# New Jacobi Elliptic Function Solutions for the Zakharov Equations

Authors: Yun-Mei Zhao
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

A generalized(G′/G)-expansion method is proposed to seek the exact solutions of nonlinear evolution equations. Being concise and straightforward, this method is applied to the Zakharov equations. As a result, some new Jacobi elliptic function solutions of the Zakharov equations are obtained. This method can also be applied to other nonlinear evolution equations in mathematical physics.


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## Body

## 1. Introduction

In recent years, with the development of symbolic computation packages like Maple and Mathematica, searching for solutions of nonlinear differential equations directly has become more and more attractive [1–7]. This is because of the availability of computers symbolic system, which allows us to perform some complicated and tedious algebraic calculation and help us find new exact solutions of nonlinear differential equations.In 2008, Wang et al. [8] introduced a new direct method called the (G'/G)-expansion method to look for travelling wave solutions of nonlinear evolution equations (NLEEs). The method is based on the homogeneous balance principle and linear ordinary differential equation (LODE) theory. It is assumed that the traveling wave solutions can be expressed by a polynomial in (G'/G), and that G=G(ξ) satisfies a second-order LODE G′′+λG′+μG=0. The degree of the polynomial can be determined by the homogeneous balance between the highest order derivative and nonlinear terms appearing in the given NPDEs. The coefficients of the polynomial can be obtained by solving a set of algebraic equations. Many literatures have shown that the (G'/G)-expansion method is very effective, and many nonlinear equations have been successfully solved. Later, the further developed methods named the generalized (G'/G)-expansion method [9], the modified (G'/G)-expansion method [10], the extended (G'/G)-expansion method [11], the improved (G'/G)-expansion method [12], and the (G'/G,1/G)-expansion method [13] have been proposed.As we know, when using the direct method, the choice of an appropriate auxiliary LODE is of great importance. In this paper, by introducing a new auxiliary LODE of different literature [8], we propose the generalized (G'/G)-expansion method, which can be used to obtain travelling wave solutions of NLEEs.In our contribution, we will seek exact solutions of the Zakharov equations [14]:
(1.1)ntt-cs2nxx=β(|E|2)xx,(1.2)iEt+αExx=δnE,
which are one of the classical models on governing the dynamics of nonlinear waves and describing the interactions between high- and low-frequency waves, where n is the perturbed number density of the ion (in the low-frequency response), E is the slow variation amplitude of the electric field intensity, cs is the thermal transportation velocity of the electron ion, and α≠0, β≠0, δ≠0 and cs are constants.Recently, many exact solutions of (1.1)-(1.2) have been successfully obtained by using the extended tanh-expansion method, the extended hyperbolic function method, the F-expansion method, the (G'/G,1/G)-expansion method [13–19], and so on.In this paper, we construct the exact solutions to (1.1)-(1.2) by using the generalized (G'/G)-expansion method. Furthermore, we show that the exact solutions are expressed by the Jacobi elliptic function.


## 2. The Generalized(G′/G)-Expansion Method

Suppose that we have a nonlinear partial differential equation (PDE) foru(x,t) in the form
(2.1)N(u,ut,ux,utt,uxt,uxx,…)=0,
where N is a polynomial in its arguments.Step 1.
By takingu(x,t)=u(ξ), ξ=x-ct, we look for traveling wave solutions of (2.1) and transform it to the ordinary differential equation (ODE)
(2.2)N(u,-cu′,u′,c2u′′,-cu′′,u′′,…)=0.Step 2.
Suppose the solutionu of (2.2) can be expressed as a finite series in the form
(2.3)u=a0+∑i=1m‍ai(f'f)i,
where a0, ai(i=1,2,…,m) are constants to be determined later; f=f(ξ) is a solution of the auxiliary LODE
(2.4)f'2=Pf4+Qf2+R,
where P, Q, and R are constants.Step 3.
Determine the parameterm by balancing the highest order nonlinear term and the highest order partial derivative of u in (2.2).Step 4.
Substituting (2.3) and (2.4) into (2.2), setting all the coefficients of all terms with the same powers of (f'/f)k(k=1,2,…) to zero, we obtain a system of nonlinear algebraic equations (NAEs) with respect to the parameters c, a0, ai(i=1,2,…,m). By solving the NAEs if available, we can determine those parameters explicitly.Step 5.
Assuming that the constantsc, a0, ai(i=1,2,…,m) can be obtained by solving the algebraic equations in Step 4, then substituting these constants and the known general solutions into (2.3), we can obtain the explicit solutions of (2.1) immediately.


## 3. Exact Solutions of the Zakharov Equations

In this section, we mainly apply the method proposed in Section2 to seek the exact solutions of the Zakharov equations.SinceE(x,t) in (1.2) is a complex function and we are looking for the traveling wave solutions, thus we introduce a gauge transformation:
(3.1)E(x,t)=φ(ξ)e[i(kx-ωt+ξ0)],n=n(ξ),ξ=x-cgt+ξ1,
where φ(x,t) is a real-valued function, cg, k, and ω are constants to be determined later, and ξ0 and ξ1 are constants. Substituting (3.1) into (1.1)-(1.2), we have
(3.2)(cg2-cs2)n′′=β(φ2)′′,(3.3)αφ′′+i(2αk-cg)φ′+(ω-αk2)φ-δnφ=0.Integrating (3.2) twice with respect to ξ, we have
(3.4)n=βcg2-cs2φ2+C1ξ+C2,cg2-cs2≠0,
where C1 and C2 are integration constants. Substituting (3.4) into (3.3), we have
(3.5)αφ′′+i(2αk-cg)φ′+(ω-αk2)φ-δ(βcg2-cs2φ2+C1ξ+C2)φ=0.In (3.5), we assume that
(3.6)cg=2αk,k1=βcg2-cs2.Then (3.5) becomes the nonlinear ODE
(3.7)αφ′′+(ω-αk2)φ-δk1φ3+δ(C1ξ+C2)φ=0.According to the homogeneous balance betweenφ′′ and φ3 in (3.7), we obtain m=1. So we assume that φ can be expressed as
(3.8)φ(ξ)=a0+a1(f'f),
where f=f(ξ) satisfies (2.4). By using (2.4) and (3.8), it is easily derived that
(3.9)φ′′=2a1(f′f)((f′f)2-Q).Substituting (3.8) and (3.9) into (3.7), the left-hand side of (3.7) becomes a polynomial in (f'/f) and ξ. Setting their coefficients to zero yields a system of algebraic equations in a0, a1, k, k1, and ω. Solving the overdetermined algebraic equations by Maple, we can obtain the following results:
(3.10)a0=0,a1=2αk1δ,ω=α(2Q+k2)+δC2,C1=0.Substituting (3.10) into (3.8), we obtain
(3.11)φ=2αk1δ(f'f).Substituting (3.11) into (3.1) and (3.4), we have the following formal solution of (1.1)-(1.2):
(3.12)E=2αk1δ(f'f)e[i(kx-ωt+ξ0)],n=2αδ(f'f)2+C2,
where ξ=x-2αkt+ξ1, ω=α(2Q+k2)+δC2, and k1=β/(4α2k2-cs2).With the aid of the appendix [20] and from the formal solution (3.12), we get the following set of exact solutions of (1.1)-(1.2).Case 1.
ChoosingP=m2, Q=-(1+m2), R=1, and f(ξ)=sn(ξ) and inserting them into (3.12), we obtain the Jacobi elliptic function solution of (1.1)-(1.2)
(3.13)E1=2αk1δcs(ξ)dn(ξ)e[i(kx-ωt+ξ0)],n1=2αδcs2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(-2(1+m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 2.
ChoosingP=m2, Q=-(1+m2), R=1, and f(ξ)=cd(ξ) and inserting them into (3.12), we obtain the Jacobi elliptic function solution of (1.1)-(1.2)
(3.14)E2=-2αk1δ(1-m2)sd(ξ)nc(ξ)e[i(kx-ωt+ξ0)],n2=2α(1-m2)2δsd2(ξ)nc2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(-2(1+m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 3.
ChoosingP=-m2, Q=2m2-1, R=1-m2, and f(ξ)=cn(ξ), we obtain
(3.15)E3=-2αk1δdc(ξ)sn(ξ)e[i(kx-ωt+ξ0)],n3=2αδdc2(ξ)sn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2-1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 4.
ChoosingP=-1, Q=2-m2, R=m2-1, and f(ξ)=dn(ξ), we obtain
(3.16)E4=-2αk1δm2cd(ξ)sn(ξ)e[i(kx-ωt+ξ0)],n4=2αm4δcd2(ξ)sn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2-m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 5.
ChoosingP=1, Q=-(1+m2), R=m2, and f(ξ)=ns(ξ), we obtain
(3.17)E5=-2αk1δcs(ξ)dn(ξ)e[i(kx-ωt+ξ0)],n5=2αδcs2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1,  ω=α(-2(1+m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 6.
ChoosingP=1, Q=-(1+m2), R=m2, and f(ξ)=dc(ξ), we obtain
(3.18)E6=2αk1δ(1-m2)sc(ξ)nd(ξ)e[i(kx-ωt+ξ0)],n6=2α(1-m2)2δsc2(ξ)nd2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(-2(1+m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 7.
ChoosingP=1-m2, Q=2m2-1, R=-m2, and f(ξ)=nc(ξ), we obtain
(3.19)E7=2αk1δdc(ξ)sn(ξ)e[i(kx-ωt+ξ0)],n7=2αδdc2(ξ)sn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2-1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 8.
ChoosingP=m2-1, Q=2-m2, R=-1, and f(ξ)=nd(ξ), we obtain
(3.20)E8=2αk1δm2cd(ξ)sn(ξ)e[i(kx-ωt+ξ0)],n8=2αm4δcd2(ξ)sn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2-m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 9.
ChoosingP=1-m2, Q=2-m2, R=1, and f(ξ)=sc(ξ), we obtain
(3.21)E9=2αk1δdc(ξ)ns(ξ)e[i(kx-ωt+ξ0)],n9=2αδdc2(ξ)ns2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2-m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 10.
ChoosingP=-m2(1-m2), Q=2m2-1, R=1, and f(ξ)=sd(ξ), we obtain
(3.22)E10=2αk1δcd(ξ)ns(ξ)e[i(kx-ωt+ξ0)],n10=2αδcd2(ξ)ns2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2-1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 11.
ChoosingP=1, Q=2-m2, R=1-m2, and f(ξ)=cs(ξ), we obtain
(3.23)E11=2αk1δds(ξ)nc(ξ)e[i(kx-ωt+ξ0)],n11=2αδds2(ξ)nc2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2-m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 12.
ChoosingP=1, Q=2m2-1, R=-m2(1-m2), and f(ξ)=ds(ξ), we obtain
(3.24)E12=-2αk1δcs(ξ)nd(ξ)e[i(kx-ωt+ξ0)],n12=2αδcs2(ξ)nd2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2-1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 13.
ChoosingP=1/4, Q=(1-2m2)/2, R=1/4, and f(ξ)=ns(ξ)±cs(ξ), we obtain
(3.25)E13=∓2αk1δds(ξ)e[i(kx-ωt+ξ0)],n13=2αδds2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(1-2m2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 14.
ChoosingP=(1-m2)/4, Q=(1+m2)/2, R=(1-m2)/4, and f(ξ)=nc(ξ)±sc(ξ), we obtain
(3.26)E14=±2αk1δdc(ξ)e[i(kx-ωt+ξ0)],n14=2αδdc2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(1+m2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 15.
ChoosingP=1/4, Q=(m2-2)/2, R=m2/4, and f(ξ)=ns(ξ)±ds(ξ), we obtain
(3.27)E15=∓2αk1δcs(ξ)e[i(kx-ωt+ξ0)],n15=2αδcs2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2-2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 16.
ChoosingP=m2/4, Q=(m2-2)/2, R=m2/4, and f(ξ)=sn(ξ)±icn(ξ), we obtain
(3.28)E16=2αk1δdn(ξ)(cn(ξ)∓isn(ξ))e[i(kx-ωt+ξ0)]sn(ξ)±icn(ξ),n16=2αdn2(ξ)(cn(ξ)∓isn(ξ))2δ(sn(ξ)±icn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(m2-2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 17.
ChoosingP=m2/4, Q=(m2-2)/2, R=m2/4, and f(ξ)=m2-1sd(ξ)±cd(ξ), we obtain
(3.29)E17=2αk1δ(m2-1cn(ξ)±m2sn(ξ)∓sn(ξ))e[i(kx-ωt+ξ0)]dn(ξ)(m2-1sn(ξ)±cn(ξ)),n17=2α(m2-1cn(ξ)±m2sn(ξ)∓sn(ξ))2δdn2(ξ)(m2-1sn(ξ)±cn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(m2-2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 18.
ChoosingP=1/4, Q=(1-2m2)/2, R=1/4, and f(ξ)=mcd(ξ)±i1-m2nd(ξ), we obtain
(3.30)E18=2αk1δmsn(ξ)(±m2∓1+im1-m2cn(ξ))e[i(kx-ωt+ξ0)]dn(ξ)(i1-m2±mcn(ξ)),n18=2αm2sn2(ξ)(±m2∓1+im1-m2cn(ξ))2δdn2(ξ)(i1-m2±mcn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(1-2m2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 19.
ChoosingP=1/4, Q=(1-2m2)/2, R=1/4, and f(ξ)=msn(ξ)±idn(ξ), we obtain
(3.31)E19=2αk1δmcn(ξ)(dn(ξ)∓imsn(ξ))e[i(kx-ωt+ξ0)]msn(ξ)±idn(ξ),n19=2αm2cn2(ξ)(dn(ξ)∓imsn(ξ))2δ(msn(ξ)±idn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(1-2m2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 20.
ChoosingP=1/4, Q=(1-2m2)/2, R=1/4, and f(ξ)=m2-1sc(ξ)±idc(ξ), we obtain
(3.32)E20=2αk1δ(m2-1dn(ξ)∓im2sn(ξ)±isn(ξ))e[i(kx-ωt+ξ0)]cn(ξ)(m2-1sn(ξ)±idn(ξ)),n20=2α(m2-1dn(ξ)∓im2sn(ξ)±isn(ξ))2δcn2(ξ)(m2-1sn(ξ)±idn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(1-2m2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 21.
ChoosingP=(m2-1)/4, Q=(m2+1)/2, R=(m2-1)/4, and f(ξ)=msd(ξ)±nd(ξ), we obtain
(3.33)E21=±2αk1δmcd(ξ)e[i(kx-ωt+ξ0)],n21=2αm2δcd2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2+1+k2)+δC2, and k1=β/(4α2k2-cs2).Case 22.
ChoosingP=m2/4, Q=(m2-2)/2, R=1/4, and f(ξ)=sn(ξ)/(1±dn(ξ)), we obtain
(3.34)E22=±2αk1δcs(ξ)e[i(kx-ωt+ξ0)],n22=2αδcs2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2-2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 23.
ChoosingP=-1/4, Q=(m2+1)/2, R=(1-m2)2/4, and f(ξ)=mcn(ξ)±dn(ξ), we obtain
(3.35)E23=∓2αk1δmsn(ξ)e[i(kx-ωt+ξ0)],n23=2αm2δsn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2+1+k2)+δC2, and k1=β/(4α2k2-cs2).Case 24.
ChoosingP=(1-m2)2/4, Q=(m2+1)/2, R=1/4, and f(ξ)=ds(ξ)±cs(ξ), we obtain
(3.36)E24=∓2αk1δns(ξ)e[i(kx-ωt+ξ0)],n24=2αδns2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2+1+k2)+δC2, and k1=β/(4α2k2-cs2).Case 25.
ChoosingP=1/4, Q=(m2-2)/2, R=m4/4, and f(ξ)=dc(ξ)±1-m2nc(ξ), we obtain
(3.37)E25=2αk1δsn(ξ)(∓m2±1+1-m2dn(ξ))e[i(kx-ωt+ξ0)]cn(ξ)(1-m2±dn(ξ)),n25=2αsn2(ξ)(∓m2±1+1-m2dn(ξ))2δcn2(ξ)(1-m2±dn(ξ))2+C2,
where ξ=x-2αkt+ξ1, ω=α(m2-2+k2)+δC2, and k1=β/(4α2k2-cs2).Case 26.
ChoosingR=m2Q2/(m2+1)2P, Q<0, P>0, and f(ξ)=-m2Q/(m2+1)Psn((-Q/(m2+1))ξ), we obtain
(3.38)E26=2αk1δ-Qm2+1cs(-Qm2+1ξ)dn(-Qm2+1ξ)e[i(kx-ωt+ξ0)],n26=-2αQδ(m2+1)cs2(-Qm2+1ξ)dn2(-Qm2+1ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2Q+k2)+δC2, and k1=β/(4α2k2-cs2).Case 27.
ChoosingR=(1-m2)Q2/(m2-2)2P, Q>0, P<0, and f(ξ)=-Q/(2-m2)Pdn((Q/(2-m2))ξ), we obtain
(3.39)E27=-2αk1δQ2-m2m2cd(Q2-m2ξ)sn(Q2-m2ξ)e[i(kx-ωt+ξ0)],n27=2αQm4δ(2-m2)cd2(Q2-m2ξ)sn2(Q2-m2ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2Q+k2)+δC2, and k1=β/(4α2k2-cs2).Case 28.
ChoosingR=m2(m2-1)Q2/(2m2-1)2P, Q>0, P<0, and f(ξ)=-(m2Q/(2m2-1)P)cn((Q/(2m2-1))ξ), we obtain
(3.40)E28=-2αk1δQ2m2-1dc(Q2m2-1ξ)sn(Q2m2-1ξ)e[i(kx-ωt+ξ0)],n28=2αQδ(2m2-1)dc2(Q2m2-1ξ)sn2(Q2m2-1ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2Q+k2)+δC2, and k1=β/(4α2k2-cs2).Case 29.
ChoosingP=1, Q=2-4m2, R=1, and f(ξ)=sn(ξ)dn(ξ)/cn(ξ), we obtain
(3.41)E29=2αk1δ(1-2m2sn2(ξ)+m2sn4(ξ))e[i(kx-ωt+ξ0)]cn(ξ)sn(ξ)dn(ξ),n29=2α(1-2m2sn2(ξ)+m2sn4(ξ)2)2δcn2(ξ)sn2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2-4m2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 30.
ChoosingP=m4, Q=2m2-4, R=1, and f(ξ)=sn(ξ)cn(ξ)/dn(ξ), we obtain
(3.42)E30=2αk1δ(1-2sn2(ξ)+m2sn4(ξ))e[i(kx-ωt+ξ0)]dn(ξ)sn(ξ)cn(ξ),n30=2α(1-2sn2(ξ)+m2sn4(ξ))2δdn2(ξ)sn2(ξ)cn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2-4)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 31.
ChoosingP=1, Q=2m2+2, R=1-2m2+m4, and f(ξ)=cn(ξ)dn(ξ)/sn(ξ), we obtain
(3.43)E31=2αk1δ(m2sn4(ξ)-1)e[i(kx-ωt+ξ0)]sn(ξ)cn(ξ)dn(ξ),n31=2α(m2sn4(ξ)-1)2δsn2(ξ)cn2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2+2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 32.
ChoosingP=A2(m-1)2/4, Q=(m2+1)/2+3m, R=(m-1)2/4A2, and f(ξ)=dn(ξ)cn(ξ)/A(1+sn(ξ))(1+msn(ξ)), we obtain
(3.44)E32=2αk1δ(m2sn2(ξ)+msn2(ξ)-1-m)e[i(kx-ωt+ξ0)]dn(ξ)cn(ξ),n32=2α(m2sn2(ξ)+msn2(ξ)-1-m)2δdn2(ξ)cn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2+1+k2)+δC2, and k1=β/(4α2k2-cs2).Case 33.
ChoosingP=A2(m+1)2/4, Q=(m2+1)/2-3m, R=(m+1)2/4A2, and f(ξ)=dn(ξ)cn(ξ)/A(1+sn(ξ))(1-msn(ξ)), we obtain
(3.45)E33=2αk1δ(m2sn2(ξ)-msn2(ξ)+m-1)e[i(kx-ωt+ξ0)]dn(ξ)cn(ξ),n33=2α(m2sn2(ξ)-msn2(ξ)+m-1)2δdn2(ξ)cn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(m2+1+k2)+δC2, and k1=β/(4α2k2-cs2).Case 34.
ChoosingP=-4/m, Q=6m-m2-1, R=-2m3+m4+m2, and f(ξ)=mcn(ξ)dn(ξ)/(msn2(ξ)+1), we obtain
(3.46)E34=2αk1δsn(ξ)(msn2(ξ)-1+2m2sn2(ξ)+m3sn2(ξ)-m2-2m)e[i(kx-ωt+ξ0)](msn2(ξ)+1)cn(ξ)dn(ξ),n34=2αsn2(ξ)(msn2(ξ)-1+2m2sn2(ξ)+m3sn2(ξ)-m2-2m)2δ(msn2(ξ)+1)2cn2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(2(6m-m2-1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 35.
ChoosingP=4/m, Q=-6m-m2-1, R=2m3+m4+m2, and f(ξ)=mcn(ξ)dn(ξ)/(msn2(ξ)-1), we obtain
(3.47)E35=2αk1δsn(ξ)(msn2(ξ)+1-2m2sn2(ξ)+m3sn2(ξ)+m2-2m)e[i(kx-ωt+ξ0)](msn2(ξ)-1)cn(ξ)dn(ξ),n35=2αsn2(ξ)(msn2(ξ)+1-2m2sn2(ξ)+m3sn2(ξ)+m2-2m)2δ(msn2(ξ)-1)2cn2(ξ)dn2(ξ)+C2,
where ξ=x-2αkt+ξ1, ω=α(-2(6m+m2+1)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 36.
ChoosingP=-(m2+2m+1)B2, Q=2m2+2, R=(2m-m2-1)/B2, and f(ξ)=(msn2(ξ)-1)/B(msn2(ξ)+1), we obtain
(3.48)E36=42αk1δmsn(ξ)cn(ξ)dn(ξ)m2sn4(ξ)-1e[i(kx-ωt+ξ0)],n36=32αm2sn2(ξ)cn2(ξ)dn2(ξ)δ(m2sn4(ξ)-1)2+C2,
where ξ=x-2αkt+ξ1, ω=α(2(2m2+2)+k2)+δC2, and k1=β/(4α2k2-cs2).Case 37.
ChoosingP=-(m2-2m+1)B2, Q=2m2+2, R=-(2m+m2+1)/B2, and f(ξ)=(msn2(ξ)+1)/B(msn2(ξ)-1), we obtain
(3.49)E37=-42αk1δmsn(ξ)cn(ξ)dn(ξ)m2sn4(ξ)-1e[i(kx-ωt+ξ0)],n37=32αm2sn2(ξ)cn2(ξ)dn2(ξ)δ(m2sn4(ξ)-1)2+C2,
where ξ=x-2αkt+ξ1,  ω=α(2(2m2+2)+k2)+δC2, and k1=β/(4α2k2-cs2).


## 4. Conclusions

In this paper, by using the generalized(G'/G)-expansion method, we have successfully obtained some exact solutions of Jacobi elliptic function form of the Zakharov equations. When the modulus of the Jacobi elliptic function m→0 or 1, the corresponding solitary wave solutions and trigonometric function solutions are also obtained. This work shows that the generalized (G'/G)-expansion method provides a very effective and powerful tool for solving nonlinear equations in mathematical physics.



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*Source: 854619-2012-12-05.xml*