# Exact Solutions of the Symmetric Regularized Long Wave Equation and the Klein-Gordon-Zakharov Equations

Authors: Isaiah Elvis Mhlanga; Chaudry Masood Khalique
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study two nonlinear partial differential equations, namely, the symmetric regularized long wave equation and the Klein-Gordon-Zakharov equations. The Lie symmetry approach along with the simplest equation and exp-function methods are used to obtain solutions of the symmetric regularized long wave equation, while the travelling wave hypothesis approach along with the simplest equation method is utilized to obtain new exact solutions of the Klein-Gordon-Zakharov equations.


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## Body

## 1. Introduction

The investigation of exact travelling wave solutions of nonlinear partial differential equations (NLPDEs) is important for the understanding of most nonlinear physical phenomena that appear in many areas of scientific fields such as plasma physics, solid state physics, fluid dynamics, optical fibers, mathematical biology, and chemical kinetics [1, 2]. A number of methods have been developed for finding travelling wave solutions to NLPDEs. These include the homogeneous balance method [3], the ansatz method [4, 5], variable separation approach [6], inverse scattering transform method [2], Bäcklund transformation [7], Darboux transformation [8], Hirota bilinear method [9], the (G′/G)-expansion method [10], the reduction mKdV equation method [11], the trifunction method [12, 13], the projective Riccati equation method [14], the sine-cosine method [15, 16], the Jacobi elliptic function expansion method [17, 18], the F-expansion method [19], the exp-function expansion method [20], dynamical system method [21–23], and Lie symmetry method [24–28].In this paper we study two nonlinear partial differential equations, namely, the symmetric regularized long wave equation and the Klein-Gordon-Zakharov equations. The Lie symmetry approach along with the simplest equation and exp-function methods are used to obtain solutions of the symmetric regularized long wave equation, while the travelling wave hypothesis approach along with the simplest equation method is utilized to obtain new exact solutions of the Klein-Gordon-Zakharov equations.


## 2. The Symmetric Regularized Long Wave Equation

We first consider that the symmetric regularized long wave equation (SRLW) as given by(1)utt-uxx+12(u2)xt-uxxtt=0
is a nonlinear evolution equation which arises in several physical applications, for example in sound waves in a plasma [29]. Exact travelling wave solutions of this equation were obtained using the (G′/G)-expansion method [29]. In the present work, Lie symmetry method along with the simplest equation method and the exp-function method are used to construct exact solutions for this equation. First the Lie point symmetries of the SRLW equation (1) are found using the Lie algorithm [25]. These Lie point symmetries are then used to transform (1) into an ordinary differential equation. The simplest equation method [30] and the exp-function method [20] are then used to construct exact solutions of the ordinary differential equation, which leads to the exact solutions of the SRLW equation.


### 2.1. Lie Point Symmetries of (1) and Symmetry Reduction

The symmetry group of the SRLW equation (1) is generated by the vector field
(2)X=ξ1(t,x,u)∂∂t+ξ2(t,x,u)∂∂x+η(t,x,u)∂∂u.
Applying the fourth prolongation of X to (1) and solving the resultant overdetermined system of linear partial differential equations, we obtain the following two translation symmetries:
(3)X1=∂∂t,X2=∂∂x.
Now taking the linear combination of these translation symmetries X1=∂/∂t and X2=∂/∂x, namely, the symmetry X=X1+νX2, where ν is a constant, leads to the two invariants
(4)z=x-νt,u=F(z).
Treating F as the new dependent variable and z as the new independent variable and then substituting the value of u into the SRLW equation (1) transform (1) into a fourth-order nonlinear ordinary differential equation:
(5)(ν2-1)F′′(z)-νF(z)F′′(z)-ν(F′(z))2+ν2F(z)′′′′=0.


### 2.2. Exact Solutions of (1) Using Simplest Equation Method

Now the simplest equation method [30, 31] is used to solve (5), and henceforth one obtains the exact solutions of the SRLW equation (1). The Bernoulli and Riccati equations will be used as the simplest equations. The Bernoulli and Riccati equations are well-known equations whose solutions can be expressed in terms of elementary functions [28].The Bernoulli equation which we use here is given by(6)H′(z)=cH(z)+dH2(z),
where c and d are constants. Its solution is given by
(7)H(z)=c{cosh⁡[c(z+C)]+sinh⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]},
where C is a constant of integration [28].For the Riccati equation(8)G′(z)=cG2(z)+dG(z)+e,
where c, d, and e are constants, the solutions to be used are
(9)H(z)=-d2c-θ2ctanh[12θ(z+C)],H(z)=-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2),
with θ2=d2-4ce>0 and C being a constant of integration [28].


#### 2.2.1. Solutions of (1) Using Bernoulli as the Simplest Equation

The solutions of the ODE (5) are considered to be in the form
(10)F(z)=∑i=0M‍𝒜i(H(z))i,
where H(z) satisfies the Bernoulli or Riccati equations, M is a positive integer that can be determined by balancing the highest order derivative term with the highest order nonlinear term [31], and 𝒜i, (i=0,1,…,M) are parameters to be determined.The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(11)F(z)=𝒜0+𝒜1H+𝒜2H2.
Substituting (11) into (5), making use of the Bernoulli equation (6), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2:
(12)-10ν𝒜22d2-120ν2𝒜2d4=0,-24ν2𝒜1d4-12ν𝒜1d2𝒜2-336ν2𝒜2d3c-18ν𝒜22dc=0,-ν𝒜0𝒜1c2-𝒜1c2-ν2𝒜1c4+ν2𝒜1c2=0,-8ν𝒜22c2-6𝒜2d2+6ν2𝒜2d2-21ν𝒜1d𝒜2c-3ν𝒜12d2-60ν2𝒜1d3c-6ν𝒜0𝒜2d2-330ν2𝒜2d2c2=0,-3𝒜1dc+3ν2𝒜1dc-4ν𝒜0𝒜2c2-4𝒜2c2+4ν2𝒜2c2-3ν𝒜0𝒜1dc-2ν𝒜12c2-16ν2𝒜2c4-15ν2𝒜1dc3=0,-2𝒜1d2-50ν2𝒜1d2c2+10ν2𝒜2dc-5ν𝒜12dc-130ν2𝒜2dc3+2ν2𝒜1d2-9ν𝒜1c2𝒜2-2ν𝒜0𝒜1d2-10ν𝒜0𝒜2dc-10𝒜2dc=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(13)𝒜0=-c2ν2+ν2-1ν,𝒜1=-12cdν,𝒜2=-12νd2.
As a result a solution of the symmetric regularized long wave equation (1) using the Bernoulli equation as the simplest equation is
(14)u(t,x)=𝒜0+𝒜1c{cosh⁡[c(z+C)]+sinh⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}+𝒜2c2{cosh⁡⁡[c(z+C)]+sinh⁡⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}2,
where z=x-νt and C is a constant of integration.


#### 2.2.2. Solutions of (1) Using Ricatti as the Simplest Equation

The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(15)F(z)=𝒜0+𝒜1G(z)+𝒜2G2(z).
Substituting (15) into (5), making use of the Ricatti equation (8), and then equating all coefficients of the function Gi to zero, we obtain an algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2. Solving the resultant algebraic equations, we obtain the following set of values:
(16)𝒜0=-8ceν2+d2ν2-ν2+1ν,𝒜1=-12cdν,𝒜2=-12c2ν.
It follows that the solutions for the symmetric regularized long wave equation (1) using the Ricatti equation as the simplest equation are
(17)u(t,x)=𝒜0+𝒜1{-d2c-θ2ctanh[12θ(z+C)]}+𝒜2{-d2c-θ2ctanh[12θ(z+C)]}2,u(t,x)=𝒜0+𝒜1{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}+𝒜2{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}2,
where z=x-νt with θ2=d2-4ce>0 and C is a constant of integration.


### 2.3. Solution of (1) Using the Exp-Function Method

In this section we use the exp-function method [20] to solve the symmetric regularized long wave equation (1). We consider solutions of (5) in the form
(18)F(z)=∑n=-bcanenz∑m=-pqbmemz,
where b, c, p, and q are positive integers to be determined and an and bm are arbitrary constants [20]. The balancing procedure of the exp-function method produces p=b and q=c. For simplicity, we set p=b=1 and q=c=1 so that (18) is reduced to
(19)F(z)=a1ez+a0+a-1e-zb1ez+b0+b-1e-z.
Substituting (19) into (5) and solving the resultant ODE, with the help of Maple, one possible set of values of the constants is
(20)a-1=-b-1ν,a0=b0(6ν2-1)ν,a1=-b024νb-1,b1=b024b-1.
As a result we obtain the solution
(21)u(t,x)=a1ex-νt+a0+a-1e-(x-νt)b1ex-νt+b0+b-1e-(x-νt).


## 2.1. Lie Point Symmetries of (1) and Symmetry Reduction

The symmetry group of the SRLW equation (1) is generated by the vector field
(2)X=ξ1(t,x,u)∂∂t+ξ2(t,x,u)∂∂x+η(t,x,u)∂∂u.
Applying the fourth prolongation of X to (1) and solving the resultant overdetermined system of linear partial differential equations, we obtain the following two translation symmetries:
(3)X1=∂∂t,X2=∂∂x.
Now taking the linear combination of these translation symmetries X1=∂/∂t and X2=∂/∂x, namely, the symmetry X=X1+νX2, where ν is a constant, leads to the two invariants
(4)z=x-νt,u=F(z).
Treating F as the new dependent variable and z as the new independent variable and then substituting the value of u into the SRLW equation (1) transform (1) into a fourth-order nonlinear ordinary differential equation:
(5)(ν2-1)F′′(z)-νF(z)F′′(z)-ν(F′(z))2+ν2F(z)′′′′=0.


## 2.2. Exact Solutions of (1) Using Simplest Equation Method

Now the simplest equation method [30, 31] is used to solve (5), and henceforth one obtains the exact solutions of the SRLW equation (1). The Bernoulli and Riccati equations will be used as the simplest equations. The Bernoulli and Riccati equations are well-known equations whose solutions can be expressed in terms of elementary functions [28].The Bernoulli equation which we use here is given by(6)H′(z)=cH(z)+dH2(z),
where c and d are constants. Its solution is given by
(7)H(z)=c{cosh⁡[c(z+C)]+sinh⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]},
where C is a constant of integration [28].For the Riccati equation(8)G′(z)=cG2(z)+dG(z)+e,
where c, d, and e are constants, the solutions to be used are
(9)H(z)=-d2c-θ2ctanh[12θ(z+C)],H(z)=-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2),
with θ2=d2-4ce>0 and C being a constant of integration [28].


### 2.2.1. Solutions of (1) Using Bernoulli as the Simplest Equation

The solutions of the ODE (5) are considered to be in the form
(10)F(z)=∑i=0M‍𝒜i(H(z))i,
where H(z) satisfies the Bernoulli or Riccati equations, M is a positive integer that can be determined by balancing the highest order derivative term with the highest order nonlinear term [31], and 𝒜i, (i=0,1,…,M) are parameters to be determined.The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(11)F(z)=𝒜0+𝒜1H+𝒜2H2.
Substituting (11) into (5), making use of the Bernoulli equation (6), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2:
(12)-10ν𝒜22d2-120ν2𝒜2d4=0,-24ν2𝒜1d4-12ν𝒜1d2𝒜2-336ν2𝒜2d3c-18ν𝒜22dc=0,-ν𝒜0𝒜1c2-𝒜1c2-ν2𝒜1c4+ν2𝒜1c2=0,-8ν𝒜22c2-6𝒜2d2+6ν2𝒜2d2-21ν𝒜1d𝒜2c-3ν𝒜12d2-60ν2𝒜1d3c-6ν𝒜0𝒜2d2-330ν2𝒜2d2c2=0,-3𝒜1dc+3ν2𝒜1dc-4ν𝒜0𝒜2c2-4𝒜2c2+4ν2𝒜2c2-3ν𝒜0𝒜1dc-2ν𝒜12c2-16ν2𝒜2c4-15ν2𝒜1dc3=0,-2𝒜1d2-50ν2𝒜1d2c2+10ν2𝒜2dc-5ν𝒜12dc-130ν2𝒜2dc3+2ν2𝒜1d2-9ν𝒜1c2𝒜2-2ν𝒜0𝒜1d2-10ν𝒜0𝒜2dc-10𝒜2dc=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(13)𝒜0=-c2ν2+ν2-1ν,𝒜1=-12cdν,𝒜2=-12νd2.
As a result a solution of the symmetric regularized long wave equation (1) using the Bernoulli equation as the simplest equation is
(14)u(t,x)=𝒜0+𝒜1c{cosh⁡[c(z+C)]+sinh⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}+𝒜2c2{cosh⁡⁡[c(z+C)]+sinh⁡⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}2,
where z=x-νt and C is a constant of integration.


### 2.2.2. Solutions of (1) Using Ricatti as the Simplest Equation

The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(15)F(z)=𝒜0+𝒜1G(z)+𝒜2G2(z).
Substituting (15) into (5), making use of the Ricatti equation (8), and then equating all coefficients of the function Gi to zero, we obtain an algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2. Solving the resultant algebraic equations, we obtain the following set of values:
(16)𝒜0=-8ceν2+d2ν2-ν2+1ν,𝒜1=-12cdν,𝒜2=-12c2ν.
It follows that the solutions for the symmetric regularized long wave equation (1) using the Ricatti equation as the simplest equation are
(17)u(t,x)=𝒜0+𝒜1{-d2c-θ2ctanh[12θ(z+C)]}+𝒜2{-d2c-θ2ctanh[12θ(z+C)]}2,u(t,x)=𝒜0+𝒜1{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}+𝒜2{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}2,
where z=x-νt with θ2=d2-4ce>0 and C is a constant of integration.


## 2.2.1. Solutions of (1) Using Bernoulli as the Simplest Equation

The solutions of the ODE (5) are considered to be in the form
(10)F(z)=∑i=0M‍𝒜i(H(z))i,
where H(z) satisfies the Bernoulli or Riccati equations, M is a positive integer that can be determined by balancing the highest order derivative term with the highest order nonlinear term [31], and 𝒜i, (i=0,1,…,M) are parameters to be determined.The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(11)F(z)=𝒜0+𝒜1H+𝒜2H2.
Substituting (11) into (5), making use of the Bernoulli equation (6), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2:
(12)-10ν𝒜22d2-120ν2𝒜2d4=0,-24ν2𝒜1d4-12ν𝒜1d2𝒜2-336ν2𝒜2d3c-18ν𝒜22dc=0,-ν𝒜0𝒜1c2-𝒜1c2-ν2𝒜1c4+ν2𝒜1c2=0,-8ν𝒜22c2-6𝒜2d2+6ν2𝒜2d2-21ν𝒜1d𝒜2c-3ν𝒜12d2-60ν2𝒜1d3c-6ν𝒜0𝒜2d2-330ν2𝒜2d2c2=0,-3𝒜1dc+3ν2𝒜1dc-4ν𝒜0𝒜2c2-4𝒜2c2+4ν2𝒜2c2-3ν𝒜0𝒜1dc-2ν𝒜12c2-16ν2𝒜2c4-15ν2𝒜1dc3=0,-2𝒜1d2-50ν2𝒜1d2c2+10ν2𝒜2dc-5ν𝒜12dc-130ν2𝒜2dc3+2ν2𝒜1d2-9ν𝒜1c2𝒜2-2ν𝒜0𝒜1d2-10ν𝒜0𝒜2dc-10𝒜2dc=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(13)𝒜0=-c2ν2+ν2-1ν,𝒜1=-12cdν,𝒜2=-12νd2.
As a result a solution of the symmetric regularized long wave equation (1) using the Bernoulli equation as the simplest equation is
(14)u(t,x)=𝒜0+𝒜1c{cosh⁡[c(z+C)]+sinh⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}+𝒜2c2{cosh⁡⁡[c(z+C)]+sinh⁡⁡[c(z+C)]1-dcosh⁡[c(z+C)]-dsinh⁡[c(z+C)]}2,
where z=x-νt and C is a constant of integration.


## 2.2.2. Solutions of (1) Using Ricatti as the Simplest Equation

The balancing procedure yieldsM=2, so the solutions of (5) are of the form
(15)F(z)=𝒜0+𝒜1G(z)+𝒜2G2(z).
Substituting (15) into (5), making use of the Ricatti equation (8), and then equating all coefficients of the function Gi to zero, we obtain an algebraic system of equations in terms of 𝒜0, 𝒜1, and 𝒜2. Solving the resultant algebraic equations, we obtain the following set of values:
(16)𝒜0=-8ceν2+d2ν2-ν2+1ν,𝒜1=-12cdν,𝒜2=-12c2ν.
It follows that the solutions for the symmetric regularized long wave equation (1) using the Ricatti equation as the simplest equation are
(17)u(t,x)=𝒜0+𝒜1{-d2c-θ2ctanh[12θ(z+C)]}+𝒜2{-d2c-θ2ctanh[12θ(z+C)]}2,u(t,x)=𝒜0+𝒜1{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}+𝒜2{sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)-d2c-θ2ctanh(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2c/θ)sinh⁡(θz/2)}2,
where z=x-νt with θ2=d2-4ce>0 and C is a constant of integration.


## 2.3. Solution of (1) Using the Exp-Function Method

In this section we use the exp-function method [20] to solve the symmetric regularized long wave equation (1). We consider solutions of (5) in the form
(18)F(z)=∑n=-bcanenz∑m=-pqbmemz,
where b, c, p, and q are positive integers to be determined and an and bm are arbitrary constants [20]. The balancing procedure of the exp-function method produces p=b and q=c. For simplicity, we set p=b=1 and q=c=1 so that (18) is reduced to
(19)F(z)=a1ez+a0+a-1e-zb1ez+b0+b-1e-z.
Substituting (19) into (5) and solving the resultant ODE, with the help of Maple, one possible set of values of the constants is
(20)a-1=-b-1ν,a0=b0(6ν2-1)ν,a1=-b024νb-1,b1=b024b-1.
As a result we obtain the solution
(21)u(t,x)=a1ex-νt+a0+a-1e-(x-νt)b1ex-νt+b0+b-1e-(x-νt).


## 3. The Klein-Gordon-Zakharov Equations

The Klein-Gordon-Zakharov (KGZ) equations [32](22a)utt-uxx+u+uv+|u|2u=0,(22b)vtt-vxx-(|u|2)xx=0,are a coupled system of nonlinear partial differential equations of two functionsu(x,t) and v(x,t). This model describes the interaction of the Langmuir wave and the ion acoustic wave in plasma. The function u(x,t) denotes the fast time scale component of electric field raised by electrons and the function v(x,t) denotes the deviation of ion density from its equilibrium. Here u(x,t) is a complex function and v(x,t) is a real function. Note that if we remove the term |u|2u, then this system reduces to the classical Klein-Gordon-Zakharov system [33](23a)utt-uxx+u+uv=0,(23b)vtt-vxx-(|u|2)xx=0.A number of studies have been conducted for this system ((23a) and (23b)) in different time space [34–38]. However, for the KGZ equations (22a) and (22b), Chen [39] considered orbital stability of solitary waves, while Shi et al. [33] employed the sine-cosine method and the extended tanh method to construct exact solutions of the KGZ equations (22a) and (22b).In this paper, we employ an entirely different approach, namely, the travelling wave variable approach along with the simplest equation method to obtain exact solutions of the KGZ equations (22a) and (22b).


### 3.1. Solution of (22a) and (22b) Using the Travelling Wave Variable Approach

The travelling wave variable approach converts the system of nonlinear partial differential equations into a system of nonlinear ordinary differential equations, which we then solve to obtain exact solutions of the system.In order to solve the KGZ equations (22a) and (22b), we first transform it into a system of nonlinear ordinary differential equations which can then be solved in order to obtain its exact solutions.We make the wave variable transformation(24)u=eiϕu(z),v=v(z),ϕ=px+rt,z=kx+dt,
where p, r, k, and d are real constants and d≠k. Using this transformation, (22a) and (22b) transform into(25a)(p2-r2+1)u+i(2rd-2pk)u′+(d2-k2)u′′+uv+u3=0,(25b)(d2-k2)v′′-(u2)′′=0.Integrating (25b) twice and taking the constants of integration to be zero, we obtain
(26)v=u2d2-k2.
Now substituting (26) into (25a), we get
(27)u′′=(r2-p2-1d2-k2)u+(d2-k2+1(d2-k2)2)u3,
which can be written in the form
(28)u′′=Pu+Qu3,
where
(29)P=r2-p2-1d2-k2,Q=d2-k2+1(d2-k2)2.
Solving (28), with the aid of Mathematica, we obtain the solution
(30)u(z)=±1Hisn(F∣ω),
where sn(F∣ω) is a Jacobian elliptic function of the sine amplitude [40],
(31)F=(P2-2Qc1-P)(z+c2)22,H=QP2-2Qc1+P,ω=-Qc1+P(P2-2Qc1+P)Qc1
is the modulus of the elliptic function with 0<ω<1. Here c1 and c2 are constants of integration. Reverting back to our original variables, we can now write the solution of our Klein-Gordon-Zakharov equations as
(32)u(x,t)=±1Hisn(F∣ω),
where
(33)F=(P2-2Qc1-P)(kx+dt+c2)22,
and ω and H are as above.Nowv(x,t) can be obtained from (26).It should be noted that the solution (32) is valid for 0<ω<1, as ω approaches zero, the solution becomes the normal sine function, sinz, and as ω approaches 1, the solution tends to the tanh function, tanhz.The profile of the solution (32) is given in Figure 1.Figure 1
Profile of solution (32).


### 3.2. Solutions of (22a) and (22b) Using the Simplest Equation Method

We consider the solutions of (27) in the form
(34)u(z)=∑i=0M‍Ai(G(z))i,
where G(z) satisfies the Bernoulli or the Riccati equation.


#### 3.2.1. Solutions of (22a) and (22b) Using Bernoulli as the Simplest Equation

We consider the Bernoulli equation(35)G′(z)=aG(z)+bG2(z),
where a and b are constants.The balancing procedure yieldsM=1, so the solution of (27) is of the form
(36)u(z)=A0+A1H(z).
Substituting (36) into (27), making use of the Bernoulli equation (35), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations:
(37)2A1k4b2-k2A13+d2A13-4A1d2k2b2+A13+2A1d4b2=0,3A1k4ab+3d2A0A12-3k2A0A12-6A1d2k2ab+3A0A12+3A1d4ab=0,p2A0d2-p2A0k2-r2A0d2+r2A0k2-A0k2+d2A03-k2A03+A0d2+A03=0,A1d2+3d2A02A1-p2A1k2+r2A1k2+p2A1d2+A1d4a2-3k2A02A1+A1k4a2+3A02A1-2A1d2k2a2-r2A1d2-A1k2=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(38)A0=a(d2-k2)2(k2-d2-1),A1=2b(d2-k2)k2-d2-1,p=r2d2-r2k2+k2-d2A02+k2A02-d2-A02d2-k2.
As a result, a solution of the Klein-Gordon-Zakharov equations (22a) and (22b), using the Bernoulli equation as the simplest equation, is
(39)u(x,t)=ei(px+rt)[+a(d2-k2)2(k2-d2-1)](-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)(2ab(d2-k2)(cosh⁡(a(kx+dt+c))+sinh⁡(a(kx+dt+c)))(2ab(d2-k2)(cosh⁡(a(kx+dt+c)))×(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c))-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)+a(d2-k2)2(k2-d2-1)],
where c is a constant of integration.


#### 3.2.2. Solutions of (22a) and (22b) Using Riccati as the Simplest Equation

We use the Riccati equation given by(40)G′(z)=aG2(z)+bG(z)+c,
where a, b, and c are constants. The balancing procedure yields M=1, so the solution of (27) is of the form
(41)u(z)=A0+A1G(z).
Similar calculations yield the following set of values:
(42)A0=b(d2-k2)2(k2-d2-1),A1=-2a(d2-k2)k2-d2-1d2-k2+1,c=-2(k2-d2-1)(d2b2-k2b2-2-2p2+2r2)4A1(d2-k2+1).
As a result the two solutions of (22a) and (22b) are
(43)u(x,t)=eiϕ[A0+A1{-b2a-θ2atanh[12θ(z+C)]}],u(x,t)=eiϕ[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0+A1{sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)-b2a-θ2atanh⁡(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0],
where ϕ=px+rt and z=kx+dt. θ is given by b2-4ac, C is a constant of integration, and A0 and A1 are as obtained above.It should be noted that by substituting the above value ofu(x,t) into (26), one can now obtain the solution for the variable v(x,t).


## 3.1. Solution of (22a) and (22b) Using the Travelling Wave Variable Approach

The travelling wave variable approach converts the system of nonlinear partial differential equations into a system of nonlinear ordinary differential equations, which we then solve to obtain exact solutions of the system.In order to solve the KGZ equations (22a) and (22b), we first transform it into a system of nonlinear ordinary differential equations which can then be solved in order to obtain its exact solutions.We make the wave variable transformation(24)u=eiϕu(z),v=v(z),ϕ=px+rt,z=kx+dt,
where p, r, k, and d are real constants and d≠k. Using this transformation, (22a) and (22b) transform into(25a)(p2-r2+1)u+i(2rd-2pk)u′+(d2-k2)u′′+uv+u3=0,(25b)(d2-k2)v′′-(u2)′′=0.Integrating (25b) twice and taking the constants of integration to be zero, we obtain
(26)v=u2d2-k2.
Now substituting (26) into (25a), we get
(27)u′′=(r2-p2-1d2-k2)u+(d2-k2+1(d2-k2)2)u3,
which can be written in the form
(28)u′′=Pu+Qu3,
where
(29)P=r2-p2-1d2-k2,Q=d2-k2+1(d2-k2)2.
Solving (28), with the aid of Mathematica, we obtain the solution
(30)u(z)=±1Hisn(F∣ω),
where sn(F∣ω) is a Jacobian elliptic function of the sine amplitude [40],
(31)F=(P2-2Qc1-P)(z+c2)22,H=QP2-2Qc1+P,ω=-Qc1+P(P2-2Qc1+P)Qc1
is the modulus of the elliptic function with 0<ω<1. Here c1 and c2 are constants of integration. Reverting back to our original variables, we can now write the solution of our Klein-Gordon-Zakharov equations as
(32)u(x,t)=±1Hisn(F∣ω),
where
(33)F=(P2-2Qc1-P)(kx+dt+c2)22,
and ω and H are as above.Nowv(x,t) can be obtained from (26).It should be noted that the solution (32) is valid for 0<ω<1, as ω approaches zero, the solution becomes the normal sine function, sinz, and as ω approaches 1, the solution tends to the tanh function, tanhz.The profile of the solution (32) is given in Figure 1.Figure 1
Profile of solution (32).


## 3.2. Solutions of (22a) and (22b) Using the Simplest Equation Method

We consider the solutions of (27) in the form
(34)u(z)=∑i=0M‍Ai(G(z))i,
where G(z) satisfies the Bernoulli or the Riccati equation.


### 3.2.1. Solutions of (22a) and (22b) Using Bernoulli as the Simplest Equation

We consider the Bernoulli equation(35)G′(z)=aG(z)+bG2(z),
where a and b are constants.The balancing procedure yieldsM=1, so the solution of (27) is of the form
(36)u(z)=A0+A1H(z).
Substituting (36) into (27), making use of the Bernoulli equation (35), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations:
(37)2A1k4b2-k2A13+d2A13-4A1d2k2b2+A13+2A1d4b2=0,3A1k4ab+3d2A0A12-3k2A0A12-6A1d2k2ab+3A0A12+3A1d4ab=0,p2A0d2-p2A0k2-r2A0d2+r2A0k2-A0k2+d2A03-k2A03+A0d2+A03=0,A1d2+3d2A02A1-p2A1k2+r2A1k2+p2A1d2+A1d4a2-3k2A02A1+A1k4a2+3A02A1-2A1d2k2a2-r2A1d2-A1k2=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(38)A0=a(d2-k2)2(k2-d2-1),A1=2b(d2-k2)k2-d2-1,p=r2d2-r2k2+k2-d2A02+k2A02-d2-A02d2-k2.
As a result, a solution of the Klein-Gordon-Zakharov equations (22a) and (22b), using the Bernoulli equation as the simplest equation, is
(39)u(x,t)=ei(px+rt)[+a(d2-k2)2(k2-d2-1)](-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)(2ab(d2-k2)(cosh⁡(a(kx+dt+c))+sinh⁡(a(kx+dt+c)))(2ab(d2-k2)(cosh⁡(a(kx+dt+c)))×(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c))-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)+a(d2-k2)2(k2-d2-1)],
where c is a constant of integration.


### 3.2.2. Solutions of (22a) and (22b) Using Riccati as the Simplest Equation

We use the Riccati equation given by(40)G′(z)=aG2(z)+bG(z)+c,
where a, b, and c are constants. The balancing procedure yields M=1, so the solution of (27) is of the form
(41)u(z)=A0+A1G(z).
Similar calculations yield the following set of values:
(42)A0=b(d2-k2)2(k2-d2-1),A1=-2a(d2-k2)k2-d2-1d2-k2+1,c=-2(k2-d2-1)(d2b2-k2b2-2-2p2+2r2)4A1(d2-k2+1).
As a result the two solutions of (22a) and (22b) are
(43)u(x,t)=eiϕ[A0+A1{-b2a-θ2atanh[12θ(z+C)]}],u(x,t)=eiϕ[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0+A1{sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)-b2a-θ2atanh⁡(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0],
where ϕ=px+rt and z=kx+dt. θ is given by b2-4ac, C is a constant of integration, and A0 and A1 are as obtained above.It should be noted that by substituting the above value ofu(x,t) into (26), one can now obtain the solution for the variable v(x,t).


## 3.2.1. Solutions of (22a) and (22b) Using Bernoulli as the Simplest Equation

We consider the Bernoulli equation(35)G′(z)=aG(z)+bG2(z),
where a and b are constants.The balancing procedure yieldsM=1, so the solution of (27) is of the form
(36)u(z)=A0+A1H(z).
Substituting (36) into (27), making use of the Bernoulli equation (35), and then equating all coefficients of the function Hi to zero, we obtain the following algebraic system of equations:
(37)2A1k4b2-k2A13+d2A13-4A1d2k2b2+A13+2A1d4b2=0,3A1k4ab+3d2A0A12-3k2A0A12-6A1d2k2ab+3A0A12+3A1d4ab=0,p2A0d2-p2A0k2-r2A0d2+r2A0k2-A0k2+d2A03-k2A03+A0d2+A03=0,A1d2+3d2A02A1-p2A1k2+r2A1k2+p2A1d2+A1d4a2-3k2A02A1+A1k4a2+3A02A1-2A1d2k2a2-r2A1d2-A1k2=0.
Solving this system, with the aid of Maple, we obtain the following values for the constants:
(38)A0=a(d2-k2)2(k2-d2-1),A1=2b(d2-k2)k2-d2-1,p=r2d2-r2k2+k2-d2A02+k2A02-d2-A02d2-k2.
As a result, a solution of the Klein-Gordon-Zakharov equations (22a) and (22b), using the Bernoulli equation as the simplest equation, is
(39)u(x,t)=ei(px+rt)[+a(d2-k2)2(k2-d2-1)](-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)(2ab(d2-k2)(cosh⁡(a(kx+dt+c))+sinh⁡(a(kx+dt+c)))(2ab(d2-k2)(cosh⁡(a(kx+dt+c)))×(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c))-bsinh⁡(a(kx+dt+c)))(k2-d2-1(-bsinh⁡(a(kx+dt+c))))1-bcosh⁡(a(kx+dt+c)))-1)+a(d2-k2)2(k2-d2-1)],
where c is a constant of integration.


## 3.2.2. Solutions of (22a) and (22b) Using Riccati as the Simplest Equation

We use the Riccati equation given by(40)G′(z)=aG2(z)+bG(z)+c,
where a, b, and c are constants. The balancing procedure yields M=1, so the solution of (27) is of the form
(41)u(z)=A0+A1G(z).
Similar calculations yield the following set of values:
(42)A0=b(d2-k2)2(k2-d2-1),A1=-2a(d2-k2)k2-d2-1d2-k2+1,c=-2(k2-d2-1)(d2b2-k2b2-2-2p2+2r2)4A1(d2-k2+1).
As a result the two solutions of (22a) and (22b) are
(43)u(x,t)=eiϕ[A0+A1{-b2a-θ2atanh[12θ(z+C)]}],u(x,t)=eiϕ[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0+A1{sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)-b2a-θ2atanh⁡(12θz)+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}[+sech(θz/2)Ccosh⁡(θz/2)-(2a/θ)sinh⁡(θz/2)}]A0],
where ϕ=px+rt and z=kx+dt. θ is given by b2-4ac, C is a constant of integration, and A0 and A1 are as obtained above.It should be noted that by substituting the above value ofu(x,t) into (26), one can now obtain the solution for the variable v(x,t).


## 4. Conclusion

In this paper we studied two nonlinear partial differential equations. Firstly, Lie symmetry approach along with the simplest equation and the Exp-function method were used to obtain travelling wave solutions of the symmetric regularized long wave equation. Secondly, the travelling wave hypothesis approach along with the simplest equation method is utilized to obtain new exact solutions of the Klein-Gordon-Zakharov equations.



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