\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 41, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/41\hfil Interior feedback stabilization]
{Interior feedback stabilization of  wave equations
 with time dependent delay}

\author[S. Nicaise, C. Pignotti\hfil EJDE-2011/41\hfilneg]
{Serge Nicaise, Cristina Pignotti}  % in alphabetical order

\address{Serge Nicaise \newline
Universit\'e de Valenciennes et du Hainaut Cambr\'{e}sis, MACS,
ISTV, 59313 Valenciennes Cedex 9, France}
\email{serge.nicaise@univ-valenciennes.fr}

\address{Cristina Pignotti \newline
Dipartimento di Matematica Pura e Applicata,
Universit\`a di L'Aquila,
Via Vetoio, Loc. Coppito, 67010 L'Aquila, Italy}
\email{pignotti@univaq.it}

\thanks{Submitted July 26, 2010. Published March 17, 2011.}
\subjclass[2000]{35L05, 93D15}
\keywords{Wave equation; delay feedback; stabilization}

\begin{abstract}
 We study the stabilization problem by interior damping
 of the wave equation with boundary or internal time-varying delay
 feedback in a bounded and smooth domain. By introducing suitable
 Lyapunov functionals exponential stability estimates are obtained
 if the delay effect is appropriately compensated by the internal
 damping.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction} \label{pbform}

 Let $\Omega\subset\mathbb{R}^n$ be an open bounded set
with  boundary $\Gamma$ of class $C^2$.
We assume that $\Gamma$ is divided into two  parts
$\Gamma_0$ and $\Gamma_1$; i.e.,
$\Gamma =\Gamma_0\cup\Gamma_1$, with
$\overline \Gamma_0 \cap\overline \Gamma_1 =\emptyset$
and $\operatorname{meas}\Gamma_0\neq 0$.

We consider the problem
 \begin{gather}
u_{tt}(x,t) -\Delta u (x,t)-a\Delta u_t(x,t)=0\quad \text{in }
\Omega\times (0,+\infty),\label{S1.1}\\
u (x,t) =0\quad \text{on }\Gamma_0\times (0,+\infty),\label{S1.2}\\
\mu u_{tt}(x,t)=-{{\partial (u+au_t)}\over {\partial \nu }}(x,t)
-k u_t (x,t-\tau (t)) \quad \text{on }\Gamma_1\times
(0,+\infty),\label{S1.3}\\
u(x,0)=u_0(x),\quad u_t(x,0)=u_1(x)\quad \text{in }
\Omega, \label{S1.4old}\\
\label{S1.4}
u_t(x,t)=f_0(x,t) \quad \text{in } \Gamma_1\times (-\tau(0), 0),
\end{gather}
where $\nu (x)$ denotes the outer unit normal vector to the point
$x\in\Gamma$ and $ {{\partial u}\over {\partial \nu}}$
is the normal derivative. Moreover, $\tau= \tau(t)$ is the time delay,
$\mu, a, k$ are real numbers, with
$\mu\ge 0, a>0$, and the initial
datum $(u_0, u_1,f_0)$ belongs to a suitable space.


It is well-known that the above model is exponentially stable
in absence of delay; that is, if $\tau(t)\equiv 0$.
We refer to
\cite{Chen,Lag,Lag83,LT,Komornik91,KomornikZuazua,Komornikbook,Zuazua}
for the more studied case $a=0,\ \mu=0$ and to
\cite{grobbelaar:94,Morgul,Doronin-Larkin_02,PSM,gerbi:08}
in the case $a, \mu> 0$.

In presence of a constant delay, when $\mu=0$ and the
condition \eqref{S1.3} is substituted by
$$ {{\partial u}\over {\partial \nu }}(x,t)=-k u_t (x,t-\tau),\quad \Gamma_1\times
(0,+\infty),$$
the system becomes unstable for arbitrarily small delays (see \cite{DATKO}).

Then, the  1-d version of the above model with $\mu=0$ in the boundary condition \eqref{S1.3} has been considered by Morg\"{u}l \cite{Morgul} who proposed a class of dynamic boundary controllers to solve the stability robustness problem.

In the case
$\mu>0$, \eqref{S1.3} is a so-called dynamic boundary condition.
Dynamic boundary conditions arise in many physical applications,
in particular they occur in elastic models. For
instance, these conditions appear in modelling dynamic vibrations of
linear viscoelastic rods and beams which have attached tip masses at
their free ends. See \cite{Andrews,conrad:98,littman:88,gerbi:08}
and the references therein for more details.
The above model without delay (e.g. $\tau=0$) has been proposed
in one dimension by Pellicer and S\`{o}la-Morales \cite{PSM}
as an alternative model for the classical spring-mass damper system,
the case $k=0$ being treated in \cite{grobbelaar:94}.
In both cases, no rates of convergence are proved.
In dimension higher than 1, we refer to  Gerbi and
Said-Houari \cite{gerbi:08}
where a nonlinear boundary feedback is even considered and
the exponential growth of the energy is proved if the initial
data are large enough.
A different  problem with a dynamic boundary condition
(without delay), motivated by the study of flows of
gas in a channel with porous walls, is analyzed in
\cite{Doronin-Larkin_02} where exponential decay is proved.

On the function $\tau$ we assume that there exist  positive
constants $\tau_0, \overline\tau$ such that
\begin{equation}\label{tau1}
0<\tau_0\le \tau(t)\le \overline\tau, \quad\forall t>0.
\end{equation}

Moreover, we assume
\begin{gather}\label{tau3}
\tau\in W^{2,\infty}([0,T]),\quad \forall T>0,\\
\label{tau2}
\tau'(t)\le d<1 \quad \forall t>0.
\end{gather}


Under the above assumptions on the time-delay function $\tau(t)$
we will prove that an exponential stability result
holds under a suitable assumption between the coefficients $a$ and $k$
(namely condition \eqref{relmubis} below).

We consider also the problem with interior delay
\begin{gather}
u_{tt}(x,t) -\Delta u (x,t)+ a_0 u_t (x,t) +a_1 u_t
(x,t-\tau(t))=0 \quad \text{in }\Omega\times
(0,+\infty)\label{1.1}\\
u (x,t) =0\quad \text{on }\Gamma\times (0,+\infty)\label{1.2}\\
u(x,0)=u_0(x),\quad  u_t(x,0)=u_1(x)\quad \text{in }\Omega\label{1.4old}
\\
u_t(x,t)=g_0(x,t) \quad \text{in }\Omega
\times (-\tau(0),0),\label{1.5}
\end{gather}
where  $\tau (t) >0$ is the time-varying delay, $a_0$ and $a_1$
are  real numbers with $a_0>0$, and the initial
datum  $(u_0, u_1,g_0)$ belongs to a suitable space.

The above  model, with $a_0>0$, $a_1>0$ and a constant delay
$\tau(t)\equiv \tau$ has been  studied
by the authors \cite{NPSicon} in the case of mixed
homogeneous Dirichlet-Neumann boundary conditions.
Assuming that
\begin{equation}\label{mu}
0\leq a_1 < a_0
\end{equation}
a stabilization result is given, by using a suitable observability
estimate. This is done by applying inequalities obtained from
Carleman estimates for the wave equation  by Lasiecka, Triggiani
and Yao in \cite{LTY} and by using compactness-uniqueness
arguments. Instability phenomena when \eqref{mu} is not satisfied
are also illustrated. We refer to \cite{DLP,Datko} for instability
examples of related problems in one dimension.

The analogous problem with boundary feedback has been  introduced
and studied by Xu, Yung, Li \cite{XYL} in one-space dimension
using a fine spectral analysis and in higher space dimension by
the authors \cite{NPSicon}.

The case of time-varying delay has been already studied   in
\cite{NVF}  in one space dimension and in general dimension, with
a possibly degenerate delay, in \cite{NPV}. Both   these papers
deal with boundary feedback. See also \cite{FNV} for abstract
problems also under the assumption of non-degeneracy of $\tau(t)$.



Here, we will give an exponential stability result  for problem
\eqref{1.1}-\eqref{1.5}  under the condition
\begin{equation}\label{relmu}
| a_1| <\sqrt{1-d}\,a_0\,,
\end{equation}
where $d$ is the constant in \eqref{tau2}.

The outline of the paper is the following.
In section \ref{BOUN} we study well-posedness and exponential
stability of the problem  \eqref{S1.1}--\eqref{S1.4}
with structural damping and boundary delay in both cases $\mu>0$
and $\mu =0$. In section \ref{INT} we analyze the problem with
internal delay feedback \eqref{1.1}--\eqref{1.5}.


\section{Boundary delay feedback} \label{BOUN}

In this section we concentrate on the problem with boundary
delay \eqref{S1.1}--\eqref{S1.4}.

Let $C_P$ be a Poincar\'e's type
constant defined as  the smallest positive constant such that
\begin{equation}\label{defcp}
\int_{\Gamma_1} |v|^2\,d\Gamma \leq C_P \int_{\Omega} |\nabla v|^2\,
dx, \; \forall  v\in H^1_{\Gamma_0}(\Omega ),
\end{equation}
where, as usual,
$$
H^1_{\Gamma_0}(\Omega )=\{u\in H^1(\Omega):u=0 \text{ on }
 \Gamma_0\}.
$$
First of all we will give a well-posedness result under the assumption
\begin{equation}\label{assWP}
 | k|\le \frac{a}{C_P}\sqrt{1-d},
\end{equation}
where $d$ is the positive constant of assumption \eqref{tau2}.
We have to distinguish the two cases $\mu>0$ and $\mu=0$.

\subsection{Well-posedness in the case of dynamic boundary condition}

First we study the well-posedness of \eqref{S1.1}--\eqref{S1.4}
for $\mu>0$.
We introduce  the auxiliary unknown
\begin{equation}\label{defz}
z(x,\rho ,t)=u_t(x, t-\tau(t)\rho ),\quad x\in\Gamma_1,\;
 \rho\in (0,1),\; t>0.
\end{equation}
Then, problem \eqref{S1.1}--\eqref{S1.4} is equivalent to
\begin{gather}
u_{tt}(x,t) -\Delta u (x,t)-a\Delta u_t(x,t)=0\quad \text{in }
\Omega\times (0,+\infty),\label{SP1bis}\\
\tau (t) z_t(x,\rho,t)+(1-\tau' (t)\rho )z_{\rho}(x,\rho, t)=0
\quad \text{in } \Gamma_1\times (0,1)\times (0,+\infty ),\label{SPzbis}\\
u (x,t) =0\quad \text{on }\Gamma_0\times
(0,+\infty),\label{SP2bis}\\
\mu u_{tt}(x,t)=-{{\partial (u+a u_t)}\over {\partial \nu }}(x,t)
-k z(x,1,t) \quad \text{on }\Gamma_1\times (0,+\infty),\label{SP3bis}\\
z(x,0,t)=u_t(x,t)\quad\text{on } \Gamma_1\times
(0,\infty),\label{SP3bisbis}
\\
u(x,0)=u_0(x)\quad \mbox{\rm and}\quad u_t(x,0)=u_1(x)\quad
\text{in }\Omega,\label{SP4bis}\\
z(x,\rho ,0)=f_0(x,-\rho \tau(0) )\quad\text{in }
\Gamma_1\times (0,1).\label{SP5bis}
\end{gather}

Let us denote
$$
U:=( u, u_t, \gamma_1 u_t, z )^T,
$$
where  $\gamma_1$ is the trace operator on
$\Gamma_1$. Then the previous problem is formally equivalent to
\begin{align*}
U':&=( u_t, u_{tt}, \gamma_1 u_{tt}, z_t )^T\\
&=\Big( u_t, \Delta u+a\Delta u_t,-\mu^{-1}\big({{\partial (u+a
u_t)}\over {\partial \nu }}(x,t)+k z(\cdot,1,\cdot)\Big),
\frac{\tau'(t)\rho-1}{\tau(t)}z_{\rho} \Big)^T.
\end{align*}
Therefore, problem \eqref{S1.1}--\eqref{S1.4} can be rewritten as
\begin{equation}\label{SMformulA}
\begin{gathered}
U'=\mathcal{A}(t) U,\\
U(0)=\big( u_0, u_1, \gamma_1 u_1, f_0(\cdot,-\cdot\tau )\big)^T,
\end{gathered}
\end{equation}
where the time varying operator $\mathcal{A}(t)$ is defined by
$$
\mathcal{A}(t)\begin{pmatrix}
u\\
v\\
v_1\\
 z
\end{pmatrix}
:= \begin{pmatrix}
v\\
\Delta (u+av)\\
-\mu^{-1}\left({{\partial (u+a v)}\over {\partial \nu }}+k
z(\cdot,1)\right)\\
\frac{\tau'(t)\rho-1}{\tau(t)} z_{\rho}
\end{pmatrix},
$$
with domain
\begin{equation}\label{SMdomain}
\begin{aligned}
\mathcal{D} (\mathcal{A}(t)):=\Big\{& (u,v,v_1,z)^T\in
  H^1_{\Gamma_0}(\Omega)^2 \times L^2(\Gamma_1)
\times L^2(\Gamma_1; H^1(0,1)):\\
&u+av \in E(\Delta,L^2(\Omega)), {
{{\partial (u+av)}\over {\partial\nu}} \in L^2( \Gamma_1)},
\\
&  v=v_1=z(\cdot,0)\text{on } \Gamma_1 \Big\},
\end{aligned}
\end{equation}
where
$$
E(\Delta,L^2(\Omega))=\{u\in H^1(\Omega): \Delta u\in L^2(\Omega)\}.
$$
Recall that for a function $u\in E(\Delta,L^2(\Omega))$,
${\partial u\over \partial\nu}$ belongs to $H^{-1/2}(\Gamma_1)$ and
the next Green formula is valid (see section 1.5 of
\cite{grisvard:85a})
\begin{equation}\label{green}
\int_{\Omega}\nabla u\nabla w
dx=-\int_{\Omega}\Delta u w dx+ \langle {{\partial u}\over
{\partial\nu}}; w \rangle_{\Gamma_1} \forall w\in
H^1_{\Gamma_0}(\Omega),
\end{equation}
 where
$\langle\cdot;\cdot\rangle_{\Gamma_1}$ means the duality pairing
between $H^{-1/2}(\Gamma_1)$ and $H^{1/2}(\Gamma_1)$.

Observe that the domain of $\mathcal{A}(t)$ is independent of
the time $t$; i.e.,
\begin{equation}\label{domini}
\mathcal{D} (\mathcal{A}(t))=\mathcal{D} (\mathcal{A}(0)),\quad t>0.
\end{equation}
$\mathcal{A}(t)$ is an unbounded operator in $\mathcal{H}$,
the Hilbert space defined by
\begin{equation}\label{SMspace}
\mathcal{H}:= H^1_{\Gamma_0} (\Omega)\times L^2 (\Omega)\times L^2
(\Gamma_1)\times  L^2(\Gamma_1\times (0,1)),
\end{equation}
equipped with the standard  inner product
\begin{equation}\label{SMscalar}
\begin{split}
\Big\langle \begin{pmatrix}
u\\
v\\
v_1
\\
z
\end{pmatrix},
\begin{pmatrix}
\tilde u\\
\tilde v\\
\tilde v_1
\\
 \tilde z
\end{pmatrix}
\Big\rangle_\mathcal{H}
&:=\int_{\Omega} \{\nabla u (x){\nabla \tilde u}(x) +v(x)\tilde v(x) \} dx\\
&\quad+ \int_{\Gamma_1} v_1(x)\tilde v_1(x) d\Gamma
+\int_{\Gamma_1}\int_0^1 z(x,\rho)\tilde z (x,\rho)\, d\rho\, d\Gamma.
\end{split}
\end{equation}
We can obtain a well-posedness result
using semigroup arguments by Kato \cite{kato:67,kato:76,pazy}.
The following result is proved in
\cite[Theorem 1.9]{kato:67}.

\begin{theorem}\label{thmkato}
Assume that
\begin{itemize}
\item[(i)] $\mathcal{D}(\mathcal{A}(0))$ is a dense subset of
$\mathcal{H}$,

\item[(ii)] $\mathcal{D}(\mathcal{A}(t))=\mathcal{D}(\mathcal{A}(0))$
for all $t>0$,

\item[(iii)] for all $t\in [0,T]$, $\mathcal{A}(t)$ generates a
strongly continuous semigroup on $\mathcal{H}$ and the family
$\mathcal{A}=\{\mathcal{A}(t): t\in [0,T]\}$ is stable with
stability constants $C$ and $m$
independent of $t$ (i.e. the semigroup $(S_t(s))_{s\geq 0}$
generated by  $\mathcal{A}(t)$ satisfies $\|S_t(s)u\|_{\mathcal{H}
}\leq C e^{m s}\|u\|_{\mathcal{H}}$, for all
$u\in \mathcal{H}$ and $s\geq 0$),

\item[(iv)] $\partial_t\mathcal{A}$ belongs to
$L^{\infty}_{*}([0,T], B(\mathcal{D}(\mathcal{A}(0)),\mathcal{H}))$,
the space of equivalent classes of essentially bounded, strongly
measurable functions from $[0,T]$ into the set
$B(\mathcal{D}(\mathcal{A}(0)),\mathcal{H})$ of bounded operators
from $\mathcal{D}(\mathcal{A}(0))$ into $\mathcal{H}$.
\end{itemize}
Then, problem
\eqref{SMformulA} has a unique
solution $U\in C([0,T], \mathcal{D}(\mathcal{A}(0)))\cap C^1([0,T],
\mathcal{H})$ for any initial
datum in $\mathcal{D}(\mathcal{A}(0))$.
\end{theorem}

Therefore, we will check  the above assumptions for  problem
\eqref{SMformulA}.

\begin{lemma}\label{lemmadensity}
$D(\mathcal{A}(0))$ is dense in $\mathcal{H}$.
\end{lemma}

\begin{proof}
Let $(f,g,g_1, h)^T\in \mathcal{H}$ be orthogonal to all
elements of $D(\mathcal{A}(0))$; that is,
\begin{align*}
0&=\Big\langle \begin{pmatrix}u\\v\\v_1\\z\end{pmatrix},
\begin{pmatrix}f\\g\\g_1\\h\end{pmatrix}\Big\rangle_{\mathcal{H}}\\
&=\int_{\Omega}\{\nabla u (x)\nabla f(x) +v(x)g(x)\} dx
+\int_{\Gamma_1}v_1g_1d\Gamma
+\int_{\Gamma_1}\int_{0}^{1}z(x,\rho)h(x,\rho)d\rho d\Gamma,
\end{align*}
for all $(u, v,v_1,z)^T\in D(\mathcal{A}(0))$.
We first take  $u=0$ and $v=0$ (then $v_1=0$)
and $z\in \mathcal{D}(\Gamma_1\times (0,1))$. As
$(0,0,0, z)^T\in D(\mathcal{A}(0))$, we obtain
\[
\int_{\Gamma_1}\int_{0}^{1}z(x,\rho)h(x,\rho)d\rho d\Gamma=0.
\]
Since $\mathcal{D}(\Gamma_1\times (0,1))$ is
dense in $L^{2}(\Gamma_1\times (0,1)$, we deduce that $h=0$.

In the same way, by taking $u=0$, $z=0$ and
$v\in\mathcal{D}(\Omega)$ (then $v_1=0$) we see that $g=0$.
Therefore, for $u=0$, $z=0$ we deduce also
$$
\int_{\Gamma_1}g_1v_1d\Gamma=0,\quad
 \forall v_1\in\mathcal{D}(\Gamma_1)
$$
and so $g_1=0$.

 The above orthogonality condition is then reduced to
\[
0=\int_{\Omega}\nabla u \nabla f dx,\quad\forall
 (u,v, v_1, z)^T\in D(\mathcal{A}(0)).
\]
By restricting ourselves to $v =0$ and $z=0$, we obtain
\[
\int_{\Omega}\nabla u (x)\nabla f(x) dx=0,\quad \forall
(u,0,0,0)^T\in D(\mathcal{A}(0)).
\]
But we easily see that $(u,0,0,0)^T\in D(\mathcal{A}(0))$
if and only if
$ u\in E(\Delta,L^2(\Omega))\cap H^1_{\Gamma_0}(\Omega)$.
This set is dense in $H^1_{\Gamma_0}(\Omega)$
(equipped with the inner product
$\langle., .\rangle_{H^1_{\Gamma_0}(\Omega)}$),
thus we conclude that $f=0$.
\end{proof}

Assuming \eqref{assWP} we will show that $\mathcal{A}(t)$
generates a $C_0$ semigroup on $\mathcal{H}$ and using the
variable norm technique of Kato from \cite{kato:76}
and Theorem \ref{thmkato}, that problem \eqref{SMformulA}
has a unique solution.

Let $\xi$ be a positive constant that satisfies
\begin{equation}\label{HPxiW}
\frac{| k| }{\sqrt{1-d}}\le{\xi}\le \frac{2a}{C_P}
-\frac{| k|}{\sqrt{1-d}}.
\end{equation}
Note that this choice of $\xi$ is possible from
assumption \eqref{assWP}.

We define on the Hilbert space $\mathcal{H}$ the time dependent inner
product
\begin{equation} \label{SMscalart}
\begin{aligned}
\Big\langle\begin{pmatrix}
u\\
v\\
v_1
\\
z
\end{pmatrix},
\begin{pmatrix}
\tilde u\\
\tilde v\\
\tilde v_1
\\
 \tilde z
\end{pmatrix}
\Big\rangle_{t}
&:=\int_{\Omega} \{\nabla u
(x){\nabla \tilde u}(x) +v(x)\tilde v(x) \} dx\\
&\quad +\mu \int_{\Gamma_1} v_1(x)\tilde v_1(x) d\Gamma
+\xi\tau(t)
\int_{\Gamma_1}\int_0^1 z(x,\rho)\tilde z (x,\rho) \,d\rho\,d\Gamma.
\end{aligned}
\end{equation}
Using this time dependent inner product and Theorem \ref{thmkato},
 we can deduce a well-posedness result.

\begin{theorem}\label{WP}
For any initial datum $U_0\in\mathcal{D}(\mathcal{A}(0))$
there exists a unique solution
$$
U\in C([0,+\infty), \mathcal{D}(\mathcal{A}(0)))\cap C^1([0,+\infty),
\mathcal{H})
$$
of system \eqref{SMformulA}.
\end{theorem}

\begin{proof}
We first observe that
\begin{equation}
\frac{\|\phi\|_t}{\|\phi\|_s}\leq
e^{\frac{c}{2\tau_0}|t-s|},\quad \forall t,s\in
[0,T],\label{ineq1}
\end{equation}
where $\phi=(u,v,v_1,z)^T$ and $c$ is a positive constant.
Indeed, for all $s,t\in [0,T]$, we have
\begin{align*}
\|\phi\|_t^2-\|\phi\|_s^2e^{\frac{c}{\tau_0}|t-s|}
&= \left(1-e^{\frac{c}{\tau_0}|t-s|}\right)
\Big\{\int_{\Omega}(|\nabla u (x)|^2+v^2)dx
+\mu\int_{\Gamma_1}v_1^2d\Gamma \Big\}\\
&\quad + \xi\left(\tau(t)-\tau(s)e^{\frac{c}{\tau_0}|t-s|}\right)
\int_{\Gamma_N}\int_0^1z^2(x,\rho)\,d\rho \,d\Gamma.
\end{align*}
We notice that $1-e^{\frac{c}{\tau_0}|t-s|}\leq0$. Moreover
$\tau(t)-\tau(s)e^{\frac{c}{\tau_0}|t-s|}\leq0$ for some $c>0$.
Indeed,
$\tau(t)=\tau(s)+\tau'(a)(t-s)$,
where $a\in(s,t)$,
and thus,
$$
\frac{\tau(t)}{\tau(s)}\leq 1+\frac{\left|\tau'(a)\right|}{\tau(s)}
|t-s|.
$$
By \eqref{tau3}, $\tau'$ is bounded on $[0,T]$ and therefore,
recalling also \eqref{tau1},
$$
\frac{\tau(t)}{\tau(s)}\leq 1+\frac{c}{\tau_0}|t-s|
\leq e^{\frac{c}{\tau_0}|t-s|},
$$
which proves \eqref{ineq1}.

Now we calculate $\langle\mathcal{A}(t)U,U\rangle_t$ for a fixed $t$.
Take $U=(u,v, v_1, z)^T\in \mathcal{D} (\mathcal{A}(t))$.
Then,
\begin{align*}
\langle \mathcal{A}(t) U, U\rangle_t
&= \Big\langle
\begin{pmatrix} v\\ \Delta (u+av)\\
-\mu^{-1}\Big(\frac{\partial (u+av)}{\partial\nu} +kz(\cdot ,1)\Big)
\\ \frac{\tau'(t)\rho-1}{\tau(t)}z_{\rho}
\end{pmatrix},
\begin{pmatrix} u\\ v\\ v_1\\ z
\end{pmatrix}
\Big\rangle_t\\
&=\int_{\Omega} \{\nabla v(x)\nabla u(x)+v(x)\Delta (u (x)+av(x))\} dx\\
&\quad -\xi\int_{\Gamma_1} \int_0^1 (1-\tau'(t)\rho)z_{\rho}(x,\rho )
z(x,\rho )d\rho d\Gamma\\
&\quad -\int_{\Gamma_1} \Big (
\frac{\partial (u+av)}{\partial\nu}(x) +kz(x,1)
\Big )v(x)\,d\Gamma.
\end{align*}
So, by Green's formula,
\begin{equation}\label{F}
\begin{split}
\langle\mathcal{A}(t)U, U \rangle_t
&= -k\int_{\Gamma_1} z(x,1) v(x) d\Gamma
-a\int_{\Omega}| \nabla v(x)|^2 dx\\
&\quad -\xi\int_{\Gamma_1}\int_0^1 (1-\tau'(t)\rho)z_{\rho}
(x,\rho )z(x,\rho ) d\rho d\Gamma.
\end{split}
\end{equation}
Integrating by parts in $\rho$, we obtain
\begin{equation}\label{F2}
\begin{split}
&\int_{\Gamma_1}\int_0^1 z_{\rho}(x,\rho) z(x,\rho)(1-\tau'(t)\rho)\,
d\rho \,d\Gamma\\
&= \int_{\Gamma_1}\int_0^1\frac{1}{2}
 \frac{\partial}{\partial\rho}z^2(x,\rho)
(1-\tau'(t)\rho)d\rho \,d\Gamma\\
&={\tau'(t)\over 2}\int_{\Gamma_1}\int_0^1
z^2(x,\rho)d\rho d\Gamma+{1\over 2}\int_{\Gamma_1}\{z^2(x,1)
(1-\tau'(t))- z^2(x,0)\} d\Gamma.
\end{split}
\end{equation}
Therefore, from \eqref{F} and \eqref{F2},
\begin{align*}
&\langle\mathcal{A}(t) U, U\rangle_t\\
&=-k \int_{\Gamma_1} z(x,1) v(x) d\Gamma
-a\int_{\Omega}| \nabla v(x)|^2 dx\\
&\quad -{\xi\over 2} \int_{\Gamma_1}\{ z^2(x,1)(1-\tau'(t))
-z^2(x,0)\}d\Gamma -\frac{\xi\tau'(t)}{2}\int_{\Gamma_1}\int_0^1
z^2(x,\rho)d\rho d\Gamma
\\
&=-k\int_{\Gamma_1} z(x,1) v(x) d\Gamma
 -a\int_{\Omega}| \nabla v(x)|^2 dx
-{\xi\over 2} \int_{\Gamma_1} z^2(x,1)(1-\tau'(t)) d\Gamma\\
&\quad + {\xi\over 2}\int_{\Gamma_1}v^2(x) d\Gamma
-\frac{\xi\tau'(t)}{2}\int_{\Gamma_1}\int_0^1
z^2(x,\rho)d\rho d\Gamma,
\end{align*}
from which, using Cauchy-Schwarz's inequality, a trace
estimate and Poincar\'e's Theorem, it follows that
\begin{equation}\label{F2bis}
\begin{split}
\langle\mathcal{A}(t) U,U  \rangle_t
&\le -\Big( a- \frac{| k| C_P}{2\sqrt{1-d}}-
{\xi\over 2}C_P \Big)
\int_{\Omega} | \nabla v(x)|^2 d x
\\
&\quad -\Big({\xi\over 2}(1-d)-\frac{| k|}{2}\sqrt{1-d}  \Big)
\int_{\Gamma_1}z^2(x,1) d\Gamma+\kappa(t)\langle U,U\rangle_t,
\end{split}
\end{equation}
where
\begin{equation}
\kappa(t)=\frac{(\tau'(t)^2+1)^{\frac{1}{2}}}{2\tau(t)}.\label{kappa(t)}
\end{equation}
Now, observe that from \eqref{HPxiW},
\begin{equation}
\langle \mathcal{A}(t)U,U\rangle_{t}-\kappa(t)\langle U,U\rangle_t
\leq 0,\label{dissipative}
\end{equation}
which means that the operator
$\tilde{\mathcal{A}}(t)=\mathcal{A}(t)-\kappa(t) I$ is dissipative.

Moreover,
\[
\kappa'(t)=\frac{\tau''(t)\tau'(t)}{2\tau(t)(\tau'(t)^2+1)^{\frac{1}{2}}}
-\frac{\tau'(t)(\tau'(t)^2+1)^{\frac{1}{2}}}{2\tau(t)^2}
\]
is bounded on $[0,T]$ for all $T>0$ (by \eqref{tau1} and
\eqref{tau3}) and we have
$$
\frac{d}{dt}\mathcal{A}(t)U
=\begin{pmatrix} 0\\0\\
\frac{\tau''(t)\tau(t)\rho-\tau'(t)(\tau'(t)\rho-1)}
{\tau(t)^2} z_{\rho}
\end{pmatrix}
$$
with $\frac{\tau''(t)\tau(t)\rho-\tau'(t)
(\tau'(t)\rho-1)}{\tau(t)^2}$ bounded on $[0,T]$.
Thus
\begin{equation}
\frac{d}{dt}\tilde{\mathcal{A}}(t)\in L^{\infty}_{*}([0,T],
B(D(\mathcal{A}(0)), \mathcal{H})),\label{Linfty}
\end{equation}
the space of equivalence classes of essentially bounded,
strongly measurable functions from $[0,T]$ into
$B(D(\mathcal{A}(0)), \mathcal{H})$.

Now, we  show that $\lambda I -\mathcal{A}(t)$ is surjective
for fixed $t>0$ and $\lambda >0$.
Given $(f,g, g_1, h)^T\in \mathcal{H}$, we seek
$U=(u,v, v_1, z)^T\in \mathcal{D}(\mathcal{A}(t)) $ solution of
$$
(\lambda I- \mathcal{A}(t)) \begin{pmatrix}
u\\v\\v_1\\z\end{pmatrix}
=\begin{pmatrix}f\\g\\g_1\\h
\end{pmatrix},
$$
that is verifying
\begin{equation}\label{max}
\begin{gathered}
\lambda u-v=f\\
\lambda v-\Delta (u+av)=g\\
\lambda v_1+\mu^{-1}\left(
\frac{\partial (u+av)}{\partial\nu}(x) +kz(x,1)\right)=g_1
\\
\lambda z+\frac{1-\tau'(t)\rho}{\tau(t)}z_{\rho}=h.
\end{gathered}
\end{equation}
Suppose that we have found $u$ with the appropriate regularity.
Then
\begin{equation}\label{numerare}
v:=\lambda u -f
\end{equation}
and we can determine $z$. Indeed, by \eqref{SMdomain},
\begin{equation}\label{F3}
z(x,0)=v(x),\quad \mbox{\rm for}\ \ x\in \Gamma_1,
\end{equation}
and, from \eqref{max},
\begin{equation}\label{F4}
\lambda z(x,\rho )+\frac{1-\tau'(t)\rho}{\tau(t)}z_{\rho} (x,\rho )
=h(x,\rho ),
\quad \text{for }  x\in \Gamma_1,\; \rho\in (0,1).
\end{equation}
Then, by \eqref{F3} and \eqref{F4}, we obtain
$$
z(x,\rho )=v(x)e^{-\lambda\rho\tau(t)}+\tau(t)
e^{-\lambda\rho\tau(t)} \int_0^{\rho} h(x,\sigma )
e^{\lambda\sigma\tau(t)} d\sigma,
$$
if $\tau'(t)=0$, and
\begin{align*}
z(x,\rho)
&=v(x)e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\rho)}\\
&\quad +e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\rho)}
\int_0^{\rho}\frac{h(x,\sigma)\tau(t)}{1-\tau'(t)\sigma}
e^{-\lambda\frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\sigma)}d\sigma,
\end{align*}
otherwise.
So, from \eqref{numerare},
\begin{equation}\label{F5}
\begin{split}
z(x,\rho )
&=\lambda u(x) e^{-\lambda\rho\tau(t)}-f(x) e^{-\lambda\rho\tau(t)}\\
&\quad +\tau(t) e^{-\lambda\rho\tau(t)} \int_0^{\rho} h(x,\sigma )
e^{\lambda\sigma\tau(t)} d\sigma,\quad
\text{on }\Gamma_1\times (0,1),
\end{split}
\end{equation}
if $\tau'(t)=0$, and
\begin{equation}\label{F5bis}
\begin{split}
z(x,\rho)&=\lambda u(x) e^{\lambda \frac{\tau(t)}{\tau'(t)}
 \ln(1-\tau'(t)\rho)}
-f(x)e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\rho)}\\
&\quad +e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\rho)}
 \int_0^{\rho}\frac{h(x,\sigma)\tau(t)}{1-\tau'(t)\sigma}
 e^{-\lambda\frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\sigma)}d\sigma,
\end{split}
\end{equation}
on $\Gamma_1\times (0,1)$ otherwise.

In particular, if $\tau'(t)=0$,
\begin{equation}\label{F5tilde}
z(x,1)=\lambda u(x) e^{-\lambda\tau(t)} +z_0(x),\quad x\in \Gamma_1,
\end{equation}
with $z_0\in L^2(\Gamma_1)$ defined by
\begin{equation}\label{F5star}
z_0(x)=-f(x)e^{-\lambda\tau(t)} +\tau(t)
 e^{-\lambda\tau(t)}\int_0^1 h(x,\sigma )
e^{\lambda\sigma\tau(t)} d\sigma,\quad x\in \Gamma_1,
\end{equation}
and, if $\tau'(t)\neq 0$,
\begin{equation}\label{F5tilde2}
z(x,1)=\lambda u(x) e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t))}
+z_0(x),\quad x\in \Gamma_1,
\end{equation}
with $z_0\in L^2(\Gamma_1)$ defined by
\begin{equation}\label{F5star2}
\begin{split}
z_0(x)&=-f(x)e^{\lambda \frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t))}\\
&\quad +e^{\lambda \frac{\tau(t)}{\tau'(t)}
 \ln(1-\tau'(t))}\int_0^{1}\frac{h(x,\sigma)\tau(t)}{1-\tau'(t)\sigma}
e^{-\lambda\frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)\sigma)}d\sigma,
\end{split}
\end{equation}
for $x\in \Gamma_1$.
Then, we have to find  $u$.
In view of the equation $\lambda
v-\Delta (u+av)=g$, we set $s=u+av$ and look at $s$. Now according
to \eqref{numerare}, we may write
\[
v=\lambda u-f=\lambda s-f-\lambda av,
\]
or equivalently
\begin{equation}\label{numerarenic2}
v= \frac{\lambda}{1+\lambda a} s-\frac{1}{1+\lambda a}f.
\end{equation}
Hence once $s$ will be found, we will get $v$ by
\eqref{numerarenic2} and then $u$ by $u=s-av$, or equivalently
\begin{equation}\label{numerarenic3}
u= \frac{1}{1+\lambda a} s+\frac{a}{1+\lambda a}f.
\end{equation}


By \eqref{numerarenic2} and \eqref{max}, the function $s$ satisfies
\begin{equation}\label{F6}
\frac{\lambda^2}{1+\lambda a} s-\Delta s=g+\frac{\lambda}{1+\lambda
a}f \quad\text{in } \Omega,
\end{equation}
with the boundary conditions
\begin{equation}\label{bcwDir}
s=0 \quad\text{on } \Gamma_0,
\end{equation}
as well as (at least formally)
\[
{{\partial s }\over {\partial \nu }} = \mu g_1-\mu \lambda v_1-k
z(\cdot,1) \quad \text{on }\Gamma_1,
\]
which becomes due to \eqref{numerarenic2}, \eqref{numerarenic3},
\eqref{F5tilde}, \eqref{F5tilde2} and the requirement that
$v_1=\gamma_1 v$ on $\Gamma_1$:
\begin{eqnarray}\label{SMbcw}
 {{\partial s }\over {\partial
\nu }} =- \frac{\lambda(k e^{-\lambda\tau(t)}+\mu \lambda)
}{1+\lambda a}s +l \quad \text{on }\Gamma_1,
\end{eqnarray}
where
\[
l= \mu g_1+ \frac{\lambda(\mu -k ae^{-\lambda\tau(t)}) }{1+\lambda
a}f -kz_0
 \quad \text{on }\Gamma_1
\]
if $\tau'(t)=0$, otherwise
\begin{equation} \label{SMbcwbis}
 {{\partial s }\over {\partial
\nu }} =- \frac{\lambda(k e^{-\lambda\frac{\tau(t)}{\tau'(t)}
\ln(1-\tau'(t))}+\mu \lambda) }{1+\lambda a}s +\tilde l \quad
\text{on }\Gamma_1,
\end{equation}
where
\[
\tilde l= \mu g_1+ \frac{\lambda(\mu -k a
e^{-\lambda\frac{\tau(t)}{\tau'(t)}\ln(1-\tau'(t)) }) }{1+\lambda
a}f -kz_0
 \quad \text{on }\Gamma_1.
\]
From  \eqref{F6}, integrating by parts, and using
\eqref{bcwDir}, \eqref{SMbcw},  \eqref{SMbcwbis}
we find the variational problem
\begin{equation}\label{SMF8}
\begin{split}
& \int_{\Omega}(\frac{\lambda^2}{1+\lambda a} sw+\nabla
s\cdot \nabla w) dx+\int_{\Gamma_1}  \frac{\lambda(k
e^{-\lambda\tau}+\mu \lambda) }{1+\lambda a}s w d\Gamma\\
&= \int_{\Omega}
(g+\frac{\lambda}{1+\lambda a}f) w dx + \int_{\Gamma_1} l w
d\Gamma\quad\forall w\in H^1_{\Gamma_0}(\Omega)
\end{split}
\end{equation}
if $\tau'(t)=0$, otherwise
\begin{equation}\label{SMF8bis}
\begin{split}
&\int_{\Omega}(\frac{\lambda^2}{1+\lambda a} sw+\nabla
s\cdot \nabla w) dx+\int_{\Gamma_1}  \frac{\lambda(k
e^{-\lambda\frac{\tau}{\tau'}\ln (1-\tau' )}
+\mu \lambda) }{1+\lambda a}s w d\Gamma \\
& = \int_{\Omega}
(g+\frac{\lambda}{1+\lambda a}f) w dx + \int_{\Gamma_1} \tilde l w
d\Gamma\quad\forall w\in H^1_{\Gamma_0}(\Omega).
\end{split}
\end{equation}
As the left-hand side  of \eqref{SMF8}, \eqref{SMF8bis}
is coercive on $H^1_{\Gamma_0} (\Omega )$,
the Lax-Milgram lemma guarantees the existence and uniqueness
of a solution $s\in H^1_{\Gamma_0} (\Omega )$
of \eqref{SMF8}, \eqref{SMF8bis}.

If we consider  $w\in \mathcal{D}(\Omega)$ in \eqref{SMF8},
\eqref{SMF8bis},   we
have that $s$ solves \eqref{F6} in $\mathcal{D}'(\Omega)$  and
thus $s=u+av\in E(\Delta,L^2(\Omega))$.


Using   Green's formula  \eqref{green} in \eqref{SMF8} and using
\eqref{F6}, we obtain
$$
\int_{\Gamma_1}  \frac{\lambda(k e^{-\lambda\tau}+\mu \lambda)
}{1+\lambda a}s w d\Gamma+ \langle {{\partial s}\over
{\partial\nu}}; w \rangle_{\Gamma_1}= \int_{\Gamma_1}  l w\,
d\Gamma,
$$
leading to \eqref{SMbcw} and then to the third equation of
\eqref{max} due to the definition of $l$ and the relations between
$u$, $v$ and $s$.
We find the same result if $\tau'(t)\neq 0$.

In conclusion, we have found
$(u,v,v_1, z)^T\in \mathcal{D}(\mathcal{A})$,
which verifies \eqref{max}, and thus
$\lambda I-\mathcal{A}(t)$ is surjective for some $\lambda >0$
and $t>0$.
Again as $\kappa(t)>0$, this proves that
\begin{equation}
\lambda I-\tilde{\mathcal{A}}(t)=(\lambda+\kappa(t))
 I-\mathcal{A}(t)\quad\text{is surjective}
\label{surjective}\end{equation}
for any  $\lambda >0$ and $t>0$.

Then, \eqref{ineq1}, \eqref{dissipative} and \eqref{surjective}
 imply that the family
$\tilde{\mathcal{A}}=\{\tilde{\mathcal{A}}(t): t\in [0,T]\}$
is a stable family of generators in $\mathcal{H}$ with stability
constants independent of $t$, by \cite[Proposition 1.1]{kato:76}.
Therefore, the assumptions (i)-(iv) of Theorem \ref{thmkato}
are satisifed by \eqref{domini}, \eqref{ineq1},
\eqref{dissipative}, \eqref{Linfty}, \eqref{surjective} and
Lemma \ref{lemmadensity}, and thus, the problem
\begin{gather*}
\tilde{U}'=\tilde{\mathcal{A}}(t)\tilde{U}\\
\tilde{U}(0)=U_0
\end{gather*}
 has a unique solution
$\tilde{U}\in C([0, +\infty), D(\mathcal{A}(0)))\cap C^1([0, +\infty),
\mathcal{H})$ for $U_0\in D(\mathcal{A}(0))$.
The requested solution of \eqref{formulA} is then given by
 $$
U(t)=e^{\beta(t)}\tilde{U}(t)
$$
with $\beta(t)=\int_0^t\kappa(s)ds$, because
\begin{align*}
U'(t)&=\kappa(t)e^{\beta(t)}\tilde{U}(t)+e^{\beta(t)}\tilde{U}'(t)\\
&= \kappa(t)e^{\beta(t)}\tilde{U}(t)+e^{\beta(t)}\tilde{\mathcal{A}}(t)\tilde{U}(t)\\
&= e^{\beta(t)}(\kappa(t)\tilde{U}(t)+\tilde{\mathcal{A}}(t)\tilde{U}(t))\\
&= e^{\beta(t)}\mathcal{A}(t)\tilde{U}(t)=\mathcal{A}(t)e^{\beta(t)}\tilde{U}(t)\\
&= \mathcal{A}(t)U(t).
\end{align*}
This concludes the proof.
\end{proof}

\begin{theorem}\label{SMWP}
Assume that \eqref{tau1}--\eqref{tau3} and \eqref{assWP} hold.
Then for any initial datum $U_0\in \mathcal{H}$ there exists
a unique solution $U\in C([0,+\infty), \mathcal{H})$
of problem  \eqref{SMformulA}.
 Moreover, if $U_0\in \mathcal{D}(\mathcal{A}(0))$, then
$$
U\in C([0,+\infty), \mathcal{D}(\mathcal{A}(0)
))\cap C^1([0,+\infty),\mathcal{H}).
$$
\end{theorem}

\subsection{Well-posedness in the case $\mu=0$}

As before, we use the auxiliary unknown \eqref{defz}.
Then, problem \eqref{S1.1}--\eqref{S1.4}, with $\mu=0$,
is equivalent to
 \begin{gather}
u_{tt}(x,t) -\Delta u (x,t)-a\Delta u_t(x,t)=0\quad \text{in }\Omega\times
(0,+\infty),\label{SP1}\\
\tau (t) z_t(x,\rho,t)+(1-\tau' (t)\rho )z_{\rho}(x,\rho, t)=0\quad \text{in } \Gamma_1\times (0,1)\times (0,+\infty ),\label{SPz}\\
u (x,t) =0\quad \text{on }\Gamma_0\times
(0,+\infty),\label{SP2}\\
{{\partial (u+a u_t)}\over {\partial \nu }}(x,t)=  -k z(x,1,t)
\quad \text{on }\Gamma_1\times
(0,+\infty),\label{SP3}\\
z(x,0,t)=u_t(x,t)\quad\text{on } \Gamma_1\times
(0,\infty),\label{SP3bis*}
\\
u(x,0)=u_0(x)\quad \mbox{\rm and}\quad u_t(x,0)=u_1(x)\quad
\text{in }\Omega,\label{SP4}\\
z(x,\rho ,0)=f_0(x,-\rho \tau(0) )\quad\text{in }
\Gamma_1\times (0,1).\label{SP5}
\end{gather}
If we denote
$U:=( u, u_t, z)^T$, then
$$
U':=( u_t, u_{tt}, z_t)^T= \left( u_t, \Delta
u+a\Delta u_t, \frac{\tau'(t)\rho-1}{\tau(t)}z_{\rho} \right )^T.
$$
Therefore, problem \eqref{SP1}--\eqref{SP5} can be rewritten as
\begin{equation}\label{formulA}
\begin{gathered}
U'=\mathcal{A}^0(t) U,\\
U(0)=\left( u_0, u_1, f_0(\cdot,-\cdot\tau(0) ) \right)^T,
\end{gathered}
\end{equation}
where the time dependent operator $\mathcal{A}^0(t)$ is defined by
$$
\mathcal{A}^0(t) \begin{pmatrix}
u\\
v\\
z
\end{pmatrix}
:= \begin{pmatrix}
v\\
\Delta (u+av)\\\frac{\tau'(t)\rho-1}{\tau(t)}
z_{\rho}
\end{pmatrix},
$$
with domain
\begin{equation}\label{Sdomain}
\begin{split}
 \mathcal{D} (\mathcal{A}^0(t))
:=\Big \{& (u,v,z)^T\in   H^1_{\Gamma_0}(\Omega)^2
\times L^2(\Gamma_1; H^1(0,1)):
 u+av \in E(\Delta,L^2(\Omega)), \\
& {{\partial (u+av)}\over {\partial\nu}}
=-k z(\cdot,1)\text{ on }\Gamma_1; v=z(\cdot,0) \text{ on }\Gamma_1
\Big\}.
\end{split}
\end{equation}

Note  that for $(u,v,z)^T\in \mathcal{D}(\mathcal{A}^0(t))$,
${{\partial (u+av)}\over {\partial\nu}}$ belongs
to $L^{2}(\Gamma_1)$ since
$z(\cdot,1)$ is in $L^{2}(\Gamma_1)$.

Finally, as above, observe that  domain of $\mathcal{A}^0(t)$
is independent of the time $t$; i.e.,
$$
\mathcal{D} (\mathcal{A}^0(t))=\mathcal{D} (\mathcal{A}^0(0)),
\quad t>0.
$$
Denote by $\mathcal{H}^0$ the Hilbert space
\begin{equation}\label{space}
\mathcal{H}^0:= H^1_{\Gamma_0} (\Omega)\times L^2 (\Omega)\times
L^2(\Gamma_1\times (0,1)),
\end{equation}
equipped with the  scalar product
\begin{equation}\label{Sscalar}
\begin{split}
\Big\langle \begin{pmatrix}
u\\
v\\
z
\end{pmatrix}, \begin{pmatrix}
\tilde u\\
\tilde v\\
\tilde z
\end{pmatrix} \Big \rangle_{\mathcal{H}^0}
&:=\int_{\Omega} \{\nabla u
(x){\nabla \tilde u}(x) +v(x)\tilde v(x) \}\,dx\\
&\quad +\int_{\Gamma_1}\int_0^1 z(x,\rho)\tilde z (x,\rho) d\rho
d\Gamma,
\end{split}
\end{equation}
Arguing analogously to the case $\mu>0$ we can deduce an
existence and uniqueness result.


\begin{theorem}\label{WPbis}
Assume that \eqref{tau1}--\eqref{tau3} and
\eqref{assWP} hold. Then, for any initial datum
$U_0\in \mathcal{H}^0$ there exists a unique solution
$U\in C([0,+\infty), \mathcal{H}^0)$ of problem \eqref{formulA}.
 Moreover, if $U_0\in \mathcal{D}(\mathcal{A}^0(0))$, then
$$
U\in C([0,+\infty), \mathcal{D}(\mathcal{A}^0(0)))
\cap C^1([0,+\infty),\mathcal{H}^0).
$$
\end{theorem}

\begin{remark} \label{rmk2.6}{\rm
This well-posedness theorem can be also deduced from the abstract
framework  of \cite{FNV} (see Theorem    2.2 in \cite{FNV}) for second order
evolution equations. On the contrary, the case $\mu>0$ is not
covered by this abstract result. }
\end{remark}

\subsection{Stability result}

Now, we  show that problem \eqref{S1.1}--\eqref{S1.4} is uniformly
exponentially stable under the assumption
\begin{equation}\label{relmubis}
| k| < \frac{a}{C_P}\sqrt{1-d}\,.
\end{equation}

We define the energy of system \eqref{S1.1}--\eqref{S1.4} as
\begin{equation}\label{defF}
F(t):= {1\over 2}\int_{\Omega}\{ u_t^2 +| \nabla u|^2\} dx
+{\xi\over 2}
\int_{t-\tau(t)}^t\int_{\Gamma_1}e^{\lambda (s-t)}u_t^2(x,s)
d\Gamma ds+\frac{\mu}{2}\int_{\Gamma_1}u_t^2(x,t)d\Gamma,
\end{equation}
where $\xi, \lambda$ are suitable positive constants.
We fix $\xi$ such that
\begin{equation}\label{xistrong}
\frac{| k| }{\sqrt{1-d}}<{\xi}<\frac{2a}{C_P}-\frac{| k|}{\sqrt{1-d}}.
\end{equation}
Note that \eqref{relmubis} ensures that this choice is possible.
Moreover, the parameter $\lambda$ is fixed satisfying
\begin{equation}\label{lambdabis}
 \lambda <\frac{1}{\overline\tau}
\big| \log \frac{| k|}{\xi\sqrt{1-d}}\big|\,.
\end{equation}
Remark that in the case of a constant delay, we can take
$\lambda=0$ and in that case $F(t)$ corresponds to the natural
energy of $(u,u_t, z)$ (up to the factor ${1\over 2}$),
 see \cite{Nic_Pig_IMAJ}.
Here the time dependence of the delay implies that our
system is no more invariant by translation
and therefore we have to replace the arguments from
\cite{Nic_Pig_IMAJ} by the use of an appropriate
Lyapunov functional.
We start with the following estimate.


\begin{proposition}\label{derivEstimabis}
Assume
\eqref{tau1}--\eqref{tau3} and \eqref{relmubis}.
Then, for any regular solution of problem \eqref{S1.1}--\eqref{S1.4}
the energy is decreasing  and, for a suitable positive constant $C$,
 we have
\begin{equation}\label{stimaderbis}
\begin{split}
F'(t)
&\le - C\Big\{\int_{\Omega} | \nabla u_t(x,t)|^2 dx +\int_{\Gamma_1}
 u^2_t(x,t-\tau (t))d\Gamma\Big\}\\
&\quad -C\int_{t-\tau(t)}^t \int_{\Gamma_1}
 e^{\lambda (s-t)} u_t^2(x,s) d\Gamma ds.
\end{split}
\end{equation}
\end{proposition}


\begin{proof}
Differentiating \eqref{defF}, we obtain
\begin{align*}
F'(t)&=\int_{\Omega} \{ u_t u_{tt} +\nabla u\nabla u_t \} dx
+{\xi\over 2}\int_{\Gamma_1} u_t^2(x,t) d\Gamma
+\mu\int_{\Gamma_1}u_t(t)u_{tt}(t)\, d\Gamma \\
&\quad -{\xi\over 2}\int_{\Gamma_1} e^{-\lambda\tau(t)}
 u_t^2(x,t-\tau(t))(1-\tau'(t)) d\Gamma \\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t
 \int_{\Gamma_1}e^{-\lambda (t-s)}u_t^2(x,s) d\Gamma ds,
\end{align*}
and then, applying Green's formula,
\begin{equation}\label{Stildebis}
\begin{split}
F'(t)&=\int_{\Omega} a u_t(x,t)\Delta u_t(x,t) dx
+\int_{\Gamma_1} u_t(t)\frac{\partial u}{\partial\nu}(t)d \Gamma
\\
&\quad - {{\xi}\over 2}\int_{\Gamma_1}e^{-\lambda\tau(t)}
 u_t^2(x,t-\tau(t))(1-\tau'(t))d\Gamma +{\xi\over 2}
 \int_{\Gamma_1}u_t^2(x,t) d\Gamma \\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t
 \int_{\Gamma_1}e^{-\lambda (t-s)}u_t^2(x,s) d\Gamma ds
 +\mu\int_{\Gamma_1}u_t(t)u_{tt}(t) d\Gamma.
\end{split}
\end{equation}
Integrating once more by parts and using the boundary conditions
we obtain
\begin{equation}\label{Stildebisbis}
\begin{split}
F'(t)&=-a
\int_{\Omega} |\nabla u_t(x,t)|^2 dx -k\int_{\Gamma_1} u_t(t)
u_t(t-\tau(t))d \Gamma \\
&\quad  - {{\xi}\over 2}\int_{\Gamma_1}e^{-\lambda\tau(t)}u_t^2
 (x,t-\tau(t))(1-\tau'(t))d\Gamma
+{\xi\over 2}\int_{\Gamma_1}u_t^2(x,t) d\Gamma \\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Gamma_1}
 e^{-\lambda (t-s)}u_t^2(x,s) d\Gamma \,ds.
\end{split}
\end{equation}
Now, applying Cauchy-Schwarz's inequality and recalling the
assumptions \eqref{tau1} and \eqref{tau2},
we obtain
\begin{align*}
F'(t)
&\le -a \int_{\Omega}| \nabla u_t(x,t)|^2dx+
{\xi\over 2}\int_{\Gamma_1}u_t^2(x,t) d\Gamma
+\frac{| k|}{2\sqrt{1-d}}\int_{\Gamma_1}u_t^2(x,t)d\Gamma\\
&\quad +\frac{| k|}{2}\sqrt{1-d}\int_{\Gamma_1}u_t^2(t-\tau(t))d\Gamma
\\
&\quad -{\xi\over 2}(1-d)e^{-\lambda\overline{\tau}}\int_{\Gamma_1}u^2_t
(x,t-\tau(t))d\Gamma
-\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Gamma_1}
e^{-\lambda (t-s)}u_t^2(x,s) d\Gamma \,ds
\\
&\le -\left(
a-\frac{| k| C_P}{2\sqrt{1-d}}-\frac{\xi}{2}C_P
\right )\int_{\Omega}|\nabla  u_t(x,t)|^2 dx \\
&\quad -\left(
e^{-\lambda\overline{\tau}}\frac{\xi}{2}(1-d)-\frac{| k| }{2}
\sqrt{1-d} \right )\int_{\Gamma_1} u_t^2(x,t-\tau(t)) d\Gamma
\\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Gamma_1}
e^{-\lambda (t-s)}u_t^2(x,s) d\Gamma ds,
\end{align*}
where in the last inequality we also use a trace estimate and
Poincar\'e's Theorem. Therefore, \eqref{stimaderbis} immediately
follows recalling \eqref{xistrong} and \eqref{lambdabis}.
\end{proof}


Now, let us define the Lyapunov functional
\begin{equation}\label{Lyapbis}
{\hat
F}(t)= {F} (t)+\gamma\left [\int_{\Omega}u(x,t)u_t(x,t)dx+\mu\int_{\Gamma_1}u(x,t)u_t(x,t)d\Gamma\right],
\end{equation}
where
$\gamma$ is a positive small constant that we will choose later on.

Note that, from Poincar\'e's Theorem, the functional $\hat F$ is
equivalent to the energy $F$, that is, for $\gamma$ small enough,
there exist two positive constant $\beta_1^0,\beta_2^0$ such that
\begin{equation}\label{equivbis}
\beta_1^0
\hat F(t)\le F(t)\le \beta_2^0
 \hat F(t),\quad \forall t\ge 0.
\end{equation}

\begin{lemma}\label{Multibis}
For any regular solution of problem  \eqref{S1.1}--\eqref{S1.4},
\begin{equation}\label{stimamultiplierbis}
\begin{split}
&\frac{d}{dt} \Big\{ \int_{\Omega} u(x,t) u_t(x,t)  dx  dt
+\mu\int_{\Gamma_1}u(x,t)u_t(x,t)d\Gamma \Big\}\\
&\le C\Big\{
\int_{\Omega}|\nabla u_t(x,t)|^2dx +\int_{\Gamma_1}u_t^2(x,t-\tau(t))
 d\Gamma\Big\}
-\frac{1}{2}\int_\Omega|\nabla u(x,t)|^2 dx,
\end{split}
\end{equation}
for a suitable positive constant $C$
(that is different from the one from Proposition \ref{derivEstimabis}).
\end{lemma}

\begin{proof}
Differentiating and integrating by parts we have
\begin{equation}\label{identitymulbis}
\begin{split}
\frac{d}{dt} \int_{\Omega} u u_t  dx
&=\int_{\Omega}u_t^2(x,t) dx +\int_{\Omega} u(\Delta u+a\Delta u_t) dx\\
&=\int_{\Omega}
u_t^2(x,t) dx-\int_\Omega | \nabla u(x,t)|^2dx
 -a\int_\Omega \nabla u(x,t)\cdot\nabla u_t(x,t) dx \\
&\quad +\int_{\Gamma_1} u(t)\frac{\partial(u+au_t)}{\partial\nu}(t)
 d\Gamma.
\end{split}
\end{equation}
From \eqref{identitymulbis}, using the boundary condition on
$\Gamma_1$, we obtain
\begin{equation}\label{identitymulbisbis}
\begin{split}
&\frac{d}{dt} \left\{\int_{\Omega} u u_t  dx
+\mu\int_{\Gamma_1}u(x,t)u_t(x,t)d\Gamma
\right\}\\
&=\int_{\Omega}u_t^2(x,t) dx +\int_{\Omega} u(\Delta u+a\Delta u_t) dx
+\mu\int_{\Gamma_1}u_t^2(x,t) d\Gamma+\mu\int_{\Gamma_1}u(x,t)u_{tt}(x,t) d\Gamma
\\
&=\int_{\Omega}
u_t^2(x,t) dx-\int_\Omega | \nabla u(x,t)|^2dx
-a\int_\Omega \nabla u(x,t)\cdot\nabla u_t(x,t) dx\\
&\quad -k\int_{\Gamma_1}u(t)u_t(t-\tau(t))  d\Gamma
+\mu\int_{\Gamma_1}u_t^2(x,t) d\Gamma.
\end{split}
\end{equation}
We can conclude by using  Young's   inequality,  a  trace estimate
and Poincar\'e's Theorem.
\end{proof}

Finally using the above results we can deduce an exponential
stability estimate.

\begin{theorem}\label{stabilinearbis}
Assume \eqref{tau1}--\eqref{tau3} and \eqref{relmubis}.
Then there exist positive constants $C_1, C_2$ such that for
any solution of problem \eqref{S1.1}-\eqref{S1.4},
\begin{equation}\label{stabil1bis}
F(t)\le C_1F(0) e^{-C_2t},\quad \forall t\ge 0.
\end{equation}
\end{theorem}

\begin{proof}
 From Lemma \ref{Multibis}, taking $\gamma$ sufficiently small
in the definition of the Lyapunov functional
$\hat F$, we have
\begin{equation}\label{stabil2}
\begin{split}
 \frac{d}{dt} {\hat F}(t)
&\le -C\Big\{ \int_{\Omega}
| \nabla u_t (x,t)|^2dx +\int_{\Gamma_1}u_t^2(x, t-\tau
(t))d x\Big\}\\
&\quad -C \int_{t-\tau(t)}^te^{-\lambda (t-s)}\int_{\Gamma_1}u_t^2(x,s)
d\Gamma\, ds
- \frac{\gamma}{2} \int_\Omega | \nabla u(x,t)|^2dx,
\end{split}
\end{equation}
for a suitable positive constant $C$ that is different from the
 one in \eqref{stimamultiplierbis}.
Poincar\'e's Theorem implying
$$
\int_{\Omega}| u_t (x,t)|^2dx+
\int_{\Gamma_1} | u_t (x,t)|^2ds\leq C_{P1}
\int_{\Omega} | \nabla u_t (x,t)|^2dx,
$$
for some $C_{P1}>0$,
 we obtain
\begin{equation}\label{stabil3}
\frac{d}{dt} {\hat F}(t)\le -C'F(t),
\end{equation}
for a suitable positive constant $C'$.
This clearly implies the exponential estimate
\eqref{stabil1bis} recalling \eqref{equivbis}.
\end{proof}

\begin{remark}\label{rmk2.10} {\rm
Note that in the case of a constant delay the exponential stability
result holds under the condition
$| k|<a/C_P$ (corresponding to \eqref{relmubis} since,
in this case, $d=0$), see  \cite{Nic_Pig_IMAJ}.
On the contrary, if this condition is no more valid, then some
instabilities may occur, we refer to \cite{Nic_Pig_IMAJ}
for some illustrations.
}
\end{remark}


\section{Internal delay feedback \label{INT}}

\subsection{Well-posedness}

First of all we formulate a well-posedness result under the assumption
\begin{equation}\label{assWP0}
| a_1|\le a_0\sqrt{1-d}\,.
\end{equation}
Let us set
\begin{equation}\label{defzbis}
z(x,\rho ,t)=u_t(x, t-\tau(t)\rho ),\quad x\in\Omega,\ \rho\in (0,1), \ t>0.
\end{equation}
Then, problem \eqref{1.1}--\eqref{1.5}
is equivalent to
 \begin{gather}
u_{tt}(x,t) -\Delta u (x,t)+a_0u_t(x,t)+ a_1z(x,1,t)=0\quad \text{in }\Omega\times
(0,+\infty)\label{P1}\\
 \tau(t) z_t(x,\rho,t)+(1-\tau'(t)\rho)z_{\rho}(x,\rho, t)=0\quad \text{in } \Omega\times (0,1)\times (0,+\infty )\label{Pz}\\
u (x,t) =0\quad \text{on }\partial\Omega\times
(0,+\infty)\label{P2}\\
z(x,0,t)=u_t(x,t)\quad\text{on } \Omega\times (0,\infty)\label{P3bis}
\\
u(x,0)=u_0(x)\quad \mbox{\rm and}\quad u_t(x,0)=u_1(x)
 \quad \text{in }\Omega\label{P4}\\
z(x,\rho ,0)=g_0(x,-\rho \tau(0) )\quad\text{in } \Omega\times (0,1).
 \label{P5}
\end{gather}
Let us denote
$U:=(u,u_t,z)^T$,
then
$$
U':=(u_t,u_{tt},z_t)^T=
\Big(u_t,\Delta u-a_0u_t-a_1 z(\cdot,1,\cdot),
\frac{\tau'(t)\rho-1}{\tau(t)}z_{\rho}\Big)^T.
$$
Therefore, problem \eqref{P1}--\eqref{P5} can be rewritten as
\begin{equation}\label{formulAI}
\begin{gathered}
U'=\mathcal{A}^1(t) U\\
U(0)=(u_0,u_1,g_0(\cdot ,-\cdot\tau(0) ))^T
\end{gathered}
\end{equation}
where the time dependent operator $\mathcal{A}^1(t)$ is defined by
$$
\mathcal{A}^1(t) \begin{pmatrix}
u\\
v\\
z
\end{pmatrix}
:=\begin{pmatrix}
v\\
\Delta u -a_0v-a_1z(\cdot ,1)\\
\frac{\tau'(t)\rho-1}{\tau(t)}
z_{\rho}
\end{pmatrix},
$$
with domain
\begin{equation}\label{domainI}
\begin{split}
\mathcal{D} (\mathcal{A}^1(t))
:=\Big \{&(u,v,z)^T\in \left (
H^2(\Omega)\cap H^1_0(\Omega)
\right )\times H^1(\Omega )\times L^2(\Omega; H^1(0,1)):\\
&v =z(\cdot,0) \text{ in }\Omega \Big \}.
\end{split}
\end{equation}
Note that the domain of $\mathcal{A}^1(t)$ is independent
of the time $t$; i.e.,
$$
\mathcal{D} (\mathcal{A}^1(t))
=\mathcal{D} (\mathcal{A}^1(0)),\quad t>0.
$$
Let us introduce the Hilbert space
\begin{equation}\label{spaceI}
\mathcal{H}^1:= H^1_0 (\Omega)\times L^2 (\Omega)
\times L^2(\Omega\times (0,1)),
\end{equation}
equipped with the inner product
\begin{equation}\label{scalarI}
\begin{split}
&\Big\langle
\begin{pmatrix}
u\\
v\\
z
\end{pmatrix},
\begin{pmatrix}
\tilde u\\
\tilde v\\
\tilde z
\end{pmatrix}
\Big\rangle_{\mathcal{H}^1}\\
&:=\int_{\Omega} \{\nabla u (x){\nabla \tilde u}(x) +v(x)\tilde v(x)
\} dx+\int_{\Omega}\int_0^1 z(x,\rho)\tilde z (x,\rho) d\rho d x.
\end{split}
\end{equation}


Next we state the well-posedness result then follows from
\cite[Theorem 2.2]{FNV} that extends
the well-posedness result of \cite{NPSicon} for
wave equations with
constant delays to an abstract second order
evolution equation with time-varying delay.

\begin{theorem}\label{WPI}
Assume \eqref{tau1}--\eqref{tau3} and \eqref{assWP0}. Then,
for any initial datum $U_0\in \mathcal{H}^1$ there exists
a unique solution $U\in C([0,+\infty), \mathcal{H}^1)$ of problem
\eqref{formulAI}.
 Moreover, if $U_0\in \mathcal{D}(\mathcal{A}^1(0))$, then
$$
U\in C([0,+\infty), \mathcal{D}(\mathcal{A}^1(0)))
\cap C^1([0,+\infty),\mathcal{H}^1).
$$
\end{theorem}

\begin{remark}\label{taudegenerate}
{\rm In \cite{NPV} the authors considered a wave equation with
boundary time-varying delay feedback without the assumption
$\tau(t)>\tau_0>0$
 of non degeneracy of $\tau$, but in less general spaces.
 We expect that a well-posedness result holds also
for problem \eqref{1.1}--\eqref{1.5} without this restriction on
$\tau$. However, we preferred to consider  non degenerate  $\tau$
in order to avoid technicalities. }
\end{remark}

\subsection{Stability result}

We will give  an exponential stability result
for problem \eqref{1.1}--\eqref{1.5}
under  assumption \eqref{relmu}.
We define the energy of system \eqref{1.1}--\eqref{1.5} as
\begin{equation}\label{S1}
E(t):= {1\over 2}\int_{\Omega}\{ u_t^2 +| \nabla u|^2\} dx +{\xi\over 2}
\int_{t-\tau(t)}^t\int_{\Omega}e^{\lambda (s-t)}u_t^2(x,s) dx\, ds,
\end{equation}
where $\xi, \lambda$ are suitable positive constants.
We will fix $\xi$ such that
\begin{gather}\label{Sxi}
2a_0-{{a_1}\over {\sqrt{1-d}}}-\xi >0,\quad
\xi-{{a_1}\over {\sqrt{1-d}}}>0, \\
\label{lambda}
 \lambda <\frac{1}{\overline\tau}
\big| \log \frac{| a_1|}{\xi\sqrt{1-d}}\big|\,.
\end{gather}

Note that assumption \eqref{relmu} guarantees the existence of  such a
constant $\xi$.
We have the following estimate.


\begin{proposition}\label{derivEstima}
Assume \eqref{tau1}-\eqref{tau3} and \eqref{relmu}.
Then, for any regular solution of problem \eqref{1.1}-\eqref{1.5}
the energy decays and there exists a positive constant $C$ such that
\begin{equation}\label{stimader}
E'(t)\le - C
\int_{\Omega} \{ u^2_t(x,t) +u^2_t(x,t-\tau (t))\}
-C\int_{t-\tau(t)}^t \int_\Omega e^{\lambda (s-t)} u_t^2(x,s) dx ds.
\end{equation}
\end{proposition}


\begin{proof}
Differentiating \eqref{S1}, we obtain
\begin{align*}
E'(t)&=\int_{\Omega} \{ u_t u_{tt} +\nabla u\nabla u_t \} dx
+{\xi\over 2}\int_\Omega u_t^2(x,t) \,dx\\
&\quad -{\xi\over 2}\int_\Omega e^{-\lambda\tau(t)}
 u_t^2(x,t-\tau(t))(1-\tau'(t)) dx \\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Omega}
e^{-\lambda (t-s)}u_t^2(x,s) dx ds,
\end{align*}
and then, applying Green's formula,
\begin{equation}\label{Stilde}
\begin{aligned}
E'(t)&= -a_0\int_{\Omega} u_t^2(x,t) dx
 -\int_{\Omega}a_1 u_t(t)u_t(x,t-\tau(t))d x
\\
&\quad - {{\xi}\over 2}\int_{\Omega}e^{-\lambda\tau(t)}
 u_t^2(x,t-\tau(t))(1-\tau'(t))dx +{\xi\over 2}
\int_{\Omega}u_t^2(x,t) dx \\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Omega}
 e^{-\lambda (t-s)}u_t^2(x,s) dx ds.
\end{aligned}
\end{equation}
Now, applying Cauchy-Schwarz's inequality and recalling
the assumptions \eqref{tau1} and \eqref{tau2},
we obtain
\begin{align*}
E'(t)
&\le -a_0 \int_{\Omega}u^2_t(x,t)dx
-a_1\int_\Omega u_t(t)u_t(t-\tau(t)) dx
+{\xi\over 2}\int_{\Omega}u_t^2(x,t) dx \\
&\quad-{\xi\over 2}(1-d)e^{-\lambda\overline{\tau}}\int_{\Omega}u^2_t(x,t-\tau(t))dx
-\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Omega}e^{-\lambda (t-s)}u_t^2(x,s) dx ds
\\
&\le -\left(a_0-\frac{| a_1| }{2\sqrt{1-d}}-\frac{\xi}{2}
\right )\int_{\Omega} u_t^2(x,t) dx \\
&\quad -\left( e^{-\lambda\overline{\tau}}\frac{\xi}{2}(1-d)
-\frac{| a_1| }{2}\sqrt{1-d} \right )\int_\Omega u_t^2(x,t-\tau(t))\, dx
\\
&\quad -\lambda {{\xi}\over 2}\int_{t-\tau(t)}^t\int_{\Omega}
e^{-\lambda (t-s)}u_t^2(x,s)\, dx \,ds
\end{align*}
from which easily follows \eqref{stimader} recalling
\eqref{Sxi} and \eqref{lambda}.
\end{proof}

Now, let us introduce the Lyapunov functional
\begin{equation}\label{Lyap}
\hat{E}(t)= {E} (t)+\gamma\int_{\Omega}u(x,t)u_t(x,t)dx,
\end{equation}
where $\gamma$ is a suitable small positive constant.

Note that, from Poincar\'e's Theorem, the functional $\hat{E}$ is
equivalent to the energy $E$, that is,  for $\gamma$ small enough,
there exist two positive constant $\beta_1,\beta_2$ such that
\begin{equation}\label{equiv}
\beta_1\hat{E}(t)\le E(t)\le \beta_2 \hat{E}(t),\quad \forall t\ge 0.
\end{equation}

\begin{lemma}\label{Multi}
For any regular solution of problem  \eqref{1.1}--\eqref{1.5},
\begin{equation}\label{stimamultiplier}
\begin{split}
&\frac{d}{dt}  \int_{\Omega} u(x,t) u_t(x,t)  \,dx \, dt\\
&\le C \int_{\Omega}[ u_t^2(x,t)+u_t^2(x,t-\tau(t))] dx
-\frac{1}{2}\int_\Omega|\nabla u(x,t)|^2 dx,
\end{split}
\end{equation}
for a suitable positive constants $C$.
\end{lemma}

\begin{proof}
Differentiating and integrating by parts
\begin{equation}\label{identitymul}
\begin{split}
\frac{d}{dt} \int_{\Omega} u u_t  dx
&=\int_{\Omega}u_t^2(x,t) dx +\int_{\Omega} u(\Delta u-a_0 u_t(t)
 -a_1u_t(t-\tau(t))\, dx\\
&=\int_{\Omega}
u_t^2(x,t) dx-\int_\Omega | \nabla u(x,t)|^2\,dx
 -\int_\Omega a_0u(t)u_t(t) dx \\
&\quad +\int_\Omega a_1u(t)u_t(t-\tau(t))\, dx.
\end{split}
\end{equation}
We can conclude by using  Young's inequality and Poincar\'e's
Theorem.
\end{proof}

Therefore, analogously to the case of boundary delay feedback,
 we can now obtain a uniform exponential
decay estimate.

\begin{theorem}\label{stabilinear}
Assume \eqref{tau1}--\eqref{tau3} and \eqref{relmu}.
Then there exist positive constants $C_1, C_2$ such that for any
 solution of problem \eqref{1.1}--\eqref{1.5},
\begin{equation}\label{stabil1}
E(t)\le C_1E(0) e^{-C_2t},\quad \forall t\ge 0.
\end{equation}
\end{theorem}

\begin{remark}{\rm
Using Lemma \ref{Multi} this stability result can be deduced with
the help of  Theorem 4.3 of  \cite{FNV}. The difference with
\cite{FNV} relies on the choice of a simpler Lyapunov functional
that renders the proof of the exponential decay more simple.}
\end{remark}

\begin{remark}{\rm
Note that in \cite{NPSicon} we have assumed that  the coefficient
$a_1$ of the delay term is positive. But this assumption is not
necessary. The results of  \cite{NPSicon} are valid, with analogous
proofs, also for $a_1$ of arbitrary sign  satisfying $| a_1|
<a_0$. }
\end{remark}

\begin{remark}{\rm
Note that in the proof of the stability estimate we did not use the
condition of non degeneracy of the delay $\tau(t) \ge\tau_0>0$. So,
 if $u$ is a regular solution of problem \eqref{1.1}--\eqref{1.5}
the exponential stability result holds for $u$
also in presence of  a possibly degenerate $\tau$,
(cf. Remark \ref{taudegenerate}).
The same is true for solutions to problem \eqref{S1.1}--\eqref{S1.4}.
}
\end{remark}

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