Modified Adomian decomposition method for solving nonlinear oscillatory system

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Author: Yahya Qaid Hasan and Liu Ming Zhu

\documentclass[12pt]{article} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{mathrsfs} \textwidth 150mm \textheight 222mm \oddsidemargin 15pt \evensidemargin 0pt \topmargin 0cm \headsep 0.3cm \renewcommand\baselinestretch{1.5} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amstext} \usepackage{amsopn} \usepackage{graphicx} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \newcommand{\ba}{\begin{array}} \newcommand{\ea}{\end{array}} \newcommand{\f}{\frac} \newcommand{\di}{\displaystyle} \newcommand{\Om}{\Omega} \newcommand{\cOm}{\overline{\Omega}} \newcommand{\ep}{\varepsilon} \newcommand{\la}{\lambda} \newcommand{\R}{{\mathbf R}} \newcommand{\ds}{\displaystyle} \begin{document} \date{} \title{ \bf\ Modified Adomian decomposition method for solving nonlinear oscillatory systems \thanks{}} \author{{Yahya Qaid Hasan \thanks {Corresponding author.Tel.:+68 13946033871; fax:+86 451 86417792 E-mail address: yahya217@yahoo.com}} , Liu Ming Zhu \\ {\small Department of Mathematics, Harbin Institute of Technology}\\ {\small Harbin, 150001, P.R.China}\\ {\small $$ }} \maketitle

\begin{abstract} In this paper, an efficient modification of Adomian decomposition method is introduced for solving nonlinear oscillatory equations of the form $$y^{''}(x)+cy^{'}(x)+\epsilon y(x)=f(x,y).$$ The proposed method can be applied to linear and nonlinear problems. The scheme is tested for some examples and the obtained results demonstrate efficiency of the proposed method.

{\bf Keywords}: Adomian decomposition method; Oscillatory equation. \\ MSC:65L10 \\ \end{abstract} \section {Introduction} The decomposition method has been shown[1-8] to solve effectively, easily and accurately a large class of linear and nonlinear, ordinary, partial, deterministic or stochastic differential equations with approximate solutions which converge rapidly to accurate solutions. \textbf{Adomian's decomposition method (ADM) [9-11] is a new and ingenious method for solving nonlinear equations of various kinds, the(ADM) has been successfully applied to solve many nonlinear equations in applied sciences see for example[12-14]. Wazwaz[15] compared the (ADM) with variational iteration method(VIM) and concluded the two methods, which accurately compute the solutions in a series form or in an exact form, are of great interest to applied sciences.} In recent years, many papers were devoted to the problem of approximate solution of nonlinear oscillatory equation [16-20]. The basic motivation of this work is to apply the modified Adomian decomposition method to the nonlinear oscillatory equations. For this reason, a new differential operator is proposed which can be used for nonlinear oscillatory equations. In addition, the proposed method is tested for some examples and the obtained results show the advantage of using this method.

\section {Modified Adomian decomposition method} Consider the non-linear oscillator equation written in the form $$y^{''}(x)+cy^{'}(x)+\epsilon y(x)=f(x,y),\eqno(1)$$ $$y(0)=a, y^{'}(0)=b,$$ where $c$ is a positive real number and $\epsilon$ is a parameter (not necessarily small). We propose the new differential operator, as below $$L(.)=e^{-mx}\frac{d}{dx}e^{-hx}\frac{d}{dx}e^{(m+h)x}(.),\eqno(2)$$ where $2m+h=c$, $m(m+h)=\epsilon $,

so, the problem (1) can be written as, $$Ly=f(x,y).\eqno(3)$$ The inverse operator $L^{-1}$ is therefore considered a two-fold integral operator, as below, $$L^{-1}(.)=e^{-(m+h)x}\int_{0}^{x}e^{hx}\int_{0}^{x}e^{mx}(.)dxdx.\eqno(4)$$ By applying $L^{-1}$ on (3), we have $$y(x)=\frac{1}{h}y^{'}(0)e^{-mx}+\frac{(m+h)}{h}y(0)e^{-mx}-\frac{1}{h}y^{'}(0)e^{-(m+h)x} -\frac{m}{h}y(0)e^{-(m+h)x} $$$$+L^{-1}f(x,y).\eqno(5)$$ The Adomian decomposition method introduce the solution $y(x)$ and the nonlinear function $f(x,y)$ by infinite series $$y(x)=\sum_{n=0}^{\infty}y_{n}(x),\eqno(6)$$ and $$f(x,y)=\sum_{n=0}^{\infty}A_{n},\eqno(7)$$ where the components $y_{n}(x)$ of the solution $y(x)$ will be determined recurrently. Specific algorithms were seen in [2,6] to formulate Adomian polynomials. The following algorithm: $$A_{0}=F(u),$$ $$A_{1}=F^{'}(u_{0})u_{1},$$ $$A_{2}=F^{'}(u_{0})u_{2}+\frac{1}{2}F^{''}(u_{0})u_{1}^{2},$$ $$A_{3}=F^{'}(u_{0})u_{3}+F^{''}(u_{0})u_{1}u_{2}+\frac{1}{3!}F^{'''}(u_{0})u_{1}^{3},\eqno(8)$$ $$.$$$$.$$$$.$$ can be used to construct Adomian polynomials, when $F(u)$ is a nonlinear function. By substituting(6)and(7) into (5), $$\sum_{n=0}^{\infty}y_{n}= \frac{1}{h}y^{'}(0)e^{-mx}+\frac{(m+h)}{h}y(0)e^{-mx}-\frac{1}{h}y^{'}(0)e^{-(m+h)x} -\frac{m}{h}y(0)e^{-(m+h)x}$$$$+L^{-1}\sum_{n=0}^{\infty}A_{n}.\eqno(9)$$ Through using Adomian decomposition method, the components $y_{n}(x)$ can be determined as $$y_{0}=\frac{1}{h}y^{'}(0)e^{-mx}+\frac{(m+h)}{h}y(0)e^{-mx}-\frac{1}{h}y^{'}(0)e^{-(m+h)x} -\frac{m}{h}y(0)e^{-(m+h)x},$$ $$y_{n+1}=L^{-1}A_{n}, n\geq 0,\eqno(10)$$ which gives $$y_{0}=\frac{1}{h}y^{'}(0)e^{-mx}+\frac{(m+h)}{h}y(0)e^{-mx}-\frac{1}{h}y^{'}(0)e^{-(m+h)x} -\frac{m}{h}y(0)e^{-(m+h)x},$$ $$y_{1}=L^{-1}A_{0},$$ $$y_{2}=L^{-1}A_{1},$$ $$y_{3}=L^{-1}A_{3},\eqno(11)$$ $$.$$$$.$$$$.$$ From (8) and (11), we can determine the components $y_{n}(x)$, and hence the series solution of $y(x)$ in (6) can be immediately obtained. For numerical purposes, the $n$-term approximant $$\Phi_{n}=\sum_{n=0}^{n-1}y_{k},\eqno(12)$$ can be used to approximate the exact solution. The approach presented above can be validated by testing it on a variety of several linear and nonlinear initial value problems.

\section {Numerical examples}

In this section, three oscillatory equations are considered and then are solved by standard and modified Adomian decomposition methods.

{\bf Example 1. The Linear Damping Oscillator Equation}

Consider the linear damping oscillator equation: $$y^{''}+2\epsilon y^{'}+y=0,\eqno(13)$$ $$y(0)=a,y^{'}(0)=0.$$

Standard Adomian decomposition method: we put $$L(.)=\frac{d^{2}}{dx^{2}}(.),$$ so $$L^{-1}(.)=\int_{0}^{x}\int_{0}^{x}(.)dxdx.$$ In an operator form, Eq.(13) becomes $$Ly=-2\epsilon y^{'}-y.\eqno(14)$$ By applying $L^{-1}$ to both sides of (14) we have $$y=y(0)+xy^{'}(0)-L^{-1}(2\epsilon y^{'}+y).$$ Proceeding as before we obtained the recursive relationship $$y_{0}=y(0)+xy^{'}(0),$$ $$y_{n+1}=-L^{-1}(y+2\epsilon y^{'}), n\geq 0,$$ and the first few components are as follows $$y_{0}=a,$$ $$y_{1}=-a\frac{x^{2}}{2},$$ $$y_{2}=2a\epsilon \frac{x^{3}}{3!}+a\frac{x^{4}}{4!},$$ $$y_{3}=-4\epsilon^{2}\frac{x^{4}}{4!}-4a\epsilon \frac{x^{5}}{5!}-a\frac{x^{6}}{6!},$$$$.$$$$.$$$$.$$ $$y=y_{0}+y_{1}+y_{2}+y_{3}+...=a-a\frac{x^{2}}{2}+2a\epsilon \frac{x^{3}}{3!} +a(-4\epsilon^{2}+1)\frac{x^{4}}{4!}-4a\epsilon \frac{x^{5}}{5!}-a\frac{x^{6}}{6!}+...$$

Modified Adomian decomposition method:we put $2m+h=2\epsilon $, $m(m+h)=1,$

it follows that $h=\pm 2i\sqrt{1-\epsilon^{2}}$,$m=\epsilon \mp i\sqrt{1-\epsilon^{2}},$ $i=\sqrt{-1}.$

Substitution of $h=-2i\sqrt{1-\epsilon^{2}}$, $m=\epsilon+i\sqrt{1-\epsilon^{2}},$ in Eq.(2) yields the operator $$L()=e^{-(\epsilon +i\sqrt{1-\epsilon^{2}})x}\frac{d}{dx}e^{2i\sqrt{1-\epsilon^{2}}x}\frac{d}{dx}e^{(\epsilon -i\sqrt{1-\epsilon^{2}})x}(),$$

so $$L^{-1}(.)=e^{(-\epsilon+i\sqrt{1-\epsilon^{2}})x}\int_{0}^{x}e^{-2i\sqrt{1-\epsilon^{2}}x}\int_{0}^{x}e^{(\epsilon +i\sqrt{1-\epsilon^{2}})x}(.)dxdx.$$

In an operator form, Eq.(13) becomes $$Ly=0.\eqno(15)$$ Now, by applying $L^{-1}$ to both sides of (15), we have $$L^{-1}Ly=0,$$ and it, implies that $$y=\frac{\epsilon-i\sqrt{1-\epsilon^{2}}}{-2i\sqrt{1-\epsilon^{2}}} a e^{(-\epsilon-i\sqrt{1-\epsilon^{2}})x}+\frac{\epsilon+ i\sqrt{1-\epsilon^{2}}}{2i\sqrt{1-\epsilon^{2}}}a e^{(-\epsilon+i\sqrt{1-\epsilon^{2}})x}$$$$\Rightarrow y=ae^{-\epsilon x}(\cos\sqrt{1-\epsilon^{2}} x+\frac{\epsilon}{\sqrt{1-\epsilon^{2}}}\sin \sqrt{1-\epsilon^{2}} x.$$

So the exact solution is easily obtained by proposed Adomian decomposition method .\\

{\bf Example 2.} Consider the following Duffing equation: $$y^{''}(x)+3y(x)-2 y^{3}(x)=\cos x\sin 2x,\eqno(16)$$ with initial conditions $$y(0)=0, y^{'}(0)=1.$$ The analytic solution of this equation is $y(x)=\sin x.$

Standard Adomian decomposition method: we put $$L(.)=\frac{d^{2}}{dx^{2}}(.),$$ so $$L^{-1}(.)=\int_{0}^{x}\int_{0}^{x}(.)dxdx.$$ In an operator form, Eq.(16) becomes $$Ly=\cos x\sin 2x+2 y^{3}-3y.\eqno(17)$$ By applying $L^{-1}$ to both sides of (17) we have $$y=y(0)+xy^{'}(0)+L^{-1}(\cos x\sin 2x)+L^{-1}(2 y^{3}-3y).$$ Proceeding as before we obtained the recursive relationship $$y_{0}=y(0)+xy^{'}(0)+L^{-1}(\cos x\sin 2x)=\frac{5x}{3}-\frac{\sin x}{2}-\frac{1}{18}\sin3x,$$ $$y_{n+1}=L^{-1}(2 A_{n}-3y_{n}), n\geq 0,\eqno(18)$$ when $A_{n}$'s are Adomian polynomials of nonlinear term $y^{3}$, as below\textbf{[6,21]} $$A_{0}=y_{0}^{3},$$ $$A_{1}=3y_{0}^{2}y_{1},$$ $$A_{2}=3y_{0}^{2}y_{2}+3y_{0}y_{1}^{2},\eqno(19)$$ $$.$$$$.$$$$.$$Substituting (19) into (18) gives the components $y_{0},y_{1}$,...

We take $$y=y_{0}+y_{1}.$$ By using Taylor series of $y=y_{0}+y_{1}$ with order 7 we get $$y=x-\frac{x^{3}}{6}-\frac{x^{5}}{15}+\frac{101x^{7}}{1260}+...$$

Modified Adomian decomposition method: we put $2m+h=0,$ $ m(m+h)=3,$

it follows that $h=\pm 2i\sqrt{3},$ $m=\mp i\sqrt{3}$, $i=\sqrt{-1}.$

Substitution of $h=-2i\sqrt{3} $, $m=i\sqrt{3},$ in Eq. (2)yields the operator $$L(.)=e^{-i\sqrt{3}x}\frac{d}{dx}e^{2i\sqrt{3}x}\frac{d}{dx}e^{-i\sqrt{3}x}(.),$$ so $$L^{-1}(.)=e^{i\sqrt{3}x}\int_{0}^{x}e^{-2i\sqrt{3}x}\int_{0}^{x}e^{i\sqrt{3}x}(.)dxdx.$$

In an operator form, Eq.(16) becomes $$Ly=2 y^{3}+\cos x\sin 2x.\eqno(20)$$ Applying $L^{-1}$ to both sides of (20) we find $$y=\frac{\sin\sqrt{3} x}{\sqrt{3}}+L^{-1}(\cos x\sin 2x)+2 L^{-1} y^{3}.$$ Proceeding as before we obtain $$y_{0}=\frac{\sin\sqrt{3} x}{\sqrt{3}}+\frac{\sin^{3}x}{3} ,$$ $$y_{n+1}=2L^{-1}A_{n}, n\geq 0.\eqno(21)$$ when $A_{n}$'s are Adomian polynomials of nonlinear term $y^{3}$,mentioned in (19), we obtain $y_{0},y_{1}$,...

By using Taylor series of$y=y_{0}+y_{1}$ we get $$y=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+...$$ Not that the Taylor series of the exact solution $y(x)=\sin x$ with order 7 is as below $$\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+...$$

So the rate of convergence of modified Adomian is faster than standard Adomian method for this problem.\\

{\bf Example 3}. Consider the non-linear equation: $$y^{''}(x)-2y^{'}(x)+y(x)+y^{2}(x)=\cos^{2}x+2\sin x,\eqno(22)$$ subject to the initial conditions $$y^{'}(0)=0, y(0)=1.$$ The analytic solution of this equation is $y(x)=\cos x.$

Standard Adomian decomposition method: we put $$L(.)=\frac{d^{2}}{dx^{2}}(.),$$ so $$L^{-1}(.)=\int_{0}^{x}\int_{0}^{x}(.)dxdx.$$ In an operator form, Eq.(22) becomes $$Ly=\cos^{2}x+2\sin x- y^{2}-y+2y^{'}.\eqno(23)$$ By applying $L^{-1}$ to both sides of (23) we have $$y=y(0)+xy^{'}(0)+L^{-1}(\cos^{2}x+2\sin x)+L^{-1}(- y^{2}-y+2y^{'}).$$ Proceeding as before we obtained the recursive relationship $$y_{0}=y(0)+xy^{'}(0)+L^{-1}(\cos^{2}x+2\sin x)=1+2x+\frac{x^{2}}{4}-2\sin x+\frac{\sin^{2}x}{4},$$ $$y_{n+1}=L^{-1}(- A_{n}-y_{n}+2y_{n}^{'}), n\geq 0,\eqno(24)$$ when $A_{n}$'s are Adomian polynomials of nonlinear term $y^{2}$, as below\textbf{[6,21]} $$A_{0}=y_{0}^{2},$$ $$A_{1}=2y_{0}y_{1},$$ $$A_{2}=2y_{0}y_{2}+y_{1}^{2},\eqno(25)$$ $$.$$$$.$$$$.$$Substituting (25) into (24) gives the components $y_{0},y_{1},y_{2},$...

We take $$y=y_{0}+y_{1}+y_{2}$$

Modified Adomian decomposition method: we put $2m+h=0,$ $ m(m+h)=1,$

it follows that $h=0,$ $m=-1$.

Substitution of $h=0, $ $m=-1,$ in Eq. (2)yields the operator $$L(.)=e^{x}\frac{d^{2}}{dx^{2}}e^{-x}(.),$$ so $$L^{-1}(.)=e^{x}\int_{0}^{x}\int_{0}^{x}e^{-x}(.)dxdx.$$

In an operator form, Eq.(22) becomes $$Ly=\cos^{2}x+2\sin x-y^{2}.\eqno(26)$$ Applying $L^{-1}$ to both sides of (26) we find $$y=e^{x}-xe^{x}+L^{-1}(\cos^{2}x+2\sin x)-L^{-1}y^{2}.$$ Proceeding as before we obtain $$y_{0}=\frac{1}{2}-\frac{11}{25}e^{x}+\frac{3}{5}xe^{x}+\cos x-\frac{3}{50}\cos 2x-\frac{2}{25}\sin 2x,$$ $$y_{n+1}=-L^{-1}A_{n}, n\geq 0.\eqno(27)$$ When $A_{n}$'s are Adomian polynomials of nonlinear term $y^{2}$ mentioned in (25), we obtain, $y_{0},y_{1},y_{2}$,....

The graph of $y=\sum_{i=0}^{2}y_{i}$ is sketched in Fig 1 and compared with the solution of the standard Adomian decomposition method.

The comparison between the results mentioned in Examples 1-3 show the power of the proposed method of this paper for these nonlinear oscillator equations. \section {Conclusion} Adomian decomposition method has been known to be powerful device for solving many functional equations as algebraic equations, ordinary and partial differential equations, integral equation and so on. In this paper, we proposed an efficient modification of the standard Adomian decomposition method for solving nonlinear oscillator equations. \textbf{In Example 1, the system was a linear systems and we derived the exact solution. For non-linear system we usually derive a very good approximations to the the solution, as in Example 3, and some times the exact solutions can be found, as in Example 2(Duffing equation)}. The study showed that the modified Adomian decomposition method is simple and easy to use and produces reliable results with few iterations used.

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Dear Editor, Please address all correspondence regarding the manuscript to me at the address indicated on the letterhead above. Thank you in advance for handling the manuscript. \ \ \ \ \ \ \ \ \ \ \ \closing{Sincerely yours, Yahya Qaid Hasan } \end{letter} \end{document}