Explicit Solutions of the Coupled mKdV Equation by the Dressing Method via Local Riemann Hilbert Problem

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ISSN Online: 2152-7393 ISSN Print: 2152-7385

Explicit Solutions of the Coupled mKdV Equation

by the Dressing Method via Local Riemann-Hilbert

Problem

Ting Su

∗

, Guohua Ding, Zhiwei Wang

Department of Mathematical and Physical Science, Henan Institute of Engineering, Zhengzhou, China

Abstract

We study the coupled mKdV equation by the dressing method via local Riemann- Hilbert problem. With the help of the Lax pairs, we obtain the matrix Riemann- Hilbert problem with zeros. The explicit solutions for the coupled mKdV equation are derived with the aid of the regularization of the Riemann-Hilbert problem.

Keywords

Coupled mKdV Equations, Riemann-Hilbert Problem, the Dressing Method

1. Introduction

The coupled mKdV equation

6 0

t xxx x

u u uvu v v uvv

− + =

− + = (1)

is an important member of the AKNS hierarchy [1]. Moreover, it has various applica-tions in mathematical and physical fields. In [2], Prof. Geng has given its quasi-periodic solution by using algebra-geometric methods. The equation can be solved by the me-thod of the inverse scattering transformation, Hirota direct meme-thod, Lax pairs nonli-nearization approach and others [3]-[6]. There are a lot of references for the topic [7]- [14].

In this paper, we study the Equation (1) with the help of the Riemann-Hilbert me-thod following [15][16]. The present paper is organized as follows. In section 2, we give the Jost solution of the spectral equation. In section 3, we discuss the analytic property of the Jost solution. In section 4, we give the Matrix Riemann-Hilbert Problem. In sec-tion 5, we obtain the soliton-solusec-tion of the coupled KdV Equasec-tion (2), and we drop

How to cite this paper: Su, T., Ding, G.H. and Wang, Z.W. (2016) Explicit Solutions of the Coupled mKdV Equation by the Dressing Method via Local Riemann-Hilbert Problem. Applied Mathematics, 7, 1789-1797.

http://dx.doi.org/10.4236/am.2016.715150

Received: August 2, 2016 Accepted: September 18, 2016 Published: September 21, 2016

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the curve of the solutions with the aid of the Matlab.

2. Jost Solution

First, we consider the coupled KdV equation

6 ,

6 .

t xxx x

u u uvu v v uvv

= −

(2)

As is well known [2], the Equation (2) can be derived as the compatibility of the sys-tem

, ,

x U t V

Ψ = Ψ Ψ = Ψ

₍₃₎

where the 2 × 2 matrices U and V of the form

3

0 1

,

0 2

u U ik Q Q

v

σ  

= − + _{= } _

 

(4)

3 2 2 3

3 3

1

2

2 xx x x

V = ikσ −k Q ikQ− σ +Q +QQ −Q Q− Q (5)

where k is an arbitrary constant spectral parameter.

When u=0,v=0, we obtain the special solution of Equation (3). For convenience,

we denote the special solution as 3

1 exp

2

F= _− ikxσ _

 . Then, the spectral Equation (3) is transformed into

[

3

]

1

, ,

2

x

H = − ik σ H +QH (6)

where, 1

H= ΨF− .

In what follows, we study the Jost solutions H±

( )

x k, of the Equation (6) satisfying

the asymptotic conditions H± →I, at x→ ±∞. Since trU =0, these boundary con-ditions guarantee that detH± =1 for all x.

In fact, the Jost functions H± are not mutually independent. They are intercon-nected by the scattering matrix S k

( )

:

( )

_{( )}

( )

1_, a k b k _{, det} _1.

H H FSF S k S k b k a k

−

− +

 − 

= =_ _ =

  (7)

3. Analysis Solutions

Let us rewrite the spectral Equation (6) with the boundary conditions in the integral form:

( ) ( )

( )( ) ( )

( ) ( )

( )( ) ( )

(

)

11 21

21 11

, , d ,

, , exp d ,

x

H x k I u H k H x k v H k ik x

ξ ξ ξ

ξ ξ ξ ξ

− _−∞ −

= +

= _ − _

∫

(8)

for the first column entries of the Jost matrix H−.

It is easy to know that the exponent in (8) decreases for Imk>0. The first column

[ ]1

(3)

analytic as well in the same domain. Then, we give a solution of Equation (6):

( )

(

[ ]1 [ ]2

)

, , .

x k H H

+ − +

Ω =

It can see that it is analytic as a whole in the upper half plane.

The analytic solution Ω+

( )

x k, can be expressed in terms of the Jost function. In

view of (7), we derive

[ ] [ ]

(

)

(

[ ]

_{( )}

[ ] [ ]

)

(

[ ] [ ]

)

( )

1 2 1 2 2 1 2 0

, e , ,

e 1

ikx

a k H H aH b k H H H H

b k

+ − + + + + + +

 

Ω = = + =  

  (9)

with

( )

0 . 1 a k S b k +   =  

 

(10)

In the same way,

( )

1 1 , , . 0 b k

H FS F S S SS a k

−

+ − − − + −

 

Ω = =_ _ =

  (11)

It follows from the above formal as that

( )

detΩ₊ x k, =a k . (12)

In what follows, we define a function −1

( )

_{x k}_, +

( )

_{x k}_,

− +

Ω = Ω  . It is obvious that

( )

_{[ ]}

( )

_{[ ]} 1 1 1 1 2 . H H − − − − ₋ +    

Ω = _ _

 

Then, 1

( )

,

x k

− −

Ω _{is a solution of the adjoint spectral problem. On the real axis}

( )

1 1 1 1 1

, , ,

x k x k FS F H FS F H

− + + − − + − −

− + + + − −

Ω = Ω  = =

and _det −1

( )

_{x k}_, _{a k}

( )

−

Ω =  , Ω+

( )

x k, has an asymptotic expansion as follows:

( )

( )1

_{( )}

2

1 1

, ,

x k I x O

k k

+ +  

Ω = + Ω _{+  }

  (13) and substitute it into the spectral Equation (6). Comparing with powers of k, we derive

( )1

3, .

Q=i_σ Ω₊ _ (14)

In order to solve the coupled KdV Equation (2), we should find the analytic solution +

Ω _.

4. Matrix RH Problem

Through tedious calculation, we obtain RH problem

( ) ( )

( )

1 1 0 0 1 , , , , 1 b x k x k F k F S S

a

− −

− + + +

 

Ω Ω = Λ Λ = =  

 

† ₍₁₅₎

with 2 2

1

a +b = , Imk=0.

It is easy to know that 1

( ) ( )

, ,

x k x k

−

− +

(4)

being given by the simple exponential function F. Moreover, it is obvious that

( )

x k, F

+

Ω → _for k→ ∞, in view of (12).

In order to obtain the soliton solution of the coupled KdV equation, we suppose that the zeros of a k

( )

and a k

( )

are simple and finite number. We know that determi-nants of the matrices Ω+ and Ω−−1 are given by a k

( )

and a k

( )

. We assume that

( )

detΩ+ kj =0, Imkj >0, j=1, 2,,N,

( )

1

detΩ−₋ k_j =0, Imk_l<0,l=1, 2,,N.

In this case, the RH problem (15) with zeros can be solved in view of its regulation. To obtain the relevant regular problem, let us introduce a rational matrix function

1

, j j ,

j j

j j j

Y Y

k k

I

k k Y Y

χ

− _{= +} − _ϒ _{ϒ =}

−



where the eigenvector Yj solves Ω+

( )

kj Yj =0. Here ϒj is the rank 1 projector

2 j j

ϒ = ϒ _{, and} Y_j = Y_j +.

In view of (11), we know that detΩ+

( )

k

(

k−kj

)

=0 near the point

j

k

. We obtain

(

1

)

det

χ

j 0

− +

Ω ≠ _{at the point} k_j. The matrix function Ω−₋1 will be regularized by the

rational function , l l l l l k k I k k

χ = − − ϒ

−

 

it is easy to know that the matrix 1 l

χ −

−

Ω _{has no zeros in} k_l.

The regularization of all the other zeros is performed similarly, and eventually we obtain the following representation for the analytic solutions:

1 1

, N N ,

ω χ χ χ

± ± −

Ω = Γ Γ = _ ₍₁₆₎

where the rational matrix function Γ

( )

x k, accumulates all zeros of the RH problem,

while the matrix functions ω_±_{solve the regular RH problem (without zeros)}

( ) ( )

( )

1 1 1

0

, , , , ,

x k x k x k F k F x k

ω

−

ω

− −

− + = Γ Λ Γ (17)

with Λ0

( )

k =I, thus Ω = Γ+

( )

x k, .

The matrix Γ will be called the dressing factor. It follows from (16) that the asymp-totic expansion for the dressing factor is written as

( )

( )1

_{( )}

2

1 1

, .

x k I x O

k k

 

Γ = + Γ _{+  }

  (18) We note that the dress matrix Γ

( )

k can be written as

( )

1 1

1 1 1 1 1 , N N N N N l l l l

k k k k

k I I

k k k k k k

I x y k k

=

 −   − 

Γ =_ − ϒ _{ } − ϒ _

− _ − _   − = − −

∑

      

( )

1 1 1

1 1

N N

k k k k

k I I

k k k k

−  −   − 

Γ =_ + ϒ  _ + ϒ 

− _ − _

 

 



(5)

1 . N l l l l l l k k

I y x

k k

=

− = +

−

∑

_ (20)

Thus, we derived 2N vectors xj and yj instead of N vectors Yj . It is ob-vious that

( ) ( )

1

k − k I

Γ Γ = _{at the point}k=k_j. To avoid divergence at k→k_j, we

should pose Γ

( )

kj yj xj =0, that is

0.

l l

l l j j

j l

k k

I y x y x

k k  ₋  − =    ₋    

 (21)

We note that the matrix Γ

( )

k can be decomposed into the following form:

( )

1

, 1 1 , , N lj jl

j l _l _j _l

l j

k I j l

k k k k

−

=

Γ = − =

− −

∑

_   _ (22)

where lj l 1 j

j l

y y

k k

= =

− 

  . Similarly,

( )

1

, 1

1

,

N

jl

j l j

k I j l

k k

−

=

Γ = −

−

∑

 (23)

where j ≡ yj . In what follows, we rewrite (13) as

( )1

_{( )}

3, .

Q=i_σ Γ x _ (24)

Let us differentiate the equation Ω+

( )

kj j =0 in x, and in view of (6), we derive

( )

,

( )

,

( )

,

( )

, 0,

2

j

x j j x j j x

k

x k j x k j i x k j x k j

+ + + +

∂ Ω + Ω = Ω + Ω =

thus, we have

3

1

.

2 j

x

j = − ikσ j

(25)

In the same way, we obtain the evolutionary equation

2 3 1 . 2 j t

j = ik σ j

(26) In this end, we establish explicitly the vector j as

2

3 0

1 1

exp ,

2 j 2 j

j = __−i k x i+ k _tσ _ j

 

  (27)

where j0 is a vector integration constant. Similarly, according to 1

( )

0

j

j Ω−₋ k =

 , we obtain the solution

2

3 0

1 1

exp ,

2 j 2 j

j = __−i k x i+ k t_σ _ j

 

   

  ₍₂₈₎

where j0 is a vector integration constant.

5. One Soliton Solution

(6)

(

)

(

)

1 2 2 2 2 2 1 exp

2 2 2

1 ,

1 exp

2 2 2

x

x i x t p

t

x

x i x t p

t

η ξ _ξη

ξ η

η ξ _ξη

ξ η  ₊ _{+ −} ₊    ₋ _ _    =      ₋ ₋ _{− −} ₊     ₋ _ _   

(29)

where, p p1, 2 are components of the constant vector 10 .

(

)

(

)

1 2 2 2 2 2 1 exp

2 2 2

1 ,

1 exp

2 2 2

x

x i x t p t

x

x i x t p t

η ξ _ξη

ξ η

η ξ _ξη

ξ η     + + − +     −     =      ₋ ₋ _{− −} ₊       −        _{  }  _    _{  }  _ (30)

where, p p 1, 2 are components of the constant vector 10 . The dress formula (19) reduced to

( )1

_{( )}

( )1

_{( )}

12 , 21 .

u= Γi x v= − Γi x (31)

At the same time, we have 1 1 1 1 k k I k k −

Γ = − ϒ

−



 , from which, we obtain

( )1

₍

₎

1 1

i

η η

Γ = − +  

 .

(32)

Denoting

(

2 2

)

z=

η

x+

ξ

−

η

t+

α

, κ ξ=

(

− +x 2ηt

)

+β , thus

exp 2

1 exp .

2 exp 2 z i i z i ϕ α β ϕ +     +   = _− _  +   _ ₋ _  

(33)

In the same way, defining

(

2 2

)

z=ηx+ ξ −η t+α,

ϕ ξ

= 

(

− +x 2

η

t

)

+

β

, thus

exp 2

1 exp .

2 exp 2 z i i z i ϕ α β ϕ +      +  = _− _  +    _ ₋ _       



(34)

Substituting (31) and (32) into (30), we have

exp exp

2 2

1 1 exp .

2

exp exp

2 2

z z i i

i i z z

ϕ ϕ α α ϕ ϕ + +     +   = _− _  + +  _ ₋ ₋ _       

Moreover, 1 1 exp

(

exp exp

)

2 z z

α α+

 

= _− _ + −

 

 _ _{, hence,}

( )

₍

₎

(

)

(

)

(

)

1

exp exp

2 _sec _.

exp exp

2

z z

i

i z z

z z i ϕ ϕ η η ϕ ϕ +  ₊   

Γ = − +   +

+  ₋ ₊ ₋       _    

(7)

(

)

ei sec

(

)

,

(

)

ei sec

(

)

.

u=

η η

+ ϕ ϕ+ z+z v= − +

η η

 − +ϕ ϕ z+z (35) Here, ξ, ξ, η and η determine the soliton velocity and amplitude, respectively, while α_, β_, α and β_{give the initial position and phase of the soliton. In what} follows, we plot the graph for u x t

( )

, in order to analyze the solutions (35). Figure 1

and Figure 2 are the imaginary part and real part of u x t

( )

, , respectively. From the

two solution curves, we can see that the difference between the real and imaginary part. In the same way, we drop the solution curves of v for Figure 3 and Figure 4.

Figure 1. The soliton solution curve of imaginary part of u x t

( )

, for η=0.01, η=0.5, 0.05

[image:7.595.240.506.195.405.2]

ξ= _, ξ=0.07, α=0.2, α =0.4, β =0.6, β=0.9, x∈ −

[

2, 2

]

, t∈ −

[

6,6

]

.

Figure 2. The soliton solution curve of real part of u x t

( )

, for η=0.01, η=0.5, ξ=0.05, 0.07

[image:7.595.238.513.458.668.2]

(8)

Figure 3. The soliton solution curve of imaginary part of v x t

( )

, for η=0.01, η=0.5, 0.05

ξ= _, ξ=0.07, α=0.2, α =0.4, β =0.6, β=0.9, x∈ −

[

2, 2

]

, t∈ −

[

6,6

]

.

Figure 4. The soliton solution curve of real part of v x t

( )

, for η=0.01, η=0.5, ξ=0.05, 0.07

ξ= , α=0.2, α=0.4, β=0.6, β=0.9, x∈ −

[

2, 2

]

, t∈ −

[

6,6

]

.

From the graphs, it is shown that u and v have the similar solution form. The differ-ence exists between the real and imaginary part. In fact, we chose different parameters, and the solution curves between the real part and imaginary part had corresponding changes.

Acknowledgements

[image:8.595.241.507.70.281.2] [image:8.595.239.508.337.549.2]

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Chi-na (Project No: 11301149), HeChi-nan Natural Science Foundation For Basic Research un-der Grant No: 162300410072, doctor Foundation (D2015001) and Young backbone teachers in Henan province.

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