# Inverse Eigenvalue Problems for Singular Rank One Perturbations of a Sturm-Liouville Operator

Authors: Xuewen Wu
Journal: Advances in Mathematical Physics
 (2021)

---

## Abstract

This paper is concerned with the inverse eigenvalue problem for singular rank one perturbations of a Sturm-Liouville operator. We determine uniquely the potential function from the spectra of the Sturm-Liouville operator and its rank one perturbations.


---

## Body

## 1. Introduction

Consider the boundary problem(1)L~y≔−y′′+qxy+αδx−x0yx0=λyon 0,1 with
(2)y′0=y1=0,where qx∈L20,1 is real valued, α∈ℝ\0, x0∈0,1, and δx are the Dirac delta function. It is well known [1] that the operator L~ is a self-adjoint operator in L20,1, which is a singular rank one perturbation of the Sturm-Liouville operator Ly≔−y′′+qxy.The goal of this paper is to deal with the inverse problem of recovering the potentialqx in (1) from the spectra of L and L~, by applying the method in [1] and the perturbation theory for linear operators [2]. Note that in [3], the boundary problem
(3)−y′′+qxy=λy,with (2) and the discontinuous conditions
(4)yx0+0=yx0−0=yx0,(5)y′x0+0=y′x0−0+αyx0can be regarded as the problem (1)-(2). The research of this paper can be a variation of Borg’s two-spectra theorem [4] for the second spectrum that is obtained by attaching the interface conditions (4) and (5) to the problem (3) and (2). Our immediate motivation is a recent research of del Rio and Kudryavtsev [5, 6], who considered the inverse problem for Jacobi matrices and recovered the original matrix from the spectra of it and its interior mass-spring perturbation; they indicated that the uniqueness for the inverse problem does not remain valid in general; however, under certain conditions, there exists at most a finite number of matrices corresponding to the two spectra.Such operatorL~ appears not only in electronics but in other areas such as the theory of diffusion processes, see the related references in [7, 8]. Some spectral and inverse spectral problems for the Sturm-Liouville operator with rank one perturbations have been investigated in [1, 3, 9–20]. In particular, Albeverio, Hryniv, and Nizhnik [3] considered the inverse eigenvalue problem for the Sturm-Liouville operator with the point potential vxy,δL2 and the perturbation δx−x0y,vL2 where vx∈L20,1 and x0=1. Later, Nizhnik [16] continued the problem with x0∈0,1. However, we consider the inverse problem for the operator (1) with the potential qx∈L20,1 and the perturbation δx−x0y,δL2. And the potential may not be determined uniquely just by the spectra; so, we employ the addition information. Moreover, the approach we use can also solve the problem with the perturbation cxy,cL2,cx∈L20,1 in the research [18]. In the paper, we establish the expression of the characteristic function of L~ which provides a necessary preliminary for treating with its inverse eigenvalue problem. The approach we use to prove our results can convert the problem (1) and (2) into three spectra inverse problem in [21, 22]. Actually, the spectra of L and L~ may not determine the potential uniquely (see Remark below for details). The key difficulty encountered is to identify the eigenvalues of two Sturm-Liouville problems defined on 0,x0 and x0,1 from the knowledge of the spectra of L and L~, for which we have to employ addition information of the number of zeros of the eigenfunctions, and the condition two spectra are disjoint.The main result asserts that, if the spectra ofL and L~ are disjoint, the potential qx can be determined uniquely by the spectra of L and L~ and the numbers of zeros, contained in 0,x0, of all eigenfunctions of L. We will state and prove it in the next section.


## 2. The Main Theorem and Proof

We describe some preliminaries which will be needed subsequently, due to [1]. One defines the scale of spaces H±1L associated to L as follows. The space H+1L is DL1/2 with the norm
(6)φH+1=L+11/2φL20,1,in which H+1 is easily seen to be complete. For H−1L, take L20,1 with the norm given by
(7)φH−1=L+1−1/2φL20,1and complete it. Note that H+1L and H−1L are dual in such a way that φ∈H−1 is associated to the function y∈H+1 given by ∫01φxyxdx. A Sobolev estimate shows that, for any y∈H+1L,
(8)∫01yxδx−x0dx2=yx02≤cy,L+1y<∞;that is, δx−x0 lies in H−1L. Hence, by Section 2 in ([1], p. 115) Simon, one can find a spectral measure dμL such that
(9)Fz=∫01δx−x0,L−z−1δx−x0dx≡∫01dμLxx−z.By (I.15) and (I.16) in ([1], p. 116), it follows that
(10)TrL−z−1−L~−z−1=αdFz/dz1+αFz.The following preliminaries are due to ([2], p. 245-250). The multiplicity index for αFz+1 is given by
(11)vζ;αFz+1=k,ifζisazeroofαFz+1oforderk,−k,ifζisapoleofαFz+1oforderk,0,forallotherζ∈ℂ.The multiplicity function for a closed operatorT is defined by
(12)v~ζ;T=0,ifζbelongstotheresolventset,m,ifζisaneigenvalueofTwithmultiplicitym,+∞,forallotherζ∈ℂ.Letλn be the eigenvalue of L, n≥0. Then,
(13)v~ζ;L=1,ζ∈λnn=0∞,0,ℂ\λnn=0∞.Using the same way as the proof of the W-A formulas in ([2], p. 248), by (10), we deduce
(14)v~ζ;L~=vζ;αFz+1+v~ζ;L=1+vζ;αFz+1,ζ∈λnn=0∞,vζ;αFz+1,ζ∈ℂ\λnn=0∞.In the following lemma, we give the spectrum and a characteristic function ofL~.Lemma 1.
The spectrum ofL~ consists of real eigenvalues. The characteristic function of L~ is
(15)Δ~λ=ΔλαFλ+i0+1,where Δλ is the characteristic function of L.Proof.
By (9),
(16)Fz=Gx0,x0,z,where Gx,y,z is Green’s function for L. From ([23], p. 15, p.29), Gx0,x0,z is meromorphic with simple poles in the points z=λni. Then, for λ∈ℝ\λnii=0∞, it follows
(17)Fλ+i0=Fλ,(18)dFλ+i0dλ=∫dμLyy−λ2=dGx0,x0,λdλ<∞.
Notice that the spectrum ofL consists of simple real eigenvalues. Combining with Theorem I.6 in [1], we see
(19)ImFλ+i0=∞,λ∈λnn=0∞,0,λ∈ℝ\λnn=0∞,Thus, by Theorem II.2 in [1] and (18), the spectrum of L~ consists of real eigenvalues, denoted by μnn=0∞.
By (14), each zero of αFz+1 is an eigenvalue of L~. Hence, the zeros of αFz+1 are real and consist of the zeros of αFλ+i0+1. Let aii=0∞ be the zeros set of αFλ+i0+1. Then, we have the multiplicity function
(20)vζ;αFz+1=1,ζ∈aii=0∞,−1,ζ∈λnii=0∞,0,forotherζ∈ℂ.
Using (14), it follows
(21)v~ζ;L~=2,ζ∈aii=0∞∩λnn=0∞,1,ζ∈aii=0∞\aii=0∞∩λnn=0∞,1,ζ∈λnn=0∞\aii=0∞∩λnn=0∞∪λnii=0∞,0,forotherζ∈ℂ.
Hence, by the definition ofv~ζ;L~, we have
(22)μnn=0∞=aii=0∞∪λnn=0∞\λnii=0∞.
Now, we prove that, forx0∈ℝ, Δ~x0=0, if and only if x0∈μnn=0∞. In fact, if x0∈μnn=0∞, it is easily seen that Δ~x0=0. We just prove that if Δ~x0=0, then x0∈μnn=0∞. Suppose that Δ~x0=0 and x0∉μnn=0∞, then x0∈λnii=0∞, i.e., x0 is a pole of Fz. By Theorem 1.1.2 in [23], all zeros of Δλ are simple; so,
(23)dΔdλx0≠0.
By (15), (16), and ([23], p. 15), Δ~λ is derivable; thus,
(24)dΔ~dλx0=dΔdλx0αFx0+i0+1<∞,which is contradictory to that x0 which is the pole of Fz. Then, x0∈μnn=0∞.
For(25)d2Δ~dλ2μn≠0,we also need to prove that x0 is an eigenvalue of L~ of multiplicity 2 if and only if x0 is a zero of Fλ of multiplicity 2. Actually, if x0 is an eigenvalue of L~ of multiplicity 2, then by (21), there is x0∈aii=0∞∩λnn=0∞, and it is easily to get
(26)Δ~x0=dΔ~dxx0=0.
Now, we suppose thatx0 is a zero of Fλ of multiplicity 2, i.e., (26) holds. By Δ~x0=0, there is Δx0=0 or αFx0+i0+1=0. Combining with
(27)dΔ~dλx0=dΔdλx0αFx0+i0+1+Δx0ddλαFx0+i0+1=0,there is Δx0=0 and αFx0+i0+1=0 which mean x0∈aii=0∞∩λnn=0∞. Then, x0 is an eigenvalue of L~ of multiplicity 2. The proof is complete.☐The main result in this paper is as follows.Theorem 2.
Letλnn=0∞ and μnn=0∞ be the spectrum of L and L~, respectively. If λnn=0∞∩μnn=0∞=∅, the potential qx can be determined uniquely by λnn=0∞,μnn=0∞ and the numbers of zeros, contained in 0,x0, of all eigenfunctions of L.Proof.
By Theorem 1.1.4 in [23], the characteristic function of L can be calculated by
(28)Δλ=∏n=0∞λnπ/2+nπ21−λλn.
In the same way of the proof of Theorem 1.1.4 in [23], by Hadamard’s factorization theorem [24], the characteristic function of L~ can be calculated by
(29)Δ~λ=∏n=0∞μnπ/2+nπ21−λμn.
From the given spectraλn,μnn=0∞, we can determine the function Fλ+i0 by
(30)Fλ+i0=1αΔ~λΔλ−1.
From (16), ([23], p. 15) and (2) in [23],
(31)Fλ+i0=−φx0,λψx0,λΔλ,where φx,λ and ψx,λ are the solutions of (3) under the initial conditions φ0,λ=1,φ′0,λ=0 and ψ1,λ=0,ψ′1,λ=1. Then, we get the zeros of φx0,λψx0,λ, denoted by γnn=0∞. It should be noted that
(32)γnn=0∞≔γn−n=0∞∪γn+n=0∞,where γn−n=0∞ and γn+n=0∞ are the spectra of the following problems
(33)L−y=−y′′+qxyon0,x0,y′0=yx0=0L+y=−y′′+qxyonx0,1,yx0=y1=0,respectively.
Ifλnn=0∞∩μnn=0∞=∅, λn is the pole of Fλ+i0, and γn is the zero of Fλ+i0. Using (18), it implies that Fλ+i0 is monotonically increasing on ℝ\λnn=0∞. By (16) and ([23], p. 29, [30]),
(34)limλ⟶λn−Fλ+i0=+∞,limλ⟶λn+Fλ+i0=−∞,limλ⟶−∞Fλ+i0=0.
Consequently, there is no zero ofFλ+i0 in −∞,λ0 and exactly one zero in λn,λn+1. That means
(35)λn<γn<λn+1.
We identifyγn−n=0∞ and γn+n=0∞ from γnn=0∞ by virtue of the number of the zeros of the eigenfunction contained in 0,x0. As is known ([25], p. 15) that the eigenfunction φx,λi has exactly i zeros in 0,1, we suppose that the eigenfunction φx,λi has m (m≤i) zeros in 0,x0. By comparison theorem in ([25], p14), there are not fewer than m zeros of the eigenfunction φx,λi+1 contained in 0,x0. From Lemma 1.3.1 in [25], the roots of φx,λ depend continuously on λ. Combining with Corollary 1.3.2 in [25], φx,λi+1 has m+1 zeros in 0,x0 only if there exists γp−∈λi,λi+1 such that φx0,γp−=0. Therefore, if φx,λi+1 has m+1 zeros in 0,x0, then γi∈γn−n=0∞; otherwise, γi∈γn+n=0∞. Then, we obtain γn+n=0∞ and γn−n=0∞ from γnn=0∞ with the information of the number of zeros of the eigenfunctions φx,λi contained in 0,x0.
Next, we prove thatλnn=0∞, γn−n=0∞, and γn+n=0∞ determine qx. By (35) and Theorem 3.2 in [21], we see that λnn=0∞, γn−n=0∞, and γn+n=0∞ can uniquely determine qx, a.e., on 0,1. The proof is therefore complete.☐Remark 3.
Based on the proof of Theorem2, we know that two disjoint spectra λnn=0∞ and μnn=0∞ determine γnn=0∞ uniquely rather than γn−n=0∞ and γn+n=0∞. Thus, in order to obtain the uniqueness of qx, we need to identify γn−n=0∞ and γn+n=0∞ from γnn=0∞. To this end, we have to employ the number of zeros of eigenfunctions, and the condition two spectra are disjoint, if not, we may not guarantee the uniqueness of qx. For example, if x0=1/2, then the eigenvalues of L and L± have the following asymptotic expressions:
(36)λn=n+12π+wn+κnn,γn−=2n+1π+w−n+κn−n,γn+=2n+2π+w+n+κn+n,where w=1/2∫01qxdx,w−=∫01/2qxdx,w+=∫1/21qxdx and κn,κn−,κn+∈l2. It is easy to see that there exists a positive integer N such that, for n>N,(37)λ2n<γn−≔γ2n<λ2n+1<γn+≔γ2n+1<λ2n+2,while we cannot identify γn−n=0∞ and γn+n=0∞ from γnn=0∞ for n≤N, which means that there will be at most a finite number of qx corresponding to two spectra γnn=0∞ and λnn=0∞ in virtue of Theorem 3.2 in [21]. Hence, we employ the number of zeros of eigenfunctions, and the condition two spectra are disjoint to guarantee the uniqueness of qx. And we have not found an example that the joint spectra determine the potential uniquely.



---

*Source: 1681873-2021-08-04.xml*