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

Authors: A. S. Ozkan; B. Keskin; Y. Cakmak
Journal: Abstract and Applied Analysis
 (2013)

---

## 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.


---

## Body

## 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:(1)ℓy∶=-y′′+q(x)y=λy,x∈I=⋃i=02(di,di+1)
subject to the boundary conditions
(2)U(y)∶=λ(y′(0)+h0y(0))-h1y′(0)-h2y(0)=0,(3)V(y)∶=λ(y′(1)+H0y(1))-H1y′(1)-H2y(1)=0
and double discontinuity conditions
(4)y(di+0)=αiy(di-0),y′(di+0)=αi-1y′(di-0)-(γiλ+βi)y(di-0),
where q(x) is real valued function in L2(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-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 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 conditions (4):
(5)(φφ′)(0,λ)=(-λ+h1λh0-h2),(ψψ′)(1,λ)=(-λ+H1λH0-H2).
These solutions are the entire functions of λ and satisfy the relation ψ(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:(6)φ(x,λ)={-λcosλx+O(λexp|τ|x),x<d1,γ1λ2[sinλx2λ-sinλ(2d1-x)2λ]+O(λexp|τ|x),d1<x<d2,γ1γ24λ2[cosλx+cosλ(2d1-x)-cosλ(2d2-x)-cosλ(2d2-2d1-x)]+O(λ3/2exp|τ|x),x>d2,(7)φ′(x,λ)={λ3/2sinλx+O(λexp|τ|x),x<d1,γ1λ22[cosλx+cosλ(2d1-x)]+O(λ3/2exp|τ|x),d1<x<d2,-γ1γ24λ5/2{sinλx-sinλ(2d1-x)+sinλ(2d2-x)+sinλ(2d2-2d1-x)}+O(λ2exp|τ|x),x>d2,whereτ=Imλ.The values of the parameterλ 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 αn of the problem L are defined as follows:
(8)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∶=∫01φ2(x,λn)dx+1ρ1(φ′(0,λn)+h0φ(0,λn))2+1ρ2(φ′(1,λn)+H0φ(1,λn))2+α1γ1φ2(d1-0,λn)+α2γ2φ2(d2-0,λn).
It is obvious that Δ(λ) is an entire function in λ and the zeros, namely, {λn} of Δ(λ) coincide with the eigenvalues of the problem L. Now, from (6) and (8), we can write
(9)Δ(λ)=-γ1γ24λ4×[sinλλ-sinλ(2d1-1)λ+sinλ(2d2-1)λ+sinλ(2d2-2d1-1)λ]+O(λ3exp|τ|).Lemma 1.
See the following.(i)
All eigenvalues of the problemL are real and algebraically simple; that is, Δ′(λn)≠0.(ii)
Two eigenfunctionsφ(x,λ1) and φ(x,λ2), corresponding to different eigenvalues λ1 and λ2, are orthogonal in the sense of
(10)∫01φ(x,λ1)φ(x,λ2)dx+1ρ1(φ′(0,λ1)+h0φ(0,λ1))(φ′(0,λ2)+h0φ(0,λ2))+1ρ2(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.Proof.
Consider a Hilbert SpaceH=L2(0,1)⊕ℂ4, equipped with the inner product
(11)〈Y,Z〉∶=∫01y(x)z(x)¯dx+1ρ1Y1Z1¯+1ρ2Y2Z2¯+α1γ1Y3Z3¯+α2γ2Y4Z4¯
for Y=(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=y1′(1)+H0y(1),Y3=γ1y(d1-0), and Y4=γ2y(d2-0)} such that
(12)T(Y)∶=(-y′′(x)+q(x)y(x)h1y′(0)+h2y(0)H1y′(1)+H2y(1)-y′(d1+0)+α1-1y′(d1-0)-β1y(d1-0)-y′(d2+0)+α2-1y′(d2-0)-β2y(d2-0)).
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 eigenvaluesλn by writting the following equations:
(13)-ψ′′(x,λ)+q(x)ψ(x,λ)=λψ(x,λ),-φ′′(x,λn)+q(x)φ(x,λn)=λnφ(x,λn).
If these equations are multiplied by φ(x,λn) and ψ(x,λ), respectively, subtracting them side by side and finally integrating over the interval [0,1], the equality
(14)[φ′(x,λn)ψ(x,λ)-ψ′(x,λ)φ(x,λn)]×(∣0d1-0+∣d1+0d2-0+∣d2+01)=(λ-λn)∫01ψ(x,λ)φ(x,λn)dx
is obtained. Add and subtract Δ(λ) in the left-hand side of the last equality, and use initial conditions (5) to get
(15)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)∫01ψ(x,λ)φ(x,λn)dx.
Rewrite this equality as
(16)Δ(λ)λ-λn=∫01ψ(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)(17)=∫01ψ(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)(18)=∫01ψ(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).
As λ→λn,
(19)Δ′(λn)=βnαn
is obtained by using the equality ψ(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 coefficients q~(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 function ψ(x,λ) can be represented by
(20)ψ(x,λ)=Δ(λ)ϰ(x,λ)-ψ′(0,λ)+h0ψ(0,λ)ρ1φ(x,λ).
Denote
(21)m(λ)∶=ψ′(0,λ)+h0ψ(0,λ)ρ1Δ(λ).
Then, we have
(22)ψ(x,λ)Δ(λ)=ϰ(x,λ)-m(λ)φ(x,λ).
The function m(λ) is called Weyl function [32].Theorem 2.
Ifm(λ)=m~(λ), then L=L~; that is, q(x)=q~(x), always everywhere in I;h=h~,H=H~, and si=s~i,i=1,2.Proof.
Let us define the functionsP1(x,λ) and P2(x,λ) as follows:
(23)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),
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 G~δ={λ:λ=k2,|k-k~n|>δ,n=1,2,…}, where δ is sufficiently small number and kn and k~n are square roots of the eigenvalues of the problem L and L~, respectively. One can easily show that the asymptotics
(24)Φ(x,λ)=O(λ-(i+3)/2exp(-|τ|x)),Φ′(x,λ)=O(λ-(i+2)/2exp(-|τ|x))
are valid for di<x<di+1, i=0,1,2, and sufficiently large |λ| in Gδ∩G~δ. Thus, the following inequalities are obtained from (6) and (24):
(25)|P1(x,λ)|≤Cδ,|P2(x,λ)|≤Cδ|λ|-1/2,λ∈Gδ∩G~δ.
According to the last inequalities and Liouville’s theorem, P1(x,λ)=A(x) and P2(x,λ)=0. Use (23) again to take
(26)φ(x,λ)=A(x)φ~(x,λ),Φ(x,λ)=A(x)Φ~(x,λ).
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(27)φ(x,λ)=C(λ)exp(-iλx)(1+o(1)),φ(x,λ)=C~(λ)exp(-iλx)(1+o(1))
are valid for λ→∞ on the imaginary axis, where
(28)C(λ)={-12λ,0<x<d1,γ14λ3/2,d1<x<d2,γ1γ28λ2,d2<x<1,C~(λ)={-12λ,0<x<d~1,γ~14λ3/2,d~1<x<d~2,γ~1γ~28λ2,d~2<x<1.
Assume that d1≠d~1 and d2≠d~2. There are six different cases for the permutation of the numbers di and di~. Without loss of generality, let 0<d1<d~1<d2<d~2<1.
From (26)-(27), we get γ1=γ~1,γ2=γ~2, and A(x)≡1, while x∈[0,d1)∪(d~1,d2)∪(d~2,1].
Moreover, we get(29)2λ-1/2(1+o(1))A(x)+γ1=o(1),
while x∈(d1,d~1). By taking limit in (29) as |λ|→∞, we condradict γ1>0. Thus, d1=d~1. Similarly, d2=d~2, and A(x)=1 in I. Hence,
(30)φ(x,λ)=φ~(x,λ),ψ′(x,λ)ψ(x,λ)=ψ~′(x,λ)ψ~(x,λ).
It can be obtained from (1), (4), and (5) that q(x)=q~(x), a.e. in I; si=s~i, i=1,2, and h=h~,H=H~. Consequently, L=L~.Theorem 3.
If{λn,αn}n≥0={λ~n,α~n}n≥0, then L=L~.Proof.
The meromorphic functionm(λ) has simple poles at λn, and its residues at these poles are
(31)Res{m(λ),λn}=ψ′(0,λn)+h0ψ(0,λn)ρ1Δ′(λn)=-βnΔ′(λn)=-1αn.
Denote Γn={λ:|λ|=(λn+ε)2}, where ε is sufficiently small number. Consider the contour integral
(32)Fn(λ)=12πi∫Γnm(η)(η-λ)dη,λ∈intΓn.
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, for λ∈Gδ. Hence, limn→∞Fn(λ)=0, and so
(33)m(λ)=∑‍∞n=01αn(λn-λ)
is obtained from residue theorem. Consequently, if λn=λ~n and αn=α~n for all n, then from (33), m(λ)=m~(λ). Hence, Theorem 2 yields L=L~.We consider the boundary value problemL1 with the condition
(34)y′(0,λ)+h0y(0,λ)=0
instead of (2) in L. Let {ηn2}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={λ~n,η~n}n≥0 and 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:
(35)Δ(λ)=C∏n=0∞‍(1-λλn),Δ1(η)=C1∏n=0∞‍(1-ηηn),
where C and C1 are constants which depend only on {λn} and {ηn}, respectively. Therefore, Δ(λ)≡Δ~(λ) and Δ1(η)≡Δ~1(η), when λn=λ~n and ηn=η~n for all n. Thus, ψ′(0,η)+h0ψ(0,η)=ψ~′(0,η)+h0ψ~(0,η). Moreover, ρ1=ρ~1 since h=h~. Consequently, the equality (21) yields m(λ)≡m~(λ). Hence, the proof is completed by Theorem 2.



---

*Source: 794262-2013-07-25.xml*