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\markboth{\hfil Boundary stabilization of a linear  systems \hfil EJDE--2001/78}
{EJDE--2001/78\hfil R. Bey, A. Heminna \& J.-P. Loh\'eac \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No.~78, pp. 1--23. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Boundary stabilization of a linear elastodynamic
  system with variable coefficients
 %
\thanks{ {\em Mathematics Subject Classifications:} 93B03, 93B05, 93D15.
\hfil\break\indent
{\em Key words:} elastodynamic system, elasticity, boundary stabilization,
  feedback.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted March 19, 2001. Published December 21, 2001.} }
\date{}
%
\author{Rabah Bey, Amar Heminna \& Jean-Pierre Loh\'eac}
\maketitle

\begin{abstract}
  We consider the boundary stabilization of a linear
  elastodynamic system, with variables coefficients,
  by \textit{natural} feedback.
  In \cite{G1,G3}, Guesmia  proved the
  boundary stabilization of the linear elastodynamic system,
  with variables coefficients, under restrictive conditions
  on the shape of the domain and on the data of the problem.
  Here, we propose using local coordinates in the boundary
  integrals to obtain stability under conditions that are
  only geometrical and less restrictive than those in
  \cite{G1,G3}.  We also extend the boundary stabilization
  result obtained in \cite{BHL}.
\end{abstract}

\newtheorem{thm}{Theorem}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{dfn}{Definition}
\newtheorem{remark}{Remark}

\newcommand{\diver}{\mathop{\rm div}}
\newcommand{\supp}{\mathop{\rm supp}}

\section*{Introduction}

Let $\Omega $ be a bounded open set of $\mathbb{R}^3 $ such that its
boundary $\Gamma $ satisfies
\begin{equation}
 \label{H1}
\mbox{$\Gamma$ is of class $\mathcal{C}^2$,
$\Gamma = \Gamma_0 \cup \Gamma_1$ with
$\mathop{\rm meas} (\Gamma_1 ) \not = 0$,
$\overline \Gamma_0 \cap \overline \Gamma_1 = \emptyset$.}
\end{equation}
  Let $a_{i j k l}$ ($i, j, k, l  = 1,2,3$) be a set
  of functions in $\mathcal{W}^{2,\infty}(\Omega \times \mathbb{R})$ such that
  $$
a_{i j kl } = a_{kl i j } =
  a_{j i kl } \,, \quad \hbox{in } \Omega \times \mathbb{R} \,,$$
  satisfying for some $\alpha > 0$ the condition
\begin{equation}
 \label{H11}
  a_{i j kl } \varepsilon_{i j } \varepsilon_{kl} \ge
  \alpha \varepsilon_{i j } \varepsilon_{i j } \,,
  \quad \acute{a_{i j kl }} \,\varepsilon_{i j } \,
  \varepsilon_{kl} \le 0 \,, \quad \hbox{in } \Omega \times \mathbb{R}
\end{equation}
  for every symmetric tensor $\varepsilon_{i j }$, where
  $\acute{}$ denotes $\partial / \partial t$ (we use the summation
  convention for repeated indices).
Given $\mathbf{x} $, a point on $\Gamma $, we denote by $\nu (\mathbf{x} )$
the normal unit vector pointing outward from $\Omega $.

 For a regular vector field $\mathbf{v} = (v_1, v_2, v_3)$, we define
$$v_{i , j } = \partial_j v_i \,, \quad
  \varepsilon_{i j } (\mathbf{v} ) = \frac{1}{2}( v_{i , j }
  + v_{j , i } ) \,, \quad \sigma_{i j } (\mathbf{v} ) =
  a_{i j kl} \varepsilon_{kl} (\mathbf{v} ) \,.
$$
  Let $A$ and $B$ be two positive constants.
  We consider the following problem introduced by Lagnese \cite{La2}:
\begin{equation}\label{E}
\begin{gathered}
  \mathbf u''- \diver  \bigl( \sigma (\mathbf u ) \bigr) = 0 \,, \quad \hbox{in } \Omega
 \times \mathbb{R}_{+} \,,
 \\
 \mathbf u = 0 \,, \quad \hbox{on } \Gamma_0 \times \mathbb{R}_{+} \,,
 \\
 \sigma (\mathbf u )\nu + A\mathbf u + B\mathbf u' = 0 \,, \quad \hbox{on } \Gamma_1
 \times \mathbb{R}_{+} \,,
 \\
 \mathbf u (0) = \mathbf u^0 \,, \quad \hbox{in } \Omega \,,
 \\
 \mathbf u'(0) = \mathbf u^1 \,, \quad \hbox{in } \Omega \,.
 \end{gathered}
\end{equation}
 Let $\mathbb{L}^2 (\Omega ) $ (resp. $\mathbb{H}^1 (\Omega ) $) be the space of vector
 fields $\mathbf{v} $ such that every component of $\mathbf{v} $ belongs to
 $\mathrm{L}^2 (\Omega ) $ (resp. $\mathrm{H}^1 (\Omega ) $).
We introduce the space $ \mathbb{H}^1_{\Gamma_0 } (\Omega ) = \{ \mathbf{v}
 \in \mathbb{H}^1 (\Omega ) \, / \, \mathbf{v} =0 \,, \, \hbox{on\ } \Gamma_0 \} $
and we assume
\begin{equation}
 \label{H2}
 (\mathbf u^0 , \mathbf u^1 ) \in \mathbb{H}^1_{\Gamma_0 } (\Omega ) \times \mathbb{L}^2 (\Omega ) \,.
\end{equation}
 Under this assumption, using semi-group theory, one can show
 that problem $(\ref{E} )$ is well-posed.

 The energy functional associated with this problem is
 $$ E(\mathbf u ,t) = \frac{1}{2} \int_\Omega \bigl( |\mathbf u' |^2 +
 \sigma (\mathbf u ) \colon \varepsilon (\mathbf u ) \bigr) \, d\mathbf{x} + \frac{1}{2}
 \int_{\Gamma_1 } A |\mathbf u |^2 \, d\Gamma \,. $$
 A boundary stabilization result for this system has been
 proved by Guesmia \cite{G} under restrictive conditions on
 the shape of $\Omega $ and on other data of the problem.
 We propose here a direct approach by using local coordinates
 in the expression of boundary integrals. Our conditions are only
 geometrical and are less restrictive than those in works of Guesmia.
 Our proof is constructive. Furthermore, the reader will observe that
 similar conditions have been introduced by Lagnese \cite{La1} for some
 anisotropic linear elastodynamic systems and by Lasieka and Triggiani
 \cite{LT} for the wave equation.

 For a vector field $\mathbf{h} = (h_1, h_2, h_3) \in
 \bigl( \mathcal{C}^1 (\overline \Omega ) \bigr)^3 $ we denote by
 $\gamma_\mathbf{h} $ a small (non-negative) number satisfying
 $$ \left |h_m \, \partial_m (a_{i j kl}) \, \varepsilon_{kl} \,
 \varepsilon_{i j } \right| \le \gamma_\mathbf{h} \, a_{i j kl}
 \, \varepsilon_{kl} \, \varepsilon_{i j } \,, $$
 for all symmetric tensor $\varepsilon_{i j }$.
 We assume that there exists a vector field
 $\mathbf{h} = (h_1, h_2, h_3) $ such that
\begin{equation}
 \label{H3}
 \mathbf{h} \in \bigl( \mathcal{C}^1 (\overline \Omega ) \bigr)^3 \,,
 \quad
 \mathbf{h} .\nu \le 0 \,, \quad \hbox{on } \Gamma_0 \,,
 \quad
 \mathbf{h} .\nu > 0 \,, \quad \hbox{on } \Gamma_1 \,,
\end{equation}
 and furthermore, there exist $\alpha_\mathbf{h} > 0 $ and $\beta_\mathbf{h}
 \in \mathbb{R} $ such that
\begin{equation}
\begin{gathered}
\forall \xi \in \bigl( \mathcal{C}^1 (\overline \Omega ) \bigr)^3,\quad
\int_\Omega \sigma_{i j } (\xi )
 h_{k, j } \xi_{i , k } \, d\mathbf{x} \ge \alpha_\mathbf{h} \int_\Omega
 \sigma (\xi ) \colon \varepsilon (\xi ) \, d\mathbf{x} + \beta_\mathbf{h}
 \int_{\Gamma_1 } |\xi |^2 \, d\Gamma \,,
 \\
 \max_{\overline \Omega } (\diver  (\mathbf{h} )) - \min_{\overline \Omega }
 (\diver  (\mathbf{h} )) < 2 \alpha_\mathbf{h} - \gamma_\mathbf{h} \,, \quad
 \min_{\overline \Omega } (\diver  (\mathbf{h} )) > 0 \,,
\end{gathered}
\label{H4}
\end{equation}
 where we can choose
 $\beta_\mathbf{h} = 0 $, if $\mathop{\rm meas} (\Gamma_0 ) \not = 0 $.

Under the above assumptions, we obtain the following result.

\begin{thm} \label{thm1}
 Assume $(\ref{H1} )$, $(\ref{H11} )$ and $(\ref{H2} )$. If there exists a
 vector field $\mathbf{h} $ satisfying $(\ref{H3} )$ and $(\ref{H4} )$, then there
 exists some constant $\omega > 0 $ such that the solution $\mathbf u $ of
 $(\ref{E} )$ satisfies
 $$ \forall t \ge 0 \,, \quad E(\mathbf u ,t) \le E(\mathbf u ,0) \exp (1-\omega t)
 \,. $$
\end{thm}

\begin{remark} \rm \label{rmq1}
 Since $\Omega $ is bounded and $\Gamma $ satisfies $(\ref{H1} )$,
 $\Gamma_1 $ is compact. Using continuity of $\mathbf{h} $ and $\nu $ given in
 $(\ref{H1} )$ and $(\ref{H3} )$, we get
 $$\exists k > 0 \, : \, \mathbf{h} .\nu  \ge k \,, \quad \hbox{\rm on }
 \Gamma_1 \,. $$
\end{remark}

\begin{remark} \rm \label{rmq2}
 This result can be applied when $\mathbf{h} (\mathbf{x} ) = \mathbf{x} - \mathbf{x}_0 $ ($\alpha_\mathbf{h} =1$,
 $\beta_\mathbf{h} =0$) and
 $$ \gamma_\mathbf{h} < 2 \,, \quad
 \Gamma_0 = \{ \mathbf{x} \in \Gamma \, / \, \mathbf{h} (\mathbf{x} ). \nu (\mathbf{x} ) \le 0 \, \}
 \,, \quad
 \Gamma_1 = \{ \mathbf{x} \in \Gamma \, / \, \mathbf{h} (\mathbf{x} ). \nu (\mathbf{x} ) > 0 \, \}
 \,. $$
 Especially, a possible case is $\Omega = U_1 \setminus U_2 $, where
 $U_1 $ is a convex open set, $U_2 $ is a closed set, star-shaped with
 respect to one of its points, $\mathbf{x}_0 $, such that
 $\{ \mathbf{x}_0 \} \subset U_2 \subset U_1 $.
(see figure \ref{fig1})
\end{remark}
 This case has been studied in \cite{G} for a particular shape
 of $\Omega $ ($\Gamma_1 $ is supposed to be close to a sphere).
 Furthermore, this remark can be extended as follows.

\begin{figure}
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1.eps}
\end{center}
\caption{An example of an open set $\Omega $.}
\label{fig1}
\end{figure}


\begin{thm} \label{thm2}
 Assume $(\ref{H1} )$, $(\ref{H11} )$, $(\ref{H2} )$ and suppose that
\begin{gather*}
 \gamma_{\mathbf{x} -\mathbf{x}_0 } < 2 \,,
 \\
 (\mathbf{x} - \mathbf{x}_0 ). \nu (\mathbf{x} ) \le 0 \,, \quad \hbox{\rm if } \mathbf{x} \in
 \Gamma_0 \,,
 \\
 (\mathbf{x} - \mathbf{x}_0 ). \nu (\mathbf{x} ) \ge 0 \,, \quad \hbox{\rm if } \mathbf{x} \in
 \Gamma_1 \,.
 \end{gather*}
 Then there exists a constant $\omega > 0 $ such that the solution
 $\mathbf u $ of $(\ref{E} )$ satisfies
 $$ \forall t \ge 0 \,, \quad E(\mathbf u ,t) \le E(\mathbf u ,0) \exp (1-\omega t)
 \,. $$
\end{thm}

 This paper is mainly devoted to the proof of Theorem 1.
 After introducing some notations and definitions (section 1),
 we deal with the well-posedness in section 2 and conclude with the
 stabilization in section 3 where the proofs of Theorems 1 and 2 are
 given.

\section{Notation}

 In this paper, we use the convention of repeated indices.
 As usual, we write
 $\mathop{\rm tr} (\tau ) = \tau_{11} + \tau_{22} + \ldots =
 \tau_{i i }$, $\mathbf{v} .\mathbf{w} = v_i w_i$, $ \sigma (\mathbf{v} )
 \colon \varepsilon (\mathbf{v} ) = \sigma_{i j } (\mathbf{v} )
 \varepsilon_{i j } (\mathbf{v} )$.

\subsection*{Geometrical notation}

 We define $\Omega $, $\Gamma $, $\nu $ as above. Since
 $\Gamma $ is of class $\mathcal{C}^2 $, for every point $\mathbf{x} $ of $\Gamma $,
 we can build a local $\mathcal{C}^2 $-diffeomorphism $\phi $ from an open
 subset $\hat \Gamma $ of $\mathbb{R}^2 $ onto some open neighbourhood of $\mathbf{x} $
 in $\Gamma $.
 Then, the following vectors
 $$ \mathbf{a}_{\alpha } (\mathbf{x} ) = \frac{\partial \phi }{\partial \xi_\alpha }
 \left( \phi^{-1}(\mathbf{x} ) \right) \,, \quad \alpha \in \{ 1,2 \} \,, $$
 are independent and generate the tangent plane to $\Gamma $ at $\mathbf{x} $,
 $ T_{\mathbf{x} }(\Gamma )$. Furthermore, we denote by $T(\Gamma )$ the
 tangent bundle (see \cite{Le} and \cite{Va}).

Then, we define \begin{itemize}

 \item The metric tensor $g$ related to $\phi $:
 $ g_{\alpha \beta } = \mathbf{a}_\alpha .\mathbf{a}_\beta $ ,
 for all $(\alpha , \beta )$ in $\{ 1,2 \}^2 $,

\item The inverse tensor of $g$: ${ \left(
 g^{\alpha \beta } \right)_{1 \le \alpha , \beta \le 2} } $.
\end{itemize}
 We denote by $\pi(\mathbf{x} )$ the orthogonal projection on $T_{\mathbf{x} }
 (\Gamma )$ and, for a given vector field $\mathbf{v} : \overline{\Omega }
 \to \mathbb{R}^3 $, we will write
\begin{gather*}
 \forall \mathbf{x} \in \Gamma \,, \quad \mathbf{v} (\mathbf{x} ) = \mathbf{v}_T (\mathbf{x} ) + v_\nu
 (\mathbf{x} )\nu(\mathbf{x} ) \,,
 \\
\mbox{with} \quad \mathbf{v}_T(\mathbf{x} ) = \pi (\mathbf{x} ) \mathbf{v} (\mathbf{x} ) \,, \quad
 v_\nu (\mathbf{x} ) = \mathbf{v} (\mathbf{x} ).\nu(\mathbf{x} ) \,.
\end{gather*}
 We denote by $\partial_T $ (resp. $\partial_\nu $) the tangential (resp.
 normal) derivative.
 If $v$ is some regular function, the transposed vector of
 $\partial_T v$ is the tangential gradient of $v$ and is denoted by
 $\nabla_T v$. We have
\begin{equation}
 \label{gra}
 \nabla v = \nabla_T v + \partial_\nu v \, \nu \,, \quad \hbox{on\ }
 \Gamma \,.
\end{equation}

\subsection*{Strain and stress}

If the vector field $\mathbf{v} $ is regular enough,  as in
 \cite{Le} and \cite{Va}, we can write
\begin{equation}
 \label{dif}
 \begin{aligned}
 {\mathrm d} \mathbf{v} = &
  \pi (\partial_T \mathbf{v}_T ) \pi + v_\nu (\partial_T \nu) +
 (\partial_\nu \mathbf{v}_T ) \overline{\nu }
 \\
 & + \nu \left( (\partial_T v_\nu ) - \overline{\mathbf{v}_T }
 (\partial_T \nu ) + (\partial_\nu v_\nu ) \overline{\nu } \right) \,,
 \quad \hbox{on } \Gamma \,,
 \end{aligned}
\end{equation}
 where $\overline{\mathbf{v} }$ (resp. $\overline{\tau }$) is the transposed
 vector (resp. matrix) of $\mathbf{v} $ (resp. $\tau $).
 The strain tensor $\varepsilon (\mathbf{v} )$ can be written on
 $\Gamma $ as follows
 $$ \varepsilon(\mathbf{v} ) = \varepsilon_T (\mathbf{v} ) + \nu \overline{\varepsilon_S
 (\mathbf{v} )} + \varepsilon_S (\mathbf{v} )\overline{\nu } +\varepsilon_{\nu } (\mathbf{v} )
 \nu \overline{\nu } \,, \quad  \hbox{on\ } \Gamma \,, $$
 with
 \begin{align*}
 2 \varepsilon_T (\mathbf{v} ) =& \pi (\partial_T \mathbf{v}_T ) \pi + \pi
 \overline{\partial_T \mathbf{v}_T } \pi + 2 v_\nu \partial_T \nu \,,
 \\
 2 \varepsilon_S (\mathbf{v} ) =& \partial_{\nu } \mathbf{v}_T + \nabla_T v_\nu
 - (\partial_T \nu ) \mathbf{v}_T \,,
 \\
 \varepsilon_{\nu } (\mathbf{v} ) =& \partial_{\nu } v_\nu \,.
 \end{align*}
 Similarly, we can write
 $$ \sigma (\mathbf{v} ) = \sigma_T (\mathbf{v} ) + \nu \overline{\sigma_S (\mathbf{v} )}
 + \sigma_S (\mathbf{v} ) \overline{\nu } + \sigma_{\nu } (\mathbf{v} ) \nu
 \overline{\nu } \,, \quad \hbox{on } \Gamma \,, $$
 where $\sigma_T (\mathbf{v} )$ is a linear symmetric operator field on the
 tangent plane, $\sigma_S (\mathbf{v} )$ is a tangent vector field and
 $\sigma_{\nu } (\mathbf{v} )$ is a scalar field.

\begin{remark} \rm \label{rmq3}
 Let $\mathbf{v} $ be in $\mathbb{H}^1 (\Omega )$. From the previous formul\ae ,
 we deduce
\begin{gather*}
 \varepsilon (\mathbf{v} ) \colon \varepsilon (\mathbf{v} ) =
 \varepsilon_T (\mathbf{v} ) \colon \varepsilon_T (\mathbf{v} )
 + 2 |\varepsilon_S (\mathbf{v} )|^2
 + |\varepsilon_{\nu } (\mathbf{v} )|^2 \,,
 \quad \hbox{\rm on } \Gamma \,,
 \\
 \sigma (\mathbf{v} ) \colon \varepsilon (\mathbf{v} ) =
 \sigma_T (\mathbf{v} ) \colon \varepsilon_T (\mathbf{v} )
 + 2 \overline{\sigma_S (\mathbf{v} )} \varepsilon_S (\mathbf{v} )
 + \sigma_{\nu } (\mathbf{v} )\varepsilon_{\nu } (\mathbf{v} ) \,,
 \quad \hbox{\rm on } \Gamma \,.
 \end{gather*}
\end{remark}

 We will consider the following vector spaces. \begin{itemize}

 \item $\mathcal{L}_s (T_{\mathbf{x} }(\Gamma ))$ is the space of
 linear symmetric operators of $T_{\mathbf{x} }(\Gamma)$,

 \item $\mathcal{L}_s (T(\Gamma ))$ is the space of symmetric
 operators of $T (\Gamma )$.
\end{itemize}

\begin{remark} \rm \label{rmq4}
 $\partial_T \nu (\mathbf{x} )$ belongs to $\mathcal{L}_s (T_{\mathbf{x} }(\Gamma ))$; its
 eigenvalues are principal curvatures of $\Gamma $ at $\mathbf{x} $.
\end{remark}


\subsection*{Some function spaces}

 Consider a tangent field
 $\mathbf{v}_T : \Gamma \to T(\Gamma ) $ with:
 $\mathbf{v}_T = v^1 \mathbf{a}_1 + v^2 \mathbf{a}_2 $.
 We will say that $\mathbf{v}_T $ belongs to $\mathrm{L}^2 (\Gamma , T(\Gamma
 ) )$ if $v^1 $ and $v^2 $ belong to $\mathrm{L}^2 (\Gamma ) $. In $\mathrm{L}^2 (\Gamma
 , T(\Gamma ))$, we define the following norm
\begin{equation}
 \label{nl2}
 \| \mathbf{v}_T \|_{\mathrm{L}^2 (\Gamma , T(\Gamma ))} = \Bigl( \int_{\Gamma }
 |\mathbf{v}_T |^2 \, d\Gamma \Bigr)^{1/2} \,,
\end{equation}
 which is equivalent to the norm: ${ \mathbf{v}_T \mapsto \left(
 \| v^1 \|_{\mathrm{L}^2 (\Gamma )}^2 + \| v^2 \|_{\mathrm{L}^2 (\Gamma )}^2
 \right)^{1/2} } $.
 Similarly, $\mathbf{v}_T $ belongs to $\mathrm{H}^1 (\Gamma , T(\Gamma )
 )$ if $v^1 $ and $v^2 $ belong to $\mathrm{H}^1 (\Gamma ) $ and we define a norm
 in $\mathrm{H}^1 (\Gamma , T(\Gamma ))$ by
\begin{equation}
 \label{nh1}
 \| \mathbf{v}_T \|_{\mathrm{H}^1 (\Gamma , T(\Gamma ))} = \bigl( \| v^1 \|_{\mathrm{H}^1
 (\Gamma )}^2 + \| v^2 \|_{\mathrm{H}^1 (\Gamma )}^2 \bigr)^{1/2} \,.
\end{equation}
 A field $\tau_T : \Gamma \to \mathcal{L}_s (T(\Gamma )) $ belongs to
 $\mathrm{L}^2 (\Gamma , \mathcal{L}_s (T(\Gamma )) ) $ if $(\tau_T \colon \tau_T
 )^{1/2} : \Gamma \to \mathbb{R} $ belongs to $\mathrm{L}^2 (\Gamma )$. We take
\begin{equation}
 \label{nll2}
 \| \tau_T \|_{\mathrm{L}^2 (\Gamma , \mathcal{L}_s (T(\Gamma )) )} =
 \| (\tau_T \colon \tau_T )^{1/2} \|_{\mathrm{L}^2 (\Gamma)} \,.
\end{equation}

\begin{remark} \rm \label{rmq5}
 If $\mathbf{v}_T \in \mathrm{H}^1(\Gamma , T(\Gamma )) $, then $\varepsilon_T(\mathbf{v}_T )
 \in \mathrm{L}^2 (\Gamma , \mathcal{L}_s (T(\Gamma )) ) $.
\end{remark}

 Another useful space is $\mathbb{V} = \{ \mathbf{v} \in \mathbb{H}^1 (\Omega ) \,
 / \, \mathbf{v}_T \in \mathrm{H}^1 (\Gamma , T(\Gamma )) \} $ with the norm
\begin{equation}
 \label{nv}
 \| \mathbf{v} \|_{\mathbb{V} } = \bigl( \| \mathbf{v} \|_{\mathbb{H}^1 (\Omega )}^2 +
 \| \mathbf{v}_T \|_{\mathrm{H}^1 (\Gamma , T(\Gamma ))}^2 \bigr)^{1/2} \,.
\end{equation}

\begin{prop} \label{prn1}
 The expression
 $$\| \mathbf{v}_T \|_1^2 = \int_{\Gamma } \left( |\mathbf{v}_T |^2 + \varepsilon_T
 (\mathbf{v}_T ) \colon \varepsilon_T (\mathbf{v}_T ) \right) \, d\Gamma \,. $$
 constitutes a norm equivalent to $(\ref{nh1} )$.
\end{prop}


\noindent {\sl Proof.}
 We only need to prove that there exists a constant $C>0 $ such
 that
 $$\| \mathbf{v}_T \|_{\mathrm{H}^1(\Gamma ,T(\Gamma ))} \leq C \| \mathbf{v}_T \|_1 \,, \quad
 \forall \mathbf{v}_T \in \mathrm{H}^1(\Gamma ,T(\Gamma )) \,. $$
 Assume that such a constant does not exist.
 Then there exits a sequence ${ \big(
 \mathbf{v}_T^{(k)} \big) }$ such that
\begin{gather*}
 \| \mathbf{v}_T^{(k)} \|_{\mathrm{H}^1(\Gamma ,T(\Gamma ))} = 1 \,,
 \quad \forall k \in \mathbb{N}\,,
 \\
 \left( \varepsilon_T ( \mathbf{v}_T^{(k)} ) \right) \to 0 \,,
 \quad \hbox{in } \mathrm{L}^2 (\Gamma , \mathcal{L}_s (T(\Gamma )) ) \,,
 \\
 \left( \mathbf{v}_T^{(k)} \right) \to 0 \,,
 \quad \hbox{in } \mathrm{L}^2 (\Gamma , T(\Gamma ) ) \,.
\end{gather*}
 Setting ${ \mathbf{v}_T^{(k)} = v^{k1} \mathbf{a}_1 + v^{k2} \mathbf{a}_2 } $
 and
 ${ v^{k \alpha }_{ ,\beta } = \frac{\partial v^{k \alpha }}
 {\partial \xi_\beta } } $, we get
$$
 \left( v^{k \alpha }_{ ,\beta } + g^{\lambda \alpha } g_{\beta \mu }
 v^{k \mu }_{ ,\lambda } \right) \to 0 \,, \quad
 \hbox{in } \mathrm{L}^2 (\Gamma ) \,.
$$
 With $\alpha = \beta $, we easily get
$ \left( v^{k1 }_{ , 1} + v^{k2}_{ , 2} \right) \to 0$, in $\mathrm{L}^2 (\Gamma )$.
 We have
 $${ g_{\alpha \eta } \left( v^{k \alpha }_{
 ,\beta } + g^{\lambda \alpha } g_{\beta \mu } v^{k \mu }_{ ,\lambda }
 \right) = g_{\alpha \eta } v^{k \alpha }_{ ,\beta } + g_{\beta \mu }
 v^{k \mu }_{ ,\eta } } \,.
$$
 This expression vanishes in $\mathrm{L}^2 (\Gamma ) $ as $k
 \to \infty $.
We use $(\beta ,\eta ) = (1,1)$, $(\beta ,\eta )
 = (2,2)$ and $(\beta ,\eta ) = (1,2)$ and get
\begin{gather*}
 \left( g_{11} v^{k1 }_{ , 1} + g_{21} v^{k2}_{ , 1}
 \right)
 \to 0 \,, \quad \hbox{in } \mathrm{L}^2 (\Gamma ) \,,
 \\
 \left( g_{22} v^{k2 }_{ , 2} + g_{21} v^{k1}_{ , 2} \right)
 \to 0 \,, \quad \hbox{in } \mathrm{L}^2 (\Gamma ) \,,
 \\
 \left( g_{11} v^{k1 }_{ , 2} + g_{22} v^{k2}_{ , 1} \right)
 \to 0 \,, \quad \hbox{in } \mathrm{L}^2 (\Gamma ) \,,
 \\
 \left( v^{k1 }_{ , 1} + v^{k2}_{ , 2} \right) \to
 0 \,, \quad \hbox{in } \mathrm{L}^2 (\Gamma ) \,.
\end{gather*}
 Set $\mathbf{w}^{(k)}_T = g \mathbf{v}^{(k)}_T = w^{k1} \mathbf{a}_1 + w^{k2} \mathbf{a}_2 $.
 From $\big( \mathbf{v}_T^{(k)} \big) \to 0$,
in $\mathrm{L}^2 (\Gamma , T(\Gamma ) )$ and
 previous computations, one can easily deduce that sequences
 ${ \left( w^{k1} \right) }$,
 ${ \left( w^{k1} \right) }$,
 ${ \left( w^{k1}_{ ,1} + w^{k2}_{ ,2} \right) }$,
 ${ \left( w^{k2}_{ ,1} + w^{k1}_{ ,2} \right) }$
 vanish in $ \mathrm{L}^2 (\Gamma ) $.
 Thanks to Korn's inequality in $\hat \Gamma $, we get that
 sequences ${ \left( w^{k1} \right) }$, ${
 \left( w^{k2} \right) }$ vanish in $ \mathrm{H}^1 (\Gamma ) $.
Hence, ${ \left( v^{k1} \right) }$,
 ${ \left( v^{k2} \right) }$ vanish in $ \mathrm{H}^1 (\Gamma ) $.
 This is impossible.
\hfill $\Box $

\section{Well-posedness}

 By using semi-group theory \cite{Pa}, we can show that problem
  $(\ref{E} )$ is well-posed.

\begin{prop} \label{prn2}
 Assume $(\ref{H11} )$. If $(\mathbf u^0 , \mathbf u^1 )$ belongs to
 $\mathbb{H}^1_{\Gamma_0 } (\Omega ) \times \mathbb{L}^2 (\Omega )$, then problem
 $(\ref{E} )$ has one and only one (weak) solution $\mathbf u $ which satisfies
$$
 \mathbf u \in \mathcal{C}^0 (\mathbb{R}_{+} , \mathbb{H}^1_{\Gamma_0 } (\Omega ) )
 \cap \mathcal{C}^1 (\mathbb{R}_{+} , \mathbb{L}^2 (\Omega ) ) \,.
$$
     If $(\mathbf u^0 , \mathbf u^1 )$ belongs to $( \mathbb{H}^2 (\Omega ) \cap \mathbb{H}^1_{\Gamma_0 }
 (\Omega ) ) \times \mathbb{H}^1_{\Gamma_0 } (\Omega )$ and if
 $$ \sigma (\mathbf u^0 ). \nu + A \mathbf u^0 + B \mathbf u^1 = 0 \,, \quad \mathrm{on\ }
 \Gamma_1 \,,$$
 then the (strong) solution of $(\ref{E} )$ satisfies
 $$ (\mathbf u ,\mathbf u' ,\mathbf u'' ) \in \mathcal{C}^0(\mathbb{R}_{+} , ( \mathbb{H}^2 (\Omega )
 \cap \mathbb{H}^1_{\Gamma_0 } (\Omega ) ) \times \mathbb{H}^1_{\Gamma_0 } (\Omega ) \times
 \mathbb{L}^2 (\Omega )) \,. $$
\end{prop}

\section{Stabilization}

 Following Komornik \cite{Ko}, we will prove here that the
 energy functional is exponentially decreasing with respect to time.
 From Lemma 5, to be proved later, it is sufficient to
 consider the case $\mathop{\rm meas}(\Gamma_0 ) \not=0 $.
We recall the following fundamental result \cite{Ko}.

\begin{lemma} \label{lma1}
 Let $E : \mathbb{R}_{+} \to \mathbb{R}_{+} $ be a non-increasing function and
 assume that there exists $ T > 0$ such that
 $$\int_t^{\infty } E(s) \, ds \le T E(t) \,, \quad \forall t \ge 0
 \,. $$
 Then we have
 $$ E(t) \le E(0) \exp \Bigl( 1 - \frac{t}{T} \Bigr) \,, \quad \forall
 \, t \ge T \,. $$
\end{lemma}


 First, we prove that the energy functional is non-increasing.


\begin{prop} \label{prn3}
 Under assumptions $(\ref{H1} )$, $(\ref{H11} )$, $(\ref{H2} )$,
 the weak solution $\mathbf u $
 of $(\ref{E} )$ is such that $\mathbf u' \sqrt{B} $ belongs to $\mathrm{L}_{{\rm loc}}^2
 (\mathbb{R}_{+} , \mathbb{L}^2 (\Gamma_1 )) $ and $\acute{a_{i j kl}} \,
 \varepsilon_{i j } \, \varepsilon_{kl}$ belongs to the space
 $\mathrm{L}_{{\rm loc}}^1(\mathbb{R}_{+} , \mathrm{L}^1 (\Omega )) $. The energy functional is
 non-increasing and satisfies
$$  E(\mathbf u ,T) - E(\mathbf u ,S)
     = - \int_S^T \int_{\Gamma_1 } B |\mathbf u' |^2\, d\Gamma\, dt
     + \frac{1}{2}\int_S^T \int_{\Omega}
 \acute{a_{i j kl}} \, \varepsilon_{i j}(\mathbf u) \,
 \varepsilon_{kl}(\mathbf u) \, d\mathbf{x}\, dt \,,
$$
for $0 \le S < T < +\infty$.
\end{prop}

 \noindent {\sl Proof.}
Assume first that $\mathbf u $ is a strong solution of $(\ref{E} )$
 (with appropriate initial data). We can write
\begin{align*}
 E'(\mathbf u ,t) =
 & \int_ {\Omega } \Bigl( \mathbf u' .\mathbf u'' +
 \sigma (\mathbf u ) \colon \varepsilon (\mathbf u' ) + \frac{1}{2}
 \acute{a_{i j kl}} \,
 \varepsilon_{i j } (\mathbf u) \,\varepsilon_{kl} (\mathbf u)
 \Bigr) \, d\mathbf{x} + \int_{\Gamma_1 } A \mathbf u . \mathbf u' \, d\Gamma
 \\
 = &  \int_{\Omega } \left( \mathbf u' . \diver  (\sigma
 (\mathbf u )) + \sigma ( \mathbf u ) \colon \varepsilon(\mathbf u' ) \right) \, d\mathbf{x}
 + \frac{1}{2} \int_{\Omega}\acute{a_{i j kl}}
 \, \varepsilon_{i j }(\mathbf u ) \,\varepsilon_{kl} (\mathbf u )\, d\mathbf{x}
 \\
 & + \int_{\Gamma_1 } A \mathbf u . \mathbf u' \, d\Gamma
 \\
 = &  \int_{\Gamma_1 } \left( \sigma (\mathbf u ) \nu
 \right) .\mathbf u' + \frac{1}{2} \int_{\Omega} \acute{a_{i j kl}}
 \, \varepsilon_{i j }(\mathbf u ) \, \varepsilon_{kl}(\mathbf u ) \, d\mathbf{x}
 \, d\Gamma + \int_{\Gamma_1 } A \mathbf u . \mathbf u' \, d\Gamma
 \\
 =&  - \int_{\Gamma_1 } B |\mathbf u' |^2 \, d\Gamma
  + \frac{1}{2} \int_{\Omega } \acute{a_{i j kl}} \,
 \varepsilon_{i j }(\mathbf u ) \, \varepsilon_{kl} (\mathbf u ) \, d\mathbf{x} \,.
\end{align*}
 We obtain the result by integrating the time variable from $S$ to $T$.
 A density argument completes the proof.
\hfill $\Box $


 In order to apply Lemma 1, we have to prove the following results.

\subsection*{Preliminary results}
In this subsection, we assume $(\ref{H1} )$ and
\begin{equation}
 \label{sf}
 \begin{gathered}
  (\mathbf u^0 , \mathbf u^1 )\in \left( \mathbb{H}^2 (\Omega ) \cap
 \mathbb{H}^1_{\Gamma_0 } (\Omega ) \right) \times \mathbb{H}^1_{\Gamma_0 } (\Omega ) \,,
 \\
  \sigma (\mathbf u^0 )\nu + A \mathbf u^0 + B \mathbf u^1 = 0 \,, \quad
 \hbox{on\ } \Gamma_1 \,,
 \end{gathered}
\end{equation}
 and consider the (strong) solution of $(\ref{E} )$.
 Let $\mathbf{h} $ be a vector field satisfying $(\ref{H3} )$ and
 $(\ref{H4} )$. For some positive constant $\beta $, define
 $$ M \mathbf u = 2(\mathbf{h} . \nabla )\mathbf u + \beta \mathbf u \,. $$
 The value of $\beta $ will be chosen later.

\begin{lemma} \label{lma2}
 The strong solution $\mathbf u $ of $(\ref{E} )$ satisfies
\begin{eqnarray*}
 \lefteqn{ \int_S^T \int_\Omega (\diver  (\mathbf{h} ) - \beta) \, (|\mathbf u'|^2 -
 \sigma (\mathbf u ) \colon \varepsilon (\mathbf u )) \, d\mathbf{x}\, dt }
 \\
 \lefteqn{+ 2 \int_S^T  \int_\Omega \sigma_{i ,j } (\mathbf u ) h_{k,j } u_{i ,k}
 \, d\mathbf{x}\, dt
 -\int_S^T \int_\Omega h_m \partial_m (a_{i
 j kl} ) \varepsilon_{kl} (\mathbf u ) \varepsilon_{i j } (\mathbf u )
 \, d\mathbf{x} \,dt }
 \\
 &=&  - \left[ \int_\Omega \mathbf u' M\mathbf u d\mathbf{x} \right]_S^T +
 \int_S^T \int_\Gamma \left((\sigma (\mathbf u ).\nu ) M\mathbf u - \mathbf{h} .\nu \sigma
 (\mathbf u ) \colon \varepsilon (\mathbf u ) \right) \, d\mathbf{x} \,dt \,.
 \\
 &&  + \int_S^T \int_\Gamma \mathbf{h} .\nu |\mathbf u'|^2 \, d\Gamma
 dt  \,.
\end{eqnarray*}
\end{lemma}

 \noindent {\sl Proof.}
 We use the multipliers method (see \cite{Ko}, \cite{Li}).
 Thanks to the first equation in $(\ref{E} )$, we may write
\begin{equation}
 \label{L21}
 \int_S^T \int_\Omega \mathbf u'' .M\mathbf u \, d\mathbf{x}\, dt
 = \int_S^T \int_\Omega
 \diver  (\sigma (\mathbf u )) .M\mathbf u \, d\mathbf{x}\, dt \,.
\end{equation}
Consider the left-hand side of the above equation.
\begin{align*}
 \int_S^T \int_\Omega \mathbf u'' .M\mathbf u \, d\mathbf{x}\, dt
 =&\Big[  \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \Big]_S^T
 - \int_S^T \int_\Omega \mathbf u' .M\mathbf u' \, d\mathbf{x}\, dt
 \\
 =& \Big[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \Big]_S^T
 \\
 & - 2 \int_S^T \int_\Omega u'_i h_j u'_{i ,
 j } \, d\mathbf{x}\, dt - \beta \int_S^T \int_\Omega |\mathbf u' |^2
 \, d\mathbf{x}\, dt \,.
 \end{align*}
 We have
\begin{align*}
 2 \int_S^T \int_\Omega u'_i h_j u'_{i ,
 j } \, d\mathbf{x}\, dt
 =&  \int_S^T \int_\Omega h_j
 \partial_j \left( |\mathbf u' |^2 \right) \, d\mathbf{x}\, dt
 \\
 =&  \int_S^T \int_\Gamma \mathbf{h} .\nu \, |\mathbf u' |^2 \, d\Gamma\, dt -
 \int_S^T \int_\Omega \diver  (\mathbf{h} ) |\mathbf u' |^2 \, d\mathbf{x}\,
 dt \,.
 \end{align*}
 Hence
\begin{equation} \label{L22}
 \begin{aligned}
\int_S^T \int_\Omega \mathbf u'' .M\mathbf u \, d\mathbf{x}\, dt
 =& \Big[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x}
 \Big]_S^T - \int_S^T \int_\Gamma \mathbf{h} .\nu \, |\mathbf u' |^2
 \, d\Gamma\, dt  \\
 & + \int_S^T \int_\Omega \left( \diver  (\mathbf{h} ) - \beta
 \right) |\mathbf u' |^2 \, d\mathbf{x}\, dt \,.
 \end{aligned}
\end{equation}
 Now, we consider the right-hand side of ($\ref{L21} $):
 \begin{align*}
\int_S^T \int_\Omega \diver  (\sigma (\mathbf u )) .M \mathbf u \,
 d\mathbf{x}\, dt
 = & 2 \int_S^T \int_\Omega \sigma_{i j , j } (\mathbf u ) h_m u_{i ,m} \,
 d\mathbf{x}\, dt  \\
 & + \beta \int_S^T \int_\Omega \mathbf u . \diver  (\sigma (\mathbf u
 )) \, d\mathbf{x}\, dt  \,.
 \end{align*}
 We have
\begin{eqnarray*}
\lefteqn{ \int_S^T \int_\Omega \sigma_{i j , j }
 (\mathbf u ) h_k u_{i ,k} \, d\mathbf{x}\, dt } \\
 &=&  \int_S^T \int_\Gamma \sigma_{i j }
 (\mathbf u ) \nu_j h_k u_{i ,k} \, d\Gamma\, dt
 - \int_S^T \int_\Omega \sigma_{i j }
 (\mathbf u ) \partial_j (h_k u_{i ,k} ) \, d\mathbf{x}\, dt
 \\
 &=&  \int_S^T \int_\Gamma \sigma (\mathbf u ) \nu . (\mathbf{h} .\nabla )
 \mathbf u \, d\Gamma\, dt - \int_S^T \int_\Omega \sigma_{i j }
 (\mathbf u ) h_{k, j} u_{i ,k} \, d\mathbf{x}\, dt
 \\
 && - \int_S^T \int_\Omega \sigma_{i j } (\mathbf u ) h_k u_{i ,j k} \,
 d\mathbf{x}\, dt  \,,
\end{eqnarray*}
 and
\begin{align*}
 - &\int_S^T \int_\Omega \sigma_{i j }
 (\mathbf u ) h_k u_{i ,j k} \, d\mathbf{x}\, dt
 \\
=& - \frac{1}{2} \int_S^T \int_\Omega h_k
 \partial_k (\sigma (\mathbf u ) \colon \varepsilon (\mathbf u )) \, d\mathbf{x}\, dt
 + \frac{1}{2}\int_S^T \int_\Omega h_m \partial_m (a_{i j kl})
 \varepsilon_{kl} (\mathbf u ) \varepsilon_{i j }(\mathbf u ) \,d\mathbf{x}\, dt
 \\
=& \frac{1}{2} \int_S^T \int_\Omega \diver  (\mathbf{h} )
 \, \sigma (\mathbf u ) \colon \varepsilon (\mathbf u ) \, d\mathbf{x}\, dt
 - \frac{1}{2} \int_S^T \int_\Gamma \mathbf{h} . \nu \,
 \sigma (\mathbf u ) \colon \varepsilon (\mathbf u ) \, d\Gamma\, dt  \,
 \\
&+  \frac{1}{2}\int_S^T \int_\Omega
 h_m \partial_m (a_{i j kl}) \varepsilon_{kl} (\mathbf u )
 \varepsilon_{i j }(\mathbf u ) \,d\mathbf{x}\, dt  \,.
\end{align*}
 Furthermore,
 $$ \int_S^T \int_\Omega \mathbf u . \diver  (\sigma (\mathbf u )) \, d\mathbf{x}\, dt = \int_S^T
 \int_\Gamma (\sigma (\mathbf u ) \nu ).\mathbf u \, d\Gamma\, dt - \int_S^T \int_\Omega
 \sigma (\mathbf u ) \colon \varepsilon (\mathbf u )  \, d\mathbf{x}\, dt \,. $$
 Hence
\begin{equation}  \label{L23}
 \begin{aligned}
 \int_S^T \int_\Omega& \diver  (\sigma (\mathbf u )) .M\mathbf u \,
  d\mathbf{x}\, dt \\
 =& \int_S^T \int_\Omega ( \diver  (\mathbf{h} ) - \beta )
 \, \sigma (\mathbf u ) \colon \varepsilon (\mathbf u ) \, d\mathbf{x}\, dt
 -  2 \int_S^T \int_\Omega \sigma_{i j }
 (\mathbf u ) h_{k, j} u_{i ,k} \, d\mathbf{x}\, dt   \\
 & +  \int_S^T \int_\Omega h_m \partial_m (a_{i j
 kl}) \varepsilon_{kl} (\mathbf u) \varepsilon_{i j } (\mathbf u) \, d\mathbf{x}\, dt
  \\
 & + \int_S^T \int_\Gamma \left( (\sigma (\mathbf u ) \nu ).M\mathbf u
 - \mathbf{h} .\nu \, \sigma (\mathbf u ) \colon \varepsilon (\mathbf u ) \right) \, d\Gamma
 dt \,.
 \end{aligned}
\end{equation}
 We deduce the desired result from $(\ref{L21} )$, $(\ref{L22} )$ and
 $(\ref{L23} )$.
\hfill $\Box $

\begin{lemma} \label{lma3}
 There exist $\beta > 0 $ and $C_1 > 0 $ such that
\begin{eqnarray*}
C_1 \int_S^T E \, dt &\le&
- \left[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \right]_S^T
 \\
 && +  \int_S^T \int_{\Gamma_0 } \mathbf{h} .\nu \left( \mu
 |\partial_\nu \mathbf u_T |^2 + (2 \mu +\lambda ) |\partial_\nu u_\nu |^2
 \right) \, d\Gamma\, dt
 \\
 && +  \int_S^T \int_{\Gamma_1 } \left(
 \Bigl( \frac{C_1}{2} - \beta \Bigr) A |\mathbf u |^2 - \beta B \mathbf u .\mathbf u' +
 \mathbf{h} .\nu \, |\mathbf u' |^2 \right) \, d\Gamma\, dt
 \\
 && - \int_S^T \int_{\Gamma_1 }\left( 2 (A\mathbf u +B\mathbf u' ).(\mathbf{h}
 .\nabla )\mathbf u + \mathbf{h} .\nu \, \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) \right)
 \, d\Gamma\, dt  \,.
\end{eqnarray*}
\end{lemma}

 \noindent {\sl Proof.}
 Lemma 2, $(\ref{H3} )$ and $(\ref{H4} )$ give
\begin{align*}
 \int_S^T \int_\Omega &\left( (\diver  (\mathbf{h} ) - \beta ) \left(
 |\mathbf u' |^2 - \sigma (\mathbf u ) \colon \varepsilon(\mathbf u ) \right) +
 (2 \alpha_{\mathbf{h}} - \gamma_{\mathbf{h}}) \, \sigma (\mathbf u ) \colon \varepsilon(\mathbf u )
 \right) \, d\mathbf{x}\, dt
 \\
 \leq & - \left[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x}
 \right]_S^T + \int_S^T \int_{\Gamma_1 } \mathbf{h} .\nu \, |\mathbf u' |^2
 \, d\Gamma\, dt
 \\
 & + \int_S^T \int_\Gamma \left( (\sigma(\mathbf u )\nu
 ).M\mathbf u - \mathbf{h} .\nu \, \sigma(\mathbf u )\colon \varepsilon(\mathbf u )\right) \,
 d\Gamma\, dt  \,.
 \end{align*}
 Since $(\ref{H4} )$ is satisfied, we can choose $\beta >0 $ such that
\begin{equation}
 \label{L30}
 \beta < \diver  (\mathbf{h} ) < 2 \alpha_{\mathbf{h}} + \beta  - \gamma_{\mathbf{h}}\,, \quad
  \hbox{in\ } \Omega \,,
\end{equation}
 and choose $C_1 > 0$ such that
\begin{equation}
 \label{L31}
 \begin{aligned}
  C_1\int_S^T E \, dt  \leq
 &  - \Big[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x}
 \Big]_S^T  \\
 & +  \int_S^T \int_{\Gamma_1 } \Bigl(
 \frac{C_1}{2} A|\mathbf u |^2 + \mathbf{h} .\nu \, |\mathbf u' |^2 \Bigr) \,
  d\Gamma\, dt  \\
 & +  \int_S^T \int_\Gamma \left( (\sigma
 (\mathbf u )\nu ).M\mathbf u - \mathbf{h} .\nu \, \sigma(\mathbf u )\colon \varepsilon(\mathbf u )
 \right) \, d\Gamma\, dt  \,.
 \end{aligned}
\end{equation}
 We write $\mathbf u = \mathbf u_T + u_\nu \nu $ and $\mathbf{h} = \mathbf{h}_T + h_\nu \nu $
 on $\Gamma $. With Dirichlet boundary condition, we get on $\Gamma_0 $
 $$ M\mathbf u = 2 (\mathbf{h} .\nabla )\mathbf u = 2 \mathbf{h} . \nu \, \partial_\nu \mathbf u \,, \quad
 \varepsilon_T (\mathbf u ) = 0 \,, \quad 2 \varepsilon_S (\mathbf u ) =
 \partial_\nu \mathbf u_T  \,. $$
 and, with our notations, using Remark 3, we get
 \begin{gather*}
 ( \sigma (\mathbf u )\nu ).M\mathbf u
 = 2\mathbf{h}.\nu \bigl({\overline{\sigma_S (\mathbf u )}}\partial_\nu
 \mathbf u_T + \sigma_\nu (\mathbf u ) \partial_\nu \mathbf u_\nu \bigr) \,, \quad
 \hbox{on } \Gamma_0 \,,
 \\
 \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) =
 \overline{\sigma_S (\mathbf u )} \partial_\nu \mathbf u_T +
 \sigma_\nu (\mathbf u ) \partial_\nu \mathbf u_\nu \,, \quad \hbox{on }
 \Gamma_0 \,,
  \\
 (\sigma (\mathbf u ) \nu ).M\mathbf u - \mathbf{h}.\nu \,
 \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) = \mathbf{h}.\nu \,
 \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) \,, \quad \hbox{on }
 \Gamma_0 \,.
 \end{gather*}
 Hence
\begin{equation}
 \label{L32}
 \int_{\Gamma_0 } \left( (\sigma(\mathbf u )\nu ).M\mathbf u - \mathbf{h} .\nu \,
 \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) \right) \, d\Gamma
 = \int_{\Gamma_0 } \mathbf{h} .\nu \, \sigma(\mathbf u ) \colon \varepsilon(\mathbf u ) \,
 d\Gamma \,.
\end{equation}
 Using the boundary conditon on $\Gamma_1 $, we get
\begin{equation}
 \label{L33}
  \int_{\Gamma_1 } (\sigma(\mathbf u )\nu ).M\mathbf u \, d\Gamma
 = -\int_{\Gamma_1 } 2 (A\mathbf u + B\mathbf u' ).(\mathbf{h} .\nabla )
 \mathbf u \,  d\Gamma - \beta \int_{\Gamma_1 } \mathbf u .(A\mathbf u
 + B\mathbf u' ) \, d\Gamma \,.
\end{equation}
 We deduce the result from $(\ref{L31} )$, $(\ref{L32} )$ and $(\ref{L33}
 )$.
\hfill $\Box $


\begin{lemma} \label{lma4}
 There exists $C_2 >0 $ such that
 $$\big| \int_\Omega \mathbf u' (t).M\mathbf u (t) \, d\mathbf{x} \big|
 \le C_2 E(\mathbf u ,t) \,, \quad \forall t \ge 0 \,. $$
\end{lemma}


 \noindent {\sl Proof.}
Given $t\ge 0$, for every $\eta > 0$, we can write
 $$
 \big| \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \big| \le \frac{\eta }{2}
 \|\mathbf u' \|_{\mathbb{L}^2 (\Omega )}^2 + \frac{1}{2\eta }
 \|M\mathbf u \|_{\mathbb{L}^2 (\Omega )}^2 \,. $$
 We have
\begin{eqnarray*}
\|M\mathbf u \|_{\mathbb{L}^2 (\Omega )}^2
 &=& \int_\Omega (|2(\mathbf{h} .\nabla )\mathbf u |^2
 + \beta^2 |\mathbf u |^2 +
 4 \beta \mathbf u .(\mathbf{h} .\nabla )\mathbf u ) \, d\mathbf{x}  \\
&=& \int_\Omega (|2(\mathbf{h} .\nabla )\mathbf u |^2 + \beta^2 |\mathbf u |^2 +
 2\beta \nabla (|\mathbf u |^2 ) \mathbf{h} ) \, d\mathbf{x}
 \\
&=& \int_\Omega (|2(\mathbf{h}. \nabla )\mathbf u |^2 + \beta (\beta -
 2 \diver  (\mathbf{h} ) ) |\mathbf u |^2 ) \, d\mathbf{x}
 + 2 \beta \int_{\Gamma_1 } \mathbf{h}. \nu \,
 |\mathbf u |^2 \, d\Gamma \,.
\end{eqnarray*}
 Setting $ R = \sup_{\overline{\Omega }} |\mathbf{h} | $,
 from $(\ref{L30} )$, we get
 $$
 \| M\mathbf u \|_{\mathbb{L}^2 (\Omega )}^2 \le 4 R^2 \int_\Omega |\nabla \mathbf u |^2 \,
 d\mathbf{x} + 2 \beta \int_{\Gamma_1 } \mathbf{h} .\nu \, |\mathbf u |^2 \,
 d\Gamma \,. $$
 With Korn's inequality, we can find the smallest positive real number
 $R_1 $ (depending on $\mathbf{h} $ and $\beta $) such that
 for all $\mathbf{v} \in \mathbb{H}^1_{\Gamma_0 } (\Gamma )$,
$$
4 R_1^2 \left( \int_\Omega \sigma (\mathbf{v} ) \colon
 \varepsilon (\mathbf{v} ) \, d\mathbf{x} + \int_{\Gamma_1 }A|\mathbf{v} |^2
 \, d\Gamma \right)
  \ge  4 R^2 \int_\Omega |
 \nabla \mathbf{v} |^2 \, d\mathbf{x} + 2 \beta \int_{\Gamma_1 } \mathbf{h} .\nu \, |\mathbf{v} |^2 \,
 d\Gamma \,.
$$
 It follows that
 $$
\Big| \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \Big|
 \le \frac{\eta }{2}
 \|\mathbf u' \|_{\mathbb{L}^2 (\Omega )}^2 + \frac{2 R_1^2 }{\eta }
 \Bigl( \int_\Omega \sigma (\mathbf u ) \colon \varepsilon (\mathbf u )
 \, d\mathbf{x} + \int_{\Gamma_1 }
 A|\mathbf u |^2 \, d\Gamma \Bigr) \,.
$$
 The choice $\eta = 2 R_1 $ gives the result with $C_2 = 2 R_1 $.
\hfill $\Box $

\begin{lemma}  \label{lma5}
 There exists $C_3 >0 $ such that, for every $\eta $ in $(0,1)$,
 $$ \int_S^T \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma\, dt \le \frac{C_3}
 {\eta } E(\mathbf u ,S) + \eta \int_S^T E(\mathbf u ,t) \, dt \,, \quad
 0 \le S < T <+ \infty \,. $$
\end{lemma}

 \noindent {\sl Proof.}
We proceed as in \cite{Ko}. We define $\mathbf{z} $, depending on
 $t$, as follows:
\begin{gather*}
 \diver  (\sigma (\mathbf{z} )) = 0 \,, \quad \hbox{in } \Omega \,,
 \\
 \mathbf{z} = \mathbf u \,, \quad \hbox{on } \Gamma \,.
\end{gather*}
 We have
 $$
 \int_{\Omega } \mathbf{z} .\diver  (\sigma (\mathbf{v} )) \, d\mathbf{x} = \int_{\Gamma_1 }
 \mathbf u. (\sigma (\mathbf{v} ) \nu ) \, d\Gamma \,, \quad \forall \mathbf{v} \in \mathbb{H}^2
 (\Omega ) \cap \mathbb{H}^1_0 (\Omega ) \,.
$$
 Using the definition of the energy functional and Proposition 3, we can
 find some positive constants $c_1$, $c_2$, $c_1'$, $c_2'$ such that
\begin{gather*}
 \int_\Omega |\mathbf{z} |^2 \, d\mathbf{x} \le c_1 \int_{\Gamma_1 }
 |\mathbf u |^2 \, d\Gamma \le c_2 E \, ;
 \\
\int_\Omega |\mathbf{z}' |^2 \, d\mathbf{x} \le c_1' \int_{\Gamma_1 }
 |\mathbf u' |^2 \, d\Gamma \le c_2' (-E') \,.
\end{gather*}
 Furthermore, we have
 $$
\int_\Omega \sigma (\mathbf{z} ) \colon \varepsilon (\mathbf u -\mathbf{z} ) \, d\mathbf{x} =
 - \int_\Omega (\mathbf u - \mathbf{z} ).\diver  (\sigma (\mathbf{z} )) \, d\mathbf{x} + \int_{\Gamma_1 }
 (\mathbf u - \mathbf{z} ).(\sigma (\mathbf{z} )\nu )\, d\Gamma = 0 \,.
 $$
 Then
 $$ \int_\Omega \sigma (\mathbf{z} ) \colon \varepsilon (\mathbf u ) \, d\mathbf{x} =
 \int_\Omega \sigma (\mathbf{z} ) \colon \varepsilon (\mathbf{z} ) \,
 d\mathbf{x}  \ge 0 \,.
$$
 From $(\ref{E} )$, we deduce
\begin{align*}
0= &  \int_\Omega \mathbf{z} . (\mathbf u'' - \diver  (\sigma (\mathbf u )) \,
 d\mathbf{x}\, dt  \\
= & \int_\Omega \mathbf{z} .\mathbf u'' \, d\mathbf{x} +
 \int_\Omega \sigma (\mathbf{z} ) \colon \varepsilon (\mathbf u ) \, d\mathbf{x} -
 \int_{\Gamma_1 } \mathbf{z} .(\sigma (\mathbf u )\nu ) \, d\Gamma
 \\
= & \int_\Omega \mathbf{z} .\mathbf u'' d\mathbf{x} +
 \int_\Omega \sigma (\mathbf{z} ) \colon \varepsilon (\mathbf u ) \, d\mathbf{x} +
 \int_{\Gamma_1 } \mathbf u .(A\mathbf u + B\mathbf u' ) \, d\Gamma  \,.
\end{align*}
 Hence
 $$
 \int_{\Gamma_1 } A |\mathbf u |^2 \, d\Gamma \le - \int_\Omega \mathbf{z}
 .\mathbf u'' \, d\mathbf{x} - \int_{\Gamma_1 } B \mathbf u . \mathbf u' \,
 d\Gamma \,. $$
 For $ 0 < S < T < \infty $, we obtain
$$
 \int_S^T \int_{\Gamma_1 } A |\mathbf u |^2 \, d\Gamma\, dt
\leq  - \Big[ \int_\Omega \mathbf{z} .\mathbf u' \, d\mathbf{x} \Big]_S^T
 + \int_S^T \int_\Omega \mathbf{z}' .\mathbf u' \, d\mathbf{x}\, dt
 - \int_S^T \int_{\Gamma_1 } B \mathbf u . \mathbf u' \, d\Gamma\, dt \,.
$$
 Let $C$ be a positive constant, large enough.
 Using the Cauchy-Schwarz inequality and the estimates obtained above,
 we can write for every $\theta > 0$,
\begin{eqnarray*}
 \int_S^T \int_{\Gamma_1 } A |\mathbf u |^2 \, d\Gamma\, dt
 & \le&
  C E(\mathbf u ,S) + C \int_S^T (-E'(\mathbf u ,t))^{1/2} (E(\mathbf u ,t)
 )^{1/2} \, dt  \\
 & & + B \int_S^T \int_{\Gamma_1 } |\mathbf u |\, |\mathbf u' | \,
 d\Gamma\, dt
 \\
 &\le&
 C E(\mathbf u ,S) + \frac{C \theta}{2} \int_S^T E(\mathbf u ,t) \, dt
 + \frac{C}{2\theta } E(\mathbf u ,S)  \\
 & & + \frac{1}{2} \int_S^T \int_{\Gamma_1 } A |\mathbf u |^2 \,
 d\Gamma\, dt +\frac{B^2}{2A}  \int_S^T \int_{\Gamma_1 } |\mathbf u' |^2 \,
 d\Gamma\, dt \,.
\end{eqnarray*}
 From Proposition 3, we get
 $$ \int_S^T \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma\, dt \le
 \left( \frac{2C}{A} + \frac{C}{A \theta } + \frac{B}{A^2} \right)
 E(\mathbf u ,S)
 + \frac{C \theta }{A} \int_S^T E(\mathbf u ,t) \, dt \,. $$
 We now choose ${\theta = \frac{A\eta }{C} } $ and obtain the
 desired result.
\hfill $\Box $


\subsection*{Proof of Theorem 1}

 We assume $(\ref{H1} )$ and that $\mathbf{h} $ satisfies $(\ref{H3} )$ and
 $(\ref{H4} )$.
 From Lemma 5, it suffices to consider the case
 $\mathop{\rm meas} (\Gamma_0) \not= 0 $.
 We first suppose $(\ref{sf} )$ and we consider the (strong)
 solution $\mathbf u $ of $(\ref{E} )$.

 The energy functional is non-increasing (Proposition 3). From
 Lemma 4, we deduce
 $$ - \Big[ \int_\Omega \mathbf u' .M\mathbf u \, d\mathbf{x} \Big]_S^T \le 2C_2 E(\mathbf u ,S)
 \,.
 $$
 Since $\mathbf{h} .\nu \le 0$ on $\Gamma_0$, Lemma 3 gives
\begin{eqnarray*}
C_1 \int_S^T E \, dt
 &\leq&  2C_2 E(\mathbf u ,S)  \\
 && + \int_S^T \int_{\Gamma_1 } \Bigl(
 \bigl( \frac{C_1}{2} - \beta \bigr) A |\mathbf u |^2 - \beta B \mathbf u
 .\mathbf u' + \mathbf{h}.\nu \, |\mathbf u' |^2 \Bigr) \, d\Gamma\, dt
 \\
 && - { \int_S^T \int_{\Gamma_1 } \left( 2 (A\mathbf u +B\mathbf u' ).(\mathbf{h}
 .\nabla )\mathbf u + \mathbf{h} .\nu \, \sigma(\mathbf u ) \colon \varepsilon(\mathbf u )\right)
 \, d\Gamma\, dt } \,.
\end{eqnarray*}
 There exists $c>0$ such that
 $$
 \big| \int_S^T \int_{\Gamma_1 }\beta \mathbf u .\mathbf u' \, d\Gamma\, dt
 \big|
 \le \int_S^T \int_{\Gamma_1 }|\mathbf u |^2 \, d\Gamma\, dt + c \int_S^T
 \int_{\Gamma_1 }|\mathbf u' |^2 \, d\Gamma\, dt \,.
 $$
 Hence, using Proposition 3, we can find $C_4 >0$ and $C_5 >0$ such that
\begin{align*}
C_1 \int_S^T E \, dt
 \le& C_4 E(\mathbf u ,S) + C_5 \int_S^T \int_{\Gamma_1 }  |\mathbf u |^2
 \, d\Gamma\, dt
 \\
 & - \int_S^T \int_{\Gamma_1 } \left( 2 (A\mathbf u +B\mathbf u' ).(\mathbf{h}
 .\nabla )\mathbf u + \mathbf{h} .\nu \, \sigma(\mathbf u ) \colon \varepsilon(\mathbf u )\right)
 \, d\Gamma\, dt \,.
\end{align*}
 From $(\ref{H11} )$, Remarks 1 and 3, we get
\begin{equation}  \label{T11}
 \begin{aligned}
  C_1 \int_S^T E \, dt
 \leq &  C_4 E(\mathbf u ,S) + C_5 \int_S^T
 \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma\, dt  \\
 & -  \int_S^T \int_{\Gamma_1 } 2 (A\mathbf u +B\mathbf u' ).
 (\mathbf{h} .\nabla ) \mathbf u \, d\Gamma\, dt  \\
 & -  k \alpha \int_S^T \int_{\Gamma_1 }
 \left(\varepsilon_T (\mathbf u ) \colon \varepsilon_T (\mathbf u ) +
 2 |\varepsilon_S (\mathbf u )|^2 + |\varepsilon_\nu(\mathbf u )|^2  \right) \,
 d\Gamma\, dt \,.
 \end{aligned}
\end{equation}
 Now, we estimate two integrals which appear on the right hand side
 (second line of the above formula). \smallskip


 \noindent {\sl Estimation of} $ \mathcal{I}_1 =
 \int_S^T \int_{\Gamma_1 } 2 A \mathbf u .(\mathbf{h} .\nabla )\mathbf u \,
 d\Gamma\, dt $.

\noindent We denote by $C$ some positive constant which is independent
 of $\mathbf u $ and large enough.
 We have $\mathbf u .(\mathbf{h} .\nabla ) \mathbf u = \frac{1}{2}
 \mathbf{h} . \nabla (| \mathbf u |^2 )$.
 Setting $\mathbf u = \mathbf u_T + u_\nu \nu $ and
 $\mathbf{h} = \mathbf{h}_T + h_\nu \nu $ on $\Gamma_1 $, we use $(\ref{gra} )$ and get
 $$
 \mathbf u .(\mathbf{h} .\nabla ) \mathbf u = \frac{1}{2} \nabla_T (|\mathbf u |^2 ) . \mathbf{h}_T
 + h_\nu \mathbf u_T .(\partial_\nu \mathbf u_T ) + h_\nu u_\nu (\partial_\nu u_\nu )
 \,, \quad \hbox{on } \Gamma_1 \,.
 $$
 Since
 $$\int_{\Gamma_1 } A \nabla_T (|\mathbf u |^2 ) . \mathbf{h}_T \, d\Gamma  =
 - \int_{\Gamma_1 } A |\mathbf u |^2 \diver _T (\mathbf{h}_T ) \, d\Gamma  \,, $$
 hence
\begin{equation}
 \label{I10}
 \big| \int_{\Gamma_1 }  A \nabla_T (|\mathbf u |^2 ) . \mathbf{h}_T \, d\Gamma
 \big| \le C \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma \,.
\end{equation}
 Using $\varepsilon_S (\mathbf u ) $ (see subsection 1.2), we can write
 $$
 h_\nu \, \mathbf u_T .\partial_\nu \mathbf u_T = h_\nu \, \mathbf u_T .(
 2\varepsilon_S (\mathbf u ) + (\partial_T \nu ) \mathbf u_T -\nabla_T u_\nu
 ) \,, \quad \hbox{on } \Gamma_1 \,.
$$
 Let $\theta $ be some positive number. We have
\begin{equation}
 \label{I11}
 \big| \int_{\Gamma_1 } 4 A h_\nu \, \mathbf u_T .\varepsilon_S (\mathbf u ) \,
 d\Gamma \big|
 \le \theta\int_{\Gamma_1 } |\varepsilon_S (\mathbf u )|^2 \, d\Gamma +
 \frac{C}{\theta } \int_{\Gamma_1 } |\mathbf u_T |^2 \, d\Gamma \,.
\end{equation}
 Since $h_\nu $ and $\partial_T \nu $ are bounded, we get
\begin{equation}
 \label{I12}
 \big| \int_{\Gamma_1 } 2 A h_\nu \, \mathbf u_T .(\partial_T \nu ) \mathbf u_T \,
 d\Gamma \big| \le C \int_{\Gamma_1 } |\mathbf u_T |^2 \, d\Gamma \,.
\end{equation}
 Now, observe
\begin{align*}
 \int_{\Gamma_1 } h_\nu \, \mathbf u_T .\nabla_T u_\nu \,
 d\Gamma
 =& - \int_{\Gamma_1 } u_\nu \diver _T (h_\nu \mathbf u_T ) \, d\Gamma
 \\
 =& - \int_{\Gamma_1 } h_\nu \, u_\nu \diver _T (\mathbf u_T ) \,
 d\Gamma - \int_{\Gamma_1 } u_\nu \nabla_T h_\nu .\mathbf u_T \, d\Gamma
 \,.
\end{align*}
 Proposition 1 implies
 $$
 \big| \int_{\Gamma_1 } 2 A h_\nu \, u_\nu \diver _T (\mathbf u_T ) \,
 d\Gamma  \big|
 \le \theta \| \mathbf u_T \|_1^2 + \frac{C}{\theta } \int_{\Gamma_1 }
 |\mathbf u |^2 \, d\Gamma \,.
 $$
 Hence
\begin{equation}
 \label{I13}
 \big| \int_{\Gamma_1 } 2 A h_\nu \, \mathbf u_T .\nabla_T u_\nu \, d\Gamma
 \big| \le
 \theta \int_{\Gamma_1 } \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T
 (\mathbf u_T ) \, d\Gamma + \frac{C}{\theta } \int_{\Gamma_1 } |\mathbf u |^2 \,
 d\Gamma \,.
\end{equation}
 We can also write
\begin{equation}
 \label{I14}
 \big| \int_{\Gamma_1 } 2 A h_\nu \, u_\nu (\partial_\nu u_\nu ) \,
 d\Gamma
 \big| \le \theta \int_{\Gamma_1 } |\partial_\nu u_\nu |^2 \, d\Gamma +
 \frac{C}{\theta } \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma \,.
\end{equation}
 Finally, $(\ref{I10} )$--$(\ref{I14} )$ give
\begin{equation}  \label{T12}
\begin{aligned}
 | \mathcal{I}_1 |  \le&
 \theta \int_S^T \int_{\Gamma_1 } \left( |\partial_\nu
 u_\nu |^2 + \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T )
  +  |\varepsilon_S (\mathbf u )|^2 \right) \, d\Gamma\, dt \\
 &+ \frac{C}{\theta } \int_S^T \int_{\Gamma_1 } |\mathbf u |^2
 \, d\Gamma\, dt \,.
\end{aligned}
\end{equation}
 We emphasize that, in $(\ref{T12} )$, $\theta $ is a positive number
 to be chosen later and $C$ is a positive constant which does not
 depend on $\mathbf u $. \smallskip


 \noindent {\sl Estimation of}
 $\mathcal{I}_2 = \int_S^T \int_{\Gamma_1 } 2 B \mathbf u' .(\mathbf{h}
 .\nabla )\mathbf u \, d\Gamma\, dt $.

\noindent Here, we use $(\ref{dif} )$ and get
\begin{align*}
 \mathbf u' .(\mathbf{h} .\nabla )\mathbf u
 = & \overline{\mathbf u'_T } (\partial_T \mathbf u_T ) \mathbf{h}_T
 + u_\nu \overline{\mathbf u'_T } (\partial_T \nu ) \mathbf{h}_T
 + h_\nu \overline{\mathbf u'_T } \partial_\nu \mathbf u_T
 \\
 & + u'_\nu (\partial_T u_\nu ) \mathbf{h}_T
 - u'_\nu \overline{\mathbf u_T } (\partial_T \nu ) \mathbf{h}_T
 + u'_\nu (\partial_\nu u_\nu ) h_\nu \,,
 \quad \hbox{on } \Gamma_1 \,.
\end{align*}
 This can be rewritten as
\begin{align*}
 \mathbf u' .(\mathbf{h} .\nabla )\mathbf u =
 & \mathbf u'_T .(\partial_T \mathbf u_T ) \mathbf{h}_T
 + ( u_\nu \mathbf u'_T - u'_\nu \mathbf u_T ).(\partial_T \nu )
 \mathbf{h}_T  \\
 & + u'_\nu \nabla_T u_\nu .\mathbf{h}_T
 + h_\nu (\mathbf u'_T .\partial_\nu \mathbf u_T
 + u'_\nu (\partial_\nu u_\nu ) )  \,,
 \quad \hbox{on } \Gamma_1 \,.
\end{align*}
 Since $\mathbf{h} $ and $\partial_T \nu $ are bounded, we get
\begin{gather}
 \label{I20}
 \big| \int_{\Gamma_1 } 2 B \mathbf u'_T .(\partial_T \mathbf u_T ) \mathbf{h}_T \, d\Gamma
 \big| \le \frac{\theta}{2} \| \mathbf u_T \|_1^2 + \frac{2C}{\theta }
 \int_{\Gamma_1 } |\mathbf u'_T |^2 \, d\Gamma \,,
\\
 \label{I21}
 \big| \int_{\Gamma_1 } 2 B ( u_\nu \mathbf u'_T - u'_\nu \mathbf u_T ).
 (\partial_T  \nu ) \mathbf{h}_T \, d\Gamma \big|
 \le C\Bigl( \int_{\Gamma_1 } |\mathbf u |^2 \,
 d\Gamma +  \int_{\Gamma_1 } |\mathbf u' |^2 \, d\Gamma \Bigr) \,.
\end{gather}
 Under the assumptions about $\Omega $, we observe that $\Gamma_1 $ is
 a compact manifold of dimension $2$. So, we can build a finite number of
 local maps $(U_1,\phi_1), \ldots ,$ $ (U_k,\phi_k)$ and an
 associated partition of unity $(\vartheta_1 , \ldots ,
 \vartheta_k )$. We have
\begin{eqnarray*}
 \int_{\Gamma_1 } 2 B u'_\nu \nabla_T u_\nu .\mathbf{h}_T \, d\Gamma =
 \sum_{j =1 }^\ell \int_{U_j } 2 B \vartheta_j u'_\nu
 \nabla_T u_\nu .\mathbf{h}_T \, d\Gamma \,.
\end{eqnarray*}
 Consider one of the $\ell $ terms of the previous sum. Omitting the
 index $j $, we denote ${\int_U 2 B
 \vartheta u'_\nu \nabla_T u_\nu .\mathbf{h}_T \, d\Gamma \, } $.
Using the notation introduced in section 1, we write
 $\mathbf{h}_T = h^1 a_1 + h^2 a_2 $. Setting $|g|= |\det (g)| $,
 $W=\phi^{-1} (U) $, we get
\begin{align*}
\int_U &2 B \vartheta u'_\nu \nabla_T u_\nu .\mathbf{h}_T \,
 d\Gamma\\
=& \int_W 2 B (\vartheta \circ \phi )( u'_\nu \circ
 \phi ) \Bigl( \frac{\partial (u_\nu \circ \phi )}{\partial \xi_1} h^1 +
 \frac{\partial (u_\nu \circ \phi )}{\partial \xi_2} h^2 \Bigr)
 |g|^{1/2} \, d\xi_1 d\xi_2 \,.
\end{align*}
 Since $\vartheta \circ \phi $ is continuous and compactly
 supported, $v_\nu = u_\nu \circ \phi $ belongs to $\mathrm{H}^{1/2} (W) $ and
  $\| v_\nu \|_{\mathrm{H}^{1/2} (W)} \le C \| u_\nu \|_{\mathrm{H}^{1/2} (U)}$.
 Let us define two subsets of $W$
 $$
 W^{+} = \{ (\xi_1 , \xi_2 )\in W \, / \, h^1 (\xi_1 , \xi_2 ) > 0 \}\,,
 \quad
 W^{-} = \{ (\xi_1 , \xi_2 ) \in W\, / \, h^1 (\xi_1 , \xi_2 ) < 0 \}
 \,.
$$
 We have
\begin{align*}
\int_W 2 B (\vartheta \circ \phi ) v'_\nu
 \frac{\partial v_\nu}{\partial \xi_1} h^1 |g|^{1/2} \, d\xi_1 d\xi_2
 =& \int_{W^{+}} 2 B (\vartheta \circ \phi ) v'_\nu
 \frac{\partial v_\nu }{\partial \xi_1} h^1 |g|^{1/2} \, d\xi_1 d\xi_2
 \\
 &+ \int_{W^{-}} 2 B (\vartheta \circ \phi ) v'_\nu
 \frac{\partial v_\nu }{\partial \xi_1} h^1 |g|^{1/2} \, d\xi_1 d\xi_2
 \,.
\end{align*}
 Setting ${ \psi = \left( (\vartheta \circ \phi ) h^1
 |g|^{1/2} \right)^{1/2} } $, in $W^{+} $, we have
\begin{eqnarray*}
\lefteqn{ \int_{W^{+}} 2 B (\vartheta \circ \phi ) v'_\nu
 \frac{\partial v_\nu }{\partial \xi_1} h^1 |g|^{1/2} \, d\xi_1 d\xi_2 }\\
 &=&  \int_{W^{+}} 2B \psi^2 v'_\nu \frac{\partial v_\nu }
 {\partial \xi_1} \, d\xi_1  d\xi_2  \\
 &=&  \int_{W^{+}} 2B \psi v'_\nu \frac{\partial (\psi
 v_\nu )} {\partial \xi_1} \, d\xi_1 d\xi_2 - \int_{W^{+}} B \psi
 \left(|v_\nu |^2 \right)' \frac{\partial \psi} {\partial \xi_1} \,
 d\xi_1 d\xi_2 \,.
\end{eqnarray*}
 Thus
 $$ \Big| \Big[ \int_{W^{+}} B \psi |v_\nu |^2 \frac{\partial \psi}
 {\partial \xi_1} \, d\xi_1 d\xi_2 \Big]_S^T \Big| \le C E(\mathbf u ,S)
 \,.
 $$
 We know that $\psi $ is compactly supported in $W$, and $\psi =
 0$, on $\partial W^{+} $. Define function
 $$G = \psi v_\nu \,, \, \hbox{ in\ } W^{+} \times \mathbb{R}_{+} \,, \quad
 G = 0 \,, \, \hbox{ in\ } (\mathbb{R}^2 \setminus W^{+} ) \times \mathbb{R}_{+} \,. $$
 We have
 $$\int_{W^{+}} 2B \psi v'_\nu \frac{\partial (\psi v_\nu )}{\partial
 \xi_1} \, d\xi_1 d\xi_2 = \int_{\mathbb{R}^2 } 2B G' \frac{\partial G}
 {\partial \xi_1} \, d\xi_1 d\xi_2 \,.
$$
 Let $\hat G $ be the Fourier transform of $G$, with respect to
 $\xi_1 $. We write
 $$\int_{W^{+}} 2B \psi v'_\nu \frac{\partial (\psi v_\nu )}{\partial
 \xi_1} \, d\xi_1 d\xi_2 = \int_{\mathbb{R}^2 } 4B i \pi \eta_1 \hat G'
 \hat G \, d\eta_1 d\xi_2 \,. $$
 This implies
 $$\int_S^T \int_{W^{+}} 2B \psi v'_\nu \frac{\partial (\psi v_\nu )}
 {\partial \xi_1} \, d\xi_1 d\xi_2 dt = \Big[ \int_{\mathbb{R}^2 } 2B
 i \pi \eta_1 |\hat G |^2 \, d\eta_1 d\xi_2 \Big]_S^T \,.
 $$
 But
 $$ \big| \int_{\mathbb{R}^2 } \eta_1 |\hat G |^2 \, d\eta_1 d\xi_2 \big|
 \le C_1 \| G \|_{\mathrm{H}^{1/2} (\mathbb{R}^2 )}^2 \le C_2 \| u_\nu \|_{\mathrm{H}^{1/2}
 (\Gamma_1 )}^2 \,. $$
 Hence, using the energy functional and Proposition 3, we get
 $$
 \Big| \int_S^T \int_{W^{+}} 2B \psi v'_\nu \frac{\partial
 (\psi v_\nu )}{\partial \xi_1} \, d\xi_1 d\xi_2 dt \Big|
 \le C E(\mathbf u , S) \,. 
 $$
 For the integral in $W^{-} $, we replace $a_1$ by
 $-a_1$, $h^1$ by $-h^1$, respectively and proceed as above. We can
 also get a similar result concerning the integral terms containing
 $h^2$.

 Finally, we obtain
\begin{equation}
 \label{I22}
 \Big| \int_S^T \int_{\Gamma_1 } 2 B u'_\nu \nabla_T u_\nu .\mathbf{h}_T \,
 d\Gamma \, dt \Big| \le C E(\mathbf u , S) \,.
\end{equation}
 Using $\varepsilon_S (\mathbf u ) $,
 $$ h_\nu \, \mathbf u'_T .\partial_\nu \mathbf u_T = h_\nu \, \mathbf u'_T .
 (2\varepsilon_S (\mathbf u ) -(\nabla_T u_\nu )
 + \partial_T \nu \mathbf u_T ) \,, \quad \hbox{on } \Gamma_1 \,, $$
 and, for $( \ref{I11} )$ and $( \ref{I12} )$, respectively,
\begin{gather}
 \label{I23}
 \big| \int_{\Gamma_1 } 4 B h_\nu \, \mathbf u'_T .\varepsilon_S (\mathbf u ) \,
 d\Gamma \big|
 \le \theta\int_{\Gamma_1 } |\varepsilon_S (\mathbf u )|^2 \, d\Gamma +
 \frac{C}{\theta } \int_{\Gamma_1 } |\mathbf u'_T |^2 \, d\Gamma \,, \\
 \label{I24}
 \big| \int_{\Gamma_1 } 2 B h_\nu \, \mathbf u'_T .(\partial_T \nu ) \mathbf u_T \,
 d\Gamma \big|
 \le C \int_{\Gamma_1 } \left( |\mathbf u'_T |^2 + |\mathbf u_T |^2 \right) \,
 d\Gamma \,.
\end{gather}
 Now compute
\begin{eqnarray*}
\lefteqn{ \int_S^T \int_{\Gamma_1 } h_\nu \, \mathbf u'_T .\nabla_T u_\nu
 \, d\Gamma\, dt }  \\
&=& \Big[ \int_{\Gamma_1 } h_\nu \, \mathbf u_T .\nabla_T u_\nu
 \, d\Gamma \Big]_S^T - \int_S^T \int_{\Gamma_1 } h_\nu \, \mathbf u_T
 .(\nabla_T u'_\nu ) \, d\Gamma\, dt \\
&=& \Big[ \int_{\Gamma_1 } h_\nu \, \mathbf u_T .\nabla_T u_\nu
 \, d\Gamma \Big]_S^T + \int_S^T \int_{\Gamma_1 } u'_\nu
 \diver _T (k \mathbf u_T ) \, d\Gamma\, dt  \,.
\end{eqnarray*}
 Since $\diver _T (h_\nu \mathbf u_T )= h_\nu \diver _T (\mathbf u_T )+ \nabla_T h_\nu .\mathbf u_T
 $, Proposition 1 gives
\begin{equation}
 \label{I25}
 \big| \int_{\Gamma_1 } 2 B u'_\nu \diver _T (h_\nu \mathbf u_T ) \, d\Gamma
 \big| \le \frac{\theta}{2} \| \mathbf u_T \|_1^2 + \frac{2C}{\theta }
 \int_{\Gamma_1 } |\mathbf u' |^2 \, d\Gamma \,.
\end{equation}
 Now consider ${ \int_{\Gamma_1 } h_\nu \, \mathbf u_T .\nabla_T
 u_\nu \, d\Gamma }$.
 Given $t>0$, let $\zeta $ be in $\mathrm{H}^1 (\Gamma_1 )$ (notice
 that $\mathrm{H}^1 (\Gamma_1 ) = \mathrm{H}^1_0 (\Gamma_1 )$) such that
 $$\zeta - \Delta_T \zeta = \diver _T (\mathbf u_T )(t) \,.
 $$
 Since $\diver _T (\mathbf u_T )(t) $ belongs to $\mathrm{H}^{-1/2} (\Gamma_1 )$, $\zeta $
 satisfies
\begin{equation}  \label{I26}
\begin{gathered}
  \| \zeta \|_{\mathrm{H}^1 (\Gamma_1 )} \le C \| \mathbf u_T (t)
 \|_{\mathrm{L}^2 (\Gamma_1 , T(\Gamma_1 ))} \,,
 \\
 \zeta \in \mathrm{H}^{3/2} (\Gamma_1 ) \, \hbox{ and } \| \zeta
 \|_{\mathrm{H}^{3/2} (\Gamma_1 )} \le C \| \mathbf u_T (t)\|_{\mathrm{H}^{1/2} (\Gamma_1 ,
 T(\Gamma_1 ))} \,.
 \end{gathered}
\end{equation}
 Then we have
\begin{eqnarray*}
\lefteqn{ \int_{\Gamma_1 } h_\nu \, \mathbf u_T .
  \nabla_T u_\nu \, d\Gamma }\\
&=&- \int_{\Gamma_1 } u_\nu \diver _T (h_\nu \mathbf u_T ) \, d\Gamma  \\
&=& - \int_{\Gamma_1 } u_\nu \nabla_T h_\nu .\mathbf u_T \, d\Gamma
 - \int_{\Gamma_1 } u_\nu h_\nu \zeta \, d\Gamma
 + \int_{\Gamma_1 } u_\nu  h_\nu \Delta_T \zeta \, d\Gamma
 \,.
\end{eqnarray*}
 First, thanks to $(\ref{I26} )$, we have
\begin{gather*}
 \big| \int_{\Gamma_1 } u_\nu \nabla_T h_\nu . \mathbf u_T \,
 d\Gamma \big| \le C \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma \,,
 \\
 \big| \int_{\Gamma_1 } u_\nu h_\nu \zeta \, d\Gamma
 \big| \le C \int_{\Gamma_1 } (|u_\nu |^2 + |\zeta |^2)  \, d\Gamma
 \le C \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma \,.
\end{gather*}
 Second
 $$ - \int_{\Gamma_1 } u_\nu h_\nu \Delta_T \zeta \, d\Gamma =
 \int_{\Gamma_1 } (-\Delta_T )^{1/4} (u_\nu h_\nu ) (-\Delta_T )^{3/4}
 \zeta \, d\Gamma \,, $$
 and, again with $(\ref{I26} )$,
 $$ \big| \int_{\Gamma_1 } u_\nu h_\nu \Delta \zeta \, d\Gamma \big|
 \le C \| u_\nu \|_{\mathrm{H}^{1/2} (\Gamma_1 )} \| \zeta \|_{\mathrm{H}^{3/2}
 (\Gamma_1 )} \le C \| \mathbf u \|^2_{\mathrm{H}^1 (\Omega )} \,.
 $$
 Hence
 $$ \big| \int_{\Gamma_1 } h_\nu \, \mathbf u_T .\nabla_T u_\nu \, d\Gamma
 \big| \le C \Bigl( \int_{\Gamma_1 } |\mathbf u |^2 \, d\Gamma + \| \mathbf u
 \|^2_{\mathrm{H}^1 (\Omega )} \Bigr) \,. $$
 Using Poincar\'e's inequality and Korn's inequality, we finally get
\begin{equation}
 \label{I27}
 \big| \int_{\Gamma_1 } 2 B h_\nu \, \mathbf u_T .\nabla_T u_\nu \, d\Gamma
 \big| \le C E(\mathbf u , t) \,.
\end{equation}
 Observing that the energy functional is non-increasing and using
 $(\ref{I25} )$, $(\ref{I27} )$, we obtain
\begin{equation}
 \label{I28}
 \begin{aligned}
  \big| \int_S^T \int_{\Gamma_1 }
 2 B h_\nu \, \mathbf u'_T .\nabla_T u_\nu \, d\Gamma\, dt \big|
\leq&
 C E(\mathbf u , S) + \theta \int_S^T \| \mathbf u_T \|_1^2 \, dt \\
 & + \frac{C}{\theta } \int_S^T \int_{\Gamma_1 } |\mathbf u' |^2 \,
 d\Gamma\, dt  \,.
 \end{aligned}
\end{equation}
 Again, we use boundedness of $\mathbf{h} $ and get
\begin{equation} \label{I29}
 \big| \int_{\Gamma_1 } 2 B h_\nu \, u'_\nu \partial_\nu u_\nu \,
 d\Gamma \big| \le \theta \int_{\Gamma_1 } |\partial_\nu u_\nu |^2
 \, d\Gamma + \frac{C}{\theta } \int_{\Gamma_1 } |\mathbf u' |^2 \, d\Gamma\,.
\end{equation}
 Finally, with $(\ref{I20} )$--$(\ref{I24} )$, $(\ref{I28} )$ and
 $(\ref{I29} )$, we obtain
 \begin{equation}  \label{T13}
 \begin{aligned}
 | \mathcal{I}_2 |
 \leq & \theta \int_S^T \int_{\Gamma_1 } \left(
   \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) +
   |\varepsilon_S (\mathbf u )|^2 + |\partial_\nu u_\nu |^2
 \right) \, d\Gamma\, dt \\
 & +  \frac{C}{\theta } \int_S^T \int_{\Gamma_1 }
 ( |\mathbf u |^2 + |\mathbf u' |^2) \, d\Gamma\, dt + C E(\mathbf u , S) \,.
 \end{aligned}
\end{equation}
 Again, we emphasize that, in $(\ref{T13} )$, $\theta $ is a positive
 number to be chosen later and $C$ is a positive constant which
 does not depend on $\mathbf u $.

 \noindent {\sl Final segment of the proof}.

 \noindent From $(\ref{T11} )$, $(\ref{T12} )$, $(\ref{T13} )$, we obtain
 two positive constants $C_6 $ and $C_7 $ such that
 \begin{equation}
 \label{T14}
 \begin{aligned}
 C_1 \int_S^T E \, dt
 \leq&  C_6 E(\mathbf u ,S) + \frac{C_7 }{\theta } \int_S^T
 \int_{\Gamma_1 } (|\mathbf u |^2 + |\mathbf u' |^2 ) \, d\Gamma\, dt
 \\
 & +  2 \theta \int_S^T \int_{\Gamma_1 }
 \left( \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) +
 |\varepsilon_S (\mathbf u )|^2 + |\partial_\nu u_\nu |^2 \right) \, d\Gamma\, dt
 \\
 & -  k \alpha \int_S^T \int_{\Gamma_1 } \left(
 \varepsilon_T (\mathbf u ) \colon \varepsilon_T (\mathbf u ) + 2 |\varepsilon_S
 (\mathbf u )|^2 + |\partial_\nu u_\nu |^2 \right) \, d\Gamma\, dt  \,.
 \end{aligned}
\end{equation}
 From the relations
 $\varepsilon_T (\mathbf u ) = \varepsilon_T (\mathbf u_T ) +
 \varepsilon_T (u_\nu \nu ) = \varepsilon_T (\mathbf u_T ) + u_\nu \partial_T
 \nu$ on $\Gamma $,  we have
\begin{eqnarray*}
\lefteqn{ \varepsilon_T (\mathbf u ) \colon \varepsilon_T (\mathbf u )}\\
 & =&
\varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) + 2
 \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (u_\nu \nu ) +
 \varepsilon_T (u_\nu \nu ) \colon \varepsilon_T (u_\nu \nu ) \,, \
 \hbox{on } \Gamma_1 \,.
\end{eqnarray*}
 Using $ | \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (u_\nu \nu ) |
 \le \theta \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) +
 (4\theta )^{-1} \varepsilon_T (u_\nu \nu ) \colon \varepsilon_T (u_\nu
 \nu ) $, we get
 $$
 \varepsilon_T (\mathbf u ) \colon \varepsilon_T (\mathbf u ) \ge (1-2\theta )
 \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) + \big( 1-
 \frac{1}{2\theta } \big) \varepsilon_T (u_\nu \nu ) \colon
 \varepsilon_T (u_\nu \nu ) \,, \quad \hbox{on } \Gamma_1 \,.
$$
 Since ${ \int_{\Gamma_1 }\varepsilon_T (u_\nu \nu ) \colon
  \varepsilon_T (u_\nu \nu )\, d\Gamma = \int_{\Gamma_1}|u_\nu |^2
 (\partial_T \nu : \partial_T \nu) \,d\Gamma \le C \int_{\Gamma_1 }
 |u_\nu |^2 \, d\Gamma }$, we deduce from $(\ref{T14} )$ that there
 exists $C_8 > 0 $ such that
\begin{eqnarray*}
\lefteqn{ C_1 \int_S^T E \, dt}\\
 &\le&  C_6 E(\mathbf u ,S) + \frac{C_8 }{\theta } \int_S^T
 \int_{\Gamma_1 } (|\mathbf u |^2 + |\mathbf u' |^2 ) \, d\Gamma\, dt\\
&&+ 2 \theta \int_S^T \int_{\Gamma_1 } \left(
 \varepsilon_T (\mathbf u_T )\colon \varepsilon_T (\mathbf u_T )
 + |\varepsilon_S (\mathbf u )|^2 + |\varepsilon_\nu (\mathbf u )|^2 \right)
 \, d\Gamma\, dt  \\
 &&- k \alpha \int_S^T \int_{\Gamma_1 } \left( (1-2\theta )
 \varepsilon_T (\mathbf u_T ) \colon \varepsilon_T (\mathbf u_T ) +
 2 |\varepsilon_S (\mathbf u )|^2 + |\varepsilon_\nu (\mathbf u )|^2
 \right) \,  d\Gamma\, dt \,.
\end{eqnarray*}
 Then, for $\theta > 0$ small enough, we can find a positive constant
 $C_9 $ such that
 $$
 C_1 \int_S^T E \, dt \le C_6 E(\mathbf u ,S) + C_9 \int_S^T
 \int_{\Gamma_1 } (|\mathbf u |^2 + |\mathbf u' |^2 ) \, d\Gamma\, dt \,.
 $$
 From Proposition 3 and Lemma 5, there exists $C_{10} > 0$ such that, for
 every $\eta > 0 $,
 $$
 C_1 \int_S^T E \, dt \le \frac{C_{10}}{\eta } E(\mathbf u ,S) +
 \eta \int_S^T E \, dt \,.
 $$
 Hence, for $\eta $ small enough, we get the theorem by applying
 Lemma 1 with $\omega = (C_1 -\eta )\eta/C_{10}$.
 Now, we can observe that all above constants do not depend on
 the strong solution $\mathbf u $ of $(\ref{E} )$. Hence, by a denseness
 argument, this result can be extended to a weak solution of
 $(\ref{E})$.
\hfill $\Box $

\subsection{Proof of Theorem 2}

 We now show that Theorem 1 can be applied with the vector
 field
 $$ \mathbf{h} (\mathbf{x} ) = (\mathbf{x} - \mathbf{x}_0 ) + \rho \widetilde{\mathbf{h} }(\mathbf{x} ) \,, $$
 where $\rho $ is some positive constant and $\widetilde{\mathbf{h} } \in
 \bigl( \mathcal{C}^1 (\overline \Omega ) \bigr)^3 $ is such that
 $$
 \widetilde{\mathbf{h} } = 0 \,, \quad \hbox{on } \Gamma_0 \,,
 \quad
 \widetilde{\mathbf{h} } = \nu \,, \quad \hbox{on } \Gamma_1 \,.
 $$
Note that $\mathbf{h} $ satisfies $(\ref{H3} ) $.
 Indeed,
 \begin{gather*}
 \mathbf{h} (\mathbf{x} ).\nu(\mathbf{x} ) = (\mathbf{x} - \mathbf{x}_0 ).\nu (\mathbf{x} ) \le 0 \,, \quad
 \hbox{if } \mathbf{x} \in \Gamma_0 \,,
 \\
 \mathbf{h} (\mathbf{x} ).\nu(\mathbf{x} ) = (\mathbf{x} - \mathbf{x}_0 ).\nu (\mathbf{x} ) + \rho > 0 \,, \quad
 \hbox{if } \mathbf{x} \in \Gamma_1 \,.
 \end{gather*}
Also note that  $\mathbf{h} $ satisfies $(\ref{H4} ) $.
 We will consider two cases.

\noindent\textbf{First case:} $\mathop{\rm meas} (\Gamma_0 ) \not = 0 $.
Here we choose $\beta_\mathbf{h} =0 $. Then
 $$ \int_\Omega \sigma_{i j } (\xi ) h_{k, j }
 \xi_{i , k }\, d\mathbf{x} =
 \int_\Omega \sigma_{i j }(\xi ) \xi_{i , j }\ d\mathbf{x}
 + \rho \int_\Omega \sigma_{i j } (\xi )
 \widetilde{h}_{k, j } \xi_{i , k }\, d\mathbf{x} \,.
 $$
 Since $\xi \in \mathbb{H}^1_{\Gamma_0 }(\Omega ) $ and $ \widetilde{\mathbf{h} } \in
 \bigl( \mathcal{C}^1 (\overline \Omega ) \bigr)^3 $, using Korn's
 inequality, we can find a constant $C(\widetilde{\mathbf{h} } ) > 0 $ such that
 $$ \big| \int_\Omega \sigma_{i j } (\xi )
 \widetilde{h}_{k, j } \xi_{i , k } \, d\mathbf{x} \big|
 \le C(\widetilde{\mathbf{h} } ) \int_\Omega \sigma (\xi ) \colon
 \varepsilon (\xi ) \, d\mathbf{x} \,, $$
 and
 $$ \int_\Omega \sigma_{i j } (\xi ) h_{k, j }
 \xi_{i , k } \, d\mathbf{x} \ge \big( 1-\rho C(\widetilde{\mathbf{h} } ) \big)
 \int_\Omega \sigma (\xi ) \colon \varepsilon (\xi ) \, d\mathbf{x} \,. $$
 We choose $ \alpha_\mathbf{h} = 1-\rho C(\widetilde{\mathbf{h} } ) $ and get
 $\alpha_\mathbf{h} > 0 $ for $\rho $ small enough.

\noindent\textbf{Second case:} $\mathop{\rm meas} (\Gamma_0 ) = 0 $.
Since $\mathop{\rm meas} (\Gamma_1 ) \not = 0 $, the map
 $$\xi \mapsto \big( \int_\Omega \sigma (\xi ) \colon
 \varepsilon (\xi ) \, d\mathbf{x} + \int_{\Gamma_1 }|\xi |^2 \, d\Gamma
 \big)^{1/2}$$
  defines an equivalent norm on $\mathbb{H}^1 (\Omega )$
 \cite{Li}. We have
 $$ \int_\Omega \sigma_{i j } (\xi ) h_{k, j }
 \xi_{i , k } \, d\mathbf{x} =
 \int_\Omega \sigma_{i j }(\xi ) \xi_{i , j } \, d\mathbf{x}
 + \rho \int_\Omega \sigma_{i j } (\xi )
 \widetilde{h}_{k, j } \xi_{i , k } \, d\mathbf{x} \,.
 $$
 Since  $\xi \in \mathbb{H}^1 (\Omega ) $ and $\widetilde{\mathbf{h} } \in \bigl(
 \mathcal{C}^1 (\overline \Omega ) \bigr)^3 $, we can find a constant
 $C(\widetilde{\mathbf{h} }) > 0 $ such that
 $$ \big| \int_\Omega \sigma_{i j } (\xi )
 \widetilde{h}_{k, j } \xi_{i , k} \, d\mathbf{x} \big|
 \leq C(\widetilde{\mathbf{h} } ) \big( \int_\Omega \sigma (\xi ) \colon
 \varepsilon (\xi ) \, d\mathbf{x} + \int_{\Gamma_1 } |\xi |^2 \ d\Gamma \big)
 \,, $$
 and
 $$ \int_\Omega \sigma_{i j } (\xi ) h_{k, j }
 \xi_{i , k } \, d\mathbf{x} \ge \big( 1-\rho C(\widetilde{\mathbf{h} } ) \big)
 \int_\Omega \sigma (\xi ) \colon \varepsilon (\xi ) \, d\mathbf{x} -
 C(\widetilde{\mathbf{h} } ) \int_{\Gamma_1 } |\xi |^2 \, d\Gamma \,.
 $$
 Then we choose $\alpha_\mathbf{h} = 1-\rho C(\widetilde{\mathbf{h}})$,
 which is positive for $\rho$ small enough, and
 $\beta_\mathbf{h} = - C(\widetilde{\mathbf{h}})$.

Now one can easily show that the other conditions in
 $(\ref{H4} )$ are satisfied if
 $$ \rho < \min \Bigl( \frac{1}{C(\widetilde{ \mathbf{h} })} ,
 \frac{2-\gamma_{\mathbf{x} -\mathbf{x}_0 } }{1+2C(\widetilde{ \mathbf{h} }) +
 { \max_{\overline \Omega } } (\diver  (\widetilde{ \mathbf{h} } ) ) -
 {\min_{\overline \Omega } } (\diver  (\widetilde{ \mathbf{h} } ) ) },
 \frac{3}{ | {\min_{\overline \Omega } }
 (\diver  (\widetilde{ \mathbf{h} } ) ) | } \Bigr) \,. $$
 This completes the proof.
\hfill $\Box $

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\noindent\textsc{Rabah Bey }(e-mail: bey@ec-lyon.fr)\\
\textsc{Jean-Pierre Loh\'eac} (e-mail: loheac@ec-lyon.fr)\\[3pt]
\textsc{E.C.L., D.M.I., C.N.R.S. U.M.R.} 5585,\\
 B.P. 163, 69131 \\
\'Ecully  Cedex, France \smallskip

\noindent\textsc{Amar Heminna} \\
\textsc{U.S.T.H.B.}, Institut de Math\'ematiques,\\
B.P. 32, El-Alia, 16111 Alger, Algeria \\
e-mail: ahemina@hotmail.com

\end{document}