# New Exact Solutions for a Generalized Double Sinh-Gordon Equation

Authors: Gabriel Magalakwe; Chaudry Masood Khalique
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study a generalized double sinh-Gordon equation, which has applications in various fields, such as fluid dynamics, integrable quantum field theory, and kink dynamics. We employ the Exp-function method to obtain new exact solutions for this generalized double sinh-Gordon equation. This method is important as it gives us new solutions of the generalized double sinh-Gordon equation.


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## Body

## 1. Introduction

It is well known that finding exact travelling wave solutions of nonlinear partial differential equations (NLPDEs) is useful in many scientific applications such as fluid mechanics, plasma physics, and quantum field theory. Due to these applications many researchers are investigating exact solutions of NLPDEs since they play a vital role in the study of nonlinear physical phenomena. Finding exact solutions of such NLPDEs provides us with a better understanding of the physical phenomena that these NLPDEs describe. Several techniques have been presented in the literature to find exact solutions of the NLPDEs. These include the homogeneous balance method, the Weierstrass elliptic function expansion method, theF-expansion method, the (G′/G)-expansion method, the Exp-function method, the tanh  function method, the extended tanh function method, and the Lie group method [1–10].In this work, we study one such NLPDE, namely, the generalized double sinh-Gordon equation:(1)utt-kuxx+2αsinh(nu)+βsinh(2nu)=0,n≥1,
which appears in many scientific applications [11–13]. It should be noted that when k=a, α=(1/2)b, and β=0, (1) becomes the generalized sinh-Gordon equation [14, 15]. Furthermore, if n=a=1 and b=2, (1) reduces to the sinh-Gordon equation [16].Many authors have studied the generalized double sinh-Gordon equation (1). Travelling waves solutions of (1) were obtained in [11] by using the tanh function method and the variable separable method. In [12] the method of bifurcation theory of dynamical system was used to prove the existence of periodic wave, solitary wave, kink and antikink wave, and unbounded wave solutions of (1). It should be noted that solutions obtained in [12] were different the ones obtained in [11]. Recently, solitary and periodic waves solutions of (1) were found in [13] by employing (G′/G)-expansion method. It is further shown in [13] that solutions obtained by using the (G′/G)-expansion method are more general than those given in [11], which were obtained by tanh   function method.In this paper, we employ an entirely different method, known as the Exp-function method, to obtain new exact solutions of the generalized sinh-Gordon equation (1). The paper is structured as follows. In Section 2, we obtain exact solutions of the generalized double sinh-Gordon equation (1) with the help of the Exp-function method. In Section 3 we present concluding remarks.


## 2. Exact Solutions of (1) Using Exp-Function Method

In this section we employ the Exp-function method to solve the generalized double sinh-Gordon equation (1). This method was introduced by He and Wu [17]. The Exp-function method results in the travelling wave solution based on the assumption that the solution can be expressed in the following form:
(2)H(z)=∑n=-cdanexp(nz)∑m=-pqbmexp(mz),
where c, d, p, and q are positive integers that can be determined and an and bm are unknown constants. According to Exp-function method, we introduce the travelling wave substitution u(x,t)=W(z), where z=x-ct. Then (1) transforms to the nonlinear ordinary differential equation:
(3)(c2-k)W′′(z)+2αsinh(nW(z))+βsinh(2nW(z))=0.
Further, using the transformation W(z)=(1/n)ln(H(z)) on (3), we obtain
(4)2(c2-k)H(z)H′′(z)-2(c2-k)H′(z)2+2αnH(z)3-2αnH(z)+βnH(z)4-βn=0.
We assume that the solution of (4) can be expressed as
(5)H(z)=acexp(cz)+⋯+a-dexp(-dz)bpexp(pz)+⋯+b-qexp(-qz).The values of c and d, p and q can be determined by balancing the linear term of the highest order with the highest order of nonlinear term in (4), that is, HH′′ and H4. By straight forward calculation, we have
(6)HH′′=c1exp[(2c+3p)z]+⋯c2exp[5pz]+⋯,H4=c3exp[4cz]+⋯c4exp[4pz]+⋯=c3exp[(4c+p)z]+⋯c4exp[5pz]+⋯,
where ci are coefficients only for simplicity. Balancing the highest order of Exp-function in (6), we have 2c+3p=4c+p, which yields c=p. Similarly, we balance the lowest order in (4) to determine values of d and q. We have
(7)HH′′=⋯+s1exp[-(2d+3q)z]⋯+s2exp[-5qz],H4=⋯+s3exp[4dz]⋯+s4exp[-4qz]=⋯+s3exp[-(4d+q)z]⋯+s4exp[-5qz],
where si are coefficients only for simplicity. Balancing the lowest order of Exp-function in (7), we have 2d+3q=4d+q, which yields d=q. For simplicity, we first set c=p=1 and d=q=1. then (5) reduces to
(8)H(z)=a1exp(z)+a0+a-1exp(-z)b1exp(z)+b0+b-1exp(-z).Inserting (8) into (4) and using Maple, we obtain
(9)1B[C4exp(4z)+C3exp(3z)+C2exp(2z)+C1exp(z)+C0+C-1exp(-z)+C-2exp(-2z)+C-3exp(-3z)+C-4exp(-4z)]=0,
where
(10)B=(b1exp(z)+b0+b-1exp(-z))4,C4=2αa13b1n-βb14n+βa14n-2αa1b13n,C3=-2a12b0b1c2+2a1a0b12c2+6αa0a12b1n-6αa1b0b12n+2a12b0b1k-2a0a1b12k+2αa13b0n-2a0a1b12k+2αa13b0n+4βa0a13n-2αa0b13n-4βb0b13n,C2=4βa-1a13n-8a12b-1b1c2+8a-1a1b12c2+8a12b-1b1k-8a-1a1b12k+2αa13b-1n-2αa-1b13n-4βb-1b13n+6αa0a12b0n+6αa02a1b1n-6αa1b02b1n-6βb02b12n+6αa-1a12b1n-6αa1b12b-1n+6βa02a12n-6αa0b0b12n,C1=-2a02b0b1c2+2a0a1b02c2+2a02b0b1k-2a0a1b02k-2a12b0b-1c2+2a-1a0b12c2-2a0a-1b12k+2αa03b1n+4βa03a1n-2αa1b03n-4βb03b1n+12a-1a1b0b1c2-12a-1a1b0b1k+12a0a1b-1b1k+6αa02a1b0n-6αa0b02b1n+12αa-1a0a1b1n-12αa1b-1b0b1n+6αa-1a12b0n-6αa0b-1b12n-6αa-1b0b12n+6αa0a12b-1n+12βa-1a0a12n-12βb-1b0b12n+2a12b0b-1k-12a0a1b-1b1c2,C0=2αa03b0n-2αa0b03n+βa04n+6αa-1a12b-1n+6αa02a1b-1n+6αa-12a1b1n+6αa-1a02b1n+12βa-1a02a1n-6αa1b-12b1n-6αa1b-1b02n-6αa-1b-1b12n-βb04n-6αa-1b02b1n-12βb-1b02b1n+8a-1a1b02c2-8a02b-1b1c2-8a-1a1b02k+8a02b-1b1k+6βa-12a12n-6βb-12b12n+12αa-1a0a1b0n-12αa0b-1b0b1n,C-1=12αa-1a0a1b-1n-12αa-1b-1b0b1n+2a-1a0b02c2-2a02b-1b0c2+2a02b-1b0k-2a-1a0b02k+2a0a1b-12c2-2a-12b0b1c2-2a0a1b-12k+2a-12b0b1k+2αa03b-1n+4βa-1a03n-2αa-1b03n-4βb-1b03n+12a-1a1b-1b0c2-12a-1a0b-1b1c2-12a-1a1b-1b0k+12a-1a0b-1b1k+6αa-1a02b0n-6αa0b-1b02n+6αa-12a1b0n+6αa-12a0b1n+12βa-12a0a1n-6αa1b-12b0n-6αa0b-12b1n-12βb-12b0b1n,C-2=2αa-13b1n+8a-1a1b-12c2+8a-12b-1b1k-8a-1a1b-12k+4βa-13a1n-4βb-13b1n-2αa1b-13n-8a-12b-1b1c2+6αa-1a02b-1n+6αa-12a0b0n-6αa0b-12b0n+6αa-12a1b-1n-6αa-1b-12b1n+6βa-12a02n-6βb-12b02n-6αa-1b-1b02n,C-3=6αa0a-12b-1n-6αa-1b-12b0n-2a-12b-1b0c2+2a-1a0b-12c2+2a-12b0b-1k-2a-1a0b-12k+2αa-13b0n+4βa0a-13n-2αa0b-13n-4βb0b-13n,C-4=βa-14n-βb-14n+2αa-13b-1n-2αa-1b-13n.
Equating the coefficients of exp(z) in (9) to zero, we obtain a set of algebraic equations:
(11)C4=0,C3=0,C2=0,C1=0,C0=0,C-1=0,C-2=0,C-3=0,C-4=0.
Solving the system (11) with the help of Maple, we obtain the following three cases.Case 1.
We have the following:(12)a-1=b-1,a0=-b0,a1=b1,β=αb02-4αb1b-14b1b-1,k=αb02n+2b-1b1c22b-1b1.Case 2.
We have the following:(13)a-1=b-1b1a1,a0=0,b0=0,α=-β(a12+b12)2a1b1,k=-2βa12b12n+βa14n+βb14n+8a12b12c28a12b12.Case 3.
We have the following:(14)a-1=-ϕb1,b-1=-ϕa1,α=-β(a12+b12)2a1b1,k=-2βa12b12n+βa14n+βb14n+2a12b12c22a12b12,
where ϕ=(-a0a12b0+a02a1b1+a1b02b1-a0b0b12)/(a1-b1)2(a1+b1)2.Substituting values from (12) into (8), we obtain
(15)H(z)=b1exp(z)-b0+b-1exp(-z)b1exp(z)+b0+b-1exp(-z).
As a result one of the solutions of (1) is given by
(16)u1(x,t)=1nln(b1exp(z)-b0+b-1exp(-z)b1exp(z)+b0+b-1exp(-z)),
where z=x-ct, β=(αb02-4αb1b-1)/4b1b-1,and k=(αb02n+2b-1b1c2)/2b-1b1.As a special case, if we chooseb0=2 and b-1=b1=1 in (16), then we get β=0, k=2αn+c2 and obtain the solution of the generalized sinh-Gordon equation as
(17)u1(x,t)=1nln(tanh2[(12)(x-ct)]),
which is the solution obtained in [14, 15].Now substituting the values from (13) (Case 2) into (8) results in the second solution of (1) as
(18)u2(x,t)=1nln(a1exp(z)+(b-1b1/a1)exp(-z)b1exp(z)+b-1exp(-z)),
with  z=x-ct, α=-β(a12+b12)/2a1b1, and k=(-2βa12b12n+βa14n+βb14n+8a12b12c2)/8a12b12.The third solution of (1) is obtained by using the values from (14) (Case 3) and substituting them into (8). Consequently, it is given by
(19)u3(x,t)=1nln(a1exp(z)+a0-b1ϕexp(-z)b1exp(z)+b0-a-1ϕexp(-z)),
where z=x-ct, ϕ=(-a0a12b0+a02a1b1+a1b02b1-a0b0b12)/(a1-b1)2(a1+b1)2, α=-β(a12+b12)/2a1b1, and k=(-2βa12b12n+βa14n+βb14n+2a12b12c2)/2a12b12.To construct more solutions of (1), we now set c=p=2 and d=q=2. Then (5) reduces to
(20)H(z)=(a2exp(2z)+a1exp(z)+a0+a-1exp(-z)+a-2exp(-2z))×(b2exp(z)+b1exp(z)+b0+b-1exp(-z)+b-2exp(-2z))-1.
Proceeding as above, we obtain the following three solutions of (1):
(21)u4(x,t)=1nln(a2exp(2z)+(a-1b1b-1)exp(z)+(a-1b0b-1)+a-1exp(-z))×(a2b-1a-1exp(z)+b1exp(z)+b0+b-1exp(-z)a2b-1a-1)-1,
where z=x-ct, α=-β(a-12+b-12)/2a-1b-1,
(22)u5(x,t)=1nln(a2exp(2z)+a1exp(z)+b0-a2exp(z)+b1exp(z)+b0),
with z=x-ct, β=α(b12+4a2b0)/4a2b0, and k=(αnb12+2a2b0c2)/2a2b0, and
(23)u6(x,t)=1nln(a2exp(2z)-b0+b-2exp(-2z)a2exp(2z)+b0+b-2exp(-2z)),
where z=x-ct, α=-(8a2b-2(c2-k)/b02n), and β=2(4a2b-2c2-4a2b-2k-b02c2+b02k)/b02n.By takingn=2, b-1=-1, b0=2, c=1, and b1=-1 in the solution (16), we have its profile given in Figure 1.Figure 1
Profile of solution (16).By takingn=3, b-2=1, b0=2, c=1, and a1=1 in the solution (23), we have its profile given in Figure 2.Figure 2
Profile of solution (23).


## 3. Concluding Remarks

In this paper we obtained new exact solutions of the generalized double sinh-Gordon equation (1) using the Exp-function method. We presented six different solutions of (1). Earlier, the tanh function, the bifurcation, and the (G′/G)-expansion methods [11–13] were employed to obtain exact solutions of (1). The solutions obtained in this paper were new and were different from the ones obtained in [11–13]. By taking special values of the constants, we also retrieved the solution of the generalized sinh-Gordon equation, which was obtained in [14, 15]. The Exp-function method is very simple and straightforward method for solving nonlinear partial differential equations. Indeed this has some pronounced merit as compared to the other methods. The correctness of the solutions obtained here has been verified by substituting them back into (1).



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