Research

# Inverse eigenvalue problems for a discontinuous Sturm-Liouville operator with two discontinuities

Yalçın Güldü

Author Affiliations

Department of Mathematics, Faculty of Sciences, Cumhuriyet University, Sivas, 58140, Turkey

Boundary Value Problems 2013, 2013:209  doi:10.1186/1687-2770-2013-209

 Received: 18 June 2013 Accepted: 26 August 2013 Published: 11 September 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we consider a discontinuous Sturm-Liouville operator with parameter-dependent boundary conditions and two interior discontinuities. We obtain eigenvalues and eigenfunctions together with their asymptotic approximate formulas. Then, we give some uniqueness theorems by using Weyl function and spectral data, which are called eigenvalues and normalizing constants for solution of inverse problem.

MSC: 34A55, 34B24, 34L05.

##### Keywords:
Sturm-Liouville problem; eigenvalues; eigenfunctions; transmission conditions; Weyl function

### 1 Introduction

It is well known that the theory of Sturm-Liouville problems is one of the most actual and extensively developing fields of theoretical and applied mathematics, since it is an important tool in solving many problems in mathematical physics (see [1-4]). In recent years, there has been increasing interest in spectral analysis of discontinuous Sturm-Liouville problems with eigenvalue-linearly and nonlinearly dependent boundary conditions [1,5-12]. Various physics applications of such problems can be found in [1,3,4,13-19] and corresponding bibliography cited therein.

Some boundary value problems with discontinuity conditions arise in heat and mass transfer problems, mechanics, electronics, geophysics and other natural sciences (see [3] also [20-29]). For instance, discontinuous inverse problems appear in electronics for building parameters of heterogeneous electronic lines with attractive technical characteristics [20,30,31]. Such discontinuity problems also appear in geophysical forms for oscillations of the earth [32,33]. Furthermore, discontinuous inverse problems appear in mathematics for exploring spectral properties of some classes of differential and integral operators.

Inverse problems of spectral analysis form recovering operators by their spectral data. The inverse problem for the classical Sturm-Liouville operator was studied first by Ambarsumian in 1929 [34] and then by Borg in 1945 [35]. After that, direct and inverse problems for Sturm-Liouville operator have been extended to so many different areas.

We consider a discontinuous Sturm-Liouville problem L with function ρ ( x )

l y : = ρ ( x ) y ( x ) + q ( x ) y ( x ) = λ y ( x ) , x [ a , δ 1 ) ( δ 1 , δ 2 ) ( δ 2 , b ] = Ω , (1)

where

ρ ( x ) = { ρ 1 2 , a x < δ 1 , ρ 2 2 , δ 1 < x < δ 2 , ρ 3 2 , δ 2 < x b ,

and ρ 1 , ρ 2 , and ρ 3 are given positive real numbers; q ( x ) L 2 [ Ω , R ] ; λ C is a complex spectral parameter; boundary conditions at the endpoints

l 1 y : = λ ( θ 1 y ( a ) θ 2 y ( a ) ) ( θ 1 y ( a ) θ 2 y ( a ) ) = 0 , (2)

l 2 y : = λ ( γ 1 y ( b ) γ 2 y ( b ) ) + ( γ 1 y ( b ) γ 2 y ( b ) ) = 0 (3)

with discontinuity conditions at two points x = δ 1 , x = δ 2

l 3 y : = y ( δ 1 + 0 ) θ 3 y ( δ 1 0 ) γ 3 y ( δ 1 0 ) = 0 , (4)

l 4 y : = y ( δ 1 + 0 ) θ 4 y ( δ 1 0 ) γ 4 y ( δ 1 0 ) = 0 , (5)

l 5 y : = y ( δ 2 + 0 ) θ 5 y ( δ 2 0 ) γ 5 y ( δ 2 0 ) = 0 , (6)

l 6 y : = y ( δ 2 + 0 ) θ 6 y ( δ 2 0 ) γ 6 y ( δ 2 0 ) = 0 , (7)

where θ i , γ i and θ j , γ j ( i = 1 , 6 , j = 1 , 2 ) are real numbers and

α 1 = | θ 3 γ 3 θ 4 γ 4 | > 0 , α 2 = | θ 5 γ 5 θ 6 γ 6 | > 0 , β 1 = | θ 1 θ 1 θ 2 θ 2 | > 0 and β 2 = | γ 1 γ 1 γ 2 γ 2 | > 0 .

In the present paper, we construct a linear operator T in a suitable Hilbert space such that problem (1)-(7) and the eigenvalue problem for operator T coincide. We investigate eigenvalues and eigenfunctions together with their asymptotic behaviors of operator T. Besides, we study some uniqueness theorems according to Weyl function and spectral data, which are called eigenvalues and normalizing constants.

### 2 Operator formulation and spectral properties

We make known the inner product in the Hilbert space H : = H 1 C 2 , where H 1 = ( L 2 ( Ω ) , , 1 ) , ℂ denotes the Hilbert space of complex numbers and a self-adjoint operator T defined on H such that (1)-(7) can be dealt with as the eigenvalue problem of operator T. We define an inner product in H by

F , G : = α 1 α 2 ρ 1 2 a δ 1 f ( x ) g ¯ ( x ) d x + α 2 ρ 1 2 δ 1 δ 2 f ( x ) g ¯ ( x ) d x + ρ 3 2 δ 2 b f ( x ) g ¯ ( x ) d x + α 1 α 2 β 1 f 1 g ¯ 1 + 1 β 2 f 2 g ¯ 2 (8)

for

F = ( f ( x ) f 1 f 2 ) H , G = ( g ( x ) g 1 g 2 ) H .

Consider the operator T defined by the domain

D ( T ) = { F H : f ( x ) , f ( x ) A C loc ( Ω ) , l f H 1 , l 3 = l 4 = l 5 = l 6 = 0 , f 1 = θ 1 f ( a ) θ 2 f ( a ) f 2 = γ 1 f ( b ) γ 2 f ( b ) }

such that T F : = ( l f , θ 1 f ( a ) θ 2 f ( a ) , ( γ 1 f ( b ) γ 2 f ( b ) ) ) for F = ( f , θ 1 f ( a ) θ 2 f ( a ) , γ 1 f ( b ) γ 2 f ( b ) ) D ( T ) and also l 3 - l 6 are satisfied for f.

Thus, we can rewrite the considered problem (1)-(7) in the operator form as T F = λ F .

Theorem 1The operatorTis symmetric inH.

Proof Let F , G D ( T ) . By two partial integrations, we get

T F , G = F , T G + α 1 α 2 ( W ( f , g ¯ , δ 1 0 ) W ( f , g ¯ , a ) ) + α 2 ( W ( f , g ¯ , δ 2 0 ) W ( f , g ¯ , δ 1 + 0 ) ) + W ( f , g ¯ , b ) W ( f , g ¯ , δ 2 + 0 ) + α 1 α 2 β 1 ( θ 1 f ( a ) θ 2 f ( a ) ) ( θ 1 g ¯ ( a ) θ 2 g ¯ ( a ) ) 1 β 2 ( γ 1 f ( b ) γ 2 f ( b ) ) ( γ 1 g ¯ ( b ) γ 2 g ¯ ( b ) ) α 1 α 2 β 1 ( θ 1 g ¯ ( a ) θ 2 g ¯ ( a ) ) ( θ 1 f ( a ) θ 2 f ( a ) ) + 1 β 2 ( γ 1 g ¯ ( b ) γ 2 g ¯ ( b ) ) ( γ 1 f ( b ) γ 2 f ( b ) ) ,

where by W ( f , g ; x ) , we denote the Wronskian of the functions f and g as

f ( x ) g ( x ) f ( x ) g ( x ) .

Since f and g satisfy the boundary conditions (2)-(3) and transmission conditions (4)-(7), we obtain

α 1 α 2 β 1 [ ( θ 1 f ( a ) θ 2 f ( a ) ) ( θ 1 g ¯ ( a ) θ 2 g ¯ ( a ) ) ( θ 1 g ¯ ( a ) θ 2 g ¯ ( a ) ) ( θ 1 f ( a ) θ 2 f ( a ) ) ] = α 1 α 2 W ( f , g ¯ , a ) × 1 β 2 [ ( γ 1 g ¯ ( b ) γ 2 g ¯ ( b ) ) ( γ 1 f ( b ) γ 2 f ( b ) ) ( γ 1 f ( b ) γ 2 f ( b ) ) ( γ 1 g ¯ ( b ) γ 2 g ¯ ( b ) ) ] = W ( f , g ¯ , b ) W ( f , g ¯ , δ i + 0 ) = α i W ( f , g ¯ , δ i 0 ) .

Thus, we have T F , G = F , T G , i.e., T is symmetric. □

Lemma 1Problem (1)-(7) can be considered as the eigenvalue problem of the symmetric operatorT.

Corollary 1All eigenvalues and eigenfunctions of problem (1)-(7) are real, and two eigenfunctions φ ( x , λ 1 ) and φ ( x , λ 2 ) , corresponding to different eigenvalues λ 1 and λ 2 , are orthogonal in the sense of

α 1 α 2 ρ 1 2 a δ 1 φ ( x , λ 1 ) φ ( x , λ 2 ) d x + α 2 ρ 2 2 δ 1 δ 2 φ ( x , λ 1 ) φ ( x , λ 2 ) d x + ρ 3 2 δ 2 b φ ( x , λ 1 ) φ ( x , λ 2 ) d x + α 1 α 2 β 1 ( θ 1 φ ( a , λ 1 ) θ 2 φ ( a , λ 1 ) ) ( θ 1 φ ( a , λ 2 ) θ 2 φ ( a , λ 2 ) ) + 1 β 2 ( γ 1 φ ( b , λ 1 ) γ 2 φ ( b , λ 1 ) ) ( γ 1 φ ( b , λ 2 ) γ 2 φ ( b , λ 2 ) ) = 0 .

We define the solutions

φ ( x , λ ) = { φ 1 ( x , λ ) , x [ a , δ 1 ) , φ 2 ( x , λ ) , x ( δ 1 , δ 2 ) , φ 3 ( x , λ ) , x ( δ 2 , b ] , ψ ( x , λ ) = { ψ 1 ( x , λ ) , x [ a , δ 1 ) , ψ 2 ( x , λ ) , x ( δ 1 , δ 2 ) , ψ 3 ( x , λ ) , x ( δ 2 , b ]

of equation (1) by the initial conditions

φ 1 ( a , λ ) = λ θ 2 θ 2 , φ 1 ( a , λ ) = λ θ 1 θ 1 , φ 2 ( δ 1 , λ ) = θ 3 φ 1 ( δ 1 , λ ) + γ 3 φ 1 ( δ 1 , λ ) , φ 2 ( δ 1 , λ ) = θ 4 φ 1 ( δ 1 , λ ) + γ 4 φ 1 ( δ 1 , λ ) , φ 3 ( δ 2 , λ ) = θ 5 φ 2 ( δ 2 , λ ) + γ 5 φ 2 ( δ 2 , λ ) , φ 3 ( δ 2 , λ ) = θ 6 φ 2 ( δ 2 , λ ) + γ 6 φ 2 ( δ 2 , λ ) , (9)

and similarly,

ψ 3 ( b , λ ) = λ γ 2 + γ 2 , ψ 3 ( b , λ ) = λ γ 1 + γ 1 , ψ 2 ( δ 2 , λ ) = γ 6 ψ 3 ( δ 2 , λ ) γ 5 ψ 3 ( δ 2 , λ ) α 2 , ψ 2 ( δ 2 , λ ) = θ 6 ψ 3 ( δ 2 , λ ) θ 5 ψ 3 ( δ 2 , λ ) α 2 , ψ 1 ( δ 1 , λ ) = γ 4 ψ 2 ( δ 1 , λ ) γ 3 ψ 2 ( δ 1 , λ ) α 1 , ψ 1 ( δ 1 , λ ) = θ 4 ψ 2 ( δ 1 , λ ) θ 3 ψ 2 ( δ 1 , λ ) α 1 , (10)

respectively.

These solutions are entire functions ofλfor each fixed x [ a , b ] and satisfy the relation

ψ ( x , λ n ) = κ n φ ( x , λ n )

for each eigenvalue λ n , where

κ n = θ 2 ψ ( a , λ n ) θ 1 ψ ( a , λ n ) β 1 .

Lemma 2Let λ = k 2 , k = σ + i ω .

Then the following integral equations and also asymptotic behaviors hold for ν = 0 , 1 :

d ν d x ν φ 1 ( x , λ ) = ( λ θ 2 θ 2 ) d ν d x ν cos k ρ 1 ( x a ) + 1 k ρ 1 ( λ θ 1 θ 1 ) d ν d x ν sin k ρ 1 ( x a ) + ρ 1 k a x d ν d x ν sin k ρ 1 ( x t ) q ( t ) φ 1 ( t , λ ) d t = ( λ θ 2 θ 2 ) d ν d x ν cos k ρ 1 ( x a ) + 1 k ρ 1 ( λ θ 1 θ 1 ) d ν d x ν sin k ρ 1 ( x a ) + O ( | k | ν + 1 e | Im k | ( x a ) ρ 1 ) , d ν d x ν φ 2 ( x , λ ) = ( θ 3 φ 1 ( δ 1 , λ ) + γ 3 φ 1 ( δ 1 , λ ) ) d ν d x ν cos k ρ 2 ( x δ 1 ) + 1 k ρ 2 ( θ 4 φ 1 ( δ 1 , λ ) + γ 4 φ 1 ( δ 1 , λ ) ) d ν d x ν sin k ρ 2 ( x δ 1 ) + ρ 2 k δ 1 x d ν d x ν sin k ρ 2 ( x t ) q ( t ) φ 2 ( t , λ ) d t = ( λ θ 2 θ 2 ) γ 3 k ρ 1 sin k ρ 1 ( δ 1 a ) d ν d x ν cos k ρ 2 ( x δ 1 ) + O ( | k | ν + 2 e | Im k | ( ( δ 1 a ) ρ 1 + ( x δ 1 ) ρ 2 ) ) , d ν d x ν φ 3 ( x , λ ) = ( θ 5 φ 2 ( δ 2 , λ ) + γ 5 φ 2 ( δ 2 , λ ) ) d ν d x ν cos k ρ 3 ( x δ 2 ) + 1 k ρ 3 ( θ 5 φ 2 ( δ 2 , λ ) + γ 6 φ 2 ( δ 2 , λ ) ) d ν d x ν sin k ρ 3 ( x δ 2 ) + ρ 3 k δ 2 x d ν d x ν sin k ρ 3 ( x t ) q ( t ) φ 3 ( t , λ ) d t = ( λ θ 2 θ 2 ) γ 3 γ 5 k 2 ρ 1 ρ 2 sin k ρ 1 ( δ 1 a ) sin k ρ 2 ( δ 2 δ 1 ) d ν d x ν cos k ρ 3 ( x δ 2 ) + O ( | k | ν + 3 e | Im k | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( x δ 2 ) ρ 3 ) ) .

Lemma 3Let λ = k 2 , k = σ + i ω .

Then the following integral equations and also asymptotic behaviors hold for ν = 0 , 1 :

d ν d x ν ψ 3 ( x , λ ) = ( λ γ 2 + γ 2 ) d ν d x ν cos k ρ 3 ( x b ) + 1 k ρ 3 ( λ γ 1 + γ 1 ) d ν d x ν sin k ρ 3 ( x b ) ρ 3 k x b d ν d x ν sin k ρ 3 ( x t ) q ( t ) ψ 3 ( t , λ ) d t = ( λ γ 2 + γ 2 ) d ν d x ν cos k ρ 3 ( x b ) + 1 k ρ 3 ( λ γ 1 + γ 1 ) d ν d x ν sin k ρ 3 ( x b ) + O ( | k | ν + 1 e | Im k | ( x b ) ρ 3 ) , d ν d x ν ψ 2 ( x , λ ) = γ 6 ψ 3 ( δ 2 , λ ) γ 5 ψ 3 ( δ 2 , λ ) α 2 d ν d x ν cos k ρ 2 ( x δ 2 ) + 1 k ρ 2 θ 6 ψ 3 ( δ 2 , λ ) θ 5 ψ 3 ( δ 2 , λ ) α 2 d ν d x ν sin k ρ 2 ( x δ 2 ) ρ 2 k x δ 2 d ν d x ν sin k ρ 2 ( x t ) q ( t ) ψ 2 ( t , λ ) d t = ( λ γ 2 + γ 2 α 2 ) ( k ρ 3 γ 5 sin k ρ 3 ( δ 2 b ) d ν d x ν cos k ρ 2 ( x δ 2 ) ) + O ( | k | ν + 2 e | Im k | ( ( δ 2 b ) ρ 3 + ( x δ 2 ) ρ 2 ) ) , d ν d x ν ψ 1 ( x , λ ) = γ 4 ψ 2 ( δ 1 , λ ) γ 3 ψ 2 ( δ 1 , λ ) α 1 d ν d x ν cos k ρ 1 ( x δ 1 ) + 1 k ρ 1 θ 4 ψ 2 ( δ 1 , λ ) θ 3 ψ 2 ( δ 1 , λ ) α 1 d ν d x ν sin k ρ 1 ( x δ 1 ) ρ 1 k x δ 1 d ν d x ν sin k ρ 1 ( x t ) q ( t ) ψ 1 ( t , λ ) d t = ( λ γ 2 + γ 2 α 2 α 1 ) × ( k 2 ρ 2 ρ 3 γ 3 γ 5 sin k ρ 3 ( δ 2 b ) sin k ρ 2 ( δ 1 δ 2 ) d ν d x ν cos k ρ 1 ( x δ 1 ) ) + O ( | k | ν + 3 e | Im k | ( ( δ 2 b ) ρ 3 + ( δ 1 δ 2 ) ρ 2 + ( x δ 1 ) ρ 1 ) ) .

The function Δ ( λ ) is called the characteristic function, and numbers { μ n } n 1 are called the normalizing constants of problem (1)-(7) such that

Δ ( λ ) = λ ( γ 1 φ ( b , λ ) γ 2 φ ( b , λ ) ) + ( γ 1 φ ( b , λ ) γ 2 φ ( b , λ ) ) , (11)

μ n : = α 1 α 2 ρ 1 a δ 1 φ 2 ( x , λ n ) d x + α 2 ρ 2 δ 1 δ 2 φ 2 ( x , λ n ) d x + ρ 3 δ 2 b φ 2 ( x , λ n ) d x + α 1 α 2 β 1 ( θ 1 φ ( a , λ n ) θ 2 φ ( a , λ n ) ) 2 + 1 β 2 ( γ 1 φ ( b , λ n ) γ 2 φ ( b , λ n ) ) 2 . (12)

Lemma 4The following equality holds for each eigenvalue λ n

α 1 α 2 Δ ˙ ( λ n ) = κ n μ n .

Proof Since

ρ ( x ) ψ ( x , λ ) + q ( x ) ψ ( x , λ ) = λ ψ ( x , λ ) , ρ ( x ) φ ( x , λ n ) + q ( x ) φ ( x , λ n ) = λ φ ( x , λ n ) ,

we get

φ ( x , λ n ) ψ ( x , λ ) ψ ( x , λ ) φ ( x , λ n ) ( | a δ 1 + | δ 1 δ 2 + | δ 2 b ) = ( λ λ n ) ρ 1 2 a δ 1 ψ ( x , λ ) φ ( x , λ n ) d x + ( λ λ n ) ρ 2 2 δ 1 δ 2 ψ ( x , λ ) φ ( x , λ n ) d x + ( λ λ n ) ρ 3 2 δ 2 b ψ ( x , λ ) φ ( x , λ n ) d x .

After that, add and subtract Δ ( λ ) on the left-hand side of the last equality, and by using conditions (2)-(7), we obtain

Δ ( λ ) + ( λ λ n ) ( θ 1 ψ ( a , λ ) θ 2 ψ ( a , λ ) ) ( λ λ n ) ( γ 1 φ ( b , λ n ) γ 2 φ ( b , λ n ) ) + ( 1 α 1 ) ( ψ ( δ 1 , λ ) φ ( δ 1 , λ n ) φ ( δ 1 , λ n ) ψ ( δ 1 , λ ) ) + ( 1 α 2 ) ( ψ ( δ 2 , λ ) φ ( δ 2 , λ n ) φ ( δ 2 , λ n ) ψ ( δ 2 , λ ) ) = ( λ λ n ) ρ 1 2 a δ 1 ψ ( x , λ ) φ ( x , λ n ) d x + ( λ λ n ) ρ 2 2 δ 1 δ 2 ψ ( x , λ ) φ ( x , λ n ) d x + ( λ λ n ) ρ 3 2 δ 2 b ψ ( x , λ ) φ ( x , λ n ) d x ,

or

α 1 α 2 Δ ( λ ) λ λ n = α 1 α 2 ρ 1 2 a δ 1 ψ ( x , λ ) φ ( x , λ n ) d x + α 2 ρ 2 2 δ 1 δ 2 ψ ( x , λ ) φ ( x , λ n ) d x + ρ 3 2 δ 2 b ψ ( x , λ ) φ ( x , λ n ) d x + α 1 α 2 β 1 ( θ 1 ψ ( a , λ ) θ 2 ψ ( a , λ ) ) ( θ 1 φ ( a , λ n ) θ 2 φ ( a , λ n ) ) + 1 β 2 ( γ 1 φ ( b , λ n ) γ 2 φ ( b , λ n ) ) ( γ 1 ψ ( b , λ ) γ 2 ψ ( b , λ ) ) .

For λ λ n , α 1 α 2 Δ ˙ ( λ n ) = κ n μ n is obtained by using the equality

ψ ( x , λ n ) = κ n φ ( x , λ n )

and (12). □

Corollary 2The eigenvalues of problemLare simple.

Lemma 5[36]

Let { α i } i = 1 p be the set of real numbers satisfying the inequalities α 0 > α 0 > > α p 1 > 0 , and let { a i } i = 1 p be the set of complex numbers. If a p 0 , then the roots of the equation

e α 0 λ + a 1 e α 1 λ + + a p 1 e α 0 λ + a p = 0

have the form

λ n = 2 π n i α 0 + Ψ ( n ) ( n = 0 , ± 1 , ) ,

where Ψ ( n ) is a bounded sequence.

Now, from Lemma 2 and (11), we can write

Δ ( λ ) Δ 0 ( λ ) = O ( k 6 e | Im k | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 ) ) ,

where Δ 0 ( λ ) = k 7 θ 2 γ 2 γ 3 γ 5 ρ 1 ρ 2 ρ 3 sin k ρ 1 ( δ 1 a ) sin k ρ 2 ( δ 2 δ 1 ) sin k ρ 3 ( b δ 2 ) .

We can see that non-zero roots, namely λ n 0 of the equation Δ 0 ( λ ) = 0 , are real and analytically simple.

Furthermore, it can be proved by using Lemma 5 that

λ n 0 = n π ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 + Ψ n , sup n | Ψ n | < . (13)

Theorem 2The eigenvalues { λ n } n 0 have the following asymptotic behavior for sufficiently largen:

λ n = λ n 4 0 + o ( 1 ) . (14)

Proof Denote

G n : = { λ : k 2 = λ , | k | | k n 0 | + δ } ,

where k n 0 = λ n 0 and δ is a sufficiently small number. The relations

| Δ 0 ( λ ) | C δ | k | 7 e | Im k | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 )

and

Δ ( λ ) Δ 0 ( λ ) = O ( k 6 e | Im k | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 ) )

are valid for λ G n .

Then, by Rouche’s theorem that the number of zeros of Δ 0 ( λ ) coincides with the number of zeros of Δ ( λ ) in G n , namely n + 4 zeros, λ 0 , λ 1 , λ 2 , , λ n + 3 . In the annulus, between G n and G n + 1 , Δ ( λ ) has accurately one positive zero, namely k n 2 : k n = λ n 0 + δ n , for n 1 . So, it follows that λ n + 4 = k n 2 . Applying to Rouche’s theorem in η ε = { k : | k k n 0 | < ε } for sufficiently small ε and sufficiently large n, we get δ n = o ( 1 ) . Finally, we obtain the asymptotic formula

λ n = λ n 4 0 + o ( 1 ) .

Denote

Δ i ( λ ) : = W ( φ i , ψ i , x ) : = φ i ψ i φ i ψ i , x Ω i ( i = 1 , 3 ¯ ) ,

which are independent of x Ω i and are entire functions such that Ω 1 = [ a , δ 1 ) , Ω 2 = ( δ 1 , δ 2 ) , Ω 3 = ( δ 2 , b ] .

It can be easily seen that

Δ ( λ ) : = Δ 3 ( λ ) = α 2 Δ 2 ( λ ) = α 1 α 2 Δ 1 ( λ ) .

□

Example Let q = 0 , a = 0 , b = π , δ 1 = π 4 , δ 2 = π 2 , θ 3 = γ 4 = 0 , γ 3 = 1 , θ 4 = 1 , θ 5 = γ 6 = 0 , γ 5 = 1 , θ 6 = 1 , γ 2 = 1 , γ 6 = 1 , γ 1 = γ 2 = 0 , θ 2 = 1 θ 1 = 1 , θ 1 = θ 2 = 0 .

Since

Δ ( λ ) = ρ 1 ρ 2 ρ 3 k 7 sin k ρ 1 δ 1 sin k ρ 2 ( δ 2 δ 1 ) sin k ρ 3 ( π δ 2 ) + O ( k 6 e | Im k | ( δ 1 ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( π δ 2 ) ρ 3 ) ) ,

the eigenvalues of the boundary value problem (1)-(7) satisfy the following asymptotic formulae:

λ n 1 = k n 1 = 4 ( n 4 ) ρ 1 + ε n , λ n 2 = k n 2 = 4 ( n 4 ) ρ 2 + ε n , λ n 3 = k n 3 = 2 ( n 4 ) ρ 3 + ε n ,

where ε n = O ( n 1 ) .

### 3 Inverse problems

In this section, we study the inverse problems for the reconstruction of the boundary value problem (1)-(7) by Weyl function and spectral data.

We consider the boundary value problem L ˜ with the same form of L but with different coefficients q ˜ ( x ) , θ ˜ i , γ ˜ i , δ ˜ j , θ ˜ j , γ ˜ j , i = 1 6 , j = 1 , 2 .

If a certain symbol α denotes an object related to L, then the symbol α ˜ denotes the corresponding object related to L ˜ .

The Weyl function Let Φ ( x , λ ) be a solution of equation (1), which satisfies the condition ( λ θ 1 θ 1 ) Φ ( a , λ ) ( λ θ 2 θ 2 ) Φ ( a , λ ) = 1 and transmissions (4)-(7).

Assume that the function χ ( x , λ ) is the solution of equation (1) that satisfies the conditions χ ( a , λ ) = β 1 1 θ 2 , χ ( a , λ ) = β 1 1 θ 1 and the transmission conditions (4)-(7).

Since W [ χ , φ ] = 1 , the functions χ and φ are linearly independent. Therefore, the function ψ ( x , λ ) can be represented by

ψ ( x , λ ) = θ 2 ψ ( a , λ ) θ 1 ψ ( a , λ ) β 1 φ ( x , λ ) + Δ ( λ ) χ ( x , λ )

or

Φ ( x , λ ) = ψ ( x , λ ) Δ ( λ ) = χ ( x , λ ) + θ 2 ψ ( a , λ ) θ 1 ψ ( a , λ ) β 1 Δ ( λ ) φ ( x , λ ) (15)

that is called the Weyl solution, and

θ 2 ψ ( a , λ ) θ 1 ψ ( a , λ ) β 1 Δ ( λ ) = M ( λ ) (16)

is called the Weyl function.

Theorem 3If M ( λ ) = M ˜ ( λ ) , then L = L ˜ , i.e., q ( x ) = q ˜ ( x ) , a.e. and θ i = θ ˜ i , γ i = γ ˜ i , i = 1 6 , δ j = δ ˜ j , θ j = θ ˜ j , γ j = γ ˜ j , j = 1 , 2 .

Proof We introduce a matrix P ( x , λ ) = [ P k j ( x , λ ) ] k , j = 1 , 2 by the formula

P ( x , λ ) ( φ ˜ Φ ˜ φ ˜ Φ ˜ ) = ( φ Φ φ Φ )

or

( P 11 ( x , λ ) P 12 ( x , λ ) P 21 ( x , λ ) P 22 ( x , λ ) ) = ( φ Φ ˜ + Φ φ ˜ φ Φ ˜ Φ φ ˜ φ Φ ˜ + φ ˜ Φ φ Φ ˜ φ ˜ Φ ) , (17)

where Φ ( x , λ ) = ψ ( x , λ ) Δ ( λ ) .

P 11 ( x , λ ) = χ ( x , λ ) φ ˜ ( x , λ ) φ ( x , λ ) χ ˜ ( x , λ ) + ( M ( λ ) M ˜ ( λ ) ) φ ( x , λ ) φ ˜ ( x , λ ) , P 12 ( x , λ ) = φ ( x , λ ) χ ˜ ( x , λ ) χ ( x , λ ) φ ˜ ( x , λ ) + ( M ˜ ( λ ) M ( λ ) ) φ ( x , λ ) φ ˜ ( x , λ ) .

Thus, if M ( λ ) M ˜ ( λ ) , then the functions P 11 ( x , λ ) and P 12 ( x , λ ) are entire in λ for each fixed x.

Denote G w = { λ : λ = k 2 , | k k δ | > w , δ = 1 , 2 , } and G ˜ w = { λ : λ = k 2 , | k k ˜ δ | > w , δ = 1 , 2 , } , where w is sufficiently small number, k δ and k ˜ δ are square roots of the eigenvalues of the problem L and L ˜ , respectively. It is easily shown that

Φ i ( ν ) ( x , λ ) C w | k | ν ( 2 + i ) e | Im λ | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 ) , x Ω i ( i = 1 , 2 , 3 ) , ν = 0 , 1 (18)

are valid for sufficiently large | λ | , where Ω 1 = [ a , δ 1 ) , Ω 2 = ( δ 1 , δ 2 ) and Ω 3 = ( δ 2 , b ] . Hence, Lemma 2 and (18) yield that

| P 11 ( x , λ ) | C w , | P 12 ( x , λ ) | C w | k | 1 for  λ Ω  and for  λ G w G ˜ w . (19)

According to (19), and Liouville’s theorem, P 11 ( x , λ ) = C ( x ) and P 12 ( x , λ ) 0 for x [ a , b ] { δ 1 , δ 2 , δ ˜ 1 , δ ˜ 2 } . By virtue of (17), we get

φ ( x , λ ) = C ( x ) φ ˜ ( x , λ ) , Φ ( x , λ ) = C ( x ) Φ ˜ ( x , λ ) . (20)

It is obvious that

W [ Φ ( x , λ ) , φ ( x , λ ) ] = Φ ( a , λ ) ( θ 1 λ θ 1 ) Φ ( a , λ ) ( θ 2 λ θ 2 ) = ψ ( a , λ ) ( θ 1 λ θ 1 ) ψ ( a , λ ) ( θ 2 λ θ 2 ) Δ ( λ ) 1 ,

and similarly, W [ Φ ˜ ( x , λ ) , φ ˜ ( x , λ ) ] 1 . Thus, we have C 2 ( x ) 1 .

Otherwise, the following asymptotic expressions hold

φ 1 ( x , λ ) = φ ˜ 1 ( x , λ ) = λ 2 e i k ( x a ) ρ 1 ( 1 + o ( 1 ) ) for  x < δ 1  and  x < δ ˜ 1 , φ 2 ( x , λ ) = φ ˜ 2 ( x , λ ) = λ 3 2 2 e i k ( ( δ 1 a ) ρ 1 + ( x δ 1 ) ρ 2 ) ( 1 + o ( 1 ) ) for  δ 1 < x < δ 2  and  δ ˜ 1 < x < δ ˜ 2 , φ 3 ( x , λ ) = φ ˜ 3 ( x , λ ) = λ 2 2 e i k ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( x δ 2 ) ρ 3 ) ( 1 + o ( 1 ) ) for  δ 2 < x  and  δ ˜ 2 < x . (21)

Without loss of generality, we assume that δ 1 < δ ˜ 1 and δ 2 < δ ˜ 2 . From (20)-(21), we get C ( x ) 1 for x [ a , δ 1 ) ( δ ˜ 1 , δ 2 ) ( δ ˜ 2 , b ] and also

1 λ 1 / 2 ( 1 + o ( 1 ) ) C ( x ) = e ( x δ 1 ) ρ 2 ( 1 + o ( 1 ) ) for  x ( δ 1 , δ ˜ 1 ) , 1 λ 1 / 2 e ( x δ 1 ) ρ 2 ( 1 + o ( 1 ) ) C ( x ) = e ( δ 2 δ 1 ) ρ 2 + ( x δ 2 ) ρ 3 ( 1 + o ( 1 ) ) for  x ( δ 2 , δ ˜ 2 ) . (22)

As | λ | in (22), we contradict C 2 ( x ) 1 . Therefore, δ 1 = δ ˜ 1 , δ 2 = δ ˜ 2 . Thus, φ ( x , λ ) φ ˜ ( x , λ ) , Φ ( x , λ ) Φ ˜ ( x , λ ) and ψ ( x , λ ) ψ ( x , λ ) ψ ˜ ( x , λ ) ψ ˜ ( x , λ ) . Hence, from equation (1) and transmission conditions (4)-(7), q ( x ) = q ˜ ( x ) , a.e., θ i = θ ˜ i , γ i = γ ˜ i , i = 3 6 , and from (9) and (10), θ i = θ ˜ i , θ j = θ ˜ j , γ j = γ ˜ j , i , j = 1 , 2 . □

Lemma 6The following representation holds

M ( λ ) = n = 0 α 1 α 2 μ n ( λ n λ ) .

Proof Weyl function M ( λ ) is a meromorphic function with respect to λ, which has simple poles at λ n . Therefore, we calculate

Re s λ = λ n M ( λ ) = θ 2 ψ ( a , λ n ) θ 1 ψ ( a , λ n ) β 1 Δ ˙ ( λ n ) .

Since κ n = θ 2 ψ ( a , λ n ) θ 1 ψ ( a , λ n ) β 1 and Δ ˙ ( λ n ) = κ n μ n α 1 α 2 ,

Re s λ = λ n M ( λ ) = α 1 α 2 μ n . (23)

Let Γ N = { λ : λ = k 2 , | k | = λ N + ε } , where ε is a sufficiently small number. Consider the contour integral I N ( λ ) = 1 2 π i Γ n M ( μ ) μ λ d μ , λ int Γ N . For λ G w ,

Δ ( λ ) | λ | 7 / 2 C w e | Im k | ( ( δ 1 a ) ρ 1 + ( δ 2 δ 1 ) ρ 2 + ( b δ 2 ) ρ 3 )

satisfies. Using this equality and (16), we get

| M ( λ ) | C w | λ | for  λ G w .

Thus, lim N I N ( λ ) = 0 . As a result, the residue theorem and (23) yield

M ( λ ) = n = 0 α 1 α 2 μ n ( λ n λ ) .

□

Theorem 4If λ n = λ ˜ n and μ n = μ ˜ n for alln, then L L ˜ , i.e., q ( x ) = q ˜ ( x ) , a.e., θ i = θ ˜ i , γ i = γ ˜ i , i = 1 6 , δ j = δ ˜ j , θ j = θ ˜ j , γ j = γ ˜ j , j = 1 , 2 . Hence, problem (1)-(7) is uniquely determined by spectral data { λ n , μ n } .

Proof If λ n = λ ˜ n and μ n = μ ˜ n for all n, then M ( λ ) = M ˜ ( λ ) by Lemma 6. Therefore, we get L = L ˜ by Theorem 3.

Let us consider the boundary value problem L 1 that we get the condition θ 2 y ( a , λ ) θ 1 y ( a , λ ) = 0 instead of condition (2) in L. Let { τ n } n 0 be the eigenvalues of the problem L 1 . It is clear that τ n are zeros of

Δ 1 ( τ ) : = θ 2 ψ ( a , τ ) θ 1 ψ ( a , τ ) .

□

Theorem 5If λ n = λ ˜ n and τ n = τ ˜ n for alln, then L ( q , δ j , θ k , γ i , γ j ) = L ( q ˜ , δ ˜ j , θ ˜ k , γ ˜ i , γ ˜ j ) , i = 1 6 , k = 3 6 , j = 1 , 2 .

Hence, the problemLis uniquely determined by the sequences { λ n } and { τ n } , except coefficients θ j and θ j .

Proof Since the characteristic functions Δ ( λ ) and Δ 1 ( τ ) are entire of order 1 2 , functions Δ ( λ ) and Δ 1 ( τ ) are uniquely determined up to multiplicative constant with their zeros by Hadamard’s factorization theorem [37]

Δ ( λ ) = C n = 0 ( 1 λ λ n ) , Δ 1 ( τ ) = C 1 n = 0 ( 1 τ τ n ) ,

where C and C 1 are constants dependent on { λ n } and { τ n } , respectively. Therefore, when λ n = λ ˜ n and τ n = τ ˜ n for all n, Δ ( λ ) Δ ˜ ( λ ) and Δ 1 ( τ ) Δ ˜ 1 ( τ ) . Hence, θ 2 ψ ( a , τ ) θ 1 ψ ( a , τ ) = θ ˜ 2 ψ ˜ ( a , τ ) θ ˜ 1 ψ ˜ ( a , τ ) . As a result, we get M ( λ ) = M ˜ ( λ ) by (16). So, the proof is completed by Theorem 3. □

### Competing interests

The author declares that he has no competing interests.

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