On inverse problem for tree of Stieltjes strings

For a given metric tree and two strictly interlacing sequences of numbers there exits a distribution of point masses on the edges (which are Stieltjes strings) such that one of the sequences is the spectrum of the spectral problem with the Neumann condition at the root of the tree while the second sequence is the spectrum of the spectral problem with the Dirichlet condition at the root. Анотація. Для заданого метричного дерева та двох послідовностей чисел, які чергуються, існує розподіл точкових мас на ребрах (котрі є стільтьєсівськими струнами) такий, що одна з послідовностей є спектром спектральної задачі з умовою Діріхле у корені, тоді як друга послідовність є спектром задачі з умовою Неймана у корені.


INTRODUCTION
Finite-dimensional spectral problems on an interval were considered in [5] (some recent results see in [16,17] and applications in [1,4]). Finitedimensional spectral problems on graphs occur in various fields of physics: mechanics of transverse vibrations of strings [6][7][8], longitudinal vibrations of point masses joined by springs [11], synthesis of electrical circuits [3,9]. By inverse problem we mean recovering the parameters of the problem using the spectra of system vibrations. Such a problem on an interval was completely solved in [5], see [18] for generalization. For a star graph the inverse problem was solved in [2,14,15,19]. In these papers conditions on sequences of numbers where obtained necessary and sufficient to be the 2 A. Dudko, V. Pivovarchik spectra of spectral problems on the whole graph and on the edges of it. It was shown that the solution of this inverse problem is unique only in the case of simple spectra.
Spectral problems on trees were considered in [12] where it was assumed that the interior vertices of the tree are free of point masses. In this paper we consider the case where point masses can be presented at the interior vertices.
In Section 2 we describe the spectral problems generated by Stieltjes string equations on a tree with Dirichlet and Neumann conditions at the root which is any of pendant vertices. Section 3 contains some auxiliary results on Stieltjes functions. In Section 4 we solve the direct problem, i.e. we show that the quotient of characteristic polynomials of the Dirichlet and Neumann problems can be expanded into branching continued fraction.
In Section 5 we solve the inverse problem of recovering values of point masses and the lengths of subintervals between them using the total lengths of the edges, the spectra of two problems: with the Dirichlet condition at the root, first, and the Neumann condition, the second (Theorem 5.1). Also we consider the inverse problem of recovering values of point masses and the lengths of subintervals between them using the numbers of masses on the edges, the spectra of two problems: with the Dirichlet condition at the root, first, and the Neumann condition, the second (Theorem 5.2).

STATEMENT OF SPECTRAL PROBLEMS
Let T be a plane metric tree with q ě 2 edges. We denote by v i the vertices, by d(v i ) their degrees, by e j the edges, by l j their lengths by n j ě 0 the numbers of point masses m n j which divide the string into the subintervals l n j at an interior vertex. An arbitrary vertex is chosen to be the root. The pendant vertices are free of masses. All the edges we direct away from the root.
The root v is the beginning of a subinterval of length l (j) 0 on each edge e j incident with the root. Each other vertex v i have one incoming edge e j ending with a subinterval of the length l (j) n j , while each outgoing edge e r begins at v i with an interval of lengths l It is assumed that the tree is stretched and the pendant vertices except the root are fixed. We consider two cases: in the first the root is fixed (Dirichlet problem) while in the second the root is free to move in the direction orthogonal to the equilibrium position of the tree (Neumann problem). The tree can vibrate in the direction orthogonal to the equilibrium position of the strings. The transverse displacement of the mass m  If an edge e j is incoming for an interior vertex v i then the displacement of the incoming end of the edge is denoted by w (j) n j +1 (t), while if an edge e r is outgoing for a vertex v i then the displacement of the outgoing end of the edge is denoted by w (r) 0 (t). Using such notation vibrations of the tree can be described by the system of equations n j ą 0. For each interior vertex v i (except of the root) with incoming edge e j and outgoing edges e r we impose the continuity conditions w (r) for all r corresponding to outgoing edges. Balance of forces at such a vertex v i implies where the sum is taken over all the outgoing edges. For an edge e j incident with a pendant vertex (except of the root) we impose the Dirichlet boundary condition: At the root we will consider the Dirichlet condition w(t) = 0, i.e. w To obtain the Neumann boundary value problem we change (2.5) for the generalized Neumann conditions w (j) for all j and l corresponding to the edges incident with the root and We exclude from consideration the case with a mass at the root. If the root is a pendant vertex then we denote by e 1 the edge incident with the root. In this case the Dirichlet condition (2.5) at the root is and the Neumann conditions (2.6), (2.7) in this case can be reduced to ) we obtain the corresponding spectral problems: Dirichlet problem. For each edge: For each interior vertex (except for the root) with incoming edge e j and outgoing edges e r we have For an edge e j incident with a pendant vertex (except of the root) we have the Dirichlet boundary condition: At the root we have the Dirichlet condition for all j corresponding to the edges incident with the root. If the root is a pendant vertex and e 1 the edge incident with the root then instead of (2.14) at the root we have Neumann problem. The Neumann problem on T consists of equations (2.10)-(2.13), of for all j and l corresponding to the edges incident with the root and of If the root is a pendant vertex and e 1 the edge incident with the root then instead of (2.16) and (2.17) at the root we have (2.18)

AUXILIARY RESULTS
Here we give some lemmas which are used in Section 4.

Definition 3.1.
A rational function f (z) is said to be a Nevanlinna function if: (i) it is analytic in the half-planes Imz ą 0 and Imz ă 0; The following lemmas are obvious.
is an S 0 -function if and only if C ą 0 and where A k ą 0 for k = 1, 2, . . . , n and

DIRECT PROBLEM
Here and in the sequel we consider a tree T rooted at a pendant vertex. The Dirichlet problem on this tree consists of equations (2.10)-(2.13) and (2.15), while the Neumann problem on it consists of (2.10)-(2.13) and (2.18).
First of all we notice that interior vertices of degree 2 do not influence the results and we can assume absence of such vertices without losses of generality. Let P be a path in the tree T involving the maximum number of masses. Obviously it starts and finishes with pendant vertices. We denote the initial vertex by v 0 and choose it as the root of the tree. The enumeration of other vertices is arbitrary. We direct the edges away from the root. Denote by e i the edge incoming into a vertex v i for all i. Then Here r is a combinatorial length of the path. Deleting v 0 and e 1 we obtain a new tree T 1 rooted at the vertex v 1 .
Since d(v 1 ) ą 2 we can divide our tree T 1 into subtrees having v 1 as the only common vertex. (We say that T 1 1 , T 1 2 ,…, T 1 d(v 1 )´1 are complementary subtrees of T 1 (see Fig. 4.1).
Denote by ϕ N (v) the characteristic polynomial of problem (2.10)-(2.13), (2.18) on the tree T and by ϕ D (v) the characteristic polynomial of problem (2.10)-(2.13), (2.15) on this tree. These polynomials are normalized such that is the characteristic polynomial of the Dirichlet problem (2.10)-(2.13), (2.15) on T 1 r and ϕ N,r(v 1 ) (z) is the characteristic polynomial of the Neumann (2.10)-(2.13), (2.18) on T 1 r and this expansion can be continued. The following result was proved in [12] (see the proof of Corollary 2.9 there). In that paper it was assumed absence of point masses at the interior vertices of the tree (l can be expanded in continued fraction .

Remark 4.2.
It is proved in [13] that the number of distinct eigenvalues of each of the problems (2.10)-(2.13), (2.15) and (2.10)-(2.13), (2.18) on a tree bearing masses on every edge is not less than the maximum number of point masses on a path in this tree.

INVERSE PROBLEM
Theorem 5.1. Suppose tµ k u n k=´n, k =0 and tν k u n k=´n, k =0 are symmetric (µ´k =´µ k , ν´k =´ν k ) and monotonic sequences of real numbers which interlace:

Proof. First of all we consider the rational function
where Φ T,v ą 0 is the form characteristic of the tree T which depends only on the form of the tree and the lengths of the edges. It can be found Let e 1 be the edge connecting v with v 1 and let l 1 be the length of this edge. Substituting z = 0 in (4.1) we obtain Due to (5.1) and Φ T,v ą 0, F´1(z) is an S 0 -function and, therefore can be presented as where a k ą 0 for k = 0, 1, . . . , n and b k ą 0 for k = 1, 2, . . . , n.
Since F´1(0) = n ř k=0 a k = Φ´1 T,v ě l 1 , we can choose the integer number and present F (z)´1 as follows: and We identify ta 0 , a 1 , . . . , a n 1´1 ,â n 1 u with the subintervals of the edge e 1 and tb 1 , b 2 , . . . , b n 1 u with the masses on it: Fig. 5.1). It is clear that F 1 (z)´1 belongs to S 0 and therefore It is known that if F 1 (z)´1 belongs to S 0 then

Let us choose nonnegative integers
Since Φ T 1 ,v 1 ą 0 and therefore, We arrange the set t(ν k ) 2 u n´n 1 k=1 as the union of disjoint sets Inverse problem 13 and choose numbers B j (j = 1, 2, . . . , d(v 1 )´1) such that (5.14) Then We will show that there exists a distribution of masses on the complimen- 2) the form characteristic of T j is Φ T j ,v 1 and the rational function (5.17) having N j simple zeros and N j simple poles is ϕ , where ϕ It is clear that due to (5.11) and (5.14) the left hand side of (5.18) is an S 0 -function.
We defineñ j by the inequalitiesñ We identify numbers a , . . . , m (1,j) n j on the edge e 1,j . In the same way as (5.9) we obtain Denote by v 1,j the second vertex incident with e 1,j and by d(v 1,j ) the degree of v 1,j Now we consider the tree T 1 j obtained by deleting the edge e 1,j from T j (see Fig. 5.1). Let v 1,j be the root of T 1 j . Substituting z = 0 into (5.15) and making use of (5.16) we obtain On the other hand, (5.19) implies and consequently, F´1 j,1 (0) = Φ´1 T 1 j ,v 1 . We continue this procedure. Finally we obtain a branching continued fraction. We identify a (j) 0, 1, . . . , n j´1 ), a n j = l (1,j) n j , b 1, 2, . . . , n 1,j ).
According to Theorem 4.1 the problems (2.10)-(2.13), (2.18) and (2.10)-(2.13), (2.15) with these masses and subintervals have the quotient where F (z) is given by (5.2). Since the equation (5.6) uniquely determines the sets ta k u n 1´1 k=0 Y tâ n 1 u and tb k u n 1 k=1 and (5.7) uniquely determines the number n 1 Statement 2) is valid. □ Theorem 5.2. Suppose tµ k u n k=´n, k =0 and tν k u n k=´n, k =0 are symmetric (µ´k =´µ k , ν´k =´ν k ) and monotonic sequences of real numbers which interlace: (j = 1, 2, . . . , d(v)´1) in arbitrary way and choose positive numbers B j in so that (5.13), (5.14) hold. The latter is possible due to (5.12). Denote by Φ T j ,v 1 the values obtained via (5.16) and regarded as the form factors of the trees T j we are constructing. Due to A j ą 0 and (5.14) the functions ( are S 0 -functions and, consequently, (5.19) holds.
Let us choose the length l j of the edge e j incident with v so that where n j is the number of the masses on e j . Then we present the functions as in (5.19) and (5.20) but with given n j instead ofñ j . Then we continue this procedure as in the proof of Theorem 5.1. □