double discontinuous inverse problems for sturm liouville operator with parameter dependent conditions

The purpose of this paper is to solve the inverse spectral problems for Sturm-Liouville operator with boundary conditions depending on spectral parameter and double discontinuities insid

Research Article

Double Discontinuous Inverse Problems for Sturm-Liouville

Operator with Parameter-Dependent Conditions

A S Ozkan, B Keskin, and Y Cakmak

Department of Mathematics, Faculty of Arts & Science, Cumhuriyet University, 58140 Sivas, Turkey

Correspondence should be addressed to A S Ozkan; asozkan58@gmail.com

Received 26 March 2013; Accepted 24 June 2013

Academic Editor: Dumitru Motreanu

Copyright © 2013 A S Ozkan et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited The purpose of this paper is to solve the inverse spectral problems for Sturm-Liouville operator with boundary conditions depending

on spectral parameter and double discontinuities inside the interval It is proven that the coefficients of the problem can be uniquely determined by either Weyl function or given two different spectral sequences

1 Introduction

Spectral problems of differential operators are studied in two

main branches, namely, direct spectral problems and inverse

spectral problems Direct problems of spectral analysis

con-sist in investigating the spectral properties of an operator

On the other hand, inverse problems aim at recovering

operators from their spectral characteristics Such problems

often appear in mathematics, physics, mechanics, electronics,

geophysics, and other branches of natural sciences

First and most important results for inverse problem of

a regular Sturm-Liouville operator were given by

Ambart-sumyan in 1929 [1] and Borg in 1946 [2] Physical applications

of inverse spectral problems can be found in several works

(see, e.g., [3–9] and references therein)

Eigenvalue-dependent boundary conditions were studied

extensively The references [10,11] are well-known examples

for problems with boundary conditions that depend linearly

on the eigenvalue parameter In [10,12], an operator-theoretic

formulation of the problems with the spectral parameter

contained in only one of the boundary conditions has

been given Inverse problems according to various spectral

data for eigenparameter linearly dependent Sturm-Liouville

operator were investigated in [13–17] Boundary conditions

that depend nonlinearly on the spectral parameter were also

considered in [18–23]

Boundary value problems with discontinuity condition

appear in the various problems of the applied sciences These

kinds of problems are well studied (see, e.g., [24–31])

In this study, we consider a boundary value problem generated by the Sturm-Liouville equation:

ℓ𝑦 := −𝑦󸀠󸀠+ 𝑞 (𝑥) 𝑦 = 𝜆𝑦, 𝑥 ∈ 𝐼 =⋃2

𝑖=0

(𝑑𝑖, 𝑑𝑖+1) (1) subject to the boundary conditions

𝑈 (𝑦) := 𝜆 (𝑦󸀠(0) + ℎ0𝑦 (0)) − ℎ1𝑦󸀠(0) − ℎ2𝑦 (0) = 0, (2)

𝑉 (𝑦) := 𝜆 (𝑦󸀠(1) + 𝐻0𝑦 (1)) − 𝐻1𝑦󸀠(1) − 𝐻2𝑦 (1) = 0

(3) and double discontinuity conditions

𝑦 (𝑑𝑖+ 0) = 𝛼𝑖𝑦 (𝑑𝑖− 0) ,

𝑦󸀠(𝑑𝑖+ 0) = 𝛼𝑖−1𝑦󸀠(𝑑𝑖− 0) − (𝛾𝑖𝜆 + 𝛽𝑖) 𝑦 (𝑑𝑖− 0) , (4) where𝑞(𝑥) is real valued function in 𝐿2(0, 1); ℎ𝑗and𝐻𝑗, 𝑗 =

0, 1, 2, are real numbers; 𝛼𝑖, 𝛾𝑖 ∈ R+, 𝛽𝑖 ∈ R, 𝑖 = 1, 2; 𝑑0 =

0, 𝑑1, 𝑑2∈ (0, 1), 𝑑3= 1; 𝜌1 := ℎ2− ℎ0ℎ1> 0, 𝜌2:= 𝐻0𝐻1−

𝐻2 > 0; and 𝜆 is a spectral parameter We denote the problem ()–(4) by𝐿 = 𝐿(𝑞, h, H, s1, s2), where h = (ℎ0, ℎ1, ℎ2), H =

(𝐻0, 𝐻1, 𝐻2), and s𝑖= (𝑑𝑖, 𝛼𝑖, 𝛾𝑖, 𝛽𝑖), 𝑖 = 1, 2

It is proven that the coefficients of the problem can

be uniquely determined by either Weyl function or given two different spectral sequences The obtained results are generalizations of the similar results for the classical Sturm-Liouville operator on a finite interval

2 Preliminaries

Let the functions𝜑(𝑥, 𝜆) and 𝜓(𝑥, 𝜆) be the solutions of (1

under the following initial conditions and the jump

condi-tions (4):

( 𝜑𝜑󸀠) (0, 𝜆) = ( −𝜆 + ℎ1

𝜆ℎ0− ℎ2) , ( 𝜓𝜓󸀠) (1, 𝜆) = ( −𝜆 + 𝐻1

𝜆𝐻0− 𝐻2)

(5)

These solutions are the entire functions of𝜆 and satisfy the relation𝜓(𝑥, 𝜆𝑛) = 𝛽𝑛𝜑(𝑥, 𝜆𝑛) for each eigenvalue 𝜆𝑛, where

𝛽𝑛= −(𝜓󸀠(0, 𝜆𝑛) + ℎ0𝜓(0, 𝜆𝑛))/𝜌1 The following asymptotics can be obtained from the integral equations given in the appendix:

𝜑 (𝑥, 𝜆) =

{ { { { { { { { {

−𝜆 cos √𝜆𝑥 + 𝑂 (√𝜆 exp |𝜏| 𝑥) , 𝑥 < 𝑑1,

𝛾1𝜆2[sin √𝜆𝑥 2√𝜆 −

sin √𝜆 (2𝑑1− 𝑥) 2√𝜆 ] + 𝑂 (𝜆 exp |𝜏| 𝑥) , 𝑑1< 𝑥 < 𝑑2,

𝛾1𝛾2

4 𝜆2[cos √𝜆𝑥 + cos √𝜆 (2𝑑1− 𝑥) − cos √𝜆 (2𝑑2− 𝑥) − cos √𝜆 (2𝑑2− 2𝑑1− 𝑥)]

+𝑂 (𝜆3/2exp|𝜏| 𝑥) , 𝑥 > 𝑑2,

(6)

𝜑󸀠(𝑥, 𝜆) =

{ { { { { { { { {

𝜆3/2sin √𝜆𝑥 + 𝑂 (𝜆 exp |𝜏| 𝑥) , 𝑥 < 𝑑1,

𝛾1𝜆2

2 [cos √𝜆𝑥 + cos √𝜆 (2𝑑1− 𝑥)] + 𝑂 (𝜆3/2exp|𝜏| 𝑥) , 𝑑1< 𝑥 < 𝑑2,

−𝛾1𝛾2

4 𝜆5/2{sin √𝜆𝑥 − sin √𝜆 (2𝑑1− 𝑥) + sin √𝜆 (2𝑑2− 𝑥) + sin √𝜆 (2𝑑2− 2𝑑1− 𝑥)}

+𝑂 (𝜆2exp|𝜏| 𝑥) , 𝑥 > 𝑑2,

(7)

where𝜏 = Im √𝜆

The values of the parameter 𝜆 for which the problem

𝐿 has nonzero solutions are called eigenvalues, and the

corresponding nontrivial solutions are called eigenfunctions

The characteristic functionΔ(𝜆) and norming constants

𝛼𝑛of the problem𝐿 are defined as follows:

Δ (𝜆) = 𝑊 [𝜑, 𝜓]

= 𝜆 (𝜑󸀠(1, 𝜆) + 𝐻0𝜑 (1, 𝜆)) − 𝐻1𝜑󸀠(1, 𝜆) − 𝐻2𝜑 (1, 𝜆)

= − 𝜆 (𝜓󸀠(0, 𝜆)+ℎ0𝜓 (0, 𝜆))+ℎ1𝜓󸀠(0, 𝜆) + ℎ2𝜓 (0, 𝜆) ,

𝛼𝑛 := ∫1

0 𝜑2(𝑥, 𝜆𝑛) 𝑑𝑥 + 1

𝜌1(𝜑󸀠(0, 𝜆𝑛) + ℎ0𝜑 (0, 𝜆𝑛))

2

+ 1

𝜌2(𝜑󸀠(1, 𝜆𝑛) + 𝐻0𝜑 (1, 𝜆𝑛))

2

+ 𝛼1𝛾1𝜑2(𝑑1− 0, 𝜆𝑛) + 𝛼2𝛾2𝜑2(𝑑2− 0, 𝜆𝑛)

(8)

It is obvious thatΔ(𝜆) is an entire function in 𝜆 and the zeros,

namely, {𝜆𝑛} of Δ(𝜆) coincide with the eigenvalues of the

problem𝐿 Now, from (6) and (8), we can write

Δ (𝜆) = −𝛾1𝛾2

4 𝜆4

× [sin √𝜆

√𝜆 −

sin √𝜆 (2𝑑1− 1)

√𝜆 +

sin √𝜆 (2𝑑2− 1)

√𝜆 +sin √𝜆 (2𝑑2− 2𝑑1− 1)

√𝜆 ] + 𝑂 (𝜆3exp|𝜏|)

(9)

Lemma 1 See the following.

(i) All eigenvalues of the problem 𝐿 are real and

alge-braically simple; that is,Δ󸀠(𝜆𝑛) ̸= 0.

(ii) Two eigenfunctions𝜑(𝑥, 𝜆1) and 𝜑(𝑥, 𝜆2),

correspond-ing to different eigenvalues𝜆1 and𝜆2, are orthogonal

in the sense of

∫1

0 𝜑 (𝑥, 𝜆1) 𝜑 (𝑥, 𝜆2) 𝑑𝑥 + 1

𝜌1(𝜑󸀠(0, 𝜆1) + ℎ0𝜑 (0, 𝜆1)) (𝜑󸀠(0, 𝜆2) + ℎ0𝜑 (0, 𝜆2)) + 1

𝜌2(𝜑󸀠(1, 𝜆1) + 𝐻0𝜑 (1, 𝜆1)) (𝜑󸀠(1, 𝜆2) + 𝐻0𝜑 (1, 𝜆2)) + 𝛼1𝛾1𝜑 (𝑑1− 0, 𝜆1) 𝜑 (𝑑1− 0, 𝜆2)

+ 𝛼2𝛾2𝜑 (𝑑2− 0, 𝜆1) 𝜑 (𝑑2− 0, 𝜆2) = 0

(10)

Proof Consider a Hilbert Space𝐻 = 𝐿2(0, 1) ⊕ C4, equipped

with the inner product

⟨𝑌, 𝑍⟩ := ∫1

0 𝑦 (𝑥) 𝑧 (𝑥)𝑑𝑥 +𝜌1

1𝑌1𝑍1 +𝜌1

2𝑌2𝑍2+ 𝛼1𝛾1𝑌3𝑍3+ 𝛼2𝛾2𝑌4𝑍4

(11)

for𝑌 = (𝑦(𝑥), 𝑌1, 𝑌2, 𝑌3, 𝑌4)𝑇, 𝑍 = (𝑧(𝑥), 𝑍1, 𝑍2, 𝑍3, 𝑍4)𝑇 ∈

𝐻

Define an operator𝑇 with the domain 𝐷(𝑇) = {𝑌 ∈ 𝐻 :

𝑦(𝑥), and 𝑦󸀠(𝑥) are absolutely continuous in 𝐼, ℓ𝑌 ∈ 𝐿2(0,

1), 𝑦(𝑑𝑖+0) = 𝛼𝑖𝑦(𝑑𝑖−0), 𝑌1= 𝑦󸀠(0)+ℎ0𝑦(0), 𝑌2= 𝑦󸀠

1(1)+

𝐻0𝑦(1), 𝑌3= 𝛾1𝑦(𝑑1− 0), and 𝑌4= 𝛾2𝑦(𝑑2− 0)} such that

𝑇 (𝑌):=((

(

−𝑦󸀠󸀠(𝑥) + 𝑞 (𝑥) 𝑦 (𝑥)

ℎ1𝑦󸀠(0) + ℎ2𝑦 (0)

𝐻1𝑦󸀠(1) + 𝐻2𝑦 (1)

−𝑦󸀠(𝑑1+ 0) + 𝛼−11 𝑦󸀠(𝑑1− 0) − 𝛽1𝑦 (𝑑1− 0)

−𝑦󸀠(𝑑2+ 0) + 𝛼−1

2 𝑦󸀠(𝑑2− 0) − 𝛽2𝑦 (𝑑2− 0)

) )

(12)

It is easily proven, using classical methods in the similar

works (see, e.g., [28]), that the operator𝑇 is symmetric in 𝐻;

the eigenvalue problem for the operator𝑇 and the problem 𝐿

coincide Therefore, all eigenvalues are real, and two different

eigenfunctions are orthogonal

Let us show the simplicity of the eigenvalues 𝜆𝑛 by

writting the following equations:

−𝜓󸀠󸀠(𝑥, 𝜆) + 𝑞 (𝑥) 𝜓 (𝑥, 𝜆) = 𝜆𝜓 (𝑥, 𝜆) ,

−𝜑󸀠󸀠(𝑥, 𝜆𝑛) + 𝑞 (𝑥) 𝜑 (𝑥, 𝜆𝑛) = 𝜆𝑛𝜑 (𝑥, 𝜆𝑛) (13)

If these equations are multiplied by 𝜑(𝑥, 𝜆𝑛) and 𝜓(𝑥, 𝜆),

respectively, subtracting them side by side and finally

inte-grating over the interval[0, 1], the equality

[𝜑󸀠(𝑥, 𝜆𝑛) 𝜓 (𝑥, 𝜆) − 𝜓󸀠(𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛)]

× (|𝑑1 −0

0 + |𝑑2 −0

𝑑1+0+ |1𝑑2+0)

= (𝜆 − 𝜆𝑛) ∫1

0 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛) 𝑑𝑥

(14)

is obtained Add and subtractΔ(𝜆) in the left-hand side of the

last equality, and use initial conditions (5) to get

Δ (𝜆) + (𝜆 − 𝜆𝑛) (𝜓󸀠(0, 𝜆) + ℎ0𝜓 (0, 𝜆))

− (𝜆 − 𝜆𝑛) (𝜑󸀠(1, 𝜆𝑛) + 𝐻0𝜑 (1, 𝜆𝑛))

− (𝜆 − 𝜆𝑛) 𝛼1𝛾1𝜓 (𝑑1− 0, 𝜆) 𝜑 (𝑑1− 0, 𝜆𝑛)

− (𝜆 − 𝜆𝑛) 𝛼2𝛾2𝜓 (𝑑2− 0, 𝜆) 𝜑 (𝑑2− 0, 𝜆𝑛)

= (𝜆 − 𝜆𝑛) ∫1

0 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛) 𝑑𝑥

(15)

Rewrite this equality as

Δ (𝜆)

𝜆 − 𝜆𝑛 = ∫

1

0 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛) 𝑑𝑥 + (𝜑󸀠(1, 𝜆𝑛) + 𝐻0𝜑 (1, 𝜆𝑛))

− (𝜓󸀠(0, 𝜆) + ℎ0𝜓 (0, 𝜆)) + 𝛼1𝛾1𝜓 (𝑑1− 0, 𝜆) 𝜑 (𝑑1− 0, 𝜆𝑛) + 𝛼2𝛾2𝜓 (𝑑2− 0, 𝜆) 𝜑 (𝑑2− 0, 𝜆𝑛)

(16)

= ∫1

0 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛) 𝑑𝑥

−(𝜓

󸀠(0, 𝜆)+ℎ0𝜓 (0, 𝜆))(𝜑󸀠(0, 𝜆𝑛)+ℎ0𝜑 (0, 𝜆𝑛))

ℎ0ℎ1− ℎ2 +(𝜑

󸀠(1, 𝜆𝑛)+𝐻0𝜑 (1, 𝜆𝑛))(𝜓󸀠(1, 𝜆)+𝐻0𝜓 (1, 𝜆))

𝐻0𝐻1− 𝐻2 + 𝛼1𝛾1𝜓 (𝑑1− 0, 𝜆) 𝜑 (𝑑1− 0, 𝜆𝑛) + 𝛼2𝛾2𝜓 (𝑑2− 0, 𝜆) 𝜑 (𝑑2− 0, 𝜆𝑛)

(17)

= ∫1

0 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆𝑛) 𝑑𝑥 +(𝜓

󸀠(0, 𝜆)+ℎ0𝜓 (0, 𝜆))(𝜑󸀠(0, 𝜆𝑛)+ℎ0𝜑 (0, 𝜆𝑛))

𝜌1 +(𝜑

󸀠(1, 𝜆𝑛)+𝐻0𝜑 (1, 𝜆𝑛))(𝜓󸀠(1, 𝜆)+𝐻0𝜓 (1, 𝜆))

𝜌2 + 𝛼1𝛾1𝜓 (𝑑1− 0, 𝜆) 𝜑 (𝑑1− 0, 𝜆𝑛) + 𝛼2𝛾2𝜓 (𝑑2− 0, 𝜆) 𝜑 (𝑑2− 0, 𝜆𝑛)

(18)

As𝜆 → 𝜆𝑛,

Δ󸀠(𝜆𝑛) = 𝛽𝑛𝛼𝑛 (19)

is obtained by using the equality𝜓(𝑥, 𝜆𝑛) = 𝛽𝑛𝜑(𝑥, 𝜆𝑛) Thus,

Δ󸀠(𝜆𝑛) ̸= 0

3 Main Results

We consider three statements of the inverse problem for the boundary value problem𝐿; from the Weyl function, from the spectral data{𝜆𝑛, 𝛼𝑛}𝑛≥0, and from two spectra {𝜆𝑛, 𝜇𝑛}𝑛≥0 For studying the inverse problem, we consider a boundary value problem ̃𝐿, together with 𝐿, of the same form but with different coefficients̃𝑞(𝑥), ̃h, ̃ H, ̃s𝑖, 𝑖 = 1, 2

Let the function𝜘(𝑥, 𝜆) denote the solution of (1) under the initial conditions 𝜘(0, 𝜆) = 𝜌1−1, 𝜘󸀠(0, 𝜆) = −𝜌1−1ℎ0

and the jump conditions (4) It is clear that the function𝜓(𝑥,

𝜆) can be represented by

𝜓 (𝑥, 𝜆) = Δ (𝜆) 𝜘 (𝑥, 𝜆) − 𝜓󸀠(0, 𝜆) + ℎ0𝜓 (0, 𝜆)

𝜌1 𝜑 (𝑥, 𝜆)

(20) Denote

𝑚 (𝜆) := 𝜓󸀠(0, 𝜆) + ℎ𝜌 0𝜓 (0, 𝜆)

1Δ (𝜆) . (21) Then, we have

𝜓 (𝑥, 𝜆)

Δ (𝜆) = 𝜘 (𝑥, 𝜆) − 𝑚 (𝜆) 𝜑 (𝑥, 𝜆) (22)

The function𝑚(𝜆) is called Weyl function [32]

Theorem 2 If 𝑚(𝜆) = ̃ 𝑚(𝜆), then 𝐿 = ̃𝐿; that is, 𝑞(𝑥) = ̃𝑞(𝑥),

always everywhere in𝐼; h = ̃h, H = ̃H, and s𝑖= ̃s𝑖, 𝑖 = 1, 2.

Proof Let us define the functions 𝑃1(𝑥, 𝜆) and 𝑃2(𝑥, 𝜆) as

follows:

𝑃1(𝑥, 𝜆) = 𝜑 (𝑥, 𝜆) ̃Φ󸀠(𝑥, 𝜆) − Φ (𝑥, 𝜆) ̃𝜑󸀠(𝑥, 𝜆) ,

𝑃2(𝑥, 𝜆) = Φ (𝑥, 𝜆) ̃𝜑 (𝑥, 𝜆) − 𝜑 (𝑥, 𝜆) ̃Φ (𝑥, 𝜆) , (23)

whereΦ(𝑥, 𝜆) = 𝜓(𝑥, 𝜆)/Δ(𝜆) If 𝑚(𝜆) = ̃𝑚(𝜆), then from

(22)-(23), 𝑃1(𝑥, 𝜆) and 𝑃2(𝑥, 𝜆) are entire functions in 𝜆

Denote𝐺𝛿 = {𝜆 : 𝜆 = 𝑘2, |𝑘 − 𝑘𝑛| > 𝛿, 𝑛 = 1, 2, } and

̃𝛿 = {𝜆 : 𝜆 = 𝑘2, |𝑘 − ̃𝑘𝑛| > 𝛿, 𝑛 = 1, 2, }, where 𝛿 is

sufficiently small number and𝑘𝑛and ̃𝑘𝑛are square roots of

the eigenvalues of the problem𝐿 and ̃𝐿, respectively One can

easily show that the asymptotics

Φ (𝑥, 𝜆) = 𝑂 (𝜆−(𝑖+3)/2exp(− |𝜏| 𝑥)) ,

Φ󸀠(𝑥, 𝜆) = 𝑂 (𝜆−(𝑖+2)/2exp(− |𝜏| 𝑥)) (24)

are valid for𝑑𝑖 < 𝑥 < 𝑑𝑖+1, 𝑖 = 0, 1, 2, and sufficiently large

|𝜆| in 𝐺𝛿 ∩ ̃𝐺𝛿 Thus, the following inequalities are obtained

from (6) and (24):

󵄨󵄨󵄨󵄨𝑃1(𝑥, 𝜆)󵄨󵄨󵄨󵄨 ≤ 𝐶𝛿, 󵄨󵄨󵄨󵄨𝑃2(𝑥, 𝜆)󵄨󵄨󵄨󵄨 ≤ 𝐶𝛿|𝜆|−1/2,

𝜆 ∈ 𝐺𝛿∩ ̃𝐺𝛿 (25) According to the last inequalities and Liouville’s theorem,

𝑃1(𝑥, 𝜆) = 𝐴(𝑥) and 𝑃2(𝑥, 𝜆) = 0 Use (23) again to take

𝜑 (𝑥, 𝜆) = 𝐴 (𝑥) ̃𝜑 (𝑥, 𝜆) ,

Φ (𝑥, 𝜆) = 𝐴 (𝑥) ̃Φ (𝑥, 𝜆) (26) Since𝑊[Φ(𝑥, 𝜆), 𝜑(𝑥, 𝜆)] = 1 and similarly 𝑊[̃Φ(𝑥, 𝜆), ̃𝜑(𝑥,

𝜆)] = 1, then 𝐴2(𝑥) = 1

On the other hand, the asymptotic expressions

𝜑 (𝑥, 𝜆) = 𝐶 (𝜆) exp (−𝑖√𝜆𝑥) (1 + 𝑜 (1)) ,

𝜑 (𝑥, 𝜆) = ̃𝐶 (𝜆) exp (−𝑖√𝜆𝑥) (1 + 𝑜 (1)) (27)

are valid for √𝜆 → ∞ on the imaginary axis, where

𝐶 (𝜆) =

{ { { { { { {

−1

2𝜆, 0 < 𝑥 < 𝑑1,

𝛾1

4𝜆3/2, 𝑑1< 𝑥 < 𝑑2,

𝛾1𝛾2

8 𝜆2, 𝑑2< 𝑥 < 1,

̃

𝐶 (𝜆) =

{ { { { { { {

−12𝜆, 0 < 𝑥 < ̃𝑑1,

̃𝛾1

4𝜆3/2, 𝑑̃1< 𝑥 < ̃𝑑2,

̃𝛾1̃𝛾2

8 𝜆2, ̃𝑑2< 𝑥 < 1.

(28)

Assume that𝑑1 ̸= ̃𝑑1and𝑑2 ̸= ̃𝑑2 There are six different cases for the permutation of the numbers𝑑𝑖and ̃𝑑𝑖 Without loss of generality, let0 < 𝑑1< ̃𝑑1< 𝑑2< ̃𝑑2< 1

From (26)-(27), we get𝛾1 = ̃𝛾1, 𝛾2 = ̃𝛾2, and𝐴(𝑥) ≡ 1, while𝑥 ∈ [0, 𝑑1) ∪ ( ̃𝑑1, 𝑑2) ∪ ( ̃𝑑2, 1]

Moreover, we get

2𝜆−1/2(1 + 𝑜 (1)) 𝐴 (𝑥) + 𝛾1= 𝑜 (1) , (29) while𝑥 ∈ (𝑑1, ̃𝑑1) By taking limit in (29) as|𝜆| → ∞, we condradict𝛾1 > 0 Thus, 𝑑1 = ̃𝑑1 Similarly,𝑑2 = ̃𝑑2, and 𝐴(𝑥) = 1 in 𝐼 Hence,

𝜑 (𝑥, 𝜆) = ̃𝜑 (𝑥, 𝜆) , 𝜓𝜓 (𝑥, 𝜆)󸀠(𝑥, 𝜆) = ̃𝜓󸀠(𝑥, 𝜆)

̃𝜓 (𝑥, 𝜆). (30)

It can be obtained from (1), (4), and (5) that𝑞(𝑥) = ̃𝑞(𝑥), a.e

in𝐼; s𝑖 = ̃s𝑖, 𝑖 = 1, 2, and h = ̃h, H = ̃ H Consequently,

𝐿 = ̃𝐿

Theorem 3 If {𝜆𝑛, 𝛼𝑛}𝑛≥0= {̃𝜆𝑛, ̃𝛼𝑛}𝑛≥0, then 𝐿 = ̃𝐿.

Proof The meromorphic function𝑚(𝜆) has simple poles at

𝜆𝑛, and its residues at these poles are

Res{𝑚 (𝜆) , 𝜆𝑛} = 𝜓󸀠(0, 𝜆𝑛) + ℎ0𝜓 (0, 𝜆𝑛)

𝜌1Δ󸀠(𝜆𝑛)

= − 𝛽𝑛

Δ󸀠(𝜆𝑛) = −

1

𝛼𝑛.

(31)

DenoteΓ𝑛 = {𝜆 : |𝜆| = (√𝜆𝑛+ 𝜀)2}, where 𝜀 is sufficiently small number Consider the contour integral

𝐹𝑛(𝜆) =2𝜋𝑖1 ∫

Γ𝑛

𝑚 (𝜂) (𝜂 − 𝜆)𝑑𝜂, 𝜆 ∈ int Γ𝑛. (32) There exists a constant 𝐶𝛿 > 0 such that Δ(𝜆) ≥

|𝜆|7/2𝐶𝛿exp|𝜏| holds for 𝜆 ∈ 𝐺𝛿 Use this inequality and (21)

to get|𝑚(𝜆)| ≤ 𝐶𝛿/|𝜆|3/2, for𝜆 ∈ 𝐺𝛿 Hence, lim𝑛 → ∞𝐹𝑛(𝜆) =

0, and so

𝑚 (𝜆) =∑∞

𝑛=0

1

𝛼𝑛(𝜆𝑛− 𝜆) (33)

is obtained from residue theorem Consequently, if𝜆𝑛 = ̃𝜆𝑛

and𝛼𝑛 = ̃𝛼𝑛for all𝑛, then from (33),𝑚(𝜆) = ̃𝑚(𝜆) Hence,

Theorem2yields𝐿 = ̃𝐿

We consider the boundary value problem 𝐿1 with the

condition

𝑦󸀠(0, 𝜆) + ℎ0𝑦 (0, 𝜆) = 0 (34) instead of (2) in 𝐿 Let {𝜂2

𝑛}𝑛≥0 be the eigenvalues of the problem 𝐿1 It is obvious that 𝜂𝑛 are zeros of Δ1(𝜂) :=

𝜓󸀠(0, 𝜂) + ℎ0𝜓(0, 𝜂)

Theorem 4 If {𝜆𝑛, 𝜂𝑛}𝑛≥0= {̃𝜆𝑛, ̃𝜂𝑛}𝑛≥0and h = ̃h, then 𝐿 =

̃𝐿.

Proof The functionsΔ(𝜆) and Δ1(𝜂) which are entire of order

1/2 can be represented by Hadamard’s factorization theorem

as follows:

Δ (𝜆) = 𝐶∏∞

𝑛=0

(1 − 𝜆

𝜆𝑛) ,

Δ1(𝜂) = 𝐶1∏∞

𝑛=0

(1 − 𝜂

𝜂𝑛) ,

(35)

where𝐶 and 𝐶1are constants which depend only on{𝜆𝑛} and

{𝜂𝑛}, respectively Therefore, Δ(𝜆) ≡ ̃Δ(𝜆) and Δ1(𝜂) ≡ ̃Δ1(𝜂),

when 𝜆𝑛 = ̃𝜆𝑛 and 𝜂𝑛 = ̃𝜂𝑛 for all 𝑛 Thus, 𝜓󸀠(0, 𝜂) +

ℎ0𝜓(0, 𝜂) = ̃𝜓󸀠(0, 𝜂) + ℎ0̃𝜓(0, 𝜂) Moreover, 𝜌1 = ̃𝜌1 since

h = ̃h Consequently, the equality (21) yields𝑚(𝜆) ≡ ̃𝑚(𝜆)

Hence, the proof is completed by Theorem2

Appendix

The solution𝜑(𝑥, 𝜆) satisfies the following integral equations

If𝑥 < 𝑑1,

𝜑 (𝑥, 𝜆) = (𝜆ℎ0− ℎ2)sin √𝜆𝑥

√𝜆 + (ℎ1− 𝜆) cos √𝜆𝑥 + 1

√𝜆∫

𝑥

0 sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡;

(A.1)

if𝑑1< 𝑥 < 𝑑2,

𝜑 (𝑥, 𝜆) = 𝜆ℎ0− ℎ2

√𝜆

× (𝛼+

1sin √𝜆𝑥 + 𝛼−

1sin √𝜆 (2𝑑1− 𝑥)) + (ℎ1− 𝜆)

× (𝛼+1cos √𝜆𝑥 + 𝛼−1cos √𝜆 (2𝑑1− 𝑥))

+(𝛾1𝜆 + 𝛽1) (ℎ1− 𝜆)

2√𝜆

× (sin √𝜆 (2𝑑1− 𝑥) − sin √𝜆𝑥) +(𝛾1𝜆 + 𝛽1) (𝜆ℎ0− ℎ2)

2𝜆

× (cos √𝜆𝑥 − cos √𝜆 (2𝑑1− 𝑥)) +(𝛾1𝜆 + 𝛽1)

2𝜆

× ∫𝑑1

0 (cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1− 𝑥 − 𝑡))

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + 1

√𝜆∫

𝑑 1

0 (𝛼1+sin √𝜆 (𝑥 − 𝑡)

+𝛼−1sin √𝜆 (2𝑑1− 𝑥 − 𝑡))

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + 1

√𝜆∫

𝑥

𝑑1sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡;

(A.2)

if𝑥 > 𝑑2,

𝜑 (𝑥, 𝜆)

= −(ℎ1− 𝜆) (𝛾1𝜆 + 𝛽1)

2√𝜆

× [𝛼+2(sin √𝜆𝑥 − sin √𝜆 (2𝑑1− 𝑥)) + 𝛼−

2( sin √𝜆 (2𝑑2− 𝑥) + sin √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

+(𝜆ℎ0− ℎ2) (𝛾1𝜆 + 𝛽1)

2𝜆

× [𝛼+2(cos √𝜆𝑥 − cos √𝜆 (2𝑑1− 𝑥)) + 𝛼−2(cos √𝜆 (2𝑑1− 𝑥)

− cos √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

+ (ℎ1− 𝜆)

× [𝛼+1(𝛼+2cos √𝜆𝑥 + 𝛼−2cos √𝜆 (2𝑑2− 𝑥)) + 𝛼−1(𝛼2+cos √𝜆 (2𝑑1− 𝑥)

+ 𝛼−2cos √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

+(𝜆ℎ0− ℎ2)

√𝜆

× [𝛼+1(𝛼2+sin √𝜆𝑥 + 𝛼2−sin √𝜆 (2𝑑2− 𝑥))

+ 𝛼1−(𝛼2+sin √𝜆 (2𝑑1− 𝑥)

−𝛼−

2sin √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

−(ℎ1− 𝜆) (𝛾1𝜆 + 𝛽1) (𝛾2𝜆 + 𝛽2)

4𝜆

× [cos √𝜆𝑥 + cos √𝜆 (2𝑑1− 𝑥)

− cos √𝜆 (2𝑑2− 𝑥)

− cos √𝜆 (2𝑑2− 2𝑑1− 𝑥)]

−(𝜆ℎ0− ℎ2) (𝛾1𝜆 + 𝛽1) (𝛾2𝜆 + 𝛽2)

4𝜆3/2

× [sin √𝜆𝑥 − sin √𝜆 (2𝑑1− 𝑥)

− sin √𝜆 (2𝑑2− 𝑥)

+ sin √𝜆 (2𝑑2− 2𝑑1− 𝑥)]

+(ℎ1− 𝜆) (𝛾2𝜆 + 𝛽2)

2√𝜆

× [𝛼+1(sin √𝜆 (2𝑑2− 𝑥) − sin √𝜆𝑥)

+ 𝛼1−(sin √𝜆 (2𝑑1− 𝑥)

+ sin √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

+(𝜆ℎ0− ℎ2) (𝛾2𝜆 + 𝛽2)

2𝜆

× [𝛼+1(cos √𝜆𝑥 − cos √𝜆 (2𝑑2− 𝑥))

− 𝛼1−(cos √𝜆 (2𝑑1− 𝑥)

− cos √𝜆 (2𝑑2− 2𝑑1− 𝑥))]

+ 𝛼2

2√𝜆∫

𝑑 1

0 [𝛼1+(sin √𝜆 (2𝑑2− 𝑥 − 𝑡)

+ sin √𝜆 (𝑥 − 𝑡)) + 𝛼1−(sin √𝜆 (2𝑑1− 𝑥 − 𝑡)

− sin √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡))]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 +𝛼2(𝛾14𝜆𝜆 + 𝛽1)

× ∫𝑑1

0 [cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1− 𝑥 − 𝑡) + cos √𝜆 (2𝑑2− 𝑥 − 𝑡)

+ cos √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡)]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 − 1

2𝛼2√𝜆

× ∫𝑑2

0 [𝛼1+(sin √𝜆 (2𝑑2− 𝑥 − 𝑡) − sin √𝜆 (𝑥 − 𝑡))

− 𝛼1−(sin √𝜆 (2𝑑1− 𝑥 − 𝑡)

+ sin √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡))]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 +(𝛾1𝜆 + 𝛽1)

4𝛼2𝜆

× ∫𝑑2

0 [cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1− 𝑥 − 𝑡)

− cos √𝜆 (2𝑑2− 𝑥 − 𝑡) + cos √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡)]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡

− 1

√𝜆∫

𝑑 2

𝑑 1

[𝛼+2sin √𝜆 (𝑥 − 𝑡) +𝛼−

2sin √𝜆 (2𝑑2− 𝑥 − 𝑡)]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 +(𝛾2𝜆 + 𝛽2)

2𝜆

× ∫𝑑1

0 [𝛼1+(cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑2− 𝑥 − 𝑡))

− 𝛼−

1(cos √𝜆 (2𝑑1− 𝑥 − 𝑡)

− cos √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡))]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡

−(𝛾1𝜆 + 𝛽1) (𝛾2𝜆 + 𝛽2)

4𝜆3/2

× ∫𝑑1

0 [sin √𝜆 (𝑥 − 𝑡) − sin √𝜆 (2𝑑1− 𝑥 − 𝑡)

− sin √𝜆 (2𝑑2− 𝑥 − 𝑡) + sin √𝜆 (2𝑑2− 2𝑑1− 𝑥 + 𝑡)]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡

−(𝛾2𝜆 + 𝛽2) 2𝜆

× ∫𝑑2

𝑑 1

[cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑2− 𝑥 − 𝑡)]

× 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡

+ 1

√𝜆∫

𝑥

𝑑2sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡,

(A.3) where𝛼±𝑖 = (1/2)(𝛼𝑖± 1/𝛼𝑖), 𝑖 = 1, 2

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