# The Application of the Homotopy Analysis Method and the Homotopy Perturbation Method to the Davey-Stewartson Equations and Comparison between Them and Exact Solutions

Authors: Hassan A. Zedan; W. Barakati; Nada Hamad
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

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.


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## Body

## 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 parameterp∈[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 functionq=q(x,y,t) which are given by (see [21])
(1)iqt+12σ2(qxx+σ2qyy)+λ|q|2q-ϕxq=0,ϕxx-σ2ϕyy-2λ(|q|2)x=0,λ=±1,σ2=±1.The caseσ=1 is called the DSI equation, while σ=i is the DSII equation. The parameter λ characterizes the 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 shrödinger (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,q(x,y,t) is the amplitude of a surface wave packet, while ϕ(x,y) 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 q separated into real and imaginary parts, that is, q=u(x,y,t)+iv(x,y,t). Consequently, the system (1) rewritten in the following form:
(2)ut=-12σ2(vxx+σ2vyy)-λ(u2+v2)v+ϕxv,vt=12σ2(uxx+σ2uyy)+λ(u2+v2)u-ϕxu,ϕxx-σ2ϕyy-2λ(u2+v2)x=0,λ=±1,σ2=±1,
with the initial condition
(3)u(x,0,t)=rsech[s(x-ct)]cos[k1x+k3t],v(x,0,t)=rsech[s(x-ct)]sin[k1x+k3t],ϕ(x,0,t)=ftanh[s(x-ct)],
where c=k2+σ2k1, r=(-2k3+k12σ2+k22)/λ, s=(2k3+k12σ2+k22)/σ2, f=2σ-λ/(1-σ2),k1, k2, and k3 are arbitrary constants.In this paper, we apply homotopy analysis method (HAM) for the above system. We rewrite system (1) in the following form
(4)ρt=σ2m(θxρx+12ρθxx)+m(θyρy+12ρθyy),θt=-12mρ(σ2ρxx+ρyy)+m2(σ2θx2+θy2)-1m(λρ2+ϕx),ϕxx-σ2ϕyy-4λρρx=0,
where we take q=ρ(x,y,t)·e-miθ(x,y,t) with the initial condition
(5)ρ(x,0,t)=rsech[s(x-ct)]cos[k1x+k3t],θ(x,0,t)=rsech[s(x-ct)]sin[k1x+k3t],ϕ(x,0,t)=ftanh[s(x-ct)],
where c=k2+σ2k1, r=(-2k3+k12σ2+k22)/λ, s=(2k3+k12σ2+k22)/σ2, f=2σ-λ/(1-σ2),k1, k2, and k3 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:
(6)∂2V1∂y2-∂2v1,0∂y2+p∂2v1,0∂y2+p[2mV1∂V2∂t+σ2(-m2V1(∂V2∂x)2+∂2V1∂x2)+p-m2V1(∂V2∂y)2+2λV13+2V1∂V3∂x(-m2V1(∂V2∂x)2+∂2V1∂x2)]=0,∂2V2∂y2-∂2v2,0∂y2+p∂2v2,0∂y2+p[-2mV1∂V1∂t+2σ2V1(∂V2∂x∂V1∂x+12V1∂2V2∂x2)+p+2V1∂V2∂y∂V1∂y(∂V2∂x∂V1∂x+12V1∂2V2∂x2)]=0,∂2V3∂y2-∂2v3,0∂y2+p∂2v3,0∂y2+p[-1σ2∂2V3∂x2+4λσ2V1∂V1∂x]=0.And the initial approximations are as follows:(7)v1,0(x,y,t)=ρ0(x,y,t)=ρ(x,0,t),v2,0(x,y,t)=θ0(x,y,t)=θ(x,0,t),v3,0(x,y,t)=ϕ0(x,y,t)=ϕ(x,0,t),(8)V1=v1,0+pv1,1+p2v1,2+p3v1,3+⋯,V2=v2,0+pv2,1+p2v2,2+p3v2,3+⋯,V3=v3,0+pv3,1+p2v3,2+p3v3,3+⋯,
where v(i,j), i=1,2,3,j=0,1,2,3,… are functions yet to be determined. Substituting (8) into (6) and arranging the coefficients of p powers, we have
(9)0=(-2v1,0t+mv1,0v2,1yy+2σ2mv2,0xv1,0xz+σ2mv2,0xxv1,0+mv2,0yyv1,0+2mv2,0yv1,0y)p+(σ2mv2,0xxv1,1+2σ2mv2,1xv1,0x+σ2mv2,1xxv1,0=s+mv2,1yyv1,1-2v1,1t+mv2,0yyv1,1=s+2σ2mv2,0xv1,1x+mv2,2yyv1,0=s+2mv2,0yv1,1y+2mv2,1yv1,0y)p2+(2mv2,2yv1,0y+2σ2mv2,0xv1,2x+2σ2mv2,1xv1,1x=s+mv2,1yyv1,2-2v1,2t+mv2,0yyv1,2=s+σ2mv2,0xxv1,2+σ2mv2,1xxv1,1=s+mv2,3yyv1,0+v2,2yymv1,1=s+2mv2,0yv1,2y+2σ2mv2,2xv1,0x=s+2mv2,1yv1,1y+σ2mv2,2xxv1,0p3)p3+⋯,0=(2mv1,0v2,0t+2v1,0v3,0x+σ2v1,0xx+2λv1,03-m2σ2v1,0v2,0x2-m2v1,0v2,0y2+v1,0yy+v1,1yy)p+(2v1,1v3,0x+v1,2yy+2mv1,0v2,1t+s+σ2v1,1xx+6λv1,02v1,1+s-2m2v1,0v2,0yv2,1y-2m2σ2v1,0v2,0xv2,1x+s+2v1,0v3,1x+2mv1,1v2,0t+s-m2σ2v1,1v2,0x2-m2v1,1v2,0y2)p2+(6λv1,02v1,2+v1,3yy-2m2σ2v1,1v2,0xv2,1x++2mv1,0v2,2t+2v1,1v3,1x++2mv1,2v2,0t-m2v1,0v2,1y2++2v1,2v3,0x+σ2v1,2xx+-m2σ2v1,0v2,1x2+2mv1,1v2,1t+-m2σ2v1,2v2,0x2-2m2v1,1v2,0yv2,1y++6λv1,0v1,12+2v1,0v3,2x+-2m2v1,0v2,0yv2,2y-2m2σ2v1,0v2,0xv2,2x+-m2v1,2v2,0y2)p3+⋯,0=(σ2v3,1yy+σ2v3,0yy-v3,0xx+4λv1,0v1,0x)p+(σ2v3,2yy-v3,1xx+4λv1,0v1,1x+4λv1,1v1,0x)p2+(σ2v3,3yy-v3,2xx+4λv1,0v1,2x+4λv1,1v1,1x++4λv1,2v1,0x)p3+⋯.
To obtain the unknowns vi,j(x,y,t), i,  j=1,2,3, we must construct and solve the following system which includes nine equations with nine unknowns, considering the initial conditions of vi,j(x,0,t)=0, i,  j=1,2,3:
(10)-2v1,0t+mv1,0v2,1yy+2σ2mv2,0xv1,0x+σ2mv2,0xxv1,0+mv2,0yyv1,0+2mv2,0yv1,0y=0,(11)σ2mv2,0xxv1,1+2σ2mv2,1xv1,0x+σ2mv2,1xxv1,0+mv2,1yyv1,1-2v1,1t+mv2,0yyv1,1+2σ2mv2,0xv1,1x+mv2,2yyv1,0+2mv2,0yv1,1y+2mv2,1yv1,0y=0,(12)2mv2,2yv1,0y+2σ2mv2,0xv1,2x+2σ2mv2,1xv1,1x+mv2,1yyv1,2-2v1,2t+mv2,0yyv1,2+σ2mv2,0xxv1,2+σ2mv2,1xxv1,1+mv2,3yyv1,0+v2,2yymv1,1+2mv2,0yv1,2y+2σ2mv2,2xv1,0x+2mv2,1yv1,1y+σ2mv2,2xxv1,0=0,(13)2mv1,0v2,0t+2v1,0v3,0x+σ2v1,0xx+2λv1,03-m2σ2v1,0v2,0x2-m2v1,0v2,0y2+v1,0yy+v1,1yy=0,(14)2v1,1v3,0x+v1,2yy+2mv1,0v2,1t+σ2v1,1xx+6λv1,02v1,1-2m2v1,0v2,0yv2,1y-2m2σ2v1,0v2,0xv2,1x+2v1,0v3,1x+2mv1,1v2,0t-m2σ2v1,1v2,0x2-m2v1,1v2,0y2=0,(15)6λv1,02v1,2+v1,3yy-2m2σ2v1,1v2,0xv2,1x+2mv1,0v2,2t+2v1,1v3,1x+2mv1,2v2,0t-m2v1,0v2,1y2+2v1,2v3,0x+σ2v1,2xx-m2σ2v1,0v2,1x2+2mv1,1v2,1t-m2σ2v1,2v2,0x2-2m2v1,1v2,0yv2,1y+6λv1,0v1,12+2v1,0v3,2x-2m2v1,0v2,0yv2,2y-2m2σ2v1,0v2,0xv2,2x-m2v1,2v2,0y2=0,(16)σ2v3,1yy+σ2v3,0yy-v3,0xx+4λv1,0v1,0x=0,(17)σ2v3,2yy-v3,1xx+4λv1,0v1,1x+4λv1,1v1,0x=0,(18)σ2v3,3yy-v3,2xx+4λv1,0v1,2x+4λv1,1v1,1x+4λv1,2v1,0x=0,From (8), if the three approximations are sufficient, we will obtain
(19)ρ(x,y,t)=limp→1V1(x,y,t)=∑k=03v1,k(x,y,t),θ(x,y,t)=limp→1V2(x,y,t)=∑k=03v2,k(x,y,t),ϕ(x,y,t)=limp→1V3(x,y,t)=∑k=03v3,k(x,y,t).To calculate the terms of the homotopy series (19) for ρ(x,y,t), θ(x,y,t), and ϕ(x,y,t), we substitute the initial conditions (5) into the system (9), and using Mathematica software, from (13), we obtain
(20)v1,1=12ry2sech[s(-ct+x)]=×([rσ22]cos[k3t+k1x]==×(k12σ2-2k3mrcos[k3t+k1x]==×sech[s(-ct+x)]==+(s(-2f+sσ2)=+=+r2(-2λ+k12m2σ2)=+=×cos2[k3t+k1x])==×(sech2[s(-ct+x)]))=×ti-s(f3((s))2k1σ2sin[k3t+k1x]+mrsech[s(-ct+x)]=xs=f×(c+k1mrσ2cos[k3t+k1x]sech[s(-ct+x)])=xs=f×sin[2(k3t+k1x)][rσ22])tanh[s(-ct+x)]=×ti+s2σ2cos[k3t+k1x]=×ti×(-1+m2r2sech2[s(-ct+x)](sin2[k3t+k1x]))=xits×tanh2[s(-ct+x)]).
From (10), we obtain
(21)v2,1==12my2(mrs2σ2sech3mrs2σ2sech3[s(-ct+x)]sin[k3t+k1x]=v2,1=-2k3tan[k3t+k1x]+2cstanh[s(-ct+x)]=v2,1=+mrσ2sech[s(-ct+x)]=v2,1=×(2k12sin[k3t+k1x]+stanh[s(-ct+x)]=v2,1=×(k1(1+3cos[2(k3t+k1x)])=v2,1=×sec[k3t+k1x]-3ssin[k3t+k1x]=v2,1=×tanh[s(-ct+x)])mrs2σ22sech3)).
From (16), we obtain
(22)v3,1=1σ2(y2sech2[s(-ct+x)](k1r2λsin[2(k3t+k1x)]=1σ21+s(-fs+2r2λcos2[k3t+k1x])tanh[s(-ct+x)]).In this manner, the other componentsv1,2(x,y,t), v2,2(x,y,t), v3,2(x,y,t), v1,3(x,y,t), v2,3(x,y,t), and v3,3(x,y,t) can be obtained from (14), (11), (17), (15), (12), and (18), respectively, and substituting these components  into (19) to obtain ρ(x,y,t), θ(x,y,t), and ϕ(x,y,t).


## 3. Application of the Homotopy Analysis Method

In order to apply the homotopy analysis method for (2), we choose the linear operator L[φi(x,y,t;p)]=∂2φi/∂y2 with the property L[c1,i+c2,iy]=0, i=1,2,3, where c1,i,c2,i are integral constants to be determined by initial conditions.Furthermore, (2) suggests to define the nonlinear operators(23)N1=σ4∂2φ2∂y2+σ2∂2φ2∂x2+2∂φ1∂t-2φ2∂φ3∂x+2λ(φ12+φ22)φ2,N2=σ4∂2φ1∂y2+σ2∂2φ1∂x2-2∂φ2∂t-2φ1∂φ3∂x+2λ(φ12+φ22)φ1,N3=∂2φ3∂y2-1σ2∂2φ3∂x2+2λσ2∂(φ12+φ22)∂x
we construct the zero-order deformation equations
(24)(1-p)L[φi(x,y,t;p)-zi,0(x,y,t)]=pħiNi[φi(x,y,t;p)],i=1,2,3.Whenp=0(25)φ1(x,y,t;0)=z1,0(x,y,t)=u0(x,y,t)φ1(x,y,t;0)=rsech[s(x-ct)]cos[k1x+k3t],φ2(x,y,t;0)=z2,0(x,y,t)=v0(x,y,t)φ1(x,y,t;0)=rsech[s(x-ct)]sin[k1x+k3t],φ3(x,y,t;0)=z3,0(x,y,t)=ϕ0(x,y,t)φ1(x,y,t;0)=ftanh[s(x-ct)].Whenp=1(26)φ1(x,y,t;1)=u(x,y,t),φ2(x,y,t;1)=v(x,y,t),φ3(x,y,t;1)=ϕ(x,y,t).Therefore, as the embedding parameterp increases from 0 to 1, φi(x,y,t;p) varies from initial guesses zi,0(x,y,t) to the solutions u(x,y,t),v(x,y,t) and ϕ(x,y,t), for i=1,2,3, respectively.Expandingφi(x,y,t;p) in Taylor series with respect to p for i=1,2,3, one has
(27)φi(x,y,t;p)=zi,0(x,y,t)+∑m=1+∞zi,m(x,y,t)pm,
where
(28)zi,m(x,y,t)=1m!∂mφi(x,y,t;p)∂pm|p=0.If the auxiliary linear operator, the initial guesses, and the auxiliary parametersħi are so properly chosen, the above series converge at p=1, and
(29)u(x,y,t)=z1,0(x,y,t)+∑m=1+∞z1,m(x,y,t),v(x,y,t)=z2,0(x,y,t)+∑m=1+∞z2,m(x,y,t),ϕ(x,y,t)=z3,0(x,y,t)+∑m=1+∞z3,m(x,y,t),
which must be one of solutions of the original nonlinear equations as proved by Liao [13]. Define the vectors
(30)z→i,n={zi,0(x,y,t),zi,1(x,y,t),…,zi,n(x,y,t)};i=1,2,3.We have themth-order deformation equations
(31)L[zi,m(x,y,t)-χmzi,m-1(x,y,t)]=ħiRi,m(z→1,m-1,z→2,m-1,z→3,m-1),i=1,2,3,
where
(32)R1,m(z→1,m-1,z→2,m-1,z→3,m-1)=σ4∂2z2,m-1∂y2+σ2∂2z2,m-1∂x2+2∂z1,m-1∂t=-2∑n=0m-1z2,n∂z3,m-1-n∂x=+2λ∑n=0m-1∑k=0n(z1,kz1,n-k+z2,kz2,n-k)z2,m-1-n,R2,m(z→1,m-1,z→2,m-1,z→3,m-1)=σ4∂2z1,m-1∂y2+σ2∂2z1,m-1∂x2-2∂z2,m-1∂t=-2∑n=0m-1z1,n∂z3,m-1-n∂x=+2λ∑n=0m-1∑k=0n(z1,kz1,n-k+z2,kz2,n-k)z1,m-1-n,R3,m(z→1,m-1,z→2,m-1,z→3,m-1)=∂2z3,m-1∂y2-1σ2∂2z3,m-1∂x2=+2λσ2∂∂x∑n=0m-1(z1,nz1,m-1-n+z2,nz2,m-1-n),
where z1, z2, and z3 are functions of x,y, and t, and
(33)χm={0,m≤11,m>1.
Now, the solutions of the mth-order deformation (31) for m≥1 become
(34)zi,m(x,y,t)=χmzi,m-1(x,y,t)+ħi∬0y[Ri,m(z→1,m-1,z→2,m-1,z→3,m-1)]dydy+c1,i+c2,iy.For simplicity, we supposeħ1=ħ2=ħ3=ħ.We consider the solutions of (2) with the initial conditions (25). We now obtain at m=1(35)z1,1(x,y,t)=12ħry2sech[s(x-ct)]=×{-[2k3+k12σ2+(2fs-2r2λ+s2σ2)=+-x×sech2[s(x-ct)]]cos[k1x+k3t]=+-2s(c-k1σ2)sin[k1x+k3t]=+×tanh[s(x-ct)]+s2σ2=+×cos[k1x+k3t]tanh2[s(x-ct)]},z2,1(x,y,t)=12ħry2sech[s(x-ct)]=×{-[2k3+k12σ2+(2fs-2r2λ+s2σ2)+==x×sech2[s(x-ct)]]sin[k1x+k3t]++x+2s(c-k1σ2)cos[k1x+k3t]++x×tanh[s(x-ct)]++x+s2σ2sin[k1x+k3t]++x×tanh2[s(x-ct)]},z3,1(x,y,t)=1σ2(ħsy2(fs-2r2λ)sech21σ2×[s(x-ct)]tanh[s(x-ct)]ħsy2(fs-2r2λ)).Obviously, forħ=-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 L[φi(x,y,t;p)]=∂2φi/∂y2 with the property [c1,i+c2,iy]=0, i=1,2,3, where c1,i,c2,i are integral constant to be determined by initial conditions.Furthermore, (4) suggests to define the nonlinear operators
(36)N1=φ1∂2φ2∂y2+2∂φ2∂y∂φ1∂y+2σ2∂φ2∂x∂φ1∂xN1=+σ2φ1∂2φ2∂x2-2m∂φ1∂t,N2=∂2φ1∂y2-m2φ1((∂φ2∂y)2+σ2(∂φ2∂x)2)N1=+σ2∂2φ1∂x2+2mφ1∂φ2∂t+2λφ13+2φ1∂φ3∂x,N3=∂2φ3∂y2-1σ2∂2φ3∂x2+4λσ2φ1∂φ1∂x.
We construct the zero-order deformation equations
(37)(1-p)L[φi(x,y,t;p)-zi,0(x,y,t)]=pħiNi[φi(x,y,t;p)],i=1,2,3.Whenp=0,
(38)φ1(x,y,t;0)=z1,0(x,y,t)=ρ0(x,y,t)φ1(x,y,t;0)=rsech[s(x-ct)]cos[k1x+k3t],φ2(x,y,t;0)=z2,0(x,y,t)=θ0(x,y,t)φ1(x,y,t;0)=rsech[s(x-ct)]sin[k1x+k3t],φ3(x,y,t;0)=z3,0(x,y,t)=ϕ0(x,y,t)φ1(x,y,t;0)=ftanh[s(x-ct)].Whenp=1,
(39)φ1(x,y,t;1)=ρ(x,y,t),φ2(x,y,t;1)=θ(x,y,t),φ3(x,y,t;1)=ϕ(x,y,t).
Therefore, as the embedding parameter p increases from 0 to 1, φi(x,y,t;p) varies from initial guesses zi,0(x,y,t) to the solutions ρ(x,y,t), θ(x,y,t), and ϕ(x,y,t), for i=1,2,3, respectively.Expandingφi(x,y,t;p) in Taylor series with respect to p for i=1,2,3, one has
(40)φi(x,y,t;p)=zi,0(x,y,t)+∑m=1+∞zi,m(x,y,t)pm,
where
(41)zi,m(x,y,t)=1m!∂mφi(x,y,t;p)∂pm|p=0.If the auxiliary linear operator, the initial guesses, and the auxiliary parametersħi are so properly chosen, the series (40) converge at p=1, has
(42)ρ(x,y,t)=z1,0(x,y,t)+∑m=1+∞z1,m(x,y,t),θ(x,y,t)=z2,0(x,y,t)+∑m=1+∞z2,m(x,y,t),ϕ(x,y,t)=z3,0(x,y,t)+∑m=1+∞z3,m(x,y,t),
which must be one of solutions of the original nonlinear equation as proved by Liao [13]. Define the vectors
(43)z→i,n={zi,0(x,y,t),zi,1(x,y,t),…,zi,n(x,y,t)};i=1,2,3.
We have the mth-order deformation equations
(44)L[zi,m(x,y,t)-χmzi,m-1(x,y,t)]=ħiRi,m(z→1,m-1,z→2,m-1,z→3,m-1),i=1,2,3,
where
(45)R1,m(z→1,m-1,z→2,m-1,z→3,m-1)=∑n=0m-1z1,n∂2z2,m-1-n∂y2+2∑n=0m-1∂z2,n∂y∂z1,m-1-n∂y=+2σ2∑n=0m-1∂z2,n∂x∂z1,m-1-n∂x-2m∂z1,m-1∂t=+σ2∑n=0m-1z1,n∂2z2,m-1-n∂x2,R2,m(z→1,m-1,z→2,m-1,z→3,m-1)=∂2z1,m-1∂y2-m2∑n=0m-1∑k=0nz1,k∂z2,n-k∂y∂z2,m-1-n∂y-m2σ2∑n=0m-1∑k=0nz1,k∂z2,n-k∂x∂z2,m-1-n∂x+σ2∂2z1,m-1∂x2+2λ∑n=0m-1∑k=0nz1,kz1,n-kz1,m-1-n+2∑n=0m-1z1,n∂z3,m-1-n∂x+2m∑n=0m-1z1,n∂z2,m-1-n∂t,R3,m(z→1,m-1,z→2,m-1,z→3,m-1)=∂2z3,m-1∂y2-1σ2∂2z3,m-1∂x2+4λσ2∑n=0m-1z1,n∂z1,m-1-n∂x,
where z1, z2, and z3 are functions of x, y, and t, and
(46)χm={0,m≤11,m>1.
Now, the solutions of the mth-order deformation (44) for m  ≥  1 become
(47)zi,m(x,y,t)=χmzi,m-1(x,y,t)+ħi∬0y[Ri,m(z→1,m-1,z→2,m-1,z→3,m-1)]dydy+c1,i+c2,iy.For simplicity, we supposeħ1=ħ2=ħ3=ħ.We consider the solutions of (4) with the initial conditions (38) and obtain for m=1(48)z1,1(x,y,t)=12ħry2sech[s(-ct+x)]=×([rσ22]cos[k3t+k1x]×(-k12σ2+2k3mrcos[k3t+k1x]sech[s(-ct+x)]+(s(2f-sσ2)+r2(2λ-k12m2σ2)×cos2[k3t+k1x])sech2[s(-ct+x)])==t+s(2k1σ2sin[k3t+k1x]+mrsech[s(-ct+x)]==t×(c+k1mrσ2cos[k3t+k1x]sech[s(-ct+x)])==i×sin[2(k3t+k1x)][rσ22])tanh[s(-ct+x)]==i+s2σ2cos[k3t+k1x]==i×(1-m2r2sech2[s(-ct+x)]sin2[k3t+k1x])==i×tanh2[s(-ct+x)]),(49)z2,1(x,y,t)=12mħy2(2k3tan[k3t+k1x]+mrσ2sech[s(-ct+x)]12mħy2=×((-3k12+3s2-4s2sech2[s(-ct+x)])++12mħx=×sin[k3t+k1x]++12mħx=+k1s(1+3cos[2(k3t+k1x)])++12mħx=×sec[k3t+k1x]tanh[s(ct-x)][rσ22])=12mħy2-2cstanh[s(-ct+x)][rσ22]),(50)z3,1(x,y,t)=-1σ2(ħy2sech2[s(-ct+x)]=-1σ2t×(k1r2λsin[2(k3t+k1x)]s-1σ2tt+s(-fs+2r2λcos2[k3t+k1x])tanh[s(-ct+x)]).
Obviously, for ħ=-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 m=2,3.Using a Taylor series, then the closed form solutions yield as follows [23]:
(51)q(x,y,t)=rsech[s(x+y-ct)]exp[i(k1x+k2y+k3t)],(52)ϕ(x,y,t)=ftanh[s(x+y-ct)],
where
(53)c=k2+σ2k1,r=-2k3+σ2k12+k22λ,s=2k3+σ2k12+k22λ,f=2σ-λ1-σ2,k1,k2, and k3 are arbitrary constants.


## 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. Tables1 and 2 show the absolute errors between the analytical solutions and the HPM solutions of the DS for the first three approximations with initial conditions (5) for q(x,y,t),ϕ(x,y,t) are very small with the present choice of t at x=50 and y=0.01, when k1=0.1, k2=0.03, k3=-0.3,σ=I, λ=1, and m=0.1. Tables 3, 4, 5 and 6 help us to compare the HAM results for the first three approximations when ħ=-1 with the analytical solution through the absolute errors. Both the analytical solutions, the HPM result, and the HAM result for q(x,y,t) and ϕ(x,y,t) are plotted in Figures 1, 2, 3, and 4.Table 1
The HPM results forq(x,y,t)  in comparison with the analytical solution with initial conditions (5).
t
|
q
e
-
q
|
-
0.6
9.861869456
×
1
0
-
18
-
0.4
1.073360934
×
1
0
-
17
-
0.2
1.161066111
×
1
0
-
17
0
1.248899278
×
1
0
-
17
0.2
1.336466228
×
1
0
-
17
0.4
1.423383001
×
1
0
-
17
0.6
1.509277063
×
1
0
-
17Table 2
The HPM results forϕ(x,y,t) in comparison with the analytical solution with initial conditions (5).
t
|
ϕ
e
-
ϕ
|
-
0.6
3.3574
×
1
0
-
15
-
0.4
1.9466
×
1
0
-
15
-
0.2
1.52954
×
1
0
-
15
0
1.81125
×
1
0
-
15
0.2
2.77765
×
1
0
-
15
0.4
5.04003
×
1
0
-
15
0.6
1.07655
×
1
0
-
15Table 3
The HAM results forq(x,y,t) in comparison with the analytical solution with initial conditions (3).
t
|
q
e
-
q
|
-
0.6
1.415900000
×
1
0
-
19
-
0.4
1.400513800
×
1
0
-
19
-
0.2
1.385294700
×
1
0
-
19
0
1.370240800
×
1
0
-
19
0.2
1.355350700
×
1
0
-
19
0.4
1.340622400
×
1
0
-
19
0.6
1.326054200
×
1
0
-
19Table 4
The HAM results forϕ(x,y,t) in comparison with the analytical solution with initial conditions (3).
t
|
ϕ
e
-
ϕ
|
-
0.6
0
-
0.4
0
-
0.2
4×10-10
0
0
0.2
0
0.4
2×10-10
0.6
0Table 5
The HAM results forq(x,y,t) in comparison with the analytical solution with initial conditions (5).
t
|
q
e
-
q
|
-
0.6
9.861809932
×
1
0
-
18
-
0.4
1.073356691
×
1
0
-
17
-
0.2
1.161063556
×
1
0
-
17
0
1.248898384
×
1
0
-
17
0.2
1.336466963
×
1
0
-
17
0.4
1.423385326
×
1
0
-
17
0.6
1.509280934
×
1
0
-
17Table 6
The HAM results forϕ(x,y,t) in comparison with the analytical solution with initial conditions (5).
t
|
ϕ
e
-
ϕ
|
-
0.6
0
-
0.4
0
-
0.2
0
0
0
0.2
0
0.4
0
0.6
5
×
1
0
-
10Comparison between the exact solution, the HPM solution, and the HAM solution forq(x,y,t).
(a)
(b)
(c)
(d)Comparison between the exact solution, the HPM solution,and the HAM solution forϕ(x,y,t).
(a)
(b)
(c)
(d)The results obtained by HPM and HAM forq(x,y,t), at t=0.2 in comparison with the exact solutions.
(a)
(b)
(c)The results obtained by HPM and HAM forϕ(x,y,t), at t=0.2 in comparison with the exact solutions.
(a)
(b)
(c)



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*Source: 326473-2013-05-12.xml*