Actual problems of modern science, education and training


Download 5.08 Mb.
bet86/87
Sana30.10.2023
Hajmi5.08 Mb.
#1734675
1   ...   79   80   81   82   83   84   85   86   87
Bog'liq
October 2022

Introduction.


Nonlinear evolution equations are widely used as models to describe complex physical phenomena in various fields of sciences, especially in fluid mechanics, solid- state physics, plasma physics and biology. The Kaup-Boussinesq system was first studied in the paper [1, 2]. In [3], multisoliton solutions were found, and the asymptotic behavior of these solutions was investigated. In papers [4,5], real finite-zone regular solutions of the Kaup-Boussinesq system are studied.
In this work, we consider the following Kaup-Boussinesq type system
v v  6uu 18u u  6vv  24vuu  6v u2  0,

t xxx xxx x xx x x x
(1)

u u  6vu  6v u  30u u2  0, x
t xxx x x x

under initial condition
v(x,t)
v0 (x), u(x,t) t 0
u0 (x),
x . (2)


t 0
The functions v0(x), u0(x) satisfy the following conditions:

(i)
u0(x) is absolutely continuous on each finite segment , ,
and the

inequalities hold
 
|u0 (x) | dx , (1 | x |)[| v0 (x) | | u0 (x) |]dx , (3)
 
(ii) the operator generated by the differential expression

d 2
2
T (0, k) :  dx2 v0 (x)  2ku0 (x)  k

has exactly 2N simple eigenvalues
k1(0),k2 (0),...,k2 N (0) .

The main aim of this work is to derive representations for the solutions v(x,t)
and u(x,t) of the Cauchy problem (1)–(3) within the inverse scattering method for the

quadratic pencil of Sturm-Liouville operators:
T (t, k) y :  y''v(x,t) y  2ku(x,t) y k 2 y  0,


х . (4)

The inverse scattering problem for the quadratic pencil of Sturm-Liouville operators in the class of “rapidly decreasing” functions was solved in the works [6,7], with periodical potential in [8,9].

The basic facts from scattering problem


In this section we give basic information about the scattering theory for the Sturm–Liouville equation with an energy-dependent potential.
Consider the following quadratic pencil of Sturm-Liouville equatins
T (0, k) y : y'' v(x) y 2ku(x) y k 2 y 0, х (5)

where
u(x)
and
v(x)
are real functions, moreover,
u(x)
is absolutely continuous on

each finite segment , ,
and the inequalities hold

 
|u x | dx  , (1 | x |)[v(x) | u(x) |]dx  . (6)

 
Under condition (6), Eq. (5) has solutions
Imk  0 and the asymptotic formulas hold


f (x, k ),


f (x, k )
regular in the half-plane

f (x, k)  eikx[1 o(1)],
f (x, k)  eikx[1 o(1)],
x   , (7)
x  . (8)

For real
k  0 , the pairs
f (x, k ),
f (x, k )
and
f (x, k),
f (x, k) (the bar over the function

here and below denotes complex conjugation) form two fundamental systems of solutions to equation (5). The following relations hold
f (x, k)  b(k) f (x, k)  a(k) f (x, k) , (9)
f (x, k)  b (k) f (x, k)  a(k) f (x, k). (10)

The functions
a(k )
and
b(k )
are defined for all
k R*  (, ) \{0}
and the following

equality is fulfilled

a(k )  
1


2ik
W f
, f,
b(k ) 
1


2ik
W f
, f,

Moreover, the function




a(k )
a(k) 2  b(k) 2  1, k R* .
admits an analytic continuation to the half-plane Imk  0

and can have at most a finite number of zeros
k1,k2 ,...,kN , besides, at

k kn , n 1,2,..., N
the following equality holds:

n n
f (x, k )  B f

(x, kn ) , (11)



where the quantities
B are independent of x . The corresponding functions
f (x, kn )


n
are the only
L2 (R)
solutions of (5) for Imk  0 and are the 'bound states'.

The set of the quantities
r (k)  b(k) , k R, k , k ,..., k

, , ,...,


(12)


a(k)
1 2 N 1 2 N

 
and



r (k)   b (k) , k R, k , k ,..., k

,  , ,...,


(13)

a(k)
1 2 N 1 2 N

 
are called the left and right scattering data of Eq. (5), respectively, here


n

B
n , n


k kn
 1, 2,..., N . (14)

Functions
r (k )
and
r (k )
are called the left and right reflection coefficients,

respectively.
We now turn to the question of constructing u(x)

and v(x)


from scattering data.



Note that scattering data (12), (13) and
F (x)
are bijectively related via transforms

N 1
F (x)  ie iknx r (k)eikxdk . (15)

n



2
n1 

To restore the coefficient functions
u(x)
and
v(x)
in equation (5) from the right

reflection coefficient
r (k )
we proceed as follows:

It is necessary to find the function
F (x)
by formula (15) and solve with respect

to K (0) (x, y)  L (x,), K(1) (x, y) L (x, ) integral equations
 1  1

 




F (x y)  K (0) (x, y) 
K (0) (x,t)F (t y)dt  0,
x
x y   ,


 




iF (x y)  K (1) (x, y) 
K (1) (x, t)F (t y)dt  0,
x
x y   .

Next, we have to define the function  (x)
equation of the Volterra type
as a solution to a nonlinear integral

in which

(x)  (t, (t))dt,
x
x ,

Φt, z Re K0 t,t m K 1 t,t sin2z 2 Re K 1 t,t sin2z 2 Im K 0 t,t cos2z
   

and we put
K x, y K 0 x, y cos x K 1 x, y sin
x.

    

After that, the coefficients equalities
u(x)
and
v(x)
of the equation (5) are determined by the

u(x) (x) ,

v(x)  u2 (x)  2 d Re K (x, x)cos (x)  Im K (x, x)sin
(x) .

dx    

It is worthy to remark that the functions
1
hn (x) 


d [ f
dk

(x, k)  B f



n
(x, k)]
(16)

k kn
k kn

are solutions of the equations
T (k
) y k 2 y, n 1, 2,..., N . For Imk  0 , using (7) and


n n
(8), we obtain the following asymptotics

n
h (x)  eiknx ,


x   , (17)


n n
h (x)  Beiknx ,
From asymptotics (7), (8), (17) and (18) we get
x   . (18)

W{h (x), f (x,k )}  2ik , W{h (x), f (x, k )} 2ik B
n n n n n n n

Download 5.08 Mb.

Do'stlaringiz bilan baham:
1   ...   79   80   81   82   83   84   85   86   87




Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling