**Theorem 2.**

*(a) If <sup>f</sup>* ∈ *<sup>L</sup>*2(0, *<sup>T</sup>*)*, then there exists a unique solution <sup>u</sup>*(*x*, *<sup>t</sup>*) *solving the system* (10)*–*(15)*, and mapping <sup>t</sup>* → *<sup>u</sup>f*(*x*, *<sup>t</sup>*) *is in C*(0, *<sup>T</sup>*; H1) ∩ *<sup>C</sup>*1(0, *<sup>T</sup>*; H)*.*

*(b) For each j* = 1, ..., *N, the mapping f* → *gj is a continuous mapping L*2(0, *<sup>T</sup>*) → F1*.*

**Proof.** On [0, *<sup>j</sup>*] with *<sup>j</sup>* <sup>≥</sup> 2, the wave will be generated by the "control" *<sup>∂</sup>*(*u<sup>f</sup> <sup>j</sup>* )(0, *t*), whereas on [0, -<sup>1</sup>] the wave is generated by the two controls *<sup>∂</sup>*(*u<sup>f</sup>* <sup>1</sup> )(0, *<sup>t</sup>*), *<sup>∂</sup>*(*u<sup>f</sup>* <sup>1</sup> )(-1, *t*) = *f*(*t*). We assume first that *<sup>f</sup>* ∈ *<sup>C</sup>*<sup>2</sup> <sup>0</sup> (0, *T*).

Let

$$p(t) := (u^f\_j)\_x(0, t), \text{ and } P(t) = -\int\_0^t p(s)ds. \tag{25}$$

Here, *p* is independent of *j* by (12). We have that *u<sup>f</sup>* is given by

$$
\mu\_1^f = \mathfrak{u}\_1^p + v^f,\text{ and for }j \ge 2,\ \mathfrak{u}\_j^f = \mathfrak{u}\_j^p. \tag{26}
$$

Note that *v<sup>f</sup>* has already been explicitly determined in (23). Thus, by (21), we have an explicit solution for *u<sup>f</sup>* if we can solve for *p*. We now prove the existence, uniqueness, and regularity of *P*.

By (21) and (26), we have for *j* ≥ 2,

$$\begin{aligned} g\_j(t) &= -P(t) + \int\_0^t w\_j(0,s)P(t-s)ds \\ &+ 2\sum\_{n\geq 1} (-1)^n \left( P(t-2n\ell\_j) + \int\_{2n\ell\_j}^t w\_j(2n\ell\_j,s)P(t-s)ds \right). \end{aligned} \tag{27}$$

For *j* = 1, we have by (21), (24), and (26) that

$$\begin{aligned} g\_1(t) &=& P(t) + \int\_0^t w\_1(0,s)P(t-s)ds \\ &+& 2\sum\_{n\geq 1} (-1)^n \left( P(t-2n\ell\_1) + \int\_{2n\ell\_1}^t w\_1(2n\ell\_1,s)P(t-s)ds \right) \\ &+& 2\sum\_{n\geq 1} \left( F(t-(2n-1)\ell\_1) + \int\_{(2n-1)\ell\_1}^t k\_1((2n-1)\ell\_1,s)F(t-s) \right). \end{aligned} \tag{28}$$

We remark that at the moment, we have not yet solved for either *P* or *gj* for any *j*. Let

$$\alpha = \min \{ \ell\_{j\prime} \mid j = 1, \dots, N \} \dots$$

We solve for *P* with an iterative argument using steps of length 2*α*. The iterations are necessary because the upper limits in the sums in (27), (28) increase with time due to reflections of the wave at the various vertices. In what follows, we label by *G*(*t*) various terms that we have already solved for, which by (24), includes *<sup>v</sup>f*(0, *<sup>t</sup>*). For *<sup>t</sup>* ≤ -<sup>1</sup> we have by unit wave speed that *P*(*t*) = 0. Suppose now *t* ∈ [-1, -<sup>1</sup> + 2*α*]. Then,

$$t - 2\ell\_j \le \ell\_1 + 2\alpha - 2\ell\_j < \ell\_{1\prime}$$

and hence *P*(*t* − *s*) = 0 for *s* ≥ 2*n<sup>j</sup>*, for all *j* with *n* ≥ 1. By (13), we have

$$\sum\_{1}^{N} \mathcal{g}\_{\bar{j}}(t) = 0\_{\prime}$$

and hence from (27) and (28) we get

$$NP(t) + \int\_{0}^{t} (\sum\_{j=1}^{N} w\_{j}(0, s)) P(t - s) ds = G(t), \ t \in [\ell\_1, \ell\_1 + 2a]. \tag{29}$$

It is easy to show that this is a Volterra equation of the second kind (VESK), and so admits a unique solution *P* with *PL*2(-1,-<sup>1</sup>+2*α*) ≤ *FL*2(0,2*α*). Furthermore, by differentiating this equation we get *pL*2(-1,-<sup>1</sup>+2*α*) ≤ *f L*2(0,2*α*).

Having solved for *P* on [0, -<sup>1</sup> + 2*α*], we now suppose *t* ∈ [-<sup>1</sup> + 2*α*, -<sup>1</sup> + 4*α*]. Thus for any *j* and any *n* ≥ 1, we have *t* − 2*n<sup>j</sup>* ≤ -<sup>1</sup> + 2*α*, so all terms in (27), (28) involving *P*(*t* − *s*), with *s* ≤ 2*n<sup>j</sup>* and *<sup>n</sup>* <sup>≥</sup> 1 , are known. Thus by (27), (28), and <sup>∑</sup>*<sup>N</sup>* <sup>1</sup> *gj*(*t*) = 0,

$$NP(t) + \int\_0^t (\sum\_{j=1}^N w\_j(0, s)) P(t - s) ds = G(T), \; t \in [\ell\_1 + 2\alpha, \ell\_1 + 4\alpha].$$

We can solve this VESK to determine uniquely *P*(*t*) for *t* ∈ [-<sup>1</sup> + 2*α*, -<sup>1</sup> + 4*α*], with the estimate *pL*2(-1,-<sup>1</sup>+4*α*) ≤ *f L*2(0,4*α*) holding. Iterating this process, we solve for the unique *P*(*t*) for *t* ∈ [0, *T*] as desired. The case for *<sup>f</sup>* ∈ *<sup>L</sup>*2(0, *<sup>T</sup>*) is then obtained by continuity. Part (a) of the Theorem follows easily from (21),(23), and (26). Part (b) of the theorem follows from Part (a) and (27) and (28).

Define

$$
\mathcal{R}\_1 f = u\_1^f(\ell\_{1\nu} t).
$$

**Proposition 1.** *For R*<sup>1</sup> *one can determine q*1*,* -<sup>1</sup>*, and N.*

**Proof.** Let *f*(*t*) = *δ*(*t*), so *F*(*t*) = −*H*(*t*). From (24) and (29) one has, for *t* < 3-1,

$$NP(t) + \int\_0^t (\sum\_{j=1}^N w\_j(0, s)) P(t - s) ds = 2H(t - \ell\_1) + 2 \int\_{\ell\_1}^t k\_1(\ell\_1, s) ds.$$

Thus, we have *P*(*t*) = <sup>2</sup> *<sup>N</sup> H*(*t* − -<sup>1</sup>) + *cont*, where *cont* denotes various continuous functions. We have by (22)

$$\begin{aligned} u\_1^f(\ell\_1, t) &= \quad v^f(\ell\_1, t) + \overline{u}(\ell\_1, t) \\ &= \quad -H(t) - \int\_0^t k\_1(0, s) ds + \frac{2}{N} H(t - 2\ell\_1) + const. \end{aligned}$$

Clearly, the discontinuity at *t* = 2-<sup>1</sup> gives us -<sup>1</sup> and *<sup>N</sup>*. That *<sup>R</sup>*2-1 0,1 determines *q*<sup>1</sup> is proven in [16].

Define the "reduced response operator" on *ej*, with *j* ≥ 2, by

$$\left(\mathring{\mathcal{R}}\_{0,j}p\right)(t) = \mathring{u}\_j^p(0,t)$$

associated to the IBVP (17)–(20). From (21), we immediately obtain

**Lemma 1.** *For j* <sup>=</sup> 2, ..., *N, and any h* <sup>∈</sup> *<sup>C</sup>*<sup>∞</sup> <sup>0</sup> (R+)*, we have*

$$\left(\mathcal{R}\_{0,j}h\right)(t) = \int\_0^t \mathcal{R}\_{0,j}(s)h(t-s)ds\_\prime$$

*with*

$$\mathcal{R}\_{0,\vec{j}}(\mathbf{s}) = -1 - 2\sum\_{n\geq 1} (-1)^n H(\mathbf{s} - 2n\ell\_{\vec{j}}) - \vec{r}\_{0\mathbf{j}}(\mathbf{s}),\tag{30}$$

*with r*˜0*j*(0) = 0. *If T is finite, the sums above are finite.*

**Proof.** Using (25), it is easy to see that

$$P(t - 2n\ell\_j) = \int\_0^t H(s - 2n\ell\_j) p(t - s) ds.$$

The lemma now follows easily from (21).

In what follows, we refer to *R*˜ 0,*j*(*s*) as the "reduced response function". For *f*(*t*) = *δ*(*t*), we denote the solution to the system (10)–(15) as *uδ*. We also use the following.

**Lemma 2.** *Let p*(*t*)=(*u<sup>δ</sup> <sup>j</sup>*)*x*(0, *t*)*. For j* = 2, ..., *N, we have*

$$p(t) = \sum\_{m \ge 1} \psi\_m \delta(t - \zeta\_m) + \theta\_m H(t - \zeta\_m) + a(t). \tag{31}$$

*Here a* ∈ F1*, and a*(*s*) = <sup>0</sup> *for s* < *<sup>ζ</sup>*1*, <sup>ζ</sup>*<sup>1</sup> = *<sup>β</sup>*<sup>1</sup> = -<sup>1</sup> *and ψ*<sup>1</sup> = 0*.*

This result holds from the proof of Theorem 2, the unit speed of wave propagation, and the properties of wave reflections off *x* = *<sup>j</sup>*, see [33]. The details are left to the reader.

The following result follows from (22), (25), and Lemma 2. The details of the proof are left to the reader.

**Corollary 1.** *Let gj*(*t*) = *u<sup>δ</sup> <sup>j</sup>*(0, *t*)*. For j* = 2, ..., *N, we have*

$$\mathfrak{g}\_j(t) = \sum\_{k \ge 1} \phi\_k H(t - \gamma\_k) + A(t). \tag{32}$$

$$\text{Here } A \in \mathcal{F}^1 \text{, and } A(\mathbf{s}) = 0 \text{ for } \mathbf{s} \prec \gamma\_1, \gamma\_1 = \ell\_1.$$

#### *4.3. Solution of Inverse Problem*

Here, we establish some notation. We recall the following notation: for *vk* we list the incident edges by {*ek*,*<sup>j</sup>* : *j* = 1, ..., Υ*k*}. Here, *ek*,1 is chosen to be the edge lying on the path from *γ*<sup>0</sup> to *vk*, and the remaining edges are labeled randomly.

Now let *<sup>k</sup>*<sup>0</sup> be some fixed interior vertex, and let *<sup>j</sup>*<sup>0</sup> satisfy 1 <sup>&</sup>lt; *<sup>j</sup>*<sup>0</sup> <sup>≤</sup> <sup>Υ</sup>*k*<sup>0</sup> . Denote by <sup>Ω</sup>*j*<sup>0</sup> *<sup>k</sup>*<sup>0</sup> the unique subtree of <sup>Ω</sup> having *vk*<sup>0</sup> as root with incident edge *ek*0,*j*<sup>0</sup> , and by *<sup>V</sup>j*<sup>0</sup> *<sup>k</sup>*<sup>0</sup> the set of its vertices, see Figure 4.

**Figure 4.** (**a**) Ω, (**b**) Subtree Ω*j*<sup>0</sup> *k*0

We define an associated response operator as follows. Let Γ*j*<sup>0</sup> *<sup>k</sup>*<sup>0</sup> = {*vk*<sup>0</sup> , *<sup>γ</sup>N*<sup>0</sup> , ...., *<sup>γ</sup>N*} be the boundary vertices on Ω*j*<sup>0</sup> *k*0 . Suppose *ϕ* = *ϕ<sup>b</sup>* solves the IBVP

$$\begin{array}{rcl}\frac{\partial^2 \varrho}{\partial t^2} - \frac{\partial^2 \varrho}{\partial x^2} + q\varrho & = & 0, \text{ x} \in \Omega\_{k\_0}^{j\_0} \backslash V\_{k\_0 \prime}^{j\_0} \text{ } t \in \times [0, T], \end{array} \tag{33}$$

$$\left.\Psi\right|\_{t=0} = \left.\varphi\_t\right|\_{t=0} = \begin{array}{ccccc} & & & \\ & & & \\ \end{array} \tag{34}$$

$$
\partial \boldsymbol{\varrho}(\boldsymbol{\upsilon}\_{\mathbf{k}}, t) \quad = \quad \partial \boldsymbol{\varrho}\_{j}(\boldsymbol{\upsilon}\_{\mathbf{k}}, t), \; j \in J(\boldsymbol{\upsilon}\_{\mathbf{k}}), \; \boldsymbol{\upsilon}\_{\mathbf{k}} \in \boldsymbol{V}\_{k\_{0}}^{j\_{0}} \; \backslash \; \Gamma\_{k\_{0} \prime}^{j\_{0}} \; t \in [0, T], \tag{35}
$$

$$\sum\_{j \in f(v\_k)} \varrho\_j(v\_k, t) \quad = \quad 0, \ v\_k \in V\_{k\_0}^{j\_0} \backslash \Gamma\_{k\_0 \prime}^{j\_0}, \ t \in [0, T], \tag{36}$$

$$\partial \varphi(v\_{k\_0}, t) \quad = \quad b(t), \; t \in [0, T], \tag{37}$$

$$
\varphi(\gamma\_{l\prime}t) \quad = \quad 0, \; l = \mathcal{N}\_{0\prime}..., \mathcal{N}\_{\prime} \; t \in [0, T]. \tag{38}
$$

Then we define an associated reduced response operator

$$(\bar{R}\_{k\_0\dot{\rho}\_0}b)(t) = \varphi\_{\dot{\rho}\_0}^b(\upsilon\_{k\_0\prime}t)\_{\prime\prime}$$

with associated response function *<sup>R</sup>*˜ *<sup>k</sup>*0,*j*<sup>0</sup> (*s*).

Suppose we determined *<sup>R</sup>*˜ *<sup>k</sup>*0,*j*<sup>0</sup> . It would follow from Proposition <sup>1</sup> that one could recover the following data: *<sup>j</sup>*<sup>0</sup> , *qj*<sup>0</sup> , and <sup>Υ</sup>*k* , where *vk* is the vertex adjacent to *vk*<sup>0</sup> in <sup>Ω</sup>*j*<sup>0</sup> *k*0 . In this section we will present an iterative method to determine the operator *<sup>R</sup>*˜ *<sup>k</sup>*0,*j*<sup>0</sup> from the (|Γ| − <sup>1</sup>)-tuple of operators, *RT*, which we know by hypothesis for some *T* > 2-. An important ingredient is the following generalization to a tree of Corollary 1.

**Lemma 3.** *Let T* > 0*, and let R<sup>T</sup> <sup>k</sup>*,*<sup>j</sup> be associated with* (33)*–*(38)*, defined by* (7) *and* (8)*. The response function for R<sup>T</sup> <sup>k</sup>*,*<sup>j</sup> has the form*

$$R\_{k,j}(s) = r\_{k,j}(s) + \sum\_{n \ge 1} \phi\_n H(s - \gamma\_n).$$

*Here, rk*,*<sup>j</sup>* ∈ F1*, and the sequence* {*γn*} *is positive and strictly increasing. If <sup>T</sup> is finite then the sums are finite.*

**Proof.** The proof follows from the proof of Corollary 1, together with the transmission and reflection properties of waves at interior vertices, and reflection properties at boundary vertices.

Fix *T* > 2-. The rest of this section shows how to recover *<sup>R</sup>*˜ *<sup>k</sup>*0,*j*<sup>0</sup> from *<sup>R</sup>T*. **Step 1**

For the first step, let *vk*<sup>0</sup> be the vertex adjacent to the root *γ*0, with associated edge labeled *e*1. By Proposition 1, we can use *R<sup>T</sup>* 0,1 to recover Υ*k*<sup>0</sup> , -1, *q*1.
