Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 326473, 12 pages http://dx.doi.org/10.1155/2013/326473 Research Article The Application of the Homotopy Analysis Method and the Homotopy Perturbation Method to the Davey-Stewartson Equations and Comparison between Them and Exact Solutions Hassan A. Zedan,1,2 W. Barakati,1 and Nada Hamad1 1 2 Mathematics Department, Faculty of Science, King Abdul Aziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia Mathematics Department, Faculty of Science, Kafr El-Sheikh University, Kafr El-Sheikh, Egypt Correspondence should be addressed to Hassan A. Zedan; hassanzedan2003@yahoo.com Received 20 November 2012; Accepted 1 February 2013 Academic Editor: Hui-Shen Shen Copyright © 2013 Hassan A. Zedan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We introduce two powerful methods to solve the Davey-Stewartson equations: one is the homotopy perturbation method (HPM) and the other is the homotopy analysis method (HAM). HAM is a strong and easy to use analytic tool for nonlinear problems. Comparison of the HPM results with the HAM results, and compute the absolute errors between the exact solutions of the DS equations with the HPM solutions and HAM solutions are obtained. 1. Introduction Nonlinear partial differential equations are useful in describing the various phenomena in disciplines. Apart from a limited number of these problems, most of them do not have a precise analytical solution, so these nonlinear equations should be solved using approximate methods. The application of the homotopy perturbation method (HPM) [1, 2] in nonlinear problems has been devoted by scientists and engineers, because this method continuously deforms a simple problem which is easy to solve into the under study problem which is difficult to solve. The homotopy perturbation method was first proposed by He [3–6]. The HPM yields a very rapid convergence of the solution series in most cases. The method does not depend on a small parameter in the equation. Using homotopy technique in topology, a homotopy is constructed with an embedding parameter π ∈ [0; 1] which is considered as a “small parameter.” No need to linearization or discretization, large computational work and round-off errors are avoided. It has been used to solve effectively, easily, and accurately a large class of nonlinear problems with approximations. These approximations converge rapidly to accurate solutions [7–10]. The goal of He’s homotopy perturbation method was to find a technique to unify linear and nonlinear, ordinary or partial differential equations for solving initial and boundary value problems. The HPM was successfully applied to nonlinear oscillators with discontinuities [4] and bifurcation of nonlinear problem [11]. In [6], a comparison of HPM and homotopy analysis method was made on a simple problem. In 1992, Liao employed the basic ideas of the homotopy in topology to propose a general analytic method for nonlinear problems, namely, homotopy analysis method (HAM) [12– 15]. This method has been successfully applied to solve many types of nonlinear problems by others [16–20]. In this paper, we consider the Davey-Stewartson (DS) equations for the function π = π(π₯, π¦, π‘) which are given by (see [21]) 1 σ΅¨ σ΅¨2 πππ‘ + π2 (ππ₯π₯ + π2 ππ¦π¦ ) + πσ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ π − ππ₯ π = 0, 2 σ΅¨ σ΅¨2 ππ₯π₯ − π2 ππ¦π¦ − 2π(σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ )π₯ = 0, π = ±1, π2 = ±1. (1) The case π = 1 is called the DSI equation, while π = π is the DSII equation. The parameter π characterizes the 2 Journal of Applied Mathematics focusing or defocusing case. The Davey-Stewartson I and II are two well-known examples of integrable equations in two space dimensions, which arise as higher dimensional generalizations of the nonlinear shroΜdinger (NLS) equation, as well as from physical considerations [22]. Indeed, they appear in many applications, for example, in the description of gravity-capillarity surface wave packets in the limit of the shallow water. Davey and Stewartson first derived their model in the context of water waves, from purely physical considerations. In the context, π(π₯, π¦, π‘) is the amplitude of a surface wave packet, while π(π₯, π¦) is the velocity potential of the mean flow interacting with the surface wave [22]. In [23], solution of DS equations by (HPM) where the amplitude of a surface wave packet π separated into real and imaginary parts, that is, π = π’(π₯, π¦, π‘) + πV(π₯, π¦, π‘). Consequently, the system (1) rewritten in the following form: 1 π’π‘ = − π2 (Vπ₯π₯ + π2 Vπ¦π¦ ) − π (π’2 + V2 ) V + ππ₯ V, 2 1 Vπ‘ = π2 (π’π₯π₯ + π2 π’π¦π¦ ) + π (π’2 + V2 ) π’ − ππ₯ π’, 2 ππ₯π₯ − π2 ππ¦π¦ − 2π(π’2 + V2 )π₯ = 0, (2) π = ±1, π2 = ±1, are arbitrary constants. After that we will apply homotopy perturbation method and homotopy analysis method. When implementing the homotopy perturbation method (HPM) and the homotopy analysis method (HAM), we get the explicit solutions of the DS equations without using any transformation method. Furthermore, we will show that considerably better approximations related to the accuracy level would be obtained. The homotopy perturbation method can be found in [1–11, 23]. The homotopy analysis method can be found in details in [12–22, 24–26] and only the main steps will be summarized here. 2. Application of the Homotopy Perturbation Method To investigate the traveling wave solution of (4), we first construct a homotopy as follows: 2 π2 V1,0 π2 π1 π V1,0 − + π ππ¦2 ππ¦2 ππ¦2 + π [2ππ1 with the initial condition π’ (π₯, 0, π‘) = π sech [π (π₯ − ππ‘)] cos [π1 π₯ + π3 π‘] , V (π₯, 0, π‘) = π sech [π (π₯ − ππ‘)] sin [π1 π₯ + π3 π‘] , − π2 π1 ( (3) π (π₯, 0, π‘) = π tanh [π (π₯ − ππ‘)] , where π = π2 + π2 π1 , π = √(−2π3 + π12 π2 + π22 )/π, π = √(2π3 + π12 π2 + π22 )/π2 , π = 2π√−π/(1 − π2 ), π1 , π2 , and π3 are arbitrary constants. In this paper, we apply homotopy analysis method (HAM) for the above system. We rewrite system (1) in the following form 1 1 ππ‘ = π2 π (ππ₯ ππ₯ + πππ₯π₯ ) + π (ππ¦ ππ¦ + πππ¦π¦ ) , 2 2 ππ‘ = −1 π 1 (π2 ππ₯π₯ + ππ¦π¦ ) + (π2 ππ₯2 + ππ¦2 ) − (ππ2 + ππ₯ ) , 2ππ 2 π ππ₯π₯ − π2 ππ¦π¦ − 4ππππ₯ = 0, (4) where we take π = π(π₯, π¦, π‘) ⋅ π−πππ(π₯,π¦,π‘) with the initial condition π (π₯, 0, π‘) = π sech [π (π₯ − ππ‘)] cos [π1 π₯ + π3 π‘] , π (π₯, 0, π‘) = π sech [π (π₯ − ππ‘)] sin [π1 π₯ + π3 π‘] , ππ2 ππ 2 π2 π1 ) + π2 (−π2 π1 ( 2 ) + ππ‘ ππ₯ ππ₯2 (5) π (π₯, 0, π‘) = π tanh [π (π₯ − ππ‘)] , where π = π2 + π2 π1 , π = √(−2π3 + π12 π2 + π22 )/π, π = √(2π3 + π12 π2 + π22 )/π2 , π = 2π√−π/(1 − π2 ), π1 , π2 , and k3 ππ ππ2 2 ) + 2ππ13 + 2π1 3 ] = 0, ππ¦ ππ₯ 2 π2 V2,0 π2 π2 π V2,0 − + π ππ¦2 ππ¦2 ππ¦2 + π[ −2 ππ1 2π2 ππ2 ππ1 1 π2 π2 ( ) + + π ππ1 ππ‘ π1 ππ₯ ππ₯ 2 1 ππ₯2 + 2 ππ2 ππ1 ] = 0, π1 ππ¦ ππ¦ 2 π2 V3,0 π2 π3 π V3,0 1 π2 π3 4π ππ1 − + π + π [− 2 + 2 π1 ] = 0. 2 2 2 ππ¦ ππ¦ ππ¦ π ππ₯2 π ππ₯ (6) And the initial approximations are as follows: V1,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) = π (π₯, 0, π‘) , V2,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) = π (π₯, 0, π‘) , (7) V3,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) = π (π₯, 0, π‘) , π1 = V1,0 + πV1,1 + π2 V1,2 + π3 V1,3 + ⋅ ⋅ ⋅ , π2 = V2,0 + πV2,1 + π2 V2,2 + π3 V2,3 + ⋅ ⋅ ⋅ , π3 = V3,0 + πV3,1 + π2 V3,2 + π3 V3,3 + ⋅ ⋅ ⋅ , (8) Journal of Applied Mathematics 3 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −10 −0.4 −10 −4 −0.2 0 0 π₯ 0.2 −2 0 π‘ 0 π₯ 0.4 The exact solution of β£π(π₯, π¦, π‘)β£ 2 π‘ HAM solution for β£πβ£ = β£π’ + ππ£β£ (a) (b) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −10 −0.4 −10 −0.4 −0 −0.2 −0.2 0 0 π₯ 0.2 0 π‘ 0 π₯ 0.2 HPM solution for β£πβ£ = β£ππ−0.1ππ β£ (c) π‘ 00.44 0.4 HAM solution for β£πβ£ = β£ππ−0.1ππ β£ (d) Figure 1: Comparison between the exact solution, the HPM solution, and the HAM solution for π(π₯, π¦, π‘). where V(π,π) , π = 1, 2, 3, π = 0, 1, 2, 3, . . . are functions yet to be determined. Substituting (8) into (6) and arranging the coefficients of π powers, we have 0 = (−2V1,0π‘ + πV1,0 V2,1π¦π¦ + 2π2 πV2,0π₯ V1,0π₯ + π2 πV2,0π₯π₯ V1,0 + πV2,0π¦π¦ V1,0 + 2πV2,0π¦ V1,0π¦ ) π + (π2 πV2,0π₯π₯ V1,1 + 2π2 πV2,1π₯ V1,0π₯ + π2 πV2,1π₯π₯ V1,0 +2πV2,0π¦ V1,1π¦ + 2πV2,1π¦ V1,0π¦ ) π2 + (2πV2,2π¦ V1,0π¦ + 2π2 πV2,0π₯ V1,2π₯ + 2π2 πV2,1π₯ V1,1π₯ + πV2,1π¦π¦ V1,2 − 2V1,2π‘ + πV2,0π¦π¦ V1,2 + π2 πV2,0π₯π₯ V1,2 + π2 πV2,1π₯π₯ V1,1 + πV2,3π¦π¦ V1,0 + V2,2π¦π¦ πV1,1 + πV2,1π¦π¦ V1,1 − 2V1,1π‘ + πV2,0π¦π¦ V1,1 + 2πV2,0π¦ V1,2π¦ + 2π2 πV2,2π₯ V1,0π₯ + 2π2 πV2,0π₯ V1,1π₯ + πV2,2π¦π¦ V1,0 + 2πV2,1π¦ V1,1π¦ + π2 πV2,2π₯π₯ V1,0 ) π3 + ⋅ ⋅ ⋅ , 4 Journal of Applied Mathematics 1 1 0.5 0.5 0 0 −0.5 −0.5 −10 −0.4 −10 −4 −2 −0.2 π₯ 0 0 0.4 0.2 π₯ π‘ 0 0 2 π‘ HAM solution for π when π = π’ + ππ£ The exact solution for π(π₯, π¦, π‘) (a) (b) 1 1 0.5 0.5 0 0 −0.5 −0.5 −10 −0.4 −10 −0.4 −0.2 π₯ 0 0 0.4 0.2 π‘ HPM solution for π when π = ππ−0.1ππ (c) −0.2 π₯ 0 0 0.2 π‘ 0.4 HAM solution for π when π = ππ−0.1ππ (d) Figure 2: Comparison between the exact solution, the HPM solution, and the HAM solution for π(π₯, π¦, π‘). 0 = (2πV1,0 V2,0π‘ + 2V1,0 V3,0π₯ + π2 V1,0π₯π₯ 2 + (6πV1,0 V1,2 + V1,3π¦π¦ − 2π2 π2 V1,1 V2,0π₯ V2,1π₯ 3 2 + 2πV1,0 − π2 π2 V1,0 V2,0π₯ + 2πV1,0 V2,2π‘ + 2V1,1 V3,1π₯ 2 − π2 V1,0 V2,0π¦ + V1,0π¦π¦ + V1,1π¦π¦ ) π 2 + 2πV1,2 V2,0π‘ − π2 V1,0 V2,1π¦ + (2V1,1 V3,0π₯ + V1,2π¦π¦ + 2πV1,0 V2,1π‘ + 2V1,2 V3,0π₯ + π2 V1,2π₯π₯ 2 + π2 V1,1π₯π₯ + 6πV1,0 V1,1 2 − π2 π2 V1,0 V2,1π₯ + 2πV1,1 V2,1π‘ − 2π2 V1,0 V2,0π¦ V2,1π¦ − 2π2 π2 V1,0 V2,0π₯ V2,1π₯ 2 − π2 π2 V1,2 V2,0π₯ − 2π2 V1,1 V2,0π¦ V2,1π¦ + 2V1,0 V3,1π₯ + 2πV1,1 V2,0π‘ 2 + 6πV1,0 V1,1 + 2V1,0 V3,2π₯ 2 2 − π2 π2 V1,1 V2,0π₯ − π2 V1,1 V2,0π¦ ) π2 − 2π2 V1,0 V2,0π¦ V2,2π¦ − 2π2 π2 V1,0 V2,0π₯ V2,2π₯ Journal of Applied Mathematics 5 0.7 0.7 0.6 0.6 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 1 2 π₯ For β£πβ£ = β£π’ + ππ£β£ −6 −5 −4 −3 −2 −1 3 4 5 0 −6 −5 −4 −3 −2 −1 6 HAM Exact 1 2 3 4 5 6 π₯ For β£πβ£ = β£ππ−0.1ππ β£ HPM Exact (a) (b) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 −2π − 3π 2 −π − π 2 0 π 2 π 3π 2 2π π₯ For β£πβ£ = β£ππ−0.1ππ β£ (c) Figure 3: The results obtained by HPM and HAM for π(π₯, π¦, π‘), at π‘ = 0.2 in comparison with the exact solutions. 2 −π2 V1,2 V2,0π¦ ) π3 + ⋅ ⋅ ⋅ , nine equations with nine unknowns, considering the initial conditions of Vπ,π (π₯, 0, π‘) = 0, π, π = 1, 2, 3: 0 = (π2 V3,1π¦π¦ + π2 V3,0π¦π¦ − V3,0π₯π₯ + 4πV1,0 V1,0π₯ ) π 2 − 2V1,0π‘ + πV1,0 V2,1π¦π¦ + 2π2 πV2,0π₯ V1,0π₯ 2 + (π V3,2π¦π¦ − V3,1π₯π₯ + 4πV1,0 V1,1π₯ + 4πV1,1 V1,0π₯ ) π + π2 πV2,0π₯π₯ V1,0 + πV2,0π¦π¦ V1,0 + 2πV2,0π¦ V1,0π¦ = 0, + (π2 V3,3π¦π¦ − V3,2π₯π₯ + 4πV1,0 V1,2π₯ + 4πV1,1 V1,1π₯ (10) π2 πV2,0π₯π₯ V1,1 + 2π2 πV2,1π₯ V1,0π₯ + π2 πV2,1π₯π₯ V1,0 + 4πV1,2 V1,0π₯ ) π3 + ⋅ ⋅ ⋅ . (9) To obtain the unknowns Vπ,π (π₯, π¦, π‘), π, π = 1, 2, 3, we must construct and solve the following system which includes + πV2,1π¦π¦ V1,1 − 2V1,1π‘ + πV2,0π¦π¦ V1,1 + 2π2 πV2,0π₯ V1,1π₯ + πV2,2π¦π¦ V1,0 + 2πV2,0π¦ V1,1π¦ + 2πV2,1π¦ V1,0π¦ = 0, (11) 6 Journal of Applied Mathematics 1 1 0.5 0.5 0 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 0 −6 −5 −4 −3 −2 −1 −0.5 1 2 3 4 5 6 −0.5 −1 −1 When π = ππ−0.1ππ When π = π’ + ππ£ HAM Exact HPM Exact (a) (b) 1 0.5 0 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 −0.5 −1 When π = ππ−0.1ππ HAM Exact (c) Figure 4: The results obtained by HPM and HAM for π(π₯, π¦, π‘), at π‘ = 0.2 in comparison with the exact solutions. 2πV2,2π¦ V1,0π¦ + 2π2 πV2,0π₯ V1,2π₯ + 2π2 πV2,1π₯ V1,1π₯ 2V1,1 V3,0π₯ + V1,2π¦π¦ + 2πV1,0 V2,1π‘ 2 + π2 V1,1π₯π₯ + 6πV1,0 V1,1 − 2π2 V1,0 V2,0π¦ V2,1π¦ + πV2,1π¦π¦ V1,2 − 2V1,2π‘ + πV2,0π¦π¦ V1,2 + π2 πV2,0π₯π₯ V1,2 + π2 πV2,1π₯π₯ V1,1 + πV2,3π¦π¦ V1,0 + V2,2π¦π¦ πV1,1 (12) 2 2 − π2 π2 V1,1 V2,0π₯ − π2 V1,1 V2,0π¦ = 0, + 2πV2,0π¦ V1,2π¦ + 2π2 πV2,2π₯ V1,0π₯ + 2πV2,1π¦ V1,1π¦ 2 V1,2 + V1,3π¦π¦ − 2π2 π2 V1,1 V2,0π₯ V2,1π₯ 6πV1,0 + π2 πV2,2π₯π₯ V1,0 = 0, 3 2πV1,0 V2,0π‘ + 2V1,0 V3,0π₯ + π2 V1,0π₯π₯ + 2πV1,0 2 2 − π2 π2 V1,0 V2,0π₯ − π2 V1,0 V2,0π¦ + V1,0π¦π¦ + V1,1π¦π¦ = 0, − 2π2 π2 V1,0 V2,0π₯ V2,1π₯ + 2V1,0 V3,1π₯ + 2πV1,1 V2,0π‘ + 2πV1,0 V2,2π‘ + 2V1,1 V3,1π₯ + 2πV1,2 V2,0π‘ (13) 2 − π2 V1,0 V2,1π¦ + 2V1,2 V3,0π₯ + π2 V1,2π₯π₯ (14) Journal of Applied Mathematics 7 2 2 + 2πV1,1 V2,1π‘ − π2 π2 V1,2 V2,0π₯ − π2 π2 V1,0 V2,1π₯ + π 2 π2 cos [π3 π‘ + π1 π₯] 2 − 2π2 V1,1 V2,0π¦ V2,1π¦ + 6πV1,0 V1,1 + 2V1,0 V3,2π₯ × (−1 + π2 π2 sech2 [π (−ππ‘ + π₯)] (sin2 [π3 π‘ + π1 π₯])) − 2π2 V1,0 V2,0π¦ V2,2π¦ − 2π2 π2 V1,0 V2,0π₯ V2,2π₯ ×tanh2 [π (−ππ‘ + π₯)]) . (20) 2 − π2 V1,2 V2,0π¦ = 0, (15) From (10), we obtain π2 V3,1π¦π¦ + π2 V3,0π¦π¦ − V3,0π₯π₯ + 4πV1,0 V1,0π₯ = 0, (16) V2,1 π2 V3,2π¦π¦ − V3,1π₯π₯ + 4πV1,0 V1,1π₯ + 4πV1,1 V1,0π₯ = 0, (17) = π2 V3,3π¦π¦ − V3,2π₯π₯ + 4πV1,0 V1,2π₯ + 4πV1,1 V1,1π₯ + 4πV1,2 V1,0π₯ = 0, 1 2 π¦ (πππ 2 π2 sech3 [π (−ππ‘ + π₯)] sin [π3 π‘ + π1 π₯] 2π − 2π3 tan [π3 π‘ + π1 π₯] + 2ππ tanh [π (−ππ‘ + π₯)] (18) + πππ2 sech [π (−ππ‘ + π₯)] × (2π12 sin [π3 π‘ + π1 π₯] + π tanh [π (−ππ‘ + π₯)] From (8), if the three approximations are sufficient, we will obtain × (π1 (1 + 3 cos [2 (π3 π‘ + π1 π₯)]) 3 × sec [π3 π‘ + π1 π₯] − 3π sin [π3 π‘ + π1 π₯] π (π₯, π¦, π‘) = lim π1 (π₯, π¦, π‘) = ∑ V1,π (π₯, π¦, π‘) , π→1 π=0 × tanh [π (−ππ‘ + π₯)]))) . 3 π (π₯, π¦, π‘) = lim π2 (π₯, π¦, π‘) = ∑ V2,π (π₯, π¦, π‘) , π→1 (19) π=0 From (16), we obtain 3 V3,1 π=0 = π (π₯, π¦, π‘) = lim π3 (π₯, π¦, π‘) = ∑ V3,π (π₯, π¦, π‘) . π→1 (21) To calculate the terms of the homotopy series (19) for π(π₯, π¦, π‘), π(π₯, π¦, π‘), and π(π₯, π¦, π‘), we substitute the initial conditions (5) into the system (9), and using Mathematica software, from (13), we obtain V1,1 1 = ππ¦2 sech [π (−ππ‘ + π₯)] 2 × (cos [π3 π‘ + π1 π₯] × (π12 π2 − 2π3 ππ cos [π3 π‘ + π1 π₯] × sech [π (−ππ‘ + π₯)] + (π (−2π + π π2 ) + π2 (−2π + π12 π2 π2 ) × cos2 [π3 π‘ + π1 π₯]) × (sech2 [π (−ππ‘ + π₯)])) − π (2π1 π2 sin [π3 π‘ + π1 π₯] + ππ sech [π (−ππ‘ + π₯)] × (π + π1 πππ2 cos [π3 π‘ + π1 π₯] sech [π (−ππ‘ + π₯)]) × sin [2 (π3 π‘ + π1 π₯)]) tanh [π (−ππ‘ + π₯)] 1 (π¦2 sech2 [π (−ππ‘ + π₯)] (π1 π2 π sin [2 (π3 π‘ + π1 π₯)] π2 + π (−ππ + 2π2 π cos2 [π3 π‘ + π1 π₯]) tanh [π (−ππ‘ + π₯)]) . (22) In this manner, the other components V1,2 (π₯, π¦, π‘), V2,2 (π₯, π¦, π‘), V3,2 (π₯, π¦, π‘), V1,3 (π₯, π¦, π‘), V2,3 (π₯, π¦, π‘), and V3,3 (π₯, π¦, π‘) can be obtained from (14), (11), (17), (15), (12), and (18), respectively, and substituting these components into (19) to obtain π(π₯, π¦, π‘), π(π₯, π¦, π‘), and π(π₯, π¦, π‘). 3. Application of the Homotopy Analysis Method In order to apply the homotopy analysis method for (2), we choose the linear operator πΏ[ππ (π₯, π¦, π‘; π)] = π2 ππ /ππ¦2 with the property πΏ[π1,π + π2,π π¦] = 0, π = 1, 2, 3, where π1,π , π2,π are integral constants to be determined by initial conditions. Furthermore, (2) suggests to define the nonlinear operators π1 = π4 ππ π2 π2 π2 π ππ + π2 22 + 2 1 − 2π2 3 + 2π (π12 + π22) π2 , 2 ππ¦ ππ₯ ππ‘ ππ₯ π2 = π4 2 ππ π2 π1 ππ 2 π π1 + π − 2 2 − 2π1 3 + 2π (π12 + π22) π1 , 2 2 ππ¦ ππ₯ ππ‘ ππ₯ π3 = 2 2 π2 π3 1 π2 π3 2π π (π1 + π2 ) − + ππ¦2 π2 ππ₯2 π2 ππ₯ (23) 8 Journal of Applied Mathematics We have the πth-order deformation equations we construct the zero-order deformation equations (1 − π) πΏ [ππ (π₯, π¦, π‘; π) − π§π,0 (π₯, π¦, π‘)] = πβπ ππ [ππ (π₯, π¦, π‘; π)] , π = 1, 2, 3. (24) πΏ [π§π,π (π₯, π¦, π‘) − ππ π§π,π−1 (π₯, π¦, π‘)] = βπ π π,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) , π = 1, 2, 3, (31) When π = 0 where π1 (π₯, π¦, π‘; 0) = π§1,0 (π₯, π¦, π‘) = π’0 (π₯, π¦, π‘) π 1,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) = π sech [π (π₯ − ππ‘)] cos [π1 π₯ + π3 π‘] , π2 (π₯, π¦, π‘; 0) = π§2,0 (π₯, π¦, π‘) = V0 (π₯, π¦, π‘) = π sech [π (π₯ − ππ‘)] sin [π1 π₯ + π3 π‘] , (25) = π4 2 π2 π§2,π−1 ππ§1,π−1 2 π π§2,π−1 + π +2 2 2 ππ¦ ππ₯ ππ‘ π−1 − 2 ∑ π§2,π π3 (π₯, π¦, π‘; 0) = π§3,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) π=0 = π tanh [π (π₯ − ππ‘)] . ππ§3,π−1−π ππ₯ π−1 π + 2π ∑ ∑ (π§1,π π§1,π−π + π§2,π π§2,π−π ) π§2,π−1−π , When π = 1 π=0 π=0 π1 (π₯, π¦, π‘; 1) = π’ (π₯, π¦, π‘) , π2 (π₯, π¦, π‘; 1) = V (π₯, π¦, π‘) , π 2,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) (26) π3 (π₯, π¦, π‘; 1) = π (π₯, π¦, π‘) . = π4 π−1 Therefore, as the embedding parameter π increases from 0 to 1, ππ (π₯, π¦, π‘; π) varies from initial guesses π§π,0 (π₯, π¦, π‘) to the solutions π’(π₯, π¦, π‘), V(π₯, π¦, π‘) and π(π₯, π¦, π‘), for π = 1, 2, 3, respectively. Expanding ππ (π₯, π¦, π‘; π) in Taylor series with respect to π for π = 1, 2, 3, one has π=0 π−1 π π=0 π=0 = where σ΅¨ π 1 π ππ (π₯, π¦, π‘; π) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ . σ΅¨σ΅¨ π! πππ σ΅¨π=0 (28) If the auxiliary linear operator, the initial guesses, and the auxiliary parameters βπ are so properly chosen, the above series converge at π = 1, and +∞ π’ (π₯, π¦, π‘) = π§1,0 (π₯, π¦, π‘) + ∑ π§1,π (π₯, π¦, π‘) , π=1 +∞ V (π₯, π¦, π‘) = π§2,0 (π₯, π¦, π‘) + ∑ π§2,π (π₯, π¦, π‘) , ππ§3,π−1−π ππ₯ + 2π ∑ ∑ (π§1,π π§1,π−π + π§2,π π§2,π−π ) π§1,π−1−π , (27) π=1 π§π,π (π₯, π¦, π‘) = − 2 ∑ π§1,π (32) π 3,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) +∞ ππ (π₯, π¦, π‘; π) = π§π,0 (π₯, π¦, π‘) + ∑ π§π,π (π₯, π¦, π‘) ππ , 2 π2 π§1,π−1 ππ§2,π−1 2 π π§1,π−1 + π −2 ππ¦2 ππ₯2 ππ‘ (29) π=1 2 π2 π§3,π−1 1 π π§3,π−1 − ππ¦2 π2 ππ₯2 + 2π π π−1 + π§2,π π§2,π−1−π ) , ∑ (π§ π§ π2 ππ₯ π=0 1,π 1,π−1−π where π§1 , π§2 , and π§3 are functions of π₯, π¦, and π‘, and 0, ππ = { 1, π≤1 π > 1. (33) Now, the solutions of the πth-order deformation (31) for π ≥ 1 become π§π,π (π₯, π¦, π‘) = ππ π§π,π−1 (π₯, π¦, π‘) +∞ π (π₯, π¦, π‘) = π§3,0 (π₯, π¦, π‘) + ∑ π§3,π (π₯, π¦, π‘) , π¦ + βπ β¬ [π π,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 )] ππ¦ ππ¦ π=1 which must be one of solutions of the original nonlinear equations as proved by Liao [13]. Define the vectors π§βπ,π = {π§π,0 (π₯, π¦, π‘) , π§π,1 (π₯, π¦, π‘) , . . . , π§π,π (π₯, π¦, π‘)} ; π = 1, 2, 3. 0 + π1,π + π2,π π¦. (34) (30) For simplicity, we suppose β1 = β2 = β3 = β. Journal of Applied Mathematics 9 We consider the solutions of (2) with the initial conditions (25). We now obtain at π = 1 We construct the zero-order deformation equations (1 − π) πΏ [ππ (π₯, π¦, π‘; π) − π§π,0 (π₯, π¦, π‘)] 1 π§1,1 (π₯, π¦, π‘) = βππ¦2 sech [π (π₯ − ππ‘)] 2 = πβπ ππ [ππ (π₯, π¦, π‘; π)] , × {− [2π3 + π12 π2 + (2ππ − 2π2 π + π 2 π2 ) π1 (π₯, π¦, π‘; 0) = π§1,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) = π sech [π (π₯ − ππ‘)] cos [π1 π₯ + π3 π‘] , − 2π (π − π1 π2 ) sin [π1 π₯ + π3 π‘] π2 (π₯, π¦, π‘; 0) = π§2,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) × tanh [π (π₯ − ππ‘)] + π 2 π2 = π sech [π (π₯ − ππ‘)] sin [π1 π₯ + π3 π‘] , 2 × cos [π1 π₯ + π3 π‘] tanh [π (π₯ − ππ‘)]} , = π tanh [π (π₯ − ππ‘)] . × {− [2π3 + π12 π2 + (2ππ − 2π2 π + π 2 π2 ) When π = 1, × sech2 [π (π₯ − ππ‘)]] sin [π1 π₯ + π3 π‘] π1 (π₯, π¦, π‘; 1) = π (π₯, π¦, π‘) , + 2π (π − π1 π2 ) cos [π1 π₯ + π3 π‘] π2 (π₯, π¦, π‘; 1) = π (π₯, π¦, π‘) , × tanh [π (π₯ − ππ‘)] π3 (π₯, π¦, π‘; 1) = π (π₯, π¦, π‘) . × tanh2 [π (π₯ − ππ‘)]} , 1 (βπ π¦2 (ππ − 2π2 π) sech2 π2 +∞ × [π (π₯ − ππ‘)] tanh [π (π₯ − ππ‘)]) . ππ (π₯, π¦, π‘; π) = π§π,0 (π₯, π¦, π‘) + ∑ π§π,π (π₯, π¦, π‘) ππ , Obviously, for β = −1, the obtained solutions are the same homotopy perturbation method in [2]; we continue to evaluate two terms of HAM. Now for (4), we choose the linear operator πΏ[ππ (π₯, π¦, π‘; π)] = π2 ππ /ππ¦2 with the property [π1,π + π2,π π¦] = 0, π = 1, 2, 3, where π1,π , π2,π are integral constant to be determined by initial conditions. Furthermore, (4) suggests to define the nonlinear operators 2 ππ ππ π π2 ππ ππ + 2 2 1 + 2π2 2 1 2 ππ¦ ππ¦ ππ¦ ππ₯ ππ₯ + π2 π3 = π§π,π (π₯, π¦, π‘) = σ΅¨ π 1 π ππ (π₯, π¦, π‘; π) σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ . σ΅¨σ΅¨ π! πππ σ΅¨π=0 (41) If the auxiliary linear operator, the initial guesses, and the auxiliary parameters βπ are so properly chosen, the series (40) converge at π = 1, has +∞ π (π₯, π¦, π‘) = π§1,0 (π₯, π¦, π‘) + ∑ π§1,π (π₯, π¦, π‘) , +∞ π (π₯, π¦, π‘) = π§2,0 (π₯, π¦, π‘) + ∑ π§2,π (π₯, π¦, π‘) , π π2 2 ππ1 − , ππ₯2 π ππ‘ π2 π3 1 π2 π3 4π ππ1 − + 2 π1 . ππ¦2 π2 ππ₯2 π ππ₯ where π=1 2 π2 π1 ππ ππ + 2ππ1 2 + 2ππ13 + 2π1 3 , ππ₯2 ππ‘ ππ₯ (40) π=1 (35) 2 π2 π1 ππ2 2 2 2 ππ2 − π π (( + π ( ) π2 = ) ) 1 ππ¦2 ππ¦ ππ₯ (39) Therefore, as the embedding parameter π increases from 0 to 1, ππ (π₯, π¦, π‘; π) varies from initial guesses π§π,0 (π₯, π¦, π‘) to the solutions π(π₯, π¦, π‘), π(π₯, π¦, π‘), and π(π₯, π¦, π‘), for π = 1, 2, 3, respectively. Expanding ππ (π₯, π¦, π‘; π) in Taylor series with respect to π for π = 1, 2, 3, one has + π 2 π2 sin [π1 π₯ + π3 π‘] + π2 π1 (38) π3 (π₯, π¦, π‘; 0) = π§3,0 (π₯, π¦, π‘) = π0 (π₯, π¦, π‘) 1 π§2,1 (π₯, π¦, π‘) = βππ¦2 sech [π (π₯ − ππ‘)] 2 π1 = π1 (37) When π = 0, × sech2 [π (π₯ − ππ‘)]] cos [π1 π₯ + π3 π‘] π§3,1 (π₯, π¦, π‘) = π = 1, 2, 3. (42) π=1 +∞ (36) π (π₯, π¦, π‘) = π§3,0 (π₯, π¦, π‘) + ∑ π§3,π (π₯, π¦, π‘) , π=1 which must be one of solutions of the original nonlinear equation as proved by Liao [13]. Define the vectors π§βπ,π = {π§π,0 (π₯, π¦, π‘) , π§π,1 (π₯, π¦, π‘) , . . . , π§π,π (π₯, π¦, π‘)} ; π = 1, 2, 3. (43) 10 Journal of Applied Mathematics We have the πth-order deformation equations πΏ [π§π,π (π₯, π¦, π‘) − ππ π§π,π−1 (π₯, π¦, π‘)] = βπ π π,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) , π = 1, 2, 3, Table 1: The HPM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (5). (44) where π 1,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) π−1 = ∑ π§1,π π=0 π−1 π2 π§2,π−1−π ππ§2,π ππ§1,π−1−π + 2 ∑ 2 ππ¦ ππ¦ π=0 ππ¦ π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 |ππ − π| 9.861869456 × 10−18 1.073360934 × 10−17 1.161066111 × 10−17 1.248899278 × 10−17 1.336466228 × 10−17 1.423383001 × 10−17 1.509277063 × 10−17 π−1 ππ§2,π ππ§1,π−1−π 2 ππ§1,π−1 − ππ₯ π ππ‘ π=0 ππ₯ + 2π2 ∑ π−1 + π2 ∑ π§1,π π=0 Table 2: The HPM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (5). π2 π§2,π−1−π , ππ₯2 π 2,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) = π−1 π π2 π§1,π−1 ππ§2,π−π ππ§2,π−1−π 2 − π ∑ ∑ π§1,π 2 ππ¦ ππ¦ ππ¦ π=0 π=0 π−1 π − π2 π2 ∑ ∑ π§1,π π=0 π=0 + π2 (45) ππ§2,π−π ππ§2,π−1−π ππ₯ ππ₯ π−1 π π2 π§1,π−1 + 2π ∑ ∑ π§1,π π§1,π−π π§1,π−1−π ππ₯2 π=0 π=0 |ππ − π| 3.3574 × 10−15 1.9466 × 10−15 1.52954 × 10−15 1.81125 × 10−15 2.77765 × 10−15 5.04003 × 10−15 1.07655 × 10−15 Table 3: The HAM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (3). π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 π−1 π−1 ππ§3,π−1−π ππ§2,π−1−π + 2 ∑ π§1,π + 2π ∑ π§1,π , ππ₯ ππ‘ π=0 π=0 π 3,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 ) = π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 2 π2 π§3,π−1 ππ§1,π−1−π 1 π π§3,π−1 4π π−1 − + 2 ∑ π§1,π , 2 2 2 ππ¦ π ππ₯ π π=0 ππ₯ |ππ − π| 1.415900000 × 10−19 1.400513800 × 10−19 1.385294700 × 10−19 1.370240800 × 10−19 1.355350700 × 10−19 1.340622400 × 10−19 1.326054200 × 10−19 where π§1 , π§2 , and π§3 are functions of π₯, π¦, and π‘, and 0, ππ = { 1, π≤1 π > 1. (46) Now, the solutions of the πth-order deformation (44) for π ≥ 1 become π§1,1 (π₯, π¦, π‘) 1 = βππ¦2 sech [π (−ππ‘ + π₯)] 2 π§π,π (π₯, π¦, π‘) = ππ π§π,π−1 (π₯, π¦, π‘) π¦ + βπ β¬ [π π,π (π§β1,π−1 , π§β2,π−1 , π§β3,π−1 )] ππ¦ ππ¦ 0 × (cos [π3 π‘ + π1 π₯] × (−π12 π2 + 2π3 ππ cos [π3 π‘ + π1 π₯] sech [π (−ππ‘ + π₯)] + π1,π + π2,π π¦. (47) For simplicity, we suppose β1 = β2 = β3 = β. We consider the solutions of (4) with the initial conditions (38) and obtain for π = 1 + (π (2π − π π2 ) + π2 (2π − π12 π2 π2 ) × cos2 [π3 π‘ + π1 π₯]) sech2 [π (−ππ‘ + π₯)]) Journal of Applied Mathematics 11 + π (2π1 π2 sin [π3 π‘ + π1 π₯] + ππ sech [π (−ππ‘ + π₯)] Table 4: The HAM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (3). × (π + π1 πππ2 cos [π3 π‘ + π1 π₯] sech [π (−ππ‘ + π₯)]) π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 × sin [2 (π3 π‘ + π1 π₯)]) tanh [π (−ππ‘ + π₯)] + π 2 π2 cos [π3 π‘ + π1 π₯] × (1 − π2 π2 sech2 [π (−ππ‘ + π₯)] sin2 [π3 π‘ + π1 π₯]) ×tanh2 [π (−ππ‘ + π₯)]) , |ππ − π| 0 0 4 ×10−10 0 0 2 ×10−10 0 (48) π§2,1 (π₯, π¦, π‘) Table 5: The HAM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (5). 1 = βπ¦2 (2π3 tan [π3 π‘ + π1 π₯] + πππ2 sech [π (−ππ‘ + π₯)] 2π π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 × ((−3π12 + 3π 2 − 4π 2 sech2 [π (−ππ‘ + π₯)]) × sin [π3 π‘ + π1 π₯] + π1 π (1 + 3 cos [2 (π3 π‘ + π1 π₯)]) × sec [π3 π‘ + π1 π₯] tanh [π (ππ‘ − π₯)]) −2ππ tanh [π (−ππ‘ + π₯)]) , |ππ − π| 9.861809932 × 10−18 1.073356691 × 10−17 1.161063556 × 10−17 1.248898384 × 10−17 1.336466963 × 10−17 1.423385326 × 10−17 1.509280934 × 10−17 (49) Table 6: The HAM results for π (π₯, π¦, π‘) in comparison with the analytical solution with initial conditions (5). π§3,1 (π₯, π¦, π‘) = −1 (βπ¦2 sech2 [π (−ππ‘ + π₯)] π2 × (π1 π2 π sin [2 (π3 π‘ + π1 π₯)] +π (−ππ + 2π2 π cos2 [π3 π‘ + π1 π₯]) tanh [π (−ππ‘ + π₯)]) . (50) Obviously, for β = −1, we obtained the same solutions as the one by the homotopy perturbation method in (20)–(22); we continue to evaluate six terms of (47) when π = 2, 3. Using a Taylor series, then the closed form solutions yield as follows [23]: π (π₯, π¦, π‘) = π sech [π (π₯ + π¦ − ππ‘)] exp [π (π1 π₯ + π2 π¦ + π3 π‘)] , (51) π (π₯, π¦, π‘) = π tanh [π (π₯ + π¦ − ππ‘)] , (52) where π = √− π = π2 + π2 π1 , π =√ 2π3 + π2 π12 π + π22 2π3 + π2 π12 + π22 , π , π1 , π2 , and π3 are arbitrary constants. 2π√−π , π= 1 − π2 (53) π‘ −0.6 −0.4 −0.2 0 0.2 0.4 0.6 |ππ − π| 0 0 0 0 0 0 5 × 10−10 4. Comparing the HPM Results and the HAM Results with the Exact Solutions To demonstrate the convergence of the HPM, the results of the numerical example are presented and only few terms are required to obtain accurate solutions. 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