Three spectra inverse Sturm – Liouville problems with overlapping eigenvalues

In the paper we show that the Dirichlet spectra of three Sturm–Liouville differential operators defined on the intervals [0, 1], [0, a] and [a, 1] for some a ∈ (0, 1) fixed, together with the knowledge of the normalizing constants corresponding to the overlapped eigenvalues, uniquely determine the potential q on [0, 1]. In this situation we also provide the algorithm for recovering the potential by using the given spectral data.


Introduction
In this paper we are concerned with the inverse problem for recovering the potential q on interval [0, 1] of a Sturm-Liouville equation lu := −u + q(x)u = λu (1.1) using the three spectra σ(L), σ(L − ) and σ(L + ) corresponding to three Sturm-Liouville operators L, L − and L + .These operators are generated in L 2 spaces by the differential expressions l defined on [0, 1], [0, a] and [a, 1], respectively, and the Dirichlet boundary conditions u(a) = 0 = u(1).
(1.4) been carried out by Gesztesy and Simon [9] under the more general situations of a ∈ (0, 1) and no-Dirichlet spectra.In particular, Gesztesy and Simon [9] proved uniqueness of the reconstructed q whenever the three spectra do not overlap and suggested a counterexample to uniqueness otherwise.They also raised an interesting assertion: We believe the analysis of the situation where σ(L − ) ∩ σ(L + ) has k-points will yield k-parameter sets of potentials q consistent with the given sets of eigenvalues (see [9, p. 91]).One of our purposes of this paper is to answer affirmatively the above assertion.More specifically, we shall show that the uniqueness result remains valid by using the three spectra provided that the normalizing constants α n corresponding to λ n are employed as the additional spectral data, when λ n ∈ σ(L − ) ∩ σ(L + ).This implies that when σ(L − ) ∩ σ(L + ) has kpoints, there then exists a k-parameter set of potentials q corresponding to the given three spectra.Our second purpose here is to reconstruct potentials in terms of the three spectra, together with the normalizing constants {α n } n∈Λ where The method used here is similar to that used solving half-inverse problem [15] and three spectra inverse problem without the overlapping eigenvalues [14].
Section 2 includes the uniqueness theorem and its proof.Section 3 gives the algorithm for the reconstruction of q in terms of given spectral data.

Uniqueness
In this section we shall prove our uniqueness result concerning with three spectra inverse Sturm-Liouville problems with overlapping eigenvalues.
Denote by u − (x, λ) and u + (x, λ) the solutions of Eq. (1.1) with the initial conditions It is known (see, e.g., [8] and [13, Lemma 3.4.2and proof of Theorem 3.4.1])that, for each x ∈ [0, 1], u ± (x, λ) are entire functions of λ and the eigenvalues {λ n } ∞ n=1 of the operator L are precisely zeros of the function ∆(λ) defined by where ∆(λ) is called the characteristic function of L, which also has the following representation: (2.5) where ∆ ± (λ) denote the characteristic functions of the operators L ± .Moreover, for any λ n ∈ σ(L), the normalizing constant, α n , associated with the λ n is defined by In virtue of the above preliminaries, we are now in a position to give the uniqueness result of this paper.Theorem 2.1.Let the Sturm-Liouville differential operators L and L ± be defined by (1.1)-(1.4).Let where Λ(⊂ N) is the associated index set.Then the potential q is uniquely determined almost everywhere on [0, 1] by the three spectra σ(L), σ(L ± ) and the normalizing constants {α n } n∈Λ .
Proof.Let us consider another operators L, L− and L+ of the same form (1.1)-(1.4)but with different coefficient q(x) on [0, 1].Denote by ∆(λ) and ∆± (λ) the characteristic functions of operators L and L± .Then by the hypotheses of Theorem 2.1, these operators and L and L ± have common spectra σ(L) and σ(L ± ) respectively; and normalizing constants α n for n ∈ Λ corresponding to the common eigenvalues We first prove that u ± (a, (2.10) Here u± = ∂u ± /∂λ.Since λ k is a simple eigenvalue of the operators L and L ± , it follows that u± (a, then we have the following equality [8]).Since α k = αk and ∆(λ) = ∆(λ) for all λ ∈ C, it follows that κ k = κk .This combined with (2.11) gives
By the discussion above, we see that u ± (a, µ ± k ) − ũ ± (a, µ ± k ) = 0 for all k ∈ N. Let us consider the function F(λ) defined as It is clear to check that F(λ) is an entire function and it has the following representation Here m − (a, λ) is called the Weyl m-function associated with the Sturm-Liouville operator The above argument implies that m ± (a, λ) ≡ m± (a, λ).By the uniqueness result of Marchenko [13], we have either q = q on [0, a] by using m − (a, λ) ≡ m− (a, λ) or q = q on [a, 1] by using m + (a, λ) ≡ m+ (a, λ).The proof is complete.

Reconstruction
By the uniqueness theorem above, we know that the solution of the inverse problem discussed by us, if exists, is unique.In this section we provide the way of recovering the potential q by using three Dirichlet spectra σ(L), σ(L ± ) and (if necessary) the normalizing constants {α n } n∈Λ corresponding to the overlapped eigenvalues λ n ∈ σ(L), where Λ = {n : λ n ∈ σ(L − ) ∩ σ(L + )}.The method used here is similar to that used solving half-inverse problem [15] and three spectra inverse problem [14] without the overlapping eigenvalues.
Based on condition that the potential q exists corresponding to our spectral data, according to [13, p. 32] we have Using (3.1) and (3.2), we obtain where ψ 1 (0) = 0, ψ 2 (0) = 0, Moreover, the K 1 (x, t) and K 2 (x, t) possess partial derivatives of the first order, and belonging to L 2 (0, a) and L 2 (a, 1), respectively, as a function of each of its variables when the other variable is fixed.From [10, Theorem 8] we see that two functional sequences sin γ − k t and sin γ + k (t − 1) ∞ k=1 are Riesz bases of L 2 [0, a] and L 2 [a, 1], respectively.This together with the fact that K 1,x (a, t) ∈ L 2 (0, a) and K 2,x (a, t) ∈ L 2 (a, 1) implies (see [13,Lemma 1.4.3]).On the other hand, letting L a denote the class of entire functions of exponential type ≤ a, which belong to L 2 (−∞, ∞) for real λ, then we have ψ 1 ∈ L a and ψ 2 ∈ L (1−a) .Now we are in a position to construct q.We begin by introducing the following lemma.