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\begin{document}

\vglue 20pt

\centerline{\bigbf Stabilization of a layered piezoelectric 3-D body}
\centerline{\bigbf by boundary dissipation}

\vglue .3in

\centerline{\bf B. Kapitonov\footnote{Sobolev Institute of Mathematics, Siberian Branch of Russian Academy of Sciences, Russia.\newline Visiting Researcher at the National Laboratory of Scientific Computation (LNCC/MCT), Brasil,\newline e-mail borisvk@lncc.br}, B. Miara\footnote{Laboratoire de Modélisation et Simulation numérique, École Superiore d'Ingénieurs en Électrotechnique et Électronique, 2 Boulevard Blaise Pascal, 93160 Noisy-le-Grand, France,\newline e-mail  miarab@esiee.fr} and G. Perla Menzala\footnote{National Laboratory of Scientific Computation LNCC/MCT, Rua Getulio Vargas 333, Quitandinha, Petropolis 25651-070, RJ, Brasil and Institute of Mathematics Federal University of Rio de Janeiro, RJ, P.O. 68530, Rio de Janeiro, RJ, Brasil, e-mail perla@lncc.br}}

\vglue .7in

\centerline{\bf Abstract}

\vglue .1in

\noindent We consider a linear coupled system of quasi-electrostatic equations which govern the evolution of a 3-D layered piezoelectric body. Assuming that a dissipative effect is effective at the boundary, we study the uniform stabilization problem. We prove that this is indeed the case, provided some geometric condition on the region and the interfaces hold. We also assume a monotonicity condition on the coefficients. As an application, we deduce exact controllability of the system with boundary control via a classical result due to D.L. Russell.

\newpage

\noindent{\bf \S1) Introduction}

\vglue .1in

The quasi-electrostatic equations are perhaps the most standard model for piezoelectricity. Here we consider the evolution problem of a piezoelectric structure whose 3-D displacement field $u = u(x,t) = (u_1,u_2,u_3)$ and scalar electric potential $\vr = \vr(x,t)$ is given by a model on a bounded domain $\Om$ of $\re^3$ with smooth boundary $\po\Om = S_0 \cup S_1$\,. Here $S_0$ and $S_1$ are two disjoint closed surfaces and the system read as follows
\begin{equation}
\left\{
\begin{aligned}
\,&u_{tt} - \vid T(u,\vr) = 0\\
\,&-\vid \D(u,\vr) = 0
\end{aligned}\right. \quad \text{in}\quad \Om\times(0,+\infty) \tag{1.1}
\end{equation}
where $u$ is the mechanical displacement and $\vr$ is the electric potential. Some classical references where such models were deduced are [2] and [3]. From now on summation convention with respect to repeated indices will be used. In this quasi-electrostatic piezoelectric system $T(u,\vr)$ is the mechanical stress, $\D$ is the electric displacement and $T$ and $\D$ satisfy the constitutive equations
\begin{align*}
&T^{ij}(u,\vr) = c_{ijk\ell}\,\ve_{k\ell}(u) + e_{kij}\, \frac{\po\vr}{\po x_k}\\
&\D^i(u,\vr) = -e_{ik\ell}\, \ve_{k\ell}(u) + d_{ij}\, \frac{\po\vr}{\po x_j}
\end{align*}
where the stress tensor $T = (T^{ij})$ and the electric displacement $\D = (\D^i)$ are related to the linearized deformation tensor $\ve(u) = (\ve_{ij}(u))$ whose components are given by $\ve_{k\ell}(u) = \dfrac 12 \left(\dfrac{\po u^\ell}{\po x_k}  + \dfrac{\po u^k}{\po x_\ell}\right)$. The 4th-order elasticity tensor $(c_{ijk\ell})$ is symmetric and positive, the 3rd-order coupling tensor $(e_{kij})$ is symmetric and the 2nd-order dielectric tensor $(d_{ij})$ is symmetric and positive.

\noindent In this article we prefer to rewrite the coupled system (1.1) in a more convenient form which will make more transparent our discussion of the so-called transmission problem. Let $A_i = [e_{k\ell i}]$, $D(x) = [d_{k\ell}(x)]$ and $A_{ij} = [a_{k\ell}^{ij}]$ be $3\times3$ matrices where $a_{k\ell}^{ij}$ is given by
$$
a_{kh}^{ij} = (1-\delta_{ih} \delta_{jk})c_{ikjh} + \delta_{ik}\delta_{jh}\, c_{ihjk}
$$
where $\delta_{ij}$ denotes the standard Kronecker delta. In these notations we have that
\begin{align*}
&\vid\left\{A_k\,\frac{\po u}{\po x_k}\right\} = \frac{\po}{\po x_i}\, \{e_{ik\ell}\,\ve_{k\ell}(u)\}\\
&\frac{\po}{\po x_i} \{A_i^*\nabla\vr\} = \frac{\po}{\po x_j}\,\left\{e_{k1j}\, \frac{\po\vr}{\po x_k}, e_{k2j}\, \frac{\po\vr}{\po x_k}, e_{k3j}\, \frac{\po\vr}{\po x_k}\right\}
\end{align*}
and
$$
\frac{\po}{\po x_i} \left\{A_{ij}\,\frac{\po u}{\po x_j}\right\} = \frac{\po}{\po x_j}\{c_{1jkh}\ve_{kh}(u), c_{2jkh} \ve_{kh}(u), c_{3jkh} \ve_{kh}(u)\}
$$
where $\nabla$ denotes the (spatial) usual gradient operator.

\noindent Using the above notation we rewrite system (1.1) in the form
\begin{equation*}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad \left\{
\begin{aligned}
\,&u_{tt} - \dfrac{\po}{\po x_i} \left\{A_{ij}\, \dfrac{\po u}{\po x_j} + A_i^* \nabla \vr\right\} = 0\qquad\qquad\qquad (1.2)\\
\,&\vid \left\{A_k\, \dfrac{\po u}{\po x_k} - D\nabla\vr\right\} = 0 \qquad\qquad\qquad\qquad\quad (1.3)
\end{aligned} \right.
\end{equation*}
in $\Om \times (0,+\infty)$. In (1.2) $A_i^*$ denotes the adjoint of $A_i$\,. As we mentioned above we are concerned with a transmission problem associated with system (1.2)-(1.3): We assume that the bounded domain $\Om = \O_0\backslash \ov{\O}_1$ where $\O_0$ and $\O_1$ are open bounded domains with $\ov{\O}_1 \subset \O_0$ where $\ov{\O}_1$ denotes the closure of $\O_1$\,, \, $\po\O_0 = S_0$ and $\po \O_1 = S_1$\,. Let us fix an integer $n > 1$ and $k=1,2,\dots,n$. For each $k$, let $B_k$ be an open subset with smooth boundary and such that $\ov{\O}_1 \subset B_k \subset \O_0$\,, \, $\ov B_k \subset B_{k+1}$\,. We set $\Om_0 = B_1\backslash \ov{\O}_1$ and $\Om_k = B_{k+1}\backslash \ov B_k$ for $k=1,2,\dots,n-1$. Also $\Om_n = \O_0\backslash \ov B_n$\,.

\noindent We consider problem (1.2)-(1.3) with initial conditions
\begin{equation}
u(x,0) = u_0(x), \quad u_t(x,0) = u_1(x) \tag{1.4}
\end{equation}
and boundary conditions
\begin{equation}
\left\{
\begin{aligned}
\,&\left(A_{ij}\,\frac{\po u}{\po x_j} + A_i^* \nabla\vr\right) \eta_i + \al(x)u_t = 0 \quad\text{on}\quad S_0 \times [0,+\infty)\\
\,&\vr = 0 \quad\text{on}\quad S_0 \times [0,+\infty)
\end{aligned} \right. \tag{1.5}
\end{equation}
\begin{equation}
\left\{
\begin{aligned}
\,&\left(A_k\,\frac{\po u}{\po x_k} - D\nabla\vr\right)\bullet \eta = 0 \quad\text{on}\quad S_1 \times [0,+\infty)\\
\,&u = 0 \quad\text{on}\quad S_1 \times [0,+\infty)
\end{aligned} \right.\qquad\qquad \tag{1.6}
\end{equation}
From now on the dot $\bullet$ denotes the usual inner product in $\re^3$.

\noindent Finally we require the transmission conditions
\begin{equation}
\left[D^{(m-1)}\nabla\vr^{(m-1)} - A_i \frac{\po u^{(m-1)}}{\po x_i}\right]\bullet \eta = \left[D^{(m)}\nabla\vr^{(m)} - A_i \frac{\po u^{(m)}}{\po x_i}\right]\bullet \eta \tag{1.7}
\end{equation}
\begin{equation}
u^{(m-1)} = u^{(m)}, \quad \vr^{(m-1)} = \vr^{(m)} \tag{1.8}
\end{equation}
\begin{equation}
\left[A_{ij}^{(m-1)}\, \frac{\po u^{(m-1)}}{\po x_j} + A_i^* \nabla \vr^{(m-1)}\right]\eta_i = 
\left[A_{ij}^{(m)}\, \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \vr^{(m)}\right] \eta_i\,. \tag{1.9}
\end{equation}
All these transmission conditions should hold at the interfaces $\Ga_m \times [0,+\infty)$,\linebreak $m=1,2,\dots,n$. From here on $\eta = (\eta_1,\eta_2,\eta_3)$ will always denote the unit normal vector pointing the exterior of $B_m$ or $\Om$ and $D^{(m)}$, $A_{ij}^{(m)}$, $\vr^{(m)}$ or $u^{(m)}$ are the restrictions of the corresponding matrices or functions on $\Om_m$\,.

Figures 1 and 2 illustrate simple such situations when $n=1$ or $n=3$.

\begin{figure}[h]
\begin{center}
\includegraphics{Graphic1.eps}
%\caption
\end{center}
\end{figure}

\noindent In order to mention the main result of this paper we give the assumptions on the coefficients or the matrices in problem (1.2)-(1.9):

\vglue .1in

\noindent{\bf HYPOTHESIS 1}

\vglue .05in

\noindent 1) We assume that the coefficients $c_{ijk\ell}$ and $d_{ij}$ (which are the cartesian components of the piezoelectric and electric permitivity tensors respectively) are $L^\infty(\Om)$ and satisfy the following assumptions

\newpage

$c_{ijk\ell} = c_{k\ell ij} = c_{jik\ell}$\,, \quad $d_{ij} = d_{ji}$

$d_{ij}\xi_j\xi_i \ge d_0|\xi|^2$ for some $d_0 > 0$ and vector $\xi = (\xi_1,\xi_2,\xi_3) \in \re^3$

$c_{ijk\ell}\,\la_{k\ell}\,\la_{ij} \ge c_0\,\la_{ij}\,\la_{ij}$ for some $c_0 > 0$ and any real symmetric tensor $[\la_{ij}]$ of order 3.

\vglue .1in

\noindent 2) The $3\times3$ matrices $A_{ij}(x) = [a_{k\ell}^{ij}(x)]$ (which satisfy $A_{ij}^* = A_{ji}$ due to the symmetry of $c_{ijk\ell}$) verify
$$
A_{ij}\,r_j\bullet r_i \ge c_1 \sum_{i=1}^3 |r_i|^2
$$
for some $c_1 > 0$ and any vector $r_i = (r_i^1, r_i^2, r_i^3) \in \re^3$.

\vglue .1in

\noindent 3) The matrices $A_i = [e_{k\ell i}]_{3\times3}$ are constant matrices with $e_{k\ell i} = e_{ki\ell}$, \, $D = [d_{k\ell}(x)]$ with $d_{ij} = d_{ij}(x)$ as well as $c_{ijk\ell} = c_{ijk\ell}(x)$ are piecewise constant functions which lose continuity only on $\Ga_1,\Ga_2,\dots,\Ga_n$\,.

\vglue .1in

\noindent 4) \, $\al(x) > 0$ \,\, $\forall\,x \in S_0$\,, \,\, $\al \in C(S_0)$.

\vglue .05in

\noindent{\bf Observation 1}. For a linear material with Saint-Venant Kirchhoff mechanic behaviour the terms $c_{ijkh}$ are given by
$$
c_{ijkh} = \la\,\delta_{ij}\delta_{kh} + \mu(\delta_{ik}\delta_{jh} + \delta_{ih}\delta_{jk})
$$
where $\la$ and $\mu$ are constants such that $\la+\mu \ge 0$ and $\mu > 0$. In this situation, item 2) of Hypothesis 1 holds with the constant $c_1 = \mu$. In fact, in this case
$$
A_{ij}\,v_j\bullet v_i = (\la+\mu) \left(\sum_{i=1}^3 v_i^i\right)^2 + \mu \sum_{i,j=1}^3 (v_i^j)^2 \ge \mu \sum_{i=1}^3 |v_i|^2.
$$
Assuming Hypothesis 1, we consider the total energy of the structure $E(t)$ associated with problem (1.2)-(1.9) is given by
\begin{equation}
E(t) = \sum_{m=0}^n \int_{\Om_m} \left\{|u_t^{(m)}|^2 + A_{ij}^{(m)} \, \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i} + D^{(m)}\nabla \vr^{(m)} \bullet \nabla\vr^{(m)}\right\} dx \tag{1.10}
\end{equation}
Formal calculations show that for every (smooth) solution of problem (1.2)-(1.9) the following identity holds
\begin{equation}
\frac{dE}{dt} + 2 \int_{S_0} \al(x)|u_t|^2\,d\Ga = 0 \qquad \forall\, t \ge 0 \tag{1.11}
\end{equation}
where the integral $\displaystyle \int_{S_0} \al|u_t|^2\,d\Ga$ means the surface integral of $\al|u_t|^2$ over the surface $S_0$\,.

\noindent Observe that when the structure is totally clamped, that is, when $S_1 =\po\Om$ then the energy is constant along a trajectory. This case was considered by B. Miara in [10]. The main result of this article shows that the total energy given by (1.10) decays exponentially to zero as $t \to +\infty$ provided suitable geometric conditions are imposed on $\Om$ and $\Ga_m$ and monotonicity assumptions on the coefficients of the system. The need for the above requirements were already noticed by J.-L. Lions in [9] in the treatment of certain transmission problems. Later on, J.E. Lagnese [7] also used those type of assumptions to prove controllability results for a class of second order hyperbolic problems.

\noindent There are a large number of contributions concerning piezoelectric equations and/or quasi-electrostatic equations (see [10], [1] and the references therein). However, as far as we know, a transmission problem for such class of equations was treated only for similar systems (see [6], [4] and the references therein). Uniform stabilization results for model (1.2)-(1.9) are interesting while studying exact controllability because give us an explicit expression of the feedback control instead of the difficult computation of an exact control. In a forthcoming article [5] we consider the case when no dissipation is included (that is, $\al \equiv 0$). In that case we obtain a ``boundary observation" inequality and use the HUM to solve the exact controllability problem.

\noindent Let us briefly describe the sections of this paper: Solvability of the initial boundary value problem (1.2)-(1.9) in the appropriate class of functions is outlined in Section 2. This is done via semigroup theory. In Section 3 we prove the exponential decay of the energy via the multiplier method. At this point, we needed to assume suitable geometric conditions on $\Om$, the interfaces $\Ga_m$ as well as monotonicity assumptions on the coefficients of the system.

\noindent We use standard notations, for example $H^r(\Om)$ or $H^s(\po\Om)$ will denote the Sobolev spaces of order $r$ and $s$ on $\Om$ and $\po\Om$ respectively. The norm of a vector $v \in \re^3$ will be denote by $|v|$. Given a real-valued function $g$ the notation $\displaystyle \int_S g\,d\Ga$ means ``the surface integral of $g$ over the surface $S$\,".

\vglue .3in

\noindent{\bf \S2) Well-posedness}

\vglue .1in

In this section we outline the function spaces where the solution pair $\{u,\vr\}$ of problem (1.2)-(1.9) is considered. In order to obtain the main results in the next section it is sufficient to work with smooth solutions.

\vglue .1in

Let $\Om$ be a bounded region as in the introduction. In $\Om$ we consider the following problem
\begin{equation*}
\qquad\qquad \left\{
\begin{aligned}
\,&\vid(D\nabla\vr) = \vid F \quad\text{in}\quad \Om_m \quad\qquad\quad m=0,1,\dots,n\qquad\qquad\qquad\qquad (2.1)\\
\,&\vr = 0 \text{ on } S_0\,, \, D\nabla\vr\bullet\eta = F\,\bullet\,\eta \text{ on } S_1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\, (2.2)\\
\,&\vr^{(m-1)} = \vr^{(m)} \quad\text{on}\quad \Ga_m\,, \,\,m=1,2,\dots,n \qquad\qquad\qquad\quad\qquad\qquad\qquad\,\, (2.3)\\
\,&D^{(m-1)} \nabla \vr^{(m-1)}\bullet \eta - D^{(m)}\nabla \vr^{(m)}\bullet \eta = F^{(m-1)}\bullet \eta - F^{(m)}\bullet\eta \text{ on } \Ga_m \qquad (2.4)
\end{aligned}\right. 
\end{equation*}
where $F=(F_k)$ is a given function belonging to $[H^1(\Om_m)]^3$. By elliptic theory, there exists a unique solution $\vr$ of (2.1)-(2.4) which we denote by $\vr = \be(F)$. Denote by $X$ the real Hilbert space of pairs $\{u,v\}$ of three-component vector-valued functions such that\linebreak $v^{(m)} \in [L^2(\Om_m)]^3$, \, $u^{(m)} \in [H^2(\Om_m)]^3$ and $u=0$ on $S_1$\,. The inner product in $X$ is given by 
\begin{align*}
\lg W,W_1\rg_X &= \sum_{m=0}^n \int_{\Om_m} \bigg\{v_1\bullet v_2 + A_{ij} \frac{\po u_1}{\po x_j} \bullet \frac{\po u_2}{\po x_i} +\\
&\quad + D\nabla\be\left(A_k \frac{\po u_1}{\po x_k}\right)\bullet \nabla\be\left(A_k \frac{\po u_2}{\po x_k}\right)\bigg\} dx
\end{align*}
whenever $W = (u_1,v_1)$ and $W_1 = (u_2,v_2)$ belong to $X$. We will denote by $||\,\cdot\,||_X$ the norm in $X$. In $X$ we define the unbounded operator $\A$ with domain $\D(\A)$ which consists of all the elements $(u,v) \in X$ such that

\pagebreak

\begin{equation}
\left\{
\begin{aligned}
\,&v^{(m)} \in [H^1(\Om_m)]^3, \quad v\big|_{S_1} = 0\\
\,&\left[A_{ij} \frac{\po u}{\po x_j} + A_i^*\nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\right]\eta_i + \al v = 0 \text{ on } S_0\\
\,&u^{(m-1)} = u^{(m)}, \quad v^{(m-1)} = v^{(m)} \text{ on } \Ga_m\\
\,&A_{ij}^{(m-1)} \frac{\po u^{(m-1)}}{\po x_j} + A_i^* \nabla\be\left(A_k \frac{\po u^{(m-1)}}{\po x_k}\right) \eta_i =\\
\,&\quad = A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \be\left(A_k \frac{\po u^{(m)}}{\po x_k}\right)\eta_i \text{ on } \Ga_m
\end{aligned} \right. \tag{2.5}
\end{equation}
$m=1,2,\dots,n$. In the domain of $\A$, the operator is given by
$$
\A(u,v) = \left(v, \frac{\po}{\po x_i} \left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\right\}\right).
$$

\noindent{\bf Lemma 2.1}. Assume Hypothesis 1 given in the introduction, then the operator $\A$ is dissipative, that is $\lg \A(W),W\rg_X \le 0$ \,\, $\forall\, W \in \D(\A)$.

\vglue .1in

\noindent{\bf Proof}. Let $W = (u,v) \in \D(\A)$, then 
\begin{align*}
\lg \A(W),W\rg_X &= \bigg\lg \left(v, \frac{\po}{\po x_i} \left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla \be \left(A_k \frac{\po u}{\po x_k}\right)\right\}\right), (u,v)\bigg\rg_X\\
&= \sum_{m=0}^n \int_{\Om_m} \bigg[\left(\frac{\po}{\po x_i} \left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla \be \left(A_k \frac{\po u}{\po x_k}\right)\right\}\right)\bullet v +\\
&+ A_{ij} \frac{\po v}{\po x_j} \bullet \frac{\po u}{\po x_i} + D\nabla \be \left(A_k \frac{\po v}{\po x_k}\right)\bullet \nabla \be \left(A_k \frac{\po u}{\po x_k}\right)\bigg]dx \tag{2.6}
\end{align*}
Using the boundary and interface conditions we get
\begin{align*}
&\sum_{m=0}^n \int_{\Om_m} \bigg[\left(\frac{\po}{\po x_i} \left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla \be \left(A_k \frac{\po u}{\po x_k}\right)\right\}\right)\bullet v\,dx = \\
&= -\sum_{m=0}^n \int_{\Om_m} A_{ij} \frac{\po u}{\po x_j} \bullet \frac{\po v}{\po x_i}\,dx - \sum_{m=0}^n \int_{\Om_m} A_i^* \nabla \be \left(A_k \frac{\po u}{\po x_k}\right)\bullet \frac{\po v}{\po x_i}\,dx -\\
&- \int_{S_0} \al|v|^2\,d\Ga. \tag{2.7}
\end{align*}
Observe that the following identities are valid
\begin{align*}
&\sum_{m=0}^n \int_{\Om_m} \left\{ D\nabla \be \left(A_k \frac{\po v}{\po x_k}\right)\bullet \nabla \be \left(A_k \frac{\po u}{\po x_k}\right) - A_i^* \nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\bullet \frac{\po v}{\po x_i}\right\} dx =\\
&= \sum_{m=0}^n \int_{\Om_m} \left(\nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\bullet \left(D\nabla\be \left(A_k \frac{\po v}{\po x_k}\right) - A_k \frac{\po v}{\po x_k}\right)\right] dx =\\
&= \sum_{m=1}^n \int_{\Ga_m} \be\left(A_k \frac{\po u}{\po x_k}\right) \left\{D^{(m-1)}\nabla\be^{(m-1)} \left(A_k \frac{\po v}{\po x_k}\right) - A_k \frac{\po v^{(m-1)}}{\po x_k}\right\}\bullet \eta\, d\Ga -\\
&- \sum_{m=1}^n \int_{\Ga_m} \left[D^{(m)} \nabla \be^{(m)} \left(A_k \frac{\po v}{\po x_k}\right) - A_k \frac{\po v}{\po x_k}\right]\bullet \eta \, d\Ga +\\
&+ \int_{S_1} \be\left(A_k \frac{\po u}{\po x_k}\right) \left[D\nabla\be\left(A_k \frac{\po v}{\po x_k}\right) - A_k \frac{\po v}{\po x_k}\right]\bullet\eta\,d\Ga -\\
&- \sum_{m=0}^m \int_{\Om_m} \be\left(A_k \frac{\po u}{\po x_k}\right)\left\{\vid\left(D\nabla\be \left(A_k \frac{\po v}{\po x_k}\right)\right) - \vid\left(A_k \frac{\po v}{\po x_k}\right) \right\} dx.
\end{align*}
From (2.1)-(2.4) it follows that
\begin{align*}
&\sum_{m=0}^n \int_{\Om_m} \bigg\{\left(D\nabla\be \left(A_k \frac{\po v}{\po x_k}\right)\bullet \left(\nabla\be \left(A_k \frac{\po u}{\po x_k}\right)\right)\right) -\\
&\qquad - \left(A_i^* \nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\bullet \frac{\po v}{\po x_i}\right)\bigg\} dx = 0 \tag{2.8}
\end{align*}
Consequently, from (2.4)-(2.8) we obtain that 
$$
\lg \A(u,v), (u,v)\rg_X = -\int_{S_0} \al|v|^2\,d\Ga \le 0.
$$
Now, we consider the adjoint operator $\A^*$. It can be verified that the domain of $\A^*$ consists of all elements $(u,v) \in X$ satisfying (2.5) except that we should change $\al(x)$ by $-\al(x)$. For $(u,v) \in \D(\A^*)$ we have
$$
\A^*(u,v) = -\left(v, \frac{\po}{\po x_i} \left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla\be\left(A_k \frac{\po u}{\po x_k}\right)\right\}\right).
$$
We can verify that $\A^*$ is dissipative. Obviously $\A$ is closed and densely defined. It follows by a well known criteria (see A. Pazy [12], Corollary I.44) that the operator $\A$ generates a strongly continuous semigroup of contractions $\{U(t)\}_{t\ge0}$\,. Furthermore, for $f = (f_1,f_2) \in \D(\A)$ we have
$$
\frac{d}{dt}\, U(t)f = \A U(t)f\, , \quad U(0)f = f
$$
and $U(t)f$ is the unique (strong) solution of problem (1.2)-(1.9). Let $f = (f_1,f_2) \in X$, \, $f^{(n)} = \big(f_1^{(n)}, f_2^{(n)}\big) \in \D(\A)$ such that $\lim\limits_{n\to+\infty} ||f-f^{(n)}||_X=0$. Then, $U(t)f^{(n)}$ satisfies the following identity
\begin{equation}
\int_0^T \left\{\bigg\lg U(t)f^{(n)}, \frac{d\psi}{dt}\bigg\rg_X + \lg U(t)f^{(n)}, \A^*\psi\rg_X\right\}dt = -\lg f^{(n)}, \psi(0)\rg_X \tag{2.9}
\end{equation}
where $\psi \in L^2(0,T;\D(\A^*))$, \, $\psi_t \in L^2(0,T;X)$ with $\psi(T)=0$. Passing to the limit in (2.9) we obtain
$$
\int_0^T \left\{\bigg\lg U(t)f, \frac{d\psi}{dt}\bigg\rg_X + \lg U(t)f, \A^*\psi\rg_X\right\}dt = -\lg f, \psi(0)\rg_{X_0} 
$$
that is, $U(t)f$ is the weak solution in $X$ of the abstract initial-boundary value problem
$$
W_t = \A(W), \quad W(0) = f
$$


\noindent{\bf \S3) Stabilization}

\vglue .1in

In this section we prove the boundary stabilization result. The proof is based on the theory of multipliers and it is motivated by the invariance of system (1.2)-(1.3) with constant coefficients relative to the one-parameter group of dilations in all variables. A good reference for the use of this technique is Komornik's book [6]. The multipliers have to be conveniently modified in such a way the extra boundary terms appearing in the identities can be estimated by appropriate bounds. Let $g = g(x)$ be an auxiliary scalar smooth function on $\ov\Om$ which we will choose later. Let us fix $t_0 > 0$ and consider the multiplier $L_1$ given by
\begin{equation}
L_1u = (t+t_0)u_t + (\nabla g\bullet\nabla)u + u \tag{3.1}
\end{equation}
where $\nabla = \left(\dfrac{\po}{\po x_1}, \dfrac{\po}{\po x_2}, \dfrac{\po}{\po x_3}\right)$,
$$
\nabla g \bullet \nabla = \frac{\po g}{\po x_1}\,\frac{\po}{\po x_1} + \frac{\po g}{\po x_2}\, \frac{\po}{\po x_2} + \frac{\po g}{\po x_3}\, \frac{\po}{\po x_3}
$$
and $u = (u_1,u_2,u_3)$. Let $L_2$ the operator
\begin{equation}
L_2\vr = (t+t_0)\vr\,\frac{\po}{\po t} - \nabla g \cdot \nabla\vr. \tag{3.2}
\end{equation}
We take the inner product (in $\re^3$) of $L_1u$ with equation (1.2) and apply the operator $L_2\vr$ to equation (1.3). Since $\{u,\vr\}$ is a (smooth) solution of (1.2)-(1.9) then adding the identities we obtain
\begin{equation}
\frac{\po F}{\po t} = \vid\,G + \frac{\po H_i}{\po x_i} + J. \tag{3.3}
\end{equation}
where
\begin{equation}
F = (t+t_0) \left[|u_t|^2 + A_{ij} \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} + D\nabla\vr\bullet\nabla\vr\right] + 2\,u_t\bullet\big[(\nabla g\bullet\nabla)u + u\big] \tag{3.4}
\end{equation}
\begin{align*}
G &= 2(t+t_0)\vr\,\frac{\po}{\po t} \left\{D\nabla\vr -A_k \frac{\po u}{\po x_k}\right\} + \nabla g D\nabla\vr\bullet\nabla\vr\\
&\quad -2(\nabla\vr\bullet\nabla g)D\nabla\vr - 2\nabla g \, A_k \frac{\po u}{\po x_k} \bullet \nabla\vr +\\
&\quad + 2(\nabla g \bullet \nabla \vr)A_k\,\frac{\po u}{\po x_k} \tag{3.5}
\end{align*}
\begin{align*}
H_i &= 2\left\{(t+t_0)u_t + (\nabla g\bullet\nabla)u+u\right\}\bullet\left\{A_{ij} \frac{\po u}{\po x_j} + A_i^* \nabla\vr\right\}\\
&\quad + \frac{\po g}{\po x_i}\, \left[|u_t|^2 - A_{pq}\, \frac{\po u}{\po x_q} \bullet \frac{\po u}{\po x_p}\right] \tag{3.6}
\end{align*}
and
\begin{align*}
J &= (\Delta g-1)A_{ij} \frac{\po u}{\po x_j}\bullet \frac{\po u}{\po x_i} - 2\,\frac{\po^2g}{\po x_p\po x_i} A_{ij} \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_p} +\\
&\quad + (3-\Delta g)|u_t|^2 + 2\,\frac{\po^2g}{\po x_i \po x_k}\,d_{ij}\, \frac{\po\vr}{\po x_j}\, \frac{\po\vr}{\po x_k} +\\
&\quad +(1-\Delta g) D\nabla\vr\bullet\nabla\vr \\
&\quad +2\nabla\vr\bullet\left\{(\Delta g-1)A_k \frac{\po u}{\po x_k} - \frac{\po^2g}{\po x_i \po x_k} A_k \frac{\po u}{\po x_i} - \left(A_k \frac{\po u}{\po x_k} \bullet \nabla\right)\nabla g\right\}.
\tag{3.7}
\end{align*}

\noindent{\bf Observation 2}. If we consider $g(x) = \dfrac 12\,|x-x_0|^2$ for some fixed $x_0 \in \re^3$ then $J \equiv 0$. In this case (3.3) will be a conservation law. However, due to the expressions of $G$ and $H_i$ we would require to have a definite sign for $\dfrac{\po g}{\po \eta}\,\cdot$ Thus, later on we will choose $g$ as an ``small" perturbation of $\dfrac 12\,|x-x_0|^2$. Let $\{u,\vr\}$ be an smooth solution of (1.2)-(1.9). Integration over $\Om_m \times (0,T)$ of identity (3.3) and summation over $m$ implies that
\begin{align*}
(T+t_0)E(T) &+ 2 \sum_{m=0}^n \int_{\Om_m} u_t^{(m)}\bullet\left\{(\nabla g\bullet\nabla)u^{(m)} + u^{(m)}\right\}dx\bigg]_{t=0}^{t=T} =\\
&= t_0\,E(0) + \sum_{m=1}^n \int_0^T \int_{\Ga_m} (\pmb{V}_{m-1} - \pmb{V}_m)d\Ga\,dt +\\
&+ \int_0^T \int_{S_0} \pmb{V}_n\,d\Ga\,dt +  \int_0^T \int_{S_1} \pmb{V}_0\,d\Ga\,dt +\\
&+ \sum_{m=0}^n \int_0^T \int_{\Om_m} J_m(x,t)dxdt \tag{3.8}
\end{align*}
where $E(t)$ is given by (1.10), $J_m = J_m(u,\vr,g)$ denotes the restriction of $J$ (in (3.7)) to the region $\Om_m$ and 
\begin{align*}
V_m &= 2\left\{(t+t_0)u_t^{(m)} + (\nabla g\bullet\nabla)u^{(m)} + u^{(m)}\right\}\bullet \left\{\bigg(A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} 
+ A_i^*\nabla\vr^{(m)}\bigg)\eta_i\right\} \\
&+ \frac{\po g}{\po\eta} \left(|u_t^{(m)}|^2 - A_{pq}^{(m)} \frac{\po u^{(m)}}{\po x_q} \bullet \frac{\po u^{(m)}}{\po x_p}\right)\\
&+ \frac{\po g}{\po\eta}\, D^{(m)}\nabla\vr^{(m)}\bullet\nabla\vr^{(m)} + 2(t+t_0)\vr^{(m)}\left(D^{(m)}\nabla \vr_t^{(m)} - A_k \frac{\po^2u^{(m)}}{\po x_k \po t}\right)\bullet\eta\\
&- 2(\nabla\vr^{(m)}\bullet\nabla g)\left(D^{(m)}\nabla \vr^{(m)} - A_k\,\frac{\po u^{(m)}}{\po x_k}\right)\bullet\eta\\
&- 2\, \frac{\po g}{\po\eta} \left(A_k \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m)}\right) \tag{3.9}
\end{align*}
Here $\dfrac{\po g}{\po\eta}$ denotes the normal derivative of $g$ at $x \in \Ga_m$ (or $S_0$, $S_1$). Next lemma tell us that the differences $V_{m-1}-V_m$ will have ``good" sign if we choose $g$ conveniently and assume a monotonicity condition on $\{A_{ij}^{(m)}\}$ and $\{D^{(m)}\}$:

%\vglue .2in

\noindent{\bf Lemma 3.1}. Let $\{u,\vr\}$ be an smooth solution of (1.2)-(1.9). Then, the identity
\begin{align*}
V_{m-1}-V_m &= -\frac{\po g}{\po\eta} \bigg\{\left(A_{ij}^{(m-1)} - A_{ij}^{(m)}\right) \frac{\po u^{(m-1)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} +\\
&+ A_{ij}^{(m)} \left[\frac{\po u^{(m)}}{\po x_j} - \frac{\po u^{(m-1)}}{\po x_j}\right]\bullet\left[\frac{\po u^{(m)}}{\po x_i} - \frac{\po u^{(m-1)}}{\po x_i}\right] +\\
&+ (D^{(m)} - D^{(m-1)}) \nabla \vr^{(m)} \bullet \nabla \vr^{(m)} +\\
&+D^{(m-1)}(\nabla\vr^{(m-1)} - \nabla\vr^{(m)})\bullet(\nabla\vr^{(m-1)} - \nabla\vr^{(m)}\bigg\} \tag{3.10}
\end{align*}
holds.

\vglue .2in

\noindent{\bf Proof}. The idea is to use the interface conditions (1.6)-(1.9).  In fact, direct calculations using (3.9) and the interfaces conditions imply that
\begin{align*}
V_{m-1}-V_m &= 2(\nabla g\bullet\nabla)(u^{(m-1)}-u^{(m)})\bullet\left(A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \vr^{(m)}\right)\eta_i\\
&- \frac{\po g}{\po\eta}\bigg\{\left(A_{pq}^{(m-1)} \frac{\po u^{(m-1)}}{\po x_q} \bullet \frac{\po u^{(m-1)}}{\po x_p}\right) - \left(A_{pq}^{(m)} \frac{\po u^{(m)}}{\po x_q} \bullet \frac{\po u^{(m)}}{\po x_p}\right)\\
&- (D^{(m-1)} \nabla \vr^{(m-1)} \bullet \nabla\vr^{(m-1)}) + (D^{(m)} \nabla \vr^{(m)} \bullet  \nabla\vr^{(m)})\\
&+2\left(A_k \frac{\po u^{(m-1)}}{\po x_k} \bullet \nabla\vr^{(m-1)}\right) - 
2\left(A_k \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m)}\right)\bigg\}\\
&-2\left(D^{(m)}\nabla \vr^{(m)} - A_k \frac{\po u^{(m)}}{\po x_k}\right)\bullet\eta \left(\nabla g\bullet \nabla \vr^{(m-1)} - \nabla g\bullet\nabla \vr^{(m)}\right) \tag{3.11}
\end{align*}
Now, we use the identity
\begin{align*}
&(\nabla g\bullet\nabla)(u^{(m-1)}-u^{(m)}) = \frac{\po g}{\po x_i} \left(\frac{\po u^{(m-1)}}{\po x_i} - \frac{\po u^{(m)}}{\po x_i}\right) =\\
&= \frac{\po g}{\po x_i}\, \eta_i \left(\frac{\po u^{(m-1)}}{\po\eta} - \frac{\po u^{(m)}}{\po\eta}\right) = \frac{\po g}{\po\eta} \left(\frac{\po u^{(m-1)}}{\po\eta} - \frac{\po u^{(m)}}{\po\eta}\right)
\end{align*}
in order to obtain
\begin{align*}
&2\left\{(\nabla g\bullet\nabla)(u^{(m-1)}-u^{(m)})\right\}\bullet\left\{\eta_i\left(A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \vr^{(m)}\right)\right\} \\
&= 2 \frac{\po g}{\po\eta} \left\{\left(\frac{\po u^{(m-1)}}{\po\eta} - \frac{\po u^{(m)}}{\po\eta}\right)\eta_i\bullet\left(A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \vr^{(m)}\right)\right\}\\
&= 2 \frac{\po g}{\po\eta} \left\{\left(\frac{\po u^{(m-1)}}{\po x_i} - \frac{\po u^{(m)}}{\po x_i}\right)\bullet\left(A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} + A_i^* \nabla \vr^{(m)}\right)\right\}\\
&= 2 \frac{\po g}{\po\eta} \bigg\{A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i}\\
&\quad\qquad\quad +A_i \frac{\po u^{(m-1)}}{\po x_i} \bullet \nabla\vr^{(m)} - A_i\, \frac{\po u^{(m)}}{\po x_i} \bullet \nabla \vr^{(m)}\bigg\} \tag{3.12}
\end{align*}
In a similar way we obtain

\vglue .2in

\begin{align*}
&\quad-2\left(D^{(m)}\nabla \vr^{(m)} - A_k \frac{\po u^{(m)}}{\po x_k}\right)\bullet\eta \left\{\nabla g\bullet\nabla\vr^{(m-1)} - \nabla g\bullet\nabla\vr^{(m)}\right\}\\
&= -2\left(D^{(m)}\nabla \vr^{(m)} - A_k \frac{\po u^{(m)}}{\po x_k}\right)\bullet\eta \left\{\nabla g \bullet \eta\left(\frac{\po\vr^{(m-1)}}{\po\eta} - \frac{\po\vr^{(m)}}{\po\eta}\right)\right\}\\
&= 2 \frac{\po g}{\po\eta} \left(D^{(m)}\nabla \vr^{(m)} - A_k \frac{\po u^{(m)}}{\po x_k}\right)\bullet(\nabla\vr^{(m)} - \nabla\vr^{(m-1)})\\
&= 2 \frac{\po g}{\po\eta} \bigg\{(D^{(m)}\nabla\vr^{(m)} \bullet \nabla\vr^{(m)}) - (D^{(m)}\nabla\vr^{(m)}\bullet\nabla\vr^{(m-1)})\\
&\qquad\qquad +A_k \frac{\po u^{(m)}}{\po x_k} \bullet \nabla \vr^{(m-1} - A_k \frac{\po u^{(m)}}{\po x_k} \bullet \nabla \vr^{(m)}\bigg\} \tag{3.13}
\end{align*}

\vglue .2in

From (3.11)-(3.13) we obtain the identity

\vglue .1in

\begin{align*}
V_{m-1}&-V_m = \frac{\po g}{\po\eta} \bigg\{2\,A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - A_{ij}^{(m)}\, \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i}\\
&- A_{ij}^{(m-1)}\, \frac{\po u^{(m-1)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - 
2\,D^{(m)}\nabla\vr^{(m)} \bullet \nabla \vr^{(m-1)}\\
&+ D^{(m)}\nabla\vr^{(m)}\bullet\nabla\vr^{(m)} - D^{(m-1)}\nabla\vr^{(m-1)}\bullet\nabla\vr^{(m-1)}\\
& + 2\,A_k\,\frac{\po u^{(m-1)}}{\po x_k} \bullet\nabla\vr^{(m)} 
 - 2\,A_k\, \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m)}\\
&+2\,A_k \, \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m-1)} - 2\,A_k\,\frac{\po u^{(m-1)}}{\po x_k} \bullet \nabla \vr^{(m-1)}\bigg\} \tag{3.14}
\end{align*}

\vglue .1in

Using the interface conditions (1.7) it follows that
$$
\left(A_k\,\frac{\po u^{(m)}}{\po x_k} - A_k\,\frac{\po u^{(m-1)}}{\po x_k}\right)\bullet\eta = \big(D^{(m)}\nabla\vr^{(m)} - D^{(m-1)}\nabla\vr^{(m-1)}\big)\bullet\eta.
$$

\noindent Therefore
\begin{align*}
&2\,\frac{\po g}{\po \eta} \bigg\{A_k\,\frac{\po u^{(m-1)}}{\po x_k} \bullet \nabla\vr^{(m)} - A_k\, \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m)} +\\
&\qquad\quad A_k\, \frac{\po u^{(m)}}{\po x_k} \bullet \nabla\vr^{(m-1)} - A_k\, \frac{\po u^{(m-1)}}{\po x_k} \bullet \nabla\vr^{(m-1)}\bigg\} =\\
&= 2\,\frac{\po g}{\po\eta} \left(A_k\,\frac{\po u^{(m)}}{\po x_k} - A_k\, \frac{\po u^{(m-1)}}{\po x_k}\right)\bullet\left(\nabla\vr^{(m-1} - \nabla\vr^{(m)}\right) =\\
&= 2\, \frac{\po g}{\po\eta} \left(A_k\, \frac{\po u^{(m)}}{\po x_k} - A_k\, \frac{\po u^{(m-1)}}{\po x_k}\right)\bullet\eta \left\{\frac{\po\vr^{(m-1)}}{\po\eta} - \frac{\po\vr^{(m)}}{\po\eta}\right\} =\\
&= 2\, \frac{\po g}{\po\eta} \left(D^{(m)}\nabla\vr^{(m)} - D^{(m-1)} \nabla \vr^{(m-1)}\right)\bullet\eta \left\{\frac{\po\vr^{(m-1)}}{\po\eta} - \frac{\po\vr^{(m)}}{\po\eta}\right\}=\\
&= 2\,\frac{\po g}{\po\eta} \left(D^{(m)}\nabla\vr^{(m)} - D^{(m-1)}\nabla\vr^{(m-1)}\right)\bullet \left\{\nabla\vr^{(m-1)} - \nabla\vr^{(m)}\right\}=\\
&= 2\,\frac{\po g}{\po\eta} \bigg\{-D^{(m)}\nabla\vr^{(m)} \bullet \nabla\vr^{(m)} + D^{(m)} \nabla\vr^{(m)} \bullet \nabla \vr^{(m-1}\\
&\qquad -D^{(m-1}\nabla\vr^{(m-1)}\bullet \nabla\vr^{(m-1)} + D^{(m-1)} \nabla\vr^{(m-1)}\bullet\nabla\vr^{(m)}\bigg\} \tag{3.15}
\end{align*}
From (3.14) and (3.15) we deduce that
\begin{align*}
V_{m-1}&-V_m = \frac{\po g}{\po\eta} \bigg\{2\,A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - A_{ij}^{(m)}\, \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i}\\
&- A_{ij}^{(m-1)}\, \frac{\po u^{(m-1)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - 
D^{(m)}\nabla\vr^{(m)} \bullet \nabla \vr^{(m)}\\
&- D^{(m-1)}\nabla\vr^{(m-1)}\bullet \nabla\vr^{(m-1)} + 2\,D^{(m-1)}\nabla\vr^{(m-1)}\bullet\nabla\vr^{(m)}\bigg\}. \tag{3.16}
\end{align*}
The conclusion of Lemma 3.1 follows from (3.16) observing the validity of the identities
\begin{align*}
A_{ij}^{(m)} \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i} &+ A_{ij}^{(m-1)}\, \frac{\po u^{(m-1)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} - 2\, A_{ij}^{(m)}\, \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i}\\
&= \left(A_{ij}^{(m-1)} - A_{ij}^{(m)}\right) \frac{\po u^{(m-1)}}{\po x_j} \bullet \frac{\po u^{(m-1)}}{\po x_i} +\\
&\quad + A_{ij}^{(m)} \left(\frac{\po u^{(m)}}{\po x_j} - \frac{\po u^{(m-1)}}{\po x_j}\right) \bullet \left(\frac{\po u^{(m)}}{\po x_i} - \frac{\po u^{(m-1)}}{\po x_i}\right)
\end{align*}
and
\begin{align*}
&D^{(m)}\nabla\vr^{(m)}\bullet \nabla\vr^{(m)} + D^{(m-1)} \nabla\vr^{(m-1)}\bullet \nabla\vr^{(m-1)} - 2D^{(m-1)}\nabla\vr^{(m-1)}\bullet\nabla\vr^{(m)} =\\
&= D^{(m-1)} \left(\nabla\vr^{(m-1)} - \nabla\vr^{(m)}\right)\bullet\left(\nabla\vr^{(m-1)}-\nabla\vr^{(m)}\right) + (D^{(m)} - D^{(m-1)})\nabla\vr^{(m)}\bullet\nabla\vr^{(m)} \qed
\end{align*}

\noindent Let us choose a convenient function $g(x)$: Let $\Phi(x)$ be a solution of the elliptic problem
\begin{equation}
\left\{
\begin{aligned}
\,&\Delta\Phi=1 \quad\text{in}\quad \Om\\
\,&\frac{\po\Phi}{\po\eta} = 2\, \frac{\text{Vol}(\Om)}{\text{area}(S_0)} \quad\text{on}\quad S_0\\
\,&\frac{\po\Phi}{\po\eta} = - \frac{\text{Vol}(\Om)}{\text{area}(S_1)} \quad\text{on}\quad S_1
\end{aligned} \right. \tag{3.17}
\end{equation}
which admits a solution $\Phi(x) \in C^2(\Om) \cap C^1(\ov\Om)$. Let $\delta > 0$, \, $x_0 \in \re^3$ (to be chosen later) and define
\begin{equation}
g(x) = \delta\,\Phi(x) + \frac 12\, |x-x_0|^2. \tag{3.18}
\end{equation}
Now we concentrate our discussion in estimating the term $\sum\limits_{m=0}^n \displaystyle \int_0^T \displaystyle \int_{\Om_m} J_m \,dxdt$ in (3.8).

\vglue .2in

\noindent{\bf Lemma 3.2}. Under the assumptions of Lemma 3.1\, Hypothesis 1 and choosing $g(x)$ as in (3.18) we have that
$$
\sum_{m=0}^n \int_0^T \int_{\Om_m} J_m \,dxdt \le \delta\,\tilde c \int_0^T E(t)\,dt
$$
for any $\delta > 0$ and some positive constant $\tilde c$ which depends only on $\Phi$ and the norms  of the matrices $A_{ij}$\,, $A_i$ and $D$.

\vglue .1in

\noindent{\bf Proof}. The index $m$ will be omitted in order to simplify notations. With our choice of $g(x)$, straightforward calculations show that $J_m$ (given by (3.7)) can be written as
\begin{align*}
J_m &= 2\delta\left\{A_k \frac{\po u}{\po x_k} - \frac{\po^2\Phi}{\po x_i \po x_k} A_k \frac{\po u}{\po x_i} - \left(A_k \frac{\po u}{\po x_k}\bullet\nabla\right)\nabla\Phi\right\}\bullet\nabla\vr\\
&+ 2\delta\left\{\left(A_{ij} \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i}\right) - \frac{\po^2\Phi}{\po x_p\po x_i} A_{ij} \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_p} + \frac{\po^2\Phi}{\po x_i \po x_k} d_{ij} \frac{\po\vr}{\po x_j}\, \frac{\po\vr}{\po x_k}\right\}\\
&- \delta\left\{|u_t|^2 + A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} + D\nabla\vr \bullet \nabla\vr\right\} \tag{3.19}
\end{align*}
Let us estimate each term on the right hand side of (3.19). We consider $v_i = \dfrac{\po^2\Phi}{\po x_p\po x_i}\,\dfrac{\po u}{\po x_p}$ and $\ve > 0$, then, we can write
\begin{align*}
-2\,A_{ij}\,\frac{\po u}{\po x_j}\bullet v_i = &-A_{ij}\left(\sqrt\ve\, \frac{\po u}{\po x_j} + \frac{1}{\sqrt\ve}\,v_j\right)\bullet\left(\sqrt\ve\, \frac{\po u}{\po x_i} + \frac{1}{\sqrt\ve}\,v_i\right)\\
& +\ve\,A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} + \ve^{-1}\, A_{ij}\, v_j\bullet v_i \le\\
&\le \ve\,A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} + \ve^{-1}\,A_{ij}\, v_j\bullet v_i \tag{3.20}
\end{align*}
because $A_{ij}$ satisfies assumption 2) in Hypothesis 1.

\noindent Let $c_3$ and $c_4$ the following numbers
$$
c_3 = \max_{\substack{x\in\ov\Om\\ i,j=1,2,3}} ||A_{ij}(x)||, \qquad c_4 = \max_{\substack{x\in\ov\Om\\ i,j=1,2,3}} \lv \frac{\po^2\Phi(x)}{\po x_i\po x_j}\rv,  
$$
where $||A_{ij}||$ denotes the norm of the matrix $A_{ij}$\,. With this notations, we have that
$$
|v_i| \le C_4 \sum_{j=1}^3 \lv\frac{\po u}{\po x_j}\rv
$$
and 
\begin{align*}
|A_{ij}v_j\bullet v_i| &\le ||A_{ij}||\,|v_j|\,|v_i| \le c_3\left(\sum_{j=1}^3 |v_j|\right)^2\\
&\le 9\,c_3\,c_4^2 \left\{\sum_{i=1}^3 \lv\frac{\po u}{\po x_i}\rv\right\}^2 \le 27\,c_3\,c_4^2 \sum_{i=1}^3 
\lv\frac{\po u}{\po x_i}\rv^2\\
&\le 27\,c_1^{-1}\, c_3\,c_4^2\,A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} \tag{3.21}
\end{align*}
where $c_1 > 0$ is as in item 2) of  Hypothesis 1.

\noindent From (3.20) and (3.21) we obtain the inequality
\begin{equation}
-2\delta\, \frac{\po^2\Phi}{\po x_p \po x_i}\, A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_p} \le \delta(\ve + 27(\ve\,c_1)^{-1}\,c_3\,c_4^2)A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} \tag{3.22}
\end{equation}
Similarly, we estimate
$$
2\delta\, \frac{\po^2\Phi}{\po x_i\po x_k}\, d_{ij}\, \frac{\po\vr}{\po x_j}\, \frac{\po\vr}{\po x_k} \le \delta\,\ve_1\, D\nabla\vr\bullet\nabla\vr + \delta\,\ve_1^{-1}\, D\psi\bullet\psi
$$
for any $\ve_1 > 0$ where $\psi_i = \dfrac{\po^2\Phi}{\po x_i\po x_k}\, \dfrac{\po\vr}{\po x_k}$\,, \,\, $\psi = (\psi_1,\psi_2,\psi_3)$ in order to obtain
\begin{equation}
2\delta\, \frac{\po^2\Phi}{\po x_i\po x_k}\, d_{ij}\, \frac{\po\vr}{\po x_j}\, \frac{\po\vr}{\po x_k} \le \delta(\ve_1 + 9(d_0\ve_1)^{-1}\,c_5c_4^2)D\nabla\vr\bullet\nabla\vr \tag{3.23}
\end{equation}
where $d_0 > 0$ is as in Hypothesis 1 and $c_5 = ||D||$.

\noindent Following the same reasoning as above we get the estimates
\begin{align*}
-2\delta\,\nabla\vr &\bullet \left(A_k \frac{\po u}{\po x_k}\bullet\nabla\right)\nabla\Phi \le \delta\left(3\ve_2\,d_0^{-1}\,c_4^2\,D\nabla\vr\bullet\nabla\vr\right) +\\
&+\delta\left(3(c_0\ve_2)^{-1}\, c_6^2 \, A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i}\right) \tag{3.24}
\end{align*}
where $\ve_2 > 0$ and $c_6 = \max\limits_{k=1,2,3} ||A_k||$.

\noindent Also
\begin{equation}
2\delta\, \nabla\vr \bullet A_k\, \frac{\po u}{\po x_k} \le \delta\left(\ve_3d_0^{-1}c_6\,D\nabla\vr\bullet\nabla\vr + 3c_6(c_0\ve_3)^{-1}\, A_{ij}\, \frac{\po u}{\po x_j}\bullet \frac{\po u}{\po x_i}\right) \tag{3.25}
\end{equation}
\begin{align*}
-2\delta\, \nabla\vr &\bullet\frac{\po^2\Phi}{\po x_i\po x_k} A_k\, \frac{\po u}{\po x_i} \le \delta\bigg(\ve_4\,d_0^{-1}\,c_6\, D\nabla\vr\bullet\nabla\vr +\\
&+27(c_0\ve_4)^{-1}\,c_6\,c_4^2\, A_{ij}\, \frac{\po u}{\po x_j}\bullet\frac{\po u}{\po x_i}\bigg) \tag{3.26}
\end{align*}
for any $\ve_3 > 0$ and $\ve_4 > 0$.

\noindent From the above estimates (3.22)-(3.26) we deduce from (3.19) that
\begin{align*}
J_m &\le \delta\left\{c_7\,A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i} + c_8\, D\nabla\vr\bullet\nabla\vr\right\}\\
&-\delta\left\{|u_t|^2 + A_{ij}\, \frac{\po u}{\po x_j}\bullet\frac{\po u}{\po x_i} + D\,\nabla\vr\bullet\nabla\vr\right\} \tag{3.27}
\end{align*}
where $c_7$ and $c_8$ are positive constants. Integration of inequality (3.27) in $\Om_m\times(0,T)$ and adding in $m$ completes the proof of Lemma 3.2 taking $\tilde c = \max\{c_7, c_8\}$. \qed

\vglue .1in

\noindent From now on we fix $\delta_0 > 0$ such that $\delta_0\tilde c < 1$ where $\tilde c$ is as in Lemma 3.2. Thus, we will work with the auxiliary function
$$
g(x) = \delta_0\,\Phi(x) + \frac 12\, |x-x_0|^2.
$$
Our next goal is to estimate the surface integrals (over $S_0 \times (0,T)$ and $S_1 \times (0,T)$ in (3.8)).

\noindent Now, we will impose geometric conditions on $\Om$ and $\Ga_m$\,.

\newpage

\noindent{\bf HYPOTHESIS  2}. There exists a point $x_0 \in \re^3$ such that
\begin{itemize}
\item[a)] $(x-x_0)\bullet\eta > -2\delta_0 \, \dfrac{\text{Vol}(\Om)}{\text{area}(S_0)}$\quad for all \quad $x \in S_0$
\item[b)] $(x-x_0)\bullet\eta \le \delta_0\, \dfrac{\text{Vol}(\Om)}{\text{area}(S_1)}$\,,  \quad for all \quad $x \in S_1$
\item[c)] $\delta_0\, \dfrac{\po\Phi}{\po\eta} + (x-x_0)\bullet\eta \ge 0$ \quad for all \quad $x \in \Ga_m$
\end{itemize}

\noindent $m=1,2,\dots,n$, where $\eta = \eta(x)$ denotes the unit outward normal to $S_0$, $S_1$ or $\Ga_m$\,.

\vglue .2in

\noindent{\bf Remark}. We note that the above assumptions on Hypothesis 2 are valid when $\delta_0=0$ for star-shaped surfaces $S_1,\Ga_1,\Ga_2,\dots,\Ga_n$ and strongly star-shaped surface $S_0$\,, i.e.
$$
(x-x_0)\bullet\eta > 0, \quad \forall\, x \in S_0\,.
$$
Moreover, if $\Ga_1,\Ga_2,\dots,\Ga_n$ are strongly star-shaped with respect to a point $x_0$\,, then the above conditions hold with $\delta_0 > 0$ for a class of domains which includes star-shaped domains.

\noindent Let $\la_0 > 0$ be such that
$$
\frac{\po g}{\po\eta} \ge \la_0|\nabla g| \quad\text{for any}\quad x \in S_0
$$
which is possible since $\dfrac{\po g}{\po\eta} = \delta_0\, \dfrac{\text{Vol}(\Om)}{\text{Area}(S_0)} + (x-x_0)\bullet\eta > 0$ on $S_0$ and $S_0$ is compact. Using the boundary conditions (1.5)-(1.6) we find that
\begin{align*}
V_n = &-\frac{\po}{\po t} \left\{\al|u|^2\right\} - \left[2\al(t+t_0) - \frac{\po g}{\po\eta}\right] |u_t|^2\\
&- \frac{\po g}{\po\eta} \left\{A_{pq}\, \frac{\po u}{\po x_q} \bullet \frac{\po u}{\po x_p} + D\,\nabla\vr\bullet\nabla\vr\right\} - 2\al\, u_t\bullet(\nabla g \bullet\nabla)u \tag{3.28}
\end{align*}
on $S_0$\,. Also
\begin{equation}
V_0 = \frac{\po g}{\po\eta} \left\{\left(A_{ij}\, \frac{\po u}{\po x_j} \bullet \frac{\po u}{\po x_i}\right) + (D\,\nabla\vr\bullet\nabla\vr)\right\} \quad\text{on}\quad S_1\,. \tag{3.29}
\end{equation}
Observe that
\begin{align*}
-2\al\,u_t &\bullet (\nabla g\bullet\nabla)u \le 2\al|u_t|\,|(\nabla g\bullet\nabla)u| \le\\
&\le \al\,\ve_0^{-1}|u_t|^2\, |\nabla g| + \al\,\ve_0|\nabla g| \left(\sum_{i=1}^3 |\nabla u_i|^2\right)\\
&\le \al\,\ve_0^{-1}|\nabla g|\,|u_t|^2 + \al\,\ve_0\,c_1^{-1}|\nabla g| A_{ij}\, \frac{\po u}{\po x_j}\bullet \frac{\po u}{\po x_i} \tag{3.30}
\end{align*}
for any $\ve_0 > 0$ where $c_1 > 0$ is as in Hypothesis 1.

\noindent Let $\ve_0 = c_1\la_0\,\ga_1^{-1}$ where $\ga_1 = \max\limits_{x\in S_0} \al(x)$. It follows from (3.28) and (3.30) that
\begin{equation}
V_n \le -\frac{\po}{\po t} \left\{\al|u|^2\right\} - \left[2\al(t+t_0) - \frac{\po g}{\po\eta} - \al\ga_1(c_0\la_0)^{-1}\,|\nabla g|\right] |u_t|^2 \tag{3.31}
\end{equation}
on $S_0$\,. We choose $t_0 > 0$ large enough so that $\left[2\al(t+t_0) - \dfrac{\po g}{\po\eta} - \al\ga_1(c_0\la_0)^{-1}|\nabla g|\right] > 0$ for any $t \ge 0$ and $x \in S_0$\,.
Also from Hypothesis 2 we obtain that
\begin{equation}
V_0 \le 0 \quad\text{on}\quad S_1\,. \tag{3.32}
\end{equation}
The following inequality can be proved by standard arguments
\begin{align*}
&2 \sum_{m=0}^n \int_{\Om_m} u_t^{(m)} \bullet \left\{(\nabla g\bullet\nabla)u^{(m)} + u^{(m)}\right\}dx + \int_{S_0} \al|u|^2\,d\Ga \le\\
&\le c_{10} \sum_{m=0}^n \int_{\Om_m} \left[|u_t^{(m)}|^2 + A_{ij}^{(m)}\, \frac{\po u^{(m)}}{\po x_j} \bullet \frac{\po u^{(m)}}{\po x_i}\right]dx \le c_{10}\,E(t) \tag{3.33}
\end{align*}
for some positive constant $c_{10}$ \bigg(which can be chosen as $c_{10} = \max\left\{2, c_1^{-1}(k + \max\limits_{x\in\ov\Om}|\nabla g(x)|)\right\}$ where $k > 0$ is such that
$$
\sum_{m=0}^n \int_{\Om_m} |u^{(m)}|^2\,dx + \int_{S_0} \al|u|^2\,d\Ga \le k \sum_{m=0}^n \int_{\Om_m} |\nabla u_i^{(m)}|^2\,dx
$$
where $u^{(m)} = \left(u_1^{(m)}, u_2^{(m)},u_3^{(m)}\right) \in [H^1(\Om_m)]^3$ with $u^{(m-1)} = u^{(m)}$ on $\Ga_m$\,, $m=1,2,\dots,n$.\bigg)

\vglue .2in

\noindent{\bf HYPOTHESIS 3}. The coefficients of  system (1.2)-(1.3) satisfy the following monotonicity conditions
\begin{align*}
&\left(A_{ij}^{(m-1)} - A_{ij}^{(m)}\right)v_j \bullet v_i \ge 0 \quad\text{for any}\quad v_i \in \re^3, \,\, 1 \le m \le n\\
&\left(D^{(m)}-D^{(m-1)}\right)v\bullet v \ge 0 \quad\text{for any}\quad v \in \re^3, \,\, 1 \le m \le n
\end{align*}

\vglue .1in

\noindent{\bf Theorem 3.1}. Let us assume Hypothesis 1, 2 and 3. Let $\{u,\vr\}$ be the unique solution of problem (1.2)-(1.9) as shown in Section 2. Then, there exists positive constants $c$ and $w$ such that 
$$
E(t) \le c\,\exp(-wt)E(0)\quad \forall\, t \ge 0
$$
where $E(t)$ is given by (1.10).

\vglue .1in

\noindent{\bf Proof}. It follows from identity (3.8) using Lemmas 3.1 and 3.2 together with (3.31), (3.32), (3.33) and Hypothesis 3 that
\begin{equation}
(T+t_0)E(T) \le (2c_{10}+t_0)E(0) + \delta_0\tilde c \int_0^T E(t)dt \tag{3.34}
\end{equation}
where $\delta_0\tilde c < 1$ (as in Lemma 3.2), holds for any $T > 0$. 

\noindent Let us denote by $h(T)$ the right hand side of (3.34). Clearly $\dfrac{h'(T)}{h(T)} \le \dfrac{\delta_0\tilde c}{T+t_0}$ therefore $h(T) \le \dfrac{(T+t_0)^p}{t_0^p}\,h(0)$ where $p = \delta_0\tilde c < 1$. Returning to (3.34) we obtain that
\begin{equation}
E(T) \le \frac{c_{11}}{(T+t_0)^{1-p}}\,E(0) \tag{3.35}
\end{equation}
where $c_{11} = (2c_{10}+t_0)t_0^{-p}$\,. Now, we choose $T = \widetilde T > 0$ large enough in (3.35) so that $T^* = \dfrac{c_{11}}{(\widetilde T+t_0)^{1-p}} < 1$. The semigroup (see Pazy [11]) property implies the conclusion of Theorem 3.1.

\vglue .3in

\noindent{\bf \S4) Application: Exact controllability}

\vglue .1in

In this section, we use the main result obtained above in order to prove exact boundary controllability to an arbitrary state of solutions of (1.2)-(1.9) where instead of the first boundary condition in (1.5) we consider 
\begin{equation}
\left(A_{ij}\,\frac{\po u}{\po x_j} + A_i^* \nabla\vr\right)\eta_i = h(x,t) \quad\text{on}\quad S_0\times[0,+\infty) \tag{4.1}
\end{equation}
where $f = (u_0,u_1)$ (in (1.4)) is an arbitrary element of the space $X$ (defined in Section \S2). The formulation of the exact boundary control for the above system is the following: Given the initial distribution $f = (u_0,u_1)$, a time $T > 0$ and a desired terminal state $g = (g_1,g_2)$, find a vector-valued function $h = h(x,t)$ such that the solution of (1.2)-(1.9) with condition (4.1) instead of the first boundary condition in (1.5), satisfies the conditions
$$
u(x,T) = g_1(x), \qquad u_t(x,T) = g_2(x).
$$
Let $\{U(t)\}_{t\ge0}$ be the semigroup associated with problem (1.2)-(1.9) (with condition (4.1) instead of the first boundary condition in (1.5)). Consider the following equation in $X$:
$$
w - U^*(T)U(T)w = f-U^*(T)g.
$$
The operator $F(T) = U^*(T)U(T)$ takes $Y$ into itself and $||F(T)|| < 1$ for any $T > T^*$ where  $T^*$ was chosen in the proof of Theorem 3.1 as $T^* = \dfrac{c_{11}}{(\widetilde T+t_0)^{1-p}}$ with $\widetilde T$ large enough so that $T^* < 1$. Thus, we can solve this equation for any $f$ and $g \in Y$ with $w$ satisfying
$$
||w|| \le C(||f||+||g||).
$$
Consequently, if we choose $w = (I-F(T))^{-1}(f-U^*(T)g)$ where $I$ denotes the identity and set
$$
V(x,t) = U(t)w - U^*(T-t)(U(T)w-g) \equiv (\tilde u, \tilde v) - (\overset{\approx}{u}, \overset{\approx}{v}).
$$
It follows that 
$$
V(x,0) = f(x), \qquad V(x,T) = g(x)
$$
and $(u,v) = V(x,t)$ is a weak solution of (1.2)-(1.9) (with condition (4.1) instead of the first boundary condition in (1.5)) with $h(x,t) =-\al(\tilde v + \overset{\approx}{v})$. We observe that by the energy identity
$$
||h||_{L^2(S_0\times(0,T))}^2 \le C\big(||f||_X^2 + ||g||_X^2\big).
$$
Thus, we arrive to the following assertion:

\vglue .2in

\noindent{\bf Theorem 4.1}. Assume that $\Om$,\, $\Ga_m$ and the coefficients satisfy the assumptions of Theorem 3.1. Then, for any $T > T^*$ given by initial data $f = (u_0,u_1) \in X$ and any $g = (g_1,g_2) \in X$ there exists a control $h(x,t) \in L^2(S_0\times(0,T))$ such that the corresponding solution of (1.2)-(1.9) (with the above mentioned modification in (1.5)) satisfies
$$
u(x,T) = g_1(x), \qquad u_t(x,T) = g_2(x).
$$
Moreover
$$
||h||_{L^2(S_0\times(0,T))} \le \widetilde C\big(||f||_X + ||g||_X\big)
$$
for some positive constant $\widetilde C$.

\vglue .3in

\noindent{\bf \S5) Conclusions}

\vglue .1in

\noindent In this work we consider a three-dimensional layered piezoelectric body with a dissipative mechanism effective at the boundary and appropriate transmission conditions at the interfaces. Using the multiplier technique we conclude that the total energy $E(t)$ decays exponentially fast as $t \to +\infty$, provided that the coefficients of the model satisfy a monotonicity condition and the domain as well as the interfaces also satisfy geometric requirements. As an application of our result we deduce exact controllability of the system with boundary control via a classical result due to D.L. Russell [13].

\vglue .3in

\noindent{\bf Acknowledgements}. We would like to express our thanks to the referees of this Journal for several suggestions which improved the final version of our manuscript. The first author (B.K.) was supported by a Grant of CNPq, Project 303981/03-2 (Brasil). The third author (G.P.M.) was partially supported by a Grant of CNPq and PRONEX (MCT, Brasil) and we acknowledge such important research support. The second author (B.M.) has been supported in part by the European Community's Human Potential Programme under contract ``Smart systems", number HPRN-CT-2002-00284.


\newpage

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