Double Discontinuous Inverse Problems for Sturm-Liouville Operator with Parameter-Dependent Conditions

Abstract

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 consist 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 Ambartsumyan 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=0𝑑_𝑖,𝑑_𝑖+1 (1) subject to the boundary conditions 𝑈𝑦∶=𝜆𝑦^′(0)+ℎ₀𝑦(0)−ℎ₁𝑦^′(0)−ℎ₂𝑦(0)𝑉𝑦𝑦=0,∶=𝜆^′(1)+𝐻₀𝑦(1)−𝐻₁𝑦^′(1)−𝐻₂𝑦(1)=0 (2)(3) and double discontinuity conditions 𝑦𝑑_𝑖+0=𝛼_𝑖𝑦𝑑_𝑖,𝑦−0_′𝑑_𝑖+0=𝛼_𝑖⁻¹𝑦^′𝑑_𝑖−0−𝛾_𝑖𝜆+𝛽_𝑖𝑦𝑑_𝑖−0, (4) where 𝑞(𝑥) is real valued function in 𝐿₂(0,1); ℎ_𝑗 and 𝐻_𝑗, 𝑗=0,1,2, are real numbers; 𝛼_𝑖,𝛾_𝑖∈ℝ⁺, 𝛽_𝑖∈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 (1)–(4) by 𝐿=𝐿(𝑞,𝐡,𝐇,𝐬₁,𝐬₂), where 𝐡=(ℎ₀,ℎ₁,ℎ₂),  𝐇=(𝐻₀,𝐻₁,𝐻₂), and 𝐬_𝑖=(𝑑_𝑖,𝛼_𝑖,𝛾_𝑖,𝛽_𝑖), 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 functions 𝜑(𝑥,𝜆) and 𝜓(𝑥,𝜆) be the solutions of (1) under the following initial conditions and the jump conditions (4): 𝜑𝜑^′(0,𝜆)=−𝜆+ℎ₁𝜆ℎ₀−ℎ₂,𝜓𝜓^′(1,𝜆)=−𝜆+𝐻₁𝜆𝐻₀−𝐻₂. (5) These solutions are the entire functions of 𝜆 and satisfy the relation 𝜓(𝑥,𝜆_𝑛)=𝜷_𝑛𝜑(𝑥,𝜆_𝑛) for each eigenvalue 𝜆_𝑛, where 𝜷_𝑛=−(𝜓^′(0,𝜆_𝑛)+ℎ₀𝜓(0,𝜆_𝑛))/𝜌₁.

The following asymptotics can be obtained from the integral equations given in the appendix:𝜑(𝑥,𝜆)=^{√−𝜆cos√𝜆𝑥+𝑂|𝜏|𝑥𝜆exp,𝑥<𝑑1,𝛾1𝜆2√sin𝜆𝑥2√𝜆−√sin𝜆2𝑑1−𝑥2√𝜆|𝜏|𝑥+𝑂𝜆exp,𝑑1<𝑥<𝑑2,}_{𝛾1𝛾24𝜆2√cos√𝜆𝑥+cos𝜆2𝑑1√−𝑥−cos𝜆2𝑑2√−𝑥−cos𝜆2𝑑2−2𝑑1𝜆−𝑥+𝑂3/2|𝜏|𝑥exp,𝑥>𝑑2,}𝜑^′(𝑥,𝜆)=^{𝜆3/2√sin|𝜏|𝑥𝜆𝑥+𝑂𝜆exp,𝑥<𝑑1,𝛾1𝜆22√cos√𝜆𝑥+cos𝜆2𝑑1𝜆−𝑥+𝑂3/2|𝜏|𝑥exp,𝑑1<𝑥<𝑑2,}_{−𝛾1𝛾24𝜆5/2√sin√𝜆𝑥−sin𝜆2𝑑1√−𝑥+sin𝜆2𝑑2√−𝑥+sin𝜆2𝑑2−2𝑑1𝜆−𝑥+𝑂2|𝜏|𝑥exp,𝑥>𝑑2,} (6)(7)

where 𝜏=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 function Δ(𝜆) and norming constants 𝛼_𝑛 of the problem 𝐿 are defined as follows: Δ(𝜆)𝜑=𝑊𝜑,𝜓=𝜆^′(1,𝜆)+𝐻₀𝜑(1,𝜆)−𝐻₁𝜑^′(1,𝜆)−𝐻₂𝜑()𝜓1,𝜆=−𝜆^′(0,𝜆)+ℎ₀𝜓(0,𝜆)+ℎ₁𝜓^′(0,𝜆)+ℎ₂𝜓(),𝜶0,𝜆_𝑛∶=¹₀𝜑²𝑥,𝜆_𝑛1𝑑𝑥+𝜌₁𝜑^′0,𝜆_𝑛+ℎ₀𝜑0,𝜆_𝑛²+1𝜌₂𝜑^′1,𝜆_𝑛+𝐻₀𝜑1,𝜆_𝑛²+𝛼₁𝛾₁𝜑²𝑑₁−0,𝜆_𝑛+𝛼₂𝛾₂𝜑²𝑑₂−0,𝜆_𝑛. (8) It is obvious that Δ(𝜆) is an entire function in 𝜆 and the zeros, namely, {𝜆_𝑛} of Δ(𝜆) coincide with the eigenvalues of the problem 𝐿. Now, from (6) and (8), we can write Δ(𝜆)𝛾=−¹𝛾₂4𝜆⁴×sin√𝜆√𝜆−sin√𝜆2𝑑₁−1√𝜆+sin√𝜆2𝑑₂−1√𝜆+sin√𝜆2𝑑₂−2𝑑₁−1√𝜆+𝑂𝜆³exp|𝜏|. (9)

Lemma 1. See the following.(i)All eigenvalues of the problem 𝐿 are real and algebraically simple; that is, Δ^′(𝜆_𝑛)≠0.(ii)Two eigenfunctions 𝜑(𝑥,𝜆₁) and 𝜑(𝑥,𝜆₂), corresponding to different eigenvalues 𝜆₁ and 𝜆₂, are orthogonal in the sense of ¹₀𝜑𝑥,𝜆₁𝜑𝑥,𝜆₂+1𝑑𝑥𝜌₁𝜑^′0,𝜆₁+ℎ₀𝜑0,𝜆₁𝜑^′0,𝜆₂+ℎ₀𝜑0,𝜆₂+1𝜌₂𝜑^′1,𝜆₁+𝐻₀𝜑1,𝜆₁𝜑^′1,𝜆₂+𝐻₀𝜑1,𝜆₂+𝛼₁𝛾₁𝜑𝑑₁−0,𝜆₁𝜑𝑑₁−0,𝜆₂+𝛼₂𝛾₂𝜑𝑑₂−0,𝜆₁𝜑𝑑₂−0,𝜆₂=0. (10)

Proof. Consider a Hilbert Space 𝐻=𝐿₂(0,1)⊕ℂ⁴, equipped with the inner product ⟨𝑌,𝑍⟩∶=¹₀𝑦(𝑥)𝑧(𝑥)1𝑑𝑥+𝜌₁𝑌₁𝑍₁+1𝜌₂𝑌₂𝑍₂+𝛼₁𝛾₁𝑌₃𝑍₃+𝛼₂𝛾₂𝑌₄𝑍₄ (11) for 𝑌=(𝑦(𝑥),𝑌₁,𝑌₂,𝑌₃,𝑌₄)^𝑇,𝑍=(𝑧(𝑥),𝑍₁,𝑍₂,𝑍₃,𝑍₄)^𝑇∈𝐻.
Define an operator 𝑇 with the domain 𝐷(𝑇)={𝑌∈𝐻∶𝑦(𝑥), and 𝑦^′(𝑥) are absolutely continuous in 𝐼,ℓ𝑌∈𝐿₂(0,1),𝑦(𝑑_𝑖+0)=𝛼_𝑖𝑦(𝑑_𝑖−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)(𝑥)𝑦(𝑥)ℎ+𝑞₁𝑦^′(0)+ℎ₂𝑦(0)𝐻₁𝑦^′(1)+𝐻₂𝑦(1)−𝑦^′𝑑₁+0+𝛼₁⁻¹𝑦^′𝑑₁−0−𝛽₁𝑦𝑑₁−0−𝑦^′𝑑₂+0+𝛼₂⁻¹𝑦^′𝑑₂−0−𝛽₂𝑦𝑑₂−0. (12) It is easily proven, using classical methods in the similar works (see, e.g., [28]), that the operator 𝑇 is symmetric in 𝐻; the eigenvalue problem for the operator 𝑇 and the problem 𝐿 coincide. Therefore, all eigenvalues are real, and two different eigenfunctions are orthogonal.
Let us show the simplicity of the eigenvalues 𝜆_𝑛 by writting the following equations: −𝜓^′′(𝑥,𝜆)+𝑞(𝑥)𝜓(𝑥,𝜆)=𝜆𝜓(𝑥,𝜆),−𝜑^′′𝑥,𝜆_𝑛+𝑞(𝑥)𝜑𝑥,𝜆_𝑛=𝜆_𝑛𝜑𝑥,𝜆_𝑛. (13) If these equations are multiplied by 𝜑(𝑥,𝜆_𝑛) and 𝜓(𝑥,𝜆), respectively, subtracting them side by side and finally integrating over the interval [0,1], the equality 𝜑^′𝑥,𝜆_𝑛𝜓(𝑥,𝜆)−𝜓^′(𝑥,𝜆)𝜑𝑥,𝜆_𝑛×∣^𝑑1₀⁻⁰+∣^𝑑2_𝑑⁻⁰₁₊₀+∣¹_𝑑2+0=𝜆−𝜆_𝑛¹₀𝜓(𝑥,𝜆)𝜑𝑥,𝜆_𝑛𝑑𝑥 (14) is obtained. Add and subtract Δ(𝜆) in the left-hand side of the last equality, and use initial conditions (5) to get Δ(𝜆)+𝜆−𝜆_𝑛𝜓^′(0,𝜆)+ℎ₀𝜓()−0,𝜆𝜆−𝜆_𝑛𝜑^′1,𝜆_𝑛+𝐻₀𝜑1,𝜆_𝑛−𝜆−𝜆_𝑛𝛼₁𝛾₁𝜓𝑑₁−0,𝜆𝜑𝑑₁−0,𝜆_𝑛−𝜆−𝜆_𝑛𝛼₂𝛾₂𝜓𝑑₂−0,𝜆𝜑𝑑₂−0,𝜆_𝑛=𝜆−𝜆_𝑛¹₀𝜓(𝑥,𝜆)𝜑𝑥,𝜆_𝑛𝑑𝑥. (15) Rewrite this equality as Δ(𝜆)𝜆−𝜆_𝑛=¹₀𝜓(𝑥,𝜆)𝜑𝑥,𝜆_𝑛+𝜑𝑑𝑥_′1,𝜆_𝑛+𝐻₀𝜑1,𝜆_𝑛−𝜓^′(0,𝜆)+ℎ₀𝜓(0,𝜆)+𝛼₁𝛾₁𝜓𝑑₁−0,𝜆𝜑𝑑₁−0,𝜆_𝑛+𝛼₂𝛾₂𝜓𝑑₂−0,𝜆𝜑𝑑₂−0,𝜆_𝑛=¹₀𝜓(𝑥,𝜆)𝜑𝑥,𝜆_𝑛−𝜓𝑑𝑥_′(0,𝜆)+ℎ₀𝜓(0,𝜆)𝜑^′0,𝜆_𝑛+ℎ₀𝜑0,𝜆_𝑛ℎ₀ℎ₁−ℎ₂+𝜑^′1,𝜆_𝑛+𝐻₀𝜑1,𝜆_𝑛𝜓^′(1,𝜆)+𝐻₀𝜓(1,𝜆)𝐻₀𝐻₁−𝐻₂+𝛼₁𝛾₁𝜓𝑑₁−0,𝜆𝜑𝑑₁−0,𝜆_𝑛+𝛼₂𝛾₂𝜓𝑑₂−0,𝜆𝜑𝑑₂−0,𝜆_𝑛=¹₀𝜓(𝑥,𝜆)𝜑𝑥,𝜆_𝑛+𝜓𝑑𝑥_′(0,𝜆)+ℎ₀𝜓(0,𝜆)𝜑^′0,𝜆_𝑛+ℎ₀𝜑0,𝜆_𝑛𝜌₁+𝜑^′1,𝜆_𝑛+𝐻₀𝜑1,𝜆_𝑛𝜓^′(1,𝜆)+𝐻₀𝜓(1,𝜆)𝜌₂+𝛼₁𝛾₁𝜓𝑑₁−0,𝜆𝜑𝑑₁−0,𝜆_𝑛+𝛼₂𝛾₂𝜓𝑑₂−0,𝜆𝜑𝑑₂−0,𝜆_𝑛. (16)(17)(18) As 𝜆→𝜆_𝑛, Δ^′𝜆_𝑛=𝛽_𝑛𝜶_𝑛 (19) is obtained by using the equality 𝜓(𝑥,𝜆_𝑛)=𝜷_𝑛𝜑(𝑥,𝜆_𝑛). Thus, Δ^′(𝜆_𝑛)≠0.

3. Main Results

We consider three statements of the inverse problem for the boundary value problem 𝐿; from the Weyl function, from the spectral data {𝜆_𝑛,𝜶_𝑛}_𝑛≥0, and from two spectra {𝜆_𝑛,𝜇_𝑛}_𝑛≥0. For studying the inverse problem, we consider a boundary value problem 𝐿, together with 𝐿, of the same form but with different coefficients 𝑞(𝑥),𝐡,𝐇,𝐬_𝑖,𝑖=1,2.

Let the function 𝜘(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 function 𝜓(𝑥,𝜆) can be represented by 𝜓()(𝜆)𝜘()−𝜓𝑥,𝜆=Δ𝑥,𝜆^′(0,𝜆)+ℎ₀𝜓(0,𝜆)𝜌₁𝜑(𝑥,𝜆). (20) Denote 𝑚(𝜆)𝜓∶=^′(0,𝜆)+ℎ₀𝜓(0,𝜆)𝜌₁Δ(𝜆). (21) Then, we have 𝜓(𝑥,𝜆)Δ(𝜆)=𝜘(𝑥,𝜆)−𝑚(𝜆)𝜑(𝑥,𝜆). (22) The function 𝑚(𝜆) is called Weyl function [32].

Theorem 2. If 𝑚(𝜆)=𝑚(𝜆), then 𝐿=𝐿; that is, 𝑞(𝑥)=𝑞(𝑥), always everywhere in 𝐼;𝐡=𝐡,𝐇=𝐇, and 𝐬_𝑖=𝐬_𝑖̃,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^′(),𝑃𝑥,𝜆₂(𝑥,𝜆)=Φ(𝑥,𝜆)𝜑(𝑥,𝜆)−𝜑(𝑥,𝜆)Φ(𝑥,𝜆), (23) where Φ(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. Denote 𝐺_𝛿={𝜆∶𝜆=𝑘²,|𝑘−𝑘_𝑛|>𝛿,𝑛=1,2,…} and 𝐺_𝛿={𝜆∶𝜆=𝑘²,|𝑘−𝑘_𝑛|>𝛿,𝑛=1,2,…}, where 𝛿 is sufficiently small number and 𝑘_𝑛 and 𝑘_𝑛 are square roots of the eigenvalues of the problem 𝐿 and 𝐿, respectively. One can easily show that the asymptotics Φ(𝑥,𝜆)=𝑂𝜆^{−(𝑖+3)/2}(−|𝜏|𝑥),Φexp_′(𝑥,𝜆)=𝑂𝜆^{−(𝑖+2)/2}exp(−|𝜏|𝑥) (24) are valid for 𝑑_𝑖<𝑥<𝑑_𝑖+1, 𝑖=0,1,2, and sufficiently large |𝜆| in 𝐺_𝛿∩𝐺_𝛿. Thus, the following inequalities are obtained from (6) and (24): 𝑃₁(𝑥,𝜆)≤𝐶_𝛿,𝑃₂(𝑥,𝜆)≤𝐶_𝛿|𝜆|^−1/2,𝜆∈𝐺_𝛿∩𝐺_𝛿. (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)𝜑(),Φ()(𝑥)Φ().𝑥,𝜆=𝐴𝑥,𝜆𝑥,𝜆=𝐴𝑥,𝜆 (26) Since 𝑊[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√(1,𝜑()=𝐶(𝜆)√𝜆𝑥(1+𝑜))𝑥,𝜆exp−𝑖𝜆𝑥(1+𝑜(1)) (27) are valid for √𝜆→∞ on the imaginary axis, where 𝐶(𝜆)=−12𝜆,0<𝑥<𝑑₁,𝛾₁4𝜆^3/2,𝑑₁<𝑥<𝑑₂,𝛾₁𝛾₂8𝜆²,𝑑₂𝐶(𝜆)=−1<𝑥<1,2𝜆,0<𝑥<𝑑₁,𝛾₁4𝜆^3/2,𝑑₁<𝑥<𝑑₂,𝛾₁𝛾₂8𝜆²,𝑑₂<𝑥<1. (28) Assume that 𝑑₁≠𝑑₁ and 𝑑₂≠𝑑₂. There are six different cases for the permutation of the numbers 𝑑_𝑖 and 𝑑_𝑖. Without loss of generality, let 0<𝑑₁<𝑑₁<𝑑₂<𝑑₂<1.
From (26)-(27), we get 𝛾₁=𝛾₁,𝛾₂=𝛾₂, and 𝐴(𝑥)≡1, while 𝑥∈[0,𝑑₁)∪(𝑑₁,𝑑₂)∪(𝑑₂,1].
Moreover, we get 2𝜆^−1/2(1+𝑜(1))𝐴(𝑥)+𝛾₁=𝑜(1), (29) while 𝑥∈(𝑑₁,𝑑₁). By taking limit in (29) as |𝜆|→∞, we condradict 𝛾₁>0. Thus, 𝑑₁=𝑑₁. Similarly, 𝑑₂=𝑑₂, and 𝐴(𝑥)=1 in 𝐼. Hence, 𝜑()=𝜑(),𝜓𝑥,𝜆𝑥,𝜆^′(𝑥,𝜆)𝜓()=𝜓𝑥,𝜆^′(𝑥,𝜆)𝜓().𝑥,𝜆 (30) It can be obtained from (1), (4), and (5) that 𝑞(𝑥)=𝑞(𝑥), a.e. in 𝐼; 𝐬_𝑖=𝐬_𝑖̃, 𝑖=1,2, and 𝐡=𝐡,𝐇=𝐇. Consequently, 𝐿=𝐿.

Theorem 3. If {𝜆_𝑛,𝜶_𝑛}_𝑛≥0={𝜆_𝑛,𝜶_𝑛}_𝑛≥0, then 𝐿=𝐿.

Proof. The meromorphic function 𝑚(𝜆) has simple poles at 𝜆_𝑛, and its residues at these poles are Res𝑚(𝜆),𝜆_𝑛=𝜓^′0,𝜆_𝑛+ℎ₀𝜓0,𝜆_𝑛𝜌₁Δ^′𝜆_𝑛𝜷=−^𝑛Δ^′𝜆_𝑛1=−𝜶_𝑛. (31) Denote Γ_𝑛={𝜆∶|𝜆|=(𝜆_𝑛+𝜀)²}, where 𝜀 is sufficiently small number. Consider the contour integral 𝐹_𝑛(𝜆)=12𝜋𝑖_Γ𝑛𝑚𝜂𝜂−𝜆𝑑𝜂,𝜆∈intΓ_𝑛. (32) There exists a constant 𝐶_𝛿>0 such that Δ(𝜆)≥|𝜆|^7/2𝐶_𝛿exp|𝜏| holds for 𝜆∈𝐺_𝛿. Use this inequality and (21) to get |𝑚(𝜆)|≤𝐶_𝛿/|𝜆|^3/2, for 𝜆∈𝐺_𝛿. Hence, lim_𝑛→∞𝐹_𝑛(𝜆)=0, and so 𝑚(𝜆)=^∞_𝑛=01𝛼_𝑛𝜆_𝑛−𝜆 (33) is obtained from residue theorem. Consequently, if 𝜆_𝑛=𝜆_𝑛 and 𝜶_𝑛=𝜶_𝑛 for all 𝑛, then from (33), 𝑚(𝜆)=𝑚(𝜆). Hence, Theorem 2 yields 𝐿=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 𝜂_𝑛 are zeros of Δ₁(𝜂)∶=𝜓^′(0,𝜂)+ℎ₀𝜓(0,𝜂).

Theorem 4. If{𝜆_𝑛,𝜂_𝑛}_𝑛≥0={𝜆_𝑛,𝜂_𝑛}_𝑛≥0 and 𝐡=𝐡, then 𝐿=𝐿.

Proof. The functions Δ(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=0𝜆1−𝜆_𝑛,Δ₁𝜂=𝐶₁^∞_𝑛=0𝜂1−𝜂_𝑛, (35) where 𝐶 and 𝐶₁ are constants which depend only on {𝜆_𝑛} and {𝜂_𝑛}, respectively. Therefore, Δ(𝜆)≡Δ(𝜆) and Δ₁(𝜂)≡Δ₁(𝜂), when 𝜆_𝑛=𝜆_𝑛 and 𝜂_𝑛=𝜂_𝑛 for all 𝑛. Thus, 𝜓^′(0,𝜂)+ℎ₀𝜓(0,𝜂)=𝜓^′(0,𝜂)+ℎ₀𝜓(0,𝜂). Moreover, 𝜌₁=𝜌₁ since 𝐡=𝐡. Consequently, the equality (21) yields 𝑚(𝜆)≡𝑚(𝜆). Hence, the proof is completed by Theorem 2.

Appendix

The solution 𝜑(𝑥,𝜆) satisfies the following integral equations.

If 𝑥<𝑑₁,𝜑(𝑥,𝜆)=𝜆ℎ₀−ℎ₂sin√𝜆𝑥√𝜆+ℎ₁−𝜆cos√+1𝜆𝑥√𝜆^𝑥0sin√𝜆(𝑥−𝑡)𝑞(𝑡)𝜑(𝑡,𝜆)𝑑𝑡; (A.1) if 𝑑₁<𝑥<𝑑₂, 𝜑(𝑥,𝜆)=𝜆ℎ₀−ℎ₂√𝜆×𝛼⁺₁sin√𝜆𝑥+𝛼⁻₁sin√𝜆2𝑑₁+ℎ−𝑥₁×𝛼−𝜆₊₁cos√𝜆𝑥+𝛼⁻₁cos√𝜆2𝑑₁+𝛾−𝑥₁𝜆+𝛽₁ℎ₁−𝜆2√𝜆×sin√𝜆2𝑑₁−𝑥−sin√+𝛾𝜆𝑥₁𝜆+𝛽₁𝜆ℎ₀−ℎ₂×√2𝜆cos𝜆𝑥−cos√𝜆2𝑑₁+𝛾−𝑥₁𝜆+𝛽₁×2𝜆_𝑑10cos√𝜆(𝑥−𝑡)−cos√𝜆2𝑑₁(𝑡)𝜑()+1−𝑥−𝑡×𝑞𝑡,𝜆𝑑𝑡√𝜆^𝑑10𝛼⁺₁sin√𝜆(𝑥−𝑡)+𝛼⁻₁sin√𝜆2𝑑₁(𝑡)𝜑()+1−𝑥−𝑡×𝑞𝑡,𝜆𝑑𝑡√𝜆^𝑥𝑑₁sin√𝜆(𝑥−𝑡)𝑞(𝑡)𝜑(𝑡,𝜆)𝑑𝑡; (A.2) if 𝑥>𝑑_2,𝜑()ℎ𝑥,𝜆=−¹−𝜆𝛾₁𝜆+𝛽₁2√𝜆×𝛼⁺₂sin√𝜆𝑥−sin√𝜆2𝑑₁−𝑥+𝛼⁻₂sin√𝜆2𝑑₂√−𝑥+sin𝜆2𝑑₂−2𝑑₁+−𝑥𝜆ℎ₀−ℎ₂𝛾₁𝜆+𝛽₁×𝛼2𝜆₊₂cos√𝜆𝑥−cos√𝜆2𝑑₁−𝑥+𝛼⁻₂cos√𝜆2𝑑₁√−𝑥−cos𝜆2𝑑₂−2𝑑₁+ℎ−𝑥₁×𝛼−𝜆₊₁𝛼⁺₂cos√𝜆𝑥+𝛼⁻₂cos√𝜆2𝑑₂−𝑥+𝛼⁻₁𝛼⁺₂cos√𝜆2𝑑₁−𝑥+𝛼⁻₂cos√𝜆2𝑑₂−2𝑑₁+−𝑥𝜆ℎ₀−ℎ₂√𝜆×𝛼⁺₁𝛼⁺₂sin√𝜆𝑥+𝛼⁻₂sin√𝜆2𝑑₂−𝑥+𝛼⁻₁𝛼⁺₂sin√𝜆2𝑑₁−𝑥−𝛼⁻₂sin√𝜆2𝑑₂−2𝑑₁−ℎ−𝑥₁−𝜆𝛾₁𝜆+𝛽₁𝛾₂𝜆+𝛽₂×√4𝜆cos𝜆𝑥+cos√𝜆2𝑑₁√−𝑥−cos𝜆2𝑑₂√−𝑥−cos𝜆2𝑑₂−2𝑑₁−−𝑥𝜆ℎ₀−ℎ₂𝛾₁𝜆+𝛽₁𝛾₂𝜆+𝛽₂4𝜆^3/2×sin√𝜆𝑥−sin√𝜆2𝑑₁√−𝑥−sin𝜆2𝑑₂√−𝑥+sin𝜆2𝑑₂−2𝑑₁+ℎ−𝑥₁−𝜆𝛾₂𝜆+𝛽₂2√𝜆×𝛼⁺₁sin√𝜆2𝑑₂−𝑥−sin√𝜆𝑥+𝛼⁻₁sin√𝜆2𝑑₁√−𝑥+sin𝜆2𝑑₂−2𝑑₁+−𝑥𝜆ℎ₀−ℎ₂𝛾₂𝜆+𝛽₂×𝛼2𝜆₊₁cos√𝜆𝑥−cos√𝜆2𝑑₂−𝑥−𝛼⁻₁cos√𝜆2𝑑₁√−𝑥−cos𝜆2𝑑₂−2𝑑₁+𝛼−𝑥₂2√𝜆^𝑑10𝛼⁺₁sin√𝜆2𝑑₂√−𝑥−𝑡+sin𝜆(𝑥−𝑡)+𝛼⁻₁sin√𝜆2𝑑₁√−𝑥−𝑡−sin𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()𝛼−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡+²𝛾₁𝜆+𝛽₁4𝜆×^𝑑1₀cos√𝜆(𝑥−𝑡)−cos√𝜆2𝑑₁√−𝑥−𝑡+cos𝜆2𝑑₂√−𝑥−𝑡+cos𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()1−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡−2𝛼₂√𝜆×^𝑑2₀𝛼⁺₁sin√𝜆2𝑑₂−𝑥−𝑡−sin√𝜆(𝑥−𝑡)−𝛼⁻₁sin√𝜆2𝑑₁√−𝑥−𝑡+sin𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()+𝛾−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡₁𝜆+𝛽₁4𝛼₂𝜆×^𝑑2₀cos√𝜆(𝑥−𝑡)−cos√𝜆2𝑑₁√−𝑥−𝑡−cos𝜆2𝑑₂√−𝑥−𝑡+cos𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()−1−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡√𝜆^𝑑2𝑑₁𝛼⁺₂sin√𝜆(𝑥−𝑡)+𝛼⁻₂sin√𝜆2𝑑₂(𝑡)𝜑()+𝛾−𝑥−𝑡×𝑞𝑡,𝜆𝑑𝑡₂𝜆+𝛽₂×2𝜆_𝑑10𝛼⁺₁cos√𝜆(𝑥−𝑡)−cos√𝜆2𝑑₂−𝑥−𝑡−𝛼⁻₁cos√𝜆2𝑑₁√−𝑥−𝑡−cos𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()−𝛾−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡₁𝜆+𝛽₁𝛾₂𝜆+𝛽₂4𝜆^3/2×^𝑑1₀sin√𝜆(𝑥−𝑡)−sin√𝜆2𝑑₁√−𝑥−𝑡−sin𝜆2𝑑₂√−𝑥−𝑡+sin𝜆2𝑑₂−2𝑑₁(𝑡)𝜑()−𝛾−𝑥+𝑡×𝑞𝑡,𝜆𝑑𝑡₂𝜆+𝛽₂×2𝜆_𝑑2𝑑1cos√𝜆(𝑥−𝑡)−cos√𝜆2𝑑₂(𝑡)𝜑()+1−𝑥−𝑡×𝑞𝑡,𝜆𝑑𝑡√𝜆^𝑥𝑑₂sin√𝜆(𝑥−𝑡)𝑞(𝑡)𝜑(𝑡,𝜆)𝑑𝑡, (A.3) where 𝛼^±_𝑖=(1/2)(𝛼_𝑖±1/𝛼_𝑖),𝑖=1,2.