An Inverse Problem for Quantum Trees with Delta-Prime Vertex Conditions

: In this paper, we consider a non-standard dynamical inverse problem for the wave equation on a metric tree graph. We assume that the so-called delta-prime matching conditions are satisﬁed at the internal vertices of the graph. Another speciﬁc feature of our investigation is that we use only one boundary actuator and one boundary sensor, all other observations being internal. Using the Neumann-to-Dirichlet map (acting from one boundary vertex to one boundary and all internal vertices) we recover the topology and geometry of the graph together with the coefﬁcients of the equations.


Introduction
This paper concerns inverse problems for differential equations on quantum graphs. Under quantum graphs or differential equation networks (DENs) we understand differential operators on geometric graphs coupled by certain vertex matching conditions. Network-like structures play a fundamental role in many problems of science and engineering. The range for the applications of DENs is enormous. Here is a list of a few.
-Structural Health Monitoring. DENs, classically, arise in the study of stability, health, and oscillations of flexible structures that are made of strings, beams, cables, and struts. Analysis of these networks involve DENs associated with heat, wave, or beam equations whose parameters inform the state of the structure, see, e.g., [1].
-Water, Electricity, Gas, and Traffic Networks. An important example of DENs is the Saint-Venant system of equations, which model hydraulic networks for water supply and irrigation, see, e.g., [2].
Other important examples of DENs include the telegrapher equation for modeling electric networks, see, e.g., [3], the isothermal Euler equations for describing the gas flow through pipelines, see, e.g., [4], and the Aw-Rascle equations for describing road traffic dynamics, see e.g., [5].
-Nanoelectronics and Quantum Computing. Mesoscopic quasi-one-dimensional structures such as quantum, atomic, and molecular wires are the subject of extensive experimental and theoretical studies, see, e.g., [6], the collection of papers in [7][8][9]. The simplest model describing conduction in quantum wires is the Schrödinger operator on a planar graph. For similar models appear in nanoelectronics, high-temperature superconductors, quantum computing, and studies of quantum chaos, see, e.g., [10][11][12].
Here, T is arbitrary positive number, q j ∈ C([a 2j−1 , a 2j ]) for all j, and f ∈ L 2 (0, T). The physical interpretation of conditions (3) and (4), and some other matching conditions was discussed in [20].
The well-posedness of this system is discussed in Section 2; it will be proved that In what follows, we refer to γ 0 as the root of Ω and f as the control.
We now pose our inverse problem. Assume an observer knows the boundary condition (5), and that (6) holds at the other boundary vertices, and that the graph is a tree. The unknowns are the number of boundary vertices and interior vertices, the adjacency relations for this tree, i.e., for each pair of vertices, whether or not there is an edge joining them, the lengths { j }, and the function q. We wish to determine these quantities with a set of measurements that we describe now. We can suppose v N is the interior vertex adjacent to γ 0 with e 1 the edge joining the two, see Figure 1. Our first measurement is then the following measurement at γ 0 : We show that from operator R 01 one can recover 1 and the degree Υ N of v N . Then by a well known argument, see [21], one can then determine q 1 . Having established these quantities, in our second step, we propose to place sensors on the edges incident to v N , and using these measurements together with R 0,1 to determine the data associated to these edges. Note that the one control remains at γ 0 . The goal is to repeat these steps until all data associated to the graph have been determined. To define the interior measurements we require more notation. For each interior vertex v k we list the incident edges by {e k,j : j = 1, ..., Υ k }. Here e k,1 is chosen to be the edge lying on the unique path from γ 0 to v k , and the remaining edges are labeled randomly, see Figure 2. Then the sensors measure We show that we do not need sensors at e k,1 , e k,Υ k . Thus the total number of sensors is 1 + ∑ N j=m+1 (Υ j − 2). It is easy to check that this number is equal to |Γ| − 1. We denote by R T the (|Γ| − 1)-tuple (R 0,1 , R N,2 , R N,3 , ....) acting on L 2 (0, T).

Results
Let be equal to the maximum distance between γ 0 and any other boundary vertex. Our main result is the following Theorem 1. Assume q j ∈ C([a 2j−1 , a 2j ]) for all j. Suppose T > 2 . Then from R T one can determine the number of interior and boundary vertices, the adjacency relations of the tree, q, and the lengths of the edges.

Discussion
We now compare this result to others in the literature. We are unaware of any works treating the inverse problem on general tree graphs with delta-prime conditions on the internal vertices. The most common conditions for internal vertices are continuity together with Kirchhoff-Neumann condition: ∑ j∈J(v k ) ∂u j (v k , t) = 0 and all references in this paragraph assume these conditions. In [21], the authors assume that controls and measurements take place at all boundary vertices but one. The authors use an iterative method called "leaf peeling", where the response operator on Ω is used first to determine the data on the edges adjacent to the boundary, and then to determine the response operator associated to a proper subgraph. In [21], the leaf peeling argument includes spectral methods that require knowing R T for all T. In [22], the methods of [21] are extended to the case where masses are placed at internal vertices, see also [23]; however these methods still require knowledge of R T for all T. Also in [22], it is proven that that for a single string of length with N attached masses and T > 2 , R T 01 is sufficient to solve the inverse problem. In particular, [23] uses a spectral variant of the boundary control method, together with the relationship between the response operator and the connecting operator. In [24,25], a dynamical leaf peeling argument is developed for a tree with no masses and with response operators at all but one boundary points, allowing for the solution of the inverse problem for finite T sufficiently large. An important ingredient in their leaf peeling is determining the response operators associated with subtrees, called "reduced response operators", from the response operator associated to the original tree. In all of these papers, it is assumed that there are no interior measurements. In [26], the iterative methods from [24,25,27] are adapted to a tree with masses placed at internal vertices, with a single control at the roof and measurements there and at internal vertices. For other works on quantum graphs, see [1,16,19,[28][29][30][31].
A special feature of the present paper is that we use only one control together internal observations. This may be useful in some physical settings where some or most boundary points are inaccessible. Another potential advantage of the method presented here is that we recover all parameters of the graphs, including its topology, from the (|Γ| − 1)-tuple response operator acting on L 2 (0, T). In previous papers, the authors recovered the graph topology from a larger number of measurements: the (|Γ| − 1) × (|Γ| − 1) matrix (boundary) response operator or, equivalently, from (|Γ| − 1) × (|Γ| − 1) Titchmarsh-Weyl matrix function. In [32], the inverse problems on a star graph for the wave equation with general self-adjoint matching conditions was solved by the (|Γ| − 1) × (|Γ| − 1) matrix boundary response operator.

Preliminaries
In what follows, we use the notations where H n (R) are the standard Sobolev spaces. We define the Heaviside function by H(t) = 1 for t > 0, and H(t) = 0 for t < 0. Then, we define H n ∈ F n as the unique solution to d n dt n H n = H; Here δ(t) denotes the Dirac delta function supported at t = 0. In this section and those that follow, we drop the superscript T from R T when convenient. Consider a star shaped graph with edges e 1 , ..., e N . For each j, we identify e j with the interval (0, j ) and the central vertex with x = 0, see Figure 3. Recall the notation q j = q| e j , and u j ( * , t) = u( * , t)| e j . Thus, we consider the system Let u f solve (10)- (15), and set For (11), it is standard that the waves have unit speed of propagation on the interval, so g j (t) = 0 for t < 1 and all j. It will be useful first to consider the vibrating string on an interval.

Representation of Solution on an Interval and Reduced Response Operator
We adapt a representation of u f (x, t) developed in [27], where only Dirichlet control and boundary conditions were considered. Fix j ∈ {1, ..., N}. We extend q j to (0, ∞) as follows: first evenly with respect to x = j , and then periodically. Thus q j (2k j ± x) = q j (x) for all positive integers k.
Define w j to be the solution to the Goursat-type problem A proof of solvability of this problem can be found in [33]. Consider the IBVP on the interval (0, j ): Let Then, the solution to (17)- (20) on e j can be written as In what follows, we only consider t ≤ T for some finite T, so all sums will be finite.
Let us now change the condition (20) toũ x ( j , t) = 0. In this case, the solution becomes To represent the solution of the wave equation on the edge e 1 in a star graph, we must account for the control at x = 1 . Thus it will also be useful to represent the solution of a wave equation on an interval when the control is on the right end. Consider the IBVP: Thus We now show that the system (10)-(15) is well-posed. Recall that F 1 was defined in (9), and g j (t) = u f j (0, t).

Theorem 2.
(a) If f ∈ L 2 (0, T), then there exists a unique solution u(x, t) solving the system (10)- (15), and mapping Proof. On [0, j ] with j ≥ 2, the wave will be generated by the "control" ∂(u f j )(0, t), whereas on [0, 1 ] the wave is generated by the two controls ∂(u Here, p is independent of j by (12). We have that u f is given by Note that v f has already been explicitly determined in (23). Thus, by (21), we have an explicit solution for u f if we can solve for p. We now prove the existence, uniqueness, and regularity of P.
By (21) and (26), we have for j ≥ 2, For j = 1, we have by (21), (24), and (26) that We remark that at the moment, we have not yet solved for either P or g j for any j. Let α = min{ j , j = 1, ..., N}.
We solve for P with an iterative argument using steps of length 2α. The iterations are necessary because the upper limits in the sums in (27), (28) increase with time due to reflections of the wave at the various vertices. In what follows, we label by G(t) various terms that we have already solved for, which by (24), includes v f (0, t). For t ≤ 1 we have by unit wave speed that P(t) = 0. Suppose now t ∈ [ 1 , 1 + 2α]. Then, t − 2 j ≤ 1 + 2α − 2 j < 1 , and hence P(t − s) = 0 for s ≥ 2n j , for all j with n ≥ 1. By (13), we have and hence from (27) and (28) we get It is easy to show that this is a Volterra equation of the second kind (VESK), and so admits a unique solution P with P L 2 ( 1 , 1 +2α) ≤ F L 2 (0,2α) . Furthermore, by differentiating this equation we get p L 2 ( 1 , 1 +2α) ≤ f L 2 (0,2α) .

Proposition 1.
For R 1 one can determine q 1 , 1 , and N.
Proof. Using (25), it is easy to see that The lemma now follows easily from (21).
In what follows, we refer toR 0,j (s) as the "reduced response function". For f (t) = δ(t), we denote the solution to the system (10)-(15) as u δ . We also use the following.
This result holds from the proof of Theorem 2, the unit speed of wave propagation, and the properties of wave reflections off x = j , see [33]. The details are left to the reader.
The following result follows from (22), (25), and Lemma 2. The details of the proof are left to the reader.

Solution of Inverse Problem
Here, we establish some notation. We recall the following notation: for v k we list the incident edges by {e k,j : j = 1, ..., Υ k }. Here, e k,1 is chosen to be the edge lying on the path from γ 0 to v k , and the remaining edges are labeled randomly. Now let k 0 be some fixed interior vertex, and let j 0 satisfy 1 < j 0 ≤ Υ k 0 . Denote by Ω  We define an associated response operator as follows. Let Γ j 0 Then we define an associated reduced response operator with associated response functionR k 0 ,j 0 (s). Suppose we determinedR k 0 ,j 0 . It would follow from Proposition 1 that one could recover the following data: j 0 , q j 0 , and Υ k , where v k is the vertex adjacent to v k 0 in Ω j 0 k 0 . In this section we will present an iterative method to determine the operatorR k 0 ,j 0 from the (|Γ| − 1)-tuple of operators, R T , which we know by hypothesis for some T > 2 . An important ingredient is the following generalization to a tree of Corollary 1.

Lemma 3.
Let T > 0, and let R T k,j be associated with (33)-(38), defined by (7) and (8). The response function for R T k,j has the form Here, r k,j ∈ F 1 , and the sequence {γ n } is positive and strictly increasing. If T is finite then the sums are finite.
Proof. The proof follows from the proof of Corollary 1, together with the transmission and reflection properties of waves at interior vertices, and reflection properties at boundary vertices.
Fix T > 2 . The rest of this section shows how to recoverR k 0 ,j 0 from R T .
Step 1 For the first step, let v k 0 be the vertex adjacent to the root γ 0 , with associated edge labeled e 1 . By Proposition 1, we can use R T 0,1 to recover Υ k 0 , 1 , q 1 .

Step 2
Consider e k 0 ,2 . In Step 2, we show how to solve forR k 0 ,2 , see Figure 5. Since v k 0 is the root of Ω 2 k 0 , the following equation is essentially a restatement of Lemma 1 to trees; the details of its proof are left to the reader.
Since we know 1 and q 1 , we can solve the wave equation on e 1 with known boundary data. We identify e 1 as the interval (0, 1 ) with v k 0 corresponding to x = 0. Then u f , restricted to e 1 , solves the following Cauchy problem, where we view x as the "time" variable: Since the function R 0,1 (s) is known, we can thus uniquely determine u f (0, t) = u f (v k 0 , t) and is determined. We now show how p and u δ 2 (v k 0 , t) can be used to determineR k 0 ,2 (s). The following equation follows from the definition of the response operators for any f ∈ L 2 : In what follows, it is convenient to extend f (t) ∈ L 2 (0, T) as zero for t < 0. By Lemma 1 and by an adaptation of Lemma 2 to general trees, we have the following expansions: Here, r k 0 ,2 ∈ F 1 and a(s) ∈ F 1 , and {ξ k } and {β n } are positive and increasing. Clearly a(s), r k 0 ,2 (s), {ψ m }, {θ m }, {φ n }, {γ n }, can all be determined by R 0,1 and R k 0 ,2 , whereas for nowr and the sets {α m }, {ξ m } are unknown. Inserting (39),(41) and (42) into (40), we get Here all sums have 1 as lower limit of summation. Proof. We mimic an iterative argument in [26]. Differentiating (43) and then matching the delta singularities, we get Since the sequences {γ n }, {ζ l }, {ξ m } are all strictly increasing, clearly we have γ 1 = ζ 1 + ξ 1 , so that φ 1 = α 1 ψ 1 , and so ξ 1 = γ 1 − ζ 1 and α 1 = φ 1 /ψ 1 . We represent that the set {φ 1 , γ 1 }, {ζ 1 , ψ 1 } determines the set {ξ 1 , α 1 } by We now match the term δ(t − γ 2 ) with its counterpart on the right hand side of (44). There are three possible cases.
Repeating this procedure as necessary, say for a total of N 2 times, we solve for {ξ 2 , α 2 }. We represent this process as We must have N 2 finite by (44) and the finiteness of the graph. Iterating this procedure, suppose for p ∈ N we have Here N p is chosen to be minimal, and so γ N p = ζ 1 + ξ p . We wish to solve for {ζ p+1 , φ p+1 }. We can again distinguish three cases: Case 1: γ (N p +1) = ζ k + ξ j , ∀j ≤ p, ∀k. Note that we know {ξ j } p 1 and {ζ k }, so these inequalities are verifiable. In this case, we must have γ (N p +1) = ζ 1 + ξ p+1 and ψ 1 α p+1 = φ (N p +1) , so we have determined α p+1 , ξ p+1 in this case.

Case 2:
There exists an integer Q and pairs {ζ i n , ξ j n } Q n=1 , with j n ≤ p, such that Note that all the numbers {ζ i n , ξ j n } have been determined, so these equations can be all verified. We can assume all pairs {ζ i n , ξ j n } satisfying (45) with j n ≤ p are listed. In this case, we have either Case 2i: φ (N p +1) = α j 1 ψ i 1 + ... + α j Q ψ i Q . It follows then that γ (N p +1) = ζ 1 + ξ p+1 , and We thus solve for ξ p+1 , α p+1 . Case 2ii: It follows then that α (N p +1) = ζ 1 + ξ p+1 , and we have to repeat this process with γ (N p +2) .
Hence, we can solve for {ξ p : p ≤ L}, {α p : p ≤ L} for any positive integer L given knowledge of R T 0,1 , R T k 0 ,2 for T = T(L) sufficiently large.
It remains to solve forr. In what follows, we setR(s) = 0 for s < 0. We use G(t) to denote various functions that we have already established to be determined by R 0,1 and R k 0 ,2 . Having already solved for {ξ n , α n }, we can eliminate from (43) the Heavyside functions to get, recalling ζ 1 = 1 , We solve this with an iterative argument. Let α = min m {ζ l+1 − ζ l }. For t < ζ 1 + α, we have for l > 1 that t − ζ l < 0 so r(t) = 0. Hence Letting r = t − ζ 1 , we get We solve this VESK to determiner(s), r < α. Now for t < ζ 1 + 2α, we have for l > 1 that t − ζ l < α, and so those terms in (46) with t − ζ l can be absorbed in G to again give We solve this VESK to determiner(s), r < 2α. Iterating this procedure, we solve forr(s) for any finite s.
Step 3 Because R k 0 ,j are determined by assumption for j = 2, ..., Υ k 0 − 1, the functions u f j (v k 0 , t) are determined. In Step 2, we showed u f 1 (v k 0 , t) is also determined. Hence by (4), u f Υ k 0 (v k 0 , t) is also determined. We can now carry out the argument in Step 2 on the remaining edges e k 0 ,3 , ..., e k 0 ,Υ k 0 incident on v k 0 to determineR k 0 ,j for all j.
Step 4 For each j = 2, ..., Υ k 0 , we use Proposition 1, to find the associated j , q j together with the valence of the vertex adjacent to v k 0 . Careful reading of Steps 2 and 3 shows that we can use R T 0,1 and R T k 0 ,j for any T > 2( 1 + j ).
Step 5 Let v k 1 , ... be the vertices adjacent to v k 0 , other than γ 0 . We now iterate Steps 2-4 for the each of these vertices. Choose for instance v k 1 . If it were a boundary vertex, this fact would be determined in Step 4, and then this algorithm goes to the next vertex, which we, for convenience, still label v k 1 . We can thus assume v k 1 is an interior vertex. Let us label an incident edge (other than e 2 := e k 0 ,2 ) as e 3 := e k 1 ,3 , see Figure 6. We wish to determineR k 1 ,3 . Mimicking Step 2, let u δ solve (1)-(6), let b(t) = ∂u δ 3 (v k 1 , t). We have the following formula holding by the definition of response operators: t 0R k 1 ,3 (s) b(t − s)ds = t 0 R k 1 ,3 (s)δ(t − s)ds.
Of course R k 1 ,3 (s) is assumed to be known. We determine b as follows. We have, from Step 2, that p(t) = ∂u δ 1 (v k 0 , t) is known. We identify e 2 as the interval (0, 2 ) with v k 1 corresponding to x = 0.
Since q 2 , 2 , and R k 0 ,2 are all known, we can thus determine b(t) = y x (0, t). The rest of the argument here is a straightforward adaptation of Steps 2-4 above. The details are left to the reader.
Step 6 Arguing as in Step 5, we determineR k,j for all other vertices adjacent to v k 0 and their associated edges. The details are left to the reader.
Steps above 6 Clearly this procedure can be iterated until all edges of our finite graph have been covered.

Conclusions
In this paper, we applied the ideas of the boundary control and leaf peeling methods to solve an inverse problem on a tree featuring non-standard, delta-prime vertex conditions on the interior. Our method required using only one boundary actuator and one boundary sensor, all other observations being internal. Using the Neumann-to-Dirichlet map (acting from one boundary vertex to one boundary and all internal vertices) we recovered the topology and geometry of the graph together with the coefficients q j of the equations. It would be interesting to see a numerical implementation of our method. It would also be interesting to adapt our methods to quantum graphs with cycles.

Conflicts of Interest:
The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the decision to publish the results.