# Exact Solutions for the Wick-Type Stochastic Schamel-Korteweg-de Vries Equation

Authors: Xueqin Wang; Yadong Shang; Huahui Di
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

We consider the Wick-type stochastic Schamel-Korteweg-de Vries equation with variable coefficients in this paper. With the aid of symbolic computation and Hermite transformation, by employing the(G′/G,1/G)-expansion method, we derive the new exact travelling wave solutions, which include hyperbolic and trigonometric solutions for the considered equations.


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## Body

## 1. Introduction

The partial differential equations arise in many physical fields like the condense matter physics, fluid mechanics, plasma physics, optics, and so on, which exhibit a rich variety of nonlinear phenomena. It is known that to find exact solutions of the partial differential equations is always one of the central themes in mathematics and physics.In [1], Lee and Sakthivel used the exp-function method to obtain some exact travelling waves solutions for the following Schamel-Korteweg-de Vries equation [2–4].(1)ut+αu1/2+βuux+δuxxx=0,where u=u(t,x) denotes the unknown function of the space variable x and time t and the parameters α,β, and δ are constants which refer to the activation trapping, the convection, and the dispersion coefficients, respectively. Equation (1) arises in number of scientific models, such as plasma physics and optical fibre. This equation describes the nonlinear interaction of ion-acoustic waves when electron trapping is present and also it governs the electrostatic potential for a certain electron distribution in velocity space. In addition, a generalized KdV equation is a special case of (1) which has been studied in a variety of mathematical physics contexts. Equation (1) incorporates the well-known KdV equation when α=0 and the Schamel equation when β=0 [5].In [6], Kangalgil used the extended (G′/G)-expansion method to obtain some hyperbolic function solutions and trigonometric functions with free parameters of (1). In [7], by using the sine-cosine method and the extended tanh method, Yang and Tang obtained the soliton-like solutions, the kink solutions, and the plural solutions of (1).When the inhomogeneities of media and nonuniformity of boundaries are taken into account in various real physical situations, the variable coefficient partial differential equations often can provide more powerful and realistic models than their constant coefficient counterparts in describing a large variety of real phenomena. Recently, the importance of taking random effects into account in modeling, analyzing, simulating, and predicting complex phenomena has been widely recognized in geophysical and climate dynamics, materials science, chemistry, biology, and other fields [8, 9]. Stochastic partial differential equations are appropriate mathematical models for complex systems under random influence in fluids.The Wick-type stochastic Schamel-Korteweg-de Vries equation with variable coefficients is one of the most important stochastic partial differential equations and it has many applications. In [10], Holden et al. gave white noise functional approach to research stochastic partial differential equations in Wick versions. In [11], Li et al. introduced a new direct method called the (G′/G,1/G)-expansion method to look for travelling wave solutions of nonlinear evolution equations. The determination of exact solutions to the variable coefficients partial differential equations is a complicated problem that challenges researchers greatly. We will use their theory and the (G′/G,1/G)-expansion method to give exact solutions of the following Wick-type stochastic Schamel-Korteweg-de Vries equation with variable coefficients:(2)ut+αt♢u1/2+βt♢u♢ux+δt♢uxxx=0,where ♢ is the Wick production on the Hida distribution space (S(Rd))∗, namely, (S(Rd))∗ is the white noise functional space which is defined in [12], and α(t),β(t), and δ(t) are white noise functionals.This paper is organized as follows. In Section2, we briefly describe some basic concepts on Wick-type and main steps of finding solutions. In Section 3, we describe the (G′/G,1/G)-expansion method. In Section 4, we use white noise analysis and Hermite transformation to obtain a number of Wick versions of hyperbolic, trigonometric, and rational solutions. The conclusions are given in the final section.


## 2. Some Basic Concepts and Main Steps

Assume thatS(Rd) and (S(Rd))∗ are the Hida test function space and the Hida disturbance space on Rd, respectively. And let hn(x) be the d-order Hermite polynomials. Put(3)ξnx=e-1/2x2hn2xπn-1!1/2,n≥1.We have that the collection ξnn≥1 constitutes an orthogonal basis of L2(R).If we denoteα=(α1,…,αd) as d-dimensional multi-indices with α1,…,αd∈N, we have that the family of tensor products ξ(α1,…,αd)=ξα1⊗⋯⊗ξαd(α∈Nd) forms an orthogonal basis for L2(Rd). Let α(i)=(α1(i),…,αd(i)) be the ith multi-index number in some fixed ordering of all d-dimensional multi-indices α=(α1,…,αd)∈Nd. We assume that this ordering has the following property:(4)i<j⟹α1i+⋯+αdi≤α1j+⋯+αdj.We define(5)ηi=ξαi=ξα1i⊗⋯⊗ξαdi,i≥1,and denote multi-indices as elements of the space (N0N)c of all sequences α=(α1,α2,…) with elements αi∈N0 and with compact support, that is, with only finitely many αi≠0. We denote J≡(N0N)c.Fixingn∈N, let (S)1n consist of those(6)x=∑αcαHαw∈⨁k=1nL2μ,where cα∈Rn,x1,k2≡∑αcα2(α!)2(2N)kα<∞,∀k∈N with cα2=cα2=∑k=1n(cα(k))2,cα=cα1,…,cαn∈Rn,α!=∏k=1∞αk!, and (2N)α=∏j(2j)αj, where μ is the white noise measure on (S∗(R),B(S∗(R))),(7)Hαw=∏i=1∞Hαiw,ηi,w∈SRd∗,α=α1,α2,…∈J.The space(S)-1n consists of all formal expansions(8)X=∑αbαHα,where bα∈Rn,X-1,-q≡∑αba2(2N)-qα<∞,∀q∈N. The family of seminorms x1,k,k∈N gives rise to a topology on (S)1n, and we can regard (S)-1n as the dual of (S)1n by the action(9)X,x=∑αbα,cαα!,where (bα,cα) is the usual inner product in Rn.ForX=∑αaαHα,Y=∑αbαHα∈(S)-1n with aα,bα∈Rn,(10)X♢Y=∑α,βaα,bβHα+β is called the Wick product of X and Y.We can prove that the spacesS(Rd),SRd∗,(S)1, and (S)-1 are closed under Wick products.ForX=∑αaαHα∈(S)-1n with aα∈Rn, the Hermite transform of X is denoted as follows:(11)HX=X~z=∑αaαzα∈Cn,when  convergent,where z=(z1,z2,…)∈Cn (the set of all sequences of complex numbers), zα=z1α1z2α2⋯znαn⋯ when α=(α1,α2,…)∈J.ForX,Y∈(S)-1n, we can find(12)X♢Y~z=X~zY~z,for all z such that X~(z) and Y~(z) exist. The product on the right-hand side of the above formula is the complex bilinear product between two elements of CN defined by(13)z11,…,zn1·z12,…,zn2=∑k=1nzk1zk2,zki∈C.LetX=∑αaαHα,∈(S)-1n. Then the vector c0=X~(0)∈Rn is called the generalized expectation of X and is denoted by E(X). Suppose that f:V→Cm is an analytic function, where V is a neighborhood of E(X). Assume that the Taylor series of f around E(X) has coefficients in Rn, and we can find the Wick version f♢(X)=H-1(f∘X~)∈(S)-1m.We define the Wick exponential ofX∈(S)-1n as follows:(14)exp♢⁡X=∑n=0∞X♢nn!.We can find that the Wick exponential has the same algebraic properties as the usual exponential with the use of the Hermite transformation. For example, exp♢{X+Y}=exp♢{X}♢exp♢{Y}.Suppose that modeling considerations lead us to consider an SPDE expressed formally asA(t,x,∂t,∇x,U,w)=0, where A is some given function and U=U(t,x,w) is unknown (generalized) stochastic process, where the operators(15)∂t=∂∂t,∇x=∂∂x1,…,∂∂xd,x=x1,…,xd∈Rd.Firstly, we interpret all products as Wick products and all functions as their Wick versions indicate this as(16)A♢t,x,∂t,∇x,U,w=0.Secondly, we take the Hermite transformation of (16). This turns Wick products into ordinary products and the equation takes the following form:(17)A~t,x,∂t,∇x,U~,z1,z2,…=0,where U~=H(U) is the Hermite transformation of U and z1,z2,… are complex number. Suppose we find a solution u=u(t,x,z) of the equation A~(t,x,∂t,∇x,u,z)=0 for each z=(z1,z2,…)∈Kq(r), where(18)Kqr=z=z1,z2,…∈CN,∑α≠0zα22Nqα<r2.Then, under certain conditions, we can take the inverse Hermite transformationU=H-1u∈(S)-1 and obtain a solution U of the original Wick equation (16). We have the following theorem, which was proved by Holden et al. [10].Theorem 1.
Supposeu(t,x,z) is a solution (in the usual strong, pointwise sense) of (17) for (t,x) in some bounded open set G⊂R×Rd, and for all z∈Kq(r), for some q,r. Moreover, suppose that u(t,x,z) and all its partial derivatives, which are involved in (17), are bounded for (t,x,z)∈G×Kq(r), continuous with respect to (t,x)∈G for all z∈Kq(r), and analytic with respect to z∈Kq(r), for all (t,x)∈G.
Then there existsU(t,x)∈(S)-1 such that u(t,x,z)=(U~(t,x))(z) for all (t,x,z)∈G×Kq(r), and U(t,x) solves (in the strong sense in (S)-1) (16) in (S)-1.


## 3. Description of the(G′/G,1/G)-Expansion Method

In this section, we describe the main steps of the(G′/G,1/G)-expansion method for finding travelling wave solutions of nonlinear evolution equations. As preparations, consider the second order linear ordinary differential equation(19)G′′ξ+λGξ=μ,and we let(20)ϕ=G′G,ψ=1G,for simplicity here and after. Using (19) and (20) yields(21)ϕ′=-ϕ2+μψ-λ,ψ′=-ϕψ.From the three cases of general solutions of the linear ordinary differential equation (19), we have the following.Case 1.
Whenλ<0, the general solution of the linear ordinary differential equation (19) is(22)Gξ=A1sinh⁡-λξ+A2cosh⁡-λξ+μλ,and we have(23)ψ2=-λλ2σ1+μ2ϕ2-2μψ+λ,where A1 and A2 are two arbitrary constants and σ1=A12-A22.Case 2.
Whenλ>0, the general solution of the linear ordinary differential equation (19) is(24)Gξ=A1sin⁡λξ+A2cos⁡λξ+μλ,and we have(25)ψ2=λλ2σ2-μ2ϕ2-2μψ+λ,where A1 and A2 are two arbitrary constants and σ2=A12+A22.Case 3.
Whenλ=0, the general solution of the linear ordinary differential equation (19) is(26)Gξ=μ2ξ2+A1ξ+A2,and we have(27)ψ2=1A12-2μA2ϕ2-2μψ,where A1 and A2 are two arbitrary constants.Now we consider a nonlinear evolution equation, say in two independent variablesx and t,(28)Pu,ut,ux,utt,uxt,uxx,…=0.In general, the left-hand side of (28) is a polynomial in u and its various partial derivatives. The main steps of the (G′/G,1/G)-expansion method are as follows.Step 1.
By coordinates transformationξ=x-Vt and with u(x,t)=u(ξ), (28) can be reduced to an ordinary differential equation on u(ξ) with (29)Pu,-Vu′,u′,V2u′′,-Vu′′,u′′,…=0.Step 2.
Suppose that the solution of the ordinary differential equation (29) can be expressed by a polynomial in ϕ and ψ as(30)uξ=∑i=0Naiϕi+∑i=1Nbiϕi-1ψ,where G=G(ξ) satisfies the second-order linear ordinary differential equation (19), ai(i=0,…,N),bi(i=1,…,N),V,λ, and μ are constants to be determined later, and aN2+bN2≠0.Step 3.
Determine the positive integerN in (30) by using the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in the ordinary differential equation (29). More precisely, we define the degree of u(ξ) as D[u(ξ)]=N, which gives rise to the degree of other expressions as follows:(31)Ddqudξq=N+q,Dupdqudξqs=Np+sq+N.Therefore, we can get the value of N in (30). In some nonlinear equations the balance number N is not a positive integer. In this case, we make the following transformations [13]:
(a) WhenN=q/p, where q/p is a fraction in the lowest terms, we let(32)uξ=vq/pξ,then substituting (32) into (29) to get a new equation in the new function v(ξ) with a positive integer balance number.
(b) WhenN is a negative number, we let(33)uξ=vNξ,then substituting (33) into (29) to get a new equation in the new function v(ξ) with a positive integer balance number.Step 4.
Substituting (30) into (29), using (21) and (23) (here Case 1 is taken as an example), the left-hand side of (29) can be converted into a polynomial in ϕ and ψ, in which the degree of ψ is not larger than one. Equating each coefficient of the polynomial to zero yields a system of algebraic equations about ai(i=0,…,N),bi(i=1,…,N),V,λ,μ,A1 and A2. Then we solve the algebraic equations with the aid of Maple or Mathematica. Substituting the values of ai(i=0,…,N),bi(i=1,…,N),V,λ,μ,A1, and A2 obtained into (30), one can obtain the travelling wave solutions expressed by the hyperbolic functions of (29).Step 5.
Similar to Steps3 and 4, substituting (30) into (29), using (21) and (25) (or (21) and (27)), we obtain the travelling wave solutions of (29) expressed by trigonometric functions (or expressed by rational functions).


## 4. Exact Solutions of Stochastic Schamel-Korteweg-de Vries Equation

In this section, we will give exact solutions of (2).Taking the Hermite transformation of (2), we can get the equation(34)u~tt,x,z+α~t,zu~1/2t,x,z+β~t,zu~t,x,zu~xt,x,z+δ~t,zu~xxxt,x,z=0,where z=(z1,z2,…)∈CN is a parameter.For the sake of simplicity, we denoteu(t,x,z)=u~(t,x,z),  α(t,z)=α~(t,z),  β(t,z)=β~(t,z), and δ(t,z)=δ~(t,z). In the following, we apply the (G′/G,1/G)-expansion method to construct the travelling wave solutions of (34). In order to obtain exact solutions of (34), we consider the transformation(35)ut,x,z=uξ,ξ=pt,zx+qt,z,where p(t,z) and q(t,z) will be determined later. Substituting (35) into (34), we have(36)utt,x,z+αt,zu1/2t,x,z+βt,zut,x,zuxt,x,z+δt,zuxxxt,x,z=u′ξptt,zx+qtt,z+αt,zu1/2ξ+βt,zuξu′ξpt,z+δt,zu′′′ξp3t,z=0.This implies(37)u′ξptt,zx=0,(38)u′ξqtt,z+αt,zu1/2ξ+βt,zuξu′ξpt,z+δt,zu′′′ξp3t,z=0.Equation (37) means pt(t,z)=0. These imply that p(t,z) is a constant; denote it by p. Suppose qt(t,z)=cα(t,z),β(t,z)=βα(t,z) and δ(t,z)=δα(t,z), where c,β, and δ are real constants. From (38), we obtain(39)qt,z=c∫0tαs,zds.Then by (38) and (39), we get the following ordinary differential equations with constant coefficients:(40)cu′ξ+pu1/2ξ+pβuξu′ξ+p3δu′′′ξ=0.Integrating (40) once, and considering the constants of integration as zero, we can find(41)cuξ+23pu3/2ξ+12pβu2ξ+p3δu′′ξ=0.When we consider the transformations(42)uξ=v2ξ,u′ξ=2vξv′ξ,u′′ξ=2v′ξ2+vξv′′ξ,where ξ=px+q(t,z), we can rewrite (41) as(43)cv2ξ+23pv3ξ+12pβv4ξ+2p3δv′ξ2+2p3δvξv′′ξ=0.For simplicity, (43) can be written as(44)v2ξ+ϑv3ξ+ιv4ξ+κv′ξ2+κvξv′′ξ=0,where ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c, and c≠0.By balancing betweenvξv′′(ξ) and v4(ξ) in (44), we get (45)N+N+2=4N⟹N=1.Consequently, we assume (46)vξ=a1ϕξ+b1ψξ,where a1 and b1 are constants to be determined later satisfying a12+b12≠0.By employing the(G′/G,1/G)-expansion method with the aid of symbolic computation, we derive new exact travelling wave solutions, which include hyperbolic, trigonometric, and rational solutions for (34); there are three cases to be discussed as follows.Case 1 (hyperbolic function solution).
Ifλ<0, substituting (46) into (44) and using (21) and (23), the left-hand side of (44) becomes a polynomial in ϕ and ψ. Setting the coefficients of this polynomial to be zero, we yield a system of algebraic equations in a1,b1,μ, and λ as follows:(47)ϕ4:a14ι+a12κ-λ6a12b12ι+3κb12σ1λ2+μ2+b14λ2ισ1λ2+μ22=0,ϕ3:a13ϑ+λ6κμa1b1-3ϑa1b12σ1λ2+μ2-8ιμλ2a1b13σ1λ2+μ22=0,ϕ3ψ:2κa1b1+ιa13b1-ιλa1b13σ1λ2+μ2=0,ϕ2:a12-4λκa12-λσ1λ2+μ2b12+λ6ιa12b12+4κb12+κμ2a12+2ιλ3b14σ1λ2+μ22+2μλ2κμb12-2ϑb13σ1λ2+μ22-4ιμ2λ3b14σ1λ2+μ23=0,ϕ2ψ:3ϑa12b1-5μκa12+λ5μκb12-ϑb13σ1λ2+μ2+2λμ6ιa12b12+κb12σ1λ2+μ2-16κμλ2b14σ1λ2+μ22=0,ϕ:6μκa1b1-3ϑa1b12σ1λ2+μ2-8ιμa1λb13σ1λ2+μ22=0,ϕψ:2a1b1+5λκa1b1+2λμ3ϑa1b12-6κμa1b1σ1λ2+μ2-4ιλ2a1b13σ1λ2+μ2+16ιμ2λ2a1b13σ1λ2+μ22=0,ψ:2μκa12-λ3μκb12-ϑb13σ1λ2+μ2-2μb12+2κμ2a12σ1λ2+μ2-4μ2λϑb13-κμb12σ1λ2+μ22-8ιμ3λ2b14σ1λ2+μ23+4ιμλ2b14σ1λ2+μ22=0,ψ0:a12κ-b12-a12μ2κ-b12λκσ1λ2+μ2+2λμb12μκ-ϑb13σ1λ2+μ22+ιλ2b14σ1λ2+μ22-4ιμ2λ2b14σ1λ2+μ23=0.Solving the above algebraic equations, we get the following solutions: when3κσ1/ιλκ2(7ι-48κ)2-3ικϑ2>0, then(48)a1=0,b1=±κλ7ι-48κ3κσ1ιλκ27ι-48κ2-3ικϑ2,λ=30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2,μ=±ϑιλ3κσ1ιλκ27ι-48κ2-3ικϑ2.From these solutions, we obtain the hyperbolic function solutions of (34).Theorem 2.
Suppose that30ιϑ2+12ιϑ2(7ι-48κ)+(7ι-48κ)2/2κ(7ι-48κ)2<0 and 3κσ1/ιλκ2(7ι-48κ)2-3ικϑ2>0. Then (34) admits an exact solution(49)ut,x,z=3κ3λ2σ17ι-48κ2ιλκ27ι-48κ2-3ικϑ2d2t,x,z,where(50)dt,x,z=A1sinh⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2ξ+A2cosh⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2ξ±ϑι3κσ1ιλκ27ι-48κ2-3ικϑ2,ξ=px+c∫0tα(s,z)ds,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ1=A12-A22, and  A1 and A2 are two arbitrary constants.Case 2 (trigonometric function solution).
Ifλ>0, substituting (46) into (44) and using (21) and (25), the left-hand side of (44) becomes a polynomial in ϕ and ψ. Setting the coefficients of this polynomial to be zero, we yield a system of algebraic equations in a1,b1,μ, and λ as follows:(51)ϕ4:3κa12+κa14+λ6ιa12b12+3κb12σ2λ2-μ2+λιb14σ2λ2-μ22=0,ϕ3:a13ϑ+λ3ϑa1b12-6κμa1b1σ2λ2-μ2-8ιμλ2b13σ2λ2-μ22=0,ϕ3ψ:2κa1b1+ιa13b1+ιλa1b13σ2λ2-μ2=0,ϕ2:a12+4κλa12+λσ2λ2-μ2b12+λ4κb12+6ιa12b12+κa12μ2+2ιλ3b14σ2λ2-μ22+2μλ2κμb12-ϑb13σ2λ2-μ22+4ιμ2λ3b14σ2λ2-μ23=0,ϕ2ψ:-5κμa12+3ϑa12b1+λϑb13-3κμb12σ2λ2-μ2-2λμ6ιa12b12+2κb12σ2λ2-μ2-4ιμλ2b14σ2λ2-μ22=0,ϕ:3ϑa1b12-6κa1b1σ2λ2-μ2+8ιμλa1b13σ2λ2-μ22=0,ϕψ:2a1b1+5κλa1b1+2μλ6μκa1b1-3ϑa1b12σ2λ2-μ2+4ιλ2a1b13σ2λ2-μ2+16ιμ2λ2a1b13σ2λ2-μ22=0,ψ:2κμa12-λϑb13-3κμb12σ2λ2-μ2+2μb12+κμ2a12σ2λ2-μ2-4λμ2ϑb13-κμb12σ2λ2-μ22+8ιμ3λ2b14σ2λ2-μ23=0,ψ0:κa12+b12+κμ2a12+κλb12σ2λ2-μ2+2μλκμb12-2ϑb13σ2λ2-μ22+ιλ2b14σ2λ2-μ22+4ιμ2λ2b14σ2λ2-μ23=0.
Solving the above algebraic equations, we get the following solutions: when3σ2/12ϑ2-100ι>0, then(52)a1=0,b1=±53σ212ϑ2-100ι,λ=12κ,μ=∓ϑκ3σ212ϑ2-100ι.From these solutions, we obtain the hyperbolic function solutions of (34).Theorem 3.
Suppose that3σ2/12ϑ2-100ι>0 and κ>0. Then (34) admits an exact solution(53)ut,x,z=75σ212ϑ2-100ιe2t,x,z,where(54)et,x,z=A1sin⁡ξ2κ+A2cos⁡ξ2κ∓2ϑ3σ212ϑ2-100ι,ξ=px+c∫0tα(s,z)ds,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ2=A12+A22, and  A1 and A2 are two arbitrary constants.Case 3 (rational function solutions).
Ifλ=0, substituting (46) into (44) and using (21) and (27), the left-hand side of (44) becomes a polynomial in ϕ and ψ. Setting the coefficients of this polynomial to be zero, we yield a system of algebraic equations in a1,b1,μ, and λ as follows:(55)ϕ4:3κa12+ιa14+3b12κ+2ιa12A12-2μA2+ιb14A12-2μA22=0,ϕ3:ϑa13+3ϑa1b12-6κμa1b1A12-2μA2-8ιμa1b13A12-2μA22=0,ϕ3ψ:3κa1b1+2ιa13b1+2ιa1b13A12-2μA2=0,ϕ2:a12+4κλa12+b12+κμ2a12A12-2μA2+λκb12A12-2μA2+2μκμb12-ϑb13A12-2μA22+4ιμ2b14A12-2μA23=0,ϕ2ψ:3ϑa12b1-5κμa12+ϑb13-12ιμa12b12-7κμb12A12-2μA2-4ιμb14A12-2μA22=0,ϕψ:2a1b1+5κλa1b1+2μ6κμa1b1-3ϑa1b12A12-2μA2+16ιμ2a1b13A12-2μA22=0,ψ:μκλa12+μb12+μ2a12+κλb12A12-2μA2+2μ2κμb12-ϑb13A12-2μA22+4ιμ3b14A12-2μA23=0,ψ0:κa12λ2=0.
Solving the above algebraic equations, we get the following solutions: when3κ(A12-2μA2)/ι<0, then(56)a1=0,b1=±-3κA12-2μA2ι,μ=∓ϑ5κ-3κA12-2μA2ι,ι=0.From the above expression of b1,μ, and ι, we can get a contradiction. In Case 3, (34) does not admit an exact solution.In order to get exact solutions of (2), we give the following conditions: (⋆) suppose α(t),β(t), and δ(t) satisfy the conditions that there exist a bounded open set G⊂R×Rd,q>0, and r>0 such that ut(t,x,z),u1/2(t,x,z)ux(t,x,z),u(t,x,z)ux(t,x,z), and uxxx(t,x,z) are uniformly bounded for all (t,x,z)∈G×Kq(r), continuous with respect to (t,x)∈G for all z∈Kq(r) and analytic with respect to z∈Kq(r), for all (t,x)∈G.Under condition (⋆) Theorem 1 implies that there exists u(t,x)∈(S)-1 such that(57)ut,x,z=Hut,x,which solves (34) for all (t,x,z)∈G×Kq(r),u(t,x) which solves (2). From (57), we can find that u(t,x) is the inverse Hermite transformation of u(t,x,z).Hence, by Theorems2 and 3, we yield the solutions of (2) as follows.Theorem 4.
Suppose that30ιϑ2+12ιϑ2(7ι-48κ)+(7ι-48κ)2/2κ(7ι-48κ)2<0 and 3κσ1/ιλκ2(7ι-48κ)2-3ικϑ2>0. Then (2) admits an exact solution(58)ut,x=3κ3λ2σ17ι-48κ2ιλκ27ι-48κ2-3ικϑ2ht,x♢2,where(59)ht,x=A1sinh♢⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2px+c∫0tαsds+A2cosh♢⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2px+c∫0tαsds±ϑι3κσ1ιλκ27ι-48κ2-3ικϑ2,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ1=A12-A22,A1 and A2 are two arbitrary constants.Theorem 5.
Suppose that3σ2/12ϑ2-100ι>0 and κ>0. Then (2) admits an exact solution(60)ut,x=75σ212ϑ2-100ιit,x♢2,where(61)it,x=A1sin♢⁡12κpx+c∫0tαsds+A2cos♢⁡12κpx+c∫0tαsds∓2ϑ3σ212ϑ2-100ι,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ2=A12+A22, and  A1 and A2 are two arbitrary constants.Since Wick versions of functions are usually difficult to evaluate, we will give some non-Wick versions of solutions of (2) in special cases.Letα(t)=f(t)+aW(t), where a is a constant. f(t) is an integral or bounded measurable function on R+.W(t) is Wiener white noise: that is, W(t)=B˙t,  Bt is a Brown motion. We have the Hermite transforms: α(t,z)=f(t)+aW~(t,z), where W~(t,z)=∑k=1∞∫0tηksdszk,z=(z1,z2,…)∈CN is parameter, and ηk(s) is defined in the second section.Usingexp♢{X}=exp{X} for nonrandom X,exp♢{Bt}=exp{Bt-1/2t2} and the definitions of sin⁡(ξ),cos⁡(ξ),sinh⁡(ξ), and cosh⁡(ξ), we have(62)sin♢⁡Bt=12exp♢⁡iBt-exp♢⁡-iBt=sin⁡Bt-12t2,cos♢⁡Bt=12exp♢⁡iBt+exp♢⁡-iBt=cos⁡Bt-12t2,sinh♢⁡Bt=12exp♢⁡Bt-exp♢⁡-Bt=sinh⁡Bt-12t2,cosh♢⁡Bt=12exp♢⁡Bt+exp♢⁡-Bt=cosh⁡Bt-12t2.Hence, by Theorems4 and 5, we yield the solutions of (2) in special cases as follows.Theorem 6.
Suppose that30ιϑ2+12ιϑ2(7ι-48κ)+(7ι-48κ)2/2κ(7ι-48κ)2<0 and 3κσ1/ιλκ2(7ι-48κ)2-3ικϑ2>0. Then (2) admits an exact solution(63)ut,x=3κ3λ2σ17ι-48κ2ιλκ27ι-48κ2-3ικϑ2mt,x2,where(64)mt,x=A1sinh⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2px+c∫0tfsds+caBt-12t2+A2cosh⁡-30ιϑ2+12ιϑ27ι-48κ+7ι-48κ22κ7ι-48κ2px+c∫0tfsds+caBt-12t2±ϑι3κσ1ιλκ27ι-48κ2-3ικϑ2,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ1=A12-A22, and  A1 and A2 are two arbitrary constants.Theorem 7.
Suppose that3σ2/12ϑ2-100ι>0 and κ>0. Then (2) admits an exact solution(65)ut,x=75σ212ϑ2-100ιnt,x2,where(66)nt,x=A1sin⁡12κpx+c∫0tfsds+caBt-12t2+A2cos⁡12κpx+c∫0tfsds+caBt-12t2∓2ϑ3σ212ϑ2-100ι,ϑ=2p/3c,ι=pβ/2c,κ=2p3δ/c,c≠0,σ2=A12+A22, and  A1 and A2 are two arbitrary constants.


## 5. Conclusion

On the earlier works given in the reference list, the authors considered the exact solutions of the Schamel-Korteweg-de Vries equation. But in this paper, we use the(G′/G,1/G)-expansion method to study the Wick-type stochastic Schamel-Korteweg-de Vries equation. We derive some new exact travelling wave solutions, which include hyperbolic and trigonometric solutions for the considered equations. The obtained solutions with free parameters may be important to expose most complex physical phenomena or to find new phenomena. It is shown in this paper that the (G′/G,1/G)-expansion method, with the help of symbolic computation like Maple or Mathematica, is direct, concise, and elementary.Compared to other methods, like the exp-function method [1], the extended (G′/G)-expansion method [6], the sine-cosine method, and the extended tanh method [7], the (G′/G,1/G)-expansion method is effective and powerful in finding exact solutions of many other nonlinear evolution equations in mathematical physics, applied mathematics, and engineering. The equations are very difficult to be solved by traditional methods.



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