\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 69, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/69\hfil Oldroyd fluids]
{Existence of solutions for an Oldroyd model of viscoelastic fluids}

\author[G. M. de Ara\'ujo, S. B. de Menezes,  A. O. Marinho,\hfil EJDE-2009/69\hfilneg]
{Geraldo M. de Ara\'ujo, Silvano B. de Menezes, Alexandro O. Marinho}  % in alphabetical order

\address{Geraldo M. de Ara\'ujo \newline
Instituto de Matem\'atica, Universidade Federal do Par\'a,
Bel\'em-PA, Brazil} 
\email{gera@ufpa.br}

\address{Silvano B. de Menezes \newline
Departamento Matem\'atica, Universidade Federal do Cear\'a,
Fortaleza-CE, Brazil} 
\email{silvano@ufpa.br}

\address{Alexandro O. Marinho \newline
Departamento de Matem\'atica, Universidade Federal do
Piau\'i - Campus Ministro Reis Veloso, Parnaiba-PI, Brazil}
\email{marinho@ufpi.edu.br}

\thanks{Submitted October 20, 2008. Published June 1, 2009.}
\thanks{Alexandro O. Marinho is partially supported by FAPEPI-Piau\'i-Brazil}
\subjclass[2000]{74H45} 
\keywords{Oldroyd equation; variable viscosity; penalty method}

\begin{abstract}
 In this paper we investigate the unilateral problem for an
 Oldroyd model of a viscoelastic fluid. Using the penalty
 method, Faedo-Galerkin's approximation, and basic result from the
 theory of monotone operators, we establish the existence of weak
 solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

It is well know that, the motion of incompressible fluids is
described by the system of Cauchy equations
\begin{equation}\label{eqI.1}
\begin{gathered}
 \frac{\partial u}{\partial t}+u_i \frac{\partial u}{\partial x_i}
 + \nabla p=\mathop{\rm div}\sigma + f\\
 \mathop{\rm div}u=0,
\end{gathered}
\end{equation}
where $u=(u_1,\dots,u_n)$ is the velocity, $p$ is the pressure in the
fluid, $f$ is the density of external forces and $\sigma$ is the
deviator of the stress tensor, that is, $\sigma$ has the purpose of
letting us consider reactions arising in the fluid during its
motion. The vector $ (u_i\frac{\partial u_j}{\partial
x_i})$, $j=1,2,\dots,n$, is denoted by $(u.\nabla)u$. The Hooke's Law
establishes a relationship between the stress tensor $\sigma$ and
the deformation tensor
$ D_{ij}(u)=\frac12\big(\frac{\partial u_i}{\partial x_j}
+\frac{\partial u_j}{\partial x_i}\big)$ and their derivatives.
Therefore is the Hooke's Law
that establishes the type of fluid. Such relationship is also called
of \textit{rheological equation} or \textit{equation of state} (see
Serrin \cite{serin} or Clifford \cite{clifford}). For example, for
an incompressible Stokes fluid the relationship has the form
\begin{equation}\label{eqI.2}
\sigma=\alpha D + \beta D^2
\end{equation}
where $\alpha$ and $\beta$ are scalar functions. If in
\eqref{eqI.2} $\alpha=2\nu$ positive constants. and
$\beta\equiv0$ we have the Newton's Law $\sigma=2\nu D$, which
substituting in \eqref{eqI.1} we obtain the equations of motion of
Newtonian fluid, which is called the Navier-Stokes equations:
$$
u'-\nu\Delta u+(u.\nabla)u+\nabla p=f,\quad \mathop{\rm div}u=0,
$$
where $\nu$ is called the kinematic coefficient of viscosity. The
Navier-Stokes model was studied from the mathematical point of view
by Leray \cite{leray1} and later by Ladyzhenskaya \cite{lady}. We
mention other deep contributions by Lions \cite{lions1}, Temam
\cite{temam}, Tartar \cite{tartar2} and many others researchers.

The model studied in this work,  introduced by Oldroyd
\cite{oldroyd,oldroyd1}, was proposed for
viscous incompressible fluids whose defining equations have the form
\begin{equation}\label{eqI.3}
\big(1+\lambda\frac{\partial}{\partial
t}\big)\sigma=2\nu\big(1+ k\nu^{-1}\frac{\partial}{\partial t}\big)D,
\end{equation}
where $\lambda,\nu,k$ are positive constants with
$\nu-\frac{k}{\lambda}>0$. In this fluid the stress
after instantaneous cessation of the motion die out like
$e^{-\lambda^{-1}t}$, while the velocities of the flow after
instantaneous removal of the stress die out like $e^{-k^{-1}t}$.

Assuming that $\sigma(0)=0$ and $D(0)=0$, we write the relationship
$(\ref{eqI.3})$ in the form of integral equation
\begin{equation}\label{eqI.4}
\sigma(x,t)=2k\lambda^{-1}D(x,t) +
2\lambda^{-1}(\nu-k\lambda^{-1})
\int_0^te^{-\frac{(t-\xi)}{\lambda}}D(x,\xi)d\xi.
\end{equation}
Thus, the equation for the motion of Oldroyd fluid  can be written
by the system of integro-differential equations
\begin{equation}\label{eqI.10}
\frac{\partial u}{\partial t} + (u.\nabla) u
- \mu \Delta u - \int_0^t \beta(t-\xi)\Delta u(x,\xi)d\xi
+ \nabla p=f,\quad x\in \Omega,\,t>0
\end{equation}
and the incompressible condition
\[
\mathop{\rm div} u=0,\quad x\in \Omega,\;t>0,
\]
with initial and boundary conditions
\[
u(x,0)=u_0,\quad x\in \Omega,\quad \text{and}\quad u(x,t)=0,\quad
x\in \Gamma,\; t\geq
0.
\]
Here, $\mu=k\lambda^{-1}>0$ and $\beta(t)=\gamma e^{-\delta t}$,
where $\gamma=\lambda^{-1}\left(\nu - k\lambda^{-1}\right)$ with
$\delta=\lambda^{-1}$. For physical details and mathematical
modelling see \cite{oskolkov,astarita,oldroyd,wilkinson}.

 The mixed problem above was investigated by Oskolkov
\cite {oskolkov}, where he proves existence of weak solution for all
$n\in\mathbb{N}$ in certain Sobolev class.

In Br\'{e}zis \cite{brezis1} we find investigation for a unilateral
problem for the case of the Navier-Stokes equations.

In the present work we consider a unilateral problem similar to
Br\'{e}zis \cite{brezis1}, adding a memory term,  that is
$-\int_0^tg(t-\sigma)\Delta u(\sigma)d\sigma$. More
precisely, in this paper we study a unilateral problem or a
variational inequality, c.f. Lions \cite{lions1}, for the operator
$$
L=\frac{\partial u}{\partial t} +
(u.\nabla) u - \mu \Delta u - \int_0^t \beta(t-\xi)\Delta
u(x,\xi)d\xi + \nabla p - f
$$
 under standard hypothesis on $f$ and
$u_{0}$. Making use of the penalty method and Galerkin's
approximations, we establish existence and uniqueness of weak
solutions.

This work is organized as follows: In Section 2, we introduce the
notation and main results. In Section 3, we  proof to the
results. Finally, in Section 4, we prove an simple result of
uniqueness.

\section{Notation and Main Results}
\label{secnot}

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$
with the boundary $\partial \Omega$ of class $C^2$. For $T>0$, we
denote by $Q_T$ the cylinder $(0,T)\times \Omega$, with lateral
boundary $\Sigma_T=(0,T)\times \partial \Omega$. By
$\langle.,.\rangle$ we will represent the duality pairing between
$X$ and $X'$, $X'$ being the
topological dual of the space $X$, and by $C$ we denote various
positive constants.
We propose the variational inequality
\begin{equation}
\begin{gathered}
u'-\mu\Delta u+(u.\nabla)u-\int_0^tg(t-\sigma)\Delta u(\sigma)d\sigma
+ \nabla p\geq f \quad \text{in } Q_T\\
\mathop{\rm div} u=0\quad \text{  in  }Q_T\\
u=0\quad\text{on }\Sigma_T\\
u(x,0)=u_{0}(x)\quad\text{in }\Omega,
\end{gathered} \label{eq1.1.1}
\end{equation}
where $g:[0,\infty)\to [0,\infty)$ is a function of
$W^{1,1}(0,\infty)$ satisfying
\begin{gather}
\frac{\mu}{2}-2\int_0^{\infty}g(s)\,ds>0; \label{H1} \\
-C_1g\leq g'\leq -C_2g, \label{H1tilde}
\end{gather}
where $C_1, C_2, C_3$ are positive constants;
\begin{equation}
g(0)>0, \label{H1bar}
\end{equation}
As an example,  $g(s)= e^{-\frac{8}{\mu}s}$ satisfies the thhree
conditions above.

To formulate problem (\ref{eq1.1.1}) we need some notation
about Sobolev spaces. We use standard natation of $L^{2}(\Omega)$,
$L^p(\Omega)$, $W^{m,p}(\Omega)$ and $C^p(\Omega)$ for functions
that are defined on $\Omega$ and range in $\mathbb R$, and the
notation $L^{2}(\Omega)^n$, $L^p(\Omega)^n$, $W^{m,p}(\Omega)^n$ and
$C^p(\Omega)^n$ for functions that range in $\mathbb R^n$. Besides,
we work also with the spaces $L^p(0,T; H^m(\Omega))$ or $L^p(0,T;
H^m(\Omega))^n$. To complete this recall on functional spaces, see for
instance, Lions \cite{lions1}.

Also we define the following spaces
$$
\mathcal{V}=\{\varphi\in \mathcal{D}(\Omega)^n:
 \mathop{\rm div} \varphi=0\},
$$
$V=V(\Omega)$ is the closure of $\mathcal{V}$ in the space $H_0^1(\Omega)^n$
with inner product and norm denoted, respectively by
$$
((u,z))=\sum^{n}_{i,j=1}\int_{\Omega}\frac{\partial u_i}
{\partial x_j}(x)\frac{\partial z_i}{\partial x_j}(x)\, dx,\quad
\|u\|^2=\sum^{n}_{i,j=1}\int_{\Omega}\Big(\frac{\partial
u_i}{\partial x_j}(x)\Big)^2 dx,
$$
$H=H(\Omega)$ is the closure of $\mathcal{V}$ in the space
$L^2(\Omega)^n$ with inner product and norm defined, respectively, by
$$
(u,v)=\sum^{n}_{i=1}\int_{\Omega}{u_i(x)v_i(x)}\,dx, \quad
|u|^2=\sum^{n}_{i=1}\int_{\Omega}{|u_i(x)|}^2\,dx
$$
and $V_2$ is the closure of $\mathcal{V}$ in $H^2(\Omega)^n$
with inner product and norm denoted, respectively by
$$
((u,z))_{V_2}=\sum^{n}_{i=1}(u_i,v_i)_{H^2(\Omega)},\quad
\|u\|^2_{V_2}=((u,u))_{V_2},
$$


\begin{remark} \label{rmk2.1} \rm %\label{rem1}
$V$, $H$ and $V_2$ are Hilbert's spaces, $V_2\hookrightarrow
V\hookrightarrow H\hookrightarrow V'$ with embedding dense and
continuous.
\end{remark}

Let $K$ be a closed and convex subset of $V\cap V_2$ with $0\in K$.
We introduce the following bilinear and the trilinear forms:
\begin{gather*}
 a(u,v)=\sum^{n}_{i,j=1}\int_{\Omega}\frac{\partial
u_i}{\partial x_j}(x)\frac{\partial v_i}{\partial
x_j}(x)\,dx=((u,v)),
\\
b(u,v,w)=\sum^{n}_{i,j=1}\int_{\Omega}u_i(x)\frac{\partial
v_j}{\partial x_i}(x)w_j(x)\,dx,
\end{gather*}
We also assume that
\begin{equation} \label{H2}
a(v,v)+b(v,\varphi,v)+\int_0^tg(t-\sigma)((v,v))d\sigma \geq0\quad
\forall \varphi\in K,\; \forall v\in V.
\end{equation}
Next we shall state the main results of this paper.

\begin{theorem} \label{thm1}
If $f\in L^{2}(0,T;H)$ and hypotheses \eqref{H2} holds, then there
exists a function $u$ such that
\begin{gather}
u\in L^{2}(0,T;V)\cap L^{\infty}(0,T;H) \label{eq1}\\
u(t)\in K\quad \text{a.e.} \label{eq2}
\\
\begin{aligned}
&\int_0^T\langle\varphi',\varphi-u\rangle+
\mu a(u,\varphi-u)+b(u,u,\varphi-u)\\
&-\Big(\int_0^tg(t-\sigma)\Delta
u(\sigma)d\sigma,\varphi-u\Big)dt\\
&\geq \int^T_0\langle
f,\varphi-u\rangle\,dt,\quad
\forall \varphi \in L^2(0,T;V),\;
 \varphi' \in L^{2}(0,T;V'),
\end{aligned}\label{eq3} \\
 \varphi(0) =0,\quad \varphi(t)\in K\text{ a.e.} \notag \\
u(0)=u_0. \notag
\end{gather}
\end{theorem}

\begin{theorem} \label{thm2} %\label{thm2.2}
Assumption \eqref{H2}, $n=2$, and
\begin{gather}
f\in L^2(0,T;V),\quad  f'\in L^2(0,T;V') \label{eq4}\\
u_0\in K. \label{eq5}
\end{gather}
Suppose also that
\begin{equation}
 (f(0),v)-\mu a(u_0,v)-b(u_0,u_0,v)
  =(u_1,v)\quad \text{for all $v\in V$  some $u_1\in V$.}
 \label{eq6}
\end{equation}
Then there exists a unique function $u$ such that
\begin{gather}\label{eq7}
u\in L^2(0,T;V\cap V_2), \quad u'\in L^2(0,T;V)\cap L^{\infty}(0,T;H) \\
\label{eq8}
u(t)\in K,\quad \forall t\in[0,T]
\\
\begin{aligned}
&(u'(t),v-u(t))+\mu a(u(t),v-u(t))+ b(u(t),u(t),v-u(t)) \\
&+ \int^T_0\int_0^tg(t-\sigma)((u(\sigma),v-u(t)))d\sigma dt \\
&\geq(f(t),v-u(t))\quad \forall  v\in K, \text{ a.e. in }t,
\end{aligned} \label{eq9}
\\
u(0)=u_0. \label{eq10}
\end{gather}
\end{theorem}

The proof of Theorems \ref{thm1} and \ref{thm2} will be given
in Section \ref{secdem} by the penalty method. It consists in
considering a perturbation of the operator $L$ adding a singular
term called penalty, depending on a parameter $\epsilon>0$. We solve
the mixed problem in $Q$ for the penalized operator and the
estimates obtained for the local solution of the penalized equation,
allow to pass to limits, when $\epsilon$ goes to zero, in order to
obtain a function $u$ which is the solution of our problem.

First of all, let us consider the penalty operator
$\beta:V\to V'$ associated to the closed convex set $K$,
c.f. Lions \cite[p. 370]{lions1}. The operator $\beta$ is monotonous,
hemicontinuous, takes bounded sets of $V$ into bounded sets of $V'$,
its kernel is $K$ and $\beta:L^2(0,T;V)\to L^2(0,T;V')$
is equally monotone and hemicontinous. The penalized problem
associated with the variational inequalities (\ref{eq3}) and
(\ref{eq9}) consists in, given $0<\epsilon<1$, find $u_{\epsilon}$
satisfying
\begin{equation}
\begin{gathered}
(u_{\epsilon}',v)+\mu a(u_{\epsilon},v)+
b(u_{\epsilon},u_{\epsilon},v)-\int_0^tg(t-\sigma)(\Delta
u_{\epsilon}(\sigma),v)d\sigma
+\frac{1}{\epsilon}(\beta (u_{\epsilon}),v)=(f,v),\\\
\forall\, v\in V, \quad
u_{\epsilon}\in L^2(0,T;V),\quad
u_{\epsilon}'\in L^2(0,T;V')\\
u_{\epsilon}(x,0)=u_{{\epsilon}_{0}}(x).
\end{gathered} \label{eq11}
\end{equation}
We suppose $n=2$. The solution of this problem is given by the
followings theorems.

\begin{theorem}  \label{thm3}
If $f\in L^{2}(0,T;H)$ and hypotheses \eqref{H1} holds, then, for each
$0<\epsilon <1$ and $u_{\epsilon_ 0}\in H$, there exists a function
$u_{\epsilon}$ with
$u_{\epsilon}\in L^2(0,T;V)\cap L^{\infty}(0,T;H)$,
$u_{\epsilon}'\in L^2(0,T;V')$ solution of \eqref{eq11}.
\end{theorem}

\begin{theorem} \label{thm4}
If $f\in L^2(0,T;V)$ and $f'\in L^{2}(0,T;V')$ and hypotheses
\eqref{H1} holds, then for each $0<\epsilon<1 $ and $u_{\epsilon_ 0}\in V$,
there exists a function $u_{\epsilon}$ with $u_{\epsilon}\in
L^{\infty}(0,T;V\cap V_2)$,
$u_{\epsilon}'\in L^2(0,T;V)\cap L^{\infty}(0,T;H)$ satisfying
\eqref{eq11}.
\end{theorem}

\section{Proof of the Results}\label{secdem}

\subsection*{Proof of Theorem \ref{thm1}}

We first prove  Theorem \ref{thm3} for the penalized problem.
We employ the Faedo-Galerkin method. We note
that the embedding
$V{\hookrightarrow}V\stackrel{\rm comp}{\hookrightarrow}H\hookrightarrow
V'$ are continuous and dense and that $V$ is compactly and densely
embedded in $H$. Let $\{w_\nu, \lambda_\nu \}$,
$\nu \in \mathbb{N}$, be solutions of the spectral problem
\begin{equation}
((w,v))=\lambda(w,v),\quad \forall v\in V.
\label{eq3.1}
\end{equation}

We consider $(w_\nu)_{\nu \in \mathbb{N}}$ a Hilbertian basis for
Faedo-Galerkin method. We represent by $V_m=[w_1,w_2,\dots,w_m]$ the
 $V$ subspace generated by the vectors $w_1,w_2,\dots,w_m$ and let
us consider
$$
u_{{\epsilon}_{m}}(t)=\sum^{m}_{j=1}{g_{j}}_{m}(t)w_j
$$
solution of approximate problem
\begin{equation}
\begin{gathered}
\begin{aligned}
&(u'_{{\epsilon}_{ m}},w_j)+\mu
a(u_{{\epsilon}_{ m}},w_j)+ b(u_{{\epsilon}_{ m}},u_{{\epsilon}_{ m}},w_j)\\
&-\int_0^tg(t-\sigma)(\Delta
u_{{\epsilon}_{m}}(\sigma),v)d\sigma+\frac{1}{\epsilon}\langle\beta
u_{{\epsilon}_{ m}},w_j\rangle\\
&= \langle f(t),w_j\rangle,\quad j=1,2,\dots m
\end{aligned}\\
 u_{{\epsilon}_{ m}}(x,0)\to u_{\epsilon}(x,0)\quad
 \text{strongly in }V.
\end{gathered} \label{eq3.2}
\end{equation}
This system of ordinary differential equations has a
solution on a interval $[0,t_m[$, $0<t_m<T$. The first estimate
permits us to extend this solution to the whole
 interval $[0,T]$.

\begin{remark} \label{rmk3.1}  \rm
To obtain a better notation, we omit the parameter $\epsilon$ in the
approximate solutions.
\end{remark}

\subsection*{First estimate}
Multiplying both sides of (\ref{eq3.2}) by $g_{j}$ and adding from
$j=1$ to $j=m$, we obtain
$$
\frac{1}{2}\frac{d}{dt}|u_m(t)|^2+\mu\|u_m{(t)}\|^2
+\int^t_0g(t-\sigma)(\nabla u_m(\sigma),\nabla u_m{(t)})d\sigma
=(f(t),u_m(t)),
$$
since $b(u_m,u_m,u_m)=0$ (see Lions \cite{lions1}) and\\
$(\beta u_m(t),u_m(t))\geq0$ because $\beta$ is monotone and $0\in
K$. Its follows that
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|u_m(t)|^2+\mu\|u_m{(t)}\|^2\\
&\leq\Big|\int_{\Omega}\nabla
u_m(x,t)\Big(\int_0^tg(s-\sigma)\nabla u_m(x,\sigma)\Big)d\sigma
dx\Big|_{\mathbb{R}}\\
&\quad+ |f(t)||u_m(t)| +
\int_{\Omega}|\nabla u_m(x,t)|_{\mathbb{R}}|g\ast\nabla
u_m(x,t)|_{\mathbb{R}}dx,
\end{aligned}
\label{eq3.3}
\end{equation}
where $\ast$ denotes the convolution in $t$.
It follows from (\ref{eq3.3}) that
\begin{align*}
&\frac{d}{dt}|u_m(t)|^2+2\mu\|u_m{(t)}\|^2 \\
& \leq 2 \int_{\Omega}|\nabla u_m(x,t)|_{\mathbb{R}} |g\ast\nabla
u_m(x,t)|_{\mathbb{R}}\,dx+2|f(t)|C\|u_m(t)\|\\
&=2\int_{\Omega}|\nabla u_m(x,t)|_{\mathbb{R}}|g\ast\nabla
u_m(x,t)|_{\mathbb{R}}\,dx
+ 2\sqrt{\frac{2}{3\mu}}C|f(t)|\sqrt{\frac{3\mu}{2}}\|u_m(t)\|\\
&=2\int_{\Omega}|\nabla u_m(x,t)|_{\mathbb{R}} |g\ast\nabla
u_m(x,t)|_{\mathbb{R}}dx
+ \frac{3\mu}{2}\|u_m(t)\|^2+\frac{2}{3\mu}C^2|f(t)|^2.
\end{align*}


\begin{remark} \label{rmk3.2} \rm
We note that from Cauchy-Schwarz inequality and Fubini's theorem we
have
$$
\|g*\nabla u_m\|_{L^2(Q)}\leq \|g\|_{L^1(0;\infty)}\|\nabla
u_m\|_{L^2(Q)}.
$$
\end{remark}

Thus, integrating $0$ to $t$ the inequality above, using the
Remark \ref{rmk3.2} and using Gronwall's inequality we obtain
$$
|u_m|_{L^{\infty}(0,T;H)}^2+
\left(\frac{\mu}{2}-2\|g\|_{L^1(0,\infty)}\right)\|u_m{}\|_{L^{2}(0,T;V)}^2\leq\frac{2}{3\mu}
C + C^2|f|_{L^{2}(0,T;H)}^2.
$$
Integrating these last inequality in $t\in[0,T]$ and using
\eqref{H1}, we have
\begin{gather}
u_m\quad \text{is bounded in }L^\infty(0,T;H)\label{eq3.5}\\
u_m\quad \text{is bounded in }L^2(0,T;V).\label{eq3.6}
\end{gather}
 From (\ref{eq3.6}), we obtain
\begin{equation}
\beta(u_m)\quad \text{is bounded in }L^2(0,T;V')\label{eq3.6.1}
\end{equation}

\subsection*{Second estimate}
By Remark \ref{rmk3.2}, we observe that,
\begin{equation}\label{eq3.7}
\text{if}\quad  \xi\in
L^2(0,T;H)\quad \text{then}\quad
\int_0^tg(t-\sigma)\xi(\sigma)d\sigma\in L^2(0,T;H).
\end{equation}
Similarly we obtain
\begin{gather}
\int_0^tg(t-\sigma)\xi(\sigma)d\sigma\in V\quad \text{ if }
 \xi(t)\in V\,,\label{eq3.8}
\\
\int_0^tg(t-\sigma)\xi(\sigma)d\sigma\in
V'\quad \text {if } \xi(t)\in V'\,.\label{eq3.8.1}
\end{gather}
 We consider
$\tilde{u}_m=u_m$, $\tilde{w}=w$ in $[0,T]$ and
$\tilde{u}_m=0$, $\tilde{w}=0$ out of $[0,T]$,
 $\widetilde{g}(\xi) =g(\xi)$ if $\xi\geq 0$ and zero if $\xi<0$.
Therefore, $\nabla \widetilde{u}_m\in L^2(\mathbb{R};H)$,
$\widetilde{w}\in L^2(\mathbb{R};V)$ and
 $\widetilde{g}\in L^1(\mathbb{R})$. This
implies
\begin{align*}
&\int_0^T\int_0^tg(t-\sigma)\left((u_m(\sigma),w(t))\right)\,d\sigma\,dt
\\
&=\int_{\mathbb{R}}\int_{\mathbb{R}}\widetilde{g}(t-\sigma)
\int_{\Omega}\nabla \widetilde{u}_m(x,\sigma)\nabla \widetilde{w}(x,t)
\,dx\,d\sigma \,dt\\
&= \int_{\mathbb{R}}\int_{\Omega}\widetilde{g}*\nabla\widetilde{u}_m(x,t)
\nabla \widetilde{w}(x,t)\,dx\,\,dt\\
&=\int_{\mathbb{R}}\int_{\Omega}\nabla
\widetilde{u}_m(x,\sigma)\widetilde{\breve{g}}*\nabla
\widetilde{w}(x,\sigma)\,dx\,d\sigma,
\end{align*}
where $\widetilde{\breve{g}}(x)=\widetilde{g}(-x)$. We observe that
(\ref{eq3.6}) implies that
\begin{equation}\label{eq3.8b}
 \int_0^T ((u_m(t),w))dt\to\int_0^T
((u(t),w))dt,\,\,\,\forall\,\,w\in L^{2}(0,T;V).
\end{equation}
From (\ref{eq3.7}), we have that $\widetilde{g}*\widetilde{w}(t)\in
V,\,\,\forall\,w\in L^2(0,T;V)$, therefore \eqref{eq3.8b} yield
$$
\int_{\mathbb{R}}\left(\nabla
\widetilde{u}_m(\sigma),\widetilde{\breve{g}}*\nabla\widetilde{w}(\sigma)
\right)dt\to \int_{\mathbb{R}}\left(\nabla
\widetilde{u}(\sigma),\widetilde{\breve{g}}*\nabla\widetilde{w}(\sigma)
\right)dt.
$$
We note that
\begin{align*}
\int_{\mathbb{R}}\left(\nabla
\widetilde{u}(\sigma),\widetilde{\breve{g}}*\nabla\widetilde{w}
(\sigma)\right)\,dt
&= \int_{\mathbb{R}}\left(\widetilde{g}*\nabla
\widetilde{u}(\sigma),\nabla\widetilde{w}(\sigma)\right)\,dt\\
&= \int_{\Omega}\int_{\mathbb{R}}\int_{\mathbb{R}}\widetilde{g}
(t-\sigma)\nabla \widetilde{u}(x,\sigma)d\sigma\nabla
\widetilde{w}(x,t)d\sigma\,dt\,dx\\
&= \int_0^T\int_0^tg(t-\sigma)\left((u(\sigma),w(t))\right)d\sigma
dt\,.
\end{align*}
Then
\begin{equation}
\int_0^T\int_0^tg(t-\sigma)\left((u_m(\sigma),w(t))\right)\,d\sigma\,dt
\to \int_0^T\int_0^tg(t-\sigma)\left((u(\sigma),w(t))\right)d\sigma
dt, \label{eq3.9}
\end{equation}
for all $w\in L^2(0,T;V)$.

Let $P_m$ be the orthogonal projection $H\mapsto V_m$; that is,
$$
P_m\varphi=\sum^{m}_{j=1}(\varphi,w_j)w_j,\quad \varphi\in H.
$$
By the  choice of $(w_{\nu})_{\nu \in \mathbb{N}}$ we have
$$
\|P_m\|_{\mathcal{L}(V,V)}\leq1\quad\text{and}\quad
 \|P_m^*\|_{\mathcal{L}(V',V')}\leq 1.
$$
We note that $P_mu_m'=u_m'$. Multiplying both sides of the
approximate equation (\ref{eq3.2}) by the vector $w_j$ and adding
from $j=1 \text{ to }j=m$, we obtain using the notations and ideas
of Lions \cite[pages 75-76]{lions1} and (\ref{eq3.7}), that
\begin{equation}
(u'_m)\quad \text{is bounded in } L^2(0,T;{V'}).
\label{eq3.10}
\end{equation}
The boundedness in (\ref{eq3.6}), (\ref{eq3.10}) and the Aubin-Lions
compactness Theorem imply that there exists a subsequence from
$(u_m)$, still denoted by $(u_m)$, such that
\begin{equation}
u_m\to u \quad \text{strongly in  $L^2(0,T;H)$
  and a. e. in $Q$}. \label{eq3.11}
\end{equation}
 Returning to the notation $u_{{\epsilon}_m}$, using
(\ref{eq3.5}), (\ref{eq3.6}) and (\ref{eq3.11}) (see Lions
\cite[pages 76-77]{lions1}),  (\ref{eq3.6.1}) and \eqref{eq3.9} we
obtain
\begin{equation}
\begin{gathered}
(u_{\epsilon}',v)+a(u_{\epsilon},v)+
b(u_{\epsilon},u_{\epsilon},v)-\int_0^tg(t-\sigma)(\Delta
u(\sigma),v)d\sigma
 +\frac{1}{\epsilon}(\zeta,v)=(f,v),\\
\forall v\in V, \quad  u_{\epsilon}\in L^2(0,T;V),\quad u_{\epsilon}'\in L^2(0,T;V')\\
 u_{\epsilon}(x,0)=u_{{\epsilon}_{0}}(x).
\end{gathered} \label{eq3.11.1}
\end{equation}

It is necessary to prove that $\zeta=\beta (u_{\epsilon})$. We make
this using the monotony of the operator $\beta$ (see Lions
\cite[Chap. 2]{lions1}). Therefore, we have proved the Theorem
\ref{thm3}.


\subsection*{Proof of Theorem \ref{thm1}}
 From (\ref{eq3.5}), (\ref{eq3.6}), (\ref{eq3.11}) and
Banach-Steinhauss theorem, it follows that there exists a subnet
$ (u_{\epsilon})_{0<\epsilon <1}$, such that it
converges to $u$ as $\epsilon \to 0$, in the weak sense . This
function satisfies (\ref{eq1}). On the other hand, we have from
\eqref{eq3.11.1} that
\begin{equation}
\beta u_{\epsilon}=\epsilon[f-u_{\epsilon}'-Au_{\epsilon}
-Bu_{\epsilon}-\int_0^tg(t-\sigma)\Delta
u(\sigma)d\sigma].  \label{eq3.26}
\end{equation}
Where $\langle Au_{\epsilon},v\rangle=a(u,v)$ and
$\langle Bu_{\epsilon},v\rangle=b(u_{\epsilon},u{_\epsilon},v)$.

Since $\int_0^tg(t-\sigma)\Delta u(\sigma)d\sigma\in\,V'$ and
$[f-u_{\epsilon}'-Au_{\epsilon}-Bu_{\epsilon}]$ is bounded, we have
\begin{equation}
\beta u_{\epsilon}\to0\quad \text{in  }\mathcal{D}'(0,T;V'). \label{eq3.27}
\end{equation}
Since
$\beta u_{\epsilon}$  is bounded in  $L^2(0,T;V')$, %\label{eq3.26-extra}
we have
\begin{equation}
\beta u_{\epsilon}\to0\;\text{  weak in }\;L^2(0,T;V').
\label{eq3.30}
\end{equation}
On the other hand we deduce from \eqref{eq3.11.1} that
\begin{equation}
 0\leq \int^T_0\langle\beta
u_{\epsilon},u_{\epsilon}\rangle\,dt\leq\epsilon\; C. \label{eq3.31}
\end{equation}
Thus
$\int^T_0\langle\beta u_{\epsilon},u_{\epsilon}\rangle
dt\to0$.  %\label{eq3.32}
We have that
$$
\int^T_0\langle\beta u_{\epsilon}-\beta\varphi,u_{\epsilon}-\varphi\rangle\,dt
\geq0,\quad \forall\varphi\text{in  } L^2(0,T;V),
$$
 because $\beta$
is a monotonous operator. Thus,
\begin{equation}
\int^T_0\langle\beta
u_{\epsilon},u_{\epsilon}\rangle\,dt-\int^T_0\langle\beta
u_{\epsilon},\varphi\rangle\,dt
-\int^T_0\langle\beta\varphi,u_{\epsilon}-\varphi\rangle\,dt\geq0.
\label{3.31extra}
\end{equation}
We have from (\ref{eq3.30}) and
(\ref{3.31extra}) that
\begin{equation}
\int^T_0\langle\beta\varphi,u(t)-\varphi\rangle\,dt\leq0.
\label{eq3.33}
\end{equation}
Taking $\varphi=u-\lambda v$, with $v\in L^2(0,T;V)$ and
$\lambda>0$, we deduce using the hemicontinuity of $\beta$ that
\begin{equation}
\beta (u(t))=0, \label{eq3.34}
\end{equation}
and this implies that $u(t)\in K$  a. e.

Next, we prove that $u$ is a solution of inequality (\ref{eq3}). Let
us consider $\mathbf{X}_{\epsilon}$ defined by
\begin{equation}
\begin{aligned}
\mathbf{X}_{\epsilon}
&=\int^T_0\langle\varphi',\varphi-u_{\epsilon}\rangle\,dt+
\int^T_0a(u_{\epsilon},\varphi-u_{\epsilon})\,dt
+ \int^T_0b(u_{\epsilon},u_{\epsilon},\varphi-u_{\epsilon})\,dt\\
&\quad +\int_0^T\int_0^tg(t-\sigma)
((u_{\epsilon}(\sigma),\varphi-u_{\epsilon}))d\sigma\,dt,
 - \int^T_0\langle f,\varphi-u_{\epsilon}\rangle\,dt,
\end{aligned} \label{eq3.43}
\end{equation}
with $\varphi \in L^2(0,T;V), \varphi' \in L^2(0,T;V'),\varphi(0)=0$,
$\varphi(t)\in K$ a.e.
It follows from (\ref{eq3.43}) that
\begin{equation}
\begin{aligned}
\mathbf{X}_{\epsilon}
&=\int^T_0\langle\varphi',\varphi\rangle\,dt
  -\int^T_0\langle\varphi',u_{\epsilon}\rangle\,dt
  +\int^T_0a(u_{\epsilon},\varphi)\,dt
  -\int^T_0a(u_{\epsilon},u_{\epsilon})\,dt\\
&\quad +\int^T_0b(u_{\epsilon},u_{\epsilon},\varphi)\,dt
  -\int^T_0b(u_{\epsilon},u_{\epsilon},u_{\epsilon})dt
 +\int_0^T\int_0^tg(t-\sigma)
 ((u_{\epsilon}(\sigma),\varphi))d\sigma\,dt\\
&\quad - \int_0^T\int_0^tg(t-\sigma)
 ((u_{\epsilon}(\sigma),u_{\epsilon}))d\sigma\,dt
 -\int^T_0\langle f,\varphi\rangle\,dt
 +\int^T_0\langle f,u_{\epsilon}\rangle\,dt.
\end{aligned}\label{eq3.44}
\end{equation}
On the other hand, taking $v=\varphi-u_{\epsilon}$ in \eqref{eq11}
and integrating in $Q_T$, we obtain that
\begin{equation}
\begin{aligned}
&-\int^T_0\langle u_{\epsilon}',\varphi\rangle\,dt
 +\int^T_0\langle u_{\epsilon}',u_{\epsilon}\rangle\,dt
 -\int^T_0a(u_{\epsilon},\varphi)\,dt
 +\int^T_0a(u_{\epsilon},u_{\epsilon})\,dt\\
&-\int^T_0b(u_{\epsilon},u_{\epsilon},\varphi)\,dt
 +\int^T_0b(u_{\epsilon},u_{\epsilon},u_{\epsilon})dt
 -\int_0^T\int_0^tg(t-\sigma) ((u_{\epsilon}(\sigma),\varphi))d\sigma\,dt\\
&+\int_0^T\int_0^tg(t-\sigma)((u_{\epsilon}(\sigma),u_{\epsilon}))  d\sigma\,dt
 -\frac{1}{\epsilon}\int^T_0\langle\beta u_{\epsilon}
 -\beta\varphi,\varphi-u_{\epsilon}\rangle\,dt\\
& +\int^T_0\langle f,\varphi\rangle\,dt
  -\int^T_0\langle f,u_{\epsilon}\rangle\,dt=0,
\end{aligned} \label{eq3.45}
\end{equation}
because $\beta\varphi=0$. Adding member to member (\ref{eq3.44}) and
(\ref{eq3.45}), we obtain
\begin{equation}
\begin{aligned}
\mathbf{X}_{\epsilon}
&= \int^T_0\langle\varphi',\varphi\rangle\,dt-\int^T_0\langle\varphi',u_{\epsilon}\rangle\,dt-\int^T_0\langle
u_{\epsilon}',\varphi\rangle\,dt\\
&\quad +\int^T_0\langle
u_{\epsilon}',u_{\epsilon}\rangle\,dt+\frac{1}{\epsilon}\int^T_0\langle\beta\varphi-\beta
u_{\epsilon},\varphi-u_{\epsilon}\rangle\,dt\geq0,
\end{aligned} \label{eq3.46}
\end{equation}
because
\begin{align*}
&\int^T_0\langle\varphi',\varphi\rangle\,dt
-\int^T_0\langle\varphi',u_{\epsilon}\rangle\,dt
-\int^T_0\langle u_{\epsilon}',\varphi\rangle\,dt
+\int^T_0\langle u_{\epsilon}',u_{\epsilon}\rangle\,dt\\
& =\int^T_0\langle\varphi'-u_{\epsilon}',\varphi-u_{\epsilon}\rangle\geq0.
\end{align*}
On the other hand, $b(u_{\epsilon},u_{\epsilon},u_{\epsilon})=0$.
 From (\ref{eq3.43})-(\ref{eq3.44}) it follows that
\begin{equation}
\begin{aligned}
\mathbf{X}_{\epsilon}
&= \int^T_0\langle\varphi',\varphi-u_{\epsilon}\rangle\,dt+
\int^T_0\!\!a(u_{\epsilon},\varphi)\,dt\\
&\quad + \int^T_0\int_0^tg(t-\sigma)((u_{\epsilon},\varphi))
d\sigma\,dt-\int^T_0\langle
f,\varphi-u_{\epsilon}\rangle\,dt\\
&\geq \int^T_0a(u_{\epsilon},u_{\epsilon})dt
 +\int^T_0b(u_{\epsilon},\varphi,u_{\epsilon})\,dt
 + \int^T_0\int_0^tg(t-\sigma)((u_{\epsilon},u_{\epsilon}))
d\sigma\,dt.
\end{aligned} \label{eq3.47}
\end{equation}
Consider
\begin{equation}
\mathbf{Y}_{\epsilon}
= \int^T_0a(u_{\epsilon},u_{\epsilon})dt+\int^T_0b(u_{\epsilon},
\varphi,u_{\epsilon})\,dt
+ \int_0^T\int_0^tg(t-\sigma)((u_{\epsilon}
,u_{\epsilon}))d\sigma\,dt. . \label{eq3.48}
\end{equation}
It follows from \eqref{H2} with $v=u-u_{\epsilon}$ that
$$
a(u-u_{\epsilon},u-u_{\epsilon})+b(u-u_{\epsilon},
\varphi,u-u_{\epsilon})+\int_0^tg(t-\sigma)((u-u_{\epsilon}
,u-u_{\epsilon}))d\sigma \geq0.
$$
 On the other hand, we can write
\begin{align*}
\mathbf{Y}_{\epsilon}
&=\int^T_0a(u_{\epsilon}-u,u_{\epsilon}-u)\,dt
 +\int^T_0b(u_{\epsilon}-u,\varphi,u_{ \epsilon}-u)\,dt\\
&\quad +\int^T_0a(u,u_{\epsilon}-u)\,dt
 +\int^T_0a(u_{\epsilon},u)\,dt+\int^T_0b(u,\varphi,u_{\epsilon}-u)\,dt\\
&\quad +\int^T_0b(u_{\epsilon} ,\varphi,u)\,dt
 +\int_0^T\int_0^tg(t-\sigma)((u_{\epsilon}-u ,u_{\epsilon}-u))d\sigma\,dt\\
&\quad +\int_0^T\int_0^tg(t-\sigma)((u,u_{\epsilon}-u))d\sigma\,dt
 +\int_0^T\int_0^tg(t-\sigma)((u_{\epsilon},u))d\sigma\,dt.
\end{align*}
 This implies
\begin{equation}
\begin{aligned}
\mathbf{Y}_{\epsilon}
&\geq \int^T_0a(u_{\epsilon},u)\,dt
 +\int^T_0a(u,u_{\epsilon}-u)\,dt
 + \int^T_0b(u,\varphi,u_{\epsilon}-u)\,dt \\
&\quad +\int^T_0b(u_{\epsilon},\varphi,u)\,dt
 + \int_0^T\int_0^tg(t-\sigma)((u ,u_{\epsilon}-u))d\sigma\,dt\\
&\quad + \int_0^T\int_0^tg(t-\sigma)((u_{\epsilon} ,u))d\sigma\,dt.
\end{aligned}\label{eq3.49}
\end{equation}
Taking $\limsup$ in (\ref{eq3.49}) we obtain
\begin{equation}
\limsup\mathbf{Y}_{\epsilon}\geq
\int^T_0a(u,u)\,dt+\int^T_0b(u,\varphi,u) \,dt
+ \int_0^T\int_0^tg(t-\sigma)((u,u))d\sigma\,dt.
 \label{eq3.50}
\end{equation}
It follows from (\ref{eq3.47}) and (\ref{eq3.50}) that
\begin{equation}
\begin{aligned}
&\limsup\Big\{\int^T_0\langle\varphi',\varphi-u_{\epsilon}\rangle\,dt
 +\int^T_0a(u_{\epsilon},\varphi)\,dt \\
&+\int_0^T\int_0^tg(t-\sigma)((u_{\epsilon}
,u))d\sigma\,dt-\int^T_0\langle f,\varphi-u_{\epsilon}\rangle\,dt\Big\}\\
& \geq\int^T_0a(u,u)\,dt+\int^T_0b(u,\varphi,u)\,dt
 + \int_0^T\int_0^tg(t-\sigma)((u ,u))d\sigma\,dt.
\end{aligned} \label{eq3.51}
\end{equation}
It follows from (\ref{eq3.51}) that
\begin{align*}
&\int^T_0\langle\varphi',\varphi-u\rangle\,dt
 +\int^T_0a(u,\varphi-u)\,dt+\int_0^Tb(u,u,\varphi-u)\,dt\\
&+\int_0^T\int_0^tg(t-\sigma)((u,\varphi-u))d\sigma\,dt\\
&\geq\int^T_0\langle f,\varphi-u\rangle\,dt
\end{align*}
for all $\varphi\in L^2(0,T;V)$, $\varphi'\in L^2(0,T;V')$,
$\varphi(0)=0$, $\varphi(t)\in K$ a.e.


\subsection*{Proof of Theorem \ref{thm2}}
We first prove Theorem \ref{thm4} for the penalized problem.
As in the proof of Theorem \ref{thm1}, we employ the Faedo-Galerkin Method.
Let $(w_{\nu})_{\nu \in \mathbb{N}}$ be a Hilbertian basis of $V$. By
$V_m=[w_1,w_2,\dots w_m]$ we represent the subspace generated by the m
first vectors of $(w_{\nu})$. Consider
$$
u_{{\epsilon}_{m} }=\sum _{j=1} ^{m} {g_{j}}_{m}w_j
$$
 solution of
approximate penalized problem
\begin{equation}
\begin{gathered}
\begin{aligned}
&(u'_{{\epsilon}_{ m}},w_j)
+\mu a(u_{{\epsilon}_{ m}},w_j)
+ b(u_{{\epsilon}_{ m}},u_{{\epsilon}_{m}},w_j)\\
&\quad -\int_0^tg(t-\sigma)(\Delta u_{\epsilon m}(\sigma),v)d\sigma
 +\frac{1}{\epsilon}\langle\beta u_{{\epsilon}_{ m}},w_j\rangle\\
& = \langle f(t),w_j\rangle,\quad j=1,2,\dots m
\end{aligned}\\
 u_{{\epsilon}_{ m}}(x,0)\to u_{\epsilon}(x,0)\quad
\text{strongly in } V.
\end{gathered} \label{eq3.53}
\end{equation}


\subsection*{First estimate}
 As in the proof of Theorem \ref{thm3}, omitting the parameter
$\epsilon$ and taking $v=u_{m}$ in the approximate equation
(\ref{eq3.53}) we obtain
\begin{gather}
(u_m)\quad \text{is bounded in  }L^{\infty}(0,T;H),\label{eq3.54}\\
(u_m)\quad \text{is bounded in  }L^2(0,T;V), \label{eq3.55}
\end{gather}

\subsection*{Second estimate}
 In both sides of  (\ref{eq3.53}) we take the derivatives
with respect $t$ and consider $v=u_m'(t)$. We obtain
\begin{equation}
\begin{aligned}
&(u_{m}''(t),u_{m}'(t))+\mu a(u_{m}'(t),u_{m}'(t))\,\\
&+ b(u_{m}'(t),u_{m}(t),u_m'(t))+b(u_{m}(t),u_{m}' (t),u_m'(t))\\
& +\frac{1}{\epsilon}((\beta u_{m}(t))',u_{m}'(t))
 +\int_0^tg'(t-\sigma) ((u_m(t),u_m'))d\sigma\\
& +g(0)((u_m(t),u_m'(t) )) +
\frac{1}{\epsilon}\left(\left(\beta
u_{m}\right)'(t),u'_m(t)\right)\\
&= (f'(t),u_{m}'(t)),
\end{aligned}\label{eq3.57extra}
\end{equation}
because
$$
\frac{d}{dt}\Big(\int_0^tg(t-\sigma)\Delta
u_m(\sigma)\,d\sigma\Big)
=g(0)\Delta u_m(t) +
\int_0^tg'(t-\sigma)\Delta u_m(\sigma)\,d\sigma.
$$
We note that
\begin{equation}\label{eq3.57extra1}
\begin{gathered}
u_m'(0)\to u_1\quad \text{strongly in } H,\\
u_m(0)\to u_0 \quad \text{strongly in } V.
\end{gathered}
\end{equation}
Indeed, \eqref{eq3.57extra1}$_1$ is obtained using (\ref{eq3.53})
with $t=0$ and (\ref{eq6}). Note that $\beta(u_0)=0$.
Then
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|u_m'(t)|^2+\mu\|u_m'(t)\|^2+
b(u_m'(t),u_m(t),u_m'(t))\\
&+\int_0^tg'(t-\sigma) ((u_m(t),u_m'))d\sigma
+g(0)\frac{1}{2}\frac{d}{dt}\|u_m(t)\|^2\\
&=(f'(t),u_m'(t)),
\end{aligned}\label{eq3.57}
\end{equation}
because  $b(u_m(t),u_m'(t),u_m'(t))=0$ and
$((\beta u_m)'(t),u_m'(t))\geq0$ (see Lions \cite[page 399]{lions1}).

\begin{remark} \label{rmk3.3} \rm % \label{rmkad}
The derivative with respect to $t$ of $(\beta(v(t)),w)$ is only
formal. The correct method is to consider the difference equation in
$t+h$ and $t$, divided by $h$ and take the limits when $h \to 0$.
Here is fundamental the operator $\beta$ to be monotonous. This
justify the formal procedure of taking the derivative with respect
to $t$, on both sides of (\ref{eq3.53}) and take $v=u_m'(t)$. See
Brezis \cite{brezis1}, Browder \cite{browder} or Lions \cite{lions2}
for details.
\end{remark}

As $n=2$, we have (see Lions \cite[page 70]{lions1})
\begin{equation}
\|u\|^2_{L^4(\Omega)}\leq C\|u\||u|,\quad \forall\,u\in
H_0^1(\Omega).\label{eq3.58}
\end{equation}
Moreover, $H_0^1(\Omega)\hookrightarrow L^2(\Omega)$; therefore,
$$
|b(u,v,u)|\leq\sum_{i,j=1}^2\int_{\Omega}|u_i(x)|
\big|\frac{\partial v_j}{\partial x_i}(x)\big||u_j(x)|
\leq\|u\|_{(L^4(\Omega))^2}\|v\|.
$$
 This and (\ref{eq3.58}) imply
\begin{equation}
\begin{aligned}
|b(u_m'(t),u_m(t),u_m'(t))|
&\leq \sqrt{\frac{\mu}{2}}\|u_m'(t)\|C\sqrt{\frac{2}{\mu}}
|u_m'(t)|\|u_m(t)\|\\
&\leq \frac{\mu}{4}\|u_m'(t) \|^2+\frac{C^2}{\mu}\|u_m(t)\|^2|u_m'(t)|^2.
\end{aligned} \label{eq3.58b}
\end{equation}
Therefore,
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|u_m'(t)|^2+\mu\|u_m'(t)\|^2 +
g(0)\frac{1}{2}\frac{d}{dt}\|u_m(t)\|^2\\
&\leq \Big|\int_0^tg'(t-\sigma)
((u_m(t),u_m'))d\sigma\Big|_{\mathbb{R}}
+ |b(u_m'(t),u_m(t),u_m'(t))|_{\mathbb{R}}\\
&\quad +\sqrt{\frac{2}{\mu}}\|f'(t)\|_{V'}\sqrt{\frac{\mu}{2}}\|u_m'(t)\|,
\end{aligned} \label{eq3.59}
\end{equation}
Therefore, from (\ref{eq3.59}), (\ref{eq3.58}) and Remark \ref{rmk3.2}
we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}|u_m'(t)|^2
 + g(0)\frac{1}{2}\frac{d}{dt}\|u_m(t)\|^2 + \mu\|u_m'(t)\|^2\\
&\leq \frac{\mu}{4}\|u_m'(t)\|^2+\frac{C^2}{2\mu}\|u_m(t)\||u_m'(t)|^2\\
&\quad + \|g'\|_{L^1(0,\infty)}\|u_m'(t)\|\|u_m(t)\|\, +
\frac{C_1}{\mu}\|f'(t)\|^2+\frac{\mu }{4} \|u_m'(t)\|^2,
\end{aligned} \label{eq3.62}
\end{equation}
Integrating (\ref{eq3.62}) from $0$ to $t$ and using the hypothesis
\eqref{H1tilde}, \eqref{H1bar} we obtain
\begin{equation}
\begin{aligned}
&|u_m'(t)|^2+\left(\frac{\mu}{2}-2\|g\|_{L^1(0,\infty)}\right)
\int^t_0\|u_m'(s)\|^2ds\\
&\leq C_2^2\|g\|_{L^1(0,\infty)}\int_0^T\|u_m(t)\|^2\,dt\,
 + C\int^t_0\|u_m(s)\|^2|u_m'(s)|^2ds
 +C\int_0^T\|f'(t)\|^2\,dt.
\end{aligned} \label{eq3.65}
\end{equation}
 From \eqref{eq3.5} and hypothesis on $f$ we obtain
\begin{equation}
|u_m'(t)|^2+\left(\frac{\mu}{2}-2\|g\|_{L^1(0,\infty)}\right)
\int^t_0\|u_m'(s)\|^2ds
\leq C+C\int^t_0\|u_m(s)\|^2|u_m'(s)|^2ds. \label{eq3.65b}
\end{equation}
Being $(u_m)$ is bounded in $L^2(0,T;V)$ we have, using Gronwall's
inequality in \eqref{eq3.65} and hypothesis $H1$, that
\begin{gather}
(u_m')\quad \text{is bounded in  }L^2(0,T;V)\label{eq3.66}\\
(u_m')\quad \text{is bounded in  }L^{\infty}(0,T;H)\label{eq3.67}.
\end{gather}


\subsection*{Third estimate}
Let $(w_{\nu})$ be the orthonormal system of $V\cap V_2$ formed by
the eigenfunctions of the Laplace operator.

As in the proof of Theorem \ref{thm3}, omitting the parameter
$\epsilon$ and taking $w_j=-\Delta u_{m}$ in the approximate
equation (\ref{eq3.53}) we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\frac{d}{dt}\|u_m(t)\|^2+\mu|\Delta u_m(t)|^2\\
& \leq|b(u_m(t),u_m(t),-\Delta u_m(t))|_{\mathbb{R}}\\
&\quad +\Big|\int_{\Omega}\Delta
u_m(x,t)\Big(\int_0^tg(t-\sigma)\Delta
u_m(x,\sigma)d\sigma\Big)\Big|_{\mathbb{R}}dx \\
&\quad + \frac{1}{\mu}|f(t)|^2 + \frac{\mu}{4}|\Delta u_m(t)|^2
\end{aligned} \label{eq3.70.1}
\end{equation}
because $\langle\beta u_m,-\Delta u_m\rangle\geq0$ (see
Haraux \cite[page 58]{haraux1}).
We note that
\begin{equation}
\begin{aligned}
|b(u_m(t),u_m(t),-\Delta u_m(t)|
& \leq\sum_{i,j=1}^2\int|u_{m_j}(t)\big|\frac{\partial
u_{m_j}}{\partial x_i}(t)\big||\Delta u_{m_j}(t)|\\
&\leq\|u_m(t)\|^2_{(L^3(\Omega))^2}\big\|\frac{\partial
u_m}{\partial x_i}(t)\big\||\Delta u_m(t)|,
\end{aligned} \label{eq3.70.2}
\end{equation}
because $H_0^1(\Omega)\hookrightarrow L^3(\Omega)$,
$H_0^1(\Omega)\hookrightarrow L^6(\Omega)$, with
$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$.

Substituting (\ref{eq3.70.2}) in (\ref{eq3.70.1}) and using the
Remark \ref{rmk3.2}, we obtain
\begin{align*}
&\frac{d}{dt}\|u_m(t)\|^2+\left(\frac{\mu}{2}-2\|g\|_{L^1(0,\infty)}
 \right)|\Delta u_m(t)|^2\\
&\leq C\|f(t)\|^2+\left(C\|u_m(t)\|^2_{(L^3(\Omega))^2}-C\right)
 \|u_m(t)\|^2.
\end{align*}
 Integrating the above inequality from $0$ to $t$,  observing that
$u_m\in L^2(0,T;V)\subset L^2(0,T;(L^3(\Omega))^2$ and using the
Gronwall's Lemma,  it follows that
\begin{gather}
u_m\quad \text{is bounded in }L^{\infty}(0,T;V)\label{eq3.70.5}\\
u_m\quad \text{is bounded in }L^2(0,T;V_2).
\end{gather}
To complete the proof of Theorem \ref{thm4}, we use the same argument
used in the proof of Theorem \ref{thm3}.

We shall now prove Theorem \ref{thm2}. From the previous
convergence, and Banach-Steinhauss theorem, it follows that there
exists a subnet $(u_{\epsilon})_{0<\epsilon <1}$, such
that it converges to $u$ as $\epsilon \to 0$, in the  sense of
previous convergence.

This function satisfies (\ref{eq7}) and (\ref{eq8}). Using the same
arguments used in Theorem \ref{thm1} we obtain that $\beta u=0$.
Therefore, $u$ satisfy (\ref{eq10}) of Theorem \ref{thm2}.

We need to show only that $u$ is a solution of inequality
(\ref{eq9}) a.e. in $t$. In fact, we have that $u_{\epsilon}$
satisfies
\begin{equation}
\begin{gathered}
(u_{\epsilon}',\widetilde{v})+\mu
a(u_{\epsilon},\widetilde{v})+b(u_{\epsilon},u_{\epsilon},\widetilde{v})+
\int_0^tg(t-\sigma)((u_{\epsilon}, \widetilde{v}))d\sigma
+\frac{1}{\epsilon}(\beta
u_{\epsilon},\widetilde{v})=(f,\widetilde{v}), \\
u_{\epsilon}(0)=u_0.
\end{gathered} \label{eq3.72}
\end{equation}
for all $\widetilde{v}\in V$.
 Then from (\ref{eq3.72}), with $\widetilde{v}=v-u_{\epsilon}$,
$v\in K$, we have
\begin{equation}
\begin{aligned}
&(u_{\epsilon}',v-u_{\epsilon})+\mu a(u_{\epsilon},v)
 +b(u_{\epsilon},u_{\epsilon},v)
 + \int_0^tg(t-\sigma)((u_{\epsilon},v))d\sigma-(f,v-u_{\epsilon})\\
& \geq\mu a(u_{\epsilon},u_{\epsilon})+
\int_0^tg(t-\sigma)((u_{\epsilon}, u_{\epsilon}))d\sigma,\quad
\forall v\in K,
\end{aligned} \label{eq3.80}
\end{equation}
because $(\beta u_{\epsilon}-\beta v,u_{\epsilon}-v)\geq0$.
Let us denote
$$
X_{\epsilon}^{v}
= (u_{\epsilon}',v-u_{\epsilon})+\mu
a(u_{\epsilon},v)+b(u_{\epsilon},u_{\epsilon},v)
+ \int_0^tg(t-\sigma)((u_{\epsilon}, v))d\sigma -(f,v-u_{\epsilon}).
$$
We obtain
\begin{equation}
 X^{v}_{\epsilon}\geq \mu
a(u_{\epsilon},u_{\epsilon})+\int_0^tg(t-\sigma)((u_{\epsilon},
u_{\epsilon}))d\sigma,\quad \forall v\in V. \label{eq3.81}
\end{equation}
Let $\psi\in C^0([0,T])$ with $\psi(t)\geq0$. Then $v\psi\in
C^0([0,T];V)$ for all $v\in V$.
$$
u_{\epsilon i}u_{\epsilon j}\to u_{i}u_j\text{  weakly in  }L^2(0,T,L^2(\Omega))
$$
It follows from (\ref{eq3.81}) that
\begin{equation}
\begin{aligned}
&\int^T_0\psi(u_{\epsilon}',v-u_{\epsilon})\,dt
+\mu\int^T_0\psi a(u_{\epsilon},v)\,dt
+ \int^T_0\psi b(u_{\epsilon},u_{\epsilon},v)\,dt\\
&+\psi\int_0^T\int_0^tg(t-\sigma)((u_{\epsilon}, v))d\sigma\,dt
 -\int^T_0\psi(f,v-u_{\epsilon})\,dt\\
&\geq\mu\int^T_0\psi a(u_{\epsilon},u_{\epsilon})\,dt
 +\int_0^tg(t-\sigma)((u_{\epsilon}, u_\epsilon))d\sigma.
\end{aligned}\label{eq3.84}
\end{equation}
Taking $\limsup$ in both side of inequality (\ref{eq3.84}) we obtain
\begin{equation}
\begin{aligned}
&\int^T_0\psi(u',v-u)\,dt
+\mu\int^T_0\psi a(u,v)\,dt
-\int^T_0\psi b(u,u,v)\,dt \\
&+\int_0^T\int_0^tg(t-\sigma)((u, v))d\sigma\,dt
 -\int^T_0\psi(f,v-u)\,dt\\
& \geq\mu\int^T_0\psi a(u,u)\,dt
+\int_0^T\int_0^tg(t-\sigma)((u, u))d\sigma\,dt,
\end{aligned} \label{eq3.85}
\end{equation}
because
$$
\limsup\mu\int^T_0\psi a(u_{\epsilon},u_{\epsilon})\,dt
\geq \liminf\mu\int^T_0\psi a(u_{\epsilon},u_{\epsilon})\,dt
\geq \mu\int^T_0\psi a(u,u)\,dt
$$
and
\begin{align*}
\limsup\int_0^T\int_0^tg(t-\sigma)((u{\epsilon}, u_{\epsilon}))d\sigma\,dt
& =\limsup\int_0^T\int_0^tg(t-\sigma)(-\Delta u_\epsilon, u_\epsilon)d\sigma\,dt\\
& =\int_0^T\int_0^tg(t-\sigma)(-\Delta u, u)d\sigma\,dt\\
& =\int_0^T\int_0^tg(t-\sigma)((u, u))d\sigma\,dt
\end{align*}
 From (\ref{eq3.85}) we obtain finally
\begin{equation}
\begin{aligned}
&(u',v-u)+\mu a(u,v-u)+ b(u,u,v-u) +\int_0^tg(t-\sigma)((u, v-u))d\sigma\\
& \geq(f,v-u)\quad \forall v\in K,\text{ a.e. in }t.
\end{aligned}\label{eq3.88}
\end{equation}

\section{Uniqueness}\label{secuni}

We now prove that when $n=2$ we have uniqueness in Theorem
\ref{thm2}. Indeed, suppose that $u_1,u_2$ are two solutions of
(\ref{eq9}) and set $w=u_2-u_1$ and $t \in (0,T)$. Taking $v=u_1$
(resp. $u_2$) in the inequality (\ref{eq9}) relative to $v_2$ (resp.
$v_1$) and adding up the results we obtain
\begin{align*}
&-\int^t_0(w',w)\,dt-\mu\int^t_0a(w,w)\,dt+\int^t_0b(u_1,u_1,w)\,dt\\
&- \int^t_0b(u_2,u_2,w)\,dt-\int_0^t\int^t_0g(t-\sigma)((w,w))\geq0.
\end{align*}
Therefore,
\begin{equation} \label{eq4.1}
\frac{1}{2}\int^t_0\frac{d}{dt}|w(t)|^2\,dt+\mu\int^t_0\|w(t)\|^2\,dt
\leq\int^t_0|b(w,u_2,w)|\,dt,
\end{equation}
because $\int_0^t\int^t_0g(t-\sigma)((w,w))\geq0$
and $b(u_2,u_2,w)-b(u_1,u_1,w)=b(w,u_2,w)$. On  the other hand,
if $n=2$, we have (see Lions \cite[page 70]{lions1})
\begin{equation}
|b(w(t),u_2(t),w(t))|\leq C \|w(t)\||w(t)|\|u_2(t)\|. \label{eq4.2}
\end{equation}
It follows from (\ref{eq4.1}) and (\ref{eq4.2}) that
$$
|w(t)|^2+\frac{\mu}{2}\int^t_0\|w(t)\|^2\,dt\leq
C \int^t_0|w(t)|^2\|u_2(t)\|^2\,dt.
$$
This implies, using Gronwall's inequality that $w=0$, because
$u_2\in L^2(0,T;V)$, therefore $u_1(t)=u_2(t)$, for all
$t\in [0,T]$.

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\end{document}
