On the Conversion of Partial Differential Equations

Syed Tauseef Mohyud-Din

HITEC University, Taxila Cantt, Pakistan

Reprint requests to S. T. M.-D.; E-mail: syedtauseefs@hotmail.com

Z. Naturforsch. 65a, 896 – 900 (2010); received June 11, 2009 / revised November 20, 2009

This paper outlines the conversion of partial differential equations (PDEs) into the correspondingordinary differential equations (ODEs) by a complex transformation which is widely used in theexp-function method. The proposed homotopy perturbation method (HPM) is employed to solve thetravelling wave solutions. Several examples are given to reveal the reliability and efficiency of thealgorithm.

Key words: Homotopy Perturbation Method; Transformation; Partial Differential Equations.

1. Introduction

Recently, He and Wu [1] developed the exp-functionmethod to seek solitary, periodic, and compacton-like solutions of nonlinear differential equations whichare widely used in nonlinear sciences [1 – 16]. Theexp-function technique [1, 17] uses a transformationη = kx+ωt, which converts the given partial differ-ential equations (PDEs) into the corresponding ordi-nary differential equations (ODEs). The basic moti-vation of the present study is the elegant coupling ofthe above transformation and He’s homotopy pertur-bation method (HPM) [2, 6 – 20] for solving PDEs.It is worth mentioning that HPM has been devel-oped by He by merging the standard homotopy andperturbation and is used for finding appropriate so-lutions of nonlinear problems of physical nature. Inour proposed algorithm, the given PDE is convertedto the corresponding ODE by using the transforma-tion η = kx + ωt. The variational iteration method(VIM) is applied to the re-formulated ODE whichgives the solution in terms of transformed variablesand the inverse transformation yields the requiredseries of solution. It is observed that the proposedcombination is very effective, convenient, and eas-ier to implement; and does not require discretization,linearization, and calculation of the so-called Ado-mian’s polynomials. Numerical results are very en-couraging.

0932–0784 / 10 / 1100–0896 $ 06.00 c© 2010 Verlag der Zeitschrift fur Naturforschung, Tubingen · http://znaturforsch.com

2. Homotopy Perturbation Method (HPM) andHe’s Polynomials

To explain He’s homotopy perturbation method, weconsider a general equation of the type

L(u) = 0, (1)

where L is any integral or differential operator. We de-fine a convex homotopy H(u, p) by

H(u, p) = (1− p)F(u)+ pL(u), (2)

where F(u) is a functional operator with known solu-tions ν0, which can be obtained easily. It is clear that,for

H(u, p) = 0, (3)

we have

H(u,0) = F(u), H(u,1) = L(u).

This shows that H(u, p) continuously traces an implic-itly defined curve from a starting point H(ν0,0) to asolution function H( f ,1). The embedding parametermonotonically increases from zero to unit as the triv-ial problem F(u) = 0 continuously deforms the origi-nal problem L(u) = 0. The embedding parameter p ∈(0,1] can be considered as an expanding parameter [2,6 – 20]. The homotopy perturbation method uses this

S. T. Mohyud-Din · On the Conversion of Partial Differential Equations 897

homotopy parameter p as an expanding parameter toobtain

u =∞

∑i=0

piui = u0 + pu1 + p2u2 + p3u3 + · · · , (4)

if p → 1, then (4) corresponds to (2) and becomes theapproximate solution of the form

f = limp→1

u =∞

∑i=0

ui. (5)

It is well known that series (5) is convergent for mostof the cases and also the rate of convergence is depen-dent on L(u); see [1, 6 – 19]. We assume that (5) has aunique solution. The comparisons of like powers of pgive solutions of various orders.

3. Examples

3.1. Example 1

Consider the following Helmholtz equation:

∂ 2u(x,y)∂ 2x2 +

∂ 2u(x,y)∂ 2y2 − u(x,y) = 0

with initial conditions

u(0,y) = y, ux(0,y) = y+ cosh(y).

The exact solution for this problem is given as

u(x,y) = yex + y+ cosh(y).

Applying the transformation η = x+ t (by setting k =ω = 1), we get the following ODE:

2d2udη2 − u = 0

with

u((η) = A, u′(η) = B,

where A and B are unknown parameters which aresubsequently determined by using the initial condi-tions. Now, we apply the convex homotopy perturba-tion method

u0 + pu1 + p2u2 + · · ·=A+Bη + p

∫ η

0

∫ η

0

[12(u0 + pu1 + p2u2 + · · ·)

]dsds.

Comparing the co-efficient of like powers of p yields

p(0) : u0(η) = A+Bη ,

p(1) : u1(η) = A+Bη +1

12Bη3 +

14

Aη2,

p(2) : u2(η) = A+Bη +1

480Bη5 +

196

Aη4

+1

12Bη3 +

14

Aη2,....

The series solution is given by

u(η) = A+β η +1

480β η5 +

196

Aη4

+1

12β η3 +

14

Aη2 + · · · .

The inverse transformation will yield

u(x,y) = A+B(x+ y)+1

480B(x+ y)5 +

196

A(x+ y)4

+1

12B(x+ y)3 +

14

A(x+ y)2 + · · ·

and the use of the initial condition gives

A =−2(2y3ey + y2 + y2 (ey)2 − 6y2ey + 24yey

− 24ey+ 12+ 12(ey)2)y[ey(48+ y4)]−1

,

B = 6[− 4y2ey + y2 + y2(ey)2 + 2y3ey

+ 8yey + 4+ 4(ey)2][ey(48+ y4)]−1

.

The solution after two iterations is given by

u(x,y) =[96yey + 8yeyx3 + 48x− 4y2eyx3 − 6y3eyx2

+ 96yeyx+ 2y3evx3 + 4y4eyx2 + 2y5eyx

+ y2e2yx3 + 2y3e2yx2 + y4e2yx+ 24yeyx2

+ 48e2yx+ 2y5ey + 4e2yx3 + y2x3 + 2y3x2

+ y4x+ x3][2ey(48+ y4)]−1

.

3.2. Example 2

Consider the following Helmholtz equation:

∂ 2u(x,y)∂ 2x2 +

∂ 2u(x,y)∂ 2y2 + 8u(x,y) = 0

with the initial conditions

u(0,y) = sin(2y), ux(0,y) = 0.

The exact solution for this problem is

u(x,y) = cos(2x)sin(2y).

898 S. T. Mohyud-Din · On the Conversion of Partial Differential Equations

(a)

(b)

Fig. 1 (colour online). Solution by the proposed algorithm (a)and exact solution (b) of Example 1.

Applying the transformation η = x+ t (by setting k =ω = 1), we get

d2udη2 + 4u = 0

with

u(η) = A, u′(η) = B.

Applying the convex homotopy perturbation method,we get

u0 + pu1+ p2u2 + · · ·=A+Bη − 4p

∫ η

0

∫ η

0(u0 + pu1+ p2u2 + · · ·)dsds.

Comparing the co-efficient of like powers of p yields

p(0) : u0(η) = A+Bη ,

p(1) : u1(η) = A+Bη − 23

Bη3 − 2Aη2,

p(2) : u2(η) = A+Bη +215

Bη5 +23

Aη4

− 23

Bη3 − 2Aη2,... .

Fig. 2 (colour online). Solution of Example 2.

The series solution is given by

u(η)=A+Bη+215

Bη5+23

Aη4−23

Bη3−2Aη2+ · · · ,

the inverse transformation will yield

u(x,y) = A+B(x+ y)+2

15B(x+ y)5 +

23

A(x+ y)4

− 23

B(x+ y)3 − 2A(x+ y)2+ · · · ,

and the use of the initial condition gives

A = 15sin(2y)(2y4 − 6y2 + 3)−16y6 + 4y8 + 45

,

B =−60ysin(2y)(−3+ 2y2)

−16y6 + 4y8 + 45.

The solution after two iterations is given by

u(x,y) = sin(2y)[− 45− 30x42y4 − 24xy5− 60y2x4

−80y3x3 − 60y4x2 + 16y3x5 + 60y4x4 + 80y5x3

+40y6x2 + 90x2 + 16y6− 4y8][− 16y6 + 4y8 + 45]−1.

3.3. Example 3

Consider the homogeneous telegraph equation

∂ 2u(x, t)∂ 2x2 =

∂ 2u(x, t)∂ 2t2 +

∂u(x, t)∂ t

− u(x,y)

with initial and boundary conditions

I.C. : u(x,0) = ex, ux(x,0) =−2ex.

B.C. : u(0, t) = e−2t , ux(0, t) = e−2t ,

The exact solution for this problem is

u(x, t) = ex−2t ,

S. T. Mohyud-Din · On the Conversion of Partial Differential Equations 899

(a) (b)

Fig. 3 (colour online). Solution by the proposed algorithm (a) and exact solution (b) of Example 3.

where A and B are unknown parameters which are sub-sequently determined by using the initial conditions.Applying the convex homotopy perturbation methodwe get

u0 + pu1+ p2u2 + · · ·

= A+Bη + p∫ η

0

∫ η

0

[(∂ 2u0

∂ s2 + p∂ 2u1

∂ s2 + · · ·)

+

(∂u0

∂ s+ p

∂u1

∂ s+ · · ·

)− (u0 + pu1+ · · ·)

]dsds.

Comparing the co-efficient of like powers of p, follow-ing approximants are obtained:

p(0) : u0(η) = A+Bη ,

p(1) : u1(η) = A+Bη − 118

Bη3 +16

Aη2 +13

Bη2,

p(2) : u2(η) = A+Bη+2

1080Bη5+

1216

Aη4+1

54Bη4

+7

54Bη3 +

127

Aη3 +16

Aη2 +13

Bη2,... .

The series solution is given by

u(η) = A+Bη +2

1080Bη5 +

1216

Aη4 +154

Bη4

+7

54Bη3 +

127

Aη3 +16

Aη2 +13

Bη2 + · · · ,

the inverse transformation would yield

u(x, t) = A+B(x− 2t)+2

1080B(x− 2t)5

+1

216A(x− 2t)4 +

154

B(x− 2t)4

+7

54B(x− 2t)3+

127

A(x− 2t)3

+16

A(x− 2t)2+13

B(x− 2t)2 + · · · ,and the use of the initial condition gives

A = 3e−2t(9+ 6t− 6t2 + 4t3)

4t4 + 27− 36t,

B = 9e−2t(3+ 2t+ 2t2)

4t4 + 27− 36t.

The solution after two iterations is given by

u(x, t) = e−2t[54+ 3x3− 72tx+ 27x2+ 2tx3 − 6t2x2

+ 2t2x3 − 8t3x2 + 8t4x+ 54x− 72t+ 8t4]· [2(4t4 + 27− 36t)

]−1.

4. Conclusion

In this paper, we applied a reliable combination ofHPM and the transformation introduced by He and Wu[1] for solving partial differential equations. It may beconcluded that the proposed coupling is very powerfuland efficient in finding the analytical solutions for awide class of PDEs.

Acknowledgement

The author is highly grateful to the referee for his/her very constructive comments.

900 S. T. Mohyud-Din · On the Conversion of Partial Differential Equations

[1] J. H. He and X. H. Wu, Chaos, Solitons, and Fractals30, 700 (2006).

[2] A. Aslanov, Z. Naturforsch. 64a, 149 (2009).[3] M. A. Abdou and A. A. Soliman, Phys. D 211, 1

(2005).[4] J. Biazar and H. Ghazvini, Comput. Math. Appl. 54,

1047 (2007).[5] D. D. Ganji, H. Tari, and M. B. Jooybari, Comput.

Math. Appl. 54, 1018 (2007).[6] J. H. He, Top. Method Nonlinear Anal. 31, 205 (2008).[7] J. H. He, Int. J. Mod. Phys. 10, 1144 (2006).[8] J. H. He, Appl. Math. Comput. 156, 527 (2004).[9] J. H. He, Int. J. Nonlinear Sci. Numer. Simul. 6, 207

(2005).[10] J. H. He, Appl. Math. Comput. 151, 287 (2004).[11] J. H. He, Int. J. Nonlinear Mech. 35, 115 (2000).[12] J. H. He, Int. J. Mod. Phys. B 22, 3487 (2008).

[13] S. T. Mohyud-Din, M. A. Noor, and K. I. Noor,Math. Prob. Eng. (2009), Article ID 234849, DOI:10.1155/2009/234849.

[14] S. T. Mohyud-Din and M. A. Noor, Math. Prob. Eng.(2007), Article ID 98602, DOI: 10.1155/2007/98602.

[15] S. T. Mohyud-Din, M. A. Noor, and K. I. Noor, Int. J.Nonlinear. Sci. Numer. Simul. 10, 223 (2009).

[16] S. T. Mohyud-Din and M. A. Noor, Z. Naturforsch.64a, 157 (2009).

[17] S. D. Zhu, Inter. J. Nonlinear Sci. Numer. Simul. 8, 461(2007).

[18] M. A. Noor and S. T. Mohyud-Din, Comput. Math.Appl. 54, 1101 (2007).

[19] A. Yildirim, Z. Naturforsch. 63a, 621 (2008).[20] E. M. E. Zayed, T. A. Nofal, and K. A. Gepreel, Z. Na-

turforsch. 63a, 627 (2008).