\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 $[0,t_m[$, $00$, 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 $b(u_2,u_2,w)-b(u_1,u_1,w)=b(w,u_2,w)$. 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