\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 117, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/117\hfil Existence and asymptotic behavior] {Existence and asymptotic behavior of global regular solutions for a 3-D Kazhikhov-Smagulov model with Korteweg stress} \author[M. Ezzoug, E. Zahrouni \hfil EJDE-2016/117\hfilneg] {Meriem Ezzoug, Ezzeddine Zahrouni} \address{Meriem Ezzoug \newline Unit\'e de recherche: Multifractals et Ondelettes, FSM, University of Monastir, 5019 Monastir, Tunisia} \email{meriemezzoug@yahoo.fr} \address{Ezzeddine Zahrouni \newline Unit\'e de recherche: Multifractals et Ondelettes, FSM, University of Monastir, 5019 Monastir, Tunisia. \newline FSEGN, University of Carthage, 8000 Nabeul, Tunisia} \email{ezzeddine.zahrouni@fsm.rnu.tn} \thanks{Submitted March 9, 2016. Published May 10, 2016.} \subjclass[2010]{35Q30, 76D03, 35B40} \keywords{Kazhikhov-Smagulov-Korteweg model; global solution; \hfill\break\indent uniqueness; asymptotic behavior} \begin{abstract} In this article, we consider a 3-D multiphasic incompressible fluid model, called the Kazhikhov-Smagulov model, with a specific Korteweg stress tensor. We prove the existence of a global unique regular solution to the Kazhikhov-Smagulov-Korteweg model provided that initial data and external force are sufficiently small. Furthermore, in the absence of external forcing, the solution decays exponentially in time to the equilibrium solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} In this article, we study a 3-D Kazhikhov-Smagulov-Korteweg (KSK) model describing the motion of a viscous incompressible mixture of two fluids having different densities. This type model can be derived from the compressible Navier-Stokes system. Let $\Omega$ be a bounded open set in $\mathbb{R}^3$ with boundary $\Gamma$ that is regular enough. We denote by $[0,T]$ the time interval, for $T>0$. The mixture of two fluids is described by the density $\rho(t,\mathbf{x}) \ge 0$, the mass velocity field $\mathbf{v}(t,\mathbf{x})$ and the pressure $p(t,\mathbf{x})$, depending on the time and space variables $(t,\mathbf{x}) \in [0,T] \times \Omega$. According to Dunn and Serrin \cite{DS} (see also Bresch et al \cite{BDL2003}), we consider the compressible Navier-Stokes system \begin{equation} \label{ModeleCompres} \begin{gathered} \frac{\partial}{\partial t} (\rho \mathbf{v}) + \operatorname{div}\big(\rho \mathbf{v} \otimes \mathbf{v}\big) = \rho \mathbf{g} + \operatorname{div}\big(\mathbf{S} + \mathbf{K}\big), \\ \frac{\partial\rho}{\partial t} + \operatorname{div}(\rho \mathbf{v}) = 0, \end{gathered} \end{equation} where $\mathbf{g}$ stands for the gravity acceleration (but it can include further external forces). The viscous stress tensor $\mathbf{S}$ and the Korteweg stress tensor $\mathbf{K}$ given by \begin{equation} \label{tensors} \begin{gathered} \mathbf{S} = (\nu \operatorname{div}\mathbf{v} - p)\mathbf{I} + 2\mu \mathbf{D}(\mathbf{v}), \\ \mathbf{K} = (\alpha \Delta\rho + \beta |\nabla\rho|^2)\mathbf{I} + \delta (\nabla\rho\otimes\nabla\rho) + \gamma D^2_x \rho, \end{gathered} \end{equation} where $\mathbf{D}(\mathbf{v}) = (\nabla \mathbf{v} + \nabla \mathbf{v}^T)/2$ is the strain tensor and $D^2_x \rho $ is the hessian matrix of the density $\rho$. The pressure $p$ and the coefficients $\alpha$, $\beta$, $\gamma$, $\delta$, $\nu$ and $\mu$ are functions of $\rho$. As in \cite{FS2001}, choosing the viscosity coefficients $\nu$ and $\mu$ constants in the viscous stress tensor $\mathbf{S}$, we have \begin{equation} \label{tensor_viscous1} \operatorname{div} \mathbf{S} = \nu \nabla(\operatorname{div}\mathbf{v}) - \nabla p + 2 \mu \operatorname{div}\big(\mathbf{D}(\mathbf{v})\big). \end{equation} In the Korteweg stress tensor $\mathbf{K}$, we consider the special case: $$ \alpha = \kappa \rho, \quad \beta = \frac{\kappa}{2}, \quad \delta = - \kappa, \quad \gamma=0, $$ for some constant $\kappa >0$, called Korteweg's constant. This choice corresponds essentially to the Korteweg's original assumptions connected with the variational theory of Van Der Waals (see \cite{K}). Therefore, the Korteweg stress tensor yields \begin{equation}\label{tensor_korteweg} \mathbf{K} = \frac{\kappa}{2} (\Delta \rho^2 - | \nabla \rho |^2) \mathbf{I} - \kappa (\nabla \rho \otimes \nabla \rho), \end{equation} and we obtain \begin{equation} \label{tensor_korteweg2} \operatorname{div} \mathbf{K} = \kappa \rho\nabla(\Delta \rho) = \kappa \nabla(\rho \Delta\rho) - \kappa \Delta\rho \nabla\rho. \end{equation} On another side, Fick's law which relates the velocity to the derivatives of the density (see \cite{KS, AnKazM}), gives \begin{equation}\label{Decomp} \mathbf{v} = \mathbf{u} - \lambda \nabla\ln(\rho), \end{equation} with a volume velocity field $\mathbf{u}$ that is solenoidal ($\operatorname{div}\,\mathbf{u} = 0$) and $\lambda>0$ is interpreted as a diffusion coefficient. Consequently, we use \eqref{Decomp} in the compressible Navier-Stokes system \eqref{ModeleCompres}, and after some calculations, we obtain the following system, that we call the Kazhikhov-Smagulov-Korteweg (KSK) model, \begin{equation} \label{Modele1} \begin{gathered} \begin{aligned} &\rho \Big(\frac{\partial\mathbf{u}}{\partial t} + (\mathbf{u}\cdot\nabla)\mathbf{u}\Big) - \mu \boldsymbol{\Delta} \mathbf{u} - \lambda \big(\nabla\rho\cdot\nabla\big)\mathbf{u} - \lambda \big(\mathbf{u}\cdot\nabla\big)\nabla\rho \\ &+ \nabla P + \frac{\lambda^2}{\rho} \Big(\Delta\rho \nabla\rho + \big(\nabla\rho\cdot\nabla\big)\nabla\rho - \frac{|\nabla\rho|^2}{\rho} \nabla\rho\Big) = \rho \mathbf{g} - \kappa \Delta\rho \nabla\rho, \end{aligned} \\ \frac{\partial\rho}{\partial t} + \mathbf{u} \cdot \nabla\rho = \lambda \Delta \rho, \\ \operatorname{div} \mathbf{u} = 0. \end{gathered} \end{equation} With $\mathcal{Q}_T = (0,T) \times \Omega$ and $\Sigma = (0,T) \times \Gamma$, the unknowns for the model \eqref{Modele1} are $\rho : \mathcal{Q}_T \to \mathbb{R}$ the density of the fluid, $\mathbf{u} : \mathcal{Q}_T \to \mathbb{R}^3$ the incompressible velocity field and $P : \mathcal{Q}_T \to \mathbb{R}$ the modified pressure. We attach to \eqref{Modele1} the following boundary and initial conditions: \begin{gather} \mathbf{u}(t,\mathbf{x}) = 0, \quad \frac{\partial\rho}{\partial\mathbf{n}} (t,\mathbf{x}) = 0, \quad (t,\mathbf{x}) \in \Sigma, \label{CondAuxBord1} \\ \mathbf{u}(0,\mathbf{x}) = \mathbf{u}_0(\mathbf{x}), \quad \rho(0,\mathbf{x}) = \rho_0(\mathbf{x}), \quad \mathbf{x} \in \Omega, \label{CondInitiales1} \end{gather} with the compatibility condition $\operatorname{div} \mathbf{u}_0 = 0$, where $\rho_0 : \Omega \to \mathbb{R}$ and $\mathbf{u}_0 : \Omega \to \mathbb{R}^3$ are given functions. We denote by $ \mathbf{n} $ the unit outward normal on the boundary $\Gamma$. Throughout this work, we assume the hypothesis \begin{equation} \label{ProprieteRho0} 0 < m \leq \rho_0(\mathbf{x}) \leq M < +\infty, \quad \mathbf{x} \in \Omega. \end{equation} Let us mention some known results about the Kazhikhov-Smagulov model without the Korteweg stress tensor. Taking $\kappa = 0$, many authors study the global existence of solution for the so-called Kazhikhov-Smagulov model. We can refer for instance to \cite{AnKazM, KS, BES, Secchi1988}. In \cite{BDV}, Beir\~{a}o da Veiga considered the same model \eqref{Modele1} without Korteweg term and proved the existence of a unique local solution for arbitrary initial data and external force and the existence of a unique global regular solution for small initial data and external force. Moreover, he proved that if $\mathbf{g} =0$, the solution decay exponentially in time to the equilibrium solution with zero velocity field. In \cite{BDV-S-V}, Beir\~{a}o da Veiga et al. have previously found the same results obtained in \cite{BDV}, in the non-viscous case for an Euler system. The aim of this work is to establish the same kind of results given in \cite{BDV} for \eqref{Modele1}. That is existence of a unique global in time regular solution of the Kazhikhov-Smagulov-Korteweg model \eqref{Modele1} for small initial data and external force. Also, we study the longtime behavior of the solution and show that it converges to a constant solution with zero velocity field. We think that the results presented here can be extended if we replace the Laplace operator by the $p$-Laplace operator $\operatorname{div}\big(|\nabla\mathbf{u}|^{p-2} \nabla\mathbf{u}\big)$, $10$, $\mathbf{g} \in L^2\big(0,T;\mathbf{L}^2(\Omega)\big)$ and $$ \widehat{\rho} = \frac{1}{|\Omega|} \int_\Omega \rho_0(\mathbf{x})\, d\mathbf{x}. $$ There exist positive constants $\gamma_1$, $\gamma_2$, $\gamma_3$ depending on $\Omega$, $\lambda$, $\mu$, $\kappa$, $M$, $m$, such that if \begin{equation}\label{HypPettitesseChap3} \begin{gathered} \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \rho_0 - \widehat{\rho} \|^2_{{H^2(\Omega)}} \leq \gamma_1, \\ \| \mathbf{g} \|^2_{{L^\infty(0,+\infty;L^2(\Omega))}} \leq \gamma_2, \end{gathered} \end{equation} then there exists a unique regular solution $(\mathbf{u}, \rho)$ of problem \eqref{Modele1}, \eqref{CondAuxBord1}, \eqref{CondInitiales1}, global in time such that \begin{gather*} \mathbf{u} \in L^2\big(0,T; \mathbf{H}^2(\Omega)\big) \cap \mathcal{C}\big([0,T]; \mathbf{V}\big), \\ \rho \in L^2\big(0,T; H^3_N\big) \cap \mathcal{C}\big([0,T]; H^2_N\big). \end{gather*} Moreover if $\mathbf{g} = \boldsymbol 0$, the solution $(\mathbf{u}, \rho)$ decays exponentially in time to the equilibrium solution $(\boldsymbol 0, \widehat{\rho})$, such that $\forall t\geq 0$, \begin{equation} \label{Asymptotic} \| \nabla\mathbf{u}(t) \|_{L^2(\Omega)}^2 + \| \rho(t) - \widehat{\rho} \|_{{H^2(\Omega)}}^2 \leq \big(\| \nabla\mathbf{u}_0 \|_{L^2(\Omega)}^2 + \| \rho_0 - \widehat{\rho} \|_{{H^2(\Omega)}}^2\big) \mathrm{e}^{-\gamma_3 t}. \end{equation} \end{theorem} \section{Proof of Theorem \ref{ThExistence2Chap3}\label{Section3}} \subsection*{Intermediate results} In this section we present some results to be used in proving Theorem \ref{ThExistence2Chap3}. First of all, integrating the convection-diffusion equation \eqref{Modele1}$_2$ over $\Omega$, we see that $$ \frac{d}{dt} \int_\Omega \rho(t,\mathbf{x}) \, d\mathbf{x} = 0, $$ and we note that the mean value of $\rho$ is conserved: $$ \int_\Omega \rho(t,\mathbf{x}) \, d\mathbf{x} = \int_\Omega \rho_0(\mathbf{x}) \, d\mathbf{x}. $$ Therefore, we set \begin{equation} \label{SuiteTranslation} \sigma = \rho - \widehat{\rho}, \end{equation} such that $\widehat{\rho} = \frac{1}{|\Omega|} \int_\Omega \rho_0(\mathbf{x}) \, d\mathbf{x}$ and $\int_\Omega \sigma(t,\mathbf{x}) \, d\mathbf{x} = 0$. Next, the KSK model \eqref{Modele1} is equivalent to find $(\mathbf{u}, \sigma)$ satisfying \begin{equation} \label{ModeleKSK2} \begin{gathered} \mathbb{P}\big(\rho \frac{\partial\mathbf{u}}{\partial t}\big) - \mu \mathbb{P}\boldsymbol{\Delta}\mathbf{u} = \mathbf{F}(\mathbf{u}, \sigma), \\ \frac{\partial\sigma}{\partial t} - \lambda \Delta \sigma = G(\mathbf{u}, \sigma), \\ \operatorname{div}\mathbf{u} = 0, \end{gathered} \end{equation} where \begin{equation} \label{F expression} \begin{gathered} \begin{aligned} \mathbf{F}(\mathbf{u}, \sigma) &= \mathbb{P}\Big(\rho \mathbf{g} - \kappa \Delta\rho \nabla\rho - \rho \big(\mathbf{u}\cdot\nabla\big)\mathbf{u} + \lambda \big(\nabla\rho \cdot\nabla\big) \mathbf{u} + \lambda \big(\mathbf{u} \cdot\nabla\big) \nabla\rho \\ &\quad - \frac{\lambda^2}{\rho} \Delta\rho \nabla\rho - \frac{\lambda^2}{\rho} \big(\nabla\rho \cdot\nabla\big) \nabla\rho + \lambda^2 \frac{|\nabla\rho|^2}{\rho^2} \nabla\rho\Big), \end{aligned}\\ G(\mathbf{u}, \sigma) = -\mathbf{u} \cdot \nabla\sigma, \end{gathered} \end{equation} Problem \eqref{ModeleKSK2} is coupled with the boundary and initial conditions \begin{gather*} \mathbf{u}(t,\mathbf{x}) = 0, \quad \frac{\partial\sigma}{\partial\mathbf{n}}(t,\mathbf{x}) = 0, \quad (t,\mathbf{x}) \in \Sigma, \\ \mathbf{u}(0,\mathbf{x}) = \mathbf{u}_0(\mathbf{x}), \quad \sigma(0,\mathbf{x}) = \sigma_0(\mathbf{x}), \quad \mathbf{x} \in \Omega, \end{gather*} where $\sigma_0(\mathbf{x}) = \rho_0(\mathbf{x}) - \widehat{\rho}$. We introduce the spaces: \begin{align*}%\label{EspaceX1Chap3} \mathcal{X}_1 =\Big\{&\bar{\mathbf{u}}: \bar{\mathbf{u}} \in L^2\big(0,T; \mathbf{H}^2(\Omega)\big) \cap \mathcal{C}\big([0,T]; \mathbf{V}\big); \frac{\partial\bar{\mathbf{u}}}{\partial t} \in L^2\big(0,T; \mathbf{H}\big); \bar{\mathbf{u}}(0) = \mathbf{u}_0; \\ &\| \bar{\mathbf{u}} \|^2_{{\mathcal{C}([0,T]; \mathbf{V})}} + \| \bar{\mathbf{u}} \|^2_{{L^2(0,T; H^2(\Omega))}} + \| \frac{\partial\bar{\mathbf{u}}}{\partial t} \|^2_{{L^2(0,T; \mathbf{H})}} \leq 2 C_4 \| \nabla\mathbf{u}_0 \|_{L^2(\Omega)}^2 \Big\} \end{align*} and \begin{align*} %\label{EspaceX2Chap3} \mathcal{X}_2 = \Big\{& \bar{\sigma}: \bar{\sigma} \in L^2\big(0,T; H^3_{N,0}\big) \cap \mathcal{C}\big([0,T]; H^2_{N,0}\big); \frac{\partial\bar{\sigma}}{\partial t} \in L^2\big(0,T; H^1(\Omega)\big); \\ & \bar{\sigma}(0) = \sigma_0;~ \| \bar{\sigma} \|^2_{\mathcal{C}([0,T];H^2(\Omega))} + \| \bar{\sigma} \|^2_{L^2(0,T;H^3(\Omega))} \leq 2 \|\sigma_0 \|^2_{{H^2(\Omega)}}; \\ &\| \frac{\partial\bar{\sigma}}{\partial t} \|^2_{{L^2(0,T;H^1(\Omega))}} \leq K_0; \; \| \bar{\sigma} - \sigma_0 \|_{{\mathcal{C}(\bar{Q}_T)}} \leq \frac{m}{2}\Big\}. \end{align*} Here $C_4$ is a positive constant depending on $\mu$, $\bar{M}$, $\bar{m}$ and we denote by $K_0$ a positive constant depending on norms of initial data $\| \nabla\mathbf{u}_0 \|_{L^2(\Omega)}$ and $\| \sigma_0 \|_{{H^2(\Omega)}}$. Now, we define the linearized problem as follows: Given $(\bar{\mathbf{u}}, \bar{\sigma}) \in \mathcal{X}_1 \times \mathcal{X}_2$ such that $\bar{\sigma}=\bar{\rho}-\widehat{\rho}$, find $(\mathbf{u},\sigma)\in\mathcal{X}_1\times\mathcal{X}_2 $ such that $\sigma=\rho-\widehat{\rho}$ satisfying \begin{equation} \label{SystemeLineaire} \begin{gathered} \mathbb{P}\big(\bar{\rho} \frac{\partial\mathbf{u}}{\partial t}\big) + \mu \mathbb{A}\mathbf{u} = \mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma}), \\ \frac{\partial\sigma}{\partial t} - \lambda~\Delta \sigma = G(\bar{\mathbf{u}}, \bar{\sigma}), \\ \operatorname{div}\mathbf{u} = 0, \\ \int_\Omega \sigma(t,\mathbf{x}) \, d\mathbf{x} = 0, \end{gathered} \end{equation} For $(\bar{\mathbf{u}}, \bar{\sigma}) \in \mathcal{X}_1 \times \mathcal{X}_2$, we define the map $$ \Phi : \mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{X}_1 \times \mathcal{X}_2, $$ such that $\Phi(\bar{\mathbf{u}}, \bar{\sigma}) = (\mathbf{u}, \sigma)$ defined by \eqref{SystemeLineaire}. Since \eqref{SystemeLineaire} is a linear problem with respect to $\mathbf{u}$ and $\sigma$, it is clear that $\Phi$ is well defined (see \cite[\S 2]{BDV}, \cite[Vol.I, Chap.1, Theorem 3.1]{LM} and \cite[Vol.II, Chap.4, Theorem 5.2]{LM}). Analogously as in \cite{BDV}, we can prove \emph{the existence of a local regular solution in time} to \eqref{Modele1} for arbitrary initial data and external force in the three-dimensional case. For this, we consider the linearized problem \eqref{SystemeLineaire} and we prove via an application of Schauder fixed point theorem, the existence of a fixed point $(\bar{\mathbf{u}}, \bar{\sigma}) \in \mathcal{X}_1 \times \mathcal{X}_2$ for the map $\Phi$, such that $$ (\bar{\mathbf{u}}, \bar{\sigma}) = (\mathbf{u}, \sigma). $$ (See \cite{BDV} for a detailed study.) To prove the main result of this article, Theorem \ref{ThExistence2Chap3}, we need some useful results. On one hand, from the estimate \eqref{ProprieteRho0} for the initial density $\rho_0$ follows a similar estimate for $\bar{\rho}$. \begin{proposition}\label{Prop1Chap3} Let $\bar{\sigma}\in\mathcal{X}_2$. Then the function $\bar{\rho}=\bar{\sigma}+\widehat{\rho}$ satisfies \begin{equation} \label{ProprieteRhoBAR} \bar{m} \equiv \frac{m}{2} \leq \bar{\rho}(t,\mathbf{x}) \leq M + \frac{m}{2} \equiv \bar{M}, \quad (t,\mathbf{x}) \in \mathcal{Q}_T. \end{equation} \end{proposition} On the other hand, the right-hand side $\mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma})$ of \eqref{SystemeLineaire}, defined by \eqref{F expression}, satisfies the following property. \begin{proposition} Let $\mathbf{g} \in L^2\big(0,T,\mathbf{L}^2(\Omega)\big)$ and $(\bar{\mathbf{u}}, \bar{\sigma}) \in \mathcal{X}_1 \times \mathcal{X}_2$. Then $\mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma})$ defined by \eqref{F expression}, satisfies \begin{equation}\label{eq1Chap3} \begin{aligned} \| \mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma}) \|^2_{L^2(\Omega)} &\leq C \Big(\|\nabla\bar{\mathbf{u}} \|^{2(1+\beta)}_{L^2(\Omega)} \| \nabla\bar{\mathbf{u}} \|^{2(1-\beta)}_{H^1(\Omega)} + \| \nabla\bar{\sigma} \|^{2(1+\beta)}_{H^1(\Omega)} \|\Delta\bar{\sigma}\|^{2(1-\beta)}_{H^1(\Omega)} \\ &\quad + \| \nabla\nabla\bar{\sigma} \|^{2\beta}_{L^2(\Omega)} \| \nabla\nabla\bar{\sigma} \|^{2(1-\beta)}_{H^1(\Omega)} \| \nabla\bar{\mathbf{u}} \|^2_{L^2(\Omega)} + \| \nabla\bar{\sigma} \|_{H^1(\Omega)}^6 \\ &\quad + \| \nabla\bar{\mathbf{u}} \|^{2\beta}_{L^2(\Omega)} \| \nabla\bar{\mathbf{u}} \|^{2(1-\beta)}_{H^1(\Omega)} \|\nabla\bar{\sigma}\|_{H^1(\Omega)}^2 + \| \mathbf{g} \|^2_{L^2(\Omega)}\Big), \end{aligned} \end{equation} where $C= C\big(\lambda, \kappa, \bar{M}, \bar{m}\big)$, and \begin{equation*} \beta = \begin{cases} 1/2 & \text{if } d = 2, \\ 1/4 & \text{if } d = 3. \end{cases} \end{equation*} \end{proposition} \begin{lemma}\label{Lemma1} Let $(\bar{\mathbf{u}}, \bar{\sigma})\in \mathcal{X}_1 \times \mathcal{X}_2$ and $\mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma})\in\mathbf{L}^2(\Omega)$ satisfy \eqref{F expression}. Then a solution $(\mathbf{u}, \sigma)$ of the linearized problem \eqref{SystemeLineaire} satisfies the following estimates: \begin{gather}\label{Inegalite0} \begin{split} &\frac{\mu}{2} \frac{d}{dt} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \frac{\mu \varepsilon_0}{2} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} + \big(\frac{3m}{4} - \frac{\varepsilon_0 M^2}{\mu} \big) \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)} \\ &\leq \big(\frac{1}{m} + \frac{\varepsilon_0}{\mu} \big) \| \mathbf{F}(\bar{\mathbf{u}}, \bar{\sigma}) \|^2_{L^2(\Omega)}, \end{split}\\ \label{Inegalite13} \begin{split} &\frac{d}{dt}\| \Delta\sigma \|^2_{L^2(\Omega)} + \lambda \|\nabla\Delta\sigma\|^2_{L^2(\Omega)} \\ &\leq C_1 \varepsilon_1 \Big(\| \nabla\bar{\mathbf{u}} \|^{2}_{H^1(\Omega)} + \|\nabla\nabla\bar{\sigma} \|^{2}_{H^1(\Omega)}\Big) + 2C_2 \varepsilon_1^{-k_d} \Big(\| \bar{\mathbf{u}} \|^{k_d+3}_{H^1(\Omega)} + \|\nabla\bar{\sigma} \|^{k_d+3}_{H^1(\Omega)}\Big), \end{split} \end{gather} where $\varepsilon_0$, $\varepsilon_1$ being arbitrary, $C_1$, $C_2$ are positive constants depending only on $\Omega$, and \begin{equation*} k_d = \begin{cases} 3 & \text{if } d = 2, \\ 7 & \text{if } d = 3. \end{cases} \end{equation*} \end{lemma} \subsection*{Global solutions} Let $(\mathbf{u}, \rho)$ be a local solution of \eqref{Modele1}, such that $\rho = \sigma + \hat{\rho}$. We will prove that this local solution is, in fact, a global solution. On the one hand, we choose $\varepsilon_0 = \frac{m\mu}{4 M^2}$ in \eqref{Inegalite0} to obtain \begin{equation*} \frac{\mu}{2} \frac{d}{dt} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \frac{m}{2} \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)} + \frac{m\mu^2}{8M^2} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} \leq \big( \frac{1}{m} + \frac{m}{4M^2}\big) \|\mathbf{F} \|^2_{L^2(\Omega)}. \end{equation*} Next, we use \eqref{eq1Chap3} for $\beta=\frac{1}{4}$ as follows: \begin{align*} \| \mathbf{F} \|^2_{L^2(\Omega)} &\leq C \Big( \| \nabla\mathbf{u} \|_{L^2(\Omega)}^{5/2} \| \nabla\mathbf{u} \|_{H^1(\Omega)}^{3/2} + \| \nabla\sigma \|^{5/2}_{H^1(\Omega)} \| \Delta\sigma \|_{H^1(\Omega)}^{3/2} \\ &\quad + \| \nabla\mathbf{u} \|_{L^2(\Omega)}^{1/2} \| \nabla\mathbf{u} \|_{H^1(\Omega)}^{3/2} \| \nabla\sigma \|_{H^1(\Omega)}^2 + \| \nabla\sigma \|_{H^1(\Omega)}^6 \\ &\quad + \| \nabla\nabla\sigma \|_{L^2(\Omega)}^{1/2} \| \nabla\nabla\sigma \|_{H^1(\Omega)}^{3/2} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \mathbf{g} \|^2_{L^2(\Omega)}\Big). \end{align*} Applying the Young inequality $\big(ab\leq \frac{a^5}{5}+\frac{4}{5}b^{5/4}\big)$, we obtain \begin{align*} \| \mathbf{F} \|^2_{L^2(\Omega)} &\leq C \Big( \big(\| \nabla\mathbf{u} \|_{L^2(\Omega)}^{5/2} + \| \nabla\sigma \|^{5/2}_{H^1(\Omega)}\big) \big(\| \nabla\mathbf{u} \|_{H^1(\Omega)}^{3/2} + \| \Delta\sigma \|_{H^1(\Omega)}^{3/2}\big) \\ &\quad + \| \mathbf{g} \|^2_{L^2(\Omega)} + \| \nabla\sigma \|_{H^1(\Omega)}^6\Big). \end{align*} Consequently, \begin{equation}\label{eq4Chap3} \begin{aligned} &\frac{\mu}{2} \frac{d}{dt} \| \nabla\mathbf{u} \|_{L^2(\Omega)}^2 + \frac{m}{2} \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)} + \frac{m\mu^2}{8M^2} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} \\ &\leq C \big(\| \nabla\mathbf{u} \|_{L^2(\Omega)}^{5/2} + \| \nabla\sigma \|^{5/2}_{H^1(\Omega)}\big) \big(\| \nabla\mathbf{u} \|_{H^1(\Omega)}^{3/2} + \| \Delta\sigma \|_{H^1(\Omega)}^{3/2}\big) \\ &\quad + C \| \nabla\sigma \|_{H^1(\Omega)}^6 + C \| \mathbf{g} \|^2_{L^2(\Omega)}, \end{aligned} \end{equation} where $C=C(\lambda, \kappa, M, m)$. On the other hand, using \eqref{Inegalite13} for $k_d=7$ and taking $\varepsilon_1 = \min\big(\frac{\lambda}{2 C_1}, \frac{m \mu^2}{32 M^2 C_1}\big)$, we obtain \begin{equation}\label{eq5Chap3} \begin{aligned} &\frac{d}{dt} \| \Delta\sigma \|^2_{L^2(\Omega)} + \frac{\lambda}{2} \| \nabla\Delta\sigma \|^2_{L^2(\Omega)} \\ &\leq \frac{m \mu^2}{32 M^2} \| \nabla\mathbf{u} \|^{2}_{H^1(\Omega)} + C \big( \| \mathbf{u} \|^{10}_{H^1(\Omega)} + \| \nabla\sigma \|^{10}_{H^1(\Omega)} \big), \end{aligned} \end{equation} where $C=C(\lambda, \mu, M, m, \Omega)$. From \eqref{eq4Chap3} and \eqref{eq5Chap3}, and recalling the equivalent norms $\|\mathbf{u}\|_{{H^2(\Omega)}}$ and $\|\mathbb{A}\mathbf{u}\|_{L^2(\Omega)}$ in $\mathbf{H}^2(\Omega)\cap\mathbf{V}$, it follows easily that \begin{equation}\label{Equation4} \begin{split} &\frac{d}{dt}\Big(\frac{\mu}{2} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big) + \frac{m}{2} \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)} \\ &+ \frac{3m\mu^2}{32M^2} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} + \frac{\lambda}{2} \| \nabla\Delta\sigma \|^2_{L^2(\Omega)} \\ &\leq C \big( \|\nabla\mathbf{u} \|^{10}_{L^2(\Omega)} + \| \Delta\sigma \|^{10}_{L^2(\Omega)} \big) + C \big(\| \nabla\mathbf{u} \|^{5/2}_{L^2(\Omega)} + \| \Delta\sigma \|^{5/2}_{L^2(\Omega)}\big) \\ &\quad\times \big(\| \mathbb{A}\mathbf{u} \|_{L^2(\Omega)}^{3/2} + \| \nabla\Delta\sigma \|_{L^2(\Omega)}^{3/2}\big) + C \| \Delta\sigma \|^6_{L^2(\Omega)} + C \| \mathbf{g} \|^2_{L^2(\Omega)}. \end{split} \end{equation} Using the Young inequality $\big(ab \leq \frac{a^4}{4}+\frac{3}{4}b^{4/3}\big)$, inequality \eqref{Equation4} is rewritten as \begin{align*} &\frac{d}{dt}\Big(\frac{\mu}{2} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big) + \frac{m}{2} \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)} \\ & + \frac{m\mu^2}{16M^2} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} + \frac{\lambda}{4} \| \nabla\Delta\sigma \|^2_{L^2(\Omega)} \\ &\leq C \Big( \| \nabla\mathbf{u} \|^{10}_{L^2(\Omega)} + \| \Delta\sigma \|^{10}_{L^2(\Omega)} + \| \mathbf{g} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^{6}_{L^2(\Omega)} \Big), \end{align*} where $C=C(\lambda, \mu, \kappa, M, m, \Omega)$. Then, put $\alpha = \min(\frac{\mu}{2},1)$ and we write the above inequality as \begin{align*} &\frac{d}{dt}\Big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big) + \frac{m}{2\alpha} \| \frac{\partial \mathbf{u}}{\partial t} \|^2_{L^2(\Omega)}\\ &+ \frac{m\mu^2}{16M^2\alpha} \| \mathbb{A} \mathbf{u} \|^2_{L^2(\Omega)} + \frac{\lambda}{4\alpha} \| \nabla\Delta\sigma \|^2_{L^2(\Omega)} \\ &\leq \frac{C}{\alpha} \big( \| \nabla\mathbf{u} \|^{2}_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big)^{4} \big( \| \nabla\mathbf{u} \|^{2}_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big) \\ &\quad + \frac{C}{\alpha} \big( \| \nabla\mathbf{u} \|^{2}_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big)^{2} \big( \| \nabla\mathbf{u} \|^{2}_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big) + \frac{C}{\alpha} \| \mathbf{g} \|^2_{L^2(\Omega)}. \end{align*} Since $\| \mathbb{A}\mathbf{u} \|_{L^2(\Omega)} \geq C_{\Omega} \| \nabla\mathbf{u} \|_{L^2(\Omega)}$ and $\| \nabla\Delta\sigma \|_{L^2(\Omega)} \geq C_{\Omega} \| \Delta\sigma \|_{L^2(\Omega)}$, it follows that for some positive constants $c_1$, $c_2$ depending on $\Omega$, $\lambda$, $\mu$, $\kappa$, $M$, $m$, we have \begin{equation}\label{Equation1} \begin{aligned} &\frac{d}{dt}\Big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big) \\ &\leq c_2 \| \mathbf{g} \|^2_{L^2(\Omega)} -\Big[c_1 - c_2 \big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big)^4 \\ &\quad - c_2 \big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big)^2\Big] \big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\big). \end{aligned} \end{equation} Integrating in time from 0 to $t < T_1$, and taking into account that $(\mathbf{u}, \sigma)\in \mathcal{X}_1\times\mathcal{X}_2$, we find for every $t\in[0,T_1)$, \begin{align*} &\| \nabla\mathbf{u}(t) \|^2_{L^2(\Omega)} + \| \Delta\sigma(t) \|^2_{L^2(\Omega)} \\ &\leq \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \Delta\sigma_0 \|^2_{L^2(\Omega)} -2 \big(C_4 \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \Delta\sigma_0 \|^2_{L^2(\Omega)}\big) \\ &\quad\times \Big[c_1 - 16 c_2 \big(C_4 \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \Delta\sigma_0 \|^2_{L^2(\Omega)}\big)^4 - 4 c_2 \big(C_4 \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} \\ &\quad + \| \Delta\sigma_0 \|^2_{L^2(\Omega)}\big)^2\Big] T_1 + c_2 \| \mathbf{g} \|^2_{L^\infty(0,T_1,L^2(\Omega))} T_1. \end{align*} Consequently, for every $t\in[0,T_1)$, \begin{equation*} \| \nabla\mathbf{u}(t) \|^2_{L^2(\Omega)} + \| \Delta\sigma(t) \|^2_{L^2(\Omega)} \leq \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \Delta\sigma_0 \|^2_{L^2(\Omega)}, \end{equation*} provided that \begin{equation} \label{HypPettitesse2Chap3} \begin{gathered} C_4 \| \nabla\mathbf{u}_0 \|^2_{L^2(\Omega)} + \| \Delta\sigma_0 \|^2_{L^2(\Omega)} < \frac{1}{2} \Big(\frac{\sqrt{\frac{c_1}{2 c_2}+1}-1}{2}\Big)^{1/2}, \\ c_2 \| \mathbf{g} \|^2_{{L^\infty(0,+\infty; L^2(\Omega))}} < \frac{7}{8} ~c_1 \Big(\frac{\sqrt{\frac{c_1}{2 c_2}+1}-1}{2}\Big)^{1/2}. \end{gathered} \end{equation} Finally, we conclude that $(\mathbf{u}, \sigma)$, such that $\sigma=\rho-\widehat{\rho}$, is a global solution of \eqref{ModeleKSK2}, and for all $T>0$, we have \begin{gather*} \mathbf{u} \in L^2\big(0,T; \mathbf{H}^2(\Omega)\big) \cap \mathcal{C} \big([0,T]; \mathbf{V}\big), \\ \rho-\widehat{\rho} \in L^2\big(0,T; H^3_{N,0}\big) \cap \mathcal{C}\big([0,T]; H^2_{N,0}\big). \end{gather*} \subsection*{Uniqueness} Let $(\mathbf{u}_1, \rho_1)$, $(\mathbf{u}_2, \rho_2)$ be two solutions of \eqref{Modele1} such that $\mathbf{u}_1(0, \mathbf{x}) = \mathbf{u}_2(0, \mathbf{x}) = \mathbf{u}_0 (\mathbf{x}) $ and $\rho_1(0, \mathbf{x}) = \rho_2(0, \mathbf{x}) = \rho_0(\mathbf{x})$. We put $\mathbf{u} = \mathbf{u}_1 - \mathbf{u}_2$ and $\rho = \rho_1 - \rho_2$. The system verified by $(\mathbf{u}, \rho)$ reads \begin{equation}\label{Modele9} \begin{gathered} \mathbb{P}\Big(\rho_1 \frac{\partial\mathbf{u}}{\partial t} \Big) + \mathbb{P}\Big(\rho \frac{\partial\mathbf{u}_2}{\partial t}\Big) + \mu \mathbb{A}\mathbf{u} = \mathbf{F}_1 - \mathbf{F}_2, \\ \frac{\partial\rho}{\partial t} + \mathbf{u}_1 \cdot \nabla\rho + \mathbf{u} \cdot \nabla\rho_2 = \lambda \Delta \rho,\\ \operatorname{div}\mathbf{u} = 0, \\ \mathbf{u}(0,\mathbf{x}) = 0, \quad \rho(0,\mathbf{x}) = 0, \end{gathered} \end{equation} where \begin{align*} \mathbf{F}_1 &\equiv \mathbf{F}(\mathbf{u}_1, \rho_1) \\ &= \mathbb{P}\Big(\rho_1~ \mathbf{g} - \kappa \Delta\rho_1 \nabla\rho_1 - \rho_1 (\mathbf{u}_1 \cdot \nabla) \mathbf{u}_1 + \lambda (\nabla\rho_1 \cdot \nabla) \mathbf{u}_1 \\ &\quad + \lambda (\mathbf{u}_1\cdot\nabla)\nabla\rho_1 - \frac{\lambda^2}{\rho_1} \Delta\rho_1 \nabla\rho_1 - \frac{\lambda^2}{\rho_1} \big(\nabla\rho_1\cdot\nabla\big)\nabla\rho_1 + \lambda^2 \frac{|\nabla\rho_1|^2}{\rho_1^2} \nabla\rho_1\Big), \\ \mathbf{F}_2 &\equiv \mathbf{F}(\mathbf{u}_2, \rho_2) \\ &= \mathbb{P}\Big(\rho_2 \mathbf{g} - \kappa \Delta\rho_2 \nabla\rho_2 - \rho_2 (\mathbf{u}_2 \cdot \nabla) \mathbf{u}_2 + \lambda (\nabla\rho_2 \cdot \nabla) \mathbf{u}_2 \\ &\quad + \lambda (\mathbf{u}_2\cdot\nabla)\nabla\rho_2 - \frac{\lambda^2}{\rho_2} \Delta\rho_2 \nabla\rho_2 - \frac{\lambda^2}{\rho_2} \big(\nabla\rho_2\cdot\nabla\big)\nabla\rho_2 + \lambda^2 \frac{|\nabla\rho_2|^2}{\rho_2^2} \nabla\rho_2\Big). \end{align*} First, taking the inner product of $\text{\eqref{Modele9}}_1$ with $\mathbf{u}$ in $\mathbf{H}$, we have $$ \Big(\mathbb{P}\big(\rho_1 \frac{\partial\mathbf{u}}{\partial t}\big), \mathbf{u}\Big) + \Big(\mathbb{P} \big(\rho \frac{\partial\mathbf{u}_2}{\partial t}\big), \mathbf{u}\Big) + \mu \Big(\mathbb{A}\mathbf{u}, \mathbf{u}\Big) = \Big(\mathbf{F}_1 - \mathbf{F}_2, \mathbf{u}\Big). $$ Then, by using the definition of operator $\mathbb{P}$, such that $$ \Big(\mathbb{P}\mathbf{u}, \mathbf{v}\Big) = \big(\mathbf{u}, \mathbf{v}\big), \quad \forall \mathbf{u} \in \mathbf{L}^2(\Omega),\; \forall \mathbf{v} \in \mathbf{H}, $$ we have $$ \Big(\rho_1 \frac{\partial\mathbf{u}}{\partial t}, \mathbf{u}\Big) = \frac{1}{2} \frac{d}{dt} \Big(\rho_1 \mathbf{u}, \mathbf{u}\Big) - \frac{1}{2} \Big(\frac{\partial\rho_1}{\partial t} \mathbf{u}, \mathbf{u}\Big). $$ Since $\rho_1$ is a solution of the convection-diffusion equation \eqref{Modele1}$_2$, we obtain \begin{align*} &\frac{1}{2} \frac{d}{dt} \Big( \rho_1 \mathbf{u}, \mathbf{u} \Big) + \mu \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} \\ &= \frac{\lambda}{2} \Big( \Delta\rho_1, \mathbf{u}^2 \Big) - \frac{1}{2} \Big( \mathbf{u}_1 \cdot \nabla\rho_1, \mathbf{u}^2 \Big) - \Big( \rho \frac{\partial\mathbf{u}_2}{\partial t}, \mathbf{u} \Big) + \Big(\mathbf{F}_1 - \mathbf{F}_2, \mathbf{u}\Big). \end{align*} By using Green's theorem and Cauchy-Schwarz and Young inequalities, we arrive at \begin{equation}\label{Inegalite6} \begin{aligned} &\frac{1}{2} \frac{d}{dt} \big( \rho_1 \mathbf{u}, \mathbf{u} \big) + \frac{\mu}{2} \| \nabla\mathbf{u} \|^2_{L^2(\Omega)} \\ &\leq \frac{\lambda}{4} \| \Delta\rho \|_{L^2(\Omega)}^2 + \Big(\frac{C}{\lambda} \| \frac{\partial\mathbf{u}_2}{\partial t} \|_{L^2(\Omega)}^2 + \frac{C\lambda^2}{2\mu} \| \nabla\rho_1 \|^2_{{L^{\infty}(\Omega)}} \\ &\quad + \frac{1}{2} \| \nabla\rho_1 \|_{{L^{\infty}(\Omega)}} \| \mathbf{u}_1 \|_{{L^{\infty}(\Omega)}}\Big) \| \mathbf{u} \|^2_{L^2(\Omega)} + \big(\mathbf{F}_1 - \mathbf{F}_2, \mathbf{u}\big). \end{aligned} \end{equation} Second, taking the inner product of \eqref{Modele9}$_2$ with $-\Delta\rho$ in $L^2(\Omega)$, we obtain \begin{equation}\label{Inegalite7} \begin{aligned} &\frac{1}{2} \frac{d}{dt} \| \nabla\rho \|_{L^2(\Omega)}^2 + \frac{\lambda}{2} \| \Delta\rho \|_{L^2(\Omega)}^2 \\ &\leq \frac{1}{\lambda} \| \mathbf{u}_1 \|_{{L^{\infty}(\Omega)}}^2 \| \nabla\rho \|_{L^2(\Omega)}^2 + \frac{1}{\lambda} \| \nabla\rho_2 \|_{{L^{\infty}(\Omega)}}^2 \| \mathbf{u} \|_{L^2(\Omega)}^2. \end{aligned} \end{equation} By adding \eqref{Inegalite6} and \eqref{Inegalite7}, it follows that \begin{equation} \label{Inegalite8} \begin{split} &\frac{d}{dt} \Big( \big( \rho_1 \mathbf{u}, \mathbf{u} \big) + \| \nabla\rho \|_{L^2(\Omega)}^2 \Big) + \mu \| \nabla\mathbf{u}\|^2_{L^2(\Omega)} + \frac{\lambda}{2} \|\Delta\rho\|_{L^2(\Omega)}^2\\ &\leq \Psi_1(t)~ \Big(m \| \mathbf{u} \|_{L^2(\Omega)}^2 + \|\nabla\rho\|_{L^2(\Omega)}^2 \Big) + 2 \big(\mathbf{F}_1 - \mathbf{F}_2, \mathbf{u}\big), \end{split} \end{equation} where $\Psi_1 \in L^1\big([0,T]\big)$ dependent on $\mathbf{u}_1$, $\mathbf{u}_2$, $\rho_1$, $\rho_2$. In particular, applying Cauchy-Schwarz and Young inequalities $\big(ab \leq \varepsilon a^2 + \frac{b^2}{\varepsilon} \big)$, the embedding $H^2(\Omega)\subset L^\infty(\Omega)$ and the equivalent norms, we obtain the inequality \begin{equation*} 2 \Big| \big(\mathbf{F}_1 - \mathbf{F}_2, \mathbf{u}\big) \Big| \leq \Psi_2(t) \Big(m \| \mathbf{u} \|^2_{L^2(\Omega)} + \| \nabla\rho \|^2_{L^2(\Omega)} \Big) + \varepsilon \Big( \| \mathbf{u} \|_{H^1(\Omega)}^2 + \| \rho \|^2_{{H^2(\Omega)}} \Big), \end{equation*} where $\Psi_2 \in L^1\big([0,T]\big)$ dependent on $\varepsilon$, $\mathbf{u}_1$, $\mathbf{u}_2$, $\rho_1$, $\rho_2$, $\mathbf{g}$, with $\varepsilon>0$ being arbitrary. Therefore, using this last estimate in \eqref{Inegalite8} and choosing $\varepsilon > 0$ such that $\varepsilon < \min\big(\mu, \frac{\lambda}{2}\big)$, we arrive at \begin{equation*} \frac{d}{dt} \Big( \big( \rho_1 \mathbf{u}, \mathbf{u} \big) + \| \nabla\rho \|_{L^2(\Omega)}^2 \Big) \leq \Big(\Psi_1(t) + \Psi_2(t)\Big) \Big(m \| \mathbf{u} \|_{L^2(\Omega)}^2 + \| \nabla\rho \|_{L^2(\Omega)}^2 \Big). \end{equation*} Since $\rho_1$ is a solution of \eqref{Modele1} satisfying the maximum principle, we have $\| \mathbf{u} \|_{L^2(\Omega)}^2 \leq m^{-1} \big(\rho_1 \mathbf{u}, \mathbf{u}\big)$ and we obtain \begin{equation*} \frac{d}{dt} \Big(\big(\rho_1 \mathbf{u}, \mathbf{u} \big) + \| \nabla\rho \|_{L^2(\Omega)}^2\Big) \leq \Big(\Psi_1(t) + \Psi_2(t)\Big) \Big(\big(\rho_1 \mathbf{u}, \mathbf{u}\big) + \| \nabla\rho \|_{L^2(\Omega)}^2 \Big). \end{equation*} Finally, from the Gronwall Lemma and from $\mathbf{u}(0)=0$, $\rho(0)=0$, we deduce the uniqueness of the solution of \eqref{Modele1}. \subsection*{Asymptotic behavior} Let us prove the inequality \eqref{Asymptotic} in Theorem \ref{ThExistence2Chap3}. Assume that $\mathbf{g} = \boldsymbol 0$. Then under hypothesis \eqref{HypPettitesse2Chap3}$_1$, the inequality \eqref{Equation1} is rewritten as \[ \frac{d}{dt}\Big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big) \leq -\frac{7}{8} c_1 \Big(\| \nabla\mathbf{u} \|^2_{L^2(\Omega)} + \| \Delta\sigma \|^2_{L^2(\Omega)}\Big). \] Consequently, since $\sigma=\rho-\widehat{\rho}$ and from Gronwall Lemma, we obtain \eqref{Asymptotic}. Finally, from this inequality \eqref{Asymptotic}, we conclude that the solution $(\mathbf{u}, \rho)$ of \eqref{Modele1}, converges to a constant solution as $t\to + \infty$: \begin{gather*} \mathbf{u}(t, \mathbf{x}) \to \boldsymbol 0 \quad \text{in } \mathbf{V}, \\ \rho(t, \mathbf{x}) \to \widehat{\rho} \quad \text{in } H^2_N. \end{gather*} The convergence is exponential in time. 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