# Applications of an Extended(G′/G)-Expansion Method to Find Exact Solutions of Nonlinear PDEs in Mathematical Physics

Authors: E. M. E. Zayed; Shorog Al-Joudi
Journal: Mathematical Problems in Engineering
 (2010)

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## Abstract

We construct the traveling wave solutions of the (1+1)-dimensional modified Benjamin-Bona-Mahony equation, the (2+1)-dimensional typical breaking soliton equation, the (1+1)-dimensional classical Boussinesq equations, and the (2+1)-dimensional Broer-Kaup-Kuperschmidt equations by using an extended(G′/G)-expansion method, where G satisfies the second-order linear ordinary differential equation. By using this method, new exact solutions involving parameters, expressed by three types of functions which are hyperbolic, trigonometric and rational function solutions, are obtained. When the parameters are taken as special values, some solitary wave solutions are derived from the hyperbolic function solutions.


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## Body

## 1. Introduction

The investigation of the traveling wave solutions of nonlinear partial differential equations (NPDEs) plays an important role in the study of nonlinear physical phenomena.  In recent years, new exact solutions may help to find new phenomena. The exact solutions have been investigated by many authors (see, e.g., [1–27]) who are interested in nonlinear physical phenomena. Many powerful methods have been presented such as the homogeneous balance method [13], the tanh method [4, 15, 24], the inverse scattering transform [1], the exp-function expansion method [2, 6, 20], the Jacobi elliptic function expansion [17], the Backlund transform [8, 9], the generalized Riccati equation [18], the modified extended Fan sub-equation method [19], the truncated Painlevé expansion [27], and the auxiliary equation method [10, 11]. More recently, the (G′/G)-expansion method [14, 22, 23, 26] has been proposed to obtain traveling wave solutions. This method is firstly proposed by the Chinese mathematicians Wang et al. [14] for which the traveling wave solutions of the nonlinear evolution equations are obtained. This method has been extended to solve difference-differential equations [28, 29]. The improved (G′/G)-expansion method has been used in [21, 25]. Recently, Gou and Zhou [5] have obtained the exact traveling wave solutions of some nonlinear PDEs using an extended (G'/G)-expansion method.In the present article, we use the extended(G′/G)-expansion method which is proposed in [5] to derive traveling wave solutions for some nonlinear PDEs in mathematical physics namely; the (1+1)-dimensional modified Benjamin-Bona-Mahony equation, the (2+1)-dimensional typical breaking soliton equation, the (1+1)-dimensional classical Boussinesq equations, and the (2+1)-dimensional Broer-Kaup-Kuperschmidt equations. The extended (G′/G)-expansion method used in this article can be applied to further equations such as difference-differential equations which can be done in forthcoming articles.


## 2. Description of an Extended(G′/G)-Expansion Method

Consider the nonlinear partial differential equation in the following form:(2.1)F(u,ut,ux,utt,uxt,uxx,…)=0,
where u=u(x,t) is unknown functions, and F is a polynomial in u(x,t) and its partial derivatives. In the following, we give the main steps for solving (2.1) using an extended (G'/G)-expansion method [5].Step 1.
The traveling wave variable(2.2)u(x,t)=u(ξ),ξ=x-Vt,
where V is a constant to be determined latter, permits us reducing (2.1) to an ODE in the form
(2.3)P(u,-Vu′,u′,V2u′′,-Vu′′,u′′,…)=0,
where P is a polynomial in u(ξ) and its total derivatives.Step 2.
Suppose the solution of (2.3) can be expressed in (G'/G) as follows:
(2.4)u(ξ)=a0+∑i=1n{ai(G'G)i+bi(G'G)i-1σ(1+1μ(G'G)2)},
where G=G(ξ) satisfies the following second-order linear ODE:
(2.5)G′′(ξ)+μG(ξ)=0,
while ai,bi(i=1,…,n) and a0 are constants to be determined, such that σ=±1 and μ≠0. The positive integer “n” can be determined by balancing the highest-order derivatives with the nonlinear terms appearing in (2.3).Step 3.
Substituting (2.4) into (2.3) and using (2.5), collecting all terms with the same powers of (G'/G)k and (G'/G)kσ(1+(1/μ)(G′/G)2) together, and equating each coefficient of them to zero, yield a set of algebraic equations for a0,ai,bi and V.Step 4.
Since the general solution of (2.5) has been well known for us, then substituting ai,bi,Vand the general solution of (2.5) into (2.4), we have the traveling wave solutions of the nonlinear partial differential equation (2.1).Remark 2.1.
It is necessary to point out that by adding the term(G'/G)kσ(1+(1/μ)(G'/G)2) into (2.4), the ansätz proposed here is more general than the ansätz in the original (G'/G)-expansion method [14]. Therefore, the extended (G'/G)-expansion method is more powerful than the original (G'/G)-expansion method [14] and some new types of traveling wave solutions and solitary wave solutions would be expected for some NPDEs. If we choose the parameters in (2.4) and (2.5) to take special values, the (G'/G)-expansion method can be recovered by our proposed method.


## 3. Applications

In this section, we will apply the extended(G′/G)-expansion method to some nonlinear PDEs in mathematical physics as follows.


### 3.1. Example 1: The (1+1)-Dimensional Modified Benjamin-Bona-Mahony Equation

We start with the following (1+1)-dimensional nonlinear dispersive modified Benjamin-Bona Mahony equation [20] written in the following form:
(3.1)ut+ux-αu2ux+uxxx=0,
where α is a nonzero positive constant. This equation was first derived to describe an approximation for surface long waves in nonlinear dispersive media. It can also characterize the hydromagnetic waves in cold plasma, acoustic waves in inharmonic crystals and acoustic-gravity waves in compressible fluids. Yusufoglu [20] has used the Exp-function method to find the traveling wave solutions of (3.1). Let us now solve (3.1) by the proposed method. To this end, we see that the traveling wave variable (2.2) permits us converting (3.1) into the following ODE:
(3.2)(1-V)u′-αu2u′+u′′′=0.
Integrating (3.2) with respect to ξ once, we get
(3.3)K+(1-V)u-13αu3+u′′=0,
where K is an integration constant. Considering the homogeneous balance between the highest-order derivatives and nonlinear terms in (3.3), we get n=1. Hence we suppose that the solution u(ξ) of (3.3) has the form
(3.4)u(ξ)=a0+a1(G′G)+b1σ(1+1μ(G′G)2),
where G=G(ξ) satisfies (2.5). Substituting (3.4) along with (2.5) into (3.3), collecting all terms with the same powers of (G′/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.5)K+3a0(1-V)-αa03-3σαa0b12=0,a1(1-V)+2μa1-αa1a02-σαa1b12=0,μαa0a12+σαa0b12=0,6μa1-μαa13-3σαa1b12=0,3b1(1+μ-V)-3αb1a02-σαb13=0,-3μαb1a12+6μb1-σαb13=0,2αa0a1b1=0.
Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.6)a1=σ6α,V=2μ+1,b1=a0=K=0.Case 2.
One has(3.7)b1=σ6μσα,V=1-μ,a1=a0=K=0.Case 3.
One has(3.8)a1=σ32α,b1=σ3μ2σα,V=12(2+μ),a0=C=0,whereσ=±1. From (3.4) and the general solution of (2.5), we deduce the traveling wave solutions of (3.1) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.9)u(ξ)=σ-6μα(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ)),
where ξ=x-(1+2μ)t and A,B are arbitrary constants while σ=±1. Case 2 gives the exact traveling wave solutions:
(3.10)u(ξ)=σ6μσασ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2],
where ξ=x-(1-μ)t. Case 3 gives the exact traveling wave solutions:(3.11)u(ξ)=σ32α{μσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]},
where ξ=x-((2+μ)/2)t. When μ>0, then Case 1 gives the exact traveling wave solutions:(3.12)u(ξ)=σ6μα(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ)).
Case 2 gives the exact traveling wave solutions(3.13)u(ξ)=σ6μσασ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2].
Case 3 gives the exact traveling wave solutions:(3.14)u(ξ)=σ3μ2α{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2](Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+1σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}.
In particular, we deduce from (3.9) that the solitary wave solutions of (3.1) are derived as follows.IfA=0,B≠0 and μ<0, then we obtain
(3.15)u(ξ)=σ-6μαcoth(-μξ),
while if A≠0,A2>B2, and μ<0, then we obtain
(3.16)u(ξ)=σ-6μαtanh(-μξ+ξ0),
where ξ0=tanh-1(B/A). Similarly, we can find more solitary wave solutions of (3.1) using (3.10)-(3.11) but we omitted them for simplicity.


### 3.2. Example 2: The (2+1)-Dimensional Typical Breaking Soliton Equation

In this subsection, we study the following the (2+1)-dimensional typical breaking soliton equation [3] in the following form:
(3.17)uxt-4uxuxy-2uxxuy+uxxxy=0,
which was first introduced by Calogero and Degasperis [3]. Tian et al. [12] have reduced new families of soliton-like solutions via the generalized tanh method which are of important significance in explaining some physical phenomena. Mei and Zhang [7] have reduced more families of new exact solutions which contain soliton-like solutions and periodic solution based on a newly generally projective Riccati equation expansion method and its algorithm. Let us now solve (3.17) by the proposed method. To this end, we see that the traveling wave variable
(3.18)u(x,y,t)=u(ξ),ξ=x+y-Vt
permits us to convert (3.17) into the following ODE:
(3.19)-Vu′′-6u′u′′+u(4)=0.
Integrating (3.19) with respect to ξ once yields
(3.20)K-Vu′-3(u′)2+u′′′=0,
where K is an integration constant. Considering the homogeneous balance between  the highest-order derivatives and nonlinear terms in (3.20) we deduce that the solution u(ξ) of (3.20) has the same form of (3.4). Substituting (3.4) along with (2.5) into (3.20), collecting all terms with the same powers of (G′/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2) and setting them to zero, we have the following algebraic equations: (3.21)K+μa1(V-2μ-3μa1)=0,-3σb12+a1(V-8μ-6a1)=0,3σb12+3μa1(a1+2)=0,b1(V-5μ-6μa1)=0,6b1(a1+1)=0.Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.22)a1=-2,V=-4μ,b1=K=0,Case 2.
One has(3.23)a1=-1,V=-μ,b1=σμσ,K=0.From (3.4) and the general solution of (2.5), we deduce the traveling wave solutions of (3.17) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.24)u(ξ)=-2-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,
where ξ=x+4μt. Case 2 gives the exact traveling wave solution:
(3.25)u(ξ)=σμσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]--μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0.
where ξ=x-μt. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.26)u(ξ)=-2μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0.
Case 2 gives the exact traveling wave solution:(3.27)u(ξ)=μ{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+σσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0
In particular, we deduce from (3.24) that the solitary wave solutions of (3.17) are derived as follows.IfA=0,B≠0, and μ<0, then we obtain
(3.28)u(ξ)=-2-μcoth(-μξ)+a0,
while if A≠0,A2>B2, and μ<0, then we obtain
(3.29)u(ξ)=-2-μtanh(-μξ+ξ0)+a0,
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.17) using (3.25) but we omitted them for simplicity.


### 3.3. Example 3: The (1+1)-Dimensional Classical Boussinesq Equations

In this subsection, we study the following (1+1)-dimensional classical Boussinesq equations [16]:
(3.30)vt+[(1+v)u]x=-13uxxx,ut+uux+vx=0.
This system has been derived by Wu and Zhang [16] for modelling nonlinear and dispersive long gravity wave traveling in two horizontal directions on shallow water of uniform depth. Let us now solve (3.30) by the proposed method. To this end, we see that the traveling wave variables (2.2) permit us converting (3.30) into the following ODEs:
(3.31)-Vv'+[(1+v)u]'+13u′′′=0,-Vu'+uu'+v'=0.
Integrating (3.31) with respect to ξ once yields(3.32)K1-Vv+(1+v)u+13u′′=0,(3.33)K2-Vu+12u2+v=0,
where K1 and K2 are integration constants. Considering the homogeneous balance between highest order derivatives and nonlinear terms in Equations (3.32), (3.33) we deduce that the solution u(ξ) which has the same form of (3.4) while, v(ξ) has the following form:
(3.34)v(ξ)=c0+c1(G'G)+d1σ(1+1μ(G'G)2)+c2(G'G)2+d2(G'G)σ(1+1μ(G'G)2).
Substituting (3.4) and (3.34) along with (2.5) into (3.32), collecting all terms with the same powers of (G'/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.35)K1+a0+σb1d1+c0(a0-V)=0,3σb1d2+a1(3c0+2μ+3)+3c1(a0-V)=0,μa1c1+σb1d1+μc2(a0-V)=0,3σb1d2+μa1(2+3c2)=0,3d1(a0-V)+b1(3+μ+3c0)=0,d2(a0-V)+a1d1+b1c1=0,b1(3c2+2)+3a1d2=0.
Similarly, substituting (3.4) and (3.34) along with (2.5) into (3.33), collecting all terms with the same powers of (G'/G)k, (G'/G)kσ(1+(1/μ)(G'/G)2) and setting them to zero, we have the following algebraic equations: (3.36)2K2+2c0+σb12+a0(a0-2V)=0,c1+a1(a0-V)=0,σb12+2μc2+μa12=0d1+b1(a0-V)=0b1a1+d2=0.Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.37)c0=-(2μ+3)3,c2=-23,a1=2σ3,V=a0,c1=d1=b1=d2=0,K1=-a0,K2=16(6+4μ+3a02).Case 2.
One has(3.38)c0=-(μ+3)3,c2=-23,b1=2σμ3σ,V=a0,d1=d2=c1=a1=0,K1=-a0,K2=16(6-2μ+3a02).Case 3.
One has(3.39)c0=-(μ+3)3,c2=-13,b1=μ3σ,d2=-σ3μσ,a1=σ3,V=a0,c1=d1=0,K1=-a0,K2=16(6+μ+3a02).wherea0 is an arbitrary constant. From (3.4), (3.34) and the general solution of (2.5), we deduce the traveling wave solutions of (3.30) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.40)u(ξ)=2σ-μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,v(ξ)=2μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(2μ+3)3,
Case 2 gives the exact traveling wave solution:
(3.41)u(ξ)=2σμ3σσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=2μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(μ+3)3.
Case 3 gives the exact traveling wave solution(3.42)u(ξ)=σ-μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+μ3σσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,(3.43)v(ξ)=-σ3-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(μ+3)3,
where ξ=x-a0t. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.44)u(ξ)=2σμ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0,v(ξ)=-2μ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-(2μ+3)3.
Case 2 gives the exact traveling wave solution
(3.45)u(ξ)=2σμ3σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+a0,v(ξ)=-2μ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-(μ+3)3.
Case 3 gives the exact traveling wave solution:(3.46)u(ξ)=μ3{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]σ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+1σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0,v(ξ)=-μ3{(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))×σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2}-(μ+3)3,
where ξ=x-a0t. In particular, we deduce from (3.40) that the solitary wave solutions of (3.30) are derived as follows:IfA=0,B≠0 and μ<0, then we obtain(3.47)u(ξ)=2σ-μ3coth(-μξ)+a0,v(ξ)=2μ3csch2(-μξ)-1;
if A≠0,A2>B2, and μ<0, then we obtain
(3.48)u(ξ)=2σμ3tanh(-μξ+ξ0)+a0,v(ξ)=-2μ3sech2(-μξ+ξ0)-1,
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.30) using (3.41)-(3.43) but we omitted them for simplicity.


### 3.4. Example 4: The (2+1)-Dimensional Broer-Kaup-Kuperschmidt Equations

In this subsection, we study the following (2+1)-dimensional Broer-Kaup-Kuperschmidt equations [19]:
(3.49)uyt-uxxy+2(uux)y+2vxx=0,vt+vxx+2(uv)x=0.
This system has been widely applied to many branches of physics like plasma physics, fluid dynamics, nonlinear optics and so on. Yomba [19] has obtained new and more general solutions of (3.49) including a series of nontraveling wave and coefficient function solutions using the modified extended Fan subequation method. Let us now solve (3.49) by the proposed method. To this end, we see that the traveling wave variables (3.18) permit us converting (3.49) into the following ODEs:
(3.50)-Vu′′-u′′′+2(uu′)′+2v′′=0,-Vv′+v′′+2(uv)′=0.
Integrating Equation (3.50) with respect to ξ once, yields(3.51)K1-Vu′′-u′′+2uu′+2v′=0,(3.52)K2-Vv+v′+2uv=0,
where K1 and K2 are integration constants. Considering the homogeneous balance between  the highest-order derivatives and nonlinear terms in (3.51) and (3.52), we deduce the solution u(ξ) which has the same form of (3.4) while, v(ξ) has the same form of (3.34). Substituting (3.4) and (3.34) along with (2.5) into (3.51), collecting all terms with the same powers of (G'/G)k, (G'/G)kσ(1+1/μ(G'/G)2), and setting them to zero, we have the following algebraic equations:
(3.53)K1-2μc1+μa1(V-2a0)=0,2σb12+2μ(2c2+a1+a12)=0,a1(V-2a0)-2c1=0,2σb12+4μc2+2μa1(1+a1)=0,2μd2+μb1(1+2a1)=0,b1(V-2a0)-2d1=0,b1(2a1+1)+2d2=0.
Similarly, substituting (3.4) and (3.34) along with (2.5) into (3.52), collecting all terms with the same powers of (G'/G)k, (G′/G)kσ(1+1/μ(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.54)K2-μc1+2σb1d1+c0(2a0-V)=0,2σb1d2+c1(2a0-V)-2μc2+2c0a1=0,2σb1d1+μc2(2a0-V)+μc1(2a1-1)=0,2σb1d2+μc2(2a1-1)=0,2b1c0+d1(2a0-V)-μd2=0,2b1c1+d2(2a0-V)+d2(2a1-1)=0,2b1c2+d2(2a1-1)=0.
Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.55)c0=-μ,c2=-1,a1=1,V=2a0,K1=K2=d1=d2=c1=b1=0.Case 2.
One has(3.56)c0=-μ4,c2=-12,d2=σ2μσ,b1=-σμσ,V=2a0,K1=K2=a1=d1=c1=0.Case 3.
One has(3.57)c0=-μ2,c2=-12,a1=12,d2=σ2μσ,b1=-σ2μσ,V=2a0,K1=K2=d1=c1=0,wherea0 is an arbitrary constant. From (3.4), (3.34), and the general solution of (2.5), we deduce the traveling wave solutions of (3.49) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.58)u(ξ)=-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,v(ξ)=μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ.
Case 2 gives the exact traveling wave solution:
(3.59)u(ξ)=-σμσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=σ2-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ4.
Case 3 gives the exact traveling wave solution
(3.60)u(ξ)=-μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))-σ2μσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=σ2-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ2,
where ξ=x-2a0t. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.61)u(ξ)=μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0,v(ξ)=-μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-μ.
Case 2 gives the exact traveling wave solution:
(3.62)u(ξ)=-σμσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+a0,v(ξ)=μ2{σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-12σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}.
Case 3 gives the exact traveling wave solution:
(3.63)u(ξ)=μ2{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2](Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))-σσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0,v(ξ)=μ2{(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))×σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-1},
where ξ=x-2a0t. In particular, we deduce from (3.58) that the solitary wave solutions of (3.49) are derived as follows.IfA=0,B≠0, and μ<0,then one obtain
(3.64)u(ξ)=-μcoth(-μξ)+a0,v(ξ)=μcsch2(-μξ);
if A≠0,A2>B2, and μ<0, then we obtain
(3.65)u(ξ)=-μtanh(-μξ+ξ0)+a0,v(ξ)=-μsech2(-μξ+ξ0),
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.49) using (3.59)-(3.60) but we omitted them for simplicity.


## 3.1. Example 1: The (1+1)-Dimensional Modified Benjamin-Bona-Mahony Equation

We start with the following (1+1)-dimensional nonlinear dispersive modified Benjamin-Bona Mahony equation [20] written in the following form:
(3.1)ut+ux-αu2ux+uxxx=0,
where α is a nonzero positive constant. This equation was first derived to describe an approximation for surface long waves in nonlinear dispersive media. It can also characterize the hydromagnetic waves in cold plasma, acoustic waves in inharmonic crystals and acoustic-gravity waves in compressible fluids. Yusufoglu [20] has used the Exp-function method to find the traveling wave solutions of (3.1). Let us now solve (3.1) by the proposed method. To this end, we see that the traveling wave variable (2.2) permits us converting (3.1) into the following ODE:
(3.2)(1-V)u′-αu2u′+u′′′=0.
Integrating (3.2) with respect to ξ once, we get
(3.3)K+(1-V)u-13αu3+u′′=0,
where K is an integration constant. Considering the homogeneous balance between the highest-order derivatives and nonlinear terms in (3.3), we get n=1. Hence we suppose that the solution u(ξ) of (3.3) has the form
(3.4)u(ξ)=a0+a1(G′G)+b1σ(1+1μ(G′G)2),
where G=G(ξ) satisfies (2.5). Substituting (3.4) along with (2.5) into (3.3), collecting all terms with the same powers of (G′/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.5)K+3a0(1-V)-αa03-3σαa0b12=0,a1(1-V)+2μa1-αa1a02-σαa1b12=0,μαa0a12+σαa0b12=0,6μa1-μαa13-3σαa1b12=0,3b1(1+μ-V)-3αb1a02-σαb13=0,-3μαb1a12+6μb1-σαb13=0,2αa0a1b1=0.
Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.6)a1=σ6α,V=2μ+1,b1=a0=K=0.Case 2.
One has(3.7)b1=σ6μσα,V=1-μ,a1=a0=K=0.Case 3.
One has(3.8)a1=σ32α,b1=σ3μ2σα,V=12(2+μ),a0=C=0,whereσ=±1. From (3.4) and the general solution of (2.5), we deduce the traveling wave solutions of (3.1) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.9)u(ξ)=σ-6μα(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ)),
where ξ=x-(1+2μ)t and A,B are arbitrary constants while σ=±1. Case 2 gives the exact traveling wave solutions:
(3.10)u(ξ)=σ6μσασ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2],
where ξ=x-(1-μ)t. Case 3 gives the exact traveling wave solutions:(3.11)u(ξ)=σ32α{μσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]},
where ξ=x-((2+μ)/2)t. When μ>0, then Case 1 gives the exact traveling wave solutions:(3.12)u(ξ)=σ6μα(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ)).
Case 2 gives the exact traveling wave solutions(3.13)u(ξ)=σ6μσασ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2].
Case 3 gives the exact traveling wave solutions:(3.14)u(ξ)=σ3μ2α{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2](Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+1σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}.
In particular, we deduce from (3.9) that the solitary wave solutions of (3.1) are derived as follows.IfA=0,B≠0 and μ<0, then we obtain
(3.15)u(ξ)=σ-6μαcoth(-μξ),
while if A≠0,A2>B2, and μ<0, then we obtain
(3.16)u(ξ)=σ-6μαtanh(-μξ+ξ0),
where ξ0=tanh-1(B/A). Similarly, we can find more solitary wave solutions of (3.1) using (3.10)-(3.11) but we omitted them for simplicity.


## 3.2. Example 2: The (2+1)-Dimensional Typical Breaking Soliton Equation

In this subsection, we study the following the (2+1)-dimensional typical breaking soliton equation [3] in the following form:
(3.17)uxt-4uxuxy-2uxxuy+uxxxy=0,
which was first introduced by Calogero and Degasperis [3]. Tian et al. [12] have reduced new families of soliton-like solutions via the generalized tanh method which are of important significance in explaining some physical phenomena. Mei and Zhang [7] have reduced more families of new exact solutions which contain soliton-like solutions and periodic solution based on a newly generally projective Riccati equation expansion method and its algorithm. Let us now solve (3.17) by the proposed method. To this end, we see that the traveling wave variable
(3.18)u(x,y,t)=u(ξ),ξ=x+y-Vt
permits us to convert (3.17) into the following ODE:
(3.19)-Vu′′-6u′u′′+u(4)=0.
Integrating (3.19) with respect to ξ once yields
(3.20)K-Vu′-3(u′)2+u′′′=0,
where K is an integration constant. Considering the homogeneous balance between  the highest-order derivatives and nonlinear terms in (3.20) we deduce that the solution u(ξ) of (3.20) has the same form of (3.4). Substituting (3.4) along with (2.5) into (3.20), collecting all terms with the same powers of (G′/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2) and setting them to zero, we have the following algebraic equations: (3.21)K+μa1(V-2μ-3μa1)=0,-3σb12+a1(V-8μ-6a1)=0,3σb12+3μa1(a1+2)=0,b1(V-5μ-6μa1)=0,6b1(a1+1)=0.Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.22)a1=-2,V=-4μ,b1=K=0,Case 2.
One has(3.23)a1=-1,V=-μ,b1=σμσ,K=0.From (3.4) and the general solution of (2.5), we deduce the traveling wave solutions of (3.17) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.24)u(ξ)=-2-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,
where ξ=x+4μt. Case 2 gives the exact traveling wave solution:
(3.25)u(ξ)=σμσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]--μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0.
where ξ=x-μt. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.26)u(ξ)=-2μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0.
Case 2 gives the exact traveling wave solution:(3.27)u(ξ)=μ{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+σσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0
In particular, we deduce from (3.24) that the solitary wave solutions of (3.17) are derived as follows.IfA=0,B≠0, and μ<0, then we obtain
(3.28)u(ξ)=-2-μcoth(-μξ)+a0,
while if A≠0,A2>B2, and μ<0, then we obtain
(3.29)u(ξ)=-2-μtanh(-μξ+ξ0)+a0,
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.17) using (3.25) but we omitted them for simplicity.


## 3.3. Example 3: The (1+1)-Dimensional Classical Boussinesq Equations

In this subsection, we study the following (1+1)-dimensional classical Boussinesq equations [16]:
(3.30)vt+[(1+v)u]x=-13uxxx,ut+uux+vx=0.
This system has been derived by Wu and Zhang [16] for modelling nonlinear and dispersive long gravity wave traveling in two horizontal directions on shallow water of uniform depth. Let us now solve (3.30) by the proposed method. To this end, we see that the traveling wave variables (2.2) permit us converting (3.30) into the following ODEs:
(3.31)-Vv'+[(1+v)u]'+13u′′′=0,-Vu'+uu'+v'=0.
Integrating (3.31) with respect to ξ once yields(3.32)K1-Vv+(1+v)u+13u′′=0,(3.33)K2-Vu+12u2+v=0,
where K1 and K2 are integration constants. Considering the homogeneous balance between highest order derivatives and nonlinear terms in Equations (3.32), (3.33) we deduce that the solution u(ξ) which has the same form of (3.4) while, v(ξ) has the following form:
(3.34)v(ξ)=c0+c1(G'G)+d1σ(1+1μ(G'G)2)+c2(G'G)2+d2(G'G)σ(1+1μ(G'G)2).
Substituting (3.4) and (3.34) along with (2.5) into (3.32), collecting all terms with the same powers of (G'/G)k, (G′/G)kσ(1+(1/μ)(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.35)K1+a0+σb1d1+c0(a0-V)=0,3σb1d2+a1(3c0+2μ+3)+3c1(a0-V)=0,μa1c1+σb1d1+μc2(a0-V)=0,3σb1d2+μa1(2+3c2)=0,3d1(a0-V)+b1(3+μ+3c0)=0,d2(a0-V)+a1d1+b1c1=0,b1(3c2+2)+3a1d2=0.
Similarly, substituting (3.4) and (3.34) along with (2.5) into (3.33), collecting all terms with the same powers of (G'/G)k, (G'/G)kσ(1+(1/μ)(G'/G)2) and setting them to zero, we have the following algebraic equations: (3.36)2K2+2c0+σb12+a0(a0-2V)=0,c1+a1(a0-V)=0,σb12+2μc2+μa12=0d1+b1(a0-V)=0b1a1+d2=0.Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.37)c0=-(2μ+3)3,c2=-23,a1=2σ3,V=a0,c1=d1=b1=d2=0,K1=-a0,K2=16(6+4μ+3a02).Case 2.
One has(3.38)c0=-(μ+3)3,c2=-23,b1=2σμ3σ,V=a0,d1=d2=c1=a1=0,K1=-a0,K2=16(6-2μ+3a02).Case 3.
One has(3.39)c0=-(μ+3)3,c2=-13,b1=μ3σ,d2=-σ3μσ,a1=σ3,V=a0,c1=d1=0,K1=-a0,K2=16(6+μ+3a02).wherea0 is an arbitrary constant. From (3.4), (3.34) and the general solution of (2.5), we deduce the traveling wave solutions of (3.30) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.40)u(ξ)=2σ-μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,v(ξ)=2μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(2μ+3)3,
Case 2 gives the exact traveling wave solution:
(3.41)u(ξ)=2σμ3σσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=2μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(μ+3)3.
Case 3 gives the exact traveling wave solution(3.42)u(ξ)=σ-μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+μ3σσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,(3.43)v(ξ)=-σ3-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ3(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-(μ+3)3,
where ξ=x-a0t. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.44)u(ξ)=2σμ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0,v(ξ)=-2μ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-(2μ+3)3.
Case 2 gives the exact traveling wave solution
(3.45)u(ξ)=2σμ3σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+a0,v(ξ)=-2μ3(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-(μ+3)3.
Case 3 gives the exact traveling wave solution:(3.46)u(ξ)=μ3{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]σ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+1σσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0,v(ξ)=-μ3{(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))×σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2}-(μ+3)3,
where ξ=x-a0t. In particular, we deduce from (3.40) that the solitary wave solutions of (3.30) are derived as follows:IfA=0,B≠0 and μ<0, then we obtain(3.47)u(ξ)=2σ-μ3coth(-μξ)+a0,v(ξ)=2μ3csch2(-μξ)-1;
if A≠0,A2>B2, and μ<0, then we obtain
(3.48)u(ξ)=2σμ3tanh(-μξ+ξ0)+a0,v(ξ)=-2μ3sech2(-μξ+ξ0)-1,
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.30) using (3.41)-(3.43) but we omitted them for simplicity.


## 3.4. Example 4: The (2+1)-Dimensional Broer-Kaup-Kuperschmidt Equations

In this subsection, we study the following (2+1)-dimensional Broer-Kaup-Kuperschmidt equations [19]:
(3.49)uyt-uxxy+2(uux)y+2vxx=0,vt+vxx+2(uv)x=0.
This system has been widely applied to many branches of physics like plasma physics, fluid dynamics, nonlinear optics and so on. Yomba [19] has obtained new and more general solutions of (3.49) including a series of nontraveling wave and coefficient function solutions using the modified extended Fan subequation method. Let us now solve (3.49) by the proposed method. To this end, we see that the traveling wave variables (3.18) permit us converting (3.49) into the following ODEs:
(3.50)-Vu′′-u′′′+2(uu′)′+2v′′=0,-Vv′+v′′+2(uv)′=0.
Integrating Equation (3.50) with respect to ξ once, yields(3.51)K1-Vu′′-u′′+2uu′+2v′=0,(3.52)K2-Vv+v′+2uv=0,
where K1 and K2 are integration constants. Considering the homogeneous balance between  the highest-order derivatives and nonlinear terms in (3.51) and (3.52), we deduce the solution u(ξ) which has the same form of (3.4) while, v(ξ) has the same form of (3.34). Substituting (3.4) and (3.34) along with (2.5) into (3.51), collecting all terms with the same powers of (G'/G)k, (G'/G)kσ(1+1/μ(G'/G)2), and setting them to zero, we have the following algebraic equations:
(3.53)K1-2μc1+μa1(V-2a0)=0,2σb12+2μ(2c2+a1+a12)=0,a1(V-2a0)-2c1=0,2σb12+4μc2+2μa1(1+a1)=0,2μd2+μb1(1+2a1)=0,b1(V-2a0)-2d1=0,b1(2a1+1)+2d2=0.
Similarly, substituting (3.4) and (3.34) along with (2.5) into (3.52), collecting all terms with the same powers of (G'/G)k, (G′/G)kσ(1+1/μ(G′/G)2), and setting them to zero, we have the following algebraic equations:
(3.54)K2-μc1+2σb1d1+c0(2a0-V)=0,2σb1d2+c1(2a0-V)-2μc2+2c0a1=0,2σb1d1+μc2(2a0-V)+μc1(2a1-1)=0,2σb1d2+μc2(2a1-1)=0,2b1c0+d1(2a0-V)-μd2=0,2b1c1+d2(2a0-V)+d2(2a1-1)=0,2b1c2+d2(2a1-1)=0.
Solving these algebraic equations by Maple or Mathematica, we obtain the following results.Case 1.
One has(3.55)c0=-μ,c2=-1,a1=1,V=2a0,K1=K2=d1=d2=c1=b1=0.Case 2.
One has(3.56)c0=-μ4,c2=-12,d2=σ2μσ,b1=-σμσ,V=2a0,K1=K2=a1=d1=c1=0.Case 3.
One has(3.57)c0=-μ2,c2=-12,a1=12,d2=σ2μσ,b1=-σ2μσ,V=2a0,K1=K2=d1=c1=0,wherea0 is an arbitrary constant. From (3.4), (3.34), and the general solution of (2.5), we deduce the traveling wave solutions of (3.49) as follows.Whenμ<0, then Case 1 gives the exact traveling wave solution:
(3.58)u(ξ)=-μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))+a0,v(ξ)=μ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ.
Case 2 gives the exact traveling wave solution:
(3.59)u(ξ)=-σμσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=σ2-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ4.
Case 3 gives the exact traveling wave solution
(3.60)u(ξ)=-μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))-σ2μσσ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+a0,v(ξ)=σ2-μ2σ(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))×σ[1-(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2]+μ2(Asinh(-μξ)+Bcosh(-μξ)Acosh(-μξ)+Bsinh(-μξ))2-μ2,
where ξ=x-2a0t. When μ>0, then Case 1 gives the exact traveling wave solution:
(3.61)u(ξ)=μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))+a0,v(ξ)=-μ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-μ.
Case 2 gives the exact traveling wave solution:
(3.62)u(ξ)=-σμσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]+a0,v(ξ)=μ2{σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-12σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}.
Case 3 gives the exact traveling wave solution:
(3.63)u(ξ)=μ2{σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2](Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))-σσσ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]}+a0,v(ξ)=μ2{(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2σσ(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))×σ[1+(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2]-(Bcos(μξ)-Asin(μξ)Acos(μξ)+Bsin(μξ))2-1},
where ξ=x-2a0t. In particular, we deduce from (3.58) that the solitary wave solutions of (3.49) are derived as follows.IfA=0,B≠0, and μ<0,then one obtain
(3.64)u(ξ)=-μcoth(-μξ)+a0,v(ξ)=μcsch2(-μξ);
if A≠0,A2>B2, and μ<0, then we obtain
(3.65)u(ξ)=-μtanh(-μξ+ξ0)+a0,v(ξ)=-μsech2(-μξ+ξ0),
where ξ0=tanh-1(B/A). Similarly, we can find more solitary solutions of (3.49) using (3.59)-(3.60) but we omitted them for simplicity.


## 4. Conclusion

In this paper, we have seen that three types of traveling wave solutions in terms of hyperbolic, trigonometric, and rational functions for the (1+1)-dimensional modified Benjamin-Bona-Mahony equation, the (2+1)-dimensional typical breaking soliton equation, the (1+1)-dimensional classical Boussinesq equations, and the (2+1)-dimensional Broer-Kaup-Kuperschmidt equations are successfully found out by using the extended(G′/G)-expansion method. The performance of this method is reliable, effective, and giving many new solutions to many other nonlinear PDEs. Finally, the solutions of the proposed nonlinear evolution equations in this paper have many potential applications in physics and engineering.



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*Source: 768573-2010-08-22.xml*