\documentclass[12pt,reqno,a4paper]{article} \usepackage[latin1]{inputenc} % accents \usepackage{a4,latexsym,amssymb,amsfonts} \usepackage{amsmath} \usepackage{amsthm} \usepackage[dvips]{epsfig} \usepackage{enumerate} \usepackage{hyperref} \usepackage[english]{babel} \usepackage{graphicx} \textheight240mm \textwidth170mm \hoffset=-9mm \pagestyle{plain} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \renewcommand{\contentsname}{Table des mati\`eres} \renewcommand{\cleardoublepage}{\clearpage} \renewcommand{\partname}{Partie} \renewcommand{\listtablename}{Liste des tables} \newcommand{\bibname}{Bibliography} \newcommand{\cqfd}{\nolinebreak \hspace*{5mm} \rule{2mm}{2mm}} \newcommand{\f}{\nolinebreak \hspace*{5mm} \rule{1mm}{1mm}} \newcommand{\ov}{\overline} \newcommand{\wi}{\widetilde} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\thesubsection {\thesection.\arabic{subsection}} \numberwithin{equation}{section} \newtheorem{thm}{\bf Theorem}[section] \newtheorem{lem}[thm]{\bf Lemma} \newtheorem{pre}{Proof} \newtheorem{prop}[thm]{\bf Proposition} \newtheorem{cor}[thm]{\bf Corollary} \newtheorem{rks}[thm]{\bf Remarks} \newtheorem{rk}[thm]{\bf Remark} \newtheorem{df}[thm]{\bf Definition} \newtheorem{example}{Example}[section] \begin{document} \title{{\Large{\bf Stokes problem with the possibility of controlling the velocity in a L-shaped domain }}} \date{ } \maketitle \centerline{{ OMAR CHAKRONE$^{1}$,} { OKACHA DIYER$^{2}$, { DRISS SBIBIH$^{2}$ }}} %\author{O. CHAKRONE, O. DIYER AND D. SBIBIH} \begin{center} $^{1}$Universit\'e Mohammed I, Faculté des sciences\\ Laboratoire LANOL, Oujda, Maroc. \end{center} \begin{center} $^{2}$Universit\'e Mohammed I, Ecole Sup\'erieure de Technologie\\ Laboratoire MATSI, Oujda, Maroc. \end{center} \begin{center}{chakrone@yahoo.fr\\ odiyer@yahoo.fr\\ sbibih@yahoo.fr} \end{center} \textbf{Abstract:} The movement is studied from a viscous and incompressible homogeneous fluid which crosses a field of the channel in the form of L, with the possibility to exert pressure of known difference between two opposite edges. We extend previous work in \cite{AB} which studies a problem of Stokes in the stationary case and with one parameter that characterizes the pressure difference between two sides in a specific domain (symmetric channel). We show existence, unicity and regularity of the solution of an evolution problem with four parameters that characterize the pressure difference between two opposite sides of our field.\\ \textbf{Keywords:}Stokes problem, Regularity of the solution, weak solution. \\ Mathematics Subject Classification. 76D05, 35Q30.\\ \section{Introduction} Let $\Omega \subset \mathbb{R}^{2}$ be a bounded domain with boundary $\partial\Omega=\Gamma=\bigcup_{i=1}^{8}\Gamma_{i}$, where $\Gamma_{1}=\{0\}\times [0,1]$, $\Gamma_{2}=\{0\}\times [1,3]$, $\Gamma_{3}=[0,1]\times \{3\}$, $\Gamma_{4}=\{1\}\times [1,3]$, $\Gamma_{5}=[1,5]\times\{1\}$, $\Gamma_{6}=\{5\}\times [0,1]$, $\Gamma_{7}=[1,5]\times\{0\}$ and $\Gamma_{8}=[0,1]\times \{0\}$, see Figure \ref{F1}. Given four real numbers $\lambda_{1}, \lambda_{2}, \lambda_{3}$ and $\lambda_{4}$, we consider the problem: $$(S_{1}) \left\{% \begin{array}{ll} \hbox{ Find } u=(u_{1},u_{2})\in V_{1} \hbox{ such that } \\ \sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial u_{i}}{\partial t}v_{i}\ + \sum\limits_{i=1}^{2}\int_{\Omega}\nabla u_{i}\nabla v_{i} =\lambda_{1}\int_{0}^{1}v_{1}(5,y)dy+ \lambda_{2}\int_{1}^{3}v_{1}(1,y)dy\\ ~~~~~~~~~~~~~~~~~~~~~~~+\lambda_{3}\int_{0}^{1}v_{2}(x,3)dx+\lambda_{4}\int_{1}^{5}v_{2}(x,1)dx & \hspace*{-1.3cm} \forall v=(v_{1},v_{2})\in V_{1},\\ u_{1}({\bf{x}},0) = a_{01}({\bf{x}}) & \hbox{ a.e. {\bf{x}} in }\Omega,\\ u_{2}({\bf{x}},0) = a_{02}({\bf{x}}) & \hbox{ a.e. {\bf{x}} in }\Omega,\\ \end{array}% \right.$$ where $V_{1}$ is the closing of $\{v=(v_{1},v_{2})\in C^{1}([0,T];H); \ div~v=\frac{\partial v_{1}}{\partial x}+ \frac{\partial v_{2}}{\partial y}=0,\ v_{i}|_{\Gamma_{1}}=v_{i}|_{\Gamma_{6}},\ v_{i}|_{\Gamma_{2}}=v_{i}|_{\Gamma_{4}},\ v_{i}|_{\Gamma_{3}}=v_{i}|_{\Gamma_{8}},\ v_{i}|_{\Gamma_{5}}=v_{i}|_{\Gamma_{7}} \mbox{ for } i=1,2\}$ in $C([0,T];H)$, $H$ is the closing of $\vartheta =\{u\in (\mathcal{D}(\Omega))^{2}; \ div~v=0\}$ in $(L^2(\Omega))^2$, where $\mathcal{D}$ consist of all functions in $C^{\infty}(\Omega)$ which have compact support in $\Omega$ and $a_{0}=(a_{01},a_{02})\in H$.\\ \vspace{4cm} \begin{figure}[h!]\vspace{-4cm} \begin{center} $\hspace*{-6cm}\vspace{1cm}\Gamma_{3}$ $$\vspace{-1cm}\hspace{-3.5cm}\Gamma_{4}$$ $$\hspace*{-2.5cm}\vspace{-4cm}\hspace{-7cm}\Gamma_{2}$$ % Requires \usepackage{graphicx} \includegraphics[width=9cm,height=8cm]{domaine.eps}\vspace{-1.4cm}\\ \vspace{-2.5cm} \hspace{1cm} $\Gamma_{5}$ \vspace*{0.8cm}\\ $\Gamma_{1}$ \hspace*{8.5cm} $\Gamma_{6}$ \vspace*{0.5cm}\\ $\hspace*{-2cm}\Gamma_{8}$ \hspace*{3cm} $\Gamma_{7}$ \\\vspace{-1.7cm} Domain $\Omega$\vspace{2cm} \caption{Vertical plane from the channel}\label{F1}\vspace{1cm} \end{center} \end{figure}\vspace{-1.6cm} As applications of this problem, we cite the different types of flows for example see \cite{ABe,DL}. This problem was studied by C. Amrouche, M. Batchi and J. Batina in the stationary case with only one parameter see \cite{AB}, we extend the preceding work to a problem of evolution with four parameters. Our aim is, in first time to prove the existence, uniqueness and regularity of the solution, second time we show the equivalence between the variational problem, where the notion of pressure does not appear explicitly, and classic problem which highlights the pressure and these differences between the opposite sides of our field. We cite also a variety of works in the stationary case see \cite{ABe,FG}.\\ This paper is organized as follows. In Section 2, we establish existence, unicity and regularity of the solution of the system $(S_{1})$ and we prove the equivalence between our variational problem and the classical problem associated. In Section 3, we give the conclusion which shows the importance of the problem to be studied. \section{Resolution of problem $(S_{1})$} \subsection{Functional spaces adapted to the problem} Let us denote by $V$ the closing of $\vartheta$ in $(H^{1}(\Omega))^{2}$. We consider the following Banach spaces $L^{2}(0,T;V)$ and $C([0,T];H)$ with the norms $\|u\|_{L^{2}(0,T;V)}=\left(\int_{0}^{T}\|u(t)\|^{2}_{V}dt\right)^{\frac{1}{2}}$ and $\|u\|_{C([0,T];H)}=\sup\limits_{t\in [0,T]}\|u(t)\|_{H}$ respectively. \subsection{Existence and uniqueness of the solution} \begin{prop} \label{prop 1} If $u\in W(a,b;V,V')$, then for all $v$ in $V$, we have $$\frac{d}{dt}(u(.),v)_{V,V}=\langle u'(.),v \rangle_{\mathcal{D'}(]a,b[),V} \mbox{ in } \mathcal{D'}(]a,b[),$$ where $W(a,b;X,Y)=\{u\in L^{2}(a,b;X);\ u'\in L^{2}(a,b;Y)\}$ is a Hilbert space with the norm $\|u\|_{W}=\left(\int_{a}^{b}(|u(t)|^{2}_{X}+|u'(t)|^{2}_{Y})dt\right)^{\frac{1}{2}}$ (see \cite{DL}). \end{prop} \begin{proof}Let $\varphi\in \mathcal{D}(]a,b[)$, we have $\int_{a}^{b}\langle u'(t),v \rangle \varphi(t) dt=\int_{a}^{b}\langle u'(t)\varphi(t),v\rangle dt$. Since $u'\in L^{2}(a,b;V')$, we deduce that the function $t\rightarrow \langle u'(t),v\rangle $ is in $L^{2}(a,b)$ for all $v\in V$.\\ In the same way, we have \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_{a}^{b}\langle u'(t)\varphi(t),v\rangle dt&=&\langle \int_{a}^{b}u'(t)\varphi(t)dt,v\rangle = -\langle \int_{a}^{b}u(t)\varphi'(t)dt,v\rangle \\ &=&-\int_{a}^{b}\langle u(t)\varphi'(t),v\rangle dt=-\int_{a}^{b}(u(t),v)\varphi'(t)dt. \end{eqnarray*} Thus $\int_{a}^{b}\langle u'(t),v\rangle \varphi(t)dt=-\int_{a}^{b}(u(t),v)\varphi'(t)dt= \int_{a}^{b}\frac{d}{dt}(u(t),v)\varphi(t)dt$, and therefore $\langle u'(.),v\rangle =\frac{d}{dt}(u(.),v)$. \end{proof} \begin{thm} \label{thm 1} If the solution of the problem $(S_{1})$ exists, it is necessarily unique. \end{thm} \begin{proof} Let $u_{1}$ and $u_{2}$ be two solutions of the problem $(S_{1})$. Put $w=u_{1}-u_{2}$, $$(\partial_{t}w,v)+((w,v))=0~~\forall v\in V_{1}\mbox{ and } w(0)=0,$$ where $(\partial_{t}w,v)=\sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial w_{i}}{\partial t}v_{i}$, $((w,v))=\sum\limits_{i=1}^{2}\int_{\Omega}\nabla w_{i}\nabla v_{i}$. Then $$\int_{0}^{T}(\partial_{t}w,v)dt+\int_{0}^{T}((w,v))dt=0.$$ Let us take $v=w\chi_{(0,s)}(t),~~s\in (0,T)$, thus $\int_{0}^{s}(\partial_{t}w,w)dt+\int_{0}^{s}((w,w))dt=0.$\\Hence $$\int_{0}^{s}(\partial_{t}w,w)dt=-\int_{0}^{s}((w,w))dt=-\int_{0}^{s}\|w(t)\|_{V}^{2}dt \leq 0.$$ Consequently $$ \frac{1}{2}\|w(s)\|_{H}^{2}-\frac{1}{2}\|w(0)\|_{H}^{2}=\frac{1}{2}\|w(s)\|_{H}^{2}\leq 0,$$ and this shows that $w=0$. \end{proof} The problem $(S_{1})$ becomes $$ \left\{% \begin{array}{ll} \hbox{ Find } u\in V_{1} \hbox{ such that }\\ \frac{\partial}{\partial t}(u,v)+((u,v))= \langle \lambda_{1}e_{1},v\rangle_{\Gamma_{ 6}}+\langle \lambda_{2}e_{1},v\rangle_{\Gamma _{4}}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~ +\langle \lambda_{3}e_{2},v\rangle_{\Gamma_{3}}+\langle\lambda_{4}e_{2},v\rangle_{\Gamma_{5}} &\forall v\in V_{1},\\ u({\bf{x}},0) = a_{0}({\bf{x}}) & \hbox{ a.e. } {\bf{x}} \hbox{ in } \Omega,\\ \end{array}% \right.$$ where $e_{1}=(1,0)$ and $e_{2}=(0,1)$.\\ We are looking for an approximate solution $(u_{m})$ of the form: \begin{equation} \label{E1} u_{m}(t)=\sum\limits_{i=1}^{m}g_{i}(t)w_{i}, \end{equation} where $\{w_{i}\}$ is a Hilbertienne basis of $H$ and $g_{i}\in C([0,T]), g_{i}(0)=g_{i0}=(u_{0},w_{i})$. We have $\forall j=1,\ldots,m $ \begin{equation}\label{E2} \begin{aligned} \frac{\partial}{\partial t}(u_{m}(t),w_{j})+((u_{m}(t),w_{j}))&= \langle\lambda_{1}e_{1},w_{j}\rangle_{\Gamma _{6}}+\langle\lambda_{2}e_{1},w_{j}\rangle_{\Gamma_{ 4}} \\ &+ \langle\lambda_{3}e_{2},w_{j}\rangle_{\Gamma _{3}}+\langle\lambda_{4}e_{2},w_{j}\rangle_{\Gamma _{5}}. \end{aligned} \end{equation} Put $V_{m}=\{w_{1},w_{2},\ldots,w_{m}\}$. By replacing \eqref{E1} in \eqref{E2}, we obtain \begin{equation}\label{E3} \frac{\partial}{\partial t}(\sum\limits_{i=1}^{m}g_{i}(t)w_{i},w_{j}) +((\sum\limits_{i=1}^{m}g_{i}(t)w_{i},w_{j}))=\gamma_{j}, \end{equation} where $\gamma_{j}=\langle\lambda_{1}e_{1},w_{j}\rangle_{\Gamma _{6}}+\langle\lambda_{2}e_{1},w_{j}\rangle_{\Gamma _{4}}+\langle\lambda_{3}e_{2},w_{j}\rangle_{\Gamma _{3}}+\langle\lambda_{4}e_{2},w_{j}\rangle_{\Gamma _{5}}$.\\ Thus \begin{equation}\label{E4} \frac{\partial}{\partial t} \left[\sum\limits_{i=1}^{m}g_{i}(t)(w_{i},w_{j}) \right]+\sum\limits_{i=1}^{m}g_{i}((w_{i},w_{j})) =\gamma_{j}. \end{equation} So $g'_{j}(t)+\frac{1}{\alpha_{j}}g_{j}(t)=\gamma_{j}$, where $\alpha_{j}$ are the eigenvalues of $\psi :H\rightarrow V , f\mapsto u$, with $u$ is the unique solution of the problem \begin{equation*} u \in V; ((u,v))=(f,v)~~~~ \forall v\in V. \end{equation*} We have the following results: $\psi(w_{i})=\alpha_{i}w_{i}$ for all $i\geq 1$, all the eigenvalues are strictly positive and $\alpha_{i}\rightarrow 0 ~~ \mbox{when}~~ i\rightarrow \infty $, $\{w_{i}\}$ is an orthogonal system in V and we have $((w_{i},w_{j}))=\frac{1}{\alpha_{i}}\delta _{ij}$, where $\delta _{ij}$ is the kronocker symbol. We obtain $$(E) \left\{% \begin{array}{ll} g'_{j}(t)+\frac{1}{\alpha_{j}}g_{j}(t)=\gamma_{j} &\forall j=1,\ldots,m ~~a.e~~ t\in (0,T), \\ g_{j}(0) = g_{j0} \end{array}% \right.$$ which admits a unique solution. Consequently $u_{m}=\sum\limits_{i=1}^{m}g_{i}(t)w_{i}$ is the solution of the approximate problem and in the same way as previously, we show the unicity of the solution $u_{m}$. \begin{thm} \label{thm 2} If $u_{0}\in V_{1}$, the approximate solution $u_{m}$ satisfies $$ \mbox{ there exist } c,c' > 0 \mbox{ such that } \|u'_{m}\|_{L^{2}(0,T;H)}\leq c \mbox{ and } \|u_{m}\|_{C([0,T];V)}\leq c'.$$ \end{thm} \begin{proof} The function $u_{m}(x,t)=\sum\limits_{i=1}^{m}g_{i}(t)w_{i}$ satisfies $$ \begin{array}{rcl} \frac{\partial}{\partial t}(u_{m},v)+((u_{m},v))&=&\langle\lambda_{1}e_{1},v\rangle_{\Gamma _{6}} +\langle\lambda_{2}e_{1},v\rangle_{\Gamma_{ 4}}+\langle\lambda_{3}e_{2},v\rangle_{\Gamma_{3}}+ \langle\lambda_{4}e_{2},v\rangle_{\Gamma_{5}}\\ &:=&b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma)~~ \forall v\in V. \end{array} $$ Then $(u'_{m},v)+((u_{m},v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma)$. For $v(t)=u'_{m}(t)=\sum\limits_{i=1}^{m}g'_{i}(t)w_{i}\in V,$ we have \begin{equation}\label{E5} (u'_{m},u'_{m})+((u_{m},u'_{m}))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},u'_{m},\Gamma)~~ for~~ a.e.~~t\in (0,T). \end{equation} Each term of \eqref{E5} is integrable, indeed $$(u'_{m},u'_{m})=\sum\limits_{i=1}^{m}|g'_{i}(t)|^{2}\in L^{1}(0,T)$$ because $g'_{j}(t)=\gamma_{j}-\frac{1}{\alpha_{j}}g_{j}(t)$ and $g'_{j}(t)\in L^{2}(0,T)~~(g_{j}\in C([0,T]))$. $$((u_{m},u'_{m}))=((\sum\limits_{i=1}^{m}g_{i}w_{i},\sum\limits_{i=1}^{m}g_{j}w_{j}))= \sum\limits_{i=1}^{m}\frac{1}{\alpha_{i}}g_{i}g'_{i}\in L^{1}(0,T).$$ $$b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},u'_{m},\Gamma)= \sum\limits_{i=1}^{m}g'_{i}b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},w_{i},\Gamma)\in L^{1}(0,T).$$ We deduce that \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_{0}^{s}\|u'_{m}(t)\|^{2}_{H}dt+\int_{0}^{s}((u_{m},u'_{m}))dt&=& \int_{0}^{s} (\langle\lambda_{1}e_{1},u'_{m}\rangle_{\Gamma_{ 6}} +\langle\lambda_{2}e_{1},u'_{m}\rangle_{\Gamma_{ 4}}\\ &+& \langle\lambda_{3}e_{2},u'_{m}\rangle_{\Gamma_{ 3}} +\langle\lambda_{4}e_{2},u'_{m}\rangle_{\Gamma_{ 5}})dt,~~\forall s\in (0,T). \end{eqnarray*} If we put $A(m,s)= \int_{0}^{s}\|u'_{m}(t)\|^{2}_{H}dt+\frac{1}{2}\|u_{m}(s)\|^{2}_{V}-\frac{1}{2}\|u_{m}(0)\|^{2}_{V}$, then we have \begin{eqnarray*} % \nonumber to remove numbering (before each equation) A(m,s) &=& \lambda_{1}\int_{0}^{s}\int_{0}^{1}u'_{m1}(5,y)(t)dydt +\lambda_{2}\int_{0}^{s}\int_{1}^{3}u'_{m1}(1,y)(t)dydt \\ &+&\lambda_{3}\int_{0}^{s}\int_{0}^{1}u'_{m2}(x,3)(t)dxdt +\lambda_{4}\int_{0}^{s}\int_{0}^{5}u'_{m2}(x,1)(t)dxdt.\\ \end{eqnarray*} So \begin{eqnarray*} % \nonumber to remove numbering (before each equation) A(m,s) &\leq& |\lambda_{1}||\int_{0}^{1}\int_{0}^{s}u'_{m1}(5,y)(t)dtdy| +|\lambda_{2}||\int_{0}^{3}\int_{0}^{s}u'_{m1}(1,y)(t)dtdy| \\ &+&|\lambda_{3}||\int_{0}^{1}\int_{0}^{s}u'_{m2}(x,3)(t)dtdx| +|\lambda_{4}||\int_{0}^{5}\int_{0}^{s}u'_{m2}(x,1)(t)dtdx|.\\ \end{eqnarray*} Thus \begin{eqnarray*} % \nonumber to remove numbering (before each equation) A(m,s) &\leq& |\lambda_{1}||\int_{0}^{1}u_{m1}(5,y)(s)dy-\int_{0}^{1}u_{m1}(5,y)(0)dy|\\ &+&|\lambda_{2}||\int_{0}^{3}u_{m1}(1,y)(s)dy-\int_{0}^{3}u_{m1}(1,y)(0)dy| \\ &+&|\lambda_{3}||\int_{0}^{1}u_{m2}(x,3)(s)dx-\int_{0}^{1}u_{m2}(x,3)(0)dx|\\ &+&|\lambda_{4}||\int_{0}^{5}u_{m2}(x,1)(s)dx-\int_{0}^{5}u_{m2}(x,1)(0)dx|.\\ \end{eqnarray*} So \begin{eqnarray*} % \nonumber to remove numbering (before each equation) A(m,s) &\leq&c_{1} [\int_{\Gamma_{6}}|u_{m1}(s)|d\sigma + \int_{\Gamma_{6}}|u_{m1}(0)|d\sigma +\int_{\Gamma_{4}}|u_{m1}(s)|d\sigma + \int_{\Gamma_{4}}|u_{m1}(0)|d\sigma\\ &+&\int_{\Gamma_{3}}|u_{m2}(s)|d\sigma + \int_{\Gamma_{3}}|u_{m2}(0)|d\sigma +\int_{\Gamma_{5}}|u_{m2}(s)|d\sigma + \int_{\Gamma_{5}}|u_{m2}(0)|d\sigma]\\ &\leq&c_{2} (\int_{\partial \Omega}|u_{m}(s)|d\sigma+\int_{\partial \Omega}|u_{m}(0)|d\sigma)\\ &\leq&c_{3}[ (\int_{\partial \Omega}|u_{m}(s)|^{2}d\sigma)^{\frac{1}{2}}+(\int_{\partial \Omega}|u_{m}(0)|^{2}d\sigma)^{\frac{1}{2}}]~~~~\mbox{ (Holder's inequality) }\\ &=&c_{3} [\|u_{m}(s)\|_{L^{2}(\partial \Omega)}+\|u_{m}(0)\|_{L^{2}(\partial \Omega)}]\\ &\leq&c_{4}[ \|u_{m}(s)\|_{V}+ \|u_{m}(0)\|_{V}]~~~~~~~~~~~~~~~~~~ (V \hookrightarrow L^{2}(\partial\Omega) \mbox{ a continuous injection }),\\ \end{eqnarray*} where $c_{i},i=1 \ldots 4$, are positive reals.\\ On the one hand \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_{0}^{s}\|u'_{m}(t)\|^{2}_{H}dt &=& A(m,s)-\frac{1}{2}\|u_{m}(s)\|^{2}_{V}+\frac{1}{2}\|u_{m}(0)\|^{2}_{V} \\ &\leq& c_{4} [\|u_{m}(s)\|_{V}+\|u_{m}(0)\|_{V}]-\frac{1}{2}\|u_{m}(s)\|^{2}_{V}+\frac{1}{2}\|u_{m}(0)\|^{2}_{V}.\\ \end{eqnarray*} Let us consider $P_{m}$ the projection of $V$ on the space $span(w_{0},w_{1},\ldots,w_{m})$, we have $u_{0}\in V$ and $u_{m}(0)=\sum\limits_{i=1}^{m}(u_{0},w_{i})w_{i}=P_{m}u_{0}$. Hence $\|u_{m}(0)\|_{V}\leq \|u_{0}\|_{V}$ because $\|P_{m}\|\leq 1$.\\ So \begin{equation}\label{E6} \int_{0}^{s}\|u'_{m}(t)\|_{H}^{2}dt\leq c_{4}[ \|u_{m}(s)\|_{V}+\|u_{0}\|_{V}]-\frac{1}{2}\|u_{m}(s)\|^{2}_{V}+\frac{1}{2}\|u_{0}\|^{2}_{V}. \end{equation} We put $\|u_{m}(s)\|_{V}=R_{m}(s)$, then we obtain \begin{equation}\label{E14} \int_{0}^{s}\|u'_{m}(t)\|_{H}^{2}dt\leq c_{4} R_{m}(s)-\frac{1}{2}(R_{m}(s))^{2}+c_{5},~~~~\mbox{ where } c_{5} \geq 0 . \end{equation} Consequently there exists $c_{6}>0$ such that $$\int_{0}^{s}\|u'_{m}(t)\|^{2}_{H}dt\leq c_{6}~~\forall~~ m .$$ On the other hand, we have \begin{equation}\label{E15} \int_{0}^{s}\|u'_{m}(t)\|_{H}^{2}dt+\frac{1}{2}\|u_{m}(s)\|^{2}_{V}-\frac{1}{2}\|u_{m}(0)\|^{2}_{V} \leq c_{4}[ \|u_{m}(s)\|_{V}+\|u_{m}(0)\|_{V}]. \end{equation} So \begin{equation}\label{E15} \frac{1}{2}\|u_{m}(s)\|^{2}_{V}-\frac{1}{2}\|u_{m}(0)\|^{2}_{V} \leq c_{4}[ \|u_{m}(s)\|_{V}+\|u_{m}(0)\|_{V}]. \end{equation} Thus \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \frac{1}{2}\|u_{m}(s)\|^{2}_{V}-c_{4} \|u_{m}(s)\|_{V} &\leq& \frac{1}{2}\|u_{m}(0)\|^{2}_{V}+c_{4} \|u_{m}(0)\|_{V} \\ &\leq& \frac{1}{2}\|u_{0}\|^{2}_{V}+c_{4}\|u_{0}\|_{V}.\\ \end{eqnarray*} Finally there exists $c_{7}>0$ such that $$\|u_{m}(s)\|_{V}\leq c_{7}~~\forall~~ m .$$ \end{proof} \begin{thm}\label{thm 3} If $u_{0}\in V$, then the solution of the problem $(S_{1})$ exists, unique and satisfies $u\in L^{\infty}(0,T;V)$, $\frac{\partial u}{\partial t}\in L^{2}(0,T;H)$. \end{thm} \begin{proof} From Theorem \ref{thm 2}, we have $\|u_{m}\|_{(C([0,T]);V)}\leq C$. Since $L^{\infty}(0,T;V)\sim (L^{1}(0,T;V'))'$ and $V'$ is reflexive, separable, we deduce that there exists $u_{m_{k}};~~u_{m_{k}}\rightharpoonup ^{*} u $ when ${m_{k}}\rightarrow \infty$ in $L^{\infty}(0,T;V)$, we show also that $\|u'_{m}\|_{L^{2}(0,T.H)}\leq c'$. Now we can extract $(u'_{m_{k}})$ such that $u'_{m_{k}}\rightharpoonup^{*} \eta $ when ${m_{k}}\rightarrow \infty$ in $L^{\infty}(0,T;H)$. We have \begin{equation}\label{E7} (u'_{m_{k}},v)+((u_{m_{k}},v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma),~~\forall~~v\in V. \end{equation} Hence \begin{equation}\label{E8} (u_{m_{k}},v')+((u_{m_{k}},v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma),~~\forall~~v\in V. \end{equation} When $m_{k}\rightarrow \infty$, we deduce from \eqref{E7} and \eqref{E8} respectively: \\ \begin{equation}\label{E9} (\eta,v')+((u,v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma),~~\forall~~v\in V. \end{equation} \begin{equation}\label{E10} (u',v)+((u,v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma),~~\forall~~v\in V. \end{equation} Then, according to \eqref{E9} and \eqref{E10} we get $(\eta,v)=(u',v)~~\forall~~v\in V$, thus $\eta=u'$.\\ When $m_{k}\rightarrow \infty$, we have \begin{equation}\label{E11} (\frac{\partial u}{\partial t},v)+((u,v))=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma),~~\forall~~v\in V. \end{equation} So $u$ is a solution of the problem $(S_{1})$ and it is unique according to Theorem \ref{thm 1}. \end{proof} \begin{thm}\label{thm 4} Let $u\in C([0,T];V)$. $u$ is a solution of $(S_{1})$ if and only if there exists $p\in L^{2}(0,T;\mathbb{R})$ such that $(u,p)$ is a solution of the problem: $(S_{2}) \left\{% \begin{array}{ll} \frac{\partial u_{1}}{\partial t}-\triangle u_{1}+\frac{\partial p}{\partial x}=0 &\hspace*{-2.5cm} \hbox{ in~~ }\Omega\times(0,T),\\ \frac{\partial u_{2}}{\partial t}-\triangle u_{2}+\frac{\partial p}{\partial y}=0 &\hspace*{-2.5cm} \hbox{ in~~ }\Omega\times(0,T),\\ div~u=0 &\hspace*{-2.5cm}\hbox{ in~~ }\Omega\times(0,T),\\ u_{i}|_{\Gamma_{ 1}}=u_{i}|_{\Gamma_{ 6}},u_{i}|_{\Gamma_{ 2}}=u_{i}|_{\Gamma_{ 4}},u_{i}|_{\Gamma_{ 3}} =u_{i}|_{\Gamma_{ 8}}, u_{i}|_{\Gamma_{ 5}}=u_{i}|_{\Gamma_{ 7}}&\hspace*{-2.5cm}\hbox{ for~~ }i=1,2,\\ \frac{\partial u_{i}}{\partial x}|_{\Gamma_{ 1}}=\frac{\partial u_{i}}{\partial x}|_{\Gamma_{ 6}}, \frac{\partial u_{i}}{\partial x}|_{\Gamma_{ 2}}=\frac{\partial u_{i}}{\partial x}|_{\Gamma_{ 4}}&\hspace*{-2.5cm}\hbox{ for ~~ }i=1,2,\\ \frac{\partial u_{i}}{\partial y}|_{\Gamma_{ 3}}=\frac{\partial u_{i}}{\partial y}|_{\Gamma_{ 8}}, \frac{\partial u_{i}}{\partial y}|_{\Gamma_{ 5}}=\frac{\partial u_{i}}{\partial y}|_{\Gamma_{ 7}}&\hspace*{-2.5cm}\hbox{ for ~~}i=1,2,\\ p|_{\Gamma_{ 6}}-p|_{\Gamma_{ 1}}=-\lambda_{1},p|_{\Gamma_{ 4}}-p|_{\Gamma_{ 2}}=-\lambda_{2},p|_{\Gamma_{ 3}} -p|_{\Gamma_{ 8}} =-\lambda_{3}, p|_{\Gamma_{ 5}}-p|_{\Gamma_{ 7}}=-\lambda_{4},\\ u_{1}({\bf{x}},0) = a_{01}({\bf{x}}) &\hspace*{-2.5cm} \hbox{ a.e.~~} {\bf{x}} \hbox{~~ in~~} \Omega,\\ u_{2}({\bf{x}},0) = a_{02}({\bf{x}}) &\hspace*{-2.5cm} \hbox{ a.e.~~} {\bf{ x }} \hbox{~~ in~~} \Omega. \end{array}% \right.$ \end{thm} \begin{proof} Let us assume that $(S_{2})$ have a solution $(u=(u_{1},u_{2}),p)$.\\ Thus $$\sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial u_{i}}{\partial t}v_{i} d{\bf{x}} -\sum\limits_{i=1}^{2}\int_{\Omega}\triangle u_{i}~v_{i} d{\bf{x}}+\int_{\Omega}\frac{\partial p}{\partial x}~v_{1} d{\bf{x}} +\int_{\Omega}\frac{\partial p}{\partial y}~v_{2} d{\bf{x}}=0, \forall v=(v_{1},v_{2}) \in V_{1},$$ so $$\sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial u_{i}}{\partial t}v_{i} d{\bf{x}} -\sum\limits_{i=1}^{2}\int_{\Omega}\triangle u_{i}~v_{i} d{\bf{x}}+\int_{\Omega}(\nabla p).v d{\bf{x}}=0, \forall v=(v_{1},v_{2}) \in V_{1}.$$ Hence $\forall v \in V_{1}$, we have $$\sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial u_{i}}{\partial t}v_{i} d{\bf{x}} +\sum\limits_{i=1}^{2}\int_{\Omega}\nabla u_{i}~\nabla v_{i} d{\bf{x}} -\sum\limits_{i=1}^{2}\int_{\partial \Omega}\frac{\partial u_{i}}{\partial n}.v_{i} d\sigma +\int_{\partial \Omega}pv.\vec{\eta}d\sigma-\int_{\Omega}p~div~vd{\bf{x}}=0,$$ where $\vec{\eta}$ is the unit outward normal to $\partial\Omega$. On the other hand, we have $$\int_{\partial \Omega}\frac{\partial u_{i}}{\partial n}.v_{i}d\sigma=\sum\limits_{j=1}^{8}\int_{\Gamma_{j}}\frac{\partial u_{i}}{\partial n}.v_{i} d\sigma=0.$$ \begin{align*} % \nonumber to remove numbering (before each equation) \int_{\partial \Omega}pv.\vec{\eta} d\sigma&=-\int_{0}^{1}p|_{\Gamma_{1}}(x,y)v_{1}(5,y)dy- \int_{1}^{3}p|_{\Gamma_{ 2}}(x,y)v_{1}(1,y)dy +\int_{0}^{1}p|_{\Gamma_{3}}(x,y)v_{2}(x,3)dx \\ &+ \int_{1}^{3}p|_{\Gamma_{4}}(x,y)v_{1}(1,y)dy+\int_{1}^{5}p|_{\Gamma_{5}}(x,y)v_{2}(x,1)dx +\int_{0}^{1}p|_{\Gamma_{6}}(x,y)v_{1}(5,y)dy \\ &- \int_{1}^{3}p|_{\Gamma_{7}}(x,y)v_{2}(x,1)dx-\int_{0}^{1}p|_{\Gamma_{ 8}}(x,y)v_{2}(x,3)dx. \end{align*} Hence \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_{\partial \Omega}pv.\vec{\eta} d\sigma &=& \int_{0}^{1}(-p|_{\Gamma_{1}}(x,y)+p|_{\Gamma_{ 6}}(x,y))v_{1}(5,y)dy +\int_{1}^{3}(-p|_{\Gamma_{2}}(x,y) \\ &+& p|_{\Gamma_{ 4}}(x,y))v_{1}(1,y)dy+\int_{0}^{1}(p|_{\Gamma_{3}}(x,y)-p|_{\Gamma_{ 8}}(x,y))v_{2}(x,3)dx \\ &+& \int_{1}^{5}(p|_{\Gamma_{5}}(x,y)-p|_{\Gamma_{7}}(x,y))v_{2}(x,1)dx \\ &=&-\lambda_{1}\int_{0}^{1}v_{1}(5,y)dy-\lambda_{2}\int_{1}^{3}v_{1}(1,y)dy -\lambda_{3}\int_{0}^{1}v_{2}(x,3)dx\\ &-&\lambda_{4}\int_{1}^{5}v_{2}(x,1)dx. \end{eqnarray*} And $\int_{\Omega}p~div~v dx=0$ because $v\in V_{1}$.\\ Now we assume that $(S_{1})$ have a solution $u$. $\forall v \in V_{1}$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \sum\limits_{i=1}^{2}\int_{\Omega}\frac{\partial u_{i}}{\partial t}v_{i}d{\bf{x}}+ \sum\limits_{i=1}^{2}\int_{\Omega}\nabla u_{i}\nabla v_{i} d{\bf{x}} &=& \lambda_{1}\int_{0}^{1}v_{1}(5,y)dy +\lambda_{2}\int_{1}^{3}v_{1}(1,y)dy\\ &+&\lambda_{3}\int_{0}^{1}v_{2}(x,3)dx + \lambda_{4}\int_{1}^{5}v_{2}(x,1)dx. \end{eqnarray*} Thus \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \frac{\partial}{\partial t}(u,v)+((u,v)) &=& \langle\lambda_{1}e_{1},v\rangle_{\Gamma_{6}}+\langle\lambda_{2}e_{1},v\rangle_{\Gamma_{4}} +\langle\lambda_{3}e_{2},v\rangle_{\Gamma_{3}}+\langle\lambda_{4}e_{2},v\rangle_{\Gamma_{ 5}} \\ &=& b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma). \end{eqnarray*} Put $\forall v\in V_{1},\ \forall s\in [0,T]$, $$ \alpha(s):=(u(s),v),\ \beta(s):=((u(s),v)),\ \gamma(s):=b(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma).$$ Thus $$ \alpha'(s)+\beta(s)-\gamma(s)=0 \mbox{ in } \mathcal{D'}(0,T).$$ So $$\int_{0}^{t}\alpha'(s)ds+\int_{0}^{t}\beta(s)ds-\int_{0}^{t}\gamma(s)ds=0~~\forall t\in (0,T).$$ As $\gamma(s)$ is independent of time $t$, we have $$\alpha(t)-\alpha(0)+\int_{0}^{t}((u(s),v))ds-tb(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma)=0.$$ $$\alpha(t)-\alpha(0)=-\int_{0}^{t}((u(s),v))ds+tb(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma).$$ Therefore $$\alpha(t)-\alpha(0)=-((\int_{0}^{t}u(s)ds,v))+tb(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma).$$ Hence $\forall t\in (0,T), \forall v\in V_{1}, (u(t)-u_{0},v)+((U(t),v))=tb(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},v,\Gamma)$, where $ U(t):=\int_{0}^{t}u(s)ds$. We have\ $$\int_{0}^{1}v_{1}(5,y)dy=\int_{\Omega}e_{1}v(x,y)\chi_{\Gamma_{ 6}}dxdy,$$ $$\int_{1}^{3}v_{1}(1,y)dy=\int_{\Omega}e_{1}v(x,y)\chi_{\Gamma_{ 4}}dxdy,$$ $$\int_{0}^{1}v_{2}(x,3)dx=\int_{\Omega}e_{2}v(x,y)\chi_{\Gamma_{ 3}}dxdy,$$ $$\int_{1}^{5}v_{2}(x,1)dx=\int_{\Omega}e_{2}v(x,y)\chi_{\Gamma_{ 5}}dxdy,$$ where $\chi_{\Gamma_{ i}}$ is the characteristic function of ${\Gamma_{i}}$, for $i=3,\ldots,6$. Thus $$(u(t)-u_{0},v)+(\nabla U(t),\nabla v)-t((\lambda_{1}\chi_{\Gamma_{ 6}}+\lambda_{2}\chi_{\Gamma _{4}})e_{1}+(\lambda_{3}\chi_{\Gamma_{ 3}}+\lambda_{4}\chi_{\Gamma _{5}})e_{2},v)=0.$$ In particular for $v\in \mathbf{\vartheta}$, we have $$(u(t)-u_{0}-\triangle U(t)-t((\lambda_{1}\chi_{\Gamma_{ 6}}+\lambda_{2}\chi_{\Gamma _{4}})e_{1}+(\lambda_{3}\chi_{\Gamma_{ 3}}+\lambda_{4}\chi_{\Gamma _{5}})e_{2},v)=0.$$ According to \cite{DL}, there exists $Q(t) \in \mathcal{D'}(\Omega)$ such that $$ u(t)-u_{0}-\triangle U(t)-t((\lambda_{1}\chi_{\Gamma_{ 6}}+\lambda_{2}\chi_{\Gamma _{4}})e_{1}+(\lambda_{3}\chi_{\Gamma_{ 3}}+\lambda_{4}\chi_{\Gamma _{5}})e_{2}=-\nabla Q(t) \in \mathcal{D'}(\Omega), \forall t\in [0,T].$$ Thus $$u(t)-u_{0}-\triangle U(t)+\nabla Q(t)=t((\lambda_{1}\chi_{\Gamma_{ 6}}+\lambda_{2}\chi_{\Gamma _{4}})e_{1}+(\lambda_{3}\chi_{\Gamma_{ 3}}+\lambda_{4}\chi_{\Gamma _{5}})e_{2},$$ where $Q \in C([0,T];L^{2}(\Omega))$.\\ Thus, $\forall \varphi\in\mathcal{D}(\Omega),\ \forall t\in [0,T]$ $$ \begin{array}{rl} \int_{\Omega}(u(t)-u_{0})\varphi({\bf{x}})d{\bf{x}}&- \int_{\Omega}U(t)\triangle\varphi ({\bf{x}}) d{\bf{x}}-\int_{\Omega}Q(t)div\varphi({\bf{x}})d{\bf{x}} \\ &= \int_{\Omega}t((\lambda_{1}\chi_{\Gamma _{6}}+\lambda_{2}\chi_{\Gamma_{4}})e_{1} +(\lambda_{3}\chi_{\Gamma _{3}}+\lambda_{4}\chi_{\Gamma_{ 5}})e_{2})\varphi({\bf{x}})d{\bf{x}}. \end{array} $$ Consequently, $ \forall \psi \in \mathcal{D}(0,T)$ \begin{align*} & -\int_{0}^{T}\int_{\Omega}t((\lambda_{1}\chi_{\Gamma _{6}}+\lambda_{2}\chi_{\Gamma_{ 4}})e_{1} +(\lambda_{3}\chi_{\Gamma _{3}}+\lambda_{4}\chi_{\Gamma_{ 5}})e_{2})\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt\\&= -\int_{0}^{T}\int_{\Omega}u(t)\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt +\int_{0}^{T}\int_{\Omega}u_{0}({\bf{x}})\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt \\ &+\int_{0}^{T}\int_{\Omega}U(t)\triangle\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt+\int_{0}^{T} \int_{\Omega}Q(t)div\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt,\\ \end{align*} and $$ \int_{0}^{T}\int_{\Omega}u_{0}({\bf{x}})\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt=0.$$ Let $\Omega_{T}=\Omega\times(0,T),$ we get \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_{0}^{T}\int_{\Omega}U(t)\triangle\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt& =& \int_{\Omega}\triangle\varphi({\bf{x}})d{\bf{x}} \int_{0}^{T}U(t)\psi'(t)dt\\ &=&-\int_{\Omega}\triangle\varphi({\bf{x}})d{\bf{x}}\int_{0}^{T}u(t)\psi(t)dt\\ &=& \int_{\Omega_{T}}u(t)\triangle\varphi({\bf{x}})\psi(t)d{\bf{x}}dt. \end{eqnarray*} Thus \begin{align*} &-\int_{0}^{T}\int_{\Omega}t((\lambda_{1}\chi_{\Gamma _{6}}+\lambda_{2}\chi_{\Gamma_{ 4}})e_{1} +(\lambda_{3}\chi_{\Gamma _{3}}+\lambda_{4}\chi_{\Gamma_{ 5}})e_{2})\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt\\ &=\int_{\Omega_{T}}((\lambda_{1}\chi_{\Gamma _{6}}+\lambda_{2}\chi_{\Gamma_{ 4}})e_{1} +(\lambda_{3}\chi_{\Gamma _{3}}+\lambda_{4}\chi_{\Gamma _{5}})e_{2})\varphi({\bf{x}})\psi(t)d{\bf{x}}dt\\ &=0, \end{align*} because $\varphi\in\mathcal{D}(\Omega)$. $$\int_{0}^{T}\int_{\Omega}Q(t)div\varphi({\bf{x}})\psi'(t)d{\bf{x}}dt= -\int_{\Omega_{T}}Q'(t)div\varphi({\bf{x}})\psi(t)d{\bf{x}}dt.$$ Put $P_{1}(t)=-Q'(t) \in \mathcal{D'}(0,T;L^{2}(\Omega))\mbox{ and } \phi(t,{\bf{x}})=\varphi({\bf{x}})\psi(t)$, we conclude that\\ $$-\int_{\Omega_{T}}u(t,{\bf{x}})\frac{\partial\phi}{\partial t}(t,{\bf{x}})d{\bf{x}}dt -\int_{\Omega_{T}}u(t,{\bf{x}})\triangle_{{\bf{x}}}\phi(t,{\bf{x}})d{\bf{x}}dt+ \int_{\Omega_{T}}P_{1}(t,{\bf{x}})div_{{\bf{x}}}\phi(t,{\bf{x}})d{\bf{x}}dt=0.$$ Since $$\int_{\Omega_{T}}P_{1}(t,{\bf{x}})div_{{\bf{x}}}\phi(t,{\bf{x}})dxdt= \int_{\partial\Omega\times(0,T)}P_{1}\phi(t,{\bf{x}})d\sigma -\int_{\Omega_{T}}\nabla P_{1}\phi(t,{\bf{x}})d{\bf{x}}dt$$ and $$\int_{\partial\Omega\times(0,T)}P_{1}\phi(t,{\bf{x}})d\sigma =0,$$ we obtain $$ -\int_{\Omega_{T}}u(t,{\bf{x}})\frac{\partial\phi}{\partial t}(t,{\bf{x}})d{\bf{x}}dt -\int_{\Omega_{T}}u(t,{\bf{x}})\triangle_{{\bf{x}}}\phi(t,{\bf{x}})d{\bf{x}}dt-\int_{\Omega_{T}}\nabla P_{1}\phi(t,{\bf{x}})d{\bf{x}}dt=0.$$ We know that the set of all finite linear combinations of the functions $\phi(t,{\bf{x}})=\varphi({\bf{x}})\psi(t)~~(\varphi\in\mathcal{D}(\Omega),\psi\in\mathcal{D}(0,T))$ is dense in $\mathcal{D}(\Omega_{T})$. Thus $$\frac{\partial u}{\partial t}(t,{\bf{x}})-\triangle u(t,{\bf{x}})+\nabla P=0,$$ where $P=-P_{1}$.\\ From the first two equations of the problem $(S_{1})$ and $(S_{2})$, we find \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \sum\limits_{i=1}^{2} \int_{\partial \Omega}\frac{\partial u_{i}}{\partial x}.v_{i} d\sigma -\int_{\partial \Omega}P v.\vec{\eta} d\sigma &=& \lambda_{1}\int_{0}^{1}v_{1}(5,y)dy+\lambda_{2}\int_{1}^{3}v_{1}(1,y)dy\\ &+& \lambda_{3}\int_{0}^{1}v_{2}(x,3)dx+\lambda_{4}\int_{1}^{5}v_{2}(x,1)dx. \end{eqnarray*} Thus $$ \langle\frac{\partial u}{\partial x}-P\vec{\eta},v\rangle=\langle\lambda_{1}e_{1},v\rangle_{\Gamma_{ 1}}+ \langle\lambda_{2}e_{1},v\rangle_{\Gamma_{ 2}} +\langle\lambda_{3}e_{2},v\rangle_{\Gamma _{3}}+\langle\lambda_{4}e_{2},v\rangle_{\Gamma_{ 4}},$$ and \begin{equation*} \langle\frac{-\partial u}{\partial x}+Pe_{1},v\rangle_{\Gamma _{1}}+\langle\frac{-\partial u}{\partial x}+Pe_{1},v\rangle_{\Gamma_{ 2}} +\langle\frac{\partial u}{\partial y}-Pe_{2},v\rangle_{\Gamma_{ 3}} +\langle\frac{\partial u}{\partial x}-Pe_{1},v\rangle_{\Gamma_{ 4}} \end{equation*} \begin{equation*} +\langle\frac{\partial u}{\partial y}-Pe_{2},v\rangle_{\Gamma _{5}}+\langle\frac{\partial u}{\partial x}-Pe_{1},v\rangle_{\Gamma_{ 6}} +\langle\frac{-\partial u}{\partial y}+Pe_{2},v\rangle_{\Gamma _{7}}\\ +\langle\frac{-\partial u}{\partial y}+Pe_{2},v\rangle_{\Gamma_{ 8}} \end{equation*} \begin{equation}\label{E13} =\langle\lambda_{1}e_{1},v\rangle_{\Gamma_{1}} +\langle\lambda_{2}e_{1},v\rangle_{\Gamma_{ 2}} +\langle\lambda_{3}e_{2},v\rangle_{\Gamma _{3}}+\langle\lambda_{4}e_{2},v\rangle_{\Gamma_{ 4}}. \end{equation} We have $u|_{\Gamma_{ 1}}=u|_{\Gamma_{ 6}}$, thus $u_{1}|_{\Gamma _{1}}=u_{1}|_{\Gamma_{ 6}}$ and $ u_{2}|_{\Gamma_{ 1}}=u_{2}|_{\Gamma _{6}}$, as $u_{i}(0,y)=u_{i}(5,y)$ for all $y\in [0,1]$, we have $ \frac{\partial u_{i}}{\partial y}(0,y)=\frac{\partial u_{i}}{\partial y}(5,y).$ Moreover we know that $div\,u=0 \mbox{ in } \Omega$ and $u\in C^{1}(\bar{\Omega})^{2}$, we conclude that $\frac{\partial u_{1}}{\partial x}(0,y)=-\frac{\partial u_{2}}{\partial y}(0,y)=-\frac{\partial u_{2}}{\partial y}(5,y)$ for all $y\in [0,1]$. Thus $\frac{\partial u_{1}}{\partial x}(0,y)=\frac{\partial u_{1}}{\partial x}(5,y)$. We consider the space $$H_{11}^{\frac{1}{2}}(\Gamma_{ 1})=\{\varphi \in L^{2}(\Gamma _{1}); \exists v \in H^{1}(\Omega),~ v|_{\Gamma_{3}}=v|_{\Gamma_{ 8}},~ v|_{\Gamma_{ 5}}=v|_{\Gamma _{7}},~ v|_{\Gamma_{ 2}}=v|_{\Gamma _{4}} ,~ v|_{\Gamma_{ 1} \cup \Gamma_{6} }=\varphi\}$$ Let $\mu\in H^{\frac{1}{2}}_{11}(\Gamma_{1})$, we put $\nu=(0,\mu_{1})^{t} $ where $\mu_{1}=\left\{ \begin{array}{ll} \mu & \hbox{on } \Gamma_{1} \cup \Gamma_{6}\\ 0 & \hbox{on } \Gamma_{2} \cup \Gamma_{3}\cup \Gamma_{4}\cup \Gamma_{5}\cup \Gamma_{7}\cup \Gamma_{8}. \end{array} \right.$ It is clear that $\nu \in (H^{\frac{1}{2}}(\Gamma))^{2} \mbox{ and } \int_{\partial\Omega}\nu.\vec{\eta} d\sigma=0$, so there exists $v\in (H^{1}(\Omega))^{2}$ such that $div\,v=0 \mbox{ in } \Omega \mbox{ and } v=\nu \mbox{ on } \Gamma$ (see \cite{AB,GR}), therefore $v \in V_{1}$. According to \eqref{E13}, we have for all $\mu \in H_{11}^{\frac{1}{2}}(\Gamma_{1})$ $$ \int_{0}^{1}\frac{\partial u_{2}}{\partial x}(0,y)\mu dy= \int_{0}^{1}\frac{\partial u_{2}}{\partial x}(5,y)\mu dy,$$ thus $$ \frac{\partial u_{2}}{\partial x}(0,y)= \frac{\partial u_{2}}{\partial x}(5,y).$$ Similarly, we have $$\frac{\partial u}{\partial x}|_{\Gamma_{2}}=\frac{\partial u}{\partial x}|_{\Gamma_{4}},~ \frac{\partial u}{\partial y}|_{\Gamma_{3}}=\frac{\partial u}{\partial y}|_{\Gamma_{8}},~ \frac{\partial u}{\partial y}|_{\Gamma_{5}}=\frac{\partial u}{\partial y}|_{\Gamma_{7}}.$$ According to \eqref{E13}, we have \begin{equation*}%\label{E12} \langle Pe_{1},v\rangle_{\Gamma_{ 1}}+\langle Pe_{1},v\rangle_{\Gamma_{ 2}} +\langle -Pe_{2},v\rangle_{\Gamma_{ 3}} +\langle -Pe_{1},v\rangle_{\Gamma_{ 4}} +\langle -Pe_{2},v\rangle_{\Gamma_{ 5}}+\langle -Pe_{1},v\rangle_{\Gamma_{ 6}} \end{equation*} \begin{equation}\label{E12} +\langle Pe_{2},v\rangle_{\Gamma_{ 7}} +\langle Pe_{2},v\rangle_{\Gamma_{ 8}} = \langle\lambda_{1}e_{1},v\rangle_{\Gamma_{1}} +\langle\lambda_{2}e_{1},v\rangle_{\Gamma_{ 2}} +\langle\lambda_{3}e_{2},v\rangle_{\Gamma_{ 3}}+\langle\lambda_{4}e_{2},v\rangle_{\Gamma_{ 4}}. \end{equation} On the other hand, let $\varepsilon_{1}=$ $\left\{ \begin{array}{ll} \varepsilon & \hbox{on } \Gamma_{ 1} \cup \Gamma_{ 6},\\ 0 & \hbox{on } \Gamma_{ 2} \cup \Gamma_{ 3}\cup \Gamma_{ 4}\cup \Gamma_{ 5}\cup \Gamma_{ 7}\cup \Gamma_{ 8}, \end{array} \right.$ where $\varepsilon \in H_{11}^{\frac{1}{2}}(\Gamma_{ 1})\mbox{ and } \tau=(\varepsilon_{1},0)^{t} $. We have $\tau\in (H^{\frac{1}{2}}(\Gamma))^{2}\mbox{ and } \int_{\Gamma}\tau.\vec{\eta} d\sigma =0 $. So, there exists $v \in (H^{1}(\Omega))^{2}$ such that $div\,v=0\mbox{ in } \Omega \mbox{ and } v=\tau \mbox{ on } \Gamma $ (see \cite{AB,GR}). In particular $v\in V_{1}$. According to \eqref{E12}, $$\int_{0}^{1}(p(0,y)-p(5,y))\varepsilon =\int_{0}^{1}\lambda_{1}\varepsilon$$ we conclude that $P|_{\Gamma_{ 1}}-P|_{\Gamma_{ 6}}=\lambda_{1}$. In the same way we show the other relations of pressures. \end{proof} \section{Conclusion} We studied a nonstationary variational problem, which has some nonlinearity in the second member. We proved the existence, uniqueness and regularity of the solution. We gave the equivalence between the weak and the classical problem that highlights the differences in pressures between two opposite edges. \begin{thebibliography}{99} \bibitem{AB} C. Amrouche, M. Batchi, J. Batina, Navier-Stokes equations with periodic boundary conditions and pressure loss, {\it Applied Mathematics Letters}. \textbf{20} (2007), 48--53. \bibitem{ABe} C. Amrouche, M.\'A.R. Bellido, Very weak solutions for the stationary Oseen and Navier-Stokes equations, {\it Comptes Rendus Mathematiques}. \textbf{348} (2010), 335--339. \bibitem{DL} {R. Dautray, J.L. Lions}, {Analyse mathématique et calcul numérique pour les sciences et les techniques}, tome3, Masson, Paris, 1985. \bibitem{FG} R. Farwig, G.P. Galdi, H. Sohr, Very weak solutions and large uniqueness class of stationary Navier-Stokes equations in bounded domains of $\mathbb{R}^{2}$, {\it Journal of Differential Equations}. \textbf{227} (2006), 564--580. \bibitem{GR} {V. Giraut, P.A. Raviart}, {Finite Element Methods for Navier-Stokes Equations}, Springer Series SCM, 1986. \end{thebibliography} \end{document}