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Partial Differential Equations And Mathematica, 1997

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ISBN 0-8493-7853-2, Kythe Prem K.
Acquiring Editor: Tim Pletscher
Senior Project Editor: Susan Fox
Cover Design: Denise Craig
Typesetting: TechType Works
Marketing Manager: Susie Carlisle
Direct Marketing Manager: Becky McEldowney
Library of Congress Cataloging-in-Publication Data
Kythe, Prem K.
Partial differential equations and Mathematica / Prem K. Kythe,
Pratap Puri, Michael R. Schaferkotter. p. cm.
Includes bibliographical references and index.
ISBN 0-8493-7853-2 (alk. paper)
1. Differential equations, Partial. 2. Mathematica (Computer file)
I. Puri, Pratap, 1938- II. Schaferkotter, Michael R., 1955- III. Title.
QA374.K97 1996 96-36946
515'.353—dc21 CIP
This book contains information obtained from authentic and highly regarded sources. Re­
printed material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use.
Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any infor­
mation storage or retrieval system, without prior permission in writing from the publisher.
CRC Press, Inc.’s consent does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from CRC Press for such copying.
Direct all inquiries to CRC Press, Inc., 2000 Corporate Blvd., N.W., Boca Raton, Florida 33431.
© 1997 by CRC Press, Inc.
No claim to original U.S. Government works
International Standard Book Number 0-8493-7853-2
Library of Congress Card Number 96-36946
Printed in the United States of America 1 2 3 4 5 6 7 8 9 0
- · - - - * — - w i d - f r e e p a p e r
I
Contents
P r e f a c e, vii
I n t r o d u c t i o n t o M a t h e m a t i c a, xiii
C h a p t e r 1: I n t r o d u c t i o n, 1
1.1. Notation and Definitions, 1
1.2. Initial and Boundary Conditions, 3
1.3. Classification of Second Order Equations, 4
1.4. Some Known Equations, 7
1.5. Superposition Principle, 17
1.6. Exercises, 19
C h a p t e r 2: M e t h o d o f C h a r a c t e r i s t i c s, 22 First Order Equations, 22
2.1. Linear Equations with Constant Coefficients, 22
2.2. Linear Equations with Variable Coefficients, 28
2.3. First Order Quasi-linear Equations, 33
2.4. First Order Nonlinear Equations, 38
2.5. Geometrical Considerations, 46
2.6. Some Theorem on Characteristics, 47 Second Order Equations, 49
2.7. Linear and Quasi-linear Equations, 49
2.8. Exercises, 56
iii
C h a p t e r 3: L i n e a r E q u a t i o n s w i t h C o n s t a n t C o e f f i c i e n t s, 6 3
3.1. Inverse Operators, 63
3.2. Homogeneous Equations, 72
3.3. Nonhomogeneous Equations, 74
3.4. Exercises, 79
C h a p t e r 4: O r t h o g o n a l E x p a n s i o n s, 8 2
4.1. Orthogonality, 82
4.2. Orthogonal Polynomials, 85
4.3. Series of Orthogonal Functions, 88
4.4. Trigonometric Fourier Series, 90
4.5. Eigenfunction Expansions, 103
4.6. Bessel Functions, 106
4.7. Exercises, 110
C h a p t e r 5: S e p a r a t i o n o f V a r i a b l e s, 1 1 7
5.1. Introduction, 117
5.2. Hyperbolic Equation, 118
5.3. Parabolic Equation, 125
5.4. Elliptic Equation, 132
5.5. Cylindrical Coordinates, 142
5.6. Spherical Coordinates, 144
5.7. Nonhomogeneous Problems, 146
5.8. Exercises, 152
C h a p t e r 6: I n t e g r a l T r a n s f o r m s, 1 6 0 Laplace Transforms, 162
6.1. Notation, 162
6.2. Basic Laplace Transforms, 164
6.3. Inversion Theorem, 183
6.4. Exercises, 189 Fourier Transforms, 195
6.5. Fourier Integral Theorems, 195
6.6. Properties of Fourier Transforms, 196
6.7. Fourier Sine and Cosine Transforms, 204
6.8. Finite Fourier Transforms, 210
6.9. Exercises, 213
C h a p t e r 7: G r e e n ’s F u n c t i o n s, 2 1 7
7.1. Definitions, 217
7.2. Parabolic Equations, 231
7.3. Elliptic Equations, 238
7.4. Hyperbolic Equations, 244
iv CONTENTS
CONTENTS
7.5. Applications, 249
7.6. Exercises, 252
C h a p t e r 8: W e i g h t e d R e s i d u a l M e t h o d s, 266
8.1. Line Integrals, 267
8.2. Variational Notation, 270
8.3. Multiple Integrals, 273
8.4. Weak Variational Formulation, 274
8.5. Galerkin Method, 279
8.6. Rayleigh-Ritz Method, 283
8.7. Choice of Test Functions, 287
8.8. Transient Problems, 292
8.9. Other Methods, 295
8.10. Exercises, 296
C h a p t e r 9: P e r t u r b a t i o n M e t h o d s, 305
9.1. Taylor Series Expansions, 306
9.2. Successive Approximations, 309
9.3. Boundary Perturbations, 311
9.4. Exercises, 314
C h a p t e r 10: F i n i t e D if fe re n c e s, 321
10.1. Finite Difference Schemes, 321
10.2. First Order Equations, 325
10.3. Second Order Equations, 328
10.4. Exercises, 343
A p p e n d i x A: G r e e n ’s I d e n t i t i e s, 350 A.l. Green’s Identities, 350 A.2. Exercises, 352
A p p e n d i x B: T a b l e s o f T r a n s f o r m P a i r s, 3$4 A p p e n d i x C: G l o s s a r y o f M a t h e m a t i c a F u A p p e n d i x D: M a t h e m a t i c a P a c k a g e s a n d
n c t i o n s, 361 N o t e b o o k s, 368
B i b l i o g r a p h y, 369
I n d e x, 373
A u t h o r s
Prem Kishore Kythe (Kulshrestha), born in India, 1930; Naturalized U.S. citizen; received Ph.D. (Math) from the Aligarh Muslim University, India, in 1961; Professor of Mathematics at the University of New Orleans since 1974; Author of A n I n t r o d u c t i o n t o B o u n d a r y E l e m e n t M e t h o d s (CRC Press, 1995); F u n d a m e n t a l S o l u t i o n s f o r D i f f e r e n t i a l O p e r a t o r s a n d A p p l i c a t i o n s (Birkhauser, 1996); Over 40 research publications in the areas of univalent functions, boundary value problems in continuum mechanics, differential equations, Laplace transform, wave theory, wave structure in rotating flows; Numerous citations in research articles and graduate text/monographs; Listed in American Men and Women of Science.
Pratap Puri, bom in India, 1938; Naturalized U.S. citizen; got Ph.D. (Math) from the Indian Institute of Technology, Kharagpur, India, in 1962; Professor of Mathematics at the University of New Orleans since 1976; Over 40 research articles published in boundary value problems in continuum mechanics and integral transforms; Numerous citations in research articles; Listed in American Men and Women of Science.
Michael Richard Schaferkotter, born in Mississippi, 1955; U.S. citi­
zen; M.S. (Math) from the Louisiana State University in Baton Rouge, 1984; Instructor in Mathematics at the University of New Orleans, 1988-1995; cur­
rently working at the Sverdrup Technology, Inc., Stennis Space Center, Mississippi. Two research articles published in topological semigroups and linear programming.
Preface
Overview
Why another introductory textbook on partial differential many are already available? The question is rightly ask is in order.
equations when so ed, and the justification
We have found that every year an increasing nu advanced courses involving boundary value problems, numerical techniques, such as finite difference, finite element methods. At the same time they are introduce ι equations as graduate students, although a few do m knowledge of the subject through other courses in en It is a pity than an opportunity to leam this subject is lost, because the students encounter textbooks that graduate level courses. Even when the textbooks are wi students, quite often that may not be the case for a maj of these textbooks, though written with quality materi on hard analysis. Our textbook, written for a two-sei at attracting junior and senior undergraduate student;; training in the subject and do not miss out on elenm simple beauty of the subject. In the pedagogical spirit avoided the extreme situation where a beginner’s cou
mber of students enter which deal mostly with element, or boundary d to partial differential inage to acquire some gineering and physics, at undergraduate level are graded strictly for xtitten for undergraduate ij ority of students. Most al, are generally based mester course, is aimed so they get an early entary techniques and of moderation we have rse is so advanced and
viii
PREFACE
severe that it is likely to break the spirit of even mature students in an attempt to cover practically everything in the subject. On the other hand, one should encourage textbooks on this subject which in pace and thought are graded to undergraduate levels.
Accordingly, the authors have striven to produce a beginner’s textbook which is mature, challenging, and instructive, and which, at the same time, is reasonable in its demands. Certainly, it is not claimed that partial differential equations can become easy and effortless. However, the authors’ combined classroom experience over a number of years justifies the effort that the subject can be made reasonably easy to understand despite its complexity, provided that the student has a thorough background in multivariate calculus and ordi­
nary differential equations. It can impart understanding and profit even to the undergraduate juniors and seniors who take it only for one semester before their graduation. The goal, then, has been to produce a textbook that provides both the basic concepts and the methods for those who will take it only for a semester, and a textbook which also provides adequate training and encour­
agement for those who plan to continue their studies in the subject itself or in applied areas. The distinctive features and the scope of the book can be determined from the table of contents.
Audience
Most of the material in this textbook, especially the first six chapters, is devel­
oped for a beginner’s course on partial differential equations. These chapters are designed primarily for junior/senior level undergraduate students in math­
ematics, physics, and engineering who have completed at least the courses on multivariate calculus and ordinary differential equations, and possess some working knowledge of Mathematica in case they opt to use its versatility in symbolic manipulation and graphics capabilities. Adequate material on other topics from mathematical analysis is provided in the text as and when needed.
The book represents a two semester course. The first six chapters can be taught at the earliest after the completion of multivariate calculus and ordinary differential equations, while the remaining part definitely requires some degree of maturity. An important consideration at this point is the need for engineering and physics majors to learn the subject at an earlier stage. In most cases they start using partial differential equations and their solutions prior to any formal training in the subject. As a result, their understanding of applied technical areas is hampered by the lack of familiarity with the theory and methods of
PREFACE
partial differential equations. It is our hope that ma engineering majors will take this course at the beginn and thus leam and enjoy other technical subjects with
thematics, physics and ing of their junior years better understanding.
The book provides a comprehensive and systemat: theory and applications that can readily be followed by at the junior or senior level.
:ic
coverage of the basic undergraduate students
Salient Features
This textbook has evolved out of lecture notes developed while teaching the course to undergraduate seniors, and graduate students in mathematics, engi­
neering, and physics at the University of New Orleans during the past three years. Much effort has gone into the organization of the subject matter in order to make the course attractive to students and the textbook easy to read. Although there is a large number of classical textbooks available on the sub­
ject, there has been a need for an introductory textbc
This publication, based on classroom experience, fulfills such a need. The
for the average student eory and techniques of
mathematical contents of the book are simple enough to understand the methodology and the fine points of th partial differential equations. The Mathematica component has been presented in detail to bring out the salient features of different methods. The chapters present a balanced two-semester course material, which can be tailored to the needs of different levels of instructions. The following table outlines some suggested curricula at the three levels.
Elementary/Juniors Beginning Seniors Beginr
Chapter 1 Chapter 3 Chapter 4 Chapter 5 Chapter 6
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5
Chapte
Chapte
Chapter 8
ok with Mathematica.
ing Graduates r 2 r 7
Chapte
Chapte
r 9
r 10
Chapter 6
X
PREFACE
Chapter 1 provides some useful definitions, classification of second order partial differential equations, some well-known equations, and the superposi­
tion principle. The method of characteristics for first and second order partial differential equations is studied in Chapter 2. This topic is usually ignored in most of the textbooks, or delayed toward the end of the book. However, it is our opinion and experience that it helps the students understand the nature of the solutions and form a guide for higher order partial differential equations, provided the topic is handled with clarity and ample geometrical presenta­
tions. Mathematica is found to be very useful in achieving this perspective. An old technique of inverse operators, borrowed from the theory of ordi­
nary differential equations, has been used in Chapter 3 to solve homogeneous and nonhomogeneous partial differential equations with constant coefficients. Chapter 4 puts together the concepts of orthogonality, orthonormality, or­
thogonal polynomials, series of orthogonal functions, trigonometric Fourier series, eigenfunction expansions, and the Bessel functions. This material is needed in Chapter 5 which deals with the method of separation of variables for boundary and initial value problems. These problems involve the wave, heat and Laplace equations, with homogeneous and nonhomogeneous bound­
ary conditions, in the Cartesian, polar cylindrical and spherical geometries. The integral transforms, especially the Laplace and Fourier transforms, are presented in Chapter 6. These techniques are powerful tools to solve different types of boundary value problems with initial conditions.
The advanced material consists of the Green’s functions (Chapter 7); weighted residual methods based on the theory of the variational calculus (Chapter 8); perturbation methods (Chapter 9) applied to problems involving partial differential equations only; and lastly the numerical methods based on finite differences, where Mathematica unfolds the intricate details and the beauty of these methods. We have decided not to include other numerical methods, like the finite element and boundary element methods, which are now fully developed into individual courses with many fine textbooks.
Although the text is rich in developing the underlying mathematical anal­
ysis with sufficient theorems and proofs, the emphasis is basically on the de­
velopment of methods. A large number of examples in every chapter presents the techniques that are representative of virtually every concept in the book. There are over 130 examples solved with meticulous detail. Besides, there are over 170 exercises, spread chapterwise throughout the book. Unlike most textbooks, the answers, hints, or sometimes detailed solution of all exercise are provided on the spot. The authors feel that this will enhance the interest of both the students and the instructors in the subject.
PREFACE
xi
The icon | www | highlights the Mathematica mat and details are provided on the spot as how Mathemat: tive situations, or the reader is directed to the correi web site, which are identified by the chapter number, example number, with platform-independent present;
rial throughout the book, ica works in the respec- sponding material on the , section number, and/or ation.
Besides a general introduction to Mathematica in to Chapter 1), which describes the Mathematica style Mathematica functions, a glossary of Mathematica fu: are provided in Appendix C.
the very beginning (prior and important concepts, notions used in the book
Although Mathematica occupies a major portion in the form of notebooks and packages available on the CRC web server as mentioned below, it opens up an opportunity to use the symbolic manipulation and graphics facilities The textbook is, however, independent of the Mathe be used with the same ease and advantage without M is not available at an institution.
matica interface and can athematica if this facility
til;
A word of caution: It has been our experience enthralled with Mathematica that they fail to learn a underlying theory. This is our hope that the instructor: entrapment by an excessive infusion of the technolog; course material.
at some students get so nalytical techniques and s will avoid this kind of ;y into their introductory
Mathematica Interface
Mathematica packages,with detailed instructions, ard Wide Web http://www.crcpress.com/books/isbr
, sequel this web site is referred to as the CRC web s Mathematica material can be also found in Append! the book. This material requires at least Mathematic interface based on Mathematica version 3.0 shall be web site soon after this version is marketed.
available via the World /0-8493-7853-2. In the erver. Details about the ix D toward the end of a 2.0. The Mathematica available on the above
At times the reader will find that Mathematica cqi is a little awkward to read. This is due to the priori situations can be found in long Mathematica session but do not exist in the Mathematica files and Noteboo1 web server.
de embedded in the text ity for typesetting. Such s presented in the book, ks available on the CRC
xii
PREFACE
Acknowledgments
We thank some of our colleagues for their encouragement. The help provided by our students, and especially by Mr. Pedro M. Jordan, is gratefully acknowl­
edged. We also thank Ms. Fleur F. Simmons for proofreading the final copy of the manuscript, and to persons at the TechType Works, Inc., Gretna, LA, for typesetting the manuscript.
Kythe, Puri, and Schaferkotter New Orleans, Louisiana
I n t r o d u c t i o n t o M a t h e m a t i *
z a
Introduction
Mathematica is a powerful mathematical progran ematica provides numeric, symbolic, and graphi one in the mathematical aspects of problem sol\ been found for Mathematica in investigating and neering, mathematics, and physics, as well as ecc Mathematica can also be used as a high level pro ematica will run on most of the major platforms, to desktop systems and laptops.
Mathematica is comprised of two parts, cal Kernel. The Kernel is the computation engine, wh The Front End takes the form of either a noteboo End) or a command line interface, allowing the us Kernel.
Mathematica Notebooks allow one to add note sions to the work in a similar fashion to a word p will be able to import and export graphics in mo tations can be prepared and also documents for e may be cut and pasted within notebooks or bet\ reuse or modify text, graphics, and calculations.
iming environment. Math- cal tools in order to assist ing. Significant uses have inalyzing problems in engi- nomics and other sciences, 'ramming language. Math- from Cray supercomputers
led the Front End and the ch does all the calculations, c interface (advanced Front er to communicate with the
s, explanations, and conclu- ocessor. In version 3.0 one >t graphic formats. Presen- lectronic publishing. Items /een notebooks in order to
xiv
INTRODUCTION TO MATHEMATICA
The Notebook Front End is a file that organizes text, graphics, and cal­
culations in cells. Many different kinds of cells comprise a Mathematica Notebook. Input cells contain Mathematica commands and may be evaluated by pressing SHIFT-RETURN or simply RETURN when the cell is selected. Text cells contain text information and are not evaluated by the Kernel. Graphic cells contain pictures of plots and graphs.
Cells can be formatted with various attributes, such as the font to use in displaying text, font size, and color. The cells may be grouped in an outline fashion in order to organize a document into sections containing titles, headings, and subheadings.
A Mathematica Notebook can be transferred from one platform to another without losing information or formatting, though some consideration should be given in order to successfully port the notebooks. For example, one should use the Uniform Style command to eliminate font variation within cells when translating a Macintosh Notebook for Windows. For more information on this subject, see the note “Notebook Conversion Tips”, which is available on MathSource (see section below). The Notebooks may also be sent via electronic mail since the notebooks are ASCII text files.
Conventions
Reserved words in the Mathematica programming environment always begin with a capital letter. The arguments of functions are delimited by brackets ([ ]), while parentheses are used to effect grouping. Lists, which are the primary data structure of Mathematica are delimited with braces ({ }) with the elements of the list separated by commas.
Certain symbols should be pointed out. The multiplication symbol is represented by * or by a space as in a * b or a b. The symbol = stands for substitution, as in t = 1, while equal is denoted by = =, as in Equal(x,i] or x = = t yields True only if x and t have the same value. The Mathematica command Not can be rendered !, as in a:! = f, which is True if x and t do not have the same value. The last input is denoted % while "/0n stands for I n [ n ], which is the input cell number n. Finally, never type the prompt I n [n] : = that begins each line. Mathematica automatically puts the I n and Out prompts. Type only the text that follows the I n prompt.
GETTING STARTED
XV
Getting Started
After the program is running either a command line i Notebook will appear. To begin, one just types and I\ characters into an input cell. With cursor in the inp key or SHIFT-RETURN keys together to evaluate the on output cell as shown below. The Mathematica Nc found on the CRC web server mentioned in the Pr< new users. The notebook provides explanation and ( started.
iterface will appear or a lathematica will put the ut cell, press the ENTER input cell and generate tebook Intro2Mma. ma, -face, is a notebook for samples in order to get
In[l]:=
1+1
Out[l]=
2
In[2]:=
x := T a b l e [ S i n [ k ],{ k,l,5 } ]//N
X
Out[3j—
{0.841471, 0.909297, 0.14112, -0.75681 In[3]:=
P l o t [ S i n [ x ],{ x,0,2 P i } ]
Out [4]=
12, -0.958924}
xvi
INTRODUCTION TO MATHEMATICA
- G r a p h i c s -
Occasionally, it will be necessary to interrupt or abort a Mathematica calculation. On most systems there is a command key sequence to interrupt and abort. For GUI (Graphical User Interface) systems, there is a menu choice, under the Action menu, that will enable an interrupt or abort of a calculation.
File Manipulation
Mathematica can be used for file manipulation on many different computers as well as non-Unix systems. The advantage is that one never has to learn the file manipulation commands of different computer systems. A few of the com­
mands for file and directory manipulation follow. See the book Mathematica (Wolfram, 1991) for other commands and examples.
In [4]:=
(* G iv e t h e c u r r e n t wo r ki ng d i r e c t o r y *) D i r e c t o r y []
Out[5]=
M a c i n t o s h HD:A p p l i c a t i o n s:M a themati ca 2.2
In[5]:=
ORDINARY DIFFERENTIAL EQUATIONS ( * L i s t a l l f i l e s i n t h e c u r r e n t wo r ki ng d i r e c t o r y *) F i l e N a m e s []
Out[6]=
{ M a t h e m a t i c a, M a themati ca K e r n e l, Ma.thLive, P a c k a g e s } In[6]:=
(* L i s t a l l p a c k a g e s c o n t a i n i n g ‘ ‘ PI e l s o f s u b d i r e c t o r i e s b e l o w t h e c u r r e n t
F i l e N a m e s ["* P l o t *.m", 2]
o t ’ ’ i n t h e two l e v - d i r e c t o r y *)
In[7]:=
{ P a c k a g e s:G r a p h i c s:C o n t o u r P l o t 3 D.m,
P a c k a g e s:G r a p h i c s:F i l l e d P l o t.m,
P a c k a g e s:G r a p h i c s:I m p l i c i t P l o t.m,
P a c k a g e s:G r a p h i c s:M u l t i p l e L i s t P l o t.m P a c k a g e s:G r a p h i c s:P a r a m e t r i c P l o t 3 D.m P a c k a g e s:G r a p h i c s:P l o t F i e l d.m,
P a c k a g e s:G r a p h i c s:P l o t F i e l d 3 D.m,
P a c k a g e s:M i s c e l l a n e o u s:W o r l d P l o t.m,
P a c k a g e s:ProgrammingExamples:P a r a m e t r i c P l o t 3 D.m}
Ordinary Differential Equations
Differential equations are used in many areas c to study processes that are continuous in space or command D So l v e computes solutions to ordinary well as systems of ordinary differential equations, differential equations.
f natural science in order time. The Mathematica differential equations, as and also first order partial
DSo l v e is a collection of algorithms that allows Mathematica to solve a wide range of equations. Mathematica can solve various types of equations
xviii
INTRODUCTION TO MATHEMATICA
including linear homogeneous and inhomogeneous equations, second order variable coefficient equations, second order non-linear equations, and first or­
der partial differential equations. A numerical approximation may be obtained using the N D S o l v e function.
In[8}:=
D S o l v e [ y'[ x ] == y [ x ],y [ x ],x ] Out[7]=
{ { y [x] - > EXC [ 1 ] } }
A question that very often appears at this stage is, “How can the results from DSolve be used in other calculations?” The answer is that the output from Solve, NSolve, DSolve, or N D S o l v e is a list in which each element is a list of rules. One can assign a name to the solution and easily check that the solution is correct.
In[9}:=
f [ x j := y [ x ]/.F i r s t [ “/.] f C x ]
O u t [ 9 ] =
EXC [ 1 ]
I n [ 1 0 ]:=
f'[ x ] = = f [ x ]
0 u t [ 1 0 ] =
T r u e
TO THE STUDENT
To the Instructor
The Mathematics Department at the University of New Orleans realized the value in using Mathematica for the teaching of Calculus. The authors recog­
nized that the symbolic, numerical, and graphical capabilities of Mathematica were well suited to augment the teaching of partial differential equations. Sub­
sequently, the first author taught a course on Ordinary Differential Equations with Mathematica, and the first two authors recently gave a course in Par­
tial Differential Equations in which Mathematica was used. The third author taught Calculus and Mathematica, and Vector Calculus using Mathematica.
The notebook concept creates a manageable interface for the student, with out all the headaches of programming I/O (input a nd output), and creating sophisticated graphics, as well as simplifying complicated algebraic expres­
sions. It should be understood by the students that, although Mathematica will provide the power to perform mathematical tasks, Me used as a crutch to solve the problems.
athematica should not be
m;
We suggest that the instructor work through t what is to be emphasized. We also ask that the in representative examples are presented in the text and exercises can be worked out in an analogous nature of a problem may require some variation ai given Mathematica code. In all cases the student explore possibilities.
It is also suggested that notebooks be downloade Web via http://www.crcpress.com/books/isbn/
made such that they cannot be erased or modified. The students should copy the originals and modify a copy. The instructor should also become aware of the hardware capabilities in terms of memory and speed. Mathematica is quite capable of using substantial amounts of memory in evaluating expressions.
ie examples to discover structor realize that only aterial. Other examples fashion. Sometimes the nd/or modification of the should be encouraged to
ed from the World Wide >0-8493-7853-2, and be
To the Student
Currently there is a revolution involving the use of computer technology to
XX
INTRODUCTION TO MATHEMATICA
facilitate the learning of mathematics as well as other sciences. The technol­
ogy, in this case takes the form of Mathematica and the associated Notebooks. With this technology, the student can easily explore many of the graphical, numerical, and symbolic aspects of any number of problems. You will find that using Mathematica and the computer to learn partial differential equations can be both exciting and frustrating at the same time! A few suggestions will follow that will help you to maximize your experience.
Although Mathematica is a programming environment, you do not need to learn how to program Mathematica. You can learn by example and easily adapt the examples to solve most of the problems. We encourage you to copy, paste, and edit whenever possible. Besides, if you have a working example that can be slightly modified to solve your problem, then there will be less chance for a typing mistake if you let the computer do the typing by cutting and pasting. We recommend that you first cut and paste, and then modify the example.
Mathematica is capable of making mistakes. So the student should be able to verify and check some results by hand. At times, you will have to do the entire calculation by hand in order to verify that Mathematica is providing you with the correct answer to your problem.
Some of you will, no doubt, be interested in learning how to program Math­
ematica, or have a question about Mathematica. The moderated newsgroup comp.soft-sys.math.mathematica is a forum which offers the opportunity to ask questions and receive answers regarding Mathematica related issues. Note that the standard rules of list netiquette apply.
MathSource
MathSource is a well-organized and easily accessible online database for Mathematica materials. For information on MathSource commands, send an email with Help Intro in the body to mathsource@wri.com
.
1
I n t r o d u c t i o n
In many mathematical modeling formulations, partial derivatives are required to represent physical quantities. These derivatives always involve more than one independent variable, generally the space variables x, y.... and the time variable t. Such formulations have one or more dependent variables, which are the unknown functions of the independent variables. The resulting equations
are called partial differential equations, which,
and/or boundary conditions, represent physical phenomena.
1.1. Notation and Definitions
Definitions about order, linearity, homogeneity, and solutions for partial dif­
ferential equations resemble those in the case of ordinary differential equations and are as follows: The order of a partial differer as the order of the highest derivative appearing in
... du du d2u d2u , d2u derivatives
dx’ dy’ dx2’ dydx dy2
t oget her wi th the initial
tial equati on is the same the equation. The partial
and —- j are sometimes denoted by ux
Uxx, uxy and uyy, or p, q, r, s and t respectively. The most general first order partial differential equation with two independent variables x and y is written in the form
F(x, y,u,p,q) = 0, p = ux,
The most general second order partial differential equation is of the form
F(x,y,u,p,q,r,s,t) = 0,
iy.
(1.1)
*xyi
t = U
yy
(1.2)
2
CHAPTER 1: INTRODUCTION
A partial differential equation is said to be linear if the unknown function u and all its partial derivatives appear in an algebraically linear form, i.e., of the first degree. For example, the equation
$11 Wjj + 2(^12 U x y “t- $22 ^y y “l·- ^1 ^ x "t” ^2 CU — f, (1.3)
where the coefficients an, <212, a22> 61,62. and c and the function / are func­
tions of x and y, is a second order linear partial differential equation in the unknown u(x,y). An operator L is a linear differential operator iff L{au + βν) = aLu + fJLv, where a and β are scalars, and u and v are any functions with continuous partial derivatives of appropriate order.
A partial differential equation Lu = 0 is said to be homogeneous, whereas Lu = g, where L is any differential operator and g φ 0 is a given function of the independent variables, is said to be nonhomogeneous. For example,
(x + 2y)ux + x2uy = cos(x2 4- y2) is a nonhomogeneous first order linear equation, whereas
(x + 2 y)ux + x2uy = 0
is homogeneous. Thus, a linear homogeneous equation is such that whenever u is a solution of the equation, then cu is also a solution where c is a constant. A function u = φ is said to be a solution of a partial differential equation if φ and its partial derivatives, when substituted for u and its partial derivatives occurring in the partial differential equation, reduce it to an identity in the independent variables. The general solution of a partial differential equation is a linear combination of all solutions of the equation with as many arbitrary functions as the order of the equation; a partial differential equation of order k has k arbitrary functions. A particular solution of a partial differential equation is one that does not contain arbitrary functions or constants.
A partial differential equation is called quasi-linear if it is linear in all the highest order derivatives of the dependent variable. For example, the most general form of a quasi-linear second order equation is
A(x, y, u, p, q)uxx + B(x, y, u, p, q)uxy + C{x, y, u, p, q)uyy+
+ f (x,V,u:p,q) = 0. (1.4)
It is assumed that the reader is familiar with the theory and methods of ordinary differential equations. Since the subject of partial differential equa­
tions is broad, we shall discuss certain well-known equations of second order in detail.
1.2. INITIAL AND BOUNDARY CONDITIONS
3
1.2. I n i t i a l and B o u n d a r y Co ndi ti on s
A partial differential equation subject to certain initial or boundary conditions is known as an initial problem. The initial conditions, also known as C a values o f the unknown function u and an appropriati at the initial point.
conditions in the form o f value or a boundary value u c h y c o n d i t i o n s, are the e number o f its derivatives
The boundary conditions fall into the following three categories:
(i) D i r i c h l e t c o n d i t i o n s (also known as boundary are the values o f the unknown function u pres< boundary d D o f the domain D under considerai
:onditions o f the first kind) eribed at each point o f the tion.
(ii) N e u m a n n c o n d i t i o n s (also known as boundary kind) are the values o f the normal derivatives prescribed at each point o f the boundary d D.
c o n d i t i o n s o f t he s e c o n d i f t he unknown f un c t i o n u
( i i i ) R o b i n c o n d i t i o n s (also known as boundary c or mixed boundary conditions) are the values the unknown function u and its normal derivativi o f the boundary d D.
The f o l l o w i n g pr o bl e ms are e x a mpl e s o f e ach
ut = kuxx, 0 < χ < I, t u{x,0) = f{x), ut(x,0) = g(x),
u( 0,t ) =Ti, u( l,t ) =T2,
u t = k uxx, 0 < χ < I, t
u ( x, 0) = f ( x ), u t ( x, 0) = g ( x ), u x ( 0,t ) = T i, u x ( l,t ) = T 3
Ut = k u x
0 < χ < I, t
u ( x,0 ) = f ( x ), u x ( x,0 ) = g ( x )
u ( 0,t ) + a u x ( 0,t ) = 0
u(l,t) + Pux(l,t) = 0,
onditions o f the third kind, 3f a linear combination o f e prescribed at each point
category, respectively:
> 0,
0 < χ < I, (1.5)
t > 0;
> 0,
0 < χ < I, t > 0;
> 0,
0 < χ < I,
t > 0.
(1.6)
(1.7)
4
CHAPTER 1: INTRODUCTION
1.3. C l a s s i f i c a t i o n o f S e c o n d O r d e r E q u a t i o n s
I f f = 0 in Eq (1.3), the most general form o f a second order homogeneous equation is
In order to show a correspondence with an algebraic quadratic equation, we replace u x by a, u y by β, u x x by a 2, u x y by α β, and u y y by β 2. Then Eq ( 1.8) reduces to a second degree polynomial in a and β\
P ( a, β ) = a n a 2 + 2α ί 2 α β + α22β 2 + b i a + ύ 2 β + c. (1.9)
It is known from analytical geometry and algebra that the polynomial equation P ( a, β ) = 0 represents a h y p e r b o l a, p a r a b o l a, or e l l i p s e according as its discriminant a\2 — a\\a 22 is positive, zero, or negative. Thus, Eq (1.8) is classified as hyperbolic, parabolic, or elliptic according as a\2 — ( i\\a 22 = 0.
An alternate approach to classify the types o f Eq (1.8) is based on the following theorem:
T h e o r e m 1.1. T h e r e l a t i o n 4 > ( x,y ) = C i s a g e n e r a l i n t e g r a l o f t h e o r d i n a r y d i f f e r e n t i a l e q u a t i o n
P r o o f. S i n c e t h e f u n c t i o n u = <j >( x,y) satisfies Eq (1-11), t he n
«11 Uxx
+ 2a12 uxy + a 22 uyy +bi ux + b 2 uy + cu = 0.
( 1.8 )
o - n d y 2 — 2a i 2 d x d y + a 22 d x 2 = 0
(1.10)
i f f U = Φ { χ,ν ) i s a p a r t i c u l a r s o l u t i o n o f t h e e q u a t i o n a n u2 + 2a i2 u x u y + a 22 u 2 = 0.
(1.11)
a n
+ a 22 — 0
(1.12)
holds for all x, y in t h e domain o f definition of u = ψ ( χ, y ) and φ ν φ 0. In order t h a t t h e relation φ ( χ, y ) = C be t h e general s ol uti on o f Eq
1.3. C L A S S I F I C A T I O N
5
(1.12), we must show t h a t the function y defined
C satisfies Eq (1.12). Suppose t h a t y = f ( x, C ) is such a function. The n
d y ^ _ d x
H e n c e, i n v i e w o f E q ( 1 - 1 1 ),
Φ χ { χ, y )
4>y{x,y)
y =f ( x,C)
aU 1 1 “ 2(112
d y
d x
+ a 22
«11 — - 2 o 12 -
i m p l i c i t l y b y φ{ χ, y) =
0-22
= 0.
y = f ( x,C )
T h u s, y = f ( x,C ) satisfies Eq (1-12).
( 1.13)
Conversely, let φ { χ.y ) = C be a general solution o f Eq (1.11). We must show t h a t for each point ( x,y )
a n Φ2Χ + 2 a u φ?. φ ν + α22 φ 2ν 0.
If we can show t h a t Eq (1.14) is satisfied for an t h e n Eq (1.14) will be satisfied for all points, i a solution o f Eq (1.14), we construct through ( x ( 1.11) where we set φ ( χ ο, y o ) = Co, and consider For all points o f t hi s curve we have
rbitrary point ( z o, y o ), ince </>(#, y) represents o, yo) an integral o f Eq t h e curve y = f { x, Co).
On
d y
d x
^a12 ( ) + a22
Ι ΨΧ \ 0 Ι Ψχ
a i i - - 7— ~ 2ai2 ——
If we set x = X o in t hi s equation, we get
a n Φ2χ { χ o, yo) + 2a 12 φ χ { χ 0, yo) Φν { ζ o, yo) + w h e r e y 0 = f ( x 0, C o ). ·
E q ( 1.1 0 ) o r ( 1.1 1 ) i s c a l l e d t h e c h a r a c t e r i s t i c e q u a t i o n o f t h e par­
t i a l differential equation (1.3) or (1.8); t h e related integrals are called c h a r a c t e r i s t i c s.
( 1.1 4 )
' «22
0.
y=f ( x,C0)
0-22 Φν{Χθ,νθ) = 0,
6
C H A P T E R 1: I N T R O D U C T I O N
Eq (1.13), regarded as a quadratic equation in d y/d x, yields two solutions:
dy _ a12 ± ^ a j 2 - a i l 022 dx a n
The expression under the radical determines the type of the differential equation (1.3) or (1.8). Thus, Eq (1.3) or (1.8) is of the hyperbolic, parabolic, or elliptic type according as a\2 — a n a 22 = 0.
E x a m p l e 1.1. The Tricomi equation ux x + x u yy + u = 0, for which a\2 — a\\a 22 = —x, is hyperbolic if x < 0, parabolic if x = 0, and elliptic if x > 0. ■
The general form of a linear second order partial differential equa­
tion in n variables χχ,... , xn is
η n
^ 2 aijUXiXj+ ^ 2 b i U X i +CU + f - 0, (1-15)
i,j = 1 i = l
wher e t h e coeffi ci ent s a ^, bk, c, a n d d a r e r ea l c o n s t a n t s or f unc t i ons of xi,... , xn. If we assume t h a t the second order partial derivatives of u are continuous, then the terms involving the highest order derivatives, i.e., those in the first summation in (1.15), can be arranged such t h a t = a j i. If we consider the quadratic form
then at a fixed point P° = (a:®,... ,x ° ) the coefficients atJ are con­
stants. This quadratic form can always be transformed by an affine
transformation into the canonical form
n
Q = Y 2 ai Wi ’
i = 1
where not all a* vanish. Then the partial differential equation (1.15) is
elliptic if all a % have the same sign; hyperbolic if all a.t except one have the same sign; ultrahyperbolic if two or more a j have different signs; and
parabolic if one or more a t vanish.
For quasi-linear second order partial differential equations, the above criteria still hold, since only the highest order terms are considered for this classification.
1.4. S O M E K N O W N E Q U A T I O N S
7
E x a m p l e 1.2. For t h e partial differential 4u y z + 4u z z = 0 in R 3, t h e quadratic form is
equation 4u xx + u yy
4 v f + v\ + 4υ21>3 + 4^3 = 0,
which, by se t t i n g 2 ν χ = w i, v 2 + 2 v 3 = w 2, and t o w 2 + w 2 = 0. Hence t h e given equation isi coefficient o f w 3 is zero. ■
E x a m p l e 1.3. Consider u x x — x 2 y u y y = « 11^22 = x 2 y > 0, so t h e partial differential eq1
0, y > 0. Here a 12 illation is hyperbolic, ι
E x a m p l e 1.4. Consider e x y u x x + u y y sinb
a i\a 22 — —e xy sinh x, and t h e partial differenti; if x < 0, parabolic i f x = 0, and elliptic i f x >
x + u = 0. Here a\2 — .1 equation is hyperbolic
0.
loai
www | T he classification o f a given second qrder partial differential
-ding t h e Mathematica E|quationType .ma found
ce.
equation into i ts t y p e can be achieved by package Equ ati onType .m, and the Notebook on t h e CRC web server mentioned in t h e Prefi
1.4. S o m e K n o w n E q u a t i o n s
The following equations appear frequently duri ical phenomena:
1. H e a t e q u a t i o n in R 1: u t = k u x x, where u denotes t h e temperature
distribution and k t h e thermal diffusivity.
2. W a v e e q u a t i o n in R 1: u t t = c 2 u x x, wher placement, e.g., o f a vibrating string from i ts e> c t h e wave speed.
3. L a p l a c e e q u a t i o n in R 2: V 2u ξ u x x + '
d e n o t e s t h e L a p l a c i a n.
v 2 — 2 v 3 = w 3, reduces parabolic because t h e
ng t h e analysis o f phys-
e u represents t h e dis­
equilibrium pos iti on, and
= 0, where V 2 = V · V
4. T r a n s p o r t ( T r a f f i c ) e q u a t i o n: u t + a ( u ) u x = 0.
8
C H A P T E R 1: I N T R O D U C T I O N
4a. T r a n s p o r t e q u a t i o n in R 1: u t + a u x = 0, where a is a constant.
5. B e r g e r ’s e q u a t i o n in R 1: u t + u u x = 0, which arises in t h e st u dy o f a stream o f particles or fluid flow wi t h zero viscosity.
6. E i k o n a l e q u a t i o n in R 2: u 2 + u 2 = 0, which arises in geometric optics.
7. P o i s s o n ’s e q u a t i o n in R n: V 2u = /, also known as t h e nonhomo­
geneous Laplace equation in R n; it arises in various field theories and ele ctr ostati cs.
8. H e l m h o l t z e q u a t i o n in R 3: ( V 2 u + k 2 ) = 0, which arises, e.g., in
underwater scattering.
9. K l e i n - G o r d o n e q u a t i o n in R 3: u t t — c 2 V 2u + r n 2 u = 0, which arises
in quantum field theory, where m denotes t h e mass.
10. T e l e g r a p h e r ’s e q u a t i o n in R 3: u t t — c 2 + a u t + m 2 u = 0, where a is t h e damping coefficient; i t arises in t h e s t u d y of electrical transmission in telegraph cables when t h e current may leak t o t h e ground.
11a. S c h r o d i n g e r e q u a t i o n in R 3: u t = i [ V 2 u + V ( x ) u ], where V { x ),
x € R?, denotes t h e potential; i t arises in quantum mechanics.
l i b. C u b i c S c h r o d i n g e r e q u a t i o n in R 3: u t = i [ V 2 u + σ ι ι |m]2], where
σ = ± 1; t h i s is a semilinear version o f 11a.
12. S i n e - G o r d o n e q u a t i o n in R 3: u t t — c 2'V 2 u + m 2 u = 0, which arises in quantum field theory.
13. Semilinear h e a t equation in R 3: Ut — fcV2u = f ( x,t,u ).
1 4 a. S e m i l i n e a r w a v e e q u a t i o n in R 3: u t t — c 2 V 2 u = f ( x:t,u ).
1 4 b. S e m i l i n e a r K l e i n - G o r d o n e q u a t i o n in R 3: U t t — c2 V 2«. + m 2 u +
7u p = 0, where 7 denotes a coupling c onstant, and p > 2 is an integer.
14c. D i s s i p a t i v e K l e i n - G o r d o n e q u a t i o n in R 3: U t t — c2V 2u + a u t +
m 2u + u p = 0.
1.4. S O M E K N O W N E Q U A T I O N S
9
= f { x,u ).
1 ( u a V u ), where k > 0 ation, and arises in the
14d. D i s s i p a t i v e s i n e - G o r d o n equation in J?3: s i n u = 0.
15. S e m i l i n e a r P o i s s o n ’s e q u a t i o n in R 3:
16. P o r o u s m e d i u m e q u a t i o n in R 3: u t = k V and a > 1 are constants; i t is a quasi-linear equ seepage flows through porous media.
17. Biharmonic equation in R 3 ·. V 4 u ξ V 2( V 4u) = 0; i t arises in elastodynamics.
18. K o r t e w e g d e V r i e s ( K d V ) e q u a t i o n in R 1: which arises in shallow water waves.
19a. E u l e r ’s e q u a t i o n s in R 3: u4 + ( u · V ) u
denotes t h e ve lo c i ty field, and p t h e pressure
19b. N a v i e r - S t o k e s e q u a t i o n s in R 3: u ( + ( u
. · V ) u + - V p = ί/V u, P
w h e r e υ denotes t h e kinematic v i sc osi ty o f a fluid.
20. M a x w e l l ’s e q u a t i o n s in R 3: E t — V χ H = 0, H t + V χ E = 0,
where E and H denotes t h e electric and t h e magnetic field, respectively; t h e y are a s ys tem of six equations in six unknowns.
Utt ~ c 2 V 2u + a u t +
Ut CU Ux Uxxx — 0,
H— V p = 0, where u P
O r i g i n s o f t h e s e a n d o t h e r e q u a t i o n s o f ma r e l a t e d t o s o m e i n t e r e s t i n g p h y s i c a l p r o b l e m s. \ t i o n o f s o m e o f t h e m a s e x a m p l e s w h i c h w i l l
t h e m a t i c a l p h y s i c s a r e Ve s h a l l p r e s e n t d e r i v a - a l s o b r i n g o u t c e r t a i n
a s p e c t s o f m a t h e m a t i c a l m o d e l i n g o f t h e s e p r o b l e m s.
E x a m p l e 1.5. ( O n e - d i m e n s i o n a l w a v e e q a s t r i n g ) Consider a stretched string o f l ength I ends. It is assumed t h a t (i) t h e string is t hi n ar no resistance t o change o f form except a change t e nsi o n To in t h e string is much larger than acting on i t so t h a t t h e l atter can be neglected equilibrium st a t e be s i t ua t e d along t h e x - a x i s. displacement o f t h e string at ti me t from i t s eq
n a t i o n f o r v i b r a t i o n s o f which is fixed at both d flexible, i.e., i t offers in length, and (ii) t h e e force due t o gravity . Let t h e string in i ts Let u ( x,t ) denote the uilibrium position. T h e
10
C H A P T E R 1: I N T R O D U C T I O N
shape of the string at a fixed t is represented in Fig. 1.1.
U
Let us further assume that the vibrations are small, which implies that the displacement u(x,t) and its derivative ux are small enough so that their squares and products can be neglected. As a result of vibrations, let a segment (Xi,X2) of the string be deformed into the
segment PQ. Then at time t the length of the arc PQ is given by
fX 2 ______
/ y/l + ul dx & χ2 - x\, (1-16)
J X\
which simply means that under small vibrations the length of the seg­
ment of the string does not change. By Hooke’s law, the tension T at each point in the string is independent of t, i.e., during the motion of the string any change in T can be neglected in comparison with the tension in equilibrium. We shall now show that the tension T is also independent of x. In fact, it is evident from Fig. 1.1 that the x-component of the resulting tension at the points P and Q must be in equilibrium, i.e.,
T(x i) cos a(xi) — T(x2) cos a(x2) = 0,
where a(x) denotes the angle between the tangent at a point x and the positive x-axis at time t. Since the vibrations are small,
cosa(x) = —, ^ „ = —, « 1,
\j 1 -|- tan a(x) -y/l + ux
whi ch i mpl ies t h a t T(x 1) « T(x2). Since xi and x.2 are arbitrary, the magnitude of T is independent of x. Hence, if To denotes the tension at equilibrium and T the tension in the vibrating string, then T « To for all x and t.
No w, t h e s u m o f t h e c o m p o n e n t s o f t e n s i o n Τ ( χ χ ) a t P and T ( x 2)
1.4. S O M E K N O W N E Q U A T I O N S
11
at Q along t h e u - a x i s must be zero, i.e.,
0 = T0 [ s i n a ( x2) - s i n a ( i i ) ]
t a n a ^ ) t a n a ( x i )
= T 0 = T0 = T 0
\J l + t a n2 a ( x 2 ) i/l + t a n2 a ( x i )
uX2
V 1 +U'2X2 V 1 + Uh
d u
<T0
d u d u d ) p 2 d x i
d u
d x
£ = 22
r x i
d 2 u
L
- - - - - -- ό τ
d x 2 ’
using (1.16). Let g ( x,t ) denote t h e external force per unit mass acting
onent o f g ( x,t ) acting
on t h e string along t h e u - a x i s. Then t h e comp on t h e segment P Q along t h e w-axis is given by
rx 2
/ g ( x,t ) d x.
J X\
L e t p ( x ) be t h e linear dens ity o f t h e string. The t h e segment P Q is
( 1.17)
~L p(x)U dx'
Hence t h e sum o f the components (1.17), (1.18 zero, i.e.,
d 2u , . , d 2 u
px 2
Jx 1
T° d ^ + S 0 M ) - P ( z )
Since x\ and x -ι are arbitrary, i t follows from (1 must be zero, which gives
, d 2u d 2 u
T h i s r e p r e s e n t s t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n t h e s t r i n g.
I f p = const, t he n (1.21) reduces t o
( 1.18) n t h e inertial force on
(1.19)
and (1.19) must be
d x = 0. (1.20)
!0) t h a t t h e integrand
(1.21)
for t h e vibrations o f
d 2u
d t 2
"XL
= + f(x,t),
dx2
(1.22)
12
C H A P T E R 1: I N T R O D U C T I O N
where a = y j T o/p, f ( x. t ) = g ( x, t )/p. In t h e absence o f external forces, Eq (1.22) becomes
d2i
,d2u
(1.23)
Ot2 dx2 ’
which is t h e wave equation for free vibrations (oscillations) o f the string. ■
A : Area in equilibrium
. U
A': Deformed area
n'
/
Lf
K
Fig. 1.2. Vibrations of a membrane.
E x a m p l e 1.6. (T w o - d i m e n s i o n a l w a v e e q u a t i o n f o r o s c i l l a t i o n s o f a m e m b r a n e ) Suppose t h a t a membrane which is a perfectly flexible t h i n s t retched s h e e t occupies a region D i n t h e x y - p l a n e in i t s equi­
librium s t a t e. Further, let t h e membrane be subjected t o a uniform t e nsi o n T applied on i ts boundary d D. This means t h a t t h e force ac t­
ing on an element d s o f t h e boundary 3 D is equal t o T d s. We shall examin e t h e transverse oscillations o f t h e membrane, which move per­
pendicular t o t h e x y - p l a n e at each point in t h e direction of t h e u - a x i s. Thus, t h e displacement u at a point ( x, y ) € D is a function o f x, y and t. Assuming t h a t t h e oscillations are small, i.e., t h e functions u, u x, and u y are so small t h a t their squares and products can be neglected, let A denote an arbitrary area o f t h e membrane ( A e D ) s i t u a t e d in equilib­
rium in t h e x y - p l a n e and bounded by a curve L. After t h e membrane is displaced from i ts equilibrium position, let t h e area A be deformed into an area A' bounded by a curve L' (see Fig. 1.2.), which at t i m e t is defined by
A! = j j ^ j l + u 2x + u 2y d x d y t t J j d x d y = A.
Thus we can neglect t h e change in A during t h e oscillations, and t he
1.4. S O M E K N O W N E Q U A T I O N S
13
te nsi on in t h e membrane remains constant and e T.
:[ u a l t o i t s i n i t i a l v a l u e
N o t e t h a t t h e t e n s i o n T which is perpendicula lies at all points in t h e tangent plane t o t h e su:: denote an element o f t h e boundary L'. Then t h e ■
du
element is T d s', and — = c o s a, where a is t h e angle between t h e a n
t e n s i o n v e c t o r T a n d t h e u - a x i s, a n d n is t h e outward normal t o the
acting on t h e element
Ence t h e component o f
ar t o t h e boundary L' face area A'. Let d s' te nsi on acting on this
boundary L. Then t h e component o f t h e tension
du
d s' in t h e direction o f t h e u - a x i s is T — d s'. He:
a n
t h e resultant force acting on t h e boundary L' al·
jng t h e n - a x i s is
T( fp
J v dn J l 9n
-IL
ds
'd2u d2u'
ι a \d x 2 + d y 2 /
by Green’s identity, where, in view o f small osciL d s' « d s, and replaced L' by L. Let g ( x,y,t ) denote an external force per unit area acting on t h e membrane along t h e u - a x i s. Then t h e t o t a l force actin g on t h e area A' is given by
LL
g ( x,y,t ) d x d y.
Let p ( x, y, t ) be t h e surface dens ity o f th e membr force at all t i mes t is
IL
(1.24)
d x d y, ations, we have taken
(1.25) m e. Then t h e inertial
(1.26)
d and boundary V, we find
■9(x
y,t)
d x d y = 0,
p(x,y,t)^dxdy.
ι a dt
Since t h e sum o f t h e inertial force and t h e t o t a l force is equal and o pp osi t e t o t h e resultant of t h e tension on the from ( 1.24)—(1.26) t h a t
f ί I" / 2u rn ( d 2U d2U
U a ~ T [ f a? + 2
or, since A i s arbitrary,
^ d2u __ f d 2u d2u
dt2
\<9ie2 dy2
T h i s i s t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n f or s m a l l b r a n e. I f t h e d e n s i t y p = const, t he n Eq (1.5 external forces reduces to
+ g ( x,y,t ).
d2u _ 2 ( d2u d2u\ d t 2 a \ d x 2 ^ d y 2 ) ’
(1.27)
Dscillations o f a mem- 7) in t h e absence of
Γ/p..
( 1.2 8 )
14
C H A P T E R 1: I N T R O D U C T I O N
Ex am p l e 1.7. (Heat transfer equation for a uniform isotropic body) Let u(x, y, z, t) denote the temperature of a uniform isotropic body at a point (χ, y, z) and time t. If different parts of the body are at different temperatures, then heat transfer takes place within the body. Consider a small surface element SS of a surface 5 drawn inside the body. Under the assumption that the amount of heat SQ passing through the ele­
ment SS in time St is proportional to St, SS, and the normal derivative
the coordinates (χ, y, z) of points in the body but is independent of the direction of the normal to the surface S, and Vn denotes the gradient
Q denote the heat flux which is the amount of heat passing through the unit surface area per unit time. Then Eq (1.29) implies that
Now, consider an arbitrary volume V bounded by a smooth surface S. Then, in view of (1.30), the amount of heat entering through the surface S in the time interval [ti, £2] is
by divergence theorem, where n is the inward normal to the surface
S. Let SV denote a volume element. The amount of heat required to change the temperature of this volume element by Su in time St is
where c(x, y, z) and p(x, y, z) are the specific heat and density of the body, respectively. Integrating (1.31) we find that the amount of heat required to change the temperature of the volume V by Su = u(x, y, z, t+ St) — u(x, y, z, t) is given by
du
SQ = —k-£— SS St — —k SS St Vnu,
dn
(1.29)
wher e k is t he t her mal conduct i vi t y of t he body whi ch depends onl y on
i n t he di rect i on of t he out ward nor mal t o t he surface el ement SS. Let
(1.30)
(1.31)
<5<?2 = [u(x,y,z,t + St) - u( x,y,z,t )] p(x,y,z)SV, (1.32)
1.4. S O M E K N O W N E Q U A T I O N S
15
We shall now assume t h a t the bod y contains heat sources, and let g(x, y, z, t) denote t h e dens ity o f such heat sources. Then t h e amount o f heat released by or absorbed in V in t h e ti me interval [£1,^2] is
ft2
Q 3 —
x,y,z,t)
Since Q2 = Q i + Q 3, we find from ( 1.3 1 ) - ( 1.3 4 )
f *//£
du _ ., _ . ,
°P~dt ~ ' ^ Vm' ~ 9yx'V
o r, s i n c e t h e v o l u m e V and t h e ti me interval [t
1)
du
cp— = V · (fcVli) + g{x,y,z,t)
dV.
t h a t
M )
( 1.3 4 )
dV =0,
t2] are arbitrary,
d
du
d
which is t h e required heat transfer equation fo
body. If c, p, and fc are c onstant, Eq (1.35) beco
du _ 2 ( d2u d2u d2u\ .
dt \ dx2 dy2 dz2 J ^
(1.35) r a uniform isotropic mes
where a = y/k/cp is known as the thermal diffusivity, and / = g/cp denotes t h e heat source (sink) function. In t h e absence o f heat sources ( i.e., when g(x,y,z,t) = 0 ), Eq (1.36) reduces heat conduction equation
du oT—ro — = a V u dt
o ( d2u d2u
12
I + -----
\ dx2 dy2
c,y,z,t),
( 1.3 6 )
t o t h e h o m o g e n e o u s
d2u dz2
In t h e case when t h e temperature distribution throughout t h e body
reaches t h e ste ady state, i.e., when t h e temperat dent o f time, Eq (1.37) reduces t o t h e Laplace eq
ire becomes indepen- uation
_ 2 d2u d2u d2u
9 Ϊ 2 + ^ 2 + ^ 2 = Φ'
F o r t h e d e r i v a t i o n o f t h e h e a t c o n d u c t i o n e q u a t i o n i n R1 x R+, con­
sider a laterally i nsulated rod o f uniform cross se c ti on w i t h area A and constant dens ity p, constant specific heat c, and constant thermal con­
du ct i v i t y fc. We shall assume t h a t t h e temperature u(x,t) is a function
(1.37)
(1.38)
16
C H A P T E R 1: I N T R O D U C T I O N
o f x and t only, t > 0, and use t h e law o f conservation o f energy to derive t h e heat conduction equation. Consider a segment P Q o f the rod, wi t h coordinates x and x + A x (Fig. 1.3). Let R denote t h e rate a t which t h e heat is accumulating on t h e segment P Q. Then, assuming t h a t there are no heat sources or sinks in t h e rod, R is given by
R = r ^ a c f A n ( (,t )
J x d t
P
β
o | — n
χ * + Δ χ
Fig. 1.3. Segment P Q on a t hi n uniform rod.
N o t e t h a t R can also b e evaluated as t h e t o t a l flux across t h e boundaries o f t h e segment P Q, which gives R = k A [ u x ( x + A x,t ) — u x ( x,t ) ]. Now, using t h e me an-val ue theorem for integrals, we get
c p A u t ( x + h A x,t ) A x = k A [ u x ( x + A x,t ) — u x ( x,t )\, 0 < h < 1.
After dividing b o t h sides by c p A A x and taking t h e limi t as A x —> 0, we get
u t ( x,t ) = a 2 u x x ( x,t ). ■
E x a m p l e 1.8. (O n e - d i m e n s i o n a l t r a f f i c S o w p r o b l e m ) Let p ( x,t ) denote t h e traffic dens ity which represents t h e number of vehicles per mile at t i m e t at an arbitrary yet fixed pos ition I o n a roadway. Let q ( x, t ) denote t h e traffic flow which is a measure o f number o f vehicles per hour passing a fixed pos iti on x. Consider a section o f t h e roadway bounded by t h e positions χ = χ χ and x = x 2, and assume t h a t there are no e x it s or entrances between t hese two positions. Then t h e number N
o f v e h i c l e s i n t h e s e g m e n t [ χ χ,χ 2\ is given by TV = I p ( x,t ) d t. The
J X\
rate o f change o f N wi t h respect t o ti me t is equal t o t h e difference between t h e number of vehicles per unit t i m e entering t h e po s it i o n at χ = Χ χ and t h a t leaving at t h e pos iti on x = x 2, i.e.,
1.5. S U P E R P O S I T I O N P R I N C I P L E
17
As in t h e case o f heat conduction (Example 1. also known as t h e integral representation o f co can be wr i tte n as
), Eq ( 1.39), which is nservation o f vehicles,
q { x i,t ) - q ( x 2) = -
rx 2
J X\
d_
d x
q { x,t ) d x =
d ί 1
dl Jx,
w h i c h a f t e r t a k i n g t h e p a r t i a l d e r i v a t i v e d/d t and noti ng t h a t x\ and x 2 are arbitrary leads difFerential equation
+ Ά =
0.
d t d x
Til·
Let u ( x,t ) denote t h e v e lo c i ty o f a vehicle, o f vehicles per hour passing a given pos iti on is vehicles t i mes t h e v e lo c i ty o f vehicles, we obtain
p ( x, t ) d x,
inside t h e l ast integral t h e required partial
t o
(1.40)
en, since t h e number i q u a l t o t h e de ns it y o f
q ( x,t ) = p ( x,t ) u ( x,t ).
I f w e a s s u m e t h a t t h e v e l o c i t y u depends only u ( p ), i.e., t h e vehicles slow down as t h e traffic d u
— < 0. T h i s i n e q u a l i t y i m p l i e s t h a t t h e t r a f f i c t h e t r a f f i c d e n s i t y, i.e., q = q ( p ), and Eq (1.40)
= n
d t d p d x
n t h e d e n s i t y p, u = <jlensity increases, then
flow depends onl y on
t he n reduces t o
d p
d t
^ + c ( p ) ^ = 0,
's
where c ( p ) = d q/d p. Eq (1.41) is a first order linear partial differential equation. ■
1.5. Superposition Principle
Let L denote a linear differential operator o f an; Th e superposition principles for homogeneous linear differential equations are represented by
(1.41)
homogeneous q u a s i -
y order and any kind, and non-homogeneous t h e following two t h e o ­
rems:
18
CHAPTER 1: INTRODUCTION
T h e o r e m 1.2. Let L u = 0 be a differential equation. Suppose u\ and U2 are two linearly independent solutions. Then c\Ui + C2 U2 is also a solution.
P R O O F. By hypotheses L u\,2 = 0. By definition L(c\Ui + C2 U2 ) = c ^ L u i + C2 L U 2 = 0. ■
T h e o r e m 1.3. I f L u = Y ^ C i f i be a non-homogeneous linear dif­
fer ential equation and i f Lgt- = f k, then Y ^ C i g i is a solution o f the above differential equation.
P R O O F. ci9i = Σ ι ci L 9i = Σ ι ci f i - Thus Σ ΐ c%9i satisfies
the differential equation. It is obvious from these two principles that if υ is a solution of an equation L u = 0 and if F a solution of L u = f, then v + F is also a solution of L u = /. A generalized superposition principle is defined as follows:
T h e o r e m 1.4. I f the functions u^, i = 1,2,, are separately the solutions o f a linear homogeneous differential equation L ( u ) — 0, then
OO
the series u = C» Ui is also a solution o f the differential equation,
i =1
provided that the derivatives appearing i n L ( u ) can be differentiated te r m - b y -te r m.
P r o o f. In fact, if the derivatives of u appearing in L ( u ) — 0 can be differentiated term-by-term, we have
dnu = y, dnUi dxmdtn- m _ 2-* 1 dxmdtn- m ’
i= 1
and since the equation L(u ) = 0 is linear and a convergent series can be added term-by-term, we can write
(
OO \ OO
= Y,C i L ( Ui) = 0.
i= 1 ) i —1
1.6. EXERCISES
19
The sufficient condition for term-by-term differentiability is the uni­
form convergence of the series Σ c.i 9"Ui
i = 1
d x md t n
1.6. Exercises
2,Uyy 0
“ I I
Auxx - 7u.
0
Wt
Classify the partial differential equation as hyperbolic, parabolic, or elliptic:
1.1.
1.2.
1.3.
1.4.
1.5.
1.6.
1.7.
1.8.
+ a u v
0, α φ 0
o u xx + 2 a u xy + u yy = 0, α φ 0
4Ut t i 2 t i x t -(- 9^3,3, 0
‘2 u x f Λ α i f 0
2UXy ^ U y y 0
8^7.7. 2 u
x y
3 u.
y v
0
Fo r w h a t v a l u e s o f x a nd y a r e t h e f ol l ow: e q u a t i o n s h y p e r b o l i c, p a r a b o l i c, o r e l l i p t i c?
l.Q. UXX XXLyy 0.
A n s. Hyperbolic for x > 0, parabolic for x =j= 0, elliptic for x < 0.
1.1 0. UXX <2 x u x y “I” y U y y = 0-
A n s. Hyperbolic for x 2 > y, parabolic for x 2 = y, elliptic for
A n s. Hyperbolic A n s. Hyperbolic A n s. Elliptic A n s. Parabolic A n s. Parabolic A n s. Hyperbolic A n s. Elliptic A n s. Hyperbolic
ng partial differential
1.1 1. u xx + 2 x u xy + (1 - y 2)u
“yy
0.
A n s. Hyperbolic for x 2 + y 2 > 1, parabolic for x 2 + y 2 = 1, elliptic for x 2 + y 2 < 1.
1.1 2. X U x x “i“ X U Xy “1“ y U y y 0.
A n s. Hyperbolic for x 2 > 4x y, parabolic for x 2 < 4x y.
x2
= 4
xy,
e l l i p t i c f o r
1.1 3. (1 + y 2) u xx + (1 + x 2) u yy = 0. A n s. Elliptic for all x and y.
20
CHAPTER 1: INTRODUCTION
1.1 4. Ί ΐ χ χ x u Xy -}- y u y y x y u y 0.
Ans. Hyperbolic for x 2 > Ay, parabolic for x 2 = Ay, elliptic for x 2 < Ay.
1.1 5. U x x "f" X Uy y 0.
Ans. Elliptic for χ Φ 0.
1.1 6. uxx — 2 s i n x u XJ, — cos2 x u y = 0.
Ans. Hyperbolic for all x.
1.1 7. y U x x + Uy y = 0.
Ans. (Tricomi) Hyperbolic for y < 0, parabolic for y = 0, and elliptic for y > 0.
1.1 8. uxx — 2 c o s x u xy — (3 + sin2 x) uyy — y uy = 0.
Ans. Hyperbolic for all x.
1.1 9. ( 1 + x 2) uxx + (1 + y 2) uyy + x u x + y uv = 0.
Ans. Elliptic for all x and y.
1.2 0. x 2uxx — y 2u yy = 0 x > 0, y > 0.
An s. Hyperbolic.
1.21. y 2uxx + x 2uyy = 0, x > 0, y > 0.
An s. Elliptic.
1.22. x uxx -I- 2xy uxy y uyy — 0.
Ans. Parabolic for all nonzero x and y.
1.2 3. ( 1 — x ) u xx — uyy — ux = 0, 0 < x < 1.
An s. Hyperbolic for x < 1, parabolic for x = 1, and elliptic for x > 1.
1.2 4. x uxx y Uyy 2yUy = 0.
Ans. Hyperbolic for all nonzero x and y.
1.2 5. x uxx *2ixyUxy “i- y Uyy -I- x u x yUy — 0.
Ans. Parabolic for all nonzero x and y.
1.6. EXERCISES
21
1.2 6. u x x y U y y 2 Uy 0.
Ans. Hyperbolic for y > 0, parabolic for y
y < 0.
WWW
All o f t h e above exercises are solved in t h e Mathemati ca Note-
erver mentioned in t h e
b ook Equ a t i o nT y pe.m a found on t h e CRC web Preface.
0, and e lliptic for
2
M e t h o d o f C h a r a c t e r i s t i c s
It is customary in modern texts on partial differential equations either to com­
pletely ignore first order partial differential equations or postpone their dis­
cussion to a much later chapter. Since first order partial differential equations are important from both physical and geometrical standpoints, their study is essential to understand the nature o f solutions and form a guide to the solutions o f higher order partial differential equations.
F i r s t Order Equati ons
First order partial differential equations occur in a variety o f situations. Some o f the common ones are traffic flow, conservation laws, Mainardi-Codazzi relations in differential geometry, and shock waves. In order to solve first order linear, quasi-linear, or nonlinear partial differential equations, the method o f characteristics is very useful. This method is explained in the next six sections. Second order equations, confined to linear equations, are discussed in §2.7. We shall limit our discussion to problems in R 2. An extension to higher dimensions, though routine, is more complicated.
2.1. Linear Eq u a ti on s with Co ns tant Coefficients
The most general form o f first order linear partial differential equations with constant coefficients is
a u x + b u y + k u = f ( x,y ). (2.1)
2.1. CONSTANT COEFFICIENTS
23
du = uxdx + Uydy.
Comparing (2.1) and (2.2) one gets the auxiliary system of equations
du
If u ( x, y) is a solution of (2.1) then
dx
a
dy = _____________________
b f(x,y) — ku
The solution of the left pair is bx — ay = c. The other pair
dx du
a f{x,y)-ku
can be reduced to an ordinary linear differential eq pendent variable and x as the independent variable by
du ku f(x, (bx — c)/a
dx ^ a a
the integrating factor for which is ekx^a.
Thi s obs ervat i on l e ads us t o i nt roduce a new dependent vari abl e v =
ue
,k x/a
, reducing Eq (2.1) to
a vx + b vy = f(x, y)ehx/a = g(
*,y)·
Note that a reduction can also be obtained by substituting v = u eky^h. This substitution will lead to avx + bvy = f (x,y) eky^b. Thus, we need to consider only the formal reduced form
aux + buy = f(x,y). The auxiliary system of equations for Eq (2.4) is
(2.2)
(2.3)
luation with u as the de- This equation is given
dx
a
dy du
b f(x,y)'
The solution of — = — is bx - ay a b y
ay + c dy
x = — 7— » a substitution of this value into —
(2.4)
(2.5)
c, which when solved for x gives
du
yields
24
CHAPTER 2: METHOD OF CHARACTERISTICS
which reduces to du = F(y,c)dy. Its solution is u = G(y, c) + ci, where GyiVic) = F{y-c)· Thus, the general solution is obtained by replacing c\ by φ(ό) and c by bx — ay, thereby yielding
u(x, y) = G(y, bx - ay) + <£(bx - ay).
Eqs (2.3) are known as the equations of t he characteristics. The sys­
tem (2.3) has two independent equations, with two solutions of the form F(x,y,u) = 0 and G(x,y,u) = 0. Each of these solutions represents a family of surfaces. The curves of intersection of these two families of sur­
faces are known as the characteristics of the partial differential equation. The projections of these curves in the (x, y)—plane are called the base char­
acteristics, which are often called characteristics for brevity when there is no ambiguity. The general solution represents a family of surfaces, which are called integral surfaces.
Thus, the equation bx — ay = c represents a family of planes. The intersection of any one of these planes with an integral surface is a curve whose projection in the (x, y)-plane will again be given by bx — ay = c, but this time this equation represents a straight line and is the base characteristic. The solution u on a base characteristic bx — ay = c is therefore given by u = G(y, c) + ci, and the general solution is the same as above.
An alternate procedure is to introduce a new set of coordinates
ξ = bx — ay, and η — bx + ay. (2.6)
This substitution reduces Eq (2.4) to
« „ = F (e,„ ), F { i,v) = m ± o m ^ o M, (2.7)
and the solution of (2.6) is
η ( ξ,η ) = φ(ξ) + 0 ( ξ,η ), (2.8)
where Οη(ξ,η) = Ρ(ξ,η). If f (x,y) = 0 in Eq (2.4), then the auxiliary system of equations is dx/a = dy/b, du = 0. The solutions of these equations are bx — ay = c, and u = c\ = φ[μ) = φ(1)χ — ay). This procedure can also be regarded as a problem in rotation of axes (see Exercise 2.22).
Note that Eq (2.8) is
u(x, y) = ς6(6χ — ay) + G ( bx — ay, bx + ay)
2.1. CONSTANT COEFFICIENTS
25
If an initial condition u ( x, ψ(χ) ) = μ{χ) is prescrib
u (χ, V’W ) = β( χ ) = 4>{bx ~ αψ(χ) ) + G(bx -
c a n b e u s e d t o d e t e r m i n e φ( χ) uniquely. Thus, tl solution u for the partial differential equation (2.1) s condition is established. (For the existence and un see end o f §2.7.)
Exa mple 2.1. Consider
2ux — 3 uy = cosx. T h e auxiliary s ys tem o f equations is
dx dy du
2 —3 c o s x
T h e first sol uti on is then given by 3x + 2y = c. dx du . 1 .
— = -------. Its solution is u = C\ Λ— sinx. No
2 cos x 2
3x + 2y = c, t h e general sol uti on becomes
aip(x),bx + αψ(χ))
e existence o f a unique abject to the above initial queness o f the solution,
sd, t hen
u = f(3x + 2 y) + — sin x
A l t e r n a t e l y, t h e s u b s t i t u t i o n ξ = 3x + 2y, η -= 3x — 2y reduces t h e equation to
1 ( ζ + v )
Ur, = — COS---------------.
η 12 6
which yields
« = /(0 + | s i n ^ ± ^
The other equation is ing t h a t c\ = /( c ) and
On replacing ξ and η by their values in x and general solution.
In t hi s problem t h e characteristics are give t ersection o f t h e planes 3x + 2y = c and the
— sin x + ci. T h e projections o f t hese curves on
y. one g e t s t h e above
n by t h e curves o f in­
integral surfaces u =
t h e ( x, y ) - p l a n e u
= 0
26
CHAPTER 2: METHOD OF CHARACTERISTICS
are t h e base characteristics. Graphs of t hese characteristics are shown in Fig. 2.1 for c = 1 and C\ = 0.
c h a r a c t e r i s t i c
Fig. 2.1.
I f a linear partial differential equation is o f t h e form P ( x. y ) u x + Q ( x,y ) u y = 0, t he n t h e base characteristics and t h e characteristics
are t h e same curves. We will now develop solutions for some specific
condi ti ons prescribed on initial curves. For example, i f u = 1 on t h e ini tial curve y = 0, then
Λ , 1 ( ■ . 3x + 2y
u = 1 + — | sin x — sm
which is an integral surface denoted in Fig. 2.2 by S\. Also, if u = x 1
o n t h e i n i t i a l c u r v e y = x, then
1 . (3a; + 2 y ) 2 1 . 3 x + 2 y
u = - sm χ H--------------------------- s m --------------,
2 25 2 5
which is another integral surface denoted by S-2 - T h e graphs o f t he integral surfaces S i and S2 and t h e characteristics are shown in Fig.
2.2. No t e t h a t an initial curve (or initial line) is a curve where an ini tial condi ti on on u is prescribed. ■
| www | A complete Mathemati ca sol uti on for t hi s example and the individual pl ot s o f t h e integral surfaces S i and 5 2 are available in the Mathemati ca N ot e book Example2.1 .ma.
It is obvious t h a t t h e sol uti on o f a first order linear partial differ­
e nti al equation represents a surface and contains an arbitrary function
2.1. CONSTANT COEFFICIENTS
27
and not an arbitrary constant. Clearly, t h e solu|t i ly o f surfaces. A unique surface is obtained if i ni tial curve, which is not a characteristic. T h e n t h e e xis ten ce and uniqueness of t h e sol uti on of are closely related t o t h e exis tence and uniquen t h e auxiliary s y s t em (2.5) o f ordinary differentia] §2-7).
characteristics
ion represents a fam- u is prescribed on an eader can observe t h a t a first order equation ess o f t h e sol uti ons o f equations (see end o f
Fig. 2.2.
Ex a mple 2.2. Consider
4 u x + U y = x 2 y.
T h e a u x i l i a r y s y s t e m o f e q u a t i o n s i s
d x d y d u
4 1 x 2 y
T h e f i r s t s o l u t i o n i s t h e n g i v e n b y x — A y = c, and t h e method o f t h e previous example, is
3a:4 — 4ca:3 ,. , 3a:4
u = Cl + i g = /( c) + -----
On replacing c by x — A y, one g e t s t h e general solution
the solution, following 4 ex 3
48
3a:4 — 4(a: — Ay ) x 3
χ4 i x 3y
28
CHAPTER 2: METHOD OF CHARACTERISTICS
2.2. L i n e a r E q u a t i o n s w i t h V a r i a b l e C o e f f i c i e n t s
Th e general form o f first order linear partial difFerential equations wi t h variable coefficients is
P ( x, y ) u x + Q (x, y ) u y + f ( x, y ) u = R ( x, y). (2.9)
Once again our at t em p t is t o eliminate t h e term in u from Eq (2.9 ). T hi s can be accomplished by s u bs titu tin g
u = v e ~ ^ x,y ^,
where ξ ( χ, y) satisfies t h e equation
P{x,y)£x{x,y) + Q{x,y) £y{x,y) = f(x,y)·
H e n c e, E q ( 2.9 ) i s f o r m a l l y r e d u c e d t o
P ( x,y ) u x + Q { x,y ) u y = R ( x,y ), (2.10)
where P, Q, R in (2.10) are not t h e same as in (2.9 ). T h e me th o d for sol vi ng t he se equations, known as Lagrange’s method, is esse nti al ly t h e same as in t h e previous se c ti on except t h a t now t h e auxiliary s ys tem o f equations is
d x dy du _ „.
~P ~ ~Q ~ ~R' ^ Λ ^
which becomes more complicated. This s y s t em has two sol uti ons o f t h e t y p e
g(x,y,u)=cu and h ( x,y,u ) = c2,
representing two families o f surfaces. The curves o f intersection o f t he se surfaces are called characteristics o f t h e equation. The projection o f a in t h e plane u = 0 is called a base characteristic.
If R( x,y ) = 0, t h e n there is no difference in t h e base characteris­
t i c s and t h e characteristics. Frequently t h e word base is o m i t t ed from t h e term base characteristic. These characteristics are clearly a o n e - parameter family o f curves. In some cases i t is convenient t o introduce a parameter, say s, in t h e auxiliary sy s t em o f equations, which are the n expressed in t h e form
2.2. VARIABLE COEFFICIENTS
E x am ple 2.3. Consider
3ux 4- 4 uy + 14(x + y) u = 6x e The first s t e p is t o find a function ζ ( χ, y ) whic 3Cr + = 14(x + y).
A p a r t i c u l a r s o l u t i o n o f t h i s e q u a t i o n i s ζ ( χ,ν ) s u b s t it u t io n u = v e ~ ^ x’y^ reduces t h e given eq
3vx + 4 vy = 6a;
with auxiliary equations
dx dy dv
Y “ T “ Qx
d x d y
T h e s o l u t i o n o f — = — i s 4a; - 3y = c\. an 3 4
is υ = x 2 + C2 - Now, as before, v can easi' x 2 + /(4a; — 3y). Let us examine t he se solutii
(χ+y)2 _
ti satisfies t h e equation
= ( x + y ) 2. The n the uati on t o
write c i = 4 x — 3 y = g ( x,y,v ) and c2 = v — x an expression o f t h e form F ( g, h ) = 0, then
F x = Fg (gx -1- g v V x) F h ( h x -t- h v v x) = 0,
which reduces t o
4F g + ( v x - 2 x ) F h - 0.
S i m i l a r l y t h e e x p r e s s i o n f o r F y yields
— 3 F g + v y F h = 0.
E l i m i n a t i n g F g and F h from t he se two equations, one gets
d s o l u t i o n o f ^ ?
3 6a;
y be found t o b e v = 3ns a l i t t l e more. I f we h ( x, y, v ), and consider
det
2a;
i.e.,
3 vx + 4 v y = 6 x. Thus, F ( g,h ) = 0 is also a solution of t h e course, we have t o replace v by u e <x+y') t o o original problem. ■
Solutions o f t h e t y p e F ( g, h ) known as g e n e r a l s o l u t i o n s.
0 o r g =
De f i n i t i o n 2.1. T w o C1 f u n c t i o n s g and h are said t o be f u n c ­
t i o n a l l y i n d e p e n d e n t i f V y x V h / 0.
29
differential equation. Of btain t h e s ol uti on o f t h e
F ( h ) or h = F ( g ) are
W e s h a l l s t a t e a n i m p o r t a n t t h e o r e m.
30
CHAPTER 2: METHOD OF CHARACTERISTICS
T heo rem 2.1. L e t g = c\ a n d h = c2 b e a n y t w o f u n c t i o n a l l y i n d e p e n d e n t s o l u t i o n s o f E q ( 2.9 ). T h e n g = F ( h ), o r h — F ( g ), o r F ( g,h ) = 0 r e p r e s e n t s t h e g e n e r a l s o l u t i o n o f E q ( 2.9 ).
C o r o l l a r y 2.1. I f g = c\ i s a s o l u t i o n o f E q ( 2.9 ), t h e n F ( g ) = 0 i s a l s o a s o l u t i o n o f E q ( 2.9 ).
Ex a m p l e 2.4. L e t u s f u r t h e r c o n s i d e r t h e r e d u c e d e q u a t i o n 3 v x + 4v y = 6x from Example 2.3. The auxiliary equations in t h e parametric form are
d x d y dv
— = — = — = d s.
3 4 6x
T h e s o l u t i o n s a r e
z = 3 s + c i, a n d y = 4 s + c 2, ( 2.1 3 )
a n d, t h e r e f o r e,
v = 9 s z + 6 c i s + C3. (2-14)
dv
T h e last solution is obtained by first su bs t it u t in g x — 3s + c i in — = d s.
6 x
T h e v a l u e s o f x, y, v in terms o f s represent t h e parametric form o f the equation o f t h e characteristics o f th e given partial differential equation. In order t o find a specific characteristic, we need t o have an initial
condition, e.g., x = X q, y = y o, v = Vo a.t s — s q. B y eli mi nati ng s from
(2.13) and (2.14) we get
4 x — 3y = 4 Οχ — 3 c2 = a and v = x 2 — c f + C3 = χ 2 + β,
f r o m w h i c h w e g e t
v = x 2 + f ( 4 x — 3y ). m
W e n o t e h e r e t h a t, e x c e p t f o r s i n g u l a r c a s e s, a u n i q u e c h a r a c t e r i s t i c w i l l, i n g e n e r a l, p a s s t h r o u g h a p o i n t i n s p a c e. T h u s, i f a c o n t i n u o u s i n i t i a l c u r v e i s p r e s c r i b e d, t h e n a u n i q u e c h a r a c t e r i s t i c w i l l p a s s t h r o u g h e v e r y p o i n t o f t h e i n i t i a l c u r v e. T h e l o c u s o f t h e s e c h a r a c t e r i s t i c s w i l l f o r m t h e i n t e g r a l s u r f a c e. H e n c e, i f t h e i n i t i a l c u r v e i s a c h a r a c t e r i s t i c i t s e l f, t h e n t h e e x i s t e n c e o f a n i n t e g r a l s u r f a c e c a n n o t b e g u a r a n t e e d ( s e e e n d o f § 2.7 ).
2.2. VARIABLE COEFFICIENTS
31
Ex a mple 2.5. Consider
ux +exuy = y, u( 0,y) = 1 jf y.
T h e auxiliary s ys tem o f equations is
dx dy du
1 ex y
T h u s, t h e s o l u t i o n s o f t h e a u x i l i a r y s y s t e m are: w h i c h y i e l d s y = ex + c. From t h e second set of y dx which results in du = (ex + c) dx. Its solutii hence
u = ex + cx + f(c), or u = ex + (y ·
N o w, i n v i e w o f t h e i n i t i a l c o n d i t i o n, u( 0, y) = which yields f ( y ) = y + 1. Hence t h e solution i
(2.15)
given by dy = ex dx, Eq (2.15) one g e t s du = on is u = ex + cx + C\;
u(x, y) = ex + (y — ex ) x + y — ex + l = l -\- y + x y — x e x.
| w w w | T h e p l o t s o f t h e s u r f a c e s y = ex + c a: for c = 1 are available in t h e Mathematica Not' T h e intersection o f t he se surfaces is a character^
E
x a mple
2.6. Consider
2y u x + uy = x, u( 0,y) = /(
The auxiliary s ys tem of equations is
dx 2V
dy
l
du
x
I n t h i s e x a m p l e, a s l i g h t l y d i f f e r e n t p r o c e d u r e w i w i l l f i r s t f i n d t h e e q u a t i o n o f t h e c h a r a c t e r i s t i c s p o i n t (xo,yo) on t h e integral surface. T h e left equations is dx = 2y dy, whose solution is t h e fa:
S ι : x = y + c.
W h e n a s u r f a c e S'i p a s s e s t h r o u g h t h e p o i n t b e c o m e s x = y 2 + x 0 — yl - If the ini tial c u( 0,y) = f ( y ), t he n t h e value o f y on t h e in
*)x + f ( y- e*).
i + f ( y — i ) = i + y,
J i d u = ex + (y — ex ) x book E x a m p l e 2.5. ma. s t i c.
y)·
1 be demonstrated. We through an arbitrary pair in t h e auxiliary mily o f surfaces
(xo,Vo), i t s equation irve is x = 0, where itial curve is given by
32
CHAPTER 2: METHOD OF CHARACTERISTICS
y 2 = yg — x o, where y o is t h e value o f y on t h e integral surface through ( x 0,y o ) at x = 0. The right pair of auxiliary equations is d u = x d y, which also represents a family o f surfaces. Let us denote t h e integral surface through (0, yo) by SY T h e curve o f intersection o f S i and 5 2 is a characteristic. The differential equation o f t hi s characteristic is given by d u = ( y 2 + x q — y l ) d y. It s solution is
At x = 0,
V 3
u = γ + {xo - y%)y + ci.
λ3
« ( 0, y o ) = ί γ + (χ ο - yo) yo + c! = f { y 0 ),
3
a n d s u b s t i t u t i n g t h e v a l u e o f y o, one gets
Cl = f { { y l - x o ) l/2 ) + | ( 2/0 - X o ) 3/2 ·
Us in g t h i s value of c\ i n u, t h e value of u at (xo- Vo) is given by
2 2
u ( x 0,y 0) = - { y l - x o f/2 - ψ ΐ + xoyo + /( ( i/o - ^ o ) 1/2)·
Since ( x o,y o ) is an arbitrary point, t h e expression for w(x0, y o ) is gen­
eralized t o
u ( x, y) = \{ y 1 - x f'2 - + x y + f { ( y 2 - χ ) 1/2)· ■
°V Ex a mple 2.7. Consider
(.x + 2y ) u x + ( y - x ) u y = y. T h e auxiliary equations are
d x dy dv
x + 2y y - x y '
The first two equations can be expressed as
dy = y - x
d x x + 2 y
2.3. QUASI-LINEAR EQUATIONS
33
Thi s is a homogeneous ordinary differential equ; A standard s u b s t it u t io n for such problems is y
a t i o n o f t h e f i r s t o r d e r. = v x, which leads t o a
first order ordinary differential equation w i t h separable variables
(1 + 2v ) d v _ d x 1 + 2 v 2 x
T h i s e q u a t i o n c a n b e s o l v e d t o g i v e _ i V 2 y
V 2 t
a n
+ l n ( a r + 2 y 2 )
T h e o t h e r s o l u t i o n c a n b e o b t a i n e d b y o b s e r v i r
t h u s y i e l d i n g u = - ( x + y ) + C2 - Hence t h e gen 3
i V i y
= Cl.
i g t h a t d x + d y = 3 d u, eral solution becomes
u = | ( * + y ) +/[ V 2 t a n 1 — ^ + l n ^ 2 + 2y 2) j.
| www | The plots o f t h e surfaces
V^tan-1 + ln(x2 4- 2y 2 ) = 1, and u=^- ( x + y ) x 3
are available in t h e Mathematica Notebook Exaiaple2.7 .ma. N o t e t h a t t h e intersection o f t he se two surfaces gives a particular characteristic.
2.3. F i r s t Order Q u a s i -l i n e a r Eq ua ti ons
If t h e coefficients P, Q, and R in Eq (2.9) are but not of u x and u y. the n t h e equation is kno t h e s e equations t h e first order derivatives occur although t h e equation need not be linear in u. shock waves o f various kinds, e.g. traffic flow, technique is t h e same as for first order linear e< point is st i l l t h e s ys tem of auxiliary equations
E x a m p l e 2.8. Consider t h e quasi-linear equation u x + u u y = 0, u ( 0,y ) = f\y ).
T h e s y s t e m o f a u x i l i a r y e q u a t i o n s i s
f u n c t i o n s o f x, y, and u, wn as quasi-linear. In anly in t h e first degree, Such equations occur in water waves. The basic l ua t i o ns. T he starti ng (2.11).
34
CHAPTER 2: METHOD OF CHARACTERISTICS
d x d y
Here d u = 0 implies u = c. Using t hi s value o f u in — = — one
1 u
y = c x + c\. Thus, t h e characteristics are given by t h e curves o f inter­
se c ti on o f t h e surfaces y — u x — c\ and u = c, and t h e general solution can b e expressed as u — g ( y — u x ). Applying t h e ini tial condition, we g e t f ( y ) = g ( y ). Hence t h e solution t o t h e problem is u = f ( y — u x ).
y
I f f ( y ) = Vi t h e s ol uti on becomes u =
1 + x
I f f ( y ) = IJ2 · the n we have u = ( y — u x ) 2, which after some algebraic simplification yields
1 + 2x y ± λ/I + 4 x y U = 2^2 '
A careful e xamination by checking t h e limit as x —> 0 shows t h a t t h e
valid sol uti on is ________
1 + 2 x y — J I + 4 x y U= 2 ^ ·'
| www | In t h e Mathemati ca Notebook Exampl e 2.8. ma, t h e following pl ot s are given:
(i) T h e graphs o f y — c x = C\ for c = 1,2,3, and C\ = 0,1, 2. These curves are characteristics and also base characteristics.
y
(ii) The graph o f u = represents an integral surface.
1 4- x
2^ _l_ 2x y λ j\ 4xy
( i i i ) T h e g r a p h o f u = -------------- — ^------------- represents another integral
surface.
Exa mple 2.9. Consider
u x + g ( u ) u v = 0, u ( 0,y ) = f ( y ).
T h e s y s t e m o f a u x i l i a r y e q u a t i o n s i s d x d y
i g ( u y
d u = 0.
As in Example 2.7, u = c, and y = g ( c ) x + c i. The general sol uti on is u = h ( y — x g ( u ) ). After applying t h e ini tial condition, we ge t f ( y ) = h( y). Hence t h e solution t o t h e problem is u = f ( y — x g ( u ) ). ■
2.3. QUASI-LINEAR EQUATIONS
35
E xample 2.1 0. Consider
UX + Uy = U 2 + 1,
T h e auxiliary equations are then
d x = dy =
u ( 0,y )
d u u 2 +1 ’
Th e solutions are y = x + c, and t a n u = x + c\. Thus, t h e general
sol uti on is u = t a n ( x + g ( y — x ) ), and t h e pa1 problem is
t a n x + f ( y — x)
XI ~ I I '
1 — f ( y — x ) t a n x
f ( y ) ·
ticular s o l ut i o n for t h e
u = y
characteristic
x - y
Fig. 2.3.
E xample 2.1 1. Consider t h e partial differential equation
4“ Uy = l,
cylinder
2
where u = 1 is prescribed on t h e ini tial curve x < 1. The auxiliary sy s t em (2.11) for t hi s part
= 0 ( x - a x i s ) for 0 < ial differential equation
36
CHAPTER 2: METHOD OF CHARACTERISTICS
Th e s ol uti on o f d x = 2y d y is t h e parabolic cylinder x = y 2 + A, while t h a t o f d u = d y is t h e plane u = y + B, where t h e parameters A and B are constant for each characteristic. For any point (X o,y o,u o ) we find t h a t A = χ o — j/οι and B = z o — y o- If t h e point (x o,y o,u o ) is taken, e.g., as t h e origin of t h e t h e coordinate sys tem, then
x — y 2, and u = y. (2.16)
T h e equation o f t h e characteristic is t h e i ntersecting curve (in t hi s case, t h e parabola) o f t h e sol uti on (2.16), as shown in Fig. 2.3.
T h e characteristic through t h e point ( 1,0,1 ) is t h e intersection o f t h e parabolic cylinder x = y 2 + 1 and t h e plane u = y + 1. Moreover, i f u = x 2 on t h e initial curve y = 0, then u = y + ( x — y 2 ) 2 represents t h e integral surface. ■
Exa mple 2.1 2. Consider
u x + 2 u u y = 1.
Th e auxiliary equations are
d x = ^ - = du.
2 u
T h e s o l u t i o n s a r e u = x + c, and u 2 = y + c\. Hence t h e general solution is
u = x + f ( u 2 — y), or u 2 = y + g ( u — x ).
I t i s i m p o r t a n t t o c h o o s e t h e a p p r o p r i a t e g e n e r a l s o l u t i o n s f o r t h e g i v e n i n i t i a l c o n d i t i o n s. T h u s, f o r e x a m p l e, i f t h e i n i t i a l l i n e ( c u r v e ) i s x = y, and t h e value o f u on t hi s ini tial line is u ( y,y ) = y, the n from t h e first sol uti on we get u = x. B ut t h e second sol uti on gives y 2 = y + ff(0), which does not yield a value for t h e function g.
I f t h e i n i t i a l l i n e i s y = x, and u ( y, y ) = y 2, t h e n t h e second solution yields
u 2 = y + ( u — x ) 2 + 2( u — x ) ± ( u — x )\/l + 4( u — x ),
w h e r e t h e p l u s s i g n c o r r e s p o n d s t o x > y > and t h e minus sign
t o x < y < i. S u bs titu tin g y = x and u = y 2 in t h e second solution, we have y 4 = y + g ( y 2 — y ). Now let z = y 2 — y, then
y = i ( l ± λ/I + 4ζ) , and y 4 - y = z 2 + 2z ± zy/1 + 4z = g( z),
2.3. QUASI-LINEAR EQUATIONS
which gives t h e solution o f th e equation. B u t if we ge t y 2 = y + f ( y A — y ), which is more diffici;
E xample 2.1 3. Consider
( y + u ) u x + y u y = x - y
F o r t h i s p r o b l e m, i t i s c o n v e n i e n t t o u s e t h e p a r a m e t r i c f o r m (2.12), i.e.,
w e u s e t h e f i r s t s o l u t i o n
I t t o r e s o l v e. ■
d x
d y
d u
d.
y + u y x - y
w h i c h c a n b e r e w r i t t e n a s
d x d y d u
* = » + "' T s = y' d~s =
T h e s o l u t i o n o f t h e m i d d l e e q u a t i o n i s y = A e and third equations yields
d ( u + x) ds
= u + x.
I t s s o l u t i o n i s u + x = B e s. Subtracting t h e third results in
d ( u — x)
= x - u - 2 y,
ds
which can be expressed as
d( u — x) ds
(u — x ) = —2 A
This equation is linear in u — x. Its solution is
u — x = C e ~ s — A e s.
y
R e p l a c i n g e s by — in t h e two solutions, we get
Λ
B
u + x = — y, and u — x + y Ά
N o t i n g t h a t B/A and C A can be replaced c2 = f ( c i ), we have
by
( u - x + y ) y = f
u + x
y
a s t h e g e n e r a l s o l u t i o n o f t h e g i v e n e q u a t i o n.
37
auxiliary equations in
x - y.
’. Ad di tion o f t h e first
iirst equation from the
C A
y
c\ and C2, and that
(2.17)
38
CHAPTER 2: METHOD OF CHARACTERISTICS
Ex a mple 2.1 4. Consider
x y u u x + ( x 2 — u 2 ) u y = x 2y.
T h e a u x i l i a r y e q u a t i o n s a r e d x
d y d u
x y u x 2 — u 2 x 2 y
F r o m
d x d u x y u x 2 y ’
we g e t x — u = c\, and t he n using t hi s solution in
d y d u
x 2 — u 2 x 2y ’
we obtai n
Thi s equation yields
, c\d u
ydy = - τ -,— ■
Uz + Cl
y2 =
2 a t a n 1 — l· C2, i f c\ = a 2, a
i U _ a , -r 2
a In- 1 'r ~
u + a
T h e g e n e r a l s o l u t i o n c a n n o w b e f o u n d.
+ c 2, i f c i = - a .
2.4. F i r s t O r d e r N o n l i n e a r E q u a t i o n s
T h e g e n e r a l f o r m o f a f i r s t o r d e r n o n l i n e a r e q u a t i o n i s F ( x,y,u,u x,u y ) = 0.
Consider t h e two-p arameter family o f surfaces
f ( x,y,u,a,b ) = 0.
T h e n
fx
+ /« «! = o, a n d fy
+ fuUy
= 0.
(2.18)
(2.19)
(2.20)
2.4. NONLINEAR EQUATIONS
Equations (2.19) and (2.20) form a set of thre parameters a and b. If a and b are eliminated one gets an equation of the type (2.18). Therefc assume that the solution of (2.18) is of type (2. envelope* of this family will also be a solution this point we state the Cauchy problem: Dete: u and its partial derivatives satisfy
e equations in the two from these equations, :ore, it is reasonable to 19). It is clear that any of equation (2.18). At mine u(x,y) such that
F(x,y,u,ux,uy) = 0,
subject to the condition u(0,y) = <fr(y). In scribed on the initial line x = 0 (y-axis). Init be prescribed on any simple curve which is Eq (2.18). Thus, for example, if the param* is x = x(s),y = y(s), then the initial data can
this
case u[x, y) is pre- ial data can, however, not a characteristic of etric form of the curve be written as
u(x{s),y(s)) = tp(s).
We will now distinguish between different kinds
(2.18).
Complete integral: A two-parameter famih f (x, y, u, a, b) = 0 is known as a complete integ
General integral: If b = g(a), where g is an an envelope of the family of solutions of the co then the envelope whose equation contains a known as the general integral corresponding to
Singular integral: If the two-parameter fa: has an envelope, then the equation of this en singular integral of (2.18).
Cauchy’s method of characteristics:
to the method of characteristics discussed earli linear partial differential equations.
Consider a first order nonlinear equation ( we will use the notation
ux = p and uy = q.
39
of solutions of equation
of solutions of the type ral.
arbitrary function and ihplete integral is found, ,n arbitrary function is the solution (2.19).
mily of solutions (2.19) Lvelope is known as the
This method is similar er for linear and quasi-
!.18). For convenience,
(2.21)
* A n e n v e l o p e o f t h e f a m i l y o f s u r f a c e s f ( x, y, u y a, 6) is a s u r f a c e w h i c h t o u c h e s s o m e m e m b e r o f t h i s f a m i l y a t e v e r y p o i n t.
40
CHAPTER 2: METHOD OF CHARACTERISTICS
I f u ( x,y ) = c is a sol uti on o f (2.18), t h e n u and i t s partial derivatives
sa t i sf y t h e equation (2.18). B u t t h e t o t a l derivative o f u is given by
d u = p d x + q d y. (2.22)
Differentiating (2.18) and (2.22) with respect t o p, we get
d x + ^ d y = 0, (2.23)
dp
and
FP + F^ = °- (2·24)
Equations (2.23) and (2.24) yield
d x = dy = p d x + q d y = d u = ^
F p F q p F p + q F q p F p + q F q
T h e s e a r e t h e c h a r a c t e r i s t i c ( o r a u x i l i a r y ) e q u a t i o n s o f E q ( 2.1 8 ). I t
c a n b e e a s i l y v e r i f i e d t h a t t h e y r e d u c e t o t h e c h a r a c t e r i s t i c e q u a t i o n s o f
a l i n e a r o r a q u a s i - l i n e a r p a r t i a l d i f f e r e n t i a l e q u a t i o n a c c o r d i n g a s E q
( 2.1 8 ) i s l i n e a r o r q u a s i - l i n e a r. T h e p a r a m e t e r t introduced in (2.25) is such t h a t
Since p is a function o f t, it follows tha t
dp d x d y
= P x ~ ^ = P y ^ q = P x F p + q x F q, ( q x P y ) · (2.27)
Differentiating Eq (2.18) w i t h respect t o x, we have
F x -f· F u u x -I- F p P x + F q q x — 0.
Us ing t h i s equation and inserting t h e value o f F p p x + F q q x in (2.27), we get
^ = - ( F x + F u u x ) = - ( F x + P F U). (2.28)
Similarly,
dq
dt
Combining equations (2.25), (2.28), and (2.29) we get
= - ( F y + qFu ). (2.29)
d x dy d u dp dq
Fp F q p F p + q F q - ( F x + p F u ) - ( F y + q F u )
= dt. (2.30)
2.4. NONLINEAR EQUATIONS
41
Thi s s ys tem o f auxiliary equations (also know equations) is used t o solve a nonlinear equation
n as t h e characteristic by Cauchy’s method.
T h e difference between t h e equations (2.25) equations (2.11) for a linear partial differentii e quations (2.25) contain p and q e x pl ic i t ly and solve t he m we need additional equations which Th e solution is found by eliminating p and (2.30) and t h e given equation. The eliminant two arbitrary c onstants and will represent a o equation. We demonstrate t h e method by t h e
Ex a mple 2.1 5. Consider
and t h e corresponding 1 equation is t h a t t h e , therefore, in order t o are included in (2.30). from t h e sol uti ons o f will, in general, contain omplete integral o f t h e following examples.
u = Apq.
The auxiliary s ys tem (2.30) for t hi s equation is
d x d y d u d%
λ = - 4 ρ ’ λ = - 8 μ' 5
Since
we get
d x dq dy
— = 4 - ^, and - £ = 4 d t d t d t
x + c\ = A q and y + c 2 =)= 4 p. (2.32)
en equation, we get t he
Su bs titu tin g t he se values o f p and q into t h e giv complete sol uti on as
u = ^ ( x + c\) ( y
+ c2).
However, i f we demand t h a t t h e solution pass t he n (2.33) may or may not yield t h e required s ince t h e s ol uti on of
d 2x dq
d t 2 dt
Aq = —
d:
dt
is x = ci + C2 e- t, we can- require t h a t u = y 2 on t h e initial curve x = 0, which correspond solution given by (2.33) fails t o yield t h e requ: t h i s si t ua t i o n we follow an alternate approach
dq
dt
-q-
(2.31)
dp dt ’
(2.33)
through a given curve, solution. For example,
be t h e ini tial c ondition 5 t o t = 0. The n the red solution. To avoid The ini tial values for
42
CHAPTER 2: METHOD OF CHARACTERISTICS
χ, y, and u can be taken t o be x o — 0, y o = v, and u o = v 2, where v is a parameter. Let t h e ini tial value o f q be qo = I — I . Thus,
\d y J o ® - ( | ) 0= 2»”= 21''
V
a n d s i n c e u o = ^ P o q o, we find t h a t t he initial value o f p is po = B ut
o
from equations (2.31) we note t h a t p and q can be solved in terms o f t as
p = A e - 4, and q = B e ~ l. (2-34)
Su b s t it u t in g t h e initial values in (2.32) and (2.34) we find t h a t
c i = 8i/, C2 = —f/2, A = v / 8, B = 2 v.
T h e n, f r o m ( 2.3 2 ) a n d t h e g i v e n e q u a t i o n, x, y and u can be expressed in terms o f v and t as
x = 8i/(e~t — 1), y = - ^ ( e - i + l ), u — i/2 e ~ 2 t. (2.35)
Th e required sol uti on can now be found by eliminating v and t from(2.35)
| www | In t h e Mathemati ca Not e book Example2.15 .ma, t h e plot o f t h e above s ol uti on represents an integral surface.
We will now consider some special cases of equation (2.18).
Exa mple 2.16. Consider
u = p x + qy + f ( p,q ), or F = p x + q y + f ( p, q ) - u = 0. (2.36)
T hi s is a special t y p e o f nonlinear equation. It always has a complete solution which can be obtained in a simple manner. T h e last two o f t h e characteristic equations are
dp = 0, and d q = 0,
which yield p = a and q = b, where a and b are arbitrary constants. S u b s t it u t in g t h e s e values in (2.36), one gets u = a x + b y + f ( a, b) as the compl ete solution. Equations o f t h e t y p e (2.36) are known as C l a i r a u t e q u a t i o n s. There are other special type s o f partial differential equations which yield t h e complete sol uti on in a relatively easy manner. ■
2.4. NONLINEAR EQUATIONS
43
Exa mple 2.17. Consider
f ( p, q) = o.
N o t e t h a t t h e characteristic equations will yield dp = 0 and dq = 0. So p = a, and solving f ( p,q) = 0 for q will give q =- g( a); t he n observing t h a t
du = adx + g(a) dy,
w e g e t
u = ax + g( a)y + c.
As an example, let f ( p, q) = p2 + q2 — 1 = 0. T h e auxiliary equations (2.30) are
(2.37)
= P, -JZ = Q’ du = dt, dp = C
dt
dt
Using dp = 0, we get p = a and q = \/l — a 2, an w i t h du = p d x + qdy yield
, dq = 0.
i t he se two combined
u= ax + ys/1 — a2 + c.
T hi s is a compl ete solution. Another complete sol dx
b y u s i n g p = a in — = p and noti ng t h a t d dt
dx
du = —, which gives a
u = — + a. a
S i m i l a r l y, du =
dy
ution can be obtained u = dt, thus g e t ti n g
Vl — a2
implies tha t
u =
y
V l ^ a 2
Thus, we can write au = x + aa, and
+ β·
uy 1 — a2 = y + αβ.
R e p l a c i n g α α a n d αβ by —c and —d, respectively, get
u2 = (x - c)2 + ( y - d)2, which is another complete solution. ■
and eliminating a we
44
CHAPTER 2: METHOD OF CHARACTERISTICS
Ex a mple 2.1 8. Consider
F ( u,p,q ) = 0. (2.38)
T h e last three terms o f t h e s ys tem (2.30) yield
dp dq , d p d q
— = d t, or — = —,
- p F u —q F u p
i.e., p = a 2 q, where a 2 is an arbitrary constant. This equation together w i t h (2.38) can be solved for p and q, and then we proceed as in t h e previous example. Thus, let F ( u,p, q ) = u 2 + p q —A = 0. The n following t h e above procedure we have p = a 2 q, which gives
q = ± -\/4 — u 2 and ρ = ± a\/4 ~ a
T h e r e f o r e, s i n c e d u = p d x + q d y, we have
or
d u = ±\/4 — u 2 ( a d x + — d y ) ,
\ a J
^ U ± ( a d x + — d y ) ,
which gives
Hence
λ/4 — u 2 V a
s i n 1 — = ± I a x -\— y + c 2 \ a
u = ±2 sin ( a x + - y + c
| w w w | I n t h e M a t h e m a t i c a N o t e b o o k E x a m p l e 2.1 8 .ma, t h e p l o t s o f u = 2s i n f a a; + - y 4 - c
V a
f o r c = 0,1,2, a n d a = 1,2 represent some particular integral surfaces.
E x a mple 2.1 9. If F ( x,y,u,p,q ) = 0 is independent o f u and can be expressed as φ ( χ,ρ ) = i p ( y, q ), t he n each o f t he se functions must be c onst an t.Th us, i f φ ( χ,ρ ) = c and i p ( y, q ) = c can be solved for p and q, t h e n a compl ete integral can be obtained. For example, consider
F{x,y,u,p,q) = p 2( 1 - x 2) - q2(4 - y 2) = 0.
2.4. NONLINEAR EQUATIONS
45
which gives
Then
p2(i — χ2) = q2(4 — y2) —
P =
\/l - x 2
a n d q =
w h e r e w e h a v e i g n o r e d t h e n e g a t i v e s o l u t i o n s. N q d y, we have
ow since d u = p d x +
a , a
d u = , d x +
V T
\j 4 - y 2
d y,
which can be integrated t o give
a ^sin-1 x + si n-1 ^ + b
u =
| www | In t h e Mathematica Not e book Example!! t h e above sol uti on for a = 1 and b = 0,1 repr integral surfaces.
.1 9.ma, t h e pl ots of i s e n t some particular
E xa mple 2.2 0. Consider
F ( x, y, u,p, q ) = 2p q y - p u - 2 a
f o r w h i c h
F x = 0, F y = 2pq, F u = - p, F p = 2qy - v,, and F q = 2py.
T h e a u x i l i a r y e q u a t i o n s a r e
d x d y d u
d u
= 0,
2 q y — u 2 p y 2 pq y — p u + 2 p q y p u + ■
T h e last pair reduces to
dp dq p q
which gives p q = a. Using t hi s value o f p q in (2.39), we get
2 a y — p u — 2 a = 0,
which yields
at.
.2 ’
= 0,
(2.39)
dp dq 4 a p 2 —p q'
p u = 2 a y — 2a, and q
2 ( a y
-a)'
The se equations after integration yield
u 2 = 4 ( a y — a ) ( x — β ). ·
4 6 C H A P T E R 2: M E T H O D O F C H A R A C T E R I S T I C S
2.5. G e o m e t r i c a l C o n s i d e r a t i o n s
B e f o r e w e discuss t h e geometrical interpretation of t h e partial difFer­
e ntial equation (2.18), i.e., F ( x,y,u,p,q ) = 0, let us recall t h e geo­
metrical interpretation of a first order ordinary differential equation y' = f ( x,y ). Here f ( x,y ) represents t h e slope o f any integral curve at t h e point ( x,y ). T hi s slope is unique at every point. If we graph f ( x,y ) = c, the n t h e curve so obtained is known as an i socline or a curve o f constant slope. Of course, t h e curve i t s e lf does not have con­
st a n t slope, but every integral curve which intersects f ( x,y ) = c has t h e sl ope c at t h e point o f intersection. Since t h e correspondence be­
tween integral curves and t h e points o f an isocline is o n e - t o - o n e, the number o f integral curves is, in general, equal t o t h e number of points on t h e isocline, i.e., there e x is t s a single infinity o f them. However, t h e ex ce pt i o n t o t hi s rule occurs when isoclines intersect at a point, in which case t h e point is a singular point o f t h e differential equation, or when t h e i socline is also a solution curve, in which case t h e isocline is a straight line wi t h slope c and t h e isocline is an envelope o f t h e integral
dy y
curves, e x ce p t for t h e equation — = — whose isoclines and the integral
d x x
curves are t h e same.
T h e si t ua t i o n for a partial differential equation is somewhat com­
plicated. In t h i s case t h e values o f p and q are not unique at a fixed point ( x,y,u ). I f an integral surface is g ( x,y,u ) = 0, the n p and q represent t h e slopes o f t h e curves of intersection o f t h e surface w i t h t h e planes u = const. Moreover, p, q, —1 represent t h e direction ratios of t h e normal t o t h e surface at t h e point ( x,y,u ). The derivatives p and q are constrained by Eq (2.18). Obviously, at a fixed point, p and q can be represented by a single parameter. Hence, there are i nfinitely many possible normals and consequently infinitely many integral sur­
faces passing through any fixed point. So, unlike t h e case o f ordinary differential equations, we cannot determine a unique integral surface by making it pass through a point.
2.6. SOME THEOREMS
47
Cauchy establ i shed t h a t a unique integral su by making it pass through a continuous t w i st e d s as an initial curve, except when the curve is a ch ferential equation. Now, t h e infinity of normals p point generates a cone known as t he normal co: ing tange nt planes t o t h e integral surfaces envel< t h e Monge cone. In t h e case o f a linear equatii degenerates into a plane since each normal is pe line.
rface can be obtained pace curve, also known
aracteristic o f t h e dif- assing t hr ough a fixed ine. The correspond- ope a cone known as on, t h e normal cone rpendicular t o a fixed
tin
Consider t h e equation a p + b q = c. Then perpendicular t o t h e direction ratios a,b, c. Thi; a fixed point. T h e Monge cone t he n degenerate planes known as t h e Monge pencil. The commo: t h e line through t h e fixed point wi t h direction is known as t h e Monge axis.
e direction p, q, — 1 is s direction is fixed at into a coaxial set of i:n axis o f t h e planes is ratios a,b,c. This line
2.6. Some Theorems on Character!
Suppose u ( x,y ) = f { x,y ) is an integral su differential equation Eq (2.18). Then t h e set of
x o,yo, u 0 = u ( x 0,y 0), p 0 =
= ( d- L )
d x J(xo,yo)
%
1)
If
which represents a plane w i t h normal ( p o, qo, — t h e point (x 0, y o, u o ), is called a p l a n e e l e m e n t. lies on S, t h e n t h e element { x Q,y o,u o,P o,q o ) sai called an i n t e g r a l e l e m e n t o f t h e surface. Let R ( x o,V o ) in t h e plane u = 0. If t h e functions in R, the n t h e element ( x o,y o,u o,P o,q o ) is called 5.
A curve Γ wi t h parametric equations x — x lies on t h e surface S i f u ( t ) = f ( x ( t ),y ( t ) ) for all If a point P o on Γ corresponds t o t h e value io of
t h e direction o f t h e tangent line is given by ( — ,^ j -, —
sties
face S o f t h e partial Jiumbers
d f\
d yJ(*o,yo)
and passing through t h e point ( x o,y o,u o ) .tisfies Eq (2.18) and is be a neighborhood of and f y are continuous a t a n g e n t e l e m e n t of
( t ),y = y ( t ),u = u ( t ) admissible values o f t. t h e parameter t, t he n d y d u\
$ t ’ d t ’ d t ) t =t o
T h i s
48
CHAPTER 2: METHOD OF CHARACTERISTICS
direction is normal t o ^p Q = ( ^ “ )to>9o = i f
d u\ ( d x\ f d u\
~dt)to=p° U J to+qo U ) i0·
Thus, a set o f five functions x ( t ),y ( t ),u ( t ),p ( t ),q ( t ), which s a t i s f y t h e c ondition
d u , x d x , . d y
d e f i n e s a s t r i p on t h e curve Γ. If t hi s strip is an integral element, t he n i t is an integral strip o f t h e partial differential equation. If t hi s integral strip at each point touches a generator o f t h e Monge cone, the n t h e integral strip is a c h a r a c t e r i s t i c s t r i p.
W e w i l l s t a t e s o m e t h e o r e m s o n t h e c h a r a c t e r i s t i c s. T h e p r o o f s c a n b e f o u n d i n t h e r e f e r e n c e s c i t e d b e l o w.
T H E O R E M 2.2. A n e c e s s a r y a n d s u f f i c i e n t c o n d i t i o n f o r a s u r f a c e t o be a n i n t e g r a l s u r f a c e o f a p a r t i a l d i f f e r e n t i a l e q u a t i o n i s t h a t a t e a c h p o i n t i t s t a n g e n t e l e m e n t s h o u l d t o u c h i t s e l e m e n t a r y c o n e ( t a n g e n t c o n e o r M o n g e c o n e ) o f t h e e q u a t i o n.
T h e o r e m 2.3. T h e f u n c t i o n F ( x, y, u,p, q ) i s c o n s t a n t a l o n g e v e r y c h a r a c t e r i s t i c s t r i p o f t h e e q u a t i o n F ( x,y,u,p,q ) = 0.
T h e o r e m 2.4. I f a c h a r a c t e r i s t i c s t r i p c o n t a i n s a t l e a s t o n e i n ­
t e g r a l e l e m e n t o f F ( x,y,u,p,q ) — 0, i t i s a n i n t e g r a l o f t h e e q u a t i o n F ( x,y,u,u x,u v ) = 0.
For t h e linear partial differential equation a p + b q = c, we have
T h e o r e m 2.5. E v e r y s u r f a c e g e n e r a t e d b y a o n e - p a r a m e t e r f a m i l y o f c h a r a c t e r i s t i c c u r v e s i s a n i n t e g r a l s u r f a c e o f t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n.
T h e o r e m 2.6. E v e r y c h a r a c t e r i s t i c c u r v e w h i c h h a s o n e p o i n t i n c o m m o n w i t h a n i n t e g r a l s u r f a c e l i e s e n t i r e l y o n t h e i n t e g r a l s u r f a c e.
2.7. LINEAR AND QUASI-LINEAR EQUATIONS
49
THEOREM 2.7. E v e r y i n t e g r a l s u r f a c e i s g e n e f a m i l y o f c h a r a c t e r i s t i c c u r v e s.
T h e o r e m 2.8. I f
D =
d x d y d s d t
d y d x d s d t
— a
d y
d t
- b — φ 0 a t
e v e r y w h e r e o n a n i n i t i a l c u r v e C, t h e n t h e i n i o n e a n d o n l y o n e s o l u t i o n. I f, h o w e v e r, D = C t h e i n i t i a l v a l u e p r o b l e m c a n n o t be s o l v e d u n l e s i c u r v e, a n d t h e n t h e p r o b l e m h a s a n i n f i n i t y o f s<
H a l v a l u e p r o b l e m h a s e v e r y w h e r e a l o n g C, C i s a c h a r a c t e r i s t i c M u t i o n s.
P r o o f s o f T h e o r e m s 2.2, 2.3 a n d 2.4 c a n b e p a g e s 6 2 - 6 4 ), a n d o f T h e o r e m s 2.5, 2.6, 2.7, ai H i l b e r t ( 1 9 6 5, p a g e s 6 4 - 6 6 ).
f o j i n d i n S n e d d o n ( 1 9 5 7, ,n d 2.8 i n C o u r a n t a n d
Se cond Or d e r Equa t i c
2.7. Li ne a r a n d Qua s i —l i n e a r Eq ua t
F o r a l i n e a r o r q u a s i - l i n e a r p a r t i a l d i f f e r e n t i a l h i g h e r o r d e r, t h e c h a r a c t e r i s t i c e q u a t i o n i s d e t e s t o r d e r t e r m s i n t h e p a r t i a l d i f f e r e n t i a l e q u a t k n o w n a s t h e p r i n c i p a l p a r t o f t h e partial differe t h e solution o f t h e characteristic equation leads first order partial differential equation, t h e solut t i c equation o f a second order partial differential coordinate transformation which when applied i der partial difFerential equation t o a simpler forn: called t h e c a n o n i c a l f o r m. Consider a second or equation
a n « » + 2 ai2 u x y + a22 u y y + F ( x, y, u, u.
■ated by a o ne- par amet er
i n s
10ns
q u a t i o n o f s e c o n d o r i r m i n e d b y t h e h i g h - 10η. T h e s e t e r m s a r e n t i a l e q u a t i o n. W h i l e t o t h e s o l u t i o n o f t h e i o n o f t h e c h a r a c t e r i s - ,1 e q u a t i o n l e a d s t o a e d u c e s t h e s e c o n d o r - . T h i s s i m p l e r f o r m i s i e r p a r t i a l d i f f e r e n t i a l
uy) 0,
(2.40)
where a n,a\2 and <222 are functions of x and y only, and F is a function
50
CHAPTER 2: METHOD OF CHARACTERISTICS
of x,y,u,ux, and uy. The different canonical forms of Eq (2.40) are Elliptic : u^ + ηηη + ΰ(ξ, η,υ,,χΐξ,υ,η) = 0,
τ τ κ I' / ~ Ur,r} ΤΙ ’> U' Url ) = i n λ -ι \
H y p e r b o l i c : ( 2.4 1 )
L υ,ζ η + υ ( ξ,η,η,ι ΐ ξ,η η ),
Par abol i c : + G( £^,u,u£,uv) =0.
In order to reduce Eq (2.40) to a canonical form, we introduce a reversible transformation
ξ = ξ{χ,ν), and η = η(χ,ν), (2.42)
with the condition that the Jacobian
J = d{x y) ~ ^xTly ~ ^ y ^ (2-43)
Using this transformation and noting that
Ux = ΙΙξ ζχ + ΊΙη Tjx,
Uy — «ξ ζy “t- Urj Tj y,
Uxx — U^ £ ζ χ 4" 2 ΐ ί ξ η ζ χ η χ + Uη η Tjx + ΊΙξ ζ χ χ Ίί η Tjx x,
UXy — U^£ ζ Χ ζ y -f- Uζ η ( ζ χ Tjy + ξ y TjX ) + Uη η ΐ]χ Tjy + Ι ί ξ £ Xy + Ι ί η TjXy ,
Uyy = ωξξ ζ y + 2Μξη T\y + Ι ί η η Tjy + Μξ £ y y + Ι Ι η Tj yy,
(2.44)
Eq (2.40) reduces t o
^ 1 1 υ,ζξ + 2A12 υ,ξη + A 22 υ,ηη + G(£, η, u, ιΐζ, u„) = ο, (2.45)
where G is a function of ξ,η, u,u^, and un, and An, A12, and A22 are functions of ξ and η, given by
^4ii = an ξχ + 2ai2 ξχ ξυ + 2 ξ%,
A i
2
= an ξχ % + a i 2 (ξχ ξυ + ηχ) + α22 ξν ην, (2.46)
-422 = a i l VI + 2 α ΐ2 ηχ ηυ + α22 tfy.
The funct i on G is linear or nonlinear according as F is linear or non­
linear.
If we now choose ξ and η such t h a t both s ati sfy t h e condition
a n ζ χ + 2ai2 ξ χ + a22 ζ 2 = 0, (2-47)
2.7. LINEAR AND QUASI-LINEAR EQUATIONS
t h e n A n = A 22 = 0. E q ( 2.4 0 ) w i l l t h e n r e d u c e t o
2A12 uiv 4- G ( £, η, i t, u ^,u v ) ■- 0.
I t c a n b e v e r i f i e d ( a f t e r s o m e t e d i o u s a l g e b r a ) t h a t
^12 — ^ 11^22
(af2 ~ <2116(22) (ξχ % +
T h i s m e a n s t h a t t h e s i g n o f t h e q u a n t i t y I
t h e r e v e r s i b l e t r a n s f o r m a t i o n ( 2.4 2 ), a n d t h e q u i f I J\ = 1. A n i m p o r t a n t c o n s e q u e n c e o f t h i s r d i f f e r e n t i a l e q u a t i o n d o e s n o t c h a n g e i t s c l a s s i f i e d t r a n s f o r m a t i o n s [see §1.3, w h e r e i t w a s p r o v e d s o l u t i o n o f E q ( 2.4 7 ), t h e n φ ( χ,ν ) — c is a s o l u t i i e q u a t i o n s ( 1.1 0 ) a n d ( 2.4 7 ) a r e c a l l e d c h a r a c t e: p a r t i a l d i f F e r e n t i a l e q u a t i o n ( 2.4 0 ). E q u a t i o n g i v e n b y
dy
dx 2 a
a\2 ± λ/a - u - a n a 2‘.>
12
I f t h e p a r t i a l d i f F e r e n t i a l e q u a t i o n i s h y p e r b o l i c l u t i o n s, r e s u l t i n g i n t w o c h a r a c t e r i s t i c s f o r t h e t i o n. I f t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n i s p a r a t o n e r e a l s o l u t i o n, a n d h e n c e o n l y o n e c h a r a c t e r i s t h e r e a r e n o r e a l s o l u t i o n s, a n d s o t h e r e a r e n o
I n o r d e r t o t r a n s f o r m t h e p a r t i a l d i f F e r e n t i a l i c a l f o r m, w e i n t r o d u c e t w o i n d e p e n d e n t v a r i η = η( χ, y), w h e r e ξ a n d η a r e s o l u t i o n s o f E q ( 2.4 9 ). I n t h e c a s e o f a p a r a b o l i c e q u a t i o n we h a v e o n l y o n e s o l u t i o n, s o η is c h o s e n a r ­
b i t r a r i l y e x c e p t t h a t i t m u s t s a t i s f y t h e c o n d i t i o n ( 2.4 3 ). I n t h e c a s e o f a n e l l i p t i c e q u a t i o n, t h e s o l u t i o n s a r e c o m p l e x c o n j u g a t e s, a n d we c a n u s e t h e r e a l a n d i m a g i n a r y p a r t s o f t h e s o l u t i o n s a s t h e n e w i n d e ­
p e n d e n t v a r i a b l e s. I t w i l l b e s h o w n i n C h a p t e r 5 t h a t c a n o n i c a l f o r m s a r e f r e q u e n t l y n e c e s s a r y t o s o l v e p a r t i a l d i f F e r e n t i a l e q u a t i o n s b y t h e m e t h o d oF s e p a r a t i o n o f v a r i a b l e s.
51
ζ -y f]x)
( 2.4 8 )
* ΐ 2—α ι1ι α 22) is i n v a r i a n t u n d e r i n t i t y i t s e l f is i n v a r i a n t ■esult is t h a t t h e p a r t i a l t i o n u n d e r n o n s i n g u l a r t h a t i f ζ = φ( χ, y) i s a o n o f E q ( 1.1 0 ) ]. B o t h r i s t i c e q u a t i o n s o f t h e 10) h a s t w o s o l u t i o n s
a
( 2.4 9 )
t h e n t h e r e a r e t w o so- p j a r t i a l d i f f e r e n t i a l e q u a - ol i c, t h e n t h e r e is o n l y t i c. I n t h e e l l i p t i c c a s e c h a r a c t e r i s t i c s.
e q u a t i o n t o i t s e a n o n - a b l e s ξ = £ ( x,y ) a n d
W e w i l l n o w d e m o n s t r a t e t h e e f f e c t i v e n e s s d u c i n g s e c o n d o r d e r p a r t i a l d i f F e re n t i a l e q u a t i o n s s o m e e x a m p l e s.
f t h i s t e c h n i q u e f o r r e - t o c a n o n i c a l f o r m s b y
E x a m p l e 2.2 1. T r a n s f o r m t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n y 2 u xx - 4 x y u xy + 4 x 2 u yy + ( x 2 + y 2) u x + u y = 0
52
CHAPTER 2: METHOD OF CHARACTERISTICS
t o i ts canonical form. This equation is parabolic and i ts characteristic equation, given by
y 2 (dy) 2 + 4x y d x d y + 4 x 2 ( d x ) 2 = 0,
has onl y one solution
d y 2 x „ o 9
— = --------, which yields 2 x + y = c.
d x y
I n t h i s c a s e t h e r e i s o n l y o n e c h a r a c t e r i s t i c c u r v e, a n d s o w e m a k e t h e s u b s t i t u t i o n
ξ = 2 x 2 + y 2, and η = x.
T h e s u b s t i t u t i o n f o r η is arbitrary in t hi s situation, t h e o n l y c ondition bei ng t h a t t h e Jacobian should be nonsingular. Thus, we have
Ux = 4 x U£ + Uv,
Uy = 2 y u £,
Ux x = 16x2 + S x Ηξ η + Uv v + 4«ξ,
u Xy — 8x y ν*ξξ Ί- 2y u u y y = 4y2 ι ΐ ζ ξ + 2 u (.
S u b s t i t u t i o n o f t h e s e v a l u e s i n t o t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n l e a d s t o t h e c a n o n i c a l f o r m
( ξ - 2 η 2) u v v + [4ξ + 4 η ( ξ - η 2 ) + 2 ν/ξ ^ 2 ^ 2] «ξ + ( ξ - η 2 ) η η = 0. ■
| www | T h e pl ots o f th e characteristic curves for t hi s example are avail­
able in t h e Mathematica Not e book Exampl e 2.2 1 .ma.
Ex a mple 2.2 2. Transform t h e partial differential equation
2 i o 2 V2 3 x 3
y U>XX 4x y UXy 4“ OX Uy y UX Uy 0
x y
t o t h e canonical form. The principal part in t hi s partial differential equation is similar t o t h e previous example, but it is hyperbolic and, therefore, i t s characteristic equation
y2 (dy)2 + 4 x y d x d y + Sx2 (dx)2 = 0
2.7. LINEAR AND QUASI-LINEAR EQUATIONS
53
has two independent solutions, namely,
— = — —, which yields 3 x2 4 d x y
a n d
d y x 2
— = —, which yields x + y = c 2. d x y
H e n c e, t h e n e w i n d e p e n d e n t v a r i a b l e s a r e
ξ = 3 a;2 4- y 2, and η = x 2 f - y z Th e partial derivatives of u wi t h respect t o t h e new variables are given
by
u x = 6xu% 4- 2i n,, u y = 2 y u ^ + 2 y u v,
u x x = 36a;2 Ηξ ζ 4- 24a;2 u ^ v 4- 4a;2 η η η -jf 6uj 4- 2u v, u x y = 1 2 x y Μξξ 4 - 16x y u ^ 4- 4a Uyy = 4y (ΐ^ξξ 4“ 2ΪΧξη 4“ Uηη) 4“ 2 (ϋξ
Su bs t it u t in g t he se values into t h e given partial get t h e canonical form
whose solution is
( ξ - ν ) ( ξ - 3 ν ) Η η = 0,
u = /( 3 a;2 + y 2) 4- g { x 2 4- y
I w w w | T h e p l o t s o f t h e c h a r a c t e r i s t i c c u r v e s f o r
b i e i n t h e M a t h e m a t i c a N o t e b o o k E x a m p l e 2.2 2
W e m e n t i o n h e r e a n i m p o r t a n t p r o p e r t y o f f e r e n t i a l e q u a t i o n s. T h e y a r e c a p a b l e o f t r a n s p i n t h e i n i t i a l d a t a a l o n g a c h a r a c t e r i s t i c. T h e C 2 are called strict solutions, whereas t hos e w: function or i ts first two derivatives are called ge: will demonstrate t hi s idea by a simple example reader is referred t o t h e t e x t s by John (1982 Hilbert (1965).
V = c i,
4~ u ^ .
differential equation we
t hi s example are avail- .ma.
hyperbolic partial dif- orting a discontinuity solutions t h a t are in ilth discontinuity in t he neralized solutions. We For more detai ls t h e and by Courant and
54
CHAPTER 2: METHOD OF CHARACTERISTICS
Ex a mple 2.2 3. Solve t h e partial differential equation
utt — u xx = 0, 0 < x < oo,
s u bj e ct t o t h e conditions
u ( 0,t ) = u( x, 0) = u t ( x, 0) = 0,
where H ( t ) is t h e Heaviside step function. B y introducing t h e charac­
t e ri s t i c coordinates ξ = x + t, η = x — t, t h e partial differential equation is reduced t o
ΙΧξη — 0.
It s s ol uti on is given by
u = /( £ ) + 9 ( v ) = f ( x + t ) + g ( x - t ).
T h e t e r m f ( x + t ) represents a wave traveling wi t h a negative veloc­
i t y coming from infinity. Since there are no sources or boundaries at infinity, it is not possible for a wave t o either emanate or be reflected from infinity. (This is also known as Sommerfeld’s radiation condition.) Therefore, t h e function f ( x + 1) must be taken t o be zero. Thus, we have from t h e boundary condition
g ( - t ) = H ( t ) e -\
w h i c h y i e l d s
u = H ( t - x ) e ~ { t ~ x ).
T h e i n i t i a l c o n d i t i o n s a r e t h e n a u t o m a t i c a l l y s a t i s f i e d. I n t h i s c a s e t h e d i s c o n t i n u i t y i n u propagates along t h e characteristic x = t.
EXAMPLE 2.2 4. T r a n s f o r m t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n V ^ X X y UXy “1“ <$X 'Uyy 0
t o t h e canonical form. In t h i s case t h e partial differential equati on is o f t h e e lliptic type. The characteristic equation is
y 2 ( dy) 2 + 4x y d x d y + 8 x 2 ( d x ) 2 = 0,
d y , , 2i £ - ( - Ϊ + 2.) —,
a n d i t s s o l u t i o n i s
2.7. LINEAR AND QUASI-LINEAR EQUATIONS
55
which yields
y 2 + 2(1 ± i ) x 2 = c i,2. In t hi s si t ua t i o n we define t h e new independent
ξ = 2 x 2 + y 2, and η = 2
T h e n, a s i n E x a m p l e s 2.2 2 a n d 2.2 3, w e g e t t h e
2 ν ( ξ - v ) K? + y - ηη) + ( ξ + η ) ^ ξ + ( ξ
I w w w | T h e p l o t s o f t h e c h a r a c t e r i s t i c c u r v e s f o r b i e i n t h e M a t h e m a t i c a N o t e b o o k E x a m p l e 2.2 4
c o o r d i n a t e s a s
c a n o n i c a l f o r m - η ) u v = 0. ■
t hi s example are avail- .ma.
It can be seen from t he se examples t h a t c more o f theoreti cal interest, also provide in s sol uti on o f t h e partial differential equations
anonical forms, though ome cases t h e general
The wel l-k nown Cauchy-Kowalewsky theore: ness and ex is t en ce o f quasi-linear partial differe: certain specific conditions. A s tate men t of t hi s pendent variables is as follows:
m guarantees t h e unique- ntial equations under theorem for two inde-
THEOREM 2.9. C o n s i d e r a q u a s i - l i n e a r s e c o n d o r d e r p a r t i a l d i f f e r ­
e n t i a l e q u a t i o n w h i c h c a n be s o l v e d f o r u x x, i.e.
= F ( x,y,u x,u y ),
w h e r e F i s a n a n a l y t i c f u n c t i o n o f x, y, u X! a n d n y i n a d o m a i n Ω C R 2. L e t t h e C a u c h y d a t a o n a c u r v e x = x q be
( 2.5 0 )
u ( x o,y ) = f ( y ), a n d u x ( x 0,y )
w h e r e f a n d g a r e a n a l y t i c f u n c t i o n s i n a ne (%o,yo)· T h e n t h e C a u c h y p r o b l e m h a s a n a n c n e i g h b o r h o o d o f t h e p o i n t ( x o,V o ) a n d t h i s s o l c l a s s o f a n a l y t i c f u n c t i o n s.
S i m p l y s t a t e d, t h i s t h e o r e m g u a r a n t e e s a u n i t h e f o r m o f a T a y l o r ’s s e r i e s i n a n e i g h b o r h o o d o f a b o v e s t a t e m e n t i s t r u e i f t h e s e c o n d o r d e r p a r t i c a n b e s o l v e d f o r u y y or u x y. A similar stateme
9(y),
ighborhood o f a poi nt lytic soluti on i n s o me i t i on is uni que i n the
que s ol uti on u ( x, y ) in t he point (xo, 2/0)· The
al differential equation i t holds for first order
56
CHAPTER 2: METHOD OF CHARACTERISTICS
partial difFerential equations. There is, however, an excepti on. IF t h e Cauchy d a t a is prescribed on a characteristic, a unique s ol uti on may not e x is t. For example, consider u x = 0. Its solution is u — f ( y ). IF t h e Cauchy da t a is u ( x, 0) = φ ( χ ), where φ ( χ ) is not constan t, t he n no s olution can be Found.
For a general st a t e m en t oF t hi s theorem for higher order partial difFerential equations, see Courant and Hilbert (1965) and Petrovskii ( 1967).
2.8. Exercises
2.1. Solve t h e equation p + q + 3u = e ~ 3x s i n ( x + 2 y ), w i t h t h e initial condi ti on u( x, 0) = 0.
A n s. u ( x, y ) = ( l/3 ) e ~ 3x [cos(:r — y ) — c o s ( x + 2y )\.
2.2. S o l v e t h e e q u a t i o n 2 p + q + 2 ( 2 x — y ) u = 6 x 2 e y ~ x .
A n s. 2y — x = c\ and uex ~ y — x 3 = c2;
general solution: F ( 2 y — x, u e x ~ y — x 3 ) = 0.
2.3. Solve t h e equation p + q\/l — y 2 = 0, wi t h initial conditions (a) u(0,y) = y, (b)u(a;,0) = x2.
A n s. ( a ) u ( x,y ) = y c o s x — y/l — y 2 s i n x; (b) u = ( si n-1 y — x ) 2.
2.4. S o l v e t h e e q u a t i o n p + q = \u ~'2, wi t h t h e ini tial c ondition
u
u ( 0,y ) = sin y.
A n s. u 3 ( x, y ) — x + s i n3(j/ - x ).
v?
2.5. F i n d t h e g e n e r a l s o l u t i o n oF t h e e q u a t i o n 2p + u q = —.
y
2|/ I n χι χχι
A n s. — = ci a n d -------------------- = c 2; a n d t h e g e n e r a l s o l u t i o n i s
y u
f u 2 y In y — x u
\y ’ u
2.6. F i n d t w o F u n c t i o n a l l y i n d e p e n d e n t s o l u t i o n s oF (y — u ) p + ( u — x ) q =
2.8. EXERCISES
57
x - y.
An s. x + y + u = c i and x 2 + y 2 + u 2 = c 2
2.7. F i n d a c o m p l e t e s o l u t i o n o f u =
1
P q
A N S. u = ± V 2 x + a ± \/2 y + b, or u2 + b =
a and b are arbitrary constants. Discuss t l | e relationship between th e two solutions.
So lu tio n. We have
F x = F y = 0 ,F U = —1, F p = - 1 /p The auxiliary equations are
2 a [ x +
y
a — 1
where
dx
dt
= —i !p\
dy
dt
M 2 d u
-1/q' Si
dp dq
T t =p'T t =q-
n e x p r e s s t h e f i r s t t w o
, , d p ^ , d q
N o t i n g t h a t — - = 1, a n d — - = 1, w e c a r
p d t q d t
■ Φ , dq _ . . . 2
equations a s d x = ----- 5·, d y = -----5-· Solving t he se we ge t p =
p i qi
1 1
and q 2 = Taking t h e p os it i ve square roots, we get
2 x + a
2 y + b
u = \/2 x + a + \/2 y + b as a solution.
An alternate approach is t o note t h a t soluticn equations can be expressed as u = A e ~ l ,p u p = A B = A i. u q = A C = A 2. Using these values in t h e given equation, we get
llll
u — — I— — —-----1——
p q A l A 2
Integrating u p = A B = A i, u q = A C — A 2, we ge t u 2 = 2 A i x + g i ( y ), and u 2 = 2 A 2 y + g 2 ( x ). Comparing t he se two values o f u 2, we find t h a t
,F q = - l/q 2.
_ 1 1 _ _
p q
ns of t h e characteristic B e t,q = C e t. Hence,
I.
2(^4ia: + A 2y + A 3) = 2 ( A i x +
y + A:
Now we wi l l discuss t h e relationship between t h e two solutions
A - 1
= V 2x + a + \f 2 y + b, and
A
- 1
■ X +
A
- y + B ),
58
CHAPTER 2: METHOD OF CHARACTERISTICS
where we have suppressed t h e subscripts and replaced A3 b y S. Define b= - a ( A - l ) + B( A- l )/A.
T h e n
φ ( χ, y, u ) = — u + \/2 x + a + \J 2 y — a ( A — 1) + B ( A — 1)/^4 = 0. Then
Λ _ _ L ________________________________________ = n
2 y/2 ^ + T i 2 ^ 2 y - α(Λ - 1) + B ( A - 1 )/A
T h u s
( A — \) λ/2 χ + a = \j 2 y — a ( A — 1) + B ( A — 1 )/A, which yields
u = \j 2 x + a + ( A — l ) V 2 x + a.
H e n c e
u = A\/2 x + a or u 2 = 2 A?x + A 2 a.
N o w s o l v i n g ( A — l ) y j 2 x + a = \j 2 y — a ( A — 1) + B ( A — l)/^4 for a and using t h e value so obtai ned in t h e expression for tt2, we get
u 2 = 2 ^ A x + — β — y + ■
S o clearly t h e second solution is t h e envelope of t he first.
2.8. Find a solution o f (1 + q 2 ) u — p x = 0, which passes through t h e curve 2 u = x 2,y = 0.
A n s. u 2 = x 2 ( y + x 2/4 ), or 4u 2 = x 2 ( 4 y 2 + x 2 ).
2.9. F i n d a s o l u t i o n o f F = p 2 + q 2 - 4u = 0, which passes through y = 0, u = x 2 + 1.
A n s. u = x 2 + ( y + l ) 2.
2.1 0. Find a solution o f F = u — p 2 + q 2 = 0, which passes through y = 0, 4u + x 2 = 0.
A n s. 4u = — { x ± y/2 y ) 2.
2.1 1. F i n d a g e n e r a l s o l u t i o n o f t h e e q u a t i o n x p — y q = x.
A n s. u = x + f ( x y ), or x y = g ( u — x ).
2.8. EXERCISES
59
2.1 2. Find a general sol uti on of t h e equation χ p + q — x u. Ans. „ = 1+19
2.1 3. Find a complete solution of the equation F A n s. (a 2 y 2 - 2c u)3//2 - a3t/3 = 3 a c ( 2 c x - u y ) Investigate i f a solution e x is t s at x = 0, for function of y which vanishes along wi t h q for A n s. Yes, and 6 = 0 gives the required solutii
p q y —p u/2 —cq = 0. b.
V h i c h t h e s o l u t i o n i s a y = 0. on.
2.1 4. Find a general solution o f t h e equation F A n s. /[ ( i t 4 x ) e y/a, (u — x ) e ~ v ^ a\ = 0. Discu; neighborhood o f x = 0.
2.1 5. Find t h e solution of (y + u ) p + yq = x — y
x =
0,u = y.
A n s. u + x = y.
2.1 6. Find a complete solution o f t h e equation F — 2 p x 2 y + 2 q x y 2 + p q — 4 u x y = 0.
A n s. u = a x 2 + b y 2 + ab.
2.1 7. F i n d a c o m p l e t e s o l u t i o n o f t h e e q u a t i o n F = 1 + u p x + u q y — u 2 = 0.
A n s. u 2 = 1 + a x 2 + b y 2.
2.1 8. F i n d t h e s o l u t i o n o f t h e e q u a t i o n F = p x
( a ) w i t h t h e i n i t i a l c o n d i t i o n u = a y 2 — b at
(b) u = a ( 1 + y 2 ) + 2 b y, at x = 1.
A n s. (a) u = a y 2 — b x 2 ·, (b)u = a ( x 2 + y 2 ) 4
2.1 9. Find a general solution o f t h e equation F
0. A l s o f i n d t h e p a r t i c u l a r s o l u t i o n s u b j e c t t b x = 1, a u = 4 4 b — y 2.
^ N S' ^ ( j ~ x 2 4 ’ x 2 ^ ) = a u ~ ^x 2 + y 2 =
N o t e t h a t b o t h g e n e r a l a n d c o m p l e t e s o l u t i o n!
2.2 0. F i n d a g e n e r a l s o l u t i o n o f t h e e q u a t i o n F A n s. y = (u - x ) f ί V - j.
p u — aq — x = 0. ss t h e s ol uti on in t h e
which passes through
4 q y - 2 u = 0, x = 1;
2bxy.
p x y + q y 2 — 2 u y —4q = t h e ini tial c ondition
ls are not unique. 2 p x + q y — u — x = 0.
60
CHAPTER 2: METHOD OF CHARACTERISTICS
2.2 1. F ind a compl ete solution o f t h e equation F = u 2 ( p 2 + q 2 + 1) — a 2 = 0.
AH +
1 + a 1
2.2 2. F i n d t h e g e n e r a l s o l u t i o n o f t h e e q u a t i o n 3 u x + 4 u y — 5 u = 10, subject t o t h e ini tial c ondition u ( x, 0) = x.
So l u t i o n. We first rotate t h e x,y axes through an angle Θ; thus,
t h e new axes ξ, η are related t o t h e old axes x, y by t h e relations
ξ = x cos Θ + y sin θ, x — ξ cos θ — η sin Θ,
η = —x sin Θ + y cos Θ, y = ξ sin θ + η c o s Θ.
H e n c e, w e h a v e
u ( x,y ) = u ( £ c o s# — ?7s i n#,£ s i n# + r?cos0) = ι ν ( ξ,η ),
d d£ d dr\ d . d . n d
a i ~ S m d r j' d d£ d dr\ d . n d n d
d ^ = &y ~di + & y Ί ϊ η = Sm <9ξ + ° ° S θ η ’
T h e g i v e n p a r t i a l d i f F e r e n t i a l e q u a t i o n t h e n b e c o m e s
d x v d z v
( 3 c o s Θ + 4 sin Θ) + (4 cos 0 - 3 sin Θ) — 5 w = 10.
σ ξ dη
dw
The coefficient oF —— in the above equation vanishes iF tan Θ = 4/3, ΰη
i.e., sin# = 4/5, and co s Θ = 3/5. Wi th these values, t h e above equati on reduces t o
dw
which has t h e general solution
w{ £,n ) = -2 + g { v ) e ^,
o r
4
3z/5H—/5
u( x, y ) = — 2 + g ( 3 y/5 — 4 x/5 ) e V .
N o t e t h a t t h i s i s t h e g e n e r a l F o r ma l s o l u t i o n oF t h e g i v e n p a r t i a l d i f F e r e n t i a l e q u a t i o n. N o w, t o f i n d t h e p a r t i c u l a r s o l u t i o n s u b j e c t
2.8. EXERCISES
61
t o t h e initial condition u ( x, 0) = x, we have and t he n t h e above equation gives
x = —2 + ς ( η) e 3x/5,
4x
η = — — when y = 0, 5
thus,
g(v) = (x + 2)e
—3 x/b /o __
(2 - 477
S u b s t i t u t i n g t h i s v a l u e o f g ( η ) into t h e abcA obtai n
/5) e 3?7/4. e general solution, we
u ( x, y) = -2 + ( 2 - ) e3,)/4 e 3 x l 5 + 4 y/5
p3/4( 3
- 2+ [2 + \( A x - 3 y )\ e"5^ 4,
which is t h e unique solution o f t h e given part: subject t o t h e given ini tial condition u ( x, 0)
ial differential equation x.
w w w I S e e t h e M a t h e m a t i c a N o t e b o o k E x e r c i:s e2.2 2 .ma.
I n p r o b l e m s ( 2.2 3 ) - ( 2.3 0 ) f i n d t h e c h a r a c t e r i s t i c: r e d u c e t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n t o i t s
2.2 3 ( a2 + x 2 ) u x x + ( a2 + y 2 ) u y y + x u x + y u y H i n t: Solve t h e characteristic equation t o di s t i t u t i o n is
ξ = l o g ( x + y o 2 4- x 2 ), and η = l o g
2.2 4 T h e T r i c o m i e q u a t i o n u y y — y u x x = 0 for Uc — tt„
. Uc — tt
A n s. υ,ξη --------
6(ξ - η )'
2.2 5 ( 1 + s i n x ) u x x — 2 cos x u x y + (1 — sin x ) u % i ( l - s i n x ) u!/ = 0.
A n s. 4U,,,, + u n = Ο,ξ,η — y ± l o g ( l + sina:). now be solved t o yield
« = /( £ ) +
s where pos sible and canonical form.
(). A n s. "f- tt^^ — 0. etermine t h a t t h e sub-
(y + \/a 2 + y 2)·
y > 0.
(1 + s i n a:)2 ~t“ ~
2c o s x
T h e r e d u c e d f o r m c a n
62
CHAPTER 2: METHOD OF CHARACTERISTICS
2.2 6 e 2 x u x x — 2e x + v u x y + e 2 y u y y + e 2 x u x + e 2 y u y = 0. A n s. u m = 0, ξ = e ~ x + e ~ y, η = e ~ x - e ~ y.
2.2 7 e 2 x u x x — 5e x + y u x y + 4 e 2 y u y y + e 2 x u x + 4 e 2 y u y = 0. A n s. ΐ ί ξ η = 0, ξ = e ~ x + e ~ y, η = 4 e ~ x + e ~ y.
2.2 8 9 u x x — I 2 u x y + 4 u y y + 12u x — 8 u y + 4 u = 0.
A n s. Z 6 u v v + 12^ 4- u = 0, u = [/( ξ ) + η 9 {ξ)]εη/β, ξ,η = 2 χ ± 3y.
2.2 9 3u x x — 7 u x y + 4 u y y + 5u x — u y + 3 u = 0.
A n s. 5 u (rt - u ( = 0, /( ξ ) + 9 ( η ) ε η/5, ξ = 2 x + y, η = x + 3 y.
2.3 0 2u x x + 6 u x y 4- Qu y y 4- 2u x 4- 3 u y — 2 u = 0.
3 3
A n s. 9 ( η ξξ - u vr t ) + 6 u v - 4 u = 0, ξ = y - - χ, η = - x.
3
L i n e a r E q u a t i o n s
w i t h C o n s t a n t C o e f f i c i e n t s
We will use the inverse operator method to solve horn geneous partial differential equations with constant c< although basically developed and frequently used ί ferential equations, becomes useful for finding gen differential equations with constant coefficients. Th general solutions o f second order partial differential coefficients and determining their particular solution conditions is also discussed in a later section. Befo differential equations with constant coefficients we w technique o f inverse operators from the theory o f or tions. This review should prove useful in discussin nonhomogeneous partial differential equations with
ogeneous and nonhomo- jefficients. This method, or solving ordinary dif- eral solutions o f partial 3 problem o f finding the equations with constant ; under auxiliary (initial) :e we discuss the partial ill first review in §3.1 the dinary differential equa- g the homogeneous and :onstant coefficients.
3.1. Inverse Operators
„ n d , 1
i t j j represents —, then — is defined as the inverse d x D
~ φ { χ ) = J φ( χ) άχ.
I f /( D ) r e p r e s e n t s a p o l y n o m i a l i n D with constan
operator o f D, i.e., coefficients, then /( D )
64
CHAPTER 3: LINEAR EQUATIONS
is a linear differential operator, and we define its inverse as 1/f ( D ). Thus,
f ( D )
f ( D )
φ( χ)
φ[χ).
Note that [/( - D )0(a:)] is not necessarily equal to φ ( χ ). However, i f
γ ζ ^ Φ ( χ ) = vH1 )' then 'φ ( χ ) contains arbitrary constants, and ψ ( χ ) =
φ ( χ ) for some value o f these arbitrary constants. In the sequel we will ignore arbitrary constants. We will list some formulas for the operator pair f ( D ) and 1
m:
1· f (D) ΤΤΤλ^(x) = Φ(χ)·
U ( D )
1
{ x ) =
1 W ) = 1
3 · [C101(*) + ο2φ2(χ)} = c i - ^ y
/2 (2?) V/i C 0 )
φ ΐ { χ ) + C2JJjyj<t>2{X)·
φ{χ) .
4.
1
-eax, provided that f ( a ) φ 0.
f { D ) f ( a )
5. /( ΰ ) φ ( χ ) ε α χ = e a x f ( D + α ) φ ( χ ).
3.1. INVERSE OPERATORS
65
COS
a x
1 f cos ι sin
1 0 · 7 7 γϋΓ\ i. a x = — — 2 Γ ’ Pr o v i d e d t h a t /( f { D 2) [ s m f ( - a 2)
- α 2 ) φ 0.
11.
1 2.
(D 2 + a 2 ) \ sin
{ “
I si
a: cos
cos \ χ sm
a x = ± -
a x
a D 2 + b D + c \ sin
2 a
c o s
c — α ω “ I < ω cos v 1 sin
ω χ =
sin
T±bu>{ ω χ cos
(c - α ω 2 )
1 3.
1
1
■ x =
/( D ) o o [ l + g ( D ) ]
= - L [ l - g ( D ) + g 2( D ) - g 3( D) +
(X o
where the terms o f degree n + 1 or higher are ignored
P roof o f F ormula 4. If φ ( χ ) = e a x, we know
D e ax = aea x, D z eax = a 2 e a x, ... D
T h u s
f ( D ) e a x = f ( a ) e a x.
I f w e t a k e y j ~ j y j as the inverse operator o f /( D ), the
f ( D ) f ( a )
eax provided /(
62ω2
+ g n ( D ) + · · · ] Xr'
hat
Therefore, the particular integral y v o f the equation by
Exa mple 3.1. Consider ( D2 + D + l ) y
1 o „ 1
n o b v i o u s l y
α ) φ 0.
f ( D ) y = A e a x is given
yP =
r,2x
D2 + D + 1 22 + 2 + l
2x. Then
1 2x
7 β ''
E x a mple 3.2. Consider (£>4 + 8) y = e x. Then
66 CHAPTER 3: LINEAR EQUATIONS
,, — i ex = 3: Px =
Vv D 4 + 8 l 4 + 8 9
An Applica tio n of F ormula 13. If φ ( χ ) = x n, then
f ( D ) — a o D n + a\D n + · · · + a n —i D + a
w h i c h g i v e s
w h e r e
f ( D ) a n ( l + g ( D ) y
g { D ) — — ( a n _ ± D + a n - 2 D 2 + · · · + ao-D” )
Q"n
and
1 1
f ( D ) an
= - ( i + g ( D ) )
-1
N o w, i n o r d e r t o f i n d t h e p a r t i c u l a r i n t e g r a l y p o f f { D ) y = A x 11, we apply t h e inverse ° f f ( D ) t o t h e ordinary difFerential equation
and get
y =
1
(.A x n )
f ( D )
= A - ± - ( l + g ( D ) ) - 1 x
A
a
A [ i - g ( D ) + ( g ( D ) ) 2 - ( g ( D ) f + · · ■]
( I n. L - 1
w h e r e t e r m s o f d e g r e e n + l a n d h i g h e r i n D are ignored in t h e above expansion on t h e right side.
3.1. INVERSE OPERATORS
67
Ex a mple 3.3. Consider (D 2 + D + 2) = .
. Then
1
V p
D 2 + D + 2
2|1 + \θ + \θ
l + | i D + i D ’
= \[ ι - (\ν + \η Λ + ( Ι ο + \*
2 ' 2'
X4
I f 1 1 9 1 9 1 q 1
= 2 l 1 - 2 D - 2 D + 4 D + 2 ° + 4 D + l - D * + 0 ( D 5 )\x\
where 0 ( D 5) means terms containing D 5 and Thus,
\D + \D 2
3 ^ 4
V p —
x — 2 x — 6 x 4- 3 x 4* 1 2 x 4- 6
x 4 — 2 x 3 — 3 x 2 + 9a:
Exa mple 3.4. Consider (D 4 + 2 D 3 + D 2 ) y
U p —
D 2 ( D 2 + 2 D + 1 ) 1)2 (1 +
- 2
D 2
1
D 2
1
(1 + D ) ~ * x ·
[ 1 - 2 D + 3 D 2 - A D 3 + 0
j y z [a:3 - 6 x 2 + 18a: - 24] 20^ 2"
= χτγχ 5 — ττχ4 + 3a;3 — 12a;2. ι
An Applica tio n o f F ormula 10. I f φ ( χ ) = we know t h a t
cos a x
2 ( cos α χ 2 Γ cc
1 sin a x \ si
sm a x
h i g h e r p o w e r s i n D. 3'
- 3a: — 9 + —
- -v-3
x 3. Then
D Y
(^)]
sin ax or cosax, then
68
CHAPTER 3: LINEAR EQUATIONS
a n d
Even i f t h e operator f ( D ) is not an even function, we can st i l l use t h e above formula; e.g.,
( D 3 + 2D 2 + 3 D + l ) cos a x = [ ( —a 2) D + 2 ( —a 2) + 3D + l ] cos a x
— [(3 - a 2) D + 1 - 2 a 2] cos a x = — (3 — a 2) a sin a x + ( l — 2a2) c o s a x.
We, therefore, notice t h a t when an operator o f the t y p e f { D ) is applied t o c o s a x or sin a x, we can set D 2 = —a 2, and reduce f ( D ) t o a linear operator o f first order. We shall use this observation t o find particular integrals in t hi s case.
If f ( D ) happens t o be o f the form / (D 2), i.e., i f t h e ordinary dif­
ferential equation is
. f c o s a x
f{D2)y=\ .
y sin ax,
then
A f c o s a x A f c o s a x
^ / ( D 2 ) \ sin a x, / ( - a 2 ) \ sin ax,
provided t h a t / ( ~ a 2 ) φ 0.
If f ( D ) is in general a polynomial containing b o t h even and odd degree terms, the n we let D 2 = —a 2, and reduce f ( D ) t o a linear operator in D o f the form a D + β, so t h a t if
f ( D ) y
Γ cc
= A\ .
I si
cos a x sin ax,
then
A f cos a x ^ f ( D ) \ sin ax.
3.1. INVERSE OPERATORS
69
The n we let D 2 = —a2, which reduces f ( D ) t o a D + β. Thus, A f cos a x ^ a D + β \ sin ax,
_ A ( a D - β ) 1 cos a x
f co i \ sir
( a D + β ) ( α ΰ — β ) \ sin a x, A ( a D - β ) ί c o s a x
a 2 D 2 — β 2 \ sin ax,
A ( —a a sin a x —
—a 2a 2 — β 2 \ a a cos a x — β ί
E x a m p l e 3.5. L e t (D 2 + l ) y = 2 s i n x.
2
The:
Vp
■ srnx.
D 2 + 1
Now, i f we let D 2 = — 1, the n D 2 + 1 = 0, and th applicable. ■
E x a m p l e 3.6. Let ( D4 + D 2 + l ) y = 2 sin
2
V p —
• s m x.
c o s a x in a x.
e a b o v e m e t h o d i s n o t
c. Then
D 4 + D 2 + 1 Now, if we l et D 2 = —1, then D 4-|- D2-l-l = 1, an
E x a m p l e 3.7. Let (D 3 + 2 D 2 + D + l ) y = 3
Vp —
- 2 2D + 2 ( - 2 2) + D + 1
cos 2x
3 „ —3 ( 3D - 7)
c o s 2x =
- 3 D - 7 - 3 9 D 2 - 49
cos 2x =
( 3 D + 7) ( 3 D —
=! (3
9 ( —22) — 49
d we get y p = 2 sin x.
3 c o s 2 x. Then
cos 2x
7)
D — 7) cos 2x
- — [6 sin 2x + 7 cos 2 x ]. ι
Οϋ
A n A p p l i c a t i o n o f F o r m u l a 6. I f φ ( χ ) = V a function o f x o f t h e t y p e x n, cos b x, sin b x or will first give a heuristic jus tifi cation for formul
D (
V e ax) = eaxD V + a V e ax = e “
( x ) e ax, where V ( x ) is :n cos b x, x ” sin b x, we , 6. No t e t h a t
D + a)V,
70
CHAPTER 3: LINEAR EQUATIONS
which yields
D 2 ( V e a x ) = D [ea x ■ ( D + a ) V ] = e a x ( D + a ) V.
S i m i l a r l y,
D n ( V e a x ) = e a x ( D + a ) n V, which, in general, gives f ( D ) ( V e a x ) = e a x f ( D + a ) V.
W e c a n, t h e r e f o r e, a s s u m e t h a t
1 ( V e a x ) = e a x — ^ ----- -V.
f ( D ) v ' f ( D + a )
W e s h a l l u s e t h i s f o r m u l a t o o b t a i n t h e p a r t i c u l a r i n t e g r a l i n t h i s c a s e, i.e., s i n c e f o r t h e o r d i n a r y d i f f e r e n t i a l e q u a t i o n f ( D ) y = φ ( χ ) = V e a x,
f τ τ π τ n 'r ^
yP =
V e ax = ea x - —— - y,
/( D ) f ( D + a )
o u r p r o b l e m r e d u c e s t o f i n d i n g —7—7-----r V. I f V is o f t h e form x n,
f ( D + a )
c o s a x or sin b x, t he n we can use formula 13 or 10 t o obtai n t h e sol u­
tion; i f V is o f t h e form x n cos b x or x n sin b x, the n we use de Moivre’s theorem and write
elbx = cos b x + i sin b x,
o r
c o s b x = y t e l b x, sin b x = S e l b x,
s o t h a t i f V = x n cos b x, we write V = <i t x n e l b x, and i f V = x'% s i n 6:r, we write V = S t x n e l b x. Thus,
f ( D + a)
3?
f ( D + a y
or
s 1
f ( D + a)
Now f ( D + a ) is another polynomial in D, and we can write f ( D + a ) = /i (.D ). We consider
1 Xneibx = _ L - Xneibx
f ( D + a ) f i { D )
a n d u s i n g t h e f o r m u l a (6) w e g e t
^ - x n e i bx = c i bx ^ x n,
f i ( D ) h { D + i b )'
x n can be evaluated by t h e use discussion also covers t h e case when f ( x ) = x n cos a x or x n sin a x.
w h e r e — — ^- - -- -- -- x n can be evaluated by t h e use o f formula 13. T hi s
h { D + ib)
3.1. INVERSE OPERATORS
E x a m p l e 3.8. Consider ( D4 + D 3 - 3 D 2 -
4ex
Up
D 4 + D 3 - 3 D 2 - D + 2 1 1
= 4
( D — l )2 D 2 + 3 D + 2
ex — 4
D 4 + D 3 1
(D — 1)
21 + 3 + 2
6 ( D — l)2 3 ( D + l - l )5
2 x 1 , 2 χ χ 2 1 2 x
- e x — τ ■ 1 = - e — = - r e 1. ■
3 D 2 3 2 3
E x a m p l e 3.9. Consider ( D3 - D 2 + 4 D - 1
Vp —
-2 sin 2a; = 2-
4) y = 2 sin 2a;. Then 1
D 3 - D 2 + 4 D - 4 D3 - Ef
I f w e p u t D2 = —4, /( D ) — 0, a n d, t h e r e f o r e, w 1
2 + 4 D - 4 e w r i t e
Vp
= 2
= 2 3 = 2 3
D 3 — D2 + 4 D - 4 1
s i n 2a; = 2 9
D 3 - D2
( D — 1) ( D2 + 4 )
1 ( 1
2 5
].
D - 2 % \( D — 1 ) ( D + 2 i )
( D - 2 i ) ( D
2 i x \ _ 2 c j.
1
________
2 ( D — 2?) ( —2 — i )
22« = _ Iq.— 1
2 - i
J2 i x
1
2 ( 22 + 1 ) D
2 ( 2 + i ) ( D - 2?)
a 6)
( 1) (using t h e formul
1~ 2- i
= — Q - 2 5
,2 i a;
D
( l ) = - - 3 ( 2 - i ) e:
,2i x
1 ^
· — 9 a;( 2 — z ) ( c o s 2a; — i sin 2a;) = — (cos
Ex a mple 3.1 0. When φ ( χ ) = e a x and / differential equation is of t h e form
71
D + 2) y = 4 e x. Then 1
- 3 D 2 - D + 2 1
sin 2 x.
J 2 i x
4 D - 4
0 2 i x
1 ) ( D + 2 i )
1 ( 1
J2ix
( D — 2 i ) V ( 2 i - l ) ( 4 i )
(e2ix · 1)
2a; — 2 sin 2 x ). ι
'a) = 0, the o r d i n a r y
f ( D ) y = A e ax
72
CHAPTER 3: LINEAR EQUATIONS
Since /( a ) = 0, ( D — a ) must be a factor of /( D ). Then f ( D ) = ( D — a ) n f i ( D ), where f i ( a ) Φ 0, and
,. — δ \ Pa x — a ^ ( ·*· r'
Vp ( D — a ) n f\( D ) ( D — a ) n \f ( D )
,ax
/i ( a ) ( D - a ) and using formula 6 we can write
4 1 A 1
i ( e a x . n = e a x — i l ) h ( a ) ( D - a y [ > h(a) D"[ )
- A c a* ( χ Τ
f l i p ) \nl·
I f f i D ) y = sin a x or c o s a x and i f /( D ) becomes zero by l et t i n g D2 = —a 2, the n i t is convenient t o consider c o s a x and sin a x as real and imaginary parts of e m x and deal wi t h t h e problem as for e a x. ·
| w w w | T h e M a t h e m a t i c a p a c k a g e I n v e r s e O p e r a t o r .m c a n b e u s e d t o s o l v e e x a m p l e s o f t h e a b o v e t y p e, a n d t h o s e i n t h e s e q u e l.
3.2. Homoge neous Eq u a t i o n s
L e t L be a linear partial difFerential operator with constant coefficients in two variables x and y. The n L u = f ( x,y ) is a partial difFerential
Q
equation wi t h constant coefficients. IF we define D x = -— and D v =
d x
d d i
— ,b o t h a t, e.g., D lx = ~, the n
m,n
L = Y ^ A x i ^ D {. (3.1)
i,j ~ 1
We shall first discuss the homogeneous case L u = 0, and limi t ourselves t o t h e case where L can b e expressed as a product of linear factors in D.
3.2. HOMOGENEOUS EQUATI
ONS
T h e o r e m 3.1. I f u\,u 2:... ,u n a r e s o l u t i c
n
i s a l i n e a r p a r t i a l d i f f e r e n t i a l o p e r a t o r, t h e n ^ o f L u = 0.
ns o f L u = 0, w h e r e L Ci Ui i s a l s o a s o l u t i o n
P r o o f. S i n c e u i, u 2,... , u n are solutions of
L u2
= L u n = 0. But the n L I ^
73
T h e o r e m 3.2. I f t h e o p e r a t o r L o f o r d e r n l i n e a r l y i n d e p e n d e n t f a c t o r s o f t h e t y p e a,i D: g e n e r a l s o l u t i o n o f L u = 0 i s g i v e n b y
L u = 0, we have L u i
n
^ C i L u i = 0. ■
n can be fact ored int o + b i D y + Ci, t h e n t h e
a x (
i — 1
P r o o f. I t w a s e s t a b l i s h e d i n § 2.1 t h a t t f b i D y + a ) u i = 0 is Ui = <j )(bi X - a t y ) e ~ Ci V/bi. T linearly independent solutions o f L u = 0 for i =
n
Theorem 3.1, u t is a sol uti on o f L u = 0.
1
T h e o r e m 3.3. I f a D x + b D y + c i s a f a c t o r o f m u l t i p l i c i t y k, t h e n t h e c o r r e s p o n d i n g s o l u t i o n i s
k - 1
Σ χ% f i ( a y ~ b x )<
— c x j a
(3.2)
e solution o f (a i D x + hen obviously m are n 1,2,... , n. Hence, by
(3.3)
2=0
A proof can be established by su bs titution.
E x a m p l e 3.1 1. Solve ( 4 D l - 1 6 D x D y + 1 5 D can be w r i t te n as (2D x — 3 D,,) ( 2 D T — 5D,.) u u = f i ( 3 x + 2y ) + f 2 ( 5 x + 2y ). ■
) u = 0. Thi s equation = 0, so t h e s ol uti on is
E x a m p l e 3.1 2. Solve (2D 2X - D x D y - 6 D\ + 4D x - 8 D y ) u = 0. T hi s equation can be written as (D x — 2 D y ) ( 2 D x + 3D y + 4 ) u = 0, so t h e solution is u = f i ( 2 x + y ) + e ~ 2 x f 2 ( 3 x — 2y\. ■
74
CHAPTER 3: LINEAR EQUATIONS
Ex a mple 3.1 3. Solve ( S D X + 7 D y ) ( 2 D x — 5 D y + 3 ) 3u = 0. In this e xampl e t h e equation is already in the factored form and i ts solution is
u = /( 7 x - 3y ) + e ~ 3 x/2 [ f 1 ( 5 x + 2y ) + x/2(5x + 2 y ) + x2/3(5x + 2y)]. ■
3.3. Nonhomogeneous Equations
I f t h e partial differential equation is L u — /( x ), then one finds t h e gen­
eral s ol uti on u c (complementary function) t o the homogeneous equa­
t i o n L u = 0 and looks for any function g ( x ) tha t satisfies L u = /( x ), where g ( x ) is also known as t h e particular solution and sometimes de­
not e d by Up. Thus, t h e solution o f t h e partial difFerential equation is u = u c + g ( x ). In t hi s section we will discuss methods for finding the particular solutions.
T h e operator technique for finding t h e particular solutions for ordi­
nary difFerential equations is applicable For t h e cases where t h e me th od oF undetermined coefficients For ordinary difFerential equations is used. T hi s technique is useFul in finding particular solutions oF partial difFer­
e nti al equations. Thus, iF f ( D x, D y ) is a linear partial difFerential oper­
ator, t he n t h e corresponding inverse operator is defined as
We will s t a t e some obvious results:
1
f ( D x,D y )
1
f i { D x, D y ) f 2 ( D x, D y )
f { D X, Dy)
4>{x,y) =
4>(x,y)
1
= Φ(χ,υ), 1
1
f 2 ( D X, Dy)
f l { D X, Dy ) 1
f l ( D X,D y )
f 2{ D X, D y )
φ { χ,ν )
f { D X, Dy )
(3.4)
Φ{χ,ν)
(3.5)
H d!,d,) [C1 φ,{ ζ · » > + C! Φ Λ τ · »>1 = Cl i ( D „ D y ) φ'( I'y)
+ c 2
f { D X, Dy )
Φ2( ζ,ν),
(3.6)
0a x + b y _
f i n - ft
Me ° X+b"’ (3.7)
f [ D x, D y ) f ( a,b )
f ( D x, ΰ ν ) φ( χ, y ) e ax+by = eax+by f ( D x + a,D y + b) φ( χ, y ), (3.8)
f ( D X,D y )
φ ( χ,ν ) e'
a x + b y _ a x + b y _
f {Dx + D y + b)
Φ(χ,ν)
3.3. NONHOMOGENEOUS EQUATIONS
75
= e by
1
f ( D x + a,D y ) 1
/( D x, D y + b)
e y φ ( χ e a x φ ( χ
<y)
y ).
f { D x,D ) cos { a x + b y ) = /( - a 2, - b 2 )
f ( D l, D l ) sin (arc + b y ) = f ( - a 2, - b 2'
Ex a m p l e 3.1 4. F o r a p a r t i c u l a r s o l u t i o n ( 3 D2 + A D x D y — D y ) u =
note t h a t
1
oi'
(3D l + 4D x D y - D y ) 1
x - 3 y
[3 + 4 ( —3) — ( —3)]
x - 3 y _
E x a m p l e 3.1 5. F o r a p a r t i c u l a r s o l u t i o n o f p a r t i a l d i f F e r e n t i a l e q u a ­
t i o n
w e h a v e
( 3 D2 — D y ) u = s i n ( a x + b y )
( 3D l - D y )
s i n ( a a; + b y ) =
( - 3 a2 - D y b c o s ( a x +
by)
E
x a mple
3.1 6. To find a particular solutic
( 3 D 2 — D y ) u = e x sin(a:
we have
1
( 3 D
2 - D y )
e x sin(a: + y ) = e
( 3 D2 + 6D X + 3 - D y )
( 3 { D x sin(a: + y )
( 3.9 )
c o s ( a x + b y ), (3.10)
s i n ( a x + b y ). (3.11)
t h e equation
—3 y
( 3.1 2 )
s i n ( a x + b y )
— 3 a 2 s i n ( a x + b y )
b 2 + 9 a 4
(3.13)
n for t h e equation
i ).
I
L)2 - Dy
s i n ( x + y)
1
76 CHAPTER 3: LINEAR EQUATIONS
(3(-l)2+6Dx + 3-Dy)
sin(x + y)
s i n ( x + y) = ex ^n2\ sin(* + v)
= e
( 6 Dx - D y) v (36 D2 - D 2)
x 7 cos(x + y)
- 3 5
= — ^ ex cos(x + y). ■ (3-14)
E x a m p l e 3.17. To solve utt — c2uxx = 0, such that u(x, 0) = e~x, ut(x, 0) = 1 + x, note that the partial differential equation can be written as (Dt — cDx)(Dt + cDx)u = 0, which gives the solution as u = f ( x + ct) + g(x — ct). This solution is known as the d’Alembert’s solution (see Eq (5.24) ). Applying the initial conditions, we get
f { x ) + g ( x ) = e ~ x, (3.15)
c f'( x ) — c g'( x ) = 1 + x. (3.16)
On integrating (3.16) with respect to x we get
f(x) -g(x) = y ) +Ci. (3.17)
Eqs (3.15) and (3.17) yield
λ/ \ 1 q. 1 . x2. cj
/( x ) _ 2 e + Y c ( x + Y ) + T ’
9 ( x ) = l e - x - ± ( x + ^ ) - c i/2.
Hence
u = i e~x(ect + e~ct) + (x + 1 )t. ·
A ge ne r a l s c heme f or i n i t i a l va l ue pr obl e ms f or t h e wave e q u a t i o n i s a s f ol l ows: Sol ve utt = c2uxx, subject to the conditions u(x, 0) = φ(χ), ut(x, 0) = φ'{χ). Then as in the above example
f(x) +g(x) = φ(χ),
c f ( x ) - c g'( x ) = ψ'{ χ ).
3.3. NONHOMOGENEOUS EQUATIONS
77
C o n s e q u e n t l y
f ( x ) = \\Φ ( χ ) + \ψ ( χ ) + 9 ( χ ) = ^[Φ(χ) ~ \Φ{χ) -
-=ι],
=ι]>
which yields
u ( x,t ) — \{ φ ( χ + ct) — φ ( χ — ct)\ + ττ- [ t + c t ) — φ ( χ — c t ) ]. 2 2c
E x am ple 3.1 8. It is interesting t o note Laplace equation by t h e above method. We w
t h a t we can solve t h e ill solve
Uxx Uyy 0,
such t h a t u ( x,Q ) = φ { χ ) and u y ( x,0) = φ'( χ )
Uy y 0 as
(.D x + i D y ) ( D x - i D y ) u and, therefore, i t s general solution is
u = f ( x + iy) + g ( x - ii J)
Applying t h e i ni t ial conditions, we get f ( x ) + g i g'{ x ) — φ'(χ). Consequently
f(x) = \ {Φ(χ) ~ ίψ(χ) +
9{χ ) - 2^ ( χ ) + ί Φ { χ ) -
Thus,
1 i
u ( x,y ) = 2 + ί ν ) + Φ(χ- i y ) ] + 2 \Φ{ή - i y ) - Φ{χ + i y ) ] ■
The final value of u is real. If φ ( χ ) = e x, solution is given by
u ( x,y ) = i [ e (x+i^ + e (x“ iy)] + ^ [ t a n _1(a; -
r ^ + i1 - y )
r ^ + ( i - y ) n i x2 + fl + v)2 J ’
= e ~ cos y ln , „ , N ,
4 l x 2 + (1 + y ) 2 J
where we have used t h e formula
1 (
tan a = i z. or a = — ln -
2
w i t h a = t a n -1 (a; — i y ) — t a n -1 (a: + i y ). ■
We can express u x x 4- 0,
χ ) = φ( χ ), and i f'( x ) —
e l ­
a n d φ'
1
1 + x 2 ’
t h e
i y ) — t a n (a: + i y ) ]
+ * )
— z
78
CHAPTER 3: LINEAR EQUATIONS
E x a m p l e 3.19. Consider
utt — c Uxx , 0 ^ x 1 f
su bje ct t o t h e conditions u ( 0, t ) = u ( l, t ) = 0, for t > 0, and u ( x, 0) = x, u t ( x,0) = 0. T h e general sol uti on is
which yi el ds f ( z ) = —g ( —z ). Also f ( l + c t ) + g ( l - c t ) = 0 is equivalent t o
which in turn gives f ( z ) = f ( z + 21). This l ast equation implies t h a t t h e function f ( x ) is a periodic function o f period 21. T h e solution, thus, reduces to
u = f ( c t + x ) - f ( c t — x ).
A p p l y i n g t h e i n i t i a l c o n d i t i o n s, w e g e t
f ( x ) - f ( - x ) = x, and f'( x ) - f'( - x ) = 0,
i.e., /'( x ) is an even function, which means t h a t f ( x ) is an odd function,
i.e., f ( x ) = —f ( —x ). Hence 2f ( x ) = x. Since f ( x ) is an odd periodic function o f period 21, i t can be expressed as a Fourier sine series. Thus
u = f ( x + c t ) + g ( x — c t ).
F r o m t h e b o u n d a r y c o n d i t i o n s w e f i n d t h a t
f ( c t ) + g { - c t ) = 0, or f ( z ) + g ( - z ) = 0,
f(ct + 1)- f{ct - 0 = 0,
which yields
/ x 1 ν'-* V- i ) r . n n, , ,, , w, ,
u ( x, t) = — ------------- [sm — ( c t + I ) - sin — ( c t - I)}
I “ ( _ l ) n + i η π
O t h e r t e c h n i q u e s f r o m o r d i n a r y d i f f e r e n t i a l e q u a t i o n s s u c h a s t h e m e t h o d o f u n d e t e r m i n e d c o e f f i c i e n t s a n d t h e v a r i a t i o n o f p a r a m e t e r s
3.4. EXERCISES
79
technique can also b e extended t o find t h e particular s ol uti on corre­
sponding t o t h e nonhomogeneous term in t h e partial differential equa­
tions.
3.4. E x e r c i s e s
Evaluate (use t h e inverse operator method o f §i
3.1. (,D - 3 ) _ 1( x 3 + 3 a:- 5 ).
A n s. — ί ( 9 α;3 + 9 x2 + 33x - 34)
27
3.2. ( D - l ) ~ 1(2x).
A n s. - 2 x.
3.3. ( D - l ) _ 1(a:2).
A n s. - ( x 2 + 2 x + 2).
3.4. ( 4 D2 - 5 D ) - I ( x2e - X).
A n s. - | ^ ( 8 1 a:2 + 234a:+ 266).
3.5. ( D2 - 3 D + 2)-1 s i n 2 x.
3 1
A n s. — c o s 2 i — — sin 2a:.
20 20
3.6. D 2( 2sin2a:).
A n s. —^ s i n 2 x..
3.7. D ~ 3 x.
x 4
An s. —.
2 4
3.8. D ~ 2 ( 3 e 3 x ).
p Z x
A n s. —.
3
3.9. D -1( 2 a: + 3 ).
A n s. a:2 + 3a:.
3.1 0. ( D 3 - D 2 ) ~ 1 ( 2 x 3 ).
/ 5 ,γ 4
ANS. - 2 ( — + — + a:3 + 3a:2 \2 0 4
3.1 1. ( D 2 + 3 D + 2 ) ~ 1( e “ ).
. 1 - 3 i :
A N S. e .
10
3.1 2. ( D2 - 3 D + 2 ) -1( 3 s i n x ).
3
A n s. — ( s i n x + 3 c o s x ).
3.1 3. ( D 2 + 3 D + 2 ) _ 1 (8 + 6e x + 2sina:).
.1):
Ans. 4 + ex + 7· (sin a: - 3 cos a:).
5
3.14. (D 5 + 2 D3 + D) ~1( 2x + sina; + cosa:).
x2
Ans. x 2 + — (cosa; — sina:).
8
Find the general solution of the following partial differential equations:
3.15. (3D2 - 2Dx Dy - 5D2y)u = 3a: + y + ex~y.
11 1 1
A n s. u = /( 5x 4- 3y) + g( x - y) + — x 3 + - x 2y + - x e x~y.
5 4 b o
3.1 6. ( Dx - 10D2x D 2y + 9Dy) u = 135sin(3a; + 2y).
A n s. f l (3a: + y) + h { x - 3y) + gi ( x + y ) + g2{x ~ y ) ~ sin(3x + 2y).
3.1 7. ( D x - 2D y f u = 125ex siny.
Ans. f i ( 2x + y) + x f
2
(‘
2
x + y ) + x 2h { 2 x + y ) - e x (2 cosy + 11 siny).
3.18. Find the particular solution for the following partial differential equations:
(a) ( D2 — Dy)u = 17ex+2/sin(a: — 2y).
A n s. — ex+y {sin(a: — 2y) + 4cos(x — 2y)}.
( b ) ( D\ + D2)u = Qxy + 25e3x+4y.
Ans. x 3y + e3x+4y.
( c ) ( Dl + D 2 - Dx )u = 37e5y cos(3x + Ay).
A n s. e5y sin(3x + 4y).
3.19. Show t h a t u = f ( a y — bx) e~cy^h is also a solution of
(iaDx + bDy + c) u = 0.
80 CHAPTER 3: LINEAR EQUATIONS
3.2 0. Find t h e general sol uti on o f 3 u x + A u y — 2 u = 1, su bje ct t o t h e initial c ondition u ( x,0 ) = x 2.
S o l u t i o n. He r e t a n# = 4/3, t h u s
d w 2 1
d i ~ 5 W~ 5 ’
3.4. EXERCISES
81
whose general solution is
or
w&v) = ~ 2 + 9 ( v ) e2i u ( x,y ) = - ^ + g ( ^ y - ^ x\ e6
r/25+8y/25
3.21. Find the general solution of u x — u y + u = s i n x.
S o l u t i o n, t a n# = —1, thus θ = 4 π/4, and d w 1 1
whose general solution is w = 1 + <7(77) e^^2
x + y
u(x,y ) = 1 + g 1 -
V2
Using t h e ini tial condition, we get s i n x = 1 t ha t
9 ( v ) = - ( s i n λ/2 η + 1) e
T h e n
u(x, y ) = 1 — (sin \/2η + 1 ) e ~ 7)/,v/2 = 1
3.22. Solve u x + u y — u = 0, subject t o t h e ini tial c ondition u ( x, 0) = h ( x ).
S o l u t i o n. H e r e t a n# = 1, t h u s θ = π/4, a general sol uti on is w = <7(77) , or
u ( x,y ) = 5( ri)ei/v/5.
The initial c ondition yields
h ( x ) = g { - x/V 2) e x/2 = g { i 7 or 5(77) = Η ( -\/2 η ) e’?/'/2. Hence
u ( x, y ) = h ( ~ - s/2r?) = h ( x - y ) e y.
3.2 3. S o l v e u t t - c 2 u
XX — 0) subject t o t h e conditions u ( x, 0) = l n ( l + x 2) and Mt(x,0) = e ~ x.
A n s.
u(x,t) = ~ [ l n { l + (x + ct)2} + l n { 1 + ( x -
1, such t h a t u ( x, 0) =
or
( y - x )/2
+ G(—x/V 2 ) e x!2, so T)/V2
+ [1 - s i n ( x + 1/)] e27.1
1
nd V 2 — = w, whose
c i) }j + - e x c o s h c i.
4
O r t h o g o n a l E x p a n s i o n s
Unlike the ordinary differential equations, the general solution o f a partial dif­
ferential equation consists o f one or more arbitrary functions. It is not easy to determine the particular form o f these functions from the prescribed boundary and initial conditions even i f the general solution is known. However, it is often possible to solve a specific boundary value or initial value problem in the form o f an infinite series o f functions known as eigenfunctions or char­
acteristic functions. This chapter is devoted to developing orthogonal series, trigonometric Fourier series, eigenfunction expansions, and Bessel functions. Orthogonal expansions are important for the method o f separation o f variables which shall be discussed in the next chapter.
4.1. O r th og on a l it y
The inner product o f two real-valued functions f i and f 2 defined on an interval a < x < b is given by
( Λ,/2) = [ f i { x ) f
2
( x) dx, (4.1)
J a
provided the integral in (4.1) exists.
D e f i n i t i o n 4.1. The functions /1 and f 2 are said to be orthogonal on the interval α < x < b i f (/1, f 2) = 0, i.e., the integral (4.1) vanishes.
4.1. ORTHOGONALITY
83
D e f i n i t i o n 4.2. A set of real-valued functi defined on the interval a < x < b is said to be an o r t l\ on the interval a < x < b if for each m and η,π ι φ
(fm,fn) = / f m( x ) f n( x ) dx = 0, 771, Π
J a
ons { f i ( x ),f2 ( x ),· · · } o g o n a l s e t o f f u n c t i o n s n,
p r o v i d e d e a c h i n t e g r a l e x i s t s. T h u s, f o r e x a m p l e, t hb o r t h o g o n a l i t y r e l a t i o n s
7Ϊ7ΓΧ
f o r t h e f u n c t i o n c o s —— a r e
f l η π χ
J o"8 —
η π χ τ η π χ , cos —— cos —-— α χ
0, η
2 ’
1, η
D e f i n i t i o n 4.3. The n o r m o f the functions /, and defined by
ll/nll = y/(fn,fn) = ( j fn(x) dx
Φ m
= τ η φ 0 (4.3)
= m = 0.
( x ) is denoted by | |/n ||,
D e f i n i t i o n 4.4. An orthogonal set o f functio; o r t h o n o r m a l s e t o f f u n c t i o n s on the interval a < n — 1,2,· · ·.
T h u s, i f { f n(x)} is an orthonormal set o f functio:
rb
( f m j n ) = [ f m { x ) f n ( x ) d x ==
J a
f 1 i f 771 = 71
where 6 m n = < is the Kronecker delta. If an orthogonal set o f
I. 0 i f m ψ n functions {/„ ( x ) } is defined on the interval a < x can always construct an orthonormal set o f functions
9n{x) =
fn(x)
ll/nll ’
In fact, in view o f (4.5),
rb fm(x) f n ( x )
*\9 m i 9 n ) — f J a
ll/mll | |/„ and he nc e | | gn || = 1 f or al l n.
d x =
a < x <
l l/m l l II /„
= 1,2,
( 4.2 )
1/2
> 0.
( 4.4 )
ms {/n ( x ) } i s c a l l e d an < 6 i f | |/n || = 1 f o r a l l
ns, t hen
Smni ( 4· ^)
< b, with | |/n || Φ 0, we 5 9 n { x ) by defining
b.
(4.6)
f m,f n ) = 6 m n, (4.7)
84
CHAPTER 4: ORTHOGONAL EXPANSIONS
E x a m p l e 4.1. The functions f n ( x ) = sin n x, n = 1,2, · · ■, make an orthogonal set o f functions on t h e interval —π < χ < π, since
( f m,f n ) = / sin m x sin n x d x
J — 7Γ
1 Γ
= — I [ c o s ( t o — n ) x — c os( t o + η ) χ ] d x
2 7 _ π
= 0, m φ n,
a n d
s i n ( m — n ) x si n( m + n ) x
m — n
m + n
( f n,f n ) = / sin2 n x d x = ^ ί (1 —c o s2n x ) d x
J —tt J — π
sin 2na;
2 n
(4.8)
Thus, 11/„|| = ν/π, and t h e orthonormal set o f functions is given by
sm n x _
9n\X) — )— ί ^ 2, ·
V71"
E x am p l e 4.2. T h e set o f functions
f 1 cos x sin x cos 2 x sin 2 x
f o r ms a n o r t h o g o n a l s e t o n t h e i n t e r v a l — π < χ < π. In Exampl e 1.9, sin 7xx
w e h a v e s e e n t h a t — i s o r t h o n o r m a l o n t h e i n t e r v a l —π < χ < π.
λ/ΤΓ
N o w, t o v e r i f y t h e o r t h o n o r m a l i t y o f o t h e r f u n c t i o n s,
c o s m x cos n x
/:
d x
\[π
^ f * [cos(m - n ) x - cos(m + n ) x ) d x = 0, for m φ n,
2 π
1
2 π
f * [cos(m — n ) x — c os( m + n ) x ] d x
f or m = n.
A l s o,
Γ -
J
— π
sin m x sin n x
Γέ ypK 1
~2π
d x
1 Γ
= — J [sin(m — n ) x + si n ( m + n ) x ] d x
c o s ( m — n)x
c o s ( m + n)x
to — n
m + n
0, f o r a l l t o, n.
4.2. ORTHOGONAL POLYNOMIALS
F i n a l l y,
£
L
1 s in n x , 1 cos na
ax = —
V2tt \β ' 1 cos n x
* \/2 π y/π
dx =
κλ/2 η 1 sin nx
7 Γ\/2 n - ^ = >) d x = 7 T ]\ =
\/2 π ) 2 π Ι - π
8 5
Ex a m p l e 4.3. T h e s e t o f f u n c t i o n s
1, c o s a;, s i n x, c o s 2 x, s i n i f o r m s a n o r t h o g o n a l s e t o n t h e i n t e r v a l —π < x
c o s m x s i n n x dx =
f ·
J —π
U l · 1"'
< π, f o r w h i c h t o + ^ ) π χ dx sin(m j- η) πχ dx = 0
for all m, η = 0,1,2, ■ · ■. The orthonormal set i$ 1 cos x sin x cos 2x sin 2at
4.2. Orthogonal Polynomials
D e finition 4.5. The weighted inner prod and g, with weight w > 0, is defined by
(/, g) w = f f ( x) g( x) w( x)
J a
= o,
= o,
1.1
act of two functions / dx. (4.9)
Some classes of orthogonal polynomials are c efined for different val­
ues of a, b, w( x) for η = 0,1,2, ■ ■ ■ as follows:
86
CHAPTER 4: ORTHOGONAL EXPANSIONS
O r t h o g o n a l P o l y n o m i a l s ___________ a ______ b___________ w ( x )
C h e b y s h e v * ( 1 s t k i n d ) T n ( x )
-1
1
(1 — χ 2 ) ~ χ!2
C h e b y s h e v ( 2 n d k i n d ) U n ( x )
-1
1
( 1 - x 2 f'2
H e r m i t e H n ( x )
— 00
00
e -*2
J a c o b i P n *'^ ( x )
-1
1
( l - x ) a ( l - x
L a g u e r r e L n ( x )
0
00
e ~ x
L e g e n d r e P n { x )
-1
1
1
| w w w | T h e M a t h e m a t i c a p a c k a g e o r t h o n o r m a l i t y .m i s a v a i l a b l e o n t h e C R C w e b s e r v e r. I t c a n b e u s e d t o v e r i f y t h e o r t h o n o r m a l i t y o f s e t s o f f u n c t i o n s.
E x a m p l e 4.4. T h e L e g e n d r e p o l y n o m i a l s P n ( x ) are t h e solutions o f t h e Legendre equation
(1 — x 2 ) y" — 2x y' + n ( n = 1) y = 0.
T he se polynomials are also called the zonal harmonics o f t h e first kind. T h e orthogonal i ty relation for the Legendre polynomials P n ( x ) is
/
i ( 0 for η φ m
Pn(x) Pm(x) dx = < 2
2 τ ι -f- 1
for n = m
We w i l l p r e s e n t a M a t h e m a t i c a s e s s i o n t o e v a l u a t e t h e i n t e g r a l
J P n ( x ) P m ( x ) d x
f or n, m = 0,... , 10.
In[l]:=
l e g e n d r e = T a b l e [ L e g e n d r e P [ η,χ ], { η, 0 , 1 0 } ];
* Also written as Tchebyscheff.
87
88
CHAPTER 4: ORTHOGONAL EXPANSIONS
-Graphics-
In[2]:=
orthogonality =
Table[Integrate[legendre[[i]] legendre[[ j ] ], {x, -1, 1}], {i, 1, 10}, {j, 1 , 10}];
MatrixForm[orthogonality]
Out [2]=
2
0
0
0
0
0
0
0
0
0
0
2
3
0
0
0
0
0
0
0
0
0
0
2
5
0
0
0
0
0
0
0
0
0
0
2
7
0
0
0
0
0
0
0
0
0
0
2
9
0
0
0
0
0
0
0
0
0
0
2
11
0
0
0
0
0
0
0
0
0
0
2
13
0
0
0
0
0
0
0
0
0
0
2
15
0
0
0
0
0
0
0
0
0
0
2
17
0
0
0
0
0
0
0
0
0
0
2
19
4.3. Series of O r t h o g o n a l F u n c t i o n s
Some important types of series expansions are obtained from orthogonal sets of functions.
D e fin ition 4.6. Let {gi {x), g2{x), ■ ■ ·} be an orthogonal set of functions on an interval a < x < b, and let a function f i x ) be repre­
sented in terms of the functions gn (x), n = 1:2,· · ·, by a convergent series
(X)
f(x) = Y^cngn( x). (4.10)
n=l
This series is called a generalized Fourier series of f ( x ), and the coeffi­
cients c„, n = 1,2,· · ·, are called the Fourier coefficients of f ( x ) with
4.3. SERIES OF ORTHOGONAL FUNCTIONS
respect t o t h e orthogonal set o f functions g n ( x )
In view o f Definition 4.4, i t is easy t o de te rn If we multiply b o t h sides o f (4.10) by g m ( x ) fc wi t h respect t o x over t h e interval [a, 6], and i ntegration, which is justified in t h e case of un obtai n
rb nb / °°
(/,5m) = / f g m d x = / l ^ c n g n ( x )
\n = 1
OO
= ^ ^ Cn (gn {x) i gm (
n = 1,2,· · ·.
ine t h e coefficients c n. r a fixed to, i ntegrate assume t e r m - b y - t e r m iform convergence, we
g m { x ) d x
tr))·
n = 1
Since in v i ew of t h e relations (4.6) and (4.7) t h a t (g n,g m ) = ||i7n ||2 for n = m, and t he n (4.1
(<h
9 m ) — filld
l ) yields
( f,g n ) = cn \\g
2
η II 5
or
_ ( f,g n ) _ i
II gn
l l 3 n l | S
/ f ( x ) g,
J a
E x a m p l e 4.5. U s i n g t h e o r t h o g o n a l s e t o f f u 4.3 a n d f o r m u l a ( 4.1 2 ), t h e r e p r e s e n t a t i o n ( 4.1 0 )
f { x ) = a 0 + Σ ( a „ cos n x + b n sin n x ),
n = 1
where, since ||5 o || = ~ ^ =, ||3n|| = ~ η = for n = 1| \/27γ ν π
are given by
1 Γ a,° = — / f ( x ) dx,
1 Γ
a n = - f i x )
π J - π
1 Γ
b n = - / f i x ) si π J - π
c o s n x d x sin n x d x.
f or n = 1,2, ■ ■
89
( 4.11)
{x ) d x.
(4.12)
nctions from Example becomes
(4.13)
2, · · ·, t h e coefficients
(4.14)
We have introduced t h e t r i g o n o m e t r i c F o u r i e r s e r i e s o f f ( x ) in t h e above example, under t h e assumption t h a t t h e series (4.13) converges
90
CHAPTER 4: ORTHOGONAL EXPANSIONS
and represents t h e function f ( x ). The coefficients ao, a„ and bn for n = 1,2,· · · are called t h e F o u r i e r c o e f f i c i e n t s, and (4.14) are known as E u l e r ’s f o r m u l a s.
4.4. T r i g o n o m e t r i c F o u r i e r S e r i e s
A f u n c t i o n f ( x ) is said t o be p e r i o d i c o f period p i f f ( x + p ) = f ( x ), c < x < c + p, where c is a constant. For example, t h e functions s i n x and c o sx have period 2π. More generally, each o f t h e functions
27ϊ7Γ£ 2τΖ7ΓΧ
s i n and c o s is periodic o f period p where n is a pos iti ve
p p
integer. Hence, i f t h e infinite series
1 ^ ( 2η π χ , . 2η π χ\ .
- α 0 + > αη c o s h bn s m (4-15)
2 Ρ P J
i s c o n v e r g e n t, t h e n i t r e p r e s e n t s a f u n c t i o n o f p e r i o d p.
T h e o r e m 4.1. ( F o u r i e r T h e o r e m I, f o r p e r i o d i c f u n c t i o n s ) L e t f ( x ) b e a s i n g l e - v a l u e d p i e c e w i s e c o n t i n u o u s p e r i o d i c f u n c t i o n o f p e ­
r i o d p o n a f i n i t e i n t e r v a l I = [c, c + p\, w h e r e c i s a c o n s t a n t. T h e n t h e s e r i e s ( 4 - 1 5 ) c o n v e r g e s t o f ( x ) a t a l l p o i n t s o f c o n t i n u i t y a n d t o
^ [ f ( x + ) + f ( x ~ ) ] (4-16)
a t t h e p o i n t s o f d i s c o n t i n u i t y ( a n d a l s o a t a l l p o i n t s o f c o n t i n u i t y ). T h e c o e f f i c i e n t s a o, a n a n d bn are g i v e n by
2 f c+P j* / \ 2η π χ
a n = ~ f ( x ) c o s d x, ( n = 0,1,2,· · · ),
P J c P
2 r c +P _ 2η π χ on = ~ f ( x ) s i n — — d x, (n = l,2,· · · ).
P J c P
w h e r e c i s a c o n s t a n t s u c h t h a t t h e i n t e r v a l [c, c + ρ] = I.
( 4.1 7 )
I f w e s e t p = 2 L and c = — L in (4.17), t he n t he se formulas become r L
Ί Γ T17TX
a„ = — J f ( x ) COS —— d x, (n = 0,1,2,· · · )
1 ί ^ 7Ί7ΓΤ
bn = j J f i x') sin — d x, (n = l,2, · · ■ ) ·
4.4. TRIGONOMETRIC FOURIER SERIES
91
E x a m p l e 4.6. Let /( x ) = | x |, —π < χ < τ (see Fig. 4.1). No t e t h a t f ( x ) is an even function, since /( —x ) = | - r| = |x| = f ( x ). Then, from (4.18), w i t h L = π,
a. o
x d x = π
1 Γ 2 f
- |x| d x = —
π J - π π Jo
- Γ f ( x ) cos n x d x = — Γ s π Jo tt Jo
2 r c o s n x x s i n n x i 71·
c o s n x d x
π L
( 4
n J o n odd
= <. η*π I 0, n even
and bn = 0 for n = 1,2, · · ■. Hence the Fourier series (4.15) is
π 4 » c os(2n — 1 ) π
7Γ \
~ ~ π 2-*
^; ( 2n — l ) 2
n = l v '
—π < χ < π. ι
F i g. 4.1. G r a p h o f /( χ ) = | x |, —π < χ < π.
T h e f o l l o w i n g M a t h e m a t i c a s e s s i o n i l l u s t r a t e s t h i s e x a m p l e.
I n [ 3 ]:=
Clear[f,a0,a,b,fourier] ; f [xj : = Abs [x]
Integrate[Abs[x],{x,-Pi,Pi}]
0 n::n o n e: Me s s a g e S e r i e s D a t a::c s a n o t f o u n d.
92
CHAPTER 4: ORTHOGONAL EXPANSIONS
O u t [ 5 ] =
Integrate[Abs[χ], {x, -Pi, Pi}]
Mathemati ca will n o t integrate | x |. Observe t h a t t h e function f ( x ) cos n x is even and f ( x ) sin n x is odd. So we proceed as follows:
I n [6]:=
aO = (2/Pi) Integrate[x,{x,0,Pi}]
O u t [ 6 ] =
Pi
I n [ 7 ]:=
a = Table[2 Integrate[x Cos[n χ],{χ,Ο,Ρί}]/Pi,{n,10}]
O u t [7]=
-4 -4 -4 -4 -4
^Pi’°’ 9ΡΪ’°’ 25 Pi’0, 49Pi ’ °’ 81Pi’°'
I n [ 8 ]:=
b = Table[(Integrate[-x Sin[n x],{x,-Pi,0}]
+ Integrate[x Sin[n χ],{χ,Ο,Ρί}])/Pi,{n,10}]
O u t [ 8 j =
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
I n [ 9 ]:=
fourier[x_] = aO/2 +
Sum[a[[n]] Cos[n x]+ b[[n]] Sin[n χ] ,{n, 1,10}]
O u t [ 9 ] =
4.4. TRIGONOMETRIC FOURIER SERIES
93
rPi 4Cos[x] 4 Cos [3 x] 4 Cos [5 x] 4
~2 Pi 9ΡΪ 25 Pi
I a [ 1 0 ]:=
Plot[{f[x].fourier[χ]},{χ,-0.5,0.5}]
C o s [ 7 x ] 4 Cos [ 9 x ]
49 Pi
81 Pi
}
Exa mple 4.7. Consider
/( * ) =
'2, x = 0,
3, 0 < x < 2,
2, x = 2,
k 1, 2 < x < 4,
such that f ( x + 4) = f(x). This function is piec< is of period 4 (see Fig. 4.2). Note that /(2-) = this example, c = 0, p = 4. Then, using (4.17) w
ewise continuous and 3 and /(2+) = 1. In
e get
a° = 2 ^ J 3dx + J I dx } = 4,
2 P4
ηπχ [ ^ n
3 cos — αχ 4- / 1 cos —
I f f2 . ηπι f 4 . ηπχ
H Z 3,m — dx + l l s m “
dx I = 0,
■ dx
3 — 2 cos ηπ — cos 2 ηπ
ηπ
0,
= <, Ji_
ηπ ’
ii’
n is even, if n is odd.
Hence t h e Fourier series is
,, s _ 4 ^ 1 . (2η - 1) π χ
n = 1
Also, no t e t h a t in view o f (4.16), for e x a m p l e,/( 4 ) = - [/( 4 + ) + /( 4 —)], i.e., 2 = ^ ( 3 + 1)· ■
94 CHAPTER 4: ORTHOGONAL EXPANSIONS
3
• ·
12 · · · · ·
;
o
CN|
1
• Tf
2 4 6 8 10
Fig. 4.2. Graph o f f { x ).
T h e f o l l o w i n g M a t h e m a t i c a s e s s i o n i l l u s t r a t e s t h i s e x a m p l e.
I n [11]:=
Clear[f,aO,a,b,f ourier];
f [x_] : = If [x==0,2, If [0<x<2,3, If [x==2,2,
If[2<x<4,1]]]]
Integrate[f[x],{x,0,4}]
On::none: Message SeriesData::csa not found.
General::intinit: Loading integration packages —
please wait.
Out [13] =
Integrate[If[x ==0, 2, If[0 < x < 2, 3, If[x == 2, 2, If[2 < x < 4, 1]]]], {x, 0, 4}]
In[14]:—
4.4. TRIGONOMETRIC FOURIER SERIES
95
aO = (1/2) (Integrate[3,{x,0,2}] + Integrate[1,{x,2,4}])
Out[14]=
In[15]:=
a = Table[1/2 (Integrate[3 Cos[n Pi x, Integrate[Cos[n Pi x/4],{x,2,4}]),{n,:
Out[15]=
{ 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 }
In[16]:=
b = Table[1/2 (Integrate[3 Sin[n Pi x Integrate[Sin[n Pi x/4],{x,2,4}]),{n,
Out[16]=
'4],{x,0,2}] + 6}]
P i 3 P i 5 P i 7 P i 9 P i H P i
In[17]:—
f o u r i e r [x_] : = aO/2 + S u m [ b [ [ n ] ] S i n [ n P l o t [ { f [ x ].f o u r i e r[χ]},{χ,-0.5,4}, P l o t R a n g e - > { { - l,5 },{ - 1,3.5 } } ]
Ou t [ 1 8 ] =
W -
^4],{ x,0,2 } ] + :L6}]
4 4
0, ,0, ,0 }
13 Pi 15 Pi 1
Pi x/2],{n,1,16}]
-1
96
CHAPTER 4: ORTHOGONAL EXPANSIONS
E x a m p l e 4.8. Consider the wave equation utt = c2uXX: where c is the wave velocity, subject to the boundary conditions u(0, t) = 0 = u(l,t ) for t > 0, and the initial conditions u(x, 0) = x, and ut (x, 0) = 0. We shall assume the d’Alembert’s solution of the wave equation (see Example 3.17 and (5.24))
u(x,t) = f ( x + ct) + g(x — ct). (4-19)
From the boundary condition u(0,t ) = 0 we find that
f(ct) + g(-ct) = 0,
or, setting ct = z, we get f ( z) = —g(—z). Using the other boundary condition u( l,t ) = 0 we have
f ( l + c t ) + g ( l - c t ) = 0,
which implies that f ( c t + 1) — f ( c t — l ) = 0, i.e., f ( z ) = f ( z + 2 l ). This last equation means that / is a periodic function of period 21. The solution thus reduces to
u(x, t) = f(ct + x) — f(ct — x).
If we apply the initial conditions we get
f{x) - f{-x) = x, and f'( x) - f'( - x ) = 0,
which means that f'( x)
is an even function, and therefore f ( x )
an odd function, i.e., f ( x ) = - f ( - x ). Hence 2 f ( x ) = x. Since f ( x ) is an odd periodic function of period 21, it can be expressed as a Fourier sine series
.. . x I ^ ( - l ) n+1 . ηπ χ
2 π ' n
71=1
whi c h yi el ds
. , I ^ ( - l ) n+1
u{x,t) = ~ Σ'
7Γ · ^
π n
n = 1
ηπίοί + χ) . ηπίσί - x) sm —;
--------- sm-------;------
21 ( — 1) ηπχ ηποί
= — > -— ----- sm —— cos —;—.
π —J n I I
n = 1
This solution can be compared with (5.23). ■
4.4. TRIGONOMETRIC FOURIER SERIES
97
E x a m p l e 4.9. Consider the wave equation the wave velocity, subject to the boundary condi
(ii) u(l,t) = 0, and the initial conditions (iii) ut(x, 0) = 0. Using the d’Alembert’s solution (4
itions (i) ux(0, t ) = 0, u(x, 0) = x, and (iv) 19) we find that
du(x, t) df(x + ct) d(x + ct) dg(x — ct) d(x — ct)
dx
d{x + ct) dx
d(x -
which, by t aki ng ct = z, in view of condition f'(z) + g'(-z) = 0, where ' ξ d/dz, i.e., f'(z) = integration with respect to z yields f (z) = g(—z
9{z) = f ( ~ z ) + ci.
Condi t i on (ii) gi ves f (l + ct) + g(l — ct) = 0, whl· of g from (4.20) becomes
).
f(l + ct) + f(ct - l ) + c i =
Condition (iii) , in view of (4.20), gives
f(x) + f(~x) +ci = x,
whereas condition (iv) gives
c/'(x) + c f ( - x ) = 0,
i.e., f'(x) = —/'( —x), which implies that /' is therefore, /(x) is an even function, i.e., f ( —x) = x — ci. If we define
Φ ( χ ) = f { x ) + j,
then we have
ipict + I) + - I) = 0,
or, by taking ct — I = v, we get
ψ(υ) + φ(ν + 21) = 0.
If we set υ = ζ + 21, then
v>(c + 4 0 = - < k c + 2/) = φ (\:
Hence, φ is a periodic function of period 41, a Let
TITT'T
φ{χ) = An sin + Bn cos
tt = c2 u x x, where c is
ct)
dx
(i) yields ^(Ο,ί) = —g’{—z) which upon + Ci, and hence
(4.20)
ch by using the value
an odd function, and, f ix). Hence, 2f (x) =
d φ(χ) + φ{—χ) = χ. ιπχ
2Τ'
98
CHAPTER 4: ORTHOGONAL EXPANSIONS
Then
u(x, t) = ψ(οί + x) + 1p{ct — X)
. r . η π. , . . ηπ, .ι
= A n | s i n — (ct + x) + s i n — (ct - x) j
_ r η π η π . . ι
+ | c o s — ( c i + x) + cos — (ct - X) j
„ . . nnct ηπχ „ mrct ηπχ
= 2 A n s m —j - cos - j - + B n cos c o s
Using t h e i ni t ia l condition (iv), we find t h a t U t ( x, 0) = 0 gives A n = 0.
TlTTCt 717Γ
Again, condition (ii) yields 2 B n cos — - cos — = 0; thus, B n = 0 if
/Ll ii
n = 2m, and B n ^ 0 i f n = 2m — 1, where m is a p os it i ve integer. Thi s gives
. . ^ „ ( 2 m — l W c i ( 2 m — \)πχ
U(X, t) = > 2 B 2 m - l C O S TT^ cos ■
m= 1
21
21
Since
we find t h a t
. v -'r>r, ( 2 m — \)πχ
i{x, 0) = > 2B m c o s —----------= x,
m= 1
21
B m =
( 2 m — \)πχ x c o s — αχ
I Jo 21
( 2 m — 1 ) π χ] 1
it
x s i n ■
21
21
( 2 m — 1 ) π 2 ( - l ) m Z
/
J o
21
o (2m — 1 ) π ( 2 m — 1 ) π χ
sin - 41
21
d x
(2m — 1 ) π ( 2 m — 1 ) 2π 2
D e f i n i t i o n 4.7. A function / defined on an interval [a,b] is said
t o be o f bounded variation on [a, b] if i t s t o t a l variation var(/) on [a, b]
i s f i n i t e, i.e.,
n
v a r (/) = s u p V ] |/( i j ) - /( f » - i ) |, ( 4.2 1 )
t h e s u p r e m u m b e i n g t a k e n o v e r a l l p a r t i t i o n s P
a = to < ti < ■ ■ ■ < t n = b (4.22)
4.4. TRIGONOMETRIC FOURIER SERIES
99
of the interval [a, b], and n is an arbitrary positi the choice of the values ti, t 2, ■ · ■ , tn- i satisfies bounded variation form a vector space, with nor:
ive integer such that .22). All functions of
II
(4
ll/ll = l/( a ) l +var(/).
A monot one f unct i on / on [a, b] is of bounded variation is
var(/) = |/( 6) - f ( a)|.
A function / of the class C1 is of bounded variati variation
ariation, and its total
on on [a, b], with total
var( f ) = [ \f'(x)\dx, J a
provi ded /' exi st s and is bounded i n (a, 6). A r esul t is: A funct i on / i s sai d t o sat i sfy t he Lips· exi st s a const ant M > 0 such that
generalization of this :<\hitz condition if there
\f(x)~f(y)\ < M\x — y\.
Obvi ously, such a f unct i on is uni forml y cont i nuo vari at i on. Some wel l - known r esul t s are:
A funct i on / is of bounded var i at i on on [a, b\ is If / and g are of bounded variation on [a,b], the bounded variation on [a,b].
If / is of bounded variation on [a, 6] and c is a function cf is also of bounded variation on [a, 6] If / is of bounded variation on [a, b], then (a) v function; (b) var(/) is an increasing function, a: at xq £ [a, 6] iff var(/) is continuous at xq. Mor bounded variation on [a, b] can be represented as functions g and h:
f (x) = g(x) + h(x), a < x
such t h a t g is nondecreasing and h is nonincreasiri / has the following properties:
(i) one-sided limits f (x+) and f ( x —) from the exist at each point;
(u) / has at most countably many discontinuitie
(iii) / is bounded and integrable over the interv
(4.23) >us and is of bounded
bounded on [a, b}. ti f + g and / g are of
eal number, then the
ar(/) is an increasing ,nd (c) / is continuous eover, a function / of sum of two monotone
<t b, (4.24)
g. Each such function interior of the interval
s in the interval, and al.
100
CHAPTER 4: ORTHOGONAL EXPANSIONS
T h e o r e m 4.2. ( F o u r i e r T h e o r e m I I ) L e t f ( x ) d e n o t e a p e r i o d i c f u n c t i o n o f p e r i o d 2 π w h o s e i n t e g r a l f r o m —π t o π e x i s t s. I f t h a t i n t e g r a l i s i m p r o p e r, l e t i t be a b s o l u t e l y c o n v e r g e n t. T h e n a t e a c h p o i n t x w h i c h i s i n t e r i o r t o a n i n t e r v a l o n w h i c h f i s o f b o u n d e d v a r i a t i o n, t h e F o u r i e r s e r i e s f o r t h e f u n c t i o n f c o n v e r g e s t o t h e a v e r a g e v a l u e ( 4 - 1 6 ).
T h e a s y m p t o t i c b e h a v i o r o f t h e Fo ur i e r c o e f f i c i e n t s o f a p e r i o d i c f u n c t i o n f ( x ) is given by t h e following theorem:
T h e o r e m 4.3. A s n —> oo, t h e F o u r i e r c o e f f i c i e n t s a n a n d bn a l ­
w a y s a p p r o a c h z e r o a t l e a s t a s r a p i d l y a s a/n w h e r e a i s a c o n s t a n t i n d e p e n d e n t o f n. I f t h e f u n c t i o n f { x ) i s p i e c e w i s e c o n t i n u o u s, t h e n e i t h e r a n o r b n, a n d i n g e n e r a l b o t h, d e c r e a s e n o f a s t e r t h a n a/n. I n g e n e r a l, i f f ( x ) a n d i t s f i r s t k — 1 d e r i v a t i v e s s a t i s f y t h e c o n d i t i o n s o f t h e F o u r i e r t h e o r e m s I a n d I I, t h e n t h e F o u r i e r c o e f f i c i e n t s a n a n d bn a p p r o a c h z e r o a s n —» oo a t l e a s t a s r a p i d l y a s a/n k + 1. M o r e o v e r, i f f ^ ( x ) i s n o t e v e r y w h e r e c o n t i n u o u s, t h e n e i t h e r a n o r bn, a n d i n g e n e r a l b o t h, a p p r o a c h z e r o n o f a s t e r t h a n a/n k + l.
T h i s t h e o r e m i m p l i e s t h a t t h e s m o o t h e r t h e f u n c t i o n / i s, t h e f a s t e r i t s F o u r i e r s e r i e s c o n v e r g e s. I t s h o u l d b e n o t e d t h a t t h e Fo ur i e r s e r i e s f o r t w o - a n d t h r e e - d i m e n s i o n a l f u n c t i o n s a r e s i m i l a r t o t h e a b o v e a n a l ­
y s i s f or o n e - d i m e n s i o n a l f u n c t i o n s.
P r o o f s o f t h e s e t h e o r e m s a r e a v a i l a b l e i n D a v i s ( 1 9 6 3 ), C h u r c h i l l a n d B r o w n ( 1 9 7 8 ), a n d Wa l ke r ( 1 9 8 8 ).
D e f i n i t i o n 4.8. L e t / b e a f u n c t i o n d e f i n e d o n t h e i n t e r v a l 0 < x < L such t h a t t h e integrals
[ f ( x ) sin d x, n = l,2,· · ·,
J O L*
e x i s t. T h e n t h e s e r i e s
o o
Σ b u s i n g (4.25)
n=l
where
2 ί ^ 777ΓΤ
bn = — J f ( x ) s m - £ - d x, (n = 1,2,· · · ), (4.26)
is c a l l e d t h e Fourier sine series
o f / On t h e i n t e r v a l 0 < x < L.
4.4. TRIGONOMETRIC FOURIER SERIES
101
No t e t h a t t h e series expansion (4.25) is ide: metric Fourier series (4.15) o f an o d d function —L < x < L, which coincides wi t h f { x ) on t h e
ntical t o t h e trigono- defined on t h e interval interval 0 < x < L.
D e f i n i t i o n 4.9. L e t / b e a f u n c t i o n d e f i n e d o n t h e i n t e r v a l 0 < x < L such t h a t t h e integrals
f L ι* / \ η π χ I j [ x ) cos a x, n = 0,l,
Jo L
e xist. Then t h e series
ao
T
4“ ^
n = 1
cos ·
η π χ
~ r
where
2 f L . η π χ
α η = γ j ( x ) cos —— α χ, [ n = 0,1
L Jo L
is called t h e F o u r i e r c o s i n e s e r i e s of / on t h e interval 0 < x < L.
N o t e t h a t t h e s e r i e s e x p a n s i o n ( 4.2 7 ) i s id^: m e t r i c Fo ur i e r s e r i e s ( 4.1 5 ) o f a n e v e n function —L < x < L, which coincides wi t h f ( x ) on t h e
E x a m p l e 4.1 0. To find t h e Fourier cosine se represents f ( x ) = x on t h e interval 0 < χ < π Then
ao
x d x = π,
1 1
π Jo
Γ
J o
2( cos?wr — 1)
2 Γ ,2
a „ = — / x cos n x α χ = π ./η π
η
π η
0, i f η is e
- Λ, i t
π η *
Hence t h e Fourier cosine series is
π 4 v ^ 1
/ TTTT COS(27l
jt ^ 2n - l ) 2 v
^ (2η - 1)
η —1 v '
(4.27)
2,···),
(4.28)
ntical t o t h e trigono- lefi ned on t h e interval interval 0 < x < L.
r i e s o f p e r i o d 2π w h i c h l e t L = π in (4.28).
x sm n x cos n x
+
n *
iiven, i s o d d.
— l ) x.
102
CHAPTER 4: ORTHOGONAL EXPANSIONS
T h e graph o f t h e right side of t hi s series is given in Fig. 4.3. No t e t h a t
at x = 0, t h e right side o f t h e series equals
π 4 ^ 1 π 4 7Γ2
2 _ π γ ( 2 η - 1 ) 2 _ 2 _?Γ Ϊ _ ’
00 1 π2
since —^ = —. At χ = π, t h e right side o f t h e above series is
6
1
, π 4 π2 obvi ou sl y equal t o — H — = π. ■
2 7Γ 8
| www | p l o t f o u r i e r .ma, available in t h e Mathematica Notebook, gen­
erates a t a bl e of t h e elements of a trigonometric Fourier series and pl ot s their graphs for a given function.
E x am p l e 4.1 1. In t h e case o f a jump discontinuity t h e Fourier series leads t o what is known as t h e Gibbs phenomenon. Consider, e.g., t h e Fourier sine series for
tt \ - / 1( 0 < ^ < 1
i-l, -1 < x < 0.
Then, t h e coefficient b n, defined by (4.23) are given by
00 4
bn = V s i n (2n + 1 ) π χ.
2n +1 v 1
n=0
4.5. EIGENFUNCTION EXPANSIONS
103
T h e p a r t i a l sums
Λ4 = £
n=0
2n + 1
sin(2n + 1
)πχ
M-
define the k harmonics, which approximate the
4.4. Notice the sharp peaks in the harmonics M40 near 0 which is the discontinuity of /(x). height or overshoot of these peaks is greater ths The width of the overshoot goes to zero as k remains at 9% both at the top and the bottom
lim max|/(x) - Mk(x)\ φ
fc—+ OO
This phenomenon does not go away even when tt is increased. ■
jump as shown in Fig. i, M2, Mio, M20, and ibbs showed that the 4n /(0+) by about 9%.
- 00, but the height such that
0.
e number of harmonics
Μ, M,
M 20
Fig. 4.4. Gibbs phenomenon
4.5. Eigenfunction Expansions
The Sturm-Liouville problems arise in the solul; problems when one uses the method of separat: method, discussed in detail in the next chapter, ful methods in solving boundary value problems
m
,
ion of boundary value ion of variables. This is one of the most use- involving partial dif-
104
CHAPTER 4: ORTHOGONAL EXPANSIONS
ferential equations. A Sturm-Li ouvi l le problem c onsists o f the S t u r m - Liouville equation
d
dx
p{x)
dy
dx
+ [q{x) + X w( x)\y = 0,
(4.29)
which is a linear second order ordinary difFerential equation defined on a given interval a < x < b and satisfies t h e boundary condi ti ons o f the form
ai y{a) + bi y'{a) = 0,
a2y(b) + b2 y'(b) = 0,
where Λ is a real parameter, and ai,a2,bi,b2 are given real c onstan ts such t h a t οχ and bi, or a2 and b2 are b o t h not zero. It is obvious t h a t (4.29)-(4.30) always has a trivial sol uti on y = 0. T h e nontrivial solutions o f t h i s problem are called t h e eigenf unct i ons φ„( χ) and t h e corresponding values o f Λ t h e eigenvalues Xn o f the problem. T he pair ( Φ η,λ η ) is known as t h e eigenpair.
T h e o r e m 4.4. Let the f unct i ons p, q, r a n dp' in Eq (4-29) be real­
valued and conti nuous on the int erval a < x < b. Let φ„ι{χ) and φη (χ) be the ei genf unct i ons o f the probl em ( 4· 29) - ( 4· 30) wit h corresponding eigenvectors Xm and Xn, respectively, such t hat Xm φ Xn. Then
f b
/ φm( x ) φ n ( x ) w ( x ) d x = 0, rn φ n, (4-31)
J a
i.e., the eigenfunctions <^m and φη are orthogonal wi t h respect to the weight f unc t i on w( x ) on the i nt erval a < x < b.
P r o o f o f t h i s t h e o r e m c a n b e f o u n d i n a n y s t a n d a r d t e x t b o o k o n o r d i n a r y d i f f e r e n t i a l e q u a t i o n s, e.g., R o s s ( 1 9 6 4 ), B o y c e a n d D i P r i m a ( 1 9 9 2 ).
D e f i n i t i o n 4.1 0. T h e b o u n d a r y c o n d i t i o n s o f t h e t y p e
y( a) = y(b), y'{ a) = y'(b) (4.32)
are known as t h e peri odi c boundary conditions. In t h i s case t h e sol uti on is o f period b — a.
T h e e i g e n f u n c t i o n e x p a n s i o n o f a n a r b i t r a r y f u n c t i o n f ( x ) in t h e interval a < x < b is given by
OO
f ( x ) = Σ ° n (4.33)
n = 1
where φ η are t h e eigenfunctions, with correspoi t h e Stu rm-Li ouvi l le (or eigenvalue) problem
4.5. EIGENFUNCTION EXPANSIONS
l ^ + x 2 y ~ 0'
s u b j e c t t o t h e b o u n d a r y c o n d i t i o n s ( 2.2 5 ), a n d d e t e r m i n e d b y ( 4.1 2 ).
Ex a m p l e 4.1 2. T h e s e t o f o r t h o g o n a l f u: c o s 2 x, s i n 2 x, · · · o f E x a m p l e 4.3 a r e t h e e i g e r v a l u e p r o b l e m
y" + Ay = 0, ν(-π)=ν(π), y'{
T h e c o r r e s p o n d i n g e i g e n v a l u e s a r e Xn = n, (n
a c t i o n s 1, cosx, s i n x, functions o f t h e eigen-
E x a m p l e 4.1 3. For t h e eigenvalue problem interval 0 < x < L, and (a) subject t o t h e Di: t i ons 3/( 0) = 0 = y( L), t h e eigenpair is
(4.34) defined on the richlet boundary condi-
ηπ
Φη (x) = sin λ ηχ, λ η = —, (η
and (b) wi t h t h e Neumann boundary condition th e eigenpair is
Φη(χ) — COS A n X, Ay]
ηπ
Τ ’
( n =
I www I e i g e n p a i r .ma, available in t h e Mathem used t o obtai n t h e eigenvalues and eigenfunction: value problem.
Tables 4.1 and 4.2 at t h e end o f t h i s chapte: t h e solution of t h e eigenvalue problem wi t h th: conditions (4.32) in t h e Cartesian and t h e polar respectively. N o t e t h a t t h e Bessel equation i coordinates
f y , ( 2 _ v ‘
dr* r dr X
y = 0, 0 < r <
105
i ding eigenvalues A„ of (4.34)
t h e coefficients cn are
π) = ν'( π ).
1,2,···)· ■
== 1,2,-·-);
3 y'( 0) = 0 = y'(L), the
0,1,2,·
atica Notebook, can be s for a given boundary
r provide t h e d a t a for ree t y p e s o f boundary cylindrical coordinates, t h e polar cylindrical
a,
v > —,
_ 2 ’
( 4.3 5 )
i s a S t u r m - L i o u v i l l e e q u a t i o n.
106
CHAPTER 4: ORTHOGONAL EXPANSIONS
4.6. Bessel F u n c t i o n s
Th e Bessel functions o f order v are the solutions o f t h e ordinary differ­
e ntial equation
χ2^ ί ί +χ% + ^χ2 - μ2^ = 0’ (4·36)
which is known as t h e Bessel equation o f order v. Its regular solutions for each v > 0 are t h e Bessel functions o f t h e first kind
n=0
JS* ( - I'I 71 /x\2 n + u
·'» ( * > = Σ „ | ( „ + „ ) 1 ( a ) ’ - C O < X < CO. ( 4.37)
The se power series solutions are obtained by t h e Frobenius method, detai ls o f which are available in any standard book on ordinary dif­
ferential equation, e.g., Ross (1964), Boyce and Di Pri m a (1992). T h e infinite series (4.37) is uniformly convergent and can be differentiated or i ntegrated t e r m - b y - t e r m. The differentiation and integration formulas are as follows:
J„(-x) = ( - 1 )" J „ ( x ),
^ [ x v Jv{x)\ = xv Jv-i{x),
^ [ x ~" J „ ( x ) ] = - x _" J „ + i ( x ),
x J'( x ) = v Jv(x) - x J v+i(x), x J'( x ) = —v J u(x) + x J„_ i ( x ),
f
Jo
(4.38)
t v d t = x v J v ( x ),
/ o
where 1 = d/d x.
In particular,
Jo(x) = -Ji(x),
and
pX
t J 0( i ) d t = x J i ( x ).
/
Jo
Jo
The integral representation for J v ( x ) is
J u ( x ) = — [ co s( x s i n# — ν θ ) ά θ, (4.39)
π Jo
4.6. BESSEL FUNCTIONS
107
1 Γ 2 f
J o ( x) = — cos(x s i n θ ) ά θ = —
π Jo π Jo
a n d, i n p a r t i c u l a r,
π/2
coi:(o; c o s#) ( 4.4 0 )
N o t e t h a t J 2k is an even function and J 2k + i an o j 0(0) = 1, but J„(0) = 0 for v > 1. In fact, J„ h v at x = 0. T h e integral representation (4.39) for all real x and v > 0.
dd function. Moreover, as a zero o f mul ti pl ici ty $hows t h a t || J„(a;)|| < 1
W W W
P l o t s o f J v { x ) and J l ( x ) for v = 0,
b e s s e l.m a in t h e Mathematica Notebook.
derivatives, i t is found re almost e ven ly spread J „ (x ) and J'v { x ) each e simple exce pt x = 0. o f t he se functions be and a'„„; the n t h e zeros interlace according t o t h e
From t h e graphs o f Jo, J\, J 2, J3 and their t h a t each J„ decays for large x, and their zeros a and interlaced. In fact, for real v t h e function^ have c ountably many real zeros, all o f which a: For nonnegative ν, let t h e n - t h pos iti ve zerc denoted by a Vi, inequalities
ctv,l < CLv- (-1,1 < Ci v,2 < Oi u+1,2 < 0 v < < c t y + i i < θί ν 2 < a i/+ i,2
i.e., e a c h J v possesses an increasing unbound^ zeros. In fact, all zeros o f J v are real for v > v not an integer, t h e number o f complex zeros c part o f ( — v ). I f t h e integer part of ( —u ) is odd on t h e imaginary axis. If v > 0, all zeros o f J and Stegun, 1965, p. 372).
B y taking t h e limit o f t h e integral representation (4.40) as x —* oo, it can be shown t h a t
lim J v { x ) = 0.
In fact, in v i ew of t h e Ri emann-Lebesgue lei:
/
7Γ Λ7Γ
F ( x ) c o s K x d x = lim / F i x ) sinK
2 /'7r/2
lim J 0 ( x ) = lim — / cos(
x —►cx» χ —>oo 7Γ J q
= - l i m - f 1
T—o o π J o
cos(x
7 ^
1,2,3 are available in
/,3 < · ■ ·
<: a' , <
V, 3
d sequence o f pos iti ve 1. B ut for v < — 1 and if J v is twice t h e integer two o f t h e s e zeros lie are real (Abramowitz
nma which s t a t e s t h a t x d x = 0, we get
os θ ) ά θ
i f )
dy = 0.
108
CHAPTER 4: ORTHOGONAL EXPANSIONS
Similarly,
1 Γ
J v[x) = — I cos(x sin# — ι/θ)άθ π J o
1 Γ
= — I [cos cos(ar sin Θ) + sin νθ sin(a; sin θ)] άθ
77 Jo
t e n d s t o z e r o as x —> oo.
E x a m p l e 4.1 4. Consider the polar cylindrical form of the Laplace operator
L = — - —
dr2 ^ r dr
i n R1. Then the corresponding eigenvalue problem is L4> + λ'2φ — 0. It can be shown that the radially symmetric eigenfunctions φ^ of the Laplace equation subject to the Dirichlet boundary condition φ( 1) = 0 on the unit disk U are
Φη = (k = 1,2,...),
s uc h t h a t
λι < Λ2 < Λ3 < · · ■
a r e t h e pos i t i ve z er os of J o - The s e e i ge nf unc t i ons f or m a c ompl e t e o r t h o g o n a l s e t, i.e., f or m φ n
( t/o ( A mr ), i/o ( A n ^ ) ) — Ι J o (Aj^r) J o { X n/r') τ d r = 0.
Jo
Then, for any function / with the norm
l l/l l 2 = f f (r) 2rdr < +00,
Jo
we have the Fourier-Bessel expansion
0 0
f ( r ) = X ] c„ J 0(A„r), (4.41)
n=1
where
( f,J o M )
^ μ 0(λ„Γ)||2 ’ (4 42)
II τ / λ mi2 J l ( X n r )
| | Jb(A„T-)| r =-----~------■■
In general, we have
4.6. BESSEL FUNCTIONS
109
THEOREM 4.4. T h e e i g e n f u n c t i o n s φ η = Χ ( λ ητ·), ( η = 1,2, · · · ), f o r m a c o m p l e t e o r t h o g o n a l s e t o f r a d i a l l y s y m m e t r i c s q u a r e i n t e g r a b l e f u n c t i o n s s u c h t h a t f o r a s q u a r e i n t e g r a b l e f u n c t i o n f t h e F o u r i e r - B e s s e l e x p a n s i o n
fij·) — ^ ' cn Jυ (Λri
n = 1
h o l d s, w h e r e
(/,Λ(ΛηΓ))
” IWA„r)||2 ’
a n d Xn a r e t h e p o s i t i v e z e r o s o f J„ f o r n = 1,2
IΛ (A,
i»0II2 = Γ J l i K
Jo
( 4.4 3 )
( 4.4 4 )
, a n d
r ) r d r
[JKK)}2
2
Jy+l(An) 2
Jl- i(A„)
T h e o r e m 4.5. T h e e i g e n f u n c t i o n s f o r t h e z e r o D i r i c h l e t c o n d i t i o n o n t h e u n i t d i s k U = {
Jo (A n r ), J u ( X n r ) c o s i'0, J „ { X
n = 1,2,· · ·, a n d u > 1, w i t h t h e e i g e n v a l u e s z e r o s o f J „. T h e s e e i g e n f u n c t i o n s f o r m a c o m p l t h e H i l b e r t s p a c e L 2 ( U ) o f a l l s q u a r e i n t e g r a b l e
( 4.4 5 )
w h i c h a r e t h e p o s i t i v e e t e o r t h o g o n a l b a s i s i n f u n c t i o n s o n U.
P r o o f s o f t h e s e t h e o r e m s c a n b e f o u n d i n W a t s o n ( 1 9 4 4 ).
For t h e t h r e e t y p e s o f b o u n d a r y c o n d i t i o n s th< i n T a b l e 4.2 a t t h e e n d o f t h i s c h a p t e r. H e n c e, a o n t h e s e t U χ ( 0,2π) has an eigenfunction exp
OO
/( Μ ) = Σ cosn6> +5„(i
n=0
where each o f t h e functions f n and g n has an (4.43).
L a p l a c e o p e r a t o r w i t h < 1 } a r e
r ) sin ν θ,
( 4.4 6 )
e e i g e n p a i r s a r e d e f i n e d f u n c t i o n f ( r, Θ) defined ansion o f t h e form
sin η θ ],
(4.47)
expansi on o f t h e f o r m
110
CHAPTER 4: ORTHOGONAL EXPANSIONS
4.7. Exercises
Show t h a t each given set of functions is orthogonal on the given interval, and determine the corresponding orthonormal set of functions:
4.1. j s i n ^ ^ j ,n = 1,2,3,·
—l <x <
4.2. | c o s ^ n J x. |, η = 0,1,2, ■ ■ · ; —l < x < l
4.3. { s i n 2 n x } , n = 1,2,3, ■ · · ; 0 < χ < π
4.4. { c o s 2 7 i:r } ,7 1 = 1,2, 3, · · ·; 0 < χ < π
4.5. {sin37ix} , η = 0,1,2, ■ · · ; - π < χ < π
4.6. {cos Snx} , η = 0,1,2,3, ■ ■ · ; —π < χ < tt
4.7. { s i n 2 7 i a;, c o s 2 n x } , n = 1,2,3, ■ · · ; |x| < π
4.8. Find th e Fourier series for the function /(x) which is assumed to have the period 2π:
t \ tt \ - / 1 i f ~ π/2 < χ <Ί ΐ/2
( a) /0 ) | _ χ ,f η τ ι 2 < χ < 3π/2
(b) f ( x ) = χ, —π < χ < π.
( c ) /( x ) = x 2, —7Γ < x < 7 Γ.
/ v 4 / 1 1 r
An s. ( a ) — I c o s x — - c o s 3 x — h - c o s o x + ·
π \ 3 5
( b ) 2 ^ s i n x - ^ s i n 2 x + ^ s i n 3 x - i s i n 4 x H- - - - -
( c ) —- - - - -4 ( c o s x — \ cos 2x + ^ cos 3x — — cos 4x +
3 V 4 9 16
4.9. Find the Fourier series of the period function /( x ) of period T:
( a ) fix)
— 1 i f — 1 < x < 0
1 i f 0 < x < 1, T = 2.
4.7. EXERCISES
(b) f ( x ) = 1 - x 2, —1 < x < 1, T = 2.
An s. ( a ) — ( βίηπχ + ^ βίη3πχ + βϊηδπχ π \ 3 5
2 4
1
(b) - -1— ~ cos 7ΓΧ — - cos 2 π χ + - cos 3 π χ — w 3 π 1 ' Λ 0
4.1 0. Find t h e trigonometric Fourier series of —4 < x < 4.
th e function /( x ) = x,
Ans. N o t e t h a t /( x ) is an odd function, wh L = 4. Then
ich implies a n = 0; and
1 / . η π χ 8
- I x sm —— a x = -------
2 Jo 4 ηπ
a 00
* -;Σ
and the series is
cos η π =
( —1 )"+ η π
-------------- s m —-
π ; n 4
η —1
8(-
4 i )"+ 1
ηπ
4.1 1. Find t h e trigonometric Fourier series o f
f i x ) = |
π, —π < χ χ, 0 < χ <
<
71’
Ans. N o t e t h a t L — π. The function / is Then
1
ao
; j f { x ) d x = ~
1 /· π
an = ~ f { x)
t f ·/ — 7Γ
c o s n x d x =
π η ^
r, f o r o d d n,
0, f o r e v e n n,
b n = l Γ/( x )
tf ./-7Γ
sin n x d x = —, (
n
111
(n = 1,2,· · · ),
t h e f u n c t i o n 0,
n e i t h e r o d d n o r e v e n.
c o s η π — 1
n = 0,l,2,· · · ).
112
CHAPTER 4: ORTHOGONAL EXPANSIONS
The series is
1
cos n x sin n x
n
4.1 2. F i n d t h e t r i g o n o m e t r i c F o ur i e r s e r i e s o f t h e f u n c t i o n
' 7Γ, —71" < X < 0,
. π
gix) = 2 ’ x = ’
x, 0 < χ < -κ.
H i n t: T h e f u n c t i o n g ( x ) is the same as f i x ) in Exercise 4.11, e xce pt at x = 0. Since t he se two functions have t h e same values at all po i n ts e xce pt a finite number (only one in t hi s case) in t h e same interval, t h e function g i x ) has t h e same Fourier series as t h a t for f { x ) in Exercise 4.11.
4.1 3. Find t h e trigonometric Fourier series o f t h e function f i x ) = a2 — x 2, 0 < x < a.
A n s.
4.1 4. S h o w t h a t
H i n t: U s e i n d u c t i o n, o r t h e r e s u l t
rir/2
0
c o s'" Θ sin" θ ά θ =
n + l 2
4.1 5. S h o w t h a t
OO
e*(t—1/0/2 = J n { x ) [ t n + ( - i ) nr n].
H i n t: Multiply the series expansions for ext!2 and e x^ 2t'>
4.7. EXERCISES
4.16. Show t h a t
c o s x
= J0(x) + 2^2(- l ) n ^2n(x),
71=1
OO
s i n x = 2 ^ ( - 1 )" J 2n + i ( x )
n=l
H i n t: set t — i in Exercise 4.15.
4.1 7. Show t h a t
cos(x sin #) = J 0 ( x ) + 2 ^ J 2n(
η — 1
oo
sin(x sin#) = 2 ^ J
2
n+i(x)
sin
n=0
113
x)
cos 2n#, 2 η θ.
H i n t: s e t t = e in Exercise 4.15.
4.1 8. Prom t h e results o f Exercise 4.17, d e du ct tha t
1 Γ
J 2n { x ) = — / c o s ( x s i n#) po s 2 n#d#, 2tt J 0
1 Γ
J
2
n+i{x)
= — I
s i n ( x s i n#) s i n 2η θ ά θ.
Jo
4.1 9. S h o w t h a t
f
J o
J v ( y ) 2 y d y =
( a 2 - v 2 ) J „ { a ) ·
H i n t: U s e t h e i d e n t i t y x J'u { x ) (x J'( x ) )' = J'( x ) [ x 2J"( x ) + x J'( x ) ]
4.2 0. Show t h a t i f A is a pos iti ve zero of J v { x ) H i n t: Set x = a r and use Exercise 4.19.
4- a 2 J'( a ) 2
then (4.45) holds.
+ λ 2 y = 0,0 < x < L, subject to the boundary conditions a i y( 0) + bi y'(0) = 0, a 2 y(L) + b2 y ’{L) = 0.
1 a■
i o n: f ( x ) = Ύ ^ ο η φη{χ) , cn - ^ ^ J f{x)<j>n(x) dx, hi = j ·, {i = 1,2).
Table 4.1: Eigenvalue P r o b l e m i n C a r t e s i a n C o o r d i n a t e s
d2v
dl
dx2
Ei genfunction expansion
Boundary conditions
At x = 0 A t x — L
A„ a r e t h e r o o t s o f
1.
D i r i c h l e t
D i r i c h l e t
s i n Anx
ai Φ 0, bi = 0
02 Φ 0, i>2 = 0
2.
D i r i c h l e t
N e u m a n n
s i n A„x
οι Φ 0, bi = 0
02 = 0, 62 Φ 0
3.
D i r i c h l e t
R o b i n
s i n A„x
Οι Φ 0, bi = 0
02 Φ 0, 62 Φ 0
4.
N e u m a n n
D i r i c h l e t
c o s Anx
Oi = 0, bi φ 0
02 Φ 0, f>2 = 0
5.
Neumann
Neumann
cosAnx
Oi = 0, bi Φ 0
o2 = 0, 62 φ 0
L
2
L
2
AnL — s i n AnL cos AnL
2λΙ
L
2
L ,
2
s i n A L = 0, i.e., Xn = ( n = 0,1,2, ■ · · )
cos λ L = 0, i.e., A„ = (n = 1,2,···)
λ + h2 t a n AL = 0 *
cos XL = 0, i.e., An = ( n = 1,2,· · · ) s i n A L = 0, i.e., A„ = (n = 0,1,2,· · · )
L ’
(2n — 1)π 2L ’
(2n — 1)π
I ’
ηπ
Τ ’
Η
>
ca
r
H
6. Neumann Robin cosAn x
ai = 0, bi φ
0 02 φ
0, b2 Φ
0
7. Robin Dirichlet sin \„ (L - x)
οι Φ
0, bi Φ
0 02 φ
0, 62 = 0
8. Robin Neumann cosA n (L — x)
οι φ 0, 61 Φ 0 02 = 0, 62 Φ 0
9. Robin Robin
Οι Φ 0, 61 φ 0 o2 ^ 0, 62 7^ 0
A„ c o s A „ x — /i i s i n A n x -
An L + s i n AnL cos A„L
2K.
AjiL — s i n AnL cos AjiL
2λΙ
A„L -I- s i n A nL.c o sA „L
2Al
(A" + /l?) {L + W + i ^ ) ~ h\
A t a n A L = h2
A c o t AL = hi
A t a n A L = — hi
A (/i i - h2)
t a n XL = —
X2 + h i h 2
o
ffi
>
TJ
H
M
ta
o
ta
ffi
O
o
o
z
>
f
m
x
TJ
>
2:
o
2;
w
* If L = -------- > 0, then Ao = 0 is an eigenvalue with φο = X.
02
** Replace L by 2L for n = 0.
1 1
* * * If L = —------— > 0, then Ao = 0 is an eigenvalue, with (
hi n 2
h i
Table 4.2: Eigenvalue Problem in Polar Cylindrical Coordinates
116
T a b l e 4.2
j· λο = 0 is also an eigenvalue for u = 0; then φο — 1 and ||φο||2 =
5
S e p a r a t i o n o f V a r i a b l e s
The method o f separation o f variables is a well-* solving ordinary differential equations. This meth almost all linear homogeneous partial differential coefficients in canonical form, and exhibits the po principle to construct the general solution o f such first order partial differential equations can always o f characteristics, the method o f separation o f vai to solve higher order partial differential equations method is to transform a partial differential equatii differential equations as the number o f independei differential equation by representing the solution o f each independent variable. After these ordinary solved, the method reduces to solving eigenvalue p the general solution as an eigenfunction expansion evaluated by using the boundary conditions and the cases the solution is written in terms o f a series o f c
stablished technique for od is easily adaptable to equations with constant wer o f the superposition equations. Since linear be solved by the method iables is usually applied . The basic idea o f this >n into as many ordinary it variables in the partial is a product o f functions differential equations are roblems and constructing where the coefficients are nitialconditions. Inmost rthogonal functions.
5.1. Intr odu cti on
Consider the partial differential equation
Oj Uxx b Uxy “1“ C U y y “1“ β Ζίχ "Ί- f Uy
+ g u = 0. (5.1)
118
CHAPTER 5: SEPARATION OF VARIABLES
The first step in the general technique is to eliminate the term with mixed partial derivatives by introducing a new set o f coordinates χ', y' (called c h a r a c t e r i s t i c c o o r d i n a t e s, see §2.7). Thus, we have
αχ ux>x> + Cl Uy'y> + e\ ux> + f l U y > + 3 i u = 0. (5.2)
We now assume a solution o f the form u = X( x') Y( yr) in (5.2), and obtain
α ι Χ" Y + CXY" + e j X' Y + f i XY' + g i XY = 0,
or, formally,
y f + « +,= », (5.3)
where Li ( Dx) is a linear differential operator in x' and L2 ( Dy) is a linear differential operator in y'. Since the first term in (5.3) is a function o f x' only and the second term is a function o f y' only, while the third term is a constant, the only way Eq (5.3) can be solved is i f each o f the first two terms is also constant, thus
LxX λ L2Y , L ,
——— = A, —~ — = μ, such that λ + μ + g i = 0. a y
W e s h a l l e x p l a i n t h i s m e t h o d b y s o m e e x a m p l e s.
5.2. H y p e r b o l i c E q u a t i o n s
E x a m p l e 5.1. T h e p r o b l e m o f a v i b r a t i n g s t r i n g i s d e f i n e d b y t h e o n e - d i m e n s i o n a l w a v e e q u a t i o n ( § 1 - 4 ). C o n s i d e r t h e b o u n d a r y v a l u e p r o b l e m
d 2 u 2 d 2 u n I f n A\
W = c a ^ · 0 < x < 1 · <5-4)
u ( 0,t ) — 0 = u ( l,t ), t > 0, (5.5)
u ( x,0 ) = f ( x ), u t ( x,0 ) = g { x ), 0 < χ < I, (5.6)
where / 6 C 1 is a given function. We seek t h e solution o f t h e form
u ( x,t ) = X ( x ) T ( t ), (5.7)
where X is a function of x onl y and T a function o f t only. We assume
here t h a t a sol uti on o f t h e form (5.7) exis ts. Sometimes t h i s method
requires some modifications, as in Example 5. carry out t h e details t o see i f t h e method works tha t
— - Y T » n n H
O t 2 ~ ’ d x 2
where t h e primes denote t h e derivative wi t h resp independent variable. T h e Eq (5.4) reduces t o
XT" = <?X"T,
or, after separating t h e variables, it becomes
5.2. HYPERBOLIC EQUATIONS
= c
,X"
X
In Eq (5.8) we have been able t o separate t h e va: s tage in t h e development o f t hi s method t h a t or abandon t h e me th od depending on whether in separating t h e variables.
riables. It is o n l y at this we may either continue or not we are successful
Let us now fix t, and let x vary over t h e inte r si t ua t i o n where X ( x ) — T ( t ) for all x and t is const. Hence, from (5.8) we can write
L U
Έ—
~ x
fc = .
Thus, t h e set o f equations (5.9) is equivalent t o t equations:
T" — k c 2 T = 0,
X" - k X = 0.
Since t h e constant k is arbitrary, i t is necessary value for Eqs (5.10) and (5.11) in order t h a t Eq general solution o f Eq (5.10) is
T ( t )
cy/kt
a — cy/kt
ci e l'vn,L + C
2
e~^wnjL for fc > 0 Cit + c2 for k = 0 k c i cos c\/—k t + C2 sin c ^ f - k t and o f Eq (5.11) is
Χ{χ)
ά χ β'Λ * + d 2 e
-\f k x
f or fc > 0
d\x + d 2 for fc = 0 _ d\ cos \/— k x + d 2 sin \f ^ - k x
119
3. We s h a l l, h o w e v e r, f or t h i s p r o b l e m. N o t e
C'T,
c t t o i t s c o r r e s p o n d i n g
( 5.8 )
/a l 0 < χ < I. T h e onl y when X ( x ) = T { t ) =
ist. (5.9)
wo ordinary differential
(5.10)
(5.11)
for fc t o have t h e same (5.9) be satisfied. The
(5.12)
for k < 0,
(5.13)
for k < 0.
120
CHAPTER 5: SEPARATION OF VARIABLES
In v i ew o f t h e boundary conditions (5.5) we must have
X ( 0 ) T ( t ) = 0 = X { l ) T ( t ) for all t > 0. (5.14)
Using t h e s e condi ti ons in (5.13) for k > 0 we get t he s ys tem o f equations X(0) = d\ + c?2 = 0,
X { 1 ) = d i e V k l = + d 2e ~'/El = 0. ^'15^
Th e sy s t em (5.15) is c onsistent, i.e., it has a nontrivial solution, iff t h e determinant o f i ts coefficients vanishes. B u t since
det
1 1 e V k l g — Viel
e - V k i _ eV k i φ 0)
a nonzero solution for X ( x ) in (5.13) for k > 0 is not possible. N e x t, for k = 0, t h e boundary conditions (5.14) imply t h a t d i = 0 and d -2 = 0. Hence there is no nonzero solution for k = 0. Finally, for k < 0, let us s e t k = —A2. Then t h e general solution (5.13) in t hi s case becomes
X ( x ) = d i cos Ax + d 2 sin Ax,
which under t h e boundary conditions (5.14) yields
X (0) = d i = 0, and X ( I ) = d i sin AI = 0.
In order t o avoid a trivi al sol uti on in t h i s case, we choose A such t h a t
T17T
AI is a pos it i ve multiple o f π, i.e., XI = η π, or A = —. The pos it i ve values o f A are chosen because t he negative multiples give t h e same eigenfunctions as t h e p os it i ve ones. This result leads t o an infinite set o f sol uti ons which are denoted by
v ( \ a ■ η π χ X n (x) = 0*2,n S i n ——,
where each solution corresponds t o t h e eigenvalue
(5.16)
T h e sol uti ons for T ( t ) for t h e choice o f k < 0, as in (5.16), are t h e n obtained from (5.12) as
m η π α ί , η π α ί
T n { t ) = C i,„ C O S — ---------h c 2,„ s i n — —.
5.2. HYPERBOLIC EQUATIONS
121
In view o f (5.7), t he n t h e infinite set o f solution Un ( x,t ) = X „ ( x ) T n ( t )
S I S
. n n c t A n C O S ;
---------h B v
I
w h e r e t h e c o n s t a n t s A n and B n are determin conditions. The eigenfunctions are contained whereas t h e eigenvalues for t h i s boundary value (5.16).
(5.17) id from t h e boundary in t h e s ol uti on (5.17), problem are given by
m
The n e x t st e p is t o obtai n t h e particular th e initial conditions (5.6). At t hi s point it one solution in (5.17) will s ati sfy (5.6). In vie principle (see §1.5), any finite sum o f t h e sol solution o f t hi s boundary value problem. We a linear combination o f t he se solutions which conditions (5.6). Even i f t hi s technique fails, infinite series o f solutions (5.17), i.e.,
oluti on which satisfies ay so happen t h a t no :kv o f t h e superposi tion utions (5.17) is also a should, therefore, find also satisfies t h e initial we can always t r y an
u ( x,t ) = X n { x ) T n ( t ) =
n = 1
n= 1
. m r c t A n COS —;
---------l·-
(For convergence o f t h e series (5.18), see Chapte: t his series expansion formally and verify t h a t tin (5.5) are s t i l l satisfied. We shall now s ati sfy the and thereby obtai n t h e sol uti on o f t hi s problem initial conditions (5.6) we get
OO
η π χ
/ λ . *ιπχ Jt/ .
i ( x, 0 ) = 2 _ ^ A n sm —— = j ( x )
n = 1
S i n c e / e C 1, t h e infinite series (5.19) is a Fou: f ( x ) can be regarded as an odd function wi t h can expand t h i s function f ( x ) on t h e interval f ( —x ) = — f ( x ) on t h e interval — I < x < 0, ar all x, where /( 0 ) = 0. The n t h e coefficients given by
fix)
where t h e last integral representation holds . η π χ
S1n —-— a r e o d d a n d t h e i r p r o d u c t is even. T h e
n n c t
s i n ■
η π χ
s i n ·
η π σ ί
η π χ
s i n ■
( 5.1 8 ) ir 4.) B u t w e c a n t a k e e b o u n d a r y c o n d i t i o n s i n i t i a l c o n d i t i o n s ( 5.6 ) . U s i n g t h e f i r s t o f t h e
( 5.1 9 )
ri e r s i n e s e r i e s. H e n c e p e r i o d 21. Thus, we 0 < χ < I such t h a t d f ( x + 21) - /( x ) for ln for n = 1,2,... are
■ η π Χ , ,-
sm —— α χ, (5.20)
because b o t h f ( x ) and n, taki ng t h e derivative
of (5.18), we get
122 CHAPTER 5: SEPARATION OF VARIABLES
OO
i r c y ^ π r i n d . . n i r c t . η π χ
u t — — 2 _ ^ n B n c o s — A n s m — j — s m ——, (5.21)
which, in view of the second of the initial conditions (5.6), gives
oo
/ n\ π 0 S T' D · η π χ f \
u t ( x, o ) = — 2 _ j sin ~ η ~ = 9 { χ ),
n = 1
where
B
” ~ m d J
Sin d x' n = 1’ 2 ’ · · · · (5 ·22)
Hence, the solution (5.18) is completely determined.
We shall now derive the d ’Alembert solution for this problem. From (5.18) we have
1 ν -' , ( η π ( χ + c t ) . η π ( χ — c t ) 1
» = 2 Σ Λ· · ( ™ I + 3m , )
n = 1 ^ '
1 ^ „ Γ η π ( χ — c t ) η π ( χ + c t ) Ί /κ no\
+ - J 2 B n\C 0 S — ^ - COS — i---------- >- \ ^ · Ι ό >
η — 1 ^
= ^ [ f ( x + ct) + f ( x ~ ct)] + \i ~ G(x + ct) + G(x - ci)]>
where
oo
t! \ A ■ n n Z
/( 2) = Σ ^ Λη s i n _ r >
n—1
a s i n ( 5.1 9 ), a n d
OO
G(z ) = Σ Βη c os ~ j - ·
n — 1
S i n c e
, l i c ^ . η π ζ 1
G'( z ) =---------- > n B n sin — = — g { z ),
c ί I c
n = 1
we obtain from (5.23) the formal solution, known as the d’Alembert solution for this problem, as
u ( x,t ) — φ(χ + ct ) + ψ( χ — ct ), (5-24)
5.2. HYPERBOLIC EQUATIONS
123
w h e r e c i s t h e w a v e v e l o c i t y, a n d
φ { x + c t ) = - [ F ( x + c t ) - G ( x ψ ( χ - c t ) = ^ [ F ( x - c t ) 4- G ( x
+ c t ) ], - c t ) ] .
A n i n t e r p r e t a t i o n o f t h e s o l u t i o n o f t h i s p r o b l e m i s a s f o l l o w s: A t e a c h p o i n t X o o f t h e string
/ \ Ji Ti C . c ,
i { X 0,t ) = } j nan cos—j—(t 4- on)
n n c x o $in —:—.
n=1
This equati on describes a harmonic motion wi t h
mrcxο Λ m l .
Th e p o i n t s whe r e s m —;— = 0, i.e., x = — (:
/ Tl
main fixed during t h e entire process; t he se poir
of t h e standi ng wave. But t h e points where (2m + 1)/
amplitudes a n sin
rn = 1,2,... , η — 1), re­
t s are called t h e n o d e s . n n c xo .
s m — -— = ± 1, i.e.,
2 n
, vibrate w i t h t h e maximum amplitude a n. T he se points
For any t t h e structure o f
are called t h e m a x i m a o f t h e standing wave, the standing wave is described by
u ( x,t ) = J 2 c n ( t ) sin
TlTTCQ'o
n=1
ι
where
Gn(t) CXn COS UJn{t -Ι- δη)> ^n
F o r t h o s e t i m e s t when c o sun(t + 6 n ) = ± 1, th i ts maximum value where t h e ve lo c i ty becomes
We shall illustrate an example by the following Mathemati ca ses­
sion.
wkcx o
m r c
~ T'
e displacement reaches zero. ■
I n [ l ]:=
L:= 1 c := 1
124
CHAPTER 5: SEPARATION OF VARIABLES
f [xJ := Sin[x] g[x_] := x~ 2+1
A[n_] := A[n] = 2/( n P i c ) N I n t e g r a t e [ g [ x ] S i n [ n P i x/L ] , { x, o, L } ]//N//C h o p B [ n _ ]:= B[n] = 2/L N I n t e g r a t e [ f [ x ] S i n [ n P i x/L ],{ x,0,L } ] //C h o p
Table[{n, A[n], B[n]}, {η,1,8}]//TableForm
O u t [ 7 ] =
{ 1.
0.6 0 7 9 2 7, 0.5 9 6 0 9 4 }
{ 2,
- 0.0 5 0 6 6 0 6,
- 0.2 7 4 8 1 }
{ 3,
0.0 6 7 5 4 7 5,
0.1 8 0 5 9 9 }
{ 4,
- 0.0 1 2 6 6 5 1,
- 0.1 3 4 7 7 8 }
{ 5,
0.0 2 4 3 1 7 1,
0.1 0 7 5 7 5 }
{ 6.
- 0.0 0 5 6 2 8 9 5,
- 0.0 8 9 5 3 4 8 }
{ 7,
0.0 1 2 4 0 6 7,
0.0 7 6 6 8 6 7 }
{ 8,
- 0.0 0 3 1 6 6 2 9,
- 0.0 6 7 0 6 8 3 }
I n [ 8 ]:=
u[x_,t_,n_] := (A[n] Sin[n Pi c t/L] + B[n] Cos[n Pi c t/L]) Sin[n Pi x/L] ;
uapprox[x_,t_] := Sum[u[χ,t,k],{k,8}]
I n [ 1 0 ]:=
graphs =
Table[Plot[uapprox[x,t],{x,0,l},
PlotRange->{-2,2},
Ticks->{{0,1},{-2,2}},
DisplayFunction->Identity],
{t,0,2,1/3}];
I n [ l l ]:=
g r a p h s a r r a y = P a r t i t i o n [ g r a p h s,2 ];
5.3. PARABOLIC EQUATIONS
125
In[12]:—
Show[GraphicsArray[graphsarray],
DisplayFunction->$DisplayFunction]
-Graphi csArray-
5.3. P a r a b o l i c E q u a t i o n s
tion
Ex a mple 5.2. Consider the one-dimensional heat conduction equa-
(5.25)
du d2u
at ” a?’ 0<x<i’
subject to the boundary conditions
u(0, t) = 0 = u(l,t ), t > (j
and the initial condition
u(x, 0) = f(x), 0 < χ < I
w h e r e / e C 1 i s a p r e s c r i b e d f u n c t i o n. I n p h y s i c r e p r e s e n t s t h e h e a t c o n d u c t i o n i n a r o d w h e n i t j:
( 5.2 6 )
( 5.2 7 )
a l t e r m s, t h i s p r o b l e m s e n d s a r e m a i n t a i n e d
126
CHAPTER 5: SEPARATION OF VARIABLES
at zero temperature while t h e initial temperature u at any point o f t h e rod is prescribed as f ( x ) · Let us assume t h e solution in t h e form
u ( x, t ) = X ( x ) T ( t ),
w h i c h a f t e r s u b s t i t u t i o n i n t o E q ( 5.2 5 ) y i e l d s t h e s e t o f e q u a t i o n s
1 ^ = ^. ( 5.2 8 )
k T X K J
A s i n E x a m p l e 5.1, t h e o n l y s i t u a t i o n w h e r e t h e s e t w o e x p r e s s i o n s c a n b e e q u a l i s f or e a c h o f t h e m t o b e c o n s t a n t, s a y e a c h e q u a l t o c. E q ( 5.2 8 ) t h e n y i e l d s t w o o r d i n a r y d i f f e r e n t i a l e q u a t i o n s
Τ' - c k T = 0, (5.29)
X"- c X = 0, (5.30)
where t h e boundary conditions (5.26) reduce to
X ( 0 ) T { t ) = 0 = X ( l ) T ( t ), or X ( 0 ) = 0 = X ( l ), (5.31)
exce pt for t h e case when t h e rod has zero initial temperature at every point. This si tuati on, being uninteresting, can be neglected. As in th e case o f Example 5.1, we notice t h a t for a nonzero sol uti on o f the problem (5.30)-(5.31) we must choose negative values o f c. Hence we set c = —A2, and find t h a t the eigenvalues c = —η 2 π 2/1 2 have t h e corresponding eigenfunctions
v f \ λ ■ η π χ
X n { x ) = A n s i n —p.
Eq (5.29) t h e n becomes
τ · + ^ τ = ο
i Z
whose general solution for each n is given by T n ( t ) = B n e ~ k n 2 ^ t/l 2. Hence, we consider an infinite series of t h e form
,( x,t ) = Σ χ η { χ ) Τ η {ί) = Y j C n S m r^ e - kn2^ t/l\ (5.32)
n = 1 n=l
5.3. PARABOLIC EQUATIONS
127
Now, we use the initial condition (5.27) in (5.32) and obtain
/ ^ . η π χ
i ( x, 0) = 2 2 C n sm ~ Ί ~ = f
n = l
w h i c h s h o w s t h a t f ( x ) can be represented as a
exten di ng / as an odd, piecewise continuous function o f period 21 with
x ),
(5.33)
Fourier sine series, by
piecewise continuous derivatives. Equation (5.33 C n as
C r
=t Lm
. η π χ . sm —-— a x
f f ( x )
J o
) gives t h e coefficients
Π',τχ
sm ■
dx, n = 1,2,... .
( 5.3 4 )
H e n c e t h e s o l u t i o n ( 5.3 2 ) i s c o m p l e t e l y d e t e r m.n e d f o r t h i s p r o b l e m. N o t e t h a t t h e s e r i e s i n ( 5.3 3 ) c o n v e r g e s s i n c e u(x,Q) does, and the expone nti al expression in (5.32) is les s t h a n 1 for each n and all t > 0 and approaches zero as t —> oo. ■
I n [ 1 3 ]:=
L:= 1 k:= 1
f [ x j := x"2
A [ n _ ]:= A[n] = 2/( n P i c ) N I n t e g r a t e [ ^ [ x ] S i n [ n P i x/L ], { x, 0, L } ]//N//C h o p;
Table[{n, A[n]}, {η,1,8}]//TableForm
O u t [ 1 7 ] =
{ 1,0.3 7 8 6 0 7 }
{ 2,- 0.3 1 8 3 1 }
{ 3,0.2 0 2 6 5 1 }
{ 4,- 0.1 5 9 1 5 5 }
{ 5,0.1 2 5 2 6 }
{ 6,- 0.1 0 6 1 0 3 }
{ 7,0.0 9 0 1 9 3 5 }
{ 8,- 0.0 7 9 5 7 7 5 }
In[18]:=
128
CHAPTER 5: SEPARATION OF VARIABLES
u[x_,t_,n_] := A[n] Sin[n Pi x/L] Exp[-k t(n Pi/L)~2] uapprox[x_,t_] := Sum[u[x,t, j] ,{j ,5}]
In[19]:=
graphs = Table[Plot[uapprox[x,t],{x,0,l}, PlotRange->{0,1},
Ticks->{{0,1},{0,1}},
DisplayFunction->Identity],
{t,0,1/3,1/24}];
In[20]:=
graphsarray = Partition[graphs,2];
Show[GraphicsArray[graphsarray], DisplayFunction->$DisplayFunction]
0 1 0 1 0 1 0 1 1: 1 1 1
0 1 0 1 0 1 0 1 - G r a p h i c s A r r a y -
We s h a l l n o w p r e s e n t a M a t h e m a t i c a s e s s i o n w i t h t h e n o n h o m o g e ­
n e o u s i n i t i a l c o n d i t i o n
f x2 0 < x < 1/2 u ( x,0) = <
V ’ \ x + 1 1/2 C r < 1.
In[21]:=
5.3. PARABOLIC EQUATIONS
129
L:= 1 k:= 1
f [x_] := x~2 g [ x _ ] := x+1
I n [ 2 5 ]:=
A [ n _ ]:= A[n] = 2/( n P i c ) ( N I n t e g r a t e { x, 0, 1/2 } ] +
N I n t e g r a t e [ g [ x ] S i n [ n P i x/L ],{ x, 1/2 T a b l e[{η, A [η], Β[η]}, {η,1,8 } ]//T a b!
O u t [ 2 6 ] =
{ 1,1.1 4 4 2 3 }
{ 2,- 1.0 6 6 7 6 }
{ 3,0.4 1 9 6 3 5 }
{ 4,- 0.1 1 9 3 6 6 }
{ 5,0.2 5 3 6 1 6 }
{ 6,- 0.3 4 6 0 3 }
{ 7,0.1 8 1 5 1 5 }
{ 8,- 0.0 5 9 6 8 3 1 }
I n [ 2 7 ]:=
u [ x _,t _,n _ ] := A[n] S i n [ n P i x/L ] Exp[ uapprox [ x _,t _ ] := S u m [ u [ x,t,j ],{ j,5 } ] I n [ 2 9 ]:=
[ f [ x ] S i n [ n P i x/L ] , , 1 } ] )//N//C h o p; sForm
•k t ( n P i/1 )"2]
graphs =
T a b l e [ P l o t [ u a p p r o x [ χ,t ],{ χ,0,1 }, P l o t R a n g e - > { 0,2.5},
Ticks->{{0,1},{0,1,2,2.5}}, DisplayFunction->Identity],
{t,0,1/3,1/24}];
In[30]:=
130
CHAPTER 5: SEPARATION OF VARIABLES
graphsarray = Partition[graphs,4];
I n [ 3 1 ]:=
Show[GraphicsArray[graphsarray] , DisplayFunction->$DisplayFunction]
-GraphicsArray-
An interesti ng si t ua t i o n arises i f t h e function f ( x ) is zero in the ini tial condi ti on (5.27), but the boundary conditions are nonhomoge­
neous.
E x am ple 5.3. Consider t h e dimensionless partial differential equa­
t i o n governing t h e plane wall transient heat conduction
ut = Uxx, o < x < 1, (5.35)
w i t h t h e boundary conditions
u( 0,t ) = 1, u ( l,i ) = 0, t > 0, (5.36)
and t h e i ni t ia l condition
u(x, 0) = 0, 0 < cc < 1. (5.37)
Since t h e homogeneous initial c ondition (5.37) does not allow us com­
p u t e t h e Fourier coefficients, as in (5.34), t h e standard technique used in t h e above examples is not directly applicable t o t hi s problem. Instead,
5.3. PARABOLIC EQUATIONS
131
we proceed as follows: First, find a partial sol uti on o f t h e problem; although there is more t h a n one way t o determine t h e particular solu­
ti on, we, e.g., take t h e st e a d y st a t e case, where t h e equati on becomes u,x x = 0, which after integrating twice has t h e general solution
ΰ ( χ ) = C\X + C2,
wi t h t h e boundary conditions ΰ ( 0 ) = 1, ΰ(1) C2 = 1, and t h e st e a d y s t a t e s ol uti on is
u ( x ) = 1 — x.
N e x t, formulate a homogeneous problem by writing u ( x, t ) as a sum o f t h e s t e a d y s t a t e sol uti on u ( x ) and a transient term v ( x, t ), i.e.,
u ( x,t ) = u ( x ) + v ( x,t )
or
v ( x,t ) = u ( x,t ) — u ( x ).
Hence t h e problem reduces t o finding v ( x,t ). Ii we s u b s t it u t e υ from
(5.38) into ( 5.35), we get
Vt ~ v x x, (5.39)
where t h e boundary conditions (5.36) and t h e initial condi ti on (5.37) reduce to
w(0, t ) = u ( 0,t ) — ΰ ( 0 ) = 0, w(l, t ) = u ( l, t ) — ΰ ( 1 ) = 0,
0. Thus, οχ = — 1,
(5.38)
and
1.
v ( x, 0) = u ( x, 0) — u ( x ) = x
N o t i c e t h a t t h e p r o b l e m ( 5.3 9 ) - ( 5.4 1 ) i s t h e s a m e a s i n E x a m p l e 5.2 w i t h k = 1, I = 1, f ( x ) = x — 1, and u replaced by v. Hence i t s general sol uti on from (5.32) is given by
v ( x,t ) = ^ ~ ^ C n e n π 4 s i n n e r,
n= 1
and t h e coefficients C n are determined from (5.34) as
C n = 2 / ( x — 1) sin η π χ dx J o
Hence,
9 00 1 v { x,t ) =
-----
nr ^ n
(5.40)
(5.41)
(5.42)
_2_
ηπ
π ■ί—' η
η = 1
e η π 4 sin ηπχ,
(5.43)
132
CHAPTER 5: SEPARATION OF VARIABLES
a n d f i n a l l y f r o m ( 5.3 8 )
s m m r x. m
( 5.4 4 )
A Mathemati ca session for t hi s example can be carried out as in t h e previous examples.
5.4. E l l i p t i c E q u a t i o n s
E x a m p l e 5.4. We shall consider t h e potenti al problem for t h e rec­
t angl e R : {0 < x < a, 0 < y < b }:
P h y s i c a l l y, t h i s p r o b l e m a r i s e s i f t h r e e e d g e s o f a t h i n i s o t r o p i c r e c t a n ­
g u l a r p l a t e a r e i n s u l a t e d a n d m a i n t a i n e d a t z e r o t e m p e r a t u r e, w h i l e t h e f o u r t h e d g e i s s u b j e c t e d t o a v a r i a b l e t e m p e r a t u r e f ( x ) until t h e s t e a dy s t a t e condi ti ons are at t ai ne d throughout R. Then t h e s t e a dy s t a t e value o f u ( x, y ) represents t h e distribution o f temperature in t h e interior of t h e plate. As before, we seek a solution o f t h e form u ( x, y ) = X ( x ) Y (y ), which, after s u b s t it u t io n into Eq (5.45) leads t o t h e s e t o f two ordinary differential equations :
where c is a c onstant, as in Example 5.2. Since t h e first three boundary condi ti ons in (5.46) are homogeneous, t h e y become
Uxx “1“ U y y 0, x, y (Ξ R,
(5.45)
s u bje ct t o t h e Dirichlet boundary conditions
u ( 0, y ) = 0 = u ( a, y ), u ( x, 0) = 0, u ( x, b ) = f ( x ). (5.46)
X" - c X = 0, Y" + c Y = 0,
(5.47)
(5.48)
X ( 0 ) = 0, X ( a ) = 0, y ( o ) = 0,
(5.49)
but t h e fourth boundary condition which is nonhomogeneous must be used separately. Now, taking c = —A2, as before, t h e s ol uti on o f (5.47)
5.4. ELLIPTIC EQUATIONS
subject to the first two boundary conditions in (5.49) leads to the eigen­
values and the corresponding eigenfunctions as
ττττ . . . ηπχ
λη = —5-, Xn{x) = sin , n
az a
whil e for t hese eigenvalues t he sol ut i ons of (5.55) sat i sfyi ng t he t hi r d boundar y condi t i on i n (5.49) are
1 3 3
Yn(y) = s i n h ^ ^, n = 1,2, a
Hence, for ar bi t r ar y const ant s Cn,n = 1,2,...,
we get
n=1
( *,!,) = £ c „ s i n ^ s i n h ^. (5.51)
g the fourth boundary
The coefficients Cn are then determined by usin condition in (5.46). Thus,
OO ,
/ i \ ,/ χ ^ η π χ · u η Έ °
u { x, o) = j ( x ) = > Cn s i n smh 1 ^
n = 1
(5.50)
. ηπχ
s m dx, n --= 1,2,...
a
wh i c h, i n v i e w o f t h e F o u r i e r s e r i e s e x p a n s i o n, y
_ . , η π ί ) 2 f a .. .
Cn smh = - f (x)
a a J 0
This solves the problem completely.
In particular, if f{x) = /0 = const, then
7Ϊ7Γ& 2/o [ l - ( - 1 )
0 < x < a, ields
C„ sinh ■
Then from (5.51), we have
ηπ
(5.52)
, , _ 2/o ^ 1 — ( —1)" Ά ΐ ί { η π χ/α ) sinh(n7r?//a)
' ^ π n sinh(n7r6/a)
n=l
A Mathematica session for the following more general boundary value problem is presented below:
o)
(5.53)
« ι ι — Uyy = 0, 0 < χ < α, o < y < b,
su bje ct t o t h e boundary conditions
u ( x,0 ) = u { x,b ) = f 2 ( x ) for 0 < i < a,
and t h e i ni t ia l conditions
u ( 0,y ) = g i { y ), u { a,y ) = g 2 { y ) for 0 < y < b.
T h e s o l u t i o n i s u ( x,y ) = u\( x,y ) + u 2 ( x,y ), where
OO
Ui(x,y) = 53 [ A n cosh(An y ) + B n sinh(A„2/)] si nA „ x,
n —1
134 CHAPTER 5: SEPARATION OF VARIABLES
2 f a
A n = - I f i { x ) s i n X „ x d x, a Jo
” s i n h A n b
w i t h λ „ = ± η π/α, and
OO
u 2 ( x,y ) = Σ ^ ο β\ι ( μ η χ ) + b n s i n h ^ n:r)] sin/xnt/,
n= 1
2 f b
a n = jj J g i ( y ) s i n μ n y d y,
b n = - ί g 2 { y ) s i n/i ny d y - a n c o s h/i na |,
sinh μ η α L b J 0 J
w i t h μ α η = ± m r/b. We shall take a = 1, 6 = 2, f i ( x ) = x 2, f 2 ( x ) = x + 2, g i ( y ) = y, and g 2 ( y ) = y + l.
I n [ 3 2 ]:=
L:= 1 M:= 2
1 [n_] := n Pi/L//N m[n_] := n Pi/M//N
2 f a
— / f 2 ( x ) s i n\n x d x — A n c o s hX n b , .a Jo
5.4. ELLIPTIC EQUATIONS
I n [ 3 6 ]:=
f l [ x j
= x ‘ 2
f 2 [x_]
= x+2
g 1 ty-]
= y
g 2 [ x j
= y + l
I n [ 4 0 ]:=
A[n_] : =
Α[η] =2/L NIntegrate[f1[x] Sin[l[n] x
B[n_] : =
B[n] = 1/Sinh[l[n] M]
(2/L NIntegrate[f2 [x] Sin[l[n] χ], -A[n] Cosh[l[n] M])//Chop;
135
], { x, 0, L } ]//C h o p;
{ x, 0, L}]
Table[{η, A[n],B[n]}, {η,1,8}]//ColumnForm
O u t [ 4 3 ] =
{ 1,0.3 7 8 6 0 7,- 0.3 6 6 7 2 2 }
{ 2,- 0.3 1 8 3 1,0.3 1 8 3 0 8 }
{ 3,0.2 0 2 6 5 1,- 0.2 0 2 6 5 1 }
{ 4,- 0.1 5 9 1 5 5,0.1 5 9 1 5 5 }
{ 5,0.1 2 5 2 6,- 0.1 2 5 2 6 }
{ 6,- 0.1 0 6 1 0 3,0.1 0 6 1 0 3 }
{ 7,0.0 9 0 1 9 3 5,- 0.0 9 0 1 9 3 5 }
{ 8,- 0.0 7 9 5 7 7 5,0.0 7 9 5 7 7 5 }
I n [ 4 1 ]:=
ul [x_,y_,n_] := (A[n] Cosh[l[n] y] + B[n] Sinh[l[n] y]) Sin[l[n] x] ulapprox[x_,y_] := Sum[ul[x,y,n],{n,8}]
I n [ 4 3 ]:=
threeDplotl=Plot3D[ulapprox[x,y],{x,0,L},{y,0,M}, DisplayFunction->Identity];
Show[threeDplotl,DisplayFunction->$DisplayFunction]
136 CHAPTER 5: SEPARATION OF VARIABLES
-GraphicsArray-
I n [ 4 5 ]:=
cvalsl:= Table[i,{i,0,1/4,1/4 4 } ] cvals2:= Table[i,{i,1/4,3/2,5/4 4 } ] contourvals:= Union[cvalsl,cvals2]
I n [ 4 8 ]:=
ul [x_,y_,n_] := (A[n] Cosh[l[n] y]+B[n] Sinh[l[n] y]) Sin[l[n] x]
I n [ 4 9 ]:=
ulapprox[x_,y_] := Sum[ul[x,y,n],{n,8}]
I n [ 5 0 ]:=
contourgraphsl= ContourPlot[ulapprox[x,y], {x,0,l},{y,0,2},
PlotPoints->40,
Contours->contourvals,
ContourShading -> False, DisplayFunction-> Identity];
5.4. ELLIPTIC EQUATIONS
In[51]:=
Show[GraphicsArray[{threeDplotl,contourgraphsl}], DisplayFunction -> $DisplayFunction]
137
1.5
0.5
-GraphicsArray-
In[52]:=
a [n_] : = a [n] =
2/M NIntegrate[gl[y] Sin[m[n] y], b[n_] := b[n] = 1/Sinh[m[n] L]
(2/M NIntegrate[g2[y] Sin[m[n] y], a[n] Cosh[m[n] L])//Chop
y, 0, M}]//Chop { y, 0, M}]-
In[54]:=
u2[x_,y_,n_] : = (a[n] Cosh[m[n] x] + b[n] Sinh[m[n] x]) * Sin[m[n] y]
u2approx[x_,y_] := Sum[u2[χ,y,n],{n,l,8
threeDplot2= Plot3D[u2approx[x,y] ,{x,(| ,L}, {y ,0 ,M} , DisplayFunction->Identity] ;
138
CHAPTER 5: SEPARATION OF VARIABLES
contourgraphs2=
ContourPlot[u2approx[χ,y], {x,0,l},{y,0,2}, PlotPoints->40, Contours->contourvals, ContourShading->False, DisplayFunction->Identity] ;
I n [ 5 7 }:=
Show[GraphicsArray[{threeDplot2,contourgraphs2}], DisplayFunction->$DisplayFunction]
-Graphi csArray-
I n [ 5 8 }:=
uapprox[x_, y_] : = ulapprox[x,y] + u2approx[x,y]; threeDplot=
Plot3D[uapprox[χ,y],{χ,0,L},{y,0,Μ}, DisplayFunction->Identity];
I n [ 6 0 ]:=
contourgraphsu=
ContourPlot[uapprox[x,y],{x,0,1},{y,0,2}, PlotPoints->40,Contours->contourvals,
5.4. ELLIPTIC EQUATIONS
139
ContourShading->False,
DisplayFunction->Identity];
Show[GraphicsArray[ {threeDplot,contcurgraphsu}], DisplayFunction->$DisplayFunction]
0 0.2 0.40.60.1
- G r a p h i c s A r r a y -
(* S u p e r p o s e the S u r f a c e G r a p h i c s plot ov e r th e C o n t o u r G r a p h i c s pl ot *)
I n [ 6 2 ]:=
m y C P =
C o n t o u r P l o t [ u a p p r o x [ χ,y ],
{x,0,L},{y,0,M},
C o n t o u r s - > U n i o n [ T a b l e [ i,{ i,0,1/4,1/44}],
T a b l e [ i,{ i,1/4,3/2,1/8}]] ,
P l o t P o i n t s - > 3 0,
A s p e c t R a t i o - > A u t o m a t i c,
P l o t R a n g e - > A l l,
C o n t o u r L i n e s - > T r u e,
C o n t o u r S h a d i n g - > T r u e,
C o n t o u r S t y l e - >
( M a p [ { H u e [#,1,R a n d o m []],T h i c k n e s s [.006]}&, R a n g e [0,1,1/12]]),
C o l o r F u n c t i o n - > H u e,
T i c k s - > { R a n g e [ 0,1 ].R a n g e [0,2] ,{ - 1.5,1.5}}, D i s p l a y F u n c t i o n - > I d e n t i t y ]
140
CHAPTER 5: SEPARATION OF VARIABLES
-Graph!csArray-
I n [ 6 3 ]:=
myContourGP= FirstOGraphicsQmyCP;
myContourGP= NOmyContourGP/. {x_AtomQ, y_AtomQ}->{x, y, I n [ 6 5 ]:=
S h o w [
{SurfaceGraphicsQmyCP,Graphics3D@myContourGP}, Axes->True,
BoxRatios->{l,1,1}, DisplayFunction->$DisplayFunction]
5.4. ELLIPTIC EQUATIONS
141
E x a m p l e 5.5. Consider the potential problem
Uxx + Uyy = 0, 0 < x < π, 0 < y < 1,
subject to the mixed boundary conditions
u(x,0) = uqcosx, u(x, 1) = uosin2x, ux(0,y) = 0 = ux(n, y).
The separ at i on of vari abl es t echni que l eads t o t he same set of or di nar y di fferent i al equat i ons as i n ( 5.47)-(5.48), i.e.,
(5.54)
(5.55)
X" + Λ2 X = 0, X'(0) = 0 = Χ'(π)
and
Υ" - λ 2Υ = 0.
Th e e i ge nva l ue s a n d t h e c or r e s po ndi ng e i ge nf unc t i ons f or ( 5.56) a r e
Ao = 0, Χ ο ( ^ ) = 1,
A „ = n 2, Xn(x) = cosnx, n =
A0 + Boy, n = 0,
An cosh ny + Bn sinh ny, n 4 1,2,__
(5.56)
(5.57)
(5.58)
and subsequently the solutions of (5.57) are Yn(y)
Hence, usi ng t he superposi t i on pr i nci pl e, we get
OO
u(x, y) = Ao + B0y + ^ [An cosh ny + Bn sinh ny] cos nx. (5.59)
71=1
Now, the first boundary condition in (5.55) leads
OO
u(x, 0) = Aq + ^ An COS 7
mx = u
71 = 1
By matching the coefficients of similar terms on both sides of (5.60) we find that ylo = 0, A\ = uq, and An = 0 for n > becomes
i{x, y) = Bay + uQ coshy cosx + ^ Bn sinlmj/ cos nx. (5.61)
n = l
to
3 cosa;.
(5.60)
2. Hence the solution
142
CHAPTER 5: SEPARATION OF VARIABLES
Similarly, using t h e second boundary condition in (5.55) we find from
(5.61) t h a t
CX)
u ( x, 1) = Bo + u o cosh 1 c o s x + ^ B n sinh n cos n x
n = l
,2 1 — c o s 2 x
= Uo sm X = Uo -----------,
from which, after comparing t h e coefficients of similar terms on b o t h sides, we get
o _ u°
Β ο - γ,
B 2 = ~:
Β χ =
u o c o s h l sinh 1
Uq
Br, = 0 for n > 3.
2 sinh 2 ’
Hence, from (5.61) t h e general solution is given by
f \ u o ,
u { x,y ) = γ υ + uo
cosh y —
c o s h 1 s i n h y
s i n h 1
s i n h 2 y cos x — Uo . _■ cos 2 x
2 si nh 2
u o
1 s i n h ( l — y ) si n h 2 u
—y H---------:——— cos x - - . , „ cos 2 x
2 smh 1 2 smh 2
(5.62)
5.5. C y l i n d r i c a l C o o r d i n a t e s
T h e thr e e- di me nsi on al Laplacian in cylindrical coordinates is
„ 2 _ d 2 I d I d 2 d 2 V = d ^ + r f r + ^ d P + d ^ · ( 5 -63)
E x a m p l e 5.6. ( C i r c u l a r d r u m ) I f a circular drum is struck in t h e center, i t s vibrations are radially symmetric. We shall solve t h e bound­
ary value problem
5.5. CYLINDRICAL COORDINATES
143
s u b j e c t t o t h e b o u n d a r y c o n d i t i o n s
u(r,0) = /( r ), r < 1, u t ( r,0 ) = 0, i t ( l,i ) = 0, t
I f we t a k e u( r,t ) = T( t ) R( r ), then Eq (5.64) reduces to the system of ordinary differential equations
rjnff
Ύ
r" + r' R + R
R
k.
Here, again, k — —A2 yields nontrivial solutions. Then the system (5.66) gives the uncoupled ordinary differential equation
or
T" + λ2Τ = 0,
d2R 1 dR 2 — + - — + \2R = 0, dr1 r dr
2 d2R dR ,2 o n ^ r 2 — r + r — + A r i l = 4), dr * dr
> 0.
(5.65)
(5.66)
which is the Bessel equation. The eigenvalues A, of Jo (A), with the corresponding eigenfunctions . of the first equation in (5.67) are Tn = cos Ant. the vibrating circular drum struck a t the center
o o
u(x,t) = Cn cos Xnt Jq(Ai
n = 1
where the coefficients Cn are (see Appendix B)
_ fp f ( r ) Jo( Xnr ) r dr
a r e t h e p o s i t i v e z e r o s -To(Anr)· The solutions Hence,the solution of is given by
Cn =
Jo[MXnr)}2rdr
Marc Kac (1966) asked the question: “Can c drum?” This means one should answer the quest of different shapes and struck in their centers ha\ (Protter, 1987). This question has been resolved Webb and Wolpert (1992) ■
w w w 1 F o r a n a n i m a t i o n n o t e b o o k, see d r u m.m a
(5.67)
(5.68)
(5.69)
(5.70)
ne hear the shape of a on whether two drums e the same eigenvalues negatively by Gordon,
144
CHAPTER 5: SEPARATION OF VARIABLES
5.6. Spherical Coordinates
Using the transformation χ = p sin φ cos Θ. y = p sin φ sin θ, z = p cos φ, where p > 0, 0 < φ < π, and 0 < θ < 2π, th e Laplacian in the spherical coordinate system becomes
2 _ 92 2 d 1 d2 I d 2 cot φ d
~ dp2 p dp p2 sin2 φ dO2 ρ2 ΰφ2 ρ2 ΰφ
E x a m p l e 5.7. ( Cooling ball ) Consider the boundary value prob­
lem
ut = V2w, 0 < yo < 1, 0 < φ < π, 0 < θ < 2π,
s u b j e c t t o t h e c o n d i t i o n s u(p, φ, 0) = f ( p, φ), u( 1, φ, t) = 0. The prob­
lem describes the temperature distribution in the interior of the unit ball dropped in cold water. The first condition implies t h a t the tem­
perature u is not uniform but depends on p and φ but not on Θ. Thus the solution can be assumed formally to be
ιι{ρ,φ,ή = β ( ρ ) Φ (0 ) Τ ( ί ), which after separating the variables gives
(5.71)
Uff
I _ ZV
p <£" + cot φΦ'
T
= - a 2 = - = λ
R ‘ ρ 2 Φ ~ T
T h e l e f t s i d e o f E q ( 5.7 2 a ) c a n b e e x p r e s s e d a s
R"
- R ’
\
R
+ OL
\
, Φ" + C O t φ Φ'
P = ----------- φ— ■
/
(5.72a)
In order t h a t this equation be satisfied, the terms on each side must be constant. Thus,
1 R" + - R'
+ ol2
R
\
μ = —
Φ" + c o t φ Φ' ~Φ '
( 5.7 2 b )
/
5.6. SPHERICAL COORDINATES
145
I t is known t h a t μ = n( n + 1) (Courant and Hilbert, 1963). Then Eq (5.72b) yields
Φ" + cot φ Φ' + n( n + 1) Φ = 0, p2 R" + 2p R 1 + a 2p2 R — n( n + 1) — R = 0.
If we set x = ap in the second equation, then it becomes
x 2R" + 2 x R'+ ( x 2 - ( n + l ) )
w h i c h u n d e r t h e t r a n s f o r m a t i o n w = λ/x R red tion
1 'i 2 ^ )
( 5.7 3 )
x2w" + i w'+ ^' χ2 — (n+ ^) and has a bounded solution w = J n+1/2 (z)· Hence
Jn+l/2(^p)
R = 0,
ices to the Bessel equa- w = 0, (5.74)
R(p)
y/ap
The eigenfunctions = π+1/2^ p n ( c A
y/ap
basis in the L2-space on the unit ball, independent of Θ, where a rnn are
positive zeros of Jn+i/2(a p)· Hence the solut distribution in the unit ball is given by
Jn+l/2iamnp)
m= l n=0
y/CXmnP
where
, using /; P 2(x) dx
2 n + l
Cm/n —
(2n + 1) 1/0 ^ β(&τητι)
f ί P ^ J n + l/2 { & m n P ) P n(c Jo Jo
a n d
β ^ η ι η ) — 2 I P J n + l/2 { ° imn P) — [> Jo
T h e f a c t o r J a mn c a n b e a b s o r b e d i n C r n n. ■
s φ) form an orthogonal
on for the temperature
■Pn (cos0), (5.75)
;θ)/(ρ,φ) sin φ dp άφ,
n + l/2 ( a m n ) ] ·
146
CHAPTER 5: SEPARATION OF VARIABLES
5.7. N o n h o m o g e n e o u s P r o b l e m s
In the above examples we have seen t h a t the method of separation of variables is applicable to steady state linear problems with homoge­
neous governing equations and three homogeneous and one nonhomo­
geneous boundary conditions. Nonhomogeneity, however, occurs from other conditions as well. For example, there may be more th a n one non­
homogeneous boundary conditions, or the governing equation may be nonhomogeneous. In order to use the method of separation of variables, a nonhomogeneous problem can be divided into finitely many simple problems with homogeneous equations an d/o r homogeneous boundary conditions. Then the solution of the given problem is obtained from the superposition of the solutions of all these simple problems.
« =/2w
0
f2(x)
« =f3(x)
« =f4(x)
= 0 X
0
+ 0 --------- X
0
/iW
0
0
+
/3W 0
+ 0
/4w
0
0
Fig. 5.1.
EXAMPLE 5.8. (with four nonhomogeneous boundary condi t i ons) Consider the steady state temperature distributions governed by Eq
(5.45) in the region R, with more th a n one nonhomogeneous boundary conditions, viz.,
u(x,0) = fi(x), u(x,b) = f 2(x), 0 < x < a,
( 5.7 6 )
u( 0,y ) = f 3(y), u( a,y ) = f 4(y), 0 < y < b.
5.7. NONHOMOGENEOUS PROBLEMS
147
This problem can be resolved as a superposition of the four problems, shown in Fig. 5.1. Hence,
u(x,y) = Ui(x,y) + u 2( x,y) + u 3( x,y) -j u4( x,y), (5.77)
where the solution of each simple problem is obtained as in Example 5.5. ■
In certain cases with mixed boundary con- separation of variables can be readily used by u( x, y) which depends on the geometry and ma problem.
l i t ions, the method of translating th e function .terial symmetry of the
Exa mple 5.9. Consider the Laplace equation (5.45) in a half-strip (see Fig. 5.2) subject to the boundary conditions
lim u{x
x —►- j - o o
u{0,y) = f ( y), uy ( x,0) = 0, uy (x,b) + P[u(x,b)
w h e r e β is known as the film coefficient.
Fig. 5.2.
This problem has more th a n one nonhomogeneou: By using the translation U( x,y ) = u( x,y) — and (5.78) reduces to
Uxx U,
yy
0,
U(0,y) = f(y) - «oo = F{y)
U(y) = 0, Uy (x,b) = f i U{x,b
l i m
£ —►4-0O
We c a n n o w a s s u m e t h a t U( x,y ) = X( x ) Y ( y ) of the two ordinary differential equations (with
X" - \2X = 0, Y" + A2Y
y) = «oo,
t i o o ] = 0,
(5.78)
s b o u n d a r y condi t i ons. , t h e pr o b l e m (5.45)
U{x,y) = 0, (5.7 9 )
which reduces the set c = A2):
= 0.
148
CHAPTER 5: SEPARATION OF VARIABLES
These two equations lead to the general solution
U(x, y) = (Ai e~Ax + Ai eAx) {Βχ cos Xy 4- B 2 sin Xy ). (5.80)
Now, since 1^,(0) = 0 and Yy (b) + β Υ (b) = 0, we obtain the eigenfunc­
tions as cos An y with the corresponding eigenvalues A„, which are the positive roots of the equation
A„ ta n Xnb = β, n = 1,2,.... (5.81)
Using th e boundary condition l i m ^ o o X{ x ) = 0, we obtain
OO
U(x, y) - Σ Cn e~AnX cos Xny + Uoo. (5.82)
n = 1
Then, in view of the nonhomogeneous boundary condition X (0) = F( y), we have
OO
F(y) = f(y) - Woo = cosXny,
n = 1
w h e r e t h e c o e f f i c i e n t s Cn are given by 2Xn
Cn =
/ [f (y) - u oo } cos Anydy. Jo
Xnb + sin Anb cos Anb (see the table in Chapter 4). Hence, the temperature distribution is
U{x,y) = u(x,y) -
^ An e~XnX cos Xny f b
= 2 > ■■■ , ■■ ,------— / [/(η) — Moo cos Χ„η αη.
Xnb + sm Anb cos Anb J 0
( 5.8 3 )
I n p a r t i c u l a r, i f f ( y ) = uq = const, the temperature distribution re­
duces to
u(x,y)—Uoo „ v"·' sinA„6 \ _ . ,
~ = 2 Σ ν ^ "· ϊ\ r i e cos Xnydy. (5.84)
uo — Wqo " A„o + sm Anb cos Anb
n = l
I n v i e w o f ( 5.8 1 ), t h e e i g e n v a l u e s A„ a r e t h e p o s i t i v e r o o t s o f t a n ξ — Bi
— = 0, where ξ = Anb, and Bi = βb is the Biot number. Three of these roots, denoted by £i, £2, and £3, are shown in Fig. 5.3. ■
5.7. NONHOMOGENEOUS PROE
COt ξ
Fig. 5.3. Distribution of the eigenvalues ξ,
E x a m p l e 5.10. We shall consider a proble nonhomogeneous in the governing equation, w: neous boundary conditions. Assume t h a t heat angular bar at a constant rate q per unit vol temperature gradient in the ^-direction, and th k of the bar is constant. Then the steady state t is governed by the equation
k {^xx Uyy) Ί~ q — 0 Let the linear and homogeneous boundary cond : ux (0,y ) = 0, u( a,y) =
uy (x, 0 ) = 0, u( x,b) =
E q u a t i o n ( 5.8 5 ), b e i n g n o n h o m o g e n e o u s, i s nol i a s s u m e t h e s o l u t i o n a s
u( x,y) = V( x,y ) +φ( χ ),
t h e n p r o b l e m ( 5.8 5 ) - ( 5.8 6 ) r e d u c e s t o t h e f ol l ow
άφ( 0 )
ά2φ q —- + - = 0 dx2 k ’
dx
0, φ(
K l + Vyy = 0,
Vx(0,y) = 0 = V( a,y), \^ ( α;,0 ) = 0, V
and
L E M S
14 9
:m w h i c h i s l i n e a r a n d i t h l i n e a r a n d h o mo g e - i s g e n e r a t e d i n a r e c t - u me, t h a t t h e r e i s n o e t h e r m a l c o n d u c t i v i t y e m p e r a t u r e d i s t r i b u t i o n
( 5.8 5 ) i t i o n s b e 0,
o. ( 5'86)
s e p a r a b l e. B u t, i f we
( 5.8 7 ) i n g t wo p r o b l e m s:
a) = 0,
(5.88)
x,b) = —φ{χ). (5.89)
150
CHAPTER 5: SEPARATION OF VARIABLES
Solution of problem (5.88) is readily obtained as
Problem (5.89) is separable and its solution is given by
ka ' Xn coshAn 6
n=0 n
where the eigenvalues A„ = ^ n ^ n = 0,1,___ The solution of
this problem is then obtained by adding (5.91) and (5.90). Note t h a t this problem can also be solved by taking
u(x,y) = V(x,y)+tp{y). (5.92)
In the case when the r ate is variable, say, q = q(x), we should use the substitution (5.87); if q = q(y), then the substitution (5.92) will make the equation in V separable. ■
Exa mple 5.11. Consider the nonhomogeneous wave equation
utt = c 2u xx + f ( x,t ), 0 < χ < I, (5.93)
with the homogeneous (Dirichlet) boundary conditions u( 0,t ) = 0 = u ( l,t ), t > 0, and the initial conditions u(x,0) = g(x), u t (x, 0) = h( x),
0 < χ < I. Using the Fourier series method, which is the same as the
method of separation of variables, we seek a solution in the form
OO
/ \ v —> , ν . 717ΓΧ , .
u ( x,t ) = 2 _ ^ u n ( t ) sin—, (5.94)
n+l
where t is regarded as a parameter. The functions /, g, h are written as Fourier series
f (x,t ) = Y ] f n { t ) s i n ^ ^, f n (t ) = j f /( ξ,ί ) sin άξ,
n + l 1 1 J° 1
g{ x) = Y ] g n s i n ^ ^, gn = j i g{£) s i n ^ d £, (5.95)
n+l 1 1 J° 1
hix) = ^ 2 hn sin hn = j [ f t ( 0 Sin άξ.
n + l 1 L Jo 1
5.7. NONHOMOGENEOUS PROBLEMS
After substituting (5.95) into (5.93) we get
OO ζ
ς {»
n = l ^
( ^) +
ι 2
Un{t) - fn(t) } si:
where ύ = d2u/d t 2. This relation is satisfied if t he series are zero, i.e., if
^n(^) "Ί"
I2
■ Un(t ) — f n{
T h e s o l u t i o n un (t) of the ordinary differential coefficients can be easily obtained under the ini
equation with constant tial conditions
0 ) = g(x) = γ un (0) sin Τ^ ψ- = ^ gn sin
U\X
71=1
OO
n-l
71 = 1
= 1
Thus, u n (0) = gn, and u t (0) = hn. Now, we define the solutions un {t ) in the form
un{t ) = U* ( t ) + U 2(t ),
w h e r e
* i ( i ) = — ί
nnc J0
s m
nnc( t — t ) I
r e p r e s e n t s t h e s o l u t i o n o f t h e n o n h o m o g e n e o u s mo g e n e o u s i n i t i a l c o n d i t i o n s, a n d
2 /. \ Τ17Γ C t 1 ,
un{t ) = 9n COS— 1 hn SI
l nirc
is the solution of the homogeneous equation with the prescribed initial conditions. Hence,
u(x,t) = j r [u\( t ) + u\(t)]
n= 1
OO r
= Σ — /
Jo
. nnc( t — τ ) . ηπ χ s i n sin —-—
l l
Σ
( ηποί 1 ηπ
I 9n
c o s — ----- 1------- sin
V l ηπο
71=1 X
151
ηπχ
n — =0,
all the coefficients of
)·
ηπχ
I ’
/ λ\ ί / \ v ^ . /_>. . ίιτϊχ \i -ν . . ηπ χ
u t ( x,0) = h { x ) = 2 _ ^ u n { 0 ) sin — = } ^ h n sin —
(t) dr
e q u a t i o n w i t h t h e h o - mret
f n( r ) dr
( 5.9 6 )
zt) \ . ηπ χ
1 s i n —-—.
152
CHAPTER 5: SEPARATION OF VARIABLES
Note t h a t the second term is the solution of the corresponding problem with / = 0 (representing a freely vibrating string with prescribed initial conditions; see Exercise 5.16). The first term represents the forced vibrations of the string under the influence of an external force. ■
5.8. E x e r c i s e s
5.1. Solve u r r + - u r = ut t, subject to the conditions u( r, 0) = u 0r,
r
u t ( r,0) = 0, and u ( a,t ) = auo, lim u ( r,t ) < +oo, where uo is a
r —>0
constant.
OO
A n s. u = J o ( a i r ) sinc^t, where
2=1
_ f “ r ( r - a ) J o ( a i r ) d r U° l o r Jq { a i r ) dr
5.2. Sol ve x 2uxy + 3y 2u = 0, such that u( x: 0) = eX!x.
A n s. u = ey2+l/x.
5.3. Solve u x x —u t = A e ~ ax, A > 0,a > 0, where u( 0,t ) = 0 = u ( L,t ) for t > 0, and u( x, 0) = f ( x ) for 0 < x < L.
A n s. u = v - A? - ( e~aL - l ) % + - 4 r e~aL, where cr α Δ v 'L a *
v = f l A n s i n ^ e - » W L
n=l
2 f L ηπχ
L j Sm~ L ~
5.4. Solve ut = a2 u xx + f ( x,t ), 0 < χ < I, with the boundary con­
ditions u ( 0,t ) = 0 = u ( l,t ) for 0 < χ < I, and the initial condition u( x, 0) = 0 for t > 0.
OO
A n s. u ( x,t ) = ^ 2 un( t ) w h e r e
n = 1
« η ( ί ) = ί en W { t - T)/l2 f n ( r ) d r,
Jo
5.8. EXERCISES
153
sin ■
mri;
5.5. Solve utt = c uxx, 0 < x < L, subject ditions u( 0,t ) = u i, u( l,t ) = u2 for 0 < are prescribed quantities, and the initial con ut (x,0) = h( x) for t > 0.
Ans. u ( x,t ) = U(x) + v( x,t ), where U( x
d e s c r i b e s t h e s t e a d y s t a t e s o l u t i o n ( s t a t i c t h e s o l u t i o n o f t h e p r o b l e m i n E x e r c i s e 5.1 0
t o
t h e b o u n d a r y c o n - ~ < I, where u\, u2 .ditions u(x, 0) = g( x),
) = u i + (u2 - U\) —, flection), and v( x) is
de:
5.6. Find the interior temperature of the cooling ball of Example 5.7, An s.
< π.
π/2 <
2\/Xm Jq Jn+1/2
Cmn —
^ fo ^"+1/2 i ^mP)2
odp
curved surface p = 1,
5.7. Determine the steady state temperature inside a solid hemisphere 0 < p < 1, 0 < φ < π/2
(a) when the base φ = π/2 is at 0° and the 0 < φ < π/2 is at 1°.
(b) when the base φ = π/2 is insulated, but the temperature on the
curved surface is /(ώ). H i n t: — v ' dz
OO
dp cos φ 1 ^
Ans. (a) u( x) = - £ p2n+1[P2n (0) - P2n+2
n=0
(b) u( x) = ^ c„p2n P 2„(cos(/)), where
Cn = (4 n +
5.8. Solve ut = ux
7 1 = 0
,π/2
-ΐ)/ m P2n(cos</>) sin φάφ. Jo
■π < χ < π, subject to the conditions u( x, 0) = f { x ), ti( — π,ί ) = « ( π,ί ), and ux (— π,ί ) = υ^( π,ί ), where f ( x ) is a periodic function of period 2π. This problem describes the heat flow inside a rod of length 2π which is shaped in the form of a closed circular ring. H i n t: Assume X ( x ) = Acosg;x + Bsi nuj x.
A n s. ωη = n; u( x,t ) = "2< (a„cos
άξ.
p) d-P
0 ON THE BASE.
( 0) ] P2n+l (.x)·
n x + bn sin nx), where
71=0
1 π 1 π
an = — f /( x ) cos n x d x, bn = — f f ( x ) sin n x d x.
7Γ 7Γ
1 5 4 C H A P T E R 5: S E P A R A T I O N O F V A R I A B L E S
5.9. S o l v e t h e p r o b l e m Ut = V2w, r < 1, 0 < 2 < 1, such t h a t u( r,z, 0) = 1, u ( l,z,t ) = 0, and u( r,0,t ) = 0 = u( r,l,t ). This problem describes the temperature distribution inside a homoge­
neous isotropic solid circular cylinder.
Ans.
OO
u{r,z,t)= Σ Cmn e_(A^ +n2,r2)t J0 (Amr) s i n n7rz,
m,n=1
where Am are the zeros of Jo, and
4 (1 — (—1)”)
c —
v-' mn.
—
Λ γγι Λ (Λ
η
5.10. Find the steady state temperature in a solid circular cylinder of radius 1 and height 1 under the conditions that the flat faces are kept at 0 ° and the curved surface at 1°.
Ans.
Jo(rarr) sin ηπζ
i ( r,z) = 4 Σ
Ι ο(ηπ) s i n n7r
n odd
5.11. Solve the steady state problem of temperature distribution in a half-cylinder 0 < τ < 1,0 < θ < π, 0 < z < 1, where the flat faces are kept at 0 ° and the curved surface at 1°.
An s.
. . 16 I m{mrr) sin ηπζ
u( r,6,z ) = —
π ^ I m(nn) βίηηπ
m,n=l
.,n m.n odd
5.12. Solve ^ ( χ ^ τ ), 0 < x < 1, t > 0, subject to the
ot ax \ ox J
conditions u( x, 0) = f (x) and u ( l,t ) = 0. H i n t: S e t 4x = r 2 a n d
solve as in Ex a mple 5.6.
OO
Ans. u( x,t ) = ^ Cne~Xnt^ Jo(An^/x), where An are the zeros of
n = 1
c _ fp /i v*) Jo(XnVx)yxdx fo[Jo{^nVx)}2Vxdx
J o a n d
5.8. EXERCISES
155
5.13. Solve u tt = c (uxx + Uyy) in the rectangle R = { ( x,y) : 0 < x <
a, 0 < y <b}, subject to the condition u = 0 for t > 0, and the initial conditions u( x, y, 0 ) g( x,y). This problem describes a vibrating Interpret the solutions « n, u\2, «21, «22, «ι: membrane a = b = 1.
A n s.
OO
u(x,y,t) ^ ' (Amn cos Amnt + sin )
o n t h e b o u n d a r y o f R = ut ( x,y,0) =
e c t a n g u l a r m e m b r a n e. , a n d U31 f o r a s q u a r e
m,n = 1 m,n o d d
w h e r e
4 ( b Γ f i \ · m7r
Π
α
g(x,y) si
.
—
B-mn ~~
τηπχ
si n
τηπχ
&&λγγιγ1 J(\ J (\ Cl
for m, n = 1, 2,...; t h e ei genval ues ar e
λγηη — C7T \/ ^ ^ ·
The solutions un, U
12
, U
2 1
,
^ 22, Wi 3, and U31 5.4.
λ . τ ηπχ . ηπχ} m n ) s i n sin —7—,
ηπν
- y dx dy,
s m ^ d x d y,
b
are represented in Fig.
Fig. 5.4.
5.14. Solve uxy — Axyu = 0, such t h a t u( 0,y) =
Δ NQ •71
-- 2+y
2
156
CHAPTER 5: SEPARATION OF VARIABLES
5.15. Solve u xx + uyy = 0, under the conditions u(0, y) = 0 = η( π, y), u( x, 0 ) = sinx, lim u( x,y) < +oo.
+ 00
A n s. u = e ~ y sinx.
5.16. Solve u xx — utt = e~a π 1 sin απχ, subject to the conditions u( x, 0 ) = 0, ut (x, 0 ) = 0, and u( 0,t ) = u ( l,t ) = 0, where a is a constant.
A n s. 0 n, fcos ant — e~a π 1 — απ sin ant) sin απχ.
αζπ*( 1 + αΔπ Δ)
5.1 7. S o l v e r 2u rr + r ur + uee = 0, such t h a t u(b, Θ) = /(#), u(r, Θ + 2π) = u ( r,6), and lim u( r,8) < +oo (circular disc problem).
τ’—+0
H i n t: Separate the variables and show t h a t the only relevant part of the solution reduces to
u(r, Θ) = c0 + Σ r a ( A( a) cos αθ + B (a) sin αθ).
N o t e t h a t u n d e r t h e g i v e n c o n d i t i o n s u(r,9) must have a Fourier series representation in Θ and therefore a = n is a positive integer.
5.18. Solve uxx + uyy = 0, under the conditions u( x, 0) = 0 = u( x, π), u( 0, y) = 0, and ΐί(π, y) = cos2 y.
A n s.
OO
u = ^ 2 Cn sinh n x sin ny,
n =l
w h e r e
2 Γ 2
Cn - —— / cos·2 y sin nydy
π smh ηπ J 0
1 - ( - 1 )" ( —1 )nn
π sinh ηπ
a n d Ci = 0.
2 n n2 — 4
5.19. Solve u r r + - u r + — u g o = 0, subject to the conditions u = 0 for
Θ = 0 or π/2, and ur = sin# at r = a.
A n s.
u =
71=1
C nr 2n s i n 2 ηθ,
5.8. EXERCISES
157
4 n ( - 1 Γ+ 1 n π ( 4 η2 — l ) o2n_1
w h e r e
5.2 0. S o l v e t h e p r o b l e m o f t r a n s v e r s e v i b r a t i o n s o f a b e a m: uu + a?uxxxx = 0, subject to the conditions u(0, t) --= u( L,t ) = uxx(0,t ) = u xx( L,t ) = ut ( x,0) = 0, and u ( x,0) = f ( x ).
χ ( 4 )
A n s. L e t u = X( x ) T( t ), then we have
X
'~ ^ τ = λ' where
it can be shown t h a t
A is a parameter. By standard arguments the relevant values of A are positive values. Let A = a 4. Then the solutions for X and T are given by
X = A cos a x + B sin a x + C cosh a x + D sin ax,
a n d
Τ = E cos a 2t + F sin a 2 X{0) = 0 means A + C = 0, and X( L ) = 0 yi
A cos a L + B sin a L + C cosh a L + D sin a L = 0,
and Xcx(0 ) = 0 implies 2 a 2A = 0, which g: Xxx{L) = 0 which yields a 2( Bs mh.aL — D i i n a L ) have a pair of two homogeneous equations:
elds
ves A = C = 0, and 0. We thus
B sinh a L — D sin a L = 0, B s i n h a L
F o r B and D to have nontrivial values, we m
sinh a L sin a L = 0,
i.e., a L = ηπ, and B = 0 and T(0) = 0 are Absorbing E in D and using the initial condition we get
D sin a L = 0. i s t have
equivalent to F = 0.
n= 1
. η π χ η ζπΗ , sm - j - cos —j j ~,
w h e r e Dn =
2 f L
l L /( i )
. η π χ . sin —— ax. L·
5.2 1. S o l v e r u, u ( a, 0) = 0
ture in a finite cylinder). A n s.
4u0
π
= 0, u ( a, z ) = u q,
u { a,h ), and lim. u ( r,z ) < +oo
_ 4wo
7Γ
Jo{^r) . ntz ^ Jo (ηπα) Sm h
u n d e r t h e c o n d i t i o n s ( S t e a d y s t a t e t e m p e r a -
158
CHAPTER 5: SEPARATION OF VARIABLES
5.22. Solve uxx — utt = e π 1 sin πχ, such t h a t u( x,0) = 0 = ut (x, 0) = u( 0,t ) = ux {l,t ).
An s. A p a r t i c u l a r s o l u t i o n i s g i v e n b y
p π 2(1 + π 2)
Now define u = v + up. Then the problem becomes
βίηπχ βίηπχ
Vxx Vtt = 0, 0) — 2ΤΪ j 2\ ’ Vt\X) 0) — j 2 T5
π (1 + π*) (1 + 7T^)
6- π 2ί
v( 0,t ) = 0, vx (l, t) = - π ( 1 + π 2 ) ·
Assume v = X( x ) T( t ). Then we have
X" T"
— = — = const.
When the constant is zero, the solution will not make any contribu­
tion to v. So we consider two cases: (i) when const = A2, and (ii) when const = —a 2. In the first case the solution is
ui = Σ * - λί (A cosh A χ + B sinh Ax), λ
and in the second case
V2 = Σ Μ 1 cos a x cos a t + A 2 cos a x sin at
a
+ A 3 sin a x cos at + A 4 sin a x sin a t ).
A p p l y i n g t h e b o u n d a r y c o n d i t i o n s t o v + v\ + we get A = 0,
B
= ο/,—
9
Γ
1
A = π 2, Αχ = A
2
=
0, a
= (n
+ - )?r, and A 3, A
4
π
ό(1
+ π Δ)
2
are to be determined from the initial conditions. Thus we have
_TT2 f · - ~ 2
e sin π χ e
u = —
π 2 (1 + π 2) π3 (1 + π 2)
4 - ^ ( A3 sin a x cos at + A 4 sin a x sin a t ).
We n o w a p p l y t h e i n i t i a l c o n d i t i o n s
s i n π χ s i n h
π 2 (1 + π 2) π3 ( 1 + π 2 )
β ί η π χ β ί η ] ι π 2χ _
w( *- ° ) = - ττ2(λ 4--7Γ2Ί - ^ sm a x = 0,
5.8. EXERCISES
E
sin πχ sinh
A3 sin a x = 0,„ ,— — +
π 2 (1 + π 2) π3 (1
•f π 2)
and
, είηπχ 8ίη1ιπ2χ
Μ χ, ο) = ·7τ-{— ^ ι 2\ + ζ -,
( 1 + π ^ ) π(1 + π*) ζ—'
sin πχ sinh
αΑ*
sin α χ
(1 + π 2) π(1 -h π 2)
Define
Γ , . (2η + 1) πχ ,
i i = / sin π χ s i n αχ =
' 2 (ύτι + 3)(2η — 1)
/■
J ο
= ί sir Jo
τ I ■ u 2 · (2« + 1)ττχ
h = sinh π χ s i n----------------
dx =
A 3 = 2
Α, = -
_π2(1 + π 2) ' π3 (1 -Ι-π2) _ | ’
h
(2η + 1)π
Λ
_(1 + π 2) π
159
= 9 ι ( χ )
α Α ί sin α χ = Ο,
= 9 2 ( χ ) ·
- 4 ( - l ) r
4 ( —1 ) η π2 c o s h π
2π 2 + (2 η + I) 2
( 1
5.2 3. S o l v e t h e P o i s s o n e q u a t i o n uxx + uyy subject to the Dirichlet boundary conditions u{ x, 0 ) = u(x, 1).
A n s.
- 1, 0 < x,y < 1, u( 0,y) = 0 = u ( l,y ) =
β ι η ^ π χ s j:
n kiry
π"1 j 3k 2 + j 2k 3
j,k o d d
6
I n t e g r a l T r a n s f o r m s
The technique of integral transforms is a powerful tool for the solution of linear partial differential equations.
A function F( x) may be transformed by the formula
provided F( s ) exists, where K( s,x ) is known as the kernel of the transform. A transform becomes useful if we can obtain f ( x ) from F( s) by some inversion formula. Some well-known integral transforms and their inversion formulas, known as the transform pairs, are given below.
1. The Fourier cosine transform /c(a) of f ( x ) is defined as
F(s) = f f ( x ) K ( s,x ) d x
J a
^ c { f { x ) } = f c { a ) = f ( x ) c o s ( x a ) d x, (6.1 a)
and its inverse is
2. The Fourier sine transform f s (a) of / (a:) is defined as
INTEGRAL TRANSFORMS
and its inverse is T
1 {/* ( “ ) } = f ( x ) = \j\J q f s ( a ) s i
3. Th e F o u r i e r c o mp l e x t r a n s f o r m F f ( x ) = f ( a ) of f ( x ) is defined as
and its inverse is
1 Γ°° ~
^ - 1{ f { a ) } = f ( x ) = - ^ = J J {,
4. Th e L a p l a c e t r a n s f o r m i s d e f i n e d as
/•OO
C{ f( t ) } = F( s) = f ( s ) = / /( )
Jo
161
(xa)da. (6.2 b)
)elxa dx, (6.3a)
ct)e~ixa da. (6.3b)
and its inverse is C~
5. Th e Me l l i n t r a n s f o r m i s d e f i n e d as
c - M o c
pOO
M { f ( x ) } = F m (s) = / f { x ) xi Jo
a n d i t s i n v e r s e i s 6. Th e Ha n k e l t r a n s f o r m o f o r d e r n i s d e f i n e d as
-j pc-\-ioo
poo
n{f(x)} = K(s) = / x f ( x ) J, Jo
wh e r e i t s i n v e r s e i s Ή
p OC
Μ-Ρ’ηίβ)} = f ( x ) = / sF( s ) Jo
T h e s e d e f i n i t i o n s a r e n o t u n i q u e, p a r t i c u l a r l y i n t h e c a s e o f F o u r i e r a n d Ha n k e l t r a n s f o r ms wh i c h a r e s o me t i me s d e f i n e d i n a d i f f e r e n t ma n n e r. I n f a c t, one
e~st dt,
1 dx,
ri(sx) dt,
( 6.4 a )
( s ) e d s. (6.4b)
(6.5a)
s)x s ds. (6.5b)
(6.6 a)
7n(sx)dt. (6.6 b)
162
CHAPTER 6: INTEGRAL TRANSFORMS
can develop an infinity of transforms. However, the six transforms defined above are frequently used. Some of the other better-known transforms are Meijer, Kontorowich-Lebedev, Mehler-Foch, Hilbert, and Laguerre. We will, however, discuss only the Laplace and Fourier transforms. Once the use of one transform is completely understood, it is a simple matter to extend one’s understanding to another transform.
In most cases a function has to satisfy Dirichlet’s conditions in order to possess an integral transform. These conditions in the interval (a,b) are (i) a function has only a finite number of extremum points in (a,b), and (ii) a function has only a finite number of finite discontinuities in (a,b) and no infinite jumps. Unless otherwise stated, it will be assumed that all the functions in the sequel satisfy Dirichlet’s conditions.
The student who lacks the knowledge of contour integration technique may omit all material in this chapter containing this technique.
L a p l a c e T r a n s f o r m s
6.1. N o t a t i o n
It is expected that the reader is familiar with the elementary theory of the Laplace transforms. The following notation is used:
pOO
C{f(t)} = F(s) = f(s)= f(t)e~st dt,
Jo
and
C-^Fis)} = C- 1 {f(s)} = f(t),
where s is the variable of the transform which is, in general, a complex variable. Note that the Laplace transform F( s ) exists for s > a, if the function f ( t ) is piecewise continuous in every finite closed interval 0 < t < b (b > 0 ), and f ( t ) is of exponential order a, i.e., there exist α, M, and io > 0 such that
e~at \f(t)\ < M f o r i > to-
W e w i l l n o w s t a t e s o m e b a s i c p r o p e r t i e s o f t h e L a p l a c e t r a n s f o r m s.
6.1. NOTATION
163
(i) £ { e atf ( t ) } = F ( s - a ),
a n d C~1 { F( s — a)} = eatf ( t ).
( i i ) £ { H ( t - a ) f ( t - a ) } = e~asF( s),
a n d C~l {e“ asF (s ) } = H( t - a ) f ( t - a).
( i i i ) Co n v o l u t i o n Th e o r e m:
CT1 {G( s ) F( s ) } = f f ( t - u) g( u) du = Jo
and C - 1 { s nF( s ) } = ~ f ( t )
dt n'
( v ) C
" { > } = /'
f ( u) du.
[ f { u ) g { t - u ) d %, ιo
m - s n- 2f ( § ) — s f ^ n~ ’ (o) dn
Hn F (
( v i ) £ { t nf ( t ) } = ( - 1 Γ —, and C'1 j ( - l ) n - j/
( v i i ) l f C { f ( x,t ) } = F ( x,s ),t h e n
c | 4 f e i ) } = ^ g i i ), ,„ d c -
dF(x,s) 1 _ d f ( x,t )
Th e l a s t t wo r e s u l t s a r e b a s e d on t h e Le i b n i z r u l e t i ve. Th e Le i b n i z r u l e s t a t e s t h a t i f g(x, t ) is an inti
$π(χ f,)
each value of x, and the partial derivative — — ex
the region under consideration, and if
dx
f { x ) = f 9(x,t)dt, J a
t hen
f'(x) = f
J a
dg{x,t )
dx
d t.
dx
dx
and a r e e x t r e me l y e f f e c - t g r a b l e f u n c t i o n o f t for
ists and is continuous in
(6.7)
164
CHAPTER 6: INTEGRAL TRANSFORMS
6.2. Basic Laplace Transforms
A table of basic Laplace transform pairs is given in Appendix B. We shall show the effectiveness of the above properties in the derivation of certain Laplace transforms.
We start with the easily established result that
£ {ea i } = —-—. (6.8)
*· } s — a
Di f f e r e n t i a t i n g b o t h s i d e s wi t h r e s p e c t t o a, we g e t
£ { i e at } = r ^ 2 ’ (6·9)
1 J (s — a) z and repeating this differentiation n times, we find that
c {tneat\ = -----—-τ—- y. (6.10 )
1 1 (s - a)n+1 v
After replacing a by ib, choosing an appropriate n, and comparing the real and imaginary parts of both sides, we can get the Laplace transforms of functions t p cos bt and t p sin bt, and then combining with Property (i), we can get the Laplace transforms of functions t peai cos bt and t peat sin bt. For example, if we choose n = 2, then we have
(6.11)
L J (5 — a ) 6
Now l et t i ng a = ib, we get which yields
£ { t2(cos bt + i sin bt ) } = ^ ^ . (6.13)
Expanding the numerator on the right side of (6.13), we get
r 9/ , s i 2(s3 + Si s2b — 3sb2 — ib3) . .
C{t (cos bt + i sin bt)} = ------------ 2 + ^2^3------------' (6-14)
6.2. BASIC LAPLACE TRANSFORMS
165
( 6.1 5 )
( 6.1 6 ) I n [ l J:=
« C a l c u l u s ‘ L a p l a c e T r a n s f o r m ‘
In[ 2]:=
L a p l a c e T r a n s f o r m [ t ~ n E x p [ a t ],t,s ]
Out[2j =
( - a + s) - 1 - n Ga mma [ 1 + n ]
In[3]:=
L a p l a c e T r a n s f o r m [ t ~ 2 E x p [ a t ],t,s ]
Out[3]=
2
( - a + s) 3
In[4]:=
L a p l a c e T r a n s f o r m [ t ~ 2 E x p [ I b t ],t,s ]
Out [4]=
2_____
( - 1 b + s) 3
In[5]:=
T h e n e q u a t i n g t h e r e a l a n d i ma g i n a r y p a r t s i n ( 6.1 4 )
2 ( s 3 — 3 sb2'
a n d
£ { i 2 c os f e i } =
C {t 2 sin&i} =
(s2 + b2)3
2(3s 2b - b3 (,s 2 + b2)3
we o b t a i n
166
CHAPTER 6: INTEGRAL TRANSFORMS
L a p l a c e T r a n s f o r m [ t"2 C o s [ b t ],t,s ]//S i m p l i f y
Out[5]=
2 s ( —3 b2 + s 2 )
( b2 + s2)3
In[6]:=
L a p l a c e T r a n s f o r m [ t ~ 2 S i n [ b t ],t,s ]//S i m p l i f y
Out [6]=
2 b ( - b2 + 3 s 2 )
( b2 + s 2 )
The Lapl ace transforms o f C {eatt 2 cos bt ) and Ceat {t 2 sin bt} can now be easily obtained.
An important Laplace inverse is
c 1 { — — !· = e r f c ^ f ’ ^6·17)
where
2 f x 2
erf (a;) = —-j= / e u du, erfc(x) = 1 — erf(a:). (6.18)
V71- Jo
In[7]:=
( * A n i m p o r t a n t f o r m u l a * )
Out[8]=
I n v e r s e L a p l a c e T r a n s f o r m [ E x p [ - a S q r t [ s ] ]/s,s,t ]
1 - E r f [
6.2. BASIC LAPLACE TRANSFC
]
2 S q r t [ t ]
RMS
167
We can derive a large number of the Laplace inverses by using Properties
(iHvii).
E x a m p l e 6.1. C
, — dy/s
7 Γ
c a n b e o b t a
n e d b y d i f f e r e n t i a t i n g
f o r m u l a ( 6.1 7 ) w i t h r e s p e c t t o a. Thus,
c - 1
,-ΟλΑ
7 ^
-a /
At
is obtained after differentiating (6.17) with resp out the negative sign on both sides. Although
e-as/s
e c t t o a and canceling the usual method of
deriving the Laplace inverse of
y/s
is by contour integration, or by
using the Laplace inverse of
-dy/s
S
(Churchill, 1972): Define - —— = y and e~a^
V =
2 s 3/2
, a n i n t e r e s t i n g m e t h o d i s a s f o l l o ws = z. Then
-“%/« _ — 2s
A
w h i c h y i e l d s S i mi l a r l y, z' =
'2 V 5
2 sy' + y + az = 0. “av/I yields
2 z'
+ ay — 0.
Taking the inverse transform of these equations
aG — F — 2tF' = 0, and a F —
w h e r e £ - 1 {y } = F( t ), and C~1{ z } = G(t ). Fro in F and G we get
we get 2t G = 0,
b i t h e s e t w o e q u a t i o n s
The solution of (6.19) is
F — — e-a2/4t
Vt
which gives
G = ^ 4 =e ~a2/4t.
2 V P
N o t e t h a t i f a = 0, then y = and Fi t ) = —= implies t h a t
γ S yf-Kt
A = Hence
V*
F(t) = - £ = e~a2/4t, G = - j L = e~a2/4t. (6.20)
VVt ν 7rt3
Γ g-aVs Ί X - 2/At
O n e c a n t h e n i n t e g r a t e £ _1 < — t=— > = —y=e a ' with respect to
1 Vs J \ixt
e~aVs
a from 0 to a and obtain C { ----------\ . In this problem we have
168 CHAPTER 6: INTEGRAL TRANSFORMS
1 /7Γ
assumed t h a t C—p = * — (see Exercise 6.11).
Vt V s
In [ 8]:=
N e e d s ["C a l c u l u s ‘ L a p l a c e T r a n s f o r m ‘"]
( * A n i m p o r t a n t L a p l a c e I n v e r s e * )
In[9]:=
f [ s _ ]:= E x p [ - a S q r t [ s ] ]/s
I n v e r s e L a p l a c e T r a n s f o r m [ f [ x ],s,t ] = = E r f c [ a/( 2 S q r t [ t ] ) ]
Out[ll]=
1 - E r f [ — ■■■ —, , ] = = E r f c [ ]
2 S q r t [ t ] 2 S q r t [ t ]
In[U]:=
6.2. BASIC LAPLACE TRANSFO] R H S:= E r f c [ a/( 2 S q r t [ t ] ) ]
I n[ 12]:=
D [ R H S, a ]
Out [ 14j =
( E a2/( 4 t ) S q r t [ P i ] S q r t [ t ] ^
I n[ 14]:=
I n v e r s e L a p l a c e T r a n s f o r m [ - D [ f [ s ],a ],s,t
Out [ 15] =
T r u e
IMS 169 ] = = - D [ R H S,a ]
E x a m p l e 6.2. £ _1 ( e _ a ^ l = —7= = e _ “2/ >- J 2 ^
ferentiating the formula in the previous example negative sign. ■
lt is obtained by dif- md canceling out the
In[l]:=
N e e d s ["C a l c u l u s ‘ L a p l a c e T r a n s f o r m'"];
( * f [ s ] i s d e f i n e d a s i n E x a m p l e 6.1 * f [ s ] S q r t [ s ]
Out[3]=
1
Ea S q r t [ s ] S q r t [ s ]
)
170
CHAPTER 6: INTEGRAL TRANSFORMS
I n [ 3 ]:=
D [ f [ s ] S q r t [ s ], a ]
O u t [ 4 ] =
_ E- ( a S q r t [ s ] )
I n [ 4 ]:=
I n v e r s e L a p l a c e T r a n s f o r m [ - D [ f [ s ] S q r t [ s ],a ],s,t ]
O u t [ 5 ] =
2 E a2/( 4 t ) S q r t [ P i t 3 ]
(1yJ~S J
> = erfc — yz
w i t h r e s p e c t t o a from 0 to a, we get
/ν · ( ^ 1 ώ = Γ β Λ ^ ώ.
Jo \ » j Jo 2 Vt
No w a f t e r c h a n g i n g t h e o r d e r o f i n t e g r a t i o n a n d t h e L a p l a c e i n v e r s i o n a n d c a r r y i n g o u t t h e i n t e g r a t i o n o n t h e l e f t s i d e, we g e t
I dx = C~l ( s-*12 - s - 3/2e"a^ ), (6.21)
/V ‘
Jo
and the right side yields
f
Jo
erfc — t= dx 2 Vt
x erfc
2 Vt
fo-
Η— I x e x !At dx ο ν π ί .
a erfc
2 Vt
° 2 J - e - “ 2/4 t + 2 1/i
we ;
S i n c e £ 1 j s 3//2| = 2 y ^ -,
r - 1 { δ - 3/2 β - α ^ } = 2 y ^ e - a 2/4i - a e r f c ^ =.
( 6.2 2 )
6.2. BASIC LAPLACE TRANSFORMS
171
In[l]:=
N e e d s ["C a l c u l u s ‘ L a p l a c e T r a n s f o r m ‘"];
( * f [ s ] i s d e f i n e d a s i n E x a m p l e 6.1 =0 f [ s ]/.a - > x S q r t [ s ]
Out[ 2]=
1
E S q r t [ s ] x s
( * C h a n g e t h e o r d e r o f o p e r a t i o n s * )
I n [ 3 ]:=
LHS = I n t e g r a t e [ f [ s ]/.a - > x ,{ x,0,a } ]
O u t [ 3 ] =
s - ( 3/2 )_ _ _ _ _ _ _ _ _ _I _ _ _ _ _ _ _
E a S q r t [ s ] s 3/2
In[4]:=
I n v e r s e L a p l a c e T r a n s f o r m [ f [ s ] ,s,t ] = = E ) r f c [ a/( 2 S q r t [ t ] ) ]
Out[4]=
1 - E r f [ -
- ] = = E r f c [ -
'2 S q r t [ t ] J u 2 S q r t [ t ]
( * D e f i n e a s u b s t i t u t i o n * )
In[5]:=
172
CHAPTER 6: INTEGRAL TRANSFORMS
t r a n s =
I n v e r s e L a p l a c e T r a n s f o r m [ f [ s ],s,t ] - > E r f c [ a/( 2 S q r t [ t ] ) ]
Out[5]=
1 - E r f [------— — ] - > E r f c [ & ]
2 S q r t [ t ] 2 S q r t [ t ]
In[6]:=
I n t e g r a t e [ E r f c [ x/( 2 S q r t [ t ] ) ],{ x,0,a } ]
Out[ 6]=
a + 2 S q r t [ t ] _ 2 S q r t [ t ] _ & ^ a
S q r t [ P i ] E a 2/( 4 t ) S q r t [ P i ] 2 S q r t [ t ]
( * C o l l e c t t h e t e r m s i n v o l v i n g a, i n o r d e r t o a p p l y t h e t r a n s f o r m a t i o n * )
In[7]:—
RHS=
C o l l e c t [ I n t e g r a t e [ E r f c [ x/( 2 S q r t [ t ] ) ],{ x,0,a } ],a ]/.t r a n s
Out[7]=
2 S q r t [ t ] 2 S q r t [ t ] & a
S q r t [ P i ] E a 2/( 4 t ) S q r t [ P i ] 2 S q r t [ t ]
( * P r e v e n t e v a l u a t i o n o f t h e I n v e r s e L a p l a c e T r a n s f o r m * )
In[ 8]:=
I L T [ X _ ] := H o l d [ I n v e r s e L a p l a c e T r a n s f o r m [ X,s,t ] ]
s e x p r:=
I L T [ I n v e r s e L a p l a c e T r a n s f o r m [ s ~ ( - 3/2 ) E x p [ - a S q r t [ s ] ] ] ]
6.2. BASIC LAPLACE TRANSFORMS
173
In[10]:=
S o l v e [
I n v e r s e L a p l a c e T r a n s f o r m [ s ~ ( - 3/2 ),s,t ] - I L T [ s e x p r ] = = R H S,I L T [ s e x p r ] ]//S i m p l i f y
Out [ 8]=
{ { H o l d [ I n v e r s e L a p l a c e T r a n s f o r m [ -
2 S q r t [ t ] a
a E r f c [ — —
E a S q r t [ s ] s3/2
, s,t ] ] - >
E a2/( 4 t ) S q r t [ P i ] 2 S q r t [ t ]
E x a mp l e 6.4. E v a l u a t e £ 1
t h a t
0- a s/s + c
1
£_1 { e ~a^\ = —^ = e ~ a I ' 2 \/π ί 3
Hence, using Property (i),
£ - i
We know from (6.20) (6.23)
I At
ct —
2 \Ζπϊ?
U s i n g t h e c o n v o l u t i o n t h e o r e m w i t h F( s ) — -
s
we get
- a y/s+Z I f t
£ _1
I — ■
J o 2\/π υ?
Note t h a t
a 1 f c a + — i-
“2/4t. (6.24)
and G(s) = e~aV a2/4udu. (6.25)
1 f c
a n d
2\/π ΐ ί 3 4\/x u 3 2 y t i 4\/π ζ ί 3 2 y u
c u + ^ - = ( V ^ + - a V t = ( V ^ - ^ ) 2 + aVS-
4u v ν ““ 2^/ΰ
If we now substitute
a .— a
1 + a n d V = h
w 2-v/ tz
2 j u — \/cH,
174
CHAPTER 6: INTEGRAL TRANSFORMS
then the integral on the right side of (6.25) can be expressed as
Hence
1
poo poo
e “ ^ / e~x2dx + e - a^ / e~v' J x 1 j y i
1
dy
-a Vs+c
j = l [ e ^ e,f c ( ^ + v S ) +
e—ay/c er£c — Vc t J j. ■ (6.26)
We will state a very useful theorem without proof.
T h e o r e m 6.1. I f G( s ) = ^ G f c ( s ) is uni f orml y convergent, then
C 1G(s) = g{t) = ] T 9fc(i),
(6.27)
where £ Gk(s) = s*(i)·
Ex a mple 6.5. Since
-1 { s - 3^ 8}
= £ - i / J —
\ S3/2
” ( _ l ) n j ” (
η) ι ς>η+3/2
i - i 1
s 2!s2 3!s
1 + ■■■ + ( -!)" 1
nis
-1)η (2λ/ί ) 2η+1
0'n!s n+3/2 o (2n + l)!
sin(2\/i),
we find t h a t this result and Property (iv) give
j s ~ 1//2e1'/'5 j = — cos(2\/i ) · ι
(6.28)
( 6.29)
6.2. BASIC LAPLACE TRANSFORMS
175
g e b r a'S y m b o l i c S u m'"]
( * An i n t e r e s t i n g u s e o f t h e p a c k a g e S y m b o l i c S u m * )
In[l]:
N e e d s ["C a l c u l u s ‘ L a p l a c e T r a n s f o r m","Al ( * C o m p u t e a n i n f i n i t e s um * )
In[ 2]:=
S u m [ s ~ ( 3/2 ) ( l/s ) ~ k/k! ,{ k,0,l n f i n i t y } f l
Out[2]=
E l/s s 3/2
( * C o m p u t e a n i n f i n i t e s u m * )
In[3]:=
S u m [ ( - 1 ) ~ k/( s ~ ( k + 3/2 ) k!),{ k,0,I n f i n i t y } ]
Out[3]=
1
£ l/s s3/2
( * C o m p u t e t h e I n v e r s e L a p l a c e T r a n s f o r i j i o f a t e r m * )
In[4]:=
I n v e r s e L a p l a c e T r a n s f o r m [ ( - l )"k/( s ~ ( k + ^/2 ) k!),s,t ]
Out [4]=
( - l ) k t 1/2 + k
3
k!G a m m a [ — + k ]
176
CHAPTER 6: INTEGRAL TRANSFORMS
( * Sum t h e i n f i n i t e s e r i e s * )
In[5]:=
l/S q r t [ P i ] S u m [ ( ( - l ) ~ k ( ( 2 S q r t [ t ] )"( 2 k + l ) ) )/( 2 k + l )!, { k,0,I n f i n i t y } ]
Out [5]=
S i n [ 2 S q r t [ t ] ]
S q r t [ P i ]
( * C o m p u t e d e r i v a t i v e a n d u s e P r o p e r t y 4 * )
In [ 6]:=
D [ l/( S q r t [ P i ] ) S i n [ 2 S q r t [ t ] ],t ]//S i m p l i f y
Out [6]=
C o s [ 2 S q r t [ t ] ]
S q r t [ P i ] S q r t [ t ]
In[7j:=
H o l d [ I n v e r s e L a p l a c e T r a n s f o r m [ R e l e a s e [ s"l s ~ ( - l/2 ) ] E x p [ l/s ],s,t ] ] = =
D [ l/( S q r t [ P i ] ) S i n [ 2 S q r t [ t ] ],t ]//S i m p l i f y
Out[7]=
H o l d [ I n v e r s e L a p l a c e T r a n s f o r m [ E ^ ^ S q r t [ s ],s,t ] ] = =
C o s [ 2 S q r t [ t ] ]
S q r t [ P i ] S q r t [ t ]
In[8]:=
( * N o t e Mma f a i l s t o c o m p u t e t h e f o l l o w i n g t r a n s f o r m * )
6.2. BASIC LAPLACE TRANSFORMS I n v e r s e L a p l a c e T r a n s f o r m [ s ~ ( - 1/2 ) Exp [ 1/s ],s,t ] Out[8]=
$ F a i l e d
1 7 7
n i l
E x a m p l e 6.6. C o n s i d e r a s e mi - i n f i n i t e x < oo, —oo < y, z < oo, which is kept at an while its face x = 0 is maintained at a timi /( f ). The problem is to find the temperature the Laplace transform to the heat conduction get Τ χχ
e d i u m b o u n d e d b y 0 < i n i t i a l z e r o t e m p e r a t u r e, e - d e p e n d e n t t e m p e r a t u r e f o r t > 0. By applying equation kTxx = Tt, we T, where T = £ { T }. The solution of this equation is
T = Ae mx + Be
( 6.3 0 )
w h e r e m =
. S i n c e T remains bounded x ~ > 00, we find t h a t
A = 0. The boundary condition a t x = 0 ib the transform domain yields B = f ( s ), where f ( s ) is the Laplace transform of /( t ). Thus, t he solution in the transform domain is
T = f(s) e~mx.
In order to carry out the inversion, we use the convolution property and Example 6.2 and get
/
Jo
x e
-f(t - T
2τ\/π kr
I f f ( s ) = 1, then the solution for T reduces tc
dr.
- I
t xe~x2^ kT ‘iT'J'Kkr
dr
This solution is the fundamental solution for tta tion for the half-space. In the special case wh is given by
T- To erfc ( ^ = )
e heat conduction equa- en f ( t ) = 1, the solution
In the above example, we have assumed ε. function whose Laplace transform is 1. The question arises: Is there such a function? We shall
178
CHAPTER 6: INTEGRAL TRANSFORMS
t r y to answer this question in a heuristic manner. Consider the step function H (f) which is defined by
f f ( t ) =
0 for t < 0 for t > 0
Fig. 6.1.
The Laplace transform of H
(f) is H(s)
= -. Then, by Property 4 of the
Laplace transforms, CH'( t ) = sH(s
) = 1. Let us examine H'( t ) closely. Obviously, it vanishes for |f| > 0 and does not exist for t — 0. From the graph of H( t ), it is clear t h a t there is a vertical jump at t = 0 (see Fig. 6.1). Therefore, it is reasonable to assume t h a t lim H'( t ) —> oo. But
t->o
since
1 fs
- / H'( t ) dt = 1, it is obvious t h a t a function like H ’(t) does
^ J—ε
not exist in the classical sense. Such a function is called generalized function or distribution. The function H'( t ) is a generally denoted by 6(t) and is known as the Dirac delta function. This function is defined such t h a t
«5(f) =
0 for |f| > 0 for f = 0,
and
oo
fm
dt
(6.31)
To make this function acceptable in the classical sense, we can mod­
ify the definition as follows:
6.2. BASIC LAPLACE TRANSFO
RMS
This definition is consistent with the classical definition of a function and automatically satisfies Equation (6.31). An important consequence of Equation (6.31) is t h a t if /( f ) is any continuous function, then
/
OO /ȣ
<5(f)/(f) dt = lim / <5(f)/(f) dt = /( 0 ).
-oo
To prove this assertion, we note t h a t by definitic
[ dt = lim [ /( f ) dt = lim
J-oo ^oJ_e2e
e—*o
where t' is a point a t which /( f ) takes its average t h a t t' € (—ε,ε), and, therefore, t' —► 0 as ε —> 0 the Dirac delta function, see §7.1.1.
179
le
2 ef ( t') = /( 0 ),
value in (—ε,ε ), such . For more details on
E x a m p l e 6.7. Consider an infinite slab bou: —oo < y, z < oo, with initial zero temperature: maintained at a constant temperature To and tb tained a t zero temperature. The problem is t o fir the slab for f > 0. Proceeding as in the above in the transform domain is given by Equation boundary conditions in the transform domain we
A + B =
s
nded by 0 < x < /, The face x = 0 is e face χ = I is main- d the temperature in Example, the solution 6.30). Applying the get
and
Aeml + B e
- m l
These two equations yield
_ ~ T 0eml To
2s s i n h ml' s
Using these values in T and simplifying, we find t h a t
ψ = T 0 sinh(/ - x) s sinh I
- A.
R e w r i t i n g t h i s s o l u t i o n a s
180
CHAPTER 6: INTEGRAL TRANSFORMS
and expanding the last factor by the binomial theorem, we get
_ ^0 e —m l m ( l - x) _ e ~ m ( l - x ) ^ j 'y ' g-2n m l
rp °° , v
0 ^ ^ m(2n/+x) g—m[(2n+2)/ — x] J
0
which on inversion yields
T = To f; L l w + _ e r t f i n i ± £ )
Q \ ών rCt π/ί
Alternately, we can use the Cauchy residue theorem and obtain a Fourier series type result. Thus
rrt ' · j r Toest sinh(Z - x)
T = > residues or -----------— 1-------
s sinh/
oo . Ί (6.32)
= T0 j^l - j - ^ — e- n2^ kt/12 sin (η π χ/I ) j. ■
E x a m p l e 6.8. Consider a solid sphere of radius a. Suppose its
initial temperature is zero and its surface is maintained a t a tempera­
ture To for t > 0. The problem is to determine the temperature of the sphere at any subsequent time. The heat conduction equation in this case is
Trr + - T r = l T t. (6.33)
r k
If we introduce a new independent variable u, such t h a t u = r T, then the heat conduction equation reduces to
Urr = \u t, (6-34)
k
which can be solved as in Example 6.2. ■
Ex a mple 6.9. Solve the wave equation
Utt = c ( 6.3 5 )
s u b j e c t t o t h e i n i t i a l c o n d i t i o n s
u = ut = 0, f o r t < 0,
6.2. BASIC LAPLACE TRANSFORMS
and the boundary conditions
u = 0 at x = 0, and ux = T
I f we a p p l y t h e L a p l a c e t r a n s f o r m t o t h e wa v e e q u a t i o n ( 6.3 5 ), we g e t
ΰχχ = c~2s2u.
I t s s o l u t i o n i s
U = A e ~ sx/c + B e sx/c.
A p p l y i n g t h e b o u n d a r y c o n d i t i o n s i n t h e t r a n s f o r m d o m a i n, we g e t
181
a t x
A + B = 0, and - A e ~ sl/c + B
S o l v i n g f o r A and B and substituting their values in the solution for u we get
Tc
, sx smh —
cosh
si '
This equation can be expressed as
Tc f esy - e~sy
esL + e
-sL
/ ’
sl/c = 2Σ
c>2
(6.36)
where y = x/c, and L = Ijc. This, after some m t h a t in Example 6.7, yields
e = - j 5 3 ( - i ) n ( β- ^ 2η+ί ) ι - ^ - e~s^
s o V
which, after inversion, gives
OO
u = Γ β - 1 )" [(i - (2n + 1 ) L + y) H( t - (2n o
(f — (2 n
+ 1 ) L — y) H( t
— (2n
Alternately,
u = Y residues of {ΰ est \
- T [ * - $ ± { - i r
a n i p u l a t i o n s i m i l a r t o
2n+l)L+y]'j
f 1 ) L + y ) ~
+ 1 ) 1 - 2/) ]. ( 6.3 7 )
81_
π 2 ^ (2η + I )2
. (2 η + 1) πχ
s m- -- - -- -——-— c
21
3 S
(2 η + 1 ) π ο ί ι
21 J'
( 6.3 8 )
182
CHAPTER 6: INTEGRAL TRANSFORMS
Dfv = c 2D2v. (6.42)
E x a m p l e 6.10. The partial differential equation for propagation of sound waves produced by the motion of a sphere of radius a in an infinite expanse of fluid is given by
r 2D2u = c2 Dr ( r2 Dr )u, (6.39)
with initial conditions
u(r, 0) = Dt u(r, 0) = 0, (6.40)
and the boundary conditions
Dru(a,t) = f(t), and u —> 0 as r —> oo. (6-41)
We first introduce a new independent variable v = ru. This substi­
tuti on reduces the partial differential equation to the standard wave equation
Applying the Laplace transform and using the second boundary condi­
tion, the solution in the transform domain is given by
v = A e ~ srl\
ΟΓ A
u = - e~sr/c. r
A p p l y i n g t h e f i r s t b o u n d a r y c o n d i t i o n i n t h e t r a n s f o r m d o m a i n, we g e t
ΰ = a C c - s ( r - a )/c
s + - r a
B y t h e c o n v o l u t i o n p r o p e r t y
1ί ^ Γ τ ) = ί e - Ht ~x)f ( x ) d x = <t>(t), k = ~. (6.43)
{s + k) J 0 a
C
T h u s t h e s o l u t i o n i s
W = £ - 1 |-----^ £ _ ^ i f ) e - s ( r - a )/c | t ( 6 4 4 )
w h i c h b y P r o p e r t y ( i i ) a n d E q u a t i o n ( 6.4 4 ) y i e l d s
6.3. INVERSION THEOREM
If /( f ) = S(t), then φ(ί) = e ct^a, and the solutii
We can derive solutions for other values of f i t propriate φ function. ■
o n b e c o m e s
r — a
6.3. I n v e r s i o n T h e o r e m
We will now establish the inversion theorem: Theorem 6.2. I f F( s ) is the Laplace trans;
λο - Η οο
•ι fC-j-lOQ
'(‘>-55j L F W e “
I n o r d e r t o p r o v e t h i s t h e o r e m, we f i r s t s t a t e a n d p r o v e a l e mma.
L e m m a 6.1. I f f ( z ) is analytic and of orde plane ϊ ϊ ζ > ■y, where η and k are real constants
r +i0 n ±
1 ΓΊ'■
f{z o) = /
2m β^οο z0
dz,
PROOF. Consider the rectangle in Fig. 6.2. Choose β > |ύ| and such t h a t Z(j lies in this rectangle. By the Cauchy integral formula
183
(6.46)
by evaluating the ap-
form of fit), then
ds.
(6.47)
:r 0 ( z k) i n the hal f - then
Sftzo > 7-
(6.48)
L
f { z )
dz = 2mf ( z o)
/ρ Z — Zq
w h e r e Γ i s t h e c o n t o u r AB C DA. L e t S denote the contour ABCD, then
m
f J i ± d z = f ^ d z -
J r Z Zq J d ^
I d a z — z q
L
(6.49)
Z - Zq
dz.
184
CHAPTER 6: INTEGRAL TRANSFORMS
Since
we have
ί
Jda z ~ zo Ja
- L
DA z ~ 2 0 l +i 0 f { z )
- i/3 z z 0
dz +
L
AD Z - Z0
f (z)
■ dz,
s Z — Zq
dz = 2ni f ( z o).
( 6.5 0 )
Real Axis
No w c o n s i d e r
L
f ( z )
F i g. 6.2.
dz as β —> cx). Obviously, β —>■ oc implies t h a t
Is z ~ z o
\z\ —» oo on S. Thus |z| > β for points on S. If we take β large enough
z0 1 1 . ..
— I < - implies t h a t
so t h a t β > 2\zo\, then \zo\ < ^ β < \\z\, or
> -. Noting t h a t |/( z ) | < M\z\~h for large 2, we
1 -
zo
get
> 1 -
I £0
1 2
m
0
1
m
<
M
(-?)
<
2 M
βΗ + 1·
I t n o w f o l l o ws t h a t
m
s z ~ z 0
L
dz
<
2M f U I
βΙο+
2 M β fc+!
( l e n g t h o f S)
6.3. INVERSION THEOREM
Thus,
Hence, from (6.50),
/( z )
f Λ
Js z -
Zo
dz = 0.
185
ρ+ίβ
J -1-10
f ( z )
7 —i/3 % ^ 0
o r
F ( s } = _ L r ‘ m
27ΓΖ J-f — ίβ s
dz.
T h e p r o o f o f T h e o r e m 6.2 f o r t h e L a p l a c e t r a n t a r y. B y t a k i n g t h e L a p l a c e i n v e r s e o f b o t h s i d e s o we h a v e
m = l
■ ■'« - s s/
s — ζ 2πί
7 - 1/3
( 6.5 1 )
s f o r m i s n o w e l e me n - f t h e a b o v e e q u a t i o n,
p+ip
/ F( z ) e zt dz. ■
J 7 — 2/3
(6.52)
Lemma 6.2. ///( z ) < C i T *, z =
where Ro,C and k are constants, then I eztf(z)\
provided t > 0, where Γ is the arc B B'C or CA!A of the circular arc with chord A B (Fig. 6.3).
< θ < π,R > Ra, dz —> 0 as R —> 0 0, and R is the radius
P r o o f. C o n s i d e r t h e i n t e g r a l o v e r t h e a r c BB'. Since for B B', we
η
have a = cos-1 , where a is the angle BOB', we get H
Similarly, I ezt f ( z ) dz —> 0 as R —> oo.
JA'A
186 CHAPTER 6: INTEGRAL TRANSFORMS
Fig. 6.3.
Let us now consider the integral over the arc B'C A'. By following the above procedure, we get
[ GZtf(z) Jbb'c
dz
< C R
CR
- k + 1
- k + 1 I —R t sin
2 C R
L
[
I
L
„ R t c o s θ I
άθ
where θ = π/2 + φ
π/2
π/2
- R t s i n
< 2 C R ~ k+l I β~Ηί φ/πάφ
n C R k
+ 1
( e — 1) —> 0 a s R —> oo.
Hence
j eztf(z)
άζ —> 0 as R —> oo, provided t > 0.
2 ρ +ί β
T h i s r e s u l t e n a b l e s u s t o c o n v e r t t h e i n t e g r a l / F ( z ) ι
2 π ί y 7 _j/3
i n t o a n i n t e g r a l o v e r t h e c o n t o u r — Γ.
izt άζ
E xa mple 6.11. E v a l u a t e d 1
6.3. INVERSION THEOREM -ay/a
by
::o n t o u r i n t e g r a t i o n. I f
m = c
-1
-a^/s
then, by the Laplace inversion theorem, we have
m ~ b l.
c+ioc „ -a y/s
e ds.
187
Fig. 6.4.
(6.53)
Consider th e Bromwich contour LABCi CDL, (Fig. 6.4). Then by Cauchy’s theorem
rc+ioo —ay/s r r
= / est ds = / F ( s ) d s + / F( s ) d s +
Jc-ioo s JLD J dC
[ F( s ) ds + [ F(i Jc
1 Jba
Now, as established in Lemma 6.2,
j F(s)ds+ f F(s)ds J ld Jam
w h e r e
-ay/s
) ds +
0,
/ F( s ) ds Jam
188
CHAPTER 6: INTEGRAL TRANSFORMS
The integral over the circle C\ can be easily shown to be <1τΛ. This can be done by taking the radius to be ε and substituting s = eete. On BA, s = t i e", and
0 — dy/ue
in/2+utex
elndu
/•it—> OO -[
I b a = / — i ^ e ~ a
Λ-.0 we
POO i
= / _ e- i aV^- ut du
Jo u
— I —e ui (cos οχ/ΰ — i sin α\/ΰ ) du
Jo u
1 2
= 2 / - e -" ‘ (cosav — isinai>)dv,
Jo V
w h e r e u = v 2. Similarly
Hence
f Z1 0 0 1 2
/ = —2 1 - e ~ v *(cosav + i s i nav) dv.
J cd Jo v
f +/ — « Γ 1,
i d ) ./ΒΛ Vo u
/° ° 1 __ 2,
I n o r d e r t o e v a l u a t e t h e i n t e g r a l / - e v si
Jo v
sin av dv.
s i n avdv, we consider the
integral / e lo
f
J 0
-υ t+iav
> 2 /**
e ~ t' 1 c o s avdv =
Jo
fOO
= $te~a2/4t e - ( v V t - i a/2 V t ) 2 d v
Jo
dv
= 5Re‘
/ 2
e ~ u d u where u = v\/t — i a/2\/t
■ia/2\/t poo p 0
/ e~u2d u + /
Jo J— ia
c - a 2/4 t
Vt
Hence
ί
Jo
' cos av dv =
i/2y/t
V ^ e ~ a2/4t
2-v/i
(6.54)
Integrating both sides of this equation with respect to a from 0 to a, we get
POO 1
/ i e - ^ s i Jo v
s m a v d v =\l ~ f e χ2/4ί dx = erf
4i io 2 2v^
6.4. EXERCISES
189
Thus
£ _ 1
0—αχ/Ί
1
2 π ί
2πί — 4 i — erf — ■= 2 2y/t
=- e r fc
2 f t
( 6.5 5 )
6.4. E x e r c i s e s
6.1. Us i n g t h e t e c h n i q u e s s h o w n a b o v e a n d £ t h e L a p l a c e t r a n s f o r m o f s i n at, cos at, eht si
n at, ebt cos at, t n eht,
a n d s i n h bt.
6.2. S h o w t h a t £
- i | e - a V 5 |
2%/π ί 3
- a /4 t
6.3. U s i n g t h e L a p l a c e t r a n s f o r m m e t h o d, s o l v e t h e p a r t i a l d i f f e r e n t i a l
e q u a t i o n ut = uzz — Mu, given t h a t u(z, 0) lim u(z, t) —» 0 for t > 0. [This problem corn
a viscous fluid on an infinite moving plate under the influence of a
-at
} =
s — a
-, derive
= 0, and u( 0, t) = uq, sponds to the flow of
constant magnetic field applied perpendicular the solution when M = 0.
A n s.
- z s i M
erfc
6.4. Using the Laplace transform method, solve the partial differential
equation in the transform domain ut = uz u( z, 0) = 0, and u ( 0,t) = u0, lim u( z,t ) —»
the solution in the transform domain in the form u = — e ^ 1 +
s
powers of k]. Invert the first two terms of thi^ expansion
. u z kz ( z2
A n s. — = erfc — ■= H — ------
«0 2y/t Μ^/πί V 2i
This problem corresponds to the flow of a viscoelastic fluid on an infinite moving plate. Obtain the exact solutic integrals by using the contour integration
to the plate.] Derive
-Z-F-y/Mt 2 Vt
+ k utzz, given t h a t 3 for t > 0. Expand
I f l
A n s. u( z,t ) = 1 -----------/ — e
ττ Λ x
Xx
X
it
n i n t e r m s of def i ni t e
dx\, where ί = fc. A
190
CHAPTER 6: INTEGRAL TRANSFORMS
6.5. Using the Laplace transform method, solve the partial differential
equation ut — u xx, with the initial condition u( x, 0) = 0 and the
boundary conditions u x (0,t) = 0, and ux ( l,t ) = 1.
» . . , · coshz-v/s
Ans. The solution in the transform domain is u = —- 77.------------= .
s6'z sinh y/s
F i n d t w o d i f f e r e n t i n v e r s e s o f t h i s s o l u t i o n, b y e x p a n d i n g t h e s o l u ­
t i o n i n a s e r i e s o f t h e t y p e s h o wn i n E x a m p l e 6.7 a n d b y t h e r e s i d u e t h e o r e m. T h u s,
OO s
u = Σ \ 2\ί Φ + ( e - {2n+1~x)2/4t + e- ( 2"+ 1+z)2/4i)
n
= 0 *·
, . . 2n + 1 - x 2n + 1 + x
— (2n + 1 — x) e r f c -------- (2n 4- 1 + x) e r f c ------
V 1 2 f t 2f t
a n d
.- £ +.- 1 - 1: S z i f
2 6 η2π 2
n= 1
6.6. Using the Laplace transform method, solve the partial differ­
ential equation uu — uxx, with the initial conditions u( x, 0 ) =
(1 — x ) 2
——, Ut { x, 0 ) = 0, and the boundary conditions u x ( 0,t ) = 1
and ux ( 1, t ) = 0.
1 2 (1 - x)2
Ans. u = —t — ------- —.
2 2
6.7. Usi ng t h e Lapl ace t r a n s f o r m me t h o d, sol ve t h e p a r t i a l di f f e r ent i a l e q u a t i o n u t = u xx, with the initial condition u( x, 0) — 0 and the boundary conditions ux (Q,t) = 0 and u ( l,t ) = 1.
cosli cc /s
A n s. T h e s o l u t i o n i n t h e t r a n s f o r m d o m a i n i s ΰ = ------ -— ■=.
s cosh y/S
F i n d t w o d i f f e r e n t i n v e r s e s o f t h i s s o l u t i o n, b y e x p a n d i n g t h e s o l u ­
t i o n i n a s e r i e s o f t h e t y p e s h o wn i n E x a m p l e 6.7 a n d b y t h e r e s i d u e t h e o r e m.
A n s. u =
0
OO
, 2n + 1 — χ _ 2n + 1 + x e r f c = b erfc
2 f t 2 yft
1)7n ( 2η -I- 1)π
_ -, _ V V _ i xn4cos(2n + l)7rx/2 (2η+1)*π2 ί/4
L> i i h + i w
6.8. Using the Laplace transform method, solve the partial differential
(1 — x ) 2
equation utt = uxx, with the initial condition u(x,0) = -------- ,
6.4. EXERCISES
191
An s.
OO o
u t ( x,0 ) = 0, a n d t h e b o u n d a r y c o n d i t i o n s u ( 0,t ) = 1, u x ( l,t ) = 0.
= Y J{ - 1)n^ { m - 2 n - 2 + x ) + H {t o
00
+ 5 ^ - 2n - 2 + x ) 2H( t - 2n
o 2
+ (t — 2n — x ) 2H( t — 2n — x)] - i ( l
6.9. Using the Laplace transform method, solve equation
« n - u tt = e π t βίηπαί,
u (x,0) = 0, Mt ( x,0 ) = 0, u( 0,t )
An s. Applying Laplace transform to the partial differential equa­
tion, we get (D2 — s 2)u = —:— ^ βίηπχ. Its
- 2 n - x)}
- 2 + x)
x) 2H{t ) -
the partial differential
S + TT2
u = Aesx + Be~sx - -
ux(l,t) = 0.
;ial differentii solution is given by
1
(s + 7t2) ( s 2 + π 2) Applying the boundary conditions we get A
7Γ
s ( Aes — Be ~ s) +
These two equations yield A = - B
( s + 7T2) (s2 + 7Γ2
T h u s,
u = —
2 s(s + 7r2) (s2 + π: π sinh sx 1
!) sinh s'
s(s + n 2) ( s2 + 7Γ2) sinh s (s + 7T2) (s 2 1 1
7T2 ( 1 + 7T2) ( s + π 2) ( 1 + 7T2) ( s 2 + 7Γ
1_______r 1_______ s
2/Ί _|_ π 2\ [ s _|_
π 2(1 + π 2) Ls + π 2 s2 + π
2 4- TT2 Ά2 7Τ2
+
s + 1
π Ι π 2(1 + 7r2)(s + π 2) ' ( 1 + τ r2) (s 2 +7r
οο
[ Σ (e(rc_2fc_1)s - e- ( 2M-i+*)s)j - ο
<■>
i ] si
sin πχ.
Β = 0, and cos π = 0.
sin π χ
+ π 2)
1 ι sinh sx 2) π 25-1 sinhs
sin πχ
1
:) π 2β
ί — -
Ls + π 2
π 2(1 + π 2) Ls + π 2 s2 + π 2
+ ■
s2 + π2
s i n π χ.
192
CHAPTER 6: INTEGRAL TRANSFORMS
On inversion we find
χ f e-7r2(t+z-2fc-l)
π l· 7γ2(1 + π 2) cos(f + x - 2k - 1) + sin(i + x - 2k - 1) 1
- 2 k -
(1 + π 2)
1 ^ r
g-7T2(t-x-2fc-l)
π tt2(1 + π 2)
cos(f — x — 2k — 1) + sin(f — x — 2k — 1) 1 i TJU OJ
+ (1 + 7Γ2)
1 2
re π * — cos π ί 4- π sin πί] sin πχ.
π 2 ( 1 + π 2)
6.10. So l v e t h e d i f f u s i o n e q u a t i o n
u t = a u xx, 0 < χ < π, t > 0,
subject to the boundary conditions u( 0,t ) = 1 — e~* and u( n,t ) = 0 for t > 0, and the initial condition u( x, 0) = 0 for 0 < χ < π.
An s. By the Laplace transform method, we get
d2U s
dx 2 a ’
with 0(0, s) = -r-^-— r, and ϋ( π, s) = 0, which has the solution
v ' s( s + 1 ) v ’
, 1 s i n h J s l a i n — x)
u( x, s) = —---- — -------------- = = --------.
s(s + 1) sin y/s/a n
T h e n t h e i n v e r s i o n f o r m u l a g i v e s
1 [c+i™ est Sinh J s/a ( n - χ) ,
u(X:t) = — v ' v ____ ’- ds,
2m J c_ ioo s( s
+ 1) sinh y s/απ
w h e r e c i s a n y p o s i t i v e c o n s t a n t. A s s u m i n g t h a t a is not of the form n ~ 2, the integrand has simple poles at s = 0,- 1, and —an2, n =
1,2, · · ■. The contour is completed by an infinite left side semicircle with 5fts = c as diameter, which is defined as the limit of a sequence of semicircles Γη t h a t cross the negative s-axis between the poles at
6,4. EXERCISES
193
—an2 and —a( n + l ) 2. The limit of the integral n —» oo. The residue at the pole s = 0 is (π — x
)/
o- t sin [(π - χ )/λΜ
s i n( 7r j y/a)
T h e r e s i d u e a t s = —on2 is given by
ι + n 2 st sinh y/s/a( n - x) s—>an2 s( s + 1) sinh y/s/απ nix
l i m
.1 s m nx
( an2 — 1)
Hence
u ( x,t) = I Z l e - ^ [ ( π - x V V a l + 2 y
π s i n (π J a ) π “
v ' v , n = 1
Note t h a t u
a s t —> oo, which gives t h
ature in the interval 0 < χ < tt.
6.11. Show t h a t
C{t*}
Γ(ρ+1)
s p + i >
where Γ(.τ) is the gamma function defined by An s. In C{ t p\ = [ t p e~st dt, let st = χ. T
3. In C{ t p} = [
Jo
pco i rc
/ t p e~s t dt =
Jo *p+1 Jo
d a r o u n d Γ η i s z e r o a s π, and at s = —1 it is
- a n £
sm n x e_ an2t
n( an2 — 1)
e steady state temper-
f°
T(x) =
Jo
xpe~x dx.
hen
x p+1 e~~x <t
ix =
6.12. Solve the nonhomogeneous Cauchy problem in R 1 x R +:
ut - auxx = f ( x,t ), x e R 1, u( x,0) = g(x), t > 0,
where g( x) is prescribed.
So l u t i o n. Using the Laplace transform we
d2u _
— - s u = - f ( x ),
Γ(ρ+1)
s p +!
get
CHAPTER 6: INTEGRAL TRANSFORMS
with the boundary conditions «.(0, s) = 0 = u(l, s). The solution of the above homogeneous equation is given by
The integrand in this solution has simple poles a t s = 0, — fc2, where fcn = ηπ/Ι,η = 1,2, ■ · · . We choose a contour of integration t h a t avoids these poles and take u > 0. Then it can be shown t h a t the residue a t the pole s = 0 is zero, while a t the poles s = —fc2 it is given by
Hence the final formal solution of the problem is given by
X
By inversion, the solution of this problem becomes
Alternately, if we use the series representation
n=0
for | a: | < 1, then we can write
OO
sinh(ly/s) = 2e~l'/° e'2nl^,
w h e r e ‘iRs is chosen such t h a t e 21 ^ < 1. Then expressing the hyperbolic functions in terms of exponentials, using the formula
6.5. FOURIER INTEGRAL THEOREMS
where z is independent of s and t (see §A.2), orders of summation and integration when solution for III < 1.
and interchanging the needed, we obtain the
F o u r i e r T r a n s f o r m s
We will not discuss the underlying theory of Fourier transforms. We will only define and discuss their properties and applications. From the definitions of th e transform pairs (6.1a,b) and (6 Fourier cosine and sine transforms and their ini
195
2a,b) we note t h a t the verses are symmetric. But the Fourier complex transform and its inverse are related in the following manner: If T f ( x ) = /( a ), then T f ( x ) = /( —a ). Various authors have defined the Fourier transform in different ways, but we
shall follow the notation used by Sneddon (19. Fourier transform pairs is given in Appendix B.
57). A table of basic
6.5. F o u r i e r I n t e g r a l T h e o r e m s
T h e o r e m 6.3 ( F o u r i e r i n t e g r a l t h e o r the Di ri chl et ’s conditions on the entire real line grable on (—00,00), then
\{ f ( x + 0) + f ( x - 0) ] = i - /“ e ~ i ax da J'
T h e o r e m 6.4 ( F o u r i e r c o s i n e t h e o r e m Di ri chl et ’s conditions on the non-negative real integrable on (0,oo), t hen
e m ). I f f ( x ) satisfies and is absolutely inte-
f ( u ) e i audu. (6.56)
. I f f ( x ) satisfies the line and is absolutely
1 2 f ° ° f ° °
~[f(x+0) + f(x — 0)] = — / da f ( u) cos(cra) cos(ax) du. (6.57)
2 ^ Jo Jo
T h e o r e m 6.5 ( F o u r i e r s i n e t h e o r e m ). I f f ( x ) satisfies the Di ri chl et ’s conditions on the non-negative real line and is absolutely
196
CHAPTER 6: INTEGRAL TRANSFORMS
integrable on (0,oo), then
1 2 r 00 /*oo
- [ f ( x + 0 ) +f ( x — 0)] = — / da f ( u ) s i n ( a u ) s m( a x ) d u. (6.58)
2 ^ Jo Jo
I f f ( x ) is continuous then ^ [ f i x + 0) + f ( x — 0)] = f ( x ). These three integrals form the basis of the Fourier transforms.*
6.6. P r o p e r t i e s o f F o u r i e r T r a n s f o r m s
We will use the following notation: Let Tf ( x) = /( a ). Then
(1) Tf [ x — a) = eiaa f (a).
(2) Ff ( ax) = - i -/(a/a).
| a|
(3) Te%axf (x) = f ( a + a).
(4) T j { x ) = f ( —a).
( 5 ) Fx nf (x) =
= ( ^ ) ·
6.6.1. F o u r i e r t r a n s f o r m s o f t h e d e r i v a t i v e s o f a f u n c ­
t i o n. As s umi ng t h a t f (x) is differentiable n times and the function and its derivatives approach zero as |x| —> oo, then it can be easily established t h a t
/dO(a) = ( - i c O/f r - 15,
where f l'p'1 is the Fourier transform of f l'p> (x), which is the p - t h deriv­
ative of f (x) for 0 < p < n.
* S e e S n e d d o n ( 1 9 5 7 ) f o r p r o o f.
then
. PROPERTIES OF FOURIER TRANSFORMS
If lim /^ ( x) = 0, and lim f ^ p\x ) = c. χ —»·οο χ—>0 y 2 ;
/^ = - Cp_ 1 + a/i P - 1)
and
/i p) =
6.6.2. C o n v o l u t i o n t h e o r e m s f or Fou|r: convolution or Faltung of f ( t ) and g(t) over (
OO
1 r°° i r c
l * g = ^ i J l'ri)9(x~'')dri = ^ i i,
,ί ( χ - ν ) 9 ( ν ) ά η. (6.61)
T h e o r e m 6.6. Let f ( a ) andg( a) be the Fovfr; andg( x), respectively. Then the inverse Fourier is
Γ
Λ { f ( a ) g ( a ) } = J f ( v) ai x - v ) d v -
PROOF. C o n s i d e r
/
OO -γ f OC j*
f(v)g(x - V) άη = - j = j f{v) dr] J 1 f °°
/
oo
/( « ) <?(
•oo
a) e~laxda
w h i c h p r o v e s t h e t h e o r e m.
T h e o r e m 6.7. L e t f ( a ) andg( a) be the Fou and g( x), respectively, then
/
OO pOC
f ( a ) g ( a ) d a = / /( -
• o o J —o o
P ro o f. Consider
J f(ot)g(a) da = J f(a) da --L= J
/
OO ι nC
.J iv)dri7 ^ L
/
oo
g( v) f ( - v) dv,
-oo
( 6.5 9 )
( 6.6 0 )
1 9 7
i e r t r a n s f o r m. T h e
oo) is defined by
ier transforms of f(x) transform of f(a) g(a)
g{a)e-l^ x~^da
/
OO
fiv)
-oo
elaxrl άη
rier transforms of f(x)
ii)g(v)dv.
(6.62)
g(η)elar|dη
)
f(a)eiar]da by Property 4.
198
CHAPTER 6: INTEGRAL TRANSFORMS
6.6.3 S o m e F o u r i e r t r a n s f o r m f o r m u l a s.
E x a m p l e 6.12. Find the Fourier transform of f ( x ) = e k^,k > 0.
Tf{x) = f(a) = -== / e~k^ e ixa dx ν 2 π J—oo
i r /*° r°°
= - == / ekxei xadx + e~kxei x adx
v 2 7 T J —
oo Jo
1/1 1 \ k V2
\/2π \k + i a —k + i a j f n { k2 + a 2)
Λγλ/2
No w b y P r o p e r t y 4, t h e F o u r i e r T r a n s f o r m o f /( x ) = /—,,2 2\
γ 7r(rC "f X )
should be /( a ) = e~k^. It is interesting as well as instructive to check if this is the case. Since f { x ) = /( —x), f ( a ) = /( —a ), we have
fcV2 1 /·°° k V2
= — ϊ
λ/27Γ J _
i r
π J -oo (^2 + a:2)
fK(k2 + x2) 7_oo \/^ 2 + z2)
OO
dx
/
O i.„ixa r
oo i.„ixa
dx+ dx
.oc(k2+x2) aX+ J0 (k2 +x 2) aX.
' /* o o j^e ~ i x a ,0 0 ^ r,g i x a
io (.k2 + x2) dx + Jo {k2 + x2) dx.
-u
- H
°° k cos xa
OO i
k cos xo;
(k 2 + x 2)
dx.
The standard method of evaluating this integral is by contour integra­
tion. The contour is the upper half-circle with radius i? and center at the origin if cc > 0 and the lower half-circle if a < 0. Its value is e~ka if a > 0, and eka if a < 0. Thus the value can be expressed as e_fc'“ L
A number of other Fourier transforms can be found by differenti-
k^/2
ating both sides of T e ~k wi t h respect to k. For
y/Tr{k2 + a 2)
p2 k 2 - a 2
e x a m p l e, t h e F o u r i e r t r a n s f o r m o f \x\e~k^ is \j — -
7Γ ( k 2 + a 2) 2
6.6. PROPERTIES OF FOURIER TRANSFORMS
199
( * L o a d t h e p a c k a g e * )
In[l]:=
N e e d s ["C a l c u l u s 1F o u r i e r T r a n s f o r m"];
In[ 2]:=
?$ F o u r i e r O v e r a l l C o n s t a n t
Out[2]=
$ F o u r i e r O v e r a l l C o n s t a n t i s t h e d e f a u o p t i o n F o u r i e r O v e r a l l C o n s t a n t ( a n o p t f o r m a n d r e l a t e d f u n c t i o n s ).
In[3]:=
$ F o u r i e r O v e r a l l C o n s t a n t
Out[3]=
1
( * R e s e t t h e c o n s t a n t * )
In[4]:=
$ F o u r i e r O v e r a l l C o n s t a n t = 1/S q r t [ 2 P
Out[4]=
1
2 S q r t [ P i ]
t s e t t i n g f o r t h e i o n t o F o u r i e r T r a n s -
In[5]:=
200
CHAPTER 6: INTEGRAL TRANSFORMS
F o u r i e r T r a n s f o r m [ E x p [ - k A b s [ x ] ],x,w ]
O u t [ 5 ] =
r 2
k S q r t [ — ]
P i
( * A l t e r n a t e c a l c u l a t i o n; t a k e s t i m e * )
In [6]:=
i n t l:= I n t e g r a t e [ E x p [ k χ ] E x p [ I x a ],x,- I n f i n i t y,0 ] i n t 2:= I n t e g r a t e [ E x p [ - k χ ] E x p [ I x a ],x,0,I n f i n i t y ]
In[8]:=
r e s u l t = 1/S q r t [ 2 P i ] ( i n t i + i n t 2 )//S i m p l i f y
G e n e r a l::i n t i n i t: L o a d i n g i n t e g r a t i o n p a c k a g e s — p l e a s e
w a i t.
O u t [ 7 ] =
k S q r t [ — ]
P i
k^ + a?·
Example 6.13. Find the Fourier transform of f ( x ) = e kx . Then Ί Γ°°
/» = - =/ e~kx2 eixa dx V J — oo
I
g — k(x2 — ixoc/k — a2/4k2 a2/4k2)
\/2 π J —qq i r°°
_ / e-k{x-ia/k) -a /4k
\/2π J —oc
1 r°°
e~OL2/4k
/ e - u 2 du
J — oo
y/2nk
1 e - a2>i k f
a2/4fc
6.6. P R O P E R T I E S O F F O U R I E R T R A N S F O R MS
y/2irk
1
y p 2 k
201
In[l}:=
N e e d s ["C a l c u l u s'F o u r i e r T r a n s f o r m"];
In[2]:=
F o u r i e r T r a n s f o r m [ E x p [ - k t 2 ~ ],t,w ]
Out[2]=
1 _____________
S q r t [ 2 ] E w2/( 4 k ) g q r t Μ
E x a m p l e 6.1 4. F i n d t h e Four i e r t r a n s f o r m a n d f ( x ) = x e ~ax for x > 0.
of f ( x ) = 0 for x < 0
1 r°°
Tf(x) = - = / f ( x )
V27T J -oo
1 Γ
m h
Jo -ax+iax
eiax dx
Xe-axeiax dx
xe
λ/2Ϊτ{ία — a) o y/2ir (i a — a)
1—/
ia - a) Jo
\ί^(%α 1
1_____ Γ
i a - a) Jo
e-ax£iax άχ
\/2 π ( ί α — a) 2
e- axelax dx
202
CHAPTER 6: INTEGRAL TRANSFORMS
E x a m p l e 6.15. Find the Fourier transform of f ( x ) = 0 if x < b and f ( x ) = β_α2χ2 if 0 < b < x.
I f 00 2
- J L · f
Jb
OO
2et x adx
- a2 ( x - i a/2 a 2) 2- a 2/4 a 2 ^
ao
p—ct/Aa
e~u~ du
/
α\/2π
( a b —i a/2 a )
1 :e~a2t i a* e r f c f ab — — ] .1
2a\/2 V 2a
« D e c l a r e .m
O u t [ l ] =
{ D e c l a r e, N e w D e c l a r e, N o n P o s i t i v e, R e a l Q }
I n [ 2 }:=
D e c l a r e [ a,P o s i t i v e ]; f [ x _ ]:= E x p [ - a ~ 2 x ~ 2 ]
i n t:= I n t e g r a t e [ f [ χ ] E x p [ I x a l p h a ],{ x,b,I n f i n i t y } ] s u b [ X _ ] := 1 - E r f [X] - > E r f c [ X//T o g e t h e r ]
S i m p l i f y [ 1/S q r t [ 2 P i ] i n t ]/.s u b [ a * ( ( - I/2 * a l p h a )/a ~ 2 + b ) ]
O u t [ 6 ] =
- I a l p h a + 2 a2 b n
E r f c [ - - - - - - - - - - - - - - - - - - - - - - - - - - - - ]
_______________ 2 a __________
2 S q r t [2] a E a l P h a 2/( 4 a 2 )
E x a m p l e 6.1 6. Solve t h e p a r t i a l di f f e r ent i a l e q u a t i o n ux + uy + k y u = f ( x ), in the domain |x| < 0, y > 0, with the boundary conditions
6.6. PROPERTIES OF FOURIER TRANSFORMS
203
u(x, 0) = 0, lim u ( x,y ) = 0, where f ( x ) is
X—*±<30
f(x) -> 0 as |x|
The partial differential equation in the domain of the Fourier trans­
form with respect to x is
ΰ(α, y)y + ( - i a + ky) u(a, y) =■
w h e r e u = T u and f ( a ) = T f. The differential equation of first order and its solution subject conditions is
iquation in ΰ is a linear to the given boundary
ΰ = eiay- ky
/2f(a) /
Jo
-i at+k
T h i s s ol ut i on on i nversi on yi el ds
u = e-ky >/2 rV
[' f ( χ - y + t)<M
Jo
N o t e t h a t b y u s i n g P r o p e r t y 1, T ^ 1 f ( a ) e ~ iai't
E x a m p l e 6.1 7. F i n d t h e s o l u t i o n o f t h e Le
Uyy = 0 in the domain |x| < oo and y > 0, w u —> 0 as |x| —> oo or as y —> oo and u( x, 0) = δ Fourier transform to the partial differential equ. we get
a2u = 0.
*yy
The appropriate solution is
= Ae ~ My.
A p p l y i n g t h e b o u n d a r y c o n d i t i o n a t y = 0 in we get u( a, 0) = A = - , Hence ΰ = , : e
o b t a i n
\/2π
u( x,y) = -
1 V
t i l
π x 2 + y2
We can now use Duhamel’s principle to obtain lem with arbitrary condition u(x, 0) = f ( x ) · T l
yf(v)
1 f °
i ( x,y) = - / π J- i
(x - η)2 + y
w h i c h i s k n o w n a s t h e P o i s s o n i n t e g r a l r e p r e s e n p r o b l e m i n t h e h a l f - p l a n e. ■
a f u n c t i o n s u c h t h a t
/( a ).
( 6.6 3 )
( 6.6 4 )
2/2 dt. (6.65)
y) - f ( x - y + t). ·
.place’s equation uxx + t h the conditions t h a t x ). After applying the i t i o n with respect to χ,
the transform domain On inverting, we
e solution to the prob- \en the solution is
dr?, (6.66)
t a t i o n f o r t h e D i r i c h l e t
204
CHAPTER 6: INTEGRAL TRANSFORMS
E x a m p l e 6.18. Find the general solution of the diffusion equation Ut = k uxx for the homogeneous initial condition u( x, 0) = f { x ), sub­
ject to the conditions t h a t lim f ( x ),u ( x,t ) —> 0. Applying Fourier
x —► i o o
transform to the partial differential equation we get
u -f k a 2u = 0,
w i t h t h e i n i t i a l c o n d i t i o n u( a, 0) = f ( a ). Hence, the solution after applying the initial condition is
ΰ(α,ί) = /( a ) e -fe"2t,
which on inversion yields
Ί ί °°
w h e r e we h a v e u s e d t h e c o n v o l u t i o n p r o p e r t y a n d E x a m p l e 6.1 3. ■
6.7. F o u r i e r S i n e a n d C o s i n e T r a n s f o r m s
We s h a l l d i s c u s s F o u r i e r s i n e a n d F o u r i e r c o s i n e t r a n s f o r m s i m u l t a n e ­
o u s l y. T h i s i s b e c a u s e m o s t o f t h e t i m e t h e y a r e u s e d s i m u l t a n e o u s l y t o s o l v e a n y p r o b l e m.
6.7.1. P r o p e r t i e s o f F o u r i e r s i n e a n d c o s i n e t r a n s f o r m s.
Hx ) = f c{a), f s f ( x ) = f s ( a), (6.67)
Fjc(x) = /(<*), Fs f s ( x) = /( a ), (6-68)
TJ(kx) = ^/ο φ, k > 0, (6.69)
Tsf{kx) = i/,φ, k > 0, (6.70)
a + b
6.7. FOURIER SINE AND COSINE TR.
Fcf(kx) c o s i i = ^ [/c, k
F cf ( k x ) s mb x = ^ + t
T
f{kx)cosbx = ^\} s
Fsf{kx) s i n bx = [/c
a + 6
a — b
k
+ f c - f s + f s -fc
6'
A N S F O R M S 2 0 5
k
-b
k
d — b
k
+ b
T c x 2 n f ( x ) = (-1) Tcx2n+1f{x) = (-1) Fsx 2nf{x) = (-1) T sx 2n+1f ( x) = (~1)”+1
d2nf c(a)
da2n
d 2 n + 1 f s (
da2n
d2nf s(a)
da2n
d2n+l
)]■
)]■
)]·
)]■
k > 0,
k > 0,
k > 0,
k > 0,
a)
c(a)
άα2η
6.7.2. Convolution theorems for Fourjier sine and cosine transforms.
T h e o r e m 6.8. Let f c(a) and gc(a) be the forms of f ( x) and g(x), respectively, and let f Fourier sine transforms of f (x) and g(x), respec
fc
1 f ° °
Λ 1ίο{α)()ο{α)\ = ~η= / g( v) [f (\x ~ v\)
V ^ T T J o
P r o o f. W e h a v e
c o s a:
c o s ax
J
f ° ° ~ 2 f ° °
f c{a)gc{a) cos axda = —= / f c(a) ο νπ Jo
2 Ζ* 0 0 f ° °
= V* Jo 9^ dV Jo
1 2 /O C
= 2 Vn J0 9^ dr]J0 fc(a)[cosa(\x - η\) i r°°
= 2
yo g(v)[f (\x- v\) + f ( x + '.
( 6.7 1 )
( 6.7 2 )
)
( 6.7 3 )
’ ( 6.7 4 )
( 6.7 5 )
( 6.7 6 )
( 6.7 7 )
( 6.7 8 )
Fourier cosine trans- la) and gs(a) be the tively. Then
f ( x + η )\d η. ( 6.7 9 )
rOC
:da / g(η) cos αη άη Jo
c o s αηάα
+ c o s a ( a; 4 - η)} da /)] άη. m
T h e o r e m 6.9.
/>X r OC
/ f c { a ) g s { a ) s i n a x d a = - 9 { η ) [ ί (\χ - η\) ~ ϊ ( χ + η ) } ά η,
( 6.8 0 )
a n d
rOO -j^ rOC
/ f s ( a ) g c ( a ) s i n a x d a = - ί ( η ) [9 (\χ - η\) - 9 { χ + η ) } ά η.
( 6.8 1 )
T h e o r e m 6.1 0.
pOO pOO pOO
/ f c ( a ) g c { a ) d a = f { v ) g { v ) d ( v ) = f s { a ) g s ( a ) d a. (6.82) Jo Jo Jo
T h e p r o o f s f o r t h e s e t h e o r e m s a r e l e f t a s e x e r c i s e s.
We wi l l n o w d e r i v e F o u r i e r s i n e a n d c o s i n e t r a n s f o r m s o f s o me f u n c ­
t i o n s.
E x a m p l e 6.19. Define
p oo poo
h = e ~ ax s i n b x d x, h = e ~ ax cos b x d x.
Jo Jo
T h e n /2 = — — - i ’i, a n d I\ = —f y. Solving for /1 and I 2, we get a a a
h = 9 b ,9, and I 2
2 0 6 C H A P T E R 6: I N T E G R A L T R A N S F O R M S
a 2 + b2 ’ a 2 + b2
If /( x ) = e-6*, then
/«=(“ ) - { I a 2 + b2'
a n d
12 a
Λ (a)
π a 2 + b2 These two results yield on inversion
/
Jo
c o s a x _ π _ h_
. da = —e ,
a 2 4- b2 26
6.7. FOURIER SINE AND COSINE TRANSFORMS
207
a n d
f
Jo
a sin α χ , π i j ο , ,2 d a = - e 6x1 a 2· + b2 2
An interesting integral is obtained by defining
m = { J
f o r 0 < x < q. 0 for 6 < x.
T h e n f c ( a ) — J — -~n a'\ Also, define g ( x ) = e V π a
2 a
5c(«) = A/-
π a 2 + a 2 ’
Thus, we have
/ f c ( a ) g c ( o i ) d a = - -
J o π J o a
s i n ab
da =
= ί ε - α η ά η = Jo
E x a m p l e 6.20. Show t h a t
a2 + a2
1 - e~ab
\ Then
POO
I f(v)g(v)dv
foo 1
Jo A2S1
si n Xa sin XbdX = — min 2
I n f ac t, i f we defi ne /( * ) :
t h e n
1 for 0 < x < a 0 for a < x
9(x) =
pOO p oo
/ f c { X ) g c ( X ) d X = / f ( x ) g ( x ) d x.
Jo Jo
T h i s, o n u s i n g ( 6.8 3 ), y i e l d s
2 f ° ° 1 .
— / sin Xa sin Xb d X = / f ( x ) g ( x ) d x = min(a, b). \
π Jo λ Jo
( 6.8 3 )
1 f o r 0 < x < b 0 for b < x,
208
CHAPTER 6: INTEGRAL TRANSFORMS
EXAMPLE 6.21. Find the Fourier cosine transform of f ( x ), where
f { x ) :
x for 0 < x < 1,
2 — x for 1 < x < 2, 0 for 2 < x < oo.
Here,
W i * ) ] = y - - ΐ/ϊ [
[/„
1
x cos x d x -\-
π
2 r x s i n a x
a
1 i Z-1 .
/s i
0 a 7o
si n a x d x
x ) c o s x d x
( 2 — x ) s i n a x 2
a
— [ sin a x d x 1 a J i -1
2 cos a — 1 — cos 2a
I n [ l ]:=
N e e d s [
"C a l c u l u s ‘ F o u r i e r T r a n s f o r m ‘","A l g e b r a'T r i g o n o m e t r y ‘"]; I n [ 2 ]:=
C l e a r [ f ];
f [ x _ ]:= I f [ 0 < x < 1, x ,I f [ 1 < x < 2, 2 - x ,0 ] ]
I n [4]:=
F o u r i e r C o s T r a n s f o r m [ f [ t ],t,w ]
O n::n o n e: M e s s a g e S e r i e s D a t a::c s a n o t f o u n d.
O n::n o n e: M e s s a g e S e r i e s D a t a::c s a n o t f o u n d.
O n::n o n e: M e s s a g e S e r i e s D a t a::c s a n o t f o u n d.
G e n e r a l::s t o p: F u r t h e r o u t p u t o f O n::n o n e
w i l l b e s u p p r e s s e d d u r i n g t h i s c a l c u l a t i o n.
O u t [ 4 ] =
6.7. FOURIER SINE AND COSINE TRANSFORMS F o u r i e r C o s T r a n s f o r m [ f [ t ],t,w ]
I n [ 5 ]:=
i n t l:= I n t e g r a t e [ x C o s [ a l p h a x ],{ x,0 i n t 2:= I n t e g r a t e [ ( 2 - x ) C o s [ a l p h a χ ],
r e s u l t = S q r t [ 2/P i ] ( i n t i + i n t 2 )//S i m p l i f y
O u t [ 7 ] =
2 a l p h a o
4 S q r t [ — ] C o s [ a l p h a ] S i n [ ]
P i 2__
a l p h a z
209
1}]
x.1,2 } ]
Example 6.2 2. We s ha l l now use t h e Four i e r t r a n s f o r m t o sol ve Exa mp l e 6.7, whi ch was e a r l i e r sol ved by t h e La pl a c e t r a n s f o r m. T h e p a r t i a l di f f e r ent i a l e q u a t i o n al ong t h e b o u n d a r y a n d i n i t i a l c ondi t i ons a r e
k Uxx — Xit ,
u = 0, u —> 0 as x —> 0 for t < 0, u == To for t > 0.
Applying Fourier sine transform to the partial differential equation, we get
+ k a 2u s = k a T 0,
a t
w h e r e u s is the Fourier sine transform of u. Its ΰ, = A e k a h -
2 To
π a
B y a p p l y i n g t h e i n i t i a l c o n d i t i o n a t t = 0, w e g e t
A + J - — = 0. V 7Γ a
He n c e
vf?
s o l u t i o n i s g i v e n b y
210
CHAPTER 6: INTEGRAL TRANSFORMS
T h u s, u ( x,t ) i s g i v e n b y
π Jo a \
6.8. F i n i t e F o u r i e r T r a n s f o r m s
W h e n t h e d o m a i n o f t h e p h y s i c a l p r o b l e m i s f i n i t e, i t i s g e n e r a l l y n o t c o n v e n i e n t t o u s e t h e t r a n s f o r m s w i t h a n i n f i n i t e r a n g e o f i n t e g r a t i o n. I n m a n y c a s e s, f i n i t e F o u r i e r t r a n s f o r m c a n b e u s e d w i t h a d v a n t a g e. We d e f i n e
a s t h e f i n i t e F o u r i e r s i n e t r a n s f o r m o f f ( x ). The function f ( x ) is then given by
( 6.8 4 )
( 6.8 5 )
Similarly, the finite Fourier cosine transform is defined as
(6.86)
a n d t h e i n v e r s e is
6.8. FINITE FOURIER TRANSFORMS 211
Example 6.23. Consider the Laplace equation for a rectangle
Uxx Uyy -- 0,
wi t h t h e b o u n d a r y condi t i ons
ω(0> y ) = u ( a, y ) = u ( x: b) = 0, and
Applying th e finite Fourier sine transform to u ( from 0 to a, we have
(6.88)
u(x, 0) = f ( x ).
x,y ) with respect t o x
Γ'
Jo
( u s ) r a ( n) = I U x x S i n ^ ^ - d x
η π
u ( 0,t ) - ( - l ) n u ( a, i)] -
u s {n,y) = 0.
Then Equation (6.8 8) becomes
d2 η 2π 2 dy 2 a2
Solving for u s ( n,y ), we get
us (n, y) = A e ™ y/a + B e ~ n ^/a.
S i n c e u s ( n,b ) = 0, we can express u s ( n,y ) as
us (n,y) = A n {e™(y- b)/a - e - n^ ~ b)/a).
Applying the boundary condition at y = 0, we get
Αη(ε~ηπΙ>/'α — envh/a) = f s (r
w h i c h, a f t e r s o l v i n g f o r A n and substituting its value in u s ( n, y ). yields
s i n h [ n 7 r ( y — b)/a]
u s ( n,y ) =
He n c e,
w h e r e
u ( x
■ » Κ Σ
s i n h ( n 7 r b/a 2 s i n h [ n7r(& — y )/a\
f s ( n
( 6.8 9 )
■u3 ( n ).
n),
a ^ sinh(n7r&/a)
M n ) = ί /( £ ) sin(n7r£/a)d£. Jo
s i n ( η π χ/a ),
212
CHAPTER 6: INTEGRAL TRANSFORMS
E x a m p l e 6.2 4. S o l v e t h e w a v e e q u a t i o n
u t t = c 2 u x x, 0 < x < l, subject to the conditions
u ( 0,t ) = g { t ) and u ( l,t ) = 0 for 0 < χ < I, u ( x, 0) = 0 for t > 0.
Taking the finite Fourier sine transform with the boundary conditions at x = 0,1, and using ( 6.8 9 ), we get
717TC2
us =
dt 2 I2 6 I
T h e g e n e r a l s o l u t i o n o f t h i s e q u a t i o n i s
. . , n - K c t „ . m r c t _ . .
u s { n ) = A cos — V B sm — h ws,p(n),
where u Slp ( n ) is the particular solution which can be obtained by the variation of parameters as
/x \ n - K c i t — t)
s,p { n ) = c J g { r ) s m d t
= c [ g ( t — t ) s i Jo
η π ο τ sin —:— d r.
W i t h t h i s c h o i c e o f u s>p ( n ), the constants A and B become zero because of the initial condition u s ( n, 0) = 0. Hence u s ( n ) = u Stp ( n ). Thus, by
( 6.8 5 )
n 00 rt
, . 2c v—v . η π χ j , . . η π ο τ ,
u( x,t ) = — > s i n —- / g ( t - τ ) sm —— d r.
1
n=l 1 J° 1
Alternately, by applying the Laplace transform, we have
d2u s 2 _
which, when solved with the boundary conditions u(0, s ) = g ( s ), u ( l, s ) = 0, gives
N , sinh{ s ( l - x )/c )
u ( x, s ) = g ( s ) . . , , ■;-----.
v sm ( s l/c )
6.9. EXERCISES
213
This can be inverted for different values of g ( t ) i.e., as s —> 0, we get
U{x,s) = g ( s ) e ~ sx/c,
which on inversion gives
u(x, t ) = H ( t — x/c ) g ( t — x/
T h i s s o l u t i o n a l s o f o l l o ws f r o m d ’A l e m b e r t gen t h e wa v e e q u a t i o n. ■
N o t e t h a t t h e f i n i t e c o s i n e t r a n s f o r m s f o r t h t i o n u can be obtained analogous to (6.89) (see
N o t e t h a t a s t —*■ oo,
c).
eral solution (5.24) of
3 derivatives of a func- Exercise 6.23).
6.9. E x e r c i s e s
6.13. F i n d t h e c o m p l e x F o u r i e r t r a n s f o r m o f o f t h e f o l l o w i n g f u n c t i o n s:
, f 0 f o r x < 0
(a) f { x ) = \_ ax
y e a x f o r x > 0.
A n s. — f - L 1------------.
y/2 π (a — ia)
J" 0 f o r x < 0,
(b) f ( x ) = I l
— e a x - e f o r x > 0 a n c b > a > 0. y x
H i n t: U s e p a r t ( a ) a n d i n t e g r a t e w i t h r e s p e c t t o a.
An s.
\/2π
I n-
i a
a — i a
0 f o r I xl > a,
( c ) f ( x ) =
An s.
f o r | x | < a.
r ( l — c o s aa).
( d ) f ( x ) = cos ax 2 and f ( x ) = sin ax2
Hint: Use Example 6.13 and define k as ia.
λ 1 / . α2 α2λ 1
An s. — = sin - — cos — and — -=
2 y/a V 4a 4a J 2 y/a
f o r I d < 1
s n
( e ) f ( x ) =
A l
f o r I d > 1.
a 2 a 2
11 ~A c o s ~7~
4 a 4 a
214
CHAPTER 6: INTEGRAL TRANSFORMS
V 2
Ans. sin a.
s i n f c r f o r | x | < 1
f o r | x | > 1. s i n( f c — a ) sin(fc + a )
Oiy/π
(f ) f ( x ) = { “
An s. —
y/2n
(g) /( * ) =
k — a 1
k + a
f or x > 0
\fx
= 0 for x < 0
Hint: Substitute a x
Ans
0 « 7γ/4
v 2 e171/2.
f o r x < 0
Ans.
e — ί π/4
>/R
0 f o r x > 0. a > 0.
(i) f ( x ) =
An s.
1
>/R'
\/M'
6.1 4. So l v e t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n (a2D 2 — D 2) u = 0, subject to the conditions u( x, 0) = f ( x ) + g( x) and D t u( x, 0) = a ( f'( x ) — g'( x) ), where u( x,t ), /( x ), g( x), u'( x,t ), f'( x ), and g'( x) all go to zero as |x| —> oo.
Ans. u = f ( x + at ) + g( x — at ).
6.1 5. S o l v e t h e f o l l o wi n g s y s t e m: D t u — v = —D x h, D t v + u = 0, D t h + c 2 D x u = 0, subject to the conditions u(x,0) = v ( x,0 ) = 0 and h ( x, 0) = k if |x| < a and /i(x, 0) = 0 if |x| > a; in addition u ( x,t ),v ( x,t ) and h ( x,t ) approach 0 as |x| —> oo. This problem is connected with shallow water waves.
Ans.
h{x,t ) = 2Ji Γ ϋ Ξ ί 2 2 > [
^ J — oc &
in(aa) r l + a 2c 2 c o s ( i\/l + a 2c 2 )
1 + a 2 c 2
c o s ( a x ) dx.
6.1 6. S o l v e t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n r 2D 2u = c?Dr ( r 2D r ) u, with initial conditions u( r,t ) = Dt u ( r,t ) = 0 for t < 0 and the boundary condition D r u( a,t ) = H( t ) f ( t ), by using Fourier tr ans­
form with respect to t.
6.9. EXERCISES
A n s. See Example 5.10. H i n t. T x -— ,
ia — b
piak(r — a )
Τ ~ λ — = —\/2 π H i t — k ( r — a ) ] e ~ b ^ ~
i a — b lution theorem.
= — \/2πH( t ) e bt and fc(r_a)l. Use the convo-
6.17. Find the Fourier sine and cosine transfor
A n s. T cf{x) = Ts f {x) =
Va + or
ms of f ( x ) = —7=e ν X
yj V a2 + a2 — a
6.18. Find the Fourier cosine transform of — [e
x
A n s. — ln
2 π V b2
a2 + a2
a
1
6.1 9. F i n d t h e Four i e r si ne t r a n s f o r m of — e °
x
. [2 _i Q
A n s. \ — ta n —.
V π a
6.2 0. Der i ve a d d i t i o n a l f or mul as for Four i e r si n f rom Ex a mp l e s 5.5 a n d 5.6 by di f f er ent i at i: r e s p e c t t o a.
6.2 1. F i n d t h e F o u r i e r s i n e a n d c o s i n e t r a n s f o r: a > 0.
λ ϊ / \ V V a 2 + a2 + a
A n s. f c(a) = ----- —, and f s(a)
V a 2 +
6.2 2. F i n d Four i e r si ne a n d cosi ne t r a n s f o r ms 0.
^ 3i arctan( a/a )
ANS./l ( a ) + i/,( a ) - 2 ^ + <>2) 3/1.
H i n t: fc(a)+ifs(a) = ; then express (a — ia) in polar
form.
λ/2(a — ia)3'
215
Va2 +
a *
ax — e~bx], Vt a,m > 0.
, SRa > 0.
a n d cosi ne t r a n s f o r ms ng or i n t e g r a t i n g wi t h
ms of f ( x ) = —j=e.
y/X
W a ·
2 + a 2 — a
o f f ( x ) = yf xe ax, a >
6.2 3. D e r i v e t h e f o r mu l a s f o r t h e f i n i t e s i n e a n d c o s i n e t r a n s f o r m s f o r t h e f i r s t, s e c o n d, a n d t h i r d d e r i v a t i v e s o f a f u n c t i o n u( x) in the interval [0,/].
216
CHAPTER 6: INTEGRAL TRANSFORMS
Ans. For the finite sine transform ' d u\ η π
dX/, , *c(").
is gi ven by (6.89),
For t h e f i ni t e cosi ne t r a n s f o r m
(£) = [ ( - 1 ) Μ 0 - « ( 0 ) ] + ^ ( η ),
( S ) = [(- i)rv(° - u,(o)] - τ “c(n
( s ) c = [(-1)v'(')-u>)]-!^[(-
- ^ ΰ,( η ).
uc(n).
l )"u ( 0 - u ( 0 ) ]
7
Green’s Functions
The solution of a given linear partial differential eqi source in the region under consideration subject to conditions is generally called a Green’s function, to generate solutions for the partial differential eqi of boundary conditions and internal sources. This portance in a variety of physical problems. For the functions one can assume the presence of an internal ary condition which results in the same effect as the source, represented by the Dirac delta function, beloi known as generalized functions or distributions. W first study certain elementary aspects of the distributi and construction of Green’s functions for different problems are carried out through various examples
| www | Refer to the Mathematica Notebook Green
lation due to a unit point homogeneous boundary This solution enables us lation subject to a range echnique is of great im- derivation of the Green’s lource or a certain bound- point source. The point igs to a class of functions th this in mind, we shall on theory. The definition types of boundary value ind exercises.
s. ma for this chapter.
7.1. D e f i n i t i o n s
Let R n denote the Euclidean n-space, and R + th numbers. Then | x — y\ defines the Euclidean distar y in R n. An open ball of radius r centered at a poi { x : \x — x 0\ < r }, and denoted by B ( x 0,r ). T the open ball B ( x 0,r ) will be denoted by S ( x Q,r
: s e t o f n o n n e g a t i v e r e a l ice b e t we e n p o i n t s x and it x q 6 R n is defined by ie boundary (surface) of
= { x : \x — x 0| = r},
218
CHAPTER 7: GREEN’S FUNCTIONS
is the surface area of the unit ball in R n. The e —
n e i g h b o r h o o d o f a s e t A c R n is Α ε = U x € a B ( x,s ). The complement of a set B with respect to a set A will be denoted by A\B, the product of the sets A and B by A χ B, and the closure of a set A by A. The characteristic
A complex-valued function / is said to belong to the class CP(Q) if it is continuous together with the derivatives D k f ( x ), |fc| < p, 0 < p < oo, in a domain Ω. The function / in the class Cp(fl), for which all derivatives D k f ( x ), |/c < p, admit continuous continuations in the closure Ω, form the class of functions CP(Q). The class C°°(f l ) consists of functions / which are infinitely differentiable on Ω, i.e., continuous partial derivatives of all orders exist. These classes are linear sets; thus, every linear combination X f + μ g, where Λ and μ are arbitrary complex numbers, also belongs to the respective class.
A function defined on R n is said to belong to the class C ^ { R n ) if it is infinitely differentiable on R n and vanishes outside some bounded region.
The s u p p o r t of a continuous function / (written supp /) is the closure of the set { x € R n : f i x ) Φ 0 }. Then the class (R n) denotes the set of functions in C p { R n ) that have compact support.*
The Hilbert space L 2 [a, bj is a complete inner product space with the norm defined by
Complex-valued functions which form a complex vector space become an inner product space if we define
(or indicator) function of the set A is defined by x a { x ) =
and is obtained from the inner product defined by
*A complex-valued function / on R n is said to have compact support if there exists a compact set K such that /(x) = 0 for each x not in K.
7.1. DEFINITIONS
219
where t e [a, b] is kept real, and the bar denotes the norm is then defined by
\ 1/2 \m\2 d t\
s i n c e f { t ) f ( t ) = \f { t )\2.
f
J a
We s h a l l n o w d i s c u s s s o me b a s i c c o n c e p t s and f u n c t i o n a l s, g e n e r a l i z e d f u n c t i o n s, a n d d i s t r i b u t i o n s r e s u l t s c a n b e e a s i l y f o u n d i n t h e l i t e r a t u r e ( s e e Fr i e dl a hi S h i l o v ( 1 9 6 4 ), Ky t h e ( 1 9 9 5, 1996), Ru b i n s t e i n ( 196S)
D e f i n i t i o n 7.1. Le t a r e a l n u mb e r
/ /C
JRn
:) φ ( χ ) ά χ = (/,</>) be
associated with each x 6 R n for every function said to be a f u n c t i o n a l on R n. The function φ is kno
£ C0°° ( R n ). Then / is nvn as the t e s t f u n c t i o n.
Th u s, e.g., t h e F o u r i e r s e r i e s o f / e C 1 [0, π], defi
2 f
f i x) = V bn s i n n x, where bn — — /
π J o
i s a f u n c t i o n a l o n R 1, with test functions in the set { useful properties of test functions are
si
(i) if φ ι ( χ ) and <fa(x) are test functions on R n, so is β ι φ ι ( χ ) + C2 0 2<x), where c\ and C2 are real nu:
(ii) if φ( χ) G Cq° ( R n ), so do all partial derivative class Cq° ( R n )·,
( i i i ) i f ^ ( x ) € C q° ( R n ) and a(x) are infinitely differentiable, then the product a( x) 0(x) belongs to the class C°° (R n ); and
(iv) if φ (xi, ■ ■ ■ ,x m ) G C%° ( R m ) and φ ( i m+i, · then φ ( x i, ■ ■ · , xm) ‘φ (xm.^i, · · ■ , xn ) £ C q (.
D e f i n i t i o n 7.2. A functional / on R n is said β φ'2 ) = λ (/, φ ι ) + μ ( }, φ2 ) for all real numbers Λ, μ
m m
No t e t h a t (/, 0 ) = 0, a n d (/, ^ α η φ η ) = ^ a n {.
71 — 1 7 1 = 1
o m p l e x c o n j u g a t e. T h e
results from the theory Proofs for most of the der (1982), Gelfand and , and Stakgold (1979)).
ned by
f { x ) sin n x d x, nx,sin2x, · · · }. Some
their linear combination riibers;
s of 0(x) belong to the
,x n) e C Z ° ( R n ~ m ), R n ).
t o b e l i n e a r if (/, λ φ ι +
:ui daU0i i2 e C ^ { R n ). >>■
220
CHAPTER 7: GREEN’S FUNCTIONS
D e f i n i t i o n 7.3. A l i n e a r f u n c t i o n a l / o n C % ° ( R n ) i s s a i d t o b e c o n ­
t i n u o u s i f t h e n u m e r i c a l s e q u e n c e {/, <pm ) —» 0 a s m —> o o, w h e r e {<^>TO( x ) } i s a n u l l s e q u e n c e i n C°° ( Rn ), i.e., s u p p <pm, m = 1,2, ■ · ·, i s c o n t a i n e d i n a s u f f i c i e n t l y l a r g e b a l l, a n d l i m m a x D fc0 m ( x ) | = 0 f o r e v e r y m u l t i - i n d e x
m —*oc> y ( = Pn
k, |fc| < n.
D e f i n i t i o n 7.4. A c o n t i n u o u s l i n e a r f u n c t i o n a l / o n C £ ° ( Rn) is said to be a d i s t r i b u t i o n. The number (/, φ ) is called the value of / at φ, or the action of / on φ.
Th e s p a c e V of all distributions on Cq° ( Rn) is a linear space. A locally integrable function /( x ) in R n generates an n-dimensional distribution / such that for all φ e C £ ° (R n )
Hence, every locally integrable function / can be regarded as a distribu­
tion. Let /i ( x ) and /2(x) be two different continuous functions. Then each generates a different distribution such that there exists a φ in C'^° (R n ) for which (/ι,φ ) Φ (/2,<A), i.e., (/i - Φ 0. Two functions f x and f 2
a r e s a i d t o b e e q u a l a l mo s t e v e r y wh e r e ( a.e.) o n a b o u n d e d d o ma i n Ω i f
/ |/i - /2 | d x = 0. Hence, two locally integrable functions that are equal
Jn
a.e. generate the same distribution. A distribution of the form (7.1), where /( x ) is locally integrable, is said to be r e g u l a r. All other distributions are called s i n g u l a r, although formula (7.1) can be used formally for such distri­
butions.
E x a m p l e 7.1. The functional (χ η,φ ) = / φ ( χ ) ά χ., Ω £ R n,
w h e r e χ η is the indicator function of the domain Ω, generates a linear, piecewise continuous and regular distribution. Note t h a t χ η { χ ) = H { x )
Rn
/» o o roc
/ f ( x Ι,·'· ,Χ η ) Φ ( Χ I,·'· ,X n ) d x i ■·■ d x n.
— o o J — o o
(7.1)
’ is the Heaviside unit function with
0, x < 0
Ω as the interval (0, o o ). ■
7.1.1. D i r a c d i s t r i b u t i o n. Let x' be a fixed point in R n. Con­
sider the functional 6 X < defined by ( δ χ >,φ ) = φ ( χ'), which assigns to
7.1. DEFINITIONS
221
each test function φ its value at x'. The functi continuous on Cg° ( R n ), and hence a distributi shall show t h a t <i>o (denoted simply by 6) is a The proof is by contradiction: Assume t h a t δ exists a locally integrable function /( x ) such t h
113
onal δχ ι is linear and on with pole x'. We singular distribution, regular. Then there
/,
/( χ ) φ ( χ ) dx = φ{ 0) for every .
The functional φ( 0) is taken as the definition of is known as the Dirac delta function. Hence,
£ C ° ° ( R n). (7.2)
ihe density 6 ( x ) which
( δ ( χ,χ ι),φ ( χ ) ) = ί δ ( χ,χ,) φ ( χ ) ώ J Rn
In the classical sense the Dirac delta function is
δ(χ, x') =f | °’
I 00,
if χ Φ x' if x = x'
c = φ ( χ'). (7.3)
defined as
(7.4)
This function is a generalized function defined by
.. 1 sin(n(a: - x
o(x, x ) = — lim
7Γ n—*oo
which, in the limit, is zero a t every point χ φ x' Thus, the Dirac delta function represents a po source point χ', i.e.,
δ(χ, x')
»oo as x —>.
0, ot her wi s e.
T h i s f unc t i on is us ed i n def i ni ng a c o n c e n t r a t e d a n d fl ui d mechani cs, a p o i n t mass i n t h e t h e o r y t i a l, a p o i n t cha r ge i n el ec t r oni cs, a n i mpul si ve o t h e r s i mi l a r s i t u a t i o n s i n physi cs a n d mechani cs
A cons equence of (7.3) defi nes a bas i c prope: whe r e χ,χ' € Ω c J?3, a n d φ e C(fi). If φ ( χ ) =
SiL6{^)iV={X ϋχ-
))
(7.5)
a n d i nf i ni t e a t x = x'. int singularity a t the
(7.6)
impulsive force in solid gravitational poten- force in acoustics and
r ty of this function as (7.7)
if
if
x'e i l x' £ Ω,
Ξ Ω Ω.
1 i n ( 7.7 ), w e g e t
( 7.8 )
222
CHAPTER 7: GREEN’S FUNCTIONS
Also,
JJJ^6(k(x,x'))dx = J J J ^6(x,x')dx. (7.9)
For ( a, b) 6 R 1 the basic property (7.7) becomes
f δ(χ, χ')φ{χ) dx = φ(χ'), (7-10)
J a
where a, b can be — oo or +oo for unbounded intervals.
Note t h a t δ( χ,χ') has the units [L- 1 ], and 6 ( t,t') has the units
[T- 1 ] if t, and t' denote time, where L denotes the unit of length
and T the unit of time. In general, δ( χ,χ') has the units such t h a t
I 6(x,x')dV = 1. Also, we often write δ(χ) for <5(x,0), and δ( χ,ί ) J J J a
f o r δ(χ) 6(t ).
T h e D i r a c d e l t a f u n c t i o n o n a r e g i o n Ω i s d e f i n e d i n t e r m s o f t h e c o m p l e t e o r t h o n o r m a l s e t o f e i g e n f u n c t i o n s f n for the region Ω, as
OO
<5ω(χ,χ') = ] Γ/„ ( χ )/„ ( χ'), (7.11)
n = 0
whe r e x a n d x' ar e t h e fiel d a n d t h e sour ce po i n t, r espect i vel y, i n Ω.
E x a m p l e 7.2. In view of Example 5.2, the spatial orthogonal eigen­
functions f n ( x ) for the one-dimensional problem u t = k u x x, —a < x < a, subject to the initial and boundary conditions u( x, 0) = F ( x ) for —a < x < a, and u ( —a,t ) = 0 = u ( a,t ) for t > 0, are given by
1 717TX
— s i n , which in complex form are
2 a a
f n = — e i n ™ l a. (7.12)
Hence the Dirac delta function in the region —a < x < a for the steady state (as t —+ oo) one-dimensional Laplace equation is represented by
Γ 1 V ° ° Ρίηπ(χ-χ')/α
r I I
δ{χ,χ') = \ 4 ^ - ° ° 6 - l ° r | x | < a, (713)
I 0, for Ixl > a.
7.1. DEFINITIONS
223
Fig. 7.1. Dirac delta function with 5,10, and 25 terms, respectively.
The graphs of the real part of this functior —α < x < a are shown in Fig. 7.1, using 5, 10, with a = 3, and x' = 0. They show t h a t the higher and narrower as n increases. ■
Example 7.3. To determine the Fourier delta function 6 ( x ), consider the function
/κ ( ζ ) = ( 2 k
I o,
|x| < κ, x\ > κ.
Note t h a t lim f K( x ) = δ ( χ ). Also, e ax f K{ x ) € «—►0
a, which implies t h a t T f K{ x ) = F K( a ) is analytic of the transform domain. Then
Ρκ(α)
i r°°
φ.*Ljax)'
' d x
1 f K 1 ■
= -!= / 7Γ- d x =
ν2π J - K 2k
V
lim F K(a) = - j — κ~*° γ 2π
b y m e a n s o f t h e m e a n - v a l u e t h e o r e m w i t h —κ
f o r t h e b a s i c i n t e r v a l a n d 25 t e r m s i n ( 7.1 3 ), p e a k b e c o me s i n f i n i t e l y
t r a n s f o r m o f t h e D i r a c
L ± { —oo, oo) for all real in the entire a- p la n e
1
2π
Xq < κ. Hence
<
224
CHAPTER 7: GREEN’S FUNCTIONS
which implies t h a t
!F6{x) = and ^ { l } = V 2 t t S ( x ). ■
γ 2π
7.1.2. Heaviside function. The Heaviside function H{ x) defines
pOC
the distribution in (#, φ) = I ψ(χ) dx, and so, in the distributional
Jo
sense (7.1), its derivative is defined by
H'{x) = 6(x). (7.14)
Note t h a t since H( x ) is not differentiable at x — 0 in the classical sense, t he distributional definition (7.14) for H'{ x) when formally integrated gives
0, x < 0 x > 0.
H(x) = J δ(χ) dx = |
7.1.3. Harmonic functions. The functions whose Laplacian is zero are known as harmonic functions.
L e m m a 1. The function 1
(7.15)
r y/(x - Zo)2 + (y- yo)2 + ( * - Zo ) 2 is harmonic in a domain that does not contain the point x o =( xo, yo,zo)·
L e m m a 2. Let u(x) be in class C2(Cl), and let its first derivatives be continuous up to the boundary S of the domain Ω. Let Xo be a fixed point in Ω, and x any other point in Ω. Then
r
where r is defined in L e m m a 1, and n is the outward normal vector to the surface S = 9Ω.
P r o o f. The proof will also establish a very basic technique in the
case when r = 0, i.e., when x = xo- Let v = -. Since this function is
r
7.1. DEFINITIONS
225
undefined when x = xo, we cannot apply Green entire domain Ω. So we indent the point Xo by a the point xo and with a small enough radius ε. L
Ωε, and Ωι = Ω\Ωε. Then both functions u and
(72(Ωχ), and Green’s identity (A.7) is valid in Ω
s identity (A.7) to the sphere Ωε centered at et 5 ε be the surface of
v = - are in the class r
l- T h u s, f r o m ( A.7)
1 d u d (1/r )
r d n dn
dS
+
II,
1 d u r d n
L e t ε —> 0. T h e s u r f a c e i n t e g r a l o v e r 5 i n ( ε. So we c o n s i d e r t h e s e c o n d i n t e g r a l o v e r S E. Si the normal n is directed inward on the boundary get
3(?)| = 3 (1/r)
Is.
dn
dr
which gives
= 47tu (xe) —> 47ru (x0)
where x e is a point on S e, which satisfies the m the surface integral over S e in (7.17) and approai
+
d ( l M
d n
d S. (7.17)
.17) is independent of nee r = ε = const, and S s (see Fig. 7.2), we
1
e) 4πε as ε —> 0,
(7.18)
ean-value theorem for .cthes xq as ε —> 0. Since
226
CHAPTER 7: GREEN’S FUNCTIONS
u € 6,2(Ω), the first derivatives of u are bounded in Ω = Ω U S. Hence, there exists a positive number K such t h a t \d u/d n\ < K. Then
f f I p. d S < ϋ i f < i S = * 4 ^ = 4 ,K e
J ) S e r d n ε J J S e ε
0 a s ε —y 0.
Thus, as ε —> 0, we get i n r
1 d u d ( l/r )\
u — d S - 4n u (xo),
5 \ r o n a n '
w h i c h g i v e s ( 7.1 6 ).
N o t e. In a finite domain Ω in F t?, with boundary Γ, the identities analogous to (A.7) and (7.16) are
J J (jiV2« — v V 2 u ) d A = J
d v d u\
u - v — d s.
d n d n J
( 7.1 9 )
a n d
“(χ°> = ά/Γ
, 1 d u < 9 1 n ( l/r ) , ,
j3 2,
V 2u d A, (7.20)
where xo = (.το, y 0 ), d A = d x d y, and s is an arc-length along the curve
Γ. The proof follows by using the technique of Lemma 2 with υ = ln -
r
and indenting the point xq by a circle Γε; then V 2v = 0, and
9 1 n ( l/r )
L
d n
d l n ( l/r )
d n
9 1 n ( l j r
d r
d s
I f
1
= - / u d s = - u ( χ ε) 2πε
ε J r e ε
—> 2n u ( x q ) as ε —> 0.
Thus,
/,
1 1 d U A
In - — d s
r d n
I f
1 1
< Λ Ί η - / d s = K ln - = 2πε l n > 0 as ε —» 0,
ε Jre ε ε
where
d u
d n
< K. The result follows as ε —> 0.
Some important properties of harmonic functions are defined by the following theorems:
7.1. DEFINITIONS
227
T h e o r e m 7.1. L e t t he f u n c t i o n u ( x ) € C 2 be
IL
d u
u — d S > 0. i s d n
I n f a c t, i f we l e t v = u in Green’s first identit
LIL
Since the volume integral (left side) in (7.22) >
{ d x ) +\d y j
d u\2 ( d u\2
+ ' d z j
d V =
harmonic in Ω. T h e n (7.21)
y (A.5), we get
i l/£ ds- <7-22>
we get (7.21). ■
T h e o r e m 7.2. T h e i n t e g r a l o f t h e n o r m a l d e f u n c t i o n o v e r t he b o u n d a r y o f t h e d o m a i n i s z e r o
r i v a t i v e o f a h a r m o n i c
IL
u p d S = 0.
s d n
T h e r e s u l t f o l l o ws b y a p p l y i n g G r e e n ’s i d e n t i t y ( A.7) t o u = v = 1.
fu·
T h e o r e m 7.3. T h e v a l u e o f a h a r m o n i c p o i n t o f a f i n i t e d o m a i n c a n be e x p r e s s e d i n t ern: v a l u e s o f i t s n o r m a l d e r i v a t i v e o n t he s u r f a c e o f
. . 1 f f ( l du 8(1/·
“<Xo) = s/A (;a;- u~ ^
This result is obtained by applying (7.16) to
N
ote that there are no second order derivat or (7.24). Thus, to ensure that these three res assume that the harmonic function u e C2(S) Ω C Ω. Then we apply (7.16) to Ω, and by t a Ω —> Ω we find t h a t the above analysis need not derivatives of u are continuous up to the bounda:
i.e.,
(7.23)
nction at an interior s of its values and the the domain, i.e.,
dS.
(7.24)
the harmonic function
ives in (7.21), (7.23), ults are valid we must Consider a domain king the limit process ssume t h a t the second ry S.
228
CHAPTER 7: GREEN’S FUNCTIONS
T h e o r e m 7.4. A f u n c t i o n u ( x ) h a r m o n i c i n t he d o m a i n Ω h a s d e r i v a t i v e s o f a l l o r d e r s i n s i d e Ω.
P r o o f. Consider an arbitrary point xo S Ω and surround this point by a domain Ω' bounded by its surface S' such t h a t Ω' U S' C Ω. Since u is harmonic in Ω, so it is harmonic in Ω' and u € C 2 ( i Y ). Then by (7.24)
Since the point xo does not lie on S', we find t h a t
1 _ 1
r V(x~ xo)2 + (y- yo)2 + (z- z0)2
i s c o n t i n u o u s a n d h a s c o n t i n u o u s d e r i v a t i v e s o f a l l o r d e r s a t t h e p o i n t x o- H e n c e t h e r i g h t s i d e o f ( 7.2 5 ) c a n b e d i f f e r e n t i a t e d a n y n u m b e r o f t i m e s u n d e r t h e i n t e g r a l s i g n. ■
T h e o r e m 7.5. T h e v a l u e o f a h a r m o n i c f u n c t i o n a t t he c e n t e r o f a s p h e r e i s e q ua l t o t h e a r i t h m e t i c m e a n o f i t s v a l u e o n t he s u r f a c e o f t h i s s p h e r e.
PROOF. Let «(xo) be harmonic inside the sphere |x — xo| = R. Then by (7.24)
/ \ 1 f f f 1 d u d ( l/r )\ Jry
“ (Xo) = 3 i i X „ ( r a;- “ — ) d s · ( 726)
9 ( Α) | d ( l/r )\ 1
Now, since = - —r, we get from (7.26)
d n ls H d r I r =R R 2 v '
u{xo)=^ I I s Ruds=^ l l <κ^Φ)^θάθάΦ··
(7.27)
T h e o r e m 7.6 ( M a x i m u m P r i n c i p l e ). A n o n - c o n s t a n t f u n c t i o n w h i c h i s h a r m o n i c i n s i d e a b o u n d e d d o m a i n Ω w i t h b o u n d a r y S a n d c o n t i n u o u s i n t he c l o s e d d o m a i n Ω = Ω U S a t t a i n s i t s m a x i m u m a n d m i n i m u m v a l u e s o n l y o n t h e b o u n d a r y o f t he d o m a i n.
7.1. DEFINITIONS
229
Jo:
P r o o f. Assume t h a t w(x) attains its maximi point xo E Ω. Draw a sphere Sp : |x — xo| =/>si|' by Theorem 7.5,
where equality holds only if u = u ( x o) = const tains its maximum value in Ω, we must have u (xq t h a t we must have the equality sign in (7.28). Tbi stant both inside and on the boundary of the spp< show t h a t u ( x ) is constant throughout Ω: Let point of Ω. Then we show t h a t u ( x ) = u (xo)· by a line L which may be a polygonal (broken) be the shortest distance between L and S. Then constant value u (xo) in the sphere Sq : |x — xo| diameter of the sphere. Suppose t h a t x i is the section of this sphere and the line L. Then we ha thus, we have u (x) = u (xo) on the sphere S\ suppose t h a t x2 is the last point at the interse and the line L. Then, u ( x ) = u (xo) on the spher< Continuing this process of constructing sphere: be covered by a finite number of these spheres, inside one of these spheres, and therefore, u (x) argument can show t h a t a harmonic function c mum value inside Ω. Hence, the function u ( x and minimum values in a closed bounded domai it attains them on the boundary of Ω, since a har: at tai n them inside Ω. ■
mm v a l u e a t a n i n t e r i o r
ch t h a t Sp c Ω. Then
C o r o l l a r y 1. I f u and U are continuous i in Ω such that u <U on S, then u <U also at
I n f ac t, t h e f unc t i on JJ — u is continuous i Ω; hence U — u > 0 on S. Then, in view of t (Theorem 7.6), we must have U — u > 0 at all
C o r o l l a r y 2. I f u and U are continuous ; in Ω for which |u| <U on S, then |u| < U also
I n f ac t, t h e t h r e e ha r moni c f unct i ons —U, u lation —U < u < U o n S. If we apply Corollary
dS = w”ax,
(7.28)
on Sp. Since u(x) at- 1 < u“ ax
, whi ch me a ns us, u (x) becomes con- lere Sp. We shall now be any other interior in the points xo and x line inside Ω, and let d by (7.28), u (x) has the = d2/4, where d is the ast point at the inter­
ne u (xi) = u (xo), and — xi| = d2/4. Again, ction of the sphere Si ■e : | x - x 2| = d2/4. the entire line L can and the point x will lie = u (xo)· An analogous annot at tai n its mini- attains its maximum η Ω, and in particular, monic function cannot
•ί Ω U S, and harmonic all points inside Ω.
η Ω, and harmonic in lie maximum principle points inside Ω. ■
η Ω U 5 and harmonic at all points inside Ω.
, and U satisfy the re- 1 twice, we find t h a t
230 CHAPTER 7: GREEN’S FUNCTIONS
-U < u < U a t all points inside Ω, or |u| < U inside Ω.
7.1.4. T h e c o n c e p t o f a G r e e n ’s f u n c t i o n. Consider the heat conduction problem in a region Ω C R 3 with the boundary surface S = U"=1Sj, defined by
V2u(x, t ) + - - f ( x, t ) = - d u ^' ^, for t > 0,
k a d t (7.29a)
du
k i - ---- 1- h i U = g ( x.,t ) o n S i, t > 0,
° Ui (7.29b)
u(x, 0) = F ( x ), (7.29c)
Q
where —— denotes the derivative with respect to the outward normal
orii
n to the boundary surface S i, the coefficients ki and hi are constant, and the function p(x, t ) is the prescribed mixed boundary condition on each S i. Note t h a t with fcj = 0 the boundary condition reduces to the Dirichlet type, and with hi = 0 to the Neumann type. In order to solve problem (7.29) we consider the auxiliary problem
V2G ( x,t;x',t/) + y 6( x,x!) 6 ( t,t') = in Ω, t > 0,
q q a d t (7.30)
h h hi u = 0 on S i, t > 0,
OTli
such t h a t
G ( x,i;x',i') = 0 t'< t, (7.31)
where <5(x, χ') = δ ( χ,χ') S ( y,y') δ ( ζ,ζ') is the Dirac delta function in
R 3 for the space coordinates x = ( x,y,z ) and x' = ( x', y', z'), and
S ( t,t ’) is the Dirac delta function for the time coordinate t > t ’. Note
t h a t the function G(x, i;x',t'), known as G r e e n ’s f u n c t i o n for problem
(7.29), satisfies the auxiliary problem (7.30) with homogeneous bound­
ary conditions and zero initial condition (7.31), and has an impulsive heat source at the space-time source point (x',t').
Definit ion 7.5. Green’s function G ( x,t;x',t') in R 3, which is a solution of problem (7.30)-(7.31), represents the temperature distribu­
tion in the region Ω which initially at zero temperature is subjected to the homogeneous boundary conditions due to an impulsive heat source of unit strength situated at the space-time point (x',t').
7.2. PARABOLIC EQUATIONS
231
the
The notation G(x, t; x', t') is composed of two The first p ar t x, t denotes the field point where sive heat source located a t the source point signifi· the point x at time t. The second part x',t' dene the impulsive heat source situated at the point x ‘ taneous (impulsive) heat at an earlier time t'. T has the physical significance of an entire space-ti: be visualized as G(effect; cause) ξ G( x,t;x',t')
p a r t s i n i t s a r g u m e n t.
effect of the impul- es the temperature at ites the cause which is generating an instan- he combined notation me process which can
Definit ion 7.6. The reciprocity relation for x', t') which satisfies the auxiliary problem (7.3C
3reen’s function G(x, i; (7.31) is defined by
G(x, t; x', t') = G(x', —i'; x, - t ).
T h e p h y s i c a l s i g n i f i c a n c e o f t h i s r e l a t i o n i s t h a t; t o a c a u s e a t x',t' for t' < t is the same as the a cause at x, — t.
G r e e n ’s f u n c t i o n G ( x, t; x',t') is singular at tl(ie source point χ' £ Ω, such t h a t
LG(x, t; x', t') = S(x, x'), (7.33)
where L is a differential operator.
(7.32)
the effect at x, t due effect at x', —t' due to
7.2. P a r a b o l i c E q u a t i o n s
Before we determine Green’s functions for paraty< solve problem (7.29). In view of the reciproc: (7.30) is written for the function G(x', - t'; x, —
VSG + i i ( x,*■)«(,,t') = 4 §
where Vq denotes the Laplacian in the variable
d2 82
d x ’2
, and the minus sign on the right side results because t has
x by x' and t by t' in
d y'2 d z'2
been replaced by —t'. Similarly, if we replace (7.29a), we get
V ^ + -/( x,,i') = -
1 d u ( x' ,t')
d t'
o l i c e q u a t i o n s, we wi l l .t y r e l a t i o n ( 7.3 2 ), E q ) a s
Ω,
x', i.e., W l
( 7.3 4 )
d2
i n Ω.
( 7.3 5 )
232
CHAPTER 7: GREEN’S FUNCTIONS
Then, if we multiply (7.34) by u and (7.35) by G, subtract, and inte­
grate the resulting equation with respect to x' over the region Ω and with respect to t' from 0 to t o = t + ε, where ε > 0 is arbitrarily small, we find t h a t
So ^ U L “ UV°G') d Q + ^ So ^ /f L ί/( χ/’ t > ) G
- ^ < 5 ( χ',χ ) < 5 ( ί',ί ) « ] d n = \J J J [Gu]?=0dQ- (7·36)
In view of Green’s identity (A.7), the first volume integral is changed to the surface integral; thus,
i i i (GV”“ - "V»G) Λ ■- έ J i £ - · g ) <*·
Also, in view of the property (7.7) of the Dirac delta distribution, the second term in the second volume integral in (7.36) is
f dt' f f f u6(x! S(t',to) dQ, = it(x, to);
Jo J J Jil
and the integrand on the right side of (7.36) in the limit as ε —> 0 becomes
[G«];“_ 0 = G u ( t o ) - G F ( x') = - G F ( x'),
t'=t0 t'=0 t'= 0
s i n c e G = G ( x,t;x',i o ) = G ( x,i;χ',i + ε ) = 0 f o r t < t 0 because
£ — io
the time for the effect precedes the time for the cause (impulse). Hence, Eq (7.36) gives
^-ΙϋΛ^)Λ,+ΠΛϋΙα
+
^ /* /» al λ'Σ//8
du d G\ .
F w— c?5i, (7.37)
s \ drii d r i i j
where G = G(x, t: x', 0). If we multiply the boundary condition
lt'=0
(7.29b) by G and (7.30) by u and subtract, we get
7.2. PARABOLIC EQUATIONS
233
where G = G ( x,t;x i',t') denotes Green’s function evaluated on the
Si
boundary surfaces S i, i = 1,2,· ■■ ,n. After substituting (7.38) into (7.37), we obtain the solution of the problem (7.29) in terms of Green’s function G ( x,t;x',t') as
u(x.,t) = J J J G(x, t\x!,ϋ ) F ( x!) d n
+ ^ J d t' J J J G(x, t, x'; t') /(:
+ a j Μ'Σ γ. J J s G ( x,i;x',l
Now we will derive Green’s functions for som Without loss of generality, we will translate th origin, i.e., we will take x = 0 and t 1 = 0, and G(x, t ). The results so obtained can then be t source point at (x', t r) by replacing x b y x - x' am however, yield G(x — x'; t — t r) which in our not
Example 7.4. Green’s function G ( x,t;x',t ) diffusion operator in R 1 satisfies the equation
d G d 2G
— α - τ τ - = o( x — x ) o ( t
d t
d x 2
A s s u m i n g t h a t G = 0 f o r t < t', we apply the L (4.40) and get
, d 2 G _ s t, s G — a ——^ = e δ ( χ — x d x z
T h e s o l u t i o n o f t h i s e q u a t i o n u n d e r t h e c o n d i t i c ± o o i s
G ( *,x,;e,i,) = 1 ^ - 4 = e - r ^,
2
w a s
w h i c h o n i n v e r s i o n g i v e s
G ( x, x'; t, t') = — ^ ^— p^x 2y/7ra( t - t>)
, t') rfO
) f ( x',t') d S i. (7.39)
e parabolic equations, e source point to the denote G(x, i; 0,0) by ranslated again to the d t by t — t'. This will, ation is G(x, t; x', t').
f o r t h e h o m o g e n e o u s
t'). (7.40)
sjplace transform to Eq
n t h a t G —» 0 as x
r = \x — χ
I,
/4 a( t —t')
234
CHAPTER 7: GREEN’S FUNCTIONS
E x a m p l e 7.5. For the diffusion operator in R n, Green’s function G(x, t ) satisfies the equation
<9G(x, t ) d t
a V2G ( x, t ) = i ( x, t ) = 6 ( x ) 6 ( t ). (7.41)
We apply the Fourier transform T x to both sides of Eq (7.41). Then, 'd G N
T x ( ^ ) - a T x ^ 2G\ = T x [ 6 ( x,t ) }.
S i n c e
T x [ S ( x,t ) ] = Τ χ\δ { χ ) ■ S( t ) ] = m (a) ■ 6 ( t ) = ■ «(*)>
(2ττ)*
where 1 is the identity function, and
T x
dG_
dt
T X[V2G ] = T x [ V 2G] = — \a\2 T X{ G\,
E q ( 7.4 1 ) i s t r a n s f o r m e d i n t o
^-t T x G ( a,t ) + a\a\2T x G ( a,t ) = . 6 ( t ),
w h i c h h a s t h e s o l u t i o n
T G ( a t ) — e- ala|2t
(27r)n/2e
If we apply the inverse Fourier transform T ~ l, we obtain G ( x,t ) = T ~ 1 [ G ( a,t ) }
— f - a\a\2t - i ( a- x) j _ - | x | 2/4at
( 2 π ) η J R n (4παί)"/2
which, on translating to the source point ( χ',t'), yields
G i x t'-χ! t') — ~ ^ ____ β - | χ - χ Ί 2/4 a ( t - t')
G ( x,i,x,i j - [ 4 7 r a ( i _ i 0 ] n/2 e
(7.42)
( 7.4 3 )
7.2. PARABOLIC EQUATION
T h e g r a p h s f o r G ( x, t ) f o r 0 < t\ < t z < t ^ a r e s h o w n i n F i g. 7.3.
t = 0.2:5
Fig. 7.3. Graphs of G(x, t ) in
Example 7.6. We will solve the Cauchy prcpblem in R 1, i.e., solve the initial value problem
Ut — (i Uxx,
u(x, 0 ) = f ( x ), —oo < X < 0 0,
whe r e f ( x ) is a bounded and continuous functl· (7.44a) is also known as the transient Fourier eqi prove t h a t the function
1 f ° °
= —===== / f ( y ) ν4τΓβΐ J —qq
which belongs to the class C°° with respect to x i
Eq (7.44a) such t h a t lim u ( x,t ) — f ( x ), —oo
M e t h o d 1: We use the method of separation of variables in the form u ( x
, t )
= X ( x ) T ( t ).
Then we find t h a t T ( t ) = e
A cos X x + B sin X x, where Χ,Α,Β are arbitrary. If we assume A = A (Λ), B = B (A), then
u\( x,t ) = [A(A) cos Ax + B ( X ) sin is a solution of Eq (7.44a). Moreover,
f
J —C
u\( x, t ) dX,
235
R 2.
t > 0,
(7.44a)
(7.44b)
on on the x-axis. Eq iation in R 1. We shall
/4 at
dy,
(7.45)
,nd t for t > 0, satisfies
< X < 00.
\x\e
- a\2t
236
CHAPTER 7: GREEN’S FUNCTIONS
with proper choice of A ( A) and B ( A), is also a solution of Eq (7.44a). Now, in view of the initial condition (7.44b)
/
OO p OO
u\( x,0 ) d X = / [A(A) cos Ax + B (\) sin Ax] dA. (7.46)
■oo J —OO
Th e f u n c t i o n f ( x ), being continuous and bounded, has the Fourier integral representation
•ι pOO p OO
f ( x ) = 7T d X f ( y ) c o s\( y - x ) d y. (7.47)
J— OO J— OO
Comparing (7.46) and (7.47), we find t h a t
1 p OO 1 poo
Α ( λ ) = — J f ( y ) cos X y d y, B { X ) = — J f ( y ) sin X y d y. Thus,
/
OO
u\( x,t ) dX
-OO
i rOC pOO
2- J dX J f (y) cos x(y — x) e~aX t
rOO
d y
r OO r OO
— / f ( y ) dV / cos X ( y — χ ) ι J — o o J — o o
1 r°°
\/4 π α ί
1 poo
= -j= = / f ( y ) e ~{y- x) /4atdy
\/4 π α ί J -oo
M e t h o d 2: We apply a Fourier transform in space (with a as the transform variable) and a Laplace transform in time (with s as the transform variable). Then Eq (7.44a) with the initial condition (7.44b) is transformed into
(s + a a 2 ) u ( a, s ) — f ( a ) = 0, (7.48)
7 / \
where ΰ = T [ u\, and / = J F [ f ]. Since u ( a,s ) = ^ -2-, the function
u ( a, s ) has a simple pole at s = — a a 2. Hence, by applying the inverse Fourier and Laplace transforms, we get the solution of the Cauchy problem (7.44) as
u ( x,t ) = - ^ = Γ ( l i m \^ ~ Γ *'* * ^ a\e s t d s } \ e ~ i a x d a. V 2 n 7-oo I L2™ Ju- i R s + a a 2 J f
7.2. PARABOLIC EQUATIONS
237
A direct inversion of the Laplace transform yields u ( a,t ) = e~0“2t/( a ), which, after the inversion of the Fourier tram
u( a,t ) e
1 . rJ_J<“>'
sform, gives d a
- ( i a x - t | a a 2t ) ^a
I f we i d e n t i f y g ( a ) = e f ( a ) g ( a ), we have
, then for the convolution of the product
1 r°° ~ 1 r°
= / f ( a ) ~ g ( a ) e -'a * d a = - ± = / V2 π J -oo v27t J- i
f ( y ) g ( x - y ) d y. (7.49)
Since f ( a ) is identified with the Fourier comp- f ( x ) in (7.44b), we find t h a t f ( y ) in (7.49) condition and
g{x - y ) =
1
-(x-y)'
V4παί
and hence by the convolution theorem we get t problem from (7.49).
If f ( x ) = 6 ( x ) in (7.44b), then the solution of the Cauchy problem (7.44) is given by
u(x, t)
y/Airat
-x
/4αί
E x a m p l e 7.7. (Schrodinger equation) If w geneous case, then the Fourier heat equation
)»(*,«) = /<
has two interpretations:
(i) If a > 0 is a real constant depending on specific heat and thermal conductivity of the medium, then u ( x, t ) determines the temperature distribution. The function f ( x, t ) on the right side describes local heat
onent of the Cauchy d at a corresponds to the initial
/4 at
ie solution of the Cauchy
t > 0.
e consider the non-homo-
x,t )
(7.50)
238
CHAPTER 7: GREEN’S FUNCTIONS
production minus absorption.
(ii) The function u ( x, t ) defines a particle density and a is the diffusion
tfx
coefficient. If a is purely imaginary such t h a t a = where m is the
mass of the quantum particle, and h = 1.054 χ 10- 2 7 erg-sec is Planck’s constant, then Eq (7.50) defines the Schrodinger equation
i n d u { x,t) + ft ^ = (7.51)
a t 2 to
I n vi ew o f ( 7.42) we have a = h/2 i m, and Green’s function for Eq (7.50) in R 1 is given by
G ( x,t ) = H ( t ) l- ^ ^ e im*2/2Ht.. (7.52)
7.3. E l l i p t i c E q u a t i o n s
We will consider the Laplace and the Helmholtz operators and derive Green’s functions for them.
Example 7.8. Assuming t h a t the source point G(x') is at the ori­
gin, Green’s function G(x) = G(x, 0) for the Laplacian V2 in R n sat­
isfies the equation
V2G(x) =<5(x). (7.53)
T h e n, i n vi ew o f t h e p r o p e r t y (7.11),
J J V 2G(x) d S = J J S { x ) d S = 1. (7.54)
We s ha l l f i r s t cons i der t h e case n = 3. Si nce t h e o p e r a t o r V2 is i nva r i ­
a n t u n d e r a r o t a t i o n of c o o r d i n a t e axes, we s ha l l seek a s ol ut i on t h a t de p e n d s onl y on r = |rc|. For r > 0, G ( r ) will satisfy the homogeneous equation V 2G = 0, i.e., in spherical coordinates
7.3. ELLIPTIC EQUATIONS
239
which has a solution G ( r ) = — |-B. If we require
r
at infinity, then B = 0, and G ( r ) = A/r. In ord' take into account the magnitude of the source (7.53) over a small sphere S e of radius ε and cen
er to determine A, we at x = 0. Integrating ter at x = 0, we obtain
t h e p o t e n t i a l t o v a n i s h
J J V 2G d x = 1,
which, by using (A.4), gives
8 G l as€ dr
JJ
J J di
d S = 1,
where d S e is the surface of the sphere S E. P presses the conservation of charge, i.e., the fl through the closed surface d S e (of area 4πε2) is t he interior of S E. Now, substituting G = A/r A = 1/(4π), and hence Green’s function for Laplace equation is
G = T ~ 1
4 π τ
( 7.5 5 )
h y s i c a l l y, E q ( 7.5 5 ) e x- αχ o f t h e e l e c t r i c f i e l d e q u a l t o t h e c h a r g e i n i n ( 7.5 5 ), we f i n d t h a t t h e t h r e e - d i m e n s i o n a l
4 π | χ ■
F o r n = 2, we h a v e
1 - 2 - ^ = 0.
r d r V’ d r J
= 'J-
w h i c h h a s a s o l u t i o n G ( r ) = C l n r + D. We set
value at infinity, and use a result similar to ('
electric charge through the boundary d C e (of le: radius ε. Then C = —1/2π, and
G(x, x') = — ln - = ~ ln —
2π r 2π x
In general, Green’s function for the free-space Laplacian in R n is given by
” >2' <758>
where 5 n ( l) = 2πη//2/Γ( η/2) is the surface ariii unity, and r = |x —x'|. Note t h a t the notation is not standard; it is sometimes defined with a the factor 1/2π. ■
(7.56)
cz + y ,
D = 0 to ensure a zero .55) for the flux of an n gth 2πε) of a circle of
(7.57)
a of a sphere of radius this Green’s function minus sign, or without
far
240
CHAPTER 7: GREEN’S FUNCTIONS
E x a m p l e 7.9. ( M e t h o d o f i m a g e s) The problem of finding Green’s function G(x, x') = G(x,y;x',y') inside some region Ω bounded by a closed curve Γ with the homogeneous boundary condition G — 0 on Γ amounts to t h a t of finding the electrostatic potential due to a point charge at the point (x', y') inside a grounded conductor in the shape of the boundary Γ. Consider, for example, the region Ω as the half­
plane y > 0. Then Green’s function is given by the point charge of strength 1/2π at (x',y') together with an equal but opposite charge (of strength —1/2π) at the point (x', —y') which is the image of the point (x',y') in the x-axis (Fig. 7.4). Thus, Green’s function for the half-plane such t h a t G = 0 on the x-axis and G —► 0 as r —> oo, where r = |x — x'| = \/( x — x')2 + (y — y')2, is given by
/ /\ 1 , (χ - x1)2 + (y - y')2
G(x,y,x ,y ) = — l n?-------- ΤΫ ΤΤΊ — i—Λ2 -
4π ( χ — x'Y 4- ( y + y') 2.
( 7.5 9 )
1/2π
o
<-x\ y')
1/2π
o
(x\ y')
0
-
1/2π
-
1/2π
o
0
-y')
(χ\ -y1)
Fig. 7.4.
Since the algebraic sum of the charges over the entire (x, $/)-plane is zero, the condition G —> 0 as r —> oo becomes possible because any nonzero residual charge will make Green’s function behave like l n r. This means t h a t we cannot determine a Green’s function for the half­
plane y > 0 subject to the conditions G y = 0 on the x-axis and G —* 0 as r —> oo, because in this case t h a t charge a t each (x', y') and (x', —y') are of the same sign and their algebraic sum is not zero. However, we can determine Green’s function for the quarter-plane x > 0, y > 0, subject to the conditions G = 0 on y = 0 and G y = 0 on x = 0. As seen in Fig. 7.4, we have the charge of strength 1/2π at each (x', y r) and (—x',y'), and the charge of strength —1/2π at each (x', —y') and
7.3. ELLIPTIC EQUATIONS
241
{—x',y'). Thus, Green’s function in this case is G( x,y;x',y')
J _ i [(x - x
')2
+ (y - y
')2
4π [(x - x
')2
+ (y + y') 2}[(
E x a m p l e 7.1 0. G r e e n ’s f u n c t i o n f o r t h e H e l m h o l t z e q u a t i o n i n R n satisfies
(V2 + M)G(r) = £(r),
(V2 - h
2
)G(r)
t h a t
w h e r e h2 = —μ, and yfji is defined such imaginary part, i.e., ^[μ = α + ίβ, with β > 0, < We will therefore take h = so t h a t h is
real negative. We will assume t h a t Green’s func symmetric. Then for χ φ 0, the function G(r) coordinates
+ μΓη-10 =
6
(r), r = | x — x'|,
(7.61)
it has a nonnegative nd β = 0 iff μ e [0, oo). real positive when μ is :tion G ( r ) is spherically itiust satisfy in spherical
dr V
dr J
If we substitute G = wr 1 (”/2), then Eq (7. equation of the Bessel type of order (n/2) — 1
d / dw\ w ( n
\2
+μ τ
dr
whose gene r al s ol ut i on ca n be w r i t t e n i n t e r ms as
w ( r ) = C l H ^/2) _ l { V J i r ) + C 2H ^/2 ) _ ^
I f μ £ [0, oo), then y/μ has positive imaginary part and the Hankel function becomes exponentially large as r —> oo, but
i/( n/2) —i ( v ^ - r) is exponentially small. Since G vanishes at r = oo, we
(n/2)-
must have C 2 = 0, and then from (7.63) we get
G = Ci H,
(i)
( n/2 ) ·
As i n E x a m p l e 7.8, we a p p l y ( A.3 ) t o ( 7.6 1 ) a n d o b t a i n
dG
/,
d r
d S = 1,
x + x
')2
+ ( y - y')
i\2
x + x
')2
+ (y + y') 2]
( 7.6 0 )
0.
( 7.6 2 )
6 2) c a n b e r e d u c e d t o a n w i t h p a r a m e t e r μ, i.e.,
w = 0,
of t h e Hanke l f unc t i ons jUr), n > 2. (7.63)
( 7.6 4 )
242 CHAPTER 7: GREEN’S FUNCTIONS
or
BC
lim r"_1S’n(l)—— = 1, (7.65)
r —*o dr
w h e r e S „ ( 1 ) i s t h e s u r f a c e a r e a o f a s p h e r e o f u n i t r a d i u s. F o r s m a l l r, we have the asymptotic expansion
Thus, substituting (7.64) into (7.65) and using (7.65) we find t h a t
i/v 7 * y n/2)-1
1 _ [(n/2) - 1]!5„(1) ~~ 4 \ 2π /
Hence, for n > 2 and μ £ [0, oo), the required Green’s function is given by
G(r,^) = \{ ^ ) { ' ^ S/2 ) - i ( v ^ r )» n ^ 2> (7·66)
or, wr i t i n g h 2 = —μ, i.e., h = —i y/μ, or y/μ = i h, we have
t f ( n/2 ) - l (\/M r ) = H ( n/2 ) - l ( i f l r ) = ^ K ( n/2) - l ( h r ), where K ( n/2) - i are the modified Bessel functions, thereby yielding 1 / h χ (n/2) —1
G [ r,~ K') = 2^ ( 2^ ) K ( n/2) - i i h r ), n > 2, (7.67)
which holds whenever —h 2 £ [0,oo), i.e., for all h with 9?/i > 0. For n = 2, Green’s functions (7.66) and (7.67) become
G { r ^ ) = l- H ^\^ r ) = ^ K 0{hr). (7.68)
For n = 3, by using H (^ 2 { z ) = - ( - ) * 2 —q 2, we get
Fig. 7.5. Green’s functions of the Helmholtz operator, with μ = 1.2: (a), (b): Real and imaginary parts of one-dimensional solution; (c), (d): Real and imaginary parts of two-dimensional solution.
244
CHAPTER 7: GREEN’S FUNCTIONS
The one-dimensional Green’s function is found directly, and it is
jpiy/Ji\x\ p-h\x\
^'Ί ν Γ —' (770)
Note t h a t if the Helmholtz equation is taken as (V2 + k 2 ) u = 0, where the wave number fc > 0 is real, then Green’s function in R 2 is given by
{ in R 2,
G ( r,k ) t =\ _ ^ 0 [ ν μ h (7.71)
I T x r - in f l ­
i t is obvious from Fig. 7.5 t h a t the real and the imaginary parts of Green’s functions of the Helmholtz equation in one- and two-dimensional cases exhibit wave structure. ■
7.4. H y p e r b o l i c E q u a t i o n s
d 2
We shall denote Green’s function for the wave operator Dc = —r — c2V2
Ot2
in R n by G n ( x,t ). I t satisfies the equation
□ c Gn ( x,i ) = <5(x,£), (7.72)
where x € R n and t e (0, t ) c R +. We will use the Laplace transform method and derive Green’s functions for n = 1,2 and 3.
E x a m p l e 7.11. In R 1, Green’s function G i { x,x'\t,t') satisfies the equation
d 2G j 2 d 2 G x
= δ ( χ — x') S ( t — t').
d t 2 d x 2
T a k i n g t h e L a p l a c e t r a n s f o r m w i t h z e r o i n i t i a l d a t a, we g e t
s2Gi - — e~st 6 ( x - x').
αχ1
A s s u m i n g t h a t G\ is finite a t x = ±oo, the solution of the above equa­
tion is
7.4. HYPERBOLIC EQUATIONS
245
w h i c h o n i n v e r s i o n g i v e s
2c
0 for c ( t — t')
f o r c ( i — £')
<:
= y c H ( c ( t - t') - \x
E x a m p l e 7.1 2. I n R 2, Green’s function G·.
d 2G 2 2 ( d 2 G 2 d 2 G 2\ .. jv\
«5- - c U? - - g g r ) = «<* - * w * - » w - 1 )·
Applying the Laplace transform we get
s G 2 - c
2 ^ 2G2 + d 2G 2
e ~ s t'6 ( x
\ d x 2 d y 2
o r u s i n g t h e a x i a l s y m m e t r y w i t h r 2 = ( x — x')
s G2 - c
_2 ( d 2G 2 | l d G
y d r 2 r d r The solution of this equation is
e~st'6{x
> 1 1 - x'\ \x - x'\
( 7.7 3 )
s a t i s f i e s t h e e q u a t i o n
- x') 6 { y - y'),
+ ( y - y
')21
- x') 6 { y - y').
G 2(r,s·, t') = — K o (
where K q is the modified Bessel function of the third kind and zero order. Since
C ~ l { K o { a s ) } - H { t ~ a
y/t2 — a 2
(see Erdelyi et al., 1954, or Abramowitz and Stegun, 1965), on inversion we get
1
G( r,t:t') = < 2 n c y/c i { t - t') 2 - r 2 1 0 for r > c ( t — t') H ( c ( t — t') — r)
27r c y/c 2( t — t') 2 — r 2
for
)
r < c(t — t')
( 7.7 4 )
246
CHAPTER 7: GREEN’S FUNCTIONS
E x a m p l e 7.13. In R 3, Green’s function G3 satisfies the equation d 2 G 3 2 ( d 2 G 3 d 2 G 3 r)2 G:i
c - — 5--------—r- +
d t 2 \ d x 2 d y 2 d z 2
= δ ( χ - x') 6 ( y — y') 8 ( z — z') 6 { t — t').
A p p l y i n g t h e L a p l a c e t r a n s f o r m we g e t
\ d x 2 d y 2 d y 2
= e ~ s t'6 { x - x') 6 ( z - z') 6 ( y - y f ),
o r u s i n g t h e a x i a l s y m m e t r y w i t h r2 = ( x — x') 2 + ( y — y') 2 + ( z — z') 2,
s 2^ 3 - c2 ( j^ r + = e~st'6(x ~ x')Hy - y')t (z - Λ
w h i c h h a s t h e s o l u t i o n ( n e g l e c t i n g t h e o t h e r s o l u t i o n e ST^c since G3 —> 0 as r —> 0 0)
G 3( r,s;t ’)
— si' p — s r j c
e e
r = lx — x'l
4 n r
O n i n v e r s i o n t h i s g i v e s
G a M;*') = 4^ * 3 ■
or
^"3 (*^) y ) z f t ) x , y f t )
s i t - t'- - y/( x - x')2 + (y - y 0 2 + (2; - z') 2)
= ^ 2--------------- 4 ^ ------------------------------- ^
The graphs of Green’s functions G\, G 2, and G3 are presented in Figs. 7.6, 7.7, and 7.8.
“1*
1
l/2c
|
J \
1
/ \
i
1 l/2 jic2t
1
1 1*1
0 ui
0
0
0
ct 0
ct 0
cr
F i g s. 7.6.
F i g. 7.7.
F i g. 7.8.
7.4. HYPERBOLIC EQUATIONS
247
In the above solutions (7.73)-(7.75), we find th;
source a t x' a t time t' affects the position of the
t if |x — x'| = c\t — t'\, i.e., when the distance field (or observation) point is c times the time, a wave propagating in all directions with veloci’ of time \t — t'\ the effect of the source is located away. The wave structure, and the non-occurren propagation, which is known as H u y g e n s ’ pr i n c. represented by Green’s functions (7.73)-(7.75)
a t the impulsive point field point x a t time f]rom the source to the The source generates ;y c. After a duration at a distance c\t — t'\ ce of diffusion in wave d b l e, in the three cases, is as follows:
JS
In R 1, the solution (7.73) shows t h a t the w stantaneously at a point source 6 ( x,t ) a t time interval — c t < x < c t, where there exist two edgi t h a t move forward with velocity c. This wave front edge and has amplitude 1/2c. Hence, w; this case. A three-dimensional representation ϋ 1 is shown in Fig. 7.9. I t can be viewed as t a t the point source and propagating as a plan front edge|x| = c t moves with the velocity c per x = 0. There does not exist a rear edge of the v
In R 2, Green’s function defined by (7.74) sho' originates instantaneously a t the point source δ( it occupies the entire circle |x| < c t (see Fig. 7 |or| = ct propagates throughout the plane with propagation exists behind the front edge a t all the wave has no rear edge. The wave diffusion < Huygens’ principle does not apply.
ave t h a t originates in- t > 0 will cover the es defined by x = ± c t observed behind the Ive diffusion occurs in of Green’s function in hat of a wave starting wave | a; | < c t whose pendicular to the plane if ave in this case.
rtvs t h a t the disturbance x, t ), and a t time t > 0 10). The wavefront at velocity c. But wave subsequent times, and iccurs in this case, and
248
CHAPTER 7: GREEN’S FUNCTIONS
I
Fig. 7.10. Wave propagation in R 2.
I n R 3, Green’s function (7.75) implies t h a t the disturbance t h a t originates a t a point source 6 ( x,t ) at time t > 0 occupies a spherical surface of radius c t and center at the origin. The wave propagates as a spherical wave with wavefront at |x| = c t and velocity c, and after the wave has passed there will be no disturbance (see Fig. 7.11). Huygens’ principle is applied in this case. The amplitude of the wave decays like r
_ 1 as the radius increases.
There is a significant difference between the two- and three-dimen­
sional cases. If a stone is dropped in a calm shallow pond, the leading water wave spreads out in a circular form with its radius increasing uniformly with time, but the water contained by this wave continues to move after its passage. This is because of the Heaviside function in the solution (7.74) which leaves a wake behind it. On the other hand, in the three-dimensional case if a shot fired suddenly at time t = t' in still air is heard only on expanding spherical surfaces with center a t the firing gun and radius c\t —1'\, where c is the sound velocity. But the air does not continue to reverberate after the passage of this wave. This is because of the presence of the Dirac delta function in the solution (7.75), which represents a sharp bang and no tail effect.
Huygens’ principle accounts for the simplicity of communications in our three-dimensional world. If it were two dimensional, the commu­
nications would have been impossible since utterances could be hardly distinguished from one another.
7.5. APPLICATIONS
I
Figs. 7.11. Wave propagation in R 3
7.5. A p p l i c a t i o n s
A n i m p o r t a n t a p p l i c a t i o n o f G r e e n ’s f u n c t i o n s f or L with homogeneous boundary conditions is in nonhomogeneous boundary value problems. Sir G(x, x'; t, t'), or G(x, x') in steady state problems the equation
L[G] = <5(x - x') 6 ( t - t ’ )
t o g e t h e r w i t h c e r t a i n p r e s c r i b e d h o mo g e n e o u s bc c a n u s e t h e l i n e a r s u p e r p o s i t i o n p r i n c i p l e t o d e t t h e n o n h o m o g e n e o u s e q u a t i o n
L [ u ( x,t ) ] = F ( x,t ),
w h i c h s a t i s f i e s t h e s a m e b o u n d a r y c o n d i t i o n s. T Ω C Ji" is given by
t)
u( x,t ) = / G ( x,x'-,t,t r) F ( x.
J n
We wi l l c o n s i d e r o n l y t w o - d i m e n s i o n a l b o u n d a r y
2 4 9
.Xi
a d i f f e r e n t i a l o p e r a t o r Bn d i n g t h e s o l u t i o n o f c e a G r e e n ’s f u n c t i o n , i s r e q u i r e d t o s a t i s f y
( 7.7 6 )
u n d a r y c o n d i t i o n s, we e r m i n e t h e s o l u t i o n o f
( 7.7 7 ) h e s o l u t i o n i n a r e g i o n
d x'. ( 7.7 8 )
value problems.
Consider the nonhomogeneous equation
^ 2u( x,y ) = F ( x,y ) i n Ω C R 2, ( 7.7 9 )
250
CHAPTER 7: GREEN’S FUNCTIONS
which is subject to certain nonhomogeneous boundary conditions on the boundary Γ of the region Ω. Then, in view of (7.19) where we take v = G, we have
U(x',y/) = J J G ( x,y;x',y') F { x,y ) d S + J ~ d s -
(7.80)
du
The boundary conditions on both u and — are, in general, not pre-
on
scribed a t all points of Γ, but those on G are prescribed precisely, i.e., the boundary conditions on G are such t h a t they annul whichever val- d u
u e s o f u or —- are unknown. For example, let u be known (but not d n
d u
o n a p o r t i o n Γ ι o f Γ; t h e n, i f u = 0 is prescribed on Γι, we must
have G = 0 on Γι, because only then will the integrals in (7.80) vanish. We will consider the following three types of boundary conditions:
(a) (Dirichlet) u is prescribed on Γ;
du
(b) (Neumann) — is prescribed on Γ;
on
du
(c) (Mixed) u is prescribed on Γι and —— is prescribed on Γο, where ^ o n
Γ ι U Γ2 = Γ.
I n e a c h c a s e G r e e n ’s f u n c t i o n G m u s t s a t i s f y t h e s a m e h o mo g e n e o u s b o u n d a r y c o n d i t i o n a s t h a t s a t i s f i e d b y u.
I n t h e c a s e w h e n F ( x, y ) = 0, Eq (7.80) with the Dirichlet boundary condition (a) becomes
f d G
u(x',y') = J u — ds, (7.81)
du
where G = 0 on Γ. But if — is prescribed on Γ (Neumann condition
(b)), Eq (7.80) becomes
u ( x\y') = - J G ^ d s, (7.82)
, d u where — = 0 on Γ. d n
E x a m p l e 7.1 4. L e t Ω b e t h e h a l f - p l a n e y > 0. Then Green’s function associated with the Dirichlet boundary condition u = 0 on the
7.5. APPLICATIONS
251
boundary y = 0 is given by (7.59). Hence, for u from (7.81) t h a t
/ / ,s j ι ,, x v | , yu, — s'^ 2
u
/ / t\ 1 f ° ° t, Λ d Γΐ ( χ - χ 1) + ‘ { x'y ) = - T * j _ J ( x ) - FyV'i (!C- x - r +
,_ j _ r/M _ i v c
4tt 7-00 (x - x') + y'
= t Γ _______μ _
π Λ o o ( x - x') 2 + y/2
(y - y O2'
( x, 0 ) = /( x ) w e f i n d
E x a m p l e 7.1 5. Let Γ b e t h e ci rcl e r — a, a
In this case Green’s function is given in Exercise
on the circle Γ, we find from (7.81) t h a t
. . ,. 1 Γ2π 9 r a 2 [ r 2 - 2 r r'c os ( 6
u ( r , θ') = — / /( 0) τ - ln — ^ i -
4 π J o d r L r 2 r/2 _ 2 r r'a 2 cosi
= J _ Λ Γ 2r - 2r'cos(fl - 0')
4 τ τ Λ L r 2 — 2 r r'
n d l e t u ( a,0 ) = f ( 0 ).
d d
7.1 5. S i n c e — = - r - o n o r
c o s (# — θ') + r
2 r r - 2 a 2r' cos( 0 - θ')
2 π
r 2 r'2 — 2 r r'a 2 c os ( 0 — θ') + a 4 -I r =.
[ 2π _ m
Jo a
a2 — r'2 ^
r,2[r2 — 2r r' cos( 0 — θ') + r
E x a m p l e 7.1 6. To f i nd t h e ha r moni c funct l · p l a n e x > 0, y > 0, subject to the boundary cond:
3φ
φ(χ, 0 ) = /( χ ), 0 < χ < oo, and ^;( 0, y ) =
n o t e t h a t i n vi ew of ( 7.80), t h e s ol ut i on is gi ven
r ( dG dφ
φ { χ · ί ) = ΐ Α φ^ ~ 9^
whe r e G is defined by Eq (7.60) (Example 7.8 the quarter-plane {x > 0, y > 0}. Then f°° r ί ) Γι
Φ(χ'ι y') = — J f i x ) [ — 1 d x
Γ
■ I [ f l - » 9'»*''9
- f ( x ) \----------------- 2 +
π J 0 L ( x - x') 2 + y ’
+^L
x'2 + ( y + y 1) 2
g(y)
( y + y') 2 i y=o
d x
- θ') + r'21
θ — θ') + Ο 4 -I r=a
α ά θ
α ά θ άθ.ι
,/2
o n φ in the qu ar te r- itions
g(y), 0 < y < oo,
by ds,
a n d Γ i s b o u n d a r y o f
J 1
r/')2 I yl 2 J
+ x') 2 + y' dy.
d x
252
CHAPTER 7: GREEN’S FUNCTIONS
Note t h a t the signs in the two integrals above result from the fact t h a t d d d d
— = — — and — = — — for the quarter-region. ■ o n o y o n o x
7.6. E x e r c i s e s
7.1. F i n d a h a r m o n i c f u n c t i o n φ in the semicircular domain r < a, 0 < θ < π, such t h a t φ { τ, 0) = 0, φ ( τ,π ) — k = φ ( α,θ ), where A; is a constant.
o ° / T \ n
Ans. Φ( γ,Θ) = Α θ + Β + I - ) [cn β ί η η θ + d n cosn#] is the
n
= 1 'O'
general solution of the Laplace equation in polar cylindrical coordi-
δ2φ 1 8 φ 1 δ 2 φ , ,
nates Η — I— —τ. The boundary conditions on θ = 0, π
or· * r o r r2 ο θ ι
a r e s a t i s f i e d i f we t a k e
Θ ^ ^
<f>(r, Θ) = b > c n ( - j sin η θ.
π \a J
n
= 1
From the last boundary condition on r = a we get
» ke S2' ■ n
k = h > c n smnw,
π ί
71 = 1
which gives
π \ π J ηπ
Thus, the required harmonic function is
k9 2k v—' /r\n sin η θ
k 0 2 k /r\
*
μ > = τ + τ Σ (;)
n
n=l
Note t h a t this solution represents the stream function for the two- dimensional flow of a je t liquid from a Borda mouthpiece (Mackie, 1 9 8 9 ).
7.6. EXERCISES
253
.2. Show t h a t Green’s function (7.42) satisfies Ans. The function G( x,t ) is locally integrabl
t h e c o n d i t i o n ( 7.4 1 ).
3 i n R n+1 a n d
[ G ( x,t ) J Rn
(4παί) η/2 JRn β
n 1 roc
~U^L‘
kl
dot = 1.
I f t > 0 , then G e C°°. Thus,
d G dt
d G
dxi
d 2G
\X\
Aat2 21
Xi 2 a t
- S i g,
G,
d a u 2
Xi
1
4 α2ί 2 2 a t
G.
H e n c e
^ - - n V 2C = ( | x| ~ 71
d t l 4 at2 21
Fl 4 at'
L e t φ € V { R n+l ). Then, using (7.84), we hav
^ - av2G’^=-<G’!l +av2^
ί """
= l i m J ΰ ( χ,ε ) φ ( χ,ε )
= l i m / ΰ ( χ,ε ) φ ( χ,ε ) ι ^ J Rn
+ l i m f G(x,e) [</>(.
£- >0 JH"
= lim / G(x, ε ) φ ( χ,0 ) ά, ^ J Rn
/4at dx
( 7.8 3 )
( 7.8 4 )
n
2i
G = 0.
+ a V 2</> ) d x d i + α,ν2φ ) dx dt
\Έ + αν'φ)
d x +
2 G^ φ ά χ ά ί
v +
ε ) — φ( χ, 0)] d x
254
CHAPTER 7: GREEN’S FUNCTIONS
which, in view of (7.83), gives
I
G ( x, t)[0(x, ε ) — φ ( χ, 0)] d x < Κ ε ί ΰ ( χ,ε ) ά χ = Κ ε.
J R n J r «
We will now show t h a t G ( x,t ), defined by (7.42), converges to 6 ( x ) as t —> 0+ in V'( R n ). Let φ { χ ) £ V. Since
[ G ( x,t ) [ 0 ( x ) - 0 (0 ) } d x < K f e - N 2/4a t\x\d x
J r" (4παί)
(4παί) Λ)
.«!| Λ Γ,ν.· 4 = 5 Λ
^ n/2 Λ
then, as t —> 0+, we have
( G ( x, ί ),0 ) = ί
G ( x,ί ) φ ( χ ) ά χ = 0(0) [ G ( x.,t ) d x +
J r" J r ™
+ [
G ( x, i ) [ 0 ( x ) - </>(0)] dx -> 0(0) = (<5,0 ),
JRn
i.e., G ( x,f ) —> 6(x) as t —» 0+ in O'{ R n ).
7.3. S e t a t = ε > 0 in (7.42) for n = 1, and consider the sequence of
functions /ε(χ) = .— e~x ^4ε. Show t h a t the sequence {/ε(χ)}
2y7r£
converges to <5(x) in i i1 as £ —> 0.
A n s. Notice from Fig. 7.2 t h a t the peak profile for Green’s function (7.42) smoothes out gradually as ί increases. Hence, the functions /e (x) vary appreciably over successively smaller intervals about the origin as ε —► 0. Thus, for any φ e C ° ° ( R 1),
/
oo 1 λοο
0
( x )/£ ( x ) d x = — - =
/ 0
( χ ) ε “ χ2/4ε d x
- oo Ζ\/7 Γ £ J— oo
- £ £ < * r
( 7.8 5 )
w h e r e we h a v e u s e d t h e M a c l a u r i n s e r i e s e x p a n s i o n f o r 0 ( x ) a n d t h e n t h e s u b s t i t u t i o n x = 2- y/i · A n e v a l u a t i o n o f t h e i n t e g r a l o n t h e r i g h t s i d e o f ( 7.8 5 ) s h o ws t h a t
l i m (/e ( x ),0 ( x ) ) = 0 ( 0 ).
7.6. EXERCISES
255
1
T h u s, t h e f u n c t i o n s f s { x ) a p p r o a c h δ ( χ ) a s ε
7.4. Show t h a t
o = l.
A n s.
/
OO
G ( x,t )
-OO
d t =
(n - 2 ) S n ( l
/
OO poo -i
G ( x,t ) d t = / ^4t d t
-oo Jo ( 2 v^rt)
I ^,[2 —n /.oo
/ u"/2 ~ 2 e ~ u d u
Jo
4πη/2
( n - 2 ) 5 n ( l )
Μ2“ η, n
> 3.
7.5. The complex conjugate of the Schrodinger equation in R 1 x R + is
du ti2 d u 2
dt 2m 9x2 ’
Find its Green’s function.
ifl
A n s. Set a = — in ( 7.4 2 ). 2m
7.6. F i n d Gr e e n ’s f unc t i on for t h e F o k k e r - P l a n
d u d i d
m = ai{ai+x''t
A n s. G r e e n ’s f unc t i on sat i sf i es
m ~ Tx { i + *) ]G(x·t;x'· *"> i(l· x')S{t·(,)·
Let χ = X e ~ t, u = v e *. Then the Fokker-Plank equation becomes
dv
J i t
d 2v
dt 6 d X 2'
dv d 2v d f = d X 2 ’
I f we s e t 2T = e2 t, t h e n t h e a b o v e e q u a t i o n
0.
x| ", n > 3, with
r(s-0
i2—n
4πη/2
:k equation in R l
r e d u c e s t o
256
CHAPTER 7: GREEN’S FUNCTIONS
which, in view of (7.45), has Green’s function
G(x,t;x',t') = - ···· H ~ f/) = e - ( » - » - (t- 1'))2/2(i-e-i<‘- ‘,)) v ’ ’ ’ ' -^2^(1 - e-2(*-*'))
7.7. F i n d t h e a xi s ymme t r i c s ol ut i on of t h e e q u a t i o n
s uch t h a t G e ( a ) = 0 and G'e ( r ) is finite at r = 0, where r > a > 0,
for the (a) A e ( r ) = —ί— e ~ r /4ε, and (b) Ae(r) = - A t e ~ r .
4 π ε π ε 2
S h o w t h a t G ( r ) = lim Ge(r) = — l n r for any fixed r = |x| > 0. ε—>o 2π
A n s. (a) Integrating the given equation with respect to r, and taking the constant of integration zero since G'e { r ) is finite at r = 0, we get
d G e T dr
= έ Η - Γ’/4'] ·
Hence
1 r 1 Γ e-z2/4£
G e{r) = — I n - - — / ------------ d z, r > a > 0,
2π a 2π J a z
w h e r e t h e i n t e g r a l o n t h e r i g h t s i d e t e n d s t o z e r o a s ε —* 0 for a fixed r > 0. Hence
G(r) = lim Ge(r) = -j-ln J.
ε—>0 Ζ7Γ a
W i t h o u t l o s s o f g e n e r a l i t y, we c a n t a k e a = 1 since r extends to the entire R 2 plane, and the result follows. P ar t (b) can be solved similarly. Note t h a t the function A e ( r ) reduces to the delta function as ε —> 0. Thus what we have shown is t h a t V2G(r) = 6 ( r ), which is true in view of the property (7.8).
7.8. Find Green’s function for the Laplacian in R 1.
A n s. F o r G r e e n ’s f u n c t i o n o f t h e o n e - d i m e n s i o n a l L a p l a c e e q u a -
d 2 G
t i o n, we n o t e t h a t ( 7.5 3 ) b e c o me s = δ ( χ ι χ>)> whose general
solution for a fixed x' is
G(x) = ~^\x — x'\ + Ax + B.
7.6. EXERCISES
257
If we require spherical symmetry, i.e., G = G and we set B = 0. Then Green’s function
- i | x - x'|.
7.9. Prove t h a t Green’s function G(x, x') for the (7.56), satisfies the following properties: (a) out the domain Ω C R 3, and (b) Green’s func G (x, x') = G (x', x).
A n s. (a) G ( x,x') = 0 on S and G ( x, x' of a small enough sphere centered at x' sin x —> x'. In view of Theorem 7.6 (maximu: t h a t G (x, x') > 0 throughout Ω. Since G(: find t h a t G (x, x') < 0 in the closed domain
Laplacian, defined by G (x, x') > 0 through- tion is symmetric, i.e.,
> 0 on the surface :e G ( x, x') —> +oo as principle), it follows ,x 0 ) | s = —1/4T t r, we Ω, which implies t h a t
ra
0 < G ( x,x') <
1
4 n r
i n s i d e Ω. ( b ) A p p l y G r e e n ’s s e c o n d i d e n t i t y v = G ( x, x2), where the integration domai domain Ω minus the two small spheres Si an tively, at χχ and x 2, each with radius ε. The:
[ i f [G (χ, χχ) V2G (x, x 2) - G (x, x 2) V 2G (χ, χχ)] d V J J J n 2
- 11.
S - S i - S i
d G ( x,x 2) G ( x,x x ) — G (x,
or
0 =
IL
+
ii
G ( χ,χ ι ) d G ^ X2^ - G ( x,x 2 G t x.x O ^ & ^ - O i x,
d n
As ε —+ 0, t h e i n t e g r a l over S\ (centered at χ χ and similarly the integral over S2 (centere G ( x 2,x x ), which proves the symmetry of Gr·
7.10. Find Green’s function for the Laplace Operator in a sphere of radius R.
A n s. I n F i g. 7.12, we n o t e t h a t t h e ρ ο ί φ x = (x,y,z ) is inside Ω, p = \O M\, p\ =
Μ χ = χ χ = (χχ, y i, ζ χ ) is the inverse point td
( | x - x'| ), t h e n A — 0,
is g i v e n b y G ( x ) =
( A.7) t o u = G (χ, χχ), η Ω2 is chosen as the d ,S2, centered, respec- m
\ 9 G (χ, χ χ )',
Χί) a„ 1J S'
d S +
d G (χ, χχ) d n
d G ( χ,χ χ )',
X2) τ:------- I d S.
d n
wi l l t e n d t o G ( χ χ, x 2) ), d a t x 2 ) wi l l t e n d t o e e n ’s f u n c t i o n.
M with coordinates (pMx |, where the point M such t h a t
ppi = R 2.
258
CHAPTER 7: GREEN’S FUNCTIONS
Let Ρ = ζ = ( ξ,η,ζ ) be an arbitrary point on the surface S of the sphere. Then, since the triangles OMP and OM\P are similar (having a common angle a at the origin), we find t h a t r/r\ = p/R, i.e.,
1 _ R l _
r p r\
H e n c e, G r e e n ’s f u n c t i o n f o r t h e s p h e r e i s
G(i,x') = ^ - = j - - -.
4 7 ΐ τ 4 π p r\
M,
7.1 1. F i n d Gr e e n ’s f unc t i on for t h e o p e r a t o r V2 — k 2, where fc > 0 is real.
A n s. Follow the method of Example 7.10 and deduce t h a t Green’s function is
— sinh(fcr) in R 1,
G ( r ) =
2k
^ K 0 ( k r ) i n i?2,
—— e t kr in R s,
4 π τ
w h e r e r = |x — x'|. Green’s functions are used in the problems of neutron diffusion.
7.12. Show t h a t Green’s function (7.71) satisfies Eq (7.61).
s i n k x
A n s. S i n c e t h e f u n c t i o n s c o s k\x\ a n d —rr-— b e l o n g t o t h e c l a s s
7.6. EXERCISES
259
JL J_ = _ JEi. JL eik\x ι = kx
d Xi |x| |a;|3 ’ d x i
V - l · Ve ifc|x| = ^ e ik^
'T· I „ l *
C ° ° ( R n ), a n d s i n c e, i n v i e w o f E x e r c i s e 7.4,
V - 7 = -4τΓ(5(χ) (by (2.l)3)),
y2eife|x|
2ifc
k2 ) ei kM,
we apply the Leibniz formula V2( a/) = /V 2a + 2Va · V/ 4- a V 2/, and find t h a t
ι / i
72 I i,2\ _ _ cifc|x| y 2 1 I 2 ( V
lil
(V2 + fc2)-
X
2i k ,i.
bl·
F l 2 i k
b l
= — 4 π 6 ( χ ).
.1 3. Sh o w t h a t G r e e n ’s f u n c t i o n f o r t h e Ca u c
d d . d . . . 1
— = — + ι —, ζ = x + i y i s G(z) = —.
a z a x o y n z
A n s. We will show t h a t G ( z ) defined above
V g ^ N ^ J _ y 2 e » f e | a:|
J \x\
— t _ eik\x\ _)_ b— eik\x\ b l kl
hy-Riemann operator
satisfies the equation
^=G(z) = = 6(x ^y)·
Let C be the boundary of a region D C R 2, and let / G C 1 ( D ) be such t h a t f ( z ) = f ( x,y ) = 0 and z e R 2\D. Let us assume t h a t the (closed) contour C is piecewise smooth and is traversed in the positive sense such t h a t the region D remains to the left. Then, since
d f
df
f
260
CHAPTER 7: GREEN’S FUNCTIONS
w e h a v e
JD ^^^dxdy= ~^c^dz-
Since the function 1/z is locally integrable in R 2, we take l/z and D = {ζ : ε < \z\ < R } (see Fig. 7.13, where C is the same as C r ). Then for all φ e V such t h a t supp φ c U r, we have, using the above result,
,5.1
"dz ζ'1
/ 1
JuR z
j r: dx dy d z
/ε<|ζ|<Λ
p d x d y z d z
= — lim f
ε-°Λ
= lim f f
^ I d x d y + l f/· - [ Y d z
d z z 2 \J c R J c J * .
= — ^ lim [ - ■ —- d z = — ^ lim i f φ { ε β %9) ά θ
^ ε ~ * ° J\z\= s 2 2 ε^Ο
/|ζ|=ε 2
= π φ ( 0 ) = < πδ, φ >
7.1 4. Us e t h e m e t h o d o f i ma g e s t o d e t e r m i n e G r e e n ’s f u n c t i o n f o r a d i s k o f r a d i u s a. HINT: The image of a charge at a point P' inside the disk has an equal but opposite charge at the inverse point Q' (Fig. 7.14). Note t h a t if the coordinates of P' are { τ', θ'), then those of Q' are { α 2/τ', θ').
7.6. EXERCISES
261
An s. In polar coordinates
1 , P P' 2π n P Q 7
whe r e C is a constant which is added to ensure t h a t G = 0 on the
boundary, we have on
boundary. Since
P P'
PQ'
= — when P is on the
1 r
t h e b o u n d a r y — l n h C = 0, which gives C
2 π a
η, η ι λ/\ 1 i a2[r2 — 2rr' cos(^ — θ') + r/Z]
G ( r,9;r,« ) = -!„ — —
r/2 L.2 _
■ COS
1 ^ a 2 [ r 2 — 2 r r'c o s ( 9
4 π r2 r/2 _ 2 r r'a 2 cos( 0 — 0') + a4
7.15. Solve the Laplace equation d 2u
1 <92«
1 <9u
9 r2 ^ r d r ^ r 2 8Θ2
s u b j e c t t o t h e c o n d i t i o n s u ( r, 0) = f ( r ) and Ans. This is the well-known problem of sol· tion in the half-plane y > 0. Taking the defined by
C,
■-= — ln —, and hence 2tr r'
{Θ
- θ') + 0') + r'2]
0,
u(r, π) = π for r > 0. ving the Laplace equa- Fourier sine transform
w(r, n) = / u ( r, Θ) sin η θ ά θ,
Jo
262
CHAPTER 7: GREEN’S FUNCTIONS
we g e t
2 d 2u d u 2 _ ,
r - r - z + r — n u — - n f ( r ).
d r 2 d r
T h i s e q u a t i o n i s s o l v e d s u c h t h a t t h e s o l u t i o n i s f i n i t e a t r = 0 a n d t e n d s t o z e r o a s r —> oo. To do this, first we determine Green’s function associated with the above problem. The solutions for the homogeneous equation are r ±n, and thus
G(r,r') — <
u ( i y, rSr.
f \ n
A ( — ι , r > r r
w h e r e A is such t h a t G (r.r') has a discontinuity of derivative of amount 1 j r'2 at r = r', i.e., A = - l/2 n r. Then
u = —n
pOO
/ G { r,r') f ( r') d r',
Jo
a n d
u ( r, n )
wh i c h g i v e s
u( r.
f®
1 r Γτ 1 00 / f \ n
’ e ) = - [ j o ^7 ^ ( 7 ) sin η θ f { r') d r'
/
OQ -
^ Σ ( ^ ) sin n0f{r')dr'
K?»(re")/(/) *'
+ /” ^ Σ ( ^") “/M*'
= “ ί ί ^ T- 7 - ^ f ( r') d r'
π I Jo τ' r — τ'ειθ
1 r' ι
+ l
-w;
r s i n t
r 2 — 2r r' cos Θ + r
f ( r') dr'
7.6. EXERCISES
263
+
pOO
Jr r'2 -·.
rsint
2r r' cos Θ + r s i n 0/( r')
= l~ [
π Jo
= - 1
π Jo i x ~ r') 2 + y 2
r 2 — 2r r' cos Θ + r “ ><r'>
f ( x ) on the entire
in Cartesian coordinates. Note t h a t if we remove the restriction on u being zero on half of the x-axis, and take x-axis, then we must add the solution for
above solution, which will give the solution under this condition as
u{ x,y ) = ^ [
m
( x — r') 2 +
f{ r') dr'j
dr'
< 0, i.e., to the
• d x'.
7.1 6. L e t a f u n c t i o n u e be defined by
ωε ( χ,2/) = — l n ( r + ε), r
Show t h a t V2ue gives a function which tenc s to the delta function S(x)6(y) as ε —> 0.
Ans. Since V2ue = 0, and ue —> — l n r as
2 π
behaves like Green’s function for the Lapla as ε —> 0. Using the requirement (7.53), t h e i
V2ue(x, y ) = δ ( χ, y ) = S { x ) 6 ( y ).
x 2 + y 2.
ε —»■ 0, we find t h a t u e ian in R 2 in the limit
7.17. Show t h a t for the wave operator
/
OO
G3(x, y, z, t; x', y', ζ', t') d z' = G 2 ( x, y, t; x', y',t').
-OO
/
o o
15(z — 2/) dz' = 1, we find from (7.75) t h a t
-OO
/
OO
G3{ x,y,z,t;x',y',z',t') dz'
•OO
, r oo <5 ( t - 1' - ^λ/( χ - χ')2 + { y - y')2 + ( z - z') A - 1 / V
_________ c_ ___________ / >
264 CHAPTER 7: GREEN’S FUNCTIONS
y/{ x - x')2 + { y - y1)2 + { z - z')2
/
OO -OO
, - o o s ( t - t' - - ^ { x - x')2 + ( y - y')2 + u2)
— / — -------- ° ------— dz'
4π J-oo i/( x - x') 2 + (y - y')2 + u2
where z — z' = u 6{t — t'—v)
r
J r /
Jrj c \!(?v 2 — r 2 w h e r e r 2 4- u2 = υ 2, r 2 = (x — x')2 + ( y — j/) 2
for t — t'< r j c
2π \/c2v2 — r 2 = G2(x,y,t-,x',y',t').
7.18. Show t h a t for t h e wave o p e r a t o r
/
OO
G2( x,y,t;x',y',t') dz' = Gi(x,t;x',t').
•OO
An s. Fol l ow t h e me t h o d i n Exer ci s e 7.19.
7.19. Use Four i e r t r a n s f o r m me t h o d t o der i ve t h e G r e e n ’s f unc t i on Gi(x,t;x',t'), for the wave operator Dc.
Ans. Apply the Fourier transform T x to the equation
d2G\ 2d2G\
6(x — x')6(t — t').
dt2 dx2
For the sake of simplicity, let us translate the source point to the origin, take t' = 0, and denote ^rx [Gi(x, i)] by Gi(a,t). Then after an application of this Fourier transform we get
^ G i ( a,t ) +c2\a\2Gi{a,t) = 1(a) -6{t), which has the solution
- , , __y v sinclali
O i ( a,t ) = H ( t ) ----:—:—.
c\a\
7.6. EXERCISES
265
Since
.F-1
s i n c | a | i
a
A-Kct
Ssc
(see Fig. 7.15), we find t h a t
H ( c t — x ) = 2c '
X\
8
Weigh ted Resi du al Met hod s
Variational formulation of boundary value problems originates from the fact that weighted variational methods provide approximate solutions of such prob­
lems. Variational methods for solving boundary value problems are based on the techniques developed in the calculus of variations. They deal with the prob­
lem of minimizing a functional, and thus reducing the given problem to the solution of Euler-Lagrange differential equations. If the functional to be min­
imized has more than one independent variable, the Euler-Lagrange equations are partial differential equations. Conversely, a boundary value problem can be formulated as a minimizing problem. The functional which corresponds to the partial differential equation is generally known as the energy function. In the case when the solution is not available in a simple form, an approxi­
mate solution such that it minimizes the energy equation can be found. The
OO
approximating function is a linear combination of the form ^ α φ ί, co = 1,
i=0
of specially chosen functions φ ί which are known as the test functions or in­
terpolation functions. The function φο satisfies the same boundary conditions as the original unknown function, while the remaining functions φ ί,ί φ 0, satisfy the homogeneous boundary conditions. The constants c t, i φ 0, are then determined by minimizing the energy function.
The weak variational formulation is defined in §8.4, and the problem of constructing an appropriate functional for a given partial differential equation is discussed there. The Galerkin and the Rayleigh-Ritz weighted residual methods are examined, with examples, in §8.5 and §8.6, and some less fre­
quently used weighted residual methods, like the collocation method, the
least-square method, and the method o f moments, are outlined in §8.9.
| www | Refer to the Mathematica Notebook g a l e r k i n. ma for this chapter.
8.1. LINE INTEGRALS 267
8.1. L i n e I n t e g r a l s
In many types o f boundary value problems, the variai: provide precise formulations which can be applied o f coordinates. We shall derive the Euler equation condition for the solution o f the following problem: which the integral
ional methods are used to in any prescribed system which is the necessary Find a function u ( x ) for
I ( u ) = / F ( x,u,u') d x
J a
t h
i s a m i n i m u m, w h e r e F is twice-differentiable wii two fixed points A ( a,c ) and B ( b,d ) in the x y - pi; Γ : χ ι— > u ( x ) (see Fig. 8.1(a)). Then c = u ( d u/d x = u'( x ) at each point Q o f Γ. Thus, the cu o f the integral in (8.1). The value o f the integral 1 we replace Γ by a new curve joining A and Β. I variation o f I with Γ, i.e., to determine as to for w has a minimum (or a maximum) value, we shall co: which are defined as follows: First we select any ciii
0(a) = 0 = 0(6)
(see Fig. 8.1(b)). Then for any value o f the parameter a, the curve Γα is, in view o f (8.2), defined by
U = U (x) = u ( x ) + α φ ( χ ), U ( a )
H e n c e i t p a s s e s t h r o u g h A and B. From (8.3) we fi
U· = ψ - = u'( x ) + α φ'(: d x
( 8.1 )
r e s p e c t t o x, u. u'. Let ine be joined by a curve a), d = u ( 6), and u' = rve Γ determines a value will, however, change i f i order to investigate the iat curve Γ the integral I iffine to a set o f curves Γα irve u = <p( x) such that
(8.2)
c, U ( b ) = d. (8.3)
nd that on the curve Γη
(8.4)
268
CHAPTER 8: WEIGHTED RESIDUAL METHODS
X
u = §(x)
Fig. 8.1(b).
For any value o f a, we thus have a curve Γ α and may form the value o f I along Γα by substituting the values from (8.3) and (8.4) as
I ( u,a ) = f F ( x, U, U') dx. J a
(8.5)
U s i n g t he di f f e re nt i a t i on f or mul a
d F d F d U
d F d U'
d a d U d a d U' d a ’
and noting from (8.3) and (8.4) that = φ ( χ ), = φ'( χ ), we find that
d a d a
8 F
d a
d_F 1d U
ΊΓ7 = Φ( χ ) ί^ 7 +Φ'(χ)§£ρ·
Hence,
I f we integrate by parts the second term in the integrand in (8.6), and use (8.2), we get
dl fb xrdF d /d F\-\,
d ^ = J a M d U - d i i d i r )\d x · (8-7)
Let us assume that there is a twice-differentiable curve, say Γ, for which the value o f I is a minimum. Then, the value o f I on Γ will be less than the value o f I on any other curve Γ α. Thus, 1 ( a ) will assume a minimum value for a = 0, since d l/d a is continuous. But, then U = u and U' = u' when a = 0. Hence, taking a = 0 in (8.7), and I'( 0) = 0, we find that
/
J a
,. . rd F d ( d F\-\ ,
(8.8)
8.1. LINE INTEGRALS
269
Since the solution φ(χ) is arbitrary, except for the c in the square brackets in (8.8) is continuous, which
onditions (8.2), the factor mplies that
dF_ d /dF\ du dx V du')
0
Id
nd
for all x € [a,b]. In fact, i f the expression withi ( 8.8) were nonzero at any point, say x q, there wou! including this point, say x\ < x q < χ?, in which by taking the function φ(χ) in Fig. 8.2, the integral except for the interval ( x i,x 2 ) where it is nonzero If the integral I, defined by (8.1), has a minimum (o sufficiently smooth curve Γ joining A and B, then u o f the differential Eq (8.9). This is known as the E i i l and its solution u[x) as the extremal. Note that the computed by recalling that u = u(x) and u1 = u'{i) o f x. After this differentiation is carried out, Eq (8.
dF
du
d2F d2F du d2F
dxdu' dudu' dx du'2
which is a second order differential equation, and arbitrary constants which must be determined froi curve passes through A and B.
S i mi l a r l y, b y f o l l o w i n g t h e a b o v e p r o c e d u r e a n d t w i c e i n t h e t h i r d t e r m a p p e a r i n g i n t h e i n t e g r a n d o f
dF dF
X
( 8.9 )
n t h e s q u a r e b r a c k e t s i n b e s o m e s m a l l i n t e r v a l t r e m a i n s n o n z e r o; t h e n, i n ( 8.9 ) w o u l d b e z e r o T h u s, w e h a v e p r o v e d: τ a m a x i m u m ) a l o n g a n y = u(x) will be a solution er-Lagrange equation, derivative d/dx in (8.9) is du/dx are functions becomes
<>)1
d2u
dx2
0,
(8.10)
its solution contains two nn the conditions that the
using integration by parts
d2F -1 dU" \
dx,
(8.11)
270 CHAPTER 8: WEIGHTED RESIDUAL METHODS
and taking the additional requirement that φ'( α ) = 0 = </>'(&), we find that the Euler equation which provides a necessary condition for the functional
I ( u ) = f F ( x,u,u',u") d x (8-12)
J a
t o be a mi ni mum i s
d F d /d F\ d 2 / d F \ n /0 1 0,
(see Exercise 8.1.).
If the integral I depends on two functions u and v, i.e., i f
I(u,v) = ί F ( x, u, v, u', ν') d x, ( 8 - 1 4 )
J a
then there are two Euler equations:
dF d ( d F\_ dF d/d F\_ du d x \ du') ’ dv d x\ dv' /
To d e r i v e ( 8.1 5 ), i n t r o d u c e t w o f u n c t i o n s φ( χ) and ψ( χ) and two parameters a and β such that U = u + αφ( χ), V = ν + βφ{χ) · Then d l/d a = 0 and 3 1/θ β = 0.
8.2. V a r i a t i o n a l N o t a t i o n
Eq (8.3) implies that the difference U — u = α φ ( χ ), or Q Q a in Fig. 8.1, is the change in U, starting from u and the value o f a — 0. Hence α φ ( χ ) = U — u is called the (first) v a r i a t i o n in u and is denoted by S u. It may also be regarded as the differential d U, since d a = a — 0 = a at a = 0; then d U/d a = φ ( χ ) so that
d U = ( ^ j J - ^ d a = φ ( χ ) α = S u. (8.16)
Also, Eq (8.4) may be rewritten as U* = u' + α φ'( χ ), or
U'- u' = α φ'( χ ). (8.17)
The term α φ'( χ ) = U' — u' is the variation o f u' and is denoted by 6 u'. It is also the differential
dU' = ( ~ j ~ ) d a = φ'( χ ) α = Su'. (8.18)
From (8.16) and (8.18) it follows that
.2. VARIATIONAL NOTATION d ( S u )
d x
(6u)' = 6(u'),
( 8.1 9 )
Va r i a t i o n o f u" and higher
i.e., the operator δ and d/d x are commutative. The derivatives o f u are defined analogously.
If u is changed to u + S u = u + a w, then the corresponding change in a function F = F ( x, u, u') at u is defined by
SF = ^ S u + ^ S u'. a u ou'
Since δ behaves like a differential operator, it has th
(i) <5(/i ± f a ) = 6 f i ± 5/2,
(ii) <5(/i f a ) = f i S f a + f i <5/i,
(m)S(h/fi) = (fi6f1- f 16f2)/fi, (iv) 6( f ) n = n(/)"_1 6f,
( v ) D ( S u ) = S ( D u ),
( v i ) δ j ^ u ( x ) d x = /ab 6 u ( x ) d x,
where ' = D = d/d x, and /1, /2 are functions of (vi) are commutative properties under differential a! above six properties ( i ) - (v i ) can be easily proved
Now, we find from (8.7) that
δ ΐ
bv
rd F d /O F
( d a ) a=o Ja \-du dx'
w h e r e t h e v a r i a t i o n s a t t h e e n d p o i n t s a r e z e r o. ( 8.1 2 ) w e h a v e
S:i:
δ ΐ
/
J a
^ d2 ( d F
ί d u d x\ d u') d x 2 V d u h
w h e r e t h e v a r i a t i o n s a r e r e s t r i c t e d b y t h e c o n d i t i o n s
F o r t h e i n t e g r a l I in (8.14) whose integrand F ( x independent variables u and v, we define the variat: so that
Su = α φ ( χ ), 6 u ’ — α φ ’ ( χ ), S v = α ψ ( χ )
2 7 1
( 8.2 0 ) e f o l l o w i n g p r o p e r t i e s:
u, u'. Note that (v) and nd integral operators. The using (8.20).
($£)]*“**’ (8'21)
milarly, for I defined by
)] 6udx, (8.22)
: φ ’ ( α ) = 0 = 0'(6).
, u, u', ν, ν') contains two ions as total differentials
δν' = α ψ'( χ ).
272
CHAPTER 8: WEIGHTED RESIDUAL METHODS
Then the definition in this case becomes
d F d F d F d F 6 F = ^ 6 U + ^ 6U' + ^ 6 V ^ SV>· ^
For the integral I in (8.14) we have cr f b \d F d f d F\i , [ b i d F d f d F\i c J . .
6I = J a [ t o - t e { w )\Sudx+J a [ - ^ - ^ { ^ )\δνάχ· (8·24)
In the parametric case, we can introduce a new parameter t as the indepen­
dent variable and regard x and u (or x, u and v ) as the dependent variables. For example, let us consider the integrand F ( x, u,u'). Using the hat to de­
note differentiation with respect to the parameter t, we have: u' = u/x, and F ( x, u, u') d x = F ( x, u, u/x ) x d t. Let x = a when t = t\, and x = b when t = t 2, and denote F ( x,u,u/x ) x by G ( x,u,x,u ). Then, the integral (8.1) can be written as
rb pt2
1 = 1 F ( x,u,u') d x = / G ( x,u,x,u ) d t,
J a J t i
and (8.21) becomes (using (8.24) with x, u, v replaced by t,x, u, respectively) cr f t 2 r d G d/d G\λ. ^ f t 2 r d G d/d G\i c ,
6 I = Jtl [ - ^ - d A w\6xdt +Jtl [ ^ - ά Λ ^ )\δη(1ί’ (8·25)
where δ χ and S u are zero at i i and t 2.
E x a m p l e 8.1. S h o w t h a t i n R 2
V ( S u ) ■ V u = ^ | V u | 2. A direct c omput at i on shows tha t
(8.26)
8.3, MULTIPLE INTEGRALS
273
8.3. M u l t i p l e I n t e g r a l s
Let F ( a,y,u,p,q ) be a twice-differentiable fun< where t h e dependent variable u is a function oi' d u/d x, q = u y = d u/d y. We shall st u dy t h e va:
ct ion o f five variables, x and y and p = u x = riation o f t h e integral
I ( u ) = J J F ( x,y,u,p,q )
d S
o v e r a s u r f a c e S which passes through a fixed determine for which surface S t h e integral I ( u ) i t h e me th od o f §8.1, we first choose any surface , such t h a t φ ( χ, y ) = 0 on t h e curve Γ, where Γ boundary curve B in t h e x y - p l a n e. Since u ( x, define a surface S a for any a by
boundary curve B and i* a minimum. Following > defined by u = φ ( χ, y ) is t h e projection o f the h ) passes through Γ, we
U ( x,y ) = u ( x, y) + α φ { χ, j/). T h e surface S a also passes through Γ. On t hi s
P = P -(“ Οίφχ , Q = Uy —
so t h a t t h e integral I is wri tten as
I = U s F (x'y ’ U’P'Q')d x
Hence
dJ_
d a
d F ± d F ± d F . 3 υ Φ + 8 Ρ Φχ + d Q ®
in
U s i n g ( A.3 ) w e f i n d f r o m ( 8.2 9 ) t h a t
I i f 8 /d F
L = o ~ J J L d U
dl _ ι
d a \
a /o r\ o
d x \ d P ) d y
II'
s i n c e U = u, P = p, Q — q when a = 0. o f d l/d a | Q by d a = a — 0 = a and write & (8.21), tha t
61 =
J J L d u d x ' ' ft*·
\ d p J d y d q
)1
( 8.2 7 )
s u r f a c e q + α φ υ,
d y.
d x d y.
( 8.2 8 )
( 8.2 9 )
%\φάχΛν'
w e m u l t i p l y t h e v a l u e u for α φ, we find, as in
S u d x d y.
(8.30)
274
CHAPTER 8: WEIGHTED RESIDUAL METHODS
If u ( x,y ) is t h e surface for which I is a minimum, t h e n 6 1 = 0 for arbitrary 6 u. Since t h e expression in t h e square brackets in (8.30) is continuous, it must be zero. T hi s gives a necessary c ondition for I t o be a minimum as
dF 8 /dF\ d /dF\ n .
du dx ( dp) dy ( dq) “ ' ^ ^
T h i s i s t h e E u l e r e q u a t i o n f o r ( 8.2 7 ), a n d a n y s u r f a c e c o r r e s p o n d i n g t o a s o l u t i o n o f t h i s e q u a t i o n i s a n e x t r e m a l.
I f F = F ( x,y,u,p,q,r,s,t ), where p and q are defined above and τ = u x x, s = u x y, and t = u y y, t he n t h e Euler equati on t o minimize
t h e integral I = J J F ( x,y,u,p,q,r,s,t ) d x d y is given by
* L ( dZ\ 92
( d F\ 92
( d F\
d u d x\d p ) d y\d q ) ^ ~ d x 2 V d r / d x d y V d s ) d y 2 \ d t )
( 8.3 2 )
E x a m p l e 8.2. T h e c o n d i t i o n t h a t t h e D i r i c h l e t i n t e g r a l
,(“)=//[ (!) +( l ) } dxdp
b e a m i n i m u m i s
d 2 u d 2 u
9 Ϊ 2 + ^ 2 = ’
w h i c h i s t h e L a p l a c e e q u a t i o n. ■
8.4. W e a k V a r i a t i o n a l F o r m u l a t i o n
T h e w e a k v a r i a t i o n f o r m u l a t i o n o f b o u n d a r y v a l u e p r o b l e m s i s d e r i v e d f r o m t h e f a c t t h a t v a r i a t i o n a l m e t h o d s f o r f i n d i n g a p p r o x i m a t e s o l u ­
t i o n s o f b o u n d a r y v a l u e p r o b l e m s, v i z., G a l e r k i n, R a y l e i g h - R i t z, c o l ­
l o c a t i o n, o r o t h e r w e i g h t e d r e s i d u a l m e t h o d s, a r e b a s e d o n t h e w e a k v a r i a t i o n a l s t a t e m e n t s o f t h e b o u n d a r y v a l u e p r o b l e m s. I n f a c t, t h e
8.4. WEAK VARIATIONAL FORMULATION
275
weak formulations are more general than the co: mulations, since even the irregular boundary con aged in the weak formulation. We shall not disci , strong formulations, but rather explain the metln tional formulation, which will in turn define the
rresponding strong for- ditions are easily man- ss the evolution of the ,od of the weak varia- underlying concept.
We shall consider a general form of a second value problem, defined by Eq (8.31), in a two with the prescribed boundary conditions
order mixed boundary dimensional region Ω
u0 on Γ ι,
and
dF
dF
~S~nx + ~E~ny = Qo on Γ 2
dp
8q
where F = F ( x,y,u,p,q ), and n x, n y are the unit vector ή normal to the boundary Γ = Γχ U t h a t Γχ Π Γ2 = 0.
di:
Γ2
For example, a special case of (8.31) is when 'd u\2 , /d u\2
1 r /d u\2 /d u\2 ]
2 1*1 (&) + feU) j"
F ~ 2 [ k l [ d x ) + u ‘\d y,
T h i s e q u a t i o n a r i s e s i n h e a t c o n d u c t i o n p r o b l e m r e g i o n w i t h f c i, a s t h e r m a l c o n d u c t i v i t i e s i n / b e i n g t h e h e a t s o u r c e ( s i n k ). H e r e
3 F _ d u 3 F
dp 1 d x ’ dq
a nd E q ( 8.3 1 ) b e c o me s
3 /, du
a ( Ou\ d / a u\
d x \ 1 d x ) d y\ 2d y )
I f ki = fc2 = 1, then we get the Poisson equ; appropriate boundary conditions.
The weak variational formulation for Eq (8. the following three steps:
STEP 1: Multiply Eq (8.31) by a test function the product over the region Ω:
\dF d ( d F\ d ( dF\i
I ~n~ I - — I — I I w d:
L du d x\ d v
, du dy
/
IL
■ dp )
d s d F' d y\d q J J
(8.33a)
(8.33b)
irection cosines of the of the region Ω such
F is defined as f u.
3 i n a t w o - d i me n s i o n a l t j he x, y directions, and
<~v J J
OU
in Ω. ation —V 2u = f with
31) can be obtained by
tv (ξξ Su) and integrate dy = 0. (8.34)
276
CHAPTER 8: WEIGHTED RESIDUAL METHODS
T h e t e s t function w is arbitrary, but i t must sati sfy t h e homogeneous essenti al boundary conditions (8.33a) on u.
S T E P 2: U s e f o r m u l a ( A.3 ) c o m p o n e n t w i s e t o t h e s e c o n d a n d t h i r d t e r m s i n ( 8.3 4 ), i n o r d e r t o t r a n s f e r t h e d i f f e r e n t i a t i o n f r o m t h e d e ­
p e n d e n t v a r i a b l e u t o t h e t e s t function w, and identify t h e t y p e o f t h e boundary condi ti ons admissible by t h e variational form:
■ d F d w d F d w d F i
* » ϊ + & a? + a F a j J d x d >
r . β j?
v
- / ( —- n x + — n y ) w d s = 0. (8.35)
J t=Γ!υΓ2 V dP d(l J
N o t e t h a t t h e f o r mu l a ( A.3 ) d o e s n o t a p p l y t o t h e f i r s t t e r m i n t h e
i n t e g r a n d i n ( 8.3 4 ).
N o t i c e t h a t t h i s s t e p a l s o y i e l d s b o u n d a r y t e r m s w h i c h d e t e r m i n e t h e n a t u r e o f t h e e s s e n t i a l a n d n a t u r a l b o u n d a r y c o n d i t i o n s f or t h e p r o b l e m. T h e g e n e r a l r u l e t o i d e n t i f y t h e e s s e n t i a l a n d n a t u r a l b o u n d ­
a r y c o n d i t i o n s f or ( 8.3 1 ) i s a s f o l l o ws: T h e e s s e n t i a l b o u n d a r y c o n d i t i o n i s p r e s c r i b e d o n t h e d e p e n d e n t v a r i a b l e ( u in t hi s c ase), i.e.,
u = uq on Γι
is t h e essenti al boundary condition for (8.31). T h e t e s t function w in t h e boundary integral (8.35) satisfies t h e homogeneous form o f t h e same boundary condition as t h a t prescribed on u. The natural boundary condi ti on arises by specifying t h e coefficients o f w and i t s derivatives in t h e boundary integral in (8.35). Thus,
d F d F
= «ο on Γ2
is t h e natural boundary c ondition in a Neumann boundary value prob­
lem. In one -di m e nsi onal problems, use integration by parts i nstead o f th e divergence formula (A.3).
In order t o equalize t h e continuity requirements on u and w, t h e differentiation in t h e divergence formula (A.3) has been transferred from F t o w. It imparts weaker continuity requirements on t h e sol uti on u in t h e variational problem t h a n in t h e original equation.
S TE P 3: Simplify t h e boundary terms by using t h e prescribed bound­
ary conditions. This will affect t h e boundary integral in (8.35) which
8.4. WEAK VARIATIONAL FORMULATION
277
is spli t into two terms, one on Γι and t h e other
Ii
r d F d w d F d w d F L d u ^ d x d p ^ d y d q
j d x d y
( d F d F\
I τ? _ -I- - -- -- - -
u r2 { d p Ux + d u
T h e i n t e g r a l o n Γ ι v a n i s h e s s i n c e w = 6 u = 0 on Γ ι. T h e natural boundary c ondition is s u b s t it u t e d in t h e integral on Γ2. The n (8.36) reduces to
f f Γ d F d w d F d w d F i , , f
J J n t d u + d x d p + d y d q J V J r
T h i s i s t h e w e a k v a r i a t i o n a l f o r m f o r t h e p r o b l e m ( 8.3 1 ). W e c a n w r i t e ( 8.3 7 ) i n t e r m s o f t h e b i l i n e a r a n d l i n e a r d i f f e r e n t i a l f o r m s a s
b ( w, u ) = l ( w ),
w h e r e
, f f \dw dF dw dF
W’U J Jn '-dx dp + dy dq
= - H w a- £ i x i y + i
o n Γ 2:
n y Sj w d s = 0. (8.36)
w q o d s = 0. (8.37)
d x d y, wqo ds.
(8.38)
(8.39)
Formula (8.38) defines t h e weak variational form for Eq (8.31) subject t o t h e boundary conditions (8.33). The quadrati w i t h t h i s variational form is given by
E x a m p l e 8.3. Consider t h e sy s t em o f Navie: a t wo- d im e ns ion al flow o f a viscous, incompre v e lo c i ty fields):
d u d u 1 d p f d 2 u
Ud ^ + V d ^ = ~'p d i + l/\d ^ M d v d v 1 d p /d 2 v
Uf o + V d ^ = +
d u d v d x ^ d y ’
functional ass oci ated
(8.40)
r - S t o k e s e quati ons for ssible fluid ( pressure-
d 2u\
d^2) ’
d 2v\
f o f ) ’
278
CHAPTER 8: WEIGHTED RESIDUAL METHODS
in a region Ω, with boundary conditions u = uo, v = vo on Γ ι, and (du du \ 1
■ w “* + a s"» J
(dv dv \ 1
"l a i"* + = *»’
on Γ2, where ( u,υ) denotes t h e v e lo c i ty field, p t h e pressure, and i x,t y t h e prescribed values o f t h e secondary variables. Let w\, W 2, w 3 be t h e t e s t functions, one for each equation, such t h a t w i and W 2 s a t i s f y t h e essential boundary c onditions on u and v, respectively, and w3 does no t s a t i sf y any essential condition. Then
//„ H
d u d u\ p d w 1 ( d w\d u d w\d u u —— I- v — — 1
(Ow\ ou aw
1 aw\i , ,
+ ~ ^ a i )\d x d v
d x d y J p d x V d x d x d y d y <
w\t x d s,
L
0 =
f f r ( d v d v\ p d w 2 ( d w 2 d v d w 2 d v \ ι , ,
J i h ( “ & + v s i > - 7 % ■ + "( - 3 7 a i + - W a i )\d x i y
- / W 2t y d s, j Γι
0 =
Then b ( ( w 1,w 2,w 3 ), (u,v ))
in
, d u d u\ f d v d v\ ( d u d v', , , ,
w\ U— + v — +w2 U— + V— +w3 u — + v — \d x d y ' d x d y J \ d x d y J \ d x d y 1
+ v
f f /dwi du dw\ du dw
2
dv dw
2
dv\
J J
QI dx dx ^ dy dy ^ dx dx ^ dy dy) X ^
l ( w i,w 2,w 3) = / ( w i i x + w 2i y )
J r 2
ds.
/ r 2 '
No t e t h a t t h e boundary integral in t h e linear form l ( w i,v ) 2, w 3 ) has no term c ontaining w 3. ·
8.5. GALERKIN METHOD
279
Ra·
We shall now discuss t h e Galerkin and t h e which are t h e two most widely used weighted re; ta i n i n g approximate numerical solutions o f boui|i< It will be found t h a t these two methods give t h e geneous boundary value problems. In fact it cai are similar for homogeneous problems.
,yleigh-Ri tz methods, s i du a l meth ods for ob- dary value problems, i a m e results for homo- be proved t h a t t he y
8.5. G a l e r k i n M e t h o d
Consider t h e boundary value problem
L u = f in Ω,
su bj e ct t o t h e boundary conditions
u = g ( s ) on Γι
d u
d n
+ k ( s ) u = h ( s )
o n Γ
w h e r e Γ = Γ i U Γ2 i s t h e b o u n d a r y o f t h e r e g i o: a p p r o x i m a t e s o l u t i o n u o f t h e form
N
U — ^ ί ψ ί ■ i = 1
An approximate solution does not, in general, sal
(8.43). The residual (error) associated w i t h an defined by
N
'•(ΰ) = L u - f = L
2=1
N o t e t h a t i f uq is an exact solution o f ( 8.4 1 ) - T he Galerkin me th od requires t h a t t h e residu respect t o t h e basis functions </>, (also called th in (8.44), i.e.,
{r: Φί) = 0.
Hence
IL
{L(u) - ϊ}φίάχάy = 0,
i = 1,
( 8.4 1 )
( 8.4 2 )
2, ( 8.4 3 )
h Ω. L e t u s c h o o s e a n
( 8.4 4 )
.t i s f y t h e s y s t e m ( 8.4 1 ) - a p p r o x i m a t e s o l u t i o n i s
-/·
( 8.4 5 )
( S
.4 3 ), t h e n r ( u o ) = 0. al be orthogonal wi t h e trial functions) used
(8.46)
,N,
(8.46a)
280
CHAPTER 8: WEIGHTED RESIDUAL METHODS
or
&!L 4>i Ltpj dx dy = J J /</>* dx dy,
w h i c h i n t h e m a t r i x f o r m i s w r i t t e n a s
[ A] { c } = { b }, (8.46b)
where
—
= Φί Lφj dx dy, h = f φί dx dy. (8.46c)
J Jci J Jq
In t h e examples given below, we shall choose different values o f N in (8.34) for t h e trial function u. There is some guidance from geometry for such choices; moreover, t h e y should sat i sf y t h e essenti al c ondi ti ons and exhi bi t t h e nature o f t h e approximation sol uti ons vis-a- v i s t h e e x a ct solutions (see §8.7 for some choices). However, t h e larger t h e N, t h e bet t er t h e approximation becomes.
E x a m p l e 8.4. Consider t he Poisson equation
2 ( 92u d2u\
- V 2u = - i — + — J = c, 0 < a: < a, 0 < y < b,
s u c h t h a t u = 0 at x = 0, a and y = 0, b. First we choose t h e first order approximate solution as
= a x y ( x — a) (y — b).
N o t e t h a t t h i s c h o i c e s a t i s f i e s a l l f o u r D i r i c h l e t b o u n d a r y c o n d i t i o n s. T h e G a l e r k i n e q u a t i o n ( 8.4 6 a ) g i v e s
Π
α
[ - 2a ( y 2 - by + x 2 — ax) - c] x y ( x - a)( y — b) dx dy = 0 which simplifies t o
Thus,
5c
a _ 2 ( a 2 + & 2 ) ·
8.5. GALERKIN METHOD
281
Hence
5 c
2( a 2 + 62)
x y { x - a ) ( y - b)
A l t e r n a t e l y, w e c a n s o l v e t h i s p r o b l e m b y c h a p p r o x i m a t e s o l u t i o n a s
~ ( 2 ) v ' . J t v x . K T n y
u\ ' = > α,-fc s i n sm —;—
y ι J a h
jik =1
N
o osi ng t h e first order
y
which is an orthogonal trigonometric series w i t h a finite number o f
(2"\
terms. No t e t h a t u\ satisfies t h e boundary conditions. Also no t e t h e
orthogonal i ty condition
f a τ η π χ η π χ , f 0,
/ s m s i n a x = <
J o a a I a/2,
T h e G a l e r k i n e q u a t i o n ( 8.4 6 a ) i n t h i s c a s e g i v e s
/7
J o J o
ν η φ η m = n.
?27t2 k 2 7r2\ . j r c x . k n y
“M — + - H S1” — —
x sin ■
Hence
k 2
ι[ V j 2
a j k 4 I a2 ' b2
- ( 1 — c o s j 7 r ) ( l — CO s/c7r)
o r
Oijk —
jfc7T2
4 c ( l — cos J7r ) ( l — cos &π)α262
πijk{b2j 2 + a 2fc2) Thus, t h i s approximate solution is
. ] πχ , f a y
s i n —— d x d y = 0. b
_ (2) _ ^ 4 a262c ( l — c o s j 7 r ) ( l — c o s f e i r )
j,k = l
j f a 4 ( a 2 k 2 + b 2j 2
I f t h e n u m b e r o f t e r m s i n e a c h s u m i s i n f i n i t e, t h e n u 2^ becomes t h e e x a ct solution uq.
. ι π χ k i τ υ
i n s i n z .
a b
282
CHAPTER 8: WEIGHTED RESIDUAL METHODS
At the center point ( a/2,b/2), we have
(2) v"' 4o262c ( l — cosj w) ( l — coskir) . j n kir
j k w ^ k 2 + b2j 2) s m T s m T ·
j,k=l
I f a = b, the n at t h e center point ( a/2, a/2 )
~(2) 4a2c ( l — c o s J7r ) ( l — coskn) . j n . k -π
u i,c = L · L · j k w4{ f + k 2) sm T sin T
j fc
a 2c r 8 8 8 ί
- ^ [ 8 + Ϊ 5 + Ϊ 5 + 8Ϊ + "Ί = U °
36.64 /a\2 ~ π4 CV2/ '
For t h e i V - t h approximation, t h e trial functions are chosen as 0jfc(x, y) f j ( x ) 9 k ( y ), where
/,· ( £ ) = x J { x - a ), g k ( y ) = y k { y - b).
T h e n t h e i V - t h a p p r o x i m a t e s o l u t i o n i s
N
u N { x,y ) = Σ a j k <t >j k { x,y ),
j,f c=i
a n d t h e r e s i d u a l i s
r = —c — Σ [f"( x)9k(y) + fj{x)g'k(yj\ ·
j\fc= 1
Hence, since t h e Galerkin method requires t h a t (<£mn, r ) = 0 for m, n
1,2,· · · ,N, we get
° = Ιο I +
fm{x)gn(y) dxdy,
w h i c h a f t e r i n t e g r a t i o n y i e l d s
N
Σ ajk [j p{j, m, a) q(k, n,b ) + k p(k, n, b) q(j, m, a)]
j,k = 1
+ cam+2 b"+2 h( m, n ) = 0, (m, n = 1,2, ■ · ■ , N),
,.6. RAYLEIGH-RITZ METHOD
where
p(j,m,a ) = aJ+m+1 q{k,n,b) = bk+n+z h( m, n
i - l 2
j + m — 1 j + m 1 2
H-
fc + n + 1 k + n + 1
(m + 1 ) ( m + 2 )(n + l ) ( n + 2)
T he coefficients a j f. for j, k = 1, 2, ■ ■ · , iV can be above s y s t em o f equations. The result for ΰ\ fou: t h i s general case.
determined from t h e hd earlier follows from
77ΓΧ
The trial functions 4 > j k ( x,y ) = s i n sin
proximation , belong t o t h e set o f orthogonal solving t h e given boundary value problem by t h e me th od (see Table 4.1, Di richl et-Diri chlet case,
kny
283
j + 1
j + m +
1 1
+
k τι 3
, used in t h e ap-
functions o btai ne d by eparation o f variables and Exercise 5.28). ■
8.6. R a y l e i g h —R i t z M e t h o d
Consider t h e Poisson equation —V 2u = /, 1 boundary c onditions u = 0 on Γχ and du/dn weak variational formulation leads t o
I(u)
= Ι Ι α{\Μ 2- ί « } “^ ·'
i th t h e homogeneous 0 on Γ2. Then, t h e
A generalization o f t h e result in (8.47) for t h e case / w i t h t h e above homogeneous boundary conditic se l f-adjoi nt and pos it i ve definite operator, leads
o f t h e s y s t em u L u = ns, where L is a linear t o t h e functional
Kn) = 5 I i { u L u — 2f u } d x
0.
( 8.4 7 )
dy.
( 8.4 8 )
T h e o r e m 8.1. I f the operator L is self-adjoint and positive defi­
nite, t hen the unique solution of Lu = / with homogeneous boundary conditions occurs at a mi ni mum value of I (u).
284
CHAPTER 8: WEIGHTED RESIDUAL METHODS
An application o f Theorem 8.1 is t h e R a y l e ig h - Ri t z method, where
L u = f b y constructi ng minimizing sequences and securing t h e approx­
i mate solutions by a limiting process based on such sequences. Thus, we choose a complete set o f linearly i ndependent basis ( t e s t ) functions φί, i = 1, ■ · ·, and t he n approximate t h e e xact sol uti on uq by taking t h e approximate solution ΰ in t h e form
where t h e c onstan ts c, are chosen such t h a t t h e functional I ( ύ ) is min­
imized at each stage. If ΰ —> uo as n —> oo, then t h e me th od yi elds a convergent solution. At each stage t h e method reduces t h e problem t o t h a t o f solving a set o f linear algebraic equations. T h e detai ls for t h e boundary value problem — V 2 u = f wi t h homogeneous boundary condi ti ons are as follows: Using (8.49) in t h e functional (8.48) we get
we find t h e direct solution o f t h e variational problem for t h e s y s t em
n
(8.49)
Ι ( ΰ ) = I ( c i, ■ · ■ ,c „ )
thus
Hence
(8.50)
and
(8.51)
(8.52)
8.6. RAYLEIGH-RITZ METHOD
285
Now, i f we choose a such t h a t I ( c i ) is a minim t h e n from (8.50) we get
n
^ ' A i j C i = hi, i = 1, ■
3 = 1
w h i c h i n t h e m a t r i x n o t a t i o n i s
m e ) = {h},
w h e r e t h e m a t r i x [^4] h a s e l e m e n t s A l j given by (8 h i given by (8.52), and { c } = [ci, ■ · · , c n ]T. No t e o f linear algebraic equations t o be solved for t h Ci, and [A] is non-si ngular i f L is positive definite.
j m ( i.e., dl/dci
= 0 ),
.53)
(8.54)
.51), { h } has e lements t h a t (8.54) is a sy s t em e unknown parameter
Th e R a y l e ig h - R i t z method can be developed, for u t h e equation (8.38), where we require t h a t geneous essential conditions only. Then t hi s pr minimizing t h e functional (8.40). In other wore}; proximate s ol uti on o f (8.38) in t h e form
alternately, by solving w s a t i s f y t h e homo- oblem is equivalent to s, we will find an ap-
un — 'y ' Cj (f>j + Φο, i = i
where t h e coefficients Cj are chosen such t h a t w = φ ί, i = 1, · · · ,n, i.e.,
ΗΦί, Un) ~ 1 ( Φΐ ) j ί = 1,· · ·
or
n
, ^ ' Cj φ j + <£o) = 1{Φι
3 = 1
t h u s,
^^Cjbfyi, φj) — 1{φί) ^(φΐ,φο).
3 = 1
T h i s e q u a t i o n i s a s y s t e m o f n linear algebraic eq Cj and has a unique solution i f t h e coefficient m; singular and thus has an inverse.
T h e functions φ ί must s ati sfy t h e following should be wel l-d efi ned such t h a t ^ φ ι,φ ^ ) φ at least t h e essential homogeneous boundary c {<Ai}?= ι must be linearly i ndependent, and (iv) t ’ complete.
3.55)
Eq (8.38) is true for
(8.56)
nations in n unknowns atrix in (8.56) is non­
requirements: (i) φ ί (ii) φ ί must s ati sfy ondition, (iii) t h e set lie set { 0j }”=1 must be
286
CHAPTER 8: WEIGHTED RESIDUAL METHODS
E x a m p l e 8.5. Consider t h e Bessel equation
x 2u" + x u' + ( x 2 — l ) u = 0, u ( l ) = 1, u ( 2) = 2.
P u t u = ν + x. Then t h e given equation and t h e boundary conditions become
x 2v" + x v' + ( χ 2 — 1)υ + x 3 = 0, u ( l ) = 0 = v ( 2 ).
I n t h e s e l f - a d j o i n t f o r m t h i s e q u a t i o n i s w r i t t e n a s
x 2 — 1
x v ” + ν' +
-ν + x = 0.
For t h e 1st approximation, we take
vi = αχφι = ai(x — l ) ( x — 2).
The n using (8.46) we ge t j f ( L v i — /) φ ι dx = 0, which gives Γ 2 χ 2 - 1
J [ 2 a i x - ( 3 — 2x)ai~\------------( x —l ) ( x — 2 ) a i +x 2]( x — l ) ( x - 2 ) d x = 0,
which, on integration, yi elds a i = —0.811, and thus,
Wi = ν i + x = —0.8 1 1 (x — \) {x — 2) + x.
T h e e x a c t s o l u t i o n i s u = c i J\{ x ) + C2 Y i { x ), where c\ = 3.60756, C2 = 0.75229. A comparison w i t h t h e e x a ct solution in t h e following tabl e shows t h a t u\ is a g ood approximation:
X
U l
Inexact
1.3
1.4703
1.4706
1.5
1.7027
1.7026
1.8
1.9297
1.9294
3.7. CHOICE OF TEST FUNCTI
DNS
ion
E x a m p l e 8.6. Consider t h e 4th-order equat
[(x + 2 l ) u"]" + b u — k x = 0, 0 ^ x < 1,
w i t h t h e b o u n d a r y c o n d i t i o n s: u { l ) = 0 = u 0, [(x + 2/)u"]'(0) = 0. We choose t h e t e s t funct
Φι(χ) = (χ — ί)2(χ2 + 2/x + 31 Φι{χ) = ( x — l)3( 3x2 + 4 l x +
'(/), ( x + 2 l ) u"( 0 ) ions
2)
2\
For t h e 1st approximation, we have u i = α ι φ\( χ ). The n f 0 ( L u i — /) φ ι ( x ) d x = 0, which gives
o j ( 2 4 i + 57.6 + 1616/4/3 1 5 + 9 q l 2/5 ) If, e.g., we take I = b = 1, and k = 3, t h e n a\ = u i = 0.011917(a; - l ) 2 ( x 2 + 2 x For t h e 2nd approximation, we take U2 = α ι φ\(.
f ( L u 2 - f ^
J o
i d x = 0, and
which, wi t h I = b = 1, k = 3, yield
f \l U2 -
Jo
83.9 1 1 a i - 67.313a2 = 1 67.213ai - 91.882o2 = 0.7li=3.
Thus a\ = 0.013743, o2 = 0.002279, and u 2 = 0.0 1 3 7 4 3 (x - l )2( x 2 + 2 x + 3) + 0.002279(
Instead o f determining t h e e xact solution, we c Thus, e.g., u i ( 0.5 ) = 0.012662, and u2(0.5) g ood results. ■
287
■F k l/3 = 0. 0.011917, and + 3).
c) + a 2«^2( x ). Then
/) φ 2 { χ ) ά χ = 0
x — l ) 3( 3 x2 + 4 x + 3).
an compare u i and u 2.
0.012964, which give
8.7. C h o i c e o f T e s t F u n c t i o n s
N o t e t h a t a suitable choice o f th e t e s t functions φ ί ( χ, y ) can be made by taki ng linear combinations o f polynomials, or trigonometric functions,
288
CHAPTER 8: WEIGHTED RESIDUAL METHODS
such t h a t t h e y sat i sf y t h e boundary conditions. For example, we can choose a s y s t em o f functions
<f>o = g, <t>i=gx, <i>2 = gy, Φζ = 9χ2, Φα = gxy,..., (8.57)
where g = g ( x, y ). It can be shown t h a t t h e s ys tem (8.57) is complete.
y
Y
1
/~\_ x
____J a
-p
0
/
-q
Fig. 8.3. Fig. 8.4.
Some practical rules for constructing the functions g ( x, y ) in (8.57) are as follows:
(i) For t h e rectangle [—a, a; —6,6]:
g ( x,y ) = {x2 - a 2) ( y 2 - 62).
(ii) For a circle o f radius r and center at origin:
g { x,y ) = r 2 - x 2 - y 2.
(iii) I f t h e boundary Γ o f a region Ω is defined by F ( x,y ) = 0, where F £ C n, t he n
g ( x,y ) = ± F { x,y ).
See (ii) above i f Γ is a circle.
(iv) For t h e case o f a convex polygon whose sides are defined by a i x + h y + c i = 0, · · · , a m x + bm y + c m = 0, we have
g ( x, y) = ± ( a 1x
+ b i y + c i ) · · ■ (a m x + b m y + cm).
See (i) above for a rectangle.
(v) T h e choice in (iv) is also suitabl e in different t y p e s o f regions bounded by curved lines; e.g., for a sector formed by t h e circles o f radii r and r/2, as in Fig. 8.3, we have
g(x,y) = (r2 - x2 - y2)(x2 -rx + y2).
8.7. CHOICE OF TEST FUNCTIONS
289
(vi) For nonconvex polygons, t h e function g{x piecewise in different parts o f t h e region, and we for any r e- entrant angles. Thus, for t h e region i:
y) must be assigned must introduce moduli Fig. 8.4,
g(x, y) = (\x\ + \y\- x - y) (x + p) (y + q ) ( l - x -2y( x + p) (y + q)(l - x) {h - y), -2(x + y) ( x + p) ( y + q)(l - x) ( h - - 2x( x +p) ( y + q)(l - x ) ( h - y)
W e c a n a l s o t a k e ( 2 n d c h o i c e )
g( x,y ) = ( x2 + y 2 - x\x\ - y\y\) ( x + p ) ( y +
In t h i s case g £ C l. A third choice is t o assign separatel y in t h e three parts o f t h e corner region
(x + p) x( h — y) ( ai + a2x + a;
in [—p, 0; 0, h]
{x + p) (y + q)(bi + b2x + b3y in [ - p,0;- Q,0 ]
(y + q)(l~ z) ( ci + C
2
X + c3y in [0,/; - q, 0],
un{x,y) = <
w h e r e a^, bk, ck {k = 1, · · · , n) are parameters whl· by the following condition on the axes x = 0 and
(x + p) xh( ai + a2x + a^x2 + ■■■) = (x + p)q(bi P(y + q)(bi + b2y H + bny m) = (y + q)yl(cy
I n v i e w o f t h e a b o v e c o n s i d e r a t i o n s, t h e t e s t f uni c a l l e d t h e s h a p e f u n c t i o n s f o r t h e r e g i o n Ω.
E x a m p l e 8.7. T o r s i o n o f a p r i s m a t i c r o d s e c t i o n o f l e n g t h 2a and width 26 is defined by
V2u = 2, u = 0 on Γ.
For the rectangle shown in Fig. 8.5, we choose
Φ(χ, y) = (a2 - x2){b2 - y2,
'(h - y )
in [0, Z; —q, 0] y ), in [—p, 0; —q, 0] in [—p, 0; 0, h\.
q ) ( l - x ) ( h - y ).
t h e f u n c t i o n s un ( x,y) o f Fig. 8.4:
y Η 1- anym)
+ ■■■ + bn y m ) + Cnym )
+
i c h m u s t b e c o n n e c t e d y = 0:
+ b2x
+ b^x2
H )
c2y + h cny m).
:t i o n s φί ( χ,y) are also
of rectangular cross-
290
CHAPTER 8: WEIGHTED RESIDUAL METHODS
and seek an approximate solution of the form
un{x,y) = (a2 - x 2)(b2 - y 2) ( Ai 4- A 2x 2 + A 3y 2 -\ h A nx 2ly 2j).
F i r s t, f o r n = 1, we use (8.46a) with / = 2 and find t h a t (with φ( χ) = (a2 — x 2)(62 — x 2))
'- I t i - p - w -') * * *
= 2 f [ [l - Ai ( a 2 - x 2) - Ai {b2 - y 2)]{a2 - x 2){b2 - y 2) dy dx
J - a J - b
128 ο ο , η i 2\ a 2i3
= -a 6 (a + b ) A\ + — a b .
4 5 9
T h u s, Αχ = 5/4 ( a 2 + b2), and
5 (a2 — x 2)(b2 — y 2)
Ul = 4 ^ T b 2 '
Y
| 1
2a
Fig. 8.5.
T he torsional moment
f a f b , , 40 G 9 a % 3
M = J _ b Ul y ~9 a 2 + b 2'
w h e r e G is t h e shear modulus, and Θ is t h e angle o f t w i s t per unit length. T h e t ange nti al stresses t z x and r z y are given by
8.7. CHOICE OF TEST FUNCTIONS
291
For a = 6, we find t h a t M = 2OG0a4/9 « 0.138 classical sol uti on is given by
ji °°
u = a x
cosh i 2 n ~ l )7ry
x 2 ou y *
π 3 (2n - 1 ) 3 c o s h
8(2a) 4G0. The exact
. ( 2 n - 1 ) π χ s i n
- - - - -- - - - - - - - -- - - - -,
w h i c h g i v e s
3 2 a 4 ^ 1
M = 2G0< —------------— > —------- — tanh
1 6 π 5 ^ (2η - l ) 5 v n = l ' '
F o r a = b, t h e e xact value o f M is 0.1406(2a very well wi t h t h e approximate value obtained a method. ■
)‘[G 9
which compares bove by t h e Galerkin
E x a m p l e 8.8. Solve V 4u = 0 on t h e rectan; 8.5) under t h e boundary conditions
d 2u
d x d y
= 0,
d 2u
d x d y
d 2u
d y 2
= 0,
=(>-£)
at
d 2u d x 2
0 at y
w h e r e c i s a c o n s t a n t. T h i s p r o b l e m p e r t a i n s
r e c t a n g u l a r p l a t e u n d e r t e n s i l e f o r c e s.
F i r s t, w e w i l l r e d u c e t h e a b o v e b o u n d a r y c o n d b o u n d a r y c o n d i t i o n s: T h e f u n c t i o n
Uq
_ y 2 \ 6 b 2 )
o b v i o u s l y s a t i s f i e s t h e g i v e n b o u n d a r y c o n d i t i o n: g r a t i n g e a c h o n e o f t h e a b o v e b o u n d a r y c o n d i t i o T h e n V 4 w = 2 c/b 2, and t h e boundary conditions
d u
0, =0 at x =
d y 2
d u d x d y
d 2 u d 2 u
d x d y ’ f a 2 = t y
(2 n — 1 ) n b
2 a
* l e [ - a, a; -6,6] ( F i g.
x = ± a,
±6,
t h e expans ion of a
to
t i ons t o homogeneous
s (it follows by inte-
ns). Set u = u q + u. become
± a,
± 6.
292
CHAPTER 8: WEIGHTED RESIDUAL METHODS
The se boundary conditions will be satisfied i f t h e following conditions are met:
du
u = 0, ——=0 ata? = ± o,
Ox
du ,
u — 0, — =0 at u = ± 6.
d y
Thus, t h e given problem reduces t o t h a t o f minimizing t h e integral I ( u ) = J J [ ( V 2u)2 - d x d y.
T h e n, b y R a y l e i g h - R i t z ( o r G a l e r k i n ) m e t h o d
( V4u n - /) 4>j d x d y = 0, j = l,...,n, (8.58)
II
ia
where u n is t h e n - t h approximate solution, which, in view o f t h e geo­
metric symmetry o f t h e rectangle, is taken as
Un = ( χ 2 - a 2 ) 2 ( y 2 - b2 ) 2 ( a i + a 2 x 2 + a 3 y 2 H ).
For n = 1, we find from (8.58) tha t
/
a pb
/ [ 2 4 a i ( y:
■a J-b
2 — b2)2 + 1 6 α ι ( 3 χ 2 — a 2 ) ( 3 y 2 — b2 )
+ 2 4 α ι ( χ 2 - a2)2 - ~^](χ2 ~ a2)2{y2 ~ b2)2 d y d x = 0, '5 4 256 b2 64 64'
/t>4 25b b4 b \
ν τ + ι ^ + τ ^ Γ 1 ^
w h i c h g i v e s a i = 0.0 4 3 2 5 3 c/a 6, an d
a 4v 2 ’
c y 2
λ 0.0 4 2 5 3 c, , 0
0
Ul = u 0 + U l = - y (1 - ^ 2) + — ------ (a: - a ) { y 2 - b 2 ) 2.
8.8. T r a n s i e n t P r o b l e m s
F o r t i m e - d e p e n d e n t p r o b l e m s t h e s e m i - d i s c r e t e f o r m u l a t i o n i s u s e d t o c h o o s e t h e b a s i s f u n c t i o n s. T h u s, f o r o n e - d i m e n s i o n a l p r o b l e m s t h e N-th
a p p r o x i m a t e s o l u t i o n i s t a k e n a s
N
u N ( x,t ) = φ ο + Σ <ζί ( ί ) Φ ί ( χ ) ’ (8.59)
j = 1
8.8. TRANSIENT PROBLEM
where, as before, t h e functions <j)j s ati sfy t h e h c ondi ti ons and φ ο is chosen as in (8.55). Then, R a y l e ig h - R i t z me th od such t h a t t h e residual is o basis functions i = 1,2, · · ■ ,N, we obtai n the differential equations in t. For example, for t h e V 2u, t hi s s ys tem is
Ν N
(Φί,Φί) = (Φί,Φι)
j
= 1 3 = 1
omogeneous boundary using t h e Galerkin or •ifthogonal t o t h e first N N first order ordinary diffusion equati on u t =
-t-
T h s
w h e r e t h e d o t d e n o t e s t h e t i m e d e r i v a t i v e, t h i s s y s t e m a r e s u b j e c t t o a n o t h e r G a l e r k i n a p i t s r e s i d u a l R = u ( x, 0) — u n ( x, 0) is orthogon functions φ j. T hi s yields t h e s y s t em o f N algebr;
i ni t ial c ondi ti ons for proximation such t h a t al t o t h e first N basis aic equations
N
Σ < * ( ° ) -
3 =
1
which is generally solved for t h e unknowns Cj(0)
by numerical methods.
E x a m p l e 8.9. Consider t h e heat conduction equation d u d 2 u 1 d u
0 < r < 1
d t d x 2 r d r
s u b j e c t t o t h e b o u n d a r y c o n d i t i o n s u r ( 0,t ) ini tial condition u ( r, 0) = l n r. For t h e first or can take t h e basis function as φχ{τ) = Co + C\r t h e coefficients co, c\, and c 2 we require t h a t
condi ti ons o f t h e problem. Thus, φι (1) = co + c c i + 2c2a = 0. B y solving t he se two equatic find t h a t Ci = 2a c o/( l — 2a), and c 2 = —Co cq = 1 —2a, t he n c i = 2a, and c 2 = — 1, and t h e
ns in terms o f co, we /( l — 2a). If we take basis function becomes φ ι ( r ) = 1 — 2a + 2a r — r 2, or φ ι ( r ) = 1 — b + b r — r 2, wi t h b = 2a. This sugges ts t h a t for t h e TV-th approximation we should choose t h e basis functions as
>j{r) = l - b j + bjr
r3 + 1
w i t h φ ο — 0. T he N - t h order approximate sol t h e se m i - d i s c re t e form as
293
rj Ί ΨΌ / 1
Φ ο ( τ ) ),
u ( l,t ) = 0, and t h e der approximation, we b c 2 r 2. To determine s a t i s f y t h e boundary dφl
+ c 2 = 0, and
d r
, J = 1,2,-- - ,JV, ution is t h e n taken in
N
uN(r,t) =
^ C j ( i ) ^ ( r )
3 =
1
294 CHAPTER 8: WEIGHTED RESIDUAL METHODS
T h e residual is given by
N Γ
Σ\έ3®Φ3
i = l *·
+ Cj(t)
(j _)_ 1)2rj 1 _
bj
T h e n for t h e Galerkin method N ,1
0 =
Σ / { h + V + 1) V 1 - 7
j=iJ o I L r
• ( l — bi + b i r — r i + 1 } r d r
N .1
= Σ / { C j i t f f i'i ’ i ) + c 3 ( t ) 9 { i J ) } ,
7 = 1 v'°
f o r i = 1,2,· · · , A'', where
a; -\ ( 1 — M ( 1 — bj) i bi + bj — bibj bibj 1 — bj
2 + 3 + 4 i + 3
1 όί bj
j + 3 i + 4 j + 4"r z + j + 4 ’
ό,: 1
+ -
= (j + i ) 5
1 — bi b + -
J + 1 j + 2 i + j + 2
bibj bj
Th e ini tial c ondition is, in general, satisfied approximately. Thi s is accomplished by requiring t h a t t h e residual
N
Λ = Σ ^ ° ) <^ τ· ) - 1ηί'
j~ i
be orthogonal t o t h e basis functions φ %{ τ ), i.e., ( ^,ϋ ) = 0 for i =
1,2,··· , iV. Thi s means t h a t
lim [ ϋ φ ί ( Γ )
r d r = 0 for j = 1,2, · ■ · , N,
s i n c e R has a logarithmic singularity at r = 0. After evaluating t hi s improper integral, we obt ai n a s ys tem o f N algebraic equations:
ς ^ ο )/( « ) = !4 ^
j = 1
4 3 6 3 ( i + 3 ):
r, i = 1,2,· · · ,J V,
8.9. OTHER METHODS
which can be solved for the unknowns Cj(0).
8.9. O t h e r M e t h o d s
There are other weighted residual methods, whii t h e sequel. The se methods are are not freque: will not present any examples. Interested rea information on t he se me th ods in Connor and ( 1980), Kantorovitch and Krylov (1958), and
ch we will mention in n t l y used, and so we i er s will find detailed Brebbia (1973), Davies ddy (1984).
Ri?
8.9.1. C o l l o c a t i o n m e t h o d. In this m eth
od, t h e trial function
is chosen t o s a t i s f y t h e boundary conditions, are determined by forcing u n t o s ati sfy t h e diffi prescribed set o f points, i.e., t h e residual r van
8.9.2. L e a s t —s q u a r e m e t h o d. T hi s methi t o t h e residual. The trial functions are chosen t conditions. T h e residual is minimized by choo such t h a t t h e functional
I ( u ) = I i ( r ( u )}2 d x d y
i s a m i n i m u m. T h u s d l/d c i = 0, i = 1, · · · , n. $ince
■ ■ ,c n ) = J J |
and L is linear, we ge t which implies
J J ^ ^ } d x d y = J J /Σ φ ί
dx
295
and t h e parameters Ci 'erential equation at a ishes at t he se points.
od is applied directly o s ati sfy t h e boundary i n g t h e parameters Cj
/} d x d y
d x d y = 0,
dy, i = 1, · ■ ■ ,n.
296
CHAPTER 8: WEIGHTED RESIDUAL METHODS
8.9.3. M e t h o d o f m o m e n t s. In t h e equation ( r, W i ) = 0, where r is t h e residual and Wi are t h e weight functions, we can use any linearly i ndependent and complete set of weight functions Wi. T he simplest choice for 1 - D problems is t h e set {1, x, x 2,x 3, ■ · · }. The n t h e successive higher moments o f t h e residual are forced t o vanish, i.e.,
< r, x 1 > = 0, i = 0,1,2,· · ·.
T h i s s c h e m e i s c a l l e d t h e m e t h o d o f m o m e n t s.
8.1 0. E x e r c i s e s
8.1. F i l l i n t h e d e t a i l s i n t h e d e r i v a t i o n o f t h e E u l e r e q u a t i o n ( 8.1 3 ).
d F | b
A n s. From (8.11), i ntegrating by parts twice and using 0 ( x ) —— =
U U I a
,,x,± ( m 1
^ d x\d U")
d l f b
= 0, we get
Λ2 / r ) P \1
d x.
_ A ( d F\
d u d x \d u') d x 2 \d u" )
8.2. F i l l i n t h e d e t a i l s i n t h e d e r i v a t i o n o f t h e E u l e r e q u a t i o n ( 8.1 5 ). A n s. I n t r o d u c e t w o f u n c t i o n s φ ( χ ) and ψ ( χ ), and two parameters a and β, respectively, such t h a t U = u + α φ ( χ ), V = ν + β ψ ( χ ). T he n d i/d a = 0 and d l/d f i — 0 lead t o (8.15).
8.3. F ind t h e geodesics for t h e following problems:
(a) On t h e ary-plane, take I = J d s = J y j l + y ’ 2 d x.
( b ) O n t h e x y - plane, take I = J ds = I y/l + r 2( d6/dr) 2 dr.
( c ) O n t h e c y l i n d e r x 2 + y 2 = a2, — oo < z < oo, take x — a cost,
y = o s i n i, and I = J ds = j \/a 2 4- (dz/dt ) 2 dt.
A n s. ( a ) S t r a i g h t l i n e s y = C1X + C 2; (b) Straight l ines r cos (fl—c i ) = c2; (c) z = c i t + C2-
8.4. A ray o f light moves between two fixed points in t h e x y - p l a n e wi t h variable v e lo c i ty v( x,y). B y Fermat’s law, i t s travel t i m e is
/
ds ί λ/ι + y'2
— = / ----------------dx. Show t h a t t h e paths for a minimum travel
v J v ti me are given by
vy" dvy l + d v = 0
V7! + y'2 dx dy
8.10. EXERCISES
297
8.5. Find t h e extremal when t h e following integral is minimized:
(a) J\y'2 + y 2 ) d x.
( b ) j (y"2 + V 2) d x.
A n s. ( a ) y = Ciex +C2e _ x; (b) y = [ c x e * ^ + C4e- 1/^ s in( y/v/2).
8.6. I f t h e end points are not fixed, derive
f b d F b f b r
/ F ( x,u,u') d x = 6 u — — +
Ja 9 u' a J a I
Γ dF_ . d u
.1), a n d l e t 6 1 denote e δ α = α, δ β = β. Set
8.7. Take <5u = α φ ( χ ) + β φ { χ ) in t h e integral t h e t o t a l differential of I at a = 0 = β, where
TT d F d / d F\ . . . .
H = — ----------— —- , and show t h a t t h e variation d l in t h i s case
o u d x \ d u' /
can be wri tten as
61
pb pb
= d a I Η φ ά χ + ά β I i f xpd x. Ja Ja
C-2
8.8. Find t h e extremal for t h e problem o f det prescribed length I joining A B and maximizin; bounded by Γ, x - a x i s and two fixed ordinate:
A n s. ( x — c i )2 + ( y — C2)2 = fc2, where c i,
Γ pass through A and B and have length I.
D e r i v e t h e v a r i a t i o n a l f o r m u l a t i o n f o r t h e b o u:
8.9 - 8.1 4 ( h e r e a, b, f, g are functions o f x; u o, fo* m o> Qo> ϊοο> u oo are constants):
8 ’9 ’ ~ έ ( α 3 ~~ ^ = = U°' a ^ {l ) ^ q° ’ ° < X < ^ (o n e _
dimensional heat conduction).
A n s.
.. . d w d u ,
b ( w,u ) = / a—— — d x,
J 0 d x d x
l ( w ) = f w f d x — w ( 0) J o
d u 1 d x
d / d u\
1 d x )
( o n e - d i m e n s i o n a l d e f o r m a t i o n o f a b a r ).
8 · 1 0 · - ώ ( α £ ) - ° ω + χ 2 = 0; u ( 0 ) = 0 ’ α
c 2e
- χ/λ/2^ c O s ( y/-\/2) +
i ( Z ) ] Sudx-
e r m i n i n g a c u r v e Γ o f g t h e a r e a A — J y d x, s.
, and k make t h e arc
ndary value problems
+ q o w ( l ).
(1) = 1, 0 < x < 1
298
CHAPTER 8: WEIGHTED RESIDUAL METHODS
An s.
,, . f 1 ( d w d u \ ... ί1 , .
blw, u)= a —— cwu dx, l(w) = — / wx dx + w( 1).
Jo \ d x d x J Jo
d ( du du\ d ( du du\ . ^ ,
* • 1 1 · c u ^ - + C i2 ^ - - C21— +C22^ - +/ = 0 in Ω wi t h
dx \ dy dx) dy\ dx dy)
boundary conditions u = Uq on Γ ι, and q n = C u - — h c i 2— ) n x +
\ dx dy)
( c2i g “ + c22 = ?o. o n Γ 2, w h e r e d j, u 0, and q 0 are pre-
' d y scribed.
An s.
, i f [ d w ( d u d u\ d w f d u d u \
B i W'U} = J L l d i { C n d i + C l 2 i i ) + d i i C2‘ T + T y '
d u ^
+ w/J d x d y,
l ( w ) = J w q n d s,
, , d u d u\ ί d u d u\
where q n = [ c n — + c 12 — I n x + I <c2i — + c22^ ] %·
-,tj insulated *3.
«or 0
kux = q0(y)
kux = - h( u- ux)
A B
F i g. 8.6.
/d2T d2T\
8.1 2. —A; I r- + -7 — - ) = / i n t h e r e g i o n Ω w i t h b o u n d a r y c o n d i t i o n s
\ dx1 dy2/
a s s h o w n i n F i g. 8.6. T h e f o l l o w i n g b o u n d a r y c o n d i t i o n s a r e p r e ­
s c r i b e d: k u x = q o ( y ) on H A; = — h ( u — Uqo) on B C; u = u o ( x )
o n A B, and d u/d n = q 0 = 0 on C D E F G H (i nsul ated), where k
i s t h e t h e r m a l c o n d u c t i v i t y o f t h e m a t e r i a l o f t h e r e g i o n Ω, h and «00 are ambient quantities, and d u/d n = —d u/d x = —u x on H A ( two-d ime ns ion al heat conduction).
8.10. EXERCISES
299
A n s.
° ‘ - I L k ( ^ i + ^ )'° d:Cdy·
f f
, / dw dT d w d T\ f
= / / k [ τ τ + t t ) d x d y - ku
J J n \dx dx dy dy J J c
dT dT \ , — nx + — nyJ ds.
bed temperature To): undary. Tx ): nx = 1, 0; on C3 — CDEFGH (insulated boundary): q = dT/dn = 0;
(y))
The boundary conditions on C\ = AB (prescifi nx = 0, ny = — 1; on C2 = BC (convective
bo
and on Ci = HA (prescribed conduction qo Thus,
b(w,u) = j j k + t t ) d x d v
J Jn \ ux ay ay j
+
n b f b
l(w) = - w(0,y)qo(y)dy + hTx / w( Jo Jo
h / w(a, y)T(a, y) dy, Jo
,y)dy.
d ( rdu 1 / dv dx 1 ί dx ^ 2 V dx
)1}
+ g = o,
d2 / d2v\ d j dv rdu 1 ^ civ Λ 21 \
dx2 V dx2) dx \ dx I dx 2 V dx ) J J
Λ , dvi
u = v = 0 at x = 0, i; — = 0.
dx\x=o
deflection bending of a beam).
Ans. Let w\ and w2 be the two test functions such that they satisfy the essential boundary v. Then
0 =
/
Jo
dw χ
J du
dx
| dx
d?w2
d2v
dx2
dx2
+ wig
dw2
+ a
dw2 dv dx dx
du
dx
Thus,
b((wi,w2), (u,v))
f l Γ dwi (
Jo ί0 dx I
du 1
adi +
2
dw 2 dv c du 1 dx dxXdx 2
l({wi,w2)) = - J (wig + w2f^ dx
: nx = —1, ny — 0.
/ = 0;
0 ],., = m“ <1” ge'
one for each equation, conditions on u and
dx,
dv
dx
+ w2f
dx
dv\2\ d2w2d2v dx)
J dx2 dx2
+ b- } ] dx,
dv\2 dx) dw2
300
CHAPTER 8: WEIGHTED RESIDUAL METHODS
8.1 4. Find t h e functional I ( u ) for t h e transverse deflection u o f a membrane stretched across a frame, in t h e shape o f a curve C, su bjected t o a pressure loading f { x,y ) per unit area. Assume t h a t t h e t e nsi o n T in t h e membrane is constant. N o t e t h a t u satisfies the equation
A n s. T h e variation o f t h e t o t a l work done by t h e force f J T is
/ / \7 2 u 6 u d x d y J J a
^ f f 6 [ V u\2 d x d y — f ~ 6 u d s,
which leads to
N o t e t h a t Exampl e 8.1 is used. Thus,
= V 2 u 6 u = — ψ δ υ,.
8.1 5. Consider t h e Poisson boundary value problem: — V 2 u = f i n Ω, wi t h t h e boundary conditions u = 0 on C\ and d u/d n = 0 on C2. Show t h a t
8.10. EXERCISES
301
Ans.
dw
dy
f f f dw du l(w) = // f w d x d y — /
J Jn Jc
t) d x d y ·
VJ
8.1 6. Us e t h e G a l e r k i n m e t h o d, a n d t h e Rj
“ k * έ Κ Ε 4 ΐ ) + / = 0 · 0 < I <
w h e r e E I is called the flexural rigidity of t
du ^ rd2u
„ ( 0 ) = o = - ( o ), e i —2
= Mo,
[ T a k e w = φί = x l+1. The exact solution is
fl 4
.y l e i g h - R i t z m e t h o d, t o , E I > 0, / = const, tie beam, with
1 ( e i ^ ) L = 0·
,· fl’ Mo 2
E l u = 24 ~ Ύ Χ + Ύ Χ
An s.
. fl d2w d 2u
bi w'a) = L E I d ^ d Z d x ·
!( «,)
-/
Jo
w f dx + w( 0)
dx
dw
E I ^
dx
x=0
dx
fou
x=0
Exact sol uti on is obtained by direct integral
8.1 7. — V 2« = 1 in Ω = {(x, ?/) : 0 < x,y < 1}
du
w du
u{ l,y ) = 0 = ti(x, 1), (0,y) ^ 0 = ^ ( z.0 ) ·
dn
Ans. If we take w = φί = (1 — x l ){l — y choice satisfies th e essential boundary cond: ural boundary conditions. Hence, we assu: solution as u\ = αφι, φχ = (1 - x 2) ( l
(2i- 1)πχ (2i - can take w = φί = cos --------- -— cos --------
exact solution is
u(x,y) = | { ( i - y 2)
( —l ) fc c o s [ ( 2/c — \)iry/2
3 2
-tj- 3
k—1
( 2 k — l ) 3cosh[(2fc — 1) π/2]
du
dn
ds.
rdw du dx dx
(I) + m 0
x=0
dw
dx
i t i o n. s u c h t h a t
,i = 1, · · · ,n, then this itions, b u t not the nat- the first approximate y 2). Alternatively, we
i = 1,· · · ,n. The
me
cosh[(2A: — 1 ) π χ/2 ] j
302
CHAPTER 8: WEIGHTED RESIDUAL METHODS
8.18. Find the iV-th approximate solution of Example 8.4 by taking
, , . . . , / \ . j n x . kny
t h e b a s i s f u n c t i o n s a s <pjk{x, y) = s i n sm ——.
d b
An s.
Σ a i k ~ t ~ ( i? +
j
|fc = l
a b n 2 ( j 2 k 2 \ 4c a b
Ka 2 b 2 ) j k n 2
f o r b o t h j, k odd.
8.1 9. Find t h e approximate sol uti on by t h e Galerkin me th od for t h e nonlinear problem Ut = u x x + e u 2 on 0 < x < 1, su bje ct t o t h e boundary conditions u(0, t ) — 0 = u ( l,i ) 4 and t h e i ni t ia l condition u ( x, 0) = 1.
H i n t. Choose <l ) j ( x) = sin j n x.
A n s.
Σ - g [ 2 ( 1 3jT i7r:>
j,k=1 J
+ ^ Σ C>"( i )/( m'n'i ) ] }> m,n = 1,2,· · · ,J V,
τ η φ η
w h e r e
. 1 — c o s i m — n + j ) w 1 — c o s i m — n — j ) n
f ( m,n,j ) = -------------------------:-------------------------------------------------;-----
m — n + j m — n — j
1 — c os( m + n + j ) - K 1 — c o s( m + n — j ) i r
m + n + j τ η + n — j
N
T o f i n d C j ( 0 ), solve ( ) — 0, where R =
j,k—l
8.2 0. U s e t h e G a l e r k i n m e t h o d t o s o l v e t h e P o i s s o n e q u a t i o n V 2w = 2 s u b j e c t t o t h e D i r i c h l e t c o n d i t i o n u = 0 along t h e boundary o f t h e square { —a < x, y < a } (Fig. 8.5).
H i n t: Use t h e basis functions <f >( x,y) = (a 2 — x 2 ) ( a 2 — y 2 ), and consider t h e approximate solution
u N (x, y) = {a2 - x 2) ( a2 - y 2) ( A i + A 2x 2 + A 3 y 3 H--------h A n x 2 l y 2:i ).
A n s. F o r = 1, w e h a v e
f f [ - 2 ( a 2 - y 2 ) A i - 2 ( a 2 - x 2 ) A x + 2 ] ( a 2 —x 2 ) ( a 2 — y 2 ) d x d y = 0. J — a J — a
8.10. EXERCISES
303
T h i s y i e l d s
. 5 5 (a2 — x2)(
Ai = s Ui = - ---------8
a2 - y2)
We must have A 2 = A 3. Then for N = 3, t u 2 — (a 2 - x 2 ) ( a 2 - y 2 ) [ A i + A
w h e r e
At =
1 2 9 5 1 4 1 6a 2'
A·? —
4 ke
i ( x2 + y 2% 25
4432a4 ’
and
u 2 ( x,y ) =
Alternately, i f we choose t h e basis functions j, k odd, then
jnx kny
$S <Pjk = cos cos ,
2α a
un
Σ
] π χ &jk c o s ——
c o
w h i c h l e a d s t o
ra pa
j,k= 1 j,k odd
f° fa Γ /j 2Tr2 k2w2\ jw
y..y ^ [5 ^ ( 4 3 · + - o r'”
a
mn
H e n c e f o r j = m and k = n
_ 128a2( - l ) j +fc~ 2 j k { j 2 + ^2)ττ4
.21. Use t h e Galerkin method t o solve t h e eij Au = 0 in t h e cylindrical polar coordinates
H i n t: Solve ( r ^\ + Xu = 0, 0 < r r dr \ dr J
J7TT
A n s. Take <j>j {r) = cos ~ ^ · For t h e first approximation, φ\ =
πΓ . 7TT , . , ,
cos —, and u 1 = αι cos —, which leads to
2α 2α
2, [ Ί ΐ ί [ Ι Ξ ( - s i n - ) l
Jo I r dr L 2α V 2a J J
c*i + Aa
15.
a2
4 + ^ ( x 2 + y2)
k-Ky
~2a'
cos ■
k-Kxy 1
*
t
]
x kny , ,
x cos — — cos -—— dx dy = 0. 2c, 2 a
/ 2
genvalue problem V 2it - for 0 < r A
<: a.
wr ι
cos — >· r dr = 0. 2a J
304
CHAPTER 8: WEIGHTED RESIDUAL METHODS
T h i s gives t h e equation for t h e eigenvalue A as
7Γ
T (,i + ^ J - A a 2
Hence,
π 2( π2 -1-4) 5.8304
Ai
4 α2 ( π 2 — 4)
5 779
T h e e x a ct value is Ai = 2—. For t h e second order approximation
7rr 3πΓ 5.792
U2 = « ι cos - — h a 2 cos ——, which gives λ2 = —.
2α 2 a a z
8.2 2. U s e t h e G a l e r k i n m e t h o d t o d e t e r m i n e t h e l o w e s t f r e q u e n c y
( f u n d a m e n t a l t o n e ) o f t h e v i b r a t i o n o f a h o m o g e n e o u s c i r c u l a r p l a t e
Ω o f r a d i u s a and center at at the origin o f cylindrical coordinates,
clamped at t h e entire edge, i.e., solve V 4u = Au su bje ct t o t h e
c onditions u( o ) = 0 = u r ( a ).
H i n t: M i n i m i z e t h e v a r i a t i o n a l p r o b l e m I ( u ) = I I V 4 u d x d y,
J J a
s u c h t h a t I u 2 d x d y = 1, subject t o t h e given conditions.
J J a Ans. Solve
d2 1 d \ /d2u 1 d u\
dr2 ^ r dr J \ dr 2 r dr J U'
x / r 2\j+1
T a k e H n = 2 _,a i 2 J ' Then, e.g., for U2, we have
/1 9 2 Aa4\ /1 4 4 Aa4\ „
( ~ 9 5~ J ^ a2 ( ~ 9 g- y ~
/1 4 4 A-;1 \ ( 9 6 Ao1 \ ,
Ql C“ 9-------- 6 - ) +<Η Τ - — ) = °'
a n d t h e e q u a t i o n f o r A i s
Q 7 Q 9
( A a4)2- - - - - - -— A a4 + 4 3 5 4 5 6 = 0,
ϋ
w h i c h h a s t h e s m a l l e r r o o t a s A =- - - - - - - - - - . U s i n g t h i s v a l u e o f
a
A i n t h e a b o v e s y s t e m o f t w o e q u a t i o n s, w e f i n d a 2 = 0.325 α ι, and
r2 \2 ( r2' 2
u 2 = a x + 0.325 ( l - ^
w h e r e ct\ can be found from the above system of two equations.
9
P e r t u r b a t i o n M e t h o d s
U y
The perturbation methods provide approximate solu and initial value problems. These methods are contain a small parameter, say ε, and the solution parameter occurs, in general, in a partial differential
L u + ε Ν u = 0,
where L is a linear partial differential operator, an or a linear differential operator which makes the cult. If ε = 0 reduces Eq (9.1) to an ordinary diff^i perturbation method will fail.
Another kind o f perturbation problems arises by In this case the parameter ε will appear in the bounc common perturbation methods discussed here are
( 1) series expansion in powers o f ε, and
(2) successive approximations.
These methods apply when the partial differential equation or the boundary is perturbed.
ame
Quite often these two methods will yield the s Frequently the perturbation solution is equivalent to corresponding integral equation. For the perturbation it is assumed that u is a continuous function o f the that the differences in the two problems (perturbed singular in character. These methods are thus app! the solution to an ideal simple problem is known
tions for boundary value sed when such problems for ε = 0 is known. This equation o f the form
(9.1)
d N is either a nonlinear s olution o f Eq (9.1) diffe­
rential equation, then the
perturbing the boundary, dary conditions. The two
e approximate solution, he iterative solution o f the method t o be successful, perturbation parameter and and unperturbed) are not ied to problems in which and, for a more realistic
306
CHAPTER 9: PERTURBATION METHODS
situation, the differential equation or the boundary conditions or the region is perturbed.
There are other more complicated methods which deal with singular per­
turbation problems. We shall, however, not discuss such methods in this book. For more information about them the reader is referred to Kevorkian (1990).
| www | Refer to the Mathematica Notebook p e r t u r b a t i o n. ma for this chap­
ter.
9.1. T a y l o r S e r i e s E x p a n s i o n s
The general scheme for this method is as follows: Consider Eq (9.1) subject to some prescribed boundary conditions and/or initial conditions. Assume that the solution o f the homogeneous equation L u = 0 subject to the same prescribed conditions is known. Then, in order to solve the given problem we shall assume that u possesses a series expansion in powers o f ε o f the form
OO
u = uo + eiti + 82
U
2
-\----- = (9.2)
n=0
If we substitute this power series for u into Eq (9.1), we get
(
OO \ / OO
Σ ε" u n J + ε Ν ί y ~'j ε ” u n = 0
n=Q J
\n = 0
Assuming that Uq satisfies the prescribed conditions and that u n, η Φ 0, satisfy homogeneous conditions, we can obtain a system o f partial differential equations by comparing coefficients o f various powers o f ε on both sides o f Eq (9.3). These partial differential equations are such that a partial differential equation in u n will depend only on ito, « ι, ■ · · , u n -\. The solutions to u o, u i,· · · ,u n-1 are known successively, i.e., one solves first for u q, which enables one to solve for u\ and so on, and one can finally solve for u n. We will demonstrate this method by some specific examples.
(9.3)
E x a m p l e 9.1. Consider
9.1. TAYLOR SERIES EXPANSIONS
307
We will consider two cases:
(a) For f ( u ) = u, let us assume t h e power se: differential equation and t h e boundary conditi 0,1,2,3, · · ·, are given by
ries ( 9.2). T h e partial ons for each u n, n =
V 2u 0 = 0, u o ( l,6 ) = g ( 0 )
V 2u x + u 0= 0, u i ( l,0 ) = O,
V Un _|_i + un = 0, un+1 ( 1, Θ
T h e g e n e r a l s o l u t i o n f o r V 2 u o = 0, b y t h e m v a r i a b l e s, i s
u o
= ^ r n ( A n cos η θ + B „
s i n
w h e r e
Σ ( Λ „ c o s η θ + B n sin η θ ) = η ( θ ).
71=0
I f g(6) is a periodic Dirichlet function, t h e n A „ mined, and t h e subsequent equations for u n can l et g(0) = c o sΘ. In t hi s case uq = r c o s#, and equation for u\ becomes
V 2u i + r c o s# = 0.
(9.8)
ential equati on o f t h e V z v + c i r n cos ρ θ + C2 r m s i n g# = 0
Since t h e particular integral for a partial differ· t y pe
is o f t h e form
vp = c n r n+2 cos ρ θ + C
22
r m+2 t h e sol uti on for u\ is given by
u i = ^ r ( l — r 2) c o s#.
8
Similarly, t h e sol uti on for u 2 is given by
«2 = ^ r ( 2 - 3 r2 + r 4) co^ Other terms can be obtained similarly.
(9.5)
0.
ithod o f separation o f η θ ), (9.6)
(9.7)
and B n can be deter­
g e solved. For example, t h e partial differential
:sin q 0,
( 9.9 )
( 9.1 0 )
308
CHAPTER 9: PERTURBATION METHODS
(b) For f ( u ) = u2, we use the series (9.2) for u in powers of ε, where ito satisfies the boundary condition u o ( l,#) = cos#, and up satisfies the homogeneous boundary condition. Thus, the partial differential equations for uo, ui, U
2
, ■ ■ ■ are given by
1 1
UQrr “I- UQr “I- ^ H0ΘΘ 0,
lllrr + ^1 r 9^100 “t” ^0 0, (9-11)
r ri
U2rr H U2r H ^«200 + 2 uoUi = 0,
r r2
and so on. The solution for uq is clearly Uq = r cos#, and the equation for u\ becomes
Uirr H Uir ~Uigg + Γ2 COS2 # = 0,
r rz whose solution is then given by
Ul = ~ r^ + Ia ^ 2 _r4)cos261· (9·12)
We can continue the process to find u2, «3, ■ ■ ■. ■
Ex a mple 9.2. To solve
urr + - u r 4 - -\uge + euug — 0, u ( l, #) = cos #, (9.13)
r r1
up to the first three terms of the power series solution, we use the series
(9.2) for u. Substituting this series into the partial differential equation
and comparing coefficients of different powers of ε on both sides, we get
1 1
UOrr “t- ~t~ ϊϊ'ΜΟΘΘ 0,
r rz
Uirr H Uir ----~Uigg + U()Uog = 0, (9-14)
r rz
1 1
U 2 r r + -U2r + Τ ^2ΘΘ + Ui UqB + U0Uig = U. r
T h e new b o u n d a r y cond i t i o n s a r e u o ( l,#) = c os#, a n d u n ( l,#) = 0, n > 1. I t is easy to see t h a t uo(l, #) = r cos #, and the partial differen­
tial equations for ui is
Uirr + i u i r + uigg = r 2 cos # sin # = ^ r 2 sin2#. (9.15)
r rz I
9.2. SUCCESSIVE APPROXIMATIONS
1 „2/„2
It s sol uti on is u i = — r (r — 1) sin 2Θ. The partial differential equation for U2 becomes
1 1
U2rr + “ ^2r "Ί"" ^^2ΘΘ
1
r r * 2 4
a n d i t s s o l u t i o n i s g i v e n b y
U2 = ^ r 3 (r2 _ X) cos361 + π ^ ^ 7"4 “ ^ co
r (r — l ) ( s i n2#i i i n# — 2c o s 2#c o s#),
—r3( r 4 — 1) cos 3#
640 v '
1
2304
s#
(r6 — l ) c o s#. ■ (9.16)
We can continue t h e process as long as we quired degree o f accuracy. There will, o f cou convergence, bu t one observes t h a t t h e coefficii small, so convergence for values of ε < 1 appear;
need t o o b t a i n t h e re- rse, be t h e question o f i ents are g e t t i n g fairly s likely.
309
9.2. S u c c e s s i v e A p p r o x i m a t i o n s
T h e general scheme for t h i s me th od is as follow; approximation t o be uo, which satisfies t h e give conditions and t h e homogeneous equation L u — imation is t h e n u i, which satisfies t h e equation given boundary and initial conditions. T h e pr t h e required degree o f accuracy is achieved. It o f difficulty is directly proportional t o t h e orde r will demonstrate t hi s me th od by solving t h e ab
is
E x a m p l e 9.3. We now solve Example 9.1(b method o f successive approximations. The partial for uo is t h e same as in (9.11), i.e.,
UQrr
= 0
~U0r + “ 2η0θθ
a n d i t s s o l u t i o n i s u o = r cos Θ. B u t t h e partial u i is
Uirr + - U l r + gg + ε τ 2 cos2 Θ = 0, U i ( l,0 ) = cos#. (9.17)
s: We assume t h e first :n boundary and initial 0. T he second approx- Xui+eTVuo = 0 and t h e Dcess is continued until obvious t h a t t h e order o f approximation. We ove examples.
') for /( u ) = u2 by t h e differential equation
differential equation for
310
CHAPTER 9: PERTURBATION METHODS
I t s s ol uti on is
1 1
u i = r c o s# + ε ^ ( 1 ~ f 4 ) + ^ (r2 — r4) c o s 2#· (9.18)
N o t i c e t h a t here u i is ac tual ly t h e sum uq + e u i o f Exampl e 9.1. ■
E x a m p l e 9.4. We now solve t h e Example 9.2 by t h e me th o d o f suc­
c essive approximations. Obviously, t h e solution for uo is uo = r cos #.
Now t h e ne x t approximation u i satisfies t h e given boundary c ondition and t h e partial differential equation
Uirr “ Uir "^ U i g g -|- S UqILqq — 0, (9.19)
or
1 1 ε Uirr H— u i r Η— » u ^ g — - r 2 sin 2# = 0. (9.20)
r H 2
T h e s ol uti on for u i is e asi ly seen t o be
u i = r c o s# + - ^ - ( r 4 — r 2 ) s i n 2#. (9.21)
24
T h e ne x t approximation u 2 once again satisfies t h e same boundary condi ti ons but t h e new partial differential equation is
1 1 ,
U2rr + - U 2r + “ o u200 + S U i U i g = 0. (9.22)
Since
£ „3/ 2 , ε2 „ 4/„ 2 1\2,
u i u i e = — — s i n 2# + —r ( r - l ) ( c o s#+ 3 c o s#) - l - - — r ( r -1) s i n 4#, 2 4 8 5 7 6
( 9.2 3 )
t h e e q u a t i o n f o r u 2 becomes 1 1
U2rr H U2r H ~U2gg
r r J
.2
= s i n 2# - - ^ r 3( r 2 - l)(cos# -I- 3 cos#) - - ^ - r 4( r 2 - l ) 2 s i n 4#. 2 48 576 (9.24)
Its solution is
ε ε2
u 2 = r cos# + — r 2 ( r 2 — l ) s i n 2# + — — r ( r4 — 1) cos#
24 1152
‘2 ‘2 3 ,j i _ ■, \ _ oa _ ,„4
r ( r — 1) cos# + - - - r (r — 1) cos3# — ——r (r — l ) c o s 3#
2304 v 7 256 v 7 640
1 „4/„6 i\ 1 „ 4/„ 4 ι \ , 1 „ 4/„ 2
r (r - 1) --------— r4( r 4 - 1) + —— r4( r 2 - 1)
\ / Ί Ο Ο Ο/Ι ' ' 1 1 c o n ' '
48384 v ' 13824 v 7 11520
si n 4#.
(9.25)
9.3. BOUNDARY PERTURBATIONS
311
It is clear t h a t t h e sol uti on (9.25) is almost t ai ne d in Example 9.2. These two solutions wo i f enough terms in t h e series solution and a approximations are taken. ■
similar t o t h e one ob- 'iild be e x a c t l y t h e same large enough number o f
9.3. B o u n d a r y P e r t u r b a t i o n s
For problems involving boundary perturbation used. T h e following examples i llustrate t h i s mi
t h e series (9.2) for u is ethod.
E x a m p l e 9.5. Solve t h e harmonic (Laplace’s) equation
Uxx “f" Uyy — 0,
such t h a t
u(e sin Luy,y) = 0, u ( w,y ) = 0, u(
a n d l i m u ( x,y ) is bounded. Assuming t h e se:
y —>oo
t h a t = 0 for all n. Moreover,
u(esinujy,y) = u(0,y) + e s i n u j y u x ( 0,y ) + - ( e
u ( e s m u y,y ) = u 0( 0,y) + e u i ( 0,y ) + u 2( 0,y ) + e s i n u i y [u0,x ( 0,y )
+ £Ui,x
(«
+ - ( s s i nuj y ) 2[uo ,XX (0, y) + ε\ + ε2^ 2,χχ(0, y) + ■ ■ ■ ] +
B y comparing (9.26) and (9.27) we find t h a t conditions
Uo(0,?/) = 0, Uo(-rr,y) = 0, and u o ( ^,0 ) = s i n x, (9.28)
whereas U\( x,y ) must s ati sfy t h e conditions
ui(0,y) =
—e~v sinuiy,
and « ι ( π, y) =
I), ui(x,
0) = 0. (9.29)
x,0) = s i n x,
:ies (9.2) for tt. we find
s i nwy) 2Uxx(0, y ) +
( 9.2 6 )
,y) +e 2u2lx(0,y) + · · ·] ^l,X£ (0j y)
(9.27)
o ( x,y ) must s a t i s f y t h e
312
CHAPTER 9: PERTURBATION METHODS
N o t e t h a t u o ( x, y ) = e ~ y s i n x. Since t h e sol uti on for u\ is complicated, we will separate it into two parts: The first part will s a t i s f y Laplace’s equation and t h e boundary conditions at χ = 0, π, while t h e second part, t hou gh sati sfyi ng Laplace’s equation, will sat i sf y homogeneous boundary condi ti ons at χ = 0, π; t he n t h e sum of b o t h parts will s ati sfy t h e c ondi ti ons at y = 0. Thus, let u i = V 1 + V 2, where V « i = 0, su bj e ct t o t h e c ondi ti ons υι(0,?/) = — e ~ y s i n u j y,v i ( n, y ) = 0. Then
Vi = [ f i x ) COS wy
+ g ( x ) sin uy] e ~ y,
/( ο ) — /( π ) — 9( π ) — 0) g( o) = - i.
The n su b s t i t u t i n g v\ into Laplace’s equation and comparing t h e coef­
ficients o f sin ωχ and cos ω χ on both sides, we get
/" + (1 - ω2)/ - 2ω9 = 0, g" + (1 — u 2 ) g + 2 u>f = 0.
,2, . „ , „ (9-31)
Let 2 = / + i g. Then, multiplying t h e second equation by i in (9.31) and adding i t t o t h e first we get
z" + (1 + 2iw — ω2)ζ = 0,
or
ζ" + (1 + ίω)2ζ = 0, (9.32)
w i t h t h e boundary conditions 2(0) = —ΐ,ζ ( π) = 0. Its s ol uti on can be expressed as
2 = Α ε { 1+ί ω) χ + Β ε ( 1 + ί ω) χ. (9.33)
On applying t h e boundary conditions, we find t h a t
_,·„-ί(1+ίω)ττ ·„ί{1+ίω)·π
-, Β = - ~ --ΐ Γ - τ---------- γττ——τ - · (9.34)
• ' __11 1 -Li./,ι 1ΤΓ *1Α 1 -UiiiJ\ίγ ' '
£—ϊ{1-\-ίω)π _ βϊ(1+ϊω)π ’ g—ΐ (1+ίω)π _ £>ΐ(1+ΐω)π ’
Now sol vi ng for t h e real and imaginary parts o f z, we find t h a t
t t N s i n x c o s hω ( π — χ ) (._ λ c o s x s i n h w ( 7 r — x ) f n οεΛ = sinh ω π ’ = s i n h o w ' ( ' ’
N e x t, w e d e t e r m i n e v 2 by t h e separation o f variables me th od w i t h ho­
mogeneous boundary condi ti ons wi t h respect t o x. T h e s ol uti on is o f t h e form
OO
v 2 = A n e ~ ny s i n n x. (9.36)
0
9.3. BOUNDARY PERTURBATIONS
313
The n adding t h e solutions (9.30) and (9.36), and pl e t e sol uti on for u\ is given by
using (9.35), the com-
ω·
y ] ^
J o i r
Ul = Σ'1·'- n y sin n x + [ cosh ω (π — x ) sin x (fos u y o
— sinh ω (π — x ) cos x sir On applying t h e c ondition u i ( x, 0) = 0, we g e t fn =
sinh ω π om (9.37)
( η - 1 ) 2 + ω 2 ( n + 1 ) 2 + ω
H e n c e t h e s o l u t i o n f o r t h e p r o b l e m i s
U i ( x,y ) = [ c o s h w ^ — x ) sin x cos wy
— s i nh ω ( π — x ) cos x sinwwl — v ' yj si
00 1 1
- Y\______ -_______________ -______1 -
•4-' 1 ( η - 1 ) 2 + ω2 (τι 4- l ) 2 + ω2 J sihh
e
sin
- v
E x a m p l e 9.6. Consider
V 2u = 0,u ( l + e s i n#, Θ) = /(#).
B y expanding u(r, Θ), r = 1 + ε sin Θ about r = 1 have
u ( l + e s i n#,0) = /(#) = u ( l, #) + e s i n f l u r ( l, #) + -
We also assume t h a t u has a series expansion in p
( 9.2 ), i.e.,
u(r, Θ) = Uo + £Ui + £2u 2 + Combining (9.40) and (9.41), we get
«©(ΐ, e) = m,
u i ( l, Θ) + s m 6uor ( l, Θ) = 0 U2( l,0) + s i n#u i r + ^ s i n 2 6 u o r r ( l
( 9.3 7 )
ω
s i n h ω π'
Ι ι ω π s i n n x
- n y
ι ω π
( 9.3 8 )
( 9.3 9 ) i n a T a y l o r s e r i e s, w e
ε 2 s i n2 9 u r r ( 1, θ)-\--------.
( 9.4 0 ) o we r s o f ε o f t h e form
(9.41)
(9.42)
Θ) = 0,
314
CHAPTER 9: PERTURBATION METHODS
and so on. T h e partial differential equation t o be satisfied by u n for all n is V 2u„ = 0. Using t h e general sol uti on and applying t h e boundary condi ti ons for tt0, we get
CX)
u o ( l, #) = /(#) = cos η θ + B n sin η θ ). (9.43)
o
I f /(#) is a periodic Dirichlet function, then
A 0 = ~ [ /(#) ά θ, A n = - / /(#) cos η θ ά θ,
J — τ τ ^ -J — π
B n = — [ /(#) sin n#c£#.
^ J — 7Γ
Case 1. If f ( 9 ) = c o s#, the n
uq = r cos Θ, Ui = — ^ r2 sin 2Θ, u 2 — ~ (r3 cos Θ — r cos Θ ).
C a s e 2. I f /(#) = s i n#, t h e n
Uq — r sin 0, 1 ' „2
(9.44)
( 9.45)
u j = - ( r c o s# - 1), (9 46)
U2 = ^ ( r sin Θ — r 3 sin 3#.
For t h e sol uti on o f t hi s example by t h e method o f successive approxi­
mations see Exercise 9.4 below. ■
9.4. E x e r c i s e s
9.1. Obtain a perturbation solution for t h e problem:
Uf Uy y “I” k U y y f ,
where u ( y, t ) = 0 for 0 < t and u (0, t ) = 1 for t > 0.
(a) B y successive approximation (2 approximations).
9.4. EXERCISES
315
(b) B y expansion in powers o f k (up t o t h e first power o f k ). H i n t. Use Laplace transforms. [Compare wi|th Exercise 6.4.1
( _ y _\ . k
A n s. u = erfc
+
(Ή
\2 y/t )
9.2. O b t a i n a s t e a d y s t a t e p e r t u r b a t i o n s o l u t i o n f o r t h e p r o b l e m:
Ul “1“ k U y y f -- k u y y y “1“ Uy y Uy , U
l i m u ( y, t ) = 0, l i m u ( y,t, k )
y —>oo fc—► 0
Ste ady s t a t e in t hi s case implies t h a t t h e in: be ignored.
H i n t: Assume a sol uti on o f t h e form u ( y, t ) assuming a perturbation sol uti on for u ( y,t ), solution for t h e roots o f t h e characteristic e< A n s.
u{y,t) = ^ e~ay+i{ult+by
where
1 . A k C k 2
a = x ( l + Q o :----- ■ = = — s
2 2\/Ι Τ Ϊ 6 ω2 V l + 16a,-2
_ 1 B k D k 2
2 ° 2\/l + 16a;2 \/l + 16a;2 ’
( 0 ,t) = e™\ u ( y,t, 0).
itial condi ti ons are t o
= F ( y ) e l u t. Instead o f assume a perturbation quation.
ao
βο
^ [ V l + lGuj2 + 1],
]/\[ V l + l 6ω2^Τ],
A — Qo + otQ + /?o — 2 a;2 ao + 4a; βο,
Β = 4ωαο + 2(JQq + 2 υ 2βο + 2a;/3q — βο, C = p a 0 + qfio, D = qa0 - ρβο,
( 2 ω2 βο — 3(1 + ao) ) A + (8a> + 2u>ao +
P =
W l + 16a;2
(2ω2βο - 3(1 + a 0 ) ) B - (2ωαο - 3βο ~ 4ω ) Α -
q =
4\/l + 16a;2
9.3. Solve V 2 u + e f ( u ) = 0,u ( l,#) = <?(#), for respectively, and g{ &) = s i n#,s i n 2#, respectr
/4 π ί
3/30) B -
( A 2 - B 2 ) VI + 16a;2
2 A B
V T + m J 2
f ( u ) = u, u 2, u + u 2, ely. Find t h e first two
316
CHAPTER 9: PERTURBATION METHODS
terms in each case. A n s. For /(#) = s i n#:
u « r sin # — —r ( l — r 2) sin #,
8
u « r s i n# -1- ^ ^ ~~ r 4) — 5r 2 ^ — ^ c o s 2 ^ |
u « r sin# 4- ^ { ^-(1 — r 4) — r( 1 — r 2) sin# — (1 — r 2) cos2#
8 L 4 3
For /(#) = s i n 2#:
u « ^(1 - r2c o s 2#) + ^ | l — r2 — ^—(1 - r 2) c o s2#
2 8 ^ 0
u ~ I ( i — r 2 cos2#) + | { ^ ( 1 - r 2) - γ ( 1 - r 2)cos2#
+ ά (1 _ + 5ο {1 ^ coa 40} ’
u w ^ ( l - r 2c o s 20) + ~ r 2 ) - ^-(1 - r 2) cos 20
+ 5 5 ( I - ’'") + S <1^ r!) “ s 4#}'
9.4. Solve u r r + - u r 4- - - ^ u g g + e u r u g = 0,u ( l, #) = c o s#. r
S o l u t i o n. Su bs t it u t in g t h e series (9.2) into t h e partial differential equation and comparing coefficients o f different powers o f ε on both sides we get
1 1
UQrr ~f~ UQr “f- ^ Uqgg 0,
1 1
Uirr + Ui^· 4“ /-y i gg "t- "UorUQQ U,
r r Δ
1 1
« 2rr + “ U2r + ~ ^ U 2gg + Mir MO0 + UQr Ui g = (J.
The new boundary conditions are u o ( l, #) = cos #, and u n ( 1, #) = 0. It is e a sy t o see t h a t
uo(l,#) = r c o s#,
U i( τ,θ ) = ^ ( r 3 — r 2 ) sin 2#,
ι / -ι ι on \
U2(r’ θ) = 480 ^ " Γ) COS^ + ( 3 5 ^ " e i"" - 2240 "Ί C° s 3 ^
9.4. EXERCISES
317
Alternately, we will now solve this problem t y the method of suc­
cessive approximations: Note t h a t the solution for u0 is r cos #. The next approximation u\ satisfies the given boundary condition and the partial differential equation
1 1
U\rr H U\r H £ u 0rU(_
T h e s o l u t i o n f o r ui is
Ui =r cos # + ε — ( r 3 — r 2) si
The next approximation u2 once again satisfies the same boundary conditions but the new partial differential equation is
1 1
U2rr H U2r ^U2gg + £UirUi
wh o s e s o l u t i o n i s
U
2
— τ cos θ + ε γ ^ ( Γ3 — r2) sin 2#
+ ε2 [ ϊ ΐ θ ( Γ ί - Γ) ” 8 β + ( έ Ι'4 -?4 ·
10=0.
n20.
= 0,
29
100Vll
9.5. Solve urr + ~ u r + X;Ugg + s u r61 = 0, tt f l, #) 4= sin #. Find the first r r 2·
t h r e e t e r m s.
So l u t i o n. S u b s t i t u t i n g t h e s e r i e s ( 9.2 ) i n t o t e q u a t i o n a n d c o m p a r i n g c o e f f i c i e n t s o f d i f f e r e n s i d e s we g e t
1 1
Uq rr + Uq r "Ί- ^ U009 — C
1 1
Uirr + ~Ulr + ^ Ui gg + UQrg
1 1
U2rr + ~ u2r + ^ U206 + Uirg
T h e n e w b o u n d a r y c o n d i t i o n s a r e u,0 ( l, #) = 0 , n > 1. Clearly, u0( l,#) = r sin#, and t f equations for u ι is
1 1
Uirr H Uir H— ^Uigg — — cc
r* nr*
2240
r 3 c o s 3#
)]
2 5 r
3 96
) s i
s i n 4#.
l e p a r t i a l d i f f e r e n t i a l t p o we r s o f ε o n b o t h
= 0,
= 0.
s i n#, a n d u „ ( l,#) = e p a r t i a l d i f f e r e n t i a l
s#.
318
CHAPTER 9: PERTURBATION METHODS
I t s s o l u t i o n is
u i = - r ( l - r ) c o s#.
O
The partial differential equation for it 2 becomes
U2rr
+ - « 27- + \^ 2θθ = x ( l - 2r) sin #, r r·*· 3
whose s ol uti on is given by
U
2
= — ^ r ( l — 4r + 3 r 2) s i n#.
O O
9.6. In Exampl e 9.6, choose /(#) = si n2 #, and solve t h e problem. Ans.
1
Uo = ~ (1 — r 2 cos2#),
ui = — ( r 3 cos # — r cos #), 1
1
U2 = - + - r cos # — —r cos 2# — - cos 3# + —r cos 4#.
9.7. Solve V 2u = 0, u ( l +e c os 0, #) = /(#), for (a) /(#) = c o s#, and
(b) /(#) = s i n#.
An s.
(a)
= r cos ft,
ui = ” 2 ( 1 + 7-2 cos2#),
U2 = ^ ( r3 cos 3# + r cos #).
(b)
u 0 = r sine/,
Ui
1
r sin 2#,
u2 = - ( r 3s i n 3# — r s i n#).
9.8. F i n d t h e e x a c t s o l u t i o n o f y u x + ( x + e u ) u y = 0,u ( x, 1) = x, by t h e meth ods o f Chapter 2, and t he n find an approximate sol uti on (up t o three terms o f a series sol uti on or three approximations by th e me th od o f successive approximation) of t hi s partial differential
equation by perturbation methods. Make ri A n s.
(1 + 2 e) u2 — 2 eux + y 2 — x 2
9.4. E X E R C I S E S
e l e v a n t c o m p a r i s o n s. - 1 = 0,
ex
+ a/[ ( 1 + x 2 - y 2) + 2e(l + x
2 — y 2) + e2x 2]
l + 2e
T h e s e r i e s s o l u t i o n i s
+ ( x - y/l + x 2 - t/2j e + ( 3 x 2 - 2y2 + 2 - 3x^/1
u = 1 + x 2 — y 2
9. F i n d t h e e x a c t s o l u t i o n o f
u.
3 1 9
+ x 2 — y 2^j ε 2 +
-— ^ + (x + eu2) uv = 0, ui 1 + 2/ y
b y t h e m e t h o d s o f C h a p t e r 2, a n d t h e n f i n d ε.η a p p r o x i m a t e s o l u t i o n ( u p t o t h r e e t e r m s o f a s e r i e s s o l u t i o n o r t h r e e a p p r o x i m a t i o n s b y t h e m e t h o d o f s u c c e s s i v e a p p r o x i m a t i o n ) o f t h e p a r t i a l d i f f e r e n t i a l e q u a t i o n b y p e r t u r b a t i o n t e c h n i q u e s.
A n s.
u2 + 2 eu2(u — x) = x 2 — 2
t a n
; = V x 2 ~ 2 t a n y + e\/x 2 — 2 ta n y ^x + ε 2 V x 2 — 2t a n ~ y ( x — y/:
1 0. So l v e V 2 u = 0, 0 < χ <7τ, y > εχ, u (0,y) = u( n,y ) = 0, and u( x,8x) = sinx. S o l u t i o n. Using (9.2), we have
- \J x 2 - 2 t a n y j
-\2
ή +■■
,0) = x,
2 t a n
0 < e <C 1, such t h a t
Hence
u{x,ex) = ^ ε"χ" — (,r,0).
o y
CHAPTER 9: PERTURBATION METHODS
Comparing powers o f ε on both sides we get
u 0(x, 0) = s i n x, u i ( x,0) + χ ^ ^ ( χ,Ο ) = 0,
u 2(x, 0) + x ^ ( x,0) + y ^ ^ ( x,0) = 0,
and condi ti ons on u n ( x, 0) can be derived similarly. All u n s ati sfy Laplace’s equation. The solution for uo is clearly given by uq = e~y sin x, and u\ satisfies t h e following conditions:
= ui ( w,y) = 0, u i ( x,0) = x s i n x.
A s s u m i n g a s o l u t i o n o f t h e f o r m
w e f i n d t h a t
T h u s i t i i s c o m p l e t e l y d e t e r m i n e d. W e c a n c o n t i n u e t h e p r o c e s s a n d d e t e r m i n e u „ f o r h i g h e r v a l u e s o f n.
71=1
10
F i n i t e D i f f e r e n c e M e t h o d s
The development o f high-speed digital and person: possible to effectively use different numerical technic and initial value problems involving partial differe different methods available, the finite difference met a straightforward structure which is derived from t also known as Taylor’s formula. We shall discuss first and second order partial derivatives, and then a solve boundary and initial value problems for seconc equations.
In finite difference methods we replace the given the boundary/initial conditions by a set o f algebraic solved by various well-known numerical techniques methods and adaptive grid generation schemes. Alth method is just one o f many numerical methods, it h methods in its simplicity o f analysis and computer c with complex geometries. It is used extensively in t fluid dynamics allowing the modeling o f complex fl with finite element method in the solution o f time-c
il computers has made it ues for solving boundary ntial equations. Among lod is widely used. It has runcated Taylor’s series, ; difference schemes for )ply them to numerically . order partial differential
differential equation and equations which are then ; including sparse matrix ough the finite difference as advantages over other odes in solving problems he area o f computational )ws. It is also used along ependent problems.
1 0.1. F i n i t e D i f f e r e n c e S c h e m e s
Consider a single-valued, finite function u ( x ) wh C ° ° ( i i 1). Then by Taylor’s theorem
ich belongs to the class
322
CHAPTER 10: FINITE DIFFERENCE METHODS
u(x + h) — u(x) + hu'{x) + u"(x) + — u'"(x) H-----,
u(x - h) = u( x) — hu'( x) + u"( x) - u"'(x) + · ■ · .
(10.2)
( 1 0.1 )
If we subtract (10.2) from (10.1), we get
u(x + h) — u( x — h) = 2 hu'( x) + 0 ( h 3).
Th e n
u'( a;) =
i ( a: + h) — u( x — /i) 2h ’
(10.3)
with a truncation error of 0 ( h 2). The approximation formula (10.3) is known as the first order central difference formula, and geometrically it represents the slope of the chord A B (Fig. 10.1). Similarly,
u'(x) =
u(x) — u(x — h) 2h ’
(10.4)
and
.. . u( x + h) - u{x) u (*) = ^ ’
(10.5)
10.1. FINITE DIFFERENCE SCHEMES
323
each with a truncation error of 0 ( h 2), are known as tb difference formulas, respectively, representing the slo] BP, respectively (Fig. 10.1).
If we add (10.1) and (10.2), we get
u(x + h) + u(x — h) = 2 u( x) 4- h2 u"( x) + 0 ( h 4), which yields the second order central difference formula
u(x + h) — 2 u( x) + u( x — h)
u"{x) =
h2
with a truncation error of 0 ( h 4). In fact, based on the the truncation error is approximately h f'"( x ) for bot
schemes, and — h ^ f ^ i x ) for the central scheme.
(10.6)
Taylor series expansion, h forward and backward
In the case of a function u( x, t ) of two independe: partition the x-axis into intervals of equal length h, an of equal length k. The (x, t)-plane is divided into eqi by the grid lines parallel to Ot, defined by Xi = ih, by the grid lines parallel to Ox, defined by yj = ih,
10.2). We wi l l u s e t h e f o l l o wi n g n o t a t i o n: L e t up = the value of the function u( x, t) at a mesh point (nodi
e backward and forward pe of the chord P A and
nt variables x and t, we d the ί -axis into intervals ual rectangles of area hk
i = 0, ±1, ±2,..., and 7 = 0, ±1, ±2,... (Fig. u( i h,j k ) = Ui j denote .e) P ( i h,j k).
t
( P
( U + 1 (ih, kj)
« - I,;
c. .
i.J
( U - i
0
h
Fi g. 10.2. G r i d l i n e s.
324
CHAPTER 10: FINITE DIFFERENCE METHODS
Then, in view of (10.6), we have the following three central difference schemes:
d2u
dx2
d2u
u((i + l ) h,j k ) - 2u( i h,j k) + u ( ( i - 1 ) j,j k )
p d x 2 i,j h2
U l,j —
= h2
Ξ δχ^ 2'3 ’ * = 1’2>· · · >n ~ !>
h 2
w i t h a t r u n c a t i o n e r r o r ——u xxxx(x, t), where x,_ i < x < xf,
( 1 0.7 )
d 2u
W
d2u p dt 2
Ui
i,j+1
2Ui j + U i j - χ
hj
SfUi, k 2
k 2
j = 1,2,... ,m — 1,
( 1 0.8 )
k 2
w i t h a t r u n c a t i o n e r r o r - — utttt{x, t'), where t j - 1 < t' < tj·, and
d U j U i + U i ^ - i j — i Ui — + i ·4~ Ui — i j — \
d x d t l p Ahk
( 1 0.9 )
h2 - k 2
wi t h a t r u n c a t i o n e r r o r —— uxxxt (x, t ) — —uxt u( x'.t'). Note that the dif­
ference operators δ2 and δ2 are the finite difference analogs of the partial
d2 d2
differential operators and respectively. Moreover, from (10.3) we have the first order central difference formula
du
dx
u((i + 1 ) h,j k ) — u( ( i — 1 ) h,j k ) h
U i + l J U i - l t j
( 1 0.1 0 )
ti1 -x J
with a truncation error ——u xxx ( x,t ), and 6
du I Ui j + \ Ui j — i
d t l p k
( 1 0.1 1 )
wi t h a t r u n c a t i o n e r r o r — — u t t t ( x, t'). The first order backward and forward difference formulas for u( x, t) can be similarly derived from (10.4) and (10.5).
10.2. FIRST ORDER EQUATIONS
The second order forward and backward difference defined by
d2u\ _ Ui+
2
,j ~ + Ui j
d x 2 I i j
K2
, * = 0,1
and
d2u dx 2
Ujj 2Uj-ij + Uj - 2J _ η ι h2 ’ ’ ’
1 0.2. F i r s t O r d e r E q u a t i o n s
Consider the first order quasi-linear partial differem
aux + buy = c, or a p + bq = c,
3 2 5
2....,n - 2, (10.12)
2...., n-2. (10.13)
s c h e m e s a r e r e s p e c t i v e l y
lial equation of the form
where a, b, and c are functions of x, y, and u only yields the auxiliary system of ordinary differential e
dx dy du
a b c
It was shown in Chapter 2 that at each point of th
(10.14) there exists a direction, known as the characteristic, along which the
solution of this partial differential equation coincide:
ordinary differential equations (10.15). We shall deiote the projection in the
(x, y)-plane of this characteristic by C. Let us assu is known at every point of the characteristic C in the is distinct from the initial curve Γ on which the initia
(10.14)
see §2.3). This equation quations
(10.15)
solution domain of Eq
with the solutions of the
me that the solution of u (χ, y )-plane such that C value of u is prescribed.
(x(l· yo)
initial curce Fig. 10.3.
326
CHAPTER 10: FINITE DIFFERENCE METHODS
Let u be prescribed on the initial curve Γ which does not coincide with any characteristic C for the partial differential equation (10.14). Let P (x p,y p ) be a point on the characteristic C which passes through a point Q (x q,Vq ) on Γ such that P and Q are close to each other, i.e., \xp — xq | is small (Fig.
10.3).
We shall denote the m-th approximation of u by u^m\ and of y by y <m^ for m = 1,2,... . Let us assume that x P is known. Then, in view of (10.3) and (10.4), the first approxi mat i on of a dy = b dx is given by
aQ ( 2/^ - Vq) = bQ (χ Ρ ~ XQ) - (10.16)
which yields , and from a du = c dx we have
aQ ( “ p - UQ) = CQ (χ Ρ ~ XQ) - (10.17)
which yields u'p . For the second approxi mat i on we have, by using average values,
aQ +2<-— ( y {p ~ Vq) = bQ +2 bp (χ Ρ ~ xq), (10.18)
(2)
which yields yP , and
aQ +2 <1— ( u {2) - itQ) = CQ +2 °p ( xP - x Q), (10.19)
which yields Up'. Subsequent (higher) approximations can be similarly ob­
tained.
Exa mple 10.1. Consider the quasi-linear partial differential equa­
tion
x 3/2 ux — UUy = u 2,
where u = 1 on the initial line y = 0, 0 < x < cc. The auxiliary system of equations is
dx dy du
x3/2 — u u2
S o l v i n g ^ we f i n d t h a t —= = A, where A is constant
χΛ'2 i r u
o n a p a r t i c u l a r c h a r a c t e r i s t i c C. Thus, a t a point ( x q, 0) on the initial
Γ
10.2. FIRST ORDER EQUATIONS
curve Γ, we know t h a t u = 1, and t h a t gives
yi el ds t h e s o l ut i o n along t h e base characteristic
1 _2______ 2_
U y/ΐ y/XQ
d y du
Similarly, sol vi ng — = —x, we have y = — ln B u, and t h e s ol uti on on —u u2
t h e c h a r a c t e r i s t i c Cq is
Thus, t h e solution along a characteristic Cq (10.21). El iminating u between Eqs (10.20) a: equation o f t h e characteristic Cq as
, Λ 2 2
y = ln 1 + - = -
y/x y/XQ
Now we will find t h e first and t h e second point P ( 1.1, y). Let x q = 1, x = 1.1. Then y Q = 0, and u q = 1.
327
A — I --------—= which
y/XQ
Cq as
(10.20)
(10.21)
is given by (10.20) or -ljid ( 10.21), we find the
(10.22)
approximations at the d x = x p — x q = 0.1,
Fi rst approxi mat i on: In v i ew of (10.18) and (10.19), from x3/2 d y =
—u d x we have x.
3/2
Q
— 0 ^ = — u q d x, wh ic
i gives
(i) ,
yp ~ 3/2 ax
A l s o f r o m x'q 2 d u — Uq d x, we have
-0.1
X q 2 ( u,p ^ — 1 ^ = Uq d x,
w h i c h y i e l d s
(0
S e c o n d a p p r o x i m a t i o n: T he equation u d x =
( 1 ) 3/2 , 3/2
UQ + up' Λί. _ XQ +XP 2
1 + 1.01
( 0.1 ) = -
( 1)3/2 + ( 1.1 ) 3/
1) = 1.01. 3//2 d y gives
y {p\
L
328
CHAPTER 10: FINITE DIFFERENCE METHODS
which yi el ds y'p — —0.097507.
Also, from χ 3^2 d u = u 2 d x, we have
which yi elds U p ^ = 1.10261.
T h e e xact values from (10.22) and (10.20) are
= - 0.0 9 7 6 9 5 3,
Up = e ~ yp = 0.91485.
Higher order approximations can be continued until t h e computed val­
ues differ from each other wi thi n t he preassigned tolerance. ■
1 0.3. S e c o n d O r d e r E q u a t i o n s
Let a region Ω in t h e (x, i ) - p l a n e be partitioned into a grid ( x l, t j ), 0 < i < n, and 0 < j < m, as in Fig. 10.2. B y replacing all derivatives in a given partial difFerential equation
by their respective difference quotients, we obtai n a finite difference equation o f t h e form
where D de note s t h e difference operator. No t e t h a t Eq (10.24) is t h e discretized form o f the given equation (10.23) such t h a t t h e solution U i j approximates u ( x, t ) at t h e grid nodes.
D e f i n i t i o n 10.1. The l o c a l t r u n c a t i o n e r r o r e i t j is t h e amount by which t h e solution U i j fails t o s ati sfy Eq (10.23), i.e.,
L u = f, for x, t € Ω
( 1 0.2 3 )
D U i j = f i j for ( x i, t j ) e Ω,
(10.24)
&ij
— D Uij fij-
( 1 0.2 5 )
10.3. SECOND ORDER EQUAT:
Defin itio n 10.2. The difference equatior consistent with the given partial differential eq
lim ei i = 0. h,k-> 0
(10.26)
(10.24) is said to be
Ration (10.23) if
Defin itio n 10.3. The discretization error j — Uij, where Uij is the exact solution of the solution of Eq (10.23) evaluated at (Xi,tj).
7ij is defined as Vij = U%j — Uij, where Uij is the exact solution of Eq (10.24), and utj is
Defin itio n 10.4. The difference scheme, defined by Eq (10.24), is said to be convergent if
lim \Vij \ — lim |Ui
h,k—*Q
h,k—*0
hj
1ί3 I
0 for (
In some cases the difference method may not b it may be consistent. There are examples dis Abbott and Basco (1990). Also, an example su, Frankel is
convergent, although ;ussing these issues in ggested by Du Fort and
U,
i+l,j
-Ui
i,j +1
-Ui
UiJ-\-\ Uij —ι
2k : ”
which is always stable, where a is a constant unless k/h —> 0 as k and h 0 (Carrier and Pe;
The concept of stability of a finite differenc< the propagation of the error Eito = V^o — o w: Uito denote the initial values for the difference Vifi are the initial values obtained from the s· difference system. For a partial differential eq solution, the difference scheme (10.24) is said tc Ei j = Vij — Uij are uniformly bounded in i as
l-E^I < M for j > J,
wher e M is a positive constant and J a positivi known as the Lax equivalence theorem, states th|il is both a necessary and sufficient condition fo finite difference problem which is consistent w: and boundary value problem.
ONS
329
tj) € Ω. (10.27)
l,
j - 1
Ui - 1 ,j h2,
but is not consi st ent arson, 1988, p. 263).
e scheme is based on i t h i ncreasi ng j, where equation (10.24), and iclution of a perturbed uation with a bounded be stable if the errors j -> oo, i.e.,
e integer. A theorem, t stability of a solution the convergence of a ith a well-posed initial
In order to illustrate the finite difference method for second order equations, we shall first consider a very simple example.
330
CHAPTER 10: FINITE DIFFERENCE METHODS
E x a m p l e 1 0.2. L e t a o n e - d i m e n s i o n a l s t e a d y s t a t e h e a t c o n d u c ­
t i o n p r o b l e m b e d e f i n e d b y
u" = - x 2, 0 < x < 1, u(0) = 1, u(l) = 2. (10.28)
A partition of the interval [0,1] into equally spaced points is given by 0 = Xo < xi < ■ ■ ■ < xn = 1,
with the step size h = ij+i — Xj = —. Now, using the forward difference
scheme (10.12) on the problem (10.28), we have
^;+ 1 ^ t = - X 2, i = 0,1,2,... ,n — 1.
In particular, say, for n = 4, we get the system of equations
16 (U2 — 2Ui + Uq) — 0,
16 (t/3 — 2U2 + Ui) = —
16{U4-2U3 + U2) = -^.
In view of the boundary conditions, we have Uo = 1, and C/4 = 2. Then the above system yields
2 Ui - U 2 = 1,
Ui-2U 2 + U3 = ~, 256
U2 - 2U3 = -
129
X ’
whi ch, by usi ng t he Gauss el i mi nat i on met hod, gives
u' = 51 = L25586' u> = §5 = L5U72· u> = 1 = 1·7367·
Alternately, since only it" occurs in problem (10.28), we can use the central difference scheme (10.7). Thus,
10.3. SECOND ORDER EQUATIONS
331
Then, with n = 4, we get the system of equatic 16 (U2 - 2UX + Ua) ■■
16 (U3 — 2U2 + Ui)
16 (Ui - 2U3 + U2)
ns
or
whi ch gives 325
16
257
2 ^ = 2 5 6 ’ Ux - 2U2 + U3 =
64
521
U2 - 2U3 = -------,
2 3 256’
393
Ui = w = L29653> u2 = i ^ = 1.53516, J.OO 25b
The exact solution is
/ Λ , 13 x
u ( x ) = 1 + — x —
t ha·
A compari son wi t h t he exact sol ut i on shows scheme gi ves a bet t er approxi mat i on for pr obk ar e shown i n t he following t abl e.
Forwar d Di fference Cent r al Differ?
0.0 1.0
0.25 1.25586
0.5 1.51172
0.75 1.76367
1.0 2.0
1.0
1.26953
1.53516
1.78516
2.0
Not e t h a t i f t he i nt erval is [o, 6], t hen we us
x — a
ξ =
t o r e d u c e t h i s i n t e r v a l t o [ 0,1].
457
^ = 256 = L7816·
t t he cent r al di fference m (10.22). The r esul t s
nee Exact
1.0
1.27051 1.53646 1.78613 2.0 >
t h e t r a n s f o r m a t i o n
332 CHAPTER 10: FINITE DIFFERENCE METHODS
10.3.1. Diffusion equations. Consider the one-dimensional dif­
fusion equation
xi% = g Ί ΐ χ χ, ( 1 0.2 9 )
where a2 denotes the diifusivity. For the grid (Xi,tj) = (i h,j k), we shall discuss the following three finite difference schemes:
Forward Difference (Explicit Scheme):
UiJ +1 Uij 2 ^ ^ i j “h Ui — IJ
k ~ a h2 ’
or
Ui,j+1 = ( 1 + r S 2) Uij, ( 1 0.3 0 )
where r = a2k/h2.
Backward Difference (Implicit Scheme):
UiJ + l Ui j 2 ^ i+l,j + l ^Ui J-fl 4* Ui—i j + \
k “ a ft* ’
or
( l - r S 2x) U i j + 1 = U i j. ( 1 0.3 1 )
Crank-Nicolson (Implicit Scheme):
Uij+1-Uij 2 62Uij 4- S2Uij+\ k h2
or
( l - r- S2x) Ui j +1 = ( l + r- 6 2x ) Uij. ( 1 0.3 2 )
Note that the Crank-Nicolson scheme is derived by averaging the finite differences at the points (i,j) and (i,j + 1) (see Exercise 1 0.2 ). Note that the above forward difference scheme ( 1 0.3 0 ) is conditionally stable iff r < 1/2, but the other two schemes (1 0.3 1 ) and ( 1 0.3 2 ) are always stable.
E x a m p l e 1 0.3. C o n s i d e r t h e b o u n d a r y v a l u e p r o b l e m
ut = uxx, 0 < x < 1, t > 0, u(0,t) = 0 = u(l,t), f o r ί > 0, ( 1 0.3 3 )
u(x, 0) = f(x), f o r 0 < x < 1.
This problem has been studied in Example 5.2 by the separation of variables method. Since the space derivative is of second order, we shall
10.3. SECOND ORDER EQUATIONS
use the central difference scheme (10.7), where the temperature at a point Xi at time tj. For shall use the forward difference scheme
ut =
UiJ-y\ Uij
k :
333
Uij = u(xi,tj) denotes the time derivative we
where k = tj+i — tj. Then Eq (10.33) is approximated by
^ ^Ui J 4- Ui—i j ] = Uij-^-i Ui
i = 1,2,... , η — 1, and j = 0,1,2 where r = k/h2, and the boundary and initial c
U0j = 0 = Unj, for j = 1,2 Ui,o = f{xi) = fi, fori = 0,1,2
After rearranging the terms in Eq (10.34), we gfet
U%j +1 = r Ui -i j + (1 — 2r) Uij + r U[+i j.
(10.34) ondi t i ons become
Thi s di fference equat i on allows us t o comput e Uij+1 from the values of U computed for earlier times. Note that the value Ul^ = fi is a prescribed value of u at time t = 0 and x = Xi-
We shal l t ake n = 4, i.e., h = 1/4. The value of r in Eq (10.35) must be chosen properly so that the solution remains stable. It has been determined that for a stable solution the value of r must be such that the coefficients of u on the right side of Eq (10.35) remains non­
negative (Smith, 1985, Ch. 3). Hence, we must have 0 < r < -. Then,
1
—. In the marginal
Oa
in view of this restriction, we must have k < rh2
case when r = 1/2, we get k = 1/32. With these values of r and k, system (10.35) reduces to
(10.35)
U%,j+1 2 U%+i,j], j — 0
Since Uoj = U$j = 0 for all j, we find For j=0:
,1,2,... .
Ui, 1 — — + t^i+Ι,θ] ,
334
CHAPTER 10: FINITE DIFFERENCE METHODS
or, successively,
Ui,i = - h, U2,i = - ( f i + f 2 ), = - f 2 ·,
For j = 1:
Ui,2 = -U2,i, U2t2 = -(Uiti + Uzti), U3t2 = -U2ti,
and so on. The values of Uij for f(x) = cos -kx are listed below in a tabular form for some successive values of t.
t
x = 0
x = 0.25
x = 0.5
x = 0.7 5
X
0
1
1
1
2
0
-1^,0
1 1
!
1/3 2
0
0
0
1/1 6
0
0
0
0
0
3/3 2
0
0
0
0
0
1/8
0
0
0
0
0
Notice that the values average out in the outer columns. This will happen if r = 1/2 is chosen. The solution for problem (10.33) for different values of t with r = 0.1 is presented in the following table:
t
x = 0
x = 0.25
x = 0.5
x = 0.75
x =
0
1
1
0
1
- 1
1/32
0
\/2
0.665685
0
V2
-0.665685
0
1/16
0
0.532548
0
-0.532548
0
3/32
0
0.426039
0
-0.426039
0
1/8
0
0.340831
0
-0.340831
0
10.3.2. Wave equation. To approximate the solution of the wave equation
Utt — c Uxx, (10.36)
we can use the difference schemes (10.30)~(10.32) discussed earlier for the diffusion equation, but more frequently used schemes are the forward difference and the Crank-Nicolson. Let (xi,tj) = (i h,j k),
(i,j = 0,1,...). Then
(a) Explicit Scheme (Forward Difference):
Uij + 1 — 2Uij + Uij-1 _ 2 Ui+ l,j ~ ^ i,j + Ui - i j
k 2 ~ c P
1 0.3. S E C O N D O R D E R E Q U A T I O N S
o r
$ U u = c p 6x Ui t j,
w h e r e p = k/h.
(b) Implicit Scheme (Crank-Nicolson):
Ui,j-\-X 2Ui^j — i C 2Ui^j
P = T L h 2
+ ■
Ui
or
t f U hj = C^ f r r f r r , ,2;
[δχ Ui,j +1 + Ui
Not e t h a t t he cent r al di fference schemes (10.7) an also used (see Exampl e 10.5).
h2
i - i ] · (10-38)
d (1 0.8) are somet i mes
Ex a m p l e 10.4. I n or der t o use a fi ni t e diffq: t he wave equat i on (10.36) wi t h t he Neumann i ni
u ( z,0) = f(x), ut (x, 0) = g(x),
we t ake U1q = f(xi) = f i. Now we will find the as follows: Under the assumption that / € C2 theorem
k2
u{xi,t0) = u(xi, 0) + kut (xi, 0) + — utt(xi, 0)
k 2
= u(xi, 0) + kg(xi) + ~ c 2f"(xi) +
0(.
c2p2
= u(xit 0) + kgi + —— [fi+i
wher e gi = g(xi), fi = f{xi), and (10.6) is used The above expansion gives
Ui, 1 — Ui 0 1 2 2 ff
9i = — ^— - - 2 C P [/*+i
2/
335
(10.37)
Ui,j — 1 “1“ Ui— l,j — 1
rence scheme to solve ial conditions, i.e., for
start-up value of l ^ i , we have by Taylor’s
-H
G(k 3) fc3)
2 fi + fi-,} + 0( k2h2 + fc2), approximate f"(xi).
t o ί
+ /i - l ] ,
336
CHAPTER 10: FINITE DIFFERENCE METHODS
where Ut o = /j. Then the start-up value U^i is given by
1
U%, i — fi + k
9i + 2° P (fi+l ~ 2/j + /i - l )
Ex a mp l e 10.5. Consider the wave equation utt = uxx, 0 < x < 1, t > 0, subject to the boundary conditions
u(0,t) = 0 = u(l,t) for t > 0,
and the initial conditions
u(x, 0) = f{x), ut {x,0)=g(x) for 0 < x < 1.
This problem is solved in Example 5.1 by the separation of variables method. We shall use the central difference schemes (10.7) and (10.8) for both uxx and utt■ Thus,
Uft
2£7i j 4- Ui—i j h? ’
+ l 2L/i j 4" Ui,j — 1
P '
Then t he wave equat i on, t he boundar y and t he i ni t i al condi t i ons reduce t o
Ui,j+1 — p2U i - i j + 2(1 — p2)Uitj + p2Ui+i,j — Ui j - i,
i = 1,2,... ,n - l, j = 0,1,2,... ,
U0,j = 0 = Unj for j = 1,2,...,
Uifi = fi, Ui,i - Uifi = kgi for i = 0,1,... ,n,
(10.39)
where p = k/h, and the forward difference scheme is used for the initial condition ut (x,0) = g{x).
Let f{x) = sinx, and g(x) = 1 — x. Then, with h = fc = 1/4, the initial conditions become
Ui,o = fi = sinx* for i = 0,1,2,3,4,
Ui,i - Uito = kgi = ^(1 - Xi) for i = 0,1,2,3,4.
10.3. SECOND ORDER EQUAT
T h e n t h e f i r s t e q u a t i o n i n ( 1 0.3 9 ) b e c o m e s
U l,j + l = U q j + U 2J — U q j - U 2J +1 — — U 2J - ^ 3,j + l = U 2 J + U i j — U s j -
S i n c e t h e b o u n d a r y c o n d i t i o n s a r e U o j = 0 = U f o r s u c c e s s i v e v a l u e s o f j = 1,2,..., w h i c h i s pre t a b l e.
ONS 337
-li
-1,
-1·
i j, w e g e t t h e s o l u t i o n s e n t e d i n t h e f o l l o w i n g
t x = 0 x = 0.2 5 x = 0.5 x
: 0.7 5 x = 1
0 0 0 0 0 0.2 5 0.2 4 7 4 0.4 2 7 2 0.3 4 2 1 0.5 0.5 0.4 7 9 4 0.5 8 9 5 0.6 7 0 7 0.7 0.7 5 0.6 8 1 6 0.7 2 3 0 - 0.0 9 2 1 - 0 1.0 0 0 0 0
0
Γ 2 8 0.6 5 7 9 5 2 6 0.9 0 2 1 .0 5 2 2 0.8 4 4 7 0 ■
E x a m p l e 1 0.6. S o l v e u u = 4 u x x, 0 < x < i n i t i a l c o n d i t i o n s u ( x, 0) = s i n π χ, u t ( x, 0) = 0 b o u n d a r y c o n d i t i o n s u ( 0,t ) = 0 = u ( l,t ) f o r t e x p l i c i t s c h e m e ( 1 0.3 7 ), a n d ( b ) t h e i m p l i c i t sch t h e e x a c t s o l u t i o n is
u ( x,t ) = sin7ra: c o s 2 7 r t, a n d d ’A l e m b e r t ’s s o l u t i o n is
u ( x, t) = i [sin(7rx + 2n t ) + s i n ( n.
T h e M a t h e m a t i c a p r o g r a m f o r ( a ) i s g i v e n b e l o w o u t p u t i s p r e s e n t e d i n a t a b u l a r f o r m.
1, t > 0, s u b j e c t t o t h e f o r 0 < x < 1, a n d t h e > 0, b y u s i n g ( a ) t h e e m e ( 1 0.3 8 ). N o t e t h a t
r — 2 n t ) ].
, a n d t h e c o r r e s p o n d i n g
I n [ l ]:=
C l e a r [ u,n,m,i,j,r,k,h ]; n:= 5 m: = 5 c := 2
h:= 1/( n - l )
338
CHAPTER 10: FINITE DIFFERENCE METHODS
k:= 1/C m - l ) r:= c k/h;
u = T a b l e [ 1,{n},{m}];
I n [ 9 ]:=
f [ i _ ]:= S i n [ P i l/( n - l ) ( i - l ) ] g [ i - ] := 0
I n [ l l ):=
Do [ u [ [ i, 1] ] = f [ i ] ; u [ [ i,2 ] ] = ( l - r * 2 ) f [ i ] + k g C i ] + r'2/2 ( f [ i + l ] + f [ i - l ] );,{ i,l,n } ];
I n [ 1 3 ]:=
Do [
u [ [ l,j ] ] = 0; u [ [ η, j ] ] = 0;,
];
I n [ 1 4 ]:=
Do [
u [ [ i,j ] ] = r ~ 2 u [ [ i - l,j - 1 ] ] + 2 ( l - r ~ 2 ) u [ [ i,j - l ] ] r ~ 2 u [ [ i + l,j - l ] ] - u [ [ i,j - 2 ] ], { i,2,n - l },
{ j > 3,m } ];
u//N//C h o p//T r a n s p o s e//T a b l e F o r m
O u t [ 1 5 ] =
0 0 0 0 0
0.707107 -0.12132 -0.665476 8.11418 -44.0196
1. -0.171573 -0.941125 7.15642 -5.54069
0.707107 -0.12132 -0.665476 0.349676 27.1931
0 0 0 0 0
10.3. SECOND ORDER EQUATIONS
339
(* The e x a c t v a l u e s f o l l o w *)
In[15]:=
T a b l e [ S i n [ P i h ( i - l ) ] C o s [ 2 P i k ( j - { i,1,n },{ j,1,m } ]//N//T a b l e F o r m
Out [20]=
) ].
0
0
0
0
0
0.707107
0
-0.707107
0
0.707107
1.
0
- 1.
0
1.
0.707107
0
-0.707107
0
0.707107
0
0
0
0
0
(* N o t e t h e v a l u e o f r a n d t h e i n s t a b i l i t y *)
It is instructive to experiment with different valm study stability of solutions and find better approxim;
Part (b) can be similarly done by modifying the and using (10.38). es for h and k in order to ations.
Mathematica code in (a)
10.3.3. Poi sso n ’ s equation. Consider Poisson’s equation in a rec­
tangle f l = { 0 < x < a, 0 < y < b } with boundary Γ:
UXX + Uyy = f(x,y ),
subject to the Dirichlet boundary condition
u = g{x,y ) on Γ.
For / = 0, Eq (10.40) becomes Laplace’s equatici simple case when a = b, with the uniformly spaced The nodes are then given by ( xm, yn ) = (m h,n k ), (see Fig. 10.4). By using the central difference s< reduces to the finite difference equation
(10.40)
(10.41)
m. We shall analyze the grid lines of size h = a/4. where m,n = 0,1,2,3,4 icheme (10.7), Eq (10.40)
δχ Um,n + fiy Um ^n — h f m
(10.42)
340
CHAPTER 10: FINITE DIFFERENCE METHODS
w h e r e f m,n = f ( x m,y n ) · T h e b o u n d a r y c o n d i t i o n ( 1 0.4 1 ) t h e n b e c o m e s
Um,n = 9 m,n = 9{.x m i V n) f o r T i l, Tl = 0, 1, 2, 3, 4.
F ig. 10.4. G r i d l i n e s on t h e s q u a re Ω.
B y r e o r d e r i n g E q (1 0.4 2 ), w e get
^Λη-1-Ι,η l,n —1 —1 h f m,n · ( 1 0.4 3 )
T h e u n k n o w n v a l u e s o f u a re a t t h e n o d e s 1,2,... ,9 (Fi g. 10.4). We sh a l l u s e t he n o t a t i o n:
Ul,! = Ui, U2,i = U2, U3ti = U3,
Ui,2 = U4, U2t2 = U5, Us, 2 = U6,
Ui,3 = U-j, U2is = U%, Us,3 = Uq.
T h e n E q ( 1 0.4 3 ) c a n b e w r i t t e n i n t h e f o r m o f a s y s t e m o f a l g e b r a i c e q u a t i o n
[A] {U} = {F}, (1 0.4 4 )
w h e r e t h e m a t r i x [A] i s a 9 x 9 s y m m e t r i c m a t r i x
' 4 - 1
0
- 1
0
0
0
0
o -
4
- 1
0
- 1
0
0
0
0
4
- 1
0
- 1
0
0
0
4
- 1
0
- 1
0
0
[A] =
4
- 1
0
- 1
0
4
0
0
- 1
4
- 1
0
4
- 1
.s y m
4 .
10.3. SECOND ORDER EQUATIONS
341
and
{ U } =
Ui Ϊ
' — h2 /i,i + go
1 + 01,0 '
u2
—h2 f 2,l +
92,0
U3
-ti2 fs,i +
93,0
Ui
-h2f h2 +
90,2
u5
>, {F} = <
- h 2 f 2
2
u6
- h 2fs,i +
04,2
u7
- h,2 fi,3 + go
3 + 01,4
U8
- h 2f 2,s-\
02,4
l t/9 J
. - h 2 f 2,3 + gi
3 + 03,4 >
The order of the mat rix [A] is ( l/h — 1) x ( l/h — vectors {[/} and { F} are ( l/h - 1). The differenc has a truncation error of 0 { h 2). If the boundary (10.41) has a unique solution, h is small and [A] is (10.44) has a unique solution. The system (10.44)
(10.46)
1). The dimensions of the method explained above value problem (10.40)- nonsingular, then system :an be easily solved.
In the case of a curved boundary Γ, consider a node in Ω, which has at least one adjacent node outside Ω (see Fig. 10.5).
R
P
,2
q
/
Fig. 10.5. Curved boundary Γ.
Let P = (xm, yn) be a point inside Ω and near coordinates of the adjacent points q and r on the
Q — (2-m+l, Vn) — (%m +
I' = Vn+1) = ( % m h,y n
the boundary Γ. Then the bbundary Γ are:
Un) i
-0Λ),
where 0 < α, β < 1. Moreover, the values of u( q) and u ( r ) are known, since
342
CHAPTER 10: FINITE DIFFERENCE METHODS
u is prescribed on the boundary Γ. By Taylor’s theorem
2 l2
u(q) = u(P) +ahux(P) + uxx(P) + 0( h3),
u(Q) = u(P) ~ h u x{P) + ^ uxx(P) + 0( h3).
Aft er eli mi nati ng ux(P) from these two expansions, we obtain
= l H q) - ( l + a M P ) + a u m +
a ( l + a)
Similarly,
2[u( r) -( l +0)u( P) +Pu(R) }
Uyy{F) ~ W + β ) + W'
Hence t he finite difference approximation for Poi sson’s equati on (10.40) de­
fined on a regi on with a curved boundary Γ is
U(Q) U(R) f l 1\ m m , U(q) , U(r) 1,2
1 + a ' l +β \a + p ) U^ + a ( l + a ) + β(1+β) 2k
(10.47)
Ex a m p l e 10.7. To find fi ni t e di fference sol ut i ons of Lapl ace’s equa­
t i on uxx + uyy = 0 on the quarter-circular region Ω = {x2 + y 2 < 1, y > 0}, subject to the boundary conditions u(x, y) = 10 on x2 +y'2 = 1, y > 0}, and u(x, i/) = 0 on0 < x < l,y = 0 and ux = 0 on x = 0, 0 < y < 1, we choose the grid with h = 1/2 (see Fig. 10.6).
Fig. 10.6.
10.4. EXERCISES
343
From the boundary conditions we have Uo,o 10 = u(q) = u(r). The only unknown values c P = (zi,2/i), and Q = (xo,yi)· From (10.43) tl centered at Q is
4t/o,i — Ui,! — U- 1,1 — Uq, 2 — Uq,
Not e t h a t t he coordi nat es of q = (%/3/i, h), and of a — β — \/3 — 1. Also, the boundary conditions from (10.48) we have
2E/o,i - U n = 5.
r = (h, %/3 h). Hence yields U\t\ = 0. Thus,
Then the difference equation (10.47) gives
U(Q) , U(R) 2U(P) , U{q)
V3 y/3
or
Uo,i U i,o ________
λ/3 y/3 V 3 - 1
V 3 - 1 \/3(\/3
2C/i,i , U(q)
y/3 ( y/3 -
whi ch yi el ds
( 1 — V 5 ) Uo,ι + 2ν/3ί7ιιι — ί By solving (10.49) and (10.50), we get
35 +
= 20 + 10V3 = 6 02317 3 ^ 3 + 1
Ui,i =
3λ/3 + 1
I www I The Notebook fd.ma can be found on
This will help not only to solve problems but also codes for difference schemes.
0 = Uito and U2,o : u are at the nodes e difference equation
0.
(10.48)
(10.49)
1)
= 0,
1)
0.
!)λ/3
= 0,
(10.50)
7.046349. ι
the CRC web server, generate Mathematica
10.4. Exercises
10.1. Use the central difference scheme, with n u" — u = —2, 0 < x < 1, u'(0)
ί Ι ί N T: 16([7i+i - 2Ui + t/j_i) - £/» = - 2 for and the boundary conditions give Ui+i = Ui,
— 4, to solve 0, u(l) = 1.
i = 1,2,3,4, h = 1/4,
and Ui = 1.
344
CHAPTER 10: FINITE DIFFERENCE METHODS
10.2. Derive the Crank-Nicolson scheme (10.32). Ans. The central difference at the point ( i,j ) is
and at the point (i, j + 1) it is given by
The result follows by taking the average of these two differences.
10.3. (a) Show that the forward difference scheme (10.30) has a trun­
cation error of 0 ( k + ft2); (b) Show that the local truncation error is of 0 ( k 2 + h4) if r = 1/6.
S o l u t i o n, (a) Note that for (X i,t j ) = (i h,j k ), we have
where Xi - χ < x < Xi and t j < t j < t j +\. Then the truncation error is given by
10.4. Show that the forward scheme (10.30) is convergent for the problem u t = a 2 u x x, 0 < x < 1, t > 0, subject to the Dirichlet boundary conditions u( 0,t ) = f ( t ), u ( l,t ) = g(t ) for t > 0, and the initial condition u(x: 0) = F(x) for 0 < x < 1.
Solution. With ( x i,t t ) = (i h,j k ) for i = 0,1,... ,n and j =
provided u tt and u xxxx are bounded.
(b) If r = 1/6, then ( ut — a2 u xx). . = 0 leads to
Ui,j
+1 ^2
k
10.4. EXERCISES
345
0,1,... , m, and with nh = 1, j m = to, w value, 0 < t < to, we have
Ui,j+i — uitj + r 6xUij,
Uifi = F{xi) = Fi, U0,j = f(tj) = fj,
Let Vi 4 = U-,j — Ui ,· be the discretization
10.3. Then
Vilj+1 = τ V i - i j + (1 — 2r ) Vi j + r Vi+in­
Un,j = 9(tj) = 9j- error, as in Definition
here t o is a prescribed
-h
4·
^2 JJ
where <Xi < and tj <t j < tj+\. Since r <1/2, we have
4 Ajfc2 + Bkh2
Wt
*| J + 1
< r | V i _ w | + ( l - 2 r ) | V i
< max Vj j + AA: 4- Bkh
0 < i < n ' 1
where
A = max | — (rc, t) |, B =
ma x
si nce bot h utt and uxxxx are assumed conti: equality gives
||%! || < 11^-11 + Ak2 + H where ||Vj|| = max |Vij, or, since ||Vo|| = 0.
0 < i < n 1 ’
ll^ll < j {Ak2 4- Bkh2) < t0 (Ak + Bh2) ,
finition of the problem
i.e., ||Vj-|| —> 0 uniformly in the domain of des as h, k —> 0 for 0 < t < to-
D.5. Deri ve t he mat r i x form of t he syst em for t he pr obl em i n Exampl e 10.5 i f t he back used.
[A]
mat r i x
equat i on
is [A] {U}
= {B
, where
- 1 4 - 2 r
—2 r
0
■ 0
0
0 "I
—r
1+2 r
—r
• 0
0
0
—r
1 + 2 r ·■
• 0
0
0
—r
1 + 2 r
—r
- sym
• 0
-2 r
1 + 2 r.
k2
— Ut t (X i,t j )
a2kh2
X^xxxx (,Xi: ),
12
ZX { X j t ) I >
nuous. The above in- kh2,
af di fference equat i ons ward scheme (10.31) is
346 CHAPTER 10: FINITE DIFFERENCE METHODS
U0,j+1
Uq j 2 h r f j + i
U i j + i
U i ti
u 2,j + l
-, { β } = <
U2J
Un — l,j + l
1
Co.
' Un,j+1 >
' U n,j + 2/ΐ Γ ^ · _ ( _ ι .
1 0.6. Sol ve t h e h e a t e q u a t i o n ut = uxx, 0 < x < 1, t > 0, with n = 4,
r = 1/2, k = 1/32, subject to the boundary and initial conditions
u(0, t) = 0 = u(l, f) for t > 0, and u(x, 0) = x.
^ 2( — l ) n _ 2 2,
Ans. Exact solution: u = } ----------sin ηπχ e ” .
' ηπ
n = l
t
x = 0
x = 0.25
x = 0.5
x = 0.75
x = 1
0
0
1/4
1/2
3/4
1
1/32
0
0.25
0.5
0.75
0
1/16
0
0.25
0.5
0.25
0
3/32
0
0.25
0.25
0.25
0
1/8
0
0.125
0.25
0.125
Ο­
10.7. Solve ut = uxx + 1, 0 < x < 1, t > 0, subject to the boundary and initial conditions u(0,t) = 0 = u(l,£) for t > 0 and the initial condition u(x, 0) = 0 for 0 < x < 1, with r = 1/2, k = 1/32, and n = 4.
Ans.
ί
x = C
ι x = 0.25
x = 0.5
x = 0.75
x = 1
0
0
0
0
0
0
1/32
0
0.03125
0.03125
0.03125
0
1/16
0
0.046875
0.0625
0.046875
0
3/32
0
0.0625
0.078125
0.0625
0
1/8
0
0.0703125
0.09375
0.0703125
0 ■
10.8. Solve the wave equation utt = uxx, 0 < χ < I, t > 0, with r = 1/2, k = 1/32, and n = 4, subject to the boundary and the initial conditions:
(a) u(0,t) = 0 = u(l,t) for t > 0, and u(a:,0) = 0, ut (x, 0) = 1 for 0 < x < 1.
(b) u(0,t) = 0 = u(l,t) for t > 0, and u(x, 0) = βίηπχ, ut (x, 0) = 0
10.4. EXERCISES
347
for 0 < X < 1.
A ns. (a)
t
x =
0
x = 0.25
x = 0.5
X = I
),75
x = 1
0
0.0
0.0
0.0
0.0
0.0
0.25
0.0
0.25
0.25
0.25
0.0
0.5
0.0
0.25
0.5
0.25
0.0
0.75
0.0
0.25
0.25
0.25
0
1.0
0.0
0.0
0.0
0.0
0.0 ■
(b)
t
x
0
x = 0.25
x = 0.5
x =
0.75
x = 1
0
0.0
0.7071
1.0
0.7C
71
0.0
0.25
0.7071
1.
0.7C
71
0.0
0.5
0.0
0.2929
0.4142
0.29
29
0.0
0.75
0.0
-0.2929
-0.4142
-0.
2929
0.0
1.0
0.0
-0.7071
- 1.0
-0.
Γ071
0.0 -
10.9.
Find the system of equations [A]
{[/} =
{F}
for the Neumann
boundary value problem
UXX + Uyy = /( X, I/),
i n Ω,
du
dn
<
where Ω is t he r ect angl e {0 < x < a, 0 < y (mh, nh) for m = 0,1,2,... ,M and n = 0 Nh = b.
A n s. Eq (10.43) for m = 0,1,2,... ,M c|nd n = 0,1,2,... ,iV leads to
Um+i,n ~ Um-i,n = 2hgm,n, n = 1,2,... ,N - 1,
Um,n
+i - Um,n
- i = 2ft
gm>N, m =
1,2,... ,M —
1,
and t he boundar y condi t i on gives
U-1,„ - UltU = 2hg0,n, n = 1,2,... , N - 1,
Um, — 1 Um,l = 2h gm,0, 771 = 1, 2, . . . , Ai 1.
At a corner node where the outward nor shall take the normal derivative as the av<
=--g(x,y) on r,
b}. Choose the nodes 1,2,... , N, such that
mal
n is undefined, we erage value of the two
348
CHAPTER 10: FINITE DIFFERENCE METHODS
normal derivatives at the two adjacent boundary nodes. Thus, the boundary conditions reduce to
U - i,o + U q,- i = U\f i + U o t i + 4/150,0,
Um,- 1 + Um+1,0 = Um, 1 + Um- 1,0 + 4/igM,o,
Um+i,n + Um,n=1 = Um- i,n + Um,n - 1 + 4hgM,N,
Uq.n +i + U- i'N = U0,n - i + Ui tN + 4hgo,N-
1 0.1 0. I n Exerci se 10.9, t ake Μ = N = 3, and g = 0. Determine the matrix [^4] and the vectors {U} and {F}.
A ns.
1
to
o
1
to
0
0
0
0
o -
4 -1
0
-2
0
0
0
0
4
0
0
-2
0
0
0
4
-2
0
-1
0
0
4
-1
0
-1
0
4
0
4
0
- 2
4
-1 0 -1 4 .
and
Ui Ϊ
' /l,l '
u 2
/2,1
u 3
/3,1
Ui
/l,2
U5
> , { F } = - h 2 <
/2,2
U6
h,i
u 7
/l,3
Us
h,3
k u J
- /2,3 -
1 0.1 1. Let Ω be the square region {0 < x,y < 1} with boundary Γ. Find the finite difference equation for the boundary value problem
uxx "Ί- u>yy -4- cu — f (x, y) in Ω,
subject to the Dirichlet boundary condition u = g(x,y) on Γ.
ANS. ( c tyUm,n Um,n — 1 "Ί- Um—l,n “t" Um,n—1 “t" -Η 1,n — fm,n'
1 0.1 2. Fi nd t he fi ni t e di fference equat i on for t he boundar y val ue pr ob­
l em
(a u x ) x + (b u y ) y = 0, in Ω = {0 < x,y < 1},
where a and b are positive functions of x, y and u. A ns. On the square grid (xm,yn) = (mh,nk) tion is given by
^ m + l/2,n C^m+ Ι,η ( ^ m - t - l/2,n + —1/2,n ) U 1
+ ^m,n+1/2 ^m,n+1 (pm,n+1/2 + ^m,n — 1/2)
10.4. EXERCISES
349
,,71+^771—1/2,71 1,71
,71 + ^771,71 —1/2 ^771,71—1
= h2 f
— ,/ 771,71 ·
the difference equa-
A
G r e e n ’ s I d e n t i t i e s
A.l. Green’s Identities
Let Ω be a finite domain in Rn bounded by a piecewise smooth orientable surface 9Ω, and let w and F be scalar functions and G a vector function in the class (7°(Ω). Then
Gradient Theorem: / VFdi l = <j nF dS,
J n JdQ
Divergence Theorem: / V ■ G dQ, = ψ n ■ G dS,
Jn JdQ
Stokes Theorem: / V x G d i!= i G t dS,
Jn Jan
where n is the out ward normal to the surface <9Ω, t is the t angent vector at a
poi nt on 3Ω, j> denotes the surface or line integral, and dS denotes the surface
or line element depending on the dimension of Ω. The divergence theorem in the above form is also known as the Gauss theorem. Stokes’ theorem is a generalization of Green’s theorem which in R2 states that if G = (Gi,G2) is a continuously differentiable vector field defined on a region containing Ω U <9Ω C R2 such that <9Ω is a smooth closed contour, then
/ ( - -7:—- ) dxi dx2 = φ Gi dxi + G2 dx2
Jn V dxi dx2 J Jan
A.l. GREEN’S IDENTITIES
351
f (VG)wdSl = - [ (Vw)Gdn+ [ nwGdS, (A.l) Jn Jil JdQ
- f (V2G)wdQ.= [ (Vw) · (VG) dQ. —
J Ω J Ω
The above theorems lead to the following two useful identities:
dG
dn
wdS, (A.2)
where — = V ■ n = ^ nXi ——, i = 1, ■ ■ · , n, is the normal derivative
(JTb (J Jb J I
operator. The i-th component of the formula (A.l) can be written in a useful form as
f dG f dw £
/ w— dQ, = - \ — + φ J n dx i J n ox i J ai
nr. w G dS.
I ni?3 l et t hef unct i onsM( x), J V( x),andP( x),x == ( x i,x2,£3) = (x,y,z) £ Ω, be the components of the vector G. Then, by the divergence theorem
r (dM dN dP\
Jn\d x + dy + dz
an
[M cos(n, χ) + N cos(n, y) + P cos(n, z)\ dS, (A.4)
(A.3)
with the direction cosines cos nx, cos ny and cos
dv dv
N = u—, and P = u—, then (A.4) yields
dy dz
Γ f du dv du dv du dv \ Jn \dx dx + dydy + dz dz )
j
JdQ
(A.5)
which is known as Green’s first identity. Moreover, i f we i nterchange u and v in (A.4), we get
Γ ί du dv du dv du dv\ f
Jn \dxdx + dy dy + dz dz) Jan
If we subtract (A.5) from (A.6), we obtain Greer
nz. If we take M = u
dv
dx’
W 0
uV νάΩ.,
j (u V 2v — v V 2u) dQ, = J ~ v dS, (A.
■7)
which is also known as Green’s reciprocity theorem. Note that Green’s identi­
ties are valid even if the domain Ω is bounded by finitely many closed surfaces;
dS — [ vW2udQ,. dn Jn
(A.6)
’s second identity:
352
APPENDIX A: GREEN’S IDENTITIES
however, in that case the surface integrals must be evaluated over all surfaces that make the boundary of Ω. If we take v = 1 in (A.5), then
idn dn
If we take u = v in (A.5), then
L & - L
V 2udCt.
J -g^dS = J (uV2u 4- |Vu|2) dCl.
( A.8 )
( A.9 )
A.2. E x e r c i s e s
A.l. Us e Gr e e n ’ s f or mul as t o s how t hat
IL
V u V w d x d y
IL
V qu w d x d y + / V'zu —— ds
r J c d n
dw
[ - ^ - V 2u w d s.
J c d n
A n s.
d x d y
J J V 2u V 2w dx dy =
+//Β"(έ*ν2“+!?ν2“) <ίΙ% = J J ^ * u ) w d x d y + l ^ u ( ^ d u - §?»
S A ^ - d^ )
= i f V 4u wd x d y + f V 2u^^ ds — [ - ^ - V 2u w d s. J J r J c o n J c d n
A.2. EXERCISES
353
A.2. Prove that uV · (h Vv) = V · (uh Vv) — V' Ans. Using V ξ \d/dx + j d/dy + kd/dz,
ι ■ hVv.
the ri ght side
d dv
+
r a /, ov\ a /, ι
= n a i (',s ) +
/. 3 „ d - /
= “ ( * s + J 3 i + k 8;) ■(·*
= uV · (hVv) = the left side.
dv
+
dv
dzJl
dv - dv\
3% + k h a ύ
A.3. Show t hat u V2i> = V · ( i t Vv) — Vu ■ Vv. Ans. Take h = 1 in A.2.
Β
T a b l e s o f T r a n s f o r m P a i r s
Some basic formulas for the pairs of the Laplace, complex (exponential) Fourier, Fourier sine, Fourier cosine, finite sine, and finite cosine transforms are provided below in tabular forms. Definitions of these transforms are given in §6.1 and §6.8.
B.l. L a p l a c e T r a n s f o r m P a i r s
f(t) F(s) = f(s)
1.
1
s > 0
s
2.
eat
1
s — a ’
s > a
3.
sin at
a
s > 0
s2 + a2 ’
4.
cos at
s
s > 0
s2 + a2'
5.
sinh at
a
s > 0
s2 — a2’
6.
cosh at
s
s > 0
s2 — a2’
7.
eat sin&i
b
(s — a)2
+ b2 ’ S
m
eat cosh bt
F( s ) = f ( s ) s
B.l. LAPLACE TRANSFORM PAj lRS
9. tn ( n = 1,2,· · · )
10. tn eat ( n = 1,2,· · · )
11. H(t — a)
1 2. H(t — a) f ( t — a)
1 3. eat f(t)
1 4. f (t )*g(t )
1 5. /<">(*)
1 6. /( a i )
1 7. /0*/( ί ) ώ
18. 6(t — a)
1 9. t/( t )
2 0 erfc —^7=
2\ft
21. /(i) with period = T
(s — a) 2 + b2 n\
, s > C
ςη+1
nl
(s - a)
n + l :
e~as F(s) =
F(s - a ) = f(s - a) F{s)G(s) = ,f{s)g(s) sn F(s) - s'1} 1 f{0) -
355
s > a
s > a
7 »
- f n~H o)
a > 0
a Va/
> ) = > )
0-a y/s
S
dt
1 — e~Ts
The c ommand « C a l c u l u s'L a p l a c e T r a n s f o i c a pa c ka g e c onc e r ni ng La p l a c e t r ansf or ms.
m‘ l oads t he Mat hemat -
*f{t) is continuous in [0, T] and periodic with perio d X, T > 0.
356
APPENDIX Β : TABLES OF TRANSFORM PAIRS
f
B.2. F o u r i e r C o s i n e T r a n s f o r m P a i r s
fix) Fc{f(x)} = fc(a)
1.
fiax) \ fc ( - )
a \aJ
2. e~ax
3. χ"1/2
9. 6{x)
1 0. H{a — x)
π a2 + a2
1
y/a
4. e~ax2 ^ c-g2/4a
y/2a
5. , y [ ^ e - aa
χ2 + ά2 V 2
6. x2 f i x) - f ci a)
s i n a x [π TT. .
7. — γ 2 ~ a
s- /"M
π
§
2 sinaa
π a
/%/x2 + a2 + a\ ^ TTy/a
^ **+«* j — '
( x2 - ^ 2) 1/2, x < a /f\3/2
12.
0, x > a
/ 7j- \ «5/Ζ
( 2 ) J o M
( a2 —χ 2) - 1/2, x > o /π\3/2
13
0, x < a
©
0/Z
Yoiaa)
14. si n ( a2x 2) ^ ( cos ( a2/4 a 2) — s i n( a2/4 a 2))
15. cos ( a2x2) ( cos ( a2/4 a 2) + si n(o:2/4 a 2))
B.3. FOURIER SINE TRANSFORM PAIRS 357
B.3. F o u r i e r S i n e T r a n s f o r m P a i r s
1.
2.
3.
4.
5.
fix)
fiax)
e~ax
T--1/2
x2 + o2
6. arctan(a/:r)
7. x2 f i x)
erfc
2\fa 9. f i x )
, a > 0
10. Hia — x)
2 2
11. xe~ax
12.
13.
V i
-a/x
x cos a x
b2 + x 2
Fs{f i x) } = /«(<*)
1 ; ( a'
a Va/
a
π a 2 + a 2
a π 2 7Γ
— e 2
7Γ 1 ~ 6~aa
2 a /»
2 1 - e~aa2 π a
- a 2f s(a) + J - 4/( 0 )
2 1 — cos aa
7Γ
7τα
4\/2a3
a
D-a2/4 a2
2 ^ 5
l · © I ( f f
3/2
j — ab
-V 2aa ^cog y/2aa + sin y/2aa)
si nh aa, a < a :d>sh aa, a > a
358
APPENDIX Β : TABLES OF TRANSFORM PAIRS
B.4. C o m p l e x F o u r i e r T r a n s f o r m P a i r s
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. 11. 12.
13.
14.
15.
16.
fix)
f {n)(x)
f(ax), a > 0 f { x - a )
S( x — a)
p-a\x\
( 1, |x| < a
\ 0, |a:| > a
f 1, M < 1
I 0, \x\ > 1
fix) * g( x)
H( x + a) — H{ x — a)
x e ~ a^, a > 0 a
x2 + a2 ax
ix2 + a2)2 cos ax
s i n ax
( cos ax, |a;| < π/2 a \ 0, |x| > π/2 a
F { f i x ) } = /( a ) i i a) n /( a )
π a2 + a 2
„- a2/i a2
c o s a
s i n a
: f i a ) g{a)
π a Ϊ2 2 i aa
π ( a 2 + a2) 2
. p- aM
- a | a |
( a + a) + δ {a - a)] ( a + a) — (5(a - a)] c o s ^ a/2 a)
B.5. FINITE SINE TRANSFORM
17.
18.
1
1 + x 2
( 1 - |x|, \x\ < 1
I o, |x| > 1
- f t - W
2J 2- a
π \ c*
sin(<
B.5. F i n i t e S i n e T r a n s f o r m P a i r s
The tables are for the interval [0, π]. If the interval is [a, b], then it can be transformed into [0, π] by
y =
π (x — a) b — a
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
f ( x )
s i n m x, m = 1,2,·
OO
^ s i n n x
n = l
π — x
Λ Η
1, ο,
η = m η φ τ η
—χ, χ < α π — χ, χ > α χ ( π — α), χ < α α( π — χ), χ > α
sinha(7r — χ)
- ( - 1 ) η+1
— [1 - ( - ΐ )
ηπ L
— cos ηα, 0 η
[ι
sinh απ
Γ ( χ )
—^ sin ηα, η 2·
2η
π ( α 2 + η 2) 2η
7r (a2 + η 2)
,. 2η —η25„ Η
-----
0
PAI RS
359
/2 )
( Β.1 )
< α < π < α < π - (—1)η ea7r]
;/( 0 ) - ( - 1 )"/( π ) ]
360
APPENDIX Β : TABLES OF TRANSFORM PAIRS
B.6. F i n i t e C o s i n e T r a n s f o r m P a i r s
The tables are for the interval [0, π]. If the interval is [a, b\, then it can be transformed into [0, π] by (B.l).
f(x)
fc(n)
1.
2.
3.
7.
cos mx, m = 1,2, ao . £2,
T r + L, a n s i n nx
* n= 1
/( π - x)
J 1, 0 < x < a
\ - 1, a < x <
9. — l n( 2s i nx/2)
10. f"( x )
1, n = to
0, η φ to
( - 1 T ~ C n τ
2, n = 0 0, n = 1,2,···
π, n = 0
A [ ("I )"-!] > n = 1,2,·
π η ζ
2π2/3, η = 0 4 ( - l ) n, n = 1,2,· · ·
2(2α — π)
η = 0
si n ηα, η = 1,2, · ■
ηπ
2 ( - l ) n e Q7r - 1 π α2 + η2 J 0, η = 0 \ 1/η, η = 1,2,
2η
—n 2 S n Η [/(0) — ( —1)" /( π ) ]
c
G l o s s a r y o f M a t h e m a t i c a F u
n o t i o n s
All All is a setting used for certain options.
Apart Apart[expr] rewrites a rational expression as a sum of terms with minimal denominators. Apart[expr, var] treats all variables other than var as constants.
Append Append[expr, elem] gives expr with elem appended.
Apply Apply[f, expr] or f @@ expr replaces the head of expr by f.
Apply[f, expr, levelspec] replaces heads in parts of expr specified by lev­
elspec.
Bessell Bessell[n, z] gives the mod­
ified Bessel function of the first kind I(n, z).
BesselJ BesselJ[n, z] gives the Bessel function of the first kind J(n,
z)·
BesselK BesselK[n, z] gives the modified Bessel function of the sec-
rvn/1 Vi-n/1 TCfn 7\
Blank _ o object that Mathematic Blank [h] with head
cai
C C[i] is tin constant of solving a d DSolve.
Cancel C common denominate]
Binomial Binomial[n, m] gives the binomial coefficient.
Blankf ] is a pattern can stand for any :a expression. _h or in stand for any expression
e default form for the i-th integration produced in lifferential equation with
)mcel[expr] cancels out ctors in the numerator and of expr.
CForm CForm[expr] prints as a C lan­
guage versi on of expr.
CharacteiristicPolynomial CharacteristicPolynomialfm, x] gives the characteristic polynomial defined by the square matrix m and the vari­
able x. The result is normally equiva­
lent to Detfm - x Identity Matrix[Length[m]]].
362
Glossary
Chop Chop[expr] replaces approxi­
mate real numbers in expr that are close to zero by the exact integer 0. Chop[expr, tol] replaces approximate real numbers in expr that differ from zero by less than tol with 0.
Clear Clear[symboll, symbol2,...] clears values and definitions for the specified symbols. Clear[\patteml\, \pattem2\,...] clears values and defini­
tions for all symbols whose names match any of the specified string pat­
terns.
Collect Collect[expr, x] collects together terms involving the same power of x. Collect[expr, {xl, x 2,...}] collects together terms that involve the same powers of xl, x2, ....
ColumnForm ColumnForm[{el, e2, ...}] prints as a column with el above e2, etc. ColumnForm[list, horiz] speci­
fies the horizontal alignment of each element. ColumnForm[list, horiz, vert] also specifies the vertical alignment of the whole column.
ComplexExpand
ComplexExpand[expr] expands expr assuming that all variables are real. ComplexExpand[expr, {xl, x 2,...}] expands expr assuming that variables matching any of the xi are complex.
Conjugate Conjugate[z] gives the complex conjugate of the complex number z.
Continuation Continuation[n] is output at the beginning of the nth line in a multiline printed expression.
ContourPlot ContourPlot[f,{x,xmin,
xmax}, {y, ymin, ymax}] generates a contour plot of f as a function of x and
y·
Cos Cos[z] gives the cosine of z.
D D[f, x] gives the partial derivative of f with respect to x. D[f, {x, n}] gives the nth partial derivative with respect to x. D[f, xl, x 2,...] gives a mixed derivative.
Denominator Denominator[expr] gives the denominator of expr.
DensityPlot DensityPlot[f, {x, xmin, xmax}, {y, ymin, ymax}] makes a density plot of f as a function of x and y.
Derivative f ’ represents the deriva­
tive of a function f of one argument. Derivativefnl, n2, ...][f] is the general form, representing a function obtained from f by differentiating nl times with respect to the first argument, n2 times with respect to the second argument, and so on.
Det Det[m] gives the determinant of the square matrix m.
Dot a.b.c or Dot[a, b, c] gives prod­
ucts of vectors, matrices and tensors.
DSolve DSolve[eqn, y[x], x] solves a differential equation for the func­
tions y[x], with independent variable x. DSolve[{eqnl, eqn2,...}, {yl[xl, ...],...}, { x l,...}] solves a list of dif­
ferential equations.
Dt Dt[f, x] gives the total derivative of f with respect to x. Dt[f] gives the total differential of f. Dt[f, {x, n}]
Glossary
363
gives the nth total derivative with respect to x. Dt[f, xl, x2,...] gives a mixed total derivative.
Ε E is the exponential constant e (base of natural logarithms), with numerical value 2.71828....
Eigensystem Eigensystem[m] gives a list {values, vectors} of the eigenval­
ues and eigenvectors of the square matrix m.
Eigenvalues Eigenvalues[m] gives a list of the eigenvalues of the square matrix m.
Eigenvectors Eigenvectors[m] gives a list of the eigenvectors of the square matrix m.
Equal lhs == rhs returns True if lhs and rhs are identical.
Erf Erf[z] gives the error function erf(z). Erf[zO, zl] gives the general­
ized error function erf(zl) - erf(zO).
Erfc Erfc[z] gives the complemen­
tary error function erfc(z) = 1 - erf(z).
Exp Exp[z] is the exponential func­
tion.
Expand Expand[expr] expands out products and positive integer powers in expr. Expand[expr, patt] avoids expanding elements of expr which do not contain terms matching the pattern patt.
False False is the symbol for the Boolean value false.
FindRoot FindRootflhs == rhs, {x,
xO}] searches for a numerical solution to the equation lhs == rhs, starting with x == ptO.
F l a t t e n nested listst level n.
F o r t r a n F prints as a expr.
Flattenflist] flattens out . Flattenflist, n] flattens to
orm FortranForm[expr] Fortran language version of
Fourier Fourierflist] finds the dis­
crete Fourier transform of a list of complex numbers.
Gamma Gammafa] is the Euler gamma function Gamma(a). Gamma[a, z] is the incomplete gamma function Gamma(a. z). Gamma[a, zO, zl] is the generalized incomplete gamma func­
tion Gamma(a, zO) - Gamma(a, zl).
Gradient! Gradient is an option for FindMiniinum, which can be used to specify ths gradient of the function whose minimum is being sought. With Gradient-> Automatic, the gradient is computed symbolically. A typical set­
ting is Gradient -> {2 x, Sign[y]}.
Greater x > y yields True if x is determined to be greater than y. xl > x2 > x3 yields True if the xi form a strictly decreasing sequence.
G r e a t e r E 1 x is detei equal to y if the xi sequence
]qual x >= y yields True if :mined to be greater than or . xl >= x2 >= x3 yields True form a non-increasing
Hold Hold[expr] maintains expr in an unevalualed form.
364
Glossary
I I represents the imaginary unit Sqrtt-1].
Im Im[z] gives the imaginary part of the complex number z.
In In[n] is a global object that is assigned to have a delayed value of the n-th input line.
I n f i n i t y Infinity is a symbol that represents a positive infinite quantity.
I n t e g r a t e Integrate[f,x] gives the indefinite integral of f with respect to x. Integrate[f,{x,xmin,xmax}] gives the definite integral.
Integrate^, {x, xmin, xmax} ,{y, ymin, ym ax}] gives a multiple integral.
InverseFourier InverseFourier[list] finds the discrete inverse Fourier trans­
form of a list of complex numbers.
Jacobian Jacobian is an option for FindRoot. Jacobian -> Automatic attempts symbolic computation of the Jacobian of the system of functions whose root is being sought. A typical setting is Jacobian -> {{2 x, Sign[y]},
{y.x}}·
Join Join[listl, list2,...] concatenates lists together. Join can be used on any set of expressions that have the same head.
LegendreP LegendrePfn, x] gives the n-th Legendre polynomial. LegendrePjn, m, x] gives the associat­
ed Legendre polynomial.
Less x < y yields True if x is deter­
mined to be less than y. xl < x2 < x3 yields True if the xi form a strictly
LessEqual x <= y yields True if x is determined to be less than or equal to y. xl <= x2 <= x3 yields True if the xi form a non-decreasing sequence.
Limit Limit[expr, x->xO] finds the limiting value of expr when x approaches xO.
L i s t P l o t ListPlot[{yl, y 2,...}] plots a list of values. The x coordinates for each point are taken to be 1,2,.... ListPlot[{{xl, yl}, {x2, y2},...}] plots a list of values with specified x and y coordinates.
ListPlot3D ListPlot3D[array] gener­
ates a three-dimensional plot of a sur­
face representing an array of height values. ListPlot3D[array, shades] gen­
erates a plot with each element of the surface shaded according to the speci­
fication in shades.
Log Log[z] gives the natural loga­
rithm of z (logarithm to base E).
Log[b, z] gives the logarithm to base b.
Map Map[f, expr] or f /@ expr applies f to each element on the first level in expr. Map[f, expr, levelspec] applies f to parts of expr specified by levelspec.
MatrixForm MatrixForm[list] prints with the elements of list arranged in a regular array.
Max Max[xl, x2, ...] yields the numer­
ically largest of the xi. Max[{xl, x2, ...}, {yl,...},...] yields the largest ele­
ment of any of the lists.
increasing sequence.
Glossa ry
365
Min Min[xl, x2,...] yields the numer­
ically smallest of the xi. Min[{xl, x2, ...}, { y l,...},...] yields the smallest element of any of the lists.
N N[expr] gives the numerical value of expr. N[expr, n] does computations to n-digit precision.
NDSolve NDSoivefeqns, y, {x, xmin, xmax}] finds a numerical solution to the differential equations eqns for the function y with the independent vari­
able x in the range xmin to xmax. NDSolve[eqns, {yl, y2,...}, {x, xmin, xmax}] finds numerical solutions for the functions yi. NDSolve[eqns, y, {x, xl, x2,...}] forces a function evalua­
tion at each of x l, x2,... The range of numerical integration is from Min[xl, x 2,...] to Max[xl, x2,...].
Needs Needs[\context'\, \file\] loads file if the specified context is not already in SPackages. Needs[\con- text'YJ loads the file specified by ContextToFilename[\context'\] if the specified context is not already in SPackages.
Negative Negativefx] gives True if x is a negative number.
NIntegrate NIntegrate[f, {x, xmin, xmax}] gives a numerical approxima­
tion to the integral of f with respect to x over the interval xmin to xmax.
NonNegative NonNegative[x] gives True if x is a non-negative number.
Not !expr is the logical NOT func­
tion. It gives False if expr is True, and True if it is False.
NSolve NSolve[eqns, vars] attempts to solve numerically an equation or set of equations for the variables vars.
Any variable in eqns but not vars is regarded as a parameter.
NSolve[eqns] treats all variables encountered as vars above. NSolve[eqns, vars, prec] attempts to solve numerically the equations for vars using prec digits precision.
NullSpace NullSpace[m] gives a list of vectors that forms a basis for the null space of the matrix m.
Number N amber represents an exact integer or an approximate real number in Read.
Off Off[symbol tag] switches off a message, so that it is no longer print­
ed. Offfs] switches off tracing mes­
sages associated with the symbol s. Off[ml, m2,...] switches off several messages. Off[ ] switches off all trac­
ing messages.
On On[symbol tag] switches on a message, so that it can be printed. On[s] switches on tracing for the sym­
bol s. On[ml, m2,...] switches on sev­
eral messages. On[ ] switches on trac­
ing for all
symbols.
Out %n or Out[n] is a global object that is ass igned to be the value pro­
duced on ;he n-th output line. % gives the last result generated. %% gives the result before last. %%...% (k times) gives the 'lc-th previous result.
Paramet Parametri tmax}] p: with x
and
i c P l o t
cPlot[{fx, fy}, {t, tmin, noduces a parametric plot y coordinates fx and fy
366
G lossary
generated as a function of t. ParametricPlot[{{fx, fy}, {gx, gy},
...}, {t, tmin, tmax}] plots several parametric curves.
P a r a m e t r i cPlot3D ParametricPIot3D[{fx, fy, fz}, {t, tmin, tmax}] produces a three-dimensional space curve parameterized by a vari­
able t which runs from tmin to tmax. ParametricPlot3D[{ix, fy, fz}, {t, tmin, tmax}, {u, umin, umax}] produces a three-dimensional surface parame­
trized by t and u.
ParametricPlot3D[{fx, fy, fz, s },...] shades the plot according to the color specification s.
ParametricPlot3D[{{fx, fy, fz}, {gx, gy, g z },...},...] plots several objects together.
P a r t expr[[i]] or Part [expr, i] gives the i-th part of expr. expr[[-i]] counts from the end. expr[[0]] gives the head o f expr. expr[[i, j,...]] orPart[expr, i,j, ...] is equivalent to expr[[i]] [ [ j ] ].... expr[[ {il, i 2,...} ]] gives a list of the parts i l, i 2,... of expr.
P a r t i t i o n Part it i onis t, n] partitions list into non-overlapping sublists of length n. Partition[list, n, d] generates sublists with offset d. Partition[list, {nl, n2, ...}, {dl, d 2,...}] partitions successive levels in list into length ni sublists with offsets di.
P i Pi is pi, with numerical value 3.14159....
P l o t Plot[f, {x, xmin, xmax}] gener­
ates a plot o f f as a function o f x from xmin to xmax. Plot[{fl, f 2,...}, {x, xmin, xmax}] plots several functions fi.
Plot3D Plot3D[f, {x, xmin, xmax}, {y, ymin, ymax}] generates a three- dimensional plot o f f as a function o f x and y. Plot3D[{f, s}, {x, xmin, xmax}, {y, ymin, ymax}] generates a three- dimensional plot in which the height o f the surface is specified by f, and the shading is specified by s.
P r i n t Printfexprl, expr2,...] prints the expri, followed by a new line (line feed).
Qui t Quit[ ] terminates a Mathematica session.
Range Range[imax] generates the list {1, 2,..., imax}. Range[imin, imax] generates the list {imin,..., imax}.
Re Re[z] gives the real part of the complex number z.
Remove Remove[symboll,...] removes symbols completely, so that their names are no longer recognized by Mathematica. Remove[\forml\, \form2\,...] removes all symbols whose names match any o f the string patterns formi.
Re pla ce Replace[expr, rules] applies a rule or list of rules in an attempt to transform the entire expression expr.
Re s t Rest[expr] gives expr with the first element removed.
Rule lhs -> rhs represents a rule that transforms lhs to rhs.
SameQ lhs === rhs yields True i f the expression lhs is identical to rhs, and yields False otherwise.
Glossary
367
Series Series[f, {x, xO, n}] generates a power series expansion for f about the point x = xO to order (χ - χΟ)Λη. Series[f, {x, xO, nx}, {y, yO, ny}] suc­
cessively finds series expansions with respect to y, then x.
Set lhs = rhs evaluates rhs and assigns the result to be the value of lhs. From then on, lhs is replaced by rhs whenever it appears. {11,12,...} = {rl, r 2,...} evaluates the ri, and assigns the results to be the values of the corresponding li.
Show Show[graphics, options] dis­
plays two- and three-dimensional graphics, using the options specified. Show[gl, g2,...] shows several plots combined. Show can also be used to play Sound objects.
Simplify Simplify[expr] performs a sequence of transformations on expr, and returns the simplest form it finds.
Sin Sin[z] gives the sine of z.
Solve Solve[eqns, vars] attempts to solve an equation or set of equations for the variables vars. Any variable in eqns but not vars is regarded as a para­
meter. Solve[eqns] treats all variables encountered as vars above. Solve[eqns, vars, elims] attempts to solve the equa­
tions for vars, eliminating the variables elims.
tn
1 to imax starts wi' imin, ima: Tablefexp jmax},.. associated
. Table[expr, {i, imin, imax}] i = imin. Table[expr, {i, x, di}] uses steps di. it, {i, imin, imax}, {j, jmin, gives a nested list. The list with i is outermost.
Times x product
Tan Tari[z] gives the tangent of z.
TeXForid TeXForm[expr] prints as a TeX language version of expr.
*y*z or x y z represents a (if terms.
r Together[expr] puts terms over a common denominator, els factors in the result.
Togethe in a sum and canc
Transpose Transpose[list] transpos­
es the first two levels in list. Transpose[list, {nl, n 2,...}] transposes list so th at the nk-th level in list is the k-th level in the result.
True True is the symbol for the Boolean value true.
VectorC) expr is a are them ι otherwis True applied expr.
: only
VectorQ[expr] gives True if list, none of whose elements selves lists, and gives False e. VectorQ[expr, test] gives if test yields True when :o each of the elements in
Sqrt Sqrt[z] gives the square root of z.
Table Table[expr, {imax}] generates a list of imax copies of expr. Table[expr, {i, imax}] generates a list of the values of expr when i runs from
D
M a t h e m a t i c a P a c k a g e s a n d N o t e b o o k s
A list o f Ma thematica packages and notebooks available on the CRC web server is given below. The numbers refer to the pages in the text o f this book.
E q u a t i o n T y p e.m, 7 E q u a t i o n T y p e.m a,7, 21 E x a m p l e 2.1.ma, 26 E x a m p l e 2.5.ma, 31 E x a m p l e 2.8.ma, 34 E x a m p l e 2.1 5.ma,42 E x a m p l e 2.1 8.ma,44 E x a m p l e 2.1 9.ma, 45 E x a m p l e 2.2 1.ma, 52 E x a m p l e 2.2 2.ma, 53 E x a m p l e 2.2 4.ma, 55 E x e r c i s e 2.2 2.ma, 61 I n v e r s e O p e r a t o r.m, 72 o r t h o n o r m a l i t y.m, 86 p l o t f o u r i e r.m, 102 e i g e n p a i r.m a, 105 b e s s e l.m a, 107 drum.ma, 143 G r e e n s.m a, 217 g a l e r k i n.ma,267 p e r t u r b a t i o n.m a, 306 f d.m a, 343
B i b l i o g r a p h y
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H. Bateman, Partial Differential Equations, Dover, Ne^
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versity Press, Cambridge, 1959.
namics, Longman Scientific 3.
f Mathematical Functions,
M. Becker, The Principles and Applications of Varia Cambridge, MA, 1964.
W. E. Boyce and R. C. DiPrima, Elementary Differentia l New York, 1992.
G. Carrier and C. Pearson, Partial Differential Equatior,
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H. S. Carslaw and J. C. Jaeger, Conduction of Heat
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R. V. Churchill, Operational Methods, 3rd ed., McGrav,
_______and J. W. Brown, Fourier Series and Boundary
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J. J. Connor and C. A. Brebbia, Finite Element Tech neers, Butterworths, London, 1973.
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H. F. Davis, Fourier Series and Orthogonal Functions 1963.
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_______and P. Puri, Unsteady MHD free-convection oscillatory flow on a porous
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J. S. Walker, Fourier Analysis, Oxford University Press, Oxford, 1988.
G. N. Watson, A Treatise on the Theory of Bessel Functions, 2nd ed., Cambridge
University Press, Cambridge, 1944.
H. F. Weinberger, A First Course in Partial Differential Equations, Xerox, Lex­
ington, MA, 1965.
R. E. Williamson, R. H. Crowell and H. F. Trotter, Calculus of Vector Functions, 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1968.
S. Wolfram, Mathematica, 2nd ed., Addison-Wesley, Redwood City, CA, 1991.
E. C. Young, Partial Differential Equations, Allyn and Bacon, Boston, 1972.
E. C. Zachmanoglou and D. W. Thoe, Introduction to Partial Differential Equations with Applications, Williams & Wilkins, Baltimore, 1976.
E. Zauderer, Partial Differential Equations of Applied Mathematics, Wiley, New York, 1983.
I n d e x
adaptive grid, 321
auxiliary system, 23, 28, 31, 40, 325
B
ball, open, 217 surface of, 217, surface area of, 218 base characteristic, 24ff, 28 beam, 157
Bessel functions, 82, 106 binomial theorem, 180 Biot number, 148 boundary conditions,
Dirichlet, 3, 105, 108, 132, 150, 159, 250, 280, 302, 339, 344, 348
essential, 276 mixed, 3, 250 natural, 250
Neumann, 3, 105, 250, 347 nonhomogeneous, 146ff of first kind, 3 of second kind, 3 of third kind, 3 periodic, 104 Robin, 3 boundary perturbations, 31 Iff bounded variation, 98ff
calculus of variations, 266
Cauchy data, 56, 237 Cauchy integral formula, 183 Cauchy-Kowalewsky theorem, 55 Cauchy problem, 39, 55, 235, 237 Cauchy residue theorem, 180 Cauchy-Riemann operator, 259 Cauchy’s theorem, 187 characteristics, 5, 24ff, 47ff, 51, 325 base, 24ff, 28 Cauchy’s method of, 39ff curves, 49 circular drum, 142 class CP(Q.), 218
C°°(ii), 218
o r C
classificati
compleme
cone, normal, 47 Monge, 47, 48 tanger t, 47, 48 contour integration, 183ff, 189, 198 convoluticn theorem, 197, 204, 215 cooling ball, 144, 153 coordinate s, characteristic 118
cylind coupling c
D
d’Alembe 213 de Moivre difference difference difference
,n), 218, on, 4ff nt, 218
221
rical, 142 onstant, 8
rt’s solution, 76, 96, 97, 122,
’s theorem, 70 equation, consistent, 328 operator, 328 scheme, central, 343
374
I ndex
Crank-Nicolson, 332, 334, 335,
344
explicit, 332, 334 implicit, 332, 334 stability of, 329 differences, finite 321ff backward, 323ff, 332 first order, 322
forward, 323ff, 330, 331, 332, 334, 336, 344 second order, 323ff, 330, 335, 336, 339
differential form, 277 differential operator, linear, 68 Dirac distribution, 220ff, 232 Dirichlet-Dirichlet case, 283 discontinuity, 53, 90 distribution, 220 regular, 220 singular, 220, 221 divergence theorem, 14 Duhamel’s principle, 203
E
eigenfunction expansion, 82, 103ff eigenfunctions, 104, 108 eigenpair, 104 eigenvalue problem, 117 eigenvalues, 104, 148ff, 155 electrostatic potential, 239 electrostatics, 8 element, integral, 47 plane, 47 tangent, 47 envelope, 39 equation(s), Berger’s, 8 Bessel’s, 106, 241, 286 biharmonic, 9
characteristic, 5, 23ff, 40, 315 cubic Schrodinger, 8 difference, 328, 339, 343, 348, 349 diffusion, 204, 332 dissipative Klein-Gordon, 8 dissipative sine-Gordon, 9
eikonal, 8
elliptic, 4, 6, 50, 132ff, 238ff Euler’s, 9, 274, 282, 296 Euler-Lagrange, 269 first order, 22ff Fokker-Planck, 255 4th order, 287 heat, 7, 346
heat conduction, 15, 16, 125ff, 177, 293 heat transfer, 14ff Helmholtz, 8, 241ff hyperbolic, 4, 6, 50, 53, 118ff, 244ff Klein-Gordon, 8 Korte de Vries (KdV), 9 Laplace, 7, 16, 203,211,252, 261,339 linear with constant coefficients, 22ff, 63ff
— with variable constants, 28ff Maxwell’s, 9 Navier-Stokes, 9, 277 nonlinear first order, 38ff parabolic, 4, 6, 50, 125ff, 23Iff Poisson’s, 8, 159, 275, 280, 283,
302, 339, 342 porous media, 9 quadratic, 4
quasi-linear first order, 2, 17, 33ff
— second order, 49ff
— nonhomogeneous, 74ff Schrodinger, 8, 255 semilinear heat, 8 semilinear Klein-Gordon, 8 semilinear Poisson’s, 9 semilinear wave, 8 sine-Gordon, 8 Sturm-Liouville, 105 telegrapher’s, 8
traffic, 7 transport, 7, 8 transient Fourier, 235 Tricomi, 6, 20 ultrahyperbolic, 6
I ndex
375
error, discretization, 329 functionally independent, 29ff
truncation, 328, 344, 345 linear, 219
Euler’s formula, 90 continuous, 220
existence, 25, 27, 30
wave, 7, 97, 180, 212, 334, 346 functional, 219
general solui geometric Gibbs pheno- gravitationil Green’s ide — first function:
field point, 247 finite differences, 321ff flow, porous media, 9 traffic, 16ff of viscous fluid, 189 of viscoelastic fluid, 189 Fourier-Bessel expansion, 108ff — seco
Fourier coefficients, 88 grid, 323, 3
— cosine series, 101
— cosine theorem, 195 H
— integral theorem, 195
— sine series, 100, 102, 127 harmonics,
— sine theorem, 195 harmonic n
— theorem I, 90 Heaviside
— theorem II, 100 2
Fourier transform, 195ff Hilbert spaci
convolution theorem, 196 Hooke’s la’
formula, 198ff Huygens’
of derivatives, 196
properties of, 196 I
fundamental solution, 177
function(s), Bessel, 82, 106, 245 impulsive
characteristic, 218 infinite seri
Dirac delta, 178ff, 217, 221ff, 248, initial condii
263 initial curvc.
Hankel, 241 inner produci
harmonic, 224ff, 251ff — spac
Heaviside, 54, 178 integral surf;
indicator, 218, 220 isocline, 46
modified Bessel, 242 monotone, 99 nondecreasing, 99
nonincreasing, 99 Jacobian, 50, 52
orthogonal, 52ff
periodic, 90, 97 K
periodic Dirichlet, 307
test, 219, 287ff kernel, 160
tion, 2 empties, 8 menon, 102 potential, 221 ntity, 13, 225, 232, 350ff identity, 227, 351 s, 217ff, 230ff, 250ff nd identity, 257, 351 49
103
lotion, 123
function, 54, 178ff, 220, 24
e, 109, 218 10
principle, 247ff
f3rce, 221 es, 82
tion, 3, 130 26ff, 31 :t, 82 e, 218
'aces, 24, 30,45, 47
376
I ndex
L
Laplace inversion theorem, 183, 187 transform, 161, 162ff, 179, 180, 183, 189ff, 233, 236, 244, 245
Laplacian, 108, 142, 231, 238, 257 line integrals, 267ff Lipschitz condition, 99 L2 space, 145
M
Maclaurin series, 254 Mainardi-Codazzi relations, 22 Mathematica functions, 36Iff
notebooks and packages list, 368 maximum principle, 228ff mesh point, 323, 340 method, Cauchy’s, 39ff collocation, 266, 395 Fourier series, 150 Galerkin, 266, 279ff Gauss elimination, 330 inverse operator, 63ff Lagrange’s, 28 least-square, 267, 295 of characteristics, 22ff of moments, 267, 296 of separation of variables, 51, 82, 103, 117ff, 235, 307, 312 of undetermined coefficients, 78 of variation of parameters, 78 perturbation, 305ff Rayleigh-Ritz, 266, 283ff weighted residual, 266ff Monge axis, 47 cone, 47 pencil, 47 multi-index, 220
N
neighborhood, 218 node, 323, 347
O
operator pair, 64 operator, difference, 328 Laplace, 108 linear, 68 orthogonal eigenfunctions, 222 expansions, 82ff functions, 82 polynomials, 86 set, 83, 88, 108 orthogonality, 82ff, 88 orthonormal, 83 set, 83
norm,218
notation, 1
parabolic cylinder, 35 partial derivatives, 1 partial differential equations, elliptic, 4, 6, 19, 50 homogeneous, 2 hyperbolic, 4, 6, 19, 50 linear, 2
nonhomogeneous, 2, 4 order, 1
parabolic, 4, 19, 50 quasi-linear, 2, 17, 33ff, 325ff particular integral, 65 solution, 2, 4, 75 ultrahyperbolic, 6 perturbation methods, 305ff point charge, 221 mass, 221 Poisson integral representation, 203 polynomials, Chebyshev 1st kind, 86 — 2nd kind, 86 Hermite, 86 Jacobi, 86 Laguerre, 86 Legendre, 86, 87 principal part, 49
I ndex
377
quadratic form, 6ff quantum field theory, 8 mechanics, 8 quasi-linear equations, 2,
17
reciprocity relations, 231 residual, 279 residue, 181
Riemann-Lebesgue lemma, 107 rotation of axes, 24, 60ff
series, Fourier, 90, 109, 156 generalized Fourier, 88 infinite, 82 Maclaurin, 254 of orthogonal functions, 88ff orthogonal, 82 Taylor’s, truncated, 321 trigonometric Fourier, 82, 89, 90ff, 111
shock waves, 33 solution, general, 2, 29, 73 particular, 2, 4, 75 source point, 230ff, 233 spherical coordinates, 144ff strip, 48
characteristic, 48 Sturm-Liouville problem, 103ff, 105 equation, 105 successive approximations, 305, 309ff superposition principle, 17ff support, 218 compact, 218
Table 4.1, 114 Table 4.2, 116
of complex Fourier transform
pairs, 358 of finii e cosine transform pairs, 360
of fini e sine transform pairs, 359 of Fourier cosine transform pairs,
356
of Fourier sine transform pairs,
357
of Laplace transform pairs, 354 of transform pairs, 354ff Taylor series expansions, 306ff Taylor’s formula, 321 Tchebyscbeff, 86 tension, 9 10 test functions, 287ff theorem, binomial, 180 Cauchy’s, 187 Cauchy-Kowalewsky, 55 Cauchy residue, 180 convolution, 197, 205, 215 de Moivre’s, 70 divergence, 14, 350 Fourier cosine theorem, 195 Fourier integral theorem, 195 Fourier sine theorem, 195 Fourier theorem I, 90 Fourier theorem II, 100 Gauss, 350 gradient, 350 Green’s, 350 Green’s reciprocity, 351 inversion, 183 Laplace inversion, 183, 187 Lax equivalence, 329 Stokes, 350 Taylor’s, 321, 342 thermal diffusivity, 7, 15 tolerance, 328 torsion, 289 total variation, 98 transforms, complex Fourier, 161, 195ff, 213, 223,234, 236, 264
finite Fourier, 21 Off finite Fourier cosine, 21 Off, 216
378
I ndex
finite Fourier sine, 21 Off, 216 Fourier cosine, 160, 195, 204ff Fourier sine, 160, 195, 204ff Hankel, 161 Hilbert, 162 integral, 160ff Kantorowich-Lebedev, 162 Laguerre, 162
Laplace, 161, 162ff, 179, 180, 183, 189ff, 233, 236ff, 244 Laplace inverse, 164ff Mehler-Foch, 162 Meijer, 161 Mellin, 161 transformation, reversible, 50ff transient problems, 292ff trigonometric Fourier series, 82, 89, 90ff
U
uniqueness, 25, 27
variable, dependent, 1 independent, 1 variation, weak, 274ff variational notation, 270ff vibrations of a string, 7, 9ff, 118, 152
of a membrane, 12ff, 142 transverse, 157 viscosity, kinematic, 9
W
wave equation, 7, 97, 180, 212, 334, 346
wave propagation, 247ff waves, shallow water, 7, 214 weak variational formulation, 274ff weighted inner product, 85 residual methods, 266ff
V
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