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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 71, pp. 1--13.\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-2011/71\hfil Compact decoupling]
{Compact decoupling for thermoviscoelasticity in irregular domains}

\author[E. M.  Ait Ben Hassi,  H. Bouslous, L. Maniar \hfil EJDE-2011/71\hfilneg]
{El Mustapha  Ait Ben Hassi,  Hammadi Bouslous, Lahcen Maniar}
 % in alphabetical order

\address{D\'epartement de Math\'ematiques \\
Facult\'e des Sciences Semlalia, \hfill\break\indent
Universit\'e Cadi Ayyad, Marrakech 40000, B.P. 2390, Maroc}
\email[E. M.  Ait Ben Hassi]{m.benhassi@ucam.ac.ma}
\email[H. Bouslous]{bouslous@ucam.ac.ma}
\email[L. Maniar]{maniar@ucam.ac.ma}


\thanks{Submitted August 24, 2010. Published May 31, 2011.}
\subjclass[2000]{34G10, 47D06}
\keywords{Thermoviscoelasticity;  semigroup compactness;
\hfill\break\indent semigroup norm continuity, essential spectrum;
 fractional power}

\begin{abstract}
 Our goal is to prove the compactness of the difference between the
 thermoviscoelasticity semigroup and its decoupled semigroup. To
 show this, we  prove the norm continuity of this difference, the
 compactness of the difference of their  resolvents  and use
 \cite[Theorem 2.3 ]{Huang}. We generalize  a result by  Liu \cite{Liu}. 
 An illustrative example of a  thermoviscoelastic system with 
 Neumann Laplacian on a Jelly Roll  domain is given.
\end{abstract}

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

\section{Introduction}

Consider the  abstract thermoviscoelastic model
\begin{gather} \label{visc1}
 \ddot w(t) + A_1 w(t) -\int_{-\infty}^0g(s) A_1 w(t+s) ds + B u(t) =0,
\quad t\geq  0, \\
\label{visc2}
 \dot u(t) + A_2 u(t) -B^{\ast}\dot w(t)=0, \quad t\geq 0, \\
 \label{visc3}
w(0)=w^0, \quad \dot w(0)=w^1,\quad u(0)= u^0,\quad
w(s)= f_0(s),\quad s\in (-\infty,0),
\end{gather}
where $g$ is a given function satisfying the
following  conditions:
\begin{gather} \label{c1}
g \in \mathcal{C}^1(-\infty,0]\cap L^1(-\infty,0),\\
\label{c2}
g(t)\geq 0 , \quad  g'(t)\geq0  \quad\text{for }t<0, \\
\label{c3} \int_{-\infty}^0g(s)ds < 1.
\end{gather}
By the decoupling technique,  we obtain the system
\begin{gather} \label{dvisc1}
 \ddot{\bar{w}}(t)+  A_1\bar{w}(t)
- \int_{-\infty}^0g(s) A_1 \bar{w}(t+s) ds
+ BA_2^{-1}B^\ast \dot{\bar{w}}(t) =0,\quad t\geq0,\\
\label{dvisc2}
 \dot{\bar{u}}(t) + A_2\bar{u}(t)
-B^{\ast}\dot{\bar{w}}(t)=0, \quad t\geq0, \\
 \label{dvisc3}
\bar{w}(0)=w^0,\quad \dot{\bar{w}}(0)=w^1, \quad
\bar{u}(0)= u^0,\quad \bar{w}(s)= f_0(s),\quad
 s\in (-\infty,0).
\end{gather}
The operators $A_1$ and $A_2$ are positive self adjoint and invertible
on two Hilbert spaces $H_1$ and $H_2$,  and $B$ is an
unbounded operator from $H_2$  to $H_1$.
Liu \cite{Liu} proved that these two systems are well posed and generate  two semigroups
$\mathcal{T}:=(T(t))_{t\geq0}$ and
$\mathcal{T}_d:=(T_d(t))_{t\geq0}$.
Assuming that $BA_2^{-\gamma}$ is compact for some
$0<\gamma<1$,  he proved that  their difference is compact.
In this paper,  proceeding as in  \cite{bbm},  we show
that $t\mapsto T(t)-T_{d}(t)$ and  $t\mapsto T(t)-S(t)$
are norm continuous for $t>0$ where $\mathcal{S}:=(S(t))_{t\geq0}$
is the semigroup generated by the first equation \eqref{dvisc1}
in the decoupled system. Consequently, under no compactness assumption,
$r_{\rm crit}(T(t))=r_{\rm crit}(T_d(t))=
r_{\rm crit}(S(t))$ for $t\geq0$ and
  $\omega_0(\mathcal{T})=\max\{\omega_{\rm crit}(\mathcal{S}),s(L)\}$.

 Assuming that $A_1^{-1/2}BA_2^{-1}$ is a compact operator
(in particular if $BA_2^{-\gamma}$ is compact) and
$\int_{-\infty}^0 g(s)s^2 ds < \infty$, we  prove the compactness
of the difference $R(\lambda,L)-R(\lambda,L_d)$  for every
 $\lambda \in \rho(L)\cap \rho(L_d)$, where $L$ and $L_d$ are
the generators of $\mathcal{T}$ and $\mathcal{T}_d$, respectively.
 Thus,  \cite[Theorem 2.3]{Huang} leads to the compactness
of $T(t)-T_{d}(t)$.

 To illustrate this generalization, we consider the
 thermoviscoelastic system
\begin{gather*} %\label{ex1}
  \ddot w -\mu\Delta w -(\lambda + \mu) \nabla\operatorname{div} w
+ \mu g_1 \ast \Delta w + (\lambda + \mu)g_1\ast \nabla
\operatorname{div} w + m\nabla  u=0\\
 \text{in } \Omega \times  (0,\infty),
\\ %\label{ex2}
\dot u+\beta u - \Delta  u - m \operatorname{div}\dot w=0\quad
  \text{in }\Omega \times  (0,\infty),
\\ %\label{ex3}
 w = 0, \frac{\partial  u}{\partial n} = 0\quad
\text{on } \Gamma \times (0,\infty),\\ % \label{ex4}
w(x,0)=w^0(x),\quad \dot w(x,0)=w^1(x),\quad  u(x,0)=  u^0(x) \quad
\text{in } \Omega ,\\\ %\label{ex5}
w(x,0) + w(x,s)=f_0(x,s)\quad\text{in }\Omega \times
 (-\infty,0)),
\end{gather*}
where $\mu,\lambda$ are positive constants. The set $\Omega$ is the
Jelley Roll, a  bounded open set proposed in \cite{B.Sim},
$$
\Omega=\{(x,y)\in \mathbb{R}^2: 1/2 <r<1\} \setminus \Gamma,
$$
where $\Gamma$ is the curve, in $\mathbb{R}^2$,  given in polar
coordinates by
$$
r(\phi)=\dfrac{\frac{3\pi}2+\arctan(\phi)}{2\pi}, \quad
-\infty<\phi<\infty.
$$
 For this system, we show that  $A_1^{-1}BA_2^{-1}$,
on the canonical modified energy Hilbert space,
is a compact operator but the operator $BA_2^{-\gamma}$
is not compact for every $0<\gamma<1$.

\section{Well-posedness}

Let $H_1$ and $H_2$ be two Hilbert spaces. The operators
$A_1 : \mathcal{D}(A_1)\subset
H_1\to H_1$ and $A_2 : \mathcal{D}(A_2)\subset H_2\to H_2$
are self adjoint and positive  (with not necessarily compact
inverses), while $B : \mathcal{D}(B) \subset H_2\to H_1$ is a closed
operator with adjoint operator $B^*$. Throughout  this paper,
we  assume the following:
\begin{gather} \label{cond}
\mathcal{D}(A_2^{1/2})\hookrightarrow \mathcal{D}(B)
\quad\text{and}\quad
\mathcal{D}(A_1^{1/2})\hookrightarrow \mathcal{D}(B^{\ast}),\\
\label{Hy}
A_2^{-1}B^{\ast}A_1^{1/2} \text{ extends to a bounded
linear operator from $H_1$ to $H_2$.}
\end{gather}

 Note that the operator $-A_2$ generates
an analytic strongly continuous semigroup $(e^{-A_2t})_{t\geq0}$.
Under assumption \eqref{cond}, $ BA_2^{-1/2} $ is a bounded operator
from $H_2$ to $H_1$ and  $BA_2^{-1/2}(BA_2^{-1/2})^{\ast}$ is a
bounded self adjoint non negative operator in $H_1$.

Setting $z(t,s)=w(t)- w(t+ s)$, $s\in (-\infty,0)$, the system
\eqref{visc1}-\eqref{visc3} can be transformed into  the  system
\begin{gather}
\label{visc11}  \ddot w(t) +k A_1 w(t) +\int_{-\infty}^0g(s)
 A_1z(t,s) ds + B u(t) =0, \quad t\geq  0, \\
\label{visc21} z_t-\dot w+z_s=0,\\
\label{visc31} z(t,0)=0,\quad t\geq0,\\
\label{visc41} \dot u(t) + A_2 u(t) -B^{\ast}\dot w(t)=0, \quad
 t\geq 0,\\
\label{visc51}
w(0)=w^0, \quad \dot w(0)=w^1,\quad u(0)= u^0,\quad
z(0,s)= f_0(s),\quad  s\in(-\infty,0),
\end{gather}
with $k=1-\int_{-\infty}^0g(s) ds$.
Set the Hilbert space
\[
\mathbb{H}=\mathcal{D}(A_1^{1/2})\times H_1
\times L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))\times H_2
\]
endowed with the norm
$$
\|( w,v, z, u)\|=\Big(k\|A_1^{1/2} w\|_{H_1}^2+\|v\|_{H_1}^2
+ \|z\|^2_{L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))}+\|u\|_{H_2}^2
\Big)^{1/2}.
$$
Here the space $L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))$
consists of $\mathcal{D}(A_1^{1/2})$-valued functions $z$
on $(-\infty,0)$ endowed with the norm
\[
\|z\|^2_{L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))}
=\int_{-\infty}^0g(s)\|A_1^{1/2} z(s)\|^2ds .
\]
 System \eqref{visc11}-\eqref{visc51} can also be written as a first
order system
\begin{gather}
\label{cauch1}  \dot w=v,\\
\label{cauch2} \dot v=- k A_1 w - \int_{-\infty}^0g(s) A_1z(t,s) ds - B
  u,\\
\label{cauch3} \dot z=v-  z_s \\
\label{cauch4} \dot u =- A_2 u + B^{\ast}v,\\
\label{cauch5} (w(0),v(0),u(0),z(0))=( w^0, w^1,f_0, u^0).
\end{gather}
We associate with the system \eqref{cauch1}-\eqref{cauch4}
the operator
\begin{gather*}
L( w, v, z, u)=\Big(v,- kA_1 w -\int_{-\infty}^0g(s) A_1  z(t,s) ds - B  u,v-  z_s
, - A_2 u + B^{\ast}v\Big),
\\
\mathcal{D}(L)= \{( w, v, z, u)\in
\mathbb{H}: v\in \mathcal{D}(A_1^{1/2}), u \in \mathcal{D}(C), kw +
\int_{-\infty}^0g(s)   z(s) ds \in \mathcal{D}(A),
\\
 z \in H^1(g,(-\infty,0),\mathcal{D}(A_1^{1/2})),  z(0)=0\},
\end{gather*}
where $H^1(g,(-\infty,0), \mathcal{D}(A_1^{1/2}))$ is the set
\[
\{z\in L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2})) :
 z_s \in  L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))\}.
\]
The decoupled system \eqref{dvisc1}-\eqref{dvisc3}
can also be transformed into
\begin{gather}
\label{dcauch1}  \dot {\bar{w}}=\bar{v},\\
\label{dcauch2} \dot{ \bar{v}}=- k A_1\bar{w} -
\int_{-\infty}^0g(s) A_1 \bar{z}(t,s) ds - BA_2^{-1}B^\ast
\bar{v}, \\
\label{dcauch3}\dot {\bar{z}}=\bar{v}-  \bar{z}_s,\\
\label{dcauch4} \dot {\bar{u}} =- A_2 \bar{u} +
 B^{\ast}\bar{v},\\
\label{dcauch5}(\bar{w}(0),\bar{v}(0,\bar{z}(0),
 \bar{u}(0))=( w^0, w^1,  f_0, u^0),
\end{gather}
to which we associate  the  operator
 $$
L_d(\bar{w},\bar{v},\bar{z},\bar{u})
=(\bar{v} ,- k A_1\bar{w} -  \int_{-\infty}^0g(s)
A_1 \bar{z}(t,s) ds - BA_2^{-1}B^\ast \bar{v},
\bar{v}- \bar{z}_s, -A_2 \bar{u} +
B^{\ast}\bar{v}),
$$
with $\mathcal{D}(L_d)= \mathcal{D}(L)$.
Also, the decoupled second order equation \eqref{dvisc1} can be
written as a  first order system
\begin{gather}
\label{dcauch1w}  \dot {\bar{w}}=\bar{v},\\
\label{dcauch2w}\dot{ \bar{v}}=- k A_1\bar{w} -
 \int_{-\infty}^0g(s) A_1 \bar{z}(t,s) ds - BA_2^{-1}B^\ast
\bar{v}, \\
\label{dcauch3w}
 \dot {\bar{z}}=\bar{v}-  \bar{z}_s,\\
\label{dcauch4w}(\bar{w}(0),\bar{v}(0),\bar{z}(0))
=( w^0, w^1,  f_0),
\end{gather}
with generating operator defined on
$\mathcal{H} :=\mathcal{D}(A_1^{1/2} )\times H_1\times
L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))$ by
\begin{gather*}
M(\bar{w},\bar{v},\bar{z}) =(\bar{v}
,- k A_1\bar{w} -  \int_{-\infty}^0g(s) A_1 \bar{z}(t,s) ds
- BA_2^{-1}B^\ast \bar{v}, \bar{v}- \bar{z}_s),
\\
\begin{split}
\mathcal{D}(M)=\big\{&(\bar{w},\bar{v},\bar{z})\in
\mathcal{H} : \bar{v}\in \mathcal{D}(A_1^{1/2}), k\bar{w} +
\int_{-\infty}^0g(s)   \bar{z}(s) ds \in \mathcal{D}(A),\\
&\bar{z} \in H^1(g,(-\infty,0),\mathcal{D}(A_1^{1/2})),
 \bar{z}(0)=0\big\}.
\end{split}
\end{gather*}

\begin{remark} \label{rmk2.1} \rm
$L$, $L_d$ and $M$ are respectively the parts in $\mathbb{H}$
and $\mathcal{H}$ of the matrix operators
\begin{gather*}
L_{-1}= \begin{pmatrix}
0&I&0&0\\
-{k(A_{1})}_{-1}&0& - G_{-1}&-B\\
0&I&-\frac{d}{ds}&0\\
0&B^{\ast}&0&-{(A_{2})}_{-1}
\end{pmatrix},\\
(L_{d})_{-1}= \begin{pmatrix}
0&I&0&0\\
-{k(A_{1})}_{-1}&-BA_2^{-1/2}(BA_2^{-1/2})^{\ast} &- G_{-1}&0 \\
0&I&-\frac{d}{ds}&0\\
0&B^{\ast}&0&-{(A_{2})}_{-1}
\end{pmatrix},\\
M= \begin{pmatrix}
0&I&0\\
-{k(A_{1})}_{-1}&-BA_2^{-1/2}(BA_2^{-1/2})^{\ast}&- G_{-1}\\
0&I&-\frac{d}{ds}
\end{pmatrix},
\end{gather*}
where  $ G_{-1} z ={(A_{1})}_{-1}^{1/2}\int_{-\infty}^0g(s)
A_1^{1/2}  z(s) ds$.
\end{remark}

 Using  Lumer Phillips theorem, the
following result can be proved  analogously as in
\cite[Theorem 2.1]{Liu}.

\begin{theorem} \label{thm2.2}
The operators $L$,  $L_d$ and $M$  generate contraction strongly
continuous semigroups $(T(t))_{t\geq0}$, $(T_{d}(t))_{t\geq0}$
and $(S(t))_{t\geq0}$ on $\mathbb{H}, \mathbb{H}$ and $\mathcal{H}$
respectively.
\end{theorem}

\section{Norm continuity of the difference between the semigroups}

To show the norm continuity of the difference between the two
semigroups, we recall the following technical lemma.

\begin{lemma}[\cite{bbm}] \label{lem3.1}
The map $t\mapsto A_2^{\alpha}e^{-A_2t}$ is norm continuous  from $(0,\infty)$ to $\mathcal{L}(H_2)$ for every $0\leq \alpha<1$.
\end{lemma}

Then we have the following result.

\begin{theorem} \label{pr}
The map $t\mapsto T(t)-T_{d}( t)$ is  norm continuous from
$(0,\infty)$ to $\mathcal{L}(\mathbb{H})$.
\end{theorem}

\begin{proof}
 Let $t>0$ and $x^0=( w^0,v^0,  f_0, u^0)\in \mathcal{D}(L)$ such
that $\|x^0\|\leq 1$.
\begin{align*}
&T(t)( w^0,v^0,  f_0, u^0)-T_{d}(t)( w^0,v^0,  f_0, u^0)\\
&= \begin{pmatrix}
 w(t)-\bar{w}(t)\\
v(t)-\bar{v}(t)\\
 z(t)-\bar{z}(t)\\
 u(t)-\bar{u}(t)
\end{pmatrix}
= \int_0^t T(t-s)\begin{pmatrix}
0\\
BA_2^{-1}B^\ast \bar{v}(s)- B\bar{u}(s)\\
0\\
0
\end{pmatrix}ds .
\end{align*}
Let $0<h<1$. Setting
 $F(s):=B\bar{u}(s)-BA_2^{-1}B^{\ast}\bar{v}(s)$,
we  check that $\|F(s+h)-F(s)\|\to0$ as
$h\to0$ uniformly in $x^0$. To this end, we have
\begin{align*}
F(s)
&= Be^{-A_2s} u^0+B\int_0^{s}e^{-A_2(s-\sigma)}
B^{\ast}\bar{v}(\sigma)d\sigma
-BA_2^{-1}B^{\ast}\bar{v}(s)\\
&= Be^{-A_2s}
u^0+BA_2^{-1}\int_0^{s}A_2e^{-A_2(s-\sigma)}
B^{\ast}\bar{v}(\sigma)d\sigma-BA_2^{-1}B^{\ast}\bar{v}(s).
\end{align*}
Using an integration by parts,
\begin{align*}
F(s)&= Be^{-A_2s} u^0+[BA_2^{-1}e^{-A_2(s-\sigma)}B^{\ast}\bar{v}
(\sigma)]_0^s -BA_2^{-1}\int_0^{s}e^{-A_2(s-\sigma)}
B^{\ast}\bar{v}'(\sigma)d\sigma\\
&\quad -BA_2^{-1}B^{\ast}\bar{v}(s)\\
&= Be^{-A_2s} u^0-BA_2^{-1}e^{-A_2s}B^{\ast}v^0-BA_2^{-1}\int_0^{s}e^{-A_2(s-\sigma)}B^{\ast}\bar{v}^{\prime }(\sigma)d\sigma\\
&= Be^{-A_2s} u^0-BA_2^{-1/2}e^{-A_2s}A_2^{-1/2}B^{\ast}v^0+kBA_2^{-1}\int_0^{s}e^{-A_2(s-\sigma)}B^{\ast}A_1\bar{w}(\sigma)d\sigma\\
&\quad +BA_2^{-1}\int_0^{s}e^{-A_2(s-\sigma)}B^{\ast}[-\int_{-\infty}^0g(\tau)A\bar{z}(s,\tau)d\tau +BA_2^{-1}B^{\ast}\bar{v}(\sigma)]d\sigma\\
&= BA_2^{-1/2}A_2^{1/2}e^{-A_2s} u^0-BA_2^{-1/2}e^{-A_2s}
 A_2^{-1/2}B^{\ast}v^0 \\
&\quad +kBA_2^{-1/2}\int_0^{s}A_2^{1/2}e^{-A_2(s-\sigma)}
 A_2^{-1}B^{\ast}A_1^{1/2}A_1^{1/2}\bar{w}(\sigma)d\sigma\\
&\quad +BA_2^{-1/2}\int_0^{s}e^{-A_2(s-\sigma)}A_2^{-1/2}
 B^{\ast}BA_2^{-1}B^{\ast}\bar{v}(\sigma)d\sigma\\
&\quad -BA_2^{-1}\int_0^{s}e^{-A_2(s-\sigma)}B^{\ast}\int_{-\infty}^0g(\tau)A\bar{
z}(s,\tau)d\tau
d\sigma\\
&= BA_2^{-1/2}A_2^{1/2}e^{-A_2s} u^0-BA_2^{-1/2}e^{-A_2s}
 A_2^{-1/2}B^{\ast}v^0\\
&\quad +kBA_2^{-1/2}\int_0^{s}A_2^{1/2} e^{-A_2(s-\sigma)}
 A_2^{-1}B^{\ast}A_1^{1/2} A_1^{1/2}\bar{w}(\sigma)d\sigma\\
&\quad +BA_2^{-1/2}\int_0^{s}e^{-A_2(s-\sigma)}A_2^{-1/2}B^{\ast}BA_2^{-1}B^{\ast}\bar{v}(\sigma)d\sigma\\
&\quad -BA_2^{-1/2}\int_0^{s}A_2^{1/2}e^{-A_2(s-\sigma)}A_2^{-1}B^{\ast}A_1^{1/2}\int_{-\infty}^0g(\tau)A_1^{1/2}\bar{
z}(s,\tau)d\tau d\sigma.
\end{align*}
Since $BA_2^{-1/2}$ is a bounded operator and $s\mapsto
 e^{-A_2s}$, $s\mapsto A_2^{1/2}e^{-A_2s}$ are  norm continuous
from $(0,\infty)$ to  $\mathcal{L}(H_2)$, then the mappings
$s\mapsto BA_2^{-1/2}A_2^{1/2}e^{-A_2s}$ and
$s\mapsto BA_2^{-1/2}e^{-A_2s}A_2^{-1/2}B^{\ast}$ are
norm continuous from $(0,\infty)$ to  $\mathcal{L}(H_1)$.

Under the assumption \eqref{Hy}  we have $A_2^{-1}B^{\ast}A_1^{1/2}$
is a bounded operator ; so there exists a constant $\alpha(s)$ such
that, $\|A_2^{-1}B^{\ast}A_1^{1/2}A_1^{1/2}\bar{
w}(\sigma)\|\leq\alpha(s)\|x^0\|$, \;\;for\; every\;
$\sigma\in[0,s]$. Thus, $s\mapsto
\int_0^{s}A_2^{1/2}e^{-A_2(s-\sigma)}A_2^{-1}
B^{\ast}A_1^{1/2}A_1^{1/2}\bar{w}(\sigma)d\sigma$
is continuous in $(0,\infty)$ uniformly in $\|x^0\|\leq1$.\\
Since $A_2^{-1/2}B^{\ast}BA_2^{-1}B^{\ast}$ is a bounded
operator, using the same argument
$$
s\mapsto\int_0^{s}e^{-A_2(s-\sigma)}A_2^{-1/2}B^{\ast}
BA_2^{-1}B^{\ast}\bar{v}(\sigma)d\sigma
$$
is  continuous in $(0,\infty)$ uniformly in $\|x^0\|\leq1$. In the
other hand
$$
\|\int_{-\infty}^0g(\tau)A_1^{1/2}\bar{z}(s,\tau)d\tau
d\sigma\|\leq \Big(\int_{-\infty}^0g(\tau)d\tau\Big)^{1/2}
\Big(\int_{-\infty}^0g(\tau)\|A_1^{1/2}\bar{
z}(s,\tau)\|^2d\tau \Big){1/2}.
$$
 So
$A_2^{-1}B^{\ast}A_1^{1/2}\int_{-\infty}^0g(\tau)A_1^{1/2}\bar{
z}(s,\tau)d\tau$ is a bounded operator uniformly in $\|x^0\|\leq1$.
As a consequence,
$s\mapsto\int_0^{s}A_2^{1/2}e^{-A_2(s-\sigma)}A_2^{-1}
B^{\ast}A_1^{1/2}(\int_{-\infty}^0g(\tau)A_1^{1/2}\bar{
z}(s,\tau)d\tau) d\sigma$ is norm continuous in $(0,\infty)$
uniformly in $\|x^0\|\leq1$.
Finally, $\|F(s+h)-F(s)\|\to0$ as
$h\to0$ uniformly in $x^0$.
We have
 \begin{align*}
&\begin{pmatrix}
 w(t)-\bar{w}(t)\\
v(t)-\bar{v}(t)\\
z(t)-\bar{z}(t)\\
 u(t)-\bar{u}(t)
\end{pmatrix}\\
&= \int_0^t T(t-s)\begin{pmatrix}
0\\
BA_2^{-1}B^\ast \bar{v}(s)- B\bar{u}(s)\\
0\\
0
\end{pmatrix} ds\\
&= {\int_0^{t+h}T(t+h-s)\begin{pmatrix}
0\\
F(s)\\
0\\
0
\end{pmatrix}ds}
-{\int_0^tT(t-s)\begin{pmatrix}
0\\
F(s)\\
0\\
0
\end{pmatrix}ds}\\
&= {\int_0^{t+h}T(s)\begin{pmatrix}
0\\
F(t+h-s)\\
0\\
0
\end{pmatrix}ds}-{\int_0^tT(s)\begin{pmatrix}
0\\
F(t-s)\\
0\\
0
\end{pmatrix}ds}\\
&= {\int_0^{t}T(s)\begin{pmatrix}
0\\
F(t+h-s)-F(t-s)\\
0\\
0
\end{pmatrix} ds}+{\int_t^{t+h}T(t+h-s)\begin{pmatrix}
0\\
F(s)\\
0\\
0
\end{pmatrix} ds}\\
&= {\int_0^{t}T(s)\begin{pmatrix}
0\\
F(t+h-s)-F(t-s)\\
0\\
0
\end{pmatrix} ds}+{\int_0^{h}T(s)\begin{pmatrix}
0\\
F(t+s)\\
0\\
0
\end{pmatrix}ds}.
\end{align*}
Hence,
\begin{align*}
&\|\begin{pmatrix}
 w(t+h)-\bar{w}(t+h)\\
v(t+h)-\bar{v}(t+h)\\
 z(t+h)-\bar{z}(t+h)\\
 u(t+h)-\bar{u}(t+h)
\end{pmatrix}
-\begin{pmatrix}
 w(t)-\bar{w}(t)\\
v(t)-\bar{v}(t)\\
 z(t)-\bar{z}(t)\\
 u(t)-\bar{u}(t)\end{pmatrix}\|\\
&\leq \|{\int_0^{t}T(s)\begin{pmatrix}
0\\
F(t+h-s)-F(t-s)\\
0\\
0
\end{pmatrix}ds}\|+\|{\int_0^{h}T(s) \begin{pmatrix}
0\\
F(t+s)\\
0\\
0
\end{pmatrix}ds}\|\\
&\leq \sup_{\tau\in[0,t]} \|T(\tau)\|{\int_0^{t}
\|F(t+h-s)-F(t-s)\|ds}
+{\int_0^{h}\|T(s)\|\|F(t+s)\| ds}.
\end{align*}
In addition, there exists a constant $N$ such that
$\sup_{s\in[0,t+1]} \|F(s)\|\leq N$ uniformly in $x^0$, and thus
\begin{align*}
&\|\begin{pmatrix}
 w(t+h)-\bar{w}(t+h)\\
v(t+h)-\bar{v}(t+h)\\
z(t+h)-\bar{z}(t+h)\\
 u(t+h)-\bar{u}(t+h)
\end{pmatrix}-\begin{pmatrix}
 w(t)-\bar{w}(t)\\
v(t)-\bar{v}(t)\\
  z(t)-\bar{z}(t)\\
  u(t)-\bar{u}(t)
\end{pmatrix}\|\\
&\leq \sup_{\tau\in[0,t+1]} \|T(\tau)\|{\int_0^{t}
\|F(t+h-s)-F(t-s)\|ds}
 +c(t)Nh.
\end{align*}
As $\|F(s+h)-F(s)\|\to 0$ as $h\to 0$
uniformly in $x_0$, we conclude that
$${\int_0^{t} \|F(t+h-s)-F(t-s)\|ds}\to 0, h\to0
$$
uniformly for $x_0\in \mathcal{D}(L)$ verifying $\|x_0\|\leq1$.
This achieves the proof.
\end{proof}

\begin{theorem}\label{thwavew}
The maps  $t\mapsto T_d(t) - S(t)$ and $t\mapsto T(t) - S(t)$ are
norm continuous from $(0,\infty)$ to  $\mathcal{L}(\mathbb{H})$.
\end{theorem}

\begin{proof}
   Let $x_0=(u^0,v^0,f_0, w^0)\in \mathbb{H}$ such
that $\|x_0\|\leq 1$ and $t>0$.
\begin{align*}
&T_d(t)x_0-(S(t)((u^0,v^0,f_0),0)\\
&=(0, 0, 0,e^{-A_2t} w^0 +
\int_0^{t}e^{-A_2(t-s)}B^{\ast}\pi_2 S(s)((u^0,v^0,f_0)ds)
\end{align*}
where
$\pi_2:\mathcal{D}(A_1^{1/2})\times H_1\times L^2(g,(-\infty,0),
\mathcal{D}(A_1^{1/2})) \to H_1$,
$(u,v,z) \mapsto v$.
Set $\Delta(t)=T_d(t)x_0-(S(t)(u^0,v^0,f_0),0)$. For $h>0$, one has
\begin{align*}
 \Delta(t+h)-\Delta(t)
&= \Big(0,0,e^{-A_2(t+h)} w^0-e^{-A_2t} w^0\\
&\quad +\int_0^{t+h}e^{-A_2(t+h-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds\\
&\quad -\int_0^{t}e^{-A_2(t-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds\Big).
\end{align*}
Then
 \begin{align*}
&{ \|\Delta(t +h)-\Delta(t)\|}\\
&= {\|e^{-A_2(t+h)} w^0-e^{-A_2t}w^0
+\int_0^{t+h}e^{-A_2(t+h-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds}\\
&\quad -
{\int_0^{t}e^{-A_2(t-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds)\|}\\
&= {\|e^{-A_2(t+h)} w^0-e^{-A_2t} w^0
+\int_0^{t}[e^{-A_2(t+h-s)}-e^{-A_2(t-s)}]B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds} \\
&\quad +{\int_t^{t+h}e^{-A_2(t+h-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds)\|} \\
&\leq \|e^{-A_2(t+h)}-e^{-A_2t}\|\|w^0\|\\
&\quad +\|\int_0^{t}[e^{-A_2(t+h-s)}-e^{-A_2(t-s)}]B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds \| \\
&\quad+\|\int_t^{t+h}e^{-A_2(t+h-s)}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)ds)\|.
\end{align*}
Hence, since $A_2^{-1/2}B^{\ast}$ is a bounded operator, we
have
\begin{align*}
&{ \|\Delta(t +h)-\Delta(t)\|}\\
&\leq \int_0^{t}\|[A_2^{1/2}e^{-A_2(t+h-s)}-A_2^{1/2}e^{-A_2(t-s)}]A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0) \| ds\\
&\quad +\int_t^{t+h}\|A_2^{1/2}e^{-A_2(t+h-s)}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0))\|ds +\|e^{-A_2(t+h)}-e^{-A_2t}\|\\
&\leq \int_0^{t}\|[A_2^{1/2}e^{-A_2(t+h-s)}-A_2^{1/2}e^{-A_2(t-s)}]A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0) \|ds \\
&\quad +\int_t^{t+h}\|A_2^{1/2}e^{-A_2(t+h-s)}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0))\|ds +\|e^{-A_2(t+h)}-e^{-A_2t}\|.
\end{align*}
Since the semigroup $(e^{-A_2t})_{t\geq0}$ is analytic, it is
immediately norm continuous. Thus
$\|e^{-A_2(t+h)}-e^{-A_2t}\|\to 0$ as $h\to0$.
In the other hand $A_2^{-1/2}B^{\ast}=
(BA_2^{-1/2})^{\ast}$ is a bounded operator, thus there exists
a constant $\delta(t)$ such that
\begin{gather*}
\|A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)\| \leq \delta(t)\|A_2^{-1/2}B^{\ast}\|
\|(u^0,v^0, f_0)\|\quad  \text{for every } s \in[0,t]\\
\|A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)\| \leq \delta(t)\|A_2^{-1/2}B^{\ast}\|\|x_0\|
\quad \text{for every } s \in[0,t]\\
\|A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)\| \leq \delta(t)\|A_2^{-1/2}B^{\ast}\|\quad
\text{for every } s \in[0,t].
\end{gather*}
Moreover
\begin{align*}
&\|\int_0^{t}[A_2^{1/2}e^{-A_2(t+h-s)}-A_2^{1/2}e^{-A_2(t-s)}]
A_2^{-1/2}B^{\ast}\pi_2 S(s)(u^0,v^0,f_0)d s\|
\\
&\leq\delta(t)\|A_2^{-1/2}B^{\ast}\|\int_0^{t}\|A_2^{1/2}
e^{-A_2(t+h-s)}-A_2^{1/2}e^{-A_2(t-s)}\|ds.
\end{align*}
It follows from Lemma \ref{lem3.1} and Lebegue theorem that
\[
\|A_2^{1/2}e^{-A_2(t+h-s)}-A_2^{1/2}e^{-A_2(t-s)}\|\to0
\]
 as $ h\to0$ and $t>s$, and
\[
{\|\int_0^{t}[A_2^{1/2}e^{-A_2(t+h-s)}
-A_2^{1/2}e^{-A_2(t-s)}]A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds \|}\to 0
\]
 as $h\to 0$ uniformly in $x_0$.
For the third term, we have
\begin{align*}
&{\|\int_t^{t+h}A_2^{1/2}e^{-A_2(t+h-s)}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds)\|}\\
&= {\|\int_0^{h}A_2^{1/2}e^{-A_2s}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds\|}.
\end{align*}
Using the similar argument as in the second term, there exists
$\beta(t)$ such that
\begin{align*}
{\|\int_0^{h}A_2^{1/2}e^{-A_2s}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds\|}&\leq
\beta(t){\int_0^{h}\|A_2^{1/2}e^{-A_2s}\|ds}\|x_0\|\\
&\leq \|x_0\|\beta(t){\int_0^{h}s^{-1/2}ds}\\
&\leq 2\beta(t)\sqrt{h}
\end{align*}
(here we used  $\|A_2^{1/2}e^{-A_2t}\|=
O(t^{-1/2})$ as $t>0$; see for example
\cite[Theorem 1.4.3]{Lun}).
Consequently,
${\|\int_t^{t+h}A_2^{1/2}e^{-A_2(t+h-s)}A_2^{-1/2}B^{\ast}\pi_2
S(s)(u^0,v^0,f_0)ds)\|}\to0$ as $h\to0$
uniformly in $x_0$. Finally,
$\|\Delta(t+h)-\Delta(t)\|\to0$ as
$h\to0$ uniformly in $x_0$.
Since, by Theorem \ref{pr}, $t\mapsto T(t)-T_{d}(t)$ is norm
continuous on $(0,\infty)$, $t\mapsto T(t)-S(t)$
is norm continuous.
\end{proof}

Theorem \ref{thwavew}  leads to the following result.

\begin{corollary}\label{cor2}
 $r_{\rm crit}(T(t))=r_{\rm crit}(T_d(t))=
r_{\rm crit}(S(t))$ for $t\geq0$ and
  $\omega_0(\mathcal{T})=\max\{\omega_{\rm crit}(\mathcal{S}_w),s(L)\}$.
\end{corollary}

\section{Compactness of the difference between the two semigroups}

We have also this  main result.
\begin{theorem} \label{thm4.1}
Assume  $A_1^{-1/2}BA_2^{-1}$ is compact in
$\mathcal{L}(H_2,H_1)$  and $\int_{-\infty}^0g(s)s^2 ds< \infty$.
Then
$ R(\lambda,L)-R(\lambda,L_d) $ is compact on $\mathbb{H}$
for every $\lambda\in \rho(L)\cap \rho(L_d)$.
\end{theorem}

\begin{proof}
We have
\begin{equation} \label{resolv1}
R(\lambda,L_d)-R(\lambda,L)
=LR(\lambda,L)[L^{-1}-L_d^{-1}]L_dR(\lambda,L_d).
\end{equation}
Let $(\varphi,\psi,\eta,\xi)\in\mathbb{H}
=\mathcal{D}(A_1^{1/2})\times H_1
\times L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))\times H_2$.
We look for $(w, v,  z, u)\in D(L)$ such that
$L(w, v,  z, u)=(\varphi,\psi,\eta,\xi)$.
Note that the equation
$L(w, v,  z, u)=(\varphi,\psi,\eta,\xi)$ is safistied
is equivalent to the system
\begin{gather*}
v=\varphi\\
- k A_1 w - \int_{-\infty}^0g(s) A_1  z(t,s) ds - B  u
=\psi  v-  z_s=\eta\\
- A_2 u + B^{\ast}v=\xi
\end{gather*}
which is equivalent to the system
\begin{gather*}
v=\varphi\\
- k A_1 w - \int_{-\infty}^0g(s) A_1  z(t,s) ds - B  u=\psi\\
z_s=\varphi-\eta\\
 - A_2 u + B^{\ast}v=\xi
\end{gather*}
which is equivalent to the system
 \begin{gather*}
- k A_1 w - \int_{-\infty}^0g(s) A_1  z(t,s) ds - B  u=\psi\\
 v=\varphi\\
 z=s\varphi -\int_0^s  \eta(\tau)d\tau\\
  u =A_2^{-1} B^{\ast}\varphi -A_2^{-1}\xi.
\end{gather*}

By assumption, $\int_{-\infty}^0g(s)s^2  ds<\infty$, and since
$\varphi\in \mathcal{D}(A_1^{1/2})$, we have
$s\varphi\in L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))$.
Using H\"older theorem,
\[
\forall s\in(-\infty,0),\quad \Big(\int_s^0
\|A_1{1/2}\eta(\tau)\|d\tau\Big)^2
\leq  -s\int_s^0
\|A_1{1/2}\eta(\tau)\|^2d\tau.
\]
We can assume that $\eta$ has compact support in $(0,\infty)$,
and then
\[
-\int_{-\infty}^0g(s) s\int_s^0
\|A_1{1/2}\eta(\tau)\|^2d\tau ds<\infty.
\]
Thus  $z\in L^2(g,(-\infty,0),\mathcal{D}(A_1^{1/2}))$.
Note that
$L(w, v,  z, u)=(\varphi,\psi,\eta,\xi)$ is equivalent to the system
\begin{gather*}
\begin{split}
 w &= - k^{-1} A_1^{-1/2}[(\int_{-\infty}^0g(s) s ds)
 A_1^{1/2}\varphi+\int_{-\infty}^0g(s) A_1^{1/2}\int_s^0
 \eta(\tau)d\tau ds ]\\
&\quad -(k A_1)^{-1}  B A_2^{-1} B^{\ast}\varphi
  +(k A_1)^{-1} BA_2^{-1}\xi-(k A_1)^{-1}\psi
\end{split}\\
 v=\varphi\\
 z=s\varphi -\int_0^s  \eta(\tau)d\tau\\
  u =A_2^{-1} B^{\ast}\varphi -A_2^{-1}\xi
\end{gather*}
which is equivalent to the system
 \begin{gather*}
\begin{split}
 w &= - (k )^{-1}(\int_{-\infty}^0g(s) s ds)\varphi
 - (kA_1)^{-1/2}\int_{-\infty}^0g(s)A_1^{1/2} \int_s^0
  \eta(\tau)d\tau ds \\
&\quad -(k A_1)^{-1}  B A_2^{-1} B^{\ast}\varphi
  +(k A_1)^{-1} BA_2^{-1}\xi-(k A_1)^{-1}\psi
\end{split}\\
 v=\varphi\\
 z=s\varphi -\int_0^s  \eta(\tau)d\tau\\
  u =A_2^{-1} B^{\ast}\varphi -A_2^{-1}\xi.
\end{gather*}
Replacing $B u$ by $BA_2^{-1}B^\ast \bar{v}$ and repeating
the above procedure for $L$, we can prove that
the equation
$L_d( \bar{w},\bar{v},\bar{z},\bar{u})=(
\varphi,\psi,\eta,\xi)$ is equivalent to the system
 \begin{gather*}
\bar{v}=\varphi\\
- k A_1\bar{w} - \int_{-\infty}^0g(s) A_1 \bar{z}(t,s) ds
- BA_2^{-1} B^{\ast}\bar{v}=\psi\\
 \bar{z}_s=\varphi-\eta\\
 - A_2 \bar{u} + B^{\ast}\bar{v}=\xi
\end{gather*}
which is equivalent to the system
 \begin{gather*}
\begin{split}
\bar{w} &= (- k)^{-1}A_1^{-1/2}\Big[\Big(\int_{-\infty}^0g(s) s ds\Big)
 A_1^{1/2}\varphi\\
&\quad -\int_{-\infty}^0g(s) A_1^{1/2} \int_0^s  \eta(\tau)d\tau ds
 + BA_2^{-1} B^{\ast}\varphi+\psi\Big]
\end{split}\\
 \bar{v}=\varphi\\
 \bar{z}=s\varphi -\int_0^s  \eta(\tau)d\tau\\
 \bar{u} =A_2^{-1} B^{\ast}\varphi -A_2^{-1}\xi
\end{gather*}
which is equivalent to the system
\begin{gather*}
\begin{split}
\bar{w} &= - k ^{-1}(\int_{-\infty}^0g(s) s ds)\varphi
 + (kA_1)^{-1/2}\int_{-\infty}^0g(s)A_1^{1/2}
 \int_0^s  \eta(\tau)d\tau ds\\
&\quad  -(k A_1)^{-1} BA_2^{-1} B^{\ast}\varphi-(k A_1)^{-1}\psi
\end{split}\\
 \bar{v}=\varphi\\
 \bar{z}=s\varphi -\int_0^s  \eta(\tau)d\tau\\
 \bar{u} =A_2^{-1} B^{\ast}\varphi -A_2^{-1}\xi.
\end{gather*}
 Therefore, by an easy computation one obtains
$$
L^{-1}-L_d^{-1}=\begin{pmatrix}0&0&(kA_1)^{-1}BA_2^{-1}
&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}
$$
which is a compact operator by assumption. The claim follows from
the equality \eqref{resolv1}.
\end{proof}

Now by the Theorem \ref{thm4.1},  Theorem \ref{pr} and
\cite[Theorem 3.2]{Huang},
 we obtain the main result of this section.

\begin{theorem} \label{thm3}
Assume that $A_1^{-1/2}BA_2^{-1}$ is compact in
$\mathcal{L}(H_2,H_1)$, \eqref{c1}-\eqref{c3} and \eqref{Hy}    hold.
Then $T(t)-T_d(t)$ is compact on $\mathbb{H}$ for all $t\geq0$.
\end{theorem}

As a consequence of this theorem, we have the following result.

\begin{corollary} \label{coro4.3}
 $r_{\rm ess}(T(t))= r_{\rm ess}(T_d(t))$ for $t\geq 0$ and
$\omega_{0}(\mathcal{T})=\max\{\omega_{\rm ess}(\mathcal{T}_d),s(L)\}$.
\end{corollary}

\begin{remark} \label{rmk4.4} \rm
The  result of Theorem \ref{thm3} has been shown in \cite{Liu}
directly, assuming the compactness of the operator $BA_2^{-\gamma}$
for some $0<\gamma<1$. It is clear that this last assumption implies
that $A_1^{-1/2}BA_2^{-1}$ is compact from $H_2$ to $H_1$.
\end{remark}

\section{Application}

 We consider the following  model for a linear viscoelastic body
$\Omega$ of Boltzmann type with thermal damping
 \begin{gather}
\begin{gathered}
  \ddot w -\mu \Delta w -(\lambda + \mu)
 \nabla \operatorname{div} w + \mu g_1 \ast \Delta w +
 (\lambda + \mu)g_1\ast \nabla \operatorname{div} w
+ m\nabla  u=0 \\
 \text{in }\Omega \times (0,\infty),
\end{gathered} \label{ex1}
\\
\label{ex2} \dot u+\beta u - \Delta  u - m \operatorname{div} \dot w=0
\quad \text{in }\Omega \times (0,\infty),
\\
\label{ex3} w = 0, \frac{\partial  u}{\partial n} = 0\quad
\text{on } \Gamma \times (0,\infty),
\\
\label{ex4}w(x,0)=w^0(x), \quad \dot w(x,0)=w^1(x), \quad
 u(x,0)=  u^0(x) \quad \text{in } \Omega ,
\\
\label{ex5}  w(x,0) + w(x,s)=f_0(x,s)\quad\text{in }
\Omega \times  (-\infty,0)),
\end{gather}
where  $\lambda,\mu>0$ the Lame's constants and $m>0$ is the
thermal strain parameter, $\beta$ is a positive constant and $g$
is a given function which satisfies the following  conditions
\begin{itemize}
\item[(C1)] $g_1 \in \mathcal{C}^1[0,\infty)\cap L^1(0,\infty)$.
\item[(C2)] $g_1(t)\geq 0 $  and $g_1'(t)\leq0$  for $t>0$,
\item[(C3)] $\int_0^{\infty}g_1(s) ds <1$.
\end{itemize}
The set $\Omega$ is the
bounded open Jelly Roll  set defined in \cite{B.Sim},
$$
\Omega=\{(x,y)\in \mathbb{R}^2: \frac12<r<1\} \setminus \Gamma,
$$
where $\Gamma$ is the curve in $\mathcal{R}^2$  given in polar coordinates by
$$
r(\phi)=\dfrac{\frac{3\pi}2+\arctan (\phi)}{2\pi}, \quad
-\infty<\phi<\infty.
$$
Note that
\begin{align*}
&\int_{-\infty}^tg_1(t-s)\Delta w(x,s)ds \\
&= \int_{-\infty}^0g_1(-s)\Delta w(x,t+s)ds\\
&= \int_{-\infty}^0g_1(-s)\Delta( w(x,t+s)-w(x,t))ds
 +\int_{-\infty}^0g_1(-s)\Delta w(x,t)ds\\
&= -\int_{-\infty}^0g_1(-s)\Delta z(x,t,s)ds
 + \int_{-\infty}^0g_1(-s)ds \Delta w(x,t).
\end{align*}
A similar expression can be establish  for $g_1*\nabla div\,w$.
In order to fit the system \eqref{ex1}-\eqref{ex5} into the setting
abstract system
\eqref{visc1}-\eqref{visc3}, we take
 \begin{gather*}
H_1=L_2(\Omega,\mathbb{R}^2),\quad
H_2=L_2(\Omega,\mathbb{R}),\\
 z(x,t,s)=w(x,t)-w(x,t+s), \quad
 g(s)= g_1(-s),
\end{gather*}
and define  the operators $A_1$, $A_2$, $B$ by
\begin{gather*}
A_1w= -\mu \Delta_{D}w -(\lambda+\mu)   \nabla\,div w, \quad
\mathcal{D}(A_1)=\mathcal{D}(\Delta_{D})=H_0^1(\Omega,\mathbb{R}^2),
\\
A_2u= (\beta I-\Delta_{N})u , \quad
\mathcal{D}(A_2)=\mathcal{D}(\Delta_{N})= H^1(\Omega,\mathbb{R}),
\\
Bu=m\nabla u, \quad \mathcal{D}(B)=H^1(\Omega,\mathbb{R}).
\end{gather*}
Here, as the domain $\Omega$ is irregular the Dirichlet Laplacian
$\Delta_{D}$ and the Neumann  Laplacian $\Delta_{N}$ are defined via
quadratic forms. More precisely,   $\Delta_{D}$  is the unique
positive self adjoint operator associated to the closed quadratic
form on  $H^1_0(\Omega)$
$$
\langle \Delta f,g\rangle=\int_\Omega \nabla f\nabla g\,dx,
$$
and $\Delta_{N}$  is the unique non negative  self adjoint operator
associated to the closed quadratic form on  $H^1(\Omega)$
$$
\langle \Delta f,g\rangle=\int_\Omega \nabla f\nabla g\,dx.
$$
It is clear that the adjoint $B^\ast$ of $B$ is
$$
B^\ast w=-m\,div \,w, \quad
\mathcal{D}(B^\ast)=\{w\in H^1(\Omega,\mathbb{R}^2):
w\cdot\overrightarrow{n} =0 \text{ in }\partial \Omega \},
$$
where $\overrightarrow{n}$ is the outward unit normal vector on the
boundary $\partial \Omega$.  Note that
\[
 \mathcal{D}(A_2^{1/2})=\mathcal{D}(B)\quad \text{and}\quad
\mathcal{D}(A_1^{1/2})\hookrightarrow
\mathcal{D}(B^\ast).
\]

We have the following facts.
\begin{itemize}
\item[(i)]  $A_2^{-1}$ is not compact on $H_2$, see  \cite{B.Sim},
but $A_1$ has a compact resolvent on $H_1$. Consequently,
$A_1^{-1/2}$ and $A_1^{-1/2}BA_2^{-1}$ are compact .

\item[(ii)]  For every $\gamma \in (0,1]$, $BA_2^{-\gamma}$ is not
compact from $H_2$ into $H_1$. In fact,  it is enough to show that
$BA_2^{-1}$ is not compact from $H_2$ to $H_1$. For this, we have
\[
A_2^{-1}B^\ast BA_2^{-1}
=m^2 A_2^{-1}(-\Delta_{N})A_2^{-1}\\
=m^2 A_2^{-1}(A_2-\beta I)A_2^{-1}\\
=m^2 (A_2^{-1}- \beta A_2^{-2}).
\]
\end{itemize}
Using the spectral mapping theorem, we have $ \sigma(A_2^{-1}B^\ast
BA_2^{-1})=m^2(\sigma(A_2)^{-1}- \beta \sigma(A_2)^{-2})$.
As in \cite{B.Sim}, $(\beta, \infty)\subset \sigma(A_2)$,  so
$A_2^{-1}B^\ast BA_2^{-1}$ is not compact on $H_2$. Consequently
$BA_2^{-1}$ is not compact from $H_2$ to $H_1$.


\begin{thebibliography}{00}

\bibitem{bbm} E. M. Ait Ben Hassi, H. Bouslous and L. Maniar;
\emph{Compact decoupling for thermoelasticity in irregular
domains}, Asymptot. Anal.  \textbf{58}  (2008),  47--56.

\bibitem{En.Nag} K. J. Engel and R. Nagel;
\emph{One-Parameter Semigroups for
Linear Evolution Equations}, Graduate Texts in Mathematics \textbf{194},
Springer-Verlag, 2000.

\bibitem{Henry} D. B. Henry, A. Perissinitto and O. Lopes;
\emph{On the essential spectrum of a semigroup
of thermoelasticity}, Nonlinear Anal., TMA \text{21} (1993), 65--75.

\bibitem{Huang}  M. Li, X.  Gu and F.  Huang;
\emph{Unbounded Perturbations of Semigroups: Compactness and Norm
Continuity}, Semigroup Forum  \textbf{65} (2002),  58--70.

\bibitem{Liu} W. J. Liu;
\emph{Compactness of the difference between the thermoviscoelastic
semigroup and its decoupled semigroup}, Rocky Mountain J. Math. \textbf{
30} (2000),  1039--1056.

\bibitem{Lun} A. Lunardi;
\emph{Analytic Semigroups and Optimal Regularity in Parabolic Problems},
 Birkh\"auser, Basel-Boston-Berlin, 1995.

\bibitem{Lopes} A. F. Neves,  H. S. Ribeiro and O.  Lopes;
\emph{On the spectrum of evolution operators generated by hyperbolic
systems},   J. Funct. Anal.  \textbf{67}  (1986), 320--344.

\bibitem{Lopes1} Lopes, Orlando;
\emph{On the structure of the spectrum of a linear time
periodic wave equation},  J. Analyse Math. \textbf{47} (1986), 55--68.

\bibitem{B.Sim} B. Simon;
\emph{The Neumann Laplacian of a Jelly Roll},
Proc. Amer. Math. Soc. \textbf{114}  (1992), 783--785.

\end{thebibliography}

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