Homotopy Perturbation Method for Solving Partial Differential Equations

Homotopy Perturbation Method for Solving Partial Differential Equations Syed Tauseef Mohyud-Din and Muhammad Aslam Noor Department of Mathematics COMSATS Institute of Information Technology Islamabad Pakistan Reprint requests to S T M-D; E-mail: syedtauseefs@hotmailcom Z Naturforsch 64a 57 7 9; received June 9 8 We apply a relatively new technique which is called the homotopy perturbation method HPM for solving linear and nonlinear partial differential equations The suggested algorithm is quite efficient and is practically well suited for use in these problems The proposed iterative scheme finds the solution without any discretization linearization or restrictive assumptions Several eamples are given to verify the reliability and efficiency of the method The fact that the HPM solves nonlinear problems without using Adomian s polynomials can be considered as a clear advantage of this technique over the decomposition method Key words: Homotopy Perturbation Method; Partial Differential Equations; Helmholtz Equations; Fisher s Equations; Initial Boundary Value Problems; Boussinesq Equations Introduction In the last two decades with the rapid development of nonlinear science there has appeared everincreasing interest of physicists and engineers in the analytical techniques for nonlinear problems It is well known that perturbation methods provide the most versatile tools available in nonlinear analysis of engineering problems see [ 9] and the references therein The perturbation methods like other nonlinear analytical techniques have their own limitations At first almost all perturbation methods are based on the assumption that a small parameter must eist in the equation This so-called small parameter assumption greatly restricts applications of perturbation techniques As is well known an overwhelming majority of nonlinear problems have no small parameters at all Secondly the determination of small parameters seems to be a special art requiring special techniques An appropriate choice of small parameters leads to the ideal results but an unsuitable choice may create serious problems Furthermore the approimate solutions solved by perturbation methods are valid in most cases only for the small values of the parameters It is obvious that all these limitations come from the small parameter assumption These facts have motivated to suggest alternate techniques such as variational iteration [8 5 8 3] decomposition [33 39] epfunction [4 4] variation of parameters [4] and iter- ative [43 44] In order to overcome these drawbacks combining the standard homotopy and perturbation method which is called the homotopy perturbation modifies the homotopy method Many problems in natural and engineering sciences are modeled by partial differential equations PDEs These equations arise in a number of scientific models such as the propagation of shallow water waves long wave and chemical reaction-diffusion models see [4 5 3 4 45 65] and the references therein A substantial amount of work has been invested for solving such models Several techniques including the method of characteristic Riemann invariants combination of waveform relaation and multi-grid periodic multi-grid wave form variational iteration homotopy perturbation and Adomian s decomposition [4 5 3 4 45 65] have been used for the solutions of such problems Most of these techniques encounter the inbuilt deficiencies and involve huge computational work He [3 8] developed the homotopy perturbation method for solving linear nonlinear initial and boundary value problems by merging two techniques the standard homotopy and the perturbation technique The homotopy perturbation method was formulated by taking the full advantage of the standard homotopy and perturbation methods and has been applied to a wide class of functional equations see [ 9] and the references therein The basic motivation of the present paper is the implementation of 93 784 / 9 / 3 57 $ 6 c 9 Verlag der Zeitschrift für Naturforschung Tübingen http://znaturforschcom

58 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations this reliable technique for solving PDEs In particular the proposed homotopy perturbation method HPM is tested on Helmholtz Fisher s Boussinesq singular fourth-order partial differential equations systems of partial differential equations and higher-dimensional initial boundary value problems The proposed iterative scheme finds the solution without any discretization linearization or restrictive assumptions and is free from round off errors The HPM gives the solution in the form of a convergent series with easily computable components Unlike the method of separation of variables which requires both initial and boundary conditions the HPM gives the solution by using the initial conditions only The fact that the proposed HPM solves nonlinear problems without using Adomian s polynomials can be considered as a clear advantage of this technique over the decomposition method The Homotopy Perturbation Method To eplain the HPM we consider a general equation of the type Lu= where L is any integral or differential operator We define a conve homotopy Hu p by Hu p= pfuplu where Fu is a functional operator with known solutions v which can be obtained easily It is clear that for we have Hu p= 3 Hu =Fu Hu =Lu This shows that Hu p continuously traces an implicitly defined curve from a starting point H v to a solution function H f The embedding parameter monotonically increases from zero to unity as the trivial problem Fu= continuously deforms the original problem Lu= The embedding parameter p ] can be considered as an epanding parameter [ 9] The HPM uses the homotopy parameter p as an epanding parameter [3 8] to obtain u = i= p i u i = u pu p u p 3 u 3 4 If p then 4 corresponds to and becomes the approimate solution of the form f = lim u = p i= u i = u u u 5 It is well known that series 5 is convergent for most of the cases and also the rate of convergence is dependent on Lu For more details about the convergence of the HPM see [ 9] and the references therein The comparisons of equal powers of p give solutions of various orders In sum according to [ ] He s HPM considers the solution u of the homotopy equation in a series of p as u= i= p i u i = u pu p u and the method considers the nonlinear term Nu as Nu= i= p i H i = H ph p H where H n are the so-called He s polynomials [ ] which can be calculated by using the formula H n u u n = n n n! p n N p i u i i= p= n = 3 Numerical Applications In this section we apply the HPM for solving PDEs In particular the proposed HPM is tested on Helmholtz Fisher s Boussinesq singular fourth-order partial differential equations systems of partial differential equations and higher-dimensional initial boundary value problems Numerical results are very encouraging 3 Eample Consider the Helmholtz equation [44] u uy y uy= with the initial conditions uy=y u y=y coshy

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 59 y Eact solution HPM ITM Absolute error 99676 9974799 9974799 777 8 8 4948 4959 4959 638 6 6 45663 45333 45333 3578 4 4 7556967 755485 755485 5 367755 367759484 367759484 7 448939 44893884 44893884 8 4 4 95888 95645 95645 43 6 6 84554 845793 845793 448 8 8 85387 85549 85549 38 4636463 45974799 45974799 6564 Table Error estimates u u p u = p u pu p u u p y u y p u y p : u y=y coshy p : u y=! y 3! 3 y p : u y= 4! 4 y 5! 5 y p 3 : u 3 y= 6! 6 y 7! 7 y p 4 : u 4 y= 8! 8 y 9! 9 y gives the solution as uy=yepcoshy Table ehibits the approimate solution obtained by using the HPM and ITM It is clear that the obtained results are in high agreement with those obtained using the eact solutions Higher accuracy can be obtained by using more terms 3 Eample Consider the Helmholtz equation [44] uy uy y 8uy= with the initial conditions uy=siny u y= u u p u = [ p 8 u pu p u ] u y u y p u y p : u y=siny p : u y= siny p : u y= 8 3 4 siny p 3 : u 3 y= 4 45 6 siny gives the series solution as u y=siny 3 4 4 45 6 and in the closed form as u y =cos siny Table ehibits the approimate solution obtained by using the HPM and ITM [5] It is clear that the obtained results are in high agreement with those obtained using the eact solutions Higher accuracy can be obtained by using more terms

6 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations y Eact solution HPM ITM Absolute error 94797 9453593 9453593 76577 35867845 35867845 35867845 3 3 4669543 4669543 4669543 4 4 4997868 49978683 49978683 5 5 45464873 454648957 454648957 44 6 6 3377359 3377333 3377333 64 7 7 6749475 675984 675984 799 8 8 9877 95749 95749 958 9 9 6 795 795 8996 378447 3787363 3787363 884 Table Error estimates 33 Eample 3 Consider the Fisher s equation of the form u t t u t ut ut = subject to the initial conditions u=β This Fisher s problem can be formulated as the integral equation ut=β u tut utdt u pu p u = u β p u p u u pu p u u pu p u dt p : u y=β p : u y=β β t p : u y= t! β 3β β t3 3 β β p 3 : u 3 y= t3 3! β 6β β 3 5β 4 t4 3 β β β t5 6 β β 3 β β t6 8 β 3 β 3 β t7 63 β 4 β 4 Table 3 Numerical results for Fisher s equation t β = β = 8 69E-6 9647E-6 557339E-6 6375E-6 4 3635E-4 6E-5 337E-4 4783E-5 6 54555E-4 5333E-4 447E-4 63379E-4 8 785E-3 54998E-4 8584E-3 934E-4 445699E-3 45544E-4 995E-3 6455E-5 gives the solution in a closed form: ut= β ept β β ept Table 3 shows the numerical results 34 Eample 4 Consider the Fisher s equation of the form u t u 6u u= subject to the initial conditions u= e This Fisher s problem can be formulated as the integral equation t ut= e u 6u udt u pu p u = e p u u p u 6u pu p u u pu p u dt

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 6 Table 4 Numerical results for Fisher s equation t = t = 4 7E-3 6996E-3 57598E- 5844E- 98949E-3 5566E-3 65E- 577E- 4 9765E- 3697E-3 393E- 47E- 6 439E- 73899E-4 5579E- 5459E- 8 8573E-3 573E-4 4359E- 58693E-3 587E-3 9777E-4 9333E- 4367E-3 p : u t= e p : u t= e t [ p : u t= e 6 5e e e t e t 3 e 3 e 6 ] e t p 3 : u 3 y= 5e 3 e 6 5 6e 5e e 3 t 3 5e e 8 7 5e 5e t 4 5e e 9 5 47e e 3 t 5 e3 e t 6 e 4e4 t 7 7 e [ 5e e 6 e e e t e t 3 e t ] gives the solution in a closed form: ut= ep 5t Table 4 shows the numerical results 35 Eample 5 Consider the generalized Fisher s equation u t = u u u 6 subject to the initial conditions u= e 3 3 This Fisher s problem can be formulated as the integral equation ut= t e 3 p u u u 6 dt 3 u pu p u = p e 3 /3 u u p u u pu p u u pu p u 6 dt p : u t= e 3 /3 p : u t= 4 e3 4 3te 3/ 8 t 4 e 3/ 7/3 p : u t= [ 37 e 3/ 49/3 496e 3/ e 3/ 363 43e 3/ 9e 3 9e 9/ t 57344e 3 e 3/ 3e 3/ t 3 79e 9/ e 3/ 8 3e 3/ 3 t 4 3584e 6 e 3/ 6 3e 3/ 4 t 5 448e 5/ e 3/ 4 3e 3/ 5 t 6 448e 5/ e 3/ 4 3e 3/ 5 t 6 3e 9 e 3/ 3e 3/ 6 t 7 e / 3e 3/ 7 t 8 3768 e 3/ 4 ] 4 e 3 4 3te 3/ 8 t

6 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations Table 5 Numerical results for the generalized Fisher s equation t = t = 4 5496E- 45437E- 845E- 97465E- 779547E- 4746E- 7494E- 839974E- 4 85E- 3376E- 347E- 93E-4 6 5375E- 9936E- 494354E- 463E- 8 996E- 538E-3 6747E- 463E- 5537E- 785833E-3 87889E- 4665E- gives the solution in a closed form: ut=/tanh[ 3/4 5/t] / /3 Table 5 shows the numerical results 36 Eample 6 Consider the singularly perturbed sith-order Boussinesq equation [4 5 3 33 4] u tt = u pu αu β u Taking α = β = and pu=3u the model equation is given as u tt = u 3u u with the initial conditions u= ak e k ae k u t = ak3 k ae k e k ae k 3 where a and k are arbitrary constants The eact solution ut of the problem is given as [33] ak epk k k ut= t aepk k k t u pu p u p 3 u 3 = ak e k ae k ak3 k ae k e k ae k 3 t u p u p u p [ 4 u 4 p 4 u 4 p 4 u 4 u 3 u p u dtdt ] dtdt p : u t= ek e p : u t= ak3 k ae k e k ae k 3 t e 4e e e 4 t p : u t= e e e e 3 e 5 t 3 e 4e e 44e 78e 44e 3 e 4 3 e 8 t 4 p 3 [ : u 3 t= e 5 e 7 e 56e 46e 56e 3 e 4 ] t 5 [ e 45 e 45e 949e 7936e 3 87378e 4 56568e 5 ] t 6 [ e 45 e 87378e 6 7936e 7 ] 949e 8 45e 9 e t 6 gives the series solution as ut= ek e ak3 k ae k e k ae k 3 t e 4e e e 4 t e e e e 3 e 5 t 3 e 4e e 44e 78e 44e 3 e 4 3 e 8 t 4 8e e e e 3 e 4 e 8 t 4 e e 56e 46e 56e 3 e 4 5 e 7 t 5 [ e 45 e 45e 949e 7936e 3 87378e 4 56568e 5 ] t 6 [ e 45 e 87378e 6 7936e 7 949e 8 45e 9 e ] t 6

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 63 i t j 4 5 8886 E-4 79667 E- 535 E- 83355 E-8 83899 E-6 47468 E-4 8 6776 E-4 436 E- 5747 E- 63378 E-8 4454 E-6 4489 E-3 6 684 E-4 3988 E- 5663 E- 684 E-8 499 E-6 393 E-3 4 6573 E-4 749 E-3 48756 E- 3843 E-8 8538 E-6 463 E-4 553446 E-4 353395 E- 5663 E- 547485 E-8 34764 E-6 835783 E-4 86398 E-4 553357 E- 5474 E- 86597 E-8 554893 E-6 37353 E-3 556 E-4 35544 E- 7779 E- 5636 E-8 3636 E-6 996 E-4 4 4353 E-4 7498 E-3 4497 E- 337 E-8 59384 E-7 966 E-5 6 668 E-4 38755 E- 478 E- 59756 E-8 37675 E-6 879 E-4 8 63945 E-4 39959 E- 557 E- 6888 E-8 39 E-6 93644 E-4 79776 E-4 78946 E- 437 E- 77684 E-8 7667 E-6 48986 E-4 Table 6 Error estimates Fig Series solution ut Table 6 ehibits the absolute error between the eact and the series solutions Higher accuracy can be obtained by introducing some more components of the series solution Figure depicts the series solution ut 37 Eample 7 Consider the singularly perturbed sith-order Boussinesq equation [4 5 3 33 4] u tt = u u u u with the initial conditions u= 5 69 sech4 6 tanh 94 3 sech4 6 6 u t = 97 The eact solution of the problem is given as [ ut= 5 ] 97 69 sech4 6 69 t u pu p u = 5 69 sech4 p p tanh 94 3 sech4 6 6 u 97 6 u p u [ u pu p u p 3 u 3 4 u 4 u p 4 4 p 4 u p ] 4 t dtdt 6 u 6 p 6 u 6 p 6 u 6 dtdt dtdt p : u = 5 69 sech4 6 sinh 94 5 p 3 sech6 6 : u t= 97 5 3793 9 94cosh 3 3 sech 6 t 6 p :u t= 395sech7 6 86 5sinh 5676644 6 664 5sinh 3 [ t 3 334sech 5 6 6 35447cosh sech 5 3 6 t

64 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations i t j 4 5 77756 E-6 36557 E-4 857869 E-3 964 E- 3383 E-8 35944 E-6 8 E-6 9984 E-5 688 E-3 7388 E- 7488 E-9 4494 E-7 6 45 E-6 99 E-4 7886 E-3 783 E- 45 E-8 798 E-6 4 E-6 337 E-4 53 E- 367 E- 34944 E-8 5749 E-6 66634 E-6 3375 E-4 4747 E- 49998 E- 39983 E-9 7859 E-6 44489 E-6 3497 E-4 4365 E- 54774 E- 35559 E-8 855935 E-6 555 E-6 39744 E-4 474 E- 4998 E- 39858 E-8 78749 E-6 4 33367 E-6 337 E-4 534 E- 36685 E- 3476 E-8 5764 E-6 6 33367 E-6 33 E-4 7858 E- 7777 E- 3695 E-8 77 E-6 8 33367 E-6 9984 E-5 33 E-3 76944 E- 788 E-9 44936 E-7 77756 E-6 38778 E-4 85833 E-3 9593 E- 3444 E-8 3854 E-6 4764cosh sech 5 ]t 4 3 6 [ 3 3cosh 3 sech 5 3 6 4 388cosh sech 5 ]t 4 3 6 Table 7 Error estimates gives the series solution as ut= 5 69 sech4 6 sinh 94 5 3 sech6 6 3 t 5 3793 97 9 94cosh 3 sech 6 t 6 395sech7 6 86 5sinh 5676644 6 664 5sinh 3 [ t 3 334sech 5 6 6 35447cosh sech 5 3 6 4764cosh sech 5 ]t 4 3 6 [3cosh 3 3 sech 5 3 6 4 388cosh sech 5 ]t 4 3 6 Table 7 ehibits the absolute error between the eact and the series solutions Higher accuracy can be obtained by introducing some more components of the series solution Figure depicts the series solution ut Fig Series solution ut 38 Eample 8 Consider the following nonlinear system of partial differential equations: u t v w y v y w = u v t w u y w y u = v w t u v y u y v = w with the initial conditions uy=e y vy=e y wy=e y v u pu p u = e y p p v p v w y p w y w p y dt v p y p v y v p y w p w p w dt p t u pu p u dt

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 65 v pv p v = e y p p u y p u p w y w p w p w p u u p dt p u y p u y dt w y p w y v pv p v u dt w pw p w = e y p p u p u v y p v y v p y dt u p y p u y u p y v p v p v dt p w pw p w dt p : u yt=e y v yt=e y w yt=e y p : u yt= te y v yt=te y w yt=te y p : u yt= t! ey v yt= t! e y w yt= t! e y p 3 : u 3 yt= t3 3! ey v 3 yt= t3 3! e y w 3 yt= t3 3! e y gives the closed form solution as uvw=e y t e yt e yt 39 Eample 9 Consider the singular fourth-order parabolic equation u t 4 4 u 4 = < < t > subject to the initial conditions u = < < u 5 = t < < and the boundary conditions u t = /5 ut= u t = 6 sint sint t > 3 sint u t= sint t > 6 u pu p u = u t p 4 4 u 4 u p 4 4 p 4 u 4 dtdt p : u t= 5 t p : u t= 5 t 3 3! p : u t= 5 t 5 5! p 3 : u 3 t= 5 t 7 7!

66 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations gives the solution as ut= 5 = 5 t t3 sint 3! t5 5! t7 7! which is the eact solution It is interesting to point out that the eact solution is obtained by using the initial conditions only Moreover the obtained solution can be used to justify the given boundary conditions 3 Eample Consider the following singular fourth-order parabolic partial differential equation in two space variables: u t 4 4 u 6! 4 y y4 4 u 6! y 4 = with the initial conditions uy= and the boundary conditions u 6 y= t 6! y6 6! u yt = 56 y6 sint 6! 6! u yt= y6 sint 6! 6! u yt = 54 4 sint u yt= 4 sint u y t = 54 4 sint u y t= 4 sint u pu p u = u t p p y y4 4 u 6! y 4 u p 4 y 4 p 4 u y 4 dtdt 4 4 u 4 u p 4 4 p 4 u 4 dtdt p : u t= 6 6! y6 t 6! p : u t= 6 6! y6 t3 6! 3! p : u t= 6 6! y6 t 5 6! 5! p 3 : u 3 t= 6 6! y6 t7 6! 7! p 4 : u 4 t= 6 6! y6 t 9 6! 9! gives the eact solution easily: u yt= t 6 6! y6 t3 6! = 6 6! y6 6! 3 Eample sint 3! t5 5! t7 7! t9 9! Consider the fourth-order singular parabolic partial differential equation u 4 t sin u = 4 < < t > with the initial conditions u= sin < < u = sin < < t

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 67 and the boundary conditions ut= ut=e t sin t > u t= u t=e t sin t > u pu p u = u t p sin 4 u 4 u p 4 4 p 4 u 4 dtdt p : u t= sin p : u t= sint t p : u t= sin! t3 3! t p 3 4 : u 3 t= sin 4! t5 5! t p 4 6 : u 4 t= sin 6! t7 7! t p 5 8 : u 5 t= sin 8! t9 9! gives the solution as ut= sin t t! t3 3! t4 4! t5 5! = sine t which is the eact solution It satisfies the boundary conditions also that we did not use in the analysis 3 Eample Consider the two-dimensional initial boundary value problem u tt = y u u yy < y < t > with the boundary conditions uyt=y e t uyt= y e t ut=y e t ut= e t and the initial conditions uy= y u t y= y u pu p u = y y t p y u u u dtdt p u y u y u y dtdt p : u yt= y y t p : u yt= y t! y t3 3! p : u yt= y t4 4! y t5 5! p 3 : u 3 yt= y t5 5! y t7 7! p 4 : u 4 yt= y t8 8! y t9 9! p 5 : u 5 yt= y t! y t! gives the series solution as uyt= y t t! t3 3! t4 4! t5 5! and in a closed form as uyt= y e t which is in full agreement with [] 33 Eample 3 t6 6! t7 7! t8 8! Consider the three-dimensional initial boundary value problem u tt = 45 u 45 y u yy 45 z u zz u < y < t <

68 S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations subject to the Neumann boundary conditions u yzt= u yzt=6y 6 z 6 sinht u y zt= u y zt=6 6 z 6 sinht u z yt= u z yt=6 6 y 6 sinht and the initial conditions uyz= u t yz= 6 y 6 z 6 u pu p u = 6 y 6 z 6 t 45 p u u u dtdt 45 p y u y u y u y z u z u z u z dtdt p u pu p u dtdt p : u yzt= 6 y 6 z 6 t p : u yzt= 6 y 6 z 6 t3 3! p : u yzt= 6 y 6 z 6 t5 5! p 3 : u 3 yzt= 6 y 6 z 6 t7 7! p 4 : u 4 yzt= 6 y 6 z 6 t9 9! gives the series solution as uyzt= 6 y 6 z 6 t t3 3! t5 5! t7 7! t9 9! = 6 y 6 z 6 sinht 34 Eample 4 Consider the two-dimensional nonlinear inhomogeneous initial boundary value problem u tt = y 5 u yu yy < y < t > with the boundary conditions uyt=y t yt 6 uyt= y t yt 6 ut= t t 6 ut= t t 6 and the initial conditions uy= u t y= [ ũ u pu p u = p p ũ ũ y y ũ y y ] dtdt ũ p ũ y p : u yt= p : u yt= y t p : u yt= yt 6 p 3 : u 3 yt= gives the solution as uyt= y t yt 6 which is in full agreement with [] 35 Eample 5 Consider the three-dimensional nonlinear initial boundary value problem u tt = t u e u e y u yy e z u zz < y < t <

S T Mohyud-Din and M A Noor The HPM for Solving Partial Differential Equations 69 subject to the Neumann boundary conditions u yzt= u y zt= u z yt= and the initial conditions uyz=e e y e z u t yz= u yzt=e u y zt=e u z yt=e u pu p u =e e y e z [ e u p u p u ] t dtdt p p u y e z u ] dtdt p [ e y u y u y z u z p u z u pu p u dtdt p : u yzt=e e y e z t t4 p : u yzt= t4 t6 36 [] A Ghorbani and J S Nadjfi Int J Nonlinear Sci Numer Simul 8 9 7 [] A Ghorbani Chaos Solitons and Fractals 7 in press [3] J H He Phys Lett A 35 87 6 [4] J H He Appl Math Comput 56 57 4 [5] J H He Int J Nonlinear Sci Numer Simul 6 7 5 [6] J H He Appl Math Comput 5 87 4 [7] J H He Int J Nonlinear Mech 35 5 [8] J H He Int J Mod Phys 44 6 [9] S T Mohyud-Din and M A Noor Math Prob Eng 7 [] M A Noor and S T Mohyud-Din Math Comput Modl 45 954 7 p : u yzt= t6 36 t8 6 gives the solution as uyzt=e e y e z t 4 Conclusions We applied the homotopy perturbation method HPM for finding the solution of a system of partial differential equations The method is applied in a direct way without using linearization transformation discretization or restrictive assumptions It may be concluded that the HPM is very powerful and efficient in finding the analytical solutions for a wide class of boundary value problems The method gives more realistic series solutions that converge very rapidly in physical problems It is worth mentioning that the method is capable of reducing the volume of the computational work as compared to the classical methods while still maintaining the high accuracy of the numerical result The fact that the HPM solves nonlinear problems without using the Adomian s polynomials is a clear advantage of this technique over the decomposition method Acknowledgement The authors are highly grateful to Dr S M Junaid Zaidi Rector CIIT for providing an ecellent research environment and facilities [] M A Noor and S T Mohyud-Din Comput Math Appl 55 953 8 [] M A Noor and S T Mohyud-Din J Math Anal Appl Th 6 6 [3] M A Noor and S T Mohyud-Din Int J Math Comput Sci 345 7 [4] M A Noor and S T Mohyud-Din Int J Mod Math 8 in press [5] M A Noor and S T Mohyud-Din Int J Nonlinear Sci Numer Simul 9 4 8 [6] M A Noor and S T Mohyud-Din Math Prob Eng 8 in press [7] M A Noor and S T Mohyud-Din Comput Math Appl 8 in press

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