# Determination of discontinuous diffusion coefficients for the heat equation on a tree-shaped network

Emmanuelle Crépeau, Lionel Rosier, Julie Valein

► To cite this version:

Emmanuelle Crépeau, Lionel Rosier, Julie Valein. Determination of discontinuous diffusion coefficients for the heat equation on a tree-shaped network. Inverse Problems, 2025, 41 (7), pp.075001. 10.1088/1361-6420/ade3f9 . hal-05464931

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Submitted on 19 Jan 2026

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# Determination of discontinuous diffusion coefficients for the heat equation on a tree-shaped network

Emmanuelle Crépeau,* Lionel Rosier † and Julie Valein ‡

May 21, 2025

## Abstract

In this paper, we study the heat equation on a tree-shaped network with a piecewise regular diffusion coefficient. By developing new Carleman estimates, we establish stability results for the identification of the diffusion coefficient. These stability estimates are derived using either internal measurements or boundary observations, offering robust insights into the inverse problem for this class of equations.

**keywords** network of partial differential equations, inverse problem, Carleman estimate, parabolic equation.

MSCcodes 35K05, 35R30, 93C20

## 1 Introduction

The purpose of this paper is to investigate certain inverse problems on a finite compact metric tree graph. In this context, each edge $e_j$ of the graph is identified with a finite interval $(0, l_j)$, and the graph contains no cycles. Although our results apply to general trees, we focus on a star-shaped network to simplify both the notation and the proofs. A star-shaped network is a tree graph in which all edges are connected to a single internal vertex. We refer to Section 4 for details concerning extension to tree-shaped networks.

The inverse problem under consideration involves determining a discontinuous diffusion coefficient $c = c(x)$ in the heat equation

$$y_t - (c(x)y_x)_x = 0$$

using either internal or boundary measurements of the solution $y(x, t)$. This problem has been extensively studied in the literature (see e.g. [13, 14, 17, 18, 27, 28, 30]). The novelty of our approach lies in extending the analysis to cases where the spatial domain is defined on a tree network. The motivation of this system comes from the diffusion of a potential $V(x, t)$ along a nerve's dentritic tree. In [29] Rall, considering tapering structure, has modelized the diffusion of the potential by a parabolic system of equations taking the following form,

$$CV_t - \frac{1}{2R_a\rho(x)}(R^2(x)V_x)_x + GV = 0$$

where $C, R_a, G$ represent the membrane capacitance, the axial resistivity, the membrane conductance respectively. Here $R(x)$ is the radius of the dentritic cable and $\rho(x) = R(x)\sqrt{1+(R'(x))^2}$. After a change of variables and a rescaling, the system can be rewritten as

$$y_t - (c(x)y_x)_x + py = 0.$$

<footnote>*Univ. Polytechnique Hauts-de-France, INSA Hauts-de-France, CERAMATHS - Laboratoire de Matériaux Céramiques et de Mathématiques, F-59313 Valenciennes, France. E-mail: emmanuelle.crepeau-jaisson@uphf.fr</footnote>

<footnote>†Université du Littoral Côte d'Opale, Laboratoire de Mathématiques Pures et Appliquées J. Liouville, B.P. 699, F-62228 Calais, France. E-mail: lionel.rosier@univ-littoral.fr</footnote>

<footnote>‡Université de Lorraine, CNRS, Inria, IECL, F-54000 Nancy, France. E-mail: julie.valein@univ-lorraine.fr</footnote>
The inverse problem of determining the potential p on a tree has already been studied, for exemple in [22] with Carleman estimates, in [2, 4] by using boundary control theory methods and in [16] for a tapered fiber. Our problem is to retrieve the radius fonction c(x).

More precisely our objective is to derive stability estimates that provide "sharp" results with respect to the choice of measurements,

More specifically, we consider a star-shaped network $\mathcal{R}$ consisting of $n+1$ edges $e_j$, each of length $l_j > 0$, $j \in [0, n] := \{k \in \mathbb{N}, 0 \le k \le n\}$, all connected at a single vertex which we designate as the origin of all edges. For any function $f: \mathcal{R} \to \mathbb{R}$ we set

$$f_j := f|_{e_j} \text{ the restriction of } f \text{ to the edge } e_j,$$

$$[f]_0 := \sum_{j=0}^{n} f_j(0) \text{ the transmission bracket at the vertex 0.}$$

For any given $T > 0$, we define $Q := \mathcal{R} \times (0, T)$. On this planar 1-D network, we consider the heat equation with a variable diffusion coefficient, given by the following system

$$\begin{cases} y_{j,t}(x,t) - (c_j(x)y_{j,x}(x,t))_x = g_j(x,t), & j \in [0, n], (x,t) \in (0, l_j) \times (0, T), \\ y_j(l_j, t) = h_j(t), & j \in [0, n], t \in (0, T), \\ y(x, 0) = y^0(x), & x \in \mathcal{R}, \end{cases} \quad (1)$$

together with the continuity condition and Kirchhoff law at the vertex 0, namely

$$y_j(0, t) = y_i(0, t) := y(0, t), \quad \forall i, j \in [0, n], t \in (0, T), \quad (2)$$

$$[cy_x(t)]_0 := \sum_{j=0}^{n} c_j(0)y_{j,x}(0, t) = 0, \quad t \in (0, T). \quad (3)$$

The diffusion coefficient c is assumed to be piecewise $C^1$, i.e. $c_i \in C^1([0, l_i])$, $i \in [0, n]$, and

$$0 < c_{\min} \le c(x) \le c_{\max}, \quad x \in \mathcal{R}. \quad (4)$$

We are studying the two following inverse problems:

* Is it possible to retrieve the diffusion coefficient, $c(x)$, for $x \in \mathcal{R}$ from internal measurements of the solution on a subdomain on all but one edges of the network and a measurement of the solution at a time $T'$?

* Is it possible to retrieve the diffusion coefficient, $c(x)$, for $x \in \mathcal{R}$ from boundary measurements of the solution on all but one of the external edges of the tree and a measurement of the solution at a time $T'$?

In the sequel, we shall use the following notations: $\mathbb{R}_+ = (0, +\infty)$, $\mathbb{R}_- = (-\infty, 0)$,

$$L^2(\mathcal{R}) := \{f : \mathcal{R} \to \mathbb{R}, f_j \in L^2(0, l_j), \forall j \in [0, n]\},$$

$$H^1(\mathcal{R}) := \{f : \mathcal{R} \to \mathbb{R}, f_j \in H^1(0, l_j), f_j(0) = f_i(0), \forall i, j \in [0, n]\},$$

$$H_0^1(\mathcal{R}) := \{f : \mathcal{R} \to \mathbb{R}, f_j \in H^1(0, l_j), f_j(l_j) = 0, f_j(0) = f_i(0), \forall i, j \in [0, n]\}.$$

For the sake of shortness, for $f \in L^1(\mathcal{R}) := \{f : \mathcal{R} \to \mathbb{R}, f_j \in L^1(0, l_j), \forall j \in [0, n]\}$, we write

$$\int_{\mathcal{R}} f dx := \sum_{j=0}^{n} \int_{0}^{l_j} f_j(x) dx.$$

The norms of the Hilbert spaces $L^2(\mathcal{R})$ and $H_0^1(\mathcal{R})$ are respectively defined by

$$\|f\|_{L^2(\mathcal{R})}^2 := \int_{\mathcal{R}} |f|^2 dx, \quad \|f\|_{H_0^1(\mathcal{R})}^2 := \int_{\mathcal{R}} |f_x|^2 dx.$$

We also denote by $L$ the operator $L := \partial_t - \partial_x(c\partial_x)$ in $Q$ and $L^*$ its formal adjoint operator, namely $L^* := -\partial_t - \partial_x(c\partial_x)$ in $Q$.

Our first main result is concerned with stability estimates with internal measurements.
**Theorem 1.** Let $T > 0$, $T' = T/2$, $\tau \in (0, T/2)$ and $\omega \subset \mathbb{R}$ be an open set with $\omega \cap (0, l_j) \neq \emptyset$ for $j \in [1, n]$. Let $\tilde{c}$ be a piecewise $C^1$ function satisfying (4) and let $h_j = 0$ for $j \in [0, n]$. Then for any $y^0 \in L^2(\mathbb{R})$, there exists $g \in C_c^\infty(\omega \times (0, T))$ such that for any piecewise $C^1$ function $c$ satisfying (4),

$$c_i(0) = \tilde{c}_i(0) \quad \forall i \in [0, n], \quad \text{and} \quad \max_{j \in [0, n]} \|c_j\|_{W^{1, \infty}(0, l_j)} \le c_{\max}^{1, \infty},$$

if $y$ and $\tilde{y}$ denote the solutions of (1), (2), and (3) associated with $c$ and $\tilde{c}$ respectively, with $g$ and $\tilde{y}_0 = y_0$, then we have

$$\|c - \tilde{c}\|_{L^2(\mathbb{R})} \le C \left( \|y_t - \tilde{y}_t\|_{L^2(\omega \times (\tau, T))} + \|y_{tt} - \tilde{y}_{tt}\|_{L^2(\omega \times (\tau, T))} \right. \\ \left. + \|y_t(T') - \tilde{y}_t(T')\|_{L^2(\omega)} + \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})} + \|y(T') - \tilde{y}(T')\|_{H^2(\omega)} \right). \quad (5)$$

Our second main result is concerned with stability estimates derived from boundary measurements.

**Theorem 2.** Let $T > 0$, $T' = T/2$, $\tau \in (0, T/2)$ and $\tilde{c}$ be a piecewise $C^1$ function satisfying (4). Then for any $y^0 \in L^2(\mathbb{R})$, there exist some functions $h_j \in C_c^1(0, T)$, $j \in [0, n]$ with $h_0 \equiv 0$, such that for any $c$ piecewise $C^1$ function satisfying (4),

$$\begin{array}{ll} c_i(0) = \tilde{c}_i(0) & \forall i \in [0, n], \\ c_i(l_i) = \tilde{c}_i(l_i) & \forall i \in [1, n], \\ \max_{j \in [0, n]} \|c_j\|_{W^{1, \infty}(0, l_j)} \le c_{\max}^{1, \infty}, & \end{array}$$

if $y$ and $\tilde{y}$ denote the respective solutions of (1), (2), and (3) associated with $c$ and $\tilde{c}$ respectively, with $g \equiv 0$ and $\tilde{y}^0 = y^0$, we have

$$\|c - \tilde{c}\|_{L^2(\mathbb{R})} \le C \left( \sum_{j=1}^{n} \left[ \|y_{tx}(l_j) - \tilde{y}_{tx}(l_j)\|_{L^2(\tau, T)} + \|y_{ttx}(l_j) - \tilde{y}_{ttx}(l_j)\|_{L^2(\tau, T)} \right. \right. \\ \left. \left. + |y_{j,x}(l_j, T') - \tilde{y}_{j,x}(l_j, T')| \right] + \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})} \right).$$

**Remark 3.** 1. It is noteworthy that only internal or boundary measurements at $n$ exterior nodes for $n + 1$ edges of the network are required. In the case of a tree network, the only needed measurements are located on all but one of the external edges of the tree (see Section 4 for more details).

2. Since the heat equation is parabolic, it is well known that internal or boundary measurements alone are not sufficient to ensure stability, see e.g. [14, 30]. Therefore, an additional measurement of the full solution at some time $T' \in (0, T)$ is required. In this work, we arbitrarily choose $T' = T/2$.

Theorem 1 provides a stability estimate for the discontinuous diffusion coefficients using measurements of the solution's derivative on open sets of $n$ edges along with measurements of the solution over the entire network of $n + 1$ edges at time $T/2$. That theorem holds under the technical assumption that the diffusion coefficients are equal at the central node. Theorem 2 establishes a similar result, based on $n$ boundary measurements of the solution's derivative under the assumption of the equality of diffusion coefficients at the central node and at $n$ exterior nodes where the measurements are taken.

The key to obtain both theorems is a global Carleman estimate. In the context of parabolic equation with discontinuous diffusion coefficient, a Carleman estimate was shown in [19] to prove the exact controllability to trajectories and in [14] in the context of inverse problem, under a condition on the monotonicity of the coefficient. Without imposing such monotonicity, a stability result for identifying the diffusion coefficient in the one-dimensional case was achieved in [13] and
Carleman estimates for parabolic operators in higher dimension are given in [25]. Moreover, [30] established the Lipschitz stability of principal coefficients for a parabolic system, assuming partial knowledge of the coefficients on an open subset of the boundary. For coupled systems, [17] addressed the inverse problem of simultaneously identifying two discontinuous diffusion coefficients in a one-dimensional parabolic system using observations of only one component. Additionally, [18] provided stability results for reaction-diffusion equations, focusing on the simultaneous determination of the diffusion coefficient and the potential using internal observations.

To bypass additional assumptions on the discontinuous diffusion coefficients, we derive a Carleman estimate in one less derivative. This approach builds on prior work: [24] introduced Carleman estimates for hyperbolic operators in $H^{-1}$ spaces, yielding Hölder stability results for inverse problems involving coefficient determination in acoustic equations with a single measurement. This idea was further explored in [20] for the controllability of parabolic systems and in [28] for the stability of discontinuous diffusion coefficients in the heat equation.

Concerning the question of uniqueness and Lipschitz stability for an inverse problem on networks by using some Carleman estimates, the only works we know consist of determining a potential, namely a coefficient of zero-th order term, for the wave equation in [8, 9, 10], for the heat and Schrödinger equations in [22], and for parabolic equations on networks with loop in [1]. Other approaches on inverse problems on graphs have been used for wave equation. We can mention the works [5, 11, 12] where they use the boundary control method to recover the tree structure, [3, 6, 7] where they develop the leaf-peeling method, an edge pruning method, to recover the potential on a tree.

To the best of our knowledge, no stability results currently exist for the identification of discontinuous diffusion coefficients in the heat equation on tree-shaped networks.

Our contribution addresses this gap. Both theorems presented in this work rely on global classical Carleman estimate in $L^2(\mathbb{R})$ for a tree-shaped network. The primary challenge lies in designing appropriate Carleman weights to handle terms at the central node arising from integrations by parts. By leveraging this $L^2(\mathbb{R})$ Carleman estimate, we derive estimates with one fewer derivative in the source term through a null controllability problem. Subsequently, we establish stability results for stationary problems under specific assumptions. Our main theorems require repeated use of these refined Carleman estimates and stability results, necessitating the preliminary construction of a control ensuring the validity of assumptions about the stationary solutions.

The paper is organized as follows. In Section 2, we provide some Carleman estimates in $L^2(\mathbb{R})$ with internal measurements, and next we derive some Carleman estimates with one fewer derivative. Finally, Section 3 is dedicated to the establishment of the stability estimates.

# 2 Carleman estimates

In this section, we will establish three Carleman estimates for both operators $L$ and $L^*$ defined on a star-shaped network $\mathbb{R}$.

## 2.1 Carleman estimates in $L^2(\mathbb{R})$ with interior observations

Let $\lambda > 0$. We define the weight functions

$$ \varphi(x, t) = \frac{e^{\lambda\beta(x)}}{t(T-t)}, \quad \eta(x, t) = \frac{e^{\lambda\bar{\beta}} - e^{\lambda\beta(x)}}{t(T-t)}, \qquad (6) $$

for all $(x, t) \in \mathbb{R} \times (0, T)$, where $\beta$ and $\bar{\beta}$ are defined as in the following lemma.

**Lemma 4.** Let $\omega_i \subset (0, l_i)$, $i \in [1, n]$, be $n$ (nonempty) open sets and let $\omega := \cup_{i \in [1, n]} \omega_i$. Then there exist $(a_0, \dots, a_n, b_1, \dots, b_n, d) \in (\mathbb{R}_-)^{n+1} \times (\mathbb{R}_+)^{n+1}$ and $\bar{\beta} > 0$ such that, defining the
weight function $\beta$ on $\mathbb{R}$ as

$$ \beta_0(x) = a_0 x + d, \text{ for } x \in [0, l_0], \qquad (7) $$

$$ \beta_j(x) = a_j x^2 + b_j x + d, \text{ for } x \in [0, l_j], j \in [1, n], \qquad (8) $$

then for all $i, j \in [0, n]$ and all $x \in [0, l_j]$, we have

$$ \beta_i(0) = \beta_j(0), \qquad (9) $$

$$ 0 < \frac{\bar{\beta}}{2} < \beta_j(x) < \bar{\beta}, \qquad (10) $$

$$ \beta'_{j}(l_j) < 0, \qquad (11) $$

$$ \beta'(x) \neq 0 \text{ on } \mathbb{R}\setminus\omega, \qquad (12) $$

and the following $(n+1) \times (n+1)$ symmetric matrix

$$ A_{\beta} := \begin{pmatrix} \beta'_{0}(0) + \beta'_{1}(0) & \beta'_{0}(0) & \cdots & \beta'_{0}(0) & -\beta'_{0}(0)[c\beta']_{0} \\ \beta'_{0}(0) + \beta'_{2}(0) & \ddots & \vdots & \vdots & -\beta'_{0}(0)[c\beta']_{0} \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ \beta'_{0}(0) + \beta'_{n}(0) & \beta'_{0}(0) & \cdots & \beta'_{0}(0)[c\beta']_{0} & -\beta'_{0}(0)[c\beta']_{0}^{2} + [c^{2}(\beta')]_{0}^{3} \end{pmatrix} \qquad (13) $$

is positive definite, i.e.

$$ \exists \alpha > 0, \forall \xi \in \mathbb{R}^{n+1}, (A_{\beta}\xi, \xi) \geq \alpha |\xi|^2. \qquad (14) $$

*Proof.* First we fix $a_0 < 0$. The continuity condition (9) follows directly from the choice of the weights in (7)-(8). Second, the matrix $A_\beta$ is positive definite if and only if all of its leading principal minors are positive. Using the definition (7)-(8), we get $\beta'_0(0) = a_0$ and $\beta'_j(0) = b_j$, $j \in [1, n]$. We choose $b_j = b_1 > 0$ for all $j \ge 1$. Thus, the symmetric matrix $A_\beta$ becomes

$$ A_{\beta} := \begin{pmatrix} a_0 + b_1 & a_0 & \cdots & a_0 & -a_0 \left( a_0 c_0(0) + b_1 \sum_{j=1}^{n} c_j(0) \right) \\ a_0 + b_1 & \ddots & \vdots & \vdots & -a_0 \left( a_0 c_0(0) + b_1 \sum_{j=1}^{n} c_j(0) \right) \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ a_0 + b_1 & a_0 & \cdots & a_0 & -a_0 \left( a_0 c_0(0) + b_1 \sum_{j=1}^{n} c_j(0) \right) \\ a_0 \left( a_0 c_0(0) + b_1 \sum_{j=1}^{n} c_j(0) \right)^2 & a_0^3 c_0^2(0) + b_1^3 \sum_{j=1}^{n} c_j^2(0) & \cdots & a_0^3 c_0^3(0) + b_1^3 \sum_{j=1}^{n} c_j^3(0) & a_0^4 c_0^4(0) + b_1^4 \sum_{j=1}^{n} c_j^4(0) \end{pmatrix}, $$

and to get the positivity of leading principal minors, we actually need

$$ (ja_0 + b_1)b_1^{j-1} > 0, \quad \forall j \in [1, n] \qquad (15) $$

$$ \mathrm{Det}(A_{\beta}) > 0. \qquad (16) $$

On the one hand, choosing $b_1 > -na_0$ yields (15), since $b_1 > 0$. On the other hand, the function $\mathrm{Det}(A_\beta) = P(a_0, b_1, c_0(0), \dots, c_n(0))$ is a polynomial in $b_1$ of degree $n+3$. More precisely, the leading order term in $b_1$ of $\mathrm{Det}(A_\beta)$ comes from the product of the diagonal terms and is given by

$$ b_1^{n+3} \sum_{j=1}^{n} c_j^2(0) $$

and choosing $b_1$ sufficiently large (positive) gives $P(a_0, b_1, c_0(0), \dots, c_n(0)) > 0$, which gives (16).

Third, we choose $a_j < 0$ such that (12) holds. Indeed, for all $j \in [1, n]$, $\beta'_j(x) = 2a_jx + b_j = 2a_jx + b_1$, and if $\omega_j \subset (p_j, q_j) \subset (0, l_j)$ we want that $-b_1/(2a_j) \in (p_j, q_j)$ which is equivalent to $-\frac{b_1}{2p_j} < a_j < -\frac{b_1}{2q_j} < -\frac{b_1}{2l_j}$. This directly implies (11).
Finally, it remains to verify (10). We clearly have $a_0l_0 + d \le \beta_0(x) \le d$ and $a_jl_j^2 + b_1l_j + d \le \beta_j(x) \le d - \frac{b_1^2}{4a_j}$ for $j \in [1, n]$. Therefore, (10) is equivalent to

$$ egin{cases} d < \bar{\beta} \\ d - \frac{b_1^2}{4a_j} < \bar{\beta}, & \forall j \in [1, n] \\ \bar{\beta} < 2(a_0l_0 + d) \\ \bar{\beta} < 2d \\ \bar{\beta} < 2(a_jl_j^2 + b_1l_j + d), & \forall j \in [1, n]. \end{cases} $$

Since $a_j < 0$ for $j \in [0, n]$, the conditions reduce to

$$ egin{cases} 1 - \frac{b_1^2}{4a_j} < \bar{\beta}, & \forall j \in [1, n] \\ \frac{\bar{\beta}}{d} < 2\left(\frac{a_0l_0}{d} + 1\right) \\ \frac{\bar{\beta}}{d} < 2\left(\frac{a_jl_j^2 + b_1l_j}{d} + 1\right), & \forall j \in [1, n]. \end{cases} $$

It is then sufficient to take $d$ large enough so that

$$ \max_{j \in [1, n]} \left(1 - \frac{b_1^2}{4da_j}\right) < \min \left\{ 2\left(\frac{a_0l_0}{d} + 1\right), \min_{j \in [1, n]} 2\left(\frac{a_jl_j^2 + b_1l_j}{d} + 1\right) \right\}, $$

and next to pick any number

$$ \nu \in \left( \max_{j \in [1, n]} \left(1 - \frac{b_1^2}{4da_j}\right), \min \left\{ 2\left(\frac{a_0l_0}{d} + 1\right), \min_{j \in [1, n]} 2\left(\frac{a_jl_j^2 + b_1l_j}{d} + 1\right) \right\} \right), $$

and to set $\bar{\beta} = \nu d$. $\square$

**Theorem 5.** Let $\tilde{\omega}_i \subset (0, l_i)$, $i \in [1, n]$, be nonempty open sets and let $\tilde{\omega} := \cup_{i \in [1, n]} \tilde{\omega}_i$. Assume that $\varphi$ and $\eta$ are as in (6) with $\beta$ as given by Lemma 4. Then there exist $\lambda_0 > 0$, $s_0 > 0$ and a positive constant $C = C(T, \mathbb{R}, \tilde{\omega}, c_{\min}, c_{\max}, \beta)$ such that the following Carleman estimate holds

$$ s\lambda^2 \iint_Q \varphi|u_x|^2 e^{-2s\eta} dxdt + s^3\lambda^4 \iint_Q \varphi^3|u|^2 e^{-2s\eta} dxdt + s^3\lambda^3 \int_0^T |\varphi(0,t)|^3|u(0,t)|^2 e^{-2s\eta(0,t)} dt \\ \le C \left[ s^3\lambda^4 \int_0^T \int_{\tilde{\omega}} \varphi^3|u|^2 e^{-2s\eta} dxdt + \iint_Q |u_t \pm (cu_x)_x|^2 e^{-2s\eta} dxdt \right. \\ \left. + s\lambda \int_0^T \varphi(0,t) |k(t)|^2 e^{-2s\eta(0,t)} dt + \frac{1}{s^2} \int_0^T |k_t(t)|^2 e^{-2s\eta(0,t)} dt \right], \quad (17) $$

for $s \ge s_0$, $\lambda \ge \lambda_0$ and $u \in D(L) := \{u \in L^2(0, T; H_0^1(\mathbb{R})), Lu \in L^2(Q)\}$ if $\pm = -$, or $u \in D(L^*) := \{u \in L^2(0, T; H_0^1(\mathbb{R})), L^*u \in L^2(Q)\}$ if $\pm = +$, where $k \in H^1(0, T)$ is as

$$ [cy_x(t)]_0 := k(t). $$

*Proof.* We only treat the case of the operator $L^* = -\partial_t - \partial_x(c\partial_x)$, the other case $(\partial_t - \partial_x(c\partial_x))$ being obtained as a direct consequence by performing the change of variable $t \to T-t$. Let $s > 0$ and $\lambda > 0$, and set $w = e^{-s\eta}u$ and $Pw = -e^{-s\eta}L^*(e^{s\eta}w)$. After computing $Pw$, we split the terms as follows:

$$ Pw = P_1w + P_2w + Rw $$
where

$$P_1 w := (cw_x)_x + s^2 \lambda^2 \varphi^2(\beta')^2 cw + s \eta_t w, \quad (18)$$

$$P_2 w := w_t - 2s\lambda\varphi c\beta'w_x - 2s\lambda^2\varphi c(\beta')^2w, \quad (19)$$

$$Rw := -s\lambda\varphi(c\beta')_x w + s\lambda^2\varphi c(\beta')^2 w. \quad (20)$$

Therefore, $P_1w + P_2w = Pw - Rw$ and

$$\iint_Q |Pw - Rw|^2 dxdt = \iint_Q |P_1w|^2 dxdt + \iint_Q |P_2w|^2 dxdt \\ + 2 \iint_Q P_1wP_2w dxdt. \quad (21)$$

The main task in the proof will be to minimize the cross-terms in $P_1wP_2w$ by positive and dominant terms looking similar to the ones in the left-hand-side of (17) and negative boundary terms that will be moved to the right-hand-side of the estimate. For the sake of clarity, we divide the proof in several steps.

### Step 1. Main calculations

We set $\langle P_1w, P_2w \rangle_{L^2(Q)} = \sum_{i,k=1}^3 I_{i,k}$ where $I_{i,k}$ is the integral over $Q$ of the product of the ith-term in $P_1w$ and the kth-term in $P_2w$. Then after some integrations by parts, we obtain

$$I_{1,1} = \iint_Q (cw_x)_x w_t dxdt = - \int_0^T w_t(0,t)[cw_x]_0 dt,$$

since $w(x,0) = w(x,T) = 0$ and $w_{j,t}(l_j, t) = 0$;

$$I_{1,2} = s\lambda^2 \iint_Q |cw_x|^2(\beta')^2 \varphi dxdt + s\lambda \iint_Q |cw_x|^2 \beta'' \varphi dxdt \\ -s\lambda \sum_{j=0}^n \int_0^T |c_j(l_j)w_{j,x}(l_j, t)|^2 \varphi_j(l_j, t) \beta_j'(l_j) dt + s\lambda \int_0^T [|cw_x|^2 \beta']_0 \varphi(0,t) dt;$$

$$I_{1,3} = 2s\lambda^2 \iint_Q |cw_x|^2(\beta')^2 \varphi dxdt + 2s\lambda^3 \iint_Q c^2ww_x(\beta')^3 \varphi dxdt \\ +2s\lambda^2 \iint_Q cww_x\varphi(c(\beta')^2)_x dxdt + 2s\lambda^2 \int_0^T w(0,t) [c^2w_x(\beta')^2]_0 \varphi(0,t) dt;$$

$$I_{2,1} = -s^2 \lambda^2 \iint_Q c|w|^2(\beta')^2 \varphi \varphi_t dxdt;$$

$$I_{2,2} = 3s^3 \lambda^4 \iint_Q |cw|^2(\beta')^4 \varphi^3 dxdt + s^3 \lambda^3 \iint_Q |w|^2(c^2(\beta')^3)_x \varphi^3 dxdt \\ +s^3 \lambda^3 \int_0^T [|c|^2(\beta')^3]_0 w^2(0,t) \varphi^3(0,t) dt;$$

$$I_{2,3} = -2s^3 \lambda^4 \iint_Q |cw|^2 \varphi^3(\beta')^4 dxdt;$$

$$I_{3,1} = -\frac{s}{2} \iint_Q |w|^2 \eta_{tt} dxdt;$$

$$I_{3,2} = s^2 \lambda^2 \iint_Q c|w|^2(\beta')^2 \eta_t \varphi dxdt - s^2 \lambda^2 \iint_Q c|w|^2(\beta')^2 \varphi \varphi_t dxdt \\ +s^2 \lambda \iint_Q |w|^2(c\beta')_x \varphi \eta_t dxdt + s^2 \lambda \int_0^T [c\beta']_0 w^2(0,t) \varphi(0,t) \eta_t(0,t) dt;$$
$$I_{3,3} = -2s^2\lambda^2 \iint_Q c|w|^2\varphi(\beta')^2\eta_t dxdt.$$

Thus, we obtain

$$\langle P_1(w), P_2(w) \rangle_{L^2(Q)} = 3s\lambda^2 \iint_Q |cw_x|^2 \varphi(\beta')^2 dxdt + s^3\lambda^4 \iint_Q |cw|^2 \varphi^3(\beta')^4 dxdt \\ - s\lambda \sum_{j=0}^{n} \int_0^T |c_j(l_j)w_{j,x}(l_j, t)|^2 \varphi_j(l_j, t)\beta'_j(l_j) dt + X + B \quad (22)$$

where $X$ is the sum of the remaining interior terms and $B$ is the sum of the trace terms at the central node.

From now on, $M > 0$ will be a generic constant depending only on the network $\mathcal{R}$, the time $T$, $c_{\min}$ and $c_{\max}$ and on $\beta_0, ..., \beta_n$, but independent of $s$ and $\lambda$.

## Step 2. Boundary terms at the central node.

**Lemma 6.** Under the hypothesis of Lemma 4 for the weight function $\beta$, the sum of the trace terms at the central node 0, denoted $B$ in (22) satisfies

$$B \geq \frac{\alpha}{2} s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt - M s \lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt \\ - \frac{M}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt \quad (23)$$

for $s$ sufficiently large and any $\lambda \ge 1$.

*Proof.* Gathering all the terms at the central node 0, we get

$$B = B_{1,1} + B_{1,2} + B_{1,3} + B_{2,2} + B_{3,2}$$

where $B_{i,j}$ comes from $I_{i,j}$.

We denote by $B_{12,2}$ the sum of $B_{1,2}$ and $B_{2,2}$, i.e.

$$B_{12,2} = s\lambda \int_0^T [|cw_x|^2\beta']_0 \varphi(0,t) dt + s^3\lambda^3 [|c|^2(\beta')^3]_0 \int_0^T |w(0,t)|^2 \varphi^3(0,t) dt.$$

Since $w = e^{-s\eta}u$ gives $w_x = s\lambda\varphi\beta'w + e^{-s\eta}u_x$, then

$$[cw_x]_0 = \sum_{j=0}^{n} c_j(0)w_{j,x}(0,t) = s\lambda\varphi(0)w(0)[c\beta']_0 + e^{-s\eta(0)}k(t), \quad (24)$$

where we wrote $\varphi(0)$, $w(0)$ and $\eta(0)$ instead of $\varphi(0, t)$, $w(0, t)$ and $\eta(0, t)$ for the sake of shortness. We can write

$$c_0(0)w_{0,x}(0) = s\lambda\varphi(0)w(0)[c\beta']_0 + e^{-s\eta(0)}k(t) - \sum_{j=1}^{n} c_j(0)w_{j,x}(0).$$
Using this expression, we thus obtain

$$
\begin{align*}
[\beta' |cw_x|^2]_0 &= \beta'_0(0) \left( s\lambda\varphi(0)w(0)[c\beta']_0 + e^{-s\eta(0)}k(t) - \sum_{j=1}^n c_j(0)w_{j,x}(0) \right)^2 \\
&\quad + \sum_{j=1}^n \beta'_j(0) |c_j(0)w_{j,x}(0)|^2 \\
&= s^2\lambda^2\varphi(0)^2\beta'_0(0)w(0)^2([c\beta']_0)^2 \\
&\quad + \beta'_0(0)e^{-2s\eta(0)}|k(t)|^2 + \sum_{j=1}^n (\beta'_0(0) + \beta'_j(0)) |c_j(0)w_{j,x}(0)|^2 \\
&\quad + 2\beta'_0(0) \sum_{i,j=1 \dots n, i < j} c_i(0)c_j(0)w_{i,x}(0)w_{j,x}(0) \\
&\quad - 2s\lambda\beta'_0(0)w(0)\varphi(0)[c\beta']_0 \sum_{j=1}^n c_j(0)w_{j,x}(0) \\
&\quad + 2s\lambda\beta'_0(0)w(0)\varphi(0)[c\beta']_0 e^{-s\eta(0)}k(t) - 2\beta'_0(0) \sum_{j=1}^n e^{-s\eta(0)}k(t)c_j(0)w_{j,x}(0).
\end{align*}
$$

Therefore,

$$
\begin{align*}
B_{12,2} = s\lambda \int_0^T \varphi(0,t) (A_\beta W(t), W(t)) \, dt &+ s\lambda \int_0^T \varphi(0,t) \beta'_0(0) e^{-2s\eta(0,t)} |k(t)|^2 \, dt \\
&+ 2s^2\lambda^2 \int_0^T \beta'_0(0) w(0,t) \varphi^2(0,t) [c\beta']_0 e^{-s\eta(0,t)} k(t) \, dt \\
&- 2s\lambda \sum_{j=1}^n \int_0^T \beta'_0(0) \varphi(0,t) e^{-s\eta(0,t)} k(t) c_j(0) w_{j,x}(0,t) \, dt,
\end{align*}
$$

with $W(t) = ((c_j(0)w_{j,x}(0,t))_{j=1...n}, s\lambda\varphi(0,t)w(0,t))$ and $A_\beta$ defined by (13).
Using Lemma 4 (see (14)), we get

$$
\begin{align*}
B_{12,2} \ge {}& \alpha s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 \, dt + \alpha s \lambda \sum_{j=1}^n \int_0^T \varphi(0,t) |c_j(0)w_{j,x}(0,t)|^2 \, dt \\
& + s \lambda \int_0^T \varphi(0,t) \beta'_0(0) e^{-2s\eta(0,t)} |k(t)|^2 \, dt + 2s^2\lambda^2 \int_0^T \beta'_0(0) w(0,t) \varphi^2(0,t) [c\beta']_0 e^{-s\eta(0,t)} k(t) \, dt \\
& - 2s\lambda \sum_{j=1}^n \int_0^T \beta'_0(0) \varphi(0,t) e^{-s\eta(0,t)} k(t) c_j(0) w_{j,x}(0,t) \, dt.
\end{align*}
$$

By Young’s inequality,

$$
\left| s^2 \lambda^2 \int_0^T \beta_0'(0) w(0,t) \varphi^2(0,t) [c\beta']_0 e^{-s\eta(0,t)} k(t) \, dt \right| \\
\le \epsilon s^3 \lambda^3 \int_0^T |w(0,t)|^2 |\varphi(0,t)|^3 \, dt + \frac{M}{\epsilon} s \lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 \, dt
$$
and

$$
\begin{aligned}
& \left| s\lambda \sum_{j=1}^{n} \int_{0}^{T} \beta'_{0}(0) \varphi(0,t) e^{-s\eta(0,t)} k(t) c_{j}(0) w_{j,x}(0,t) dt \right| \\
& \leq \epsilon s\lambda \sum_{j=1}^{n} \int_{0}^{T} \varphi(0,t) |c_{j}(0) w_{j,x}(0,t)|^{2} dt + \frac{M}{\epsilon} s\lambda \int_{0}^{T} \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^{2} dt,
\end{aligned}
$$

for any $\epsilon > 0$.

Moreover, by (24) and the fact that $\varphi(x,0)w(x,0) = \varphi(x,T)w(x,T) = 0$, we have by integration by parts

$$
\begin{aligned}
|B_{1,1}| &= \left| s\lambda \int_0^T \varphi(0,t) w(0,t) w_t(0,t) [c\beta']_0 dt + \int_0^T e^{-s\eta(0,t)} w_t(0,t) k(t) dt \right| \\
&\leq \frac{s\lambda}{2} \left| \int_0^T \varphi_t(0,t) [c\beta']_0 w(0,t)^2 dt \right| \\
&\quad + \left| -\int_0^T w(0,t) e^{-s\eta(0,t)} k_t(t) dt + s \int_0^T w(0,t) k(t) \eta_t(0,t) e^{-s\eta(0,t)} dt \right|.
\end{aligned}
$$

Using Young's inequality and the facts that for $t \in (0, T)$,

$$ T^2 |\varphi(0, t)| \geq 1, \qquad (25) $$

$$ |\varphi_t(0, t)| \leq T |\varphi(0, t)|^2 \leq T^3 |\varphi(0, t)|^3, \qquad (26) $$

(since $\beta(0) \ge 0$) and

$$ |\eta_t(0,t)| \le \frac{T}{t^2(T-t)^2} e^{\lambda\bar{\beta}} \le \frac{T}{t^2(T-t)^2} e^{2\lambda\beta(0)} = T|\varphi(0,t)|^2 = T|\varphi(0,t)|^{3/2} |\varphi(0,t)|^{1/2} \quad (27) $$

(see (10)), we obtain

$$
|B_{1,1}| \le M \left( (s\lambda T^3 + sT^2 + s^2 T^6) \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt \right. \\
\left. \qquad + \frac{1}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt + s \int_0^T e^{-2s\eta(0,t)} |\varphi(0,t)| |k(t)|^2 dt \right).
$$

Using again (24), we have

$$
\begin{aligned}
B_{1,3} = 2s^2\lambda^3 & \int_0^T w(0,t)^2\varphi(0,t)^2 c_0(0)(\beta_0'(0))^2 [c\beta']_0 dt \\
& - 2s\lambda^2 \sum_{j=1}^n \int_0^T c_0(0)c_j(0)(\beta_0'(0))^2 w(0,t)\varphi(0,t)w_{j,x}(0,t) dt \\
& + 2s\lambda^2 \sum_{j=1}^n \int_0^T (\beta_j'(0))^2 c_j(0)^2 w_{j,x}(0,t)w(0,t)\varphi(0,t) dt \\
& + 2s\lambda^2 \int_0^T (\beta_0'(0))^2 c_0(0)w(0,t)\varphi(0,t)e^{-s\eta(0,t)}k(t) dt,
\end{aligned}
$$
and then, using Young's inequality and (25), we obtain

$$|B_{1,3}| \le M \left( s^2 \lambda^3 (T^2 + T^4) \int_0^T |w(0,t)|^2 |\varphi(0,t)|^3 dt + \lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt \right. \\ \left. + \lambda \sum_{j=1}^n \int_0^T |c_j(0)w_{j,x}(0,t)|^2 \varphi(0,t) dt \right).$$

Finally, by (27), we can prove that

$$|B_{3,2}| \le MTs^2\lambda \int_0^T |w(0,t)|^2 |\varphi(0,t)|^3 dt.$$

As $B = B_{12,2} + B_{1,1} + B_{1,3} + B_{3,2} \ge B_{12,2} - |B_{1,1}| - |B_{1,3}| - |B_{3,2}|$, we then obtain for $s \ge 1$, $\lambda \ge 1$ and $\epsilon > 0$,

$$B \ge ((\alpha - \epsilon)s^3\lambda^3 - M(s\lambda T^3 + sT^2 + s^2T^6 + s^2\lambda^3(T^2 + T^4) + Ts^2\lambda)) \int_0^T |w(0,t)|^2 |\varphi(0,t)|^3 dt \\ + (\alpha s\lambda - M\lambda - \epsilon s\lambda) \sum_{j=1}^n \int_0^T \varphi(0,t) |c_j(0)w_{j,x}(0,t)|^2 dt + s\lambda \int_0^T \varphi(0,t) \beta_0'(0) e^{-2s\eta(0,t)} |k(t)|^2 dt \\ - \left(\frac{M}{\epsilon}s\lambda + Ms + M\lambda\right) \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt - \frac{M}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt. \quad (28)$$

Consequently, by taking $s$ sufficiently large and $\epsilon$ small enough, we get (23). $\square$

### Step 3. Interior terms.

**Lemma 7.** The sum $X$ of the remaining interior terms of (22) satisfies

$$|X| \le M \left( \left( s^3 \lambda^3 + \frac{1}{\epsilon} s \lambda^4 \right) \iint_Q |w|^2 \varphi^3 dxdt + (s\lambda + \epsilon s\lambda^2) \iint_Q |w_x|^2 \varphi dxdt \right)$$

for some suitable constant $M = M(T) > 0$ and all $\epsilon > 0$, $\lambda \ge 1$ and $s \ge 1$.

*Proof.* Gathering all remaining interior terms of (22), we get

$$X = X_{1,2} + X_{1,3} + \tilde{X}_{2,1} + X_{2,2} + X_{3,1} + \tilde{X}_{3,2},$$

where

$$|X_{1,2}| = \left| s\lambda \iint_Q |cw_x|^2 \beta'' \varphi dxdt \right| \le Ms\lambda \iint_Q |w_x|^2 \varphi dxdt,$$

$$|\tilde{X}_{2,1}| = \left| -2s^2\lambda^2 \iint_Q c|w|^2(\beta')^2\varphi\varphi_t dxdt \right| \le MTs^2\lambda^2 \iint_Q |w|^2 |\varphi|^3 dxdt,$$

(by an estimation similar to (26)),

$$|X_{2,2}| = \left| s^3 \lambda^3 \iint_Q |w|^2 (c^2(\beta')^3)_x \varphi^3 dxdt \right| \le Ms^3 \lambda^3 \iint_Q |w|^2 |\varphi|^3 dxdt,$$

$$|X_{3,1}| = \left| \frac{s}{2} \iint_Q |w|^2 \eta_{tt} dxdt \right| \le MT^2 s \iint_Q |w|^2 |\varphi|^3 dxdt,$$
and

$$
\begin{aligned}
|\tilde{X}_{3,2}| &= |s^2 \lambda^2 \iint_Q c|w|^2 (\beta')^2 \varphi \eta_t \,dxdt + s^2 \lambda \iint_Q |w|^2 (c\beta')_x \varphi \eta_t \,dxdt| \\
&\le (MT s^2 \lambda^2 + MT s^2 \lambda) \iint_Q |w|^2 |\varphi|^3 \,dxdt
\end{aligned}
$$

since

$$
|\eta_{tt}(x, t)| \leq M T^2 |\varphi(x, t)|^3
$$

and

$$
|\eta_t(x,t)| \leq MT |\varphi(x,t)|^2. \tag{29}
$$

Moreover, using Young's inequality, the fact that

$$
T^2 |\varphi(x,t)| \ge 1 \quad \forall (x,t) \in \mathcal{R} \times (0,T), \qquad (30)
$$

and $\lambda > 1$, we obtain

$$
\begin{aligned}
|X_{1,3}| = & \left| 2s\lambda^3 \iint_Q c^2 w w_x (\beta')^3 \varphi \,dxdt + 2s\lambda^2 \iint_Q c w w_x \varphi (c(\beta')^2)_x \,dxdt \right| \\
& \leq \frac{MT^4}{\epsilon} s\lambda^4 \iint_Q |w|^2 |\varphi|^3 \,dxdt + M\epsilon s\lambda^2 \iint_Q |w_x|^2 \varphi \,dxdt
\end{aligned}
$$

for any $\epsilon > 0$, which concludes the proof of Lemma 7.

<figure><img src="image_2.png" /></figure>

Combining (21), (22), Lemma 6 and Lemma 7, we obtain

$$
\begin{aligned}
& \iint_Q |P_1 w|^2 dxdt + \iint_Q |P_2 w|^2 dxdt + 6s\lambda^2 \iint_Q |cw_x|^2 (\beta')^2 \varphi dxdt \\
& + 2s^3 \lambda^4 \iint_Q |cw|^2 \varphi^3 (\beta')^4 dxdt - 2s\lambda \sum_{j=0}^n \int_0^T \varphi_j(l_j, t) |c_j(l_j) w_{j,x}(l_j, t)|^2 \beta_j'(l_j) dt \\
& + \frac{\alpha}{2} s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt \le \iint_Q |Pw - Rw|^2 dxdt \\
& + M \left( \left(s^3 \lambda^3 + \frac{1}{\epsilon} s \lambda^4\right) \iint_Q |w|^2 \varphi^3 dxdt + (s\lambda + \epsilon s \lambda^2) \iint_Q |w_x|^2 \varphi dxdt \right) \\
& + Ms\lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt + \frac{M}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt.
\end{aligned}
\tag{31}
$$

Using (4), (11) and (12), we then obtain

$$
\begin{aligned}
& \iint_Q |P_1 w|^2 dxdt + \iint_Q |P_2 w|^2 dxdt + s\lambda^2 \int_0^T \int_{\mathbb{R}\setminus\omega} |w_x|^2 \varphi dxdt + s^3\lambda^4 \int_0^T \int_{\mathbb{R}\setminus\omega} |w|^2 \varphi^3 dxdt \\
& + s^3\lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt \\
\le & M \left( \iint_Q |Pw - Rw|^2 dxdt + \left(s^3\lambda^3 + \frac{1}{\epsilon}s\lambda^4\right) \iint_Q |w|^2 \varphi^3 dxdt + (s\lambda + \epsilon s\lambda^2) \iint_Q |w_x|^2 \varphi dxdt \right. \\
& \left. \qquad + s\lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt + \frac{1}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt \right).
\end{aligned}
$$
By (20), we have

$$ \begin{aligned} \iint_Q |Pw - Rw|^2 dxdt &\le 2 \iint_Q |Pw|^2 dxdt + 2 \iint_Q |Rw|^2 dxdt \\ &\le 2 \iint_Q |Pw|^2 dxdt + Ms^2 \lambda^4 \iint_Q |w|^2 \varphi^3 dxdt. \end{aligned} \quad (32) $$

Consequently, we have

$$ \begin{aligned} & \iint_Q |P_1 w|^2 dxdt + \iint_Q |P_2 w|^2 dxdt + s\lambda^2 \iint_Q |w_x|^2 \varphi dxdt + s^3 \lambda^4 \iint_Q |w|^2 \varphi^3 dxdt \\ & \qquad + s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt \\ &\le M \left( \iint_Q |Pw|^2 dxdt + \left(s^3 \lambda^3 + s^2 \lambda^4 + \frac{1}{\epsilon} s \lambda^4\right) \iint_Q |w|^2 \varphi^3 dxdt \right. \\ & \qquad \left. + (s\lambda + \epsilon s\lambda^2) \iint_Q |w_x|^2 \varphi dxdt + s\lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt + \frac{1}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt \right. \\ & \qquad \left. + s\lambda^2 \int_0^T \int_\omega |w_x|^2 \varphi dxdt + s^3 \lambda^4 \int_0^T \int_\omega |w|^2 \varphi^3 dxdt \right). \end{aligned} $$

Taking $\epsilon$ small enough, $\lambda$ sufficiently large and then $s$ sufficiently large (which depends to $\lambda$ and $\epsilon$), we arrive to

$$ \begin{aligned} & \iint_Q |P_1 w|^2 dxdt + \iint_Q |P_2 w|^2 dxdt + s\lambda^2 \iint_Q |w_x|^2 \varphi dxdt + s^3 \lambda^4 \iint_Q |w|^2 \varphi^3 dxdt \\ & \qquad + s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |w(0,t)|^2 dt \\ &\le M \left( \iint_Q |Pw|^2 dxdt + s\lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt + \frac{1}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt \right. \\ & \qquad \left. + s\lambda^2 \int_0^T \int_\omega |w_x|^2 \varphi dxdt + s^3 \lambda^4 \int_0^T \int_\omega |w|^2 \varphi^3 dxdt \right). \end{aligned} \quad (33) $$

### Step 4. Back to the variable u.

The facts that $u = e^{s\eta}w$ and $\eta_x = -\lambda\beta'\varphi$ give that

$$ \begin{aligned} & s\lambda^2 \varphi e^{-2s\eta} |u_x|^2 \le C \left( s\lambda^2 \varphi |w_x|^2 + s^3 \lambda^4 |\varphi|^3 |w|^2 \right) \\ & s\lambda^2 \varphi |w_x|^2 \le C \left( s\lambda^2 \varphi e^{-2s\eta} |u_x|^2 + s^3 \lambda^4 |\varphi|^3 e^{-2s\eta} |u|^2 \right). \end{aligned} \quad (34) $$

and
These estimates combined with (33) leads to

$$
\begin{align*}
& \iint_Q |P_1 w|^2 dxdt + \iint_Q |P_2 w|^2 dxdt + s\lambda^2 \iint_Q e^{-2s\eta} |u_x|^2 \varphi dxdt \\
& \qquad + s^3 \lambda^4 \iint_Q e^{-2s\eta} |u|^2 \varphi^3 dxdt + s^3 \lambda^3 \int_0^T |\varphi(0,t)|^3 |u(0,t)|^2 e^{-2s\eta(0,t)} dt \\
& \le M \left( \iint_Q e^{-2s\eta} |u_t + (cu_x)_x|^2 dxdt + s\lambda \int_0^T \varphi(0,t) e^{-2s\eta(0,t)} |k(t)|^2 dt \right. \\
& \qquad \left. + \frac{1}{s^2} \int_0^T e^{-2s\eta(0,t)} |k_t(t)|^2 dt + s^3 \lambda^4 \int_0^T \int_\omega e^{-2s\eta} |u|^2 \varphi^3 dxdt \right. \\
& \qquad \left. + s\lambda^2 \int_0^T \int_\omega e^{-2s\eta} |u_x|^2 \varphi dxdt \right).
\end{align*}
$$

To prove the Carleman estimate (17), it remains to absorb the last term in the previous inequality. To do that, following [19, Proof of (100)], we can show that, for $\omega \subset \tilde{\omega}$,

$$
s\lambda^2 \int_0^T \int_\omega \varphi |u_x|^2 e^{-2s\eta} dxdt \le M \left( s^3 \lambda^4 \int_0^T \int_{\tilde{\omega}} \varphi^3 |u|^2 e^{-2s\eta} dxdt + \iint_Q |u_t + (cu_x)_x|^2 e^{-2s\eta} dxdt \right),
$$

which leads to the Carleman estimate (17).

2.2 Carleman estimates with one less derivative, with interior obser-
vations

**Theorem 8.** Let us consider the following problem:

$$
\left\{
\begin{array}{ll}
q_t - (cq_x)_x = f_x, & \text{in } Q, \\
q_j(l_j, t) = 0, & \forall j \in [0, n], \forall t \in (0, T), \\
q_i(0, t) = q_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[cq_x(t)]_0 = 0, & \forall t \in (0, T), \\
q(x, 0) = q_0(x), & \forall x \in \mathbb{R}.
\end{array}
\right.
\qquad (35)
$$

Let $\omega_i \subset (0, l_i)$, $i \in [1, n]$, be n nonempty open sets, and let $\omega := \cup_{i \in [1, n]} \omega_i$. Assume that $\varphi$ and $\eta$ are as in (6), and that $\beta$ is as given by Lemma 4. Assume that $q_0 \in L^2(\mathbb{R})$, $f \in L^2(0, T, H^1(\mathbb{R}))$ with $f(0, t) = 0$ for a.e. $t \in (0, T)$. Let $s_0, \lambda_0$ be as in Theorem 5, pick $\lambda \ge \lambda_0$, then there exist $\bar{s} \ge s_0 > 0$ and a positive constant $C = C(T, \mathbb{R}, \omega, c_{min}, c_{max})$ such that the following Carleman estimate holds

$$
\iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \le C \left( \iint_Q s^2 \lambda^2 \varphi^2 e^{-2s\eta} |f|^2 dxdt + \int_0^T \int_\omega s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right), \quad (36)
$$

for $s \ge \bar{s}$ and $q$ satisfying (35).

The proof relies on the original proof of Imanuvilov and Yamamoto [23], see also [24], to
obtain a Carleman estimate for a parabolic operator with one less derivative in the source term.
For simplicity, we follow the proofs given in [20] and [28].

*Proof.* Using integrations by parts and the fact that $f(0,t) = 0$, we see that the function $q$ is solution of (35) in $L^2(Q)$ if and only if it satisfies for any $g \in L^2(Q)$

$$
\iint_Q qgdxdt = - \iint_Q f z_x dxdt + \int_{\mathbb{R}} q_0 z(0) dx, \quad (37)
$$
where $z \in D(L^*)$ is the solution of

$$
\begin{cases}
-z_t - (cz_x)_x = g, & \text{in } Q, \\
z_j(l_j, t) = 0, & \forall j \in [0, n], \\
z_i(0, t) = z_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[cz_x(t)]_0 = 0, & \forall t \in (0, T), \\
z(x, T) = 0, & \forall x \in \mathbb{R}.
\end{cases}
$$

Let $s$ and $\lambda$ be as in Theorem 5 and let us introduce the following fourth order problem with unknown

$$
p \in P_0 := \{u \in C^2(\bar{Q}); \ u_j(l_j, t) = 0, \ u_i(0, t) = u_j(0, t), \ \forall i, j \in [0, n], \ [cu_x(t)]_0 = 0 \ \forall t \in [0, T]\},
$$

$$
\left\{
\begin{array}{ll}
L^*(e^{-2s\eta}Lp) + s^3\lambda^4\varphi^3 e^{-2s\eta}p\mathbb{1}_{\omega} = s^3\lambda^4\varphi^3 e^{-2s\eta}q, & \text{in } Q, \\
(e^{-2s\eta}Lp)_j(l_j, t) = 0 & \forall j \in [0, n], \forall t \in (0, T), \\
(e^{-2s\eta}Lp)_i(0, t) = (e^{-2s\eta}Lp)_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[c(e^{-2s\eta}Lp)_x(t)]_0 = 0, & \forall t \in (0, T), \\
(e^{-2s\eta}Lp)(x, 0) = (e^{-2s\eta}Lp)(x, T) = 0, & \forall x \in \mathbb{R},
\end{array}
\right.
\quad (38)
$$

where $q \in L^2(Q)$ is solution of (35).

We define on $P_0^2$ the bilinear form $a$ and on $P_0$ the linear form $b_q$ as follows,

$$
\left\{
\begin{array}{ll}
a(p, p') & := \iint_Q e^{-2s\eta} Lp Lp' dxdt + \int_0^T \int_\omega s^3 \lambda^4 \varphi^3 e^{-2s\eta} pp' dxdt, \\
b_q(p') & := \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} qp' dxdt.
\end{array}
\right.
$$

It is clear that $a$ is a positive bilinear symmetric form on $P_0^2$. As $P_0 \subset D(L)$, we can apply the Carleman estimate (17) to $p$ to obtain

$$
\iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |p|^2 dxdt \leq Ca(p,p), \text{ for all } p \in P_0. \quad (39)
$$

Thus $(P_0, \|\cdot\|_a)$, where $\|\cdot\|_a := \sqrt{a(\cdot, \cdot)}$, is a normed space.

Let $P := \bar{P}_0^{\|\cdot\|_a}$ be the completion of $P_0$ for the norm $\|\cdot\|_a$. Then $P$ is a Hilbert space for the scalar product $a(\cdot, \cdot)$ and all functions in $P$ satisfy inequality (39). Furthermore, thanks to Cauchy-Schwarz inequality and (39) we obtain,

$$
|b_q(p)| \le C^{1/2} \left( \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} \|p\|_a. \quad (40)
$$

Thus $b_q$ is a continuous linear form on $P$.

Then we can apply the Lax-Milgram Theorem and the following variational problem possesses one and only one solution $p \in P$:

$$
a(p, p') = b_q(p'), \forall p' \in P. \tag{41}
$$

Using some test functions $p' \in P_0$, we obtain after integrations by parts that the solution $p$ also solves the first equation in (38) in the distributional sense.

Let us set $\hat{u} := -s^3\lambda^4\varphi^3 e^{-2s\eta}\mathbb{1}_{\omega}p$, $\hat{z} = e^{-2s\eta}Lp$. Using appropriate test functions in (41), we see that $\hat{z} \in D(L^*)$ and $\hat{z}$ is solution of the null controllability problem,

$$
\left\{
\begin{array}{ll}
-\hat{z}_t - (c\hat{z}_x)_x = \hat{u} + s^3\lambda^4\varphi^3 e^{-2s\eta}q, & \text{in } Q, \\
\hat{z}_j(l_j, t) = 0, & \forall j \in [0, n], \\
\hat{z}_i(0, t) = \hat{z}_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[c\hat{z}_x(t)]_0 = 0, & \forall t \in (0, T), \\
\hat{z}(x, T) = 0, \hat{z}(x, 0) = 0, & \forall x \in \mathbb{R},
\end{array}
\right.
$$
and using (37) applied to $\hat{z}$ we have,

$$ \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt = - \iint_Q f \hat{z}_x dxdt - \int_0^T \int_\omega q \hat{u} dxdt. \quad (42) $$

With (40) we have

$$ a(p, p) = b_q(p) \le C^{1/2} \left( \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} a(p, p)^{1/2}. $$

Thus

$$ a(p, p) = \iint_Q e^{2s\eta} |\hat{z}|^2 dxdt + \int_0^T \int_\omega s^{-3}\lambda^{-4}\varphi^{-3}e^{2s\eta}|\hat{u}|^2 dxdt \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt. \quad (43) $$

We will now give an estimate of the integral term $\iint_Q s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}c|\hat{z}_x|^2 dxdt$. Let $Q_n := \mathbb{R} \times (t_n, T_n)$ where the numbers $0 < t_n < T_n < T$, $n \in \mathbb{N}$, will be defined later. Let

$$ I_1 := \iint_{Q_n} s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}c|\hat{z}_x|^2 dxdt. $$

First, we make an integration by parts to obtain

$$ I_1 = \underbrace{\iint_{Q_n} s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}[-(c\hat{z}_x)_x]\hat{z} dxdt}_{I_2} + \underbrace{\iint_{Q_n} s^{-2}\lambda^{-2}(\varphi^{-2}e^{2s\eta})_x(-c\hat{z}_x)\hat{z} dxdt}_{I_3} $$

As $-(c\hat{z}_x)_x = \hat{u} + s^3\lambda^4\varphi^3e^{-2s\eta}q + \hat{z}_t$ we have

$$ I_2 = \underbrace{\iint_{Q_n} s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}\hat{z}\hat{u} dxdt}_{I_4} + \underbrace{\iint_{Q_n} s\lambda^2\varphi\hat{z}q dxdt}_{I_5} + \underbrace{\iint_{Q_n} s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}\hat{z}\hat{z}_t dxdt}_{I_6}. $$

First of all, we have $(\varphi^{-2}e^{2s\eta})_x = -2\lambda\varphi^{-2}(1+s\varphi)\beta'e^{2s\eta}$, thus for $s$ sufficiently large we have,

$$ |(\varphi^{-2}e^{2s\eta})_x| \le C\lambda s\varphi^{-1}e^{2s\eta}. $$

Thus we get by using (43)

$$ \begin{aligned} |I_3| & \le C \iint_{Q_n} s^{-1}\lambda^{-1}\varphi^{-1}e^{2s\eta}c|\hat{z}||\hat{z}_x|dxdt \\ & \le \frac{1}{2} \iint_{Q_n} s^{-2}\lambda^{-2}\varphi^{-2}e^{2s\eta}c|\hat{z}_x|^2dxdt + C \iint_{Q_n} e^{2s\eta}|\hat{z}|^2dxdt \\ & \le \frac{1}{2}I_1 + C \iint_Q s^3\lambda^4\varphi^3e^{-2s\eta}|q|^2dxdt. \end{aligned} \quad (44) $$

By using Minkowski estimate and (43), we obtain

$$ \begin{aligned} |I_4| & \le C \iint_{Q_n} e^{2s\eta}|\hat{z}|^2 dxdt + C \iint_{Q_n} s^{-4}\lambda^{-4}\varphi^{-4}e^{2s\eta}|\hat{u}|^2 dxdt \\ & \le C \iint_{Q_n} e^{2s\eta}|\hat{z}|^2 dxdt + C \int_0^T \int_\omega s^{-3}\lambda^{-4}\varphi^{-3}e^{2s\eta}|\hat{u}|^2 dxdt \\ & \le C \iint_Q s^3\lambda^4\varphi^3e^{-2s\eta}|q|^2dxdt. \end{aligned} \quad (45) $$
By using Minkowski estimate and (43), we obtain

$$ 
\begin{aligned}
|I_5| & \le C \iint_{Q_n} e^{2s\eta} |\hat{z}|^2 dxdt + C \iint_{Q_n} s^2 \lambda^4 \varphi^2 e^{-2s\eta} |q|^2 dxdt \\
& \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt.
\end{aligned} 
\quad (46) $$

First of all, we make an integration by parts in time:

$$ I_6 = -\frac{1}{2} \iint_{Q_n} s^{-2} \lambda^{-2} (\varphi^{-2} e^{2s\eta})_t |\hat{z}|^2 dxdt + \left[ \frac{1}{2} \int_{\mathbb{R}} s^{-2} \lambda^{-2} \varphi^{-2} e^{2s\eta} |\hat{z}|^2 dx \right]_{t_n}^{T_n} $$

But we have

$$ 
\begin{aligned}
|(\varphi^{-2} e^{2s\eta})_t| &= |(-2\varphi^{-3}\varphi_t + 2s\varphi^{-2}\eta_t)e^{2s\eta}| \\
&\le C\varphi^{-3}(|\eta_t| + |\varphi_t|)(1+s\varphi)e^{2s\eta} \le Cse^{2s\eta},
\end{aligned} 
$$

since $|\varphi_t| + |\eta_t| \le C_T \varphi^2$. Then, we get with (43),

$$ 
\begin{aligned}
\left| \frac{1}{2} \iint_{Q_n} s^{-2} \lambda^{-2} (\varphi^{-2} e^{2s\eta})_t |\hat{z}|^2 dxdt \right| & \le C \iint_Q s^{-1} \lambda^{-2} e^{2s\eta} |\hat{z}|^2 dxdt \\
& \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt.
\end{aligned} 
\quad (47) $$

With estimates (44), (45), (46) and (47), we infer that

$$ |I_1| \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt + \left| \left[ \frac{1}{2} \int_{\mathbb{R}} s^{-2} \lambda^{-2} \varphi^{-2} e^{2s\eta} |\hat{z}|^2 dx \right]_{t_n}^{T_n} \right| \quad (48) $$

To choose the times $t_n$ and $T_n$, we need the following lemma.

**Lemma 9.** Let $T > 0$, $r > 1$ and $\zeta : (0, T) \to \mathbb{R}$ be a measurable function with $\zeta \in L^1(0, T)$. Then there exists a decreasing sequence $(t_n)_{n \ge 0}$ in $(0, T)$ with $t_n \to 0$, such that $t_n^r |\zeta(t_n)| \to 0$.

*Proof.* To prove Lemma 9, we have to show the following statement:

$$ \forall C > 0, \forall \epsilon > 0, \exists t \in (0, \epsilon) \quad t^r |\zeta(t)| < C. $$

If the statement is false, we may pick some numbers $C > 0$ and $\epsilon \in (0, T)$ such that for all $t \in (0, \epsilon)$, we have $t^r |\zeta(t)| \ge C$. This yields $\int_0^\epsilon |\zeta(t)| dt \ge C \int_0^\epsilon t^{-r} dt = \infty$, contradicting the assumption $\zeta \in L^1(0, T)$. $\square$

We apply Lemma 9 twice to the function $\zeta(t) := \int_{\mathbb{R}} e^{2s\eta} |\hat{z}|^2 dx$ with $r = 2$. Note that $\zeta \in L^1(0, T)$ by (43). We may pick two sequences $(t_n)_{n \ge 0}$, $(T_n)_{n \ge 0}$, with $0 < t_n < T_n < T$, $t_n \searrow 0$, $T_n \nearrow T$ and

$$ t_n^2 |\zeta(t_n)| + (T - T_n)^2 |\zeta(T_n)| \to 0 \text{ as } n \to \infty. $$

Since $\varphi(x, t_n)^{-2} \le C t_n^2$ and $\varphi(x, T_n)^{-2} \le C(T - T_n)^2$, we infer that as $n \to +\infty$

$$ \left| \left[ \frac{1}{2} \int_{\mathbb{R}} s^{-2} \lambda^{-2} \varphi^{-2} e^{2s\eta} |\hat{z}|^2 dx \right]_{t_n}^{T_n} \right| \to 0. $$

Letting $n \to \infty$ in (48), we conclude that

$$ \iint_Q s^{-2} \lambda^{-2} \varphi^{-2} e^{2s\eta} c |\hat{z}_x|^2 dxdt \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt. \quad (49) $$
With (42), (43) and (49), we have

$$
\begin{align*}
\iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt &= - \iint_Q f \hat{z}_x dxdt - \int_0^T \int_\omega q \hat{u} dxdt \\
&\le \left( \iint_Q s^{-2} \lambda^{-2} \varphi^{-2} e^{2s\eta} |\hat{z}_x|^2 dxdt \right)^{1/2} \left( \iint_Q s^2 \lambda^2 \varphi^2 e^{-2s\eta} |f|^2 dxdt \right)^{1/2} \\
&\quad + \left( \int_0^T \int_\omega s^{-3} \lambda^{-4} \varphi^{-3} e^{2s\eta} |\hat{u}|^2 dxdt \right)^{1/2} \left( \int_0^T \int_\omega s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} \\
&\le C \left( \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} \left( \left( \iint_Q s^2 \lambda^2 \varphi^2 e^{-2s\eta} |f|^2 dxdt \right)^{1/2} \right. \\
&\qquad \left. + \left( \int_0^T \int_\omega s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} \right).
\end{align*}
$$

Thus we obtain the desired Carleman inequality (36).

<figure><img src="image_3.png" /></figure>

2.3 Carleman estimates with one less derivative, with boundary ob-
servations

By following the same lines as in [28], we shall obtain a new Carleman inequality for boundary
observations. We will use again the Carleman estimate with distributed observation in L²(ℝ).

**Theorem 10.** Let us consider the following problem:

$$
\left\{
\begin{array}{ll}
q_t - (cq_x)_x = f_x, & \text{in } Q, \\
q_j(l_j, t) = 0, & \forall j \in [0, n], \\
q_i(0, t) = q_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[cq_x(t)]_0 = 0, & \forall t \in (0, T), \\
q(x, 0) = q_0(x), & \forall x \in \mathbb{R}.
\end{array}
\right.
\qquad (50)
$$

Extend the network $\mathbb{R}$ in another network $\hat{\mathbb{R}}$ with the edges $[0, \hat{l}_j]$, $j \in [0, n]$, where $0 < l_j < \hat{l}_j$ for $j \in [1, n]$ and $\hat{l}_0 = l_0$, and fix an open subset $\omega \subset \hat{\mathbb{R}} \setminus \mathbb{R}$ intersecting each interval $(l_j, \hat{l}_j)$ for $1 \le j \le n$. Assume that $\varphi$ and $\eta$ are as in (6), and that $\beta$ is as given by Lemma 4 for $\hat{\mathbb{R}}$ and $\omega$. Assume that $q_0 \in L^2(\mathbb{R})$, $f \in L^2(0, T, H^1(\mathbb{R}))$ with $f(0, t) = 0$ for a.e. $t \in (0, T)$. Let $s_0, \lambda_0$ be as in Theorem 5. Pick $\lambda \ge \lambda_0$. Let $V_\gamma := \cup_{1 \le i \le n} (l_i - \epsilon, l_i)$ for some $\epsilon \in (0, \inf_{1 \le i \le n} l_i)$. Then there exist $\bar{s} \ge s_0 > 0$ and a positive constant $C = C(T, \mathbb{R}, c_{min}, c_{max})$ such that if $q \in L^2(Q) \cap L^2(0, T, H^2(V_\gamma))$ solves (50), then the following Carleman estimate holds for $s \ge \bar{s}$

$$
\iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \le C \left( \iint_Q s^2 \lambda^2 \varphi^2 e^{-2s\eta} |f|^2 dxdt \right. \\
\left. \qquad + \int_0^T \sum_{j=1}^n s \lambda \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |f_j(l_j, t)| + c_j(l_j) q_{j,x}(l_j, t) |^2 dt \right).
$$

*Proof.* We still follow the proof given in [28, A.2.3]. Let $\hat{Q} = (0, T) \times \hat{\mathbb{R}}$. We extend $q$ and $f$ on $\hat{Q}$ by setting $f=0$ and $q=0$ in $\hat{Q} \setminus Q$. We set $q_0(x) = 0$ for $x \in \hat{\mathbb{R}} \setminus \mathbb{R}$. We also extend $c$ smoothly such that $0 < c_{min} \le c(x) \le c_{max} < \infty$ in $\hat{\mathbb{R}}$. We are then in the same situation as in
the proof of Theorem 8 with $\mathbb{R}$ replaced by $\hat{\mathbb{R}}$. The function $q$ satisfies the system

$$ 
\left\{ 
\begin{array}{l}
\hat{L}q := q_t - (cq_x)_x = (f_x)\mathbb{1}_{\mathbb{R}} + \sum_{j=1}^{n}(c_jq_{j,x})(l_j)\delta_{l_j}, \quad \text{in } \hat{Q}, \\
q_j(\hat{l}_j, t) = 0, \quad \forall j \in [0, n], \\
q_i(0, t) = q_j(0, t), \quad \forall i, j \in [0, n], \forall t \in (0, T), \\
[cq_x(t)]_0 = 0, \quad \forall t \in (0, T), \\
q(x, 0) = q_0(x), \quad \forall x \in \hat{\mathbb{R}}.
\end{array}
\right. 
\quad (51) 
$$

Then (37) is replaced by

$$ \iint_{\hat{Q}} qgdxdt = - \iint_{\hat{Q}} fz_xdxdt + \sum_{j=1}^{n} \int_{0}^{T} (f_j(l_j, t) + c_j(l_j)q_{j,x}(l_j, t))z_j(l_j, t)dt + \int_{\hat{\mathbb{R}}} q_0z(0)dx \quad (52) $$

where $z \in D(\hat{L}^*)$ is a solution to

$$ 
\left\{ 
\begin{array}{ll}
-z_t - (cz_x)_x = g, & \text{in } \hat{Q}, \\
z_j(\hat{l}_j, t) = 0, & \forall j \in [0, n], \\
z_i(0, t) = z_j(0, t), & \forall i, j \in [0, n], \forall t \in (0, T), \\
[cz_x(t)]_0 = 0, & \forall t \in (0, T), \\
z(x, T) = 0, & \forall x \in \hat{\mathbb{R}}.
\end{array}
\right. 
$$

We modify the calculations in the proof of Theorem 8 by taking into account the boundary terms. Since $q=0$ and $f=0$ in $\hat{Q} \setminus Q$, then (42) is replaced by

$$ \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt = - \iint_Q f \hat{z}_x dxdt + \sum_{j=1}^{n} \int_0^T (f_j(l_j, t) + c_j(l_j)q_{j,x}(l_j, t)) \hat{z}_j(l_j, t) dt. \quad (53) $$

Then we get (43) and (49) with $\hat{Q}$ instead of $Q$. To obtain the desired Carleman inequality, we just need to prove that

$$ \left| \sum_{j=1}^{n} \int_{0}^{T} (f_j(l_j, t) + c_j(l_j)q_{j,x}(l_j, t)) \hat{z}_j(l_j, t) dt \right| \\ \le C \left( \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt \right)^{1/2} \left( \sum_{j=1}^{n} \int_{0}^{T} s\lambda \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} (f_j(l_j, t) + c_j(l_j)q_{j,x}(l_j, t))^2 dt \right)^{1/2} . \quad (54) $$

As for any $j \in [1, n]$ and for any $t \in (0, T)$, we have $(\varphi^{-1}e^{2s\eta}\hat{z}^2)(., t)_{|(0,l_j)} \in W^{1,1}(0, l_j)$, and as for any function $\mu \in W^{1,1}(0, l_j)$, $|\mu(l_j)| \le C(\int_0^{l_j} |\mu_x|dx + \int_0^{l_j} |\mu|dx)$, we have by using (43) and (49)

$$ 
\begin{aligned}
& \int_0^T \sum_{j=1}^n s^{-1} \lambda^{-1} \varphi_j^{-1}(l_j, t) e^{2s\eta_j(l_j, t)} \hat{z}_j^2(l_j, t) dt \\
& \le C \left( \iint_Q s^{-1} \lambda^{-1} \varphi^{-1} e^{2s\eta} \hat{z}^2 dxdt + \iint_Q s^{-1} \lambda^{-1} |(\varphi^{-1} e^{2s\eta} \hat{z}^2)_x| dxdt \right) \\
& \le C \iint_Q s^{-1} \lambda^{-1} \varphi^{-1} e^{2s\eta} |\hat{z}|^2 dxdt + Cs^{-1} \lambda^{-1} \iint_Q |(\varphi^{-1} e^{2s\eta})_x| \hat{z}^2 dxdt \\
& \qquad + Cs^{-1} \lambda^{-1} \iint_Q \varphi^{-1} e^{2s\eta} |\hat{z} \hat{z}_x| dxdt \\
& \le C \iint_Q e^{2s\eta} |\hat{z}|^2 dxdt + Cs^{-1} \lambda^{-1} \iint_Q \varphi^{-1} e^{2s\eta} |\hat{z} \hat{z}_x| dxdt \\
& \le C \iint_Q e^{2s\eta} |\hat{z}|^2 dxdt + Cs^{-2} \lambda^{-2} \iint_Q \varphi^{-2} e^{2s\eta} |\hat{z}_x|^2 dxdt \\
& \le C \iint_Q s^3 \lambda^4 \varphi^3 e^{-2s\eta} |q|^2 dxdt
\end{aligned}
$$

which gives (54) and completes the proof of the Carleman inequality. $\square$

<figure><img src="image_4.png" /></figure>
## 3 Stability results

Let $y$ and $\tilde{y}$ be the solutions of

$$ egin{cases} y_t - (cy_x)_x = g\mathbb{1}_\omega, & \forall (x,t) \in \mathbb{R} \times (0,T), \\ y_j(l_j, t) = h_j(t), & \forall j \in [0, n], t \in (0,T), \\ y_j(0, t) = y_i(0, t), & \forall i, j \in [0, n], t \in (0,T), \\ [cy_x(t)]_0 = 0, & \forall t \in (0,T), \\ y(x, 0) = y^0(x), & \forall x \in \mathbb{R}, \end{cases} \quad (55) $$

and

$$ egin{cases} \tilde{y}_t - (\tilde{c}\tilde{y}_x)_x = g\mathbb{1}_\omega, & \forall (x,t) \in \mathbb{R} \times (0,T), \\ \tilde{y}_j(l_j, t) = h_j(t), & \forall j \in [0, n], t \in (0,T), \\ \tilde{y}_j(0, t) = \tilde{y}_i(0, t), & \forall i, j \in [0, n], t \in (0,T), \\ [\tilde{c}\tilde{y}_x(t)]_0 = 0, & \forall t \in (0,T), \\ \tilde{y}(x, 0) = \tilde{y}^0(x), & \forall x \in \mathbb{R}, \end{cases} \quad (56) $$

respectively. Let

$$ \xi := c - \tilde{c}, \quad \text{and} \quad v := (y - \tilde{y})_t. $$

Then $v$ satisfies the following problem

$$ egin{cases} v_t - (cv_x)_x = (\xi \tilde{y}_{tx})_x, & \forall (x,t) \in \mathbb{R} \times (0,T), \\ v_j(l_j, t) = 0, & \forall j \in [0, n], t \in (0,T), \\ v_j(0, t) = v_i(0, t), & \forall i, j \in [0, n], 0 < t < T, \\ [cv_x(t)]_0 = -[\xi \tilde{y}_{tx}(t)]_0, & \forall t \in (0,T), \\ v(x, 0) = (cy_x^0)_x - (\tilde{c}\tilde{y}_x^0)_x, & \forall x \in \mathbb{R}. \end{cases} \quad (57) $$

### 3.1 Stability results with interior observations

In this subsection we derive stability theorem 1 with interior observations. Our method is based on ideas borrowed from [17, 28].

#### 3.1.1 Stability for a stationary problem

We first investigate the $L^2(\mathbb{R})$-stability of the solution $c(x)$ of the following stationary problem

$$ egin{cases} -(cu_x)_x = f, & \text{in } \mathbb{R}, \\ u_j(0) = u_i(0), & \forall i, j \in [0, n], \\ [cu_x]_0 = 0, & \end{cases} \quad (58) $$

where $u \in H^1(\mathbb{R})$ and $f \in L^2(\mathbb{R})$.

Let $\tilde{c}(x)$ solve

$$ egin{cases} -(\tilde{c}\tilde{u}_x)_x = \tilde{f}, & \text{in } \mathbb{R}, \\ \tilde{u}_j(0) = \tilde{u}_i(0), & \forall i, j \in [0, n], \\ [\tilde{c}\tilde{u}_x]_0 = 0, & \end{cases} \quad (59) $$

where $\tilde{u} \in H^1(\mathbb{R})$ and $\tilde{f} \in L^2(\mathbb{R})$, and let $\xi := c - \tilde{c}$.

For any function $q: \mathbb{R} \to \mathbb{R}$, we set $q_s(x) := q(x)e^{-s\eta(x, T')}$ where $\eta$ is as in (6) and $T' \in (0, T)$. At several places, to simplify the notation, we shall write $\eta(T')$ instead of $\eta(x, T')$. Moreover in this section $\eta$ (resp. $\varphi$) denotes $\eta(T')$ (resp. $\varphi(T')$).

We begin with the following stability result:

**Lemma 11.** Let $\delta > 0$ and $T' \in (0, T)$. Assume that the functions $c$ and $\tilde{c}$ are piecewise $C^1$, that they satisfy (4), and that they are such that

$$ (H_c) \quad c_i(0) = \tilde{c}_i(0) \quad \forall i \in [0, n]. $$
Let $\beta$ be as in Lemma 4. Assume further that

$$(H_1) \quad |\beta' \tilde{u}_x| \ge \delta > 0 \quad \text{in } \mathbb{R} \setminus \omega^0,$$

$$(H_2) \quad \forall j \in [0, n], u_j \in C^2([0, l_j]) \text{ and } \tilde{u}_j \in C^2([0, l_j]),$$

$$(H_3) \quad \|\tilde{u}\|_{L^\infty(\mathbb{R})} + \|\tilde{u}_x\|_{L^\infty(\mathbb{R})} \le \delta^{-1},$$

for some open set $\omega^0$ with $\omega^0 \subset \omega$.

Let $s_0, \lambda_0$ be associated with the set $\omega^0$ as in Theorem 5. Pick any $\lambda \ge \lambda_0$. Then there exists $s_1 \ge s_0$ such that for any $s \ge s_1$ we have the following estimate (with $\xi = c - \tilde{c}$)

$$s^2 \delta^2 \int_{\mathbb{R} \setminus \omega^0} |\xi|^2 e^{-2s\eta} dx \le C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx + C s \int_{\omega^0} |\xi|^2 e^{-2s\eta} dx + C(s, c_{max}) \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2. \quad (60)$$

*Proof.* By (58)-(59), we have

$$\begin{cases} -(c_s u_x)_x - s\eta_x c_s u_x = f_s \\ -(\tilde{c}_s \tilde{u}_x)_x - s\eta_x \tilde{c}_s \tilde{u}_x = \tilde{f}_s. \end{cases}$$

Thus

$$\|f_s - \tilde{f}_s\|_{L^2(\mathbb{R})}^2 = \int_{\mathbb{R}} [(c_s u_x - \tilde{c}_s \tilde{u}_x)_x + s\eta_x(c_s u_x - \tilde{c}_s \tilde{u}_x)]^2 dx.$$

As $(a+b)^2 \ge a^2 + 2ab$ we obtain,

$$\|f_s - \tilde{f}_s\|_{L^2(\mathbb{R})}^2 \ge s^2 \int_{\mathbb{R}} |\eta_x(c_s u_x - \tilde{c}_s \tilde{u}_x)|^2 dx + 2s \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)_x (\eta_x(c_s u_x - \tilde{c}_s \tilde{u}_x)) dx.$$

We estimate each term in the right hand side of the previous inequality. We first have a look at

$$B_1 := \int_{\mathbb{R}} |\eta_x(c_s u_x - \tilde{c}_s \tilde{u}_x)|^2 dx = \int_{\mathbb{R}} \lambda^2 \varphi^2(\beta')^2 (c_s u_x - \tilde{c}_s \tilde{u}_x)^2 dx.$$

As $c_s u_x - \tilde{c}_s \tilde{u}_x = \xi_s \tilde{u}_x + c_s(u_x - \tilde{u}_x)$ and $(a+b)^2 \ge \frac{a^2}{2} - b^2$, we have

$$\begin{aligned} B_1 &= \lambda^2 \int_{\mathbb{R}} \varphi^2(\beta')^2 (\xi_s \tilde{u}_x + c_s(u_x - \tilde{u}_x))^2 dx \\ &\ge \lambda^2 \int_{\mathbb{R}} \varphi^2(\beta')^2 \left( \frac{1}{2} \xi_s^2 \tilde{u}_x^2 - c_s^2 (u_x - \tilde{u}_x)^2 \right) dx. \end{aligned}$$

With Hypothesis (H$_1$), $|\beta_x \tilde{u}_x| \ge \delta$ on $\mathbb{R} \setminus \omega^0$, we infer that

$$\begin{aligned} B_1 &\ge \frac{\lambda^2 \delta^2}{2} \int_{\mathbb{R} \setminus \omega^0} \varphi^2 \xi_s^2 dx - C(\beta, \lambda, c_{max}) \int_{\mathbb{R}} \varphi^2 e^{-2s\eta} |u_x - \tilde{u}_x|^2 dx \\ &\ge \frac{\lambda^2 \delta^2 e^{\lambda \bar{\beta}}}{2(T'(T-T'))^2} \int_{\mathbb{R} \setminus \omega^0} \xi_s^2 dx - C(\beta, c_{max}, s, \lambda, T') \int_{\mathbb{R}} |u_x - \tilde{u}_x|^2 dx. \end{aligned} \quad (61)$$

On the other hand,

$$\begin{aligned} B_2 &:= \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)_x (\eta_x(c_s u_x - \tilde{c}_s \tilde{u}_x)) dx \\ &= \frac{1}{2} \sum_{j=0}^{n} [(c_{j,s} u_{j,x} - \tilde{c}_{j,s} \tilde{u}_{j,x})^2 \eta_{j,x}]_0^{l_j} - \frac{1}{2} \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)^2 \eta_{xx} dx \\ &= -\frac{\lambda}{2} \sum_{j=0}^{n} (c_{j,s} u_{j,x} - \tilde{c}_{j,s} \tilde{u}_{j,x})^2 (l_j) \beta'_j(l_j) \varphi_j(l_j) + \frac{\lambda}{2} \sum_{j=0}^{n} (c_{j,s} u_{j,x} - \tilde{c}_{j,s} \tilde{u}_{j,x})^2 (0) \beta'_j(0) \varphi(0) \\ &\quad - \frac{1}{2} \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)^2 \eta_{xx} dx. \end{aligned}$$
First of all, we have $\beta'_j(l_j) < 0$ for all $j \in [0, n]$, by (11). Thus the first term in the right hand side of the previous equation is nonnegative. We now prove that the second term in this expression is also nonnegative.

Let $B_3 := \sum_{j=0}^{n} (c_{j,s}u_{j,x} - \tilde{c}_{j,s}\tilde{u}_{j,x})^2(0)\beta'_j(0)\varphi(0)$. We have that $c_{j,s}(0) = \tilde{c}_{j,s}(0)$ for all $j \in [0, n]$ (see $(H_c)$), and hence $[c_s(u_x - \tilde{u}_x)]_0 = [c_s u_x - \tilde{c}_s \tilde{u}_x]_0 = 0$, so that

$$c_{0,s}(0)(u_{0,x} - \tilde{u}_{0,x})(0) = -\sum_{j=1}^{n} c_{j,s}(0)(u_{j,x} - \tilde{u}_{j,x})(0).$$

It follows that

$$
\begin{aligned}
B_3 &= \varphi(0) \left( \beta'_0(0) \left( \sum_{j=1}^{n} c_{j,s}(0)(u_{j,x} - \tilde{u}_{j,x})(0) \right)^2 + \sum_{j=1}^{n} (c_{j,s})^2(0)(u_{j,x} - \tilde{u}_{j,x})^2(0)\beta'_j(0) \right) \\
&= \varphi(0) \left( \sum_{j=1}^{n} (\beta'_0(0) + \beta'_j(0))(c_{j,s}(0)(u_{j,x} - \tilde{u}_{j,x})(0))^2 \right. \\
&\qquad \left. + 2\beta'_0(0) \sum_{\substack{i,j=1, i<j \\ i,j< n}}^{n} c_{i,s}(0)(u_{i,x} - \tilde{u}_{i,x})(0)c_{j,s}(0)(u_{j,x} - \tilde{u}_{j,x})(0) \right) \\
&= \varphi(0)(A_{\beta}W, W),
\end{aligned}
$$

with $A_{\beta}$ denoting the matrix defined in Lemma 4, and $W = ((c_{j,s}(0)(u_{j,x} - \tilde{u}_{j,x})(0))_{j=1...n}, 0)$. As the matrix $A_{\beta}$ is positive definite, we infer that $B_3 \ge 0$. It follows that, by $(H_2)$,

$$
\begin{aligned}
B_2 &\geq -\frac{1}{2} \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)^2 \eta_{xx} dx \\
&\geq -C(\lambda, \beta, T') \int_{\mathbb{R}} (c_s u_x - \tilde{c}_s \tilde{u}_x)^2 dx \\
&\geq -C \int_{\mathbb{R}} (c_s(u_x - \tilde{u}_x) + \xi_s \tilde{u}_x)^2 dx \\
&\geq -C \|\tilde{u}_x\|_{\infty}^2 \int_{\mathbb{R}} \xi_s^2 dx - C(s, c_{max}) \int_{\mathbb{R}} (u_x - \tilde{u}_x)^2 dx.
\end{aligned}
\qquad (62)
$$

With (61), (62) and $(H_3)$ we obtain

$$
\begin{aligned}
& \|f_s - \tilde{f}_s\|_{L^2(\mathbb{R})}^2 \\
& \geq s^2 \frac{\lambda^2 \delta^2 e^{\lambda \bar{\beta}}}{2(T'(T-T'))^2} \int_{\mathbb{R} \setminus \omega^0} |\xi_s|^2 dx - C(s, \beta, \lambda, T', c_{max}) \int_{\mathbb{R}} (u_x - \tilde{u}_x)^2 dx - \frac{Cs}{\delta^2} \int_{\mathbb{R}} \xi_s^2 dx.
\end{aligned}
\qquad (63)
$$

Thus we obtain (60) for $s$ large enough.

<figure><img src="image_5.png" /></figure>

The following lemma gives an $L^2$-estimate of $\xi$ on a neighborhood of $\omega^0$.

**Lemma 12.** Let $c, u, f, \tilde{c}, \tilde{u}, \tilde{f}$ be as in Lemma 11. Assume further that

$(H_4)$ There exist some open intervals $\omega_i^1 \subset (0, l_i)$, $i \in [1, n]$, such that:

1. $\omega^0 \subset \omega^1 := \cup_{i \in [1, n]} \omega_i^1 \subset \omega$,

2. $\tilde{u}_{|\partial\omega^1} = 0$,

3. $\tilde{u} \in C^2(\omega^1)$ and $|\tilde{u}_x|^2 - \tilde{u}\tilde{u}_{xx} \ge \delta > 0$ in $\omega^1$.
Assume in addition that $\|c\|_{W^{1,\infty}(\omega_1)} \le c_{max}^{1,\infty}$. Then there exist $C > 0$ and $C(c_{max}^{1,\infty}) > 0$ such that

$$ \int_{\omega^1} |\xi|^2 dx \le C \|f - \tilde{f}\|_{L^2(\omega^1)}^2 + C(c_{max}^{1,\infty}) \|u - \tilde{u}\|_{H^2(\omega^1)}^2. \quad (64) $$

*Proof.* In $\omega^1$, we have

$$ -(\xi \tilde{u}_x)_x = -(c \tilde{u}_x - c \tilde{u}_x)_x = f - \tilde{f} + (c(u - \tilde{u})_x)_x. \quad (65) $$

Thus

$$ \int_{\omega^1} -\xi \tilde{u} (\xi \tilde{u}_x)_x dx = \int_{\omega^1} (f - \tilde{f} + (c(u - \tilde{u})_x)_x) \xi \tilde{u} dx. \quad (66) $$

On the other hand, since $\tilde{u}_{|\partial\omega^1} = 0$, we obtain by an integration by parts that

$$ \int_{\omega^1} -\xi \tilde{u} (\xi \tilde{u}_x)_x dx = \int_{\omega^1} \xi^2 \left( (\tilde{u}_x)^2 - \frac{1}{4} (\tilde{u}^2)_{xx} \right) dx. $$

Thanks to $(H_4)$, 3., we have $(\tilde{u}_x)^2 - \frac{1}{4}(\tilde{u}^2)_{xx} \ge \frac{\delta}{2}$. This yields

$$ \int_{\omega^1} \xi^2 dx \le C(\delta) \int_{\omega^1} (f - \tilde{f} + (c(u - \tilde{u})_x)_x) \xi \tilde{u} dx, $$

and hence, using Cauchy-Schwarz inequality and $(H_3)$,

$$ \begin{aligned} \int_{\omega^1} \xi^2 dx &\le C(\delta) \int_{\omega^1} (f - \tilde{f} + (c(u - \tilde{u})_x)_x)^2 dx \|\tilde{u}\|_{L^\infty(\omega^1)}^2 \\ &\le C(\delta) \|f - \tilde{f}\|_{L^2(\omega^1)}^2 + C(c_{max}^{1,\infty}, \delta) \|u - \tilde{u}\|_{H^2(\omega^1)}^2, \end{aligned} $$

which ends the proof of Lemma 12.

<figure><img src="image_6.png" /></figure>

We are in a position to state a stationary stability result:

**Proposition 13.** Let $c, \tilde{c}$ piecewise $C^1$ satisfying (4) and $\max_{j \in [0,n]} \|c_j\|_{W^{1,\infty}(0,l_j)} \le c_{max}^{1,\infty}$. Under hypothesis $(H_c), (H_1), (H_2), (H_3)$ and $(H_4)$, there exist $C > 0$, $s_0 > 0$ such that for all $s \ge s_0$, there exist $C(s) > 0$ and $C(s, c_{max}, c_{max}^{1,\infty}) > 0$ with

$$ s^2 \int_{\mathbb{R}} |\xi|^2 e^{-2s\eta} dx \le C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx + C(s) \|f - \tilde{f}\|_{L^2(\omega^1)}^2 \\ + C(s, c_{max}, c_{max}^{1,\infty}) \left( \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2 + \|u - \tilde{u}\|_{H^2(\omega^1)}^2 \right). $$

*Proof.* We have by using Lemma 11 and Lemma 12

$$ \begin{aligned} s^2 \int_{\mathbb{R}} |\xi|^2 e^{-2s\eta} dx &\le s^2 \int_{\mathbb{R}\setminus\omega^0} |\xi|^2 e^{-2s\eta} dx + s^2 \int_{\omega^1} |\xi|^2 e^{-2s\eta} dx \\ &\le C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx + C(s) \int_{\omega^0} |\xi|^2 e^{-2s\eta} dx + C(s, c_{max}) \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2 \\ &\quad + C(s) \int_{\omega^1} |\xi|^2 dx \\ &\le C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx + C(s, c_{max}) \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2 + C(s) \int_{\omega^1} |\xi|^2 dx \\ &\le C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx + C(s, c_{max}) \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2 \\ &\quad + C(s) \|f - \tilde{f}\|_{L^2(\omega^1)}^2 + C(s, c_{max}^{1,\infty}) \|u - \tilde{u}\|_{H^2(\omega^1)}^2. \end{aligned} $$

<figure><img src="image_7.png" /></figure>
### 3.1.2 Stability for the evolution problem

Assume given $y^0 \in L^2(\mathbb{R})$ and $g \in C_c^\infty(\omega \times (0, T))$ (conveniently chosen). Let $y$ (resp. $\tilde{y}$) denote the solution of (55) (resp. (56)) with $h_j(t) = 0$ for $j \in [0, n]$ and $t \in (0, T)$, and $\tilde{y}^0(x) = y^0(x)$ for $x \in \mathbb{R}$. We aim at deriving stability estimates for $c - \tilde{c}$ from measurements of derivatives of $y - \tilde{y}$ for $(x, t) \in \omega \times (0, T)$ and for $(x, t) \in \mathbb{R} \times \{T/2\}$.

Notice first that there is no loss of generality in assuming that $\omega = \cup_{j=1}^n (p_j, q_j)$ with $0 < p_j < q_j < l_j$ for $1 \le j \le n$ by shrinking $\omega$ if nedded. In what follows, the function $\beta$ is those given in Lemma 4 for the open set $\omega = \cup_{j=1}^n (p_j, q_j)$.

As we shall use the stability estimates in the stationary case, we need first to prove the existence of some control input $g$ such that the assumptions $(H_1), (H_2), (H_3)$ and $(H_4)$ are satisfied for $\tilde{u} = \tilde{y}(., T/2)$. This is done in the following proposition.

**Proposition 14.** Let $\tilde{c}$ be a piecewise $C^1$ function satisfying (4) and $(H_c)$. Let $T > 0$ and $\tilde{y}^0 \in L^2(\mathbb{R})$. Pick $T' = T/2$. Then there exist two open sets $\omega^0 \subset \omega^1 \subset \omega$, a function $g \in C_c^\infty(\omega \times (0, T))$ such that the solution $\tilde{y}$ of (56) associated with $\tilde{y}_2^0(x) := \tilde{y}^0(x) - 2$ and $h_j(t) = -2$ for $j \in [0, n]$ and $t \in (0, T)$ satisfies for some $\delta > 0$

$$|\beta' \tilde{y}_x(., T')| \ge \delta > 0 \quad \text{in } \mathbb{R} \setminus \omega^0; \qquad (67)$$

$$\tilde{y}_j(., T') \in C^2([0, l_j]) \quad \forall j \in [0, n]; \qquad (68)$$

$$\|\tilde{y}(., T')\|_{L^\infty(\mathbb{R})} + \|\tilde{y}_x(., T')\|_{L^\infty(\mathbb{R})} \le \delta^{-1}; \qquad (69)$$

$$\tilde{y}(., T')|_{\partial\omega^1} = 0; \qquad (70)$$

$$|\tilde{y}_x(., T')|^2 - \tilde{y}(., T')\tilde{y}_{xx}(., T') \ge \delta > 0 \text{ in } \omega^1. \qquad (71)$$

*Proof.* We proceed as in [17]. In a first step, we construct a function $\rho(x)$ satisfying conditions very similar to (67)-(71). In a second step, we show that a control input $g$ can be designed so that the solution $\tilde{y}$ of (56) is close enough to $\rho$ for the norm $\|u\|_{L^\infty(\mathbb{R})} + \|u_x\|_{L^\infty(\mathbb{R})} + \|u_{xx}\|_{L^\infty(\mathbb{R})}$ at time $t = T'$ (approximate controllability).

**Step 1.** We need the following lemma:

**Lemma 15.** Let $\tilde{c}$ be a piecewise $C^1$ function satisfying (4) and $(H_c)$. Then there exist two open sets $\omega^0 \subset \omega^2 \subset \omega$ and a function $\rho \in H^1(\mathbb{R})$ such that for some $\hat{\delta} > 0$

$$|\beta' \rho_x| \ge \hat{\delta} > 0 \quad \text{in } \mathbb{R} \setminus \omega^0; \qquad (72) \qquad \rho_j(l_j) = -2 \quad \forall j \in [0, n]; \qquad (77)$$

$$\rho_j \in C^2([0, l_j]) \quad \forall j \in [0, n]; \qquad (73) \qquad (\tilde{c}_j \rho_{j,x})_x(l_j) = 0 \quad \forall j \in [0, n]; \qquad (78)$$

$$\|\rho\|_{L^\infty(\mathbb{R})} + \|\rho_x\|_{L^\infty(\mathbb{R})} \le \hat{\delta}^{-1}; \qquad (74) \qquad (\tilde{c}_j \rho_{j,x})_x(0) = 0 \quad \forall j \in [0, n]; \qquad (79)$$

$$\rho|_{\partial\omega^2} = 0; \qquad (75) \qquad \rho(0) = -1; \qquad (80)$$

$$|\rho_x|^2 - \rho \rho_{xx} \ge \hat{\delta} > 0 \quad \text{in } \omega^2; \qquad (76) \qquad [\tilde{c} \rho_x]_0 = 0. \qquad (81)$$

*Proof of Lemma 15:* We first construct $\rho_j = \rho|_{[0,l_j]}$ for any given $j \in [0, n]$. Since $\beta'_j(x) \ne 0$ for $x \in [0, l_j] \setminus (p_j, q_j)$ and since $[0, l_j] \setminus (p_j, q_j)$ is compact, we can find two numbers $p_j^0, q_j^0$ with $0 < p_j < p_j^0 < q_j^0 < q_j < l_j$ such that

$$|p_j^0 - p_j| + |q_j^0 - q_j| < \frac{1}{10}(q_j - p_j) \qquad (82)$$

and

$$\beta'_j(x) \ne 0 \quad \forall x \in [0, l_j] \setminus (p_j^0, q_j^0).$$

Pick some numbers $p_j^2 \in (p_j, p_j^0)$ and $q_j^2 \in (q_j^0, q_j)$. Set

$$\rho_j(x) := r_j(x - p_j^2)(q_j^2 - x) \quad \forall x \in [p_j, q_j]$$
where $r_j > 0$ is chosen so that $\rho_j(p_j) > -1$ and $\rho_j(q_j) > -2$. We note that we have $\rho_{j,x}(x) > 0$ for $x \in [p_j, \frac{p_j^2+q_j^2}{2})$ and $\rho_{j,x}(x) < 0$ for $x \in (\frac{p_j^2+q_j^2}{2}, q_j]$. Since $p_j^0 < \frac{p_j^2+q_j^2}{2} < q_j^0$ by (82), we infer that $\rho_{j,x}(x) > 0$ for $x \in [p_j, p_j^0]$ and $\rho_{j,x}(x) < 0$ for $x \in [q_j^0, q_j]$.

Pick $\varepsilon_j \in (0, \min(p_j, l_j - q_j))$. We set

$$ \rho_j(x) := -1 + k_j \int_0^x \frac{ds}{\tilde{c}_j(s)} \quad \text{for } 0 \le x \le \epsilon_j, $$

$$ \rho_j(x) := -2 + k_j \int_x^{l_j} \frac{ds}{\tilde{c}_j(s)} \quad \text{for } l_j - \varepsilon_j \le x \le l_j, $$

where $k_j > 0$ is chosen (small enough) so that $\rho_j(\epsilon_j) < \rho_j(p_j)$ and $\rho_j(l_j - \epsilon_j) < \rho_j(q_j)$. Then $\rho_j(l_j) = -2$, $\rho_j(0) = -1$, and $(\tilde{c}_j \rho_{j,x})_x(l_j) = (\tilde{c}_j \rho_{j,x})_x(0) = 0$. Next, we can extend $\rho_j$ as a function in $C^2([0, l_j])$ so that $|\beta'_j \rho_{j,x}| > 0$ in $[0, l_j] \setminus (p_j^0, q_j^0)$. On the other hand, $\rho_j(p_j^2) = \rho_j(q_j^2) = 0$, and noticing that $\rho_{j,xx} < 0$ in $(p_j, q_j)$ and that $\rho_{j,x}(p_j^2) \ne 0$, $\rho_{j,x}(q_j^2) \ne 0$, we see that $|\rho_{j,x}|^2 - \rho_j \rho_{j,xx} > 0$ in $[p_j^2, q_j^2]$.

For $\rho_0$, we set $k_0 := - \sum_{j=1}^n k_j < 0$ and

$$ \rho_0(x) := -1 + k_0 \int_0^x \frac{ds}{\tilde{c}_0(s)} \quad \text{for } 0 \le x \le \epsilon_0, $$

$$ \rho_0(x) := -2 - k_0 \int_x^{l_0} \frac{ds}{\tilde{c}_0(s)} \quad \text{for } l_0 - \varepsilon_0 \le x \le l_0, $$

where $\epsilon_0 \in (0, l_0/2)$ is chosen (small enough) so that $\rho_0(\epsilon_0) > \rho_0(l_0 - \epsilon_0)$. Next we extend $\rho_0$ as a function in $C^2([0, l_0])$ so that $\rho_{0,x} < 0$ in $[0, l_0]$.

Setting

$$ \omega^0 := \cup_{j=1}^n (p_j^0, q_j^0) \quad \text{and} \quad \omega^2 := \cup_{i=1}^n (p_j^2, q_j^2), $$

we see that (72)-(81) hold. We note that $\rho_{0,x}(0) < 0$ while $\rho_{j,x}(0) > 0$ for $j \in [1, n]$. This completes the proof of Lemma 15.

**Step 2.** It follows from Theorem 5 that the following unique continuation property

$$ z \in L^2(0, T, H_0^1(\mathbb{R})) \\ \left. \begin{array}{ll} z_t + (\tilde{c}z_x)_x = 0 & \text{in } \mathbb{R} \times (0, T) \\ [\tilde{c}z_x(t)]_0 = 0 & \text{for } t \in (0, T), \\ z = 0 & \text{in } \omega \times (0, T) \end{array} \right\} \Rightarrow z = 0 \quad \text{in } \mathbb{R} \times (0, T) $$

holds for any $T > 0$. By duality, this yields the approximate controllability in $L^2(\mathbb{R})$ and for any $T > 0$ of the system

$$ \left\{ \begin{array}{ll} u_t - (\tilde{c}u_x)_x = g\chi_\omega, & \forall (x,t) \in \mathbb{R} \times (0, T), \\ u_j(l_j, t) = 0, & \forall j \in [0, n], t \in (0, T), \\ u_j(0, t) = u_i(0, t), & \forall i, j \in [0, n], 0 < t < T, \\ [\tilde{c}u_x(t)]_0 = 0, & \forall t \in (0, T), \\ u(x, 0) = 0, & \forall x \in \mathbb{R}, \end{array} \right. \quad (83) $$

with control input $g \in C_c^\infty(\omega \times (0, T))$.

We denote by $\tilde{U}(u^0, g, h)$ the solution of the system

$$ \left\{ \begin{array}{ll} u_t - (\tilde{c}u_x)_x = g\chi_\omega, & \forall (x,t) \in \mathbb{R} \times (0, T), \\ u_j(l_j, t) = h_j(t), & \forall j \in [0, n], t \in (0, T), \\ u_j(0, t) = u_i(0, t), & \forall i, j \in [0, n], t \in (0, T), \\ [\tilde{c}u_x(t)]_0 = 0, & \forall t \in (0, T), \\ u(x, 0) = u^0(x), & \forall x \in \mathbb{R}, \end{array} \right. \quad (84) $$
where $h = (h_j)_{j \in [0,n]}$.

Let $\tilde{A}$ denote the operator $\tilde{Au} := -\partial_x(\tilde{c}\partial_x u)$ with domain $D(\tilde{A}) := \{u \in H_0^1(\mathbb{R}), \partial_x(\tilde{c}\partial_x u) \in L^2(\mathbb{R}), [\tilde{c}u_x]_0 = 0\} \subset L^2(\mathbb{R})$. It is easily seen that the operator $\tilde{A}$ is selfadjoint, nonnegative, and sectorial (see [21]), and hence $-\tilde{A}$ generates a semigroup of contractions $(e^{-t\tilde{A}})_{t \ge 0}$ which is analytic.

Let

$$C(\mathbb{R}) := \{u : \mathbb{R} \to \mathbb{R}; u_j \in C([0, l_j]) \ \forall j \in [0, n], u_j(0) = u_i(0) \ \forall i, j \in [0, n]\}$$

and

$$C_0(\mathbb{R}) := \{u \in C(\mathbb{R}); u_j(l_j) = 0 \text{ for } j \in [0, n]\}$$

be endowed with the norm $\| \cdot \|_{L^\infty(\mathbb{R})}$. We note that $H_0^1(\mathbb{R}) \subset C_0(\mathbb{R})$.

Let $\tilde{Au} := -\partial_x(\tilde{c}\partial_x u)$ denote the operator with domain

$$D(\tilde{A}) := \{u \in H_0^1(\mathbb{R}); \partial_x(\tilde{c}\partial_x u) \in C_0(\mathbb{R}), [\tilde{c}u_x]_0 = 0\} \subset C_0(\mathbb{R}).$$

It is easily seen (see e.g. [15]) that $-\tilde{A}$ generates a semigroup of contractions in $C_0(\mathbb{R})$.

Note that $u \in D(\tilde{A})$ implies that $u_j \in C^2([0, l_j])$ for all $j \in [0, n]$, and that

$$\|u^0\|_{D(\tilde{A})} := \|u^0\|_{L^\infty(\mathbb{R})} + \|\tilde{A}u^0\|_{L^\infty(\mathbb{R})} \sim \|u^0\|_{L^\infty(\mathbb{R})} + \|u_x^0\|_{L^\infty(\mathbb{R})} + \|u_{xx}^0\|_{L^\infty(\mathbb{R})}. \quad (85)$$

From classical semigroup theory, there exists $C = C(T) > 0$ such that for all $u^0 \in D(\tilde{A})$, we have

$$\|\tilde{U}(u^0, 0, 0)(t)\|_{D(\tilde{A})} \le C\|u^0\|_{D(\tilde{A})} \quad \forall t \in [0, T].$$

Let $\tilde{y}^0 \in L^2(\mathbb{R})$ and $T' = T/2$. We note that $e^{-T'\tilde{A}}\tilde{y}^0 \in D(\tilde{A}^2) \subset D(\tilde{A})$. We set

$$\rho_2 := \rho + 2 - \tilde{U}(\tilde{y}^0, 0, 0)(T') = \rho + 2 - e^{-T'\tilde{A}}\tilde{y}^0.$$

Then $\rho_2 \in D(\tilde{A})$, by (73) and (77)-(81). Let $\varepsilon > 0$ be given. We pick $\eta \in (0, T')$ such that

$$\|\tilde{U}(\rho_2, 0, 0)(\eta) - \rho_2\|_{D(\tilde{A})} \le \varepsilon. \quad (86)$$

By approximate controllability of (83) in time $T' - \eta$, we may pick a control input $g \in C_c^\infty(\omega \times (0, T' - \eta))$, that we may extend by 0 for $t \ge T' - \eta$, such that

$$\|\tilde{U}(0, g, 0)(T' - \eta) - \rho_2\|_{L^2(\mathbb{R})} \le \varepsilon.$$

On the other hand, we notice that for some constant $C' > 0$ and all $v^0 \in L^2(\mathbb{R})$

$$\|\tilde{U}(v^0, 0, 0)(\eta)\|_{D(\tilde{A})} \le C'\|v^0\|_{L^2(\mathbb{R})}. \quad (87)$$

Indeed, both the map $v^0 \in L^2(\mathbb{R}) \to \tilde{U}(v^0, 0, 0)(\eta) = e^{-\eta\tilde{A}}v^0 \in D(\tilde{A}^2)$ and the embedding $D(\tilde{A}^2) \subset D(\tilde{A})$ are continuous. Using (87) with $v^0 = \tilde{U}(0, g, 0)(T' - \eta) - \rho_2$ and the semigroup property yield

$$\|\tilde{U}(0, g, 0)(T') - \tilde{U}(\rho_2, 0, 0)(\eta)\|_{D(\tilde{A})} \le C'\varepsilon,$$

and hence, with (86),

$$\|\tilde{U}(0, g, 0)(T') - \tilde{\rho}\|_{D(\tilde{A})} \le (C' + 1)\varepsilon. \quad (88)$$

Let

$$\tilde{y} := \tilde{U}(\tilde{y}^0 - 2, g, -2) = \tilde{U}(\tilde{y}^0, 0, 0) + \tilde{U}(0, g, 0) - 2$$

for $\tilde{U}(-2, 0, -2) = -2$. Then

$$\|\tilde{y}(T') - \rho\|_{D(\tilde{A})} = \|\tilde{U}(0, g, 0)(T') - \rho_2\|_{D(\tilde{A})} \le (C' + 1)\varepsilon. \quad (89)$$
For any $j \in [1, n]$, we have $\rho_j(p_j) < 0$, $\rho_j(q_j) < 0$ and $\min_{x \in [p_j^0, q_j^0]} \rho_j(x) > 0$. Using (89), we see that for $\varepsilon$ small enough we still have $\tilde{y}(p_j, T') < 0$, $\tilde{y}(q_j, T') < 0$ and $\min_{x \in [p_j^0, q_j^0]} \tilde{y}(x, T') > 0$. Therefore we can pick $p_j^1 \in (p_j, p_j^0)$ and $q_j^1 \in (q_j^0, q_j)$ such that $\tilde{y}(p_j^1, T') = \tilde{y}(q_j^1, T') = 0$ and $\tilde{y}(x, T') > 0$ for $x \in (p_j^1, q_j^1)$. Set $\omega^1 = \cup_{j=1}^n (p_j^1, q_j^1)$. Then (67)-(71) hold true for $\varepsilon$ and $\delta$ small enough. The proof of Proposition 14 is complete. $\square$

We are ready to prove the main stability result for the evolution problem with interior observations, namely Theorem 1.

**Proof of Theorem 1.** Let $y^0 \in L^2(\mathbb{R})$, and let $\tilde{y}^0 = y^0$. Let $g \in C_c^\infty(\omega \times (0, T))$ be as given in Proposition 14. Pick $h_j(t) = 0$ for $j \in [0, n]$ and $t \in (0, T)$. Let $y$ (resp. $\tilde{y}$) denote the solution to (55) (resp. (56)). Pick $T' = \frac{T}{2}$ and $\tau \in (0, T')$ with $g \equiv 0$ on $\omega \times (0, \tau)$. Let $A$ and $\mathcal{A}$ be the same operators as previously but corresponding to the coefficient $c$. Since $-A$ is the infinitesimal generator of an analytic semigroup with $0 \in \rho(A)$, we have by [26, Theorem 6.13] that for all $t > 0$, $e^{-tA} : L^2(\mathbb{R}) \to D(A^4)$ continuously. In particular, setting $Y(t) := e^{-tA}y^0$, we have

$$Y \in C([\tau, T], D(A^4)), \quad Y_t \in C([\tau, T], D(A^3)), \quad \text{and} \quad Y_{tt} \in C([\tau, T], D(A^2)).$$

Recall that $D(A^2) \subset D(A)$. Let $\xi := c - \tilde{c}$. Note that $\xi(0) = 0$. It follows from (85) and the fact that $\xi(0) = 0$ that the maps $u^0 \in D(A) \to u_x^0 \in L^\infty(\mathbb{R})$ and $u^0 \in D(A) \to \xi u_x^0 \in H^1(\mathbb{R})$ are continuous.

We infer that $Y_{tx}, Y_{ttx} \in C([\tau, T], L^\infty(\mathbb{R}))$, and that $\xi Y_{ttx} \in C([\tau, T], H^1(\mathbb{R}))$. Note that $y(t) = Y(t) + U(0, g, 0)(t)$, where $U$ is defined as above with $\tilde{c}$ replaced by $c$. Note that $C_c^2(\omega) \subset D(A)$, so that $g \in C_c^2([0, T], D(A))$ and $U(0, g, 0) \in C^2([0, T], D(A))$. It follows that $y_{tx}, y_{ttx} \in C([\tau, T], L^\infty(\mathbb{R}))$, and that $\xi y_{ttx} \in C([\tau, T], H^1(\mathbb{R}))$. Similarly, we have that $\tilde{y}_{tx}, \tilde{y}_{ttx} \in C([\tau, T], L^\infty(\mathbb{R}))$, and that $\xi \tilde{y}_{tx} \in C([\tau, T], H^1(\mathbb{R}))$.

Replacing $y^0$ (resp. $\tilde{y}^0$) by $y(\tau)$ (resp. $\tilde{y}(\tau)$), we can assume that

$$y_{tx}, y_{ttx}, \tilde{y}_{tx}, \tilde{y}_{ttx} \in C([0, T], L^\infty(\mathbb{R})), \quad \xi \tilde{y}_{tx} \in C([0, T], H^1(\mathbb{R})).$$

Note that it could now happen that $y^0 \neq \tilde{y}^0$. As (5) is invariant by uniform translation of both $y$ and $\tilde{y}$ by $-2$, we can pick $h_j(t) = -2$ (instead of 0) for all $t \in (0, T)$ and all $j \in [0, n]$, so that the function $\tilde{y}$ is now the one given by Proposition 14, while $y$ is now the solution to (55) with initial data $y^0 - 2$, with boundary data $h_j(t) = -2$ for $t \in (0, T)$ and $j \in [0, n]$, and still with the same function $g \in C_c^\infty(\omega \times (0, T))$ as above. Let $v = (y - \tilde{y})_t$. Then $v$ satisfies (57), and we have

$$\begin{aligned} \int_{\mathbb{R}} |v(T')|^2 e^{-2s\eta(T')} dx &= \int_0^{T'} \int_{\mathbb{R}} 2vv_t e^{-2s\eta} dxdt + \int_0^{T'} \int_{\mathbb{R}} |v|^2 \frac{\partial}{\partial t} (e^{-2s\eta}) dxdt \\ &= \int_0^{T'} \int_{\mathbb{R}} 2vv_t e^{-2s\eta} dxdt - 2s \int_0^{T'} \int_{\mathbb{R}} |v|^2 \eta_t e^{-2s\eta} dxdt. \end{aligned} \quad (90)$$

Let us denote by $A_1$ the term at the left hand side and by $A_2$ and $A_3$ the two right members of the previous equation.

1. Upper bounds for $A_1$.

We know that $|\eta_t| \le C\varphi^2$ (see (29)), thus

$$|A_3| \le C_s \int_0^{T'} \int_{\mathbb{R}} |v|^2 \varphi^2 e^{-2s\eta} dxdt \le C_s \iint_Q |v|^2 \varphi^3 e^{-2s\eta} dxdt \quad (91)$$

where we used (30). Furthermore, we note that $\xi_i(0) = 0$ for all $i \in [0, n]$ (due to $(H_c)$), so that $[cv_x(t)]_0 = -[\xi \tilde{y}_{tx}(t)]_0 = 0$ and $(\xi \tilde{y}_{tx})(0, t) = 0$ so we can apply the Carleman inequality given in Theorem 8 and get
$$s \iint_Q |v|^2 \varphi^3 e^{-2s\eta} dxdt \le C \left( s \int_0^T \int_\omega |v|^2 \varphi^3 e^{-2s\eta} dxdt + \iint_Q |\xi \tilde{y}_{tx}|^2 \varphi^2 e^{-2s\eta} dxdt \right). \quad (92)$$

First of all, we note that (omitting variables $x, t$)

$$\varphi^2 e^{-2s\eta} = \varphi^2(T')e^{-2s\eta(T')} \left(\frac{T'(T-T')}{t(T-t)}\right)^2 e^{-2s\eta(T')} \left(\frac{T'(T-T')}{t(T-t)}-1\right). \quad (93)$$

Picking $T' = T/2$, we easily obtain that, for $s, \lambda$ large enough such that $s\eta(T') > 1$, we have $\left(\frac{T'(T-T')}{t(T-t)}\right)^2 e^{-2s\eta(T')} \left(\frac{T'(T-T')}{t(T-t)}-1\right) \le 1$. Thus with (93) we have the following inequality:

$$\varphi^2 e^{-2s\eta} \le \varphi^2(T')e^{-2s\eta(T')}. \quad (94)$$

Then we get, since $\tilde{y}_{tx} \in C([0, T], L^\infty(\mathbb{R}))$,

$$\iint_Q |\xi \tilde{y}_{tx}|^2 \varphi^2 e^{-2s\eta} dxdt \le C \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx.$$

Then, with (91) and (92), we obtain

$$|A_3| \le C \left[ \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx + s \int_0^T \int_\omega |v|^2 \varphi^3 e^{-2s\eta} dxdt \right]. \quad (95)$$

Using Cauchy-Schwarz inequality and (30), we have

$$|A_2| \le C \left( \iint_Q s|v|^2 \varphi^3 e^{-2s\eta} dxdt + \iint_Q s^{-1}\varphi^{-1} e^{-2s\eta} |v_t|^2 dxdt \right). \quad (96)$$

Observe that $v_t$ satisfies

$$\left\{
\begin{array}{ll}
(v_t)_t - (cv_{xt})_x = (\xi \tilde{y}_{ttx})_x, & \forall (x, t) \in \mathbb{R} \times (0, T), \\
v_{j,t}(l_j, t) = 0, & \forall j \in [0, n], t \in (0, T), \\
v_{j,t}(0, t) = v_{i,t}(0, t), & \forall i, j \in [0, n], t \in (0, T), \\
[cv_{tx}(t)]_0 = -[\xi \tilde{y}_{ttx}(t)]_0 = 0, & \forall t \in (0, T).
\end{array}
\right.
\quad (97)$$

Applying Carleman estimate given in Theorem 8 to $v_t$, we have by using (94), (30) and the fact that $\tilde{y}_{ttx} \in C([0, T], L^\infty(\mathbb{R}))$

$$\begin{aligned}
& \iint_Q s^{-1} \varphi^{-1} e^{-2s\eta} |v_t|^2 dxdt \\
& \le C s^{-4} \iint_Q (s\varphi)^3 e^{-2s\eta} |v_t|^2 dxdt \\
& \le C s^{-4} \left[ \int_0^T \int_\omega s^3 |v_t|^2 \varphi^3 e^{-2s\eta} dxdt + \iint_Q |\xi \tilde{y}_{ttx}|^2 (s\varphi)^2 e^{-2s\eta} dxdt \right] \\
& \le C \left[ \int_0^T \int_\omega s^{-1} |v_t|^2 \varphi^3 e^{-2s\eta} dxdt + \int_{\mathbb{R}} |\xi|^2 s^{-2} \varphi^2(T') e^{-2s\eta(T')} dx \right]. \quad (98)
\end{aligned}$$

Thus, with (92), (95)-(96) and (98), we obtain the following upper bounds for $A_1$ for $s$ sufficiently large:

$$A_1 \le C \left[ \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx + \int_0^T \int_\omega (s|v|^2 + s^{-1}|v_t|^2) \varphi^3 e^{-2s\eta} dxdt \right]. \quad (99)$$
## 2. Lower bounds for $A_1$.

First we have

$$v(T') = (cy_x - \tilde{c}\tilde{y}_x)_x(T'). \quad (100)$$

Thanks to the study of the stationary case and Proposition 13, we can write

$$ 
\begin{aligned}
A_1 &= \int_{\mathbb{R}} |(cy_x - \tilde{c}\tilde{y}_x)_x(T')|^2 e^{-2s\eta(T')} dx \\
&\geq C s^2 \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx - C(s) \left( \|y_t(T') - \tilde{y}_t(T')\|_{L^2(\omega^1)}^2 \right. \\
&\qquad \left. + \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})}^2 + \|y(T') - \tilde{y}(T')\|_{H^2(\omega^1)}^2 \right).
\end{aligned} 
\quad (101) $$

Thus, combining (101) with (99), we obtain

$$ 
\begin{aligned}
s^2 \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx - C(s)(\|y_t(T') - \tilde{y}_t(T')\|_{L^2(\omega^1)}^2 \\
+ \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})}^2 + \|y(T') - \tilde{y}(T')\|_{H^2(\omega^1)}^2) \\
\leq C \left[ \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx + \int_0^T \int_{\omega} (s|v|^2 + s^{-1}|v_t|^2) \varphi^3 e^{-2s\eta} dx dt \right].
\end{aligned} 
\quad (102) $$

Thus, for $s$ sufficiently large, we have the stability estimate

$$ 
\begin{aligned}
s^2 \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx \leq C \int_0^T \int_{\omega} \varphi^3(s|y_t - \tilde{y}_t|^2 + s^{-1}|y_{tt} - \tilde{y}_{tt}|^2) e^{-2s\eta} dx dt \\
+ C(s)(\|y_t(T') - \tilde{y}_t(T')\|_{L^2(\omega^1)}^2 + \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})}^2 + \|y(T') - \tilde{y}(T')\|_{H^2(\omega^1)}^2),
\end{aligned} 
\quad (103) $$

which completes the proof of Theorem 1. $\square$

## 3.2 Stability results with boundary observations

We derive, in the case of boundary measurements, results similar to those obtained in the previous section. In a first time, as for the proof of the Carleman result with one less derivative, we extend the network $\mathbb{R}$ in another network $\hat{\mathbb{R}}$ and we follow the previous case. We make the following assumptions, ($H_c$) and

$$ (H_{Ext}) \quad c_i(l_i) = \tilde{c}_i(l_i), \quad \forall i \in [1, n]. $$

### 3.2.1 Stability for the stationary problem

We easily prove the following result, with the same notations as in Section 3.1.1.

**Proposition 16.** Let $c, \tilde{c}$ piecewise $C^1$ satisfying (4) and $\max_{j \in [0, n]} \|c_j\|_{W^{1, \infty}(0, l_j)} \leq c_{max}^{1, \infty}$. Under hypotheses $(H_1)$ (with $\omega^0 = \emptyset$), $(H_2)$, $(H_3)$, $(H_c)$ and $(H_{Ext})$, there exist $C > 0$, $s_0 > 0$ such that for all $s \geq s_0$, there exist $C(s)$, $C(s, c_{max}) > 0$,

$$ 
\begin{aligned}
s^2 \int_{\mathbb{R}} \varphi^2 |\xi|^2 e^{-2s\eta} dx \leq C \int_{\mathbb{R}} |f - \tilde{f}|^2 e^{-2s\eta} dx \\
+ C(s, c_{max}) \left( \|u - \tilde{u}\|_{H^1(\mathbb{R})}^2 + \sum_{j=1}^n e^{-2s\eta(l_j, T')} |u_{j,x}(l_j) - \tilde{u}_{j,x}(l_j)|^2 \right).
\end{aligned} 
$$

We only need to take care of the boundary term appearing in $B_2$ in Lemma 11. Using the fact that $\beta'_0(l_0) < 0$ and that $\beta'_j(l_j) > 0$ for $j \in [1, n]$ (since $\beta$ is as given by Lemma 4 for $\mathbb{R}$ and $\omega \subset \hat{\mathbb{R}} \setminus \mathbb{R}$), we can obtain the result.
### 3.2.2 Stability for the evolution problem

We also prove the main stability result for the evolution problem with boundary measurements, namely Theorem 2.

**Proof of Theorem 2.** Let $y^0 \in L^2(\mathbb{R})$. First, we pick $\tau \in (0, T')$ where $T' = T/2$. We take $h_j(t) = 0$ for $0 \le t \le \tau$, so that $y(\tau) \in D(A^4)$. As in the proof of Theorem 10, we extend the network $\mathbb{R}$ in another network $\hat{\mathbb{R}}$, and we fix an open set $\omega \subset \cup_{j=1}^n(0, \hat{l}_j)$ which intersects each interval $(l_j, \hat{l}_j)$ for $j \in [1, n]$. We also extend $c$ and $\tilde{c}$ smoothly such that $0 < c_{min} \le c(x) \le c_{max} < \infty$ in $\mathbb{R}$. We can extend $y(\tau)$ (resp. $\tilde{y}(\tau)$) as a function in $D(A^4)$ (resp. $D(\tilde{A}^4)$) for $c$ and $\hat{\mathbb{R}}$ (resp. for $\tilde{c}$ and $\hat{\mathbb{R}}$). Applying Proposition 14 to $\hat{\mathbb{R}}$ and $\tilde{y}(\tau)$, we can find an input $g \in C_c^\infty(\omega \times (\tau, T))$ such that $|\tilde{y}_x(., T)| \ge \delta > 0$ in $\mathbb{R}$ and $\|\tilde{y}(., T')\|_{L^\infty(\mathbb{R})} + \|\tilde{y}_x(., T')\|_{L^\infty(\hat{\mathbb{R}})} \le \delta^{-1}$. We pick a function $\psi \in C^\infty(\mathbb{R})$ such that $\psi(t) = 1$ for $t \le T'$ and $\psi(t) = 0$ for $t \ge 3T/4$, and we set $h_j(t) = \psi(t)\tilde{y}_j(l_j, t)$ for $t \in [\tau, T]$ and $j \in [1, n]$, and $h_0(t) = 0$ for $t \in [0, T]$. We claim that $h_j \in C_c^1([0, T])$ for $j \in [1, n]$. Indeed, $0 = h'_j(\tau^-) = (\tilde{c}_j\tilde{y}_{j,x})_x(l_j, \tau) = h'_j(\tau^+)$ thanks to (56) applied with $x = l_j$ and $t = \tau^-, t = \tau^+$. On the other hand, $h_j(t) = 0$ for $t \in [0, T] \setminus (\tau, 3T/4)$ and $j \in [1, n]$.

Next, we use the same type of proof as previously for the case of interior observation. We have three terms $A_1, A_2$ and $A_3$ to estimate as in (90).

By using the same estimate as in (91), we get

$$|A_3| \le C s \int_0^T \int_{\mathbb{R}} |v|^2 \varphi^3 e^{-2s\eta} dx dt. \quad (104)$$

By applying the Carleman estimate given in Theorem 10 (since $[cv_x(t)]_0 = -[\xi \tilde{y}_{tx}(t)]_0 = 0$ due to $(H_c)$), we obtain

$$s \iint_Q |v|^2 \varphi^3 e^{-2s\eta} dx dt \le C \left( \iint_Q \varphi^2 e^{-2s\eta} |\xi \tilde{y}_{tx}|^2 dx dt \right. \\ \left. + \int_0^T \sum_{j=1}^n s^{-1} \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{tx}(l_j, t) + c(l_j) v_x(l_j, t)|^2 dt \right). \quad (105)$$

As in the previous section, we can get with $T' = T/2$ and (94)

$$|A_3| \le C \left[ \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx \right. \\ \left. + \int_0^T \sum_{j=1}^n s^{-1} \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{tx}(l_j, t) + c(l_j) v_x(l_j, t)|^2 dt \right]. \quad (106)$$

We clearly have (96) and (97). Therefore by applying the Carleman estimate given in Theorem 10 to $v_t$, we obtain

$$\iint_Q s^{-1} \varphi^{-1} e^{-2s\eta} |v_t|^2 dx dt \le C s^{-4} \iint_Q (s\varphi)^3 e^{-2s\eta} |v_t|^2 dx dt \\ \le C \left[ s^{-2} \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx \right. \\ \left. + s^{-3} \int_0^T \sum_{j=1}^n \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{ttx}(l_j, t) + c(l_j) v_{xt}(l_j, t)|^2 dt \right]. \quad (107)$$
With (106) and (107), we obtain an upper bound for $A_1$ for $s$ sufficiently large:

$$A_1 \le C \left[ \int_{\mathbb{R}} |\varphi(T')|^2 e^{-2s\eta(T')} |\xi|^2 dx \right. \\ \left. + s^{-3} \int_0^T \sum_{j=1}^n \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{ttx}(l_j, t) + c(l_j) v_{xt}(l_j, t)|^2 dt \right. \\ \left. + \int_0^T \sum_{j=1}^n s^{-1} \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{tx}(l_j, t) + c(l_j) v_x(l_j, t)|^2 dt \right]. \quad (108)$$

We now search for lower bounds for $A_1$. First we have that

$$v(T') = (cy_x - \tilde{c}\tilde{y}_x)_x(T'). \quad (109)$$

Thanks to the study of the stationary case and Proposition 16, we can write

$$A_1 = \int_{\mathbb{R}} |(cy_x - \tilde{c}\tilde{y}_x)_x(T')|^2 e^{-2s\eta(T')} dx \\ \ge C s^2 \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx \\ - C(s) \left( \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})}^2 + \sum_{j=1}^n e^{-2s\eta(l_j, T')} |y_{j,x}(l_j, T') - \tilde{y}_{j,x}(l_j, T')|^2 \right) . \quad (110)$$

Thus with (108), (110) and ($H_{Ext}$), we get for $s$ sufficiently large

$$s^2 \int_{\mathbb{R}} e^{-2s\eta(T')} |\xi|^2 dx \\ \le C(s) \left[ \|y(T') - \tilde{y}(T')\|_{H^1(\mathbb{R})}^2 + \sum_{j=1}^n e^{-2s\eta(l_j, T')} |y_{j,x}(l_j, T') - \tilde{y}_{j,x}(l_j, T')|^2 \right. \\ \left. + s^{-3} \int_0^T \sum_{j=1}^n \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{ttx}(l_j, t) + c(l_j) v_{xt}(l_j, t)|^2 dt \right. \\ \left. + \int_0^T \sum_{j=1}^n s^{-1} \varphi_j(l_j, t) e^{-2s\eta_j(l_j, t)} |\xi(l_j) \tilde{y}_{tx}(l_j, t) + c(l_j) v_x(l_j, t)|^2 dt \right].$$

This completes the proof of Theorem 2, since $\xi(l_j)\tilde{y}_{tx}(l_j) = 0$ for $j \in [1, n]$ (see ($H_{Ext}$)). $\square$

# 4 Conclusion

As it was said in the introduction, these results can be extended to general tree-shaped networks: with boundary measurements at all exterior nodes except one, the root of the tree, or with internal measurements at all exterior edges of the tree except one. The key point is in the choice of the weights in the Carleman estimates. More precisely, we obtain, by integration by parts, instead of one matrix $A_\beta$ defined in Lemma 4, several matrices of the same kind, corresponding to each interior node. Then, we take second order polynomial functions like (8) for external edges and affine functions like (7) for all remaining edges. The construction is made by induction in a similar way as in [9]. The main difficulty here is just to use clear and understandable notations.

However the case of a general network with a cycle remains an open question.
# Acknowledgements

EC was partially supported by MathAmSud-22-MATH-08 SCIPinPDEs.

LR was partially supported by the ANR project ANR-24-CE40-5470 and the COFECUB project COFECUB-20232505780P.

JV was partially supported by the ANR project TRECOS ANR-20-CE40-0009.

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