Ambarzumian-type problems for discrete Schr\"{o}dinger operators

We discuss the problem of unique determination of the finite free discrete Schr\"{o}dinger operator from its spectrum, also known as Ambarzumian problem, with various boundary conditions, namely any real constant boundary condition at zero and Floquet boundary conditions of any angle. Then we prove the following Ambarzumian-type mixed inverse spectral problem: Diagonal entries except the first and second ones and a set of two consecutive eigenvalues uniquely determine the finite free discrete Schr\"{o}dinger operator.


Introduction
The Jacobi matrix is a tridiagonal matrix defined as where n ∈ N, a k > 0 for any k ∈ {1, 2, . . . , n − 1} and b k ∈ R for any k ∈ {1, 2, . . . , n}. When a k = 1 for each k ∈ {1, 2, . . . , n − 1}, the Jacobi matrix (1.1) defines the finite discrete Schrödinger operator. Inverse spectral problems aim to recover {a k } n−1 k=1 and {b k } n k=1 from spectral information. Ambarzumian-type problems focus on inverse spectral problems for free discrete Schrödinger operators, i.e. a k = 1 and b k = 0 for every k, or similar cases when b k = 0 for some k.
In this paper, we first revisit the classical Ambarzumian problem for the finite discrete Schödinger operator in Theorem 3.3, which says that the spectrum of the free operator uniquely determines the operator. Then we provide a counterexample, Example 3.4, which shows that knowledge of the spectrum of the free operator with a nonzero boundary condition is not sufficient for unique recovery. In Theorem 3.5, we observe that a non-zero boundary condition along with the corresponding spectrum of the free operator is needed for the uniqueness result. However, in Theorem 3.6, we prove that for the free operator with Floquet boundary conditions, the set of eigenvalues including multiplicities is sufficient to get the uniqueness up to transpose.
We also answer the following mixed Ambarzumian-type inverse problem positively in Theorem 4.3.
The paper is organized as follows. In Sect. 2 we fix our notations. In Sect. 3 we consider the problem of unique determination of the finite free discrete Schrödinger operator from its spectrum, with various boundary conditions, namely any real constant boundary condition at zero, and Floquet boundary conditions of any angle. In Sect. 4 we prove the above mentioned Ambarzumian-type mixed inverse spectral problem.

Notations
Let us start by fixing our notation. Let S n represent the discrete Schrödinger matrix of size n × n Let us also introduce the following matrices: Let us denote the free discrete Schrödinger matrix of size n × n by F n : and F n (θ ) are defined accordingly. In addition, let p n (x) be the characteristic polynomial of F n with zeroes λ 1 , . . . , λ n and let q n (x) be the characteristic polynomial of S n with zeroes μ 1 , . . . , μ n . The discrete Schrödinger matrix S n has n distinct simple eigenvalues, so we can order the eigenvalues as λ 1 < λ 2 < · · · < λ n and μ 1 < μ 2 < · · · < μ n . This property is valid for any Jacobi matrix and can be found in [19], which provides an extensive study of Jacobi operators.
Throughout the paper we will discuss our results via matrices using the notations introduced above, but let us note that the matrices S n (b, B) and S n (θ ) can be represented as difference equations. Remark 2.1 Given S n , let us consider the discrete Schrödinger matrix where all b k 's are the same as in S n except b 1 and b n are replaced by b 1 + b and b n + B respectively for b, B ∈ R, i.e. S n + b(δ 1 , ·)δ 1 + B(δ n , ·)δ n . (2.1) The discrete Schrödinger matrix (2.1) is given by the difference expression with the boundary conditions In order to get a unique difference expression with these boundary conditions for a given discrete Schrödinger matrix, we can think of the first and the last diagonal entries of the matrix S n as the boundary conditions at 0 and n + 1 respectively. Therefore S n (b, B) denotes the discrete Schrödinger matrix with boundary conditions b at 0, B at n + 1.
If we consider the difference expression (2.2) with the Floquet boundary conditions then we get the matrix representation S n (θ ).

Ambarzumian problem with various boundary conditions
Firstly let us obtain the first three leading coefficients of q n (x). This is a well-known result, but we give a proof in order to make this section self-contained.

Lemma 3.1
The characteristic polynomial q n (x) of the discrete Schrödinger matrix S n has the form Proof The characteristic polynomial q n (x) is given by det(xI n − S n ). Let us consider Liebniz' formula for the determinants is an n × n matrix and sgn is the sign function of permutations in the permutation group S n . If we use (3.1) with the identity permutation, then det(xI n −S n ) becomes n i=1 (x − b i ), so we get that q n (x) is a monic polynomial and the coefficient of The coefficient of the x n−2 term of q n (x) is formed from the sum of all disjoint pairs of a i . Hence we obtain 1≤i< The only other permutation that will yield an x n−2 term is a transposition. However, if |i − j| ≥ 1, then the product will be zero. Thus we are looking for transpositions where |i − j| = 1. There are n − 1 of these, namely (1, 2), where r n−3 (x) is a polynomial of degree at most n − 3. Since the signature of a transposition is negative, we derive −x n−2 for each product. Summing over all n − 1 permutations and adding to 1≤i< j≤n b i b j yields our desired result.

Corollary 3.2
The characteristic polynomial p n (x) of the free discrete Schrödinger matrix F n has the form where P n−2 (x) is a polynomial of degree at most n − 2.
Now let us give a proof of Ambarzumian problem with Dirichlet-Dirichlet boundary conditions, i.e. for the matrix F n (0, 0) = F n in our notation. Proof In order for the two matrices to have all the same eigenvalues, they must have equal characteristic polynomials. Comparing the results from Lemma 3.1 to Corollary 3.2, we must have This leads us to conclude that which only occurs when all the b i are zero, i.e. S n = F n , since b i ∈ R for each i ∈ {1, · · · , n}.
A natural question to ask is whether or not we get the uniqueness of the free operator with non-zero boundary conditions. After Theorem 3.3, one may expect to get the uniqueness of a free discrete Schrödinger operator from a spectrum with nonzero boundary condition at 0. However, this is not the case because of the following counterexample: The matrices A and B have the same characteristic polynomial x 3 − 2x 2 − 2x + 2, so they share the same spectrum.
This example shows that Theorem 3.3 is a special case, so in order to get the uniqueness of a rank-one perturbation of the free operator, we also need to know the non-zero boundary condition along with the spectrum.  F n (b, 0). Now, we approach the Ambarzumian problem with Floquet boundary conditions. Let us recall that S n (θ ) and F n (φ) denote a discrete Schrödinger operator and the free discrete Schrödinger operator with Floquet boundary conditions for the angles 0 ≤ θ < 1 and 0 ≤ φ < 1, respectively.
The following theorem shows that with Floquet boundary conditions, the knowledge of the spectrum of the free operator is sufficient for the uniqueness up to transpose. Theorem 3.6 Suppose that S n (θ ) shares all of its eigenvalues with F n (φ), including multiplicity, for 0 ≤ θ, φ < 1. Then b 1 = · · · = b n = 0 and θ ∈ {φ, 1 − φ}, i.e. S n (θ ) Proof Let us define D[k, l] as the following determinant of a (l − k + 1) × (l − k + 1) matrix for 1 ≤ k < l ≤ n: Let us consider the characteristic polynomial of S n (θ ) by using cofactor expansion on the first row: Then by using cofactor expansions on the first row of the determinant in the second term and on the first column of the determinant in the third term we get Now let's use cofactor expansion on the first column of the determinant in the third term. We see that the determinant in the fourth term is the determinant of an upper triangular matrix. Therefore, Finally, observing that the determinant in the third term is that of a lower triangular matrix, we get At this point note that D[k, l] is the characteristic polynomial of the following discrete Schrödinger matrix Therefore using Lemma 3.1 and Eq. (3.4), we obtain where f n−3 is a polynomial of degree at most n − 3, independent of θ . Using the same steps for F n (φ), we obtain where g n−3 is a polynomial of degree at most n − 3, which is independent of φ. Comparing Eqs. (3.5) and (3.6), like we did in the Proof of Theorem 3.3, we can conclude that the diagonal entries {b i } n i=1 of S n (θ ) must be zero. Note that the expression consisting of the first three terms in the right end of (3.4), [3, n] − D[2, n − 1] is independent of θ . In addition, we observed that b 1 = · · · = b n = 0. Therefore using the equivalence of the characteristic polynomials of S n (θ ) and F n (φ), we obtain e 2πiθ + e −2πiθ = e 2πiφ + e −2πiφ , which can be written using Euler's identity as 2 cos(2πθ) = 2 cos(2πφ). (3.7) Equation (3.7) is valid if and only if θ differs from φ or −φ by an integer. Since 0 ≤ θ, φ < 1, the only possible values for θ are φ and 1 − φ. This completes the proof.

An Ambarzumian-type mixed inverse spectral problem
Let us recall that S n,m denotes the following n × n discrete Schrödinger matrix for 1 ≤ m ≤ n: and F n denotes the free discrete Schrödinger matrix of size n × n. In this section our goal is to answer the following Ambarzumian-type mixed spectral problem positively for the m = 2 case.
Inverse Spectral Problem If S n,m and F n share m consecutive eigenvalues, then do we get b 1 = · · · = b m = 0, i.e. S n,m = F n ? When m = 1, this problem becomes a special case of the following result of Gesztesy and Simon [10]. For a Jacobi matrix given as (1.1), let us consider the sequences {a k } and {b k } as a single sequence b 1 , a 1 , b 2 , a 2 , · · · , a n−1 , b n = c 1 , c 2 , · · · , c 2n−1 , i.e. c 2k−1 :=b k and c 2k :=a k .  Proof Following the notations of Theorem 4.1, c 2 , · · · , c 2n−1 are known for S n,1 . Hence by letting k = 1 in Theorem 4.1, we get S n,1 = F n . Now let us prove the m = 2 case. Let λ 1 < λ 2 < · · · < λ n denote the eigenvalues of F n , and letλ 1 <λ 2 < · · · <λ n denote the eigenvalues of S n,2 . Proof We start by proving the following claim.
Using cofactor expansion on the first row for the first term and the first column for the second term, we get Finally, using cofactor expansion on the first column of the second term, we get Note that the cofactor expansions we derived so far are given in [10] in a more general setting for Jacobi matrices. One just needs to look at (2.4) in [10] and keep in mind that (2.5) in [10] takes place.
First let us show that b 1 and b 2 have the same sign. If n is even and k = n/2, then λ k = −λ k+1 . Hence b 1 = b 2 = 0. If n is odd and k = (n − 1)/2 or k = (n + 1)/2, then one of the eigenvalues λ k or λ k+1 is zero, so b 2 is undefined. For all other values of k, two consecutive eigenvalues λ k and λ k+1 and hence b 1 and b 2 have the same sign.
Without loss of generality, let us assume both λ k and λ k+1 are negative and b 1 ≤ b 2 . Let us define the matrix M n (t) with the real parameter t as follows: Note that the kth eigenvalue of M n (−b 2 ) is greater than or equal toλ k , since M n (−b 2 ) ≥ S n,2 . Let us also note that M n (0) = F n . Let us denote the kth eigenvalue of M n (t) by λ k (t) and the corresponding eigenvector by X (t), normalized as ||X (t)|| = 1. Since M n (t) is a smooth function of t around 0 and λ k is a simple eigenvalue of F n , we get that λ k (t) is a simple eigenvalue of M n (t), λ k (0) = λ k , and λ k (t) and X (t) are smooth functions of t around 0 (see Theorems 9.7 and 9.8 in [14]). Let us also observe that M n (t) is self-adjoint, ||X (t)|| = 1 and M n (t)X (t) = λ k (t)X (t). Therefore the Hellmann-Feynman Theorem (Theorem 1.4.7 in [17]) implies where X (t) T = [X 1 (t), X 2 (t), . . . , X n (t)]. Since X (t) is a non-zero eigenvector of the tridiagonal matrix M n (t), at least one of X 1 (t) and X 2 (t) is non-zero. Therefore by Eq. (4.4), there exists an open interval I ⊂ R containing 0 such that λ k (t) < 0 for t ∈ I , i.e. λ k (t) is decreasing on I . This implies existence of 0 < t 0 < − b 2 satisfying This contradicts with our assumption that λ k =λ k . Therefore only the first case of the claim is true, i.e. b 1 = b 2 = 0 and hence S n,2 = F n .