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

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.

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

𝑢_𝑡+6𝑢𝑢_𝑥+𝑢_𝑥𝑥𝑥=0 (1.1) have been of some interest since 150 years. Some generalizations of this last equation have been studied recently. For instance, the equation

𝑢_𝑡+𝑘₁𝑡^𝑛𝑢𝑢_𝑥+𝑘₂𝑡^𝑚𝑢_𝑥𝑥𝑥=0, (1.2) 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:

𝑢_𝑡+𝛼(𝑡)𝑢𝑢_𝑥+𝑘𝛼(𝑡)𝑢_𝑥𝑥𝑥=𝐹(𝑡), (1.3) 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]:

𝑢_𝑡+𝛼𝑢𝑢_𝑥+𝛽𝑢_𝑥𝑥𝑥=𝐹(𝑡)(𝛼,𝛽=const), (1.4) which results from (1.3) by taking 𝛼=const and 𝑘=𝛽/𝛼.

We suppose that the solution to (1.3) has the form

𝑢(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑣(𝑥,𝑡). (1.5) Therefore, (1.3) reduces to

𝑣_𝑡+𝛼(𝑡)𝑣𝑣_𝑥+𝛼(𝑡)𝐹(𝑡)𝑑𝑡𝑣_𝑥+𝑘𝛼(𝑡)𝑣_𝑥𝑥𝑥=0. (1.6) Now we consider the transformation

𝑣=𝑉(𝜉),𝜉=𝜆𝑥+ℎ(𝑡)𝑑𝑡, (1.7) 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

ℎ(𝑡)𝑉^′(𝜉)+𝛼(𝑡)𝜆(𝑉(𝜉)+𝑓(𝑡))𝑉^′(𝜉)+𝑘𝜆³𝛼(𝑡)𝑉^′′′(𝜉)=0, (1.8) where

𝑓=𝑓(𝑡)=𝐹(𝑡)𝑑𝑡. (1.9)

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

𝐹(𝑢,𝑢_𝑥,𝑢_𝑡,𝑢_𝑥𝑥,𝑢_𝑥𝑡,𝑢_𝑡𝑡,…)=0 (2.1) can be found using the expression

∑𝑢(𝜉)=^{𝑑𝑛=−𝑐}𝑎_𝑛exp(𝑛𝜉)∑^𝑞_𝑛=−𝑝𝑎_𝑚,exp(𝑚𝜉) (2.2) where 𝑐, 𝑑, 𝑝, and 𝑞 are positive integers which could be freely chosen; 𝑎_𝑛 and 𝑏_𝑛 are unknown constants to be determined. According to this, we suppose that solutions to (1.8) can be expressed in the form

∑𝑣(𝜉)=^{𝑑𝑛=−𝑐}𝑎_𝑛exp(𝑛𝜉)∑^𝑞_𝑚=−𝑝𝑏_𝑚=𝑎exp(𝑚𝜉)^−𝑐exp(−𝑐𝜉)+⋯+𝑎_𝑑exp(𝑑𝜉)𝑏_−𝑝exp(−𝑝𝜉)+⋯+𝑏_𝑞,exp(𝑞𝜉) (2.3) where 𝑐, 𝑑, 𝑝, and 𝑞 are positive integers which are unknown to be determined later; 𝑎_𝑛 and 𝑏_𝑚 are unknown constants. We have the following two cases.

2.1. Case 1: 𝑝=𝑐=1 and 𝑑=𝑞=1

In this case, the trial solution to (1.8) becomes

𝑎𝑉(𝜉)=¹exp(𝜉)+𝑎₀+𝑎₋₁exp(−𝜉)exp(𝜉)+𝑏₀+𝑏₋₁.exp(−𝜉) (2.4) 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)𝑎₁𝑏²₀, 𝑏₋₁=𝑏²₀/4, 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

𝑎𝑢(𝑥,𝑡)=¹(𝑏₀+2𝑒^𝜉)²+24𝑏₀𝑘𝜆²𝑒^𝜉(𝑏₀+2𝑒^𝜉)²+𝐹(𝑡)𝑑𝑡, (2.5) where

𝜉=𝜉(𝑥,𝑡)=𝜆𝑥+ℎ(𝑡)𝑑𝑡=𝜆𝑥−𝜆𝛼(𝑡)(𝑎₁+𝑘𝜆²+𝐹(𝑡)𝑑𝑡)𝑑𝑡, (2.6) and 𝑎₁, 𝑏₀, and 𝜆 are arbitrary real or complex numbers.

2.2. Case 2: 𝑝=𝑐=2 and 𝑑=𝑞=2

In this case, the trial solution to (1.8) becomes

𝑎𝑉(𝜉)=²exp(2𝜉)+𝑎₁exp(𝜉)+𝑎₀+𝑎₋₁exp(−𝜉)+𝑎₋₂exp(−2𝜉)exp(2𝜉)+𝑏₁exp(𝜉)+𝑏₀+𝑏₋₁exp(−𝜉)+𝑏₋₂.exp(−2𝜉) (2.7) As in the first case, we obtain an algebraic system. Solving it gives

𝑎₀=−−2𝑎₁𝑏₁(𝑎₂+3𝑘𝜆²)+𝑎₂(𝑎₂𝑏²₁+6(𝑏²₁−𝑏₀)𝑘𝜆²)+𝑎²₁6𝑘𝜆²,𝑎₋₁=𝑎₁−𝑎₂𝑏₁432𝑘³𝜆⁶(−𝑎₁𝑏₁(27𝑎₂𝑘𝜆²+2𝑎²₂+72𝑘²𝜆⁴)+(𝑎₂+6𝑘𝜆²)(12𝑎₂𝑏²₁𝑘𝜆²+𝑎²₂𝑏²₁+72𝑏₀𝑘²𝜆⁴)+𝑎²₁(𝑎₂+9𝑘𝜆²𝑎)),₋₂=16912𝑘⁴𝜆⁸𝑎₂(𝑎₁−𝑎₂𝑏₁)²(8𝑎₂𝑏²₁𝑘𝜆²−2𝑎₁𝑏₁(𝑎₂+4𝑘𝜆²)+𝑎²₂𝑏²₁+𝑎²₁+48𝑏₀𝑘²𝜆⁴𝑏),₋₁=1432𝑘³𝜆⁶(𝑎₁−𝑎₂𝑏₁)(9𝑎₂𝑏²₁𝑘𝜆²−𝑎₁𝑏₁(2𝑎₂+9𝑘𝜆²)+𝑎²₂𝑏²₁+𝑎²₁+72𝑏₀𝑘²𝜆⁴𝑏),₋₂=16912𝑘⁴𝜆⁸(𝑎₁−𝑎₂𝑏₁)²(8𝑎₂𝑏²₁𝑘𝜆²−2𝑎₁𝑏₁(𝑎₂+4𝑘𝜆²)+𝑎²₂𝑏²₁+𝑎²₁+48𝑏₀𝑘²𝜆⁴),ℎ(𝑡)=−𝜆𝛼(𝑡)(𝑎₂+𝑓(𝑡)+𝑘𝜆²). (2.8) From (1.5), (1.7), and (2.7) we may verify that to this set of values corresponds the solution

1𝑢(𝑥,𝑡)=(𝑎₁−𝑎₂𝑏₁+12𝑘𝜆²𝑒^𝜉)²(𝑎²₁𝑎₂+𝑎₁(−2𝑎²₂𝑏₁+24𝑎₂𝑘𝜆²𝑒^𝜉+144𝑘²𝜆⁴𝑒^𝜉)+𝑎₂(𝑎²₂𝑏²₁−24𝑎₂𝑏₁𝑘𝜆²𝑒^𝜉+144𝑘²𝜆⁴𝑒^𝜉(𝑒^𝜉−𝑏₁+)))𝐹(𝑡)𝑑𝑡, (2.9) where

𝜉=𝜉(𝑥,𝑡)=𝜆𝑥+ℎ(𝑡)𝑑𝑡=𝜆𝑥−𝜆𝛼(𝑡)(𝑎₂+𝑘𝜆²+𝐹(𝑡)𝑑𝑡)𝑑𝑡, (2.10) 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

𝑉(𝜉)=𝑎₀+^𝑚_𝑖=1𝜎^𝑖−1(𝜉)(𝑎_𝑖𝜎(𝜉)+𝑏_𝑖𝜏(𝜉)), (3.1) where 𝑎₀, 𝑎₁, 𝑏₁,… are constants and 𝜎(𝜉) and 𝜏(𝜉) satisfy the system

𝜎^′(𝜉)=𝜖𝜎(𝜉)𝜏(𝜉),𝜏^′(𝜉)=𝑅+𝜖𝜏²(𝜉)−𝜇𝜎(𝜉). (3.2) In (3.2), 𝜖=±1 and 𝑅 and 𝜇 are certain constants. These equations have following solutions.

Case 1. When 𝜖=−1 and 𝑅≠0, 𝜎₁√(𝜉)=𝑅sech(𝑅𝜉)𝜇sech(√𝑅𝜉)+1,𝜏₁√(𝜉)=𝑅tanh(√𝑅𝜉)𝜇sech(√,𝜎𝑅𝜉)+1₂√(𝜉)=𝑅csch(𝑅𝜉)𝜇csch(√𝑅𝜉)+1,𝜏₂√(𝜉)=𝑅coth(√𝑅𝜉)𝜇csch(√.𝑅𝜉)+1 (3.3)

Case 2. When 𝜖=1 and 𝑅≠0, 𝜎₃√(𝜉)=𝑅sec(𝑅𝜉)𝜇sec(√𝑅𝜉)+1,𝜏₃√(𝜉)=𝑅tan(√𝑅𝜉)𝜇sec(√,𝜎𝑅𝜉)+1₄√(𝜉)=𝑅csc(𝑅𝜉)𝜇csc(√𝑅𝜉)+1,𝜏₄√(𝜉)=𝑅cot(√𝑅𝜉)𝜇csc(√.𝑅𝜉)+1 (3.4)

Case 3. When 𝑅=𝜇=0, 𝜎₅𝐶(𝜉)=𝜉,𝜏₅1(𝜉)=.𝜖𝜉 (3.5) In this last case, we seek solutions to (1.8) in the form 𝑉(𝜉)=^𝑚_𝑖=0𝑎_𝑖𝜏^𝑖(𝜉), (3.6) 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: 𝜏²𝜇(𝜉)=−𝜖[𝑅−2𝜇𝜎(𝜉)+²−1𝑅𝜎²(𝜉)]. (3.7)

3.1. Periodic and Soliton Solutions

Periodic and soliton solutions are obtained when 𝑅≠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 𝜎^𝑖(𝜉)𝜏^𝑗(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

𝑉(𝜉)=𝑎₀+𝑎₁𝜎(𝜉)+𝑏₁𝜏(𝜉)+𝜎(𝜉)(𝑎₂𝜎(𝜉)+𝑏₂𝜏(𝜉)). (3.8) We equate each coefficient of the polynomial to zero. This will give an overdetermined system of algebraic equations involving the parameters 𝑎_𝑖, 𝑏_𝑖(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)𝑏²₁𝑅³𝛼𝜆−𝑏²₁𝑅³𝛼𝜖²𝜆=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+2𝑏²₁𝑅²𝛼𝜆𝜇𝜖²+𝑎₀𝑎₁𝑅²𝛼𝜆𝜖+𝑎₁𝑓𝑅²𝛼𝜆𝜖+𝑎₁𝑅²ℎ𝜖+2𝑏₁𝑏₂𝑅³𝑏𝛼𝜆−₂₁𝑅²𝛼𝜆𝜇=0.(vi)−24𝑎₂𝑘𝑅³𝛼𝜆³𝜖⁴+12𝑎₁𝑘𝑅²𝛼𝜆³𝜇𝜖⁴+16𝑎₂𝑘𝑅³𝛼𝜆³𝜖²−𝑏²₁𝑅𝛼𝜆𝜇²𝜖²−2𝑏²₂𝑅³𝛼𝜆𝜖²+𝑏²₁𝑅𝛼𝜆𝜖²−6𝑎₁𝑘𝑅²𝛼𝜆³𝜇𝜖²+6𝑏₁𝑏₂𝑅²𝛼𝜆𝜇𝜖²+𝑎²₁𝑅²𝛼𝜆𝜖+2𝑎₀𝑎₂𝑅²𝛼𝜆𝜖+2𝑎₂𝑓𝑅²𝛼𝜆𝜖+2𝑎₂𝑅²ℎ𝜖+𝑏²₂𝑅³𝛼𝜆−2𝑏₁𝑏₂𝑅²𝛼𝜆𝜇=0.(vii)6𝑏₁𝑘𝑅⁴𝛼𝜆³𝜖⁵−8𝑏₁𝑘𝑅⁴𝛼𝜆³𝜖³−𝑎₀𝑏₁𝑅³𝛼𝜆𝜖²−𝑏₁𝑓𝑅³𝛼𝜆𝜖²−𝑏₁𝑅³ℎ𝜖²+2𝑏₁𝑘𝑅⁴𝛼𝜆³𝜖+𝑎₀𝑏₁𝑅³𝛼𝜆+𝑏₁𝑓𝑅³𝑏𝛼𝜆+₁𝑅³ℎ=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:𝑢₁(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1−sin(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₂(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1−cos(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₃(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1−cosh(√.−𝑅(𝜆𝑥+𝐻(𝑡))) (3.9)(ii)𝑎₂=𝑏₁=𝑏₂=0, 𝑎₁=−6𝑘𝜆², ℎ(𝑡)=−𝜆𝛼(𝑡)(𝑎₀+∫𝐹(𝑡)𝑑𝑡−𝑅𝑘𝜆²), 𝜇=1:𝑢₄(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1+sin(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₅(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1+cos(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₆(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−6𝑘𝜆²𝑅1+cosh(√.−𝑅(𝜆𝑥+𝐻(𝑡))) (3.10)(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: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(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₈(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))) (3.11)(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: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(√,𝑢𝑅(𝜆𝑥+𝐻(𝑡)))₁₁(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. (3.12) 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: 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))) (3.13) 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 solution 𝑢11(𝑥,𝑡), −3≤𝑥≤4 and −1≤𝑡≤1.1.

Figure 2 

Graphic of solution 𝑢12(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:𝑢₁₃(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀+6𝑘𝜆²𝑅(1+sin(√−𝑅(𝜆𝑥+𝐻(𝑡))))cos²(√.−𝑅(𝜆𝑥+𝐻(𝑡))) (3.14)(vi)𝑎₁=𝑏₁=0, 𝑎₂=6𝑘𝜆²/𝑅, 𝑏₂=6𝑘𝜆²/√−𝑅, ℎ(𝑡)=−𝜆𝛼(𝑡)(𝑎₀+∫𝐹(𝑡)𝑑𝑡+𝑅𝑘𝜆²), 𝜇=0:𝑢₁₄(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀+6𝑘𝜆²𝑅(1−sin(√−𝑅(𝜆𝑥+𝐻(𝑡))))cos²(√.−𝑅(𝜆𝑥+𝐻(𝑡))) (3.15)(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: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,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))) (3.16)(3.17)(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: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,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))) (3.18)(3.19)

3.2. Rational Solutions

We seek rational solutions to (1.8) (𝑅=𝜇=0) in the form given by (3.6) with 𝑚=2,

𝑉(𝜉)=²_𝑖=0𝑎_𝑖𝜏^𝑖(𝜉), (3.20) where

𝜏^′(𝜉)=𝜏²(𝜉). (3.21) Note that this last equation has the general solution

1𝜏(𝜉)=−.𝜉+𝐶 (3.22) 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)𝜆𝑎²₁+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 gives 𝑎₀=𝑎₀, 𝑎₁=0, 𝑎₂=−12𝑘𝜆² and

ℎ(𝑡)=−𝜆𝛼(𝑡)(𝐹(𝑡)𝑑𝑡+𝑎₀). (3.23) A rational solution of (1.8) is given by

𝑉(𝜉)=𝑎₀−12𝑘𝜆²(𝜉+𝐶)². (3.24) According to (1.5), (1.6), and (1.7) we obtain the following rational solution to (1.3):

𝑢₁₉(𝑥,𝑡)=𝐹(𝑡)𝑑𝑡+𝑎₀−12𝑘𝜆²(𝜆𝑥−∫(𝜆𝛼(𝑡)(∫𝐹(𝑡)𝑑𝑡+𝑎₀))𝑑𝑡+𝐶)². (3.25)

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].