# Exact Solutions for a Third-Order KdV Equation with Variable Coefficients and Forcing Term

Authors: Alvaro H. Salas S; Cesar A. Gómez S
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

The general projective Riccati equation method and the Exp-function
method are used to construct generalized soliton solutions and periodic solutions to special
KdV equation with variable coefficients and forcing term.


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## Body

## 1. Introduction

It is well  known that in an early phase of the development of the solitons theory, there were already many applications in physics and engineering. In particular, traveling waves as solutions of the KdV equation(1.1)ut+6uux+uxxx=0
have been of some interest since 150 years. Some generalizations of this last equation have been studied recently. For instance, the equation(1.2)ut+k1tnuux+k2tmuxxx=0,
where k1, k2 are arbitrary constants, which have applications in physics, has been analyzed in [1] from the point of view of its exact solutions. The search of explicit solutions to nonlinear partial differential equations (NLPDEs) using analytic methods is not an easy task. However, the use of computational methods facilitates this work. Some powerful computational methods such as the tanh method [2], the generalized tanh method [3, 4], the extended tanh method [5–10], the improved tanh-coth method [11–13], the Exp-function method [14–18], the modified Exp-function method [19], the Cole-Hopf transformation [20], the projective Riccati equation method (PREM) [21, 22], the generalized projective Riccati equations method [23–25], the extended hyperbolic function method [26], and many other methods have been developed in this direction. The PREM and the Exp-function method have been used in a satisfactory form to solve some NLPDEs [15–17, 24, 25, 27–30]. In this paper, we use this last two methods to obtain soliton and periodic solutions to the following special KdV equation with variable coefficients and forcing term:(1.3)ut+α(t)uux+kα(t)uxxx=F(t),
where F(t) is an external forcing function varying with time t, k is a constant, and α=α(t) is a function of t, α(t)≠0. Equation  (1.3)  is a generalization of the following equation [15, 17, 31]:(1.4)ut+αuux+βuxxx=F(t)(α,β=const),
which results from (1.3) by taking α=const and k=β/α.We suppose that the solution to (1.3) has the form(1.5)u(x,t)=∫F(t)dt+v(x,t).
Therefore, (1.3) reduces to(1.6)vt+α(t)vvx+α(t)∫F(t)dtvx+kα(t)vxxx=0.
Now we consider the transformation(1.7)v=V(ξ),ξ=λx+∫h(t)dt,
where λ is a constant and h(t) is an unknown function of t to be determined later. Substituting (1.7) into (1.6), we obtain(1.8)h(t)V′(ξ)+α(t)λ(V(ξ)+f(t))V′(ξ)+kλ3α(t)V′′′(ξ)=0,
where(1.9)f=f(t)=∫F(t)dt.


## 2. The Exp-Function Method

Recently   He and Wu [14] have introduced the Exp-function method to solve nonlinear equations. In particular, the Exp-function method is a useful tool for solving nonlinear equations with high nonlinearity. The method has been used in a satisfactory way by other authors to solve a great variety of nonlinear wave equations [14–19]. The Exp-function method is very simple and straightforward  and is based on a priori assumption that traveling wave solutions to a nonlinear partial differential equation in the form(2.1)F(u,ux,ut,uxx,uxt,utt,…)=0
can be found using the expression(2.2)u(ξ)=∑n=-cdanexp(nξ)∑n=-pqamexp(mξ),
where c, d, p, and q are positive integers which could be freely chosen; an and bn are unknown constants to be determined. According to this, we suppose that solutions to (1.8) can be expressed in the form(2.3)v(ξ)=∑n=-cdanexp(nξ)∑m=-pqbmexp(mξ)=a-cexp(-cξ)+⋯+adexp(dξ)b-pexp(-pξ)+⋯+bqexp(qξ),
where c, d, p, and q are positive integers which are unknown to be determined later; an and bm are unknown constants. We have the following two cases.


### 2.1. Case1: p=c=1 and d=q=1

In this case, the trial solution to (1.8) becomes(2.4)V(ξ)=a1exp(ξ)+a0+a-1exp(-ξ)exp(ξ)+b0+b-1exp(-ξ).
Substituting (2.4) into (1.8) we obtain a polynomial equation in the variable η=exp(ξ). Equating to zero the coefficients of all powers of η yields a set of algebraic equations. Solving it with the aid of a computer, we get a0=b0(a1+6kλ2), a-1=(1/4)a1b02,   b-1=b02/4, and  h(t)=-λα(t)(a1+f(t)+kλ2), and,  from (1.5), (1.7), and (2.3), one solution to (1.3) is given by(2.5)u(x,t)=a1(b0+2eξ)2+24b0kλ2eξ(b0+2eξ)2+∫F(t)dt,
where(2.6)ξ=ξ(x,t)=λx+∫h(t)dt=λx-∫λα(t)(a1+kλ2+∫F(t)dt)dt,
and a1, b0, and λ are arbitrary real or complex numbers.


### 2.2. Case2: p=c=2 and d=q=2

In this case, the trial solution to (1.8) becomes(2.7)V(ξ)=a2exp(2ξ)+a1exp(ξ)+a0+a-1exp(-ξ)+a-2exp(-2ξ)exp(2ξ)+b1exp(ξ)+b0+b-1exp(-ξ)+b-2exp(-2ξ).
As in the first case, we obtain an algebraic system. Solving it gives(2.8)a0=--2a1b1(a2+3kλ2)+a2(a2b12+6(b12-b0)kλ2)+a126kλ2,a-1=a1-a2b1432k3λ6(-a1b1(27a2kλ2+2a22+72k2λ4)+(a2+6kλ2)(12a2b12kλ2+a22b12+72b0k2λ4)+a12(a2+9kλ2)),a-2=16912k4λ8a2(a1-a2b1)2(8a2b12kλ2-2a1b1(a2+4kλ2)+a22b12+a12+48b0k2λ4),b-1=1432k3λ6(a1-a2b1)(9a2b12kλ2-a1b1(2a2+9kλ2)+a22b12+a12+72b0k2λ4),b-2=16912k4λ8(a1-a2b1)2(8a2b12kλ2-2a1b1(a2+4kλ2)+a22b12+a12+48b0k2λ4),h(t)=-λα(t)(a2+f(t)+kλ2).
From (1.5), (1.7), and (2.7) we may verify that to this set of values corresponds the solution(2.9)u(x,t)=1(a1-a2b1+12kλ2eξ)2(a12a2+a1(-2a22b1+24a2kλ2eξ+144k2λ4eξ)+a2(a22b12-24a2b1kλ2eξ+144k2λ4eξ(eξ-b1)))+∫F(t)dt,
where(2.10)ξ=ξ(x,t)=λx+∫h(t)dt=λx-∫λα(t)(a2+kλ2+∫F(t)dt)dt,
and a1, a2, b1,  and λ are arbitrary real or complex numbers.


## 2.1. Case1: p=c=1 and d=q=1

In this case, the trial solution to (1.8) becomes(2.4)V(ξ)=a1exp(ξ)+a0+a-1exp(-ξ)exp(ξ)+b0+b-1exp(-ξ).
Substituting (2.4) into (1.8) we obtain a polynomial equation in the variable η=exp(ξ). Equating to zero the coefficients of all powers of η yields a set of algebraic equations. Solving it with the aid of a computer, we get a0=b0(a1+6kλ2), a-1=(1/4)a1b02,   b-1=b02/4, and  h(t)=-λα(t)(a1+f(t)+kλ2), and,  from (1.5), (1.7), and (2.3), one solution to (1.3) is given by(2.5)u(x,t)=a1(b0+2eξ)2+24b0kλ2eξ(b0+2eξ)2+∫F(t)dt,
where(2.6)ξ=ξ(x,t)=λx+∫h(t)dt=λx-∫λα(t)(a1+kλ2+∫F(t)dt)dt,
and a1, b0, and λ are arbitrary real or complex numbers.


## 2.2. Case2: p=c=2 and d=q=2

In this case, the trial solution to (1.8) becomes(2.7)V(ξ)=a2exp(2ξ)+a1exp(ξ)+a0+a-1exp(-ξ)+a-2exp(-2ξ)exp(2ξ)+b1exp(ξ)+b0+b-1exp(-ξ)+b-2exp(-2ξ).
As in the first case, we obtain an algebraic system. Solving it gives(2.8)a0=--2a1b1(a2+3kλ2)+a2(a2b12+6(b12-b0)kλ2)+a126kλ2,a-1=a1-a2b1432k3λ6(-a1b1(27a2kλ2+2a22+72k2λ4)+(a2+6kλ2)(12a2b12kλ2+a22b12+72b0k2λ4)+a12(a2+9kλ2)),a-2=16912k4λ8a2(a1-a2b1)2(8a2b12kλ2-2a1b1(a2+4kλ2)+a22b12+a12+48b0k2λ4),b-1=1432k3λ6(a1-a2b1)(9a2b12kλ2-a1b1(2a2+9kλ2)+a22b12+a12+72b0k2λ4),b-2=16912k4λ8(a1-a2b1)2(8a2b12kλ2-2a1b1(a2+4kλ2)+a22b12+a12+48b0k2λ4),h(t)=-λα(t)(a2+f(t)+kλ2).
From (1.5), (1.7), and (2.7) we may verify that to this set of values corresponds the solution(2.9)u(x,t)=1(a1-a2b1+12kλ2eξ)2(a12a2+a1(-2a22b1+24a2kλ2eξ+144k2λ4eξ)+a2(a22b12-24a2b1kλ2eξ+144k2λ4eξ(eξ-b1)))+∫F(t)dt,
where(2.10)ξ=ξ(x,t)=λx+∫h(t)dt=λx-∫λα(t)(a2+kλ2+∫F(t)dt)dt,
and a1, a2, b1,  and λ are arbitrary real or complex numbers.


## 3. General Projective Riccati Equation Method

The projective  Riccati equation method was introduced initially in [21] and generalizations of this method have been used in a satisfactory way by several authors to solve nonlinear partial equations [22–25]. Using this last method [24, 25, 27–30], we seek solutions to (1.8) in the form(3.1)V(ξ)=a0+∑i=1mσi-1(ξ)(aiσ(ξ)+biτ(ξ)),
where a0, a1, b1,… are constants and σ(ξ) and τ(ξ) satisfy the system(3.2)σ′(ξ)=ϵσ(ξ)τ(ξ),τ′(ξ)=R+ϵτ2(ξ)-μσ(ξ).
In (3.2),  ϵ=±1 and R and μ are certain constants. These equations have following solutions.Case 1.
Whenϵ=-1 and R≠0,
(3.3)σ1(ξ)=Rsech(Rξ)μsech(Rξ)+1,τ1(ξ)=Rtanh(Rξ)μsech(Rξ)+1,σ2(ξ)=Rcsch(Rξ)μcsch(Rξ)+1,τ2(ξ)=Rcoth(Rξ)μcsch(Rξ)+1.Case 2.
Whenϵ=1 and R≠0,
(3.4)σ3(ξ)=Rsec(Rξ)μsec(Rξ)+1,τ3(ξ)=Rtan(Rξ)μsec(Rξ)+1,σ4(ξ)=Rcsc(Rξ)μcsc(Rξ)+1,τ4(ξ)=Rcot(Rξ)μcsc(Rξ)+1.Case 3.
WhenR=μ=0,
(3.5)σ5(ξ)=Cξ,τ5(ξ)=1ϵξ.
In this last case, we seek solutions to (1.8) in the form
(3.6)V(ξ)=∑i=0maiτi(ξ),
where τ′(ξ)=τ2(ξ).
For any pair(σ(ξ),τ(ξ)) of functions given by (3.3) or (3.4) the following equation holds:
(3.7)τ2(ξ)=-ϵ[R-2μσ(ξ)+μ2-1Rσ2(ξ)].


### 3.1. Periodic and Soliton Solutions

Periodic and soliton solutions are obtained whenR≠0 and ϵ=±1 and it  corresponds  to the first  two cases. Substituting (3.1), along with (3.2) and (3.7) into the left hand of (1.8) and collecting all terms with the same power in σi(ξ)τj(ξ), we get a polynomial in the variables σ=σ(ξ) and τ=τ(ξ). We may choose m=2. Thus, we seek solutions to (1.8) in the form(3.8)V(ξ)=a0+a1σ(ξ)+b1τ(ξ)+σ(ξ)(a2σ(ξ)+b2τ(ξ)).
We equate each coefficient of the polynomial to zero. This will give an overdetermined system of algebraic equations involving the parameters ai, bi(i=0,…,m), λ and μ, R, and the unknown function h(t). Having determined these parameters, we  may determine V(ξ), and using (1.5) we obtain an exact solution u(x,t) in a closed form. The corresponding system reads.(i)
b12R3αλ-b12R3αϵ2λ=0.(ii)
6a1kRαλ3ϵ4-6a1kRαλ3μ2ϵ4+48a2kR2αλ3μϵ4-3b1b2Rαλμ2ϵ2+3b1b2Rαλϵ2-18a2kR2αλ3μϵ2+4b22R2αλμϵ2+3a1a2R2αλϵ-b22R2αλμ=0.(iii)
24a2kRαλ3ϵ4-24a2kRαλ3μ2ϵ4-2b22Rαλμ2ϵ2+2b22Rαλϵ2+2a22R2αλϵ=0.(iv)
6b1kαλ3μ4ϵ5+6b1kαλ3ϵ5-96b2kRαλ3μ3ϵ5-12b1kαλ3μ2ϵ5+96b2kRαλ3μϵ5+36b2kRαλ3μ3ϵ3-36b2kRαλ3μϵ3-3a2b1Rαλμ2ϵ2-3a1b2Rαλμ2ϵ2+3a2b1Rαλϵ2+3a1b2Rαλϵ2+8a2b2R2αλμϵ2-a2b2R2αλμ=0.(v)
24b2kαλ3μ4ϵ5+24b2kαλ3ϵ5-48b2kαλ3μ2ϵ5-4a2b2Rαλμ2ϵ2+4a2b2Rαλϵ2=0, -6a1kR3αλ3ϵ4+5a1kR3αλ3ϵ2-3b1b2R3αλϵ2+2b12R2αλμϵ2+a0a1R2αλϵ+a1fR2αλϵ+a1R2hϵ+2b1b2R3αλ-b12R2αλμ=0.(vi)
-24a2kR3αλ3ϵ4+12a1kR2αλ3μϵ4+16a2kR3αλ3ϵ2-b12Rαλμ2ϵ2-2b22R3αλϵ2+b12Rαλϵ2-6a1kR2αλ3μϵ2+6b1b2R2αλμϵ2+a12R2αλϵ+2a0a2R2αλϵ+2a2fR2αλϵ+2a2R2hϵ+b22R3αλ-2b1b2R2αλμ=0.(vii)
6b1kR4αλ3ϵ5-8b1kR4αλ3ϵ3-a0b1R3αλϵ2-b1fR3αλϵ2-b1R3hϵ2+2b1kR4αλ3ϵ+a0b1R3αλ+b1fR3αλ+b1R3h=0.(viii)
24b2kR4αλ3ϵ5-24b1kR3αλ3μϵ5-28b2kR4αλ3ϵ3+28b1kR3αλ3μϵ3-2a1b1R3αλϵ2-2a0b2R3αλϵ2-2b2fR3αλϵ2+2a0b1R2αλμϵ2+2b1fR2αλμϵ2-2b2R3hϵ2+2b1R2μhϵ2+5b2kR4αλ3ϵ-5b1kR3αλ3μϵ+a1b1R3αλ+a0b2R3αλ+b2fR3αλ-a0b1R2αλμ-b1fR2αλμ+b2R3h-b1R2μh=0.(ix)
-12b1kR2αλ3ϵ5+36b1kR2αλ3μ2ϵ5-96b2kR3αλ3μϵ5+8b1kR2αλ3ϵ3-32b1kR2αλ3μ2ϵ3+92b2kR3αλ3μϵ3-a0b1Rαλμ2ϵ2-b1fRαλμ2ϵ2-3a2b1R3αλϵ2-3a1b2R3αλϵ2+a0b1Rαλϵ2+b1fRαλϵ2+4a1b1R2αλμϵ2+4a0b2R2αλμϵ2+4b2fR2αλμϵ2-b1Rμ2hϵ2+b1Rhϵ2+4b2R2μhϵ2+3b1kR2αλ3μ2ϵ-11b2kR3αλ3μϵ+a2b1R3αλ+a1b2R3αλ-a1b1R2αλμ-a0b2R2αλμ-b2fR2αλμ-b2R2μh=0.(x)
-48b2kR2αλ3ϵ5-24b1kRαλ3μ3ϵ5+144b2kR2αλ3μ2ϵ5+24b1kRαλ3μϵ5+28b2kR2αλ3ϵ3+12b1kRαλ3μ3ϵ3-100b2kR2αλ3μ2ϵ3-12b1kRαλ3μϵ3-2a1b1Rαλμ2ϵ2-2a0b2Rαλμ2ϵ2-2b2fRαλμ2ϵ2-4a2b2R3αλϵ2+2a1b1Rαλϵ2+2a0b2Rαλϵ2+2b2fRαλϵ2+6a2b1R2αλμϵ2+6a1b2R2αλμϵ2-2b2Rμ2hϵ2+2b2Rhϵ2+6b2kR2αλ3μ2ϵ+a2b2R3αλ-a2b1R2αλμ-a1b2R2αλμ=0.In the equations above,f=f(t)=∫F(t)dt, α=α(t), and h=h(t). Solving the previous system with the aid of a computer, we obtain many solutions to (1.3). These solutions may be obtained from (1.5) and  are given by (3.9)–(3.19). In these formulas, H(t)=∫h(t)dt  and λ, a0 are arbitrary parameters. In a formula containing R, we suppose that R>0 and if the expression -R appears, we choose R<0. If a formula involves -R and 1-μ2 (see, e.g.,  (3.17)) we consider that R<0 and |μ|≤1.First Group.ϵ=1:.
(i)
a2=b1=b2=0, a1=6kλ2, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=-1:(3.9)u1(x,t)=∫F(t)dt+a0-6kλ2R1-sin(R(λx+H(t))),u2(x,t)=∫F(t)dt+a0-6kλ2R1-cos(R(λx+H(t))),u3(x,t)=∫F(t)dt+a0-6kλ2R1-cosh(-R(λx+H(t))).(ii)
a2=b1=b2=0, a1=-6kλ2, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=1:(3.10)u4(x,t)=∫F(t)dt+a0-6kλ2R1+sin(R(λx+H(t))),u5(x,t)=∫F(t)dt+a0-6kλ2R1+cos(R(λx+H(t))),u6(x,t)=∫F(t)dt+a0-6kλ2R1+cosh(-R(λx+H(t))).(iii)
a1=-6kλ2μ, b1=0, a2=6kλ2(μ2-1)/R, b2=6kλ2R(1-μ2)/R, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=μ:(3.11)u7(x,t)=∫F(t)dt+a0-6kλ2R(1+μcos(R(λx+H(t)))-1-μ2sin(R(λx+H(t))))μ2+2μcos(R(λx+H(t)))+cos2(R(λx+H(t))),u8(x,t)=∫F(t)dt+a0-6kλ2R(1+μcosh(-R(λx+H(t)))-μ2-1sinh(-R(λx+H(t))))(μ+cosh(-R(λx+H(t))))2,u9(x,t)=∫F(t)dt+a0-6λ2kR(1+μsin(R(λx+H(t)))-1-μ2cos(R(λx+H(t))))1+μ2+2μsin(R(λx+H(t)))-cos2(R(λx+H(t))).(iv)
a1=-6kλ2μ, b1=0, a2=6kλ2(μ2-1)/R, b2=-6kλ2R(1-μ2)/R, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=μ:(3.12)u10(x,t)=∫F(t)dt+a0-6kλ2R(1+μcos(R(λx+H(t)))+1-μ2sin(R(λx+H(t))))μ2+2μcos(R(λx+H(t)))+cos2(R(λx+H(t))),u11(x,t)=∫F(t)dt+a0-6kλ2R(1+μcosh(-R(λx+H(t)))+μ2-1sinh(-R(λx+H(t))))(μ+cosh(-R(λx+H(t))))2.
Figure 1 shows solution u11(x,t) for the choices μ=1.3, R=-2.3, k=2.3, λ=1.9, a0=1, F(t)=sin(t), α(t)=t, -3≤x≤4, and -1≤t≤1.1:
(3.13)u12(x,t)=∫F(t)dt+a0-6kλ2R(1+μsin(R(λx+H(t)))+1-μ2cos(R(λx+H(t))))1+μ2+2μsin(R(λx+H(t)))-cos2(R(λx+H(t))).
Figure 2 shows solution u12(x,t) for the choices μ=0.3, R=2.3, k=2.3, λ=1.9, a0=1, F(t)=sin(t), α(t)=t, -3≤x≤4, and -1≤t≤1.1.Figure 1
Graphic of solutionu11(x,t), -3≤x≤4 and -1≤t≤1.1.Figure 2
Graphic of solutionu12(x,t), -3≤x≤4 and -1≤t≤1.1.Second Group.ϵ=-1:.
(v)
a1=b1=0, a2=6kλ2/R, b2=-6kλ2/-R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=0:(3.14)u13(x,t)=∫F(t)dt+a0+6kλ2R(1+sin(-R(λx+H(t))))cos2(-R(λx+H(t))).(vi)
a1=b1=0, a2=6kλ2/R, b2=6kλ2/-R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=0:(3.15)u14(x,t)=∫F(t)dt+a0+6kλ2R(1-sin(-R(λx+H(t))))cos2(-R(λx+H(t))).(vii)
a1=6kλ2μ, a2=6kλ2(1-μ2)/R, b1=0, b2=6kλ2R(μ2-1)/R,  h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2),μ=μ:(3.16)u15(x,t)=∫F(t)dt+a0+6kλ2μRμ+cosh(R(λx+H(t)))+6kλ2R(1-μ2+μ2-1sinh(R(λx+H(t))))(μ+cosh(R(λx+H(t))))2,(3.17)u16(x,t)=∫F(t)dt+a0+6kλ2μRμ+cos(-R(λx+H(t)))+6kλ2R(1-μ2+1-μ2sin(-R(λx+H(t))))μ2+2μcos(-R(λx+H(t)))+cos2(-R(λx+H(t))).(viii)
a1=6kλ2μ, a2=-6kλ2(1-μ2)/R, b1=0, b2=-6kλ2R(μ2-1)/R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=μ:(3.18)u17(x,t)=∫F(t)dt+a0+6kλ2μRμ+cosh(R(λx+H(t)))+6kλ2R(1-μ2-μ2-1sinh(R(λx+H(t))))(μ+cosh(R(λx+H(t))))2,(3.19)u18(x,t)=∫F(t)dt+a0+6kλ2μRμ+cos(-R(λx+H(t)))+6kλ2R(1-μ2-1-μ2sin(-R(λx+H(t))))μ2+2μcos(-R(λx+H(t)))+cos2(-R(λx+H(t))).


### 3.2. Rational Solutions

We seek rational solutions to (1.8) (R=μ=0) in the form given by (3.6) with m=2,(3.20)V(ξ)=∑i=02aiτi(ξ),
where(3.21)τ′(ξ)=τ2(ξ).
Note that  this last equation has the general solution(3.22)τ(ξ)=-1ξ+C.
We now substitute (3.20) into (1.8), and using the relation τ′(ξ)=τ2(ξ) we obtain an equation whose left-hand side is a polynomial in the variable τ=τ(ξ). We equate each coefficient of this polynomial to zero and we get the following algebraic system.(i)
h(t)a1+α(t)λ(∫F(t)dt)a1+α(t)λa0a1=0.(ii)
α(t)λa12+2h(t)a2+2α(t)λ(∫F(t)dt)a2+2α(t)λa0a2=0.(iii)
6ka1λ3+3a1a2λ=0.(iv)
24ka2λ3+2a22λ=0.Solving this system givesa0=a0, a1=0,  a2=-12kλ2 and(3.23)h(t)=-λα(t)(∫F(t)dt+a0).
A rational solution of (1.8) is given by(3.24)V(ξ)=a0-12kλ2(ξ+C)2.
According to (1.5), (1.6), and (1.7) we obtain the following rational solution to (1.3):(3.25)u19(x,t)=∫F(t)dt+a0-12kλ2(λx-∫(λα(t)(∫F(t)dt+a0))dt+C)2.


## 3.1. Periodic and Soliton Solutions

Periodic and soliton solutions are obtained whenR≠0 and ϵ=±1 and it  corresponds  to the first  two cases. Substituting (3.1), along with (3.2) and (3.7) into the left hand of (1.8) and collecting all terms with the same power in σi(ξ)τj(ξ), we get a polynomial in the variables σ=σ(ξ) and τ=τ(ξ). We may choose m=2. Thus, we seek solutions to (1.8) in the form(3.8)V(ξ)=a0+a1σ(ξ)+b1τ(ξ)+σ(ξ)(a2σ(ξ)+b2τ(ξ)).
We equate each coefficient of the polynomial to zero. This will give an overdetermined system of algebraic equations involving the parameters ai, bi(i=0,…,m), λ and μ, R, and the unknown function h(t). Having determined these parameters, we  may determine V(ξ), and using (1.5) we obtain an exact solution u(x,t) in a closed form. The corresponding system reads.(i)
b12R3αλ-b12R3αϵ2λ=0.(ii)
6a1kRαλ3ϵ4-6a1kRαλ3μ2ϵ4+48a2kR2αλ3μϵ4-3b1b2Rαλμ2ϵ2+3b1b2Rαλϵ2-18a2kR2αλ3μϵ2+4b22R2αλμϵ2+3a1a2R2αλϵ-b22R2αλμ=0.(iii)
24a2kRαλ3ϵ4-24a2kRαλ3μ2ϵ4-2b22Rαλμ2ϵ2+2b22Rαλϵ2+2a22R2αλϵ=0.(iv)
6b1kαλ3μ4ϵ5+6b1kαλ3ϵ5-96b2kRαλ3μ3ϵ5-12b1kαλ3μ2ϵ5+96b2kRαλ3μϵ5+36b2kRαλ3μ3ϵ3-36b2kRαλ3μϵ3-3a2b1Rαλμ2ϵ2-3a1b2Rαλμ2ϵ2+3a2b1Rαλϵ2+3a1b2Rαλϵ2+8a2b2R2αλμϵ2-a2b2R2αλμ=0.(v)
24b2kαλ3μ4ϵ5+24b2kαλ3ϵ5-48b2kαλ3μ2ϵ5-4a2b2Rαλμ2ϵ2+4a2b2Rαλϵ2=0, -6a1kR3αλ3ϵ4+5a1kR3αλ3ϵ2-3b1b2R3αλϵ2+2b12R2αλμϵ2+a0a1R2αλϵ+a1fR2αλϵ+a1R2hϵ+2b1b2R3αλ-b12R2αλμ=0.(vi)
-24a2kR3αλ3ϵ4+12a1kR2αλ3μϵ4+16a2kR3αλ3ϵ2-b12Rαλμ2ϵ2-2b22R3αλϵ2+b12Rαλϵ2-6a1kR2αλ3μϵ2+6b1b2R2αλμϵ2+a12R2αλϵ+2a0a2R2αλϵ+2a2fR2αλϵ+2a2R2hϵ+b22R3αλ-2b1b2R2αλμ=0.(vii)
6b1kR4αλ3ϵ5-8b1kR4αλ3ϵ3-a0b1R3αλϵ2-b1fR3αλϵ2-b1R3hϵ2+2b1kR4αλ3ϵ+a0b1R3αλ+b1fR3αλ+b1R3h=0.(viii)
24b2kR4αλ3ϵ5-24b1kR3αλ3μϵ5-28b2kR4αλ3ϵ3+28b1kR3αλ3μϵ3-2a1b1R3αλϵ2-2a0b2R3αλϵ2-2b2fR3αλϵ2+2a0b1R2αλμϵ2+2b1fR2αλμϵ2-2b2R3hϵ2+2b1R2μhϵ2+5b2kR4αλ3ϵ-5b1kR3αλ3μϵ+a1b1R3αλ+a0b2R3αλ+b2fR3αλ-a0b1R2αλμ-b1fR2αλμ+b2R3h-b1R2μh=0.(ix)
-12b1kR2αλ3ϵ5+36b1kR2αλ3μ2ϵ5-96b2kR3αλ3μϵ5+8b1kR2αλ3ϵ3-32b1kR2αλ3μ2ϵ3+92b2kR3αλ3μϵ3-a0b1Rαλμ2ϵ2-b1fRαλμ2ϵ2-3a2b1R3αλϵ2-3a1b2R3αλϵ2+a0b1Rαλϵ2+b1fRαλϵ2+4a1b1R2αλμϵ2+4a0b2R2αλμϵ2+4b2fR2αλμϵ2-b1Rμ2hϵ2+b1Rhϵ2+4b2R2μhϵ2+3b1kR2αλ3μ2ϵ-11b2kR3αλ3μϵ+a2b1R3αλ+a1b2R3αλ-a1b1R2αλμ-a0b2R2αλμ-b2fR2αλμ-b2R2μh=0.(x)
-48b2kR2αλ3ϵ5-24b1kRαλ3μ3ϵ5+144b2kR2αλ3μ2ϵ5+24b1kRαλ3μϵ5+28b2kR2αλ3ϵ3+12b1kRαλ3μ3ϵ3-100b2kR2αλ3μ2ϵ3-12b1kRαλ3μϵ3-2a1b1Rαλμ2ϵ2-2a0b2Rαλμ2ϵ2-2b2fRαλμ2ϵ2-4a2b2R3αλϵ2+2a1b1Rαλϵ2+2a0b2Rαλϵ2+2b2fRαλϵ2+6a2b1R2αλμϵ2+6a1b2R2αλμϵ2-2b2Rμ2hϵ2+2b2Rhϵ2+6b2kR2αλ3μ2ϵ+a2b2R3αλ-a2b1R2αλμ-a1b2R2αλμ=0.In the equations above,f=f(t)=∫F(t)dt, α=α(t), and h=h(t). Solving the previous system with the aid of a computer, we obtain many solutions to (1.3). These solutions may be obtained from (1.5) and  are given by (3.9)–(3.19). In these formulas, H(t)=∫h(t)dt  and λ, a0 are arbitrary parameters. In a formula containing R, we suppose that R>0 and if the expression -R appears, we choose R<0. If a formula involves -R and 1-μ2 (see, e.g.,  (3.17)) we consider that R<0 and |μ|≤1.First Group.ϵ=1:.
(i)
a2=b1=b2=0, a1=6kλ2, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=-1:(3.9)u1(x,t)=∫F(t)dt+a0-6kλ2R1-sin(R(λx+H(t))),u2(x,t)=∫F(t)dt+a0-6kλ2R1-cos(R(λx+H(t))),u3(x,t)=∫F(t)dt+a0-6kλ2R1-cosh(-R(λx+H(t))).(ii)
a2=b1=b2=0, a1=-6kλ2, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=1:(3.10)u4(x,t)=∫F(t)dt+a0-6kλ2R1+sin(R(λx+H(t))),u5(x,t)=∫F(t)dt+a0-6kλ2R1+cos(R(λx+H(t))),u6(x,t)=∫F(t)dt+a0-6kλ2R1+cosh(-R(λx+H(t))).(iii)
a1=-6kλ2μ, b1=0, a2=6kλ2(μ2-1)/R, b2=6kλ2R(1-μ2)/R, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=μ:(3.11)u7(x,t)=∫F(t)dt+a0-6kλ2R(1+μcos(R(λx+H(t)))-1-μ2sin(R(λx+H(t))))μ2+2μcos(R(λx+H(t)))+cos2(R(λx+H(t))),u8(x,t)=∫F(t)dt+a0-6kλ2R(1+μcosh(-R(λx+H(t)))-μ2-1sinh(-R(λx+H(t))))(μ+cosh(-R(λx+H(t))))2,u9(x,t)=∫F(t)dt+a0-6λ2kR(1+μsin(R(λx+H(t)))-1-μ2cos(R(λx+H(t))))1+μ2+2μsin(R(λx+H(t)))-cos2(R(λx+H(t))).(iv)
a1=-6kλ2μ, b1=0, a2=6kλ2(μ2-1)/R, b2=-6kλ2R(1-μ2)/R, h(t)=-λα(t)(a0+∫F(t)dt-Rkλ2), μ=μ:(3.12)u10(x,t)=∫F(t)dt+a0-6kλ2R(1+μcos(R(λx+H(t)))+1-μ2sin(R(λx+H(t))))μ2+2μcos(R(λx+H(t)))+cos2(R(λx+H(t))),u11(x,t)=∫F(t)dt+a0-6kλ2R(1+μcosh(-R(λx+H(t)))+μ2-1sinh(-R(λx+H(t))))(μ+cosh(-R(λx+H(t))))2.
Figure 1 shows solution u11(x,t) for the choices μ=1.3, R=-2.3, k=2.3, λ=1.9, a0=1, F(t)=sin(t), α(t)=t, -3≤x≤4, and -1≤t≤1.1:
(3.13)u12(x,t)=∫F(t)dt+a0-6kλ2R(1+μsin(R(λx+H(t)))+1-μ2cos(R(λx+H(t))))1+μ2+2μsin(R(λx+H(t)))-cos2(R(λx+H(t))).
Figure 2 shows solution u12(x,t) for the choices μ=0.3, R=2.3, k=2.3, λ=1.9, a0=1, F(t)=sin(t), α(t)=t, -3≤x≤4, and -1≤t≤1.1.Figure 1
Graphic of solutionu11(x,t), -3≤x≤4 and -1≤t≤1.1.Figure 2
Graphic of solutionu12(x,t), -3≤x≤4 and -1≤t≤1.1.Second Group.ϵ=-1:.
(v)
a1=b1=0, a2=6kλ2/R, b2=-6kλ2/-R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=0:(3.14)u13(x,t)=∫F(t)dt+a0+6kλ2R(1+sin(-R(λx+H(t))))cos2(-R(λx+H(t))).(vi)
a1=b1=0, a2=6kλ2/R, b2=6kλ2/-R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=0:(3.15)u14(x,t)=∫F(t)dt+a0+6kλ2R(1-sin(-R(λx+H(t))))cos2(-R(λx+H(t))).(vii)
a1=6kλ2μ, a2=6kλ2(1-μ2)/R, b1=0, b2=6kλ2R(μ2-1)/R,  h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2),μ=μ:(3.16)u15(x,t)=∫F(t)dt+a0+6kλ2μRμ+cosh(R(λx+H(t)))+6kλ2R(1-μ2+μ2-1sinh(R(λx+H(t))))(μ+cosh(R(λx+H(t))))2,(3.17)u16(x,t)=∫F(t)dt+a0+6kλ2μRμ+cos(-R(λx+H(t)))+6kλ2R(1-μ2+1-μ2sin(-R(λx+H(t))))μ2+2μcos(-R(λx+H(t)))+cos2(-R(λx+H(t))).(viii)
a1=6kλ2μ, a2=-6kλ2(1-μ2)/R, b1=0, b2=-6kλ2R(μ2-1)/R, h(t)=-λα(t)(a0+∫F(t)dt+Rkλ2), μ=μ:(3.18)u17(x,t)=∫F(t)dt+a0+6kλ2μRμ+cosh(R(λx+H(t)))+6kλ2R(1-μ2-μ2-1sinh(R(λx+H(t))))(μ+cosh(R(λx+H(t))))2,(3.19)u18(x,t)=∫F(t)dt+a0+6kλ2μRμ+cos(-R(λx+H(t)))+6kλ2R(1-μ2-1-μ2sin(-R(λx+H(t))))μ2+2μcos(-R(λx+H(t)))+cos2(-R(λx+H(t))).


## 3.2. Rational Solutions

We seek rational solutions to (1.8) (R=μ=0) in the form given by (3.6) with m=2,(3.20)V(ξ)=∑i=02aiτi(ξ),
where(3.21)τ′(ξ)=τ2(ξ).
Note that  this last equation has the general solution(3.22)τ(ξ)=-1ξ+C.
We now substitute (3.20) into (1.8), and using the relation τ′(ξ)=τ2(ξ) we obtain an equation whose left-hand side is a polynomial in the variable τ=τ(ξ). We equate each coefficient of this polynomial to zero and we get the following algebraic system.(i)
h(t)a1+α(t)λ(∫F(t)dt)a1+α(t)λa0a1=0.(ii)
α(t)λa12+2h(t)a2+2α(t)λ(∫F(t)dt)a2+2α(t)λa0a2=0.(iii)
6ka1λ3+3a1a2λ=0.(iv)
24ka2λ3+2a22λ=0.Solving this system givesa0=a0, a1=0,  a2=-12kλ2 and(3.23)h(t)=-λα(t)(∫F(t)dt+a0).
A rational solution of (1.8) is given by(3.24)V(ξ)=a0-12kλ2(ξ+C)2.
According to (1.5), (1.6), and (1.7) we obtain the following rational solution to (1.3):(3.25)u19(x,t)=∫F(t)dt+a0-12kλ2(λx-∫(λα(t)(∫F(t)dt+a0))dt+C)2.


## 4. Conclusions

In this paper, by using the projective Riccati equation and the Exp-function methods, with the help of a symbolic computation engine, we obtain exact solutions for a generalized KdV equation with forcing term (1.3). The methods certainly works well for a large class of very interesting nonlinear equations. The main advantage of these methods is their capability of greatly reducing the size of computational work compared to existing techniques such as the pseudospectral method, the inverse scattering method, Hirota's bilinear method, and the truncated Painlevé expansion. The Exp-function method gives us more general solutions with some free parameters than the projective Riccati equation method. It also has other interesting applications. For instance, the Exp-function method may be applied not only to differential equations  but also to differential-difference equations or Stochastic equations [32–35].



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