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The mKdV equation with the initial value problem is studied numerically by means of the homotopy perturbation method. The analytical approximate solutions of the mKdV equation are obtained. Choosing the form of the initial value, the single solitary wave, two solitary waves and rational solutions are presented, some of which are shown by the plots.

Partial differential equations widely describe many phenomena in the world. Although many mathematicians and physicists presented various methods to find the explicit solutions of the partial differential equations, it is a difficult and important task to build the solutions of initial and boundary value problem. Recently, the homotopy perturbation method (HPM) have been applied into many problems [

The initial value problem of mKdV equation is as following:

{ u t + 6 u 2 u x + u x x x = 0 , u ( x , 0 ) = f ( x ) . (1)

The mKdV equation arises in many different fields, such as shallow water model, plasma science, biophysics and so on. A Darboux transformation was developed for generating dark multi-soliton solutions of the mKdV equation [

This paper is arranged as follows: In Section 2, by using HPM, we obtain the analytical approximate solution of Equation (1). In Section 3, by taking the form of the initial value, some exact solutions of mKdV equation are obtained. And some pictures are given to show the structure of the obtained solutions. Finally, some conclusions and discussions are given in Section 4.

In order to obtain the analytical approximate solution of Equation (1), we consider the one-parameter family of Equation (1) as follows

( u − u 0 ) t + p ( 6 u 2 u x + u x x x ) = 0, (2)

where the parameter p ∈ [ 0,1 ] and u 0 = f ( x ) .

If p = 0 , we meet u = u 0 .

If p = 1 , we come back to the original problem (1). Let the solution u ( x , t ) of the system (2) be written in the form of an infinite series,

u ( x , t ) = ∑ i = 0 ∞ u i ( x , t ) p i . (3)

Then u ( x , t ) = ∑ i = 0 ∞ u i ( x , t ) is a series solution of Equation (1).

Substituting Equation (3) into Equation (2), and equating the coefficients of p , p 2 , ⋯ , we have

u 1 , t + 6 u 0 2 u 0 , x + u 0 , x x x = 0 , (4)

u 2 , t + 6 u 0 2 u 1 , x + u 1 , x x x + 12 u 0 u 1 u 0 , x = 0 , (5)

u 3 , t + 6 u 0 2 u 2 , x + 6 u 1 2 u 0 , x + u 2 , x x x + 12 u 0 u 1 u 1 , x + 12 u 0 u 2 u 0 , x = 0 , (6)

and so on. Solving Equations (4), (5) and (6), one can obtain

u 1 ( x , t ) = − ( 6 u 0 2 u 0 , x + u 0 , x x x ) t , (7)

u 2 ( x , t ) = 1 2 ( 144 u 0 3 u 0 , x 2 + 36 u 0 4 u 0 , x x + 12 u 0 2 u 0 , x x x x + 72 u 0 , x 2 u 0 , x x + 36 u 0 u 0 , x x 2 + 60 u 0 u 0 , x u 0 , x x x + u 0 , x x x x x x ) t 2 , (8)

u 3 ( x , t ) = − 1 6 ( 900 u 0 , x u 0 , x x x 2 + 6480 u 0 4 u 0 , x 3 + 864 u 0 , x 5 + 504 u 0 u 0 , x x x u 0 , x x x x + 2376 u 0 3 u 0 , x x u 0 , x x x + 6264 u 0 2 u 0 , x 2 u 0 , x x x + 1512 u 0 3 u 0 , x u 0 , x x x x + 3888 u 0 5 u 0 , x u 0 , x x + 7992 u 0 2 u 0 , x u 0 , x x 2 + 144 u 0 u 0 , x u 0 , x x x x x x + 324 u 0 u 0 , x x u 0 , x x x x x + 1404 u 0 , x u 0 , x x u 0 , x x x x + 324 u 0 , x 2 u 0 , x x x x x + 108 u 0 4 u 0 , x x x x x + 216 u 0 6 u 0 , x x x + 1296 u 0 , x x 2 u 0 , x x x + 18 u 0 2 u 0 , x x x x x x x + 9504 u 0 u 0 , x 3 u 0 , x x + u 0 , x x x x x x x x x ) t 3 .

(9)

Hence, we obtain the solution of Equation (1)

u ( x , t ) = f ( x ) + u 1 ( x , t ) + u 2 ( x , t ) + u 3 ( x , t ) + ⋯ ,

where u 1 ( x , t ) , u 2 ( x , t ) and u 3 ( x , t ) are given by Equations (7), (8) and (9) respectively.

In this section, we will study the single soliton, two-soliton and rational solutions of mKdV equation.

Consider the following case:

{ u t + 6 u 2 u x + u x x x = 0 , u ( x , 0 ) = − 2 k exp ( k x ) exp ( 2 k x ) + 1 .

From the above section, we can have

u 0 ( x , t ) = − 2 k exp ( k x ) exp ( 2 k x ) + 1 ,

u 1 ( x , t ) = − 2 k 4 exp ( k x ) ( exp ( 2 k x ) − 1 ) t ( exp ( 2 k x ) + 1 ) 2 ,

u 2 ( x , t ) = − k 7 exp ( k x ) ( exp ( 4 k x ) − 6 exp ( 2 k x ) + 1 ) t 2 ( exp ( 2 k x ) + 1 ) 3 ,

u 3 ( x , t ) = − k 10 exp ( k x ) ( exp ( 6 k x ) − 23 exp ( 4 k x ) + 23 exp ( 2 k x ) − 1 ) t 3 3 ( exp ( 2 k x ) + 1 ) 4 ,

u ( x , t ) = − 2 k exp ( k x ) exp ( 2 k x ) + 1 − 2 k 4 exp ( k x ) ( exp ( 2 k x ) − 1 ) ( exp ( 2 k x ) + 1 ) 2 t − k 7 exp ( k x ) ( exp ( 4 k x ) − 6 exp ( 2 k x ) + 1 ) ( exp ( 2 k x ) + 1 ) 3 t 2 − k 10 exp ( k x ) ( exp ( 6 k x ) − 23 exp ( 4 k x ) + 23 exp ( 2 k x ) − 1 ) 3 ( exp ( 2 k x ) + 1 ) 4 t 3 + ⋯ .

Using Taylor series, one can obtain the exact solution

u ( x , t ) = − 2 k exp ( k ( x − k 2 t ) ) exp ( 2 k ( x − k 2 t ) ) + 1 . (10)

In this case, we take

f ( x ) = 4 exp ( x ) exp ( 2 x ) + 1 .

Then from the above section, one can have

u 0 ( x , t ) = 4 exp ( x ) exp ( 2 x ) + 1 ,

u 1 ( x , t ) = 4 exp ( x ) ( exp ( 2 x ) + 1 ) 4 ( exp ( 6 x ) + 73 exp ( 4 x ) − 73 exp ( 2 x ) − 1 ) t ,

u 2 ( x , t ) = 2 exp ( x ) ( exp ( 2 x ) + 1 ) 7 ( exp ( 12 x ) + 2158 exp ( 10 x ) + 2863 exp ( 8 x ) − 26236 exp ( 6 x ) + 2863 exp ( 4 x ) + 2158 exp ( 2 x ) + 1 ) t 2 ,

u 3 ( x , t ) = 2 exp ( k x ) 3 ( exp ( 2 x ) + 1 ) 10 ( exp ( 18 x ) + 58951 exp ( 16 x ) + 225620 exp ( 14 x ) − 1999268 exp ( 12 x ) − 6147250 exp ( 10 x ) + 6147250 exp ( 8 x ) + 1999268 exp ( 6 x ) − 225620 exp ( 4 x ) − 58951 exp ( 2 x ) − 1 ) t 3 ,

u ( x , t ) = 4 exp ( x ) exp ( 2 x ) + 1 + 4 exp ( x ) ( exp ( 2 x ) + 1 ) 4 ( exp ( 6 x ) + 73 exp ( 4 x ) − 73 exp ( 2 x ) − 1 ) t + 2 exp ( x ) ( exp ( 2 x ) + 1 ) 7 ( exp ( 12 x ) + 2158 exp ( 10 x ) + 2863 exp ( 8 x ) − 26236 exp ( 6 x ) + 2863 exp ( 4 x ) + 2158 exp ( 2 x ) + 1 ) t 2

+ 2 exp ( k x ) 3 ( exp ( 2 x ) + 1 ) 10 ( exp ( 18 x ) + 58951 exp ( 16 x ) + 225620 exp ( 14 x ) − 1999268 exp ( 12 x ) − 6147250 exp ( 10 x ) + 6147250 exp ( 8 x ) + 1999268 exp ( 6 x ) − 225620 exp ( 4 x ) − 58951 exp ( 2 x ) − 1 ) t 3 + ⋯ .

Using Taylor series, one can obtain the exact solution

u ( x , t ) = 4 ( exp ( t − x ) + 3 exp ( 27 t − 3 x ) + 3 exp ( 29 t − 5 x ) + exp ( 55 t − 7 x ) ) 1 + 4 exp ( 2 t − 2 x ) + 6 exp ( 28 t − 4 x ) + 4 exp ( 54 t − 6 x ) + exp ( 56 t − 8 ) . (11)

Here, our goal is to find the rational solution of mKdV equation. To do this, we consider the form of the initial value as follows:

f ( x ) = 2 I x − a

Due to the above section, it is obtained

u 0 ( x , t ) = 2 I x − a ,

u 1 ( x , t ) = − 36 I t ( x − a ) 4 ,

u 2 ( x , t ) = 432 I t 2 ( x − a ) 7 ,

u 3 ( x , t ) = − 5184 I t 3 ( x − a ) 10 ,

u ( x , t ) = 2 I x − a − 36 I ( x − a ) 4 t + 432 I ( x − a ) 7 t 2 − 5184 I ( x − a ) 10 t 3 + ⋯ .

From the knowledge of Taylor series, one can get the exact solution

u ( x , t ) = 2 I [ ( x − a ) 3 − 6 t ] ( x − a ) [ ( x − a ) 3 + 12 t ] ,

which is singular at x = a or ( x − a ) 3 + 12 t = 0 .

In summary, we successfully apply homotopy perturbation method to the mKdV equation with the initial value problem and obtain the analytical approximate solution of the mKdV equation. Using the form of the initial value, the single solitary wave, two solitary waves and rational solutions of the mKdV are obtained. Here, we get the two solitary waves solution without using bilinear forms, Wronskian, etc. In our later works, we will focus on the form of the initial value that can create the two solitary waves solutions.

This work was supported by the National Natural Science Foundation of China under Grant No. 11305048, the Science and Technology Research Key Project of Education Department of Henan Province under Grant No. 13A110329, the Basic and Frontier Research Program of Henan Province under Grant No. 132300410223, the Doctor Foundation of Henan Polytechnic University under Grant No. B2011-006, and the Key Teacher Foundation of Henan Polytechnic University (Grant 2014).

Dong, Z.Z. and Wang, F. (2018) Construction of Solitary Wave Solutions and Rational Solutions for mKdV Equation with Initial Value Problem by Homotopy Perturbation Method. Open Access Library Journal, 5: e4383. https://doi.org/10.4236/oalib.1104383