m=0
(2.4-112)
Note that the second term on the right side of Equation (2.4-112) is equal to the right side of Equation (2.4-106). Therefore, Equation (2.4-112) is the general solution of Equation (2.4-98a). The solution can be written in a closed form by observing that
~
m=0
v
which is Equation (2) in Gupta and Douglas (1967). Before summarizing the Frobenius method, we note that Equation (2.4-98a) can be transformed to a much simpler equation involving constant coefficients. Replacing c A by —— transforms the equation to
2
We summarize Frobenius's method of finding a solution near a regular singular point as follows. We assume that the solution is of the form
oo
y = Zcnxn+r'
n=0
c 0*°
(2.4-121)
We differentiate the series term by term and substitute the resulting expressions in the differential equation. On setting the coefficient of the lowest power of x to zero, we obtain a quadratic (for a
118
ADVANCED MATHEMATICS
second order equation) equation in r. We denote the two solutions by rj and r2. We now consider the following three cases. (a) rj and r2 are distinct and do not differ by an integer. By comparing powers of x, we obtain a recurrence formula that allows us to obtain cj, c 2 , c 3 ,... in terms of c 0 and r\ (or r 2 ). The two linearly independent solutions are
oo
yi = X ^ ( r ^ x 1 1 ^
n=0
(2.4-122a)
yi = X
n=0
cn(r2)x"+r2
(2.4-122b)
(b)
?i = r2. In this case, one solution yj is given by Equation (2.4-122a). To obtain the other solution, we assume a solution of the form
oo
y = y i i n x + £ bnxn+r>
n=0
(2.4-123)
We proceed as in case (a) to obtain b n . (c) rj and r 2 differ by an integer. Let us assume that r t > r 2 . The solution y 1 can be obtained as in (a). We try to compute the coefficients c s with the value of r = r 2 . If all the coefficients can be computed as in Examples 2.4-5 and 6, the second linearly independent solution y 2 is obtained. If, in the computation of c s , we have to divide by zero as in Example 2.4-4, we assume a solution in the form of Equation (2.4-123) and proceed to calculate b n . Note that in case (b) one of the solutions always has a i n x term whereas in case (c) this is not the case (see Examples 2.4-5 and 6). If one solution is known, we can obtain the second solution by the method of variation of parameters and this is explained in the next section. 2.5 METHOD OF VARIATION OF PARAMETERS
We use the method of variation of parameters to find a second linearly independent solution of Equation (2.1-2a) if one solution is known. Let yj be a solution of Equation (2.1-2a) and we assume a second solution to be given by y(x) = u ( x ) y i ( x ) On differentiating and substituting y, y1, and y" in Equation (2.1-2a), we obtain (2.5-1)
In the previous section, it is stated that the general solution of a second order O.D.E. is a linear combination of two linearly independent solutions. To determine the two constants, we need to impose two conditions. In many physical problems, the conditions are imposed at the boundaries and these problems are boundary value problems. Many of the second order boundary value problems can be stated as follows (r y')'+ (q + A,p) y = 0, subject to a i y ( a ) + a 2 y'(a) = 0 b 1 y(b) + b 2 y'(b) = 0 (2.6-lb) (2.6-lc) a<x<b (2.6-la)
SERIES SOLUTIONS AND SPECIAL FUNCTIONS
727
where r, p, and q are continuous real functions of x, A , is a constant parameter (possibly complex), aj, a2, bj, and b2 are constants. The system defined by Equations (2.6-la to c) is the Sturm Liouville problem. Equation (2.6-la) can be written as r y " + r ' y ' + (q + Ap)y = 0 (2.6-ld)
We assume X to be real and it can be positive, zero, or negative. We consider these three cases separately. (a) X,<0. For convenience, we set A. = - m 2 and the solution of Equation (2.6-3a) is y = c1emx + c2e-mx where Cj and c 2 are constants. To satisfy Equations (2.6-3b, c), we require 0 = C!+c2 0 = m (c, emn-c2 e~mn) (2.6-5a) (2.6-5b) (2.6-4)
Equation (2.6-10b) implies that either C5 is zero which leads to a trivial solution or that cos nn is zero which provides the non-trivial solution. The cos function has multiple zeros and cos n7t is zero if n = (2s+l)/2, s = 0, 1,2, ... (2.6-11)
The system defined by Equations (2.6-3a to c) has an infinite number of eigenvalues and they are given by
Xs
= f2^1) '
s = 0' 1 ' 2
'-
(2.6-12)
The corresponding eigenfunctions are
ys = sinf^s^J-jx
(2.6-13)
Note that the eigenvalues are real, positive, and discrete. Such properties are associated for example with discrete energy levels in quantum mechanics. • Next we discuss the general properties of the Sturm Liouville problem.
124
ADVANCED
MATHEMATICS
The Eigenvalues Are Real Suppose X is complex and this implies that y is also complex. Taking the complex conjugate of Equations (2.6-la to c) and noting that r, p, q, aj, a2, bj and b 2 are real, we obtain (ry 1 )' + ( q + I p ) y = 0 a 1 y(a) + a 2 y ' ( a ) = 0 b1y(b)+b2y'(b) = 0 where y is the complex conjugate of y. We multiply Equation (2.6-la) by y and Equation (2.6-14a) by y and subtract one from the other to yield (X-X)pyy = y(ry')'-y(ry')' On integrating, we obtain (2.6-15) (2.6-14a) (2.6-14b) (2.6-14c)
if i is odd. These polynomials are the Legendre polynomials and are denoted by P^ (x). The constants c 0 (or q ) are chosen such that P^ (1) is unity. In general.1).7-1
Legendre polynomials
A relatively easy method of computing P^ (x) is to use Rodrigues' formula which can be written as P^ (X) = .d)
P.3x) (2.i ) ( s + i + l)/(s+2)(s+l)
(2.7-11)
From Equation (2. The first few Legendre polynomials are shown in Figure 2.J — A _ ( x 2 . for any integer i . The case we considered earlier is k = 0 {Jt = 0) and the even solution is a constant. we observe that if Z is a non-negative integer. the even solution (a polynomial) is valid for all values of x and.U) PpU) *
i\
'
P*(XX
P|(x J // yf
l \ / \
///
.7-12a.l(
FIGURE 2. P2 (X) = 1 (3x2 .7-13)
. P^x) = x P3 (x) = 1 (5x2 . if i is even. c^+2 is zero and so are CZ+A' ci+6> •• ^nus tnc mfiniie series becomes a polynomial and the solution is valid for all values of x.i 2 . the odd solution is a polynomial.128
ADVANCED
MATHEMATICS
c s + 2 = c s ( s 2 + s .7-12c. In particular.l /
2ZJt\ dxZ
(2.i ) / ( s + 2 ) ( s + l ) = c s ( s . the degree of the polynomial is i .7-11).7-1 and are P 0 (x) = 1 .b) (2.
The orthogonal property can be written as ( P i (x) Pm(x) dx = — 1 — 8e m J-l 2i + l where 8^ m is the Kronecker delta.7-1) can be obtained by the method of variation of parameters. and the integrating factor IF. 20.7-20) (2.2xP/J + i ( i + 1) Pj] + (1-x 2 ) (u"P i + 2u'P<J) . we can calculate ?2.7-15) ( i + l)Pj + 1 (x) = (2i + l ) x P i ( x ) .2xu'P i = 0 Since P^ is a solution of Legendre' s equation. The second linearly independent solution of Equation (2.6-2c. From Equations (2. we obtain u [(1-x 2 ) PJ . Equation (2.E. is (2. P3.7-18) simplifies to (l-x2)Piu"+uI[2(l-x2)P/J-2xPi] = 0 On writing u' = v.7-17) (2...7-18)
.
l
= X l * pi
i=o
to
(2-7-14)
V l-2xt+t 2
Another method of determining P^ (x) is to use the recurrence formula which can be written as (2.7-20) is a first order O. Equation (2.SERIFS SOLUTIONS AND SPECIAL FUNCTIONS
129
The Legendre polynomials can also be obtained by using the generating function (l-2xt+t ) Expanding this function in powers of t yields
oo
.i P i _ 1 ( x ) Knowing Po and Pj. we deduce that the Legendre polynomials are orthogonal with respect to weight one. 21).7-16)
On differentiating Q^ (x) twice and substituting the resulting expressions in Equation (2.)-x/(l-x2)] = 0 dx A * Equation (2. An important property of the Legendre polynomials is the orthogonal property..7-19) can be written as 4 x + 2v[(Pl/P. We denote the solution by Q^ (x) and assume that Q i (x) = u(x)P i (x) (2.7-19) (2.
.7-1) with k = i ( i + 1).D.
7'-43) is J v (x). We consider the two cases where 2v is an odd or an even integer separately. If v is not an integer or zero. r 2 = .
(2. \
/
/
^
/
I
I
\/TT
-3* -2* -I*
W*\
I
3
r 'i ! -*
! ! !
. For v (* 0 or an integer). the other solution is obtained by considering the other root of the indicial equation [Equation (2. B must be zero.7-2 Gamma function Thus one solution of Equation (2. the general solution of Equation (2. If v is zero.7-63)
(2.v ) is an integer. In Example
. this implies that 2v (rl=v. J_v (x) is singular at the origin.* •
i i n
FIGURE 2.v /[(p!)r(p-v+l)]
Note that whereas J v (x) has no singularity at the origin.7-43) is y = AJ v (x) + BJ_ v (x) where A and B are constants. That is to say.4-3. the two linearly independent solutions are not obtained in a straight forward manner as described earlier for non-integral values of v. If the solution is finite at the origin. the two roots are coincident and this case has been considered in Example 2. the other solution is J_v (x) and is written as
oo
J_v(x) = X (-D P (x/2) 2 p . If the two roots differ by an integer.7-46b)].SERIES SOLUTIONS AND SPECIAL FUNCTIONS
ZiZ
! U *\
]
I
I
'
|r(u)l
I
I 2.7-64)
We recall that if the two roots of the indicial equation are coincident or differ by an integer.
J v (x) and J_v (x) are not linearly independent.J_v (x)] / sin vn (2.v + l) is ±°° for p < v .4-5.7-43) is given by Equation (2.7-68) are zero.7-62b.4-84d)] and we have one solution starting with c 0 and the other starting with cj. (2.7-63) can be written as
oo
L V W = X (-D q+V (x/2) 2q+V /[(q + v)!(q!)]
q=0
(2. the Bessel function of the second kind Y v (x) is defined as Y v (x) = [Jv (x) cos v% . We recall that if 2v is one. Instead.7-64 or 69). J v (x) and J_v (x) are obtained in the case where 2v is an odd integer. we obtain J 1 / 2 ( x ) a n ^ -L1/2 00.7-69)
. B is zero. we deduce that Y v (x) exists in the limit as v tends to an integer. If v is not an integer. we deduce that J_ V W = ( . we solved the case where 2v is one. This method yields (2. we conclude that both the numerator and the denominator on the right side of Equation (2. the other linearly independent solution can be obtained by the method of variation of parameters.7-65)
Comparing Equations (2. J x[J v (x)] 2
(2-7-67)
The solution y2 is usually not considered.138
ADVANCED
MATHEMATICS
2.v . v is an integer and from Equation (2. By applying l'Hopital's rule. Writing q = p . the general solution is y = AJ v (x) + BY v (x) where A and B are constants. Equation (2.7-68)
From Equation (2. Thus. The function Y v (x) has a in(x/2) term (see Problem 19b) and is singular at the origin. Since J v (x) is known. If y is finite at the origin.Similarly.D V J v ( x ) If v is an integer. the solution of Equation (2. q is not necessarily zero [see Equation (2. If 2v is an even integer. but if v is an integer only Equation (2. By assigning to c 0 and Cj the values given earlier.7-69) is valid.7-63) we note that the series starts from p = v since F ( p . 65).7-66).7-66)
yi = J v 00 I d]L—.
b).7-71a) (2.7-4.7-3 and Yo and Yl in Figure 2.7-43) instead of J v (x) and Y v (x).
JuU)
FIGURE 2. we obtain
Jv(x) = i-(Ht1) + Hf) Yv(x) = i ( H f .
We recall that there are circumstances when it is preferable to work with exp (± ix) rather than with sin x and cos x.7-7 lb)
The functions Jo and Jj are shown in Figure 2.7-70b)
are linearly independent and the general solution of Equation (2. From Equations (2. Equally. there are circumstances when it is preferable to choose Hankel functions (Bessel functions of the third kind) of order V as solutions of Equation (2.7-43)
is a linear combination of H^.7-70a. and H^ . Table 2. are defined by H ^ } = J v (x) + iY v (x) H f = Jv(x)-iYv(x) The functions H^ and H^ (2.7-1 lists some properties of Bessel functions. These Hankel functions H^ and H*.< } )
(2.7-70a) (2.7-3
Bessel functions of the first kind
.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
US.
the two linearly independent solutions are I v (x) and I_ v (x). We seek a series solution as described previously for Bessel's equation. p. 1963.7-81) The function K v (x) is linearly independent of I v (x) and the limit as v tends to an integer is defined. 524)
d4 Y d2 Y
—*A.i 7 l v / 2 J v (ix) (2. is real.7-77) for all values of v (including integral values) are I v (x) and K v (x). The function I v (x). That is to say. defined by Equation (2. A simplified form of the equation governing the linear stability of a Newtonian fluid flowing between two parallel walls can be written as (Rosenhead. Kamke (1959) and Murphy (1960) list several such possibilities.7-79)
Comparing Equations (2. we define a new function K v as (2.7-3. the solution needs to be given in the form of real variables and J v (ix) and J_ v (ix) are not in a suitable form.7-78)
Equation (2. If v is an integer.7-77) are J v (ix) and J_ v (ix).7-79).7-78) is exactly Equation (2.b)
Note that we have already established that the solution of Equation (2. Example 2. The properties of all Bessel functions are given in Watson (1966). Usually.7-77) is J v (ix) and to obtain the real part of the solution we multiply J v (ix) by a complex constant (i~ V ).7-82)
. The solution we obtain can be denoted by I v (x) as follows I v (x) = ]T (x/2) 2 p + v /[(p!) r ( p + v + 1)]
p=0
(2.-iri — ^ . we deduce that I v (x) = T v J v (ix) = e .v 2 ) y = 0 dz dz
(2.142 2
ADVANCED MATHEMATICS
z2 ^ + z ^ + (z 2 . they are Bessel functions with purely imaginary argument.7-62b. 79). If v is not an integer. The two linearly independent solutions of Equation (2.7-43) with z being replaced by x and the linearly independent solutions of Equation (2.= 0 dr| 4 dr| 2
(2. Many second order linear differential equations can be transformed to Bessel's equation (or to another standard equation) by a suitable substitution.7-80a.
146
ADVANCED
MATHEMATICS
for each e > 0 whenever N > N e and for all x in the interval. we obtain
an
j ~ =
s
(b
N
p(x)[-2fy s + 2 y s X c n y n ] d x Ja n=l and using the orthogonal property of yn. to the mean value of f(x) as x approaches x 0 from the left and from the right. If D —> 0 as N —> °°. The function y n can be continuous though f (x) can be discontinuous. the set of
functions (y l5 y2. s = l .) is complete and the space they span is a Hilbert space.[f(x 0 + ) + f (x o _)].X c n y n ] dx =
Ja n=l Ja
X c n y n +(X cnyn) ]dx
n=l n=l (2.8-1 lb)
( p(x)[-2fy s + 2c s y s 2 ]dx
Ja
(2. = 0. we require that the integral I p (x) [f (x) ~2J c n y n ] dx be a minimum. If for
rb
every f (x) for which j
•'*
2
£
converges m
p (x) f dx is finite.8-8a).8-10a.
2
that is to say.
. the sum ^ c n y n converges in the mean to
n=l
f (x). Let Ja
(b
D=
N
2
(h
P(x)[f2-2f
N
N
2
P « [ f .8-1 la)
On differentiating. . 2. Note that in this case the requirement is that the integrals exist and the function can have a jump discontinuity in the interval. X c n v n
n = l
the mean to f (x). the sum ^
n=l
cnvn
c o n v erges
to ^.. we write f^-=
acs
(2.8-llc)
Minimizing D implies that c s is given by Equation (2.b)
D is a function of c n and D is a minimum if | 5 . The approximation in this case is in the
N
least square sense.
N
If f(x) has a jump discontinuity at x 0 . Another approximation which is widely used in the treatment of experimental data is the least square a p p r o x i m a t i o n . Using the least square criterion.. (2.. The series is the Fourier series and the coefficient c n is the Fourier coefficient.. We demonstrate that this is the present case..
n a =1
71 I
f(x)cosnxdx J-n .8-14c)
The coefficients a n and b n given by Equations (2.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
MZ
Trigonometric Fourier Series The trigonometric functions sin x and cos x are periodic with period 2TC. b] to [-n .aQ and not a 0 so that the formula for a n [Equation (2. c) are special cases of Equation (2. If f (x) is defined in the interval [a.a0 + 2^ [a n cosnx + b n sinnx]
2
(2. the Fourier series [ p ( x ) = l ] . That is to say
oo
f(x) = i.
The functions sin x and cos x form an orthogonal basis in the interval [-71. The orthogonal property of cos x and sin x can be written as (2.c.8-13a)
r
(2.8-12)
I sinnxcosmxdx = 0.a ) transforms the interval [a.7t] with unit weight fK f 2 dx is bounded.8-8a).8-14a)
n=l
. b]. J-n
r
r
for all m and n
(2. If f(x) is periodic and of period 2TC and I J-n converges in the mean to f (x).8-14b)
b n = JJ-n
f(x)sinnxdx
(2. A periodic function f(x) of period T is defined by Equation (1.1-14).8-14b.d)
I sinnxsinmxdx = I cosnxcosmxdx = n 8 n m J-n J-n where 8 n m is the Kronecker delta. Note that we have written j . For simplicity. 7t].8-14b)] is
. we assume f (x) to be defined in the interval [-%. then the transformation x* = 27t[x-(a + b ) / 2 ] / ( b . n].8-13b.n
(2.
8-21)
The product of two even (or odd) functions is an even function and the product of an even and an odd function is an odd function.8-23b)
(2. we obtain
oo
(2. I
n
f (X) sin nx dx
(2.8-24a)
fn
b n = 2.8-23a)
an = 2.8-25a. If f (x) is an even function.150
ADVANCED MATHEMATICS
The functions cos x. x 2n (n is an integer) are even functions and the functions sin x. x^ 2n + 1^ are odd functions.8-14atoe) reduce to
oo
f (x) * 1 a0 + £
n=l
an cos nx
(2.8-24b)
Jo
Note that we have made use of the fact that fn 0. If f(x) is an even function f'(0) = 0 and if f(x) is odd f(0) = 0 (2.8-23c)
f(x) » £
n=l
b n sinnx
(2. it is possible to define f (x) to be even or odd. Mathematically.b) if f(x) is even
In many situations.8-22) (2. I f(x)cosnxdx n Jo
bn = 0 Similarly. all coefficients b n are zero and Equations (2. it is known that f (x) has a period of 2% but is defined only in the interval 0 < x < 7C. if f (x) is odd
I
J-n
f(x)dx = I 21 f(x)dx. if f (x) is an odd function. Jo
r7 C
(2. but usually the physics of the problem dictates whether f (x) is even or odd (see Chapter 5).
.
we can write
. that is to say. This means that
t x e x
oo
f x
I
/x 0
f(x)dx = i . If f(x) has a Fourier series. Under the conditions stated earlier.
ADVANCED MATHEMATICS
f(x). but not in the interval -K < x < 0.|
/x 0
aodx+^T I
n = l /x 0
(a n cosnx + b n sinnx)dx
(2.
-27T
-7T
0
7T
27T
x
FIGURE 2.U2. we deduce that y
l— -
sTo(2s+l)2
= — 8
(2. f (x) can be represented by Equations (2.8-28).8-29)
If f (x) has a Fourier series.8-14a to c). if f'(x) is piecewise continuous and f (—TC) = f (n).8-2
Even function
Note that both series on the right side of Equations (2.8-30)
where x 0 is arbitrary. At x = 0.8-17. the series can be differentiated term by term. the series can be integrated term by term. 28) represent the same function in the interval 0 < x < K. the function is continuous and from Equation (2.
b)
at x = 0 and x = JL. The deflection w of a uniform beam of length Z with an elastic support under a given external load p is given by E I ^ + k w = p(x) dx 4 where E I is the flexural rigidity and k is the modulus of the elastic support.
W
mini
FIGURE 2. Assume that the load is constant and is applied on the interval Jt /3 < x < 2Z13.8-40b)
Example 2.8-41)
d_w _ dx 2
0
(2.U6
ADVANCED MATHEMATICS
cosh n = sinhJL
l + 2
V
— 1 —
(2.
Beam on elastic foundation under an external load
.8-4
.8-42a. If the beam is supported at the ends.8-41) is given in von Karman and Biot(1940).8-4. The derivation of Equation (2. Obtain the deflection w (x). as shown in Figure 2. the boundary conditions are
2
w =
(2.8-4.
3-8a).8-65d. 61a.8-66b)
w(x) = ^ 71
I Jo
sin aa cos ax a ( a 4 E I + k)
da
(2.9-1) yields
. we obtain
(2.9-1. b) in Equation (2. Example 2. that is to say.8-65a.9-1)
The origin (x = 0) is an irregular singular point. This is illustrated in the next example.8-64a) (2.8-51f. we can obtain a formal series solution which is a good approximation for small values of x.f)
J-oo
Combining Equations (2. 2.c)
J-oa
B(a) = I
p(x)sinaxdx = p 0 I sin ax dx = 0
J-a
r
(2. Find a power series that satisfies the equation x 3 y" + (x2 + x ) y ' .8-64b)
A(a) = I
r
r
p(x)cosaxdx = 2p 0 I cos ax dx = (2p0 sin cca)/a
JO
r
(2.9 ASYMPTOTIC SOLUTIONS
We have obtained convergent series solution in the neighborhood of x 0 (= 0) if x 0 is an ordinary or regular singular point. In some cases.e.8-67)
Examples of Legendre-Fourier and Bessel-Fourier series are given in Chapter 5.8-66a) (2. x (x +x)/x and x (-l)/x do not have a Taylor series about x = 0.3-8a to c) in Equation (2. there is no method of generating a convergent series solution.8-64a to 65f) yields C(a) = 2 p 0 s i n a a / [ a ( a 4 E I + k)] D(a) = 0 Substituting Equations (2.8-62b) yields
/•oo
(2.1SZ
ADVANCED
MATHEMATICS
(a 4 EI + k)C(oc) = A (a) (a 4 EI + k)D(a) = B(a) From Equations (2.b. g.y = 0 (2. b).8-66a. We seek a formal series solution as given by Equation (2. Substituting Equations (2. If x 0 is an irregular singular point.
9-3c. We can increase the value of n as long as the value of the n th term is less than the value of the (n .l ) s + 1 (s-1)! x s
(2.2 c 2 = 2c 2
xs:
(s-l)(s-2)cs_1 + (s-l)cs_1 + s c s . we find that y (0.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
oo
/££
]T [(r + 2)(r+l)c r + 2 x r + 3 + (r+l)c r + 1 x r + 2 + ( r + l ) c r + 1 x r + 1 .c r x r ] = 0
r=0
(2.
The power series s n (x) represents f(x) asymptotically as x—>0 if x ~ n [ f ( x ) .( s . Since the asymptotic series is divergent.9-4)
Can this divergent series [Equation (2. From Equation (2.f)
CJ+2C2-C2 = 0 => c 2 = . the first two terms are sufficient and y(0. (s-1) xs + .9-3g.Cj [10"1 .]= cl £
s=l
( .. Using the divergent series.s n ] —> 0..c s = 0
=> c s = .1) to be accurate to two decimal places.h)
The formal series that satisfies the differential equation is
oo
c 1 [ x .9-3e.6 x 10" 4 + 2-4 x 10~4 + .
.
n
The divergent series s n (x) [= X
r=0
crxF]
is an asymptotic series expansion of a function
f(x).1) = 0.l ) t h term if the value of the n* term is greater than that of the (n .9-6)
The error in approximating a function by a convergent series decreases as the number of terms increases.10" 2 + 2 x 10~3 .9-2)
Comparing powers of x.1) c s _!
(2.9-5)
We note that the magnitude of the terms decreases as the order increases and if we require y (0. for asymptotic series.d) (2..x 2 + 2x 3 + ( .99 Cj.. it is seen that the approximation improves as the magnitude of x decreases for a fixed value of n.1) .9-3b) (2.9-6).l) t h term.l ) s + 1 2-3-4 . for all n > 0 as x—>0 (2. the error may increase as n increases. We are certain of attaining the required accuracy by taking a sufficient number of terms. we obtain x°: x1: x2: x3: c0 = 0 cj-cj = 0 => cj is arbitrary (2. ] (2. Usually. only a few terms are required...9-4)] be used to calculate the values of y for small values of x ? The answer is yes.9-3a) (2.l) t h term.c x 2c 2 + 2c2 + 3c 3 -C3 = 0 => c 3 = .1) can be written as y (0. We stop at the (n .
we obtain
oo oo
u' = ie ix X c n x. In applied mathematics. On differentiating.9-13a)
.4-44)] for large values of x. Example 2.n -e i x £ n c ^ . Even if a convergent series is available. A second order differential equation is in its normal form if the first derivative (dy/dx) is not present. We first transform Equation (2.9-9)
For large values of x.4-44) yields x 2 (u" v + 2u' v1 + u v") + x (u1 v + u v') + x 2 u v = 0 or x2vu"+ u'(2x2v'+xv) + u(x2v"+xv'+x2v) = 0 (2.4-44) to its normal form. one often neither knows nor cares if the series converges or not [Van Dyke (1975)]. we look for a solution as x —> °°.9-8a) (2. To achieve this. in this case. we write y = u(x)v(x) Differentiating and substituting the resulting expressions in Equation (2.6-3a)] and the solution is e lx .164
ADVANCED
MATHEMATICS
Asymptotic series are widely used in the solution of differential equations and for evaluating integrals. This suggests that we seek a solution of the form
oo
u = e ix £ c n x" n
n=0
(2.9-8b) becomes u" + u ( l + l/4x 2 ) = 0 (2. therefore we expand in reciprocal powers of x." " 1
n=0 n=l
(2.9-11) (2.9-12)
Note that.9-2. Equation (2.9-7)
We now impose the condition 2x 2 v'+ xv = 0 The solution is v = x~ 1/2 Equation (2. it is sometimes profitable to consider an asymptotic series as shown in the next example.9-8b) (2.9-10) (2.9-11) is approximately the simple harmonic equation [Equation (2. Solve Bessel's equation of order zero [Equation (2.
From the two examples we have considered. In Example 2. r = -l/2 (2. that the series representations of w r and wj [Equations (2.9-36)
as x —> °°.n/4)]
(2.4-53)] and.b)
Further details on asymptotic expansions can be found in Cesari (1963).9-2. we need to consider only the first seven terms of the series (four for w r and three for Wj) and we obtain J o (6) = 0.
. d)] are divergent. where X(x) = ix. using the ratio test. However. we deduce.cos x)] [wr cos (x .15064 (2.9-16c.7t/4) + wf sin (x . Since Poincare's pioneering work.9-15i). X = r = 0 and it is not common for an asymptotic solution to be simply a power series.SERIES SOLUTIONS AND SPECIAL FUNCTIONS -1/2
762
Jo ~
V JL
[wr (cos x + sin x) + Wj (sin x .9-35)
c n x" n
(2.9-37a. and Wasow (1965). • An asymptotic series is often more useful than a convergent series.9-33a) (2. we can postulate that the asymptotic solution of a differential equation can be of the form y . Nayfeh (1973).'
From Equation (2. to determine J 0 (6) accurate to five places of decimals.9-33b)
* MWv A. we obtain J o (6) = 0. considerable progress has been made in the understanding of the asymptotic series.9-1.{exp [X(x)]} x r £
n=0
(2.9-34)
The function Jo (x) can also be represented by a convergent series [Equation (2.15067 and this value is accurate to four decimal places only. Usually. after adding the first twenty one terms. Van Dyke (1964). exponential functions are involved as in Example 2.
9-39a. Example 2. = . we have the Reynolds number which can be small (Stokes flow) or large (boundary-layer flow).. t m = tQ + etj +. 39a. A particle of unit mass is thrown vertically upwards (y-direction) with an initial velocity v 0 .9-41a) (2... The equation of motion (Newton's second law of motion) is y = -[g + ey2] (2. If the air resistance at speed v is assumed to be ev 2 .. as y (t m ) = y 0 (t0 + etj + . We take the origin at the surface of the earth.. = v 0 At time t m .b)
We have identified e to be a small parameter and we start by expanding all functions of interest in powers of e..9-40a) in Equations (2.9-40b)
..) + .9-3. That is y (t) = y o (t) + eyi(t) + ..9-38.9-38)
where g is the gravitational acceleration and the dot denotes differentiation with respect to time.9-41c) (2. In fluid dynamics. we consider its reciprocal.. the particle is at rest and this is expressed.. (2.9-40a) (2.) + ey l (t0 + ztl + . we obtain respectively y ' o + e y j + .170
ADVANCED MATHEMATICS
Parameter Expansion
Many physical problems involve a parameter which can be small or large. The next example illustrates the method of parameter expansion. We seek a solution in powers of £ and often the series solution is an asymptotic series... Substituting Equation (2.)] yo(O) + e yi (O) + . determine the maximum height y m reached by the particle at time t m .[g + e ( y 2 + 2ey 0 y j +.. where e is a small positive constant. The initial conditions are y(0) = 0.. y(0) = v0 (2...9-42a) (2.. using Taylor's expansion.9-41b) (2. We consider the case when the parameter is small and we denote it by 8. If the parameter is large. b). = 0 y o (O) + ey !«)) + .
B.4 and Example 2. If x < (1/e). If we are required to evaluate exp (-ex) for large values of x. the rate equations for components A.. Thus the approximation of exp ( . i) The function exp (-ex) can be approximated as
oo
exp (-ex) = ^
n=0
(-l) n (ex) n /n! = 1-ex + e 2 x 2 + . (2.
(2. ( e « l ) by the first three terms. it is not possible to obtain the analytic solution and the quasi-steady state approximation is introduced.14-14 to 16) in dimensionless form. we deduce from Equation (1. is valid for x < ( l / e ) but not for x » ( l / e ) . as an asymptotic expansion. the solution in powers of e is often singular (see Section 2.b)
Several methods have been developed to extend the validity of the solution. we need to include more terms as the value of x increases.b)
The infinite series is convergent and.14-2. but if x » (1/e). the approximation is no longer valid and the magnitude of the first three terms can exceed one while exp (-ex) < 1 for all x > 0.9-54a)] is not useful (see Example 2. Follow Bowen et al. In Example 1. Discuss the validity of this approximation.) Except for the first term.14-15) is set to zero. We next introduce several examples involving a singular perturbation.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
775
The regular perturbation (straight forward expansion) method as described in the preceding example may generate a solution which is not valid throughout the region of interest (usually at infinity or at the origin). Example 2.9-54a.9-4. (1963) and write Equations (1. ii) The function Vx + e can be approximated as VxT~e = Vx" (1 + e / x ) m = VxT (1 + e/2x . Show that the solution by the regular perturbation is singular. we consider the first three terms as shown in Equation (2. In the case of a convergent series. the approximation is valid. This is discussed in the next example. This implies that the left side in Equation (1.9-4). In this case. iii) If the coefficient of the highest derivative in a differential equation is e.9-56)
. (e = 0)..9-2). There is a limit to the number of terms that can be included.9-55a.14-15) that cB = ( k j / k 2 ) c A (2. In more complicated cases. and C involved in first order reactions are solved. Setting dB/dt = 0..e x ) .e 2 /8x 2 + . the convergent series [Equation (2. The basic idea is to use more than one scale.9-54b). we have a singular perturbation.. all the other terms are singular at the origin.
The coefficient of dc B /dt is e. The solutions given by Equations (2. Expanding c B in powers of 8. Note that in Equations (2.9-67b. We use two different time scales. the limit of e lim
e->0 t*->0 *
is one. We further note that the zeroth approximation
CIO
= e~l*>
CBO
=°
(2. one for the inner solution and one for
. We obtain a solution near t* = 0. This zeroth order solution cannot be improved because c B is not uniformly valid. t * 0. we note that Equation (2.9-68)
Comparing Equations (2. On reversing the
taking the limit as e —> 0 first and then as t* —> 0. c B and c B are known quantities and are defined in Equations (2.) [e" 1 . the limit of e~l
-t*/F
limiting process.9-62b) can be written as c B .9-64d. d) respectively. This implies that lim e" l * /£ (2.b)
Equation (2. This implies that the quasi-steady state method is valid for the zeroth approximation and not for higher approximations. f).l * / e *
t*->0 e->0
We further note that e~l changes rapidly near t* = 0. The Stokes solution of the flow past a sphere is uniformly valid but the higher approximation is not valid and is known as the Whitehead paradox.. A thorough discussion on the Whitehead paradox is given in Van Dyke (1975). At time t ~ 0. f)] and no arbitrary constant can be introduced so as to satisfy the initial conditions. 68). On is zero. It is the presence of the term e -t*/f in Equation (2. d. 67b) are not uniformly valid and do not satisfy the initial condition [cB(0) = 0].e(1 + e + . This suggests that to obtain a uniformly valid solution we need to seek the solution in two separate regions.9-67b) is a valid approximation in the case 8 —> 0. which is the inner region or boundary layer where 8 is important.176
ADVANCED MATHEMATICS
The approximate solutions for c A and c B are c*A « e~l .9-67a. c B « ee~ l (2.9-56.9-70)
e. which is the outer region.9-60b)] reduces to a system of algebraic equations [Equations (2. and a solution for t* » 0.9-69a..9-62b) that contributes to the singularity.e" 1
/£]
(2. A similar situation exists in fluid mechanics.b)
is uniformly valid.9-64b. the differential equation [Equation (2.9-64b. the approximation is not valid.
E°:
T£..9-73. by our choice of T..-<*
dc®*
(29-74a)
The initial conditions can be applied in the inner region and they are 4i}o*(O) = c®*(0) = 0 The solution of Equation (2..9-74c. the coefficient of dc B /dT is one and not e. The time scale for the inner region is T = t*/e Using Equation (2.9-67a)] is uniformly valid and we need to solve only Equation (2.9-72)
Note that. Equation (2. we obtain
.9-71).9-74a) that satisfies Equation (2.9-72) and comparing powers of e.9-74c) is <$* = 0 Combining Equations (2. c A [Equation (2. This perturbation method is the matched asymptotic expansion method (boundary layer method) and was introduced by Prandtl to resolve d'Alembert's paradox in fluid mechanics (see Chapter 3).9-73) in Equation (2.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
177
the outer solution. 74b) yields (2. 71.9-60b) by the method of matched asymptotic expansion. The independent variable for the inner region is chosen such that the coefficient of the highest derivative is not e. We denote the inner solution by ci and it is expanded as
of = c < » *+ <+.9-66a.
Substituting Equation (2. In the present example.d)
.2.9-75) (2.9-60b) becomes ^ = -CB
+
(2.9-71)
EC A
(2.
t
* (o) *
(2. .9-80a. the arbitrary constant is determined by using a matching principle.9-79) is identically satisfied. 78.c)
On solving Equation (2.e ~ T
The inner solution c^J can be written as
(2. The solution which is uniformly valid is the composite solution and is denoted by CA .b. we obtain
c£j* = l . The simplest one was proposed by Prandtl and can be expressed as
Km 4 ° * = Jim c£o)*
T—>oo B
t*_>0
(2. Equation (2. The outer solution Cg' is given by Equation (2.9-76c) subject to Equation (2.b..9-67b) is valid for
t » 0. As mentioned earlier. no condition can be imposed on the outer solution.. +"i o
r / o \*~] i
ci j and [ci J denote the outer limit (T —> °°) of the inner solution and the inner limit (t* —> 0) of the outer solution respectively.e
T
T—>~
l)
= lim ee~ l
t*—>0
t*
= E
(2.9-67b.
For the outer region. This solution is given by (2. Equation (2. If the outer solution is obtained by solving a differential equation.t * .9-83a)
.9-76a.178
ADVANCED MATHEMATICS
dc®* —^-+41J
dT
ci
= e " e l = 1-ET + . the outer solution is determined by an algebraic equation and there is no arbitrary constant..9-77)
c£°* = E(l-e" T ) + .9-74d). Note that since all conditions are at time t = 0.9-82) Combining Equations (2. 81a) yields c£c)* = £ ( l .9-81a.b)
[
/•«. There exists a region where both the outer and the inner solutions are valid and are equal.c)
The matching principle can be interpreted as follows.1
(2.9-79)
B
In the present example. 80c.9-78)
is the time scale and the solution given by Equation (2.T ) + £ e .£ + O ( £ 2 ) (2.e . That is to say lim £ ( l .9-79) implies (2.9-67b)..
k is the thermal conductivity of the fin.
Sheppard and Eisenklam (1983) have considered the dispersion of an ideal gas through porous masses of solid particles. Deduce that the indicial equation of 2d y . If T denotes the fin temperature and T A denotes the air temperature. 2^ dy " x z — '.
where C is the Laplace transform of the concentration. obtain C in a power series of x up to x5.+ (x . show from an energy balance that y (= T . 13b. at x = b .x z ) — . as shown in Figure 2. Npo is the dispersion number. y remains finite y = TB-TA
.0 (b . and p is the Laplace transform variable. Obtain one series solution which is non-singular at the origin and verify that it is e x . Use the method of variation of parameters to show that the second solution is I (e~ x /x) dx. find the second series solution.x) ^ + (b . p = h / k s i n a . By expanding e~x in powers of x. and a is the half angle of the vertex of the triangular fin. The gas is assumed to be compressible and after some simplifications (details are given in the paper). The outer radius of the fin is b. The appropriate boundary conditions are at x = 0. x0 is a residence time. the equation to be solved is
N D O (l + Bx)^-|-(l+Bx+|N D O ) d ^ + [S-(l+Bx)3/2(T0)(p)]c = 0
\X A.T A ) satisfies the equation x (b . On expanding all quantities in powers of x up to x5.182
ADVANCED
MATHEMATICS
1 lb.x) y = 0 dx2 dx where x is the distance from the rim of the fin. B is related to the permeability of the porous mass. Show that the origin is an ordinary point. h is the heat transfer coefficient.xy = 0 dx dx 2 is r 2 = 0. 12a. Jenson and Jeffreys (1963) have considered the problem of the temperature distribution in a transverse fin of triangular cross-section.a .2x) ^ .P-13b.
Obtain a series solution for y if a = 5 cm. and T B = 373 K. The potential u at any point P is given by u = e/rj . where a positive charge e is placed at point A.e / r 2 ^ y(7c) + p 2 y1 (TC) = 0
.P-15b. a distance a from the origin. of the Sturm-Liouville problem dx 2 a i y ( O ) + oc2y'(O) = 0 .P-13b 14a. as shown in Figure 2.SERIES SOLUTIONS AND SPECIAL FUNCTIONS
J83. A negative charge . Show that the origin (x = 0) is a regular singular point. a 2 P 2 ) t a n 7 t *^ ~ " ^ ( P i a 2 ~ a i P2) 15b.
—/TCL
__!___ j°
FIGURE 2. (3 = 380 W/mK. a distance .a from the origin. are given by (al Pj + A . Transverse cooling fin Show that the eigenvalues A.e is placed at B.
Tg is the temperature of the pipe and a is the radius of the pipe. Consider the following electrostatic problem. T A = 288 K. b = 15 cm.
A (a) = 1 / ( 1 + a 2 )
Jo 1 + a 2 2
(ii) I Jo
cos a d a
1+a2
= -S2e
Answer: B(a) = a / ( l + a 2 )
29a. Differentiate the series term by term and substitute the resulting expression in the differential equation.
. Calculate the radius of \r=0 j convergence of the series and comment on the validity of the series as a solution of the equation. the cos terms.
Show that the equation
X
2di dx 2
+ ( 3 x
_1)dy+y dx
=
o
has an irregular singular point at the origin. Answer: a0 = c/n 2 as = ks7tc 2 T 3 /[a 2 + k 2 s 2 7C 2 T 2 ] bs = c T 2 a s / [ a 2 + k 2 s V T 2 ]
a* = sV-n 2 T 2
28 a.188. and the sin terms. Assume that y (t) is also periodic of period 2T and write its Fourier series expansion. if f(t) is periodic of period 2T and is equal to ct/2T in the interval (0. 2T). Determine the Fourier integral representation of e" x f(x) = 0 Deduce that x<0 x>0
r
(i) I
doc
= 2E.
ADVANCED
MATHFMATICS
Find the Fourier series of f(t).
I~ \
Find a power series I ^ c r x I t n a t satisfies the equation.
Answer. Determine the Fourier coefficients by comparing the coefficients of the constant term.
Determine the uniformly valid solution to order e. We introduce two time scales To and T] and write
t = T o + 8 Tj
Substitute t in the solution and determine Tj such that the secular term is eliminated.190
ADVANCED MATHEMATICS
y (0) = 1 . To obtain a uniformly valid solution. Answer: Tl = 3T o /8co 2
.
^
d t
=0
t=0
Show that the solution to order e is y = cos cot+ e [(cos cot-cos 3 cot)/32 co + (3tsin cot)/8co] The presence of the term 3t sin cot (secular term) renders the solution invalid for large values of t. we can use the method of multiple scales.
The theory of differential equations has been extended within the domain of complex variables. However. therefore. Applied mathematicians. Wessel.1-3) (3.1 INTRODUCTION
The inadequacy of the real number system (rational and irrational numbers) in solving algebraic equations was known to mathematicians in the past. and asymptotic expansions. The concept of a function was subsequently extended to complex functions of the type w = f(z) where z (= x + i y) is the independent variable.CHAPTER 3
COMPLEX VARIABLES
3. The concept of complex variables is a powerful and a widely used tool in mathematical analysis. were solved by accepting V-T as a possible number. Euler was the first to introduce the symbol i for i^A with the basic property i2 = . when d is zero. It was not until around 1800 that sound footing was given to the complex number system by Gauss. called an "imaginary" (as opposed to real) number. Complex integral calculus has found a wide variety of applications in evaluating integrals. which could not be solved in the domain of real numbers. no actual meaning could be assigned to the expression V^T. Real roots are special cases.) He also established the relationships between complex numbers and trigonometric functions. Gauss proved that every algebraic equation with real coefficients has complex roots of the form c + i d. so as to obtain meaningful solutions to simple equations such as x2+ 1 = 0 (3. it appears that equations. This notation. has had its own shortcomings. This usage still prevails in the literature.1-1)
For quite sometime. inverting power series. however.1-2)
.1 (Electrical engineers use j to denote i^T. It therefore became necessary to extend the real number system. It was. Argand proposed a graphical representation of complex numbers. forming infinite products. and Argand. (3. in those times.
As in vector algebra.
^
X
2 = x-Ly
FIGURE 3.2-1. a complex number can be represented by a point in a plane. This representation is the Argand diagram and is shown in Figure 3. it implies
. Just as a real number can be represented by a point on a line.192
ADVANCED
MATHEMATICS
physicists. We can also regard z as an ordered pair of real numbers.2 BASIC PROPERTIES OF COMPLEX NUMBERS
We can write a complex number z as z = x + iy where x and y are real numbers. If z^ (=Xj + i y i ) and Z2 (=X2 + iy2) are equal. It is indispensable for students in mathematical. and engineering sciences to have some knowledge of the theory of complex analysis. and engineers make extensive use of complex variables. (3.2-1
Argand diagram
Two complex numbers are equal if and only if (iff) their real and imaginary parts are equal. we write z as (x. physical. The numbers x and y are the real and imaginary parts of z respectively and are denoted by Re (z) and Im(z). y).> z=x+Ly
A
N.2-1)
y . 3.
.3-1.z o | > 8 represents the exterior of the circle. The curve C is simple if it does not cross itself. the curve C is a closed curve. It can also be expressed as |z| = 1 The set of all points z which satisfy the inequality |z-zo| < § (3.3-2b) (3. It consists of all points z lying inside but not on the circle of radius 5 with the center at zQ.3-1 6 neighborhood of z 0
Similarly.3-2a)
If z (a) and z (b) are equal.3-3b)
is called the 5 neighborhood of the point zQ.
*~
X
FIGURE 3. It is in fact the unit circle.COMPLEX VARIABLES
799
defines a curve C that joins the point z(a) = x(a) + iy(a) to the point z(b) = x(b) + iy(b) (3.3-3a)
is a simple closed curve. For example z = cost + i sin t (0<t<27i) (3.3-4) (3. the inequality | z . This is illustrated in Figure 3.
The conditions x > 0 and x < 0 define the right half plane and the left half plane respectively. S is an open set. (3.3-6b). none or all of its boundary points is a region. it is called an exterior point of S if there exists a neighborhood of z 0 which does not contain points of S. Similarly. Example 3. Let S be a set of complex numbers and let z. the set of all points z (= x + i y) such that y > 0 is the upper half plane.b)
The real functions u and v denote the real and imaginary parts of w. The set of all points in the plane that do not lie on | z | = 1 is an open set which is not connected. be a complex variable. The open set | z | < 1 is connected and so is the annulus 1 < | z | < 2. The unit circle does not belong to the set. Then find the value of f (1 + i). S is closed. Express the function w = f(z) = z 2 + 2 z . Just as the variable z is decomposed into real and imaginary parts.2Q0
ADVANCED
MATHEMATICS
A point z 0 is said to be an interior point of a set S whenever there is some neighborhood of z 0 that contains only points of S . A set that is formed by taking the union of a domain and its boundary is a closed
region. Clearly an open set contains none of its boundary points. We write w = f(z) (3. An open set that is connected is a domain.3 z in the form of Equation (3. This is because we cannot join a point inside the unit circle to a point outside the unit circle without crossing the unit circle. we have the lower half plane.b)
. A domain together with some.3-1. which varies in S.
Finally. If z 0 is neither an interior point nor an exterior point of S. y) (3. A function f defined on a set S of complex numbers assigns to each z in S a unique complex number w. We now define functions of a complex variable. We write w = f (z) = u (x. If every point of a set S is an interior point of S. w can be decomposed into a real and an imaginary part. The set S is the domain of definition of f(z) and the set of all images f (z) is the range of f (z). We note that a complex valued function of a complex variable is a pair of real valued functions of two real variables.3-7a. z 2 in S can be joined by a polygonal line that lies entirely in S. An open set S is a connected set if every pair of points z i . The closure S of a set S is the closed set consisting of all points in S as well as the boundary of S.3-5)
The number w is the image of z under f. If a set S contains all its boundary points . y) + i v (x. for y < 0.3-6a. it is a boundary point of S.
. such as the rules for differentiating a constant.c .3-22b)
The function f is not differentiable every where except at the origin where f'(0) is zero irrespective of 9.3-22a) (3.3-2lc) f'(z 0 ) = z 0 + z 0 f(z0) = z o . =— and ^— exist and satisfy the equations dx 3y dx dy 9u _ 3v du _ 9v H 3 23aM
Proof: Since f(z) is differentiable at any point z 0 in D. integer power of z. If f (z) is analytic for all finite values of z. Theorem 1 A necessary condition for the function f (z) to be analytic in a domain D is that the four partial j • <. e > 2 l
cos 9 + sin 9 . • All the familiar rules.i S i n . f(z) is an entire function.3-21b)
(3. We thus define a broad class of functions. we obtain respectively from Equation (3. sum. Functions which are differentiable at a single point are not of great interest. if z approaches z 0 along the real axis (9 = 0). The basic criterion for analyticity of a complex function f(z) is given by the Cauchy-Riemann conditions. then f'( z o) must exist. derivatives ^r— . and along the imaginary axis (0 = n/2).du du dv . ^— . product and quotient of differentiable functions as well as the chain rule of differential calculus of real variables hold in the case of complex variables. Synonyms for analytic are regular and holomorphic. The function f(z) is analytic at z 0 if its derivative exists at each point z in some neighborhood of z 0 .2Q4
ADVANCED
MATHEMATICS
•
=
^ + Zo ( c ° S 2 e . difference. Also the limit does not exist.z o (3. Points where f (z) ceases to be analytic are singular points. if f (z) is defined and has a derivative at every point in D.. dv . For example. z 0 + z 0 (cos 29 . . A function f (z) is analytic in a domain D.i sin 29)
(3.3-21c)
Thus f '(z 0 ) depends on 9 and is not unique.
The functions u and v of the analytic function are also conjugate functions (harmonic conjugates). Hence f(z) is analytic. If a harmonic function u (x. •
Let f (z) be an analytic function. The solution of Laplace's equation is a harmonic function.
(3. we deduce that both u and v satisfy the equations ^ + ^ = 0
(3. y) through the Cauchy-Riemann relations. The analytic function f (z) can then be determined.3-5.3 x y 2
(3. and their partial derivatives satisfy the Cauchy-Riemann equations and are continuous functions of x and y. v.3-42a. We will see in Section 3. y) = x 3 .b) (3. b).3-45)
.3-43a.3-44a.b.d)
ay2 axz
From Equations (3. b) are Laplace's equations which will be solved in Chapter 5. Assume that the mixed second derivatives of u and v exist and are equal.4 that this function is ez. y) is given in some domain D.3-43b.3-29a.COMPLEX VARIABLES
202
The functions u. From Equations (3. we have
fiax3 y ax ay
ax2
^
=^ =-^
ax2 ay z
ay2
+ ^ = 0
(3.3-44a)
(3. d). we can determine the harmonic conjugate v (x. Both the real and imaginary parts of an analytic function are harmonic functions. Show that
u(x. Example 3.3-44b)
a x2 a y 2
Equations (3. The derivative f'(z) = Y~ + i^= e x cosy + i e x s i n y (3.3-42c)
= e x (cos y + i sin y) which is identical to the given function.3-43c.
we note that < > | and \|/ satisfy the Cauchy-Riemann relations and are therefore harmonic functions.3-63)
* • = _ .3-61a. 62a to d). The derivative of O yields both velocity components. and will be discussed further in this chapter.b)
vy .3-28b.3-26. • The concept of a complex potential is widely used in hydrodynamics. we deduce dv 3v (3. Applications of complex potential in electrostatics are given in Ferraro (1956).| f
(3. we have ^=|^+i^=vx-ivv dz dx dx x y Similarly using Equation (3.
. ! • + ! ? .
(3.3-61a) 3 vY 3v
it =-i?
We introduce two functions (j) (x.211
ADVANCED
MATHEMATICS
From Equations (3.3-64a.3-62a to d). They are conjugate functions. we have (3.3-29a.3-59.3-62c. | £ = . 60). we can obtain both the potential (Re O) and the stream function (Im O).3-65a.3-62a.b)
Using the complex function 3>.d)
From Equations (3. Equations (3.3 . 60.b) (3. The combinationty+ i \|/ is an analytic function and the complex potential is given by O(z) = < > | + iY Differentiating O(z) with respect to z and using Equations (3. v x -iv y
dz 8y dy x y
(3. b). 62a to d). 62a to d).61b >
Substituting v x for u and vy for (-v).
vx = £ = J f . the components v x and vy satisfy the Cauchy-Riemann relations. y) such that
( 3 . b) are identical to Equations (3. y) and \|/ (x. The function < ) | is the potential and \|/ is the stream function. This can also be verified by combining Equations (3. That is.3-59.
4-20b) (3. That is to say. then Arg z is zero and the definition of Ln z is identical to the /8n in the theory of real variables.COMPLEX VARIABLES
2JZ
x
= e"
(3. we can write i n z = Lnz±2n7ti (n is an integer) (3.4-21) is a multiple-valued function and since the argz can differ by multiples of 2K.4-22a) (3. Some of the properties of Jin z (Ln z) are:
. b) yields i n z = i n l z l + iargz (3.4-18) (3. we obtain Izl = eu argz = v Combining Equations (3. Ln z is defined as Lnz = i n l z l + iargz -7t<argz<7i.4-23)
If z is a real positive number.4-16)
Since arg z can differ by multiples of 2n.4-17) (3. We then have the principal value of Jin z and we denote it by Ln z.4-19) (3.4-17) becomes z = eu (cos v + i sin v) From Equation (3. 20a.4-19).4-22b)
The function i n z as defined by Equation (3.4-18.4-20a) (3.4-21) (3. we restrict the definition to the principal values of arg z (Arg z). Equation (3.4-15)
then u = inx We now extend the definition of Jin to the complex variables and write z = ew It follows that we can define Jin z as w = inz Separating w into its real and imaginary parts (w = u + iv). (3.
is l a always equal to 1? From Equation (3.i) = l i n 2 .4-27) (3.4-27) and noting that 1 can be written as e2nm. The principal values of z a and ocz are obtained by substituting i n z by Ln z in Equations (3.4-29a) (3.
Since i n z and Ln z differ only by an arbitrary multiple by 27ti.4-30a) (3. (iii) (-i) 1 .4-28b) (3.4-3.c) (i) i n e. If a is a complex number.4-3la) (3.^ i (iii) Taking the principal value. Find the values of (i) i n e = i n I eI + i arg e = 1 + i (0 + 2nK) = 1 + 2n7ti (ii) Ln (1 .4-26) (3.i ) .4-31d) (3.i I + i Arg (1 .4-26. ^(inz) =1 If a is any complex number. we have (-i)i = e i L n ^ _ e i (-iic/2) _ e n/2 Example 3.4-31b) (3. we define z a to be z a = exp (a i n z) Similarly the function ocz is defined as a z = exp (z i n a) (3.4-28a) (3.4-30b. (ii) L n ( l .4-29b)
. (3. Example 3. 27). we have l a = exp [a i n e 2raii ] = exp [a (27tni)] = exp [-27inp + 27tni y] = exp [-2n.5-25b)
In general z a and ocz are multiple-valued functions since i n z is a multiple-valued function.np] {cos 27Cny+ i sin 27tn'y} (3.COMPLEX VARIABLES
219.i) = i n 11 .4-31 c) (3.4-2.
To begin with. Note that if n is zero (consider only the principal value) l a is equal to 1.b
(3. then
I [f(t) + g(t)] dt = I f ( t ) d t + I g(t)dt Ja Ja Ja
.220
ADVANCED
MATHEMATICS
To obtain Equation (3.2. A definite integral of a complex function of a complex variable is defined on the curve joining the two end-points of the integral in the complex plane. This two dimensional aspect has an effect on complex integration. we stated that a complex plane is required to display complex numbers. Ja Ja Several properties of real integrals are carried over to complex integrals. Let f(t) = u(t) + iv(t). if f (t) and g(t) are complex functions.5-3c)
where a (= p + i y) is a complex constant.4-24f) and we have expressed the complex number a as y + i p .5-2)
u(t)dt + i | v(t)dt Ja la
fb fb Note that both I u(t)dt and I v(t) dt are real. For example. a<t<b (3.c .5-1)
and we assume that u (t) and v (t) are continuous functions of t. We define
r
f(t)dt = Ja
fb
tb
(3.
. Thus in genera] l a is not equal to 1.4-3lb). P and y are real. we consider the definite integral of a complex valued function of a real variable t over a given interval a < t < b.5 COMPLEX INTEGRATION
In Section 3.5-3b)
af(t)dt = a
Ja Ja
f (t) dt
(3.5-3a)
f(t) dt = Ja Ja
f(t)dt + Jc
f(t)dt
(3. 3.b . we have made use of Equation (3.
0<t < 1 (3.5-5) and if we let (-C) be the curve that traces the same set of points in the reverse order.b)
L = I |z'(t)|dt Ja
(3.5-10)
If C is given by Equation (3.5-1 la)
A curve C that is constructed by joining a finite number of smooth curves end to end is called a contour (or path). Arrows indicate orientation
Curve C is smooth if z (t) is continuous and non-zero in the interval. A formula for representing the line segment joining two points z t and z 2 in a complex plane is z = zx + t (z2 . -b<t<-a (3.C) is given by z(t) = x(-t) + iy(-t).b
(3.5-9a.zx). curve (.y(a)]
V
^--
I
-^ z = ( a ) = z ( b )
c2
FIGURE 3.222
ADVANCED
MATHEMATICS
y " z(b)=[x(b). the differential arc length is given by
ds = V[x'(t)]2 + [y'(t)f dt = |z'(t)|dt
and the length L of the curve C is given by
.y(b)]
/
(
(
-
z(a) = [x(o).5-1 lb)
.5-1
Simple (Cj) and simple closed (C2) curves in the complex plane. If C is smooth.
In the theory of definite integrals of real functions. For complex integrals. we have taken the moduli of all complex quantities and we are dealing with real quantities.5-14b) is replaced by the length of the interval of integration (= b . b).5-2.5-12)
Differentiating z(t) from Equation (3. Contour integration in the real two-dimensional plane will be considered in Chapter 4. It was stressed in Section 3.5-14a)
Ja
(ii) I f(z)dz c
Ja
< J|f(z)|dz < ML c (3. yields
I f(z)dz = I (u + iv) (dx + idy) c c = I (udx-vdy) + i I (v dx + u dy) c c
(3.2 that complex numbers are not ordered and inequalities have meaning only when associated with real numbers.5-13a)
(3.5-3a to d).a.5-12).5-5). we can replace the real variable t by the complex variable z in Equations (3. we have the following inequalities:
(i)
f (t) dt
<
|f(t)|dt
(3. the length L in Equation (3. where a and b are the limits of integration).5-14b)
where M is the upper bound of lf(z)l and L is the length of the contour C.5-14a. and decomposing f into its real and imaginary parts. That is to say. Evaluate the integral I ( z .5-13b)
The contour integral has similar properties to those of integrals of a complex function of a real variable. Note that in Equations (3. Example 3.z o ) n d z
c
.COMPLEX VARIABLES
22J
The integral of f (z) along a curve C joining the points z(a) to z(b) is a contour (line) integral and is written as
I f(z)dz = I f(z)z'(t)dt c
(3. substituting the resulting expression in Equation (3.
0) to the origin is the mid-point of the segment. It follows from Equation (3.b)
I di c •
< 4V2"
(3.
. as shown in Figure 3. As z varies along the segment.COMPLEX VARIABLES
225
y^
(O+L)
^_
0
-^
(I+LO)
»~
X
FIGURE 3.5-19)
We note that a simple closed contour C divides the complex plane into two domains.5-14b) that (3.5-20a. and the other domain is unbounded and is the exterior of C.5-2
Path of integration
The closest point on the segment joining the points (0. The upper bound M of — is given by z4 M = lV2 j = 4 The length L of the segment is v2 . One domain is bounded and is referred to as the interior of C. i) and (1. its distance from the origin I z I varies and its minimum value is l / V z . as shown in Figure 3. The distance from the origin to that point is 1 /V2 .5-3. The maximum value of — is l / ( l / V 2 j and is V2 .5-2.
A consequence of Cauchy's theorem is the concept of path independence.3-29a.jp) dx dy c s
(3.5-23a)
I (v dx + u dy) = \\ (|^ . We have
| (u dx . That is to say I f(z)dz = I ( u d x .4). Proof: The integral can be written as in Equation (3.5-23a. b) are zero. Let z\ and Z2 be two arbitrary points on C and let them divide C into two arcs Cj and C2 as shown in Figure 3.|H) dx dy
c s
(3.5-21) holds.5-5. then I f(z)dz = 0 c The integral round a closed contour is also denoted by (j) f (z) dz.5-13b). Since f(z) is analytic. It follows that Equation (3.5-22) (3.5-23b)
where S is the domain enclosed by C. b) and the right side of Equations (3. Consider the integral round a closed contour C.j £ .5-21)
We transform each of the integrals on the right side of Equation (3.b)
.5-24a.COMPLEX VARIABLES
227
Cauchy's Theorem
If f (z) is analytic in a simply connected domain D and if C is a simple closed contour that lies in D. We have
I f(z)dz= I f(z)dz+ I f(z)dz = 0
c C! c2
(3.5-22) to a double integral using the two-dimensional Stokes (Green's) theorem (see Section 4.v d y ) + i | c c c (vdx + udy) (3.v dy) = II (. u and v satisfy the Cauchy-Riemann relations (Equations 3.
Evaluate I z 2 dz along each of the straight lines OA.5-5 We deduce
Path independence
I f(z)dz = . Thus the value of the integral is independent of the path and depends only on the end points. x(t) = t. x (t) = 1. y (t) = t. OB and AB as illustrated in Figure 3.^
C.5-6.C 2 .5-1 lb) are along OA: along OB: along AB: x (t) = t. c2
f(z)te = I f(z)dz -c2
(3.5-26c.5-25b) implies that the integral of f(z) from zj to z 2 is the same whether we integrate along Cj or .l c.b) (3.
FIGURE 3. The parametric equations of the lines (Equation 3. 0<t< 1 0<t< 1 0<t< 1 (3.5-25a. y (t) = t.f)
.5-26a.5-26e. Example 3.228
ADVANCED MATHEMATICS
22
I
^
.5-4. Equation (3. y (t) = 0.d) (3.b)
Note that .C 2 is the arc obtained by changing the orientation of C2 and is the curve joining z^ and z 2 .
the doubly connected domain has been cut by Lj and L2 and the resulting domain is simply connected. • A multiply connected domain can be transformed to a simply connected domain by making suitable cuts.5-29a)
=
3 + (" l
+
¥ ) " (~f + ¥ ) = °
(3. one pair of cuts is sufficient and for a triply connected region two pairs of cuts are needed so as to obtain a simply connected region. the area enclosed by them are on the left.5-7
Transformation of a doubly connected domain to a simply connected one
.5-29b.5-7. As we move along the curves.230
ADVANCED
MATHEMATICS
I z2dz = I
C OA
z2dz+|
AB
z2dz+|
BO
z2dz
(3.c)
verifying Cauchy's theorem. For a doubly connected region. The curve enclosing the simply connected domain D is C L j C ^ I ^ C . By Cauchy's theorem I f(z)dz = I f ( z ) d z + |
C L!
f(z)dz+|
C[
f(z)dz+|
L2
f(z)dz = 0
(3.b)
CLjQL^C
FIGURE 3. Note the orientation of the curves in Figure 3. In Figure 3.5-30a.5-7.
do not intersect each other. n) inside another simple closed contour C.5-31) then becomes
I f(z)dz = I f(z)dz
(3. Let the function f(z) be analytic in domain D which contains all contours and the region between them.COMPLEX VARIABLES
231
We note that Lj and L 2 are in opposite direction (that is.5-32)
Equation (3.5-5. Suppose we have n simple closed contours which we denote as C. (3. Via Cauchy's z—a theorem
J 2"a
c
I Az_
= 0
(3. Equation (3. by convention. Evaluate I -^z— where C is a simple closed curve. their limits of integration are interchanged) and so their contributions will cancel. have no common points. the point z = a is inside C. Example 3. the function —±— is analytic everywhere. the contours C. Since the point z = a is outside C. are oriented in the same direction.I f(z)dz c c. We then write
I f(z)dz = X I f(z)dz J j=i j
C
(3. (j = 1.5-33)
Cj
All the contours C and C.5-32) can be generalized to the case where there is more than one curve inside C..5-30b) becomes I f(z)dz = . usually in the anticlockwise direction which. is the positive direction. That is to say.5-31)
We note that C is oriented anticlockwise and Cj is oriented clockwise.. Consider the following two
J
Z — 3.. Equation (3. .5-34)
.
c cases: (i) (ii) (i) the point z = a is outside C. The regions interior to each contour C. 2. We reverse the orientation of C\ and denote (-Cj) by Cj.
we have
I -dZ. f (z) is analytic and using Equation (3. we deduce
. 37c).c) 0<6<27C (3.5-33).
Jo
£e lfl
Jo
From Equations (3.5-37a.5-35)
FIGURE 3.I -dz_
I z-a c I c.232
ADVANCED
MATHEMATICS
(ii)
In this case. On C and Cj and the region enclosed by the two curves. z-a
(3. The integral then becomes _ _dz_ _ i_e±_ i 2TI i (3. we write z = a + eei6.5-8. the function is singular at z = a and we enclose it by a circle Q of radius e with centre at the point of singularity as shown in Figure 3. b.5-8
Contour integral with point z = a inside C
To evaluate the integral along Cj.5-35.= .5-36)
d6
=
de
=
J z~a c.
we have verified Equation (3.5-39)
J z~zo
c
In Example 3.COMPLEX VARIABLES
221
I -&.5-38)
Cauchy's Integral Formula
Let f(z) be analytic in a simply connected domain D and let C be a simple closed positively oriented contour that lies in D.
Example 3.= 2%\
I z —a c
(3.
Integral Formulae for Derivatives
If f (z) is analytic in D. The values of these derivatives at a point z 0 in D are given by the formulae
f'(Zo) =
^/(7vdz
.5-40C)
In Equations (3.5-6.5-5. we have 1 M _ dz = 2 n i f ( z n ) (3. all derivatives exist. it has derivatives of all orders in D which are also analytic functions in D. C is any simple closed path in D which encloses zQ. For any point zQ which lies interior to C. Find the value of the integral I
e
coshz
dz
c
.5-40b)
f (n) (z 0 ) = ^
(
f(Z)
. We omit the proofs of these results but consider their applications.5-40a toe). We also note that if one derivative exists. (z-z0)
< 3 '5"40a)
(3. dz
(3. This is only true for complex variables.5-39) for the case f (z) equals to one.
234
ADVANCED
MATHEMATICS
where contour C is a square whose sides are: x = ±4 and y = ±4, described in the positive direction. The region D, the contour C and the point z 0 are shown in Figure 3.5-9. From Equation (3.5-40b), we identify f(z) = e z coshz z0 = n (3.5-4 la) (3.5-41b)
y '•
D
'C
7T~4
x
FIGURE 3.5-9
Integration around a square
The conditions for Equation (3.5-40b) are satisfied and it follows that
f
eZcoshz
dz = in - ^ - (ezcoshz)
(3.5-42)
1 (z-nf
dz2
c Carrying out the differentiation, we obtain
2
— (ez cosh z) = 2 e z (sinh z + cosh z) dz 2 = 2e 27r atz = 7 C
(3.5-43a)
(3.5-43b,c)
COMPLEX VARIABLES
231
Combining Equations (3.5-42, 43b) yields
[
eZ c o s h z
dz = 2irce271
(3.5-44)
c
{z~n)3
Example 3.5-7. Let f (z) be analytic inside and on the circle of radius R with its centre at the origin. Let z 0 (= r 0 ei0°) be any point inside C as shown in Figure 3.5-10. Show that
27C
JQ R 2 -2r 0 cos(e 0 - ¥ ) + r5
p (R^)f(R^)
/
(3.5-45)
y
L/^
ro
V z'
o
FIGURE 3.5-10
Point z 0 and its inverse point z1
Obtain the real and imaginary parts of f (z 0 ). Since z 0 is inside C, using the Cauchy's integral formula (Equation 3.5-39), we have
f(z } = v °;
^ f
2TII J z-z0
ffldz
(3.5.46)
c
We define the inverse point Z\ of z 0 with respect to C to be
226 2 2
ADVANCED MATHEMATICS
zj = | - = f- ei9° zo o
(3.4-47a,b)
(3.5-48) The point z\ is outside C and from Cauchy's theorem (Equation 3.5-21), we have
Substituting Equations (3.5-52a, b) in Equation (3.5-5Id) and equating the real and imaginary parts, we obtain
u(r0,90) = £ f
Z71
Jo
2 ^ ° ^ \ d¥ R - 2 R r0 cos (6 0 - \|/j + r^
(3,-53a,
v(r0,9j = JL I'"
111
te-4)^*>
Rz - 2 R r0 cos (6 0 - \|/j + r£
d¥
(3.5.53b)
Jo
Equations (3.5-5Id, 53a, b) are Poisson's integral formulae and are important in potential theory. We note that the values of f at any point inside C can be determined from the values of f on C. The functions u and v are harmonic functions, that is to say, are solutions of Laplace's equation, and we deduce that the solution of Laplace's equation is determined by the values of the function at the boundary only.
Morera's Theorem
If f is a continuous function in a simply connected domain D and if
I f(z)dz = 0 c
for every closed contour C in D, f (z) is analytic in D.
(3.5-54)
Goursat proved Cauchy's theorem (Equation 3.5-21) requiring only the existence and not the continuity of f'(z). Morera's theorem implies that f(z) is analytic as a consequence of Cauchy's theorem. Maximum Modulus Principle If f is a continuous analytic function and is not a constant in a closed bounded domain D, then I f (z) I has its maximum value on the boundary C and not at an interior point. If M is the maximum value of lf(z)l on C,
2i£
ADVANCED MATHEMATICS
I f(z)l < M for all z in D If f (z) is constant, lf(z)l < M for all z in D
(3.5-55a)
(3.5-55b)
The maximum modulus principle is true for harmonic functions but not for any smooth real valued functions of two real variables. 3.6 SERIES REPRESENTATIONS OF ANALYTIC FUNCTIONS
In this section, the equivalence between analytic functions and power series is explored. The concept of sequences, series and power series of complex numbers are in many cases similar to those of real variables. Sequences and Series Let Zj, Z2,..., zn be a sequence of complex numbers. A sequence {zn} converges to zQ if lim z n = z0
n-> oo
The sum of a convergent series of complex numbers can be found by computing the sum of its real and
oo
imaginary parts. A series is absolutely convergent if X | z | is convergent.
n=l
Example 3.6-2. Discuss the convergence of the geometric series
oo
X zn = 1 +z + z 2 + ... + z n + ...
n=0
n
(3.6-7)
240
ADVANCED
MATHEMATICS
The partial sum s n is given by l-zn+1
1 - z
sn
=
(3.6-8)
If I z I < 1, we deduce
lim
n —> oo
sn
= 7T7
i
z,
(3-6"9)
The series £ z n converges to —^—. 1 -z The function —-— is analytic inside the circle I z I < 1 and has the representation
1 Z
oo
j
1
-
= X zn
n=0 "
(3.6-10)
1-Z
If I z I > 1 , the series diverges.
Comparison Test
oo
Let X M be a convergent series with real non-negative terms. If, for all n greater than N,
n=0
lz n l < M n the series X z n also converges absolutely.
(3.6-1 la)
Ratio Test
oo
Let X z be a complex series and
n=0
n
lim J-^fL = L
n -> oo | Zn I
(3.6-12)
If L < 1, the series converges absolutely and if L > 1, the series diverges. No conclusion can be drawn if L is one. Note that in both tests we use the absolute value of zn since complex numbers are not ordered.
COMPLEX VARIABLES
241
Example 3.6-3. Determine the convergence of the series
X
n=l
—.
2n
Using the ratio test, we have
J z ^
l z nl
=
| n + l+ i | 2 _ ^ _ . i t ( n + l ) 2 + l ]
2 n + 1 |n + i | 2
2
6 _, 3
n2+l
Taking the limit as n tends to infinity yields ^ Since the limit is less than one, the series converges.
A series of the form
oo
Z
n=0
c n ( z - z / = co + C l (z-z o )+ ... +c n (z-z 0 ) n + ...
(3.6-14)
where z is a complex variable, z 0 , c 0 , Cj, ... are complex numbers is a power series. By a change of origin, we can set z 0 to be zero. In Chapter 2, we have shown that every power series of a real variable has a radius of convergence R. This result can be extended to complex variables. Every series has a radius of convergence R (0 < R < «>) and the series converges absolutely if I z - z 0 I < R and diverges if Iz - zQ I > R. On the circle of convergence (Iz - z 0 I = R), the series may converge at some points and may diverge at other points. When R is zero, the series converges only at z 0 and when R is infinity, the series converges for all values of z. The radius of convergence depends on the absolute values of l c n l . converges to the limit L, the radius of convergence R is given by If the sequence y |cnl
R =f
An alternative equation for R is 1 = lim - ^ ± 1 .
(3.6-15)
(3.6-16)
if the limit exists. A power series represents a continuous function at every point inside its circle of convergence.
242
ADVANCED MATHEMATICS
~
(-l)nZ2n
Example 3.6-4. Determine the radius of convergence of the series the function it represents. From Equation (3.6-14), we identify 0,
c =
2 J -—
n=0
22n(n|)2
and determine
if n is odd (3.6-17a,b) .f . , if n is even
(-l)n/2 —-— 2 n [(n/2)!] 2 For odd values of n
V l c nl = 0
For even values of n lim V ' c n l
n->oc
(3.6-18a)
= lim
1
=0
(3.6-18b,c)
n ->~ 2 [ ( n / 2 ) ! ] 2 / n
Thus R is infinity and the series converges absolutely for all values of z. In Chapter 2, we have defined Bessel functions of a real variable. Replacing x by z, we find that the series represents the complex Bessel function of order zero.
Taylor Series
We now consider the expansion of an analytic function f (z) as a power series. Let f (z) be analytic everywhere inside the circle C with centre at z 0 and radius R. At each point z inside C,
f(z)= X ^ - f M z - z o ) n
n=0 n!
(3.6-19)
That is, the power series converges to f (z) when I z - z 0 I < R. We first prove the theorem when z 0 is the origin. Let z be any point inside the circle C of radius R, as shown in Figure 3.6-1. Let C\ be a circle with radius Rj < R, and let £, denote a point lying on Cj. Since z is interior to Cj and f(z) is analytic, we have, using the Cauchy's integral formula
COMPLEX VARIABLES
243
f(z) = JL f ^ J
27C1 ] ^_z
(3.6.20)
c,
Note that Cj has to be positively oriented.
y '•
FIGURE 3.6-1
Illustration for the proof of Taylor series
We expand —-— in powers of — (< 1), as
-L-
= Ifl- ^r1
(3.6-2la)
=
1
i + i + i \ 2 + ... + /z\n-l
+
( z/ ^)
(3.6-21b)
=
J - + - L z + - L z 2 + ... t - L z ^ ' + z"
1
1
(3.6-21c)
Multiplying each term by ——r and integrating around the circle, we have
ZJll
If M denotes the maximum value of If (^) I on C\, we write the absolute value of the remainder as follows
rn M2TCRI MR, it \n
Rn(z)
< I
L- = - — L
—
(3.6-25)
2TC ( R i - r j R ;
( R l " r ) lRl/
Since (r/Ri) is less than one, it follows from Equation (3.6-25) that lim Rn = 0
n-> °°
(3.6-26)
We proved that f(z) has a power series representation given by Equation (3.6-23a) with R n tending to zero as n tends to infinity. That is to say, f (z) has an infinite series representation. The proof was restricted to the case where z 0 is the origin. This infinite series is the Maclaurin series which is a special case of the Taylor series. To extend the proof to the case where z 0 is not the origin, we need to shift the origin to z 0 . We define a function g(z) to be g(z) = f(z + z 0 ) (3.6-27)
Since f (z) is analytic in I z - z 0 I < R, g (z) is analytic in I (z + z 0 ) - z 0 I < R, which is I z I < R. Thus g (z) has a Maclaurin series expansion which is written as
Note that in the case of complex variables, if f (z) is analytic in I z - z 0 I < R, the Taylor series represents the function, the remainder term Rn tends to zero as n tends to infinity. In the calculus of real variables, the remainder term R n (Equation 1.2-12) does not have this property. In the theory of complex variables, an analytic function has a power series representation and the power series is analytic. If f (z) is analytic in a domain containing z 0 , and zj is the nearest singular point to z 0 , the Taylor series (Equation 3.6-31) is convergent in the domain I z - z 0 I < I Zj - z 0 I. Since the Taylor series is convergent for an analytic function, term by term differentiation and integration are permissible. The radii of convergence of the differentiated and integrated series are the same as that of the original series. Example 3.6-5. Expand 1 / ( I + z 2 ) in a Taylor (Maclaurin) series about the origin. Determine its radius of convergence. The singularities of — - — are 1+z2 z = i, z = -i (3.6-32)
Since — - — is analytic in the domain I z I < 1, the radius of convergence of the series is one. 1 + z2 The radius of convergence can also be deduced from Equation (3.6-15) and is found to be one, as expected. Note that, in the case of real variables, the radius of convergence can be deduced from the series expansion only. The function l / ( l + x2) has no singularity along the real line. Laurent Series If the function f(z) is not analytic at a point zQ, it does not have a Taylor series about z 0 . Instead it can be represented by a power series with both positive and negative powers of (z - z 0 ). This series is a Laurent series. Laurent's theorem can be stated as follows. Let D be the annular region bounded by two concentric circles C o and C^ with centre z 0 and radii Rj and R.2 respectively, as shown in Figure 3.6-2. Let f(z) be analytic within D and on C o and C^. At every point z inside D, f(z) can be represented by a Laurent series which can be written as
f(z) = X a n (z-z 0 ) n + X
n=0
n—
(3.6-35a)
n = l ( z _ Z ( ) )n
where
an
= T^
I
f(
^ ^ , .
n = 0, 1, 2, ...
(3.6-35b)
bn = - L (
f &
^
,
n = 1, 2, ...
(3.6-350
The integrals around Co and C^ are to be taken in the anticlockwise direction. The proof of this theorem is similar to that of Taylor's theorem. We surround the point z by a circle y as shown in Figure 3.6-2. Let £ be any point on the curves C o , Q or y. From Equation (3.5-33), we have
[tJL^-ilGLK- (1^=0 J S-z J S-z J S-z
C, Co Y
(3.6-36)
where C\, C o and y are considered to be in the anticlockwise direction.
COMPLEX VARIABLES
247
yt
\.
D
y
•
X
FIGURE 3.6-2
Illustration for the proof of convergence of Laurent series
Using Equation (3.5-39), we write the last integral as
27U f (z) = [ ^ ^
Y
(3.6-37)
Combining Equations (3.6-36, 37) yields
f(z) =
^
[ m«k _ J , ( £ ^ i
^_z 2711 J c0 ^_z
0.6-38)
27C1 J c,
On C], 11, I > I z I and we have the same situation as in the case of the Taylor series. Equations (3.6-21 to 26) can be carried over and we have
1
I f (5) d^ _ y
c,
,
,n
r 3 6 _ 39 v
27Ti J ^ 7
"ntt)^12"^
(
}
where the an are given by Equation (3.6-35b).
248
ADVANCED
MATHEMATICS
On Co, I z I > 11,1 and, in this case, we write
-^- =
£-z
=
L
(z-z o )-^-z o )
= _JL_ L M " 1
(z-z0) L
+
(3 . 6 . 4 oa,b)
(z-z0)
_ i _ + j^-zj
(z-z0) (z-z0)2
+ fe z z Q T +
(z-zo)n
fe-z/
(z-zo)n(z-^)
(36.4Oc)
Multiplying Equation (3.6-40c) by -~
f(^) and integrating around C o , we obtain
(3.6-41)
where b n are given by Equation (3.6-35c). The remainder Tn is given by
Tn = -L ^ _ (
27ti (z-z o ) n j
Co
fe-fff^
(z-^)
(3 . 6 .42)
Let K be the maximum value of I f(^) | on Co. For any £ on Co, we have
Comments 1. If f(z) is analytic in a region l z - z o l < R , the coefficients b n in the Laurent series are zero. The Laurent series reduces to the Taylor series about the point z 0 . The Laurent series in a specified annulus Rj < Iz - z 0 I < R/> is unique. That is to say, if, by any method, we have obtained a series representation for an analytic function in the given annulus, that series is the Laurent series. In the examples that follow, we shall derive a representation of f (z) by methods other than via Equations (3.6-35a to c). Since f(z) is analytic in the annulus, the contours Cj and C o in Equations (3.6-39, 45) can be replaced by any circle C lying in the annulus. That is to say, the contour integrals can be taken around the same curve C, as long as C lies in the annulus Rj < Iz - z 0 I < R2.
2.
3.
Example 3.6-6. Expand f(z) =
z+3
(3.6-46)
(z2-z-2jz in powers of z in the following regions (i) (ii) within the unit circle about the origin, within the annular region between concentric circles about the origin having radii 1 and 2 respectively, exterior to the circle of radius 2.
Note that the series in Equations (3.6-48b, 49b, 50b) converge in the region indicated. In (i), (ii) and (iii), we have written f (z) in a form such that when we expand the appropriate expression as a binomial series, the expansion is valid. The series we have obtained are the Laurent series expanded about the origin which is a singular point of f (z). The other two singular points of f (z) are at z = -1 and z = 2. The domain is divided into regions such that in each region the function is analytic. Example 3.6-7. Prove that
oo
cosh(z+ i ) = a where
,2n
o
+ I an|zn+-L|,
Iz I> 0
(3.6-5 la)
an = - L I
%
cos n6 cosh (2 cose) d0
(3.6-5 lb)
JO
The function cosh (z + — ) is analytic for all non-zero finite values of z. Therefore it can be expanded in a Laurent series at any point z about the origin in the region 0 < I z I < °°. Equation (3.6-35a) becomes
00 00
6-58b)
(3.^ .) I xz + y z /
(3. for all values of y and. we obtain < > | + i \ | / = a 1 (x + iy) + b l ( x " 1 ^ (3. 56d) yields
an = . In Example 3. incompressible flow past an infinite stationary circular cylinder of radius a.6-60)
.6-59a)
x2 + y2
¥ = Y fa! . we deduce $ = atx+
blX
+ i y fal .6-8. there is no singularity and the complex potential O(z) is analytic and can be represented by a Laurent series. incompressible flow. The centre of the cylinder is taken to be the origin.6-57)
We have shown in Example 3. We start by choosing the simplest Laurent series given by O(z) = a l Z + ^L (3. irrotational.. Deduce the complex potential for a two-dimensional.3-6 that the potential < j ) and the stream function \|/ are given by the real and imaginary parts of <E>(z) respectively.252
ADVANCED
MATHEMATICS
Combining Equations (3. There is no flow in the region I z I < a and singularities may be present in this region. Separating O(z) into its real and imaginary parts.6-59b)
The cylinder is a streamline and we can assume that \|/ is zero on the cylinder.L
cosh (2 cos e) cos n9 d6
(3.6-59b). we have shown that an analytic function can be a suitable complex potential for an irrotational..6-5 lb)
Jo
Example 3.6-58b). we obtain
0=L .6-58a)
U +y) = a l x + -~~
xz + yz From Equation (3. from Equation (3. This implies that \j/ = 0 for x 2 + y 2 = a2.M
(3.3-6. In the region a < I z I < °o.6-55c.^ .r )
\ x z + y 1}
(3.
we obtain as (x2 + y2) tends to infinity vM = &\ From Equations (3. 6-59a).COMPLEX VARIABLES
251
Far away from the cylinder. z 0 is an isolated singular point. If z 0 is a singular point but f(z) is analytic in the region 0 < Iz .6-60.6.6-64) (3.21 < R. The point z = 2 is an isolated singular point.4-1. 3.6-63) (3. The origin is not an isolated singular point. 62). The function 1 / (z .z / + X b n (z-z o r n
n=0 n=l
(3.3-62a.6-57) can be written as ^ ) = v0O(z + | . The function i n z is singular at the origin but also along the negative part of the real axis as illustrated in Figure 3.) (3. There are many other singular points near the origin. That is to say.6-35b.zQ I < R for some positive R. 0) From Equations (3. If z 0 is an isolated singular point in the annulus 0 < Iz .2) has a singular point at z = 2 and is analytic in the region 0 < I z . b n are given by Equations (3. the flow is undisturbed by the cylinder and we assume the velocity distribution to be v = (v^.6-62) (3.
. f (z) is analytic in a region in which the point z = 2 has been removed.35a)
where the coefficients an.6-61)
The function O is analytic in the annulus a < I z I < « and satisfies all the boundary conditions and is the complex potential for the present flow. c).7 RESIDUE THEORY
We have defined the singular point (singularity) zQ of f(z) to be the point at which f(z) ceases to be analytic. f (z) has a Laurent series representation which can be written as
oo oo
f(z) = X a n ( z .z 0 I < R. we deduce t>! = a 2 v 00 Equation (3.
Discuss the nature of the singularities of
(i)
cosz-1 . We can also obtain this result using 1'Hopital's rule. (3.. the expansion of (cosz . The coefficient bj is the residue and is denoted as Res (z 0 ) or Res [f (z).254
ADVANCED
MATHEMATICS
We consider three types of singularities (i) If all the coefficients b n are zero.. the function &&%.l ) / z 2 can be deduced to be
cosz
~1 z2
=
_ l +^ _ ^ + 2 4! 6!
( 37-3) { }
We define the value of (cos z . the Laurent series reduces to the Taylor series. We define lim f(z) = aQ
z->z 0
(3. z 0 ] (iii) If all the coefficients b n do not vanish.
.7-1. If m is one. The singular point is a removable singular point. j s n o t defined at the origin.7-1)
For example. b m ) are non-zero and the coefficients b n vanish for all n > m.7-2)
The series in Equation (3. The singular point z 0 is a pole of order m if (bj. . (ii) If all but a finite number of the coefficients b n are zero. . the function is analytic everywhere... We can also use l'Hopital's rule to deduce that its value at the origin is (-1/2).
(iii)
M ^ z5
at the origin. z2
(ii)
e 1/z2 .7-2) is convergent and we define f (0) to be one..^ + | ^ . (i) The expansion of cos z in powers of z is known. The origin is a removable singular point.l ) / z 2 at the origin to be (-1 IT). z 0 is a pole. z 0 is a simple pole. Note that the function is not defined at the origin. The origin is a removable singularity. In so doing. z 0 is an essential singular point. The function has a series representation f(z) = s i | z = 1 .
Example 3.
we have
I f(z)dz = X I f ( z ) d z
Jc j=i JCJ
(3. 12b).7-12a)
Jq
(3.. .7-11)
Consider the integral I f(z) dz for a fixed value of i... positively oriented. Cauchy's residue theorem states that k f(z)dz = 2ni £ Res[f(z). .7-11. z n contained in the interior of C. Each circle Cj (j = 1.7-10)
Proof. Let C\. The residue bj as defined by Equation
JCi
(3.6-35c) is given by bj = . within and on which a function f (z) is analytic except at a finite number of singular points z l5 Z2.. with their centers at the isolated singular points z 1.
. the function l/f(z) has a pole of order 8.. • If all the singularities of f(z) in the finite complex plane are poles.. as shown in Figure 3.L
27tl
I f(z)dz
(3. . z 2 .COMPLEX VARTARLES
257
The function f(z) has a zero of order 8.zJ Jc n=l [ (3. .. Cauchy's Residue Theorem Let C be a simple closed contour. f(z) is a meromorphic function.. k) lies inside C and exterior to the other circles.7-10) is obtained by combining Equations (3. From Equation (3. C2.5-33).. Therefore. .7-1. be k circles each positively oriented..7-12b)
27uib1 = I f(z)dz
Equation (3.. z^ respectively. Q.
f(z) has a simple pole at z.7-13)
From Equation (3.z i ) n
n=0
+
. which implies that h(z) has a simple zero at zj [h(Zj) = O].7-13).7-15. that is to say
f(z) = S j l
(3. X
FIGURE 3. we deduce that bj is given by bj = lim (z-Zj)f(z)
Z—>Zj
(3.7-1
Illustration for the proof of Cauchy's residue theorem
Some Methods of Evaluating the Residues (i) Simple pole at n
The Laurent series of f(z) about Zj can be written as
f(*)= £ a n ( z .228
ADVANCED
MATHEMATICS
y
i
_
^.7-14)
If the function f (z) is given in the form of a rational function.7-14) can be written as
. Equation (3.^ (z~zi)
0.
7-17b)
(z-zj)2
= (z-Zirmg(z) where g (z) is analytic and non-zero at Zj (see Equations 3.7-17a).ta J^L kzj£ld
dz 1 "" 1
If f (z) is as given by Equation (3. it is possible to decompose f (z) into its partial fractions. By comparing coefficients of powers of (z .7-16c)
f(z) = £ a n (z-z i ) n + . we can expand g(z) and h(z) in their Taylor series.+ .7-17a. it has a Taylor series which can be written as
g(z)=
I^r^-Zi)"
n=0
n-
(3-7-18)
Comparing Equations (3. Since g (z) is analytic.7-15).^
n=0 (Z-Zi)
+
.. .
.16a)
(3.. 18). 19b )
(m-1)!
. we find that the residue b } is given by
(m-l)( \
b> =
isdyr
z ^^i
("-19a)
(3 .A .7-16b)
= lim / i^\ *->* \[h(z)-h(z i )]/(z-z i )/ = ^ L h'(Zi) The derivative h'(zj) is non-zero because h(z) has only a simple zero at zy (ii) Pole of order m > 2 The Laurent series of f (z) about Zj is
(3. 7 . In many cases.K m
1 z ^ Zi
fef^M
[h(z)-h(zi)J
(3. b.COMPLEX VARIABLES
259
b.7-8a to g).7-17a)
(3. we can determine bj.zj) with Equation (3. .7. +-*»(Z-Zi)m
(3.
7-41)
where F (cos 9. sin 9) is a rational function of cos 9 and sin 9.7-42)
.7-39) and using Equation (3. and C is the curve defining the surface of the cylinder. Differentiating Equation (3. sin e) d6 o
(3.V o o (z + a 2 /z) (3.7-40a)
(3. we write z = eie The trigonometric functions cos 0 and sin 9 can be written as (3. We consider a few examples. we have shown that Equation (3. Calculate the force if O = .7-39)
In Example 3. This surprising result is known as d'Alembert's paradox and is discussed at length by Batchelor (1967).7-40b)
The residue is zero and this means that there is no force acting on the cylinder.6-8.iF y =
T[
Jc
tefdz
( 3 -7"38)
where p is the density of the fluid. we obtain
F X~ ' ^= ^ l V ~ I' " ^+ ^ ^
v2 = i p ^ 2 7 U i SR.7-39) represents the complex potential for a flow past a circular cylinder of radius a with center at the origin. Triginometric Integrals To evaluate integrals of the form
J
f27t
F (cos 6.7-38).COMPLEX VARIABLES
263
F *. • The integral of real variables can be evaluated using the residue theorem. O is the complex potential.es
(3.
7-52a) (3. but the limits on the right side of Equation (3. we write.COMPLEX VARIABLES
265
Combining Equations (3.°°. °°) is defined by
r
Jo
f(x)dx = lim
a -*°°
r
Jo
f(x)dx
(3. 49b) results in
I = i ^ M = E = -TM= 2 Va 2 -b 2 Va 2 -b 2
Improper Integrals of Rational Functions
(3. for example. the improper integral of f over [0.7-54)
. we have
f(x)dx =
lim
I f(x)dx
(3.b)
We recall from the theory of functions of real variables that if f (x) is a continuous function on 0 < x < oo.7-5 lb)
J—
a ^-°°A
When both limits exist.7-51a)
provided the limit exists.7-53)
J-a
It may happen that the limit in Equation (3.7-53) exists. 0]. the function f(x) defined by f(x) = x The limit (3.7-44b. Consider. for an integrable function on the whole real line (—°°.7-5Oa.7-52b)
JO
= I
J-oo
f(x)dx +
The value of the improper integral is computed as
r
J-c
f(x)dx = lim
a ^°°
r
f(x)dx
(3. Similarly if f(x) is continuous on (. °°)
r
f(x) dx = J-oo
a
f
lim ^ /
a
r
f(x) dx + lim f(x) dx a ^°° Jo f(x)dx (3.7-52a) may not exist. 46b.
we have
.
f°°
Consider the integral I
J— oo
f (x) dx where (3. The contour associated with the complex integration must therefore also include the real line. This is defined by Equation (3.7-55)
does not exist. From now on. We denote the semi-circle by F and we choose R to be large enough so that the semi-circle encloses all the poles of f (z).7-2..) of an integral over the interval (-oo. Note that the integral on the left side involves the real line.7-53) if the limit exist.v. The contour we choose is a semi-circle.> ~ 2
(3.
The integral is then given by
J
r°°
k
[ f(x)dx = 2ni nX Res[f(z).7-56)
This leads us to define the Cauchy principal value (p. Q (x) has no real zeros.R
. the principal value of I
J-oo
f (x) dx is implied whenever the integral appears. But
lim
I xdx = 0
/-a
r
(3. The degree of P (x) is at least two less than that of Q (x). zj J-R Jr n= 1
(3.7-57)
(i) (ii) (iii) (iv)
f(x) = P ( x ) / Q ( x ) P (x) and Q(x) are polynomials.266
ADVANCED
MATHEMATICS
T
lim
a—>°o/0
a2
xdx = lim =a .. z ] = 1
-oo
(3.
k
f(x)dx+| f(z)dz = 27ti X Res[f(z). centered at the origin with radius R in the upper half plane as shown in Figure 3..7-59)
. z^ are the poles of f (z) that lie in the upper half plane.7-58)
where z l5 z 2 .. By the residue theorem. c»).
7-61)
Contour integral for evaluating infinite integrals
From condition (iv).7-61) (3. e—>0 (3.7-60)
I f(z)dz = I f(Re ie )iRe ie d6 Jr Jo
(3.7-58). we deduce that for sufficiently large R I z f (z) I < 8 for all points of F.7-59) reduces to Equation
.7-60). Equation (3.
0
>
1
R
(3.7-58) implies the integral along F to be zero and this will be established next.7-62b)
Jo
< en (3. we obtain (3. and theby integral around F (3.7-62c) Equation is obtained using inequality As R—>(3.7-2
.7-59).7-60)
(3.7-62b) oo. By substituting z = Reie into the second integral on the left side of Equation (3.COMPLEX VARIABLES
267
Equation (3.7-62a)
r
< £
d9
(3. is zero.
I
-R
FIGURE 3.
The contour is indented by making a semi-circle of radius e. The contour integral given in Equation (3. zn]
n= 1
(3.7-3
Contour integral with a pole on the real axis
It was shown earlier that in the limit as R —> °°. we can write
(3. We denote the small semi-circle by y. we have z = a + eeie and the contour integral becomes (3. I f (z) dz —> 0. where a is real. Suppose f (z) has a pole at z = a. We now consider the integral around y.7-59) is now modified as follows
J
f ( x ) d x + | f(z)dz+ I
f(x)dx+(
f(z)dz = 2ni X Res[f(z).7-3.7-81)
I f(z)dz = I f(a + e e i e ) e i e i e d 0 Jy Jn Since f(z) has a simple pole at z = a. On y.7-80)
-R
Jy
A+e
Jr
-R
0
R
FIGURE 3.7-82)
.COMPLEX VARIABLES
271
semi-circles in the upper half plane to remove the poles from the real axis. centered at z = a in the upper half plane as shown in Figure 3.
272
ADVANCED
MATHEMATICS
f(z) = g ( z ) / ( z . In the limit. ..7-4. Example 3.7-80) becomes
J
f oo
k
! f(x)dx = 2ni 2 Res[f(z). one at z = -a denoted by Ji.7-72).7-86)
If there is more than one simple pole on the real axis and if we denote these poles by aj.7-82). We illustrate this by considering an example.7-83)
where g(z) is analytic and can be expanded as a Taylor series about the point a. 84) into Equation (3. at -a and a. where C is the indented contour to be defined later. 83) that g(a) is the residue of f(z) at a and is denoted by Ra. as R —> °o and e —> 0.7-9. that is to say. The
Jc a z -z z
function f(z) [ = l / ( a 2 . The function g(z) can be approximated by g(z) = g(a) + 0(e) Substituting Equations (3.z 2 ) ] has two poles on the real axis. a2.z] + 7tiRa
-oo
n=l
(3.
.7-85a)
(3. the sum of the residues at aj. We indent the contour by making two small semi-circles of radius e. The contour C is shown in Figure 3.7-84)
I f(z) dz -> i I g(a) de Jy Jn -> -iicg(a)
(3.~ a2-x2 TLjsrna^ a (3. . The same modification can be applied to Equation (3.7-85b)
We note from Equations (3. we obtain. and the other at z = a denoted by y2. (3.7-83. as £ —> 0.. ^ ••• > a i > we replace R a in Equation (3. Equation (3.7-87)
=
a is real
I We consider the integral I
1 2
eiz
2
dz.7-14. Show that cosjLdx ) . a^ .7-86) by the sum of R .a )
(3.
^ .7-85b).7-92b. The function e i z / (a 2 . Equation (3. so the right side of Equation (3.7-91) simplifies to
I
Jyi a .x z
= i7c[R a
+
R_J
(3.d x .7-96)
The residue R a is given by
(3.7-93b)
f eiz
lim ^-rdz =-i7rR_ a (3.7-88) is zero. 93b).a and a. = lim -a z _^_ a e~ia = ^ From Equations (3. (a-z)(a + z)
(3.4r
i p~la /
de
j
n
(3J-92a)
(3.7-92b)
-» " ^ ^ Using Equation (3.z
f-ziz
eiz
. Equation (3.7-88) simplifies to
. Similarly we have
I eiz
Urn ^ 0 J Y2 .274
ADVANCED
MATHEMATICS
In the limit as e —> 0.7-95) where R a is the residue at z = a.7-94)
^o
jYl a2-z2
Equation (3.7-93a)
(3.7-94) is a special case of Equation (3. we find that the residue R_a at z = . we obtain /Z+f)£lZ.d z a2_z2 =-i7tRa (3.7-97a)
. e—> 0.z 2 ) has no other poles except at .a is given by R . a .7-14)._<*. In the limiting case R—> < * > .^ .
the mapping is one to one. we obtain
(3.7-98b) into its real and imaginary parts. we regard one complex plane to be z [= (x. From Equations (3.7-98b)
=
(3.8
_sinjL_ dx = 0
(3.8-2a. y)] and the other to be w [= (u. b).b) (3. v) plane there corresponds one and only one point in the (x.7-99a)
.7-97b)
. If to each point in the (u.^ d x J_ooa 2 -x 2
= i^[e-ia-eia]
2a
(3. The equation of T is written as w(t) = f(z) = u(t) + iv(t) (3._ooa 2 -x 2
I fl2 Yx2 y_oo a —
3.3-6a. we have I _cos_x_ d x ILsin_a
a
(3.7-96). We examine the geometric properties of this mapping. we consider the function f defined by Equation (3.3-5) to be a mapping from one subset of a complex plane to another. The tangent to C at P is given by
. y) plane and vice versa.7-98a)
= J (sin a) Separating Equation (3. The equation of a curve C in the (x. v) plane. we have defined a curve in the (x.3. v) plane. y) plane by introducing a parameter t.COMPLEX VARIABLES
275
eia = -~^ Substituting the values of R a and R_a into Equation (3. In Section 3. v)] and f establishes the correspondence between the (x.7-99b)
CONFORMAL MAPPING
In this section.8-1)
Suppose the curve C passes through a point P at which t takes the value tQ. y) plane is written as z(t) = x(t) + iy(t) The curve C is transformed to a curve T in the (u. y) plane and the (u.
8-3)
t
lo
The tangent to the curve T in the (u. Using the Cauchy-Riemann conditions [Equations (3.276
ADVANCED
MATHEMATICS
d* at
t
lo
= <£ dt
t
lo
+i % at
(3.8-5) 3v dy dy J L dt -
Equation (3.8-7d)
ax -^ dz
The condition that the determinant be non-zero is equivalent to the condition that — be non-zero.8-5) expresses the transformation of tangent vectors from the z-plane to the w-plane. it ensures that non-zero tangent vectors in one plane are transformed to non-zero tangent vectors in the other plane. as at at
"dul dt dv.8-4)
£o
to
[o
Using the chain rule "-^ and ^ can be written. v) plane at the point [u (tQ). v (t 0 )] is (3.8-7c)
(3. If it is non-zero.dt J
r^L 3x 3v \_dx
^L 1 [ dx 1 9y dt (3. The condition for a unique solution is that the determinant du dx J = 9v dx 3v 9y du dy # 0 (3. in matrix form.8-7b)
= ^
=
+
i | ^
ax
2
(3. dz
. b)]. .3-29a. J can be written as
a_u a_v
J
a u a_v
~ ax a y
_ /3u\ 2
ay ax
/av) 2
(3-8"7a)
(3.8-6)
The determinant J is the Jacobian.
Further discussions are given in Chapter 4. the mapping is conformal. but the sense is not the same. y0). The Jacobian J is defined as
. in a conformal mapping. we briefly review the case of two variables. v 0 ). U
FIGURE 3. The curves Fj and F 2 intersect at (u0 . both the magnitude and sense of angles are preserved. measured from Fj and F 2 .
dz
y A
VJ
uo. Suppose Cj and C 2 are mapped by f to Fj and F 2 respectively in the w-plane. Let Uj and u 2 be two differentiable functions of two variables xj and x2. as shown in Figure 3. the mapping is isogonal.
(c)
4^ i s non-zero. y 0 ). f is single valued.COMPLEX VARIABLES
2ZZ
The curves Cj and C 2 in the z-plane intersect at (x 0 .yo)
^
V
La
r.8-1.8-1
Conformal mapping
The Jacobian J introduced earlier plays an important role in the theory of transformations. Here. measured from Cj to C 2 as illustrated in Figure 3. Let A9 be the angle between the tangent vectors to Cj and C 2 at (x 0 . v0) and let Aa be the angle between the tangent vectors to Fi and F 2 at (u 0 . The conditions that f represents a conformal mapping are (a) (b) f is analytic. If the magnitude of A0 is equal to the magnitude Aa.
1 ^ X 1 +.8-1. If A0 is equal to Aa. Thus.
8-18) (3. v) in the w-plane is mapped to a point (x + Bj.
^2
= 0
bl b2
(3.
.8-13)
If Xi and x 2 are also functions of yj and y2. the double integral of any function (j) over R is given by I I (J)(x 1 .8-2. B2) as shown in Figure 3.U2) = 5
d(yi>Y2)
K U 2 ) a(xl. x 2 (u!.y + B2) (3.8-17) into its real and imaginary parts. Equation (3. If we take A to be unity.8-17) (3. Separating Equation (3. we have
a(Ul. we obtain (u.u 2 )] | J| duj du 2 R
R-
(3.COMPLEX VARIABLES
al
279. The Jacobian J gives the magnification of the area due to the transformation. u 2 ) plane.8-15)
If < ] ) is unity. y) by (Bj. We now consider some transformations in the complex plane.v) = (x + B 1 . Linear Transformation Consider the transformation w = Az + B where A and B are complex constants.8-16)
Equation (3.x 2 )dx 1 dx 2 = I I ^ [ x j f u j . we obtain the area of the closed region. This corresponds to a displacement of (x.8-18) shows that a point (u.8-16) becomes u + i v = (x + i y) + (Bj + i B2) where B is written as B j + i B 2 .x2) 3 ( X 1' X 2) ^(yi'Y2)
8
j
If a closed region R in the (xl5 x2) plane is mapped into a closed region R' in the (uj. ^ ) . y + B 2 ) in the z-plane. B^ and B 2 are real constants.
8-2
Translation
If we take B to be zero and we write w.
.
1
6
~
I
0
1
8
1
» •
I
2
I
4
I
6
I
8
^
FIGURE 3.b) (3. Equation (3.c)
Equation (3.8-19a. A and z in their polar form as w = pe'*.8-21a) represents a contraction (a < 1) or an expansion (a > 1) of the radius vector r by the factor a. Equation (3. The shape of all the figures is preserved.8-21a. Combining these two cases. z = reie (3.8-16) represents a displacement.8-16) becomes p e j * = are**0""9* We deduce p = ar. < > | = a +9 (3. and a rotation.280
ADVANCED
MATHEMATICS
V I
v "
w
8 -
'
8 -
z
2 1
0 2 4
i. we deduce that the linear transformation given by Equation (3.8-3.b. a magnification. This transformation is illustrated in Figure 3.8-21b) is a rotation through an angle a.8-20) A = aeia.
8-4. as shown in Figure 3.v = 3.8-16. the distance between two points has been magnified by the factor V2" and all the lines have undergone a rotation of n/4. Reciprocal Transformation Consider the transformation w = 1 (3. 23).8-28a.
u + v=l.8-28c)
.
u-v =. .B
y\
c1/
R
\ ^
\
R1 >v
^
0
A
X
Nv
/
VY/4 o1
FIGURE 3.8-29) (3. We identify A and B to be A = (1+i) = n e i7tM B = (l-2i) The transformation is as described earlier.
v i *
B'
c
.8-27a-d)
respectively in the w-plane. We note that all the points in the z-plane are displaced by (1.2) in the w-plane.b) (3.
u+v=-l
(3.282
ADVANCED
MATHEMATICS
u .8-4 Mapping of a rectangle in the z-plane to another rectangle in the w-plane The transformation can also be deduced by comparing Equations (3. The rectangle R in the z-plane is mapped into rectangle R' in the w-plane.l .
the inverse mapping is also a reciprocal transformation.8-30)
From Equation (3.COMPLEX VARIABLES
283.8-5. we obtain the identity transformation. The equation
.8-5
Reciprocal transformation
On reflecting about the real axis.
r
FIGURE 3.8-32)
On applying the transformation twice in succession.b)
y z. That is to say.8-29). is given by z.e" i e (3. We recall that if z 0 is a point in the complex plane.5-7. we obtain z = -1 (3.= ± e i e zo r (3.8-31a. in this case.
/
1%/
\
I
Xe
\
r
^7
1
z. This is illustrated in Figure 3. The reciprocal transformation maps circles or straight lines into straight lines and circles. The process of inversion was introduced in Example 3. its inverse point Zj. with respect to the unit circle. from Equation (3. we deduce that the reciprocal transformation consists of an inversion with respect to the unit circle and a reflection about the real axis. z\ is transformed to w.
Writing z in its polar form.8-30). = J . We note that.8-29) becomes w = 1. Equation (3.
c ± 0) is mapped into a circle passing through the origin in the w-plane.8-34.8-33) can also be written as azz + bz + bz + c = 0 where a and c are real constants and b is a complex constant. b. But the points inside the circle in the z-plane are mapped to points outside the circle in the w-plane. the circle not passing through the origin in the z-plane (a ^ 0.
(3. d are complex constants and (ad . the straight line passing through the origin in the z-plane (a = c = 0) is mapped into a straight line passing through the origin in the w-plane.8-33)
(3. Bilinear Transformation The bilinear (Mobius) transformation is given by
w = ^±4
cz + d where a. we deduce the following (i) the straight line not passing through the origin in the z-plane (a = 0. Equation (3.284
ADVANCED
MATHEMATICS
a (x2+ y 2 ) + bxx + b2y + c = 0 represents a circle if a is non-zero and a straight line if a is zero. center at the origin in the w-plane.be) is non-zero. c = 0) is mapped into a straight line not passing through the origin in the w-plane.8-36) can be written as
(3. 35).8-29) into Equation (3. Substituting Equation (3. center at the origin in the z-plane. c. (3. the circle passing through the origin in the z-plane (a ^ 0.8-35)
(ii)
(iii)
(iv)
Note that the unit circle.8-36)
. we obtain cww + b w + bw + a = 0 From Equations (3. Equation (3. if a is non-zero and c is zero. is mapped into a unit circle.8-34). c * 0) is mapped into a circle not passing through the origin in the w-plane. we have a circle passing through the origin. we have a straight line.8-34)
If a is zero.
we obtain w = eie(z-zn) .8-3.8-47) (3.i + iz) (5-3i)-z(l+i) P-84JJ
Example 3. v _ .i.8-45a.z 0 ) (3.b/a by z 0 and . with the condition that Im (zQ) is positive.i) in the z-plane into (0.b)
. 45a.0> (z-z0) The real axis of the z-plane is mapped into w = e ' 6 (X I ^ (x . The bilinear transformation can be written as
(z .b/a corresponds to w = 0. .
We denote . we obtain 2 ( l .8-2. 1. we have Iwl = loci = 1 Combining Equations (3. the point z = . The point zQ is then mapped to the origin which is inside the unit circle.8-44). and the lower plane (z 0 ) is mapped into the outside of the unit circle.d/c by z 0 . From Equation (3. 2 .8-44)
(z~zo)
The origin in the z-plane is mapped into a point on the unit circle in the w-plane.zn)
w = a )—=&•
(3. Find the bilinear transformation that maps the upper half plane Im (z) > 0 into the unit circle I w I < 1. i) in the w-plane. we deduce (i) (ii) the point z = . Using Equation (3. we obtain
hzHln!) _
(-l)(i-w)
(i + i-z)(2)
(l+2i)(2-i-z)
('-*42J
On simplifying Equation (3. From Equation (3. Determine the bilinear transformation that maps the points (1 + i.8-46) (3. b).286
ADVANCED
MATHEMATICS
Example 3.8-42).8-41).d/c corresponds to w = °o.8-44.8-36).
The Schwarz-Christoffel transformation is the required transformation and is given by iw.8-6.8-47). Determine the complex
The motion of the ideal fluid is due to a source of strength m at
potential <&.8-49) may be considered to be a constant which can be absorbed in K.°o.8-48)
Thus the real axis of the z-plane is mapped into the unit circle in the w-plane. (3.8-51)
.3-6. which would allow the determination of the velocity and of the streamlines. its interior angle is zero and we may take the corresponding point a in the w-plane to be infinity. Equation (3. and we wish to map its boundary into the real axis (v = 0) of the w-plane. 0) respectively in the w-plane. We map the vertices A. If one vertex of the polygon is at infinity.. C.= K ( w . and (<*>... 0).ES
287
From Equation (3. B. (1. and D to the points (.l ) 1 / 2 (3.
Schwarz-Christoffel Transformation
Suppose we have a polygon with interior angles o^. The semi-rectangular channel ABCD is shown in Figure 3. (X2 .a 1 ) 1 . In this example. as shown in Example 3.a 2 ) 1 . An ideal fluid is bounded by a semi-infinite rectangle given by x>0. l>y>0 (3.COMPLEX VARIARI. It has already been shown that a point on the upper z-plane is mapped into a point inside the unit circle in the w-plane.8-4. (-1.0)..8-49) becomes &£.c t 2 / 7 C . we obtain _ eie(x-z0) e~ie(x-z0)
1
(3.a^) in Equation (3.a i / 7 t ( w . a2 are the real values of w (= u) corresponding to the vertices of the polygon. Equation (3.8-46) is the required transformation.= K ( w + l ) 1 / 2 ( w .8-50) -. Example 3.8-49)
where K is a constant and a1. 0). The factor (w .
we have w = ae
i 9
(3. ) .8-60)
+—
a.m i n | w + i sinh%\ = . we find that the singular points are z = ±l Consider a circle of radius a with center at the origin in the z-plane.
(3.m i n (w 2 + sinh 2 1) = .8-55a.m i n [w .m i n (cosh2rcz + sinh2 £•) ' (3.8-57b) (3.
as shown in Figure 3.8-57a)
(3.8-61)
. p.8-58a) (3.i s i n h 2 .8-58c)
Joukowski Transformation
The Joukowski transformation plays an important role in aerodynamics. we have w = cosh (TCZ) The source in the z-plane at ^——.8-57c) = i sinh & in the w-plane.8-59). b) into Equation (3.8-59) is zero.8-58b) (3.8-53). It transforms an aerofoil into a circle.8-6.8-56)
w = cosh[|-(l +i)l
= cosh^ c o s h ^ + sinh^ s i n h ^ = isinh| The source at S = ^ ' in the z-plane is mapped into a source at S'
(3. 1965. 210] 0> = . In the w-plane. The complex potential < E > due to a source of strength m at i sinh ~ and a wall along the real axis is [Milne-Thomson. The transformation can be written as w = z+ 1 The singular points of the transformation are the points at which ^(3. From Equation
(3.is mapped into a source at (3.COMPLEX VARIABLES
289
Substituting Equations (3.
which is equivalent to saying that the circle does not pass through the singular points.8-7
Mapping of a unit circle into a segment of the real line
Example 3.8-62a. v=0 (3. b) become u = 2cos6. The center of Cj and C 3 is at the
.8-64a.8-7. A circle passing through one singular point and enclosing the other singular point in the z-plane is mapped into an aerofoil in the w-plane.
r TT
71 "x
-2
2
*u
FIGURE 3. we obtain
u
= 'a
+1
' cos9.290
ADVANCED
MATHEMATICS
Separating Equation (3.b)
The circle passing through both singular points in the z-plane is mapped into a segment of the real axis in the w-plane. b) yields j^r +j ^ = 1 (a2+l) (a2-l) (3. If the radius of the circle is one (a = 1). Equations (3.8-63)
A circle in the z-plane is mapped into an ellipse in the w-plane.8-62a. as shown in Figure 3.8-59)]. and C3 in the z-plane into the w-plane under the Joukowski transformation [Equation (3.8-5. as long as the radius of the circle is not one.
v = l l J l i l sine
a
(3.8-62a.b)
a
Eliminating B from Equations (3. Discuss the transformation of the circles Cj.8-61) into its real and imaginary parts. C2.
8-65a. It also intersects C^ at the singular point A and C 3 at the point B [=(2. can be written as u2 v2 1
h+ V =1
The equation for C2 is given by z = \ + |eie
(38-66)
(3. an ellipse [Equation (3. C2.0)]. b)] and can be written as -2 < u < 2. in this example.b)
The circle C3 is mapped into F 3 . v=0 (3.8-8
Mapping of the circles Cj. The circle C2 passes through one singular point (z = -1) and encloses the other singular point (z = 1). The circles C\.
y i
v i. 0) and its radius is 3/2.8-8.COMPLEX VARIABLES
297
origin and their radii are one and two respectively.
FIGURE 3. and C 3 from the z-plane to the w-plane
The unit circle Cj is mapped into Fj. The center of Ci is on the real axis at (112. and C3 are shown in Figure 3.864a.8-67)
. a segment of the real axis in the w-plane [Equation (3. C 2 .8-63)] and.
5-7. C O . and s are real. If we need to solve Laplace's equation subject to a boundary condition in a complicated domain D. Determine all three values of z 1/3 and plot them in the Argand diagram. we can map the complicated region into a simpler region. there exists a transformation that can map it into a unit disk (I z I < 1). Unfortunately it is not shown how the transformation can be obtained. for any simply connected domain which is not the whole complex plane.8-7Oa. Riemann has shown that. Write z = 4 Y 2 + 4 V 2 i in polar form. we can transform the domain to a simpler one D1 and solve the problem in D'. Using a conformal transformation to the w-plane. If the geometry of the region of interest is complicated. such as the unit circle. v)]} is also harmonic satisfying the same boundary conditions [Equations (3. b)] at the transformed boundary. we have mentioned the importance of Laplace's equation and have shown that an analytic solution is a solution of Laplace's equation. the solution of Laplace's equation is known for any domain.
PROBLEMS la.
TI* = I f(s)e. We can then invert back to D.3. we have shown that if the value of a harmonic function is given on a circle of radius R. the transformed function K (u. In Example 3. v) {= § [x (u.
In Section 3. In all physical problems.8-70a. By using a conformal transformation. We have applied this technique in Example 3. The complex viscosity T] of a linear viscoelastic liquid is given by
2a. since we can map the domain into a unit circle. In theory. its value at any other point is known. y (u.
. the problem of imposing the boundary conditions can be demanding. Let <|) be a harmonic function satisfying conditions at the boundary. the solution has to satisfy certain boundary conditions. v).8-4.i ( O s ds
Jo
where f.b)
d<b where §0 is a real constant and — is the derivative of § with respect to the normal to the boundary curve. The main problem is to determine the suitable transformation from D to D'.COMPLEX VARIABLES
293.= 0 dn (3. such as (j) = <|>0 or dd> -p.
what are the stream lines in this case? Do they represent the flow due to a source at the origin? Answer: arctan (y / x) 7b. If f (z) is an analytic function. If f (z) is the complex potential O. Show that the equation
6b. Obtain the harmonic conjugate v (x.3-26). The stream lines are given by \}/ = constant. y) = inVx 2 + y2
is a harmonic function.294
ADVANCED
MATHEMATICS
If we write n* = Ti'-iri" f(s) = Ke~s/X
where K and X are constants. identify the potential < j ) and the stream function \|/.
u(x. calculate r|' and r|". Discuss the continuity and differentiability of the following functions (i) (ii) 5a. use the Cauchy-Riemann conditions to obtain two other expressions for f'(z 0 ). Determine and plot the curve given by z. f(z) = z 2 f(z) = 1 (iii) (iv) f(z) = z 2 + | f(z)=|z|
l+X2®2
Starting form the definition of f'(z 0 ) given by Equation (3.1 4a. Answer: ^K ^ (oKX2
l+X2(a2
3b. show that
—7 + — r If (z) I = 4 f (z) 3x z dyz
. The quantities r\ and r\" are the dynamic viscosity and dynamic rigidity respectively. y) and the function f (z). one in terms of u only and the other in terms of v only.
dz Calculate lim
z->0
sinz
z
along the real and imaginary axes. If C is the closed curve (Cj .
lib. This is shown in Figure 3.
c
what condition(s) in the Cauchy's theorem is (are) not satisfied?
y "
-I t
x
FIGURE 3. Is the origin a singular point? 9a. 10a. show that -"4 = 0. z ) is an analytic function.
Cj is the upper semi-circle of unit radius.1 .
Evaluate I
zdz along the semi-circles Cj and C2. centered at the origin traversed in the clockwise direction from ( . If f ( z . 0) to (1.COMPLEX VARIABLES
295
8a.
Determine the real and imaginary parts of f(z) = ^
Li
Are the Cauchy-Riemann conditions satisfied? Compute f'(z).P-llb
Integration around semi-circles
. what is the value of I z dz ? Is the integral zero? If not. 0) and C 2 is the image of Cj about the real axis.C2).P-llb.
What is the value of T if the radius of C is changed from 1 to 3/2 ? 14a.J — = l + z + .296
ADVANCED
MATHEMATICS
12a. . for which of the following < E > is T zero? (i) (ii) O = z O = i log z
(iii)
* =
^
If F is not zero. z 0 is not enclosed by C z 0 is enclosed by C Answer: 0 Answer: 27iicosz 0
The circulation T round a closed curve C in a two-dimensional flow is given by
r
= Re I ^ dz
I dz
C
where O is the complex potential.. . 1 -z |z|<l \62ni
. If C is the unit circle with center at the origin.. . evaluate T.
Evaluate the integral I —sin z
AZ ^ w here C is a simple closed curve for the following
1 (z-z 0 ) 2
C
cases (i) (ii) 13b. + z n + . Evaluate the integral I (cos 2 z + 2cosh 4 z)
1
C
l
z
where C is any simple closed curve that encloses the origin. Show that the expansion of —^— about the origin is . Answer: 15b.
Locate the singularities of the following functions and discuss their nature (i) (ii) e1/z ez/z3 (iii) (iv) zsin(l/z) (cos z)/(z-7c) 2
22a.
Find the Laurent series of (z-l)(z-3) (i) (ii) in powers of z.
Locate the poles of the following functions and specify their order rs z(z-2) ( z + i r ( z 2 + 4) . in powers of ( z . are there points at which the series about the point (0. 17a.. Determine the radius of convergence of the new series. ez cos z z
21a.
Expand Ln (
\ for | z | > 1. Sketch the circles of convergence of the two series.
z
19a. Find the Taylor series of sin z about the point Z = K/2.
... What is its radius of
18a. Is the origin an isolated singular point? To describe the behavior of a function f(z) at infinity. Do they coincide? If not.
e 2z
Obtain the Laurent series of (z-1)3 convergence?
about the point
z = 1. we make the transformation £ = l / z and the point at infinity in the z-plane is mapped to the origin in the ^-plane.Expand the same function about the point (0.
The function cosec (1 /z) has singularities at the origin and at points on the real axis given by z = 1 /nn . 1/2) is convergent while the series about the origin is not convergent? This process of extending the region of convergence is called the analytic continuation of the function. Discuss the behavior of the following functions at infinity (i) (ii) ez z1/2 (iii) (iv) zsin(l/z) z/[(z-l)(z-3)]
23b. 16a. Note that this point lies inside the circle of convergence and is not on the line joining the origin to the singular point.1 ) . 20a. 1/2).
7. i. y. V . a and q are vectors. j and k are unit vectors along the x. A vector is also defined as a tensor of order one. It is a tensor of order zero and temperature is an example of a scalar.. T_Y.
TABLE 4.. . is represented by nine numbers (components).A. The extra stress tensor in fluid mechanics is an example of a second order tensor and can be denoted by T. z axes respectively. T__. Third and fourth order tensors will also be introduced..
^y Aj£
J
x. we briefly review some of the known and useful results of vector algebra and of vector calculus..2-3) (4.. velocity is a vector and.2-1. in a three-dimensional space. or time t. when referred to rectangular Cartesian axes..2-2) (4. In Table 4. we will be dealing mainly with vectors and second order tensors. x v .. . x.2 VECTORS
In this section.
A..CHAPTER 4
VECTOR AND TENSOR ANALYSIS
4. v.2-4) (4.2-5) (4.2-6)
.2-1) (4. y. A vector is defined as a quantity that has both magnitude and direction. it is specified by its three components (v x .2-1 Results of vector analysis
a+v=v+a v + (a + q) = (v + a) + q c1v = vc1 (Cj + c 2 ) Y = c 1 v + c 2 v a « v = v « a (dot or scalar product) a • v = | a | | v | cos 0 (0 is the angle between a and v) (4. and % " . A scalar has only magnitude and is completely characterized by one number. For example. v . it is described by three numbers (components). Cj and c 2 are scalars and cp and \)/ are differentiable scalar functions of position x. z... In a three-dimensional space. A second order tensor. v z ). v y . x. 4. t v .1 INTRODUCTION
In this chapter.
If v is a velocity and C is a closed curve then I.3-14b)
I =
+
=
5 41> -1 3
Line Integral of a Vector Function Let v [x (s). The tangent to C at the point P is by definition the line joining P to a neighboring point P 1 as P' approaches P. Evaluate I = j
JA
(4. then di / I = dt/ dr
E(S + A S ) . y (s). Thus X is by definition given by T = lim As->0 = ^ ds If C is defined in terms of t instead of s. 13b). is known as the circulation around C. z (s)] be a vector field defined at all points of a smooth curve C and T be the unit tangent to C.£ ( S )
(4. Example 4.3-17a) (4. 12c. we obtain 3 3~~3 (4. If F ( = v ) is a force.3-2.3-7.3-16)
/" d T
The scalar line integral of v along C is then defined as I = f Y'Tds Jc = f Y»dr Jc where dr = T d s . then I represents the work done by F in moving a particle along C.3-17b)
Y»dr
(4. c.3-14a) (4.VECTOR AND TENSOR ANALYSIS
$02.3-15b)
As
(4. b.
Combining Equations (4.3-15a) (4. 8a. in fluids mechanics.3-18)
.
It then follows that y = 2t 2 since y = 2x2.3-22b. The more practice one has. 2.VECTOR AND TENSOR ANALYSIS
Ml
The path of integration is shown in Figure 4. 2).3-4
Line integral along a parabola
The vector position r of any point on the parabola may be written as r = t i + 2 t 2 2 + 2k (4. fBF-dx = f
J A J Q
[ t i .c)
The evaluation of vector line integrals.3-4 with point A = (0. such as
. the luckier one gets. The point A corresponds to t = 0 and the point B corresponds to t = 1.2 j + 4 t 2 k ] » [ i + 4tj]dt
-
(4.3-22a)
= f1 [t-8t]dt =-l
JO 2
(4. Such parametrization is usually done via educated guessing. 2) and point B = (1.3-21)
This has been obtained by choosing x = t.
AZ
)
•
y
/
FIGURE 4. 0.
3-23b)
can be done by integrating each component separately.o. 0. y.3-23a)
(4. This is illustrated in the next example.2TTb)l
(t=27T) '
I I
/o(N x
y
FIGURE 4. 0) to the point (a.3-4. 2rcb) Jc Jc on the circular helix illustrated in Figure 4. a sin t. bt) (4. z) = (a cos t. Example 4.3-5
Line integral along a helix
.m
ADVANCED MATHEMATICS
I = f Y ds Jc or I = f vxdr Jc
(4.3-5.3-24)
(a. Evaluate I £ ds and I £ x d£ from the point (a. 0. given by X = (x.
If further f (x. y) is 1/(1 + x). y. we integrate f(x. x = a and x = b.3-31).1 = l-(x-l)2 (4. y) = 1.x 2 + 2x . z) dz J dy dx
)
(4. A thin plate is bounded by the parabola y = 2x . Thus. In a single integral the integration is taken along a curve. y. The equation of the parabola may be written as y = 1 . The area of the base of the cylinder is A. Thus a triple integral can be defined as
* rh fv(x) / /•q(x.y)dydx
(4. then
f
f f(x. the area A over which the integration is to be performed is the area bounded by the curves y = u(x).3-29a)
Ja Jc
J a J c
rb
_ J a.3-31)
Ja. Thus an alternative notation is /•b i-v(x) re
f(x. In a double integral the integration is taken over an area. [yu(x) (. in Equation (4.3-28b). If z = f (x. y) dy dx = f
f <p(x) V|/ (y) dy dx
(4. y = v(x).314
ADVANCED MATHEMATICS
of integration is not important. y) to z = q (x.x 2 and y = 0. We can extend the process of double integration to triple or higher integration.3-30)
If f (x. y) then the double integral is the volume of the cylinder formed by lines parallel to the z-axis and bounded by z = 0 and z = f. y). y) is separable and may be written as a product of a function of x and a function of y.y)
In Equation (4. y) resulting in a function of x and y./p(x.3-32a) (4. Determine its mass if the density at any point (x.y) \
I =
I
I
f(x.
(p(x)dx
l[f
d
J L.3-5.:
y(y)dy
(4.3-32b)
. z) with respect to z between the limits z = p(x.y)dydx=
J a Ju(x) JJ A
f(x.3-29b)
Thus in this case the repeated integral becomes the product of two single integrals. Example 4.3-30) is the area A. say g (x. then the double integral in Equation (4. The triple integral is thus reduced to a double integral and can be evaluated as shown before.
3-34)
y=2x-x 2
y
FIGURE 4.3.VECTOR AND TF.NSOR ANALYSIS
575
The parabola intersects y = 0 at x=0 x=2 (4.o) **
Shape of the flat plate.J — dy dx Jo [Jy=O 1 +x J
Note that the double integral results from the fact that we consider a plate of unit thickness.3-6.3-33b)
The shape of the plate is shown in Figure 4.3-6
y^o
\(2. we integrate with respect to y first from y = 0 to y = 2x .35a)
.3-34). Since the density is independent of y.
(4. Then we integrate with respect to x from x = 0 to x = 2. The integral with respect to y is indicated by the arrow
In Equation (4. The mass M of the plate of unit thickness is given by
f2 f2x-x2 1
M =I I . we obtain on integrating with respect to y
M=
r2r^q2x-x2dx
JO L 1 + x J o
(4.x 2 as can be seen from Figure 4.3-6.3-33a) (4.
= _ d(u. The limits of integration can be inferred from the diagram.3-39)
where A is the area of integration in the xy-plane and A' is the corresponding area in the uv-plane. then y = y (u. y(u. It is always useful to sketch the region of integration. z(u.b)
Similarly for a triple integral. z) dx dy dz =jj j [f(x(u. w). v. v. w).| y 3 / 2 ] 0 (4. The Jacobian J is given by dx 3 C V) 3u J = g ^ .3-37b)
=\
= 1 (l-Vy)dy
Jo = [y . v)] IJ I du dv
A A'
(4.VECTOR AND TENSOR ANALYSIS
3JJ
We can interchange the order of integration. w are defined by
. If u and v are two new variables such that x = x (u. v).3-41)
V V where V is the region of integration in the xyz-space and V' is the corresponding region in the uvwspace. v. then A is given by
A=/o1[/^dTy
/•I
(43"3?a)
(4. y. y) dx dy = jj f [x(u. • A change of variables may sometimes make the evaluation of a double integral easier. v) 3y_ 3u dx dv _ 3y 3v
(4. v) (4.b)
jj f(x. y(u. v) . we have
Iff f(x.3-37c. v.d)
In this example we have seen that it is easier to integrate with respect to y first and then with respect to x. w)) I J | du dv dw] (4. The variables u.3-40a.3-38a.
3-49c. If we keep the value of v fixed and let u vary.x 2 . y = y (u.y 2 x = a sin 9 cos < > |. then P will trace out a curve C u as shown in Figure 4.3-50c.320
ADVANCED MATHEMATICS
= &&£. y. v) (4.3-50a) (4. in the xyz-space is mapped into a rectangle.3-49a)
Equation (4.3-4%) (4. Similarly by fixing the value of u and letting v vary. z = a cos 0 (4. in the 0<|)-plane. y) or in parametric form as x = x (u.| (a 2 + b 2 } [ _ 1 ( c o s J \ l * = 4rcabc
( a 2 + h2 + C 2 }
e s i n 2 e + 2 CQS e ) l " + 2c 21 J0 L
cose 3
I Joj
(4. e) defines a mapping (projection) of a region A in the xyz-space into a region A' in the uv-plane.e) (4. 0 < < Let P be a point with vector position r on a general surface S given by Equations (4.3-9. z) = constant or z = f(x. Thus the equation of the surface of a sphere of radius a and center at the origin may be written as x2 + y 2 + z 2 = a2 z = ± Va 2 .d. y = a sin 0 sin (j).3-48e) (4.3-49c. d. v).3-49c to e). > | < 2n. z = z (u. a curve C v will be traced out.3-5!)
.2-10) by
n=f^4
(4. 0 < 0 < K. The tangent to Cu is given by ~~ ( = ! " ) and the tangent to C v is — ( = I V ) .e)
We note that the surface of a sphere of radius a. A unit normal n to the surface S is a vector which is perpendicular to the tangents to C u and C v and is given via Equation (4.3-48f)
Surfaces The equation of a surface S may be written as < p (x. v).3-50b) (4.d.
z(s)] satisfy Equation (4. n is also given by
a=
(4. Similarly we can deduce that grad(p»T v =O where T v is the unit tangent to C v .3-53. y(s). [x(s). From Equations (4.3-49a).3-54)
_grad(L I grad < p|
(43
. we deduce that grad cp is perpendicular to both T_u and T v and thus is a normal to the surface S. 54).3-52)
From the definition of grad (p and the tangent to C u .3-9
Curves C u and C v on surface S
The curve Cu can also be given by Equation (4.3-52) can be written as gradcp«T u =O where T u is the unit tangent to C u .55)
. On differentiating with respect to s and using the chain rule we obtain |cpdx+3«pdy+3cpdz=0 dx ds dy ds dz ds (4. we find that Equation (4. That is to say.3-1) and since C u lies on the surface S.3-53)
(4.VECTOR AND TENSOR ANALYSIS
321
Z
-i
\
>/
oJ^ / x r
•
y
FIGURE 4.
The surfaces we shall consider are two-sided surfaces.3-51).3-10 is an example of a one-sided surface.3-50c to e) —^.3-56a)
. Find the unit normal n to the surface of a sphere of radius a. 55). It is obtained by twisting a strip of paper once and gluing the ends together.a sin 0 k 30 (4.3-10
Mobius strip
A surface is said to be smooth if its unit normal exists and is continuous everywhere on the surface. A Mobius strip illustrated in Figure 4. An insect can crawl on that strip and reach all points on the strip without ever having to cross an edge! Thus the strip has only one side. there are surfaces which have one side only.322
ADVANCED
MATHEMATICS
If r u and r v are interchanged in Equation (4. The equation of the surface of the sphere is given by Equations (4. Example 4.3-51. Therefore. If S is a closed surface. In such a case we cannot designate a positive side.3-8.= a cos 0 cos §i + a cos 0 sin <|)j . The union of a finite number of smooth surfaces forms a simple surface. However. using Equations (4. centered at the origin. the sign of n is reversed. n is chosen to be positive if it points outwards. we obtain one circle. we need to establish a convention to label one side of S to be positive. If we cut the strip along the centre line.
FIGURE 4.
z)dS s
(4. sin 0 sin §.3-56b)
a= f^4
_ (a 2 sin 2 9 cos <|).J-J/UJ
a^sin 0 = (sin 0 cos ((). then the surface element dS is given by dS= | r u x r v l d u d v (4.3-61).3-61)
In Equation (4.3-57c)
=§
(4.3-50a) and cp is given by (p = x 2 + y 2 + z2 = a2 = constant grad (p = 2x i + 2y j + 2z k = 2r From Equation (4. it has to satisfy the equation of the surface. z is not an independent variable. a 2 sin 2 0 sin §. z) is a scalar function.VECTOR AND TENSOR ANALYSIS
321
— = .3. cos 0) (4.a sin 9 sin §i + a sin 6 cos (j)j
(4. a2sin 0 cos 0)
— —
(4. That is to say.3-60b) (4.57a)
{H.
Surface and Volume Integrals The surface integral is an extension of the double integral to an integration over a surface S.3-55) we have n = -p^j= I (4.3-57d)
The equation of the surface of the sphere is also given by Equation (4.3-59b)
since I r I = a.3-59a) (4.3-58) (4.y.3-60a) (4. If S is given in parametric form. If cp (x. it is given by Equation (4. y.3-49b).3-62)
. then the surface integral of cp over S is denoted by
I =J|(p(x.
4-3. Substituting Equations (4.4-23)] for this simple geometry.4-30)
where T is a unit tangent to C and n a unit outward normal to the surface S.4-26b)
h J v-nds =h jf (-^f+ ^ c s2
The normal n to the curve C is given by
dxdy
(4.4-29)
= //& + ^l d x d y
A
( 4 -4-23)
where we have replaced S2 by A. We have thus deduced Gauss' theorem in two dimensions [Equation (4. we obtain
(4.4-26a. the line integrals along AB will cancel. The direction of AB from C to Cj is opposite to the direction from Ci to C. 28) and dividing by h. we obtain
f (vx dy . then
<f v • T ds = f f curl v • n dS
C S
(4. Thus we only need to evaluate the integrals along C and C\.vy dx) = ff ( ^ + ^ ) dxdy s2 c
(4.4-27. Therefore.4-25b).4-27)
(4.
Stokes' Theorem
If the vector field v and curl v are defined everywhere on a simple open surface S bounded by a curve C. b) into Equation (4. We then go around Ci in the positive direction once and leave C\ at B to rejoin C at A and proceed along C in the positive direction until completion of the circuit. This is illustrated in Figure 4.4-28) Combining Equations (4.VECTOR AND TENSOR ANALYSIS
-
i i i
dS c = h ds where ds is the line element along the circle that encloses the surface S2. then at a point A on C we make a cut and draw a curve from A to a point B on Ci. We have to ensure that the direction of integration is chosen
. If the surface S is enclosed by two curves C and Ci.
We assume the boundary condition to be (p = f(x. everywhere except at P. We want to determine the value of (p at a point P.5-11) which gives the value of (p on S is known as the Dirichlet condition.18 that we need to construct a Green's function G that satisfies the homogeneous equation. y.18 for ordinary differential equations.G ! ) d S = / / / U2G^
S V
V 2 < p ) d V
(4. the potential (p satisfies the equation V2 9 = p (x. we have Laplace's equation. inside a volume V enclosed by a surface S. which is sufficient to ensure a unique solution to Equation (4. In some problems.5-12)
Replacing \|/ by G in Equation (4. V G = 0. such as electrostatics. that is to say G=0 on S (4. ~.5-13)
.5-10) using the method of Green's functions which was introduced in Section 1.5-10) is known as Poisson's equation and if p = 0. which for incompressible fluids (p = constant) simplifies to div v = V • v = 0 (4.is given at the boundary and this condition is the Neumann condition.5-10)
Equation (4.4-22) we have
/ / ( ^ . We choose P to be the origin. in the present case. We propose to solve the inhomogeneous Equation (4.5-8b) is the equation of continuity in fluid dynamics.5-10).5-9a. z) on S (4. We recall from Section 1.
ADVANCED MATHFMATICS
where £• = — + v • grad is the material or substantial derivative.£M
. which is a homogeneous equation. y. We also assume that G satisfies the homogeneous boundary condition. z) (4.5-11)
The boundary condition given by Equation (4. L X at Equation (4.b)
Solution of Poisson's Equation
In many applications.
5-15b)
(4.1
(4.5-14) as e — » 0.5-15e)
(4.1 £
(4. Equation (4.5-13) becomes (4. G is singular at the origin. As in Example 4.5-17a)
.4-1.> 0 To evaluate the right side of Equation (4.5-12) and cp satisfies Equations (4. we need to isolate the origin by enclosing it with a sphere of radius e. In V .VECTOR AND TENSOR ANALYSIS
339
Since G and V G are not defined at the origin. the region enclosed by S and S e .5-15c)
dr
//[^-<#s=-f ff M^lkH d e
se
L J
(4.5-15d)
= -J
I
j 9 + £ ^ 1 sin 0 d(j) d0
(4.5-14)
s
se
v-ve
Near the origin we assume ~ to be the dominant term of G. To evaluate I £ we note that p is finite everywhere and let its upper bound in V e be M.5-16)
where I £ is the contribution from V e . G satisfies Laplace's equation.V e . On S e we have r=e (4. Then in V e we have Ip|<M (4. we work in terms of spherical coordinates. surface S e and volume V e as in Example 4. a s e . 11).5-15f)
= -47U(p(0).5-10.4-1.5-15a)
| . Equation (4.= -fdn G . we write
f f f G p dV = (((
V-Ve V
G p dV .
we generate a relationship between x and y.5-17c) (4. y) (4.5-17b) (4. we obtain
< p < 0 ) = if/// G p d V + // f ^ d S ]
.5-2 la) (4. b). 15f.5-18)
G
Ie<M|7ie3l ^ |7tMe2 Thus as e -^ 0.5-17d) (4. On eliminating t between them.5-14. Combining Equations (4. then x(t + T) = x(t) y(t + T) = y(t) (4.5-20a) (4. y) and g(x.5-21b)
. following Chapters 5 and 6 on partial differential equations. If we can solve Equations (4. Non-Existence of Periodic Solutions Many dynamical systems to be covered in Chapter 10 are governed by a non-linear autonomous system 4 r = f (x. y) are continuous functions with continuous partial derivatives.5-17e) (4. We can plot x versus y and the obtained curve is the path or trajectory.5-19) gives the value of 9 at P in the form of an integral involving the Green's function G. Ie -> 0. The system is said to be autonomous because f and g do not depend explicitly on time t. V S
( 4 5 -19)
Equation (4.5-20b)
where f (x. 18).5-20a. If x and y are periodic and of period T. We will be in a position to construct G.^C
Ve
ADVANCED
MATHEMATICS
= ^rte3 =^
(4. we obtain x and y as functions of t. y) dt ^ = g (x. 16. The xy-plane is known as the phase plane.
g t) d t
=0
( 4 '5-22b)
(4. then the left side of Equation (4. The closed curve will be traversed once as t increases from t 0 to t 0 + T.5-22c) cannot be zero. Thus if 5— + ^ p does not change sign in the phase space.4-23) we set vx = f and vy = g.
(ii)
Thus laws (i) and (ii) can be expressed as
.5-22c) follows from using Equations (4. In other words there is no periodic solution.m. This experimental observation is usually stated as Neumann's law and Lenz's law. If ^— + ~
dx
dy
is of one sign only in the phase plane. So this leads to a contradiction which implies that there is no closed curve.5-20a. b). then an additional electromotive force (e.) is set up in the circuit and is of magnitude ~-. Equation (4. that is • = —+~
dx
dy
is either positive or negative
throughout the phase plane.5-22b) is obtained by introducing the variable t. Thus a periodic solution corresponds to a closed path in the phase plane.f. (i) Neumann's law: if the magnetic flux N through a closed circuit varies with time.
Maxwell's Equations
Faraday discovered that if a closed circuit is being moved across a magnetic field or if the circuit is placed in a varying magnetic field. Other criteria for determining the existence or non-existence of periodic solutions are given in Cesari (1971). This is Bendixson's negative criterion. b) does not have a periodic solution.VECTOR AND TENSOR ANALYSIS
m
The path in the phase plane will then be a closed curve.5-20a. then we can write
// (fx~ + If) dx dy = §(f dy -gdx)
A C
( 4 -5-22a)
= f T ( f *. Lenz's law: the current induced in the circuit opposes the change in N. If in Equation (4.5-22c)
Equation (4.5-20a. for every t 0 . Integrating once around the closed curve C corresponds to integrating from t = t 0 to t = t 0 + T. then the system given by Equations
OX 0y
(4. It should be pointed out that the non-linear Equations (4. b) are usually difficult to solve exactly and only approximate solutions can be obtained. a current is generated in the loop. Thus it is of interest to have analytical criteria to determine the existence of periodic solutions.
jf ?= «n dS s
(4.5-25)
where B is the magnetic induction.M2
ADVANCED MATHEMATICS
e-m.6 GENERAL CURVILINEAR COORDINATE SYSTEMS AND HIGHER ORDER TENSORS (4.f.f.5-28)
Cartesian Vectors and Summation Convention A vector y is represented by its components which depend on the choice of the coordinate system. around a closed circuit is equal to the change in potential in going around the circuit once. and N are measured in the same system of units.5-27) holds for every S. we have e.m.d N
(4.2 = at Equation (4. If we change the coordinate system. = (t E » T d s (4.5-23 to 25)
£ E • T ds = . we obtain by combining Equations (4.m. Since S is a fixed surface. it follows that curlE = . we write
(4.[ [ ?= • n dS s c
Applying Stokes' theorem.f. the components will generally change. Since the e. = . Thus there is a relationship between the components of v in
.5-27)
Since Equation (4.m. The vector v however is independent of the coordinate system.f.5-28) is one of Maxwell's equations in electromagnetic field theory. The total normal magnetic induction N across any surface S is
N = [ [ B»ndS
S
(4.5-23)
provided both the e. 4.5-24)
c
where E is the electric field.5-26)
jf s
n»curlEdS = .
z) system as shown in Figure 4. y.6-2a)
and
k
= ^x i * I +
Vy
I * I + ^z £ * I
(4. It is the object of this section to establish such relationships. z) obtained by rotating the (x. z) system.VECTOR AND TENSOR ANALYSIS
Ml
one coordinate system and the components of the same vector v in another coordinate system. Vector v has components vx.
v = vxi + vyj + v z k = vxi + vyj + v z k
(4. v y and vz are obtained through forming dot products with unit vectors j_ respectively. z) system and components v x . y. vy. b) with the unit vector i . v z in the (x. vz in the (x. y.6-2c)
v z = v x I • k + vy J • k + vz k • k
. y.6-1. vx= v x i » i + v y j / i + v z k • i Similarly.6-1 Vectors v and r in two coordinate systems The component v x is obtained by forming the dot product of Equations (4.6-la. We wish to relate those components.b)
z
i
^
X.
7
/ PCx.6-la.
vy
(4.y. and this can be achieved as follows. z) to another Cartesian coordinate system (x.z)
/I
/
I
: h
FIGURE 4. v . y. We start by considering a transformation from one Cartesian coordinate system (x.6-2b) (4.
The three equations given in matrix form in Equation (4. y. the indices in £mn notation has the following advantages (i) are written as numbers instead of as x.j = k«k = 1 Inverting the set of Equations (4. x 2 . v 2 . j . Note that the indices in (x 1 .k = i .b) (4.6-6b)
= % inm
. i )
cos (i. 3 ) and (5. we can extend it to an infinite dimensional space.e. z) will be relabelled as (x 1 .6-3d. j ..344
ADVANCED
MATHEMATICS
Note that in the rectangular Cartesian system i»j_ = j _ . In this notation. are the direction cosines and can be represented in a more compact form by £mn. v v z ) will be denoted by (vj. x 3 ) and (x *. i ) £2l = J ' i = |1| | l | c o s ( J . 1? 8.6-2a to c). z and this
in the case of an extension to an n dimensional space.
(ii)
Similarly a vector v with components (v x .6-5c. 2 > < 3 3) respectively. 6.b. x 2 . k) will be denoted by (6. where m takes the values 1. 2 and 3. The nine quantities £ \ \. v z ) can be represented by v m . . The coordinates (x.k = O i .6-4). i) is the cosine of the angle between the unit vectors i and i . k) and (i.6-3a. we would run out of letters. x 3 ) respectively. In Equation (4.d)
where £n = I « i = |1| 111 cos ( I . v 2 = vy and v 3 = vz. the notation is more compact. where both indices m and n take the values 1.f)
K\
vy =
I hi hi hi \ /M
£2l £22 £23 vy (4. vy. 2 and 3.i = j . j .c) (4. the unit vectors (i.6-5a.. z) and (x.6-6a) (4. v t = vx.6-4) can now be written as
3
vm
= X
n=l
^n ^nm
(4.6-4)
\vj
\ £3l £32 £33 I \v z /
(4. Thus x 2 is not x squared and we shall denote x 2 squared as (x 2 ) 2 with a bracket round x2. x 2 . we obtain (4. v 3 ). The components (v x . where n can be greater than 26. 5. Indeed. y. 2. £21. y. x 3 ) are written as superscripts and the reason for this notation will be explained later.
b).6-10)
. 2 or 3. v 2 . From the definition of £mn.6-7) can be replaced by any other letter except n (or p). 5r
(4. Thus the three equations represented by Equation (4. Equations (4. In our case. and Equation (4.6-6a) are obtained by assigning the value of m = 1. Thus in Equation (4.6-6a). r (we cannot use 5.6-6b). The index n is called the dummy (repeated) index and the right side of Equation (4. x ^ x 3 ) coordinate system is an orthonormal coordinate system §n-I
r
= 8nr
(4. we must apply the same value to m wherever m occurs.6-7).6-7)
The right side of Equation (4. m can take the values of 1. which is the same as the summation over all the possible values of n as implied by Equation (4.b) can be written as ^=vm£-m = %^n (4. Similarly the free index m in Equation (4.6-6a) is a summation over all the possible values of n as indicated by the £ sign.VECTOR AND TENSOR ANALYSIS
345
In Equation (4. and is equal to 0 if n ^ r. whenever an index occurs twice and twice only in an expression it implies summation over all the possible values of that index.6-8)
Since the (x '.6-9)
where 8nr is the Kronecker delta and is equal to 1 if n = r. we can replace it by other letter. v 3 ) can be expressed in terms of (v l 5 v 2 . we should not replace the dummy index n (or p) by the free index m. making use of the properties of £pm.6-la. The components (vj. known as the Einstein summation convention.6-7). unless stated otherwise. or by repeating the process used in obtaining Equation (4. 2 and 3 in turn.5r= v
n
6
n
. According to this convention. p say. Once a value of m is chosen.6-6b).6-5a. we obtain vm5m. n or 5_ m due to the summation convention). we have &
m
-S
r
= ^rm
(4. We shall adopt the second procedure to demonstrate the elegance and conciseness of the summation convention. the index n is a dummy index and the expression on the right side of Equation (4.6-7) is a summation over all the possible values of p.6-la. Since n is a dummy suffix.6-6b) can equally well be written as
vm=
Vpm
(4.6-6b) implies summation over all possible values of n. Equations (4. v 3 ) by inverting Equation (4. the index m is known as the free index and it can take any of its possible values. To obey the summation convention. The summation sign occurs so frequently that it is useful to adopt a convention.b)
Forming the dot product with 5.
b). it is not generally permissible to do so in a general coordinate transformation.6-12b)
^rmxm = xr
Here we have freely written the indices as superscripts and subscripts. is associated with 6. (4.6-13a)
Note that £Tm can also be obtained via Equation (4. from Ox*x 2 x 3 to Ox 1 x 2 x 3 is a linear transformation. Although this is permissible in our present coordinate transformation. That is to say
xm =
xP|pm
(4. b) imply that dxT dxm = 8x m dlr
tAr^ (4-6'13b)
. the transformation.6-1 lb)
On summing the right side of Equation (4. m. r and the second index. Substituting Equations (4. which is r.6-12a) (4.6-13a. is associated with 8 m of the unbarred coordinate system.6-9.6-12a) by replacing the dummy index p by r. then the components of the vector position OP (= r) will transform from the barred to the unbarred system or vice-versa.6-12a. Because the direction cosines £rm are constants. we obtain ^rmvm = % 5nr = vr (4-6-1 la) (4. and the coordinates of P are (x1. 1 lb).6-7.6-13c)
. as we shall discuss in the next section. to yield r)xm 'nn=|=7 Equations (4. 10) into Equation (4. as defined by Equations (4. x 3) relative to the Ox 1 x 2 x 3 and Ox 1 x 2 x 3 coordinate systems respectively. due to the definition of 5 nr . If P is any point in space.6-12b) that
(4. x2. that the only surviving term is v r .6-8). x3) and (x *. we find.6-1 la) for all possible values of n.346
ADVANCED
MATHEMATICS
Note that the first index in £rm. Further we note from Equation (4. x 2. according to Equations (4.
the index m "cancels" and we have the superscript r on both sides of the equation. we obtain
fdxm
axr
vr = 3xr
dxm
V m
(4.6-1 lb). Similarly the spherical polar coordinate system is chosen for solving flow past a sphere. we are left with a subscript r and on the left side we also have only a subscript r. Similarly in Equation (4. General Curvilinear Coordinate Systems The rectangular Cartesian coordinate system is not always the most convenient coordinate system to use in solving problems. q3) as shown in Figure 4.6-14a). We now consider a general curvilinear coordinate system (q1.6-14b) should be written as vr = — dxm vm (4. The laminar flow of a fluid in a circular pipe is solved using a cylindrical polar coordinate system. b)] and thus there is no need to make a distinction between subscripts (covariant components) and superscripts (contravariant components). components that transform according to Equation (4. In Equation (4. The superscript m is associated with Ox 1 x 2 x 3 and the subscript r with Ox 1 x 2 x 3 . m is a dummy index. both laws of transformation are equivalent [Equations (4. We can consider them as "cancelling" each other. we need to distinguish between covariant and contravariant components. The velocity is then a function of the radial position r only and not of two variables x1 and x2.b)
Vm
Equations (4. b) describe two types of transformation laws. Components that transform according to Equation (4. b) into Equation (4. we regard m as a superscript and r as a subscript) and dxT dx r once as a subscript in vm.6-14a. Substituting Equations (4. The choice of coordinate system depends on the geometry of the problem.6-14b) are known as contravariant components and the indices are written as superscripts. q2. But for Cartesian components.6-14a) are called covariant components (vm) and the indices are written as subscripts.6-14a.6-2. it occurs once as a dxm Bxm superscript in —=— (in the expression—=—. Equation (4.6-15). Thus in proper tensor notation.VECTOR AND TENSOR ANALYSIS
341
This is only true in the case of a Cartesian transformation.6-13a.6-15)
Note the symmetry in the notation.6-14a. On the right side.
. Thus in a general coordinate transformation.
the tangent and the normal coincide.
. In the particular case of the rectangular Cartesian coordinate system.3-55) and. which explains why Equation (4.b) Base vectors defined as tangents to coordinate curves are covariant base vectors.6-3. They are given by Equation (4.6-1. in the present notation. For example the base vector 5 3 is normal to the surface x^ = constant.6-3 Generalized coordinate system and base vectors
The base vectors g m are given by (4.c)
Note that both g m and g n are not necessarily unit vectors.b.6-19a. Example 4. The base vectors 8 m can also be considered as being normal to the planes x m = constant.6-13c) is valid. Base vectors defined as normal to coordinate surfaces q n = constant are contravariant base vectors and are denoted by g n as illustrated in Figure 4.VECTOR AND TENSOR ANALYSIS
349
V3
q1 FIGURE 4.6-20a. that is to say. they are written as g n = V q n = gradq" = ^ 6^ (4. This is not generally the case. it is parallel to the x3-axis. Obtain the covariant and contravariant base vectors for the spherical polar coordinate system.
the components v s are defined as .6-34a)
(4. 20)] and the chain rule.352
ADVANCED
MATHEMATICS
Since in the general case.J At->o At
(4.6-33) are contravariant components. covariant and contravariant base vectors are different. we can no longer take dot products. b) yields ^ =
v m
(4Mla)
(4. to establish the relation between components v m and v m . as follows
«. in the q m coordinate system. See also Equation (4. are defined as
vm
Aa m _ jjm _ J _ At-»o At 8q m = -|-
(4. As far as the summation convention is concerned.6-32a.6-35a)
.6-15).6-33)
Components of vectors which transform according to Equation (4.= a ? ^
-S. General coordinate systems are not necessarily orthogonal and g j • g 2 is not necessarily zero.6-19. Obtain the law of transformation of the velocity components v\ The velocity components v m .6-30a..6-8).6-2.b)
It now follows that _ da n v n = ^— v m dq m (4. repeated indices should appear as subscript-superscript pairs.6-34b)
On transforming to the q s coordinate system. Those relationships are obtained via the definitions [Equations (4. as in Equation (4. one has to pay special attention to the position of the indices. In the general case.s . Example 4. Aq s v = lim .6-3!b)
^ I n
= vng
n
(4.fg
Substitution in Equation (4.
6-38).6-37a)
= ^ 1 vm dqm Thus the components v s transform as contravariant components.6-36)
(4. we obtain v^lim ^ ^ At->o dqm At
(4.6-37b)
We further note from Equation (4.6-36) into (4. In Equations (4. using the chain rule.6-40a) (4. That is v = vmgm = vngn The relation between g n and g m can be deduced. as follows gra = Vqm = ^ Y qn (4.6-38)
= ^ . 27).VECTOR AND TENSOR ANALYSIS
353
= ^ According to the chain rule Aq s = ^5_L Aq m dqm Substituting Equation (4.6-35a). we have expressed the vector v in terms of the covariant base vectors g m . b). •
(4.6-30a. We could equally have expressed v in terms of the contravariant base vectors g n .6-36) that the components Aqm also transform as contravariant components.6-26. we obtain ^ v
m
(4. but we still write the indices as superscripts because Aqm are contravariant components.6-39) (4. as can be seen from Equation (4.6-35b)
(4.6-40b) into Equation (4.gn dq n ~ Substituting Equation (4. Aqm are contravariant components and the indices are written as superscripts.6-40b)
g
n
= vn g «
(4. Although the coordinates q m are not components of a vector. The transformation from q m to q n is arbitrary.6-41)
.
q2 and q3. the components u n are given by (4. we have seen that V the quantity V ^ v is a tensor of order two and is known as the velocity gradient. In Example 4. It is a _ (p is a vector but (p is a scalar.6-3.6-43b) (4. A vector has both a magnitude and a direction and its components transform according to Equations (4. In fluid mechanics.6-14a)]. One index is sufficient to specify its components.6-3.6-42) are known as covariant components [see.6-45a) (4. where cp is a scalar function of q1. also Equation (4.6-33. transform as covariant components of a vector. A scalar is a tensor of order zero and its numerical value at a point remains invariant when the coordinate system is transformed.
. Example 4. we have 9cp 9q m (4. q 3) coordinate system. Thus tensor of order one.6-44) Applying the chain rule to Equation (4.354
ADVANCED
MATHEMATICS
We then deduce
% = f ^ vm
(4. Tensors of Arbitrary Order
(4-6-45b)
So far we have considered only scalars and vectors. Show that the components of V(p. 42) when the coordinate system is transformed.6-44). q 2 . Let u = V cp then we define u m as um = | ^ In the (q 1 .6-42)
Components of vectors which transform according to Equation (4.6-43a)
8q m = ^ »m Thus the components u n transform as covariant components.
6-33. One recognizes the 3qP 3 q m 9qP quantity — — (in the unbarred coordinate system) as representing the Kronecker delta.6-46). We need two indices to specify the components of a second order tensor. It thus follows that 9q l 9q
T™
| r1^vS
=8
pu'=
U p
(46-5Oa-b)
In Equation (4. 42). we are at liberty to change the free index m to p and the dummy index n to s.6-46). which is another tensor of order two. The stress tensor. Equation (4. be written as —^ — .6-47)
l £ ». is equal to the sum of the velocity gradient and its transpose.VECTOR AND TENSOR ANALYSIS
355
the rate of deformation tensor. while a quantity such as —~— is not in general 8 \. q 2 . This can be written as u m = T m n v" (4. maps linearly the surface force on the surface of a deformable continuum to the unit normal on the surface. we obtain 3qP dqm 9q r = 3 q m dqn
s
(4. Similarly. " ^ | ^ V
da m Multiplying both sides of Equation (4. In a three-dimensional space. Equation (4. according to the chain rule. 3q p
r
a dV ^u' q
= T-
a^ a^ v
( 4 -6"49)
—5— would represent 5 s t . 2 and 3 and T has nine components.6-48. q 3) coordinate system. The quantities T m n are the components of a second order tensor which we denote as T.6-47) becomes (4.6-46)
In Equation (4. The components of u are written as covariant components and the components of v are written in contravariant form.can.6-46) becomes um = Tmn v n Using Equations (4.6-48) by —-—. In the (q *. we then obtain up = T ps vs (4. which is a tensor of order two. m and n can take the values 1.
dar d a m Note that —^. Thus a tensor of order two transforms a vector u linearly to another vector v. we have adopted the summation convention by writing the dummy index as a subscript-superscript pair.6-51)
.
.
y
Contravariant components T m n transform according to
T mn
=
3<L^ <*iil T rs
dqr dqs
(4. The components Cjjk^ need four indices and are the components of a fourth order tensor.6-42).6-52) are known as covariant components. q3) and (q l.S
T
m n
f4 6-53^
^
}
dq p dqs Components that transform according to Equation (4. q 3 )in Equation (4. The law of transformation for a fourth order tensor can be obtained by generalizing Equations (4. q 2 .6-55) Tensors of order higher than two also exist and are frequently used. Note the similarity to Equation (4. An example of a tensor of order three is the permutation tensor. the infinitesimal strain tensor and the elastic tensor respectively. The constitutive equation of a linear elastic material may be written as tij = Ciju yu (4.6-57a)
3qP 3q r 3q s 3q> ""
. y k ^ a n d cyk^ are the stress tensor.6-54)
For second order tensors. which will be defined in the next section. in addition to covariant and contravariant components. we can have mixed components T ™ which transform according to (4. A tensor that maps linearly a second order tensor to another second order tensor is a tensor of order four. it becomes T
1 P
3a
=
S
m
ZH
.6-53 to 55) as follows
T -SiSiH!5lT i w
prSt
(4.6-56)
where tjj. q2.356
ADVANCED MATHEMATICS
Comparing Equations (4. _
D
da n °4
-A. we deduce
dq m Ml
T
(4 6-52)
Interchanging (ql.6-52). 51).6-50b.
Metric and Permutation Tensors Equation (4. The commutative law does not hold and g j g • is in general not equal to g : g j . ds 2 .6-58) can also be written as I = uigjvJ gj = um g m v n g n = =
ui VJI
(4. If P and Q are two neighboring points with vector positions r and r + dr with coordinates (q1) and (qJ + dqi) respectively. T = uv = u®v Equation (4.f)
i I j = um v n I
m
I "
Tij
l i I j = Tmn I m I "
A second order tensor T is also known as a dyad and the notation adopted in Equations (4.6-58a.6-60C) (4.6-60a)
dqmdqn
= T*k ' TT
dqm dqn
(4. The second order tensor T may be defined as the dyadic product of vectors u and v.57c)
A second order tensor may also be defined as the juxtaposition of two vectors u and v. between P and Q is ds 2 = dr«dr (4. If they are equal.b)
(4.6-60d)
= gm'gndqmdqn = gmn dq m dq n
.VECTOR AND TENSOR ANALYSIS
357_
Tprst =
3 £ M_ _Bql M_ T ijk^ dq[ 9q j dqk dq£
(4.6-59c. Tensors of higher order can likewise be defined. The component Tjj is not necessarily equal to the component Tjj. The juxtaposition of the two vectors u and v is also known as the outer product of the two vectors.b) (4. then the square of the distance.6-57b)
v
rSt
*tmLWL%!_Ti
a q 1 3q r 3q S 3q l
jW
(46.6-59a to f) is called the dyadic notation.d) (4.6-59e.6-19a) defines the covariant base vector g m .6-60b)
(4. T is a symmetric tensor.6-59a.
6-67c. 3 appear in the clockwise direction .6-68)
.b.c)
°
l
(4.6-65d) (4.6-65b)
(4.6-65e)
In Equation (4.b) (4. The relationship between gmj and gJn can be established by combining Equations (4. and is defined as 0.6-63c. if the indices 1.d) (4.1 . The convention is framed so as to be compatible with the rules of transformation.f)
= e312 =
= e213 = .VECTOR AND TENSOR ANALYSIS
359
It can be seen from Equation (4.6-65c)
— — hnr 3x r p r ^ 9x r
(4.6-65e). we have written the Kronecker delta as a mixed tensor so as to conform to the convention that an index appearing as a subscript (superscript) on one side of the equation should also appear as a subscript (superscript) on the other side of the equation.6-64b) that gJn components transform as contravariant components. if any two indices are equal ejjk = 1.6-67e. The permutation tensor is an example of a third order tensor which in an orthonormal coordinate system is denoted by ejjk.
e ll2 = e l22 = e123 e321
(4.6-67a. The fundamental reason for writing the Kronecker delta as a mixed tensor is because it transforms as a mixed second order tensor.!
Let w be the vector product of two vectors u and v. 2. 3 appear in the anticlockwise direction Thus. Wj is given by
wi
= e ijk u j v k
(4. 64c) and the result is
in
dxP dxP dq) dqn
tAtzcs
Smjg
a^^a^^
=M ^
dqm = dqm = -^1 dqm = K 9x r
(4-6"65a)
*£L
3x r
(4. 2. if the indices 1.6-66a. Then in an orthonormal coordinate system. for example.
6-71b)
= Bta^ff«'v'
Multiplying both sides of Equation (4. so as to avoid confusion with the Cartesian components.6-71a. we deduce that
(4-6"73a)
(4. We put a bar over the components of the vectors in the present coordinate system. Equation (4. u m and v n ) to the orthonormal coordinate system using Equations (4.6-70)
Transforming (w^.6-71b) by ——.6-73b)
e pst
= *«» ivit^
= ^
eAnn
dq£
dqm
dqn
(4-6"74)
which is the law of transformation of covariant components. then the determinant of D can be written as Igl =eijkdildj2dk3 (4-6-69)
We extend the definition of the permutation tensor to a general curvilinear coordinate system and we denote it by e^ mn . 42). Equation (4.26Q
ADVANCED
MATHEMATICS
If D is a (3 x 3) matrix.6-73a.6-72) becomes dqe dqm dqn
wp = e ^ a x V ^ ^ U s V t
= epst u s v t From Equations (4. Alternatively e^mn may be written as
eAnn
(4. identifying the coordinates qm as x m and q n as qn. Thus w"^ is expressed in a general barred coordinate system as
W£
= e£mn " *" ^ "
(4.
(4. we do not distinguish between covariant and contravariant components.mn f £ »• flf V
dqe 3x s 3x l
(4.6-75)
.6-33. b).6-70) becomes
f 7 » r = e. with elements dy. we obtain
9xP dqe 3xr
r
dqe dqm dqn
£^mn
s
t
v (4-6-72)
^ H
w
j V 1T7 TT u
dxP 3q £ dxP dx s 3 x l Using the chain rule on the left side of the equation and noting that in an orthonormal coordinate system.
The components of the metric tensor are given by grr = g r • g r = 1 See = I e * I e = g«x|> = g<t>-g<t>
= f2
(4. its dual g rs and its determinant g for the spherical polar coordinate system.6-4.b) X r2 (4.f)
r2sin2e
All the other gjj are zero since the spherical polar coordinate system is orthogonal.VECTOR AND TENSOR ANALYSIS
361
where g is the determinant of the metric tensor g m n .g r = l
gee =
(4. Contravariant and Physical Components
(4.6-77e. grr = g r .b) (4. The contravariant form of the permutation tensor is
E1* = j= e ijk
(4. we have (4.6-1. Calculate the metric tensor gjj.d)
gG. Via the metric tensor it is possible to establish a relationship between these two types of components and this is done as follows v = vm g m = vn g n Forming the dot product with g r .6-79)
It has been noted that the covariant and contravariant base vectors do not in general have the same dimensions and it is not surprising that the covariant and contravariant components of a vector also do not have the same dimensions.6-78e.d) (4.6-77c.b)
.6-78a. =^TLVY ~* r2sin29 The determinant g is given by g = r4sin20 Covariant.6-77a. The base vectors have been obtained in Example 4.6-80a.6-76)
Example 4.6-78c.f)
(4.gG
=
g** = g * ' g < .
6-86a)
.6-82a) (4. This process is known as lowering and raising indices.362
Vm
ADVANCED MATHEMATICS § m ' §r =
v"
In * Ir
(4.6-83a. Similarly for higher order tensors we have Trs = gnr gms T n m
Tnm = gnr gms T f s
(4. Equation (4. The normalized covariant base vectors are given by
Kn^lfVvf^
| S nl °nn
(4.6-82C)
T™ = g nr g ms g pk gqi TJ*
In a space in which a metric is defined. The physical components of a vector are expressed in terms of normalized base vectors.6-82b) (4. it is possible to transform covariant components to contravariant components and vice versa.6-8lc) shows the transformation from contravariant components to covariant components. This makes it necessary to define the so called physical components.6-85) implies that for each n. b) there is no summation and the index n is contained in brackets to designate physical components.6-84a. yields (vn--^l)gn = O (4.
vn
_ ^n^
&nn
(4. c). noting that m is a dummy index.b)
Note that in Equations (4. It was pointed out in Example 4.6-84C)
=>nn
Combining Equations (4.6-83a.6-8la)
v m 8™ = v n gm vr = gnr v n
(4.6-8 lc)
Equation (4. A vector v may be written as v = v =
v
m
g
m
= v (n) g ( n )
V-^M
(4.6-1 that g r and g Q do not have the same magnitude and they have different dimensions.6-81b) (4.6-84a.6-85)
Since the vectors g n are base vectors and are linearly independent.b) (4.
g n m in Equation (4.6-87) into Equation (4. we obtain Vg™~ v n = v (n) (4. b) there is no summation over n. The index n occurs three times! The contravariant component vn can be written as
vn
= gnm
Vm
(4. dqm In Example 4. in terms of normalized covariant base vectors.6-2. In Equation (4.6-88). In the framework of an orthogonal coordinate system. Obtain the contravariant. the physical components V(n) can be obtained via the metric tensor. at contravariant components of v are vr = ^ = f The
(4.6-89).6-90b)
= V g 1 ^ Vg^ T m
= Vg^Vg^T^
(4.6-5.6-89)
Similarly the physical components of higher order tensors are defined. For an orthogonal coordinate system we have V)
=
^ ^
^
^
(4. we obtain v(n) = ^ g" m v m (4-6-88)
In an orthogonal coordinate system. they may not be identical since Equation (4. we have shown that -^— transforms as contravariant components.6-87)
Substituting Equation (4. We could equally have chosen to define physical components via normalized contravariant components.6-9 la.6-89) could not be obtained from Equation (4. Indeed.b)
. b.VECTOR AND TENSOR ANALYSIS
v(n)
363
or
= ^ n
v"
(4.6-86b)
That is to say.6-90c)
So far. there is no summation over n.6-90a) (4. Note that in Equation (4. both are identical. for a non-orthogonal system.6-86b).6-86a.6-88) represents three non-zero components for each value of n. combining Equations (4. In the case of a non-orthogonal coordinate system. Example 4. we have defined the physical components.6-6la. 88). covariant and physical components of the velocity vector v of a particle in the spherical polar coordinate system.
contravariant and physical components.6-95e. In a space in which a metric tensor exists.d)
» = ^»* = 7±eH
( 4 -6"96e'f)
In Examples 4. contravariant or physical components.d)
u«> = g<* u^ = — . then on the left side we must also have a mixed component. the components of a tensor refer to physical components. whereas the r. a tensor can be represented in terms of covariant. there is no distinction between covariant. as shown in Examples 4. That is to say. Each expression in the equation should be a tensor component of the same kind and order. 0 components.6-5 and 6. all have the same dimension. In the rectangular Cartesian coordinate system.6-95c. 9 and (j) physical components.6-5 and 6. • The laws of physics are independent of the coordinate system and they should be written in tensorial form.b) (4.6-96c.
.1 — ^ v r 2 sin 2 0 5(t> The physical components are
(4. The metric tensor is the Kronecker delta. They can be transformed from one to the other by the process of raising or lowering the indices.VECTOR AND TENSOR ANALYSIS
365
ue
= g e e u e = -L ^
(4. if on the right side of the equation we have a mixed component which is covariant of order m and contravariant of order n. Many authors omit the word physical when referring to physical components of a tensor. Finally the components of a tensor have to be measured in terms of certain units and it is desirable to express all the components in terms of the same physical dimensions. The quantities that enter into the equations expressing these laws should be in covariant or contravariant components. we note that the r covariant and contravariant components do not have the same dimension as the 9. Thus it is safe to assume that unless otherwise stated. covariant of order m and contravariant of order n.f)
u(r) = Vg" ur = jjp > ) = V g ^ u0 = 1 H
(4.6-96a.
but functions of (q1. q 2 .7-3)
Note that 5 p .6-19b). we obtain
h~-JTs-™ + vmjf
dqJ dqJ dqJ Using Equation (4.7
COVARIANT DIFFERENTIATION
We have shown in Example 4.6-26) is a linear transformation.6-30a).6-55. we have dvJ 3qj = 3 3qs (dql -i —-*. We shall presently show that if v1 is the contravariant component of a vector v.7-1b) 3 q s dq£ 3qJ 9q s 3q ' 3qi 3v
i
Comparing Equations (4. — — is a covariant component of a
dq*
vector. On transforming from (q 1 .6-26.6-3 that if cp is a scalar function. q 3 ).366
ADVANCED
MATHEMATICS
4.7-la)
(4. q 2. q 3 ) to (q *. .^ S dqJ dq m 3qJ
p P
(4J"2)
(4. q2.
dv1
aqj
does
Consider the transformation given by Equations (4. That is to say. In general the partial deviative of the components of a vector is not a tensor. transforms as a component of a tensor only if the transformation dqi 3q s dq £ given by Equation (4.v [dq1 dqs --L3qJ
(An
. 27). q 3 ) .6-3 lb)
. the base vector in the rectangular Cartesian coordinate system is a constant. 7-lb). we note that
transforms as a mixed component if
9qj
32qi dvi — = 0. not transform as tensor. The transformation of the covariant base vector 5 p to g ^ is given by Equation (4. From Equation (4. (4. This is because the base vectors are in general not constants. we find ^ = . On taking the partial deviative of a vector there is a contribution from the base vectors and this has to be taken into account.
7-9)
The covariant derivative of higher order tensors are
.7-5a)
=
3v£
_3qJ
+v m
m
a V dq e] -4-
g
(4.7-5c. allowing us to factor out g £. it is given by -. = — + Jv m dq) lm jj 'J
(4. The covariant derivative of v^ with respect to q-i is denoted by v^ • or v^ •. denoted by [rs.3 .7-6)
m
j{
3 q m 3qJ ^?
is known as the Christoffel symbol of the second kind and is also denoted by T£ •. d). we have replaced the dummy index m in the first term on the right side of the equation by I. From Equations (4.7-5a to b).7-7)
The notation correctly suggests that \e ^ represents a mixed component of a second order tensor.g£
(4.7-8)
The covariant derivative of the covariant component v^ is given by
w£
d\e i = —• 'J aqj
Imj \i j !
v
(4.7-5c)
= [v'Jg/
(4. t] (or r r s t ). The quantity (4.7-5b)
3q m 9qJ 3x p J ~ '
(4. is defined as I K t] = Su
r s
(4. 3qj and { z .VECTOR AND TENSOR ANALYSIS
367
dv dwm m 32xP -=: = — r g m + v m
:
dq£ .7-5d)
In going from Equation (4. do not transform as tensors. lm J)
The Christoffel symbol of the first kind. £ I I \
vl .
(iv) On performing the transformation given by Equations (4. does the Christoffel symbol
\3q m 3qJ
/
transform as a mixed third order tensor. (iii) The metric tensors gy. no summation is implied. £ is a dummy index and the dot product of the two second order tensors is another second order tensor.k = Tij>k + S ij(k (ii) (4. Rules of Covariant Differentiation (i) The covariant derivative of the sum (or difference) of two tensors is the sum (or difference) of their covariant derivatives (Tij + Sij). 27).7-13d)
In Equations (4.7-16a) (4. g^m and the Kronecker delta 8|j are constants with respect to covariant differentiation Sij.7-16b).6-26.7-15)
The covariant derivative of a dot (or outer) product of two tensors is equal to the sum of the two terms obtained by the dot (or outer) product of each tensor with the covariant derivative of the other tensor (Ti£ S^j).c)
A consequence of this result is that the metric and Kronecker tensors can be put outside the covariant differentiation sign.7-13a to d). k = °rs. since the metric tensors are constants.j 3
(4.k = 0 (4.k) S mj (4.k + (Ti£. the Christoffel symbol transforms as
ti\ __ j M a q ^ a q ^ ^ ^ L
\m j) Is t/ 3q r 3 q m 3qj 9 q t 9q m 3 . In the rectangular Cartesian coordinate system.b.7-14)
Note that only for the linear transformation
= 0 .k = g ^ .VECTOR AND TENSOR ANALYSIS
369
1
= U"^?
(4.k = Ti£ Smj.7-16a). That is to say
. we have formed the outer product of two second order tensors resulting in a fourth order tensor.k = T i i s \ k + (TiAk) S ^j ( T i^ Smj). In Equation (4. all the Christoffel symbols are zero.7-17a.7-16b)
Note that in Equation (4.
From Equations (4.7-1. Example 4.b)
{/J-4« I I !(«**> = -"in2e
( e 8 H «"!«*»>•* i 9 j = " 2 g e e !<«••> = .6-57c) and is thus zero in all coordinate systems. we can always set up a rectangular Cartesian coordinate system and the Christoffel symbols are zero. 0. Since the coordinate system is orthogonal. R " 1 ^ is a fourth order tensor and transforms according to Equation (4. The only non-zero Christoffel symbols are
(9re! = -2«rr|rteee) = -r
(4.VECTOR AND TENSOR ANALYSIS
321
We can interchange the order of covariant differentiation if R " 1 ^
= 0. calculate v r r + v e expression in physical components.S i n 9 c ° S e {/ j " 2 g++ I M = t
( 4 -7-21c-d)
«™°*> (4-7"21g'h) <4-™iJ)
< 4 -7-21M)
0
j/ 6 )-2«**lr( 8 «). we make use of Equations (4.7-7. Let v* be the contravariant components of the velocity vector v.7-2. consequently the Riemann Christoffel tensor is zero. Obtain v' j in + v^ x . we have
Example 4. In a Euclidean space. R 1 1 1 ^ is zero and the order of covariant differentiation is not important as long as the components of the tensor have continuous second partial derivatives. Rewrite this
. We conclude that in a Euclidean space.c < * e
spherical polar coordinate system.7-21a. 21a to 1).6-77a to f. Calculate the Christoffel symbols of the second kind for the spherical polar coordinate system (r. §). The metric tensors are given by Equations (4. The Riemann-
Christoffel tensor is a property of the space and is independent of the vector v. 78a to f). That is to say.7-13a to d).
7-2%) (4. The sum v* i is independent of the coordinate system and is a scalar or an invariant.28)
3qr 3qJ
Setting j = i and summing Vi '' ^ ! M aq'dq4 = 8 sr v r>s = vss = v^j
vr
(4. and Curl The operator V in a general curvilinear coordinate system can be defined as V=griL dqr If cp is a scalar function gradcp = Vcp = g r ^SL . Grad. Div. we have shown that — J — transforms as a covariant tensor. c) are obtained by using the usual chain rule.a qr In Example 4.7.6-55)
v
i
'J
V
3ql v r
'S
(4 7.r
(4.v l g t ~ ~ dqr ~ ~ dqr = gr!S
Vs.b)
(4.7. That is to say. we obtain the same value in the barred and in the unbarred coordinate system.6-3. The indices i and s are dummy and can be interchanged freely.7-32c.30)
(4.7-3 la.7-32a.29a)
'S (4. and the property of the Kronecker delta respectively.7-29d)
Equations (4.vs g s = gr | .b) (4.7-29c) (4.d)
= I' It
yt .VECTOR AND TENSOR ANALYSIS
373_
The component v* • is a mixed tensor and will transform according to Equation (4. 3qr If v is a vector 1 v = g r f .r
.7-29b.
7-34c) (4.1 .7-34a)
Ir#Is It
(4. This confirms the statement made earlier that the operation of taking the grad of a tensor raises the order of the tensor. which in the V notation is written as V .7-33a) (4.
The divergence of a tensor of order n is a tensor of order n .7-33b) (4. y = gr — »vsg = v s r gr • g s = v s r 8rs = vss The divergence of a second order tensor is defined as Z " l = gr|~7-Tstgs gt .b)
(4.7-33c) (4. since t is the only free index.7-33d)
(4. Thus the divergence of a vector is a scalar and the divergence of a tensor of order two is a vector. In Example 4.dq r ~ =
T S t . then Vi = ^ dql
V 2 cp can be written as
(4. we have calculated the divergence of a vector.214
ADVANCED MATHEMATICS
The components vS) r and v 1 r are the covariant and mixed components respectively of a second order tensor.7-36)
.7-34d)
= Tst>r 8rs g t = T st>s g t The component T s t
s
is the contravariant component of a vector.7-34b) (4.7-35a. The Laplacian of a scalar function cp is given by p = div (grad <p) = V • (V cp) V2 < If we denote grad cp by v.r
(4.7-3.
b. 2a to d) yields ul = . Since tensors of order zero are scalars and are independent of direction.9 ISOTROPIC. 4. To find out if a tensor is isotropic or not. x 3 ) and rotate the axes to obtain a new coordinate system (x 1 . the components of these tensors are identical in all rectangular Cartesian coordinate systems. we can write "m = ^mn u n (4-9-1)
ii)
Let [x 1) be the coordinate axes obtained by rotating the (x*) system through n rad about the x3-axis. and (u ') are the components of the same vector in the (x 1) coordinate system.U j . u3 = u3 (4. u2 = u2 = 0 (4.9-3a.c. i) All tensors of order zero are isotropic. their properties are independent of direction. 0 is the only isotropic tensor of order one. b). u2 = . that is to say. u3) are the components of a tensor of order one (a vector) in the (x1) coordinate system.8-14)
The left side of Equation (4. we deduce that for u to be isotropic Uj = Uj = 0.
(4. 13).u 2 .c)
From Equations (4. x 3 ).b. ^22 = ~ 1 '
f
33=1'
the other 4 j = 0.9-2a.c.8-12. the tensor is isotropic. x 2 .d)
.8-11) becomes p ^ + vJ v1 ti = -TCJ1 j + p g1 (4.9-3a. Below we list the isotropic tensors of order zero to four.b. u 2 . x2.9-1.8-14) is often written as —— ——I where 2_ is known as the substantial (material) derivative. OBJECTIVE TENSORS AND TENSOR-VALUED FUNCTIONS
Isotropic Tensors Many materials are isotropic. then •*H=-1.d)
Combining Equations (4.9-4a. If in the new coordinate system the components are identical. Thus if these properties are described by tensors. we express its components in a rectangular Cartesian coordinate system (x1. It is the time derivative following a material element. If (Uj. Equation (4. they are isotropic.382
ADVANCED MATHEMATICS
Combining Equations (4. Here we consider only Cartesian coordinate systems.
iv) The permutation tensor e} • k is an isotropic tensor of third order. iii) The Kronecker delta 8j: is isotropic.is a component of a second order tensor in the (x*) coordinate system and T r s is a component of the same tensor in the (x J) coordinate system.9-8)
ijkersk
=
5
ir
5
js-
5
is
5
jr
( 4 -9"9)
v)
Any fourth order isotropic tensor c} • k £ may be expressed as
cHk£
= ^5ij5k^
+
^8ik5j^
+ v5
i^5jk
(4. incompressible Newtonian fluid. Any second order isotropic tensor can be expressed in terms of the Kronecker delta. = ' r i ' s j S i j = ^ri^si = S r s Thus the Kronecker delta transforms into itself and is thus an isotropic tensor. there is no non-zero isotropic vector. A useful relation between e^ j k and 8 r s is
e
(4'9"7a)
(4.9-5)
Letting Tj: be the Kronecker delta.c)
(4. For a Newtonian fluid.
.9-6) becomes Tr.9-10)
where X. Example 4. In fluid mechanics. the isotropic part of the stress tensor 7U-• can be written as rc[j0) = . we obtain u3 = 0 Thus 0 is the only isotropic tensor. that is to say.9-1. Obtain the constitutive equation of an isotropic. then
T = I • I • T-(A 9-6\
(4. If T.9-7b. Equation (4.VECTOR AND TENSOR ANALYSIS
383
By rotating the axes about the x^axis through n radians. \i and v are scalars. the deviatoric stress tensor x_ depends linearly on the rateof-deformation y .p S j j where p is a scalar.
Consider two observers. Since the second observer is both translating and rotating relative to the first one. Quantities which are indifferent to the motion of the observers are known as objective quantities.9-12a) and others (Bird et al. Q is orthogonal at all times. The coefficient (v + |i) is known as the viscosity of the fluid.
Objective Tensors
The constitutive equation of a material should be independent of the motion of the material. one at rest (coordinate system x = x1).9-13)
The vector c (t) in Equation (4. Some authors adopt the positive sign in Equation (4.. The matrix Q (t) denotes the rotation of the second observer relative to the first one and the elements of Q are the £-.is symmetric. Such transformation is known as a transformation of frames of reference.9-13) denotes the translation of the second observer relative to the first observer.9-12c) (4.)Vij
Equation (4.8j. and the other in relative motion (coordinate system x*1 = x*). both c and Q are functions of time t.
.
VYki5jk]
= ± [ X y k k 6 i j + ^Y i j + VY ji ] = ± ( v + M. Also j . Note that in the
present transformation.384
ADVANCED MATHEMATICS
The constitutive equation of a Newtonian fluid may be written as (4.9-12b) (4. Alternatively we may state that the constitutive equation should be the same for all observers. 1987) adopt the negative sign.9-12d)
^yi. That is to say. we can relate these two systems by x* = c(t) + Q(t)«x (4. irrespective whether they are at rest or in motion. Y kk is zero.9-11) Since the fluid is isotropic.9-10) *ij = i ^ j S ^ = ±[XYkk5iJ +^k^
+
+ vfii/^jYw
+
(4-9-12a) (4. the direction cosines of the axes x** relative to x*.9-12d) follows from Equation (4.9-12c) since the fluid is incompressible. we obtain using Equation (4.
Note that due to the rotation of the axes. i) Velocity vector v Differentiating Equation (4. Equally a second order tensor T is an objective tensor if T* = Q « T » Q t We now examine the objectivity of some tensors. but to an observer standing on the road we are travelling at a finite velocity. v is not an objective vector.9-17) (4. we seem to be at rest. If we denote the components of a vector u relative to the x1 coordinate system as u1 and the components of the same vector u relative to the x** coordinate system as u*'.9-13). This observation is a common experience. ii) The line element (ds)2 From Equation (4. we obtain v* = c (t) + Q (t) • v + Q (t) • x (4.e)
.9-14)
Since c (t) and Q (t) do not vanish at all times.Qdx = dxt dx = (ds)2 (4.9-18c) (4. we have dx* = Q ' d x (ds*)2 = dx*t dx* = dx*«dx* = dxtQt.9-13).VECTOR AND TENSOR ANALYSIS
381
Objective tensors are thus tensors which are invariant under a change of frame of reference.9-18a.b) (4.9-18d.9-15) (4. the components u*1 transform to components uJ under the usual tensor transformation laws. Sitting in a moving bus and watching the passenger sitting opposite to us.9-16) (4. then if u* = Q«u u is an objective tensor (vector).
I 3 = det T .9-26)
is its characteristic equation.9-24)
The trace of tensor T (tr T ) is the sum of the diagonal elements.VECTOR AND TENSOR ANALYSIS
287
=
Q . Thus y is an objective tensor and is an admissible quantity in a constitutive equation. The scalar product of u and u (u1 Uj) is an invariant. The equation of motion holds only relative to an inertial frame of reference.is a scalar and is an eigenvalue of T . Q t + i_JQ. The equation of motion is not objective. y .1 2 X + 1 3 = 0 where I2 = tr T . This can be expressed as T-u = Xu where A . I2 = l [ ( t r I ) 2 . These invariants arise naturally when we consider the eigenvalues and eigenvectors of a second order tensor. The functions Ij.
Tensor-Valued Functions
As indicated earlier.9-23d)
(4.t r ( T 2 ) ] . For second order tensors. A non-zero vector u is said to be an eigenvector of a second order tensor T if the product T • u is parallel to u.] = 0 On expanding the determinant.
. They have an important role in tensor-valued functions. we have three principal invariants.9-26) (4. the invariants of a tensor are scalar quantities which remain unchanged when the coordinate system is transformed.9-24) is det[T-^I.Qtj
(4.9-23e)
= Q-Y'Q1"
Equation (4. Not all equations in physics are objective.9-23e) follows since Q is orthogonal.X3 + I x X2 . I 2 and I 3 are known as the principal invariants of T . The condition for the existence of a non-zero solution to Equation (4. (4. and Equation (4. we obtain .9-25) (4. because velocity and acceleration are not objective quantities. If the motion of the earth can be neglected then a frame of reference fixed on the surface of the earth can be considered to be an inertial frame.
.9-28c)
Further if cp is a scalar function of T . . which in turn is a function of A.b) (4.2 I 2 III = t r T 3 = l"(6I 3 + 2 l J .9-30)
. + a n T n where a 0 . Then I = A. We also need to consider tensor-valued functions of T and we write S = F (T) or Sjj = Fjj ( T k e ) (4.j.9-27e.b)
If S can be expressed as a polynomial in T .. we have S = ao|: + a1T + a2T2+a3T3+.. (4. X2 and Xg as its diagonal elements.. . That is to say. a 1 ? .f)
If T is a symmetric tensor. That is to say.9-27c. X2 anc* ^3 are its eigenvalues. if X-y. T satisfies Equation (4. Every matrix satisfies its own characteristic equation.9-31) (4. T can be transformed to a diagonal matrix with Xy.d) (4. then its eigenvalues are real and it is diagonalisable.I 2 T +I3J = 0 Expanding F as a polynomial in T .9-28a)
1 2 = XXX2 + Xfa + X2X?) 13 = \{kfa
(4. The Cayley-Hamilton theorem can be stated as follows.j + X2 + X3 (4. then (p is a function of the invariants of T .388
ADVANCED MATHEMATICS
Another set of invariants is defined as I = tr T = Ij II = tr T 2 = I 2 .T 3 + I 1 T 2 .9-28b) (4.6 I 1 I 2 ) (4. a n are constants. X-^ and X3 in the case of a symmetric tensor T .9-26) and substituting T for X yields . then the Cayley-Hamilton theorem can be used to simplify the representation of S .9-27a.9-29a.
that is to say. \_ . we have S = Q 5 Q
t =
QF(T)Qf
(4.
|
= F(I)
(4. all powers of T higher than two in Equation (4. We note that T 2 may be written in terms of T .9-30) by T .9-13). I_ and I 1? I 2 and I 3 . Thus Equation (4.b) (4. Similarly.9-32) is a representation of S without requiring that it can be expressed as a polynomial in T . Since F is the same in both coordinate systems.9-30)]. T . on rotating the axes. we find that T
can be replaced by
T . and the three invariants of T .9-32)
In continuum mechanics.9-36) and the inverse of T I T "M if it exists.
(4. I premultiplying Equation (4. F is the same in all Cartesian coordinate systems.VECTOR AND TENSOR ANALYSIS
289
Using the Cayley-Hamilton theorem [Equation (4. and S and T are symmetric. T . |3j. On
. we obtain (4. Thus.9-35)
If F is isotropic.T 2 + Ii T . and P 2 are functions of the invariants of T .9-31) simplifies to
§ = P0I+PiT + P2T2
where Po.9-31) can be replaced by T 2 .9-33c)
T =QTQt where Q is the orthogonal matrix defined in Equation (4.1 2 I + I 3 T 4 = 0 Thus an alternative representation of S is | = Tol+ Yll +Y-11'1 (4. Equation (4.9-35) defines an isotropic second order tensor-valued function F .9-37) .9-34)
Combining the two sets of equations. it is not uncommon to restrict attention to isotropic materials and F is then an isotropic function.9-33a.
(4. we deduce
Q F ( T ) Q1" = F I Q T QtJ
Equation (4.
.1 has been dropped since we are considering the deviatoric stress. x is a tensor-valued function of y and if we assume that x can be expanded as a power series of y we obtain Equation (4.9-38a)
+ ¥ 6 ( T ? T 2 + T2T]) + V7(T1II+
T^T1)
+ V8
(T5T1+
lllfj
(4. A Stokesian fluid is a fluid whose deviatoric stress tensor t depends on the rateof-deformation y . T 1 and T 2. A Stokesian fluid is isotropic. t r ( j j ) . I 2 and I 3 . Equation (4. t r ( | ? T 2 ) .9-39)
representation of t . Obtain a representation for T ..9-38b)
The quantities \|/ 0 ..9-32) which. t r f T i j j ) .390
ADVANCED MATHEMATICS
where y 0 .9-2.9-39) is an exact (4. The results for a function of an arbitrary number of tensors can be found in Truesdell and Noll (1965). T2) = V 0 I + V i T 1 + V2T2 + V3T? + V 4 T2 + V 5 ( T 1 T 2 + T2|i) is an isotropic
(4. Yj and y j are functions of 1^. The ten principal invariants are
tr(Ti). S = F j T j . the first invariant of y is zero because the
. If the fluid is incompressible. Example 4. we have increased the number of functions from three to eight.
(i = 1. T 2 and their products. in this case. is written as I = P1Y + P2Y 2 The term P o . The number of arguments of each function has increased from three to ten.2) tr(T?jj)
tr(TiT2). In the case where F function of two symmetric tensors. t r f j j ) .
In extending from one tensor to two tensors.. \|/g are functions of the invariants of T j . Since both x and y are symmetric. The function F can be a function of more than one tensor.
Rivlin obtained Equation (4. Thus. v is a conservative field. deduce that curl v = 0. Are they better prophets than Lord Kelvin? PROBLEMS 1 a.
Ifv = 2yzj_-xyj+xzk (i) (v«V)cp. (iv) (v»V)V<p
y«(V<p). cp = ax + by + cz cp = ax2 + 2bxy + 2cz2 (p = f(r). Determine the acceleration potential cp.9-39). If v x a = q x a. It is thus not surprising that Equation (4. If the magnitude of a vector a (t) is a constant.
5 a. 1967) did not believe vectors would be of the slightest use to any creature. 3 a. determine the relation between v and q. Determine V cp in each of the following cases (i) (ii) (iii) 4b. 2a. the third invariant is also zero. in shear flows. such as shear flows. we observe that Stokes proposed his constitutive equation in 1845. calculate the following (iii) (vxVJcp. (ii)
and cp=xyz. It is perhaps appropriate to close this chapter by observing that Lord Kelvin (Crowe.9-39) is known as the constitutive equation of a Reiner-Rivlin fluid. show that a (t) is perpendicular to da ——. r2 = x 2 + y 2 + z 2 Answer: r f'(r)/r
Show that Y (v • v) = 2 (v • V) v + 2 v x curl v If v = V cp. Hamilton stated the Cayley-Hamilton theorem in 1853 for a special class of matrices. As a matter of historical interest. that is to say.VECTOR AND TENSOR ANALYSIS
391
equation of continuity has to be satisfied. but it was only in 1945 that Reiner combined both results to obtain Equation (4. If v is a conservative field. show that the acceleration defined by a = -== + (v • V) v at is also a conservative field.
. In some flows.9-39) two years later without requiring the polynomial approximation. Can we imagine a present-day physicist not using vectors at all? Many engineers believe tensors are of no use. generalized by Cayley in 1858. Pj and P2 can be functions of the second invariant of y .
(1960) obtained the equation of continuity for a binary mixture by considering a volume element Ax Ay Az fixed in space.1 j k . The volume V is enclosed by a surface S. 12a.a d S
s
O O T
where v = x i + y j + z k and S is the surface of a sphere of radius a. (i) v = xi + 4yj + zk . Using the
Jv
mass balance and the divergence theorem to transform the surface integral to a volume integral.4-40) is equivalent to Equation (4. Bird et al. The output of A across the surface at J\ is I n A • n dS. v = xi + y j + (z . 13b. the volume V is bounded by the coordinate planes and the plane 2x + y + 2z = 6 in the positive octant. Answer: * ^ a
Verify the divergence theorem in each of the following cases. by evaluating both the volume and surface integrals.3-75) is identical to Equation (4. dV. V is the region occupied by the circular cylinder x 2 + y 2 < 1. The time rate of change of mass of A in V is ^ — I pA dV. deduce the equation
. n A (= p A v A) is the mass flux vector and n is the unit outward normal to S. We can equally consider an arbitrary volume V fixed in space.394
ADVANCED
MATHEMATICS
(i)
(ii)
| J ( x 2 + y2)dS
s J I (xi + y j + z k ) » n dS s
where n is the unit normal. The rate of production of A by chemical reaction in volume V is I r.4-23). where pA is the density of A. Use the divergence theorem to evaluate
Answer: (i) * * Oa
t
(ii) 24 a3
ff v .
(ii)
14a. 15a.3-41).
Show that Equation (4. Show that Equation (4. 16b. z = ±l. n is the unit outward normal to S.
Write down the set of direction cosines which corresponds to this rotation. Calculate the covariant and contravariant base vectors of the elliptical coordinate system. 0 < t. x 2 . 6) when referred to Oxyz. y. find x. Explain your result. The set of axes Oxyz.6-53) and the fact that in a rectangular Cartesian coordinate system the metric tensor is the Kronecker delta 5ij. by using the results obtained in Problem 19a. Using the summation convention. T|. Let Tjj be a covariant tensor of order two. z = 3. o 1 [x~
2
_s
2 0
i
2 0
o
1
y
z
If the coordinates of a point P are x =1. z. 5. z) is x 1 = cosh 4 cos r). 3. x3 = z . Calculate the sum of the products ujVi and UjVj. 4) and (4. -n <r\ <K. obtain their components when referred to O x y z .
. x 2 = sinh \ sin r). 18a. Calculate Vx 2 + y 2 + z 2 and V x 2 + y 2 + z 2 . <«. Verify that uiv i = u i v i . v<j> is a quantity whose tensorial properties are not known. This process of setting a superscript equal to a subscript is known as the process of contraction. But the inner product Tij v<j> is known to be a covariant vector (tensor of order
22a. If the vectors u and v have components (2.6-55). z) coordinate system considered in Problem 19a. The transformation from the rectangular Cartesian coordinate system (x J . Calculate the metric tensor gij and its dual g rs for the (£. Let m = n in Equation (4. 21a. r\. the direction of rotation being from the x-axis to the y-axis. Hence.VECTOR AND TENSOR ANALYSIS
231
17a. Is it an orthogonal coordinate system? 20b.
-°°<Z<°°.
The set of axes Ox y z is obtained by rotating the set of axes Oxyz through an angle of 60° about the z-axis.
19a. show that
"xi
y
z
=
r i
2
n. x 3 ) to the elliptical coordinate system (£. Ox y z are as defined in Problem 17a.
Determine the geometrical shapes of the ^ and r\ coordinate curves. Also obtain gij by using the transformation given by Equation (4. y = 2. show that T nn is a scalar.
0) to the Cartesian coordinate system is given by x 1 = r cos 9.P-23a
Four-roll mill
.
Obtain the physical components of the velocity in the polar coordinate system.
V 2 = CX 2
where c is a constant. That is to say. Show that v<j> is a contravariant vector. The Taylor (1934) four-roll mill consists of four infinitely long cylinders immersed in a fluid. the fluid is drawn in at A and C and is expelled at B and D.P-23a. cr sin 0 cos 0]
o | f 'o
—A 2
c
—".C X 1 . as shown in Figure 4. Each cylinder rotates in the direction opposite to the two adjacent cylinders.396
ADVANCED MATHEMATICS
one). The axes of the cylinders pass through the corners of a square as shown in Figure 4. The velocity components are
Vl = .cos 2 0).
FIGURE 4. The stagnation point O is taken to be the origin of a rectangular Cartesian coordinate system Ox 1 x 2 . v<j> transforms as given by Equation (4. 23 a.P-23a. x 2 = r sin 0.6-33). Answer: [cr (sin2 0 . This method of ascertaining whether a quantity is a tensor or not is known as the quotient law. The resulting flow field is a point of stagnation at O. The transformation from the polar coordinate system (r.
the volume integral may be written as
/•I rn fin I I I u u r z sin 6 dr d0 d<b
h k Jo Show that the volume integral is equal to ^JL j . In modelling a vibrating jet. may be written as
V« =
v(0)
2 dJ+Ul(r'z)COS0
= -^(rui)sin0
v(z) = ~ w ( z ) The covariant component of the rate-of-strain tensor y. where u is a unit vector. Appendix E. x 2 = r sin 0. (1987). in the cylindrical polar coordinate system (r. In the sperical polar coordinate system (see Example 4. 0.
le el
Ir e l
The transformation from the Cartesian coordinate system (x 1 . 0. z) system is orthogonal and the metric tensors are gn = g zz = 1.i
+
Show that the physical components of j (denoted by A j in the original paper) are
. 26b. (1993) assumed that the physical components of the velocity field. geg = r2.is defined as
Yij = v i.VECTOR AND TENSOR ANALYSIS
397
24b. x 3 ) to the cylindrical coordinate system (r. it is required to evaluate the volume integral of the dyad u u.
The (r.j
v j. z) is x 1 = r cos 0. z). Chan Man Fong et al. Determine them.4-1). 25 a. x 2 .
In Bird et al. z) are / \ and / \. x 3 = z. Show that the only non-zero Christoffel symbols of the second kind corresponding to the cylindrical polar coordinate system (r. 0. over a unit sphere. 0.
I'
34b. I2) is a function of the first two principal invariants of a second order tensor T.• i.• J is objective.• + V. Working with Cartesian components. Verify that the Christoffel symbols cancel. v* are the velocity components and
Y.T rs)
.+ ai^^ 6 rs.J J.
By considering the transformation of coordinates given in Problem 17a.
(n)
•ij 3Y ij v r?i = —— + vss3Y —ij
(2)
dv 1 y . show that
dty
dty
=
dty 1
\
3T^
a!7 6 . The derivative defined in (ii) is known as the Oldroyd contravariant upper convected derivative.] 36a. 37b. Show that
35b. The rate-of-deformation tensor y has been shown to be objective. (ii) 6 E
Compute the principal invariants of y for the two flows. show that if Ty is an isotropic tensor.• = rU V.S j SJ
3v j v -ii • u. The rate-of-deformation tensor y for (i) a simple shear flow and (ii) an uniaxial elongational flow are given respectively by (i) TO 1 0" Y= T 1 0 0 L0 0 0y = e
(ii)
.
The scalar function ty (I\.
at
ax s
dxs
dxs
[Note that in (ii) we have ordinary partial derivatives and not covariant derivatives as in (i).VECTOR AND TENSOR ANALYSIS
222
1 j " TCij yij dV = j v^ijnJdS
V S
where V is the volume enclosing the surface S.
(i)
-j—
= -^— + v s y1J
s
is not objective.H °° 0-10
[0 0 2 Answer: (i) 2 "f. 8jj. it is necessarily of the form A.
.
E. is one in which there is only one independent variable and all derivatives are ordinary derivatives.). which we denote by x 1 and x 2 .E.CHAPTER 5
PARTIAL DIFFERENTIAL EQUATIONS I
5. we have considered ordinary differential equations (O.E. x 2 ).D. A P. A partial differential equation (P.1 INTRODUCTION
In the first two chapters.D. we consider only two independent variables . c is the concentration. Most processes that are of interest to engineers and scientists take place in a two or three-dimensional space. Other examples of P.E. we shall see that. The equation describing the vibrations of a string can be written as
—T
= c
^
(5-1-D
at 2
ax2
where y is the displacement of the string from its equilibrium position. we have seen that to determine the velocity field of an incompressible and irrotational flow. The diffusion of a material in a homogeneous medium is governed by the equation ^ = « ^ (5-1-2)
2
at
ax2
where a is a constant.) is one in which there are two or more independent variables and the derivatives that occur in it are partial derivatives. and their solutions will be given in this and the next chapter. and c is a constant. We denote the unknown function by u (Xj. in mechanics. The number of independent variables is usually more than one and the equations governing these processes are partial differential equations. can be written as
. In Chapter 10. In Chapter 3. x is the coordinate of a point on the string.D. we have to solve Laplace's equation. An O.E. which is a partial differential equation. t and x are time and position as in Equation (5. For simplicity. t is the time. time may also be involved.1-1). Hamilton's equations of motion are given by a set of partial differential equations. Frequently.D.D.
the coefficients of u and its derivatives are only functions of the independent variables Xj and x 2 ..E. that is to say. -.1-3)
where f is a function. . 2) are of second order and are linear.+ a 3 (x 1? x 2 )u = g ( x 1 > x 2 ) If g is zero.402
ADVANCED MATHEMATICS
f( Xl . Equations (5.^-. The order of the equation is defined to be the order of the highest order partial derivative in the equation. the equation is linear.u.^._ . is an equation in which there is no term which is a function of the independent variables only.1-1.= °
C5-2"1)
where &i and a2 are given functions. ..+ u r . (5.) = 0
dXj dx 2 9xj
(5.1-5)
Just as for ordinary differential equations.D. A first order linear n o n homogeneous partial differential equation can be written as a ^ x j . " (5.=0
dXj dx 2
. An example of a non-linear equation is 3u (du \ n v— + 3— = 0
dxj dx 2 /
.1-6)
In this section. x 2 ) ^ .x 2 . Method of Characteristics Consider the linear homogeneous equation which is written as
al(xl'x2)^~+ a
2(xl'x2)^. it is non-linear.^-. a homogeneous P. An example is 3u 3u _ j . If f is linear only in the highest order partial derivatives. 5.1-4)
If the equation is neither linear nor quasi-linear. we present several methods of solving first order partial differential equations. the equation is quasi-linear.2 FIRST ORDER EQUATIONS (5.. If f is linear in u and its derivatives.
.+ a 2 (x l 5 x 2 ) ^ . the equation is homogeneous.
3-1.x2)
(5. I".2-2a) (5.x 2 ) a2(x1. . u is a function of one variable s and we can write du ds 3u dx.
= a 1 (x 1 .
.2-5)
Along Fj. Geometrically. b. Solve the equation
g. 3xj ds 3u dx9 3x2 ds (5.2-1.PARTIAL DIFFERENTIAL EQUATIONS I
403
The solution of Equation (5. can be written as ^
al
. we need to know the value of the function at a given point.2-4) is u = C where C is a constant. additional conditions need to be prescribed. defined in parametric form by [see Equations (4. we prescribe the value of the function u along a curve.= 0
9xj
given that
l
(5.^
a2
(5-2-6)
The solution has an arbitrary constant and the characteristics form a one-parameter family of curves. u is a constant and Fj is a characteristic curve.2-7)
dx2
. Example 5.+ x. Let Fj be a curve in the (x1? x 2 ) plane. we regard u as a surface in the (xj. F 1? F 2 . For a first order ordinary differential equation. b) which.2-2b)
=
Along Fj. 3) yields
411 = 0
ds The solution of Equation (5.2-3)
Combining Equations (5.2-2a. x2) plane and the surface is generated by a family of curves.
-j± dx ^2. that is to say.2-4)
(5. For a partial differential equation.. The characteristics are obtained by solving Equations (5. 15)]
dxi
.2-2a.2-1) in the (x ls x2) plane.
(5. To obtain a solution. we need to impose an initial condition. by eliminating s.2-1) is a function u that satisfies Equation (5..
The values of u in the region bounded by the characteristics passing through the origin and (0. x 2 ) is any point in R.2-9a. We denote this region by R which is shaded in Figure 5.404
ADVANCED
MATHEMATICS
u (0. b)]
£ =' •
517 = x>
The solution is
% .2-11). u is a constant and we write u = C The parabola will intersect y at the point (5. which is x j . the value of u is constant. Let (xj. x 2 ) = 1 + x 2 . On each of the parabolas.2-15)
u = 1 + 1 (2x 2 -xf)
(5.10 >
Combining Equations (5. x 2 ) is given by Equation (5.2-14a.
0 < x2 < 2
(5. 13. 2) can be determined.2 .
X2
=2
fe-^l2)
(5.2-2a.x>
( 5 -2-9-b)
(5 . 14b) yields C = 1 + 1 (2x 2 -xj 2 ) Since (x"j.2-8.2-8) and combining Equations (5.b)
The value of u along y is given by Equation (5.2-1.
(5.2-8)
The equations for the characteristics are [Equations (5. The equation of the characteristic that passes through (xj. we can replace it by (x 1? x 2 ) and the solution in R is (5.2-13) (5.2-1.2-1.2-12)
xx = 0.2 x 2 = Xj 2 -2x 2 = K On this parabola. The values of u are given along the x2-axis in the interval 0 < x 2 < 2 and we have denoted this line segment by y in Figure 5.2-11)
The characteristics are parabolas and are shown in Figure 5. b) yields
xf-2x 2 = K
where K is a constant. x 2 ) be any point in R.2-16)
.
+ f 2 (x 1 . we obtain 3O 9xj
+
(5. the conditions must be consistent.
FIGURE 5.u) = 0 Differentiating with respect to xx and x2.PARTIAL DIFFERENTIAL EQUATIONS I
401
0
X.2-1
Characteristics F of the differential equation and the line y along which initial data are given
• Note that y is not a segment of a characteristic F.x 2 . u ) j ^ . otherwise there is no solution.x2. That is to say. u ) Suppose the solution can be written in an implicit form as O(Xl.2-1%)
. since u is constant on F.2-17)
(5.2-18)
3O 8u_ 9u 3x2
= Q
(5. u must also be constant on y. If y is part of F.2-19a)
i * + ^ ^L 3x2 3u 8x2
=
o
(5. This method can be applied to a quasi-linear equation which can be written as f 1 ( x 1 .= g ( x l 5 x 2 . x 2 . u ) ^ .
22a.2-20a.2-20b)
du
^dx 2
= -jrL / jr dx 2 du
Substituting Equations (5. b) can be written as dx -^J.! * / £
dXj dXj
(5. and u. in this case.2-21) is now a linear equation in O.x 2 . f| and f2 are also functions of u and that the characteristics depend on u. The characteristics can also be given in parametric form and Equations (5.2-17).2-22b) yields the values of u along the characteristics and Equation (5.= f j ( x . b) into Equation (5. Note that. The characteristics of Equation (5. x 2 . However.2-23a)
g-
=g(Xl.2-20a)
(5. x 2 . If g is zero. we obtain
3cb
f
1
3o
z
3o
=° (5.406
ADVANCED MATHEMATICS
Assuming that ^-— is not identically zero.u) (5.2. unless g is zero. . only Xj and x 2 are independent variables and u is the dependent variable).x2.
.2-21) are given by
^
fl
=
^1
f2
=
dL
§
(5. j s not zero. We consider all three as independent variables (in the original problem.2-21)
du
l^-+f2 5^ + g ^
dXj dx 2
Equation (5.b)
The characteristics of Equation (5.2-17) is solved. Solving Equation (5.2-23b) (5. The initial values of u are given along the curve y and the solution (u) is defined along the characteristics that pass through y.= .u)
(5.2-23c)
Note that in this case ^U. it now involves three variables x t .2-22a). we deduce
all
p. as shown in Figure 5.2-22a. u ) dxo -^= f 2 (x 1 .2-17) are obtained by solving Equation (5.2-2. u is a constant along the ds characteristics.
I-1) implies that ^
+
A(pvx) = O
(5.
.2-24)
Our driving experience leads us to believe that vx is a decreasing function of p. Model the movement of cars at a traffic light We model the automobile traffic on the road by considering the flow to be one-dimensional and the road to be the x-axis. we have a traffic jam and v x reduces to zero.2-26a.2-2
Characteristics Tj of the differential equation and the initial data curve y
Example 5.PARTIAL DIFFERENTIAL EQUATIONS I
401
U
/
r
'
FIGURE 5.d)
where L is a typical length. A simple function that describes this situation is vx = v m d . If p is zero.2-24. a block).p / p c ) We non-dimensionalize Equations (5. we can travel at the maximum allowable speed v m and if p reaches a critical value p c .b. 25) by introducing (5.2-25)
u=X
vm
< J = -^-.e. which can be the distance between two traffic lights (i. We denote the local density (number of automobiles per unit length) by p (x) and the velocity by v x .c. The principle of the conservation of the number of cars (Equation A.
Pc
t =V '
L
^= T
L
(5.2-2.
2-27a)
(5. we deduce (5. The road in front of the traffic light is free of vehicles and behind the traffic light the automobiles are bumper-tobumper.2-28)
We now assume that at the origin (£.b)
The speed of propagation c of the state c is given by c = -ji From Equations (5.2 .2-32)
. = 0) there is a traffic light which turns green at T = 0. The density a initially can be written as a (£. This can be shown by differentiating c along a characteristic curve.3Oa >
(5.2-30b)
dt _J = i _ 2 o
We note that the slopes of the characteristics change sign at a equals to a half.2-27b)
||
+
(l-2o)g-0
(5. b) yields
(5.8-29a. £ > 0
The characteristics of Equation (5.2-3la. The factor (1 . a is a constant (g = 0).2a) can be interpreted as the speed of the propagation of the state a. Also along each characteristic. b).2-3la. We have da = 0 = ~ dx + | | d£ (5.2-27a.2-24.2-28) are given by
f
or
=
T^
( 5 .0) = { ' ^< (5.4Q8
ADVANCED
MATHEMATICS
Equations (5.b)
1 0. 25) now become
a7 + ^ M ) = 0
u = (l-o) Combining Equations (5.
0
0
£
FIGURE 5.2-3. In II. the slopes of the characteristics vary from one to minus one and this implies (Equation 5. and III as shown in Figure 5. For negative values of £. in I and in. That is to say. the values of a are zero and one respectively.2-28).2-3
Characteristics and path of vehicles (. The x-axis corresponds to a value of o equals to one half. T) upper half plane into three regions which we denote by I.) for the continuum traffic model
.
(5. a is zero and we deduce from Equation (5. The characteristics Fj and F2 divide the (£...2-27b. u > c.2-33b) is obtained from Equation (5. For positive values of %. the slopes are -1 and we denote by T2 the characteristic that passes through the origin. 33b). We now draw the characteristics which are shown in Figure 5.2-33a)
= (l-2o) Equation (5.2-3. we find that the velocity of the vehicles is greater than the speed of propagation of o. II.2-33b)
Comparing Equations (5.2-30b) that the slopes of the characteristics are one.PARTIAL DIFFERENTIAL EQUATIONS I
409
(5. On the characteristics. The initial conditions are given along the ^-axis. Let Fj be the characteristic that passes through the origin. the value of o is a constant.2-3Ob) that the values of a vary from zero to one.
410
ADVANCED MATHEMATICS
Let D be any point on a characteristic in II, as illustrated in Figure 5.2-3. Integrating Equation (5.2-30b) and imposing the condition that the characteristic passes through the origin yields £ = (1 -2a) x From Equation (5.2-34), we deduce that the value of a on the characteristic through D is o= ^(l-|) (5.2-35) (5.2-34)
Having obtained a description of a, we now consider the path of an individual automobile. In this model, the car at the origin takes off at maximum speed since the state a associated with it is always zero (along Fj). The car initially at A (£ = - a ) in Figure 5.2-3 is associated with the condition a = 1. That is to say, it cannot move. It has to wait a finite time for the car density a to become less than one, before it can move. This is associated with point B in the figure. The time it has to wait can be obtained from Equation (5.2-36b) and is x = a (5.2-37)
From now on as it moves forward in space and time, the value of a is given by Equation (5.2-35). The speed u of the car is given by Equation (5.2-27b) and is written as
(5.2-38a,b)
Note that, in Equation (5.2-30b), -— refers to the slope of the characteristics which are defined by the dx J P.D.E. [Equation (5.2-28)]. In Equation (5.2-38b), - ^ refers to the speed of the car and is defined dx by the trajectory of the car, as shown by the broken line in Figure 5.2-3. Note also that, in the case of the first car, a = 0 and u = c. That is to say, the slope of the characteristic is also the velocity of the car. Integrating Equation (5.2-38b) yields
Equation (5.2-39) is the equation of a parabola and is the path of the vehicle shown as a broken line in Figure 5.2-3. The time required for the vehicle to reach the traffic light (£ = 0) from B can be obtained from Equation (5.2-40) and is given by x = 4a (5.2-41)
The last vehicle in the block (£, = -1) crosses the traffic light after five units of time (5a) and if the light changes in less than five units of time, the car has to wait for the next change. The typical path of a vehicle starting at point A can be summarized as follows. After the light turns green, it remains stationary (AB) for a finite time (a), after which it starts to accelerate to cross the traffic light after a further 4a units of time at point C. It continues to accelerate until it reaches its maximum speed asymptotically. • The quasi-linear equation considered in this example can be used to describe the movement of glaciers and floods. Further details are given in Rhee et al. (1986); see Problem 3b.
Lagrange's Method
Lagrange's method is similar to the method of characteristics. Equation (5.2-17) is now also solved via Equations (5.2-22a, b). The latter equations are now referred to as the auxiliary (or subsidiary) equations. They can also be written as (5.2-42a,b)
dxj
fj
dxj
fj
Instead of defining the solution along the characteristics, in Lagrange's method the solutions of Equations (5.2-42a, b) are written in implicit form as O 1 ( x 1 , x 2 , u ) = Cj O 2 ( x b x 2 , u) = C 2 where C^ and C 2 are the arbitrary constants of integration. The functions Oj and O 2 are integrals of the auxiliary equations and are solutions of Equation (5.2-17). This can be shown by differentiating ®i to yield (5.2-43a) (5.2-43b)
From Equations (5.2-20a, b), we deduce that Oj is a solution of Equation (5.2-17). Similarly, O 2 is a solution and a general solution can be written as F(ObO2) = 0 (5.2-46) where F is an arbitrary function. The system of Equations (5.2-22 a , b) might not be easy to integrate. A possible method of solving the equations involves writing Equations (5.2-22 a, b) as follows dx 9 du X,dx,+Lidx ? + vdu dx, -7+ = -j2- = — = l * 2 where we used the following relation
(5.2-47a,b,c)
r = J =
b d
a
r
Xa + lie
Xb + |id
—
(5.2-48a,b)
We now choose the functions X, \x, and v such that Xf 1 +^if 2 + v g = 0 Since we are now dividing by zero, we have to require that A,dxj+Lidx 2 +vdu = 0 Comparing the exact differentials [Equations (5.2-44, 50)], we identify
X =
(5.2-49)
(5.2-50)
™±, dxj
n=^ I , 3x2
v - ^ I du
(5.2-51a,b,c)
Since X, L i and v are known functions of x l5 x 2 and u, we can integrate Equations (5.2-5 la toe) to obtain O j . A different choice for the set {X, Li, v} in Equation (5.2-49) will provide a different function O 2 .
PARTIAL DIFFERENTIAL EQUATIONS I
47i
The general solution will be given by Equation (5.2-46) which implicitly determines u as a function of Xj and x2. The arbitrary function F is determined by the initial conditions. Different choices of 3>j will generate different functions F, but the initial conditions will ensure that the solution is unique. Example 5.2-3. Chromatography is widely used as a method of separation of chemical species and references are given in Rhee et al. (1986). The simplest case is the chromatography of a single solute. Suppose that the solution flows along the z-direction through a bed with void fraction e at a constant superficial velocity v0. If c is the concentration in the fluid phase and n is the concentration in the stationary phase, a mass balance [Rhee et al. (1986) or Bird et al. (I960)] yields (5.2-52) We assume a linear relation between n and c and write n = Kc where K is a constant. Combining Equations (5.2-52, 53) yields (5.2-53)
Our aim is now to determine the form of the function f. That is to say, we want to know if f is an exponential function, a trigonometric function, etc. To determine the form of the function, we use Equations (5.2-55a, b). From Equation (5.2-55b), we determine that the unknown function f has to be a Heaviside function. The solution is given by Equation (5.2-61). The reader should verify that this equation satisfies the other boundary condition. From Equations (5.2-55a, b), we deduce that c = c0H(t-pz) Transformation Method We illustrate this method by considering a first order partial differential equation which can be written as
al
(5.2-61)
( x l' X2> 57~ + a 2 (xi» X2) a~
+ a 3 ( x l' X2) u =
f
(xl> X2>
(5.2-62)
We make a change of independent variables and write £ = ^(x1,x2), Using the chain rule, we obtain r| = r((x 1 ,x 2 ) (5.2-63a,b)
|3L
9xj
=
f£ 1 1
9q dxl
+
|£p .
dr\ 8x1
(5.2-64a)
|3L . | » | L
dx2 dc, dx 2
+
| i |!L
dT) 3x 2
(5.2-64b)
Substituting Equations (5.2-64a, b) into Equation (5.2-62) yields
PARTIAL DIFFERENTIAL EQUATIONS I
415
(5.2-65) We now choose £ and r\ such that we reduce the P.D.E. to an equation which involves only one derivative. One possibility is to choose % = xx and to require the coefficient of 3— to be zero. That is to say (5.2-66a)
a'!^7 + a
^=0
(5-2"66b)
This allows us to obtain r\ by solving
P- = ^
dx 2 a2 Along each of the characteristics given by Equation (5.2-67), T| is constant. Equation (5.2-65) simplifies to
&1fe,n)
(5.2-67)
| | + a 3fe,TI) u = ffe,TI)
u =
(5.2-68a)
(5.2-68b)
or
^L
3^
+
£i
aj
1
ax
Equation (5.2-68b) can be solved by introducing an integrating factor I defined as
I = exp I 52- d^
Equation (5.2-68b) can now be written as
(5.2-69)
4-(uI) = —I
3^ aj
(5.2-70)
On integrating, we obtain the solution to the problem. u = I
1
— Id£ + F(r|) I) a l
(5.2-71)
416
ADVANCED MATHEMATICS
where F (r\) is an arbitrary function of r), which is determined from the initial (boundary) conditions. The solution in terms of x{ and x 2 is obtained via Equations (5.2-66a, b). The transformation from (xj, x2) to (£, r|) is valid as long as the Jacobian
ii
8xj J = 9xj is non-zero.
ii
9x 2 (5.2-72) dx2
Example 5.2-4. Solve the problem considered in Example 5.2-3 by the present method. Equations (5.2-54) can be written as
0 1
§7 + v° If = °
<52-73)
<«-74>
(5.2-75a,b,c)
We identify aj to be a, a 2 to be v 0 , ^ to be t, and T| is obtained by solving
Example 5.2-5. Solve the following equation using the three methods described in this section (5.2-79) subject to the condition u=2 on x 2 = xf (5.2-80)
In this example, Equations (5.2-22a, b) are dx, dx 9
=
du
•at= ~H
x 2 = Kxj
2^2
(52-81a'b)
From Equation (5.2-8la), we deduce that the characteristics are (5.2-82)
where K is a constant. The characteristics are straight lines through the origin, with slope K. Combining Equations (5.2-8 lb, 82), we obtain
*± = —
X! 2Kxj 2
(5.2-83)
Solving Equation (5.2-83), we obtain along each characteristic u = KxJ + C where C is a constant. The constant C depends on the characteristic; that is to say, it is a function of K (= X2/X1). Using Equation (5.2-82), Equation (5.2-84) can be written as u = x 1 x 2 + f(x 2 /x 1 ) Imposing the condition given by Equation (5.2-80) yields 2 = xf + fCXi) We deduce
f(xj) = 2-Xj 3 (5.2-87)
The two solutions of the auxiliary equations are given by Equations (5.2-82, 92). The general solution [Equation 5.2-46)] can be written as
Xlx2-u
= g(x2/Xl)
(5.2-93)
where g is an arbitrary function. Using Equation (5.2-80), we find, as in the previous method, that g is given by g(x 2 /xj) = ( x 2 / X l ) 3 - 2 It follows that Equation (5.2-88) is the solution of Equations (5.2-79, 80). We now use the transformation method. We introduce two new variables £, and r\ and choose ^ to be Xj. The variable T| is given by Equation (5.2-66b) and in this example, it is (5.2-95) The characteristics of Equation (5.2-95) are ^ = ^ x2 x2 (5.2-96) (5.2-94)
Reverting to the original variables (replacing £ by xj and r\ by x 2 /x!), Equation (5.2-103) becomes Equation (5.2-71). As expected, all three methods yield the same solution.
420
ADVANCED MATHEMATICS
5.3
SECOND ORDER LINEAR EQUATIONS
Second order linear partial differential equations are frequently encountered in the applications of mathematics. They are classified into three types: hyperbolic, parabolic, and elliptic. Each type describes a different physical phenomenon. Hyperbolic equations describe wave phenomena, parabolic equations describe diffusion processes, and elliptic equations describe equilibrium conditions. We next deduce the canonical form of each type via a transformation of independent variables.
Classification
The general second order linear partial differential equation in two independent variables (xp x2) can be written as d u n^T
dXj
0
where a^, a12, a22, bj, b 2 , and c are functions of Xj and x2. We simplify Equation (5.3-1) by changing the variables (xj, x2) to (£, T|), in an attempt to reduce the number of higher derivatives. This transformation is given by Equations (5.2-63a, b). From Equation (5.2-64a), we have
The transformation (Xj, x 2 ) to (^, r|) is not yet defined. As mentioned earlier, the aim of the transformation is to simplify Equation (5.3-1). We examine the possibility of defining a d2u transformation such that the coefficient of vanishes. This does not affect the generality of the
a2u
equation. That is to say, Equation (5.3-3) with the coefficient of = 0 will be equivalent to
a^2
Equation (5.3-1). An example is provided by Equations (5.4-6b, 21). That is to say, we seek the conditions under which it is possible to have
J&t
\dx 1 /
a12ii ii + ajiif - 0
axj ax2 \ax 2 /
(5.3-4,
The solution of Equation (5.3-4) provides the desired transformation for £. Assuming that 3x 2 is non-zero, Equation (5.3-4) can be written as
Consider the following cases, (i) ax| > 4 a n a 2 2 In this case, we have two distinct roots and this is the hyperbolic case. Consider the curve E, = constant On differentiating, we obtain ^ - dx, + ^ - dx 9 = 0 3xj l dx 2 l It follows that (5.3-8) (5.3-7)
ii/il
9x] 3x 2
=
_d^
dxj
(5.3-9)
Substituting Equation (5.3-9) into Equation (5.3-6) yields dx^ d Xl
=
a12 + V a 1 2 - 4 a n a 2 2 2an
(5.3-10)
We have two possible solutions; that is to say, we have two possible curves (t, = constant), one corresponding to the positive root and the other to the negative root. Noting that the coefficient of is identical to that of — - if we replace S, by r|, we choose one root for
the curve % (= constant) and the other root for r\ (= constant), since r\ is also a function of Xj and x 2 . These two curves are the characteristics of the partial differential equation. For different values of the constants, these characteristics span a two-dimensional space. For a hyperbolic equation, we have two real characteristics (^ = constant, r\ = constant) and they are obtained by solving the equations dxo_ dxl dx^
dxj
=
a12 + V a 1 2 - 4 a 1 1 a 2 2 2all a12-Va122-4ana22
2all
(5.3-1 la)
=
(5 3-llb)
respectively.
PARTIAL DIFFERENTIAL EQUATIONS I
423.
With the coefficients of — - and
as
^2
equal to zero, we obtain the canonical form of a
hyperbolic equation which is written as
J ^
+ P l
^
+
P 2 ^ + Yu - o
(5.3-12)
where P l 5 P 2 , Y > a n d o are functions of t, and TJ. Note that Equation (5.1-1) is hyperbolic. In Equation (5.3-6), we have assumed ajj to be non-zero; but if a ^ is zero, we can still solve for = — / = — from Equation (5.3-5) and finally obtain Equation (5.3-12). If both ajj and a22 are zero, the equation is already in the canonical form. (ii) a12 = 4 a n a 2 2 This is the parabolic case and the two characteristics are coincident. We choose £ (= constant) to be a characteristic and the coefficient of 32u to be zero. The coefficient of
Equation (5.1-2) is a parabolic equation, (iii) a 12 < 4 a u a 2 2 This is the elliptic case and the characteristics are complex. Since the coefficients a;; are real, the characteristics are complex conjugates [Equations (5.3-1 la, b)] and we write £ = a + if* ri = a - i p where a and P are real. Writing Equation (5.3-4) in terms of a and P and equating the real and imaginary parts to zero yields (5.3-16a) (5.3-16b)
3a\2
all + a12
8a 3a
+ a22 dXj dx 2
(3a\2 (d$\2 3p 3p /3p\2
= all + a12 + a 22
dXj/
\"X2/
\dxi/
" x i "X2
\3 X 2/
(5.3-17a)
2ail^L^+ai2(^^+*L
iL
+2
a 2 2 ^ ^ ,o
8x 2 9x 2
(5.3-Hb)
3xj 3xj
\8xj 3x 2 3x 2 8xj
Rewriting Equations (5.3-3) in terms of a and P and using Equations (5.3-17a, b), we find
2
9 9
that the coefficient of
is zero and the coefficients of
and
are equal. The
3a 3P
canonical form of an elliptic equation is ^ | 3a2
+
3a 2
3p2
^ |
+
P
5
^ 5a
+
P
6
^
+
y2u=a
2
(5.3-18)
3p2
3p
where P 5 , P^, y 2 , and a 2 are functions of a and p. Laplace's equation is an elliptic equation.
PARTIAL DIFFERENTIAL EQUATIONS I
425
As usual, we require the transformations from (x l5 x2) to (£, r\) to be non-singular; that is to say, the Jacobian J, given by Equation (5.2-72), is non-zero. In the elliptic case, we replace the complex functions £ and r\ by the real quantities a and (3 and this is implied in all subsequent discussions. Let the coefficients of—-, , and be denoted by a 11? a 1 2 , and a 2 2 respectively. By
direct computation from Equations (5.2-72, 3-3), we obtain
a 12
-4aila22
=
j 2 ( a 12
- 4ana22)
(5.3-19)
The Jacobian J is real, its square is positive, and Equation (5.3-19) implies that the sign of a l2 ~ 4 a l l a 2 2 is preserved on a transformation of coordinates. That is to say, the type (hyperbolic, parabolic, or elliptic) of a P.D.E. is independent of the coordinate system. The quantity a l2 ~ 4a 1 1 a 2 2 can be a function of (x1? x 2 ), its sign can depend on the region of the (xj, x 2 ) plane. The type of an equation may change on moving from one region to another. Example 5.3-1. Reduce the equation ^ 3xj x2^=0 3x 2 (5.3-20)
In both cases, the singularities correspond to the xj-axis (x2 = 0). That is to say, both hyperbolic and elliptic equations have singularities on the Xj-axis and it is across this axis that Equation (5.3-20)
4 METHOD OF SEPARATION OF VARIABLES
This method relies on the possibility of writing the unknown function u (xj. We can extend this method to more than two variables. the partial differential equation reduces to two ordinary differential equations for fl and f2. The y component of the equation of motion is
. 5. the boundary and initial conditions will have to change according to the region in the (Xj. x2) plane. Wave Equation An elastic string is tied at its ends (x = 0. as shown in Figure 5. it is at P 0 Q 0 .4-1.4-1. In a given problem. This will be addressed later. x 2 ) as the product of two functions f1 (Xj) f 2 (x 2 ). At equilibrium. We illustrate this method by solving the following equations. x = L).PARTIAL DIFFERENTIAL EQUATIONS I
427
changes from elliptic to hyperbolic. In many cases. The tension T acts along the tangents at P and Q.
y Jk
\s
I
iS/v^ + S^
0
Po
Qo
L
x
FIGURE 5. We assume that the tension T in the string is uniform and the density of the string is p.4-1
Vibrating string
The angles these tangents make with the x-axis are \|/ and y + 8y. as shown in Figure 5. Consider a small element PQ of the string of length 5s. The displacement from the equilibrium position is y.
n2.2Y
(5. This condition is written as y (0. so there is no displacement at these two points. The conditions given by Equations (5. we need to impose additional conditions. where x and t are independent variables.4-10) by XT yields 1 d 2 T = c 2 d2X = constant t t T dt 2 X dx 2 . -£
dt
= g(x)
t=0
(5. Equation (5. We choose the constant to be . we obtain (5. 0) = f(x).4-11)
Since the left side of Equation (5. b) are the initial (Cauchy) conditions.4-11) represents two ordinary linear equations which can be written as follows
d2T 9
5_L+n2T = 0 dt 2
. t) = y (L.4-10)
The function XT is not identically zero. These are expressed as y(x.4-8a.4-9)
X^
dt 2
= c 2 ^fT
dx 2
(5.4-12a) (5. The string is tied at its ends. b) are the boundary conditions and those given by Equations (5.* A *" (5.b)
The vibrations depend on the initial shape of the string and the speed at which the string is released.b)
where f (x) and g (x) are given functions of x.4-12b)
2
^A+n_^x = 0 c2 dx2
. so as to satisfy the boundary conditions. We now seek a solution of the form y = X(x)T(t) Differentiating and substituting into Equation (5.4-11) is a function of t only and the right side a function of x only.4-8a.PARTIAL DIFFERENTIAL EQUATIONS I
429
To complete the problem. each side is equal to a constant.4-6b). dividing Equation (5. t) = 0 (5.4-7a.4-7a.
ri = x .PARTIAL DIFFERENTIAL EQUATIONS I
431
The right sides of Equations (5. In this case.E.3-12)]. We note that we first impose the boundary conditions and then the initial conditions.'s and on integrating.4-6b) to its canonical form [Equation (5.4-21)
. If the string is released from rest [g(x) = 0].c t) where F and G are arbitrary functions. b)] £ = x + ct. Its period in x is (2L) and in t.4-20a. x 2 (= t).4-22a) (5. It is necessary to use the superposition principle.4-6b) becomes 82v -^JL.D. the eigenfunctions are the trigonometric functions and we obtain a Fourier series.b)
From Equation (5. is decomposed in two O. On differentiating y partially with respect to t. The P. The coefficients of the expansion are then determined.3-1 la.4-19a)
Jo
L
Bs = . This procedure is necessary because the initial conditions are given as functions and not as constants. the coefficients B s are zero for all s.2 S7CC J
(L g (X) Sin (Sffii)
VL /
dx
(5. we obtain (5. it is I—).3-3) with xl (= x).c t (5.This is accomplished by introducing the characteristics.4-18a.= 0 The solution is y = F ( 0 + G Cn) = F (x + c t) + G (x . b) are the Fourier series of f (x) and g(x) and the Fourier coefficients are given by
As = f
L
f(x)sinfesx)
dx
(5. that is to say. which are [Equations (5.E. to expand the functions given in the initial conditions in terms of appropriate eigenfunctions.4-22b) (5. we find that Equation (5.4-19b)
The solution y is periodic both in x and t. we obtain constants and not arbitrary functions.
D'Alembert's Solution
D'Alembert proposed a solution to the wave equation by transforming Equation (5.D.
on integrating Equation (5. we obtain arbitrary functions and these functions are determined by using the initial conditions.4-28b)
The value of t is not fixed and Equations (5. for any a.c t) is negative. we observe that initially the shape of the string is given by f (x). they are reflected. then at time t the point x that corresponds to the same value of TI is Xj + ct. F is also periodic with period 2L.4-29) (5.4-7a. We note in this case that.4-28b) becomes F (a) + G (2L . From Equation (5. To do this.4-27)
We need to extend the range of the interval when (x + c t) exceeds L and when (x . Thus f (x .ct) is a pattern moving to the right at a speed c without changing shape.c t ) = 0 F(L + ct) + G ( L .PARTIAL DIFFERENTIAL EQUATIONS I
.4-28a. we consider the simpler case where the string is released from rest. The functions f (x) and g (x) are defined only in the interval 0<x<L (5. if initially (t = 0) the point xj corresponds to a constant value r|. it is seen that this implies that g is zero.
. b)] and from Equation (5. one moving to the right and the other to the left. The transformation rj [= (x . The solution can now be extended for all values of (x . and at the end points (x = 0 and x = L).4-30) shows that G is periodic with period 2L.4-31)
Consider a function f(r|). We deduce that. From Equation (5.c t ) = 0 If we replace L + ct by o. Similarly. we need to impose the boundary conditions [Equations (5. Equation (5.c t)] represents a displacement to the right at a speed c.4-22b).4-16).4-30)
Equation (5. f(r|) is constant.a) = 0 (5.4-31). G(-a) = G(2L-a) (5. the original pattern breaks up into two similar patterns but only half the size (amplitude) of the initial one.4-28a) (5.c t) and (x + c t).4-21). For a fixed value of TJ.4-8b). Afterwards (t > 0).
4J3
This solution is d'Alembert's solution and is equivalent to Equation (5.4-31) represents the superposition of two waves. They are periodic and of period 2L. To examine further the physical significance of the solution. Equation (5. 29) are valid for any t and a.4-2. Equation (5. That is to say.4-26c) is then written as y = X[f(x
+ ct) +
f(x-ct)]
(5. we deduce F(ct) + G ( .)] to the left as shown in Figure 5. One pattern moves to the right and the other [f (£.
4-7a.4-26c) simplifies to yx+ct
y(x.4-3. t) depends only on the interval (x . This is the domain of dependence and it is shown in Figure 5. 8a.
. Motivated by physical considerations. It can be seen from Equation (5.4-26c) that the domain of dependence is the same in the case f ^ 0. boundary and initial conditions [Equations (5.c t) and (x + c t).t) = ± I
Jx-ct
g(C)dC
(5.4-32)
The value of y at (x. Equation (5. These conditions are sufficient and necessary to determine a unique solution. b. b)] have been imposed.4-2
Displacement of an initial pattern
Next we consider the case where f is zero and g is non-zero.434
ADVANCED MATHEMATICS
y"
t =o
1
1
1
^~
X
y
J
t >o
X
FIGURE 5.
In this case. c is the temperature and a 2 is the thermal diffusivity.IV-1) simplifies to ^A at
=
jc.1-2) also describes the conduction of heat in an isotropic medium.1-2) which is reproduced here for convenience
i£ = a 2 — 3x 2 9t
(5-1-2)
Equation (5.PARTIAL DIFFERENTIAL EQUATIONS I
431
lx. &p& is the diffusivity. For simplicity. we write Equation (5. t is the time and x is the position. Under appropriate conditions.^ ^ A 3x2
(5. Equation (A.
.t)
L
A
•
x-ct
FIGURE 5.4-33)
where c A is the concentration of A. Consider the diffusion of chemical species A in a binary system of A and B.4-3 Diffusion Equation
x+ct
Domaine of influence
x
Diffusion is a process by which matter is transported from one part of a system to another.4-33) as Equation (5.
The surfaces are kept at zero concentration at all times.1-2) subject to Equations (5.4-39)
Bsexp(-s27i2a2t/L2]sin^^-
(5.4-34b.
s = l . c) constitute an eigenvalue problem..
n
= ^ a . c(O. we find that the partial differential equation leads to two ordinary differential equations.4-36a) is T n = C n e"11'1 where C n is a constant. we deduce An = 0 ..436
ADVANCED
MATHEMATICS
We consider the diffusion out of a plane sheet of finite thickness L. 34b. 2. Using the principle of superposition. t) = X(x) T(t) (5.4-34a) (5. The initial and boundary conditions can be written as c(x. The solution is
Xn
=Ancos~
+ Bnsin^f
(5-4-37)
Imposing the boundary conditions.0) = f(x).b. the general solution of Equation (5. c) is c = ^
s=l
(5.4-36b)
— ' + —X = 0 dx 2 a2
where n2 is a positive constant which is determined by the boundary conditions. They are
d j + n2T = 0
2 9
(5. Initially the distribution of the diffusing substance is f(x).4-38a. Equations (5.t) = 0 .
(5. c(L.4-36b.c)
We seek a solution of the form c (x. .4-36a)
(5.4-40)
The Fourier coefficients B s are obtained from the initial conditions and they are given by
. 0<x<L t>0 (5.4-35)
Proceeding as in the case of the wave equation.c)
The solution of Equation (5. t) = 0 .4-34b.
y) = T (L. x = L) are kept at zero temperature and the wall (y = 0) is kept at a temperature f (x). It decays more rapidly with increasing a in agreement with physical expectation.4-43)
(5.4-45a. decaying exponentially in time.PARTIAL DIFFERENTIAL EQUATIONS I
437
Bs
= f Wo
f (x) sin S ? 2 L d x L
(5. Equation (5. As y tends to infinity. y) = 0 T (x. y = 0. this is the first problem considered in detail by Fourier. T —> 0 as y —> 00 (5. y—>oo (5.4-44). as in the case of a vibrating string. and dividing by XY (* 0) yields
.4-45c.d)
The two dimensional heat equation is given by (see Appendix HI)
Kma*lil + tl)
where T is the temperature and a2 is the thermal diffusivity. x = L. as pointed out in Carslaw and Jaeger (1973). the temperature tends to zero.c. unlike the case of a vibrating string. At steady state.4-41)
The solution is sinusoidal in x. Laplace's Equation We consider the steady temperature in a semi-infinite rectangular solid. substituting the resulting expression into Equation (5.4-43) reduces to ^ 1 +^ 1 = 0 dx2 dy2
(5.4-44)
The walls (x = 0.4-42a. The boundary conditions are T (0. 0) = f (x).d)
We employ the method of separation of variables and write T = X(x)Y(y) (5. Historically.4-46)
Differentiating.b) (5. The solid is bounded by the planes x = 0.b.
Co. and Do are constants. 0 and R are given by 0 = Ao0 + B o R = C o i n r + Do where Ao.5-4. If the function has no singularity and is periodic in 0. Equation (5. F is given by
F
m*0
(5. and B m are constants. The solution /En r is the fundamental solution and is singular at the origin. Am. 9). Using the principle of superposition.5-13a) (5. If m is zero.PARTIAL DIFFERENTIAL EQUATIONS I
447
cosmB = cos m (9 + 2 s %) where s is an integer. we make a change of variable and write a = inr (i = / r T ) (5. Ao.5-15)
Equation (5.5-9) is the Euler (unidimensional) equation and its solution is R = Cr m + Dr~ m .5-16) (5.5-13b)
= Eo + X LAm cos m6 + B m sin mBj r m
m=l
(5.5-11)
This implies that m must be an integer. The solution of Equation (5.5-8a).5-14)
where E o . Bo. 8. where C and D are constants. To solve Equation (5.5-8a) becomes
a
2d^R+adR+(a2_m2)R da 2 da
= 0
(55_1?)
. we have to solve Equations (5.
(5. and Co are zero.5-12)
(5. the constants D.5-4) is Z = Hcosnz + Gsinnz where H and G are constants. In the three-dimensional case (m # 0).
z) = i. and n are to be determined from the boundary conditions.442
ADVANCED MATHEMATICS
Equation (5. Its solution is R = LIm(nr) + MKm(nr) (5. K.5-20a. The constant M must then be zero.5-18) becomes (5. The coefficients A n m and B n m are given by
. G.5-2 la)
=\ X
An0 sin n z
f^|
+
£
Km
cos m 6
+ Bnm s i n **] ^ nz
^
|
(5.5-22)
The right side of Equation (5. The general solution is (5.5-21b) yields
CO OO
f (0. n) = 0 u(a.5-17) is the Bessel equation with complex argument. L and M are constants. we assume the boundary conditions to be u (r.5-20c) to Equation (5. 6. there is no loss of generality in assuming that R(a) is one and Equation (5. z) = f(8.5-18)
where I m and K m are the modified Bessel functions of the first and second kind respectively.n = l
[A nm cos m6 + B n m sin me] sin nz
(5. 6.5-19)
Im(na)
The constants A. b) imply that H is zero and that n is an integer. If the radius of the cylinder is a.5-21b)
Applying Equation (5. To complete the problem. The function K m has a singularity at the origin. 0) = u (r.b) (5. z) Equations (5. H.5-22) is the (double) Fourier series of f (6. z).5-20a. We assume that the function remains finite at the origin.] £ A n0 sin nz + ^
n=l m.5-20c)
u = X G n s i n n z ^ | + X £ Gn sin nz[Amcosm9 + Bm sin me]
n=l 0 n=l m=l
m
^
|
(5. B.
and L is a constant. There are an infinite number of them and we denote them as Xnm (knl. 0) (5. If the height is L.5-23b)
n Jo Jo
We have assumed the height of the cylinder to be n for convenience. z)sinnz cosmG d0 dz
(5.5-24a)
d*R dr 2
+ r
dR dr
+
( n 2 r 2 _ m 2) R
=
o
(5 .2n
f (6.5-24b)
Equation (5.PARTIAL DIFFERENT!AT. nm ) = 0 (5.5-27a.. and u tends to zero as z tends to infinity. EQUATIONS I
44?
A n m = -2.5-27a) implies that n a are the zeros of J m . 0.5-28a.5-3a.5-23a)
K2 JO JO
= \
I
I
f
(0' z> s i n
n z sin m e d e d z
(5.I
rn
Bnm
I
.5-8b) remains unchanged. The remaining conditions are assumed to be u (a. 8) become 5-^-n2Z = 0 dz 2
r2
(5. 9. b) has to be changed. u (r.5-26) (5. the solution cannot be sinusoidal in z as given here. Xn2. . Equations (5.5-4. 0) = f (r. The sign of n 2 in Equations (5.) J m (na) = J m (A. Equation (5. We have already assumed that u is finite as r tends to zero. it can be scaled to n by introducing z 1= ^j..b)
. If the cylinder is semi-infinite. and u tends to zero as z tends to infinity. The solution of Equation (5.5-25)
Equation (5.5-24b) is the Bessel equation and the solution that remains finite at the origin is R = LJm(nr) where J m is the Bessel function of the first kind.b) (5.5-24a) that satisfies the condition Z tends to zero as z tends to infinity is Z = Ke~ n z where K is a constant. z) = 0 .
f . Equation (5. The singular points of the equation are at x (= ± 1) which corresponds to 0 (= 0. If the solution is finite for all values of 0.x 2 ) ^ .5-37d)
By writing X as n (n + 1).5-39)
Note that we have replaced X by n ( n + 1) and n is an integer.. The solution is 0 = AP n (x) + BQ n (x) (5. 0) = f (0) Combining Equations (5.5-37a) is the Euler equation and its solution is R = Cr n + ^ n+l
(5. Equation (5.7t). 41) yields
oo
(5.5-42) shows that f(0) is expressed as a Fourier-Legendre series and the coefficients A n can be determined using the orthogonal properties of Legendre polynomials. D must be zero.5-40. B must be zero.5-43)
.5-41)
f(6) = £
n=0
A nanPn(cos0)
(5.5-40)
Suppose the boundary condition on the sphere of radius a is specified as u (a.5-38)
where A and B are constants and P n and Qn are the Legendre functions of the first and second kind respectively. The coefficients A n are given by
An =
(2n+l) I 2an Jo
f(e)
pn(cos
0 ) sin e d e
(5. If R is finite at the origin.5-37d) becomes the standard Legendre equation discussed in Chapter 2.2 x — + XQ = 0
dx dx
(5.5-42)
Equation (5. The general solution is
oo
u
= X A n r n P n (cos0)
n=0
(5.446
ADVANCED MATHEMATICS
( l . and n is an integer. P n are the Legendre polynomials.
the differential equations form eigenvalue problems which generate eigenfunctions. 37a. The functions to be chosen depend on the coordinate system. The solution of Equation (5.5-7. Each type of equation requires different conditions. Therefore. The initial conditions are expanded in terms of these eigenfunctions in the form of Fourier series. In the general case.5-35a.6 BOUNDARY AND INITIAL CONDITIONS
The three types of second order linear partial differential equations have been derived in a physical context and the prescribed boundary and initial conditions have been based on physical considerations. hyperbolic functions. and we have to solve Equations (5. In general. from the initial conditions. both u and -^— are at given initially and at the boundaries only u is given. b). The initial value of -=— can be determined from Equation (5. In Example 3. The method of separation of variables consists of writing the unknown function of two or more independent variables as a product of two or more functions. and orthogonal functions considered in Chapter 2. In addition. In the method of separation of variables. the undetermined functions are expressed as Fourier series.5-44)
The coefficients C n and D n are determined from the boundary conditions. Together with the boundary conditions. The solution is then given as
u
= X [Cnrn + -^j") P n( c o s e ) rn+ / n=o I
(5. the solution is — and it is the fundamental solution. These two types of equations describe evolutionary processes and. When n is zero.5-37b) is the associated Legendre function. the solution of a partial differential equation involves undetermined functions rather than undetermined constants. only u needs to be known initially. The partial differential equation is transformed to two or more ordinary differential equations. The solution of Equation (5. For convenience. In the case of a parabolic equation. The region can be enclosed by surfaces of spheres of radii a and b with the possibility that b tends to infinity. we have shown that the solution of Laplace's
.6-1. the condition at the origin cannot be imposed and D is not necessarily zero. The constants generated by the solutions of the ordinary differential equations are identified as the coefficients of the Fourier series and they can be determined. we denote the dependent variable by u. u is a function of all three variables. Laplace's equation characterizes the steady state and u is prescribed on the surfaces that enclose the region of interest. For a hyperbolic equation.1-2). 5.PARTIAL DIFFERENTIAL EQUATIONS I
441
We have now obtained the solution inside the sphere. Possible base functions for the series are trigonometric functions. If the region of interest is outside the sphere. the values of u are given at the boundaries.5-35a) involves sin m§ and cos m(j) and we require the function to be single valued. These are shown in Figure 5. such that each of these functions is a function of one variable only. m has to be an integer. The problem can be completed by imposing appropriate boundary conditions. we can integrate forward to determine the values of u at a later stage.
6-4) (5. Solve the equation ^ = 0 dxdt subject to u (x. This example violates this rule and the problem is not well posed.6-6a._ ^ AS (5. t) =
g 2 (t)
(5.6-1) is transformed to X"+n2X = 0 Y"-n2Y = 0 The solution that satisfies Equations (5.6-2.b) (5. conditions at y = 0. b) is u = sinhny sin nx 3— n . as a result of having imposed the wrong type of boundary conditions.PARTIAL DIFFERENTIAL EQUATIONS I
449
This example was first considered by Hadamard.6-5) is u = F(x) + G(t) Imposing Equations (5. u (0.6-3b)
As n tends to infinity.6-2a. In most cases. The solution of Equation (5. 1) = f2(x) u (1. we have imposed initial conditions.6-5) is a hyperbolic equation and we have prescribed boundary conditions. 0) = f2 (x) . we expect that a small change in the initial conditions leads to a small change to the corresponding solution.6-8a) (5.6-6a to d) yields F(x) + G(0) = fj(x) (5.6-3a) (5. both u and = — tend to zero but the solution u oscillates with increasing dy amplitude.6-7)
. Instead.6-6c. The solution does not approach zero as the initial conditions tend to zero. as opposed to the required initial conditions. Equation (5. t) =
gl
(5. that is to say.d)
(t) . Note that we have to solve an elliptic equation and that we did not impose conditions on the curve enclosing the region required.
Equation (5. Using the method of separation of variables and writing u as X(x)Y(y). Example 5.6-5)
u (x.
6-8b) (5. 5. Example 5. In Section 4. that is to say. and are therefore not arbitrary. b). Chan Man Fong et al.7-1a)
. the equations governing the motion are
IT-1-i!
-v 3t 3f
(5.7-1.. This is another example of an ill posed problem. (1993) considered the transient flow of a thin layer of a Maxwell fluid on a rotating disk. We illustrate this method by considering a few examples. If u is given on part of the boundary and ^— on the remaining part of the boundary.6-8a.6-8c) (5._ " . small changes in initial and boundary conditions as well as in the coefficients of the equations lead to small changes in the solution. Hadamard has proposed the following conditions for a well posed problem: (i) (ii) (iii) the existence of a solution.6-8d)
From Equations (5..4 satisfy Hadamard's conditions.5. In most physical problems. the solution is unique. we deduce that both f j (x) . For some non-homogeneous problems. the boundary conditions can be imposed arbitrarily.450
ADVANCED MATHEMATICS
F(x) + G ( l ) = f2(x) F(O) + G(t) = g^t) F(l) + G(t) = g2(t)
(5.7 NON-HOMOGENEOUS PROBLEMS
The method of separation of variables was used to solve homogeneous partial differential equations with homogeneous boundary conditions.
The equations solved in Section 5. we have defined Dirichlet's problem (u is given on the boundary) and Neumann's problem (^— is given on the on boundary). the condition imposed at one end (x = 0) is independent of the condition at the other end (x = 1). we on have Robin's problem.F(x) are constants and this implies that fj(x) and f2(x) can differ only by a constant.
N
.F(x) and f2(x) . we can introduce an auxiliary function and the problem can be reduced to a homogeneous problem. In dimensionless form.
d)
Equation (5.7-5)
Note that this choice introduces one extra degree of freedom.7-2a.
(5. ^ ) (5. Substituting Equation (5.tj) = f^zj + g ^ ) (5. Equations (5.7-3a. Xj is a dimensionless number characterizing the relaxation time of the fluid.7-6a) so as to remove the inhomogeneity.7-6b)
As a result of having introduced an extra degree of freedom. tj). The initial and boundary conditions are f (z .7-6a)
dt]
at!
dz2
(5.7-la.b)
where rj is the dimensionless radial distance.c.b. 7). The functions f and x are related to the dimensionless radial velocity u 0 and the dimensionless shear stress T(rz) by u 0 = rj f ( z . 3d) imply
ff(l. t l ) = O
We assume that fj satisfies boundary conditions (5.4)
(5.7-4) and separating the resulting equation into an equation for f j and another one for g. 0) = f (0. we are allowed to choose Equation (5.7-3c. namely
(5. That is to say. b) yields ^ 3tf • * . We introduce another function and write ffz. tj) = T(1. we obtain
A
—r dz = -1 (5. 3z2 . tj) = 0 Combining Equations (5.7-5) into Equation (5.7-7)
.7-lb. 3tj * .7-4) is non-homogeneous and the method of separation of variables cannot be applied directly.PARTIAL DIFFERENTIAL EQUATIONS I
451
where tj and z are the dimensionless time and height respectively.7. one function on the left side is replaced by two functions on the right side. T (rz) = r ^ l z . 0) = T(Z .
PARTIAL DIFFERENTIAL EQUATIONS I
451
K3 = 0 a = —-—. n is an integer
(5.7-17)
[An + B n ] s i n a n z
(5.b)
KK-K)
We can now substitute An and B n into Equations (5.^ .7-16)
where A n and B n are constants (Fourier coefficients). and bnSubstituting Equation (5. 7 . The coefficients A n and B n are found to be
An
. we obtain
x = l . a n is given by Equation (5.7-15a) (5.7-16. b) by sin a n z and cos a n z and integrating with respect to z between 0 and 1.7-9.7-15b).7-3a.7-lb) and solving the resulting equation subject to Equation (5.7-16) into Equation (5. by multiplying both sides of Equations (5. 11.7-18b)
"n
The Fourier coefficients An and B n are determined in the usual manner.7-18a.b " tl ]sina n z
n=0
(5.7-3d). I^i!
(5 .z = £
n=0
(5. 2&Z&.ai > tl +B n e. b) on f and x yields A_ . and an and b n are given by Equations (5. a n . 14a to d. that is to say.19 a. 15a.+ z + ]T [A n e. d) with a.7-15b)
From Equations (5. and b being replaced by ocn.7-18a)
Z_l = £
n=0
anAn + b
nBn|cosa^
J
(5.7-14c. We next consider an example where the boundary condition is non-homogeneous. we find that f can be written as
f = . 17) and f and t are given in the form of an infinite series. b) and using the principle of superposition.z + X M " e ~ a " t + b " B " e ~ b n 1 cos a n z
n=0 L "n J Applying the initial conditions given by Equations (5.
. a.
aA\-K)
B n .
= 0 . The diffusion of oxygen into blood is important in surgery.454
ADVANCED MATHEMATICS
Example 5.7-1
Cross section of a wetted-wall column
The diffusion equation can be written as [Equation (A.7-1. Hershey et al. the following assumptions are made: (i) the system is at steady state 3 .IV-1)] (5. A film of blood of thickness L flows down along the z-axis and is in direct contact with a rising stream of oxygen. It is illustrated in Figure 5.7-2.
0
L
y /
x
^
/
f / f f
Oxygen
Blood
i*—Column ^
wall
/ / /
z?
FIGURE 5. c is the concentration of oxygen in the blood. It is given by TT = ^ + v«gradc Dt at where y is the velocity of the blood. and — is the substantial (material) derivative. In this problem. (1967) have modelled this process as a wetted-wall column. (5.7-20b)
.7-20a) where «0OB is the diffusion coefficient which is assumed to be constant.
It is equally possible to use Laplace transforms to reduce a partial differential equation in two independent variables to an ordinary differential equation. B2. we define the Laplace transform of u with respect to t by
.00
L[u(x.| x + £x 3
(5. We recall that the Laplace transform of a function f (t) was defined by Equation (1. t)] = U(s. Likewise.7-41) and using Equation (5. if u (x. This is not surprising because.t)dt
JO
(5.17-1). we have used Laplace transforms to simplify an ordinary differential equation to an algebraic equation.7-47).8
LAPLACE TRANSFORMS
In Chapter 1. c) are obtained by making the following substitutions respectively x = cos 6 (5.7-46a)
y = l . t) is a function of two independent variables x and t. including non-homogeneous problems. the solution is arbitrary to the extent of a constant.7-46b)
Similarly Bj. ••• can be calculated. for a Neumann problem.8-1)
. The method of integral transforms. The constant A o is arbitrary. can be used to solve partial differential equations. The boundary condition given by Equation (5. The coefficients B n are known and so the An (n > 1) can be determined. (5.7-48)
We deduce that An = ^ for all n > 1.7-38a) can now be written as
00
3f
r=1
=ZBnPn(cos9)
n=0
(5-7-47)
Differentiating T from Equation (5.7-45b. we obtain
00 00
£
n=0
n
A n P n (cos 9) = £
n=0
B n Pn (cos 0)
(5.460
ADVANCED MATHEMATICS
Equations (5. x) = I e" st u(x. introduced next.7-49)
5.
.8-5a. Wimmers et al. t) = c 0 . Combining Equations (5.S-oa. We illustrate the method of Laplace transforms by solving a few partial differential equations. 2 dC AB — 7 + [dr2 r dr .S + KJL.IV-3).k c A where k is the rate constant. 3) yields (5.8-5b)]. (1984) studied the diffusion of a gas from a bubble into a liquid in which a chemical reaction occurs. Example 5. and r is the radial distance.b. t) dt). the concentration of the absorbed gas in the liquid phase is found."^AB T 7 r T ~ ~ [r z dr \ dr }\ We make the substitution C = £ and Equation (5.8-3)
We denote the Laplace transform of c A by C (= I e~st c A (r.PARTIAL DIFFERENTIAL EQUATIONS I
461
Other properties of Laplace transforms are listed in Chapter 1. by simplifying Equation (A. to be
^^J-Lffr2^]*^
(5.8-7) ro d 2 C .8-2)
where c A is the concentration of gas A in the liquid.8-6b) reduces to the following equation with constant coefficients (5.c) (5. .b)
(* R
. c A (r. p.J9 I (. 0) = 0 ( r > R ) . We assume that the rate equation is first order and RA is given by RA = . Taking the Laplace transform of Equation (5. If the bubble is at rest and has a constant radius R. RA is the production rate of A..8-2. c A — > 0 as r — > °° for all t (5. t is the time.8-1. J S A B is the diffusion coefficient of solute A in solvent B.8-4) and using the initial conditions [Equation (5. we obtain " 1 d (r2 dCW _ (s + k) C .8-4) The boundary and initial conditions are c A (R.
k is zero and Equation (5. and evaluating the integral using the Cauchy residue theorem [Equation (3.J . then letting the contour tend to infinity. There is a general inversion formula which can be written as
f(t) = L.7-10)]. This is illustrated in Figure 5. These functions are defined by
erf x = 3=
e~u du
(5.8-17)
• In using the method of Laplace transforms.8-1. The method we have used is to read the inversion from the table of transforms. we obtain cA = ^ erfc ( '~R ) (5. The integration is usually done by taking L to be a finite line. closing the contour. we need to invert the transform F (s) so as to obtain the solution f (t).8-2. see Problem 9b in Chapter 4. The constant y is chosen so that all singularities of F(s) are to the left of L. One method of obtaining a closed contour C is by joining the line L to an arc of a circle centered at the origin and of radius R.8-12b) reduces to
C
= ^ o exp . as shown in Figure 5.8-18)
where s (= x + iy) is a complex variable and the path of integration is along a straight line L in the complex s-plane.1 L ^ «®AB
(r-R)
(5-8-16)
From the table of Laplace transforms.PARTIAL DIFFERENTIAL EQUATIONS I
463
where erfc is the complement of the error function (erf).8-15a)
^ Jo
erfcx = 1-erfx = -L in I
Jx
e~u du
(5.8-15b)
(To show that — (
Vif Jo
e~u du is one.
.1 [F(s)] = ±r
Jy-ioo
estF(s)ds
(5.)
If no chemical reaction occurs.
s) =e-^/Vs~ Using Equation (5.i V i ).8-57)
Note that the second integral on the left side of Equation (5. The values of A and B are found to be A = —±= ( e ^ + e " ^ ) .PARTIAL DIFFERENTIAL EQUATIONS I
421
The solutions satisfying Equations (5. 2 Vs -Vs B = ~= 2 Vs (5.8-5la to c) are C = A e~zVi. we deduce that Vs~ [Ae"^ + B C e ^ .8-60)
Jy_ioo V T
We note that the integrand has a branch point at s = 0 and we proceed to evaluate the integral as in Example 5. we invert C and obtain 1 1 c (x. we determine C (0. where A and B are constants.8-56) is zero because C is continuous at z= 1.8-59)
e s x e~"^ e %.8-56)
Combining Equations (5. 56) and letting e —> 0. we integrate Equation (5. b. = B(e i V i + e.i v
2111
(5. s) and this is given by C(0.8-54a) (5. On Lj and hj.8-54a. ds
(5.8-53) and obtain [iC-1 -si Cdz = 8(z-l)dz =-1 (5.8-3).8-55) z>1 z<l (5.8-2.8-18).8-58a. the square roots of s can respectively be written as
. The only non-zero contribution to the integral is along Lj and L2 (Figure 5. We further assume that C is continuous at z = 1 and this implies that A = B(e 2 V 5 +l) (5. 0) = .8-54b)
The derivative of C at z = 1 is not continuous and to evaluate it.e " ^ ) ] = 1 (5.b)
To determine the concentration at ground level (z = 0).
and y = 0 as shown in Figure 5.. If u is the potential.9-6)
.. In the semi-infinite space bounded by the planes x = 0. Example 5.9-4)
Its inverse is given by
oo
f(x) = ^ F C O + X F c n ^ 8 1 1 ^
Z
<5-9"5)
n=l
*-
The Fourier sine and cosine transforms can be used to solve ordinary and partial differential equations. u satisfies the equation 9 u +
3x 2 3y 2
9 u = _p
. The procedure is similar to that employed when Laplace transforms are adopted. (5..9-1. We solve an electrostatic problem using the method of Fourier transforms.9-1.. we can define its finite Fourier cosine transform as
Fcn
= ? I f00cosli?L
Jo
dx
(5. we deduce that the inverse is
CO
(5. x = %. there is a uniform distribution of charge of density (p/47i).9-3)
n= l
Similarly if f (x) is an even periodic function of period 2L.9-lb)
We can regard b n as the finite Fourier sine transform of f (x). In keeping with the notation used in Laplace transforms.9-la). we write
Fsn
= f
L
fOOsin^dx
Jo
L
(5.PARTIAL DIFFERENTIAL EQUATIONS I
475
bn
= f I fOOsinQp dx L Jo L
(5.9-2)
From Equation (5.
Similarly the Fourier cosine transform and its inverse are defined by
Fc(cc) = I f(x)cosaxdx Jo f(x) = -2.9-21a)
f(x) = J ^ I Fc (a) cos ax da Jo
Note that both sets of Equations (5. Sedahmed (1986) studied the rate of mass transfer at a cathode. some authors define the Fourier cosine transform and its inverse as
f°°
Fc(cc) = 4^
11
I f(x)cosaxdx Jo
(5. The mass diffusion equation is determined by simplifying Equation (A.9-21b)
^ oo
f(x) = — I I f(x)cosaxdx cos ax da n Jo Jo
(5. For the sake of symmetry.9-24a)
.PARTIAL DIFFERENTIAL EQUATIONS I
479
The inverse of the Fourier sine transform Fs(oc) is given by Equation (5.IV-1) with R^ = 0 to yield ^ dt = J9 *?dy2 (5.I F c (a)cosaxda Jo
r
(5. b) lead to
p oo
(5.9-20a)
(5.
V 7E
Example 5. b.9-19b)] can be defined with the factor v — in front of the integral.9-19a). 21a. The initial and boundary conditions are c = Cj for y = 0 at t > 0
(5.9-20a.9-20b)
The definitions given by some authors differ from those given here by having different factors in front of the integral.9-16b)] and the Fourier sine transform [Equation (5.9-2.9-23)
where c is the concentration and J9 is the diffusivity.9-22)
Similarly the Fourier transform [Equation (5.
t) = 2. we obtain c* (y. Using the inversion formula.9-16a.9-19a)]. b)] is used. 0<x<a \ Fs(ct) J-fl-cos(aa)] (5. we find that the inverse of Cs is c* = erfc(y/2fjS)T) which is the solution quoted by Sedahmed (1986).9-1 and 2.9-31)
To obtain c .PARTIAL DIFFERENTIAL EQUATIONS I
481
r
s
= i.9-1 Fourier sine transform f(x) 1.( l .a a
tJ9)
(5.9-33)
0.9-1). the Fourier transform [Equations (5. a < x < °° j
e" x xe" x erfc ax x~a.sin ay da
(5.e . If the region of interest is the whole infinite space (-«> to <»).9-32)
Jo
From the table of Fourier sine transforms (see Table 5. we have used the sine transform because the values of u and c are given at the origin. TABLE 5. • In Examples 5.I
71
^—
a
. r is the gamma function
al-a
. 0 < a < l a/(l+a2) 2a/(l + a2)2 X [1 _ eXp (-a 2 /4a 2 )] -^cosNr(l-a) V2 . We choose the cosine transform if the first derivative is given at the origin. This is shown in the next example. we can either look up in a table of sine transforms or use the inversion formula [Equation (5.
490
ADVANCED
MATHEMATICS
f°° = Jo exp [-iy In (a/r)]
COsh(y9)
dy
(5. 32c. • So far. This process is illustrated in the next example.10-32c)
ycosh(y0o)
Equation (5. and t. In general. we have considered partial differential equations in two independent variables. r.e"^ 111 ^]dy! 2%1 \ Jo ycoshy6 0 j
(5. if we start with a partial differential equation involving three independent variables. Stastna et al.4-27) noting that cosh is an even function.10-34b)
ycoshy90 ye [Sin ( y i n (a/r)] dy (5. we find
lim f
£—>0 } i e
(f)m
cosm9
dm = f
Jo
exp [iy In (a/r)]
COshy9
dy
(5.10-3.10-30b.10-34c)
1
1
+
1 j
n
cosh
Jo
ycoshy90
We have the solution in the form of an integral.10-32b) is obtained by using Equation (3. The diffusivities are assumed to be constant and we denote them by
. Example 5.i . Similarly.1) transforms to reduce it to an ordinary differential equation. we need to apply two transforms to reduce it to an ordinary differential equation.10-28) yields
T = .10-34a)
(
=
0 OO
in+ I Jo
y 9
[2i sin (y In (a/r)] dy
(5. we have reduced a partial differential equation to an ordinary differential equation.10-3. We need to invert (n . Similarly. an equation in n independent variables needs (n . Symmetry is assumed and there is no dependence on 0.10-33)
mcosm80
ycoshy0o
Substituting Equations (5. (1991) considered the diffusion of a cylindrical drop through a membrane. the concentration is c 2 and it is a function of z. In the membrane K2. Applying the integral transform method.[in + f" C O S h y 9 [e*ln<a'r> . A circular cylindrical drop Kj of radius a and height h diffuses into a circular membrane K2 of infinite radius and of thickness Z as shown in Figure 5. 33) into Equation (5. It is assumed that Kj is filled with a penetrant of concentration c 1 which is a function of z and t only.1) times so as to obtain the solution.
10-48)
r
c2 = Jo
=
aC 2 J 0 (ar)da 2 a ^ f^^j^f^jr^n-l * '° '° nTl
exp{.10-40. 5. We need to introduce an auxiliary function and the method can be complicated. The solution is often expressed as an infinite series.. s)f(t)dt (5.10-49b)
.. (2n .. then c 1 and c 2 can be determined from Equations (5. the method of separation of variables cannot be applied directly. 49) respectively. we find that
(5. we obtain
(5.
|
2i
which is the solution given by Stastna et al. If the equations are nonhomogeneous.X) n . _ \ .. .e ) ap cosh |3i
From the table of Laplace transforms. (1991).494
ADVANCED
MATHEMATICS
C2 = F ( s ) a J ' ( a a ) s i n h P ( Z . The method of integral transforms can be applied to linear differential equations (ordinary or partial) with arbitrary boundary and initial conditions.10-49a)
f [2 I L
2 /2n .11-1)
. \2i / J
( t .n(2n-l)(z--e)ttdx
2i Using Equation (5.. If f(t) is specified.1 \2 . .a +7T
(5. .1) (z .11 SUMMARY
The method of separation of variables is suitable for homogeneous equations with homogeneous boundary conditions.
— '— dx da
(5. The problems solved by the method of separation of variables can be solved by the method of integral transforms..10-2).T ) <O7> sm
. . The various transforms F(s) of a function f(t) introduced in this chapter can be written as
rb
F(s) = Ja K(t.10-47)
s.
In the next chapter. s) e~ st e~ ist sin (st) cos (st) sin £2Ei cos^i
J—/
t J n (st)
ts~l
. can be complicated. that is to say.PARTIAL DIFFERENTIAL EQUATIONS I
495
where K (t. Other kernels than those discussed in this chapter have been considered by Latta in the handbook edited by Pearson (1974). finding f(t) given F(s). TABLE 5. The methods of separation of variables and of integral transforms are applicable to linear equations only. it might involve evaluating an integral which cannot be written in a closed form.11-1 lists the kernel corresponding to various transformations. s) is the kernel. Numerical methods can be used to evaluate the integral. Extensive tables of integral transforms are given by Erdelyi et al. the solution is obtained immediately. Otherwise. Table 5. (1954). we shall consider methods which can be used to solve non-linear equations. If f(t) can be read from a table. The inverse process.11-1 Kernels of various transforms Transforms Laplace Fourier Fourier sine Fourier cosine Fourier finite sine Fourier finite cosine Hankel Mellin a 0 -<*> 0 0 0 0 0 0 b °° oo °° °° L L oo °° K (t.
. As a first approximation..x 2 )
Answer: 5 x i / x 2
.= 2x l U given that u(0. They can be written.
1 given that u(xj. p and vx are given by P = Po(1+X).0) = Xj du du given that u = 5 on x 2 = x^
2 Answer: ( x j . Solve the following equations
(i)
|H..496
ADVANCED MATHEMATICS
PROBLEMS 1 a. The direction of flow is taken to be the x-axis and the flow velocity vx is assumed to be a function of x and t. in the absence of a pressure gradient. x 2 ) = (sinhx 2 )/x 2 Answer: -L sinh (X 2 + x 2 )
2b. Obtain p and v x if initially (t = 0)..••-. v x = v o ( l +x) Answer: v o (l + x ) / (1 + tv Q ) po(l+x)/(l+tvo)2 3b. as
3vv 8vY —— + v —— = 0
at
x
dx
where p is the density and vx is the velocity.
du
n
du
"
(m)
x 2 ^ .2 X l x 2 ^ .
The equations governing one-dimensional compressible flow are the equations of continuity and motion given in Appendix I and II.+ |H-
= 0
. the shear stress x between the glacier and its bed is given by t = pgha
where p 0 and v0 are constants.. We model the downhill movement of a glacier as the sliding of a block of height h down a plane surface of constant slope a.
4v m J 2 (2. Chiappetta and Sobel (1987) have modeled the combustion-gas sampling probe as a hemisphere whose flat surface is kept at a constant temperature T c . 0)] dr u where k is the thermal conductivity and h is the heat transfer coefficient. ^ ) = T (r. The system is steady and the temperature T at any point inside the hemisphere satisfies Laplace's equation. By considering only the first term of the infinite series. estimate the time required for the velocity at the centre line to be reduced by 90%. which in spherical polar coordinate. 23b. Note that the boundary conditions are not homogeneous and we introduce u defined by u = T-Tc Solve for u and then.4 ) 2 v t J 0 (0) y Answer: — n L ^ [2. T is finite everywhere in the hemisphere. as a model for a turbulent flow in a channel.4)] 2 22b. Show that the transformation (Cole-Hopf transformation)
.(a. 0) = h [T r .T (a. Burgers (1948) has proposed the equation 9v 3v d v —+ v— = v—3t dx 3x z where v is a positive constant.PARTIAL DIFFERENTIAL EQUATIONS I
505
Solve for vzt. obtain T. Gas at constant temperature TQ > TQ is flowing over the probe.4)e-( 2 . can be written as (see Appendix HI)
dr I
dr /
sin 0 30
36 J
2
2
where a is the radius of the hemisphere.f ) = T c . The boundary conditions are on the flat surface: T (r.
dT on the curved surface: k -^.4J 1 (2. .
we have used the method of characteristics to solve first order partial differential equations and the wave equation (d'Alembert's solution). we have used the method of separation of variables and transform methods to solve partial differential equations.b)
. a method which has been encountered in Chapter 5. We now describe Riemann's method of solving hyperbolic equations. y) (6. Discontinuous initial data are carried along the characteristics.E. y ) .CHAPTER 6
PARTIAL DIFFERENTIAL EQUATIONS II
6.2-2a.D. and parabolic. We recall that the characteristics are the curves along which information is propagated.D.2-la. and for each type we discuss one method. and a similarity variable is introduced to solve parabolic equations.D.1 INTRODUCTION
In Chapter 5.'s.E.'s. hyperbolic.b)
+a(x'y)a^~ + b<Xy)^p
We assume that the prescribed conditions are on a curve y and that they can be written as u = g(x. y) (6. and Laplace's equation are all important equations in classical physics. the hyperbolic equations are solved by the method of characteristics. y)u = p(x. Here. elliptic. It is also shown that hyperbolic equations have real characteristics.2 METHOD OF CHARACTERISTICS
In Chapter 5.E.E. The canonical form of a linear hyperbolic equation is 32 L(u) = ^ d r) 3 +c(x. the diffusion (heat) equation. The method of Green's function which was employed in Chapter 1 to solve non-homogeneous O.D. We consider also non-linear equations.'s can be classified in three types.'s is now extended to solve elliptic P. In this chapter. In this chapter. 6. The equations we considered in Chapter 5. we seek the solution of Schrodinger's equation which is one of the most important equations in quantum mechanics. P. ^ = h(x. we introduce additional methods which can be used to solve P. namely the wave equation.
512
ADVANCED MATHEMATICS
where ^— is the normal derivative of u on y. on We also assume that y is not a characteristic of Equations (6.2-la, b), as shown in Figure 6.2-1.
y
l
y0 •
°V—•*—
1
^
x0
*
FIGURE 6.2-1
Characteristics Tx and T 2 and curve y along which conditions are prescribed
We now determine the value of u at a point P (x 0 , y0) which is a point of intersection of two characteristics. The solution at P is determined by the domain of dependence A (see also Figure 5.4-3). The adjoint operator L* of the operator L defined by Equations (6.2-la, b) is [see Equation (6.5-1) for the general case]
L* W
Equation (6.2-9) expresses the value of u at P in terms of v and the sum of integrals. By choosing v appropriately, we can reduce some of the integrals to zero. Following Riemann, we impose on v (the Riemann function) the following conditions
514
ADVANCED
MATHEMATICS
(a)
v satisfies the adjoint equation, that is to say L*(v) = 0 (6.2-10a)
(b)
along x = constant (Fj) j p = av (6.2-10b)
(c)
along y = constant (F 2 ) | ^ = bv (6.2-10c)
(d)
at the point P v(xo,yo) = 1 (6.2-lOd)
Substituting Equations (6.2-10a to d) into Equation (6.2-9) yields
u(P) = v(Q)u(Q)-J [(auv-u^-)dy-(buv + v|^)dx]+j j vpdxdy
Y A
(6.2-11)
Equation (6.2-11) is not symmetrical in the sense that it only involves the point Q and the derivative 3— but not R and ^—. To obtain a more symmetric expression, we consider the identity
The function u and its first partial derivatives on y are given by Equations (6.2-2a, b), R and Q are on y; that is, u(R) and u(Q) are known, p is given in A, and if the Riemann function v is known, u(P) can be evaluated. The problem of obtaining a solution to Equations (6.2-la, b, 2a, b) has now been transformed to finding the Riemann function v satisfying Equations (6.2-10a to d). Once v is found, u(P) can be obtained from Equation (6.2-14). Note that Equation (6.2-14) can be written in a general form as
u(P) = J K£,P)p($)dS
(6.2-15)
where the kernel K depends on the operator and the boundary conditions, and is independent of p. The Riemann method is similar to the Green's function method [see Equation (1.18-1)]. Note that we have assumed that each characteristic intersects y at no more than one point. This condition is necessary, otherwise the problem might have no solution. Example 6.2-1. Solve the following equation using Riemann's method
Equation (6.2-29) arises in the transmission of electrical impulses in a long cable with distributed capacitance, inductance, and resistance. Note that if a = P = 0 (no loss of current and no resistance), Equation (6.2-26) reduces to the wave equation [Equation (5.1.1)]. We can simplify Equation (6.2-29) by assuming that u is of the form
518
ADVANCED MATHEMATICS
u(x, t) = 0(t) w(x, t) In terms of 6 and w, Equation (6.2-29) becomes
(6.2-31)
(6.2-32)
U2
d*2
at ' at
'
at2
dt
/
Replacing u (x, t) by two functions 9 and w in Equation (6.2-31) allows us to impose an additional condition on say 9 so as to simplify Equation (6.2-32). We choose to set the coefficient of 3w 3t 2d0~ + a 9 = o dt The solution of Equation (6.2-33) is G = C0e-at/2 where C o is a constant. Substituting Equation (6.2-34) into Equation (6.2-32) yields
a w 2a w "! ,
c
We note from Equations (6.2-36a, c) that t = 0 corresponds to £, = TJ. We evaluate the value of w at a point P(£ 0 ,r| 0 ). The point P, the characteristics, and the curve y on which w and its derivatives are prescribed by Equations (6.2-38a to c) are shown in Figure 6.2-2.
V
% p
I
•—y*
/
r, /
/
FIGURE 6.2-2
Characteristics r x and T 2 of the telegraph equation
The Riemann function v has to satisfy Equations (6.2-10a to d) and the operator L defined by Equation (6.2-37) is self adjoint. The function v is a function of £, and r) and depends also on t,0 and r\ 0 . We write v as v (£, rj; £ 0 , r\ 0 ). The characteristics Fj and T2 are given by J^: ^ = ^0 ; T2:
TI=TI0
(6.2-39a,b)
The conditions that v has to satisfy can be written as J^L. + Kv = 0 (6.2-40a)
dfcdTi -[v(^,Tio;^,Tio)] = 0 9^ (6.2-40b)
520
ADVANCED
MATHEMATICS
^[v(£0,Ti; $0,TI0)] = 0
V(^0,TI0; ^0,TI0)
(6.2-40C) (6.2-40d)
= 1
Rather than solving Equation (6.2-40a) right away, we first examine the form that v might take. We expand v about (^0, r\ 0) in a Taylor series and we obtain
Be,
O|l
z
d£,
+ ^ ( T 1 - T 1 O ) 2 ^ - + ( ^ - ^ ) ( T 1 - T V ) ) - ^ - + ...
(6.2-41)
All the derivatives are evaluated at the point (^0,f\0). and 3Tin a^ n - ^ = -KvU _ = -K
Comparing Equations (6.2-61, 5.4-31, 32), we realize that, in the present example, u is damped. • An extension of the method described here to higher order problems is given in Courant and Hilbert (1966). Abbott (1966) has described and solved several problems of engineering interest by the method of characteristics. In Chapter 8, a numerical method for solving hyperbolic equations based on characteristics is presented. 6.3 SIMILARITY SOLUTIONS
In many physical problems, the solution does not depend on the independent variables separately but on a combination of the independent variables. In this case, a similarity solution exists and the existence of a similarity solution is usually associated with the process of diffusion. We describe the method of obtaining a similarity solution by considering a few examples. Example 6.3-1. Look for a similarity solution for the diffusion problem obtained by considering a generalized form of Equation (A.IV-1)
^ = A[D(c)|^|,
dt dx L dxj c(0, t) = c 0 ,
X>0, t>0
c(x, 0) = c 2
(6.3-1)
(6.3-2a,b,c)
lim c (x, t) = c, ,
We introduce a similarity variable by writing
ri = xat^
where a and p are constants. We assume c to be a function of r\ only. Its derivatives are given by ^ dt
=
We now substitute, wherever possible, the combination of x and t by T| and Equation (6.3-5) becomes (6.3-6) Both sides of Equation (6.3-6) are functions of Tj only if (x 2 /t) is a function of TJ. We choose r| to be
Note that the value of a has been chosen so that the first term on the right side of Equation (6.3-6) vanishes. We now check that the initial and boundary conditions are consistent with the choice of TJ. From Equation (6.3-7), we deduce x = 0=»T| = 0, x—>OO=>T|—>oo, t = O=>T|—>«> (6.3-lOatof)
If D is a function of c, Equation (6.3-9) is a non-linear equation and we can solve it numerically. It is generally easier to solve an O.D.E. than a P.D.E. Thus, looking for a similarity solution can simplify the problem. On the curves (TJ = x/Vt = constant), c(r|) is a constant, hence the name
similarity solution.
To simplify the problem, we assume D to be constant and
PARTIAL DIFFERENTIAL EQUATIONS II
521
c\ = c 2 = 0 Equation (6.3-9) becomes
2
(6.3-12a,b)
^4+-l-Ti^ = 0 dr ,2 2D dr| The solution is
(6.3-13) ^ J
P
c = c3
2
exp(-£ /4D)d^ + c 4 (6.3-14)
Jo
where C3 and C4 are constants. Note that the lower limit of integration in Equation (6.3-14) is arbitrary. That is to say, had we integrated between the limits 1 and T|, the constants C3 and C4 would have been adjusted accordingly. However, since the boundary conditions refer to r\ = 0, the obvious choice for the lower limit is zero. Imposing the initial and boundary conditions [Equations (6.3-2a to c, 10a to f, 12a, b)] yields
We assume T| to be of a more general form than Equation (6.3-3) and write Tl = y / $ 0 0 where t, is to be specified later. Equation (6.3-20) suggests that \|/ can be written as V = v 00 (x)^(x)f(Ti) Computing the derivatives of \|/, we have (6.3-23) (6.3-22)
^
dx
=
l^^f+v
dx ^
^
"dx
f + v
4
« ^
=
^
f
+v
^ f - v ^ ^ f
°° dx °° dx drj
(6.3-24a,b)
(6.3-24c,d)
°°^ dr| 3x = ^ ^ dx dr] ^
dx ^
^ = v ^ , 3y °°dr| '
^ dxdy
^ ^ 1 £ dx dr| 2
?!?.^i!f,
3y 2 ^ dri2
^
3y 3
=^4
£ &x\
(6.3-24e,f,
Substituting Equations (6.3-24a to f) into Equation (6.3-20), we obtain after some simplification
i!l +fejL(gv ) f ^ I + ( ^ ^ ) h - f f f] = 0
dr)3
(6.3-25)
L v dx
vs
°°;
dr|2
V v dx /[
Ur|j J
For the existence of similarity solutions, we require that |^(^vj = a (6.3-26a)
PARTIAL DIFFERENTIAL EQUATIONS II
527
V IT=P
where a and P are constants. Equation (6.3-25) can now be written as
<6'3'26b)
i!l + af4+P[l-(£fl = °
dr(3 dT]2 L ldTl/J
(6-3-27)
Equation (6.3-27) is the Falker Skan equation. Various values of a and (3 are associated with various outer flow situations. The following cases have been examined. Blasius Flow In this case, we choose a =1 p = 0 (6.3-28a,b)
where the prime denotes differentiation with respect to T|. Equation (6.3-30) is a third order non-linear O.D.E. and its solution, subject to the three boundary conditions, can be obtained numerically. Stagnation Flow Substituting the values of a = 1, p = 1 (6.3-32a,b)
The system defined by Equations (6.3-35, 36a to c) can be solved numerically. The solution of various flows is discussed in Rosenhead (1963). Example 6.3-3. Stefan problems are concerned with the phase change across a moving boundary. We consider the phase change from a solid to a liquid. The boundary between the solid [x < s (t)] and the liquid [x > s (t)] is given by x = s(t) and is shown in Figure 6.3-1. We assume that there is no motion and that the temperature T, in both phases, satisfies a simplified form of Equation (A.III-1) (6.3-37)
pc? I = k^ 1
dt 9xz
(6.3-38)
where p is the density, c is the specific heat, and k is the thermal conductivity. We assume that they are constants and have the same values in both phases. At the interface, the temperature is at the melting point T m .
3-46a. For the interface. 45a. and is written as
^ + J . b).3-47)
(6.E..b)
Note that the same TJ was obtained in Example 6.3-43a.3-48)
dr| 2
2D
dr)
where D = k/pc.D.
£2.3-42).
t a 2 /ai
T = —3—
t a 3 /a]
(6.3-38) is reduced to an O. We introduce a set of new variables defined by Ti = —^—.3-43a.3-44)
One possible solution is
^
= 0. = 1
(6. The interface (x = s) is now given by r\ = s 0 The conditions at the interface are (6.530
ADVANCED MATHEMATICS
We now combine the variables x. and T in such a way as to reduce the number of variables.b)
The ratios OC2/CC1 and aL^/CLy can be obtained from Equation (6. leading to 1_^= a1 2^-21 a{ a j (6. we assume that s is given by s = s o ff where s0 is a constant.T I ^ = 0
(6.T m Equation (6.3-45a. We define f(Tl) = T .3-50) (6.3-49)
. x= T (6. b. t. we obtain T\ = -X.3-1.b)
Combining Equations (6.
52a) and f becomes
(6. These methods can be extended to more than two independent variables.b)
Equations (6.3-55). b) imply that.55)
The interface (s0) is obtained by solving Equation (6.3-5 lb) yields r| > s0 (6. we have the liquid phase and the temperature is at the melting point (f = 0).3-54a)
JoSo e"^ /4D d$
f = 0.PARTIAL DIFFERENTIAL EQUATIONS II
531
f(s») = o'
ffl!.3-53)
P e-^/4Dd^
f=_l+l2 . The solution of Equation (6. we have introduced three methods of obtaining similarity variables. at x = 0. 0<"n<s 0 (6. (6. the temperature is one degree below the melting point and beyond the interface.
. Crank (1984) has discussed Stefan problems in detail. To complete the problem. In this section.2-52a.3-54b)
L!oesJ/4D[S°e-^/4D^= j 2k Jo
(6>3.Imposing Equation (6.3-5la.3-48) is
f = CJ + CQ f e-^ / 4 D d^
JO The constants C o and Cj are determined from Equations (6.1 . we need to define the domain which we assume to be semi-infinite (0 < x < oo). Further details can be found in Ames (1972) and in Bluman and Cole (1974). The additional boundary conditions we impose are f(0) = ./-1^
lim fCn) = 0
T)
>oo
( 6 -3-5ia-b)
where sQ is in the solid phase and s 0 is in the liquid phase.3-52a.
it is stated that G has to satisfy the following conditions (a) V G = 0. The conditions which G has to satisfy in the case of ordinary differential equations are stated in Chapter 1. Note that some authors include the factors (1/471. that the solution of Poisson's equation is known if the Green's function for that problem is available.4-la) (6. z ) . We now have to compute G. y). In particular. u = f (x. if V 2 u = p (x. 4).4-lb)
where S is the surface enclosing V. 1/2TC) in the definition of G. y).4-5a)
. it is shown. In Chapter 4.4-3a) (6.532
ADVANCED
MATHEMATICS
6. y. the Dirichlet problem defined by V 2 u = p (x. as an application of Green's theorem.4-2. These conditions can be extended to partial differential equations. in a plane.4-3b)
where C is the curve enclosing the domain A. in V on S (6.
U(P) =
2K \\ G P d A + I f | f
-A C
ds
(64"4)
G is the two-dimensional Green's function. u = f(x. These factors are then absent in Equations (6. This method is widely used to solve elliptic partial differential equations.4
GREEN'S FUNCTIONS
Dirichlet Problems
Green's functions were used to solve boundary value problems in Chapter 1. Similarly. z) . In Chapter 4. in A on C (6. everywhere in V except at P (6. y. The value of u at a point P is given by
u < P )= ^[///GpdV+//flM
_V S
(6-4-2>
where G is the Green's function for the problem.
x o ) 2 + ( y . G is symmetric. y 0 . G behaves as the fundamental solution of Laplace's equation [Equations (5. except for simple domains such as a half space or a circle. The function G is the potential of a unit charge placed at (x0. The Green's function G (x. y. Upper half plane Consider the upper half plane z > 0.dS = An
se
I y^. x.4-1 lb) (6. The potential at (x. x 0 . To maintain the plane z = 0 at zero potential. z) due to a unit charge at (x 0 .y o ) 2 + ( z + z 0 ) 2 In the two-dimensional case (y > 0). ZQ) and the plane z = 0 is kept at zero potential. y0. y 0 . y0. It is generally not easy to calculate G. Since the Laplacian is self-adjoint. that is G (x.4-10)
This can be deduced from physical considerations.ds = 27t
(6.4-1 lc)
.4-5a' or 9a). z) (6. Each is of radius e with e —> 0. .z 0 ) . y. y0. z 0 . zQ) is given by G = -i--1 r r r = V(x-xo)2+(y-yo)2+(z-zo)2 r' = V ( x .4-5c')
(6.4-9c')
where S e and C e are respectively the spherical surface and circumference that enclose the point P. y. y0. z 0 ) = G (x 0 . We next deduce G for simple geometries by the method of images.534
ADVANCED MATHEMATICS
where r is the radial distance from P. z). 39)] and G satisfies Equations (6. z. Condition (c) for the three and two-dimensional cases can respectively be replaced by
I I ^p. x0. y. we place a unit charge of opposite sign at (x0. z. y.5-13b. y 0 .4-1 la) (6. G is given by (6. Note that near P. z 0 ) due to a unit charge at (x. zQ) is the same as the potential at (xQ.
and P' can be written respectively as OQ = r[sin9 cos()). P. On introducing spherical polar coordinates (r.4-1 If Q is any point. The inverse point P' is defined such that OP* OP' = a2 (6. the vector positions of Q. as shown in Figure 6.4-12)
Consider a sphere of radius a with a unit negative charge placed at P. 0. ())).4-1. cos0] (5.4-15a)
.4-14)
r2
where rj and r 2 are the distances from Q to P' and P respectively. The point P is at a distance b from O. sinG sin(J). the centre of the sphere. G is given by G = ±
Image system for a sphere and a circle
+ v^brj
(6.PARTIAL DIFFERENTIAL EQUATIONS II
5£5_
G = ^in[(x-x o ) 2 +(y-y o ) 2 ]-i-in[(x-x o ) 2 +(y + yo)2]
Sphere and circle
(6.4-13)
o
p
1
P1
FIGURE 6. The image system that produces a zero potential on the surface of the sphere is a charge a/b placed at the inverse point P' of P.
everywhere in V except at P
(6. At steady state. the temperature satisfies Equation (5. modifying the boundary condition is not sufficient and we need to impose additional conditions.4-40). By definition V G = 0 everywhere except at P. it is stated that the solution to a Neumann problem is arbitrary to the extent of a constant. We impose the following conditions on the Green's function for the Neumann problem (N)
2
(a)
V N = A (constant).7-3.4-40)
Equation (6. One possible physical interpretation is to consider the problem of heat conduction. the boundary condition must satisfy the left side of Equation (6.4-38). denote the surface of this sphere by S e and the volume by V e . n = 0) leading to the impossibility of determining c m n . In Example 5.4-1 to determine G. in Example 6-4-3. the function G2 will not be uniquely defined. It follows that. One possibility is to modify condition (a) [Equation (6.4-4lc) (6.4-5a or 9a)]. implies
J J ^ . Enclose P by a sphere of radius e. Thus.PARTIAL DIFFERENTIAL EQUATIONS II
547
//^ dS = /// v 2 ° d V
S V 2
<S4-38)
Let P be the singular point. for a Neumann problem the eigenfunctions V mn are of the form cosnmx cosnfty and one of the eigenvalues 'kmn can be zero (m = 0. the heat flux across the boundary must be zero and this is expressed by the left side of Equation (6.4-41b) (6.4-38) where G is the temperature. if we use the method adopted in Example 6. from Equation (6. Also.4-38). V G is zero and. we obtain (6.4-5c') contradicts Equation (6.nearP
. on S on N--l/r.4-41d)
= A + 4TT8(X-£0) (b) (c) ^ = 0 . In the volume V .4-44) which is Laplace's equation. To maintain a steady temperature.V e .dS = 0
(6. Note that if G satisfies Laplace's equation everywhere in V.4-4la) (6.4-39)
2
s
se
Imposing the condition 3G/8n = 0 on S.
PARTIAL DIFFERENTIAL EQUATIONS II
543. In order for the fluid not to cross the line y = 0. in hydrodynamics.x 0 ) 2 + (y .4-47)
Neumann problems are usually associated with hydrodynamics where the usual boundary condition is that the normal velocity (3u /dn.x 0 ) 2 + (y + y 0 ) 2 ] On differentiating partially with respect to y. Example 6.4-46) in two-dimensions is
u(P) = -L I I Np dS . u is the potential) at the boundary is zero.I Nh ds
_S C
(6.4-4.
where the constant term is A Jjj
V
u d V.
. This is the difference between hydrodynamics and electrostatics. This term can be ignored since the solution to a Neumann
problem is arbitrary to the extent of a constant. 1967) N = 1 i n [(x .4_49)
=0 y=0
(6. If N is the potential.4-46) and can be calculated once N is known. Further discussions on the image system in hydrodynamics can be found in Milne-Thomson (1967). We illustrate this method in the next example.x 0 ) 2 + (y . we obtain (6. The analog of Equation (6. b) is given by Equation (6.4-44a. Find the Neumann function N for the plane y > 0.y 0 ) 2
+
(y±zo)
(x . the image of a source is another source and not a sink. Neumann's functions can also be constructed by the method of eigenfunction expansion as explained in the next example. A source is placed at (x 0 .4-48b)
^L =
dy It follows that ^
dy
<±zi&
(x .x 0 ) 2 + (y + y 0 ) 2 ] = 1 i n [(x .4-48a) (6. N is given by (Milne-Thomson. The solution of Equations (6.x 0 ) 2 + (y + y 0 ) 2
(6. We can determine N by the method of images. y 0 ).x 0 ) 2 + (y .y 0 ) 2 ] + 1 i n [(x . we need to place another source of equal strength at (x 0 .4-50)
• Note that.-y 0 ).y 0 ) 2 ] [(x .
61) yields
(6.
ADVANCED
MATHEMATICS
Q
P
y0
—i
1
^
~XQ
XQ
X
R
y°
s
FIGURE 6. That is to say.4-60)
Joo
Note that the contour integral is taken in the counter clockwise sense.4-61)
. so that the integral along the y-axis is from °° to 0. On the x-axis
f = -~
an ay Combining Equations (6.4-2
Image system of a mixed boundary value problem
From Equations (6.4-57a to d).546
.4-60.4-59)
where C is the contour enclosing the positive quadrant. we deduce that u (P) is given by
u (P) = ^
[ (f H . along the positive x and y axes. From Equations (6.4-59) can be written as u(P) = J 2K
f^-dx-l
JO
dn
Ghdy
(6.Gh) ds c
(6. we deduce that Equation (6.47).4-4.
x 20 ) = G(x 1 0 . At about the same time. It was the mathematician Jordan who saw the connection between the theory proposed by Heisenberg and the theory of matrices. x 2 . x 2 0 . x 1 0 . It has led to a number of applications. 10) can be replaced by G via Equation (6.5-9. His theory is based on the possibility that matter can have particle and wave properties. electron microscopes. and elliptic equations but the method involving Riemann's functions is restricted to hyperbolic equations. To formulate the laws of quantum mechanics on an acceptable basis.5-15)
The function G in Equations (6. have to be applied. from an apparently different starting point and using the classical continuous formalism. both G and G* are equal. New devices based on the principles of quantum mechanics are still being developed. such as lasers. G is symmetric in (xi. de Broglie suggested that
. The method of Green's functions is applicable to hyperbolic.
The importance of quantum mechanics needs no stressing. Newton's laws have to be replaced by the laws of quantum
mechanics. They fail to describe the motion of bodies traveling at high speeds. 6. silicon chips. at speeds approaching the velocity of light. Schrodinger.5-11) and.PARTIAL DIFFERENTIAL EQUATIONS II
557
G ( x 1 . For micro particles. in selfadjoint cases.6 QUANTUM MECHANICS
Limitations of Newtonian Mechanics Newton's law of motion are adequate to describe the motion of bodies on a macroscopic scale usually encountered in everyday life. Note that the heat (diffusion) equation is not self-adjoint whereas the Laplace and wave equations are self-adjoint. parabolic.x 2 ) and (xio>x2o)-
(6. he was not aware of the existence of the matrices. provided an alternative foundation for quantum mechanics. This led to the publication of the famous Born-Heisenberg-Jordan (1926) paper. the laws of relativity. such as atoms and their constituents. Heisenberg proposed the theory of matrix mechanics. x 2 ) that is to say. x 1 . Greenberg (1971). that is to say. In 1924. and Zauderer (1983). Only hyperbolic equations have two real characteristics. though at that time. and non-linear optics. two approaches were adopted in the 1920's. It provides explanations for a variety of physical phenomena ranging from the structure of the atoms to the beginning of the universe. Courant and Hilbert (1966). Further details on Green's functions can be found in Morse and Feshbach (1953). as developed by Einstein at the turn of the twentieth century. In this case.
6-3)
Schrodinger Equation It was stated earlier that Schrodinger based his theory on the possibility of matter having both particle and wave properties. < & is the potential energy. the wave properties have to be considered for microscopic particles only. The function \j/ is normalized such that
.b)
(ii)
Heisenberg's uncertainty principle It is not possible to simultaneously measure the momentum p and the position x to arbitrary precision. t ) | the probability density of the particle being at x at time t.6-1)
(6. has to be small and the wave properties can be neglected. The wave is
function \|/ is complex and the observable quantity is not \|/ but |\|/| .T T ~ V2V|/ + Oi)/ = i h . (i) Electromagnetic energy occurs in discrete quantities. t) obeys a diffusion equation which can be written as . He proposed that the wave function V|/(x. for macroscopic particles (p is large). the matrix and the wave mechanics were thought to be unrelated until Schrodinger showed the equivalence of the two formalisms.^ 2m at
2
(6. Thus.6-2a. The relationship between the momentum p of the particle and the wavelength A. we can deduce that. h = h/2n. For a short period. ApAx = h (6.6-4)
where m is the mass of the particle. The quantity | \ | / ( x .552
ADVANCED
MATHEMATICS
atomic particles might behave as waves. the wavelength A. If Ap and Ax are the errors in the measurements of p and x respectively. The energy E is related to the frequency v by E = h v = h co
O = 2TZV. From Equation (6. We state some of the basic assumptions of quantum mechanics. G
(6. of the "particle wave" is p = h/X where h is Planck's constant.6-1). where h is the Planck's constant. and V
is the Laplacian.
2
2
U+
]
u ["25T V
° ( 2 ° U'
=
^ dt
= E
(6. We write \|/(x. the solution is of the form given by (6. That is to say.6-6)
\-t.t)|2dx = 1
V
(6.— V u + Ou = Eu 2m f = C e-[Ei/n where C is a constant. The wave function \|/(x.6-4) and assuming a static potential O (x). Equally. does not have a visible orbit as a particle but its presence can be detected when it hits a target and acts as a particle. Its position is given by a probability density and it is this probability density which is associated with a wave.6-8a) is the time-independent Schrodinger equation.b) (6.t) depends on the potential <P and if O is time independent. Equation (6.6-9) (6.6-7a. Equation (6. The property which is manifested depends on the situation under consideration.6-8b)
. which is an electromagnetic wave.6-8a) (6. b) lead to -h2 2 . a photon. we have (6. Note that it is not claimed that the particle has become a wave but rather that its position is not known with certainty. Equations (6.6-10a. \}/ can be considered to be a wave of probabilities.b)
where E is a constant representing the total energy (kinetic + potential).t) = u ( x ) f ( t ) Substituting \|/ into Equation (6.6-7a.6-5)
where V is the whole space. t) can be written as V|/ = u (x) e-iEt/n The probability density P r is given by P r = |v|/| 2 = | u ( x ) | 2 and is independent of time.PARTIAL DIFFERENTIAL EQUATIONS II
553
| |v|/(x. Thus. we have a coexistence of wave and particle properties. These waves are abstract waves in the same sense that crime waves are abstract waves. The solution \|/(x.6-4) can be solved by the method of separation of variables.
5-38).6-43)
(iii)
If n > i .t i l l d .42)
(ii)
P7n = ( -l) n H=Jl21p"
* (i+n)! is a multiple of PJ? and we need to consider only positive n. For H to be single valued. Equation (6. If n is non-zero.x2)] G = 0
dx L dxJ
(6. n has to be an integer.7-11)].6-38b)
where n is another separation constant.6-40) is the associated Legendre equation [Equation (2.2 AJ^_[ix2_l)*]
2* (i!) We note the following properties of PJ? (i) n and Jt are integers.
The function F can now be written as
.6-40)
If n is zero (axial symmetry).6-38b) is H = Aein* + Be-in* where A and B are arbitrary constants.7-1)] and the solution is given by Equation (5.6-39)
-f. [Note that the term inside the square bracket is of order 2/2 and its (/6 + n) derivative is zero if n > i ] . dxi +n
(66.6-40) is the standard Legendre equation [Equation (2. The solution of Equation (6.7-33)] and the solution can be written as G(x) = CP{(x) where C is a constant.[d .n 2 /(l .] + [i (i + 1) .x2) iS. This shows that P»
(6. Z is positive [see Equation (2.x 2 ) " . The associated Legendre functions PJ? (x) are defined by (6.558 2
ADVANCED MATHEMATICS
^ + n dC
2
H
= 0
(6.6-41)
Pj?(x) = .6-38a) is transformed to the standard form by setting x = cos 0 and it becomes (6. Equation (6. Equation (6.6-42) that PJ? is zero. we deduce from Equation (6.
Since O is time independent and is discontinuous. Equation (6. We next consider an example which leads to an application.560
ADVANCED MATHEMATICS
the energy level observed in high resolution measurements can be accounted for by including the spin effects.6-55b) (6.6-8a) in the regions x < 0 and x > 0 separately. the solution of Equation (6.E) Ih (6.
x<0
x>0
(6.6-52a.2
<\2
= E u !
x < 0
(6.6-53b) can be written as u = Ce^ 2 X + D e ^ 2 X where C and D are constants and \^ is given by %2 = V2m (O0 .6-3. In this example.6-54a)
where A and B are constants and A-j is given by Xj = (V2mE )lh Since O 0 > 0.^ — ^ + %u = E\x.6-8a) can be written for the given O as _«_il 2m dx 2
ft2
t. 2m dx 2 The solution of Equation (6. The conditions requiring u and du / dx to be continuous at the origin yield the following
.6-54b)
The condition that u tends to zero as x tends to infinity implies that C is zero. the potential has a jump discontinuity at the origin. we need to solve Equation (6.6-53b)
(6. Solve the one-dimensional Schrodinger equation for a potential given by O(x) =
f0.6-53a)
H2
.b)
where O 0 (constant) is greater than the total energy E. Example 6.6-55a) (6. This potential function can represent the motion of a charged particle between two electrodes kept at different voltages.
\%. Interested readers can consult Eisberg and Resnick (1985).6-53a) is u = AeiXl" + B e " a i X
x>0
(6.
This phenomenon can be observed experimentally and has found an application in the tunnel diode which is used in modern electronics.6-56a.b)
In quantum mechanics.b) (6. we obtain A = -£. there is a finite probability that the particle may penetrate the classically excluded region and this phenomenon is called penetration (tunneling).V 2 m ( O 0 .6-61) (6. it is not possible to state that O 0 is definitely greater than E and that the particle cannot enter the region x > 0. if the potential energy is greater that E.
.2 « / i / V 2 m ( O 0 .-iA.6-60a. For small values of x. b).1/A.6-56a) (6. For details.b)
The probability P r decreases rapidly with increasing x.6-55b. we deduce from Equations (6. If the depth of penetration is Ax.O 0 . The probability P r of finding the particle in the region x > 0 is Pr = | u | 2 = | D | 2 e " 2 ^ x (6. The particle does not enter the region x > 0.E (6. 58a.6-58a.6-56b)
(6. there is a finite probability of finding a particle in this region.PARTIAL DIFFERENTIAL EQUATIONS II
561
A+B = D i?4(A-B) = . That is to say.? i 2 D On solving Equations (6.b)
The constant D can be obtained by applying the normalization condition.6-3) and is Ap = h/Ax . Note that the statement O 0 is greater than E should be subjected to the uncertainty principle. the kinetic energy must be negative which is physically impossible.(Xl + i X2).E ) The uncertainty in the momentum Ap can be obtained from Equation (6.6-57a. see Eisberg and Resnick (1985). b) that Ax .E ) The total energy E is proportional to p 2 /2m and it follows that AE . This is not predicted in classical mechanics where the total energy E is constant. In quantum mechanics. 2Xl B = -E-(A..6-59a. 2 ) 2XX
(6.
1-1)
In Equation (7. mathematical modelling is now more realistic and a new field of numerical simulation has been opened up.150) is correct to at most one significant figure.1 INTRODUCTION
Many engineering problems can not be solved exactly by analytical techniques.149 = 0. We will look. f (0. This type of error is known as loss of significance. We can now use iterative methods with much greater speed. for example. at Newton's method of approximating the solution of an algebraic or transcendental equation.150 radians. at the Gaussian elimination method for solving linear systems of equations and at the Euler and Runge-Kutta methods for solving initial-value problems. in one form or another. Since the arrival of computers.CHAPTER 7
NUMERICAL METHODS
7. Many non-linear equations which were intractable in the past can now be solved approximately by numerical methods.
. In some cases.150-0. the potential of these methods has been realized and the entire character of numerical methods has changed. This type of error is known as round off error and is unavoidable. the number of significant digits will be reduced. the value of sin (0. We first briefly examine the possible sources of error that may occur in the solutions obtained by numerical methods. Losing all significant figures during machine computations is not unheard of.150) is correct to three significant figures but f (0. Numerical methods. In this chapter. If we subtract two almost equal numbers. As a result of these possibilities.s i n x We now evaluate f(x) when x is equal to 0. We can also solve large systems of equations numerically. The knowledge of numerical methodology is essential for determining approximate solutions. we present different methods which will be useful to applied science students.1-2a) (7.001 (7. we generate errors. have been studied for several centuries. It is therefore necessary to check the numbers during the calculations. Most numbers have an infinite decimal representation and for computational purposes they have to be rounded to a finite number of decimal places.1-2b) (7. During the process of calculation. Successful simulations of complex processes in science and engineering are now being achieved. we consider the function f (x) given by f(x) = x . As an example.1-2a).150) = 0.
1-3b)
The value of sin (0. 7. We need to establish that the sequence converges to the solution. In computational mathematics.000562 (7. In this sense.1-4) gives a better approximation to f (0.^x3 o Using Equation (7. numerical analysis is an art and not a science.1-3a) (7. It is often possible to provide an estimate of such errors..14944 and it can be seen that Equation (7. Alternatively we will also say that x is a zero of the function f. we can consider only finite sums. As mentioned earlier.2-1) for a given f. we consider infinite processes.I X 3 + . iterative methods are used to solve equations numerically. This equation could be algebraic or transcendental.2-1). For example.2 SOLUTIONS OF EQUATIONS IN ONE VARIABLE
We begin our discussion by considering equations of the form f(x) = 0 (7.2-1)
where x and f (x) are real or complex.x .( x . If x is a real or complex number satisfying Equation (7. In classical analysis.
. This results in an error called the truncation error. We need to find values of the variable x that satisfy Equation (7.150) correct to five significant figures is 0. we can approximate sin x by its Taylor series and f (x) is approximated as f(x) . Ideally we should always perform an error analysis before accepting a particular solution which was generated numerically. Next we present four different methods of solving Equation (7. we can rely on intuition and past experience and hope that the results obtained are reliable and useful. In such a case.1-4) (7.1-2b). ) o .150) than Equation (7.2-1).1-3a).570
ADVANCED
MATHEMATICS
it is possible to rewrite the function to be evaluated in such a way that the loss of significant figures does not occur.150) = 0. such as infinite sums. we say that x is a root of the equation. A more detailed discussion on error analysis can be found in Hildebrand (1956) and Elden and Wittmeyer-Koch (1990). Often this is not possible as the problem is too complicated. These methods generate a sequence of numbers. we find that f (0..
with f(a) and f(b) having opposite signs. the required root lies in the interval (a. If f(xj) and f(a) have the same sign. One starts by letting Xj be the mid-point of [a.
f(a) ^ r ^
f*x. b). The method consists of a repeated halving of the interval [a. there exists at least one number x such that a < x < b for which f (x) = 0. b] with f(a) and f(b) having opposite signs. b]. defined on the closed interval [a. as illustrated in Figure 7.b). We then proceed to find the mid-point of (xj.2-1
Identification of the root (•) of the equation via the bisection method
. xj) and in this case x2 is the mid-point of (a. If xj satisfies Equation (7. b].2-1). until the difference between two consecutive values of xj is within the required accuracy. f(xj) has the same sign as either f(a) or f(b). the required root lies in the interval (xj. If not. We repeat this process of halving until a satisfactory value of the root is obtained.)-T--V 0 -J L \
a x.NUMERICAL METHODS
£77
Bisection Method (Internal Halving Method) This method is based upon the use of the intermediate mean-value theorem which states that for a continuous function f. Similarly if f (xj) and f (b) have the same sign. xj is the required root. xj). That is to say.
f(x)
i.2-1. \
nr-*
jb x
FIGURE 7.
1 0 = 0 (7. f (xj)].365. TABLE 7.5. Note that as k increases the difference x k .3651 f(xk) 2. The equation of the secant line is
.2-1 Values of x^ and f(x^) k 1 2 3 4 5 6 7 8 9 10 11 12 13 xk 1. The values of x k and f (x k ) are given in Table 7. In this method. at least one root lies in the interval (1.37 -1. we see that the root accurate to three decimal places is 1.3437 1.3652 1. It is not necessary that x 0 < Xj or that f(xQ) and f(xj) have different signs.2) and x1? the mid-point of (1.3655 1.162 -0.3671 1. x j .3632 1.b)
We note that f (1) is negative (-5) and that f (2) is positive (14).25 1.2-2a.029 -0. f (X])] which will intersect the x-axis at some point X2. Secant Method Consider the graph of the function f as shown in Figure 7.848 -0.2-1. That is to say. Solve the following equation via the bisection method f(x) = x 3 + 4 x 2 .048 -0. we consider two points [xQ.5.797 0.0038 0.x k _ 1 decreases and f(x k ) approaches zero.102 0. f (x 0 )] and [xj.3125 1.5) which is found to be positive (2.2).0127 0.2-1.25.5 1. We draw the secant line. is 1.0045 0.37) and the root lies between 1 and 1.375 1.572
ADVANCED
MATHEMATICS
Example 7.351 -0. it is not necessary that the root lies in the interval [xQ.0003 -0. The second approximation X2 is 1.3661 1.0017
From Table 7.3593 1. We continue this process.3651 1. We calculate f (1. Thus. that is the line passing through points [x 0 .2-1 for various values of k. f (x 0 )] and [xj.2-2.
2-31) do not converge to the root.2-32c) (7.365 in ten iterations. However.2 .484 x 4 = 0. the iteration formula is (7. is greater than one and condition (iii) is violated.2-6 ten iterations are needed to achieve the same accuracy. computed via Equation (7.365) but they are oscillating. Other forms of g(x) can be chosen.30 )
In this case.2-32a to d). g(x).621 (7. We now examine the properties of the iteration function g (x). we obtain xi = 2 x 2 = 0. For Newton's method. in the interval (1.2-28) converges to the solution 1.2-34). An alternative form for Equation (7.2-31) Again.2-25) is given by
g(x) =
2xi ± ixi ± I 0
3x 2 + 8x
(? 2 3 5 )
It is shown in Example 7. •
. it is faster to compute g (x) as given in Equation (7.2-3 that the root is obtained after three iterations whereas in Example 7. The sequences generated by Equation (7. we find that | g'(x)|.2-32a) (7. we note that the x^ are not converging to the solution (1.2-2) is
X
= ^xT4)
< 7 .2-27) compared to that in Equation (7. 1. We can verify that g (x) satisfies the conditions for the convergence of the iteration.5).2-35). starting with an initial value equal to 1.NUMERICAL METHODS
_ _ _ _
579
We note that the iteration given by Equation (7. The function g (x) and g'(x) are given by
g(x)
g-00
= roTT4)
= ~^±^
(x2 + 4x) 2
<7-2-33>
(7 .2-32b) (7. 2 _34)
From Equation (7.833 x 3 = 2.2-32d)
From Equations (7.
.3-la. 1) 2) 3) 4) There are n (not necessarily distinct) real or complex roots. 4) are compatible provided V i = an (7.3-5a) (7.! x""1 + . then necessarily f(x) = ( x .b) with real or complex coefficients.. We note the following facts regarding Equations (7.580
ADVANCED MATHEMATICS
For human computers. Newton's method is used for faster convergence.3-la. n > 0.x o ) q ( x ) + R where q (x) = bn_! x""1 + bn_2 xn"2 + ...b)
Newton's Method
To use Newton's method for computing the root of an equation [Equation (7. 3. they occur as conjugate pairs. for calculating f (xfc) and f' (x k ) where f is a polynomial. Equations (7. + aj x + a0 = 0 where a n * 0. If xQ is a root of Equation (7.3-la. (7. known as Horner's method.x o ) g ( x ) where g (x) is a polynomial of degree (n . 7.b).3-3)
.3-4) (7. If all the coefficients are real and complex roots exist. Once a reasonably good approximation is obtained.1). b. Newton's method should be avoided if f' (x) is close to zero near the root.b) we can write f(x) = ( x . To avoid dividing by zero. + t>! x + b 0 and R is a constant. We want to find the roots (real or complex) of the polynomial equation f (x) = a n xn + a n .3-2) (7.3-la. Recall that for any polynomial as given in Equations (7. If n is odd and all coefficients are real. We describe a method.2-11)].3 POLYNOMIAL EQUATIONS
We now apply the discussion of the previous section to the special case where f (x) is a polynomial of degree n.3-1 a. we need to evaluate f(xjc) and f'(x k ). there is at least one real root. it is recommended to start with the bisection or secant method.
we consider the reciprocal variable — .3-21a.b)
As previously noted.
i
Xj
Example 7. and consequently Equation (7. we find f (0.b) (7. f(0) = 6 f (0) = 16 (7. 1].NUMERICAL METHODS
583
We now briefly outline a method for locating intervals containing the zeros of Equation (7. It thus follows that to approximately determine all the real zeros of Equation (7. it suffices to search in the intervals [-1.°°). For example. Furthermore if x is a zero of f(x).0).1].3-17c)
= ^k+an-lx = -Lf(x) x
-
+ a l x n " 1 + a O^]
We note that changing the variable x to 1/x allows us to look for solutions on the intervals [-1. x ^ 0.3-22a)
.l ) = 47.0) and (0.3-19a.3-18 to 21b). »). and x. From Equations (7. Similarly if f (xp f (xA < 0. 1/x is a zero of f. the zeros of f are in the interval (0. f (_1) = 47. 1 e (-oo.3-3.1] evaluating both
•s
f and f to determine sign changes.3-1) has a zero between -L and —.b)
[ 1. f has a zero between x. Also
f (i) = a n(i) n +
Vl(x-r1 + . in order to consider the finite intervals [-1.3-17a) (7.3-20a. and if x e [-1.3-17b) (7.125 (7. The converse is also true.e [1. Locate the roots of the polynomial equation f(x) = 16x 4 -40x 3 + 5x2 + 20x + 6 = 0 (7.0) and (0. we deduce that there is no change of sign in the interval — Consequently. Recall that if f(xj) f (xp < 0.b) (7.3-1) where f(x) is a polynomial of degree greater than 2.-l).3-18a.3-1).^ 0
+
(7. to identify sign changes.1]. 0).0) and (0.5) = 0. f has a zero between Xj and Xj.40x + 16 We note that f ( .1] instead of on the original interval (-<». To determine them we evaluate f at several points. f (1) = f (X) = 6x4 + 20x3 + 5x2 . J. We note that if x € (0.
.5M
ADVANCED
MATHEMATICS
f (0. 2 .4-2b) Equation
where A = (ay) is the coefficient matrix.1). 7.5. —L) = ( — .4-la. 3 .4-2a)
or
Ax = b
(7. + a l n x n = bj
a21 x l + a 22 X2 +
-
+ a2n x n = b 2
(7.8516 f (1) = 7
(7. The quartic can be written as a product of two quadratic expressions...b.75) = -0.75) and at least one other in the interval (0. ^ ) .c) is
n
Xaijxj
i=i
=b i '
i = l .
(7.We have identified two approximate roots. 0. .b. The rows of A are linearly independent.4-2b) has a unique solution if the inverse of A exists.3-22c)
Thus f has at least one zero in the interval (0. 2) and at least one other in the interval ( l . n
(7. The rank of A is n.c)
anlxl+an2x2
+-
+ a nn x n
= bn
A more compact form of writing Equations (7.75.4 SIMULTANEOUS LINEAR EQUATIONS
We now describe methods of solving systems of linear equations which are written as a n x 1 + a 1 2 x 2 + . 1) 2) 3) 4) 5) The determinant of A is not zero.4-la.
. One of them is known and the other is obtained by division. The homogenous system (A x = 0) has only the trivial solution (x = 0). The columns of A are linearly independent. It then follows that f has at least one zero in the interval l—^— . The conditions for the existence of A"1 can be stated in the following alternate forms..3-22b) (7. . and x and b are column matrices.
. . x2. We then choose the new coefficient of x 2 in the second row as pivot and reduce the coefficient of x 2 to zero in row three as well as in the subsequent rows.
Gaussian Elimination Method The basic idea behind this method is to convert the n x n coefficient matrix of the given system to an upper triangular matrix. multiplying a row by a non-zero constant. We then operate on the augmented matrix A g .. > v n } are linearly independent. Vectors {v 2 . v n } is a set of vectors in Rn. The row operations described earlier are to be applied to column b as well.. v 2 .4-1.
If A" exists. adding a multiple of one row to another row. where b is the (n+l) th column of A g . Similarly we reduce the coefficients of xj in all subsequent rows to zero. We note that the following operations on a system of linear equations do not change the solution of the system 1) 2) 3) interchanging two rows. We continue this process until we have only one unknown xn in the n th row. . v n } span Rn. We reduce the matrix A to a triangular form by the following elementary operations.. Vectors {vj. v 2 . It is economical to consider the matrix A and the column b simultaneously.j. we can determine the solutions for x n . otherwise A is said to be a singular matrix.
We now discuss a standard method of solving the set of Equations (7. . Recall also that if {vj. We choose a n as a pivot and keep the first row unchanged. We multiply the first row by an appropriate factor and add it to the second row so as to reduce the coefficient of xi in the second row to zero. The matrix A g is now a rectangular matrix with n rows and (n+1) columns.. v 2 .4-1). xj. -•• . we say that the matrix A is non-singular. By backward substitution.4-2b) has a unique solution iff A is non-singular. Solve the following system of equations via the Gaussian elimination method. v n } form a basis for Rn. . which is obtained by combining A and b .
. Thus the linear system (7. Example 7.NUMERICAL METHODS
585. . We can solve for xn.. the following statements are equivalent. 1) 2) 3) Vectors {vj. .. . v 2 ..
9
and this allows us to determine X3. a pivot row must be selected.9 . However.4-5b)
If. we would not be able to proceed.4-4b)
(7. That is. since the coefficient a 22 is zero.4.4-4c)
(7. x 2 and Xj provided the vector b is known.9 -9 -10 4 " (row 2 .4-5b) as "1 A(2) = 2 4 " (7.5 x row 1) (7. interchanging rows 2 and 3 allows us to eliminate the coefficient of x 2 in such a way as to proceed towards the computation of the required triangular matrix.4-4d)
The following observations concerning this method are important. we were to choose the second row as the pivot row.4-5a)
Choosing the first row as a pivot row yields " A(1) = 1 0 0 2 0 .r o w 1) (row 3 . At each stage in the Gaussian elimination procedure.1 0 0 0 . Appropriate multiples of the pivot row are added to other rows to perform the desired elimination. we write Equation (7.5c)
0 .
. We illustrate a problem which could arise by considering the following non-singular matrix 1 A = 1 5 2 2 .NUMERICAL METHODS
587
X2 = . in the usual manner. The selection of this pivot row is very important.1 Substituting the values of x 2 and X3 in row 1 yields X ! + 2 ( .l ) + 3(2) = 5 and the value of Xj is xx = 1 •
(7.5 1 10 4 " (7.
We interchange the third row with the first row and write the augmented matrix as 5 Ag = 2 1 2 5 2 . this strategy will always work. After the first column of zeros has been determined.
.4-6a. as described.588
ADVANCED MATHEMATICS
This example illustrates the need for adopting an appropriate pivoting strategy. If the matrix is non-singular.4-2. Example 7. (i) (ii) (iii) Scan the first column to find the largest absolute value | a j j |.3 0
l 6
(7.c)
(7. scan the second column below the first row to find the largest absolute value | a i 2 1. i * 1.b. One such strategy is as follows. Solve the following system of equations using the Gaussian elimination method.4-6d)
4
23
Choosing the first row as the pivot row. The row in which this element is found now becomes the second row.3 x 3 = 0 Here the largest absolute value in the first column occurs in the third row. Xj + 5x 2 + 4x 3 = 23 2xj + i-x 2 + x 3 = 6 5x x + 2 x 2 . Exchange this row with the first row. we proceed to generate zeros in the first column.
(iv)
This process is continued until the matrix has been transformed into an upper triangular matrix.
4-7a.5 (7. yielding
Xl
= 1.6 0 4. x 2 = 100
(7.5 7.b) is X! = . Performing the elimination.4-10a. When the matrix is singular.1 r o w 1) J (row 3 .-*.b)
(7. we obtain X2 and finally.b) is
.NUMERICAL METHODS
589
5
2 .4-8a.01x1 + x 2 = 2 The exact solution of Equations (7. we do not have a unique solution.4-9a.4-8a.4_6f)
0 0
The last row identifies X3 as 7.4-6e)
46
46
°
10
10
23
We now look at the second column and note that ~-
> .6 23 2.3 0
3 22
A(g1} =
° " T o To
ft
6
(row 2 . We then interchange the second and
third rows leaving the first row unchanged.
x3 = 3
(7.5/2.. as stated earlier.b. Substituting this value of X3 in the second row.
x2 = 2 .b)
The exact solution of Equations (7. we find the new matrix to be " 5 A<2)
=
2 . We examine the case of ill-conditioning by considering the following two systems of equations
x l + X2 =
1 (7. we obtain x^ .4-9a.b)
(i) Xj + 1.9 9 .5.c)
The second kind of problem that can arise in the solution of a system of linear equations deals with singular or ill-conditioned (nearly singular) matrices.1 row 1)
(7.01x2 = 2 and x t + x2 = 1 (ii) 1.3 0 " 4. from the first row.
11x1 = 0 iff x = 0.
. we assign a real number which we call a norm and denote it by II x II. We define the closeness of two matrices by the condition number which in turn is expressed in terms of the norm of the matrices.4-1 la. we find that Equations (7. The determinant of each of the coefficient
L l iJ
matrices is nearly zero.001
1
ri
which are.590
ADVANCED MATHEMATICS
X! = 100. A norm shares certain properties with absolute values. the solution (-99. A norm satisfies the following properties.4-14a. 100) is changed to (100.01 '
r
I
I
1
1. x2 * 2 (7. -99). in a sense.4-12a.b) can have an incorrect approximate solution.b)
The equations are also satisfied approximately by xx « 0 . The coefficient matrices in the three previous cases were
'i
1. close to the singular matrix
ii
. 1) 2) 3) llxll > 0 for all x .b)
Note that for a small change in the value of the coefficient in the second equation.4-13a.001 The exact solution here is X! = 1 .01
ii
1 '
r
I
li
1.4-12a. I IcxI I = I c III x I I for any real constant c. To each vector (or matrix) x.
x2 = .9 9
(7. from one quadrant to another quadrant! Consider the system of equations
Xj + x 2 = 2
1. II x + y II < II x II + II y II (triangle inequality). x2 = 1
(7. That is to say. Ill-conditioning is thus in a sense a measure of the closeness of the coefficient matrix to a singular matrix. Vector and matrix norms are defined as follows.b)
Here.001 Xj+ x 2 = 2.b)
(7. The above examples are cases of ill-conditioning.
4-15 a.4-3. irrespective of the direction. The familiar Euclidean norm. The L p norm is defined as Lp = I I x lip = ( X l x i l P ) 1 / P for p > l where Xi are the components of x.. + xl The I IxI I x norm is defined as 11x11^= max |Xj|
~ l<i<n
(7. The maximum value of the column sum is IIAII1. the Lj norm is the sum of the absolute values of the components of x . as defined by Equation (7..4-19a. The Lj and L^ norms of a matrix are defined as
n
Lj = HAH. II j norm. is obtained by setting p = 2. For p = 1. To calculate II A. is obtained by adding the absolute values of the elements of each column of A. In this case. I I ^ we interchange the roles of columns and rows and repeat the process described earlier in the determination of IIAII j . Calculate IIAII j and IIAII^ if
.
591
The first condition states that all non-zero vectors have a positive norm.4-18a.b)
L^ = HAIL = max X
~ l<i<n j = 1
a ii
J
(7. The second property implies that the norm of a vector is invariant.b)
The II A. = max V
l<j<n
i = 1
aH
J
(7. which is the length of the vector x. as a measure of x.NUMERICAL METHODS
.4-16)
(7. It is easier to compute I I x II ^ than I I x II2 and it is not surprising that I IxI I ^ is preferred to I I x II2 in numerical analysis. The third property is the usual triangle inequality.4-15b) becomes llxll 2 = Vx} + x^ + .b)
(7.4-18b). Example 7.4-17)
The I IxI I ^ chooses the maximum absolute value of x. Equation (7.
If K ( A ) is large we say that the matrix is illconditioned.oi i ' ]
The Lj and L ^ norms are
n
< 7 -4"27)
Li = HAH.1 1 -" -"
(7. We can then expect that a small change in b will produce a large change in x resulting in loss of accuracy.01. Note that K ( A ) is undefined for a singular matrix. The resulting n x 2 n matrix is in the case of the matrix A given by Equation (7. Example 7.c)
n
^
= UAH
= °°
= max T
1 ^ 2
j = 1
a-J
= max(2.b.4-28a. We separate the matrices A and 1 by a dotted line.4-26c)
Equation (7. we use the Gauss-Jordan method which proceeds as follows.. The other elements in the last row are zeros. 2) = 2. We then perform the Gaussian elimination process from the n t h row upwards until A has been transformed to I and I is then transformed to A" . Determine K ( A ) using the Lj and L ^ norms where
^ .01
(7. r • i T
1
I_ • A"1
. 2. To the matrix A .NUMERICAL METHODS
593
or
118x11 i II 5b II -Jff * HA II II A' 1 I I . This quantity IIAII II A" II is called the condition number or COND ( A ) and is denoted by K ( A ) . we adjoin the identity matrix I and we obtain a n x 2n matrix which we write as y A : I_ j .01
J
(7.[ i. We use the Gaussian elimination method to reduce A to the usual triangular form.b. = max Y
i^j<2
i= 1
a-.4-4.oi
i :
o i ]
(7-4"3Oa)
." II with the relative change of vector b. We divide the resulting last row by an appropriate number so that the element of the last row and last column is one.4-26c) shows that a relative change in the solution vector is given by the product of IIAII II A.4-27).c)
To obtain the inverse of A.= max(2.01) = 2. performing simultaneously the same operations on I. Below we illustrate the method
1
:
1
0
"
[ A ^ ] = [i.4-29a. The choice of norm is not important.
4-38)
IIA-BII = 0. the coefficient matrix A is written as the difference of two matrices B o and C o . when the number of variables is large and when the coefficient matrix is sparse (has many zero entries). Equation (7. Sparse matrices of high order are of frequent occurrence. iterative techniques are preferable in terms of computer storage and time requirements. Iterative Method When the dimension of the system of linear equations to be solved is small.4-40)
. When using an iterative method.4-37) (7.NUMERICAL METHODS
.
$95
We conclude this section by stating the nearest singular matrix theorem which helps to circumvent the cumbersome problem of finding A" . However.4-40) by the inverse of B o we obtain (7. An example is when partial differential equations are solved numerically (see Chapter 8). we have K(A) > 2. we choose B to be "1 B =
i
1
i
(7. If A and B are a non-singular and a singular matrix respectively
^BT
£
"^'"
(7'4"36>
Using this result in the above example.4-2b) is now written as Box = b + Cox Premultiplying Equation (7.01.} — = 102
(7.4-39a.01 So that II A"1 II > .01 x 102 which is of the same order of magnitude as that obtained earlier. the Gauss elimination method is useful.b)
With II A II = 2.
with the largest coefficient
xl
= l ~ IX2 + IX3
= 7 + 7 x l + f X3
( lSt r o w )
( 3rd r o w )
(7.b.
Jacobi's method
In Jacobi's method. That is to say. Example 7. Substituting these values into the right side of the set of equations generates new approximations that are closer to the required value. to solve for the variable with the largest coefficient.c)
~^X2
+
^X3
=
"^
We first write Equations (7.4-5. we can generate a sequence of values of x from Equation (7. Each component of x on the right side of Equation (7. We first consider the method of Jacobi and then the Gauss-Seidel method. choosing whenever possible. c) for one variable in each row. b.4-42)
A judicious choice of BQ and CQ will lead to a sequence of x k which will converge to the required solution x. § 0 is chosen to be a diagonal matrix.4-4 lb) as follows
x k+l
= ixk
+
£
(7. We begin with some initial approximate values for each variable.4-43a. These new values are then substituted in the right side of each equation to generate successive approximations until the difference of two successive approximations of each variable is within the desired degree of accuracy. solve the system of equations
o X i 4" X<2 — Xo — O
2 x 1 + x 2 + 9x3 = 12
xl
(7. we solve each equation for only one of the variables.4-44a)
(7. Using Jacobi's iteration method.4-42) may be taken equal to zero if no information is available.4-4 lb)
Starting from an initial guess x 0 .4-43a.596
ADVANCED
MATHEMATICS
x = BQ1 b + B o! (: O X = Bx + £
(7. as shown in the next example.4-44b)
X2
.4-4la) (7.
is an eigenvalue of A and x is the corresponding eigenvector Ax^ = lx_ (7. If A. In this section. A real matrix Q is orthogonal iff Q"1 = Q f If A is a real symmetric matrix.5-4) (7. eigenvalues arise in connection with vibration problems in mechanical engineering. We say that a matrix A is similar to another matrix B if there exists a non-singular matrix P such that A = P"'BP If A. in this case. we briefly summarize the basic results and definitions. Most of the algorithms for estimating eigenvalues of a symmetric matrix A make use of similarity transformations and are therefore carried out in two stages. If A is triangular or diagonal.5
EIGENVALUE PROBLEMS
Problems involving the determination of eigenvalues and eigenvectors often arise in science and engineering. The general problem of finding all eigenvalues of a non-symmetric matrix is much more difficult as it easily leads to stability problems with respect to perturbations. there exists an orthogonal matrix Q such that Q"1 A Q = C (7. the matrix is reduced to a suitable form and.600
ADVANCED MATHEMATICS
7. in the second stage. is an eigenvalue of A with associated eigenvector x.5-1)
Suppose that matrix A is similar to B .5-2) (7. the diagonal entries of A are the eigenvalues of A. Thus the diagonal entries of C. then A is also an eigenvalue of B with P x as the associated eigenvector of B . Before proceeding with the method. For example. the method of determining the eigenvalues is executed. In the first stage. are the eigenvalues of
A-
.5-3)
where C is a diagonal matrix. we study the problems of calculating the eigenvalues of a square matrix. in discussing the stability of an aircraft in aeronautical engineering and in quantum mechanics.
5-8)
. the Hermitian transpose of A is the transpose of the complex conjugate of A and is denoted by A .5-7) (7. If A = AH (7. -0 4 5J
0 3 4 2 0 0 0 2 15 _ 0 0 0 5-l_
We next describe a method of transforming a real symmetric matrix to a tridiagonal matrix. there exists a real orthogonal matrix Q such that Q A Q is diagonal. The diagonal elements of A are then its eigenvalues.5-5)
the eigenvalues of A are all real and A is known as a Hermitian matrix.5-6)
The only entries in A that could be non-zero are the elements along the diagonal. A matrix U is a unitary matrix iff U" 1 = U H A matrix A (= a^) is tridiagonal if a^ = 0 whenever I i . Let v be a unit vector (column matrix).
Householder Algorithm
It was stated earlier that for any real symmetric matrix A.j I > 1 (7. Consider the matrix Q = I . the subdiagonal and the super diagonal. If A is a matrix with complex entries. Examples of tridiagonal matrices are 3 10] .NUMERICAL METHODS
601
The eigenvalues of a real symmetric matrix are real numbers and their corresponding eigenvectors are mutually orthogonal.1 2 0 0 0" 2 13 0 0
124 . The methods proposed so far do not always converge. Attempts at diagonalizing A have not always been successful. Instead we transform A to a tridiagonal form via the method of Householder which is economical and reliable.2 v vf (7. and the columns of Q are the eigenvectors.
Example 7.5-9a) (7.5-9b.5-10b) (7.2 v k v J (7.2 v v f ) = I-4vv + 4 v v t vyt = I (7.5-9b). the matrix A n 2 will be a tridiagonal matrix. Reduce the matrix A to a tridiagonal matrix by the method of Householder.5-10d)
We choose v k such that its first k elements are zero.5-11)
We choose v j to be 0" v} = v2 _V3_ V2 + V3 = 1 (7.2) operations.5-9c) follows from Equation (7.2 v y t ) ( I . After (n .c)
t
Equation (7.5-12b) (7.5-10a) (7. since v is a unit vector. we perform the following sequence of transformations Ao = A Ak = Q k A k _ ! Q £ Qk = I . The product v ' v is a scalar.5-1. Note that v v' is a symmetric matrix and its transpose is v v ' . if A is given by
all a 12 a 13
A =
a12 a13
&22 a23 a23 a33
(7. we note QQ1" = ( ^ .5-12a)
. To transform a matrix A to a tridiagonal form. We illustrate the method by the following example.5-lOc)
XkXk =
l
(7.602
ADVANCED MATHEMATICS
To verify that Q is orthogonal.
The orthogonal matrix Qj is now constructed (Equation 7.5-26)
Then
v. + a i 2 n
(7..1 [l + *•»<*•»")]
ah (sign a12) vij =
J2S v
(7. If the matrix A is a (n x n) matrix (n > 2).5-15a to f).. The choice of v 1 2 ls different from Vjj (j > 3) because we want to reduce the matrix A to a tridiagonal form [see also Equations (7..5-22. we have considered a ( 3 x 3 ) matrix.NUMERICAL METHODS
605
We choose the positive sign in Equation (7.5-22). we start with v j given by
l\
= (°'vl2'v13'-'vln)
(7-5-24)
Then from Equation (7. Let Sj be given by
Sl
= a122 + a123 + .22 . A j will then be of the form
. Substituting v 2 and v3 into Equation (7.5-10c)..5-23) and it follows that we take the negative sign in Equation (7. 0 x x (7.5-25)
0
x
x
x
where the x denote non-zero entries. .5-27a)
(7..5-1.5-27b) is valid only for j > 3.5-10c) and the operations defined by Equations (7. 23)]. Qj will be of the form 1 0 0 Ql = 0 x x 0 x x .. • In Example 7. the matrix A (1) is tridiagonal.5-10a.5-27b)
>
J^3
Note that Equation (7.. b) can be performed..
... We write v 2 as a row vector
y j = (0.." A .
. the first row and column are of the desired form.d)
v13 = ..5-30)
.5-31a.5-27a...f)
..
0
X
Aj =
0
x
x
. . Use the Householder method to find a tridiagonal matrix similar to the matrix "1 A = 2 2 . We now proceed to reduce the second row and column to the form of a tridiagonal matrix.
6(V273 )
From Equation (7..1 2
S as defined by Equation (7.1 1 3 2 2 1 -1 2 2 1 . b) (7...b)
V12
= VH1+1]
=
Vl
(7.5-32c..J L = .5-26) is given by S^ = (-l) 2 + 2 2 + 2 2 = 9 The Vj: are given by Equations (7. as shown in the next example.
x
(7. v2n)
(7.5-29)
We then proceed to construct Q 2 and obtain A 2 as described earlier.1 (7..b)
(7. Example 7. The process is continued until we obtain A n 2 which will be of the tridiagonal form.5-2. Ql is
= -^f
6
(7.^ 6(V2/3) 6 v14 = . v 23 .5-10c).5-32e. 0.606
ADVANCED MATHEMATICS
X
X X
X
X
0
.5-32a.5-28)
0
x
x
x
In Aj.
7778 -0. the elements which have been reduced to zero in the first column and row at the first iteration remain zero at the second and subsequent iterations. Once the matrix has been reduced to a tridiagonal matrix.6Q8
ADVANCED
MATHEMATICS
The orthogonal matrix Q 2 is given by
?2 =1 "
2l2l\
(7. the most efficient method of obtaining all the eigenvalues is the QR method.6000
0 0 (7.7857 0
0 -0.8000
The tridiagonal matrix A 2 is obtained by the similarity transformation A2 = Q 2 A j Q | (7.5-39a) 0 1 0 0 0 -0. = 0 0
3 3. The main idea is based upon the following result of linear algebra. In general. This property made Householder's method very reliable. The QR Algorithm We now discuss the powerful QR algorithm for computing eigenvalues of a symmetric matrix.9899
1 0 0 0 A 2 is then calculated and is 1 3 A.6000 -0.1414 0 0 0 (7.5-41 a)
=
Q2QIAQ1QJ
(7.5-39b) -0.
.0222 -0.5-40) -0. all the zeros created in previous iterations are not destroyed in subsequent iterations.1414 0.7857 3.5-41c)
= P A Pf •
We note that in the Householder method.9899 -0.5-41b)
(7.
5-45).5-42)
Since rotation matrices are orthogonal matrices. we deduce that c =
an
(7.5-45)
(7.5-43d. there exists an upper (right) triangular matrix R and an orthogonal matrix Q such that A = QR (7. P2 =
r i
0
o
c
o i
-s .5-46b)
V a?i + a ? 2
With this choice of c and s and denoting Pj A by Aj .5-44b)
We set the element in the second row and first column of P l A to be zero.5-43a. we use rotation matrices to transform A to the triangular matrix R.NUMERICAL METHODS
609
If A is an (n x n) real matrix. P3 =
r c
0
o
1
-s
0
_ 0
0
1 J
[ 0
s
c J
[
s
0
c
_
(7.c) where c = cos 0 and s = sin 0 (7. that is sajj-ca12 = 0 From Equation (7.b.5-46a)
V Al + a?2
s =
ai2
(7. a rotation matrix may be one of the following
c
Pj = s
-s
c
o I
0 .e)
Let A be a (3 x 3) tridiagonal matrix and form the product of A with the rotation matrix P j c PjA= s -s c 0 1 f an 0
a i2
a 12
a22
0
a23
(7. In the case of a (3 x 3) matrix.5-44a)
0
0
1 J _ | 0
a 23 a 33 _
=
san-ca12 0
sa 1 2 +ca 2 2 a 23
ca 2 3 a33
(7. we write
.
given a collection of experimental data points [x k .NUMERICAL METHODS
_677
R = gn-l A n-2 = P n _! P n . . n . Similarly we choose P r so as to reduce the element aj. we address the following problem. many of the elements are already zero.5-53a) (7.. of any matrix A r (r = 0. 7. b.1) will be of the form given in Equation (7.5-52a... Lagrange. Lagrange Interpolation This method is based on choosing a polynomial p n (x) of degree n for which p n ( x k ) = f(x k ). and Newton.
The terms c and s are given by c = . f (x k )]. the c is along the diagonal elements and the s is off diagonal.
(r) . In a tridiagonal matrix... 2 . . .. . k = 0. b.is . The elements of P r are zero every where except (i)
. a[[ s =
(7.. we can successively reduce the non-diagonal elements in the lower triangular section of the matrix to zero.s ..b)
The matrices P r (r = 1.2. n . A standard text on the eigenvalues of a matrix was written by Wilkinson (1965).5-43a. and the reduction to an upper triangular matrix is fast.
P-j is s and Pj. (11)
along the diagonal where they are one with the exception of P^ where they are c . x n ). xn) or a curve that passes through the data points. An upper triangular matrix R is then obtained and the eigenvalues are the diagonal elements. c).1. l . We note that in Equations (7. n.
_(r) . . . c) except that they will be (n x n) matrices.2 ) to zero.5-53b)
jl
Vafi + afi
By this process..5-43a. The points x k are sometimes called the nodes. This problem has been studied by well known mathematicians such as Gauss. . 1. The earliest method requires one to fit a polynomial (an interpolation function) that approximates the function f over (x 0 .6-1)
..6 INTERPOLATION
In this section. .g i A
(7. (7. We wish to estimate the value of f (x) for some x in the interval (x 0 . k = 0.
20) = 1. 1.w> {1 ^ ^ (7. we have to decide at the outset of the degree of the polynomial to be considered.
(x-x Q )(x. We could consider all the given points but the calculations become lengthy.20. we only have to calculate the higher differences. The above conditions of continuity. The points closest to 1.05-1.12-1.05-1. 1.12 than Newton's formula. The points at which the low degree polynomials are joined are termed knots.15 and 1.
.10)(U2-1.15)(1. (x-xo)(x-x2)(x-x3) (x o -xi)(x o -*2)(*o. Using Lagrange's method. In the recent past. Using Newton's method. Spline Functions A high degree polynomial interpolation method is not satisfactory when the number of data points becomes too large. slope and curvature at the knots can be easily satisfied. piecewise polynomial approximations have become prominent. or when the data points are associated with a function whose derivatives are large or do not exist.12-1.10. we have to perform the calculations again. Such piecewise low degree polynomials are called splines.20) = (1.05830
n n
_. We write n M =
P 3 W
(X-Xl)(x-X2)(x~x3) f/x ^ . If subsequently we wish to improve the approximation by using a higher degree polynomial._nw (L°2470) +
n
.6-36b)
We note that Lagrange' s formula produces a more accurate value for V1.x 3) (x^x^-x^-x/^ +
(X-XQ)(X-XI)(X-X3)
f(x } + 2>
(x 2 -x 0 )(x 2 -x 1 )(x 2 -x 3 ) Thus
n P 1O . although we have used fewer points in Lagrange's method.10)(1. it is more convenient to approximate the function by a piecewise polynomial function in a small interval.05-1.05.NUMERICAL METHODS
612
For Lagrange's formula.15)(1.12 are: 1. First degree polynomials (straight lines) are continuous but not smooth functions.6-35)
3(L12)
(1. Instead of trying to approximate a function over the entire interval by a single polynomial of a high degree. we approximate Vx by a third degree polynomial involving four points. Cubic splines are the most commonly used splines. Second degree polynomials (parabolas) allow us to impose continuity and slope conditions at the knots but the curvature changes abruptly at the knots. A spline function is a function consisting of polynomial pieces joined together with certain smoothness conditions. since we cannot make use of the values already computed.X l )(x-x 2 ) (x 3 -x 0 )(x3-x 1 )(x 3 -x 2 )
3l
(7. Their first derivatives are discontinuous at the knots.
.
i = 0..6-38a)
(7. .. Thus over the whole interval (x 0 .6-38b)
= xi+l~xi
(7. !
i
i i i
i
i
. fj). Instead of using a high degree polynomial to pass through the (n + 1) points.6-1 We write Si(x) as Si(x) = aA + bi (x Xi )
Approximating f (x) by a set of cubic splines
+ q (x . 1. slope and curvature conditions. Each Sj(x) joins only two points Xj andx i + 1 . we choose a set of cubic splines Sj(x).1)
(7. This is illustrated in Figure 7. x n )..
x-^. (x n .
VI
y
f (x)
f^r^C.6-37.(x i+1 ) = S i + 1 (x i + 1 ) Combining Equations (7.. (xj. bj. f0). fn) be the given data points.. Continuity at the knots requires S.. the function is approximated by n cubic splines. 38a) yields aj + bihi + Cih?+ dihf = a i + 1 where
hi
(7. x i + 2
xn X
FIGURE 7.6-1.6-38c)
.62Q
ADVANCED
MATHEMATICS
Let (x 0 .
!
I
i
' •
I
i
i
i i *
•
! _ •
l
i
i
l
i
0
X-.6-37)
where the coefficients aj.Xj)2 + di(x-
Xi )3. cj and dj are to be determined from the definition of cubic splines and the following continuity. (n .
6-39a) becomes ai = fj The continuity of the slopes at the knots imposes S|(x i + 1 ) = S. 43a.NUMERICAL METHODS
62/
At the knots. Equation (7.. b) yields c0 = 0 Cn-i+Sd^h. Cj and dj).6-40a.
. 121 and 144.6-42b)
Equations (7.6-40b) (7. Use cubic splines to determine an interpolation formula for the function Vx . bj. 39b. Using Equation (7. the values of S.. Si(xi) = f. the conditions (7.6-39b) (7. We need to impose two additional conditions which are S"Q (x0) = 0 S.6-37. + 1 (x i + 1 ) On differentiating Sj and Sj + 1 . b) are applied at the interior points and the two end points (x 0 .6-40a) bj + 2 c i h i + 3djhf = b i + 1 Similarly the continuity of the curvature at the knots leads to s"(x i + 1 ) = S" +I (x i + 1) Ci + Sdjhj = c i + 1 (7. x n ) have to be excluded. 42a.6-43b) (7.6-43a) (7. 40b._i(x n ) = 0 Combining Equations (7.6-38b. (x) are equal to f (x).6-37). b) form a set of 4n equations to be solved for the 4n unknowns (aj.! = 0 (7.6-42a) (7. The values of the function are 10.6-4. We choose the three knots to be 100. we obtain from Equation (7. 11 and 12.6-41b) (7.6-40a) (7. Consider three points (knots).6-39a)
The derivatives of f are not known. Example 7.6-41a) (7. 41b.
we use Sj. we use S o . For the interval 10 < x < 11. yj}. b.6-56a) (7.6-46a. b. b) into Equations (7. Suppose we have a set of n data points {XJ.6-54b)
= Z(a
i=l
+ bxi-yi)2wi
The best values of a and b imply that we choose a and b such that f(a. = 3b = °
Combining Equations (7. for each i). and a set of weights {Wj}. We want to obtain the best values of a and b to fit the data.b) = X efwj
i=l
(7. The deviation at each point is 8j = a + b x j .6-54a) (7.6-44a. The weights express our confidence in the accuracy of the points. If we think that they are all equally accurate.6-57b)
If all the points are equally weighted (w. b. That is of of " .6-53)
f(a.6-56a.y i The weighted sum of the squares of the deviation is (7. b) can be rewritten respectively as
n n n
(7-6-55a'b)
(7.
Substituting Equations (7. __ . 52a.6-56b)
aX
i=i
n
wi
+ b X xiwi = X viwi
i=i
n
(7. we have
. 55a. b) is a minimum. b. Least Squares Approximation The concept of least squares approximation was introduced in Chapter 2. we obtain S o and Sj. 51a. we set Wj = 1 for all i.NUMERICAL METHODS
623. b) yields 2 X ( a + bxi-yi)wi = 0 2 X ( a + b x i . and for the interval 11 < x < 12.6-57a)
i=i
n
a X xiwi
i=i
+ b
X x?wi
i=i
=
X x iyi w i
i=i
(7. we start with the simplest case of a linear polynomial (y = ax + b).yi) x i w i = ° Equations (7. Here. = 1. b).6-54a. where n is greater than 2. We assume that the x's are exact. 50a._ .
N
r.
we obtain
f'(x0) -
f(X° + h)2-hf(X0"h)
(7 . 12).x 0 ) is the distance between the points xj and x 0 .2-11. Since
Similarly if we approximate f(x) by a polynomial of second degree p2(x) and differentiate p 2 (x). x 0 + h).7-3b)
behaves well in the interval. 3b). we find that the error is of the order h. The error can be reduced by taking h to be smaller.7-3a) may be written as
(7.7 " 4)
Equation (7. It is also referred to as the central difference formula.7-4) is known as the three-point formula as it involves the three points (x 0 .h) and at (x 0 + h ) . provided f f is generally not known. (Xo)
. since the derivative at x 0 depends on the value of f at (x 0 .7-lb)
where h (= xj . ffro-no-ffro)
(77. It is found to be (see Problem 16a)
.7-2. From Equations (1. A four-point formula is obtained by considering a third degree polynomial.h.f " ( x o + 0h) where 0 < 9 < 1.^^-^-tfV+eh)
By comparing Equations (7. Equation (7. A formula for f' (x 0 ) is then given approximately by
f . as the derivative at x 0 depends on the forward point xj.
(7.7-3a)
f'^.7-2) is known as the two-point formula since it involves two points XQ and x^.)-f(x 0 )
h
(7. It is also known as the forward difference formula.2)
The formula given by Equation (7.626
ADVANCED
MATHEMATICS
=
f(x. x 0 . we cannot estimate the error. we have
2
f(x o + h) = f(x o ) + hf'(x o ) + | .
We apply the formulas given earlier to each subinterval and the desired result is the sum of the integrals of each subinterval. If the interval [a. .7-11)
In Equation (7.7-12b)
rb
The integral I f (x) dx can be written as
Ja
.7-10d)
f (x) dx « ^
•'Xo
[f (x0) + 3 f (Xl) + 3 f (x2) + f (x3)]
(7. We denote the break-points by xj.630
ADVANCED
MATHEMATICS
J
'
xo
I(x-x1)(x-x9)
(x-xn)(x-x9) ". This should be resisted. the formulas given so far will not generate good result. x 3 ).7-10d) is the well known Simpson's rule.10b)
~ ^ [ 2(x-x1)(x-x2)dx . . b] is large. we subdivide the interval [a. If we approximate f(x) by a third degree polynomial p 3 (x).. N (7. we write xj = a + ih. . 1.7-12a) (7.. The points Xj and x 2 are in the interval (x0.)
f(X°)+
.^ f Vx o )(x-x 2 )dx + ^ ) [ 2(x_x )(x_x ) d x
2h 2 Jx0
« | [ f ( x o ) + 4f(x1) + f(x2)] Formula (7. in the present section. b] into N subintervals of equal length h. Equally. . h = ^ i = 0. the limits of integration are (xQ. As pointed out in the previous section.7-11). we obtain the 3/8 rule which is given by
2 h 2 J Xo
(7. for example) over the subintervals. x l5 x 2 and x 3 are equidistant points and the distance between two consecutive points is h. (x-xft)(x-x.1
f
I (-h)(-2h)
(h)('h)
h 2 JXo
f(x ' ) +
(2h)(h)
HdX
<77 . We may be tempted to approximate f(x) by a high degree polynomial. it is better to divide the interval into subintervals and approximate the function by a low degree polynomial (cubic splines. that is to say. X3) such that the points x 0 .7-lOc) (7.
6-10)
fh f(x) dx can then be approximated by The integral I
/a
.7-19) can then be written as I = Th + C h 2 where C is a constant and is assumed to be independent of h. we obtain an improved value of the integral which we denote by T ( 2 ) .. We denote the value obtained with a gap h/2 by T^^. h/2.NUMERICAL METHODS
633
The error involved in evaluating the integral depends on h.7-21)
= ^(Th/2-Th) 3h 2
(7-7-22)
Substituting C into Equation (7. This method of obtaining an improved value of the integral from two approximate values is known as the Romberg method (extrapolation method). we may write I = T h + O(h 2 ) (7.7-23)
We can repeat this process of halving the interval.Thus I is given by I = T h / 2 + Ch 2 /4 From Equations (7. We recall that the Lagrange interpolation formula for the polynomial p n (x) can be written as
PnW = X f ( x i ) 4 ( x )
i=0
(7. We now recalculate the integral using the same Trapezoidal rule but halving h. we obtain
C
(7. .7-19)
We now assume that the error which is of O(h2) is proportional to h2..7-20. Thus T ( 2 ) (=1) is given by
T(2)
= T h/2 + ^ ( T h / 2 .7-21).. using intervals of h. h/4. It is widely used in computer programs.T h )
(7.7-20)
(7.7-14d)] by T h . If we denote the
rb
exact value of I f(x) dx by I and the value obtained using the Trapezoidal formula [Equation (7. the length of the gap. Equation (7. that is to say. h/2 n successively until two improved values of the integral are within the desired degree of accuracy. 21).
=
A 0 [a 0 + a 1 z 0 + a 2 z2 + a3z3] + A 1 [a 0 + a 1 z 1 + a 2 z2 + a3z3]
(7. we transform the interval [a. b] to [. We choose Xj such that Equation (7.7-26) We consider the case where f(z) is approximated by p 3 (z).l ^) We now have to determine the points Zj which lie in the interval [-1. 1].7-24C)
(7.7-24d)
Note that f(Xj) are the values of f at the points Xj and are constants. It can be written as p 3 (z) = ao + a 1 z + a 2 z 2 + a 3 z 3 We now choose z 0 .7-24d) is "exact" or "best" in a sense to be defined later. That is to say p 3 (z)dz = p 3 (z o )A o + p 3 (z 1 )A 1 (7. A o and Aj such that Equation (7.1. We now assume that the Xj are not given.7-24a)
fb
~ Sf(xi) ^iWdx i=0 Ja « Z
i=0
f ( x i) A i
(7. For simplicity. The Aj 1= I /8j(x) dx are known as the weights. z 1.7-29)
.634
ADVANCED
MATHEMATICS
C
f(x)dx Ja
f
p n (x)dx Ja n [b = X f(x i )i i (x)dx i=0 Ja (7.7-27)
Carrying out the required integration.7-24b) (7. They can therefore be taken
I rb
\
Ja. So far the points Xj are given and are equally spaced.7-26) is "exact".i f(z)dz . This can be achieved by writing _ _2x-(a + b) (b-a) {l. 1] such that
c
J.
\
j
out of the integral sign.7-28) (7.7-28) becomes 2ao+^2. Equation (7.X f ( z i ) A i i=o (7.
33).d)
The first solution (Ao = 0) implies.
z0
= °>
zi=zo'
zl=-zo
(7. 30c. This is illustrated in the next example. a1( a2 and a3 and.7-30c) (7.b.7-30b or d). A1 = l . Form Equation (7.7-35)
Equation (7.7-30b) (7. a ls a2 and a3. z0.7-29) holds for all a0.7-30d)
| = Aozo + A l z l 0 = A o zJ + AjzJ
The set of Equations (7. we deduce A0 = A! Combining Equations (7.7-32a. from Equation (7.L r ) J-i \ V3 / W3 j (7.b. Of the four possible solutions. V3 z\=-}= V3 (7.7-30a) (7.7-35) is the Gaussian two point formula and it involves evaluating the function at two points.7-26). the only valid solution is the last one.7-30b). Thus the number of evaluations of the integrand is the same as in the Trapezoidal rule. A1. that either A1 or z1 is zero and Equation (7. Zj) and we have exactly four equations to solve.7-34a .NUMERICAL METHODS
635
Equation (7.7-31) are A0 = 0. we obtain A0=l.7-33)
Substituting Equations (7.7-30d) yields
Aozo(zl"zo)
= °
(7.
.c. by comparing the coefficients of a0.7-31)
The possible solutions of Equation (7. we have approximated the function f (x) by a third degree polynomial and we expect better accuracy. we obtain respectively 2 = A o +A! 0 =
AQZQ+AJZJ
(7.c.7-30c) cannot be satisfied.7-30a.d) (7. zo=--L. Multiplying Equation (7.7-30a to d) involves four unknowns (Ao.d) into Equation (7. but in this case.7-34a.7-30b) by z \ and subtracting the resulting expression from Equation (7. we obtain 1 f(z)dZ = f( ^ r ) + f / .
32103 We now halve the gap and the interval is now 0.7-41b)
The value we obtain for the integral using only two points is as accurate as Simpson's rule using seven points.b)
Comparing T ' ' with the values obtained in Example 7.y + 1.3
I
V7dx = 0.(1 + 1.07238 + 1.7-4la)
(7. To use the Gaussian quadrature. we have To
15
(7.7-23) T ( 2 ) = 0.7-36a. Using Equation (7.l5z+ 1.7-3.7-35.7-9d) and the values given in Table 7. •
. 1.14017) = 0.15
(7.40) yields
/•I.7-2. we have x = O.7-4.15 dz
(7.7-38a. though the number of points considered is less.y + 1.32149
/ o .
We use the trapezoidal rule to evaluate Tjj using h = 0.7-2. l 5 / . 1].b)
= 0^5.15 I Vo.636
ADVANCED
MATHEMATICS
rl. From Equation (7.14017) = 0. we find the Romberg method gives an answer nearer to the exact values.b)
The improved value T1-2-1 is given from Equation (7. From Equation (7. we have (7..7.32137 + 1(0.15
= 0.15 + .(l + 2 x 1.15z+1.7-39)
I
Vx~dx = 0.15.3] to [—1. Evaluate I
Vx dx using the Romberg method and the Gaussian method. l 5 / . we have T 0 3 = 03.7-25).14d) and Table 7.7-40)
Combining Equations (7.7-37a.32137
(7.32148 (7.3
r—
Example 7.15 By direct substitution.3.32103) = 0. / o .32137-0. we have to change the interval from [1.
.8-2c.D.. . z (0) = 0.E.b) (7. whereas y" = f ( x . see Hilderbrand (1956) or Stroud and Secrest (1966).1).'s) with all conditions specified at one value of the independent variable is called an initial value problem. with the conditions specified at different points represents a boundary value problem.8-4a)
.E. In this case. INITIAL VALUE PROBLEMS
The order and the degree of an ordinary differential equation (O.D.7-26) is "exact" if f (x) is a polynomial of degree (2n .8-1)
is a two point boundary value problem whereas if the conditions to be satisfied were y(a) = Y 0 . we have 2n unknowns (z 0 .d)
it would be an initial value problem.. However if the conditions are specified at more than one point.8-2a.8 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS.E. y' (0) = 2. . y (b) = Y : (7.) have been discussed in Chapter 1.D. z" = z + 1 with y (0) = 1. . An O. which are discussed in Chapter 2.8-3a) (7.8-3c to f)
is an initial value problem of a second order system of differential equations. z (0) = 1 0<x<l (7. For example. A o .D.1) has 2n constants which exactly corresponds to the number of unknowns. An_i) and we have to choose them such that Equation (7. For further details. In the case where time is the independent variable. Similarly y" = ( z 2 + l ) y . 1 . . The determination of Zj and Aj is simplified by considering Legendre polynomials..NUMERICAL METHODS
637
We can generalize the two-point formula to an n-point formula. all conditions may be specified at t = 0. (or a system of O. y ' ) . a<x<b (7. 7.8-3b) (7. the differential equation a 2 y" + a 1 y l + a o y = g(x) subject to y (a) = Y o . This is possible because a polynomial of degree (2n . y . z n . the O.E. y'(a) = Y 2 (7.
c)
is a boundary value problem (Sturm-Liouville problem)..8-5a)] into an integral equation. we obtain
yOO = y o + I f(t.8-6) is known as an integral equation.8-8)
.8-5a) subject to the initial condition given by Equation (7.8-5a) (7. b] I f (x.. z) I < L I y ...f (x. 2. In this section. we can determine y^ and subsequently y 2 .8-5b).8-6) by considering
yk+1 to = y0 + I f & yfc) d t ' Ja
k = 0. we seek a solution of Equation (7.
(7. First Order Equations A first order initial value problem can be written as ^ = f(x.8-5b)
y(a) = y 0
Formally integrating Equation (7. Equation (7.6M
ADVANCED MATHEMATICS
a 1 y(a) + a 2 y'(a) = c : .8-7)
Since y 0 is given. The iteration will converge if (a) (b) f(x.
bj y(b)+ b 2 y'(b) = c 2
(7.. only initial value problems will be considered. -oo<y<oo there is a constant L.8-6) and we cannot evaluate such an integral.8-6)
Note that the unknown variable y is inside the integral sign in Equation (7. . 2-20). such that for any two numbers y.2-21).8-4b. y) is continuous in the interval a < x < b . 1.z I Condition (7. .8-6. As suggested by the iteration given by Equation (7.y)dt Ja
(7. y) (7. y) . There is a similarity between Equations (7. (7.8-8) is the Lipschitz condition. z and any x e [a. We have transformed the differential equation [Equation (7.
we obtain Yi+l = yi + hf(Xi.8-9) is Euler's formula. Equation (7. it is not suitable for numerical computation because it requires the evaluation of many integrals. which is of length h. b] into subintervals [xv xj +1 ] and assume that in each subinterval.8-9)
yiL
/yU)
i
1—i
0 X. Figure 7. b) using Equation (7.8-5a).8-5a. y) is constant. Euler's method
We subdivide the interval [a. Integrating Equation (7. f (x.. y i ) where y(x. • : approximate value of yi+l
. Although it ensures that if f satisfies the conditions stated earlier a solution is generated.
X
FIGURE 7.8-1
Euler's method.8-7). Its accuracy is not very good.NUMERICAL METHODS
639
Picard's method involves solving Equations (7.
!
^
X|+.8-1 illustrates this method. x : exact value of yj+x. Other numerical methods are available and are considered next. (7.) is denoted by y.
8-23).083
0 0.yi) + f ( x i + 1 . y) (= y'.8-2
Values of xj and improved values of yj
i 0 1 2 3
Xi
f(xj. are given in Table 7. we obtain an improved value of yj + 1 .8-23)
We determine y i+ j using Equation (7. We assume that yl+\ is given by y i+1 = y i + a k 1 + b k 2 (7. we find that Heun's method gives the exact solution to three decimal places.8-23). We derive the formula for the Runge-Kutta method of order two by generalizing Equation (7.8-1 and 2.3
Comparing the last column of Tables 7. yj) by the average value of f (x i? yt) and f (xj + i. Solve the initial value problem of Example 7. They are more accurate than Euler's method and do not involve finding derivatives as in Taylor's method. We can then calculate f (x.
TABLE 7. Use the same value of h.8-2. Equation (7. It represents an improvement over Euler's method. to determine the value of y at x i + 1 . we only used the value of f (x.1 0.8-9) now becomes Yi+l = yi + |[f(Xi.020 1. We can improve the accuracy of Euler's method by replacing f (xj.1 0.8-24)
.8-9). yj) 0 0.3091
f(xi+1. y-l+\). The values of yj.8-23).046 1. y i + 1 ) ] (7. Combining Table 7.
Runge-Kutta methods
The Runge-Kutta methods are extensions of Heun's method. we generate Table 7. y) at selected points on each subinterval.NUMERICAL METHODS
645
Heun's method
In Euler's method [Equation (7.yi+1) 0.1 0.8-9)].8-1 via Heun's method. We only need to evaluate the function f (x.202 0. Vj+i).202 0. and finally using Equation (7. the slope of y) at the point X.8-1.4244
yi+1 1. obtained by using Euler's method.005 1.3091 0. +1 .8-1 and Equation (7.8-3.2 0. Example 7.
204990
1.148125 0.8-5.0 0. once the computation can be started. f2 and f3 so as to be able to calculate y 4 .J.1155 1.10000 0. we need to know f0. for example. The methods considered previously are self-starting methods as they require only the values of (XJ. yj. This is the disadvantage of this method. Several multistep formulas have been proposed [see. The Adams-Bashforth method requires only one computation of f per step. This is illustrated in Examples 7. yj using the Runge-Kutta method i xj kx k2 k3 k4 yi+1 (7. We use the Runge-Kutta method of order four to calculate y^ .11575 0.This type of formula is known as a multistep formula.20496
0.14896 0. Example 7. such as RungeKutta methods.1 0. TABLE 7. A common practice is to start with a Runge-Kutta method and once the values of f at a sufficient number of points have been generated. Gerald and Wheatley (1994)]. In Problem 17b.8-3 Values of Xj. f\. The calculations are given in Table 7.8-33). yj) to determine yj +1 . prior to applying Equation (7.166450 0.8-5 and 7. (xi-2> Yi-2) a n d ( x i-3' Yi-3).j). yi-3).184740 -
0.4496
. we indicate the derivation of the Adams-Moulton formula. To get the method started.8-34a) (7.131575 0.16642 0.NUMERICAL METHODS
647
Note that in Equation (7.3
0. Vj_2. Take h to be 0. y^) to calculate y i + 1 but also (XJ.18566
0. to determine ( y ^ . as can be seen from Equation (7.2 0. Solve the initial value problem y' = 2x + y y(0) = 1 using the Adams-Bashforth method. we need to use other methods. but it involves less computation than the Runge-Kutta methods of order four. a multistep methods is used.115000 0.V3.13155 0.8-33). To use the Adams-Bashforth method. we not only need to know (xj.8-33).1.8-34b)
0 1 2 3
0.2642 1.8-3. whereas the Runge-Kutta method of order four requires four evaluations of f at each step.
9f0] (7. we have Y5 = Y4 + ^ [55f4 .39)
.8-38a) (7.8395] = 1.1878 + 48. we have Y4 = y 3 + ^ [55f3 ..6735-9] (7.59f3 + 37f2 .8-35) becomes y 4 = 1.8-37a) (7.11.
yi)
= kj/h
(7. Higher Order or Systems of First Order Equations Any r t h order differential equation can be written as a system of equations. f3 have previously been evaluated. since f ^ .8-36)
Equation (7.9fj] (7.8-3. the third order differential equation r first order
^+y^i+h)2+y
dx 3 can be written as dx 2 \dx /
=g(x)
(7.648
ADVANCED MATHEMATICS
From Equation (7.. • We next extend the methods employed to solve first order equations to higher order equations.728-98.891 +61. we need to perform only one calculation.4496 + ^ = 1.8-33). As stated earlier.f3. For example.8-38c) [112...5754. we obtain f0 . we need to evaluate only f4. at each step.8-37b)
= 1.8..9462
We can similarly proceed to calculate yg. From Equation (7.8-35)
From Table 7.147-120.6754 Continuing.8-38b) (7.6754 +Q-L [136..8-30b). we have f(xj. Note that to determine y 5 .59f2 + 37fx .
NUMERICAL METHODS
659
0. dx 2 dx y(a) = A .8
1
FIGURE 7. the conditions here are given at more than one point and the formulas given in Section 7.
Shooting Method
This method is widely used.9
BOUNDARY VALUE PROBLEMS
Boundary value problems have been discussed in Chapter 2.9-lb.8 cannot be applied. We assume that
. We described two methods which can be used to solve boundary value problems.9-lc) and replacing it by imposing a condition at x = a.2
0.c)
We convert the boundary value problem to an initial value problem by dropping the boundary condition given by Equation (7.8-3
Deformation of D' versus capillary number Ca for various values of radius R
7.8
• * * •
*
#
•
*
R=5 •
R:7
a
0.+ p ( x ) ^ +q(x)y = r(x).9-la) (7. We consider a second order linear equation ^ .4
Ca
0. It consists of transforming the boundary value problem to an initial value problem.4
*
^
.6
0. y(b) = B a<x<b (7.
I——
°0
0. Unlike initial value problems.
it is unlikely that the boundary condition given by Equation (7.9-la to c). we can compute yj and y 2 as in the linear case and we use Equation (7.y 2 ( b ) (7.9-lc) will be satisfied. Since oci is a guessed value of y'(a). With y'(a) given by Equation (7.9-4.9-la). and the value of y'(a) that will satisfy Equation (7.9-lc) will still not be satisfied.y > ( a ) ]
< 7 -9-7)
In Equation (7. ( a ) + S^l k < a ) . all the terms on the right side are known.9-3)
where c 1 and c 2 are constants.9-lb) and y must also satisfy this condition. from these two values of a (ocj and a 2) it is possible to obtain the exact value of y'(a). we obtain
y ' w = y .9-1. 2) form an initial value problem and we can integrate the differential equation using one of the formulas given in the previous section.9-5) (7. 3) yields 1 = cx + c 2 Imposing condition (7. The above method is applicable only for linear equations. Let y 1 be the solution obtained by using (Xj and y 2 be the solution with a 2 . Combining Equations (7.9-lc). Equation (7. That is to say y = c1y1+c2y2 (7. we can compute y 3
.9-3) is not valid. is a solution of Equation (7. 5).9-lc) is known.9-la.9-7).9-2)
Equations (7. a linear combination of yj and y 2 is also a solution of Equation (7. we cannot apply the principle of superposition and Equation (7. Unless we are lucky. In the linear case.9-7).9-3) and using Equations (7. For non-linear equations. we compute the solution y that satisfies Equations (7.660
ADVANCED
MATHEMATICS
y'(a) = (*!
(7.9-la).9-4)
Cl
(7-9"6a'b)
Differentiating y in Equation (7.9-6a.9-1 b. Both y 1 andy 2 satisfy Equation (7. However. we have B = c i y i ( b ) + c 2 y 2 (b) From Equations (7. b). Since the system is linear. b.9-7) to obtain an improved value of y'(a) which we denote as y3(a). We now guess another value of y'(a) (say a2) and we proceed with the integration. we obtain _ " B-y2(b) yi(b)-y2(b) ' _ Yi(b)-B °2 " y l ( b ) . as illustrated in Figure 7. Note that y3(a) is not the exact value of y'( a ).Using y 3 (a).
. Equation (7.9-7)..x / 5 ) y y(l) = 2.9-10b. We need to determine a such that Equation (7.9-8) can also be obtained as follows.
y '
V2lb)"
v*V
'
V|(b)
"
/
/
^
*
1
i i
i i
[ J
i . b) can be solved by the secant method [Equation (7.9-8)
Equation (7.l (7.9-1.9-9a.9-9a. we obtain
Shooting method
yi+2(a) = y.b)
Equations (7.9-8). Example 7. The solution y is a function of y'(a) (= a) and the solution can be written as y(oc. x).NUMERICAL METHODS
667
in the same way as yj and y2 were computed.
(7.
I
0
1
o
1
b x
^
FIGURE 7. y(3) = . We repeat this process until successive solutions yj and y i+1 are within the required degree of accuracy.9-1 c) is satisfied. Solve the boundary value problem y" = x + ( l .2-6)] and the result is Equation (7.c)
.B = 0 (7.9-lc) can be written as g(cc) = y(cc. ^ ^ l b ) k i ( a ) -y[ (-1
i = L 2.9-1 On generalizing Equation (7.9-10a) (7. 1 ) . (a) + y ..
we need to solve an initial value problem.9-11).y.^)
dx 2
x dx/
(7.9-13c) by the initial condition
Pdx
a
=a
(7. b.9-15b. We need to obtain a good approximation to a via Equation (7.NUMERICAL METHODS
661
To obtain g'(ock). b) and then g'(a) is obtained.9-11) to obtain an improved value of a. we obtain
£'(a. a) = 0 £ ' ( a . d) with respect to a yields £(cc. The iteration is repeated until two iterates yield values within the desired degree of accuracy. a) = 1 (7. 16a. b) can be solved and ^ ( a . d) define an initial value problem. b. which in turn implies finding g'(a k ).9-16a) (7. From Equation (7.9-13a.9-16b)
The initial value system defined by Equations (7. d). Then on differentiating £.9-14)
The prime denotes differentiation with respect to x and this notation will be kept from now on.9-14) yields
^'(a. we note that g'(a) is £(a.9-13d)
Equations (7.c)
We replace Equation (7.x) = ~[y'(a. b) is obtained. Equations (7.
y(b) = B
(7.by ^(oc. This can be substituted in Equation (7. We demonstrate this by considering the boundary value problem given by
4=f(x.x) = ^ ^
dy da = ^ dy
+
+
i L 3Z1
dy' da '
(7.9-13a) with respect to a and using Equation (7. We denote ^ .9-15b.9-13.9-15a)
^ dy
(7. with respect to x and interchanging the order of differentiation. Differentiating Equation (7. x). b) are the associated variational equations which need to be solved simultaneously with Equations (7. 16a.9-13b.9-12).9-13a.)
y(a) = A .
.x)]
(7.9-15b)
Similarly differentiating Equations (7.9-13b.
9-la). b)
Y
xj
=
(7-9-23)
Substituting Equations (7. We denote a by xQ and b by x n + 1 .9-17c). 23) into Equation (7.9-19a to d. Further. The value of a 0 chosen is 0. 22a . b] into (n + 1) equal intervals.9-22b.y j . We again consider the boundary value problem given by Equations (7. the number of initial conditions to be guessed is also high.9-la to c). Finite Difference Method In transforming a two-point boundary value problem to an initial value problem. We use the central difference scheme and the derivatives at x. each of length h. (= x0 + jh) are given by
d y |
dx d*y dx2
xj
=
^lHM=aiCil
2h 2h VjH-l-^j + yj-l h2
(7 .
•
A method widely used to solve boundary value problems is the method of finite differences which is considered next.324. The shooting method can then be very laborious. We divide the interval [a. if we did not make a good guess of the initial conditions. If the order of the differential equation is high. In this method. which is close enough to the value given by Equation (7.2 and after six iterations the value of y (1) was found to converge to 0. An alternative numerical method of solving boundary value problems is the method of finite differences.9-21c) (7.9-21b) (7. we have replaced the boundary condition(s) at one end by guessed initial condition(s) at the other end. 1 )]
h
+ q(xj)yj
= r(Xj)
(7. 9 .2 y j + yj.1] + 2t[p(x j )(y j + 1 . we obtain
^ [ y j + l .9-2 Id)
The initial value problem given by Equations (7. we replace the derivatives by finite differences. the convergence might be slow. 21 a to d) can be solved using the RungeKutta method with an initial value of a (= a 0 ).9-24)
.NUMERICAL METHODS
^ 5
y2(-l) = « y3(-0 = 0 y4(-!) = 1
(7.
2. Equation (7. Catalytic monoliths are widely used in the automobile industry to control the emissions from the vehicle exhaust systems.9-4.2528 0. n) resulting in a set of non-linear algebraic (or transcendental) equations.9-22b. )
(7. Thus by discretizing Equation (7. This set of equations are solved by iteration techniques. b).2 and 7.
ADVANCED MATHEMATICS
-(2+h2) 1
1 -(2+h2)
0 1
?1 Y2 =
° 0 (7. Numerical solution 0.668
.
^.5211 0.9-2 Values of y .9-32). Example 7. Nowadays numerical software is available which implements the formulas given in this chapter. However.75 0. This is illustrated in the next example. y 2 and y3 and their values are given in Table 7. ^ ^ . ^ .
.9-33) can be written out for each value of j (j = 1.9-33)
The boundary conditions are given by Equations (7.9-28a) will be non-linear.50 0. (1993) modeled a catalytic combustion in a monolith reactor.9-32)
o
l
-(2+h 2 ) J |_y3j
L-sinhl_
Solving Equation (7.25 0.5214 0.1-2^ + ^-1 = ^ 4 . For a given f. we still have to use our judgment in choosing the appropriate method. . we obtain yj. The monolith usually consists of a number of cells through which the exhaust gases flow...9-26a. Spence et al.4. using Equations (7.2526 0.8223
For non-linear equations. Equation (7.8226 Exact solution 0. TABLE 7. 23).9-2. we obtain. . some of which are described in Sections 7.9-13a). We consider only a single cell and a schematic diagram of such a cell is shown in Figure 7.9-2 together with the values of the exact solution.
though
.b) (7. D c . as shown in Figure 7. x 2 = T B (x)/T B0 .9-39)
The shooting method is found to work only for P e < 1 and P e > 500. b) form a standard boundary value problem for a known w(x). Pe is the Peclet number.9-38b)
The boundary value problem defined by Equations (7.9-37a. The reasons for the failure are as follows. For large values of P e . = PeJk{2yi-w-e0) d£
(7. w = T i (x)/T B 0 .9-39) is almost exact and the method works.9-36d) (7. T m . Tj are the temperatures at the outer wall. and yc are dimensionless parameters. X J = C B ( X ) / C O . 37a. L is the length of the cell. 0Q = T O / T B O . y = T m (x)/T B0 . F.9-36a) is written as a system of first order equation as follows ^L d^
= y
(7.9-36a.d)
= x2(0) = 1
Note that Equation (7.9-36a) is almost zero and the starting value of yj(O) given by Equation (7. TB(x) is the bulk gas temperature and T B0 is TB(0). As an initial guess. J D . For small values of P e .=0 d £ ^=1
Xl (0)
(7.9-2.1) /w]}"
where £ = x/L. J H . The outside wall temperature (0O) is given and Equations (7. J k . The associated boundary conditions are ^ = ^ =0 d £ i. The complementary function of Equation
r
/
1
(7. the mean solid wall temperature. and the temperature at the inner wall.6ZQ
ADVANCED
MATHEMATICS
~-
= -JD[xl(^)-aw^)]
(7.9-36b to 38b) can be solved by the shooting method.9-37c. the errors remain small and the method works.9-38a)
^2. CB(x) is the fuel concentration in the bulk gas and Co is CB(0). Equation (7. y j (0) is assumed to be
Y l (0)
= [w(O) + e o (O)]/2
(7. TQ.9-36e)
°=JH[x2^)-w(^)]+Jk[y^)-w^)]-rDcexp[yc(w-1)/w]aw a w = xj {1 + (DC/JD) exp [yc (w .9-36d) is an algebraic equation and allows us to determine w(0).9-36a) is a linear combination of exp [± ^V2P e Jk J and. round off errors propagate as multiples of these exponentials. in the shooting method.9-36c) (7. the left side of Equation (7.
Using Equation (7. We illustrate this situation further by considering the following example.10-1.10-3) numerically.l ]
+ C2a2 1
(7. (1993) have used alternative methods to overcome this instability problem. The exact solution is y = e~2x We note that y is a decreasing function of x. The solution is of the form y.l = 0 The values of a j
2
(7. we have seen that due to round-off errors.10-2)
(7-10-3)
Instead of solving Equation (7. The round-off errors (or truncation errors) amplify as the integration proceeds and the magnitude of the errors exceeds the solution.10-lb)
(7.10-5)
From Equation (7. we deduce that a j 2 are the roots of a2 + 2 h a .10-3). the shooting method does not work when the Peclet number is in the range of 1 and 500.2 h y j
(7.10-5).10-4)
[«2 + 2 h a 2 . we discuss the problem of instability in more details. the finite difference equation is yj+i-yj-i = .10-la)
(7.9-4. In the next section.= -2y
dx y(0) = 1 Examine the stability of the system.10 STABILITY
In Example 7.10-4) into Equation (7.NUMERICAL METHODS
]
577
unreliable. Spence et al.= C 1 a J 1 + C 2 a ^ Substituting Equation (7.10-6)
are
.9-22b).
7. we seek an analytic solution. This is an example of instability. Example 7.l ] = 0
(7. Write the finite difference equation for the system
P. we obtain Cjoc^"1 [ 0 4 + 2 1 1 0 4 .
10-1 b). If Equation (7. When x. is a decreasing function of
X J-
In an actual computation.10-11) becomes the dominant term and the solution yj oscillates. choose another one.
.10-10a) (7.-.( 1 +2h) The solution is given approximately by yj = C 1 ( l . It is multiplied by an exponentially increasing function of x. and noting that lim (1 -2h) 2 x J / 2 h = e"2xJ h->0 lim (1+2h) 2 x J / 2 h = e2xJ h->0 The solution y.£Z2
____
ADVANCED
MATHEMATICS
al2
= -2h ± V l + 4 h 2
(7.10-8a) (7. the solution would be oscillatory for some values of X.exceeds a certain critical value. Instability depends on the differential equation and the method used.2 h a2 = . If one method is unstable. there is round-off error and the value of C 2 will not be exactly zero.2 h ) j + C 2 ( . but will be small.10-9) (7. can be written as ^ = qe" 2 *) +C 2 (-l) j e 2 x J (7.10-7)
Since h is assumed to be small.10-10b) (7.10-11) (7. This shows that the difference equation is unstable. The roots are then approximately given by «1 = l . we find C 2 to be zero and y..10-3) were solved numerically.l ) j ( l + 2 h ) j Replacing j by Xj (Xj=jh). greater than a certain value. This arises because of the error in the initial condition and of replacing the first order differential equation by a two-point difference equation.10-8b)
Applying boundary condition (7. the second term (which is the error term) on the right side of Equation (7. we expand V1 + 4h in powers of h.
7148 The cubic x 3 .111.4ac . Answer: 0. Answer: 1. calculate xj a and xjt.
Obtain the three roots of x 3 . b.8608
. One root x^ of a quadratic equation a x 2 + bx + c = 0 is xl = [-b + V b 2 . show that xj can be approximated by xla = . b = 111 and c = 0. By expanding V b .4 a c ] / 2 a . Use (i) the secant method and (ii) Newton's method to find the root. Calculate xx if a = 1. x = x l a .2541.4 a c ) ] For the given values of a. 2]. Is the answer reliable? Note that Xj is given by the difference of two almost equal numbers and this could result in loss of significant figures. 1.c / b By multiplying the numerator and denominator by b + V b .. and x = x^.618 Deduce that an iterative formula for finding the cube root of a real number A is
x k+l
3a..4ac . show that x^ can be written as xlb = [-2c/(b + V b 2 .
PROBLEMS la. Answer: -2. Test the convergence in each case. Evaluate a x ^ + b x + c for x = Xj.11.1 has a root in the interval [1.2x .NUMERICAL METHODS
673. 0.4 x = 0 has a root in the interval [0. 2a. Comment on the results.1149. 5a.107
Use the formula to calculate the cube root of 30.
4a
= xk-(x^-A)/3x£ Answer: 3. The equation e x .4x + 1 by the fixed point iteration method.1]. Obtain the root (i) by the method of bisection and (ii) by Newton's method. c.
h. 6).6-10). x2 = x 0 + h.
The solution of the differential equation
subject to the condition
y(*i) = yj /* x i+i
is
y ( x i + 1 ) = yi + i = y . y^) by fj. Calculate
• it
i j ( x 0 ) and i j (x 0 ) and verify Equations (7.. Xj and Xj_l5 y i+1 is given by yi+i = y i + ^ ( 5 f i + i + 8fi-fi_1) 18 a...676
ADVANCED
MATHEMATICS
15b. In the derivation of the Adams-Bashforth formulae. Can we replace the QR method by the Gaussian elimination in the calculation of eigenvalues?
16a. show that Vj+1 is given respectively by yi+i = y1+^-(23fi-i6fI_1 yi+i = y i + ^ ( 5 5 f i . If the point XJ+J is included. x ^ .7-5.i
+
5f i _ 2 )
37fi-2-9fi_3)
+
These formulae are the Adams-Bashforth formulae [Equation (7. By approximating f by a polynomial of degree 2 (through 3 points) and of degree 3 (through 4 points).8-33)].
17b. and x3 = x 0 + 2h. .5 9 f i . Xj +1 . + I
f(x. Denote f (XJ. y) by a polynomial p n (x). deduce the expressions P2( x ) and P3(x) for equidistant points x 0 . .
From Equation (7.y)dx
Approximate f(x. x1? x 2 . and x 3 . Choose x^ = x 0 . Show that if f is approximated by a quadratic expression passing through xj + 1 . The Gaussian elimination method transforms a matrix to an upper diagonal matrix. we obtain the Adams-Moulton formulae.5-1) have the same eigenvalues. we have interpolated f through the points Xj. and not through the point Xj+i. . with a constant stepsize h. Show that the following functions are not cubic splines
. in the form given by Equation (7.6-10). taking the interpolation points to be Xj.
Show that the matrices A and B defined by Equation (7.
and
a2
such that
I
[f (x) .y (x)] dx
is a minimum (least squares
approximation). 21a.. The sales of the product for three different prices are given in the following table Price ($) 1. The measured value has been reported to be 0.8
Compute the cubic splines that pass through the three given points.NUMERICAL METHODS
677
Il-24x+18x2-4x3. The function f(x) = e x .50 330
Find the least squares regression line and estimate the demand when the price is $1.25 375 1.
/•I
2
Obtain
a0.1 0 I n _ i
Use this formula to calculate I4 retaining (i) three significant and (ii) five significant figures. (i) f(x) = -54 + 72x-30x2 + 4 x 3 . Answer: 349 22b.
. 0 < x < l is approximated by y(x) = a o + a 1 x + a 2 x 2 .40. Show that the recurrence formula for evaluating the integral I n defined by
*
In = n . (ii) f(x) = -35 + 51x-22x2 + 3x3. 19a.00 Demand | 450 1. Expand f(x) in a Taylor series about x = 0.5 up to the term x 2 .
l<x<2 2<x<3 l<x<2 2<x<3
The viscosity r| of water at various temperatures T is given in the following table T("C) 10 20 30
Ti(cp) I 1.
&\.8904 cp. 13-31x + 2 3 x 2 .3
L0
0. 20b. The demand for a certain product is a linear function of the price. Compare this series with y (x). Evaluate the viscosity at 25°C.5 x 3 .
Answer: -0. + 1 . Show that. under certain conditions which should be stated. and (b) the finite difference method. 27 a. In Problem 17b. analytically and numerically.l . y(4) = 3 0<x<4
by the method of finite differences. we obtain a multistep formula which can be written as
. using the Runge-Kutta method. accurate to four decimal places. Use the Adams-Bashforth method to obtain y at x = 1 with h = 0.NUMERICAL METHODS
679
Work to five decimal places. 0 1 (T-20) = 0 dx2 The boundary conditions are T(0) = 4 0 . T(10) = 200
Find the analytical solution and solve the problem numerically using (a) the shooting method. The steady one-dimensional heat equation can be written as
2
— .5). we can integrate from Xj_j to x i+1 instead of from Xj to x.1.11587. dx 2 dx y(0) = . Solve the boundary value problem ^ .0 . is 1. Choose an appropriate value of h.+2(2-x)^=2(2-x). given that y(0.1). 29a. Compare the results obtained by the three methods. compare the accuracy and amount of computations required associated with the two methods.3784 28b. Is h = 1 appropriate? 30b. A vibrating system with a periodic forcing term is given by ^ + 64y = 16cos8t dt 2 y(0) = y'(0) = 0 Find y (0.
Which of the two methods gives the correct solution? Discuss the stability of the two methods. y(O) = l
by the formula obtained in this problem and by Euler's formula.680
ADVANCED MATHEMATICS
y i + i = y i _i + 2hfi
where h = Xj+j .
.
Solve the initial value problem ^ = -2y+l.Xj.
the value of u(x i? yj+k) by u i j + 1 . We shall also consider the method of finite elements.'s.
Vj )
= u i + l i j = u i. Many equations cannot be solved exactly and we have to be satisfied with approximate solutions. yj)by u^. Using computers.. We divide the xy-plane into a grid consisting of (n x m) rectangles with sides Ax = h and Ay = k as shown in Figure 8. we have u( X i +h. we consider the unknown function u to be a function of two variables x and y.'s. The solution is then an approximate solution.
.CHAPTER 8
NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
8.D.E. we can sum only a finite number of terms. non-homogeneity can be transformed to homogeneity via auxiliary functions as shown in Section 5. the method of separation of variables can generally be applied only to linear homogeneous equations with homogeneous boundary conditions. u i + s j + t denotes the value of u(xj+sh. we have discussed several analytical methods for solving P. In this chapter. 8. Using Taylor's theorem. and we have to integrate them numerically. the infinite Fourier series solution obtained in Chapter 5 has the appearance of an exact solution. in many cases. numerical methods are the most appropriate methods of solving some P. y^+tk). These methods cannot be used to solve all P.'s.E. and that of u(x i +h.2-la.D. Similarly. the value of u(Xj+h. For instance. the Fourier coefficients are expressed as integrals and some of them cannot be evaluated analytically.'s.D. For example. yj+k) by u i + 1 j + 1 .D.E.j +
h
d ^.j
where Rn is the remainder term.b)
i.
+
1>J
h 2 d2 Y a ^
+ R
"
(8.7. Also. but. we shall extend the method of finite differences described in Chapter 7 to P.2 FINITE DIFFERENCES
For simplicity.y|) by u i + 1 j .2-1. Even some of the exact solutions given in Chapter 5 are in reality approximate solutions. We denote the value of u(x i . where s and t are integers.1 INTRODUCTION
In Chapters 5 and 6.E. Under favorable conditions.
b) and the temperature is positive.3-1. This example can be associated with the heat transfer in a rod of unit length with its two ends kept at zero degrees. j + l
+ u
i .j
+
( u i+l. Such an analysis can be found in a more advanced text. the explicit method is valid only when 0 < r < 1/2. Chapter 5] is
u (x. Crank-Nicolson Implicit Method Crank and Nicolson (1947) proposed that
3x2
be approximated by the mean value of
i. b.3-13) and these values are also tabulated in Table 8.3-13)
Proceeding as in case (a).j
i-l.3-14)
Equation (8.3-15)
.3-8a.(2n + I ) 2 7C2t] 7i2n=0 (2n+l)2
(8. such as the one by Richtmyer and Morton (1967).3-5) becomes
u i.j
i.3-1. stability. b).j
(8. it can be seen that the values of u obtained in case (b) cannot be the solution of Equations (8.2 u i . A convergence.Y — ^ sin [(2n + 1) Ttx] exp [. the temperature can never be negative and we conclude that the values of u in column (b) are unacceptable. This example shows the importance of the choice of the value of r and the explicit method is not valid for all values of r. The values of u obtained in case (a) are in qualitative agreement with the analytical solution whereas the values of u obtained in case (b) are not.j-2ui. we can compute the values of Uj l .j + l . 7a. t) = -£.3-1. The analytical solution [see Problem 1 la.686
ADVANCED MATHEMATICS
In case (b) r=l Equation (8.j
ui. and compatibility analysis is necessary.j = f (ui+l. the temperature distribution is given by Equations (8.j + u i-l. Uj 2 . ••• from Equation (8. Initially.j)
(8. In the present example.3-4b) is now replaced by
dx2
i. Equation (8.U +l)
(8. From the physics of the problem.j + i
+ u
+l-ui. We next consider an implicit method which is valid for a wider range of values of r. From Table 8.3-12)
+ l = -ui.3-14) predicts that u is an exponentially decreasing function of t.j
32u and
3 x 2
d2u .j
+ u
i + l. 8a.
1 4 . we can solve. Solve Equation (8.3-2.4
x 5 = 0.01 and solving Equation (8.3-7a.6 0.5.1 0 0 .6921
1. and d [Equations (8. A can be written as " 4 . the value of r is unity. In this case.4 0.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
689
From the initial and boundary conditions.1. we can obtain the values of Uj at time t 2 .01 0.3
N x 4 = 0.3956 0.5 only. and the analytical solution at the point x = 0. at time t j . From the symmetry of the problem. b. 20a tog).0 0.02.2 1.3-2 Numerical (N) and analytical (A) values of u N tj
xo = o
N Xi=0. Assume a to be unity and choose h = 0. Example 8.1 0 0 -1 4 .8 d = 1. we can calculate dj at time t o (=O) and solving Equation (8. we find that initially the dj are " 0.1 0 A = 0 . These values can be used to compute dj at time tj and from Equation (8.6 -
(8. TABLE 8. We can now calculate dj at time t = 0.1 and k = 0.3-24)
For the given A.1989 0. b).3-2la).3-21a).6461
1.7381 0.7691 0.8 0.2
N x 3 = 0.01.01. 24)].4 " 0.3789
0.1 4 0 0 0 .1936
0..3-23. This process can be repeated to obtain the values of uj at time t3.3-21a).0 0.2 0.1
N x 2 = 0.6 .0 0.3-18b.5834 0. we obtain the values of u at time t = 0. 8a. t m .5400
0.3-23)
From Equations (8. We can repeat this process and obtain the values of u at subsequent times. we need to calculate up to x = 0.. using the CrankNicolson formula.7743 0.3-21a) to obtain Vj which are the values of u at points Xj at time t = 0. using the symmetry property (v6 = v 4 ). we obtain vj which are the values of u. Table 8. With the given choice of h and k. Equation (8.02
0..5 N A
0.6809
. by the elimination method.3-2 gives the values of u at various times t..2 0 0 0 1 4
(8.0 0 0
0.3-1) subject to Equations (8.
[Equation (8.3-25)
(8.2.3-5)]
u o..3-27)
We note that u^ • is outside our region and can be eliminated via Equation (8. but the implicit method does.3-1) as shown later.
Derivative Boundary Conditions
So far we have assumed that u is given on the boundary.j)
(8.3-3a) by ^ = Pu. From Tables 8. there is no restriction on r.5. it is seen that the Crank-Nicolson formula provides more accurate values of u than the explicit method for r = 0. in the heat conduction problem.690
ADVANCED
MATHEMATICS
We noted that for r = 1.j
+ i = (1-2r)uo. we use the central difference form of du/dx Equation (8. • Compared to the explicit method [Equation (8. t>0 d x x=0 where (3 is a constant.3-26) becomes (UlJ-u_1. The Crank-Nicolson formula is convergent for all values of r. if the material is insulated.3-26)
(8. in the Crank-Nicolson method. the explicit method did not generate a meaningful solution. though the smaller the value of r the better the accuracy. the time interval can be larger. however. To obtain better accuracy. j = 0.2-3c)] and = 0 (8. the difference equation at the origin (x 0 = 0) is given by [Equation (8. . If we employ the explicit formula. Equation (8.3-1 and 2.j
+ r(ul.3-28)
.3-25) corresponds to the case J3 = 0. 1.3-18a)] involves more computation. and this generally compensates for the increase in computational effort at each grid point. the Crank-Nicolson method [Equation (8. We now replace the boundary condition (8.j) = 2hpu0j .. it could be that the derivative of u is given on the boundary. In some cases.j + u-l. there is no heat flow across the surface and the boundary condition is ^ on where n is the unit outward normal to the surface. For example.3-5)].
3-30)
where y is a constant.j + ul.j
+ 2r[Un-l.r v j = (1 .3-20a to g). b) yields ( 1 + r + r h p ) v o . the values of u at subsequent times can be computed.j-2hPuo.3-29a) (8.r .3-34a.j]
(8.w _ 1 = 2hpw 0 Combining Equations (8.3-31) and using the initial condition. We have two equations to determine v0 (= u 0 .3-31)
From Equation (8. Uj 0 are known and from Equation (8. we deduce that
u n. we eliminate u_j : to obtain
u o.j-O+hy)un.3-27) is true for all times and we have
VJ-V^J
(8. If the Crank-Nicolson formula is used.j
+ r(ul.3-33a) (8.3-33b)
w 1 . 28).3-32)
= 2hpv 0
(8. the vector v has two new elements (v0 and vn) .3-18a) . Using the same technique as at the origin. we obtain from Equation (8.+ l) and vn (= u n .j
+ l = u n.( l + h p ) u O J ]
From the initial condition u 0 0 .
(8.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
697
Combining Equations (8.j
+ l = (1-2r)uo.r ) wQ + rw.r v j = rw_j + 2 (1 . the values of u at later times at the origin can be computed.+ 1) .j)
(8. 33a. Suppose that at the other end (xn = 1 ) the boundary condition is given by ^
d x
= yu.3-29b)
= uOJ + 2 r [ u 1 J .
x=l
t>0
(8.3-32.3-29b).hPr) w 0 + rwj = c 0 The boundary condition at the other end (xn =1) can be treated similarly. Equation (8.
. Note that v0 and vn are not given and we work with Equations (8. and similarly vector c is increased by two elements (c0 and c n ).3-27. b)
The matrix A has been augmented by two columns and two rows.r v _ j + 2 ( 1 + r ) v o .3-19a to e) instead of with Equations (8.
4-11) Note that. j . .j
=
(k) i U i.4-10a)
o
--4
1 B =
i l l i 1
-4 1 '.4). Section 4.13 )
.4-10b)
Q
"
o
l
-4 l
1 -4
If the size of A is small. Section 7. and given that it is sparse. .4 u i .j+l+ui.j-l -4ui. From Equation (8.j3
< 8 . In this chapter.R. j + u i.4-3b). it is more economical to adopt the method of iteration. <8-4-12)
To improve convergence.j
=
(k) " U i. j ]
. j + U i . we see that a possible iteration formula is (8.R.4-8b) can be solved by the method of Gaussian elimination.). We use the same notation as in Chapter 7 and the k th iterate is written as u.j +f
(k) (k+l) (k) (k+l) (k) [ui+l.j-1"
Equation (8.O.o ..j +ui. But if A is large. and
Ui. as in the Gauss-Seidel method. . method is essentially a modified version of the Gauss-Seidel method (see.4-12) is modified to
(k+l) u i. we have used the new values (k + 1 iterate) of u : i .
0
"
(8.l .j + 4
(k) (k+i) (k) (k+i) (k) [ u i + l . we describe the method of successive over relaxation (S. .j + l + u i .NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
$99
"1 i
I A = B I '.O. Equation (8.4-11) can also be written as
(k+l) U i. (8.4 . Equation (8.j+ui-l.l . This method is discussed in Chapter 7. The S.
d) 9y 2 0<x<3. .4-20a)
.75
(8. For the Dirichlet O (coop) is given by problem in a rectangular region.4-19)
Taking into account the boundary conditions.8 0 . u(0.O. u 2 j . .4-16a.4-1.700
ADVANCED MATHEMATICS
The optimum value of G O depends on the mesh size and lies between 1 and 2. we deduce that G0Op is given approximately by co op = 1.25 .25.9 0 ] Solving Equation (8. Example 8. the optimum value of G coop = 4
(8. we obtain
V][
(8.4-18a. 3) = 9 0 .75 v 4 = u 2>2 = 43.4-8a) directly (Gaussian elimination).1 6 0 .y) = 7 0 . see Hageman and Young (1981).4-15)
We divide the region into 9 unit squares.4-18c.1 0 . y) = 0 (8.4-7a to h). we need to assume the values of the zeroth iterative of the interior points.R.4_14)
2 + ^4-(cos J + cosJ^r
For details on co op . From Equation (8.b. method. Equation (8.
We now solve the same set of equations by the S.0)=10.4-17)
=u
u
=41. Uj 2 '
U<U U = U
and U2
2) c a n ^ e written as
[u 2 k i + 7 0 + u i k 2 + 10
U + °'289
-4uU
]
(8. u(x. The values of u at the four interior points are given by Equations (8.c. One possibility is to assume that the value of all the interior points is the same and equal to the average value of the boundary points. Solve the equation U i + ^ J i = 0. .4-14). 0<y<3 (8.
v 2 = u 2 1 = 23.b) (8. 3x 2 subject to u(x.156 (8. u (3.d)
v 3 = Uj 2 = 61. To start the iterative process.4-13) for the four unknowns
(Uj j . The transpose of the column vector x is given by LT = [ .
0<x<l.5 42.331 43.4-1.4 u £ ]
(8.4-2.039 24. we assume that the value of u at the interior points is equal to the average value at the boundary points.750
Neumann Problem In this case.4-20b)
(8.4-22a. This will introduce points which are outside the region of interest and these fictitious outside points can be eliminated as previously shown in the parabolic case.240 43.4-20d)
To start the iteration.4 u ® ]
(8.407 63.752 k=3 41.252 23. Example 8.289 [ u ^ + 70 + 90 + u £ * 1 } . Solve the equation — + — = 0. 0<y<l (8.2 u2(2 k=0 42.761 61. The average value in this case is 42. the values of u at the boundary points are not given and have to be determined at the same time as the interior points.5.277 23.4u(jk)2]
-S^ =u£
+
0.5 k=l 41.055 20.289 [u(^+1) + 90 + u * + 1 ) . The next example shows how this is done. The values of Ujj at subsequent iterations are given in Table 8.5 42.4-20c)
u ( j k2+1} = u ^ + 0. We can proceed in the same way as in the case of parabolic equations and write the normal derivative in a finite difference form.689 61.4-1 Values of u j : using the S. TABLE 8.R.784 43.703 k=2 41.b)
.O. method ug uu u2)i "1.289 [u ( £ \ u £
+
10 .757 43. It can be seen that the convergence is rapid.4-21)
ax 2
dy2
subject to the conditions ^ = ^=0 dx x=0 3 y y=0 (8.755 k=4 41.197 60.5 42.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
70/
u £ 1 } = u™ + 0.
4-3 that the domain of dependence of Equation (8.5-11) is exactly that of Equation (8.b)
From Table 8. we observe that there is complete agreement between the analytical and numerical results. This concept is discussed in Chapter 5. Another way of showing that Equation (8.5-11) provides an exact solution of the problem is to consider the domain of dependence of the solution.h + tp .5-la).t ) + f (x + t) Using Equation (8. we have used the method of characteristics to solve hyperbolic equations.f ( x i .j-ui.5-1.5-15b) (8. Section 4) of Equations (8.j-l
= f ( x i + i . In Chapters 5 and 6. In Figure 8.15d) is exactly the left side of Equation (8.5-13a. Equation (8.5-11).5.5-1 and 5.5-1.5-5a) is a stable algorithm for solving Equation (8.
u=0
(8.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
709
t = 0. Thus.5-11) is
ui +
(8. Noting that the characteristics in this case have gradients ± 1. This method has the advantage that discontinuities in the initial data are propagated along the characteristics. D'Alembert's solution (Chapter 5.t j ) + f(x i + 1 +t j ) + f(x i _ 1 -t j ) + f(x i _ 1 +t j ) .5-15d)
We note that the right side of Equation (8. The parameter r is the Courant parameter and it has been shown by Courant et al. (1928) that if r is less than or equal to one.5-8a). Section 4.5-15a)
= f (xj + h .t j _ 1 ) .tp + f (xj + h + tp + f (xj .
. • We now show that Equation (8.5-11) is an exact difference equation of the problem.t j + h)-f(xi + tj-h) = f(xi + h + tj) + f ( x i .j+ui-l. we observe by comparing Figures 8.5-15c) (8.f ( x i .f ( x i + tj_1) (8.5-la to e) with g (x) = 0 can be written as u = f ( x . we have marked the pivotal triangular domains which determine the value of Uj .h .t j ) = ui>j + 1 (8. the right side of Equation (8. This result is due to r being unity (h = k).h .5-14)
l.5-14).5.+ 1. this method is generally more suitable for cases of discontinuities in the given data.tp + f (xj .
b) and is found to be x R = 0. qp = qQ = 0 p p = 0.5-3.5-32c.2.5-34a. The characteristics are given by r.b) (8. In this problem.045.0) respectively as shown in Figure 8.2. = 1 dx + 41 = m_ = .5-32a.:
+
#L = m . We choose the points P and Q to be (0. Solve the problem stated in Example 8. y R = 0.2 .4 (8.4
^
x
FIGURE 8.5-2.5-3
Numerical solution of the wave equation by the method of characteristics
The point (x R . e) and are Up = 0. y R ) is determined from Equations (8.3.d)
T :
The initial values of u are given along the x-axis which is not a characteristic.i -
r_ /W*
/ * *
iZ_
0 .0) and (0.5-8d.5-3.2
_v<
.b.1 dx
(8. p Q = 0. u Q = 0.714
ADVANCED
MATHEMATICS
Example 8.1 (8.f)
t
.e.5-1 by the method of characteristics.3 Q
1
. and q at P and Q can be computed from Equations (8.
.1.5-33a.1 P
1
.105.5-33d.5-26a.b)
This can also be seen from Figure 8. The values u. y is replaced by t.c) (8. p .
1 (8. 8. we obtain p R = 0.6-1 shows an example of such a situation. This is usually the case where the geometry is simple.5-37)
From Table 8. q R = -0.6 IRREGULAR BOUNDARIES AND HIGHER DIMENSIONS
So far.5-36a.b) (8.5-35b)
The approximate value of u R is given approximately by Equation (8.
y
-•
S
P
Q^S
__\c
X
FIGURE 8.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
7L5
Equations (8. we have considered only the case where the grid points lie on the boundaries.5-35a) (8. we note that this is exactly the value obtained by the finite difference method.5-1. The values of u at other points in the xt-plane can similarly be obtained. it is not always possible to arrange for the grid points to be on the boundaries.3.5-3Ob) and is found to be u R = 0.6-1
Irregular boundary
.5-28a. For curved boundaries.5-35a.075 (8. Figure 8. b). b) reduce to PR-°-4-qR = 0 °-2~PR-qR = ° On solving Equations (8.
t j .k
. Some of the methods described in this chapter can be used for non-linear equations. Laplace's equation in three dimensions for a cubical grid of size h 3 is
u i+l.2u i.6-10)
As in the two-dimensional case. k
+ u i.j.k + l. 8.2 u i.( i
+
e2)Up]
= Q
(g 6
_g)
k 2 e 2 ( i + e2)
The difference equation for the other points closest to the boundary are deduced in the same way. Other methods are available and are described in more advanced books.q + e^up] h2el(i + el)
+
[uR + e 2 u T .k
{ u i.6-10) can be solved by the methods described earlier.7 NON-LINEAR EQUATIONS
One of the advantages of numerical methods is that many of the methods based on linear equations with constant coefficients can be carried over directly to non-linear equations.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
711
[iiQ + e ^ s .2u i. we deduce that
u i.j. Equation (8. j . k . however it involves a larger number of points and the direct method is not recommended.j. the difference equation is complicated and it is preferable to use the finite elements method which will be described later. k + u i.k-l
= ( )
h2
h2
h2 (8.k.k + u i.6-9)
From Equation (8. For Neumann conditions.j.l .7-2a)
k
h2
J
.2 u i > i
J > JL
+U
i-u]
(8 . The difference form of the quasi-linear equation
3T = srl D ( x . j .k
+ u i. j . k) is the average of the values of u at the six neighboring points that surround it.k
= ^[ui+l.j.l ]
(8. j . U i ) h + l J . ui.k + u i .j.j
+ l.j.j + l.j.l .k.j.j-i.6-9).k + u i-l.t i U ) d
can be written as
Ui 'J + 1 ~ U i ' J
(8-7-l)
= D( X i .k+l + u i .j. the value of u at the point (i. The derivation of the difference equation for three and higher dimensional problems is straightforward.k
+ u i .
Implicit formulae which are unconditionally stable have also been derived.
0<x<l
(8. More recently. it has been widely used to obtain approximate solutions to continuum problems. 1].7-1). It must be pointed out that one does not usually solve one-dimensional problems by this method. We introduce the basic concepts of the finite element method by considering one-dimensional problems. Zienkiewicz and Morgan (1983) have shown that a generalized finite element method can be defined which includes all the numerical methods and it is left to the user to choose the optimum method. These formulae can be found in more advanced books on numerical solutions of P.8 FINITE ELEMENTS
The finite element method was developed to study the stresses in complex discrete structures. we have only one independent variable and the equation we need to solve is an ordinary differential equation.b)
By an appropriate scaling.
. The battle between finite differences and finite elements is over. We consider the differential equation
.8-2a. For regular boundaries and simple equations. the finite element method is the obvious choice. Several formulae have been proposed which are more complicated than Equation (8.
One-Dimensional Problems
In a one-dimensional problem. the method of finite differences is preferable because finite difference equations are easier to set up.2
^ L = _g(x).722
ADVANCED MATHEMATICS
The accuracy of Equation (8. t:. 8 .D.7-2b) can be improved by replacing D (Xj. The reason for this choice will become apparent after we have introduced the method. dx 2 subject to u (0) = A.E.8-1)
u (1) = B
(8.7-2b) but which give results closer to the solution of Equation (8. we can always transform any arbitrary interval into an interval of [0. Uj :) by an average value of D over the points considered. The one-dimensional case is used to ease the way to the understanding of two and higher dimensional problems. It was pointed out earlier that if the boundary is irregular.
we obtain an expression of the form I = F(a1)a2 g (8. In Chapters 9 and 10. They can be polynomials.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
721
Variational method
In the finite difference method. we give examples of such variational formulations. it is shown that the function y (x) that yields the extreme values of the integral I. we look for a function that extremizes an integral.8-1).
u .5)
(88-6)
srHnr) .8-3)
^1 _ A.8-4) yields
( °. i l
9y dx 9y' Consider the integral
=0
(8. They are usually chosen such that the boundary conditions are satisfied. or other functions whose properties are known.g r ^ ^ ( i u -H]-°
We approximate u by
m
Simplifying Equation (8. Instead of looking for a function that satisfies the differential equation. are constants and the Nj(x) are the shape (trial) functions.8-5).X
i= l
aiNiW
(8-8"7)
where a.s u ] d x
Applying the condition given by Equation (8.8-6). we replace the derivative by a finite difference. defined by
I = I f (x. In Chapter 9. we reformulate the problem into a variational problem. y')dx Ja is
(8. This means that the solution of Equation (8.8-8)
. y.8-5). we obtain Equation (8.8-1) is the function that extremizes the integral I defined by Equation (8. On substituting Equation (8. trigonometric functions.8-4)
i = / o te) 2 .8-7) into Equation (8. In the finite element method.
1] into a number of subintervals and each subinterval is called an element.2. we choose the weight function to be the approximate solution (w = u).
Galerkin method
Not all problems can be formulated as extremum problems. Equation (8.b)
Jo
In the Galerkin method.8-1). . On the first interval. then u is an exact solution of Equation (8. If R is identically zero.8-9). In the method of finite elements.8-1 la) becomes /"xo I N : [ajN^' + a 2 N 2 ' + g] dx = 0 JO
(8. we obtain the coefficients a. We approximate the solution u in the interval 0 < x < l by u(x).8-12a)
. We now describe a method which provides a weak solution to Equation (8. In each element.. and on substituting them into Equation (8. we subdivide the interval [0. we approximate u by Equation (8..8-7) provide sufficient accuracy. and the extreme values of I are given by dF ^ . The function u generally does not satisfy Equation (8. This method is known as the RayleighRitz method (see Example 9.11-2).8-1). 1] into subintervals (elements) and on each element.8-7). . In the finite element method. We illustrate this method in Example 8. Suppose we subdivide the interval [0. we obtain an approximate solution for Equation (8.8-1). The function u is a weak solution if the integral of R with respect to a weight function w (x) is zero.1] into [0. x 0 ) and (x 0 .8-9)
From Equation (8.724
ADVANCED MATHEMATICS
It can now be considered as a function of a.
i = 1.8-7).8-1) and we write ~ + g(x) = R dx 2 (8. Since each element is shorter than the whole interval.8-10)
where R is the residual. That is to say
C
Jo
Td2~ 1 w(x) —^ + g(x) dx =
Ldx2 J
(l
w(x)R(x)dx = 0 (8.m
(8. 1] and assume that two terms in Equation (8.8-7).8-7). we can expect that a better accuracy can be obtained for the same number of terms in Equation (8. u is approximated by Equation (8.8-1 la.8-1. we subdivide the interval [0.= 0.
8-39)
which simplifies the computation of Gj.g (c ) [aj + bjx + cjy ] A «^-g(c)A
(8. 42c).8-42b). Similarly.g ( c ) [aj + bjx + Cjy ] I I dx dy (8.8-40a. we can obtain a sparse matrix K. x and y can be approximated as the average values at the nodes. then g can be approximated by its values at the centre of mass of the triangular element and we denote it by g(c).b)
. Example 8.8-2. Thus using Equations (8.NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS
731
= [bjbj + c j Cj] (area of the element) = Afbjbj + CjCj] Using Equation (8. and G . x and y are approximated as x « x = i.(xj + Xj + x k ) y « y = j (Yi + yj + y k ) The approximate value of Gj is thus G| . K is a (3 x 3) matrix for each element.8-42a) (8.8-4ia. is a vector with m components. Equation (8. y We note that since each triangular element has three nodes.8-38c.8-26a to c) into Equation (8.8-24a).8-42b) (8.8-42c) is obtained by substituting Equations (8. . That is to say. We then have to assemble them to obtain the global K. then the global matrix K. If the domain has m nodes.8-38b) (8. ^ is a (m * m) matrix and the global G / > . Solve the equation
.8-38c)
= I I g ^ j + bjx + cjy] dxdy
(8.b) (8. we can compute Kjj and gj for each element E.8-31d) can be written as
Gi
(8.8-42c)
Equation (8. and G is a vector with three components. By numbering the nodes appropriately. •>. If the triangular element is small enough.
y (b)) respectively. where its value is 0. and for computing approximate solutions of differential equations.1-1. In differential calculus we learned to determine the extreme values (maxima or minima) of a function. the function f (x) = x 2 has a minimum at x = 0. If the coordinates of A and B are (a. Calculus of variations is an extension of the above concept.
(9-'-»
y(b)
y(a) ^ ^ . y (a)) and (b.1-1
Curves joining two points A and B
. a
J
*s/Uy
b
x
FIGURE 9. For example.1 INTRODUCTION
Calculus of variations was introduced by Johann Bernouilli two centuries ago and is now widely used in optimization. control theory. f (x) is greater than zero everywhere else.CHAPTER 9
CALCULUS OF VARIATIONS
9. Suppose we wish to determine the shortest plane curve joining two points A and B in the xy-plane as shown in Figure 9. then the length s of the curve y = y (x) joining A and B is given by
s=f
J a
V r + (l) 2 d x
y .
1-1) is referred to as a functional. where ( ' ) denotes differentiation with respect to x and Ax = x . In variational calculus we need to determine a function..1-1 shows two such curves. we assign a number s to each function y (x) and the rule refers to an integration with respect to x from a to b. irrespective of the sign of Ax is f'(x o ) = O (9. Hence Q represents an extremum.x 0 . we review the theory of extreme values in differential calculus. Figure 9.. Figure 9. We need to determine the function r B that will maximize the integral. The reaction rates rA. we wish to find the extreme values of a functional.2-2a.2-1)
retains the same sign for all small values of Ax.
9. Expanding the right side of Equation (9. It can be seen from Equation (9. we can obtain the corresponding value of s and we have to determine the curve y(x) which yields the minimum value of s.2-1 illustrates this for the situations where x 0 is at P (both Af and Ax change sign) or at Q (only Ax changes sign). In kinetics we can consider the reactions »B — > C.2-1) about x 0 yields f (x) . rB and r c may be functions of temperature and concentration. In this example. irrespective of the sign of Ax. The integral on the right side of Equation (9.b)
. tj) is given by
P
1=
JO
r B dt
(9.2
FUNCTION OF ONE VARIABLE
In Chapter 1.1-1).f (x0) = Af = f' (x 0 ) Ax + 1 f "(x 0 ) (Ax)2 + .2-3) (9. where a functional can loosely be defined as a rule which assigns a real number to a function. In engineering we encounter similar variational problems..1-2)
Equation (9. A— We may wish to determine the reaction temperature that will produce the maximum amount of B in a fixed time interval from 0 to tj.1-2) is similar to Equation (9. such that an integral is an extremum. The production I of B in the time interval (0. Prior to considering variational calculus.2-2) that the condition for Af to preserve its sign. y 0 and y.740
ADVANCED MATHEMATICS
For each curve y (x). if it exists. That is to say. it is stated that a function of one variable f(x) has an extreme value in the neighborhood of x 0 if Af given by Af = f(x o + Ax)-f(x o ) (9.
.. x 2 . X) = f (x1? x 2 .n (9. m
(9.4-3b)
dX
From the (n + 1) equations [Equations (9.4 CONSTRAINED EXTREMA AND LAGRANGE MULTIPLIERS
We have assumed earlier that Xj. . xn) subject to the condition g ( X l . . . x2. and the constraints are given by g j ( x 1 .. If we have m constraints. Thus if Xj represents the length of a rod and x 2 is the temperature. In practice.. There might be a relationship between the Xj.4-3a. then x^ is a function of x2. it is a maximum along one line and a minimum along another line. . instead of one.. x 2 .. .. The Lagrangian becomes m L = f (xj. Xj and x 2 are not completely independent...4-4)
(9. 2 .. x n )
j=l
j = l .x n ) = 0 We now consider the function L. x n . if Xj is the cost of a product. It is preferable to use the method of Lagrange multipliers. 2 . . The extremum of L is then given by |^=0. we can solve for Xj = Xj and X.. x 2 . . in the neighborhood of the origin. . .. .CALCULUS OF VARIATIONS
745
The extreme value of f is at the origin where Dj and f are zero and the quadratic expression is not definite. .... sometimes called a Lagrangian. .. In theory it is possible to eliminate all dependent variables and retain only the independent variables. . .. . .. xn are all independent and can take any arbitrary value..4-2) (9. xn) + Xg (xj. this can be very cumbersome and the process of elimination can be very time consuming. The origin is a saddle point. For example. . .4-1)
— =0
(9. and is given by L (xj.. x 2 . x 2 .4-3a) (9. . Suppose we wish to find the extremum of f (xj. This might not necessarily be the case.b)]. then Xj cannot be negative. xn) + ]T. 9..x n ) = 0 . .. i = l . Xj gj (x lf x 2 . f is greater than zero if Xj and x 2 are of the same sign and less than zero if Xj and x 2 are of opposite sign.4-5)
... x n ) where X is the Lagrange multiplier. x 2 .
Substituting Equation (9.5 EULER-LAGRANGE EQUATIONS
We now deduce the conditions for the functional (integral) I given by 1=
I" f(x. the point (x1? x2) has to be on the circle of unit radius with the centre located at the origin.CALCULUS OF VARIATIONS
747
Note that.5-3)
Integrating the second term by parts yields
/•b
. On expanding the function f [x.5-5). in this example.
(9.5-4) into Equation (9.y' + eTi(x)]dx-l f[x. (Xj. x 2 ) can be any point and thus f = XjX2 has a saddle point at the origin. The points a and b are fixed and we also assume y (a) and y (b) to be fixed.5-4)
Since y (a) and y (b) are fixed. y')dx
(9. according to the Taylor expansion [Equation (1. we obtain
i-b
AI = I
eri |^.dx
(9. The increment AI is given by
i-b rb
AI = I f[x. y^(x)]dx
Ja Ja
(9.5-2)
where e is a small quantity and r\ is an arbitrary continuous function of x.7-7)] and substituting into Equation (9. we deduce that AI preserves its sign for all arbitrary functions erj (x) if
. y^ + er|'(x)] in powers of e. r\ (a) = r\ (b) = 0.y o + ETi(x). we obtain
. (Xj.3-3.
fb
en' — dx = [ETI-^-1 -
eri-jU-^. We can consider a neighboring curve yj = yQ + er| (x) as shown in Figure 9.5-1)
to have an extreme value.5-3).5-2).
/a
y. Let the function y 0 (x) represent an extreme value for I.b
AI=
[eT] K^.a L ^y dx \dy'
]dx + O(82)
. yo(x).+ eif . x2) cannot be the origin.5-5)
From Equation (9.1-1.±IK
. 9. y o + er|(x). In the present problem.^ + O(e2) dx
(9. In Example 9.
If y does not satisfy Equations (9.5-11) reduces to | = 0 (9. y = y (b) (9. b)]. y is not an admissible solution and the functional I does not have an extreme value. dx
-=0
dxdy'
dy'
a y 8y'
Hint: Note that
can be written as g [x.fte 1^
(9.. Example 9. y' (x)] and compute
Thus to determine the extremum of the functional I. at x = b.5-12a. y' of f are present and Equation (9. we need to solve a second-order differential equation [Equation (9. a2f
y
.CALCULUS OF VARIATIONS
749
Thus the condition for ^
d£ e=o
= 0 for all r\ (x) is also given by Equation (9.5-6).6-2). b). We now consider these special cases. f is a function of x and y only and it follows % =0 (9.6-1. 9. y. subject to the boundary conditions at x = a.5-11)]. The variation
~
d£
is also known as the Gateaux variation. a2f Oy') 2
n
.5-6).5-11) is then simplified. y (x).5-12a.6 SPECIAL CASES Explicitly
Function f Does Not Depend on y'
In this case. Find the extreme values of
. y = y (a). not all the arguments x. The solution y = y (x) has to satisfy the boundary conditions [Equations (9.6-2)
From Equation (9. we obtain the relation y = y (x).5-11) -^-. we have
af By
a2f
--y
.5-12a. Expanding Equation (9.6-1)
ay
Equation (9. b)
In practice.
it satisfies Equation (9.1719
Comparing Equations (9. suppose we have a straight line joining the points (1.6-3)
The boundary conditions are (a) (b) y(l) = l.6-7e. which yields 2xy-l=0 (9.6-4a) (9.6-6) [y = (l/2x)] does not satisfy Equation (9.2
1 = 1=
[XJ Jl I 4x2
Udx = 2x) Ji
-L dx
4x
(9. The extremum of I is then given by Equation (9.e)
= . The equation of such a line is y = l(3-x) Substituting this y into I.x ) 2 .b. 1/4). 9d).6-2). y(l)= 1/2. I has no extremum.6-7d.6-4a) and. However. it is seen that ^ is a minimum.6-6) (9.750
ADVANCED
MATHEMATICS
I"
1=1
2
(xy z -y)dx (9. in this case. y(2)=l y(2) = l/4 (9.1733
For the purpose of comparison.6-4b) and y = (l/2x) gives the extreme value of I.6-7a.2 .c)
(9.6-8)
f2
I = I2 = 1 { ^ ( 3 .
(9.c.1 i n 2 = -0.b. In this case
.l (3-x) jdx = .d)
.1 1 =-0.6-5)
The solution of Equation (9. 1/2) to (2.6-4b)
In this problem f = xy2-y which is not an explicit function of y1. we obtain (9.6-9a.
the function y = (l/2x) minimizes I.b)
I /T^2?
where c 2 is a constant.6-15a.c ? x 2 + c 2
(9.5-12a. f =
dx
(9.12)
y. Of all the functions that satisfy Equation (9. Equation (9.J .6-11) dy' From Equation (9.6-14)
Vl-c 2 x 2
Integrating yields
y = I
C]X
dx = . we obtain
y' =
hi
(9.CALCULUS OF VARIATIONS
757
This means that there is no function (curve) y (x) that extremizes I and satisfies Equation (9. here we always have a solution.
Cl
.b)
^
x
Solving for y'.6-11). we introduce two arbitrary constants which are determined by the end conditions [Equations (9. we can solve for y' and y' can be further integrated to obtain y.6-2.6-13a.V1 . b)]. Unlike the previous case. Since we have integrated twice. Function f Does Not Depend on y Explicitly
In this case.6-11) yields
\ V 1 + ^—^ (v')2— 1 is not an explicit function of
xVl+(y')2
(9.6-4b).6-4a).6.5-6) simplifies to
f(!M = o
dx \dy J
(96-io)
which on integrating yields = constant = Cj (9. Equation (9. Example 9. Find the extremum of
i= C H ^ L
In this example.
As usual.6-22). If the curve passes through the origin (0.6-18) (9.752
ADVANCED
MATHEMATICS
We may rewrite Equation (9.c 2 x 2 ) (9.y l ^ r l = o
L dy The solution of the above equation is f .= constant = Cj dy'
(9. Function f Does Not Depend on x Explicitly Since f is not an explicit function of x (9.6-21)
(9.6-16) Cj=c2=l Substituting the values of Cj and c 2 into Equation (9. 1). 0) and the point (1. 1).b)
I"0
Equation (9.6-20) by y1. we find that the resulting expression is an exact differential which may be written as
-f. The constants c1 and c 2 are determined by the boundary conditions.[ f .6-16). a2f .c 2 ) 2 = ( l .y —T7 = ° 8ydy 0y')2
( 9 6 -19)
.6-20)
Multiplying Equation (9. we can solve for y' and this can be integrated so as to obtain the required y as a function of x.6-16)
and this represents a family of circles with centres along the y-axis.6-22)
From Equation (9.6-15b) as c2 ( y .5-11) simplifies to df By .^ . we obtain from Equation (9. we obtain (y-l)2 + x2=l which is the equation of unit circle with centre at (0.y ' .6-17a.. the two integration constants are determined from the boundary conditions.
. a2f _ y T ^ T .n£*ns (9.
Find the extremum of
. while falling under gravity.26)
^(l+(y')2)1/2y-l/2 V2g
il. Integrating y' yields (9.6-25a.6-30)
.b)
Jo
Since f
=
v
^ Jo V 7
(9.(y1)2 0 + (y')2)~m
y~m
= Cl
(9.6. 6-24)
t =
fa ds = J L f3 Vl+(y') 2 dx
(9. we have to determine the path that a particle will take. so that the time taken is a minimum. It is usually accepted that this problem marks the beginning of calculus of variations. If we take one of the points to be the origin of the Cartesian coordinate system.1-1. Equation (9. x the horizontal axis. using Equations (9. b) is. Then the time t needed for the particle to fall from the origin (0. we can then solve for y'.
^
f
Vl + (y')2
dx
(9. we obtain I =ciy1/2(l+(y')2)1/2 Squaring both sides of the above expression.
8y'
y T " 2 ( i + (/>2)-'/2
V2g
(9 .CALCULUS OF VARIATIONS
7J1
Example 9. and y the vertical axis. 6 .6-22) becomes
(i + (y') 2 ) 1 / 2 y'm
. then the velocity v at height y is computed through the conservation of energy (potential + kinetic = constant) to be v = /2gy (9. It was first proposed by Johann Bernouilli in 1696. 0) to a point (a.6-23)
This example is known as the brachistochrone problem.6-24)
where g is the gravity. In this example.6-3. 27)
Substituting f and dy'
and absorbing V2g into the constant c 1 .6-28)
On simplifying the above expression.6-29)
I A / -^— dy = I dx ] J V l-c2y
(9.
that the work done by a conservative force is independent of the path taken. If we take (p.6-44)
(9. y') dx = I ^ dx = ())b .6-45)
Then the work done by a from one point to another is independent of the path and is only a function of the difference of potential between the two points. Find the extremum of
.5-12a. then (9.c)
«-. The above result is equivalent to the statement in mechanics.6-40) is identically satisfied.-(£-£)
If Equation (9.6-40)
Equation (9. then the expression given by Equation (9.6-40) is identically satisfied.b)].6-36) is an exact differential.6-42a.6-43)
(9.6-40) gives y as a function of x and there is no integration constant.6-4.CALCULUS OF VARIATIONS
755
The condition for I to be an extremum is dp da (9. Example 9.6-41)
The integral becomes I = | f (x. q) to be the components of a vector a in a twodimensional space. and there exists a function < ] > such that dd> (9. such that > | a = ± grad <
(9.<j)a Ja Ja ax Thus I is independent of the path taken and all curves yield the same value of I.b. y. This is similar to the case where f is not an explicit function of y'. Thus an extremum of I will exist only if y satisfies the end conditions [Equations (9. If Equation (9. then curl a = 0 So a is a conservative force and there exists a potential <|).
6-48a.6-40) is identically satisfied. y. there is no extremum.
In general. . 0) and (1. .b)
Thus the condition is 2y = 2x or x-y = 0 (9.
f1
1= (y + xy')dx=
fl
[ydx + xdy] =
fl
d (xy) = [xy]^ = y (1) (9. I does not depend on which curve y (x) is taken. So in this case there is no extremum.6-5.6-49)
In this example.6-50atoe)
Jo
•
Jo
Jo
Therefore.o (9. 9.256
ADVANCED
MATHEMATICS
1=1 (y2 + x 2 y')dx
(9. then we have to solve the second order differential Equation (9. y ^n\ Thus we find the extremum of
. 1). then the first end condition is satisfied. q = x. . such as y ".6-40) with p = y2 and q = x2 (9. . and y'. then the curve y = x will give the extremum of I.7 EXTENSION TO HIGHER DERIVATIVES
We extend the theory to f being an explicit function of higher derivatives of y. so Equation (9. y ' " . Find the extremum of
r1
1=1 (y + xy')d* .6-47a. if f is an explicit function of all three variables x. Example 9. If the second end condition is modified to (1.6-46)
Jo
The requirement for an extremum of I is given by Equation (9. p = y.b)
If the end conditions are represented by the points (0. In this case.5-11) and the two end conditions will determine the two arbitrary constants.2). but the second one is not.
8-1)
L dy Ja
Using Equation (9. y". The y-coordinate of the curve will have its ends at any point along the ordinates x = a and x = b as shown in Figure 9. If the value of y is fixed at one of the end points (say at x = a) and is free at the other end. since y (a) and y (b) are not fixed.y. We obtain Equations (9.5-6. y " . . are given at x = a and x = b.7-3)
dy
dx U y '
dxMay"
dx3 (ay111
""
dxn \ 3y (n) j
This equation is essentially a generalization of Equation (9. y1. 4). In deducing Equation (9..8-1) simplifies to -^ =0 (9. are assumed to be satisfied. but the values of y and its derivatives at the end points not specified. but we can now no longer impose the boundary conditions r| (a) = r\ (b) = 0.5-6).8 TRANSVERSALITY (MOVING BOUNDARY) CONDITIONS
We reconsider the problem of finding the extremum of I [see Equation (9. then the condition given by Equation (9.7-2)
where e is small and r\ (x) is an arbitrary function of x and proceeding as in Section 9. we can show that the condition for I to be an extremum is (9. we have assumed that y. y " \ . Thus the conditions for I to be an extremum are now given by Equations (9. then by considering another function y = y o + eii(x) (9. We proceed as before by considering two curves y 0 and y 0 + £T|. as required. . we obtain Equations (9. Instead we need to deduce two new conditions.5. y ( n ) ) d x
(9.8-2)
V x=b
. 9.8-1). y'.. The alternating signs are a result of several integrations by parts of the derivatives of T|. 8-1).7-3). .CALCULUS OF VARIATIONS
757
I =
/a
f(x.5-3.5-5. The extremum curve has a greater degree of freedom and we cannot determine the two arbitrary constants in the solution of the Euler-Lagrange equation from the end conditions.7-1)
If y = y o ( x ) is the function that will make I an extremum. We impose a new transversality condition Tl-2-7 =0 (9.5-1)] with the end points x = a and x = b fixed. 6).8-1. The existence and continuity of y and its derivatives.
8-14a) (9. Imposing the end conditions y(0)=l. k1 = ( a . we can assume that y' is a constant.2 a + 2] [2a-2] = 0 (9.8-13a. We suppose y (1) = a. we find that ko=l.8-1 la. At least one of the roots must be real and since c1 is an arbitrary constant.8-9)
The function y and the integral I are given by y=(a-l)x+l (9.8-7)
I = 70
dx = 1
(9.2 a + 2] Jo
(9.8-6).8-14b)
The solution is a = 1 and I = 1 which is the solution given by Equation (9. The condition for I to be an extremum is still given by Equation (9. That is ^ = 0 or 2 [ a 2 .b)
We regard a as a variable and determine the values of a that will make I a minimum.8-8)
Let us determine I when the values of y are specified at both ends. we obtain automatically the value of y(l) which renders I an extremum.
.b) y(l) = a (9. Thus y = k 1 x + k0 where kj and kQ are arbitrary.l ) (9.CALCULUS OF VARIATIONS
759
y=1 Thus the extremum of I is given by
(9.8-12)
I = [a2 .2 a + 2] I dx = [ a 2 .b) (9.8-4).8-10a. which is a cubic equation in y'. Using the transversality condition.
260
ADVANCED MATHEMATICS
9.9
CONSTRAINTS
So far we have been concerned with finding the extremum of (b Ja with y satisfying certain end conditions.9-3)
J* = I g(x. y')dx
(9. y o + e 1 r | 1 + 8 2 r i 2 . y.5-1)
J = I g (x. algebraic equations or differential equations. Consider a neighboring curve to y 0 given by y = y o (x) + e 1 r| 1 (x) + e 2 r| 2 (x) (9. y o + e 1 r ) j + e 2 r i 2 ) d x
(9. We define I* and J* as
/•b
I
=1
/a
f(x. We need to find the extremum of I Equation (9. in many problems y has to satisfy additional conditions and these conditions may take the form of integral relations.9-4b).9-4a. y. In extremizing I we can allow £j to be completely arbitrary as before and then adjust e2 so that the condition given by Equation (9. y')dx = constant = C
(9. The functions Tj^x) and T|2(x) are arbitrary functions. we have introduced two parameters £l and £ 2 .
= C
(9. y'Q+elr\[+e2ri'2)dx Ja.9-lb) is satisfied. subject to the constraint
1=
f(x.b)
I* and J* can be considered as functions of £j and £ 2 and the extremum of I* is given by subject to the constraint given by £j = e 2 = 0. then they are known as isoperimetric constraints which we shall consider next.9-la.9-2)
In this case. Suppose we have to find the extremum of I as given by Equation (9. This problem is equivalent to the case of finding the extremum of a function of two
. If the supplementary conditions are of an integral type.5-1). However.b)
We assume that y o (x) is the function that extremizes I. y o + EJTIJ+ e 2 ri 2 .
9-12)
Jo
and the length s is given by Equation (9.9-11)
Example 9.M y + W l + t y 1 ) 2 ) ] ^
(9. we obtain
.9-1.1-1).9-1 The area A = I ydx
i
*
Area (shaded) enclosed by a curve of length s (9.9-1).9-10).
y
o
FIGURE 9.762
ADVANCED
MATHEMATICS
and
gdx = C
(9. Given a curve of fixed length s joining the points (0.9-13)
dy dx [ dy Taking the partial derivatives. 0) as shown in Figure (9. remembering that y and y' are to be treated as independent variables.^ f . Using Equation (9. the extremum is given by
-My + x V l + t y f l . find the extremum of the shaded area enclosed by the curve and the x-axis. 0) and (1.
9-25a)
I
\2X>
X
J
o
= 2A.and c 2 = ± V A
(9.9-27a) (9. y ) (9. Find the extremum of
.9-26)
Thus the required curve is the semi-circle with centre (1/2.9-2. then we have to choose the Lagrange multiplier as a function of x. We can now express X in terms of
(9.1-1).1-1) yields
s = XI dx — jo y?i 2 -(x-i/2) 2
This integral is of a standard form and its value is
(9. X .0) and radius 1/2. y. -j .9-27b)
Example 9. y') = constant or y1 = g ( x .1/4) ) and radius X. with solutions Cj = }. ±V X2.1/2 and the equation of the curve is given by y2+(x-l/2)2= 1 (9. instead of a constant. Differentiating Equation (9.9-24)
s^fsin-1^^]1
(9.9-23)
To obtain X. If s = K/2.C l ) 2 = l 2 . • If the constraint is of algebraic type or in the form of a differential equation.siiT1( -J-)
[since sin"1 (-x) = -sin" 1 (x)]. The constraints can be of the form f (x.9-25b)
s and the required curve is a circle with centre
(1/2.9-21) with respect to x and substituting y' into Equation (9.764
ADVANCED MATHEMATICS
cf + ( l . we impose the isoperimetric constraint given by Equation (9. This result can also be obtained from geometrical considerations.
9-30)
Subject to the constraint yi-y' = o The Lagrangian L now is given by L = (l+(yi)2) + ?i(y1-y') = 0 (9. yi' yi» . y 2 (b) = b 2 . y 1 (b) = b 1 .. y 2 (a) = a 2 . y n (a) = an y n (b) = b n (9.10-2a... y 2 .10-3a.CALCULUS OF VARIATIONS
265
1 = 1 [l+(y") 2 ]dx
(9. ... and then we shall be able to complete the above example. We write yi = y" Equation (9.9-32) (9.10 SEVERAL DEPENDENT VARIABLES We generalize to the case of several dependent variables....• y i i ) d x
(9.9-28) becomes (9. . . . . y n . 9. y n .c) (9.c)
. . We need to determine the n functions y 1. yn that extremize the functional I given by
i=| }a
f(x.b. y2. .9-29)
1 = 1 [l+(yi)2]dx
JO
(9.9-28)
Jo
We reduce the integrand to the usual form in terms of y'. The function f depends on n dependent variables y J .. . y2.9-31)
Thus the problem of having a higher derivative in the integrand has been transformed to one involving two dependent variables y and y^ subject to a constraint.10-1)
We assume the end points are fixed and the conditions are y 1 (a) = a 1 . . yi.b. We first extend the theory of calculus of variations to the case of more than one dependent variables.
.5-9). yoi+ejTi!. . .10-6)
The extremum of I is then given by
ft* aT
aei
=
(9. as Tli —
+
^ T
L
dx = 0
( 9 .9-2. yon+EnV yii+eiTll.10 " 8 )
Ja [
^i
hi
The conditions for I to be an extremum are obtained by integrating the second term in Equation (9.)dx
(9...io-7)
£i=0
Equation (9. again following the development leading to Equation (9. Suppose y 01 . y o2 . 11) become
. .766
ADVANCED
MATHEMATICS
We proceed as in the case of one variable.10-5a...b)
!*(%) =
/a
f (x. Equations (9. yOn are the functions that extremize I and consider the functions y i = yoi + e i r l i ( x ) ' i=1 n (9.
dy{ dx \ dy[ j
i=l. Complete Example 9.n
(9.10-10) (9. y.10-8) by parts and the result is
iL__l(_*U0. The extremum of I is now given by
*_A(ik\ = 8y dx 1 dy' I ^-±(^)
0
(9.9-32). .10-11)
=0
3Yl
dx [ dy[ J
Using Equation (9..10-4)
where the r^ are arbitrary functions.2. satisfying the conditions Tii(a) = Tii(b) = 0 The functional I* (£j) is given by (9..10-1.10-9)
Example 9.10-10.. n +e n Ti.. .10-7) can be written. ..
.. A particle P of mass m lies on a smooth horizontal table and is connected to another particle Q of mass 3m by an inextensible string which passes through a smooth hole in the table as shown in Figure 9..10-1.10-33)
*n
velocities are q . 0) as shown.2. The components of the velocity of P are (r. q2. be the origin.10-35)
. The polar coordinates of P are (r. If the particle P is projected from rest with a speed V8ag along the table at right angle to the string when it is at a distance a from the hole. q .e-r) (9. The generalized coordinates of the dynamical system are q1.10-36) (9.10-37) (9. r 0). determine the ensuing motion of P and Q.CALCULUS OF VARIATIONS
J69
1
=I
L d t
(9.r). . qn and the generalized
*1 *2
*j
(9. q .10-34) are known as Lagrange's equations. Taking the level of the table to be the zero-level of potential energy.. Thus the condition for I to be an extremum is
^L-A/^M dt\9q'j Bq1
o.. L is then a function of q1 and q .
i=l.10-32)
in an extremum..10-4. The dot denotes differentiation with respect to time. . Let O. the position of the hole. Since the string is inextensible. The kinetic energy of the system is K = i. the potential energy of particle Q is . Example 9. L is the Lagrangian and is given by L = K-q> where K is the kinetic energy and c p is the generalized potential energy. where k is the length of string.n
(9. ....m ( r 2 + r 2 0 2 ) + 3 m ( f 2 ) The potential energy cp is q >= -3mg(i-r) TheLagangian L becomes L= \ m(f2 + r202) + 2lllr2+3mg(. Q will move up or down with speed r.10-34)
The n equations given by Equation (9.3 mg(/8 .
10-43) may now be written as
2dr^2)
=
^3^~
3 g
(9.10-45)
which can be integrated to yield
2 r 2
•9
= _ ^ . we obtain
4
-r'=8a_g_3g r3
(9.10-42)
Substituting the value of 0 into Equation (9.10-46.3 g r + C2 r2
4a p
(9. r* is negative (-2g) [see Equation (9. 47). this does not imply that the radial acceleration at t = 0 is zero. we find that r
r. r* is positive. r = 0 . C 2 is calculated to be C 2 = 7ag From Equations (9.10-46)
where C 2 is a constant.2=7ag_2*jg_3gr
(9. That is to say.10-47) is given by
(9104g)
2
r2
2
We can deduce from Equation (9. we deduce that r = 0.10-43)
The term r can be written as
" =i
<h
= i < f >$ ' ^
• 2t
(f2)
(9.10-48). r" will be zero when r3 = 8a 3 /3. At r = 2a.10-43).10-43)] and
.10-44a. the particle Q will initially more up and r will increase. Consequently.& .10-38b). when r = 2a and Q will come momentarily to rest. From the initial conditions r = a.d)
Equation (9. From Equation (9. that initially since r = a. although the radial velocity is zero initially (at time t = 0).c.b.CALCULUS OF VARIATIONS
721
6 = ^
(9.
Geometrically it means that we are considering a surface instead of a curve.2a). That is to say. because of the lack of friction.772
ADVANCED
MATHEMATICS
so Q will fall. u X i (x!. The value of u (xj.x 2 ). 2). = =— (i = 1. let us consider a function u (xj. The particle P aV^gas^ has a radial velocity given by Equation (9.11-1
^
Region R bounded by curve C
We need to find the function u which extremizes I over the region R and satisfies the boundary condition
.10-48) and a tangential velocity — ~ . For simplicity. The radial position of P lies therefore between a and 2a.a) and (Z .'
Figure 9. we allow the cycle to repeat when r is again equal to a.x 2 )] dxj dx2
(9..
Q
FIGURE 9. 9. Thus Q will oscillate between the heights ( i . u X2 (Xj. x 2 ) which is a function of two independent variables Xj and x 2 .x 2 ).11 SEVERAL INDEPENDENT VARIABLES The integrand might be a function of two or more independent variables.
R
u(Xj. x2) is given on C.11-1)
where u x .11-1. and R is a region in the XjX2-plane bounded by a curve C as shown in
1 O A. We now have to find the extremum of I = l | f [xv\2.
The system given by Equations (9.dv: \ where y-.1. The equation of continuity is given by -^i. Note that in this example. as
g
+
^_
p f l
=
0.11-44.= v H = 0
dXj
M
(9.are the velocity components.
1 = 1. Show that the equations governing the slow flow of a generalized Newtonian fluid are equivalent to extremizing a functional. Remember the convention. x^ is the deviatoric stress tensor.11-46)
m e rate
°f deformation tensor. = 1 / 1 y.-44)
where p is the pressure. p is the density and fj are the external applied forces (usually only gravity). neglecting inertia. ni is the component of the unit normal to the surface S t .. requiring summation over repeated indices. y.\ = ^rL +^is I I . we consider different boundary conditions on different parts of the enclosing surface. v. / 3v. (9. The equations of motion. can be written. ) ^ .1 1 ( 1 1 . 46) subject to the boundary conditions stated above will be shown to be equivalent to finding the extremum of
.11-45a..CALCULUS OF VARIATIONS
779
(9. when referred to a rectangular Cartesian coordinate system O x ^ X g .2.. i = = — * .b)
The constitutive equation of a generalized Newtonian fluid may be written a Tij = .(p ns + x.
Let V be the flow region (volume) bounded by two non-overlapping surfaces S v and S t . and
is the second invariant of y.11-43)
where V . j
Example 9.3
(9.n:)] is given on S t . The velocity is given on S v and the traction (surface) force tj [= .11-2.
This integral has to be minimized and we need to compute ^ .1 l-68a. (1987).r k)
J
KR
(9.CALCULUS OF VARIATIONS
Z21
Since we have defined 7 and v z as functions of \a\.q a where C o =
°
(9. da Differentiating yields 4 1 = C n (n+l)[2a(n+l)] n »2(n + l ) . 75). b.C 1 = 0 da Hence RAF ) 1/n R(l-k)< l4 *> /n n 2^7/ 2(n+l) .b)
using vz. given by Equations (9.11-73) .11-74a. (9-H-75) (9.Q 1 . For smaller values of k and n.11-72) will be zero when integrated from a = .1-76a. This technique is known as the Rayleigh-Ritz method.5 and n > 0. the first term in both integrals of Equation (9.
Cl
rcLRm(l-k)n
(n+l)(l
+
2
2
= £ ± £ KAP R2(l -k 2 ) .b)
a =
We can now calculate the volumetric flow rate Q.l to a = 1. In Bird et al. Integration yields
I = CopaCn + l f l ^ .
.
<2n+1)
2n)[nR(l-k)] n + 1
'
The integral I is now a function of a. the value of Q for various values of n and k are compared with the exact solution. since they are odd functions of a. the approximate solution gets worse and could probably be improved by considering more terms in the trial solution. " . which is given by
^/ R ^-«fer (i -c. This is followed by expressing the unknown functions as a series of simple functions that satisfy the boundary conditions. The above approximate solution is within 2% of the exact solution for k > 0. The unknown coefficients are determined by satisfying the variational principle. • The method we have used to solve the present boundary-value problem consists of rewriting the governing differential equations and boundary conditions as a variational problem.25.
. only along the straight lines x = a. x = b). yj.8 to the case where the functional depends on n functions. y 2 . The quantity 3— is given by
d8j
df
<*ei
J.. [ df dy:
+
j=1
df dy': ]
(9.12-2a.y 2 )
(9.12-4a)
dy.
i = l. y n . Using Leibnitz's rule.786
ADVANCED
MATHEMATICS
Although simple problems.12-1)
Jp
The end points P and Q are not fixed nor are the values of y 1.10-7) becomes
* /Q
|L=
l^dx +f ^
. We assume that P and Q lie on two surfaces respectively given by
x
= <l>(yi>y2)'
x = \|/(y 1 .10.12 TRANSVERSALITY CONDITIONS WHERE THE FUNCTIONAL DEPENDS ON SEVERAL FUNCTIONS
We extend the theory considered in Section 9.f | * = 0 . but now the end points are free to move on the surfaces given by Equation (9. more complex problems require the use of an approximate solution.. y . We further relax the condition that the end points can move only along the ordinates (that is.aej
|^ = 2
3 y j 9£j
=2
j=1
^^+r^^ :r
dy.b)
where P and Q are functions of 8j.b)
We proceed as in Section 9. b). such as this flow problem. Equation (9. can be solved exactly via the equations of continuity and motion.y 2 . We allow the end points to move along given curves or surfaces and we label the end points as P and Q.12-3a.2
(9. For simplicity we derive the transversality conditions for the case of two functions y 1 and y 2 .
9.dB{ 9yj d^
v
(9-12"4b)
dx '
. y2)dx (9.12-2a. Variational methods are widely used to obtain approximate solutions to boundary value problems in continuum mechanics. y 2 at these points. We aim to find the extremum of
rQ
1= f(x. . y i ..
12-4c)
Substituting Equation (9. we obtain
0=
-^ I
dx +
li
+ f ^ -f^-
(9.2
(9.12-5). That is to sa y . Substituting Equation (9. Therefore.10-4)] are related for example b y an expression such as y ^ + y2 = c.CALCULUS OF VARIATIONS
787
& F 3f ty d df dy: i+ _ ^ _ i = 2.12-5)
J P a£i J9 Yj dx a y .a £ i j p
^ Q a^ P
using the convention of summation over repeated indices.d / df 1_ 9 ^ dx ^ 3 y j
(9.12-7)
Next.12-6) where d^ is the Kronecker delta. we deduce the extremum conditions to be
K_±(K\ 3Yi dx [ dy[ J
= 0. to be functions of x and e.12-2a. we differentiate to obtain
.12-8)
(9. dx a y j 3e. we deduce the transversality conditions. if equations y\ and y 2 [see Equation (9. It can be seen from Equation (9.12-2a. The transversality conditions are to be imposed at the points P and Q which are on the surfaces given b y Equation (9.12-9)
Since y^ is a function of x and e. then a change in £j automatically induces a change in e 2 and we can indeed consider the y .
i=l. b). except at P and Q. The transversality conditions may now be written as
2 2
"iL^L + iL^ +f 3Ql = 0 dy\ de dy'2 de 5e _ '^^L+^_^l+faP] dy[ 9e 3y 2 9e de
= 0
(9. | j [a yj . y 2 and x are related through Equation (9. derivatives with respect to £j are now replaced b y derivatives with respect to e.10-4) that (9. b) and we can regard y j and y2 to be functions of x and e instead of considering them individual^ as functions of £j and e 2 .12-6) into Equation (9. Thus y i .12-3) and simplifying. j=i p Y j ^
dy.12-4c) into Equation (9. At an y point in space. y i and y2 are independent.
b)
The extremum condition will be given by Equation (9.12-22)
S =I
JO
Vl+(y') 2 dx
(9.12-24a. the transversality conditions at Q are
"iL 3 ( f _£ . y n .
i=1
"
(9.12-7) with y^ = y.. The transversality conditions then reduce to
"iL+?l(f-y'it] .12-25a.b)
With f given by Equation (9. y 2 .12-2!)
Example 9. Thus if f is a function of yj.."
L3y dy \ 3y' j
Q
(912. then the end points P and Q will be free to move on the curves.2-18)
Similar expression for the condition at P can be established. 0). \ j=i J3yj j J
Q
=0 . ] ]
_3yi 3y..12-23)
Since y does not appear explicitly in the integrand.12-19a. we have
?
y
= Co
=>
y' = k 1 (constant)
(9. Equation (9. The equation of the circle is (x-4)2 + y 2 = 9 The distance from the origin to any point Q on the circle is (9. .12-7) simplifies to — dx — dy' = 0 => ^ = constant 3y' (9. x = \j/(y) (9.b)
Vl+y'2
. £ . Find the shortest distance from the origin to a point on the circle of radius 3 and centre (4.CALCULUS OF VARIATIONS
789
The above derivations can be extended to more than two functions.12-1..12-23).20)
"jL+?lf_y'iLl
3y' dy dy' J P
=o
(9. If f is a function of only one dependent variable y. respectively given by x = 9(y)..
.. we obtain
2
kl
\
--*1J-
V l + kf -
kl
=0
(9. . In this section.k. y n . consider two types of constraints: an integral type and an algebraic type. y = 0) yields y = kjx (9.9.c)
Substituting the value of y into Equation (9.12-22).12-27a. We shall. Q lies on the circle given by Equation (9. y = 0 and the point Q is on the x-axis.0).. A. as in Section 9.
The solution is kj = 0. ..12-20). The condition for the extremum of I subject to the isoperimetric constraint is
.. From Equation (9. . Thus
f
dy
.13-1)
subject to
Jj = I gj (x.. ..12-22) as well as on the line given by Equation (9. in addition to the end points conditions.12-28)
V1 +kf
(x-4)
Vl+kf . yj.790
ADVANCED
MATHEMATICS
Integrating Equation (9. we established the conditions for the extremum of a functional which depends on only one function.b. y ^ d x
(9. * L =^L
dy (x-4)
=
dil±
(x-4)
(9. .13 SUBSIDIARY CONDITIONS WHERE THE FUNCTIONAL DEPENDS ON SEVERAL FUNCTIONS In Section 9.. y n .13-2a.. Then the minimum distance is 1.. y^) dx = constant = Cj . The function was subjected to certain supplementary conditions.12-26).12-20). X2.. we consider the case of a functional which depends on several functions.
j = i. we make use of Equation (9. yj. .. y l 5 y 2 ..12-25b) and using the end condition at the origin (x = 0. 9..b)
We use the method of Lagrange multipliers and introduce ?ij.12-26)
To determine the slope k j ... . We now look for an extremum of the functional
1=
/a
f(x. Ja The end conditions are prescribed. y } . k
(9..9. we deduce that Q is the point (1.
11-44b) (9. z0) and (xj.796
ADVANCED
MATHEMATICS
or
xy-xy = c^zy-zy)
(9. Silastic implants are widely used in plastic surgery (Atkinson. The balloon is then filled with saline over a period of several weeks until the skin has been sufficiently stretched so as to cover an adjacent defect area.13-1.11-44a) (9. y0. Zj) are given.
.13-43b) can be integrated to yield inCx-CjZ) = i n y + /8nc 2 or or x-CjZ = c 2 y x-c2y-c1z = 0 (9. The arbitrary constants Cj and c 2 can be determined if the fixed points (x0.13-44c)
Equation (9. Formulate the problem of silastic implant and determine the optimum design of the implant.13-42b)
where Cj is a constant. 1988).13-44c) is the equation of a plane passing through the origin. The intersection of the plane given by Equation (9. The plastic surgeon needs to know the optimum design of the implant so as to cover the damaged area for a given volume of saline.13-3. Let A be the area of the base of the implant which is in the xy-plane. yj. Example 9.13-25) is the arc of the great circle. This problem is similar to the Plateau problem considered in Section 9.11. Equation (9. as shown in Figure 9.13-43a)
•
Z
•
V
or
l—
= 5y
(9.13-43b)
x-CjZ
Equation (9.13-44c) and the sphere given by Equation (9. The technique consists of inserting a deflated silastic balloon underneath the skin.13-42b) may be written as yCx-Cjz) = y(x-cjz) •
X-Ci
(9.
11-24) is given by
S=j|
A
[ l + h 2 + h2] 1/2 dxdy
(9..13-47a)
. I) = S-IV (9. dy
The volume V of saline is
V = I I h (x.13-45)
. the surgeon will perform the operation and advance the skin to cover the defect area. Once the skin has been sufficiently expanded.13-1
Geometry of an implant
The base of the implant is PQRP and the defect area to be covered is PTRP. then the surface area S according to Equation (9. y) dx dy = Vo
A
(9. h "
= 5
dh -.
•
.*
/!
\
t
FIGURE 9. . x dx
.13-46a. 9h where h =^-.CALCULUS OF VARIATIONS
797
y
R
IMPLANT 1
y
HEALTHY «MN
A
KCALTHV SKIN
^tm^
V^TS^
I
I
. y) be the height of the expanded skin above the base. This isoperimetric problem can be solved by introducing the Lagrange multiplier and is equivalent to finding the extremum of L(h.b)
To ensure that the defect area will be covered we need to know the minimum surface S for a given volume V. Let h (x.
and V o can be obtained. We consider an isotropic body of volume V with non-homogeneous thermal conductivity k(Xj) which is unknown. y) respectively. X = 2. Bj = 0. it can be substituted into Equation (9. For other values of the geometric parameters and X. large values of (3 imply that more importance is given to achieving the
.4632. Bj and Q : A{ = 1. This problem has been considered by Meric (1985).CALCULUS OF VARIATIONS
799
The parameter X can be determined in terms of V o . d. For the values of a = oil = d = 1.8186.2192.13-46) and the relation between A. Also the values of E and the required values of E for various conditions are tabulated in Atkinson's paper. Thus varying the values of X is equivalent to varying the values of V o .13-55)
Small values of (3 imply that more weight is given to attaining the desired temperature distribution T d . We need to determine the optimal distribution of k(Xj) such that it satisfies a desired distribution kd(Xj) and a desired temperature distribution Td(Xj) in the body. ocj and X. Atkinson has found the following values for A 1 .13-1) is given by It L = l ? \ (1+h^j
m
dx
(9. Bj and Cj are obtained numerically for various values of a. p\ d and a j . (3. The length of the curve L on the inflated balloon corresponding to a horizontal (y = constant) section PiP 2 (Figure 9. Cj = 0.5. At the other extreme. (3 = 1.13-4. This is a problem in the manufacture of composite insulating materials. Mathematically we are required to minimize the so-called performance index J defined by (9. The extension E produced by the inflation of the balloon in the x-direction for a given y is E = L-(x!-x2) The balloon must be inflated to a volume such that E is sufficient to cover the defect.13-54)
J = | [(T-T d ) 2 + p(k-k d ) 2 ]dV
where T is the temperature and (3 is a weighting factor.13-53)
where the coordinates of Pj and P 2 are (x1? y) and (x2.
(9. The shape of the implant is determined by the values of a.85. Solve the problem of optimization of thermal conductivities of isotropic solids. since once h has been obtained in term of X. The values of Ai. the reader is referred to the original paper. Example 9.
..2. it costs (T .. Then determine the minimum surface area. We are required to find the straight line V = mT + c which will "best" approximate the data. 4a. A processing operation produces twenty cups per minute at 420 K.. if we denote the error at the point Vj by e. Find the radius r and the height h so that the surface area of the can is a minimum. i = 1.. determine the profit as a function of T. use the method of Lagrange multipliers. By considering E 2 as a
i=l
(ii)
function of m and c. then ej = (V(Tj) . Use the following methods (i) determine the surface area of the can in terms of r and h and substitute r or h in terms of V and hence write the surface area as a function of one variable only. Tj). show that for E equations
n n
to be a minimum. Answer: h = 2r 3a. If the cups are sold at one dollar
each. n. The constraint is the fixed volume V. A cylindrical can holds Vm 3 of soup. A volume temperature profile is represented by n experimental points (Vj.Vj).6). m and c must satisfy the
m^Tj
i=l
+ nc =
Xvi
i=l
mlT^cXTj^TjVj
i=l i=l i=l
The above equations are known as normal equations (see Section 7. That is. What is the most profitable temperature to run the machine? Answer: T = 450K 2a.CALCULUS OF VARIATIONS
$01
PROBLEMS la. An extra two cups per minute can be produced for each degree above 420 K. At a given temperature T.420) 2 15 + -———— dollars per minute to operate the machine. Evaluate the integral
I = I ( y2 + x 2 y')dx Jo
.Vj) = (mTj + c . n The "best" line can be defined such that E = ^ e 2 is a minimum.
0) to (1. Answer: cot (() = a cos (0 + c2) 7a. Answer: y = x 5a. 9.804
ADVANCED MATHEMATICS
along the following curves joining the points (0. z = a cos (|)
Let 0 = F((|>) be a curve lying on the surface of the sphere joining two points A and B. The length of the curve is given by /•B rB
I ds = I Vdx 2 + dy 2 + dz 2 JA JA
Show that in the spherical polar coordinate system
/•B
. d) and is rotated about the x-axis. 1): (i) (ii) (iii) y=x y = x2 y = 1/2 (x + x 2 ) Answer: 0. 6b. Find the curve y (x) which extremizes the functional
. y = a sin § sin 0 .667 Answer: 0.700 Answer: 0. y. cosh
Any point (x.
Answer: y = c.B
I ds = a I V 1 + (F'(<(>))2 sin 2 ^ d^
JA JA By solving the Euler-Lagrange equation.675
Find the curve that will extremize I. determine the function F (())) that minimizes the length of the curve. deduce if I is a maximum or a minimum. The surface area S thus generated is given by
/•b
S = 27C I y V l + ( y ' ) 2 Ja
dx (x — c ) —
cl
Find the curve y (x) which will extremize S. z) on the surface of a sphere of radius a can be expressed in a spherical polar coordinate system (r. c). (b. (])) as x = a sin § cos 0 . From the values of I determined using (i to iii). The curve y = y (x) passes through the points (a.
V 1 + (y ) dx. The differential equation is known as the
dxf
V i=l
I
Schrodinger equation (see Section 6.4 .806
ADVANCED MATHEMATICS
10a. If C A is the concentration of A. (p is the potential.T)dx
Jo
. then the relation between C A and T is given by a kinetic equation of the form
dC A
_
e(C
T)
~dT " g ( C A '
T)
The production of A is given by
1=
f(CA. Consider the irreversible reaction A — > B in a tubular reactor of length Z.
Show that extremizing the functional
1 = 111
R
I In" K + ^2 + ^3) + ^ dx l dx2 dx3
subject to the condition
(II 2
I I I \|/ dxl dx 2 dx 3 = 1
R leads to k2 2
V
~2nT
V+9¥ = . 12b.
Find the shortest distance from the origin (0.6).
/•P / —^
Hint: the distance s = I x = 4 + 2y = <|>(y). The constraint is known as the normalization condition. k and m are constants. y (0) = 0 and the point P is on the line Answer: ^-V~5
lib. 0) to any point P on the line 2y = x .^
/ 3
d2
\
The operator V 2 is the Laplacian
=^
. We wish to determine the best operating temperature T as a function of the axial position x along the tube so as to maximize the production of B.
In Example 9.V 3 C A and the boundary conditions are: C A = 1 at x = 0 and T = T o at x = i . the dependent variables are C A and T. then vz = vz ^ ^ / ( a Ap) has to satisfy the following equations of motion V vz = . we have to minimize the production of A.11-3. Answer: T = 2 sinh 2x + V T cosh 2x + C (2 cosh 2x + V T sinh 2x) C = [ T 0 . we have to extremize I subject to the kinetic equation.CALCULUS OF VARIATIONS
807
Thus to maximize the production of B. that is to say. The flow is still an axial flow but the interface between the two liquids is not circular and the axial velocity v z is a function of r and 0. such as crude oil. The boundary conditions are
. The equation of constraint is in the form of a differential equation and has to be satisfied at each point along the tube. The Lagrange multiplier is thus a function of x and is not a constant. Write down the Lagrange function L. = 9r 2 1 r 3r 1 3 ^ r 2 dQ2 1 7pis the
where r\: is the viscosity of liquid i (i = 1. g = T . If v z is non-dimensionalized and Ap is the pressure drop. V Laplacian. we used a variational method to solve the axial flow of a power-law fluid between two co-axial circular cylinders. In this example. In the present problem we consider the axial flow of two immiscible fluids in a circular pipe of radius a and length Z. Show that the Euler-Lagrange equations are
3C A
9C A
2
dx
2 /
Determine T if f = C A + T . Bentwich has shown that the power required to transport a required volume of liquid.( 2 s i n h 2 i +V3~ cosh 2Jt)] /(2 cosh 21 +V3~ sinh 2 i ) 13b.1
2_
in phase (liquid) 1
r^/r^V vz = -1
2 _
in phase (liquid) 2
2 Tp. This flow situation was investigated by Bentwich (1976). 2). is reduced if a small volume of less viscous liquid is added to the crude.
Note that for one phase flow v z is given by: v z = 1 .808
ADVANCED MATHEMATICS v = 0 on the surface of the pipe which is taken to be at r = — = 1. (j)nm = Jn (A. ^ a At the interface Vzl 1 = V Z | 2 9v z
ni^rT
3vz
j^^rT
2
where n is the outward normal to the surface.
. the above expansion of vz corresponds to a Fourier series expansion. Show that the above boundary value problem corresponds to extremizing the functional
Rj
Rj+R. have to satisfy. On substituting the expression for v z in the double integral.2
where Rj denote the surface occupied by liquid i. We now require to extremize I with respect to Ao and A n m (properties of Jn are given in Section 2. Find the equations that extremize I and write down the algebraic equations that A o and A ^ .7 2 For a fixed 7. 1 and 2 indicate the evaluation is to be made at the interface in liquid 1 and 2 respectively. An approximate solution can be expressed as
N
Vz
M
= A0 % + X
n=0 m=l
X
Anm
V
where ())0 = 1 .7 2 . Jn is the Bessel function of the first kind of order n.nm7) cosn0. A n m are constants which are to be determined.7). we obtain I as a function of Ao and A n m . ^ n m are the roots of the equations Jn (knm) = 0 and the coefficients A o .
2-la. Let the vector position of particle i be denoted by r j relative to an origin O. Students used to deterministic concepts find it hard to think in terms of probability. The present chapter does not intend to provide such a course. each particle is defined by three coordinates. unlike the Cartesian coordinates need not all have the same physical dimension. The choice of the generalized coordinates depends on the geometry of the problem under consideration. then the two generalized coordinates are. a topic which is gaining widespread popularity. ql=r. as shown in Figure 10. q 2 .b)
If the particle is constrained to move on a flat surface of the cylinder. It will be at an elementary level. q2 = 6 (10. we shall extend the analytical mechanics introduced in Chapter 9. Next. Our aim is more modest.1 INTRODUCTION
It is not uncommon nowadays to require students in engineering to take a course in statistical mechanics. the only two generalized coordinates are ql=z. we shall consider probability. If there are m constraints. the number of generalized coordinates will be reduced to 3N .CHAPTER 10
SPECIAL TOPICS
10. say the surface z = h.2-1. If there is no constraint between the particles. We may regard the system to be in a 3N dimensional space and to have 3N degrees of freedom. Hamiltonian mechanics will be introduced.m.b)
. This can also be used to pave the way for a course in chaos. The q1 are independent variables and are known as generalized coordinates.2-2a. We denote these 3N quantities by q1. In a three-dimensional space. q2 = 6 (10...2-1. In the first few sections. We plan to provide the necessary background for the students to be able to follow with ease a more demanding course in statistical mechanics. . Finally. we need 3N quantities to describe the configuration of the system. some topics in thermodynamics and Brownian motion will be examined. but we shall emphasize the interpretation of probability.. The generalized coordinates. The idea of an ensemble average will be discussed.2 PHASE SPACE Consider a system consisting of N particles. q 3N . as shown in Figure 10. 10. Thus if a particle is forced to move on the curved surface of a cylinder of radius a.
. ..m) degrees of freedom and we have n generalized coordinates which are related to the N vector positions r l 5 r 2 . .. t) (10.2-3b)
.2-3a)
I N = rj^Cq^q 2 . r N .q n . .2-1
Motion of a particle on the curved and flat surface of a circular cylinder
We now consider the system to have n (= 3N . . t) where t is the time. — ..6) r j = r ^ q ^ q 2 . The relationship can be expressed as (see Section 4..q n .810
ADVANCED MATHEMATICS
z
I I I
' . The velocity of particle i can be obtained by differentiating r j and we obtain
(10. _1_* y
X
FIGURE 10..
i l _ 9r. as they refer to a particle i.
(10. L is given by L = i-m(x2 + y2 + z2)-(p It follows that ^ = mx = px dx In Equation (10.b) (10.2-5b)
In Example 9. dr. I
(10. z).k dr.2-5a)
_. then in Cartesian coordinates (x. the indices denoting generalized coordinates are written as superscripts. Following the notation introduced in Chapter 4. p x is the x-component of the momentum. Consider a single particle of mass m moving in a potential field which depends on the position only.2-7b).2-4b). adopt a different notation.2-7a. we have deduced Lagrange's equation which is
^-A/iM = aq1 dt \aq J j
0
(9. y.10-34)
where L = K . dr.b)
3qJ
9t
In Equation (10. The kinetic energy of the system is given by
K =
2 £
1
m
i . there is less of a compelling need to distinguish between subscripts and superscripts.10-3. Note that the indices on the vectors Vj and r i appear as subscripts. we have adopted the summation convention.2-6)
. •.SPECIAL TOPICS
811
Vj = £ j
= — q
]
+ ^
(10.i ' . Some authors dealing with analytical mechanics. They employ subscripts throughout. This suggests that a generalized momentum associated with the q1 coordinate can be defined as (10. As they consider mostly scalar quantities such as kinetic and potential energies.2-4a.cp is the Lagrangian and (p is the generalized potential.
p n . .. Let q be the displacement of the particle from its equilibrium position at any time t. .2-8)
dq If the coordinate qj does not appear explicitly in L (qJ may be present in L). A particle of mass m is attached to the end of a linear spring of natural length /Co and of negligible mass.2-9)
. q 1 . The Lagrangian is given by L = \ [m(q)2-G(q)2] The equation of motion is (10. The dynamical state of the system is thus defined by the 2n quantities. qn and the momenta (mass x velocity) by the n generalized momenta p 1 . The mass is displaced from its equilibrium by a distance q 0 along the length of the spring and is then released from rest. which defines a space of 2n dimensions.. qn. The kinetic energy K = ±-m(q) 2 The potential energy is 9 = i-G(q) 2 where G is the modulus of the spring.. Determine the subsequent motion.2-1..812 P1 = .2-12) (10.7 7
ADVANCED MATHEMATICS (10. q-i is known as a cyclic (or ignorable) coordinate. The configuration of the system is specified by the n generalized coordinates q1. The motion of the system is thus described by a path (trajectory or orbit) in this 2ndimensional space....2-10) (10. p 1 . .10-34) simplifies to = pJ = constant aqj Thus the generalized momentum of a cyclic coordinate is a constant... Equation (9. Example 10. . . .2-11) (10. The spring is fixed at the other end and lies on a smooth horizontal table. known as the phase space. p n . .
. q1. p n ). (qn. In the phase space.. If we have n oscillators not necessarily all with the same energy.
f
FIGURE 10. the variables are (q1. t) system.. p 1 ).. It is desirable to transform from the (q1. the variables are (q1. .3-2)
Differentiating Equation (10. t).2-2
The trajectory of an oscillator
1 0 . 3 HAMILTON'S EQUATIONS OF MOTION In Lagrangian dynamics.3-1). pJ).L ( q J .. we have
. q J . pi) corresponds to an oscillator.814
ADVANCED
MATHEMATICS
describe the same ellipse. This can be accomplished by the use of the Legendre transformation.. we have 2n variables (q 1 . qj. The differential of H is given by
ffi =
j = l.3-1)
9H dt
+
3H d q j 3qJ
+
3H d p J 9pJ
(10. t) system to the (q1.n
(10. p 1 .t) = p m q m . t ) . and we sum over m. q1.. Each set of (q1.2. The starting point on the ellipse depends on the initial conditions. A new function H known as the Hamiltonian is defined as H(pJ.
we deduce that K is a quadratic function in q J .2-8).^ d q j . .3-5)
(10.
N
.3-3C)
= qmdpm--^dqJ-^ aqj at Equation (10. Then from Equation (10.— aqj 3L 3H
(10. .4 1 d q j . we deduce qj = — aPJ pj = .^-\ dqj + q m dpm .3-4b.ar = ^
(103-6c)
Equations (10. We also assume that the system is conservative. From Equations (9.3-6a to c) are known as Hamilton's equations of motion.2-8) may be written as
.p j dqj .2-3).3-6a)
(10.^
(10.3-3c) is obtained using Equation (10.^
dqJ . They are a set of first order equations.3-6b) " .2-5b). 5).c)
(10.2-8). then < p is a function of qj but not of q J .^ 9qJ 3q J 9t
(10. 10.3-4a)
(10.3-4c) into Equation (10. We now consider the case where t does not occur explicitly in Equation (10. It then follows that Equation (10.3-3b)
\
aqJj
aqj
at
(10.^ at Comparing Equations (10. we have
*h=*m aqJ dt[aqJ|
= ft (Pj) = P j
Substituting Equation (10. whereas Lagrange's equations are second order equations. Note also the symmetry in pj and qj.3-2.SPECIAL TOPICS
8H
dH = pmdqm + q m d p m .3-3a)
= fpi .3-3c) yields dH = q j dpj .10-34. Such a situation may arise if the constraints are time independent. | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
". . .recommended for the teacher and researcher as well as for graduate students. In fact, [it] has a place on every mathematician's bookshelf." —American Mathematical Monthly
Linear Algebra and Its Applications, Second Edition presents linear algebra as the theory and practice of linear spaces and linear maps with a unique focus on the analytical aspects as well as the numerous applications of the subject. In addition to thorough coverage of linear equations, matrices, vector spaces, game theory, and numerical analysis, the Second Edition features student-friendly additions that enhance the book's accessibility, including expanded topical coverage in the early chapters, additional exercises, and solutions to selected problems.
Beginning chapters are devoted to the abstract structure of finite dimensional vector spaces, and subsequent chapters address convexity and the duality theorem as well as describe the basics of normed linear spaces and linear maps between normed spaces.
Further updates and revisions have been included to reflect the most up-to-date coverage of the topic, including:
The QR algorithm for finding the eigenvalues of a self-adjoint matrix
The Householder algorithm for turning self-adjoint matrices into tridiagonal form
The compactness of the unit ball as a criterion of finite dimensionality of a normed linear space
Additionally, eight new appendices have been added and cover topics such as: the Fast Fourier Transform; the spectral radius theorem; the Lorentz group; the compactness criterion for finite dimensionality; the characterization of commentators; proof of Liapunov's stability criterion; the construction of the Jordan Canonical form of matrices; and Carl Pearcy's elegant proof of Halmos' conjecture about the numerical range of matrices.
Clear, concise, and superbly organized, Linear Algebra and Its Applications, Second Edition serves as an excellent text for advanced undergraduate- and graduate-level courses in linear algebra. Its comprehensive treatment of the subject also makes it an ideal reference or self-study for industry professionals.
Synopsis
This introduction to linear algebra by world-renowned mathematician Peter Lax is unique in its emphasis on the analytical aspects of the subject as well as its numerous applications. The book grew out of Dr. Lax's course notes for the linear algebra classes he taught at New York University. Geared to graduate students as well as advanced undergraduates, it assumes only limited knowledge of linear algebra and avoids subjects already heavily treated in other textbooks. While it discusses linear equations, matrices, determinants, and vector spaces, it also includes a number of exciting topics, such as eigenvalues, the Hahn-Banach theorem, geometry, game theory, and numerical analysis.
About the Author
Peter D. Lax, PhD, is Professor Emeritus of Mathematics at the Courant Institute of Mathematical Sciences at New York University. Dr. Lax is the recipient of the Abel Prize for 2005 "for his groundbreaking contributions to the theory and application of partial differential equations and to the computation of their solutions". * A student and then colleague of Richard Courant, Fritz John, and K. O. Friedrichs, he is considered one of the world's leading mathematicians. He has had a long and distinguished career in pure and applied mathematics, and with over fifty years of experience in the field, he has made significant contributions to various areas of research, including integratable systems, fluid dynamics, and solitonic physics, as well as mathematical and scientific computing. | 677.169 | 1 |
Details about Multivariable Mathematics:
Multivariable Mathematics combines linear algebra and multivariable mathematics in a rigorous approach. The material is integrated to emphasize the recurring theme of implicit versus explicit that persists in linear algebra and analysis. In the text, the author includes all of the standard computational material found in the usual linear algebra and multivariable calculus courses, and more, interweaving the material as effectively as possible, and also includes complete proofs. * Contains plenty of examples, clear proofs, and significant motivation for the crucial concepts. * Numerous exercises of varying levels of difficulty, both computational and more proof-oriented. * Exercises are arranged in order of increasing difficulty.
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Patterns, ratios, equality, algebraic functions, and variables are some of the concepts covered in this printable book for elementary students. You'll find a variety of materials to encourage your young students to learn math | 677.169 | 1 |
First Course in Computational Algebraic Geometry is designed for young students with some background in algebra who wish to perform their first experiments in computational geometry. Originating from a course taught at the African Institute for Mathematical Sciences, the book gives a compact presentation of the basic theory, with particular emphasis on explicit computational examples using the freely available computer algebra system, Singular. Readers will quickly gain the confidence to begin performing their own experiments. | 677.169 | 1 |
Popular Textbooks
Details about Number Power 4: Geometry:
Number Power is the first choice for those who want to develop and improve their math skills! Number Power 4: Geometry introduces lines, angles, triangles, other plane figures, and solid figures.
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Rent Number Power 4: Geometry 1st edition today, or search our site for other textbooks by Robert Mitchell. Every textbook comes with a 21-day "Any Reason" guarantee. Published by McGraw-Hill Education.
Need help ASAP? We have you covered with 24/7 instant online tutoring. Connect with one of our Geometry tutors now. | 677.169 | 1 |
John Napier
Free Course
This free course looks at Scotsman John Napier, best known to for his treatise on Protestant religion. However, it was his interest in a completely different subject that radically altered the course of mathematics. After 40 years of dabbling in maths, he revealed his table of logarithms in the early 17th century.
After studying this course, you should be able to:
understand the significance of John Napier's contributions to mathematics
give examples of the factors that influenced Napier's mathematical workJohn Napier
Introduction
This course provides an overview of John Napier and his work on logarithms. It discusses his approach to this lasting invention and looks at the key players who worked with him, including Briggs, Wright and Kepler.
This OpenLearn course provides a sample of Level 2 study in Mathematics [Tip: hold Ctrl and click a link to open it in a new tab. (Hide tip)] 8th July 2011
Last updated on: Tuesday, 15 | 677.169 | 1 |
I'm having difficulty with a course that pulls quite a bit from real analysis, which I haven't taken. (But, I have taken the intro to intro to real analysis courses.) Is there a book or online learning regimen that you would recommend?
I'm currently taking my second semester of real analysis and we use Rudin. I wouldn't recommend it as an intro; it's incredibly non-explicit about what it's doing. Expect a lot of "it follows similarly" (lies) and "left as an excersize for the reader." We're working on constructing the Lebesgue Measure, and each page takes easily an hour to understand enough to work with, nevermind comfortably. Coming from a fairly successful math major who finishes their degree this semester, it's tough work.
My first semester was taught out of A Friendly Intro to Real Analysis, which is a lot, well, friendlier. It's doesn't construct the reals or go into measure spaces or anything like that, but it handles up to multi-variable calculus. I don't know what parts you want to learn, but for the basics, I'd go with that.
I loved Rudin when we used it for my real analysis class, but we never got to Lebesgue integration. I tried to read that chapter after the class and could never really get through it. Glad I'm not the only one who thinks it was the real weak point of the book.
Elementary Analysis: The Theory of Calculus by Kenneth Ross is a good introductory text which serves as a smooth transition from lower division (i.e. calculus) mathematics to upper division analysis. Rudin is a standard and very good, but requires more "mathematical maturity" and is more difficult, with exercises ranging from very reasonable to extremely difficult. Both have been used for intro analysis classes at UC Berkeley, for what that's worth.
"This is a dangerous book Understanding Analysis is so well-written and the development of the theory so well-motivated that exposing students to it could well lead them to expect such excellence in all their textbooks. It might not be a good idea to create such expectations. You might not want to adopt this text unless you're comfortable teaching from a book in which the exposition will nearly always be clearer than your lectures. Understanding Analysis is perfectly titled; if your students read it, that's what's going to happen."
I learned from Rosenlicht about 25 years ago and it was already a Dover reprint at that time. The professor's comment was "the value of epsilon hasn't changed for hundreds of years so why do we need a new expensive book?"
In learning mathematics it can be very helpful to have different books on the same subject: You get more examples, slightly different explanations, and more problems to work. Dover books, even the marginal ones, can be very helpful in this regard because they are so cheap.
I have it. It is very good, and the price is excellent. It might not be as clear as Abbot's or Ross' books but it is still pretty clear, some parts you may have to do some google searching to make that last tick in your brain you need to put you over the edge on a concept but other than that it is a pretty smooth read. I wouldn't bother with Rudin to start, first off it is $100, second off, it is pretentiously terse and instead of spending hours trying to understand Rudin, you could just cut the bullshit and understand the same concepts straight away with cheaper and more effective textbooks. Use Rudin later as a reference.
Tldr; Great book, can't beat the price tag either. Rudin is overrated, especially as an intro.
Its accessible and has all the major elements of good intro to real analysis. Also, it's free online in PDF. You can get a paperback version for about $10 though if you want a physical book. If you get through this one then move on to Rudin.
Rudin is pretty much the standard for a real course in real analysis, but I wouldn't use it for a first introduction. You probably don't need something so in-depth. I got my first intro to analysis with "A Course in Modern Analysis and its Applications" by Graeme Cohen - I've never heard of anyone else using this book; I just found it at a bookstore, but it's very clear and well written.
One trouble is, I'm not sure what my level is. Of the things the professor calls "real analysis," I've seen some things before, and some are new to me. One of the intro to Intro to Analysis courses was taught by the Intro to Analysis prof, who may or may not have covered some of the material. In truth, I'm not sure what real analysis is.
I'm with you there. I'm a high school student with a vested interest and a fair amount of books because I have an aptitude for teaching myself. I was operating on the book having an introduction to proofs. And analysis I've been told that it's a superset of calculus in that it analyses continuous sets. Other than that, don't really know.
I assumed that the book might be a bit elementary for you because of it's intro to proof, something I'm guessing you have a solid ground in, and I figured the rest of the book would take an elementary approach, that's why I got it as a self-teaching book.
I thought it was pretty reasonable. Rudin is not really appropriate for most undergrads. It's do-able, but it requires a lot of mathematical maturity. If you've time, one thing you can attempt to do is work through Rudin and really try to understand all the details(which will probably involve filling several notebooks full of notes)
Thanks! Unfortunately, it is a graduate student course, but I needed another elective to graduate. I'll start on an easier one and eventually work through Rudin I guess. In theory, I may have it less painfully in grad school the second time around :)
I have both Rudin and Serge Lang's Real Analysis book, and I am reading the latter at the moment. I find Lang's to be more friendly. I guess you could try Lang's "undergraduate analysis," but so far every book I've read specifically aimed at undergraduates involves lossy compression of ideas, which only makes me more confused. | 677.169 | 1 |
Word problems are the most difficult part of any math course ?- and the most important to both the SATs and other standardized tests. This book teaches proven methods for analyzing and solving any type of math word problem. more...
A self-teaching guide to basic arithmetic, covering whole numbers, fractions, percentages, ratio and proportion, basic algebra, basic geometry, basic statistics and probability You'll be able to learn more in less time, evaluate your areas of strength and weakness and reinforce your knowledge and confidence. more...
Preempt your anxiety about PRE-ALGEBRA! Ready to learn math fundamentals but can't seem to get your brain to function? No problem! Add Pre-Algebra Demystified , Second Edition, to the equation and you'll solve your dilemma in no time. Written in a step-by-step format, this practical guide begins by covering whole numbers, integers, fractions, decimals,... more...
Your solution to MATH word PROBLEMS! Find yourself stuck on the tracks when two trains are traveling at different speeds? Help has arrived! Math Word Problems Demystified , Second Edition is your ticket to problem-solving success. Based on mathematician George Polya's proven four-step process, this practical guide helps you master the basic procedures... more...
Stack the odds in your favor for mastering probability Don't leave your knowledge of probability to chance. Instead, turn to Probability Demystified , Second Edition, for learning fundamental concepts and theories step-by-step. This practical guide eases you into the subject of probability using familiar items such as coins, cards, and dice. As... more... | 677.169 | 1 |
Math Skills Center
Visit us in the CMC
The Math Skills Center is located on the 2nd floor of the Center for Mathematics & Computing (between the Math Department and the computer labs and classrooms).
The Math Skills Center supports students who are taking math and math-related courses. It's also a great place to study, to do homework, and to meet for one-on-one tutoring.
The Math Skills Center provides a number of services:
drop-in tutoring
one-on-one tutoring (see Russ for more information)
algebra and pre-calculus review
pre-calculus software and VCR tapes on statistics and calculus
preparation software for GRE, LSAT and MCAT entrance exams
lending library and magazine rack
The mission of the Math Skills Center is to support all Carleton students in any mathematics or math-related course they are taking in which they are experiencing difficulty, either with the mathematical concepts or with the mathematical tools needed to succeed in the course. Our mission is to "level the playing field" by giving students who enter Carleton without strong mathematics backgrounds the tools they need to succeed here at Carleton.
We believe that by providing a physical space and a staff of tutors actively moving about the lab, engaging students in order to assist them as they work on problems, we can create those "learning moments" that allow students to assimilate the material more easily and retain it longer.
Rather than serving as a space where students can come to "get the right answer," we strive to provide a space where group activity, discussion, and inquiry are valued, and where students come to inquire, reflect, and gain insight into a problem; in other words, we offer a place where math comes alive as students actively engage in the learning process. | 677.169 | 1 |
Description
The fun and easy way to learn pre-calculus
Getting ready for calculus but still feel a bit confused? Have no fear. Pre-Calculus For Dummies is an un-intimidating, hands-on guide that walks you through all the essential topics, from absolute value and quadratic equations to logarithms and exponential functions to trig identities and matrix operations.
With this guide's help you'll quickly and painlessly get a handle on all of the concepts — not just the number crunching — and understand how to perform all pre-calc tasks, from graphing to tackling proofs. You'll also get a new appreciation for how these concepts are used in the real world, and find out that getting a decent grade in pre-calc isn't as impossible as you thought.
Updated with fresh example equations and detailed explanations
Tracks to a typical pre-calculus class
Serves as an excellent supplement to classroom learning
If "the fun and easy way to learn pre-calc" seems like a contradiction, get ready for a wealth of surprises in Pre-Calculus For Dummies!
About the authors
Yang Kuang, PhD, is a professor of mathematics at Arizona State University. He currently serves on the calculus committee where he and other members discuss what and how to teach calculus to students majoring in math and physical sciences. Elleyne Kase is a professional writer.
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Zina Sverdlik
??? This book does not load. Just spent 12.99 on it...... fix the problem or i want a refund ..
Bret Hart
Will not load I purchased this book for 9.99 and it doesn't open. Every other google play book I own works exceptionally. Want a refund
User reviews
Zina Sverdlik February 26, 2014
??? This book does not load. Just spent 12.99 on it...... fix the problem or i want a refund .. | 677.169 | 1 |
Find a Martins AdditionsA solid fundamental understanding of arithmetic is very important when moving away from the basic concepts we use every day, to the more abstract concepts of algebra. It's also just as important to learn the language of mathematics, so we are prepared to communicate the higher, more abstract concepts in algebra. Java was the first programming language I really got involved in | 677.169 | 1 |
good except that pictures and diagrams don't always match up with the page that you are on. For example a diagram would be on page four or five and the written text would be on page three. This sometimes makes the problems hard to solve. That is why I am giving it four stars instead of five. | 677.169 | 1 |
The purpose of this book is to introduce the basic ideas of mathematical proof to students embarking on university mathematics. The emphasis is on helping the reader in understanding and constructing proofs and writing clear mathematics. This is achieved by exploring set theory, combinatorics and number theory, topics which include many fundamental ideas which are part of the tool kit of any mathematician. This material illustrates how familiar ideas can be formulated rigorously, provides examples demonstrating a wide range of basic (...) methods of proof, and includes some of the classic proofs. The book presents mathematics as a continually developing subject. Material meeting the needs of readers from a wide range of backgrounds is included. Over 250 problems include questions to interest and challenge the most able student as well as plenty of routine exercises to help familiarize the reader with the basic ideas (...) Williams | 677.169 | 1 |
Details about Calculus:
Teaches the techniques of differential and integral calculus that students are likely to encounter in undergraduate courses in their majors and in subsequent professional activities. This work provides an understanding of the basic concepts of calculus. It assumes that students have completed high school algebra.
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Rent Calculus 8th edition today, or search our site for other textbooks by Laurence D. Hoffmann Calculus tutors now. | 677.169 | 1 |
Soquel PrealgebraLouise A.
...It is a new language and yet based on our arithmetic. Algebra simply uses an "x" instead of a number. Algebra 2 is the time in the development of our curriculum that takes the basic skills and puts them into context. | 677.169 | 1 |
As the name suggests, our website is for students of any grade and all earnest Math learners To prepare for Standardized Tests:
• Students who take various standardized tests such as SAT, GRE, GMAT, PCAT and other MBA and MCA tests in which Math is one important skill tested will find here relevant and useful math material on math subjects such as:
Probability: Matrices How will math-for-all-grades explain the math subjects and math lessons? OR What is the method of discussing the math subjects and lessons?
• A coherent practical anecdote is presented to kindle interest and effortlessly lead the student into the actual topic.
Header-Links and a Terse Recap of all Math Concepts
• Links of the all of the math lessons under each major math topic are provided to take the student directly to the web page dealing with the lesson on clicking.
For example, under basic algebra > factorization, links of all major types of factorization are displayed, on clicking which, the student will be taken to the web page explaining the related lesson, say factoring trinomial perfect squares.
• A terse overview of all the math concepts in every math lesson is provided for two reasons
One, to give a bird's eye view of the concepts to be discussed.
Another, more importantly, to serve a ready and quick recap of the salient concepts learnt in the specific math lesson.
Huge Number of Examples and Problems are solved
math-for-all-grades believes the key to mastery over math is to solve tens and scores of problems.
One Picture is worth a Thousand Words
• Wherever possible, appropriate images are included to illustrate the otherwise abstruse and uninteresting method of reading text – a necessary and indispensable mode of learning math.
Resources to Add Value to your Math Learning
• On the Navigation bar of the Home page of this website
there are buttons displayed under the Title: Math related. It contains:
o Math Glossary:
This gives a succinct definition of all important math terms and concepts in each math lesson.
The math glossary is exhibited topic wise. Alphabetical arrangement is also due for construction
o Math Formulas:
Math formulas will be arranged Subject-Wise. Math formulas on Algebra, Arithmetic, Trigonometry, Coordinate Geometry, Probability, Statistics, and Vectors will be presented for ready recap of the numerous important formulas in all math lessons in each math subject. | 677.169 | 1 |
Now in its third edition, Mathematical Concepts in the Physical Sciences, 3rd Edition provides a comprehensive introduction to the areas of mathematical physics. It combines all the essential math concepts... | 677.169 | 1 |
progress report. sent weekly to parents, with recommended revision lessons to ensure concepts are mastered before moving on to the next level of difficulty.
Systematic and Structure Approach
Questions are organised by types, enabling students to identify a pattern in solving heuristic questions. Students are trained to tackle different types of heuristic problem sums systematically through a step by step approach presented in question hints and solutions. A series of video tutorials is also included for selected concepts, accessible in the Video Resources section.
Gradual and Comprehensive Learning Solution
Three questions a day ensure workload is kept to a minimal, yet sufficient for students to master heuristic problem solving skills. One concept is scheduled daily, allowing students to cover the entire syllabus over the year. Daily practice is provided for students to develop their problem solving skills on a consistent basis.
Critical Performance Tracking
All student activities are tracked and daily progress report are sent to parents providing an analysis of areas of weaknesses.
For Students, Parents and Tutors
Through the use of video examples and step by step hints, PSLEMath Online is able to assist teachers and tutors in imparting problem solving skills toANS Traffic Android (GCC)
ANS Traffic for Android (GCC) is a special application combining traffic information and navigation, and helps you to plan your routes quickly and effectively when driving in any of the eight GCC countries (United Arab Emirates, Saudi Arabia, Oman, Qatar, Bahrain, Kuwait, Jordan, Lebanon).
It's a highly user-friendly application that can make drivi…ANS Traffic Android (SAF)
ANS Traffic for Android (SAF) is a special combination of traffic and navigation applications that plan your routes quickly and effectively when driving in Southern Africa and even help you to save time (countries included: Botswana, Burundi, Kenya, Lesotho, Malawi, Mozambique, Namibia, Rwanda, South Africa, Swaziland, Uganda).
It's a highly user-f… | 677.169 | 1 |
Iterated Function Systems in the Classroom
Charles Waiveris
The title may appear daunting, but the exercises, which can be presented to students from middle school to graduate school, are not. The exercises center on creating fractal images in the xy-plane with free. easy-to-use software and questions appropriate to the level of the student. Numerous images are shown.
This is available to members of NCTM. Please log in now to view this article. If you are interested in a NCTM membership join now. | 677.169 | 1 |
The Virginia
Algebra Resource Center (VARC) website is designed to help Algebra I teachers
realize some of the potential of the Internet.VARC finds, screens, catalogs, and cross-references information that
encourages the application of the Algebra I Standards of Learning and
emphasizes the inter-relatedness of each Algebra I standard to other
mathematical concepts.
Activities
posted to the Virginia Algebra Resource Center are correlated to the Algebra I
Standards of Learning.The activities
are referenced by standard and by algebra topic.The Algebra Resource Center also catalogs general mathematics
resources and interactive sites that are available on the Internet.An online forum for questions and answers
from students, teachers, and parents is provided as well as listings of
mathematics publications, organizations, research articles and books, and
software.The student homework help
page has links to a variety of Internet resources. | 677.169 | 1 |
1,001 practice opportunities to score higher in statistics 1,001 Statistics Practice Problems For Dummies takes you beyond the instruction and guidance offered in Statistics For Dummies to give you a more hands-on understanding of statistics. The practice problems offered range in difficulty, including detailed explanations and walk-throughs.... more...
At a time of rapid demographic change and amidst the many educational challenges facing the US, this critical new collection presents mathematics education from a culturally responsive perspective. It tackles the most crucial issues of teaching mathematics to an ethnically diverse school population, including the political dimension of mathematics... more...
This book offers ideas, blue-prints and actions that can help raise public awareness of the importance of mathematical sciences in our contemporary society. It covers national experiences, exhibitions, mathematical museums, and other popularization activities. 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...
CRYSTAL?Alberta was established to research ways to improve students? understanding and reasoning in science and mathematics. To accomplish this goal, faculty members in Education, Science, and Engineering, as well as school teachers joined forces to produce a resource bank of innovative and tested instructional materials that are transforming teaching... more... | 677.169 | 1 |
a one or two-semester Applied Calculus course, this innovative text features a graphing calculator approach, incorporating real-life applications and such technology as graphing utilities and Excel® spreadsheets to help students learn mathematical skills that they will use in their lives and careers. The texts overall goal is to improve learning of basic calculus concepts by involving students with new material in a way that is different from traditional practice. The development of conceptual understanding coupled with a commitment to make calculus meaningful to the student are guiding forces. The material involves many applications of real situations through its data-driven, technology-based modeling approach. The ability to correctly interpret the mathematics of real-life situations is considered of equal importance to the understanding of the concepts of calculus.
CALCULUS CONCEPTS, Fifth Edition, presents concepts in a variety of forms, including algebraic, graphical, numeric, and verbal. Targeted toward students majoring in liberal arts, economics, business, management, and the life and social sciences, the text's focus on technology along with its use of real data and situations make it a sound choice to help students develop an intuitive, practical understanding of concepts.
Additional Product Information
Features and Benefits
Each chapter opens with a real-life situation and several questions about the situation that can be answered using the concepts and skills to be covered.
Each section incorporates a brief concept development narrative, interspersed with Quick Examples that highlight specific skills as well as formal examples that illustrate the application of the skills and concepts in a real-world setting.
A NAVG (numeric, algebraic, verbal, and graphical) compass icon indicates places in the text where a concept is demonstrated through multiple representations. This feature helps students recognize connections between different representations, and is particularly helpful for students who use alternative learning styles.
A Concept Inventory listed at the end of each section gives students a brief summary of the major ideas developed in the section.
A Concept Review activity section at the end of each chapter provides practice with techniques and concepts. Complete answers to the Concept Review activities are included in the answer key located at the back of the text.
Concept Check is an end-of-chapter checklist that describes the main concepts and skills taught in the chapter and identifies sample odd-numbered activities corresponding to each item. Students can complete these representative activities to help them assess their understanding of the chapter content and identify the areas on which they need to focus their study.
A Chapter Summary reviews and connects major chapter topics and concepts, and further emphasizes their practical importance.
Activities, such as online Projects and Writing Across the Curriculum, reinforce the authors' innovative approach. Spreadsheet and Graphing Calculator Activities give students the opportunity to use technology as they learn difficult calculus concepts.
Section Activities at the end of each section reinforce concepts and allow students to explore topics using, for the most part, actual data in a variety of real-world settings. These activities, designed to encourage students to communicate in written form, include questions and interpretations pertinent to the data. The activities do not mimic the examples in the chapter discussion, requiring students to think more independently. Possible answers to odd-numbered activities are given at the end of the book.
End-of-chapter Projects are optional group assignments that allow students to practice composing reports and giving oral presentations.
What's New
Many of the book's examples and activities are new. In addition, many data sets have been revised to incorporate more recent data.
The concept of limits is introduced early in Chapter 1 and used throughout the discussion of models in the remainder of that chapter. The concept is also used to help students understand differentiation and integration.
Formerly presented in a self-contained chapter, coverage of sine models has been incorporated throughout the text in optional sections and activities.
Differential equations and slope fields are introduced in a pair of optional sections located at the end of the integration chapters.
The text has been carefully rewritten so that narrative sections are as clear and concise as possible.
While a real-world context is still used as the platform for most of the discussion, some of the less critical details of these contextual descriptions are now presented to the side of the primary narrative in Notes, allowing students to focus on key ideas without potentially getting distracted.
Definitions and other important mathematical elements are highlighted in boxes for easy reference, and selected mathematical and interpretation skills are illustrated in Quick Examples.
Each section's activity set has been reworked to incorporate an orderly development of the skills and concepts presented in that section.
Many activities have been rewritten to be more student-friendly, and others have been replaced with up-to-date applications. Even-numbered activities are similar to, but not necessarily identical to, odd-numbered activities.
All solutions have been reworked and answers have been rewritten to be concise. Activities requiring essay-style answers are clearly markedDiploma
(ISBN-10: 0538735619 | ISBN-13: 9780538735612)
PowerLecture DVD
(ISBN-10: 0538735406 | ISBN-13: 9780538735407)
This DVD provides the instructor with dynamic media tools for teaching. Create, deliver, and customize tests (both print and online) in minutes with ExamView® Computerized Testing, featuring algorithmic equations. There's also a link to the Solution Builder online solutions manual, making it easy to build solution sets for homework or exams.
This manual contains fully worked-out solutions to all of the odd-numbered exercises in the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer courseGo beyond the answers--see what it takes to get there and improve your grade! This manual provides worked-out, step-by-step solutions to the odd-numbered problems in the text, giving you the information you need to truly understand how these problems are solved.
If your instructor has chosen to package Enhanced WebAssign with your text, this manual will help you get up and running quickly with the Enhanced WebAssign system so you can study smarter and improve your performance in class | 677.169 | 1 |
Maple includes more than 5,000 mathematical functions and a variety of options for high-performance computing. Maple 16 makes available new techniques and tools in the Clickable Math collection, such as "drag-to-solve" and "smart popups." Drag-to-solve allows users to tackle problems by dragging terms to the desired location and save each step in the calculation. Smart popups let users adjust a single part of a highlighted calculation and preview the answer. It also notifies the user which mathematical identities can be used, if a subexpression can be factored, and shows the plot.
"Teaching Calculus with Maple: A Complete Kit" was developed over five years by Jack Weiner, professor emeritus of mathematics at University of Guelph in Ontario, Canada. The kit has been field- tested in classrooms with 15 to 600 students.
Notes for the teachers with annotation and demonstrations using Maple;
Automatically graded homework assignments for Maple T.A. users, including feedback and hints, as well as assignments for users who don't have access to Maple T.A.; and
Extra homework assignments from Stewart's Calculus.
"Using Maple as the foundation of each lecture has resulted in fun, extremely interactive classes. We're reaching more students now than ever before," said Weiner. "In addition, the use of Maple T.A. to reinforce learning by providing lots of practice and immediate feedback gives students more confidence, which ultimately leads to greater success."
"Teaching Calculus with Maple: A Complete Kit" is free for schools that use Maple or Maple T.A | 677.169 | 1 |
Kaufmann and Schwitters have built this text's reputation on clear and concise exposition, numerous examples, and plentiful problem sets. This traditional text consistently reinforces the following common thread: learn a skill; practice the skill to help solve equations; and then apply what you have learned to solve application problems. This simple, straightforward approach has helped many students grasp and apply fundamental problem solving skills necessary for future mathematics courses. Algebraic ideas are developed in a logical sequence, and in an easy-to-read manner, without excessive vocabulary and formalism.
The open and uncluttered design helps keep students focused on the concepts while minimizing distractions. Problems and examples reference a broad range of topics, as well as career areas such as electronics, mechanics, and health, showing students that mathematics is part of everyday life.
The text's resource package—anchored by Enhanced WebAssign, an online homework management tool—saves instructors time while also providing additional help and skill-building practice for students outside of class.
Additional Product Information
Features and Benefits
Kaufmann/Schwitters' straightforward, 3-step problem-solving approach is introduced and reinforced throughout the textbook. The steps are: learn the algebraic skill; use the skill to help solve equations; apply the skill to solve the application problem.
Learning Objectives can be found at the beginning of each section and are mapped to the Problem Sets and the Chapter Summary.
Made up of predominately true/false questions, a Concept Quiz immediately precedes the problem set in each section and allows students to check their understanding of the mathematical concepts and definitions introduced in the section before moving on to their homework.
Problem Sets, a focal point of every revision, contain a wide variety of skill-development exercises. Problems have been added, deleted, and reworded based on users' suggestions.
Thoughts Into Words problems are included in every end-of-section problem set. These problems are designed to give students an opportunity to express in written form their thoughts about various mathematical ideas.
Further Investigations are included in many of the problem sets. These problems offer more challenging exercises and lend themselves to small group work. They also provide the instructor with flexibility in incorporating advanced mathematical topics depending on specific course needs.
The grid format of the Chapter Summaries makes reviewing fast and easy.
What's New
NEW Study Skill Tips appear at the beginning of each chapter to encourage best study practices throughout the course. A thought-provoking question related to the presented Study Skill Tip encourages students to think more about their current study habits or their past experiences with math.
NEW section 0.1 has been added to provide as an opportunity for students to review fractions so that they are prepared for the intermediate algebra course. Mastery Sets appear throughout the section to help students gauge how well they know the content presented.
NEW Sample Problems have been added to the each Objective in the Chapter Summary. Sample Problems provide students an opportunity to try problems similar to the presented Examples for each Objective.
NEW Apply Your Skill Examples present real-life applications so students can see the relevance of math in everyday life.
Alternate Formats
Choose the format that best fits your student's budget and course goals
Efficacy and Outcomes
Reviews
"We currently use this text and LOVE IT! It is very appropriate for our students. We do not hear complaints that the text is too difficult to understand/read/comprehend." - Stacey Moore, Wallace State Community College- Hanceville
— Stacey Moore
"It's a very nice text that is instructor-friendly. It's easy to use in the classroom - the classroom examples are priceless. WebAssign is very nice with this. Also, it's full of examples and exercises - it's the best Algebra for College Students text I've seen." -Stacey Ernstberger, LaGrange College
— Stacey ErnstbergerThe Complete Solutions Manual provides worked-out solutions to all of the problems in the text.
Solutions Builder
(ISBN-10: 128586400X | ISBN-13: 9781285864006)
This online instructor database offers complete worked solutions to all exercises in the text, allowing you to create customized, secure solutions printouts (in PDF format) matched exactly to the problems you assign in class. For more information, visit
Mastering Mathematics: How to Be a Great Math Student
(ISBN-10: 0534349471 | ISBN-13: 9780534349479)
Providing solid tips for every stage of study, Mastering Mathematics stresses the importance of a positive attitude and gives students the tools to succeed in their math course. This practical guide will help students: avoid mental blocks during math exams, identify and improve areas of weakness, get the most out of class time, study more effectively, overcome a perceived "low math ability," be successful on math tests, get back on track when feeling "lost," and much more!5864042 | ISBN-13: 9781285864044)
NEW! Get a head-start. The Student Workbook contains all of theJerome E. KaufmannKaren L. Schwitters | 677.169 | 1 |
Graph Theory
Overview and graduate students of topology, algebra, and matrix theory. Fundamental concepts and notation and elementary properties and operations are the first subjects, followed by examinations of paths and searching, trees, and networks. Subsequent chapters explore cycles and circuits, planarity, matchings, and independence. The text concludes with considerations of special topics and applications and extremal theory. Exercises appear throughout the text.
Read an Excerpt
Graph Theory
Dover Publications, Inc.
For years, mathematicians have affected the growth and development of computer science. In the beginning they helped design computers for the express purpose of simplifying large mathematical computations. However, as the role of computers in our society changed, the needs of computer scientists began affecting the kind of mathematics being done.
Graph theory is a prime example of this change in thinking. Mathematicians study graphs because of their natural mathematical beauty, with relations to topology, algebra and matrix theory spurring their interest. Computer scientists also study graphs because of their many applications to computing, such as in data representation and network design. These applications have generated considerable interest in algorithms dealing with graphs and graph properties by both mathematicians and computer scientists.
Today, a study of graphs is not complete without at least an introduction to both theory and algorithms. This text will attempt to convince you that this is simply the nature of the subject and, in fact, the way it was meant to be treated.
Section 1.1 Fundamental Concepts and Notation
Graphs arise in many settings and are used to model a wide variety of situations. Perhaps the easiest way to adjust to this variety is to see several very different uses immediately. Initially, let's consider several problems and concentrate on finding models representing these problems, rather than worrying about their solutions.
Suppose that we are given a collection of intervals on the real line, say C = {I1, I2, ..., Ik}. Any two of these intervals may or may not have a nonempty intersection. Suppose that we want a way to display the intersection relationship among these intervals. What form of model will easily display these intersections?
One possible model for representing these intersections is the following: Let each interval be represented by a circle and draw a line between two circles if, and only if, the intervals that correspond to these circles intersect. For example, consider the set
C = {[-4, 2], [0, 1], (-8, 2], [2, 4], [4, 10)}.
The model for these intervals is shown in Figure 1.1.1.
Next, we consider the following old puzzle. Suppose there are three houses (call them h1, h2 and h3) and three utility companies (say gas (g), water (w) and electricity (e)). Our problem is to determine if it is possible to connect each of the three houses to each of the three utilities without crossing the service lines that run from the utilities to the houses. We model this puzzle by representing each house and each utility as a circle and drawing a line between two circles if there is a service line between the corresponding house and utility. We picture this situation in Figure 1.1.2. A solution to this problem would be a drawing in which no lines crossed. The drawing of Figure 1.1.2 is not a solution to the problem, but merely an attempt at modeling the problem.
In our third problem, suppose you are the manager of a company that has four job openings (say j1, j2, j3 and j4) and five applicants a1, ..., a5 and that some of these applicants are qualified for more than one of your jobs. How do you go about choosing people to fill the jobs so that you will fill as many openings as possible? We picture such a situation in Figure 1.1.3. Again, each job and each applicant can be represented as a circle. This time, a line is drawn from a circle representing an applicant to each of the circles representing the jobs for which the applicant is qualified. A solution to this problem would be a set of four lines joining distinct jobs to distinct applicants, that is, one line joins each job to a distinct applicant. For example, the lines joining j1 and a2, j2 and a1, j3 and a4 and j4 and a5 constitute a solution to this problem. Since lines only join jobs to applicants, this is clearly the maximum number of lines possible. Can you find another solution? The real problem is how can we find solutions in general?
Despite the fact that these problems seem very different, we have used a similar type of diagram to model them. Such a diagram is called a graph. Formally, a graph G = (V, E) is a finite nonempty set V of elements called vertices, together with a set E of two element subsets of V called edges. In our example diagrams, each circle is a vertex and each line joining two vertices is an edge. If the context is not clear, we will denote V or E by V (G) or E (G), respectively, to show they come from the graph G. In Figure 1.1.2, the vertices are h1, h2, h3, g, w, e and the edges are
For simplicity, we will usually denote edges by consecutively listing the vertices at either end. For example, the edge {h1, g} would be denoted h1g or gh1.
One of the beauties of graphs is that they may be thought of in many ways: formally as set systems, geometrically as the diagrams we have presented and algebraically, as we shall see later. Such diverse representations afford us an opportunity to use many tools in studying graphs and to apply graph models in many ways. To do this effectively, of course, we need to build more terminology and mathematical machinery.
Given a graph G = (V, E), the number of vertices in V is called the order of G and the number of edges in E is called the size of G. They shall be denoted as | V | and | E |, respectively. The interval graph of Figure 1.1.1 has order 5 and size 6. If a graph G has order p and size q, we say G is a (p, q) graph. Two vertices that are joined by an edge are said to be adjacent, as are two edges that meet at a vertex. If two vertices are not joined by an edge, we say they are nonadjacent or independent. Similarly, two edges that do not share a common vertex are said to be independent. The set of all vertices adjacent to a vertex v is called the neighborhood of v and is denoted N (v). An edge between vertices u and v is said to have u (or v) as an end vertex. Further, the edge is said to be incident with v (or with u) and v is said to dominate u (also, u dominates v). The number of edges incident with a vertex v is called the degree of v and is denoted deg v or by degG v if we wish to emphasize that this occurs in the graph G. The minimum degree and maximum degree of a vertex in the graph G are denoted by δ(G) and Δ(G), respectively. A graph in which each vertex has degree r is called an r-regular graph (or simply regular). We now present the theorem traditionally called The First Theorem of Graph Theory.
Consequently, any graph contains an even number of vertices of odd degree.
Proof. Since each edge has exactly two end vertices, the sum of the degrees counts each edge exactly twice. Thus, the sum is obtained. Since 2q is even, an even number of vertices of odd degree must then be present in the sum.
We have considered three problems thus far. The drawing of Figure 1.1.1 is a solution to the first problem, and we have an idea of what a solution for one example of the third problem looks like. But the drawing given for the utilities problem does not provide a solution to that problem. This does not mean there is no solution, only that our drawing fails to provide one. What if we try other drawings (see Figure 1.1.4)? One of the interesting features of these drawings is the freedom we have to shape them. There are no restrictions on the size of the vertices or on the length or even the shape of the edges. These drawings are very much free-form. We are also free to choose an entirely different representation for our graph, for example the set representation we used in defining graphs. But this freedom also presents us with some difficulties. If a graph is presented in different ways, how can we determine if the presentations really represent the same graph?
Mathematicians use the term isomorphism to mean the "fundamental equality" of two objects or systems. That is, the objects really have the same mathematical structure, only nonessential features like object names might be different. For graphs, "fundamentally equal" means the graphs have essentially the same adjacencies and nonadjacencies. To formalize this concept further, we say two graphs G1 and G2 are isomorphic if there exists a 1-1 and onto function f: V (G1) ->V (G2) such that xy [member of] E (G1) if, and only if, f (x) f (y) [member of] E (G2) (that is, f preserves adjacency and nonadjacency). We use the function f to express the correspondence between vertices that are "essentially the same" in the two graphs. The function f is called an isomorphism.
Example 1.1.1 The two drawings of the house-utilities graph are again shown in Figure 1.1.5. An isomorphism between G1 and G2 is determined by the function f: V (G1) ->V(G2) where:
f(a) = x, f(b) = r, f(c) = y, f(d) = s, f(e) = z, f(g) = t.
An isomorphism from G2 to G1 is given by f-1, the inverse of f.
Can you find other isomorphisms from G1 to G2?
A subgraph of G is any graph H such that V (H) [subset or equal to] V (G) and E (H) [subset or equal to] E (G); we also say G contains H. If H is a subgraph of G and V (H) = V (G), we say that H is a spanning subgraph of G. A more restricted but often very useful idea is the following: Given a subset S of V (G), the subgraph induced by S, denoted , is that graph with vertex set S and edge set consisting of those edges of G incident with two vertices of S.
The graphs in Figure 1.1.6 illustrate these ideas. Since V (H) = V (G), H is a spanning subgraph of G. Also, I is an induced subgraph of G since all edges of G with both end vertices in V (I) are contained in I. However, J is not an induced subgraph of G since the edge from 1 to 5 is in G, but is not in J.
Several natural and useful variations on graphs will be helpful. The first is the idea of a multigraph, that is, a graph with (possibly) multiple edges between vertices. A pseudograph allows edges that begin and end at the same vertex (called a loop). If we think of the edge between two vertices as an ordered pair rather than a set, a natural direction from the first vertex of the pair to the second can be associated with the edge. Such edges will be called arcs (to maintain the historical terminology), and graphs in which each edge has such a direction will be called directed graphs or digraphs. For digraphs, the number of arcs directed away from a vertex v is called the outdegree of v (denoted od v) and the number of arcs directed into a vertex v is the indegree of v (denoted id v). Often, for emphasis, we denote the arc directed from u to v as u ->v. In a digraph, we define the degree of a vertex v to be deg v = id v + od v. If u ->v is an arc of the digraph, we say that u dominates v and that v is dominated by u. Sometimes we say u is adjacent to v or v is adjacent from u.
Clearly, we can produce even more variations such as pseudodigraphs, multidigraphs and pseudomultidigraphs. Although these will not play as significant a role in our study of graphs, at times they will be useful. In this text we will be sure the reader understands the kind of graph under consideration, and the term graph will always be as we defined it: finite order, without loops, multiple edges or directed edges.
Section 1.2 Elementary Properties and Operations
A quick inspection of a road map of the southern states shows several important cities and the interstates that connect them. We model a portion of this map in Figure 1.2.1.
It seems natural to think of a graph (or digraph) when trying to model a road system. Vertices represent cities, and edges (or arcs) represent the roads between the cities. In tracing the route from one location to another, we would traverse some sequence of roads, beginning at some starting point, and finally reaching our destination. For example, we could travel from Atlanta to Nashville by first leaving Atlanta and traveling along I75 to Chattanooga, then following I24 to Nashville. Such a model leads us naturally to formally define the following concepts.
Let x and y be two vertices of a graph G (not necessarily distinct vertices). An x - y walk in G is a finite alternating sequence of vertices and edges that begins with the vertex x and ends with the vertex y and in which each edge in the sequence joins the vertex that precedes it in the sequence to the vertex that follows it in the sequence. For example, in the graph of Figure 1.2.1, one b - n walk is
b, I59, c2, I75, a, I20, b, I59, c2, I24, n
while another is
b, I20, a, I85, c1, I85, a, I75, c2, I24, n.
The number of edges in a walk is called the length of the walk. Note that repetition of vertices and edges is allowed. Ordinarily, we will use a more compact notation for a walk by merely listing the vertices involved, noting that the edge between consecutive vertices (at least in a graph or digraph) is then implied. We will only use the full notation for walks in multigraphs, where there is a choice of edges and this choice is important, or for emphasis of the edges involved. An x - y walk is closed if x = y and open otherwise. Two walks are equal if the sequences of vertices and edges are identical.
Several stronger types of walks are also important. An x - y trail is an x - y walk in which no edge is repeated, and an x - y path is an x - y walk in which no vertex is repeated, except possibly the first and the last (if the path is closed). Clearly, a path is also a trail. We consider a single vertex as a trivial path (walk or trail). In the graph of Figure 1.2.1, we see that a, b, a, c2, b, is an a - b walk of length 4, while a, b, c2, a, c1 is an a - c1 trail of length 4, and a, c2, n, m is an a - m path of length 3. | 677.169 | 1 |
All our rights reserved. No part of this publication may be reproduced, stored in a retrieval
system or transmitted, in any form or by any means, electronic, mechanical, photocopying,
recording or otherwise without the prior permission of the Association of Business Executives
(ABE).
The learner will: Be able to use algebraic methods to solve business problems.
Assessment Criteria
The learner can:
Indicative Content
3.1 Solve equations using
algebraic methods.
3.1.1 Solve linear and simultaneous equations.
3.1.2 Solve quadratic equations using factorisation and
formulae.
3.2 Solve equations using roots or
logarithms.
3.2.1 Solve and simplify equations using roots or
logarithms.
3.3 Determine the equation of a
straight line.
3.3.1 Determine the equation of a straight line through
two points and also when given one point and its
gradient.
3.3.2 Determine the gradient and intercepts on the x or
y axes for a straight line.
Learning Outcome 4
The learner will: Be able to construct and use graphs, charts and diagrams in a business
context.
Assessment Criteria
The learner can:
Indicative Content
4.1 Draw charts and diagrams
derived from tabular data.
4.1.1 Draw charts and diagrams derived from tabular
data: e.g. bar charts, pie charts, scatter diagrams.
4.2 Plot graphs applying the
general rules and principles of
graphical construction.
4.2.1 Plot graphs applying the general rules and
principles of graphical construction, including choice,
range and scale of axes.
4.3 Plot and interpret
mathematical graphs.
4.3.1 Plot and interpret mathematical graphs for simple
linear, quadratic, exponential and logarithmic equations.
4.3.2 Identify points of importance on graphs e.g. points
of maximum and minimum; points of intercept with the x
and y axes.
Learning Outcome 5
The learner will: Understand and be able to apply the laws of probability to find solutions to
business problems.
Assessment Criteria
The learner can:
Indicative Content
6.1 Explain and apply the laws of
probability.
6.1.1 Recognise outcomes which are equally likely, not
equally likely or subjective.
6.1.2 Use appropriate formulae to determine
probabilities for complementary, mutually exclusive,
independent and conditional events.
6.1.3 Determine probabilities, using a sample space,
two way table or tree diagram.
6.2 Calculate the expected value
of an outcome.
6.2.1 Use probabilities to calculate the expected value
of an outcome.
6.3 Determine probabilities using
the normal distribution.
6.3.1 Determine probabilities using the normal
distribution, by making use of standard normal
distribution tables.
6.3.2 Represent normal probabilities as areas under the
standard normal distribution curve.
or Annual Depreciation =
( ) ( )
life Useful
life useful of end at Value - asset of Cost
Reducing balance method:
D = B(1 ÷ i)
n
where: D = depreciated value at the end of the n
th
time period
B = original value at beginning of time period
i = depreciation rate (as a proportion)
n = number of time periods (normally years).
STRAIGHT LINE
A linear function is one for which, when the relationship is plotted on a graph, a straight line is
obtained.
The expression of a linear function, and hence the formula of a straight line, takes the
following form:
y = mx + c
Note that: c = the y intercept (the point where the line crosses the y axis)
m = the gradient (or slope) of the line
QUADRATIC EQUATION
A quadratic equation of the form ax
2
+ bx + c = 0 can be solved using the following formula:
x =
a 2
ac 4 b b
2
÷ ± ÷
The number of columns indicates the actual amounts involved. Consider the number 289.
This means that, in this number, there are:
two hundreds (289)
eight tens (289)
nine units (289).
Before we look at other number systems, we can note that it is possible to have negative
numbers – numbers which are less than zero. Numbers which are more than zero are
positive. We indicate a negative number by a minus sign (÷). A common example of the use
of negative numbers is in the measurement of temperature – for example, ÷20°C (i.e. 20°C
below zero).
We could indicate a positive number by a plus sign (+), but in practice this is not necessary
and we adopt the convention that, say, 73 means +73.
Roman Number System
We have seen that position is very important in the number system that we normally use.
This is equally true of the Roman numerical system, but this system uses letters instead of
figures. Although we invariably use the normal denary system for calculations, Roman
numerals are still in occasional use in, for example, the dates on film productions, tabulation,
house numbers, etc.
Here are some Roman numbers and their denary equivalent:
XIV 14 XXXVI 36
IX 9 LVlll 58
Binary System
As we have said, our numbering system is a base ten system. By contrast, computers use
base two, or the binary system. This uses just two symbols (figures), 0 and 1.
In this system, counting from 0 to 1 uses up all the symbols, so a new column has to be
started when counting to 2, and similarly the third column at 4.
C. ARITHMETIC
We will now revise some basic arithmetic processes. These may well be familiar to you, and
although in practice it is likely you will perform many arithmetic operations using a calculator,
it is important that you know how to do the calculations manually and understand all the rules
that apply and the terminology used.
When carrying out arithmetic operations, remember that the position of each digit in a
number is important, so layout and neatness matter.
Addition (symbol +)
When adding numbers, it is important to add each column correctly. If the numbers are
written underneath each other in the correct position, there is less chance of error.
So, to add 246, 5,322, 16, 3,100 and 43, the first task is to set the figures out in columns:
246
5,322
16
3,100
43
Total 8,727
Note that each row is added up as well as each column, thus providing a full cross-check.
Subtraction (symbol ÷)
Again, it is important to set out the figures underneath each other correctly in columns and to
work through the calculation, column by column, starting with the units column.
So, to subtract 21 from 36, the first task is to set the sum out:
36
÷21
Total 15
As with addition, you must work column by column, starting from the right – the units column.
Now consider this calculation:
34
÷18
Total 16
Where it is not possible to subtract one number from another (as in the units column where
you cannot take 8 from 4), the procedure is to "borrow" one from the next column to the left
and add it to the first number (the 4). The one that you borrow is, in fact, one ten – so adding
that to the first number makes 14, and subtracting 8 from that allows you to put 6 into the
total. Moving on to the next column, it is essential to put back the one borrowed and this is
added to the number to be subtracted. Thus, the second column now becomes 3 ÷ (1 + 1), or
3 ÷ 2, which leaves 1.
Multiplication (symbol ×)
The multiplication of single numbers is relatively easy – for example, 2 × 4 = 8. When larger
numbers are involved, it is again very helpful to set out the calculation in columns:
246
× 23
Division (symbol ÷)
In division, the calculation may be set out in a number of ways:
360 ÷ 24, or 360/24, or most commonly
24
360
Just as there were three names for the various elements involved in a multiplication
operation, so there are three names or terms used in division:
the dividend is the number to be divided – in this example, it is 360
the divisor is the number by which the dividend is to be divided – here it is 24 and
the answer to a division operation is called the quotient – here the quotient for our
example is 15.
Division involving a single number divisor is relatively easy – for example, 16 ÷ 2 = 8. For
larger numbers, the task becomes more complicated and you should use a calculator for all
long division.
The Rule of Priority
Calculations are often more complicated than a simple list of numbers to be added and/or
subtracted. They may involve a number of different operations – for example:
8 × 6 ÷ 3 +
2
8
D. DEALING WITH NEGATIVE NUMBERS
Addition and Subtraction
We can use a number scale, like the markings on a ruler, as a simple picture of addition and
subtraction. Thus, you can visualise the addition of, say, three and four by finding the position
of three on the scale and counting off a further four divisions to the right, as follows:
This illustrates the sum 6 ÷ 4 = 2.
Negative numbers can also be represented on the number scale. All we do is extend the
scale to the left of zero:
÷6 ÷5 ÷4 ÷3 ÷2 ÷1 0 1 2 3 4 5 6
We can now visualise the addition and subtraction of negative numbers in the same way as
above. Consider the sum (÷3) + 4. Start at ÷3 on the scale and count four divisions to the
right:
÷4 ÷3 ÷2 ÷1 0 1 2 3 4
Therefore, (÷3) + 4 = 1.
(Note that, here, we have shown the figure of "minus 3" in brackets to avoid confusion with
the addition and subtraction signs.)
The order of the figures of these calculations does not matter, so we can also write:
(÷3) + 4 = 4 + (÷3) = 1
We can illustrate the second sum as follows:
E. FRACTIONS
A fraction is a part of a whole number.
When two or more whole numbers are multiplied together, the product is always another
whole number. In contrast, the division of one whole number by another does not always
result in a whole number. Consider the following two examples:
12 ÷ 5 = 2, with a remainder of 2
10 ÷ 3 = 3, with a remainder of 1
The remainder is not a whole number, but a part of a whole number – so, in each case, the
result of the division is a whole number and a fraction of a whole number. The fraction is
expressed as the remainder divided by the original divisor and the way in which it is shown,
then, is in the same form as a division term:
12 ÷ 5 = 2, with a remainder of 2 parts of 5, or two divided by five, or
5
2
Each part of the fraction has a specific terminology:
the top number is the numerator and
the bottom number is the denominator.
Thus, for the above fractions, the numerators are 2 and 1, and the denominators are 5 and 3.
When two numbers are divided, there are three possible outcomes.
(a) The numerator is larger than the denominator
In this case, the fraction is known as an improper fraction. The result of dividing out an
improper fraction is a whole number plus a part of one whole, as in the two examples
above. Taking two more examples:
9
42
= 4 whole parts and a remainder of 6 parts of 9, or 4
9
6
24
30
= 1 whole part and a remainder of 6 parts of 24, or 1
24
6
(b) The denominator is larger than the numerator
In this case, the fraction is known as a proper fraction. The result of dividing out a
proper fraction is only a part of one whole and therefore a proper fraction has a value of
less than one.
Examples of proper fractions are:
5
1
,
3
2
,
7
4
,
5
4
, etc.
(c) The numerator is the same as the denominator
In this case, the expression is not a true fraction. The result of dividing out is one whole
part and no remainder. The answer must be 1. For example:
5
5
= 5 ÷ 5 = 1
Fractions are often thought of as being quite difficult. However, they are not really hard as
long as you learn the basic rules about how they can be manipulated. We shall start by
looking at these rules before going on to examine their application in performing arithmetic
operations on fractions.
Basic Rules for Fractions
Cancelling down
Both the numerator and the denominator in a fraction can be any number. The
following are all valid fractions:
5
2
,
50
20
,
93
47
,
9
6
Which whole number will divide into both 20 and 50? There are several: 2, 5, and 10.
However the rule states that we use the largest whole number, so we should use 10.
Dividing both parts of the fraction by 10 reduces it to:
5
2
Note that we have not changed the value of the fraction, just the way in which the parts
of the whole are expressed. Thus:
50
20
=
5
2
A fraction which has not been reduced to its lowest possible terms is called a vulgar
fraction.
Cancelling down can be a tricky process since it is not always clear what the largest
whole number to use might be. For example, consider the following fraction:
231
165
In fact, both terms can be divided by 33, but you are very unlikely to see that straight
away. However, you would probably quickly see that both are divisible by 3, so you
could start the process of cancelling down by dividing by three:
3 231
3 165
÷
÷
=
77
55
This gives us a new form of the same fraction, and we can now see that both terms can
be further divided – this time by 11:
11 77
11 55
÷
÷
=
7
5
So, large vulgar fractions can often be reduced in stages to their lowest possible terms.
Changing the denominator to required number
The previous process works in reverse. If we wanted to express a given fraction in a
different way, we can do that by multiplying both the numerator and denominator by the
same number. So, for example,
5
2
could be expressed as
20
8
by multiplying both terms by 4.
If we wanted to change the denominator to a specific number – for example, to express
½ in eighths – the rule is to divide the required denominator by the existing
denominator (8 ÷ 2 = 4) and then multiply both terms of the original fraction by this
result:
4 2
4 1
×
×
=
8
4
The rule for converting this is to divide the numerator by the denominator and place
any remainder over the denominator. Thus we find out how many whole parts there are
and how many parts of the whole (in this case, eighths) are left over:
8
35
= 35 ÷ 8 = 4 and a remainder of three = 4
8
3
This result is known as a mixed number – one comprising a whole number and a
fraction.
Changing a mixed number into an improper fraction
This is the exact reverse of the above operation. The rule is to multiply the whole
number by the denominator in the fraction, then add the numerator of the fraction to
this product and place the sum over the denominator.
For example, to write 3
3
2
as an improper fraction:
(a) first multiply the whole number (3) by the denominator (3) = 3 × 3 = 9, changing
the three whole ones into nine thirds
(b) then add the numerator (2) to this = 2 + 9 = 11, thus adding the two thirds to the
nine thirds from the first operation
(c) finally place the sum over the denominator =
3
11
Adding and Subtracting with Fractions
(a) Proper fractions
Adding or subtracting fractions depends upon them having the same denominator. It is
not possible to do these operations if the denominators are different.
Where the denominators are the same, adding or subtracting is just a simple case of
applying the operation to numerators. For example:
7
3
+
7
2
=
7
5
8
5
÷
8
1
=
8
4
=
2
1
Where the denominators are not the same, we need to change the form of expression
so that they become the same. This process is called finding the common denominator.
Consider the following calculation:
5
2
+
10
3
We cannot add these two fractions together as they stand – we have to find the
common denominator. In this case, we can change the expression of the first fraction
by multiplying both the numerator and denominator by 2, thus making its denominator
10 – the same as the second fraction:
10
4
+
10
3
=
10
7
The lowest common denominator for these three fractions is 20. We must then convert
all three to the new denominator by applying the rule explained above for changing a
fraction to a required denominator:
5
3
+
4
1
÷
10
1
=
20
12
+
20
5
÷
20
2
=
20
15
=
4
3
The rule is simple:
find the lowest common denominator – and where this is not readily apparent, a
common denominator may be found by multiplying the denominators together
change all the fractions in the calculation to this common denominator
then add (or subtract) the numerators and place the result over the common
denominator.
(b) Mixed numbers
When adding mixed numbers, we deal with the integers and the fractions separately.
The procedure is as follows:
first add all the integers together
then add the fractions together as explained above, and then
add together the sum of the integers and the fractions.
Consider the following calculation:
3 + 4
4
1
+ 5
2
1
+ 2
4
3
Here, it is not possible to subtract the fractions even after changing them to the
common denominator:
6
3
÷
6
5
To complete the calculation we need to take one whole number from the integer in the
first mixed number, convert it to a fraction with the same common denominator and add
it to the fraction in the mixed number. Thus:
4
2
1
÷ 3 + |
.
|
We often see the word "of" in a calculation involving fractions – for example, ¼ of
£12,000. This is just another way of expressing multiplication.
(b) Dividing a fraction by a whole number
In this case, we simply multiply the denominator by the whole number and place the old
numerator over the product (which is now the new denominator). For example:
4
3
÷ 5 =
5 4
3
×
=
20
3
3
2
÷ 5 =
5 3
2
×
=
15
2
(c) Multiplying one fraction by another
In this case, we multiply the numerators by each other and the denominators by each
other, and then reduce the result to the lowest possible terms. For example:
4
3
×
3
2
=
3 4
2 3
×
×
=
12
6
=
2
1
(e) Dealing with mixed numbers
All that we have said so far in this section applies to the multiplication and division of
proper and improper fractions. Before we can apply the same methods to mixed
numbers, we must complete one further step – convert the mixed number to an
improper fraction. For example:
5
2
1
×
11
4
=
2
11
×
11
4
=
22
44
= 2
4
4
3
÷
2
1
=
4
19
×
1
2
=
4
38
= 9
2
1
(It is always helpful to set out all the workings throughout a calculation. Some
calculations can be very long, involving many steps. If you make a mistake somewhere,
it is much easier to identify the error if all the workings are shown. It is also the case
that, in an examination, examiners will give "marks for workings", even if you make a
mistake in the arithmetic. However, if you do not show all the workings, they will not be
able to follow the process and see that the error is purely mathematical, rather than in
the steps you have gone through.)
Cancellation can also be used in the multiplication of fractions. In this case it can be
applied across the multiplication sign – dividing the numerator of one fraction and the
denominator of the other by the same number. For example:
4
1 3
×
4 12
1
=
16
1
Remember that every cancelling step must involve a numerator and a denominator.
Thus, the following operation would be wrong:
1 4
3
×
3 12
1
=
3
3
Note, too, that you cannot cancel across addition and subtraction signs.
However, you can use cancellation during a division calculation at the step when,
having turned the divisor upside down, you are multiplying the two fractions. So,
repeating the calculation involving mixed numbers from above, we could show it as
follows:
4
4
3
÷
2
1
=
2 4
19
×
1
1 2
=
2
19
= 9
2
1
F. DECIMALS
Decimals are an alternative way of expressing a particular part of a whole. The term
"decimal" means "in relation to ten", and decimals are effectively fractions expressed in
tenths, hundredths, thousandths, etc.
Unlike fractions though, decimals are written completely differently and this makes carrying
out arithmetic operations on them much easier.
Decimals do not have a numerator or a denominator. Rather, the part of the whole is shown
by a number following a decimal point:
the first number following the decimal point represents the number of tenths, so 0.1 is
one-tenth part of a whole and 0.3 is three-tenths of a whole
the next number to the right, if there is one, represents the number of hundredths, so
0.02 is two-hundredths of a whole and 0.67 is six-tenths and seven-hundredths (i.e. 67
hundredths) of a whole
the next number to the right, again if there is one, represents the number of
thousandths, so 0.005 is five-thousandths of a whole and 0.134 is 134 thousandths of a
whole
and so on up to as many figures as are necessary.
We can see the relationship between decimals and fractions as follows:
0.1 =
10
1
, 0.3 =
10
3
, 1.6 = 1
10
6
G. PERCENTAGES
A percentage is a means of expressing a fraction in parts of a hundred – the words "per cent"
simply mean "per hundred", so when we say "percentage", we mean "out of a hundred". The
expression "20 per cent" means 20 out of 100. The per cent sign is %, so 20 per cent is
written as 20%. Some examples:
4% means four-hundredths of a whole, or four parts of a hundred =
100
4
28% means twenty-eight-hundredths of a whole, or twenty-eight parts of a hundred =
100
28
2. Change the following percentages into fractions and reduce them to their lowest
possible terms
(a) 2% (b) 8%
(c) 7½% (d) 18%
3. Calculate the following:
(a) 15% of £200 (b) 25% of 360°
(c) 12½% of 280 bottles (d) 27% of £1,790
(e) 17.5% of £138
4. Calculate the following:
(a) 60 as a percentage of 300 (b) 25 as a percentage of 75
(c) £25 as a percentage of £1,000 (d) 30 as a percentage of 20
Now check your answers with those given at the end of the chapter.
I. FURTHER KEY CONCEPTS
In this last section we shall briefly introduce a number of concepts which will be used
extensively later in the course. Here we concentrate mostly on what these concepts mean,
and you will get the opportunity to practice their use in later chapters. However, we do
include practice in relation to the expression of numbers in standard form.
Indices
Indices are found when we multiply a number by itself one or more times. The number of
times that the multiplication is repeated is indicated by a superscript number to the right of
the number being multiplied. Thus:
2 × 2 is written as 2
2
(the little "2" raised up is the superscript number)
2 × 2 × 2 is written as 2
3
4
2y
can be further cancelled by dividing both the numerator and denominator by 2 to
give
2
y
or we could simply say we can cancel
4x
2xy
by dividing both parts by 2x.
We can always cancel like terms in the numerator and denominator.
Indices, Powers and Roots
You will remember that indices are found when we multiply the same number by itself several
times. The same rules apply in algebra, but we can investigate certain aspects further by
using algebraic notation.
To recap:
x × x is written as x
2
(referred to as "x squared")
x × x × x is written as x
3
(referred to as "x cubed")
x × x × x × x is written as x
4
(referred to as "x to the power four" or "x raised to the
fourth power")
and so on.
In, for example, x
5
, x is termed the base and 5 is called the exponent or index of the power.
Note two particular points before we go on:
x = x
1
There is in fact a general rule that, when we have two expressions with the same base,
multiplication is achieved by adding the indices. Thus:
x
m
× x
n
= x
m + n
(Rule 1)
(b) Division of indices
When we divide expressions with the same base, we find we can achieve this by
subtracting the indices. For example:
3
6
x
x
=
x x x
x x x x x x
× ×
× × × × ×
In general, we have (x
m
)
n
= x
mn
(Rule 3)
(e) Roots
Indices do not have to be whole numbers. They can be fractional.
Let us consider x
½
(i.e. x raised to the power half). We know from Rule (3) that:
(x
½
)
2
= x
½ × 2
= x
Thus, x
½
when multiplied by itself gives x. However, this is exactly the property that
defines the square root of x. Therefore, x
½
is the square root of x which you will
sometimes see denoted as:
2
x or, more commonly, as . x
Therefore, x
½
÷ . x (The symbol "÷" means "identical with".)
Similarly, (
3
1
x )
3
= x. Thus,
3
1
x is the cube root of x which is denoted by the symbol
3
x .
Similarly, x
¼
is the fourth root of x denoted by
4
x .
Thus, in general, we can say that:
n
1
x is the n
th
root of x or
n
x .
Note that in all these expressions the number to the left of the root sign – for example
the 3 in
3
x – is written so that it is level with the top of the root sign. A number written
level with the foot of the root sign – for example, x 3 – would imply a multiple of the
square root of x – in this example, 3 times the square root of x.
We shall often need to use square roots in later work, particularly in statistics and in
stock control.
(f) Roots of powers
There is one more type of index you may meet – an expression such as:
or
3
5
x , or in general
n
m
x .
To interpret these, remember that the numerator of a fractional index denotes a power
and that the denominator denotes a root.
Thus,
2
3
x is (x
3
)
½
, i.e. the square root of (x cubed) or
2 3
x . Alternatively, it is (x
½
)
3
, i.e.
the cube of (the square root of x) or ( x )
3
.
Let us consider its value when x = 4:
2
3
4 =
2 3
4 = 4 4 4 × × = 64 = 8 (since 8 × 8 = 64)
or
2
3
4 = ( ) 4
3
= 2
3
= 2 × 2 × 2 = 8
Therefore it does not matter in which order you perform the two operations, and we
have:
n
m
x =
n m
x = (
n
x )
m
x = 15
You should have noticed that the object of such operations is to isolate a single unknown, x,
on one side of the equals sign.
Questions for Practice 2
Simplify the following expressions, writing "no simpler form" if you think any cannot be further
simplified:
1. x ÷ 7 = 21
2. x ÷ 3 = 26
3. x + 9 = 21
4.
5
x
= 11
5.
8
x
= 7
6. 3x = 30
Now check your answers with those given at the end of the chapter.
= £200
This explains the principle of substitution. In the next chapter you will get much more practice
in this.
Rearranging Formulae
Finally, we should note that a formula can be rearranged to find a different unknown quantity.
So for example, when considering simple interest, we may wish to know how long it would
take to earn a specified amount of interest from a loan made to someone else, at a particular
rate of interest.
The way in which the formula is expressed at present does not allow us to do that, but by
using the rules we have discussed in relation to simple equations, we can rearrange the
formula so that it does.
Again, we start by restating the original formula:
I = P ×
100
r
× n
Note that we can also express this as:
I =
100
n r P
To find n, we need to rearrange the formula so that n is isolated on one side of the equation.
We can do this by dividing both sides by P ×
100
r
or, as it may also be expressed,
100
r P
:
|
.
|
Compound Interest
Let us now look at how compound interest differs from simple interest.
If an amount of money is invested and the interest is added to the investment, then the
principal increases and so the interest earned gets bigger each year.
Suppose you begin with £1,000 and the interest is 8% per year. The interest in the first year
is 8% of £1,000 which is £80. Adding this to the original principal, the amount invested at the
end of the first year is £1,080.
Now, in the second year, the interest is 8% of £1,080 which is £86.40. The amount accrued
at the end of the second year is £1,166.50.
You can continue working out the amounts in this way. However, to find the amount at the
end of any year, you are finding 100% + 8% = 108% of the amount at the beginning of the
year. So you can multiply by 1.08 each time.
There is a standard formula which can be used to calculate compound interest, as follows:
A =
n
100
r
1 P |
.
|
D. DEPRECIATION
Depreciation is a measure of the wearing out, consumption or other loss of value of an asset
arising from use, passage of time, or obsolescence through technology or market changes.
For example, a car loses value over time and is worth less each year.
This is an important concept in business where certain types of asset need to be accurately
valued in order to help determine the overall value of the business. The amount by which an
asset depreciates each year may also have important consequences.
There are several methods for calculating depreciation, but the most commonly used are the
straight-line method (using fixed instalments) and the reducing-balance method (applying a
fixed percentage). It is acceptable to use different methods for different categories of asset,
as long as the method chosen is applied consistently from year to year.
Straight-Line Depreciation
In the straight-line method, depreciation is calculated by dividing the cost of the asset by the
number of years the asset is expected to be used in the business. We can express this as a
formula as follows:
annual depreciation =
life useful
asset of cost
Thus, over the period of years selected, the value of the asset will be reduced to zero.
In the case of certain types of asset, such as cars, it may be anticipated that the asset will be
used for say three years and then sold. In this case, the amount of depreciation is the
difference between the cost of the asset and its anticipated value at the end of its useful life,
divided by the number of years over which it will be used. The formula would then be:
annual depreciation =
( ) ( )
life useful
life useful of end at value - asset of cost
year.
Now check your answers with those given at the end of the chapter.
E. PAY AND TAXATION
Pay and taxation are complex issues and it is not our intention here to explore them in any
detail. Rather, we shall concentrate on the principles of their calculation, using only a basic
construction of how pay is made up and the way in which the taxation system works.
Again, we shall be applying a number of the basic numerical concepts we have considered
previously – including the distinction between fixed and variable values, percentages and
simple formulae. This is also an area where it is very important to adopt clear layout when
working through calculations.
Make Up of Pay
All employees receive payment for their labour. At its most basic, this is paid in the form of
either a salary or wages.
Salaries are generally paid monthly and are based on a fixed annual amount. Thus:
monthly pay =
12
salary annual
For example, an employee with an annual salary of £18,000 would receive £1,500 every
month (£18,000 ÷ 12).
Wages are generally paid weekly and are usually based on a fixed rate for working a certain
number of hours. For example, many wage earners are required to work a fixed number of
hours in a week, and they are paid for these hours at a basic hourly rate.
Thus, an office clerk is paid £5.30 per hour for a 40 hour week. What is her weekly wage?
weekly wage = rate of pay × hours worked
= £5.30 × 40 = £212
An employee can often increase a basic wage by working longer than the basic week – i.e.
by doing overtime. A higher hourly rate is usually paid for these additional hours, the most
common rates being time and a half (i.e. 1½ times the basic hourly rate) and double time
(twice the basic hourly rate).
Consider a factory worker who works a basic 36 hour week for which she is paid a basic rate
of £5.84 per hour. In addition, she works 5 hours overtime at time and a half and 3 hours
overtime at double time. Calculate her weekly wage.
People who are employed as salespersons or representatives, and some shop assistants,
are often paid a basic salary or wage plus a percentage of the value of the goods they have
sold. In some cases, their basic salary or wage can be quite small, or even non-existent, and
the commission on their sales forms the largest part of their pay.
For example, a salesman earns a basic salary of £8,280 per year plus a monthly commission
of 5% on all sales over £5,000. What would be his income for a month in which he sold
goods to the value of £9,400?
basic monthly pay =
12
salary annual
=
12
280 , 8 £
= £690
He earns commission on (£9,400 ÷ £5,000) worth of sales:
commission = 5% of (£9,400 ÷ £5,000) = 0.05 × £4,400 = £220
Total monthly pay = basic pay + commission
= £690 + £220 = £910
Questions for Practice 6
1. Calculate the weekly wage of a factory worker if she works 42 hours per week at a rate
of pay of £5.25 per hour
2. If an employee's annual salary is £18,750 how much is she paid per month?
3. A basic working week is 36 hours and the weekly wage is £306. What is the basic
hourly rate?
4. A trainee mechanic works a basic 37.5 hour week at an hourly rate of £4.90. He is paid
overtime at time and half. How much does he earn in a week in which he does 9 hours
overtime?
5. An employee's basic wage is £6.20 per hour, and she works a basic 5-day, 40-hour
week. If she works overtime during the week, she is paid at time and a half. Overtime
worked at the weekend is paid at double time. Calculate her wage for a week when she
worked five hours overtime during the week and four hours overtime on Saturday.
6. An insurance representative is paid purely on a commission basis. Commission is paid
at 8% on all insurance sold up to a value of £4,000 per week. If the value of insurance
sold exceeds £4,000, he is paid a commission of 18% on the excess. Calculate his
total pay for the 4 weeks in which his sales were £3,900, £4,500, £5,100 and £2,700.
Now check your answers with those given at the end of the chapter.
This means that, for a person with a total taxable pay of £20,000 in a year, the total tax
liability would be:
£20,000 × 20% = £4,000
For a person with a total taxable pay of £45,000 in a year, the total tax liability would be:
£36,000 × 20% £7,200
(£45,000 ÷ £36,000) × 40% = £3,600
£10,800
There are many possible allowances and non-taxable items which affect the level of taxable
pay. Here we shall consider just two.
Personal allowances – the amount of money an individual is allowed to earn each year
before he or she starts to pay tax. This amount (sometimes called free-pay) depends
on the individual's circumstances – for example, whether they are single or married,
number of children, etc. If your personal allowances are greater than your actual pay,
then you would pay no income tax at all.
The level of personal allowances are generally set as an annual amount, but this can
be converted into monthly or weekly amounts of "free-pay".
Pension contributions – the amounts paid by an individual into a pension scheme, such
as an employer's occupational pension scheme, are not taxable. Therefore, pension
contributions, where they are deducted from pay by the employer, may also be
deducted from gross pay before calculating tax. (Note that this does not necessarily
apply to any other types of deductions from pay. Where pension contributions are not
paid through the employer, an adjustment may be made to a person's personal
allowances to allow for them.)
Now we can develop a framework for calculating the amount of tax payable annually, using
the tax bands set out here.
gross taxable pay = gross pay ÷ non-taxable items
net taxable pay = gross taxable pay ÷ personal allowances
tax liability = first £36,000 of net taxable pay × 20%
plus remainder of net taxable pay × 40%
Consider the following examples, using the bands and tax rates set out here.
Example 1:
How much annual tax will be paid by someone who earns £18,600 per annum and has
personal allowances of £5,800? What is the monthly tax deduction?
There are no non-taxable items specified, so:
gross taxable pay = gross pay ÷ non-taxable items
= £18,600
net taxable pay = gross taxable pay ÷ personal allowances
= £18,600 ÷ £5,800 = £12,800
tax liability = £12,800 × 20% = £2,560
The monthly income tax deducted will be £2,560 divided by 12, i.e. £213 rounded down.
Questions for Practice 2
1. £31
2. £5,950
3. £1,560
4. £24.72
5. £304
6. There are three parts to this question.
(a) They would be charged £3 for each cash transaction. Therefore, they would be
changing:
£107 into NZD = 299.39 NZD
£97 into AUD = 238.43 AUD
£117 into HKD = 2299.05 HKD.
(b) (i) £570
(ii) £5.70
(c) Total cost is the £900 worth of foreign currency plus the commission of £9 for the
cash transactions and £5.70 for the travellers cheques = £914.70. Spread
between the four members of the family, cost per person is £228.68.
Now we shall look at some harder examples!
(a) Where the unknowns are denominators
Consider the following equations:
x
4
÷
y
3
= 4 (1)
x
3
–
y
7
= 2 (2)
Here we can treat
x
1
and
y
1
as the unknowns and then solve the problem in the usual
way.
To eliminate
y
1
, multiply (1) by 7 and (2) by 3 and then subtract:
C. QUADRATIC EQUATIONS
All the equations we have studied so far have been of a form where all the variables are to
the power of 1 and there is no term where x and y are multiplied together – i.e. xy does not
feature in the equation. (The types of equation studied so far are known as linear equations
since, as we shall see in the next chapter, if we draw a graph of the outcome of the equation
for all the possible values of the unknowns, the result is a straight line.)
By contrast, a quadratic equation is one that contains the square of the unknown number, but
no higher power. For example, the following are quadratic expressions in x:
4x
2
+ 7x
2x
2
÷ x + 1
2
1
x
2
÷ 2
We can also have quadratic expressions in x
2
. For example, 3x
4
+ 2x
2
+ 5 is a quadratic in x
2
,
since it may be also be written as:
3(x
2
)
2
+ 2(x
2
) + 5, or
3y
2
+ 2y + 5, where y stands for x
2
Solving Quadratic Equations with no Terms in x
In particular circumstances, there is a quicker and easier method than the factor method of
solving a quadratic equation.
Consider the following equation:
25x
2
÷ 36 = 0
Using the factor method, this would be solved as follows:
(5x ÷ 6)(5x + 6) = 0
5x = 6, or 5x = ÷6
x = 1.2 or ÷1.2.
There is, however, a simpler method of obtaining this answer and you should use it
whenever there is no term in x in the given equation:
25x
2
÷ 36 = 0
25x
2
= 36
x
2
=
25
36
Solving Quadratic Equations by Formula
Since the method just used for solving a quadratic equation applies to all cases, even where
no real roots can be found, it can be used to establish a formula.
Consider the equation ax
2
+ bx + c = 0, where a, b and c may have any numerical value.
Rearranging the equation we get:
ax
2
+ bx + c = 0
ax
2
+ bx = ÷c
x
2
+
a
b
x = ÷
a
c
At this point, we need to "complete the square", which means adding or subtracting the
expression that is needed in order to be able to factorise the LHS. In this case, it is:
2
a 2
b
|
.
|
\
|
Remember that this needs to be added to both sides of the equation to keep it balanced.
x
2
+
a
b
x +
2
a 2
b
|
.
|
This is an important formula. When using it, always write it down first, then write the values of
a, b, c in the particular case, paying great attention to the signs. It can be used to solve any
quadratic equation, although you should use the factor method whenever possible.
Example 1:
Consider the following equation:
5x
2
÷ 15x + 11 = 0
x =
a 2
ac 4 b b
2
÷ ± ÷
Remember that the square of a number is always positive, and that no real square root can
be found for a negative number.
Therefore, if the term b
2
÷ 4ac is negative, no real roots of the equation can be found.
For example:
2x
2
+ 4x + 5 = 0
Applying the formula:
x =
a 2
ac 4 b b
2
÷ ± ÷
Relationship between Logarithms and the Exponential Number
From this discussion of natural logarithms and the exponential number, you may have
identified that a relationship exists between the two. The natural logarithm function f(x) = e
x
is
the inverse of the exponential function f(y) = ln(x). Notice that the variables x and y have
been reversed. This is not a mistake! When we take the natural logarithm of an exponential
function the result is the exponent, so if y = e
x
, then ln(y) = x.
Use your calculator to confirm this natural logarithmic and exponential relationship. First,
think of a number for y and then calculate the natural logarithm of this number to find x (i.e.
using the formula y = e
x
). Then, using this value for x, calculate the value of y using the
exponential function y = e
x
. The resulting value of y should be the same as the initial number
you chose for y, thus demonstrating that an inverse relationship exists between the natural
logarithm function f(x) = e
x
and the exponential function f(y) = ln(x). Thus, we can conclude
that:
ln(e
x
) = x
Laws of Logs and Exponents
Numbers expressed in logarithmic or exponential form obey the same rules applicable to
powers and indices, which have already been described in Chapter 2.
Here we give the various rules of logarithms (which can be equally applied to common
logarithms (i.e. to the base 10) and natural logarithms (to the base e)). Since logarithms are
basically powers, there should be some logic to the following rules. Thus, for any base:
- log(p × q) = log p + log q
- log |
.
|
p
n
= n log p] we can rearrange the second part of the expression:
log(x) + log(y
2
)
Using the rule [log(p × q) = log p + log q] we can rearrange the expression:
log(xy
2
)
Example 2:
Simplify the following logarithm expression into a single log term:
3ln(A) ÷ 4ln(B)
Using the rule [log
x = 130
Therefore, the cost of tea = £1.30 per kilo, and the cost of sugar = £0.50 per kilo.
3. Let: x = the original income from advertisements (in £)
y = the original income from sales (in £).
Thus, in the first year:
x + y = 670 (i)
In the second year:
(x + 12
2
1
%
x) + (y ÷ 16
3
2
%
y) = 670 ÷ 12.50
(x +
8
1
x) + (y ÷
6
1
y) = 657
2
1
8
9
x +
6
5
y = 657
2
1
Multiplying this by 24 to eliminate the fractions, we get:
27x + 20y = 15,780 (ii)
To eliminate y from the pair of equations, multiply (i) by 20:
20x + 20y = 13,400
and subtract this from (ii) to get:
7x = 2,380
x = 340
Substituting for x in (i):
340 + y = 670
y = 330
Thus, the original income from:
advertisements = £340
sales = £330.
4. Let: x = the average speed (in miles per hour)
y = the distance (in miles)
We can use a formula for the time taken:
speed
distance
= time taken
The points should be plotted with a small x using a sharp pencil.
To plot the co-ordinate (1,1) you count one unit across (along the x-axis) and then count one
unit up (along the y-axis). Be careful not to get these in the wrong order or when you come to
plot the co-ordinate (7,4) you will have plotted the point (4,7), which is somewhere
completely different!
(Remember that if the x co-ordinate is negative you count to the left and if the y co-ordinate
is negative you count downwards.)
Once you have plotted all the given points, you should join them all up.
If your graph is a linear graph, all the points will be in a straight line and you can join them
using a ruler. (In fact, to draw a linear graph, you only need to plot two points which, when
joined and extended across the whole scale of the x-axis, give the complete graph for all
values of the independent variable.)
However, if your graph is a quadratic graph, then the plot line will form a curve. The points
should be joined with a smooth curve. The easiest way of doing this is to trace through the
points with a pencil, but without touching the paper, to find the shape of the curve. When you
are satisfied that the path of the curve is smooth, draw the curve through the points in one
movement.
The full graph derived from the previous data on the height of a plant and extended over nine
days is given in Figure 5.3.
Figure 5.3: Graph of height of plant against time
Alternatively, as this is a simple problem, you could tabulate just the first and the last lines,
working the rest in your head as you go along.
We can also find a single function of x by substituting that value directly into the given
equation, for example:
f(x) x
2
2x
Replace x by 3 and find the value:
f(3) 3
2
2 3 15
Replace x by 2 and find the value
f(2) (2)
2
2(2) 0
You should have found the points: (3, 7), (0, 2), (2, 8).
Now draw your two axes to cover the range of values for x specified (although it is good
practice to extend them beyond this – leaving some spare room at each end of each axis),
mark the scale, plot your points and join them up. Then compare your result with Figure 5.6.
Figure 5.6: Graph of y 2x + 2
Characteristics of Straight-Line Graphs
If you consider the graphs we have drawn as examples, and those you drew in the Questions
for Practice, we can start to identify a number of characteristics of straight-line graphs.
The general equation for a straight line is y mx c where m and c are constants which may
have any numerical value, including 0. The value of m is known as the coefficient of x, and
plays a very important part in calculations concerning a straight-line graph.
Lines parallel to an axis
Consider the case where m 0. This means y c.
Whatever value x has, y is always equal to c. The graph of y c must, then, be a line
parallel to the x-axis at a distance c from it. Note that y 0 is the x-axis itself.
In the same way, x c is a line parallel to the y-axis at a distance c from it. x 0 is the
y-axis.
Plot these seven points on the graph and then join the points with a smooth curve, as shown
in Figure 5.10. From the graph you will see that function y x
2
is minimised where y 0 and
x 0 (i.e. the co-ordinates where y x
2
is minimised are (0,0)).
Figure 5.10: Graph of y x
2
Equations with x as the Denominator
These equations (hyperbolic equations) take the form of y
x
a
where a is a constant.
They result in a characteristic curve, the direction of which may vary.
Example 1:
Let's look at the case where a 1:
y
x
1
Just as we saw that the shape of the quadratic graphs changed according to whether the
coefficient of x
2
is positive or negative, so the shape of graphs developed from equations
where x is the denominator change:
x
a
(i.e. where both the a and x are positive) gives a graph which slopes down from left
to right
x
a
gives a graph which slopes down from left to right
x
a
gives a graph which slopes up from right to left
x
a
gives a graph which slopes up from right to left.
These are illustrated in Figure 5.15.
Figure 5.15: Shape of graphs from the equation
x
a
Reading from the graph, when y 0, x 1.
Questions for Practice 5
1. Use a scientific calculator to construct a table of values for y log
10
(x) when x 0,5,
10, 15, 20 and 25 (give your answers to 2 decimal places). Plot these values on a
graph.
2. The sales S (in thousands of units) of a product after it has been on the market for t
months is given by the equation: S 100(1 e
0.1625t
).
(a) Construct a table of values for S for t 0,5, 10, 15, 20 and 25 (give S values to
the nearest whole number).
(b) Use the table to draw a graph of S against t and from your graph estimate the
value of S when t is very large.
Now check your answers with those given at the end of the chapter.
Now, consider the five slopes illustrated in Figure 5.18.
In the five shapes, the slope of the line(s) increases as we go from (i) through to (v).
Figure 5.18: Gradient of a slope
In (i), the gradient (s) is zero (no slope).
In the other cases, the gradient is defined as the height of the right-angled
triangle divided by the base. Therefore
in (ii), the gradient of s is 1/2 0.5
in (iii), the gradient of s is 2/2 1
in (iv), the gradient of s is 4/2 2
in (v), the gradient of s is 8/2 4
The gradient of the line s is
2
2
1.
Gradient of a Straight Line
We can easily calculate the gradient of a straight line joining two points from the co-ordinates
of those points.
Firstly, we shall examine the method by reference to plotting the line on a graph.
Example 1:
Plot the points (1,3) and (4,8) and find the gradient of the line joining them.
The procedure would be as follows:
(a) Plot the two points on a graph and join them to give a sloping line, as shown in Figure
5.20.
Figure 5.20: Gradient of the line connecting points (1, 3) and (4, 8)
This time we shall work out the lengths of the base and the height from the co-ordinates of
the points.
base 2 (1) 3
height 4 (3) 7
We can check that this is correct by reference to the plotted triangle on the graph.
Note that the slope is negative.
Therefore the gradient
base
height
3
7
2
3
1
Note that these lengths are directed lengths: AC is positive if the movement from A to C is in
the same direction as O to x, and negative if the movement from A to C is in the opposite
direction to O to x. In the same way, CB is positive if the movement from point C to B is in the
same direction as O to y. (The point at O is the origin, i.e. (0,0).)
Taking the lines in Figure 5.22, we find the following:
In line (i), AC and CB are both positive, hence the gradient
AC
CB
is positive.
In line (ii), AC is negative, CB is positive, and the gradient is negative.
We can further see that AC is the difference between the co-ordinates along the x-axis and
CB is the difference between the co-ordinates along the y-axis. Therefore, we could calculate
the gradient where the points are, say, (5,1) and (7,2) as follows:
5 7
1 2
2
1
Rearranging the equation m
x
c y
we get:
mx y c
y mx c
This is the general equation of any straight line. Note that c is the intercept that the line
makes with the y-axis. If the line cuts the y-axis below the origin, its intercept is negative, c.
Using this method, given any straight-line graph, its equation can be found immediately.
Also, given any linear equation, we can find the gradient of the line (m) immediately. For
example:
Find the gradient of a line with the equation 6x 3y 19.
We need to put the equation into the general form y mx c:
y
3
19 x 6
We can work out the gradient of the line on this graph, based on two co-ordinates. One of
those co-ordinates must be (0, 0), since when there is no production, no variable costs will
be incurred. If we know that one other point on the graph is (9, 3), then the gradient (which is
m) is:
m
0 9
0 3
9
3
3
1
We can see that this line is parallel to the x-axis, with an intercept at 2. This is represented by
the equation:
b 2 (where b is fixed costs)
Businesses are not only interested in costs as either fixed or variable, but in their total costs.
We can say that:
total costs variable costs fixed costs
If total costs y, and using the notation from above, we get the following equation:
y a b
Substituting from the equations established for a and b, we get:
y
3
1
x 2
This is a standard linear equation. We could work out a table of values for it (which we shall
skip here) and draw a graph. This has been done, again on the same graph as in Figure
5.26, in Figure 5.27.
Note that the line for total costs is in fact a combination of the lines for variable and fixed
costs. Its intercept is at the level of fixed costs and its gradient is that of variable costs.
Break-even Analysis
We can further develop the costs graph by including revenue on it.
Revenue is the total income from sales of the units produced. If we assume that each unit
sells for £1, we can say that y x where y is the income from sales (in £) and x is the number
of units sold.
Adding the graph of this to the costs graph shown in Figure 5.27, we get Figure 5.29. (Note
that we are assuming that all units produced are sold.)
Figure 5.29: Costs and revenue graph
The warehousing cost is linear – if the amount purchased in each order is doubled, then the
warehousing costs are doubled as there are, on average, twice as many items in the
warehouse. This shown in Figure 5.31.
Figure 5.31: Warehousing costs
The cost of purchasing is not linear. The cost per annum depends on the number of orders
placed, as it is a fixed amount per order.
For example, assume that there are 250 working days each year:
(a) Sprockets purchased in batches of 1,000, i.e. a batch every day:
cost £100 250 £25,000
(b) Sprockets purchased in batches of 2,000, i.e. a batch every other day:
cost £100 125 £12,500
(c) Sprockets purchased in batches of 4,000, i.e. a batch every fourth day:
cost £100 62.5 £6,250.
Each time the amount purchased is doubled, the cost per annum halves. The graph is shown
in Figure 5.32.
Size of purchased lot
Cost
Stock
level
There are four factors that influence the changes in a time series over time. These are the
trend, seasonal variations, cyclical fluctuations, and irregular or random fluctuations.
However, you only need to concern yourself with the trend.
The trend is the change in general level over the whole time period and is often referred to as
the secular trend. You can see in Figure 5.34 that the trend is definitely upwards, in spite of
the obvious fluctuations from one quarter to the next. A trend can thus be defined as a clear
tendency for the time series data to travel in a particular direction in spite of inter-period
fluctuations (be they large or small).
If we were to draw a simple linear trend line through the time series data shown in Figure
5.34, then we would get a trend line similar to that shown in Figure 5.35. Based on this trend
line, we can clearly see that lost days from work (per quarter) have increased over the
period, from approximately 25 days lost per quarter in 2003 to approximately 50 days lost per
quarter in 2007.
1. Plot the time series on a historigram.
2. Using a ruler, draw a linear trend line through the time series plotted in Question 1 and
comment on the trend.
Now check your answers with those given at the end of the chapter.
The graph is shown on the next page. Reading off from it, we get:
(a) Profit is maximised (£17,000) at 4,000 units of output, i.e. when x 4.
(b) At an output of between 1,000 and 7,000 units (i.e. between x 1 and x 7),
profits generated are in excess of £8,000 (i.e. y 8 or above).
As it stands, this data does not allow us to draw any useful conclusions about the group of
employees. In order to do this, we need to classify the data in some way. One way would be
to classify the individual cases by sex:
Again, although the data has now acquired some of the characteristics of useful information,
it is still a collection of data about individuals. In statistics though, we are concerned with
things in groups rather than with individuals. In comparing the height of Frenchmen with the
height of Englishmen, we are concerned with Frenchmen in general and Englishmen in
general, but not with Marcel and John as individuals. An insurance company, to give another
example, is interested in the proportion of men or women who die at certain ages, but it is not
concerned with the age at which John Smith or Mary Brown, as individuals, will die.
We need to transform the data into information about the group, and we can do this by
tabulation. The number of cards in each group, after classification, is counted and the results
presented in a table.
Another question might have been "What is the wage distribution of the employees?"
The answer can be given in another simple table, Table 6.3.
Table 6.3: Wage distribution
Wages group Number of employees
£40 but less than £60 105
£60 but less than £80 510
£80 but less than £100 920
£100 but less than £120 1,015
£120 but less than £140 300
£140 but less than £160 150
Total 3,000
* Note: 40 – 59.99 is the same as "40 but less than 60" and similarly for the other columns.
This table is much more informative than are the two previous simple tables, although it
is more complicated. We could have complicated it further by dividing the groups into
male and female employees, or into age groups.
Later in this chapter, we will look at a list of the rules you should try to follow in
compiling statistical tables. At the end of that list you will find a table relating to our
3,000 employees, which you should study as you read the rules.
Secondary Statistical Tabulation
So far, our tables have merely classified the already available figures – the primary statistics.
However, we can go further than this and do some simple calculations to produce other
figures – secondary statistics.
As an example consider again Table 6.2, and this time calculate how many employees there
are on average per workshop. This is obtained by dividing the total (3,000) by the number of
workshops (5), and the table appears as follows:
Table 6.5: Distribution of employees by workshop
Workshop Number employed
A 600
B 360
C 660
D 840
E 540
Total 3,000
Average number of
employees per workshop:
The proportions given here are also secondary statistics. In commercial and business
statistics it is more usual to use percentages than proportions, and so in Table 6.6 these
would be 3.5%, 17%, 30.7%, 33.8%, 10% and 5%.
Of course secondary statistics are not confined to simple tables. They are used in complex
tables too, as in the example presented in Table 6.7.
Table 6.7: Inspection results for a factory product in two successive years
Machine no.
Year 1 Year 2
Output No. of
rejects
% of
rejects
Output No. of
rejects
% of
rejects
1 800 40 5.0 1,000 100 10.0
2 600 30 5.0 500 100 20.0
3 300 12 4.0 900 45 5.0
4 500 10 2.0 400 20 5.0
Total 2,200 92 4.2 2,800 265 9.5
Average per
machine
Note (a): Total number employed in each workshop as a percentage of the total workforce.
Note (b): Total number in each wage group as a percentage of the total workforce.
This table can be called a "twofold" table as the workforce is broken down by wage and
workshop.
Questions for Practice 1
These questions are typical of those likely to be set in an examination.
1. The human resources/personnel manager of a firm submits the following report on
labour and staff to the managing director:
"The total number of employees is 136, of whom 62 are male. Of these
men, 43 are industrial labourers and the rest are staff. The age distribution
of the 43 is 'under 18' (seven), '18–45' (22) and the others are over 45. Our
total industrial labour force is 113, of whom 25 are 'under 18' and 19 are
'over 45'. The total number of 'under-18s' on the site is 29 (10 male, 19
female). There are 23 'over-45s', of whom only 4 are male staff."
The managing director, quite rightly, thinks that this is gibberish and instructs you to
present it to him in a tabular form which he can readily understand. Show the table you
would produce.
Simple Frequency Distributions
A useful way of preparing a frequency distribution from raw data is to go through the records
as they stand and mark off the items by the "tally mark" or "five-bar gate" method. First look
at the figures to see the highest and lowest values so as to decide the range to be covered
and then prepare a blank table. Then mark the items on your table by means of a tally mark.
To illustrate the procedure, Table 6.11 shows the procedure for the data in Table 6.10 after all
20 items have been entered.
Table 6.11: Time taken for one employee to perform task
Length of time
(minutes)
Tally marks Frequency
3.4 | | 2
3.5 | | | 3
3.6 | | | | 4
3.7 | | | | 5
3.8 | | | | 4
3.9 | | 2
Total 20
Grouped frequency distributions are only needed when there is a large number of values
and, in practice, would not have been required for the small amount of data in our example.
Table 6.13 shows a grouped frequency distribution used in a more realistic situation, when an
ungrouped table would not have been of much use.
Table 6.13: Age distribution of employees in an office
Age group
(years)
Number
of employees
16 to < 21 10
21 to < 26 17
26 to < 31 28
31 to < 36 42
36 to < 41 38
41 to < 46 30
46 to < 51 25
51 to < 56 20
56 to < 60 10
Total 220
2. (a) The complete table is as follows:
Number of Stoppages of Work Arising from
Industrial Disputes on Construction Sites
Cause
Year 1 – Year 6 Year 7 – Year 12
1 day
More than
1 day
Total 1 day
More than
1 day
Total
Pay 74.47 602.53 677 29.47 296.53 326
Demarcation 6.66 104.34 111 52 807 859
Working
conditions
18 126 144 16 93 109
Unofficial* 78.37 764.63 843 144.03 3,143.97 3,288
All 177.5 1,597.5 1,775 241.5 4,340.5 4,582
* Note that we have given the figures as worked out from the information
supplied. Obviously, though, fractional numbers of stoppages cannot occur, and
you should round off each figure to the nearest whole number.
The subtotals of the columns will then be 177; 1,598; 241; and 4,341.
(b) The overall number of stoppages in the later period was more than 2½ times
greater than in the earlier period. In both periods, the number of unofficial
stoppages was the highest category. This was particularly remarkable in the later
period, when they amounted to more than twice the number of all other
stoppages put together. Demarcation disputes rose in number dramatically,
although stoppages over pay and working conditions both declined. About 95% of
stoppages in the later period lasted more than one day, compared with 90% in
the previous period, with those concerning working conditions being less likely to
last more than one day.
Frequency Dot Diagram
This is a simple form of graphical representation for the frequency distribution of a discrete
variate. (In statistics, the term "variate" is used when you might have expected the term
"variable". Whilst the two terms do not mean exactly the same thing, roughly speaking, at this
level, they can be used interchangeably. Loosely, a variate is a quantity that can have any of
a set of numerical values. Do not worry too much about the difference between these two
terms.)
A horizontal scale is used for the variate and a vertical scale for the frequency. Above each
value on the variate scale we mark a dot for each occasion on which that value occurs.
Thus, a frequency dot diagram of the information in Table 7.1 would be as shown in Figure
7.1.
Figure 7.1: Frequency dot diagram to show length
of time taken by operator to complete a given task
Frequency Polygon
Instead of drawing vertical bars as we do for a frequency bar chart, we could merely mark
the position of the top end of each bar and then join up these points with straight lines. When
we do this, the result is a frequency polygon, as in Figure 7.3.
Figure 7.3: Frequency polygon to show length
of time taken by operator to complete a given task
The last two groups have been lumped together as one. A wrong form of histogram, using
heights instead of areas, for the amended data would look like Figure 7.5.
Figure 7.5: Histogram showing age distribution of employees in an office
(amended data)
Often it happens, in published statistics, that the last group in a frequency table is not
completely specified. For example, the last few groups in our age distribution table may have
been specified as follows:
Table 7.4: Age distribution of employees in an office (further amended extract)
Age group
(years)
Number
of employees
40 to < 45 30
45 to < 50 25
50 and over 30
Before you begin, you might want to arrange the data in numerical order, but is not a required
step as ordering can be done later. But what is necessary is that you separate each number
into a stem and a leaf. Since these are two digit numbers, the tens digit is the stem and the
units digit is the leaf (e.g. for the number 79, the stem is "7" and the leaf is "9").
Next, group the numbers with the same stems (e.g. 79 and 77 are in the same group as both
have a stem of 7). Then, list the stems in numerical order (in our example, the stems range
from 5 to 8 as the lowest value is 50 and the highest value is 87) in a column. In a second
column list the leaves to the right of the appropriate stem values. If your leaf values are not in
increasing order, order them now.) Finally, add a title to your diagram. If you have
successfully followed these steps, then you should have finished up with a stem and leaf
diagram identical to the one shown in Figure 7.10. As you will see, for the stem value of 5 we
have leaf values of 0, 7 and 9, which correspond to the original values of 50, 57 and 59.
B. PICTOGRAMS
Pictograms are the simplest method of presenting information visually. These diagrams are
variously called "pictograms", "ideograms", or "picturegrams" – the words all refer to the
same thing. Their use is confined to the simplified presentation of statistical data for the
general public.
Pictograms consist of simple pictures which represent quantities. There are two types and
these are illustrated in the following examples. The data we will use is shown in Table 7.8.
Table 7.8: Cruises organised by a shipping line between Year 1 and Year 3
Year Number of
cruises
No. of passengers
carried
1 200 100,000
2 250 140,000
3 300 180,000
Limited Form
We could represent the number of cruises by showing pictures of ships of varying size, as in
Figure 7.11.
Figure 7.11: Number of cruises Years 1-3
(Source: Table 7.8)
Each matchstick man is the same height and represents 20,000 passengers, so there can be
no confusion over size.
These diagrams have no purpose other than generally presenting statistics in a simple way.
They are not capable of providing real accurate detail. Consider Figure 7.13 – it is difficult to
represent a quantity of less than 10m barrels – so we are left wondering what "[" means.
Figure 7.13: Imports of crude oil
This figure is also an example of a percentage component bar chart, i.e. the information is
expressed in percentages rather than in actual numbers of visitors.
If you compare several percentage component bar charts, you must be careful. Each bar
chart will be the same length, as they each represent 100%, but they will not necessarily
represent the same actual quantities – for example, 50% might have been 1 million, whereas
in another year it may have been nearer to 4 million and in another to 8 million.
Horizontal Bar Chart
A typical case of presentation by a horizontal bar chart is shown in Figure 7.17. Note how a
loss is shown by drawing the bar on the other side of the zero line.
Figure 7.17: Horizontal bar chart for the So & So Company Ltd to show profits made
by branches in Year 1 and Year 2
Pie charts and bar charts are especially useful for "categorical" variables as well as for
numerical variables. The example in Figure 7.17 shows a categorical variable – i.e. the
different branches form the different categories, whereas in Figure 7.15 we have a numerical
variable, namely, time. Figure 7.17 is also an example of a multiple or compound bar chart as
there is more than one bar for each category.
European
51%
Commonwealth
23%
USA
22%
O
t
h
e
r
s
While scatter diagrams are used to study possible relationships between two variables, it is
important to stress that these diagrams cannot prove that one variable causes the other.
However, they may indicate the existence and strength of an association between two
variables. This association is known as correlation.
Thus, the relationship shown in Figure 7.18, where both variables are increasing together, is
described as positive correlation (both variables are positively correlated with each other).
In contrast, if one variable increases while the other decreases, the relationship is described
as negative correlation. For example, we would expect the efficiency of the machines in this
manufacturing company to decline as they get older, suggesting that the efficiency and age
of machinery are negatively correlated.
If the scatter diagram does not show any relationship between the variables then no
correlation exists. For example, we would not expect any correlation between the cost of raw
materials and employee output.
These three types of correlation are shown in Figure 7.19.
Sources of revenue in Ruritania for year ending 31 March
5. Illustrate by means of a bar chart the following data relating to the number of units sold
of a particular kind of machine:
Year Home sales Commonwealth sales Foreign sales
1 6,500 5,000 5,800
2 6,250 5,400 5,700
3 5,850 6,600 5,500
6. The unit price of a commodity and its demand over a period of time are shown in the
following table:
Price
(£ per item)
Number of items sold
(000)
1.20 41
2.39 38
3.30 34
4.31 29
5.27 26
6.29 30
7.26 25
8.24 22
9.18 19
10.16 15
Draw a scatter diagram of the data and comment on the relationship between price and
the amount sold.
Now check your answers with those given at the end of the chapter.
(called "x bar") and the number of items by
the letter n. The calculation of the arithmetic mean can then be described by formula:
n
Σx
x
or
Σx
n
1
x
The latter expression is customary in statistical work. Applying it to the example above, we
have:
175 1225
7
1
x ) ( cms.
You will often find the arithmetic mean simply referred to as "the mean" when there is no
chance of confusion with other means (which we are not concerned with here).
The Mean of a Simple Frequency Distribution
When there are many items (that is when n is large), the arithmetic can be eased somewhat
by forming a frequency distribution. Consider the following example.
Example 1:
Table 8.2: Height distribution data
Height in cms
(x)
No. of men at
this height
(f)
Product
(fx)
160 1 160
163 1 163
168 1 168
180 2 360
187 2 374
Total Σf 7 Σ(fx) 1225
You should have obtained the following answers:
the total numbers of items, Σf 100
the total product, Σ(fx) 713
the arithmetic mean,
100
713
x 7.13
Make sure that you understand everything we have covered so far. Revise it if necessary,
before going on to the next paragraph. It is most important that you do not get muddled about
calculating arithmetic means.
The Mean of a Grouped Frequency Distribution
Suppose now that you have a grouped frequency distribution. In this case, you will
remember, we do not know the actual individual values of x, only the groups in which they lie
(unless you have a stem and leaf diagram, which we will discuss later). How then can we
calculate the arithmetic mean? The answer is that we cannot calculate the exact value of
x ,
but we can make an approximation sufficiently accurate for most statistical purposes. We do
this by assuming that all the values of x in any group are equal to the midpoint (or mid-value)
of that group.
The procedure is very similar to that for a simple frequency distribution (which is why we
stressed the need for revision) and is shown in this example:
So, using the same formula (and letting the mid-value x), we can calculate the arithmetic
mean of the grouped frequency distribution in Table 8.4 as follows:
50
660 1
Σf
Σfx
x
,
33.2
Provided that Σf (i.e. n) is not less than about 50 and the number of groups is not less than
about 12, the arithmetic mean thus calculated is sufficiently accurate for all practical
purposes.
There is one pitfall to be avoided when using this method – if all the groups should not have
the same class interval, be sure that you get the correct mid-values! Table 8.5 is part of a
larger table with varying class intervals, to illustrate the point:
Table 8.5: Data for grouped frequency distribution
Group Mid-value
(x)
0 to < 10 5
10 to < 20 15
20 to< 40 30
40 to < 60 50
60 to < 100 80
That is Σ(x – x
o
) 35 cms or Σd 35 cms.
We can make use of this property and use it as a short cut for finding
x .
We start by choosing some value of x as an assumed mean. We try to choose it near to
where we think the true mean ( x ) will lie, and we always choose it as the midpoint of
one of the groups when we are dealing with a grouped frequency distribution.
In the example just given, the total deviation (d) does not equal zero, so 170 cannot be
the true mean. As the total deviation is positive, we must have underestimated in our
choice of x
o
, so the true mean is higher than 170. As there are seven readings, we
need to adjust x
o
upwards by one seventh of the total deviation, i.e. by (35) 7 5.
Therefore the true value of
All we can really say is that "70 to < 80" is the modal group (the group with the largest
frequency).
You may be tempted to say that the mode is 75, but this is not true, nor even a useful
approximation in most cases. The reason is that the modal group depends on the method of
grouping, which can be chosen quite arbitrarily to suit our convenience. The distribution could
have been set out with class intervals of five instead of 10, and would then have appeared as
follows (only the middle part is shown, to illustrate the point):
Table 8.14: Revised data for determining the mode of a grouped frequency distribution
Group Frequency
60 to < 65 16
65 to < 70 22
70 to < 75 21
75 to < 80 19
80 to < 85 12
85 to < 90 8
The total frequency is 206 and therefore the median is the 103
rd
item which, from the
cumulative frequency column, must lie in the "60 to < 70" group. But exactly where in the "60
to < 70" group? Well, there are 92 items before we get to that group and we need the 103
rd
Calculating the Median from a Stem and Leaf Diagram
Similarly, we can use a stem and leaf diagram to calculate the median value of a distribution
of grouped data as the original data is preserved. Again, let us consider our previous
example of a stem and leaf diagram which shows the distribution of marks awarded to an ice
skater in a competition (Figure 8.6). Again, remember that the stem values represent the
units of a mark awarded and the leaf values represent the tenths of a mark awarded.
Figure 8.6: Stem and leaf diagram to show the marks awarded to a competition skater
Stem Leaf
0 0
1
2
3 5 9
4 0 1 1 3
5 0 5 8
224.5 (1.424 50)
224.5 71.2 153.3 153 shares to the nearest whole number.
(N.B. You may have chosen a different x
o
, but the result will be the same.)
(b) Half of the total frequency is 280.5 and therefore the median lies in the "100 –
149" group and it is the 58.5
th
(280.5 222) item within it.
Median 99.5 1 . 126
110
50 5 . 58
126 shares to the nearest whole number.
(c) The modal group is the "50 – 99" group and the estimated mode is given by
drawing this group, together with the surrounding class groups, on a histogram.
This is shown on the next page.
Reading from this, the estimated mode 83 shares.
Σf
Σfx
29.1
i.e. the arithmetic mean age of manufacturing employees is 29.1 years.
(b) The median age of manufacturing employees is 27 years.
(c) In our example, there are two 8 leaves on the 1 stem (i.e. two data points of value
18) and two 5 leaves on the 2 stem (i.e. two data points of value 25). Thus, the
data set is bimodal with the mode age of manufacturing employees being 18
years and 25 years.
This is the same ogive that we drew in Chapter 8 when finding the median of a grouped
frequency distribution.
You will notice that we have added the relative cumulative frequency scale to the right of the
graph. 100% corresponds to 206, i.e. the total frequency. It is then easy to read off the values
of the variate corresponding to 25%, 50% and 75% of the cumulative frequency, giving the
lower quartile (Q
1
), the median and the upper quartile (Q
3
) respectively.
Q
1
= 46.5
median = 63 (as found previously)
Q
3
= 76
The difference between the two quartiles is the interquartile range and half of the difference
is the semi-interquartile range or quartile deviation:
Quartile deviation =
2
Q Q
1 3
÷
, 5 . 51
4
206
= and
75% of the total frequency, i.e. 154.5
and then read from the ogive the values of the variate corresponding to 51.5 and 154.5 on
the cumulative frequency scale (i.e. the left-hand scale). The end result is the same.
Calculation of the Quartile Deviation
The quartile deviation is not difficult to calculate in situations where the graphical method is
not acceptable. Remember that graphical methods are never quite as accurate as
calculations.
We shall again use the same example, and the table of values is reproduced for convenience
as Table 9.2:
Table 9.2: Data for finding the quartile deviation of a grouped frequency distribution
Group Frequency
(f)
Cumulative
frequency
(F)
0 to < 10 4 4
10 to < 20 6 10
20 to < 30 10 20
30 to < 40 16 36
40 to < 50 24 60
50 to < 60 32 92
60 to < 70 38 130
70 to < 80 40 170
80 to < 90 20 190
90 to < 100 10 200
100 to < 110 5 205
110 to < 120 1 206
Total 206
10
24
½ 15
×
= 40 + 6.458
= 46.458
Similarly, the upper quartile will be the 154½
th
item which is in the "70 to < 80" group and is
the (154½ ÷ 130) = 24½
th
item within it.
So, the upper quartile = 70 +
10
40
½ 24
×
= 70 + 6.125
= 76.125
Remember that the units of the quartiles and of the median are the same as those of the
variate.
The quartile deviation =
2
Q Q
1 3
÷
=
2
458 . 46 125 . 76 ÷
=
2
667 . 29
= 14.8335
= 14.8 to 1 decimal place.
The quartile deviation is unaffected by an occasional extreme value. However it is not based
on the values of all the items in the distribution and, to this extent, it is less representative
than the standard deviation (which we discuss later). In general, when a median is the
appropriate measure of location, then the quartile deviation should be used as the measure
of dispersion.
Deciles and Percentiles
It is sometimes convenient, particularly when dealing with wages and employment statistics,
to consider values similar to the quartiles, but which divide the distribution more finely. Such
partition values are deciles and percentiles. From their names you will probably have
guessed that the deciles are the values which divide the total frequency into tenths and the
percentiles are the values which divide the total frequency into hundredths. Obviously it is
only meaningful to consider such values when we have a large total frequency.
The deciles are labelled D
1
, D
2
... D
9
. For instance, the second decile (D
2
) is the value below
which 20% of the data lies, and the sixth decile (D
6
) is the value below which 60% of the data
lies.
The percentiles are labelled P
1
, P
2
... P
99
. Again we can say that, for example, P
5
is the value
below which 5% of the data lies and P
64
is the value below which 64% of the data lies.
Using the same example as previously, let us calculate as an illustration the third decile D
3
.
The method follows exactly the same principles as the calculation of the median and
quartiles.
8 . 61 206
100
30
= ×
We are therefore looking for the value of the 61.8
th
item. A glance at the cumulative
frequency column shows that the 61.8
th
item lies in the "50 to < 60" group, and is the (61.8 ÷
60) = 1.8
th
item within it. So:
D
3
= 50 +
10
32
8 . 1
×
= 50 +
32
18
= 50.6 to one decimal place.
Therefore 30% of our data lies below 50.6.
We could also have found this result graphically – check that you agree with the calculation
by reading D
3
from the graph in Figure 9.1. You will see that the calculation method enables
us to give a more precise answer than is obtainable graphically.
C. STANDARD DEVIATION
The most important of the measures of dispersion is the standard deviation. Except for the
use of the range in statistical quality control and the use of the quartile deviation in wages
statistics, the standard deviation is used almost exclusively in statistical practice.
The standard deviation is defined as the square root of the variance, and so we need to know
how to calculate the variance first.
The Variance and Standard Deviation
We start by finding the deviations from the mean, and then squaring them, which removes
the negative signs in a mathematically acceptable fashion. The variance is then defined as
the mean of the deviations from the mean, squared:
variance =
n
) x x ( Σ
2
÷
, where n is the total frequency.
Consider the following example in Table 9.3.
\
|
÷ =
Formula (c)
The choice of the formula to use depends on the information that you already have and the
information that you are asked to give. In any calculation, you should aim to keep the
arithmetic as simple as possible and the rounding errors as small as possible.
If you already know
x
and it is a small integer, there is no reason why you should not
use formula (a).
If you already know
x
but it is not an integer (as in the previous example), then formula
(b) is the best to use.
If you do not know
x , then you should use formula (c) – particularly if you are not
asked to calculate
x .
When you are dealing with data in the form of a simple or grouped frequency distribution,
then the formula for calculating the standard deviation needs to be expressed slightly
differently:
Σf
) x x f Σ
σ
2
÷
=
(
Formula (d)
or
|
|
.
|
\
|
×
It is always subtracted from the approximate variance to get the corrected variance. In
our case the correction is:
÷(20/30)
2
= ÷0.4444.
(c) The corrected variance is thus:
2.7333 ÷ 0.4444 = 2.2889
(d) The standard deviation is then the class interval times the square root of the corrected
variance:
SD =
The assumed mean should be chosen from a group with the most common interval and
c will be that interval. If the intervals vary too much, we revert to the basic formula.
Characteristics of the Standard Deviation
In spite of the apparently complicated method of calculation, the standard deviation is the
measure of dispersion used in all but the very simplest of statistical studies. It is based on all
of the individual items. It gives slightly more emphasis to the larger deviations but does not
ignore the smaller ones and, most important, it can be treated mathematically in more
advanced statistics.
D. THE COEFFICIENT OF VARIATION
Suppose that we are comparing the profits earned by two businesses. One of them may be a
fairly large business with average monthly profits of £50,000, while the other may be a small
firm with average monthly profits of only £2,000. Clearly, the general level of profits is very
different in the two cases, but what about the month-by-month variability? We can compare
the variability of the two firms by calculating the two standard deviations.
Let us suppose that both standard deviations come to £500. Now, £500 is a much more
significant amount in relation to the small firm than it is in relation to the large firm. So
although they have the same standard deviations, it would be unrealistic to say that the two
businesses are equally consistent in their monthly profits.
To overcome this difficulty, we can express the standard deviation as a percentage of the
mean in each case, using the following formula:
When the items are not symmetrically dispersed on each side of the mean, we say that the
distribution is skewed or asymmetric.
A distribution which has a tail drawn out to the right is said to be positively skewed, while one
with a tail to the left is negatively skewed.
Two distributions may have the same mean and the same standard deviation but they may
be differently skewed. This will be obvious if you look at one of the skew distributions in
Figure 9.3 and then look at the same one through from the other side of the paper!
Figure 9.3: Skewed distributions
What does skewness tell us? It tells us that we are to expect a few unusually high values in a
positively skewed distribution or a few unusually low values in a negatively skewed
distribution.
If a distribution is symmetrical, the mean, mode and median all occur at the same point, i.e.
right in the middle. But in a skew distribution, the mean and the median lie somewhere along
Variate
Frequency
Variate
Frequency
(a)
For most distributions, except for those with very long tails, the following approximate
relationship holds:
mean ÷ mode = 3(mean ÷ median)
The more skewed the distribution, the more spread out are these three measures of location,
and so we can use the amount of this spread to measure the amount of skewness. The most
usual way of doing this is to calculate Pearson's first coefficient of skewness:
Pearson's first coefficient of skewness =
SD
mode mean ÷
However as we have seen the mode is not always easy to find, and so we can use an
equivalent formula for Pearson's second coefficient of skewness:
Pearson's second coefficient of skewness =
SD
median) ean m ( 3 ÷
You need to be familiar with these formulae as the basis of calculating the skewness of a
distribution (or its "coefficient of skewness", as it is also called). When you do the calculation,
remember to get the correct sign (+ or ÷) when subtracting the mode or median from the
mean. You will get negative answers from negatively skewed distributions, and positive
answers for positively skewed distributions.
The value of the coefficient of skewness is between ÷3 and +3, although values below ÷1
and above +1 are rare and indicate very skewed distributions.
Examples of variates with positive skew distributions include size of incomes of a large group
of workers, size of households, length of service in an organisation and age of a workforce.
Negative skew distributions occur less frequently. One such example is the age at death for
the adult population in the UK.
Variate
Frequency
You are required to:
(a) Derive by graphical methods the median, lower and upper quartile values for Year
1 and Year 11.
(b) Calculate for the two distributions the median values and explain any difference
between these results and those obtained in (a).
(c) Estimate for each year the number of people whose assets are less than (i)
£2,500, and (ii) £25,000.
(d) Write a brief report on the data, using your results where relevant.
4. Items are produced to a target dimension of 3.25 cm on a single machine. Production
is carried out on each of three shifts, each shift having a different operator – A, B and
C. It is decided to investigate the accuracy of the operators. The results of a sample of
100 items produced by each operator are as follows:
Accuracy of operators
Operator
Dimension of item (cm)
3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29
A 1 6 10 21 36 17 6 2 1
B 0 0 7 25 34 27 6 1 0
C 3 22 49 22 4 0 0 0 0
= 2.2727 class interval units
= £1,363.6
= £1,364 to the nearest £.
The standard deviation measures the dispersion or spread of a set of data. It does this
by taking every reading into account. We take the deviation of each reading from the
mean and square it to remove any negative signs in a mathematically acceptable
fashion. We then add all the squared deviations, find their mean and finally take the
square root.
The lower quartile Q
1
is the value below which 25% of the data lies, i.e. it is the
value of the 25th item.
Q
1
= 3.235 +
01 . 0
21
) 17 25 (
×
÷
= 3.235 +
21
01 . 0 8 ×
= 3.2388 cm.
Similarly
Q
3
= 3.255 +
01 . 0
17
) 74 75 (
×
÷
= 3.255 +
01 . 0
17
1
× = 3.2556 cm.
The semi-interquartile range =
2
Q Q
1 3
÷
=
2
0168 . 0
= 0.008 cm to three decimal
places.
This tells us something about the spread of the middle 50% of the data. It could
be of use to the manufacturer for predicting the variation in A's production, so that
possible wastage when items are made too far from the target dimension could
be estimated.
5. Coefficient of skewness =
deviation standard
median) 3(mean ÷
=
5
) 11 10 ( 3 ÷
= ÷0.6
Thus we can say that the distribution is slightly negatively skewed – i.e. asymmetric
with the peak of the curve being to the right.
= 0.167
As a third and final example, imagine a box containing 100 beads of which 23 are black and
77 white. If we pick one bead out of the box at random (blindfold and with the box well
shaken up), what is the probability that we will draw a black bead? We have 23 chances out
of 100, so the probability is:
P(B) =
100
23
= 0.23
Probabilities of this kind, which can be assessed from our prior knowledge of a situation, are
also called a priori probabilities.
In general terms, we can say that if an event E can happen in h ways out of a total of n
possible ways, then the probability of that event occurring (called a success) is given by:
P(E) =
outcomes possible of number total
occurring E of ways possible of number
=
n
h
In the preceding examples, we introduced the concept that some outcomes are equally likely,
while others are not equally likely. In probability, when there are the same chances for more
than one outcome, we say that the outcomes are equally likely to occur. For example:
If someone tosses a coin, the chances of getting a heads or tails are the same, i.e. all
outcomes are equally likely. The probability of getting a heads is P(H) = 0.5 and the
probability of getting a tails is P(T) = 0.5.
If someone throws a die, the chances of getting a six are no different to getting any
other number, i.e. all outcomes are equally likely. The probability of getting a one is
P(1) = 0.167, the probability of getting a two is P(2) = 0.167, etc.
But not all outcomes are equally likely to occur. In the example about the box containing 100
beads, of which 23 are black and 77 white, the probability of getting a black bead is P(B) =
0.23, whereas the probability of getting a white bead is P(R) = 0.77. Thus, the chances of
getting a red bead or a black bead are not the same, i.e. all outcomes for this particular event
are not equally likely.
Empirical Probabilities
Often it is not possible to give a theoretical probability of an event.
For example, what is the probability that an item on a production line will fail a quality control
test? This question can be answered either by measuring the probability in a test situation or
by relying on previous results (i.e. empirically). If 100 items are taken from the production line
and tested, then:
probability of failure, P(F) =
tested items of no. total
fail which items of no.
are complementary, i.e. they are mutually exclusive with a total
probability of 1. Thus:
P(A) + P( A ) = 1
This relationship between complementary events is useful, as it is often easier to find
the probability of an event not occurring than to find the probability that it does occur.
Using this formula, we can always find P(A) by subtracting P( A ) from 1.
C. THE TWO LAWS OF PROBABILITY
In this section we will define the two laws of probability, namely the addition law and the
multiplication law, and discuss the situations when each can be used.
Addition Law for Mutually Exclusive Events
Consider again the example of throwing a die. You will remember that:
P(6) =
= 0.5
This illustrates that
P(1 or 2 or 3) = P(1) + P(2) + P(3)
This result is a general one and it is called the addition law of probabilities for mutually
exclusive events. It is used to calculate the probability of one of any group of mutually
exclusive events.
It is stated more generally as:
P(A or B or ... or N) = P(A) + P(B) + ... + P(N)
where A, B ... N are mutually exclusive events.
If all possible mutually exclusive events are listed, then it is certain that one of these
outcomes will occur. For example, when the die is tossed there must be one number showing
afterwards.
P(1 or 2 or 3 or 4 or 5 or 6) = 1
Using the addition law for mutually exclusive events, this can also be stated as
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
Again this is a general rule. The sum of the probabilities of a complete list of mutually
exclusive events will always be 1.
Example:
An urn contains 100 coloured balls. Five of these are red, seven are blue and the rest are
white. One ball is to be drawn at random from the urn. What is the probability that it will be
red?
Probability that ball is red P(R) =
100
5
= 0.05
What is the probability that it will be blue?
Probability that ball is blue P(B) =
52
4
= = ÷ +
(b) Let event B be "the card is a spade" and event A be "the card is the ace of diamonds".
Here, the conditions are not met by both events occurring at the same time (i.e. they
are not non-mutually exclusive events), so the probability is given by:
P(spade or ace of diamonds) = P(A) + P(B)
P(A) =
of the customers buy at least one kind of bread.
Multiplication Law for Independent Events
Consider an item on a production line. This item could be defective or acceptable. These two
possibilities are mutually exclusive and represent a complete list of alternatives.
Assume that:
probability that it is defective, P(D) = 0.2
probability that it is acceptable, P(A) = 0.8
Now consider another facet of these items. There is a system for checking them, but only
every tenth item is checked. This is shown as:
probability that it is checked, P(C) = 0.1
probability that it is not checked, P(N) = 0.9
Again these two possibilities are mutually exclusive and they represent a complete list of
alternatives. An item is either checked or it is not.
Consider the possibility that an individual item is both defective and not checked. These two
events can obviously both occur together so they are not mutually exclusive. They are,
however, independent. That is to say, whether an item is defective or acceptable does not
affect the probability of it being tested.
There are also other kinds of independent events. If you toss a coin once and then again a
second time, the outcome of the second test is independent of the results of the first one.
The results of any third or subsequent test are also independent of any previous results. The
probability of heads on any test is 0.5 even if all the previous tests have resulted in heads.
To work out the probability of two independent events both happening, you use the
multiplication law for independent events. This can be stated as:
P(A and B) = P(A) × P(B) if A and B are independent events.
Again this result is true for any number of independent events. So:
P(A and B ... and N) = P(A) × P(B) ... × P(N)
02 . 0
1,000
20
=
For the item taken from the second batch:
probability that it is defective: P(D
2
) =
005 . 0
10,000
50
=
Since these two probabilities are independent:
P(D
1
and D
2
) = P(D
1
) × P(D
2
) = 0.02 × 0.005 = 0.0001
Example 2:
A card is drawn at random from a well shuffled pack of playing cards. What is the probability
that the card is a heart? What is the probability that the card is a three? What is the
probability that the card is the three of hearts?
Probability of a heart: P(H) =
52
13
=
4
1
Probability of a three: P(3) =
52
4
=
13
1
Probability of the three of hearts, since the suit and the number of a card are independent:
P(H and 3) = P(H) × P(3) =
The result of each throw is independent, so:
probability of no six in all three throws =
216
125
6
5
6
5
6
5
= × ×
Since no sixes and one or more sixes are mutually exclusive and cover all possibilities:
probability of one or more sixes =
216
91
216
125
1 = ÷
Multiplication Law for Dependent Events
To work out the probability of two dependent events both happening, you use the
multiplication law for dependent events. This can be stated as:
P(A and B) = P(A) × P(B|A)
where: A and B are dependent events and the vertical line is a shorthand way of writing
"given that".
Example 1:
For example, suppose there are 10 marbles in a bag, and 3 are defective. Two marbles are
to be selected one after the other without replacement. What is the probability of selecting a
defective marble followed by another defective marble?
Probability that the first marble selected is defective: P(A)
10
3
=
Probability that the second marble selected is defective: P(B)
9
2
=
Thus, P(A and B)
7%
9
2
10
3
= × =
This means that if this experiment were repeated 100 times, in the long run 7 experiments
would result in defective marbles on both the first and second selections.
Example 2:
Select one card at random from a deck of cards and find the probability that the card is an 8
and a diamond.
We know that there are four '8' cards (i.e. P(8)
) B ( n
B) and n(A
where n = the number of possible outcomes.
Similarly, the conditional probability of event B occurring, given that A also occurs, is
calculated by the following formula:
P(B|A) =
n(A)
B) and n(A
where n = the number of possible outcomes.
Consider the case of 100 candidates attempting their driving test for the first time. Suppose
80 of the candidates pass and 50 of them have had 10 or more lessons. Of those who have
had 10 or more lessons, 45 of them passed. What is the probability of passing at the first
attempt if you have had only 6 lessons?
We can construct a table which identifies the possible outcomes as follows in Table 10.1.
Table 10.1: A partially completed contingency table
Passed Failed Total
10 plus lessons 45 50
Less than 10 lessons
Total 80 100
We will call event A "passed" and event B "having had less than 10 lessons".
Make sure you understand the construction of this table:
We have listed the possible outcomes of event A along the top – these outcomes are
either to have passed or failed the test. Thus, the columns will show the number of
outcomes for either possibility.
We have listed the possible outcomes of event B along the side – these are either to
have had 10 or more lessons, or less than 10 lessons. Thus, the rows will show the
number of outcomes for either possibility.
We have entered in the information that we know: that there are a total of 100 possible
outcomes; that the total number of outcomes in which event A was "passed" is 80; that
the total number of outcomes in which event B is "had 10 or more lessons" is 50; and
that the total number of outcomes in which event A was "passed" and event B was "had
10 or more lessons" is 45.
We can complete the rest of the table by subtraction as in Table 10.2.
Table 10.2: Contingency table
Passed Failed Total
10 plus lessons 45 5 50
Less than 10 lessons 35 15 50
Total 80 20 100
= 0.7
We could consider the question: "what is the probability of failing at the first attempt if you
have had more than 10 lessons?". Here event A is "failed" and event B is "having had more
than 10 lessons". The calculation is:
P(A|B) =
n(B)
B) and n(A
=
50
5
= 0.1
Questions for Practice 1
See if you can work out the answer to the following questions using the preceding table:
"What is the probability of failing at the first attempt if you have only had 8 lessons?"
Remember to be clear about which is event A and event B, then read off the total number of
outcomes for event B and for event A and B together, and then apply the formula.
Now check your answers with those given at the end of the chapter.
When the second ball is selected, there are only seven balls left and the mix of red and white
balls will be determined by the selection of the first ball – i.e. the probability of a red or a
white ball being selected is conditional upon the previous event. Thus:
If the first ball selected was red, there will be four red and three white balls left. The
probabilities for the selection of the second ball are then:
P(R) =
7
4
, and P(W) =
7
3
If the first ball selected was white, there will be five red and two white balls left. The
probabilities for the selection of the second ball are then:
P(R) =
7
5
, and P(W) =
7
2
This process of working out the different conditional probabilities can be very complicated
and it is easier to represent the situation by means of a tree diagram as in Figure 10.1. The
probabilities at each stage are shown alongside the branches of the tree.
Figure 10.1: Example 1 – tree diagram
The diagram is constructed as follows, working from left to right.
At the start we take a ball from the bag. This ball is either red or white so we draw two
branches labelled R and W, corresponding to the two possibilities. We then also write
on the branch the probability of the outcome of this simple experiment being along that
branch.
We then consider drawing a second ball from the bag. Whether we draw a red or a
white ball the first time, we can still draw a red or a white ball the second time. So we
mark in the two possibilities at the end of each of the two branches of our existing tree
2nd Ball 1st Ball
Start
R
R
W
W
W
R
8
5
We have given the first ball drawn a subscript of 1 – for example, red first = R
1
. We shall give
the second ball drawn a subscript of 2.
The probabilities in the first stage will be:
P(R
1
) =
6
3
balls of no. total
balls red of no.
=
P(W
1
) =
6
2
(P(B
1
) =
6
1
The possibilities in stage 2 may now be mapped on the tree diagram. When the second ball
is drawn, the probability of it being red, white or blue is conditional on the outcome of the first
event. Thus, note there was only one blue ball in the bag, so if we picked a blue ball first then
we can have only a red or a white second ball. Also, whatever colour is chosen first, there
are only five balls left as there is no replacement.
Start
R1
W1
B1
6
3
2
1
6
3
=
(c) both a score greater than 3 and an odd number – one possibility: P =
6
1
G. EXPECTED VALUE
Probabilities may also be used to assist the decision-making process in business situations.
Before a new undertaking is started, for example, it may be possible to list or categorise all
the different possible outcomes. On the basis of past experience it may also be possible to
attribute probabilities to each outcome. These figures can then be used to calculate an
expected value for the income from the new undertaking.
Example:
Suppose a company is considering opening a new retail outlet in a town where it currently
has no branches. On the basis of information from its other branches in similar towns, it can
categorise the possible revenue from the new outlet into categories of high, medium or low.
Table 10.6 shows the probabilities that have been estimated for each category. You will note
that the probabilities sum to 1.
Table 10.6: Probabilities of likely revenue returns
Likely Revenue Probability
High 0.4
Medium 0.5
Low 0.1
Consider now a continuous variable in which you have been able to make a very large
number of observations. You could compile a frequency distribution and then draw a
frequency bar chart with a very large number of bars, or a histogram with a very large
number of narrow groups. Your diagrams might look something like those in Figure 10.6.
Figure 10.6: Frequency bar chart and histogram
The "normal" or "Gaussian" distribution was discovered in the early eighteenth century and is
probably the most important distribution in the whole of statistical theory. This distribution
seems to represent accurately the random variation shown by natural phenomena – for
example:
heights of adult men from one race
weights of a species of animal
distribution of IQ levels in children of a certain age
weights of items packaged by a particular packing machine
life expectancy of light bulbs.
The typical shape of the normal distribution is shown in Figure 10.8. You will see that it has a
central peak (i.e. it is unimodal) and that it is symmetrical about this centre. The mean of this
distribution is shown as "m" on the diagram, and is located at the centre. The standard
deviation, which is usually denoted by "σ", is also shown.
Figure 10.8: The normal distribution
Further Probability Calculations
It is possible to calculate the probability of an observation being in the shaded area shown in
Figure 10.12, using values from the tables. This represents the probability that x is between
m ÷ 0.7σ and m + 1.5σ: i.e.
P(m ÷ 0.7σ < x < m + 1.5σ)
Figure 10.12: Probability of an observation being in a particular area
(ii) One die must show a six and the others must not show a six. Any one of
the three dice could be the one that shows a six.
Number of ways the die showing six can be chosen = 3
probability of six on one die =
|
.
|
\
|
6
1
probability of not obtaining six on the other two dice =
2
6
5
|
.
|
\
|
Therefore:
P(one six) = 3 ×
|
.
|
\
|
6
1
×
2
6
5
|
.
|
\
|
=
72
25
(iii) Probability of obtaining at least one six
= 1 ÷ probability of not obtaining any sixes
= 1 ÷ probability that all three dice show 1, 2, 3, 4 or 5
= 1 ÷
0.009881 to 4 sig. figs.
(ii) Probability that the three will be all of any one of the four colours
= probability that all three are red
+ probability that all three are blue
+ probability that all three are green
+ probability that all three are white
=
ABE - Introduction to Quantitative Methods
Description
Following the introduction of the new Qualifications and Credit Framework (QCF) in the UK, ABE has taken the opportunity to thoroughly revise its suite of qualifications in order to provide the mos...
Following the introduction of the new Qualifications and Credit Framework (QCF) in the UK, ABE has taken the opportunity to thoroughly revise its suite of qualifications in order to provide the most relevant and focused learning experience for students.
All accredited colleges can, for a limited time, download ABE's new and revised QCF study manuals below. These study manuals apply to exams beginning in December 2011. The study manuals will be posted below as and when they become available so we thank you for your patience whilst we finalise the outstanding texts. | 677.169 | 1 |
Careers in Mathematics
Mathematics Majors have a great variety of career options in business, education, academia, and government. The professional societies offer a wealth of information on their websites.
Here are some of the many disciplines that Mathematics graduates find challenging and rewarding employment opportunities in:
Operations Researchers use statistics, linear algebra, stochastic modelling and analytical skills to find optimal solutions of complex decision-making problems. Financial modeling, marketing, manufacturing, simulation, and public policy are often involved. Start with INFORMS (Institute for Operations Research and Management Societies).
Actuary - use statistics to predict changes in the marketplace; most commonly applied to finance, investment, insurance, and pension funds. See, for example, Be An Actuary from the Society of Actuaries. Related fields include Economics andFinancial Mathematics.
Math Teacher - there are lots of options here, including primary, secondary and college. A teaching credential is usually required to teach in a public school, and a Masters degree to teach in community college. See, for example, Teach For America.
Machine Learning deals with learning from data, with the intent of being able to make intelligent predictions based on new data after having received some set of training data. It relies heavily on statistical data analysis, classification, probability theory, graph theory, algorithms, and analysis. Start with AAAI, IEEE Careers, and the ACM Job Center
Mathematical Biologists use a combination of deterministic (e.g., differential equations) and stochastic theories to develop mathematical descriptions of processes in living organisms. Mathematical biologists work at the interface of multiple disciplines, notably mathematics, biology, computation, physics, and bio-engineering and often work in teams with specialists in other fields. Mathematical Biologists are often Computational Biologists and vice-versa. For more information see the Society for Mathematical Biology.
Computational Biologists use mathematical models to perform biological experiments in-silico. Bioinformaticians develop and utilize tools for the storage and retrieval of biological data, often using advanced statistical and computational techniques. For more information see the International Society for Computational Biology, and bioinformatics.org
Astrodynamicists perform satellite orbit mission design for NASA and various commercial organizations. Tasks include design and maintenance of satellite orbits, maneuver design, and the specification of satellite orbits to meet specific scientific observational requirements. They may work with Space Systems Analysts who evaluate telemetry data in real time from satellites to determine satellite health, develop satellite command loads, and develop and implement fault recovery procedures in the event of operational failures. Both of these jobs require the types of strong analytical skills that Mathematicians strive to develop. On-the-job-training is usually provided since these skills are not taught in universities. See JPL Career Launch or American Institute of Aeronautics and Astronautics (AIAA) for more information.
Many mathematicians go on to become successful Business consultants,Attorneys, or Physician. It is the analytical skills that you take with you to these new disciplines, though it is often possible to leverage mathematical knowledge as well, particularly operations research, financial mathematics, mathematical modeling, statistics, in these areas.
Math Professors are probably most familiar to you as teachers but that is just part of their job. Other responsibilities include doing research and service to the community. The amount of time available to do research will depend on the type of college you teach at. In community colleges there is little to no time for research. In a comprehensive university (like CSUN) faculty spend the bulk of their time devoted to instruction, typically with only one or two days per week available for research. In a research oriented university (like UCLA) the roles are reversed, with faculty typically teaching one to three classes per year. Thus depending on your particular interests you should be able to find a school that allows the balance that fits your needs. Math Professors may be pure or applied mathematicians, applied mathematical scientists, specialists in mathematics educationor work in one of the disciplines. They often do interdisciplinary research because other researchers value their ability to clearly analyze and explain problems. See mathjobs.org
What Employers Really Want
In a recent survey by the AACU (Assoc. of Amer. Coll. & Univ.) major employers indicated that your major is only secondary. What they really want includes: | 677.169 | 1 |
Mathematicians Delight
An introduction to mathematics which starts with simple arithmetic and algebra and proceeds through to graphs, logarithms, trigonometry to calculus and imaginary numbers. The author, who is internationally renowned for his innovative teaching methods, offers insights into the pleasures of mathematics that will appeal to readers of all backgrounds | 677.169 | 1 |
The purpose of the Entry Ticket: Intro. to Quadratic Functions is to activate students' prior knowledge about working with functions. I start by having students work on the Entry Ticket as soon as they enter the class – as the year has progressed it has become more and more automatic that students take out their binders and get to work on the Entry Ticket rather than milling around or socializing. This also frees up a couple of quick minutes for me to take care of housekeeping (attendance, etc.) and not waste valuable instructional time.
This entry ticket reviews work students completed earlier in the year with linear functions. The entry ticket accomplishes two goals:
Providing a real-world context helps students understand and apply concepts (Bagel World is a local hot spot for high school students and pretty much every one else that lives in town!)
Providing students with opportunities to practice skills that I am going to ask them to apply to a new situation.
I tell students that the questions on the entry ticket review ideas we are going to build on in class today. The entry ticket focuses on a linear relationship, while we will focus on quadratic relationships in class today.
About 5 minutes into class, I ask students to turn-and-talk with a partner about the Entry Ticket. I prompt them to share how they solved the problem. I encourage students to identify the algebraic rules used to solve each problem. After a few minutes of partner work, we will review the Entry Ticket as a class. I plan to start by asking groups to share any discrepancies/errors and how to correct them.
After discussing the Entry Ticket, I turn my attention to the agenda for the day. We review the objectives as a class. I explain how this lesson's objective fits into the bigger objectives of the unit: to support students who have difficulty seeing the big picture and/or shifting back and forth between the gestalt and the details of lessons and units. I typically have students write down the homework assignment during this time and I hand out copies of the homework, asking students to file the homework in their binders.
After the day's agenda has been reviewed, the class shifts to the middle of the lesson.
Resources
After reviewing the objectives for the day's lesson, I cue students to open to a new piece of paper in their notebook. Many of my students continue to benefit from an explicit directive to take notes. Then, I let students know that I want them to take notes as they watch a 2-minute video. I write the following focus prompt on a whiteboard: Based on the video, describe different characteristics of contexts that involve quadratic functions.
I then show this as an introduction to contexts involving quadratic functions:
After showing the video I ask students to take an additional two minutes to respond to the prompt. We then have a 3-5 minute class discussion where I give students a chance to share their ideas. In order to differentiate instruction, I lead the conversation patiently. It is important to allow students sufficient time to integrate their ideas. It is beneficial to provide multiple opportunities for students to express themselves (written and verbal).
After the second video is complete, I ask students to complete the following Think-Pair-Share protocol:
Since I believe students need to be taught skills of conversation and language use in school, I next have students work independently on the vocabulary for this unit. I explicitly teach students the difference between brick words and mortar words and how to use them in an academic conversation.
To accomplish these goals efficiently, I have students work on iPad Minis using a math dictionary to look up the definitions for terms. As an alternative, students could use an online math dictionary. I recommend Wolfram Alpha as it provides excellent knowledge, is accurate and also provides good visual examples for many terms.
Instructional Note: With some classes, I might choose to provide students with the definitions of the words (see: ______ to allow students to focus more on generating multiple representations of meaning for each word to develop a deeper understanding of the vocabulary terms. I have included two versions of the vocabulary for teachers - one with definitions and one without.
Today, as the students wrap up, I have students file the work in a vocabulary section of their notebook and remind them vocabulary is part of the Notebook Check. We will be using the vocabulary later in the unit for a writing exercise on creating their own functions.
Technology Note: The use of iPad Miinis is supported by a technology grant for Modeling with Mathematics and Universal Design for Learning in the Math Classroom by the Hardscrabble Education Fund.
I review the PowerPoint Slides: Intro. to Quadratic Functions to provide students with direct instruction and examples to practice. I start at a basic level, but I expect to move quickly. My goal is to deepen student understanding of the features of quadratic functions.
I begin an introduction to the basic form of a quadratic function. I show students a number of graphs and equations as examples. The next slide focuses on the value of the coefficient of the leading term, a, in a quadratic equation. I include the situation when a=0 because I want students to make the connection between quadratic functions and linear functions. A linear function is related to a quadratic in so far as it can be considered a quadratic with a value of a = 0.
Then, I discuss two examples of graphing quadratic functions with students. I chose two examples that can factor without having to complete the square. I want to focus on the basic ideas necessary to graph a quadratic function.
After graphing the two functions, the class then shifts to determining the domain and range of quadratic functions.
Resources (1)
Resources
As a means of introducing students to identifying the domain of a quadratic function, I ask students to watch a LearnZillion video that demonstrates how to determine the domain and range of a quadratic function from its graph. Before starting the video, I put the words domain and range on the whiteboard and ask students to brainstorm things they remember about the two terms.
After the video, I give students the opportunity to ask questions they have about the domain of a quadratic function. Then, I show students an example of a quadratic function on the board. As a class, we determine the domain and range of the function from its graph.
To give students the opportunity to practice on their own, I display graphs of quadratic functions (and their equations) on the SmartBoard. I may well use the examples from the PowerPoint Slides. As we look at the functrions, I plan to complete the following discussion protocol with students:
Think: 3-5 minutes working in pairs (writing down their own responses)
Pair: 2-3 minutes talking with another pair (this makes students work in groups of 3-5) comparing their answers and asking clarifying questions of each other).
Resources (1)
Resources
To conclude today's lesson, I will ask students to complete an Exit Ticket: Introduction to Quadratic Functions. The Exit Ticket uses a similar context as the entry ticket. I designed the task to help students reflect on the characteristics of linear and quadratic functions. I will ask my students to work on their own, so that I can use the Exit Ticket as an informal assessment. On the Exit Ticket I ask students to explain their reasoning. I want them to take time at the end of class to integrate their thinking.
Tonight's Homework: Introduction to Quadratic Functions assignment asks students to find 3 examples of quadratic functions in real life. I ask students to identify examples that were not included in the class videos. I provide them with an idea organizer to complete. I provide this resource to help the students focus their ideas and choose supporting examples. This homework supports a focus on using academic language.
Thanks so much for the positive feedback Nicole. I could not find the original video, but I think the second video will meet the intent of the introduction. I deleted the old link and added a second example video that connects parabolas to Super Mario Brothers.
Thank you so much for sharing these awesome lesson ideas and explicitly explaining your teaching strategies. I really appreciate it and love your real world applications. This will be so helpful when working with my students, especially my remedial students.
Do you have another link to the video you used referencing contexts of quadratic functions. That link no longer works and would love to see what you had! | 677.169 | 1 |
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POLYMATH 5.
0 - OVERVIEW
POLYMATH is a proven computational system, which has been specifically created for educational or professional use. The various POLYMATH programs allow the user to apply effective numerical analysis techniques during interactive problem solving on personal computers. Results are presented graphically for easy understanding and for incorporation into papers and reports. Students, engineers, mathematicians, scientists, or anyone with a need to solve problems will appreciate the efficiency and speed of problem solution. The POLYMATH main, task selection window is shown below.
The main options available in this window are the following: LIN: Linear Equations Solver. Enter (in matrix form) and solve a new system of up to 64 simultaneous linear equations. NLE: Nonlinear Equations Solver. Enter and solve a new system of up to 200 nonlinear algebraic equations. DEQ: Differential Equations Solver. Enter and solve a new system of up to 200 ordinary differential equations. REG: Data analysis and Regression. Enter, analyze, regress, and plot set of up to 350 data points. calc: Calculator. Enter and evaluate explicit expressions with a variety of intrinsic functions. units: Unit Conversion. Convert selected units into desired units. const: Useful Constants. Find selected scientific and engineering constants
setup: Parameter Settings. problem title. and system time is shown in the status bar... Each equation (row) is indicated by a number. Modify parameters of numerical solution algorithms Additional information such as the current filename. and each column by a variable name (x1. Systems containing up to 64 linear equations can be entered. x2.. You can use the cursor to highlight one of the cells. All equations are checked for correct syntax and other errors upon entry.
Nonlinear Equations Solver
The purpose of this program is to allow you to solve systems of nonlinear algebraic equations. b). move the highlight to the appropriate cell and either start typing in a number or erase and retype the value already in the cell. The data entry screen is shown below. Multiple roots are given for a single equation. xn.
. After you finished entering the data the system of equations can be solved by selecting "Solve" from the drop-down "Program" menu or by bringing the cursor to the blue arrow on the screen and pressing the left mouse button. To enter or modify a value in a data cell. The system may contain up to 200 nonlinear simultaneous (implicit) and supporting auxiliary (explicit) equations. This is the current cell. You can move the highlight by using the arrow keys or the left mouse click. Only real roots (non-complex) can be found. Column (variable) names can be changed by selecting a column or a cell and then choosing the "Variable name.. The elements of the matrix of coefficients and the vector of constants must be entered. The matrix of coefficients and the vector of coefficients are displayed..." option from the "Edit" drop-down menu or from the menu activated with the right mouse click.
Linear Equations Solver
The purpose of this program is to allow you to solve systems of linear algebraic equations where the set of linear equations is inputted in a matrix-vector form.
have been entered.incomplete (missing definition. ready to be solved. initial state etc.
The options and information available on this screen are the following: Add NLE: Input a new nonlinear (implicit) algebraic equation Add EE: Input a new explicit algebraic equation Remove: Delete the highlighted equation Edit: Edit/Alter the highlighted equation Comments: This check-box hides/unhides the comments column in the equations entry table.). Green light . : Solve the system of equations.
.The equation input/modification window is shown below. : Selecting this option opens a window showing a list of defined and undefined variables : Move the highlighted equation downward : Move the highlighted equation upward : Problem setup state. optionally users can benefit from it by adding comments and remarks. including initial guesses for the variables.complete. Solve with: Select the preferred solution algorithm from 4 choices. Red light . Note that this option is displayed only when a valid set of equations. The comments column is not used by POLYMATH for the numerical calculations.
The options and information available on this screen are the following: Add DE: Input a new differential equation Add EE: Input a new explicit algebraic equation Remove: Delete the highlighted equation Edit: Edit/Alter the highlighted equation
. The equation input/modification window is shown below. The system may contain up to 200 differential and supporting explicit algebraic equations.) are given. Detailed additional information is provided with regard to the following subjects: • Entering the equations and initial estimates into the program • Variables and expressions • Solution algorithms • Solution report • Troubleshooting
Differential Equations Solver
The purpose of this program is to help you solve systems of first-order ordinary differential equations. missing initial values etc.At the bottom of the window the number of nonlinear and auxiliary equations is listed and problem setup related warning messages (undefined variables.
polynomial and nonlinear regression techniques). Table: When this option is marked a table containing the values of all the variables as function of independent variable will be created during integration. various curves and equations to the data using multiple linear. Red light . At the bottom of the window the number of nonlinear and auxiliary equations is listed and problem setup related warning messages (undefined variables. This optional feature allows users to add comments. initial state etc. missing initial values etc. Initial Value: Initial value of the independent variable.complete. The data is stored in a column-wise fashion where every column is associated with a name (variable) and can be addressed separately. The comments column is not used by POLYMATH for the numerical calculations. Tabular results of solution of a set of ordinary differential equations will be stored also in the data table. differentiated. has been entered. The stored data can be regressed (link to Regression) (meaning fitting a straight line. Solve with: Select the preferred solution algorithm identified in the pull down menu from among 6 choices including two for stiff problems. Indep Var: Name of the independent variable. Green light . Note that this option is displayed only when a valid set of equations. The data table can be accessed directly by selecting the "REG" option when creating a new file. analyzed (link to data analysis) (meaning interpolated. Final Value: Final value of the independent variable. Report: When this option is marked a report containing initial. including initial and final values of the variables. list of the equations in the system and some additional information will be shown after the integration is finished.) are given. a plot of the integrated values of the differential variables versus the independent variable will be created during integration.
.
Data Table for Regressions.). final. manipulation and storage of numerical data. maximal and minimal values of all the variables. : Integrate the system of equations. ready to be solved. integrated and various statistics are calculated) and plotted. : Selecting this option opens a window showing a list of defined and undefined variables : Move the highlighted equation downward : Move the highlighted equation upward : Problem setup state. Graph: When this option is marked.Comments: This check-box can hide/unhide the comments column in the equations entry table.incomplete (missing definition. and General Plotting
The data table is used for input. Curve Fitting.
. transformation functions are used to linearize a nonlinear regression model... for example.
Linear & Polynomial Regression
This part of the program will fit a polynomial of the form: P(x) = a0 + a1*x + a2*x^2 + .ENTERING AND MANIPULATING DATA
The data entry and manipulation window (Data Table) is shown above.1). an by minimizing the sum of squares of the deviations between the calculated y or P(x) and the data for y. select the appropriate option from the bottom of the data table. an are regression parameters to a set of N tabulated values of x (independent variable) versus y (dependent variable). The window for selection of the type of the curve to be fitted is shown below.. . The highest degree allowed for a polynomial is N .1 (thus n >= N . It is much like a spreadsheet. The rest are data cells. which may be up to 300 rows by 30 columns (default values.. . a 1st order polynomial must be selected. . a1.
. Column names can be changed. The program calculates the coefficients a0. The top row of cells contains the column names (the default names being C01 to C30).. + an*x^n where a0. a1. Column definitions can be very useful when. To carry out regression with analysis or plotting of the data. Contents of a column can be defined as function of previously defined columns. To carry out linear regression.. the total number of rows and columns can be changed in the setup dialog box).
the free parameter is set to zero in the regression model (a0=0).The options available in this window are the following: Dependent Variable: Select the dependent variable for regression. Polynomial Differentiation: If this parameter is marked. a graph showing the calculated curve (or points) and the data points is prepared and displayed. Through origin: If this option is marked. Polynomial Degree: Select the degree of the polynomial (indicated by 'n' in equation above). Independent Variable: Select the independent variable for regression (indicated by 'x' in the regression equation above). Residuals:
. Graph: If this option is marked. select the '1/Linear' polynomial for linear regression. Polynomial Integration: If this parameter is marked. the derivative of the polynomial at the specified point is calculated. the integral of the polynomial between the specified boundaries is calculated.
... Store Model in Column: Store the regression model and the calculated parameters in the next available empty column. To carry out a multiple linear regression.If this option is marked.. Note that the number of data points must be greater than n+1 (thus N >= n+1). xn (independent variables) versus y (dependent variable). Report: If this option is marked... an by minimizing the sum of squares of the deviations between the calculated and the data for y..
. . . x2. . select the "Multiple linear" option in the regression solver window. + an*xn where a0.. a1.. a graph showing the deviation between the data and the calculated values of the dependent variable (error. an are regression parameters. x2. to a set of N tabulated values of x1.. residuals) points is prepared and displayed. a report showing the regression model the numerical values and confidence intervals of the parameters and some additional statistical and other information are presented and displayed.... xn) = a0 + a1*x1 + a2*x2 + . The window of options selection for multiple linear regression is shown below..
Multiple Linear Regression
This part of the program will fit a linear function of the form: y(x1. a1. . Solve (blue arrow): Carry out the regression and the additional calculations. The program calculates the coefficients a0.
The options available in this window are the following: Dependent Variable: Select the one of the column names as the dependent variable for regression. Store Model in column: Store the regression model and the calculated parameters in the next available empty column. a report showing the regression model the numerical values and confidence intervals of the parameters and some additional statistical and other information are
. the free parameter is set to zero in the regression model (a0 = 0). . Through origin: If this option is marked..xn' in the regression equation above).. residuals) points is prepared and displayed. Note that holding down the Cntr key while then pressing left button of the mouse will select more than one variable. Residuals: If this option is marked. Report: If this option is marked. a graph showing the calculated points and the data points is prepared and displayed. x2. a graph showing the deviation between the data and the calculated values of the dependent variable (error. . Independent Variables: Select the independent variables for regression (indicated by other column names 'x1. Graph: If this option is marked.
…. …. a graph showing the calculated points and the data points is prepared and displayed. an initial guess for all the parameters should be provided in this area. am) where a0. ….
Enter Model: Type in a new regression model equation or edit an existing equation. …. a0. an are regression parameters to a set of N tabulated values of x1. The model entry communication box for entering the nonlinear model is shown below.presented and displayed Solve (blue arrow): Carry out the regression and the additional calculations. xn. a1. Residuals:
. a1. Enter Initial Guess for Model Parameters: After entering the model. x2.
Multiple Nonlinear Regression
This part of the program will fit a nonlinear function of the form: y = f (x1. a2. xn (independent variables) versus y (dependent variable). Graph: If this option is marked. x2. Note that the number of data points must be greater than m + 1 (thus N >= m + 1).
differentiate. integrate and plot data for which no regression model is available. type in an equation containing numerical constants.
Data Analysis
Tabulated data often cannot be represented satisfactorily by a regression model or it can be difficult to find such a model. The options available in this window are the following. The purpose of this part of the program to help you interpolate. The program will identify the valid column names and will assign all the other names appearing in the equation as parameter names. press the "Enter Model" button. On the right hand side. and parameter names. a report showing the regression model the numerical values and confidence intervals of the parameters and some additional statistical and other information is presented and displayed. The window for selection of the type of the data analysis to be carried out is shown below. Store Model in column: Store the regression model and the calculated parameters in the next available empty column. a graph showing the deviation between the data and the calculated values of the dependent variable (error. You are requested to provide initial estimates for all the parameters. Solve (blue arrow): Carry out the regression with either the Levenberg-Marquardt or other algorithm and perform the additional calculations.
. column names. Report: If this option is marked. On the left-hand side. select the name of the dependent variable (it should be the name of an existing data column).If this option is marked. To enter a model equation. residuals) points is prepared and displayed.
Solve: Carry out the analysis
Graph Preparation and Editing
.Interpolation: Calculation of the value of the dependent variable for a specified value of the independent variable Differentiation: Calculation of the derivative of the dependent variable for a specified value of the independent variable Integration: Calculation of the integral of the dependent variable for a specified region of the independent variable Independent variable: Select the independent variable/s for the analysis Dependent variable: Select the dependent variable for the analysis Solve with: Select the solution method.
The resulting graph shown below has many options. Once the graph has been edited. The window for creation of a graph in the Data Table give a selection of the column variable for plotting as shown in the following window. spreadsheets.
. or desktop publishing packages.Graphical presentations can be created from the Data Table or are given as an option from several of the programs. it can be copied to other applications such as word processors.
Legend box: The legend box can be "dragged" to a different location. The following options are available: Max Y-axis: Change the upper bound on the Y (vertical) axis. Scatter connected: Show the curve connecting the calculated and/or stored data points.
. if necessary. Auto scale: Determines whether the program automatically changes the scale as users remove/add function curves. Title: Add or change graph title. Curves and Functions: Option to remove variables and to add function plots to the graph. Subtitle: Add or change graph subtitle. Min Y-axis: Change the lower bound on the Y (vertical) axis. "X-axis" and "Y-axis" dropdown menus or by pressing the right button on the mouse. Min X-axis: Change the lower bound on the X (horizontal) axis. using the mouse.The graph editing options can be reached using either the "Curves". remove them if they are shown. Inverse grid lines: Show grid lines if they are not shown. Using the right button the menu options are displayed as shown below. Max X-axis: Change the upper bound on the X (horizontal) axis. Draw points: Show the calculated and/or stored data points.
change the width of the various lines and curves and change the size of the points drown.In addition to the options presented here there are options available to change the format of the labels in the X and Y axes. These options can be reached using the drop-down menus.
. | 677.169 | 1 |
This is a vital document, a contract between student and
instructor. Read it thoroughly! This syllabus contains class-specific
information. It is supplemented by the MATH
ONLINE SYLLABUS
which contains information relevant to ALL of Prof. Keely's math online courses.
You must read carefully and abide by both documents.
Course Description:
The math 090-095 sequence is designed for the student who is
prepared to take algebra at an accelerated pace. This sequence can instead be
taken in a normal-paced three-term sequence by taking math 089-091-093 (Algebra
I-II-III). The 089-091-093 sequence covers the same material overall, but fewer
topics/chapters per term than the accelerated 090-095 sequence. By opting to
take the accelerated sequence you should be prepared for the added workload and
fast-paced delivery. See Clark Mathematics Division's
Should I take
the math 089-091-093 or the math 090-095 sequence? flyer for more information.
This is the first course in a two-term sequence of
developmental algebra intended to improve your critical thinking abilities and
prepare you for college level mathematics courses. This course focuses on linear
functions, graphs, and equations. We will study algebraic expressions, linear equations,
linear inequalities, an introduction to graphs and functions, lines, systems of linear
equations, integer exponents, and polynomials.
This is the second course in a two-term sequence of
developmental algebra intended to improve your critical thinking abilities and
prepare you for college level mathematics courses. This course focuses on
algebraic processes related to polynomial, rational, radical, quadratic,
exponential, and logarithmic functions and their graphs.
Prerequisites:
Both math 090/095 online courses require a basic knowledge of
computers and the internet.
Prerequisite to Math 090: Pre-Algebra (Clark's math 030) passed with a grade of at
least "C" or qualifying score on the placement test. Prerequisite to Math 095: Elementary Algebra or Algebra II (Clark's math
090 or 091) passed
with a grade of at least "C" or qualifying score on the placement test.
Course Materials:
MyMathLab access to the online text and testing center is REQUIRED. MyMathLab is an
online course delivery system provided by the textbook publisher. In MyMathLab you can read the
complete textbook online,
watch video lectures, work interactive tutorial
exercises, and take tests.
See box below for purchasing options, registration directions, and MyMathLab course
ID code.
ATTENTION NEW
MYMATHLAB USERS: Once you open the package you may not be able to get a
refund. Don't open the access kit until you are sure that you want to take this
class.
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unless you really want one. If you do opt to buy the hard copy text new, it comes bundled with MyMathLab
access. If you buy the hard copy text used you will have to purchase the
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tests are conducted in MyMathLab. See
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College Algebra
College Algebra is accessible to both students who will take higher-level math courses after algebra and students for whom algebra will be their last college math course. The material is both comprehensive and rigorous and provides examples that apply to everyday life. Each chapter provides historical contexts on algebraic concepts, technology notes, and application examples. | 677.169 | 1 |
Algebraic geometry
It can be seen as the study of solution sets of systems of polynomials.
When there is more than one variable, geometric considerations enter and are important to understand the phenomenon.
One can say that the subject starts where equation solving leaves off, and it becomes at least as important to understand the totality of solutions of a system of equations as to find some solution; this does lead into some of the deepest waters in the whole of mathematics, both conceptually and in terms of technique.
June 11, 2009 — Hirzebruch's problem at the interface of topology and algebraic geometry has occupied mathematicians for more than 50 years. A professor of mathematics at the Ludwig-Maximilians-Universitaet in ... read more
Sep. 16, 2013 — Researchers have found high school students in the United States achieve higher scores on a standardized mathematics test if they study from a curriculum known as integrated ... read more
Aug. 3, 2009 — As mathematics continues to become an increasingly important component in undergraduate biology programs, a more comprehensive understanding of the use of algebraic models is needed by the next ... read more
Jan. 18, 2016 — Over centuries, humans have tried to discover a Theory of Everything. Possible candidates for this cachet, such as String Theory and Grand Unified Theory, require higher dimensions or ... read more
Aug. 16, 2006 — Studying complex systems, such as the movement of robots on a factory floor, the motion of air over a wing, or the effectiveness of a security network, can present huge challenges. Mathematician ... read more
June 20, 2011 — A new study finds important differences in math curricula across US states and school districts. The findings suggest that many students across the country are placed at a disadvantage by less ... read more
Mar. 4, 2009 — Greek mathematics is considered one of the great intellectual achievements of antiquity. It has been decisive to the academic and cultural development of Western civilization. The three Roman authors | 677.169 | 1 |
Calc III is generally pretty easy. LA tends to be more proof based but I don't know what the course would look like with differential equations included. I took them at the same time and found them both pretty easy, but I prefer proof based math over calculation.
"Multivariate Calculus" is most calc IIIs, but most defined "Multivariate Calculus" courses are the equivalent of calculus 4 through X depending on how deep they take the discipline.
My school's multivariate calculus course (requires Linear, Calc 3, and Differential Eq) is effectively conceptual math proofing of the general forms of calculus 1/2/3 theorums and is of a Junior/Senior level comprehension of the program.
At most US high school's, precalculus does not include linear algebra. It is usually trig with some study of different types of functions. Linear Algebra is helpful but not needed for Multivariate Calculus.
It depends... Multi-variable is a definite next step once you've done derivative and integral single-variable calculus, but you only really need indefinite integrals to take differential equations. As for linear algebra, the math is really easy. It's literally algebra, as in adding and multiplying numbers. However, it is conceptually a leap. It is more theoretical and not as directly related to your previous knowledge. I think it's only slightly harder than multivariable calc, but that's just me. I find diff eq just as much advanced calculus as I do multivariable calc (or close), so you can't go wrong with either.
I'm of the opposite camp. I took multi first, and some of the vector calculus seemed really unmotivated to me. I have a lot of trouble learning things when I can't pick up on why anyone would be doing it. That was 3 years ago and I had literally no mathematical maturity, but you get my drift.
If the linear algebra is treated relatively rigorously, as I suspect it will be (the only LA + DE composite class I've ever heard of is at UC Berkeley), it may be easier to you. I haven't taken DE (not required for the major at my Uni and I've never had the right schedule), but I've been told most entry-level courses are fairly easy (and don't really build on things you learn in multi). You should ask someone who's taken both or either, that'll be your best bet.
Depends on your department for the specifics, but at least where I go, calc 3 and linear algebra complement each other pretty well. The ability to work with vectors is important for multivariate calculus.
Solving certain kinds of differential equations can depend on the concept of a partial derivative, which lives in the domain of multivariable calculus. I've never heard of LA and DE being combined in a single course before, though, so I'd go with existentialhero's answer.
techniques from linear algebra are often necessary for solving vector ODE's. i've seen some treatments of PDE's also require heavy linear algebra. the algebra of linear transformations also helps expedite proofs in DE's in general. it's becoming norm here in california. | 677.169 | 1 |
Find a Woodinville review basic organizational knowledge of expressions and equations that include: order of operations, odd and even numbers, and use of signed numbers, and real numbers. The following topics in each category are covered:
Pre Algebra: number problems, multiples, factors, primes, divisibility an... | 677.169 | 1 |
Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied.
Here is a detailed and comprehensive presentation of linear algebra based on axiomatic treatment of linear spaces. The text maintains a balance between modern algebraic interests and traditional linear algebra. Revised, with new material and exercises | 677.169 | 1 |
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This is a Stand Alone Instructinal Resource (StAIR) designed to teach and test students on their understanding of different triangle classifications. It could be used as initial instruction, or as an effective review tool at the end of a unit/lesson on triangle classification Triangles StAIR to your Bookmark Collection or Course ePortfolio
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"This is a thorough, self-contained introductory textbook for training undergraduate students in basic mathematical and statistical methods that are important in biological sciences. Students are introduced to topics ranging from probability and statistics to matrix theory and calculus, with a brief introduction to modeling using difference and differential equations. Two unique features of this textbook are the inclusion of real-world biological data to motivate particular methods and the use of MATLAB for computational purposes."--Linda J. S. Allen, Texas Tech University
"This spectacular book develops the reader's ability to quantitatively analyze problems arising in biology, and illustrates the great utility of mathematical models and computing to provide answers to key questions. The biological examples and student projects are excellent."--Carlos Castillo-Chavez, Arizona State University
"This is a good introductory text for life sciences undergraduates who do not have a strong background in mathematics and need to familiarize themselves with core math concepts and their applications to biology. The problems and examples are well chosen, and the book is written in a style that is clear and makes it easy for students to use on their own."--Joceline Lega, University of Arizona
There are no other books quite like this one on the market. Other texts on the subject do not have nearly the amount of statistics and probability that this one has, nor do they do as much to help build practical MATLAB skills. The abundance of data-driven examples, exercises, and student projects also sets this book apart from its competitors."Trachette L. Jackson, University of Michigan | 677.169 | 1 |
About the Book
Bibliographic Details
Title: Visualization in Mathematics, Reading and ...
Publisher: Springer
Publication Date:
Binding: Hardcover
Book Condition: New
Edition:
Book Type: Hardcover
Description:
Hardcover. 106 pages. Dimensions: 9.3in. x 6.2in. x 0.5in.Science education at school level worldwide faces three perennial problems that have become more pressing of late. These are to a considerable extent interwoven with concerns about the entire school curriculum and its reception by students. The rst problem is the increasing intellectual isolation of science from the other subjects in the school curriculum. Science is too often still taught didactically as a collection of pre-determined truths about which there can be no dispute. As a con- quence, many students do not feel any ownership of these ideas. Most other school subjects do somewhat better in these regards. For example, in language classes, s- dents suggest different interpretations of a text and then debate the relative merits of the cases being put forward. Moreover, ideas that are of use in science are presented to students elsewhere and then re-taught, often using different terminology, in s- ence. For example, algebra is taught in terms of x, y, z in mathematics classes, but students are later unable to see the relevance of that to the meaning of the universal gas laws in physics, where p, v, t are used. The result is that students are c- fused and too often alienated, leading to their failure to achieve that extraction of an education from a scheme of instruction which Jerome Bruner thought so highly desirable. This item ships from multiple locations. Your book may arrive from Roseburg,OR, La Vergne,TN. Bookseller Inventory # 9789048188154 | 677.169 | 1 |
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Talking Point: The need for change in maths education
By DAPHNE JACKSON
Mathematics has been selected as one of the three core subjects in the
National Curriculum for schoolchildren betwen the ages of 5 and 16. As such,
it has to fill a variety of functions. For example, when supplemented by
A-level or AS-level courses, or by those leading to vocational qualifications,
school mathematics courses have to provide a foundation for a wide array
of users, including engineers, scientists, computer programmers, economists
and designers. The courses must also provide a satisfying experience for
potential mathematicians, the most gifted of whom tend to develop at an
early age. And then there is a strongly perceived need to upgrade the elementary
mathematical capability of the population as a whole.
The difficulty caused by the plethora of requirements is that changes
in mathematics education which may benefit one particular group may be seen
as adverse for others. It is important to be aware of some of the conflicts
which can arise, particularly at a time when the transition from the dual
examinations of O-level and CSE to GCSE has already taken place – although
its effects are only slowly becoming evident – and GCSE courses themselves
are being revised or replaced by syllabuses and examinations which meet
the requirements of the National Curriculum.
One requirement which will remain is the need to ensure that pupils
of all abilities have adequate opportunities to show what they know, understand,
and can do. In mathematics, this is mainly achieved by offering papers at
several levels of difficulty: foundations, intermediate and higher. However,
such a strategy places restrictions on the grades which can be achieved.
It is a cause for concern that a significant percentage of pupils who were
entered for the intermediate and higher levels were ungraded.
Another development is the extensive use of coursework, which will be
a compulsory part of GCSE maths from 1991. In principle, this is intended
to allow pupils to demonstrate their initiative, and apply their mathematical
skills to problem-solving.
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As a result of these changes, the portfolio of skills and knowledge
acquired by pupils up to the age of 16 is likely to be significantly different
from that of some years ago. Pupils are likely to perform less well in algebraic
manipulation, computation and aspects of number, as indicated by a recent
report from the School Examinations and Assessment Council (SEAC)*. However,
they are likely to show improved abilities in geometry, the use of measures,
and some aspects of probability and statistics; they should also be more
skilled at investigative mathematics.
Furthermore, if more pupils are to stay in full-time education beyond
16, then from 1994 onwards those who have achieved levels 7 to 10 in the
National Curriculum should enter advanced studies, but some of these pupils
will have taken only intermediate papers at GCSE. Experienced teachers have
cast doubt on the feasibility of this. But if it is possible in English,
we must make it possible in mathematics. It follows that ensuring continuity
between subjects studied at the 5 to 16 stage, and those followed subsequently,
requires a rethink of A-level and AS-level syllabuses and examinations.
In the past, a degree of uniformity has been achieved through the so-called
'Inter Board Common Core'. Negotiated in the early 1980s, the core emphasises
calculus, trigonometry and functions. However, this set of mathematical
topics is seen by some as reflecting too strongly the needs of engineering
and physics some years ago.
Calculus should certainly be seen as part of our cultural heritage.
But it is difficult to argue that trigonometry is more important to engineers
than probability. Similarly, physicists now make substantial use of discrete
mathematics and mathematical modelling, as do environmental scientists and
social scientists.
I suggest, therefore, that the proper way to introduce a much needed
degree of uniformity into mathematics courses is to reduce the number of
different syllubuses on offer. At present there are over 40 A-level syllabuses
in mathematics and statistics. The mathematics committee of the SEAC would
like to reduce this to five or six, of which one might omit some elements
of the present 'common core'. We do not see the need for a 'harder' mathematics
course along the lines of most of the present syllabuses, usually called
'further maths'.
The role of AS-level syllabuses and examinations has not yet been fully
established and accepted, especially in mathematics. Of the candidates entered
for AS-level examinations in 1989, 27 per cent took mathematics, and 62
per cent of these candidates were 17 or under. Only 60 per cent of mathematics
candidates achieved grades A to E, whereas 74 per cent achieved these grades
in the A-level examination that year. Note that teachers are using the AS-level
as a one-year qualification.
A possible solution to the problem of achieving a steady progression
through to post-16 mathematics may lie in the use of modular schemes, similar
to the one recently devised by the Mathematics in Education and Industry
Project, set up in 1966. This scheme consists of 22 modules in pure mathematics,
mechanics, and statistics. A student who passes three appropriately linked
modules can receive an AS-level award in mathematics.
With six modules, a student can gain an A-level award, and A-level and
AS-level further awards are also possible. The present 'common core' is
clearly contained in three pure mathematics modules. By no means all combinations
are allowed, so that the number of routes through the scheme is consistent
with our desire to reduce the number of titles.
Obviously, we hope that schemes of this kind will carry students along
step by step, even if their initial perception is that mathematics is a
difficult subject. At the same time, attempting as many as nine modules
should present a suitably demanding challenge to an able student.
It is important that colleagues in higher education do all they can
to support teachers of mathematics in the schools. Running workshops for
gifted young mathematicians should be considered as one way of doing this.
Above all, teachers in higher education must be aware of developments at
both the 5 to 16 and 16 to 19 levels in schools, and must revise their first
year syllabuses and teaching modules accordingly.
* Assessment Matters No 3, APU Mathematics Monitoring 1984-88.
Daphne Jackson is head of the department of physics at the University
of Surrey, and director of the Physics Centre of the Computers in Teaching
Initiative. She is a member of the School Examinations and Assessment Council,
and chairs the Council's mathematics committee. | 677.169 | 1 |
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Videos Will Help Students "Ace" Math
02/01/97
Ace-Math, an award-winning video tutorial series, is suited for students trying to grasp fundamental mathematical concepts, parents who want to help their child with their homework, or people who need to brush up on math skills for a specialized license or test.
There are nine separate series, each with many individual videos: Basic Mathematical Skills, Pre-Algebra, Algebra I, Algebra II, Advanced Algebra, Trigonometry, Calculus, Geometry, and Probability and Statistics. Each series except Algebra I consists of 30-minute videotapes explaining various concepts. Algebra I has 16 hour-long videos.
For only $29.95, Ace-Math purchasers get a 30-minute tape with the right to make two back-up copies. This lets educators keep the tape in the learning center and let students check out a copy to take home -- with the added security of another back-up copy!
These innovative tapes have been purchased by institutions such as NASA, the U.S. Coast Guard and IBM, and are in use at institutions such as the Los Angeles Public Library and New York Public Library.Video Resources Software, Miami, FL, (888) ACE-MATH,
This article originally appeared in the 02 | 677.169 | 1 |
...
Show More with graphs, real data, and linear algebraic manipulations in these various disciplines. The authors' informal tone and exclusion of unnecessary technical terminology are well-suited to the non-science major to whom the book is targeted.Enhancing the text's practical approach are a thorough integration of graphing technology, high-quality, real-world applications, and an emphasis on modeling. Functions are introduced by the "Rule of Four"—graphically, numerically, symbolically, and verbally. The text also features frequent use of data tables to display functions and an early introduction to realistic multivariate problems.Functions and Change introduces concepts through applications (in context) instead of through theory. Many exercises are conceptual, requiring thought and interpretation, not just quantitative answers.A wide diversity of relevant applications reveals mathematics as an integral part of nature, science, and society. In addition to offering more student appeal, the real-world context allows students to check their answers against their own intuition.Key Idea boxes provide reinforcement by highlighting important concepts as they are presented.In response to user feedback, Another Look review sections, accompanied by Enrichment Exercises, appear at the end of each section in the first six chapters, to enrich algebraic presentation and symbol manipulation and provide alternative treatments of the material.Skill-Building exercises precede each exercise set in the text and additional exercises feature further practice in setting up formulas and a wider variety of non-science applications.An icon has been added to help students identify ongoing problems, such as applications that are revisited in a more sophisticated way each time they appear.Areas of new or enhanced content include the basic concept of a function; treatment of linear regression by hand calculations; exponential regression exercises; a section on logarithmic functions; a section on composition of functions and piecewise-defined functions; an expanded and self- contained section on quadratic functions; expanded coverage of polynomial and rational functions, including treatment of asymptotes; and expanded coverage of periodic functions and trigonometry | 677.169 | 1 |
Monday, December 20, 2010
Computer graphics programming uses linear algebra so heavily, you could basically say it's based on it. Yet, many people — perhaps most — who develop an interest in graphics programming don't have a background in Linear Algebra. At most universities, it is taught as an upper-level (300 or 400) mathematics course, which means that the majority of students who aren't majoring or minoring in math or certain hard sciences, typically don't take them. Even if you have studied it, if you've gone a period of time without using it, you very likely have forgotten it.
Yet, if you want to go beyond a certain level in graphics programming, you need to understand it.
Today, I stumbled across a free course in Linear Algebra, and it looks to be quite good. It's dense but, hey, this is higher math we're talking about here, so there's not much that can be done to simplify it without making it incorrect. But the price is right, the content is good, and what's even better is that the course doesn't assume much in the way of specific prior knowledge.
Make no bones about it, this is hard stuff to learn, but if you've got a reason to learn it (like a drive to create computer graphics), it helps a lot. Linear algebra in the classroom taught in an abstract manner bored me to tears. Having a real reason and being able to do things with it makes it much more rewarding and fun | 677.169 | 1 |
This app covers the following topics applicable to Multivariable Calculus, Advanced Calculus, and Vector Calculus:
- Evaluate any numeric expression, or substitute a value for a variable - Plot 2D or 3D functions of your choice - Determine the limit of a function as it approaches a specific value or values - Differentiate any single or multivariable function - Find the critical points and saddle points of a function - Calculate the gradient of a function - Identify the local extrema of a function - Find the single, double, or triple integral of a function - Determine the dot or cross product of two vectors - Calculate the divergence or curl of a vector field
The Wolfram Multivariable Calculus Course Assistant is powered by the Wolfram|Alpha computational knowledge engine and is created by Wolfram Research, makers of Mathematica—the world's leading software system for mathematical research and education.
The Wolfram Calculus Course Assistant draws on the computational power of Wolfram|Alpha's supercomputers over a 3G, 4G, or Wi-Fi connection.
What's New in Version 1.2
Screenshots
Customer Reviews
Missing a few key features
by
Mathguy2718
The app should include plots of vector fields. Also the limit activity should include a 3d plot and contour plot of the function to help students see the relationships between the plots and the limits. There are a lot of good features in this app, but it could definitely be improved.
This is awesome application. Just need new more updates.
by
Sherap6
I am taking Multivariable Calculus course in this summer semester.
Meh!
by
Tooslicky
No Spherical or Cylindrical Support. Furthermore it does not always work with more advanced triple integrals.
Lastly be aware that while the user can make manual conversions to spherical or cylindrical it will not always produce a correct integral or even interpret it, especially if your attempting a work around for sphereical | 677.169 | 1 |
Math modeling handbook now available
April 23, 2014
This is the cover of the free math modeling handbook published by the Society for Industrial and Applied Mathematics this month. Credit: SIAM
Math comes in handy for answering questions about a variety of topics, from calculating the cost-effectiveness of fuel sources and determining the best regions to build high-speed rail to predicting the spread of disease and assessing roller coasters on the basis of their "thrill" factor. How does math do all that?
That is the topic of a free handbook published by the Society for Industrial and Applied Mathematics (SIAM) this month: "Math Modeling: Getting Started and Getting Solutions."
Finding a solution to any of the aforementioned problems—or the multitude of other unanswered questions in the real world—will likely involve the creation, application, and refinement of a mathematical model. A math model is a mathematical representation of a real-world situation intended to gain a qualitative or quantitative understanding in order to predict future behavior. Such predictions allow us to come up with novel findings, enable scientific advances, and make informed decisions.
The handbook provides instructions and a process for building mathematical models using a variety of examples to answer wide-ranging questions.
The inspiration for the handbook came from Moody's Mega Math (M3) Challenge, a high school applied math contest organized by SIAM. Despite the tremendous success of the nine-year-old Challenge, which is currently available to 45 U.S. states and Washington, D.C., organizers found that many participating students—high school juniors and seniors—were having trouble coming up with approaches and solutions to the open-ended realistic problems posed by the contest. Participants expressed their frustration in post-contest surveys and emails.
"We have been enthusiastic about the high level of insight and analysis demonstrated by participants in the Challenge, especially the winning teams," says M3 Challenge Project Director Michelle Montgomery. "However, it became clear to us that, given the lack of modeling courses in most high school curricula, many of the participants did not have access to basic resources necessary to create a successful model. We came up with the handbook to give every participant these tools."
This type of thinking created an "aha" moment, so to speak, for handbook authors Karen Bliss, Katie Fowler, and Ben Galluzzo, long-time Challenge judges who have been part of the contest's problem development team for the past two years.
"All students, especially those interested in STEM disciplines, need as much practice in solving open-ended problems as possible, but they often do not get many chances to do that in school,"says Fowler, who is an associate professor of mathematics at Clarkson University. "Math modeling skills allow students to approach problems they initially may feel are outside of their comfort zone, and we want to give them the confidence to tackle them."
Further motivated by a series of SIAM-National Science Foundation (NSF) workshops on the topic of math modeling across the curriculum, the trio began work on a modeling guide. What started as a pamphlet with step-by-step guidance about the modeling process grew into a 70-page, full color handbook, with a companion document that makes connections to the Common Core State Standards as well as easy-to-use reference cards for those who want to get straight to the crux of modeling. The guide is suitable for teachers as well as high school and undergraduate students interested in learning how to model.
"Math modeling is challenging, but it's also surprisingly accessible. The guidebook is designed to remove perceived roadblocks by presenting modeling as a highly-creative iterative process in which multiple approaches—to the same problem—can lead to meaningful results," says Galluzzo, an assistant professor of mathematics at Shippensburg University.
The handbook, as well as the Challenge itself, has another, more pressing goal: motivating our younger generation to pursue higher education and careers in science and math. "SIAM does a big service to the math community at large by giving high school students the opportunity to see how math is more than just a series of formulas and rote memorization," says Bliss, an assistant professor of mathematics at Quinnipiac University. "Students at all levels have the means to produce highly creative solutions to interesting problems. Seeing that math can be a powerful tool for solving truly important problems through M3 Challenge participation might be just enough to encourage a student to study math or another STEM discipline in college."
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book contains 24 illustrated math problem sets based on a weekly series of space science problems. Each set of problems is contained on one page. The problems were created to be authentic glimpses of modern science and engineering issues, often...(View More) involving actual research data. Learners will use mathematics to explore problems that include basic scales and proportions, fractions, scientific notation, algebra, and geometry.(View Less) | 677.169 | 1 |
Mathematics: Graphic Methods Books
Roland's Flora of Nova Scotia is the most comprehensive book ever published on the province's plants. It is an indispensable resource for anyone interested in what grows where in Nova Scotia. With easy-to-read descriptions, foolproof keys and complete,…
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The time has now come when graph theory should be part of the education of every serious student of mathematics and computer science, both for its own sake and to enhance the appreciation of mathematics as a whole. This book is an in-depth account of graph…
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This book stresses the connection between, and the applications of, design theory to graphs and codes. Beginning with a brief introduction to design theory and the necessary background, the book also provides relevant topics for discussion from the theory of… | 677.169 | 1 |
Hockessin Microsoft ExcelJohn B.
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2009 Assessment Report 2009 GENERAL COMMENTS
The number of students who sat for the 2009 examination was 7223. Almost 15% of students scored 90% or more of the available marks, compared with 16% in 2008, and just over 1% received full marks, compared with 3% in 2008. It was encouraging to see that the use of the integrand, with 'dx', was not a problem for students this year. However, some students commenced finding the rule for an inverse function by writing f -1 = x = or similar, presumably in an attempt to indicate that they were finding f -1(x) by solving x = f(y) for y. Many students were unable to correctly evaluate simple arithmetic calculations involving decimals and fractions. In Questions 1a., 2b., 4, 5, 6, 7 and 10a., answers were often either left unsimplified or incorrectly evaluated when simplification was attempted. The use of brackets in expressions where more than one term is multiplied or subtracted from another was a problem for a significant number of students. This was particularly evident in Questions 1b. and 2b. Students are strongly advised to reread a question upon completion to ensure they have done all that is required. In Question 2b., many students applied the quotient rule but neglected to substitute x = π to complete their answer. In Question 3, most students knew what to do to find the rule of the inverse function f -1(x) but neglected to state the domain of f -1. Students need to be aware that they must show working in questions worth more than one mark. Failure to show appropriate working will result in marks not being awarded. The number of marks allocated to a question is usually an indication of the number of steps/concepts required. Question 1a. Marks 0 % 16 1 x × + log e ( x ) x
= 1 + log e ( x )
The quotient rule was appropriately attempted by most students. However, too many students did not include the brackets on the first line and/or incorrectly cancelled out the 2x + 2 in the denominator with the 2x + 2 in the numerator. A majority of students did not substitute x = π correctly. | 677.169 | 1 |
Accelerated 6th Grade
CPM offers two options for acceleration to Algebra 1 in the 8th grade. Both pathways utilize the Core Connections, Courses 1-3 (CC1-3) middle grades courses, so no special books will need to be ordered. In the few cases where the pacing guide uses a lesson from the second book out of sequence, those pages may be downloaded and copied for the students (see files below). This approach gives schools flexibility from year to year as the number of students following the accelerated pathway fluctuates.
The first option, which is recommended by the CPM mentor teachers and authors, begins acceleration in the 6th grade. Students complete CC1 and half of CC2 in 6th grade, then complete CC2 and all of CC3 in 7th grade. They take CC Algebra in 8th grade. This plan offers the most flexibility and ease for students to opt in or out of acceleration with respect to the content and cognitive demand of these courses. If the pace is too demanding in the 6th grade, students can resume a regular 7th grade course with a head start. If students opt in at 7th grade, it will be easier for them to catch up on the content from the first part of CC2.
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The 6th Grade Pathway (three books in two years) to Algebra in the 8th grade: | 677.169 | 1 |
Links to articles about society's conflicting expectations for the goals (ends) and pedagogy (means) of school mathematics. Concerns about the relation of skills to understanding, about connections with other...
An interactive mathematics course offered via the Internet to students throughout the Hawaiian islands. Cathi Sanders, a teacher at Punahou School in Honolulu, teaches this course through ESchool, a pilot project of the...
The mission of the Math Machines organization is to "improve the quality of mathematical education, enhance the transfer of mathematical thinking into other classes, and increase students' ability to apply rigorous...
This math skills review was written for first year college chemistry students who have weak math backgrounds, but is useful for all science students in high school and college. Topics covered include algebraic...
This site includes a number of mathematical lessons that relate directly to physics topics.
Each link provides a short example that would provide helpful supplemental materials in mathematics or science education... | 677.169 | 1 |
The Technology Executive's Definitive Guide to Modernization Strategy in the Age of Disruption, a book is for corporate technology executives with the concern for business continuity in a climate of high disruption and digital transformation.
iOS 10
The Unofficial Guide To iOS 10 - Everything You Need To Know About iOS 10 - Advanced Tips And Tricks And New iOS 10 Features For The iPhone 6s, iPhone 6s Plus And iPad!
The latest version of Apple's mobile operating system, iOS 10, is the biggest change since iOS made its debut in 2007. It brings a new era to the way you are able to interact with your favorite device.
The subjects treated in this book are parts of the works related to importance of ship stability assessment. The book is approaching the problems of ships stability loss through the study of operational aspects as well as dynamic ship behaviour in severe sea conditions offering a picture of some stability failure modes that presently are not covered by any regulations or criteriaThis short, illustrated book is about the observations and insights of Buckminster Fuller. Though he's best remembered for the geodesic dome and Buckyballs, Fuller offered many thought-provoking ideas about evolution, history, technology and existence itself, including the observation that created the title for this book.
Discover the classic techniques of lock picking to defeat any of the most common door locks. Learn how locks work and what makes them vulnerable. Practice your new found skills and get relevant info on tools, lock types and also padlocks, cuffs and plastic wrist tiesThis is a concise article detailing mathematical equations showing when and where chaos occurs in linear algebraic equations and systems of linear simultaneous algebraic equations. The approach used is very simple and easy to understand by students.This concise article of forty pages takes you on a short tour on how to handle matrices using MATLAB. Topics covered include how to generate matrices in MATLAB, different operations on matrices, and how to handle matrix, vector, and scalar quantities. Various MATLAB functions associated with matrices are also explored.
This concise article takes you on a short tour on how to plot graphs using MATLAB. The presentation covers both two-dimensional and three-dimensional graphs. This article is taken form the bestselling book "MATLAB for Beginners: A Gentle Approach." | 677.169 | 1 |
Geometry Lessons in the Waldorf School – Volume 2
Freehand Form Drawing and Basic Geometric Construction in Grades 4 and 5
Ernst Schuberth
Softbound
Includes CD-ROM with additional exercises and color plates.
$18.00
This book is the second volume* of mathematician and Waldorf teacher Ernst Schuberth's Geometry Lessons in the Waldorf School. With an abundance of black and white drawings and a clear, descriptive text, Schubreth covers the free form drawing and basic constructions taught in the Waldorf school fourth and fifth grades.
A real bonus is the accompanying CD Rom that has offers additional exercises and color plates for the teacher to study. | 677.169 | 1 |
Math for the Trades43
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About the Book
This book serves readers who want to assess and improve the math skills they need to succeed at work. A complete review of arithmetic, algebra, geometry, and word problems ensures improvement of these essential math skills. With over 200 on-the-job practice questions, it is a necessary tool for employees who need math skills to complete their jobs with ease. | 677.169 | 1 |
Foundation for Advancements in Science and Education
Since its inception, the mission of this independent non-profit public charity has "evolved continuously, encompassing challenging new issues and exploring innovative applications of information technology." In particular, FASE sells The Eddie Files,
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GCSE Maths Revision - Steven Britton, Top Grade Tutoring
GCSE Maths Master is a DVD available in two levels, higher level and foundation level, designed to make revision faster and simpler, with worked examples and exam tips. There is also a non-calculator skills DVD available, suitable for revising for non-calculatorGirls' Angle
Girls' Angle is a math club for girls and a supportive community for all girls and women engaged in the study, use and creation of mathematics. By mentoring, publishing the bimonthly Bulletin, and solving math problems on commission, networking professional
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Granular Volcano Group - Sebastien Dartevelle
The site aims to describe and understand granular flows, fluid dynamic, supercomputer modeling, grain features, and behaviors in volcanology. Includes descriptions of frictional, kinetic, and collisional rheologies of granular matter, and definitions
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Graphing Stories - Dan Meyer and the BuzzMath Team
Videos, each 102 seconds in length, of stories for students to graph using the freely downloadable accompanying student template. Select from among constant, decreasing, increasing, linear, parabolic, periodic, piecewise, and step graphs; story subjects
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Guaranteach - Alasdair Trotter and George Tattersfield
Thousands of very short video lessons, recorded by dozens of different professional teachers whose methods and styles of teaching math vary widely. The diverse perspectives and "bite-size" chunks in Guaranteach's video library let viewers stay focused
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Handheld Geometry - Nevil Hopley
A dynamic geometry site offering lessons and challenges for handheld devices and graphing calculators, including the Cabri™ Jr. Application for the Texas Instruments TI83+ and TI84+, and the TI-Nspire™. The author offers to make users' contributions
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HegartyMaths - Colin Hegarty
Hundreds of narrated screencasts covering the UK's Key Stage 3, GCSE and A-Level curriculums, along with many past paper solution walkthroughs. Instructional videos — most for students; some for teachers — also available from the HegartyMaths
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The Ignite Show - Georgia Public Broadcasting
TV shows and blog that "raise respect and awareness of educators and education through the voices of teachers, parents and students" in student-centered classrooms. Half hour-long videos that feature math teachers include episodes starring 2008 California
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Interior Point Methods - Steven Wright
Research into algorithms for linear programming which approach the solution through the interior of the feasible polygon, rather than moving around the boundary from vertex to vertex, as simplex methods do. Background; thumbnail sketch; zipped Postscript
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Is All About Math - Julio Cesar de la Yncera
Main site for a mathematical Video PodCast that will cover a big range of topics in college-level mathematics accessible to the majority of the students with some mathematical inclinations. The Video PodCast will also try to be inspirational for the non-mathematically
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iSchoolPolymath - Sarah Prendergast
Video libraries, syllabi, Pinterest inspiration pages, and other materials for math courses taught at iSchoolPolymath, a public high school in New York City. Prendergast's Algebra I course incorporates gamification, hands-on projects, problem solving,
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Jo Boaler
From the author of What's Math Got to Do with It?: How Parents and Teachers Can Help Children Learn to Love Their Least Favorite Subject. Watch a YouTube clip of Boaler teach using Inquiry Based Learning method, or click through her presentation entitled
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Josh Hertel
Hertel serves the University of Wisconsin-La Crosse as an assistant professor in the mathematics department. He completed his Ph.D. in Mathematics Education at Illinois State University with the freely downloadable dissertation "Investigating the Purpose
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Karismath - CLSO Learning Systems, Inc.
Math topics, including Algebra, are presented visually, through lessons (Flash animations that explain a target concept), demos (Flash animations that explain every exercise template to be used) and worksheets (PDF format) that can be downloaded. Each
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Khan Academy - Salman Khan
Salman Khan has recorded thousands of YouTube videos on a variety of topics, largely math and finance. The Khan Academy, which Khan founded, is a not-for-profit organization with the mission of providing a high quality education to anyone, anywhere. To
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Learning Math is Fun - Kevin Cornell
Learn the order of operations by watching a video of the PEMDAS song. Before becoming an elementary school principal, singer "Mister C" taught 5th grade science and served as a math coach. See also Cornell's science site at
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Learning Wave Communications - Human Relations Media
Learning Wave is a producer of videos, CD-ROMs and other supplementary materials for math education including interactive games such as Absurd Math. Under the name HRM Video it has created programs such as middle school math mysteries, applications-based
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Factoring Polynomials (Lesson 26 of 61)
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
1.97 MB | 6 pages
PRODUCT DESCRIPTION
This is a math test prep lesson that explains how to factor polynomials by understanding how to determine and divide by greatest common factors, factoring by grouping, and factoring the difference of squares as part of the Algebra material that many state exams cover.
This is one of 61 lessons available in the workbook titled The Essentials of High School Math from Willow Tree Publishing. Each book contains an answer key to all lessons, but an answer key for just this lesson is not available | 677.169 | 1 |
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Read More through modeling. Its intent is to demonstrate the relevance and practicality of discrete mathematics to all students. The fifth edition includes a more thorough and linear presentation of logic, proof types and proof writing, and mathematical reasoning. This enhanced coverage will provide students with a solid understanding of the material as it relates to their immediate field of study and other relevant subjects. The inclusion of applications and examples to key topics has been significantly addressed to add clarity to every subject. True to the fourth edition, the text specific web site supplements the subject matter in meaningful ways, offering additional material for students and instructors. Discrete math is an active subject with new discoveries made every year. The continual growth and updates to the web site reflect the active nature of the topics being discussed. This text is designed for a one- or two-term introductory discrete mathematics course to be taken by students in a wide variety of majors, including computer science, mathematics, and engineering. College Algebra is the only explicit prerequisiteCustomer Reviews
Discrete Mathematics and Its Applications
by Kenneth H Rosen
Very difficult material
The mathematics in this book were very discrete and much too abstract for a beginner to understand...
Ed P
Sep 26, 2013
avoid the international edition
although the description said the contents are the same, the pages and questions are out of order. I was unable to use this book for my class.
Hbird
Oct 4, 2012
Beware
Beware of international editions. The one my son received did not have the same questions in it as the american one and therefore we had to buy the american edition as his instructor required these questions be anwsered as part of his assignment.
Phillip R
Oct 4, 2012
excellent seller fast shipping love to do business again
James H
Mar 24, 2011
Great book
Through and well explained. In addition to the math an excellent foundation for Computer Science | 677.169 | 1 |
Technical MathematicsIndividuals preparing for technical, engineering technology, or scientific careers will benefit from the major projects and integrated use of calculators in this 3rd edition that allow them to solve problems in much the same manner as they will on the job! Thorough coverage of precalculus topics ranging from algebra and geometry to trigonometry and analytic geometry place an emphasis on how these topics are used in specific occupations. Written in an easy-to-understand manner, this comprehensive book provides numerous application-oriented exercises and examples that will help users learn to use mathematics and technology in situations related to their future work.
Preface
xv
The Real Number System
1
(61)
Some Sets and Basic Laws of Numbers
2
(14)
Basic Operations with Real Numbers
16
(15)
Exponents and Roots
31
(11)
Significant Digits and Rounding Off
42
(7)
Scientific and Engineering Notation
49
(13)
Review
59
(2)
Test
61
(1)
Algebraic Concepts and Operations
62
(49)
Addition and Subtraction
63
(5)
Multiplication
68
(6)
Division
74
(7)
Solving Equations
81
(14)
Applications of Equations
95
(16)
Review
107
(2)
Test
109
(2)
Geometry
111
(59)
Lines, Angles, and Triangles
112
(12)
Other Polygons
124
(6)
Circles
130
(7)
The Area of Irregular Shapes
137
(7)
Geometric Solids
144
(12)
Similar Geometric Shapes
156
(14)
Review
162
(5)
Test
167
(1)
Building Design
168
(2)
Functions and Graphs
170
(60)
Relations and Functions
171
(14)
Operations on Functions; Composite Functions
185
(8)
Rectangular Coordinates
193
(5)
Graphs
198
(9)
Calculator Graphs and Solving Equations Graphically
207
(11)
Introduction to Modeling
218
(12)
Review
226
(2)
Test
228
(2)
An Introduction to Trigonometry
230
(42)
Angles, Angle Measure, and Trigonometric Functions
231
(5)
Values of the Trigonometric Functions
236
(6)
The Right Triangle
242
(8)
Trigonometric Functions of Any Angle
250
(9)
Applications of Trigonometry
259
(13)
Review
265
(4)
Test
269
(1)
Chip Away
270
(2)
Systems of Linear Equations and Determinants
272
(42)
Linear Equations
273
(9)
Graphical and Algebraic Methods for Solving Two Linear Equations in Two Variables
282
(7)
Algebraic Methods for Solving Three Linear Equations in Three Variables | 677.169 | 1 |
This site is devoted to learning mathematics through practice. Many dozens of practice problems are provided in Precalculus,...
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This site is devoted to learning mathematics through practice. Many dozens of practice problems are provided in Precalculus, Calculus I - III, Linear Algebra, Number Theory, and Abstract Algebra. The last two subject areas -- referred to as "books" on the site -- are under construction. To each topic within a book (for example, Epsilon and Delta within Calculus I) there is a "module" of approximately 20 to 30 problems. Each module also includes a help page of background material. The modules are interactive to some extent and often provide suggestions when wrong answers are enteredOW -- Calculus on the Web to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material COW -- Calculus on the Web
Select this link to open drop down to add material COW -- Calculus on the Web to your Bookmark Collection or Course ePortfolio
This site consists of examples, exercises, games, and other learning activities associated with the textbook, Discrete...
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This site consists of examples, exercises, games, and other learning activities associated with the textbook, Discrete Mathematics: Mathematical Reasoning and Proof with Puzzles, Patterns and Games by Doug Ensley and Winston Crawley. Requires Adobe Flash player Discrete Math Resources to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Discrete Math Resources
Select this link to open drop down to add material Discrete Math Resources to your Bookmark Collection or Course ePortfolio
This is a graduate level reference work with 8,000 entries illuminating nearly 50,000 mathematical notions. If you are...
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This is a graduate level reference work with 8,000 entries illuminating nearly 50,000 mathematical notions. If you are looking for the definition of a term, this site most likely has it.Each entry is actually a small survey article on the subject in question, written by an expert. Articles list further references, so that deeper study after a quick refresher is possible Encyclopaedia of Mathematics to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Encyclopaedia of Mathematics
Select this link to open drop down to add material Encyclopaedia of Kunkel's Mathematics Lessons to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Kunkel's Mathematics Lessons
Select this link to open drop down to add material Kunkel's Mathematics Lessons to your Bookmark Collection or Course ePortfolio
of Miletus to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Thales of Miletus
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The material presented in the following pages are for middle school students, high school students, college students, and all...
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The material presented in the following pages are for middle school students, high school students, college students, and all who are interested in mathematics. You will find interactive programs that you can manipulate and a lot of animation that helps you to grasp the meaning of mathematical ideas MER with Java to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Manipula Math with Java
Select this link to open drop down to add material Manipula Math with Java History of Pi to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material A History of Pi
Select this link to open drop down to add material A History of Pi to your Bookmark Collection or Course ePortfolio
Highlights of Calculus is a series of short videos that introduces the basic ideas of calculus — how it works and why it is...
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Highlights of Calculus is a series of short videos that introduces the basic ideas of calculus — how it works and why it is important. The intended audience is high school students, college students, or anyone who might need help understanding the subject.In addition to the videos, there are summary slides and practice problems complete with an audio narration by Professor Strang. You can find these resources to the right of each video.This resource is also available on Highlights for High School.About the InstructorProfessor Gilbert Strang is a renowned mathematics professor who has taught at MIT since 1962. Read more about Prof. StrangAcknowledgementsSpecial thanks to Professor J.C. Nave for his help and advice on the development and recording of this program.The video editing was funded by the Lord Foundation of Massachusetts to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material Calculus
Select this link to open drop down to add material Calculus Table: Mathematical Modeling
Select this link to open drop down to add material Calculus of the Dinner Table: Mathematical Modeling to your Bookmark Collection or Course ePortfolio | 677.169 | 1 |
Find a Pico Heights, CAThe math sections measure a student's ability to reason quantitatively, solve mathematical problems, and interpret data presented in graphical form. These sections focus on four areas of mathematics that are typically covered in the first three years of American high school education: Arithmetic | 677.169 | 1 |
84374730547197977
Publication Year:
2008
ISBN-13:
9780547197975
Language:
English
Author:
Vernon C. Barker, Joanne Lockwood, Richard N. Aufmann
Educational Level:
College Audience
ISBN:
9780547197975
Detailed item info
Synopsis
Based on the best-selling series by the Aufmann team, this hardcover text for the intermediate algebra course adheres to the formula that has made the Aufmann developmental texts so reliable for both students and instructors. The text's clear writing style, emphasis on problem-solving strategies, and proven Aufmann Interactive Method--in an objective-based framework--offer guided learning for both lecture and self-paced courses. The completely integrated learning system is organized by objectives. Each chapter begins with a list of learning objectives, which are woven throughout the text, in Exercises, Chapter Tests, and Cumulative Reviews, as well as through the print and multimedia ancillaries. The result is a seamless, easy-to-follow learning system | 677.169 | 1 |
04866469 Source Book in Mathematics (Dover Books on Mathematics)
This work presents, in English translation, the great discoveries in mathematics from the Renaissance to the end of the nineteenth century. You are able to read the writings of Newton, Leibniz, Pascal, Riemann, Bernoulli, and others, exactly as the world saw them for the first time. Succinct selections from 125 different treatises and articles, most of them unavailable elsewhere in English, offer a vivid, firsthand story of the growth of mathematics. The articles are grouped in five sections: I. The Field of Number. Twenty-four articles trace developments from the first steps in printed arithmetic through selected number systems, to the early phases of modern number theory. II. The Field of Algebra. Eighteen articles on algebra include writings by Fermat, John Wallis, Newton, Leibniz, Abel, Galois, etc. III. The Field of Geometry. Thirty-six articles on geometry span 500 years. Here are the early writers such as Fermat, Desargues, Pascal, and Descartes; and some of the men who revived the study in the nineteenth century and developed non-Euclidian forms: Lobachevsky, Bolyai, Riemann, and others. IV. The Field of Probability. Selections from Fermat, Pascal, De Moivre, Legendre, Chebyshev, and Laplace discuss crucial topics in the early history of this branch of modern mathematics. V. The Field of Calculus, Functions, and Quaternions. The development of the calculus, function theory, and quaternions is covered from early sources of calculus to important advances relating to the commutative law in quaternions and Ausdehnungslehre. This section contains works of Bessel, Mobius, W. R. Hamilton, Leibniz, Berkeley, Cauchy, Fermat, and six other pioneering mathematicians. Each article is preceded by a biographical-historical Introduction | 677.169 | 1 |
College Algebra, Math 163, Notes
This is My Personal College Algebra Study Notes.
Hi everyone,
I am publishing this hub to help anyone who is or will take College Algebra. Information in this hub is strictly from the notes I took from lectures and personal studying. This hub is not intended to teach someone College Algebra, but to aid people who need to refer to information covered in College Algebra. I want everyone reading this hub to know that this hub doesn't completely cover every material in the textbook I used or in every College Algebra class. Also, there may be errors in this hub, so if you find any, please leave me a comment to fix them so this hub won't misguide anyone studying the subject.
What is College Algebra?
According to the CCBC (Community College of Baltimore County) mathematics course description, College Algebra explores the nature and scope of college mathematics through the study of functions. Topics include the study of polynomial, rational, radical, piece-wise defined, and absolute value functions and their graphs and applications as well as modeling with these functions. Additional topics include complex numbers, the binomial theorem, inverse functions, operations with functions, and exponential and logarithmic functions and their graphs and applications. | 677.169 | 1 |
Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied.
The constructive approach to mathematics has recently enjoyed a renaissance. This was caused largely by the appearance of Bishop's Foundations of Constructive Analysis, but also by the proliferation of powerful computers, which stimulated the development of constructive algebra for implementation purposes. In this book, the authors present the fundamental structures of modern algebra from a constructive point of view. Beginning with basic notions, the authors proceed to treat PID's, field theory (including Galois theory), factorisation of polynomials, noetherian rings, valuation theory, and Dedekind domains.Read more...
Abstract:
Presents the fundamental structures of modern algebra from a constructive point of view. This book contains basic notions, and also covers PID's, field theory (including Galois theory), factorisation of polynomials, noetherian rings, valuation theory, and Dedekind domains.Read more... | 677.169 | 1 |
Known for its continued excellence, CALCULUS OF A SINGLE VARIABLE: EARLY TRANSCENDENTAL FUNCTIONS, Sixth Edition, offers instructors and students innovative teaching and learning resources. The Larson team always has two main objectives for text revisions: to develop precise, readable materials for students that clearly define and demonstrate concepts and rules of calculus; and to design comprehensive teaching resources for instructors that employ proven pedagogical techniques and save time. The Larson/Edwards Calculus program offers a solution to address the needs of any calculus course and any level of calculus student. Every edition from the first to the sixth of CALCULUS: EARLY TRANSCENDENTAL FUNCTIONS has made the mastery of traditional calculus skills a priority, while embracing the best features of new technology and, when appropriate, calculus reform ideas.
Additional Product Information
Features and Benefits
Applications-Carefully chosen applied exercises and examples are included throughout to address the question, "When will I use this?" These applications are pulled from diverse sources, such as current events, world data, industry trends, and more, and relate to a wide range of interests.
Writing About Concepts-Writing exercises at the end of each section are designed to test students' understanding of basic concepts in each section, encouraging them to verbalize and write answers and promote technical skills that will be invaluable in their future careers.
Theorems-Theorems provide the conceptual framework for calculus. Theorems are clearly stated and separated from the rest of the text by boxes for quick visual reference. Key proofs often follow the theorem and can be found on LarsonCalculus.com.
Definitions-As with theorems, definitions are clearly stated using precise, formal wording and are separated from the text by boxes for quick visual reference.
Explorations-Explorations provide unique challenges to study concepts that have not yet been formally covered in the text. They allow students to learn by discovery and introduce topics related to ones presently being studied. Exploring topics in this way encourages students to think outside the box.
Historical Notes and Biographies-Historical notes provide students with background information on the foundations of calculus. The Biographies introduce students to the people who created and contributed to calculus.
Technology-Throughout the book, technology boxes show students how to use technology to solve problems and explore concepts of calculus. These tips also point out some of the pitfalls of using technology.
Section Projects-Projects appear in selected sections and encourage students to explore applications related to the topics they are studying. They provide an interesting and engaging way for students to work and investigate ideas collaboratively.
Putnam Exam Challenges-Putnam Exam Questions appear in selected sections. These actual Putnam Exam Questions will challenge students and push them to the limits of their understanding of calculus.
What's New
LarsonCalulus.com-This robust companion website offers multiple tools and resources. Access to these features is free. Students can watch videos explaining concepts or proofs from the book, explore examples, view three-dimensional graphs, download articles from math journals, and much more!
Interactive Examples-Examples throughout the book are accompanied by interactive examples at LarsonCalculus.com. These interactive examples use Wolfram's free CDF Player and allow students to explore calculus by manipulating functions or graphs, and observing the results.
Proof Videos-Students can watch videos of co-author Bruce Edwards as he explains the proofs of the theorems in CALCULUS OF A SINGLE VARIABLE: EARLY TRANSCENDENTAL FUNCTIONS, Sixth Edition, at LarsonCalculus.com.
How Do You See It?-The "How Do You See It?" feature in each section presents a real-life problem that students solve by visual inspection using the concepts learned in the lesson. This exercise is excellent for classroom discussion or test preparation.
Remark-These hints and tips reinforce or expand on concepts, help students learn how to study mathematics, caution students about common errors, address special cases, or show alternative or additional steps to a solution of an example.
Exercise Sets-The exercise sets have been carefully and extensively examined to ensure they are rigorous and relevant and include all topics our users have suggested. The exercises have been reorganized and titled so students can better see the connections between examples and exercises. Multi-step, real-life exercises reinforce problem-solving skills and mastery of concepts by giving students the opportunity to apply the concepts in real-life situations.
Appendix A (Proofs of Selected Theorems) now appears in video format on LarsonCalculus.com. The proofs also appear in text form at CengageBrain.com.
Alternate Formats
Choose the format that best fits your student's budget and course goals
Instructor Supplements
This comprehensive instructor DVD includes resources such as electronic versions of the Instructor's Resource Guide and Printed Test Bank, complete pre-built PowerPoint lectures, all art from the text in both jpeg and PowerPoint formats, ExamView® algorithmic computerized testing software, JoinIn™ content for audience response systems (clickers), and a link to Solution Builder.
Solution Builder
(ISBN-10: 1285778324 | ISBN-13: 9781285778327)
This online instructor database offers complete worked-out solutions to all exercises in the text, allowing you to create customized, secure solutions printouts (in PDF format) matched exactly to the problems you assign in class.
This manual contains worked-out solutions to all exercises in Chapters 1-6 to all exercises in Chapters 7-9, and requires no set up from instructors For instructors, CourseMate includes Engagement Tracker, a first-of-its kind tool that monitors student engagement.If you are not certain this is the correct access code for your course, please contact your Cengage Learning Consultant.
List Price = $125.00
| College Bookstore Wholesale Price = $93.75
Student SupplementsIf you are not certain this is the correct access code for your course, please contact your instructor.
List Price = $125.00
| College Bookstore Wholesale Price = $93.75Bruce H. Edwards
Dr. Bruce H. Edwards is Professor of Mathematics at the University of Florida. Professor Edwards received his B.S. in Mathematics from Stanford University and his Ph
Dr. Bruce H. Edwards is Professor of Mathematics at the University of Florida. Professor Edwards received his B.S. in Mathematics from Stanford University and his Ph | 677.169 | 1 |
Product Description
This Prentice Hall All-in-One Student Workbook accompanies the sold-separately Prentice Hall Mathematics Course 2 Textbook. "Daily notetaking pages" provide the lesson objective, "key concepts," additional fill-in-the-blank "example" problems, and quick check problems. The practice problem pages follow the same lesson instructions as the textbook, but provide different problems for each chapter. The Guided Problem Solving pages lead students through a step-by-step solution to an application problem. Vocabulary pages contain a variety of activities to increase reading and math understanding, including graphic organizers and review puzzles. The Daily Notebooking Pages are all first (in order of lesson) and are then followed by a section with the Practice, Guiding Problem, and Vocabulary worksheets. The Practice & Guided Problem worksheets alternate in lesson order, with the vocabulary worksheets all at the end of that chapter's section. Course 2. Grade 7. Version A | 677.169 | 1 |
05343775operative Group Problem Solving in Physics
COOPERATIVE GROUP PROBLEM SOLVING IN PHYSICS is a teaching guide designed for use in any introductory algebra- or calculus-based physics courses. Built on a solid foundation of research from cognitive psychology, education, and physics education, this book enables instructors to integrate group problem solving into their classroom. Clear and concise directions guide instructors in the manipulation of classroom setup, content, and structure of the course. The techniques, tested and refined over a decade by many professors and thousands of students, successfully lead students to a better understanding of physics. The authors offer specific preparation techniques and teaching tools, along with real, context-rich problems designed to encourage students to develop an organized, logical, problem-solving strategy. COOPERATIVE GROUP PROBLEM SOLVING can be used as the major focus of your course or as a supplement in combination with other teaching tools | 677.169 | 1 |
James Stewart's CALCULUS: EARLY TRANSCENDENTALS texts are widely renowned for their mathematical precision and accuracy, clarity of exposition, and outstanding examples and problem sets. Millions of students worldwide have explored calculus through Stewart's trademark style, while instructors have turned to his approach time and time again. In the Eighth Edition of SINGLE VARIABLE CALCULUS: EARLY TRANSCENDENTALS, Stewart continues to set the standard for the course while adding carefully revised content. The patient explanations, superb exercises, focus on problem solving, and carefully graded problem sets that have made Stewart's texts best-sellers continue to provide a strong foundation for the Eighth Edition. From the most unprepared student to the most mathematically gifted, Stewart's writing and presentation serve to enhance understanding and build confidence.
Features and Benefits
Four carefully crafted diagnostic tests in algebra, analytic geometry, functions, and trigonometry appear at the beginning of the text. These provide students with a convenient way to test their pre-existing knowledge and brush up on skills they need to successfully begin the course. Answers are included, and students who need to improve will be referred to points in the text or on the book's website where they can seek help.
Stewart's writing style speaks clearly and directly to students, guiding them through key ideas, theorems, and problem-solving steps, and encouraging them to think as they read and learn.
Every concept is supported by thoughtfully worked examples—many with step-by-step explanations—and carefully chosen exercises. The quality of this pedagogical system is what sets Stewart's texts above others.
Examples are not only models for problem solving or a means of demonstrating techniques; they also encourage students to develop an analytic view of the subject. To provide further insight into mathematical concepts, many of these detailed examples display solutions that are presented graphically, analytically, and/or numerically. Margin notes expand on and clarify the steps of the solution.
The text's clean, user-friendly design provides a clear presentation of calculus. The art program, with its functional and consistent use of color, helps students identify and review mathematical concepts more easily.
The topic of Differential Equations is unified by the theme of modeling. Qualitative, numerical, and analytic approaches are given equal consideration.
Stewart draws on physics, engineering, chemistry, biology, medicine, and social science to motivate students and demonstrate the power of calculus as a problem-solving tool in a wide variety of fields.
Stewart's presentation repeatedly provides answers to the question, "When will I use this?" You'll find many examples of how calculus is used as a problem-solving tool in fields such as physics, engineering, chemistry, biology, medicine, and the social sciences.
Stewart's text offers an extensive collection of more than 8,000 quality exercises. Each exercise set is carefully graded, progressing from skill-development problems to more challenging problems involving applications and proofs. The wide variety of types of exercises includes many technology-oriented, thought-provoking, real, and engaging problems.
Conceptual exercises encourage the development of communication skills by explicitly requesting descriptions, conjectures, and explanations. These exercises stimulate critical thinking and reinforce the concepts of calculus.
A wealth of engaging projects reinforce concepts. "Writing Projects" ask students to compare present-day methods with those of the founders of calculus. "Discovery Projects" anticipate results to be discussed later. "Applied Projects" feature content that engages student interest and demonstrates the real-world use of mathematics. "Laboratory Projects" anticipate results to be discussed later or encourage discovery through pattern recognition.
Comprehensive review sections follow each chapter and further support conceptual understanding. A "Concept Check" and "True/False Quiz" allow students to prepare for upcoming tests through ideas and skill-building.
"Strategies" sections (based on George Polya's problem-solving methodology) help students select what techniques they'll need to solve problems in situations where the choice is not obvious, and help them develop true problem-solving skills and intuition.
Historical and biographical margin notes enliven the course and show students that mathematics was developed to help explain and represent natural phenomena.
Additional exercises have been added to the existing end-of-chapter collections of more challenging exercises called "Problems Plus." These sections reinforce concepts by requiring students to apply techniques from more than one chapter of the text, and by patiently showing them how to approach a challenging problem.
What's New
New examples have been added (see Examples 6.1.5, and 11.2.5, for instance), and the solutions to some of the existing examples have been amplified.
Several new application based problems in the book have been added to help the students strengthen the understanding of concepts and make the leap to discovering the impact of Calculus in its various applications.
Two new projects have been added: The project, Controlling Red Blood Cell Loss During Surgery, describes the ANH procedure, in which blood is extracted from the patient before an operation and is replaced by saline solution. This dilutes the patient's blood so that fewer red blood cells are lost during bleeding and the extracted blood is returned to the patient after surgery. The project, Planes and Birds: Minimizing Energy, asks how birds can minimize power and energy by flapping their wings versus gliding.
Each section of the text is discussed from several viewpoints. The Instructor's Guide contains suggested time to allot, points to stress, text discussion topics, core materials for lecture, workshop/discussion suggestions, group work exercises in a form suitable for handout, and suggested homework assignments.
Cengage Learning Testing Powered by Cognero is a flexible, online system that allows you to author, edit, and manage test bank content; create multiple test versions in an instant; and deliver tests from your LMS, your classroom or wherever you want. This is available online via your SSO account at login.cengage.com., as well as summary and focus questions with explained answers. The Study Guide also contains "Technology Plus" questions, and multiple-choice "On Your Own" exam-style questions.
Complement your text and course content with study and practice materials. Cengage Learning Calculus CourseMate brings course concepts to life with interactive learning, study, and exam preparation tools that support the printed textbook. Watch student comprehension soar as your class works with the printed textbook and the textbook specific website. Calculus goes beyond the book to deliver what you need!
IMPORTANT: This printed access card includes access to Enhanced WebAssign for only one term. When purchasing, Enhanced WebAssign engages students and helps them develop a deeper understanding of course concepts in Physics. If you are not sure that this is the correct access code for your course, please contact your Cengage Learning ConsultantIf you are not certain this is the correct access code for your course, please contact your Cengage Learning Consultant Enhanced WebAssign engages students and helps them develop a deeper understanding of course concepts in Physics. If you are not sure that this is the correct access code for your course, please contact your Cengage Learning Consultant.
Student Supplements and summary and focus questions with explained answers. The Study Guide also contains "Technology Plus" questions and multiple-choice "On Your Own" exam-style questions.
The more you study, the greater your success. You can make the most of your study time by accessing everything you need to succeed in one place—online with CourseMate. You can use CourseMate to read the textbook, take notes, watch videos and take practice quizzesIf you are not certain this is the correct access code for your course, please contact your instructor If you are not sure that this is the correct access code for your course, please contact your instructor.
Meet the Author
Author Bio
James Stewart
The late James Stewart received his M.S. from Stanford University and his Ph.D. from the University of Toronto. He did research at the University of London and was influenced by the famous mathematician George Polya at Stanford
The late James Stewart received his M.S. from Stanford University and his Ph.D. from the University of Toronto. He did research at the University of London and was influenced by the famous mathematician George Polya at Stanford | 677.169 | 1 |
Overview
has appended a section on Differential Geometry.
Related Subjects
Table of Contents
I Introduction: Vectors and Tensors.- Three-Dimensional Euclidean Space.- Directed Line Segments.- Addition of Two Vectors.- Multiplication of a Vector v by a Scalar ?.- Things That Vectors May Represent.- Cartesian Coordinates.- The Dot Product.- Cartesian Base Vectors.- The Interpretation of Vector Addition.- The Cross Product.- Alternative Interpretation of the Dot and Cross Product. Tensors.- Definitions.- The Cartesian Components of a Second Order Tensor.- The Cartesian Basis for Second Order Tensors.- Exercises.- II General Bases and Tensor Notation.- General Bases.- The Jacobian of a Basis Is Nonzero.- The Summation Convention.- Computing the Dot Product in a General Basis.- Reciprocal Base Vectors.- The Roof (Contravariant) and Cellar (Covariant) Components of a Vector.- Simplification of the Component Form of the Dot Product in a General Basis.- Computing the Cross Product in a General Basis.- A Second Order Tensor Has Four Sets of Components in General.- Change of Basis.- Exercises.- III Newton's Law and Tensor Calculus.- Rigid Bodies.- New Conservation Laws.- Nomenclature.- Newton's Law in Cartesian Components.- Newton's Law in Plane Polar Coordinates.- The Physical Components of a Vector.- The Christoffel Symbols.- General Three-Dimensional Coordinates.- Newton's Law in General Coordinates.- Computation of the Christoffel Symbols.- An Alternative Formula for Computing the Christoffel Symbols.- A Change of Coordinates.- Transformation of the Christoffel Symbols.- Exercises.- IV The Gradient, the Del Operator, Covariant Differentiation, and the Divergence Theorem.- The Gradient.- Linear and Nonlinear Eigenvalue Problems.- The Del Operator.- The Divergence, Curl, and Gradient of a Vector Field.- The Invariance of ? · v, ? × v, and ?v.- The Covariant Derivative.- The Component Forms of ? · v, ? × v, and ?v.- The Kinematics of Continuum Mechanics.- The Divergence Theorem.- Differential Geometry.- Exercises. | 677.169 | 1 |
Studies in Advanced Mathematics
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A Computational Introduction
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Stability, Symbolic Dynamics, and Chaos, 2nd Edition
Several distinctive aspects make Dynamical Systems unique, including:treating the subject from a mathematical perspective with the proofs of most of the results includedproviding a careful review of background materialsintroducing ideas through examples and at a level accessible to a beginning…
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The study of composition operators lies at the interface of analytic function theory and operator theory. Composition Operators on Spaces of Analytic Functions synthesizes the achievements of the past 25 years and brings into focus the broad outlines of the developing theory. It provides a…
The Heat Equation and the Atiyah-Singer Index Theorem
This book treats the Atiyah-Singer index theorem using the heat equation, which gives a local formula for the index of any elliptic complex. Heat equation methods are also used to discuss Lefschetz fixed point formulas, the Gauss-Bonnet theorem for a manifold with smooth boundary, and the…
Mathematics and Applications
Wavelets is a carefully organized and edited collection of extended survey papers addressing key topics in the mathematical foundations and applications of wavelet theory. The first part of the book is devoted to the fundamentals of wavelet analysis. The construction of wavelet bases and the fast…
Vibration and Damping in Distributed Systems, Volume II discusses asymptotic methods, including equations with variable coefficients, asymptotic estimates of eigenfrequencies of membranes and plates, WKB approximations and the wave propagation method of Keller and Rubinow, which are developed and…
Vibration and Damping in Distributed Systems, Volume I provides a comprehensive account of the mathematical study and self-contained analysis of vibration and damping in systems governed by partial differential equations. The book presents partial differential equations techniques for the…
An Introduction to Operator Algebras is a concise text/reference that focuses on the fundamental results in operator algebras. Results discussed include Gelfand's representation of commutative C*-algebras, the GNS construction, the spectral theorem, polar decomposition, von Neumann's double…
Several Complex Variables and the Geometry of Real Hypersurfaces covers a wide range of information from basic facts about holomorphic functions of several complex variables through deep results such as subelliptic estimates for the ?-Neumann problem on pseudoconvex domains with a real analytic…
Ever since the groundbreaking work of J.J. Kohn in the early 1960s, there has been a significant interaction between the theory of partial differential equations and the function theory of several complex variables. Partial Differential Equations and Complex Analysis explores the background and…
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CR Manifolds and the Tangential Cauchy Riemann Complex provides an elementary introduction to CR manifolds and the tangential Cauchy-Riemann Complex and presents some of the most important recent developments in the field. The first half of the book covers the basic definitions and background… | 677.169 | 1 |
ALEX Lesson Plans
Title: Family Ties: Parabolas
Description:
ThisStandard(s): [MA2015] AM1 (9-12) 11: (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Understand the importance of using complex numbers in graphing functions on the Cartesian or complex plane. [N-CN9] (Alabama)
Subject: Mathematics (9 - 12), or Technology Education (9 - 12) Title: Family Ties: Parabolas Description: This
Thinkfinity Lesson Plans
Title: There Has to Be a System for This Sweet Problem
Description:
In this Illuminations lesson, students use problem-solving skills to find the solution to a multi-variable problem that is solved by manipulating linear equations. The problem has one solution, but there are multiple variations in how to reach that solution.
Standard(s): [MA2015] AL1 (9-12) 21: Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. [A-REI7]
Subject: Mathematics Title: There Has to Be a System for This Sweet Problem Description: In this Illuminations lesson, students use problem-solving skills to find the solution to a multi-variable problem that is solved by manipulating linear equations. The problem has one solution, but there are multiple variations in how to reach that solution. Thinkfinity Partner: Illuminations Grade Span: 9,10,11,12 | 677.169 | 1 |
Just Plain Algebra
Date: 01/07/98 at 23:57:39
From: Ryun Patenaude
Subject: Just plain algebra
I try hard but I just don't get algebra. Do you have any advice or any
programs you might recommend?
Thank you,
Ryun Patenaude
Date: 01/12/98 at 12:30:24
From: Doctor Joe
Subject: Re: Just plain algebra
Hi Ryun,
The first thing you must learn in algebra is this golden rule:
Don't panic and always have a clear mind.
At your age, the type of algebra you encounter (correct me if I'm
wrong; you might be an expert in higher algebra such as group theory,
linear spaces, homology and topos theory) should be arithmetic
operations, the solving of algebraic equations in a unknown, usually
x, and most difficult of all, word problems that involve the
formulation of an algebraic equation.
Follow the following steps. I hope they are useful but they are by no
means exhaustive:
I am focusing on the aspect of word-problems that involve the
formulation of an algebraic equation.
Step 1:
Read the question carefully and find/underline the unknown quantity
that is involved in the question. Note that this unknown quantity
will be the one particular quantity that other unknowns depend on.
Example:
There are 3 pieces of wire. The length of the first is 20 percent of
the length of the second, and the length of the third is 100 percent
of the length of the second.
In this question, clearly the length of the second piece of wire is
the desired unknown on which the others depend.
Step 2.
Let x (or any symbol you like) be the unknown quantity.
Step 3.
Define the other quantities in terms of the unknown.
This may prove to be the most difficult step. My advice is to try to
imagine you already know the value of x. Then your job becomes finding
the other quantities as if you know what x is, and you don't need to
simplify those expressions in terms of x.
Step 4.
Form the final equation. Usually, this comes in the form of a total or
a final statement in the question.
Example:
In the previous example, if it is further given that the 3 pieces of
wire total 23 cm in length, then the equation will be
0.2x + x + 1.1x = 23
Step 5.
Simplify expressions in x (this you can practice by doing more
simplification of expression exercises).
Step 6.
Add, subtract, multiply or divide by suitable numbers on both sides of
the equation one step at a time.
Look at the following examples and you'll know what I mean by Step 6.
It is more meaningful this way:
Example:
Suppose we have the equation:
2x + 3 = 4 - 5x
Ask yourself: Isn't it more systematic if we group things of the
same type together? (i.e. the unknowns with the unknowns and the
known with the known).
How do we make this happen? We see a 3 on the left; suppose we
subtract 3 from the quantities on both sides? The equality still
holds, so we have
(2x + 3) - 3 = (4 - 5x) - 3
Then, 2x + 3 - 3 = 4 - 5x -3
2x = 4 - 3 - 5x
2x = 1 - 5x
Likewise, add 5x on both sides,
2x + 5x = 1 - 5x + 5x
7x = 1
Now, to eliminate the outstanding quantity 7 and make x stand on its
own (so to speak), we multiply by 1/7 on both sides:
(1/7)*7x = 1/7 * 1
(1/7 * 7) x = 1/7
1 * x = 1/7
x = 1/7
I hope this helps you understand algebra.
-Doctor Joe, The Math Forum
Check out our web site! | 677.169 | 1 |
Full text
The above pdf files were produced by Texshop on a Mac. In case you have trouble seeing the equations (e.g. you just see a lot of dots) you might try the following alternate Acrobat version of the entire text.
Hard copies
You can buy a printed paper copy of this material from Copiesmart on University Boulevard. | 677.169 | 1 |
This course will discuss the basic notions
of Euclidean and Non-Euclidean geometry, in two dimensions.
We will focus on the three main examples:
the Euclidean Plane, the Projective Plane, and the Hyperbolic Plane.
Each will take up approximately one-third of the course material.
In the discussion of each of these three geometries,
we will begin with the synthetic approach,
via an axiomatic treatment.
Then we will turn to the introduction of coordinates in each geometry.
Finally we will discuss changes of coordinates in each geometry,
leading to the groups of automorphisms of each geometry.
Tesselations or tilings in each geometry will cap each unit.
This course was taught using a modified Moore method.
Each day the course met, I handed out a list of definitions
and theorems to be proved.
During the next class period, students were called on in turn
to present solutions to the problems and proofs of the theorems.
If a student was called on to present a particular problem,
he or she was responsible for writing the solution to hand in.
At the end of the class period, I handed another sheet of
definitions, problems, and theorems.
I have taught this year-long course every other year
since 1987.
It varies slightly from one incarnation to the next.
The study of Projective Geometry has its roots
in the art and science of the late middle ages.
The basic objects of study in Projective Geometry
is the same as in all geometries: points, lines, planes, conics, etc.
However Projective Geometry systematically introduces
the concept of infinity into the geometric universe,
thereby ``compactifying'' the ordinary space R^n (or C^n)
by adding points at infinity.
This first semester course covers the basic theory of Projective Geometry
over fields, concentrating on the complex numbers
(although much of the first part of the theory works over any field).
The prerequisites are a good understanding of linear algebra
and a little abstract algebra (notion of a ring and ideal is enough).
Don't be scared by the high number:
M566 should be plenty as a prerequisite.
We start by covering the linear theory of projective space
(points, lines, planes, etc.)
and then some of the non-linear aspects:
conic curves, quadric surfaces, cubic curves, etc.
Fall Syllabus:
Affine Geometry and Projective Geometry
Definition of Projective Space
Homogeneous Coordinates in Projective Space
Linear Subspaces, Points, Lines, Planes, Collinearity Conditions
Schubert Conditions, Changes of Coordinates
Conics and Quadrics
Curves in the plane, degree
Classification of conics
Quadric surfaces and hypersurfaces, classification by rank
Varieties in projective space
Homogeneous polynomials, the homogeneous coordinate ring
The Ideal of a projective variety
Rational functions
Cubic plane curves, classification, singular cubics
Smooth cubics, Weierstrass form, the group law
Cubic surfaces and the 27 lines
We have used "Undergraduate Algebraic Geometry", by Miles Reid,
London Mathematical Society Student Text No. 12, Cambridge University Press
as a text for the fall semester.
In 1993 we used "Algebraic Geometry" by Joe Harris,
Graduate Texts in Mathematics No. 133, Springer-Verlag.
Spring Syllabus:
This has varied somewhat,
depending on the interests and level of the students.
In 1996 we read chapters of
"Basic Algebraic Geometry I: Varieties in Projective Space",
by I. R. Shafarevich, Springer-Verlag.
In 1994 we continued in Joe Harris' book in the second semester.
In 1992 I taught a course in Algebraic Surfaces, with no text.
This year-long sequence is the basic graduate algebra course
in our department. The Algebra Qualifying Examination is based
on these two semester courses. Several texts have been used
in the past few years, including Herstein, Lang, and Hungerford.
In 1993-4 I used: | 677.169 | 1 |
An extension of the Easy Peasy All-in-One Homeschool
Algebra 1
Note: CK12 is offering free tutoring done by college students. We are starting with an EP group with them where you can post your questions any time Here's the info: You will need a create a free CK12 account then join the Easy Peasy Tutoring Group here. (You may be asked to enter the code, heiv7 .) Click on Q&A to ask a question. When you ask, include the question you are having trouble with. You should get an answer within 24 hours. (Remember in an online forum, never give out personal information.)
Course Description: Students will engage in real world and hands-on problem solving while using their developing skills in algebra. Students will learn new material through animations, videos, reading, and guided practice. The topics covered in this course include: real numbers, algebraic expressions, graphing to solve inequalities and absolute value, graphing to solve linear equations, systems of equations, factoring polynomial equations, relations and functions, quadratic equations, radical and rational expressions and equations, and probability. Students will also do timed PSAT practice questions.
Remember that the topic text is always available for reminder, review, help, or to learn the lesson if that is easier for you than the video.
Day 3
Do the warm up, presentation, worked examples as necessary and practice for solving equations.
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
Day 4
If you want to move more quickly, DO NOT do the review on the same day as the lesson. The point of putting the review on the next day is to try to force the information into your longer term memory instead of using it only in your short term memory enabling you to easily forget it and leave it behind. If you want to move on, after a review day, you can start the next lesson. You could do the whole thing or just watch the presentation. Then the next day you could use the text and worked examples. You can find what groove works best Connect Four with equations. Click on two-step equations. If you aren't sure of how to get the right answers, go back and choose one-step equations to practice easier problems before coming back to two-step equations. (Here's a walk through of one step equations.)
Almost every other day is a review day. It gives two days to learn each new lesson. Make sure you get 100% on the lesson before you move on. There will also be another review activity on these days to have more practice on other topics from the course.
Early in the course these days will go by quickly because it is probably easyRead the lesson on properties of addition and do the guided practice. Then check your answers. (right below it)
Day 11
Do the warm up and presentation for absolute value. Use the worked examples and topic text to help you.
Complete the practice.
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
Review as directed if you missed any. Then you may retake the review at this point for a new score. I don't need to keep saying it do I? You know the rules? You may add a half point for any newly correct answers if you can correctly solve them now and haven't just cheated.
Read the lesson on properties of multiplication and do the guided practice. Then check your answers.
Complete the design a roller coaster project. Make sure you read the grading rubric before you begin. (There are links on the page, but they don't work. Here's one link with some insight into roller coasters and functions, but you can also do some of your own research if you need to. Problem three says to use glogster for your poster, but you can present your information however you like. Glogster is no longer free. If you would like to try another online poster site, here's one, Prezi.)
Sample schedule
Day 1: Complete number one. Look at all of the links on the page.
Day 2: Design your roller coaster for number 2 and outline paper.
Day 3: Write your paper.
Day 4: Create your poster and present it. Keep in mind that there is a grade for presentation.
This is just a sample schedule. If you need lots of time to write, then you should design your roller coaster on the first day to leave two days for writing. If you don't finish in four days, you will lose lots of points. You will score poorly. You'll also have to finish over the weekend! No getting behind.
Use the rubric to score your project out of 8.
Add 10 points for completing on time.
Add 5 points for your paper having a clear intro, at least three points and a conclusion. Take off a point for each of those things that are missing.
Do the warm up and presentation for parallel lines. Use the worked examples and topic text to help you.
Complete the practice.
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
STOP
Time for a report card and portfolio/records updating I hope you are following the rules. You may add a half point for any newly correct answers if you can correctly solve them now and haven't just cheated.
Try to get at least 4 girls/women and four boys/men, including yourself.
Complete parts 1 and 2.
Day 56
Complete your height project from Day 55.
Take graph paper and put a project title on it.
Make two tables of data, one for the boys and one for the girls.
Graph the three equations for each set of data. You can make two graphs (one for boys and one for girls) and use different colors for the three different equation lines.
Use your graphs to predict the heights of your parents (or other adults you measured). Measure their real heights and see how close they came.
Write a paragraph summary of what you did, what seems to be the best equation for predicting height and why.
Score up to 5 points for two tables and three equations.
Score up to 5 points for your paragraph. Make sure you followed all of the directions.
Record your score out of 10.
You can include this in your portfolio.
Day 57
The SAT is a test you'll take in 11th grade. It is required by colleges. You will need a good score to show the college of your choice that you will be a good student. A good score also shows that you've been learning something and not just home playing video games.
In 10th grade you can take a practice test called the PSAT. Some schools give full scholarships to students who score very high. That could save your parents $100,000! So do your best 🙂
When you take the PSAT or SAT, you need to know how to play the game. It's a bit of a game and knowing the rules will help you win.
You get one point for each correct answer. You get zero points for anything left blank. You lose a quarter of a point if you get one wrong. So it's not a good idea to just guess. If you can eliminate at least one answer, then your odds of guessing the right one increase and statistically speaking, it's in your favor to guess. If you can eliminate two or more of the answers, then you really should guess at the answer. Of course it's best to know the answer!
NOTE! As of 2015 the PSAT is eliminating the penalty for a wrong answer. So it is best to guess if you don't know! Don't leave any answers blank.
NOTE! As of 2016 the SAT is eliminating the penalty for a wrong answer. That means you should never leave an answer blank. Just guess if you don't know.
Here's another reason not to cheat. The truth has a way of making itself known. If your PSAT/SAT scores don't match up with your grades, everyone will know something is up. Just copying answers won't help you learn anything and it will eventually show if you aren't learning.
Day 58
The PSAT and the SAT are also timed tests. You have to stop when time is up. Give it a try.
Give yourself twenty minutes to complete this test. Read the directions below before you begin.
Get the timer ready and then begin. If you have time left over, go back and look over your answers to make sure you are happy with them.
Don't flip out if you feel like you don't know something. Use your brain. It talks about scatter plots and trend lines. Maybe you have no idea what those are. But look at the question. It says estimate. You know what that is. It has a graph. You can estimate where the "trend" is headed. Look at the answers. Only one is reasonable in relation to the graph.
Review as directed if you missed any. Then you may retake the review at this point for a new score.
Take the quiz on solving absolute value equations. The access code is always written in the sentence under the title. (That link wasn't working. Here's the alternate quiz. Still score up to five points for up to five correct answers.)
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
Day 79
Do the review from solving systems of linear equations by elimination.
Record your score out of 5.
Review as directed if you missed any. Then you may retake the review at this point for a new score.
Take the quiz on solving systems of equations using the elimination method. The access code is under the title. Read the fine print. (This quiz isn't working right now, so here's an alternate quiz. Take it repeatedly if necessary. Do five problems and get a point for each "x" and "y" solved for correctly.)
Record your score out of 10.
Day 80
Do the warm up and presentation for rate problems. Use the worked examples and topic text to help you.
Complete the practice.
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
Do these three sets of review exercises. Work to get a perfect score. Use the links to go over the material again if necessary. Topics: Non-linear functions, perpendicular lines, solving and graphing linear inequalities in two variablesSelect three rectangles to measure. After measuring the length and width of each, use the Pythagorean Theorem to calculate the length of the diagonal. Then measure the actual distance of the diagonal. How accurate were your calculations compared to your measurements?
Choose at least two bigger rectangles like a table or door. (Don't just do three books.) Make sure you are able to measure the diagonal. You could use string to measure and then measure the string. (This activity idea is from the NROC course.)Find the equation of the line. (The slope is rise/run. Say, "Rise over run, we're gonna have fun." 'Cause math is fun like that.) Read the instructions and make sure you start the game. I tried to play without starting the game and it didn't work. Also, you must play multiple levels. No stopping after the first easy level "We Can Work It Out." (From NROC algebra course: If it takes Bob 3 hours to paint a room and it takes Jeff 5 hours to paint the same room, how many hours would it take if the two painters work together? Many students will attempt to take an average and come up with an answer of 4 hours, but 4 hours doesn't make sense. Why would it take longer than Bob working alone? Surely the two painters would complete the room in less than 3 hours. Using the algebra work model, we discover that the room should be painted in slightly less than two hours.)
Review as directed if you missed any. Then you may retake the review at this point for a new score.
Take the quiz on the properties of numbers. You might want to copy down the key before you begin! The number that is a symbol is Pi. Remember that the access code is in the sentence under the title. (This quiz isn't working. Here's an alternative quiz. No grade.)
Day 164
Do the warm up and presentation for deductive reasoning. Use the worked examples and topic text to help you.
Complete the practice.
Record your score out of 6 for the practice problems. You get a point for any you get correct before it shows you the answer.
Figure your course grade. Enter on your fourth quarter grading sheet your total score for each quarter. Divide by the total score from all four quarters. That can be your grade, but I also think you can award up to half of the grade for completing the daily assignments. Then you would take the grade you just calculated, divide it in half and add it to 50, or whatever grade you deem appropriate. Example of the scoring calculation: | 677.169 | 1 |
The following unit, which has been
planned and ready to be implemented into an eighth grade Algebra I class,
deals with systems of equations. Students will see how two functions
with the same variables will be able to work together and solve for the two
unknown variables. The students will have to use many procedures to
come up with the answers. Graphing, adding, subtracting, solving for
a variable, and using multiples will be just some of the skills that they
should have already mastered to work through this unit.
This unit is probably one of the more
life-applying units that the students will have. In turn the teacher
will be able to answer the age-old question, "When are we ever going to have
to use this?" with more ease and plenty of examples. The
material will pull in the students because there will many different and
exciting activities that will show students that they don't have to answer
one problem the same way. Numerous methods will be applied to the same
question but will all result in the same answer. This lesson
plan will fit into the curricular recommendations made by the NCTM/Illinois
Learning Goals in that students will continue to expand their knowledge in
algebra and working with and understanding functions in more complex ways.
The use of graphs, calculators, and other manipulatives will help the student
to see what exactly a problem is doing and how to better solve it.
By the end
of this unit a student should be able to recognize a system of equations
and be able to solve it using either a graph, substituting for one variable,
eliminating a variable through addition, subtraction and multiplication.
The students should also be able to be given a word problem and create a
system on their own. The unit project will get them ready to be able
to create problems on their own. This is done to help students work
and manipulate with systems. | 677.169 | 1 |
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