Solution of Nonlinear Partial Differential Equations by Mixture Adomian Decomposition Method and Sumudu Transform

Abstract

This chapter is fundamentally centering on the application of the Adomian decomposition method and Sumudu transform for solving the nonlinear partial differential equations. It has instituted some theorems, definitions, and properties of Adomian decomposition and Sumudu transform. This chapter is an elegant combination of the Adomian decomposition method and Sumudu transform. Consequently, it provides the solution in the form of convergent series, then, it is applied to solve nonlinear partial differential equations.

1. Introduction

Many of nonlinear phenomena are a necessary part in applied science and engineering fields [1]. The wide use of nonlinear partial differential equations is the most important reason why they have drawn mathematician’s attention. Despite this, they are not easy to find an answer, either numerically or theoretically. In the past, active study attempts were given a large amount of attention to the study of getting exact or approximate solutions of this kind of equations.

Therefore, it becomes increasingly important to be familiar with all traditional and recently developed methods for solving partial differential equations. For some examples of the traditional methods, such as, the separation of variables method, the method of characteristics, the $\sigma$-expansion method, the integral transforms and Hirota bilinear method [2, 3, 4, 5]. Moreover, the recently developed methods like, Adomian decomposition method (ADM) [1, 6, 7, 8, 9], He’s semi – inverse method, the tanh method, the sinh – cosh method, the homotopy perturbation method (HPM) [3, 4, 10, 11], the differential transform method (DTM), the variational iteration method (VIM) [1, 5, 12], and the weighted finite difference.

In this chapter, our presentation will be based on applying the new method, namely the Adomian Decomposition Sumudu Transform Method (ADSTM) for solving the nonlinear partial differential equations. This method is an elegant combination of the Sumudu transform method and decomposition method.

2. Sumudu Transform

A long time ago, differential equations wared a necessary part in all aspects of applied sciences and engineering fields. In this chapter we need to develop a new technique for help us to obtain the exact and approximate solutions of these differential equations.

Watugula [13] introduced a new integral transform and called it as Sumudu transform, which is defined as:

$$F\left(u\right)=S\left[f\left(t\right)\right]=\underset{0}{\overset{\infty }{\int }}\frac{1}{u}{e}^{\left(-\frac{t}{u}\right)}f\left(t\right)\mathit{dt};$$E1

Watugula [13] applied this transforms to the solution of ordinary differential equations. Because of its useful properties, the Sumudu transforms helps in solving complex problems in applied sciences and engineering mathematics. Henceforward, is the definition of the Sumudu transforms and properties describing the simplicity of the transform.

Definition 1: The Sumudu transform of the function $f\left(t\right)$ is defined by:

$$F\left(u\right)=S\left[f\left(t\right)\right]=\underset{0}{\overset{\infty }{\int }}\frac{1}{u}{e}^{\left(-\frac{t}{u}\right)}f\left(t\right)\mathit{dt}$$E2

Or,

$$F\left(u\right)=S\left[f\left(t\right)\right]=\underset{0}{\overset{\infty }{\int }}f\left(ut\right)e^{-t}\mathit{dt}$$E3

For any function$f\left(t\right)$ and $-{\tau }_1<u<{\tau }_2$.

3. The relation between Sumudu and Laplace transform

The Sumudu transform $F_s\left(u\right)$ of a function $f\left(t\right)$ defined for all real numbers $t\ge 0$. The Sumudu transform is essentially identical with the Laplace transform.

Given an initial $f\left(t\right)$ its Laplace transform $G\left(u\right)$ can be translated into the Sumudu transform $F_s\left(u\right)$ of $f$ by means of the relation;

$$F\left(u\right)=\frac{G\left(\frac{1}{u}\right)}{u},\text{and}\mathrm{it}’\mathrm{s}\text{inverse},G\left(s\right)=\frac{F_s\left(\frac{1}{s}\right)}{s}$$

Theorem 1: Let $f\left(t\right)$ with Laplace transform $G\left(s\right)$, then, the Sumudu transform $F\left(u\right)$ of $f\left(t\right)$ is given by, $F\left(u\right)=\frac{G\left(\frac{1}{u}\right)}{u}$.

Proof:

Form definition ($1$.1.1) we get:

$F\left(u\right)=\underset{0}{\overset{\infty }{\int }}{e}^{-t}f\left(ut\right)\mathit{dt}$, If we set $w=ut$ and $\mathit{dt}=\frac{\mathit{dw}}{u}$ then:

$$F\left(u\right)=\underset{0}{\overset{\infty }{\int }}{e}^{\left(-\frac{w}{u}\right)}f\left(w\right)\frac{\mathit{dw}}{u}=\frac{1}{u}\underset{0}{\overset{\infty }{\int }}{e}^{\left(-\frac{w}{u}\right)}f\left(w\right)\mathit{dw}$$

By definition of the Laplace transform we get: $F\left(u\right)=\frac{G\left(\frac{1}{u}\right)}{u}$.

Theorem 2: It deals with the effect of the differentiation of the function $f\left(t\right)$, $k$ times on the Sumudu transform $F\left(u\right)$ if $S\left[f\left(t\right)\right]=F\left(u\right)$ then:

1. $S\left[f^{\prime }\left(t\right)\right]=\frac{1}{u}\left[F\left(u\right)-f\left(0\right)\right]$

2. $S\left[f^{″}\left(t\right)\right]=\frac{1}{u^2}\left[F\left(u\right)\right]-\frac{1}{u^2}f\left(0\right)-\frac{1}{u}{f}^{\prime }\left(u\right)$

3. $S\left[f^{\left(n\right)}\left(t\right)\right]=\frac{1}{u^n}\left[F\left(u\right)\right]-\frac{1}{u^n}\sum _{k=0}^{n-1}{u}^k{f}^{\left(k\right)}\left(0\right)=u^{-n}\left[F\left(u\right)-\sum _{k=0}^{n-1}{u}^k{f}^{\left(k\right)}\left(0\right)\right]$

Where $f^{\left(0\right)}\left(0\right)=f\left(0\right),f^{\left(k\right)}\left(0\right),k=1,2,3,\cdots ,n-1$ are the nth-order derivatives of the function $f\left(t\right)$ evaluated at, $t=0$.

Proof:

i. Using integration by parts,

ii. $\begin{array}{l}S\left[f^{\prime }\left(t\right)\right]={\left[\frac{1}{u}exp\left(-\frac{t}{u}f\left(t\right)\right)\right]}_0^{\infty }+\frac{1}{u}\underset{0}{\overset{\infty }{\int }}\frac{1}{u}exp\left(-\frac{t}{u}\right)f\left(t\right)\mathit{dt}=-\frac{1}{u}f\left(0\right)+\frac{1}{u}F\left(u\right)\\ S\left[f^{\prime }\left(t\right)\right]=\frac{1}{u}\left[F\left(u\right)-f\left(0\right)\right]\end{array}$.

Using integration by parts;

$$S\left[f^{″}\left(t\right)\right]={\left[\frac{1}{u}{e}^{\left(-\frac{t}{u}\right)}{f}^{\prime }\left(t\right)\right]}_0^{\infty }+\frac{1}{u}\underset{0}{\overset{\infty }{\int }}\frac{1}{u}{e}^{\left(-\frac{t}{u}\right)}{f}^{\prime }\left(t\right)\mathit{dt}$$

From (i) $\begin{array}{c}=-\frac{1}{u}{f}^{\prime }\left(0\right)+\frac{1}{u}S\left[f^{\prime }\left(t\right)\right]\\ S\left[f^{″}\left(t\right)\right]=\frac{1}{u^2}\left[F\left(u\right)\right]-\frac{1}{u^2}f\left(0\right)-\frac{1}{u}{f}^{\prime }\left(0\right)\end{array}$.

iii. By definition the Laplace transform for $f^{\left(n\right)}\left(t\right)$ is given by

$$G_n\left(s\right)=s^{n}G\left(s\right)-\sum _{k=0}^{n-1}{s}^{n-\left(k+1\right)}{f}^{\left(k\right)}\left(0\right)$$

By using the relation between Sumudu and Laplace transform;

$$G_n\left(\frac{1}{u}\right)=\frac{G\left(\frac{1}{u}\right)}{u^n}-\sum _{k=0}^{n-1}\frac{f^{\left(k\right)}\left(0\right)}{u^{n-\left(k+1\right)}}$$

Since $F_n\left(u\right)=\frac{G_n\left(\frac{1}{u}\right)}{u^n}$, we get:

$$\begin{array}{l}u{F}_n\left(u\right)=\frac{uF\left(u\right)}{u^n}-\sum _{k=0}^{n-1}\frac{f^{\left(k\right)}\left(0\right)}{u^{n-k}{u}^{-1}}\\ F_n\left(u\right)=\frac{F\left(u\right)}{u^n}-\sum _{k=0}^{n-1}\frac{f^{\left(k\right)}\left(0\right)}{u^{n-k}}\\ F_n\left(u\right)=u^{-n}F\left(u\right)-\sum _{k=0}^{n-1}{u}^{-n}{u}^k{f}^{\left(k\right)}\left(0\right)\\ S\left[f^{\left(n\right)}\left(t\right)\right]=F\left(u\right)=u^{-n}\left[F\left(u\right)-\sum _{k=0}^{n-1}{u}^k{f}^{\left(k\right)}\left(0\right)\right]\end{array}$$

4. Adomian decomposition method

Many of nonlinear phenomena are a necessary part in applied science and engineering fields. Nonlinear equations are noticed in a different type of physical problems [1], such as fluid dynamics, plasma physics, solid mechanics, and quantum field theory.

The wide use of these equations is the most important reason why they have drawn mathematician’s attention. Despite this, they are not easy to find an answer, either numerically or theoretically.

In the past, active study attempts were given a large amount of attention to the study of getting exact or approximate solutions of this kind of equations. It is significant to note that several powerful methods have been advanced for this purpose.

The Adomian decomposition method will be used in this chapter and in other chapters to deal with nonlinear equations. The Adomian decomposition method proves to be powerful, effective and successfully used to handle most types of linear or nonlinear ordinary or partial differential equations, and linear or nonlinear integral equations.

In the following, the Adomian scheme for calculating a wide variety of forms of nonlinearity.

5. Calculation of Adomian polynomials

It is well known that the Adomian decomposition method suggests the unknown linear function $u$ may be represented by the decomposition series;

$$u=\sum _{n=0}^{\infty }{u}_n,$$E4

Where the components $u_n,n\ge 0$ can be elegantly computed in a recursive way. However, the nonlinear term $F\left(u\right)$, such as $u^2,u^3,u^4,sinu,e^u,u{u}_x,{u_x}^2,$ etc., can be expressed by an infinite series of the so- called Adomian polynomials $A_n$ given in the form;

$$F\left(u\right)=\sum _{n=0}^{\infty }{A}_n\left(u_0,u_1,u_2,\dots ,u_n\right).$$E5

The Adomian polynomials $A_n$ for the nonlinear term $F\left(u\right)$ can be evaluated by using the following expression;

$$A_n=\frac{1}{n!}\frac{d^n}{d{\lambda }^n}{\left[F\left(\sum _{i=0}^n{\lambda }^i{u}_i\right)\right]}_{\lambda =0},n=0,1,2,\dots$$E6

Assuming that the nonlinear function is $F\left(u\right)$, therefore, by using (6), Adomian polynomials are given by;

$$\begin{array}{l}{A}_0=F\left(u_0\right),\\ A_1=u_1{F}^{\prime }\left(u_0\right),\\ A_2=u_2{F}^{\prime }\left(u_0\right)+\frac{1}{2!}{u_1}^2{F}^{″}\left(u_0\right),\\ A_3=u_3{F}^{\prime }\left(u_0\right)+u_1{u}_2{F}^{″}\left(u_0\right)+\frac{1}{3!}{u_1}^3{F}^{‴}\left(u{}_0\right).\end{array}$$E7

Other polynomials can be generated in a similar manner.

Substituting (7) into (5) gives;

$$\begin{array}{ll}F\left(u\right)& =A_0+A_1+A_2+A_3+\dots =F\left(u_0\right)+\left(u_1+u_2+u_3+\dots \right)F^{\prime }\left(u_0\right)\\ & +\frac{1}{2!}\left({u_1}^2+2u_1{u}_2+{u_2}^2+\dots \right)F^{″}\left(u_0\right)+\frac{1}{3!}\left({u_1}^3+3{u_1}^2{u}_2+3{u_1}^2{u}_3+\dots \right)F^{‴}\left(u_0\right)+\dots \\ & =F\left(u_0\right)+\left(u-u_0\right)F^{\prime }\left(u_0\right)+\frac{1}{2!}{\left(u-u_0\right)}^2{F}^{″}\left(u_0\right)+.\dots \end{array}$$

The last expansion confirms a fact that the series in $A_n$ polynomials is a Taylor series about a function $u_0$ and not about a point as is usually used.

In the following, we will calculate Adomian polynomials for several forms of nonlinearity.

5.1 Nonlinear polynomials

$$\text{If}F\left(u\right)=u^2$$

The polynomials can be found as follows:

$$\begin{array}{l}{A}_0=F\left(u_0\right)={u_0}^2,A_1=u_1{F}^{\prime }\left(u_0\right)=2u_0{u}_1,A_2=u_2{F}^{\prime }\left(u_0\right)+\frac{1}{2!}{u_1}^2{F}^{″}\left(u_0\right)=2u_0{u}_2+{u_1}^2,\\ A_3=u_3{F}^{\prime }\left(u_0\right)+u_1{u}_2{F}^{″}\left(u_0\right)+\frac{1}{3!}{u_1}^3{F}^{‴}\left(u{}_0\right)=2u_0{u}_3+2u_1{u}_2.\end{array}$$

And so on. Proceeding as before, we find $u^3,u^4,u^5,\dots$, etc.

5.2 Nonlinear derivatives

Case1.$F\left(u\right)={\left(u_x\right)}^2$

$$A_0={{u_0}_x}^2,A_1=2u_{0x}{u}_{1_x},A_2=2u_{0x}{u}_{2x}+{{u_1}_x}^2,A_3=2u_{0x}{u}_{3x}+2u_{1x}{u}_{2x}.$$

And so on. In a similar, we get ${u_x}^3,{u_x}^4,{u_x}^5,.\dots$, etc.

Case 2. $F\left(u\right)=u{u}_x=\frac{1}{2}{L}_x\left(u^2\right)$

The $A_n$ polynomials in this case given by;

$$\begin{array}{l}{A}_0=F\left(u_0\right)=u_0{u}_{0_x},A_1=\frac{1}{2}{L}_x\left(2u_0{u}_1\right)=u_{0x}{u}_1+u_0{u}_{1_x},\\ A_2=\frac{1}{2}{L}_x\left(2u_0{u}_2+{u_1}^2\right)=u_{0_x}{u}_2+u_0{u}_{2_x}{u}_0+u_1{u}_{1_x},\\ A_3=\frac{1}{2}{L}_x\left(2u_0{u}_3+2u_1{u}_2\right)=u_{0_x}{u}_3+u_{1_x}{u}_2+u_{2_x}{u}_1+u_{3_x}{u}_0.\end{array}$$

And so on.

5.3 Trigonometric nonlinearity

$$\text{If}F\left(u\right)=sinu$$

The Adomian polynomials for this form nonlinearity are given by;

$$A_0=sin{u}_0,A_1=u_{1}cos{u}_0,A_2=u_{2}cos{u}_0-\frac{1}{2!}{u_1}^{2}sin{u}_0,A_3=u_{3}cos{u}_0-u_1{u}_{2}sin{u}_0-\frac{1}{3!}{u_1}^{3}cos{u}_0.$$

And so on. In a similar way, we find $F\left(u\right)=cosu$.

5.4 Hyperbolic nonlinearity

$$\text{If}F\left(u\right)=sinhu$$

The $A_n$ polynomials in this case are given by;

$$A_0=sinh{u}_0,A_1=u_{1}cosh{u}_0,A_2=u_{2}cosh{u}_0+\frac{1}{2!}{u_1}^{2}sinh{u}_0,A_3=u_{3}cosh{u}_0+u_1{u}_{2}sinh{u}_0+\frac{1}{3!}{u_1}^{3}cosh{u}_0.$$

And so on. In a parallel manner, Adomian polynomials can be calculated for $F\left(u\right)=coshu$.

5.5 Exponential nonlinearity

$$\text{If}F\left(u\right)=e^u$$

The Adomian polynomials in this form of nonlinearity are given by;

$$A_0=e^{u_0},A_1=u_1{e}^{u_0},A_2=\left(u_2+\frac{1}{2!}{u_1}^2\right)e^{u_0},A_3=\left(u_3+u_1{u}_2+\frac{1}{3!}{u_1}^3\right)e^{u_0}.$$

And so on. Proceeding as a before, we find $F\left(u\right)=e^{-u}$.

5.6 Logarithmic nonlinearity

$$\text{If}F\left(u\right)=lnu,u>0$$

The $A_n$ polynomials for logarithmic nonlinearity are given by;

$$A_0=ln{u}_0,A_1=\frac{u_1}{u_0},A_2=\frac{u_2}{u_0}-\frac{1}{2}\frac{{u_1}^2}{{u_0}^2},A_3=\frac{u_3}{u_0}-\frac{u_1{u}_2}{{u_0}^2}+\frac{1}{3}\frac{{u_1}^3}{{u_0}^3}.$$

And so on. In a similar way, we find $F\left(u\right)=ln\left(1+u\right),-1<u\le 1$.

6. A New algorithm for calculating Adomian polynomials (The alternative algorithm for calculating Adomian polynomials)

It is well known about the main disadvantage of the calculating Adomian polynomials $A_n$, that it is a difficult method to perform calculation so called nonlinear terms. There is an alternative algorithm to reduce the demerits of formula introduced before, which depends mainly on algebraic, trigonometric identities and on Taylor expansions.

In the alternative algorithm which is a simple and reliable may be employed to calculate Adomian Polynomials $A_n$.

The new algorithm will be clarified by discussing the following suitable forms of nonlinearity.

6.1 Nonlinear polynomials

$$\text{If}F\left(u\right)=u^2$$

We first set,

$$u=\sum _{n=0}^{\infty }{u}_n.$$E8

Substituting (8) into $F\left(u\right)=u^2$ gives;

$$F\left(u\right)={\left(u_0+u_1+u_2+u_3+u_4+\dots \right)}^2.$$E9

Expanding the expression at the right- hand side gives;

$$F\left(u\right)={u_0}^2+2u_0{u}_1+2u_0{u}_2+{u_1}^2+2u_0{u}_3+2u_1{u}_2+.\dots$$E10

The expansion in (10) can be rearranged by grouping all terms with the sum of the subscripts of the components is the same. This means that we can rewrite (10) as;

$$F\left(u\right)=\underset{A_0}{\underbrace{{u_0}^2}}+\underset{A_1}{\underbrace{2u_0{u}_1}}+\underset{A_2}{\underbrace{2u_0{u}_2+{u_1}^2}}+\underset{A_3}{\underbrace{2u_0{u}_3+2u_1{u}_2}}+\dots .$$E11

This gives Adomian polynomials for $F\left(u\right)=u^2$ by;

$$A_0={u_0}^2,A_1=2u_0{u}_1,A_2=2u_0{u}_2+{u_1}^2,A_3=2u_0{u}_3+2u_1{u}_2.$$

And so on. Proceeding as before, we get $u^3,u^4,u^5,.\dots$, etc.

6.2 Nonlinear derivatives

Case 1.$\text{If}F\left(u\right)={u_x}^2.$

We first set;

$$u_x=\sum _{n=0}^{\infty }{u}_{nx}.$$E12

Substituting (12) into $F\left(u\right)={u_x}^2$ giving;

$$F\left(u\right)={\left(u_{0x}+u_{1x}+u_{2x}+u_{3x}+u_{4x}+\dots \right)}^2.$$E13

Squaring the right – hand side gives;

$$F\left(u\right)={{u_0}_x}^2+2u_{0x}{u}_{1x}+2u_{0x}{u}_{2x}+{{u_1}_x}^2+2u_{0x}{u}_{3x}+2u_{1x}{u}_{2x}+.\dots$$E14

Grouping the terms as discussed above, we find;

$$F\left(u\right)=\underset{A_0}{\underbrace{{{u_0}_x}^2}}+\underset{A_1}{\underbrace{2u_{0x}{u}_{1x}}}+\underset{A_2}{\underbrace{2u_{0x}{u}_{2x}+{{u_1}_x}^2}}+\underset{A_3}{\underbrace{2u_{0x}{u}_{3x}+2u_{1x}{u}_{2x}}}+.\dots$$E15

Adomian polynomials are given by;

$$A_0={{u_0}_x}^2,A_1=2u_{0x}{u}_{1_x},A_2=2u_{0x}{u}_{2x}+{{u_1}_x}^2,A_3=2u_{0x}{u}_{3x}+2u_{1x}{u}_{2x}.$$

Case 2.$F\left(u\right)=u{u}_x$

Note that this form of nonlinearity appears in advection problems and in nonlinear Burgers equations. We first set;

$$u=\sum _{n=0}^{\infty }{u}_n,u_x=\sum _{n=0}^{\infty }{u}_{\mathit{nx}}.$$E16

Substituting (16) into $F\left(u\right)=u{u}_x$ yields;

$$F\left(u\right)=\left(u_0+u_1+u_2+u_3+u_4+\dots \right)\times \left(u_{0x}+u_{1x}+u_{2x}+u_{3x}+u_{4x}+\dots \right);$$E17

Multiplying the two factors gives;

$$\begin{array}{c}F\left(u\right)=u_0{u}_{0x}+u_{0x}{u}_1+u_0{u}_{1x}+u_{0x}{u}_2+u_{1x}{u}_1+u_{2x}{u}_0+u_{0x}{u}_3+u_{1x}{u}_2+\\ +u_{2x}{u}_1+u_{3x}{u}_0+u_{0x}{u}_4+u_0{u}_{4x}+u_{1x}{u}_3+u_1{u}_{3x}+u_2{u}_{2x}+\dots \end{array}.$$E18

Proceeding with grouping the terms are obtained;

$$F\left(u\right)=\underset{A_0}{\underbrace{u_0{u}_{0x}}}+\underset{A_1}{\underbrace{u_{0x}{u}_1+u_0{u}_{1x}}}+\underset{A_2}{\underbrace{u_{0x}{u}_2+u_{1x}{u}_1+u_{2x}{u}_0}}+\underset{A_3}{\underbrace{u_{0x}{u}_3+u_{1x}{u}_2+u_{2x}{u}_1+u_{3x}{u}_0}}\dots$$E19

Consequently, the Adomian polynomials are given by;

$$\begin{array}{l}{A}_0=u_0{u}_{0_x},A_1=u_{0x}{u}_1+u_0{u}_{1_x},A_2=u_{0_x}{u}_2+u_0{u}_{2_x}{u}_0+u_1{u}_{1_x},\\ A_3=u_{0_x}{u}_3+u_{1_x}{u}_2+u_{2_x}{u}_1+u_{3_x}{u}_0.\end{array}$$

Proceeding as before, we find $F\left(u\right)=u^2{u}_x$.

6.3 Trigonometric nonlinearity

$$\text{If}F\left(u\right)=sinu$$

First, we should be separate $A_0=F\left(u_0\right)$ from other terms. To achieve this goal, we first substitute;

$$u=\sum _{n=0}^{\infty }{u}_n;$$E20

Into $F\left(u\right)=sinu$ to obtain;

$$F\left(u\right)=sin\left[u_0+\left(u_1+u_2+u_3+u_4+\dots \right)\right].$$E21

To separate $A_0$, recall the trigonometric identity;

$$sin\left(\theta +\varphi \right)=sin\theta cos\varphi +cos\theta sin\varphi .$$E22

This means that;

$$F\left(u\right)=sin{u}_{0}cos\left(u_1+u_2+u_3+u_4+\dots \right)+cos{u}_{0}sin\left(u_1+u_2+u_3+u_4+\dots \right).$$E23

Separating $F\left(u_0\right)=sin{u}_0$ from other factors and using Taylor expansion for, $cos\left(u_1+u_2+u_3+u_4+.\dots \right)$ and, $sin\left(u_1+u_2+u_3+u_4+.\dots \right)$ gives;

$$\begin{array}{c}F\left(u\right)=sin{u}_0\left(1-\frac{1}{2!}{\left(u_1+u_2+\dots \right)}^2+\frac{1}{4!}{\left(u_1+u_2+\dots \right)}^4-\dots \right)+\\ +cos{u}_0\left(\left(u_1+u_2+\dots \right)-\frac{1}{3!}{\left(u_1+u_2+\dots \right)}^3+\dots \right)\end{array},$$E24

So that;

$$F\left(u\right)=sin{u}_0\left(1-\frac{1}{2!}\left({u_1}^2+2u_1{u}_2+\dots \right)+\dots \right)+cos{u}_0\left(\left(u_1+u_2+\dots \right)-\frac{1}{3!}{u_1}^3+\dots \right).$$E25

The last expansion can be rearranged by grouping all terms with the same sum of subscripts. This leads to;

$$\begin{array}{l}F\left(u\right)=\underset{A_0}{\underbrace{sin{u}_0}}+\underset{A_1}{\underbrace{u_{1}cos{u}_0}}+\underset{A_2}{\underbrace{u_{2}cos{u}_0-\frac{1}{2!}{u_1}^{2}sin{u}_0}}+\\ +\underset{A_3}{\underbrace{u_{3}cos{u}_0-u_1{u}_{2}sin{u}_0-\frac{1}{3!}{u_1}^{3}cos{u}_0}}+\dots \end{array}$$E26

This completes the calculation of the Adomian polynomials for nonlinear operator $F\left(u\right)=sinu$, therefore we write;

$$\begin{array}{l}{A}_0=sin{u}_0,A_1=u_{1}cos{u}_0,A_2=u_{2}cos{u}_0-\frac{1}{2!}{u_1}^{2}sin{u}_0,\\ A_3=u_{3}cos{u}_0-u_1{u}_{2}sin{u}_0-\frac{1}{3!}{u_1}^{3}cos{u}_0.\end{array}$$

And so on. In a similar way, we find $F\left(u\right)=cosu$.

6.4 Hyperbolic nonlinearity

If $F\left(u\right)=sinhu$ then, we first substitute

$$u=\sum _{n=0}^{\infty }{u}_n;$$E27

Into $F\left(u\right)=sinhu$ to obtain;

$$F\left(u\right)=sinh\left[u_0+\left(u_1+u_2+u_3+u_4+\dots \right)\right].$$E28

To calculate $A_0$, recall the hyperbolic identity;

$$sinh\left(\theta +\varphi \right)=sinh\theta cosh\varphi +cosh\theta sinh\varphi .$$E29

Accordingly, Eq. (29) becomes;

$$F\left(u\right)=sinh{u}_{0}cosh\left(u_1+u_2+u_3+u_4+\dots \right)+cosh{u}_{0}sinh\left(u_1+u_2+u_3+u_4+\dots \right).$$E30

Separating $F\left(u_0\right)=sinh{u}_0$ from other factors and using Taylor expansion for $cosh\left(u_1+u_2+u_3+u_4+.\dots \right)$ and $sinh\left(u_1+u_2+u_3+u_4+.\dots \right)$ gives;

$$\begin{array}{ll}F\left(u\right)& =sinh{u}_0\left(1+\frac{1}{2!}{\left(u_1+u_2+\dots \right)}^2+\frac{1}{4!}{\left(u_1+u_2+\dots \right)}^4+\dots \right)\\ & +cosh{u}_0\left(\left(u_1+u_2+\dots \right)+\frac{1}{3!}{\left(u_1+u_2+\dots \right)}^3+\dots \right)\\ & =sinh{u}_0\left(1+\frac{1}{2!}\left({u_1}^2+2u_1{u}_2+\dots \right)+\dots \right)+cosh{u}_0\left(\left(u_1+u_2+\dots \right)+\frac{1}{3!}{u_1}^3+\dots \right)\end{array}$$

By grouping all terms with the same sum of subscripts we find

$$F\left(u\right)=\underset{A_0}{\underbrace{sinh{u}_0}}+\underset{A_1}{\underbrace{u_{1}cosh{u}_0}}+\underset{A_2}{\underbrace{u_{2}cosh{u}_0+\frac{1}{2!}{u_1}^{2}sinh{u}_0}}+\underset{A_3}{\underbrace{u_{3}cosh{u}_0+u_1{u}_{2}sinh{u}_0-\frac{1}{3!}{u_1}^{3}cosh{u}_0}}+\dots$$

Consequently, the Adomian polynomials for $F\left(u\right)=sinhu$ are given by;

$$\begin{array}{l}{A}_0=sinh{u}_0,A_1=u_{1}cosh{u}_0,A_2=u_{2}cosh{u}_0+\frac{1}{2!}{u_1}^{2}sinh{u}_0,\\ A_3=u_{3}cosh{u}_0+u_1{u}_{2}sinh{u}_0+\frac{1}{3!}{u_1}^{3}cosh{u}_0.\end{array}$$

Similarly as before, we find $F\left(u\right)=coshu$.

6.5 Exponential nonlinearity

$$\text{If}F\left(u\right)=e^u.$$

Substituting

$$u=\sum _{n=0}^{\infty }{u}_n;$$E31

Into $F\left(u\right)=e^u$ gives;

$$F\left(u\right)=e^{\left(u_0+u_1+u_2+u_3+u_4+\dots \right)}.$$E32

Or equivalently;

$$F\left(u\right)=e^{u_0}\times e^{\left(u_1+u_2+u_3+u_4+\dots \right)}.$$E33

Keeping the term $F\left(u_0\right)=e^{u_0}$ and using Taylor expansion for the other factors we obtain;

$$F\left(u\right)=e^{u_0}\times \left(1+\left(u_1+u_2+u_3+\dots \right)+\frac{1}{2!}{\left(u_1+u_2+u_3+\dots \right)}^2+\dots \right).$$E34

By grouping all terms with an identical sum of subscripts we find

$$F\left(u\right)=\underset{A_0}{\underbrace{e^{u_0}}}+\underset{A_1}{\underbrace{u_1{e}^{u_0}}}+\underset{A_2}{\underbrace{\left(u_2+\frac{1}{2!}{u_1}^2\right)e^{u_0}}}+\underset{A_3}{\underbrace{\left(u_3+u_1{u}_2+\frac{1}{3!}{u_1}^3\right)e^{u_0}}}+.\dots$$E35

It then follows that;

$$A_0=e^{u_0},A_1=u_1{e}^{u_0},A_2=\left(u_2+\frac{1}{2!}{u_1}^2\right)e^{u_0},A_3=\left(u_3+u_1{u}_2+\frac{1}{3!}{u_1}^3\right)e^{u_0}.$$

And so on. Proceeding as a before, we find $F\left(u\right)=e^{-u}$.

6.6 Logarithmic nonlinearity

$$\text{If}F\left(u\right)=lnu,u>0$$

Substituting

$$u=\sum _{n=0}^{\infty }{u}_n;$$E36

Into $F\left(u\right)=lnu$ gives;

$$F\left(u\right)=ln\left(u_0+u_1+u_2+u_3+u_4+\dots \right).$$E37

Eq. (34) can be written as;

$$F\left(u\right)=ln\left(u_0\left(1+\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right)\right).$$E38

Using the identity $ln\left(\alpha \beta \right)=ln\alpha +ln\beta$, Eq. (38) becomes;

$$F\left(u\right)=ln\left(u_0\right)+ln\left(1+\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right).$$E39

Separating $F\left(u_0\right)=ln\left(u_0\right)$ and using Taylor expansion of the remaining term, we obtain;

$$F\left(u\right)=ln\left(u_0\right)+\left\{\begin{array}{l}\left(\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right)-\frac{1}{2}{\left(\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right)}^2+\frac{1}{3}{\left(\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right)}^3\\ -\frac{1}{4}{\left(\frac{u_1}{u_0}+\frac{u_2}{u_0}+\frac{u_3}{u_0}+\dots \right)}^4+\dots \end{array}\right\}$$E40

Proceeding as before, Eq. (40) can be rewritten as;

$$F\left(u\right)=\underset{A_0}{\underbrace{ln\left(u_0\right)}}+\underset{A_1}{\underbrace{\frac{u_1}{u_0}}}+\underset{A_2}{\underbrace{\frac{u_2}{u_0}-\frac{1}{2}\frac{{u_1}^2}{{u_0}^2}}}+\underset{A_3}{\underbrace{\frac{u_3}{u_0}-\frac{u_1{u}_2}{{u_0}^2}+\frac{1}{3}\frac{{u_1}^3}{{u_0}^3}}}+.\dots$$E41

Based on this, the Adomian polynomials are given by;

$$A_0=ln\left(u_0\right),A_1=\frac{u_1}{u_0},A_2=\frac{u_2}{u_0}-\frac{1}{2}\frac{{u_1}^2}{{u_0}^2},A_3=\frac{u_3}{u_0}-\frac{u_1{u}_2}{{u_0}^2}+\frac{1}{3}\frac{{u_1}^3}{{u_0}^3}.$$

And so on. In a like manner, we obtain $F\left(u\right)=ln\left(1+u\right),-1<u\le 1$.

7. Adomian decomposition Sumudu transform method for solving nonlinear partial differential equations

In this section, we will concentrate our study on the nonlinear PDEs. There are many nonlinear partial differential equations which are quite useful and applicable in engineering and physics.

The nonlinear phenomena that appear in the many scientific fields’ such as solid state physics, plasma physics, fluid mechanics and quantum field theory can be modeled by nonlinear differential equations. The significance of obtaining exact or approximate solutions of nonlinear partial differential equations in physics and mathematics is yet an important problem that needs new methods to develop new techniques for obtaining analytical solutions. Several powerful mathematical methods are used for this purpose. We, propose a new method, namely Adomian Decomposition Sumudu Transform Method (ADSTM) for solving nonlinear equations. This method is a combination of Sumudu transform and decomposition method which was introduced by D. Kumar, J. Singh and S. Rathore.

(ADSTM) provides the solution for nonlinear equations in the form of convergent series. This forms the motivation for us to apply (ADSTM) for solving nonlinear equations in understanding different physical phenomena.

To illustrate the basic idea of this method, we consider a general non-homogeneous partial differential equation with the initial conditions of the form:

$$\begin{array}{c}DU\left(x,t\right)+RU\left(x,t\right)+NU\left(x,t\right)=g\left(x,t\right)\\ U\left(x,0\right)=h\left(x\right),U_t\left(x,0\right)=f\left(x\right).\end{array};$$E42

Where $D$ is the second order linear differential operator $D=\frac{{\partial }^2}{\partial t^2}$, $R$ is linear differential operator of less order than $D$, $N$ represent the general nonlinear operator and $g\left(x,t\right)$ is the source term.

Taking the Sumudu transform of both sides of Eq. (42), we get:

$$S\left[DU\left(x,t\right)\right]+S\left[RU\left(x,t\right)\right]+S\left[N\left(x,t\right)\right]=S\left[g\left(x,t\right)\right];$$E43

Using the differentiation property of the Sumudu transform and given initial conditions, we have:

$$S\left[U\left(x,t\right)\right]=u^{2}S\left[g\left(x,t\right)\right]+h\left(x\right)+uf\left(x\right)-u^{2}S\left[RU\left(x,t\right)+NU\left(x,t\right)\right].$$E44

If we apply the inverse operator $S^{-1}$ to both sides of Eq. (44), we obtain:

$$U\left(x,t\right)=G\left(x,t\right)-S^{-1}\left[u^{2}S\left[RU\left(x,t\right)+NU\left(x,t\right)\right]\right].$$E45

Where $G\left(x,t\right)$ represents the term arising from the source term and the prescribed initial conditions. Now, apply the Adomain decomposition method:

$$U\left(x,t\right)=\sum _{n=0}^{\infty }{U}_n\left(x,t\right);$$E46

The nonlinear term can be decomposed as:

$$NU\left(x,t\right)=\sum _{n=0}^{\infty }{A}_n\left(U\right);$$E47

For some Adomain polynomials $A_n\left(U\right)$ that are given by:

$$A_n\left(U_0,U_1,U_2,\dots ,U_n\right)=\frac{1}{n!}\frac{d^n}{d{\lambda }^n}{\left[N\left(\sum _{n=0}^{\infty }{\lambda }^n{U}_n\right)\right]}_{\lambda =0},n=0,1,2,.\dots$$

Substituting Eq. (46) and Eq. (47) in Eq. (45), we get:

$$\sum _{n=0}^{\infty }{U}_n\left(x,t\right)=G\left(x,t\right)-S^{-1}\left[u^{2}S\left[R\sum _{n=0}^{\infty }{U}_n\left(x,t\right)+\sum _{n=0}^{\infty }{A}_n\left(U\right)\right]\right].$$E48

Accordingly, the formal recursive relation is defined by:

$$\begin{array}{l}{U}_0\left(x,t\right)=G\left(x,t\right),\\ U_{k+1}\left(x,t\right)=-S^{-1}\left[u^{2}S\left[R{U}_k+A_k\right]\right].k\ge 0.\end{array}$$E49

The Adomian decomposition Sumudu transform method will be illustrated by discussing the following examples.

Example 1: Consider the following nonlinear partial differential equation:

$$U_t+{\mathit{UU}}_x=0;$$E50

With the initial condition:

$$U\left(x,0\right)=x.$$E51

Taking the Sumudu transform of both sides of Eq. (50) and using the initial condition, we have:

$$S\left[U\left(x,t\right)\right]=x-uS\left[{\mathit{UU}}_x\right].$$E52

Applying $S^{-1}$ to both sides of Eq. (52) implies that:

$$U\left(x,t\right)=x-S^{-1}\left[uS\left[{\mathit{UU}}_x\right]\right];$$E53

Following the technique, if we assume an infinite series of the form (53), we obtain:

$$\sum _{n=0}^{\infty }{U}_n\left(x,t\right)=x-S^{-1}\left[uS\left[\sum _{n=0}^{\infty }{A}_n\left(U\right)\right]\right].$$E54

Where $A_n\left(U\right)$ are Adomian polynomials that represent the nonlinear terms.

The first few components of the Adomian polynomials are given by;

$$\begin{array}{l}{A}_0\left(U\right)=U_0{U}_{0_x},\\ A_1\left(U\right)=U_0{U}_{1_x}+U_1{U}_{0_x},\\ ....\end{array}$$E55

This gives the recursive relation:

$$\begin{array}{l}{U}_0\left(x,t\right)=x,\\ U_{k+1}\left(x,t\right)=-S^{-1}\left[uS\left[A_k\right]\right],k\ge 0.\end{array}$$E56

The first few components are given by:

$$\begin{array}{l}{U}_0\left(x,t\right)=x,\\ U_1\left(x,t\right)=-S^{-1}\left[uS\left[A_0\right]\right]=-xt,\\ U_2\left(x,t\right)=-S^{-1}\left[uS\left[A_1\right]\right]=x{t}^2,\\ U_3\left(x,t\right)=-S^{-1}\left[uS\left[A_2\right]\right]=-x{t}^3.\end{array}$$E57

And so on. The solution in a series form is given by:

$$U\left(x,t\right)=x\left(1-t+t^2-t^3+\dots \right);$$E58

And in a closed form of:

$$U\left(x,t\right)=\frac{x}{1+t}.$$E59

Example 2: Consider the following nonlinear partial differential equation:

$$U_t+{\mathit{UU}}_x=x+x{t}^2;$$E60

With the initial condition:

$$U\left(x,0\right)=0.$$E61

Proceeding as in Example 1, Eq. (60) becomes:

$$\sum _{n=0}^{\infty }{U}_n\left(x,t\right)=xt+\frac{x{t}^3}{3}-S^{-1}\left[uS\left[\sum _{n=0}^{\infty }{A}_n\left(U\right)\right]\right].$$E62

The modified decomposition method admits the of a modified recursive relation given by:

$$\begin{array}{l}{U}_0\left(x,t\right)=xt,\\ U_1\left(x,t\right)=\frac{x{t}^3}{3}-S^{-1}\left[uS\left[A_0\right]\right]\\ U_{k+1}\left(x,t\right)=-S^{-1}\left[uS\left[A_k\right]\right],k\ge 1.\end{array}$$E63

Consequently, we obtain:

$$\begin{array}{l}{U}_0\left(x,t\right)=xt,\\ U_1\left(x,t\right)=\frac{x{t}^3}{3}-S^{-1}\left[uS\left[x{t}^2\right]\right]=0\\ U_{k+1}\left(x,t\right)=0,k\ge 1.\end{array}$$E64

In few of Eq. (64), the exact solution is given by:

$$U\left(x,t\right)=xt.$$E65

Example 3: Consider the nonlinear partial differential equation:

$$U_{\mathit{tt}}+{U_x}^2+U-U^2=t{e}^{-x};$$E66

With the initial condition

$$U\left(x,0\right)=0,U_t\left(x,0\right)=e^{-x}.$$E67

By taking Sumudu transform for (66) and using (67) we obtain:

$$S\left[U\left(x,t\right)\right]=u^3{e}^{-x}+u{e}^{-x}-u^{2}S\left[{U_x}^2-U^2+U\right].$$E68

Applying $S^{-1}$ to both sides of (68) we obtain;

$$U\left(x,t\right)={\mathit{te}}^{-x}+\frac{1}{6}{t}^3{e}^{-x}-S^{-1}\left[u^{2}S\left[{U_x}^2-U^2+U\right]\right].$$E69

Substituting;

$$U\left(x,t\right)=\sum _{n=0}^{\infty }{U}_n\left(x,t\right);$$E70

And the nonlinear terms of;

$${U_x}^2=\sum _{n=0}^{\infty }{A}_n,U^2=\sum _{n=0}^{\infty }{B}_n.$$E71

Into (69) gives;

$$\sum _{n=0}^{\infty }{U}_n\left(x,t\right)={\mathit{te}}^{-x}+\frac{1}{6}{t}^3{e}^{-x}-S^{-1}\left[u^{2}S\left(\sum _{n=0}^{\infty }{A}_n+\sum _{n=0}^{\infty }{U}_n\left(x,t\right)-\sum _{n=0}^{\infty }{B}_n\right)\right]$$E72

This gives the modified recursive relation;

$$\begin{array}{l}{U}_0\left(x,t\right)=t{e}^{-x},\\ U_1\left(x,t\right)=\frac{1}{6}{t}^3{e}^{-x}-{L_t}^{-1}\left(A_0+U_0-B_0\right)\\ U_{k+1}\left(x,t\right)=-{L_t}^{-1}\left(A_k+U_k-B_k\right),k\ge 1.\end{array}$$E73

The first few of the components are given by;

$$\begin{array}{l}{U}_0\left(x,t\right)=t{e}^{-x},\\ U_1\left(x,t\right)=\frac{1}{6}{t}^3{e}^{-x}-{L_t}^{-1}\left(A_0+U_0-B_0\right)=0,\\ U_{k+1}\left(x,t\right)=0,k\ge 1.\end{array}$$E74

The solution in a closed form is given by;

$$U\left(x,t\right)=t{e}^{-x}.$$E75

Example 4: Consider the following nonlinear partial differential equation,

$$U_{tt}+U^2-{U_x}^2=0;$$E76

With the initial conditions

$$U\left(x,0\right)=0,U_t\left(x,0\right)=e^x.$$E77

By taking Sumudu transform for (76) and using (77) we obtain:

$$S\left[U\left(x,t\right)\right]=u{e}^x+u^{2}S\left[{U_x}^2-U^2\right].$$E78

By applying the inverse Sumudu transform of (78), we get:

$$U\left(x,t\right)=t{e}^x+S^{-1}\left[u^{2}S\left[{U_x}^2-U^2\right]\right];$$E79

This assumes a series solution of the function $U\left(x,t\right)$is given by:

$$U\left(x,t\right)=\sum _{n=0}^{\infty }{U}_n\left(x,t\right);$$E80

Using (80) into (79), we get:

$$\sum _{n=0}^{\infty }{U}_n\left(x,t\right)=t{e}^x+S^{-1}\left[u^{2}S\left[\sum _{n=0}^{\infty }{A}_n\left(U\right)-\sum _{n=0}^{\infty }{B}_n\left(U\right)\right]\right].$$E81

Where $A_n\left(U\right)$ and $B_n\left(U\right)$ are Adomian polynomials that represents nonlinear terms.

The few components of the Adomian polynomials are given as follows:

$$\begin{array}{l}{A}_0\left(U\right)={U_{0_x}}^2,A_1\left(U\right)=2U_{0_x}{U}_{1_x},\\ B_0\left(U\right)={U_0}^2,B_1\left(U\right)=2U_0{U}_1,\end{array}$$E82

And so on. From the above equations we obtain:

$$\begin{array}{l}{U}_0\left(x,t\right)=t{e}^x,\\ U_{k+1}\left(x,t\right)=S^{-1}\left[u^{2}S\left[A_k-B_k\right]\right],k\ge 0.\end{array}$$E83

The first few terms of $U_n\left(x,t\right)$ follows immediately upon setting:

$$\begin{array}{l}{U}_1\left(x,t\right)=S^{-1}\left[u^{2}S\left[A_0-B_0\right]\right]=S^{-1}\left[u^{2}S\left[{U_{0_x}}^2-{U^2}_0\right]\right]=0\\ U_{k+1}\left(x,t\right)=0,k\ge 1.\end{array}$$E84

Therefore the solution obtained by ADSTM is given as follows:

$$U\left(x,t\right)=t{e}^x.$$

8. Nonlinear physical models

Now we will, concentrate our study on the linear and nonlinear particular applications that appear in applied science. The wide use of these physical significant problems is the most important reason why they have drawn mathematician’s attention in recent years.

Nonlinear partial differential equations have witnessed remarkable improvement. Nonlinear problems appear in the many scientific fields’ such as gravitation, chemical reaction, fluid dynamics, dispersion, nonlinear optics, plasma physics, acoustics, and others. Several approaches have been used such as the Adomian decomposition method, the variational iteration method, and the characteristics method and perturbation techniques to examine these problems.

(ADSTM) gives the solution of nonlinear equations in the form of convergent series. The main advantage of this method is its potentiality of combining two powerful methods for deriving exact and approximate solution of nonlinear equations. This forms the motivation for us to apply (ADSTM) for solving nonlinear equations in understanding different physical phenomena.

The following section offers the effectiveness of the Adomian decomposition Sumudu transform method (ADSTM) in solving nonlinear physical models.

Example 5: Consider the following inhomogeneous advection problem:

$$U_t+{\mathit{UU}}_x=2t+x+t^3+x{t}^2;$$E85

With the initial condition:

$$U\left(x,0\right)=0.$$E86

Following discussion presented above, we obtain the recursive relation;

$$\begin{array}{l}{U}_0\left(x,t\right)=t^2+xt+\frac{t^4}{4}+\frac{xt}{3},\\ U_{k+1}\left(x,t\right)=-S^{-1}\left[uS\left[A_k\right]\right],k\ge 0.\end{array}$$E87

This gives;

$$\begin{array}{l}{U}_0\left(x,t\right)=t^2+xt+\frac{t^4}{4}+\frac{x{t}^3}{3},\\ U_1\left(x,t\right)=-\frac{t^4}{4}-\frac{x{t}^3}{3}-\frac{2}{15}x{t}^5-\frac{7}{72}{t}^6-\frac{1}{63}x{t}^7-\frac{1}{98}{t}^8.\end{array}$$E88

It is easily observed that two noise term appears in the components $U_0\left(x,t\right)$ and $U_1\left(x,t\right)$. By canceling these terms from $U_0\left(x,t\right)$, the remaining non-canceled term of $U_0\left(x,t\right)$ may provide the exact solution.

The exact solution is given by;

$$U\left(x,t\right)=t^2+xt.$$

Example 6: Consider the following nonlinear Klein – Gordon equation:

$$U_{\mathit{tt}}-U_{\mathit{xx}}+U^2=x^2{t}^2;$$E89

Subject to the initial conditions:

$$U\left(x,0\right)=0,U_t\left(x,t\right)=x.$$E90

Following the discussion presented above, we find a recursive relation;

$$\begin{array}{l}{U}_0\left(x,t\right)=xt+\frac{1}{12}{x}^2{t}^4,\\ U_{k+1}\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_k\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[u^{2}S\left[A_k\right]\right],k\ge 0.\end{array}$$E91

So the Adomian polynomials $A_n$ are given as follows;

$$\begin{array}{l}{A}_0={U_0}^2,\\ A_1=2U_0{U}_1,\\ A_2=2U_0{U}_2+{U_1}^2.\end{array}$$

And so on. Using modified recursive relation Eq. (91) can be rewritten in the scheme;

$$\begin{array}{l}{U}_0\left(x,t\right)=xt,\\ U_1\left(x,t\right)=\frac{1}{12}{x}^2{t}^4+S^{-1}\left[u^{2}S\left[{\left(U_0\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[u^{2}S\left[A_0\right]\right],\\ U_{k+1}\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_k\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[u^{2}S\left[A_k\right]\right],k\ge 1.\end{array}$$E92

This lead to;

$$\begin{array}{l}{U}_0\left(x,t\right)=xt,\\ U_1\left(x,t\right)=\frac{1}{12}{x}^2{t}^4+S^{-1}\left[u^{2}S\left[{\left(U_0\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[u^{2}S\left[A_0\right]\right]=0,\\ U_{k+1}\left(x,t\right)=0,k\ge 1.\end{array}$$E93

Therefore, the exact solution is given by;

$$U\left(x,t\right)=xt.$$

Example 7: Consider the following Sine-Gordon equation with the given initial conditions:

$$U_{\mathit{tt}}\left(x,t\right)-U_{\mathit{xx}}\left(x,t\right)=sinU;$$E94

Subject to the initial conditions;

$$U\left(x,0\right)=\frac{\pi }{2},U_t\left(x,t\right)=0.$$E95

Using the recursive scheme yields;

$$\begin{array}{l}{U}_0\left(x,t\right)=\frac{\pi }{2},\\ U_{k+1}\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_k\right)}_{\mathit{xx}}\right]\right]+S^{-1}\left[u^{2}S\left[A_k\right]\right],k\ge 0.\end{array}$$E96

The first few the Adomian polynomials for $sinU$ are given as;

$$\begin{array}{l}{A}_0=sin{U}_0,\\ A_1=U_{1}cos{U}_0,\\ A_2=U_{2}cos{U}_0-\frac{1}{2!}{U_1}^{2}sin{U}_0,\\ A_3=U_{3}cos{U}_0-U_1{U}_{2}sin{U}_0-\frac{1}{3!}{U_1}^{3}cos{U}_0.\end{array}$$E97

Combining (96) and (97) leads to;

$$\begin{array}{l}{U}_0\left(x,t\right)=\frac{\pi }{2},\\ U_1\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_0\right)}_{\mathit{xx}}\right]\right]+S^{-1}\left[u^{2}S\left[A_0\right]\right]=\frac{t^2}{2!},\\ U_2\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_1\right)}_{\mathit{xx}}\right]\right]+S^{-1}\left[u^{2}S\left[A_1\right]\right]=0,\\ U_3\left(x,t\right)=S^{-1}\left[u^{2}S\left[{\left(U_2\right)}_{\mathit{xx}}\right]\right]+S^{-1}\left[u^{2}S\left[A_2\right]\right]=-\frac{1}{240}{t}^6.\\ \end{array}$$E98

And so on. The series solution is;

$$U\left(x,t\right)=\frac{\pi }{2}+\frac{t^2}{2!}-\frac{1}{240}{t}^6+.\dots$$

Example 8: Consider the following one – dimensional Burgers equation:

$$U_t=U_{\mathit{xx}}-{\mathit{UU}}_x;$$E99

Subject to the initial conditions:

$$U\left(x,0\right)=x.$$E100

Following the discussion presented above, we find a recursive relation;

$$\begin{array}{l}{U}_0\left(x,t\right)=x,\\ U_{k+1}\left(x,t\right)=S^{-1}\left[\mathit{uS}\left[{\left(U_k\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[\mathit{uS}\left[A_k\right]\right],k\ge 0.\end{array}$$E101

Using the Adomian polynomials we obtain;

$$\begin{array}{l}{U}_0\left(x,t\right)=x,\\ U_1\left(x,t\right)=S^{-1}\left[a\mathit{uS}\left[{\left(U_0\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[\mathit{uS}\left[A_0\right]\right]=-xt,\\ U_2\left(x,t\right)=S^{-1}\left[a\mathit{uS}\left[{\left(U_1\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[\mathit{uS}\left[A_1\right]\right]=x{t}^2,\\ U_3\left(x,t\right)=S^{-1}\left[a\mathit{uS}\left[{\left(U_2\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[\mathit{uS}\left[A_2\right]\right]=-x{t}^3.\end{array}$$E102

Summing these iterates gives the series solution;

$$U\left(x,t\right)=x\left(1-t+t^2-t^3+\dots \right);$$E103

Consequently, the exact solution is given by;

$$U\left(x,t\right)=\frac{x}{1+t}.$$

Example 9: Consider the following nonlinear Schrodinger equation:

$$i{U}_t+U_{\mathit{xx}}-2{\left|U\right|}^{2}U=0;$$E104

Subject to the initial condition:

$$U\left(x,0\right)=e^{ix}.$$E105

Following the discussion presented above, we find;

$$\begin{array}{l}{U}_0\left(x,t\right)=e^{ix},\\ U_1\left(x,t\right)=S^{-1}\left[i\mathit{uS}\left[{\left(U_0\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[2\mathit{iuS}\left[A_0\right]\right]=-3it{e}^{ix},\\ U_2\left(x,t\right)=S^{-1}\left[i\mathit{uS}\left[{\left(U_1\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[2\mathit{iuS}\left[A_1\right]\right]=\frac{1}{2!}{\left(3it\right)}^2{e}^{ix},\\ U_3\left(x,t\right)=S^{-1}\left[i\mathit{uS}\left[{\left(U_2\right)}_{\mathit{xx}}\right]\right]-S^{-1}\left[2\mathit{iuS}\left[A_2\right]\right]=-\frac{1}{3!}{\left(3it\right)}^3{e}^{ix}.\end{array}$$E106

In a few of (106), the series solution is given by;

$$U\left(x,t\right)=e^{ix}\left(1-\left(3it\right)+\frac{1}{2!}{\left(3it\right)}^2-\frac{1}{3!}{\left(3it\right)}^3+\dots \right);$$E107

The exact solution is;

$$U\left(x,t\right)=e^{i\left(x-3t\right)}$$

Example 9: Consider the following homogeneous nonlinear KdV equation:

$$U_t-6{\mathit{UU}}_x+U_{\mathit{xxx}}=0;$$E108

Subject to the initial condition;

$$U\left(x,0\right)=6x.$$E109

Following the discussion presented above, we find a recursive relation;

$$\begin{array}{l}{U}_0\left(x,t\right)=6x,\\ U_{k+1}\left(x,t\right)=-S^{-1}\left[\mathit{uS}\left[{\left(U_k\right)}_{\mathit{xxx}}\right]\right]+S^{-1}\left[6\mathit{uS}\left[A_k\right]\right],k\ge 0.\end{array}$$E110

That gives the first few the components by;

$$\begin{array}{l}{U}_0\left(x,t\right)=6x,\\ U_1\left(x,t\right)=-S^{-1}\left[\mathit{uS}\left[{\left(U_0\right)}_{\mathit{xxx}}\right]\right]+S^{-1}\left[6\mathit{uS}\left[A_0\right]\right]=6^{3}xt,\\ U_2\left(x,t\right)=-S^{-1}\left[b\mathit{uS}\left[{\left(U_1\right)}_{\mathit{xxx}}\right]\right]+S^{-1}\left[\mathit{auS}\left[A_1\right]\right]=6^{5}x{t}^2,\\ U_3\left(x,t\right)=-S^{-1}\left[\mathit{buS}\left[{\left(U_2\right)}_{\mathit{xxx}}\right]\right]+S^{-1}\left[6\mathit{uS}\left[A_2\right]\right]=6^{7}x{t}^3.\end{array}$$E111

In a few of (111), the series solution is given by;

$$U\left(x,t\right)=6x\left(1+\left(36t\right)+{\left(36t\right)}^2+{\left(36t\right)}^3+\dots \right);$$E112

The exact solution is;

$$U\left(x,t\right)=\frac{6x}{1-36t},\left|36t\right|<1.$$

9. Conclusion

In this chapter, we have combined the Adomian decomposition method and Sumudu transform to solve some of the nonlinear partial differential equations. This method has advantages of converting to the exact or approximate solutions and can easily handle a wide class of nonlinear differential equations.