Exact Solutions for Nonlinear Wave Equations by the Exp-Function Method

Abstract

This paper elucidates the main advantages of the exp-function method in finding exact solutions of nonlinear wave equations. By the aid of some mathematical software, the solution process becomes extremely simple and accessible.

1. Introduction

One of the most important aspects in nonlinear science is how to solve an exact solution of a nonlinear equation. Recently many different methods have appeared, among which the homotopy perturbation method [1–4], the tanh-method [5], the sinh-method [6, 7], and the F-expansion method [8–11] have caught much attention; however, all these methods are valid for some special kinds of nonlinear equations. It is therefore very much needed to find a universal approach to nonlinear equations; this is very challenging indeed, and the exp-function method [12–15] meets this requirement. The exp-function method itself is mathematically beautiful and extremely accessible to nonmathematicians. The use of the method requires no special knowledge of advanced calculus, and it is especially effective for solitary solutions.

2. Exp-Function Method

The exp-function method was first proposed by He and Wu [16], and we consider a general partial differential equation (PED) in the form 𝑃𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑦,𝑢_𝑥𝑥,𝑢_𝑡𝑡,𝑢_𝑦𝑦=0 (1) to pick out the main solution process and its advantages.

Use a transformation [16] 𝜉=𝑘𝑥+𝜔𝑡+𝑙𝑦, (2) where 𝑘,𝜔, and 𝑙 are unknown constants and should be determined later. By (2), we can convert (1) to the following nonlinear ordinary differential equation: 𝐺𝑢,𝑢^′,𝑢^′′,𝑢^′′′,…=0. (3) According to the exp-function method, we assume that its solution can be expressed in the following form [16, 17]: 𝑢(𝜉)=∑^𝑑_𝑛=−𝑐𝑎_𝑛exp(𝑛𝜉)∑^𝑞_𝑚=−𝑝𝑏_𝑛(),exp𝑚𝜉 (4) where 𝑐,𝑑,𝑝,𝑞 are positive integers that could be freely chosen. To determine the value of 𝑐 and 𝑝, we balance the linear term of highest order of (3) with the highest order of the nonlinear term. Similarly for determining the value of 𝑑 and 𝑞, we balance the lowest orders of linear and nonlinear terms in (3). By substituting (4) into (3), collecting terms of the same term of exp(𝑖𝜉), and equating the coefficient of each power of exp to zero, we can get a set of algebraic equations for determining unknown constants.

3. Exact Solution for Nonlinear Wave Equation

In order to illustrate the basic solution process of the exp-function method, we use the Burgers-Huxley equation as an example, which can be expressed as [18] 𝑢_𝑡+𝑢_𝑥𝑥+3𝑘𝑢𝑢_𝑥+𝑐𝑢+𝑢²+𝑢³=0, (5) where 𝑢=𝑢(𝑥,𝑡) is an unknown function, 𝑢_𝑡,𝑢_𝑥 are the partial derivatives of 𝑢(𝑥,𝑡) with respect to 𝑡 and 𝑥, respectively, and 𝑘 and 𝑐 are arbitrary constants.

According to the exp-function method [15–17], we introduce a complex variation 𝜉 defined as 𝜉=𝑘𝑥+𝜔𝑡. (6) Equation (5) thus becomes an ordinary differential equation as 𝑘²𝑢^′′+(3𝑢+𝜔)𝑢^′+𝑐𝑢+𝑢²+𝑢³=0. (7) We suppose that the solution of (7) can be expressed as 𝑢(𝜉)=𝑎_𝑐exp(𝑐𝜉)+⋯+𝑎_−𝑑exp(−𝑑𝜉)𝑏_𝑝exp𝑝𝜉+⋯+𝑏_−𝑞.exp−𝑞𝜉 (8) Thus we have 𝑢^′′=𝛾₁exp𝑐+3𝑝𝜉𝛾₂(),𝑢exp4𝜉₃=𝑐₃exp(3𝑐𝜉)+⋯𝑐₄=𝑐exp3𝑝𝜉+⋯³exp3𝑐+𝑝𝜉𝑐₄.exp4𝑝𝜉 (9) Balancing highest order of exp-function in (9), we have 3𝑐+𝑝=𝑐+3𝑝, which leads to the result 𝑝=𝑐. Similarly we balance the lowest orders of linear and nonlinear terms in (5) to determine values of 𝑑 and 𝑞, and we can get 𝑑=𝑞. For simplicity, we set 𝑝=𝑐=1 and 𝑞=𝑑=1; then (8) reduces to 𝑢(𝜉)=𝑎₁exp(𝜉)+𝑎₀+𝑎₋₁exp(−𝜉)exp(𝜉)+𝑏₀+𝑏₋₁().exp−𝜉 (10) Substituting (10) in to (5), we have 1𝐴𝐸₃exp(3𝜉)+𝐸₂exp(2𝜉)+𝐸₁exp(𝜉)+𝐸₀+𝐸₋₁exp(−𝜉)+𝐸₋₂exp(−2𝜉)+𝐸₋₃exp(−3𝜉)=0, (11) where =[𝑏₀+exp(𝜉)+𝑏₋₁exp(−𝜉)]³, 𝐸₃=𝑎³₁+𝑎²₁+𝑐𝑎₁,𝐸₂=𝑐𝑎₀−𝑎₀𝑎₁−𝜔𝑎₀+3𝑎₀𝑎²₁+4𝑎²₁𝑏₀+𝑘²𝑎₀+2𝑎₁𝑏₀𝑐+𝜔𝑎₁𝑏₀−𝑘²𝑎₁𝑏₀,𝐸₀=𝑎³₀+𝑎²₀𝑏₀−7𝑎₀𝑎₋₁+6𝑎₀𝑎₁𝑎₋₁+11𝑎₀𝑎₁𝑏₋₁+2𝑏₀𝑎₁𝑎₋₁+2𝑐𝑎₀𝑏₋₁+2𝑐𝑏₀𝑎₋₁−3𝜔𝑏₀𝑎₋₁+𝑐𝑎₀𝑏²₀−6𝑘²𝑎₀𝑏₋₁+3𝑘²𝑎₋₁𝑏₀+3𝑘²𝑎₋₁𝑏₋₁𝑏₀+2𝑐𝑎₁𝑏₋₁𝑏₀+3𝜔𝑎₁𝑏₋₁𝑏₀,𝐸₁=𝑐𝑎₋₁−4𝑎₁𝑎₋₁−2𝜔𝑎₋₁+3𝑎₁𝑎²₀+3𝑎²₁𝑎₋₁+7𝑎²₁𝑏₋₁+4𝑘²𝑎₋₁−2𝑎²₀+𝑘²+𝑐𝑎₁𝑏²₀+5𝑎₀𝑎₁𝑏₀+2𝑐𝑎₁𝑏₋₁−𝜔𝑎₀𝑏₀+2𝜔𝑎₁𝑏₋₁+2𝑐𝑎₀𝑏₀,𝐸₋₁=3𝑎²₀𝑎₋₁+3𝑎₁𝑎²₋₁+4𝑎²₀𝑏₋₁−5𝑎²₋₁+𝑎₋₁𝑏²₀𝑘²+4𝑎₁𝑏²₋₁𝑘²−𝑎₀𝑎₋₁𝑏₀+8𝑎₁𝑎₋₁𝑏₋₁+(2𝑐−2𝜔)𝑎₋₁𝑏₋₁+𝑐𝑎₋₁𝑏²₀+𝑐𝑎₁𝑏²₋₁−4𝑎₋₁𝑏²₋₁𝑘²−𝜔𝑎₋₁𝑏²₀+2𝜔𝑎₁𝑏²₋₁+2𝑐+𝜔−𝑘²𝑎₀𝑏₋₁𝑏₀,𝐸₋₂=3𝑎₀𝑎²₋₁−2𝑎²₋₁𝑏₀+𝑘²𝑎₀𝑏²₋₁+5𝑎₀𝑎₋₁𝑏₋₁+(𝑐+𝜔)𝑎₀𝑏²₋₁+2𝑐−𝑘²−𝜔𝑎₋₁𝑏₋₁𝑏₀,𝐸₋₃=𝑎³₋₁+𝑎²₋₁𝑏₋₁+𝑐𝑎₋₁𝑏²₋₁. (12) Setting the coefficients of exp(𝑖𝜉),(𝑖=0,±1,±2,±3) to zero, we have 𝐸₃=0,𝐸₂=0,𝐸₁𝐸=0₀𝐸=0,₋₃=0,𝐸₋₂=0,𝐸₋₁=0. (13) With the help of some mathematical software, we can solve the solutions of the algebraic equations.

Case 1. Consider 𝑎₁=√1−4𝑐−12,𝑏₀=−𝑎₀3𝑎²₁−𝑎₁+𝑘²+𝑐−𝜔𝑎₁−𝑘²+4𝑎₁,+2𝑐+𝜔𝜔=𝑐𝑎₀−𝑎₀𝑎₁+3𝑎₀𝑎²₁+4𝑎²₁𝑏₀+𝑎₀𝑘²+2𝑎₁𝑏₀𝑐−𝑎₁𝑏₀𝑘²𝑎₀−𝑎₁𝑏₀,𝑎₋₁=𝑏₋₁√1−4𝑐−12. (14) This implies the following exact solution: 𝑢(𝜉)=√1−4𝑐−12exp(𝜉)+𝑎₀+𝑏₂√1−4𝑐−12()×(𝜉)−exp−𝜉exp3𝑎₀𝑎²₁−𝑎₀𝑎₁+𝑎₀𝑘²+𝑐𝑎₀−𝜔𝑎₀2𝑐𝑎₁+𝜔𝑐−𝑎₁𝑘²+4𝑎²₁+𝑏₂exp(−𝜉)⁻¹, (15) where +𝜉=𝑘𝑥𝑐𝑎₀−𝑎₀𝑎₁+3𝑎₀𝑎²₁+4𝑎²₁𝑏₀+𝑎₀𝑘²+2𝑎₁𝑏₀𝑐−𝑎₁𝑏₀𝑘²𝑎₀−𝑎₁𝑏₀𝑡, (16)𝑎₀,𝑏₂ are parameters, 𝑏₂≠0, and 𝑘 is a free real number.

Case 2. Consider 𝑎₁=12,𝑏₀𝑎=0,₂𝑏=−²2,𝑏₂=𝑏₂,𝑎₀=√22+−16𝑏₂𝑘²+192𝑏₂+2𝑐𝑏₂+8𝑏₂𝜔,𝜔=𝑘²1+𝑐+4. (17) This case gives another exact solution as follows: 𝑢()=1𝑥,𝑡2exp𝑘𝑥+𝑘²1+𝑐+4𝑡+√2210𝑏₂𝑐−8𝑏₂𝑘²+192𝑏₂+2𝑏₂−𝑏₂exp−𝑘𝑥−𝑘²1+𝑐+4𝑡×exp𝑘𝑥+𝑘²1+𝑐+4𝑡+𝑏₂exp−𝑘𝑥−𝑘²1+𝑐+4𝑡⁻¹, (18) where 𝑏₂,𝑘, are nonzero free parameters.

Case 3. Consider 𝑎₁√=−1−4𝑐+12,𝑎₀𝑏=0,₀=0,𝑏₂=𝑏₂,𝑎₂√=−1+1−4𝑐2𝑏₂. (19) This results in the following exact solution: 𝑢(𝜉)=−√1−4𝑐+12(𝜉)−√exp1+1−4𝑐2𝑏₂()×(𝜉)exp−𝜉exp+𝑏₂exp(−𝜉)⁻¹, (20) where 𝑏₂ is nonzero free parameter.

4. Conclusion

By some mathematical software, the solution process is extremely simple and abundant solutions are predicted [19–21]. The exp-function method is a universal tool for nonlinear equations and can be easily extended to fractional calculus [22–27].

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This project is sponsored by the National Natural Science Foundation of China (NSFC, Grants U1204703 and U1304614), the Key Scientific and Technological Project of Henan Province (122102310004), the Fundamental Research Funds for the Central Universities (HUST: 2012QN087 and 2012QN088), and the Innovation Scientists and Technicians Troop Construction Projects of Zhengzhou City (10LJRC190).