134520092106

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Journal of Applied Mathematics, Islamic Azad University of Lahijan Vol.6, No.21, Summer 2009

47

Application of Homotopy Perturbation Method for

Higher- Order Boundary Valve Problems

H. Jafari a , H. Hoseinzadeh a , S. Seyedein b a Departement of mathematics and computer science, University of mazandaran ,Babolsar, Iran

b Islamic Azad University, Ghaemshah Branch

Abstract

This paper deals with the higher – order boundary valve problems by using the direct

homotopy perturbation method and indirect homotopy perturbation method .This met-hod

based on an embedding parameter and provides the approximate solution without

discretization and computing of the adomian polynomials.In these paper we construct

approximate polynomials to find approximate solution of higher-order boundary value

problems. Numerical camparisions are made between direct homotopy perturbation method

and indirect homotopy perturbation method.

Keywords: Homotopy Pertubution Method, Forth Order Boundary Valve Problems, Approximate Solution,

Analitycal Solution.

1 Introduction

In this paper, we consider the special 2m-order boundary value of the from

10),()()2(<<= x y x f x y m

With the boundary conditions

(2 ) (2 )

2 2(0) , (1) 0, 1, 2, ... 1. j j

j j y A y B j m= = = − (2)

Where y(x) and f(x,y) are assumed real and as many as times differentiable as required

for ]1,0[∈ x and j

A2 and j

B2 , )1(,...,1,0 −= m j are real finit constants Djidjedi, Twizell

and boutayeb [3], more over the constants )1,...(2,1,0,2 −= m j A j describe the even

order derivatives at the boundary 0= x .

(1)

*Corresponding authorE-mail address: [email protected]

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In this paper, we use the direct HPM and indirect HPM for solving two boundary

value problems and carry out the comparison between them. Scott has solved the forth order boundary value broblem [14] for very large valve of c

with the orthonormalization process And also Noor and mohyud-Din [1] have solvedthis problem by using variational iteration method. It is well known that the forth-order

boundary value problems arise in the mathematical modeling of viscoelastic and

inelastic flows [12], deformation of beams [15] and plate deflection theory [2].several

numerical and analytical methods including finit difference method [5], Adomian

decompotion method [13, 16], differential transform method [4],variational iteration

method [1], have been developed for solving general fourth-order boundary value

problem.

This paper is organized as follows. In section 2, we introduce the analysis of the

homotopy perturbation method.In section 3 we present numerical results

2 Homotopy Perturbation Method

In recent years, the application of perturbation techniques in nonelinear problems

has been devoted by scientists and engineers [6, 7, 8]. The most perturbation methods

are based on the assumption that a small parameter exists too over-strict to find wide

application. Therefore, many new techniques have been proposed to eliminate the small

parameter assumption, such as homotopy perturbation method [9, 10]. We consider the

following none-linear differential equation

( ) ( ) ( ), , L u N u f r r + = ∈Ω

(3) with the boundary conditions

( , ) 0, ,u

B u r n

∂= ∈ Γ

∂ (4)

where L is linear operator, while N is none linear operator, B is boundary operator, Γ is

the boundary of domain Ω and )(r f is known analytic function HPM technique defines

the homogony [ ] R pr v →×Ω 1,0:),( Which satisfies

[ ] [ ] 0)()()()()()1(),( 0 =−++−−= r f v N v LPu Lv L p pv H (5)

or

[ ]0 0( , ) ( ) ( ) ( ) ( ) ( ) 0, H v p L v L u PL u P N v f r = − + + − = (6)

where Ω∈r and [ ]1,0∈P is an embedding parameter, 0u is an initial approximation

of Eq. (3) ,which satisfies the boundary condition, from Eq. (5) and (6) we have

,0)()()0,( 0 =−= u Lv Lv H 0)()()()1,( =−+= r f u N v Lv H (7)

the changing process of p from zero tounity is just of ).( pr v from )(0 r u to )(r u .

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In topology,this is called deformation, and ),()( 0u Lv L − )()()( r f v N v L −+ are homotopic.

The basic assumption is that the solution of Eq. (3) and (4) can be expressed as a

power series in p

...2

2

10 +++= v p pvvv (8)

The approximate solution of Eq. (3), therefore, can be readily obtained.

1

......210

→

+++==

p

vvvv Limu(9)

The convergence of the series (9) has been proved in Him [11].

3 Numerical Resudts

In this section first we solve two boundary valve problem with direct homotopy

perturbation method then we solve them with indirect homotopy perturbation method.

Example 1

we first consider second-order boundary value problem

0)()( =−′′ x y x y (10)

With boundary conditions.

367879.0)1(,1)0( == y y (11)

The analytic solution is given by xe x y−

=)( .First, we transform the boundary value

problem to following integral equation.

∫ −+−=x

dt t yt x x x y0

)()(1)( (12)

Then, substituting ...2

2

10 +++= y p py y y in to(12).We obtain the following equation.

dt y p py yt x p x py py y x

.....))((1......0

2

2

10210 +++−−−=+++ ∫ (13)

So, the homotopy solution 210 y y y y ++= for Eq.( 10 ) is given by

!5!4!3!21)(

5432 x x x x

x x y −+−+−= (14)

Now, we solve the problem by indirect homotopy perturbation method.We can rewritethe second-order boundary value problem (10)–(11) as system of differential equations

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,, 21

1 ydx

dy y y ==

1

2

ydx

dy

= (15)

With A y y == )0(,1)0( 21 which can be written as a system of integral equaticns:

dt t y A y

dt t y y

x

x

)(

)(1

012

021

∫

∫

+=

+=

(16)

Using Eq.(5) and (8) for Eq (10), we have

∫

∫

++++=+++

+++=+++

x

o

x

dt y p py y p A y p py y

dt y p py y p y p py y

...)(...

...)(...

12

2

111022

2

2120

022

2

212012

2

1110

(17)

Comparing the coefficients of like powers of p, adding all terms: 171110 ... y y y y +++=

we have

!7!6!5!4!3!21)(

765432 Ax x Ax x Ax x

AX x y +++++++= (18)

Using the boundary condition at x=1, we get A= -0.999981. In table 1, we list results

obtain by direct homotopy perturbation method and indirect homotopy perturbation

methed.

Table 1

Numerical solution for Example 1

Example 2

We consider forth - order boundary valve problem. This problem is special form of

forth-order boundary value problem.

12

1)()1()( 2)4(

−+−′′+= cx xcy yc x y (19)

with boundary value conditions

1)0(,1)0( =′= y y

x Analitcal solution Direct HPM Indirect HPM

0.1 0.904837418 036 0.904837416667 0.904839318608

0.2 0.818730753078 0.818730666667 0.818734573182

0.3 0.740818230682 0.74081725 0.7408239971010.4 0.670330046036 0.670314666667 0.670327824126

0.5 0.606530659713 0.656510416667 0.606540455263

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).1cosh(1)1(),1sinh(5 / 1)1( +=′+= y y (20)

Where .10,10.10,1 86=c Here, we suppose 1=c .

The exact solution for this problem is ).sinh(211)( 2 x x x y ++= (21)

In view of the homotopy perturbation method, we can construct the following

homotopy for Eq(19):

)]()()1[(12

1)( 2)4(

xcy x yc pcx x y −′′++−= (22)

Subsituting ...2

2

10 +++= y p py y y in to (22), we obtain the following set of linear

equations:

M

,0)0(,0)0(,0)0(,0)0(,2:

,0)0(,0)0(,0)0(,0)0(,2:

,0)0(,0)0(,0)0(,0)0(,2:

,)0(,)0(,1)0(,1)0(,2

11:

333322

)4(

3

3

222211

)4(

2

2

111100

)4(

1

1

0000

2)4(

0

0

=′′′=′′=′=−′′=

=′′′=′′=′=−′′=

=′′′=′′=′=−′′=

=′′′=′′=′=+−=

y y y y y y y p

y y y y y y yP

y y y y y y y p

B y A y y y x y p

Solving the above equations, we obtain

6432

0!6

1

!4

1

!3!21)( x x x

B x

A x x y +−+++=

)!10

1

!8

1

!7!6!5

1

!4

1()

!8

1

!6

1

!5!4(2)( 10876548654

1 x x x B

x A

x x x x x B

x A

x y +−+++−+−+=

So, the homotopy solution 10 uuu += +… for Eq. (19) is given by...

134405040720720120601212621)(

8766554432 x Bx Ax x x Bx Ax x Bx Ax

x x y+−−−−++−+++=

.

10

...3628800

+−x

(23)

Using boundary at 1= x , we find B A, .

Now, we solve it by indirect homotopy perturbation method. We rewrite the forth-order

boundary value problem (19) and (20) as the system of differential equations:

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1212,

,,

213

44

3

32

21

1

−+−==

===

x y ydxdy y

dxdy

ydx

dy y

dx

dy y y

(24)

Where B y A y y y ==== )0(,)0(,1)0(,1)0( 4321 which can be written as a

system of integral equations:

dt t t yt y B ydt t y A y

dt t y ydt t y y

x x

x x

12

1)()(2,)(

)(1,)(1

2

0 1340 43

30 0

221

−+−+=+=

+=+=

∫∫

∫ ∫(25)

Using (5) and (8) for Eq. (25), we have

∫

∫

++−+++−+=+++

++++=+++

x

x

dt y p py y y p py yt p B y p py y

dt yP py y p yP py y

012

2

111032

2

3130

2

42

2

4140

12

2

210 0212

2

1110

....)(..).(212

1...

...).(1...

M

Comparing the coefficients of like powers p, adding all terms: 171110 ...... y y y y ++= we

have

2016016802520240240601201212621)(

87766554432 x Bx x Ax x Bx x x Ax Bx Ax

x x y ++−+−+−−++++=

Using the boundary conditions at x=1 .We get A=1/00043, B=0/998376.

In table 2, we list results obtained by direct homotopy perturbation method and

indirect homotopy perturbation method.

4 conclusions

This article deals with the numerical solution of higher-order boundary value

problems using direct homotopy perturbation method and indirect homotopy

perturbation method. These techniques were tested on two examples, and was seen to

produce Satisfactory results. The results show that direct homotopy perturbaticn method

is very strightforword, and we achieve the solution in less procedure and calculation.

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Table 2

Comparison of numerical errors

References

[1] M.A. Noor, S.T. Mohyud-Din, Variational iteration technique for solving higher order boundary value

problems, Appl. Math. Comput. (2006), doi: 10.2016/ j.amc.2006.12.07.[2] M.M. Chawla, C.P. Katti, Finite difference methods for two-point boundary-Value problems

involving higher order differential equations, BIT 19 (1979) 27-33.

[3] Djidjedi K, Twizell EH, Boutayeb A, Numerical methods for special non linear boundary value

problems of order m2 . J Comput Appl math 1993, 47:35-45.

[4] V.S. Erturk, S. Momani, Comparing numerical method for solving forth-order boundary value

problems, Appl. Math. Comput. (2006), doi:10.1016/j.amc.2006.

[5] E. Doedel, Finit difference method for nonlinar two-point bondary-value problems, SIAM J. Number.

Anal. 16 (1979) 173-185[6] H. Jafari, M. Zabihi and M. Saidy, Application of Homotopy-perturbation method for solving Gas

Dynamics Equation, Applied Mathematical Sciences, Vol. 2 no. 48 (2008) 2393 - 2396.[7] Z. Z. Ganji, D. D. Ganji, H. Jafari and M. Rostamian, Application of the homotopy perturbation

method for coupled system of partial differential equatons with time fractional derivatives,topological

method in nonelinear analysis,Vol.31, No.2 (2008) 341-348

[8] H. Jafari, J. Sadeghi, M. Zabihi and A. R. Amani, Application of Homotopy perturbation method

for two coupled scalar fields, The Icfai University Journal of Computational Mathematics, Vol. 1(3)

(2008) 56-66

[9] J.H. He, a coupling method of homotopy technique and perturbation technique for nonlinear problem,

Int. j. Non-linear Meth.35 (1) (2000) 37-43.

[10] J.H. He, Homotopy perturbation technique, Comput. Math. Appl. Meth. Eng.178 (1999) 257-262.

[11] He Ji-H. homotopy perturbation method: a new nonlinear analytical technique, Appl math comput

2003.135:73-9. 3-185.

[12] S.M. Momani, Some problem in non-Newtonian fluid mechanics, Ph.D. thesis, Wasle University,

United Kingdom, 1991.[13] S. Momani, K.Moadi, A reliable algorithm for solving forth-order boundary value problems, Appl.

Math. Comput. 22 (3) (2006) 185-197.

[14] M.Scott, some special problems, private communication, 2006.

[15] T.F.Ma,j. Silva, iteration solution for a beam equation with nonlinear boundary Conditions of third

order, Appl. Math. Comput. 159(1) (2004) 11-18.

[16] A.M. Wazwaz, The numerical solution of special forth-order boundary value problems by the

modified decompotion method, Int. J.Comput. Math. 79 (3) (2002) 345-356.

x Analitcal solution Errors[Direct HPM] Errors[Indirect HPM]

0.0 1.0000000 0.0000 0.00000.1 1.1051667500 7.4E-5 1.8E-6

0.2 1.2213360025 2.5E-4 6.4E-6

0.3 1.3495202934 4.6E-4 1.2E-5

0.4 1.4907523258 6.5E-4 1.7E-5

0.5 1.6460953055 6.6E-4 2.1E-5

0.6 1.8166535821 7.5E-4 2.2E-5

0.7 2.0035837018 6.1E-4 1.9E-5

0.8 2.20810559822 3.8E-4 1.2E-5

0.9 2.4315167257 1.3E-4 4.5E-6

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