# New Exact Traveling Wave Solutions for Two Nonlinear

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2012 International Conference on Computer Technology and Science (ICCTS 2012) IPCSIT vol. 47 (2012) © (2012) IACSIT Press, Singapore DOI: 10.7763/IPCSIT.2012.V47.66

New Exact Traveling Wave Solutions for Two Nonlinear Evolution Equations
Qinghua Feng+
School of Science, Shandong University of Technology, Zhangzhou Road 12, Zibo, Shandong, China, 255049
Abstract. In this paper, a generalized sub-ODE method is proposed to construct exact solutions of two
nonlinear equations. As a result, some new exact traveling wave solutions for them are found.
Keywords:sub-ODE method, traveling wave solution, exact solution, nonlinear equation,

1. Introduction
In scientific research, seeking the exact solutions of nonlinear equations is a hot topic. Many approaches have been presented so far. Some of these approaches are the homogeneous balance method [1,2], the hyperbolic tangent expansion method [3,4], the trial function method [5], the tanh-method [6-8], the nonlinear transform method [9], the inverse scattering transform [10], the Backlund transform [11,12], the Hirotas bilinear method [13,14], the generalized Riccati equation [15,16], the theta function method [17-19], the sine-Ccosine method [20], the Jacobi elliptic function expa-nsion [21,22], the complex hyperbolic function method [23-25], and so on..
In this paper, we proposed a sub-ODE method to construct exact traveling wave solutions for NLEES. The rest of the paper is organized as follows. In Section 2, we describe the sub-ODE method for finding traveling wave solutions of nonlinear evolution equations, and give the main steps of the method. In the subsequent sections, we will apply the method to find exact traveling wave solutions of the Boussinesq equation and (2+1) dimensional Boussinesq equation. In the last Section, some conclusions are presented.

2. Description of the Bernoulli Sub-ODE method
In this section we present the solutions of the following ODE:

G '+ λG = μG2 ,

(2.1)

where λ ≠ 0,G = G(ξ ) When μ ≠ 0 , Eq. (2.1) is the type of Bernoulli equation, and we can obtain the solution as

G= 1 ,

(2.2)

μ + deλξ

λ

where d is an arbitrary constant. Suppose that a nonlinear equation, say in two or three independent variables x, y and t , is given by

+ Corresponding author. Tel.: +86-13561602410 E-mail address: [email protected]

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P(u, ut,ux , uy , utt , uxt , uyt , uxx , uyy ......) = 0

(2.3)

where u = u(x, y, t) is an unknown function, P is a polynomial in u = u(x, y, t) and its various partial derivatives, in which the highest order derivatives and nonlinear terms are involved. By using the solutions of Eq. (2.1), we can construct a serials of exact solutions of nonlinear equations:.
Step 1.We suppose that

u(x, y,t) = u(ξ ),ξ = ξ (x, y,t)

(2.4)

the traveling wave variable (2.4) permits us reducing Eq. (2.3) to an ODE for u = u(ξ )

P(u,u ',u '',......) = 0

(2.5)

Step 2. Suppose that the solution of (2.5) can be expressed by a polynomial in G as follows:

u(ξ

)

=

αmGm

+

α Gm−1 m−1

+

......

(2.6)

where G = G(ξ ) satisfies Eq. (2.1), and αm ,αm−1... are constants to be determined later, αm ≠ 0 . The
positive integer m can be determined by considering the homogeneous balance between the highest order derivatives and nonlinear terms appearing in (2.5).
Step 3. Substituting (2.6) into (2.5) and using (2.1), collecting all terms with the same order of G together, the left-hand side of Eq. (2.5) is converted into another polynomial in G . Equating each coefficient of this
polynomial to zero, yields a set of algebraic equations for αm ,αm−1,...λ, μ .
Step 4. Solving the algebraic equations system in Step 3, and by using the solutions of Eq. (2.1), we can construct the traveling wave solutions of the nonlinear evolution equation (2.5).
In the subsequent sections we will illustrate the proposed method in detail by applying it to Boussinesq equation and (2+1) dimensional Boussinesq equation.

3. Application for Boussinesq Equation
In this section, we will consider the following Boussinesq equation:

utt + α uxx + β (u 2 )xx + γ ux(4) = 0,α < 0
Suppose that

(3.1)

u(ξ ),ξ = k(x − ct)

(3.2)

where the constants c, k can be determined later. By using (3.2), (3.1) is converted into an ODE

(α + c2 )u ''+ β (u2 ) ''+ γ k3u(5) = 0

(3.

3)

Integrating (3.3) twice, and take the integration constant for zero, then we have

(α + c2 )u + βu2 + γ k3u ''' = 0

(3.

4)

355

Suppose that the solution of (3.4) can be expressed by a polynomial in G as follows:

∑ m
u(ξ ) = aiGi
i=0

(3.5)

where ai are constants, and G = G(ξ ) satisfies Eq. (2.1). Balancing the order of u2 and u ''' in Eq. (3.5), we
obtain that 2m = m + 3 ⇒ m = 3 .So Eq. (3.5) can be rewritten as

u(ξ ) = a3G3 + a2G 2 + a1G + a0 , a3 ≠ 0

where a3, a2 , a1, a0 are constants to be determined later.
Substituting (3.5) into (3.3) and collecting all the terms with the same power of G together, the left-hand side of Eq. (3.3) is converted into another polynomial in G . Equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows:

G0 : β a02 + α a0 + c2a0 = 0 G1 : 2β a0a1 + α a1 + c2a1 + γ k 3a1μ3 = 0 G2 : −7γ k 3λa1μ 2 + 2β a0a2 + (c2 + α )a2 + 8γ k 3a2μ3 + β a12 = 0 G3 : −38γ k 3a2μ 2λ + (c2 + α )a3 + 2β a0a3 + 2β a1a2 +12γ k3λ 2a1μ + 27γ a3k 3μ3 = 0 G4 : 54γ a2μk3λ2 + β a22 + 2β a1a3 −111γ a3λk3μ2 − 6γ a1k3λ3 = 0 G5 :144γ a3μk3λ2 + 2β a2a3 − 24γ a2k3λ3 = 0 G6 : −60γ a3k3λ3 + β a32 = 0 Solving the algebraic equations above, yields: Case 1:

a = 60γ k3λ3 ,a = a = a = 0,k = k, c = −α , λ = λ, μ = 0

3

β

210

Substituting (3.6) into (3.5), we have

u (ξ ) = 60γ k3λ3 G3 , ξ = k(x − −α t)

1

β

(3.6)
(3. 7)

Combining with Eq. (2.2) and we can obtain the traveling wave solutions of (3.1) as follows:

u (x, t) = 60γ k 3λ3 [de−λk (x− −αt) ]3

1

β

where d is an arbitrary constant. Case 2:

(3.8)

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a = 60γ k3λ3 ,a = a = a = 0,k = k, c = − −α , λ = λ, μ = 0

3

β

210

Substituting (3.6) into (3.5), we have

u (ξ ) = 60γ k3λ3 G3 , ξ = k(x + −αt)

1

β

Combining with Eq. (2.2) and we can obtain the traveling wave solutions of (3.1) as follows:

(3.9)
(3. 10)

u (x, t) = 60γ k 3λ3 [de−λk (x+ −αt) ]3

1

β

where d is an arbitrary constant

4. Application for (2+1) dimensional Boussinesq Equation
In this section, we will consider the following (2+1) dimensional Boussinesq equation:

(3.11)

utt − uxx − uyy − (u2 )xx − uxxxx = 0

(4.1)

Suppose that

u(x, y,t) = u(ξ ),ξ = kx + ly + mt + d
(4.2)

where l, k, m, d are constants that to be determined later.
By (4.2), (4.1) is converted into an ODE

(m2 − k2 − l2)u ''− 2k2(u '2 + uu '') − k4u '''' = 0

(4.3)

Integrating (4.3) once we obtain

(m2 − k2 − l2)u '− 2k2uu '− k4u ''' = g

(4.4)

where g is the integration constant. Suppose that the solution of (4.4) can be expressed by a polynomial in G as follows:

∑ m
u(ξ ) = aiGi
i=0

(4.5)

where ai are constants, and G = G(ξ ) satisfies Eq.(2.1).
Balancing the order of uu ' and u ''' in Eq.(4.4), we have 2m +1 = m + 3 ⇒ m = 2 .So Eq. (4.5) can be rewritten as

u(ξ ) = a2G2 + a1G + a0 , a2 ≠ 0

(4.6)

where a2 , a1, a0 are constants to be determined later.

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Substituting (4.6) into (4.4) and collecting all the terms with the same power of G together, equating each coefficient to zero, yields a set of simultaneous algebraic equations as follows:

G0 : −g = 0

G1 : (k 2 + l 2 − m2 )a1λ + k 4a1λ3 + 2k 2a0a1 = 0

G2 : −7k 4μa1λ 2 + 8k 4a2λ3 + 2(k 2 + l 2 − m2 )a2λ +(m2 − k 2 − l 2 )μa1 − a12 + 2k 2a12λ + 4k 2a0a2λ − 2k 2a0a1μ = 0

G3 : 2(m2 − k 2 − l 2 )a2μ − 4k 2a0a2μ + 6k 2a1a2λ −38a2μk 4λ 2 − 2k 2a12μ +12a1λk 4μ 2 = 0

G4 : −6k 2a1a2μ + 54a2λk 4μ 2 − 6a1k 4μ3 + 4k 2a22λ = 0

G5 : −24a2k 4μ3 − 4k 2a22μ = 0

Solving the algebraic equations above, yields:

a = −6k 2μ 2 , a = −6k 2μλ, a = − 1 l2 + k 2 − m2 + λ 2k 4 k = k,l = l, m = m, d = d

2

1

02

k2

where k,l, m, d are arbitrary constants.

(4.7)

Substituting (4.7) into (4.6), we get that

u(ξ ) = −6k 2μ 2G2 − 6k 2μλG − 12 l2 + k 2 −km2 2 + λ 2k 4 ξ = kx + ly + mt + d Combining with Eq. (2.2), we can obtain the traveling wave solutions of (4.1) as follows:

(4.8)

u(ξ ) = −6k 2μ 2 ( 1 )2 − 6k 2μλ( 1 )

μ + deλξ

μ + deλξ

λ

λ

− 12 l 2 + k 2 −km2 2 + λ 2k 4

Remark : Our result (4.9) is new exact traveling wave solutions for Eq. (4.1).

(4.9)

5. Conclusions
In the present work, we propose a new sub-ODE method, and then test its power by finding some new traveling wave solutions of Boussinesq equation and (2+1) dimensional Boussinesq equation. This method is one of the most effective approaches handling nonlinear evolution equations. One can see the method is concise and effective. Also this method can be used to many other nonlinear problems.

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