Boundary Control: BC-Method

Abstract

With this chapter we start the discussion on how to solve in full generality the inverse problems for Schrödinger operators on metric graphs.

19.1 Inverse Problems: First Look

With this chapter we start the discussion on how to solve in full generality the inverse problems for Schrödinger operators on metric graphs. When considering Ambartsumian type theorems in Chaps. 14 and 15, we have already pointed out (see the introductory comments to Chap. 14) that the solution of the inverse problem means recovering all three members of the triple

• the metric graph $\Gamma$,

• the potential q,

• the vertex conditions.

We do not discuss how to reconstruct the magnetic potential since it can be eliminated, leading to different vertex conditions (see Chap. 16). On the contrary, we are going to consider spectral data dependent on the magnetic fluxes through the cycles in the graph, thus allowing a non-destructive investigation of quantum graphs in real world experiments. The fact that spectral and transport properties of nano-systems depend on the magnetic fluxes $\Phi _j$ is well-known for physicists as the Aharonov-Bohm effect [9, 110, 113, 459, 470, 485, 508]. More precisely, we shall use spectral data for the magnetic fluxes equal to 0 and $\pi$. These spectral data correspond to the standard Schrödinger operators on $\Gamma$ with zero magnetic potential and possibly extra signing conditions (3.43) introduced on every cycle.

The proven Ambartsumian type theorems allow us to solve the inverse spectral problem in certain very specific cases. In general, a single spectrum is not enough to solve the inverse problem. For example, a potential on an interval is determined by the two spectra corresponding to different boundary conditions at one of the endpoints [375]. In the case of metric graphs one may extend the set of spectral data by adding the spectra of the problems obtained by amending vertex conditions at different vertices. We do not pursue this direction, since the knowledge of all vertices trivializes reconstruction of the metric graph and the problem is highly overdetermined for large graphs.

Our set of spectral data will contain the M-functions associated with a relatively small set of vertices, to be called the contact set. One should imagine that contact vertices are used to approach the graph. For example, in the case of trees the contact set can be chosen to coincide with all degree one vertices. On one hand, drawing an arbitrary tree on a sheet of paper the degree one vertices naturally form the graph’s boundary. On the other hand, the M-function’s diagonal entry associated with any degree one vertex determines the potential on the corresponding edge. This reconstruction can be carried out using the Boundary Control method described in this chapter. As a result we end up with the M-function associated with a smaller tree. Repeating the procedure we solve the inverse problem step-by-step. This procedure is described in Chap. 20.

In general the contact set should be allowed to contain higher degree vertices, since there are graphs without degree one vertices. The following assumption will be assumed in the rest of the book.

Assumption 19.1

The contact set $\partial \Gamma$ is a non-empty subset of the vertex set that contains all degree one vertices.

To guarantee the unique solvability of the inverse problem the contact vertices should be well-distributed inside the graph $\Gamma$ taking into account its topology. Without any knowledge of the graph’s structure it is hard to formulate explicit conditions on how the contact vertices should be placed. The number of required contact vertices may reduced if one considers the M-functions depending on the magnetic fluxes through the cycles. This will allow us to reconstruct the M-function for a spanning tree associated with the original metric graph $\Gamma$. We call the corresponding method Magnetic Boundary Control as it uses ideas from the classical Boundary Control method, but the spectral data are magnetic flux dependent.

Using the MBC-method different approaches having local and global characters are combined. The local approach we are going to use is the Boundary Control method (BC-method) due to Belishev [67, 70,71,72], who formulated and developed this approach bringing ideas from control theory to the area of inverse problems. Local approaches to inverse problems have been used earlier independently by Gopinath and Sondhi [241, 242] and by Blagoweshchenskii [91], but it was Belishev, who turned BC-method into a standard tool to solve inverse problems, with the help of numerous collaborators and colleagues, in particular: S. Avdonin, D. Korikov, Ya. Kurylev, L. Oksanen, L. Pestov, and A. Vakulenko. BC-method takes particular simple form in one dimension, where it is closely related to the solution of the inverse problem using the asymptotics of the M-function, suggested recently by Simon et al. [238, 441, 472], the relations are well-described in [460]. As the name suggests, the BC-method uses ideas from control theory to solve inverse problems using boundary observations. The Laplace transform connects the response operator appearing in the BC-method to the graph’s M-function (see (19.12)). The procedure of opening the cycles has global character and is based entirely on the connection between the M-functions for the graph and its spanning tree.

In this chapter we give a comprehensive introduction to the Boundary Control method and discuss how it can be used to solve the inverse problem for the star graph. This approach emerged from our papers [37, 44].

Before we proceed let me mention that the inverse problems for operators on metric graphs have been discussed using alternative sets of spectral data. It is not always straightforward to establish connections between these approaches. The method of spectral mapping was used by V.A.Yurko and collaborators, let me mention the most important publications: [223, 224, 509,510,511,512,513,514,515,516,517,518,519,520,521,522,523,524]. M. Belishev with collaborators employed the BC-method, in particular its variant used to reconstruct Riemann surfaces, to solve inverse problems for graphs [68, 69, 73,74,75, 280]. See also [98,99,100,101, 109, 122, 151, 331, 426, 427, 429, 430, 447, 447, 458, 507].

19.2 How to Use BC-Method for Graphs

With the Schrödinger operator $L_{q} = - \frac {d^2}{dx^2} + q$ on $[0, \infty )$ one associates the wave equation

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle \frac{\partial^2}{\partial t^2} u + L_{q} u = 0, \; \, x \in \Gamma, \; t > 0, \\[3mm] \displaystyle u (x,0) = \frac{\partial}{\partial t} u (x,0) \equiv 0, \\[3mm] \displaystyle u (0,t) = f (t). \end{array} \right. \end{aligned} $$

(19.1)

The function $f$ is called the boundary control. Solving the wave equation one obtains a certain differentiable function $u^f (x,t)$. The linear operator

(19.2)

is called the response operator, and contains all information that an observer placed at the origin can possibly obtain sending waves to $[0, \infty )$ and collecting their response. In the theory of one-dimensional inverse problems it is proven that the response operator determines the potential $q$ (see Sect. 19.4). The reconstruction procedure is local in the sense that in order to reconstruct the potential on the interval $[0, \ell ]$ one needs to know the response operator only for all $t \le T = 2 \ell .$ Since the propagation speed is equal to 1, the time $T = 2 \ell$ is precisely the time needed for the wave to travel from the boundary point $x = 0$ to $x = \ell$ and back. It is clear that this result is optimal since the response operators for any $T' < T$ are independent of the form of the potential on the interval $x >T'/2.$ A precise solution of the inverse problem following BC-method is described in Sect. 19.4.

The described properties of the BC-method show that it may be applied to Schrödinger operators on graphs in order to recover the potential on the edges having degree one vertex as one of the endpoints. No serious modification is required, since the wave evolution on a metric graph has finite speed of propagation. Let us assume that the boundary control is applied at some degree one vertex. Then for small values of time $t$ the waves initiated by the boundary control may reach only a small neighbourhood of the vertex. More precisely the wave function may be different from zero only for points $x$ at distances less than or equal to $t$ from the vertex. If $T$ is less than double the length of the pendant edge, then the response operator ${\mathbf {R}}$ coincides with the response operator for the half-axis with the same potential on the interval $[0, T/2).$ This principle can be extended to compare response operators for two arbitrary quantum graphs with equal potentials on one of the pendant edges: the corresponding entries of the response operators are identical for sufficiently small values of t.

19.3 The Response Operator and the M-Function

Assume that a magnetic Schrödinger operator $L_{q,a}^{{\mathbf {S}}} (\Gamma )$ is given. On the metric graph $\Gamma$ we select any non-empty contact set of vertices $\partial \Gamma$ satisfying Assumption 19.1. As before we assume standard vertex conditions on $\partial \Gamma$ to facilitate our presentation. The vertex conditions at the internal vertices are arbitrary (and are given in (17.5)).

For a given continuous function u, let $\vec {u}^\partial$ be the vector whose entries are the values of $u$ at the contact vertices $\partial \Gamma$. Similarly, the vector of extended normal derivatives $\partial \vec {u}^\partial$ will have the coordinates:

$$\displaystyle \begin{aligned} \partial u(V^m) = \sum_{x_j \in V^m} \partial u(x_j), \quad V^m \in \partial \Gamma.\end{aligned}$$

These vectors have dimension $M_\partial = \# \partial \Gamma$, the number of vertices in the contact set.

Consider the wave equation on $\Gamma$,

$$\displaystyle \begin{aligned} {} \frac{\partial^2}{\partial t^2} u (x,t) + \left(i \frac{\partial}{\partial x} + a(x) \right)^2 u (x,t) + q(x) u (x,t) = 0, \; \; x \in \Gamma, \; t > 0, \end{aligned} $$

(19.3)

with zero initial data

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle u (x,0) = 0, \\ \displaystyle \frac{\partial}{\partial t} u (x,0) = 0, \end{array} \right. \end{aligned} $$

(19.4)

subject to the matching conditions (17.5) at all internal vertices, and to the continuity condition and boundary control

$$\displaystyle \begin{aligned} {} \vec{u}^\partial (t) = \vec{f} (t), \end{aligned} $$

(19.5)

at the contact vertices.

The boundary control$\vec {f}$ is a vector valued function with values in $\mathbb C^{M_\partial }.$ If the boundary control is not a smooth function, then one has to consider weak solutions to the wave equation. On the other hand if $\vec {f}$ is at least two times continuously differentiable and both potentials are sufficiently continuous,

$$\displaystyle \begin{aligned} {} \begin{array}{c} \displaystyle \vec{f} \in C^2 (\mathbb R_+, \mathbb C^D), \quad \vec{f}(0) = \vec{f}'(0) = 0, \\[3mm] \displaystyle a \in C^2 (\Gamma \setminus {\mathbf{V}}), q \in C (\Gamma \setminus {\mathbf{V}}), \end{array} \end{aligned} $$

(19.6)

then the function $w (x,t)$ satisfies Eq. (19.3) in the conventional sense.

The solution to this wave equation will be denoted by $u^{\vec {f}}.$ We are going to study the properties of this solution in Sect. 19.4. In particular, since the wave equation has a finite speed of propagation the solution $w (x,t)$ will be equal to zero outside the $t$-neighbourhood of the contact set.

The dynamical response operator${\mathbf {R}}$ is then given by the equality

(19.7)

The dynamical response operator is a natural generalisation of the Dirichlet-to-Neumann map and therefore is sometimes referred to as the dynamical Dirichlet-to-Neumann map, since it connects the Dirichlet and Neumann boundary data for any solution to the wave equation on $\Gamma .$ This operator originally defined on functions satisfying conditions (19.6) can be extended to the set of $L_2^{\mathrm {loc}}$-functions by continuity. The dynamical response collects all information that an observer may obtain about the quantum graph via boundary measurements. As we shall immediately see it is closely related to the graph’s M-function and the scattering matrix.

Assume that the boundary control has compact support as a function of time

$$\displaystyle \begin{aligned} \vec{f} \in C_0^\infty ( (0, \infty); \mathbb C^{M_\partial}). \end{aligned} $$

(19.8)

Then for sufficiently large $t$ (to the right of the support of $\vec {f}$) the evolution is described by the wave equation on $\Gamma$ with the Dirichlet conditions on the contact set, hence the energy is preserved. It follows in particular that the $L_2 (\Gamma )$ norm of $u$ is uniformly bounded, and the Laplace transform can be used to solve the original wave equation (19.1), yielding

$$\displaystyle \begin{aligned} \hat{u} (x,s) = \int_0^\infty e^{-st} u(x,t) dt, \; \; \; \mbox{Re} \, s > 0. \end{aligned} $$

(19.9)

The function $\hat {u}$ is a solution of the following differential equation

$$\displaystyle \begin{aligned} s^2 \hat{u} (x,s) + \left(i \frac{\partial}{\partial x} + a(x) \right)^2 \hat{u} (x,t) + q(x) \hat{u} (x,t) = 0, \; \; x \in \Gamma, \; \mbox{Re} \, s > 0, \end{aligned} $$

(19.10)

satisfying the matching conditions (17.5) at all internal vertices, continuous on $\partial \Gamma$ and satisfying the condition

$$\displaystyle \begin{aligned} \hat{\vec{u}}^\partial (s) = \hat{\vec{f}} (s) \end{aligned} $$

(19.11)

on the contact set, where $\hat {\vec {f}} (s)$ is the Laplace transform of $\vec {f}.$

The extended normal derivatives of the function $\hat {w}(x,s)$ on the contact set may be calculated using the graph M-function (17.7). The result is

$$\displaystyle \begin{aligned} \partial \hat{\vec{u}}^\partial = {\mathbf{M}}_\Gamma (-s^2) \hat{\vec{u}}^\partial.\end{aligned}$$

We have proven the following remarkable formula

$$\displaystyle \begin{aligned} {} \widehat{\left({\mathbf{R}} \vec{f} \right)} (s) = {\mathbf{M}}_\Gamma (-s^2) \hat{\vec{f}} (s), \end{aligned} $$

(19.12)

which implies that the M-function and the dynamical response operator are in one-to-one correspondence (this was noticed first in [37]). The connection between the scattering matrices and M-functions was established in the previous chapter (see (18.40)).

In what follows we are going to switch between the three equivalent sets of spectral data:

• the Titchmarsh-Weyl M-function ${\mathbf {M}}_\Gamma (\lambda )$;

• the scattering matrix ${\mathbf {S}}_\Gamma (\lambda )$;

• the dynamical response operator ${\mathbf {R}} = {\mathbf {R}} (\Gamma ).$

Problem 85

Calculate the response operator for the Laplacian on the interval $[0,1]$ if the contact set is given by

• both endpoints;

• the left endpoint assuming Dirichlet condition at the other endpoint;

• the left endpoint assuming Neumann condition at the right endpoint.

Problem 86

Check that formula (19.12) connecting the M-function and the response operator is valid for the interval ${\mathbf {I}} = [0,1]$with the contact set $\partial {\mathbf {I}} = \{0\}$ and Neumann condition at $x=1$.

19.4 Inverse Problem for the One-Dimensional Schrödinger Equation

In this section we describe how to use the BC-method to reconstruct the potential in the one-dimensional Schrödinger equation. As we already explained earlier this reconstruction is local in the sense that the knowledge of the response operator for relatively small values of the time parameter t allows one to reconstruct the potential $q$ on a part of the interval $[0,\infty )$ close to the point $x=0$, where control is applied. Here we are going to follow [36, 37, 40,41,42], see also recent review papers [43, 460], where relations between the BC method and other local inverse methods are clarified.

The method works under rather weak assumptions on the potential, such as $q \in L_{1, \mathrm {loc}} [0, \infty )$ as described in [460], but we restrict our presentation to continuous potentials in order to make the presentations transparent. In this case all equations are satisfied in the classical sense and there is no need to work with weak solutions. We essentially follow [460] in our presentation of the BC-method.

Consider the wave equation on the interval $[0, \infty )$ with boundary control at the point $x = 0$:

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle - \frac{\partial^2}{\partial x^2} u (x,t) + q(x) u (x,t) = - \frac{\partial^2}{\partial t^2} u (x,t), \\[2mm] \displaystyle u(x,t) = 0, \quad t < x \; \; \mbox{(causality condition)}, \\[2mm] \displaystyle u(x,0) = \frac{\partial}{\partial t} u (x,0) = 0, \\[2mm] \displaystyle u(x,0) = f(t) . \end{array} \right. \end{aligned} $$

(19.13)

We will firstly be interested in the solution of this problem for sufficiently small $t \leq T$.

The solution possesses the following representation:

$$\displaystyle \begin{aligned} {} \displaystyle u^f (x,t) = \left\{ \begin{array}{ll} \displaystyle f(t-x) + \int_{x}^t w(x,s) f(t-s) ds, & x \leq t, \\[2mm] 0, & t \leq x, \end{array} \right. \end{aligned} $$

(19.14)

where $w (x,t)$ is the unique solution to the Goursat problem (Fig. 19.1)

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle \left(- \frac{\partial^2}{\partial x^2} + q(x) \right) w = - \frac{\partial^2}{\partial t^2} w, \\[5mm] \displaystyle w (0, t) = 0, \\[1mm] \displaystyle w (x, x) = - \frac{1}{2} \int_{0}^x q(y) dy. \end{array} \right. \end{aligned} $$

(19.15)

Fig. 19.1

Goursat problem

Formula (19.14) is nothing else than the Duhamel principle telling that the solution at any point $(x,t)$ can be written as a linear combination of the boundary controls at $\tau \in [0, t-x]:$

$$\displaystyle \begin{aligned} u^f (x,t) = f(t-x) + \int_0^{t-x} w(x, t - \tau) f (\tau) d \tau, \; \; x \leq t. \end{aligned} $$

(19.16)

The boundary control at $\tau > t-x$ cannot influence the value of the wave function at $(x,t)$ due to the finite propagation speed. The following definition introduces the control operator connecting the boundary control $f$ to the solution of the wave equation.

Definition 19.2 (Control Operator)

The operator $W^T$ on $L_2 (0,T)$ defined by

$$\displaystyle \begin{aligned} {} \left( W^T f \right) (x) = f(T-x) + \int_x^T w (x,\tau ) f(T- \tau) d \tau, \end{aligned} $$

(19.17)

where $w(x,s)$ solves the Goursat problem (19.15) is called the control operator.

The control operator is invertible and bounded on $L_2 (0,T)$, since it can be inverted by solving a Volterra equation of the second kind. The inverse operator solves the Boundary Control problem:

Given a fixed time $T> 0$ and a function $g \in L_2 (0,T),$ find a boundary control $f \in L_2 (0, T)$ such that

$$\displaystyle \begin{aligned} {} \left( W^T f \right) (x) = g(x) \; \; \mbox{for all } x \in (0, T). \end{aligned} $$

(19.18)

The representation (19.14) allows one to calculate the response operator ${\mathbf {R}}^T$ (already introduced in the previous section by (19.2))

$$\displaystyle \begin{aligned} ({\mathbf{R}}^T f) (t) = \frac{\partial}{\partial x} u(0,t),\end{aligned}$$

where $u$ is the unique solution to the BC problem (19.13)

$$\displaystyle \begin{aligned} {} \left( {\mathbf{R}}^T f \right) (t) = - f'(t) + \int_0^t r(t-\tau) f(\tau) d \tau, \end{aligned} $$

(19.19)

where

$$\displaystyle \begin{aligned} r(t) = \frac{\partial}{\partial x} w (0, t). \end{aligned} $$

(19.20)

Here we used the fact that the kernel $w$ is differentiable as a solution to the Goursat problem with a continuous potential.

Definition 19.3 (Connecting Operator)

The operator on $L_2 (0,T)$ defined by

$$\displaystyle \begin{aligned} C^T = W^{T*} W^T \end{aligned} $$

(19.21)

is called the connecting operator.

The connecting operator can also be defined using its quadratic form

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \langle C^T f, g \rangle_{L_2 [0,T]} & = & \displaystyle \langle (W^T)^* W^T f, g \rangle_{L_2[0,T]} = \langle W^T f, W^T g \rangle_{L_2[0,T]} \\[3mm] & = & \displaystyle \langle u^f ( \cdot, T), u^g (\cdot, T) \rangle_{L_2[0,T]}. \end{array} \end{aligned} $$

(19.22)

Observe that $C^T$ is boundedly invertible and positive, since $W^T$ is boundedly invertible,

$$\displaystyle \begin{aligned} C^T > 0 \; \, \mbox{in }L_2 (0,T).\end{aligned}$$

In what follows we are going to prove two important properties concerning the control operator forming the core of the BC-method.

Lemma 19.4

The connecting operator $C^T$ admits the following representation:

$$\displaystyle \begin{aligned} {} \left( C^T f \right) (t) = f(t) + \int_0^T \left[ p(2T-t-s) - p (\vert t-s \vert) \right] f(s) ds, \end{aligned} $$

(19.23)

where

$$\displaystyle \begin{aligned} {} p (t) = \frac{1}{2} \int_0^{t} r (s) ds. \end{aligned} $$

(19.24)

Proof

To prove this representation it will be convenient to introduce the following rather simple linear operators:

• the operator of odd continuation $S^T$:

  $$\displaystyle \begin{aligned} \left( S^T f \right) (t) = \left\{ \begin{array}{ll} f(t), & 0 \leq t \leq T, \\[3mm] - f(2T-t), & T < t \leq 2T; \end{array} \right.\end{aligned}$$

• the operator extracting the odd part $Q_{2T}$:

  $$\displaystyle \begin{aligned} \left( Q_{2T} f \right) (t) = \frac{1}{2} \left[ f(t) - f(2T-t) \right];\end{aligned}$$

• the operator of restriction $N^T$:

  $$\displaystyle \begin{aligned} N^T f = f \vert_{[0,T]} ;\end{aligned}$$

• the integration operator $J_{2T}$:

  $$\displaystyle \begin{aligned} \left( J_{2T} f \right) (t) = \int_0^t f(s) ds, \; \; 0 \leq t \leq 2T.\end{aligned}$$

It is easy to check that

$$\displaystyle \begin{aligned} {} \left( S^T \right)^* = 2 N^T Q_{2T}. \end{aligned} $$

(19.25)

To prove representation (19.23), consider arbitrary functions $f, g \in C_0^\infty [0,T]$. Let $f_- = S^T f$ and set

$$\displaystyle \begin{aligned} w^{f,g} (s,t) := \int_0^T u^{f_-} (x,s) \overline{u^g} (x,t) dx . \end{aligned} $$

(19.26)

Our goal is to calculate this function for $s = t = T$ since formula (19.22) implies

$$\displaystyle \begin{aligned} {} w^{f,g} (T,T) = \langle C^T f, g \rangle_{L_2 [0,T]}. \end{aligned} $$

(19.27)

To calculate $w^{f,g}(T,T)$ we first show that the function $w^{f,g}$ satisfies the inhomogeneous wave equation and then use d’Alembert’s formula to get its solution. Integration by parts gives the following equalities:

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \left[ \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial s^2} \right] w^{f,g} (s,t) & = & \displaystyle \int_0^T \left[ u^{f_-} (x,s) \overline{u^{g}_{tt}} (x,t) - u^{f_-}_{ss} \overline{u^{g}} (x,t) \right] dx \\[3mm] & = & \displaystyle \int_0^T \left[ u^{f_-} (x,s) \left( \overline{u^g_{xx}} (x,t) - q(x) \overline{u^g} (x,t) \right) \right. \\ & & \displaystyle \; \; \; \; \; \; \; \; \; \; \hspace{5mm} \left. - \left( u^{f_-}_{xx} (x,s) - q(x) u^{f_-} (x,s) \right) \overline{u^g} (x,t) \right] dx \\[3mm] & = & \displaystyle \int_0^T \left[ u^{f_-} (x,s) \overline{u^g_{xx}} (x,t) - u^{f_-}_{xx} \overline{u^g} (x,t) \right] fx \\[3mm] & = & \displaystyle \left[ u^{f_-} (x,s) \overline{u^g_{x}} (x,t) - u^{f_-}_{x} \overline{u^g} (x,t) \right] \vert_{x=0}^T \\[3mm] & = & \displaystyle - f_- (s) \overline{ ({\mathbf{R}}^T g )} (t) + ( {\mathbf{R}}_{2T} f_-) (s) \overline{g} (t), \end{array} \end{aligned} $$

(19.28)

where we used that

$$\displaystyle \begin{aligned} u_{tt}^{f} = u_{xx}^f - q(x) u^f \quad u_{tt}^{g} = u_{xx}^g - q(x) u^g ,\end{aligned}$$

and that

$$\displaystyle \begin{aligned} u^{f_-} (T,s) = u^g (T,t) \equiv 0,\end{aligned}$$

since $f,g \in C_0^{\infty } [0,T].$

Summing up, the function $w^{f,g}$ is a solution to the inhomogenous wave equation

$$\displaystyle \begin{aligned} w^{f,g}_{tt} - w^{f,g}_{ss} = - f_- (s) \overline{ ({\mathbf{R}}^T g )} (t) + ( {\mathbf{R}}_{2T} f_-) (s) g(t) \end{aligned} $$

(19.29)

in the region $0 \leq s \leq 2T, \; \, 0 \leq t \leq T$ with zero initial conditions

$$\displaystyle \begin{aligned} w^{f,g} (s,0) = w^{f,g}_t (s,0) = 0. \end{aligned} $$

(19.30)

The solution is given by d’Alembert’s formula [182], which we use for $t= s = T$

$$\displaystyle \begin{aligned} w^{f,g} (T,T) = - \frac{1}{2} \int_0^T d \eta \int_{\eta}^{2T-\eta} d \xi \left[ f_- (\xi) \overline{ ({\mathbf{R}}^T g )} (\eta) - ( {\mathbf{R}}_{2T} f_-) (\xi) \overline{g} (\eta) \right] \end{aligned} $$

(19.31)

Taking into account that $\int _{\eta }^{2T-\eta } f_- (\xi ) d \xi = 0$ ($f_-$ is an odd function with respect to $\xi = T$) we have

$$\displaystyle \begin{aligned} {} w^{f,g} (T,T) = \frac{1}{2} \int_0^T d \eta \Big( \int_{\eta}^{2T-\eta} d \xi ( {\mathbf{R}}_{2T} f_-) (\xi) \Big) \overline{g} (\eta) . \end{aligned} $$

(19.32)

On the other hand we have

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \int_{\eta}^{2T-\eta} ( {\mathbf{R}}_{2T} f_-) (\xi) d \xi & = & \displaystyle \left( J_{2T} {\mathbf{R}}_{2T} f_- \right) (2T-\eta) - \left( J_{2T} {\mathbf{R}}_{2T} f_- \right) (\eta) \\[3mm] & = & \displaystyle - 2 \left( Q_{2T} J_{2T} {\mathbf{R}}_{2T} f_- \right) (\eta), \end{array}\end{aligned}$$

and expression (19.32) takes the form (using (19.25))

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle w^{f,g} (T,T) & = & \displaystyle - \int_0^T \left(N^T Q_{2T} J_{2T} {\mathbf{R}}_{2T} S^T f \right) (\eta) \overline{g} (\eta) d \eta \\[3mm] & = & \displaystyle - \frac{1}{2} \langle (S^T)^* J_{2T} {\mathbf{R}}_{2T} S^T f, g \rangle_{L_2[0,T]}. \end{array} \end{aligned} $$

(19.33)

Remembering (19.22) and (19.27) we obtain

$$\displaystyle \begin{aligned} {} C^T = - \frac{1}{2} (S^T)^* J_{2T} {\mathbf{R}}_{2T} S^T, \end{aligned} $$

(19.34)

since $C_0^\infty [0,T]$ is dense in $L_2 [0,T].$

Using representation (19.19) we obtains formula (19.23) for the connecting operator. □

Problem 87

Using change of variables check that (19.34) implies representation (19.23) for the connecting operator.

Let $y$ be the solution of the Sturm-Liouville problem

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle - y'' (x) + q(x) y(x) = 0, \\[3mm] y(0) = 0, \; y'(0) = 1. \end{array} \right. \end{aligned} $$

(19.35)

Consider the boundary control problem (19.18): find the control function $f^T$ such that

$$\displaystyle \begin{aligned} {} \left( W^T f^T \right) (x) = \left\{ \begin{array}{ll} y(x), & x \leq T, \\[3mm] 0, & x > T. \end{array} \right. \end{aligned} $$

(19.36)

One may prove that on such boundary control the connecting operator has very simple form.

Lemma 19.5

Let$f^T$be the boundary control leading to the solution of the Sturm-Liouville problem$y$for$x < T$, defined by formula (19.36). Then the connecting operator$C^T$maps$f^T$to$T-x$

$$\displaystyle \begin{aligned} {} \left( C^T f^T \right) (x) = T-x, \; \; x \in [0, T]. \end{aligned} $$

(19.37)

Proof

Consider arbitrary function $g \in C_0^\infty (0,T)$—smooth function with compact support inside $(0, T).$ In our calculations we are going to use that the wave equation has finite speed of propagation and therefore

$$\displaystyle \begin{aligned} w^g (x,t) = 0 , \; \; \mbox{and} \; \; \frac{\partial}{\partial t} w^g (x,t) = 0 \; \mbox{provided} \; x > t .\end{aligned}$$

In particular we have:

• $w^g (x,0) = 0 , \; x > 0 ;$

• $\frac {\partial }{\partial t} w^g (x,0) = 0 , \; x > 0;$

• $w^g (T,t) = 0 , \; T > t ;$

• $\frac {\partial }{\partial t} w^g (T,t) = 0 , \; T > t.$

Then we may perform the following calculations, mostly using integration by parts

(19.38)

Here we used that $w^g$ satisfies the wave equation (19.13) and $y$—the Sturm-Liouville equation (19.35). Since the function $g$ is arbitrary, we get the operator equality (19.37). □

Problem 88

Consider formula (19.38) and check all steps.

We are finally ready to describe the solution of the inverse problem using the BC-method.

Algorithm to solve the inverse problem using BC-method

• (1) Reconstruct the kernel$r(\tau ), \; 0 < \tau < 2T$of the response operator${\mathbf {R}}_{2T}$ assuming that it is given by (19.19)

  $$\displaystyle \begin{aligned} \left( {\mathbf{R}}_{2T} f \right) (t) = - f'(t) + \int_0^t r(t-\tau) f(\tau) d \tau.\end{aligned}$$

• (2) Calculate the connecting operator using formula (19.23)

  $$\displaystyle \begin{aligned} \left( C^T f \right) (t) = f(t) + \int_0^T \left[ p(2T-t-s) - p (\vert t-s \vert) \right] f(s) ds,\end{aligned}$$

  where

  $$\displaystyle \begin{aligned} p (t) = \frac{1}{2} \int_0^{t} r (s) ds, \quad t \in [0, 2 T],\end{aligned}$$

  is determined by the kernel of the response operator.

• (3) Invert the response operator, i.e. solve Eq. (19.37)

  $$\displaystyle \begin{aligned} \left( C^T f^T \right) (x) = T-x, \; \; x \in [0, T]\end{aligned}$$

  to find the boundary control function $f^T (x)$ leading to the solution $y(x)$ on the interval $x \in [0, T]$ as a result of the boundary control. Note that the boundary control function $f^T (\cdot )$ depends on $T$, i.e. to get the linear function $T-x$ as the result of connecting operator, different boundary controls dependent on $T$ should be applied.

• (4) Calculate the solution$y$ using its relation with the control function $f^T$ via (19.17)

  $$\displaystyle \begin{aligned} \left( W^T f^T \right) (x) = f^T(T-x) + \int_x^T w (x,\tau ) f^T(T- \tau) d \tau,\end{aligned}$$

  leading to

  $$\displaystyle \begin{aligned} y(T) = \Big( W^T f^T \Big) (T-0) = f^T ({T} - {T}+0) = f^T (+0) .\end{aligned}$$

  The value of $y(T)$ is roughly equal to the initial value of the boundary control function $f^T$ that has to be applied to get $y$ on the interval $[0,T].$

• (5) Calculate the potential$q$ using that $y$ is a solution to the Schrödinger equation

  $$\displaystyle \begin{aligned} {} q(T) = \frac{\displaystyle y'' (T) }{\displaystyle y(T)} = \frac{\displaystyle \frac{d^2}{dT^2} f^T (+0)}{\displaystyle f^T(+0)}. \end{aligned} $$

  (19.39)

  Note that in formula (19.39) one needs to take the limit first and then differentiate the control function $f^T (+0)$ with respect to $T.$

We summarise our studies as

Theorem 19.6

The response operator${\mathbf {R}}^T$for the Schrödinger differential expression$- \frac {d^2}{dx^2} + q(x)$on$[0, \infty )$with locally integrable potential$q$determines the unique potential on the interval$[0, T/2]$.

The advantage of the described method is that solution to the inverse problem is essentially reduced to reconstruction of the kernel of an integral operator and inversion of another integral operator. The rest is just the integration and differentiation. The nature of this method is local: to recover potential close to $x = 0$ one needs to know the response operator ${\mathbf {R}}^T$ for small values of $T.$

An alternative approach to inverse spectral problems in one dimension based on A-amplitude was developed by B. Simon and collaborators [238, 441, 443, 472]. This approach is based on the analysis of the M-function. The two approaches have already been compared in [43, 460].

19.5 BC-Method for the Standard Laplacian on the Star Graph

In this section we study the boundary response operator for the Laplacian on any equilateral star graph with standard vertex conditions at the middle vertex. The case of Laplacian (zero potential) is important, since as we shall see later on, the singularities in the kernel of the response operator for the Schrödinger evolution are determined precisely by the response operator for the Laplacian. Let us denote by $\ell$ the (common) length of the edges and by $N$ the number of edges. Every edge is glued by one endpoint to the central vertex and the opposite endpoints belong to the contact set. It will be convenient to identify functions $u$ from $L_2 (\Gamma )$ with the vector-valued functions $\vec {u} \in L_2 ([0,\ell ], \mathbb C^N)$, so that the central vertex corresponds to $x= 0$.

To calculate the dynamical response operator we need to find the unique solution to the wave equation

$$\displaystyle \begin{aligned} {} \frac{\partial^2 }{\partial t^2} \vec{u} (x,t) - \frac{\partial^2 }{\partial x^2} \vec{u} (x,t) = 0, \; \; x \in (0, \ell), \, t \in (0,T), \end{aligned} $$

(19.40)

satisfying standard vertex conditions at the origin, subject to the boundary control

$$\displaystyle \begin{aligned} {} \vec{u} (\ell, t) = \vec{f} (t), \end{aligned} $$

(19.41)

and with zero initial data (19.4)

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle \vec{u} (x,0) = 0, \\ \displaystyle \frac{\partial }{\partial t} \vec{u} (x,0) = 0. \end{array} \right. \end{aligned} $$

(19.42)

It is clear that solutions to the differential equation can be written as a combination of d’Alembert waves

$$\displaystyle \begin{aligned} {} \vec{u} (x,t) \equiv \vec{u}^{\vec{f}} (x,t) = \vec{b} ( t+x) + \vec{a} (t - x), \end{aligned} $$

(19.43)

where $\vec {b}$ and $\vec {a}$ denote, respectively, the waves going toward the central vertex and coming from it. The boundary control initiates waves on the edges $E_n, \, n=1,2, \dots , N$, which reach the central vertex at the time $t = \ell$. Therefore for sufficiently small $t$ ($t < \ell$) the solution is given by just one traveling wave

$$\displaystyle \begin{aligned} {} \vec{u} (x,t) = \vec{f} (t+x - \ell), \; \; t < \ell. \end{aligned} $$

(19.44)

The argument is chosen in a special way in order to satisfy the boundary control (19.41):

$$\displaystyle \begin{aligned} \vec{u} (\ell, t) = \vec{f} (t+\ell-\ell) = \vec{f} (t), \quad t < \ell.\end{aligned}$$

For such relatively small values of $t$ the value of $\vec {u}$ on each interval $E_n$ is determined by the corresponding component $(\vec {f})_n$ of the boundary control function.

For $t$ slightly larger than $\ell$ (more precisely for $\ell < t < 2 \ell$) the solution on $E_n$ in addition to the wave initiated by $(\vec {f})_n$ contains a wave going away from the central vertex

$$\displaystyle \begin{aligned} {} \vec{u} (x,t) = \vec{f} (t+x-\ell) + \vec{a} (t-x). \end{aligned} $$

(19.45)

The incoming wave remains the same since no wave coming from the central vertex may turn back inside the edge but the time $t < 2 \ell$ is not enough to reach the central vertex and return back to any of the boundary vertices. It turns out that the outgoing wave $\vec {a}$ can be taken equal to $S_{\mathbf {v}}^{\mathrm {st}} \vec {f} (t-x-\ell )$ leading to the solution

$$\displaystyle \begin{aligned} \vec{u} (x,t) = \vec{f} (t+x - \ell) + S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t-x-\ell), \quad \ell < t < 2 \ell.\end{aligned}$$

Here $S_{\mathbf {v}}^{\mathrm {st}}$ is the vertex scattering matrix corresponding to the standard conditions (3.41). The intuition behind this formula should be rather clear: the waves coming to the central vertex penetrate to the other edges with the amplitudes equal to the entries of the vertex scattering matrix $S_{\mathbf {v}}^{\mathrm {st}}.$ Let us check directly from the definition that this formula gives the correct solution. First of all the combination of d’Alambert waves always give a solution to the wave equation. The boundary control (19.41) is satisfied since the second term in the solution is identically equal to zero

It remains to check that standard conditions are satisfied at the central vertex. The matrix $S_{\mathbf {v}}^{\mathrm {st}}$ possesses the representation

$$\displaystyle \begin{aligned} S_{\mathbf{v}}^{\mathrm{st}} = - {\mathbf{I}} + 2 P_{(1,1, \dots,1)}, \end{aligned} $$

(19.46)

where $P_{(1,1, \dots ,1)}$ is the orthogonal projector on the vector $(1,1, \dots , 1) \in \mathbb C^N.$ We get

$$\displaystyle \begin{aligned} \vec{u} (0,t) = \vec{f} (t-\ell) + S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t-\ell) = 2 P_{(1,1, \dots,1)} \vec{f} (t-\ell) ,\end{aligned}$$

implying that all coordinates in the vector $\vec {w} (0,t)$ are equal. To check the Kirchhoff condition on the normal derivatives we calculate

$$\displaystyle \begin{aligned} \begin{array}{ccl} \partial_n \vec{u} (0,t) & = & \displaystyle \frac{\partial}{\partial x} \Big( \vec{f} (t+x - \ell) + S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t-x-\ell) \Big) \vert_{x= 0} \\[3mm] & = & \displaystyle 2 ({\mathbf{I}} - P_{(1,1, \dots,1)}) \vec{f} (t-\ell) \\[3mm] & = & \displaystyle P_{(1,1, \dots,1)}^\perp \vec{f} (t-\ell) , \end{array}\end{aligned}$$

where $P_{(1,1, \dots ,1)}^\perp = 1 - P_{(1,1, \dots ,1)}$ is the projector on the subspace orthogonal to $(1,1, \dots , 1)$ in $\mathbb C^N$. It follows that the sum of derivatives is zero.

On the next interval $2 \ell < t < 3 \ell$ the waves initiated by the boundary control for $0 < t < \ell$ have enough time not only to reach the central vertex, but also to return back and reflect from the contact vertices. The solution is given by

$$\displaystyle \begin{aligned} {} \vec{u} (x,t) = \vec{f} (t+x - \ell) + S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t-x-\ell) - S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t+x-3\ell) , \quad 2 \ell < t < 3 \ell. \end{aligned} $$

(19.47)

To check that this formula is correct there is no need to consider the central vertex since the third term is identically equal to zero there

$$\displaystyle \begin{aligned} \vec{f} (t+0-3\ell) \equiv 0, \quad t < 3 \ell.\end{aligned}$$

On the other hand, the second and third terms cancel each other on the boundary:

$$\displaystyle \begin{aligned} S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t-\ell-\ell) - S_{\mathbf{v}}^{\mathrm{st}} \vec{f} (t+\ell-3\ell) \equiv 0.\end{aligned}$$

In fact formula (19.47) can be applied for any $0 < t < 3 \ell$ since the third term is identically equal to zero on $t < 2\ell$ and the second term on $t < \ell .$

Continuing this procedure it is straightforward to obtain solution to the boundary control problem for arbitrary $t > 0$

$$\displaystyle \begin{aligned} \vec{u}(x,t) = \left\{ \begin{array}{ll} \displaystyle \sum_{m=0}^{n-1} (-1)^m \Big( (S_{\mathbf{v}}^{\mathrm{st}} )^{m}\vec{f} (t+x- (2m+1)\ell)& \\ +( S_{\mathbf{v}}^{\mathrm{st}})^{m+1} \vec{f} (t-x - (2m+1)\ell)\Big)& \\[5mm] \displaystyle + (-1)^n(S_{\mathbf{v}}^{\mathrm{st}})^n \vec{f} (t+x- (2n+1)\ell) ,&\, 2 n \ell < t < (2n+1) \ell; \\[5mm] \displaystyle \sum_{m=0}^{n} (-1)^m \Big( (S_{\mathbf{v}}^{\mathrm{st}} )^{m}\vec{f} (t+x- (2m+1)\ell) &\\ +( S_{\mathbf{v}}^{\mathrm{st}})^{m+1} \vec{f} (t-x - (2m+1)\ell)\Big)& \, 2 (n+1) \ell < t < (2n+2) \ell. \end{array} \right. \end{aligned} $$

(19.48)

Note that the formula may be simplified taking into account that $\big (S_{\mathbf {v}}^{\mathrm {st}} \big )^2 = {\mathbf {I}}$ eliminating all even powers of the vertex scattering matrix and substituting odd powers with just $S_{\mathbf {v}}^{\mathrm {st}}.$

Let us calculate now the dynamical response operator for $T \in (0, 3 \ell )$ using formula (19.47)

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \left( {\mathbf{R}}^T \vec{f} \right) (t) & = & \displaystyle \partial_n \vec{u} (\ell, t) = - \frac{\partial}{\partial x} \vec{u} (\ell, t) \\[3mm] & = & \displaystyle - \frac{d}{dt} \vec{f} (t) + 2 S_{\mathbf{v}}^{\mathrm{st}} \frac{d}{dt} \vec{f} (t-2\ell) \\[3mm] & = & \displaystyle - \vec{f}'(t) + 2 S_{\mathbf{v}}^{\mathrm{st}} \vec{f}'(t-2 \ell) .\end{array} \end{aligned} $$

(19.49)

The response operator can be seen as a convolution operator with the generalised kernel

$$\displaystyle \begin{aligned} {} - \delta'(t) + 2 S_{\mathbf{v}}^{\mathrm{st}} \delta'(t-2 \ell). \end{aligned} $$

(19.50)

We see that the kernel of the response operator is singular and the singularities occur at the time delays corresponding to the time needed for the wave to travel from the contact set to the central vertex and back. It is important to note that the second and third terms in the solution determine the same singularity in the kernel, hence coefficient 2 in the formula. It follows that the response operator determines the distance to the nearest vertex in the case of standard conditions.

19.6 BC-Method for the Laplacian on the Star Graph with General Vertex Conditions

Our goal now is to obtain an explicit formula for the solution of the boundary control on the star graph assuming most general vertex conditions at the central vertex. We again use vertex notations $\vec {u} \in L_2 ([0,\ell ], \mathbb C^N)$ implying that the vertex conditions (3.21) can be written as

$$\displaystyle \begin{aligned} {} i (S - I) \vec{u}(0) = (S +I) \vec{u}'(0). \end{aligned} $$

(19.51)

To calculate the dynamical response operator we need to find the unique solution to the wave equation (19.40) satisfying vertex conditions (19.51), subject to the boundary control (19.41) and zero initial data (19.42). The same ideas can be applied

• the solution to the wave equation is given as a sum of d’Alembert waves (19.43);

• for $0 < t < \ell$ only one travelling wave is present implying that the solution is identical to the solution (19.44) for the case of standard vertex conditions;

• for larger values of $t$ the solution is obtained step by step by calculating waves reflected from the central vertex or from the contact set.

Hence we start by calculating the solution for $\ell < t < 2 \ell$

$$\displaystyle \begin{aligned} {} \vec{w} (x,t) = \vec{f} (t+x-\ell) + \vec{a} (t-x). \end{aligned} $$

(19.52)

Our immediate aim is to calculate the function $\vec {a}$ from the vertex conditions (19.51). It will be more convenient to write these conditions using Hermitian matrices as it was done in Sect. 3.4:

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{l} \displaystyle P_{-1} \vec{u} (0) = 0, \\[3mm] \displaystyle ({\mathbf{I}}-P_{-1} ) \vec{u}' (0) = A ({\mathbf{I}} -P_{-1}) \vec{u}, \end{array} \right. \end{aligned} $$

(19.53)

where $P_{-1}$ is the projection on the eigensubspace of $S$ corresponding to the eigenvalue $-1$ and

$$\displaystyle \begin{aligned} A = ({\mathbf{I}}- P_{-1}) i \frac{S-{\mathbf{I}}}{S+{\mathbf{I}}} ({\mathbf{I}}-P_{-1}).\end{aligned}$$

The boundary values at the origin of the solution given by (19.52) are

$$\displaystyle \begin{aligned} \begin{array}{l} \vec{u}(0,t) = \vec{f} (t-\ell) + \vec{a} (t), \\[2mm] \vec{u}^{\prime}_x (0,t) = \vec{f}'(t-\ell) - \vec{a}' (t). \end{array} \end{aligned} $$

(19.54)

Substitution into vertex conditions (19.53) yields

$$\displaystyle \begin{aligned} \left\{ \begin{array}{l} \displaystyle P_{-1} \left( \vec{f}(t-\ell) + \vec{a} (t) \right) = 0, \\[2mm] \displaystyle ({\mathbf{I}}-P_{-1} ) \left( \vec{f}'(t-\ell) - \vec{a}' (t) \right) = A ({\mathbf{I}}-P_{-1}) \left( \vec{f} (t-\ell) + \vec{a} (t) \right). \end{array} \right. \end{aligned} $$

(19.55)

These equations can easily be solved. The only difficulty is that the signs in front of $a'$ on different sides of the second equation are different. Here are the explicit solutions:

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle P_{-1} \vec{a} (t) = - P_{-1} \vec{f} (t-\ell), \\[3mm] \displaystyle ({\mathbf{I}}-P_{-1}) \vec{a} (t) = ({\mathbf{I}}-P_1) \vec{f} (t-\ell) - 2 A e^{-At} \int_{\ell}^t e^{A\tau} ({\mathbf{I}}-P_{-1}) \vec{f} (\tau - \ell) d\tau, \end{array} \end{aligned}$$

(19.56)

where we used notation (3.31) for the high energy limit $S_v (\infty )$ of the vertex scattering matrix $S_{\mathbf {v}} (k)$ given by (3.20). The integral from $\ell$ to $t$ should be interpreted so that it is equal to zero whenever $t \leq \ell .$

Having calculated $\vec {a}$ we obtain the solution to the wave equation satisfying vertex conditions for $t < 2 \ell$

$$\displaystyle \begin{aligned} \begin{array}{ccl} {} \displaystyle \vec{u} (x,t) & = & \displaystyle \vec{f} (t+x-\ell) + S_{\mathbf{v}} (\infty) \vec{f} (t-x-\ell) \\ & & \displaystyle -2 A e^{-A(t-x-\ell)} \int_0^{t-x-\ell} e^{A\tau} ({\mathbf{I}}- P_{-1}) \vec{f} (\tau) d \tau.\\ \end{array} \end{aligned} $$

(19.57)

Note that we decided to change the argument in the convolution integral so that the solution $\vec {w} (x,t)$ is given as a linear combination of $\vec {f} (\tau ), \; 0 \leq \tau \leq t-x-\ell .$ Here $x + \ell$ is precisely the delay time needed for a wave to travel from the contact point ($\ell$) to the central vertex (point $0$) and back to point $x.$

It should be clear for the reader that our next step is to obtain an explicit formula for the solution of the wave equation for $t \in (2 \ell , 3 \ell ).$ The central vertex can be treated in the same way as before and we obtain precisely the same formula for the wave $\vec {a}.$ The only difference is that we have to take into account reflection of this wave from the contact set $x = \ell .$ Reflection due to the Dirichlet condition results in multiplication by $-1.$ It is easy to check that the following function satisfies not only the vertex conditions at the central vertex, but also boundary control for $t \in (2 \ell , 3 \ell )$:

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \vec{u} (x,t) & = & \displaystyle \vec{f} (t+x-\ell) \\[3mm] && \displaystyle + S_{\mathbf{v}}(\infty) \vec{f} (t-x-\ell)\\[3mm] &&- 2 A e^{-A(t-x-\ell)} \int_0^{t-x-\ell} e^{A\tau} ({\mathbf{I}}-P_{-1}) \vec{f} (\tau) d \tau\\[3mm] && \displaystyle - S_{\mathbf{v}}(\infty) \vec{f} (t+x-3\ell)\\[3mm] &&+ 2 A e^{-A(t+x-3\ell)} \int_0^{t+x-3\ell} e^{A\tau} ({\mathbf{I}}- P_{-1}) \vec{f} (\tau) d \tau. \\ \end{array} \end{aligned} $$

(19.58)

As before the formula can be used for any $t < 3 \ell$ since the second and the third terms vanish for $t < \ell$, while the fourth and fifth terms—for $t < 2 \ell .$

To check that this function satisfies the boundary control consider

$$\displaystyle \begin{aligned} \begin{array}{ccl} \vec{u} (\ell, t) & = & \vec{f} (t) + S_{\mathbf{v}} (\infty) \vec{f} (t-2 \ell) - 2A e^{-A(t-2 \ell)} \int_0^{t-2 \ell} e^{A\tau} ({\mathbf{I}}-P_{-1}) \vec{f}(\tau) d \tau \\[3mm] & & - S_{\mathbf{v}} (\infty) \vec{f} (t-2 \ell) + 2A e^{-A(t-2 \ell)} \int_0^{t-2 \ell} e^{A\tau} ({\mathbf{I}}-P_{-1}) \vec{f}(\tau) d \tau \\[3mm] & = & \vec{f}(t). \end{array} \end{aligned}$$

Checking the vertex condition at $x = 0$ one should just take into account that the last two terms in (19.58) vanish at $x = 0$ for $t < 3 \ell$:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \begin{array}{cl} & - S_{\mathbf{v}}(\infty) \vec{f} (t+x-3\ell) + 2 A e^{-A(t+x-3\ell)} \int_0^{t+x-3\ell}\\[3mm]& \quad \times e^{A\tau} ({\mathbf{I}}- P_{-1}) \vec{f} (\tau) d \tau \vert_{x= 0, t < 3 \ell } \\[3mm] = & - S_{\mathbf{v}}(\infty) \vec{f} (t-3\ell) + 2 A e^{-A(t-3\ell)} \int_0^{t-3\ell} e^{A\tau} ({\mathbf{I}}- P_{-1}) \vec{f} (\tau) d \tau \vert_{t < 3 \ell} \\[3mm] \equiv & 0, \end{array}\end{array} \end{aligned} $$

implying that all formulas are identical to just considered in the case $\ell < t < 2 \ell$.

It is clear that the process can be continued further adding more and more waves as we have already done for the standard Laplacian. For any finite $T$ the formula for the solution for $t < T$ will contain a finite number of terms. For the solution of the inverse problem it will be enough to have $T = 3 \ell .$ Let us analyse obtained solution (19.58) given by the five terms:

• The first term $\vec {f} (t+x-\ell )$ represents the waves initiated by the boundary control. One may think about it as a free wave, since in the case of infinite interval the solution is given just by this term.

• The second term $S_{\mathbf {v}}(\infty ) \vec {f} (t-x-\ell )$ represents the sum of the wave with precisely the same profiles as the coordinate functions in the initial wave $\vec {f} (t+x-\ell )$ multiplied by the entries of $S_{\mathbf {v}} (\infty )$, but going in the opposite direction. Its exact value at any point $x$ at the time $t$ is determined by the value of the control function at the moment $t-x-\ell .$

• The third term $-2 A e^{-A(t-x-\ell )} \int _0^{t-x-\ell } e^{A\tau } ({\mathbf {I}}-P_{-1}) \vec {f} (\tau ) d \tau$ is also a wave traveling in the outward direction, but its amplitude at the point $x$ and time $t$ is determined by the values of the control function in the whole interval $[0, t-x-\ell ].$ The integral kernel is a smooth function.

• The last two terms represent the outgoing wave that has been reflected by the contact set and transformed into an incoming wave with the opposite amplitude.

The dynamical response operator for $T \in (0, 3 \ell )$ is given by

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \left( {\mathbf{R}}^T \vec{f} \right) (t) & = & \displaystyle \partial_n \vec{u} (\ell, t) = - \frac{\partial}{\partial x} \vec{w} (\ell, t) \\[2mm] & = & \displaystyle - \frac{d}{dt} \vec{f} (t) + 2 S_{\mathbf{v}}(\infty) \frac{d}{dt} \vec{f} (t-2\ell) - 4 A ({\mathbf{I}}-P_{-1}) \vec{f} (t-2\ell)) \\[3mm] && \displaystyle + 4 A^2 e^{-A(t-2 \ell)} \int_{-\infty}^{t-2\ell} e^{A \tau} ({\mathbf{I}}-P_{-1}) \vec{f} (\tau) d \tau \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} & = & \displaystyle - \vec{f}'(t) + 2 S_v (\infty) \vec{f}'(t-2 \ell) - 4 A ({\mathbf{I}}-P_{-1}) \vec{f} (t-2\ell)) \\[3mm] && \displaystyle + 4 A^2 e^{-A(t-2 \ell)} \int_{-\infty}^{t-2\ell} e^{A \tau} ({\mathbf{I}}-P_{-1}) \vec{f} (\tau) d \tau, \end{array} \end{aligned} $$

(19.59)

where the third term appears as the result of the differentiation of the integral. The generalised kernel $r (t-\tau )$ of the response operator is

$$\displaystyle \begin{aligned} {} r (t) &= - \delta'(t) + 2 S_{\mathbf{v}}(\infty) \delta'(t-2 \ell) - 4 A ({\mathbf{I}}-P_{-1}) \delta (t-2 \ell) \\ & \quad + 4A^2 e^{-A(t-2 \ell)} ({\mathbf{I}}-P_{-1}) \theta (t-2 \ell), \end{aligned} $$

(19.60)

where $\theta$ is the Heaviside function. The kernel $4A^2 e^{-A(t-2 \ell )} ({\mathbf {I}}-P_{-1}) \theta (t-2 \ell )$ is locally $L_2$, therefore we have the following classification of the singularities in the kernel:

• the initial wave $\vec {f} (t+x-\ell )$ determines the $\delta '$ singularity $- \delta '(t)$;

• the reflected wave $S_{\mathbf {v}} (\infty ) \vec {f} (t-x-\ell )$ determines the delayed $\delta '$ and $\delta$ singularities $2 S_{\mathbf {v}}(\infty ) \delta '(t-2 \ell )$ and $-4 A ({\mathbf {I}}-P_{-1}) \delta (t-2 \ell )$.

In addition there is an integral operator with the bounded kernel $4A^2 e^{-A(t-2 \ell )} ({\mathbf {I}}-P_{-1}) \theta (t-2 \ell ).$

These properties of the dynamical response operator will be very important for our future analysis, especially for the solution of the inverse problem. Let us remember that the kernel of the dynamical response operator contains the delayed $\delta '$ and $\delta$ singularities, of course provided $P_{-1} \neq {\mathbf {I}}.$ The delay is equal to the time needed for the wave to travel from the contact vertex to the central vertex and back. It follows that in general examining the response operator we may determine the distance to the nearest vertex, since it is not really important that the graph we consider is a star graph. The only possible obstacle is that the reflection coefficient from the nearest vertex could be identically zero.