# Analyzing “Homotopy Perturbation Method for Solving Fourth-Order Boundary Value Problem"

Authors: Afgan Aslanov
Journal: Mathematical Problems in Engineering
 (2011)

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## Abstract

We analyze a previous paper by S. T. Mohyud-Din and M. A. Noor (2007) and show the mistakes in it. Then, we demonstrate a more efficient method for solving fourth-order boundary value problems.


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## Body

## 1. Problem

Let  us consider the fifth-order boundary problem of the type(1.1)u(4)(x)=f(u,u′,u′′,u′′′)+g(x)
with the  boundary conditions(1.2)u(0)=α1,u′(0)=α2,u(1)=β1,u′(1)=β2,
where f and g are  continuous functions and α1, α2, β1, and β2 are real constants.The homotopy perturbation method (HPM) is employed in [1] for solving such problems. The purpose of this paper is to point out the mistakes in paper [1] and demonstrate more efficient method for solving the problems of type (1.1)-(1.2).First of all, we show the mistakes.(1)  In Example3.2 the approximate solution
(1.3)uapprox(x)=512+480x+224.00000000226055x2+…
of the problem
(1.4)u(4)(x)=u(x)+u′′(x)+ex(x-3),u(0)=1,u′(0)=0,u(1)=0,u′(1)=-e
(see formula (3.19) in [1]) has no relationship with the exact solution uexact(x)=(1-x)ex.  And “error” is at least uapprox(0)-uexact(0)=512-e=509.28.It is strange that the authors demonstrated the reliability of the errors in the table (see Table 3.2 in [1]).(2)  In  Example 3.3 the approximate solution(1.5)uapprox(x)=11.3472-262.827x-3.40768x3+…
of the problem(1.6)u(4)=sinx+sin2x-(u′′(x))2,u(0)=0,u′(0)=1,u(1)=sin1,u′(1)=cos⁡1
(see formula (3.28) in [1]) has no relationship with the exact solution u=sinx, since uapprox(0)=11.3472, uapprox′(0)=-262.827.


## 2. Homotopy Perturbation Method

The basic ideas of the standard HPM were given by He [2, 3], and a new interpretation of HPM was given by He [4]. We introduce a new reliable procedure for choosing the initial approximation in HPM. To do so, we consider the following general nonlinear differential equation(2.1)Lu+Nu=f(u,x)
with some initial boundary conditions, where L and N are, respectively, the linear and nonlinear operators.According to HPM, we construct a homotopy which satisfies the following relations:(2.2)H(u,p)=Lu-Lv0+pLv0+p[Nu-f(u,x)]=0,
where p∈[0,1] is an embedding parameter and v0 is an initial approximation. When we put p=0 and p=1 in (2.2), we obtain(2.3)H(u,0)=Lu-Lv0,H(u,1)=Lu+Nu-f(u,x),
respectively. In topology, this is called deformation and Lu-Lv0 and Lu+Nu-f(u,x) are called homotopics.The solution of (2.2) is expressed as(2.4)u(x)=u0(x)+pu1(x)+p2u2(x)+….
Hence, the approximate solution of (1.5) can be expressed as(2.5)u(x,t)=u0(x)+u1(x)+u2(x)+….
Mohyud-Din  and Noor [1] tried to rewrite the problem as a system of integral equations and then HPM applied for each equation. In the use of HPM, what we are mainly concerned about are  the auxiliary operator L and the initial guess v0. We take L=(d/dx4)(·), and(2.6)v0=α1+α2x+Ax3+Bx4+L-1(g(x)),
where A and B are yet to be determined. Then using the boundary conditions u(1)=β1, u′(1)=β2 we determine A and B.


## 3. Applications

Here we apply the HPM to solve correctly the problems in [1].Example 3.1 (see [1, Example 3.2]).
We have(3.1)u(4)(x)=u(x)+u′′(x)+ex(x-3)
with boundary conditions
(3.2)u(0)=1,u′(0)=0,u(1)=0,u′(1)=-e.
We construct a homotopy which satisfies the relation
(3.3)u(4)(x)-v0(4)(x)+p[v0(4)(x)-u(x)-u′′(x)-ex(x-3)]=0,
where
(3.4)v0=1+Ax2+Bx3+L-1(ex(x-3)).
Now substituting (3.4) into (3.3), we obtain
(3.5)u0(4)+pu1(4)+p2u2(4)+⋯-v0(4)(x)+p[v0(4)(x)-u0-pu1-p2u2-⋯-u0′′-pu1′′-p2u2′′-⋯-ex(x-3)]=0,
and, equating the coefficients of a like powers of p, we get a system of equations:
(3.6)u0(4)(x)-v0(4)(x)=0,u0(0)=1,u0′(0)=0,u0′′(0)=A,u0′′′(0)=B,u1(4)+v0(4)(x)-u0-u0′′-ex(x-3)=0,u1(0)=0,u1′(0)=0,u1′′(0)=0,u1′′′(0)=0,u2(4)-u1-u1′′=0,u2(0)=0,u2′(0)=0,u2′′(0)=0,u2′′′(0)=0,u3(4)-u2-u2′′=0,u3(0)=0,u3′(0)=0,u3′′(0)=0,u3′′′(0)=0,….
Solving (3.6), we get
(3.7)u0=1+Ax2+Bx3+L-1(ex(x-3))=1+Ax2+Bx3+6x-7ex+xex+52x2+23x3+7=6x-7ex+Ax2+Bx3+xex+52x2+23x3+8,u1=20+x5(120B+110)+x4(112A+1324)+x6(1360A+1144)+x7(1840B+11260)+1614x3+1216x2+ex(2x-20)+18x,u2=L-1(u1+u1′′)=L-1(32x-36ex+Ax2+A6x4+Bx3+A360x6+B10x5+B840x7+4xex+292x2+133x3+34x4+215x5+1144x6+11260x7+36)=52+x6(1360A+29720)+x7(1840B+132520)+x8(110080A+12240)+x9(B30240+122680)+x10(A1814400+1725760)+x11(B6652800+19979200)+1640x3+1244x2+ex(4x-52)+48x+32x4+415x5,….
Using only three-term approximation, we have
(3.8)u=u0+u1+u2=6x-7ex+Ax2+Bx3+xex+52x2+23x3+8+20+x5(120B+110)+x4(112A+1324)+x6(1360A+1144)+x7(1840B+11260)+1614x3+1216x2+ex(2x-20)+18x+52+x6(A360+29720)+x7(B840+132520)+x8(A10080+12240)+x9(B30240+122680)+x10(A1814400+1725760)+x11(B6652800+19979200)+1640x3+1244x2+ex(4x-52)+48x+32x4+415x5.
Now it follows from conditions u(1)=0, u′(1)=-e that A=-0.46717 and B=-0.38354 and, therefore,
(3.9)u=u0+u1+u2=80+72x-79ex+7xex+32.033x2+9.2832x3+2.0027x4+0.34749x5+4.4627×10-2x6+5.0392×10-3x7+4.0008×10-4x8+3.1409×10-5x9+1.1204×10-6x10+4.2558×10-8x11,
or, in power series form,
(3.10)u=1-0.467x2-0.38347x3-0.1223x4-0.01918x5-6.762×10-3x6-9.132×10-4x7+4.0008×10-4x8+3.1409×10-5x9.
Higher accuracy level can be attained by evaluating some more terms of u(x).Example 3.2 (see [1, Example 3.3]).
We have(3.11)u(4)(x)=sinx+sin2x-(u′′(x))2
with boundary conditions
(3.12)u(0)=0,u′(0)=1,u(1)=sin1,u′(1)=cos⁡1
(the exact solution of the problem is u=sinx).
We construct a homotopy which satisfies the relation(3.13)u(4)(x)-v0(4)(x)+p[v0(4)(x)+(u′′(x))2-sinx-sin2x]=0,
where
(3.14)v0=x+Ax2+Bx3+L-1(sinx+sin2x).
Substituting (3.14) into (3.13), we obtain
(3.15)u0(4)+pu1(4)+p2u2(4)+⋯-v0(4)(x)+p[v0(4)(x)+(u0′′+pu1′′+p2u2′′+⋯)2-sinx-sin2x]=0,
and, equating the coefficients of a like powers of p, we get a system of equations:
(3.16)u0(4)(x)-v0(4)(x)=0,u0(0)=1,u0(0)=0,u0′(0)=1,u0′′(0)=A,u0′′′(0)=B,u1(4)+v0(4)(x)+(u0′′)2-sinx-sin2x=0,u1(0)=0,u1′(0)=0,u1′′(0)=0,u1′′′(0)=0,u2(4)+2u0′′u1′′=0,u2(0)=0,u2′(0)=0,u2′′(0)=0,u2′′′(0)=0,u3(4)+(u1′′)2+2u0′′u2′′=0,u3(0)=0,u3′(0)=0,u3′′(0)=0,u3′′′(0)=0,….
Solving (3.16) we get
(3.17)u0=x+Ax2+Bx3+L-1(sinx+sin2x)=x+Ax2+Bx3+132+sinx-132cos⁡2x+16x3-116x2-x+148x4,u1=-11260Ax7-x8(15040A+11680B)-x9(16048B-190720A)+x10(A113400+B100800-1181440)-16A2x4-140B2x6-110ABx5+O(x11),u2=x14(17567560A2+130270240AB+123284800A-11441440B2+12522520B)+x13(121621600A2-175405400AB+11853280A-1739200B2+4343243200B)+x11(-234989600A2+115400AB+13158400B2)+x10(17226800A2+14200BA)+x12(-1249480A2-133326400AB+1907200A+144352B2)+x9(13780A2+1336B3)+145A3x6+4105A2Bx7+9560AB2x8.
Using only three-term approximation we have
(3.18)u=x+Ax2+Bx3+132+sinx-132cos⁡2x+16x3-116x2-x+148x4-11260Ax7-x8(15040A+11680B)-x9(16048B-190720A)+x10(1113400A+1100800B-1181440)-16A2x4-140B2x6-110ABx5+x9(13780A2+1336B3)+145A3x6+4105A2Bx7+9560AB2x8.
Now by using the conditions u(1)=sin1, u′(1)=cos1, we have a system of equations of degree three. Solving this system numerically (applying some standard computer programs) we have that A=5.8611×10-3, B=-0.17455 and the series solution
(3.19)u=x+5.861×10-3x2-0.17455x3-6.3333×10-6x4+8.4356×10-3x5+2.0161×10-3x6-2.0329×10-4x7-9.2803×10-5x8+O(x9).


## 4. Conclusion

In this paper we have used the homotopy perturbation method for finding the solution of fourth-order linear and nonlinear boundary value problems. We presented a simple way to chooseL and v0 when we use the homotopy perturbation method. In most cases, our simple choice yields very good approximation of exact solution.



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*Source: 296796-2011-06-16.xml*