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Republic of Iraq
Ministry of Higher Education and Scientific Research
University of Baghdad
College of Education for Pure Science / Ibn Al-Haitham
Department of Mathematics
Semi-Analytical Iterative Methods for Non-Linear
Differential Equations
A Thesis
Submitted to the College of Education for Pure Science \ Ibn Al-Haitham,
University of Baghdad as a Partial Fulfillment of the Requirements for the
Degree of Master of Science in Mathematics
By
Mustafa Mahmood Azeez
B.Sc., University of Baghdad, 2010
Supervised By
Asst. Prof. Dr. Majeed Ahmed Weli
2017 AD 1438 AH
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๏ดพ11๏ดฟูุฑูุน ุงููู ุงูุฐูู ุขู ููุง ู ููู ูุงูุฐูู ุฃูุชูุง ุงูุนูู ุฏุฑุฌุงุช
ูู ุญ ุงูุฑ ู ุญู ุงูุฑ ุงููู ุณู ุจ
ู ุตุฏู ุงููู ุงูุนุธู
๏ดพ11๏ดฟุงุงููุฉ โ ุงูู ุฌุงุฏูุฉุณูุฑุฉ
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ูุฏุงุก ุงูุฅ
ุงูุฑุญูู ""ุจุณู ูููุง ุงูุฑุญู ู
ุนู ูู ููู ุงุนู ููุง ูุณูุฑู) (ู ูุฑุณููู ูุงูู ุคู ููู ููุงู
ู "ุตุฏู ูููุง ุงูุนุธู"
ุงููุญุธุงุช ูุงูุชุทูุจ ..ูุงููุทูุจ ุงูููุงุฑ ุฅุงู ุจุทุงุนุชู .. ุจูุฑุฑ ุฅููู ุงููุทูุจ ุงูููู ุฅุงู
..ูุงู ุชุทูุจ ุงูุฌูุฉ ุฅุงู ุจุฑุคูุชู ..ูุงู ุชุทูุจ ุงุขูุฎุฑุฉ ุฅุงู ุจุนูู ..ุฅุงู ุจุดูุฑ
ุฅูู ูุจู ุงูุฑุญู ุฉ ูููุฑ ุงูุนุงูู ูู ..ู ุงุฃูู ุงูุฉ ููุตุญ ุงุฃูู ุฉุบ ุงูุฑุณุงูุฉ ูุฃุฏ ุฅูู ู ู ุจู
"ูู ูุณูู ุชุนุงูู ุนููุตูู ูููุง " ุฏุณูุฏูุง ู ุญู
ูุงูู ู ู ู ูุฏ ูู ุทุฑูู ุงูุนูู ..ุฃุณุงุชูุชู ุงุฃููุงุถู ุงูู ุจูุงุฑู ุงูุนูู ูุงูู ุนุฑูุฉ
ุตุฏูุงุก ุงุงููู ูุงุฃู ูุงูู ,ุฏ ู ุฌูุฏ ุงุญู ุฏ .ู .ุฃ ุงููุงุถูุฅูู ุงุณุชุงุฐู
ูู ูู ู ุดูุงุฑู ุงูุชุนููู ู ูู ุง ุฑุงู ููู ู ู ุฏุนู
ู ุตุทูู ู ุญู ูุฏ
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Supervisorโs Certification
I certify that this thesis was prepared under my supervision at the department
of mathematics, college of education for pure science / Ibn Al-Haitham,
university of Baghdad as a partial fulfillment of the requirements for the
degree of master of science in mathematics.
Signature:
Name: Asst. Prof. Dr. Majeed Ahmed Weli
Date: 11 / 9 / 2017
In view of the available recommendation, I forward this thesis for debate by
the examining committee.
Signature:
Name: Asst. Prof. Dr. Majeed Ahmed Weli
Head of Department of Mathematics
Date: 11 / 9 / 2017
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Committee Certification
We certify that we have read this thesis entitled "Semi-Analytical Iterative
Methods for Non-Linear Differential Equations" and as examining committee,
we examined the student (Mustafa Mahmood Azeez) in its contains and what is
connected with it, and that in our opinion it is adequate for the partial fulfillment of
the requirement for the degree of Master of Science in Mathematics.
(Member and Supervisor)
Signature:
Name: Asst. Prof. Dr. Majeed A. Weli
Date: 12 / 12 / 2017
(Member)
Signature:
Name: Asst. Prof. Dr. Osama H. Mohammed
Date: 12 /12 / 2017
(Chairman)
Signature:
Name: Prof. Dr. Saad N. Ali
Date: 11 / 12 / 2017
(Member)
Signature:
Name: Lecturer Dr. Ghada H. Ibrahim
Date: 11 / 12 / 2017
Approved by the University Committee of graduate studies.
Signature:
Name: Prof. Dr. Khalid F. Ali
The dean of Education College for Pure Science/ Ibn AL Haitham University of
Baghdad
Date: 12 / 12 / 2017
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ACKNOWLEDGEMENTS
I thank God Almighty our creator for his generosity and great good for helping
me in my scientific career and finishing my current thesis. I would like to thank
my supervisor Dr. Majeed Ahmed Weli who supervised this thesis, and for his
time, patience, motivation and encourage me all the time.
Last but not least. I would like to thank all department staff, my family and dear
friends who have had teach, encourage and support that provided me to finish
my thesis.
Mustafa Mahmood
2017
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Abstract
The main aim of this thesis is to use two semi-analytical iterative
methods to find the analytical solutions for some important problems in
physics and engineering.
The first objective is to use an iterative method which is proposed by
Temimi and Ansari namely (TAM) for finding the analytical solutions for
Riccati, pantograph, and beam deformation equations. Also, the TAM has
been applied for 1D, 2D and 3D nonlinear Burgersโ equations and systems
of equations to get the analytic solutions for each of them. The convergence
of the TAM has been investigated and successfully proved for those
problems.
The second objective is to use an iterative technique based on Banach
contraction method (BCM) which is easy to apply in dealing with nonlinear
terms. The BCM is used to solve the same physical and engineering
problems mentioned above. In addition, we have been presented several
comparisons among the TAM, BCM, variational iteration method (VIM) and
Adomian decomposition method (ADM) with some conclusions and future
works. The presented methods are efficient and have high accuracy. The
TAM and BCM do not require any restrictive assumptions in the nonlinear
case. In addition, both methods do not rely on using any additional
restrictive assumptions as in the ADM or VIM or any other iterative
methods.
The software used for the calculations in this presented thesis is
MATHEMATICAยฎ 10.0.
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AUTHORโS PUBLICATIONS
Journal Papers
1. M. A. Al-Jawary, M. M. Azeez, Efficient Iterative Method for Initial and
Boundary Value Problems Appear in Engineering and Applied Sciences,
International Journal of Science and Research, 6 (6) (2017) 529- 538.
2. M. A. AL-Jawary, M. M. Azeez, G. H. Radhi, Analytical and numerical
solutions for the nonlinear Burgers and advection-diffusion equations by
using a semi-analytical iterative method, Computers & Mathematics with
Applications (under review).
3. M. A. AL-Jawary, M. M. Azeez, G. H. Radhi, Banach Contraction
Method for Some Physical and Engineering Problems, (under
preparation).
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Contents
Subject Page
No.
Introduction i
Chapter 1: Primary Concepts 1
1.1 Introduction 1
1.2 Preliminaries 2
1.2.1 Initial Value Problems 2
1.2.2 Boundary Value Problems 2
1.3 Ordinary Differential Equations (ODEs) 4
1.3.1 Riccati differential equation 5
1.3.2 Analytical methods available in literatures to solve the
Riccati differential equation
5
1.3.3 The Adomian decomposition method 5
1.3.4 Error analysis 8
1.3.5 Variational Iteration Method 11
1.3.6 Pantograph differential equation 15
1.3.7 VIM for solving the pantograph differential equation 16
1.3.8 Beam deformation equation 17
1.3.9 VIM for solving the nonlinear beam deformation problem 18
1.4 Partial Differential Equations (PDEs) 22
1.4.1 Burgersโ equation 22
1.4.2 Exact and numerical solutions for nonlinear Burgersโ
equation by ADM
24
1.4.3 The VIM to solve the 2D Burgersโ equation 30
Chapter 2: Semi-Analytical Iterative Method for Solving Some
Ordinary and Partial Differential Equations
33
2.1 Introduction 33
2.2 The basic idea of the TAM 34
2.3 The convergence of the TAM 35
2.4 Application of the TAM with convergence for the ODE and
PDE
36
2.4.1 Exact and numerical solutions for nonlinear Riccati
equation by TAM
36
2.4.2 TAM for solving linear pantograph equation 42
2.4.3 Solving the nonlinear beam equation by using the TAM 45
2.4.4 Solving the nonlinear Burgersโ equation by using TAM 50
2.4.5 Solving the nonlinear system of Burgersโ equations by
the TAM
64
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Chapter 3: Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
87
3.1 Introduction 87
3.2 Contractions: Definition and Example 88
3.3 Banach Contraction Principle 89
3.3.1 Solution of differential equations using Banach
contraction principle
89
3.4 The basic idea of the BCM 91
3.5 The BCM to solve Riccati, pantograph and beam deformation
equations
92
3.5.1 Exact and numerical solutions for nonlinear Riccati
equation by BCM
92
3.5.2 BCM for solving linear pantograph equation 96
3.5.3 Solving the nonlinear beam equation by using BCM 97
3.5.4 Solving the nonlinear Burgersโ equation by using the
BCM
101
3.5.5 Solving the nonlinear system of Burgersโ equations by
using BCM
108
Chapter 4: Conclusions and Future Works 117
4.1 Introduction 117
4.2 Conclusions 117
4.3 Future Works 121
References 122
Appendices 130
Appendix A: Code of Mathematica for chapter one (ADM) 130
Appendix B: Code of Mathematica for chapter one (VIM) 134
Appendix C: Code of Mathematica for chapter two (TAM) 136
Appendix D: Code of Mathematica for chapter three (BCM) 137
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List of Tables
Tableโs name Pageโs
No. Table 1.1: The maximal error remainder: ๐๐ธ๐ ๐ by the ADM,
where ๐ = 1, โฆ ,6. 11
Table 1.2: The maximal error remainder: ๐๐ธ๐ ๐ by the VIM,
where ๐ = 1, โฆ ,6. 14
Table 1.3: Results of the absolute errors for VIM. 21
Table 1.4: The maximal error remainder: ๐๐ธ๐ ๐ by the ADM. 30
Table 1.5: The maximal error remainder: ๐๐ธ๐ ๐ by the VIM. 32
Table 2.1: The maximal error remainder: ๐๐ธ๐ ๐ by the TAM. 41
Table 2.2: Results of the absolute errors for TAM. 48
Table 2.3: The maximal error remainder: ๐๐ธ๐ ๐ by the TAM,
where ๐ = 1, โฆ ,4.
56
Table 3.1: The maximal error remainder: ๐๐ธ๐ ๐ by the BCM,
where ๐ = 1, โฆ ,6.
95
Table 3.2: Results of the absolute errors for BCM. 100
Table 3.3: The maximal error remainder: ๐๐ธ๐ ๐ by the BCM. 104
Table 4.1: Comparative results of the maximal error remainder:
๐๐ธ๐ ๐for TAM, BCM, VIM and ADM, where ๐ = 1, โฆ ,6. 118
Table 4.2: Comparative results for the absolute errors of the TAM,
BCM, VIM and HPM.
119
Table 4.3: Comparative results the maximal error remainder:
๐๐ธ๐ ๐ for TAM, BCM, VIM and ADM, where ๐ = 1, โฆ ,4. 120
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List of Figures
Figureโs name Pageโs
No.
Figure 1.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by
ADM.
11
Figure 1.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by
VIM.
14
Figure 1.3: Beam on the elastic bearing. 18
Figure 1.4: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by
ADM.
30
Figure 1.5: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by
V1M.
32
Figure 2.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by
TAM.
41
Figure 2.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by
TAM.
56
Figure 3.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by
BCM.
95
Figure 3.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by
BCM.
104
Figure 4.1: Comparison of the maximal error remainder for TAM,
BCM, VIM and ADM.
118
Figure 4.2: Comparison of the maximal error remainder for TAM,
BCM, VIM and ADM.
120
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LIST OF SYMBOLS AND ABBREVIATIONS
๐ด๐: Adomian Polynomials
๐ด๐ท๐: Adomian Decomposition Method
๐ต๐ถ๐: Banach Contraction Method
๐ต๐๐: Boundary Value Problem
๐ธ๐ : Error Remainder
๐ป๐ด๐: Homotopy Analysis Method
๐ผ๐๐: Initial Value Problem
Re: Reynolds number
๐๐ธ๐ : Maximum Error Remainder
๐๐ท๐ธ๐ : Ordinary Differential Equations
๐๐ท๐ธ๐ : Partial Differential Equations
๐๐ด๐: Temimi- Ansari Method
๐๐ผ๐: Variational Iteration Method
๐๐ผ๐ธ: Volterra Integral Equation
๐: Lagrangeโs multiplayer
๐ฟ: Variation
|๐๐|: Absolute Error
โ โ: Norm
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Introduction
i
INTRODUCTION
The ordinary and partial differential equations play an important role in
many problems that appears in different fields of engineering and applied
sciences. The past few decades have seen a significant progress to implement
analytical, approximate methods for solving linear and nonlinear differential
equations. For examples: Adomian decomposition method (ADM) [68],
variational iteration method (VIM) [4], homotopy perturbation method
(HPM) [63], differential transform method (DTM) [78] and many others.
Although these methods provided or given some useful solutions, however,
some drawbacks have emerged such as the calculation of the Adomian
polynomials for the nonlinear problems in ADM, calculate the Lagrange
multiplier in the VIM and the calculation became more complicated after
several iterations, homotopy construction and solving the corresponding
equations in HPM.
In this thesis, some of the ordinary and partial differential equations that
appear in the problems of chemistry, physics, engineering and other applied
sciences will be solved by using iterative methods.
One of these equations is the Riccati equation, which is an initial value
problem of nonlinear ordinary differential equation, which plays a significant
role in many fields of applied science such as random processes, optimal
control, diffusion problems, network synthesis and financial mathematics
[26].
The other one is the pantograph equation which is originated from the
work of Ockendon and Tayler on the collection of current by the pantograph
head of an electric locomotive [38]. The pantograph equations are appeared in
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Introduction
ii
modeling of various problems in engineering and sciences such as biology,
economy, control and electrodynamics [16].
Moreover, the beam deformation equation will be solved, which is a
nonlinear boundary value problem (BVP) which is frequently used as
mathematical models in viscoelastic, inelastic flows and deformation of
beams [1].
Furthermore, the other equation that will be solved is the Burgersโ
equation which is a partial differential equation and firstly introduced by
Harry Bateman in 1915 [27] and it was subsequently processed to become as
Burgersโ equation [45]. Burgersโ equation has a lot of importance in
engineering and physical fields, especially in problems that have the form of
nonlinear equations and equation systems. The applications of Burgersโ
equations by mathematical scientists and researchers has become more
commonplace and interesting, it has been known that this equation describes
different types of phenomena such as modeling of dynamics, heat conduction,
acoustic waves, turbulence and many others [5, 10, 36, 45, 56 , 61, 77]. In
recent decades, some scientists and researchers used some methods and
techniques to solve these types of problems [22, 39, 42, 44, 71] which will be
discussed in this thesis and solved in reliable iterative methods.
Recently, Temimi and Ansari have introduced a semiโanalytical iterative
method namely (TAM) for solving nonlinear problems [28]. The main feature
of the TAM is very effective and reliable, does not require any restrictive
assumptions to deal with nonlinear terms, time saver, and has a higher
convergence and accuracy. The TAM was inspired from the homotopy
analysis method (HAM) [70] and it is one of the famous iterative methods
that used for solving nonlinear problems [65]. Moreover, the TAM is
successfully used to solve many differential equations, such as Duffing
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Introduction
iii
equations [51], kortewegโde vries equations [18], chemistry problems [52]
and nonlinear thin film flow problems [53].
In addition, Sehgal and Bharucha-Reid proved the probabilistic version
of the classical Banach contraction method (BCM) in 1972 [3]. The BCM
characterized as one of the developments of Picardโs method where the ease
of application clearly observed which makes it distinct from the other known
iterative methods. The BCM used to solve various nonlinear functional delay
equations including partial differential equations (PDE), ordinary differential
equations (ODE), system of ODEs and PDEs and algebraic equations [75].
The main advantages of the BCM, it does not require any external
assumptions to deal with the nonlinear terms like the TAM. It avoided the
large computational work or calculated the Lagrange's multipliers in the VIM.
This thesis has been arranged as follows: in chapter one, the introductory
concepts about the non-linear differential equations (ordinary and partial) and
some analytic and approximate methods will be introduced to solve some
scientific applications, such as ADM and VIM. Chapter two presents the basic
idea of the TAM as well as an overview of the convergence of TAM, which is
based on the Banach fixed-point theory. Furthermore, the TAM is
successfully implemented to solve the Riccati equations, pantograph, and
beam deformation equations. In addition, the TAM will be implemented to
solve the Burgers and system equations in 1D, 2D and 3D to illustrate the
efficiency and accuracy of the proposed method. In chapter three, the basic
concepts of the BCM will be presented, and the application of the BCM for
the same problems solved by TAM will be introduced. Finally, in Chapter
four the conclusions and future works will be given.
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CHAPTER 1
PRIMARY
CONCEPTS
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Chapter One Primary Concepts
1
Chapter 1
Primary Concepts
1.1 Introduction
The importance of the non-linear differential equations highlights when
dealing with the problems that are occurring in different fields such as
physical and chemical sciences and engineering. Many of these problems are
of a nonlinear form. It is known that there is a difficulty in to solving these
nonlinear problems and require to use efficient methods to get the analytic,
approximate or numerical solutions.
Moreover, the problems that will be solved in this thesis are
mathematical models contain non-linear ODEs such as Riccati equation,
pantograph equation, beam deformation equation and nonlinear PDEs such as
Burgersโ equation, that appeared in many engineering and applied sciences.
This chapter contains four sections. In section two, some definitions and
theorems that will be used in the next chapters will be given. In section three,
some types of ODEs are presented, such as Riccati, pantograph and beam
deformation equations. In addition, some analytic and approximate methods
will be introduced to solve some scientific applications, such as ADM and the
VIM. Finally, in section four, the Burgersโ equations and system of equations
in 1D, 2D, and 3D will be introduced. Also, the 1D, 2D Burgersโ equations
will be given in literature by ADM and VIM.
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Chapter One Primary Concepts
2
1.2 Preliminaries
1.2.1 Initial Value Problems [6]:
Differential equations are often divided into two classes, ordinary and
partial, according to the number of independent variables. A more meaningful
division, however, is between initial value problems, which usually model
time-dependent phenomena, and boundary value problems, which generally
model steady-state systems.
A differential equation (DE) that has given conditions allows us to find
the specific function that satisfies a given DE rather than a family of
functions. These types of problems are called initial value problems (IVP). In
physics or other sciences, frequently amounts to solving an initial value
problem. For example, an initial value problem is a differential equation
๐ฃโฒ(๐ฅ) = ๐(๐ฅ, ๐ฃ(๐ฅ)), with ๐: ๐ โ ๐ ร ๐ ๐ โ ๐ ๐,
where ๐ is an open set of ๐ ร ๐ ๐, together with a point in the domain of ๐,
(๐ฅ0, ๐ฃ0) โ ๐ called the initial condition. A solution to an initial value problem
is a function ๐ฃ that is a solution to the differential equation and satisfies
๐ฃ(๐ฅ0) = ๐ฃ0.
1.2.2 Boundary Value Problems [6]:
When imposed an ordinary or a partial differential equation the
conditions are given at more than one point and the differential equation is of
order two or greater, it is called a boundary value problem (BVP). For
example, if the independent variable is time over the domain [0,1], a
boundary value problem would specify values for ๐ฃ(๐ฅ) at both ๐ฅ =
0 and ๐ฅ = 1, whereas an initial value problem would specify a value of
๐ฃ(๐ฅ) and ๐ฃโฒ(๐ฅ) at time ๐ฅ = 0. In addition, there are four types of boundary
value problems or boundary conditions: Dirichlet, Neunmann, Mixed, Robin
boundary conditions.
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Chapter One Primary Concepts
3
The boundary value problems have a greater importance in various
scientific fields. Many of researchers and scientists experiencing in their
scientific researchers have a problem in finding the desired solution for those
boundary value problems. These problems come with boundary conditions,
which are satisfying the solution. In some cases, these problems are
formulated from BVP to Volterra integral equation. The integral equation
make the reaching to the solution simple and easy.
Definition 1.1: [6]
An integral equation is called Volterra integral equation (VIE) if limits
of integration are functions of x rather than constants. The first kind Volterra
integral equations, is:
๐(๐ฅ) = ฮป โซ ๐(๐ฅ, ๐ก)๐ฃ(๐ก)๐๐ก
๐ฅ
๐
, (1.1)
where the unknown function ๐ฃ(๐ฅ) appears only inside integral sign, and the
Volterra integral equations of the second kind is
๐ฃ(๐ฅ) = ๐(๐ฅ) + ฮป โซ ๐(๐ฅ, ๐ก)๐ฃ(๐ก)๐๐ก,
๐ฅ
๐
(1.2)
where the unknown function ๐ฃ(๐ฅ) appears inside and outside the integral sign.
Definition 1.2: [57]
Let (๐, ๐) be a metric space. Then a mapping ๐น: ๐ โ ๐, is said to be
Lipschitz if there exists a real number ๐ โฅ 0, such that
๐(๐น(๐ฅ), ๐น(๐ฆ)) โค ๐๐(๐ฅ, ๐ฆ) for all ๐ฅ, ๐ฆ in ๐.
In addition ๐น is called a contraction mapping on ๐ if ๐ < 1.
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Chapter One Primary Concepts
4
Theorem 1.1 (Banach Fixed Point Theorem): [69]
Let (๐, ๐) be a non-empty complete metric space with a contraction
mapping ๐น: ๐ โ ๐. Then ๐น admits a unique fixed-point x* โ ๐
F(x*) = x*
Theorem 1.2: [57]
Let ๐น be a mapping of a complete metric space (๐, ๐) into itself such
that ๐น๐ is a contraction mapping of ๐ for some positive integer ๐. Then ๐น has
a unique fixed point in ๐.
Proof: See [57].
1.3 Ordinary Differential Equations (ODEs)
An ordinary differential equation is a relation containing one real
independent variable ๐ฅ โ ๐ = (โโ, โ), the real dependent variable v, and
some of its derivatives ( ๐ฃโฒ, ๐ฃโฒโฒ, ๐ฃโฒโฒโฒ, โฆ , ๐ฃ๐), ๐ฃโฒ =๐๐ฃ
๐๐ฅ. ODEs arise in many
contexts of mathematics and science. Various differentials, derivatives, and
functions become related to each other via equations, and thus an ODE is a
result that describes dynamically changing phenomena, evolution, or
variation. Also, in specific mathematical fields, include geometry and
analytical mechanics, and scientific fields include much of physics, chemistry
and biology [4].
In this section, we have examined three specific problems in physical
and engineering sciences that are governed by non-linear ODEs and they are
solved by using ADM and VIM.
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Chapter One Primary Concepts
5
1.3.1 Riccati differential equation
The Riccati differential equation is named after the Italian noble man
Count Jacopo Francesco Riccati (1676-1754) [26]. This equation has many
applications such as network synthesis, financial mathematics [12], control
the boundary arising in fluid structure interaction [31]. Consider the following
nonlinear Riccati differential equation [20].
๐ฃโฒ(๐ฅ) = ๐(๐ฅ) + ๐(๐ฅ)๐ฃ(๐ฅ) + ๐(๐ฅ)๐ฃ2(๐ฅ), ๐ฃ(๐ฅ0) = ๐ผ, ๐ฅ0 โค ๐ฅ โค ๐๐ , (1.3)
where ๐(๐ฅ), ๐(๐ฅ) and ๐(๐ฅ) are continuous functions, ๐ฅ0, ๐๐ and ฮฑ are
arbitrary constants, and ๐ฃ(๐ฅ) is unknown function.
A substantial amount of research work has been done to develop the
solution for Riccati differential equation. The most used methods are ADM
[68], HAM [34]. Taylor matrix method [59] and Haar wavelet method [79],
HPM [82] combination of Laplace Adomian decomposition and Pade
approximation methods [64], and many other methods available in the
literature.
1.3.2 Analytical methods available in literatures to solve the
Riccati differential equation
In this subsection, the ADM and VIM will be used to solve the nonlinear
Riccati differential equation.
1.3.3 The Adomian decomposition method
In the early 1980s, George Adomian had introduced the ADM [22, 23,
24]. This method plays an important role in applied mathematics. The
importance of the ADM is due to its strength and effectiveness, and it can
easily deal with many types of ODEs, PDEs (linear and nonlinear), integral
equations and the other types of equations [7]. It consists of decomposing the
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Chapter One Primary Concepts
6
unknown function ๐ฃ(๐ฅ) of any equation into a sum of an infinite number of
components defined by the decomposition series.
๐ฃ(๐ฅ) = โ ๐ฃ๐(๐ฅ)
โ
๐=0
or equivalently
๐ฃ(๐ฅ) = ๐ฃ0(๐ฅ) + ๐ฃ1(๐ฅ) + ๐ฃ2(๐ฅ) + โฏ.
Where the linear components ๐ฃ๐(๐ฅ), ๐ โฅ 0, are evaluated in recursive
manner. The decompositions method concerns itself with finding the
components ๐ฃ0, ๐ฃ1, ๐ฃ2, ... etc.
The zero component is identified by all terms that are not included under
the integral sign. Consequently, those components ๐ฃ๐(๐ฅ), ๐ > 1 of the
unknown function ๐ฃ(๐ฅ) are completely determined by setting the recurrence.
Consider the following nonlinear differential equation
๐ฟ๐ฃ + ๐๐ฃ = ๐(๐ฅ), (1.4)
where ๐ฟ and ๐ are linear and nonlinear operators, respectively, and ๐(๐ฅ) is
the source inhomogeneous term. The ADM introduces for Eq. (1.4) in the
form
๐ฃ0(๐ฅ) = ๐(๐ฅ),
๐ฃ๐+1(๐ฅ) = โซ (โ ๐ฃ๐(๐ก))๐๐ก, ๐ โฅ 0
โ
๐=0
๐ฅ
0
(1.5)
For the nonlinear solution ๐ฃ(๐ฅ), and the infinite series of polynomials
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Chapter One Primary Concepts
7
๐ฃ(๐ฅ) = โ ๐ด๐(๐ฃ0, ๐ฃ1, โฆ , ๐ฃ๐)
โ
๐=0
,
where the components ๐ฃ๐(๐ฅ) of the solution ๐ฃ(๐ฅ) will be determined,
recurrently and ๐ด๐ are the Adomian polynomials which are obtained from the
definitional formula [24]
๐ด๐ =1
๐!
๐๐
๐๐๐[๐(โ ๐๐๐ฃ๐)
๐
๐=0
]๐=0, ๐ = 1,2, โฆ
The formulas of the first several Adomian polynomials from ๐ด0 to ๐ด5, will be
listed below as given in [24]
๐ด0 = ๐(๐ฃ0),
๐ด1 = ๐ฃ1๐โฒ(๐ฃ0),
๐ด2 = ๐ฃ2๐โฒ(๐ฃ0) +1
2!๐ฃ1
2๐โฒโฒ(๐ฃ0),
๐ด3 = ๐ฃ3๐โฒ(๐ฃ0) + ๐ฃ1๐ฃ2๐โฒโฒ(๐ฃ0) +1
3!๐ฃ1
3๐โฒโฒโฒ(๐ฃ0),
๐ด4 = ๐ฃ4๐โฒ(๐ฃ0) + (1
2!๐ฃ2
2 + ๐ฃ1๐ฃ3)๐โฒโฒ(๐ฃ0) +1
2!๐ฃ1
2๐ฃ2๐โฒโฒโฒ(๐ฃ0)
+1
4!๐ฃ1
4๐โฒโฒโฒโฒ(๐ฃ0),
๐ด5 = ๐ฃ5๐โฒ(๐ฃ0) + (๐ฃ2๐ฃ3 + ๐ฃ1๐ฃ4)๐โฒโฒ(๐ฃ0) + (1
2!๐ฃ1๐ฃ2
2 +1
2!๐ฃ1
2๐ฃ3)๐โฒโฒโฒ(๐ฃ0)
+1
3!๐ฃ1
3๐ฃ2๐โฒโฒโฒโฒ(๐ฃ0) +1
5!๐ฃ1
5๐โฒโฒโฒโฒโฒ(๐ฃ0),
โฆ
and so on.
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Chapter One Primary Concepts
8
Note 1.1:
It is worth mentioning that we can use the transformation formula of
converting multiple integrals to single [6]
โซ โซ โฆ โซ ๐ฃ(๐ฅ๐)๐๐ฅ๐ โฆ ๐๐ฅ1
๐ฅ
0
=1
(๐ โ 1)!โซ (๐ฅ โ ๐ก)๐โ1๐ฃ(๐ก)๐๐ก. (1.6)
๐ฅ
0
๐ฅ
0
๐ฅ
0
1.3.4 Error analysis
We used the appropriate function of the maximal error remainder [35,
54] to assess the accuracy of the approximate solution.
๐ธ๐ ๐(๐ฅ) = ๐ฟ(๐ฃ) โ ๐(๐ฃ) โ ๐. (1.7)
The maximal error remainder is
๐๐ธ๐ ๐ = max0โค๐ฅโค1
|๐ธ๐ ๐(๐ฅ)|. (1.8)
The ADM will be implemented to solve the Riccati differential equation
Example 1.1 [74]:
Let us consider the Riccati equation of the form.
๐ฃโฒ = โ๐ฃ2 + 1, with initial condition ๐ฃ(0) = 0. (1.9)
Eq. (1.9) can be converted to the following Volterra integral equation
๐ฃ(๐ฅ) = ๐ฅ โ โซ ๐ฃ2(๐ก)๐๐ก, (1.10)๐ฅ
0
The Adomain method proposes that the solution ๐ฃ(๐ฅ) can be expressed by the
decomposition series:
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Chapter One Primary Concepts
9
๐ฃ(๐ฅ) = โ ๐ฃ๐(๐ฅ)
โ
๐=0
,
and the nonlinear term ๐ฃ2, be equated to:
๐ฃ2 = โ ๐ด๐.
โ
๐=0
Therefore, Eq. (1.10) will be:
โ ๐ฃ๐(๐ฅ) = ๐ฅ โ โซ (โ ๐ด๐
โ
๐=0
(๐ก))๐๐ก, (1.11)๐ฅ
0
โ
๐=0
The following recursive relation:
๐ฃ0 = ๐ฅ,
๐ฃ๐+1 = โ โซ (โ ๐ด๐
โ
๐=0
(๐ก)) ๐๐ก, ๐ฅ
0
๐ โฅ 0. (1.12)
Now, we find Adomian polynomial ๐ด๐.
๐ด0 = ๐(๐ฃ0) = ๐ฃ02,
๐ด1 = ๐ฃ1 ๐โฒ(๐ฃ0) = 2๐ฃ0๐ฃ1,
๐ด2 = ๐ฃ2๐โฒ(๐ฃ0) +1
2!๐ฃ1
2๐โฒโฒ(๐ฃ0) = 2๐ฃ0๐ฃ2 + ๐ฃ12,
๐ด3 = ๐ฃ3๐โฒ(๐ฃ0) + ๐ฃ1๐ฃ2๐โฒโฒ(๐ฃ0) +1
3!๐ฃ1
3๐โฒโฒโฒ(๐ฃ0) = 2๐ฃ0๐ฃ3 + 2๐ฃ1๐ฃ2,
โฎ
Consequently, by applying the Mathematicaโs code, see appendix A, we get
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Chapter One Primary Concepts
10
๐ฃ0 = ๐ฅ,
๐ฃ1 = โ โซ ๐ด0(๐ก)๐๐ก = โ๐ฅ3
3,
๐ฅ
0
๐ฃ2 = โ โซ ๐ด1(๐ก)๐๐ก =2๐ฅ5
15,
๐ฅ
0
๐ฃ3 = โ โซ ๐ด2(๐ก)๐๐ก = โ17๐ฅ7
315,
๐ฅ
0
๐ฃ4 = โ โซ ๐ด3(๐ก)๐๐ก =62๐ฅ9
2835,
๐ฅ
0
๐ฃ5 = โ โซ ๐ด4(๐ก)๐๐ก = โ1382๐ฅ11
155925,
๐ฅ
0
๐ฃ6 = โ โซ ๐ด5(๐ก)๐๐ก =21844๐ฅ13
6081075,
๐ฅ
0
Then, we have
๐ฃ(๐ฅ) = ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
17๐ฅ7
315+
62๐ฅ9
2835โ
1382๐ฅ11
155925+
21844๐ฅ13
6081075.
The obtained series solution in Eq. (1.9) can be used to calculate the
maximal error remainder for show the highest accuracy level that we can
achieve. It can be clearly seen in table 1.1 and figure 1.1 that by increasing the
iterations the errors will be reduced.
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Chapter One Primary Concepts
11
Table 1.1: The maximal error remainder: ๐๐ธ๐ ๐ by the ADM, where ๐ =1, โฆ ,6.
n ๐ด๐ฌ๐น๐ by ADM
1 6.65556 ร 10โ5
2 3.76891 ร 10โ7
3 1.96289 ร 10โ9
4 9.72067 ร 10โ12
5 4.65513 ร 10โ14
6 2.15106 ร 10โ16
Figure 1.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by ADM.
1.3.5 Variational Iteration Method
The method used to solve the Riccati differential equation is called the
VIM [20]. The VIM established by Ji-Huan He in 1999 is now used to deal
with a wide variety of linear and nonlinear, homogeneous and inhomogeneous
equations [40, 41]. The obtained series can be employed for numerical
purposes if exact solution is not obtainable. To illustrate the basic concepts of
the VIM, we consider the following nonlinear equation [39, 43]:
๐ฟ๐ฃ(๐ฅ) + ๐๐ฃ(๐ฅ) = ๐(๐ฅ), ๐ฅ > 0, (1.13)
1 2 3 4 5 6
10 15
10 13
10 11
10 9
10 7
10 5
n
ME
Rn
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Chapter One Primary Concepts
12
where, ๐ฟ is a linear operator, ๐ a nonlinear operator and ๐(๐ฅ) is the source of
the inhomogeneous term. The VIM introduces functional for Eq. (1.13) in the
form:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ ๐(๐ )(๐ฟ๐ฃ๐(๐ ) + ๐๏ฟฝ๏ฟฝ๐(๐ ) โ ๐(๐ ))๐๐ , (1.14)๐ฅ
0
where ๐ is a general Lagrangian multiplier which can be identified optimally
via the variational theory, and ๏ฟฝ๏ฟฝ๐ as a restricted variation. The Lagrange
multiplier ๐ may be constant or a function and it is given by the general
formula [8, 9]:
๐(๐ ) = (โ1)๐1
(๐ โ 1)!(๐ โ ๐ฅ)๐โ1. (1.15)
However, for fast convergence, the function ๐ฃ0(๐ฅ) should be selected by
using the initial condition for ODE as follows:
๐ฃ0(๐ฅ) = ๐ฃ(0), for first order.
๐ฃ0(๐ฅ) = ๐ฃ(0) + ๐ฅ๐ฃโฒ(0), for second order.
๐ฃ0(๐ฅ) = ๐ฃ(0) + ๐ฅ๐ฃโฒ(0) +1
2!๐ฅ2๐ฃ"(0), for third order.
โฆ
and so on.
The successive approximations ๐ฃ๐+1, ๐ โฅ 0 of the solution ๐ฃ๐(๐ฅ) will
be readily upon using selective function ๐ฃ0(๐ฅ). Consequently, the solution is
given by
๐ฃ(๐ฅ) = ๐๐๐๐โโ
๐ฃ๐(๐ฅ).
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Chapter One Primary Concepts
13
In what follows, the VIM will be implemented to solve the Riccati
differential equation.
Example 1.2 [74]:
Rewrite the example 1.1 and solve it by VIM.
๐ฃโฒ = โ๐ฃ2 + 1, with initial condition ๐ฃ(0) = 0. (1.16)
By using Eq. (1.14), we get
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ [๐(๐ )(๐ฃ๐โฒ + ๐ฃ๐
2(๐ ) โ 1)]๐ฅ
0
๐๐ , (1.17)
By using the formula in Eq. (1.15) leads to ฮป = โ1. Substituting this value of
the Lagrange multiplier ฮป = โ1 into the functional Eq. (1.17) gives the
iteration formula:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) โ โซ (๐ฃ๐โฒ + ๐ฃ๐
2(๐ ) โ 1)๐ฅ
0
๐๐ . (1.18)
The initial approximation will be:
๐ฃ0(๐ฅ) = ๐ฃ(0) = 0,
By using equation (1.18), the following successive approximations will be
obtained:
๐ฃ0 = 0,
๐ฃ1 = ๐ฃ0 โ โซ (๐ฃ0โฒ + ๐ฃ0
2(๐ ) โ 1)๐ฅ
0
๐๐ = ๐ฅ,
๐ฃ2 = ๐ฃ1 โ โซ (๐ฃ1โฒ + ๐ฃ1
2(๐ ) โ 1)๐ฅ
0
๐๐ = ๐ฅ โ๐ฅ3
3,
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Chapter One Primary Concepts
14
๐ฃ3 = ๐ฃ2 โ โซ (๐ฃ2โฒ + ๐ฃ2
2(๐ ) โ 1)๐ฅ
0
๐๐ = ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
๐ฅ7
63,
By continuing in this way till n = 6, we have
๐ฃ6(๐ฅ) = ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
17๐ฅ7
315+
62๐ฅ9
2835โ
1382๐ฅ11
155925+
20404๐ฅ13
6081075
โ766609๐ฅ15
638512875+
4440998๐ฅ17
10854718875โ
3206482๐ฅ19
24105934125+ โฏ , (1.19)
Table 1.2 and Figure 1.2 illustrate the convergence of the solution
through the use of the maximal error remainder which can be clearly seen, the
error will be reduced by increasing the number of iterations.
Table 1.2: The maximal error remainder: ๐๐ธ๐ ๐ by the VIM, where ๐ =1, โฆ ,6.
n ๐ด๐ฌ๐น๐ by VIM
1 0.01
2 6.65556 ร 10โ5
3 2.65463 ร 10โ7
4 7.56708 ร 10โ10
5 1.67808 ร 10โ12
6 3.04791 ร 10โ15
Figure 1.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by VIM.
1 2 3 4 5 6
10 13
10 10
10 7
10 4
n
ME
Rn
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Chapter One Primary Concepts
15
1.3.6 Pantograph differential equation
The pantograph is a type of delay differential equations. In 1971 Taylor
and Okondon developed first the mathematical model of the pantograph [46].
Pantograph equation used in many applications, such as industrial
applications [49], studies based on biology, economy, control and
electrodynamics [58]. In addition, the nonlinear dynamic systems and
population models. Consider the following generalized pantograph equation
[11].
๐ฃ(๐)(๐ฅ) = โ โ ๐๐๐(๐ฅ)๐ฃ(๐)(๐ผ๐๐ฅ + ๐ฝ๐) + ๐(๐ฅ)
๐โ1
๐=0
๐ฝ
๐=0
, (1.20)
with initial conditions
โ ๐๐๐๐ฃ(๐)(0) = ๐๐
๐โ1
๐=0
, ๐ = 0, 1, โฆ , ๐ โ 1,
where ๐๐๐(๐ฅ) and ๐(๐ฅ) are analytical functions, ๐ฝ โฅ 1, ๐๐๐ , ๐๐ , ๐ผ๐ and ๐ฝ๐ are
real or complex constants.
Pantograph equation was solved by many authors either analytically or
numerically. For instance, Yusuf oglu [81] proposed an efficient algorithm for
solving generalized pantograph equations with the linear functional argument,
the authors of investigated an exponential approximation to obtain an
approximate solution of generalized pantograph-delay differential equations
[73]. High-order pantograph equations with initial conditions have been also
studied by Taylor method [62], ADM [14], DTM [78] and HPM [16].
Recently, Doha et al proposed and developed a new Jacobi rational-Gauss
collocation method for solving the generalized pantograph equations on a
semi-infinite domain [15].
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Chapter One Primary Concepts
16
1.3.7 VIM for solving the pantograph differential equation
Example 1.3 [11]:
Let us deal with the following pantograph differential equation,
๐ฃโฒโฒ =3
4๐ฃ + ๐ฃ (
๐ฅ
2) โ ๐ฅ2 + 2,
with the initial conditions ๐ฃ(0) = 0, ๐ฃ โฒ(0) = 0. (1.21)
By using Eq. (1.14), we get:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ [๐(๐ ) (๐ฃ๐โฒโฒ โ
3
4๐ฃ๐ โ ๐ฃ๐ (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ . (1.22)
Using the formula in Eq. (1.15) leads to ฮป = (๐ โ ๐ฅ). Substituting this value of
the Lagrange multiplier ฮป = (๐ โ ๐ฅ) into the functional Eq. (1.22) gives the
iteration formula:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ [(๐ โ ๐ฅ) (๐ฃ๐โฒโฒ โ
3
4๐ฃ๐ โ ๐ฃ๐ (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ ,
The initial approximation will be:
๐ฃ0(๐ฅ) = ๐ฃ(0) + ๐ฅ๐ฃโฒ(0) = 0, (1.23)
By using Eq. (1.23), the following successive approximations will be obtained
๐ฃ0 = 0,
๐ฃ1 = ๐ฃ0 + โซ [(๐ โ ๐ฅ) (๐ฃ0โฒโฒ โ
3
4๐ฃ0 โ ๐ฃ0 (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ = ๐ฅ2 โ๐ฅ4
12,
๐ฃ2 = ๐ฃ1 + โซ [(๐ โ ๐ฅ) (๐ฃ1โฒโฒ โ
3
4๐ฃ1 โ ๐ฃ1 (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ = ๐ฅ2 โ13๐ฅ6
5760,
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Chapter One Primary Concepts
17
๐ฃ3 = ๐ฃ2 + โซ [(๐ โ ๐ฅ) (๐ฃ2โฒโฒ โ
3
4๐ฃ2 โ ๐ฃ2 (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ ,
= ๐ฅ2 โ91๐ฅ8
2949120 ,
๐ฃ4 = ๐ฃ3 + โซ [(๐ โ ๐ฅ) (๐ฃ3โฒโฒ โ
3
4๐ฃ3 โ ๐ฃ3 (
๐
2) โ ๐ 2 + 2)]
๐ฅ
0
๐๐ ,
= ๐ฅ2 โ17563๐ฅ10
67947724800 ,
and so on. Then:
๐ฃ(๐ฅ) = ๐ฅ2 โ small term,
This series converges to the exact solution [11],
๐ฃ(๐ฅ) = ๐๐๐๐โโ
๐ฃ๐(๐ฅ) = ๐ฅ2.
1.3.8 Beam deformation equation
In the last two decades with the rapid development of applied science,
there has appeared mathematical model of the beam deformation equation
which is used in structural engineering and fluid mechanics.
Recently, many analytical methods are used to solve nonlinear beam
deformation equation, such as HPM [55], VIM [1], and ADM [25].
According to the classical beam theory in [1], the function ๐ฃ = ๐ฃ(๐ฅ)
represents the configuration of the deformed beam. The length of the beam is
๐ฟ = 1, where ๐ฅ = 0 at the left side and ๐ฅ = 1 at the right side and ๐ = ๐(๐ฅ) is
a load which causes the deformation [1], as it is shown in figure 1.3.
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Chapter One Primary Concepts
18
Let us consider the nonlinear beam deformation problem as a general
fourth order boundary value problem of the form [1],
๐ฃ(4)(๐ฅ) = ๐(๐ฅ, ๐ฃ, ๐ฃโฒ, ๐ฃ'', ๐ฃ'''), (1.24)
with the boundary conditions:
๐ฃ(๐) = ๐ผ1, ๐ฃโฒ(๐) = ๐ผ2,
๐ฃ(๐) = ๐ฝ1, ๐ฃโฒ(๐) = ๐ฝ2,
where ๐ is a continuous function on [a,b] and the parameters ๐ผ1,
๐ผ2, ๐ฝ1 and ๐ฝ2 are finite real arbitrary constants. Eq. (1.24) used as
mathematical models in engineering.
Figure 1.3: Beam on the elastic bearing.
In this subsection, the following problem of the beam deformation will be
solved by using the VIM.
1.3.9 VIM for solving the nonlinear beam deformation problem
Example 1.4 [1]:
Let us consider the following form of the beam deformation equation,
๐ฃ(4) = ๐ฃ2 + ๐(๐ฅ),
subject to the boundary conditions
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Chapter One Primary Concepts
19
๐ฃ(0) = 0, ๐ฃโฒ(0) = 0, ๐ฃ(1) = 1, ๐ฃโฒ(1) = 1. (1.25)
Where
๐(๐ฅ) = โ๐ฅ10 + 4๐ฅ9 โ 4๐ฅ8 โ 4๐ฅ7 + 8๐ฅ6 โ 4๐ฅ4 + 120๐ฅ โ 48,
By using Eq. (1.14), we have:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ [๐(๐ ) (๐ฃ๐(4)
โ ๐ฃ๐2 โ ๐(๐ ))]
๐ฅ
0
๐๐ . (1.26)
By using the formula in Eq. (1.15) leads to ฮป = (๐ โ๐ฅ)3
3!. Substituting this value
of the Lagrange multiplier ฮป = (๐ โ๐ฅ)3
3! into the functional Eq. (1.26) gives the
iteration formula:
๐ฃ๐+1(๐ฅ) = ๐ฃ๐(๐ฅ) + โซ [(๐ โ ๐ฅ)3
3!(๐ฃ๐
(4)โ ๐ฃ๐
2 โ ๐(๐ ))]๐ฅ
0
๐๐ . (1.27)
Let us assumed that an initial approximation has the form:
๐ฃ0(๐ฅ) = ๐๐ฅ3 + ๐๐ฅ2 + ๐๐ฅ + ๐, (1.28)
where a, b, c, and d are unknown constants to be further determined.
By using the iteration formula (1.27), the following first-order approximation
is obtained:
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Chapter One Primary Concepts
20
๐ฃ1 = ๐ฃ0 + โซ [(๐ โ ๐ฅ)3
3!(๐ฃ0
(4)โ ๐ฃ0
2 โ ๐(๐ ))]๐ฅ
0
๐๐ ,
= ๐ + ๐๐ฅ + ๐๐ฅ2 + ๐๐ฅ3 + (โ2 +๐2
24) ๐ฅ4 + (1 +
๐๐
60) ๐ฅ5
+ (๐2
360+
๐๐
180) ๐ฅ6 + (
๐๐
420+
๐๐
420) ๐ฅ7
+ (โ1
420+
๐2
1680+
๐๐
840) ๐ฅ8 +
๐๐๐ฅ9
1512+ (
1
630+
๐2
5040) ๐ฅ10
โ๐ฅ11
1980โ
๐ฅ12
2970+
๐ฅ13
4290โ
๐ฅ14
24024, (1.29)
Imposing the boundary conditions (1.25), into ๐ฃ1, results in the following
values:
๐ = โ0.00687154, ๐ = 2.00593, ๐ = 0, ๐ = 0. (1.30)
The following first-order approximate solution is then achieved:
๐ฃ1 = 2.005929514398968๐ฅ2 โ 0.006871538792438514๐ฅ3 โ 2๐ฅ4
+ ๐ฅ5 + 0.00001413881948623746๐ฅ8
โ 0.000009116284704424509๐ฅ9
+ 0.001587310955961384๐ฅ10 โ๐ฅ11
1980โ
๐ฅ12
2970+
๐ฅ13
4290
โ๐ฅ14
24024. (1.31)
Similarly, the following second-order approximation may be written:
๐ฃ2 = ๐ฃ1 + โซ [(๐ โ ๐ฅ)3
3!(๐ฃ1
(4)โ ๐ฃ1
2 โ ๐(๐ ))]๐ฅ
0
๐๐ , (1.32)
Imposing the boundary conditions, (1.25), into ๐ฃ2, yields
๐ = โ8.27907 ร 10โ7, ๐ = 2.000000763, ๐ = 0, ๐ = 0. (1.33)
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Chapter One Primary Concepts
21
The following second-order approximate solution is obtained:
๐ฃ2 = 2.0000007627556786๐ฅ2 โ 8.279069998454606 ร 10โ7๐ฅ3 โ 2๐ฅ4
+ ๐ฅ5 + 1.816085295168468 ร 10โ9๐ฅ8 โ โฏ. (1.34)
In order to check the accuracy of the approximate solution, we calculate
the absolute error, |๐๐| = |๐ฃ(๐ฅ) โ ๐ฃ๐(๐ฅ)| where ๐ฃ(๐ฅ) = ๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2 is
the exact solution and ๐ฃ๐(๐ฅ) is the approximate solution. In table 1.3 the
absolute error of VIM with n = 1, 2 is presented.
Table 1.3: Results of the absolute errors for VIM.
x |๐๐| for VIM |๐๐| for VIM
0 0 0
0.1 5.24236 ร 10โ5 6.79965 ร 10โ9
0.2 1.82208 ร 10โ4 2.3887 ร 10โ8
0.3 3.48134 ร 10โ4 4.62946 ร 10โ8
0.4 5.09092 ร 10โ4 6.90558 ร 10โ8
0.5 6.24721 ร 10โ4 8.72068 ร 10โ8
0.6 6.57823 ร 10โ4 9.58101ร 10โ8
0.7 5.81138 ร 10โ4 9.01332 ร 10โ8
0.8 3.92709 ร 10โ4 6.67054 ร 10โ8
0.9 1.45715 ร 10โ4 2.80674 ร 10โ8
1.0 3.3447 ร 10โ8 0
It can be seen clearly from table 1.3, that by increasing the iterations, the
errors will decreasing.
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Chapter One Primary Concepts
22
1.4 Partial Differential Equations (PDEs)
PDE is an equation that contains the dependent variable, and its partial
derivatives. In the ordinary differential equations, the dependent variable ๐ฃ =
๐ฃ(๐ฅ) depends only on one independent variable x. Unlike the ODEs, the
dependent variable in the PDEs, such as ๐ฃ = ๐ฃ(๐ฅ, ๐ก) or ๐ฃ = ๐ฃ(๐ฅ, ๐ฆ, ๐ก), must
depend on more than one independent variable. If ๐ฃ = ๐ฃ(๐ฅ, ๐ก), then the
function ๐ฃ depends on the independent variable ๐ฅ, and on the time variable
๐ก [7]. In addition, it is well known that most of the phenomena that arise in
mathematical, chemical, physics and engineering fields can be described by
partial differential equations. For example, the heat flow and the wave
propagation phenomena, physical phenomena of fluid dynamics and many
other models described by partial differential equations.
In this section, we have examined the Burgersโ equations governed by
nonlinear PDEs and will be introduced by using the ADM and the VIM.
1.4.1 Burgersโ equation
The Burgers equation was first presented by Bateman [27] and treated
later by Johannes Martinus Burgers (1895-1981) then it is widely named as
Burgersโ equation [45]. The introduction of the applications of Burgersโ
equations by mathematical scientists and researchers has become more
important and interesting, especially in problems that have the form of
nonlinear equations and systems of equations. It has been known that this
equation describes different types of phenomena such as modeling of
dynamics, heat conduction, acoustic waves and many others [5, 10, 36, 45,
61]. In addition, the 2D and 3D Burgersโ equations have played an important
role in many physical applications such as modeling of gas dynamics and
shallow water waves [37, 48].
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Chapter One Primary Concepts
23
We consider the 1D, 2D and (1+3)-D Burgersโ equations as follows [13,
76, 80],
๐๐ฃ
๐๐ก+ ๐ฃ
๐๐ฃ
๐๐ฅ= ๐ (
๐2๐ฃ
๐๐ฅ2) , 0 โค ๐ฅ โค 1, ๐ก > 0. (1.35)
๐๐ฃ
๐๐ก+ ๐ฃ
๐๐ฃ
๐๐ฅ+ ๐ฃ
๐๐ฃ
๐๐ฆ= ๐ (
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2) , 0 โค ๐ฅ, ๐ฆ โค 1, ๐ก > 0. (1.36)
๐๐ฃ
๐๐ก+ ๐ฃ
๐๐ฃ
๐๐ฅ= ๐ (
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2+
๐2๐ฃ
๐๐ง2) , 0 โค ๐ฅ, ๐ฆ, ๐ง โค 1, ๐ก > 0. (1.37)
In addition, the system of coupled Burgerโs equation,
๐๐ฃ
๐๐กโ 2๐ฃ
๐๐ฃ
๐๐ฅ+
๐(๐ฃ๐ข)
๐๐ฅ=
1
๐ ๐(
๐2๐ฃ
๐๐ฅ2),
๐๐ข
๐๐กโ 2๐ข
๐๐ข
๐๐ฅ+
๐(๐ฃ๐ข)
๐๐ฅ=
1
๐ ๐(
๐2๐ข
๐๐ฅ2) , 0 โค ๐ฅ โค 1, ๐ก > 0, (1.38)
and the system of 2D Burgersโ equation,
๐๐ฃ
๐๐ก+ ๐ฃ
๐๐ฃ
๐๐ฅ+ ๐ข
๐๐ฃ
๐๐ฆ=
1
๐ ๐(
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2),
๐๐ข
๐๐ก+ ๐ฃ
๐๐ข
๐๐ฅ+ ๐ข
๐๐ข
๐๐ฆ=
1
๐ ๐(
๐2๐ข
๐๐ฅ2+
๐2๐ข
๐๐ฆ2) , 0 โค ๐ฅ, ๐ฆ โค 1, ๐ก > 0, (1.39)
and the system of 3D Burgersโ equation,
๐๐ฃ
๐๐ก+ ๐ข
๐๐ฃ
๐๐ฅ+ ๐ฃ
๐๐ฃ
๐๐ฆ+ ๐ค
๐๐ฃ
๐๐ง=
1
๐ ๐(
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2+
๐2๐ฃ
๐๐ง2),
๐๐ข
๐๐กโ ๐ข
๐๐ข
๐๐ฅโ ๐ฃ
๐๐ข
๐๐ฆโ ๐ค
๐๐ข
๐๐ง=
1
๐ ๐(
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2+
๐2๐ฃ
๐๐ง2),
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Chapter One Primary Concepts
24
๐๐ค
๐๐ก+ ๐ข
๐๐ค
๐๐ฅ+ ๐ฃ
๐๐ค
๐๐ฆ+ ๐ค
๐๐ค
๐๐ง=
1
๐ ๐(
๐2๐ฃ
๐๐ฅ2+
๐2๐ฃ
๐๐ฆ2+
๐2๐ฃ
๐๐ง2),
0 โค ๐ฅ, ๐ฆ, ๐ง โค 1, ๐ก > 0, (1.40)
where k = 1
๐ ๐ is an arbitrary constant (Re: Reynolds number) for simulating
the physical phenomena of wave motion, and thus determines the behavior of
the solution. If the viscosity is k = 0, then the equation is called the equation
inviscid Burgersโ. It is the lowest approximation of the one-dimensional
system.
Recently, a number of researchers have solved the Burgersโ equations.
For example, Marinca and Herisaun proposed the optimal homotopy
asymptotic method (OHAM) [66], Daftardar-Jafari method (DJM) in [32]. In
addition, considerable attention has been given to ADM for solving Burgersโ
equation [5]. Also, Laplace decomposition method (LDM) [60]. Moreover,
HPM was proposed by He [47], and many others [10, 61, 76].
1.4.2 Exact and numerical solutions for nonlinear Burgersโ
equation by ADM
In this subsection, the ADM is used to get an exact solution for the 1D
Burgersโ equation and numerical solution for the 2D Burgersโ equation. As
given in [7, 30].
Example 1.5 [33]:
Let us consider the following 1D Burgersโ equation.
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ = ๐ฃ๐ฅ๐ฅ, with initial condition ๐ฃ(๐ฅ, 0) = 2๐ฅ. (1.41)
Applying the integration on both sides of Eq. (1.41) from 0 to ๐ก and
using the initial condition at ๐ฃ(๐ฅ, 0) = 2๐ฅ, we have
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Chapter One Primary Concepts
25
๐ฃ(๐ฅ, ๐ก) = 2๐ฅ + โซ (โ๐ฃ๐ฃ๐ฅ+๐ฃ๐ฅ๐ฅ)๐๐ก. (1.42)๐ก
0
By using the decomposition series for the linear term ๐ฃ(๐ฅ, ๐ก) and the
series of Adomian polynomials for the nonlinear term ๐ฃ๐ฃ๐ฅ give
โ ๐ฃ๐(๐ฅ, ๐ก) = 2๐ฅ โ โซ (โ ๐ด๐
โ
๐=0
) ๐๐ก๐ก
0
+ โซ (โ ๐ฃ๐๐ฅ๐ฅ
โ
๐=0
) ๐๐ก. (1.43)๐ก
0
โ
๐=0
Identifying the zeroth component ๐ฃ0(๐ฅ, ๐ก) by the term that arise from the
initial condition and following the decomposition method, we obtain the
recursive relation
๐ฃ0(๐ฅ, ๐ก) = 2๐ฅ,
๐ฃ๐+1(๐ฅ, ๐ก) = โ โซ (โ ๐ด๐
โ
๐=0
) ๐๐ก +๐ก
0
โซ (โ ๐ฃ๐๐ฅ๐ฅ
โ
๐=0
) ๐๐ก,๐ก
0
๐ โฅ 0. (1.44)
The Adomian polynomials for the nonlinear term ๐ฃ๐ฃ๐ฅ have been derived in
the form
๐ด0 = ๐ฃ0๐ฅ๐ฃ0,
๐ด1 = ๐ฃ0๐ฅ๐ฃ1 + ๐ฃ1๐ฅ๐ฃ0,
๐ด2 = ๐ฃ0๐ฅ๐ฃ2 + ๐ฃ1๐ฅ๐ฃ1 + ๐ฃ2๐ฅ๐ฃ0,
๐ด3 = ๐ฃ0๐ฅ๐ฃ3 + ๐ฃ1๐ฅ๐ฃ2 + ๐ฃ2๐ฅ๐ฃ1 + ๐ฃ3๐ฅ๐ฃ0,
๐ด4 = ๐ฃ0๐ฅ๐ฃ4 + ๐ฃ1๐ฅ๐ฃ3 + ๐ฃ2๐ฅ๐ฃ2 + ๐ฃ3๐ฅ๐ฃ1,
โฎ
Consequently, we obtain:
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Chapter One Primary Concepts
26
๐ฃ0(๐ฅ, ๐ก) = 2๐ฅ,
๐ฃ1(๐ฅ, ๐ก) = โ โซ ๐ด0๐๐ก + โซ (๐ฃ0๐ฅ๐ฅ)๐๐ก = โ4๐ก๐ฅ, ๐ก
0
๐ก
0
๐ฃ2(๐ฅ, ๐ก) = โ โซ ๐ด1๐๐ก + โซ (๐ฃ1๐ฅ๐ฅ)๐๐ก = 8๐ก2๐ฅ, ๐ก
0
๐ก
0
๐ฃ3(๐ฅ, ๐ก) = โ โซ ๐ด2๐๐ก + โซ (๐ฃ2๐ฅ๐ฅ)๐๐ก = โ16๐ก3๐ฅ, ๐ก
0
๐ก
0
๐ฃ4(๐ฅ, ๐ก) = โ โซ ๐ด3๐๐ก + โซ (๐ฃ3๐ฅ๐ฅ)๐๐ก = 32๐ก4๐ฅ, ๐ก
0
๐ก
0
โฎ
Summing these iterates gives the series solution
๐ฃ(๐ฅ, ๐ก) = โ ๐ฃ๐(๐ฅ, ๐ก),
โ
๐=0
= 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 + 32๐ฅ๐ก4 โ 64๐ฅ๐ก5 + 128๐ฅ๐ก6 โ 256๐ฅ๐ก7
+ 512๐ฅ๐ก8 โ โฏ,
This series converges to the exact solution [33]:
๐ฃ(๐ฅ, ๐ก) =2๐ฅ
1 + 2๐ก ,
= 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 + 32๐ฅ๐ก4 โ 64๐ฅ๐ก5 + 128๐ฅ๐ก6
โ 256๐ฅ๐ก7 + 512๐ฅ๐ก8 โ 1024๐ฅ๐ก9 + 2048๐ฅ๐ก10 โ โฏ.
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Chapter One Primary Concepts
27
Example 1.6 [80]:
Consider the following 2D Burgersโ equation,
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ = ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ), ๐ > 0, (๐ฅ, ๐ฆ) โ [0,1] ร [0,1], (1.45)
with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
Applying the integration on both sides of Eq. (1.45) from 0 to ๐ก and
using the initial condition at ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ), we have
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ)
+ โซ (โ๐ฃ๐ฃ๐ฅ โ ๐ฃ๐ฃ๐ฆ + ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ)) ๐๐ก. (1.46)๐ก
0
By using the decomposition series for the linear term ๐ฃ(๐ฅ, ๐ฆ, ๐ก) and the
series of Adomian polynomials for the nonlinear term ๐ฃ๐ฃ๐ฅ, ๐ฃ๐ฃ๐ฆ give
โ ๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
= ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ) โ โซ (โ ๐ด๐
โ
๐=0
) ๐๐ก ๐ก
0
โ โซ (โ ๐ต๐
โ
๐=0
) ๐๐ก๐ก
0
+ โซ (โ ๐(๐ฃ๐๐ฅ๐ฅ +
โ
๐=0
๐ฃ๐๐ฆ๐ฆ)) ๐๐ก, (1.47)๐ก
0
identifying the zeroth component ๐ฃ0(๐ฅ, ๐ฆ, ๐ก) by the term that arises from the
initial condition and following the ADM, we obtain the recursive relation
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
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Chapter One Primary Concepts
28
๐ฃ๐+1(๐ฅ, ๐ฆ, ๐ก) = โ โซ (โ ๐ด๐
โ
๐=0
) ๐๐ก๐ก
0
โ โซ (โ ๐ต๐
โ
๐=0
) ๐๐ก +๐ก
0
โซ (โ ๐(๐ฃ๐๐ฅ๐ฅ +
โ
๐=0
๐ฃ๐๐ฆ๐ฆ)) ๐๐ก๐ก
0
๐ โฅ 0. (1.48)
The Adomian polynomials for the nonlinear term ๐ฃ๐ฃ๐ฅ, ๐ฃ๐ฃ๐ฆ have been derived
in the form
๐ด0 = ๐ฃ0๐ฅ๐ฃ0,
๐ด1 = ๐ฃ0๐ฅ๐ฃ1 + ๐ฃ1๐ฅ๐ฃ0,
๐ด2 = ๐ฃ0๐ฅ๐ฃ2 + ๐ฃ1๐ฅ๐ฃ1 + ๐ฃ2๐ฅ๐ฃ0,
๐ด3 = ๐ฃ0๐ฅ๐ฃ3 + ๐ฃ1๐ฅ๐ฃ2 + ๐ฃ2๐ฅ๐ฃ1 + ๐ฃ3๐ฅ๐ฃ0,
and
๐ต0 = ๐ฃ0๐ฆ๐ฃ0,
๐ต1 = ๐ฃ0๐ฆ๐ฃ1 + ๐ฃ1๐ฆ๐ฃ0,
๐ต2 = ๐ฃ0๐ฆ๐ฃ2 + ๐ฃ1๐ฆ๐ฃ1 + ๐ฃ2๐ฆ๐ฃ0,
๐ต3 = ๐ฃ0๐ฆ๐ฃ3 + ๐ฃ1๐ฆ๐ฃ2 + ๐ฃ2๐ฆ๐ฃ1 + ๐ฃ3๐ฆ๐ฃ0,
consequently, we obtain:
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
๐ฃ1(๐ฅ, ๐ฆ, ๐ก) = โ โซ ๐ด0๐๐ก โ โซ ๐ต0๐๐ก + โซ ๐(๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฆ๐ฆ)๐๐ก,๐ก
0
๐ก
0
๐ก
0
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Chapter One Primary Concepts
29
= ๐ก(โ8๐2๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐๐๐๐ (2๐๐ฅ)๐๐๐ (2๐๐ฆ)2๐ ๐๐(2๐๐ฅ)
+ 2๐๐ ๐๐(2๐๐ฅ)๐ ๐๐(2๐๐ฆ)),
๐ฃ2(๐ฅ, ๐ฆ, ๐ก) = โ โซ ๐ด1๐๐ก โ โซ ๐ต1๐๐ก + โซ ๐(๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฆ๐ฆ)๐๐ก,๐ก
0
๐ก
0
๐ก
0
= โ2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + 32๐4๐ก2๐2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + โฏ,
๐ฃ3(๐ฅ, ๐ฆ, ๐ก) = โ โซ ๐ด2๐๐ก โ โซ ๐ต2๐๐ก + โซ ๐(๐ฃ2๐ฅ๐ฅ + ๐ฃ2๐ฆ๐ฆ)๐๐ก,๐ก
0
๐ก
0
๐ก
0
= 16๐4๐ก3๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ256
3๐6๐ก3๐3๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + โฏ,
๐ฃ4(๐ฅ, ๐ฆ, ๐ก) = โ โซ ๐ด3๐๐ก โ โซ ๐ต3๐๐ก + โซ ๐(๐ฃ3๐ฅ๐ฅ + ๐ฃ3๐ฆ๐ฆ)๐๐ก,๐ก
0
๐ก
0
๐ก
0
=2
3๐4๐ก4๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 64๐6๐ก4๐2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + โฏ,
then
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+2
3๐4๐ก4๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + 16๐4๐ก3๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+ 32๐4๐ก2๐2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ โฏ.
The obtained series solution in Eq. (1.45) can be used to calculate the
maximal error remainder to show the highest accuracy level that we can
achieve. This can be clearly seen in table 1.4 and figure 1.4.
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Chapter One Primary Concepts
30
Table 1.4: The maximal error remainder: ๐๐ธ๐ ๐ by the ADM.
n ๐ด๐ฌ๐น๐ by ADM
1 1.61705 ร 10โ2
2 2.98775 ร 10โ5
3 4.9464 ร 10โ8
4 7.1962 ร 10โ11
Figure 1.4: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by ADM.
1.4.3 The VIM to solve the 2D Burgersโ equation
In this subsection, the VIM is used to get a numerical solution for the 2D
Burgersโ equation. As given in [50].
Example 1.7 [80]:
Rewrite the example 1.6 and solve it by VIM,
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ = ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ), (1.49)
with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
By using Eq. (1.14), we get:
1.0 1.5 2.0 2.5 3.0 3.5 4.0
10 8
10 5
10 2
n
ME
Rn
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Chapter One Primary Concepts
31
๐ฃ๐+1(๐ฅ, ๐ฆ, ๐ก)
= ๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
+ โซ [๐(๐ )(๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ + ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฅ
๐ก
0
+ ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฆ โ ๐(๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฅ๐ฅ
+ ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฆ๐ฆ))] ๐๐ . (1.50)
Using Eq. (1.15) leads to ฮป = โ1. Substituting this value of the Lagrange
multiplier into the functional Eq. (1.50) gives the following iteration formula:
๐ฃ๐+1(๐ฅ, ๐ฆ, ๐ก)
= ๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
โ โซ (๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ + ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฅ
๐ก
0
+ ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฆ โ ๐(๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฅ๐ฅ
+ ๐ฃ๐(๐ฅ, ๐ฆ, ๐ )๐ฆ๐ฆ)) ๐๐ . (1.51)
By applying Mathematicaโs code (see appendix B), we get the following
initial approximation
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
By using Eq. (1.51) the obtained successive approximations will be
๐ฃ0 = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
๐ฃ1 = ๐ฃ0 โ โซ (๐ฃ0๐ + ๐ฃ0๐ฃ0๐ฅ
+ ๐ฃ0๐ฃ0๐ฆโ ๐(๐ฃ0๐ฅ๐ฅ
+ ๐ฃ0๐ฆ๐ฆ)) ๐๐ ,
๐ก
0
= ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ ๐ก(8๐2๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+ 2๐๐๐๐ (2๐๐ฅ)๐๐๐ (2๐๐ฆ)2๐ ๐๐(2๐๐ฅ) โ 2๐๐ ๐๐(2๐๐ฅ)๐ ๐๐(2๐๐ฆ)),
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Chapter One Primary Concepts
32
๐ฃ2 = ๐ฃ1 โ โซ (๐ฃ1๐ + ๐ฃ1๐ฃ1๐ฅ
+ ๐ฃ1๐ฃ1๐ฆโ ๐(๐ฃ1๐ฅ๐ฅ
+ ๐ฃ1๐ฆ๐ฆ)) ๐๐ ,
๐ก
0
= ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+ 32๐4๐ก2๐2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+ 40๐3๐ก2๐๐๐๐ (2๐๐ฅ)๐๐๐ (2๐๐ฆ)2๐ ๐๐(2๐๐ฅ) โ โฏ,
continuing in this way till n = 6, we have
๐ฃ6 = ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+2
3๐4๐ก4๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + โฏ, (1.52)
Table 1.5 and Figure 1.5 illustrate the convergence of the solution
through the use of the maximal error remainder. Clearly seen, the errors will
be decreased as the number of iterations will be increased.
Table 1.5: The maximal error remainder: ๐๐ธ๐ ๐ by the VIM.
n ๐ด๐ฌ๐น๐ by VIM
1 1.61705 ร 10โ2
2 2.52036 ร 10โ5
3 3.56783 ร 10โ8
4 4.53593 ร 10โ11
Figure 1.5: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by V1M.
1.0 1.5 2.0 2.5 3.0 3.5 4.010 11
10 8
10 5
10 2
n
ME
Rn
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CHAPTER 2
SEMI-ANALYTICAL
ITERATIVE METHOD FOR
SOLVING SOME ORDINARY
AND PARTIAL DIFFERENTIAL
EQUATIONS
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
33
Chapter 2
Semi-Analytical Iterative Method for Solving Some
Ordinary and Partial Differential Equations
2.1 Introduction
Temimi and Ansari have proposed recently a semi-analytical iterative
method namely (TAM) to solve various linear and nonlinear ODEs and PDEs
[28]. For examples, solving nonlinear ordinary differential equations [29], the
Korteweg-de Vries Equations [18], Duffing equations [51], the nonlinear thin
film flow problems [53], and some chemistry problems [52].
The main advantages of the TAM, it does not required any restrictive
assumptions for nonlinear terms since the ADM required the so-called
Adomian polynomial, avoided the large computational work or using any
other parameters or restricted assumptions that appear in other iterative
methods such as VIM and HPM. In addition, the TAM for solving non-linear
ODEs and PDEs have successfully implemented.
This chapter is organized as follows: in section two, the basic idea of the
TAM is presented. In section three, an overview of the convergence of the
TAM based on the Banach fixed-point theorem will be given. In section four,
some types of non-linear ODEs equations will be solved by TAM, namely the
Riccati equation, pantograph equation and beam deformation equation.
Moreover, the Burgersโ equations will be solved and the convergence will be
proved.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
34
2.2 The basic idea of the TAM
We start by pointing out that the nonlinear differential equation can be
written in operator form as:
๐ฟ(๐ฃ(๐ฅ)) + ๐(๐ฃ(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฅ โ ๐ท,
with boundary conditions ๐ต(๐ฃ, ๐๐ฃ
๐๐ฅ ) = 0, ๐ฅ โ ๐. (2.1)
Where x denotes the independent variable, ๐ฃ(๐ฅ) is an unknown function,
๐(๐ฅ) is a known function, ๐ is the boundary of the domain ๐ท, ๐ฟ is a linear
operator, ๐ is a nonlinear operator and ๐ต is a boundary operator. The main
requirement here is that ๐ฟ is the linear part of the differential equation, but it
is possible to take some linear parts and add them to ๐ as needed. The method
works in the following steps, starts by assuming that ๐ฃ0(๐ฅ) is an initial guess
of the solution to the problem [28].
๐ฟ(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ต(๐ฃ0, ๐๐ฃ0๐๐ฅ
) = 0. (2.2)
To generate the next iteration of the solution, we solve the following
problem:
๐ฟ(๐ฃ1(๐ฅ)) + ๐(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ต(๐ฃ1, ๐๐ฃ1๐๐ฅ
) = 0. (2.3)
Thus, we have a simple iterative procedure which is effectively the
solution of a linear set of problems i.e.
๐ฟ(๐ฃ๐+1(๐ฅ)) + ๐(๐ฃ๐(๐ฅ)) + ๐(๐ฅ) = 0, ๐ต(๐ฃ๐+1, ๐๐ฃ๐+1๐๐ฅ
) = 0. (2.4)
It is noted that each of the ๐ฃ๐(๐ฅ), are solutions to Eq. (2.1). Thus,
evaluating more approximate terms provides better accuracy.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
35
2.3 The convergence of the TAM
The Banach fixed-point theorem also known as the contraction mapping
principle is an important tool in the theory of metric spaces. It guarantees the
existence and uniqueness of fixed points of certain self-maps of metric spaces.
Also, provides a constructive method to find those fixed points. The theorem
is named after Stefan Banach (1892โ1945), and was first stated by him in
1922 [69]. In this section, some basic concepts and the main theorem of the
convergence will be presented.
Theorem 2.1: [33]
Suppose that (๐, โ โ๐) and (๐, โ โ๐) be Banach spaces and T: X โ Y, is
a contraction nonlinear mapping, that is
โ ๐ฃ, ๐ฃโ โ ๐ ; โ๐(๐ฃ) โ ๐(๐ฃโ)โ๐ โค ๐โ๐ฃ โ ๐ฃโโ๐, 0 < ๐ < 1.
According to theorem 1.1, having the fixed point ๐ฃ, that is ๐(๐ฃ) = ๐ฃ, the
sequence generated by the TAM will be regarded as
๐ฃ๐ = ๐(๐ฃ๐โ1), ๐ฃ = ๐๐๐๐ โ โ
๐ฃ๐,
and suppose that ๐ฃ0 โ ๐ต๐(๐ฃ) where ๐ต๐(๐ฃ) = {๐ฃโ โ ๐, โ๐ฃโ โ ๐ฃโ < ๐} then
we have the following statements:
1. โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
2. ๐ฃ๐ โ ๐ต๐(๐ฃ),
3. ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
36
2.4 Application of the TAM with convergence for the ODE
and PDE
In this section, some types of non-linear ODEs equations will be solved
by TAM, namely the Riccati equation, pantograph equation and beam
deformation equation. Moreover, the Burgersโ equations will be solved and
the convergence will be proved.
2.4.1 Exact and numerical solutions for nonlinear Riccati
equation by TAM
The TAM will be implemented to solve the Riccati differential equation
which is a nonlinear ODE of first order. So, to verify the accuracy of the
TAM in solving this kind of problems, the following examples will be
discussed.
Example 2.1 [19]:
Let us consider the following Riccati differential equation,
๐ฃโฒ = โ ๐ฅ โ โ 3๐ฅ + 2โ 2๐ฅ๐ฃ โ โ ๐ฅ๐ฃ2, with initial condition ๐ฃ(0) = 1. (2.5)
By applying the TAM by first distributing the equation as,
๐ฟ(๐ฃ) = ๐ฃโฒ, ๐(๐ฃ) = โ2โ 2๐ฅ๐ฃ + โ ๐ฅ๐ฃ2 and ๐(๐ฅ) = โโ ๐ฅ + โ 3๐ฅ.
Thus, the initial problem will be
๐ฟ(๐ฃ0(๐ฅ)) = โ ๐ฅ โ โ 3๐ฅ, ๐ฃ0(0) = 1. (2.6)
By integrating both sides of Eq. (2.6) from 0 to x, we obtain
โซ ๐ฃ0โฒ (๐ก) โ ๐ก
๐ฅ
0
= โซ (โ ๐ก โ โ 3๐ก) โ ๐ก, ๐ฃ0(0) = 1.๐ฅ
0
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
37
Therefore, we have
๐ฃ0(๐ฅ) =1
3+ โ ๐ฅ โ
โ 3๐ฅ
3.
The second iteration can be given as
๐ฟ(๐ฃ1(๐ฅ)) + ๐(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ1(0) = 1. (2.7)
By using simple manipulation, one can solve Eq. (2.7) as follows:
โซ ๐ฃ1โฒ (๐ก) โ ๐ก
๐ฅ
0
= โซ (โ ๐ก โ โ 3๐ก + 2โ 2๐ก๐ฃ0(๐ก) โ โ ๐ก๐ฃ02(๐ก)) โ ๐ก, ๐ฃ1(0) = 1.
๐ฅ
0
Then, we obtain
๐ฃ1(๐ฅ) =1
14+
8โ ๐ฅ
9+
โ 4๐ฅ
18โ
โ 7๐ฅ
63.
We turn the function ๐ฃ1(๐ฅ) by using Maclaurin series expansion to
exponential function, we get
๐ฃ1(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2โ
๐ฅ3
6โ
23๐ฅ4
24โ
209๐ฅ5
120.
Applying the same process, we get the second iteration ๐ฃ2(๐ฅ),
๐ฟ(๐ฃ2(๐ฅ)) + ๐(๐ฃ1(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ2(0) = 1.
Then, we have
๐ฃ2(๐ฅ) =11
9720+
195โ ๐ฅ
196+
โ 2๐ฅ
126โ
โ 3๐ฅ
243โ
โ 5๐ฅ
630+
โ 6๐ฅ
486+
โ 8๐ฅ
3528โ
5โ 9๐ฅ
6804
+โ 12๐ฅ
6804โ
โ 15๐ฅ
59535.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
38
We can write the function ๐ฃ2(๐ฅ) in the following series form,
๐ฃ2(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2+
๐ฅ3
6+
๐ฅ4
24+
๐ฅ5
120+
๐ฅ6
720โ
79๐ฅ7
5040โ
3919๐ฅ8
40320
โ116479๐ฅ9
362880.
By continuing in this way, we will get a series of the form:
๐ฃ(๐ฅ) = ๐๐๐๐โโ
๐ฃ๐(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2+
๐ฅ3
6+
๐ฅ4
24+
๐ฅ5
120+ โฏ,
This series converges to the following exact solution [19].
๐ฃ(๐ฅ) = โ ๐ฅ = 1 + ๐ฅ +๐ฅ2
2+
๐ฅ3
6+
๐ฅ4
24+
๐ฅ5
120+
๐ฅ6
720+ โฏ.
Suppose that T: [0,1] โ ๐ , then ๐ฃ๐ = ๐(๐ฃ๐โ1) and 0 โค ๐ฅ โค 1.
According to the theorem 2.1 for nonlinear mapping T, a sufficient
condition for convergence of the TAM is strictly contraction T, the exact
solution is ๐ฃ = ๐ฃ(๐ฅ) = โ ๐ฅ, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ1
3+ โ ๐ฅ โ
โ 3๐ฅ
3โ โ ๐ฅโ,
โ๐ฃ1 โ ๐ฃโ = โ1
14+
8โ ๐ฅ
9+
โ 4๐ฅ
18โ
โ 7๐ฅ
63โ โ ๐ฅโ โค ๐ โ
1
3+ โ ๐ฅ โ
โ 3๐ฅ
3โ โ ๐ฅโ
= ๐โ๐ฃ0 โ ๐ฃโ,
But, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ = 1
2,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.19551 < 1,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
39
Then, we have
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
โ๐ฃ2 โ ๐ฃโ = โ11
9720+
195โ ๐ฅ
196+
โ 2๐ฅ
126โ
โ 3๐ฅ
243โ
โ 5๐ฅ
630+
โ 6๐ฅ
486+
โ 8๐ฅ
3528โ
5โ 9๐ฅ
6804+
โ 12๐ฅ
6804โ
โ 15๐ฅ
59535โ โ ๐ฅโ โค ๐โ๐ฃ1 โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ = ๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ =1
2,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.0580412 < 1,
Thus, we get
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get:
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0, and ๐๐๐๐ โ โ
๐๐ = 0, we drive
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ = โ ๐ฅ, which is the exact solution [19].
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
40
Example 2.2:
Let us recall example 1.1
๐ฃโฒ = โ๐ฃ2 + 1, with initial condition ๐ฃ(0) = 0. (2.8)
By applying the TAM by first distributing the equation as
๐ฟ(๐ฃ) = ๐ฃโฒ, ๐(๐ฃ) = ๐ฃ2 and ๐(๐ฅ) = โ1
Thus, the initial problem will be
๐ฟ(๐ฃ0(๐ฅ)) = 1, ๐ฃ0(0) = 0, (2.9)
By integrating both sides of Eq. (2.9) from 0 to x, we have
โซ ๐ฃ0โฒ (๐ก) โ ๐ก
๐ฅ
0
= โซ (1) โ ๐ก, ๐ฃ0(0) = 0,๐ฅ
0
So, we get
๐ฃ0(๐ฅ) = ๐ฅ.
The second iteration can be given as
๐ฟ(๐ฃ1(๐ฅ)) + ๐(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ1(0) = 0, (2.10)
By using simple manipulation, one can solve Eq. (2.10) as follows
โซ ๐ฃ1โฒ (๐ก) โ ๐ก
๐ฅ
0
= โซ (โ๐ฃ02(๐ก) + 1) โ ๐ก, ๐ฃ1(0) = 0.
๐ฅ
0
Then, we obtain
๐ฃ1(๐ฅ) = ๐ฅ โ๐ฅ3
3.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
41
Applying the same process to get the second iteration ๐ฃ2(๐ฅ),
๐ฟ(๐ฃ2(๐ฅ)) + ๐(๐ฃ1(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ2(0) = 0,
We have
๐ฃ2(๐ฅ) = ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
๐ฅ7
63.
Continuing in this way till n = 6, but for brevity, they are not listed.
The obtained series solution in Eq. (2.8) can be used to calculate the
maximal error remainder to show the highest accuracy level that we can
achieve. This can be clearly seen in table 2.1 and figure 2.1
Table 2.1: The maximal error remainder: ๐๐ธ๐ ๐ by the TAM.
n ๐ด๐ฌ๐น๐ by TAM
1 6.65556 ร 10โ5
2 2.65463 ร 10โ7
3 7.56708 ร 10โ10
4 1.67806 ร 10โ12
5 2.97852 ร 10โ15
6 1.24033 ร 10โ16
Figure 2.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by TAM.
1 2 3 4 5 6
10 15
10 12
10 9
10 6
n
ME
Rn
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
42
2.4.2 TAM for solving linear pantograph equation
In this subsection, the TAM will be applied for second order linear
pantograph equation which is used in many applications, such as industrial
applications, studies based on biology, economy and others. The following
example presents a solution for pantograph equation by using the TAM.
Example 2.3 [11]:
Resolve example 1.3 by using TAM.
๐ฃโฒโฒ =3
4๐ฃ + ๐ฃ (
๐ฅ
2) โ ๐ฅ2 + 2,
with initial conditions ๐ฃ(0) = 0 , ๐ฃโฒ(0) = 0. (2.11)
To solve Eq. (2.11) by TAM, distribute the equation as follows
๐ฟ(๐ฃ) = ๐ฃ", ๐(๐ฃ) = โ3
4๐ฃ โ ๐ฃ(
๐ฅ
2) and ๐(๐ฅ) = ๐ฅ2 โ 2.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ)) = โ๐ฅ2 + 2, ๐ฃ0(0) = 0, ๐ฃ0โฒ(0) = 0. (2.12)
Integrate both sides of Eq. (2.12) twice from 0 to x and substitute the initial
conditions, we get
โซ โซ ๐ฃ0โฒโฒ(๐ก)โ ๐กโ ๐ก
๐ฅ
0
๐ฅ
0
= โซ โซ (โ๐ก2 + 2)โ ๐กโ ๐ก๐ฅ
0
๐ฅ
0
, ๐ฃ0(0) = 0, ๐ฃ0โฒ(0) = 0.
Then, we obtain
๐ฃ0(๐ฅ) = ๐ฅ2 โ๐ฅ4
12.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
43
The second iteration can be carried as
๐ฟ(๐ฃ1(๐ฅ)) + ๐(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ1(0) = 0, ๐ฃ1โฒ(0) = 0. (2.13)
By integrating both sides of Eq. (2.13) twice from 0 to ๐ฅ, we obtain
โซ โซ ๐ฃ1โฒโฒ(๐ก) โ ๐กโ ๐ก
๐ฅ
0
๐ฅ
0
= โซ โซ (โ๐ก2 + 2 +3
4๐ฃ0(๐ก) + ๐ฃ0 (
๐ก
2)) โ ๐กโ ๐ก,
๐ฅ
0
๐ฅ
0
with initial conditions ๐ฃ1(0) = 0, ๐ฃ1โฒ(0) = 0.
Thus, we get
๐ฃ1(๐ฅ) = ๐ฅ2 โ13๐ฅ6
5760.
The next iteration is
๐ฟ(๐ฃ2(๐ฅ)) + ๐(๐ฃ1(๐ฅ)) + ๐(๐ฅ) = 0 , ๐ฃ2(0) = 0, ๐ฃ2โฒ(0) = 0.
Thus, we have
๐ฃ2(๐ฅ) = ๐ฅ2 โ91๐ฅ8
2949120 .
The next iteration is
๐ฟ(๐ฃ3(๐ฅ)) + ๐(๐ฃ2(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ3(0) = 0, ๐ฃ3โฒ(0) = 0.
Then, we get
๐ฃ3(๐ฅ) = ๐ฅ2 โ17563๐ฅ10
67947724800 .
By continuing in this way, we will get:
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
44
๐ฃ(๐ฅ) = ๐ฅ2 โ small term,
This series converges to the exact solution [11]:
๐ฃ(๐ฅ) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ) = ๐ฅ2.
Suppose that ๐: [0,1] โ ๐ , then ๐ฃ๐ = ๐(๐ฃ๐โ1) and 0 โค ๐ฅ โค 1.
Following the same procedure as for the Riccati equation, the exact solution
for pantograph equation is ๐ฃ = ๐ฃ(๐ฅ) = ๐ฅ2, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ๐ฅ2 โ๐ฅ4
12โ (๐ฅ2)โ,
โ๐ฃ1 โ ๐ฃโ = โ๐ฅ2 โ13๐ฅ6
5760โ (๐ฅ2)โ โค ๐ โ๐ฅ2 โ
๐ฅ4
12โ (๐ฅ2)โ = ๐โ๐ฃ0 โ ๐ฃโ,
But, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ = 1
4,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.00169271 < 1,
Then, we have
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
โ๐ฃ2 โ ๐ฃโ = โ๐ฅ2 โ91๐ฅ8
2949120โ (๐ฅ2)โ โค ๐โ๐ฃ1 โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ =
๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ =1
4,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.00120267 < 1,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
45
Thus, we get
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get:
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐๐ โ โ
๐๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ = ๐ฅ2, which is the exact solution [11].
2.4.3 Solving the nonlinear beam equation by using the TAM
In this subsection, the TAM will be implemented to solve fourth order
nonlinear beam deformation problem which is a boundary value problems that
are considered one of more important equations because they are widely used
in different scientific specializations. These boundary value problems appear
in applied mathematics, engineering, several branches of physics and others.
Example 2.4:
Let us recall example 1.4
๐ฃ(4) = ๐ฃ2 โ ๐ฅ10 + 4๐ฅ9 โ 4๐ฅ8 โ 4๐ฅ7 + 8๐ฅ6 โ 4๐ฅ4 + 120๐ฅ โ 48,
with boundary conditions ๐ฃ(0) = ๐ฃโฒ(0) = 0, ๐ฃ(1) = ๐ฃโฒ(1) = 1. (2.14)
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Chapter Two Semi-Analytical Iterative Method for Solving
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46
By applying the TAM for Eq. (2.14), we get
๐ฟ(๐ฃ) = ๐ฃ(4), ๐(๐ฃ) = โ๐ฃ2,
๐(๐ฅ) = ๐ฅ10 โ 4๐ฅ9 + 4๐ฅ8 + 4๐ฅ7 โ 8๐ฅ6 + 4๐ฅ4 โ 120๐ฅ + 48.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0, ๐ฃ0(0) = ๐ฃ0โฒ(0) = 0, ๐ฃ0(1) = ๐ฃ0โฒ(1) = 1. (2.15)
By integrating both sides of (2.15) from 0 to ๐ฅ four times, one can obtain
โซ โซ โซ โซ ๐ฃ0(4)(๐ก) โ ๐ก โ ๐ก โ ๐ก โ ๐ก
๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
= โซ โซ โซ โซ โ๐(๐ก) โ ๐ก โ ๐ก โ ๐ก โ ๐ก,๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
Then, we have
๐ฃ0(๐ฅ) = ๐ฃ0(0) + ๐ฅ๐ฃ0โฒ (0) +
๐ฅ2
2๐ฃ0
โฒโฒ(0) + ๐ฅ3
6๐ฃ0
โฒโฒโฒ(0)
โ โซ โซ โซ โซ ๐(๐ก) โ ๐ก โ ๐ก โ ๐ก โ ๐ก, ๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
(2.16)
Imposing the boundary conditions (2.15), into Eq. (2.16), and by applying
Mathematicaโs code, see appendix C, the results are as follows.
๐ฃ0(๐ฅ) =718561๐ฅ2
360360+
4019๐ฅ3
540540โ 2๐ฅ4 + ๐ฅ5 โ
๐ฅ8
420+
๐ฅ10
630โ
๐ฅ11
1980โ
๐ฅ12
2970
+๐ฅ13
4290โ
๐ฅ14
24024.
The second iteration can be carried through and we have
๐ฟ(๐ฃ1(๐ฅ)) + ๐(๐ฃ0(๐ฅ)) + ๐(๐ฅ) = 0,
๐ฃ1(0) = ๐ฃ1โฒ(0) = 0, ๐ฃ1(1) = ๐ฃ1โฒ(1) = 1. (2.17)
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Chapter Two Semi-Analytical Iterative Method for Solving
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47
By integrating both sides of (2.17) from 0 to ๐ฅ four times, one can obtain
โซ โซ โซ โซ ๐ฃ1(4)(๐ก) โ ๐ก โ ๐ก โ ๐ก โ ๐ก
๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
= โซ โซ โซ โซ (๐ฃ02(๐ก) โ ๐(๐ก)) โ ๐ก โ ๐ก โ ๐ก โ ๐ก,
๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
Then, we have
๐ฃ1(๐ฅ) = ๐ฃ1(0) + ๐ฅ๐ฃ1โฒ (0) +
๐ฅ2
2๐ฃ1
โฒโฒ(0) + ๐ฅ3
6๐ฃ1
โฒโฒโฒ(0)
+ โซ โซ โซ โซ (๐ฃ02(๐ก) โ ๐(๐ก)) โ ๐ก โ ๐ก โ ๐ก โ ๐ก,
๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
(2.18)
Similarly, by imposing the boundary conditions (2.17), into Eq. (2.18), the
results are as follows:
๐ฃ1(๐ฅ) =4720792684282308505367359๐ฅ2
2360410309588661890560000+
18553248327867926839๐ฅ3
1180205154794330945280000โ 2๐ฅ4 +
๐ฅ5 โ3107407679๐ฅ8
218163673728000+
2887896659๐ฅ9
294520959532800+
7018307521๐ฅ10
1472604797664000โ
22553๐ฅ11
4281076800+
4019๐ฅ12
3210807600โ
718561๐ฅ14
1818030614400โ
4019๐ฅ15
3718698984000+
1799641๐ฅ16
4958265312000โ
10117717๐ฅ17
85587287385600โ
68671๐ฅ18
655004750400+
127713533๐ฅ19
1941434080185600+
11676571๐ฅ20
7550021422944000โ
2086087๐ฅ21
186529941037440+
11827๐ฅ22
3321402084000โ
19๐ฅ23
49945896000+
๐ฅ24
61856071200โ
๐ฅ25
111891780000โ
1153๐ฅ26
672022030680000+
289๐ฅ27
112699346760000โ
2857๐ฅ28
5522267991240000โ
191๐ฅ30
1525197826152000โ
๐ฅ31
38914512436800+
๐ฅ32
498105759191040.
The next iteration is
๐ฟ(๐ฃ2(๐ฅ)) + ๐(๐ฃ1(๐ฅ)) + ๐(๐ฅ) = 0,
with boundary conditions ๐ฃ2(0) = ๐ฃ2โฒ(0) = 0, ๐ฃ2(1) = ๐ฃ2โฒ(1) = 1. (2.19)
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48
By solving Eq. (2.19), we get
๐ฃ2(๐ฅ) = 1.9999999795270846๐ฅ2 + 2.745035293882598 ร 10โ8๐ฅ3 โ
2๐ฅ4 + ๐ฅ5 โ 2.817792217763172 ร 10โ8๐ฅ8 +
2.079400216725163 ร 10โ8๐ฅ9 + 9.392717549543088 ร
10โ9๐ฅ10 + ๐[๐ฅ]11. (2.20)
In order to check the accuracy of the approximate solution, we calculate
the absolute error, where ๐ฃ(๐ฅ) = ๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2 is the exact solution and
๐ฃ๐(๐ฅ) is the approximate solution. In table 2.2 the absolute error of TAM with
n = 1, 2 is presented.
Table 2.2: Results of the absolute errors for TAM.
๐ |๐๐| for TAM |๐๐| for TAM
0 0 0
0.1 1.02627 ร 10โ7 1.77279 ร 10โ10
0.2 3.47659 ร 10โ7 5.99375 ร 10โ10
0.3 6.41401 ร 10โ7 1.1028 ร 10โ9
0.4 8.93924 ร 10โ7 1.53129 ร 10โ9
0.5 1.02777 ร 10โ6 1.752 ร 10โ9
0.6 9.93141 ร 10โ7 1.68284 ร 10โ9
0.7 7.86445 ร 10โ7 1.32351 ร 10โ9
0.8 4.64662 ร 10โ7 7.76486 ร 10โ10
0.9 1.471 ร 10โ7 2.44241 ร 10โ10
1.0 2.22045 ร 10โ16 1.11022 ร 10โ16
It can be seen clearly from table 2.2, that by increasing the iterations, the
errors will decreasing.
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49
In order to discuss the convergence for TAM, suppose that ๐: [0,1] โ ๐ , then
๐ฃ๐ = ๐(๐ฃ๐โ1) and 0 โค ๐ฅ โค 1.
The convergence issue can be done as follows:
Since, the exact solution is ๐ฃ = ๐ฃ(๐ฅ) = ๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ๐ฃ0 โ (๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2)โ,
โ๐ฃ1 โ ๐ฃโ = โ๐ฃ1 โ (๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2)โ โค ๐โ๐ฃ0 โ (๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2)โ
= ๐โ๐ฃ0 โ ๐ฃโ,
But, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ = 1
3,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.99894 < 1,
Then, we have
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
โ๐ฃ2 โ ๐ฃโ = โ๐ฃ2 โ (๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2)โ โค ๐โ๐ฃ1 โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ =
๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ฅ ๐ [0,1], 0 < ๐ < 1, when ๐ฅ = 1
3,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.99472 < 1,
Thus, we get
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
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โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐๐ โ โ
๐๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ = ๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2, which is the exact
solution [1].
2.4.4 Solving the nonlinear Burgersโ equation by using TAM
This subsection produces an implementation for the TAM to solve the
first order nonlinear Burgersโ equation and finding the exact solutions for
different types of Burgersโ equations in 1D, 2D and (1+3)-D.
Example 2.5:
Resolve example 1.5 by using TAM [33].
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ = ๐ฃ๐ฅ๐ฅ, with initial condition ๐ฃ(๐ฅ, 0) = 2๐ฅ. (2.21)
By applying TAM for Eq. (2.21), we get
๐ฟ(๐ฃ) = ๐ฃ๐ก , ๐(๐ฃ) = ๐ฃ๐ฃ๐ฅ โ ๐ฃ๐ฅ๐ฅ and ๐(๐ฅ, ๐ก) = 0.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ, ๐ก)) = 0, ๐ฃ0(๐ฅ, 0) = 2๐ฅ. (2.22)
By using a simple manipulation, one can solve Eq. (2.22) as follows:
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Chapter Two Semi-Analytical Iterative Method for Solving
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โซ ๐ฃ0๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, 0) = 2๐ฅ.
Then, we get
๐ฃ0(๐ฅ, ๐ก) = 2๐ฅ.
The second iteration can be carried through and is given as
๐ฟ(๐ฃ1(๐ฅ, ๐ก)) + ๐(๐ฃ0(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ฃ1(๐ฅ, 0) = 2๐ฅ. (2.23)
By integrating both sides of Eq. (2.23) from 0 to t, we get
โซ ๐ฃ1๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= โซ (โ๐ฃ0(๐ฅ, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ก) + ๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ก)) โ ๐ก๐ก
0
, ๐ฃ1(๐ฅ, 0) = 2๐ฅ.
Thus,
๐ฃ1(๐ฅ) = 2๐ฅ โ 4๐ฅ๐ก.
The next iteration is
๐ฟ(๐ฃ2(๐ฅ, ๐ก)) + ๐(๐ฃ1(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ฃ2(๐ฅ, 0) = 2๐ฅ.
Then, we have
๐ฃ2(๐ฅ) = 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ16๐ฅ๐ก3
3.
The next iteration is
๐ฟ(๐ฃ3(๐ฅ, ๐ก)) + ๐(๐ฃ2(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ฃ3(๐ฅ, 0) = 2๐ฅ.
Then, we get
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Chapter Two Semi-Analytical Iterative Method for Solving
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๐ฃ3(๐ฅ, ๐ก) = 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 +64๐ฅ๐ก4
3โ
64๐ฅ๐ก5
3+
128๐ฅ๐ก6
9
โ256๐ฅ๐ก7
63.
By continuing in this way, we will get series of the form:
๐ฃ(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ก) = 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 + 32๐ฅ๐ก4 โ โฏ,
This series converges to the exact solution [33]:
๐ฃ(๐ฅ, ๐ก) =2๐ฅ
1 + 2๐ก= 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 + 32๐ฅ๐ก4 โ 64๐ฅ๐ก5 + โฏ.
To check the convergence of the proposed method, let us suppose that T: ๐ ร
[0,1
2) โ ๐ 2, then ๐ฃ๐ = ๐(๐ฃ๐โ1) and ๐ก โค
๐
2 , 0 < ๐ < 1.
According to the theorem 2.1 for nonlinear mapping T, a sufficient
condition for convergence of the TAM is strictly contraction T, the exact
solution is ๐ฃ = ๐ฃ(๐ฅ, ๐ก) =2๐ฅ
1+2๐ก, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ2๐ฅ โ2๐ฅ
1 + 2๐กโ,
โ๐ฃ1 โ ๐ฃโ = โ2๐ฅ โ 4๐ฅ๐ก โ2๐ฅ
1 + 2๐กโ โค ๐ โ2๐ฅ โ
2๐ฅ
1 + 2๐กโ = ๐โ๐ฃ0 โ ๐ฃโ,,
But, โ ๐ก ๐ [0,๐
2) , 0 < ๐ < 1, when ๐ก =
1
4, ๐ฅ = 1,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.5 < 1,
Then, we get
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Chapter Two Semi-Analytical Iterative Method for Solving
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53
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
Also,
โ๐ฃ2 โ ๐ฃโ = โ2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ16๐ฅ๐ก3
3โ
2๐ฅ
1+2๐กโ โค ๐โ๐ฃ1 โ ๐ฃโ โค
๐๐โ๐ฃ0 โ ๐ฃโ = ๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ก ๐ [0,๐
2) , 0 < ๐ < 1, when ๐ก =
1
4, ๐ฅ = 1,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.353553 < 1,
Then, we have
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐๐ โ โ
๐๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ = 0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ =2๐ฅ
1+2๐ก , which is the exact solution [33].
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Example 2.6:
Let us recall example 1.6
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ = ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ), (2.24)
with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
Applying the TAM by first distributing the equation as
๐ฟ(๐ฃ) = ๐ฃ๐ก , ๐(๐ฃ) = ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ โ ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ) and ๐(๐ฅ, ๐ฆ, ๐ก) = 0.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) = 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ). (2.25)
By using simple manipulation, one can solve Eq. (2.25) as follows:
โซ ๐ฃ0๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
hence, we get
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
The second iteration can be carried through and given as
๐ฟ(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
with initial condition ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ). (2.26)
By integrating both sides of Eq. (2.26) from 0 to t, we get
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โซ ๐ฃ1๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ (โ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ฆ, ๐ก) โ ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)๐ฃ0๐ฆ(๐ฅ, ๐ฆ, ๐ก)๐ก
0
+ ๐(๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ฃ0๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก))) โ ๐ก,
with initial condition ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ). (2.27)
Thus,
๐ฃ1(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ1
2๐๐ก๐ ๐๐(4๐๐ฅ) +
1
2๐๐ก๐ ๐๐(4๐๐ฆ) โ
4๐2๐ก๐๐ ๐๐(2๐๐ฅ โ 2๐๐ฆ) โ 4๐2๐ก๐๐ ๐๐(2๐๐ฅ + 2๐๐ฆ) โ
1
2๐๐ก๐ ๐๐(4๐๐ฅ + 4๐๐ฆ).
The next iteration is
๐ฟ(๐ฃ2(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
with initial condition ๐ฃ2(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ). (2.28)
Then, we have
๐ฃ2(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+1
2(โ๐๐ ๐๐(4๐๐ฅ) + ๐๐ ๐๐(4๐๐ฆ) โ 8๐2๐๐ ๐๐(2๐๐ฅ โ 2๐๐ฆ)
โ 8๐2๐๐ ๐๐(2๐๐ฅ + 2๐๐ฆ) โ ๐๐ ๐๐(4๐๐ฅ + 4๐๐ฆ))๐ก + โฏ.
By continuing in this process till n = 4, we can get
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๐ฃ4(๐ฅ, ๐ฆ, ๐ก)
= ๐๐๐ (2๐๐ฆ)๐ ๐๐ (2๐๐ฅ) +1
2(โ๐๐ ๐๐ (4๐๐ฅ) + ๐๐ ๐๐ (4๐๐ฆ)
โ 8๐2๐๐ ๐๐ (2๐๐ฅ โ 2๐๐ฆ) โ 8๐2๐๐ ๐๐ (2๐๐ฅ + 2๐๐ฆ) โ ๐๐ ๐๐ (4๐๐ฅ
+ 4๐๐ฆ))๐ก + โฏ. (2.29)
Table 2.3 and Figure 2.2 illustrate the convergence of the solution
through the use of the maximal error remainder. Clearly seen, the errors will
be decreased as the number of iterations will be increased.
๐๐ธ๐ ๐ = ๐๐๐ฅ0โค๐ฅโค10โค๐ฆโค1
|๐ธ๐ ๐(๐ฅ, ๐ฆ)| , ๐ = 1, โฆ , ๐, when ๐ = 0.1, t =0.0001.
Table 2.3: The maximal error remainder: ๐๐ธ๐ ๐ by the TAM, where ๐ =1, โฆ ,4.
n ๐ด๐ฌ๐น๐ by TAM
1 5.99981 ร 10โ3
2 2.89069 ร 10โ6
3 1.12531 ร 10โ8
4 4.54397 ร 10โ11
Figure 2.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by TAM.
1.0 1.5 2.0 2.5 3.0 3.5 4.0
10 10
10 8
10 6
10 4
10 2
n
ME
Rn
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Example 2.7 [76]:
Let us consider the following 2D Burgersโ equation.
๐ฃ๐ก = ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ , with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ. (2.30)
By using the TAM, we get the following:
๐ฟ(๐ฃ) = ๐ฃ๐ก , ๐(๐ฃ) = โ๐ฃ๐ฃ๐ฅ โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ and ๐(๐ฅ, ๐ฆ, ๐ก) = 0.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) = 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ. (2.31)
By using a simple manipulation, one can solve Eq. (2.31) as follows:
โซ ๐ฃ0๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ.
Then, we get
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ.
The second iteration can be carried through and is given as
๐ฟ(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
with initial condition ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ. (2.32)
By integrating both sides of Eq. (2.32) from 0 to t, we get
โซ ๐ฃ1๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ (๐ฃ0(๐ฅ, ๐ฆ, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ฃ0๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก)) โ ๐ก๐ก
0
,
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with initial condition ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ.
Thus,
๐ฃ1(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ + (๐ฅ + ๐ฆ)๐ก.
The next iteration is
๐ฟ(๐ฃ2(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0, ๐ฃ2(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ.
Then, we have
๐ฃ2(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 +1
3(๐ฅ + ๐ฆ)๐ก3.
By continuing in this way, we will get series of the form:
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 +
(๐ฅ + ๐ฆ)๐ก3 + (๐ฅ + ๐ฆ)๐ก4 + โฏ.
This series converges to the exact solution [76]:
๐ฃ(๐ฅ, ๐ฆ, ๐ก) =๐ฅ + ๐ฆ
1 โ ๐ก,
= (๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 + (๐ฅ + ๐ฆ)๐ก3 + (๐ฅ + ๐ฆ)๐ก4
+ (๐ฅ + ๐ฆ)๐ก5 + โฏ.
In order to prove the convergence of TAM for 2D problem, let us assume that
T: ๐ 2 ร [0,1) โ ๐ 3, then ๐ฃ๐ = ๐(๐ฃ๐โ1), 0 โค ๐ก < 1.
By following similar steps as for example 2.5, since the exact solution is
๐ฃ = ๐ฃ(๐ฅ, ๐ฆ, ๐ก) =๐ฅ+๐ฆ
1โ๐ก, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ๐ฅ + ๐ฆ โ๐ฅ + ๐ฆ
1 โ ๐กโ,
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โ๐ฃ1 โ ๐ฃโ = โ๐ฅ + ๐ฆ + (๐ฅ + ๐ฆ)๐ก โ๐ฅ + ๐ฆ
1 โ ๐กโ โค ๐ โ๐ฅ + ๐ฆ โ
๐ฅ + ๐ฆ
1 โ ๐กโ
= ๐โ๐ฃ0 โ ๐ฃโ,
But, โ ๐ก ๐ [0,1), when ๐ก = 1
2, ๐ฅ = 1, ๐ฆ = 1,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.5 < 1,
Then, we get
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
Also,
โ๐ฃ2 โ ๐ฃโ = โ(๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 +1
3(๐ฅ + ๐ฆ)๐ก3 โ
๐ฅ + ๐ฆ
1 โ ๐กโ
โค ๐โ๐ฃ1 โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ = ๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ก ๐ [0,1), when ๐ก = 1
2, ๐ฅ = 1, ๐ฆ = 1,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.456435 < 1,
Thus,
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get:
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
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Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐๐ โ โ
๐๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ = 0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ =๐ฅ+๐ฆ
1โ๐ก , which is the exact solution [76].
Example 2.8 [76]:
Let us consider, the following (1+3)-D Burgersโ equation with the following
initial condition.
๐ฃ๐ก = ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง,
with initial condition ๐ฃ(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง. (2.33)
Now, applying the proposed method by first distributing the equation as
๐ฟ(๐ฃ) = ๐ฃ๐ก , ๐(๐ฃ) = โ๐ฃ๐ฃ๐ฅ โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ โ ๐ฃ๐ง๐ง and ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0.
Thus, the initial problem which needs to be solved is
๐ฟ(๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) = 0, ๐ฃ0(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง. (2.34)
By using a simple manipulation, one can solve Eq. (2.34) as follows:
โซ ๐ฃ0๐ก(๐ฅ, ๐ฆ, ๐ง, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง.
Then, we get
๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง.
The second iteration can be carried through and is given as
๐ฟ(๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
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with initial condition ๐ฃ1(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง. (2.35)
By integrating both sides of Eq. (2.35) from 0 to t, we get
โซ ๐ฃ1๐ก(๐ฅ, ๐ฆ, ๐ง, ๐ก) โ ๐ก๐ก
0
= โซ (๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ฆ, ๐ง, ๐ก) + ๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ง, ๐ก)๐ก
0
+ ๐ฃ0๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ง, ๐ก) + ๐ฃ0๐ง๐ง(๐ฅ, ๐ฆ, ๐ง, ๐ก)) โ ๐ก,
with initial condition ๐ฃ1(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง.
Thus,
๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก.
The next iteration is
๐ฟ(๐ฃ2(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
with initial condition ๐ฃ2(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง.
Then, we have
๐ฃ2(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2
+ 1
3(๐ฅ + ๐ฆ + ๐ง)๐ก3.
The next iteration is
๐ฟ(๐ฃ3(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฃ2(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
with initial condition ๐ฃ3(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง.
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Then, we get
๐ฃ3(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 + (๐ฅ + ๐ฆ + ๐ง)๐ก3
+2
3(๐ฅ + ๐ฆ + ๐ง)๐ก4 +
1
3(๐ฅ + ๐ฆ + ๐ง)๐ก5.
By continuing in this way, we will get series of the form:
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 + (๐ฅ + ๐ฆ + ๐ง)๐ก3
+ (๐ฅ + ๐ฆ + ๐ง)๐ก4 + (๐ฅ + ๐ฆ + ๐ง)๐ก5 + โฏ,
This series converges to the exact solution [76]:
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ฅ + ๐ฆ + ๐ง
1 โ ๐ก,
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 + (๐ฅ + ๐ฆ + ๐ง)๐ก3
+ (๐ฅ + ๐ฆ + ๐ง)๐ก4 + (๐ฅ + ๐ฆ + ๐ง)๐ก5 + (๐ฅ + ๐ฆ + ๐ง)๐ก6 + โฏ.
For convergence issue, suppose that T: ๐ 3 ร [0,1) โ ๐ 4, then ๐ฃ๐ = ๐(๐ฃ๐โ1)
and 0 โค ๐ก < 1, similar procedure can be followed, the exact solution is
๐ฃ = ๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ฅ+๐ฆ+๐ง
1โ๐ก, therefore, we have
โ๐ฃ0 โ ๐ฃโ = โ๐ฅ + ๐ฆ + ๐ง โ๐ฅ + ๐ฆ + ๐ง
1 โ ๐กโ,
โ๐ฃ1 โ ๐ฃโ = โ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก โ๐ฅ + ๐ฆ + ๐ง
1 โ ๐กโ
โค ๐ โ๐ฅ + ๐ฆ + ๐ง โ๐ฅ + ๐ฆ + ๐ง
1 โ ๐กโ = ๐โ๐ฃ0 โ ๐ฃโ,
But, โ ๐ก ๐ [0,1) , when ๐ก =1
3, ๐ฅ = 1, ๐ฆ = 1, ๐ง = 1,
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โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.333333 < 1,
Then, we get
โ๐ฃ1 โ ๐ฃโ โค ๐ โ๐ฃ0 โ ๐ฃโ,
Moreover,
โ๐ฃ2 โ ๐ฃโ = โ(๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 +1
3(๐ฅ + ๐ฆ +
๐ง)๐ก3 โ๐ฅ+๐ฆ+๐ง
1โ๐กโ โค ๐โ๐ฃ1 โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ = ๐2โ๐ฃ0 โ ๐ฃโ,
Since, โ ๐ก ๐ [0,1) , when ๐ก = 1
3, ๐ฅ = 1, ๐ฆ = 1, ๐ง = 1,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 โ โ
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐ = 0.293972 < 1,
Thus,
โ๐ฃ2 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
Similarly, we have
โ๐ฃ3 โ ๐ฃโ โค ๐3โ๐ฃ0 โ ๐ฃโ,
By continuing in this way, we get
โ๐ฃ๐ โ ๐ฃโ โค ๐๐โ๐ฃ0 โ ๐ฃโ,
Therefore,
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐๐ โ โ
๐๐ = 0, then
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๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ =๐ฅ+๐ฆ+๐ง
1โ๐ก,which is the exact solution[76].
2.4.5 Solving the nonlinear system of Burgersโ equations by the
TAM
Systems of partial differential equations have attracted much attention in
studying evolution equations describing wave propagation, in investigating
the shallow water waves [37, 48], examining the chemical reactionโdiffusion
model [48]. Moreover, systems of Burgersโ equations have a lot of
importance in engineering and physical fields, engineering, chemistry, finance
and in different fields in science [5, 10, 36, 45, 61]. To solve many
researchers have been interested it by various techniques such as the lattice
Boltzmann method (LBM) [21], LADM [17], ADM [72] and the HAM [2]. In
this section, the TAM will be implemented to solve the Burgersโ equation and
systems of Burgersโ equations in 1D, 2D and 3D.
Example 2.9:
Let us consider, the following system of coupled Burgersโ equation [50],
๐ข๐ก โ ๐ข๐ฅ๐ฅ โ 2๐ข๐ข๐ฅ + (๐ข๐ฃ)๐ฅ = 0,
๐ฃ๐ก โ ๐ฃ๐ฅ๐ฅ โ 2๐ฃ๐ฃ๐ฅ + (๐ข๐ฃ)๐ฅ = 0, ๐ฅ โ [โ๐, ๐], ๐ก > 0,
Subject to the initial conditions:
๐ข(๐ฅ, 0) = ๐ ๐๐(๐ฅ), ๐ฃ(๐ฅ, 0) = ๐ ๐๐(๐ฅ). (2.36)
To solve the system of Eq. (2.36) by using the TAM, we have
๐ฟ1(๐ข) = ๐ข๐ก , ๐1(๐ข) = โ๐ข๐ฅ๐ฅ โ2๐ข๐ข๐ฅ + (๐ข๐ฃ)๐ฅ and ๐(๐ฅ, ๐ก) = 0,
๐ฟ2(๐ฃ) = ๐ฃ๐ก , ๐2(๐ฃ) = โ๐ฃ๐ฅ๐ฅ โ2๐ฃ๐ฃ๐ฅ + (๐ข๐ฃ)๐ฅ and ๐(๐ฅ, ๐ก) = 0.
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Thus, the initial problem which needs to be solved is
๐ฟ1(๐ข0(๐ฅ, ๐ก)) = 0, ๐ข0(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
๐ฟ2(๐ฃ0(๐ฅ, ๐ก)) = 0, ๐ฃ0(๐ฅ, 0) = ๐ ๐๐(๐ฅ). (2.37)
By using simple manipulation, one can solve Eq. (2.37) as follows:
โซ ๐ข0๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ข0(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
โซ ๐ฃ0๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, 0) = ๐ ๐๐(๐ฅ).
Therefore, we have
๐ข0(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ),
๐ฃ0(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ).
The second iteration can be carried through and is given as
๐ฟ1(๐ข1(๐ฅ, ๐ก)) + ๐1(๐ข0(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ข1(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
๐ฟ2(๐ฃ1(๐ฅ, ๐ก)) + ๐2(๐ฃ0(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ฃ1(๐ฅ, 0) = ๐ ๐๐(๐ฅ). (2.38)
By integrating both sides of Eq. (2.38) from 0 to t, we get
โซ ๐ข1๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= โซ [๐ข0๐ฅ๐ฅ(๐ฅ, ๐ก) + 2๐ข0(๐ฅ, ๐ก)๐ข0๐ฅ(๐ฅ, ๐ก)๐ก
0
โ (๐ข0(๐ฅ, ๐ก)๐ฃ0(๐ฅ, ๐ก))๐ฅ
] โ ๐ก , ๐ข1(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
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โซ ๐ฃ1๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= โซ [๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ก) + 2๐ฃ0(๐ฅ, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ก)๐ก
0
โ (๐ข0(๐ฅ, ๐ก)๐ฃ0(๐ฅ, ๐ก))๐ฅ
] โ ๐ก , ๐ฃ1(๐ฅ, 0) = ๐ ๐๐(๐ฅ).
Thus, we get
๐ข1(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ),
๐ฃ1(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ).
The next iteration is
๐ฟ1(๐ข2(๐ฅ, ๐ก)) + ๐1(๐ข1(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ข2(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
๐ฟ2(๐ฃ2(๐ฅ, ๐ก)) + ๐2(๐ฃ1(๐ฅ, ๐ก)) + ๐(๐ฅ, ๐ก) = 0, ๐ฃ2(๐ฅ, 0) = ๐ ๐๐(๐ฅ). (2.39)
Once again, by taking the integration to both sides of problem (2.39), we have
โซ ๐ข2๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= โซ [๐ข1๐ฅ๐ฅ(๐ฅ, ๐ก) + 2๐ข1(๐ฅ, ๐ก)๐ข1๐ฅ(๐ฅ, ๐ก)๐ก
0
โ (๐ข1(๐ฅ, ๐ก)๐ฃ1(๐ฅ, ๐ก))๐ฅ
] โ ๐ก , ๐ข2(๐ฅ, 0) = ๐ ๐๐(๐ฅ),
โซ ๐ฃ2๐ก(๐ฅ, ๐ก) โ ๐ก๐ก
0
= โซ [๐ฃ1๐ฅ๐ฅ(๐ฅ, ๐ก) + 2๐ฃ1(๐ฅ, ๐ก)๐ฃ1๐ฅ(๐ฅ, ๐ก)๐ก
0
โ (๐ข1(๐ฅ, ๐ก)๐ฃ1(๐ฅ, ๐ก))๐ฅ
] โ ๐ก , ๐ฃ2(๐ฅ, 0) = ๐ ๐๐(๐ฅ).
Therefore, we get
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๐ข2(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ),
๐ฃ2(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ).
By continuing in this way, we will get series of the form:
๐ข(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ก),
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ
1
6๐ก3๐ ๐๐(๐ฅ) + โฏ,
๐ฃ(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ก),
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ
1
6๐ก3๐ ๐๐(๐ฅ) + โฏ.
This series converges to the exact solution [50],
๐ข(๐ฅ, ๐ก) = โ โ๐ก๐ ๐๐(๐ฅ),
๐ฃ(๐ฅ, ๐ก) = โ โ๐ก๐ ๐๐(๐ฅ).
The convergence, suppose that ๐ข๐ = ๐(๐ข๐โ1) and ๐ฃ๐ = ๐(๐ฃ๐โ1), 0 โค ๐ก < 1.
๐ข = ๐๐๐๐ โ โ
๐ข๐, ๐ฃ = ๐๐๐๐ โ โ
๐ฃ๐.
Similar procedure can be followed, the exact solution is ๐ข = ๐ข(๐ฅ, ๐ก) =
โ โ๐ก๐ ๐๐(๐ฅ), and ๐ฃ = ๐ฃ(๐ฅ, ๐ก) = โ โ๐ก๐ ๐๐(๐ฅ), we have
โ๐ข0 โ ๐ขโ = โ๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ,
โ๐ฃ0 โ ๐ฃโ = โ๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ.
Also,
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โ๐ข1 โ ๐ขโ = โ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ โค ๐1โ๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ
= ๐1โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ = โ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ โค ๐2โsin(๐ฅ) โ โ โ๐ก sin(๐ฅ)โ
= ๐2โ๐ฃ0 โ ๐ฃโ.
But, โ ๐ก ๐ [0,1), when ๐ก = 1
4, ๐ฅ =
๐
2 ,
โ๐ข1 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.130203 < 1,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.130203 < 1.
Then, we get
โ๐ข1 โ ๐ขโ โค ๐1โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ.
Also,
โ๐ข2 โ ๐ขโ = โ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ โค ๐1โ๐ข1 โ
๐ขโ โค ๐1๐1โ๐ข0 โ ๐ขโ = ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ = โ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ โ โ๐ก๐ ๐๐(๐ฅ)โ โค ๐2โ๐ฃ1 โ
๐ฃโ โค ๐2๐2โ๐ฃ0 โ ๐ฃโ = ๐22โ๐ฃ0 โ ๐ฃโ.
Since, โ ๐ก ๐ [0,1), when ๐ก = 1
4, ๐ฅ =
๐
2 ,
โ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1
2 โ โโ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.105226 < 1,
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โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2
2 โ โโ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.105226 < 1.
Then, we have
โ๐ข2 โ ๐ขโ โค ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ โค ๐22โ๐ฃ0 โ ๐ฃโ.
Similarly, we have
โ๐ข3 โ ๐ขโ โค ๐13โ๐ข0 โ ๐ขโ,
โ๐ฃ3 โ ๐ฃโ โค ๐23โ๐ฃ0 โ ๐ฃโ.
By continuing in this way, we get:
โ๐ข๐ โ ๐ขโ โค ๐1๐โ๐ข0 โ ๐ขโ,
โ๐ฃ๐ โ ๐ฃโ โค ๐2๐โ๐ฃ0 โ ๐ฃโ.
Therefore,
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ โค ๐๐๐๐ โ โ
๐1๐ โ๐ข0 โ ๐ขโ = 0 , and ๐๐๐
๐ โ โ๐1
๐ = 0, then
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ =0, then ๐๐๐๐ โ โ
๐ข๐ = ๐ข = โ โ๐ก๐ ๐๐(๐ฅ), which is the exact
solution.
Similarly, we get
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐2๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐
๐ โ โ๐2
๐ = 0, then
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๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ = โ โ๐ก๐ ๐๐(๐ฅ), which is the exact
solution [50].
Example 2.10 [30]:
Let us consider, the following system 2D Burgersโ equation.
๐ข๐ก + ๐ข๐ข๐ฅ + ๐ฃ๐ข๐ฆ โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ = 0,
๐ฃ๐ก + ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ = 0,
Subject to the initial conditions:
๐ข(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ, ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ. (2.40)
To solve the system of Eq. (2.40) by using the TAM
๐ฟ1(๐ข) = ๐ข๐ก , ๐1(๐ข) = ๐ข๐ข๐ฅ + ๐ฃ๐ข๐ฆ โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ and ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
๐ฟ2(๐ฃ) = ๐ฃ๐ก , ๐2(๐ฃ) = ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ and ๐(๐ฅ, ๐ฆ, ๐ก) = 0.
Thus, the initial problem which needs to be solved is
๐ฟ1(๐ข0(๐ฅ, ๐ฆ, ๐ก)) = 0, ๐ข0(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
๐ฟ2(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) = 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ. (2.41)
By using a simple manipulation, one can solve Eq. (2.41) as follows:
โซ ๐ข0๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ข0(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
โซ ๐ฃ0๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= 0, ๐ฃ0(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ.
Then, we get
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๐ข0(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ),
๐ฃ0(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ โ ๐ฆ).
The second iteration can be carried through and is given as
๐ฟ1(๐ข1(๐ฅ, ๐ฆ, ๐ก)) + ๐1(๐ข0(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0, ๐ข1(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
๐ฟ2(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐2(๐ฃ0(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0, ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ. (2.42)
By integrating both sides of Eq. (2.42) from 0 to t, we get
โซ ๐ข1๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ [โ๐ข0(๐ฅ, ๐ฆ, ๐ก)๐ข0๐ฅ(๐ฅ, ๐ฆ, ๐ก) โ ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)๐ข0๐ฆ(๐ฅ, ๐ฆ, ๐ก)๐ก
0
+ ๐ข0๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ข0๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก)] โ ๐ก , ๐ข1(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
โซ ๐ฃ1๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ [โ๐ข0(๐ฅ, ๐ฆ, ๐ก)๐ฃ0๐ฅ(๐ฅ, ๐ฆ, ๐ก) โ ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)๐ฃ0๐ฆ(๐ฅ, ๐ฆ, ๐ก)๐ก
0
+ ๐ฃ0๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ฃ0๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก)] โ ๐ก , ๐ฃ1(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ.
Thus, we have
๐ข1(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ) โ 2๐ก๐ฅ,
๐ฃ1(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ โ ๐ฆ) โ 2๐ก๐ฆ.
The next iteration is
๐ฟ1(๐ข2(๐ฅ, ๐ฆ, ๐ก)) + ๐1(๐ข1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
72
๐ข2(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
๐ฟ2(๐ฃ2(๐ฅ, ๐ฆ, ๐ก)) + ๐2(๐ฃ1(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ก) = 0,
๐ฃ2(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ. (2.43)
Once again, by taking the integration on both sides of problem (2.43), we
have
โซ ๐ข2๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ [โ๐ข1(๐ฅ, ๐ฆ, ๐ก)๐ข1๐ฅ(๐ฅ, ๐ฆ, ๐ก) โ ๐ฃ1(๐ฅ, ๐ฆ, ๐ก)๐ข1๐ฆ(๐ฅ, ๐ฆ, ๐ก)๐ก
0
+ ๐ข1๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ข1๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก)] โ ๐ก , ๐ข2(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ,
โซ ๐ฃ2๐ก(๐ฅ, ๐ฆ, ๐ก) โ ๐ก๐ก
0
= โซ [โ๐ข1(๐ฅ, ๐ฆ, ๐ก)๐ฃ1๐ฅ(๐ฅ, ๐ฆ, ๐ก) โ ๐ฃ1(๐ฅ, ๐ฆ, ๐ก)๐ฃ1๐ฆ(๐ฅ, ๐ฆ, ๐ก)๐ก
0
+ ๐ฃ1๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก) + ๐ฃ1๐ฆ๐ฆ(๐ฅ, ๐ฆ, ๐ก)] โ ๐ก , ๐ฃ2(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ.
Therefore, we get
๐ข2(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ) โ 2๐ก๐ฅ + 2(๐ฅ + ๐ฆ)๐ก2 โ4๐ก3๐ฅ
3,
๐ฃ2(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ โ ๐ฆ) โ 2๐ก๐ฆ + (2๐ฅ โ 2๐ฆ)๐ก2 โ4๐ก3๐ฆ
3.
By continuing in this way, we will get a series of the form:
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
73
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ฆ, ๐ก) ,
= (๐ฅ + ๐ฆ) โ 2๐ฅ๐ก + 2(๐ฅ + ๐ฆ)๐ก2 โ 4๐ฅ๐ก3 + 4(๐ฅ + ๐ฆ)๐ก4 โ 8๐ฅ๐ก5
+ 8(๐ฅ + ๐ฆ)๐ก6 โ 16๐ฅ๐ก7 + 16(๐ฅ + ๐ฆ)๐ก8 โ โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก),
= (๐ฅ โ ๐ฆ) โ 2๐ฆ๐ก + (2๐ฅ โ 2๐ฆ)๐ก2 โ 4๐ฆ๐ก3 + (4๐ฅ โ 4๐ฆ)๐ก4 โ 8๐ฆ๐ก5
+ (8๐ฅ โ 8๐ฆ)๐ก6 โ 16๐ฆ๐ก7 + (16๐ฅ โ 16๐ฆ)๐ก8 โ โฏ.
This series converges to the exact solution [30]:
๐ข(๐ฅ, ๐ฆ, ๐ก) =๐ฅ โ 2๐ฅ๐ก + ๐ฆ
1 โ 2๐ก2,
= (๐ฅ + ๐ฆ) โ 2๐ฅ๐ก + 2(๐ฅ + ๐ฆ)๐ก2 โ 4๐ฅ๐ก3 + (4๐ฅ + 4๐ฆ)๐ก4 โ 8๐ฅ๐ก5
+ (8๐ฅ + 8๐ฆ)๐ก6 โ 16๐ฅ๐ก7 + (16๐ฅ + 16๐ฆ)๐ก8 โ 32๐ฅ๐ก9 + โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ก) =๐ฅ โ 2๐ฆ๐ก โ ๐ฆ
1 โ 2๐ก2,
= (๐ฅ โ ๐ฆ) โ 2๐ฆ๐ก + (2๐ฅ โ 2๐ฆ)๐ก2 โ 4๐ฆ๐ก3 + (4๐ฅ โ 4๐ฆ)๐ก4 โ 8๐ฆ๐ก5
+ (8๐ฅ โ 8๐ฆ)๐ก6 โ 16๐ฆ๐ก7 + (16๐ฅ โ 16๐ฆ)๐ก8 โ 32๐ฆ๐ก9 + โฏ.
For convergence issue, suppose that ๐ข๐ = ๐(๐ข๐โ1), ๐ฃ๐ = ๐(๐ฃ๐โ1), 0 โค ๐ก < 1,
๐ข = ๐๐๐๐ โ โ
๐ข๐, ๐ฃ = ๐๐๐๐ โ โ
๐ฃ๐ .
Since, the exact solution is ๐ข = ๐ข(๐ฅ, ๐ฆ, ๐ก) =๐ฅโ2๐ฅ๐ก+๐ฆ
1โ2๐ก2, and ๐ฃ = ๐ฃ(๐ฅ, ๐ฆ, ๐ก) =
๐ฅโ2๐ฆ๐กโ๐ฆ
1โ2๐ก2 , therefore, we have
โ๐ข0 โ ๐ขโ = โ๐ฅ + ๐ฆ โ (๐ฅ โ 2๐ฅ๐ก + ๐ฆ
1 โ 2๐ก2)โ,
โ๐ฃ0 โ ๐ฃโ = โ๐ฅ โ ๐ฆ โ (๐ฅ โ 2๐ฆ๐ก โ ๐ฆ
1 โ 2๐ก2)โ.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
74
Also,
โ๐ข1 โ ๐ขโ = โ๐ฅ + ๐ฆ โ 2๐ก๐ฅ โ (๐ฅ โ 2๐ฅ๐ก + ๐ฆ
1 โ 2๐ก2)โ
โค ๐1 โ๐ฅ + ๐ฆ โ (๐ฅ โ 2๐ฅ๐ก + ๐ฆ
1 โ 2๐ก2)โ = ๐1โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ = โ๐ฅ โ ๐ฆ โ 2๐ก๐ฆ โ (๐ฅโ2๐ฆ๐กโ๐ฆ
1โ2๐ก2 )โ โค ๐2 โ๐ฅ โ ๐ฆ โ (๐ฅโ2๐ฆ๐กโ๐ฆ
1โ2๐ก2)โ =
๐2โ๐ฃ0 โ ๐ฃโ.
But, โ ๐ก ๐ [0,1), when ๐ก = 1
4, ๐ฅ = 1, ๐ฆ = 1,
โ๐ข1 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.75 < 1,
and when ๐ก = 1
2, ๐ฅ = 0, ๐ฆ = 1,
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.666667 < 1.
Then, we get
โ๐ข1 โ ๐ขโ โค ๐1 โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ โค ๐2 โ๐ฃ0 โ ๐ฃโ.
Also,
โ๐ข2 โ ๐ขโ = โ๐ฅ + ๐ฆ โ 2๐ก๐ฅ + 2(๐ฅ + ๐ฆ)๐ก2 โ4๐ก3๐ฅ
3โ (
๐ฅโ2๐ฅ๐ก+๐ฆ
1โ2๐ก2 )โ โค
๐1โ๐ข1 โ ๐ขโ โค ๐1๐1โ๐ข0 โ ๐ขโ = ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ = โ๐ฅ โ ๐ฆ โ 2๐ก๐ฆ + (2๐ฅ โ 2๐ฆ)๐ก2 โ4๐ก3๐ฆ
3โ (
๐ฅโ2๐ฆ๐กโ๐ฆ
1โ2๐ก2 )โ โค
๐2โ๐ฃ1 โ ๐ฃโ โค ๐2๐2โ๐ฃ0 โ ๐ฃโ = ๐22โ๐ฃ0 โ ๐ฃโ.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
75
Since, โ ๐ก ๐ [0,1), when ๐ก = 1
4, ๐ฅ = 1, ๐ฆ = 1,
โ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1
2 โ โโ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.228218 < 1,
and when ๐ก = 1
2, ๐ฅ = 0, ๐ฆ = 1,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2
2 โ โโ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.625485 < 1.
Thus,
โ๐ข2 โ ๐ขโ โค ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ โค ๐22โ๐ฃ0 โ ๐ฃโ.
Similarly, we have
โ๐ข3 โ ๐ขโ โค ๐13โ๐ข0 โ ๐ขโ,
โ๐ฃ3 โ ๐ฃโ โค ๐23โ๐ฃ0 โ ๐ฃโ.
By continuing in this way, we have
โ๐ข๐ โ ๐ขโ โค ๐1๐โ๐ข0 โ ๐ขโ,
โ๐ฃ๐ โ ๐ฃโ โค ๐2๐โ๐ฃ0 โ ๐ฃโ.
Therefore,
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ โค ๐๐๐๐ โ โ
๐1๐โ๐ข0 โ ๐ขโ = 0 , and ๐๐๐
๐ โ โ๐1
๐ = 0, then
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ =0, then ๐๐๐๐ โ โ
๐ข๐ = ๐ข =๐ฅโ2๐ฅ๐ก+๐ฆ
1โ2๐ก2, which is the exact solution,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
76
Similarly, we have
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐2๐โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐
๐ โ โ๐2
๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ = 0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ =๐ฅโ2๐ฆ๐กโ๐ฆ
1โ2๐ก2 , which is the exact solution
[30].
Example 2.11:
Let us consider, the following system of 3D Burgersโ equation,
๐ข๐ก โ ๐ข๐ข๐ฅ โ ๐ฃ๐ข๐ฆ โ ๐ค๐ข๐ง โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ โ ๐ข๐ง๐ง = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
๐ฃ๐ก + ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ + ๐ค๐ฃ๐ง + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
๐ค๐ก + ๐ข๐ค๐ฅ + ๐ฃ๐ค๐ฆ + ๐ค๐ค๐ง + ๐ค๐ฅ๐ฅ + ๐ค๐ฆ๐ฆ + ๐ค๐ง๐ง = โ(๐ฅ, ๐ฆ, ๐ง, ๐ก),
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
โ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)), (2.44)
Subject to the initial conditions:
๐ข(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
๐ฃ(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
๐ค(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง.
To solve the system of Eq. (2.44) by using the TAM, we have
๐ฟ1(๐ข) = ๐ข๐ก , ๐1(๐ข) = โ๐ข๐ข๐ฅ โ ๐ฃ๐ข๐ฆ โ ๐ค๐ข๐ง โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ โ ๐ข๐ง๐ง,
โ ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
77
๐ฟ2(๐ฃ) = ๐ฃ๐ก , ๐2(๐ฃ) = ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ + ๐ค๐ฃ๐ง + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง,
โ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐ฟ3(๐ค) = ๐ค๐ก , ๐3(๐ค) = ๐ข๐ค๐ฅ + ๐ฃ๐ค๐ฆ + ๐ค๐ค๐ง + ๐ค๐ฅ๐ฅ + ๐ค๐ฆ๐ฆ + ๐ค๐ง๐ง,
โโ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)).
Thus, the initial problem which needs to be solved is
๐ฟ1(๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐ข0(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
๐ฟ2(๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐ฃ0(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
๐ฟ3(๐ค0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐ค0(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง. (2.45)
By using simple manipulation, one can solve Eq. (2.45) as follows:
โซ ๐ข0๐ก(๐ฅ, ๐ฆ, ๐ง, ๐ก) โ ๐ก๐ก
0
= โซ [โ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))] โ ๐ก๐ก
0
,
๐ข0(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
โซ ๐ฃ0๐ก(๐ฅ, ๐ฆ, ๐ง, ๐ก) โ ๐ก๐ก
0
= โซ [โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))] โ ๐ก๐ก
0
,
๐ฃ0(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
โซ ๐ค0๐ก(๐ฅ, ๐ฆ, ๐ง, ๐ก) โ ๐ก๐ก
0
= โซ [โ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))] โ ๐ก๐ก
0
,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
78
๐ค0(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง.
Then, we get
๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= 1 โ โ ๐ก +๐ฅ
2+ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2+ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2+ โ ๐ก๐ง
โ1
2โ 2๐ก๐ง,
๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= 1 โ โ ๐ก +๐ฅ
2โ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2โ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2โ โ ๐ก๐ง
โ1
2โ 2๐ก๐ง,
๐ค0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= โ ๐ก โ๐ฅ
2+ โ ๐ก๐ฅ +
1
2โ 2๐ก๐ฅ โ
๐ฆ
2+ โ ๐ก๐ฆ +
1
2โ 2๐ก๐ฆ โ
๐ง
2+ โ ๐ก๐ง
+1
2โ 2๐ก๐ง.
We turn the function ๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก), ๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก) and ๐ค0(๐ฅ, ๐ฆ, ๐ง, ๐ก) by using
Taylor series expansion to exponential function, we get
๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (๐ฅ + ๐ฆ + ๐ง) โ ๐ก +1
2(โ1 โ ๐ฅ โ ๐ฆ โ ๐ง)๐ก2 +
1
6(โ1 โ 3๐ฅ โ 3๐ฆ
โ 3๐ง)๐ก3 +1
24(โ1 โ 7๐ฅ โ 7๐ฆ โ 7๐ง)๐ก4 +
1
120(โ1 โ 15๐ฅ โ 15๐ฆ
โ 15๐ง)๐ก5 + ๐[๐ก]6,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
79
๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ1 โ 2๐ฅ โ 2๐ฆ โ 2๐ง)๐ก +1
2(โ1 โ 3๐ฅ โ 3๐ฆ
โ 3๐ง)๐ก2 +1
6(โ1 โ 5๐ฅ โ 5๐ฆ โ 5๐ง)๐ก3 +
1
24(โ1 โ 9๐ฅ โ 9๐ฆ
โ 9๐ง)๐ก4 +1
120(โ1 โ 17๐ฅ โ 17๐ฆ โ 17๐ง)๐ก5 + ๐[๐ก]6,
๐ค0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (1 + ๐ฅ + ๐ฆ + ๐ง) + (1 + 2๐ฅ + 2๐ฆ + 2๐ง)๐ก
+1
2(1 + 3๐ฅ + 3๐ฆ + 3๐ง)๐ก2 +
1
6(1 + 5๐ฅ + 5๐ฆ + 5๐ง)๐ก3
+1
24(1 + 9๐ฅ + 9๐ฆ + 9๐ง)๐ก4 +
1
120(1 + 17๐ฅ + 17๐ฆ + 17๐ง)๐ก5
+ ๐[๐ก]6.
The second iteration can be carried through and is given as
๐ฟ1(๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐1(๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
๐ข1(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
๐ฟ2(๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐2(๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
๐ฃ1(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
๐ฟ3(๐ค1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐3(๐ค0(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + โ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
๐ค1(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง. (2.46)
By integrating both sides of Eq. (2.46) from 0 to t, we turn the
function ๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก), ๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก) and ๐ค1(๐ฅ, ๐ฆ, ๐ง, ๐ก) by using Taylor series
expansion to exponential function, we get
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
80
๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(โ2 โ ๐ฅ โ ๐ฆ โ ๐ง)๐ก2 +
1
6(โ3
โ 5๐ฅ โ 5๐ฆ โ 5๐ง)๐ก3 +1
24(โ2 โ 13๐ฅ โ 13๐ฆ โ 13๐ง)๐ก4 +
1
120(9
โ 23๐ฅ โ 23๐ฆ โ 23๐ง)๐ก5 + ๐[๐ก]6,
๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ1 โ 2๐ฅ โ 2๐ฆ โ 2๐ง)๐ก +1
2(โ1 โ 3๐ฅ โ 3๐ฆ
โ 3๐ง)๐ก2 +1
6(โ1 โ 5๐ฅ โ 5๐ฆ โ 5๐ง)๐ก3 +
1
24(โ1 โ 9๐ฅ โ 9๐ฆ
โ 9๐ง)๐ก4 +1
120(โ1 โ 17๐ฅ โ 17๐ฆ โ 17๐ง)๐ก5 + ๐[๐ก]6,
๐ค1(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (1 + ๐ฅ + ๐ฆ + ๐ง) + (1 + 2๐ฅ + 2๐ฆ + 2๐ง)๐ก +1
2(1 + 3๐ฅ + 3๐ฆ
+ 3๐ง)๐ก2 +1
6(1 + 5๐ฅ + 5๐ฆ + 5๐ง)๐ก3 +
1
24(1 + 9๐ฅ + 9๐ฆ + 9๐ง)๐ก4
+1
120(1 + 17๐ฅ + 17๐ฆ + 17๐ง)๐ก5 + ๐[๐ก]6.
The next iteration is
๐ฟ1(๐ข2(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐1(๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
๐ข2(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
๐ฟ2(๐ฃ2(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐2(๐ฃ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
๐ฃ2(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
๐ฟ3(๐ค2(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + ๐3(๐ค1(๐ฅ, ๐ฆ, ๐ง, ๐ก)) + โ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 0,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
81
๐ค2(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง.
Similarly, we have
๐ข2(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2 +
1
6(โ4 โ 3๐ฅ
โ 3๐ฆ โ 3๐ง)๐ก3 +1
24(โ15 โ 23๐ฅ โ 23๐ฆ โ 23๐ง)๐ก4 +
1
120(โ16
โ 75๐ฅ โ 75๐ฆ โ 75๐ง)๐ก5 + ๐[๐ก]6,
๐ฃ2(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2 +
1
6(โ2
โ 3๐ฅ โ 3๐ฆ โ 3๐ง)๐ก3 +1
24(โ7 โ 11๐ฅ โ 11๐ฆ โ 11๐ง)๐ก4
+1
120(โ14 โ 31๐ฅ โ 31๐ฆ โ 31๐ง)๐ก5 + ๐[๐ก]6,
๐ค2(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= (1 + ๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2 +
1
6(2
+ 3๐ฅ + 3๐ฆ + 3๐ง)๐ก3 +1
24(7 + 11๐ฅ + 11๐ฆ + 11๐ง)๐ก4 +
1
120(14
+ 31๐ฅ + 31๐ฆ + 31๐ง)๐ก5 + ๐[๐ก]6.
By continuing in this way, we will get series of the form:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2 +
1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 + โฏ,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
82
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2
+1
6(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก3 + โฏ,
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ค๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= (1 + ๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 + โฏ.
This series converges to the exact solution:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง),
= (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 +
1
24(๐ฅ + ๐ฆ + ๐ง)๐ก4 + โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง),
= (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2
+1
6(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก3 +
1
24(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก4 + โฏ,
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง),
= (1 + ๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 +
1
24(๐ฅ + ๐ฆ + ๐ง)๐ก4 + โฏ.
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
83
Finally, the convergence can be done in the following steps, suppose that
๐ข๐ = ๐(๐ข๐โ1), ๐ฃ๐ = ๐(๐ฃ๐โ1) and ๐ค๐ = ๐(๐ค๐โ1), 0 โค ๐ก โค 1.
๐ข = ๐๐๐๐ โ โ
๐ข๐ , ๐ฃ = ๐๐๐๐ โ โ
๐ฃ๐ , ๐ค = ๐๐๐๐ โ โ
๐ค๐ .
Since, the exact solution is
๐ข = ๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง),
๐ฃ = ๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง),
๐ค = ๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง).
Therefore, we have
โ๐ข0 โ ๐ขโ = โ๐ข0 โ โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ,
โ๐ฃ0 โ ๐ฃโ = โ๐ฃ0 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ,
โ๐ค0 โ ๐คโ = โ๐ค0 โ (1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))โ.
Also,
โ๐ข1 โ ๐ขโ = โ๐ข1 โ โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ โค ๐1โ๐ข0 โ โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ
= ๐1โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ = โ๐ฃ1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ โค ๐2โ๐ฃ0 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ
= ๐2โ๐ฃ0 โ ๐ฃโ,
โ๐ค1 โ ๐คโ = โ๐ค1 โ (1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))โ
โค ๐3โ๐ค0 โ (1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))โ = ๐3โ๐ค0 โ ๐คโ .
But, โ ๐ก ๐ [0,1], when ๐ก = 1, ๐ฅ = 1, ๐ฆ = 1, ๐ง = 1,
โ๐ข1 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.82038 < 1,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
84
โ๐ฃ1 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.673461 < 1,
โ๐ค1 โ ๐คโ
โ๐ค0 โ ๐คโโค ๐3 = 0.673461 < 1.
Then, we get
โ๐ข1 โ ๐ขโ โค ๐1โ๐ข0 โ ๐ขโ,
โ๐ฃ1 โ ๐ฃโ โค ๐2โ๐ฃ0 โ ๐ฃโ,
โ๐ค1 โ ๐คโ โค ๐3โ๐ค0 โ ๐คโ.
Also,
โ๐ข2 โ ๐ขโ = โ๐ข2 โ โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ โค ๐1โ๐ข1 โ ๐ขโ โค ๐1๐1โ๐ข0 โ ๐ขโ
= ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ = โ๐ฃ2 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)โ โค ๐2โ๐ฃ1 โ ๐ฃโ โค ๐2๐2โ๐ฃ0 โ ๐ฃโ
= ๐22โ๐ฃ0 โ ๐ฃโ,
โ๐ค2 โ ๐คโ = โ๐ค2 โ (1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง))โ โค ๐3โ๐ค1 โ ๐คโ โค ๐3๐3โ๐ค0 โ ๐คโ
= ๐32โ๐ค0 โ ๐คโ.
Since, โ ๐ก ๐ [0,1], when ๐ก = 1, ๐ฅ = 1, ๐ฆ = 1, ๐ง = 1,
โ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1
2 โ โโ๐ข2 โ ๐ขโ
โ๐ข0 โ ๐ขโโค ๐1 = 0.807042 < 1,
โ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2
2 โ โโ๐ฃ2 โ ๐ฃโ
โ๐ฃ0 โ ๐ฃโโค ๐2 = 0.672965 < 1,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
85
โ๐ค2 โ ๐คโ
โ๐ค0 โ ๐คโโค ๐3
2 โ โโ๐ค2 โ ๐คโ
โ๐ค0 โ ๐คโโค ๐3 = 0.672965 < 1.
Then, we get
โ๐ข2 โ ๐ขโ โค ๐12โ๐ข0 โ ๐ขโ,
โ๐ฃ2 โ ๐ฃโ โค ๐22โ๐ฃ0 โ ๐ฃโ,
โ๐ค2 โ ๐คโ โค ๐32โ๐ค0 โ ๐คโ.
Similarly, we have
โ๐ข3 โ ๐ขโ โค ๐13โ๐ข0 โ ๐ขโ,
โ๐ฃ3 โ ๐ฃโ โค ๐23โ๐ฃ0 โ ๐ฃโ,
โ๐ค3 โ ๐คโ โค ๐33โ๐ค0 โ ๐คโ.
By continuing in this way, we get:
โ๐ข๐ โ ๐ขโ โค ๐1๐โ๐ข0 โ ๐ขโ,
โ๐ฃ๐ โ ๐ฃโ โค ๐2๐โ๐ฃ0 โ ๐ฃโ,
โ๐ค๐ โ ๐คโ โค ๐3๐โ๐ค0 โ ๐คโ.
Therefore,
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ โค ๐๐๐๐ โ โ
๐1๐ โ๐ข0 โ ๐ขโ = 0 , and ๐๐๐
๐ โ โ๐1
๐ = 0, then
๐๐๐๐ โ โ
โ๐ข๐ โ ๐ขโ =0, then ๐๐๐๐ โ โ
๐ข๐ = ๐ข = โ ๐ก(๐ฅ + ๐ฆ + ๐ง), which is the exact
solution,
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Chapter Two Semi-Analytical Iterative Method for Solving
Some Ordinary and Partial Differential Equations
86
Similarly, we have
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ โค ๐๐๐๐ โ โ
๐2๐ โ๐ฃ0 โ ๐ฃโ = 0 , and ๐๐๐
๐ โ โ๐2
๐ = 0, then
๐๐๐๐ โ โ
โ๐ฃ๐ โ ๐ฃโ =0, then ๐๐๐๐ โ โ
๐ฃ๐ = ๐ฃ = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง), which is the exact
solution,
Also,
๐๐๐๐ โ โ
โ๐ค๐ โ ๐คโ โค ๐๐๐๐ โ โ
๐3๐ โ๐ค0 โ ๐คโ = 0 , and ๐๐๐
๐ โ โ๐3
๐ = 0, then
๐๐๐๐ โ โ
โ๐ค๐ โ ๐คโ =0, then ๐๐๐๐ โ โ
๐ค๐ = ๐ค = 1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง), which is the
exact solution.
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CHAPTER 3
BANACH CONTRACTION
METHOD FOR SOLVING SOME
ORDINARY AND PARTIAL
DIFFERENTIAL EQUATIONS
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
87
Chapter 3
Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
3.1 Introduction
Various analytic and semi-analytic iterative methods have been used to
solve many variant problems in different fields of the applied sciences have
significantly improved the iterative techniques. One of these methods is
Banach contraction method (BCM) based on using Banach contraction to
provide the required solution for different nonlinear kinds of functional
equations [75]. The BCM characterized as one of the developments of
Picardโs method where the ease of application clearly observed which makes
it distinct from the other known iterative methods.
In this chapter, the BCM will be used to solve some important problems
in physics and engineering which already solved by TAM in chapter two. The
first part of this chapter will be presented the analytic solutions for Riccati,
pantograph, and beam deformation equations. Moreover, the BCM will be
used to solve 1D, 2D and 3D of Burgersโ equations and systems of equations.
The BCM does not required any additional restrictions to deal with nonlinear
terms.
The work in this chapter is organized as follows: In section two, the
basic idea of the BCM will be presented. Section three discusses the
application of the BCM for the Riccati, pantograph, and beam deformation
equations. In section four, the BCM will be applied to solve Burgersโ
equations and system of equations in 1D, 2D and 3D.
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
88
3.2 Contractions: Definition and Example
Definition 3.1: [3]
Let (๐, ๐) be a metric space. Then a mapping ๐: ๐ โ ๐,
(a) A point ๐ฅ โ ๐ is called a fixed point of ๐ if ๐ฅ = ๐(๐ฅ).
(b) ๐ is called contraction if there exists a fixed constant ๐ < 1 such that
๐(๐(๐ฅ), ๐(๐ฆ)) โค ๐๐(๐ฅ, ๐ฆ), for all ๐ฅ, ๐ฆ in ๐.
A contraction mapping is also known as Banach contraction.
Example 3.1 [3]:
(a) Consider the usual metric space (๐ , ๐). Define
๐(๐ฅ) =๐ฅ
๐+ ๐, for all ๐ฅ โ ๐ .
Then, ๐ is contraction on ๐ if ๐ > 1 and the solution of the equation,
๐ฅ โ ๐(๐ฅ) = 0 is ๐ฅ =๐๐
๐โ1.
(b) Consider the Euclidean metric space (๐ 2, ๐). Define
๐(๐ฅ, ๐ฆ) = (๐ฅ
๐+ ๐,
๐ฆ
๐+ ๐), for all (๐ฅ, ๐ฆ) โ ๐ 2.
Then, ๐ is contraction on ๐ 2 if ๐, ๐ > 1. Now, solving the equation,
๐(๐ฅ, ๐ฆ) = (๐ฅ, ๐ฆ) for a fixed point, we get is ๐ฅ =๐๐
๐โ1 and ๐ฆ =
๐๐
๐โ1.
Using induction, one can easily get the following concerning iterates of a
contraction mapping.
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
89
If ๐ is a contraction mapping on a metric space (๐, ๐) with contraction
constant ๐, then ๐๐ (where the superscript represents the ๐th iterate of ๐) is
also a contraction on ๐ with constant ๐๐.
3.3 Banach Contraction Principle
In 1922 Banach published his fixed point theorem also known as
Banachโs Contraction Principle uses the concept of Lipschitz mapping. Also,
it has long been used in analysis to solve various kinds of differential and
integral equations [67].
Theorem 3.1 (Banach Contraction Principle): [75]
Let (๐, ๐) be a complete metric space and ๐: ๐ โ ๐ be a contraction
mapping. Then ๐ has a unique fixed point ๐ฅ0 and for each ๐ฅ โ ๐, we have
๐๐๐๐โโ
๐๐(๐ฅ) = ๐ฅ0,
Moreover for each ๐ฅ โ ๐, we have
๐(๐๐(๐ฅ), ๐ฅ0) โค ๐๐
1 โ ๐๐(๐(๐ฅ), ๐ฅ).
3.3.1 Solution of differential equations using Banach contraction
principle
Let ๐(๐ฅ, ๐ฃ) be a continuous real-valued function on [๐, ๐] ร [๐, ๐]. The
Cauchy initial value problem is to find a continuous differentiable function
๐ฃ on [๐, ๐] satisfying the differential equation [67],
๐ฃโฒ(๐ฅ) = ๐(๐ฅ, ๐ฃ(๐ฅ)), ๐ฃ(๐ฅ0) = ๐ฃ0 . (3.1)
Consider the Banach space ๐ถ[๐, ๐] of continuous real-valued functions with
supremum norm defined by
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
90
โ๐ฃโ = sup{ |๐ฃ(๐ฅ)| โถ ๐ฅ โ [๐, ๐]}.
Integrate both sides of Eq. (3.1), we obtain an integral equation
๐ฃ(๐ฅ) = ๐ฃ0 + โซ ๐(๐ก, ๐ฃ(๐ก)) ๐๐ก.๐ฅ
๐ฅ0
(3.2)
The problem (3.1) is equivalent the problem solving the integral equation
(3.2). We define an integral operator ๐น: ๐ถ[๐, ๐] โ ๐ถ[๐, ๐] by
๐น๐ฃ(๐ฅ) = ๐ฃ0 + โซ ๐(๐ก, ๐ฃ(๐ก)) ๐๐ก.๐ฅ
๐ฅ0
Thus, a solution of Cauchy initial value problem (3.1) corresponds with a
fixed point of ๐น. One may easily check that if ๐น is contraction, then the
problem (3.1) has a unique solution.
Now our purpose is to impose certain conditions on ๐ under which the
integral operator ๐น is Lipschitz with ๐ < 1.
Theorem 3.2: [67]
Let ๐(๐ฅ, ๐ฃ) be a continuous function of ๐ท๐๐(๐) = [๐, ๐] ร [๐, ๐] such
that ๐ is Lipschitz with respect to ๐ฃ, that is there exists ๐ > 0 such that
|๐(๐ฅ, ๐ฃ) โ ๐(๐ฅ, ๐ข)| โค ๐|๐ฃ โ ๐ข|, for all ๐ฃ, ๐ข โ [๐, ๐] and for ๐ฅ โ [๐, ๐].
Suppose (๐ฅ0, ๐ฃ0) โ ๐๐๐ก(๐ท๐๐(๐)). Then for sufficiently small โ > ๐, there
exists a unique solution of the problem (3.1).
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
91
3.4 The basic idea of the BCM
Let us consider the general functional equation [75],
๐ฃ = ๐(๐ฃ) + ๐ (3.3)
where ๐(๐ฃ) is a nonlinear operator and ๐ is a known function.
Define successive approximations as
๐ฃ0 = ๐,
๐ฃ1 = ๐ฃ0 + ๐(๐ฃ0),
๐ฃ2 = ๐ฃ0 + ๐(๐ฃ1),
๐ฃ3 = ๐ฃ0 + ๐(๐ฃ2),
โฎ
By continuing, we have
๐ฃ๐ = ๐ฃ0 + ๐(๐ฃ๐โ1), ๐ = 1,2,3, โฆ . (3.4)
If ๐๐ is contraction for operator some positive integer k, then ๐(๐ฃ) has
a unique fixed-point and hence the sequence defined by Eq. (3.4) is
convergent in view of theorem 1.2, and the solution of Eq. (3.3) is given by
๐ฃ = ๐๐๐๐ โ โ
๐ฃ๐ (3.5)
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
92
3.5 The BCM to solve Riccati, pantograph and beam
deformation equations
In this section, three types of non-linear equations namely Riccati
equation, pantograph equation, and beam deformation equation will be solved
by the BCM.
3.5.1 Exact and numerical solutions for nonlinear Riccati
equation by BCM
The BCM will be implemented to solve the Riccati differential equation
which is a nonlinear ODE of first order, the following examples will be
solved.
Example 3.2:
Let us recall example 2.1:
๐ฃโฒ = โ ๐ฅ โ โ 3๐ฅ + 2โ 2๐ฅ๐ฃ โ โ ๐ฅ๐ฃ2, with initial condition ๐ฃ(0) = 1. (3.6)
Eq. (3.6) can be converted to the following Voltera integral equation
๐ฃ(๐ฅ) = ๐(๐ฅ) + ๐(๐ฃ), (3.7)
where
๐(๐ฅ) =1
3+ โ ๐ฅ โ
โ 3๐ฅ
3 and ๐(๐ฃ) = โซ (2โ 2๐ก๐ฃ โ โ ๐ก๐ฃ2) ๐๐ก.
๐ฅ
0
By implementing the BCM for Eq. (3.7), we have
๐ฃ๐(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (2โ 2๐ก๐ฃ๐โ1(๐ก) โ โ ๐ก๐ฃ๐โ12 (๐ก)) ๐๐ก,
๐ฅ
0
๐ = 1,2, โฆ .
Then, we get
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
93
๐ฃ0(๐ฅ) = ๐(๐ฅ) =1
3+ โ ๐ฅ โ
โ 3๐ฅ
3,
๐ฃ1(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (2โ 2๐ก๐ฃ0(๐ก) โ โ ๐ก๐ฃ02(๐ก)) ๐๐ก,
๐ฅ
0
=1
14+
8โ ๐ฅ
9+
โ 4๐ฅ
18โ
โ 7๐ฅ
63,
The function ๐ฃ1(๐ฅ) can be written by using Taylor series expansion,
๐ฃ1(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2โ
๐ฅ3
6โ
23๐ฅ4
24โ
209๐ฅ5
120โ
1639๐ฅ6
720โ
12161๐ฅ7
5040
โ87863๐ฅ8
40320โ
625969๐ฅ9
362880โ
4425479๐ฅ10
3628800+ ๐[๐ฅ]11,
๐ฃ2(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (2โ 2๐ก๐ฃ1(๐ก) โ โ ๐ก๐ฃ12(๐ก)) ๐๐ก,
๐ฅ
0
= 11
9720+
195โ ๐ฅ
196+
โ 2๐ฅ
126โ
โ 3๐ฅ
243โ
โ 5๐ฅ
630+
โ 6๐ฅ
486+
โ 8๐ฅ
3528โ
5โ 9๐ฅ
6804
+โ 12๐ฅ
6804โ
โ 15๐ฅ
59535,
Similarly, by using Taylor series expansion, we have
๐ฃ2(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2+
๐ฅ3
6+
๐ฅ4
24+
๐ฅ5
120+
๐ฅ6
720โ
79๐ฅ7
5040โ
3919๐ฅ8
40320
โ116479๐ฅ9
362880โ
2735039๐ฅ10
3628800+ ๐[๐ฅ]11,
Continuing in this way, we get a series of the form:
๐ฃ(๐ฅ) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ) = 1 + ๐ฅ +๐ฅ2
2+
๐ฅ3
6+
๐ฅ4
24+
๐ฅ5
120+ โฏ, (3.8)
This series converges to the following exact solution [19]:
๐ฃ(๐ฅ) = โ ๐ฅ.
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
94
Example 3.3 [74]:
Rewrite the example 1.1 and solve it by BCM.
๐ฃโฒ = โ๐ฃ2 + 1, with initial condition ๐ฃ(0) = 0. (3.9)
Eq. (3.9) can be converted to the following Voltera integral equation
๐ฃ(๐ฅ) = ๐(๐ฅ) + ๐(๐ฃ), (3.10)
where
๐(๐ฅ) = ๐ฅ and ๐(๐ฃ) = โ โซ (๐ฃ2) ๐๐ก.๐ฅ
0
By applying the BCM for Eq. (3.10), we have
๐ฃ๐(๐ฅ) = ๐ฃ0(๐ฅ) โ โซ (๐ฃ๐โ12 (๐ก)) ๐๐ก,
๐ฅ
0
๐ = 1,2, โฆ .
Thus, we have
๐ฃ0(๐ฅ) = ๐(๐ฅ) = ๐ฅ,
๐ฃ1(๐ฅ) = ๐ฃ0(๐ฅ) โ โซ(๐ฃ02(๐ก)) ๐๐ก = ๐ฅ โ
๐ฅ3
3,
๐ฅ
0
๐ฃ2(๐ฅ) = ๐ฃ0(๐ฅ) โ โซ (๐ฃ12(๐ก)) ๐๐ก =
๐ฅ
0
๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
๐ฅ7
63,
๐ฃ3(๐ฅ) = ๐ฃ0(๐ฅ) โ โซ (๐ฃ22(๐ก)) ๐๐ก,
๐ฅ
0
= ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
17๐ฅ7
315+
38๐ฅ9
2835โ
134๐ฅ11
51975+
4๐ฅ13
12285โ
๐ฅ15
59535,
By continuing in this manner till ๐ = 6, we have
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
95
๐ฃ6(๐ฅ) = ๐ฅ โ๐ฅ3
3+
2๐ฅ5
15โ
17๐ฅ7
315+
62๐ฅ9
2835โ
1382๐ฅ11
155925+
21844๐ฅ13
6081075
โ909409๐ฅ15
638512875+
543442๐ฅ17
986792625โ
77145154๐ฅ19
371231385525+ โฏ, (3.11)
The obtained series solution in Eq. (3.9) can be used to calculate the
maximal error remainder for show the highest accuracy level that we can
achieve. This can be clearly seen in table 3.1 and figure 3.1
Table 3.1: The maximal error remainder: ๐๐ธ๐ ๐ by the BCM,
where ๐ = 1, โฆ ,6.
n ๐ด๐ฌ๐น๐ by BCM
1 6.65556 ร 10โ5
2 2.65463 ร 10โ7
3 7.56708 ร 10โ10
4 1.67808 ร 10โ12
5 3.04791 ร 10โ15
6 3.46945 ร 10โ18
Figure 3.1: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 6 by BCM.
1 2 3 4 5 6
10 15
10 11
10 7
n
ME
Rn
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
96
3.5.2 BCM for solving linear pantograph equation
In this subsection, the BCM will be applied for second order linear
pantograph equation of order two. The following example presents a solution
for pantograph equation by using the BCM.
Example 3.4:
Resolve example 1.3 by using BCM [11].
๐ฃโฒโฒ =3
4๐ฃ + ๐ฃ (
๐ฅ
2) โ ๐ฅ2 + 2, (3.12)
with initial conditions ๐ฃ(0) = 0 , ๐ฃโฒ(0) = 0.
Integrate both sides of Eq. (3.12) twice from 0 to ๐ฅ and using the initial
condition at ๐ฅ = 0, we get
๐ฃ(๐ฅ) = ๐ฅ2 โ๐ฅ4
12+ โซ โซ (
3
4๐ฃ(๐ก) + ๐ฃ (
๐ก
2)) ๐๐ก๐๐ก,
๐ฅ
0
๐ฅ
0
(3.13)
By using the formula in Eq. (1.6), we get
๐ฃ(๐ฅ) = ๐ฅ2 โ๐ฅ4
12+ โซ (๐ฅ โ ๐ก) (
3
4๐ฃ(๐ก) + ๐ฃ (
๐ก
2)) ๐๐ก
๐ฅ
0
, (3.14)
By applying the BCM for Eq. (3.14), we get
๐ฃ(๐ฅ) = ๐(๐ฅ) + ๐(๐ฃ),
where
๐(๐ฅ) = ๐ฅ2 โ๐ฅ4
12 and ๐(๐ฃ) = โซ (๐ฅ โ ๐ก) (
3
4๐ฃ(๐ก) + ๐ฃ (
๐ก
2)) ๐๐ก
๐ฅ
0
,
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Chapter Three Banach Contraction Method for Solving Some
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97
๐ฃ๐(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (๐ฅ โ ๐ก) (3
4๐ฃ๐โ1(๐ก) + ๐ฃ๐โ1 (
๐ก
2)) ๐๐ก,
๐ฅ
0
๐ = 1,2, โฆ .
Thus, by applying the Mathematicaโs code, see appendix D, we have
๐ฃ0(๐ฅ) = ๐(๐ฅ) = ๐ฅ2 โ๐ฅ4
12,
๐ฃ1(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (๐ฅ โ ๐ก) (3
4๐ฃ0(๐ก) + ๐ฃ0 (
๐ก
2)) ๐๐ก =
๐ฅ
0
๐ฅ2 โ13๐ฅ6
5760,
๐ฃ2(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (๐ฅ โ ๐ก) (3
4๐ฃ1(๐ก) + ๐ฃ1 (
๐ก
2)) ๐๐ก
๐ฅ
0
= ๐ฅ2 โ91๐ฅ8
2949120,
๐ฃ3(๐ฅ) = ๐ฃ0(๐ฅ) + โซ (๐ฅ โ ๐ก) (3
4๐ฃ2(๐ก) + ๐ฃ2 (
๐ก
2)) ๐๐ก
๐ฅ
0
,
= ๐ฅ2 โ17563๐ฅ10
67947724800,
By continuing in this way, we get
๐ฃ(๐ฅ) = ๐ฅ2 โ small term, (3.15)
This series converges to the exact solution [11]:
๐ฃ(๐ฅ) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ) = ๐ฅ2.
3.5.3 Solving the nonlinear beam equation by using BCM
In this subsection, the BCM will be implemented to solve fourth order
nonlinear beam deformation problem.
Example 3.5:
Resolve example 1.4 by using BCM [1].
๐ฃ(4) = ๐ฃ2 โ ๐ฅ10 + 4๐ฅ9 โ 4๐ฅ8 โ 4๐ฅ7 + 8๐ฅ6 โ 4๐ฅ4 + 120๐ฅ โ 48,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
98
with the boundary conditions ๐ฃ(0) = ๐ฃโฒ(0) = 0, ๐ฃ(1) = ๐ฃโฒ(1) = 1. (3.16)
By integrating both sides of Eq. (3.16) four times from 0 to ๐ฅ, we obtain
๐ฃ(๐ฅ) = ๐ฃ0(๐ฅ)
+ โซ โซ โซ โซ (๐ฃ2 โ ๐ก10 + 4๐ก9 โ 4๐ก8 โ 4๐ก7 + 8๐ก6 โ 4๐ก4๐ฅ
0
๐ฅ
0
๐ฅ
0
๐ฅ
0
+ 120๐ก โ 48) ๐๐ก ๐๐ก ๐๐ก ๐๐ก . (3.17)
Let us assume the following initial component,
๐ฃ0(๐ฅ) = ๐๐ฅ3 + ๐๐ฅ2 + ๐๐ฅ + ๐,
By using Eq. (1.6), we obtain the following
๐ฃ(๐ฅ) = ๐ฃ0(๐ฅ)
+ โซ(๐ฅ โ ๐ก)3
6(๐ฃ2 โ ๐ก10 + 4๐ก9 โ 4๐ก8 โ 4๐ก7 + 8๐ก6 โ 4๐ก4
๐ฅ
0
+ 120๐ก โ 48) ๐๐ก, (3.18)
By implementing the BCM:
๐ฃ๐(๐ฅ) = ๐ฃ0(๐ฅ)
+ โซ(๐ฅ โ ๐ก)3
6(๐ฃ๐โ1
2 (๐ก) โ ๐ก10 + 4๐ก9 โ 4๐ก8 โ 4๐ก7 + 8๐ก6 โ 4๐ก4๐ฅ
0
+ 120๐ก โ 48) ๐๐ก, ๐ = 1,2,3, โฆ. (3.19)
Then, we get
๐ฃ0(๐ฅ) = ๐(๐ฅ) = ๐๐ฅ3 + ๐๐ฅ2 + ๐๐ฅ + ๐,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
99
๐ฃ1(๐ฅ) = ๐ฃ0(๐ฅ)
+ โซ(๐ฅ โ ๐ก)3
6(๐ฃ0
2(๐ก) โ ๐ก10 + 4๐ก9 โ 4๐ก8 โ 4๐ก7 + 8๐ก6 โ 4๐ก4๐ฅ
0
+ 120๐ก โ 48) ๐๐ก,
= ๐ + ๐๐ฅ + ๐๐ฅ2 + ๐๐ฅ3 โ 2๐ฅ4 +๐2๐ฅ4
24+ ๐ฅ5 +
1
60๐๐๐ฅ5 +
๐2๐ฅ6
360
+1
180๐๐๐ฅ6 +
1
420๐๐๐ฅ7 +
1
420๐๐๐ฅ7 โ
๐ฅ8
420+
๐2๐ฅ8
1680
+1
840๐๐๐ฅ8 +
๐๐๐ฅ9
1512+
๐ฅ10
630+
๐2๐ฅ10
5040โ
๐ฅ11
1980โ
๐ฅ12
2970+
๐ฅ13
4290
โ๐ฅ14
24024, (3.20)
Imposing the boundary conditions (3.16), into ๐ฃ1(๐ฅ), we get
๐ = โ0.00687154, ๐ = 2.00593, ๐ = 0, ๐ = 0. (3.21)
The following first-order approximate solution is achieved:
๐ฃ1(๐ฅ) = 2.005929514398968๐ฅ2 โ 6.871538792438514 ร 10โ3๐ฅ3
โ 2๐ฅ4 + ๐ฅ5 + 1.413881948623746 ร 10โ5๐ฅ8
โ 9.116284704424509 ร 10โ6๐ฅ9 + 1.587310955961384
ร 10โ3๐ฅ10 โ๐ฅ11
1980โ
๐ฅ12
2970+
๐ฅ13
4290โ
๐ฅ14
24024. (3.22)
๐ฃ2(๐ฅ) = ๐ฃ0(๐ฅ)
+ โซ(๐ฅ โ ๐ก)3
6(๐ฃ1
2(๐ก) โ ๐ก10 + 4๐ก9 โ 4๐ก8 โ 4๐ก7 + 8๐ก6 โ 4๐ก4๐ฅ
0
+ 120๐ก โ 48) ๐๐ก,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
100
= ๐ + ๐๐ฅ + ๐๐ฅ2 + ๐๐ฅ3 โ 2๐ฅ4 +๐2๐ฅ4
24+ ๐ฅ5 +
1
60๐๐๐ฅ5 +
๐2๐ฅ6
360
+1
180๐๐๐ฅ6 +
1
420๐๐๐ฅ7 +
1
420๐๐๐ฅ7 โ
๐ฅ8
420+
๐2๐ฅ8
1680
+1
840๐๐๐ฅ8 โ
๐๐ฅ8
420+
๐3๐ฅ8
20160+
๐๐๐ฅ9
1512โ
๐๐ฅ9
756+ โฏ. (3.23)
Once again, by imposing the boundary conditions (3.16) in the second
component ๐ฃ2(๐ฅ), the obtained coefficients will be:
๐ = โ8.27907 ร 10โ7, ๐ = 2, ๐ = 0, ๐ = 0.
In order to check the accuracy of the approximate solution, we calculate
the absolute error, where ๐ฃ(๐ฅ) = ๐ฅ5 โ 2๐ฅ4 + 2๐ฅ2 is the exact solution and
๐ฃ๐(๐ฅ) is the approximate solution. In table 3.2 the absolute error of BCM with
๐ = 1, 2. It can be seen clearly from table 3.2 by increasing the iteration, the
error will be decreasing.
Table 3.2: Results of the absolute errors for BCM.
x |๐๐| for BCM |๐๐| for BCM
0 0 0
0.1 5.24236 ร 10โ5 6.79965 ร 10โ9
0.2 1.82208 ร 10โ4 2.3887 ร 10โ8
0.3 3.48134 ร 10โ4 4.62946 ร 10โ8
0.4 5.09092 ร 10โ4 6.90558 ร 10โ8
0.5 6.24721ร 10โ4 8.72068 ร 10โ8
0.6 6.57823 ร 10โ4 9.58101 ร 10โ8
0.7 5.81138 ร 10โ4 9.01332 ร 10โ8
0.8 3.92709 ร 10โ4 6.67054 ร 10โ8
0.9 1.45715 ร 10โ4 2.80674 ร 10โ8
1.0 3.3447 ร 10โ8 0
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
101
3.5.4 Solving the nonlinear Burgersโ equation by using the BCM
In this subsection, we implement the BCM to solve the first order
nonlinear Burgersโ equation and finding the exact solutions for different types
of Burgersโ equations in 1D, 2D and (1+3)-D.
Example 3.6:
Rewrite the example 1.5 and solve it by BCM [33].
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ = ๐ฃ๐ฅ๐ฅ, with initial condition ๐ฃ(๐ฅ, 0) = 2๐ฅ. (3.24)
Integrating both sides of Eq. (3.24) from 0 to ๐ก and using the initial condition
at ๐ก = 0, we get
๐ฃ(๐ฅ, ๐ก) = 2๐ฅ + โซ (โ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ) ๐๐ก,๐ก
0
(3.25)
Applying the BCM for Eq. (3.25), we get
๐ฃ(๐ฅ, ๐ก) = ๐(๐ฅ, ๐ก) + ๐(๐ฃ),
where
๐(๐ฅ, ๐ก) = 2๐ฅ and ๐(๐ฃ) = โซ (โ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ) ๐๐ก.๐ก
0
๐ฃ๐(๐ฅ, ๐ก) = ๐ฃ0(๐ฅ, ๐ก) + โซ (โ๐ฃ๐โ1(๐ฅ, ๐ก)(๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ
+ (๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ๐ฅ
) ๐๐ก,๐ก
0
๐ = 1,2, โฆ. (3.26)
Then, we have
๐ฃ0 = ๐(๐ฅ, ๐ก) = 2๐ฅ,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
102
๐ฃ1 = ๐ฃ0 + โซ (โ๐ฃ0๐ฃ0๐ฅ + ๐ฃ0๐ฅ๐ฅ) ๐๐ก = ๐ก
0
2๐ฅ โ 4๐ก๐ฅ,
๐ฃ2 = ๐ฃ0 + โซ (โ๐ฃ1๐ฃ1๐ฅ + ๐ฃ1๐ฅ๐ฅ) ๐๐ก = ๐ก
0
2๐ฅ โ 4๐ก๐ฅ + 8๐ก2๐ฅ โ16๐ก3๐ฅ
3,
By continuing in this way, the result is
๐ฃ(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ก) ,
= 2x โ 4xt + 8xt2 โ 16xt3 + 32xt4 โ 64xt5 + โฏ, (3.27)
This series converges to the following exact solution [33]:
๐ฃ(๐ฅ, ๐ก) =2๐ฅ
1 + 2๐ก= 2๐ฅ โ 4๐ฅ๐ก + 8๐ฅ๐ก2 โ 16๐ฅ๐ก3 + 32๐ฅ๐ก4 โ 64๐ฅ๐ก5 + โฏ.
Example 3.7:
Recall example 1.6:
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ = ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ), (3.28)
with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
By taking the integration on both sides of equation (3.28) from 0 to ๐ก, we
obtain
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ)
+ โซ (โ๐ฃ๐ฃ๐ฅโ๐ฃ๐ฃ๐ฆ + ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ)) ๐๐ก,๐ก
0
(3.29)
Now, by applying the BCM for Eq. (3.29), we have
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ก) + ๐(๐ฃ),
where
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
103
๐(๐ฅ, ๐ฆ, ๐ก) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ) and
๐(๐ฃ) = โซ (โ๐ฃ๐ฃ๐ฅโ๐ฃ๐ฃ๐ฆ + ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ)) ๐๐ก.๐ก
0
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
= ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)
+ โซ (โ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฅ โ ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฆ
๐ก
0
+ ๐(๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฅ๐ฅ + ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฆ๐ฆ)) ๐๐ก, ๐ = 1,2, โฆ. (3.30)
So, we get
๐ฃ0 = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ),
๐ฃ1 = ๐ฃ0 + โซ (โ๐ฃ0๐ฃ0๐ฅ โ ๐ฃ0๐ฃ0๐ฆ + ๐(๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฆ๐ฆ)) ๐๐ก,๐ก
0
= ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + ๐ก(โ8๐2๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
โ 2๐๐๐๐ (2๐๐ฅ)๐๐๐ (2๐๐ฆ)2๐ ๐๐(2๐๐ฅ) + 2๐๐ ๐๐(2๐๐ฅ)๐ ๐๐(2๐๐ฆ)),
๐ฃ2 = ๐ฃ0 + โซ (โ๐ฃ1๐ฃ1๐ฅ โ ๐ฃ1๐ฃ1๐ฆ + ๐(๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฆ๐ฆ)) ๐๐ก,๐ก
0
= ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
โ 8๐2๐ก๐๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) + โฏ,
continue till n = 4, we get
๐ฃ4 = ๐ฃ0 + โซ (โ๐ฃ3๐ฃ3๐ฅ โ ๐ฃ3๐ฃ3๐ฆ + ๐(๐ฃ3๐ฅ๐ฅ + ๐ฃ3๐ฆ๐ฆ)) ๐๐ก,๐ก
0
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
104
= ๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ 2๐2๐ก2๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ)
+2
3๐4๐ก4๐๐๐ (2๐๐ฆ)๐ ๐๐(2๐๐ฅ) โ โฏ,
The obtained series solution in Eq. (3.28) can be used to calculate the
maximal error remainder to check the accuracy of the obtained approximate
solution. It can be clearly seen in table 3.3 and figure 3.2 that by increasing
the iterations the errors will be reduced.
Table 3.3: The maximal error remainder: ๐๐ธ๐ ๐ by the BCM.
n ๐ด๐ฌ๐น๐ by BCM
1 1.61705 ร 10โ2
2 2.52036 ร 10โ5
3 3.56783 ร 10โ8
4 4.5361 ร 10โ11
Figure 3.2: Logarithmic plots of ๐๐ธ๐ ๐versus ๐ is 1 through 4 by BCM.
Example 3.8:
Recall example 2.7:
๐ฃ๐ก = ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ , with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ. (3.31)
By taking the integration for both sides of Eq. (3.31) from 0 to ๐ก, we obtain
1.0 1.5 2.0 2.5 3.0 3.5 4.010 11
10 8
10 5
10 2
n
ME
Rn
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
105
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ + โซ (๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ) ๐๐ก,๐ก
0
(3.32)
Implementing the BCM for Eq. (3.32), we get
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ก) + ๐(๐ฃ),
where
๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ and ๐(๐ฃ) = โซ (๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ) ๐๐ก.๐ก
0
Thus, we have
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
= ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)
+ โซ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฅ + ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฅ๐ฅ
๐ก
0
+ ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)๐ฆ๐ฆ) ๐๐ก, ๐ = 1,2, โฆ. (3.33)
Thus, we get
๐ฃ0 = ๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ,
๐ฃ1 = ๐ฃ0 + โซ (๐ฃ0๐ฃ0๐ฅ + ๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฆ๐ฆ) ๐๐ก ๐ก
0
= ๐ฅ + ๐ฆ + (๐ฅ + ๐ฆ)๐ก,
๐ฃ2 = ๐ฃ0 + โซ (๐ฃ1๐ฃ1๐ฅ + ๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฆ๐ฆ) ๐๐ก,๐ก
0
= ๐ฅ + ๐ฆ + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 +1
3(๐ฅ + ๐ฆ)๐ก3,
By continuing in this way, we have a series of the form:
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
106
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก) ,
= (๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 + (๐ฅ + ๐ฆ)๐ก3 + โฏ, (3.34)
This series is convergent to the following exact solution [76]:
๐ฃ(๐ฅ, ๐ฆ, ๐ก) =๐ฅ + ๐ฆ
1 โ ๐ก= (๐ฅ + ๐ฆ) + (๐ฅ + ๐ฆ)๐ก + (๐ฅ + ๐ฆ)๐ก2 + (๐ฅ + ๐ฆ)๐ก3 + โฏ.
Example 3.9:
Recall example 2.8:
๐ฃ๐ก = ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง,
with initial condition ๐ฃ(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง. (3.35)
Applying the integration on both sides of Eq. (3.35) from 0 to ๐ก and using the
initial condition at ๐ก = 0, we get
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง + โซ (๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง)๐ก
0
๐๐ก. (3.36)
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง and ๐(๐ฃ) = โซ (๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง) ๐๐ก.๐ก
0
Now, applying the BCM for Eq. (3.36), we have
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) + ๐(๐ฃ),
where
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง and ๐(๐ฃ) = โซ (๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง) ๐๐ก.๐ก
0
Thus, we get
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
107
๐ฃ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= ๐ฃ0(๐ฅ, ๐ฆ, ๐ง, ๐ก)
+ โซ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ง, ๐ก)(๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ง, ๐ก))๐ฅ
+ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ง, ๐ก))๐ฅ๐ฅ
๐ก
0
+ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ง, ๐ก))๐ฆ๐ฆ
+ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ง, ๐ก))๐ง๐ง
) ๐๐ก, ๐ = 1,2,3, โฆ .
Thus, we have
๐ฃ0 = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ฅ + ๐ฆ + ๐ง,
๐ฃ1 = ๐ฃ0 + โซ (๐ฃ0๐ฃ0๐ฅ + ๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฆ๐ฆ + ๐ฃ0๐ง๐ง) ๐๐ก,๐ก
0
= ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก,
๐ฃ2 = ๐ฃ0 + โซ (๐ฃ1๐ฃ1๐ฅ + ๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฆ๐ฆ + ๐ฃ1๐ง๐ง) ๐๐ก,๐ก
0
= ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 +1
3(๐ฅ + ๐ฆ + ๐ง)๐ก3,
and so on, therefore we get the following series:
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 + (๐ฅ + ๐ฆ + ๐ง)๐ก3
+ โฏ,
This series converges to the exact solution [76]:
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ฅ + ๐ฆ + ๐ง
1 โ ๐ก,
= ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก + (๐ฅ + ๐ฆ + ๐ง)๐ก2 + (๐ฅ + ๐ฆ + ๐ง)๐ก3 + โฏ.
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
108
3.5.5 Solving the nonlinear system of Burgersโ equations by
using BCM
In this subsection, the BCM will be implemented to solve the first order
nonlinear Burgersโ equation for the exact solutions for different types of
systems of Burgersโ equations in 1D, 2D and 3D.
Example 3.10:
Rewrite the example 2.9 and solve it by BCM [50].
๐ข๐ก โ ๐ข๐ฅ๐ฅ โ 2๐ข๐ข๐ฅ + (๐ข๐ฃ)๐ฅ = 0,
๐ฃ๐ก โ ๐ฃ๐ฅ๐ฅ โ 2๐ฃ๐ฃ๐ฅ + (๐ข๐ฃ)๐ฅ = 0,
Subject to the initial conditions:
๐ข(๐ฅ, 0) = ๐ ๐๐(๐ฅ), ๐ฃ(๐ฅ, 0) = ๐ ๐๐(๐ฅ). (3.37)
By solving Eq. (3.37), we get
๐ข(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) + โซ (๐ข๐ฅ๐ฅ + 2๐ข๐ข๐ฅ โ (๐ข๐ฃ)๐ฅ) ๐๐ก,๐ก
0
๐ฃ(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) + โซ (๐ฃ๐ฅ๐ฅ + 2๐ฃ๐ฃ๐ฅ โ (๐ข๐ฃ)๐ฅ) ๐๐ก,๐ก
0
(3.38)
To solve the system of Eq. (3.38) by using of the BCM, we have
๐ข(๐ฅ, ๐ก) = ๐(๐ฅ, ๐ก) + ๐1(๐ข), ๐ฃ(๐ฅ, ๐ก) = ๐(๐ฅ, ๐ก) + ๐2(๐ฃ),
where
๐(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) and ๐1(๐ข) = โซ (๐ข๐ฅ๐ฅ + 2๐ข๐ข๐ฅ โ (๐ข๐ฃ)๐ฅ)๐ก
0
๐๐ก,
๐(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ) and ๐2(๐ฃ) = โซ (๐ฃ๐ฅ๐ฅ + 2๐ฃ๐ฃ๐ฅ โ (๐ข๐ฃ)๐ฅ) ๐๐ก.๐ก
0
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
109
Thus
๐ข๐(๐ฅ, ๐ก) = ๐ข0(๐ฅ, ๐ก)
+ โซ ((๐ข๐โ1(๐ฅ, ๐ก))๐ฅ๐ฅ
+ 2๐ข๐โ1(๐ฅ, ๐ก)(๐ข๐โ1(๐ฅ, ๐ก))๐ฅ
๐ก
0
โ (๐ข๐โ1(๐ฅ, ๐ก)๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ
) ๐๐ก, ๐ = 1,2, โฆ .
๐ฃ๐(๐ฅ, ๐ก) = ๐ฃ0(๐ฅ, ๐ก)
+ โซ ((๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ๐ฅ
+ 2๐ฃ๐โ1(๐ฅ, ๐ก)(๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ
๐ก
0
โ (๐ข๐โ1(๐ฅ, ๐ก)๐ฃ๐โ1(๐ฅ, ๐ก))๐ฅ
) ๐๐ก, ๐ = 1,2, โฆ . (3.39)
Then,
๐ข0 = ๐(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ),
๐ฃ0 = ๐(๐ฅ, ๐ก) = ๐ ๐๐(๐ฅ),
๐ข1 = ๐ข0 + โซ (๐ข0๐ฅ๐ฅ + 2๐ข0๐ข0๐ฅ โ (๐ข0๐ฃ0)๐ฅ) ๐๐ก๐ก
0
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ),
๐ฃ1 = ๐ฃ0 + โซ (๐ฃ0๐ฅ๐ฅ + 2๐ฃ0๐ฃ0๐ฅ โ (๐ข0๐ฃ0)๐ฅ) ๐๐ก๐ก
0
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ),
๐ข2 = ๐ข0 + โซ (๐ข1๐ฅ๐ฅ + 2๐ข1๐ข1๐ฅ โ (๐ข1๐ฃ1)๐ฅ) ๐๐ก,๐ก
0
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ),
๐ฃ2 = ๐ฃ0 + โซ (๐ฃ1๐ฅ๐ฅ + 2๐ฃ1๐ฃ1๐ฅ โ (๐ข1๐ฃ1)๐ฅ) ๐๐ก,๐ก
0
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ),
Continuing in this way, we get:
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
110
๐ข(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ก) ,
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ
1
6๐ก3๐ ๐๐(๐ฅ) +
1
24๐ก4๐ ๐๐(๐ฅ)
โ โฏ,
๐ฃ(๐ฅ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ก) ,
= ๐ ๐๐(๐ฅ) โ ๐ก๐ ๐๐(๐ฅ) +1
2๐ก2๐ ๐๐(๐ฅ) โ
1
6๐ก3๐ ๐๐(๐ฅ) +
1
24๐ก4๐ ๐๐(๐ฅ)
โ โฏ.
This series converges to the exact solution [50]:
๐ข(๐ฅ, ๐ก) = โ โ๐ก๐ ๐๐(๐ฅ),
๐ฃ(๐ฅ, ๐ก) = โ โ๐ก๐ ๐๐(๐ฅ).
Example 3.11:
Rewrite the example 2.10 and solve it by BCM [30].
๐ข๐ก + ๐ข๐ข๐ฅ + ๐ฃ๐ข๐ฆ โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ = 0,
๐ฃ๐ก + ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ = 0,
Subject to the initial conditions:
๐ข(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ, ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ. (3.40)
Integrating both sides of Eq. (3.40) from 0 to ๐ก and using the initial conditions
at ๐ก = 0, we get
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ + โซ (โ๐ข๐ข๐ฅ โ ๐ฃ๐ข๐ฆ + ๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ) ๐๐ก,๐ก
0
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ โ ๐ฆ + โซ (โ๐ข๐ฃ๐ฅ โ ๐ฃ๐ฃ๐ฆ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ) ๐๐ก๐ก
0
, (3.41)
To solve the system of Eq. (3.41) by using of the BCM
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
111
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ก) + ๐1(๐ข), ๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ก) + ๐2(๐ฃ)
where
๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ and ๐1(๐ข) = โซ (โ๐ข๐ข๐ฅ โ ๐ฃ๐ข๐ฆ + ๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ) ๐๐ก,๐ก
0
๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ โ ๐ฆ and ๐2(๐ฃ) = โซ (โ๐ข๐ฃ๐ฅ โ ๐ฃ๐ฃ๐ฆ + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ) ๐๐ก๐ก
0
.
Thus
๐ข๐(๐ฅ, ๐ฆ, ๐ก)
= ๐ข0(๐ฅ, ๐ฆ, ๐ก)
+ โซ (โ๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก)(๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฅ
๐ก
0
โ ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)(๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฆ
+ (๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฅ๐ฅ
+ (๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฆ๐ฆ
) ๐๐ก, ๐ = 1,2, โฆ,
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก)
= ๐ฃ0(๐ฅ, ๐ฆ, ๐ก)
+ โซ (โ๐ข๐โ1(๐ฅ, ๐ฆ, ๐ก)(๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฅ
๐ก
0
โ ๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก)(๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฆ
+ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฅ๐ฅ
+ (๐ฃ๐โ1(๐ฅ, ๐ฆ, ๐ก))๐ฆ๐ฆ
) ๐๐ก, ๐ = 1,2, โฆ. (3.42)
Therefore, we have
๐ข0 = ๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ,
๐ฃ0 = ๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ โ ๐ฆ,
๐ข1 = ๐ข0 + โซ (โ๐ข0๐ข0๐ฅ โ ๐ฃ0๐ข0๐ฆ + ๐ข0๐ฅ๐ฅ + ๐ข0๐ฆ๐ฆ) ๐๐ก๐ก
0
= ๐ฅ + ๐ฆ โ 2๐ฅ๐ก,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
112
๐ฃ1 = ๐ฃ0 + โซ (โ๐ข0๐ฃ0๐ฅ โ ๐ฃ0๐ฃ0๐ฆ + ๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฆ๐ฆ) ๐๐ก๐ก
0
= ๐ฅ โ ๐ฆ โ 2๐ฆ๐ก,
๐ข2 = ๐ข0 + โซ (โ๐ข1๐ข1๐ฅ โ ๐ฃ1๐ข1๐ฆ + ๐ข1๐ฅ๐ฅ + ๐ข1๐ฆ๐ฆ) ๐๐ก,๐ก
0
= ๐ฅ + ๐ฆ โ 2๐ฅ๐ก + 2(๐ฅ + ๐ฆ)๐ก2 โ4๐ฅ๐ก3
3,
๐ฃ2 = ๐ฃ0 + โซ (โ๐ข1๐ฃ1๐ฅ โ ๐ฃ1๐ฃ1๐ฆ + ๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฆ๐ฆ) ๐๐ก,๐ก
0
= ๐ฅ โ ๐ฆ โ 2๐ฆ๐ก + (2๐ฅ โ 2๐ฆ)๐ก2 โ4๐ฆ๐ก3
3,
Continuing in this way, we get the following forms:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ฆ, ๐ก) ,
= ๐ฅ + ๐ฆ โ 2๐ฅ๐ก + 2(๐ฅ + ๐ฆ)๐ก2 โ 4๐ฅ๐ก3 + 4(๐ฅ + ๐ฆ)๐ก4 โ 8๐ฅ๐ก5
+ โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก) ,
= ๐ฅ โ ๐ฆ โ 2๐ฆ๐ก + (2๐ฅ โ 2๐ฆ)๐ก2 โ 4๐ฆ๐ก3 + (4๐ฅ โ 4๐ฆ)๐ก4 โ 8๐ฆ๐ก5
+ โฏ.
These forms converges to the exact solutions [30]:
๐ข(๐ฅ, ๐ฆ, ๐ก) =๐ฅ โ 2๐ฅ๐ก + ๐ฆ
1 โ 2๐ก2,
= ๐ฅ + ๐ฆ โ 2๐ฅ๐ก + 2(๐ฅ + ๐ฆ)๐ก2 โ 4๐ฅ๐ก3 + 4(๐ฅ + ๐ฆ)๐ก4 โ 8๐ฅ๐ก5 + โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ก) =๐ฅ โ 2๐ฆ๐ก โ ๐ฆ
1 โ 2๐ก2,
= ๐ฅ โ ๐ฆ โ 2๐ฆ๐ก + (2๐ฅ โ 2๐ฆ)๐ก2 โ 4๐ฆ๐ก3 + (4๐ฅ โ 4๐ฆ)๐ก4 โ 8๐ฆ๐ก5 + โฏ.
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
113
Example 3.12:
Recall example 2.11:
๐ข๐ก โ ๐ข๐ข๐ฅ โ ๐ฃ๐ข๐ฆ โ ๐ค๐ข๐ง โ ๐ข๐ฅ๐ฅ โ ๐ข๐ฆ๐ฆ โ ๐ข๐ง๐ง = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
๐ฃ๐ก + ๐ข๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ + ๐ค๐ฃ๐ง + ๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ + ๐ฃ๐ง๐ง = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก),
๐ค๐ก + ๐ข๐ค๐ฅ + ๐ฃ๐ค๐ฆ + ๐ค๐ค๐ง + ๐ค๐ฅ๐ฅ + ๐ค๐ฆ๐ฆ + ๐ค๐ง๐ง = โ(๐ฅ, ๐ฆ, ๐ง, ๐ก),
where
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง) โ โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)),
โ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง) + โ ๐ก(1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง)), (3.43)
Subject to the initial conditions:
๐ข(๐ฅ, ๐ฆ, ๐ง, 0) = ๐ฅ + ๐ฆ + ๐ง,
๐ฃ(๐ฅ, ๐ฆ, ๐ง, 0) = โ(๐ฅ + ๐ฆ + ๐ง),
๐ค(๐ฅ, ๐ฆ, ๐ง, 0) = 1 + ๐ฅ + ๐ฆ + ๐ง.
Integrating both sides of in Eq. (3.43) from 0 to ๐ก and using the initial
conditions at ๐ก = 0, we get
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= 1 โ โ ๐ก +๐ฅ
2+ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2+ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2+ โ ๐ก๐ง
โ1
2โ 2๐ก๐ง + โซ (๐ข๐ข๐ฅ + ๐ฃ๐ข๐ฆ + ๐ค๐ข๐ง + ๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ + ๐ข๐ง๐ง) ๐๐ก,
๐ก
0
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
114
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= 1 โ โ ๐ก +๐ฅ
2โ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2โ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2โ โ ๐ก๐ง
โ1
2โ 2๐ก๐ง + โซ (โ๐ข๐ฃ๐ฅ โ ๐ฃ๐ฃ๐ฆ โ ๐ค๐ฃ๐ง โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฆ๐ฆ โ ๐ฃ๐ง๐ง) ๐๐ก,
๐ก
0
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก)
= โ ๐ก โ๐ฅ
2+ โ ๐ก๐ฅ +
1
2โ 2๐ก๐ฅ โ
๐ฆ
2+ โ ๐ก๐ฆ +
1
2โ 2๐ก๐ฆ โ
๐ง
2+ โ ๐ก๐ง +
1
2โ 2๐ก
+ โซ (โ๐ข๐ค๐ฅ โ ๐ฃ๐ค๐ฆ โ ๐ค๐ค๐ง โ ๐ค๐ฅ๐ฅ โ ๐ค๐ฆ๐ฆ โ ๐ค๐ง๐ง) ๐๐ก,๐ก
0
(3.44)
To solve the system of Eq. (3.44) by using of the BCM, we have
๐ข0 = 1 โ โ ๐ก +๐ฅ
2+ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2+ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2+ โ ๐ก๐ง โ
1
2โ 2๐ก๐ง,
๐ฃ0 = 1 โ โ ๐ก +๐ฅ
2โ โ ๐ก๐ฅ โ
1
2โ 2๐ก๐ฅ +
๐ฆ
2โ โ ๐ก๐ฆ โ
1
2โ 2๐ก๐ฆ +
๐ง
2โ โ ๐ก๐ง โ
1
2โ 2๐ก๐ง,
๐ค0 = โ ๐ก โ๐ฅ
2+ โ ๐ก๐ฅ +
1
2โ 2๐ก๐ฅ โ
๐ฆ
2+ โ ๐ก๐ฆ +
1
2โ 2๐ก๐ฆ โ
๐ง
2+ โ ๐ก๐ง +
1
2โ 2๐ก ,
By writing the functions ๐ข1, ๐ฃ1, ๐ค1 by using Taylor series expansions, we get
๐ข1 = (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(โ2 โ ๐ฅ โ ๐ฆ โ ๐ง)๐ก2 +
1
6(โ3 โ 5๐ฅ
โ 5๐ฆ โ 5๐ง)๐ก3 +1
24(โ2 โ 13๐ฅ โ 13๐ฆ โ 13๐ง)๐ก4 +
1
120(9
โ 23๐ฅ โ 23๐ฆ โ 23๐ง)๐ก5 + ๐[๐ก]6,
๐ฃ1 = (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2
+1
2(โ1 โ ๐ฅ โ ๐ฆ โ ๐ง)๐ก3 +
1
24(โ12 โ 13๐ฅ โ 13๐ฆ โ 13๐ง)๐ก4
+1
40(โ13 โ 19๐ฅ โ 19๐ฆ โ 19๐ง)๐ก5 + ๐[๐ก]6,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
115
๐ค1 = (1 + ๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
2(1 + ๐ฅ + ๐ฆ + ๐ง)๐ก3 +
1
24(12 + 13๐ฅ + 13๐ฆ + 13๐ง)๐ก4
+1
40(13 + 19๐ฅ + 19๐ฆ + 19๐ง)๐ก5 + ๐[๐ก]6. (3.45)
In a similar way, we have
๐ข2 = (๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2 +
1
6(โ4 โ 3๐ฅ โ 3๐ฆ
โ 3๐ง)๐ก3 +1
24(โ15 โ 23๐ฅ โ 23๐ฆ โ 23๐ง)๐ก4 +
1
120(โ16 โ 75๐ฅ
โ 75๐ฆ โ 75๐ง)๐ก5 + ๐[๐ก]6,
๐ฃ2 = (โ๐ฅ โ ๐ฆ โ ๐ง) + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2 +
1
6(โ2 โ 3๐ฅ
โ 3๐ฆ โ 3๐ง)๐ก3 +1
24(โ7 โ 11๐ฅ โ 11๐ฆ โ 11๐ง)๐ก4 +
1
120(โ14
โ 31๐ฅ โ 31๐ฆ โ 31๐ง)๐ก5 + ๐[๐ก]6,
๐ค2 = (1 + ๐ฅ + ๐ฆ + ๐ง) + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(2 + 3๐ฅ + 3๐ฆ + 3๐ง)๐ก3 +
1
24(7 + 11๐ฅ + 11๐ฆ + 11๐ง)๐ก4
+1
120(14 + 31๐ฅ + 31๐ฆ + 31๐ง)๐ก5 + ๐[๐ก]6. (3.46)
By continuing, we have the following formulas:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 + โฏ,
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ฃ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= โ๐ฅ โ ๐ฆ โ ๐ง + (โ๐ฅ โ ๐ฆ โ ๐ง)๐ก +1
2(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก2
+1
6(โ๐ฅ โ ๐ฆ โ ๐ง)๐ก3 + โฏ,
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Chapter Three Banach Contraction Method for Solving Some
Ordinary and Partial Differential Equations
116
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐๐๐ โ โ
๐ค๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) ,
= 1 + ๐ฅ + ๐ฆ + ๐ง + (๐ฅ + ๐ฆ + ๐ง)๐ก +1
2(๐ฅ + ๐ฆ + ๐ง)๐ก2
+1
6(๐ฅ + ๐ฆ + ๐ง)๐ก3 + โฏ.
Those forms converge to the following exact solutions:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ ๐ก(๐ฅ + ๐ฆ + ๐ง),
๐ฃ(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โโ ๐ก(๐ฅ + ๐ฆ + ๐ง),
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 1 + โ ๐ก(๐ฅ + ๐ฆ + ๐ง).
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CHAPTER 4
CONCLUSIONS AND FUTURE
WORKS
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Chapter Four Conclusions and Future Works
117
Chapter 4
Conclusions and Future Works
4.1 Introduction
The main objective of this thesis has been achieved by solving some of
non-linear ODE and PDE by different iterative methods. This purpose has
been obtained by implementing two iterative methods, which are so-called the
semi-analytic method of TAM and BCM. In addition, the comparison
between the proposed methods and other methods such as ADM and VIM
will be presented. Also, some conclusions, and future works will be given.
4.2 Conclusions
From the present study one can conclude the following:
1. The two iterative methods are efficient and reliable to find the exact
solutions for Riccati and pantograph equations, different types of Burgersโ
equations and equations systems in 1D, 2D and 3D. The efficiency and
accuracy of the TAM has been proved by studying the convergence.
2. The two proposed methods did not require any resulted assumption to deal
with the nonlinear terms unlike ADM was need the so-called Adomain
polynomial in nonlinear case. In addition, the VIM required the Lagrange
multiplier.
3. We take example for nonlinear Riccati equation it was solved in chapter 1,
chapter 2 and chapter 3
๐ฃโฒ = โ๐ฃ2 + 1, with initial condition ๐ฃ(0) = 0. (4.1)
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Chapter Four Conclusions and Future Works
118
When comparing the results of TAM and BCM with those of the ADM
and VIM, the numerical solutions obtained by BCM are more accurate.
The maximal error remainders will be decreased when there is more
obtaining of iterations which are clarified in the figure 4.1 and table 4.1.
Table 4.1: Comparative results of the maximal error remainder: ๐๐ธ๐ ๐for
TAM, BCM, VIM and ADM, where ๐ = 1, โฆ ,6.
n ๐ด๐ฌ๐น๐ by TAM ๐ด๐ฌ๐น๐ by BCM ๐ด๐ฌ๐น๐by VIM ๐ด๐ฌ๐น๐by ADM
1 6.65556 ร 10โ5 6.65556 ร 10โ5 0.01 6.65556 ร 10โ5
2 2.65463 ร 10โ7 2.65463 ร 10โ7 6.65556 ร 10โ5 3.76891 ร 10โ7
3 7.56708 ร 10โ10 7.56708 ร 10โ10 2.65463 ร 10โ7 1.96289 ร 10โ9
4 1.67806 ร 10โ12 1.67808 ร 10โ12 7.56708 ร 10โ10 9.72067 ร 10โ12
5 2.97852 ร 10โ15 3.04791 ร 10โ15 1.67808 ร 10โ12 4.65513 ร 10โ14
6 1.24033 ร 10โ16 3.46945 ร 10โ18 3.04791ร 10โ15 2.15106 ร 10โ16
Figure 4.1: Comparison of the maximal error remainder for TAM, BCM,
VIM and ADM.
4. We take example for nonlinear beam equation it was solved in chapter 1,
chapter 2 and chapter 3
๐ฃ(4) = ๐ฃ2 โ ๐ฅ10 + 4๐ฅ9 โ 4๐ฅ8 โ 4๐ฅ7 + 8๐ฅ6 โ 4๐ฅ4 + 120๐ฅ โ 48,
with the boundary conditions ๐ฃ(0) = ๐ฃโฒ(0) = 0, ๐ฃ(1) = ๐ฃโฒ(1) = 1. (4.2)
1 2 3 4 5 6
10 20.020.03
n
ME
Rn
ADM
VIM
BCM
TAM
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Chapter Four Conclusions and Future Works
119
In comparison the results by the TAM and BCM with the some existing
methods such as VIM and HPM [1], it is observed in general that the
approximate solutions obtained by the TAM converge faster without any
restricted assumptions and possesses high-order accuracy. As given in
table 4.2, we compare the absolute error of TAM, with n = 2 together with
the VIM, and the HPM [1].
Table 4.2: Comparative results for the absolute errors of the TAM, BCM,
VIM and HPM.
x |๐๐| for TAM |๐๐| for BCM |๐๐| for VIM |๐๐|for HPM[1]
0 0 0 0 0
0.1 1.77279 ร 10โ10 6.79965 ร 10โ9 6.79965 ร 10โ9 1.896 ร 10โ7
0.2 5.99375 ร 10โ10 2.3887 ร 10โ8 2.3887 ร 10โ8 6.4171 ร 10โ7
0.3 1.1028 ร 10โ9 4.62946 ร 10โ8 4.62946 ร 10โ8 1.18180 ร 10โ6
0.4 1.53129 ร 10โ9 6.90558 ร 10โ8 6.90558 ร 10โ8 1.6405 ร 10โ6
0.5 1.752 ร 10โ9 8.72068 ร 10โ8 8.72068 ร 10โ8 1.8703 ร 10โ6
0.6 1.68284 ร 10โ9 9.58101ร 10โ8 9.58101 ร 10โ8 1.7815 ร 10โ6
0.7 1.32351ร 10โ9 9.01332 ร 10โ8 9.01332 ร 10โ8 1.3816 ร 10โ6
0.8 7.76486 ร 10โ10 6.67054 ร 10โ8 6.67054 ร 10โ8 7.958 ร 10โ7
0.9 2.44241ร 10โ10 2.80674 ร 10โ8 2.80674 ร 10โ8 2.437 ร 10โ7
1.0 1.11022 ร 10โ16 0 0 6.0 ร 10โ10
It can be observed clearly from table 4.2, the absolute error for TAM is lower
than BCM, VIM and HPM.
5. We take example for nonlinear 2D Burgersโ equation it was solved in
chapter 1, chapter 2 and chapter 3
๐ฃ๐ก + ๐ฃ๐ฃ๐ฅ + ๐ฃ๐ฃ๐ฆ = ๐(๐ฃ๐ฅ๐ฅ + ๐ฃ๐ฆ๐ฆ), (4.3)
with initial condition ๐ฃ(๐ฅ, ๐ฆ, 0) = ๐ ๐๐(2๐๐ฅ)๐๐๐ (2๐๐ฆ).
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Chapter Four Conclusions and Future Works
120
The TAM has also achieved the rapid convergence of our approximate
solution by the sequence of the curves of error remainder functions. Also
it does not require any restricted assumptions like the ADM and VIM,
which are clarified in the mentioned figure 4.2 and table 4.3.
Table 4.3: Comparative results the maximal error remainder: ๐๐ธ๐ ๐ for
TAM, BCM, VIM and ADM, where ๐ = 1, โฆ ,4.
n ๐ด๐ฌ๐น๐ by TAM ๐ด๐ฌ๐น๐ by BCM ๐ด๐ฌ๐น๐by VIM ๐ด๐ฌ๐น๐ by ADM
1 5.99981 ร 10โ3 1.61705 ร 10โ2 1.61705 ร 10โ2 1.61705 ร 10โ2
2 2.89069 ร 10โ6 2.52036 ร 10โ5 2.52036 ร 10โ5 2.98775 ร 10โ5
3 1.12531 ร 10โ8 3.56783 ร 10โ8 3.56783 ร 10โ8 4.9464 ร 10โ8
4 4.54397 ร 10โ11 4.5361 ร 10โ11 4.53593 ร 10โ11 7.1962 ร 10โ11
Figure 4.2: Comparison of the maximal error remainder for TAM, BCM,
VIM and ADM.
6. The analytic solutions of these different problems are excellent as
compared to the results of other iterative methods. In nonlinear equations
that emerge much of the time to express phenomenon the proposed
methods are effective and strong enough to give the reliable results.
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Chapter Four Conclusions and Future Works
121
4.3 Future works
In this section some of future works will be suggested
1) Solving the EmdenโFowler equation by using TAM
๐ฃโฒโฒ +๐
๐ฅ ๐ฃโฒ + ๐(๐ฅ)๐(๐ฃ) = 0, ๐ฃ(0) = ๐ฃ0, ๐ฃโฒ(0) = 0, (4.4)
where ๐(๐ฅ) and ๐(๐ฃ) are some given functions of ๐ฅ and ๐ฃ, respectively,
and ๐ is called the shape factor.
2) Using BCM for solving a Thomas-Fermi equation
๐ฃโฒโฒ =๐ฃ
32
โ๐ฅ, ๐ฃ(0) = 1, ๐ฃ(โ) = 0, (4.5)
3) Solving the Blasius equation by using TAM
๐ฃโฒโฒโฒ +1
2๐ฃ๐ฃโฒโฒ = 0, ๐ฃ(0) = 0, ๐ฃโฒ(0) = 1, ๐ฃโฒโฒ(0) = ๐ผ, ๐ผ > 0. (4.6)
4) Solving the sine-Gordon equation by using TAM
๐ฃ๐ก๐ก โ ๐2๐ฃ๐ฅ๐ฅ + ๐ผ sin(๐ฃ) = 0, ๐ฃ(๐ฅ, 0) = ๐(๐ฅ), ๐ฃ๐ก(๐ฅ, 0) = ๐(๐ฅ), (4.7)
where ๐ and ๐ผ are constants. It is clear that this equation adds the
nonlinear term sin(๐ฃ) to the standard wave equation.
5) Using BCM for solving the telegraph equation
๐ฃ๐ฅ๐ฅ = ๐๐ฃ๐ก๐ก + ๐๐ฃ๐ก + ๐๐ฃ, ๐ฃ(๐ฅ, 0) = ๐(๐ฅ), ๐ฃ๐ก(๐ฅ, 0) = ๐(๐ฅ), (4.8)
where ๐ฃ = ๐ฃ(๐ฅ, ๐ก) is the resistance, and ๐, ๐ and c are constants related to
the inductance, capacitance and conductance of the cable respectively.
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REFERENCES
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References
122
REFERENCES
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[2] A. K. Alomari, M. S. M. Noorani, R. Nazar, "The Homotopy Analysis
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Burgersโ Equations", Applied Mathematical Sciences 2 (40), 1963โ
1977, (2008).
[3] A. Latif, "Banach Contraction Principle and Its Generalizations",
Springer International Publishing Switzerland 2014, 33โ46, (2014).
[4] A. M. Wazwaz, "The variational iteration method for solving linear and
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[5] A. M. Wazwaz, "Partial Differential Equations: Methods and
Applications", (2002).
[6] A. M. Wazwaz, "Linear and nonlinear integral equations", Springer
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APPENDICES
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Appendices
130
Appendix A: Code of Mathematica for chapter one (the ADM)
**************************************************************
Code of example (1.1)
**************************************************************
(* v'=-v2 +1, v(0)=0 *)
tt=AbsoluteTime[];
zz=SessionTime[];
v0=x;
v1= -Integrate[v0^2/. xt,{t,0,x}]
-x3/3
v2= -Integrate[2v0*v1/. xt,{t,0,x}]
2x5/15
v3= -Integrate[2v0*v2+v1^2 /. xt,{t,0,x}]
-17x7/315
v4= -Integrate[2v0*v3+2v1*v2 /. xt,{t,0,x}];
v5= -Integrate[2v0*v4+2v1*v3+v2^2 /. xt,{t,0,x}];
v6= -Integrate[2v0*v5+2v2*v3+2v1*v4 /. xt,{t,0,x}];
v=v0+v1+v2+v3+v4+v5+v6
x-x3/3+2x5/15-17x7/315+62x9/2835-1382x11/155925+ 21844x13/6081075
r1=Abs[D[v0+v1,{x,1}]+(v0+v1)^2-1];
Plot[r1,{x,0,1}]
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Appendices
131
r2=Abs[D[v0+v1+v2,{x,1}]+(v0+v1+v2)^2-1];
Plot[r2,{x,0,1}]
r3=Abs[D[v0+v1+v2+v3,{x,1}]+(v0+v1+v2+v3)^2-1];
Plot[r3,{x,0,1}]
r4=Abs[D[v0+v1+v2+v3+v4,{x,1}]+(v0+v1+v2+v3+v4)^2-1];
Plot[r4,{x,0,1}]
0.2 0.4 0.6 0.8 1.0
0.1
0.2
0.3
0.4
0.5
0.2 0.4 0.6 0.8 1.0
0.05
0.10
0.15
0.20
0.2 0.4 0.6 0.8 1.0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
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Appendices
132
r5=Abs[D[v0+v1+v2+v3+v4+v5,{x,1}]+(v0+v1+v2+v3+v4+v5)^2-1];
Plot[r5,{x,0,1}]
r6=Abs[D[v0+v1+v2+v3+v4+v5+v6,{x,1}]+(v0+v1+v2+v3+v4+v5+v6)^2-
1];
Plot[r6,{x,0,1}]
x ={0, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.1};
r1;r2;r3;r4;r5;r6;
Y1=Max[r1]
0.2 0.4 0.6 0.8 1.0
0.005
0.010
0.015
0.020
0.025
0.030
0.2 0.4 0.6 0.8 1.0
0.002
0.004
0.006
0.008
0.2 0.4 0.6 0.8 1.0
0.0005
0.0010
0.0015
0.0020
0.0025
0.0030
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Appendices
133
0.0000665556
Y2=Max[r2]
3.76891ร10-7
Y3=Max[r3]
1.96289ร10-9
Y4=Max[r4]
9.72067ร10-12
Y5=Max[r5]
4.65513ร10-14
Y6=Max[r6]
2.15106ร10-16
Y={Y1,Y2,Y3,Y4,Y5,Y6};
n={1,2,3,4,5,6};
ListLog-
Plot[{{1,Y1},{2,Y2},{3,Y3},{4,Y4},{5,Y5},{6,Y6}},JoinedTrue,PlotRan
geAll,FrameTrue,
AesTrue,FrameLabel{Row[{Style["n",FontSlantItalic]}],Row[{Style[
"MERn",FontSlantItalic]}]},PlotMarkers{Automatic,15}]
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Appendices
134
Appendix B: Code of Mathematica for chapter one (the VIM)
**************************************************************
Code of example (1.7)
**************************************************************
(* D[v,{t,1}]+v*D[v,{x,1}]+ v*D[v,{y,1}]= k (D[v,{x,2}]+D[v,{y,2}]),
v(x,y,0)= sin(2ฯx)cos(2ฯy) *)
tt=AbsoluteTime[];
zz=SessionTime[];
v0= sin(2ฯx)cos(2ฯy);
v1=v0-Integrate[(D[v0,{t,1}]+v0*D[v0,{x,1}]+ v0*D[v0,{y,1}]-k
(D[v0,{x,2}]+D[v0,{y,2}]),{t,0,t}];
v2=v1-Integrate[(D[v1,{t,1}]+v1*D[v1,{x,1}]+ v1*D[v1,{y,1}]-k
(D[v1,{x,2}]+D[v1,{y,2}]),{t,0,t}];
v3=v2-Integrate[(D[v2,{t,1}]+v2*D[v2,{x,1}]+ v2*D[v2,{y,1}]-k
(D[v2,{x,2}]+D[v2,{y,2}]),{t,0,t}];
v4=v3-Integrate[(D[v3,{t,1}]+v3*D[v3,{x,1}]+ v3*D[v3,{y,1}]-k
(D[v3,{x,2}]+D[v3,{y,2}]),{t,0,t}];
r1=Abs[D[v1,{t,1}]+v1*D[v1,{x,1}]+ v1*D[v1,{y,1}]-k
(D[v1,{x,2}]+D[v1,{y,2})];
r2=Abs[D[v2,{t,1}]+v2*D[v2,{x,1}]+ v2*D[v2,{y,1}]-k
(D[v2,{x,2}]+D[v2,{y,2})];
r3=Abs[D[v3,{t,1}]+v3*D[v3,{x,1}]+ v3*D[v3,{y,1}]-k
(D[v3,{x,2}]+D[v3,{y,2})];
r4=Abs[D[v4,{t,1}]+v4*D[v4,{x,1}]+ v4*D[v4,{y,1}]-k
(D[v4,{x,2}]+D[v4,{y,2})];
k =0.1;
t =0.0001;
x ={0, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.1};
y ={0, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.1};
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Appendices
135
r1;r2;r3;r4;
Y1=Max[r1]
0.0161705
Y2=Max[r2]
0.0000252036
Y3=Max[r3]
3.56783ร10-8
Y4=Max[r4]
4.53593ร10-11
Y={Y1,Y2,Y3,Y4};
n={1,2,3,4};
ListLog-
Plot[{{1,Y1},{2,Y2},{3,Y3},{4,Y4}},JoinedTrue,PlotRangeAll,Frame
True,AesTrue,FrameLabel{Row[{Style["n",FontSlantItalic]}],Row[
{Style["MERn",FontSlantItalic]}]},PlotMarkers{Automatic,15}]
1.0 1.5 2.0 2.5 3.0 3.5 4.010 11
10 8
10 5
10 2
n
ME
Rn
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Appendices
136
Appendix C: Code of Mathematica for chapter two (the TAM)
**************************************************************
Code of example (2.4)
**************************************************************
(* v''''=v2-x10+4x9-4x8-4x7+8x6-4x4+120x-48, v(0)= v'(0)=0, v(1)= v'(1)=1 *)
(* L(v)= v'''', N(v)= v2 and f(x)=-(-x10+4x9-4x8-4x7+8x6-4x4+120x-48) *)
tt=AbsoluteTime[];
zz=SessionTime[];
v01[x_]=v0[x]/.First@DSolve[{v0''''[x] -x10+4x9-4x8-4x7+8x6-4x4+120x-48,
v0[0]0, v0'[0]0, v0[1]=1, v0'[1]=1}, v0, x];
Expand[v01[x]]
718561x2/360360+4019x3/540540-2x4+x5-x8/420+x10/360-x11/1980-
x12/2970+x13/4290-x14/24024
v11[x_]=v1[x]/.First@DSolve[{v1''''[x] (v01[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v1[0]0, v1'[0]0, v1[1]=1, v1'[1]=1}, v1, x];
v22[x_]=v2[x]/.First@DSolve[{v2''''[x] (v11[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v2[0]0, v2'[0]0, v2[1]=1, v2'[1]=1}, v2, x];
v33[x_]=v3[x]/.First@DSolve[{v3''''[x] (v22[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v3[0]0, v3'[0]0, v3[1]=1, v3'[1]=1}, v3, x];
v44[x_]=v4[x]/.First@DSolve[{v4''''[x] (v33[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v3[0]0, v4'[0]0, v4[1]=1, v4'[1]=1}, v4, x];
v55[x_]=v5[x]/.First@DSolve[{v5''''[x] (v44[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v5[0]0, v5'[0]0, v5[1]=1, v5'[1]=1}, v5, x];
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Appendices
137
v66[x_]=v6[x]/.First@DSolve[{v6''''[x] (v55[x])^2 -x10+4x9-4x8-4x7+8x6-
4x4+120x-48, v6[0]0, v6'[0]0, v6[1]=1, v6'[1]=1}, v6, x];
x={0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1};
v=x5-2x4+2x2;
r1=Abs[v-v11[x]]
{ 0, 1.02627ร10-7, 3.47659ร10-7, 6.41401ร10-7, 8.93924ร10-7, 1.02777ร10-6,
9.93141ร10-7, 7.86445ร10-7, 4.64662ร10-7, 1.471ร10-7, 2.22045ร10-16 }
r2=Abs[v-v22[x]];
r3=Abs[v-v33[x]];
r4=Abs[v-v44[x]];
r5=Abs[v-v55[x]];
r6=Abs[v-v66[x]];
Appendix D: Code of Mathematica for chapter three (the BCM)
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Code of example (3.4)
**************************************************************
(* v''=3/4*v+v(x/2)-x2+2, v(0)=0, v'(0)=0 *)
tt=AbsoluteTime[];
zz=SessionTime[];
v0= x2-x4/12;
a0=v0 /. xx/2;
v1= v0+Integrate[(x-t)*((3/4*v0+a0)/. xt),{t,0,x}]
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Appendices
138
x2-13x6/5760
a1=v1 /. xx/2;
v2= v0+Integrate[(x-t)*((3/4*v1+a1)/. xt),{t,0,x}]
x2-91x8/2949120
a2=v2 /. xx/2;
v3= v0+Integrate[(x-t)*((3/4*v2+a2)/. xt),{t,0,x}]
x2-17563x10/67947724800
a3=v3 /. xx/2;
v4= v0+Integrate[(x-t)*((3/4*v3+a3)/. xt),{t,0,x}]
x2-13505947x12/9184358065766400
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ุงูู ุณุชุฎูุต
ุงูุญููู ุฅููุฌุงุฏ ุงููุฏู ุงูุฑุฆูุณู ู ู ูุฐู ุงูุฑุณุงูุฉ ูู ุงุณุชุฎุฏุงู ุงุซูุชูู ู ู ุทุฑุงุฆู ุชูุฑุงุฑูุฉ ุดุจู ุชุญููููุฉ
ุงูุชุญููููุฉ ูุจุนุถ ุงูู ุณุงุฆู ุงูู ูู ุฉ ูู ุงูููุฒูุงุก ูุงูููุฏุณุฉ.
ู ู ูุจู ุงูุชู ูู ู ูุงุฃููุตุงุฑู ุฃุงู ุงูุฐู ุงูุชุฑุญูุงุฃูููู ุฉ ุงูุชูุฑุงุฑู ูุทุฑููุฉุง ุงุณุชุฎุฏุงู ูู ุงุฃููู ุงููุฏู
ูุฐุฑุงุน ุงูู ุชุฐุจุฐุจ. ุงุชุดูู ู ุจุงูุชูุบุฑุงูุ ุฑููุงุชูุ ูู ุนุงุฏุงูุชุงูุญููู ุงูุชุญููููุฉ ุฅููุฌุงุฏ (TAM) ููู ุงูู
( ุนูู ู ุนุงุฏุงูุช ูุงูุธู ุฉ ู ุนุงุฏุงูุช ุจุฑุฌุฑุฒ ุงูุบูุฑ ุฎุทูุฉ ุงุฃูุญุงุฏูุฉ ุงูุจุนุฏ TAMูุฏ ุชู ุชุทุจูู ุงูู ) ุงูุถุงุ
( ูุฏ ุชู TAMูุงูุซูุงุฆูุฉ ุงุงูุจุนุงุฏ ูุงูุซุงูุซูุฉ ุงุงูุจุนุงุฏ ููุญุตูู ุนูู ุงูุญููู ุงูุชุญููููุฉ ููู ู ููุง. ุงูุชูุงุฑุจ ููู )
ุงูุชุญูู ู ูู ูุฅุซุจุงุชู ุจูุฌุงุญ ูุชูู ุงูู ุณุงุฆู.
ูุณูู ุงูุชู( BCMุงููุจุงุถ ุจูุงุฎ )ุทุฑููุฉ ุฃุณุงุณ ุนูู ุชูุฑุงุฑูุฉ ุชูููุฉ ุงุณุชุฎุฏุงู ูู ุงูุซุงูู ูุงููุฏู
( ูุญู ููุณ ุงูู ุณุงุฆู BCM. ุชู ุงุณุชุฎุฏุงู ุงูู )ุงูุฎุทูุฉ ุบูุฑ ุงูู ุตุทูุญุงุช ู ุน ุงูุชุนุงู ู ูู ุชุทุจูููุง
( TAMุจูู ุงูู )ุงูู ูุงุฑูุงุช ู ู ุงูุนุฏูุฏ ูุฏู ูุง ุฐููุ ุฅูู ูุจุงุฅูุถุงูุฉุงูููุฒูุงุฆูุฉ ูุงูููุฏุณูุฉ ุงูู ุฐููุฑุฉ ุฃุนุงูู.
( ู ุน ุจุนุถ ADM) ุฃุฏูู ูุงู ุชููููุทุฑููุฉ ( ูVIM) ุทุฑููุฉ ุงูุชุบุงูุฑูุฉ ุงูุชูุฑุงุฑูุฉ( ูุงูBCMูุงูู )
( ูุงูู TAMุนุงููุฉ. ุงูู )ูุชู ุชูู ุฏูุฉ ููุคุฉ ุงูู ูุฏู ุฉุงูุทุฑุงุฆู .ุงุงูุณุชูุชุงุฌุงุช ูุงุฃูุนู ุงู ุงูู ุณุชูุจููุฉ
(BCM ุงู ุชุชุทูุจุงู ุงูุฉ ุงูุชุฑุงุถุงุช ู ููุฏุฉ ูู ุงูุญุงูุฉ )ูุงู ุงูุทุฑููุชูู ุงู ุฐููุ. ุจุงุฅูุถุงูุฉ ุงูู ุบูุฑ ุฎุทูุฉ
( ุฃู VIM( ุฃูุงูู )ADMูู ุงูู )ุงูุญุงู ููุชุนุชู ุฏุงู ุนูู ุงุณุชุฎุฏุงู ุฃู ุงูุชุฑุงุถุงุช ู ููุฏุฉ ุงุถุงููุฉ ูู ุง
ุฃูุฉ ุทุฑุงุฆู ุชูุฑุงุฑูุฉ ุฃุฎุฑู.
.10 ยฎู ุงุซูู ุงุชููุงุงูุจุฑูุงู ุฌ ุงูู ุณุชุฎุฏู ููุญุณุงุจุงุช ูู ูุฐู ุงูุฑุณุงูุฉ ุงูู ูุฏู ุฉ ูู
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ุฌู ููุฑูุฉ ุงูุนุฑุงู
ูุฒุงุฑุฉ ุงูุชุนููู ุงูุนุงูู ูุงูุจุญุซ ุงูุนูู ู
ุฌุงู ุนุฉ ุจุบุฏุงุฏ
ูููุฉ ุงูุชุฑุจูุฉ ููุนููู ุงูุตุฑูุฉ / ุงุจู ุงูููุซู
ูุณู ุงูุฑูุงุถูุงุช
ุฉ ููู ุนุงุฏุงูุช ูุงุฆู ุชูุฑุงุฑูุฉ ุดุจู ุชุญูููุทุฑ
ุฎุทูุฉุงูุบูุฑ ุงูุชูุงุถููุฉ ุงูุฎุทูุฉ ู
ุฑุณุงูุฉ
ุ ุฌุงู ุนุฉ ุจุบุฏุงุฏ ุงุจู ุงูููุซู / ู ูุฏู ุฉ ุฅูู ูููุฉ ุงูุชุฑุจูุฉ ููุนููู ุงูุตุฑูุฉ
ูู ุงูุฑูุงุถูุงุชู ุงุฌุณุชูุฑ ุนููู ูุฌุฒุก ู ู ู ุชุทูุจุงุช ููู ุฏุฑุฌุฉ
ู ู ูุจู
ู ุตุทูู ู ุญู ูุฏ ุนุฒูุฒ ุจุฅุดุฑุงู
ู ุฌูุฏ ุฃุญู ุฏ ููู .ุฏ. ู . ุฃ
ู 2017 ฺพ 1438