# Exponential Stability of the System of Transmission of the Wave Equation with a Delay Term in the Boundary Feedback

Salah-Eddine Rebiai

► To cite this version:

Salah-Eddine Rebiai. Exponential Stability of the System of Transmission of the Wave Equation with a Delay Term in the Boundary Feedback. 25th System Modeling and Optimization (CSMO), Sep 2011, Berlin, Germany. pp.276-285, 10.1007/978-3-642-36062-6_28. http-01347547

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Distributed under a Creative Commons Attribution 4.0 International License
Exponential stability of the system of
transmission of the wave equation with a delay
term in the boundary feedback

Salah-Eddine Rebiai

Laboratoire des Techniques Mathématiques, Faculté des Sciences,
Université de Batna, 05000 Batna, Algeria
rebiai@hotmail.com

**Abstract.** We consider a system of transmission of the wave equation with Neumann feedback control that contains a delay term and that acts on the exterior boundary. First, we prove under some assumptions that the closed-loop system generates a $C_0$-semigroup of contractions on an appropriate Hilbert space. Then, under further assumptions, we show that the closed-loop system is exponentially stable. To establish this result, we introduce a suitable energy function and use multiplier method together with an estimate taken from [3] (Lemma 7.2) and compactness-uniqueness arguments.

**Keywords:** Wave equation, transmission problem, time delays, boundary stabilization, exponential stability.

# 1 Introduction

It is by now well-known that certain infinite-dimensional second-order systems are not robust with respect to arbitrarily small delays in the damping. This lack of stability robustness was first shown to hold for the one-dimensional wave equation ([2]). Later, further examples illustrating this phenomenon were considered in [1]: the two-dimensional wave equation with damping introduced through Neumann-type boundary conditions on one edge of a square boundary and the Euler-Bernoulli beam equation in one dimension with damping introduced through a specific set of boundary conditions on the right end point.

Recently, Xu et al [9] established sufficient conditions that guarantee the exponential stability of the one-dimensional wave equation with a delay term in the boundary feedback. Nicaise and Pignotti [5] extended this result to the multi-dimensional wave equation with a delay term in the boundary or internal feedbacks. The same type of result was obtained by Nicaise and Rebiai [6] for the Schrödinger equation.

Motivated by the references [9], [5] and [6]; we investigate in this paper the problem of exponential stability for the system of transmission of the wave equation with a delay term in the boundary feedback.

Let $\Omega$ be an open bounded domain of $\mathbb{R}^n$ with a boundary $\Gamma$ of class $C^2$ which consists of two non-empty parts $\Gamma_1$ and $\Gamma_2$ such that $\overline{\Gamma_1} \cap \overline{\Gamma_2} = \emptyset$. Let $\Gamma_0$
with $\overline{\Gamma}_0 \cap \overline{\Gamma}_1 = \overline{\Gamma}_0 \cap \overline{\Gamma}_2 = \emptyset$ be a regular hypersurface of class $C^2$ which separates $\Omega$ into two domains $\Omega_1$ and $\Omega_2$ such that $\Gamma_1 \subset \partial\Omega_1$ and $\Gamma_2 \subset \partial\Omega_2$. Furthermore, we assume that there exists a real vector field $h \in (C^2(\overline{\Omega}))^n$ such that:
(H.1) The Jacobian matrix $J$ of $h$ satisfies

$$ \int_{\Omega} J(x)\zeta(x) \cdot \zeta(x) d\Omega \geq \alpha \int_{\Omega} |\zeta(x)|^2 d\Omega, $$

for some constant $\alpha > 0$ and for all $\zeta \in L^2(\Omega; \mathbb{R}^n)$;

(H.2) $h(x) \cdot \nu(x) \leq 0$ on $\Gamma_1$;

(H.3) $h(x) \cdot \nu(x) \geq 0$ on $\Gamma_0$.

where $\nu$ is the unit normal on $\Gamma$ or $\Gamma_0$ pointing towards the exterior of $\Omega$ or $\Omega_1$.

Let $a_1, a_2 > 0$ be given. Consider the system of transmission of the wave equation with a delay term in the boundary conditions:

$$ y''(x, t) - a(x)\Delta y(x, t) = 0 \qquad \text{in } \Omega \times (0, +\infty), \qquad (1) $$

$$ y(x, 0) = y^0(x), y'(x, 0) = y^1(x, 0) \qquad \text{in } \Omega, \qquad (2) $$

$$ y_1(x, t) = 0 \qquad \text{on } \Gamma_1 \times (0, +\infty), \qquad (3) $$

$$ \frac{\partial y_2(x, t)}{\partial \nu} = -\mu_1 y_2'(x, t) - \mu_2 y_2'(x, t - \tau) \qquad \text{on } \Gamma_2 \times (0, +\infty), \qquad (4) $$

$$ y_1(x, t) = y_2(x, t), \qquad \text{on } \Gamma_0 \times (0, +\infty), \qquad (5) $$

$$ a_1 \frac{\partial y_1(x, t)}{\partial \nu} = a_2 \frac{\partial y_2(x, t)}{\partial \nu} \qquad \text{on } \Gamma_0 \times (0, +\infty), \qquad (6) $$

$$ y_2'(x, t - \tau) = f_0(x, t - \tau) \qquad \text{on } \Gamma_2 \times (0, \tau). \qquad (7) $$

where:

$$ -a(x) = \begin{cases} a_1, & x \in \Omega_1 \\ a_2, & x \in \Omega_2 \end{cases} $$

$$ -y(x, t) = \begin{cases} y_1(x, t), & (x, t) \in \Omega_1 \times (0, +\infty) \\ y_1(x, t), & (x, t) \in \Omega_2 \times (0, +\infty) \end{cases} $$

- $\frac{\partial}{\partial \nu}$ is the normal derivative.

- $\mu_1$ and $\mu_2$ are positive real numbers.

- $\tau$ is the time delay

- $y^0, y^1, f_0$ are the initial data which belong to suitable spaces.

In the absence of delay, that is $\mu_2 = 0$, Liu and Williams [4] have shown that the solution of (1)-(6) decays exponentially to zero in the energy space $H_{\Gamma_1}^1(\Omega) \times L^2(\Omega)$ provided that

$$ a_1 > a_2 \qquad (8) $$

and $\{\Omega, \Gamma_0, \Gamma_1, \Gamma_2\}$ satisfies (H.1), (H.2), (H.3), and (H.4) $h(x) \cdot \nu(x) \geq \gamma > 0$.
The purpose of this paper is to investigate the stability of problem (1) – (7) in the case where both $\mu_1$ and $\mu_2$ are different from zero. To this end, assume as in [5] that

$$ \mu_1 > \mu_2. \tag{9} $$

and define the energy of a solution of (1) – (7) by

$$ E(t) = \frac{1}{2} \int_{\Omega} \left[ |y'(x,t)|^2 + a(x) |\nabla(y(x,t)|^2 \right] dx + \frac{\xi}{2} \int_{\Gamma_2} \int_{0}^{1} |y(x,t - \tau \rho)|^2 d\rho \, d\sigma(x), \tag{10} $$

where

$$ a_2 \tau \mu_2 < \xi < a_2 \tau (2\mu_1 - \mu_2), \tag{11} $$

We show that if {$\Omega, \Gamma_0, \Gamma_1, \Gamma_2$} satisfies (H.1), (H.2) and (H.3), then there is an exponential decay rate for $E(t)$. The proof of this result combines multipliers technique and compactness-uniqueness arguments.

The main result of this paper can be stated as follows.

**Theorem 1.** Assume (H1), (H.2), (H.3), (8) and (9). Then there exist constants $M \ge 1$ and $\omega > 0$ such that

$$ E(t) \leq M e^{-\omega t} E(0). $$

Theorem 1 is proved in Section 3. In Section 2, we investigate the well-posedness of system (1) – (7) using semigroup theory.

## 2 Well-posedness of problem (1) – (7)

Inspired from [5] and [6], we introduce the auxiliary variable $z(x, \rho, t) = y(x, t - \tau \rho)$. With this new unknown, problem (1) – (7) is equivalent to

$$ y''(x, t) - a(x) \Delta y(x, t) = 0 \quad \text{in } \Omega \times (0, +\infty), \tag{12} $$

$$ y(x, 0) = y^0(x), y'(x, 0) = y^1(x) \quad \text{in } \Omega, \tag{13} $$

$$ y(x, t) = 0 \quad \text{on } \Gamma_1 \times (0, +\infty), \tag{14} $$

$$ \frac{\partial z(x, \rho, t)}{\partial t} + \frac{1}{\tau} \frac{\partial z(x, \rho, t)}{\partial \rho} = 0 \quad \text{on } \Gamma_2 \times (0, +\infty) \tag{15} $$

$$ \frac{\partial y_2(x, t)}{\partial \nu} = -\mu_1 y_2'(x, t) - \mu_2 z(x, 1, t) \quad \text{on } \Gamma_2 \times (0, +\infty), \tag{16} $$

$$ y_1(x, t) = y_2(x, t) \quad \text{on } \Gamma_0 \times (0, +\infty), \tag{17} $$

$$ a_1 \frac{\partial y_1(x, t)}{\partial \nu} = a_2 \frac{\partial y_2(x, t)}{\partial \nu} \quad \text{on } \Gamma_0 \times (0, +\infty), \tag{18} $$

$$ z(x, 0, t) = y'(x, t) \quad \text{on } \Gamma_2 \times (0, +\infty) \tag{19} $$

$$ z(x, \rho, 0) = f_0(x, -\tau \rho) \quad \text{on } \Gamma_2 \times (0, 1) \tag{20} $$

Now, we endow the Hilbert space

$$ \mathcal{H} = H_{\Gamma_1}^1(\Omega) \times L^2(\Omega) \times L^2(\Gamma_2; L^2(0, 1)) $$
with the inner product

$$ \left\langle \begin{pmatrix} u \\ v \\ z \end{pmatrix} ; \begin{pmatrix} \bar{u} \\ \bar{v} \\ \bar{z} \end{pmatrix} \right\rangle = \int_{\Omega} (a(x)\nabla u(x)\nabla \bar{u}(x) + v(x)\bar{v}(x)) dx + \xi \int_{\Gamma_2} \int_0^1 z(x, \rho)\bar{z}(x, \rho) d\rho d\sigma(x) $$

and define a linear operator in $\mathcal{H}$ by

$$ D(A) = \{(u, v, z)^T \in H^2(\Omega_1, \Omega_2, \Gamma_1) \times H_{\Gamma_1}^1(\Omega) \times L^2(\Gamma_2; H^1(0, 1)); \\ \frac{\partial u}{\partial \nu} = -\mu_1 v - \mu_2 z(\cdot, 1), v = z(\cdot, 0) \text{ on } \Gamma_2 \} \quad (21) $$

$$ A \begin{pmatrix} u \\ v \\ z \end{pmatrix} = \begin{pmatrix} v \\ a(x)\Delta u \\ -\tau^{-1} \frac{\partial z}{\partial \rho} \end{pmatrix} \quad (22) $$

The spaces used for the definition of $\mathcal{H}$ and $D(A)$ are

$$ H_{\Gamma_1}^1(\Omega) = \{u \in H^1(\Omega) : u = 0 \text{ on } \Gamma_1\} \\ H^2(\Omega_1, \Omega_2, \Gamma_1) = \{u_i \in H^2(\Omega_i) : u = 0 \text{ on } \Gamma_i, u_1 = u_2 \text{ and } a_1 \frac{\partial u_1}{\partial \nu} = a_2 \frac{\partial u_2}{\partial \nu} \text{ on } \Gamma_0\} $$

Then we can rewrite (12) – (20) as an abstract Cauchy problem in $\mathcal{H}$

$$ \left\{ \begin{array}{l} \frac{d}{dt} Y(t) = AY(t) \\ Y(0) = Y_0 \end{array} \right. \qquad (23) $$

where

$$ Y(t) = (y, y', z)^T \text{ and } Y_0 = (y_0, y_1, f_0(\cdot, -\tau))^T $$

**Proposition 1.** The operator $A$ defined by (21) and (22) generates a strongly continuous semigroup on $\mathcal{H}$. Thus, for every $Y_0 \in \mathcal{H}$, problem (23) has a unique solution $Y$ whose regularity depends on the initial datum $Y_0$ as follows:

$$ Y(\cdot) \in C([0, +\infty); \mathcal{H}) \text{ if } Y_0 \in \mathcal{H}, \\ Y(\cdot) \in C([0, +\infty); D(A)) \cap C^1([0, +\infty); \mathcal{H}) \text{ if } Y_0 \in D(A). $$

$$ \textit{Proof.} \text{ Let } Y = \begin{pmatrix} u \\ v \\ z \end{pmatrix} \in D(A). \text{ Then} $$

$$ \langle AY, Y \rangle = \int_{\Omega} a(x) \nabla u(x) \cdot \nabla v(x) dx + \int_{\Omega} (a(x) \Delta u(x)) v(x) dx - \frac{\xi}{\tau} \int_{\Gamma_2} \int_0^1 z_{\rho}(x, \rho) z(x, \rho) d\rho d\Gamma \quad (24) $$
Applying Green's first theorem, we obtain

$$
\begin{align}
\int_{\Omega} (a(x)\Delta u(x))v(x)dx &= a_1 \int_{\Gamma_1} v(x) \frac{\partial u(x)}{\partial \nu}d\Gamma - a_1 \int_{\Omega_1} \nabla u(x) \cdot \nabla v(x)dx + \nonumber \\
&\quad a_2 \int_{\Gamma_2} v(x) \frac{\partial u(x)}{\partial \nu}d\Gamma - a_2 \int_{\Omega_2} \nabla u(x) \cdot \nabla v(x)dx \nonumber \\
&= a_2 \int_{\Gamma_2} v(x)\{-\mu_1v(x) - \mu_2z(x,1)\}d\Gamma - \int_{\Omega} a(x)\nabla u(x) \cdot \nabla v(x)dx \tag{25}
\end{align}
$$

Integrating by parts in $\rho$, we get

$$
\int_{\Gamma_2} \int_0^1 z_{\rho}(x, \rho) z(x, \rho) d\rho d\Gamma = \frac{1}{2} \int_{\Gamma_2} \{z^2(x, 1) - z^2(x, 0)\} d\Gamma \quad (26)
$$

Inserting (25) and (26) into (24) results in

$$
\begin{align*}
\langle AY, Y \rangle = -a_2\mu_1 \int_{\Gamma_2} v^2(x) d\Gamma - a_2\mu_2 \int_{\Gamma_2} v(x) z(x, 1) d\Gamma - \\
\frac{\xi}{2\tau} \int_{\Gamma_2} z^2(x, 1) d\Gamma + \frac{\xi}{2\tau} \int_{\Gamma_2} v^2(x) d\Gamma
\end{align*}
$$

from which follows using the Cauchy-Schwarz inequality

$$
\langle AY, Y \rangle \le -(a_2\mu_1 - \frac{a_2\mu_2}{2} + \frac{\xi}{2\tau}) \int_{\Gamma_2} v^2(x) d\Gamma - (\frac{\xi}{2\tau} - \frac{a_2\mu_2}{2}) \int_{\Gamma_2} z^2(x, 1) d\Gamma \quad (27)
$$

(27) implies that

$\langle AY, Y \rangle \le 0$

Thus A is dissipative.

Now we show that for a fixed $\lambda > 0$ and $(g, h, k)^T \in \mathcal{H}$, there exists $Y = (u, v, z)^T \in D(A)$ such that

$$
(\lambda I - A) Y = (g, h, k)^T
$$

or equivalently

$$
\lambda u - v = g \tag{28}
$$

$$
\lambda v - a(x) \Delta u = h \tag{29}
$$

$$
\lambda z + \frac{1}{\tau} z_{\rho} = k \qquad (30)
$$

Suppose that we have found $u$ with the appropriate regularity, then we can
determine $z$. Indeed, from (21) and (30) we have

$$
\begin{cases}
z_{\rho}(x, \rho) = -\lambda \tau z(x, \rho) + \tau k(x, \rho) \\
z(x, 0) = v(x)
\end{cases}
$$
The unique solution of the above initial value problem is

$$z(x, \rho) = e^{-\lambda \tau \rho} v(x) + \tau e^{-\lambda \tau \rho} \int_{0}^{\rho} e^{\lambda \tau s} k(x, s) ds$$

and in particular

$$z(x, 1) = \lambda e^{-\lambda \tau} u(x) + z_0(x), \quad x \in \Gamma_2$$

where

$$z_0(x) = -e^{-\lambda \tau} g(x) + \tau e^{-\lambda \tau} \int_{0}^{1} e^{\lambda \tau s} k(x, s) ds$$

By (28) and (29), the function $u$ satisfies

$$\lambda^2 u - a(x) \Delta u = h + \lambda g \quad (31)$$

Problem (31) can be reformulated as

$$\int_{\Omega} (\lambda^2 u - a(x) \Delta u) w dx = \int_{\Omega} (h + \lambda g) w dx, \quad w \in H_{\Gamma_1}^{1}(\Omega) \quad (32)$$

Using Green's first theorem and recalling (21), we express the right-hand side of (32) as follows

$$\int_{\Omega} (\lambda^2 u - a(x) \Delta u) w dx = \int_{\Omega} (\lambda^2 uw + a(x) \nabla u \cdot \nabla w) dx + a_2 \int_{\Gamma_2} \{\mu_1(\lambda u - g)w + \mu_2(\lambda e^{-\lambda \tau} u(x) + z_0(x))w\} d\Gamma$$

Therefore (32), can be rewritten as

$$\int_{\Omega} (\lambda^2 uw + a(x) \nabla u \cdot \nabla w) dx + a_2 \int_{\Gamma_2} (\mu_1 + \mu_2 e^{-\lambda \tau}) \lambda u w d\Gamma = \int_{\Omega} (h + \lambda g) w d\Gamma \\ + a_2 \mu_1 \int_{\Gamma_2} g w d\Gamma - a_2 \mu_2 \int_{\Gamma_2} z_0 w d\Gamma, \quad \forall w \in H_{\Gamma_1}^1(\Omega). \quad (33)$$

Since the left-hand side of (33) is coercive on $H_{\Gamma_1}^1(\Omega)$, the Lax-Milgram Theorem guarantees the existence and uniqueness of a solution $y \in H_{\Gamma_1}^1(\Omega)$ of (31). If we consider $w \in D(\Omega)$ in (28), then $y$ is a solution in $D'(\Omega)$ of

$$\lambda^2 u - a(x) \Delta u = h + \lambda g \quad (34)$$

and thus $\Delta u \in L^2(\Omega)$.

Combining (33) together with (34), we obtain after using Green's first theorem

$$a_2 \int_{\Gamma_2} (\mu_1 + \mu_2 e^{-\lambda \tau}) \lambda u w d\Gamma + a_2 \int_{\Omega} \frac{\partial u}{\partial \nu} w d\Gamma = a_2 \mu_1 \int_{\Gamma_2} g w d\Gamma - a_2 \mu_2 \int_{\Gamma_2} z_0 w d\Gamma$$

which implies that

$$\frac{\partial u(x)}{\partial \nu} = -\mu_1 v(x) - \mu_2 z(x, 1)$$

So, we have found $(u, v, z)^T \in D(A)$ which satisfies (28) – (30). Thus, by the Lumer-Phillips Theorem (see for instance [8], Theorem 1.4.3), generates a strongly continuous semigroup of contractions on $\mathcal{H}$.
### 3 Proof of Theorem 1

We prove Theorem 1 for smooth initial data. The general case follows by a standard density argument.

We proceed in several steps.

**Step 1.**

Since

$$E(t) = \frac{1}{2} \|(y, y', z)\|_{H}^{2}$$

Then, we deduce from the proof of Proposition 1 that $E(t)$ is non-increasing and

$$\frac{d}{dt} E(t) \leq -C \int_{\Gamma_2} \{y^2(x, t) + y'^2(x, t)\} d\Gamma \quad (35)$$

where

$$C = \min\{a_2\mu_1 - \frac{a_2\mu_2}{2} + \frac{\xi}{2\tau}, \frac{\xi}{2\tau} - \frac{a_2\mu_2}{2}\}$$

**Step 2.**

Set

$$E(t) = \mathcal{E}(t) + E_d(t)$$

where

$$\mathcal{E}(t) = \frac{1}{2} \int_{\Omega} \{a(x) |\nabla y(x, t)|^2 + |y'(x, t)|^2\} dx$$

and

$$E_d(t) = \frac{\xi}{2\tau} \int_{\Gamma_2} \int_{0}^{1} |y'(x, t - \tau \rho)|^2 d\rho d\Gamma$$

$E_d(t)$ can be rewritten via a change of variable as

$$E_d(t) = \frac{\xi}{2\tau^2} \int_t^{t+\tau} \int_{\Gamma_2} y'^2(x, s-\tau) d\Gamma ds \quad (36)$$

From (36), we obtain

$$E_d(t) \leq C_1 \int_0^T \int_{\Gamma_2} y'^2(x, s-\tau) d\Gamma ds \quad (37)$$

for $0 \le t \le T$ and $T$ large enough.

**Step 3.**

By applying energy methods (multiplier $2h \cdot \nabla y + (divh - \alpha)y$) (see the appendix) to problem (1) – (7), we obtain for all $T > 0$.

$$\int_{0}^{T} \mathcal{E}(t)dt \leq C_{2}(\mathcal{E}(0) + \mathcal{E}(T)) + C_{3} \int_{0}^{T} \int_{\Gamma_{2}} \left\{ \left( \frac{\partial y(x, t)}{\partial \nu} \right)^{2} + y'^{2}(x, t) \right\} d\Gamma dt + C_{4} \int_{0}^{T} \int_{\Gamma_{2}} |\nabla_{\sigma} y(x, t)|^{2} d\Gamma dt + C_{5} \int_{0}^{T} \int_{\Omega} |y(x, t)|^{2} d\Omega dt \quad (38)$$
where $\nabla_{\sigma} y$ is the tangential gradient of $y$.

**Step 4.**

We eliminate the tangential gradient from (38) by using the following estimate due to Lasiecka and Triggiani (Lemma 7.2 in [3])

$$ \int_{\epsilon}^{T-\epsilon} \int_{\Gamma_2} |\nabla_{\sigma} y(x, t)|^2 d\Gamma dt \le C_6 \left\{ \begin{aligned} & \int_0^T \int_{\Gamma_2} \left\{ \left( \frac{\partial y(x, t)}{\partial \nu} \right)^2 + y'^2(x, t) \right\} d\Gamma dt + \\ & \|y\|_{L^2(0,T;H^{1/2+\delta}(\Omega))}^2 \end{aligned} \right\} $$

where $\epsilon$ and $\delta$ are arbitrary positive constants. We obtain

$$ \int_{0}^{T} \mathcal{E}(t)dt \le C_2(\mathcal{E}(0) + \mathcal{E}(T)) + C_7 \int_{0}^{T} \int_{\Gamma_2} \left\{ \left( \frac{\partial y(x, t)}{\partial \nu} \right)^2 + y'^2(x, t) \right\} d\Gamma dt + C_8 \|y\|_{L^2(0,T;H^{1/2+\delta}(\Omega))}^2 \quad (39) $$

**Step 5.**

We differentiate $\mathcal{E}(t)$ with respect to $t$ and apply Green's first theorem. We obtain

$$ \frac{d}{dt} \mathcal{E}(t) = a_2 \int_{\Gamma_2} y'(x, t) \frac{\partial y(x, t)}{\partial \nu} d\Gamma dt \quad (40) $$

From (40), we get via the Cauchy-Schwarz inequality

$$ \mathcal{E}(0) \le \mathcal{E}(T) + \frac{a_2}{2} \int_0^T \int_{\Gamma_2} \left\{ y'^2(x, t) + \left( \frac{\partial y(x, t)}{\partial \nu} \right)^2 \right\} d\Gamma dt \quad (41) $$

Insertion of (41) into (39) yields

$$ \int_{0}^{T} \mathcal{E}(t)dt \le 2C_2\mathcal{E}(T) + C_9 \int_{0}^{T} \int_{\Gamma_2} \left\{ \left( \frac{\partial y(x, t)}{\partial \nu} \right)^2 + y'^2(x, t) \right\} d\Gamma dt + C_8 \|y\|_{L^2(0,T;H^{1/2+\delta}(\Omega))}^2 \quad (42) $$

**Step 6.**

Since $E(t)$ is non-increasing and $E(t) = \mathcal{E}(t) + E_d(t)$, then (42) together with (37) implies that

$$ TE(T) \le 2C_2\mathcal{E}(T) + C_9 \int_0^T \int_{\Gamma_2} \left\{ \left( \frac{\partial y(x, t)}{\partial \nu} \right)^2 + y'^2(x, t) \right\} d\Gamma dt + C_8 \|y\|_{L^2(0,T;H^{1/2+\delta}(\Omega))}^2 + TC_1 \int_0^T \int_{\Gamma_2} y'^2(x, t-\tau) d\Gamma dt \quad (43) $$

Thus invoking again the identity $E(t) = \mathcal{E}(t) + E_d(t)$ and recalling the boundary condition (4), we obtain from (43)

$$ E(T) \le C_{10} \int_{0}^{T} \int_{\Gamma_2} \{y'^2(x, t) + y'^2(x, t-\tau)\} d\Gamma dt + C_{11} \|y\|_{L^2(0,T;H^{1/2+\delta}(\Omega))}^2 \quad (44) $$
for $T$ large enough.

**Step 7.**

We drop the lower order term on the right-hand side of (44) by a compactness-uniqueness argument to obtain

$$E(T) \leq C_{12} \int_{0}^{T} \int_{\Gamma_{2}} \{y^{\prime 2}(x, t) + y^{\prime 2}(x, t - \tau)\} d\Gamma dt \quad (45)$$

**Step 8.**

The estimate (45) together with (35) yields

$$E(T) \leq \frac{C_{13}}{1 + C_{13}} E(0) \quad (46)$$

The desired conclusion follows now from (46) since the system (1) – (7) is invariant by translation.

## References

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2. Datko, R., Lagnese, J., Polis, M.P.: An example on the effect of time delays in boundary feedback stabilization of wave equations. SIAM J. Control Optim. 24, 152-156 (1986)

3. Lasiecka, I., Triggiani, R.: Uniform stabilization of the wave equation with Dirichlet or Neumann feedback control without geometrical conditions. Appl. Math. Optim. 25, 189-244 (1992)

4. Liu, W., Williams, G.H.: The exponential stability of the problem of transmission of the wave equation. Bull. Austra. Math. Soc. 97, 305-327 (1998)

5. Nicaise, S., Pignotti, C.: Stability and instability results of the wave equation with a delay term in the boundary or internal feedbacks. SIAM J. Control Optim. 45, 1561-1585 (2006)

6. Nicaise, N., Rebiai, S.E.: Stabilization of the Schrödinger equation with a delay term in boundary feedback or internal feedback. Portugal. Math. 68, 19-39 (2011).

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Appendix: Sketch of Proof of (38)

We multiply both sides of (1) by $2h \nabla y + (divh - \alpha)y$ and integrate over $(0, T) \times \Omega$.
We obtain

$$
\begin{align*}
& 2 \int_0^T \int_\Omega a(x) J \nabla y \cdot \nabla y d\Omega dt + \alpha \int_0^T \int_\Omega \{y'^2 - a(x) |\nabla y|^2\} d\Omega dt = \\
& - \int_\Omega \{2y'h \nabla y + (divh - \alpha)y'y\}_0^T d\Omega - \int_0^T \int_\Omega a(x) y \nabla y \cdot \nabla (divh - \alpha) d\Omega dt + \\
& a_1 \int_0^T \int_{\Gamma_1} \left| \frac{\partial y_1}{\partial \nu} \right|^2 h \cdot \nu d\Gamma dt - (a_1 - a_2) \int_0^T \int_{\Gamma_0} |\nabla y_1|^2 h \cdot \nu d\Gamma dt - \\
& \frac{(a_1 - a_2)^2}{a_2} \int_0^T \int_{\Gamma_0} \left| \frac{\partial y_1}{\partial \nu} \right|^2 h \cdot \nu d\Gamma dt + \int_0^T \int_{\Gamma_2} |y_2'|^2 h \cdot \nu d\Gamma dt + \\
& 2a_2 \int_0^T \int_{\Gamma_2} \left| \frac{\partial y_2}{\partial \nu} \right|^2 h \cdot \nabla y_2 d\Gamma dt - a_2 \int_0^T \int_{\Gamma_2} |\nabla y_2|^2 h \cdot \nu d\Gamma dt + \\
& a_2 \int_0^T \int_{\Gamma_2} \left| \frac{\partial y_2}{\partial \nu} \right|^2 (divh - \alpha) d\Gamma dt \tag{47}
\end{align*}
$$

after using the boundary conditions (3) and (5). Identity (47) is used together
with (H.1), (H.2), (H.3) and (8) to obtain estimate (38).