New Soliton Solutions to the Initial Value Problem for the Two Component Short Pulse Equation

(1)

ISSN Online: 2327-4379 ISSN Print: 2327-4352

DOI: 10.4236/jamp.2019.71002 Jan. 9, 2019 13 Journal of Applied Mathematics and Physics

New Soliton Solutions to the Initial Value

Problem for the Two-Component

Short Pulse Equation

Yan Luo

College of Science, University of Shanghai for Science and Technology, Shanghai, China

Abstract

We formulate a matrix Riemann-Hilbert problem to the initial value problem for the two-component system proposed by Matsuno. By solving the asso-ciated Riemann-Hilbert problem, we can get the soliton solutions of the two-component system. One and two soliton solutions are investigated in detail.

Keywords

Riemann-Hilbert Problem, Two-Component Short Pulse Equation, Soliton Solutions

1. Introduction

The SP (short pulse) equation

( )

3

1 _,

6

xt _xx

u = +u u (1) where u u x t=

( )

, _{represents the magnitude of the electric field and subscripts}

x and t appended to u denote partial differentiation, has been proposed as a model equation describing the propagation of ultrashort optical pulses in non-linear media [1]. A numerical analysis shows that as the pulse length shortens, the SP equation becomes a better approximation to the solution of the Maxwell equation when compared with the prediction of the NLS equation [2]. Then Matsuno proposed a novel multi-component generalization of the SP equa-tion [3], which generalizes the SP equation describing the propagation of ul-tra-short pulses in optical fibers. Here we consider the multi-component sys-tem [3].

How to cite this paper: Luo, Y. (2019) New Soliton Solutions to the Initial Value Problem for the Two-Component Short Pulse Equation. Journal of Applied Ma-thematics and Physics, 7, 13-22.

https://doi.org/10.4236/jamp.2019.71002

Received: December 5, 2018 Accepted: January 6, 2019 Published: January 9, 2019

http://creativecommons.org/licenses/by/4.0/

(2)

DOI: 10.4236/jamp.2019.71002 14 Journal of Applied Mathematics and Physics

2

, ,

1

1 _, _{1,2, ,}

2

n i xt i i x j

j _x

u u u u i n

=

 

= + _ _ =



∑

 (2)

with reduced two-component short pulse system if letting n = 2, which given by

(

)

(

)

2 2

1 2 1 2

xt _{x x}

u u u v u

v v u v v

 _{= +} _ ₊ _

  

 

 = + _ + _ 

(3)

where u u= 1 and v u= 2. Then we make the transformation

( )

, ,

2 2

u v u v u v

i

+ −

 

→ _ _ and this system can be written as

(

)

(

)

1 2 1 2

xt x x

u u uvu

v v uvv

 _{= +}

 

 _{= +}



(4)

We named Equation (4) as 2SPE (two-component system). Obviously if we put u v= _{, then 2SPE reduces to the SP equation. The integrability, soliton}

solu-tions, and other features of Equation (1) and Equation (4) common to the com-pletely integrable PDEs (partial differential equations) have been studied from various points of view [3] [4] [5] [6]. And the semi-discretization of a mul-ti-component SP equation was studied in [7], super extensions in [8].

Here, in this paper, we consider the initial value problem for the 2SPE (4) with the initial value data u x t

(

, =0

)

=u x0

( )

∈S R

( )

, v x t

(

, =0

)

=v x0

( )

∈S R

( )

and S R

( )

denotes the Schwartz space. The present paper is devoted to analyze the IVP (initial value problem) for the Equation (4) by Riemann-Hilbert ap-proach.

Organization of the paper: In Section 2, we formulate the associated Rie-mann-Hilbert problem by performing the similar spectral analysis to the Equa-tion (4) as [9]. In Section 3, to obtain the soliton solutions of the Equation (4), we need analyze the residue conditions of the Riemann-Hilbert problem. Espe-cially, the one and two soliton solutions are investigated in detail.

2. Riemann-Hilbert Problem for the Two-Component System

The 2SPE admits a WKI (Wadati-Konno-Ichikawa)-type Lax pair as follow:

(

)

(

)

, , , , x

t

X x t k T x t k

ϕ ϕ

=  

=

 (5) where

1 1 3 2

0 1

,

0

2 4 2

u

ik i

X ikX T uvX

v

ikσ σ

 

= = + _{− } _

  (6) with

1 3 2

1 1 0 0

, , .

1 0 1 0

x x

u i

X

v σ σ i

−

     

=_ _ =_ _ =_ _

− _ − _ _ _

(3)

2.1. Spectral Analysis for

k

= 0

Introducing

(

)

₀

(

)

₄ 3

, , , , e

t ikx

ik

x t k x t k σ

ϕ

µ

 ₊ 

 

 

= (8) Then we can get the Lax pair of

_µ

0

0 0 0 0

3 1

0 0 0 0

3 2

,

1 _,

4

x

t

ik V

V ik

µ σ µ µ

 − _ _=

 

 

− =

 _ _



(9)

where

0 0

1 2

0 ₂ ₂ ₂

, 0

2 2 2

x x

x

ik_uv ik_uvu u u

V ik V

v ik_uvv v ik_uv

 ₋ 

 

 

= _ _ =  

 

  _ ₊ ₋ _

 

(10)

We define two eigenfunctions of 0

(

_{, ,}

)

j x t k

µ

(

)

( )

₍

₎

( )

₍

₎

(

)

( )

₍

₎

( )

₍

₎

3 3

0 0 0

1 1 1

0 0 0

2 1 2

, , e , , e , , d

x _{ik x y} _{ik x y}

ik x y ik x y x

x t k I V y t k y t k y

σ σ

µ µ

− − −

−∞

+∞ ₋ ₋ ₋

 _{= +}

 

 = −



∫

(11)

Proposition 1: The functions

{ }

0 2 1 j

µ are bounded and analytic as:

(

)

(

)

0 0

1 D D2, 1 , 2 D D1, 2 .

µ ∈ µ ∈ ₍₁₂₎ where 0

(

)

2, 1

j D D

µ

  ∈

  means that the 1-th column µ0j₁ and 2-th column 0

2 j µ

 

  of

µ

0j are bounded and analytic in D2 and D1, respectively,

{

}

1 Im 0

D = k∈ k> and D2 =

{

k∈Imk<0 .

}

Proposition 2: The functions 0

(

_{, ,}

)

j x t k

µ

_{have the expansions as:}

(

)

( )

0 _{, ,} 0 2 _, _0.

0 j

iu

x t k I k O k k

iv

µ = +_ _ + →

 

(13)

2.2. Spectral Analysis for

k

=

∞

Define

( )

1 1

1

, ,

2 1 ₁

x

m v m

G x t

m m

u

 ₋ − 

 

+  

= _ _

−

 

 

(14)

(

, ,

)

_x

(

,

)

1 d

)

₄ 2,

t

p x t k x m x t x

k

∞

′ ′

= −

_∫

− − (15) where m= +1 u vx x.

Introduce the following transformation:

(

x t k, ,

)

G x t

( )

, ehσ3

(

x t k, , e

)

hσ3e hσ3eikp x t k(, , )σ3

ϕ = − µ −− −+

(16)

(4)

(

)

( )

, d ,

4 1

x u v_{x xx} u v_{xx x}

h x t x

m m

− _−∞

− _′ _′

=

+

∫

(17)

(

)

( )

, d ,

4 1

x xx xx x

x

u v u v

h x t x

m m

+∞ +

− _′ _′

=

+

∫

(18)

(

)

( )

, d .

4 1

x xx xx x

u v u v

h h h x t x

m m

+∞

+ − _−∞

− _′ _′

= + =

+

∫

(19)

We find the Lax pair equation of

µ

(

x t k, ,

)

[

]

[

33

]

21 , ,

x x

t t

ikp V

µ σ µ µ

 − =

 

− =

 (20) where

(

)

(

)

1

1 0

4 2

1

0

4 2

x _{x xx}

x x

x _{xx x}

x x

m m _{u v}

mv mv

V

m m _{u v}

mu mu

 ₊ 

 ₋ ₊ 

 

= −

 + 

 ₋ 

 

 

(21)

(

)

(

)

(

)

(

)

(

)

(

)

2 3

0

1 1 ₁ 1 1

0

4 4

1 1

0 1

4 ₁ ₁

0

1 0

4 2

1

0

4 2

x x

x

x x

x

x _{x xx}

x x

x _{xx x}

x x

u V

v

ik m ik m

m u v m uv

v

m _m _{u v} _m _uv

u

m m _{u v}

mv mv

m m _{u v}

mu mu

σ  

 

= _ − _ − ⋅ _ _

   

 ₋ ₊ ₊ 

 ₋ 

 

+  

 + + − 

 

 

 ₊ 

 ₋ ₊ 

 

⋅

 ₊ 

 ₋ 

 

 

(22)

Then, define two eigenfunctions of

µ

j

(

x t k, ,

)

:

( ) ( )

₍

₎

( ) ( )

₍

₎

( ) ( )

₍

₎

( ) ( )

₍

₎

3 3

, , , , , , , ,

1 1 1

, , , , , , , ,

2 1 2

e , , e , , d

x _{ik p x t k p y t k} _{ik p x t k p y t k}

ik p x t k p y t k ik p x t k p y t k x

I V y t k y t k y

σ σ

µ µ

− −

   

   

−∞

+∞ _ − _ _ − _

 = +  

 = −



∫

(23)

{ }

µ

j ₁2 are bounded and analytic as:

(

)

(

)

1 D D2, 1 , 2 D D1, 2 .

µ

∈

µ

∈

(24) Proposition 4: The functions

µ

j

(

x t k, ,

)

also satisfies

(

, ,

)

1 , .

j x t k I O _k k

µ = +  _{ } → ∞

 

(25) Proposition 5: (Symmetry property) In this paper, we consider the function

( )

,

(5)

satisfy the following symmetry property:

( )

j k j k

µ − =µ

₍₂₆₎

2.3. Spectral Function

s

(

k

)

The eigenfunctions

µ

1

(

x t k, ,

)

and

µ

2

(

x t k, ,

)

are related

(

)

(

)

( , , ) 3

_{( )}

( , , ) 3

1 x t k, , 2 x t k, , eikp x t kσ s k e ikp x t kσ

µ ₌µ −

(27) where the matrix s k

( )

is independent of

( )

x t, and has the form

( )

a k

( ) ( )

_{( )}

b k

_{( )}

s k

c k d k

 

=  

  (28)

The function a k

( )

and d k

( )

can be computed by

( )

(

[ ] [ ]

)

( )

(

[ ] [ ]

)

1 1 2 2 2 1 1 2

det det

a k d k

µ µ

 =

 

=

 (29)

where det

( )

X means the determinate of matrix X.

Hence, form the analytic properties of the functions µj, we know that d k

( )

and a k

( )

are analytic in D1 and D2, respectively.

2.4. The Relation between

µ

j

(

x t k

, ,

)

and

µ

0j

(

x t k

, ,

)

µ

j

(

x t k, ,

)

and

µ

0j

(

x t k, ,

)

are related

(

)

( ) (

)

(

)

( ) (

)

3 3

3 3 3

1 0

1 1

1 0

2 2

, , e , , , e ,

, , e , , , e e ,

h ikd

h ikd h

x t k G x t x t k

σ σ

σ σ σ

µ

− −

− +

− − −

− −

 =

 

=

 (30)

where d− =

∫

_−∞x

(

m x t

( )

′, −1 d

)

x′ and d _x

(

m x t

( )

, 1 d

)

x

+∞

+ =

∫

′ − ′.

Proposition 7: The spectral functions a k

( )

and d k

( )

have the following asymptotic behaviors as k→0_:

( )

2

2 3

2

2 3

e 1

2

e 1

2

h

d

a k ikd k O k

d

d k ikd k O k

−

 ₌  ₋ ₋ ₊ 

  

  



 

 ₌ ₊ ₋ ₊

 

 _ _



(31)

where d d= −+d+.

2.5. The Riemann-Hilbert Problem

Let us define

(

)

[ ] [ ]

_{( )}

[ ]

( )

[ ]

1 2

2 1 1

1 1

2 2 2

, , ,

,

k D d k

M x t k

k D a k

µ µ

 

∈

_ _

 

= 

 

 _∈

 

 

 



(6)

Making coordinate transformation

( )

, _x

(

( )

, 1 d

)

( )

,

y x t = −x

∫

+∞ m x t′ − x′= −x d x t+

(33) and defining

(

, ,

)

(

( )

, , ,

)

M y t k =M x y t t k ₍₃₄₎

Then, M y t k

(

, ,

)

_{satisfies the Riemann-Hilbert problem as follows:}

(

, ,

)

(

, ,

) (

, , ,

)

,

M y t k+ =M y t k J y t k k− ∈

(35)

(

, ,

)

, ,

M y t k = Ι k→ ∞

₍₃₆₎

where

(

)

₄ 3

( )

₄ 3

0

, , e e .

t t

ik ky ik ky

k k

J y t k σ J k σ

 ₋  ₋  ₋ 

   

   

= ₍₃₇₎

To obtain the solution u x t

( )

, and v x t

( )

, of the Equation (4), we need the asymptotic behavior of M x t k

(

, ,

)

as k→0_,

(

, ,

)

e h 3 1

(

, ,

)

d iu

( )

2 eh 3

M x t k G x t k I k O k

iv d

σ σ

− +

− −

+

   

= _ + _ _+ _

−

 

 

(38)

Hence, we can get the solution of Equation (4)

( )

(

( )

)

( )

(

( )

)

, , ,

u x t u y x t t

v x t v y x t t

 =

 

=

 (39)

where

( )

(

)

(

)

( )

(

)

(

)

( )

(

)

(

)

1

11 0

1

2 12

0

1

2 21

0

, ,0 , , 1

, lim

, ,0 , ,

e , lim

, ,0 , ,

e , lim

k

h

k

h

k

M y t M y t k x y t y

ik

M y t M y t k u y t

ik M y t M y t k v y t

ik

−

→

−

→

−

→

  _⋅  ₋

 

 _{= +}

 

  _⋅ 

 _⋅ ₌  

 

 _ _

⋅

 _ _

⋅ =

 

(40)

3. Solitons

In order to get the soliton solutions of 2SPE, we assume that d k

( )

has N sim-ple zero points

{ }

₁ 1

N j j

k ₌ ∈D _and a k

( )

has N simple zero points

{ }

₁ 2 N j j

k ₌ ∈D _.

Then, there exist some constant Cj and Cj such that,

(

)

(

)

(

)

(

)

2

2 2

4

2 1

2 4

1 2

Res , , e , ,

j j j

t ik y

k

k k j j

t ik y

k

j j

k k

M y t k C M y t k

 

 − 

 

 

=

 

 

− _ − _

 

=



 _{ } ₌ _{ }

    

  

 _{ } ₌ _{ }

   



(41)

(7)

(

)

(

)

(

)

(

)

2 2 2 4

2 ₁ 1

2 4

1 ₁ 1

e 0

, , ₁ , ,

e 1

, , ₀ , ,

j j j j t ik y k N j j j j t ik y k N _j j j _j C

M y t k M y t k

k k

C

M y t k M y t k

k k    ₋      =    ₋      =     _ _{ =} ₊ _ _        _{ } ₋     _{ }     _ _ =_{ }+ _ _ −   

∑

(42)

3.1. One Soliton Solution

Take N =1 into Equation (42), evaluating at k k= 1 and k k= 1 of the 1-th

and 2-th column of M y t k

(

, ,

)

, then solving the algebra system for

(

)

(

)

11 , , 1 , 21 , , 1

M y t k M y t k and M12

(

y t k M, , 1

)

, 22

(

y t k, ,1

)

, we can get

(

)

(

)

(

)

1 ₂ 1 ₂

1 1

1 2

1

1 ₂ 1 ₂

1 1 1 11 1 2 2 4 4 1 1

1 1 1 1

2 4 1 1 1 12 1 2 2 4 4 1 1

1 1 1 1

2 1 21 1 1 , , e e 1 e , , e e 1 e , , t t

ik y ik y

k k

t ik y

k

t t

ik y ik y

k k

ik y

M y t k

C C

k k k k

C k k

M y t k

C C

k k k k

C

M y t k

     −  −  −             −           −  −  −          − = − ⋅ − − − = − ⋅ − − =

(

)

2 1

1 2 1 2

1 1

1 2 1 2

1 1 4 1 1 2 2 4 4 1 1

1 1 1 1

22 1

2 2

4 4

1 1

1 1 1 1

e e 1 1 , , e e 1 t k t t

ik y ik y

k k

t t

ik y ik y

k k

k k

C C

k k k k

M y t k

C C

k k k k

   −           −  −  −               −  −  −                            ₋     ₋ _⋅  − −   ₌    ₋ _⋅  − −  (43)

By the symmetry condition, k1=ib. And for simplify choosing k1= −ib

and 2 20

1 2 eic y

C ₌ ib +

, 2 2 0

1 2 e ic y

C ₌ ib − +

; b, c and y0 are real constants; denoting

1 2 2

1

4

t

k y i

k

ϕ

=  − =

ϕ

  , and 2 4

t by

b

ϕ = + , then we get one soliton solution

( )

(

( )

)

₍

₎

( )

(

( )

)

₍

₎

(

)

2 2 0 2 2 2 0 2 0 2 1 1

, , , e

cosh 2 2

1 1

, , , e

cosh 2 2

1 tanh 2 2 1

ic h

u x t u y x t t

b y

v x t v y x t t

b y

x y y

(8)

3.2. Two-Soliton Solutions

Take N=2_{into Equation (42), we can get}

( )

1 2 2 2

1 2

1 ₂ 2 ₂

1 2 1 ₂ 1 2 2 4 4 1 2

12 11 1 11 2

1 2

2 2

4 4

1 2

22 21 1 21 2

1 2

2 2

4

1 2

11 12 1

1 e e e e 1 e e 1 t t

ik y ik y

k k

t t

ik y ik y

k k

t

ik y ik

k

C C

M k M k M k

k k k k

C C

M k M k M k

k k k k

C C

M k M k

k k      −   −               −   −              − _ − _ −   = + − − = + + − − = + + −

( )

2 ₂ 2

1 2 2 2

1 2 4 12 2 2 2 2 4 4 1 2

21 22 1 22 2

1 2 e e t y k t t

ik y ik y

k k

M k k k

C C

M k M k M k

k k k k

   −              − _ − _ − _ − _                 ₋    = +  − −  (45)

1) k2= −k1

Letting k1= +a bi , k2 = − +a bi and choosing k1= −a bi , k2 = − −a bi

and denoting

ϕ

1=a y z

(

−

)

and

ϕ

2=b y z

(

+

)

, where

₍

₎

2 2 4 t z a b =

+ , repeat-ing the process of one soliton solution, then we can get the two soliton solutions.

Parameter selection case one: Choosing

(

)

2

1 2bi eic

C a bi

a

= ⋅ + ⋅ ,

(

)

2

1 2bi e ic

C a bi

a

−

= ⋅ − ⋅

_,

₍

₎

2

2 2bi eic

C a bi

a

= ⋅ − ⋅ ,

(

)

2

2 2bi e ic

C a bi

a

−

= ⋅ + ⋅

_{, where}_c

is an arbitrary constant, we have

( )

1 2

2 2 2 2 2 2

2 1

2 2 2 1 2 1

2 2 2 2 2 2

2 1

2 2 2 1 2 1

2 2 2 2

2

sin 4 sinh 4 2

,

cosh 2 sin 2

cosh 2 cos 2 sinh 2 sin 2

2

, e

cosh 2 sin 2

cosh 2 cos 2 sinh 2 sin 2

2 , e cosh 2 ic h ic h b a ab x y t y

a

a b a b

a b

ab u y t

a b a b

a b

ab v y t

a b a b

ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ + − +  −  = + _ + _ + _ + _  +  − = ⋅ _ _ + _ + _ + − = ⋅

+ + 2 2

1 sin 2ϕ

       _ _  _ _     (46)

Remark 8: The Result (46) is coincide with the breather-solution of SP equation as

[6], since the parameter chosen satisfies the symmetry condition of SP equation as [9]. Parameter selection case two: Choosing

(

)

2

1 2b eic

C a bi

a

= ⋅ + ⋅ ,

(

)

2

1 2b e ic

C a bi

a

−

= ⋅ − ⋅

_,

₍

₎

2

2 2b eic

C a bi

a

= ⋅ − ⋅ ,

(

)

2

2 2b e ic

C a bi

a

−

= ⋅ + ⋅

_{, then, we}

can get

( )

1 2

2 2 2 2 2 2

2 1

2 2 2 1 2 1

2 2 2 2 2 2

2 1

2 2 2 1 2 1

2 2 2 2

2

sin 4 sinh 4 2

,

sinh 2 sin 2

sinh 2 cos 2 cosh 2 sin 2

2

, e

sinh 2 sin 2

sinh 2 cos 2 cosh 2 sin 2

2 , e sinh 2 ic h ic h b a ab x y t y

a

a b a b

a b

ab u y t

a b a b

a b

ab v y t

a b a b

ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ + − +  +  = − _ − _ + _ − _  +  − = ⋅ _ _ + _ − _ + − = ⋅

+ − 2 2

1 sin 2ϕ

(9)

2) Two pure image zeros

If choosing k ib1= 1, k1= −ib1, k2 =ib2, k2 = −ib2, where b1>0, b2 >0,

and denoting 3 1 1

4

t b y

b

ϕ = + _, ₃ ₂

2

4

t b y

b

ϕ = + , then letting the parameter be

2 1 2 e1 ic

C = ib , 2

1 2 e1 ic

C ₌ ib − _, 2

2 2 e2 ic

C = ib , 2

2 2 e2 ic

C ₌ ib − _{, we have}

( )

(

)

(

)

(

)

(

)

(

) (

)

(

)

(

)

(

)

( )

(

)

(

)

(

)

3 3 3 3 3 3

3 3 3 3

3 3 3

4 1 2

1 2

4

2 2 1 2 4 4 4 4 1 2

2 1

2 3

1 2 1 2

2 2

2 2 2 2 4 2 2

1 2 1 2 1 2

4 2 2

1 2 1 2 2

2 2 1 2 2 2

1 2 1 2

2 ,

4

e e e e

e e e

2

, e

e e e

ic h

i

x y t y b b

b b

b b _b _b

b b b b

b b b b b b

i

u y t b b

b b

b b _b

b b

ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

− − − − − −

− − − −

+

− − − −

= + +

 ₋ 

 + + + 

 + + 

⋅ 

 + − − + + + 

 

 

= ⋅ +

−

+ +

⋅

(

) (

)

(

) (

)

(

)

(

)

(

)

( )

₍

₎

(

)

(

)

(

) (

)

(

) (

)

(

)

(

)

3 3 3

3 3 3 3

3 3 3 3 3 3

3 3 3

2 2

2 2 1

2 2

2 2 2 2 4 2 2

1 2 1 2 1 2

4 2 2

1 2 1 2 2

2 2 1 2 2 2 2 2

1 2 2 1

2 1 2

2

2 2 2 2 4 2 2

1 2 1 2 1 2

e e

e e e

2

, e

e e e e e

e e e

ic h

b b b

b b b b b b

i

v y t b b

b b

b b _b _b _b _b

b b

b b b b b b

ϕ ϕ

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕ ϕ

− −

− − − −

− +

− − − − − −

− − − −

 

 + + 

 

 + − − + + + 

 

 

= ⋅ +

−

+ + +

+ ⋅

+ − − + + +

(

ϕ3

)

2

                       

 



 



 



 



 



 



 



(48)

Remark 9: To our knowledge, the result (47) and result (48) are new.

4. Conclusion

In this article, we consider IVP for the 2SPE with initial value in Schwartz space. We begin with the Lax pair of 2SPE and then we formulate a Riemann-Hilbert problem in new coordinate (y, t), which implies that we can get the parametric form of soliton solutions in terms of the solution of the associated Rie-mann-Hilbert problem. After that, we can get the soliton solution by analyzing the residue condition of the Riemann-Hilbert problem. Last we obtain the soli-ton solutions, in particular the new breather-solution (47) and two solisoli-ton solu-tion (48) were obtained.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this pa-per.

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(10)

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