# New Exact Solutions of Ion-Acoustic Wave Equations by (G′/G)-Expansion Method

Authors: Wafaa M. Taha; M. S. M. Noorani; I. Hashim
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

The (G′/G)-expansion  method is used to study ion-acoustic waves equations in plasma physic for the first time.  Many new exact traveling wave solutions of the Schamel equation,  Schamel-KdV (S-KdV), and the two-dimensional modified KP (Kadomtsev-Petviashvili) equation with square root nonlinearity are constructed. The traveling wave solutions obtained via this method are expressed by  hyperbolic functions, the trigonometric functions, and the rational functions. In addition to solitary waves solutions, a variety of special solutions like kink shaped, antikink shaped, and bell type solitary solutions are obtained when the choice of parameters is taken at special values. Two- and three-dimensional plots are drawn to illustrate the nature of solutions. Moreover, the solution obtained via this method is in good agreement with previously obtained solutions of other researchers.


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## Body

## 1. Introduction

The ion-acoustic solitary wave is one of the fundamental nonlinear waves phenomena appearing in fluid dynamics [1] and plasma physics [2]. To allowing for the trapping of some of the electrons on ion-acoustic waves, Schamel proposed a modified equation for ion-acoustic waves [3] given by
(1)ut+u1/2ux+δuxxx=0,
where u is the wave potential and δ is a constant, this equation describing the ion-acoustic wave, where the electrons do not behave isothermally during their passage of the wave in a cold-ion plasma. Then, combining the equations of Schamel and the KdV equation, one obtains the so-called one-dimensional form of the Schamel-KdV (S-KdV) equation equation [4, 5]:
(2)ut+(αu1/2+βu)ux+δuxxx=0,δβ≠0,
where β, α, and δ are constants. This equation is established in plasma physics in the study of ion acoustic solitons when electron trapping is present, and also it governs the electrostatic potential for a certain electron distribution in velocity space. Note that we obtain the KdV equation when α=0 and the Schamel equation when β=0 for δ=1. Due to the wide range of applications of (2), it is important to find new exact wave solutions of the Schamel-KdV (S-KdV) equation. Another equation arising in the study of ion-acoustic waves is the so-called modified Kadomtsev-Petviashvili (KP) equation given by [6]
(3)(ut+αu1/2ux+βuxxx)x+δuyy=0.Equation (3) was firstly derived by Chakraborty and Das [7]; the modified KP equation containing a square root nonlinearity is a very attractive model for the study of ion-acoustic waves in a multispecies plasma made up of non-isothermal electrons in plasma physics.In the literature, the KP equation is also known as the two-dimensional KdV equation [8].It has lately become more interesting to obtain exact analytical solutions to nonlinear partial differential equations such as the one arising from the ion-acoustic wave phenomena, by using appropriate techniques. Several important techniques have been developed such as the tanh-method [9, 10], sine-cosine method [11, 12], tanh-coth method [13], exp-function method [14], homogeneous-balance method [15, 16], Jacobi-elliptic function method [17, 18], and first-integral method [19, 20] to solve analytically nonlinear equations such as the above ion-acoustic wave equations.Moreover, in the standard tanh method developed by Malfliet in 1992 [21], the tanh is used as a new variable. Since all derivatives of a tanh are represented by tanh itself, the solution obtained by this method may be solitons in terms of sech2 or may be kinks in terms of tanh. We believe that the (G′/G)-expansion method is more efficient than the tanh method. Moreover, the tanh method may yield more than one soliton solution, a capability which the tanh method does not have. The sine-cosine method yields a solution in trigonometric form. The Exp-function method leads to both generalized solitary solution and periodic solutions. The homogeneous-balance method is a generalized tanh function method for many nonlinear PDEs. The first integral method, which is based on the ring theory of commutative algebra, was first proposed by Feng. There is no general theory telling us how to find its first integrals in a systematic way; so, a key idea of this approach to find the first integral is to utilize the division theorem. The traveling wave solutions expressed by the (G′/G)-expansion method, which was first proposed by Wang et al. [22], transform the given difficult problem into a set of simple problems which can be solved easily to get solutions in the forms of hyperbolic, trigonometric, and rational functions. The main merits of the (G′/G)-expansion method over the other methods are as follows. (i)
Higher-order nonlinear equations can be reduced to ODEs of order greater than 3.(ii)
There is no need to apply the initial and boundary conditions at the outset. The method yields a general solution with free parameters which can be identified by the above conditions.(iii)
The general solution obtained by the(G′/G)-expansion method is without approximation.(iv)
The solution procedure can be easily implemented in Mathematica or Maple.In fact, the(G′/G)-expansion method has been successfully applied to obtain exact solution for a variety of NLPDE [23–34].In this paper, the(G′/G)-expansion method is used to study ion-acoustic waves equations in plasma physic for the first time. We obtain many new exact traveling wave solutions for the Schamel equation, S-KdV, and the two-dimensional modified KP equation. The traveling wave solutions obtained via this method are expressed by hyperbolic functions, the trigonometric functions, and the rational functions. In addition to solitary waves solutions, a variety of special solutions like kink shaped, antikink shaped, and bell type solitary solutions are obtained when the choice of parameters is taken at special values. Two- and three-dimensional plots are drawn to illustrate the nature of solutions. Moreover, the solution obtained via this method is in good agreement with previously obtained solutions of other researchers.Our paper is organized as follows: in Section2, we present the summary of the (G′/G)-expansion method, and Section 3 describes the applications of the (G′/G)-expansion method for Schamel equation, S-KdV equation, and modified KP equation, and lastly, conclusions are given in Section 4.


## 2. Summary of the(G′/G)-Expansion Method

In this section, we describe the(G′/G)-expansion method for finding traveling wave solutions of NLPDE. Suppose that a nonlinear partial differential equation in two independent variables, x and t, is given by
(4)p(u,ut,ux,uxt,utt,uxx,…)=0,
where u=u(x,t) is an unknown function, P is a polynomial in u=u(x,t) and its various partial derivatives, in which highest order derivatives and nonlinear terms are involved.The summary of the(G′/G)-expansion method can be presented in the following six steps.Step 1.
To find the traveling wave solutions of (4), we introduce the wave variable:
(5)u(x,t)=u(ζ),ζ=(x-ct),
where the constant c is generally termed the wave velocity. Substituting (5) into (4), we obtain the following ordinary differential equations (ODE) in ζ (which illustrates a principal advantage of a traveling wave solution; i.e., a PDE is reduced to an ODE):
(6)p(u,cu′,u′,cu′′,c2u′′,u′′,…)=0.Step 2.
If necessary, we integrate (6) as many times as possible and set the constants of integration to be zero for simplicity.Step 3.
Suppose that the solution of nonlinear partial differential equation can be expressed by a polynomial in(G′/G) as
(7)u(ζ)=∑i=0m‍ai(G′G)i,
where G=G(ζ) satisfies the second-order linear ordinary differential equation
(8)G′′(ζ)+λG′(ζ)+μG(ζ)=0,
where G′=dG/dζ, G′′=d2G/dζ2, and ai, λ, and μ are real constants with am≠0. Here, the prime denotes the derivative with respect to ζ. Using the general solutions of (8), we have(9)
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.The above results can be written in simplified forms as(10)(G′G)={-λ2+λ2-4μ2tanh{λ2-4μ2ζ},λ2-4μ>0,-λ2+4μ-λ22tan{4μ-λ22ζ},λ2-4μ<0,(c2c1+c2ζ)-λ2,λ2-4μ=0.Step 4.
The positive integerm can be accomplished by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (6) as follows: if we define the degree of u(ζ) as D[u(ζ)]=m, then the degree of other expressions is defined by
(11)D[dqudζq]=m+q,D[ur(dqudζq)s]=mr+s(q+m).
Therefore, we can get the value ofm in (7).Step 5.
Substituting (7) into (6), useing general solutions of (8), and collecting all terms with the same order of (G′/G) together, then setting each coefficient of this polynomial to zero yields a set of algebraic equations for ai, c, λ, and μ.Step 6.
Substituteai, c, λ, and μ obtained in Step 5 and the general solutions of (8) into (7). Next, depending on the sign of the discriminant A=λ2-4μ, we can obtain the explicit solutions of (4) immediately.


## 3. Applications of the(G′/G)-Expansion Method

### 3.1. Schamel Equation

In order to find the solitary wave solution of (1), we use the transformations
(12)u(x,t)=v2(x,t),v(x,t)=v(ζ),ζ=kx-ct.
Then, (1) becomes
(13)-cvv′+kv2v′+δk3(vv′′′+3v′v′′)=0.
Integrating (13) with respect to ζ and setting the integration constant equal to zero, we have
(14)-c2v2+k3v3+k3δ(v′)2+k3δvv′′=0.
According to the previous steps, using the balancing procedure between v3 with vv′′ in (14), we get 3m=2m+2 so that m=2. Now, assume that (14) has the following solution:
(15)v(ζ)=a0+a1(G′G)+a2(G′G)2,a2≠0,
where a0, a1, and a2 are unknown constants to be determined later. Substituting (15) along with (8) into (14) and collecting all terms with the same order of (G′/G), the left hand side of (14) is converted into a polynomial in (G′/G). Equating each coefficient of the resulting polynomials to zero yields a set of algebraic equations for a0, a1, a2, δ, λ, c, k, and μ as follows:
(16)(G′G)6:13ka23+10k3δa22=0,(G′G)5:18k3δa22λ+12k3δa1a2+ka1a22=0,(G′G)4:6k3δa0a2+16k3δa22μ+8k3δa22λ2-12ca22+21k3δa12+ka0a22+ka12a2=0,(G′G)3:14k3δa22λμ+5k3δa12λ+18k3δa1a2μ+9k3δa1λ2a2+2ka0a1a2-ca1a2+13ka13+2k3δa0a1+10k3δa0a2λ=0,(G′G)2:2k3δa12λ2+4k3δa12μ+8k3δa0a2μ-12ca12-ca0a2+15k3δa1a2λμ+ka0a12+4k3δa0a2λ2+6k3δa22μ2+3kδa0a1λ+ka02=0,(G′G)1:2k3a0a1μ+k3δa0a1λ2+6k3δa0a2λμ+ka02-ca0a1+6k3a1a2δμ2+3k3δλμa12=0,(G′G)0:k3δa12μ2-12ca02+13ka03+k3δλμa0a1+2k3δμ2a0a2=0.
On solving the above set of algebraic equations by Maple, we have
(17)a0=-30μδk2,a1=-30k2δλ,a2=-30k2δ,c=δk3(4λ2-16μ).
Now, (15) becomes
(18)v(ζ)=-30μδk2-30k2δλ(G′G)-30k2δ(G′G)2.
Substituting the general solution of (8) into (18), we obtain the three types of traveling wave solutions depending on the sign of A=λ2-4μ.IfA>0, we have the following general hyperbolic traveling wave solutions of (1):(19)v(x,t)=-30μδk2-30k2δλ×[-λ2+A2ffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30k2δ[-λ2+A2ffffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2,where c1 and c2 are arbitrary constants.IfA<0, we have the following general trigonometric function solutions of (1):(20)v(x,t)=-30μδk2-30k2δλ×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{A/2}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30k2δ[-λ2+A2ffffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.IfA=0, we have the following general rational function solutions of (1):
(21)v(x,t)=-30μδk2-30k2δλ[-λ2+(c2c1+c2ζ)]-30k2δ[-λ2+(c2c1+c2ζ)]2,
where ζ=kx-δk3(4λ2-16μ)t.Writingu(x,t)=v2(x,t) and setting c2=μ=0 and λ=2 in (19), we reproduce the result of Khater and Hassan [35] (see their Equation (4.7)),
(22)u(x,t)=4(900k4δ2sech4{ζ}),
where ζ=kx-16δk3t.Note that Khater and Hassan [35] obtained only hyperbolic solutions, but in this work, we found two additional types of solutions, that is, trigonometric and rational solutions.


### 3.2. S-KdV Equation

To find the general exact solutions of (2), we first write u(x,t)=v2(x,t) to transform (2) into
(23)vvt+(αv+βv2)vx+δvvxxx=0.
Assume the traveling wave solution of (23) in the form
(24)v(x,t)=V(ζ),ζ=k(x-ct).
Hence, (23) becomes
(25)-cVV′+(αV2+βV3)V′+k2δ(VV′′′+3V′V′′)=0.
Suppose that the solution of (25) can be expressed by a polynomial in (G′/G) as
(26)V(ζ)=∑i=0m‍ai(G′G)i,ai≠0
and G(ζ) satisfies (8). The homogeneous balance between the highest order derivative VV′′′ and the nonlinear term V3V′ appearing in (25) yields m=1, and hence, we take the following formal solution:
(27)V(ζ)=a0+a1(G′G),
where the positive integers a0 and a1 are to be determined later. Substituting (27) along with (8) into (25), collecting all the terms with the same power of (G′/G), and equating each coefficient to zero yield a set of simultaneous algebraic equations for a0, a1, c, k, α, β, and δ as follows:
(28)(G′G)5:-βa14-12δk2a12=0,(G′G)4:-3βa13a0-βa14λ-27δk2a12λ-αa13-6δk2a0a1=0,(G′G)3:-3βa13λa0-19δk2a12λ2-12δk2a0a1λ-20δk2a12μ-αa13λ-βa14μ+ca12-2αa12a0-3βa12a02=0,(G′G)2:ca0a1-3βa12λa02-2αa12λa0+ca12λ-26δk2a12λμ-8δk2a0a1μ-7δk2a0a1λ2-3βa13μa0-βa1a03-αa1a02-αa13μ=0,(G′G)1:-αλa1a02+ca0a1λ-8δk2a12μ2-7δμk2a12λ2-3βμa12a02+ca12μ-βλa1a03-8δλμk2a0a1-2αμa12a0=0,(G′G)0:cμa0a1-αμa1a02-βμa1a03-3δλk2a12μ2-δμλ2k2a0a1-2δμ2k2a0a1=0.
The above system admits the following sets of solutions:
(29)a0=0,a1=4α5βλ,μ=0,c=-16α275β,k=±21/-75δβαλ,(30)a0=-4α5β,a1=-4α5βλ,μ=0,c=-16α275β,k=±21/-75δβαλ,(31)a0=5βa1λ-4α10β,μ=25β2a12λ2-16α2100β2a12,c=-16α275β,k=±β-12δa1.
Now, substituting (29)-(30) into (27) gives, respectively,
(32)V1(ζ)=4α5βλ(G′G),V2(ζ)=-4α5β-4α5βλ(G′G),V3(ζ)=5βa1λ-4α10β+a1(G′G).
When substituting the general solutions (9) into (32), we obtain the following three types of traveling wave solutions:Case
A
>
0: (hyperbolic type)
(33)v1(x,t)=-2α5β+2α5β(c1sinh{(λ/2)ζ}+c2cosh{(λ/2)ζ}c1cosh{(λ/2)ζ}+c2sinh{(λ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,v2(x,t)=-2α5β-2α5β(c1sinh{(λ/2)ζ}+c2cosh{(λ/2)ζ}c1cosh{(λ/2)ζ}+c2sinh{(λ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,v3(x,t)=-2α5β+2α5β×(c1sinh{(2α/5βa1)ζ}+c2cosh{(2α/5βa1)ζ}c1cosh{(2α/5βa1)ζ}+c2sinh{(2α/5βa1)ζ}),ζ=±β-12δa1x+16α275βt.If we setc2=0 and write u(x,t)=v2(x,t), then the above solutions can be written as
(34)u1(x,t)=4α225β2[-1±tanh(α5-3βδ[x+16α275βt])]2,(35)u2(x,t)=4α225β2[1±tanh(α5-3βδ[x+16α275βt])]2,(36)u3(x,t)=4α225β2[-1±2tanh(α5-3βδ[x+16α275βt])]2.
Note that (35) is exactly the same solution of Khater and Hassan [35] as given in their first equation of (3.9) with ζ0=0. Similarly we can obtain the second solution of (36) in Hassan [5] if we set c1=0 in our solution (35). The solution (35) represents kink shaped solitary and antikink shaped solitary solutions (depending upon the choice of sign) which are shown graphically in Figure 1 for the case c1=1:(a) 2D profile of (35): kink shaped solitary (u+, blue line), anti-kink shaped solitary (u−, black line). (b) Corresponding 3D plots when + sign is taken and when −ve sign is taken, with α=1, β=-1, and δ=1.
(a)
(b)Case
A
<
0: (trigonometric type)
(37)V1(x,t)=-2α5β+2αi5β(-c1sin{(iλ/2)ζ}+c2cos{(iλ/2)ζ}c1cos{(iλ/2)ζ}+c2sin{(iλ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,V2(x,t)=-2α5β-2iα5β×(-c1sin{(iλ/2)ζ}+c2cos{(iλ/2)ζ}c1cos{(iλ/2)ζ}+c2sin{(iλ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,V3(x,t)=-2α5β+2iα5β×(-c1sin{(2iα/5βa1)ζ}+c2cos{(2iα/5βa1)ζ}c1cos{(2iα/5βa1)ζ}+c2sin{(2iα/5βa1)ζ}),ζ=±β-12δa1x+16α275βt.But ifc2=0 and u(x,t)=v2(x,t), then trigonometric type solution becomes
(38)u1(x,t)=4α225β2[-1±itan(α53βδ[x+16α275βt])]2,u2(x,t)=4α225β2[1±itan(α53βδ[x+16α275βt])]2,u3(x,t)=4α225β2[-1±2itan(α53βδ[x+16α275βt])]2.Case
A
=
0: (rational type)
(39)u1(x,t)=4α225β2[-1+2(c2c1+c2ζ)]2,u2(x,t)=4α225β2[1+2(c2c1+c2ζ)]2,u3(x,t)=[2α5β+a1(c2c1+c2ζ)]2.As mentioned before, the(G′/G)-expansion method gives more general types of solutions than that found by Khater and Hassan [35] and Hassan [5].


### 3.3. The Modified Two-Dimensional KP (Kadomtsev-Petviashvili) Equation

The modified KP equation (3) containing a square root nonlinearity is a very attractive model for the study of ion-acoustic waves in plasma physic [8]. We will obtain more general exact solutions of the modified KP equation. In order to find the traveling wave solution of (3), we let
(40)v(x,y,t)=v(ζ),ζ=(x+ky-ct).
Now, taking u(x,y,t)=v2(x,y,t), (3) becomes
(41)(-c+δk2)vv′′+(-c+δk2)v′2+αv2v′′+2αvv′2+βvv′′′′+4βv′v′′′+3β(v′′)2=0,
where k, c, β, δ, and α are constants and the prime denotes differentiation with respect to ζ. Integrating (41) with respect to ζ and setting the integration constant equal to zero, we obtain
(42)(-c+δk2)vv′+αv2v′+3βv′v′′+βvv′′′=0.
Balancing v2v′ with vv′′′ gives m=2. Therefore, we can write the solution of (42) in the form
(43)v(ζ)=a0+a1(G′G)+a2(G′G)2,
where a0, a1, and a2 are constants to be determined later. Substituting (43) along with (8) into (42) and collecting all terms with the same order of (G′/G), the left hand sides of (42) are converted into a polynomial in (G′/G). Setting each coefficient of each polynomial to zero, we derive a set of algebraic equations for a0, a1, a2, δ, λ, α, β, c, k, and μ as follows:
(44)(G′G)7:-60βa22-2αa23=0,(G′G)6:-2αa23λ-5αa1a22-60βa1a2-150a22λ=0,(G′G)5:-12βa12-5αa1a22λ-124βa22λ2-2k2δa22-24βa0a2-4αa0a2-144βa1a2λ+2ca22-4αa12a2-2αa23μ=0,(G′G)4:-196βa22λμ-6αa0a1a2-111βa1a2λ2-4αa0a22λ-32βa22λ3-αa13-2k2δa22λ+2ca22λ-54βa0a2λ-5αa1a22μ-4αa12a2λ+3ca1a2-6βa0a1-3k2δa1a2=0,(G′G)3:2ca0a2-76βa22μ2-4αa0a22μ-k2δa12-27βa1λ3a2-74βa22λ2μ+ca12-6αa0a1a2λ+3ca1a2λ-40βa0a2μ-38βa0a2λ2-19βa12λ2-2αa0a12-4αa12a2μ-168βa1a2λμ-3k2δa1a2λ+2ca22μ-2αa02a2-20βa12μ-αa13λ-12βa0a1λ-2k2δa0a2=0,(G′G)2:2ca0a2λ-4βa12λ3-8βa0a1μ-3k2δa1a2μ-αa13μ-52βa0a2λμ+ca0a1+3ca1a2μ-6αa0a1a2μ-k2δa12λ-60βa1a2μ2-αa02a1-2αa0a12λ-2k3δa0a2λ-2αa02a2λ-7βa0a1λ2-26βa12λμ+ca12λ-57βa1λ2a2μ-8βa0a2λ3-k2δa0a1-54βa22λμ2=0,(G′G)1:-2αa02μ+2ca0a2μ-αa02a1λ-2k2δa0a2μ-8βa12μ2+ca0a1λ-16βa0a2μ2-8βa0a1λμ-βa0a1λ3-14βa0a2λ2μ-36βλμ2a1a2-12βa22μ3-k2δa0a1λ-k2δa12μ+ca12μ-2αa0a12μ-7βμλ2a12=0,(G′G)0:-6βμ3a1a2+ca0a1μ-αμa02a1-3βλμ2a12-βμλ2a0a1-6βλμ2a0a2-2βa0a1μ2-k2δμa0a1=0.Solving this system by Maple gives two sets of solutions.Case 1.
We have(45)a0=-30βμα,a1=-30βλα,a2=-30βα,c=-16βμ+4βλ2+k2δ.
Substituting the above case and the general solution (8) into (43) and according to (42), we obtain three types of traveling wave solutions of (3) as follows.IfA>0, we have the hyperbolic type(46)v(x,y,t)=-30βμα-30βλα×[-λ2+A2fffaf×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30βα[-λ2+A2fffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2.In particular, if c1≠0, c2=0, λ>0, and μ=0, then u(x,y,t) becomes
(47)u(x,y,t)=225β2λ44α2sech4{λ2ζ},ζ=x+ky-(4βλ2+k2δ)t.
If A<0, we have the trigonometric type
(48)v(x,y,t)=-30βμα-30βλα×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30βα[-λ2+A2fffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.
So, the traveling wave solutions of (3) in this case are
(49)u(x,y,t)=225β2λ44α2sec4{-λ22ζ},ζ=x+ky-(4βλ2+k2δ)t.Case 2.
We have(50)a0=-5β(λ2+2μ)α,a1=-30βλα,a2=-30βα,c=16βμ-4βλ2+k2δ.
If A>0, we have the hyperbolic type(51)v(x,y,t)=-5β(λ2+2μ)α-30βλα×[-λ2+A2fff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30βα[-λ2+A2fffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2.However, if c1≠0, c2=0, λ>0, and μ=0, then u(x,y,t) becomes
(52)u(x,y,t)=25β2λ44α2[2-3sech2{λ2ζ}]2,ζ=x+ky-(-4βλ2+k2δ)t.
If A<0, we have the trigonometric type
(53)v(x,y,t)=-5β(λ2+2μ)α-30βλα×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30βα[-λ2+A2fffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.
In the particular case when c1≠0, c2=0, λ>0, and μ=0, u(x,y,t) becomes
(54)u(x,y,t)=25β2λ44α2[2-3sec2{-λ22ζ}]2,ζ=x+ky-(-4βλ2+k2δ)t.
If A=0, we have the rational type
(55)v(x,y,t)=-5β(λ2+2μ)α-30k2δλ[-λ2+(c2c1+c2ζ)]-30k2δ[-λ2+(c2c1+c2ζ)]2,
where ζ=x+ky-(16βμ-4βλ2+k2δ)t.We remark that our results in (47) and (52), when c1≠0, c2=0, λ>0, and μ=0, match those of Khater et al. [6] (2.19) when a=1. In Figure 2, we plot the bell type solitary for 2D profile and the corresponding 3D plot of (47) for parameters α=2, β=0.4 and δ=0.1, λ=1, k=1, and t=0.5.Bell type solitary (a) 2D profile and (b) corresponding 3D plot of (47) for parameters α=2, β=0.4 and δ=0.1, λ=1, k=1, and t=0.5.
(a)
(b)


## 3.1. Schamel Equation

In order to find the solitary wave solution of (1), we use the transformations
(12)u(x,t)=v2(x,t),v(x,t)=v(ζ),ζ=kx-ct.
Then, (1) becomes
(13)-cvv′+kv2v′+δk3(vv′′′+3v′v′′)=0.
Integrating (13) with respect to ζ and setting the integration constant equal to zero, we have
(14)-c2v2+k3v3+k3δ(v′)2+k3δvv′′=0.
According to the previous steps, using the balancing procedure between v3 with vv′′ in (14), we get 3m=2m+2 so that m=2. Now, assume that (14) has the following solution:
(15)v(ζ)=a0+a1(G′G)+a2(G′G)2,a2≠0,
where a0, a1, and a2 are unknown constants to be determined later. Substituting (15) along with (8) into (14) and collecting all terms with the same order of (G′/G), the left hand side of (14) is converted into a polynomial in (G′/G). Equating each coefficient of the resulting polynomials to zero yields a set of algebraic equations for a0, a1, a2, δ, λ, c, k, and μ as follows:
(16)(G′G)6:13ka23+10k3δa22=0,(G′G)5:18k3δa22λ+12k3δa1a2+ka1a22=0,(G′G)4:6k3δa0a2+16k3δa22μ+8k3δa22λ2-12ca22+21k3δa12+ka0a22+ka12a2=0,(G′G)3:14k3δa22λμ+5k3δa12λ+18k3δa1a2μ+9k3δa1λ2a2+2ka0a1a2-ca1a2+13ka13+2k3δa0a1+10k3δa0a2λ=0,(G′G)2:2k3δa12λ2+4k3δa12μ+8k3δa0a2μ-12ca12-ca0a2+15k3δa1a2λμ+ka0a12+4k3δa0a2λ2+6k3δa22μ2+3kδa0a1λ+ka02=0,(G′G)1:2k3a0a1μ+k3δa0a1λ2+6k3δa0a2λμ+ka02-ca0a1+6k3a1a2δμ2+3k3δλμa12=0,(G′G)0:k3δa12μ2-12ca02+13ka03+k3δλμa0a1+2k3δμ2a0a2=0.
On solving the above set of algebraic equations by Maple, we have
(17)a0=-30μδk2,a1=-30k2δλ,a2=-30k2δ,c=δk3(4λ2-16μ).
Now, (15) becomes
(18)v(ζ)=-30μδk2-30k2δλ(G′G)-30k2δ(G′G)2.
Substituting the general solution of (8) into (18), we obtain the three types of traveling wave solutions depending on the sign of A=λ2-4μ.IfA>0, we have the following general hyperbolic traveling wave solutions of (1):(19)v(x,t)=-30μδk2-30k2δλ×[-λ2+A2ffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30k2δ[-λ2+A2ffffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2,where c1 and c2 are arbitrary constants.IfA<0, we have the following general trigonometric function solutions of (1):(20)v(x,t)=-30μδk2-30k2δλ×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{A/2}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30k2δ[-λ2+A2ffffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.IfA=0, we have the following general rational function solutions of (1):
(21)v(x,t)=-30μδk2-30k2δλ[-λ2+(c2c1+c2ζ)]-30k2δ[-λ2+(c2c1+c2ζ)]2,
where ζ=kx-δk3(4λ2-16μ)t.Writingu(x,t)=v2(x,t) and setting c2=μ=0 and λ=2 in (19), we reproduce the result of Khater and Hassan [35] (see their Equation (4.7)),
(22)u(x,t)=4(900k4δ2sech4{ζ}),
where ζ=kx-16δk3t.Note that Khater and Hassan [35] obtained only hyperbolic solutions, but in this work, we found two additional types of solutions, that is, trigonometric and rational solutions.


## 3.2. S-KdV Equation

To find the general exact solutions of (2), we first write u(x,t)=v2(x,t) to transform (2) into
(23)vvt+(αv+βv2)vx+δvvxxx=0.
Assume the traveling wave solution of (23) in the form
(24)v(x,t)=V(ζ),ζ=k(x-ct).
Hence, (23) becomes
(25)-cVV′+(αV2+βV3)V′+k2δ(VV′′′+3V′V′′)=0.
Suppose that the solution of (25) can be expressed by a polynomial in (G′/G) as
(26)V(ζ)=∑i=0m‍ai(G′G)i,ai≠0
and G(ζ) satisfies (8). The homogeneous balance between the highest order derivative VV′′′ and the nonlinear term V3V′ appearing in (25) yields m=1, and hence, we take the following formal solution:
(27)V(ζ)=a0+a1(G′G),
where the positive integers a0 and a1 are to be determined later. Substituting (27) along with (8) into (25), collecting all the terms with the same power of (G′/G), and equating each coefficient to zero yield a set of simultaneous algebraic equations for a0, a1, c, k, α, β, and δ as follows:
(28)(G′G)5:-βa14-12δk2a12=0,(G′G)4:-3βa13a0-βa14λ-27δk2a12λ-αa13-6δk2a0a1=0,(G′G)3:-3βa13λa0-19δk2a12λ2-12δk2a0a1λ-20δk2a12μ-αa13λ-βa14μ+ca12-2αa12a0-3βa12a02=0,(G′G)2:ca0a1-3βa12λa02-2αa12λa0+ca12λ-26δk2a12λμ-8δk2a0a1μ-7δk2a0a1λ2-3βa13μa0-βa1a03-αa1a02-αa13μ=0,(G′G)1:-αλa1a02+ca0a1λ-8δk2a12μ2-7δμk2a12λ2-3βμa12a02+ca12μ-βλa1a03-8δλμk2a0a1-2αμa12a0=0,(G′G)0:cμa0a1-αμa1a02-βμa1a03-3δλk2a12μ2-δμλ2k2a0a1-2δμ2k2a0a1=0.
The above system admits the following sets of solutions:
(29)a0=0,a1=4α5βλ,μ=0,c=-16α275β,k=±21/-75δβαλ,(30)a0=-4α5β,a1=-4α5βλ,μ=0,c=-16α275β,k=±21/-75δβαλ,(31)a0=5βa1λ-4α10β,μ=25β2a12λ2-16α2100β2a12,c=-16α275β,k=±β-12δa1.
Now, substituting (29)-(30) into (27) gives, respectively,
(32)V1(ζ)=4α5βλ(G′G),V2(ζ)=-4α5β-4α5βλ(G′G),V3(ζ)=5βa1λ-4α10β+a1(G′G).
When substituting the general solutions (9) into (32), we obtain the following three types of traveling wave solutions:Case
A
>
0: (hyperbolic type)
(33)v1(x,t)=-2α5β+2α5β(c1sinh{(λ/2)ζ}+c2cosh{(λ/2)ζ}c1cosh{(λ/2)ζ}+c2sinh{(λ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,v2(x,t)=-2α5β-2α5β(c1sinh{(λ/2)ζ}+c2cosh{(λ/2)ζ}c1cosh{(λ/2)ζ}+c2sinh{(λ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,v3(x,t)=-2α5β+2α5β×(c1sinh{(2α/5βa1)ζ}+c2cosh{(2α/5βa1)ζ}c1cosh{(2α/5βa1)ζ}+c2sinh{(2α/5βa1)ζ}),ζ=±β-12δa1x+16α275βt.If we setc2=0 and write u(x,t)=v2(x,t), then the above solutions can be written as
(34)u1(x,t)=4α225β2[-1±tanh(α5-3βδ[x+16α275βt])]2,(35)u2(x,t)=4α225β2[1±tanh(α5-3βδ[x+16α275βt])]2,(36)u3(x,t)=4α225β2[-1±2tanh(α5-3βδ[x+16α275βt])]2.
Note that (35) is exactly the same solution of Khater and Hassan [35] as given in their first equation of (3.9) with ζ0=0. Similarly we can obtain the second solution of (36) in Hassan [5] if we set c1=0 in our solution (35). The solution (35) represents kink shaped solitary and antikink shaped solitary solutions (depending upon the choice of sign) which are shown graphically in Figure 1 for the case c1=1:(a) 2D profile of (35): kink shaped solitary (u+, blue line), anti-kink shaped solitary (u−, black line). (b) Corresponding 3D plots when + sign is taken and when −ve sign is taken, with α=1, β=-1, and δ=1.
(a)
(b)Case
A
<
0: (trigonometric type)
(37)V1(x,t)=-2α5β+2αi5β(-c1sin{(iλ/2)ζ}+c2cos{(iλ/2)ζ}c1cos{(iλ/2)ζ}+c2sin{(iλ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,V2(x,t)=-2α5β-2iα5β×(-c1sin{(iλ/2)ζ}+c2cos{(iλ/2)ζ}c1cos{(iλ/2)ζ}+c2sin{(iλ/2)ζ}),ζ=±21/-75δβαλx+16α275βt,V3(x,t)=-2α5β+2iα5β×(-c1sin{(2iα/5βa1)ζ}+c2cos{(2iα/5βa1)ζ}c1cos{(2iα/5βa1)ζ}+c2sin{(2iα/5βa1)ζ}),ζ=±β-12δa1x+16α275βt.But ifc2=0 and u(x,t)=v2(x,t), then trigonometric type solution becomes
(38)u1(x,t)=4α225β2[-1±itan(α53βδ[x+16α275βt])]2,u2(x,t)=4α225β2[1±itan(α53βδ[x+16α275βt])]2,u3(x,t)=4α225β2[-1±2itan(α53βδ[x+16α275βt])]2.Case
A
=
0: (rational type)
(39)u1(x,t)=4α225β2[-1+2(c2c1+c2ζ)]2,u2(x,t)=4α225β2[1+2(c2c1+c2ζ)]2,u3(x,t)=[2α5β+a1(c2c1+c2ζ)]2.As mentioned before, the(G′/G)-expansion method gives more general types of solutions than that found by Khater and Hassan [35] and Hassan [5].


## 3.3. The Modified Two-Dimensional KP (Kadomtsev-Petviashvili) Equation

The modified KP equation (3) containing a square root nonlinearity is a very attractive model for the study of ion-acoustic waves in plasma physic [8]. We will obtain more general exact solutions of the modified KP equation. In order to find the traveling wave solution of (3), we let
(40)v(x,y,t)=v(ζ),ζ=(x+ky-ct).
Now, taking u(x,y,t)=v2(x,y,t), (3) becomes
(41)(-c+δk2)vv′′+(-c+δk2)v′2+αv2v′′+2αvv′2+βvv′′′′+4βv′v′′′+3β(v′′)2=0,
where k, c, β, δ, and α are constants and the prime denotes differentiation with respect to ζ. Integrating (41) with respect to ζ and setting the integration constant equal to zero, we obtain
(42)(-c+δk2)vv′+αv2v′+3βv′v′′+βvv′′′=0.
Balancing v2v′ with vv′′′ gives m=2. Therefore, we can write the solution of (42) in the form
(43)v(ζ)=a0+a1(G′G)+a2(G′G)2,
where a0, a1, and a2 are constants to be determined later. Substituting (43) along with (8) into (42) and collecting all terms with the same order of (G′/G), the left hand sides of (42) are converted into a polynomial in (G′/G). Setting each coefficient of each polynomial to zero, we derive a set of algebraic equations for a0, a1, a2, δ, λ, α, β, c, k, and μ as follows:
(44)(G′G)7:-60βa22-2αa23=0,(G′G)6:-2αa23λ-5αa1a22-60βa1a2-150a22λ=0,(G′G)5:-12βa12-5αa1a22λ-124βa22λ2-2k2δa22-24βa0a2-4αa0a2-144βa1a2λ+2ca22-4αa12a2-2αa23μ=0,(G′G)4:-196βa22λμ-6αa0a1a2-111βa1a2λ2-4αa0a22λ-32βa22λ3-αa13-2k2δa22λ+2ca22λ-54βa0a2λ-5αa1a22μ-4αa12a2λ+3ca1a2-6βa0a1-3k2δa1a2=0,(G′G)3:2ca0a2-76βa22μ2-4αa0a22μ-k2δa12-27βa1λ3a2-74βa22λ2μ+ca12-6αa0a1a2λ+3ca1a2λ-40βa0a2μ-38βa0a2λ2-19βa12λ2-2αa0a12-4αa12a2μ-168βa1a2λμ-3k2δa1a2λ+2ca22μ-2αa02a2-20βa12μ-αa13λ-12βa0a1λ-2k2δa0a2=0,(G′G)2:2ca0a2λ-4βa12λ3-8βa0a1μ-3k2δa1a2μ-αa13μ-52βa0a2λμ+ca0a1+3ca1a2μ-6αa0a1a2μ-k2δa12λ-60βa1a2μ2-αa02a1-2αa0a12λ-2k3δa0a2λ-2αa02a2λ-7βa0a1λ2-26βa12λμ+ca12λ-57βa1λ2a2μ-8βa0a2λ3-k2δa0a1-54βa22λμ2=0,(G′G)1:-2αa02μ+2ca0a2μ-αa02a1λ-2k2δa0a2μ-8βa12μ2+ca0a1λ-16βa0a2μ2-8βa0a1λμ-βa0a1λ3-14βa0a2λ2μ-36βλμ2a1a2-12βa22μ3-k2δa0a1λ-k2δa12μ+ca12μ-2αa0a12μ-7βμλ2a12=0,(G′G)0:-6βμ3a1a2+ca0a1μ-αμa02a1-3βλμ2a12-βμλ2a0a1-6βλμ2a0a2-2βa0a1μ2-k2δμa0a1=0.Solving this system by Maple gives two sets of solutions.Case 1.
We have(45)a0=-30βμα,a1=-30βλα,a2=-30βα,c=-16βμ+4βλ2+k2δ.
Substituting the above case and the general solution (8) into (43) and according to (42), we obtain three types of traveling wave solutions of (3) as follows.IfA>0, we have the hyperbolic type(46)v(x,y,t)=-30βμα-30βλα×[-λ2+A2fffaf×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30βα[-λ2+A2fffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2.In particular, if c1≠0, c2=0, λ>0, and μ=0, then u(x,y,t) becomes
(47)u(x,y,t)=225β2λ44α2sech4{λ2ζ},ζ=x+ky-(4βλ2+k2δ)t.
If A<0, we have the trigonometric type
(48)v(x,y,t)=-30βμα-30βλα×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30βα[-λ2+A2fffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.
So, the traveling wave solutions of (3) in this case are
(49)u(x,y,t)=225β2λ44α2sec4{-λ22ζ},ζ=x+ky-(4βλ2+k2δ)t.Case 2.
We have(50)a0=-5β(λ2+2μ)α,a1=-30βλα,a2=-30βα,c=16βμ-4βλ2+k2δ.
If A>0, we have the hyperbolic type(51)v(x,y,t)=-5β(λ2+2μ)α-30βλα×[-λ2+A2fff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]-30βα[-λ2+A2fffffff×(c1sinh{(A/2)ζ}+c2cosh{(A/2)ζ}c1cosh{(A/2)ζ}+c2sinh{(A/2)ζ})]2.However, if c1≠0, c2=0, λ>0, and μ=0, then u(x,y,t) becomes
(52)u(x,y,t)=25β2λ44α2[2-3sech2{λ2ζ}]2,ζ=x+ky-(-4βλ2+k2δ)t.
If A<0, we have the trigonometric type
(53)v(x,y,t)=-5β(λ2+2μ)α-30βλα×[-λ2+A2ffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]-30βα[-λ2+A2fffffff×(-c1sin{(A/2)ζ}+c2cos{(A/2)ζ}c1cos{(A/2)ζ}+c2sin{(A/2)ζ})]2.
In the particular case when c1≠0, c2=0, λ>0, and μ=0, u(x,y,t) becomes
(54)u(x,y,t)=25β2λ44α2[2-3sec2{-λ22ζ}]2,ζ=x+ky-(-4βλ2+k2δ)t.
If A=0, we have the rational type
(55)v(x,y,t)=-5β(λ2+2μ)α-30k2δλ[-λ2+(c2c1+c2ζ)]-30k2δ[-λ2+(c2c1+c2ζ)]2,
where ζ=x+ky-(16βμ-4βλ2+k2δ)t.We remark that our results in (47) and (52), when c1≠0, c2=0, λ>0, and μ=0, match those of Khater et al. [6] (2.19) when a=1. In Figure 2, we plot the bell type solitary for 2D profile and the corresponding 3D plot of (47) for parameters α=2, β=0.4 and δ=0.1, λ=1, k=1, and t=0.5.Bell type solitary (a) 2D profile and (b) corresponding 3D plot of (47) for parameters α=2, β=0.4 and δ=0.1, λ=1, k=1, and t=0.5.
(a)
(b)


## 4. Conclusion

The(G′/G)-expansion was applied to solve the model of ion-acoustic waves in plasma physics where these equations each contain a square root nonlinearity. The (G′/G)-expansion has been successfully used to obtain some exact traveling wave solutions of the Schamel equation, Schamel-KdV (S-KdV) equation, and modified KP (Kadomtsev-Petviashvili) equation. Moreover, the reliability of the method and the reduction in the size of computational domain give this method a wider applicability. This fact shows that our algorithm is effective and more powerful for NLPDE. In all the general solutions (22), (35), (47), and (52), we have the additional arbitrary constants c1, c2, λ, and μ. We note that the special case c1≠0, c2=0, λ>0, and μ=0 reproduced the results of Khater and Hassan [35], Hassan [5] and Khater et al. [6]. Many different new forms of traveling wave solutions such as the kink shaped, antikink shaped, and bell type solitary solutions were obtained. Finally, numerical simulations are given to complete the study.Moreover, all the methods have some limitations in their applications. In fact, there is no unified method that can be used to handle all types of nonlinear partial differential equations (NLPDE). Certainly, each investigator in the field of differential equations has his own experience to choose the method depending on form of the nonlinear differential equation and the pole of its solution. So, the limitations of the(G′/G)-expansion method used a rise only when the equation has the traveling wave and becomes powerful in finding traveling wave solutions of NLPDE only.In our future works, we can extend our method by introducing a more generalized ansätzG′2=d2G2+d3G3+d4G4, where G=G(ζ), to solve Schamel equation, Schamel-KdV (S-KdV) equation, and modified Kadomtsev-Petviashvili (KP) equation.



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*Source: 810729-2013-11-04.xml*