efficient method for solving some types of partial
TRANSCRIPT
Republic of Iraq
Ministry of Higher Education
and Scientific Research
University of Baghdad
College of Education for Pure Science,
Ibn Al-Haitham
Efficient Method for Solving Some Types of
Partial Differential Equations
A Thesis
Submitted to the Department of Mathematics, College of Education for
Pure Science- Ibn Al-Haitham, University of Baghdad as Partial
Fulfillment of Requirements for the Degree of
Master of Science in Mathematics
By
Myasar Obaid Enadi
Supervised By
Prof. Dr. Luma Naji Mohammed Tawfiq
AC 1440 AH 2019
ุจุณู ุงููู ุงูุฑุญู ู ุงูุฑุญูู
ููุงู ุฑุจ ุฃูุฒุนูู ุฃู ุฃุดูุฑ ูุนู ุชู ุงูุชู ุฃู ุนู ุช ุนูู ูุนูู ูุงูุฏู ูุฃู ๏ดพ19๏ดฟ ู ูู ุนุจุงุฏู ุงูุตุงูุญูู ุฃุนู ู ุตุงูุญุง ุช ุฑุถุงู ูุฃุฏุฎููู ุจุฑุญู ุช
ุตุฏู ุงููู ุงูุนุธูู
(19)ุงูุขูุฉ ุณูุฑุฉ ุงููู ู
ูุฏูุง ู ุญู ุฏ ูุงูุตูุงุฉ ูุงูุณูุงู ุนูู ุฃุดุฑู ุงูู ุฑุณููู ุณ ุงูุญู ุฏ ููู ุนูู ูุนู ู ุงูุชู ูุง ุชุนุฏ ููุง ุชุญุตู
ูุฃูู ุงูุทูุจูู ุงูุทุงูุฑูู
ูู ููุงูุฉ ุฑุณุงูุชู ุงูุฏู ุซู ุฑุฉ ุฌูุฏู ุงูู ุชูุงุถุนุฉ
ุงูุญูุงุฉ ู ุตุงุนุจ ู ู ุฃุฌููู ุชููู ุงูุฐูู ุฅูู
ู ู ุฃุฌููู ุชุณุชุญู ุงูุญูุงุฉ ุงูุฐูู ุฅูู
"ุงูุนุงุฆูุฉ ุงููุฑูู ุฉ "ุญุจูู ูู ุงูููุจ ููุงููุง ุงูููุงุก ูุงููุจุถ ุณููุฅูู ู ู
ูู ู ูุงุฌู ู ุญู ุฏ ุชูููู"ุงูุฏูุชูุฑ " ุงูุฃุณุชุงุฐุฑุณุงูุชู ุนูู ุงูู ุจุฅุดุฑุงููุชุฅูู ู ู ุดุฑู
ููุชูุฌููุงุชูุง ุงูุนูู ูุฉ ูุนู ุงููุจูุฑุตุจุฑูุง ูุญุฑูู ูุฅููุงุฆูุง ุญููุง ุงูุงูุชู ูู ุชููู
ุฎุงูุต ุูู ุง ุฃุชูุฌู ุจ ุงูุนู ู ูุฌุงุฒุจุดูู ูุจูุฑ ูู ุฅุชู ุงู ู ุง ุงูุชู ุณุงูู ุช ุงูุชู ูุง ุชูุฏุฑ ุจุซู ู ู
ุดูุฑู ู ุชูุฏูุฑู
ุฅูู ุจูุงุฑู ุงูุนูู ูุงูู ุนุฑูุฉ ุฃุณุงุชุฐุชู ุงูุฃูุงุถู
ูู ูุชุฑุฉ ุฏุฑุงุณุชูุฅูู ูู ู ู ุณุงุนุฏูู
ุงูุจุงุญุซ
ุฏุงุกูุงูุฅู
To โALLAHโ and to his prophet โMohammedโ, my praise and thanks are due
for their blessings without which this research would not have been achieved
without their blessings. It is pleasure to express my deep appreciation to my
supervisor Prof. Dr. Luma Naji Mohammed Tawfiq for her invaluable advice,
assistance, cooperation, and support throughout the course of preparing my thesis.
I dedicate this work to my dear parents and my wife for their financial and
moral support throughout the study period, and thank them so much.
I would like to thank everyone who helped me and stayed by my side
throughout the study period from my family, friends and teachers.
Researcher
Acknowledgements
Journal Papers
1. Enadi, M. O., and Tawfiq, L.N.M., 2019, New Technique for Solving
Autonomous Equations, Ibn Al-Haitham Journal for Pure and Applied Science,
32(2), p: 123-130, Doi: 10.30526/32.2.2150
2. Enadi, M. O., and Tawfiq, L.N.M., 2019, Solving Systems of Non-linear
Partial differential Equations by Using Coupled Method, International Journal
of Modern Mathematical Sciences, 17(1), p: 68-77.
3. Enadi, M. O., and Tawfiq, L.N.M., 2019, New Approach for Solving Three
Dimensional Space Partial Differential Equation, Baghdad Science Journal,
16(3), p: 786-792, Doi: http://dx.doi.org/10.21123/bsj.2019.16.3(Suppl.).0780.
4. Enadi, M. O., and Tawfiq, L.N.M., 2019, Using New Coupled Method for
Solving System of Nonlinear Wave Equation, Computers and Mathematics with
Applications, Elsevier, Under review.
Authorโs Publications
Symbol Definition
ADM Adomian Decomposition Method
An Adomian Polynomials
BCs Boundary Conditions
BVP Boundary Value Problem
CuTBs Cubic Trigonometric Bspline
DTM Differential Transform Method
Eq. Equation
Eqs. Equations
Figs. Figures
HAM Homotopy Analysis method
Hn He Polynomials
HPM Homotopy Perturbation Method
iD-PDEs
i=1,2,3,4 i dimensional Partial differential equations, i=1,2,3,4
ICs Initial Conditions
IVP Inatial Value Problem
KdV Korteweg-deVries
LT Laplace Transformation
MHPM Modified Homotopy Perturbation Method
Non-Linear Linear and Nonlinear
NT New Transformation
NTHPM New transform homotopy perturbation method
List of Symbols and Abbreviations
ODEs Ordinary Differential Equations
PDEs Partial Differential Equations
RLW Real Like Wave
t Time
VIM Variational Iteration Method
๐ New transform for the function f
ฮฉ Omega
๐ New Transformation
๐ค Gamma
โ โ Norm
Subject Page
No.
Abstract I
Introduction 1
1 Chapter One : Preliminaries 6
1.1 Introduction 6
1.2 Overview of Differential Equations 6
1.3 Some Basic Concepts of the Homotopy Perturbation
Method
9
1.4 New Transformation 15
1.4.1 Definition of New Transformation 15
1.4.2 The General Properties of the New
Transformation
16
1.4.3 The Advantages of the New Transformation 17
2 Chapter Two : Couple New Transform with HPM to Solve
Some Types of Partial Differential Equations
20
2.1 Introduction 20
2.2 New Transformation โ Homotopy Perturbation Method 20
2.3 Solve Linear PDEs by NTHPM 21
2.4 Illustrative Application 24
2.5 Solve Nonlinear PDE by NTHPM 27
2.6 Applications 29
2.7 Solve Automatous Equation by NTHPM 35
2.8 Illustrative Examples 38
2.9 Convergence of the Solution for Linear Case 42
2.10 Convergence of the Solution for Nonlinear Case 45
Contents
3 Chapter Three : Solving System of Partial Differential
Equations by New Couple Method
3.1 Introduction 49
3.2 Solving System for 2Equation Nonlinear PDEs by
NTHPM
50
3.3 Illustrative Examples for System of 1D-PDEs 54
3.4 Illustrative Examples for System of 2D-PDEs 60
3.5 Solving System for 3Equations Nonlinear 1D-PDEs 69
3.6 Convergence for the Series Solution 80
4 Chapter Four: Application Model 84
4.1 Introduction 84
4.2 Formulation Mathematical Model 84
4.3 Solving Model Equation by the NTHPM 89
4.4 Experiment Application 92
5 Chapter Five : Conclusions and Future Work 101
5.1 Conclusions 101
5.2 Future Works 103
References 105
I
In this thesis, a new method based on a combined form of the new transform
with homotopy perturbation method is proposed to solve some types of partial
differential equations, for finding exact solution in a wider domain. It can be used
to solve the problems without any discretization, or resorting to the frequency
domain or restrictive assumptions and it is free from round-off errors. This method
is called the new transform homotopy perturbation method.
In this thesis, we focuse on some basic concepts of the partial differential
equations.The first objective is implement suggested method in order to solve
some types of PDEs with initial condition such that: Klein-Gordan equation, wave-
like equations, autonomous equation, system of two or three non-linear equations,
Burgers' equations, coupled Hirota Satsuma KdV type II, and RLW equation.
Finally, The proposed method is used to solve application model which is soil
moisture equation where traditional HPM leads to an approximate solution. The
second aim which is the convergence of the series solution is studied, the series
solution converge to to the exact form is proved.Some examples are provided to
illustrate the reliability and capability of the suggested method.The practical results
Abstract
II
show that the proposed method is efficient tool for solving those types of partial
differential equations.
1
Many phenomena that arise in mathematical physics and engineering
fields can be described by partial differential equations (PDEs). In physics for
example, the heat flow and the wave propagation phenomena are well
described by PDEs [45,55]. So, it is a useful tool for describing natural
phenomena of science and engineering models. Most of engineering
problems are nonlinear PDEs, and it is difficult to solve them analytically.
The obtaining of the exact solution of nonlinear PDEs in physics and
mathematics is still a significant problem that needs new efficient
implemented methods to get exact solutions. Various powerful mathematical
methods have been proposed for obtaining exact and approximate analytic
solutions. Some of the classic analytic methods are perturbation techniques
[8] and Hirotaโs bilinear method [46]. Perturbation techniques were generated
useful solutions in describing both quantitative and qualitative properties of
the problem, which is an advantage compared to numerical solutions.
However, some drawbacks were obvious for complex equations due to either
such parameters cause a divergence of solutions as the quantities
increase/decrease, or the non-existence of small or large perturbation
Introduction
Introduction
2
parameters. In problems where these quantities do not exist, the parameter
has to be artificially introduced which may lead to incorrect results [17].
Perturbation techniques are therefore found to be mainly suitable for weakly
nonlinear problems.
In recent years, many researchers have paid attention to study the
solutions of non-linear PDEs by using various methods. Among these are the
Adomian Decomposition Method (ADM) [2,38,39], tanh method, Homotopy
Perturbation Method (HPM) [43], Homotopy Analysis Method (HAM) [35],
the Differential Transform Method (DTM) [9], Cubic Trigonometric B-
Spline Method [29,30], Laplace Decomposition Method [26,43], Variational
Iteration Method (VIM) [37, 56], parallel processing [39,40,49] and semi
analytic technique [41, 42, 48, 50].
In this thesis we suggested a new method based on combine two efficient
methods to get exact solution for some types of PDEs such autonomous
equation which describes the appearance of the stripe pattern in two
dimensional systems. Moreover, this equation was applied to a number of
problems in variety systems, e.g., Rayleigh-Benard convection, Faraday
instability, nonlinear optics, chemical reactions and biological systems. The
approximate solutions of the autonomous equation were presented by
Introduction
3
differential transformation method [1], reduce differential transformation
[31].
The system of PDEs arises in in many areas of mathematics, engineering
and physical sciences. These systems are too complicated to be solved
exactly so it is still very difficult to get exact solution for most problems. A
vast class of analytical and numerical methods has been proposed to solve
such problems. But many systems such as system of high dimensional
equations, the required calculations to obtain its solution in some time may
be too complicated. Recently, many powerful methods have been presented,
such as the coupled method [27,30]. Herein we solved such systems by
proposed method and we get exact solution without using computer
programming and calculating.
This thesis is organized as follows:
In Chapter one, a brief review of basic definitions and concepts relate to the
work is introduced. It includes an overview of PDEs and their types.
Chapter two contains the implementation of proposed method based on
coupled two efficient methods such Homotopy Perturbation Method (HPM)
and new transform defined by Luma and Alaa in [51], that we will say the
New Transform Homotopy Perturbation Method (NTHPM) for obtaining
exact solutions to some types of PDEs such as: Klein-Gordan equation,
Introduction
4
wave-like equations, autonomous equation, 3D-PDEs, and other 3rd order
PDEs.
Chapter three contains the implementation of NTHPM for solving some types
of system of PDEs. The efficiency of the proposed method is verified by the
examples. The convergence of series solution to the exact analytic solution
function is proved.
In chapter four the proposed method are successfully implemented to solve
1D, 2D, and 3D soil moisture model equation to determine the moisture
content in soil.
Finally, in chapter five the conclusions and future works are given.
Preliminaries
6
1.1. Introduction
This chapter includes some basic definitions and concepts related to the
problems for this thesis. An overview of differential equations and their types is
introduced. In addition, we review some traditional techniques such as HPM for
solving partial differential equations, for comparison with the proposed approach
NTHPM illustrated by examples.
1.2. Overview of Differential Equations
Differential equations are used in different field of science and engineering. It's a
relation involving an unknown function (or functions say dependent variables) of
one or several independent variables and their derivatives with respect to those
variables. Many real phenomena in various fields such as engineering, physical,
biological and chemical are modeled mathematically by using differential equations
[47, 48, 50, 60]. Commonly, most real science and engineering processes including
more than one independent variable and the corresponding differential equations
are called partial differential equations (PDEs). However, the PDEs have been
reduced to ordinary differential equations (ODEs) using simplified assumptions.
Chapter One
Preliminaries
Preliminaries Chapter One
7
Where ODE is a differential equation for a function of single independent variable
[47].
The order of a PDE is the order of the highest partial derivative that appears in
the equation. A PDE s are classified as homogeneous or inhomogeneous. A PDE of
any order is called homogeneous if every term of the PDE contains the dependent
variable or one of its derivatives; otherwise, it is called an inhomogeneous PDE
[55].
In research field, there are several types of PDEs which depends on the
application that are used. Each application has its own special governing equations
and properties that should considered individually.
A PDE is called linear if the power of the dependent variable and each partial
derivative contained in the equation is one and the coefficients of the dependent
variable and the coefficients of each partial derivative are constants or independent
variables. However, if any of these conditions is not satisfied, the equation is called
nonlinear [55]. Also, it can consider a semi linear, if it is linear in partial
derivative only. In addition, it can consider a quasi linear, if it is linear in the first
partial derivatives or it is nonlinear in dependent variable [60].
The general form of quasi linear 2nd order inhomogeneous PDE with two
independent variables can write as [Hoffman, 2001]:[55]
๐ด๐ข๐ฅ๐ฅ + ๐ต๐ข๐ฅ๐ฆ + ๐ถ๐ข๐ฆ๐ฆ + ๐ท๐ข๐ฅ + ๐ธ๐ข๐ฆ + ๐น๐ข = ๐บ (1.1)
Where A, B, C, D, E, and F are the coefficients and the inhomogeneous term G
may depend on x and y. The above equation (1.1) can be classified to:
Preliminaries Chapter One
8
1) Elliptic, when ๐ต2 โ 4๐ด๐ถ < 0;
2) Parabolic, when ๐ต2 โ 4๐ด๐ถ = 0;
3) Hyperbolic, when ๐ต2 โ 4๐ด๐ถ > 0.
A solution of a PDE is a function satisfies the equation under discussion and
satisfies the given conditions as well. In order to find the solution of PDEs, initial
and / or boundary conditions used to solve PDEs, so the PDEs with initial
conditions (ICs) is said to be initial value problem (IVPs), the PDEs with boundary
conditions (BCs) is said to be boundary value problems (BVPs), but the PDEs with
initial and boundary conditions is said to be initial-boundary value problems, the
boundary conditions (BCs) which can be classify into three types:[60]
1) Dirichlet boundary condition: numerical values of the function are specific of
the boundary of the region.
2) Neumann boundary condition: specifies the values that the derivative of a
solution to take on the boundary of the domain.
3) Mixed boundary conditions: defines a BVP in which the solution of the given
equation is required to satisfy different boundary conditions on disjoint parts of the
boundary of the domain where the condition is stated. In effect, in a mixed BVP,
the solution is required to satisfy the Dirichlet or Neumann boundary conditions in
a mutually exclusive way on disjoint parts of the boundary.[60]
Preliminaries Chapter One
9
An elliptic partial differential equation with mixed boundary conditions is called
a Robbins problem [60].
Some types of differential equation, such as nonlinear differential equation
cannot be easy to solve, so we use numerical or approximate solution. In this thesis
we suggest an efficient method to solve important types of PDEs to get exact
solution.
We are beginning with Homotopy Perturbation Method (HPM).
1.3. Some Basic Concepts of the Homotopy Perturbation Method
The HPM was first proposed by He J. Huan in 1999 [13] for solving
differential and integral equations, non-linear and has been successfully applied to
solve non-linear differential equations, and other fields for more details see [15]. It
is a combine of traditional perturbation method with homotopy method and it
suggested to overcome the difficulty arising in calculating Adomian polynomials.
This method has many advantages such as it is applied directly to the nonlinear
problems without linearizing the problem. In this section, some basic concepts of
this method have been explained.
Definition 1.1 [20]
Let X and Y are two topological spaces. Two continuous functions ๐: ๐ โ ๐ and
๐: ๐ โ ๐ are said to be homotopic, denoted by โ ๐ , if a continuous function
๐ป: ๐ ร [0,1] โ ๐ ,such that:
๐ป(๐ฅ, ๐ฃ) = ๐(๐ฅ), โ๐ฅ โ ๐
Preliminaries Chapter One
10
๐ป(๐ฅ, 1) = ๐(๐ฅ), โ๐ฅ โ ๐
In this case, ๐ป is said to be a homotopy.
Now, to illustrate definition (1.1), consider the following examples.
Example 1.1[20]
Let ๐and ๐ be any topological spaces, ๐ be the identity function and ๐ be the zero
function, then define ๐ป: ๐ ร [0,1] โ ๐ by:
๐ป(๐ฅ, ๐) = ๐ฅ(1 โ ๐), โ๐ฅ โ ๐, โ ๐ โ [0,1]
Then ๐ป is a continuous function and
๐ป(๐ฅ, 0) = ๐ฅ = ๐(๐ฅ), โ๐ฅ โ ๐
๐ป(๐ฅ, 1) = 0 = ๐(๐ฅ), โ๐ฅ โ ๐
Therefore ๐ โ ๐ .
Remark
Let ๐: ๐ โ ๐ and ๐: ๐ โ ๐ be continuous functions. Define
๐ป: ๐ ร [0,1] โ ๐ by
๐ป(๐ฅ, ๐) = (1 โ ๐)๐(๐ฅ) + ๐๐(๐ฅ), โ ๐ฅ โ ๐ ,โ ๐ โ [0,1]
Then, ๐ป(๐ฅ, 0) = ๐(๐ฅ), โ๐ฅ โ ๐
and,๐ป(๐ฅ, 1) = ๐(๐ฅ), โ๐ฅ โ ๐
Therefore โ ๐ .
Now, to illustrate the basic idea of the HPM, we consider the following
nonlinear differential equation:
Preliminaries Chapter One
11
A(u) f (x), x(1.2)
where A is differential operator, f is a known function of x. The operator A can
generally speaking be divided into two operators L and N, where L is a linear
operator, and N is a nonlinear operator. Therefore equation (1.2) can be rewritten as
follows:
L(u) N(u) f (x) 0
According to [16], we can construct a homotopy H:[0,1]which satisfies
the homotopy equation:
๐ป(๐ฃ, ๐) = (1 โ ๐)[๐ฟ(๐ฃ) โ ๐ฟ(๐ข0)] + ๐[๐ด(๐ฃ) โ ๐(๐ฅ)] = 0
Or
๐ป(๐ฃ, ๐) = ๐ฟ(๐ฃ(๐ฅ)) โ ๐ฟ(๐ข0(๐ฅ)) + ๐๐ฟ(๐ข0(๐ฅ)) + ๐[๐(๐ฃ(๐ฅ)) โ ๐(๐ฅ)] = 0 (1.3)
where p[0,1], u0 is an initial approximation solution of equation (1.2).
Obviously, from equation (1.3) we have:
H (v, 0) L(v) L(u0)
H (v,1) A(v) f (x) 0
The changing process of p from zero to unity is just that of v(x, p) from u0(x) to
u(x).
Therefore, L(v) L(u0) A(v) f (x), x
and u0(x) u(x) , x
Assume that the solution of equation (1.2) can be written as a power series in p as
follows:
Preliminaries Chapter One
12
๐ฃ(x, p) = โ p๐โ๐=0 vi(x) = ๐ฃ0 + ๐๐ฃ1 + ๐2๐ฃ2 + โฏ (1.4)
Setting p =1 in equation (1.4), can get:
๐ข(๐ฅ) = โ v๐(x)
โ
๐=0
= lim๐โ1
๐ฃ = ๐ฃ0 + ๐ฃ1 + ๐ฃ2 + โฏ (1.5)
This is the solution of equation (1.2)
To illustrate the efficiency of method, consider the following examples.
Example 1.2[55]
Consider 2nd order linear homogeneous Klein-Gordan equation.
๐ข๐ก๐ก = ๐ข๐ฅ๐ฅ + ๐ข๐ฅ + 2๐ข , โโ < ๐ฅ < โ , ๐ก > 0
Subject to the IC: ๐ข(๐ฅ, 0) = ๐๐ฅ , ๐ข๐ก(๐ฅ, 0) = 0
Using the HPM we have
๐ป(๐ฃ, ๐) = (1 โ ๐)(๐ฃ๐ก๐ก โ ๐ข0๐ก๐ก) + ๐(๐ฃ๐ก๐ก โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฅ โ 2๐ฃ)
= ๐ฃ๐ก๐ก โ ๐ข0๐ก๐ก + ๐(๐ข0๐ก๐ก โ ๐ฃ๐ฅ๐ฅ โ ๐ฃ๐ฅ โ 2๐ฃ) = 0
๐0: ๐ฃ0๐ก๐ก โ ๐ข0๐ก๐ก = 0
๐ฃ0๐ก = ๐ข0๐ก โน ๐ฃ0 = ๐ข0 = ๐๐ฅ
๐1: ๐ฃ1๐ก๐ก = (๐ฃ0๐ฅ๐ฅ + ๐ฃ0๐ฅ + 2๐ฃ0 โ ๐ข0๐ก๐ก) = ๐๐ฅ + ๐๐ฅ + 2๐๐ฅ = 4๐๐ฅ
๐ฃ1๐ก = 4๐ก๐๐ฅ โน ๐ฃ1 = 4๐ก2
2!๐๐ฅ
๐2: ๐ฃ2๐ก๐ก = (๐ฃ1๐ฅ๐ฅ + ๐ฃ1๐ฅ + 2๐ฃ1) = 4๐ก2
2!๐๐ฅ + 4
๐ก2
2!๐๐ฅ + 8
๐ก2
2!๐๐ฅ = 16
๐ก2
2!๐๐ฅ
๐ฃ2๐ก = 16๐ก3
3!๐๐ฅ โน ๐ฃ2 = 16
๐ก4
4!๐๐ฅ
Preliminaries Chapter One
13
๐3: ๐ฃ3๐ก๐ก = (๐ฃ2๐ฅ๐ฅ + ๐ฃ2๐ฅ + 2๐ฃ2) = 16๐ก4
4!๐๐ฅ + 16
๐ก4
4!๐๐ฅ + 32
๐ก4
4!๐๐ฅ = 64
๐ก4
4!๐๐ฅ
๐ฃ3๐ก = 64๐ก5
5!๐๐ฅ โน ๐ฃ3 = 64
๐ก6
6!๐๐ฅ
And so on, to get
๐ข(๐ฅ, ๐ก) = lim๐โ1
๐ฃ = ๐ฃ0 + ๐ฃ1 + ๐ฃ2 + ๐ฃ3 + โฏ
๐ข(๐ฅ, ๐ก) = ๐๐ฅ + 4๐ก2
2!๐๐ฅ + 16
๐ก4
4!๐๐ฅ + 64
๐ก6
6!๐๐ฅ + โฏ
๐ข(๐ฅ, ๐ก) = ๐๐ฅ (1 +(2๐ก)2
2!+
(2๐ก)4
4!+
(2๐ก)6
6!+ โฏ ) = ๐๐ฅ ๐๐๐ โ(2๐ก)
Example 1.3[60]
.PDE order nonlinear rdfollowing 3onsider the C
๐ข๐ก +1
2๐ข๐ฅ
2 = ๐ข๐ฅ๐ฅ๐ก , โโ < ๐ฅ < โ , ๐ก > 0
Subject to the initial condition (IC): ๐ข(๐ฅ, 0) = ๐ฅ
Using the HPM we have:
๐ป(๐ฃ, ๐) = (1 โ ๐)(๐ฃ๐ก โ ๐ข0๐ก) + ๐ (๐ฃ๐ก +1
2(๐ฃ2)๐ฅ โ ๐ฃ๐ฅ๐ฅ๐ก)
= ๐ฃ๐ก โ ๐ข0๐ก + ๐ (๐ข0๐ก +1
2(๐ฃ2)๐ฅ โ ๐ฃ๐ฅ๐ฅ๐ก) = 0
๐0: ๐ฃ0๐ก โ ๐ข0๐ก = 0 , ๐ฃ0 = ๐ฅ
๐1: ๐ฃ1๐ก = โ (๐ข0๐ก +1
2(๐ฃ0
2)๐ฅ โ ๐ฃ0๐ฅ๐ฅ๐ก) , ๐ฃ1 = โ๐ฅ๐ก
๐2: ๐ฃ2๐ก = โ (1
2(2๐ฃ0๐ฃ1)๐ฅ โ ๐ฃ1๐ฅ๐ฅ๐ก) , ๐ฃ2 = ๐ฅ๐ก2
Preliminaries Chapter One
14
๐3: ๐ฃ3๐ก = โ (1
2(2๐ฃ0๐ฃ2 + ๐ฃ1
2)๐ฅ โ ๐ฃ2๐ฅ๐ฅ๐ก) , ๐ฃ3 = โ๐ฅ๐ก3
And so on, to get:
๐ข(๐ฅ, ๐ก) = โ ๐๐๐ฃ๐(๐ฅ, ๐ก) = ๐ฅ โ(โ๐ก)๐ = ๐ฅ
1 + ๐ก
โ
๐=0
โ
๐=0
This is exact solution
Example 1.4[60]
.PDE order linear rdfollowing 3onsider the C
๐ข๐ก + ๐ข๐ฅ = 2๐ข๐ฅ๐ฅ๐ก , โโ < ๐ฅ < โ , ๐ก > 0
Subject to the IC: ๐ข(๐ฅ, 0) = ๐โ๐ฅ
Using HPM we have:
๐ป(๐ฃ, ๐) = (1 โ ๐)(๐ฃ๐ก โ ๐ข0๐ก) + ๐(๐ฃ๐ก + ๐ฃ๐ฅ โ 2๐ฃ๐ฅ๐ฅ๐ก)
= ๐ฃ๐ก โ ๐ข0๐ก + ๐(๐ข0๐ก + ๐ฃ๐ฅ โ 2๐ฃ๐ฅ๐ฅ๐ก) = 0
๐0: ๐ฃ0๐ก โ ๐ข0๐ก = 0 , ๐ฃ0 = ๐โ๐ฅ
๐1: ๐ฃ1๐ก = 2๐ฃ0๐ฅ๐ฅ๐ก โ ๐ฃ0๐ฅ โ ๐ข0๐ก , ๐ฃ1 = ๐ก๐โ๐ฅ = ๐1(๐ก)๐โ๐ฅ
๐2: ๐ฃ2๐ก = 2๐ฃ1๐ฅ๐ฅ๐ก โ ๐ฃ1๐ฅ , ๐ฃ2 = (2๐ก +๐ก2
2) ๐โ๐ฅ = ๐2(๐ก)๐โ๐ฅ
๐3: ๐ฃ3๐ก = 2๐ฃ2๐ฅ๐ฅ๐ก โ ๐ฃ2๐ฅ , ๐ฃ3 = (4๐ก + 2๐ก2 +๐ก3
3!) ๐โ๐ฅ = ๐3(๐ก)๐โ๐ฅ
And so on, where ๐๐(๐ก) is a polynomial which has the following form:
๐๐(๐ก) = 2๐โ1๐ก + โ ๐๐๐๐ก๐
๐
๐=0
; ๐๐2 , ๐๐3, โฆ โฆ โฆ , ๐๐๐ > 0
By induction, we get:
Preliminaries Chapter One
15
๐ฃ๐ = ๐๐(๐ก)๐โ๐ฅ
And
๐ฃ๐+1 = ๐ฟ๐กโ1[2๐ฃ๐๐ฅ๐ฅ๐ก โ ๐ฃ๐๐ฅ] = ๐ฟ๐ก
โ1[2๐ฟ๐ก๐๐(๐ก) + ๐๐(๐ก)]๐โ๐ฅ
= [2๐๐(๐ก) + ๐ฟ๐กโ1๐๐(๐ก)]๐โ๐ฅ โก ๐๐+1(๐ก)๐โ๐ฅ
It is easy to see that ๐๐(๐ก) โ +โ (๐ โ +โ) for any ๐ก > 0. Therefore, the infinite
series:
โ ๐ฃ๐
โ
๐=0
= [1 + โ ๐๐(๐ก)
โ
๐=1
] ๐โ๐ฅ
Is divergent.
We note that in example 1.3, the PDE is nonlinear and the method gave the exact
solution, in example 1.4, the PDE is linear but the method miss fire to get the exact
solution. For these reasons we suggest efficient method based on coupled new
transformation with HPM and denoted by NTHPM. Now, firstly introduce the new
transformation proposed by luma-Alaa [51].
1.4. New Transformation
In this section, a new integral transformation is introduced. The
domain of the new transformation (NT) is wider than of the domain of
other transformation; therefore, it is more widely used to solve problems.
Definition 1.2 [51]
The new transformation of a function f(t) is defined by:
Preliminaries Chapter One
16
๐(๐ข) = ๐{๐(๐ก)} = โซ ๐โ๐ก๐ (๐ก
๐ข) ๐๐ก
โ
0
, (1.6)
Where u is a real number, for those values of u which the improper integral is
finite. A list of the NT for common functions is presented in the Table (1.1).
Table 1.1: New transformation for some common functions[51]
f(t) ๐(๐ข) = ๐{๐(๐ก)} ๐ท๐
n=0,1,โฆ, nt ๐!
๐ข๐ ๐ข โ 0
, a>0 at au(a+1)/ u0
ate ๐ข
๐ขโ๐
uR\[0,a] a0 uR\[a,0]
a<0
sin(at) ๐๐ข
๐2 + ๐ข2 ๐ข โ 0
cos(at) ๐ข2
๐2 + ๐ข2 ๐ข โ 0
sinh(at) โ๐๐ข
๐2 โ ๐ข2 |๐ข| > |๐|
cosh(at) โ๐ข2
๐2 โ ๐ข2 |๐ข| > |๐|
)a-t)=H(a-t)=u(t(au au-e u>0
(t-a) /uau-e u>0
1.4.1. The General Properties of the New Transformation
If the new transformation ๐{๐} and ๐{๐} of the functions f (t) and g(t) are well-
defined and a, b are constants, then the following properties are hold:
1. Linearity property: ๐{๐๐(๐ก) + ๐๐(๐ก)} = ๐๐{๐(๐ก)} + ๐๐{๐(๐ก)} (1.7)
Preliminaries Chapter One
17
2. Convolution property: (๐ โ ๐)(๐ก) = โซ ๐(๐)๐(๐ก โ ๐)๐๐๐ก
0 (1.8)
3. ๐{๐ก๐} = ๐!
๐ฃ๐ , ๐ฃ โ 0 , ๐ = 0,1,2,3, โฆ (1.9)
4. Differentiation property: ๐{๐โฒ} = ๐ฃ(๐{๐} โ ๐(0)) (1.10)
For more details see [51].
1.4.2. The Advantages of the New Transform
The NT has many interesting properties which make it rival to the Laplace
Transform (LT). Some of these properties are:
1. The domain of the NT is wider than or equal to the domain of LT as illustrate in
Table (1.2). This feature makes the NT more widely used for problems.
2. Depending on [51], the NT has the duality with LT, therefore, the NT can be
solve all the problems which be solved by LT.
3. The unit step function in the t-domain is transformed to unity in the u-domain.
4. The differentiation and integration in the tโdomain are equivalent to
multiplication and division of the transformed function F(u) by u in the uโ
domain.
5. By Linear property (1.7), we have that for any constant aR, ๐{๐} = ๐๐{1} =
๐, and hence, ๐โ1{๐} = ๐, that is, we donโt have any problem when we dealing
with the constant term( the constant with respect to the parameter u).
Preliminaries Chapter One
18
Table 1.2: The domain of Laplace and new transformation[51]
In the next chapter, we will use a combination of new transformation (NT)
and the HPM to solve types of PDEs and get exact solution without needing
computer calculations.
f(t) Laplace New Transform
[ ๐ (f)](s) Domain [ ๐ (f)](u) Domain
tn n=0,1,2,โฆ n!/sn+1 s>0 n!/un u0
ta a>0 (a+1)/sa+1 s>0 (a+1)/ua u0
eat 1/(s-a) s>a u/(u-a)
uR/[0,a] if
a0 uR/[a,0]
if a<0
sin(at) a/(s2+a2) s>0 au/(u2+a2) u0
cos(at) s/(s2+a2) s>0 u2/(u2+a2) u0
sinh(at) a/(s2-a2) s>|a| au/(u2-a2) |u|>|a|
cosh(at) s/(s2-a2) s>|a| u2/(u2-a2) |u|>|a|
ua(t)=u(t-a)=H(t-
a) e-as/s s>0 e-au u>0
(t-a) e-as s>0 e-au/u u>0
ln(at) a>0 (ln(a/s)-ฮณ) /s s>0 ln(a/u)-ฮณ u>0
๐พ = โ โซ ๐โ๐ก๐๐๐ก ๐๐ก
โ
0
โ 0.5772 โฆ
Couple New Transform
with HPM to Solve Some
Types of Partial
Differential Equations
20
2.1. Introduction
In this chapter, we will introduce an approach in obtaining the exact
solution of non-linear partial differential equations. The new approach based
on combine two methods, new transform with HPM and denoted by NTHPM.
Then applied it to solve some important model equations. The exact solutions
of these equations are compared to the HPM. The comparisons show the
efficiency of the proposed NTHPM against the other methods. The method is
strongly and powerful to treatment the nonlinear term of nonlinear equations.
2.2. New TransformationโHomotopy Perturbation Method
The NTHPM is a new method to solve differential equation; it
successfully applied to solve types of PDEs. This method is powerful to obtain
the exact solution without using computer calculating. The method suggested
firstly by Tawfiq and Jabber in 2018 [18] to solve groundwater equation. The
NTHPM has many merits and has many advantages over the HPM and ADM.
In the present work, the suggested method is used to solve of 1D, 2D, and 3D;
Chapter Two
Couple New Transform with HPM to Solve
Some Types of Partial Differential Equations
Chapter Two NTHPM to Solve Types of PDEs
21
2nd and 3rd order non-linear PDEs equations, and comparison has been made to
the results obtained by the HPM and NTHPM.
2.3. Solve Linear PDEs by NTHPM
In this section, we will use a combination of new transform (NT) and the
HPM to get the new transform that has played an important role because its
theoretical interest also in such method that allows to solve in the simplest
form; it used have to accelerate the convergence of power series.
To illustrate the ideas of NTHPM to find the exact solution of linear three
dimensions 2nd order PDEs of the form:
๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ + ๐ข๐ง๐ง = ๐ผ๐ข๐ก ; ๐ฅ, ๐ฆ, ๐ง โ ๐ & ๐ก > 0 (2.1)
with initial condition (IC): ๐ข(๐ฅ, ๐ฆ, ๐ง, 0) = ๐(๐ฅ, ๐ฆ, ๐ง); ฮฑ is constant.
Firstly, rewrite equation (2.1) as following:
๐ฟ[๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก)] + ๐ [๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก)] = ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) (2.2)
where L: is the linear differential operator (๐ฟ = ๐ผ๐
๐๐ก), ๐ : is the remainder of
the linear operator, g(๐ฅ, ๐ฆ, ๐ง, ๐ก) is the inhomogeneous part.
We construct a Homotopy as:
๐ป(๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก)) โ ๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ง, 0))] +
๐ [๐ด[๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก)] โ ๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)] = 0 (2.3)
Where p[0, 1] is an embedding parameter and A defined as A= L+ R.
Chapter Two NTHPM to Solve Types of PDEs
22
It is clear that, if p =1, then the homotopy equation (2.3) is converted to the
differential equation (2.2).
Substituting equation (2.2) into equation (2.3) and rewrite it as:
๐ฟ(๐ข) โ ๐ฟ(๐) โ ๐๐ฟ(๐ข) + ๐๐ฟ(๐) + ๐๐ฟ(๐ข) + ๐๐ (๐ข) โ ๐ ๐ = 0
Then
๐ฟ(๐ข) โ ๐ฟ(๐) + ๐[๐ฟ(๐) + ๐ (๐ข) โ ๐] = 0 (2.4)
Since f(x, y, z) is independent of the variable t and the linear operator L
dependent on t so, L(f(x, y, z)) = 0, i.e., the equation (2.4) becomes:
๐ฟ(๐ข) + ๐๐ (๐ข) โ ๐ ๐ = 0 (2.5)
According to the classical perturbation technique, the solution of the equation
(2.5) can be written as a power series of embedding parameter p, as follows:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
(2.6)
The convergence of series (2.6) at p =1 is discussed and proved in [8,25],
which satisfies the differential equation (2.2).
The final step is determining the parts un (n= 0,1, 2,โฆ) to get the solution
u(๐ฅ, ๐ฆ, ๐ง, ๐ก).
Here, we couple the NT with HPM as follows:
Taking the NT (with respect to the variable t) for the equation (2.5) to get:
๐{๐ฟ(๐ข)} + ๐ ๐{๐ (๐ข)} โ ๐ ๐{๐} = 0 (2.7)
Now by using the differentiation property of NT (property 4) and equations
(2.2), (2.7) becomes:
Chapter Two NTHPM to Solve Types of PDEs
23
๐ฃ๐ผ๐{๐ข} โ ๐ฃ๐ผ๐(๐ฅ) + ๐ ๐{๐ (๐ข)} โ ๐ ๐{๐} = 0 (2.8)
Hence:
๐{๐ข} = ๐(๐ฅ, ๐ฆ, ๐ง) โ ๐๐{๐ (๐ข)}
๐ฃ๐ผ+ ๐
๐{๐}
๐ฃ๐ผ (2.9)
Taking the inverse of the NT on both sides of equation (2.9), to get:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐(๐ฅ, ๐ฆ, ๐ง) โ ๐๐โ1 {๐{๐ (๐ข(๐ฅ,๐ฆ,๐ง,๐ก))}
๐ฃ๐ผ} + ๐๐โ1 {
๐{๐(๐ฅ,๐ฆ,๐ง,๐ก)}
๐ฃ๐ผ} (2.10)
Then substituting equation (2.6) into equation (2.10) to obtain:
โ๐๐๐ข๐
โ
๐=0
= ๐(๐ฅ, ๐ฆ, ๐ง) โ ๐๐โ1 {๐ {๐ (โ ๐๐๐ข๐
โ๐=0 )}
๐ฃ๐ผ} + ๐ ๐โ1 {
๐{๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)}
๐ฃ๐ผ} (2.11)
By comparing the coefficient of powers of p in both sides of the equation
(2.11), we have:
๐ข0 = ๐(๐ฅ, ๐ฆ, ๐ง)
๐ข1 = โ๐โ1 {
๐{๐ [๐ข0]}
๐ฃ๐ผ} + ๐โ1 {
๐{๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)}
๐ฃ๐ผ}
๐ข2 = โ๐โ1 {
๐{๐ [๐ข1]}
๐ฃ๐ผ}
๐ข3 = โ๐โ1 {
๐{๐ [๐ข2]}
๐ฃ๐ผ} (2.12)
โฎ
๐ข๐+1 = โ๐โ1 {
๐{๐ [๐ข๐]}
๐ฃ๐ผ}
Chapter Two NTHPM to Solve Types of PDEs
24
2.4. Illustrative Application
Here, the NTHPM may be used to solve the 2nd order-PDE with initial
condition as following:
Example 2.1[60]
Let us consider the following 3D - PDE
๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ + ๐ข๐ง๐ง = ๐ผ๐ข๐ก ; ๐๐๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ ๐ & ๐ก > 0
Subject to IC: ๐ข(๐ฅ, ๐ฆ, ๐ง, 0) = ๐(๐ฅ, ๐ฆ, ๐ง) = 5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง),
where a, b, c and ๐ผ are constants. According to the equation (2.12) the power
series of p get as follows:
๐0: ๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)
๐1: ๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ(5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก
๐ผ) (๐2 + ๐2 + ๐2)
๐2: ๐ข2(๐ฅ, ๐ฆ, ๐ง, ๐ก) = (5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก2
2 ๐ผ2) (๐2 + ๐2 + ๐2)2
๐3: ๐ข3(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ(5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก3
3! ๐ผ3) (๐2 + ๐2 + ๐2)3
๐4: ๐ข4(๐ฅ, ๐ฆ, ๐ง, ๐ก) = (5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก4
4! ๐ผ4) (๐2 + ๐2 + ๐2)4
๐5: ๐ข5(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ(5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก5
5! ๐ผ5) (๐2 + ๐2 + ๐2)5
โ
๐๐: ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = (โ1)๐(5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง)) (๐ก๐
๐! ๐ผ๐) (๐2 + ๐2 + ๐2)๐
Thus, we get the following series form:
Chapter Two NTHPM to Solve Types of PDEs
25
(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
= โ(โ1)๐5 ๐ ๐๐(๐๐ฅ) ๐ ๐๐(๐๐ฆ) ๐ ๐๐(๐๐ง) (๐ก๐
๐! ๐ผ๐) (๐2 + ๐2 + ๐2)๐
โ
๐=0
So, the closed form of the above series is:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = 5 sin(๐๐ฅ) sin(๐๐ฆ) sin(๐๐ง) ๐โ๐ก๐ผ(๐2+๐2+๐2)
This gives an exact solution of the problem.
Example 2.2[60]
Consider the following 3D-PDE
๐ข๐ฅ๐ฅ + ๐ข๐ฆ๐ฆ + ๐ข๐ง๐ง + ๐๐ฅ+๐ฆ = ๐ผ๐ข๐ก ; ๐๐๐ ๐ฅ, ๐ฆ, ๐ง ๐๐ ๐ & ๐ก > 0
with IC: ๐ข(๐ฅ, ๐ฆ, ๐ง, 0) = ๐(๐ฅ, ๐ฆ, ๐ง) = ๐, where d is constants.
From equation (2.12), we get the power series of p as follows:
๐0: ๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐
๐1: ๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ก
๐ผ(๐๐ฅ+๐ฆ)
๐2: ๐ข2(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ก2
2๐ผ2(๐๐ฅ+๐ฆ)
๐3: ๐ข3(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ก3
6๐ผ3(๐๐ฅ+๐ฆ)
๐4: ๐ข4(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ก4
24๐ผ4(๐๐ฅ+๐ฆ)
โ
๐๐: ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐ก๐
๐!๐ผ๐(๐๐ฅ+๐ฆ)
Thus, the following series form is obtained:
Chapter Two NTHPM to Solve Types of PDEs
26
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) =
โ
๐=0
๐ +โ๐ก๐+1
(๐ + 1)! ๐ผ๐+1(๐๐ฅ+๐ฆ)
โ
๐=0
Therefore, the closed form of the above series is:
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐๐ฅ+๐ฆ (๐๐ก
๐ผ โ 1) + ๐
Example 2.3[55]
Consider another linear homogeneous Klein-Gordan equation
๐ข๐ก๐ก = ๐ข๐ฅ๐ฅ + ๐ข๐ฅ + 2๐ข , โโ < ๐ฅ < โ , ๐ก > 0
Subject to the ICs:
๐ข(๐ฅ, 0) = ๐๐ฅ , ๐ข๐ก(๐ฅ, 0) = 0
Taking new transformation on both sides, subject to the IC, to get:
๐[๐ข(๐ฅ, ๐ฆ, ๐ก)] = ๐๐ฅ +1
๐ฃ2๐[๐ข๐ฅ๐ฅ + ๐ข๐ฅ + 2๐ข ]
Taking inverse of new transformation, we get:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐๐ฅ + ๐โ1 [1
๐ฃ2๐[๐ข๐ฅ๐ฅ + ๐ข๐ฅ + 2๐ข]]
by HPM, we get:
๐ข(๐ฅ, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
Using equation (2.11) in equation (2.10), to get:
Chapter Two NTHPM to Solve Types of PDEs
27
โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
= ๐๐ฅ
+ ๐(๐โ1 [1
๐ฃ2๐ [โ๐๐๐ข๐๐ฅ๐ฅ(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
+โ๐๐๐ข๐๐ฅ(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
+ 2โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
]])
๐0: ๐ข0(๐ฅ, ๐ก) = ๐๐ฅ
๐1: ๐ข1(๐ฅ, ๐ก) =(2๐ก)2
2!๐๐ฅ
๐2: ๐ข2(๐ฅ, ๐ก) =(2๐ก)4
4! ๐๐ฅ
๐3: ๐ข3(๐ฅ, ๐ก) =(2๐ก)6
6!๐๐ฅ
:
๐ข(๐ฅ, ๐ก) = ๐๐ฅ (1 +(2๐ก)2
2!+(2๐ก)4
4!+(2๐ก)6
6!+ โฏ) = ๐๐ฅ ๐๐๐ โ(2๐ก)
2.5. Solve Nonlinear PDE by NTHPM
In the NTHPM can be used for solving various types of nonlinear PDEs. To
illustrate the basic idea of suggested method, we consider general nonlinear
PDEs with the initial conditions of the form:
๐ฟ๐ข(๐ฅ, ๐ก) + ๐ ๐ข(๐ฅ, ๐ก) + ๐๐ข(๐ฅ, ๐ก) = ๐(๐ฅ, ๐ก) , (2.13)
Subject to ICs: ๐ข(๐ฅ, 0) = โ(๐ฅ) , ๐ข๐ก(๐ฅ, 0) = ๐(๐ฅ) .
Chapter Two NTHPM to Solve Types of PDEs
28
where L is the 2nd order linear differential operator ๐ฟ =๐2
๐๐ก2 , R is the linear
differential operator of less order than L ; N represents the general nonlinear
differential operator and ๐(๐ฅ, ๐ก)is the source term.
Taking the new transformation on both sides of equation (2.13) to get:
๐[๐ฟ๐ข(๐ฅ, ๐ก)] + ๐[๐ ๐ข(๐ฅ, ๐ก)] + ๐[๐๐ข(๐ฅ, ๐ก)] = ๐[๐(๐ฅ, ๐ก)] , (2.14)
Using the differentiation property of the new transform, we have:
๐[๐ข(๐ฅ, ๐ก)] = โ(๐ฅ) +๐(๐ฅ)
๐ฃ+1
๐ฃ2๐[๐(๐ฅ, ๐ก) โ ๐ ๐ข(๐ฅ, ๐ก) โ ๐๐ข(๐ฅ, ๐ก)] (2.15)
Operating with the inverse of new transformation on both sides of equation
(2.15) gives:
๐ข(๐ฅ, ๐ก) = ๐บ(๐ฅ, ๐ก) โ ๐โ1 [1
๐ฃ2๐[๐ ๐ข(๐ฅ, ๐ก) โ ๐๐ข(๐ฅ, ๐ก)]] (2.16)
where G(x, t) represents the term arising from the source term and the
prescribed initial conditions. Now, we apply the HPM.
๐ข(๐ฅ, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
(2.17)
The nonlinear term can be decomposed as:
๐๐ข(๐ฅ, ๐ก) = โ๐๐๐ป๐(๐ข)
โ
๐=0
(2.18)
for He's polynomials Hn(u) (see [48-49]) that are given by:
๐ป๐(๐ข0, ๐ข1, โฆโฆ , ๐ข๐) =1
๐!
๐๐
๐๐๐[๐(โ ๐๐๐ข๐
โ๐=0 )], ๐ = 0,1,2,3, โฆ (2.19)
Chapter Two NTHPM to Solve Types of PDEs
29
Substituting equation (2.17), (2.18) and (2.19) in equation (2.16) to get:
โ ๐๐๐ข๐(๐ฅ, ๐ก)โ๐=0 = G(x, t) โ p (๐โ1 [
1
v2๐[Rโ pnun(x, t)
โn=0 +Nโ pnHn(u)
โn=0 ]]) (2.20)
Comparing the coefficient of powers of p, to get:
๐0: ๐ข0(๐ฅ, ๐ก) = ๐บ(๐ฅ, ๐ก)
๐1: ๐ข1(๐ฅ, ๐ก) = โ๐โ1 [
1
๐ฃ2๐[๐ ๐ข0(๐ฅ, ๐ก) + ๐ป0(๐ข)]],
๐2: ๐ข2(๐ฅ, ๐ก) = โ๐โ1 [
1
๐ฃ2๐[๐ ๐ข1(๐ฅ, ๐ก) + ๐ป1(๐ข)]],
๐3: ๐ข3(๐ฅ, ๐ก) = โ๐โ1 [
1
๐ฃ2๐[๐ ๐ข2(๐ฅ, ๐ก) + ๐ป2(๐ข)]],
and so on.
2.6. Applications
In this section, NTHPM is applied for solving various types of nonlinear
wave-like equations with variable coefficients.
Example 2.4[10]
Consider the following 2D- nonlinear wave-like equations with variable
coefficients.
๐ข๐ก๐ก =๐2
๐๐ฅ๐๐ฆ(๐ข๐ฅ๐ฅ๐ข๐ฆ๐ฆ) โ
๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ๐ข๐ฅ๐ข๐ฆ) โ ๐ข
with the ICs: ๐ข(๐ฅ, ๐ฆ, 0) = ๐๐ฅ๐ฆ , ๐ข๐ก(๐ฅ, ๐ฆ, 0) = ๐๐ฅ๐ฆ
Taking new transformation on both sides, subject to the IC, we get:
๐[๐ข(๐ฅ, ๐ฆ, ๐ก)] = ๐๐ฅ๐ฆ +๐๐ฅ๐ฆ
๐ฃ+
1
๐ฃ2๐ [
๐2
๐๐ฅ๐๐ฆ(๐ข๐ฅ๐ฅ๐ข๐ฆ๐ฆ) โ
๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ๐ข๐ฅ๐ข๐ฆ) โ ๐ข ]
Taking inverse of new transform, we have:
Chapter Two NTHPM to Solve Types of PDEs
30
๐ข(๐ฅ, ๐ฆ, ๐ก) = (1 + ๐ก)๐๐ฅ๐ฆ โ ๐โ1 [1
๐ฃ2๐ [
๐2
๐๐ฅ๐๐ฆ(๐ข๐ฅ๐ฅ๐ข๐ฆ๐ฆ) โ
๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ๐ข๐ฅ๐ข๐ฆ) โ ๐ข]]
By HPM, we have:
๐ข(๐ฅ, ๐ก) = โ ๐๐๐ข๐(๐ฅ, ๐ก)โ๐=0
Substation equation (2.13) in the equation (2.14), to get:
โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
= (1 + ๐ก)๐๐ฅ๐ฆ
+ ๐(๐โ1 [1
๐ฃ2๐ [โ๐๐๐ป๐(๐ข)
โ
๐=0
โโ๐๐๐พ๐(๐ข)
โ
๐=0
โโ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
]])
Where ๐ป๐(๐ข) and ๐พ๐(๐ข) are the He's polynomials having the value ๐ป๐(๐ข) =
๐2
๐๐ฅ๐๐ฆ(๐ข๐ฅ๐ฅ๐ข๐ฆ๐ฆ) and ๐พ๐(๐ข) =
๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ๐ข๐ฅ๐ข๐ฆ)
The first few components of ๐ป๐(๐ข) and ๐พ๐(๐ข) are given by:
๐ป0(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ข0๐ฅ๐ฅ๐ข0๐ฆ๐ฆ) =
๐2
๐๐ฅ๐๐ฆ[(1 + ๐ก)2๐ฅ2๐ฆ2๐2๐ฅ๐ฆ]
๐ป1(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ข1๐ฅ๐ฅ๐ข0๐ฆ๐ฆ + ๐ข0๐ฅ๐ฅ๐ข1๐ฆ๐ฆ)
=๐2
๐๐ฅ๐๐ฆ[โ2(1 + ๐ก) (
๐ก2
2!+๐ก3
3!) ๐ฅ2๐ฆ2๐2๐ฅ๐ฆ]
๐ป2(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ข2๐ฅ๐ฅ๐ข0๐ฆ๐ฆ + ๐ข1๐ฅ๐ฅ๐ข1๐ฆ๐ฆ + ๐ข0๐ฅ๐ฅ๐ข2๐ฆ๐ฆ)
And so on
๐พ0(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ(๐ข0๐ฅ๐ข0๐ฆ)) =
๐2
๐๐ฅ๐๐ฆ[(1 + ๐ก)2๐ฅ2๐ฆ2๐2๐ฅ๐ฆ]
๐พ1(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ(๐ข1๐ฅ๐ข0๐ฆ + ๐ข0๐ฅ๐ข1๐ฆ))
Chapter Two NTHPM to Solve Types of PDEs
31
=๐2
๐๐ฅ๐๐ฆ[โ2(1 + ๐ก) (
๐ก2
2!+๐ก3
3!) ๐ฅ2๐ฆ2๐2๐ฅ๐ฆ]
๐พ2(๐ข) =๐2
๐๐ฅ๐๐ฆ(๐ฅ๐ฆ(๐ข2๐ฅ๐ข0๐ฆ + ๐ข1๐ฅ๐ข1๐ฆ + ๐ข0๐ฅ๐ข2๐ฆ))
:
Comparing the coefficients of various powers of p, to get:
๐0: ๐ข0(๐ฅ, ๐ฆ, ๐ก) = ((1 + ๐ก)๐๐ฅ๐ฆ
๐1: ๐ข1(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 [
1
๐ฃ2๐[๐ป0(๐ข) + ๐พ0(๐ข) โ ๐ข0(๐ฅ, ๐ฆ, ๐ก)]]
= โ(๐ก2
2!+๐ก3
3!) ๐๐ฅ๐ฆ
๐2: ๐ข2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 [
1
๐ฃ2๐[๐ป1(๐ข) + ๐พ1(๐ข) โ ๐ข1(๐ฅ, ๐ฆ, ๐ก)]]
= (๐ก4
4!+๐ก5
5!) ๐๐ฅ๐ฆ
๐3: ๐ข3(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 [
1
๐ฃ2๐[๐ป2(๐ข) + ๐พ2(๐ข) โ ๐ข2(๐ฅ, ๐ฆ, ๐ก)]]
= โ(๐ก6
6!+๐ก7
7!) ๐๐ฅ๐ฆ
And so on
Therefore the solution is given by:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ข0(๐ฅ, ๐ฆ, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ก) + ๐ข2(๐ฅ, ๐ฆ, ๐ก) + ๐ข3(๐ฅ, ๐ฆ, ๐ก) + โฏ
= ๐๐ฅ๐ฆ (1 + ๐ก โ๐ก2
2!โ๐ก3
3!+๐ก4
4!+๐ก5
5!โ๐ก6
6!โ๐ก7
7!+โฏ)
So, ๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐๐ฅ๐ฆ(cos ๐ก + sin ๐ก)
Chapter Two NTHPM to Solve Types of PDEs
32
Example 2.5[10]
Consider the following nonlinear 1D- equation with variable coefficients.
๐ข๐ก๐ก = ๐ข2๐2
๐๐ฅ2(๐ข๐ฅ๐ข๐ฅ๐ฅ๐ข๐ฅ๐ฅ๐ฅ) + ๐ข๐ฅ
2๐2
๐๐ฅ2(๐ข๐ฅ
3) โ 18๐ข5 + ๐ข
with ICs: ๐ข(๐ฅ, 0) = ๐๐ฅ, ๐ข๐ก(๐ฅ, 0) = ๐๐ฅ
By applying NTHPM, we get:
โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
= (1 + ๐ก)๐๐ฅ
+ ๐(๐โ1 [1
๐ฃ2๐ [โ๐๐๐ป๐(๐ข)
โ
๐=0
โโ๐๐๐พ๐(๐ข)
โ
๐=0
+ 18โ๐๐๐ฝ๐(๐ข)
โ
๐=0
โโ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
]])
Where ๐ป๐(๐ข), ๐พ๐(๐ข)and ๐ฝ๐(๐ข) are He's polynomials. First few components of
He's polynomials are given by:
๐ป0(๐ข) = ๐ข02 ๐2
๐๐ฅ2(๐ข0๐ฅ๐ข0๐ฅ๐ฅ๐ข0๐ฅ๐ฅ๐ฅ) = 9(1 + ๐ก)
5๐5๐ฅ
๐ป1(๐ข) = 2๐ข0๐ข1๐2
๐๐ฅ2(๐ข0๐ฅ๐ข0๐ฅ๐ฅ๐ข0๐ฅ๐ฅ๐ฅ) + ๐ข0
2 ๐2
๐๐ฅ๐๐ฆ(๐ข1๐ฅ๐ข0๐ฅ๐ฅ๐ข0๐ฅ๐ฅ๐ฅ +
๐ข0๐ฅ๐ข1๐ฅ๐ฅ๐ข0๐ฅ๐ฅ๐ฅ + ๐ข0๐ฅ๐ข0๐ฅ๐ฅ๐ข1๐ฅ๐ฅ๐ฅ) = 45(1 + ๐ก)4 (
๐ก2
2!+๐ก3
3!) ๐5๐ฅ
:
and
๐พ0(๐ข) = (๐ข0๐ฅ)2 ๐2
๐๐ฅ2[(๐ข0๐ฅ)
3] = (1 + ๐ก)2๐2๐ฅ๐2
๐๐ฅ2[(1 + ๐ก)3๐3๐ฅ]
= 9(1 + ๐ก)5๐5๐ฅ
Chapter Two NTHPM to Solve Types of PDEs
33
๐พ1(๐ข) = 2๐ข0๐ฅ๐ข1๐ฅ๐2
๐๐ฅ2(๐ข0๐ฅ)
3 + 3(๐ข0๐ฅ)2 ๐2
๐๐ฅ2[(๐ข0๐ฅ)
2๐ข1๐ฅ]
= 45(1 + ๐ก)4 (๐ก2
2!+๐ก3
3!) ๐5๐ฅ
:
and
๐ฝ0(๐ข) = (๐ข0)5 = (1 + ๐ก)5๐5๐ฅ
๐ฝ1(๐ข) = 5(๐ข0)4๐ข1 = 5(1 + ๐ก)
4 (๐ก2
2!+๐ก3
3!) ๐5๐ฅ
:
Comparing the coefficients of various powers of p, we get:
๐0: ๐ข0(๐ฅ, ๐ก) = (1 + ๐ก)๐๐ฅ
๐1: ๐ข1(๐ฅ, ๐ก) = ๐โ1 [1
๐ฃ2๐[๐ป0(๐ข) + ๐พ0(๐ข) โ 18๐ฝ0(๐ข) + ๐ข0(๐ฅ, ๐ฆ, ๐ก)]]
= (๐ก2
2!+๐ก3
3!) ๐๐ฅ
๐2: ๐ข2(๐ฅ, ๐ก) = ๐โ1 [1
๐ฃ2๐[๐ป1(๐ข) + ๐พ1(๐ข) โ 18๐ฝ1(๐ข) + ๐ข1(๐ฅ, ๐ฆ, ๐ก)]]
= (๐ก4
4!+๐ก5
5!) ๐๐ฅ
and so on, therefore the solution is given by:
๐ข(๐ฅ, ๐ก) = ๐ข0(๐ฅ, ๐ก) + ๐ข1(๐ฅ, ๐ก) + ๐ข2(๐ฅ, ๐ก) + ๐ข3(๐ฅ, ๐ก) + โฏ
= ๐๐ฅ (1 + ๐ก +๐ก2
2!+๐ก3
3!+๐ก4
4!+
๐ก5
5!+๐ก6
6!+๐ก7
7!+โฏ) = ๐๐ฅ+๐ก
This is the exact solution.
Chapter Two NTHPM to Solve Types of PDEs
34
Example 2.6[10]
Consider the following nonlinear 1D -3rd order wave-like equation with
variable coefficients.
๐ข๐ก๐ก = ๐ฅ2 ๐
๐๐ฅ(๐ข๐ฅ๐ข๐ฅ๐ฅ) โ ๐ฅ
2(๐ข๐ฅ๐ฅ2 ) โ ๐ข , 0 < ๐ฅ < 1 , ๐ก > 0
With ICs: ๐ข(๐ฅ, 0) = 0, ๐ข๐ก(๐ฅ, ๐ฆ, 0) = ๐ฅ2
By applying NTHPM, we get:
โ๐๐๐ข๐(๐ฅ, ๐ก)
โ
๐=0
= ๐ฅ2๐ก
+ ๐(๐โ1 [1
๐ฃ2๐ [๐ฅ2โ๐๐๐ป๐(๐ข)
โ
๐=0
โ ๐ฅ2โ๐๐๐พ๐(๐ข)
โ
๐=0
โโ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
]])
Where ๐ป๐(๐ข) and ๐พ๐(๐ข) are He's polynomials. First few components of He's
polynomials are given by:
๐ป0(๐ข) =๐
๐๐ฅ(๐ข0๐ฅ๐ข0๐ฅ๐ฅ) = 4๐ก
2
๐ป1(๐ข) =๐
๐๐ฅ(๐ข1๐ฅ๐ข0๐ฅ๐ฅ + ๐ข0๐ฅ๐ข1๐ฅ๐ฅ) = โ8
๐ก4
3!
๐ป2(๐ข) =๐2
๐๐ฅ(๐ข2๐ฅ๐ข0๐ฅ๐ฅ + ๐ข1๐ฅ๐ข1๐ฅ๐ฅ + ๐ข0๐ฅ๐ข2๐ฅ๐ฅ)
:
And
๐พ0(๐ข) = (๐ข0๐ฅ๐ฅ)2 = 4๐ก2
๐พ1(๐ข) = 2๐ข0๐ฅ๐ฅ๐ข1๐ฅ๐ฅ = โ8๐ก4
3!
๐พ2(๐ข) = (๐ข12)๐ฅ๐ฅ + 2(๐ข0)๐ฅ๐ฅ(๐ข2)๐ฅ๐ฅ
:
Chapter Two NTHPM to Solve Types of PDEs
35
Comparing the coefficients of various powers of p, to get:
๐0: ๐ข0(๐ฅ, ๐ก) = ๐ฅ2๐ก
๐1: ๐ข1(๐ฅ, ๐ก) = ๐โ1 [1
๐ฃ2๐[๐ฅ2๐ป0(๐ข) โ ๐ฅ
2๐พ0(๐ข) โ ๐ข0(๐ฅ, ๐ก)]] = โ๐ฅ2 ๐ก
3
3!
๐2: ๐ข2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 [
1
๐ฃ2๐[๐ฅ2๐ป1(๐ข) โ ๐ฅ
2๐พ1(๐ข) โ ๐ข1(๐ฅ, ๐ก)]] = ๐ฅ2 ๐ก
5
5!
:
Therefore, the solution is given by:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ข0(๐ฅ, ๐ฆ, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ก) + ๐ข2(๐ฅ, ๐ฆ, ๐ก) + ๐ข3(๐ฅ, ๐ฆ, ๐ก) + โฏ
= ๐ฅ2 (๐ก โ๐ก3
3!+๐ก5
5!โโฏ) = ๐ฅ2 sin ๐ก
This is the exact solution.
2.7. Solve Autonomous Equation by NTHPM
In this section the proposed method will be used to solve one of the most
important of amplitude equations is the autonomous equation which describes
the appearance of the stripe pattern in two dimensional systems. Moreover,
this equation was applied to a number of problems in variety systems, e.g.,
Rayleigh-Benard convection, Faraday instability, nonlinear optics, chemical
reactions and biological systems [60]. The approximate solutions of the
autonomous equation were presented by differential transformation method
[18], reduce differential transformation [8]. Here a reliable couple NTHPM is
applied for solving autonomous equation.
Chapter Two NTHPM to Solve Types of PDEs
36
To illustrate the ideas of NTHPM, firstly rewrite the initial value problem
in autonomous equation in the form:
๐ข๐ก(๐ฅ, ๐ก) = ๐๐ข๐ฅ๐ฅ(๐ฅ, ๐ก) + ๐1๐ข(๐ฅ, ๐ก) โ ๐2๐ข๐(๐ฅ, ๐ก) (2.20๐)
With the IC: ๐ข (๐ฅ, ๐ก) = ๐(๐ฅ) (2.20๐)
where ๐1 and ๐2 are real numbers and c and n are positive integers.
Taking new transformation on both sides of the equation (2.20a) and using the
linearity property of the new transformation gives:
๐{๐ข๐ก(๐ฅ, ๐ก)} = ๐๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} + ๐1๐{๐ข(๐ฅ, ๐ก)} โ ๐2๐{๐ข๐(๐ฅ, ๐ก)} (2.21)
By applying the differentiation property of new transform, we have
๐ฃ๐{๐ข(๐ฅ, ๐ก)} โ ๐ฃ๐ข(๐ฅ, 0)
= ๐๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} + ๐1๐{๐ข(๐ฅ, ๐ก)} โ ๐2๐{๐ข๐(๐ฅ, ๐ก)} (2.22)
Thus, we get:
(๐ฃ โ ๐1)๐{๐ข(๐ฅ, ๐ก)} = ๐ฃ๐(๐ฅ) + ๐๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ ๐2๐{๐ข๐(๐ฅ, ๐ก)}
๐{๐ข(๐ฅ, ๐ก)} =๐ฃ๐(๐ฅ)
๐ฃโ๐1+
๐
๐ฃโ๐1๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ
๐2
๐ฃโ๐1๐{๐ข๐(๐ฅ, ๐ก)}
Taking the inverse of new transformation on equation (2.22), we obtain:
๐ข(๐ฅ, ๐ก) = ๐โ1 {๐ฃ๐(๐ฅ)
๐ฃโ๐1} + ๐โ1 (
๐
๐ฃโ๐1๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ
๐2
๐ฃโ๐1๐{๐ข๐(๐ฅ, ๐ก)}) (2.23)
Chapter Two NTHPM to Solve Types of PDEs
37
In the HPM, the basic assumption is that the solutions can be written as a
power series in p:
๐ข(๐ฅ, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ก) (2.24)
โ
๐=0
and the nonlinear term N(u) = ๐ข๐, ๐ > 1 can be presented by an infinite
series as:
๐(๐ข) = โ ๐๐๐ป๐(๐ฅ, ๐ก) (2.25)โ๐=0
where p โ [0,1] is an embedding parameter. Hn(u) is He polynomials. Now,
substituting (2.24) and (2.25) in (2.23), to get:
โ๐๐๐ข๐(๐ฅ, ๐ก) =
โ
๐=0
๐โ1 {๐ฃ๐(๐ฅ)
๐ฃ โ ๐1}
+ ๐๐โ1 (๐
๐ฃ โ ๐1๐ {โ๐๐
โ
๐=0
๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ๐2
๐ฃ โ ๐1๐ {โ๐๐๐ป๐
โ
๐=0
} )
Comparing the coefficient of powers of p, the following are obtained.
๐0: ๐ข0(๐ฅ, ๐ก) = ๐โ1 {
๐ฃ๐(๐ฅ)
๐ฃ โ ๐1}
๐1: ๐ข1(๐ฅ, ๐ก) = ๐โ1 (
๐
๐ฃ โ ๐1๐{๐ข0๐ฅ๐ฅ(๐ฅ, ๐ก)} โ
๐2๐ฃ โ ๐1
๐{๐ป0})
๐2: ๐ข2(๐ฅ, ๐ก) = ๐โ1 (
๐
๐ฃ โ ๐1๐{๐ข1๐ฅ๐ฅ(๐ฅ, ๐ก)} โ
๐2๐ฃ โ ๐1
๐{๐ป1})
๐3: ๐ข3(๐ฅ, ๐ก) = ๐โ1 (๐
๐ฃ โ ๐1๐{๐ข2๐ฅ๐ฅ(๐ฅ, ๐ก)} โ
๐2๐ฃ โ ๐1
๐{๐ป2})
Chapter Two NTHPM to Solve Types of PDEs
38
Proceeding in this same manner, the rest of the components un(x, t) can be
completely obtained and the series solution is thus entirely determined.
Finally, we approximate the analytical solution ๐ข(๐ฅ, ๐ก) by truncated series:
๐ข(๐ฅ, ๐ก) = lim๐โโ
(โ๐ข๐(๐ฅ, ๐ก)
๐
๐=0
)
The above series solutions generally converge very rapidly.
2.8. Illustrative Examples
In this section, some non-linear autonomous equations with IC are
presented to show the advantages of the proposed method.
Example 2.7 [1]
Consider linear 1D, 2nd order autonomous equation.
๐ข๐ก(๐ฅ, ๐ก) = ๐ข๐ฅ๐ฅ(๐ฅ, ๐ก) โ 3๐ข(๐ฅ, ๐ก) (2.26)
With IC: ๐ข(๐ฅ, ๐ก) = ๐2๐ฅ
Taking the new transformation on both sides of equation (2.26), we have
๐{๐ข๐ก(๐ฅ, ๐ก)} = ๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ 3๐{๐ข(๐ฅ, ๐ก)}
By applying the differentiation property of new transformation, we get:
๐ฃ๐{๐ข(๐ฅ, ๐ก)} โ ๐ฃ๐ข(๐ฅ, 0) = ๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} โ 3๐{๐ข(๐ฅ, ๐ก)}
Thus, we have:
(๐ฃ + 3)๐{๐ข(๐ฅ, ๐ก)} = ๐ฃ๐2๐ฅ + ๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)}
Chapter Two NTHPM to Solve Types of PDEs
39
๐{๐ข(๐ฅ, ๐ก)} =๐ฃ๐2๐ฅ
๐ฃ+3+
1
๐ฃ+3๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} (2.27)
Taking the inverse new transformation on equation (2.27), to get:
๐ข(๐ฅ, ๐ก) = ๐โ1 {๐ฃ๐2๐ฅ
๐ฃ+3} + ๐โ1 (
1
๐ฃ+3๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)})
Now, applying the HPM, we get:
โ๐๐๐ข๐(๐ฅ, ๐ก) =
โ
๐=0
๐2๐ฅโ3๐ก + ๐๐โ1 (1
๐ฃ + 3๐ {โ๐๐
โ
๐=0
๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)})
Comparing the coefficients of powers of p, we have:
๐0: ๐ข0(๐ฅ, ๐ก) = ๐2๐ฅโ3๐ก
๐1: ๐ข1(๐ฅ, ๐ก) = ๐โ1 (
1
๐ฃ + 3๐{๐ข0๐ฅ๐ฅ(๐ฅ, ๐ก)}) = ๐
โ1 (1
๐ฃ + 3๐{4๐2๐ฅ๐โ3๐ก})
= 4๐2๐ฅ๐โ1 (๐ฃ
(๐ฃ + 3)2) = 4๐ก๐2๐ฅโ3๐ก
๐2: ๐ข2(๐ฅ, ๐ก) = ๐โ1 (
1
๐ฃ+3๐{๐ข1๐ฅ๐ฅ(๐ฅ, ๐ก)}) = ๐
โ1 (1
๐ฃ+3๐{16๐ก๐2๐ฅ๐โ3๐ก})
= ๐โ1 (1
๐ฃ+3โ16๐2๐ฅ๐ฃ
(๐ฃ+3)2) = ๐โ1 (8๐2๐ฅ โ
2!๐ฃ
(๐ฃ+3)2+1) = 8๐ก2๐2๐ฅโ3๐ก
๐3: ๐ข3(๐ฅ, ๐ก) = ๐โ1 (1
๐ฃ + 3๐{๐ข2๐ฅ๐ฅ(๐ฅ, ๐ก)})
= ๐โ1 (1
๐ฃ+3๐{32๐2๐ฅ๐ก2๐โ3๐ก}) = 32๐2๐ฅ๐โ1 (
1
๐ฃ+3โ
2!๐ฃ
(๐ฃ+3)3)
= 32๐2๐ฅ๐โ1 (2! ๐ฃ
(๐ฃ + 3)4) =
32
3๐2๐ฅ๐โ1 (
3! ๐ฃ
(๐ฃ + 3)3+1) =
32
3๐ก3๐2๐ฅโ3๐ก
Chapter Two NTHPM to Solve Types of PDEs
40
And so on, therefore the solution u(x, t) is given by:
๐ข(๐ฅ, ๐ก) = ๐2๐ฅโ3๐ก (1 + 4๐ก + 8๐ก2 +32๐ก3
3+โฏ)
= ๐2๐ฅโ3๐ก (1 + (4๐ก) +(4๐ก)2
2!+(4๐ก)3
3!+ โฏ) = ๐2๐ฅ+๐ก
Example 2.8 [1]
Consider nonlinear 1D, 2nd order autonomous equation.
๐ข๐ก(๐ฅ, ๐ก) = 5๐ข๐ฅ๐ฅ(๐ฅ, ๐ก) + 2๐ข(๐ฅ, ๐ก) + ๐ข2(๐ฅ, ๐ก) (2.28)
With the IC: ๐ข (๐ฅ, ๐ก) = ๐ฝ
where ๐ฝ is arbitrary constant. Taking the new transformation on both sides of
equation (2.28), we have:
๐{๐ข๐ก(๐ฅ, ๐ก)} = 5๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} + 2๐{๐ข(๐ฅ, ๐ก)} + ๐{๐ข2(๐ฅ, ๐ก)}
๐ฃ๐{๐ข(๐ฅ, ๐ก)} โ ๐ฃ๐ข(๐ฅ, 0) = 5๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} + 2๐{๐ข(๐ฅ, ๐ก)} + ๐{๐ข2(๐ฅ, ๐ก)}
(๐ฃ โ 2)๐{๐ข(๐ฅ, ๐ก)} = ๐ฃ๐ฝ + 5๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} + ๐{๐ข2(๐ฅ, ๐ก)}
๐{๐ข(๐ฅ, ๐ก)} =๐ฃ๐ฝ
๐ฃ โ 2+
5
๐ฃ โ 2๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} +
1
๐ฃ โ 2๐{๐ข2(๐ฅ, ๐ก)} (2.29)
Taking the inverse new transformation on equation (2.29), we obtain:
๐ข(๐ฅ, ๐ก) = ๐โ1 {๐ฃ๐ฝ
๐ฃโ2} + ๐โ1 (
5
๐ฃโ2๐{๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} +
1
๐ฃโ2๐{๐ข2(๐ฅ, ๐ก)})
Now, applying the HPM, we get:
Chapter Two NTHPM to Solve Types of PDEs
41
โ๐๐๐ข๐(๐ฅ, ๐ก) = ๐ฝ
โ
๐=0
๐2๐ก + ๐๐โ1 (5
๐ฃ โ 2๐ {โ๐๐
โ
๐=0
๐ข๐ฅ๐ฅ(๐ฅ, ๐ก)} +1
๐ฃ โ 2๐ {โ๐๐๐ป๐
โ
๐=0
})
Comparing the coefficients of powers of p, we have:
๐0: ๐ข0(๐ฅ, ๐ก) = ๐ฝ๐2๐ก
๐1: ๐ข1(๐ฅ, ๐ก) = ๐โ1 (
5
๐ฃ โ 2๐{๐ข0๐ฅ๐ฅ(๐ฅ, ๐ก)} +
1
๐ฃ โ 2๐{๐ป0(๐ข)})
๐2: ๐ข2(๐ฅ, ๐ก) = ๐โ1 (
5
๐ฃโ2๐{๐ข1๐ฅ๐ฅ(๐ฅ, ๐ก)} +
1
๐ฃโ2๐{๐ป1(๐ข)})
๐3: ๐ข3(๐ฅ, ๐ก) = ๐โ1 (5
๐ฃ โ 2๐{๐ข2๐ฅ๐ฅ(๐ฅ, ๐ก)} +
1
๐ฃ โ 2๐{๐ป2(๐ข)})
:
First, compute An to the nonlinear part N(u) , we have:
๐(๐ข) = ๐(โ๐๐๐ป๐
โ
๐=0
)
= ๐ข02 + ๐ (2๐ข0๐ข1) + ๐
2(2๐ข0๐ข2 + ๐ข12) + ๐3(2๐ข0๐ข3 + 2๐ข1๐ข2) + โฏ
So,
๐ป0 = ๐ข02
๐ป1 = 2๐ข0๐ข1
๐ป2 = 2๐ข0๐ข2 + ๐ข12
and so on.
Moreover, the sequence of the parts un is:
๐ข0(๐ฅ, ๐ก) = ๐ฝ๐2๐ก
Chapter Two NTHPM to Solve Types of PDEs
42
๐ข1(๐ฅ, ๐ก) = ๐โ1 (0 +
1
๐ฃโ2๐{๐ฝ2๐4๐ก})
= ๐โ1 (1
๐ฃโ2โ๐ฝ2๐ฃ
๐ฃโ4) = ๐ฝ2๐โ1 (
๐ฃ
(๐ฃโ2)(๐ฃโ4))
= ๐ฝ2๐โ1 (๐ฃ
2(๐ฃโ4)โ
๐ฃ
2(๐ฃโ2)) = ๐ฝ2 [
1
2๐4๐ก โ
1
2๐2๐ก] =
๐ฝ2
2๐2๐ก(๐2๐ก โ 1)
๐ข2(๐ฅ, ๐ก) = ๐โ1 (0 +
1
๐ฃโ2๐ {2๐ฝ๐2๐ก โ
๐ฝ2
2๐2๐ก(๐2๐ก โ 1)})
= ๐โ1 (1
๐ฃโ2๐{๐ฝ3(๐6๐ก โ ๐4๐ก)}) = ๐โ1 (
๐ฝ3
๐ฃโ2(๐ฃ
๐ฃโ6โ
๐ฃ
๐ฃโ4))
= ๐ฝ3๐โ1 (๐ฃ
(๐ฃโ2)(๐ฃโ6)โ
๐ฃ
(๐ฃโ2)(๐ฃโ4) )
= ๐ฝ3๐โ1 ((๐ฃ
4(๐ฃโ6)โ
๐ฃ
4(๐ฃโ2)) โ (
๐ฃ
2(๐ฃโ4)โ
๐ฃ
2(๐ฃโ2)) )
= ๐ฝ3 [(1
4๐6๐ก โ
1
4๐2๐ก) โ (
1
2๐4๐ก โ
1
2๐2๐ก)] = ๐ฝ3 [
1
4๐6๐ก โ
1
2๐4๐ก +
1
4๐2๐ก]
= ๐ฝ3
4๐2๐ก(๐4๐ก โ 2๐2๐ก + 1) =
๐ฝ3
4๐2๐ก(๐2๐ก โ 1)2
๐ข3(๐ฅ, ๐ก) =๐ฝ4
8๐2๐ก(๐2๐ก โ 1)3
Therefore the solution u(x, t) is given by:
๐ข(๐ฅ, ๐ก) = ๐2๐ก (๐ฝ + ๐ฝ
2
2(๐2๐ก โ 1) +
๐ฝ3
4(๐2๐ก โ 1)2 +
๐ฝ4
8(๐2๐ก โ 1)3 +โฏ)
= 2๐ฝ๐2๐ก
2+ ๐ฝ(1โ๐2๐ก)
Chapter Two NTHPM to Solve Types of PDEs
43
2.9. Convergence of the Solution for Linear Case
Now, we need to show the convergence of series form to the exact form for
3D- PDEs.
Lemma 2.1 If f be continues function then
๐
๐๐กโซ๐(๐ก โ ๐)๐๐
๐ก
0
= ๐(๐ก)
Proof
Suppose that
โซ๐(๐ฅ)๐๐ฅ = ๐น(๐ฅ) + ๐
Assume that ๐ฅ = ๐ก โ ๐ then ๐๐ฅ = โ๐๐ then
๐
๐๐กโซ ๐(๐ก โ ๐)๐๐๐ก
0= โ
๐
๐๐กโซ ๐(๐ฅ)๐๐ฅ0
๐ก=
๐
๐๐กโซ ๐(๐ฅ)๐๐ฅ๐ก
0=
๐
๐๐ก[๐น(๐ฅ)|0
๐ก ] =
๐
๐๐ก[๐น(๐ก) โ ๐น(0)] =
๐
๐๐ก๐น(๐ก) โ
๐
๐๐ก๐น(0) = ๐(๐ก)
๐ ๐, ๐
๐๐กโซ ๐(๐ก โ ๐)๐๐๐ก
0= ๐(๐ก)
Lemma 2.2 Let ๐ is new transformation. Then
๐
๐๐ก(๐โ1 {
1
๐ข ๐ {๐(๐, ๐ก)}}) = ๐(๐, ๐ก) , where ๐ = (๐ฅ, ๐ฆ, ๐ง)
Proof
Using properties 2 and 3 of NT, and lemma (2.1), we have:
๐
๐๐ก(๐โ1 {
1
๐ข ๐ {๐(๐, ๐ก)}}) =
๐
๐๐ก(๐โ1 {
1
๐ข ๐ {1}๐ {๐(๐, ๐ก)}}) =
๐
๐๐ก(๐โ1{๐ {1 โ
๐(๐, ๐ก)}}) =๐
๐๐ก(1 โ ๐(๐, ๐ก)) =
๐
๐๐ก(โซ ๐(๐, ๐ก โ ๐)๐๐
๐ก
0) = ๐(๐, ๐ก)
Chapter Two NTHPM to Solve Types of PDEs
44
Theorem 2.1 (Convergence Theorem)
If the series form given in equation (2.6) with p = 1, i.e.,
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
(2.30)
is convergent. Then the limit point converges to the exact solution of equation
(2.1), where un (n = 0, 1, โฆ) are calculated by NTHPM, i.e.,
๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐โ1 {๐ +1
๐ฃ๐ผ ๐ {โ๐ [๐ข0]}}
๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐โ1 {1
๐ฃ๐ผ๐{๐ [๐ข๐โ1]}} , ๐ > 1
}
Proof
Suppose that equation (2.30) converges to the limit point say as:
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
Now, from right hand side of equation (2.1) we have:
๐ผ๐๐ค
๐๐ก= ๐ผ
๐
๐๐ก โ ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ๐=0 = ๐ผ
๐
๐๐ก [๐ข0 + ๐ข1 + โ ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ๐=2 ]
= ๐ผ ๐
๐๐ก [๐โ1 {๐ +
1
๐ฃ๐ผ ๐ {โ๐ [๐ข0]}} โโ๐โ1 {
1
๐ฃ๐ผ[๐{๐ [๐ข๐โ1]}]}
โ
๐=2
]
= ๐ผ ๐๐
๐๐กโ ๐ [๐ข0] โ
๐
๐๐ก (โ๐โ1 {
1
๐ฃ[๐{๐ [๐ข๐]}]}
โ
๐=2
)
= 0 โ ๐ [๐ข0] โโ๐
๐๐ก (๐โ1 {
1
๐ฃ[๐{๐ [๐ข๐]}]})
โ
๐=0
(2.31)
By lemma (2.2), the equation (2.31) becomes:
Chapter Two NTHPM to Solve Types of PDEs
45
๐ผ๐๐ค
๐๐ก= โโ๐ [๐ข๐]
โ
๐=0
= โ ๐ [โ๐ข๐
โ
๐=0
] = โ๐ ๐ค = ๐ค๐ฅ๐ฅ +๐ค๐ฆ๐ฆ +๐ค๐ง๐ง
Then w(x, y, z, t) satisfies equation (2.1). So, it is exact solution.
2.10. Convergence of the Solution for Nonlinear Case
Now, we must prove the convergence of solution of equation (2.13) to the
exact solution when we used the NTHPM, the solution is given in equation
(2.32), where un,, (n= 0, 1, โฆ), are calculated by new transformation, i.e.,
๐ข(๐ฅ, ๐ฆ, ๐ง, ๐ก) = ๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ง, ๐ก) + โฏ =โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
(2.32)
๐ข0 = ๐(๐ฅ, ๐ฆ, ๐ง)
๐ข๐ = โ๐โ1 {
๐{๐ [๐ข๐โ1] + ๐ด๐โ1}
๐ฃ} , ๐ โฅ 1
} (2.33)
and An, (n = 0, 1, โฆ), are defined as
๐ด๐ = ๐ข๐๐๐ข0
๐๐ง+ ๐ข๐โ1
๐๐ข1
๐๐ง+โฏ+ ๐ข0
๐๐ข๐
๐๐ง= โ ๐ข๐
๐๐ข๐โ๐
๐๐ง
๐๐=0 (2.34)
Now we proof the convergence in the following theorem.
Theorem 2.2 (Convergence Theorem)
If the series (2.32) which was calculated by NTHPM, is convergent then
the limit point converges to the exact solution for the equation (2.13). Suppose
that the limit point is:
๐ค(๐ฅ, ๐ฆ, ๐ง, ๐ก) = โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
Now, from left hand side of equation (2.13), we have:
Chapter Two NTHPM to Solve Types of PDEs
46
๐๐ค
๐๐ก=
๐
๐๐ก โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=0
= ๐
๐๐ก [๐ข0(๐ฅ, ๐ฆ, ๐ง, ๐ก) +โ๐ข๐(๐ฅ, ๐ฆ, ๐ง, ๐ก)
โ
๐=1
]
= ๐
๐๐ก [๐โ1{๐} โโ๐โ1 {
๐{๐ [๐ข๐โ1] + ๐ด๐โ1}
๐ฃ}
โ
๐=1
]
=๐
๐๐ก [๐โ1{๐} โโ๐โ1 {
๐{๐ [๐ข๐]}
๐ฃ}
โ
๐=0
โโ๐โ1 {๐{๐ด๐}
๐ฃ}
โ
๐=0
]
= ๐๐
๐๐กโ
๐
๐๐ก โ๐โ1 {
๐{๐ [๐ข๐]}
๐ฃ}
โ
๐=0
โ๐
๐๐กโ๐โ1 {
๐{๐ด๐}
๐ฃ}
โ
๐=0
= ๐๐
๐๐กโโ
๐
๐๐ก[๐โ1 {
๐{๐ [๐ข๐]}
๐ฃ}]
โ
๐=0
โโ๐
๐๐ก[๐โ1 {
๐{๐ด๐}
๐ฃ}]
โ
๐=0
(2.35)
By lemma (2.2) and equation (2.35), we get:
๐๐ค
๐๐ก= 0 โ โ๐ [๐ข๐]
โ
๐=0
โโ๐ด๐
โ
๐=0
(2.36)
However, from equation (2.34), we have:
โ๐ด๐
โ
๐=0
=โโ๐ข๐ ๐๐ข๐โ๐๐๐ง
๐
๐=0
โ
๐=0
= ๐ข0๐๐ข0๐๐ง
+ ๐ข0๐๐ข1๐๐ง
+ ๐ข1๐๐ข0๐๐ง
+ ๐ข0๐๐ข2๐๐ง
+ ๐ข1๐๐ข1๐๐ง
+ ๐ข2๐๐ข0๐๐ง
+ ๐ข0๐๐ข3๐๐ง
+ ๐ข1๐๐ข2๐๐ง
+ ๐ข2๐๐ข1๐๐ง
+ ๐ข3๐๐ข0๐๐ง
+โฏ
= ๐ข0 (๐๐ข0
๐๐ง+๐๐ข1
๐๐ง+๐๐ข2
๐๐ง+๐๐ข3
๐๐ง+โฏ) + ๐ข1 (
๐๐ข0
๐๐ง+๐๐ข1
๐๐ง+๐๐ข2
๐๐ง+๐๐ข3
๐๐ง+โฏ) +
๐ข2 (๐๐ข0
๐๐ง+๐๐ข1
๐๐ง+๐๐ข2
๐๐ง+๐๐ข3
๐๐ง+โฏ) + ๐ข3 (
๐๐ข0
๐๐ง+๐๐ข1
๐๐ง+๐๐ข2
๐๐ง+๐๐ข3
๐๐ง+โฏ) +โฏ
Chapter Two NTHPM to Solve Types of PDEs
47
= (๐ข0 + ๐ข1 + ๐ข2 + ๐ข3 +โฏ) (๐๐ข0
๐๐ง+๐๐ข1
๐๐ง+
๐๐ข2
๐๐ง+๐๐ข3
๐๐ง+โฏ) =
(โ ๐ข๐โ๐=0 ) (โ
๐๐ข๐
๐๐งโ๐=0 ) = (โ ๐ข๐
โ๐=0 ) (
๐
๐๐งโ ๐ข๐โ๐=0 ) (2.37)
Then substitute equation (2.37) in equation (2.36) to obtain:
๐๐ค
๐๐ก= โ๐ [โ ๐ข๐
โ๐=0 ] โ (โ ๐ข๐
โ๐=0 ) (
๐
๐๐งโ ๐ข๐โ๐=0 ) = โ๐ [๐ค] โ ๐ค
๐๐ค
๐๐ง
๐กโ๐๐ ๐๐ค
๐๐ก = [
๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2+๐2๐ค
๐๐ง2] โ ๐ค
๐๐ค
๐๐ง
Then w(x, y, z, t) is satisfy equation (2.13). So, its exact solution.
Solving System of Partial
Differential Equations by
New Couple Methods
49
3.1. Introduction
The system of PDEs arises in many areas of mathematics, engineering
and physical sciences. These systems are too complicated to be solved
exactly so it is still very difficult to get closed-form solutions for most
problems. A vast class of analytical and numerical methods has been
proposed to solve such problems. Such as the ADM [24, 57], VIM [6, 54],
HPM [3, 7, 8, 59], HAM [5] and DTM [28]. But many systems such that
system of high dimensional equations, the required calculations to obtain
its solution in some time may be too complicated. Recently, many powerful
methods have been presented, such as the coupled method [10, 32, 52].
In this chapter, new coupled method based on HPM and new transform
NTHPM, is presented to solve systems of PDEs. The efficiency of the
NTHPM is verified by some examples.
Chapter Three
Solving System of Partial Differential
Equations by New Couple Method
MTHPSolving System of PDEs by N Chapter Three
50
3.2. Solving System of Nonlinear PDEs by NTHPM
This section consist the procedure of the NTHPM to solve system of
nonlinear PDEs. Firstly, writes the system of nonlinear PDEs as follows:
๐ฟ[๐ข(๐ฅ, ๐ฆ, ๐ก)] + ๐ [๐ข(๐ฅ, ๐ฆ, ๐ก)] + ๐[๐ข(๐ฅ, ๐ฆ, ๐ก)] = ๐1(๐ฅ, ๐ฆ, ๐ก)
๐ฟ[๐ค(๐ฅ, ๐ฆ, ๐ก)] + ๐ [๐ค(๐ฅ, ๐ฆ, ๐ก)] + ๐[๐ค(๐ฅ, ๐ฆ, ๐ก)] = ๐2(๐ฅ, ๐ฆ, ๐ก) (3.1๐)
With IC:
๐ข(๐ฅ, ๐ฆ, 0) = ๐(๐ฅ, ๐ฆ)
๐ค(๐ฅ, ๐ฆ, 0) = ๐(๐ฅ, ๐ฆ) (3.1๐)
where all ๐ฅ, ๐ฆ ๐๐ ๐ , L is a linear differential operator (๐ฟ =๐
๐๐ก), R is a
remained of the linear operator, N is a nonlinear differential operator and
๐1(๐ฅ, ๐ฆ, ๐ก), ๐2(๐ฅ, ๐ฆ, ๐ก) are the nonhomogeneous part.
We construct a homotopy u(x, p): ๐ ๐ร [0, 1] โ R, using the homotopy
perturbation technique which satisfies
๐ป(๐ข(๐ฅ, ๐ฆ, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐ข(๐ฅ, ๐ฆ, 0))] + ๐[๐ด(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ
๐1(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ป(๐ค(๐ฅ, ๐ฆ, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐ค(๐ฅ, ๐ฆ, 0))] + ๐[๐ด(๐ฃ(๐ฅ, ๐ฆ, ๐ก)) โ
๐1(๐ฅ, ๐ฆ, ๐ก)] = 0 (3.2)
Where ๐ โ [0,1] is an embedding parameter and the operator A defined as:
๐ด = ๐ฟ + ๐ + ๐.
Obviously, if p = 0, the system (3.2) becomes:
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) = ๐ฟ(๐ข(๐ฅ, ๐ฆ, 0)), and ๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) = ๐ฟ(๐ค(๐ฅ, ๐ฆ, 0)).
MTHPSolving System of PDEs by N Chapter Three
51
It is clear that, if ๐ = 1 then the homotopy system (3.2) convert to the main
system (3.1). In topology, this deformation is called homotopic.
Substitute equation (3.1b) in system (3.2) and rewrite it as:
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) โ ๐๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ฟ(๐(๐ฅ, ๐ฆ)) +
๐๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐๐1(๐ฅ, ๐ฆ, ๐ก) = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) โ ๐๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ฟ(๐(๐ฅ, ๐ฆ)) +
๐๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐๐2(๐ฅ, ๐ฆ, ๐ก) = 0
Then
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐[๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) +
๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐1(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐[๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) +
๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐2(๐ฅ, ๐ฆ, ๐ก)] = 0 (3.3)
Since ๐(๐ฅ, ๐ฆ) and ๐(๐ฅ, ๐ฆ) are independent of the variable ๐ก and the linear
operator L dependent on t so, ๐ฟ(๐(๐ฅ, ๐ฆ)) = 0, ๐ฟ(๐(๐ฅ, ๐ฆ)) = 0, i.e., system
(3.3) becomes:
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐[๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐1(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐[๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐2(๐ฅ, ๐ฆ, ๐ก)] = 0 (3.4)
According to the classical perturbation technique, the solution of system
(3.4) can be written as a power series of embedding parameter p, in the
form:
MTHPSolving System of PDEs by N Chapter Three
52
๐ข(๐ฅ, ๐ฆ, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
๐ค(๐ฅ, ๐ฆ, ๐ก) = โ๐๐๐ค๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
(3.5)
For most cases, the series form in (3.5) is convergent and the convergent
rate depends on the nonlinear operator ๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) and ๐(๐ค(๐ฅ, ๐ฆ, ๐ก)).
Taking the NT (with respect to the variable t) for the system (3.4) to get:
๐{๐ฟ(๐ข)} + ๐ ๐{๐ (๐ข) + ๐(๐ข) โ ๐1} = 0
๐{๐ฟ(๐ค)} + ๐ ๐{๐ (๐ค) + ๐(๐ค) โ ๐2} = 0 (3.6)
Now by using the differentiation property of NT and IC in (3.1b), so
system (3.6) becomes:
๐ฃ๐{๐ข} โ ๐ฃ๐(๐ฅ, ๐ฆ) + ๐ ๐{๐ (๐ข) + ๐(๐ข) โ ๐1} = 0
๐ฃ๐{๐ค} โ ๐ฃ๐(๐ฅ, ๐ฆ) + ๐ ๐{๐ (๐ค) + ๐(๐ค) โ ๐2} = 0
Hence:
๐{๐ข} = ๐(๐ฅ, ๐ฆ) + ๐1
๐ฃ๐{๐1 โ ๐ (๐ข) โ ๐(๐ข)}
๐{๐ค} = ๐(๐ฅ, ๐ฆ) + ๐1
๐ฃ๐{๐2 โ ๐ (๐ค) โ ๐(๐ค)} (3.7)
By taking the inverse of NT on both sides of system (3.7), to get:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ) + ๐ ๐โ1 {1
๐ฃ๐{๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐(๐ข(๐ฅ, ๐ฆ, ๐ก))}}
๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ) + ๐ ๐โ1 {1
๐ฃ๐{๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐(๐ค(๐ฅ, ๐ฆ, ๐ก))}} (3.8)
Then, substitute system (3.5) in system (3.8) to get:
MTHPSolving System of PDEs by N Chapter Three
53
โ๐๐๐ข๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ข๐
โ
๐=0
) โ ๐(โ๐๐๐ข๐
โ
๐=0
)}}
โ๐๐๐ค๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ค๐
โ
๐=0
) โ ๐(โ๐๐๐ค๐
โ
๐=0
)}} (3.9)
The nonlinear part can be decomposed, as will be explained later, by
substituting system (3.5) in it as:
๐(๐ข) = ๐(โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
) = โ๐๐๐ป๐
โ
๐=0
๐(๐ค) = ๐(โ๐๐๐ค๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
) = โ๐๐๐พ๐
โ
๐=0
Then system (3.9) becomes:
โ๐๐๐ข๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ข๐
โ
๐=0
) โโ๐๐๐ป๐
โ
๐=0
}}
โ๐๐๐ค๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ค๐
โ
๐=0
) โโ๐๐๐พ๐
โ
๐=0
}} (3.10)
By comparing the coefficient with the same power of p, in both sides of the
system (3.10) we have:
๐ข0 = ๐(๐ฅ, ๐ฆ), ๐ค0 = ๐(๐ฅ)
๐ข1 = ๐โ1 {1
๐ฃ๐{๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ข0) โ ๐ป0}} , ๐ค1 = ๐
โ1 {1
๐ฃ๐{๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ค0) โ ๐พ0}}
๐ข2 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข1) + ๐ป1}}, ๐ค2 = โ๐
โ1 {1
๐ฃ๐{๐ (๐ค1) + ๐พ1}}
๐ข3 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข2) + ๐ป2}}, ๐ค3 = โ๐
โ1 {1
๐ฃ๐{๐ (๐ค2) + ๐พ2}}
:
MTHPSolving System of PDEs by N Chapter Three
54
๐ข๐+1 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข๐) + ๐ป๐}} , ๐ค๐+1 = โ๐
โ1 {1
๐ฃ๐{๐ (๐ค๐) + ๐พ๐}}
According to the series solution in system (3.5), when at p=1 we can get
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ข0(๐ฅ, ๐ฆ, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ก) + โฏ = โ ๐ข๐(๐ฅ, ๐ฆ, ๐ก)โ๐=0
๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐ค0(๐ฅ, ๐ฆ, ๐ก) + ๐ค1(๐ฅ, ๐ฆ, ๐ก) + โฏ = โ ๐ค๐(๐ฅ, ๐ฆ, ๐ก)โ๐=0
3.3. Illustrative Examples for System of 1D-PDEs
In this section, the NTHPM can be used to solve system of 1D,
nonlinear PDEs.
Example 3.1 [7]
Consider the following system of 1D, nonhomogeneous nonlinear PDEs.
๐๐ข
๐๐ฅโ ๐ค
๐๐ข
๐๐ก+ ๐ข
๐๐ค
๐๐ก= โ1 + ๐๐ฅ sin(๐ก)
๐๐ค
๐๐ฅ+๐๐ข
๐๐ก
๐๐ค
๐๐ฅ+๐๐ค
๐๐ก
๐๐ข
๐๐ฅ= โ1 โ ๐โ๐ฅ cos(๐ก)
Subject to IC: ๐ข(0, ๐ก) = sin(๐ก) , ๐ค(0, ๐ก) = cos(๐ก)
๐ข0 = sin(๐ก) , ๐ค0 = cos(๐ก)
๐ข1 = ๐โ1 {
1
๐ฃ๐ {๐ค0
๐๐ข0๐๐ก
โ ๐ข0๐๐ค0๐๐ก
โ 1 + ๐๐ฅ sin(๐ก)}}
= ๐โ1 {1
๐ฃ๐{๐๐ฅ sin(๐ก)}}
= ๐โ1 {1
๐ฃโ๐ฃ sin(๐ก)
๐ฃ โ 1} = sin(๐ก)๐โ1 {
๐ฃ
๐ฃ โ 1โ 1} = ๐๐ฅ sin(๐ก) โ sin(๐ก)
MTHPSolving System of PDEs by N Chapter Three
55
๐ค1 = ๐โ1 {
1
๐ฃ๐ {โ
๐๐ข0๐๐ก
๐๐ค0๐๐ฅ
โ๐๐ค0๐๐ก
๐๐ข0๐๐ฅ
โ 1 โ ๐โ๐ฅ cos(๐ก)}}
= ๐โ1 {1
๐ฃ๐{โ1 โ ๐โ๐ฅ cos(๐ก)}}
= ๐โ1 {โ1
๐ฃโ๐ฃ cos(๐ก)
๐ฃ(๐ฃ + 1)} = โ๐ฅ + cos(๐ก)๐โ1 {
โ๐ฃ
๐ฃ + 1+ 1}
= โ๐ฅ + ๐โ๐ฅ cos(๐ก) โ cos(๐ก)
๐ข2 = ๐โ1 {
1
๐ฃ๐ {(๐ค1
๐๐ข0๐๐ก
+ ๐ค0๐๐ข1๐๐ก) โ (๐ข1
๐๐ค0๐๐ก
+ ๐ข0๐๐ค1๐๐ก)}}
= ๐โ1 {1
๐ฃ๐{โ2 + ๐๐ฅ + ๐โ๐ฅ โ ๐ฅ cos(๐ก)}}
= ๐โ1 {โ2
๐ฃ+
1
๐ฃ โ 1+
1
๐ฃ + 1โcos(๐ก)
๐ฃ2}
= ๐โ1 {โ2
๐ฃ+
๐ฃ
๐ฃ โ 1โ 1 โ
๐ฃ
๐ฃ + 1+ 1 โ
cos(๐ก)
๐ฃ2}
= โ2๐ก + ๐๐ฅ โ ๐โ๐ฅ โ๐ฅ2
2!cos(๐ก)
๐ค2 = โ๐โ1 {
1
๐ฃ๐ {(
๐๐ข1๐๐ก
๐๐ค0๐๐ฅ
+๐๐ข0๐๐ก
๐๐ค1๐๐ฅ
) + (๐๐ค1๐๐ก
๐๐ข0๐๐ฅ
+๐๐ค0๐๐ก
๐๐ข1๐๐ฅ)}}
= ๐โ1 {1
๐ฃ๐{โcos(๐ก) โ ๐โ๐ฅ cos2(๐ก) โ ๐๐ฅ sin2(๐ก)}}
= โ๐โ1 {โcos(๐ก)
๐ฃโ๐ฃ cos2(๐ก)
๐ฃ(๐ฃ + 1)โ๐ฃ sin2(๐ก)
๐ฃ(๐ฃ โ 1)}
MTHPSolving System of PDEs by N Chapter Three
56
= ๐โ1 {cos(๐ก)
๐ฃ+ cos2(๐ก) (
โ๐ฃ
(๐ฃ + 1)+ 1) + sin2(๐ก) (
๐ฃ
(๐ฃ โ 1)โ 1)}
= ๐ฅ cos(๐ก) โ ๐โ๐ฅ cos2(๐ก) + cos2(๐ก) + ๐๐ฅ sin2(๐ก) โ sin2(๐ก)
= ๐ฅ cos(๐ก) โ ๐โ๐ฅ cos2(๐ก) + ๐๐ฅ sin2(๐ก) + cos(2๐ก)
and so on, therefore
๐ข(๐ฅ, ๐ก) = โ ๐ข๐(๐ฅ, ๐ก) = sin(๐ก) + ๐๐ฅ sin(๐ก) โ sin(๐ก) + โฏ = ๐๐ฅ sin(๐ก)โ
๐=0
๐ค(๐ฅ, ๐ก) = โ ๐ค๐(๐ฅ, ๐ก) = cos(๐ก) + ๐โ๐ฅ cos(๐ก) โ cos(๐ก) โ ๐ฅ +โฏ =โ
๐=0
๐โ๐ฅ cos(๐ก)
This is the exact solution.
Example 3.2 [3]
Consider a system of 1D, 3rd order nonlinear KdV equations (type1).
๐๐ข
๐๐กโ๐3๐ข
๐๐ฅ3โ ๐ข
๐๐ข
๐๐ฅโ๐ค
๐๐ค
๐๐ฅ= 0
๐๐ค
๐๐ก+
๐3๐ข
๐๐ฅ3+ ๐ข
๐๐ค
๐๐ฅ= 0
Subject to IC: ๐ข(๐ฅ, 0) = 3 โ 6 tanh2 (๐ฅ
2) , ๐ค(๐ฅ, 0) = โ3๐โ2 tanh2 (
๐ฅ
2)
We have the following terms:
๐0: ๐ข0(๐ฅ, ๐ก) = 3 โ 6 tanh2 (๐ฅ
2) , ๐0: ๐ค0(๐ฅ, ๐ก) = โ3๐โ2 tanh
2 (๐ฅ
2)
๐1: ๐ข1 = ๐โ1 {
1
๐ฃ๐{
๐3๐ข0
๐๐ฅ3+ ๐ข0
๐๐ข0
๐๐ฅ+๐ค0
๐๐ค0
๐๐ฅ}}
MTHPSolving System of PDEs by N Chapter Three
57
๐ข1 = ๐โ1 {
1
๐ฃ๐{(โ6 ๐ก๐๐โ3(
๐ฅ
2)๐ ๐๐โ2(
๐ฅ
2) + 12 ๐ก๐๐โ(
๐ฅ
2) ๐ ๐๐โ4(
๐ฅ
2)) + (โ18 tanh (
๐ฅ
2) ๐ ๐๐โ2(
๐ฅ
2) +
36 ๐ก๐๐โ3(๐ฅ
2)๐ ๐๐โ2(
๐ฅ
2)) + (โ18 ๐ก๐๐โ3 (
๐ฅ
2) ๐ ๐๐โ2(
๐ฅ
2))}}
๐ข1 = ๐โ1 {1
๐ฃ๐ {12 tanh3(
๐ฅ
2) sech2(
๐ฅ
2) + 12 tanh(
๐ฅ
2) sech4(
๐ฅ
2)
โ 18 tanh(๐ฅ
2)sech2(
๐ฅ
2)}}
= ๐โ1 {1
๐ฃ๐ {12 tanh(
๐ฅ
2) sech2(
๐ฅ
2) โ 18 tanh(
๐ฅ
2)sech2(
๐ฅ
2)}}
= (โ6 tanh (๐ฅ
2) sech2 (
๐ฅ
2)) ๐ก
๐1: ๐ค1 = โ๐โ1 {
1
๐ฃ๐ {๐3๐ข0๐๐ฅ3
+ ๐ข0๐๐ค0๐๐ฅ
}}
๐ค1 = โ๐โ1 {1
๐ฃ๐{(โ6๐โ2 ๐ก๐๐โ3(
๐ฅ
2)๐ ๐๐โ2(
๐ฅ
2) + 12๐โ2 ๐ก๐๐โ(
๐ฅ
2) ๐ ๐๐โ4(
๐ฅ
2)) +
(โ9๐โ2 ๐ก๐๐โ (๐ฅ
2) ๐ ๐๐โ2(
๐ฅ
2) + 18๐โ2 ๐ก๐๐โ3(
๐ฅ
2)๐ ๐๐โ2(
๐ฅ
2))}}
๐ค1 = โ๐โ1 {
1
๐ฃ๐{12๐โ2 tanh3(
๐ฅ
2) sech2(
๐ฅ
2) + 12๐โ2 tanh(
๐ฅ
2) sech4(
๐ฅ
2) โ
9๐โ2 tanh(๐ฅ
2)sech2(
๐ฅ
2)}}
= โ๐โ1 {1
๐ฃ๐{12๐โ2 tanh(
๐ฅ
2) sech2(
๐ฅ
2) โ 9๐โ2 tanh(
๐ฅ
2)sech2(
๐ฅ
2)}}
= (โ3๐โ2 tanh (๐ฅ
2) sech2 (
๐ฅ
2)) ๐ก
and so on, therefore
MTHPSolving System of PDEs by N Chapter Three
58
๐ข = 3 โ 6 tanh2 (๐ฅ
2) โ (6 tanh (
๐ฅ
2) sech2 (
๐ฅ
2)) ๐ก + โฆโฆโฆ
๐ค = โ3๐โ2 tanh2 (๐ฅ
2) โ (3๐โ3 tanh (
๐ฅ
2) sech2 (
๐ฅ
2)) ๐ก + โฆโฆโฆ
The above series closed to exact solution:
๐ข = 3 โ 6 tanh2 (๐ฅ+๐ก
2) , ๐ค = โ3๐โ2 tanh2 (
๐ฅ+๐ก
2)
Example 3.3 [3]
Consider a system of 1D, 3rd order nonlinear KdV equations (type 2).
๐๐ข
๐๐กโ๐3๐ข
๐๐ฅ3โ 2๐ค
๐๐ข
๐๐ฅโ ๐ข
๐๐ค
๐๐ฅ= 0
๐๐ค
๐๐กโ ๐ข
๐๐ข
๐๐ฅ= 0
Subject to ICs: ๐ข(๐ฅ, 0) = โ tanh (๐ฅ
โ3) , ๐ค(๐ฅ, 0) = โ
1
6โ
1
2tanh2 (
๐ฅ
โ3)
Now, we solve the system by using our suggested method to get:
๐0: ๐ข0(๐ฅ, ๐ก) = โ tanh (๐ฅ
โ3) , ๐0: ๐ค0(๐ฅ, ๐ก) = โ
1
6โ1
2tanh2 (
๐ฅ
โ3)
๐1: ๐ข1 = ๐โ1 {
1
๐ฃ๐{๐3๐ข0๐๐ฅ3
+ 2๐ค0๐๐ข0๐๐ฅ
+ ๐ข0๐๐ค0๐๐ฅ
}}
= ๐โ1 {1
๐ฃ๐{(โ
4
3โ3tanh2(
๐ฅ
โ3)sech2(
๐ฅ
โ3) +
2
3โ3sech4(
๐ฅ
โ3)) + (
1
6โ3sech2(
๐ฅ
โ3) +
1
2โ3tanh2(
๐ฅ
โ3)sech2(
๐ฅ
โ3)) + (
1
โ3tanh2 (
๐ฅ
โ3) sech2(
๐ฅ
โ3))}}
MTHPSolving System of PDEs by N Chapter Three
59
๐ข1 = ๐โ1 {
1
๐ฃ๐ {
2
3โ3tanh2(
๐ฅ
โ3) sech2(
๐ฅ
โ3) +
2
3โ3sech4(
๐ฅ
โ3)
+1
3โ3sech2(
๐ฅ
โ3)}}
= ๐โ1 {1
๐ฃ๐{
2
3โ3sech2(
๐ฅ
โ3) +
1
3โ3sech2(
๐ฅ
โ3)}}
= (1
โ3sech2 (
๐ฅ
โ3)) ๐ก
๐1: ๐ค1 = ๐โ1 {
1
๐ฃ๐{๐ข0
๐๐ข0
๐๐ฅ}}
๐ค1 = โ๐โ1 {
1
๐ฃ๐ {(
1
โ3tanh (
๐ฅ
โ3) sech2(
๐ฅ
โ3))}}
= (1
โ3tanh(
๐ฅ
โ3) sech2(
๐ฅ
โ3)) ๐ก
and so on, therefore
๐ข = โ tanh (๐ฅ
โ3) + (
1
โ3sech2 (
๐ฅ
โ3)) ๐ก + โฆโฆโฆ
๐ค = โ1
6โ1
2tanh2 (
๐ฅ
โ3) + (
1
โ3tanh(
๐ฅ
โ3) sech2(
๐ฅ
โ3)) ๐ก + โฆโฆโฆ
The above series closed to exact solution:
๐ข = โ tanh (โ๐ก + ๐ฅ
โ3) , ๐ค = โ
1
6โ1
2tanh2 (
โ๐ก + ๐ฅ
โ3)
MTHPSolving System of PDEs by N Chapter Three
60
3.4. Illustrative Examples for System of 2D-PDEs
In this section, the NTHPM will be used to solve system of 2D,
nonlinear PDEs.
Example 3.4 [7]
Consider the following system of 2D, nonhomogeneous nonlinear PDEs.
๐๐ข
๐๐กโ ๐ค
๐๐ข
๐๐ฅโ๐๐ค
๐๐ก
๐๐ข
๐๐ฆ= 1 โ ๐ฅ + ๐ฆ + ๐ก
๐๐ค
๐๐กโ ๐ข
๐๐ค
๐๐ฅโ๐๐ข
๐๐ก
๐๐ค
๐๐ฆ= 1 โ ๐ฅ โ ๐ฆ โ ๐ก
Subject to ICs:
๐(๐ฅ, ๐ฆ) = ๐ข(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ โ 1 , ๐(๐ฅ, ๐ฆ) = ๐ค(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ + 1
๐ฟ[๐ข(๐ฅ, ๐ฆ, ๐ก)] =๐๐ข(๐ฅ,๐ฆ,๐ก)
๐๐ก , ๐ [๐ข(๐ฅ, ๐ฆ, ๐ก)] = 0 ,
๐[๐ข(๐ฅ, ๐ฆ, ๐ก)] = โ๐ค๐๐ข
๐๐ฅโ๐๐ค
๐๐ก
๐๐ข
๐๐ฆ , ๐(๐ฅ, ๐ฆ, ๐ก) = 1 โ ๐ฅ + ๐ฆ + ๐ก
๐[๐ค(๐ฅ, ๐ฆ, ๐ก)] = โ๐ข๐๐ค
๐๐ฅโ๐๐ข
๐๐ก
๐๐ค
๐๐ฆ , ๐(๐ฅ, ๐ฆ, ๐ก) = 1 โ ๐ฅ โ ๐ฆ โ ๐ก
First, compute the nonlinear of ๐(๐ข) to get:
๐ค๐๐ข
๐๐ฅ= โ๐ป(1,๐)
โ
๐=0
(๐ข, ๐ค) , ๐๐ค
๐๐ก
๐๐ข
๐๐ฆ= โ๐ป(2,๐)
โ
๐=0
(๐ข, ๐ค)
Also, compute the nonlinear of ๐(๐ค) to get:
๐ข๐๐ค
๐๐ฅ= โ ๐พ(1,๐)
โ๐=0 (๐ข,๐ค) ,
๐๐ข
๐๐ก
๐๐ค
๐๐ฆ= โ ๐พ(2,๐)
โ๐=0 (๐ข, ๐ค)
๐ป(1,0) = ๐ค0๐๐ข0
๐๐ฅ , ๐ป(2,0) =
๐๐ค0
๐๐ก
๐๐ข0
๐๐ฆ
MTHPSolving System of PDEs by N Chapter Three
61
๐ป(1,1) = ๐ค1๐๐ข0
๐๐ฅ+๐ค0
๐๐ข1
๐๐ฅ , ๐ป(2,1) =
๐๐ค1
๐๐ก
๐๐ข0
๐๐ฆ+๐๐ค0
๐๐ก
๐๐ข1
๐๐ฆ
And so on
๐พ(1,0) = ๐ข0๐๐ค0๐๐ฅ
, ๐พ(2,0) =๐๐ข0๐๐ก
๐๐ค0๐๐ฆ
๐พ(1,1) = ๐ข1๐๐ค0๐๐ฅ
+ ๐ข0๐๐ค1๐๐ฅ
, ๐พ(2,1) =๐๐ข1๐๐ก
๐๐ค0๐๐ฆ
+๐๐ข0๐๐ก
๐๐ค1๐๐ฆ
And so on.
Moreover, we have the sequence of ๐ข๐ , ๐ค๐ as:
๐ข0 = ๐ฅ + ๐ฆ โ 1 , ๐ค0 = ๐ฅ โ ๐ฆ + 1
๐ข1 = ๐โ1 {
1
๐ฃ๐ {๐ค0
๐๐ข0๐๐ฅ
+๐๐ค0๐๐ก
๐๐ข0๐๐ฆ
+ 1 โ ๐ฅ + ๐ฆ + ๐ก}}
= ๐โ1 {1
๐ฃ๐{2 + ๐ก}} = ๐โ1 {
2
๐ฃ+
1
๐ฃ2} = 2๐ก +
๐ก2
2!
๐ค1 = ๐โ1 {
1
๐ฃ๐ {๐ข0
๐๐ค0๐๐ฅ
+๐๐ข0๐๐ก
๐๐ค0๐๐ฆ
+ 1 โ ๐ฅ โ ๐ฆ โ ๐ก}}
= ๐โ1 {1
๐ฃ๐{โ๐ก}} = ๐โ1 {โ
1
๐ฃ2} = โ
๐ก2
2!
๐ข2 = ๐โ1 {
1
๐ฃ๐ {(๐ค1
๐๐ข0๐๐ฅ
+ ๐ค0๐๐ข1๐๐ฅ) + (
๐๐ค1๐๐ก
๐๐ข0๐๐ฆ
+๐๐ค0๐๐ก
๐๐ข1๐๐ฆ)}}
= ๐โ1 {1
๐ฃ๐ {โ
๐ก2
2!โ ๐ก}} = ๐โ1 {โ
1
๐ฃ3โ
1
๐ฃ2} = โ
๐ก3
3!โ๐ก2
2!
MTHPSolving System of PDEs by N Chapter Three
62
๐ค2 = ๐โ1 {
1
๐ฃ๐ {(๐ข1
๐๐ค0๐๐ฅ
+ ๐ข0๐๐ค1๐๐ฅ
) + (๐๐ข1๐๐ก
๐๐ค0๐๐ฆ
+๐๐ข0๐๐ก
๐๐ค1๐๐ฆ
)}}
= ๐โ1 {1
๐ฃ๐{โ2 + ๐ก +
๐ก2
2!}} = ๐โ1 {โ
2
๐ฃ+1
๐ฃ2+1
๐ฃ3} = โ2t +
๐ก2
2!+๐ก3
3!
๐ข3 = ๐โ1 {1
๐ฃ๐ {(๐ค2
๐๐ข0๐๐ฅ
+ ๐ค1๐๐ข1๐๐ฅ
+ ๐ค0๐๐ข2๐๐ฅ
) + (๐๐ค2๐๐ก
๐๐ข0๐๐ฆ
+๐๐ค1๐๐ก
๐๐ข1๐๐ฆ
+๐๐ค0๐๐ก
๐๐ข2๐๐ฆ
)}}
= ๐โ1 {1
๐ฃ๐{โ2 โ ๐ก + ๐ก2 +
๐ก3
3!}} = ๐โ1 {โ
2
๐ฃโ1
๐ฃ2+2
๐ฃ3+1
๐ฃ4}
= โ2๐ก โ๐ก2
2!+๐ก3
3!+๐ก4
4!
๐ค3 = ๐โ1 {1
๐ฃ๐ {(๐ข2
๐๐ค0๐๐ฅ
+ ๐ข1๐๐ค1๐๐ฅ
+ ๐ข0๐๐ค2๐๐ฅ
) + (๐๐ข2๐๐ก
๐๐ค0๐๐ฆ
+๐๐ข1๐๐ก
๐๐ค1๐๐ฆ
+๐๐ข0๐๐ก
๐๐ค2๐๐ฆ
)}}
= ๐โ1 {1
๐ฃ๐{๐ก โ
๐ก3
3!}} = ๐โ1 {
1
๐ฃ2โ1
๐ฃ4} =
๐ก2
2!โ๐ก4
4!
๐ข4 =๐ก2
2!+๐ก3
3!โ๐ก4
4!โ๐ก5
5!
๐ค4 = 2๐ก โ๐ก2
2!โ๐ก3
2+๐ก4
4!+๐ก5
5!
๐ข5 = 2t +๐ก2
2!โ2๐ก3
3โ๐ก4
12+๐ก5
60+๐ก6
6!
๐ค5 = โ๐ก2
2!+๐ก4
12โ๐ก6
6!
๐ข6 = โ๐ก2
2!โ๐ก3
3!+๐ก4
12+๐ก5
60โ๐ก6
6!โ๐ก7
7!
MTHPSolving System of PDEs by N Chapter Three
63
๐ค6 = โ2๐ก +๐ก2
2!+5๐ก3
3!โ๐ก4
12โ๐ก5
30+๐ก6
6!+๐ก7
7!
And so on. Therefore
๐ข(๐ฅ, ๐ฆ, ๐ก) = โ ๐ข๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ โ 1 + 2๐ก +๐ก2
2!โ๐ก2
2!โ๐ก3
3!โ 2๐ก โโ
๐=0
๐ก2
2!+๐ก3
3!+๐ก4
4!+โฏ = ๐ฅ + ๐ฆ + ๐ก โ 1
๐ค(๐ฅ, ๐ฆ, ๐ก) = โ ๐ค๐(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ โ ๐ฆ + 1 โ๐ก2
2!โ 2๐ก +
๐ก2
2!+
๐ก3
3!+๐ก2
2!โ๐ก4
4!+โ
๐=0
2๐ก โ๐ก2
2!โ๐ก3
2+๐ก4
4!+๐ก5
5!+โฏ = ๐ฅ โ ๐ฆ โ ๐ก + 1
This is the exact solution.
Example 3.5 [22]
Consider the following 2D, nonlinear system of Burgers' equations
๐๐ข
๐๐ก+ ๐ข
๐๐ข
๐๐ฅ+๐ค
๐๐ข
๐๐ฆ=
1
๐ (๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2), (3.11)
๐๐ค
๐๐ก+ ๐ข
๐๐ค
๐๐ฅ+๐ค
๐๐ค
๐๐ฆ= (
๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2),
Subject to the ICs:
๐ข(๐ฅ, ๐ฆ, 0) = ๐ฅ + ๐ฆ , ๐ค(๐ฅ, ๐ฆ, 0) = ๐ฅ โ ๐ฆ ; (๐ฅ, ๐ฆ, ๐ก) โ ๐ 2 ร [0 ,1
โ2).
To solve equation (3.11) by using NTHPM, we construct the following
homotopy
MTHPSolving System of PDEs by N Chapter Three
64
๐ป(๐ข, ๐) = (1 โ ๐) (๐๐ข
๐๐กโ๐๐ข0
๐๐ก) + ๐ (
๐๐ข
๐๐ก+ ๐ข
๐๐ข
๐๐ฅ+ ๐ค
๐๐ข
๐๐ฆโ
1
๐ (๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2)) = 0
๐ป(๐ค, ๐) = (1 โ ๐) (๐๐ค
๐๐กโ๐๐ค0
๐๐ก) + ๐(
๐๐ค
๐๐ก+ ๐ข
๐๐ค
๐๐ฅ+ ๐ค
๐๐ค
๐๐ฆโ
1
๐ (๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2)) = 0
Or equivalent:
๐๐ข
๐๐ก=
๐๐ข0
๐๐กโ ๐ (๐ข
๐๐ข
๐๐ฅ+๐ค
๐๐ข
๐๐ฆโ
1
๐ (๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2) +
๐๐ข0
๐๐ก)
๐๐ค
๐๐ก=
๐๐ค0
๐๐กโ ๐ (๐ข
๐๐ค
๐๐ฅ+๐ค
๐๐ค
๐๐ฆโ
1
๐ (๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2) +
๐๐ค0
๐๐ก)
Applying NT on both sides of above system, to get:
๐ {๐๐ข
๐๐ก} = ๐{
๐๐ข0๐๐ก
โ ๐ (๐ข๐๐ข
๐๐ฅ+ ๐ค
๐๐ข
๐๐ฆโ1
๐ (๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2) +
๐๐ข0๐๐ก)}
๐ {๐๐ค
๐๐ก} = ๐ {
๐๐ค0
๐๐กโ ๐ (๐ข
๐๐ค
๐๐ฅ+๐ค
๐๐ค
๐๐ฆโ
1
๐ (๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2) +
๐๐ค0
๐๐ก)}
Now by using the differentiation property of NT we obtain:
๐ฃ๐{๐ข} = ๐ฃ๐ข(๐ฅ, ๐ฆ, 0) + ๐ {๐๐ข0
๐๐กโ ๐ (๐ข
๐๐ข
๐๐ฅ+๐ค
๐๐ข
๐๐ฆโ
1
๐ (๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2) +
๐๐ข0
๐๐ก)}
๐ฃ๐{๐ค} = ๐ฃ๐ค(๐ฅ, ๐ฆ, 0) + ๐ {๐๐ค0
๐๐กโ ๐ (๐ข
๐๐ค
๐๐ฅ+ ๐ค
๐๐ค
๐๐ฆโ
1
๐ (๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2) +
๐๐ค0
๐๐ก)}
By applying inverse of NT we have:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ข(๐ฅ, ๐ฆ, 0) + ๐โ1 {1
๐ฃ๐{
๐๐ข0
๐๐กโ ๐ (๐ข
๐๐ข
๐๐ฅ+ ๐ค
๐๐ข
๐๐ฆโ (
๐2๐ข
๐๐ฅ2+๐2๐ข
๐๐ฆ2) +
๐๐ข0
๐๐ก)}}
๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐ค(๐ฅ, ๐ฆ, 0) + ๐โ1 {1
๐ฃ๐{
๐๐ค0
๐๐กโ ๐ (๐ข
๐๐ค
๐๐ฅ+๐ค
๐๐ค
๐๐ฆโ (
๐2๐ค
๐๐ฅ2+๐2๐ค
๐๐ฆ2) +
๐๐ค0
๐๐ก)}} (3.12)
MTHPSolving System of PDEs by N Chapter Three
65
Suppose the series solution has the form:
๐ข = ๐ข0 + ๐๐ข1 + ๐2๐ข2 +โฏ
๐ค = ๐ค0 + ๐๐ค1 + ๐2๐ค2 +โฏ (3.13)
Substituting the system (3.12) into the system (3.13) and comparing
coefficients of the terms with the identical powers of p, this lead to
๐0 = {๐ข0(๐ฅ, ๐ฆ, ๐ก) = ๐ข(๐ฅ, ๐ฆ, 0) + ๐
โ1 {1
๐ฃ๐{
๐๐ข0
๐๐ก}}
๐ค0(๐ฅ, ๐ฆ, ๐ก) = ๐ค(๐ฅ, ๐ฆ, 0) + ๐โ1 {
1
๐ฃ๐ {
๐๐ค0
๐๐ก}}
๐1 =
{
๐ข1(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐ {โ
๐๐ข0
๐๐กโ ๐ข0
๐๐ข0
๐๐ฅโ ๐ฃ0
๐๐ข0
๐๐ฆ+ (
๐2๐ข0
๐๐ฅ2+๐2๐ข0
๐๐ฆ2)}}
๐ค1(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {1
๐ฃ๐ {โ
๐๐ค0
๐๐กโ ๐ข0
๐๐ค0
๐๐ฅโ ๐ค0
๐๐ค0
๐๐ฆ+ (
๐2๐ค0
๐๐ฅ2+๐2๐ค0
๐๐ฆ2)}}
๐2 =
{
๐ข2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐ {โ๐ข0
๐๐ข1๐๐ฅ
โ ๐ข1๐๐ข0๐๐ฅ
โ ๐ค0๐๐ข1๐๐ฆ
โ ๐ฃ1๐๐ข0๐๐ฆ
+ (๐2๐ข1๐๐ฅ2
+๐2๐ข1๐๐ฆ2
)}}
๐ค2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {1
๐ฃ๐ {โ๐ข0
๐๐ค1๐๐ฅ
โ ๐ข1๐๐ค0๐๐ฅ
โ ๐ค0๐๐ค1๐๐ฆ
โ ๐ค1๐๐ค0๐๐ฆ
+ (๐2๐ค1๐๐ฅ2
+๐2๐ค1๐๐ฆ2
)}}
And so on, we have
๐๐ =
{
๐ข๐(๐ฅ, ๐ฆ, ๐ก) = ๐
โ1 {1
๐ฃ๐{โ(โ๐ข๐
๐๐ข๐โ๐โ1
๐๐ฅโ ๐ค๐
๐๐ข๐โ๐โ1
๐๐ฆ)
๐โ1
๐=0
+ (๐2๐ข๐โ1
๐๐ฅ2+๐2๐ข๐โ1
๐๐ฆ2)}} , ๐ = 1,2, โฆ
๐ค๐(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐{โ(โ๐ข๐
๐๐ค๐โ๐โ1
๐๐ฅโ ๐ค๐
๐๐ค๐โ๐โ1
๐๐ฆ)
๐โ1
๐=0
+ (๐2๐ค๐โ1
๐๐ฅ2+๐2๐ค๐โ1
๐๐ฆ2)}} , ๐ = 1,2, โฆ
From above system, we obtain the following recurrent relations:
MTHPSolving System of PDEs by N Chapter Three
66
๐ข๐(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐{โ(โ๐ข๐
๐๐ข๐โ๐โ1
๐๐ฅโ ๐ค๐
๐๐ข๐โ๐โ1
๐๐ฆ)
๐โ1
๐=0
+ (๐2๐ข๐โ1
๐๐ฅ2+๐2๐ข๐โ1
๐๐ฆ2)}}
๐ค๐(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {1
๐ฃ๐{โ(โ๐ข๐
๐๐ค๐โ๐โ1
๐๐ฅโ ๐ค๐
๐๐ค๐โ๐โ1
๐๐ฆ)
๐โ1
๐=0
+ (๐2๐ค๐โ1
๐๐ฅ2+๐2๐ค๐โ1
๐๐ฆ2)}}
Starting with
๐ข0 = ๐ฅ + ๐ฆ , ๐ค0 = ๐ฅ โ ๐ฆ
Then we get the following results:
๐ข1(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐{โ
๐๐ข0
๐๐กโ ๐ข0
๐๐ข0
๐๐ฅโ ๐ค0
๐๐ข0
๐๐ฆ+ (
๐2๐ข0
๐๐ฅ2+๐2๐ข0
๐๐ฆ2)}}
= ๐โ1 {1
๐ฃ๐{0 โ (๐ฅ + ๐ฆ)(1) โ (๐ฅ โ ๐ฆ)(1) + 0}} = ๐โ1 {โ
2๐ฅ
๐ฃ} = โ2๐ฅ๐ก
๐ค1(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐ {โ
๐๐ค0
๐๐กโ ๐ข0
๐๐ค0
๐๐ฅโ๐ค0
๐๐ค0
๐๐ฆ+ (
๐2๐ค0
๐๐ฅ2+๐2๐ค0
๐๐ฆ2)}}
๐ค1 = ๐โ1 {1
๐ฃ๐{0 โ (๐ฅ + ๐ฆ)(1) โ (๐ฅ โ ๐ฆ)(โ1) + 0}} = ๐โ1 {โ
2๐ฆ
๐ฃ} = โ2๐ฆ๐ก
๐ข2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐{โ๐ข0
๐๐ข1
๐๐ฅโ ๐ข1
๐๐ข0
๐๐ฅโ ๐ค0
๐๐ข1
๐๐ฆโ ๐ค1
๐๐ข0
๐๐ฆ+ (
๐2๐ข1
๐๐ฅ2+๐2๐ข1
๐๐ฆ2)}}
๐ข2 = ๐โ1 {1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(โ2๐ก) โ (โ2๐ฅ๐ก)(1) โ (๐ฅ โ ๐ฆ)(0) โ (โ2๐ฆ๐ก)(1) + 0}}
= ๐โ1 {1
๐ฃ๐{2๐ฅ๐ก + 2๐ฆ๐ก + 2๐ฅ๐ก + 2๐ฆ๐ก}} = ๐โ1 {
4๐ฅ
๐ฃ2+4๐ฆ
๐ฃ2}
= 2๐ฅ๐ก2 + 2๐ฆ๐ก2
MTHPSolving System of PDEs by N Chapter Three
67
๐ค2(๐ฅ, ๐ฆ, ๐ก) = ๐โ1 {
1
๐ฃ๐{โ๐ข0
๐๐ค1๐๐ฅ
โ ๐ข1๐๐ค0๐๐ฅ
โ ๐ค0๐๐ค1๐๐ฆ
โ ๐ค1๐๐ค0๐๐ฆ
+ (๐2๐ค1๐๐ฅ2
+๐2๐ค1๐๐ฆ2
)}}
๐ค2 = ๐โ1 {1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(0) โ (โ2๐ฅ๐ก)(1) โ (๐ฅ โ ๐ฆ)(โ2๐ก) โ (โ2๐ฆ๐ก)(โ1) + 0}}
= ๐โ1 {1
๐ฃ๐{2๐ฅ๐ก + 2๐ฅ๐ก โ 2๐ฆ๐ก โ 2๐ฆ๐ก}} = ๐โ1 {
4๐ฅ
๐ฃ2โ4๐ฆ
๐ฃ2}
= 2๐ฅ๐ก2 โ 2๐ฆ๐ก2
๐ข3 = ๐โ1 {1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(2๐ก2) โ (โ2๐ฅ๐ก)(โ2๐ก) โ (2๐ฅ๐ก2 + 2๐ฆ๐ก2)(1) โ (๐ฅ โ ๐ฆ)(2๐ก2)
โ (โ2๐ฆ๐ก)(0) โ (2๐ฅ๐ก2 โ 2๐ฆ๐ก2)(1) + 0}}
= ๐โ1 {1
๐ฃ๐{โ2๐ฅ๐ก2 โ 2๐ฆ๐ก2 โ 4๐ฅ๐ก2 โ 2๐ฅ๐ก2 โ 2๐ฆ๐ก2 โ 2๐ฅ๐ก2 + 2๐ฆ๐ก2 โ 2๐ฅ๐ก2 + 2๐ฆ๐ก2}}
= ๐โ1 {1
๐ฃ๐{โ12๐ฅ๐ก2}} = ๐โ1 {
โ24๐ฅ
๐ฃ3} = โ4๐ฅ๐ก3
๐ค3 = ๐โ1 {1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(2๐ก2) โ (โ2๐ฅ๐ก)(0) โ (2๐ฅ๐ก2 + 2๐ฆ๐ก2)(1) โ (๐ฅ โ ๐ฆ)(โ2๐ก2)
โ (โ2๐ฆ๐ก)(โ2๐ก) โ (2๐ฅ๐ก2 โ 2๐ฆ๐ก2)(โ1) + 0}}
= ๐โ1 {1
๐ฃ๐{โ2๐ฅ๐ก2 โ 2๐ฆ๐ก2 โ 2๐ฅ๐ก2 โ 2๐ฆ๐ก2 + 2๐ฅ๐ก2 โ 2๐ฆ๐ก2 โ 4๐ฆ๐ก2 + 2๐ฅ๐ก2 โ 2๐ฆ๐ก2}}
= ๐โ1 {1
๐ฃ๐{โ12๐ฆ๐ก2}} = ๐โ1 {
โ24๐ฆ
๐ฃ3} = โ4๐ฆ๐ก3
MTHPSolving System of PDEs by N Chapter Three
68
๐ข4 = ๐โ1 {1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(โ4๐ก3) โ (โ2๐ฅ๐ก)(2๐ก2) โ (2๐ฅ๐ก2 + 2๐ฆ๐ก2)(โ2๐ก) โ
(โ4๐ฅ๐ก3)(1) โ (๐ฅ โ ๐ฆ)(0) โ (โ2๐ฆ๐ก)(2๐ก2) โ (2๐ฅ๐ก2 โ 2๐ฆ๐ก2)(0) โ (โ2๐ฆ๐ก3)(1)}} =
๐โ1 {1
๐ฃ๐{4๐ฅ๐ก3 + 4๐ฆ๐ก3 + 4๐ฅ๐ก3 + 4๐ฅ๐ก3 + 4๐ฆ๐ก3 + 4๐ฅ๐ก3 + 4๐ฆ๐ก3 + 4๐ฆ๐ก3}} =
๐โ1 {1
๐ฃ๐{16๐ฅ๐ก3 + 16๐ฆ๐ก3}} = ๐โ1 {
96๐ฅ
๐ฃ4+96๐ฆ
๐ฃ4} = 4๐ฅ๐ก4 + 4๐ฆ๐ก4
๐ค4 = ๐โ1 {
1
๐ฃ๐{โ(๐ฅ + ๐ฆ)(0) โ (โ2๐ฅ๐ก)(2๐ก2) โ (2๐ฅ๐ก2 + 2๐ฆ๐ก2)(0) โ (โ4๐ฅ๐ก3)(1) โ
(๐ฅ โ ๐ฆ)(โ4๐ก3) โ (โ2๐ฆ๐ก)(โ2๐ก2) โ (2๐ฅ๐ก2 โ 2๐ฆ๐ก2)(โ2๐ก) โ (โ4๐ฆ๐ก3)(โ1)}}
= ๐โ1 {1
๐ฃ๐{4๐ฅ๐ก3 + 4๐ฅ๐ก3 + 4๐ฅ๐ก3 โ 4๐ฆ๐ก3 โ 4๐ฆ๐ก3 + 4๐ฅ๐ก3 โ 4๐ฆ๐ก3 โ 4๐ฆ๐ก3}} =
๐โ1 {1
๐ฃ๐{16๐ฅ๐ก3 โ 16๐ฆ๐ก3}} = ๐โ1 {
96๐ฅ
๐ฃ4โ96๐ฆ
๐ฃ4} = 4๐ฅ๐ก4 โ 4๐ฆ๐ก4
๐ข5 = โ8๐ฅ๐ก5 , ๐ค5 = โ8๐ฆ๐ก
5
๐ข6 = 8๐ฅ๐ก6 + 8๐ฆ๐ก6 , ๐ค6 = 8๐ฅ๐ก
6 โ 8๐ฆ๐ก6
And so on, the solution of the system (3.11) can be obtained by setting ๐ =
1, i.e.,
๐ข(๐ฅ, ๐ฆ, ๐ก) = lim๐โ1
๐ข๐(๐ฅ, ๐ฆ, ๐ก) = ๐ข0 + ๐ข1 + ๐ข2 + ๐ข3 + ๐ข4 + ๐ข5 + ๐ข6 +โฏ
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ + ๐ฆ โ 2๐ฅ๐ก + 2๐ฅ๐ก2 + 2๐ฆ๐ก2 โ 4๐ฅ๐ก3 + 4๐ฅ๐ก4 + 4๐ฆ๐ก4 โ 8๐ฅ๐ก5 + 8๐ฅ๐ก6 + 8๐ฆ๐ก6
๐ข(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ + ๐ฆ)(1 + 2๐ก2 + 4๐ก4 + 8๐ก6 +โฏ) โ 2๐ฅ๐ก(1 + 2๐ก2 + 4๐ก4 + 8๐ก6 +โฏ)
๐ข(๐ฅ, ๐ฆ, ๐ก) =๐ฅ + ๐ฆ
1 โ 2๐ก2โ
2๐ฅ๐ก
1 โ 2๐ก2=๐ฅ + ๐ฆ โ 2๐ฅ๐ก
1 โ 2๐ก2
MTHPSolving System of PDEs by N Chapter Three
69
Also by similar way, we get:
๐ค(๐ฅ, ๐ฆ, ๐ก) = lim๐โ1
๐ฃ๐(๐ฅ, ๐ฆ, ๐ก) = ๐ค0 +๐ค1 +๐ค2 +๐ค3 +๐ค4 +๐ค5 +๐ค6 +โฏ
๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐ฅ โ ๐ฆ โ 2๐ฆ๐ก + 2๐ฅ๐ก2 โ 2๐ฆ๐ก2 โ 4๐ฆ๐ก3 + 4๐ฅ๐ก4 โ 4๐ฆ๐ก4 โ 8๐ฆ๐ก5 + 8๐ฅ๐ก6 โ 8๐ฆ๐ก6
๐ค(๐ฅ, ๐ฆ, ๐ก) = (๐ฅ โ ๐ฆ)(1 + 2๐ก2 + 4๐ก4 + 8๐ก6 +โฏ) โ 2๐ฆ๐ก(1 + 2๐ก2 + 4๐ก4 + 8๐ก6 +โฏ)
๐ค(๐ฅ, ๐ฆ, ๐ก) =๐ฅ โ ๐ฆ
1 โ 2๐ก2โ
2๐ฆ๐ก
1 โ 2๐ก2=๐ฅ โ ๐ฆ โ 2๐ฆ๐ก
1 โ 2๐ก2
3.5. Solving System for 3Equations Nonlinear 1D-PDEs
In this section, the procedure of NTHPM will be used to solve system
of 1D, for 3 equations nonlinear PDEs.
Firstly the system is written as follows:
๐ฟ[๐ข(๐ฅ, ๐ฆ, ๐ก)] + ๐ [๐ข(๐ฅ, ๐ฆ, ๐ก)] + ๐[๐ข(๐ฅ, ๐ฆ, ๐ก)] = ๐1(๐ฅ, ๐ฆ, ๐ก)
๐ฟ[๐ค(๐ฅ, ๐ฆ, ๐ก)] + ๐ [๐ค(๐ฅ, ๐ฆ, ๐ก)] + ๐[๐ค(๐ฅ, ๐ฆ, ๐ก)] = ๐2(๐ฅ, ๐ฆ, ๐ก) (3.14)
๐ฟ[๐ง(๐ฅ, ๐ฆ, ๐ก)] + ๐ [๐ง(๐ฅ, ๐ฆ, ๐ก)] + ๐[๐ง(๐ฅ, ๐ฆ, ๐ก)] = ๐3(๐ฅ, ๐ฆ, ๐ก)
Subject to IC:
๐ข(๐ฅ, ๐ฆ, 0) = ๐(๐ฅ, ๐ฆ)
๐ค(๐ฅ, ๐ฆ, 0) = ๐(๐ฅ, ๐ฆ)
๐ง(๐ฅ, ๐ฆ, 0) = โ(๐ฅ, ๐ฆ)
Where all ๐ฅ, ๐ฆ โ ๐ , L is a linear differential operator (๐ฟ =๐
๐๐ก), R is a
remained of the linear operator; N is a nonlinear differential operator and
๐1, ๐2, ๐3 is the nonhomogeneous part.
MTHPSolving System of PDEs by N Chapter Three
70
We construct a homotopy as: u(x, p): ๐ ๐ร [0, 1] โR,
๐ป1(๐ข(๐ฅ, ๐ฆ, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐ข(๐ฅ, ๐ฆ, 0))] + ๐[๐ด(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐1(๐ฅ, ๐ฆ, ๐ก)] =
0
๐ป2(๐ค(๐ฅ, ๐ฆ, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐ค(๐ฅ, ๐ฆ, 0))] + ๐[๐ด(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐2(๐ฅ, ๐ฆ, ๐ก)] =
0
๐ป3(๐ง(๐ฅ, ๐ฆ, ๐ก), ๐) = (1 โ ๐)[๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐ง(๐ฅ, ๐ฆ, 0))] + ๐[๐ด(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐3(๐ฅ, ๐ฆ, ๐ก)] = 0
Where ๐ โ [0,1] is an embedding parameter and the operator A defined as:
๐ด = ๐ฟ + ๐ + ๐.
Obviously, if p = 0, then the above system becomes
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) = ๐ฟ(๐ข(๐ฅ, ๐ฆ, 0)), ๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) = ๐ฟ(๐ค(๐ฅ, ๐ฆ, 0)) and ๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) = ๐ฟ(๐ง(๐ฅ, ๐ฆ, 0)).
Substitute ICs in above system and rewrite it as:
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) โ ๐๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) +
๐๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐๐1(๐ฅ, ๐ฆ, ๐ก) = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) โ ๐๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) +
๐๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐๐2(๐ฅ, ๐ฆ, ๐ก) = 0
๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(โ(๐ฅ, ๐ฆ)) โ ๐๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) + ๐๐ฟ(โ(๐ฅ, ๐ฆ)) + ๐๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) +
๐๐ (๐ง(๐ฅ, ๐ฆ, ๐ก)) + ๐๐(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐๐3(๐ฅ, ๐ฆ, ๐ก) = 0
Then
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐[๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ
๐1(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐[๐ฟ(๐(๐ฅ, ๐ฆ)) + ๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ
๐2(๐ฅ, ๐ฆ, ๐ก)] = 0 (3.15)
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๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐ฟ(โ(๐ฅ, ๐ฆ)) + ๐[๐ฟ(โ(๐ฅ, ๐ฆ)) + ๐ (๐ง(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ
๐3(๐ฅ, ๐ฆ, ๐ก)] = 0
Since ๐(๐ฅ, ๐ฆ), ๐(๐ฅ, ๐ฆ) and โ(๐ฅ, ๐ฆ) are independent of the variable ๐ก and the
linear operator L dependent on t so, ๐ฟ(๐(๐ฅ, ๐ฆ)) = 0, ๐ฟ(๐(๐ฅ, ๐ฆ)) = 0, and
๐ฟ(โ(๐ฅ, ๐ฆ)) = 0; i.e., the system (3.15) becomes:
๐ฟ(๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐[๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐1(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ฟ(๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐[๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ค(๐ฅ, ๐ฆ, ๐ก)) โ ๐2(๐ฅ, ๐ฆ, ๐ก)] = 0
๐ฟ(๐ง(๐ฅ, ๐ฆ, ๐ก)) + ๐[๐ (๐ง(๐ฅ, ๐ฆ, ๐ก)) + ๐(๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐3(๐ฅ, ๐ฆ, ๐ก)] = 0 (3.16)
According to the classical perturbation technique, the solution of the above
system can be written as a power series of embedding parameter p, as the
form
๐ข(๐ฅ, ๐ฆ, ๐ก) = โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
๐ค(๐ฅ, ๐ฆ, ๐ก) = โ๐๐๐ค๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
(3.17)
๐ง(๐ฅ, ๐ฆ, ๐ก) = โ๐๐๐ง๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
For most cases, the series form in (3.17) is convergent and the convergent
rate depends on the nonlinear operator.
Taking the NT (with respect to the variable t) for the system (3.16) to get:
๐{๐ฟ(๐ข)} + ๐ ๐{๐ (๐ข) + ๐(๐ข) โ ๐} = 0
๐{๐ฟ(๐ค)} + ๐ ๐{๐ (๐ค) + ๐(๐ค) โ ๐} = 0 (3.18)
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๐{๐ฟ(๐ง)} + ๐ ๐{๐ (๐ง) + ๐(๐ง) โ ๐} = 0
Now by using the differentiation property of NT and IC, the above system
becomes:
๐ฃ๐{๐ข} โ ๐ฃ๐(๐ฅ, ๐ฆ) + ๐ ๐{๐ (๐ข) + ๐(๐ข) โ ๐1} = 0
๐ฃ๐{๐ค} โ ๐ฃ๐(๐ฅ, ๐ฆ) + ๐ ๐{๐ (๐ค) + ๐(๐ค) โ ๐2} = 0
๐ฃ๐{๐ง} โ ๐ฃโ(๐ฅ, ๐ฆ) + ๐ ๐{๐ (๐ง) + ๐(๐ง) โ ๐3} = 0
Hence:
๐{๐ข} = ๐(๐ฅ, ๐ฆ) + ๐1
๐ฃ๐{๐1 โ ๐ (๐ข) โ ๐(๐ข)}
๐{๐ค} = ๐(๐ฅ, ๐ฆ) + ๐1
๐ฃ๐{๐2 โ ๐ (๐ค) โ ๐(๐ค)}
๐{๐ง} = โ(๐ฅ, ๐ฆ) + ๐1
๐ฃ๐{๐3 โ ๐ (๐ง) โ ๐(๐ง)}
By taking the inverse of new transform on both sides of the above system,
to get:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ) + ๐ ๐โ1 {1
๐ฃ๐{๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ข(๐ฅ, ๐ฆ, ๐ก)) โ ๐(๐ข(๐ฅ, ๐ฆ, ๐ก))}}
๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐(๐ฅ, ๐ฆ) + ๐ ๐โ1 {1
๐ฃ๐{๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ค(๐ฅ, ๐ฆ, ๐ก)) โ
๐(๐ค(๐ฅ, ๐ฆ, ๐ก))}} (3.19)
๐ง(๐ฅ, ๐ฆ, ๐ก) = โ(๐ฅ, ๐ฆ) + ๐ ๐โ1 {1
๐ฃ๐{๐3(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ง(๐ฅ, ๐ฆ, ๐ก)) โ ๐(๐ง(๐ฅ, ๐ฆ, ๐ก))}}
Then, substitute system (3.17) in the system (3.19) to get:
โ๐๐๐ข๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ข๐
โ
๐=0
) โ ๐(โ๐๐๐ข๐
โ
๐=0
)}}
โ๐๐๐ค๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ค๐
โ
๐=0
) โ ๐(โ๐๐๐ค๐
โ
๐=0
)}} (3.20)
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โ๐๐๐ง๐ =
โ
๐=0
โ(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐3(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ง๐
โ
๐=0
) โ ๐ (โ๐๐๐ง๐
โ
๐=0
)}}
The nonlinear part can be decomposed, as will be explained later, by
substituting system (3.17) in it as:
๐(๐ข) = ๐(โ๐๐๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
) = โ๐๐๐ป๐
โ
๐=0
๐(๐ค) = ๐(โ๐๐๐ค๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
) = โ๐๐๐พ๐
โ
๐=0
๐(๐ง) = ๐(โ๐๐๐ง๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
) = โ๐๐๐ฝ๐
โ
๐=0
Then the system (3.20) becomes:
โ๐๐๐ข๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ข๐
โ
๐=0
) โโ๐๐๐ป๐
โ
๐=0
}}
โ๐๐๐ค๐ =
โ
๐=0
๐(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ค๐
โ
๐=0
) โโ๐๐๐พ๐
โ
๐=0
}} (3.21)
โ๐๐๐ง๐ =
โ
๐=0
โ(๐ฅ, ๐ฆ) + ๐๐โ1 {1
๐ฃ๐ {๐3(๐ฅ, ๐ฆ, ๐ก) โ ๐ (โ๐๐๐ง๐
โ
๐=0
) โโ๐๐๐ฝ๐
โ
๐=0
}}
By comparing the coefficient with the same power of p, in both sides of the
system (3.21) we have:
๐ข0 = ๐(๐ฅ, ๐ฆ), ๐ค0 = ๐(๐ฅ, ๐ฆ), ๐ง0 = โ(๐ฅ, ๐ฆ)
๐ข1 = ๐โ1 {
1
๐ฃ๐{๐1(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ข0) โ ๐ป0}},
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๐ค1 = ๐โ1 {
1
๐ฃ๐{๐2(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ค0) โ ๐พ0}}
๐ง1 = ๐โ1 {
1
๐ฃ๐{๐3(๐ฅ, ๐ฆ, ๐ก) โ ๐ (๐ง0) โ ๐ฝ0}}
๐ข2 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข1) + ๐ป1}},
๐ค2 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ค1) + ๐พ1}}
๐ง2 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ง1) + ๐ฝ1}}
๐ข3 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข2) + ๐ป2}},
๐ค3 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ค2) + ๐พ2}}
๐ง3 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ง2) + ๐ฝ2}}
And so on, therefor
๐ข๐+1 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ข๐) + ๐ป๐}}
๐ค๐+1 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ค๐) + ๐พ๐}}
๐ง๐+1 = โ๐โ1 {
1
๐ฃ๐{๐ (๐ง๐) + ๐ฝ๐}}
According to the series solution in system (3.17), when at p=1, we can get:
๐ข(๐ฅ, ๐ฆ, ๐ก) = ๐ข0(๐ฅ, ๐ฆ, ๐ก) + ๐ข1(๐ฅ, ๐ฆ, ๐ก) + โฏ = โ๐ข๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
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๐ค(๐ฅ, ๐ฆ, ๐ก) = ๐ค0(๐ฅ, ๐ฆ, ๐ก) + ๐ค1(๐ฅ, ๐ฆ, ๐ก) + โฏ =โ๐ค๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
๐ง(๐ฅ, ๐ฆ, ๐ก) = ๐ง0(๐ฅ, ๐ฆ, ๐ก) + ๐ง1(๐ฅ, ๐ฆ, ๐ก) + โฏ = โ๐ง๐(๐ฅ, ๐ฆ, ๐ก)
โ
๐=0
Example 3.6 [7]
Consider the following nonlinear system of inhomogeneous PDEs.
๐๐ข
๐๐กโ๐๐ง
๐๐ฅ
๐๐ค
๐๐กโ1
2
๐๐ง
๐๐ก
๐2๐ข
๐๐ฅ2= โ4๐ฅ๐ก
๐๐ค
๐๐กโ๐๐ง
๐๐ก
๐2๐ข
๐๐ฅ2= 6๐ก
๐๐ง
๐๐กโ๐2๐ข
๐๐ฅ2โ๐๐ค
๐๐ฅ
๐๐ง
๐๐ก= 4๐ฅ๐ก โ 2๐ก โ 2
Subject to the ICs:
๐ข(๐ฅ, 0) = ๐ฅ2 + 1 , ๐ค(๐ฅ, 0) = ๐ฅ2 โ 1 , ๐ง(๐ฅ, 0) = ๐ฅ2 โ 1
๐ข0(๐ฅ, 0) = ๐ฅ2 + 1 , ๐ค0(๐ฅ, 0) = ๐ฅ
2 โ 1 , ๐ง0(๐ฅ, 0) = ๐ฅ2 โ 1
๐ข1 = ๐โ1 {
1
๐ฃ๐ {๐๐ง0๐๐ฅ
๐๐ค0๐๐ก
+1
2
๐๐ง0๐๐ก
๐2๐ข0๐๐ฅ2
โ 4๐ฅ๐ก}} = ๐โ1 {1
๐ฃ๐{โ4๐ฅ๐ก}}
= ๐โ1 {โ4๐ฅ
๐ฃ2} = โ2๐ฅ๐ก2
๐ค1 = ๐โ1 {
1
๐ฃ๐ {(
๐๐ง0๐๐ก
๐2๐ข0๐๐ฅ2
+ 6t)}} = ๐โ1 {1
๐ฃ๐{6๐ก}} = ๐โ1 {
6
๐ฃ2} = 3๐ก2
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๐ง1 = ๐โ1 {
1
๐ฃ๐{(
๐2๐ข0๐๐ฅ2
+๐๐ค0๐๐ฅ
๐๐ง0๐๐ก
+ 4๐ฅ๐ก โ 2๐ก โ 2)}}
= ๐โ1 {1
๐ฃ๐{4๐ฅ๐ก โ 2๐ก}} = ๐โ1 {
4๐ฅ
๐ฃ2โ2
๐ฃ2} = 2๐ฅ๐ก2 โ ๐ก2
๐ข2 = ๐โ1 {
1
๐ฃ๐{(
๐๐ง1๐๐ฅ
๐๐ค0๐๐ก
+๐๐ง0๐๐ฅ
๐๐ค1๐๐ก) +
1
2(๐๐ง1๐๐ก
๐2๐ข0๐๐ฅ2
+๐๐ง0๐๐ก
๐2๐ข1๐๐ฅ2
)}}
= ๐โ1 {1
๐ฃ๐{16๐ฅ๐ก โ 2๐ก}} = ๐โ1 {
16๐ฅ
๐ฃ2โ2
๐ฃ2} = 8๐ฅ๐ก2 โ ๐ก2
๐ค2 = ๐โ1 {
1
๐ฃ๐ {๐๐ง1๐๐ก
๐2๐ข0๐๐ฅ2
+๐๐ง0๐๐ก
๐2๐ข1๐๐ฅ2
}} = ๐โ1 {1
๐ฃ๐{8๐ฅ๐ก โ 4๐ก}}
= ๐โ1 {8๐ฅ
๐ฃ2โ4
๐ฃ2} = 4๐ฅ๐ก2 โ 2๐ก2
๐ง2 = ๐โ1 {
1
๐ฃ๐ {(
๐2๐ข1๐๐ฅ2
+ (๐๐ค1๐๐ฅ
๐๐ง0๐๐ก
+๐๐ค0๐๐ฅ
๐๐ง1๐๐ก)}}
= ๐โ1 {1
๐ฃ๐{8๐ฅ2๐ก โ 4๐ฅ๐ก}}
= ๐โ1 {8๐ฅ2
๐ฃ2โ4๐ฅ
๐ฃ2} = 4๐ฅ2๐ก2 โ 2๐ฅ๐ก2
๐ข3 = 3๐ก4 โ 6๐ฅ๐ก2 + 12๐ฅ2๐ก2
๐ค3 = 8๐ฅ2๐ก2 โ 4๐ฅ๐ก2
๐ง3 = 8๐ฅ3๐ก2 โ 4๐ฅ2๐ก2
๐ข4 = โ5๐ก4 + 16๐ฅ๐ก4 + 24๐ฅ3๐ก2 โ 12๐ฅ2๐ก2
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๐ค4 = 16๐ฅ3๐ก2 โ 8๐ฅ2๐ก2
๐ง4 = 8๐ก3 + 4๐ฅ๐ก4 โ 2๐ก4 + 16๐ฅ4๐ก2 โ 8๐ฅ3๐ก2
๐ข5 = 60๐ฅ2๐ก4 โ 10๐ก4 + 48๐ฅ4๐ก2 โ 24๐ฅ3๐ก2 + 8๐ก3
๐ค5 = 16๐ก3 + 32๐ฅ๐ก4 โ 16๐ก4 + 32๐ฅ4๐ก2 โ 16๐ฅ3๐ก2
๐ง5 = 64๐ฅ๐ก3 โ 8๐ก3 โ 32๐ฅ2๐ก4 โ 16๐ฅ๐ก4 + 2๐ก4 + 32๐ฅ5๐ก2 โ 16๐ฅ4๐ก2
And so on, thus
๐ข(๐ฅ, ๐ก) = โ ๐ข๐(๐ฅ, ๐ก) = ๐ฅ2 + 1 โ 2๐ฅ๐ก2 + 8๐ฅ๐ก2 โ ๐ก2 + 3๐ก4 โ 6๐ฅ๐ก2 +โ
๐=0
12๐ฅ2๐ก2 โ 5๐ก4 +โฏ = ๐ฅ2 โ ๐ก2 + 1
๐ค(๐ฅ, ๐ก) = โ ๐ค๐(๐ฅ, ๐ก) = ๐ฅ2 โ 1 + 3๐ก2 + 4๐ฅ๐ก2 โ 2๐ก2 + 8๐ฅ2๐ก2 โโ
๐=0
4๐ฅ๐ก2 +โฏ = ๐ฅ2 + ๐ก2 โ 1
๐ง(๐ฅ, ๐ก) = โ ๐ง๐(๐ฅ, ๐ก) = ๐ฅ2 โ 1 + 2๐ฅ๐ก2 โ ๐ก2 + 4๐ฅ2๐ก2 โ 2๐ฅ๐ก2 +โ
๐=0
8๐ฅ3๐ก2 โ 4๐ฅ2๐ก2 +โฏ = ๐ฅ2 โ ๐ก2 โ 1
Example 3.7 [3]
Consider the generalized coupled Hirota Satsuma KdV type II.
๐๐ข
๐๐กโ1
2
๐3๐ข
๐๐ฅ3+ 3๐ข
๐๐ข
๐๐ฅโ 3
๐
๐ฅ(๐ค๐ง) = 0
๐๐ค
๐๐ก+๐3๐ข
๐๐ฅ3โ 3๐ข
๐๐ค
๐๐ฅ= 0
๐๐ง
๐๐ก+
๐3๐ง
๐๐ฅ3โ 3๐ข
๐๐ง
๐๐ฅ= 0
MTHPSolving System of PDEs by N Chapter Three
78
Subject to IC:
๐ข(๐ฅ, 0) = โ1
3+ 2 tanh2(๐ฅ) , ๐ค(๐ฅ, 0) = tanh(๐ฅ) , ๐ง(๐ฅ, 0) =
8
3tanh(๐ฅ)
๐1: ๐ข1 = ๐โ1 {
1
๐ฃ๐ {
1
2
๐3๐ข0
๐๐ฅ3โ 3๐ข0
๐๐ข0
๐๐ฅ+ 3
๐
๐๐ฅ(๐ค0๐ง0)}}
= ๐โ1 {1
๐ฃ๐{(8 tanh3(๐ฅ )sech2( ๐ฅ) โ 16 tanh(๐ฅ) sech4( ๐ฅ)) + (4 tanh ( ๐ฅ) sech2(๐ฅ) โ
24 tanh3(๐ฅ )sech2( ๐ฅ)) + (16 tanh(๐ฅ) sech2(๐ฅ
2))}}
๐ข1 = ๐โ1 {
1
๐ฃ๐{โ16 tanh3(๐ฅ) sech2( ๐ฅ) + 20 tanh( ๐ฅ) sech2( ๐ฅ)
โ 16 tanh(๐ฅ )sech4(๐ฅ)}}
= ๐โ1 {1
๐ฃ๐{โ16 tanh(๐ฅ) sech2( ๐ฅ) + 20 tanh(๐ฅ )sech2(๐ฅ)}}
= (4 tanh(๐ฅ) sech2(๐ฅ))๐ก
๐1: ๐ค1 = ๐โ1 {
1
๐ฃ๐{โ
๐3๐ข0
๐๐ฅ3+ 3๐ข0
๐๐ค0
๐๐ฅ}}
= โ๐โ1 {1
๐ฃ๐{(โ4 tanh2(๐ฅ )sech2( ๐ฅ) + 2 sech4( ๐ฅ)) + (โ sech2(๐ฅ) +
6 tanh2(๐ฅ )sech2( ๐ฅ))}}
๐ค1 = โ๐โ1 {
1
๐ฃ๐{2 tanh2(๐ฅ) sech2( ๐ฅ) + 2 sech4( ๐ฅ) โ sech2(๐ฅ)}}
MTHPSolving System of PDEs by N Chapter Three
79
= โ๐โ1 {1
๐ฃ๐{2 sech2( ๐ฅ) โ sech2(๐ฅ)}} = (sech2(๐ฅ))๐ก
๐1: ๐ง1 = ๐โ1 {
1
๐ฃ๐{โ
๐3๐ง0
๐๐ฅ3+ 3๐ข0
๐๐ง0
๐๐ฅ}}
๐ง1 = โ๐โ1 {
1
๐ฃ๐ {(โ
32
3tanh2(๐ฅ )sech2( ๐ฅ) +
16
3sech4( ๐ฅ)) +
(โ8
9sech2(๐ฅ) +
16
3tanh2(๐ฅ )sech2( ๐ฅ))}}
๐ง1 = โ๐โ1 {
1
๐ฃ๐ {16
3tanh2(๐ฅ) sech2( ๐ฅ) +
16
3sech4( ๐ฅ) โ
8
3sech2(๐ฅ)}}
= โ๐โ1 {1
๐ฃ๐{
16
3sech2( ๐ฅ) โ
8
3sech2(๐ฅ)}} = (
8
3sech2(๐ฅ)) ๐ก
And so on, so
๐ข = โ1
3+ 2 tanh2(๐ฅ) + (4 tanh(๐ฅ) sech2(๐ฅ))๐ก + โฆโฆโฆ
๐ค = tanh(๐ฅ) + (sech2(๐ฅ))๐ก + โฆโฆโฆ
๐ง =8
3tanh(๐ฅ) + (
8
3sech2(๐ฅ)) ๐ก + โฆโฆโฆ
The above series closed to exact solution as
๐ข = โ tanh(๐ก + ๐ฅ) , ๐ค = โ1
6โ1
2tanh2(๐ก + ๐ฅ) , ๐ง =
8
3tanh(๐ก + ๐ฅ)
MTHPSolving System of PDEs by N Chapter Three
80
3.6. Convergence for the Series Solution
In this section the convergence of NTHPM for systems of nonlinear
PDEs is presented. The sufficient condition for convergence of the
method is addressed. Since mathematical modeling of numerous
scientific and engineering experiments lead to system of equations, it is
worth trying new methods to solve these systems. Here we show the
series solution for systems of previous sections closed to exact solution.
Definition 3.1 [33]
A Banach space is a complete, normed, vector space.
All norms on a finite-dimensional vector space are equivalent. Every finite-
dimensional normed space over R or C is a Banach space [50].
Definition 3.2 [34]
Let X is a set and let ๐: ๐ โ ๐ be a function that maps X into itself. Such
a function is often called an operator. A fixed point of ๐ is an element x โ
X, for which f(x) = x .
Definition 3.3 [34]
Let (๐, ๐) be a metric space. A mapping ๐ โถ ๐ โ ๐ is a contraction
mapping, or contraction, if there exists a constant ๐, with 0 โค ๐ < 1, such
that ๐(๐(๐ฅ), ๐(๐ฆ)) โค ๐๐(๐ฅ, ๐ฆ), ๐ฅ , ๐ฆ โ ๐ .
MTHPSolving System of PDEs by N Chapter Three
81
Theorem 3.1 [46]
A contractive function ๐ on a Banach space ๐ has a unique fixed point
๐ฅโ in R2. see [46]
Theorem 3.2 (Sufficient Condition of Convergence)
If X and Y are Banach spaces and ๐ โถ ๐ โ ๐ is a contractive nonlinear
mapping, that is
โ ๐ค , ๐คโ โ ๐ ; โ๐(๐ค) โ ๐(๐คโ)โ โค ๐พโ๐ค โ ๐คโโ , 0 < ๐พ < 1 .
Then according to Banach's fixed-point theorem, N has a unique fixed-
point u, that is ๐(๐ข) = ๐ข .
Assume that the sequence generated by NTHPM can be written as:
๐ค๐ = ๐(๐ค๐โ1), ๐ค๐โ1 = โ๐ค๐
๐โ1
๐=0
, ๐ = 1,2,3,โฆโฆ
And suppose that ๐0 = ๐ค0 โ ๐ต๐(๐ค) ๐คโ๐๐๐ ๐ต๐(๐ค) = {๐คโ โ ๐| โ๐คโ โ
๐คโ < ๐}, then we have
i. ๐ค๐ โ ๐ต๐(๐ค),
ii. lim๐โโ
๐๐ = ๐ค
Proof
(i) By inductive approach, for ๐ = 1, we have
โ๐1 โ๐คโ = โ๐(๐0) โ ๐(๐ค)โ โค ๐พโ๐ค0 โ๐คโ
MTHPSolving System of PDEs by N Chapter Three
82
Assume that โ๐๐โ1 โ๐คโ โค ๐พโ๐ค๐โ1 โ๐คโ
โค ๐พ2โ๐ค๐โ2 โ๐คโ
โค ๐พ3โ๐ค๐โ3 โ๐คโ
โค ๐พ๐โ1โ๐ค0 โ๐คโ
As induction hypothesis, then
โ๐๐ โ๐คโ = โ๐(๐๐โ1) โ ๐(๐ค)โ โค ๐พโ๐ค๐โ1 โ๐คโ โค ๐พ๐โ๐ค0 โ๐คโ
Using (i), we have
โ๐๐ โ๐คโ โค ๐พ๐โ๐ค0 โ๐คโ โค ๐พ
๐๐ < ๐ ๐๐ โ ๐ต๐(๐ค)
Because of โ๐๐ โ๐คโ โค ๐พ๐โ๐ค0 โ๐คโ and
lim๐โโ
๐พ๐ = 0, lim๐โโ
โ๐๐ โ๐คโ = 0,
that is
lim๐โโ
๐๐ = ๐ค
Application Model
84
4.1. Introduction
To illustrate the importance of suggested method, we use it to solve
soil moisture equation that is equation describe rate of moisture content in
soil. The term moisture content is used in hydrogeology, soil sciences and
soil Mechanics. Here a model of moisture content is derived and used to
build up the one, two and three dimensional space 2nd order nonlinear
homogenous PDE. Then NTHPM is used to solve this equation. The
derivation of formulation model is illustrated in section 4.2. Basic idea
for suggested method that be used to solve the model equation is
introduced in section 4.3. Illustrating applicability is presented in section
4.4. While a convergence of the solution is proved in section 4.5.
4.2. Formulation Mathematical Model [22]
Moisture content is the quantity of water contained in a soil called soil
moisture. The saturated zone is one in which the space is occupied by
water. In the unsaturated zone only part of the space is occupied by water.
There is no moisture in the dry soil, so the value of moisture content is
1 when the porous medium is fully saturated by water and its value is 0 in
Chapter Four
Application Model
lication ModelpAp Chapter Four
85
the unsaturated porous medium. So, the range of moisture content is [0,
1]. The region of the unsaturated soil is called as unsaturated zone. In
typical soil profiles some distance separates the earthโs surface from the
water table, which is the upper limit of completely water-saturated soil. In
this inverting zone the water saturation varies between 0 and 1 the rest of
the pore space normally being occupied by air. Water flow in this
unsaturated zone is complicated by the fact that the soilโs permeability to
water depends on its water saturation. In many practical situations the
flow of water through soil is unsteady because the moisture content
changes as a function of time and it is slightly saturated because all the
spaces are not completely filed with flowing liquid.
In the formulation model, we assume that the diffusivity coefficient
will be small enough constant and regarded as a perturbation parameter
because of they are equivalent to their average value over the whole range
of moisture content. Additionally, the permeability of the medium is
varied directly to the square of the moisture content.
Darcyโs law gives the motion of water in the isotropic homogeneous
medium as:
๐ = โ๐พ โโ (4.1)
lication ModelpAp Chapter Four
86
Where V is volume flux of moisture content, K is coefficients of aqueous
the conductivity, is a total moisture potential and โ is the gradient,
i.e., โ = (๐
๐๐ฅ,
๐
๐๐ฆ,
๐
๐๐ง) .
In any unsaturated porous media, continuity equation governs the motion
of water flow given as (depending on [23]):
๐(๐๐ ๐)
๐๐ก= โโ. ๐ (4.2)
where S : bulk density for medium on dry weight basis, : the moisture
content in any position (x, y, z) on a dry weight basis, and M : mass of
flux for moisture in any t โฅ 0.
From (4.1) and (4.2) we get:
๐(๐๐ ๐)
๐๐ก= โโ. ๐ = โโ. (๐. ๐) = ๐ป. (๐. ๐พ. โ๐) (4.3)
Where indicate the flux density of the medium.
Now, depending on [21] we have ๐ = ๐ โ ๐๐ง . Hence,
๐(๐๐ ๐)
๐๐ก= ๐ป. (๐. ๐พ. โ(๐ โ ๐๐ง)) (4.4)
Where indicates the pressure (capillary) potential, g is the gravitation
constant, z is an elevation of the unit of mass for water above a consistent
datum (which is the level of saturation) and the positive direction of the
z-axis is the same as that of gravity. We find
lication ModelpAp Chapter Four
87
๐๐ ๐๐
๐๐ก= ๐ป. (๐. ๐พ. โ๐) โ ๐ป. (๐. ๐พ. ๐ป(๐๐ง)) (4.5)
Since โ(๐๐ง) = (๐(๐๐ง)
๐๐ฅ,
๐(๐๐ง)
๐๐ฆ,
๐(๐๐ง)
๐๐ง) = (0,0, ๐) and divided both sides of
equation (4.5) by S, we get:
๐๐
๐๐ก= ๐ป. (
๐
๐๐ . ๐พ. โ๐) โ ๐ป. (
๐
๐๐ . ๐พ. (0,0, ๐)) (4.6)
Consider and are single valued function, so equation (4.6) can be
written as
๐๐
๐๐ก= ๐ป. (
๐
๐๐ . ๐พ.
๐๐
๐๐โ๐) โ ๐ป. (
๐
๐๐ . ๐พ. (0,0, ๐)) (4.7)
According to [21], we have ๐ท =๐
๐๐ . ๐พ.
๐๐
๐๐ is called the diffusivity
coefficient, which is constant as we assumed, so we get:
๐๐
๐๐ก= ๐ป. (๐ท. โ๐) โ ๐ป. (
๐
๐๐ . ๐พ. (0,0, ๐)) (4.8)
Let Da is the average value of D over the whole range. According to [22]
we have ๐พ โ ๐2, i.e., ๐พ โ ๐พ0. ๐2,where K0 is a constant. Hence, equation
(4.8) becomes
๐๐
๐๐ก= ๐ป. (๐ท๐. โ๐) โ ๐ป. (
๐
๐๐ . ๐พ0. ๐2. (0,0, ๐)) (4.9)
So, equation (4.9) get:
lication ModelpAp Chapter Four
88
๐๐
๐๐ก+
๐
๐๐ 2๐. ๐พ0. ๐
๐๐
๐๐ง= ๐ท๐(๐ป. ๐ป๐) (4.10)
Suppose ๐พ1 =๐
๐๐ . 2๐. ๐พ0 and substituting it in equation (4.10) we get:
๐๐
๐๐ก+ ๐พ1. ๐.
๐๐
๐๐ง= ๐ท๐(๐ป. ๐ป๐) (4.11)
Hence, the final form of the equation is
๐๐
๐๐ก+ ๐พ1. ๐.
๐๐
๐๐ง= ๐ท๐ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2+
๐2๐
๐๐ง2] (4.12)
We can simplify equation (4.12), by supposing:
๏ฟฝ๏ฟฝ =๐ฅ
๐1 , ๏ฟฝ๏ฟฝ =
๐ฆ
๐1 , ๐ง =
๐ง
๐1 (4.13)
It is clear that:
๐๐
๐๐ฅ=
1
๐1
๐๐
๐๏ฟฝ๏ฟฝ ,
๐๐
๐๐ฆ=
1
๐1
๐๐
๐๏ฟฝ๏ฟฝ ,
๐๐
๐๐ง=
1
๐1
๐๐
๐๐ง
๐2๐
๐๐ฅ2=
1
๐12
๐2๐
๐๏ฟฝ๏ฟฝ2 ,
๐2๐
๐๐ฆ2=
1
๐12
๐2๐
๐๏ฟฝ๏ฟฝ2 ,
๐2๐
๐๐ง2=
1
๐12
๐2๐
๐๐ง2
(4.14)
Substituting equation (4.14) into equation (4.12), we get:
๐๐
๐๐ก+ ๐
๐๐
๐๐ง=
๐ท๐
๐12 [
๐2๐
๐๏ฟฝ๏ฟฝ2+
๐2๐
๐๏ฟฝ๏ฟฝ2+
๐2๐
๐๐ง2] (4.15)
Let ๐ผ =๐ท๐
๐12 , then equation (4.15) will be:
๐๐
๐๐ก+ ๐
๐๐
๐๐ง= ๐ผ [
๐2๐
๐๏ฟฝ๏ฟฝ2+
๐2๐
๐๏ฟฝ๏ฟฝ2+
๐2๐
๐๐ง2] (4.16)
For simplification, the original symbols will be used instead of the
symbols in equation (4.16), i.e.,
lication ModelpAp Chapter Four
89
๐๐
๐๐ก+ ๐
๐๐
๐๐ง= ๐ผ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2+
๐2๐
๐๐ง2] (4.17)
4.3. Solving Model Equation by the NTHPM
The main purpose of this section is discuss the new implementation of
the combine NT algorithm with the HPM to solve the suggested model
equation (4.17).
To explain the NTHPM, firstly writes the nonlinear PDE (4.17) as follow:
๐ฟ[๐(๐, ๐ก)] + ๐ [๐(๐, ๐ก)] + ๐[๐(๐, ๐ก)] = ๐(๐, ๐ก) (4.18)
Subject to IC: ๐(๐, 0) = ๐(๐)
Where ๐ โ ๐ ๐ , L is a linear differential operator (๐ฟ =๐
๐๐ก) , R is a
remained of the linear operator, ๐ is a nonlinear differential operator and
๐(๐, ๐ก) is the nonhomogeneous part.
We construct a Homotopy as: (๐, ๐): ๐ ๐ ร [0,1] โ ๐ , which satisfies
๐ป(๐(๐, ๐ก), ๐) = (1 โ ๐) โ [๐ฟ(๐(๐, ๐ก)) โ ๐ฟ(๐(๐, 0))] +
๐ [๐ด ((๐(๐, ๐ก))) โ ๐(๐, ๐ก)] = 0 (4.19)
Where ๐ โ [0,1] is an embedding parameter and the operator A defined
as ๐ด = ๐ฟ + ๐ + ๐.
Obviously, if ๐ = 0, equation (4.19) becomes ๐ฟ(๐(๐, ๐ก)) โ ๐ฟ(๐(๐, 0)).
lication ModelpAp Chapter Four
90
It is clear that, if ๐ = 1 then the homotopy equation (4.19) convert to the
main differential equation (4.18). Substitute IC in equation (4.19) and
rewrite it as:
๐ฟ(๐(๐, ๐ก)) โ ๐ฟ(๐(๐)) โ ๐๐ฟ(๐(๐, ๐ก)) + ๐๐ฟ(๐(๐)) + ๐๐ฟ(๐(๐, ๐ก)) +
๐๐ (๐(๐, ๐ก)) + ๐๐(๐(๐, ๐ก)) โ ๐๐(๐, ๐ก) = 0
Then
๐ฟ(๐(๐, ๐ก)) โ ๐ฟ(๐(๐)) + ๐[๐ฟ(๐(๐)) + ๐ (๐(๐, ๐ก)) + ๐(๐(๐, ๐ก)) โ
๐(๐, ๐ก)] = 0 (4.20)
Since ๐(๐) is independent of the variable ๐ก and the linear operator ๐ฟ
dependent on t so, ๐ฟ(๐(๐)) = 0, i.e., (4.20) becomes:
๐ฟ(๐(๐, ๐ก)) + ๐[๐ (๐(๐, ๐ก)) + ๐(๐(๐, ๐ก)) โ ๐(๐, ๐ก)] = 0 (4.21)
According to the classical perturbation technique, the solution of equation
(4.21) can be written as a power series of embedding parameter p, in the
form
๐(๐, ๐ก) = โ ๐๐๐๐(๐, ๐ก)
โ
๐=0
(4.22)
For most cases, the series form (4.22) is convergent and the convergent
rate depends on the nonlinear operator ๐(๐(๐, ๐ก)).
lication ModelpAp Chapter Four
91
Taking the NT (with respect to the variable ๐ก) for the equation (4.21), we
have:
๐{๐ฟ(๐)} + ๐๐ {๐ (๐) + ๐(๐) โ ๐} = 0 (4.23)
Now by using the differentiation property of LT, so equation (4.23),
becomes:
๐ฃ๐{๐} โ ๐(๐) + ๐๐ {๐ (๐) + ๐(๐) โ ๐} = 0
Hence:
๐{๐} = ๐(๐) + ๐1
๐ฃ๐{๐ โ ๐ (๐) โ ๐(๐)} (4.24)
By taking the inverse of new transformation on both sides of equation
(4.24), to get:
๐(๐, ๐ก) = ๐(๐)
+ ๐ ๐โ1 {1
๐ฃ๐{๐(๐, ๐ก) โ ๐ (๐(๐, ๐ก)) โ ๐(๐(๐, ๐ก))}} (4.25)
Then substitute equation (4.22) in equation (4.25) to obtain:
โ ๐๐๐๐
โ
๐=0
= ๐(๐) + ๐๐โ1 {1
๐ฃ๐ {๐(๐, ๐ก) โ ๐ (โ ๐๐๐๐
โ
๐=0
) โ ๐ (โ ๐๐๐๐
โ
๐=0
)}} (4.26)
The nonlinear part can be decomposed, as will be explained later, by
substituting equation (4.22) in it as:
๐(๐) = ๐ (โ ๐๐๐๐(๐, ๐ก)
โ
๐=0
) = โ ๐๐๐ป๐
โ
๐=0
Then equation (4.26) becomes:
lication ModelpAp Chapter Four
92
โ ๐๐๐๐
โ
๐=0
= ๐(๐) + ๐๐โ1 {1
๐ฃ๐ {๐(๐, ๐ก) โ ๐ (โ ๐๐๐๐
โ
๐=0
) โ โ ๐๐๐ป๐
โ
๐=0
}} (4.27)
By comparing the coefficient with the same power of p, in both sides of
the equation (4.27) we have:
๐0 = ๐(๐)
๐1 = ๐โ1 {1
๐ฃ๐{๐(๐, ๐ก) โ ๐ (๐0) โ ๐ป0}}
๐2 = โ๐โ1 {1
๐ฃ๐{๐ (๐1) + ๐ป1}}
๐3 = โ๐โ1 {1
๐ฃ๐{๐ (๐2) + ๐ป2}}
โฎ
๐๐+1 = โ๐โ1 {1
๐ฃ๐{๐ (๐๐) + ๐ป๐}}
According to the series solution in equation (4.22), and p converges to 1,
we get:
๐(๐, ๐ก) = ๐0(๐, ๐ก) + ๐1(๐, ๐ก) + โฏ = โ ๐๐(๐, ๐ก)
โ
๐=0
(4.28)
4.4. Experiment Application
In this section, the suggested method will be used to solve the one;
two and three-dimensions model equation, with appropriate initial
condition:
lication ModelpAp Chapter Four
93
Problem 4.1
Here suggested method that can used to solve the 2D model equation
(1D space) i.e., equation governs the motion of water flow in 1D-
horizontal:
๐๐
๐๐ก+ ๐
๐๐
๐๐ฅ= ๐ผ [
๐2๐
๐๐ฅ2]
๐(๐ฅ) = ๐(๐ฅ, 0) =๐พ+๐ฝ+(๐ฝโ๐พ)e๐
๐๐+1
Where ๐ =๐พ
๐ผ(๐ฅ โ ๐), , and are parameters dimension
We have
๐ฟ[๐(๐, ๐ก)] =๐๐(๐ฅ,๐ก)
๐๐ก , ๐ [๐(๐, ๐ก)] = โ ๐ผ [
๐2๐
๐๐ฅ2] ,
๐[๐(๐, ๐ก)] = ๐๐๐
๐๐ฅ and ๐(๐, ๐ก)
First, compute ๐ป๐ to the nonlinear part ๐(๐) we get:
๐(๐) = ๐ (โ ๐๐๐๐
โ
๐=0
) = (โ ๐๐๐๐
โ
๐=0
) (๐
๐๐ง[โ ๐๐๐๐
โ
๐=0
]) = (โ ๐๐๐๐
โ
๐=0
) (โ ๐๐๐๐๐
๐๐ฅ
โ
๐=0
)
= ๐0
๐๐0
๐๐ฅ+ ๐ (๐0
๐๐1
๐๐ฅ+ ๐1
๐๐0
๐๐ฅ) + ๐2 (๐0
๐๐2
๐๐ฅ+ ๐1
๐๐1
๐๐ฅ+ ๐2
๐๐0
๐๐ฅ)
+ ๐3 (๐0
๐๐3
๐๐ฅ+ ๐1
๐๐2
๐๐ฅ+ ๐2
๐๐1
๐๐ฅ+ ๐3
๐๐0
๐๐ฅ) + โฏ
lication ModelpAp Chapter Four
94
So,
๐ป0 = ๐0๐๐0
๐๐ฅ
๐ป1 = ๐1๐๐0
๐๐ฅ+ ๐0
๐๐1
๐๐ฅ
๐ป2 = ๐2๐๐0
๐๐ฅ+ ๐1
๐๐1
๐๐ฅ+ ๐0
๐๐2
๐๐ฅ
๐ป3 = ๐3๐๐0
๐๐ฅ+ ๐2
๐๐1
๐๐ฅ+ ๐1
๐๐2
๐๐ฅ+ ๐0
๐๐3
๐๐ฅ
And so on.
From IC, we can get
๐ป0 = โ2 ๐พ2 ๐๐ (๐พ+๐ฝ+(๐ฝโ๐พ)๐๐)
๐ผ (๐๐+1)3
๐ป1 = ๐ก2 ๐พ3๐ฝ ๐๐ (๐พ+๐ฝโ4๐พ ๐๐+(๐พโ๐ฝ)๐2๐)
๐ผ2 (๐๐+1)4
๐ป2 = โ๐ก2 ๐พ4๐ฝ2 ๐๐ (๐พ+๐ฝโ(11๐พ+3๐ฝ)๐๐+(11๐พโ3๐ฝ)๐2๐+(๐ฝโ๐พ)๐3๐)
๐ผ3 (๐๐+1)5
๐ป3 = ๐ก3 ๐พ5๐ฝ3 ๐๐ (๐พ+๐ฝโ(26๐พ+10๐ฝ)๐๐+66๐พ๐2๐โ(26๐พโ10๐ฝ)๐3๐+(๐พโ๐ฝ)๐4๐)
3 ๐ผ4 (๐๐+1)6
And so on.
Moreover, the sequence of parts ๐๐ is:
๐0 =๐พ+๐ฝ+(๐ฝโ๐พ)๐๐
(๐๐+1)
๐1 = ๐ก2 ๐พ2๐ฝ ๐๐
๐ผ(๐๐+1)2
๐2 = โ๐ก2 ๐พ3๐ฝ2 ๐๐ (1โ๐๐)
๐ผ2 (๐๐+1)3
lication ModelpAp Chapter Four
95
๐3 = ๐ก3 ๐พ4๐ฝ3 ๐๐ (1โ4 ๐๐+๐2๐)
3 ๐ผ3 (๐๐+1)4
๐4 = โ๐ก4 ๐พ5๐ฝ4 ๐๐ (1โ11 ๐๐+11๐2๐โ๐3๐)
12 ๐ผ4 (๐๐+1)5
And so on.
Substituting the above values in series form (4.28), hence the solution of
the problem is close to the form:
๐(๐ฅ, ๐ก) =๐พ+๐ฝ+(๐ฝโ๐พ)e
๐โ ๐พ๐ฝ๐ผ
๐ก
e๐โ
๐พ๐ฝ๐ผ ๐ก
+1
Problem 4.2
Here suggested method will be used to solve the 3D model equation (2D
space) i.e., equation governs the motion of water flow in horizontal and
vertical:
๐๐
๐๐ก+ ๐
๐๐
๐๐ฅ= ๐ผ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2]
๐(๐ฅ, ๐ฆ) = ๐(๐ฅ, ๐ฆ, 0) =๐พ+๐ฝ+(๐ฝโ๐พ)e๐
๐๐+1
Where ๐ =๐พ
๐ผ(๐ฅ + ๐ฆ โ ๐), ๐ผ, ๐ฝ and ๐ are parameters dimension
We have
๐ฟ[๐(๐, ๐ก)] =๐๐(๐ฅ,๐ฆ,๐ก)
๐๐ก , ๐ [๐(๐, ๐ก)] = โ ๐ผ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2] ,
lication ModelpAp Chapter Four
96
๐[๐(๐, ๐ก)] = ๐๐๐
๐๐ฅ and ๐(๐, ๐ก) = 0
First, compute ๐ป๐ to the nonlinear part ๐(๐) we get:
๐(๐) = ๐ (โ ๐๐๐๐
โ
๐=0
) = (โ ๐๐๐๐
โ
๐=0
) (๐
๐๐ง[โ ๐๐๐๐
โ
๐=0
]) = (โ ๐๐๐๐
โ
๐=0
) (โ ๐๐๐๐๐
๐๐ฅ
โ
๐=0
)
= ๐0
๐๐0
๐๐ฅ+ ๐ (๐0
๐๐1
๐๐ฅ+ ๐1
๐๐0
๐๐ฅ) + ๐2 (๐0
๐๐2
๐๐ฅ+ ๐1
๐๐1
๐๐ฅ+ ๐2
๐๐0
๐๐ฅ)
+ ๐3 (๐0
๐๐3
๐๐ฅ+ ๐1
๐๐2
๐๐ฅ+ ๐2
๐๐1
๐๐ฅ+ ๐3
๐๐0
๐๐ฅ) + โฏ
So,
๐ป0 = ๐0๐๐0
๐๐ฅ
๐ป1 = ๐1๐๐0
๐๐ฅ+ ๐0
๐๐1
๐๐ฅ
๐ป2 = ๐2๐๐0
๐๐ฅ+ ๐1
๐๐1
๐๐ฅ+ ๐0
๐๐2
๐๐ฅ
๐ป3 = ๐3๐๐0
๐๐ฅ+ ๐2
๐๐1
๐๐ฅ+ ๐1
๐๐2
๐๐ฅ+ ๐0
๐๐3
๐๐ฅ
And so on.
From IC, we can get:
๐ป0 = โ ๐พ2 ๐๐ (๐พ+๐ฝ+(๐ฝโ๐พ)๐๐)
๐ผ (๐๐+1)3
๐ป1 = ๐ก ๐พ3๐ฝ ๐๐ (๐พ+๐ฝโ4๐พ ๐๐+(๐พโ๐ฝ)๐2๐)
2 ๐ผ2 (๐๐+1)4
๐ป2 = โ๐ก2 ๐พ4๐ฝ2 ๐๐ (๐พ+๐ฝโ(11๐พ+3๐ฝ)๐๐+(11๐พโ3๐ฝ)๐2๐+(๐ฝโ๐พ)๐3๐)
8 ๐ผ3 (๐๐+1)5
lication ModelpAp Chapter Four
97
๐ป3 = ๐ก3 ๐พ5๐ฝ3 ๐๐ (๐พ+๐ฝโ(26๐พ+10๐ฝ)๐๐+66๐พ๐2๐โ(26๐พโ10๐ฝ)๐3๐+(๐พโ๐ฝ)๐4๐)
48 ๐ผ4 (๐๐+1)6
And so on.
Moreover, the sequence of parts ฮธn is:
๐0 =๐พ+๐ฝ+(๐ฝโ๐พ)๐๐
(๐๐+1)
๐1 = ๐ก ๐พ2๐ฝ ๐๐
๐ผ(๐๐+1)2
๐2 = โ๐ก2 ๐พ3๐ฝ2 ๐๐ (1โ๐๐)
4๐ผ2 (๐๐+1)3
๐3 = ๐ก3 ๐พ4๐ฝ3 ๐๐ (1โ4 ๐๐+๐2๐)
24 ๐ผ3 (๐๐+1)4
๐4 = โ๐ก4 ๐พ5๐ฝ4 ๐๐ (1โ11 ๐๐+11๐2๐โ๐3๐)
192 ๐ผ4 (๐๐+1)5
And so on.
Substituting the above values in series form (4.28), hence the solution of
the problem is close to the form:
๐(๐ฅ, ๐ฆ, ๐ก) =๐พ+๐ฝ+(๐ฝโ๐พ)e
๐โ ๐พ๐ฝ2๐ผ
๐ก
e๐โ
๐พ๐ฝ2๐ผ๐ก
+1
Problem 4.3
Here suggested method will be used to solve the 4D model equation (3D
space):
๐๐
๐๐ก+ ๐
๐๐
๐๐ฅ= ๐ผ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2+ +
๐2๐
๐๐ง2]
lication ModelpAp Chapter Four
98
๐(๐ฅ, ๐ฆ, ๐ง) = ๐(๐ฅ, ๐ฆ, ๐ง, 0) =๐พ+๐ฝ+(๐ฝโ๐พ)e๐
๐๐+1
Where ๐ =๐พ
๐ผ(๐ฅ + ๐ฆ + ๐ง โ ๐), ๐ผ, ๐ฝ and ๐ are parameters dimension
We have
๐ฟ[๐(๐, ๐ก)] =๐๐(๐ฅ,๐ฆ,๐ง,๐ก)
๐๐ก , ๐ [๐(๐, ๐ก)] = โ ๐ผ [
๐2๐
๐๐ฅ2+
๐2๐
๐๐ฆ2+
๐2๐
๐๐ง2] ,
๐[๐(๐, ๐ก)] = ๐๐๐
๐๐ฅ and ๐(๐, ๐ก) = 0
First, compute ๐ป๐ to the nonlinear part ๐(๐) we get:
๐(๐) = ๐ (โ ๐๐๐๐
โ
๐=0
) = (โ ๐๐๐๐
โ
๐=0
) (๐
๐๐ง[โ ๐๐๐๐
โ
๐=0
]) = (โ ๐๐๐๐
โ
๐=0
) (โ ๐๐๐๐๐
๐๐ฅ
โ
๐=0
)
= ๐0
๐๐0
๐๐ฅ+ ๐ (๐0
๐๐1
๐๐ฅ+ ๐1
๐๐0
๐๐ฅ) + ๐2 (๐0
๐๐2
๐๐ฅ+ ๐1
๐๐1
๐๐ฅ+ ๐2
๐๐0
๐๐ฅ)
+ ๐3 (๐0
๐๐3
๐๐ฅ+ ๐1
๐๐2
๐๐ฅ+ ๐2
๐๐1
๐๐ฅ+ ๐3
๐๐0
๐๐ฅ) + โฏ
So,
๐ป0 = ๐0๐๐0
๐๐ง
๐ป1 = ๐1๐๐0
๐๐ง+ ๐0
๐๐1
๐๐ง
๐ป2 = ๐2๐๐0
๐๐ง+ ๐1
๐๐1
๐๐ง+ ๐0
๐๐2
๐๐ง
And so on.
From IC, we can get
lication ModelpAp Chapter Four
99
๐ป0 = โ2 ๐พ2 ๐๐ (๐พ+๐ฝ+(๐ฝโ๐พ)๐๐)
3 ๐ผ (๐๐+1)3
๐ป1 = ๐ก2 ๐พ3๐ฝ ๐๐ (๐พ+๐ฝโ4๐พ ๐๐+(๐พโ๐ฝ)๐2๐)
9 ๐ผ2 (๐๐+1)4
๐ป2 = โ๐ก2 ๐พ4๐ฝ2 ๐๐ (๐พ+๐ฝโ(11๐พ+3๐ฝ)๐๐+(11๐พโ3๐ฝ)๐2๐+(๐ฝโ๐พ)๐3๐)
27 ๐ผ3 (๐๐+1)5
๐ป3 = ๐ก3 ๐พ5๐ฝ3 ๐๐ (๐พ+๐ฝโ(26๐พ+10๐ฝ)๐๐+66๐พ ๐2๐โ(26๐พโ10๐ฝ)๐3๐+(๐ฝโ๐พ)๐4๐)
243 ๐ผ4 (๐๐+1)6
And so on.
Moreover, the sequence of parts ฮธn is:
๐0 =๐พ+๐ฝ+(๐ฝโ๐พ)๐๐
(๐๐+1)
๐1 = ๐ก2 ๐พ2๐ฝ ๐๐
3๐ผ(๐๐+1)2
๐2 = โ๐ก2 ๐พ3๐ฝ2 ๐๐ (1โ๐๐)
9๐ผ2 (๐๐+1)3
๐3 = ๐ก3 ๐พ4๐ฝ3 ๐๐ (1โ4 ๐๐+๐2๐)
81 ๐ผ3 (๐๐+1)4
๐4 = โ๐ก4 ๐พ5๐ฝ4 ๐๐ (1โ11๐๐+11๐2๐โ๐3๐)
972 ๐ผ4 (๐๐+1)5
And so on.
Substituting the above values in series form (4.28), hence the solution of
the problem is close to the form:
๐(๐ฅ, ๐ฆ, ๐ง, ๐ก) =๐พ+๐ฝ+(๐ฝโ๐พ)e
๐โ ๐พ๐ฝ3๐ผ
๐ก
e๐โ
๐พ๐ฝ3๐ผ๐ก
+1
lication ModelpAp Chapter Four
100
From problems 1, 2, 3, we can see that the proposed NTHPM is
applied to find the exact solution of the nonlinear 2nd order (2, 3, 4 โ D)
model equations does not require any restrictive assumptions to deal with
nonlinear terms.
102
5.1. Conclusions
In this thesis, the combination of new transform suggested by Luma
and Alaa with HPM method is proposed to get exact solution of some
types of non-linear, non-homogenous PDEs; 1D, 2D, and 3D; 2nd or 3rd
order such as Klein-Gordan equation, wave-like equations, autonomous
equation, system of two or three non-linear equations, 2D-Burgers'
equations, coupled Hirota Satsuma KdV type II, and RLW equation.
Finally, the suggested method is used to solve application model such soil
moisture equation where traditional HPM leads to an approximate
solution. So, the results reveal that NTHPM is a powerful method for
solving those types of PDEs with initial conditions. The basic idea
described in this thesis is strong enough to be employed to solve other
types of equations. The advantage of suggested method is capability of
combining two powerful methods for obtaining exact solutions for those
types of PDEs, where the HPM was disability to get the exact solution for
the same problems and solved its approximately.
Chapter Five
Conclusions and Future Work
Chapter Five Conclusions and Future Work
103
Moreover, the research finding how the solution of PDE by MTHPM
provide the agreement with real life problems. The method attacks the
problem in a direct way and in a straightforward fashion without using
linearization, or any other restrictive assumption that may change the
physical behavior of the model under discussion.
The experimental results show that the NTHPM is computationally
efficient for solving those types of problems and can easily be
implemented without computer.
The suggested method is free of unnecessary mathematical complexities.
The fast convergence and simple applicability of NTHPM provid
excellent foundation for using these functions in analytical solution of
variety problems.
The obtained results show that our proposed method has several
advantages such like being free of using Adomian polynomials when
dealing with the nonlinear terms like in the ADM and being free of using
the Lagrange multiplier as in the VIM.
The results reveal that the presented methods are reliable, effective, very
accurate and applicable to solve other nonlinear problems.
The advantage of NTHPM is its capability of combining two powerful
methods for obtaining exact.
Chapter Five Conclusions and Future Work
104
5.2. Future Works
Based on the results of the proposed method and its illustrative, the
following future works may suggest:
1- Using NTHPM for solving high dimensions PDEs.
2- Use NTHPM to solve nonlocal problems.
3- Use NTHPM to solve integral and integro-differential equations.
4- Use NTHPM to solve system of integral or integro-differential
equations.
5- Use NTHPM to solve differential equations with fractional orders.
106
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ุนูู ุงูุชุฑุงู ุตูุบุฉ ุชุญููู ุฌุฏูุฏ ู ุน ุทุฑููุฉ ุชุณุชูุฏูู ูุฐู ุงูุฑุณุงูุฉ ุงูุชุฑุญูุง ุทุฑููุฉ ุฌุฏูุฏุฉ
ุงูุญู ุงูู ุถุจูุท ูุฅูุฌุงุฏูุญู ุจุนุถ ุงููุงุน ุงูู ุนุงุฏูุงุช ุงูุชูุงุถููุฉ ุงูุฌุฒุฆูุฉ ุงูุงุถุทุฑุงุจ ุงูููู ูุชูุจู
ุงู ุงุนุงุฏุฉ ุชุฑุชูุจ ุชุฑุฏุฏ ุณูู ุชูููู ูู ุงุณุชุฎุฏุงู ุงูุทุฑููุฉ ูุญู ุงูู ุณุงุฆู ุฏูู ุฃู ูู ู ุฌุงู ุงูุณุน.
ุชูููุฏ ุจูุฑุถูุงุช ู ูู ุญุฑ ู ู ุงุฎุทุงุก ุงูุชุฏููุฑ. ูุฐู ุงูุทุฑููุฉ ุชุณู ู ุทุฑููุฉ ุงูุชุญููู ุงูุงูู ุฌุงู ุงู
ุงูุฌุฏูุฏ ููุงุถุทุฑุงุจ ุงูููู ูุชูุจู.
ุณูููู ุงูุชุฑููุฒ ุนูู ุจุนุถ ุงูู ูุงููู ุงูุงุณุงุณูุฉ ููู ุนุงุฏูุงุช ุงูุชูุงุถููุฉ ุงููุฏู ุงูุงูู ูู ุงูุฑุณุงูุฉ
.ุงูุฌุฒุฆูุฉ
ุญู ุจุนุถ ุงููุงุน ู ู ุงูู ุนุงุฏูุงุช ุงูู ุจุงูุฅุถุงูุฉูู ุชุทุจูู ุงูุทุฑููุฉ ุงูู ูุชุฑุญุฉ ุงููุฏู ุงูุซุงูู
ุงูุชูุงุถููุฉ ุงูุฌุฒุฆูุฉ ุฐุงุช ุงูุดุฑูุท ุงูุงุจุชุฏุงุฆูุฉ ู ุซู
Klein-Gordan equation, wave-like equations, autonomous equation,
system of two or three non-linear equations, Burgers' equations, coupled
Hirota Satsuma KdV type II, and RLW equation.
ุงุฎูุฑุง ุงุณุชุฎุฏู ุช ูุญู ูู ูุฐุฌ ุชุทุจููู ู ุซู ู ุนุงุฏูุฉ ุฑุทูุจุฉ ุงูุชุฑุจุฉ ุญูุซ ุงู ุญููุง ุจุทุฑููุฉ ุงูุงุถุทุฑุงุจ
ุตูุบุฉุงูููู ูุชูุจู ุงูุชูููุฏูุฉ ูุคุฏู ุงูู ุงูุญู ุงูุชูุฑูุจู. ุงูุถุง ุงุซุจุชูุง ุชูุงุฑุจ ู ุชุณูุณูุฉ ุงูุญู ุงูู ุงู
ู ูุซูููุฉ ู ูุงุจููุฉ ุงูุทุฑููุฉ ุงูู ูุชุฑุญุฉ. ุงูุถุง ุชุถู ูุช ุจุนุถ ุงูุงู ุซูุฉ ุงูุชู ุชูุถุญ. ุฉุงูู ุถุจูุท
ุงููุชุงุฆุฌ ุงูุนู ููุฉ ุงุซุจุชุช ุงู ุงูุทุฑููุฉ ุงูู ูุชุฑุญุฉ ุงุฏุงุฉ ููุคุฉ ูุญู ุชูู ุงูุงููุงุน ู ู ุงูู ุนุงุฏูุงุช ุงูุชูุงุถููุฉ
ุงูุฌุฒุฆูุฉ.
ุงูู ุณุชุฎูุต
ุฌู ููุฑูุฉ ุงูุนุฑุงู
ูุฒุงุฑุฉ ุงูุชุนููู ุงูุนุงูู ูุงูุจุญุซ ุงูุนูู ู
ุจุบุฏุงุฏ ุฌุงู ุนุฉ
/ุงุจู ุงูููุซู ููุนููู ุงูุตุฑูุฉ ูููุฉ ุงูุชุฑุจูุฉ
ููุคุฉ ูุญู ุจุนุถ ุฃููุงุน ุงูู ุนุงุฏูุงุช ุงูุชูุงุถููุฉ ุงูุฌุฒุฆูุฉ ุฉููุทุฑ
ุฑุณุงูุฉ ุฌุงู ุนุฉ ุจุบุฏุงุฏ ,ุงุจู ุงูููุซู -ููุนููู ุงูุตุฑูุฉ ุงูุชุฑุจูุฉ ูููุฉ ุฏู ุฉ ุฅูููู ู
ููู ูุดูุงุฏุฉ ู ุงุฌุณุชูุฑ ุนููู ุฌุฒุก ู ู ู ุชุทูุจุงุช ููู ูู ุงูุฑูุงุถูุงุช
ู ูู ูุจูู
ู ูุณุฑ ุนุจูุฏ ุนูุงุฏู
ุฑุงููุฅุดูุจ ุฃ . ุฏ. ูู ูู ูุงุฌูู ู ุญูู ูุฏ ุชููููู
ู 2019 ูู 1440