\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 62, pp. 1--8.\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-2009/62\hfil Navier-Stokes equation with slip-like boundary] {Navier-Stokes equation with slip-like boundary condition} \author[K. Hashizume, T. Koyama, M. \^{O}tani \hfil EJDE-2009/62\hfilneg] {Ken'ichi Hashizume, Tetsuya Koyama, Mitsuharu \^{O}tani} % in alphabetical order \address{Ken'ichi Hashizume \newline Hiroshima Institute of Technology, 2-1-1, Miyake, Saeki-ku, Hiroshima, Japan, 731-5193 } \email{khshzm@cc.it-hiroshima.ac.jp} \address{Tetsuya Koyama \newline Hiroshima Institute of Technology, 2-1-1, Miyake, Saeki-ku, Hiroshima, Japan, 731-5193 } \email{tkoyama@cc.it-hiroshima.ac.jp} \address{Mitsuharu \^{O}tani \newline Department of Applied Physics, School of Science and Engineering, Waseda University, 3-4-1, Okubo Tokyo, Japan, 169-8555 } \email{otani@mn.waseda.ac.jp} \thanks{Submitted August 22, 2008. Published May 4, 2009.} \subjclass[2000]{35Q30} \keywords{Navier-Stokes equation; time discretition; slip-like boundary} \begin{abstract} The aim of this note is to investigate a time-discretized 2-dimensional Navier-Stokes equation with a slip-like boundary condition, which arises in the melting ice problem. We prove the existence and uniqueness of a weak solution. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} Consider an ice plate, placed upright, whose vertical face is exposed to the air and melting. So this face is covered by the layer of flowing water, and the shapes of the ice and the water-layer vary as time $t$ goes on. Therefore, in the water region, this system can be described by Navier-Stokes equations with two free boundaries of the ice-water interface $\Gamma_1$ and the water-air interface $\Gamma_2$, whose movements would depend on the unknown functions. However, as a first step of analysis, we here consider the discretized Navier-Stokes equation in the time variable $t$ with the discretizing parameter $\tau > 0$ in the fixed domain $\Omega$ with given interfaces $\Gamma_1$ and $\Gamma_2$. Experiments for this kind of problems can be found in \cite{t1} and mathematical treatments for problems similar to ours are discussed by several authors (see e.g. \cite{f1,k1}). \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(54,35)(-4,-3) \put(0,30){\line(1,0){42}} \put(14,32){$\Gamma_4$} %%(8,32) \put(41.8,29.1){$\to$} \put(46,27){$y$} \put(0,0){\line(1,0){40}} \put(30,-4){$\Gamma_3$} \put(0,-2){\line(0,1){32}} \put(40,0){\line(0,1){30}} \qbezier(15,30)(25,15)(25,0) \qbezier(20,30)(35,15)(35,0) \put(28.5,22){$\Gamma_2$} \put(27.5,10){$\Omega$} \put(20,6){$\Gamma_1$} \put(-0.85,-4){$\downarrow$} \put(-4,-4){$x$} \end{picture} \end{center} \caption{The water region and its boundaries} \end{figure} Fix the $x$-axis vertically and downward, the $y$-axis in the direction of the thickness and outward, and the $z$-axis orthogonally to the $x$ and $y$ axes. The ice-water interface and the water-air interface are represented by $y=l(x,z)$ and $y=d(x,z)$ respectively. Further suppose that the size of ice plate in $z$-direction is so large that we can regard $l$ and $d$ as constant in $z$. So our problem can be formulated in the following 2-dimensional setting. Define the domain $\Omega$ which is occupied by water by $$ \Omega = \{(x,y): 0< x < 1, \; l(x) < y < d(x)\}, $$ where $l, d \in C^{0,1}([0,1])$; %\label{$l1$} that is, $l$ and $d$ are Lipschitsz continuous on $[0,1]$ and \begin{equation} 0 \leq l(x) < d(x) \leq 1 \quad \text{for all } 0\leq x \leq 1. \label{l2} \end{equation} Hence $\Omega$ is of class $C^{0,1}$. Define the ice-water interface $\Gamma_1$, the water-air interface $\Gamma_2$, the lower boundary $\Gamma_3$, and the upper boundary $\Gamma_4$ by \begin{gather*} \Gamma_1 = \{(x,y): 0\leq x \leq 1, \; y = l(x) \}, \\ \Gamma_2 = \{(x,y): 0\leq x \leq 1, \; y = d(x) \}, \\ \Gamma_3 = \{(x,y): x = 1, \; l(1) \leq y \leq d(1) \}, \\ \Gamma_4 = \{(x,y): x = 0, \; l(0) \leq y \leq d(0) \} \end{gather*} respectively. Our objective is to study the equations: \begin{equation} \begin{gathered} \frac{1}{\tau}(\mathbf{u} - \mathbf{u}_0) + (\mathbf{u}\cdot\nabla) \mathbf{u} + \frac{1}{\rho}\nabla p -\nu \Delta \mathbf{u} = \mathbf{g} \quad \text{ in } \Omega, \\ \mathop{\rm div} \mathbf{u} =0 \quad \text{in } \Omega, \end{gathered} \label{eP} \end{equation} for the fixed discretizing parameter $\tau>0$ with the boundary conditions \begin{gather} \mathbf{u} = \mathbf{0} \quad\text{on } \Gamma_1, \label{B1}\\ U_Y = V_Y = 0 \quad \text{on } \Gamma_2, \label{B2}\\ v = 0 \quad \text{on } \Gamma_3, \label{B3}\\ \mathbf{u} = \mathbf{0} \quad \text{on } \Gamma_4. \label{B4} \end{gather} \begin{figure}[ht] \begin{center} \setlength{\unitlength}{1mm} \begin{picture}(50,30)(0,0) \qbezier(10,30)(25,20)(25,-2) \put(20,20){\line(2,-3){15}} \put(20,20){\line(3,2){20}} \put(39.8,33.3){\line(-1,0){1.8}} \qbezier(39.8,33.3)(39.5,32.4) (39.2,31.5) \put(42,30){$Y$} \put(28,25.4){\line(-1,0){1.8}} \qbezier(28,25.)(27.7,24.5) (27.4,23.6) \put(23,25){$\mathbf{n}$} \put(26.5,10.2){\line(0,1){1.8}} \qbezier(26.5,10.2)(25.6,10.5)(24.7,10.8) \put(25,13){$\mathbf{t}$} \put(35,-2.5){\line(0,1){1.8}} \qbezier(35,-2.5) (34.6,-2.2) (33.2,-1.9) \put(37,-2.5){$X$} \end{picture} \end{center} \caption{The local coordinates} \end{figure} Here, the velocity vector $\mathbf{u} = (u,v)$ and the pressure $p$ are unknown functions of $(x,y)$. The initial velocity $\mathbf{u}_0$, the gravity force $\mathbf{g}$, the density $\rho$, and the kinematic viscosity $\nu$ are given data. The unit time $\tau$ is to be determined later. Put $U=\mathbf{u}\cdot\mathbf{t}$, $V=\mathbf{u}\cdot\mathbf{n}$, where $\mathbf{n}$ designates the outer unit normal vector of $\Gamma_2$ and $\mathbf{t}$ designates the downward unit tangential vector of $\Gamma_2$. Denote by $(X,Y)$ the local coordinate with directions $\mathbf{t}$ and $\mathbf{n}$. The original slip boundary condition is stated as $$ U_Y + V_X = 0 \quad \text{on } \Gamma_2, $$ (see \cite{s1}) and condition \eqref{B2} is its linearized version. In the original problem, both $\Gamma_1$ and $\Gamma_2$ move after the unit time $\tau$. But in our setting, the interfaces stay invariant. \section{Main results} Set \begin{gather*} \mathbf{V} = \{ \mathbf{u} \in \left(H^1(\Omega)\right)^2 : \mathop{\rm div}\mathbf{u} = 0, \; \mathbf{u}=\mathbf{0} \text{ on $\Gamma_1$ and $\Gamma_4$,} v=0 \text{ on } \Gamma_3 \}, \\ \mathbf{H} = \{ \mathbf{u} \in (L_2(\Omega))^2 : \mathop{\rm div}\mathbf{u} = 0 \}, \\ P_\sigma \text{is the orthogonal projection from } (L_2(\Omega))^2 \text{ onto } \mathbf{H}, \\ \mathbf{L}_4 = \{ \mathbf{u} \in \left(L_4(\Omega)\right)^2 : \mathop{\rm div}\mathbf{u} = 0 \} \end{gather*} and let $(\cdot,\cdot)$ and $|\cdot|$ denote the inner product and the norm of the space $\mathbf{H}$. Define a bounded positive bilinear form $a(\cdot,\cdot)$ on $\mathbf{V}$ by \[ a(\mathbf{u},\delta\mathbf{u}) := \int_\Omega (\nabla\,u\cdot\nabla\,\delta u + \nabla\,v\cdot\nabla\,\delta v) \,d\mathbf{x} \] for $\mathbf{u}=(u,v)$, $ \delta\mathbf{u}=(\delta u, \delta v) \in \mathbf{V}$. Also define a trilinear form $b(\cdot, \cdot, \cdot)$ on $(\mathbf{L}_4)^2 \times \mathbf{V}$ by \[ b(\tilde{\mathbf{u}},\mathbf{u},\delta\mathbf{u}) := \int_\Omega (\tilde{u} u_x \delta u + \tilde{u} v_x \delta v + \tilde{v} u_y \delta u + \tilde{v} v_y \delta v) \, d\mathbf{x} \] for $\tilde{\mathbf{u}}=(\tilde{u}, \tilde{v}) \in \mathbf{L}_4$, $\mathbf{u}=(u, v) \in \mathbf{V}$, $\delta\mathbf{u}=(\delta u, \delta v) \in \mathbf{L}_4$, where $\mathbf{x}=(x,y)$. We note that H\"older's inequality gives \begin{equation}\label{est:b} |b(\tilde{\mathbf{u}},\mathbf{u},\delta\mathbf{u})| \leq |\tilde{\mathbf{u}}|_4 |\nabla\mathbf{u}| |\delta\mathbf{u}|_4 \quad \text{for } \tilde{\mathbf{u}} \in \mathbf{L}_4, \; \mathbf{u} \in \mathbf{V}, \; \delta\mathbf{u} \in \mathbf{L}_4. \end{equation} Here $|\cdot |_4$ denotes the norm of $\mathbf{L}_4$ and \[ |\nabla \mathbf{u}| = |(|\nabla u|, |\nabla v|)|, \quad |\nabla u| =\Big( \left|\frac{\partial u}{\partial x}\right|^2 + \left|\frac{\partial u}{\partial y}\right|^2\Big)^{1/2}. \] In this paper, we are concerned with weak solutions of \eqref{eP} with boundary conditions \eqref{B1}-\eqref{B4} in the following sense: $\mathbf{u}\in \mathbf{V}$ is said to be {\it a weak solution} of \eqref{eP} with boundary conditions \eqref{B1}-\eqref{B4}, if the following relation holds. \begin{equation} \Big(\frac{1}{\tau} (\mathbf{u} - \mathbf{u}_0) - \mathbf{g}, \delta\mathbf{u}\Big) + b(\mathbf{u}, \mathbf{u}, \delta\mathbf{u}) + \nu a(\mathbf{u}, \delta\mathbf{u}) = 0 \quad \text{for all } \delta \mathbf{u} \in \mathbf{V}. \label{Pw} \end{equation} We remark that if a sufficiently smooth function $\mathbf{u}$, say in $(C^2(\bar{\Omega}))^2 \cap \mathbf{V}$, satisfies \eqref{Pw}, then $\mathbf{u}$ should satisfy equation \eqref{eP} and boundary condition \eqref{B2} on $\Gamma_2$. In fact, let $\mathbf{u} \in ( C^2(\bar{\Omega}) )^2 \cap \mathbf{V}$ and let $\mathbf{f} = - \frac{1}{\nu} \left(\frac{1}{\tau}(\mathbf{u}-\mathbf{u}_0) - \mathbf{g} + P_\sigma (\mathbf{u}\cdot\nabla) \mathbf{u} \right) \in \mathbf{H}$, then \eqref{Pw} gives \begin{equation*} a(\mathbf{u}, \delta\mathbf{u}) = (\mathbf{f},\delta\mathbf{u}) \quad \text{for all } \delta\mathbf{u}\in \mathbf{V}. \end{equation*} Here we note that since $v \equiv 0$, $\mathop{\rm div} \mathbf{u} = u_x + v_y \equiv 0 $ on $\Gamma_3$, $v_y \equiv 0$ and hence $u_x \equiv 0$ on $\Gamma_3$. Consequently, integration by parts yields \begin{equation}\label{ident:a} \begin{aligned} a(\mathbf{u},\delta\mathbf{u}) &= (\mathbf{f},\delta\mathbf{u}) \\ &= \int_\Omega (-\Delta u \delta u -\Delta v \delta v) dx + \int_{\Gamma_2} (u_Y \delta u + v_Y \delta v) dS + \int_{\Gamma_3} u_x \delta u dS \\ &= \int_\Omega (-\Delta u \delta u -\Delta v \delta v) dx + \int_{\Gamma_2} (U_Y \delta U + V_Y \delta V) dS, \end{aligned} \end{equation} where $\delta U$ and $\delta V$ are $\mathbf{t}$ and $\mathbf{n}$ components of $\delta\mathbf{u}$ on $\Gamma_2$. If we take $\delta\mathbf{u}\in (C_0^\infty(\Omega))^2 \cap \mathbf{V}$, then the term of the integration on $\Gamma_2$ in \eqref{ident:a} vanishes, whence follows \[ (\mathbf{f}, \delta\mathbf{u}) = ( - \Delta \mathbf{u}, \delta\mathbf{u}) \quad \text{for all } \delta\mathbf{u}\in (C_0^\infty(\Omega))^2 \cap \mathbf{V}. \] This says that $\mathbf{f} = - P_\sigma \Delta \mathbf{u}$ in the sense of distribution. Hence $\mathbf{f} = - P_\sigma \Delta \mathbf{u}$ holds a.e. in $\Omega$, which implies that $\mathbf{u}$ gives a solution of \eqref{eP}. Furthermore, plugging this relation into \eqref{ident:a}, we get \[ \int_{\Gamma_2} (U_Y \delta U + V_Y \delta V) dS = 0 \quad \text{for any } \delta\mathbf{u} \in\mathbf{V}, \] whence easily follows that $\mathbf{u}$ should satisfy \eqref{B2}. Our main result is stated as follows. \begin{theorem} \label{thm1} Let $\mathbf{u}_0\in\mathbf{V}$ and $\mathbf{g}\in \mathbf{H}$. There exists a positive number $\tau_0 = \tau_0(|\mathbf{g}|, |\nabla\mathbf{u}_0|)$ such that for all $\tau \in (0,\tau_0]$, \eqref{Pw} admits a unique weak solution $\mathbf{u}\in\mathbf{V}$. \end{theorem} \section{Proof of main theorem} By the Gagliardo-Nirenberg estimate and Poincare's lemma, there is a constant $K_1$ which depends only on $\Omega$ such that \begin{equation}\label{ineq:GNL4} |\mathbf{u}|_4 \leq K_1 |\nabla \mathbf{u}|^{1/2} |\mathbf{u}|^{1/2} \quad \text{for all }\mathbf{u}\in\mathbf{V}. \end{equation} Also, by the Sobolev embedding theorem, there is a constant $K_2$ such that \begin{equation}\label{ineq:L4} |\mathbf{u}|_4 \leq K_2 |\nabla \mathbf{u}| \quad \text{for all }\mathbf{u}\in\mathbf{V}. \end{equation} \begin{lemma} \label{lem1} Let $\mathbf{u}_0, \tilde{\mathbf{u}} \in \mathbf{V}$ and $\mathbf{g} \in \mathbf{H}$. Define $\tau_1 = \tau_1(\nu, K_1, |\tilde{\mathbf{u}}|_4)$ by \begin{equation}\label{tau1} \tau_1 = \begin{cases} \frac{16}{27} K_1^{-4} \nu^3 |\tilde{\mathbf{u}}|_4^{-4} & \text{for } \tilde{\mathbf{u}} \neq \mathbf{0}, \\ \text{any positive number} & \text{for } \tilde{\mathbf{u}} = \mathbf{0}, \end{cases} \end{equation} where $K_1$ is the constant appearing in \eqref{ineq:GNL4}. Then, for all $\tau \in (0, \tau_1]$, the problem \begin{equation} \quad\left(\frac{1}{\tau}(\mathbf{u} - \mathbf{u}_0) - \mathbf{g}, \delta\mathbf{u}\right) %%%\quad) + b(\tilde{\mathbf{u}}, \mathbf{u}, \delta\mathbf{u}) + \nu a(\mathbf{u}, \delta\mathbf{u}) = 0 \quad \text{for all } \delta\mathbf{u}\in \mathbf{V} \label{Pwu} \end{equation} has a unique weak solution $\mathbf{u}\in \mathbf{V}$. \end{lemma} \begin{proof} We show that the bilinear form $$ N(\mathbf{u}, \delta\mathbf{u}) := \nu a(\mathbf{u}, \delta\mathbf{u}) + \frac{1}{\tau}(\mathbf{u},\delta\mathbf{u}) + b(\tilde{\mathbf{u}}, \mathbf{u}, \delta\mathbf{u}) $$ is bounded and coercive on $\mathbf{V}$. In fact, the estimate $$ |b(\tilde{\mathbf{u}}, \mathbf{u}, \delta\mathbf{u})| \leq |\tilde{\mathbf{u}}|_4 |\nabla \mathbf{u}| |\delta\mathbf{u}|_4 \leq K_2 |\tilde{\mathbf{u}}|_4 |\nabla \mathbf{u}| |\nabla\delta\mathbf{u}| $$ assures the boundedness of $ N(\cdot,\cdot)$ for any $\tau > 0$ and also the coerciveness of $ N(\cdot,\cdot)$ for any $\tau > 0$ when $\tilde{\mathbf{u}} = \mathbf{0}$. For the case where $\tilde{\mathbf{u}} \neq \mathbf{0}$, (\ref{ineq:GNL4}) and Young's inequality give \begin{align*} |b(\tilde{\mathbf{u}}, \mathbf{u}, \mathbf{u})| &\leq |\tilde{\mathbf{u}}|_4 |\nabla \mathbf{u}| |\mathbf{u}|_4\\ &\leq |\tilde{\mathbf{u}}|_4 K_1 |\nabla \mathbf{u}|^{\frac32} |\mathbf{u}|^{1/2}\\ &\leq |\tilde{\mathbf{u}}|_4 K_1 \Big(\frac{3}{4} \eta |\nabla \mathbf{u}|^2 + \frac{1}{4} \eta^{-3}|\mathbf{u}|^2 \Big) \quad \forall \eta > 0. \end{align*} Hence choosing $\eta = \frac{2 \nu}{3 K_1 |\tilde{\mathbf{u}}|_4}$, we get \begin{align*} N(\mathbf{u},\mathbf{u}) &= \nu a(\mathbf{u}, \mathbf{u}) + \frac{1}{\tau}(\mathbf{u},\mathbf{u}) + b(\tilde{\mathbf{u}}, \mathbf{u}, \mathbf{u})\\ &\geq \Big( \nu - |\tilde{\mathbf{u}}|_4 \frac{3}{4} \eta K_1 \Big) |\nabla\mathbf{u}|^2 + \Big( \frac{1}{\tau} - |\tilde{\mathbf{u}}|_4 \frac{1}{4} \eta^{-3} K_1 \Big) |\mathbf{u}|^2 \\ & = \frac{\nu}{2} |\nabla\mathbf{u}|^2 + \Big( \frac{1}{\tau} - \frac{27 K_1^4 |\tilde{\mathbf{u}}|_4^4}{32 \nu^3} \Big) |\mathbf{u}|^2 \\ & \geq \frac{\nu}{2} |\nabla\mathbf{u}|^2 + \frac{1}{2\tau} |\mathbf{u}|^2 \quad \forall \tau \in (0, \tau_1], \end{align*} which implies the coerciveness of $N(\cdot,\cdot)$. On the other hand, $$ \delta\mathbf{u} \mapsto \Big(\frac{1}{\tau}\mathbf{u}_0 + \mathbf{g}, \delta\mathbf{u} \Big) $$ is a bounded linear functional on $\mathbf{V}$. Therefore, the Lax-Milgram theorem assures the existence of a unique weak solution of \eqref{Pwu}. \end{proof} Define an operator $F: \mathbf{L}_4 \to \mathbf{L}_4$ by $F(\tilde{\mathbf{u}})= \mathbf{u}$, where $\mathbf{u}$ is the unique weak solution of \eqref{Pwu}, whose existence is assured in Lemma \ref{lem1}. Then $F$ satisfies the following a priori estimates. \begin{lemma} \label{lem2} Let $\tilde{\mathbf{u}} \in \mathbf{L}_4$ and let $ 0< \tau \leq \tau_1^* := \min (1, \tau_1), $ then $\mathbf{u} = F(\tilde{\mathbf{u}})$ satisfies \begin{gather} |\mathbf{u}-\mathbf{u}_0|_4 \leq K_0 := 2 K_2 \Big( \frac{|\mathbf{g}|^2}{\nu} + \frac{|\nabla\mathbf{u}_0|^2}{2} \Big)^{1/2}, \label{est:u4}\\ |\nabla \mathbf{u}| \leq \frac{K_0}{K_2}. \label{est:nablau} \end{gather} \end{lemma} \begin{proof} By substitution $\delta\mathbf{u} = \mathbf{u}-\mathbf{u}_0$ in \eqref{Pwu}, we have \begin{equation} 0 = \frac{1}{\tau}|\mathbf{u}-\mathbf{u}_0|^2 + \nu a(\mathbf{u}, \mathbf{u} - \mathbf{u}_0) + b(\tilde{\mathbf{u}}, \mathbf{u}, \mathbf{u}-\mathbf{u}_0) - (\mathbf{g},\mathbf{u}-\mathbf{u}_0). \label{energyestimate1} \end{equation} Here, in view of the definition of $\tau_1$, we get \begin{align*} b(\tilde{\mathbf{u}}, \mathbf{u}, \mathbf{u}-\mathbf{u}_0) & \leq |\tilde{\mathbf{u}}|_4 |\nabla \mathbf{u}| K_1 |\nabla \mathbf{u} - \nabla \mathbf{u}_0|^{1/2} |u - u_0|^{1/2} \\ & \leq \frac{\nu}{4} |\nabla \mathbf{u}|^2 + \frac{1}{\nu}|\tilde{\mathbf{u}}|_4^2 K_1^2 |\nabla \mathbf{u} - \nabla \mathbf{u}_0| |u - u_0| \\ & \leq \frac{\nu}{4} |\nabla \mathbf{u}|^2 + \frac{\nu}{4} |\nabla \mathbf{u} - \nabla \mathbf{u}_0|^2 + \frac{1}{\nu^3}|\tilde{\mathbf{u}}|_4^4 K_1^4 |u - u_0|^2 \\ & \leq \frac{\nu}{4} |\nabla \mathbf{u}|^2 + \frac{\nu}{4} |\nabla \mathbf{u} - \nabla \mathbf{u}_0|^2 + \frac{16}{27} \frac{1}{\tau_1} |u - u_0|^2. \end{align*} Furthermore we note that \begin{gather*} \begin{aligned} a(\mathbf{u}, \mathbf{u} - \mathbf{u}_0) &= \frac12 a((\mathbf{u}+\mathbf{u}_0) + (\mathbf{u}-\mathbf{u}_0), \mathbf{u} - \mathbf{u}_0)\\ & = \frac12|\nabla\mathbf{u}|^2 + \frac12|\nabla\mathbf{u} - \nabla\mathbf{u}_0|^2 - \frac12|\nabla\mathbf{u}_0|^2, \end{aligned} \\ | (\mathbf{g}, \mathbf{u} - \mathbf{u}_0 ) | \leq \frac{1}{4 \tau} | \mathbf{u} - \mathbf{u}_0|^2 + \tau |\mathbf{g}|^2. \end{gather*} Therefore, by \eqref{energyestimate1}, we have \begin{align*} & \frac{1}{\tau} | \mathbf{u} - \mathbf{u}_0|^2 + \frac{\nu}{2} (|\nabla \mathbf{u}|^2 + |\nabla \mathbf{u} - \nabla \mathbf{u}_0|^2) \\ & \leq \ \frac{\nu}{2} |\nabla \mathbf{u}_0|^2 + \frac{\nu}{4} (|\nabla \mathbf{u}|^2 + |\nabla \mathbf{u} - \nabla \mathbf{u}_0|^2) + \left( \frac{16}{27} + \frac14 \right) ~\frac{1}{\tau} ~|\mathbf{u} -\mathbf{u}_0|^2 + \tau |\mathbf{g}|^2. \end{align*} Hence we get \begin{align*}%\label{estimate1} & \frac{K_2^{-2}~\nu}{4} |\mathbf{u} - \mathbf{u}_0|^2_4 + \frac{\nu}{4} |\nabla \mathbf{u}|^2 \\ & \leq \frac{\nu}{4}\left(|\nabla \mathbf{u} - \nabla \mathbf{u}_0|^2 + |\nabla \mathbf{u}|^2 \right) + \frac{17}{108} \frac1\tau |\mathbf{u} - \mathbf{u}_0|^2 \leq \tau |\mathbf{g}|^2 + \frac{\nu}{2} |\nabla\mathbf{u}_0|^2, \end{align*} whence follows \begin{gather*} |\mathbf{u} - \mathbf{u}_0|_4 \leq \Big(\frac{4 K_2^2}{\nu} (|\mathbf{g}|^2 + \frac{\nu}{2}|\nabla \mathbf{u}_0|^2) \Big)^{1/2} = K_0, \\ |\nabla \mathbf{u}| \leq \Big(\frac{4}{\nu}|\mathbf{g}|^2 + 2|\nabla \mathbf{u}_0|^2 \Big)^{1/2} = \frac{K_0}{K_2}. \end{gather*} Thus, \eqref{est:u4} and \eqref{est:nablau} are verified. \end{proof} \begin{lemma} \label{lem3} Set \begin{equation} \tau_0 := \frac{27}{16\cdot 17} \tau_0^*, \quad \tau_0^* := \min\Big(1, \, \frac{16}{27}~\frac{\nu^3}{K_1^4}~ \frac{1}{\left( K_0+|\mathbf{u}_0|_4 \right)^4} \Big). \label{tau0} \end{equation} Then, for any $\tau \in (0, \tau_0]$, $F(\cdot)$ becomes a contraction mapping from \[ \mathbf{L}_4^{K_0} := \{ \mathbf{v} \in \mathbf{L}_4: |\mathbf{v} - \mathbf{u}_0|_4 \leq K_0 \} \] into itself, where $K_0$ is the constant given in \eqref{est:u4}. \end{lemma} \begin{proof} We first claim that $F$ maps $\mathbf{L}_4^{K_0}$ into itself for all $\tau \in (0, \tau_0^*]$. In fact, for any $\tilde{\mathbf{u}} \in \mathbf{L}_4^{K_0}$, Lemma \ref{lem2} assures that $F(\tilde{\mathbf{u}}) \in \mathbf{L}_4^{K_0}$, provided that $0< \tau \leq \tau_1^* := \min (1, \tau_1)$ with $\tau_1 = \frac{16}{27} \nu^3 K_1^{-4}~|\tilde{\mathbf{u}}|_4^{-4}$ for $\tilde{\mathbf{u}} \neq \mathbf{0}$ and $\tau_1 < +\infty$ for $\tilde{\mathbf{u}} = \mathbf{0}$. Since $\tilde{\mathbf{u}} \in \mathbf{L}_4^{K_0}$ implies $|\tilde{\mathbf{u}}|_4 \leq |\mathbf{u}_0|_4 + K_0 $, we find \begin{equation*} \tau_1 = \tau_1(|\tilde{\mathbf{u}}|_4) = \frac{16}{27} \nu^3~K_1^{-4}~|\tilde{\mathbf{u}}|_4^{-4} \geq \frac{16}{27} \nu^3~K_1^{-4} \frac{1}{\left( K_0+|\mathbf{u}_0|_4 \right)^4} \geq \tau_0^*. \end{equation*} Thus we conclude that $F(\tilde{\mathbf{u}}) \in \mathbf{L}_4^{K_0}$ for all $\tau \in (0,\tau_0^*]$. \\ \quad Now we are going to show that $F(\cdot)$ is a contraction. Let $\tilde{\mathbf{u}_i} \in \mathbf{L}_4^{K_0}$, $\mathbf{u}_i = F(\tilde{\mathbf{u}}_i)$ ($i=1,2$), then as in the proof of Lemma \ref{lem2}, it is easy to see that $\mathbf{u}_1 - \mathbf{u}_2$ satisfies \begin{equation} \label{est:diff} \frac{1}{\tau} |\mathbf{u}_1 - \mathbf{u}_2|^2 + b(\tilde{\mathbf{u}}_1, \mathbf{u}_1, \mathbf{u}_1 - \mathbf{u}_2) - b(\tilde{\mathbf{u}}_2, \mathbf{u}_2, \mathbf{u}_1 - \mathbf{u}_2) + \nu |\nabla \mathbf{u}_1 - \nabla \mathbf{u}_2|^2 = 0. \end{equation} We here note that \begin{equation} \label{est:b:diff} \begin{aligned} &|b(\tilde{\mathbf{u}}_1, \mathbf{u}_1, \mathbf{u}_1 - \mathbf{u}_2) - b(\tilde{\mathbf{u}}_2, \mathbf{u}_2, \mathbf{u}_1 - \mathbf{u}_2)| \\ & = |b(\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2, \mathbf{u}_1, \mathbf{u}_1 - \mathbf{u}_2) + b(\tilde{\mathbf{u}}_2, \mathbf{u}_1 - \mathbf{u}_2, \mathbf{u}_1 - \mathbf{u}_2)| \\ & \leq |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4 |\nabla\mathbf{u}_1| |\mathbf{u}_1 - \mathbf{u}_2|_4 + |\tilde{\mathbf{u}}_2|_4 |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2| |\mathbf{u}_1 - \mathbf{u}_2|_4 \end{aligned} \end{equation} holds. Therefore, by \eqref{est:nablau} and \eqref{ineq:GNL4}, we get \begin{equation}\label{est:b:diff1} \begin{split} & |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4 |\nabla\mathbf{u}_1| |\mathbf{u}_1 - \mathbf{u}_2|_4 \\ & \leq |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4 K_2^{-1} K_0 K_1 |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2|^{1/2} |\mathbf{u}_1 - \mathbf{u}_2|^{1/2} \\ & \leq \frac{K_2^{-2} \nu}{4} |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4^2 + \frac{1}{\nu} K_0^2 K_1^2 |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2|~|\mathbf{u}_1 - \mathbf{u}_2| \\ & \leq \frac{K_2^{-2} \nu}{4} |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4^2 + \frac{\nu}{4} |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2|^2 + \frac{1}{\nu^3} K_0^4 K_1^4 |\mathbf{u}_1 - \mathbf{u}_2|^2. \end{split} \end{equation} Since $|\tilde{\mathbf{u}}_2|_4 \leq |\mathbf{u}_0|_4 + K_0$, \eqref{ineq:GNL4} and Young's inequality yield \begin{equation} \label{est:b:diff2} \begin{aligned} & |\tilde{\mathbf{u}}_2|_4 |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2| |\mathbf{u}_1 - \mathbf{u}_2|_4 \\ & \leq \left(|\mathbf{u}_0|_4 + K_0 \right) K_1 |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2|^{\frac32} |\mathbf{u}_1 - \mathbf{u}_2|^{1/2} \\ & \leq \frac34 \frac{\nu}{4} |\nabla\mathbf{u}_1 - \nabla\mathbf{u}_2|^2 + \frac{\nu^{-3}}{4} 4^3 \left(|\mathbf{u}_0|_4 + K_0 \right)^4 K_1^4 |\mathbf{u}_1 - \mathbf{u}_2|^2. \end{aligned} \end{equation} Then, in view of \eqref{est:diff}-\eqref{est:b:diff2}, we obtain \begin{align*} &\frac{1}{\tau} |\mathbf{u}_1 - \mathbf{u}_2|^2 + \nu |\nabla \mathbf{u}_1 - \nabla \mathbf{u}_2|^2\\ & \leq \frac{K_2^{-2} \nu}{4} |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4^2 + \frac{\nu}{2}|\nabla \mathbf{u}_1 - \nabla \mathbf{u}_2|^2 + \tilde{K} |\mathbf{u}_1 - \mathbf{u}_2|^2, \\ \tilde{K} & = \frac{1}{\nu^3} \left(K_0^4 K_1^4 + 16 ( |\mathbf{u}_0|_4 + K_0)^4 K_1^4 \right). \end{align*} Since $\tau \in (0,\tau_0]$ assures that $\frac{1}{\tau} \geq \frac{K_1^4}{\nu^3}\, 17 \, \left(|\mathbf{u}_0|_4 + K_0 \right)^4 \geq \tilde{K}$, we obtain \[ \frac{K_2^{-2} \nu}{2} |\mathbf{u}_1 - \mathbf{u}_2|_4^2 \leq \frac{\nu}{2} |\nabla \mathbf{u}_1 - \nabla \mathbf{u}_2|^2 \leq \frac{K_2^{-2} \nu}{4} |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4^2; \] i.e., $|F(\tilde{\mathbf{u}}_1) - F(\tilde{\mathbf{u}}_2)|_4 \leq \frac{1}{\sqrt{2}} |\tilde{\mathbf{u}}_1 - \tilde{\mathbf{u}}_2|_4$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1}] It is clear that $\mathbf{L}_4^{K_0}$ is a closed convex subset of $\mathbf{L}_4$. 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