World Applied Sciences Journal 14 (12): 1786-1791, 2011 ISSN 1818-4952 © IDOSI Publications, 2011
An Efficient Numerical Method for Solving Linear and Nonlinear Partial Differential Equations by Combining Homotopy Analysis and Transform Method
1Muhammad Asif Gondal, 2A.S. Arife, 1Majid Khan and 1Iqtadar Hussain
1Department of Science and Humanities, National University of Computer and Emerging Science, Islamabad, Pakistan 2Faculty of Science, Department of Mathematics, National Centre for Research, Egypt
Abstract: In this article, we propose a reliable combination of Homotopy analysis method (HAM) and Laplace decomposition method (LDM) to solve linear and nonlinear partial differential equations effectively with less computation. The proposed method is called homotopy analysis transform method (HATM). This study represents significant features of HATM and its capability of handling linear and nonlinear partial differential equations.
Key words : Homotopy analysis method • laplace decomposition method • linear and nonlinear partial
differential equations • homotopy analysis transform method
INTRODUCTION nonlinear partial differential equations in this paper. It is worth mentioning that the proposed method is an A nonlinear phenomenon appears in a wide variety elegant combination of the homotopy analysis method of scientific applications such as plasma physics, solid and Laplace transform. The advantage of this proposed state physics, fluid dynamics and chemical kinetics. A method is its capability of combining two powerful broad range of analytical and numerical methods have methods for obtaining rapid convergent series partial been used in the analysis of these scientific models. differential equations. To the best of author knowledge Mathematical modeling of many physical systems leads no such attempt has been made to combine homotopy to nonlinear ordinary and partial differential equations analysis method and Laplace transform for solving in various fields of physics and engineering. An partial differential equations. This paper considers the effective method is required to analyze the effectiveness of the homotopy analysis transform mathematical model which provides solutions method in solving linear and nonlinear equations. conforming to physical reality. Common analytic The paper is organized as follows. In Section 2, the procedures linearize the system and assume that basic concepts of HAM is presented. Section 3 contains nonlinearities are relatively insignificant. Such basic idea of HATM. Section 4, contains applications of assumptions sometimes strongly affect the solution with HATM. The conclusions are given in last Section. respect to the real physics of the phenomenon. Thus seeking solutions of nonlinear ordinary and partial BASIC IDEA OF HAM differential equations are still significant problem that needs new techniques to develop exact and approximate In HAM, a system can be writtenas: solutions. Various powerful mathematical techniques such as Adomian decomposition method [1, 2], N[u(x,t)]= 0,i = 1,2,3.... (1) Homotopy analysis method [3-5], differential transform method [6, 7], Homotopy perturbation method [8, 9], where N is a nonlinear operator, u(x,t) is unknown Variational iterative method [10, 11], Laplace ≠ decomposition method [12-15], modified Laplace function of x and t, u0(x,t) is the initial guess, 0 an decomposition method [16, 17] and homotopy auxiliary parameter and L is a auxiliary linear operator. perturbation transform method [18] to obtain exact and Also, q∈[0,1] is an embedding parameter. We can approximate analytical solutions. construct a Homotopy as follows We use the homotopy analysis method combined with the Laplace transform for solving linear and (1−φ q)L[ (x,t;q) − u0 (x,t)] = q N[ φ(x,t;q)] (2)
Corresponding Author: Majid Khan, Department of Science and Humanities, National University of Computer and Emerging, Science, Islamabad, Pakistan 1786 World Appl. Sci. J., 14 (12): 1786-1791, 2011
Eq. (1), is so-called zero-order deformation Lu(x,t)++= Ru(x,t) Nu(x,t) f(x,t) (11) equation. When q = 0, the zero-order deformation equation become where L is a linear operator, N is a nonlinear operator and ƒ(x,t) is a source function. The initial conditions are φ=(x,t;0) u0 (x,t) (3) also as
when q = 1, since ≠ 0 , we get final solution u(x,0)= g(x),ut (x,0)= f(x) (12) expression as follows Applying the Laplace transforms ℘, we obtain φ=(x,t;1) u(x,t) (4) ℘+℘+℘=℘Lu(x,t) Ru(x,t) Nu(x,t) f(x,t) (13) The embedding parameter q increases from 0 to 1.
Using Taylor's theorem, φi(x,t;q) can be expanded in a Therefore, we have power series of q as follows 2 s℘− u(x,t) su(x,0) − u(x,0)t ∞ , m =℘f(x,t) −℘ Ru(x,t) −℘Nu(x,t) φ=+(x,t;q) u0m (x,t)∑ u (x,t)q (5) m1= Or h(x) f(x) 1 m ℘u(x,t) = + +℘f(x,t) 1∂φ (x,t;q) 22 = sss um (x,t) m (6) (14) m!∂ q = 11 q0 −℘Ru(x,t) −℘Nu(x,t) ss22 where u0(x,t) is a initial guess, Li is a auxiliary linear operator and ≠ 0 is the non-zero auxiliary parameter. Now we embed the HAM in Laplace transform The power series in equation (5) is converges at q = 1. method. Hence we may write any equation in the form Therefore, we obtain N[u(x,t)]= 0,i = 1,2,3 (15) ∞ φ=+ (x,t;q) u(x,t)0m∑ u (x,t) (7) h(x) f(x) 1 1 m1= φ = + +℘ −℘φ N[ (x,t;q)] 22f(x,t) 2 R[ (x,t;q)] sss s (16) 1 According to the equation (6), the governing − ℘N[ φ (x,t;q)] −℘φ [ (x,t;q)] s2 equation of ui(x,t) can be derived from the zero-order deformation Eq.(2). Using m times differentiating with where N is a nonlinear operator, u(x,t) is unknown respective to q from the zero-order deformation function of x and t, ≠ 0 an auxiliary parameter and L equation (2) and setting q=1, we have the so-called an auxiliary linear operator. Also, q∈[0,1] is an mth-order deformation equation as: embedding parameter. Thus, we can construct such a Homotopy such as L[um (x,t)−χm u 0 (x,t)] = Rm( u m1− (x,t)) (8)
(1− q) ℘φ [ (x,t;q) − u0 (x,t)] = q N[ φ (x,t;q)] (17) where which is a zero-order deformation equation. When 1∂φm1− N[ (x,t;q)] q = 0, the zero-order deformation equation become R (u (x,t)) = (9) m m1− −∂m1− (m 1)! q q0= φ=(x,t;0) u0 (x,t) (18) and 1, m> 1 when q = 1, since i ≠ 0 we get an analytical χ=m (10) 0, m≤ 1 approximate solution as follows
HOMOTOPY ANALYSIS φ=(x,t;1) u(x,t) (19) TRANSFORM METHOD The embedding parameter q increases from 0 to 1.
We consider a general nonlinear, non-homogenous Using Taylor's theorem, φi(x,t;q) can be expanded in a partial differential equation power series of q as follows 1787 World Appl. Sci. J., 14 (12): 1786-1791, 2011
∞ m Applying Laplace transform, we get φ=+(x,t;q) u0m (x,t)∑ u (x,t)q (20) m1= 11 where ℘− − u(x,t) u(x,0)2 ut (x,0) 1∂φm (x,t;q) ss = (29) um (x,t) m (21) 11 1 m!∂ q −℘yu22 + xu = 0 q0= 2 xx yy s2 2
If u (x,t) guesses the initial condition, L is the i,0 i Using given the initial condition Eq. (28) becomes auxiliary linear operator, ≠ 0 is the non-Zero auxiliary parameter then the power series in equation (20) is 11 1 ℘ −℘22 + converges at q = 1. Therefore, we have: u(x,t)2 y uxx x uyy s2 2 (30) 2222 ∞ xyxy++ −+= φ=+(x,t;q) u(x,t) u (x,t) (22) 2 0 0m∑ ss m1=
Form Eq. (30), we define a nonlinear operator as According to the Eq. (6), the governing equation of ui(x,t) can be derived from the zero-order deformation ∂φ22 ∂φ φ =℘φ −1 ℘ 22 (x,t;q)+ (x,t;q) Eq. (2) Using m times differentiating with respective to N[ (x,t;q)] (x,t;q)22 y x 2 q from the zero-order deformation equation (2) and sx∂∂ y(31) 2222 setting q=1, we have the so-called m-th-order xyxy++ −+=2 0 deformation equation as ss
Using the above definition, we construct the ℘[u (x,t) −χ u (x,t)] = R( u (x,t)) (23) m m 0 m m1− zeroth-order deformation equation
Using the last equation the series solution is given by (1− q) ℘φ [ (x,t;q) − u0 (x,t)] = q N[ φ (x,t;q)] (32)
u (x,t)=χ u(x,t) +℘ −1 R u (x,t) (24) m m 0 ( m() m1− ) With initial conditions, where q∈[0,1] is an embedding parameter, is a non-zero auxiliary where function, ℘ is Laplace transformation operator, u0(x,t) is an initial guess of u(x,t) and φ(x,t;q) is unknown 1∂φm1− N[ (x,t;q)] function. When q = 0 and q = 1 we have R (u (x,t)) = (25) i,m i , m− 1 −∂m1− (m 1)! q q0= φ(x,t;0) = u (x,t), φ= (x,t;1) u(x,t) (33) and 0 1, m> 1 χ= m (26) Expanding φ(x,t;q) in Taylor series with respect to 0, m≤ 1 q, we obtain APPLICATION ∞ m φ=+(x,t;q) u0m (x,t)∑ u (x,t)q (34) In order to elucidate the solution procedure of the m1= homotopy Analysis transform method, we solve three where examples in this sections which shows the effectiveness 1∂φm1− (x,t;q) um (x,t)= (35) and generalizations of our proposed method. (m−∂ 1)! q m1−
Example 1: Consider the two-dimensional initial The above series is convergent at q = 1, then values problem
∞ 11 22 φ=+(x,t;q) u(x,t)0m u (x,t) (36) utt = y uxx + x uyy ,x ><> 0,y 1,t 0 (27) ∑ 22 m1= with the initial conditions The Eq. (36) must be one of the solutions of the original nonlinear equation (1), which is proved by Liao 2 2 22 u(x,y,0)=+ x y,u(x,y,0)t =−+ (x y) (28) [3]. Now, we define the vector 1788 World Appl. Sci. J., 14 (12): 1786-1791, 2011
um1−−= { u 0 (x,t),u 2 (x,t),...... um1 (x,t)} (37)
The mth order deformation equation is
−1 um =χmm u−−1 +℘ R(m u)m1 (38) where 2 2 2222 1 22∂ um1−− ∂ um 1 xyxy++ Rm (u m1−− )=℘−℘ u m1 y + x +−χ−() 1 m + (39) s222 ∂∂ x y ss2
Using the Mathematica 5.2 package, we obtain the solution as
u (x,y,t)=−−+ (t2 2t)(x22y) (40) 1 2
t u (x,y,t)=() t3 − 4 t2 − 12 t −++ 12t 24 24)(x22 + y (41) 2 24 The solution is given by tttttt234567 u(x,y,t)=−+−+−+−+ 1 t ...... (x 22 +y) (42) 2 3! 4! 5! 6! 7!
u(x,y,t)= e− t (x22 + y) (43) where = −1
Example 2: Consider the two dimensional nonlinear inhomogeneous initial value problems
15 u=+++><> 2x22 2y (xu 2 yu 2 ),x 0,y 1,t 0 (44) tt 2 xx yy with the initial conditions
u(x,y,0)= 0,u(x,y,0)t = 0 (45)
Applying Laplace transforms, we get 2x22+ 2y 15 ℘u(x,t) − −℘xu22 + yu = 0 (46) s 2s2 xx yy We define a nonlinear operator as
2 ∂φ 2 ∂φ 22+ φ =℘φ −1 ℘ (x,t;q)+ (x,t;q)− 2x 2y (47) N[ (x,t;q)] (x,t;q)2 xy sx∂∂ y s
The m-th order deformation equation is
−1 um =χmm u−−1 +℘ R(m u)m1 (48) where 12m1−−∂∂ u u m1 ∂∂u u 2x22+ 2y =℘ − ℘ k m1k−− + k m1k− − − −χ Rm (u m1−− ) u m1 2 x∑∑y ()1 m (49) 5sk0= ∂∂ x xk= 0 ∂∂ y y s
Using the Mathematica 5.2 package, we obtain the solution as
2 22 u(x,y,t)1 =−+ t (x y)
2 22 u(x,y,t)2 =−+ ( 1)t (x +y) (50) 1789 World Appl. Sci. J., 14 (12): 1786-1791, 2011
2 2 2 422 u(x,y,t)3 =−+++ t (1 2 t y)(x +y) where = −1, which is exact solution obtained as …….. special case of homotopy perturbation transform method [19]. The solution is given by CONCLUSION u(x,y,t)=+ (x2 y 22 ) t ++ (x 2 y 26 ) t +...... 51) The main aim here is to provide the series solution of linear and nonlinear partial differential where = −1. equations using Homotopy Analysis Transform Method (HATM). The solution obtained with the help Example 3 : Consider the inhomogeneous problem [19] of HATM is more general as compared to HPTM, ADM and VIM solution. We can easily recover all u+= uu 0 (52) tx results of HPTM, ADM and VIM by assuming = −1. The analysis given here shows further confidence with the initial condition on HATM.
u(x,0)= − x (53) REFERENCE
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