Open Access

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

Boundary Value Problems20132013:209

DOI: 10.1186/1687-2770-2013-209

Received: 18 June 2013

Accepted: 26 August 2013

Published: 11 September 2013

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 [14]). 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, 512]. Various physics applications of such problems can be found in [1, 3, 4, 1319] 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 [2029]). 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq1_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ1_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equa_HTML.gif
and ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq2_HTML.gif, ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq3_HTML.gif, and ρ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq4_HTML.gif are given positive real numbers; q ( x ) L 2 [ Ω , R ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq5_HTML.gif; λ C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq6_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ2_HTML.gif
(2)
l 2 y : = λ ( γ 1 y ( b ) γ 2 y ( b ) ) + ( γ 1 y ( b ) γ 2 y ( b ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ3_HTML.gif
(3)
with discontinuity conditions at two points x = δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq7_HTML.gif, x = δ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq8_HTML.gif
l 3 y : = y ( δ 1 + 0 ) θ 3 y ( δ 1 0 ) γ 3 y ( δ 1 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ4_HTML.gif
(4)
l 4 y : = y ( δ 1 + 0 ) θ 4 y ( δ 1 0 ) γ 4 y ( δ 1 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ5_HTML.gif
(5)
l 5 y : = y ( δ 2 + 0 ) θ 5 y ( δ 2 0 ) γ 5 y ( δ 2 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ6_HTML.gif
(6)
l 6 y : = y ( δ 2 + 0 ) θ 6 y ( δ 2 0 ) γ 6 y ( δ 2 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ7_HTML.gif
(7)
where θ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq9_HTML.gif, γ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq10_HTML.gif and θ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq11_HTML.gif, γ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq12_HTML.gif ( i = 1 , 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq13_HTML.gif, j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq14_HTML.gif) 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equb_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq15_HTML.gif, where H 1 = ( L 2 ( Ω ) , , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq16_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ8_HTML.gif
(8)
for
F = ( f ( x ) f 1 f 2 ) H , G = ( g ( x ) g 1 g 2 ) H . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equc_HTML.gif
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 ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equd_HTML.gif

such that T F : = ( l f , θ 1 f ( a ) θ 2 f ( a ) , ( γ 1 f ( b ) γ 2 f ( b ) ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq17_HTML.gif for F = ( f , θ 1 f ( a ) θ 2 f ( a ) , γ 1 f ( b ) γ 2 f ( b ) ) D ( T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq18_HTML.gif and also l 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq19_HTML.gif- l 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq20_HTML.gif are satisfied for f.

Thus, we can rewrite the considered problem (1)-(7) in the operator form as T F = λ F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq21_HTML.gif.

Theorem 1 The operator T is symmetric in H.

Proof Let F , G D ( T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq22_HTML.gif. 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 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Eque_HTML.gif
where by W ( f , g ; x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq23_HTML.gif, we denote the Wronskian of the functions f and g as
f ( x ) g ( x ) f ( x ) g ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equf_HTML.gif
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equg_HTML.gif

Thus, we have T F , G = F , T G https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq24_HTML.gif, i.e., T is symmetric. □

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

Corollary 1 All eigenvalues and eigenfunctions of problem (1)-(7) are real, and two eigenfunctions φ ( x , λ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq25_HTML.gif and φ ( x , λ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq26_HTML.gif, corresponding to different eigenvalues λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq27_HTML.gif and λ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq28_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equh_HTML.gif
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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equi_HTML.gif
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ9_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ10_HTML.gif
(10)

respectively.

These solutions are entire functions of λ for each fixed x [ a , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq29_HTML.gif and satisfy the relation
ψ ( x , λ n ) = κ n φ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equj_HTML.gif
for each eigenvalue λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq30_HTML.gif, where
κ n = θ 2 ψ ( a , λ n ) θ 1 ψ ( a , λ n ) β 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equk_HTML.gif

Lemma 2 Let λ = k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq31_HTML.gif, k = σ + i ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq32_HTML.gif.

Then the following integral equations and also asymptotic behaviors hold for ν = 0 , 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq33_HTML.gif:
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 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equl_HTML.gif

Lemma 3 Let λ = k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq31_HTML.gif, k = σ + i ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq32_HTML.gif.

Then the following integral equations and also asymptotic behaviors hold for ν = 0 , 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq33_HTML.gif:
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 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equm_HTML.gif
The function Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif is called the characteristic function, and numbers { μ n } n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq35_HTML.gif 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 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ11_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ12_HTML.gif
(12)
Lemma 4 The following equality holds for each eigenvalue λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq30_HTML.gif
α 1 α 2 Δ ˙ ( λ n ) = κ n μ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equn_HTML.gif
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equo_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equp_HTML.gif
After that, add and subtract Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equq_HTML.gif
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 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equr_HTML.gif
For λ λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq36_HTML.gif, α 1 α 2 Δ ˙ ( λ n ) = κ n μ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq37_HTML.gif is obtained by using the equality
ψ ( x , λ n ) = κ n φ ( x , λ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equs_HTML.gif

and (12). □

Corollary 2 The eigenvalues of problem L are simple.

Lemma 5 [36]

Let { α i } i = 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq38_HTML.gif be the set of real numbers satisfying the inequalities α 0 > α 0 > > α p 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq39_HTML.gif, and let { a i } i = 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq40_HTML.gif be the set of complex numbers. If a p 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq41_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equt_HTML.gif
have the form
λ n = 2 π n i α 0 + Ψ ( n ) ( n = 0 , ± 1 , ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equu_HTML.gif

where Ψ ( n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq42_HTML.gif 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 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equv_HTML.gif

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq43_HTML.gif.

We can see that non-zero roots, namely λ n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq44_HTML.gif of the equation Δ 0 ( λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq45_HTML.gif, 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 | < . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ13_HTML.gif
(13)
Theorem 2 The eigenvalues { λ n } n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq46_HTML.gif have the following asymptotic behavior for sufficiently large n:
λ n = λ n 4 0 + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ14_HTML.gif
(14)
Proof Denote
G n : = { λ : k 2 = λ , | k | | k n 0 | + δ } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equw_HTML.gif
where k n 0 = λ n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq47_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equx_HTML.gif
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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equy_HTML.gif

are valid for λ G n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq48_HTML.gif.

Then, by Rouche’s theorem that the number of zeros of Δ 0 ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq49_HTML.gif coincides with the number of zeros of Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif in G n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq50_HTML.gif, namely n + 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq51_HTML.gif zeros, λ 0 , λ 1 , λ 2 , , λ n + 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq52_HTML.gif. In the annulus, between G n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq50_HTML.gif and G n + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq53_HTML.gif, Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif has accurately one positive zero, namely k n 2 : k n = λ n 0 + δ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq54_HTML.gif, for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq55_HTML.gif. So, it follows that λ n + 4 = k n 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq56_HTML.gif. Applying to Rouche’s theorem in η ε = { k : | k k n 0 | < ε } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq57_HTML.gif for sufficiently small ε and sufficiently large n, we get δ n = o ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq58_HTML.gif. Finally, we obtain the asymptotic formula
λ n = λ n 4 0 + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equz_HTML.gif
Denote
Δ i ( λ ) : = W ( φ i , ψ i , x ) : = φ i ψ i φ i ψ i , x Ω i ( i = 1 , 3 ¯ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equaa_HTML.gif

which are independent of x Ω i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq59_HTML.gif and are entire functions such that Ω 1 = [ a , δ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq60_HTML.gif, Ω 2 = ( δ 1 , δ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq61_HTML.gif, Ω 3 = ( δ 2 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq62_HTML.gif.

It can be easily seen that
Δ ( λ ) : = Δ 3 ( λ ) = α 2 Δ 2 ( λ ) = α 1 α 2 Δ 1 ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equab_HTML.gif

 □

Example Let q = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq63_HTML.gif, a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq64_HTML.gif, b = π https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq65_HTML.gif, δ 1 = π 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq66_HTML.gif, δ 2 = π 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq67_HTML.gif, θ 3 = γ 4 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq68_HTML.gif, γ 3 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq69_HTML.gif, θ 4 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq70_HTML.gif, θ 5 = γ 6 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq71_HTML.gif, γ 5 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq72_HTML.gif, θ 6 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq73_HTML.gif, γ 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq74_HTML.gif, γ 6 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq75_HTML.gif, γ 1 = γ 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq76_HTML.gif, θ 2 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq77_HTML.gif θ 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq78_HTML.gif, θ 1 = θ 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq79_HTML.gif.

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 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equac_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equad_HTML.gif

where ε n = O ( n 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq80_HTML.gif.

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 ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq81_HTML.gif with the same form of L but with different coefficients q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq82_HTML.gif, θ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq83_HTML.gif, γ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq84_HTML.gif, δ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq85_HTML.gif, θ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq86_HTML.gif, γ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq87_HTML.gif, i = 1 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq88_HTML.gif, j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq14_HTML.gif.

If a certain symbol α denotes an object related to L, then the symbol α ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq89_HTML.gif denotes the corresponding object related to L ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq81_HTML.gif.

The Weyl function Let Φ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq90_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq91_HTML.gif and transmissions (4)-(7).

Assume that the function χ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq92_HTML.gif is the solution of equation (1) that satisfies the conditions χ ( a , λ ) = β 1 1 θ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq93_HTML.gif, χ ( a , λ ) = β 1 1 θ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq94_HTML.gif and the transmission conditions (4)-(7).

Since W [ χ , φ ] = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq95_HTML.gif, the functions χ and φ are linearly independent. Therefore, the function ψ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq96_HTML.gif 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equae_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ15_HTML.gif
(15)
that is called the Weyl solution, and
θ 2 ψ ( a , λ ) θ 1 ψ ( a , λ ) β 1 Δ ( λ ) = M ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ16_HTML.gif
(16)

is called the Weyl function.

Theorem 3 If M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq97_HTML.gif, then L = L ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq98_HTML.gif, i.e., q ( x ) = q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq99_HTML.gif, a.e. and θ i = θ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq100_HTML.gif, γ i = γ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq101_HTML.gif, i = 1 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq88_HTML.gif, δ j = δ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq102_HTML.gif, θ j = θ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq103_HTML.gif, γ j = γ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq104_HTML.gif, j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq14_HTML.gif.

Proof We introduce a matrix P ( x , λ ) = [ P k j ( x , λ ) ] k , j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq105_HTML.gif by the formula
P ( x , λ ) ( φ ˜ Φ ˜ φ ˜ Φ ˜ ) = ( φ Φ φ Φ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equaf_HTML.gif
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ17_HTML.gif
(17)
where Φ ( x , λ ) = ψ ( x , λ ) Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq106_HTML.gif.
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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equag_HTML.gif

Thus, if M ( λ ) M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq107_HTML.gif, then the functions P 11 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq108_HTML.gif and P 12 ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq109_HTML.gif are entire in λ for each fixed x.

Denote G w = { λ : λ = k 2 , | k k δ | > w , δ = 1 , 2 , } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq110_HTML.gif and G ˜ w = { λ : λ = k 2 , | k k ˜ δ | > w , δ = 1 , 2 , } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq111_HTML.gif, where w is sufficiently small number, k δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq112_HTML.gif and k ˜ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq113_HTML.gif are square roots of the eigenvalues of the problem L and L ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq81_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ18_HTML.gif
(18)
are valid for sufficiently large | λ | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq114_HTML.gif, where Ω 1 = [ a , δ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq60_HTML.gif, Ω 2 = ( δ 1 , δ 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq115_HTML.gif and Ω 3 = ( δ 2 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq116_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ19_HTML.gif
(19)
According to (19), and Liouville’s theorem, P 11 ( x , λ ) = C ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq117_HTML.gif and P 12 ( x , λ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq118_HTML.gif for x [ a , b ] { δ 1 , δ 2 , δ ˜ 1 , δ ˜ 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq119_HTML.gif. By virtue of (17), we get
φ ( x , λ ) = C ( x ) φ ˜ ( x , λ ) , Φ ( x , λ ) = C ( x ) Φ ˜ ( x , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ20_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equah_HTML.gif

and similarly, W [ Φ ˜ ( x , λ ) , φ ˜ ( x , λ ) ] 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq120_HTML.gif. Thus, we have C 2 ( x ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq121_HTML.gif.

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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ21_HTML.gif
(21)
Without loss of generality, we assume that δ 1 < δ ˜ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq122_HTML.gif and δ 2 < δ ˜ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq123_HTML.gif. From (20)-(21), we get C ( x ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq124_HTML.gif for x [ a , δ 1 ) ( δ ˜ 1 , δ 2 ) ( δ ˜ 2 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq125_HTML.gif 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ22_HTML.gif
(22)

As | λ | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq126_HTML.gif in (22), we contradict C 2 ( x ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq127_HTML.gif. Therefore, δ 1 = δ ˜ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq128_HTML.gif, δ 2 = δ ˜ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq129_HTML.gif. Thus, φ ( x , λ ) φ ˜ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq130_HTML.gif, Φ ( x , λ ) Φ ˜ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq131_HTML.gif and ψ ( x , λ ) ψ ( x , λ ) ψ ˜ ( x , λ ) ψ ˜ ( x , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq132_HTML.gif. Hence, from equation (1) and transmission conditions (4)-(7), q ( x ) = q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq99_HTML.gif, a.e., θ i = θ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq100_HTML.gif, γ i = γ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq101_HTML.gif, i = 3 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq133_HTML.gif, and from (9) and (10), θ i = θ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq100_HTML.gif, θ j = θ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq134_HTML.gif, γ j = γ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq104_HTML.gif, i , j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq135_HTML.gif. □

Lemma 6 The following representation holds
M ( λ ) = n = 0 α 1 α 2 μ n ( λ n λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equai_HTML.gif
Proof Weyl function M ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq136_HTML.gif is a meromorphic function with respect to λ, which has simple poles at λ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq30_HTML.gif. 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equaj_HTML.gif
Since κ n = θ 2 ψ ( a , λ n ) θ 1 ψ ( a , λ n ) β 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq137_HTML.gif and Δ ˙ ( λ n ) = κ n μ n α 1 α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq138_HTML.gif,
Re s λ = λ n M ( λ ) = α 1 α 2 μ n . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equ23_HTML.gif
(23)
Let Γ N = { λ : λ = k 2 , | k | = λ N + ε } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq139_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq140_HTML.gif, λ int Γ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq141_HTML.gif. For λ G w https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq142_HTML.gif,
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equak_HTML.gif
satisfies. Using this equality and (16), we get
| M ( λ ) | C w | λ | for  λ G w . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equal_HTML.gif
Thus, lim N I N ( λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq143_HTML.gif. As a result, the residue theorem and (23) yield
M ( λ ) = n = 0 α 1 α 2 μ n ( λ n λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equam_HTML.gif

 □

Theorem 4 If λ n = λ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq144_HTML.gif and μ n = μ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq145_HTML.gif for all n, then L L ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq146_HTML.gif, i.e., q ( x ) = q ˜ ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq147_HTML.gif, a.e., θ i = θ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq100_HTML.gif, γ i = γ ˜ i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq148_HTML.gif, i = 1 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq88_HTML.gif, δ j = δ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq102_HTML.gif, θ j = θ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq103_HTML.gif, γ j = γ ˜ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq104_HTML.gif, j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq14_HTML.gif. Hence, problem (1)-(7) is uniquely determined by spectral data { λ n , μ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq149_HTML.gif.

Proof If λ n = λ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq144_HTML.gif and μ n = μ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq150_HTML.gif for all n, then M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq97_HTML.gif by Lemma 6. Therefore, we get L = L ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq98_HTML.gif by Theorem 3.

Let us consider the boundary value problem L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq151_HTML.gif that we get the condition θ 2 y ( a , λ ) θ 1 y ( a , λ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq152_HTML.gif instead of condition (2) in L. Let { τ n } n 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq153_HTML.gif be the eigenvalues of the problem L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq151_HTML.gif. It is clear that τ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq154_HTML.gif are zeros of
Δ 1 ( τ ) : = θ 2 ψ ( a , τ ) θ 1 ψ ( a , τ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equan_HTML.gif

 □

Theorem 5 If λ n = λ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq144_HTML.gif and τ n = τ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq155_HTML.gif for all n, then L ( q , δ j , θ k , γ i , γ j ) = L ( q ˜ , δ ˜ j , θ ˜ k , γ ˜ i , γ ˜ j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq156_HTML.gif, i = 1 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq88_HTML.gif, k = 3 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq157_HTML.gif, j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq14_HTML.gif.

Hence, the problem L is uniquely determined by the sequences { λ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq158_HTML.gif and { τ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq159_HTML.gif, except coefficients θ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq160_HTML.gif and θ j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq11_HTML.gif.

Proof Since the characteristic functions Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif and Δ 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq161_HTML.gif are entire of order 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq162_HTML.gif, functions Δ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq34_HTML.gif and Δ 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq161_HTML.gif 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_Equao_HTML.gif

where C and C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq163_HTML.gif are constants dependent on { λ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq164_HTML.gif and { τ n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq159_HTML.gif, respectively. Therefore, when λ n = λ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq165_HTML.gif and τ n = τ ˜ n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq155_HTML.gif for all n, Δ ( λ ) Δ ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq166_HTML.gif and Δ 1 ( τ ) Δ ˜ 1 ( τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq167_HTML.gif. Hence, θ 2 ψ ( a , τ ) θ 1 ψ ( a , τ ) = θ ˜ 2 ψ ˜ ( a , τ ) θ ˜ 1 ψ ˜ ( a , τ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq168_HTML.gif. As a result, we get M ( λ ) = M ˜ ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-209/MediaObjects/13661_2013_Article_457_IEq97_HTML.gif by (16). So, the proof is completed by Theorem 3. □

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Sciences, Cumhuriyet University

References

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