Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 794262, pages http://dx.doi.org/10.1155/2013/794262 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 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 = ⋃ (𝑑𝑖 , 𝑑𝑖+1 ) (1) 𝑖=0 subject to the boundary conditions 𝑈 (𝑦) := 𝜆 (𝑦 (0) + ℎ0 𝑦 (0)) − ℎ1 𝑦 (0) − ℎ2 𝑦 (0) = 0, (2) 𝑉 (𝑦) := 𝜆 (𝑦 (1) + 𝐻0 𝑦 (1)) − 𝐻1 𝑦 (1) − 𝐻2 𝑦 (1) = (3) and double discontinuity conditions 𝑦 (𝑑𝑖 + 0) = 𝛼𝑖 𝑦 (𝑑𝑖 − 0) , 𝑦 (𝑑𝑖 + 0) = 𝛼𝑖−1 𝑦 (𝑑𝑖 − 0) − (𝛾𝑖 𝜆 + 𝛽𝑖 ) 𝑦 (𝑑𝑖 − 0) , (4) where 𝑞(𝑥) is real valued function in 𝐿 (0, 1); ℎ𝑗 and 𝐻𝑗 , 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 − ℎ0 ℎ1 > 0, 𝜌2 := 𝐻0 𝐻1 − 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 s𝑖 = (𝑑𝑖 , 𝛼𝑖 , 𝛾𝑖 , 𝛽𝑖 ), 𝑖 = 1, 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 SturmLiouville operator on a finite interval 2 Abstract and Applied Analysis 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): These solutions are the entire functions of 𝜆 and satisfy the relation 𝜓(𝑥, 𝜆 𝑛 ) = 𝛽𝑛 𝜑(𝑥, 𝜆 𝑛 ) for each eigenvalue 𝜆 𝑛 , where 𝛽𝑛 = −(𝜓 (0, 𝜆 𝑛 ) + ℎ0 𝜓(0, 𝜆 𝑛 ))/𝜌1 The following asymptotics can be obtained from the integral equations given in the appendix: 𝜑 −𝜆 + ℎ1 ), ( ) (0, 𝜆) = ( 𝜆ℎ0 − ℎ2 𝜑 𝜓 −𝜆 + 𝐻1 ( ) (1, 𝜆) = ( ) 𝜆𝐻0 − 𝐻2 𝜓 (5) −𝜆 cos √𝜆𝑥 + 𝑂 (√𝜆 exp |𝜏| 𝑥) , { { { { { { sin √𝜆𝑥 sin √𝜆 (2𝑑1 − 𝑥) { { − ] + 𝑂 (𝜆 exp |𝜏| 𝑥) , { {𝛾1 𝜆 [ √ 𝜆 2√𝜆 𝜑 (𝑥, 𝜆) = { { 𝛾1 𝛾2 { { { 𝜆 [cos √𝜆𝑥 + cos √𝜆 (2𝑑1 − 𝑥) − cos √𝜆 (2𝑑2 − 𝑥) − cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥)] { { { { { 3/2 { +𝑂 (𝜆 exp |𝜏| 𝑥) , 𝜆3/2 sin √𝜆𝑥 + 𝑂 (𝜆 exp |𝜏| 𝑥) , { { { { { { 𝛾1 𝜆2 { 3/2 { { { [cos √𝜆𝑥 + cos √𝜆 (2𝑑1 − 𝑥)] + 𝑂 (𝜆 exp |𝜏| 𝑥) , 𝜑 (𝑥, 𝜆) = { { 𝛾𝛾 { { − 𝜆5/2 {sin √𝜆𝑥 − sin √𝜆 (2𝑑1 − 𝑥) + sin √𝜆 (2𝑑2 − 𝑥) + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥)} { { { { { { { +𝑂 (𝜆 exp |𝜏| 𝑥) , 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 𝛼𝑛 of the problem 𝐿 are defined as follows: Δ (𝜆) = − = 𝜆 (𝜑 (1, 𝜆) + 𝐻0 𝜑 (1, 𝜆)) − 𝐻1 𝜑 (1, 𝜆) − 𝐻2 𝜑 (1, 𝜆) (𝜑 (0, 𝜆 𝑛 ) + ℎ0 𝜑 (0, 𝜆 𝑛 )) 𝜌1 𝑥 > 𝑑2 , 𝑥 < 𝑑1 , 𝑑1 < 𝑥 < 𝑑2 , (7) 𝑥 > 𝑑2 , sin √𝜆 (2𝑑2 − 2𝑑1 − 1) ] + 𝑂 (𝜆3 exp |𝜏|) √𝜆 Lemma See the following (i) All eigenvalues of the problem 𝐿 are real and algebraically simple; that is, Δ (𝜆 𝑛 ) ≠ (ii) Two eigenfunctions 𝜑(𝑥, 𝜆 ) and 𝜑(𝑥, 𝜆 ), corresponding to different eigenvalues 𝜆 and 𝜆 , are orthogonal in the sense of + (𝜑 (1, 𝜆 𝑛 ) + 𝐻0 𝜑 (1, 𝜆 𝑛 )) 𝜌2 (6) 𝛾1 𝛾2 𝜆 sin √𝜆 sin √𝜆 (2𝑑1 − 1) sin √𝜆 (2𝑑2 − 1) − + ×[ √𝜆 √𝜆 √𝜆 + = − 𝜆 (𝜓 (0, 𝜆)+ℎ0 𝜓 (0, 𝜆))+ℎ1 𝜓 (0, 𝜆) + ℎ2 𝜓 (0, 𝜆) , 𝑑1 < 𝑥 < 𝑑2 , (9) Δ (𝜆) = 𝑊 [𝜑, 𝜓] 𝛼𝑛 := ∫ 𝜑2 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 + 𝑥 < 𝑑1 , ∫ 𝜑 (𝑥, 𝜆 ) 𝜑 (𝑥, 𝜆 ) 𝑑𝑥 (𝜑 (0, 𝜆 ) + ℎ0 𝜑 (0, 𝜆 )) (𝜑 (0, 𝜆 ) + ℎ0 𝜑 (0, 𝜆 )) 𝜌1 + (𝜑 (1, 𝜆 ) + 𝐻0 𝜑 (1, 𝜆 )) (𝜑 (1, 𝜆 ) + 𝐻0 𝜑 (1, 𝜆 )) 𝜌2 + + 𝛼1 𝛾1 𝜑 (𝑑1 − 0, 𝜆 𝑛 ) + 𝛼2 𝛾2 𝜑 (𝑑2 − 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 + 𝛼1 𝛾1 𝜑 (𝑑1 − 0, 𝜆 ) 𝜑 (𝑑1 − 0, 𝜆 ) + 𝛼2 𝛾2 𝜑 (𝑑2 − 0, 𝜆 ) 𝜑 (𝑑2 − 0, 𝜆 ) = (10) Abstract and Applied Analysis Proof Consider a Hilbert Space 𝐻 = 𝐿 (0, 1) ⊕ C4 , equipped with the inner product Δ (𝜆) = ∫ 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 𝜆 − 𝜆𝑛 1 ⟨𝑌, 𝑍⟩ := ∫ 𝑦 (𝑥) 𝑧 (𝑥)𝑑𝑥 + 𝑌1 𝑍1 𝜌 1 + 𝑌2 𝑍2 + 𝛼1 𝛾1 𝑌3 𝑍3 + 𝛼2 𝛾2 𝑌4 𝑍4 𝜌2 Rewrite this equality as (11) + (𝜑 (1, 𝜆 𝑛 ) + 𝐻0 𝜑 (1, 𝜆 𝑛 )) − (𝜓 (0, 𝜆) + ℎ0 𝜓 (0, 𝜆)) 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 ∈ 𝐿 (0, 1), 𝑦(𝑑𝑖 + 0) = 𝛼𝑖 𝑦(𝑑𝑖 − 0), 𝑌1 = 𝑦 (0) + ℎ0 𝑦(0), 𝑌2 = 𝑦1 (1) + 𝐻0 𝑦(1), 𝑌3 = 𝛾1 𝑦(𝑑1 − 0), and 𝑌4 = 𝛾2 𝑦(𝑑2 − 0)} such that + 𝛼1 𝛾1 𝜓 (𝑑1 − 0, 𝜆) 𝜑 (𝑑1 − 0, 𝜆 𝑛 ) + 𝛼2 𝛾2 𝜓 (𝑑2 − 0, 𝜆) 𝜑 (𝑑2 − 0, 𝜆 𝑛 ) = ∫ 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 (𝜓 (0, 𝜆)+ℎ0 𝜓 (0, 𝜆))(𝜑 (0, 𝜆 𝑛 )+ℎ0 𝜑 (0, 𝜆 𝑛 )) −𝑦 (𝑥) + 𝑞 (𝑥) 𝑦 (𝑥) ℎ1 𝑦 (0) + ℎ2 𝑦 (0) ( 𝑇 (𝑌) := ( 𝐻1 𝑦 (1) + 𝐻2 𝑦 (1) 𝛼1−1 𝑦 𝛼2−1 𝑦 −𝑦 (𝑑1 + 0) + (−𝑦 (𝑑2 + 0) + − ℎ0 ℎ1 − ℎ2 ) ) (𝜑 (1, 𝜆 𝑛 )+𝐻0 𝜑 (1, 𝜆 𝑛 ))(𝜓 (1, 𝜆)+𝐻0 𝜓 (1, 𝜆)) (𝑑1 − 0) − 𝛽1 𝑦 (𝑑1 − 0) + (𝑑2 − 0) − 𝛽2 𝑦 (𝑑2 − 0)) (12) + 𝛼1 𝛾1 𝜓 (𝑑1 − 0, 𝜆) 𝜑 (𝑑1 − 0, 𝜆 𝑛 ) 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) 𝜑 (𝑥, 𝜆 𝑛 ) = 𝜆 𝑛 𝜑 (𝑥, 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 𝐻0 𝐻1 − 𝐻2 + 𝛼2 𝛾2 𝜓 (𝑑2 − 0, 𝜆) 𝜑 (𝑑2 − 0, 𝜆 𝑛 ) (17) = ∫ 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 + + (𝜓 (0, 𝜆)+ℎ0 𝜓 (0, 𝜆))(𝜑 (0, 𝜆 𝑛 )+ℎ0 𝜑 (0, 𝜆 𝑛 )) 𝜌1 (𝜑 (1, 𝜆 𝑛 )+𝐻0 𝜑 (1, 𝜆 𝑛 ))(𝜓 (1, 𝜆)+𝐻0 𝜓 (1, 𝜆)) 𝜌2 + 𝛼1 𝛾1 𝜓 (𝑑1 − 0, 𝜆) 𝜑 (𝑑1 − 0, 𝜆 𝑛 ) + 𝛼2 𝛾2 𝜓 (𝑑2 − 0, 𝜆) 𝜑 (𝑑2 − 0, 𝜆 𝑛 ) (18) [𝜑 (𝑥, 𝜆 𝑛 ) 𝜓 (𝑥, 𝜆) − 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 )] 𝑑 −0 × (|0 As 𝜆 → 𝜆 𝑛 , 𝑑 −0 + |𝑑2 +0 + |1𝑑2 +0 ) (14) Δ (𝜆 𝑛 ) = 𝛽𝑛 𝛼𝑛 = (𝜆 − 𝜆 𝑛 ) ∫ 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 is obtained Add and subtract Δ(𝜆) in the left-hand side of the last equality, and use initial conditions (5) to get − (𝜆 − 𝜆 𝑛 ) (𝜑 (1, 𝜆 𝑛 ) + 𝐻0 𝜑 (1, 𝜆 𝑛 )) − (𝜆 − 𝜆 𝑛 ) 𝛼2 𝛾2 𝜓 (𝑑2 − 0, 𝜆) 𝜑 (𝑑2 − 0, 𝜆 𝑛 ) = (𝜆 − 𝜆 𝑛 ) ∫ 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆 𝑛 ) 𝑑𝑥 (19) is obtained by using the equality 𝜓(𝑥, 𝜆 𝑛 ) = 𝛽𝑛 𝜑(𝑥, 𝜆 𝑛 ) Thus, Δ (𝜆 𝑛 ) ≠ Main Results Δ (𝜆) + (𝜆 − 𝜆 𝑛 ) (𝜓 (0, 𝜆) + ℎ0 𝜓 (0, 𝜆)) − (𝜆 − 𝜆 𝑛 ) 𝛼1 𝛾1 𝜓 (𝑑1 − 0, 𝜆) 𝜑 (𝑑1 − 0, 𝜆 𝑛 ) (16) (15) We consider three statements of the inverse problem for the boundary value problem 𝐿; from the Weyl function, from the spectral data {𝜆 𝑛 , 𝛼𝑛 }𝑛≥0 , and from two spectra {𝜆 𝑛 , 𝜇𝑛 }𝑛≥0 For studying the inverse problem, we consider a boundary ̃ together with 𝐿, of the same form but with value problem 𝐿, ̃ H, ̃ s̃𝑖 , 𝑖 = 1, different coefficients 𝑞̃(𝑥), h, Let the function 𝜘(𝑥, 𝜆) denote the solution of (1) under the initial conditions 𝜘(0, 𝜆) = 𝜌1−1 , 𝜘 (0, 𝜆) = −𝜌1−1 ℎ0 Abstract and Applied Analysis and the jump conditions (4) It is clear that the function 𝜓(𝑥, 𝜆) can be represented by 𝜓 (𝑥, 𝜆) = Δ (𝜆) 𝜘 (𝑥, 𝜆) − 𝜓 (0, 𝜆) + ℎ0 𝜓 (0, 𝜆) 𝜑 (𝑥, 𝜆) 𝜌1 (20) Denote 𝑚 (𝜆) := 𝜓 (0, 𝜆) + ℎ0 𝜓 (0, 𝜆) 𝜌1 Δ (𝜆) (21) Then, we have 𝜓 (𝑥, 𝜆) = 𝜘 (𝑥, 𝜆) − 𝑚 (𝜆) 𝜑 (𝑥, 𝜆) Δ (𝜆) (22) The function 𝑚(𝜆) is called Weyl function [32] ̃ that is, 𝑞(𝑥) = 𝑞̃(𝑥), ̃ Theorem If 𝑚(𝜆) = 𝑚(𝜆), then 𝐿 = 𝐿; ̃ ̃ always everywhere in 𝐼; h = h, H = H, and s𝑖 = ̃s𝑖 , 𝑖 = 1, Proof Let us define the functions 𝑃1 (𝑥, 𝜆) and 𝑃2 (𝑥, 𝜆) as follows: ̃ (𝑥, 𝜆) − Φ (𝑥, 𝜆) 𝜑̃ (𝑥, 𝜆) , 𝑃1 (𝑥, 𝜆) = 𝜑 (𝑥, 𝜆) Φ ̃ (𝑥, 𝜆) , 𝑃2 (𝑥, 𝜆) = Φ (𝑥, 𝜆) 𝜑̃ (𝑥, 𝜆) − 𝜑 (𝑥, 𝜆) Φ (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 𝐺𝛿 = {𝜆 : 𝜆 = 𝑘2 , |𝑘 − 𝑘𝑛 | > 𝛿, 𝑛 = 1, 2, } and ̃𝛿 = {𝜆 : 𝜆 = 𝑘2 , |𝑘 − ̃𝑘𝑛 | > 𝛿, 𝑛 = 1, 2, }, where 𝛿 is 𝐺 sufficiently small number and 𝑘𝑛 and ̃𝑘𝑛 are square roots of ̃ respectively One can the eigenvalues of the problem 𝐿 and 𝐿, easily show that the asymptotics Φ (𝑥, 𝜆) = 𝑂 (𝜆−(𝑖+3)/2 exp (− |𝜏| 𝑥)) , Φ (𝑥, 𝜆) = 𝑂 (𝜆−(𝑖+2)/2 exp (− |𝜏| 𝑥)) (24) are valid for 𝑑𝑖 < 𝑥 < 𝑑𝑖+1 , 𝑖 = 0, 1, 2, and sufficiently large ̃𝛿 Thus, the following inequalities are obtained |𝜆| in 𝐺𝛿 ∩ 𝐺 from (6) and (24): −1/2 𝑃1 (𝑥, 𝜆) ≤ 𝐶𝛿 , 𝑃2 (𝑥, 𝜆) ≤ 𝐶𝛿 |𝜆| , (25) ̃𝛿 𝜆 ∈ 𝐺𝛿 ∩ 𝐺 According to the last inequalities and Liouville’s theorem, 𝑃1 (𝑥, 𝜆) = 𝐴(𝑥) and 𝑃2 (𝑥, 𝜆) = Use (23) again to take 𝜑 (𝑥, 𝜆) = 𝐴 (𝑥) 𝜑̃ (𝑥, 𝜆) , ̃ (𝑥, 𝜆) Φ (𝑥, 𝜆) = 𝐴 (𝑥) Φ (26) ̃ 𝜆), 𝜑(𝑥, ̃ Since 𝑊[Φ(𝑥, 𝜆), 𝜑(𝑥, 𝜆)] = and similarly 𝑊[Φ(𝑥, 𝜆)] = 1, then 𝐴 (𝑥) = On the other hand, the asymptotic expressions 𝜑 (𝑥, 𝜆) = 𝐶 (𝜆) exp (−𝑖√𝜆𝑥) (1 + 𝑜 (1)) , ̃ (𝜆) exp (−𝑖√𝜆𝑥) (1 + 𝑜 (1)) 𝜑 (𝑥, 𝜆) = 𝐶 (27) are valid for √𝜆 → ∞ on the imaginary axis, where { < 𝑥 < 𝑑1 , − 𝜆, { { { { 𝛾 { { 3/2 𝐶 (𝜆) = { 𝜆 , 𝑑1 < 𝑥 < 𝑑2 , { { { 𝛾𝛾 { { { 𝜆2 , 𝑑2 < 𝑥 < 1, { { < 𝑥 < 𝑑̃1 , − 𝜆, { { { { { { 𝛾̃1 3/2 { 𝜆 , 𝑑̃1 < 𝑥 < 𝑑̃2 , ̃ (𝜆) = 𝐶 { { { { { 𝛾̃ 𝛾̃ { { { 𝜆2 , 𝑑̃2 < 𝑥 < { (28) Assume that 𝑑1 ≠ 𝑑̃1 and 𝑑2 ≠ 𝑑̃2 There are six different cases for the permutation of the numbers 𝑑𝑖 and 𝑑̃𝑖 Without loss of generality, let < 𝑑1 < 𝑑̃1 < 𝑑2 < 𝑑̃2 < From (26)-(27), we get 𝛾1 = 𝛾̃1 , 𝛾2 = 𝛾̃2 , and 𝐴(𝑥) ≡ 1, while 𝑥 ∈ [0, 𝑑1 ) ∪ (𝑑̃1 , 𝑑2 ) ∪ (𝑑̃2 , 1] Moreover, we get 2𝜆−1/2 (1 + 𝑜 (1)) 𝐴 (𝑥) + 𝛾1 = 𝑜 (1) , (29) while 𝑥 ∈ (𝑑1 , 𝑑̃1 ) By taking limit in (29) as |𝜆| → ∞, we condradict 𝛾1 > Thus, 𝑑1 = 𝑑̃1 Similarly, 𝑑2 = 𝑑̃2 , and 𝐴(𝑥) = in 𝐼 Hence, 𝜓 (𝑥, 𝜆) 𝜓̃ (𝑥, 𝜆) = 𝜓̃ (𝑥, 𝜆) 𝜓 (𝑥, 𝜆) 𝜑 (𝑥, 𝜆) = 𝜑̃ (𝑥, 𝜆) , (30) It can be obtained from (1), (4), and (5) that 𝑞(𝑥) = 𝑞̃(𝑥), a.e ̃ H = H ̃ Consequently, in 𝐼; s𝑖 = ̃s𝑖 , 𝑖 = 1, 2, and h = h, ̃ 𝐿 = 𝐿 ̃ ,𝛼 ̃ Theorem If {𝜆 𝑛 , 𝛼𝑛 }𝑛≥0 = {𝜆 𝑛 ̃ 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 Res {𝑚 (𝜆) , 𝜆 𝑛 } = 𝜓 (0, 𝜆 𝑛 ) + ℎ0 𝜓 (0, 𝜆 𝑛 ) 𝜌1 Δ (𝜆 𝑛 ) 𝛽 = − 𝑛 =− 𝛼𝑛 Δ (𝜆 𝑛 ) (31) Denote Γ𝑛 = {𝜆 : |𝜆| = (√𝜆 𝑛 + 𝜀) }, where 𝜀 is sufficiently small number Consider the contour integral 𝐹𝑛 (𝜆) = 𝑚 (𝜂) 𝑑𝜂, ∫ 2𝜋𝑖 Γ𝑛 (𝜂 − 𝜆) 𝜆 ∈ int Γ𝑛 (32) There exists a constant 𝐶𝛿 > such that Δ(𝜆) ≥ |𝜆|7/2 𝐶𝛿 exp |𝜏| holds for 𝜆 ∈ 𝐺𝛿 Use this inequality and (21) to get |𝑚(𝜆)| ≤ 𝐶𝛿 /|𝜆|3/2 , for 𝜆 ∈ 𝐺𝛿 Hence, lim𝑛 → ∞ 𝐹𝑛 (𝜆) = 0, and so ∞ 𝑚 (𝜆) = ∑ 𝑛=0 𝛼𝑛 (𝜆 𝑛 − 𝜆) (33) Abstract and Applied Analysis ̃ is obtained from residue theorem Consequently, if 𝜆 𝑛 = 𝜆 𝑛 ̃ 𝑛 for all 𝑛, then from (33), 𝑚(𝜆) = 𝑚(𝜆) ̃ and 𝛼𝑛 = 𝛼 Hence, ̃ Theorem yields 𝐿 = 𝐿 × (sin √𝜆 (2𝑑1 − 𝑥) − sin √𝜆𝑥) + We consider the boundary value problem 𝐿 with the condition 𝑦 (0, 𝜆) + ℎ0 𝑦 (0, 𝜆) = × (cos √𝜆𝑥 − cos √𝜆 (2𝑑1 − 𝑥)) (34) + instead of (2) in 𝐿 Let {𝜂𝑛2 }𝑛≥0 be the eigenvalues of the problem 𝐿 It is obvious that 𝜂𝑛 are zeros of Δ (𝜂) := 𝜓 (0, 𝜂) + ℎ0 𝜓(0, 𝜂) 𝑑1 × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + Proof The functions Δ(𝜆) and Δ (𝜂) which are entire of order 1/2 can be represented by Hadamard’s factorization theorem as follows: ∞ 𝑛=0 𝜆 ), 𝜆𝑛 ∞ 𝜂 Δ (𝜂) = 𝐶1 ∏ (1 − ) , 𝜂𝑛 𝑛=0 𝑑1 + ∫ (𝛼1 sin √𝜆 (𝑥 − 𝑡) √𝜆 +𝛼1− sin √𝜆 (2𝑑1 − 𝑥 − 𝑡)) × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 (35) + 𝑥 ∫ sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡; √𝜆 𝑑1 (A.2) where 𝐶 and 𝐶1 are constants which depend only on {𝜆 𝑛 } and ̃ ̃ (𝜂), {𝜂𝑛 }, respectively Therefore, Δ(𝜆) ≡ Δ(𝜆) and Δ (𝜂) ≡ Δ ̃ when 𝜆 𝑛 = 𝜆 𝑛 and 𝜂𝑛 = 𝜂̃𝑛 for all 𝑛 Thus, 𝜓 (0, 𝜂) + ̃ 𝜂) Moreover, 𝜌1 = 𝜌̃1 since ℎ0 𝜓(0, 𝜂) = 𝜓̃ (0, 𝜂) + ℎ0 𝜓(0, ̃ ̃ h = h Consequently, the equality (21) yields 𝑚(𝜆) ≡ 𝑚(𝜆) Hence, the proof is completed by Theorem if 𝑥 > 𝑑2, 𝜑 (𝑥, 𝜆) =− Appendix (ℎ1 − 𝜆) (𝛾1 𝜆 + 𝛽1 ) 2√𝜆 × [𝛼2+ (sin √𝜆𝑥 − sin √𝜆 (2𝑑1 − 𝑥)) The solution 𝜑(𝑥, 𝜆) satisfies the following integral equations If 𝑥 < 𝑑1 , sin √𝜆𝑥 + (ℎ1 − 𝜆) cos √𝜆𝑥 𝜑 (𝑥, 𝜆) = (𝜆ℎ0 − ℎ2 ) √𝜆 + 𝑥 ∫ sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡; √𝜆 if 𝑑1 < 𝑥 < 𝑑2 , 𝜑 (𝑥, 𝜆) = (𝛾1 𝜆 + 𝛽1 ) 2𝜆 × ∫ (cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1 − 𝑥 − 𝑡)) ̃ , 𝜂̃ } and h = h, ̃ then 𝐿 = Theorem If {𝜆 𝑛 , 𝜂𝑛 }𝑛≥0 = {𝜆 𝑛 𝑛 𝑛≥0 ̃ 𝐿 Δ (𝜆) = 𝐶∏ (1 − (𝛾1 𝜆 + 𝛽1 ) (𝜆ℎ0 − ℎ2 ) 2𝜆 𝜆ℎ0 − ℎ2 √𝜆 + 𝛼2− ( sin √𝜆 (2𝑑2 − 𝑥) + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] (A.1) + (𝜆ℎ0 − ℎ2 ) (𝛾1 𝜆 + 𝛽1 ) 2𝜆 × [𝛼2+ (cos √𝜆𝑥 − cos √𝜆 (2𝑑1 − 𝑥)) + 𝛼2− (cos √𝜆 (2𝑑1 − 𝑥) − cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] × (𝛼1+ sin √𝜆𝑥 + 𝛼1− sin √𝜆 (2𝑑1 − 𝑥)) + (ℎ1 − 𝜆) + (ℎ1 − 𝜆) × [𝛼1+ (𝛼2+ cos √𝜆𝑥 + 𝛼2− cos √𝜆 (2𝑑2 − 𝑥)) × (𝛼1+ cos √𝜆𝑥 + 𝛼1− cos √𝜆 (2𝑑1 − 𝑥)) + (𝛾1 𝜆 + 𝛽1 ) (ℎ1 − 𝜆) 2√𝜆 + 𝛼1− (𝛼2+ cos √𝜆 (2𝑑1 − 𝑥) + 𝛼2− cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] Abstract and Applied Analysis + 𝑑1 (𝜆ℎ0 − ℎ2 ) √𝜆 × ∫ [cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1 − 𝑥 − 𝑡) + cos √𝜆 (2𝑑2 − 𝑥 − 𝑡) × [𝛼1+ (𝛼2+ sin √𝜆𝑥 + 𝛼2− sin √𝜆 (2𝑑2 − 𝑥)) + cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡)] + 𝛼1− (𝛼2+ sin √𝜆 (2𝑑1 − 𝑥) × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 − −𝛼2− sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] (ℎ − 𝜆) (𝛾1 𝜆 + 𝛽1 ) (𝛾2 𝜆 + 𝛽2 ) − 4𝜆 𝑑2 × ∫ [𝛼1+ (sin √𝜆 (2𝑑2 − 𝑥 − 𝑡) − sin √𝜆 (𝑥 − 𝑡)) − 𝛼1− (sin √𝜆 (2𝑑1 − 𝑥 − 𝑡) × [cos √𝜆𝑥 + cos √𝜆 (2𝑑1 − 𝑥) + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡))] − cos √𝜆 (2𝑑2 − 𝑥) − cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥)] − (𝜆ℎ0 − ℎ2 ) (𝛾1 𝜆 + 𝛽1 ) (𝛾2 𝜆 + 𝛽2 ) 4𝜆3/2 × [sin √𝜆𝑥 − sin √𝜆 (2𝑑1 − 𝑥) × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + (𝛾1 𝜆 + 𝛽1 ) 4𝛼2 𝜆 𝑑2 × ∫ [cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑1 − 𝑥 − 𝑡) − cos √𝜆 (2𝑑2 − 𝑥 − 𝑡) − sin √𝜆 (2𝑑2 − 𝑥) + cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡)] + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥)] + (ℎ1 − 𝜆) (𝛾2 𝜆 + 𝛽2 ) 2√𝜆 × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 − 𝑑2 + ∫ [𝛼 sin √𝜆 (𝑥 − 𝑡) √𝜆 𝑑1 +𝛼2− sin √𝜆 (2𝑑2 − 𝑥 − 𝑡)] × [𝛼1+ (sin √𝜆 (2𝑑2 − 𝑥) − sin √𝜆𝑥) × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + 𝛼1− (sin √𝜆 (2𝑑1 − 𝑥) + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] + + (𝛾2 𝜆 + 𝛽2 ) 2𝜆 𝑑1 (𝜆ℎ0 − ℎ2 ) (𝛾2 𝜆 + 𝛽2 ) 2𝜆 × ∫ [𝛼1+ (cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑2 − 𝑥 − 𝑡)) − 𝛼1− (cos √𝜆 (2𝑑1 − 𝑥 − 𝑡) × [𝛼1+ (cos √𝜆𝑥 − cos √𝜆 (2𝑑2 − 𝑥)) − cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡))] − 𝛼1− (cos √𝜆 (2𝑑1 − 𝑥) − cos √𝜆 (2𝑑2 − 2𝑑1 − 𝑥))] + 𝛼2 𝑑1 + ∫ [𝛼1 (sin √𝜆 (2𝑑2 − 𝑥 − 𝑡) 2√𝜆 × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 − + 𝑑1 − sin √𝜆 (2𝑑2 − 𝑥 − 𝑡) (sin √𝜆 (2𝑑1 − 𝑥 − 𝑡) + sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡)] − sin √𝜆 (2𝑑2 − 2𝑑1 − 𝑥 + 𝑡))] × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + 𝛼2 (𝛾1 𝜆 + 𝛽1 ) 4𝜆 (𝛾1 𝜆 + 𝛽1 ) (𝛾2 𝜆 + 𝛽2 ) 4𝜆3/2 × ∫ [sin √𝜆 (𝑥 − 𝑡) − sin √𝜆 (2𝑑1 − 𝑥 − 𝑡) + sin √𝜆 (𝑥 − 𝑡)) 𝛼1− 2𝛼2 √𝜆 × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 − (𝛾2 𝜆 + 𝛽2 ) 2𝜆 Abstract and Applied Analysis 𝑑2 × ∫ [cos √𝜆 (𝑥 − 𝑡) − cos √𝜆 (2𝑑2 − 𝑥 − 𝑡)] 𝑑1 × 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡 + 𝑥 ∫ sin √𝜆 (𝑥 − 𝑡) 𝑞 (𝑡) 𝜑 (𝑡, 𝜆) 𝑑𝑡, √𝜆 𝑑2 (A.3) where 𝛼𝑖± = (1/2)(𝛼𝑖 ± 1/𝛼𝑖 ), 𝑖 = 1, References ă [1] V A Ambartsumyan, Uber eine frage der eigenwerttheorie, Zeitschrift făur Physik, vol 53, pp 690695, 1929 [2] G Borg, “Eine umkehrung der Sturm–Liouvilleschen eigenwertaufgabe bestimmung der differentialgleichung durch die eigenwerte,” Acta Mathematica, vol 78, pp 1–96, 1946 [3] M Kac, “Can one hear the shape of a drum?” The American Mathematical Monthly, vol 73, no 4, pp 1–23, 1966 [4] A V Likov and A Yu Mikhailov, The Theory of Heat and Mass Transfer, Gosnergoizdat, 1963 [5] O N Litvinenko and V I Soshnikov, The Theory of Heteregeneous Lines and Their Applications in Radio Engineering, Radio, Moscow, Russia, 1964 [6] J R 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[27] B Keskin, A S Ozkan, and N Yalc¸in, ? ?Inverse spectral problems for discontinuous Sturm- Liouville operator with eigenparameter dependent boundary conditions, ” Communications de la Facult´e... A S Ozkan and B Keskin, “Spectral problems for SturmLiouville operator with boundary and jump conditions linearly dependent on the eigenparameter,” Inverse Problems in Science and Engineering,