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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 147, pp. 1--25.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/147\hfil Quasistatic evolution of damage]
{Quasistatic evolution of damage in an elastic-viscoplastic material}
\author[K. L. Kuttler\hfil EJDE-2005/147\hfilneg]
{Kenneth L. Kuttler}

\address{Kenneth Kuttler\hfill\break
Department of Mathematics,
Brigham Young University,
Provo, UT 84602, USA}
\email{klkuttle@math.byu.edu}

\date{}
\thanks{Submitted September 16, 2005. Published December 12, 2005.}
\subjclass[2000]{74D10, 74R99, 74C10, 35K50, 35K65, 35Q72, 35B05}
\keywords{Existence and uniqueness; damage; comparison theorems;
\hfill\break\indent  elastic viscoplastic materials}

\begin{abstract}
 The mathematical theory of quasistatic elastic viscoplastic models
 with damage is studied. The existence of the unique local weak
 solution is established by using approximate problems and a priori
 estimates. Pointwise estimates on the damage are obtained using a
 new comparison technique which removes the necessity of including
 a subgradient term in the equation for damage.
\end{abstract}

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

\section{Introduction}\label{intro}

This work deals with quasistatic evolution of the macroscopic
mechanical state of an elastic viscoplastic body and the development of
microscopic material damage which results from internal compression or
tension. The damage of the material is caused by the opening and growth of
micro-cracks and micro-cavities which lead to the decrease in the load
carrying capacity of the body and, eventually, to the possible failure of
the system in which the body is situated. The model for the stress used here
is given as a solution to an initial value problem.
\begin{equation*}
\mathbf{\sigma }'=\mathcal{A}(\zeta \mathbf{\varepsilon }
(\mathbf{u}))'+G(\mathbf{\sigma
,\varepsilon }(\mathbf{u}),\zeta ),\quad \mathbf{\sigma }
(0)=\mathbf{\sigma }_0.
\end{equation*}
Without the damage parameter $\zeta $, this is the model of elastic
viscoplastic material. For a discussion of the mathematical theory of these
models, see \cite{ion93}and also \cite{HSbook}. In this formula for the
stress the damage parameter has values between 0 and 1. The above formula
for the stress differs from \cite{CFHS} by allowing the damage to affect the
elastic part of the stress and not just the viscoplastic part.

The novel idea of modelling material damage by the introduction of
the damage field originated in the works of Fr\'{e}mond
\cite{FN95,FN96,FR-book} and was motivated by the evolution of
damage in concrete structures. These ideas have been extended
recently in \cite{Ang1,AKRS02,BF,bon04,CFKSV,CFHS,FKNS98,FKS99,HSS,ksq05}
and in the references therein. Additional results and references can be
found in the two new monographs \cite{SST-book,SHS-book}. In this
approach the damage
field $\zeta $ varies between one and zero at each point in the body. When
$\zeta =1$ the material is damage-free, when $\zeta =0$ the material is
completely damaged, and for $0<\zeta <1$ it is partially damaged.

The evolution of the damage field is usually described by a parabolic
inclusion with a damage source function which depends on the mechanical
compression or tension. The reason it is an inclusion and not an equation is
that a subgradient is included in the model to force the damage parameter to
remain within the desired interval. It is interesting to find conditions on
the damage source function which remove the necessity for using this
subgradient term in the model.

I will show in this paper that the subgradient is not necessary when
physically reasonable conditions are made on the damage source function
which are sufficient to show the damage parameter remains in the desired
interval. This makes possible considerable improvements in the regularity of
the solutions although this aspect of the the elastic viscoplastic problem
will be postponed for another paper. The goal in this paper is to consider
the weak solutions under minimal regularity and compatibility conditions for
the data. The argument which allows pointwise estimates on the damage is
most impressive in the context of very weak solutions. It is based on a
parabolic comparison principle which is easy to prove for classical
solutions but is not obvious for the weak solutions discussed here.

The main result is Theorem \ref{23mayt2} which is an existence and
uniqueness theorem. It is seen that the equation for damage is solved in the
classical sense because all the derivatives in the partial differential
equation exist but the balance of momentum equation is only solved weakly.
In later papers, more regularity will be obtained. Also, other types of
mechanical situations will be considered such as problems with contact,
wear, friction and adhesion.

\section{The model}\label{model}

The body which occupies a domain $\Omega \in \mathbb{R}^{d}$ ($d=1,2,3$)
with outer surface $\partial \Omega =\Gamma $ assumed to be sufficiently
smooth, at least $C^{2,1}$ which means the second derivatives of the
parameterizations defining $\partial \Omega $ are Lipshitz continuous.
Volume forces of density $\mathbf{f}_B$ act in
$\Omega _T=\Omega \times (0,T)$, for $T>0$.

Denote by $\mathbf{u}$ the displacement field, $\mathbf{\sigma }$ the stress
tensor, and $\mathbf{\varepsilon }(\mathbf{u})$ the small or linearized
strain tensor. Let $\zeta $ denote the \textit{damage field}, which is
defined in $\Omega _T$ and measures the fractional decrease in the
strength of the material, to be described shortly. Integrating the
differential equation, for the stress, $\mathbf{\sigma }$ is the solution of
the integral equation
\begin{equation}
\mathbf{\sigma }(t)=\zeta (t)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}(t))-\zeta _0\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)-\mathbf{\sigma }
_0+\int_0^{t}G(\mathbf{\sigma ,\varepsilon }(\mathbf{u}
),\zeta )ds  \label{constit}
\end{equation}
where $\mathbf{u}_0$ is an initial displacement and $\mathbf{\sigma }_0$
is an initial stress.

Assume $\mathcal{A}=\{ \mathcal{A}_{ijkl}(\mathbf{x})\} $
 satisfies the usual symmetries
\begin{equation*}
\mathcal{A}_{ijkl}(\mathbf{x})=\mathcal{A}_{klij}(\mathbf{
x}),\mathcal{A}_{ijkl}(\mathbf{x})=\mathcal{A}
_{jikl}(\mathbf{x}).
\end{equation*}
Also it is always assumed
\begin{equation*}
\mathcal{A}(\mathbf{x})\mathbf{\tau \mathbf{\cdot }\tau }\geq
m_{\mathcal{A}}|\mathbf{\tau }|_{\mathbb{S}_d}^{2},\;\text{
for all }\mathbf{\ \tau }\in \mathbb{S}_d.
\end{equation*}
Here and in the rest of the paper, $\mathbf{x}$ will denote a material point.

As a result of the tensile or compressive stresses in the body, micro-cracks
and micro-cavities open and grow and this, in turn, causes the load bearing
capacity of the material to decrease. This reduction in the strength of an
isotropic material is modelled by introducing the damage field $\zeta =\zeta
(\mathbf{x},t)$ as the ratio
\begin{equation*}
\zeta =\zeta (\mathbf{x},t)=\frac{E_{eff}}{E}
\end{equation*}
between the effective modulus of elasticity $E_{eff}$ and that of the
damage-free material $E$. It follows from this definition that the damage
field should only \ have values between 0 and 1.

Following the derivation in Fr\'{e}mond and Nedjar \cite{FN95,FN96} (see
\cite{FR-book} for full details, and also \cite{SST-book}), the evolution of
the microscopic cracks and cavities responsible for the damage is described
by the differential inclusion
\begin{equation}
\zeta '-\kappa \,\Delta \zeta \in \phi (\mathbf{\varepsilon }(
\mathbf{u}),\zeta )-\partial I_{[0,1]}(\zeta ).  \label{dam}
\end{equation}
However, in this paper, I will show that the subgradient term is not
necessary provided physically reasonable assumptions are made on the source
term, $\phi (\mathbf{\varepsilon }(\mathbf{u}),\zeta )$. This assumption is
essentially that whenever $\zeta \geq 1,\phi (\mathbf{\varepsilon }
(\mathbf{u}),\zeta )\leq 0$. This makes perfect sense
because there should be no way the source term for damage to produce damage
greater than 1. Thus in this paper the damage is governed by the equation
\begin{equation*}
\zeta '-\kappa \,\Delta \zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\zeta )
\end{equation*}
rather than the inclusion (\ref{dam}). Here, the prime denotes the time
derivative, $\Delta $ is the Laplace operator, $\kappa >0$ is the damage
diffusion constant, $\phi $ is the damage source function. There have been
many different formulas proposed for $\phi $ but in this paper I will only
assume the following Lipshitz continuity of $\phi $.
\begin{equation}
|\phi (\mathbf{\varepsilon }_1,\zeta _1)-\phi (
\mathbf{\varepsilon }_2,\zeta _2)|\leq K(|
\mathbf{\varepsilon }_1-\mathbf{\varepsilon }_2|+|\zeta
_1-\zeta _2|)\label{16auge1}
\end{equation}
This may seem restrictive but one can give good physical reasons for making
this assumption \cite{kut05}. In addition, it is shown in this reference
that in the elastic case the above assumption can be completely eliminated
in the presence of suitable compatibility conditions on the initial data and
other assumptions which allow the use of elliptic regularity theorems.
Probably similar considerations will eventually apply to this elastic
viscoplastic problem but at present this is not known.

The classical form of the problem is:
Find a displacement field $\mathbf{u}:\Omega _T\to \mathbb{R}^{d}$,
a stress field $\mathbf{\sigma }:\Omega _T\to \mathbb{S}_d$, and
a damage field $\zeta :\Omega _T\to \mathbb{R}$, such
that
\begin{gather*}
-\mathop{\rm div}\mathbf{\sigma }=\mathbf{f}_B\quad \text{in }\Omega _T,
\\
\mathbf{\sigma }(t)=\zeta (t)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}(t))-\zeta _0\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)-\mathbf{\sigma }
_0+\int_0^{t}G(\mathbf{\sigma ,\varepsilon }(\mathbf{u}
),\zeta )ds\quad \text{in }\Omega _T,
\\
{\zeta }'-\kappa \Delta \zeta =\phi (\mathbf{\varepsilon }
(\mathbf{u}),\zeta )\quad \text{in }\Omega _T,
\\
\partial \zeta /\partial n=0\text{on }\partial \Omega
\times (0,T),
\\
\mathbf{u}=\mathbf{0}\text{ on }\Gamma _{D}\times (0,T),\;
\mathbf{\sigma n}=\mathbf{f}_N\text{ on }\Gamma _N
\times (0,T),
\\
\zeta (0)=\zeta _0
\end{gather*}
First consider a truncated problem which depends on the
truncation operator
$\eta _{\ast }$ which is a nondecreasing $C^{3}$ function satisfying
\begin{equation}
\eta _{\ast }(\zeta )\equiv
\begin{cases}
b & \text{if }  \zeta >1+\epsilon , \\
\zeta & \text{if }  \zeta _{\ast }\leq \zeta \leq 1, \\
a & \text{if }  \zeta <\zeta _{\ast }/2.
\end{cases} \label{2.4}
\end{equation}
where $b\geq 1$, $1>\zeta _{\ast }>0$, and $0<a$. Note that as
long as $\zeta \in [ \zeta _{\ast },1] $ it makes no difference whether one
writes $\zeta $ or $\eta _{\ast }(\zeta )$. The purpose for
using $\eta _{\ast }$ is to allow the study of global solutions. Then,
starting with initial condition $\zeta _0$ such that
$\zeta _{\ast }<\zeta _0\leq 1$ I will establish pointwise estimates
which show that $\zeta $
remains in an interval on which $\eta _{\ast }(\zeta )=\zeta $.
Replacing $\zeta $ with $\eta _{\ast }(\zeta )$, yields the
classical form of the truncated problem.

\textbf{Problem P.}  Find a displacement field
$\mathbf{u}:\Omega_T\to \mathbb{R}^{d}$, a stress field
$\mathbf{\sigma }:\Omega_T\to \mathbb{S}_d$, and a damage
field $\zeta :\Omega_T\to \mathbb{R}$, such that
\begin{gather}
-\mathop{\rm div}\mathbf{\sigma }=\mathbf{f}_B\text{ in }\Omega _T,
\label{k1}
\\
\mathbf{\sigma }(t)=\eta _{\ast }(\zeta )(
t)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}(t)
)-\eta _{\ast }(\zeta _0)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}_0)-\mathbf{\sigma }
_0+\int_0^{t}G(\mathbf{\sigma ,\varepsilon }(\mathbf{u}
),\eta _{\ast }(\zeta ))ds\text{ in }\Omega _T,
\label{k2}
\\
{\zeta }'-\kappa \Delta \zeta =\phi (\mathbf{\varepsilon }
(\mathbf{u}),\eta _{\ast }(\zeta ))\text{
in }\Omega _T,\;\zeta (0)=\zeta _0,  \label{k3}
\\
\partial \zeta /\partial n=0\text{ on }\partial \Omega \times (
0,T),  \label{k4}
\\
\mathbf{u}=\mathbf{0}\text{ on }\Gamma _{D}\times (0,T),\;
\mathbf{\sigma n}=\mathbf{f}_N\text{ on }\Gamma _N\times (
0,T).  \label{k5}
\end{gather}
Here, $\phi $ satisfies (\ref{16auge1}). I will also assume a Lipschitz
condition on the function $G$ which might depend on $\mathbf{x}$ although
this dependence is not shown explicitly.
\begin{gather}
|G(\mathbf{\sigma }_1\mathbf{,\varepsilon }_1,\zeta
_1)-G(\mathbf{\sigma }_2\mathbf{,\varepsilon }_2,\zeta
_2)|\leq K(|\sigma _1-\sigma _2|_{
\mathbb{S}_d}+|\mathbf{\varepsilon }_1-\mathbf{\varepsilon }
_2|_{\mathbb{S}_d}+|\zeta _1-\zeta _2|)
\label{3septe4}
\\
G(\mathbf{0,0},0)\in L^{2}(\Omega ;\mathbb{S}_d)
\equiv Q  \label{3septe5}
\end{gather}
where $\mathbb{S}_d$ is the space of symmetric $d\times d$ matrices with
the usual notion of inner product. {Also }$\Gamma _{D}$ and $\Gamma _N$
are disjoint subsets of $\partial \Omega $ whose union equals $\partial
\Omega $ and $\Gamma _{D}$ has positive surface measure.

\section{Abstract formulation, existence and uniqueness}
\label{weaksolutions}

I will make use of the following two theorems found in Lions \cite{lio69}
and Simon \cite{sim87}, respectively.

\begin{theorem}\label{t6.1}
Let $p\geq 1,\;q>1,\;X_1\subseteq X_2\subseteq X_3$ with
compact inclusion map $X_1\to X_2$ and continuous inclusion map
$X_2\to X_3$, and let
\begin{equation*}
S_{R}=\{\mathbf{u}\in L^{p}(0,T;X_1):\ \mathbf{u}'\in
L^{q}(0,T;X_3),\;\|\mathbf{u}\|_{L^{p}(0,T;X_1)}+\|\mathbf{u}'\|_{L^{q}(0,T;X_3)}<R\}.
\end{equation*}
Then $S_{R}$ is precompact in $L^{p}(0,T;X_2)$.
\end{theorem}

\begin{theorem}\label{t6.2}
Let $X_1,X_2$ and $X_3$ be as above and let
\begin{equation*}
S_{RT}=\{\mathbf{u}:\|\mathbf{u}(t)\|_{X_1}+\|\mathbf{u}'\|_{L^{q}(0,T;X_3)}
\leq R,\quad t\in [0,T]\},
\end{equation*}
for some $q>1$. Then $S_{RT}$ is precompact in $C(0,T;X_2)$.
\end{theorem}

Let $H\equiv (L^{2}(\Omega ))^{d}$, $H_1\equiv (H^{1}(\Omega))^{d}$,
and
\begin{equation*}
V\equiv \{\mathbf{v}\in H_1:\mathbf{v}=\mathbf{0}\quad \mbox{ on }\;\Gamma
_{D}\}.
\end{equation*}
It follows from Korn's inequality that an equivalent norm on $V$ is
\begin{equation*}
\|\mathbf{u}\|_{V}\equiv |\mathbf{
\varepsilon }(\mathbf{u})|_{Q}
\end{equation*}
and I will use this as the norm on $V$. Also,
$E\equiv H^{1}(\Omega)$.

Denote by $\mathcal{V},\mathcal{E},\mathcal{H}$, and $\mathcal{Y}$ the
spaces
\begin{equation*}
L^{2}(0,T;V),\quad L^{2}(0,T;E),\quad L^{2}(0,T;H),
\quad L^{2}(0,T;L^{2}(\Omega )),
\end{equation*}
respectively. Since $V$ is dense in $H$ one can identify $H$ with its dual
$H'$ and write
\begin{equation*}
V\subseteq H=H'\subseteq V'.
\end{equation*}
Also let $Y\equiv L^{2}(\Omega )$ and in a similar way
\begin{equation*}
E\subseteq Y=Y'\subseteq E'.
\end{equation*}

I will use the standard notation for the dual spaces and duality pairings.
Recall that $\mathbf{\sigma }$ satisfies the  identity
\[
\mathbf{\sigma }(t)=\eta _{\ast }(\zeta )(
t)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}(t)
)-\eta _{\ast }(\zeta _0)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}_0)
-\mathbf{\sigma }_0+\int_0^{t}G(\mathbf{\sigma ,\varepsilon }
(\mathbf{u}),\eta _{\ast }(\zeta ))ds\,.
\]
For fixed $\zeta \in \mathcal{Y}$ and $\mathbf{\tau }\in L^{2}(
0,T;Q)$, define $\Psi _{\zeta \mathbf{\tau }}:L^{2}(
0,T;Q)\to L^{2}(0,T;Q)$ by
\[
\Psi _{\zeta \mathbf{\tau }}(\sigma )(t)
\equiv \eta _{\ast }(\zeta )(t)\mathcal{A}\mathbf{\tau }
-\eta _{\ast }(\zeta _0)\mathcal{A}\mathbf{\varepsilon }
(\mathbf{u}_0)\\
-\mathbf{\sigma }_0+\int_0^{t}G(\mathbf{\sigma ,\tau },\eta
_{\ast }(\zeta ))ds
\]

\begin{lemma}\label{19augl1}
The operator $\Psi _{\zeta \mathbf{\tau }}$ has a unique fixed point in
$L^{2}(0,T;Q)$.
\end{lemma}

\begin{proof}
Let $\sigma _{i},i=1,2$ be two elements of $L^{2}(0,T;Q)$.
Then since $G$ is Lipschitz continuous, (\ref{3septe4}) holds,
and
\begin{align*}
|\Psi _{\zeta \mathbf{\tau }}(\sigma _1)(t)
-\Psi _{\zeta \mathbf{\tau }}(\sigma _2)(t)|_{Q}
&=
\big|\int_0^{t}(G(\mathbf{\sigma }_1\mathbf{,\tau },\eta
_{\ast }(\zeta ))-G(\mathbf{\sigma }_2\mathbf{
,\tau },\eta _{\ast }(\zeta )))ds\big|\\
&\leq K\int_0^{t}|\sigma _1(s)-\sigma _2(s)|_{Q}ds.
\end{align*}
Therefore, letting $\lambda >0$,
\[
\int_0^{T}e^{-\lambda t}|\Psi _{\zeta \mathbf{\tau }}(\sigma
_1)(t)-\Psi _{\zeta \mathbf{\tau }}(\sigma
_2)(t)|_{Q}^{2}dt
\leq \int_0^{T}e^{-\lambda t}(K\int_0^{t}|\sigma
_1(s)-\sigma _2(s)|_{Q}ds)^{2}dt.
\]
Using Jensen's inequality,
\begin{align*}
\int_0^{T}e^{-\lambda t}|\Psi _{\zeta \mathbf{\tau }}(\sigma
_1)(t)-\Psi _{\zeta \mathbf{\tau }}(\sigma
_2)(t)|_{Q}^{2}dt
&\leq \int_0^{T}K^{2}te^{-\lambda t}\int_0^{t}|\sigma _1(
s)-\sigma _2(s)|^{2}dsdt \\
&=K^{2}\int_0^{T}\int_{s}^{T}te^{-\lambda t}dt|\sigma _1(
s)-\sigma _2(s)|^{2}ds \\
&\leq K^{2}\frac{1+T\lambda }{\lambda ^{2}}\int_0^{T}e^{-\lambda s}|
\sigma _1(s)-\sigma _2(s)|^{2}ds.
\end{align*}
Note that
\[
\|f\|_{\lambda }^{2}\equiv \int_0^{T}e^{-\lambda
s}|f|_{Q}^{2}ds
\]
is an equivalent norm on $L^{2}(0,T;Q)$ is $\|\cdot\|_{\lambda }$
and that the above inequality shows that for $\lambda $ large enough,
$\Psi_{\zeta \mathbf{\tau }}$ is a contraction mapping on
$L^{2}(0,T;Q)$ with respect to $\|\cdot \|_{\lambda }$.
Therefore, it has a unique fixed point in $L^{2}(0,T;Q)$.
This proves the lemma.
\end{proof}

Denote this fixed point by
\begin{equation}
S(\zeta ,\mathbf{\tau })=\sigma .  \label{19auge4}
\end{equation}
Thus
\begin{equation}  \label{7septe1}
\begin{aligned}
S(\zeta ,\mathbf{\tau })(t)=&\eta _{\ast }(
\zeta )(t)\mathcal{A}\mathbf{\tau }-\eta _{\ast }(
\zeta _0)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}
_0) \\
&-\mathbf{\sigma }_0+\int_0^{t}G(S(\zeta (s),
\mathbf{\tau }(s))\mathbf{,\tau },\eta _{\ast }(
\zeta (s)))ds
\end{aligned}
\end{equation}
The next lemma shows the dependence of $S(\zeta ,\mathbf{\tau })$
on $\mathbf{\tau }$ and $\zeta $.

\begin{lemma} \label{19augl2}
The following inequalities hold for $C$ and $\delta $ independent
of $\mathbf{\tau }_{i}$ in $L^{2}(0,T;Q)$ and  $\zeta $.
\begin{gather}
\|S(\zeta ,\mathbf{\tau }_1)-S(\zeta ,
\mathbf{\tau }_2)\|_{L^{2}(0,t;Q)}\leq
C\|\mathbf{\tau }_1-\mathbf{\tau }_2\|
_{L^{2}(0,t;Q)},  \label{3septe1}
\\
|S(\zeta ,\mathbf{0})|_{Q}\leq C,  \label{3septe2}
\\
(S(\zeta (t),\mathbf{\tau }(t)),
\mathbf{\tau }(t))_{Q}\geq \delta |\mathbf{\tau }
(t)|_{Q}^{2}-C-C\int_0^{t}|\mathbf{\tau }(
s)|_{Q}^{2}ds,  \label{3septe3}
\\
\begin{aligned}
&(S(\zeta (t),\mathbf{\tau }_1(t)
)-S(\zeta (t),\mathbf{\tau }_2(t)
),\mathbf{\tau }_1(t)-\mathbf{\tau }_2(
t))_{Q} \\
&\geq \delta |\mathbf{\tau }_1(t)-\mathbf{\tau }
_2(t)|_{Q}^{2}-C\int_0^{t}|\mathbf{\tau }
_1(s)-\mathbf{\tau }_2(s)|_{Q}^{2},
\end{aligned}\label{3septe7}
\\
\begin{aligned}
|S(\zeta _1(t),\mathbf{\tau }(t)
)-S(\zeta _2(t),\mathbf{\tau }(t))|_{Q}^{2}
&\leq  C\big(\int_{\Omega }|\eta _{\ast }(\zeta _1(t)
)-\eta _{\ast }(\zeta _2(t))|
^{2}|\mathbf{\tau }(t)|_{\mathbb{S}_d}^{2}dx\big)\\
&\quad +C\big(\int_0^{t}|\eta _{\ast }(\zeta _1(s)
)-\eta _{\ast }(\zeta _2(s))|_{Y}^{2}ds\big)\,.
\end{aligned}
\label{3septe9}
\end{gather}
\end{lemma}

\begin{proof}
Let
\begin{gather*}
\sigma _{i}(s)=\eta _{\ast }(\zeta )(
s)\mathcal{A}\mathbf{\tau }_{i}-\eta _{\ast }(\zeta _0)
\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}_0),\\
-\mathbf{\sigma }_0+\int_0^{s}G(\mathbf{\sigma }_{i}\mathbf{,\tau }
_{i},\eta _{\ast }(\zeta ))dr =S(\zeta (
s),\mathbf{\tau }_{i}(s)).
\end{gather*}
Then since $\eta _{\ast }$ is bounded, an inequality of the following form
in which $C$ is independent of $\mathbf{\tau }_{i}$ holds.
\begin{equation*}
|\sigma _1(s)-\sigma _2(s)|
_{Q}\leq C(|\mathbf{\tau }_1(s)-\mathbf{\tau }
_2(s)|+\int_0^{s}(|\sigma _1(r)-\sigma _2(r)|_{Q}+|\mathbf{\tau }
_1(r)-\mathbf{\tau }_2(r)|_{Q})dr)
\end{equation*}
Now Gronwall's inequality implies that after adjusting $C$,
\begin{equation*}
|\sigma _1(s)-\sigma _2(s)|
_{Q}\leq C\big(|\mathbf{\tau }_1(s)-\mathbf{\tau }
_2(s)|+\int_0^{s}|\mathbf{\tau }_1(
r)-\mathbf{\tau }_2(r)|_{Q}dr\big).
\end{equation*}
This implies (\ref{3septe1}).
Next we consider (\ref{3septe2}). From the definition of
$S(\zeta ,\mathbf{\tau })$,
\begin{equation*}
S(\zeta (t),\mathbf{0})=-\eta _{\ast }(\zeta
_0)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}_0)-
\mathbf{\sigma }_0+\int_0^{t}G(S(\zeta ,\mathbf{0}),
\mathbf{0,}\eta _{\ast }(\zeta ))ds
\end{equation*}
Now from (\ref{3septe5}) and the boundedness of $\eta _{\ast }$,
\begin{align*}
&|S(\zeta (t),\mathbf{0})|_{Q}\\
&\leq |(\eta _{\ast }(\zeta _0)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}_0)+\mathbf{\sigma }_0)
|_{Q}+\int_0^{t}|G(S(\zeta ,\mathbf{0}),
\mathbf{0,}\eta _{\ast }(\zeta ))|_{Q}ds \\
&\leq |(\eta _{\ast }(\zeta _0)\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)+\mathbf{\sigma }
_0)|_{Q}+\int_0^{t}K(|S(\zeta ,\mathbf{0}
)|_{Q}+2)ds+\int_0^{t}|G(\mathbf{0,0,}
0)|_{Q}ds
\end{align*}
and so by Gronwall's inequality and (\ref{3septe4}),
\begin{equation*}
|S(\zeta (t),\mathbf{0})|_{Q}\leq
(|(\eta _{\ast }(\zeta _0)\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)+\mathbf{\sigma }
_0)|_{Q}+2+\int_0^{T}|G(\mathbf{0,0,}0)
|_{Q}ds)e^{KT}\equiv C
\end{equation*}
Consider (\ref{3septe3}). From the identity solved by $S(\zeta ,
\mathbf{\tau })$,
\begin{align*}
(S(\zeta (t),\mathbf{\tau }(t)),
\mathbf{\tau }(t))_{Q}
&\geq m_{\mathcal{A}}\zeta _{\ast }|\mathbf{\tau }(t)|
_{Q}^{2}-|(\eta _{\ast }(\zeta _0)\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)+\mathbf{\sigma }
_0)|_{Q}|\mathbf{\tau }(t)|_{Q} \\
&\quad -\int_0^{t}|G(S(\zeta ,\mathbf{\tau }),\mathbf{
\tau },\eta _{\ast }(\zeta ))|_{Q}ds|
\mathbf{\tau }(t)|_{Q}\,.
\end{align*}
So,
\begin{equation*}
(S(\zeta (t),\mathbf{\tau }(t)),
\mathbf{\tau }(t))_{Q}\geq 3\delta |\mathbf{\tau }
(t)|_{Q}^{2}-C-|\mathbf{\tau }(t)
|_{Q}\int_0^{t}|G(S(\zeta ,\mathbf{\tau }),
\mathbf{\tau },\eta _{\ast }(\zeta ))|_{Q}ds\,.
\end{equation*}
Now from (\ref{3septe5}), (\ref{3septe4}), (\ref{3septe1}), and
(\ref{3septe2}) and adjusting constants as needed,
\begin{align*}
&(S(\zeta (t),\mathbf{\tau }(t)),\mathbf{\tau }(t))_{Q}\\
&\geq 2\delta |\mathbf{\tau }(t)|
_{Q}^{2}-C-K|\mathbf{\tau }(t)|
_{Q}\int_0^{t}(|S(\zeta (s),\mathbf{\tau }
(s))|_{Q}+|\mathbf{\tau }(s)
|_{Q}+2)ds \\
&\quad -K|\mathbf{\tau }(t)|_{Q}\int_0^{t}|
G(\mathbf{0,0,}0)|_{Q}ds \\
&\geq \delta |\mathbf{\tau }(t)|
_{Q}^{2}-C-C\int_0^{t}|\mathbf{\tau }(s)|
_{Q}^{2}ds-C\int_0^{t}|S(\zeta (s),\mathbf{\tau }
(s))|_{Q}^{2}ds \\
&\geq \delta |\mathbf{\tau }(t)|
_{Q}^{2}-C-C\int_0^{t}|\mathbf{\tau }(s)|
_{Q}^{2}ds-C\int_0^{t}|\mathbf{\tau }(s)|
_{Q}^{2}ds-\int_0^{t}|S(\zeta (s),0)
|_{Q}^{2}ds \\
&\geq \delta |\mathbf{\tau }(t)|
_{Q}^{2}-C-C\int_0^{t}|\mathbf{\tau }(s)|
_{Q}^{2}ds.
\end{align*}
Next consider (\ref{3septe7}). From the assumptions on $G$ and the
definition of $S$, along with (\ref{3septe1}),
\begin{align*}
&(S(\zeta (t),\mathbf{\tau }_1(t)
)-S(\zeta (t),\mathbf{\tau }_2(t)
),\mathbf{\tau }_1(t)-\mathbf{\tau }_2(
t))_{Q} \\
&\geq m_{\mathcal{A}}\zeta _{\ast }|\mathbf{\tau }_1(t)-
\mathbf{\tau }_2(t)|_{Q}^{2}-|\mathbf{\tau }
_1(t)-\mathbf{\tau }_2(t)|_{Q} \\
&\quad\times \int_0^{t}K\big(|S(\zeta (s),\mathbf{\tau }
_1(s))-S(\zeta (s),\mathbf{\tau }
_2(s))|_{Q}+|\mathbf{\tau }_1(
s)-\mathbf{\tau }_2(s)|_{Q}\big) \\
&\geq \delta |\mathbf{\tau }_1(t)-\mathbf{\tau }
_2(t)|_{Q}^{2}-C\int_0^{t}\big(|\mathbf{\tau
}_1(s)-\mathbf{\tau }_2(s)|_{Q}^{2}\big).
\end{align*}
It only remains to prove (\ref{3septe9}).
\begin{align*}
&S(\zeta _1(t),\mathbf{\tau }(t))
-S(\zeta _2(t),\mathbf{\tau }(t))\\
&=(\eta _{\ast }(\zeta _1(t))-\eta _{\ast
}(\zeta _2(t)))\mathcal{A}\mathbf{\tau }(t)\\
&\quad +\int_0^{t}(G(S(\zeta _1,\mathbf{\tau }),
\mathbf{\tau },\eta _{\ast }(\zeta _1))-G(
S(\zeta _2,\mathbf{\tau }),\mathbf{\tau },\eta _{\ast }(
\zeta _2)))ds.
\end{align*}
Therefore,
\begin{align*}
&|S(\zeta _1(t),\mathbf{\tau }(t)
)-S(\zeta _2(t),\mathbf{\tau }(t))|_{Q}^{2}\\
&\leq C\big(\int_{\Omega }|\eta _{\ast
}(\zeta _1(t))-\eta _{\ast }(\zeta
_2(t))|^{2}|\mathbf{\tau }(t)
|_{\mathbb{S}_d}^{2}dx\big) \\
&\quad +C\big(\int_0^{t}\int_{\Omega }(|S(\zeta _1,\mathbf{
\tau })-S(\zeta _2,\mathbf{\tau })|_{\mathbb{S}
_d}^{2}+|\eta _{\ast }(\zeta _1(s))-\eta
_{\ast }(\zeta _2(s))|^{2})\,dx\,ds\big)
\end{align*}
Now by Gronwall's inequality and adjusting the constants,
\begin{align*}
|S(\zeta _1(t),\mathbf{\tau }(t)
)-S(\zeta _2(t),\mathbf{\tau }(t)
)|_{Q}^{2}
&\leq C\big(\int_{\Omega }|\eta _{\ast }(\zeta _1(
t))-\eta _{\ast }(\zeta _2(t))
|^{2}|\mathbf{\tau }(t)|_{\mathbb{S}_d}^{2}dx\big)\\
&\quad +C\big(\int_0^{t}\int_{\Omega }|\eta _{\ast }(\zeta
_1(s))-\eta _{\ast }(\zeta _2(s)
)|^{2}dx\,ds\big)
\end{align*}
This proves the lemma.
\end{proof}

Before continuing with the abstract formulation, here is a summary of the
assumptions on the functions involved in the model and the data.
\begin{align}
&\mathcal{A}(\mathbf{x})\mathbf{\tau \mathbf{\cdot }\tau }\geq
m_{\mathcal{A}}|\mathbf{\tau }|_{\mathbb{S}_d}^{2},\quad
\text{for all }\mathbf{\tau }\in \mathbb{S}_d.  \label{k11} \\
&\text{The mapping $\mathbf{x\to }\mathcal{A}(\mathbf{x})$
is measurable and bounded.}  \label{k13} \\
&\mathcal{A}(\mathbf{x})\text{ is symmetric.}
\end{align}
Here, $m_{\mathcal{A}}$ is a positive constant.

The \textit{damage source function} $\phi :\Omega \times \mathbb{S}
_d\times \mathbb{R}\to \mathbb{R}$ is Lipschitz and satisfies:
\begin{align}
&|\phi (\mathbf{x,\varepsilon }_1,\eta _{\ast }(\zeta
_1))-\phi (\mathbf{x,\varepsilon }_2,\eta _{\ast
}(\zeta _2))|\leq L_{\phi }(|
\mathbf{\ \varepsilon }_1-\mathbf{\varepsilon }_2|+|\eta
_{\ast }(\zeta _1)-\eta _{\ast }(\zeta _2)
|)\nonumber  \\
&\mbox{ for all }\mathbf{\varepsilon }_1,\mathbf{\varepsilon }
_2\in \mathbb{S}_d,\;\;\zeta _1,\zeta _2\in \mathbb{R},\quad
\mbox{a.e. } \mathbf{x}\in \Omega .  \label{k20}\\
&\mbox{The function  }\;\mathbf{x\to }\phi (\mathbf{\
x,\varepsilon },\,\zeta )\text{ is measurable.}  \label{k21} \\
&\hbox{The mapping  }\mathbf{x}\to \phi (\mathbf{x}
,0,0)\hbox{  belongs to  }L^{2}(\Omega ). \\
&\phi (\mathbf{x,\varepsilon ,}\eta _{\ast }(\zeta )
)\text{ is bounded}
\end{align}
Here, $L_{\phi }>0$ is the Lipschitz constant. I will suppress the
dependence of these functions on $\mathbf{x}$. Also it will eventually be
assumed that for $0<\zeta _{\ast }<1$,
\begin{equation}
\phi (\mathbf{\varepsilon ,}\zeta )\leq 0\quad \text{if}\quad \zeta \geq
1,\;\phi (\mathbf{\varepsilon },\zeta _{\ast })\geq 0
\label{18june1}
\end{equation}
The first of these assumptions states the source term for damage is
nonpositive whenever $\zeta =1$. This makes perfect physical sense because
it says the damage cannot be made to exceed 1. The omission of the second
condition will not be fully explored in this paper. Based on an analogy with
the elastic case, it is likely that if one leaves it out, the result will be
local rather than global solutions to the problem.

As for the initial data and forcing function, the assumptions made in this
paper are listed here.
The body force and surface traction are assumed to satisfy
\begin{equation}
\mathbf{f}_B\in C([0,T];H),\;\mathbf{f}_N\in
C\big([ 0,T] ;L^{2}(\Gamma _N)^{d}\big),
\label{k23}
\end{equation}
and $\mathbf{f}\in \mathcal{V}'$ is defined by
\begin{equation}
\langle \mathbf{f}(t),\mathbf{v}(t)\rangle
_{V',V}=(\mathbf{f}_B(t),\mathbf{v}(
t))_{H}+(\mathbf{f}_N(t),\mathbf{v}(
t))_{L^{2}(\Gamma _N)^{d}}.  \label{k25}
\end{equation}
Thus
\begin{equation}
\mathbf{f}\in C([0,T];V')\label{16auge3}
\end{equation}
The initial conditions satisfy
\begin{equation}
\zeta _0\in E,\quad \zeta _0(\mathbf{x})\in (\zeta _{\ast
},1],\quad 1>\zeta _{\ast }>0  \label{a24maye15}
\end{equation}
However, $\zeta _0\in E$ will be used initially.
Now, $L:E\to E'$ is defined by
\begin{equation}
\langle L\zeta ,\xi \rangle \equiv \int_{\Omega }\nabla \zeta \mathbf{
\mathbf{\cdot }}\nabla \xi \,dx.  \label{k25.5}
\end{equation}
Letting $\mathbf{w}\in \mathcal{V}$ and $\tau \in \mathcal{E}$, multiply (
\ref{k1}) by $\mathbf{w}$ and integrate by parts. Using the boundary
conditions for $\mathbf{u}$ this yields a variational formulation for (\ref
{k1}) which is of the form
\begin{equation}
\int_{\Omega }\sigma _{ij}\mathbf{\varepsilon }(\mathbf{w})
_{ij}dx=\int_{\Omega }\mathbf{f}_B\cdot \mathbf{w}dx+\int_{\Gamma _N}
\mathbf{f}_N\cdot \mathbf{w}d\alpha  \label{19auge1}
\end{equation}
where
$\sigma =S(\zeta ,\mathbf{\varepsilon }(\mathbf{u}))$.

Now multiply (\ref{k3}) by $\tau $ and integrate by parts. With the boundary
condition for $\zeta $, this yields the variational formulation,
\begin{equation}
\zeta '+\kappa L\zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\eta _{\ast }(\zeta )),\quad
\zeta (0)=\zeta _0  \label{19auge2}
\end{equation}
Now define for $\zeta \in \mathcal{Y}$, $A:\mathcal{Y\times V\to V}'$ by
\begin{equation}
\langle A(\zeta ,\mathbf{u}),\mathbf{w}\rangle
\equiv \int_0^{T}\int_{\Omega }S(\zeta ,\mathbf{\varepsilon }(
\mathbf{u}))_{ij}\mathbf{\varepsilon }(\mathbf{w})
_{ij}dx\,dt  \label{19auge3}
\end{equation}
The abstract version of Problem P is to find $\zeta \in \mathcal{E},\zeta
'\in \mathcal{E}'$ and $\mathbf{u}\in \mathcal{V}$ such
that
\begin{gather}
\zeta '+\kappa L\zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\eta _{\ast }(\zeta )),\;\zeta (
0)=\zeta _0,  \label{19auge7}
\\
A(\zeta ,\mathbf{u})=\mathbf{f}\text{ in }\mathcal{V}'.
\label{19auge9}
\end{gather}
I will denote this problem as $P_{V}$. It turns out that $P_{V}$ is too
difficult to study directly so I will consider a simpler problem and then
obtain the solution to $P_{V}$ as a fixed point. Fix $\zeta _1\in \mathcal{
Y}$. Then $P_{V\zeta _1}$ denotes the following problem. Find $\zeta \in
\mathcal{E},\zeta '\in \mathcal{E}'$ and $\mathbf{u}\in \mathcal{V}$
 such that
\begin{gather}
\zeta '+\kappa L\zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\eta _{\ast }(\zeta )),\;\zeta (
0)=\zeta _0\in E,  \label{19auge11}
\\
A(\zeta _1,\mathbf{u})=\mathbf{f}\text{ in }\mathcal{V}
'.  \label{19auge13}
\end{gather}
For $\lambda $ a positive constant, define new dependent variables,
$\zeta_{\lambda }$ and $\mathbf{u}_{\lambda }$ by
\begin{equation*}
\zeta _{\lambda }(t)e^{\lambda t}=\zeta (t),\quad
\mathbf{u}_{\lambda }(t)e^{\lambda t}=\mathbf{u}(t).
\end{equation*}

\begin{lemma} \label{22augl1}
 For $\zeta _1\in \mathcal{Y}$ there exists a unique
solution to Problem $P_{V\zeta _1}$ which satisfies
$\zeta ,\zeta '\in \mathcal{Y}$.
\end{lemma}

\begin{proof} There exists a unique solution, $\mathbf{u}$ to
\ref{19auge13} if and only if there exists a unique solution to
\begin{equation}
e^{-\lambda (\cdot )}A(\zeta _1,\mathbf{u}_{\lambda
}e^{\lambda (\cdot )})=e^{-\lambda (\cdot )}
\mathbf{f}\text{ in }\mathcal{V}'.  \label{2septe3}
\end{equation}
This is equivalent to
\begin{equation}
\int_0^{T}\int_{\Omega }e^{-\lambda t}S(\zeta ,e^{\lambda t}\mathbf{
\varepsilon }(\mathbf{u}_{\lambda }))_{ij}\mathbf{
\varepsilon }(\mathbf{w})_{ij}dx\,dt=\int_0^{T}\langle
e^{-\lambda (\cdot )}\mathbf{f,w}\rangle dt
\label{6dece1}
\end{equation}
for all $\mathbf{w}\in \mathcal{V}$. Now recall the definition of $S$ in
terms of a fixed point of an operator found in (\ref{7septe1}). Using this
definition, (\ref{6dece1}) occurs if and only if
\begin{align*}
&\int_0^{T}\int_{\Omega }Big(\eta _{\ast }(\zeta _1)
(t)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}_{\lambda
})-e^{-\lambda t}\eta _{\ast }(\zeta _0)\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_0)-e^{-\lambda t}\mathbf{
\sigma }_0  \\
&+\int_0^{t}e^{-\lambda t}G(S(\zeta _1,e^{\lambda s}
\mathbf{\varepsilon }(\mathbf{u}_{\lambda }))\mathbf{,}
e^{\lambda s}\mathbf{\varepsilon }(\mathbf{u}_{\lambda }),\eta
_{\ast }(\zeta _1))ds\Big)_{ij}\mathbf{\varepsilon }
(\mathbf{w})_{ij}dx\,ds \\
&=\int_0^{T}\langle e^{-\lambda (\cdot )}\mathbf{f,w}
\rangle dt
\end{align*}
To simplify the notation denote the left side of the above equation by
\begin{equation*}
\int_0^{T}\langle N_{\lambda }(t,\mathbf{u}_{\lambda }),
\mathbf{w}\rangle dt
\end{equation*}
Then $N_{\lambda }:\mathcal{V\to V}'$ given by
\begin{equation*}
\langle N_{\lambda }\mathbf{u,w}\rangle \equiv
\int_0^{T}\langle N_{\lambda }(t,\mathbf{u}),\mathbf{w}
\rangle dt
\end{equation*}
is obviously hemicontinuous and bounded. I will now show that if $\lambda $
is large enough, then $N_{\lambda }$ is also monotone and satisfies an
inequality of the form
\begin{equation*}
\langle N_{\lambda }\mathbf{u}_1-N_{\lambda }\mathbf{u}_2,\mathbf{u}
_1-\mathbf{u}_2\rangle \geq \delta \|\mathbf{u}_1-
\mathbf{u}_2\|_{\mathcal{V}}^{2}
\end{equation*}
where $\delta >0$ and does not depend on the $\mathbf{u}_{i}$. Let $\mathbf{u
}_1,\mathbf{u}_2$ be two elements of $\mathcal{V}$. Then from Lemma
\ref{19augl2},
\begin{align*}
&\langle N_{\lambda }\mathbf{u}_1-N_{\lambda }\mathbf{u}_2,\mathbf{u}
_1-\mathbf{u}_2\rangle\\
&\geq \zeta _{\ast }m_{\mathcal{A}}|
|\mathbf{u}_1-\mathbf{u}_2\|_{\mathcal{V}}^{2}
-K\int_0^{T}e^{-\lambda t}\int_0^{t}(|S\Big(\zeta
_1,e^{\lambda s}\mathbf{\varepsilon }(\mathbf{u}_1)\Big)\\
&-S(\zeta _1,e^{\lambda s}\mathbf{\varepsilon }(\mathbf{u}
_2))|+|e^{\lambda s}(\mathbf{\varepsilon }
(\mathbf{u}_1)-\mathbf{\varepsilon }(\mathbf{u}_2))|)ds
|\mathbf{\varepsilon }(\mathbf{u}_1(t))-
\mathbf{\varepsilon }(\mathbf{u}_2(t))|dt
 \\
&\geq \zeta _{\ast }m_{\mathcal{A}}\|\mathbf{u}_1-\mathbf{u}
_2\|_{\mathcal{V}}^{2}-C\int_0^{T}e^{-\lambda
t}\int_0^{t}e^{\lambda s}\big|\mathbf{\varepsilon }(\mathbf{u}_1(s))\\
&\quad -\mathbf{\varepsilon }(\mathbf{u}_2(s))|ds|\mathbf{\varepsilon }(
\mathbf{u}_1(t))-\mathbf{\varepsilon }(\mathbf{u}_2(t))\big|dt
\end{align*}
Using Holder's inequality and Jensen's inequality, the last term is
dominated by
\begin{align*}
&C\Big(\int_0^{T}|\mathbf{\varepsilon }(\mathbf{u}_1(
t))-\mathbf{\varepsilon }(\mathbf{u}_2(t) )|^{2}dt\Big)^{1/2}\\
&\times \Big(\int_0^{T}e^{-2\lambda t}\Big(
\int_0^{t}e^{\lambda s}|\mathbf{\varepsilon }(\mathbf{u}
_1(s))-\mathbf{\varepsilon }(\mathbf{u}
_2(s))|ds\Big)^{2}dt\Big)^{1/2} \\
&\leq C\Big(\int_0^{T}|\mathbf{\varepsilon }(\mathbf{u}
_1(t))-\mathbf{\varepsilon }(\mathbf{u}
_2(t))|^{2}dt\Big)^{1/2}\\
&\quad\times
\Big(\int_0^{T}e^{-2\lambda t}(\frac{e^{\lambda t}}{\lambda }-
\frac{1}{\lambda })\Big(\int_0^{t}e^{\lambda s}|\mathbf{
\varepsilon }(\mathbf{u}_1(s))-\mathbf{
\varepsilon }(\mathbf{u}_2(s))|^{2}ds\Big)dt\Big)^{1/2} \\
&\leq C\Big(\int_0^{T}|\mathbf{\varepsilon }(\mathbf{u}
_1(t))-\mathbf{\varepsilon }(\mathbf{u}
_2(t))|^{2}dt\Big)^{1/2}\\
&\quad\times  \Big(\int_0^{T}(\frac{e^{-\lambda t}}{\lambda })(
\int_0^{t}e^{\lambda s}|\mathbf{\varepsilon }(\mathbf{u}
_1(s))-\mathbf{\varepsilon }(\mathbf{u}
_2(s))|^{2}ds)dt\Big)^{1/2} \\
&=\frac{C}{\sqrt{\lambda }}\Big(\int_0^{T}|\mathbf{\varepsilon }
(\mathbf{u}_1(t))-\mathbf{\varepsilon }(
\mathbf{u}_2(t))|^{2}dt\Big)^{1/2}\\
&\quad\times \Big(\int_0^{T}(\int_0^{t}e^{-\lambda (t-s)}|\mathbf{
\varepsilon }(\mathbf{u}_1(s))-\mathbf{
\varepsilon }(\mathbf{u}_2(s))|
^{2}ds)dt\Big)^{1/2} \\
&=\frac{C}{\sqrt{\lambda }}\Big(\int_0^{T}|\mathbf{\varepsilon }
(\mathbf{u}_1(t))-\mathbf{\varepsilon }(
\mathbf{u}_2(t))|^{2}dt\Big)^{1/2}\\
&\quad\times\Big(
\int_0^{T}|\mathbf{\varepsilon }(\mathbf{u}_1(s)
)-\mathbf{\varepsilon }(\mathbf{u}_2(s))
|^{2}\int_{s}^{T}e^{-\lambda (t-s)}dt\,ds\Big)^{1/2} \\
&\leq \frac{C}{\sqrt{\lambda }}\Big(\int_0^{T}|\mathbf{\varepsilon }
(\mathbf{u}_1(t))-\mathbf{\varepsilon }(
\mathbf{u}_2(t))|^{2}dt\Big)^{1/2}
  \Big(
\int_0^{T}|\mathbf{\varepsilon }(\mathbf{u}_1(s)
)-\mathbf{\varepsilon }(\mathbf{u}_2(s))
|^{2}\frac{1}{\lambda }ds\Big)^{1/2} \\
&=\frac{C}{\lambda }\Big(\int_0^{T}|\mathbf{\varepsilon }(
\mathbf{u}_1(t))-\mathbf{\varepsilon }(\mathbf{u}
_2(t))|^{2}dt\Big).
\end{align*}
Then letting $\delta =m_{\mathcal{A}}\zeta _{\ast }/2$, it follows that for
$\lambda $ large enough, the desired inequality holds.

It follows that there exists a unique solution, $\mathbf{u}$ to
(\ref{19auge13}). Now using this $\mathbf{u}$ in the equation of
(\ref{19auge11}), one notes that the right side of the equation
 is Lipschitz in $\zeta $
and so it follows by standard results there exists a unique $\zeta $ solving
(\ref{19auge11}) which satisfies $\zeta ,\zeta '\in \mathcal{Y}$.
The way this can be done is to consider this equation with the $\zeta $ in
the right side replaced with $\widehat{\zeta }$, a fixed element of
$\mathcal{Y}$. Then since the operator on the left comes as a subgradient of
a convex lower semicontinuous functional, there exists a solution having the
desired regularity. \cite{bre73} Then one shows a high enough power of the
map taking $\widehat{\zeta }$ to $\zeta $ is a contraction. The unique fixed
point is the desired solution. This proves the lemma.
\end{proof}

Now I will continue the consideration of problem $P_{V}$ which is listed
here again.

\noindent \textbf{Problem $P_{V}$.} Find a displacement field $\mathbf{u}:
[ 0,T] \to V$ and a damage field $\zeta $ such that
\begin{equation}
A(\zeta ,\mathbf{u})=\mathbf{f}\quad \text{ in }\mathcal{V}',  \label{k29}
\end{equation}
where
\begin{gather}
\langle A(\zeta ,\mathbf{u}),\mathbf{w}\rangle
\equiv \int_0^{T}\int_{\Omega }S\big(\zeta ,\mathbf{\varepsilon }(
\mathbf{u})\big)_{ij}\mathbf{\varepsilon }(\mathbf{w})
_{ij}dx\,dt \nonumber
\\
\zeta '+\kappa L\zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\ \eta _{\ast }(\zeta )),\qquad \zeta
(0)=\zeta _0.  \label{k30}
\end{gather}
To simplify notation, let
\begin{equation*}
|\cdot |=|\cdot |_{L^{2}(\Omega )},\quad \Vert \cdot \Vert =\Vert \cdot
\Vert _{E}.
\end{equation*}
For $\zeta \in \mathcal{Y}$, let $\mathbf{u}_{\zeta }\in \mathcal{V}$ denote
the unique solution of the problem
\begin{equation}
A(\zeta ,\mathbf{u}_{\zeta })=\mathbf{f}\quad \text{ in }
\mathcal{V}'.  \label{22jane1}
\end{equation}
The following is a fundamental convergence result.

\begin{lemma}
\label{22janl1}If $\zeta _{n}\to \zeta $ in $\mathcal{Y}$ as
$n\to \infty $, then $\mathbf{u}_{\zeta _{n}}\to \mathbf{u}
_{\zeta }$ in $\mathcal{V}$.
\end{lemma}

\begin{proof}  Recall (\ref{7septe1}), listed here for convenience,
\begin{align*}
S(\zeta ,\mathbf{\varepsilon }(\mathbf{u}))(
t)&=\eta _{\ast }(\zeta )(t)\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u})-\eta _{\ast }(\zeta
_0)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}_0)\\
&\quad-\mathbf{\sigma }_0+\int_0^{t}G(S(\zeta (s),
\mathbf{\varepsilon }(\mathbf{u})(s))
\mathbf{,\varepsilon }(\mathbf{u}),\eta _{\ast }(\zeta
(s)))ds
\end{align*}
Then let
\[
m(\zeta ,\mathbf{\varepsilon }(\mathbf{u}))(
t)=-\eta _{\ast }(\zeta _0)\mathcal{A}\mathbf{
\varepsilon }(\mathbf{u}_0)
-\mathbf{\sigma }_0+\int_0^{t}G(S(\zeta (s),
\mathbf{\varepsilon }(\mathbf{u})(s))
\mathbf{,\varepsilon }(\mathbf{u}),\eta _{\ast }(\zeta
(s)))ds.
\]
For short, let $\mathbf{u}_{\zeta _{n}}=\mathbf{u}_{n}$ and $\mathbf{u}
_{\zeta }=\mathbf{u}$. Then
\begin{align*}
&\big|m(\zeta ,\mathbf{\varepsilon }(\mathbf{u})
)(t)-m(\zeta _{n},\mathbf{\varepsilon }(
\mathbf{u}_{n}))(t)\big|_{Q}^{2} \\
&=\Big|\int_0^{t}\Big(G(S(\zeta _{n}(s),
\mathbf{\varepsilon }(\mathbf{u}_{n})(s))
\mathbf{,\varepsilon }(\mathbf{u}_{n}),\eta _{\ast }(
\zeta _{n}(s)))  \\
&\quad -G(S(\zeta (s),\mathbf{\varepsilon }
(\mathbf{u})(s))\mathbf{,\varepsilon }
(\mathbf{u}),\eta _{\ast }(\zeta (s))
)\Big)ds\Big|_{Q}^{2} \\
&\leq C\int_0^{t}\Big(|S(\zeta _{n}(s),\mathbf{
\varepsilon }(\mathbf{u}_{n})(s))-S(
\zeta (s),\mathbf{\varepsilon }(\mathbf{u})(
s))|_{Q}^{2} \\
&\quad +|\mathbf{\varepsilon }(\mathbf{u}_{n})(
s)-\mathbf{\varepsilon }(\mathbf{u})(s)
|_{Q}^{2}+|\eta _{\ast }(\zeta _{n}(s)
)-\eta _{\ast }(\zeta (s))|
_{Y}^{2}\Big)ds \\
&\leq C\int_0^{t}|\eta _{\ast }(\zeta _{n}(s)
)-\eta _{\ast }(\zeta (s))|
_{Y}^{2}ds+C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}
_{n})(s)-\mathbf{\varepsilon }(\mathbf{u})
(s)|_{Q}^{2}ds \\
&\quad +C\int_0^{t}|S(\zeta _{n}(s),\mathbf{
\varepsilon }(\mathbf{u}_{n})(s))-S(
\zeta _{n}(s),\mathbf{\varepsilon }(\mathbf{u})
(s))|_{Q}^{2}ds \\
&\quad +C\int_0^{t}|S(\zeta _{n}(s),\mathbf{
\varepsilon }(\mathbf{u})(s))-S(\zeta
(s),\mathbf{\varepsilon }(\mathbf{u})(s))|_{Q}^{2}ds.
\end{align*}
Now from Lemma \ref{19augl2}, (\ref{3septe1}) and adjusting constants, this
is dominated by
\begin{align*}
&C\int_0^{t}|\eta _{\ast }(\zeta _{n}(s))
-\eta _{\ast }(\zeta (s))|
_{Y}^{2}ds+C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}
_{n})(s)-\mathbf{\varepsilon }(\mathbf{u})
(s)|_{Q}^{2}ds \\
&+C\int_0^{t}|S(\zeta _{n}(s),\mathbf{
\varepsilon }(\mathbf{u})(s))-S(\zeta
(s),\mathbf{\varepsilon }(\mathbf{u})(
s))|_{Q}^{2}ds.
\end{align*}
Now using (\ref{3septe9}) of Lemma \ref{19augl2} this is dominated by
\begin{align*}
&C\int_0^{t}|\eta _{\ast }(\zeta _{n}(s))
-\eta _{\ast }(\zeta (s))|
_{Y}^{2}ds+C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}
_{n})(s)-\mathbf{\varepsilon }(\mathbf{u})
(s)|_{Q}^{2}ds \\
&+C\int_0^{t}\int_{\Omega }|\eta _{\ast }(\zeta _{n}(
s))-\eta _{\ast }(\zeta (s))|
^{2}|\mathbf{\varepsilon }(\mathbf{u}(s))
|_{\mathbb{S}_d}^{2}dx\,ds \\
&+C\int_0^{t}\int_0^{s}|\eta _{\ast }(\zeta _{n}(
r))-\eta _{\ast }(\zeta (r))|
_{Y}^{2}drds
\end{align*}
which, after adjusting the constants, implies
\begin{equation}
\begin{aligned}
&|m(\zeta ,\mathbf{\varepsilon }(\mathbf{u}))
(t)-m(\zeta _{n},\mathbf{\varepsilon }(\mathbf{u}
_{n}))(t)|_{Q}^{2}\\
&\leq C\int_0^{t}|\eta _{\ast }(\zeta _{n}(s))
-\eta _{\ast }(\zeta (s))|
_{Y}^{2}ds+C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}
_{n})(s)-\mathbf{\varepsilon }(\mathbf{u})
(s)|_{Q}^{2}ds  \notag \\
&\quad +C\int_0^{t}\int_{\Omega }|\eta _{\ast }(\zeta _{n}(
s))-\eta _{\ast }(\zeta (s))|
^{2}|\mathbf{\varepsilon }(\mathbf{u}(s))
|_{\mathbb{S}_d}^{2}dx\,ds.  \label{7septe2}
\end{aligned}
\end{equation}
Consequently,
\begin{equation}
\begin{aligned}
&|m(\zeta ,\mathbf{\varepsilon }(\mathbf{u}))
(t)-m(\zeta _{n},\mathbf{\varepsilon }(\mathbf{u}
_{n}))(t)|_{Q}\\
&\leq C\Big(\int_0^{t}|\eta _{\ast }(\zeta _{n}(s)
)-\eta _{\ast }(\zeta (s))|
_{Y}^{2}ds\Big)^{1/2}
+C\Big(\int_0^{t}|\mathbf{\varepsilon }
(\mathbf{u}_{n})(s)-\mathbf{\varepsilon }(
\mathbf{u})(s)|_{Q}^{2}ds\Big)^{1/2}  \\
&\quad +C\Big(\int_0^{t}\int_{\Omega }|\eta _{\ast }(\zeta
_{n}(s))-\eta _{\ast }(\zeta (s)
)|^{2}|\mathbf{\varepsilon }(\mathbf{u}(
s))|_{\mathbb{S}_d}^{2}dx\,ds\Big)^{1/2}.
\end{aligned}\label{7septe3}
\end{equation}
Then from $A(\zeta _{n},\mathbf{u}_{n})=\mathbf{f}$ and
$A(\zeta ,\mathbf{u})=\mathbf{f}$ it follows
\begin{align*}
0 &=\int_0^{t}(\eta _{\ast }(\zeta _{n})\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_{n})-\eta _{\ast }(\zeta
)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}),\mathbf{
\varepsilon }(\mathbf{u}_{n})-\mathbf{\varepsilon }(
\mathbf{u}))_{Q}ds \\
&\quad+\int_0^{t}(m(\zeta _{n},\mathbf{\varepsilon }(\mathbf{
u}_{n}))-m(\zeta ,\mathbf{\varepsilon }(\mathbf{u}
)),\mathbf{\varepsilon }(\mathbf{u}_{n})-\mathbf{
\varepsilon }(\mathbf{u}))_{Q}
\end{align*}
and so from the above estimate in (\ref{7septe3}),
\begin{align*}
&\int_0^{t}(\eta _{\ast }(\zeta _{n})\mathcal{A}
\mathbf{\varepsilon }(\mathbf{u}_{n})-\eta _{\ast }(\zeta
)\mathcal{A}\mathbf{\varepsilon }(\mathbf{u}),\mathbf{
\varepsilon }(\mathbf{u}_{n})-\mathbf{\varepsilon }(
\mathbf{u}))_{Q}ds \\
&\leq \int_0^{t}|m(\zeta _{n},\mathbf{\varepsilon }(
\mathbf{u}_{n}))-m(\zeta ,\mathbf{\varepsilon }(
\mathbf{u}))|_{Q}|\mathbf{\varepsilon }(
\mathbf{u}_{n})-\mathbf{\varepsilon }(\mathbf{u})|
_{Q}ds \\
&\leq \int_0^{t}\Big[ C\Big(\int_0^{s}|\eta _{\ast }(
\zeta _{n}(r))-\eta _{\ast }(\zeta (r)
)|_{Y}^{2}dr\Big)^{1/2}
+C\Big(\int_0^{s}|\mathbf{
\varepsilon }(\mathbf{u}_{n})(r)-\mathbf{
\varepsilon }(\mathbf{u})(r)|
_{Q}^{2}dr\Big)^{1/2}  \\
&\quad +\Big(\int_0^{s}\int_{\Omega }|\eta _{\ast }(\zeta
_{n}(r))-\eta _{\ast }(\zeta (r)
)|^{2}|\mathbf{\varepsilon }(\mathbf{u}(
r))|_{\mathbb{S}_d}^{2}dx\,dr\Big)^{1/2}\Big]
|\mathbf{\varepsilon }(\mathbf{u}_{n}(s))-
\mathbf{\varepsilon }(\mathbf{u}(s))|_{Q}ds.
\end{align*}
Considering the left side of this inequality and manipulating the right side
some more, one obtains an inequality of the following form.
\begin{align*}
&\int_0^{t}\zeta _{\ast }m_{\mathcal{A}}\|\mathbf{u}_{n}-
\mathbf{u}\|_{V}^{2}ds\\
&\leq C\int_0^{t}\int_{\Omega }|
\eta _{\ast }(\zeta _{n}(s))-\eta _{\ast }(
\zeta (s))|^{2}|\mathbf{\varepsilon }(
\mathbf{u}(s))|_{\mathbb{S}_d}^{2}dx\,ds\\
&\quad +C\int_0^{t}\int_0^{s}|\eta _{\ast }(\zeta _{n}(
r))-\eta _{\ast }(\zeta (r))|
_{Y}^{2}+\|\mathbf{u}_{n}(r)-\mathbf{u}(
r)\|_{V}^{2}drds \\
&\quad +C\int_0^{t}\int_0^{s}\int_{\Omega }|\eta _{\ast }(\zeta
_{n}(r))-\eta _{\ast }(\zeta (r)
)|^{2}|\mathbf{\varepsilon }(\mathbf{u}(
r))|_{\mathbb{S}_d}^{2}dxdrds \\
&\quad +\frac{\zeta _{\ast }m_{\mathcal{A}}}{2}\int_0^{t}\|\mathbf{u
}_{n}-\mathbf{u}\|_{V}^{2}ds.
\end{align*}
Therefore, after adjusting constants and using Gronwall's inequality,
\begin{equation}
\begin{aligned}
\int_0^{t}\|\mathbf{u}_{n}-\mathbf{u}\|_{V}^{2}ds
&\leq C\int_0^{t}\int_0^{s}|\eta _{\ast }(\zeta
_{n}(r))-\eta _{\ast }(\zeta (r)
)|_{Y}^{2}dr\,ds  \notag \\
&\quad +C\int_0^{t}\int_{\Omega }|\eta _{\ast }(\zeta _{n}(
s))-\eta _{\ast }(\zeta (s))|
^{2}|\mathbf{\varepsilon }(\mathbf{u}(s))
|_{\mathbb{S}_d}^{2}dx\,ds.
\end{aligned}\label{7septe7}
\end{equation}
If the conclusion of the lemma is not true, then there exists
$\varepsilon>0 $ and $\zeta _{n}\to \zeta $ in $\mathcal{Y}$
but $\|\mathbf{u}_{n}-\mathbf{u}\|_{\mathcal{V}}\geq \varepsilon $.
Taking a subsequence, one can assume that the convergence of $\zeta _{n}$ to
$\zeta $ is pointwise a.e. But now an application of the dominated
convergence theorem in (\ref{7septe7}) yields a contradiction because the
right side of the above inequality converges to 0. This proves the lemma.
\end{proof}

Now define the operator $\Phi :\mathcal{Y}\to \mathcal{Y}$ as
follows. Let $\zeta \in \mathcal{Y}$, then $\Phi (\zeta )$ is the solution
of
\begin{equation}
\Phi (\zeta )'+\kappa L\Phi (\zeta )=\phi
(\mathbf{\varepsilon }(\mathbf{u}_{\zeta }),\eta _{\ast
}(\Phi (\zeta ))),\quad \Phi (\zeta
)(0)=\zeta _0.  \label{22jane4}
\end{equation}

\begin{lemma}
\label{22janl2} The operator $\Phi $ is continuous.
\end{lemma}

\begin{proof}
This is clear from the preceding lemma and routine Gronwall
inequality arguments exploiting the Lipschitz continuity of $\phi $.
\end{proof}

\begin{lemma} \label{22janl3}
$\Phi (\mathcal{Y})$ lies in a compact and
convex subset of $\mathcal{Y}$.
\end{lemma}

\begin{proof}
Let $\zeta \in \mathcal{Y}$. Then, it follows from (\ref
{22jane4}) and the boundedness assumption on $\phi $ that
\begin{align*}
& \frac{1}{2}|\Phi (\zeta )(t)|
_{L^{2}(\Omega )}^{2}-\frac{1}{2}|\zeta _0|_{L^{2}(\Omega
)}^{2}+\kappa \int_0^{t}\|\Phi (\zeta )(
s)\|_{H^{1}(\Omega )}^{2}ds   \\
&  \leq C+\kappa \int_0^{t}|\Phi (\zeta )
(s)|_{L^{2}(\Omega )}^{2}ds+\int_0^{t}|\Phi
(\zeta )(s)|_{L^{2}(\Omega )}^{2}ds,
\end{align*}
and so by Gronwall's inequality there is a positive constant $C$,
independent of $\zeta $, such that
\begin{equation*}
|\Phi (\zeta )(t)|_{L^{2}(\Omega
)}^{2}+\|\Phi (\zeta )\|_{\mathcal{H}
_1}^{2}\leq C.
\end{equation*}
It follows now from (\ref{22jane4}) that $\Vert \Phi (\zeta )
'\Vert _{\mathcal{E}'}\leq C$, for a positive constant $C$
which is independent of $\zeta $. Therefore, there exists another constant
$C $ such that
\begin{equation*}
\|\Phi (\zeta )\|_{\mathcal{E}
}^{2}+\|\Phi (\zeta )'\|_{
\mathcal{E}'}^{2}\leq C,
\end{equation*}
for all $\zeta \in \mathcal{Y}$, and the conclusion follows now from
Theorem \ref{t6.1}.
\end{proof}

The following lemma will be used to prove the uniqueness part in the next
theorem.

\begin{lemma}\label{6febl1}
Let $y,y'\in \mathcal{Y}$, $y(0)=0$, and
assume that $y\in L^{2}(0,T;H^{2}(\Omega ))$ and
it satisfies ${\partial y}/{\partial n}=0$ on $\partial \Omega $. Then
\begin{equation*}
\int_0^{t}(y',-\Delta y)_{L^{2}(\Omega )}ds\geq 0.
\end{equation*}
\end{lemma}

\begin{proof} Let $L:D(L)\subseteq \mathcal{Y}\to \mathcal{Y}$
 be defined by (\ref{k25.5}), where
$D(L)\equiv\{ z\in \mathcal{Y}:Lz\in \mathcal{Y}\}$.
Note that $L$ was defined above as $L:\mathcal{E}\to \mathcal{E}'$.
Then $L$ is a maximal monotone operator and $Ly=-\Delta y$ for
$y\in D(L)$. Also, since $C_0^{\infty }(\Omega )$ is dense in
$L^{2}(\Omega )$, it follows that $D(L)$ is dense in
$\mathcal{Y}$. Let
\begin{equation*}
y_{\varepsilon }\equiv (I+\varepsilon L)^{-1}y,
\end{equation*}
for a small positive $\varepsilon $. Thus, $y_{\varepsilon }'=(I+\varepsilon L)^{-1}y'\in D(L)$ and
so it is routine to verify that
\begin{equation*}
\int_0^{t}(y_{\varepsilon }',(-\Delta y_{\varepsilon
}))_{L^{2}(\Omega )}ds\geq 0.
\end{equation*}
Also, since $D(L)$ is dense in $\mathcal{Y}$, it follows from
standard results on maximal monotone operators \cite{bre73} that, as
$\varepsilon \to 0$,
\begin{gather*}
-\Delta y_{\varepsilon }=Ly_{\varepsilon }=L(I+\varepsilon L)
^{-1}y =(I+\varepsilon L)^{-1}Ly\to Ly=-\Delta y, \\
(I+\varepsilon L)^{-1}y' = y_{\varepsilon }'\to y'\quad
\text{ weakly in }\mathcal{Y}.
\end{gather*}
Therefore,
\begin{equation*}
0\leq \lim_{\varepsilon \to 0}\int_0^{t}(y_{\varepsilon
}',-\Delta y_{\varepsilon })_{L^{2}(\Omega
)}ds=\int_0^{t}(y',-\Delta y)_{L^{2}(\Omega )}ds.
\end{equation*}
This proves the lemma.
\end{proof}
Finally, here is the existence and uniqueness theorem for Problem $P_{V}$.

\begin{theorem}\label{22jant1}
Let $\zeta _0\in E$ and $\mathbf{f}\in L^{\infty }(0,T;V')$.
Then there exists a unique solution to the system \eqref{k29}
 and \eqref{k30} which satisfies
\begin{equation*}
\zeta '\in \mathcal{Y},\quad L\zeta \in L^{2}(
0,T;L^{2}(\Omega )),\quad \zeta \in L^{2}(0,T;H^{2}(
\Omega )),\quad \mathbf{u}\in L^{\infty }(0,T;V).
\end{equation*}
\end{theorem}

\begin{proof} The existence of a solution to (\ref{k29}) and (\ref{k30})
which satisfies $\zeta ,\zeta '\in \mathcal{Y}$ and $\mathbf{u}\in
\mathcal{V}$ follows from the Schauder fixed-point theorem.
Consider the equation for $\mathbf{u}$. From Lemma \ref{19augl2} applied to
$\mathbf{u}\mathcal{X}_{[ t-h,t+h] }$,
\begin{align*}
\int_{t-h}^{t+h}\langle \mathbf{f}(t),\mathbf{u}\rangle ds
&=\int_{t-h}^{t+h}(S(\zeta ,\mathbf{
\varepsilon }(\mathbf{u})),\mathbf{\varepsilon }(
\mathbf{u}))ds \\
&\geq \delta \int_{t-h}^{t+h}|\mathbf{\varepsilon }(\mathbf{u}
(s))|_{Q}^{2}ds-\int_{t-h}^{t+h}C\int_0^{s}
|\mathbf{\varepsilon }(\mathbf{u}(r))|^{2}dr\,ds-2hC
\end{align*}
and so since $\mathbf{f}\in L^{\infty }(0,T;V')$, this
implies
\begin{equation*}
\frac{\delta }{2}\int_{t-h}^{t+h}|\mathbf{\varepsilon }(\mathbf{u
}(s))|_{Q}^{2}ds\leq
2hC+\int_{t-h}^{t+h}C\int_0^{s}|\mathbf{\varepsilon }(\mathbf{u
}(r))|^{2}drds.
\end{equation*}
Now divide by $2h$ and apply the fundamental theorem of calculus to obtain
that for a.e. $t$,
\begin{equation*}
|\mathbf{\varepsilon }(\mathbf{u}(t))|
_{Q}^{2}\leq C+C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}
(s))|^{2}ds.
\end{equation*}
Then an application of Gronwall's inequality yields
\begin{equation}
|\mathbf{\varepsilon }(\mathbf{u}(t))|
_{Q}^{2}\leq C\text{ a.e.}  \label{5septe3}
\end{equation}
which shows that $\mathbf{u}\in L^{\infty }(0,T;V)$.

The regularity of $\zeta $ follows from $\zeta '\in \mathcal{Y}$
which implies $\zeta +L\zeta \in \mathcal{Y}$ and then standard regularity
results imply that $\zeta \in L^{2}(0,T;H^{2}(\Omega ))$. See \cite{gri85}.

It remains to verify the uniqueness of the solution. Suppose then that
$(\zeta _{i},\mathbf{u}_{i})$, for $i=1,2$, are two solutions
with the specified regularity. Then,
\begin{equation*}
\frac{1}{2}|\zeta _1(t)-\zeta _2(t)
|_{Y}^{2}+\kappa \int_0^{t}|\nabla (\zeta _1-\zeta
_2)(s)|^{2}ds\leq C\int_0^{t}\big(|
|\mathbf{u}_1-\mathbf{u}_2\|_{V}^{2}+|\zeta
_1-\zeta _2|_{Y}^{2}\big)ds
\end{equation*}
Hence Gronwall's inequality yields
\begin{equation}
|\zeta _1(t)-\zeta _2(t)|
_{Y}^{2}+\int_0^{t}|\nabla (\zeta _1-\zeta _2)
(s)|^{2}ds\leq C\int_0^{t}\|\mathbf{u}
_1(s)-\mathbf{u}_2(s)\|_{V}^{2}ds
\label{23maye1}
\end{equation}
Also, from the equation for $\mathbf{u}$
\begin{equation}
\int_0^{t}(S(\zeta _1,\mathbf{\varepsilon }(\mathbf{u}
_1))-S(\zeta _2,\mathbf{\varepsilon }(\mathbf{u
}_2)),\mathbf{\varepsilon }(\mathbf{u}_1)-
\mathbf{\varepsilon }(\mathbf{u}_2))_{Q}ds=0.
\label{5septe1}
\end{equation}
Now recall Lemma \ref{19augl2}. Two of the formulas established there were
\begin{align*}
&|S(\zeta _1(t),\mathbf{\tau }(t)
)-S(\zeta _2(t),\mathbf{\tau }(t))|_{Q}^{2}\\
&\leq C\Big(\int_{\Omega }|\eta _{\ast }(\zeta _1(
t))-\eta _{\ast }(\zeta _2(t))
|^{2}|\mathbf{\tau }(t)|_{\mathbb{S}_d}^{2}dx\Big)
 +C\Big(\int_0^{t}|\eta _{\ast }(\zeta _1(s)
)-\eta _{\ast }(\zeta _2(s))|_{Y}^{2}ds\Big)
\end{align*}
and
\begin{align*}
&(S(\zeta (t),\mathbf{\tau }_1(t)
)-S(\zeta (t),\mathbf{\tau }_2(t)
),\mathbf{\tau }_1(t)-\mathbf{\tau }_2(
t))_{Q} \\
&\geq \delta |\mathbf{\tau }_1(t)-\mathbf{\tau }
_2(t)|_{Q}^{2}-C\int_0^{t}|\mathbf{\tau }
_1(s)-\mathbf{\tau }_2(s)|_{Q}^{2}ds.
\end{align*}
The first of these inequalities implies
\begin{align*}
&\big|(S(\zeta _1(t),\mathbf{\tau }(
t))-S(\zeta _2(t),\mathbf{\tau }(
t)),\mathbf{\varepsilon })_{Q}\big|
\\
&\leq C\Big(\int_{\Omega }|\eta _{\ast }(\zeta _1(
t))-\eta _{\ast }(\zeta _2(t))
|^{2}|\mathbf{\tau }(t)|_{\mathbb{S}
_d}^{2}dx\Big)^{1/2}|\mathbf{\varepsilon }|_{Q} \\
&\quad +C\Big(\int_0^{t}|\eta _{\ast }(\zeta _1(s)
)-\eta _{\ast }(\zeta _2(s))|
_{Y}^{2}ds\Big)^{1/2}|\mathbf{\varepsilon }|_{Q}  \\
&\leq C|\mathbf{\varepsilon }|_{Q}\Big(\|\zeta
_1(t)-\zeta _2(t)\|_{L^{\infty
}(\Omega )}|\mathbf{\tau }(t)|
_{Q}+\Big(\int_0^{t}|\zeta _1(s)-\zeta _2(
s)|_{Y}^{2}ds\Big)^{1/2}\Big)\label{5septe2}
\end{align*}
Using these estimates in (\ref{5septe1}),
\begin{align*}
0 &=\int_0^{t}(S(\zeta _1(s),\mathbf{
\varepsilon }(\mathbf{u}_1(s)))-S(
\zeta _1(s),\mathbf{\varepsilon }(\mathbf{u}_2(
s))),\mathbf{\varepsilon }(\mathbf{u}_1(
s))-\mathbf{\varepsilon }(\mathbf{u}_2(s)
))_{Q}ds \\
&\quad +\int_0^{t}(S(\zeta _1,\mathbf{\varepsilon }(\mathbf{
u}_2))-S(\zeta _2,\mathbf{\varepsilon }(
\mathbf{u}_2)),\mathbf{\varepsilon }(\mathbf{u}
_1(s))-\mathbf{\varepsilon }(\mathbf{u}
_2(s)))_{Q}ds \\
&\geq \int_0^{t}\Big(\delta |\mathbf{\varepsilon }(\mathbf{u
}_1)(s)-\mathbf{\varepsilon }(\mathbf{u}
_2)(s)|_{Q}^{2}-C\int_0^{s}|\mathbf{
\varepsilon }(\mathbf{u}_1)(r)-\mathbf{
\varepsilon }(\mathbf{u}_2)(r)|
_{Q}^{2}dr\Big)ds \\
&\quad -C\int_0^{t}|\mathbf{\varepsilon }(\mathbf{u}_1(
s))-\mathbf{\varepsilon }(\mathbf{u}_2(s)
)|_{Q}\Big(\|\zeta _1(s)-\zeta
_2(s)\|_{L^{\infty }(\Omega )
}|\mathbf{\varepsilon }(\mathbf{u}_2)(s)
|_{Q}  \\
&\quad  +\Big(\int_0^{s}|\zeta _1(r)-\zeta
_2(r)|_{Y}^{2}dr\Big)^{1/2}\Big)ds.
\end{align*}
Now letting $r\in (3/2,2)$, so that $H^{r}(\Omega )$ imbeds compactly
into $L^{\infty }(\Omega )$, and using (\ref{5septe3}) this implies
after adjusting constants, an inequality of the form
\begin{equation}
\begin{aligned}
&\int_0^{t}\|\mathbf{u}_1-\mathbf{u}_2\|_{V}^{2}ds\\
&\leq C\int_0^{t}\int_0^{s}\|\mathbf{u}_1(r)
\mathbf{-u}_2(r)\|
_{V}^{2}drds+C\int_0^{t}\|\mathbf{u}_1(s)-
\mathbf{u}_2(s)\|_{V}\|\zeta
_1(s)-\zeta _2(s)\|_{H^{r}(
\Omega )}ds  \notag \\
&\quad +C\int_0^{t}\|\mathbf{u}_1(s)-\mathbf{u}
_2(s)\|_{V}\Big(\int_0^{s}|\zeta
_1(r)-\zeta _2(r)|_{Y}^{2}dr\Big)^{1/2}ds
\end{aligned} \label{5septe4}
\end{equation}
 From (\ref{23maye1}),
\begin{align*}
&\int_0^{t}\|\mathbf{u}_1-\mathbf{u}_2\|_{V}^{2}ds\\
&\leq C\int_0^{t}\int_0^{s}\|\mathbf{u}_1(r)
\mathbf{-u}_2(r)\|_{V}^{2}drds+C\int_0^{t}\|\mathbf{u}_1(s)-
\mathbf{u}_2(s)\|_{V}
 \|\zeta _1(s)-\zeta _2(s)\|_{H^{r}(\Omega )}ds \\
&\quad +C\int_0^{t}\|\mathbf{u}_1(s)-\mathbf{u}_2(s)\|_{V}
\Big(\int_0^{s}\int_0^{r}\|\mathbf{u}_1(p)-\mathbf{
u}_2(p)\|^{2}dp\,dr\Big)^{1/2}ds
\\
&\leq C\int_0^{t}\int_0^{s}\|\mathbf{u}_1(r)
\mathbf{-u}_2(r)\|_{V}^{2}drds+\frac{1}{4}
\int_0^{t}\|\mathbf{u}_1(s)-\mathbf{u}_2(s)\|_{V}^{2}ds \\
&\quad +C\int_0^{t}\|\zeta _1(s)-\zeta _2(s)\|_{H^{r}(\Omega )}^{2}ds \\
&\quad +\frac{1}{4}\int_0^{t}\|\mathbf{u}_1(s)-
\mathbf{u}_2(s)\|
_{V}^{2}ds+C\int_0^{t}\int_0^{s}\int_0^{r}\|\mathbf{u}
_1(p)-\mathbf{u}_2(p)\|^{2}dp\,dr\,ds
\end{align*}
Now an application of Gronwall's inequality yields
\begin{equation}
\int_0^{t}\|\mathbf{u}_1-\mathbf{u}_2\|
_{V}^{2}ds\leq C\int_0^{t}\|\zeta _1(s)-\zeta
_2(s)\|_{H^{r}(\Omega )}^{2}ds\,.
\label{5septe6}
\end{equation}
It follows from (\ref{23maye1}) that
\begin{equation}
|\zeta _1(t)-\zeta _2(t)|
_{Y}^{2}+\int_0^{t}|\nabla (\zeta _1-\zeta _2)
(s)|^{2}ds\leq C\int_0^{t}\|\zeta
_1(s)-\zeta _2(s)\|_{H^{r}(
\Omega )}^{2}ds \,. \label{5septe5}
\end{equation}
The equations for $\zeta _1$ and $\zeta _2$ imply that
\begin{align*}
&\int_0^{t}((\zeta _1'-\zeta _2')
,-\Delta (\zeta _1-\zeta _2))_{L^{2}(\Omega
)}ds+\kappa \int_0^{t}|\Delta (\zeta _1-\zeta _2)
|_{L^{2}(\Omega )}^{2}ds \\
& \leq \int_0^{t}|(\phi (\mathbf{\varepsilon }
(\mathbf{u}_1),\eta _{\ast }(\zeta _1))
-\phi (\mathbf{\varepsilon }(\mathbf{u}_2),\eta _{\ast
}(\zeta _2)),\Delta (\zeta _1-\zeta
_2))_{L^{2}(\Omega )}|ds \\
& \leq C\int_0^{t}(\|\mathbf{u}_1-\mathbf{u}
_2\|_{V}+|\zeta _1-\zeta _2|_{Y})
|\Delta (\zeta _1-\zeta _2)|_{L^{2}(\Omega)}ds.
\end{align*}
It follows from Lemma \ref{6febl1} that the first term is nonnegative, thus
from (\ref{5septe6}),
\begin{align*}
\frac{\kappa }{2}\int_0^{t}|\Delta (\zeta _1-\zeta_2)|_{Y}^{2}ds
&\leq C\int_0^{t}\big(\|\mathbf{u}_1-\mathbf{u}_2\|_{V}^{2}+|\zeta
_1-\zeta _2|_{Y}^{2}\big)ds \\
&\leq C\big[ \int_0^{t}\|\zeta _1-\zeta _2|
|_{H^{r}(\Omega )}^{2}+|\zeta _1-\zeta
_2|_{Y}^{2}\big] ds
\end{align*}
Then using regularity results, adjusting constants and using the compactness
of the imbedding of $H^{2}(\Omega )$ into $H^{r}(\Omega)$,
\begin{align*}
\int_0^{t}\|\zeta _1-\zeta _2\|_{H^{2}(
\Omega )}^{2}ds
&\leq C\big[ \int_0^{t}\|\zeta
_1-\zeta _2\|_{H^{r}(\Omega )}^{2}+|
\zeta _1-\zeta _2|_{Y}^{2}\big] ds \\
&\leq C\int_0^{t}|\zeta _1-\zeta _2|_{Y}^{2}ds+\frac{1}{2
}\int_0^{t}\|\zeta _1-\zeta _2\|
_{H^{2}(\Omega )}^{2}ds.
\end{align*}
Therefore, an inequality of the following form holds
\begin{equation*}
\int_0^{t}\|\zeta _1-\zeta _2\|_{H^{2}(
\Omega )}^{2}ds\leq C\int_0^{t}|\zeta _1-\zeta _2|
_{Y}^{2}ds.
\end{equation*}
 From (\ref{5septe5}) and the above inequality,
\begin{align*}
&|\zeta _1(t)-\zeta _2(t)|_{Y}^{2}+\int_0^{t}\|\zeta _1-\zeta _2\|
_{E}^{2}ds\\
&\leq C\int_0^{t}\|\zeta _1(s)-\zeta _2(
s)\|_{H^{r}(\Omega )}^{2}ds+\int_0^{t}|\zeta _1-\zeta _2|_{Y}^{2}ds   \\
&\leq  C\int_0^{t}|\zeta _1(s)-\zeta _2(s)|_{Y}^{2}ds
\end{align*}
and by Gronwall's inequality, $\zeta _1=\zeta _2$.  From this it
follows immediately from Lemma \ref{22augl1} that
$\mathbf{u}_1=\mathbf{u}_2$ and this proves uniqueness.
\end{proof}

\section{Removing $\eta _{\ast }$}

\label{localsolutions}

This section considers how to remove $\eta _{\ast }$ and involves only the
assumptions
\begin{gather}
\zeta _0(\mathbf{x})\in [ \zeta _{\ast },1]
\label{23maye4}
\\
\phi (\mathbf{\varepsilon },\zeta _{\ast })\geq 0,
\quad \phi (\mathbf{\varepsilon },1)\leq 0.  \label{23maye5}
\end{gather}
It is based on some fundamental comparison theorems which apply to
semilinear parabolic equations which are interesting for their own sake.

\begin{definition} \rm
Let $\Omega $ be an open set. Then $\Omega $ has the interior
ball condition  at $\mathbf{x}\in \partial \Omega $
if there exists $\mathbf{z}\in \Omega $ and $r>0$ such that
 $B(\mathbf{z},r)\subseteq \Omega $ and
$\mathbf{x}\in \partial B(\mathbf{z},r)$.
\end{definition}

With these definitions, the following is a special case of a famous lemma
by Hopf, \cite{eva93}.

\begin{lemma}\label{hopf}
Let $\Omega $ be a bounded open set and suppose
$\mathbf{x}_0\in \partial \Omega $ and $\Omega $ has the interior
ball condition at $\mathbf{x}_0$ with the ball being $B(\mathbf{z},r)$.
Suppose for $u\in C^{2}(\Omega )\cap C^{1}(\overline{\Omega })$
\begin{equation}
\Delta u\geq 0\quad \text{in }\Omega .  \label{22septe7}
\end{equation}
Then if $u(\mathbf{x}_0)=\max \{ u(\mathbf{x}):\mathbf{x}\in
\overline{\Omega }\} $ and $u(\mathbf{x})<u(\mathbf{x}_0)$ for
$\mathbf{x}\in \Omega $, it follows
\begin{equation}
\frac{\partial u}{\partial n}(\mathbf{x}_0)>0
\label{22septe8}
\end{equation}
where $\mathbf{n}$ is the exterior unit normal to the ball at the point
$\mathbf{x}_0$.
\end{lemma}

\begin{lemma}\label{11mayl2}
If $\Omega $ has $C^{2,1}$ boundary then every point of
$\partial \Omega $ has the interior ball condition. In addition, there exist
at each point of $\partial \Omega $ arbitrarily small balls tangent to
$\partial \Omega $ such that the exterior unit normal to the ball at that
point coincides with the exterior unit normal to $\Omega $.
\end{lemma}

 From now on, assume the boundary of $\Omega $ is $C^{2,1}$. Suppose the
following holds for a measurable function $f$.
\begin{gather}
f :(0,T)\times \Omega \times \mathbb{R}\to \mathbb{R},
\label{11maye2} \\
|f(t,\mathbf{x},\zeta )-f(t,\mathbf{x},\xi )
|\leq K|\zeta -\xi |,  \label{11maye3} \\
f(t,\mathbf{x},\zeta )\leq -2\varepsilon <0\text{ if }\zeta
\geq b,  \label{11maye4} \\
f(\cdot ,\cdot ,0)\in L^{2}(0,T;L^{2}(\Omega)),  \label{11maye5}
\end{gather}
Also let $\Omega _T\equiv (0,T)\times \Omega ,B_T\equiv(-T,2T)\times
(\Omega +B(\mathbf{0},1))$. In order to take a convolution,
$f$ is extended to $\widehat{f}$ as follows
\begin{equation*}
\hat{f}(t,\mathbf{x},\zeta )\equiv
\begin{cases}
f(t,\mathbf{x},\zeta )&\text{if }(t,\mathbf{x})\in
[ 0,T] \times \Omega \\
-2\varepsilon & \text{if }(t,\mathbf{x})\in B_T\setminus
\Omega _T \\
0&\text{if }(t,\mathbf{x})\notin B_T
\end{cases}
\end{equation*}
If $\zeta \in \mathcal{Y}$,
\begin{equation*}
\hat{\zeta}(t,\mathbf{x})\equiv
\begin{cases}
\zeta (t,\mathbf{x}) &\text{if }(t,\mathbf{x})\in
[ 0,T] \times \Omega \\
0 &\text{otherwise}.
\end{cases}
\end{equation*}
Now define
\begin{equation*}
f_{n}(t,\mathbf{x},\zeta )\equiv \int_{\mathbb{R}^{d+2}}\hat{f}
(t-s,\mathbf{x-y},\zeta -\xi )\psi _{n}(s,\mathbf{y},\xi
)ds\,dy\,d\xi
\end{equation*}
where $\psi _{n}$ is a mollifier having support in $B(\mathbf{0,}\frac{
1}{n})\subseteq \mathbb{R}^{d+2}$. Thus $f_{n}\in C^{\infty }(
\mathbb{R}^{d+2})$.

\begin{lemma} \label{11mayl1}
Let $f_{n}$ be defined above. Then if $n$ is large enough and
$(t,\mathbf{x})\in (0,T)\times \Omega $,
\begin{equation}
f_{n}(t,\mathbf{x},\zeta )\leq -\varepsilon \quad \text{if }\zeta
\geq b.  \label{11maye1}
\end{equation}
For $\zeta \in \mathcal{Y}$ and $\ \delta >0$ given and $\zeta _1\in
\mathcal{Y}$ $\ $arbitrary, it follows that for all $n$ sufficiently large,
depending only on $\delta $ and $\zeta $,
\begin{equation}
\Big(\int_0^{t}\int_{\Omega }|f_{n}(s,\mathbf{x},\zeta
_1)-f(s,\mathbf{x},\zeta )|^{2}dx\,ds\Big)
^{1/2}\leq \delta +\sqrt{2}K(\int_0^{t}|\zeta _1-\zeta
|_{Y}^{2}ds)^{1/2}.  \label{11maye6}
\end{equation}
\end{lemma}

\begin{proof} Since the integral is from 0 to $t$, change
$\zeta_1(s)$ to equal $\zeta (s)$ for $s>t$. Then
\begin{align*}
&\Big(\int_0^{t}\int_{\Omega }|f_{n}(s,\mathbf{x},\zeta
_1(s,\mathbf{x}))-f(s,\mathbf{x},\zeta (s,
\mathbf{x}))|^{2}dx\,ds\Big)^{1/2} \\
&\leq \Big(\int_0^{T}\int_{\Omega }|f_{n}(s,\mathbf{x}
,\zeta _1(s,\mathbf{x}))-f(s,\mathbf{x},\zeta
(s,\mathbf{x}))|^{2}dx\,ds\Big)^{1/2}
\\
&=\Big(\int_0^{T}\int_{\Omega }\Big|\int_{\mathbb{R}^{d+2}}
\big(\hat{f}(s-r,\mathbf{x-y},\zeta _1(s,\mathbf{x})-\xi) \\
&\quad  -f(s,\mathbf{x},\zeta (s,\mathbf{x}
))\big)\psi _{n}(r,\mathbf{y},\xi )dsdyd\xi
\Big|^{2}dx\,ds\Big)^{1/2}
\end{align*}
By Minkowski's inequality, the above expression is bounded by
\begin{equation}
\begin{aligned}
&\int_{B(\mathbf{0,}\frac{1}{n})}\psi _{n}(r,
\mathbf{y},\xi )
\Big(\int_0^{T}\int_{\Omega }|\hat{f}(s-r,\mathbf{x-y}
,\zeta _1(s,\mathbf{x})-\xi )\\
&-f(s,\mathbf{x},\zeta (s,\mathbf{x}))|^{2}dx\,ds\Big)^{1/2}dr\,dy\,d\xi
\\
&\leq \int_{B(\mathbf{0,}\frac{1}{n})}\psi _{n}(r,
\mathbf{y},\xi )\cdot
\Big(\int_{\mathbb{R}^{d+1}}|\hat{f}(s-r,\mathbf{x-y},\hat{
\zeta}_1(s,\mathbf{x})-\xi )\\
&\quad -\hat{f}(s,\mathbf{x},\hat{\zeta}(s,\mathbf{x}))|^{2}dx\,ds\Big)
^{1/2}dr\,dy\,d\xi
\end{aligned}\label{11maye9}
\end{equation}
Now consider the inner integral for $(r,\mathbf{y},\xi )\in
B(\mathbf{0},\frac{1}{n})$.
\begin{align*}
&\Big(\int_{\mathbb{R}^{d+1}}\Big|\hat{f}\big(s-r,\mathbf{x-y},\hat{
\zeta}_1(s,\mathbf{x})-\xi \big)-\hat{f}\big(s,\mathbf{x}
,\hat{\zeta}(s,\mathbf{x})\big)\Big|^{2}dx\,ds\Big)^{1/2}
\\
&\leq \Big(\int_{[ r,T+r] \times \Omega +\mathbf{y}}\Big|
f\big(s-r,\mathbf{x-y},\hat{\zeta}_1(s,\mathbf{x})-\xi
\big)-f\big(s-r,\mathbf{x-y},\hat{\zeta}(s,\mathbf{x})
\big)\Big|^{2}dx\,ds\Big)^{1/2} \\
&\quad +\Big(\int_{\mathbb{R}^{d+1}}\Big|\hat{f}\big(s-r,\mathbf{x-y},\hat{
\zeta}(s,\mathbf{x})\big)-\hat{f}\big(s-r,\mathbf{x-y},
\hat{\zeta}(s-r,\mathbf{x-y})\big)\Big|^{2}dx\,ds\Big)
^{1/2} \\
&\quad +\Big(\int_{\mathbb{R}^{d+1}}\Big|\hat{f}\big(s-r,\mathbf{x-y},\hat{
\zeta}(s-r,\mathbf{x-y})\big)-\hat{f}\big(s,\mathbf{x},
\hat{\zeta}(s,\mathbf{x})\big)\Big|^{2}dx\,ds\Big)^{1/2}
\\
&\leq \sqrt{2}K\Big(\int_{[ r,T+r] \times \Omega +\mathbf{y}
}|\hat{\zeta}_1(s,\mathbf{x})-\hat{\zeta}(s,
\mathbf{x})|^{2}dx\,ds\Big)^{1/2}\\
&\quad +\sqrt{2}K\Big(\int_{[r,T+r] \times \Omega +\mathbf{y}}|\xi |^{2}dx\,ds\Big)
^{1/2} \\
&\quad +K\Big(\int_{\mathbb{R}^{d+1}}|\hat{\zeta}(s,\mathbf{x})
-\hat{\zeta}(s-r,\mathbf{x-y})|^{2}dx\,ds\Big)^{1/2} \\
&\quad +\Big(\int_{\mathbb{R}^{d+1}}|\hat{f}(s-r,\mathbf{x-y},\hat{
\zeta}(s-r,\mathbf{x-y}))-\hat{f}(s,\mathbf{x},
\hat{\zeta}(s,\mathbf{x}))|^{2}dx\,ds\Big)^{1/2}\,.
\end{align*}
Using continuity of translation, in $L^{2}(\mathbb{R}^{d+1})$
and the above convention that $\zeta _1=\zeta $ on $[t,T]$,
this is dominated by
\begin{align*}
&\delta +\sqrt{2}K\Big(\int_{[r,T+r]\times \Omega +\mathbf{y}
}|\hat{\zeta}_1(s,\mathbf{x})-\hat{\zeta}(s,
\mathbf{x})|^{2}dx\,ds\Big)^{1/2} \\
&\leq \delta +\sqrt{2}K\Big(\int_0^{t}\int_{\Omega }|\zeta
_1(s,\mathbf{x})-\zeta (s,\mathbf{x})|
^{2}dx\,ds\Big)^{1/2}
\end{align*}
provided $n$ is large enough. Therefore, from (\ref{11maye9}),
\begin{align*}
&\Big(\int_0^{t}\int_{\Omega }|f_{n}(s,\mathbf{x},\zeta
_1)-f(s,\mathbf{x},\zeta )|^{2}dx\,ds\Big)
^{1/2}\\
&\leq \delta +\sqrt{2}K\Big(\int_0^{t}\int_{\Omega }|\zeta
_1(s,\mathbf{x})-\zeta (s,\mathbf{x})|
^{2}dx\,ds\Big)^{1/2}
\end{align*}
This proves (\ref{11maye6}). It remains to verify (\ref{11maye1}) for
$(t,\mathbf{x})\in \Omega _T$. Recall
$B_T\equiv (-T,2T)\times (\Omega +B(\mathbf{0},1))$ and
so
\begin{align*}
(t,\mathbf{x})-B_T &=(t-2T,t+T)\times (
\mathbf{x}-\Omega +B(\mathbf{0},1))\\
&\supseteq (-T,T)\times (\mathbf{x}-\Omega +B(
\mathbf{0},1))
\end{align*}
Then letting $\zeta \geq b$,
\begin{align*}
f_{n}(t,\mathbf{x},\zeta )
&\equiv \int_{\mathbb{R}^{d+2}}\hat{f}
(t-r,\mathbf{x-y},\zeta -\xi )\psi _{n}(r,\mathbf{y},\xi
)dr\,dy\,d\xi  \\
&=\int_{\mathbb{R}^{d+2}}\big(\hat{f}(t-r,\mathbf{x-y},\zeta -\xi
)-\hat{f}(t-r,\mathbf{x-y},\zeta )\big)\psi
_{n}(r,\mathbf{y},\xi )dr\,dy\,d\xi  \\
&\quad +\int_{\mathbb{R}^{d+2}}\hat{f}(t-r,\mathbf{x-y},\zeta )\psi
_{n}(r,\mathbf{y},\xi )dr\,dy\,d\xi  \\
&\leq K\int_{B(\mathbf{0},\frac{1}{n})}|\xi |\psi
_{n}(r,\mathbf{y},\xi )dr\,dy\,d\xi  \\
&\quad +\int_{B(\mathbf{0},\frac{1}{n})\cap ((t,\mathbf{x}
)-B_T)\times \mathbb{R}}\hat{f}(t-r,\mathbf{x-y},\zeta
)\psi _{n}(r,\mathbf{y},\xi )dr\,dy\,d\xi  \\
&\leq \frac{K}{n}+(-2\varepsilon )\int_{B(\mathbf{0},\frac{
1}{n})\cap ((t,\mathbf{x})-B_T)\times
\mathbb{R}}\psi _{n}(r,\mathbf{y},\xi )dr\,dy\,d\xi
\end{align*}
Letting $n$ be such that $1/n<\min (\frac{T}{2},1)$, it follows
\begin{align*}
B(\mathbf{0},\frac{1}{n})\cap ((t,\mathbf{x})
-B_T)\times \mathbb{R} &\supseteq  \\
B(\mathbf{0},\frac{1}{n})\cap (-T,T)\times (
\mathbf{x}-\Omega +B(\mathbf{0},1))\times \mathbb{R}
&\supseteq B(\mathbf{0},\frac{1}{n})
\end{align*}
and so for such $n$, the above conditions imply
\begin{equation*}
 f_n(t,\mathbf{x},\zeta) \leq \frac{K}{n}+(-2\varepsilon ).
\end{equation*}
Now choosing $n$ still larger, we obtain this is larger than
$-\varepsilon $. It suffices to choose
\begin{equation*}
n>\max (\frac{1}{\min (\frac{T}{2},1)},\frac{K}{
\varepsilon }).
\end{equation*}
This proves the lemma.
\end{proof}

The next lemma is fairly routine and gives conditions under which weak
solutions are actually classical solutions which are smooth enough to apply
the reasoning of Lemma \ref{hopf}.

\begin{lemma}\label{7decl1}
Let $\zeta _0\in C_{c}^{\infty }(\Omega )$ and
let $\zeta $ be the weak solution to
\begin{equation*}
\zeta '+\kappa L\zeta =f_{n}(\cdot ,\cdot ,\zeta ),\quad
\zeta (0)=\zeta _0.
\end{equation*}
where $f_{n}$ is described above. Then $\zeta $ is $C^{1}$ in $t$
and $C^{2}$ in $\mathbf{x}$.
\end{lemma}

\begin{proof}
I will give a brief argument for the sake of completeness.
Using standard theory of maximal monotone operators, it is routine to obtain
that this solution satisfies $\zeta '\in \mathcal{Y}$ and
$L\zeta\in \mathcal{Y}$. Then by elliptic regularity theorems
applied pointwise, it follows $\zeta \in L^{2}(0,T;H^{2}(\Omega ))$ and
$\frac{\partial \zeta _{n}}{\partial n}=0$ on $\partial \Omega $. Thus
\begin{equation}
\zeta '-\kappa \Delta \zeta =f_{n}(\cdot ,\cdot ,\zeta
),\quad \zeta (0)=\zeta _0  \label{12maye1}
\end{equation}
Multiplying both sides by $\zeta $ and integrating from 0 to $t$ yields
after using the Lipschitz continuity of $f$ in the last variable an estimate
for $|\zeta (t)|_{Y}$ which is independent of $t$.
Multiplying both sides by $-\Delta \zeta $ and integrating from 0 to $t$
then gives an estimate for $\|\zeta (t)|
|_{E}+\int_0^{T}\|\Delta \zeta (t)|
|^{2}dt$. Multiplying both sides by $\zeta '$ and
integrating gives an estimate for
$\|\zeta '||_{\mathcal{Y}}+\|\zeta (t)\|_{E}$.
One can also obtain the solution to (\ref{12maye1}) as a limit as
$\varepsilon \to 0$ of solutions of systems of the form
\begin{equation*}
(1+\varepsilon L)\zeta '+\kappa L\zeta =f_{n}(
\cdot ,\cdot ,\zeta ),\quad
 (1+\varepsilon L)\zeta (0)=(1+\varepsilon L)\zeta _{0n}.
\end{equation*}
obtaining similar estimates to those just mentioned for this regularized
system. These solutions have $L\zeta '\in \mathcal{Y}$ and so
$-\Delta \zeta '\in \mathcal{Y}$. Multiplying by
$-\Delta \zeta'$ and integrating yields eventually an estimate of the form
\begin{equation*}
\int_0^{t}\|\zeta '\|_{E}^{2}ds+|
\Delta \zeta (t)|_{Y}\leq C
\end{equation*}
for $C$ independent of $\varepsilon ,$which is preserved when passing to the
limit as $\varepsilon \to 0$. Thus using elliptic regularity
theorems applied for a.e. $t$, the solutions to (\ref{12maye1}) satisfy
$\zeta \in L^{\infty }(0,T;H^{2}(\Omega )),\zeta
'\in \mathcal{E}$. Now from (\ref{12maye2a}) and the assumption
$\partial \Omega $ is $C^{2,1}$ it follows $\zeta \in L^{2}(
0,T;H^{3}(\Omega ))$.

Next differentiating the equation (\ref{12maye1}) with respect to $t$ yields
\begin{equation*}
-\kappa L\zeta _{0n}+f_{n}(0,\cdot ,\zeta _{0n}(\cdot ))\in E
\end{equation*}
because $\zeta _0\in C_{c}^{\infty }(\Omega )$ and so $L\zeta
_0\in Y$. This will be the new initial condition for $\xi \equiv \zeta
'$ and we note that because of the regularity of $\zeta _0$ this
initial condition is in $E$ as just claimed. Thus there exists a unique
solution, $\xi $, to
\begin{gather*}
\xi '+\kappa L\xi = f_{n,1}(\cdot ,\cdot ,\zeta )
+f_{n,3}(\cdot ,\cdot ,\zeta )\xi , \\
\xi (0)=\kappa L\zeta _0+f_{n}(0,\cdot ,\zeta
_0(\cdot ))
\end{gather*}
which has the same regularity as $\zeta $. Thus
$\zeta '\in L^{\infty }(0,T;H^{2}(\Omega ))\cap L^{2}(
0,T;H^{3}(\Omega ))$ and $\zeta ''\in \mathcal{E}$.
By Theorem \ref{t6.2} this implies $\zeta '\in C(0,T;H^{r}(\Omega ))$
where $r>3/2$. Since the
dimension is no larger than 3, this shows
$t\to \zeta (t, \mathbf{x})$ is $C^{1}$.
 Now (\ref{12maye1}) and the fact just
shown that $\zeta '\in L^{\infty }(0,T;H^{2}(\Omega ))$ and
elliptic regularity shows that $\zeta \in L^{\infty}(0,T;H^{4}(\Omega ))$.
Therefore, by Theorem \ref{t6.2} this shows
$\zeta \in C([0,T];H^{q}(\Omega ))$ where $q>7/2$. It follows the
partial derivatives
of $\zeta $ up to order 2 are in $H^{r}(\Omega )$ where $r>3/2$.
Since the dimension is no larger than 3, this implies all these partial
derivatives are continuous. To summarize, $\zeta (t,\mathbf{x})$
is $C^{1}$ in $t$ and $C^{2}$ in $\mathbf{x}$. This proves the lemma.
\end{proof}

One could continue in this manner and using the Sobolev embedding theorems
obtain the solution to (\ref{12maye1}) is in
$C^{\infty }([0,T]\times \overline{\Omega })$ provided the boundary was
$C^{\infty }$ but it is not necessary. The purpose for this was only to
obtain solutions which are sufficiently smooth to carry out the estimate of
the following lemma which is based on the Hopf lemma.

\begin{lemma}\label{12mayl1}
Let $f$ satisfy \eqref{11maye2}-\eqref{11maye5} and suppose
\begin{equation}
\zeta '+\kappa L\zeta =f(\cdot ,\cdot ,\zeta ),\quad
\zeta(0)=\zeta _0\in Y  \label{12maye5}
\end{equation}
where $\zeta _0(\mathbf{x})\leq b$ and $L$ is the operator
defined above mapping $E$ to $E'$ as
\begin{equation}
\langle L\zeta ,\xi \rangle \equiv \int_{\Omega }\nabla \zeta
\cdot \nabla \xi dx  \label{2maye7a}
\end{equation}
Then \ $\zeta (t)(\mathbf{x})\leq b$ a.e. $\mathbf{x}$.
\end{lemma}

\begin{proof}
Let $\zeta _{0n}\in C_{c}^{\infty }(\Omega )$
such that $|\zeta _{0n}-\zeta _0|_{Y}\to 0$ and
$\zeta _{0n}(\mathbf{x})\leq b$. Also let $f_{n}$ be defined as
above so $f_{n}$ is $C^{\infty }$. Let $\zeta _{n}$ be the solution to
\begin{equation}
\zeta _{n}'+\kappa L\zeta _{n}=f_{n}(\cdot ,\cdot ,\zeta
_{n}),\quad  \zeta _{n}(0)=\zeta _{0n}.  \label{12maye2a}
\end{equation}
By Lemma \ref{7decl1}, $\zeta _{n}$ is $C^{1}$ in $t$ and $C^{2}$ in
$\mathbf{x}$.

First we show $\zeta _{n}\leq b$. Suppose the maximum value of $\zeta _{n}$
on $[0,T]\times \overline{\Omega }$ is achieved at
$(t_0,\mathbf{x}_0)$. If $t_0=0$ nothing else needs to be done
because it is assumed $\zeta _0\leq b$. Suppose then that $t_0>0$. If
$\zeta _{n}(t_0,\mathbf{x}_0)<b$, we are done again since
this implies what was to be shown. Suppose then that
$\zeta _{n}(t_0,\mathbf{x}_0)\geq b$. First suppose
$\mathbf{x}_0\in \Omega $.
Then by the second derivative test, it must follow
$\Delta \zeta _{n}(t_0,\mathbf{x}_0)\leq 0$.
Therefore, from (\ref{12maye1}),
\begin{equation*}
\zeta _{n}'(t_0,\mathbf{x}_0)=\kappa \Delta \zeta
_{n}(t_0,\mathbf{x}_0)+f_{n}(t_0,\mathbf{x}
_0,\zeta _{n}(t_0,\mathbf{x}_0))<-\varepsilon <0
\end{equation*}
which is a contradiction to the maximum occurring at
$(t_0,\mathbf{x}_0)$. The only remaining case to consider is
$\mathbf{x}_0\in \partial \Omega $. Here we will use Lemma \ref{11mayl1}.
Consider the
interior balls tangent to $\partial \Omega $ at $\mathbf{x}_0$. If any of
these balls has $\Delta \zeta _{n}\geq 0$ on that ball, then by Lemma
\ref{hopf} $\frac{\partial \zeta }{\partial n}(t_0,\mathbf{x}_0)>0$
which does not occur because in fact
$\frac{\partial \zeta }{\partial n}(t_0,\mathbf{x}_0)=0$.
Therefore, in every such ball, there
are points, $\mathbf{x}_1$ where
$\Delta \zeta _{n}(t_0,\mathbf{x}_1)<0$. It follows by continuity
of $f_{n}$ there is one of these
balls small enough that for $\mathbf{x}_1$ in it,
$\ f_{n}(t_0,\mathbf{x}_1,\zeta _{n}
(t_0,\mathbf{x}_1))<-\frac{\varepsilon }{2}$.
Therefore, passing to a limit, it follows
\begin{align*}
\zeta _{n}'(t_0,\mathbf{x}_0)
&=\lim_{\mathbf{x}_1\to \mathbf{x}_0}\zeta _{n}'(t_0,\mathbf{x}_1)\\
&=\lim_{\mathbf{x}_1\to \mathbf{x}_0}(\kappa \Delta \zeta
_{n}(t_0,\mathbf{x}_1)+f_{n}(t_0,\mathbf{x}
_1,\zeta _{n}(t_0,\mathbf{x}_1)))\leq -
\frac{\varepsilon }{2}<0
\end{align*}
which is another contradiction. This proves $\zeta _{n}\leq b$. In fact,
this shows $\zeta _{n}<b$ at points $(t,\mathbf{x})$ where
$t>0$.

Now consider the case where $f$ is not regularized. Using (\ref{12maye5})
and Lemma \ref{11mayl1} and letting $\delta >0$ be given, the following is
valid for all $n$ large enough.
\begin{align*}
&\frac{1}{2}|\zeta (t)-\zeta _{n}(t)|
_{Y}^{2}-\frac{1}{2}|\zeta _0-\zeta _{0n}|_{Y}^{2}
\\
&\leq \int_0^{t}((f_{n}(s,\mathbf{x},\zeta _{n})
-f(s,\mathbf{x},\zeta )),\zeta _{n}-\zeta )\\
&\leq \int_0^{t}|f_{n}(s,\mathbf{x},\zeta _{n})
-f(s,\mathbf{x},\zeta )|_{Y}|\zeta _{n}-\zeta
|ds \\
&\leq \Big(\int_0^{t}|f_{n}(s,\mathbf{x},\zeta _{n})
-f(s,\mathbf{x},\zeta )|_{Y}^{2}ds\Big)^{1/2}
\Big(\int_0^{t}|\zeta _{n}-\zeta |^{2}ds\Big)^{1/2} \\
&\leq \Big(\delta +K(\int_0^{t}|\zeta _{n}-\zeta |
_{Y}^{2}ds)^{1/2}\Big)\Big(\int_0^{t}|\zeta _{n}-\zeta
|^{2}ds\Big)^{1/2} \\
&\leq \delta ^{2}+(K^{2}+1)\int_0^{t}|\zeta _{n}-\zeta
|_{Y}^{2}ds
\end{align*}
and so by Gronwall's inequality,
\begin{equation*}
\max \{ |\zeta _{n}(t)-\zeta (t)|_{Y}:t\in [0,T]\}
\leq (|\zeta _0-\zeta _{0n}|_{Y}^{2}+2\delta ^{2})e^{2(K^{2}+1)T}.
\end{equation*}
Thus there exists an increasing sequence, $\{ n_{k}\} $ such that
\begin{equation*}
\max \{ |\zeta _{n_{k}}(t)-\zeta (t)|_{Y}:t\in [0,T]\}
 \leq (|\zeta_0-\zeta _{0n_{k}}|_{Y}^{2}
+2\frac{1}{k^{2}})e^{2(K^{2}+1)T}.
\end{equation*}
Taking a further subsequence, if necessary,
\begin{equation*}
\max \{ |\zeta _{n_{k}}(t)-\zeta _{n_{k+1}}(
t)|_{Y}:t\in [0,T]\} \leq \frac{1}{2^{k}}
\end{equation*}
and so as in the usual proof of completeness of $L^{p}$, it follows that
$\zeta _{n_{k}}(t)(\mathbf{x})\to \zeta
(t)(\mathbf{x})$ a.e. $\mathbf{x}$. But
$\zeta_{n_{k}}(t)(\mathbf{x})\leq b$ and so
$\zeta(t)(\mathbf{x})\leq b$ a.e. This proves the lemma.
\end{proof}

The next corollary involves weakening the assumption that
$f(t,\mathbf{x},\zeta )\leq -2\varepsilon $ when
$\zeta \geq b$ to $f(t,\mathbf{x},\zeta )\leq 0$ when $\zeta \geq b$.

\begin{corollary}\label{12mayc1}
Let $f$ satisfy
\begin{gather*}
f :(0,T)\times \Omega \times \mathbb{R}\to \mathbb{R},\\
|f(t,\mathbf{x},\zeta )-f(t,\mathbf{x},\xi )|\leq K|\zeta -\xi |, \\
f(t,\mathbf{x},\zeta ) \leq  0\quad \text{if }\zeta \geq b, \\
f(\cdot ,\cdot ,0) \in  L^{2}(0,T;L^{2}(\Omega)),
\end{gather*}
and suppose $\zeta _0\in Y$ is such that $\zeta _0(\mathbf{x})\leq b$.
Then the solution, $\zeta $, to
\begin{equation*}
\zeta '+\kappa L\zeta =f(\cdot ,\cdot ,\zeta ),\quad
\zeta(0)=\zeta _0
\end{equation*}
satisfies $\zeta (t)(\mathbf{x})\leq b$ a.e.
\end{corollary}

\begin{proof}
Let $\zeta _{\varepsilon }$ be the solution to
\begin{equation*}
\zeta _{\varepsilon }'+\kappa L\zeta _{\varepsilon }
=f(\cdot,\cdot ,\zeta _{\varepsilon })-\varepsilon ,\quad
\zeta _{\varepsilon}(0)=\zeta _0.
\end{equation*}
Thus from Lemma \ref{12mayl1}, $\zeta _{\varepsilon }(t)(
\mathbf{x})\leq b$ a.e. $\mathbf{x}$. Furthermore,
$\zeta_{\varepsilon }\to \zeta $ uniformly in $C([0,T];Y)$ and so as
in the above, a subsequence has the property that for
each $t$ $\zeta _{\varepsilon }(t)(\mathbf{x})
\to \zeta (t)(\mathbf{x})$ a.e. $\mathbf{x}$.
Thus $\zeta (t)(\mathbf{x})\leq b$ a.e. $\mathbf{x}$.
\end{proof}

A similar set of arguments implies the following corollary.

\begin{corollary}\label{23mayc1}
Let $f$ satisfy
\begin{gather*}
f :(0,T)\times \Omega \times \mathbb{R}\to \mathbb{R},\\
|f(t,\mathbf{x},\zeta )-f(t,\mathbf{x},\xi )
|\leq K|\zeta -\xi |, \\
f(t,\mathbf{x},\zeta )\geq 0\quad \text{if }\zeta \leq \zeta
_{\ast }<1, \\
f(\cdot ,\cdot ,0)\in L^{2}(0,T;L^{2}(\Omega)),
\end{gather*}
and suppose $\zeta _0\in Y$ is such that
$\zeta _0(\mathbf{x})\geq \zeta _{\ast }$. Then the solution, $\zeta $ to
\begin{equation*}
\zeta '+\kappa L\zeta =f(\cdot ,\cdot ,\zeta ),\quad
\zeta(0)=\zeta _0
\end{equation*}
satisfies $\zeta (t)(\mathbf{x})\geq \zeta _{\ast} $ a.e.
\end{corollary}

Next the above comparison results are used to eliminate $\eta _{\ast }$
under the assumption
\begin{equation}
\phi (\mathbf{\varepsilon },\zeta )\leq 0\quad \text{if }\zeta \geq
1,\quad
\phi (\mathbf{\varepsilon },\zeta )\geq 0\quad \text{if }\zeta
\leq \zeta _{\ast }  \label{13maye2}
\end{equation}
For example, a formula which has been proposed for $\phi $ in \cite{kut05}
is
\begin{equation}
\phi (\mathbf{\varepsilon },\zeta )=-(\frac{(1-m_{\zeta })\zeta }{
1-m_{\zeta }\zeta }(\lambda _{u}^{+}\Phi _{q^{\ast }}(\mathbf{\
\varepsilon }^{+})+\lambda _{u}^{-}\Phi _{q^{\ast }}(\mathbf{\varepsilon }
^{-}))-\lambda _{w})_{+}+H(\zeta )  \label{13maye7}
\end{equation}
in which $\mathbf{\varepsilon }^{+}$ and $\mathbf{\varepsilon }^{-}$ are the
positive and negative parts of the symmetric matrix, $\mathbf{\varepsilon }$
and $H(\zeta )$ is a Lipschitz function on $[\delta ,1]$ for
each $\delta >0$ which vanishes when $\zeta =0$ and $\lambda_{w}$
 is a positive parameter. The function $H$ represents self mending of
the material as might take place in a bone. Now letting
$f(t,\mathbf{x},\zeta )\equiv \phi (\mathbf{\varepsilon }
(\mathbf{u}(t,\mathbf{x})),\eta _{\ast }(\zeta ))$, it follows $f$
satisfies the conditions
needed for Corollary \ref{12mayc1}. If $H(\zeta _{\ast })$ is
large enough, then letting $f(t,\mathbf{x},\zeta )\equiv \phi (
\mathbf{\varepsilon }(\mathbf{u}(t,\mathbf{x})),\eta _{\ast}(\zeta ))$
it follows $f$ satisfies the conditions for Corollary \ref{23mayc1}.

Now recall Theorem \ref{22jant1} listed here for convenience.

\begin{theorem}
Let $\zeta _0\in E$ and $\mathbf{f}\in L^{\infty }(0,T;V')$.
Then there exists a unique solution to the system \eqref{k29}
and \eqref{k30} which satisfies
\begin{equation*}
\zeta '\in \mathcal{Y},\quad
L\zeta \in L^{2}(0,T;L^{2}(\Omega )),\quad
\zeta \in L^{2}(0,T;H^{2}(\Omega )),\quad
\mathbf{u}\in L^{\infty }(0,T;V).
\end{equation*}
\end{theorem}

Define $A'(\mathbf{u},\zeta )\in V'$ by
replacing every occurrence of $\eta _{\ast }(\zeta )$ with
$\zeta $ in the definition of $A(\mathbf{u},\zeta )$.

\begin{theorem}\label{23mayt2}
Let $\zeta _0\in E,\zeta _0(\mathbf{x})\in[\zeta _{\ast },1]$,
and $\mathbf{f}\in L^{\infty }(0,T;V')$. Also suppose \eqref{13maye2}.
 Then there exists a unique solution to
\begin{gather}
A'(\mathbf{u},\zeta )=\mathbf{f}\quad \text{ in }
\mathcal{V}',  \label{23maye7}
\\
\zeta '+\kappa L\zeta =\phi (\mathbf{\varepsilon }(
\mathbf{u}),\zeta ),\qquad \zeta (0)=\zeta _0,
\label{23maye8}
\\
\zeta '\in \mathcal{Y},\quad L\zeta \in L^{2}(
0,T;L^{2}(\Omega )),\quad \zeta \in L^{2}(0,T;H^{2}(
\Omega )),\quad \mathbf{u}\in L^{\infty }(0,T;V).
\label{23maye9}
\end{gather}
This solution satisfies $\zeta (t)(\mathbf{x})\in [\zeta _{\ast },1]$ a.e.
for each $t$.
\end{theorem}

\begin{proof} Letting $f(t,\mathbf{x},\zeta )\equiv \phi (
\mathbf{\varepsilon }(\mathbf{u}(t,\mathbf{x})),\eta _{\ast
}(\zeta ))$ and $(\zeta ,\mathbf{u})$ be the unique solution to
Theorem \ref{22jant1}, Corollaries \ref{12maye1} and \ref{23maye1} imply
$\zeta (t)(\mathbf{x})\in [\zeta _{\ast },1]$ for a.e. $\mathbf{x.}$
Therefore, the solution to Theorem \ref{22jant1} is the solution to
 Theorem \ref{23mayt2}.
\end{proof}

This is the main theorem of the paper. Note that it gives global solutions
to the problem of damage based on the assumption (\ref{13maye2}). If only
the fist half of (\ref{13maye2}) holds it can be shown that a local solution
to the problem in which the elastic viscoplastic part has $\zeta $ instead
of $\eta _{\ast }(\zeta )$ is obtained. This requires the proof that $\zeta $
is continuous with values in a suitable Sobolev space. It follows from
compatibility conditions on the initial data and further estimates. This has
been carried out in \cite{kut05} for the elastic case. However, for this
elastic viscoplastic model, it remains to be established.

\subsection*{Acknowledgments}
This problem was suggested to me by Professor Via\~{n}o when I\ visited the
University of Santiago de Compostela. The support for this visit was greatly
appreciated. I would also like to thank the referee who made many helpful
suggestions.

\begin{thebibliography}{99}
\bibitem{Ang1}  T. A. Angelov, \emph{On a rolling problem with damage and
wear,} Mech. Res. Comm., \textbf{26} (1999), 281--286.

\bibitem{AKRS02}  K. T. Andrews, K. L. Kuttler, M. Rochdi and M. Shillor,
\emph{One-dimensional dynamic thermoviscoelastic contact with damage}, J.
Math. Anal. Appl., \textbf{272} (2002), 249--275.

\bibitem{BF}  E. Bonetti and M. Fr\'{e}mond, \emph{Damage theory:
microscopic effects of vanishing macroscopic motions}, Comp. Appl. Math.,
\textbf{22} (3)(2003), 313--333.

\bibitem{bon04}  E. Bonetti and G. Schimperna, \emph{Local existence for
Fr\'{e}mond's model of damage in elastic materials}, Comp.Mech. Thermodyn.
16 (2004) no 4, 319-335.

\bibitem{bre73}  Brezis H, ``Op\'{e}rateurs maximaux monotones et
semigroupes de contractions dans les espaces de Hilbert," Math Studies, 5,
North Holland, 1973.

\bibitem{CFKSV}  M. Campo, J. R. Fern\'{a}ndez, K. L. Kuttler, M. Shillor
and J. M. Via\~{n}o, \emph{Numerical analysis and simulations of a dynamic
contact problem with damage}, preprint.

\bibitem{CFHS}  O. Chau, J. R. Fern\'{a}ndez, W. Han and M. Sofonea, \emph{A
frictionless contact problem for elastic-viscoplastic materials with normal
compliance and damage}, Comput. Meth. Appl. Mech. Engrg., \textbf{191}
(2002), 5007--5026.
%\bibitem{Ciarlet}  P.G. Ciarlet, The finite element method for elliptic
%problems, in: (P.G. Ciarlet and J.L. Lions Eds.), Handbook of Numerical
%Analysis, Vol. II (North Holland, 1991) 17--352.

%\bibitem{DL}  G. Duvaut and J. L. Lions, ``Inequalities in Mechanics and
%Physics,'' Springer-Verlag, Berlin, 1976.

\bibitem{eva93}  L.C. Evans \textit{Partial Differential Equations,}
Berkeley Mathematics Lecture Notes. 1993.

\bibitem{FN95}  M. Fr\'{e}mond and B. Nedjar, \emph{Damage in concrete: the
unilateral phenomenon}, Nuclear Engng. Design, \textbf{156} (1995), 323--335.

\bibitem{FN96}  M. Fr\'{e}mond and B. Nedjar, \emph{\ Damage, gradient of
damage and principle of virtual work,} Intl. J. Solids Structures, \textbf{
33 } (8) (1996), 1083--1103.

\bibitem{FR-book}  M. Fr\'{e}mond, ``Non-smooth Thermomechanics,'' Springer,
Berlin, 2002.

\bibitem{FKNS98}  M. Fr\'{e}mond, K. L. Kuttler, B. Nedjar and M. Shillor,
\emph{One-dimensional models of damage,} Adv. Math. Sci. Appl., \textbf{8}
(1998), 541--570.

\bibitem{FKS99}  M. Fr\'{e}mond, K. L. Kuttler and M. Shillor, \textit{\
Existence and uniqueness of solutions for a one-dimensional damage model,}
J. Math. Anal. Appl., \textbf{229} (1999), 271--294.

\bibitem{gri85}  P. Grisvard, ``Elliptic problems in non-smooth domains,''
Pittman, 1985.

\bibitem{HSS}  W. Han, M. Shillor and M. Sofonea, \emph{Variational and
numerical analysis of a quasistatic viscoelastic problem with normal
compliance, friction and damage,} J. Comput. Appl. Math., \textbf{137}
(2001), 377--398.

\bibitem{ion93}  I. R. Ionescu, M. Sofonea, Functional and Numerical Methods
in Viscoplasticity, Oxford University Press, Oxford, 1993.

\bibitem{HSbook}  W. Han and M. Sofonea, ``Quasistatic contact problems in
viscoelasticity and viscoplasticity,'' American Mathematical Society-Intl.
Press, 2002.

\bibitem{kut05}  K.L.Kuttler and M. Shillor, \textit{Quasistatic Material
Damage in an Elastic Body,} in preparation.

\bibitem{ksq05}  K. L. Kuttler, M. Shillor, \textit{Quasistatic Evolution of
Damage in an Elastic Body}. To appear in Nonlinear Analysis Real world
applications.

\bibitem{lio69}  J. L. Lions, ``Quelques M\'{e}thodes de R\'{e}solution des
Probl\`{e}mes aux Limites Non Lin\'{e}aires,'' Dunod, Paris, 1969.

\bibitem{SST-book}  M. Shillor, M. Sofonea, and J. J. Telega ``Models and
Analysis of Quasistatic Contact," Lecture Notes in Physics \textbf{655},
Springer, Berlin, 2004.

\bibitem{SHS-book}  M. Sofonea, W. Han, and M. Shillor, ``Analysis and
Approximation of Contact problems with Adhesion or Damage,'' Pure and
Applied Mathematics 276, Chapman and Hall/CRC Press. Boca Raton Florida,
2005.

\bibitem{sim87}  J. Simon, \emph{Compact sets in the space
$L^{p}(0,T;B)$,} Ann. Mat. Pura. Appl., \textbf{146} (1987), 65-96.

\end{thebibliography}

\end{document}