Magnetic Boundary Control I: Graphs with Several Cycles

Abstract

This is the first chapter devoted to the Magnetic Boundary Control method (MBC-method).

This is the first chapter devoted to the Magnetic Boundary Control method (MBC-method). It appears that this method can effectively be applied to graphs having several independent cycles, while graphs with just one cycle may require special attention and are described in the following chapter. The MBC-method is based on the following idea: using the dependence of the M-function on the magnetic fluxes one may recover the M-function for the graph on the same set of edges but with some of the cycles being opened. In some sense the new graph is closer to a tree than the original graph. For example, consider an arbitrary graph $\Gamma$ and some contact vertex $V^0$ having sufficiently large degree $d_0 \geq 3$. Let $\Gamma '$ be the metric graph obtained from $\Gamma$ by splitting the vertex $V^0$ into $d_0$ degree one vertices. We say that the vertex $V^0$ is dissolved. The M-function for $\Gamma$ known for some different values of the magnetic fluxes determines the M-function for $\Gamma '$. Then the classical BC-method can be used to recover the potential on the pendant edges in $\Gamma '$. Peeling these edges away as described in Chap. 20 we reduce the inverse problem to a smaller graph. For some graphs, by repeating the procedure the inverse problem is reduced to the inverse problem on a tree and therefore can be solved completely, while for other graphs the procedure terminates, leaving a major part of the graph unknown.

We develop here ideas proposed in [334, 336], it might be interesting to see connections with [453].

22.1 Dissolving the Vertices

Let us study how to determine the M-function when one of the vertices is dissolved.

Definition 22.1

We say that the metric graph $\Gamma _1$ is obtained from a metric graph $\Gamma$ by dissolving a certain vertex $V^0$ in $\Gamma$ if:

• the metric graphs $\Gamma$ and $\Gamma _1$ share the same set of edges $\{ E_n \}_{n=1}^N$;

• the endpoints connected at $V^0$ in $\Gamma$ form degree one vertices in $\Gamma _1$;

• all other vertices in $\Gamma$ and $\Gamma _1$ coincide.

See Fig. 22.1 where the dissolving procedure is presented schematically. The green area represents the part of the graph which is not affected by the procedure. The degree four vertex $V^0$ is substituted with four degree one vertices $V^1, \dots , V^4$.

Fig. 22.1

Dissolving a vertex

We are going to exclude the case where dissolution of $V^0$ disconnects the original graph. Then the number of cycles in the original graph $\Gamma$ and in the new graph $\Gamma _1$ differ by $d_0 -1$:

(22.1)

In other words, dissolution of $V^0$ breaks precisely $d_0 -1$ cycles in the original graph. This number does not depend on whether the vertex $V^0$ is situated well inside the graph or on its periphery (see the two graphs presented in Fig. 22.2).

Fig. 22.2

Breaking cycles by dissolving vertices

Our goal is to compare the M-functions

$$\displaystyle \begin{aligned} \mathbf M (\lambda) := \mathbf M_\Gamma (\lambda) \quad \mbox{and} \quad \mathbf M_1 (\lambda):= \mathbf M_{\Gamma_1} (\lambda) \end{aligned}$$

corresponding to $\Gamma$ and $\Gamma _1$ respectively. In our context these functions depend not only on the spectral parameter $\lambda$ but also on the magnetic fluxes. Let us denote by $\vec {\Phi }$ and $\vec {\Phi }^1$ the vectors collecting all fluxes for $\Gamma$ and $\Gamma _1$, respectively. Every cycle in $\Gamma _1$ corresponds to a certain cycle in $\Gamma$, hence one may naturally assume that the entries in the vector $\vec {\Phi }^1$ correspond to certain $\beta _1 (\Gamma _1)$ entries in the vector $\vec {\Phi }$. The remaining $\beta _1 (\Gamma ) - \beta _1 (\Gamma _1) = d_0-1$ entries in $\vec {\Phi }$ correspond to the cycles that are broken under the dissolution of $V^0 .$ It will be convenient to denote the corresponding fluxes by $\vec {\Phi }^2$, so that we have:

$$\displaystyle \begin{aligned} {} \vec{\Phi} = (\vec{\Phi}^1, \vec{\Phi}^2). \end{aligned} $$

(22.2)

We assume from now on that the fluxes $\vec {\Phi }^1$ through the preserved cycles are fixed and omit indicating the dependence of the M-functions on $\vec {\Phi }^1$.

We denote by $V^1, \dots , V^{d_0}$ the pendant vertices in $\Gamma _1$ coming from the vertex $V^0$ in $\Gamma$ and let $C_j$ be a path connecting $V^{d_0}$ to $V^j, \, j =1,2, \dots , d_0-1$. The paths on $\Gamma _1$ correspond to the cycles in $\Gamma$, that are broken under the dissolution. The corresponding fluxes are

$$\displaystyle \begin{aligned} \Phi_j = \int_{C_j} a(y) dy = \int_{V^{d_0}}^{V^j} a(y) dy, \quad j =1,2, \dots, d_0-1. \end{aligned} $$

(22.3)

These fluxes form the vector $\vec {\Phi }^2$. It will be convenient to view $\vec {\Phi }^2$ as an element of $\mathbb R^{d_0}$ despite that only $d_0-1$ of its coordinates may be non-zero:

$$\displaystyle \begin{aligned} \vec{\Phi}^2= (\Phi_1, \Phi_2, \dots, \Phi_{d_0-1}, 0 ). \end{aligned}$$

To reconstruct the M-function for $\Gamma _1$ it is enough to consider the fluxes equal to $0$ and $\pi$, therefore we introduce the notations

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \mu_j := e^{i \Phi_j}, \quad j =1,2, \dots, d_0; \\ \displaystyle \boldsymbol\mu = (\mu_1, \mu_2, \dots, \mu _{d_0}) = e^{i \vec{\Phi}}; \end{array} \end{aligned} $$

(22.4)

and consider the M-functions depending on the indices $\mu _j$ instead of the phases $\Phi _j$. To get the corresponding spectral data it is enough to consider the standard operators with zero magnetic potential and additional signing conditions (3.43) introduced on some of the cycles. These operators will be denoted $L_q^{\mathrm {sign}} (\Gamma )$ and called signed Schrödinger operators.

Our first step is to establish the relation between the diagonal element

$$\displaystyle \begin{aligned} \mathbf M^{00} (\lambda, \boldsymbol\mu) =: \mathbb M (\lambda, \boldsymbol\mu) \end{aligned}$$

associated with the vertex $V^0$ and the diagonal $d_0 \times d_0$ block of $\mathbf M_1 (\lambda , \boldsymbol \mu )$ associated with the degree one vertices in $\Gamma _1$ (coming from $V^0$). We shall find an explicit relation between the scalar Herglotz-Nevanlinna function $\mathbb M (\lambda , \boldsymbol \mu )$ and the $d_0 \times d_0$ matrix valued Herglotz-Nevanlinna function $\mathbb M_1 (\lambda , \boldsymbol \mu ) := \left \{ \mathbf M_1^{ij} (\lambda , \boldsymbol \mu ) \right \}_{i,j = 1}^{d_0}$.

The dependence of $\mathbb M_1 (\lambda , \boldsymbol \mu )$ upon $\boldsymbol \mu$ is trivial:

(22.5)

To see this, let us eliminate the magnetic potential starting from $V^{d_0}$ by using the transformation

$$\displaystyle \begin{aligned} f(x) \mapsto g(x) = e^{ -i \int_{V^{d_0}}^x a(y) dy} f(x). \end{aligned}$$

Under this transformation we have

$$\displaystyle \begin{aligned} f(V^j) = e^{i \Phi_j} g(V^j) = \mu_j g(V^j) ,\end{aligned}$$

implying

$$\displaystyle \begin{aligned} \left( \begin{array}{c} f(V^1) \\ \vdots \\ f(V^{d_0}) \end{array} \right) = \mathrm{diag}\, \{\mu_j \} \; \left( \begin{array}{c} g(V^1) \\ \vdots \\ g(V^{d_0}) \end{array} \right), \end{aligned}$$

which leads to (22.5).

The diagonal entry $\mathbb M (\lambda , \boldsymbol \mu )$ is equal to the sum of all entries in $\mathbb M_1(\lambda , \boldsymbol \mu )$:

(22.6)

This formula determines the M-function for any signed operator on $\Gamma$ through the M-function for $\Gamma _1$.

The key idea behind the reconstruction of $\mathbb M_1$ from $\mathbb M$ is to use formula (17.37), which expresses each of these two Herglotz-Nevanlinna functions through the normal derivatives of the Dirichlet eigenfunctions, i.e. the eigenfunctions satisfying Dirichlet conditions at $V^0$ in $\Gamma$ and at $V^1, \dots , V^{d_0}$ in $\Gamma _1$. These eigenfunctions simply coincide since the Dirichlet condition does not feel whether pendant vertices are glued together or not.

Let $\psi _n^{\mathrm {D}}$ denote the eigenfunction corresponding to zero fluxes through the broken cycles. These eigenfunctions can be chosen real-valued since they satisfy standard and Dirichlet vertex conditions and the fluxes in $\vec {\Phi }^1$ are all either $0$ or $\pi$. Then the normal derivatives of the Dirichlet eigenfunctions for non-zero fluxes are given by

$$\displaystyle \begin{aligned} \mu_j \partial \psi_n^{\mathrm{D}} (V^j) ,\end{aligned}$$

implying in particular that the normal derivative at $V^0$ is

$$\displaystyle \begin{aligned} \sum_{j=1}^{d_0} \mu_j \partial \psi_n^{\mathrm{D}} (V^j). \end{aligned} $$

(22.7)

It follows that the singularity of $\mathbb M(\lambda , \boldsymbol \mu )$ is of the form

(22.8)

where we have used that $\partial \psi _n^{\mathrm {D}}$ are real-valued.

Introducing the notation $a_j := \partial \psi _n^{\mathrm {D}} (V^j)$, we are faced with the following trivial problem:

• Determine $a_j$ if the numbers

  $$\displaystyle \begin{aligned} \left( \pm a_1 \pm a_2 \pm \dots \pm a_{d_0-1} + a_{d_0} \right)^2 \end{aligned}$$

  are known for all possible combinations of the signs.

It is clear that this reconstruction is possible only up to the multiplication of all $a_j$ by $-1$, which corresponds to the multiplication of the corresponding eigenfunctions by $-1$.

The sum of the squares can be obtained by averaging over all possible signs:

$$\displaystyle \begin{aligned} \sum_{i=1}^{d_0} a_j^2 = \frac{1}{2^{d_0-1}} \sum_{ \boldsymbol\mu \in (\{1,-1 \}^{d_0-1}, 1) } \left( \mu_1 a_1 + \mu_2 a_2 + \dots + \mu_{d_0-1} a_{d_0-1} + a_{d_0} \right)^2. \end{aligned} $$

(22.9)

Hence we are able to determine the following combinations of the $a_j$’s

(22.10)

We recover the products by averaging a second time

(22.11)

The product $a_k a_l = a_l a_k$ appears in the double sum precisely $2^{d_0-1}$ times, while all other products cancel since $\mu _i \mu _j$ attains $+1$ and $-1$ equally many times.

If at least three of the coefficients are nonzero, then the squares $a_j^2$ are determined as

$$\displaystyle \begin{aligned} {} a_i^2 = \frac{ (a_i a_j) \, (a_i a_l)}{(a_j a_l)}, \quad \mbox{provided} \quad a_j, a_l\neq 0 . \end{aligned} $$

(22.12)

We are able to recover one nonzero $a_j$ up to a sign, but then all other non-zero coefficients are determined from the products $a_j a_i$. We conclude that if the squared sums $\left ( \sum _{j=1}^{d_0} \mu _j a_j \right )^2$ are known for all $\boldsymbol \mu$ of the form $\boldsymbol \mu \in (\{1,-1\}^{d_0-1},1)$, then the coefficients $a_j$ are determined up to a common sign.

It follows that the diagonal element $\mathbb M (\lambda , \boldsymbol \mu )$ known for all $\boldsymbol \mu \in (\{1,-1\}^{d_0-1},1)$ determines the vector

$$\displaystyle \begin{aligned} \partial {\vec{\psi_n^{\mathrm{D}}}}_2 := \left(\partial \psi_n^{\mathrm{D}} (V^1), \partial \psi_n^{\mathrm{D}} (V^2), \dots, \partial \psi_n^{\mathrm{D}} (V^{d_0}) \right) , \end{aligned}$$

up to the common sign, hence the singular part of $\mathbb M_1 (\lambda , \vec {0})$ is determined, which as before allows us to reconstruct it up to the constant matrix $\mathbb A$, yielding

$$\displaystyle \begin{aligned} {} \mathbb M_1 (\lambda, \vec{0}) = \mathbb A + \sum_{\lambda_n^{\mathrm{D}} (\Gamma_1)} \frac{\lambda - \lambda'}{(\lambda_n^{\mathrm{D}} - \lambda) (\lambda_n^{\mathrm{ D}} - \lambda')} \left\langle \vec{\psi_n^{\mathrm{D}}}_2 , \cdot \right\rangle \partial \vec{\psi_n^{\mathrm{D}}}_2. \end{aligned} $$

(22.13)

To determine $\mathbb A$ we remember that the M-function possesses the asymptotics

$$\displaystyle \begin{aligned} \mathbb M_1(- s^2, \vec{0}) = - s\; I_{d_0} + o(1), \quad s \rightarrow \infty, \end{aligned}$$

(see (21.5)). We are now ready to prove the main result of this section:

Theorem 22.2

Let $\Gamma$ be a pendant free metric graph with contact set including the vertex $V^0$ , and let $\Gamma _1$ be the metric graph obtained from $\Gamma$ by dissolving the vertex $V^0$ . Assume that

• (1) the graph $\Gamma _1$ , and hence also the graph $\Gamma$ , is connected;

• (2) the degree$d_0$of the contact vertex$V^0$is at least three:$d_0 \geq 3$.

Let$L_{q,a}^{\mathrm {st}}$be the standard magnetic Schrödinger operator. Consider the M-functions for$\Gamma$and$\Gamma _1$dependent on the spectral parameter$\lambda$and the magnetic fluxes through the cycles$\vec {\Phi } = (\vec {\Phi }^1, \vec {\Phi }^2)$, where, following (22.2), $\vec {\Phi }^1$collects the fluxes corresponding to the cycles that are preserved under the dissolution of$V^0$.

Assume in addition two generically satisfied assumptions:

• (a) the spectrum of the Dirichlet operator on $\Gamma _1$ is simple;

• (b) for each Dirichlet eigenfunction $\psi _n^{\mathrm {D}}$ on $\Gamma _1$ the normal derivatives at the pendant vertices (coming from $V^0$ ) possess the following property:

  • among the normal derivatives either all derivatives are zero, or at least three derivatives are different from zero.

Then for any fixed$\vec {\Phi }^1 \in \{0, \pi \}^{\beta _1(\Gamma _1)}$the$|\partial \Gamma | \times |\partial \Gamma |$matrix valued M-function$\mathbf M_\Gamma (\lambda , \vec {\Phi })$taken for all possible values of$\vec {\Phi }^2 \in \{ 0, \pi \}^{d_0-1}$determines the$(|\partial \Gamma | + d_0-1) \times (| \partial \Gamma | + d_0-1)$matrix valued M-function$\mathbf M_{\Gamma _1} (\lambda , \vec {\Phi }^1)$.

Proof

We are going to assume that the magnetic fluxes $\vec {\Phi }^1$ through the cycles in $\Gamma _1$ are fixed, and will omit indication that the M-functions and the eigenfunctions depend on $\vec {\Phi }^1$.

Let us present the M-function for $\Gamma _1$ in the following block form separating the preserved and pendant vertices

$$\displaystyle \begin{aligned} \mathbf M_1 (\lambda) = \left( \begin{array}{cc} \mathbb M^{00}_1 (\lambda) & \mathbb M^{01}_1 (\lambda) \\ \mathbb M^{10}_1 (\lambda) & \mathbb M^{11}_1 (\lambda) \end{array} \right), \end{aligned} $$

(22.14)

where the quadratic $d_0 \times d_0$ block $\mathbb M^{00}_1$ corresponds to the pendant (originating from $V^0$) vertices in $\Gamma _1$ and $\mathbb M^{11}_1$ is the quadratic $(|\partial \Gamma | -1) \times (|\partial \Gamma | -1)$ block corresponding to the preserved vertices $V^j$ from $\Gamma$, $V^j \neq V^0$.

The first diagonal block $\mathbb M^{00}_1 (\lambda )$ coincides with the matrix $\mathbb M_1$ already reconstructed above. The second diagonal block $\mathbb M^{11}_1 (\lambda )$ coincides with the corresponding block in the M-function for $\Gamma$.

It remains to reconstruct the non-diagonal blocks having the singularities determined by $\partial \psi _n^{\mathrm {D}} (V^i) \partial \psi _n^{\mathrm {D}} (V^j)$ with $i = 1,2, \dots , d_0$ and $V^j$ being one of the preserved vertices. Knowing the corresponding singularity in the 0j-entry of the original M-function

$$\displaystyle \begin{aligned} {} \mathbf M^{0j} (\lambda, \vec{\Phi}^2) \underset{\lambda \rightarrow \lambda_n^{\mathrm{D}}}{\sim} \frac{1}{\lambda_n^{\mathrm{D}} - \lambda} \partial \psi_n^{\mathrm{D}} (V^j) \left( \sum_{i=1}^{d_0} e^{i \Phi_i} \partial \psi_n^{\mathrm{D}} (V^i) \right) , \end{aligned} $$

(22.15)

allows us to reconstruct $\partial \psi _n^{\mathrm {D}} (V^j)$.

Hence the blocks $\mathbb M^{01}_1$ and $\mathbb M^{10}_1$ are determined using formula (17.37) and taking into account the asymptotics (21.5). □

The theorem implies that the M-function for $\Gamma _1$ can be recovered, provided the M-functions of all signed operators on $\Gamma$ are known.

Theorem 22.2 can be proved for any fixed $\vec {\Phi }^1$ not necessarily from $\{0, \pi \}^{\beta _1(\Gamma _1)} .$ The reason we restrict our statements to $\vec {\Phi }^1 \in \{0, \pi \}^{\beta _1(\Gamma _1)}$ is that only those values of $\Phi _j$ will be used when we shall dissolve further vertices.

The assumption that at least three normal derivatives are non-zero can be weakened, one may require instead that the eigenfunction $\psi _n^{\mathrm {D}}$ has non-zero normal derivatives at one of the preserved vertices.

22.2 Geometric Ideas Behind the MBC-Method: First Examples

In this section we discuss how to apply the MBC-method to solve inverse problems for metric graphs. As before we assume that the original graphs have no pendant edges. We start by presenting examples when the whole graph can be reconstructed starting from one vertex. We then continue by discussing what may prevent complete reconstruction of the graph.

Example 22.3

Consider the graph presented in Fig. 22.3 and assume that the contact set consists of the single vertex $V$. Dissolving the vertex $V$ and peeling away the pendant vertices we arrive at a smaller graph. Repeating the procedure by dissolving the vertices $V'$ and $V''$ the inverse problem is reduced to a tree with all pendant vertices in the contact set (see the upper sequence in Fig. 22.3). The MBC-method allows us to solve the inverse problem for this graph.

Fig. 22.3

The whole graph may be reconstructed by the MBC-method

The inverse problem for this graph can be solved by dissolving the vertices $V$, $V^*$, and $V^{**}$ instead (see the lower sequence in Fig. 22.3). The resulting graph is the cycle with 3 contact points—the inverse problem is again solvable by dismantling the cycle into three intervals (see Sect. 21.2).

This example shows that the MBC-method allows us to solve the inverse problem for rather complicated graphs with arbitrary number of cycles and very few contact points.

Example 22.4

Consider the graph presented in Fig. 22.4 and assume that the contact set is given by the vertex $V$. Dissolving the vertex $V$ and peeling away the pendant vertices we arrive at a graph with three contact vertices, each having degree 2. Theorem 22.2 cannot be applied to such vertices; the fat edges form a wall separating already reconstructed edges from the rest of the graph. Note that the graph in Fig. 22.4 is obtained from the graph in Fig. 22.3 by removing two internal edges.

Fig. 22.4

Reconstruction is terminated by degree two vertices

Example 22.5

Figure 22.5 presents another graph with a single contact vertex $V$. After dissolving V  and removing the pendant edges we get the graph with three vertices. We may dissolve only the vertex $V'$ as the remaining two contact vertices have degree two. This leads to a graph with three contact vertices: two degree two vertices and one bottleneck vertex $V''$—dissolution of this vertex would disconnect the graph (see Definition 22.8). The inverse problem for the remaining graph cannot be solved by dismantling it, since the corresponding trees are not independent. Note that the original graph in this example is again a slight modification of the graph presented in Fig. 22.3.

Fig. 22.5

Reconstruction is terminated by the bottleneck

It is not surprising that not all pendant-free graphs may be reconstructed starting from a single contact vertex—the described procedure may terminate immediately or after a few steps. As Examples 22.4 and 22.5 show, there are two reasons for the termination

• degree two contact vertices;

• bottlenecks.

Problem 94

Find new examples of graphs that can be reconstructed by dissolving vertices, starting from a single contact vertex.

22.3 Infiltration Domains, Walls and Bottlenecks

Let us have a closer look at how graphs or at least parts of them may be reconstructed using the method presented in Theorem 22.2.

Definition 22.6

Let $\Gamma$ be a finite compact pendant free metric graph with contact set $\partial \Gamma$. Consider any single contact vertex $V^j \in \partial \Gamma$ and apply the MBC-method by dissolving $V^j$ and the new contact vertices appearing after peeling away the pendant edges. We repeat this procedure until it terminates or the whole graph $\Gamma$ is recovered without involving the other original contact vertices from $\partial \Gamma \setminus V^j$. The maximal subgraph $D_j \subset \Gamma$ recovered in this way is called the infiltration domain. The MBC-method determines not only the metric subgraph $D_j$ but also the potential $q$ on it.

Of course we do not exclude the case where the infiltration domain coincides with the whole original graph $\Gamma$, but we are also interested in the mechanisms preventing this.

Our first observation is that the reconstructed domain may depend on the order in which the vertices are dissolved (see Example 22.3). The infiltration domain using $V, V^*$, and $V^{**}$ is smaller than the infiltration domain obtained by dissolving $V$, $V'$, and $V''$.

In what follows we are going to choose the largest possible infiltration domains—this will be our convention for the rest of this chapter.

For metric graphs it is natural to modify the notion of the set complement as follows.

Definition 22.7

Let $\Gamma _1= (E_1, V^1)$ be a subgraph of the metric graph $\Gamma = ( E, V).$ Then graph’s complement$\Gamma \setminus \Gamma _1 =: \Gamma _2$ is the metric graph on the edge set $E_2 = E \setminus E_1$ and the vertex set $V^2 = \Big \{ V^m (\Gamma ) \setminus V^m (\Gamma _1) \Big \}_{V^m \in V^\Gamma }.$

In other words, the complement graph is built from all edges in $\Gamma$ that are not edges in $\Gamma _1$: the connections between the edges are inherited from $\Gamma$, that is two edges in $\Gamma \setminus \Gamma _1$ are connected at a vertex if they were connected at the same vertex in $\Gamma$. The corresponding equivalence class simply loses all endpoints of edges that belong to $\Gamma _1.$ Note that graph’s complement may have non-trivial intersection with the original graph: it consists of all vertices that belong to both $\Gamma _1$ and $\Gamma \setminus \Gamma _1$. These vertices form the boundary of $\Gamma _1$ with respect to the original graph $\Gamma$:

$$\displaystyle \begin{aligned} \delta_\Gamma \Gamma_1 = \Gamma_1 \bigcap \Big( \Gamma \setminus \Gamma_1 \Big). \end{aligned} $$

(22.16)

It consists of all vertices in $\Gamma$ that belong to both $\Gamma _1$ and $\Gamma _2$. Another way to characterise the subgraph’s boundary is

$$\displaystyle \begin{aligned} \delta_\Gamma \Gamma_1 = \Big\{ V^m \in \Gamma_1: \; \deg_{\Gamma_1} (V^m) < \deg_\Gamma (V^m) \Big\}, \end{aligned} $$

(22.17)

where $\deg _\Gamma$ and $\deg _{\Gamma _1}$ denote the degrees of the vertices with respect to $\Gamma$ and to $\Gamma _1$ respectively.

Let an infiltration domain $D_j$ be determined. Then for every vertex from the boundary $\delta _\Gamma D_j$ at least one of the two topological conditions required by Theorem 22.2 fails to be satisfied:

• (1) Dissolution of the vertex disconnects the complement graph.

  Consider the example presented in Fig. 22.6: $V^0$ is the original contact vertex, $V^1$ is a new contact vertex; dissolving $V^1$ the complement graph falls into two disconnected components, the infiltration domain $D_0$ is marked by red with $V^1$ being its unique boundary vertex.

  Fig. 22.6

  

  Bottleneck prevents dissolution (see Definition 22.8)

• (2) The degree of the vertex in the complement graph equals 2.

  Consider the example presented in Fig. 22.7. The infiltration domain $D_0$ corresponding to the original contact vertex $V^0$ is marked in red. The four boundary vertices $V^1, V^2, V^3, V^4$ have degree two (with respect to the graph complement to $D_0$) and therefore cannot be dissolved. The edges attached to these vertices (marked by fat red curves) can be seen as a wall surrounding the infiltration domain. We do not assume that walls belong to infiltration domains.

  Fig. 22.7

  

  Degree two vertices prevent dissolution

Degree two vertices are excluded in the original graph (they can be removed since we assume standard vertex conditions), but such vertices may appear after the reduction.

Definition 22.8

A vertex $V^0$ in a connected graph is called a bottleneck vertex if dissolving this vertex the graph becomes disconnected.

This notion is close to that of bridges in a graph—edges whose removal makes graph disconnected. Bottleneck vertices for metric graphs play a role analogous to the bridges in discrete graphs. Consider the graph presented in Fig. 22.8. Only the vertex $V^1$ is a bottleneck.

Fig. 22.8

Graph with a bottleneck vertex

One should not think that the resulting graph always has just two connected components, see for example the initial graph in Fig. 22.6.

Definition 22.9

Let $D$ be an infiltration domain in a metric graph $\Gamma$. Then the domain’s wall$W_j$ is the union of all edges in the complement to $D$ connected by at least one of the endpoints to the boundary $\partial D$ (with respect to the original graph $\Gamma$).

Consider the graph in Fig. 22.7. The infiltration domain is marked in red and is given by the 4-star. The boundary is given by the pendant vertices in the star—the vertices $V^1, V^2, V^3, V^4$. The wall is marked by thick red lines and form a single cycle connected via 5 other vertices to the rest of the graph.

With these definitions one may say that every infiltration domain is separated from the rest of the graph by its wall and the set of bottleneck vertices. Note that the two edges connected to degree two bottleneck vertices belong to the wall.

Bottlenecks connecting more than two components in the original graph always remain if the graph is reduced starting from a single contact vertex and therefore prevent further expansion of infiltration domains (see Fig. 22.6). On the other hand bottlenecks connecting just two components in the original graph are not dangerous and disappear, provided the degree of such vertices with respect to each of the graph components is not 2 (see Fig. 22.9).

Fig. 22.9

Degree two bottlenecks may allow dissolution

One should not imagine that the infiltration domains are always surrounded by walls as in Fig. 22.7—such a picture is suitable for planar graphs only. In fact any metric graph may serve as a wall; the only restriction is that every edge should be connected by one of the ends to a degree two vertex. Let us return to the graph presented in Fig. 22.3, as shown in Fig. 22.10 this graph serves as a wall for the infiltration domain on a larger graph. One places degree two dummy vertices in the middle of each edge in the original graph and connects all these vertices by a star. The middle vertex in the star serves as a contact vertex for the new larger graph.

Fig. 22.10

Any metric graph could be a wall

Example 22.10

Consider the graph $K_{3,3}$—the complete bipartite graph on three vertices presented in Fig. 22.11. Assume without loss of generality that $V^4$ is a contact vertex. Dissolving the vertex and peeling the three pendant edges we get three new contact vertices $V^1, V^3 ,$ and $V^5$. The procedure stops since all these vertices have degree two with respect to the remaining graph. It follows that the infiltration domain is just this 3-star. What is interesting is that the whole remaining graph forms the wall for the infiltration domain. The wall can be seen as a watermelon graph on three parallel edges with extra contact vertices in the middle of the edges. We see that the wall not only contains cycles but also that these cycles are not independent.

Fig. 22.11

The infiltration domain for the graph $K_{3,3}$

We summarise our studies of infiltration domains as

Observation 22.11

Consider any infiltration domain and its wall in a finite graph $\Gamma .$ The following possibilities may occur

• (1) the infiltration domain coincides with $\Gamma$ , so the wall is empty (e.g. Fig. 22.3 (upper figure));

• (2) the infiltration domain and its nonempty wall cover$\Gamma$(e.g. Fig.22.11);

• (3) the infiltration domain and its nonempty wall form a proper subgraph of$\Gamma$(e.g. Fig.22.7);

• (4) the infiltration domain does not cover$\Gamma$but the wall is empty (e.g. Fig.22.6).

One may continue these studies by investigating how two infiltration domains and their walls may be situated in relation to each other. See Fig. 22.12 for illustrations. Note that two subgraphs are considered disjoint if they share no more than a finite number of points.

Fig. 22.12

Graphs with disjoint infiltration domains and with (a) a common wall; (b) partially common walls; (c) disjoint walls; (d) disjoint walls

22.4 Solution of the Inverse Problem Via the MBC-Method

The idea behind solving the inverse problem for general graphs is to find a sufficient number of contact vertices so that the corresponding infiltration domains cover, or almost cover, the original graph. In other words we are going to assume first that the skeleton

$$\displaystyle \begin{aligned} \mathbb S : = \Gamma \setminus \left( \bigcup_{V^j \in \partial \Gamma} D_j \right) \end{aligned} $$

(22.18)

is empty or sufficiently thin. Remember that graph complement is understood in the sense of Definition 22.7. We shall present theorems providing sufficient conditions for graph reconstruction, the theorems will be ordered so that the skeleton gets less thin. Naturally in each new theorem stronger assumptions on the spectrum or eigenfunctions will be required. As before we distinguish topological assumptions (enumerated by numbers) from spectral ones (enumerated by letters). The spectral assumptions are generically satisfied.

The first result (see Theorem 22.12 below) may seem rather straightforward, but to prove it rigorously a few important points must be clarified:

• In our preliminary discussions introducing infiltration domains and proving reconstructability of the potential it was always assumed that the topology is known, hence one has to describe how topology of infiltration domains can be recovered.

• In the case of several contact points one has to determine how to couple together the corresponding infiltration domains that may (and should) have common vertices.

Theorem 22.12

Let $\Gamma$ be a finite compact pendant free metric graph without degree two vertices and loops, and let $L_{q,a}^{\mathrm {st}}$ be the corresponding standard Schrödinger operator. Assume that

• (1) the contact set$\partial \Gamma$is chosen so that the infiltration domains corresponding to each$V^j \in \partial \Gamma$cover the original graph$\Gamma$,

  $$\displaystyle \begin{aligned} {} \bigcup_{V^j \in \partial \Gamma} D_j = \Gamma, \end{aligned} $$

  (22.19)

  i.e. the skeleton $\mathbb S$ is empty,

  $$\displaystyle \begin{aligned} \mathbb S = \emptyset. \end{aligned} $$

  (22.20)

Assume in addition the following generically satisfied assumption:

• (a) the Dirichlet eigenfunctions on connected subgraphs ¹ of $\Gamma$ do not vanish identically on any edge.

Then the M-function associated with the contact set and known for all possible signings (magnetic fluxes$\Phi _i = 0, \pi , \, i = 1,2, \dots , \beta _1$) determines the graph$\Gamma$and potential$q$.

Proof

Two properties of graph M-functions in the case of standard vertex conditions will be used:

• (1) Let the metric graph $\Gamma$ be fixed and two contact sets $\partial \Gamma$ and $\partial ' \Gamma$ be given, so that $\partial ' \Gamma \subset \partial \Gamma$; then the M-function associated with $\partial \Gamma$ determines the M-function associated with $\partial ' \Gamma$ (see Sect. 17.3 on hierarchy of M-functions, in particular Theorem 17.9).

• (2) Let the metric graph $\Gamma$ and the contact set $\partial \Gamma$ be fixed; then the corresponding M-function determines the distances between the contact vertices (see Step 1 below).

Only the second statement needs a proof. This connection has already been established for metric trees (see Theorem 20.9); the same ideas can be applied to graphs with cycles as follows. □

Step 1: M-function and Distances Between Contact Points

We will establish a relation between the M-function and the distances between the contact points. As was the case with trees the simplest way to establish such a connection is via the dynamical response operator. Definition 20.7 can be generalised as follows:

Definition 22.13

Let $\mathbf R^T$ be the dynamical response operator associated with the metric graph $\boldsymbol {\Gamma }$ and the contact set $\partial \Gamma .$ Let $V^i$ and $V^j$ be any two vertices from $\partial \Gamma .$ Then the travelling time $t (V^i, V^j )$ between the vertices is given by

$$\displaystyle \begin{aligned} {} t (V^i, V^j ) = \sup \; \big\{ T: R^T_{V^i, V^j} \equiv 0 \big\}, \end{aligned} $$

(22.21)

where $R^T_{V^i, V^j}$ denotes the entry of the matrix operator $\mathbf R^T$ associated with the vertices $V^i$ and $V^j$.

Then Lemma 20.8 is modified as follows:

Lemma 22.14

Consider the Schrödinger equation on a finite metric graph $\Gamma$ with standard vertex conditions. Then the travelling time between any two contact vertices $V^i$ and $V^j$ is equal to the distance $\mathrm {dist}\; (V^i, V^j)$ between the vertices.

Proof of Lemma 22.14

The proof is almost identical to the proof of Lemma 20.8, since in determining the travelling time, one checks how the front of the wave initiated at the contact vertex $V^j$ spreads along the graph. If the path connecting $V^j$ to $V^i$ is unique, then the front of the wave evolves as if it were travelling along a certain tree cut from $\Gamma$: the tree contains the path connecting $V^j$ with $V^i$ and arbitrarily short but non-zero pendant edges adjusted to each vertex on the path. The entry of the response operator associated with the vertices $V^j$ and $V^i$ contains a $\delta '$ term delayed by the length of the shortest path. The waves reflected from the vertices and/or coming along any other path are further delayed and do not contribute to this singularity. Then Lemma 20.8 implies that the travelling time coincides with the distance between the edges.

Consider the case in which there are several shortest paths connecting the two vertices. The same argument as above applies to each path. It follows that the kernel of the $V^i,V^j$ entry of the response operator contains the sum of delayed $\delta '$ terms coming from each of the shortest paths. The amplitude for each term is equal to the product of the transmission coefficients from the vertices along the corresponding path. In the case of standard vertex conditions, all transmission coefficients are positive (see (3.41)), hence contributions from different shortest paths cannot cancel each other and the travelling time again equals the distance. □

Step 2: Recovery of the Infiltration Domains

We start by dissolving the contact vertex $V^1$ leading to the metric graph $\Gamma _1$ with $d_1 = \deg V^1$ pendant edges. Choosing any of the pendant vertices, we recover the length and the potential on the corresponding edge, which we denote by $E_1$. The length is reconstructed in the same way as for trees since for small times the nearest vertex acts as if it were a part of a star graph (see Sect. 20.2.2). Let us denote by $V'$ the vertex to which the pendant edge $E_1$ is attached. Note that $V'$ cannot coincide with $V^1$ since $\Gamma$ has no loops. Now the edge $E_1$ can be peeled away. The contact set for the new graph contains $V'$ and all $d_1-1$ pendant vertices.

Now we turn to the second pendant vertex and reconstruct the corresponding pendant edge $E_2$ (i.e. its length and the potential on it). The second edge is connected to the vertex $V'$ if and only if the travelling time between the second pendant vertex and $V'$ is equal to the length of $E_2$. If this is the case then the edges $E_1$ and $E_2$ were parallel. Otherwise we denote by $V^2$ the new vertex to which $E_2$ is connected. We can peel away the edge $E_2$.

Repeating this procedure $d_1$ times, one gets a new graph $\Gamma _2$. All newly-labeled vertices turn into contact vertices for $\Gamma _2$. The number of contact vertices is between $1$ (all peeled pendant edges are parallel) and $d_1$ (no two peeled pendant edges are parallel). If $\Gamma$ is a watermelon, then $\Gamma _1$ is a star graph and the reconstruction is accomplished since $\Gamma _2$ is trivial as a metric graph (one vertex, no edges). If $\Gamma _2$ contains pendant vertices we repeat this procedure until we obtain a pendant free graph.

The dissolution-peeling procedure is applied again to the smaller graph $\Gamma _2$. The only difference is that the graph may have not one but several contact vertices. Therefore each time when a pendant edge is peeled away, one has to compare its length to the distance to any of the contact vertices: if the length and the distance are equal, then the edge is attached to that particular vertex, otherwise one introduces a new vertex to which the edge is attached.

This procedure stops when either the whole graph $\Gamma$ is recovered or all contact vertices are either degree two or bottlenecks with respect to the unrecovered part of $\Gamma$. The corresponding recovered subgraph of $\Gamma$ is the infiltration domain $D_1$. (We reiterate that $D_1$ may depend on the order the vertices are dissolved.)

Step 3: Connecting Different Infiltration Domains Together

Starting from different contact vertices $V^1, \dots , V^{|\partial \Gamma |} \in \partial \Gamma$, corresponding infiltration domains $D_j$ are recovered. Under condition (22.19) some of these domains should have common vertices. One has to understand how the $D_j$’s are connected to each other. One should not exclude the case when $D_i \subset D_j$ for a certain $i \neq j$.

Assume that $D_1$ is recovered. It cannot be excluded that some other original contact vertices belong to $D_1$, therefore each time pendant edges are peeled away when reconstructing $D_1$, one should not only compare the length of the edge to be peeled to the distance between the pendant vertex and any other recovered vertex in $D_1$, but also to all original contact vertices in $\Gamma$: if the length and the distance are equal, then the vertex the pendant edge is attached to should be identified with the already known contact vertex.

The same procedure should be applied when any other infiltration domain is recovered: peeling away pendant edges one compares the length of the edge to the distance to all recovered vertices and identifies the vertex the edge is connected to in the case of equality. In this way connections of new infiltration domains to already recovered domains are also established.

Note that when reconstructing infiltration domains, we do not pay attention to synergy effects that may come from the interaction between neighbouring infiltration domains. This is in order to make the formulation of the theorem more transparent.

Under condition (22.19) the graph $\Gamma$ is completely recovered when all infiltration domains are determined and it is known how they are glued together.

In the above theorem the skeleton is assumed to be empty, which guarantees direct reconstruction of $\Gamma$ from $D_j$. In the following theorem the skeleton may be non-empty but is assumed to be the smallest possible—every edge in the skeleton connects the infiltration domains.

Theorem 22.15

Let $\Gamma$ be a finite compact pendant free metric graph without degree two vertices and loops and let $L_{q,a}^{\mathrm {st}}$ be the corresponding standard Schrödinger operator. Assume that

• (1) the contact set$\partial \Gamma \ni V^j$is chosen so that the union of infiltration domains corresponding to vertices in$\partial \Gamma$contains all vertices in the original graph$\Gamma$:

  $$\displaystyle \begin{aligned} {} \bigcup_{V^j \in \partial \Gamma} D_j \supset \bigcup_{V^m \in \mathbf V} V^m. \end{aligned} $$

  (22.22)

Then it holds that every edge in the skeleton $\mathbb S = \Gamma \setminus \left ( \bigcup _{V^j \in \partial \Gamma } D_j \right )$ connects two of its contact vertices.

Assume in addition the following generically satisfied assumptions:

• (a) the Dirichlet eigenfunctions on the connected subgraphs ² of $\Gamma$ do not vanish identically on any edge;

• (b) the spectra of the Dirichlet operators on the edges forming the skeleton are disjoint.

Then the M-function associated with the contact set$\partial \Gamma$and known for all possible signings (magnetic fluxes$\Phi _i = 0, \pi , \, i = 1,2, \dots , \beta _1$) determines the graph$\Gamma$and potential$q$.

Proof

Assume that the contact set $\partial \Gamma$ is fixed so that all assumptions of the theorem are fulfilled. Repeating the proof of Theorem 22.12 we conclude that all infiltration domains (including the potential on them) and their connections are recovered. It remains to determine the skeleton $\mathbb S$ and the potential on it (Fig. 22.13). Removing all infiltration domains one obtains the M-function for the skeleton associated with all skeleton contact vertices coinciding with those vertices in the skeleton which are simultaneously boundary vertices for a certain infiltration domain:

$$\displaystyle \begin{aligned} {} \partial \mathbb S = \left( \bigcup_{V^j \in \partial \Gamma} \partial D_j \right) \cap \mathbb S. \end{aligned} $$

(22.23)

Under condition (22.22) every vertex in $\mathbb S$ is a contact vertex.

Fig. 22.13

The skeleton (marked by thick black lines) may contain cycles and non-closed paths

If the original graph and hence the skeleton contain no parallel edges, then uniqueness of the skeleton and potential on it follows directly from Theorem 21.8.

To reconstruct the skeleton in the general case we need to modify the proofs of Theorems 21.6 and 23.6 and hence of Theorem 21.8. We shall see possible multiple edges in $\mathbb S$ as watermelon graphs $\mathbb W_j$ connecting two vertices. The Dirichlet eigenvalues on the edges give singularities of the M-function for the skeleton. Each eigenfunction determines precisely two singularities since the Dirichlet spectra on the edges are disjoint (assumption b)) and loops are not allowed. In this way we obtain the Dirichlet spectra of each watermelon graph $\mathbb W_j$. Similar to (21.8) the M-functions are determined up to constant matrices $\mathbf A_j$

$$\displaystyle \begin{aligned} {} \mathbf M_j (\lambda) = \mathbf A_j + \sum_{\lambda_n^{\mathrm{D}} (\mathbb W_j)} \frac{\lambda- \lambda'}{(\lambda_n^{\mathrm{D}} - \lambda) (\lambda_n^{\mathrm{D}} - \lambda')} \langle \partial \psi_n^{\mathrm{D}} \vert_{\partial \mathbb S}, \cdot \rangle_{\ell_2(\partial \mathbb S)} \partial \psi_n^{\mathrm{D}} \vert_{\partial \mathbb S}. \end{aligned} $$

(22.24)

In this formula we do not require knowledge of the normal derivatives $\partial \psi _n^{\mathrm {D}}$, instead we just sum the corresponding singularities in $\mathbf M_{\mathbb S}$. To determine $\mathbf A_j$ from the asymptotics given by (21.11) we need to know the number of parallel edges in each watermelon graph $\mathbb W_j$. We check whether the singularities of $\mathbf M_{\mathbb W_j}$ depend on any magnetic flux: if $\mathbf M_{\mathbb W_j}$ depends on $n$ fluxes, then the degrees of the vertices in $\mathbb W_j$ are equal to $N_j = n+1$. Modifying formula (21.10) we obtain the matrices $\mathbf A_j$ and therefore accomplish reconstruction of $\mathbf M_{\mathbb W_j}$.

If no parallel edges are present, then we are done, since we know that the M-function for a single interval determines its length and the potential on it. It remains to prove that the M-function for a watermelon graph determines the edge lengths and the potential on it.

Suppose $\mathbb W_j$ is a watermelon graph formed by $N_j$ edges connecting the vertices $V^1$ and $V^2$, which we assume to be contact vertices. If $N_j = 2$, then the edges form a cycle of discrete length two. The reconstruction follows from Theorem 23.4, case (2) from Chap. 23. Therefore in what follows we assume that the number of edges in the watermelon graph is at least three. Let us remove one of the two vertices, say $V^2$, from the set of contact vertices. Dissolving the remaining vertex $V^1$ we get the M-function for the star graph and can therefore determine the lengths of all edges in the watermelon and the potential on it. This completes the proof of the theorem since the skeleton is completely recovered. □

Assumptions (a) and (b) on the spectrum and eigenfunctions are generically satisfied in the following sense: assume that potential q on a copy of the real line $\mathbb R$ is fixed. Then choosing the edges $E_n$ arbitrarily and in this way fixing the potential $q$ on $\Gamma$ leads almost surely to a quantum graph with the conditions above satisfied. More precisely, the set of endpoints for which some of Dirichlet eigenfunctions vanish identically on certain edges is meagre. This can be proved following [82] since it is assumed that the graphs do not have loops and the number of subgraphs involved is finite.

Analysing the proof we see that the assumptions are too restrictive and may be weakened without actually changing the proof. We decided not to include such weaker but cumbersome assumptions in order to make formulation of the main theorem clearer—these assumptions are generically satisfied anyway. The most obvious extensions of the theorem are as follows:

• Instead of assumption (a) it is enough to assume that at least three normal derivatives of such eigenfunctions are nonzero at each vertex that is being dissolved. This is equivalent to the assumption that the eigenfunctions are non-zero on at least three edges emanating from each dissolving vertex.

• Instead of assumption (b) it is enough to assume that the spectra of the Dirichlet operators are disjoint, provided the corresponding edges are either neighbours or are separated by a single edge. In other words one may allow common eigenvalues for the Dirichlet operators on edges lying far away in the skeleton.

In the following theorem the skeleton is permitted to be even bigger; nevertheless the infiltration domains still cover a major part of the original graph. To compensate for the relaxation a new assumption (2) is introduced: it is not generically satisfied and has topological nature. In what follows it will be convenient to consider single intervals as a special case of star graphs.

Theorem 22.16

Let $\Gamma$ be a finite compact pendant free metric graph without degree two vertices and loops and let $L_{q,a}^{\mathrm {st}}$ be the corresponding standard Schrödinger operator. Assume that

• (1) the contact set $\partial \Gamma$ is chosen so that the union of infiltration domains corresponding to the vertices from $\partial \Gamma$ and their walls $W_j$ cover the original graph $\Gamma$

  $$\displaystyle \begin{aligned} \bigcup_{V^j \in \partial \Gamma} \left(D_j \cup W_j \right) \supset \Gamma \supset \bigcup_{V^m \in \mathbf V} V^m; \end{aligned} $$

  (22.25)

• (2) the skeleton contains no cycles of discrete length less than or equal to 4.

Then the skeleton$\mathbb S = \Gamma \setminus \left ( \bigcup _{V^j \in \partial \Gamma } D_j \right )$is a union of star graphs joined at skeleton contact vertices$\partial \mathbb S$.

Assume in addition the following generically satisfied assumptions:

• (a) the Dirichlet eigenfunctions on the connected subgraphs ³ of $\Gamma$ do not vanish identically on any edge;

• (b) the spectra of the Dirichlet operators on the star graphs forming the skeleton are disjoint.

Then the M-function associated with the contact set$\partial \Gamma$and known for all possible signings (magnetic fluxes$\Phi _i = 0, \pi , \, i = 1,2, \dots , \beta _1$) determines the graph$\Gamma$and potential$q$.

Proof

Our first step is to prove that the skeleton is formed by star graphs connected at contact vertices. Consider formula (22.23) for the skeleton contact set. Every edge in the skeleton belongs to a wall, and therefore at least one of its endpoints is contained in $\bigcup _{V^j \in \partial \Gamma } \partial D_j$. This vertex must be from $\partial \mathbb S$, for otherwise the edge does not belong to the skeleton. Summing up, every edge in the skeleton has at least one endpoint from the contact set $\partial \mathbb S$. Introducing Dirichlet conditions at the skeleton contact points dissolves it into a set of star graphs and single edges, which are treated as star graphs as well.

Repeating the arguments used in the proof of Theorem 22.15, we conclude that the M-function for $\Gamma$ determines the M-function for $\mathbb S$ associated with all contact vertices from $\partial \mathbb S$. We are going to use Theorem 21.7 to reconstruct the skeleton, so let us check that all conditions in the theorem are satisfied:⁴

• we have already proven that the contact vertices dismantle the skeleton into a set of star graphs—simplest trees;

• standard vertex conditions (which are a special form of delta couplings) are assumed at all vertices;

  • (1) cycles of discrete length 2 are forbidden in the skeleton, hence no star graph in $\mathbb S$ has two pendant vertices coming from the same contact vertex;

  • (2) cycles of discrete length 2, 3 and 4 are forbidden in the skeleton, hence no two star graphs have more than one common vertex;

    • (a) it is assumed that the spectra of the Dirichlet operators on the star graphs forming the skeleton are disjoint.

We see that all conditions of Theorem 21.7 are satisfied for the skeleton dismantled by the contact vertices into star graphs. Hence the corresponding M-function determines the skeleton and the potential on it. This completes the reconstruction of $\Gamma$ and $q$. □

As we already pointed out, Assumptions (a) and (b) are generically satisfied, while Assumptions (1) and (2) are related to the topology of $\Gamma$ and the choice of $\partial \Gamma$. Assumption (2) can be weakened as follows: reconstructing the skeleton we have not used dependence of the corresponding M-function on the magnetic fluxes associated with the skeleton. One may allow parallel edges and parallel stars. We leave this as a problem for the reader.

Problem 95

Study whether Assumption (3) can be weakened ensuring reconstruction of $\Gamma$ and $q$ applying the MBC-method to the skeleton.

Another possible generalisation concerns vertex conditions: it is enough to assume that the vertex conditions at the skeleton inner vertices are generalised delta couplings (required by Theorem 21.7).

Matryoshka-Type Structure of Reconstructable Graphs

It is clear that the graphs described by the three theorems above do not exhaust the whole family of graphs reconstructable via the MBC-method. Our goal here is to indicate how such a family can be described. We are going to focus on topological properties of the graphs assuming that generically satisfied spectral conditions are always fulfilled. It will be convenient to see any metric graph as a pair $(\Gamma , \partial \Gamma )$ consisting of the metric graph $\Gamma$ and the contact set $\partial \Gamma$.

The families of graphs covered by Theorems 22.12, 22.15, and 22.16 have one common feature: the inverse problem for the corresponding skeleton has already been solved (or can be easily solved). One may determine an inductive procedure characterising pairs that are reconstructable by our methods. This procedure brings to mind Russian matryoshka dolls.

Assume that we have already characterised a family $\mathcal F_0$ of reconstructable pairs, for example given by Theorem 22.16. Then we may also reconstruct all pairs whose skeletons belong to the original family $\mathcal F_0$. Denoting the new family by $\mathcal F_1$ we may repeat our argument and obtain families $\mathcal F_2$, $\mathcal F_3$, $\dots$.

Not all pairs are reconstructable: bottlenecks and degree two vertices make it impossible to reconstruct certain graphs in the same way that they prevent further growth of infiltration domains. In fact every time when the infiltration domain does not cover the whole graph we have an example of a metric graph with one contact point that does not belong to the family (see Figs. 22.4, 22.5, and 22.7). We provide here a few more examples:

• All graphs with bottlenecks of degree three and higher and contact sets lying on one side of the bottleneck are not reconstructable (see Fig. 22.6).

• Consider the graph presented in Fig. 22.14 with contact vertices $V^1, V^2, V^3$: the subgraph $\Gamma _0$ marked by blue colour is connected to the rest via degree two vertices and therefore cannot be reconstructed. Adding further peripheric vertices does not help to solve the problem.

  Fig. 22.14

  

  Degree two vertices prevent reconstruction

It is clear that the family $\mathcal F = \bigcup _n \mathcal F_n$ of all reconstructable pairs possesses a certain monotonicity property:

Increasing the contact set the pair remains in the family:

$$\displaystyle \begin{aligned} \left. \begin{array}{l} (\Gamma, \partial \Gamma ) \in \mathcal F, \\[3mm] \partial \Gamma \subset \partial' \Gamma \end{array} \right\} \Rightarrow (\Gamma, \partial' \Gamma) \in \mathcal F. \end{aligned} $$

(22.26)

Moreover Theorem 21.8 implies that every pair with the maximal possible contact set (consisting of all vertices) is reconstructable, provided the graph has no loops or parallel edges. Note that the assumption concerning parallel edges may be removed as was done in the proof of Theorem 22.15.

On the other hand fixing the contact set and making the metric graph smaller does not necessarily guarantee constructability:

$$\displaystyle \begin{aligned} \left. \begin{array}{l} (\Gamma, \partial \Gamma ) \in \mathcal F, \\[3mm] \Gamma' \subset \Gamma \end{array} \right\} \not\Rightarrow (\Gamma', \partial \Gamma) \in \mathcal F. \end{aligned} $$

(22.27)

Consider for example the graphs presented in Figs. 22.3 and 22.4: the second graph is obtained from the first one by removing two edges.