INVERSE SPECTRAL PROBLEMS FOR DISCONTINUOUS STURM-LIOUVILLE OPERATOR WITH EIGENPARAMETER DEPENDENT BOUNDARY CONDITIONS

In this study, Sturm—Liouville problem with discontinuities in the case when an eigenparameter linearly appears not only in the differential equation but it also appears in both of the boundary conditions is investigated.

Spectral problems for Sturm-Liouville operator with eigenvalue dependent boundary conditions were studied extensively.[2] and [3] are well known example for works with boundary conditions depend on eigenvalue parameter linearly.Moreover, [5], [16] and [18] are also interested in linearly conditions.Nonlinearly conditions were considered in [4], [7], [9]- [12] and [17].These works have been on Hilbert and Pontryagin space formulations, expansion theory, direct and inverse spectral theory.In [2] and [20] an operator-theoretic formulation of the problems without discontinuity conditions (4) and with spectral parameter contained in only one of the boundary conditions has been given.Boundary value problems with discontinuities inside the interval are extensively studied ( [6], [13]) These kinds of problems are often appear in mathematics, mechanics, physics, geophysics and other branches of natural properties.Discontinuous inverse problems also appear in electronics for constructing parameters of heterogeneous electronic lines with desirable technical characteristics [14], [24] and [19].It must be noted that some special cases of the considered problem (1)-( 4) arise after an application of the method of separation of variables to the varied assortment of physical problems.For example, some boundary-value problems with transmission conditions arise in heat and mass transfer problems (see, for example, [21]), in vibrating string problems when the string loaded additionally with point masses (see, for example, [22]) and in di¤raction problems (see, for example, [23]).
Inverse problem for discontinious Sturm-Liouville operator with Dirichlet conditions were given in [15].In the present paper, not only di¤erential equation, but also both of the boundary conditions of the problem L contain spectral parameter.In this case, inverse problem according to the Weyl functions [8] and the spectral data, i.e. (i): the sets of eigenvalues and norming constants; (ii): two di¤erent eigenvalues sets, is studied.

Operator Treatment.
Let the inner product in the Hilbert Space H = L 2 (0; ) C 2 be de…ned by De…ne an operator T acting in H with the domain D(T ) = fF 2 H : F 1 (x) and F 0 1 (x) are absolutely continuous in [0; d) [ (d; ]; `F1 2 L 2 (0; ); by two partial integration, we get hT F; Gi hF; Use the domain of the operator T and 1 > 0; 2 > 0 to see that, hT F; Gi = hF; T Gi : So T is symmetric.
Corollary 1.All eigenvalues of the operator T (or the problem L) are real and two eigenfunctions '(x; 1 ); '(x; 2 ) corresponding to di¤ erent eigenvalues 1 and Let us denote the solutions of (1) by '(x; ) and (x; ) satisfying the initial conditions respectively and the jump conditions (4): These solutions satisfy the relation Theorem 2.2.The following asymptotics hold for su¢ cienly large jkj where, Proof.It is clear that the functions '(x; k) and (x; k) satisfy the following integral equations; Since the proof of the equalities ( 6) and ( 7) are similar, let us prove only (7).Divide both sides of last integral equality by k 2 and put If we use the Picard's iteration method to above integral equations, we get where C = ( + + j j)(jH 0 j + jH 1 j + jH 2 j + 1) Last inequality gives (x; k) = O k 2 exp j j ( x) : From this equality and integral equations of (x; k); we get equality (7) 3. Properties of Spectrum.
In the present section, properties of eigenvalues, eigenfunctions, and the resolvent operator of the problem L are investigated.
It is obvious that the characteristic function ( ) of the problem L is as follows The roots of ( ) = 0 coincide with the eigenvalues of problem L: We de…ne norming constants by Lemma 3.1.The eigenvalues of the problem L are simple and seperated.
Proof.Let us write the following equations, 00 (x; )+q(x) (x; ) = (x; ) ; ' 00 (x; n )+q(x)' (x; n ) = n ' (x; n ) : If we multiply the …rst equation by ' (x; n ) ; second by (x; ) and subtract, after integrating from 0 to we obtain Let ' (x; n ) be an eigenfunction and use initial conditions (5); to get Pass through the limit as !n and using the equality (x; n ) = n ' (x; n ) ; to get 0 ( n ) = n n .It is obvious that 0 ( n ) 6 = 0.So eigenvalues of the problem L are simple.Since, ( ) is an entire function of ; the zeros of ( ) are seperated.So lemma is proven.
One can easily prove the following theorem using same methods in [1].
Theorem 3.2.The operator T (or problem L ) has a discrete spectrum.Moreover, the resolvent operator of T is de…ned as follows ; R (T ) := (T I) x t '(t; ) (x; ); t x The following theorem gives knowledge about the behaviour at in…nity of the eigenvalues, eigenfunctions and normalizing numbers of L: Theorem 3.3.The eigenvalues n ; eigenfunctions '(x; n ) and normalizing numbers n of the problem L have the following asymptotic estimates for large n; where Proof.Using ( 6) in ( 8), we get, for su¢ ciently large value of j j : Denote where is su¢ ciently small and k 0 n are the zeros of 0 ( ) except 0: Since, j 0 ( )j C exp( 52 exp j j ) and j ( ) 0 ( )j = O( 2 exp j j ) for 2 G n and large values n; using the Rouche's theorem, we establish that, the functions 0 ( ) and ( ) have the same number of zeros inside the contour G n : Consequently, in the annulus between G n and G n+1 , has precisely one zero, namely k 2 n .Therefore, for the eigenvalue n ; the equality n+3 = k 2 n is true.On the other hand, by using again the Rouche's theorem in " := : k 0 n < " ; for su¢ ciently small "; we get the asymptotic formulae ) is valid for large n.Finally, the equality This fact proves the equality (10).By using ( 10) in ( 6) and ( 11) in (9) we get ( 11) and ( 12) , respectively.

Inverse Problems.
In the present section, we study the inverse problem recovering the boundary value problem L from its spectral data.We consider three statements of the inverse problem of the reconstruction of the boundary-value problem L fom the Weyl function, from the spectral data f n ; n g n 0 and from two spectra f n ; n g n 0 .
The function (x; k) and M (k) are called the Weyl solution and theWeyl function, respectively for the boundary value problem L: The Weyl function is meromorphic in with simple poles in the points n : Relations ( 13) and ( 15) yield To and inversion formula It is easy to see that the where is su¢ ciently small number and k n and e k n are square roots of eigenvalues of problem L and e L; respectively. ( )(x; k) C jkj 3 exp( j j x) is valid for k 2 G ; = 0; 1; Thus we get from (17).According to the last inequalities, if M (k) = M (k) then from well known Liouville's theorem P 11 (x; k) = A(x); P 12 (x; k) 0: Using (18), we take From Proof.It follows from ( 14) the residues of M ( ) at n are  Let us consider the boundary value problem L 1 which is the problem that we take the condition y 0 (0; k) h 0 y(0; k) = 0 instead of the condition (2) in L and f 2 n g n 0 be the eigenvalues of the problem L 1 .

Theorem 4 . 3 .
If k n = e k n and n = ~ n for n = 0; 1; ::: , then L = e L; i. e., q(x) = e q(x); a.e. and d = e d; = e ; h i = e h i ; H i = e H i ; i = 0; 1; 2: Thus, the problem L is uniquely de…ned by spectral data.Proof.If k n = e k n and n = ~ n for n = 0; 1; :::, then from Lemma2, we get M ( ) = f M ( ).Hence, Theorem6 gives L = e L: study the inverse problem, we agree that together with L we consider a boundary value problem e L of the same form but with di¤erent coe¢ cients e