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The travelling solitary wave solutions to the higher order Korteweg-de Vries equation are obtained by using tanh-polynomial method. The method is effective and concise, which is also applied to various partial differential equations to obtain traveling wave solutions. The numerical simulation of the solutions is given for completeness. Numerical results show that the tanh-polynomial method works quite well.

Nonlinear wave phenomena appear in various scientific and engineering fields [

u t + u x x x x x + γ u u x x x + β u x u x x + α u 2 u x = 0 (1)

where the subscripts denote the partial derivatives of x and t. The γ, β and α are constant parameters; u represents a real scalar function u(x, t).

Consider a given evolution equation with independent variables x and t in the form are given as

P ( u , u x , u t , u x t , u x x , u t t , ⋯ ) = 0 . (2)

By using the wave variable ξ = x + κ t and substituting into Equation (2), we obtain the following ordinary differential equation

Q ( u , u ξ , u ξ ξ , u ξ ξ ξ , ⋯ ) = 0 . (3)

where u ξ , u ξ ξ , ⋯ denotes the derivative with respect to the same sole variable ξ .

The fact that the solutions of many nonlinear equations can be expressed as a finite series of tanh-function motivates us to seek for the solutions

u ( x , t ) = U ( ξ ) = ∑ i = 0 m a i tanh ( ξ ) i (4)

where ξ is a function about x and t, a 0 , a 1 , a 2 , ⋯ are constant parameters. m can be obtained by balancing the derivative term of the highest order with the nonlinear term as the follow

O ( d p u d ξ p ) = m + p , p = 1 , 2 , 3 , ⋯

and

O ( u q d p u d ξ p ) = ( q + 1 ) m + p , q = 0 , 1 , 2 , ⋯ ; p = 1 , 2 , 3 , ⋯

Usually m is a positive integer, however, once in a while, the value of m is a negative or a fraction, the other kinds of expression will introduced. In the following, we illustrate the method by using it to solve the higher-order Korteweg-de Vries equations.

Putting the variable ξ = x + κ t into Equation (1) and we find

k u ξ + u ξ ξ ξ ξ ξ + γ u u ξ ξ ξ + β u ξ u ξ ξ + α u 2 u ξ = 0 ; (5)

which is an ordinary differential equation. By using the above method, m = 2 is obtained by balancing the derivative term of the highest order with the nonlinear term.

u ( ξ ) = a 0 + a 1 tanh ( ξ ) + a 2 tanh ( ξ ) 2 (6)

substituting (6) into (5) yields a set of algebraic equations for a_{0}, a_{1}, a_{2}, γ, β and α. Collecting all terms with the same power of tanh ( ξ ) together, equating each coefficient to zero, we obtain a set of simultaneous algebraic equations as follows:

16 a 1 + α a 0 2 a 1 + 2 a 1 a 2 β − 2 a 1 a 0 β + a 1 κ = 0

2 α a 1 2 a 0 + 272 a 2 + 2 α a 0 2 a 2 − 2 a 1 2 β + 4 a 2 2 β − 2 a 1 2 γ − 16 a 0 a 2 γ + 2 a 2 κ = 0

− 136 a 1 − α a 0 2 a 1 + α a 1 3 + 6 α a 0 a 1 a 2 − 14 a 1 a 2 β + 8 a 0 a 1 γ − 18 a 1 a 2 γ − a 1 κ = 0

− 2 α a 1 2 a 0 − 1232 a 2 − 2 α a 0 2 a 2 + 4 α a 1 2 a 2 + 4 α a 2 2 a 0 + 4 a 1 2 β − 20 a 2 2 β + 8 a 1 2 γ + 40 a 0 a 2 γ − 16 a 0 a 2 γ − 2 a 2 κ = 0

240 a 1 − α a 1 3 − 6 α a 0 a 1 a 2 + 5 α a 2 2 a 1 + 22 a 1 a 2 β − 6 a 0 a 1 γ + 48 a 1 a 2 γ = 0

1680 a 2 − 4 α a 1 2 a 2 − 4 α a 2 2 a 0 + 2 α a 2 3 − 2 a 1 2 β + 28 a 2 2 β − 6 a 1 2 γ − 24 a 0 a 2 γ + 40 a 2 2 γ = 0

− 120 a 1 − 5 α a 2 2 a 1 − 10 a 1 a 2 β − 30 a 1 a 2 γ = 0

− 720 a 2 − 2 α a 2 3 − 12 a 2 2 β − 24 a 2 2 γ = 0

Solving the algebraic equations, we get the results:

1) The first case a 1 = 0

a 0 = 4 γ − − 136 α − 2 α a 2 β + 16 γ 2 − α κ α , a 2 = − 6 β − 12 γ − ( 6 β + 12 γ ) 2 − 1440 α 2 α ,

u 1 ( x , t ) = a 0 + a 2 tanh ( x + κ t ) 2 (7)

a 0 = 4 γ + − 136 α − 2 α a 2 β + 16 γ 2 − α κ α , a 2 = − 6 β − 12 γ + ( 6 β + 12 γ ) 2 − 1440 α 2 α

u 2 ( x , t ) = a 0 + a 2 tanh ( x + κ t ) 2 (8)

a 0 = 4 γ − − 136 α − 2 α a 2 β + 16 γ 2 − α κ α , a 2 = − 6 β − 12 γ + ( 6 β + 12 γ ) 2 − 1440 α 2 α

u 3 ( x , t ) = a 0 + a 2 tanh ( x + κ t ) 2 (9)

a 0 = 4 γ + − 136 α − 2 α a 2 β + 16 γ 2 − α κ α , a 2 = − 6 β − 12 γ − ( 6 β + 12 γ ) 2 − 1440 α 2 α ,

u 4 ( x , t ) = a 0 + a 2 tanh ( x + κ t ) 2 (10)

2) The second case

a 0 = 840 + 10 a 2 β + 15 a 2 γ 16 γ , a 1 = 10 a 0 γ + 6 a 2 ( 2 β + 3 γ − α a 0 ) − 1320 α , a 2 = − 3 γ + 3 γ 2 − 64 α 2 α

u 5 ( x , t ) = a 0 + a 1 tanh ( x + κ t ) + a 2 tanh ( x + κ t ) 2 (11)

a 0 = 840 + 10 a 2 β + 15 a 2 γ 16 γ , a 1 = − 10 a 0 γ + 6 a 2 ( 2 β + 3 γ − α a 0 ) − 1320 α , a 2 = − 3 γ − 3 γ 2 − 64 α 2 α

u 6 ( x , t ) = a 0 + a 1 tanh ( x + κ t ) + a 2 tanh ( x + κ t ) 2 (12)

a 0 = 840 + 10 a 2 β + 15 a 2 γ 16 γ , a 1 = 10 a 0 γ + 6 a 2 ( 2 β + 3 γ − α a 0 ) − 1320 α , a 2 = − 3 γ − 3 γ 2 − 64 α 2 α

u 7 ( x , t ) = a 0 + a 1 tanh ( x + κ t ) + a 2 tanh ( x + κ t ) 2 (13)

a 0 = 840 + 10 a 2 β + 15 a 2 γ 16 γ , a 1 = − 10 a 0 γ + 6 a 2 ( 2 β + 3 γ − α a 0 ) − 1320 α , a 2 = − 3 γ + 3 γ 2 − 64 α 2 α

u 8 ( x , t ) = a 0 + a 1 tanh ( x + κ t ) + a 2 tanh ( x + κ t ) 2 (14)

Equations (7)-(14) are the solutions in the tanh ( x + κ t ) form for higher order Korteweg-de Vries Equation (1). The numerical simulations u 1 ( x , t ) and u 2 ( x , t ) are shown with α = − 1 , β = 2 , γ = 0.4 , κ = 1 , the range of x and t are x ∈ [ − 1 , 1 ] and t ∈ [ − 1 , 1 ] in

Some new analytical solutions of the higher-order Korteweg-de Vries Equation (1) are obtained by successfully employing tanh-function method in this paper, which can be employed to discuss some interest physical phenomena, such as two-layer fluid, steady-state solitary waves in a fluid, three-layer fluid with a constant buoyancy frequency in an each layer. This tanh-function method is based on a previous work [

The authors declare no conflicts of interest regarding the publication of this paper.

Xiang, C.H. and Wang, H.L. (2019) Travelling Solitary Wave Solutions to Higher Order Korteweg-de Vries Equation. Open Journal of Applied Sciences, 9, 354-360. https://doi.org/10.4236/ojapps.2019.95029