New Soliton and Periodic Solutions for Nonlinear Wave Equation in Finite Deformation Elastic Rod. 1 Introduction

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1 ISSN print), online) International Journal of Nonlinear Science Vol.15013) No.,pp New Soliton and Periodic Solutions for Nonlinear Wave Equation in Finite Deformation Elastic Rod Peng Guo, Guixin Wan, Xiaoyun Wang, Xiaowei Sun School of Mathematics and Physics, Lanzhou Jiaotong University, Lanzhou , China Received 6 October 01, accepted 1 January 013) Abstract: In this paper, we consider nonlinear wave equation in finite deformation elastic rod, and we obtain abundant new exact solutions for it by using the G /G)-expansion method. The obtained solutions include hyperbolic function solutions, trigonometric function solutions and rational solutions. It is shown that G /G)-expansion method provides a powerful mathematical tool for solving a great many nonlinear partial differential equations in mathematical physics. Keywords: the G /G)-expansion method; travelling wave solutions; nonlinear wave; finite deformation elastic rod 1 Introduction During the last decades, investigations of exact travelling wave solutions to nonlinear evolution equations NLEEs) play an important role in the study of complex physical and mechanical phenomena. Many effective approaches for obtaining exact solutions of NLEEs have been presented, such as the tanh-function expansion method, the Jacobi elliptic function expansion method, the homogeneous balance method, F-expansion method, the exp-function expansion method and others [1-8]. Many exact solutions, including solitary wave solutions, shock wave solutions, and periodic wave solutions of NLEEs were successfully obtained. But because of the complexity of the nonlinear evolution equations, it is difficult for us to determine the solutions for the problems. Recently, Mingliang Wang et al. proposed a new method called the G /G)- expansion method to look for travelling wave solutions of NLEEs [9]. By using the G /G)-expansion method, the authors in [9-11] have successfully obtained hyperbolic function solutions, trigonometric function solutions and rational solutions of some important NLEEs. In Ref.[1], Zhifang Liu et al. derived the nonlinear wave equation in finite deformation elastic rod as follows: u t u c 0 x = x [3E ρ u + E ρ u3 + ν R u t u v 0 )], 1) x where R, c 0= E/ρ), v 0= ν/ρ), ν, E and ρ are the cross-section area of the rod, the square of the linear elastic longitudinal wave velocity, the square of the shear wave velocity, Poisson ratio, the Young s modulus and the density of the rod, respectively. Eq.1) is the double nonlinear wave equation with respect to the axial displacement gradient finite deformation elastic rod. It is shown that when the longitudinal wave propagates, the shear wave also propagates as a result of the transverse Poisson effects. Recently, travelling wave solutions for this equation were constructed by using Jacobi elliptic function expansion method [1-16]. The goal of this work is to implement the G /G)-expansion method to obtain more new exact travelling wave solutions of nonlinear wave equation in finite deformation elastic rod. Description of the G /G)-expansion method The generalized G /G)-expansion method can be summarized as follows [9-11]. Suppose that a given nonlinear partial differential equation has the form P u, u t, u x, u tt, u xx, ) = 0. ) Corresponding author. address: guopenglzjtu@16.com Copyright c World Academic Press, World Academic Union IJNS /716

2 P. Guo et al: New Soliton and Periodic Solutions for Nonlinear Wave 183 Step 1. When we seek travelling wave solutions of Eq.), the first step is to introduce the wave transformation: ux, t) = u), = x ct, 3) where c is a real constant. Under the transformation Eq.3), Eq.) becomes an ordinary differential equation Step. Take the solutions of Eq.4) in the more general form u) = a 0 + Qu, u, u, ) = 0. 4) m i=1 [a i G ) G) )i + b i G ) G) ) i ], 5) where a 0, a i and b i i = 1,,, m) are constants to be determined later. The integer m in Eq.5) can be determined by balancing the highest order nonlinear terms and the highest order linear terms of u) in Eq.4). G = G) satisfies the second-order linear ordinary differential equation G + λg + µg = 0, 6) where λ and µ are constants. The explicit expressions for the general solution of Eq.6) are as follows: when λ 4µ > 0, G) = c 1 e λ+ λ 4µ ) + c e λ λ 4µ ) ; when λ 4µ = 0, G) = c 1 + c )e λ ) ; when λ λ 4µ < 0, G) = e ) 4µ λ + c sin 4µ λ ). Step 3. Substitute Eq.5) into Eq.4) and collect all terms with the same order of G /G together. The left-hand side of Eq.4) is converted into a polynomial in G /G. Then, let each coefficient of this polynomial to be zero to derive a set of over-determined algebraic equations for a 0, a i, b i i = 1,,, m), λ, µ and c. Step 4. Solve the algebraic equations obtained in Step 3 with the aid of a computer algebra system such as Mathematica or Maple) to determine these constants. Then, substituting a 0, a i, b i i = 1,,, m), c and the solutions of Eq.6) into Eq.5), we can obtain the exact travelling wave solutions of Eq.). 3 Soliton and periodic solutions for nonlinear wave equation in finite deformation elastic rod Choose the travelling wave transformation Eq.3). Substituting Eq.3) into Eq.1), integrating it with respect to twice, and letting the integrating constant to be zero, we have u + β 1 u + β u + β 3 u 3 = 0, 8) where β 1 = c c 0) ν R c v0 ), β 3c 0 = ν R c v0 ), β c 0 3 = ν R c v0 9) ). According to Step, we get m = 1. Therefore, we can write the solution of Eq.8) in the form u) = a 0 + a 1 G ) G) ) + b 1 G ) G) ) 1. 10) Substituting Eq.10) into Eq.8), collecting the coefficients of G /G) i i = 0, 1,, 3,) and letting it be zero, we obtain the system a 0 β 1 + a 0β + a 1 b 1 β + a 3 0β 3 + 6a 0 a 1 b 1 β 3 + b 1 λ + a 1 λµ = 0, b 3 1β 3 + b 1 µ = 0, b 1β + 3a 0 b 1β 3 + 3b 1 λµ = 0, b 1 β 1 + a 0 b 1 β + 3a 0b 1 β 3 + 3a 1 b 1β 3 + b 1 λ + b 1 µ = 0, a 1 β 1 + a 0 a 1 β + 3a 0a 1 β 3 + 3b 1 a 1β 3 + a 1 λ + a 1 µ = 0, a 3 1β 3 + a 1 = 0, a 1β + 3a 0 a 1β 3 + 3a 1 λ = 0. 7) IJNS homepage:

3 184 International Journal of Nonlinear Science, Vol.15013), No., pp Solving this system by Mathematica, we obtain a 0 = β β 18β 3µ β, a 1 =, b 1 = 0, λ = 18β 3µ), 11) 3β 3 β 3 9β 3 a 0 = β + β 18β 3µ, a 1 = 0, 3β 3 b 1 = 1 [β β 3 β 1 µ)µ 9β β 3β 1 + µ) β 3 9β 1β β 3 ) β 18β 3µ], β3 9β 1 β 3 β + β β λ = 18β 3 µ) 3 β + β 18β 3µ)9β 1 β 3 β + β β 18β 3 µ) β β 3 β 1 µ) 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ β3 3, 1) a 0 = β β 18β 3µ, a 1 = 0, 3β 3 b 1 = 1 [β β 3 β 1 µ)µ 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ], β3 9β 1 β 3 β β β λ = 18β 3 µ) 3β + β 18β 3µ) 9β 1 β 3 + β + β β 18β 3 µ) β β 3 β 1 µ) 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ β3 3, 13) a 0 = β3 9β 1 β β 3 36β β 3 µ 108β3 µ, a 1 =, β 3 b 1 = [β3 9β 1β β 3 ) 3888β 3 β 3β 1β 3 )µ µ3 ] β 7 3 µ4, λ = β3 + 9β 1 β β 3 54µ β 3, 14) a 0 = β 1β + 1β µ, a 1 =, 1β 3 µ β 3 b 1 = [β 1 β + 16β 1µ β + 9β 3µ) + 16µ 13β + 18β 3µ)] µ4, λ = β β 1 16µ) 6µ β 3, 15) where µ and c are arbitrary constants. By using Eq.11)-Eq.15), Eq.10) can be written as u) = β β 18β 3µ G ) ), 16) 3β 3 β 3 G) u) = β + β 18β 3µ 3β 3 1 [β β 3 β 1 µ)µ 9β β 3β 1 + µ) β 3 9β 1β β 3 ) β 18β 3µ] G ) G) ) 1, 17) IJNS for contribution: editor@nonlinearscience.org.uk

4 P. Guo et al: New Soliton and Periodic Solutions for Nonlinear Wave 185 u) = β β 18β 3µ 3β 3 1 [β β 3 β 1 µ)µ 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ] G ) G) ) 1, 18) u) = β3 9β 1 β β 3 36β β 3 µ 108β3 µ u) = β 1β + 1β µ 1β 3 µ β 3 G ) G) ) [β3 9β 1β β 3 ) 3888β3 β 3β 1β 3 )µ + 338β3 3µ3 ] β3 7 G ) µ4 G) ) 1, 19) G ) β 3 G) ) [β 1 β + 16β 1µ β + 9β 3µ) + 16µ 13β + 18β 3µ)] 37348β3 3 G ) µ4 G) ) 1. 0) Substituting the general solutions of Eq.6) into Eq.16), we can obtain the travelling wave solutions of Eq.1) as follows: When λ 4µ > 0, u) = β β 18β 3µ λ 4µ [ c λ 4µ λ 4µ c 1 )cosh + c + c 1 )sinh ) λ ]. 1) 3β 3 β 3 λ 4µ λ 4µ c + c 1 )cosh + c c 1 )sinh When λ 4µ = 0, u) = β β 18β 3µ c 3β 3 β 3 c 1 + c λ ). ) When λ 4µ < 0, u) = β β 18β 3µ [ c 4µ λ 4µ λ cos c 1 sin ) λ ]. 3) 3β 3 β 3 4µ λ 4µ λ + c sin Substituting the general solutions of Eq.6) into Eq.17), we can obtain the travelling wave solutions of Eq.1) as follows: When λ 4µ > 0, u) = β + β 18β 3µ 1 [β 4 3β β 3 β 1 µ)µ 9β β 3β 1 + µ) β 3 9β 1β β 3 ) β 18β 3µ] [ When λ 4µ = 0, u) = β + β 18β 3µ 1 3β 3 9 When λ 4µ < 0, u) = β + β 18β 3µ 1 3β 3 9 λ 4µ c λ 4µ c 1 )cosh c + c 1 )cosh + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ ) λ ] 1. 4) [β β 3 β 1 µ)µ 9β β 3β 1 + µ) β 3 9β 1β β 3 ) β 18β 3µ] c c 1 + c λ ) 1. 5) [β β 3 β 1 µ)µ 9β β 3β 1 + µ) β 3 9β 1β β 3 ) β 18β 3µ] [ c 4µ λ cos c 1 sin 4µ λ + c sin 4µ λ 4µ λ ) λ ] 1. 6) IJNS homepage:

5 186 International Journal of Nonlinear Science, Vol.15013), No., pp Substituting the general solutions of Eq.6) into Eq.18), we can obtain the travelling wave solutions of Eq.1) as follows: When λ 4µ > 0, u) = β β 18β 3µ 1 [β 4 3β β 3 β 1 µ)µ 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ] [ When λ 4µ = 0, u) = β β 18β 3µ 1 3β 3 9 When λ 4µ < 0, u) = β β 18β 3µ 1 3β 3 9 λ 4µ c λ 4µ c 1 )cosh c + c 1 )cosh + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ ) λ ] 1. 7) [β β 3 β 1 µ)µ 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ] c c 1 + c λ ) 1. 8) [β β 3 β 1 µ)µ 9β β 3β 1 + µ) + β 3 9β 1β β 3 ) β 18β 3µ] [ c 4µ λ cos c 1 sin 4µ λ + c sin 4µ λ 4µ λ ) λ ] 1. 9) Substituting the general solutions of Eq.6) into Eq.19), we can obtain the travelling wave solutions of Eq.1) as follows: When λ 4µ > 0, u) = β3 9β 1 β β 3 36β β 3 µ λ 4µ 108β3 µ [ c λ 4µ λ 4µ c 1 )cosh + c + c 1 )sinh ) λ β 3 λ 4µ λ 4µ ] c + c 1 )cosh + c c 1 )sinh [β3 9β 1β β 3 ) 3888β3 β 3β 1β 3 )µ + 338β3 3µ3 ] β3 7µ4 When λ 4µ = 0, λ 4µ [ c c 1 )cosh c + c 1 )cosh λ 4µ + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ ) λ ] 1. 30) u) = β3 9β 1 β β 3 36β β 3 µ c 108β3 µ β 3 c 1 + c λ ) When λ 4µ < 0, [β3 9β 1β β 3 ) 3888β3 β 3β 1β 3 )µ + 338β3 3µ3 ] β3 7 µ4 c 1 + c λ ) 1. 31) u) = β3 9β 1 β β 3 36β β 3 µ 108β3 µ [ β 3 c 4µ λ cos c 1 sin 4µ λ + c sin 4µ λ 4µ λ [β3 9β 1β β 3 ) 3888β3 β 3β 1β 3 )µ + 338β3 3µ3 ] β3 7µ4 [ c 4µ λ cos c 1 sin 4µ λ + c sin c ) λ ] 4µ λ 4µ λ ) λ ] 1. 3) IJNS for contribution: editor@nonlinearscience.org.uk

6 P. Guo et al: New Soliton and Periodic Solutions for Nonlinear Wave 187 Substituting the general solutions of Eq.6) into Eq.0), we can obtain the travelling wave solutions of Eq.1) as follows: When λ 4µ > 0, u) = β 1β + 1β µ 1β 3 µ When λ 4µ = 0, u) = β 1β + 1β µ 1β 3 µ When λ 4µ < 0, u) = β 1β + 1β µ 1β 3 µ λ 4µ [ β 3 c λ 4µ c 1 )cosh c + c 1 )cosh + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ ) λ ] [β 1 β + 16β 1µ β + 9β 3µ) + 16µ 13β + 18β 3µ)] 37348β3 3µ4 λ 4µ [ c λ c 1 )cosh 4µ λ + c + c 1 )sinh 4µ ) λ λ 4µ λ 4µ ] 1. 33) c + c 1 )cosh + c c 1 )sinh c β 3 c 1 + c λ ) [β 1 β + 16β 1µ β + 9β 3µ) + 16µ 13β + 18β 3µ)] 37348β3 3 µ4 c 1 + c λ ) 1. 34) [ β 3 c 4µ λ cos c 1 sin 4µ λ + c sin 4µ λ 4µ λ ) λ ] [β 1 β + 16β 1µ β + 9β 3µ) + 16µ 13β + 18β 3µ)] 37348β3 3µ4 [ c 4µ λ 4µ λ cos c 1 sin ) λ 4µ λ 4µ λ ] 1. 35) + c sin 4 Soliton and periodic solutions for truncated nonlinear wave equation If Eq.1) is truncated in Ou 3 ), that is β 3 u 3 = 0 in Eq.8). Then Eq.8) reduce to u + β 1 u + β u = 0. 36) According to Step, we get m =. Therefore, we can write the solution of Eq.36) in the form u) = a 0 + a 1 G ) G) ) + b 1 G ) G) ) 1 + a G ) G) ) + b G ) G) ). 37) Substituting Eq.37) into Eq.36), collecting the coefficients of G /G) i i = 0, 1,, 3, 4,) and letting it be zero, we obtain the system b + a 0 β 1 + a 0β + a 1 b 1 β + a b β + b 1 λ + a 1 λµ + a µ = 0, b β + 6b µ = 0, b 1 b β + 10b λµ + b 1 µ = 0, bβ 1 + b 1β + a 0 b β + 4b λ + 8b µ + 3b 1 λµ = 0, b 1 β 1 + a 0 b 1 β + a 1 b β + 6b λ + b 1 λ + b 1 µ = 0, a 1 β 1 + a 0 a 1 β + a b 1 β + a 1 λ + a 1 µ + 6a λµ = 0, a β 1 + a 1β + a 0 a β + 3a 1 λ + 4a λ + 8a µ = 0, a 1 + a 1 a β + 10a λ = 0, 6a + a β = 0. c IJNS homepage:

7 188 International Journal of Nonlinear Science, Vol.15013), No., pp Solving this system by Mathematica, we obtain a 0 = β 1 6µ, a 1 = 6 β1 + 4µ, b 1 = 0, a = 6 β1, b = 0, λ = + 10β 1µ + 8µ, 38) β β β 5β µ a 0 = 6µ, a 1 = 6 4µ β1, b 1 = 0, a = 6 β1, b = 0, λ = + 10β 1µ + 8µ, 39) β β β 5β µ a 0 = 3β 1 44β 1 µ 156µ, a 1 = 0, b 1 = β 1 4µ)β 1 + 6µ) 3β 1 + 6µ) β 5β µ) β 5β µ), a = 0, b = 3β 1µ 44β 1 µ 156µ 3, λ = β 1 4µ), 40) β 5β µ) a 0 = β 1 6µ, a 1 = 0, b 1 = 6 µ β 1 + 4µ), β β a = 0, b = 6µ, β λ = µβ 1 4µ), 41) a 0 = 6µ, a 1 = 0, b 1 = 6 µ β 1 4µ), β β a = 0, b = 6µ, β λ = µβ 1 4µ), 4) a 0 = 0, a 1 = 0, b 1 = 7µ 3 β, a = 0, b = 6µ β, λ = µ, 43) where µ and c are arbitrary constants. By using Eq.38)-Eq.43), Eq.37) can be written as u) = β 1 6µ 6 β1 + 4µ G ) β β G) ) 6 G ) β G) ), 44) u) = 6µ 6 4µ β1 G ) β β G) ) 6 G ) β G) ), 45) u) = 3β 1 44β 1 µ 156µ β 5β µ) β 1 4µ)β 1 + 6µ) 3β 1 + 6µ) β 5β µ) G ) G) ) 1 + 3β 1µ 44β 1 µ 156µ 3 G ) β 5β µ) G) ), 46) u) = β 1 6µ 6 µ β 1 + 4µ) G ) β β G) ) 1 6µ G ) β G) ), 47) µ β 1 4µ) u) = 6µ β 6 u) = β 7µ 3 β G ) G) ) 1 6µ β G ) G) ), 48) G ) G) ) 1 6µ β G ) G) ). 49) Substituting the general solutions of Eq.6) into Eq.44), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = β 1 6µ 6 β1 + 4µ λ 4µ [ c λ 4µ c 1 )cosh + c + c 1 )sinh β β λ 4µ c + c 1 )cosh + c c 1 )sinh 6 λ 4µ [ c c 1 )cosh β c + c 1 )cosh λ 4µ + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ λ 4µ λ 4µ ) λ ] ) λ ]. 50) IJNS for contribution: editor@nonlinearscience.org.uk

8 P. Guo et al: New Soliton and Periodic Solutions for Nonlinear Wave 189 When λ 4µ = 0, When λ 4µ < 0, u) = β 1 6µ 6 β1 + 4µ c β β c 1 + c λ ) 6 c β c 1 + c λ ). 51) u) = β 1 6µ 6 β1 + 4µ [ c 4µ λ cos c 1 sin β β 4µ λ + c sin 6 [ c cos β 4µ λ 4µ λ 4µ λ ) λ ] c 1 sin 4µ λ 4µ λ 4µ λ ) λ ]. 5) + c sin Substituting the general solutions of Eq.6) into Eq.45), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = 6µ 6 4µ β1 λ 4µ [ β β When λ 4µ = 0, c λ 4µ c 1 )cosh λ 4µ c + c 1 )cosh 6 λ 4µ [ c c 1 )cosh β c + c 1 )cosh + c + c 1 )sinh + c c 1 )sinh λ 4µ + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ ) λ λ 4µ ] λ 4µ λ 4µ ) λ ]. 53) u) = 6µ 6 4µ β1 c β β c 1 + c λ ) 6 c β c 1 + c λ ). 54) When λ 4µ < 0, u) = 6µ 6 4µ β1 [ c 4µ λ cos c 1 sin β β 4µ λ + c sin 6 [ c cos β 4µ λ 4µ λ 4µ λ ) λ ] c 1 sin 4µ λ 4µ λ 4µ λ ) λ ]. 55) + c sin Substituting the general solutions of Eq.6) into Eq.46), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = 3β 1 44β 1 µ 156µ β 5β µ) [ λ 4µ + 3β 1µ 44β 1 µ 156µ 3 λ 4µ [ β 5β µ) When λ 4µ = 0, u) = 3β 1 44β 1 µ 156µ β 5β µ) β 1 4µ)β 1 + 6µ) 3β 1 + 6µ) β 5β µ) c λ 4µ c 1 )cosh λ c + c 1 )cosh 4µ + c + c 1 )sinh + c c 1 )sinh c c 1 )cosh c + c 1 )cosh λ 4µ λ 4µ λ 4µ + c + c 1 )sinh λ 4µ + c c 1 )sinh ) λ ] 1 β 1 4µ)β 1 + 6µ) 3β 1 + 6µ) β 5β µ) c 1 + c λ ) 1 c + 3β 1µ 44β 1 µ 156µ 3 β 5β µ) λ 4µ λ 4µ c ) λ ]. 56) c 1 + c λ ). 57) IJNS homepage:

9 190 International Journal of Nonlinear Science, Vol.15013), No., pp When λ 4µ < 0, u) = 3β 1 44β 1 µ 156µ β 1 4µ)β 1 + 6µ) 3β 1 + 6µ) β 5β µ) β 5β µ) [ c 4µ λ cos c 1 sin 4µ λ + c sin + 3β 1µ 44β 1 µ 156µ 3 [ c cos β 5β µ) 4µ λ 4µ λ 4µ λ ) λ ] 1 c 1 sin 4µ λ + c sin 4µ λ 4µ λ ) λ ]. 58) Substituting the general solutions of Eq.6) into Eq.47), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = β 1 6µ 6 µ β 1 + 4µ) λ 4µ [ c λ c 1 )cosh 4µ + c + c 1 )sinh β β λ 4µ c + c 1 )cosh When λ 4µ = 0, When λ 4µ < 0, 6µ λ 4µ [ β c c 1 )cosh c + c 1 )cosh λ 4µ + c c 1 )sinh + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ λ 4µ λ 4µ ) λ ] 1 ) λ ]. 59) u) = β 1 6µ 6 µ β 1 + 4µ) c β β c 1 + c λ ) 1 6µ c β c 1 + c λ ). 60) 4µ λ u) = β 1 6µ 6 µ β 1 + 4µ) [ c 4µ λ cos c 1 sin ) λ β β 4µ λ 4µ λ ] 1 + c sin 6µ [ c 4µ λ cos c 1 sin β 4µ λ + c sin 4µ λ 4µ λ ) λ ]. 61) Substituting the general solutions of Eq.6) into Eq.48), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = 6µ 6 µ β 1 4µ) λ 4µ [ c λ c 1 )cosh 4µ + c + c 1 )sinh β β λ 4µ c + c 1 )cosh When λ 4µ = 0, 6µ λ 4µ [ β c c 1 )cosh c + c 1 )cosh λ 4µ + c c 1 )sinh + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ λ 4µ λ 4µ ) λ ] 1 ) λ ]. 6) u) = 6µ 6 µ β 1 4µ) c β β c 1 + c λ ) 1 6µ c β c 1 + c λ ). 63) IJNS for contribution: editor@nonlinearscience.org.uk

10 P. Guo et al: New Soliton and Periodic Solutions for Nonlinear Wave 191 When λ 4µ < 0, u) = 6µ 6 µ β 1 4µ) [ c 4µ λ cos c 1 sin β β 4µ λ 6µ [ β + c sin c cos 4µ λ 4µ λ 4µ λ ) λ ] 1 c 1 sin 4µ λ + c sin 4µ λ 4µ λ ) λ ]. 64) Substituting the general solutions of Eq.6) into Eq.49), we can obtain the travelling wave solutions of Eq.36) as follows: When λ 4µ > 0, u) = 7µ 3 λ 4µ [ c λ 4µ c 1 )cosh β λ 4µ c + c 1 )cosh 6µ λ 4µ [ c c 1 )cosh β When λ 4µ = 0, When λ 4µ < 0, + c + c 1 )sinh + c c 1 )sinh c + c 1 )cosh λ 4µ λ 4µ λ 4µ ) λ ] 1 + c + c 1 )sinh λ 4µ + c c 1 )sinh λ 4µ λ 4µ ) λ ]. 65) 7µ 3 c u) = β c 1 + c λ ) 1 6µ c β c 1 + c λ ). 66) u) = 7µ 3 [ c 4µ λ cos β 4µ λ 4µ λ c 1 sin ) λ 4µ λ ] 1 + c sin 6µ [ c 4µ λ cos c 1 sin β 4µ λ + c sin 4µ λ 4µ λ ) λ ]. 67) 5 Conclusions In this article, the G /G)-expansion method has been successfully implemented to find new traveling wave solutions for nonlinear wave equation in finite deformation elastic rod. We obtain some new traveling wave solutions include hyperbolic function solutions, trigonometric function solutions, rational solutions. The results show that the proposed method is reliable and effective and gives more solutions. This method can be also applied to other kinds of NLEEs. Acknowledgments This work was supported by the National Natural Science Foundation of China under Grant no and the Natural Science Foundation of Gansu Province of China under Grant no. 1014RJZA046. IJNS homepage:

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