Spectral Determinants and an Ambarzumian Type Theorem on Graphs

We consider an inverse problem for Schrödinger operators on connected equilateral graphs with standard matching conditions. We calculate the spectral determinant and prove that the asymptotic distribution of a subset of its zeros can be described by the roots of a polynomial. We verify that one of the roots is equal to the mean value of the potential and apply it to prove an Ambarzumian type result, i.e., if a specific part of the spectrum is the same as in the case of zero potential, then the potential has to be zero.


Introduction
Quantum graphs arise naturally as simplified models in mathematics, physics, chemistry, and engineering [3]. Ambarzumian's theorem in inverse spectral theory refers to a setting when a differential operator can be reconstructed from at most one spectrum due to the presence of a constant eigenfunction. The original theorem from 1929 states for q ∈ C[0, π] that if the eigenvalues of −y + q(x)y = λy y (0) = y (1) = 0 (1.1) are λ n = n 2 π 2 (n ≥ 0), then q = 0 [2]. Neumann boundary conditions are crucial here for otherwise the statement is not true. Eigenvalues other than zero are used only through eigenvalue asymptotics to get 1 0 q = 0; hence a subsequence λ r = r 2 π 2 +o(1) of them is sufficient to reach the same conclusion even if q ∈ L 1 (0, π). On finite intervals inverse eigenvalue problems have a This work was supported by the Hungarian NKFIH Grant SNN-125119. 24 Page 2 of 11 M. Kiss IEOT vast literature. In general we mention Horváth [18] and the fundamental paper of Borg [7] as well as the works referenced by and referencing these. For Ambarzumian's theorem a recent stability result is found in Horváth [19]. Boman et al. [6] proved that an interval with zero potential can be identified by its spectrum even if we only know a priori that the underlying space is a graph with appropriate boundary conditions. Let us turn to the list of extensions when the underlying space is a fixed graph. On a tree with edges of equal length knowing the smallest eigenvalue 0 exactly and a specific part of the spectrum approximately is enough for recover to the zero potential (Carlson and Pivovarchik [8]). On a tree with different edge lengths this is still true (see Lemma 4.4 of Law and Yanagida [25]), however, the required set of eigenvalues is given by an existence proof.
An ultimate generalization of Ambarzumian type scenarios with a fixed underlying space is found by Davies [12]. In an abstract framework which contains arbitrary graphs with arbitrary edge lengths as particular special cases he was able to recover to the zero potential by determining the mean value of the potential from heat trace asymptotics. In the graph setting there are also trace formulas, for example (8.14) of Bolte et al. [5] that could be used to reach the same conclusion. In the context of inverse eigenvalue problems this is not entirely satisfactory however, as spectral informations are the eigenvalues themselves and it is not clear how to apply trace formulas when only a small part of the spectrum is known. We look for instead a minimal set of eigenvalues which determines the mean value of the potential and give an answer to that question if all edge lengths are equal.
For a summary on differential operators on graphs, see Pokornyi and Borovskikh [30] and Kuchment [23].
In this paper we consider a connected graph G(V, E) with edges of equal length. The graph can contain loops and multiple edges. We parametrize each edge with x ∈ (0, 1). This gives an orientation on G. We consider a Schrödinger operator with potential q j (x) ∈ L 1 (0, 1) on the edge e j and with Neumann (or Kirchhoff) boundary conditions (sometimes called standard matching conditions), i.e., solutions are required to be continuous at the vertices and, in the local coordinate pointing outward, the sum of derivatives is zero. More formally, consider the eigenvalue problem on e j for all j with the conditions  The spectral determinant or alternatively functional determinant or characteristic function of the problem (1.2)-(1.4) is a meromorphic function whose zeros coincide with its spectrum. Spectral determinants have been subject to continuous attention in the theoretical physics literature for the last 20 years [1,9,11,14,15,17,20,28,31]. As a tool for proving our Ambarzumian type results, we give a formula for the spectral determinant of Schrödinger operators (see (2.9)) and for the asymptotic distribution of some of its zeros. It is already known that for finite connected graphs, the main term of the eigenvalue asymptotics can be obtained from Weyl's law and the next terms depend on complicated combinations of 1 0 q j , j = 1, . . . , |E| (Möller and Pivovarchik [27], p. 213). We express some of these combinations as the roots of a polynomial (see (2.11)) and prove that for any connected, equilateral graph there is a root equal to the mean value of the potential. For the convenience of the reader, we emphasize this result in the context of spectral determinants: Theorem 1.1' Consider a connected graph G(V, E) with edges of equal length. The spectral determinant of Schrödinger operators on G with standard matching conditions has a sequence of roots which asymptotically differ by the mean value of the potential from the corresponding sequence of roots of the spectral determinant of the free Schrödinger operator. Precisely, the problem It is important to note at this point that the presence of the potential does not simply shift the spectrum of the Laplacian. This will be clear from (2.11). A related result is Theorem 2.1 of de Verdière [13]. One might also expect that a variant of Theorem 1.1 holds in the case of possibly different edge lengths though it seems hard to to describe the asymptotic distribution of the eigenvalues in question.
For our Ambarzumian type result, we need an improved version of this theorem: Among these eigenvalues at least one has the asymptotics (1). Moreover, if G is a bipartite graph, the same is true for k instead of 2k, i.e., there are exactly |E| − |V | + 2 eigenvalues (counting multiplicities) such that λ = (kπ) 2 + O(1) and at least one has the asymptotics λ = (kπ Although G is a directed graph, if we reverse an edge e j and change the potential to q j (1 − x) on it, we get an eigenvalue problem with the same eigenvalues and eigenfunctions (which are reversed with respect to the new direction on e j ).
Remark. For q = 0 and λ = 2k 2 π 2 (k ∈ Z + ) one eigenfunction is cos 2kπx and in each cycle there is a Dirichlet eigenfunction ± sin 2kπx on its edges (depending on the direction) and 0 everywhere else (Kurasov [24], Theorem 2(4)). In the bipartite case we can assume that V is a disjoint union of V 1 and V 2 , and that every edge points from V 1 to V 2 . Then for k odd, we can take cos kπx on all edges; besides, there are Dirichlet eigenfunctions, ± sin kπx alternately on the edges of a cycle and 0 elsewhere. Hence in both cases there are |E| − |V | + 2 independent eigenfunctions. We shall prove that the multiplicity is not greater. Remark. For a tree |E| − |V | + 2 = 1, and a tree is bipartite, hence Theorem 1.2 is a generalization of Theorem 1.2 in Carlson and Pivovarchik [8].
If the graph represents an electrical circuit in which each edge has a unit resistance, the effective resistance of an edge can be computed (or in mathematics, defined) by terms of the graph Laplacian. A result of Kirchhoff [22] is that the effective resistance of an edge e can be expressed as the number of spanning trees containing e divided by the number of all spanning trees. Interesting examples are the complete graph (if |V | > 2) or a graph with one point and a loop, which corresponds to the case of periodic boundary conditions treated in Cheng et al. [10]. Figure 1 shows a non edge-transitive example (Göbel [16]).

The Proof
Denote by c j (x, λ) the solution of (1.2) which satisfies the conditions c j (0, λ)− 1 = c j (0, λ) = 0 and by s j (x, λ) the solution of (1.2) which satisfies the conditions s j (0, λ) = s j (0, λ) − 1 = 0. Each y j (x, λ) may be written as a linear combination Then y j (0, λ) = A j (λ) is the same on each outgoing edge; hence we choose to index the functions A(λ) by vertices, and then if e j starts from v. If the eigefunctions are normalized, i.e., j y j (x, λ) [8].
where in the first sum v j denotes the starting point of e j ; and for all e j ∈ E(G), if e j points from u to v. The matrix of this homogeneous linear system of equations has the form • A is like an adjacency matrix; a vu = 1 √ λ c j (1, λ), the sum is taken on edges pointing from u to v; • B and C are like incidence matrices; The determinant of the matrix M is the so-called spectral determinant of the problem (1.2)-(1.4).

Example 1. Consider a single vertex with a loop. Then
Example 2. Consider a star graph with root r and vertices u, v and w. Let e 1 , e 2 and e 3 point from u, v and w to r, respectively. We choose to index rows and columns by r, u, v, w, e 1 , e 2 , e 3 , in that order. Then the matrix [29]).
where the sum is taken for all spanning trees τ of G.
Proof Proof. In the Leibniz formula every nonzero term (in fact there is only one) contains exactly one element from X, hence it is enough to prove this for s = 1. Let the indices of the nonzero element in X be uv. First we prove that the determinant of M 1 is independent of v. Indeed, there is a path in the tree between two arbitrary vertices, hence we can add to (or subtract from) row u the rows corresponding to the vertices of that path. Similarly, the determinant does not depend on u. Reversing an edge in G does not change the determinant; using this and by adding rows (and corresponding columns) we can assume that τ is a path. Then taking u = |V |, v = 1 and expanding the determinant from left to the right we get ((−1) |V |) ) |V |−1 = 1. M. Kiss IEOT disjoint union of V 1 and V 2 such that all edges connect V 1 to V 2 , then let us multiply by (−1) the rows in R corresponding to V 1 and the columns in R T corresponding to V 2 , respectively. This multiplies the determinant by (−1) |V | leaving the nonzero element of X unchanged for (uv) ∈ E(G). Hence the statement follows from the previous lemma.
Then consider the eigenvalue problem −y + tq j (x)y = λy with the boundary conditions (1.3)-(1.4) for t ∈ [0, 1]. The corresponding operator T (t) forms a holomorphic family and its eigenvalues λ n (t) and normalized eigenfunctions g n (t) can be represented by holomorphic functions of t (applying Theorems VII-1.8 and II-6.1 from Kato's book [21]). λ n (t) = g n (t), T (t)g n (t) =