# Comment on “Adomian Decomposition Method for a Class of Nonlinear Problems”

Authors: S. Dalvandpour; A. Motamedinasab
Journal: ISRN Applied Mathematics
 (2013)

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## Abstract

Sánchez Cano in his paper“Adomian Decomposition Method for a Class of Nonlinear Problems” in application part pages 8, 9, and 10 had made some mistakes in context; in this paper we correct them.


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## Body

## 1. Introduction

Adomian [1, 2] proposed a powerful method for solving nonlinear functional equation. The technique uses a decomposition of the nonlinear operator as a series of functions; each term of this series is a generalized polynomial called Adomian polynomial.We will see that using the Adomian decomposition method together with some properties of the nested integral [3, 4] the solution of nonlinear ordinary differential equations system is obtained.


## 2. Correct Equations

In page 8, he showed that the solutionxn(t)is given by
(1)x(t)=x0(t)+x1(t)+x2(t)+⋯+xn(t)+⋯=∝+∫0tf(u)du+aβt+[-a2∝t22!-a21Γ(3)hhhhhhhhhhhhhhhhhhhhhhh×∫0t(t-u)2f(u)dut22!]-a3βt33!+[a4∝t44!+a41Γ(5)hhhhhhhhhhhh×∫0t(t-u)4f(u)dut94!]+⋯.
By rearranging he obtained
(2)x(t)=∝[1-(at)22!+(at)44!+⋯(-1)n(at)2n2n!+⋯]+β[(at)-(at)33!+(at)55!iiiiiiiiiii+⋯(-1)n(at)2n+1(2n+1)!+⋯]+∫0t[(a(t-u))0Γ(1)-(a(t-u))2Γ(3)iiiiiiiiiiii+⋯(-1)n(a(t-u))2nΓ(2n+1)+⋯]f(u)du.
And similarly for yn(t) we will have
(3)y(t)=y0(t)+y1(t)+y2(t)+⋯+yn(t)+⋯=β-∝at-a1Γ(2)×∫0t(t-u)f(u)du-a2βt22!+a3∝t33!+a31Γ(4)×∫0t(t-u)3f(u)du+a4βt44!+⋯.
By rearranging he obtained
(4)y(t)=β[1-(at)22!+(at)44!+⋯(-1)n(at)2n2n!+⋯]-∝[(at)-(at)33!+(at)55!+⋯(-1)n(at)2n+1(2n+1)!+⋯]-∫0t[(a(t-u))1Γ(2)-(a(t-u))3Γ(4)iiiiiiiii+⋯(-1)n(a(t-u))2n+1Γ(2n)+⋯]f(u)du.
Continuing in this fashion, he concluded the following formulas:
(5)xn(t)=(-1)n[anatnn!+an1Γ(n+1)∫0t(t-u)nf(u)du],yn(t)=(-1)n+1[anatnn!+an1Γ(n+1)∫0t(t-u)nf(u)du].
But the correct formulas are given by
(6)x2n-1(t)=(-1)n+1(a2n-1βt2n-1(2n-1)!),x2n(t)=(-1)n[a2n∝t2n(2n)!+an1Γ(2n+1)iiiiiiiiiiiiiii×∫0t(t-u)2nf(u)dua2n∝t2n(2n)!],y2n-1(t)=(-1)n[a2n-1∝t2n-1(2n-1)!+an1Γ(2n)iiiiiiiiiiiiiii×∫0t(t-u)2n-1f(u)dua2n-1∝t2n-1(2n-1)!],y2n(t)=(-1)n(a2nβt2n(2n)!).
Writing (2) and (4) as a single integral, he had
(7)x(t)=∝cos(at)+βsin(at)+∫0tcos(a(t-u))f(u)du=∝cos(at)+βsin(at)+f(t)*cos(at).
Similarly,
(8)y(t)=βcos(at)-∝sin(at)-∫0tsin(a(t-u))f(u)du=βcos(at)-∝sin(at)+f(t)*sin(at).
In page 9, he uses
(9)dxdt-ay(t)=f(t),dyat+ax(t)=0iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii(a∈R,a≠0).
With x(0)=α and y(0)=β, he obtained
(10)d2ydt2+a2y(t)=f(t),y(0)=α,y´(0)=β.
In fact, the correct solution is given by
(11)dxdt-ay(t)=f(t)⟶dxdt=ay(t)+f(t),dydt+ax(t)=0⟶d2ydt2+adxdt=0,hhhhhhhhh⇓d2ydt2+a2y(t)=-af(t),y(0)=β,y´(0)=α.
In page 10, he showed two cases for the solutions x(t) and y(t) using (2), (4), and f(t)=cos(t).Case 1 (a≠1).
In this case, he obtained the solutions(12)x(t)=αcos(at)+βsin(at)+11-a2sin(t),y(t)=βcos(at)-αsin(at)+a1+a2cos(t).But the correct method is the following:
(13)x(t)=∝cos(at)+βsin(at)+∫0tcos(a(t-u))f(u)du=⟶∫0tcos(a(t-u))f(u)du=12[∫0tcos(at-au+u)cos(at-au-u)du]=[12(1-a)sin(at-au+u)iiiiii+12(-1-a)sin(at-au-u)]0t=11-a2sint-a1-a2sinat⟶x(t)=∝cos(at)+βsin(at)+11-a2sint-a1-a2sinat,x(t)=αcosat+(β-a1-a2)sinat+11-a2sint.
Similarly,
(14)y(t)=(β-a1-a2)cosat-αsinat+a1-a2cost.Case 2 (a=1).
He arrived at the formulasx(t) and y(t) as follows:
(15)x(t)=∝cos(t)+βsin(t)+12tcost,y(t)=(∝-12)cost+βsint-12tsint.
But the correct formulas are given by
(16)x(t)=∝cos(t)+βsin(t)+∫0tcos((t-u))f(u)du=⟶∫0tcos((t-u))cosudu=12[∫0tcost+cos(t-2u)du]=12[tcost-12sin(t-2u)]0t=12tcost+12sint⟹x(t)=αcost+βsint+12tcost+12sint⟹x(t)=αcost+(β+12)sint+12tcost.Similarly,(17)y(t)=βcost-αsint-12tsint.



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*Source: 806056-2013-10-07.xml*