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A FIRST COURSEIN
INTEGRAL EQUATIONS
Second Edition
Solutions Manual
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N E W J E R S E Y • L O N D O N • S I N G A P O R E • BE IJ ING • S H A N G H A I • H O N G K O N G • TA I P E I • C H E N N A I
World Scientific
A FIRST COURSEIN
INTEGRAL EQUATIONS
Second Edition
Abdul-Majid WazwazSaint Xavier University, USA
Solutions Manual
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Published by
World Scientific Publishing Co. Pte. Ltd.5 Toh Tuck Link, Singapore 596224USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE
Library of Congress Cataloging-in-Publication DataWazwaz, Abdul-Majid. A first course in integral equations. Solutions manual / by Abdul-Majid Wazwaz (Saint Xavier University, USA). -- Second edition. pages cm Includes bibliographical references and index. ISBN 978-9814675154 (pbk. : alk. paper) 1. Integral equations--Problems, exercises, etc. I. Title. QA431.W36 2015b 515'.45--dc23 2015008782
British Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.
Copyright © 2015 by World Scientific Publishing Co. Pte. Ltd.
All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.
For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.
Printed in Singapore
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THIS BOOK IS DEDICATED TO
My wife, our son, and our three daughtersfor supporting me in all my endeavors
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Contents
Preface ix
1 Introductory Concepts 1
1.2 Classification of Linear Integral Equations . . . . . . . . . . 1
1.3 Solution of an Integral Equation . . . . . . . . . . . . . . . 2
1.4 Converting Volterra Equation to an ODE . . . . . . . . . . 4
1.5 Converting IVP to Volterra Equation . . . . . . . . . . . . . 7
1.6 Converting BVP to Fredholm Equation . . . . . . . . . . . 11
1.7 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Fredholm Integral Equations 15
2.2 Adomian Decomposition Method . . . . . . . . . . . . . . . 15
2.3 The Variational Iteration Method . . . . . . . . . . . . . . . 22
2.4 The Direct Computation Method . . . . . . . . . . . . . . . 25
2.5 Successive Approximations Method . . . . . . . . . . . . . . 29
2.6 Successive Substitutions Method . . . . . . . . . . . . . . . 33
2.8 Homogeneous Fredholm Equation . . . . . . . . . . . . . . . 35
2.9 Fredholm Integral Equation of the First Kind . . . . . . . . 39
3 Volterra Integral Equations 41
3.2 Adomian Decomposition Method . . . . . . . . . . . . . . . 41
3.3 The Variational Iteration Method . . . . . . . . . . . . . . . 54
3.4 The Series Solution Method . . . . . . . . . . . . . . . . . . 57
3.5 Converting Volterra Equation to IVP . . . . . . . . . . . . . 63
3.6 Successive Approximations Method . . . . . . . . . . . . . . 67
3.7 Successive Substitutions Method . . . . . . . . . . . . . . . 75
3.9 Volterra Equations of the First Kind . . . . . . . . . . . . . 79
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viii Contents
4 Fredholm Integro-Differential Equations 854.3 The Direct Computation Method . . . . . . . . . . . . . . . 854.4 The Adomian Decomposition Method . . . . . . . . . . . . 904.5 The Variational Iteration Method . . . . . . . . . . . . . . . 944.6 Converting to Fredholm Integral Equations . . . . . . . . . 96
5 Volterra Integro-Differential Equations 1015.3 The Series Solution Method . . . . . . . . . . . . . . . . . . 1015.4 The Adomian Decomposition Method . . . . . . . . . . . . 1035.5 The Variational Iteration Method . . . . . . . . . . . . . . . 1055.6 Converting to Volterra Equations . . . . . . . . . . . . . . . 1075.7 Converting to Initial Value Problems . . . . . . . . . . . . . 1105.8 The Volterra Integro-Differential Equations of the First
Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6 Singular Integral Equations 1176.2 Abel’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . 1176.3 Generalized Abel’s Problem . . . . . . . . . . . . . . . . . . 1226.4 The Weakly Singular Volterra Equations . . . . . . . . . . . 1226.5 The Weakly Singular Fredholm Equations . . . . . . . . . . 130
7 Nonlinear Fredholm Integral Equations 1337.2 Nonlinear Fredholm Integral Equations . . . . . . . . . . . . 133
7.2.1 The Direct Computation Method . . . . . . . . . . . 1337.2.2 The Adomian Decomposition Method . . . . . . . . 1417.2.3 The Variational Iteration Method . . . . . . . . . . . 148
7.3 Nonlinear Fredholm Integral Equations of the FirstKind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
7.4 Weakly-Singular Nonlinear Fredholm Integral Equations . . 153
8 Nonlinear Volterra Integral Equations 1578.2 Nonlinear Volterra Integral Equations . . . . . . . . . . . . 157
8.2.1 The Series Solution Method . . . . . . . . . . . . . . 1578.2.2 The Adomian Decomposition Method . . . . . . . . 1638.2.3 The Variational Iteration Method . . . . . . . . . . . 168
8.3 Nonlinear Volterra Integral Equations of the First Kind . . 1708.3.1 The Series Solution Method . . . . . . . . . . . . . . 1708.3.2 Conversion to a Volterra Equation of the Second
Kind . . . . . . . . . . . . . . . . . . . . . . . . . . . 1728.4 Nonlinear Weakly-Singular Volterra Equation . . . . . . . . 173
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Preface
This Solutions Manual is provided as a supplement to accompany the textA First Course in Integral Equations, Second Edition. It is intended:• to develop readers’ skills for each linear or nonlinear exercise provided inthe textbook;• to illustrate the use of the newly developed methods along with the tra-ditional methods used in the text;• to help readers master integral equations concepts so they can solve reg-ular level and challenging exercises.
This Solutions Manual provides completely explained solutions to allexercises of the textbook. The manual can be used by a wide variety of au-dience in various fields of mathematics, physics, chemistry and engineering.Because the text A First Course in Integral Equations, Second Edition con-tains the newly developed methods and some of the classical methods, wehave aimed to make the Solutions Manual self-explanatory, instructional,and useful. The fully-worked solutions are explained in a systematic wayfocusing on the needs, suggestions and expectations of the readers. Thefollowing distinguishing features make this Solutions Manual significant:• it provides a fully worked solution for each problem in the text;• it applies the new methods along with some of the classic methods formost problems using the same strategies, notations, and terminologies usedin the text A First Course in Integral Equations, Second Edition;• it illustrates the solutions in a systematic fashion consistent with the ma-terial presented in the text;• it assists in helping readers gain knowledge and understanding of themathematical methods as they work through the problems and study thesolutions;• it addresses all questions and suggestions made by students and scholarswho used the first edition;• it complements the material and the worked examples of the textbook.
ix
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x Preface
The text A First Course in Integral Equations, Second Edition presentsa variety of valuable methods and applications which cannot be found inany other book. For this reason, and based upon a vast request by readers,this Solutions Manual has been made to provide detailed explanations andillustrations for solving each problem of the second edition.
I would like to acknowledge my wife for her genuine patience and I wishto thank her for sincerely encouraging me in all my endeavors, includingthe writing of the second edition of the textbook as well as the solutionsmanual. I would also like to acknowledge our son and three daughters fortheir support and encouragement.
Saint Xavier University Abdul-Majid WazwazChicago, IL 60655 e-mail: [email protected] 2015
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Chapter 1
Introductory Concepts
1.2 Classification of Linear Integral Equations
Exercises 1.2
1. Fredholm, linear, nonhomogeneous2. Volterra, linear, nonhomogeneous3. Volterra, nonlinear, nonhomogeneous4. Fredholm, linear, homogeneous5. Fredholm, linear, nonhomogeneous6. Fredholm, nonlinear, nonhomogeneous7. Fredholm, nonlinear, nonhomogeneous8. Fredholm, linear, nonhomogeneous9. Volterra, nonlinear, nonhomogeneous10. Volterra, linear, nonhomogeneous11. Volterra integro-differential equation, nonlinear12. Fredholm integro-differential equation, linear13. Volterra integro-differential equation, nonlinear14. Fredholm integro-differential equation, linear15. Volterra integro-differential equation, linear
16. u(x) = 1 +
∫ x
0
4u(t)dt
17. u(x) = 1 +
∫ x
0
3t2u(t)dt
18. u(x) = 4 +
∫ x
0
u2(t)dt
1
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2 Chapter 1. Introductory Concepts
19. u′(x) = 1 +
∫ x
0
4tu2(t)dt, u(0) = 2
20. u′(x) = 1 +
∫ x
0
2tu(t)dt, u(0) = 0
21. Volterra–Fredholm integral equation, nonlinear, nonhomogeneous22. Volterra–Fredholm integro-differential equation, linear, nonhomoge-neous23. Volterra–Fredholm integro-differential equation, nonlinear, nonhomo-geneous24. Volterra (singular) integral equation, nonlinear, nonhomogeneous
1.3 Solution of an Integral Equation
Exercises 1.3
1. Substituting u(x) = x+ 124 in the right hand side yields
RHS = x+
∫ 14
0
(t+
1
24
)dt
= x+ [ 12 t2 + 1
24 t]140
= x+ 124
= u(x)= LHS.
2. Substituting u(x) = x in the right hand side yields
RHS = 23x+
∫ 1
0
xt2dt
= 23x+ [ 13xt
3]10
= 23x+ 1
3x= u(x)= LHS.
3. Substituting u(x) = 2x in the right hand side yields
RHS = x+
∫ 1
0
4xt3dt
= x+ [xt4]10= 2x= u(x)= LHS
4. Substituting u(x) = sinx in the right hand side yields
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1.3. Solution of an Integral Equation 3
RHS = x−∫ x
0
(x− t) sin tdt
= x− [−x cos t− sin t+ t cos t]x0= sinx= u(x)= LHS.
5. Substituting u(x) = coshx in the right hand side yields
RHS = 2 coshx− x sinhx− 1 +
∫ x
0
t cosh tdt
= 2 coshx− x sinhx− 1 + [t sinh t− cosh t]x0= coshx= u(x)= LHS.
6. Substituting u(x) = x in the right hand side yields
RHS = x+ 15x
5 −∫ x
0
t4dt
= x+ 15x
5 − [ 15 t5]x0
= x= u(x)= LHS.
7. Substituting u(x) = x2 in the right hand side yields
RHS = 2x− x4 +
∫ x
0
4t3dt
= 2x− x4 + [t4]x0= 2x= u
′(x)
= LHS.
8. Substituting u(x) = sinx in the right hand side yields
RHS = x cosx− 2 sinx+
∫ x
0
t sin tdt
= x cosx− 2 sinx+ [sin t− t cos t]x0= − sinx= u
′′(x)
= LHS.
9. Substituting u(x) = 3 in the left hand side yields
LHS =
∫ x
0
3 (x− t)2 dt
= [−(x− t)3]x0
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4 Chapter 1. Introductory Concepts
= x3
= RHS.
10. Substituting u(x) = 32 in the left hand side yields
LHS =
∫ x
0
3
2(x− t)
12 dt
= [−(x− t) 32 ]x0
= x32
= RHS.
11. Substituting u(x) = 2 cosx− 1 into both sides yields
2 cosx− 1 =
∫ x
0
(x− t)(2 cos t− 1)dt
f(x) = x2
12. Substituting u(x) = ex into both sides yields
ex = f(x)
∫ x
0
(x− t)(et)dt
f(x) = 1 + x
13. Substituting u(x) = e−x3
into both sides yields
e−x3
= 1− α∫ x
0
3t2(e−)3
dt,
α = 1
14. Substituting u(x) = sinx into both sides yields
sinx = f(x)− 1
∫ π2
0
t sin t dt
f(x) = sinx
1.4 Converting Volterra Equation to an ODE
Exercises 1.4
1.
∫ x
0
3(x− t)2u(t)dt
2. 2xex3
− ex2
+
∫ x2
x
textdt
3.
∫ x
0
4(x− t)3u(t)dt
4. 4 sin 5x− sin 2x+
∫ 4x
x
cos(x+ t)dt
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1.4. Converting Volterra Equation to an ODE 5
5. Differentiating both sides three times, we obtain
u′(x) = 1 +
∫ x
0
2(x− t)u(t)dt
u′′(x) =
∫ x
0
2u(t)dt
u′′′
(x) = 2u(x)dtSubstituting x = 0 in the original equation and the first two derivatives,we find u(0) = u
′(0) = 1, u
′′(0) = 0
Hence, the equivalent initial value problem is
u′′′
(x) = 2u(x), u(0) = u′(0) = 1, u
′′(0) = 0
6. Differentiating both sides two times, we obtain
u′(x) = ex −
∫ x
0
u(t)dt
u′′(x) = ex − u(x)
Substituting x = 0 in the original equation and the first derivative, we findu(0) = u
′(0) = 1
Hence, the equivalent initial value problem isu′′(x) + u(x) = ex, u(0) = u
′(0) = 1
7. Differentiating both sides two times, we obtain
u′(x) = 1 +
∫ x
0
u(t)dt
u′′(x) = u(x)
Substituting x = 0 in the original equation and the first derivative, we findu(0) = 0, u
′(0) = 1
Hence, the equivalent initial value problem isu′′(x)− u(x) = 0, u(0) = 0, u
′(0) = 1
8. Differentiating both sides two times, we obtain
u′(x) = 1 + sinx+
∫ x
0
u(t)dt
u′′(x) = cosx+ u(x)
Substituting x = 0 in the original equation and the first derivative, we findu(0) = −1, u
′(0) = 1
Hence, the equivalent initial value problem isu′′(x)− u(x) = cosx, u(0) = −1, u′(0) = 1
9. Differentiating both sides two times, we obtain
u′(x) = 3 + 10x+ u(x) +
∫ x
0
2u(t)dt
u′′(x) = 10 + u
′(x) + 2u(x)
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6 Chapter 1. Introductory Concepts
Substituting x = 0 in the original equation and the first derivative, we find
u(0) = 2, u′(0) = 5
Hence, the equivalent initial value problem isu′′(x)− u
′(x)− 2u(x) = 10, u(0) = 2, u
′(0) = 5
10. Differentiating both sides two times, we obtain
u′(x) = 6 + 5u(x)−
∫ x
0
6u(t)dt
u′′(x) = 5u
′(x)− 6u(x)
Substituting x = 0 in the original equation and the first derivative, we findu(0) = −5, u
′(0) = −19
Hence, the equivalent initial value problem isu′′(x)− 5u
′(x) + 6u(x) = 0, u(0) = −5, u
′(0) = −19
11. Differentiating both sides we obtainu′(x) = sec2x− u(x)
Substituting x = 0 in the original equation we findu(0) = 0Hence, the equivalent initial value problem isu′(x) + u(x) = sec2x, u(0) = 0
12. Differentiating both sides three times, we obtain
u′(x) = 1 + 5x+ 3u(x) +
∫ x
0
[6− 5(x− t)]u(t)dt
u′′(x) = 5 + 3u
′(x) + 6u(x)− 5
∫ x
0
u(t)dt
u′′′
(x) = 3u′′(x) + 6u
′(x)− 5u(x)
Substituting x = 0 in the original equation and the first two derivatives,
we find u(0) = 1, u′(0) = 4, u
′′(0) = 23
Hence, the equivalent initial value problem isu′′′
(x)− 3u′′(x)− 6u
′(x) + 5u(x) = 0,
u(0) = 1, u′(0) = 4, u
′′(0) = 23
13. u′′′
(x)− 4u(x) = 24x,u(0) = u
′(0) = 0, u
′′(0) = 2
14. uiv(x)− u(x) = 0,u(0) = u
′(0) = 0, u
′′(0) = 2, u
′′′(0) = 0
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1.5. Converting IVP to Volterra Equation 7
1.5 Converting IVP to Volterra Equation
Exercises 1.5
1. We first sety′(x) = u(x)
Integrating both sides from 0 to x we findy(x)− 1 =
∫ x0u(t)dt
Substituting in the original equation we find
u(x) = −1−∫ x
0
u(t)dt
2. We first sety′(x) = u(x)
Integrating both sides from 0 to x, we findy(x) =
∫ x0u(t)dt
Substituting in the original equation we find
u(x) = x+
∫ x
0
u(t)dt
3. We first sety′(x) = u(x)
Integrating both sides from 0 to x we findy(x) =
∫ x0u(t)dt
Substituting in the original equation we find
u(x) = sec2(x)−∫ x
0
u(t)dt
4. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice givesy′(x)− y′(0) =
∫ x0u(t)dt
y(x)− y(0)− xy′(0) =∫ x0
∫ x0u(t)dtdt
or equivalentlyy(x) = 1 +
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = −1−∫ x
0
(x− t)u(t)dt
5. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(0) =
∫ x
0
u(t)dt
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March 12, 2015 12:19 book-9x6 9571-Root page 8
8 Chapter 1. Introductory Concepts
y(x)− y(0)− xy′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy(x) = 1 + x+
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = 1 + x+
∫ x
0
(x− t)u(t)dt
6. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(0) =
∫ x
0
u(t)dt
y(x)− y(0)− xy′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy(x) = 1 + x+
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = −11− 6x−∫ x
1
[5 + 6(x− t)]u(t)dt
7. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(1) =
∫ x
1
u(t)dt
or equivalentlyy′(x) = 1 +
∫ x1u(t)dt
Substituting in the original equation we find
u(x) = −1−∫ x
1
u(t)dt
8. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(0) =
∫ x
0
u(t)dt
y(x)− y(0)− xy′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy(x) = x+
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = −1 + 4x+
∫ x
0
[2(x− t)− 1]u(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 9
1.5. Converting IVP to Volterra Equation 9
9. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(0) =
∫ x
0
u(t)dt
y(x)− y(0)− xy′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy(x) =
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = sinx−∫ x
0
(x− t)u(t)dt
10. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice gives
y′(x)− y′(0) =
∫ x
0
u(t)dt
y(x)− y(0)− xy′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy(x) = 1− x+
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = x− sinx+ xex − ex −∫ x
0
[(x− t)ex − sinx]u(t)dt
11. We first sety′′′
(x) = u(x)Integrating both sides from 0 to x three times gives
y′′(x)− y′′(0) =
∫ x
0
u(t)dt
y′(x)− y′(0)− xy′′(0) =
∫ x
0
∫ x
0
u(t)dtdt
y(x)− y(0)− xy′(0)− 12x
2y′′(0) =
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdt
or equivalentlyy(x) = 2 + x2 + 1
2
∫ x0
(x− t)2u(t)dtSubstituting in the original equation we find
u(x) = 2x− x2 +
∫ x
0
[1 + (x− t)− 1
2(x− t)2
]u(t)dt
12. We first sety′′′
(x) = u(x)Integrating both sides from 0 to x three times gives
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March 4, 2015 14:44 book-9x6 9571-Root page 10
10 Chapter 1. Introductory Concepts
y′′(x)− y′′(0) =
∫ x
0
u(t)dt
y′(x)− y′(0)− xy′′(0) =
∫ x
0
∫ x
0
u(t)dtdt
or equivalentlyy′(x) = x+
∫ x0
(x− t)u(t)dtSubstituting in the original equation we find
u(x) = −3x− 4
∫ x
0
(x− t)u(t)dt
13. We first setyiv(x) = u(x)Integrating both sides from 0 to x three times gives
y′′′
(x)− y′′′(0) =
∫ x
0
u(t)dt
y′′(x)− y′′(0)− xy′′′(0) =
∫ x
0
∫ x
0
u(t)dtdt
y′(x)− y′(0)− xy′′(0)− 1
2x2y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdt
y(x)− y(0)− xy′(0)− 12x
2y′′(0)− 1
6x3y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdtdt
and by changing multiple integrals to single integrals and by substitutingin the original equation we find
u(x) = 2 + x− 1
2x2 − 1
6x3 −
∫ x
0
[2(x− t) +
1
6(x− t)3
]u(t)dt
14. We first setyiv(x) = u(x)Integrating both sides from 0 to x three times gives
y′′′
(x)− y′′′(0) =
∫ x
0
u(t)dt
y′′(x)− y′′(0)− xy′′′(0) =
∫ x
0
∫ x
0
u(t)dtdt
y′(x)− y′(0)− xy′′(0)− 1
2x2y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdt
y(x)− y(0)− xy′(0)− 12x
2y′′(0)− 1
6x3y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdtdt
and by changing multiple integrals to single integrals and by substitutingin the original equation we find
u(x) = 1− 1
2x2 +
1
3!
∫ x
0
(x− t)3u(t)dt
15. We first setyiv(x) = u(x)
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March 12, 2015 12:19 book-9x6 9571-Root page 11
1.6. Converting BVP to Fredholm Equation 11
Integrating both sides from 0 to x three times gives
y′′′
(x)− y′′′(0) =
∫ x
0
u(t)dt
y′′(x)− y′′(0)− xy′′′(0) =
∫ x
0
∫ x
0
u(t)dtdt
y′(x)− y′(0)− xy′′(0)− 1
2x2y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdt
y(x)− y(0)− xy′(0)− 12x
2y′′(0)− 1
6x3y′′′
(0) =
∫ x
0
∫ x
0
∫ x
0
∫ x
0
u(t)dtdtdtdt
and by changing multiple integrals to single integrals and by substitutingin the original equation we find
u(x) = 2ex − 1− x−∫ x
0
(x− t)u(t)dt
1.6 Converting BVP to Fredholm Equation
Exercises 1.6
1. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice we findy′(x) = y
′(0) +
∫ x0u(t)dt
y(x) = y(0) + xy′(0) +
∫ x0
(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary conditionat x = 1 givesy′(0) = −
∫ 1
0(1− t)u(t)dt
Accordingly, we obtainy(x) = −x
∫ 1
0(1− t)u(t)dt+
∫ x0
(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = sinx+ 4x
∫ x0
(1− t)u(t)dt+ 4x∫ 1
x(1− t)u(t)dt− 4
∫ x0
(x− t)u(t)dtThis gives
u(x) = sinx+
∫ 1
0
K(x, t)u(t)dt
K(x, t) =
{4t(1− x) 0 ≤ t ≤ x4x(1− t) x ≤ t ≤ 1
2. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice we findy′(x) = y
′(0) +
∫ x0u(t)dt
y(x) = y(0) + xy′(0) +
∫ x0
(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary condition
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March 4, 2015 14:44 book-9x6 9571-Root page 12
12 Chapter 1. Introductory Concepts
at x = 1 we obtainy′(0) = −
∫ 1
0(1− t)u(t)dt
Accordingly, we obtainy(x) = −x
∫ 1
0(1− t)u(t)dt+
∫ x0
(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = 1 + 2x2
∫ x0
(1− t)u(t)dt+ 2x2∫ 1
x(1− t)u(t)dt− 2x
∫ x0
(x− t)u(t)dtThis gives
u(x) = 1 +
∫ 1
0
K(x, t)u(t)dt,
K(x, t) =
{2xt(1− x) 0 ≤ t ≤ x2x2(1− t) x ≤ t ≤ 1
3. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice we findy′(x) = y
′(0) +
∫ x0u(t)dt
y(x) = y(0) + xy′(0) +
∫ x0
(x− t)u(t)dtSubstituting x = 1 in the last equation and using the boundary conditionat x = 1 we obtainy′(0) = −1−
∫ 1
0(1− t)u(t)dt
Accordingly, we obtainy(x) = 1− x− x
∫ 1
0(1− t)u(t)dt+
∫ x0
(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = (2x− 1) + x
∫ x0
(1− t)u(t)dt+ x∫ 1
x(1− t)u(t)dt+
∫ x0
(x− t)u(t)dtThis gives
u(x) = (2x− 1) +
∫ 1
0
K(x, t)u(t)dt,
K(x, t) =
{t(1− x) 0 ≤ t ≤ xx(1− t) x ≤ t ≤ 1
4. We first sety′′(x) = u(x)
Integrating both sides from 0 to x twice we findy′(x) = y
′(0) +
∫ x0u(t)dt
y(x) = y(0) + xy′(0) +
∫ x0
(x− t)u(t)dtSubstituting x = 1 in the first equation and using the boundary conditionx = 1 we obtainy′(0) = −
∫ 1
0(1− t)u(t)dt
Accordingly, we obtainy(x) = 1− x
∫ 1
0u(t)dt+
∫ x0
(x− t)u(t)dtSubstituting in the original equation, we obtainu(x) = x− 1 + x
∫ x0u(t)dt+ x
∫ 1
xu(t)dt−
∫ x0
(x− t)u(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 13
1.7. Taylor Series 13
This gives u(x) = (x− 1) +
∫ 1
0
K(x, t)u(t)dt,
K(x, t) =
{t 0 ≤ t ≤ xx x ≤ t ≤ 1
1.7 Taylor Series
Exercises 1.7
1. f(x) = e2x 2. f(x) = e−3x
3. f(x) = ex − 1 4. f(x) = cos(2x)5. f(x) = sin(3x) 6. f(x) = sinh(2x)7. f(x) = cosh(2x) 8. f(x) = cosh(3x)− 19. f(x) = 1 + cos(2x) 10. f(x) = 1 + sinx
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March 4, 2015 14:44 book-9x6 9571-Root page 15
Chapter 2
Fredholm IntegralEquations
2.2 Adomian Decomposition Method
Exercises 2.2
1. We set
u0(x) =13
3x
Hence, we find
u1(x) = −1
4
∫ 1
0
13
3xt2dt
u1(x) = − 13
3× 12x
and
u2(x) =13
3× 144x
and so on. Accordingly, we find
u(x) =13
4x
(1− 1
12+
1
144+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = 4x
2. We set
u0(x) = x3 − 1
5x
Hence, we find
15
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March 4, 2015 14:44 book-9x6 9571-Root page 16
16 Chapter 2. Fredholm Integral Equations
u1(x) =
∫ 1
0
x
(t4 − 1
5t2)dt =
2
15x
and
u2(x) =2
45x
and so on. Accordingly, we find
u(x) = x3 − 15x+
2
15x
(1 +
1
3+
1
9+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = x3
3. We setu0(x) = x2
Hence, we find
u1(x) =
∫ 1
0
xt3dt =1
4x
and
u2(x) =1
12x
and so on. Accordingly, we find
u(x) = x2 +1
4x
(1 +
1
3+
1
9+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = x2 + 3
8x
4. We setu0(x) = ex
Hence, we find
u1(x) = e−1∫ 1
0
etdt
u1(x) = 1− e−1andu2(x) = e−1 − e−2u3(x) = e−2 − e−3and so on. Accordingly, we findu(x) = ex + 1−
(e−1 − e−1 + e−2 − e−2 + · · ·
)u(x) = ex + 1
5. We setu0(x) = x+ sinxHence, we find
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March 12, 2015 12:19 book-9x6 9571-Root page 17
2.2. Adomian Decomposition Method 17
u1(x) = −x∫ π
2
0
(t+ sin t)dt
u1(x) = −(1 +π2
8)x
and
u2(x) = (1 +π2
8)π2
8x
and so on. Accordingly, we find
u(x) = x+ sinx− (1 +π2
8)
(1− π2
8+π4
64+ · · ·
)x
And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sinx
6. We setu0(x) = x+ cosxHence, we find
u1(x) = −2x
∫ π6
0
(t+ cos t)dt
u1(x) = −(1 +π2
36)x
and
u2(x) = (1 +π2
36)π2
36x
and so on. Accordingly, we find
u(x) = x+ cosx− (1 +π2
36)
(1− π2
36+
π4
362+ · · ·
)x
And by evaluating the infinite geometric series at the right hand side, wefindu(x) = cosx
7. We setu0(x) = cos(4x) + 1
4xHence, we find
u1(x) = −x∫ π
8
0
(cos 4t+1
4t)dt
u1(x) = −1
4(1 +
π2
2× 82)x
and
u2(x) =1
4(1 +
π2
2× 82)
π2
2× 82x
and so on. Accordingly, we find
u(x) = cos 4x+1
4x− 1
4
(1 +
π2
2× 82
)(1− π2
2× 82+
π4
4× 84+ · · ·
)x
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March 4, 2015 14:44 book-9x6 9571-Root page 18
18 Chapter 2. Fredholm Integral Equations
Hence, we findu(x) = cos 4x
8. We setu0(x) = sinhx− e−1xHence, we find
u1(x) =
∫ 1
0
x t (sinh t− e−1t)dt
u1(x) =2
3e−1x
and
u2(x) =2
9e−1x
and so on. Accordingly, we find
u(x) = sinhx− e−1x+2
3e−1x
(1 +
1
3+
1
9+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sinhx
9. We setu0(x) = 2e2x + (1− e2)xHence, we find
u1(x) = x
∫ 1
0
(2e2t + (1− e2)t
)dt
u1(x) = −1
2(1− e2)x
and
u2(x) = −1
4(1− e2)x
and so on. Accordingly, we find
u(x) = 2e2x + (1− e2)x− 1
2(1− e2)x
(1 +
1
2+
1
4+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = 2e2x
10. We setu0(x) = 1 + sec2xHence, we find
u1(x) = −∫ π
4
0
(1 + sec2t
)dt
u1(x) = −(1 +π
4)
and
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March 4, 2015 14:44 book-9x6 9571-Root page 19
2.2. Adomian Decomposition Method 19
u2(x) = (1 +π
4)π
4and so on. Accordingly, we find
u(x) = 1 + sec2x− (1 +π
4)
(1− π
4+π2
16+ · · ·
)And by evaluating the infinite geometric series at the right hand side, wefindu(x) = sec2x
11. We setu0(x) = sinxHence, we find
u1(x) = esin−1x
∫ 1
−1sin tdt
u1(x) = 0andun(x) = 0, for n ≥ 1Accordingly, we findu(x) = sinx
12. We setu0(x) = tanxHence, we find
u1(x) = −etan−1x
∫ π3
−π3tan t dt
u1(x) = 0andun(x) = 0, for n ≥ 1Accordingly, we findu(x) = tanx
13. We setu0(x) = tan−1xHence, we find
u1(x) =1
2(ln 2− π
2)x+ x
∫ 1
0
tan−1dt
u1(x) =1
2(ln 2− π
2)x+ x[t(tan−1t)− 1
2ln(1 + t2)]10
and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = tan−1x
14. We set
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March 4, 2015 14:44 book-9x6 9571-Root page 20
20 Chapter 2. Fredholm Integral Equations
u0(x) = coshxHence, we findu1(x) = (sinh 1)x+ (e−1 − 1)−
∫ 1
0(x− t) cosh tdt
u1(x) = (sinh 1)x+ (e−1 − 1)− [x sinh t− t sinh t+ cosh t]10and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = coshx
15. We set
u0(x) =1
1 + x2Hence, we find
u1(x) = (eπ4 − e−π4 )x− x
∫ 1
−1
1
1 + t2etan
−1tdt
Using integration by substitution, we set v = tan−1t, we obtainu1(x) = (e
π4 − e−π4 )x− x[etan
−1t]1−1and, henceun(x) = 0, for n ≥ 1Accordingly, we find
u(x) =1
1 + x2
16. We set
u0(x) =1√
1− x2Hence, we find
u1(x) = (eπ6 − 1)x− x
∫ 12
0
1√1− t2
esin−1tdt
Using integration by substitution, we set v = sin−1t, we obtain u1(x) =
(eπ6 − 1)x− x[esin
−1t]120
and, henceun(x) = 0, for n ≥ 1accordingly, we find
u(x) =1√
1− x2
17. We set
u0(x) =1
1 + x2Hence, we find
u1(x) =π2
32x− x
∫ 1
0
1
1 + t2tan−1t dt
Using integration by substitution, we set v = tan−1t, we obtain
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March 4, 2015 14:44 book-9x6 9571-Root page 21
2.2. Adomian Decomposition Method 21
u1(x) =π2
32x− x[
(tan−1t)2
2]10
and, henceun(x) = 0, for n ≥ 1Accordingly, we find
u(x) =1
1 + x2
18. We setu0(x) = cos−1xHence, we find
u1(x) = −πx+ x
∫ 1
−1cos−1tdt
u1(x) = −πx+ x[t(cos−1t)−√
1− t2]1−1and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = cos−1x
19. We setu0(x) = xtan−1xHence, we find
u1(x) = (π
4− 1
2)x− x
∫ 1
0
t tan−1dt
u1(x) = (π
4− 1
2)x− 1
2x[(1 + t2)tan−1t− t]10
and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = xtan−1x
20. We setu0(x) = xsin−1x+ 1Hence, we find
u1(x) = −(π
8+ 1)x+ x
∫ 1
0
(tsin−1t+ 1)dt
u1(x) = (π
8+ 1)x+
1
4x[(2t2 − 1)sin−1t+ t
√1− t2]10
and, henceun(x) = 0, for n ≥ 1Accordingly, we findu(x) = xsin−1x+ 1
21. We set
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March 4, 2015 14:44 book-9x6 9571-Root page 22
22 Chapter 2. Fredholm Integral Equations
u0(x) = sin x1+sin x + x− π
2x
u1(x) =∫ π
2
0xu0(t) dt = −x+ π
2x+ π2
8 x−π3
16xBy cancelling the noise terms, we findu(x) = sin x
1+sin x
22. We setu0(x) = sin x
1+cos x − x ln 2
u1(x) =∫ π
2
0xu0(t) dt = x ln 2− π2
8 x ln 2By cancelling the noise terms, we findu(x) = sin x
1+cos x
23. We setu0(x) = sec2 x
1+tan x − x ln 2
u1(x) =∫ π
4
0xu0(t) dt = x ln 2− π2
32x ln 2By cancelling the noise terms, we findu(x) = sec2 x
1+tan x
24. We setu0(x) = 1 + sinx− x− π2
8 x
u1(x) =∫ π
2
0xtu0(t) dt = x+ π2
8 x+ · · ·By cancelling the noise terms, we findu(x) = 1 + sinx
25. We setu0(x) = 1 + sinx+ cosx− 2x− π
2x+ π2 + π2
8
u1(x) =∫ π
2
0(x− t)u0(t) dt = 2x+ π
2x−π2 −
π2
8 + · · ·By cancelling the noise terms, we findu(x) = 1 + sinx+ cosx
26. We setu0(x) = x sinx− x− 2 + π
u1(x) =∫ π
2
0(x− t)u0(t) dt = x+ 2− π + · · ·
By cancelling the noise terms, we findu(x) = x sinx
2.3 The Variational Iteration Method
Exercises 2.3
1. Differentiating both sides givesu′(x) = 3x2 − 1
5 +∫ 1
0tu(t) dt, u(0) = 0
u0(x) = 0
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March 4, 2015 14:44 book-9x6 9571-Root page 23
2.3. The Variational Iteration Method 23
u1(x) = x3 − 15x
u2(x) = x3 − 115x
u3(x) = x3 − 145x
...un(x) = x3 − 1
5···3n−1x, n ≥ 1u(x) = x3
2. Differentiating both sides givesu′(x) = ex − 1 +
∫ 1
0tu(t) dt, u(0) = 1
u0(x) = 1u1(x) = ex − 1
2x
u2(x) = ex − 16x
u3(x) = ex − 118x
...un(x) = ex − 1
2···3n−1x, n ≥ 1u(x) = ex
3. Differentiating both sides givesu′(x) = 2
3 +∫ 1
0tu(t) dt, u(0) = 0
u0(x) = 0u1(x) = 2
3x
u2(x) = 89x
u3(x) = 2627x
...un(x) = 3n−1
3n x, n ≥ 1u(x) = x
4. Differentiating both sides givesu′(x) = 4x3 + 2x− 5
12 +∫ 1
0tu(t) dt, u(0) = 0
u0(x) = 0u1(x) = x2 + x4 − 5
12x
u2(x) = x2 + x4 − 536x
u3(x) = x2 + x4 − 5108x
...un(x) = x2 + x4 5
12·3n−1x, n ≥ 1u(x) = x2 + x4
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March 12, 2015 12:19 book-9x6 9571-Root page 24
24 Chapter 2. Fredholm Integral Equations
5. Differentiating both sides givesu′(x) = ex + 2− 3
4 +∫ 1
0tu(t) dt, u(0) = 1
u0(x) = 1u1(x) = ex + 13
8 x
u2(x) = ex + 2732x
u3(x) = ex + 133128x
...u(x) = ex
6. Differentiating both sides givesu′(x) = −e−x + 2 + 3
2 +∫ 0
−1 tu(t) dt, u(0) = 1u0(x) = 1u1(x) = e−x + 5
4x
u2(x) = e−x + 98x
u3(x) = e−x + 1716x
...u(x) = e−x
7. Differentiating both sides givesu′(x) = 1− 1
6x+ 2∫ 1
0tu(t) dt, u(0) = 1
u0(x) = 1u1(x) = 1 + x+ 5
12x2
u2(x) = 1 + x+ 4148x
2
u3(x) = 1 + x+ 185192x
2
...u(x) = 1 + x+ x2
8. Differentiating both sides givesu′(x) = ex − 2x+ 2x
∫ 1
0tu(t) dt, u(0) = 1
u0(x) = 1u1(x) = ex − 1
2x2
u2(x) = ex − 18x
2
u3(x) = ex − 132x
2
...un(x) = ex − 1
2·4n−1x2, n ≥ 1 u(x) = ex
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March 4, 2015 14:44 book-9x6 9571-Root page 25
2.4. The Direct Computation Method 25
2.4 The Direct Computation Method
Exercises 2.4
1. We setu(x) = xex + (α− 1)xwhere
α =
∫ 1
0
u(t) dt
Accordingly, we find
α =
∫ 1
0
(tet + (α− 1)t
)dt
This givesα = 1Substituting for α in the first equation we findu(x) = xex
2. We set
u(x) = x2 − 25
12x+ 1 + αx
where
α =
∫ 1
0
tu(t) dt
Accordingly, we find
α =
∫ 1
0
t
(t2 − 25
12t+ 1 + αt
)dt
This gives
α =1
12Substituting for α in the first equation we findu(x) = x2 − 2x+ 1
3. We setu(x) = x sinx+ (α− 1)xwhereα =
∫ π2
0u(t) dt
Accordingly, we find
α =
∫ π/2
0
(t sin t+ (α− 1)t) dt
This givesα = 1Substituting for α in the first equation we findu(x) = x sinx
4. We set
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March 4, 2015 14:44 book-9x6 9571-Root page 26
26 Chapter 2. Fredholm Integral Equations
u(x) = e2x − 1
4(e2 + 1)x+ αx
where
α =
∫ 1
0
tu(t) dt
Accordingly, we find
α =
∫ 1
0
t
(e2t − 1
4(e2 + 1)t+ αt
)dt
This gives
α =1
4(e2 + 1)
Substituting for α in the first equation we findu(x) = e2x
5. We setu(x) = sec2x+ (α− π
4 )where
α =
∫ π4
0
u(t) dt
Accordingly, we find
α =
∫ π4
0
(sec2t+ (α− π
4
)dt
This givesα = 1 + π
4
Substituting for α in the first equation we findu(x) = 1 + sec2x
6. We set
u(x) = sin 2x+ (α− 1
2)x
where
α =
∫ π4
0
u(t) dt
Accordingly, we find
α =
∫ π4
0
(sin 2t+ (α− 1
2)t
)dt
This gives
α =1
2Substituting for α in the first equation we findu(x) = sin 2x
7. We set
u(x) = x2 − 1
3x− 1
4+ α(x+ 2)
where
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March 4, 2015 14:44 book-9x6 9571-Root page 27
2.4. The Direct Computation Method 27
α =
∫ 1
0
u(t) dt
Accordingly, we find
α =
∫ 1
0
(t2 − 1
3t− 1
4+ α(t+ 2)
)dt
This gives
α =1
18Substituting for α in the first equation we find
u(x) = x2 − 5
18x− 5
36
8. We setu(x) = sinx+ cosx+ (α− π
2)x
where
α =
∫ π2
0
tu(t) dt
Accordingly, we find
α =
∫ π2
0
t(
sin t+ cos t+ (α− π
2)t)dt
This gives
α =π
2Substituting for α in the first equation we findu(x) = sinx+ cosx
9. We setu(x) = secx tanx+ (1− α)xwhere
α =
∫ π3
0
u(t) dt
Accordingly, we find
α =
∫ π3
0
(sec t tan t+ (1− α)t) dt
This givesα = 1Substituting for α in the first equation we findu(x) = secx tanx
10. We set
u(x) = x2 − 1
6x− 1
24+
1
2α(1 + x)− 1
2β
where
α =
∫ 1
0
u(t) dt
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March 4, 2015 14:44 book-9x6 9571-Root page 28
28 Chapter 2. Fredholm Integral Equations
β =
∫ 1
0
tu(t) dt
Accordingly, we find
α =
∫ 1
0
(t2 − 1
6t− 1
24+
1
2α(1 + t)− 1
2β
)dt
β =
∫ 1
0
t
(t2 − 1
6t− 1
24+
1
2α(1 + t)− 1
2β
)dt
This gives
α =1
3
β =1
4Substituting for α and β in the first equation we findu(x) = x2
11. We set
u(x) = sinx+1
4(α− 1)x
where
α =
∫ π2
0
tu(t) dt
Accordingly, we find
α =
∫ π2
0
t
(sin t+
1
4(α− 1)t
)dt
This givesα = 1Substituting for α in the first equation we findu(x) = sinx
12. We setu(x) = 1 + α lnx+ βwhere
α =
∫ 1
0+u(t) dt
β =
∫ 1
0+ln t u(t) dt
Accordingly, we find
α =
∫ 1
0+(1 + α ln t+ β) dt
β =
∫ 1
0+ln t (1 + α ln t+ β) dt
This gives
α =1
2
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March 4, 2015 14:44 book-9x6 9571-Root page 29
2.5. Successive Approximations Method 29
β = 0Substituting for α and β in the first equation we find
u(x) = 1 +1
2lnx
13. u(x) = x3
14. u(x) = 1 +π
4sec2x
2.5 Successive Approximations Method
Exercises 2.5
1. We selectu0(x) = 0Substituting in the original equation we find
u1(x) =11
12x
Substituting u1(x) in the original equation we obtain
u2(x) =11
12x+
1
4
∫ 1
0
11
12xt2dt
so that
u2(x) =122 − 1
122x
Proceeding as before we find
u3(x) =123 − 1
123x
...
un(x) =12n − 1
12nx
Hence, u(x) = limn→∞ un(x) = x
2. We selectu0(x) = 0Substituting in the original equation we find
u1(x) =6
7x3
Substituting u1(x) in the original equation we obtain
u2(x) =6
7x3 +
5
7
∫ 1
0
6
7x3t4dt
so that
u2(x) =72 − 1
72x3
Proceeding as before we find
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March 4, 2015 14:44 book-9x6 9571-Root page 30
30 Chapter 2. Fredholm Integral Equations
u3(x) =73 − 1
73x
...
un(x) =7n − 1
7nx3
Hence, u(x) = limn→∞ un(x) = x3
3. We selectu0(x) = 0Substituting in the original equation we find
u1(x) =13
3x
u1(x) in the original equation we obtain
u2(x) =13
3x− 1
4
∫ 1
0
13
3xt2dt
so that
u2(x) =143
3× 12x
Proceeding as before we find
u3(x) =1729
3× 122x
...
un(x) = 4x+(−1)n+1
3× 12n−1x, n ≥ 1
Hence, u(x) = limn→∞ un(x) = 4x
4. We selectu0(x) = 0Substituting in the original equation we findu1(x) = 1Substituting u1(x) in the original equation we obtain
u2(x) = 1 +
∫ 1
0
xdt
so thatu2(x) = 1 + xProceeding as before we find
u3(x) = 1 +3
2x
u4(x) = 1 +7
4x
...
un(x) = 1 +2n − 2
2n−1x, n ≥ 1
Hence, u(x) = limn→∞ un(x) = 1 + 2x
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March 4, 2015 14:44 book-9x6 9571-Root page 31
2.5. Successive Approximations Method 31
5. We selectu0(x) = 0Substituting in the original equation we findu1(x) = sinxSubstituting u1(x) in the original equation we obtain
u2(x) = sinx+
∫ π2
0
sinx sin t cos tdt
so that
u2(x) =3
2sinx
Proceeding as before we find
u3(x) =7
4sinx
...
un(x) =2n − 1
2n−1sinx, n ≥ 1
Hence, u(x) = limn→∞ un(x) = 2 sinx
6. We selectu0(x) = 0Substituting in the original equation we find
u1(x) = −1
2+ sec2x
Substituting u1(x) in the original equation we obtain
u2(x) = −1
2+ sec2x+
1
2
∫ π4
0
(−1
2+ sec2t)dt
so thatu2(x) = − π
16+ sec2x
Proceeding as before we find
u3(x) = − π2
128+ sec2x
...
un(x) = − πn−1
2× 8n−1+ sec2x, n ≥ 1
Hence, u(x) = limn→∞ un(x) = sec2x
7. We selectu0(x) = 0Substituting in the original equation we find
u1(x) = −1
4+ secx tanx
u1(x) in the original equation we obtain
u2(x) = −1
4+ secx tanx+
1
4
∫ π3
0
(−1
4+ sec t tan t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 32
32 Chapter 2. Fredholm Integral Equations
so thatu2(x) = − π
4× 12+ secx tanx
Proceeding as before we find
u3(x) = − π2
4× 122+ secx tanx
...
un(x) = − πn−1
4× 12n−1+ secx tanx
Hence, u(x) = limn→∞ un(x) = secx tanx
8. We selectu0(x) = 0Substituting in the original equation we findu1(x) = coshx+ (1− e−1)xSubstituting u1(x) in the original equation we obtain
u2(x) = coshx+ (1− e−1)x−∫ 1
0
xt(cosh t+ (1− e−1)t)dt
so that
u2(x) = coshx− 1
3(1− e−1)x
Proceeding as before we find
u3(x) = coshx+1
9(1− e−1)x
...
un(x) = coshx+(−1)n+1
3n−1(1− e−1)x, n ≥ 1
Hence, u(x) = limn→∞ un(x) = coshx
9. We selectu0(x) = 0Substituting in the original equation we findu1(x) = ex − (sinh 1)xSubstituting u1(x) in the original equation we obtain
u2(x) = ex − (sinh 1)x+1
2
∫ 1
−1x(et − (sinh 1)t)dt
so thatu2(x) = ex
Proceeding as before we findu3(x) = ex
...un(x) = ex
Hence, u(x) = limn→∞ un(x) = ex
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March 4, 2015 14:44 book-9x6 9571-Root page 33
2.6. Successive Substitutions Method 33
10. We selectu0(x) = 0Substituting in the original equation we findu1(x) = 1
4x+ sinxSubstituting u1(x) in the original equation we obtain
u2(x) = 14x+ sinx− x
∫ π2
0
(1
4t+ sin t)dt
so that
u2(x) = sinx− π2
128x
Proceeding as before we find
u3(x) = sinx+π4
128× 32x
...
un(x) = sinx+ (−1)n+1 π2n−2
2n+1 × (4)2n−2x, n ≥ 1
Hence, u(x) = limn→∞ un(x) = sinx
2.6 Successive Substitutions Method
Exercises 2.6
1. Using the successive substitutions method, and noting that f(x) =116 x, λ = 1
4 , K(x, t) = xt, we find
u(x) =11
6x+
1
4
∫ 1
0
11
6xt2dt+
1
16
∫ 1
0
∫ 1
0
11
6xt1
2t2 dt1dt+ · · ·so that
u(x) =11
6x+
11
72x+
11
864x+ · · ·
=11
6x[1 +
1
12+
1
144+ · · ·] = 2x
by finding the sum of the infinite geometric series.
2. Using the successive substitutions method, and noting thatf(x) = 1, λ = − 1
4 , K(x, t) = cosx, we find
u(x) = 1− 1
4
∫ π2
0
cosxdt+1
16
∫ π2
0
∫ π2
0
cosx cos t dt1dt+ · · ·so thatu(x) = 1− π
8cosx+
π
32cosx+ · · ·
= 1− π
8cosx[1− 1
4+
1
16+ · · ·]
= 1− π
10cosx
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March 4, 2015 14:44 book-9x6 9571-Root page 34
34 Chapter 2. Fredholm Integral Equations
by finding the sum of the infinite geometric series.
3. Using the successive substitutions method, and noting thatf(x) = 7
12x+ 1, λ = 12 , K(x, t) = xt, we find
u(x) =7
12x+ 1 +
1
2
∫ 1
0
xt(7
12t+ 1)dt+
1
4
∫ 1
0
∫ 1
0
xt2t21(7
12t1 + 1) dt1dt+ · · ·
so that
u(x) = 1 +7
12x+
25
72x
(1 +
1
6+
1
36+ · · ·
)u(x) = 1 +
7
12x+
5
12x = 1 + x
4. Using the successive substitutions method, and noting thatf(x) = cosx, λ = 1
2 , K(x, t) = sinx, we find
u(x) = cosx+1
2
∫ π2
0
sinx cos tdt+1
4
∫ π2
0
∫ π2
0
sinx sin t cos t1 dt1dt+ · · ·so that
u(x) = cosx+1
2sinx+
1
4sinx+
1
8sinx+ · · ·
= cosx+1
2sinx[1 +
1
2+
1
4+ · · ·] = cosx+ sinx
5. Using the successive substitutions method, and noting thatf(x) = 7
8x2, λ = 1
2 , K(x, t) = x2t, we find
u(x) =7
8x2 +
1
2
∫ 1
0
x2t(7
8t2)dt+
1
4
∫ 1
0
∫ 1
0
7
8x2t3t31 dt1dt+ · · ·
so that
u(x) =7
8x2 +
7
64x2 +
7
512x2 + · · ·
u(x) =7
8x2(
1 +1
8+
1
64+ · · ·
)= x2
6. Using the successive substitutions method, and noting thatf(x) = 9
10x3, λ = 1
2 , K(x, t) = x3t, we find
u(x) =9
10x3 +
1
2
∫ 1
0
9
10x3t4dt+
1
4
∫ 1
0
∫ 1
0
9
10x3t4t41 dt1dt+ · · ·
so that
u(x) =9
10x3 +
9
100x3 +
9
1000x3 + · · ·
u(x) =9
10x3(
1 +1
10+
1
100+ · · ·
)= x3
7. Using the successive substitutions method, and noting thatf(x) = sinx, λ = 1
2 , K(x, t) = cosx, we find
u(x) = sinx+1
2
∫ π2
0
cosx sin tdt+1
4
∫ π2
0
∫ π2
0
cosx cos t sin t1 dt1dt+ · · ·
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March 4, 2015 14:44 book-9x6 9571-Root page 35
2.8. Homogeneous Fredholm Equation 35
so that
u(x) = sinx+1
2cosx+
1
4cosx+
1
8cosx+ · · ·
= sinx+1
2cosx[1 +
1
2+
1
4+ · · ·] = sinx+ cosx
8. Using the successive substitutions method, and noting thatf(x) = 1, λ = 1
2 , K(x, t) = sinx, we find
u(x) = 1 +1
2
∫ π2
0
sinxdt+1
4
∫ π2
0
∫ π2
0
sinx sin t dt1dt+ · · ·so thatu(x) = 1 +
π
4sinx+
π
8sinx+
π
16sinx+ · · ·
= 1 +π
4sinx[1 +
1
2+
1
4+ · · ·]
= 1 + π2 sinx
9. Using the successive substitutions method, and noting thatf(x) = 1, λ = 1
2 , K(x, t) = sec2x, we find
u(x) = 1 +1
2
∫ π4
0
sec2xdt+1
4
∫ π4
0
∫ π4
0
sec2x sec2t dt1dt+ · · ·so thatu(x) = 1 +
π
8sec2x+
π
16sec2x+ · · ·
= 1 +π
8sec2x+ [1 +
1
2+
1
4+ · · ·]
= 1 +π
4sec2x
10. Using the successive substitutions method, and noting thatf(x) = 1, λ = 1
5 , K(x, t) = secx tanx, we find
u(x) = 1 +1
5
∫ π3
0
secx tanxdt+1
25
∫ π3
0
∫ π3
0
secx tanx sec t tan t dt1dt+ · · ·so thatu(x) = 1 +
π
15secx tanx+
π
75secx tanx+ · · ·
u(x) = 1 +π
15secx tanx
(1 +
1
5+
1
25+ · · ·
)u(x) = 1 +
π
12secx tanx
2.8 Homogeneous Fredholm Equations
Exercises 2.8
1. Using the direct computation method we find
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March 4, 2015 14:44 book-9x6 9571-Root page 36
36 Chapter 2. Fredholm Integral Equations
u(x) = 2αλwhere
α =
∫ 1
0
tu(t)dt
Substituting for u(t) from the above equation we find
α =
∫ 1
0
2αλt dt
Integrating and solving for λ where α is a constant we obtainλ = 1Hence, u(x) = 2α
2. Using the direct computation method we findu(x) = 4αλxwhere
α =
∫ 1
0
u(t)dt
Substituting for u(t) from the above equation we find
α =
∫ 1
0
4αλt dt
Integrating and solving for λ where α is a constant we obtain
λ =1
2Hence, u(x) = 2αx
3. Using the direct computation method we findu(x) = αλxwhere
α =
∫ 1
0
et u(t)dt
Substituting for u(t) from the above equation we find
α =
∫ 1
0
αλtet dt
Integrating and solving for λ where α is a constant we obtainλ = 1Hence, u(x) = αx
4. Using the direct computation method we findu(x) = αλ cosxwhere
α =
∫ π2
0
sin t u(t)dt
Substituting for u(t) from the above equation we find
α =
∫ π2
0
αλ sin t cos tdt
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March 4, 2015 14:44 book-9x6 9571-Root page 37
2.8. Homogeneous Fredholm Equation 37
Integrating and solving for λ where α is a constant we obtainλ = 2u(x) = 2α cosx
5. Expanding sin(x+ t) yields
u(x) =2
πλ
∫ π
0
[sinx cos t+ cosx sin t]u(t)dt
Using the direct computation method we find
u(x) =2
πλ[α sinx+ β cosx]
where
α =
∫ π
0
cos tu(t)dt
β =
∫ π
0
sin tu(t)dt
Substituting for u(t) from the above equation we find
α =2
πλ
∫ π
0
cos t[α sin t+ β cos t]dt
β =2
πλ
∫ π
0
sin t[α sin t+ β cos t]dt
Integrating and solving for λ where α and β are constants we obtainα = λββ = λαHence, we findλ = ±1β = ±αAccordingly, u(x) = ± 2
πα(sinx± cosx)
6. Expanding cos(x− t) yields
u(x) =2
πλ
∫ π
0
[cosx cos t+ sinx sin t]u(t)dt
Using the direct computation method we find
u(x) =2
πλ[α cosx+ β sinx]
where
α =
∫ π
0
cos tu(t)dt
β = λ
∫ π
0
sin tu(t)dt
Substituting for u(t) from the above equation we find
α =2
πλ
∫ π
0
cos t[α cos t+ β sin t]dt
β =2
πλ
∫ π
0
sin t[α cos t+ β sin t]dt
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March 4, 2015 14:44 book-9x6 9571-Root page 38
38 Chapter 2. Fredholm Integral Equations
Integrating and solving for λ where α and β are constants we obtainα = λαβ = λβHence, we findλ1 = λ2 = 1
Accordingly, u(x) =2
π(α sinx+ β cosx)
7. Using the direct computation method we findu(x) = αλ secxwhere
α =
∫ π3
0
tan tu(t)dt
Substituting for u(t) from the above equation we find
α =
∫ π3
0
αλ sec t tan t dt
Integrating and solving for λ where α is a constant we obtainλ = 1Hence we find u(x) = α secx
8. Using the direct computation method we findu(x) = αλsec2xwhere
α =
∫ π4
0
u(t)dt
Substituting for u(t) from the above equation we find
α =∫ π
4
0αλsec2t dt
Integrating and solving for λ where α is a constant we obtainλ = 1Hence we find u(x) = αsec2x
9. Using the direct computation method we findu(x) = αλsin−1xwhere
α =
∫ 1
0
u(t)dt
Substituting for u(t) from the above equation we find
α =
∫ 1
0
αλsin−1t dt
Integrating and solving for λ where α is a constant we obtain
λ =2
π − 2
Hence we find u(x) =2
π − 2αsin−1x
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March 4, 2015 14:44 book-9x6 9571-Root page 39
2.9. Fredholm Integral Equation of the First Kind 39
10. Using the direct computation method we find
u(x) = αλ(3− 3
2x)
where
α =
∫ 1
0
t u(t)dt
Substituting for u(t) from the above equation we find
α =
∫ 1
0
αλ(3t− 3
2t2) dt
Integrating and solving for λ where α is a constant we obtainλ = 1
Hence we find u(x) = α(3− 3
2x)
2.9 Fredholm Integral Equation of the First
Kind
Exercises 2.9
1. 13x =
∫ 1
0xtu(t)dt
Using the method of regularization, we obtain
uε(x) = 13εx−
xε
∫ 12
0tuε(t) dt
Using the direct computation method we find
uε(x) = ( 13ε −
αε )x
whereα =
∫ 1
0tuε(t) dt
This in turn givesα = 1
3+9ε
u(x) = limε→0 uε(x) = x
2. 34x =
∫ 1
0xt2u(t)dt
Using the method of regularization, we obtainuε(x) = 3
4εx−xε
∫ 1
0t2uε(t) dt
Using the direct computation method we finduε(x) = ( 3
4ε −αε )x
whereα =
∫ 1
0t2uε(t) dt
This in turn givesα = 1
4+16ε
u(x) = limε→0 uε(x) = 3x
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March 4, 2015 14:44 book-9x6 9571-Root page 40
40 Chapter 2. Fredholm Integral Equations
3. 12e−x =
∫ 12
0et−xu(t)dt
Using the method of regularization, we obtain
uε(x) = 12εx−
xε
∫ 12
0et−xuε(t) dt
Using the direct computation method we finduε(x) = ( 1
2ε −αε )e−x
whereα =
∫ 12
0exp−tuε(t) dt
This in turn givesα = 1
2+4ε
u(x) = limε→0 uε(x) = e−x
4. 25x
2 =∫ 1
−1 x2t2u(t)dt
Using the method of regularization, we obtainuε(x) = 2
5εx2 − x2
ε
∫−1 x
2t2uε(t) dtProceeding as before we findα = 4
10+25ε
u(x) = limε→0 uε(x) = x2
5. π2 cosx =∫ π0
cos(x− t)u(t)dtUsing the method of regularization, we obtainuε(x) = π
2ε cosx− απε cosx− β
ε sinxwhereα =
∫ π0
cos(t)uε(t) dtβ =
∫ π0
sin(t)uε(t) dtThis in turn gives
α =π2
2π + 4ε, β = 0
u(x) = limε→0 uε(x) = cosx
6. − π2 cosx =
∫ π0
sin(x− t)u(t)dtUsing the method of regularization, we obtainuε(x) = − π
2ε cosx− απε sinx+ β
ε cosxwhereα =
∫ π0
cos(t)uε(t) dt
β =∫ π0
sin(t)uε(t) dtThis in turn gives
α = − π2ε
π2 + 4ε2, β =
π3
2π2 + 8ε2u(x) = limε→0 uε(x) = sinx
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March 4, 2015 14:44 book-9x6 9571-Root page 41
Chapter 3
Volterra IntegralEquations
3.2 Adomian Decomposition Method
Exercises 3.2
1. Using the Adomian decomposition method, we set
u0(x) = 4x+ 2x2
Hence, we find
u1(x) = −∫ x
0
(4t+ 2t2)dt
u1(x) = −2x2 − 2
3x3
and
u2(x) =2
3x3 +
1
6x4
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
41
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March 4, 2015 14:44 book-9x6 9571-Root page 42
42 Chapter 3. Volterra Integral Equations
u(x) = 4x+ 2x2 − 2x2 − 2
3x3 +
2
3x3 + · · ·
u(x) = 4x
2. Using the Adomian decomposition method, we set
u0(x) = 1 + x− x2
Hence, we find
u1(x) =
∫ x
0
(1 + t− t2)dt
u1(x) = x+1
2x2 − 1
3x3
and
u2(x) =1
2x2 +
1
6x3 − 1
12x4
u3(x) =1
6x3 +
1
24x4 − 1
60x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1 + 2x
3. Using the Adomian decomposition method, we set
u0(x) = 1
Hence, we find
u1(x) = −∫ x
0
dt
u1(x) = −x
and
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March 4, 2015 14:44 book-9x6 9571-Root page 43
3.2. Adomian Decomposition Method 43
u2(x) =1
2!x2
u3(x) = − 1
3!x3
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1− x+1
2!x2 − 1
3!x3 + · · ·
u(x) = e−x
4. Using the Adomian decomposition method, we set
u0(x) = x
Hence, we find
u1(x) =
∫ x
0
t(x− t)dt
u1(x) =1
3!x3
and
u2(x) =1
5!x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = x+1
3!x3 +
1
5!x5 + · · ·
u(x) = sinhx
5. Using the Adomian decomposition method, we set
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March 4, 2015 14:44 book-9x6 9571-Root page 44
44 Chapter 3. Volterra Integral Equations
u0(x) = 3x
Hence, we find
u1(x) = −9
∫ x
0
3t(x− t)dt
u1(x) = − 1
3!(3x)3
and
u2(x) =1
5!(3x)5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 3x− 1
3!(3x)3 +
1
5!(3x)5 + · · ·
u(x) = sin 3x
6. Using the Adomian decomposition method, we set
u0(x) = 1
Hence, we find
u1(x) = −4
∫ x
0
(x− t)dt
u1(x) = − 1
2!(2x)2
and
u2(x) =1
4!(2x)4
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
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March 4, 2015 14:44 book-9x6 9571-Root page 45
3.2. Adomian Decomposition Method 45
Accordingly, we obtain
u(x) = 1− 1
2!(2x)2 +
1
4!(2x)4 + · · ·
u(x) = cos 2x
7. Using the Adomian decomposition method, we set
u0(x) = 1 + x
Hence, we find
u1(x) = −∫ x
0
(1 + t)(x− t)dt
u1(x) = − 1
2!x2 − 1
3!x3
and
u2(x) =1
4!x4 +
1
5!x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) =
(1− 1
2!x2 +
1
4!x4 + · · ·
)+
(x− 1
3!x3 +
1
5!x5 + · · ·
)u(x) = cosx+ sinx
8. Using the Adomian decomposition method, we set
u0(x) = 1− x
Hence, we find
u1(x) = −∫ x
0
(1− t)(x− t)dt
u1(x) = − 1
2!x2 +
1
3!x3
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March 4, 2015 14:44 book-9x6 9571-Root page 46
46 Chapter 3. Volterra Integral Equations
and
u2(x) =1
4!x4 − 1
5!x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) =
(1− 1
2!x2 +
1
4!x4 + · · ·
)−(x− 1
3!x3 +
1
5!x5 + · · ·
)u(x) = cosx− sinx
9. Using the Adomian decomposition method, we set
u0(x) = 1 + x
Hence, we find
u1(x) =
∫ x
0
(1 + t)(x− t)dt
u1(x) =1
2!x2 +
1
3!x3
and
u2(x) =1
4!x4 +
1
5!x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1 + x+1
2!x2 +
1
3!x3 + · · ·
u(x) = ex
10. Using the Adomian decomposition method, we set
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March 4, 2015 14:44 book-9x6 9571-Root page 47
3.2. Adomian Decomposition Method 47
u0(x) = 1− x
Hence, we find
u1(x) =
∫ x
0
(1− t)(x− t)dt
u1(x) =1
2!x2 − 1
3!x3
and
u2(x) =1
4!x4 − 1
5!x5
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1− x+1
2!x2 − 1
3!x3 + · · ·
u(x) = e−x
11. Using the Adomian decomposition method, we set
u0(x) = 2
Hence, we find
u1(x) =
∫ x
0
2(x− t)dt
u1(x) = 2
(1
2!x2)
and
u2(x) = 2
(1
4!x4)
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
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March 4, 2015 14:44 book-9x6 9571-Root page 48
48 Chapter 3. Volterra Integral Equations
Accordingly, we obtain
u(x) = 2
(1 +
1
2!x2 +
1
4!x4 + · · ·
)u(x) = 2 coshx
12. Using the Adomian decomposition method, we set
u0(x) = 1 + x
Hence, we find
u1(x) =
∫ x
0
(1 + t)dt
u1(x) = x+1
2!x2
and
u2(x) =1
2!x2 +
1
3!x3
u3(x) =1
3!x3 +
1
4!x4
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1 + 2
(x+
1
2!x2 +
1
3!x3 + · · ·
)u(x) = 2ex − 1
13. Using the Adomian decomposition method, we set
u0(x) = 1− 1
2x2
Hence, we find
u1(x) = −∫ x
0
(1− 1
2t2)(x− t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 49
3.2. Adomian Decomposition Method 49
u1(x) = − 1
2!x2 +
1
4!x4
and
u2(x) = +1
4!x4 +
1
6!x6
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1 + 2
(− 1
2!x2 +
1
4!x4 + · · ·
)u(x) = 2 cosx− 1
14. Using the Adomian decomposition method, we set
u0(x) = 1 +1
2x2
Hence, we find
u1(x) =
∫ x
0
(1 +1
2t2)(x− t)dt
u1(x) =1
2!x2 +
1
4!x4
and
u2(x) =1
4!x4 +
1
6!x6
and so on. Substitute the components obtained in the decomposition
u(x) = u0(x) + u1(x) + u2(x) + · · ·
Accordingly, we obtain
u(x) = 1 + 2
(1
2!x2 +
1
4!x4 + · · ·
)u(x) = 2 coshx− 1
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50 Chapter 3. Volterra Integral Equations
15. Using the modified decomposition method we set
u0(x) = cosx
Hence, we find
u1(x) = (1− esin x)x+ x
∫ x
0
esin t cos tdt
Using the substitution v = sin t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = cosx
16. Using the modified decomposition method we set
u0(x) = sec2x
Hence, we find
u1(x) = (1− etan x)x+ x
∫ x
0
etan tsec2tdt
Using the substitution v = tan t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = sec2x
17. Using the modified decomposition method we set
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3.2. Adomian Decomposition Method 51
u0(x) = coshx
Hence, we find
u1(x) =x
2(1− esinh x) +
x
2
∫ x
0
esinh t cosh tdt
Using the substitution v = sinh t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = coshx
18. Using the modified decomposition method we set
u0(x) = sinhx
Hence, we find
u1(x) =1
10(e− ecosh x) +
1
10
∫ x
0
ecosh t sinh tdt
Using the substitution v = cosh t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = sinhx
19. Using the modified decomposition method we set
u0(x) = x3
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52 Chapter 3. Volterra Integral Equations
Hence, we find
u1(x) = −x5 + 5
∫ x
0
t4dt
Using the substitution v = cosh t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = x3
20. Using the modified decomposition method we set
u0(x) = secx tanx
Hence, we find
u1(x) = (e− esec x) +
∫ x
0
esec t sec t tan tdt
Using the substitution v = sec t, we obtain
u1(x) = 0
and consequently
uk(x) = 0 for k ≥ 1
Accordingly, we find
u(x) = secx tanx
21. Using the noise terms phenomenon we set
u0(x) = 8x− 4x3
Hence, we find
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3.2. Adomian Decomposition Method 53
u1(x) = 4x3 − x5
By cancelling 4x3, we find
u(x) = 8x
22. Using the noise terms phenomenon we set
u0(x) = 8x2 − 2x5
Hence, we find
u1(x) = 2x5 − 27x
8
By cancelling 2x5, we find
u(x) = 8x2
23. Using the noise terms phenomenon we set
u0(x) = sec2 x− tanx
Hence, we find
u1(x) = tanx+ ln(cosx)
By cancelling tanx, we find
u(x) = sec2 x
24. Using the noise terms phenomenon we set
u0(x) = 1 + x+ x2 − 14x
4 − 15x
5 − 16x
6
Hence, we find
u1(x) = 14x
4 + 15x
5 + 16x
6 + other terms
By cancelling the noise terms, we find
u(x) = 1 + x+ x2
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54 Chapter 3. Volterra Integral Equations
25. Using the noise terms phenomenon we set
u0(x) = x sinx+ x cosx− sinx
Hence, we find
u1(x) = sinx− x cosx + other terms
By cancelling the noise terms, we find
u(x) = x sinx
26. Using the noise terms phenomenon we set
u0(x) = cosh2 x− 14 sinh(2x)− 1
2x
Hence, we find
u1(x) = 14 sinh(2x) + 1
2x + other terms
By cancelling the noise terms, we find
u(x) = cosh2 x
3.3 The Variational Iteration Method
Exercises 3.3
1. Differentiating both sides with respect to x givesu′(x) = u(x)
Using the correction functional givesu0(x) = 1,
u1(x) = 1−∫ x0
(u′
0(t)− u0(t))dt
= 1 + x,
u2(x) = 1 + x−∫ x0
(u′
1(t)− u1(t))dt
= 1 + x+ 12!x
2
u3(x) = 1 + x+ 12!x
2 + 13!x
3,and so on. The solution in a series form is given by
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3.3. The Variational Iteration Method 55
u(x) = 1 + x+ 12!x
2 + 13!x
3 + · · ·
that converges to the exact solution
u(x) = ex.
2. Differentiating both sides with respect to x givesu′(x) = 1 +
∫ x0u(t) dt
Using the correction functional givesu0(x) = 0,
u1(x) = 0−∫ x0
(u′
0(t)− 1−∫ t0u0(r) dr
)dt
= x,
u2(x) = x−∫ x0
(u′
1(t)− 1−∫ t0u1(r) dr
)dt
= x+ 13!x
3
u3(x) = x+ 13!x
3 + 15!x
5,
and so on. The solution in a series form is given byu(x) = x+ 1
3!x3 + 1
5!x5 + · · ·
that converges to the exact solutionu(x) = sinhx.
3. Differentiating both sides with respect to x givesu′(x) = 3− 9
∫ x0u(t) dt
Using the correction functional givesu0(x) = 0,
u1(x) = 0−∫ x0
(u′
0(t)− 3 + 9∫ t0u0(r) dr
)dt
= 3x,
u2(x) = 3x−∫ x0
(u′
1(t)− 3 + 9∫ t0u1(r) dr
)dt
= 3x− 13! (3x)3
u3(x) = (3x)x− 13! (3x)3 + 1
5! (3x)5,and so on. The solution in a series form is given byu(x) = (3x)x− 1
3! (3x)3 + 15! (3x)5 + · · ·
that converges to the exact solutionu(x) = sin(3x).
4. Differentiating both sides with respect to x gives
u′(x) = −4∫ x0u(t) dt
Using the correction functional gives
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56 Chapter 3. Volterra Integral Equations
u0(x) = 1,
u1(x) = 1−∫ x0
(u′
0(t) + 4∫ t0u0(r) dr
)dt
= 1− 2x2,
u2(x) = 1− 2x2 +∫ x0
(u′
1(t) + 4∫ t0u1(r) dr
)dt
= 1− 12! (2x)2 + 1
4! (2x)4
and so on. The solution in a series form is given by
u(x) = 1− 12! (2x)2 + 1
4! (2x)4 + · · ·that converges to the exact solutionu(x) = cos(2x).
5. Using the correction functional givesu0(x) = 1,u1(x) = 1 + x− 1
2!x2
u2(x) = 1 + x− 12!x
2 − 13!x
3 + 14!x
4
and so on. The solution in a series form is given by
u(x) = (x− 13!x
3 + 15!x
5 + · · ·) + (1− 12! (2x)2 + 1
4! (2x)4 + · · ·)that converges to the exact solutionu(x) = sinx+ cosx.
6. Using the correction functional givesu0(x) = 1,u1(x) = 1− x− 1
2!x2
u2(x) = 1− x− 12!x
2 + 13!x
3 + 14!x
4
and so on. The solution in a series form is given by
u(x) = (1− 12! (2x)2 + 1
4! (2x)4 + · · ·)− (x− 13!x
3 + 15!x
5 + · · ·)that converges to the exact solutionu(x) = cosx− sinx.
7. Differentiating both sides with respect to x gives
u′(x) = −x+ cosx+ sinx+∫ x0
(x− t)u(t) dt
Using the correction functional gives
un+1(x) = un(x)−∫ x0
(u′
n(t) + t− cost− sin t−∫ r0
(t− r)un(r) dr)dt
This givesu0(x) = 0,u1(x) = xu2(x) = x− 1
3!x3 + 1
5!x5
and so on. The solution in a series form is given by
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3.4. The Series Solution Method 57
u(x) = (x− 13!x
3 + 15!x
5 + · · ·)that converges to the exact solutionu(x) = sinx.
8. Using the correction functional givesu0(x) = 1,u1(x) = 1 + 1
2!x2
u2(x) = 1 + 12!x
2 + 14!x
4
and so on. The solution in a series form is given by
u(x) = 1 + 12!x
2 + 14!x
4 + · · ·that converges to the exact solutionu(x) = coshx.
3.4 The Series Solution Method
Exercises 3.4
1. Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = 2x+ 2x2 − x3 +
∫ x
0
( ∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side that involve termsof the form tn, n ≥ 0 yields
a0+a1x+a2x2+a3x
3+ · · · = (a0+2)x+(2+ 12a1)x2+( 1
3a2−1)x3+ · · ·Equating the coefficients of like powers of x in both sides we finda0 = 0a1 = 2a2 = 3ak = 0 for k ≥ 3Accordingly, we findu(x) = 2x+ 3x2
2. Substituting u(x) by the series
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58 Chapter 3. Volterra Integral Equations
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = 1 + x− 2
3x3 − 1
2x4 + 2
∫ x
0
t
( ∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 yields
a0 + a1x+ a2x2 + a3x
3 + · · · = 1 + x+ a0x2 + ( 2
3a1 −23 )x3
+( 12a2 −
12 )x4 + · · ·
Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1a2 = 1ak = 0 for k ≥ 3
Accordingly we find
u(x) = 1 + x+ x2
3. Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation and using the Taylor expansion for sinx,lead to
∞∑n=0
anxn = 1 + 2x− 2
3!x3 −
∫ x
0
( ∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side that involve termsof the form tn, n ≥ 0 yields
a0 + a1x+ a2x2 + a3x
3 + · · · = 1 + (2− a0)x− 1
2a1x
2
−(1
3a2 +
2
3!)x3 − 1
4a3x
4 · · ·
Equating the coefficients of like powers of x in both sides we find
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3.4. The Series Solution Method 59
a0 = 1a1 = 1
a2 = − 1
2!
a3 = − 1
3!
a4 =1
4!
Accordingly, we find
u(x) =
(1− 1
2!x2 +
1
4!x4 − · · ·
)+
(x− 1
3!x3 +
1
5!x5 − · · ·
)and in a closed form
u(x) = cosx+ sinx
4. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = 1 + x+
1
2!x2 +
1
3!x3 −
∫ x
0
((x− t)
∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0 + a1x+ a2x2 + a3x
3 + · · · = 1 + x+ (1
2− 1
2a0)x2
+(1
6− 1
6a1)x3 − 1
12a2x
4 + · · ·
Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1a2 = 0a3 = 0ak = 0 for k ≥ 3
Accordingly, we find
u(x) = 1 + x
5. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = −1−
∫ x
0
( ∞∑n=0
antn
)dt
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60 Chapter 3. Volterra Integral Equations
Evaluating the regular integrals at the right hand side as discussed abovegives
a0 +a1x+a2x2 +a3x
3 + · · · = −1−a0x−1
2a1x
2− 1
3a2x
3− 1
4a3x
4−· · ·
Equating the coefficients of like powers of x in both sides we finda0 = −1a1 = 1
a2 = − 1
2!
a3 =1
3!
a4 = − 1
4!
Accordingly, we find
u(x) = −(
1− x+1
2!x2 − 1
3!x3 + · · ·
)and in a closed form
u(x) = −e−x
6. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = 1− 2
∫ x
0
( ∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0 + a1x+ a2x2 + a3x
3 + · · · = 1− 2a0x− a1x2 −2
3a2x
3 − 1
2a3x
4 − · · ·Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = −2a2 = 2
a3 = −4
3...Accordingly, we find
u(x) = 1− (2x) +(2x)2
2!− (2x)3
3!+ · · ·
and in a closed form
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3.4. The Series Solution Method 61
u(x) = e−2x
7. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = 1 + xex −
∫ x
0
(t
∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0+a1x+a2x2+a3x
3+· · · = 1+x+(1− 1
2a0)x2+(
1
2!− 1
3a1)x3−· · ·
Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 1
a2 =1
2!
a3 =1
3!...
Accordingly, we find
u(x) = 1 + x+1
2!x2 +
1
3!x3 + · · ·
and in a closed formu(x) = ex
8. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = x+
∫ x
0
((x− t)
∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0 + a1x+ a2x2 + a3x
3 + · · · = x+1
2a0x
2 +1
6a1x
3 +1
12a2x
4 + · · ·
Equating the coefficients of like powers of x in both sides we finda0 = 0a1 = 1a2 = 0
a3 =1
3!...
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62 Chapter 3. Volterra Integral Equations
Accordingly, we find
u(x) = sinhx
9. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = 1− 1
2x2 −
∫ x
0
((x− t)
∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0+a1x+a2x2+a3x
3+· · · = 1−(1
2+
1
2a0)x2− 1
6a1x
3− 1
12a2x
4+· · ·
Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = 0a2 = −1a3 = 0
a4 =1
12...
Accordingly, we find
u(x) = 2
(1− 1
2!x2 +
1
4!x4 − · · ·
)− 1
and in a closed form
u(x) = 2 cosx− 1
10. Substituting u(x) by the series as shown above yields
∞∑n=0
anxn = 1− x−
∫ x
0
((x− t)
∞∑n=0
antn
)dt
Evaluating the regular integrals at the right hand side as discussed abovegives
a0+a1x+a2x2+a3x
3+ · · · = 1−x− 1
2a0x
2− 1
6a1x
3− 1
12a2x
4+ · · ·
Equating the coefficients of like powers of x in both sides we finda0 = 1a1 = −1
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3.5. Converting Volterra Equation to IVP 63
a2 = − 1
2!
a3 =1
3!
a4 =1
4!...
Accordingly, we find
u(x) =
(1− 1
2!x2 +
1
4!x4 − · · ·
)−(x− 1
3!x3 +
1
5!x5 − · · ·
)and in a closed form
u(x) = cosx− sinx
11. We findu(x) = sinhx12. We findu(x) = sinx
3.5 Converting Volterra Equation to IVP
Exercises 3.5
1. Differentiate both sides using Leibniz rule we obtain
u′(x) = −3u(x)
The equivalent initial value problem is then
u′(x) + 3u(x) = 0, u(0) = 1
Solving this equation and using the initial condition, we find
u(x) = e−3x
2. Differentiate both sides twice using Leibniz rule we obtain
u′(x) =
∫ x
0
u(t) dt
u′′(x) = u(x)
The equivalent initial value problem is then
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64 Chapter 3. Volterra Integral Equations
u′′(x)− u(x) = 0, u(0) = 1, u
′(0) = 0
The characteristic equation is
r2 − 1 = 0Solving this equation and using the initial conditions, we find
u(x) = coshx
3. Differentiate both sides twice using Leibniz rule we obtain
u′(x) = −1−
∫ x
0
u(t) dt
u′′(x) = −u(x)
The equivalent initial value problem is then
u′′(x) + u(x) = 0, u(0) = 1, u
′(0) = −1
The characteristic equation is
r2 + 1 = 0Solving this equation and using the initial conditions, we find
u(x) = cosx− sinx
4. Differentiate both sides using Leibniz rule we obtain
u′(x) = 1 + u(x)
The equivalent initial value problem is then
u′(x)− u(x) = 1, u(0) = 0
Solving this equation and using the initial conditions, we find
u(x) = ex − 1
5. Differentiate both sides twice using Leibniz rule we obtain
u′(x) = 1 +
∫ x
0
u(t) dt
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March 4, 2015 14:44 book-9x6 9571-Root page 65
3.5. Converting Volterra Equation to IVP 65
u′′(x) = u(x)
The equivalent initial value problem is then
u′′(x)− u(x) = 0, u(0) = 1, u
′(0) = 1
The characteristic equation is
r2 − 1 = 0Solving this equation and using the initial conditions, we find
u(x) = ex
6. Differentiate both sides four times using Leibniz rule we obtain
u′(x) =
1
2
∫ x
0
(x− t)2u(t) dt
u′′(x) =
∫ x
0
(x− t)u(t)dt
u′′′
(x) =
∫ x
0
u(t)dt
uiv(x) = u(x)The equivalent initial value problem is thenuiv(x)− u(x) = 0, u(0) = 1, u
′(0) = u
′′(0) = u
′′′(0) = 0
The characteristic equation isr4 − 1 = 0 with roots given by r = ±1, ±i
Accordingly we findu(x) = A cosx+B sinx+ C coshx+D sinhxUsing the initial conditions yields
u(x) =1
2(cosx+ coshx)
7. Differentiate both sides four times using Leibniz rule we obtain
u′(x) = 1 +
1
2
∫ x
0
(x− t)2u(t) dt
u′′(x) =
∫ x
0
(x− t)u(t)dt
u′′′
(x) =
∫ x
0
u(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 66
66 Chapter 3. Volterra Integral Equations
uiv(x) = u(x)The equivalent initial value problem is then
uiv(x)− u(x) = 0, u(0) = u′′(0) = u
′′′(0) = 0, u
′(0) = 1
The characteristic equation is
r4 − 1 = 0 with roots given by r = ±1, ±iAccordingly we find
u(x) = A cosx+B sinx+ C coshx+D sinhxUsing the initial conditions yields
u(x) = 12 (sinx+ sinhx)
8. Differentiate both sides twice using Leibniz rule we obtain
u′(x) = 2x+
∫ x
0
u(t) dt
u′′(x) = 2 + u(x)
The equivalent initial value problem is thenu′′(x)− u(x) = 2, u(0) = 0, u
′(0) = 0
The characteristic equation of the homogeneous part isr2 − 1 = 0This gives uc(x) = A coshx+B sinhxThe particular solution is obtained by substituting up(x) = C where weobtain up(x) = −2
Using the initial conditions in u(x) = uc(x) + up(x),we find
u(x) = 2 coshx− 2
9. Differentiate both sides twice using Leibniz rule we obtain
u′(x) = 1 +
1
2x2 −
∫ x
0
u(t) dt
u′′(x) = x− u(x)
The equivalent initial value problem is then
u′′(x) + u(x) = x, u(0) = 0, u
′(0) = 1
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March 12, 2015 12:19 book-9x6 9571-Root page 67
3.6. Successive Approximations Method 67
The characteristic equation of the homogeneous part is
r2 + 1 = 0
This gives uc(x) = A cosx+B sinx
The particular solution is obtained by substituting
up(x) = C +Dx , where we obtain up(x) = x
Using the initial conditions in u(x) = uc(x) + up(x), we find
u(x) = x
10. Differentiate both sides twice using Leibniz rule we obtain
u′(x) = 1− 2x+
1
2x2 − 1
3x3 −
∫ x
0
u(t) dt
u′′(x) = −2 + x− x2 − u(x)
The equivalent initial value problem is then
u′′(x) + u(x) = −2 + x− x2, u(0) = 0, u
′(0) = 1
The characteristic equation of the homogeneous part is
r2 + 1 = 0This gives uc(x) = A cosx+B sinxThe particular solution is obtained by substituting
up(x) = C +Dx+ Ex2, where we obtain up(x) = x− x2Using the initial conditions in u(x) = uc(x) + up(x), we find
u(x) = x− x2
3.6 Successive Approximations Method
Exercises 3.6
1. We select
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March 4, 2015 14:44 book-9x6 9571-Root page 68
68 Chapter 3. Volterra Integral Equations
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1
Substituting u1(x) in the original equation we obtain
u2(x) = 1−∫ x
0
dt
so that
u2(x) = 1− x
Proceeding as before we find
u3(x) = 1− x+1
2!x2
u4(x) = 1− x+1
2!x2 − 1
3!x3
...
Accordingly,
u(x) = limn→∞ un(x) = e−x
2. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1
Substituting u1(x) in the original equation we obtain
u2(x) = 1− 9
∫ x
0
(x− t)dt
so that
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March 4, 2015 14:44 book-9x6 9571-Root page 69
3.6. Successive Approximations Method 69
u2(x) = 1− (3x)2
2!
Proceeding as before we find
u3(x) = 1− (3x)2
2!+
(3x)4
4!
...
Accordingly,
u(x) = limn→∞ un(x) = cos(3x)
3. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1 + 2x
Substituting u1(x) in the original equation we obtain
u2(x) = 1 + 2x+ 4
∫ x
0
(x− t)(1 + 2t)dt
so that
u2(x) = 1 + 2x+(2x)2
2!+
(2x)3
3!
...
Accordingly,
u(x) = limn→∞ un(x) = e2x
4. We select
u0(x) = 0
Substituting in the original equation, we find
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March 4, 2015 14:44 book-9x6 9571-Root page 70
70 Chapter 3. Volterra Integral Equations
u1(x) = 1− 1
4x
Substituting u1(x) in the original equation we obtain
u2(x) = 1− 1
4x+
1
16
∫ x
0
(x− t)(1− 1
4t)dt
so that
u2(x) = 1− 1
4x+
( 14x)2
2!−
( 14x)3
3!
...
Accordingly,
u(x) = limn→∞ un(x) = e−14x
5. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 2
Substituting u1(x) in the original equation we obtain
u2(x) = 2− 2
∫ x
0
(x− t)dt
so that
u2(x) = 2(1− 1
2!x2)
u3(x) = 2(1− 1
2!x2 +
1
4!x4)
...
Accordingly,
u(x) = limn→∞ un(x) = 2 cosx
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March 4, 2015 14:44 book-9x6 9571-Root page 71
3.6. Successive Approximations Method 71
6. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1
Substituting u1(x) in the original equation we obtain
u2(x) = 1− 2
∫ x
0
tdt
so that
u2(x) = 1− x2
u3(x) = 1− x2 +1
2!x4
...
Accordingly,
u(x) = limn→∞ un(x) = e−x2
7. We select
u0(x) = x
Substituting in the original equation, we find
u1(x) = x+1
3!x3
Substituting u1(x) in the original equation we obtain
u2(x) = x+
∫ x
0
(x− t)(t+t3
3!)dt
so that
u2(x) = x+1
3!x3 +
1
5!x5
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March 4, 2015 14:44 book-9x6 9571-Root page 72
72 Chapter 3. Volterra Integral Equations
...
Accordingly,
u(x) = limn→∞ un(x) = sinhx
8. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1
Substituting u1(x) in the original equation we obtain
u2(x) = 1−∫ x
0
(x− t)dt
so that
u2(x) = 1− 1
2!x2
u3(x) = 1− 1
2!x2 +
1
4!x4
...
Accordingly,
u(x) = limn→∞ un(x) = cosx
9. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1 + x
Substituting u1(x) in the original equation we obtain
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March 4, 2015 14:44 book-9x6 9571-Root page 73
3.6. Successive Approximations Method 73
u2(x) = 1 + x−∫ x
0
(x− t)(1 + t)dt
so that
u2(x) = 1 + x− 1
2!x2 − 1
3!x3
u3(x) = 1 + x− 1
2!x2 − 1
3!x3 +
1
4!x4 +
1
5!x5
u3(x) =
(1− 1
2!x2 +
1
4!x4 − · · ·
)+
(x− 1
3!x3 +
1
5!x5 − · · ·
)Accordingly,
u(x) = limn→∞ un(x) = cosx+ sinx
10. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1− x
Substituting u1(x) in the original equation we obtain
u2(x) = 1− x−∫ x
0
(x− t)(1− t)dt
so that
u2(x) = 1− x− 1
2!x2 +
1
3!x3
u3(x) = 1− x− 1
2!x2 +
1
3!x3 +
1
4!x4 − 1
5!x5
u3(x) =(1− 1
2!x2 + 1
4!x4 − · · ·
)−(x− 1
3!x3 + 1
5!x5 − · · ·
)Accordingly,
u(x) = limn→∞ un(x) = cosx− sinx
11. We select
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March 4, 2015 14:44 book-9x6 9571-Root page 74
74 Chapter 3. Volterra Integral Equations
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 2− x
Substituting u1(x) in the original equation we obtain
u2(x) = 2− x+
∫ x
0
(2− t)dt
so that
u2(x) = 2 + x− 1
2!x2
u3(x) = 2 + x+1
2!x2 − 1
3!x3
u4(x) = 2 + x+1
2!x2 +
1
3!x3 − 1
4!x4
Accordingly,
u(x) = limn→∞ un(x) = 1 + ex
12. We select
u0(x) = 0
Substituting in the original equation, we find
u1(x) = 1− x− 1
2!x2
Substituting u1(x) in the original equation we obtain
u2(x) = 1− x− 1
2!x2 +
∫ x
0
(x− t)(1− t− 1
2!t2)dt
so that
u2(x) = 1−(x+ 1
2!x2 + 1
3!x3)
u3(x) = 1−(x+
1
2!x2 +
1
3!x3 +
1
4!x4)
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March 4, 2015 14:44 book-9x6 9571-Root page 75
3.7. Successive Substitutions Method 75
Accordingly,
u(x) = limn→∞ un(x) = 1− sinhx
3.7 Successive Substitutions Method
Exercises 3.7
1. Substituting λ = 1, f(x) = x, and K(x, t) = 1 into the formula yields
u(x) = x+
∫ x
0
tdt+
∫ x
0
∫ t
0
t1dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = x+1
2!x2 +
1
3!x3 + · · ·
and in a closed form, u(x) is given by
u(x) = ex − 1
2. Substituting λ = 1, f(x) = 12!x
2, and K(x, t) = 1 into the formula yields
u(x) =1
2!x2 +
∫ x
0
1
2!t2dt+
∫ x
0
∫ t
0
1
2!t21dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 12!x
2 +1
3!x3 +
1
4!x4 + · · ·
and in a closed form, u(x) is given by
u(x) = ex − x− 1
3. Substituting λ = −1, f(x) =1
3!x3, and K(x, t) = (x−t) into the formula
yields
u(x) =1
3!x3−
∫ x
0
(x−t) 1
3!t3dt+
∫ x
0
∫ t
0
(x−t)(t−t1)1
3!t31dt1dt+· · ·
Accordingly we obtain u(x) in a series form
u(x) =1
3!x3 − 1
5!x5 +
1
7!x7 + · · ·
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March 4, 2015 14:44 book-9x6 9571-Root page 76
76 Chapter 3. Volterra Integral Equations
and in a closed form, u(x) is given by
u(x) = x− sinx
4. Substituting λ = 1, f(x) = 13!x
3, and K(x, t) = (x− t) into the formulayields
u(x) =1
3!x3+
∫ x
0
(x−t) 1
3!t3dt+
∫ x
0
∫ t
0
(x−t)(t−t1)1
3!t31dt1dt+· · ·
Accordingly we obtain u(x) in a series form
u(x) =1
3!x3 +
1
5!x5 +
1
7!x7 + · · ·
and in a closed form, u(x) is given by
u(x) = −x+ sinhx
5. Substituting λ = −1 , f(x) =1
2!x2, and K(x, t) = (x − t) into the
formula yields
u(x) =1
2!x2−
∫ x
0
(x−t) 1
2!t2dt+
∫ x
0
∫ t
0
(x−t)(t−t1)1
2!t21dt1dt+· · ·
Accordingly we obtain u(x) in a series form
u(x) =1
2!x2 − 1
4!x4 +
1
6!x6 + · · ·
and in a closed form, u(x) is given by
u(x) = −1− cosx
6. Substituting λ = −1, f(x) = 1− 1
2!x2, and K(x, t) = 1 into the formula
yields
u(x) = 1− 1
2!x2 −
∫ x
0
(1− 1
2!t2)dt+
∫ x
0
∫ t
0
(1− 1
2!t21)dt1dt+ · · ·
Accordingly we obtain
u(x) = 1− x+ (1
2!x2 − 1
2!x2) + (
1
3!x3 − 1
3!x3) + · · ·
Hence, u(x) is given by
u(x) = 1− x
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March 4, 2015 14:44 book-9x6 9571-Root page 77
3.7. Successive Substitutions Method 77
7. Substituting λ = 2, f(x) = 1, and K(x, t) = 1 into the formula yields
u(x) = 1 + 2
∫ x
0
dt+
∫ x
0
∫ t
0
dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 1 + 2x+(2x)2
2!+
(2x)3
3!+ · · ·
and in a closed form, u(x) is given by
u(x) = e2x
8. Substituting λ = 1, f(x) = 3−2x, andK(x, t) = 1 into the formula yields
u(x) = 3− 2x+
∫ x
0
(3− 2t) dt+
∫ x
0
∫ t
0
(3− 2t)(3− 2t1)dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 3 + x+1
2!x2 +
1
3!x3 + · · ·
and in a closed form, u(x) is given by
u(x) = 2 + ex
9. Substituting λ = −1, f(x) = 2 + 12!x
2, and K(x, t) = (x − t) into theformula yields
u(x) = 2 +1
2!x2 −
∫ x
0
(x− t) (2 +1
2!t2)dt
+
∫ x
0
∫ t
0
(x− t)(t− t1)(2 +1
2!t21)dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 2− 1
2!x2 +
1
4!x4 + · · ·
and in a closed form, u(x) is given by
u(x) = 1 + cosx
10. Substituting λ = −1, f(x) = 1 − x +1
2!x2, and K(x, t) = (x − t) into
the formula yields
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March 4, 2015 14:44 book-9x6 9571-Root page 78
78 Chapter 3. Volterra Integral Equations
u(x) = 1− x+1
2!x2 −
∫ x
0
(x− t) (1− t+1
2!t2)dt
+
∫ x
0
∫ t
0
(x− t)(t− t1)(1− t1 +1
2!t21)dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 1−(x− 1
3!x3 +
1
5!x5 + · · ·
)and in a closed form, u(x) is given by
u(x) = 1− sinx
11. Substituting λ = 1, f(x) = 2 − 1
2!x2, and K(x, t) = (x − t) into the
formula yields
u(x) = 2− 1
2!x2 −
∫ x
0
(x− t) (2− 1
2!t2)dt
+
∫ x
0
∫ t
0
(x− t)(t− t1)(2− 1
2!t21)dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) = 1 +
(1 +
1
2!x2 +
1
4!x4 + · · ·
)and in a closed form, u(x) is given byu(x) = 1 + coshx
12. Substituting λ = 1, f(x) =1
2!x2, and K(x, t) = (x− t) into the formula
yields
u(x) = 12!x
2+
∫ x
0
(x−t) (1
2!t2)dt+
∫ x
0
∫ t
0
(x−t)(t−t1)(1
2!t21)dt1dt+ · · ·
Accordingly we obtain u(x) in a series form
u(x) =
(1
2!x2 +
1
4!x4 +
1
6!x6 + · · ·
)and in a closed form, u(x) is given byu(x) = −1 + coshx
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March 4, 2015 14:44 book-9x6 9571-Root page 79
3.9. Volterra Equations of the First Kind 79
3.9 Volterra Equations of the First Kind
Exercises 3.9
1. Noting that K(x, t) = 5 + 3x− 3t, hence K(x, x) = 5 6= 0
Differentiating both sides of the equation with respect to x yields
10x+ 3x2 = 5u(x) +
∫ x
0
3u(t)dt
or equivalently
u(x) = 2x+3
5x2 − 3
5
∫ x
0
u(t)dt
Using the modified decomposition method, we set
u0(x) = 2x
which gives
u1(x) =3
5x2 − 3
5
∫ x
0
2t dt
Consequently, we find
uk(x) = 0 for k ≥ 1
Accordingly, the exact solution is
u(x) = 2x
2. Noting that K(x, t) = et−x, hence K(x, x) = 1 6= 0
Differentiating both sides of the equation with respect to x yields
e−x − xe−x = u(x)−∫ x
0
et−xu(t)dt
or equivalently
u(x) = e−x − xe−x +
∫ x
0
et−xu(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 80
80 Chapter 3. Volterra Integral Equations
Using the modified decomposition method, we set
u0(x) = e−x
which gives
u1(x) = −xe−x +
∫ x
0
e−x dt
Consequently, we find
uk(x) = 0 for k ≥ 1
Accordingly, the exact solution is
u(x) = e−x
3. Noting that K(x, t) = 1 + x− t, hence K(x, x) = 1 6= 0
Differentiating both sides of the equation with respect to x yields
2ex − 1 = u(x) +
∫ x
0
u(t)dt
or equivalently
u(x) = 2ex − 1−∫ x
0
u(t)dt
Using the modified decomposition method, we set
u0(x) = ex
which gives
u1(x) = ex − 1−∫ x
0
et dt
Consequently, we find
uk(x) = 0 for k ≥ 1
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March 4, 2015 14:44 book-9x6 9571-Root page 81
3.9. Volterra Equations of the First Kind 81
Accordingly, the exact solution is
u(x) = ex
4. Noting that K(x, t) = 2− x+ t, hence K(x, x) = 2 6= 0
Differentiating both sides of the equation with respect to x yields
2 sinhx− coshx+ 1 = 2u(x)−∫ x
0
u(t)dt
or equivalently
u(x) = sinhx− 1
2coshx+
1
2+
1
2
∫ x
0
u(t)dt
Using the modified decomposition method, we set
u0(x) = sinhx
which gives
u1(x) =1
2coshx+
1
2+
1
2
∫ x
0
sinh tdt
Consequently, we find
uk(x) = 0 for k ≥ 1
Accordingly, the exact solution is
u(x) = sinhx
5. Noting that K(x, t) = 4 + 3x− 3t, hence K(x, x) = 4 6= 0
Differentiating both sides of the equation with respect to x yields
4 cosx+ 3 sinx = 4u(x) + 3
∫ x
0
u(t)dt
or equivalently
u(x) = cosx+3
4sinx− 3
4
∫ x
0
u(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 82
82 Chapter 3. Volterra Integral Equations
Using the modified decomposition method, we set
u0(x) = cosx
which gives
u1(x) =3
4sinx− 3
4
∫ x
0
cos tdt
Consequently, we find
uk(x) = 0 for k ≥ 1
Accordingly, the exact solution is
u(x) = cosx
6. Noting that K(x, t) = 1 + x− t, hence K(x, x) = 1 6= 0, x < π2
Differentiating both sides of the equation with respect to x yields
sec2x+ tanx = u(x) +
∫ x
0
u(t)dt
or equivalently
u(x) = sec2x+ tanx−∫ x
0
u(t)dt
Using the modified decomposition method, we set
u0(x) = sec2x
which gives
u1(x) = tanx−∫ x
0
sec2tdt
Consequently, we find
uk(x) = 0 for k ≥ 1
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March 4, 2015 14:44 book-9x6 9571-Root page 83
3.9. Volterra Equations of the First Kind 83
Accordingly, the exact solution is
u(x) = sec2x
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March 4, 2015 14:44 book-9x6 9571-Root page 85
Chapter 4
FredholmIntegro-DifferentialEquations
4.3 The Direct Computation Method
Exercises 4.3
1. u′(x) = 1
6 + 536x−
∫ 1
0xtu(t)dt, u(0) = 1
6
We first set
α =
∫ 1
0
tu(t)dt
so that the given equation can be rewritten as
u′(x) =
1
6+ (
5
36− α)x
Integrating both sides from 0 to x and using the given condition gives
u(x) =1
6(1 + x) + (
5
36− α)
x2
2
Substituting this expression of u(x) into the first equation for α yields
85
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March 4, 2015 14:44 book-9x6 9571-Root page 86
86 Chapter 4. Fredholm Integro-Differential Equations
α =
∫ 1
0
t
(1
6(1 + t) + (
5
36− α)
t2
2
)dt
so that
α =5
36
This gives
u(x) =1
6(1 + x)
2. u′(x) = 1
21x−∫ 1
0xtu(t)dt, u(0) = 1
6
We first set
α =
∫ 1
0
tu(t)dt
so that the given equation can be rewritten as
u′(x) =
1
21x− αx
Integrating both sides from 0 to x and using the given condition give
u(x) =1
6+ (
1
42− α
2)x2
Substituting this expression of u(x) into the first equation for α yields
α =
∫ 1
0
t
(1
6+ (
1
42− α
2)t2)dt
so that
α =5
63
This gives
u(x) =1
6− 1
63x2
3. u′′(x) = − sinx+ x−
∫ π/20
xtu(t)dt, u(0) = 0, u′(0) = 1
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March 4, 2015 14:44 book-9x6 9571-Root page 87
4.3. The Direct Computation Method 87
We first set
α =
∫ π/2
0
tu(t)dt
so that the given equation can be rewritten as
u′′(x) = − sinx+ (1− α)x
Integrating both sides from 0 to x twice and using the given conditionsgives
u′(x) = cosx+
1− α2!
x2
and
u(x) = sinx+1− α
3!x3
Substituting this expression of u(x) into the first equation for α yields
α =
∫ π/2
0
t
(sin t+
1− α3!
t3)dt
so that
α = 1
This gives
u(x) = sinx
4. u′′(x) = 9
4 −13x+
∫ 1
0(x− t)u(t)dt, u(0) = u
′(0) = 0
We first set
α =
∫ 1
0
u(t)dt
and
β =
∫ 1
0
tu(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 88
88 Chapter 4. Fredholm Integro-Differential Equations
so that the given equation can be rewritten as
u′′(x) = 9
4 −x3 + αx− β
Integrating both sides from 0 to x twice and using the given conditionsgives
u′(x) = 9
4x−16x
2 + α2 x
2 − βx
and
u(x) = (9
8− β
2)x2 + (
α
6− 1
18)x3
Substituting this expression of u(x) into the first equation for α yields
α =1
3, β =
1
4
This gives
u(x) = x2
5. u′(x) = 2sec2x tanx− x+
∫ π/40
xu(t)dt, u(0) = 1
We first set
α =
∫ π/4
0
u(t)dt
so that the given equation can be rewritten as
u′(x) = 2sec2x tanx+ (α− 1)x
Integrating both sides from 0 to x and using the given condition gives
u(x) = sec2x+ α−12 x2
Substituting this expression of u(x) into the first equation for α yields
α = 1
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March 4, 2015 14:44 book-9x6 9571-Root page 89
4.3. The Direct Computation Method 89
This gives
u(x) = sec2x
6. Integrating both sides from 0 to x and using the given condition gives
u(x) = 1− (5 + β)x+ (α2 − 3)x2
where
α =∫ 1
−1 u(t) dt, β =∫ 1
−1 tu(t) dt
Substituting u(x) into α and β and by solving the resulting equationswe find α = 0, β = −2 hence we find
u(x) = 1− 3x− 3x2
7. Integrating both sides from 0 to x and using the given condition gives
u(x) = sinx+ (α2 −12 )x2 + (1− β)x
where
α =∫ π
2
0u(t) dt, β =
∫ π2
0tu(t) dt
Proceeding as before we find α = 1, β = 1 hence we find
u(x) = sinx
8. Integrating both sides from 0 to x and using the given condition gives
u(x) = sinx+ cosx(α2 − 1x2 + (π2 − β)xwhereα =
∫ π2
0u(t) dt, β =
∫ π2
0tu(t) dt
Proceeding as before we find α = 2, β = π2 hence we find
u(x) = sinx+ cosx
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March 4, 2015 14:44 book-9x6 9571-Root page 90
90 Chapter 4. Fredholm Integro-Differential Equations
4.4 The Adomian Decomposition Method
Exercises 4.4
1. u′(x) = sinhx+ 1
8 (1− e−1)x− 1
8x
∫ 1
0
tu(t)dt, u(0) = 1
Integrating both sides from 0 to x and using the given condition gives
u(x) = coshx+1
16(1− e−1)x2 − 1
16x2∫ 1
0
tu(t)dt
Following the Adomian decomposition method, we set
u0(x) = coshx+1
16(1− e−1)x2
u1(x) = − 116x
2
∫ 1
0
t
(cosh t+
1
16(1− e−1)t2
)dt
so that
u1(x) = − 1
16(1− e−1)x2 − 1
16× 64(1− e−1)x2
Cancelling the noise term of u0 that appears in u1, and justifying thatthe remaining non-canceled term in u0 justifies the equation, we thereforefind
u(x) = coshx
2. u′(x) = 1− 1
3x+ x
∫ 1
0
tu(t)dt, u(0) = 0
Integrating both sides from 0 to x and using the given condition gives
u(x) = x− 1
6x2 +
1
2x2∫ 1
0
tu(t)dt
Following the Adomian decomposition method, we set
u0(x) = x− 1
6x2
u1(x) = 12x
2
∫ 1
0
t
(t− 1
6t2)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 91
4.4. The Adomian Decomposition Method 91
so that
u1(x) =7
48x2
and continuing in the same manner we find
u2(x) =7
48× 8x2
u(x) = u0(x) + u1(x) + u2(x) + · · ·
u(x) = x− 16x
2 + 748x
2(1 + 1
8 + · · ·)
we therefore find
u(x) = x
3. u′(x) = xex + ex − x+ x
∫ 1
0
u(t)dt, u(0) = 0
Integrating both sides from 0 to x and using the given condition gives
u(x) = xex − 1
2x2 +
1
2x2∫ 1
0
u(t)dt
Following the Adomian decomposition method, we set
u0(x) = xex − 1
2x2
u1(x) = 512x
2
and continuing in the same manner we find
u2(x) =5
72x2
u(x) = u0(x) + u1(x) + u2(x) + · · ·
u(x) = xex − 12x
2 + 512x
2(1 + 1
6 + 136 + · · ·
)we therefore find
u(x) = xex
4. u′(x) = x cosx+ sinx− x+ x
∫ π2
0
u(t)dt, u(0) = 0
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March 4, 2015 14:44 book-9x6 9571-Root page 92
92 Chapter 4. Fredholm Integro-Differential Equations
Integrating both sides from 0 to x and using the given condition give
u(x) = x sinx− 1
2x2 +
1
2x2∫ π
2
0
u(t)dt
Following the Adomian decomposition method, we set
u0(x) = x sinx− 1
2x2
u1(x) =1
2x2
Cancelling the noise terms between u0(x) and u1(x) and justifying thatthe non-canceled term of u0(x) satisfies the integral equation gives
u(x) = x sinx
5. u′′(x) = − sinx+ x− x
∫ π2
0
tu(t)dt, u(0) = 0, u′(0) = 1
Integrating both sides from 0 to x twice and using the given conditionsgive
u(x) = sinx+1
3!x3 − 1
3!x3∫ π
2
0
tu(t)dt
Following the Adomian decomposition method, we set
u0(x) = sinx+1
3!x3
u1(x) = − 1
3!x3 − π5
160× (3!)2x3
Cancelling the noise terms between u0(x) and u1(x) and justifying thatthe non-canceled term of u0(x) satisfies the integral equation gives
u(x) = sinx
6. u′′′
(x) = 6 + x− x∫ 1
0
u′′(t)dt u(0) = −1, u
′(0) = 1, u
′′(0) = −2
Integrating both sides from 0 to x three times and using the given con-ditions give
u(x) = −1 + x− x2 + x3 + 14!x
4 − 1
4!x4∫ 1
0
u′′(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 93
4.4. The Adomian Decomposition Method 93
Following the Adomian decomposition method, we set
u0(x) = −1 + x− x2 + x3 +1
4!x4
u1(x) = − 1
4!x4 − 1
4!× 3!x4
Cancelling the noise terms between u0(x) and u1(x) and justifying thatthe non-canceled term of u0(x) satisfies the integral equation gives
u(x) = x3 − x2 + x− 1
7. u′′′
(x) = − cosx+ x+
∫ π2
0
xu′′(t)dt, u(0) = u
′′(0) = 0, u
′(0) = 1
Integrating both sides from 0 to x three times and using the given con-ditions give
u(x) = sinx+ 14!x
4 +1
4!x4∫ π
2
0
u′′(t)dt
Following the Adomian decomposition method, we set
u0(x) = sinx+1
4!x4
u1(x) = − 1
4!x4 +
π3
4!× 48x4
Cancelling the noise terms between u0(x) and u1(x) and justifying thatthe non-canceled term of u0(x) satisfies the integral equation gives
u(x) = sinx
8. Integrating both sides from 0 to x and using the given condition we find
u(x) = sinx+ cosx− x2 + π2x+
∫ π2
0(x
2
2 − xt)u(t)dt
Using the modified decomposition method, we set
u0(x) = sinx+ cosx
u1(x) = 0
u(x) = sinx+ cosx
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March 4, 2015 14:44 book-9x6 9571-Root page 94
94 Chapter 4. Fredholm Integro-Differential Equations
9. Integrating both sides from 0 to x and using the given condition we find
u(x) = cosx− x2
2 + π2x− x+
∫ π2
0(x
2
2 − xt)u(t)dt
Using the modified decomposition method, we set
u0(x) = cosx
u1(x) = 0
u(x) = cosx
10. Integrating both sides from 0 to x and using the given condition we find
u(x) = sinx− cosx+ 2xπ2x+∫ π
2
0(x
2
2 − xt)u(t)dt
Using the modified decomposition method, we set
u0(x) = sinx− cosx
u1(x) = 0
u(x) = sinx− cosx
4.5 The Variational Iteration Method
Exercises 4.5
1. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− sin t− t cos t+ 1−∫ π
2
0un(r) dr
)dt
Using u0(x) = 0 into the correction functional gives the following suc-cessive approximationsu0(x) = 0
u1(x) = u0(x)−∫ x0
(u′
0(t)− sin t− t cos t+ 1−∫ π
2
0u0(r) dr
)dt
= x sinx− xu2(x) = u1(x)−
∫ x0
(u′
1(t)− sin t− t cos t+ 1−∫ π
2
0u1(r) dr
)dt
= (x sinx− x) + (x− π2
8 x)
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March 4, 2015 14:44 book-9x6 9571-Root page 95
4.5. The Variational Iteration Method 95
u3(x) = (x sinx− x) + (x− π2
8 x) + (π2
8 x− · · ·)u(x) = x sinx
2. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− cos t+ t sin t− 2−∫ π0un(r) dr
)dt
Using u0(x) = 0 into the correction functional gives the following suc-cessive approximationsu0(x) = 0u1(x) = x cosx+ 2xu2(x) = (x cosx+ 2x) + (π2x− 2x)
u3(x) = (x cosx+ 2x) + (π2x− 2x) + (π4
2 x− π2x) + · · ·
u(x) = x cosx
3. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 2 sec2 x tanx+ π2
32 −∫ π
4
0un(r) dr
)dt
Using u0(x) = 1 into the correction functional gives the following suc-cessive approximationsu0(x) = 1
u1(x) = sec2 x+ π4x−
π2
32x
u2(x) = sec2 x+ (x+ π4x)− (π4x+ π2
32x)u3(x) = sec2 x+ x+ · · ·u(x) = x+ sec2 x
4. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 3 + 12t−∫ 1
0run(r) dr
)dt
Using u0(x) = 1 into the correction functional gives the following suc-cessive approximationsu0(x) = 1u1(x) = 1− 6x2 + 7
2x
u2(x) = 1− 6x2 + 196 x
u3(x) = 1− 6x2 + 5518x
u(x) = 1 + 3x− 6x2
5. The correction functional for this equation is given by
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March 4, 2015 14:44 book-9x6 9571-Root page 96
96 Chapter 4. Fredholm Integro-Differential Equations
un+1(x) = un(x)−∫ x0
(u′
n(t)− et + 1−∫ 1
0run(r) dr
)dt
Using u0(x) = 1 into the correction functional gives the following suc-cessive approximationsu0(x) = 1
u1(x) = ex − 12x
u2(x) = ex − 16x
u3(x) = ex − 118x
un(x) = ex − 12×3n−1x
u(x) = ex
6. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 1912 − 2t+ 3t2 −
∫ 1
0un(r) dr
)dt
Using u0(x) = 1 into the correction functional gives the following suc-cessive approximationsu0(x) = 1
u1(x) = 1 + x2 − x3 − 712x
u2(x) = 1 + x2 − x3 − 1924x
u3(x) = 1 + x2 − x3 − 4348x
u(x) = 1− x+ x2 − x3
4.6 Converting to Fredholm Integral
Equations
Exercises 4.6
1. u′(x) = −x sinx+ cosx+ (1− π
2 )x+
∫ π2
0
xu(t)dt, u(0) = 0
Integrating both sides from 0 to x and using the given condition give
u(x) = x cosx+ (1− π2 )x
2
2 + x2
2
∫ π2
0
u(t)dt
or equivalently
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March 4, 2015 14:44 book-9x6 9571-Root page 97
4.6. Converting to Fredholm Integral Equations 97
u(x) = x cosx+ (1− π
2)x2
2+ α
x2
2
where
α =
∫ π2
0
u(t)dt
Substituting for u(x) from the above equation yields
α = −(1− π2 )
so that
u(x) = x cosx
2. u′′(x) = −ex + 1
2x+
∫ 1
0
xtu(t)dt, u(0) = 0, u′(0) = −1
Integrating both sides from 0 to x twice and using the given conditionsgive
u(x) = −ex + 1 + x3
12 + x3
6
∫ 1
0
tu(t)dt
or equivalently
u(x) = −ex + 1 + x3
12 + αx3
6
where
α =
∫ 1
0
tu(t)dt
Substituting for u(x) from the above equation yields
α = − 12
so that
u(x) = 1− ex
3. u′′(x) = − sinx+ cosx+ (2− π
2)x−
∫ π2
0
xtu(t)dt,
u(0) = −1, u′(0) = 1
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March 4, 2015 14:44 book-9x6 9571-Root page 98
98 Chapter 4. Fredholm Integro-Differential Equations
Integrating both sides from 0 to x twice and using the given conditionsgive
u(x) = sinx− cosx+ (2− π
2)x3
6− x3
6
∫ π2
0
tu(t)dt,
or equivalently
u(x) = sinx− cosx+ (2− π
2)x3
6− x3
6α,
where
α =
∫ π2
0
tu(t)dt,
Substituting for u(x) from the above equation yields
α = (2− π
2)
so that
u(x) = sinx− cosx
4. u′(x) =
7
6− 11x+
∫ 1
0
(x− t)u(t)dt, u(0) = 0
Integrating both sides from 0 to x and using the given condition give
u(x) = 76x−
112 x
2 + 12x
2
∫ 1
0
u(t)dt− x∫ 1
0
tu(t)dt,
or equivalently
u(x) = 76x−
112 x
2 + α 12x
2 − xβ,
where
α =
∫ 1
0
u(t)dt,
and
β =
∫ 1
0
tu(t)dt,
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March 4, 2015 14:44 book-9x6 9571-Root page 99
4.6. Converting to Fredholm Integral Equations 99
Substituting for u(x) from the above equation yields
α = −1
β = − 56
so that
u(x) = 2x− 6x2
5. u′(x) = 1
4x+ cos 2x−∫ π
4
0
xu(t)dt, u(0) = 0
Integrating both sides from 0 to x and using the given condition give
u(x) = 18x
2 + 12 sin 2x− 1
2x2
∫ π4
0
u(t)dt,
or equivalently
u(x) = 18x
2 + 12 sin 2x− 1
2αx2
where
α =
∫ π4
0
u(t)dt
Substituting for u(x) from the above equation yields
α = 14
so that
u(x) = 12 sinx
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March 4, 2015 14:44 book-9x6 9571-Root page 101
Chapter 5
VolterraIntegro-DifferentialEquations
5.3 The Series Solution Method
Exercises 5.3
1. u′(x) = 1− 2x sinx+
∫ x
0
u(t)dt, u(0) = 0
Using the series method, noting that a0 = 0 by using the given condition,we findu(x) = a1x+ a2x
2 + a3x3 + a4x
4 + · · ·and henceu′(x) = a1 + 2a2x+ 3a3x
2 + 4a4x3 + · · ·
Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = 1, a2 = 0, a3 = − 1
2! , a4 = 0, a5 = 14!
Accordingly, u(x) in a series form is given byu(x) = x(1− 1
2!x2 + 1
4!x4 − · · ·)
and in a closed formu(x) = x cosx
2. u′(x) = −1 + 1
2x2 − xex −
∫ x
0
tu(t)dt, u(0) = 0
Using the series method, noting that a0 = 0 by using the given condition,
101
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March 4, 2015 14:44 book-9x6 9571-Root page 102
102 Chapter 5. Volterra Integro-Differential Equations
we findu(x) = a1x+ a2x
2 + a3x3 + a4x
4 + · · ·and henceu′(x) = a1 + 2a2x+ 3a3x
2 + 4a4x3 + · · ·
Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = −1, a2 = − 1
2! , a3 = − 13! , a4 = − 1
4!
Accordingly, u(x) in a series form is given byu(x) = −(x+ 1
2!x2 + 1
3!x3 + 1
4!x4 + · · ·)
and in a closed formu(x) = 1− ex
3. u′′(x) = 1− x(cosx+ sinx)−
∫ x
0
tu(t)dt, u(0) = −1, u′(0) = 1
Using the series method, noting that a0 = −1, a1 = 1 by using the givenconditions, we findu(x) = −1 + x+ a2x
2 + a3x3 + a4x
4 + · · ·and henceu′(x) = 1 + 2a2x+ 3a3x
2 + 4a4x3 + · · ·
u′′(x) = 2a2 + 6a3x+ 12a4x
2 + 20a5x3 + · · ·
Substituting into both sides of the equation, using the Taylor expansions ofsinx and cosx, integrating the right hand side and equating the coefficientsof like terms of x in both sides givea0 = −1, a1 = 1, a2 = 1
2! , a3 = − 13! , a4 = − 1
4!
Accordingly, u(x) in a series form is given byu(x) = −(1− 1
2!x2 + 1
4!x4 − · · ·) + (x− 1
3!x3 + 1
5!x5 + · · ·)
and in a closed formu(x) = − cosx+ sinx
4. u′′(x) = −8− 1
3 (x3 − x4) +
∫ x
0
(x− t)u(t)dt, u(0) = 0, u′(0) = 2
Using the series method, noting that a0 = 0, a1 = 2 by using the givenconditions, we findu(x) = 2x+ a2x
2 + a3x3 + a4x
4 + · · ·and henceu′(x) = 2 + 2a2x+ 3a3x
2 + 4a4x3 + · · ·
u′′(x) = 2a2 + 6a3x+ 12a4x
2 + 20a5x3 + · · ·
Substituting into both sides of the equation, integrating the right hand sideand equating the coefficients of like terms of x in both sides givea0 = 0, a1 = 2, a2 = −4, a3 = a4 = · · · = 0Accordingly, u(x) is given byu(x) = 2x− 4x2
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March 4, 2015 14:44 book-9x6 9571-Root page 103
5.4. The Adomian Decomposition Method 103
5. u′′(x) = 1
2x2 − x coshx−
∫ x
0
tu(t)dt, u(0) = 1, u′(0) = −1
Using the series method, noting that a0 = 1, a1 = −1 by using the givenconditions, we findu(x) = 1− x+ a2x
2 + a3x3 + a4x
4 + · · ·and henceu′(x) = −1 + 2a2x+ 3a3x
2 + 4a4x3 + · · ·
u′′(x) = 2a2 + 6a3x+ 12a4x
2 + 20a5x3 + · · ·
Substituting into both sides of the equation, using the Taylor expansions ofcoshx, integrating the right hand side and equating the coefficients of liketerms of x in both sides givea0 = 1, a1 = −1, a2 = 0, a3 = − 1
3! , a4 = 0, a5 = − 15!
Accordingly, u(x) in a series form is given byu(x) = 1− (x+ 1
3!x3 + 1
5!x5 + · · ·)
and in a closed formu(x) = 1− sinhx
5.4 The Adomian Decomposition Method
Exercises 5.4
1. u′′(x) = 1 + x− 1
3!x3 +
∫ x
0
(x− t)u(t)dt, u(0) = 1, u′(0) = 2
Applying L−1 to both sides and using the initial conditions we findu(x) = 1 + 2x+ 1
2!x2 + 1
3!x3 − 1
5!x5 + L−1
(∫ x0
(x− t)u(t)dt)
whereL−1(.) =
∫ x0
∫ x0
(.)dtdtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtain
u0(x) = 1 + 2x+1
2!x2 +
1
3!x3 − 1
5!x5
u1(x) = L−1(∫ x
0
(x− t)(1 + 2t+1
2!t2 +
1
3!t3 − 1
5!t5)dt
)so thatu1(x) = 1
4!x4 + 1
60x5 + 1
6!x6 + · · ·
Accordingly u(x) in a series form is given by
u(x) = x+
(1 + x+
1
2!x2 +
1
3!x3 +
1
4!x4 +
1
5!x5 + · · ·
)and in a closed formu(x) = x+ ex
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March 12, 2015 12:19 book-9x6 9571-Root page 104
104 Chapter 5. Volterra Integro-Differential Equations
2. u′′(x) = −1− 1
2!x2 +
∫ x
0
(x− t)u(t)dt, u(0) = 2, u′(0) = 0
Applying L−1 to both sides and using the initial conditions we findu(x) = 2− 1
2!x2 − 1
4!x4 + L−1
(∫ x0
(x− t)u(t)dt)
whereL−1(.) =
∫ x0
∫ x0
(.)dtdtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 2− 1
2!x2 − 1
4!x4
u1(x) = L−1(∫ x
0
(x− t)(2− 1
2!t2 − 1
4!t4)dt
)so thatu1(x) = 1
12x4 − 1
6!x6 − 1
8!x8
Accordingly u(x) in a series form is given by
u(x) = 1 +
(1− 1
2!x2 +
1
4!x4 − 1
6!x6 + · · ·
)and in a closed formu(x) = 1 + cosx
3. u′(x) = 2 +
∫ x
0
u(t)dt, u(0) = 2
Applying L−1 to both sides and using the initial conditions we findu(x) = 2 + 2x+ L−1
(∫ x0u(t)dt
)whereL−1(.) =
∫ x0
(.)dtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 2 + 2xu1(x) = x2 + 1
3x3
Accordingly u(x) in a series form is given by
u(x) = 2
(1 + x+
1
2!x2 +
1
3!x3 + · · ·
)and in a closed formu(x) = 2ex
4. u′(x) = 1−
∫ x
0
u(t)dt, u(0) = 1
Applying L−1 to both sides and using the initial conditions we findu(x) = 1 + x− L−1
(∫ x0u(t)dt
)where
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March 4, 2015 14:44 book-9x6 9571-Root page 105
5.5. The Variational Iteration Method 105
L−1(.) =∫ x0
(.)dtUsing the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 1 + xu1(x) = − 1
2!x2 − 1
3!x3
Accordingly u(x) in a series form is given by
u(x) =
(1− 1
2!x2 +
1
4!x4 + · · ·
)+
(x− 1
3!x3 +
1
5!x5 + · · ·
)and in a closed formu(x) = cosx+ sinx
5. u(iv)(x) = −x+ 12!x
2 −∫ x
0
(x− t)u(t)dt,
u(0) = 1, u′(0) = −1, u
′′(0) = 0, u
′′′(0) = 1
Applying L−1 to both sides and using the initial conditions we findu(x) = 1− x+ 1
3!x3 − 1
5!x5 + 1
6!x6 − L−1
(∫ x0
(x− t)u(t)dt)
where
L−1(.) =
∫ x
0
∫ x
0
∫ x
0
∫ x
0
(.)dtdtdtdt
Using the Adomian decomposition method where we setu(x) = u0(x) + u1(x) + u2(x) + · · ·Hence we obtainu0(x) = 1− x+ 1
3!x3 − 1
5!x5 + 1
6!x6
u1(x) = L−1(∫ x
0
(x− t)(1− t+1
3!t3 − 1
5!t5 +
1
6!t6)dt
)so thatu1(x) = − 1
6!x6 + 1
7!x7 + · · ·
Accordingly u(x) in a series form is given by
u(x) = 1−(x− 1
3!x3 +
1
5!x5 − 1
7!x7 + · · ·
)and in a closed formu(x) = 1− sinx
5.5 The Variational Iteration Method
Exercises 5.5
1. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t) + t− 12 t
2 −∫ t0un(r) dr
)dt.
We find the following successive approximations:u0(x) = 0,
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March 4, 2015 14:44 book-9x6 9571-Root page 106
106 Chapter 5. Volterra Integro-Differential Equations
u1(x) = u0(x)−∫ x0
(u′
0(t)− 2et + 1−∫ t0u0(r) dr
)dt
= −2− x+ 2ex,
u2(x) = u1(x)−∫ x0
(u′
1(t)− 2et + 1−∫ t0u1(r) dr
)dt,
= x+ x2 + 12x
3 + 13!x
4 + 14!x
5 + · · ·,u3(x) = u2(x)−
∫ x0
(u′
2(t)− 2et + 1−∫ t0u2(r) dr
)dt
= x+ x2 + 12x
3 + 13!x
4 + 14!x
5 + 15!x
6 · · ·.This gives the exact solutionu(x) = xex.
2. The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 2 cos t+ 12 t
2 −∫ t0un(r) dr
)dt.
We find the following successive approximations:u0(x) = 0,
u1(x) = u0(x)−∫ x0
(u′
0(t)− 2 cos t+ 12 t
2 −∫ t0u0(r) dr
)dt
= 2 sinx− 13!x
3,
u2(x) = u1(x)−∫ x0
(u′
1(t)− 2 cos t+ 12 t
2 −∫ t0u1(r) dr
)dt
= 2x− 13!x
3 + 15!x
5 + · · ·.
This gives the exact solutionu(x) = x+ sinx.
3. Proceeding as before we find the following successive approximations:u0(x) = −1,
u1(x) = −1 + x+ 12x
2 − 13!x
3 − 14x
4,
u2(x) = −1 + x+ 12x
2 − 14x
4 + · · ·
This gives the exact solutionu(x) = x− cosx.
4. Proceeding as before we find the following successive approximations:u0(x) = 1,
u1(x) = 1 + x+ 12!x
2 − 13!x
3,
u2(x) = 1 + x+ 12!x
2 + 14!x
4 + 16!x
6 + · · ·.
This gives the exact solutionu(x) = x+ coshx.
5. Proceeding as before we find the following successive approximations:
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March 4, 2015 14:44 book-9x6 9571-Root page 107
5.6. Converting to Volterra Equations 107
u0(x) = 1 + x,
u1(x) = 1 + 2x+ 12!x
2 + 13!x
3 + · · ·,
u2(x) = 1 + 2x+ 12!x
2 + 13!x
3 + 14!x
4 + 15!x
5 + · · ·.
This gives the exact solutionu(x) = x+ ex.
6. Proceeding as before we find the following successive approximations:u0(x) = 1− x,
u1(x) = 1− x− 12!x
2 + 13!x
3 + 14!x
4 + · · ·,
u2(x) = 1− x− 12!x
2 + 13!x
3 + 14!x
4 − 15!x
5 + · · ·.
This gives the exact solutionu(x) = cosx− sinx.
5.6 Converting to Volterra Equations
Exercises 5.6
1. u′′(x) = 1 +
∫ x
0
(x− t)u(t)dt, u(0) = 1, u′(0) = 0
Integrating both sides twice from 0 to x, using the given conditions andconverting the resulting multiple integrals to a single integral we obtain
u(x) = 1 + 12!x
2 + 13!
∫ x0
(x− t)3u(t)dtUsing the Adomian decomposition method, (or any other method), wherewe set
u0(x) = 1 +1
2!x2
u1(x) =1
4!x4 +
1
6!x6
Accordingly u(x) in a series form is given by
u(x) = 1 +1
2!x2 +
1
4!x4 +
1
6!x6 + · · ·
and in a closed form
u(x) = coshx
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March 4, 2015 14:44 book-9x6 9571-Root page 108
108 Chapter 5. Volterra Integro-Differential Equations
2. u′(x) = 1−
∫ x
0
u(t)dt, u(0) = 0
Integrating both sides from 0 to x, using the given condition and con-verting the resulting multiple integral to a single integral we obtain
u(x) = x−∫ x0
(x− t)u(t)dtUsing the series solution method, (or any other method), where we set
u(x) = a0 + a1x+ a2x2 + · · ·
in both sides of the resulting integral equation we find
a0 = 0, a1 = 1, a2 = 0, a3 = − 13! , a4 = 0, a5 = 1
5! , · · ·
Accordingly, u(x) in a series form is given by
u(x) = x− 1
3!x3 +
1
5!x5 + · · ·
and in a closed form
u(x) = sinx
3. u′′(x) = x+
∫ x
0
(x− t)u(t)dt, u(0) = 0, u′(0) = 1
Integrating both sides twice from 0 to x, using the given conditions andconverting the resulting multiple integrals to a single integral we obtain
u(x) = x+ 13!x
3 + 13!
∫ x0
(x− t)3u(t)dtUsing the successive approximation method, (or any other method), wherewe set
u0(x) = 0
so that
u1(x) = x+1
3!x3
u2(x) = x+1
3!x3 +
1
5!x5 +
1
7!x7
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March 4, 2015 14:44 book-9x6 9571-Root page 109
5.6. Converting to Volterra Equations 109
un(x) =∑k=2nk=1
x2k−1
(2k−1)!
Accordingly u(x) in a closed form
u(x) = sinhx
4. u′(x) = 2− 1
2!x2 +
∫ x
0
u(t)dt, u(0) = 1
Integrating both sides from 0 to x, using the given condition and con-verting the resulting multiple integral to a single integral we obtain
u(x) = 1 + 2x− 13!x
3 +∫ x0
(x− t)u(t)dtUsing the series solution method, (or any other method), where we set
u(x) = a0 + a1x+ a2x2 + · · ·
in both sides of the resulting integral equation we find
a0 = 1, a1 = 2, a2 = 12! , a3 = 1
3! , a4 = 14! , a5 = 1
5! , · · ·
Accordingly, u(x) in a series form is given by
u(x) = x+(1 + x+ 1
2!x2 + 1
3!x3 + · · ·
)and in a closed form
u(x) = x+ ex
5. u′(x) = 1−
∫ x
0
u(t)dt, u(0) = 1
Integrating both sides from 0 to x, using the given condition and con-verting the resulting multiple integral to a single integral we obtain
u(x) = 1 + x−∫ x0
(x− t)u(t)dtUsing the series solution method, (or any other method), where we set
u(x) = a0 + a1x+ a2x2 + · · ·
in both sides of the resulting integral equation we find
a0 = 1, a1 = 1, a2 = − 12! , a3 = − 1
3! , a4 = 14! , a5 = 1
5! , · · ·
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110 Chapter 5. Volterra Integro-Differential Equations
Accordingly, u(x) in a series form is given by
u(x) =(1− 1
2!x2 + 1
4!x4 + · · ·
)+(x− 1
3!x3 + 1
5!x5 + · · ·
)and in a closed form
u(x) = cosx+ sinx
6. u′′(x) = 1 + x+
∫ x
0
(x− t)u(t)dt, u(0) = u′(0) = 1
Integrating both sides twice from 0 to x, using the given conditions andconverting the resulting multiple integrals to a single integral we obtain
u(x) = 1 + x+ 12!x
2 + 13!x
3 + 13!
∫ x0
(x− t)3u(t)dtUsing the Adomian decomposition method, (or any other method), wherewe set
u0(x) = 1 + x+1
2!x2 +
1
3!x3
u1(x) =1
4!x4 +
1
5!x5 + · · ·
Accordingly u(x) in a series form is given by
u(x) = 1 + x+1
2!x2 + +
1
3!x3 +
1
4!x4 +
1
5!x5 + · · ·
and in a closed form
u(x) = ex
5.7 Converting to Initial Value Problems
Exercises 5.7
1. u′(x) = ex −
∫ x
0
u(t)dt, u(0) = 1
Differentiating both sides with respect to x and using Leibniz rule weobtain the nonhomogeneous differential equation
u′′(x) + u(x) = ex, u(0) = 1, u
′(0) = 1
Solving the initial value problem we find
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March 4, 2015 14:44 book-9x6 9571-Root page 111
5.7. Converting to Initial Value Problems 111
u(x) = A cosx+B sinx+ 12ex
Using the initial conditions we find
u(x) = 12 (cosx+ sinx+ ex)
2. u′(x) = 1−
∫ x
0
u(t)dt, u(0) = 0
Differentiating both sides with respect to x and using Leibniz rule weobtain the homogeneous differential equation
u′′(x) + u(x) = 0, u(0) = 0, u
′(0) = 1
Solving the initial value problem we find
u(x) = A cosx+B sinx
Using the initial conditions we find
u(x) = sinx
3. u′′(x) = −x− 1
2!x2 +
∫ x
0
(x− t)u(t)dt, u(0) = 1, u′(0) = 1
Differentiating both sides twice with respect to x and using Leibnizrule we obtain the nonhomogeneous differential equation
u(iv)(x)− u(x) = −1, u(0) = u′(0) = 1, u
′′(0) = 0, u
′′′(0) = −1
Solving the initial value problem we find
u(x) = A cosx+B sinx+ C coshx+D sinhx+ 1
Using the initial conditions we find
u(x) = 1 + sinx
4. u′′(x) = 1− 1
2!x2 +
∫ x
0
(x− t)u(t)dt, u(0) = 2, u′(0) = 0
Differentiating both sides twice with respect to x and using Leibnizrule we obtain the nonhomogeneous differential equation
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March 4, 2015 14:44 book-9x6 9571-Root page 112
112 Chapter 5. Volterra Integro-Differential Equations
u(iv)(x)− u(x) = −1, u(0) = 2, u′(0) = 0, u
′′(0) = 1, u
′′′(0) = 0
Solving the initial value problem we find
u(x) = A cosx+B sinx+ C coshx+D sinhx+ 1
Using the initial conditions we find
u(x) = 1 + coshx
5. u′′(x) = − 1
2!x2 − 2
3x3 +
∫ x
0
(x− t)u(t)dt, u(0) = 1, u′(0) = 4
Differentiating both sides twice with respect to x and using Leibnizrule we obtain the nonhomogeneous differential equation
u(iv)(x)− u(x) = −1− 4x, u(0) = 1, u′(0) = 4, u
′′(0) = u
′′′(0) = 0
Solving the initial value problem we find
u(x) = A cosx+B sinx+ C coshx+D sinhx+ 1 + 4x
Using the initial conditions we find
u(x) = 1 + 4x
6. u′′(x) = −x− 1
8x2 +
∫ x
0
(x− t)u(t)dt, u(0) =1
4, u′(0) = 1
Differentiating both sides twice with respect to x and using Leibnizrule we obtain the nonhomogeneous differential equation
u(iv)(x)− u(x) = − 14 , u(0) = u
′(0) = 1, u
′′(0) = 0, u
′′′(0) = −1
Solving the initial value problem we find
u(x) = A cosx+B sinx+ C coshx+D sinhx+ 14
Using the initial conditions we find
u(x) = 14 + sinx
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March 4, 2015 14:44 book-9x6 9571-Root page 113
5.8. The Volterra Integro-Differential Equations of the First Kind 113
7. u′(x) = 1 + sinx+
∫ x
0
u(t)dt, u(0) = −1
Differentiating both sides with respect to x and using Leibniz rule weobtain the nonhomogeneous differential equation
u′′(x)− u(x) = cosx, u(0) = −1, u
′(0) = 1
Solving the initial value problem we find
u(x) = Aex +Be−x − 12 cosx
Using the initial conditions we find
u(x) = 14ex − 3
4e−x − 1
2 cosx
5.8 The Volterra Integro-Differential
Equations of the First Kind
Exercises 5.8
1. Differentiating both sides leads to
u′ = cosx− sinx− 1−∫ x0u′(t) dt
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− cos t+ sin t+ 1 +∫ t0u′
n(r) dr)dt.
We find the following successive approximations:
u0(x) = 1,
u1(x) = cosx+ sinx− x,
u2(x) = 1− 12x
2 + 14!x
4 + · · ·,
u3(x) = 1− 12x
2 + 14!x
4 − 16!x
6 + · · ·,
This gives the exact solution
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March 4, 2015 14:44 book-9x6 9571-Root page 114
114 Chapter 5. Volterra Integro-Differential Equations
u(x) = cosx.
2. Differentiating both sides leads to
u′ = −1−∫ x0u′(t) dt
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t) + 1 +∫ t0u′
n(r) dr)dt.
We find the following successive approximations:
u0(x) = 1,
u1(x) = 1− x,
u2(x) = 1− x+ 12x
2,
u3(x) = 1− x+ 12!x
2 − 13!x
3 + · · ·,
This gives the exact solution
u(x) = e−x.
3. Differentiating both sides leads to
u′ = sinhx+ coshx−∫ x0u′(t) dt
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t) +− sinh t− cosh t+∫ t0u′
n(r) dr)dt.
We find the following successive approximations:
u0(x) = 0,
u1(x) = coshx+ sinhx− 1,
u2(x) = x,
u3(x) = x+ 13!x
3 + 15!x
5 + · · ·,
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March 4, 2015 14:44 book-9x6 9571-Root page 115
5.8. The Volterra Integro-Differential Equations of the First Kind 115
This gives the exact solution
u(x) = sinhx.
4. Differentiating both sides leads to
u′ = 3ex − 2−∫ x0
(u+ u′(t)) dt
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 3et + 2 +∫ t0(un(r) + u
′
n(r)) dr)dt.
We find the following successive approximations:
u0(x) = 1,
u1(x) = 1 + x+ 12x
2 + · · ·,
u2(x) = 1 + x+ 12x
2 + 13!x
3 + · · ·,
u3(x) = 1 + x+ 12!x
2 + 13!x
3 + 14!x
4 + · · ·,
This gives the exact solution
u(x) = ex.
5. Differentiating both sides leads to
u′ = cosx+ x+ 12x
2 −∫ x0
(u+ u′(t)) dtThe correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− cos t− t− 12 t
2 +∫ t0(un(r) + u
′
n(r)) dr)dt.
We find the following successive approximations:
u0(x) = 1,
u1(x) = 1 + x+ · · ·,
u2(x) = 1 + x− 12x
2 + 14!x
4 + · · ·,
u3(x) = 1 + x− 12x
2 + 14!x
4 − 16!x
6 + · · ·,
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March 4, 2015 14:44 book-9x6 9571-Root page 116
116 Chapter 5. Volterra Integro-Differential Equations
This gives the exact solution
u(x) = x+ cosx.
6. Differentiating both sides leads to
u′ = 12 coshx+ 3
2 sinhx− 12 −
12
∫ x0
(u+ u′(t)) dtThe correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− cos t− t− 12 t
2 +∫ t0(un(r) + u
′
n(r)) dr)dt.
We find the following successive approximations:
u0(x) = 1,
u1(x) = 1 + 12x
2 + 112x
3 + 116x
4 + · · ·,
u2(x) = 1 + 12x
2 + 132x
4 + · · ·,
u3(x) = 1 + 12!x
2 + 14!x
4 16!x
6 + · · ·,
This gives the exact solution
u(x) = coshx.
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March 4, 2015 14:44 book-9x6 9571-Root page 117
Chapter 6
Singular IntegralEquations
6.2 Abel’s Problem
Exercises 6.2
1. π(x+ 1) =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = π(x+ 1) into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =d
dx
∫ x
0
t+ 1√x− t
dt
Using the substitution y = x− t so that dt = −dy we find
u(x) =d
dx
∫ x
0
x− y + 1
y12
dy
Integrating the right hand side yields
u(x) =d
dx
(2(x+ 1)y1/2 − 2
3y3/2
)|y=xy=0
117
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March 4, 2015 14:44 book-9x6 9571-Root page 118
118 Chapter 6. Singular Integral Equations
u(x) = 2√x+ 1√
x
This result can also be obtained by integrating the integral involving tby using Appendix B.
2. π2 (x2 − x) =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = π2 (x2 − x) into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =1
2
d
dx
∫ x
0
t2 − t√x− t
dt
Using the substitution y = x− t so that dt = −dy we find
u(x) =1
2
d
dx
∫ x
0
(x− y)2 − (x− y)
y12
dy
Integrating the right hand side yields
u(x) =√x( 4
3x− 1)
This result can also be obtained by integrating the integral involving tby using Appendix B.
3. 1 + x+ x2 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = 1 + x+ x2 into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =1
π
d
dx
∫ x
0
1 + t+ t2√x− t
dt
Using the substitution y = x− t so that dt = −dy we find
u(x) =1
π
d
dx
∫ x
0
1 + (x− y) + (x− y)2
y12
dy
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March 4, 2015 14:44 book-9x6 9571-Root page 119
6.2. Abel’s Problem 119
Integrating the right hand side yields
u(x) = 1π√x
(1 + 2x+ 83x
2)
This result can also be obtained by integrating the integral involving tby using Appendix B.
4. 38πx
2 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = 38πx
2 into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =3
8
d
dx
∫ x
0
t2√x− t
dt
Using the substitution t = xsin2θ so that dt = 2x sin θ cos θdθ we find
u(x) = x32
This result can also be obtained by integrating the integral involving tby using Appendix B.
5. 43x
32 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = 43x
32 into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =4
3π
d
dx
∫ x
0
t32
√x− t
dt
Using the substitution t = xsin2θ so that dt = 2x sin θ cos θdθ we find
u(x) = x
This result can also be obtained by integrating the integral involving tby using Appendix B.
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March 4, 2015 14:44 book-9x6 9571-Root page 120
120 Chapter 6. Singular Integral Equations
6. 815x
52 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = 815x
52 into the formula
u(x) =1
π
d
dx
∫ x
0
f(t)√x− t
dt
we find
u(x) =8
15π
d
dx
∫ x
0
t52
√x− t
dt
Using the substitution t = xsin2θ so that dt = 2x sin θ cos θdθ we find
u(x) = 12x
2
This result can also be obtained by integrating the integral involving tby using Appendix B.
7. x3 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = x3, f(0) = 0, f′(x) = 3x2, α = 1
2 , sin(απ) = 1 intothe formula of the generalized Abel’s integral equation
u(x) =1
π
∫ x
0
f′(t)√x− t
dt
we find
u(x) =1
π
∫ x
0
3t2√x− t
dt
Using the substitution y = x − t so that dt = −dy or using AppendixB we find
u(x) =16
5πx
52
8. x4 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = x4, f(0) = 0, f′(x) = 4x3, α = 1
2 , sin(απ) = 1 intothe formula of the generalized Abel’s integral equation
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March 4, 2015 14:44 book-9x6 9571-Root page 121
6.2. Abel’s Problem 121
u(x) =1
π
∫ x
0
f′(t)√x− t
dt
we find
u(x) =1
π
∫ x
0
4t3√x− t
dt
Using the substitution y = x − t so that dt = −dy or using AppendixB we find
u(x) =128
35πx
72
9. x+ x3 =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = x+x3, f(0) = 0, f′(x) = 1+3x2, α = 1
2 , sin(απ) =1 into the formula of the generalized Abel’s integral equation
u(x) =1
π
∫ x
0
f′(t)√x− t
dt
we find
u(x) =1
π
∫ x
0
1 + 3t2√x− t
dt
Using the substitution y = x − t so that dt = −dy or using AppendixB we find
u(x) =2√x
π(1 +
8
5x2)
10. sinx =
∫ x
0
1√x− t
u(t)dt
Substituting f(x) = sinx, f(0) = 0, f′(x) = cosx into the formula of
the generalized equation
u(x) =1
π
∫ x
0
f′(t)√x− t
dt
Using the approximation cosx ≈ 1, we find
u(x) ≈ 2π
√x
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March 4, 2015 14:44 book-9x6 9571-Root page 122
122 Chapter 6. Singular Integral Equations
6.3 Generalized Abel’s Problem
Exercises 6.3
1. Notice that α = 16 , f(x) = 36
55x116 . Therefore, we set
u(x) = sin(απ)π
ddx
∫ x0
3655 t
116
(x−t)56dt = x.
2. Notice that α = 13 , f(x) = 243
440x113 . Therefore, we set
u(x) = sin(απ)π
ddx
∫ x0
243440 t
113
(x−t)23dt = x3.
3. Notice that α = 34 , f(x) = 24x
14 . Therefore, we set
u(x) = sin(απ)π
ddx
∫ x0
24t14
(x−t)14dt = 6.
4. Notice that α = 23 , f(x) = 3πx
13 + 9x
43 , u(x) = π + 4x.
5. Notice that α = 56 , f(x) = 36
7 x76 , u(x) = x.
6. Notice that α = 23 , f(x) = 27x
73 + 9x
43 , u(x) = 4x+ 14x2.
6.4 The Weakly Singular Volterra Equations
Exercises 6.4
1. u(x) =√x− πx+ 2
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · =√x− πx
+2
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
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March 4, 2015 14:44 book-9x6 9571-Root page 123
6.4. The Weakly Singular Volterra Equations 123
u0(x) =√x− πx
yields the first component
u1(x) = 2∫ x0
1√x−t (√t− πt)dt
so that
u1(x) = πx− 83π
2x32
by using Appendix B.
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) =√x
2. u(x) =3
8πx2 + x
32 −
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = 3
8πx2 + x
32
−∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = 38πx
2 + x32
yields the first component
u1(x) = −∫ x0
1√x−t (t
3/2 − 3
8πt2)dt
so that
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March 4, 2015 14:44 book-9x6 9571-Root page 124
124 Chapter 6. Singular Integral Equations
u1(x) = − 38πx
2 − 8120πx
5/2
Canceling the noise term between u0(x) and u1(x) yields the exactsolution
u(x) = x3/2
3. u(x) = 12 −√x+
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = 12 −√x
+
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = 12 −√x
yields the first component
u1(x) =√x− 1
2πx
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) = 12
4. u(x) =√x− 1
2πx+
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · =√x− 1
2πx
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March 4, 2015 14:44 book-9x6 9571-Root page 125
6.4. The Weakly Singular Volterra Equations 125
+
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) =√x− 1
2πx
yields the first component
u1(x) = 12πx−
23πx
3/2
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) =√x
5. u(x) = x5/2 − 516πx
3 +
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = x5/2 − 516πx
3
+
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = x5/2 − 516πx
3
yields the first component
u1(x) = 516πx
3 − 27πx
7/2
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) = x5/2
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March 4, 2015 14:44 book-9x6 9571-Root page 126
126 Chapter 6. Singular Integral Equations
6. u(x) = x3 + 3233x
7/2 −∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = x3 + 3233x
7/2
−∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = x3 + 3233x
7/2
yields the first component
u1(x) = − 3233x
7/2 − 35132πx
4
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) = x3
7. u(x) = 1 + x− 2√x− 4
3x3/2 +
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = 1 + x− 2√x− 4
3x3/2
+
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
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March 4, 2015 14:44 book-9x6 9571-Root page 127
6.4. The Weakly Singular Volterra Equations 127
u0(x) = 1 + x− 2√x− 4
3x3/2
yields the first component
u1(x) = 2√x+ 4
3x3/2 − πx− 1
2πx2
Canceling the noise terms between u0(x) and u1(x) yields the exactsolution
u(x) = 1 + x
8. u(x) = 1 + 2√x−
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = 1 + 2√x
−∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = 1 + 2√x
yields the first component
u1(x) = −2√x− πx
Canceling the noise terms between u0(x) and u1(x) yields the exactsolution
u(x) = 1
9. u(x) = x2 + 1615x
52 −
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
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March 4, 2015 14:44 book-9x6 9571-Root page 128
128 Chapter 6. Singular Integral Equations
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = x2 + 1615x
52
−∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = x2 + 1615x
52
yields the first component
u1(x) = − 1615x
5/2 − 13πx
3
Canceling the noise term between u0(x) and u1(x) yields the exact so-lution
u(x) = x2
10. u(x) = 2π
√x+ 15
16x2 − x− x5/2 +
∫ x
0
1√x− t
u(t)dt, I = [0, 2]
Using the Adomian decomposition method, we substitute
u(x) =∑∞n=0 un(x)
into both sides of the integral equation, hence we obtain
u0(x) + u1(x) + u2(x) + · · · = 2π
√x+ 15
16x2 − x− x5/2
+
∫ x
0
1√x− t
(u0(t) + u1(t) + · · ·) dt
Setting the zeroth component by
u0(x) = 2π
√x+ 15
16x2 − x− x5/2
yields the first component
u1(x) = x+ x5/2 − 43x
3/2 − 516πx
3
Canceling the noise terms between u0(x) and u1(x) yields the exactsolution
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March 4, 2015 14:44 book-9x6 9571-Root page 129
6.4. The Weakly Singular Volterra Equations 129
u(x) = 2π
√x+ 15
16x2
11. We use the recurrence relation
u0(x) = 1 + x,
u1(x) = − 43x
34 − 16
21x74 +
∫ x0
1+t
(x−t)14u(t) dt = 0.
This gives the exact solution
u(x) = 1 + x.
12. We use the recurrence relation
u0(x) = x+ x2,
u1(x) = − 94x
43 − 243
140x103 +
∫ x0
t+t2
(x−t)14u(t) dt = 0.
This gives the exact solution
u(x) = x+ x2.
13. We use the recurrence relation
u0(x) = 1 + 3x2.
Proceed as before, to get the exact solution
u(x) = 1 + 3x2.
14. We use the recurrence relation
u0(x) = 5− x.
Proceed as before, to get the exact solution
u(x) = 5− x.
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March 4, 2015 14:44 book-9x6 9571-Root page 130
130 Chapter 6. Singular Integral Equations
6.5 The Weakly Singular Fredholm Equations
Exercises 6.5
1. We first decompose f(x) into two parts as
f0(x) = x2
f1(x) = − 1615x
52 + 2
√x−15 + 8x
√x−1
15 + 16x2√x−115
We then use the modified recurrence relation
u0(x) = x2,
u1(x) = − 1615x
52 + 2
√x−15 + 8x
√x−1
15 + 16x2√x−115 +
∫ 1
0u(t)√x−tu0(t) dt = 0.
This gives the exact solution by
u(x) = x2.
2. We first decompose f(x) into two parts as
f0(x) = 10
f1(x) = 20(x− 1)12 − 20(x+ 1)
12
We then use the modified recurrence relation
u0(x) = 10,
u1(x) = 20(x− 1)12 − 20(x+ 1)
12 +
∫ 1
−1u0(t)
(x−t)12dt = 0.
This gives the exact solution by
u(x) = 10.
3. We first decompose f(x) into two parts as
f0(x) = 10x
f1(x) = −9x53 + 6(x− 1)
23 + 9x(x− 1)
23
We then use the modified recurrence relation
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March 4, 2015 14:44 book-9x6 9571-Root page 131
6.5. The Weakly Singular Fredholm Equations 131
u0(x) = 10x,
u1(x) = −9x53 + 6(x− 1)
23 + 9x(x− 1)
23 +
∫ 1
0u(t)
(x−t)13u0(t) dt = 0.
This gives the exact solution byu(x) = 10x.
4. We first decompose f(x) into two parts as
f0(x) = 3 + 10x− 9x13
f1(x) = − 452 x
43 + 33
2 (x− 1)13 + 45
2 x(x− 1)13
We then use the modified recurrence relation
u0(x) = 3 + 10x− 9x13 ,
u1(x) = − 452 x
43 + 33
2 (x− 1)13 + 45
2 x(x− 1)13 +
∫ 1
0u0(t)
(x−t)23dt = 9x
13 + · · ·.
Cancelling the noise term gives the exact solution byu(x) = 3 + 10x.
5. We first decompose f(x) into two parts as
f0(x) = x+ x2 − 165 x
54 (1 + 8
9x)
f1(x) = 845 (x− 1)
14 (7 + 22x+ 16x2)
We then use the modified recurrence relation
u0(x) = x+ x2 − 165 x
54 (1 + 8
9x),
u1(x) = 845 (x−1)
14 (7+22x+16x2)+
∫ 1
0u(t)
(x−t)34dt = 16
5 x54 (1+ 8
9x)+ · · ·.
Cancelling the noise term gives the exact solution byu(x) = x+ x2.
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March 4, 2015 14:44 book-9x6 9571-Root page 133
Chapter 7
Nonlinear FredholmIntegral Equations
7.2 Nonlinear Fredholm Integral Equations
7.2.1 The Direct Computation Method
Exercises 7.2.1
1. u(x) = 1 +1
2λ
∫ 1
0
u2(t)dt
This equation can be rewritten as
u(x) = 1 +1
2λα
where
α =∫ 1
0u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(1 +
1
2λα
)2
dt
Integrating the definite integral yields
λ2α2 + (4λ− 4)α+ 4 = 0
133
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March 4, 2015 14:44 book-9x6 9571-Root page 134
134 Chapter 7. Nonlinear Fredholm Integral Equations
Solving the quadratic equation for α gives
α =2(1− λ)± 2
√1− 2λ
2λ2
Substituting for α in the above equation for u(x) we find
u(x) =1±√
1− 2λ
λ
Examining the solution u(x) we obtained leads to
λ = 0 is a singular point
λ =1
2is a bifurcation point
2. u(x) = 1− λ∫ 1
0
u2(t)dt
This equation can be rewritten as
u(x) = 1− λα
where
α =∫ 1
0u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(1− λα)2dt
Integrating the definite integral yields
λ2α2 − (1 + 2λ)α+ 1 = 0
Solving the quadratic equation for α gives
α =(1 + 2λ)± 2
√1 + 4λ
2λ2
Substituting for α in the above equation for u(x) we find
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March 4, 2015 14:44 book-9x6 9571-Root page 135
7.2. Nonlinear Fredholm Integral Equations 135
u(x) = −1±√
1 + 4λ
2λ
Examining the solution u(x) we obtained leads to
λ = 0 is a singular point
λ = −1
4is a bifurcation point
3. u(x) = 1 + λ
∫ 1
0
tu2(t)dt
This equation can be rewritten as
u(x) = 1 + λα
where
α =∫ 1
0tu2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
t (1 + λα)2dt
Integrating the definite integral yields
λ2α2 + (2λ− 2)α+ 1 = 0
Solving the quadratic equation for α gives
α =(1− λ)±
√1− 2λ
λ2
Substituting for α in the above equation for u(x) we find
u(x) =1±√
1− 2λ
λ
Examining the solution u(x) we obtained leads to
λ = 0 is a singular point
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March 4, 2015 14:44 book-9x6 9571-Root page 136
136 Chapter 7. Nonlinear Fredholm Integral Equations
λ =1
2is a bifurcation point
4. u(x) = 1 + λ
∫ 1
0
t2u2(t)dt
This equation can be rewritten as
u(x) = 1 + λα
where
α =∫ 1
0t2 u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
t2 (1 + λα)2dt
Integrating the definite integral yields
λ2α2 + (2λ− 3)α+ 1 = 0
Solving the quadratic equation for α gives
α =(3− 2λ)±
√9− 12λ
2λ2
Substituting for α in the above equation for u(x) we find
u(x) =3±√
9− 12λ
2λ
Examining the solution u(x) we obtained leads to the conclusion that
λ = 0 is a singular point
λ =3
4is a bifurcation point
5. u(x) = 1 + λ
∫ 1
0
t3u2(t)dt
This equation can be rewritten as
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March 4, 2015 14:44 book-9x6 9571-Root page 137
7.2. Nonlinear Fredholm Integral Equations 137
u(x) = 1 + λα
where
α =∫ 1
0t3u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
t3 (1 + λα)2dt
Integrating the definite integral yields
λ2α2 + (2λ− 4)α+ 1 = 0
Solving the quadratic equation for α gives
α =(2− λ)± 2
√1− λ
λ2
Substituting for α in the above equation for u(x) we find
u(x) =2± 2
√1− λ
λ
Examining the solution u(x) we obtained leads to
λ = 0 is a singular point
λ = 1 is a bifurcation point
6. u(x) = 2− 4
3x+
∫ 1
0
xt2 u2(t)dt
This equation can be rewritten as
u(x) = 2 + (α− 4
3)x
where
α =∫ 1
0t2 u2(t)dt
Substituting for u(x) from the above equation we obtain
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March 4, 2015 14:44 book-9x6 9571-Root page 138
138 Chapter 7. Nonlinear Fredholm Integral Equations
α =
∫ 1
0
t2(
2 + (α− 4
3)t
)2
dt
Integrating the definite integral yields
α =4
3
Substituting for α in the above equation for u(x) we find
u(x) = 2
7. u(x) = sinx− π8 +
1
2
∫ π/2
0
u2(t)dt
This equation can be rewritten as
u(x) = sinx+ ( 12α−
π8 )
where
α =∫ 1
0u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(sin t+ (
1
2α− π
8)
)2
dt
Integrating the definite integral yields
α =π
4
Substituting for α in the above equation for u(x) we find
u(x) = sinx
8. u(x) = cosx− π8 +
1
2
∫ π/2
0
u2(t)dt
This equation can be rewritten as
u(x) = cosx+ ( 12α−
π8 )
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March 4, 2015 14:44 book-9x6 9571-Root page 139
7.2. Nonlinear Fredholm Integral Equations 139
where
α =∫ 1
0u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(cos t+ (
1
2α− π
8)
)2
dt
Integrating the definite integral yields
α =π
4
Substituting for α in the above equation for u(x) we find
u(x) = cosx
9. u(x) = x− 18 +
1
2
∫ 1
0
tu2(t)dt
This equation can be rewritten as
u(x) = x+ ( 12α−
18 )
where
α =∫ 1
0tu2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
t
(t+ (
1
2α− 1
8)
)2
dt
Integrating the definite integral yields
α =1
4,
67
12
Substituting for α in the above equation for u(x) we find
u(x) = x, x+8
3
10. u(x) = x2 − 110 +
1
2
∫ 1
0
u2(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 140
140 Chapter 7. Nonlinear Fredholm Integral Equations
This equation can be rewritten as
u(x) = x2 + ( 12α−
110 )
where
α =∫ 1
0u2(t)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(t2 + (
1
2α− 1
10)
)2
dt
Integrating the definite integral yields
α =1
5,
43
15
Substituting for α in the above equation for u(x) we find
u(x) = x2, x2 +4
3
11. u(x) = x− 56 +
∫ 1
0
(u(t) + u2(t)
)dt
This equation can be rewritten as
u(x) = x+ (α− 56 )
where
α =∫ 1
0
(u(t) + u2(t)
)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
((t+ (α− 5
6) + (t+ (α− 5
6))2)dt
Integrating the definite integral yields
α =5
6,−1
6
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March 4, 2015 14:44 book-9x6 9571-Root page 141
7.2. Nonlinear Fredholm Integral Equations 141
Substituting for α in the above equation for u(x) we find
u(x) = x, x− 1
12. u(x) = x− 1 + 34
∫ 1
0
(2t+ u2(t)
)dt
This equation can be rewritten as
u(x) = x+ ( 34α− 1)
where
α =∫ 1
0
(2t+ u2(t)
)dt
Substituting for u(x) from the above equation we obtain
α =
∫ 1
0
(2t+ (t+ (
3
4α− 1))2
)dt
Integrating the definite integral yields
α =4
3,
16
9
Substituting for α in the above equation for u(x) we find
u(x) = x, x+ 13
7.2.2 The Adomian Decomposition Method
Exercises 7.2.2
1. u(x) = 1 + λ
∫ 1
0
tu2(t)dt, λ ≤ 1
2
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the procedure of the Adomian polynomials discussed before, we set
u0(x) = 1
Accordingly, the first components is given by
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March 4, 2015 14:44 book-9x6 9571-Root page 142
142 Chapter 7. Nonlinear Fredholm Integral Equations
u1(x) = λ∫ 1
0tdt =
1
2λ
u2(x) = λ∫ 1
0t(2u0(t)u1(t))dt = 1
2λ2
u3(x) = 58λ
3
Consequently the solution in a series form is
u(x) = 1 +1
2λ+
1
2λ2 +
5
8λ3 + · · ·
2. u(x) = 1 + λ
∫ 1
0
t3u2(t)dt, λ ≤ 1
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the procedure of the Adomian polynomials discussed before, we set
u0(x) = 1
Accordingly, the first components is given by
u1(x) = λ∫ 1
0t3dt =
1
4λ
u2(x) = λ∫ 1
0t3(2u0(t)u1(t))dt = 1
8λ2
Consequently the solution in a series form is
u(x) = 1 +1
4λ+
1
8λ2 + · · ·
3. u(x) = 2 sinx− π8 +
1
8
∫ π/2
0
u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = 2 sinx
Accordingly, the first components is given by
u1(x) = −π8 +1
8
∫ π/2
0
4sin2tdt = 0
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March 4, 2015 14:44 book-9x6 9571-Root page 143
7.2. Nonlinear Fredholm Integral Equations 143
u2(x) = 0
u3(x) = 0
Consequently the solution is
u(x) = 2 sinx
4. u(x) = 2 cosx− π8 +
1
8
∫ π/2
0
u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = 2 cosx
Accordingly, the first component is given by
u1(x) = −π8 +1
8
∫ π/2
0
4cos2tdt = 0
u2(x) = 0
u3(x) = 0
Consequently the solution is
u(x) = 2 cosx
5. u(x) = secx− x+ x∫ π/40
u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = secx
Accordingly, the first component is given by
u1(x) = −x+ x∫ π/40
sec2tdt = 0
u2(x) = 0
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March 4, 2015 14:44 book-9x6 9571-Root page 144
144 Chapter 7. Nonlinear Fredholm Integral Equations
u3(x) = 0
Consequently the solution is
u(x) = secx
6. u(x) = 32x+
3
8
∫ 1
0
xu2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the procedure of the Adomian polynomials discussed before, we set
u0(x) = 32x
Accordingly, the first component is given by
u1(x) = 932x
u2(x) = 27256x
Consequently the solution in a series form is given by
u(x) = 32x+ 9
32x+ 27256x+ · · ·
so that
u(x) ≈ 2x
7. u(x) = x2 − 1
12+
1
2
∫ 1
0
tu2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = x2
Accordingly, the first components is given by
u1(x) = − 1
12+
1
2
∫ 1
0
t5dt = 0
u2(x) = 0
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March 4, 2015 14:44 book-9x6 9571-Root page 145
7.2. Nonlinear Fredholm Integral Equations 145
Consequently the solution is given by
u(x) = x2
8. u(x) = x− π
8+
1
2
∫ 1
0
1
1 + u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = x
Accordingly, the first component is given by
u1(x) = −π8
+1
2
∫ 1
0
1
1 + t2dt = 0
u2(x) = 0
Consequently the solution is given by
u(x) = x
9. u(x) = x− 1 +2
π
∫ 1
−1
1
1 + u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = x
Accordingly, the first component is given by
u1(x) = −1 +2
π
∫ 1
−1
1
1 + t2dt = 0
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = x
10. u(x) = x− 1
4ln 2 +
1
2
∫ 1
0
t
1 + u2(t)dt
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March 4, 2015 14:44 book-9x6 9571-Root page 146
146 Chapter 7. Nonlinear Fredholm Integral Equations
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = x
Accordingly, the first component is given by
u1(x) = −1
4ln 2 +
1
2
∫ 1
0
t
1 + t2dt, = 0
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = x
11. u(x) = sinx+ cosx− π + 2
8+
1
4
∫ π/2
0
u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = sinx+ cosx
Accordingly, the first component is given by
u1(x) = −π + 2
8+
1
4
∫ π/2
0
(1 + sin 2t)dt− 0.
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = sinx+ cosx
12. u(x) = sinhx− 1 +
∫ 1
0
(cosh2t− u2(t)
)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = sinhx
Accordingly, the first component is given by
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March 4, 2015 14:44 book-9x6 9571-Root page 147
7.2. Nonlinear Fredholm Integral Equations 147
u1(x) = 0
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = sinhx
13. u(x) = cosx+ 2−∫ 1
0
(1 + sin2t+ u2(t)
)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = cosx
Accordingly, the first component is given by
u1(x) = 0
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = cosx
14. u(x) = secx− x+
∫ 1
0
x(u2(t)− tan2t
)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the modified decomposition method discussed before, we set
u0(x) = secx
Accordingly, the first component is given by
u1(x) = 0
Consequently the solution, after justifying that u0(x) justifies the equa-tion, is given by
u(x) = secx
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March 4, 2015 14:44 book-9x6 9571-Root page 148
148 Chapter 7. Nonlinear Fredholm Integral Equations
7.2.3 The Variational Iteration Method
Exercises 7.2.3
1. u(x) =3
4x+
∫ 1
0
xtu2(t) dt
Differentiating both sides of this equation with respect to x yields
u′(x) = 3
4 +∫ 1
0tu2(t)dt, u(0) = 0.
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 34 −
∫ 1
0ru2n(r) dt
)dt
where we used λ = −1 for first-order integro-differential equations.This gives the following successive approximations
u0(x) = 0,u1(x) = 0.75x,u2(x) = 0.8906x,u3(x) = 0.9483x,u4(x) = 0.9748x,
... .This gives the exact solution by u(x) = x
2. u(x) = x2 − 1
6x+
∫ 1
0
xtu2(t) dt
Differentiating both sides of this equation with respect to x yields
u′(x) = 2x− 1
6 +∫ 1
0tu2(t)dt, u(0) = 0.
The correction functional for this equation is given by
un+1(x) = un(x)−∫ x0
(u′
n(t)− 2t− 16 −
∫ 1
0ru2n(r) dt
)dt
This gives the following successive approximations
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March 12, 2015 12:19 book-9x6 9571-Root page 149
7.3. Nonlinear Fredholm Integral Equations of the First Kind 149
u0(x) = 0,u1(x) = x2 − 0.1667x,u2(x) = x2 − 0.0597x,u3(x) = x2 − 0.0230x,u4(x) = x2 − 0.0090x,
... .
This gives the exact solution by u(x) = x2
3. u(x) = x2 − 1
8x+
∫ 1
0
xtu2(t) dt
Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the following successive approximations
u0(x) = 0,u1(x) = x3 − 0.1250x,u2(x) = x3 − 0.0377x,u3(x) = x3 − 0.0122x,u4(x) = x3 − 0.0040x,
... .
This gives the exact solution by u(x) = x3
4. u(x) = x− 1
5x2 +
∫ 1
0
x2t2u2(t) dt
Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the exact solution by u(x) = x+ x2
5. u(x) = x+34
105x2 +
∫ 1
0
x2t2u2(t) dt
Differentiating both sides of this equation with respect to x and pro-ceeding as before, we obtain the exact solution by u(x) = x.
7.3 Nonlinear Fredholm Integral Equations
of the First Kind
Exercises 7.3
1. We set the transformation
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March 4, 2015 14:44 book-9x6 9571-Root page 150
150 Chapter 7. Nonlinear Fredholm Integral Equations
v(x) = u2(x), u(x) =√v(x)
This gives
vε(x) = 1ε ( 49
60 − α)x
where
α =∫ 1
0tvε(t) dt
Proceeding as in the examples, we find
α = 4960(1+3ε)
This gives
u(x) = 7√20
√x
Also
u(x) = x+ x2
2. We set the transformation
v(x) = u3(x), u(x) = 3√v(x)
This gives
vε(x) = 1ε ( 1
120 − α)x2
where
α =∫ 1
0t2vε(t) dt
Proceeding as in the examples, we find
α = 1120(1+5ε)
This gives
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March 4, 2015 14:44 book-9x6 9571-Root page 151
7.3. Nonlinear Fredholm Integral Equations of the First Kind 151
u(x) = 3
√x2
24
Also
u(x) = x− x3
3. We set the transformation
v(x) = u3(x), u(x) = 3√v(x)
This gives
vε(x) = 1ε (1− α)e2x
where
α =∫ 1
0e−6tvε(t) dt
Proceeding as in the examples, we find
α = 1−e−4
1−e−4+4ε
This gives
u(x) = 3
√4e2x
1−e−4
Also
u(x) = e2x
4. We set the transformation
v(x) = u4(x), u(x) = 4√v(x)
This gives
vε(x) = 1ε (1− α)ex
where
α =∫ 1
0e−4tvε(t) dt
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March 4, 2015 14:44 book-9x6 9571-Root page 152
152 Chapter 7. Nonlinear Fredholm Integral Equations
Proceeding as in the examples, we find
α = 1−e−3
(1−e−3+3ε)
This gives
u(x) = 4
√3ex
1−e−3
Also
u(x) = ex
5. We set the transformation
v(x) = u2(x), u(x) =√v(x)
This gives
vε(x) = 1ε ( 1
4 − α)x2
where
α =∫ 1
0tvε(t) dt
Proceeding as in the examples, we find
α = 14(1+4ε)
This gives
u(x) = x
Also
u(x) = lnx
6. We set the transformation
v(x) = u2(x), u(x) =√v(x)
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March 12, 2015 12:19 book-9x6 9571-Root page 153
7.4. Weakly-Singular Nonlinear Fredholm Integral Equations 153
This gives
vε(x) = 1ε ( 2
125 − α)x2
where
α =∫ 1
0t2vε(t) dt
Proceeding as in the examples, we find
α = 2125(1+5ε)
This gives
u(x) =√25 x
Also
u(x) = x lnx
7.4 Weakly-Singular Nonlinear Fredholm
Integral Equations
Exercises 7.4
1. We decompose f(x) into
f0(x) = x
f1(x) = − 1615x
52 + 2
5
√x− 1(1 + 4
3x+ 83x
2)
Consequently, we set the modified recurrence relation as
u0(x) = x
u1(x = f1(x) +∫ 1
0u2(t)√|x−t|
dt = 0
The exact solution is u(x) = x.
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March 12, 2015 12:19 book-9x6 9571-Root page 154
154 Chapter 7. Nonlinear Fredholm Integral Equations
2. We decompose f(x) into
f0(x) = x
f1(x) = − 3235x
72 + 2
7
√x− 1(1 + 6x+ 8x2 + 16x3)
Consequently, we set the modified recurrence relation as
u0(x) = x
u1x = f1(x) +∫ 1
0u3(t)√|x−t|
dt = 0
The exact solution is u(x) = x.
3. We decompose f(x) into
f0(x) = 1− x
f1(x) = −2√x(1− 4
3x+ 815x
2) + 1615
√x− 1(1− 2x+ x2)
Consequently, we set the modified recurrence relation as
u0(x) = 1− x
u1x = f1(x) +∫ 1
0u2(t)√|x−t|
dt = 0
The exact solution is u(x) = x.
4. We decompose f(x) into
f0(x) = 1 + x
f1(x) = − 32x
23 (1 + 6
5x+ 920x
2) + 140 (x− 1)
23 (123 + 90x+ 27x2)
Consequently, we set the modified recurrence relation as
u0(x) = 1 + x
u1x = f1(x) +∫ 1
0u2(t)
(|x−t|)13dt = 0
The exact solution is u(x) = 1 + x.
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March 12, 2015 12:19 book-9x6 9571-Root page 155
7.4. Weakly-Singular Nonlinear Fredholm Integral Equations 155
5. We decompose f(x), then we use
u0(x) = 4√
cosx
u1x = f1(x) +∫ π
2
0u4(t)
(| sin x−sin t|)13dt = 0
The exact solution is u(x) = 4√
cosx.
6. We decompose f(x), then we use
u0(x) =√
sinx
u1x = f1(x) +∫ π
2
0u2(t)
(| cos x−cos t|)14dt = 0
The exact solution is u(x) =√
sinx.
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March 4, 2015 14:44 book-9x6 9571-Root page 157
Chapter 8
Nonlinear VolterraIntegral Equations
8.2 Nonlinear Volterra Integral Equations
8.2.1 The Series Solution Method
Exercises 8.2.1
1. u(x) = x2 + 110x
5 − 12
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = x2 +
1
10x5 − 1
2
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields
a2 = 1, ak = 0, k 6= 2
157
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March 4, 2015 14:44 book-9x6 9571-Root page 158
158 Chapter 8. Nonlinear Volterra Integral Equations
Accordingly, we find
u(x) = x2
2. u(x) = x2 + 112x
6 − 12
∫ x
0
tu2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = x2 +
1
12x6 − 1
2
∫ x
0
t
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields
a2 = 1, ak = 0, k 6= 2
Accordingly, we find
u(x) = x2
3. u(x) = 1− x2 − 13x
3 +
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = 1− x2 − 1
3x3 +
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields
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March 4, 2015 14:44 book-9x6 9571-Root page 159
8.2. Nonlinear Volterra Integral Equations 159
a0 = a1 = 1, ak = 0, k ≥ 2
Accordingly, we find
u(x) = 1 + x
4. u(x) = 1− x+ x2 − 23x
3 − 15x
5 +
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = 1− x+ x2 − 2
3x3 − 1
5x5 +
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields
a0 = a2 = 1, ak = 0, k 6= 0, 2
Accordingly, we find
u(x) = 1 + x2
5. u(x) = x2 + 114x
7 − 12
∫ x
0
u3(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = x2 +
1
14x7 − 1
2
∫ x
0
( ∞∑n=0
antn
)3
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers of
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160 Chapter 8. Nonlinear Volterra Integral Equations
x in both sides yields
a2 = 1, ak = 0, k 6= 2
Accordingly, we find
u(x) = x2
6. u(x) = 12 + e−x − 1
2e−2x −
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn =
1
2+ e−x − 1
2e−2x −
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor expansions of the exponentialfunctions and equating the coefficients of like powers of x in both sides yields
ak = (−1)kk! , k = 0, 1, 2, ...
Accordingly, we find
u(x) = e−x
7. u(x) = 1− 32x
2 − x3 − 14x
3 +
∫ x
0
u3(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = 1− 3
2x2 − x3 − 1
4x3 +
∫ x
0
( ∞∑n=0
antn
)3
dt
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8.2. Nonlinear Volterra Integral Equations 161
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0 and equating the coefficients of like powers ofx in both sides yields
a0 = a1 = 1, ak = 0, k ≥ 2
Accordingly, we find
u(x) = 1 + x
8. u(x) = sinx− 12x+ 1
4 sin 2x+
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = sinx− 1
2x+
1
4sin 2x+
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor series expansion of the trigono-metric functions and equating the coefficients of like powers of x in bothsides yields
a0 = 0
a1 = 1
a2 = 0
a3 = − 13!
a2k = 0, a2k+1 = (−1)k 1(2k+1)! , k = 0, 1, 2, ....
Accordingly, we find
u(x) = sinx
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162 Chapter 8. Nonlinear Volterra Integral Equations
9. u(x) = cosx− 12x−
14 sin 2x+
∫ x
0
u2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
∞∑n=0
anxn = cosx− 1
2x− 1
4sin 2x+
∫ x
0
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor series expansion of the trigono-metric functions and equating the coefficients of like powers of x in bothsides yields
a0 = 1
a1 = 0
a2 = − 12!
a3 = 0
a2k = (−1)k 1(2k)! , a2k+1 = 0, k = 0, 1, 2, ....
Accordingly, we find
u(x) = cosx
10. u(x) = ex + 12x(e2x − 1
)−∫ x
0
xu2(t)dt
Substituting u(x) by the series
u(x) =
∞∑n=0
anxn
into both sides of the equation leads to
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8.2. Nonlinear Volterra Integral Equations 163
∞∑n=0
anxn = ex +
1
2x(e2x − 1
)−∫ x
0
x
( ∞∑n=0
antn
)2
dt
Evaluating the regular integrals at the right hand side that involveterms of the form tn, n ≥ 0, using the Taylor expansions of the exponentialfunctions and equating the coefficients of like powers of x in both sides yields
ak = 1k! , k = 0, 1, 2, ...
Accordingly, we find
u(x) = ex
8.2.2 The Adomian Decomposition Method
Exercises 8.2.2
1. u(x) = 3x+1
24x4 − 1
18
∫ x
0
(x− t)u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-
ing the procedure of the modified decomposition method discussed before,we set
u0(x) = 3x
Accordingly, the first component is given by
u1(x) =1
24x4 − 1
18
∫ x
0
(x− t)u20(t)dt
so that
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = 3x
2. u(x) = 2x− 1
2x4 +
1
4
∫ x
0
u3(t)dt
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164 Chapter 8. Nonlinear Volterra Integral Equations
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-
ing the procedure of the modified decomposition method discussed before,we set
u0(x) = 2x
Accordingly, the first component is given by
u1(x) = −1
2x4 +
1
4
∫ x
0
u20(t)dt
so that
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = 2x
3. u(x) = sinx+1
8sin 2x− 1
4x+
1
2
∫ x
0
u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-
ing the procedure of the modified decomposition method discussed before,we set
u0(x) = sinx
Accordingly, the first component is given by
u1(x) =1
8sin 2x− 1
4x+
1
2
∫ x
0
u20(t)dt
so that
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = sinx
4. u(x) = x2 +1
5x5 −
∫ x
0
u2(t)dt
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8.2. Nonlinear Volterra Integral Equations 165
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-
ing the procedure of the modified decomposition method discussed before,we set
u0(x) = x2
Accordingly, the first component is given by
u1(x) =1
5x5 −
∫ x
0
u20(t)dt
so that
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = x2
5. u(x) = x+
∫ x
0
(x− t)u3(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we set
u0(x) = x
Accordingly, the first components are given by
u1(x) =1
20x5
u2(x) =1
720x9
Hence
u(x) ≈ x+ 120x
5 + 1720x
9 + · · ·
6. u(x) = 1 +
∫ x
0
(x− t)2u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we set
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166 Chapter 8. Nonlinear Volterra Integral Equations
u0(x) = 1
Accordingly, the first components are given by
u1(x) =1
3x3
u2(x) =1
90x6
Hence
u(x) ≈ x+ 13x
3 + 190x
6 + · · ·
7. u(x) = 1 +
∫ x
0
(x− t)2u3(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we set
u0(x) = 1
Accordingly, the first components are given by
u1(x) =1
3x3
u2(x) =1
60x6
Hence
u(x) ≈ x+ 13x
3 + 160x
6 + · · ·
8. u(x) = x+
∫ x
0
(x− t)2u2(t)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we set
u0(x) = x
Accordingly, the first components are given by
u1(x) =1
6x5
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8.2. Nonlinear Volterra Integral Equations 167
u2(x) =1
756x9
Hence
u(x) ≈ x+ 16x
5 + 1756x
9 + · · ·
9. u(x) = 1 +
∫ x
0
(t+ u2(t)
)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we setu0(x) = 1
Accordingly, the first components are given by
u1(x) = 2x
u2(x) =5
2x2
Hence
u(x) ≈ 1 + 2x+ 52x
2 + · · ·
10. u(x) = 1 +
∫ x
0
(t2 + u2(t)
)dt
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and fol-
lowing the Adomian decomposition method discussed before, we set
u0(x) = 1
Accordingly, the first components are given by
u1(x) = x+ 13x
3
u2(x) = x2 +1
3x3 +
1
6x4
Hence
u(x) ≈ 1 + x+ x2 23x
3 + 16x
4 + · · ·
11. u(x) = secx+ tanx+ x−∫ x
0
(1 + u2(t)
)dt, x <
π
2
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168 Chapter 8. Nonlinear Volterra Integral Equations
Using the decomposition u(x) =∑∞n=0 un(x) into both sides and follow-
ing the procedure of the modified decomposition method discussed before,we set
u0(x) = secx
Accordingly, the first component is given by
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = secx
12. u(x) = tanx− 14 sin 2x− 1
2x+
∫ x
0
1
1 + u2(t)dt, x <
π
2
Using the decomposition
u(x) =∑∞n=0 un(x)
into both sides and following the procedure of the modified decompositionmethod discussed before, we set
u0(x) = tanx
Accordingly, the first component is given by
u1(x) = 0
Consequently, the other components vanish, hence
u(x) = tanx
8.2.3 The Variational Iteration Method
Exercises 8.2.3
1. u(x) = x− 120x
5 +∫ x0
(x− t)u3(t) dt
Differentiating both sides of this equation, and using Leibniz rule, wefind
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March 4, 2015 14:44 book-9x6 9571-Root page 169
8.2. Nonlinear Volterra Integral Equations 169
u′(x) = 1− 1
4x4 +
∫ x0u3(t) dt, u(0) = 0
The correction functional is
un+1(x) = un(x)−∫ x0
(u′
n(ξ)− 1 + 14ξ
4 −∫ ξ0u3n(r) dr
)dξ
This will give the following successive approximations
u0(x) = 0,u1(x) = x− 1
20x5,
u2(x) = 2x− 120x
5 + ( 15x
5 − 1480x
9) + 120800x
13 + · · · ,u3(x) = 2x− 1
480x9 + ( 1
480x9 + 1
20800x13)− 1
20800x13 + · · · ,
... .Cancelling the noise terms gives the exact solution by u(x) = x.
2. u(x) = x2 − 156x
8 +∫ x0
(x− t)u3(t) dt
Differentiating both sides of this equation, and using Leibniz rule, wefind
u′(x) = 2x− 1
7x7 +
∫ x0u3(t) dt, u(0) = 0
The correction functional is
un+1(x) = un(x)−∫ x0
(u′
n(ξ)− 2ξ + 17ξ
7 −∫ ξ0u3n(r) dr
)dξ
This will give the following successive approximations
u0(x) = 0,u1(x) = x2 − 1
56x8,
u2(x) = x2 − 310192x
14 + · · · ,u3(x) = x2 + · · · ,
... .
Cancelling the noise terms gives the exact solution by u(x) = x2.
3. u(x) = x+ x2 − 112x
4 − 110x
5 − 130x
6 +∫ x0
(x− t)u2(t) dt
Proceeding as before gives the following successive approximations
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170 Chapter 8. Nonlinear Volterra Integral Equations
u0(x) = 0,u1(x) = x+ x2 − 1
12x4 − 1
10x5 − 1
30x6,
u2(x) = x+ x2 + noise terms,... .
Cancelling the noise terms gives the exact solution by u(x) = x+ x2.
4. u(x) = 1 + x− 12x
2 − 13x
3 − 112x
4 +∫ x0
(x− t)u2(t) dt
Proceeding as before and by cancelling the noise terms, we obtain theexact solution by u(x) = 1 + x.
5. u(x) = 1 + x− 13x
3 − 14x
4 − 130x
5 +∫ x0
(x− t)2u2(t) dt
Proceeding as before and by cancelling the noise terms, we obtain theexact solution by u(x) = 1 + x.
8.3 Nonlinear Volterra Integral Equations of
the First Kind
8.3.1 The Series Solution Method
Exercises 8.3.1
1. Using the Taylor series of the left side, and substituting the series formof u(x) yields
112x
4− 190x
6+ 11260x
8+· · · =∫ x0
(x−t)[a0 + a1t+ a2t
2 + a3t3 + · · ·
]2dt
Integrating the right side, and equating the coefficients of like powersof x
a0 = 0, a1 = ±1, a2 = 0, a3 = ∓ 13! , · · ·
Consequently, the exact solution is given by
u(x) = ± sinx
2. Using the Taylor series of the left side, and substituting the series formof u(x) yields
12x
2 + 13x
3 + 16x
4 + 115x
5 + · · · =∫ x0
(x− t)[a0 + a1t+ a2t
2 + a3t3 + · · ·
]2dt
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8.3. Nonlinear Volterra Integral Equations of the First Kind 171
Integrating the right side, and equating the coefficients of like powersof x
a0 = ±1, a1 = ±1, a2 = ± 12! , a3 = ± 1
3! , · · ·
Consequently, the exact solution is given by
u(x) = ±ex
3. Using the Taylor series of the left side, and substituting the series formof u(x) yields
x+ 32x
2+ 32x
3+ 98x
4+ 2740x
5+· · · =∫ x0
[a0 + a1t+ a2t
2 + a3t3 + · · ·
]3dt
Integrating the right side, and equating the coefficients of like powersof x
a0 = 1, a1 = 1, a2 = 12! , a3 = 1
3! , · · ·
Consequently, the exact solution is given by
u(x) = ±ex
4. Using the Taylor series of the left side, and substituting the series formof u(x) yields12x
2− 13x
3+ 115x
5− 2315x
7+· · · =∫ x0
(x−t)[a0 + a1t+ a2t
2 + a3t3 + · · ·
]2dt
Integrating the right side, and equating the coefficients of like powersof x
a0 = ±1, a1 = ∓1, a2 = ∓ 12! , a3 = ± 1
3! , · · ·
Consequently, the exact solution is given by
u(x) = ±(cosx− sinx)
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March 4, 2015 14:44 book-9x6 9571-Root page 172
172 Chapter 8. Nonlinear Volterra Integral Equations
8.3.2 Conversion to a Volterra Equation of the SecondKind
Exercises 8.3.2
1. Differentiating both sides, and using Leibniz rule gives
v(x) = x6 + 17x
7 −∫ x0v(t) dt
This gives
v0(x) = x6, v1(x) = 0
This in turn gives
u(x) = ±x3
2. Differentiating both sides, and using Leibniz rule gives
v(x) = x8 + 19x
9 −∫ x0v(t) dt
This gives
v0(x) = x8, v1(x) = 0
This in turn gives
u(x) = ±x4
3. Differentiating both sides, and using Leibniz rule gives
v(x) = x12 + 113x
14 −∫ x0v(t) dt
This gives
v0(x) = x12, v1(x) = 0
This in turn gives
u(x) = ±x4
4. Differentiating both sides, and using Leibniz rule gives
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March 4, 2015 14:44 book-9x6 9571-Root page 173
8.4. Nonlinear Weakly-Singular Volterra Equation 173
v(x) = 32e
2x − 12 −
∫ x0v(t) dt
This gives
v0(x) = e2x, v1(x) = 0
This in turn gives
u(x) = ±ex
5. Proceeding as before gives
v0(x) = x2e2x, v1(x) = 0
This in turn gives
u(x) = ±xex
5. Proceeding as before gives
v0(x) = x2e2x, v1(x) = 0
This in turn gives
u(x) = ±xex
6. Proceeding as before gives
v0(x) = e−2x, v1(x) = 0
This in turn gives
u(x) = ±e−x
8.4 Nonlinear Weakly-Singular Volterra
Equation
Exercises 8.4
1. Decompose f(x) into two parts, then set:
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March 4, 2015 14:44 book-9x6 9571-Root page 174
174 Chapter 8. Nonlinear Volterra Integral Equations
u0(x) = x3, u1(x) = − 7293080x
283 +
∫ x0
A0(t)
(x2−t2)13dt = 0
This gives
u(x) = x3
2. Decompose f(x) into two parts, then set:
u0(x) = x4, u1(x) = − 2312048πx
12 +∫ x0
u3(t)√x2−t2 dt = 0
This gives
u(x) = x4
3. Decompose f(x) into two parts, then set:
u0(x) = 5√
1 + x4, u1(x) = −π2 (1 + 38x
4) +∫ x0
u5(t)√x2−t2 dt = 0
This gives
u(x) = 5√
1 + x4
4. Decompose f(x) into two parts, then set:
u0(x) = 4√
1 + x+ x3, u1(x) = −(π2 + x+ 23x
3) +∫ x0
u4(t)√x2−t2 dt = 0
This gives
u(x) = 4√
1 + x+ x3
5. Decompose f(x) into two parts, then set:
u0(x) = 5√
cosx, u1(x) = −2√
sinx+∫ x0
u5(t)√sin x−sin t dt = 0
This gives
u(x) = 5√
cosx
6. Decompose f(x) into two parts, then set:
u0(x) = 4√
sinx, u1(x) = 2√
cosx− 1 +∫ x0
u4(t)√cos x−cos t dt = 0
This gives
u(x) = 4√
sinx