About System of Dirac Differential Equation Which has Discontinıity Point in Semi Axis = Yarı Eksende Süreksizlik Noktalarına Sahip Dirac Diferansiyel Denklemler Hakkında

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C.Ü. Fen-Edebiyat Fakültesi

Fen Bilimleri Dergisi (2003)Cilt 24 Sayı 2

About System of Dirac Differential Equation Which has Discontinıity Point in Semi Axis

S.Gulyaz, Y.Cakmak and R.Kh. Amirov

Department of Mathematics, Cumhuriyet University Sivas Turkey

Received: 02.10.2004, Accepted: 17.11.2004

Abstract: In article, existence of transformation operators was proved for a class of Dirac operators

which have discontinuity conditions inside of some properties of kernel of this transformation operator was investigated.

AMS Subject Classification: 34A55, 34B24, 34L05

Key words: Dirac operators, transformation operator, discontinuity

Yarı Eksende Süreksizlik Noktalarına Sahip Dirac Diferansiyel Denklemler Hakkında

Özet: Bu çalışmada, aralıkta süreksizlik koşullarına sahip Dirac Operatörlerin bir sınıfı için çevirme

operatörünün varlığı ispatlandı ve bu operatörün çekirdeğinin bazı özellikleri incdelendi.

Anahtar Kelimeler: Dirac operatör, çevirme operatörü, süreksizlik

1.Introduction

For solving of the inverse problems for Dirac differential operators as regular as singular the transformation operators have a special place.

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direct and inverse problems of spectral theory can be investigated for Dirac operators. Analogous problem [14] have been investigated in finite interval.

Boundary value problems with discontinuity conditions arise in different branches of mathematics, mechanics, radio, electronics, geophysics and other fields of natural science and technology. For example, discontinuous conditions inside an interval are connected with discontinuous or nonsmooth properties of media ([1], [2]). Inverse problem of this type are connected with the investigation of discontinuous solutions of some nonlinear equations in mathematical physics.

For the classical Sturm-Liouville operators, Schrödinger equation and hyperbolic equations, direct and inverse problems are studied fairly completely (See [3], [5] and references there in).

For Dirac differential equation, direct and inverse problems have been investigated enough. (See [3], [4], [6-8], [14] and references there in). The presence of discontinuity conditions inside an interval introduces qualitative changes in the investigation of such problems. Some aspect of direct and inverse problems for differential operators with discontinuity conditions were studied in [9-14].

Let’s get system of Dirac differential equations with canonical form in semi axis, ∞ < < = Ω + (x)y y, 0 x dx dy B λ (1.1) where , , , and is a parameter.       − = 0 1 1 0 B _      − = Ω ) ( ) ( ) ( ) ( ) ( x p x q x q x p x _      = ) ( ) ( ) ( 2 1 x y x y x y p(x),q(x)∈L₂

( )

0,∞ λ

In this study, an expression of boundary value problem generated by boundary conditions 0 ) 0 ( = y (1.2) and ) 0 ( ) 0 (a− =Ay a+ y (1.3)

have been given, where 0< a<+∞, _      = α α 1 0 0 A , is real number. α ≠1

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2. Representation of Solution

Let Y ( λ be a solution of matrice equation (1.1) corresponding to the case of satisfying Y ( unite matrice) and discontinuity conditions (1.3).

) , 0 0 0 ) ( ≡ Ω x ₀(0,λ)=I I

In this case, for the functionY₀(x,λ), it is obvious that

     +∞ < <       − + < < = ₊ ₋ ₋ ₋ ₋ − x a e e a x e x Y _Bx _B _a _x Bx , 1 0 0 1 0 , ) , ( ₍₂ ₎ 0 λ λ λ α α λ (2.1) where       + = + _α α α 1 2 1 and       − = − _α α α 1 2 1 .

Let Y( λ be a solution of matrice equation (1.1) satisfying the initial condition Y and discontinuity conditions (1.3).

) , x = ) , 0 ( λ I

Let’s show that the function Y(x,λ) as

∫

∞ − + = x Bt_dt e t x K x Y x Y( ,λ) ₀( ,λ) ( , ) λ (2.2)

where is a seconder matrice function. It is obvious that the solution Y satisfies the integral equation

) , ( tx K (x,λ)

∫

∞ − _Ω + = x dt t Y t B t Y x Y x Y x Y( , ) ( , ) ( , ) 1( , ) ( ) ( , ) 0 0 0 λ λ λ λ λ (2.3)

to satisfy the of function Y given as the type of (2.2), it is necessary that the equality ) , (x λ

∫

∞ − ∞ − ∞ −       + Ω = x t Bt x Bt_dt _Y _x _Y _t _B _t _Y _t _K _x _t _e _ds _dt e t x K( , ) λ ( ,λ) 1( ,λ) ( ) ₀( ,λ) ( , ) λ 0 0 (2.4)

must be satisfied. On the constrary that if the matrice function satisfies the equality (2.4), the function Y satisfies the equation (2.2).

) , ( tx K ) , (x λ

We shall transform the right hand side of the equality (2.4) such that it will similar to left hand side of this equality. First, let’s assume the following expressions,

[

K x t BK x t B

]

t x K ( , ) ( , ) 2 1 ) , ( = ± ±

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) , ( ) , ( ) , (x t K x t K x t K = ₊ − ₋

[

BK x t K x t B

]

K x t B t x BK ( , ) ( , ) ( , ) 2 1 ) , ( ₊ + = − =−

[

BK x t K x t B

]

K x t B t x BK ( , ) ( , ) ( , ) 2 1 ) , ( ₋ − = + =

If we case the equality

       < < < <       − + < < < = − − − − − − − − + − − − x t a e x a t e e a x t e t Y x Y t x B t x a B t x B t x B , , 1 0 0 1 0 , ) , ( ) , ( ) ( ) 2 ( ) ( ) ( 1 0 0 λ λ λ λ α α λ λ

and the expression of the functions and if we transform right hand side of the equality (2.4) such that it will similar to expression in left hand side, we get the following system of integral equations for the matrice function and :

) , ( tx K_± ) , ( tx K₊ K₋( tx, ) I. for x<t<a, ξ ξ ξ ξ K t x d B t x B t x K t x x

∫

+ − + − Ω + −      + Ω − = 2 / ) ( ) , ( ) ( 2 2 1 ) , ( ξ ξ ξ ξ K t x d B t x K t x x

∫

+ + − =− Ω + − 2 / ) ( ) , ( ) ( ) , ( II. for ax> , 2a−x<t<x, ξ ξ ξ ξ ξ ξ ξ ξ α B K t x d B K t x d t x K t x a t x a

∫

+ − + + + + =− Ω + − − Ω + − 2 / ) ( 2 / ) ( ) , ( ) ( ) , ( ) ( ) , ( ξ ξ ξ ξ ξ ξ ξ ξ α B K t x d B K t x d t x K t x a t x a

∫

+ + + + + − =− Ω + − − Ω + − 2 / ) ( 2 / ) ( ) , ( ) ( ) , ( ) ( ) , ( III. for ax> , x<t <3x−a, ξ ξ ξ ξ ξ ξ ξ ξ α ξ ξ ξ ξ α α d x t K B d x a t K B d x t K B t x B t x K t x a a x t a t x a

∫

+ − + − − − + + + + + − + Ω − − − + Ω       − − − + Ω −       + Ω − = 2 / ) ( 2 / ) 2 ( 2 / ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( 2 2 ) , (

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ξ ξ ξ ξ α ξ ξ ξ ξ ξ ξ ξ ξ α d x a t K B d x t K B d x t K B t x K a x t a t x a t x a

∫

+ − − − + + + + + − − − + Ω       − − − + Ω − − + Ω = 2 / ) 2 ( 2 / ) ( 2 / ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( ) , ( ) ( ) , ( IV. for x> , a 3x−a<t<+∞, ξ ξ ξ ξ ξ ξ ξ ξ α ξ ξ ξ ξ α α d x t K B d x a t K B d x t K B t x B t x K t x a a x t a t x a

∫

+ − + − − − + + + + + − + Ω − − − + Ω       − − − + Ω −       + Ω − = 2 / ) ( 2 / ) 2 ( 2 / ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( 2 2 ) , ( ξ ξ ξ ξ α ξ ξ ξ ξ ξ ξ ξ ξ α α d x t K B d x t K B d x a t K B a x t B t x K t x a t x a a x t a

∫

+ + + + + + − − − − − − + Ω − − + Ω − − − + Ω       − −       − + Ω       − − = 2 / ) ( 2 / ) ( 2 / ) 2 ( ) , ( ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 2 2 1 0 0 1 2 ) , (

when we write successive approximations for each situations, I. for x<t<a,       + Ω − = + 2 2 1 ) , ( 0 _x _t _B x t K , 0( , )₌0 − x t K a t x I .for < <       + Ω − = + 2 2 1 ) , ( ) 0 ( _x _t _B x t K , (0)( , )₌0 − x t K , 2 , 1 , ) , ( ) ( ) , ( 2 ) ( ) 1 ( ) ( K = − + Ω − =

_∫

+ − − + x t B K t x d n K t x x n n _ξ _ξ _ξ _ξ , 2 , 1 , ) , ( ) ( ) , ( 2 ) ( ) 1 ( ) ( K = − + Ω − =

∫

+ − + − x t B K t x d n K t x x n n _ξ _ξ _ξ _ξ 2 , for . x a a x t x II > − < < K , 2 , 1 , 0 ) , ( , 0 ) , ( 0 ) , ( , 0 ) , ( ) ( ) ( ) 0 ( ) 0 ( = = = = = − + − + n t x K t x K t x K t x K n n a x t x a x III .for > , < <3 −2   + Ω − = + ) 0 ( α x t (0) ₌

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∫

+ − − + − + + + − + Ω − − − + Ω − = 2 ) ( ) 1 ( 2 ) ( ) 1 ( ) ( ) , ( ) ( ) , ( ) ( ) , ( t x a n t x a n n d x t K B d x t K B t x K ξ ξ ξ ξ ξ ξ ξ ξ α

∫

+ − + + − − − − + − + + − − + Ω − − − − + Ω       − − − − + Ω − = 2 ) ( ) 1 ( 2 ) 2 ( ) 1 ( 2 ) ( ) 1 ( ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( ) , ( t x a n a x t a n t x a n n d x t K B d x a t K B d x t K B t x K ξ ξ ξ ξ ξ ξ ξ ξ α ξ ξ ξ ξ α +∞ < < − >a x a t x IV. for , 3 2       + Ω − = + + 2 2 ) , ( ) 0 ( _x _t _B x t K α       − + Ω − = − − 2 2 2 ) , ( ) 0 ( _x _t _B t x a K α

∫

+ − − + − − − − + − + + + − + Ω − − − − + Ω       − − − − + Ω − = 2 ) ( ) 1 ( 2 ) 2 ( ) 1 ( 2 ) ( ) 1 ( ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( ) , ( t x a n a x t a n t x a n n d x t K B d x a t K B d x t K B t x K ξ ξ ξ ξ ξ ξ ξ ξ α ξ ξ ξ ξ α

∫

+ − + + − − − − + − + + − − + Ω − − − − + Ω       − − − − + Ω − = 2 ) ( ) 1 ( 2 ) 2 ( ) 1 ( 2 ) ( ) 1 ( ) ( ) , ( ) ( ) 2 , ( ) ( 1 0 0 1 ) , ( ) ( ) , ( t x a n a x t a n t x a n n d x t K B d x a t K B d x t K B t x K ξ ξ ξ ξ ξ ξ ξ ξ α ξ ξ ξ ξ α

Now, let’s obtain that written successive approximations are convergent for the situation I. and also get that the evaluation for the vector function K( tx, ):

a t x< < 0 ) , ( , 2 2 1 ) , ( (0) ) 0 ( _ ₌      + Ω − = ₋ + K x t t x B t x K

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( )

( ) 2 2 1 ) , ( ) 0 ( _x _t _dt _B x t _dt _t _dt _x K x x x σ ≤ Ω =       + Ω =

∫

∞ ∞ ∞ + ξ ξ ξ ξ K x t d B B x t d B t x K t x x t x x       + Ω Ω = Ω =

∫

+ + + − 2 ) ( ) , ( ) ( ) , ( 2 ) ( ) 0 ( 2 ) ( ) 1 ( ! 2 ) ( ) , ( 2 ) 1 ( _x _t _dt x K x σ ≤

∫

∞ −

Analogously, for n=k, the truth of the following inequalities; )! 1 2 ( ) ( ) , ( ) 1 2 ( ) 2 ( + ≤ + ∞ +

∫

K x t dt _k x k x k σ )! 2 2 ( ) ( ) , ( ) 2 2 ( ) 1 2 ( + ≤ + ∞ + −

∫

K x t dt _k x k x k σ

is obtained. Let’s show that similar inequalities are satisfied for also n= k+1: Since

(

,

)

, ) ( ) , ( (2 1) 2 ) ( ) 2 2 ( _x _t _B _ξ _K _ξ _t _x _ξ _d_ξ K k t x x k ₌₋ _Ω + ₊ ₋ − + + +

∫

(

)

(

)

( )

∫

∫ ∫

∫

∞ + + ∞ ∞ + − ∞ + − ∞ − ∞ + − + ∞ + + + = Ω + = Ω = − + Ω = − + Ω ≤ x k k x k x k x x k t x x x k k x d d d K dtd x t K dt d x t K dt t x K )! 3 2 ( ) ( ) ( ) ( 2)! (2k 1 - , ) ( , ) ( , ) ( ) , ( ) 3 2 ( ) 2 2 ( ) 1 2 ( ) 1 2 ( 2 ) 1 2 ( 2 ) ( ) 2 2 ( σ ξ ξ σ ξ ξ τ τ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ ξ Similarly, )! 4 2 ( ) ( ) , ( (2 4) ) 3 2 ( + ≤ + ∞ + −

∫

K x t dt _kk x x k σ

We get the following evaluations that series K x t dt n x n

∫

∞ = ∞ ± 0 ) ( ₍ _, ₎

[ ]

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∫

∞ − ≤ x x e dt t x K( , ) σ( ) 1.

Let’s assume that Ω function is differentiable. In this case when the expression of the function Y substitute in the equation (1.1):

) (x ) , (x λ y y dx dy B + =λ       + =       + Ω +       +

∫

∞ − ∞ − ∞ − x Bt x Bt x Bt dt e t x K x Y dt e t x K x Y x dt e t x K x Y dx d B λ λ λ λ λ λ λ ) , ( ) , ( ) , ( ) , ( ) ( ) , ( ) , ( 0 0 0

∫

∞ − ∞ − ∞ − + = Ω + Ω +       + ′ x Bt x Bt x Bt dt e t x K x Y dt e t x K x x Y x dt e t x K dx d B x Y B λ λ λ λ λ λ λ λ ) , ( ) , ( ) , ( ) ( ) , ( ) ( ) , ( ) , ( 0 0 0 here since BY₀′(x,λ)=λY₀(x,λ), +       + +

_∫

∫

∞ − − − − − − − a x Bt a x x a Bt x a x Bt_dt _K _x _t _e _dt _K _x _t _e _dt e t x K dx d B 2 3 2 3 2 2 ) , ( ) , ( ) , ( λ λ λ

∫

∞ − ∞ − ₌ Ω + Ω + x Bt x Bt_dt _K _x _t _e _dt e t x K x x Y x) ( ,λ) ( ) ( , ) λ λ ( , ) λ ( ₀

[

]

dt e t x K dt e t x K x x Y x dt e x t x K B e a x x K e x a x K e a x x K e x x K e x a x K B Bt x Bt x Bt x a x B x a B a x B Bx x a B λ λ λ λ λ λ λ λ λ λ ∞ − ∞ − − ∞ − − − − − − − − −

∫

= Ω + Ω + + ∂ ∂ + + − − + − + + − − + − − − − ) , ( ) , ( ) ( ) , ( ) ( ) , ( ) 0 2 3 , ( 3 ) 0 2 , ( ) 0 2 3 , ( 3 ) , ( ) 0 2 , ( 0 ) 2 3 ( ) 2 ( ) 2 3 ( ) 2 ( is obtained.

When partial integral is applied one time in the integral which is right hand side of last equality

[

]

[

]

=     ₊_Ω ∂ ∂ + Ω + − − − + − − − − − − − + − − ∞ − − − − −

∫

x K x t e dt x t x K B x Y x e a x x K a x x K B e x x BK e x a x K x a x K B Bt x a x B Bx x a B λ λ λ λ λ ) , ( ) ( ) , ( ) , ( ) ( ) 0 2 3 , ( ) 0 2 3 , ( 3 ) , ( ) 0 2 , ( ) 0 2 , ( 0 ) 2 3 ( ) 2 (

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[

]

[

]

Be dt x t x K Be a x x K a x x K Be x x K Be x a x K x a x K Bt x a x B Bx x a B λ λ λ λ − ∞ − − − − −

∫

∂ _∂ − − − − + − − − − − − − + − = ) , ( ) 0 2 3 , ( ) 0 2 3 , ( 3 ) , ( ) 0 2 , ( ) 0 2 , ( ) 2 3 ( ) 2 (

from this last equality,

B t t x K t x K x x t x K B ∂ ∂ − = Ω + ∂ ∂ ( , ) ) , ( ) ( ) , ( ) 1 B x x K x x K x x x BK a x ) , ( ) , ( ) ( ) , ( 0 for ) 2 − = Ω + − < <

[

]

[

K x a x K x a x

]

B x x a x K x a x K B a x ) 0 2 , ( ) 0 2 , ( ) ( ) 0 2 , ( ) 0 2 , ( for ) 3 − − − + − − = = Ω + − − − + − <

[

]

[

]

[

K x x a K x x a

] [

K x x a K x x a

]

B B B x a x K x a x K x x a x K x a x K B B x x K x x x BK a x ) 0 2 3 , ( ) 0 2 3 , ( ) 0 2 3 , ( ) 0 2 3 , ( 3 , ) 0 2 , ( ) 0 2 , ( ) 5 . 2 ( ) ( 1 0 0 1 ) 0 2 , ( ) 0 2 , ( , ) , ( ) ( ) , ( for ) 4 − − − + − = − − − + − − − − + − − = = Ω       − + − − − + − − = Ω + − > − + α α

Thus we have been proved following theorem: Theorem: Let’s say that

∫

∞ +∞ < Ω x dx x)

( then each solution Y satisfying

conditions (1.2)-(1.3) of the equation (1.1) has the expression (2.2) and moreover ) , (x λ

∫

∞ − ≤ x x e dt t x K( , ) σ( ) 1_where

∫

∞ Ω = x dt t x) ( ) ( σ .

If the function is differentiable then the function satisfies the conditions of (2.5).

) (x

σ K( tx, )

Reference

[1] O.N.Litvinenko and V.I. Soshnikov, The Theory of Heterogeneous Lines and their Applications in Radio Engineering, Radio, Moscow, 1964(in Russian).

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[4] V.A. Marchenko, Strum-Liouville Operatorsand their Applications, Naukova Dumka, Kiev, 1977; English transl. Birkhauser, 1986

[5] L.P. Nizhnik, Inverse Scattering Problems for Hyperbolic equations, Naukova Dumka, Kiev, 1977(in Russian).

[6] M.G. Gasimov, The Inverse problem of of scattering theory for Dirac equations system of the order 2n. Tr. MMO, v.19, 1968, p.41-112 (in Russian).

[7] M.G. Gasimov, T.T.Jabiyev, The detemination of Dirac Differential equation system by the spectrum. The difficulties of summer school on spectrum theory of operators and the theory of group Representation. Baku, ‘Elm’, 1975, p.46-71 (in Russian).

[8] I.M. Guseinov, On the Representation of Lost solutions for Dirac’s equation system with discontinuous coefficients. Transactions of AS Azerbaijan, No.5, 1999, p.41-45. [9] O.H.Hald, Discontinuous Inverse eigenvalue problems, Commun. Pure Appl. Math. 37, 539-577, 1984.

[10] D.Shepelsky, The inverse problem of reconstruction of the medium’s cumdıctivity in a class of discontinuous and increasing functions, in: Spectral Operator Theory and Related Topics: Adv. İn Sov. Math., v.19, AMS, Prov. p.209-232, 1994.

[11] M.Kobayashi, A uniqueness prof for discontinuous inverse Strum-Liouville problems with symmetric potentials, Inverse Problems, v.5, No.5, p.767-781, 1989. [12] R.Kh.Amirov and V.A.Yurko, On differential operators with singularity and discontinuity conditions inside an interval, Ukrainian Math. Jour. v.53, N0.11, p.1751-1769, 2001.

[13] V.A.Yurko, Integral transforms connected with differential operators having singularities inside the interval, Integral Transform Special Func., 5, No.3, 4, p. 309-322, 1997.

[14] R.Kh.Amirov, On Dirac Differential equation system which have discontinuity in interval, Ukrainian Math. Jour.(in appear)