# A Uniqueness Theorem for Bessel Operator from Interior Spectral Data

Authors: Murat Sat; Etibar S. Panakhov
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Inverse problem for the Bessel operator is studied. A set of values of eigenfunctions at some internal point and parts of two spectra are taken as data. Uniqueness theorems are obtained. The approach that was used in investigation of problems with partially known potential is employed.


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## Body

## 1. Introduction

Inverse spectral analysis involves the problem of restoring a linear operator from some of its spectral parameters. Currently, inverse problems are being studied for certain special classes of ordinary differential operators. The simplest of these is the Sturm-Liouville operatorLy=-y′′+q(x)y. For the case where it is considered on the whole line or half line, the Sturm-Liouville operator together with the function q(x) has been called a potential. In this direction, Borg [1] gave important results. He showed that, in general, one spectrum does not determine a Sturm-Liouville operator, so the result of Ambarzumyan [2] is an exception to the general rule. In the same paper, Borg showed that two spectra of a Sturm-Liouville operator determine it uniquely. Later, Levinson [3], Levitan [4], and Hochstadt [5] showed that when the boundary condition and one possible reduced spectrum are given, then the potential is uniquely determined. Using spectral data, that is, the spectral function, spectrum, and norming constant, different methods have been proposed for obtaining the potential function in a Sturm-Liouville problem. Such problems were subsequently investigated by other authors [4–6]. On the other hand, inverse problems for regular and singular Sturm-Liouville operators have been extensively studied by [7–15].The inverse problem for interior spectral data of the differential operator consists in reconstruction of this operator from the known eigenvalues and some information on eigenfunctions at some internal point. Similar problems for the Sturm-Liouville operator and discontinuous Sturm-Liouville problem were formulated and studied in [16, 17].The main goal of the present work is to study the inverse problem of reconstructing the singular Sturm-Liouville operator on the basis of spectral data of a kind: one spectrum and some information on eigenfunctions at the internal point.Consider the following singular Sturm-Liouville operatorL satisfying (1)–(3):
(1)Ly=-y′′+[ℓ(ℓ+1)x2+q(x)]y=λy,0<x<1
with boundary conditions,
(2)y(0)=0,(3)y′(1,λ)+Hy(1,λ)=0,
where q(x) is a real-valued function and q∈L2(0,1), λ spectral parameter, ℓ∈ℕ0,  H∈ℝ. The operator L is self adjoint on the L2(0,1) and has a discrete spectrum {λn}.Let us introduce the second singular Sturm-Liouville operatorL~ satisfying
(4)L~y=-y′′+[ℓ(ℓ+1)x2+q~(x)]y=λy,0<x<1,
subject to the same boundary conditions (2), (3), where q~(x) is a real-valued function and q~∈L2(0,1). The operator L~ is self adjoint on the L2(0,1) and has a discrete spectrum {λ~n}.


## 2. Main Results

Before giving some results concerning the Bessel equation, we should give its physical properties. The total energy of the particle is given byE=p2/2M=ℏ2k2/2M=k2, where p is its initial or final momentum, and k the corresponding wave number, ℏ planck constant, M particle's mass, and E energy. The reduced radial Schrödinger equation for the partial wave of angular momentum ℓ then reads [18]
(5)d2dr2Ψ1(k,r)+(k2-ℓ(ℓ+1)r2)Ψ1(k,r)=V(r)Ψ1(k,r).
When V=0, the above equation reduces to the classical Bessel equation in the form
(6)d2dr2Ψ1(k,r)+(k2-ℓ(ℓ+1)r2)Ψ1(k,r)=0.
This equation has the solution Jℓ(r), called the Bessel function.Eigenvalues of the problem (1)–(3) are the roots of (3). This spectral characteristic satisfies the following asymptotic expression [19, 20]:
(7)λn=(n+ℓ2)2π2+∫01q(x)dx-l(l+1)+an,
where the series ∑n=1∞‍an2<∞. Next, we present the main results in this paper. When b=1/2, we get the following uniqueness Theorem 1.Theorem 1.
If for everyn∈ℕ one has
(8)λn=λ~n,yn′(1/2)yn(1/2)=y~n′(1/2)y~n(1/2)
then
(9)q(x)=q~(x)a.eontheinterval(0,1).In the caseb≠1/2, the uniqueness of q(x) can be proved if we require the knowledge of a part of the second spectrum.Let{m(n)}n∈ℕ be a sequence of natural numbers with a property
(10)m(n)=nσ(1+ɛn),0<σ≤1,ɛn⟶0.Lemma 2.
Let{m(n)}n∈ℕ be a sequence of natural numbers satisfying (10) and b∈(0,1/2) are so chosen that σ>2b. If for any n∈ℕ(11)λm(n)=λ~m(n),ym(n)′(b)ym(n)(b)=y~m(n)′(b)y~m(n)(b)
then
(12)q(x)=q~(x)a.eon(0,b].Let{l(n)}n∈ℕ and {r(n)}n∈ℕ be a sequence of natural numbers such that
(13)l(n)=nσ1(1+ɛ1,n),0<σ1≤1,ɛ1,n⟶0,(14)r(n)=nσ2(1+ɛ2,n),0<σ2≤1,ɛ2,n⟶0
and let μn be the eigenvalues of (1), (2), and (15) and μ~n be the eigenvalues of (4), (2), and (15)
(15)y′(1,λ)+H1y(1,λ)=0,H≠H1.
Using Mochizuki and Trooshin's method from Lemma 2 and Theorem 1, we will prove that the following Theorem 3 holds.Theorem 3.
Let{l(n)}n∈ℕ and {r(n)}n∈ℕ be a sequence of natural numbers satisfying (13) and (14), and 1/2<b<1 are so chosen that σ1>2b-1,σ2>2-2b. If for any n∈ℕ one has
(16)λn=λ~n,μl(n)=μ~l(n),yr(n)′(b)yr(n)(b)=y~r(n)′(b)y~r(n)(b)
then
(17)q(x)=q~(x)a.eon(0,1).


## 3. Proof of the Main Results

In this section, we present the proofs of main results in this paper.Proof of Theorem1.
Before proving Theorem1, we will mention some results, which will be needed later. We get the initial value problems
(18)-y′′+[ℓ(ℓ+1)x2+q(x)]y=λy,0<x<1,(19)y(0)=0,(20)-y~′′+[ℓ(ℓ+1)x2+q~(x)]y~=λy~,0<x<1,(21)y~(0)=0.
As known from [18], Bessel's functions of the first kind of order v=ℓ-1/2 are
(22)Jv(x)=∑k=0∞‍(-1)kxv+2k2v+2kk!Γ(v+k+1)
and asymptotic formulas for large argument
(23)Jv(x)=2πx{cos[x-vπ2-π4]+O(1x)},Jv′(x)=-2πx{sin[x-vπ2-π4]+O(1x)}.
It can be shown [19] that there exists a kernel H(x,t)(H~(x,t)) continuous in the triangle 0≤t≤x≤1 such that by using the transformation operator every solution of (18), (19) and (20), (21) can be expressed in the form [8, 21],
(24)y(x,λ)=x(λ)vJv(λx)+∫0xH(x,t)t(λ)vJv(λt)dt,(25)y~(x,λ)=x(λ)vJv(λx)+∫0xH~(x,t)t(λ)vJv(λt)dt,
respectively, where the kernel H(x,t)  (H~(x,t)) is the solution of the equation
(26)∂2H(x,t)∂x2+ℓ(ℓ+1)x2H(x,t)=∂2H(x,t)∂t2+(ℓ(ℓ+1)t2+q(t))H(x,t)
subject to the boundary conditions
(27)2dH(x,x)dx=q(x),limt→0H(x,t)tv-1/2=0,[Jv′(t,λ)=O(tv-1/2)].
After the transformations
(28)ξ=14(x+t)2,η=14(x-t)2,H(x,t)=(ξ-η)-v+1/2U(ξ,η),
we obtain the following problem:
(29)∂2U∂ξ∂η-14(ξ-η)∂U∂ξ+14(ξ-η)∂U∂η=14ξηq(ξ+η)U,U(ξ,ξ)=0,∂U∂ξ+αξU=14q(ξ)ξv-1,α=-v+12.
This problem can be solved by using the Riemann method [21].
Multiplying (18) by y~(x,λ) and (20) by y(x,λ), subtracting and integrating from 0 to 1/2, we obtain
(30)∫01/2(q(x)-q~(x))y(x,λ)y~(x,λ)dx=[y~(x,λ)y′(x,λ)-y(x,λ)y~′(x,λ)]|01/2.
The functions y(x,λ) and y~(x,λ) satisfy the same initial conditions (19) and (21), that is,
(31)y~(0,λ)y′(0,λ)-y(0,λ)y~′(0,λ)=0.
Let
(32)Q(x)=q(x)-q~(x),(33)K(λ)=∫01/2Q(x)y(x,λ)y~(x,λ)dx.
If the properties of y(x,λ) and y~(x,λ) are considered, the function K(λ)  is an entire function.
Therefore the condition of Theorem1 implies
(34)y~(12,λn)y′(12,λn)-y(12,λn)y~′(12,λn)=0
and hence
(35)K(λn)=0,n∈ℕ.
In addition, using (24) and (33) for 0<x<1,
(36)|K(λ)|≤M1λv,
where M is constant.
Introduce the function(37)W(λ)=y′(1,λ)+Hy(1,λ).
By using the asymptotic forms of y and y′, we obtain
(38)W(λ)=λsin(λ-vπ2-π4)+O(1).
The zeros of W(λ) are the eigenvalues of L and hence it has only simple zeros λn because of the seperated boundary conditions. From (38), W(λ) is an entire function of order 1/2 of λ. Since the set of zeros of the entire function W(λ) is contained in the set of zeros of K(λ), we see that the function
(39)Ψ(λ)=K(λ)W(λ)
is an entire function on the parameter λ. From (36), (38), and (39), we get
(40)|Ψ(λ)|=O(1λv+1/2).
So, for all λ, from the Liouville theorem,
(41)Ψ(λ)=0,
or
(42)K(λ)=0.
It was proved in [19] that there exists absolutely continuous function H~~(x,τ) such that we have
(43)y(x,λ)y~(x,λ)=12{∫1+cos2[λx-vπ2-π4]+∫0xH~~(x,τ)×cos2[λτ-vπ2-π4]dτ},
where
(44)H~~(x,t)=2[H(x,x-2τ)+H~(x,x-2τ)]+2[∫-x+2τxH(x,s)H~(x,s-2τ)ds+∫-xx-2τH(x,s)H~(x,s+2τ)ds].
We are now going to show that Q(x)=0 a.e. on (0,1/2]. From (33), (43) we have
(45)12∫01/2Q(x){∫1+cos2[λx-vπ2-π4]+∫0xH~~(x,τ)×cos2[λτ-vπ2-π4]dτ}dx=0.
This can be written as
(46)∫01/2Q(x)dx+∫01/2cos2[λτ-vπ2-π4]×[Q(τ)+∫01/2Q(x)×H~~(x,τ)dx∫01/2]dτ=0.
Let λ→∞ along the real axis, by the Riemann-Lebesgue lemma, one should have
(47)∫01/2Q(x)dx=0,∫01/2cos2[λτ-vπ2-π4]×[Q(τ)+∫τ1/2Q(x)H~~(x,τ)dx]dτ=0.
Thus from the completeness of the functions cos, it follows that
(48)Q(τ)+∫τ1/2Q(x)H~~(x,τ)dx=0,0<x<12.
But this equation is a homogeneous Volterra integral equation and has only the zero solution. Thus we have obtained
(49)Q(x)=q(x)-q~(x)=0,
or
(50)q~(x)=q(x)
almost everywhere on (0,1/2]. Therefore Theorem 1 is proved.Theorem 4.
To prove thatq(x)=0 on [1/2,1) almost everywhere, we should repeat the above arguments for the supplementary problem
(51)Ly=-y′′+[ℓ(ℓ+1)(1-x)2+q(1-x)]y,0<x<1
subject to the boundary conditions
(52)y(1)=0,y′(0,λ)+Hy(0,λ)=0.Consequently(53)q(x)=q~(x)a.eontheinterval(0,1).Next, we show that Lemma2 holds.Proof of Lemma2.
As in the proof of  Theorem1 we can show that
(54)G(ρ)=∫0bQ(x)y(x,λ)y~(x,λ)dx=[y~(x,λ)y′(x,λ)-y(x,λ)y~′(x,λ)]|x=b,
where ρ=λ=reiθ and Q(x)=q(x)-q~(x). From the assumption
(55)ym(n)′(b)ym(n)(b)=y~m(n)′(b)y~m(n)(b)
together with the initial condition at 0 it follows that,
(56)G(ρm(n))=0,n∈ℕ.
Next, we will show that G(ρ)=0 on the whole ρ plane. The asymptotics (23) imply that the entire function G(ρ) is a function of exponential type ≤2b.
Define the indicator of functionG(ρ) by
(57)h(θ)=limr→∞supln|G(reiθ)|r.
Since |Imλ|=r|sinθ|, θ=argλ from (23) it follows that
(58)h(θ)≤2b|sinθ|.
Let us denote by n(r) the number of zeros of G(ρ) in the disk {|ρ|≤r}. According to [22] set of zeros of every entire function of the exponential type, not identically zero, satisfies the inequality
(59)limr→∞infn(r)r≤12π∫02πh(θ)dθ,
where n(r) is the number of zeros of G(ρ) in the disk |ρ|≤r. By (58),
(60)12π∫02πh(θ)dθ≤bπ∫02π|sinθ|dθ=4bπ.
From the assumption and the known asymptotic expression (7) of the eigenvalues λn we obtain
(61)n(r)≥2∑(πn/σ)[1+O(1/n)]<r‍1=2πσr(1+o(1)),r⟶∞.
For the caseσ>2b,
(62)limr→∞n(r)r≥2πσ>4bπ=2b∫02π|sinθ|dθ≥12π∫02πh(θ)dθ.
The inequalities (59) and (62) imply that G(ρ)=0 on the whole ρ plane.
Similar to the proof of Theorem1, we have
(63)q(x)=q~(x)a.eontheinterval(0,b].
This completes the proof of Lemma 2.Now we prove that Theorem3 is valid.Proof of Theorem3.
From(64)λr(n)=λ~r(n),yr(n)′(b)yr(n)(b)=y~r(n)′(b)y~r(n)(b),
where {r(n)}n∈ℕ satisfies (14) and σ2>2-2b. Similar to the proof of Lemma 2, we get
(65)q(x)=q~(x)a.eon[b,1).
Thus, it needs to be proved that q(x)=q~(x) a.e on (0,b]. The eigenfunctions yn(x,λn) and y~n(x,λn) satisfy the same boundary condition at 1. It means that
(66)yn(x,λn)=ξny~n(x,λn)
on [b,1] for any n∈ℕ where ξn are constants.
Letρn=λn, sn=μn. From (54) and (66) we obtain
(67)G(ρn)=0,n∈ℕ,G(sln)=0,n∈ℕ.
We are going to show that inequality (59) fails and consequently, the entire function of exponential type G(ρ) vanishes on the whole ρ-plane. The ρn and sn have the same asymptotics (7). Counting the number of ρn and sn located inside the disc of radius r, we have
(68)1+2πr[1+O(1n)]
of ρn's and
(69)1+2πrσ1[1+O(1n)].
of sn's.
This means that(70)n(r)=2+2π[r(σ1+1)+O(1n)],limr→∞n(r)r=2π(σ1+1).
Repeating the last part of the proof of Lemma 2, and considering the condition σ1>2b-1, we can show that G(ρ)=0 identically on the whole ρ-plane which implies that
(71)q(x)=q~(x)a.eon(0,b]
and consequently
(72)q(x)=q~(x)a.eon(0,1).
Hence the proof of Theorem3 is completed.



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