\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil A system of differential inclusions \hfil EJDE--2000/43} {EJDE--2000/43\hfil Chunpeng Wang \& Jingxue Yin \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.~{\bf 2000}(2000), No.~43, pp.~1--8. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Uniqueness of solutions to a system of differential inclusions \thanks{ {\em Mathematics Subject Classifications:} 35K50, 35K65, 35A05, 35D99. \hfil\break\indent {\em Key words:} differential inclusions, degeneracy, uniqueness. \hfil\break\indent \copyright 2000 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted May 15, 2000. Published June 12, 2000.} } \date{} % \author{ Chunpeng Wang \& Jingxue Yin } \maketitle \begin{abstract} In this paper we study the uniqueness of solutions to the initial and Dirichlet boundary-value problem of differential inclusions $$\Delta u_i+\nabla\cdot\stackrel{\rightarrow}{B_i} (u_1,u_2,\dots,u_N)\in \frac{\partial F_i(u_i)}{\partial t}, \quad i=1,2,\dots,N,$$ where $\stackrel{\rightarrow}{B_i}(s_1,s_2,\dots,s_N)$ is an $n$-dimensional vector continuously differentiable on ${\mathbb R}^N$, and $F_i(u_i)=\{w_i:u_i=A_i(w_i)\}$, $i=1,2,\dots,N$ with $A_i(s)$ continuously differentiable functions on ${\mathbb R}$ and $A'_i(s)\geq 0$. \end{abstract} \newcommand\iint{\int\hskip-2mm\int} \newtheorem{theorem}{Theorem} \section{Introduction}\label{sec1} This paper concerns with the system of differential inclusions $$\Delta u_i+\nabla\cdot\stackrel{\rightarrow}{B_i} (u_1,u_2,\dots,u_N)\in \frac{\partial F_i(u_i)}{\partial t}, \quad (x,t)\in Q_T, \quad i=1,2,\dots,N, \eqno(1.1)$$ where $\Omega$ is a bounded domain in ${\mathbb R}^n$ with smooth boundary $\partial\Omega$, $Q_T=\Omega\times(0,T)$, with $T>0$, $n$ and $N$ are positive integers, $\stackrel{\rightarrow}{B_i}(s_1,s_2,\dots,s_N)$ is an $n$-dimensional vector continuously differentiable on ${\mathbb R}^N$, and $$F_i(u_i)=\{w_i:u_i=A_i(w_i)\}, \quad i=1,2,\dots,N$$ with $A_i(s)$ continuously differentiable functions on ${\mathbb R}$ and $A'_i(s)\geq 0$. Note that if $A_i(s)$ is strictly increasing, then $F_i(u_i)$ is single-valued, and (1.1) becomes equality. However, we are interested in the case when some or all $A_i(s)$'s are only nondecreasing, and so the $F_i(u_i)$'s are interval-valued functions. System (1.1) arises from mathematical models describing the nonlinear diffusion phenomena which exist in nature extensively. An important classical case of (1.1) is that with $\stackrel{\rightarrow}{B_i}=\stackrel{\rightarrow}{0}$ and $N=1$. In this case (1.1) can be changed to $$\frac{\partial w}{\partial t}=\Delta A(w).$$ Br\'{e}zis and Crandall \cite{B1} proved the uniqueness of bounded measurable solutions for the Cauchy problem of the equation, where the nonlinear function $A(s)$ is assumed to be only non-decreasing. In other words, if $A(s)$ is differentiable, then $$A'(s)\geq 0\,;$$ namely, the equation is permitted to be strongly degenerate. Thereafter some authors tried to extend the uniqueness results to the equation with convection, i.e., $$\frac{\partial w}{\partial t}=\Delta A(w) +\nabla\cdot\stackrel{\rightarrow}{B}(w).$$ However, in most of those works, the nonlinear function $A(s)$ is assumed to be strictly increasing. In other words, the equation is weakly degenerate, see for example \cite{D1,C1,W2,Y1}. In this paper we study the uniqueness of solutions of the initial and Dirichlet boundary-value problem of (1.1). The initial-boundary conditions are $$\displaylines{ \hfill u_i=0, \quad (x,t)\in \partial\Omega\times[0,T], \quad i=1,2,\dots,N, \hfill \llap{(1.2)} \cr \hfill F_i(u_i)(x,0)=\{f_i(x)\},\quad x\in \Omega, \quad i=1,2,\dots,N. \hfill\llap{(1.3)} }$$ \paragraph{Definition} For $i=1,2,\dots,N$, let $f_i$'s be bounded and measurable functions. $(u_1,u_2,\dots,u_N)$ is called a solution of the initial and Dirichlet boundary-value problem (1.1)--(1.3), if the $u_i$'s are bounded and measurable functions and there exist bounded measurable functions $w_i\in F_i(u_i)$ such that for arbitrary test function $\varphi$ in $C^\infty({\overline{Q_T}})$ with value zero for $x\in\partial\Omega$ and for $t=T$, the following integral equalities hold \begin{eqnarray*} && \iint_{Q_T} \left(u_i\Delta\varphi-\stackrel{\rightarrow}{B_i}(u_1,u_2,\dots,u_N) \cdot\nabla\varphi+w_i\frac{\partial\varphi}{\partial t}\right)\,dx\,dt \\ &&+\int_\Omega f_i(x)\varphi(x,0)dx=0, \quad i=1,2,\dots,N. \end{eqnarray*} The main result of this paper is the following theorem. \begin{theorem} The initial and Dirichlet boundary-value problem (1.1)--(1.3) has at most one solution. \end{theorem} The method of the proof is inspired by Br\'{e}zis and Crandall \cite{B1}. Here what we consider is not the Cauchy problem but the initial and Dirichlet boundary-value problem, so we adopt the self-adjoint operators with homogeneous Dirichlet boundary condition to prove the uniqueness instead of the self-adjoint operators on the whole space. Moreover, the problem which we consider is a system of differential inclusions with convection, so we must overcome some other technical difficulties. \section{Proof of the main theorem} \label{sec2} We first introduce a family of operators. The $L^2$ theory for elliptic equations (see, e.g., \cite{G1} ) implies that for each $\lambda>0$ and $f\in H^{-1}(\Omega)$, the Dirichlet problem $$\displaylines{ \hfill -\Delta u+\lambda u=f,\quad x\in\Omega,\hfill\llap{(2.1)} \cr \hfill u=0,\quad x\in\partial\Omega,\hfill \llap{(2.2)} }$$ has a unique solution $u\in H^1_0(\Omega)$. For $0<\lambda<1$, we define the operator $$T_\lambda:H^{-1}(\Omega)\rightarrow H^1_0(\Omega), \quad f\mapsto u,$$ where $u$ is the unique solution to (2.1)--(2.2). It is easy to see that $T_\lambda$ is self-adjoint, namely, for arbitrary $f,g\in H^{-1}(\Omega)$, $$\langle f,T_\lambda g\rangle=\langle g,T_\lambda f\rangle$$ holds, where $\langle \cdot,\cdot\rangle$ represents the dual product between $H^{-1}(\Omega)$ and $H^1_0(\Omega)$. Specially, for $f\in L^2(\Omega)$ and $g\in H^{-1}(\Omega)$, we have $$\langle f,T_\lambda g\rangle=\int_\Omega f T_\lambda g dx.$$ In addition, for arbitrary $f\in L^2(\Omega)$, the $L^2$ theory for elliptic equations also implies $T_\lambda f\in H^2(\Omega)\cap H^1_0(\Omega)$ and $$\|T_\lambda f\|_{H^2(\Omega)}\leq C_0 \|f\|_{L^2(\Omega)}, \eqno (2.3)$$ here $C_0$ is a constant depending only on $n$ and $\Omega$, but independent of $\lambda$. \paragraph{Proof of Theorem 1.} Let $(u_1,u_2,\dots,u_N)$ and $(\hat{u}_1,\hat{u}_2,\dots,\hat{u}_N)$ be two solutions to (1.1)--(1.3). For $i=1,2,\dots,N$, the bounded measurable functions in $F_i(u_i)$ and $F_i(\hat{u}_i)$ satisfying the definition are denoted by $w_i$ and $\hat{w}_i$ correspondingly. For $i=1,2,\dots,N$, we set $$\displaylines{ v_i=u_i-\hat{u}_i, \quad z_i=w_i-\hat{w}_i, \cr \stackrel{\rightarrow}{H_i}= \stackrel{\rightarrow}{B_i}(u_1,u_2,\dots,u_N) -\stackrel{\rightarrow}{B_i}(\hat{u}_1,\hat{u}_2,\dots,\hat{u}_N). }$$ The definition of solutions implies that $z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}~(i=1,2,\dots,N)$ are all bounded measurable functions, and for arbitrary test function $\varphi$, namely, $\varphi\in C^\infty({\overline{Q_T}})$ with $\varphi=0$ for $x\in\partial\Omega$ and for $t=T$, the integral equalities $$\iint_{Q_T} \left(v_i\Delta\varphi-\stackrel{\rightarrow}{H_i} \cdot\nabla\varphi+z_i\frac{\partial\varphi}{\partial t}\right)\,dx\,dt=0, \quad i=1,2,\dots,N \eqno (2.4)$$ hold. Let $\psi\in C^\infty([0,T])$ with $\psi(T)=0$ and $k\in C_0^\infty(\Omega)$. Then we see that $T_\lambda k\in H^2(\Omega)\cap H^1_0(\Omega)$. By an approximate process, we may choose $\psi T_\lambda k$ as a test function. Letting $\varphi=\psi T_\lambda k$ in (2.4), we get $$\iint_{Q_T} \left(\lambda\psi v_i T_\lambda k -\psi v_i k-\psi\stackrel{\rightarrow}{H_i} \cdot\nabla T_\lambda k +\frac{\partial\psi}{\partial t} z_i T_\lambda k\right)\,dx\,dt=0.$$ Using integration by parts and the self-adjointness of $T_\lambda$, we get $$\iint_{Q_T} \left(\lambda\psi k T_\lambda v_i -\psi k v_i+\psi k T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i}) -\psi k\frac{\partial T_\lambda z_i}{\partial t}\right)\,dx\,dt=0.$$ Owing to the arbitrariness of $\psi$ and $k$, we see that $$\frac{\partial T_\lambda z_i}{\partial t}= \lambda T_\lambda v_i-v_i+T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i}) \eqno (2.5)$$ in the sense of distribution. It follows that $\frac{\partial T_\lambda z_i}{\partial t}\in L^2(Q_T)$ and $T_\lambda z_i\in H^2(\Omega)\cap H^1_0(\Omega)$. Let $\psi\in C^\infty([0,T])$ with $\psi(T)=0$. By an approximate process, we may choose $\psi T_\lambda z_i$ as a test function. Letting $\varphi=\psi T_\lambda z_i$ in (2.4), we get $$\iint_{Q_T} \left(\lambda\psi v_i T_\lambda z_i-\psi v_i z_i -\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i +\frac{\partial\psi}{\partial t} z_i T_\lambda z_i +\psi z_i \frac{\partial T_\lambda z_i}{\partial t}\right)\,dx\,dt=0. \eqno (2.6)$$ Combining (2.5) with (2.6), we see that \begin{eqnarray*} \iint_{Q_T} \Big(\lambda\psi v_i T_\lambda z_i +\lambda\psi z_i T_\lambda v_i-2\psi v_i z_i +\psi z_i T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i})&& \\ -\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i +\frac{\partial\psi}{\partial t} z_i T_\lambda z_i\Big)\,dx\,dt&=&0\,. \end{eqnarray*} Using integration by parts and the self-adjointness of $T_\lambda$, for $i=1,2,\dots, N$, we get $$\iint_{Q_T} \Big(2\lambda\psi v_i T_\lambda z_i-2\psi v_i z_i -2\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i +\frac{\partial\psi}{\partial t} z_i T_\lambda z_i\Big)\,dx\,dt=0. \eqno (2.7)$$ Let $$g_{i\lambda}(t)=\int_\Omega z_i T_\lambda z_i dx, \quad t\in[0,T], \quad i=1,2,\dots,N.$$ Now we prove that $g_{i\lambda}(t)$ converges to zero on $[0,T]$ uniformly as $\lambda\rightarrow 0$ for $i=1,2,\dots,N$. First, we show that $g_{i\lambda}(t)$ is absolutely continuous. From (2.7), we get $$\iint_{Q_T}\left(2\lambda\psi v_i T_\lambda z_i-2\psi v_i z_i -2\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i -\psi\frac{\partial (z_i T_\lambda z_i)}{\partial t}\right)\,dx\,dt=0.$$ From the arbitrariness of $\psi$, we see that \begin{eqnarray*} g'_{i\lambda}(t) &=&\frac{d}{dt}\int_\Omega z_i T_\lambda z_i dx =\int_\Omega\frac{\partial (z_i T_\lambda z_i)}{\partial t} \,dx\\ &=&2\lambda\int_\Omega v_i T_\lambda z_i dx -2\int_\Omega v_i z_i dx -2\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i dx, \quad\hbox{a.e.}~t\in [0,T]. \end{eqnarray*} Since $z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}$ are all bounded measurable functions and (2.3) holds, we get that $g'_{i\lambda}(t)\in L^1(0,T)$. Thus $g_{i\lambda}(t)$ is absolutely continuous. Next, we show that $g_{i\lambda}(0+0) \equiv\lim_{t\rightarrow 0^+}g_{i\lambda}(t)=0$. Let $$\psi_\varepsilon(t)=\int^{+\infty}_t \alpha_\varepsilon(s-\varepsilon)ds, \quad \alpha_\varepsilon(s)=\frac1\varepsilon \alpha\left(\frac{s}{\varepsilon}\right),$$ where $\alpha(s)$ denotes the kernel of one-dimensional mollifier, namely, $\alpha$ is in the space $C_0^\infty(-\infty,+\infty)$, $\alpha\geq0$, supp$\alpha=[-1,1]$ and $\int_{-1}^1\alpha(s)\,ds=1$. Thus $\psi_\varepsilon\in C^\infty([0,T])$ and $\psi_\varepsilon(T)=0$ for sufficiently small $\varepsilon>0$. Letting $\psi=\psi_\varepsilon$ in (2.7), we get $$\iint_{Q_T} (2\lambda\psi_\varepsilon v_i T_\lambda z_i -2\psi_\varepsilon v_i z_i -2\psi_\varepsilon\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i -\alpha_\varepsilon(t-\varepsilon) z_i T_\lambda z_i)\,dx\,dt=0.$$ The dominated convergence theorem implies \begin{eqnarray*} g_{i\lambda}(0+0) &=& \lim_{\varepsilon\rightarrow0^+}\int^{2\varepsilon}_0 \alpha_\varepsilon(t-\varepsilon)g_{i\lambda}(t)dt \\ &=& \lim_{\varepsilon\rightarrow0^+}\iint_{Q_T} \alpha_\varepsilon(t-\varepsilon) z_i T_\lambda z_i \,dx\,dt\\ &=& 2\lambda\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T} \psi_\varepsilon v_i T_\lambda z_i \,dx\,dt -2\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T} \psi_\varepsilon v_i z_i \,dx\,dt\\ &&-2\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T} \psi_\varepsilon\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i \,dx\,dt\,. \end{eqnarray*} Since $z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}$ are all bounded measurable functions and (2.3) holds, we get that $g_{i\lambda}(0+0)=0$. Finally, we prove that $g_{i\lambda}(t)$ converges to zero on $[0,T]$ uniformly as $\lambda\rightarrow 0$. It follows easily from the above arguments that \begin{eqnarray*} g_{i\lambda}(t) &=&g_{i\lambda}(0+0)+\int^t_0 g'_{i\lambda}(s)ds\\ &=& 2\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds -2\int_0^t\int_\Omega v_i z_i \,dx\,ds\\ && -2\int_0^t\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i \,dx\,ds\,. \end{eqnarray*} Since $w_i$ and $\hat{w}_i$ are bounded measurable and $A_i$ and $\stackrel{\rightarrow}{B_i}$ are continuously differentiable, there exist three positive constants $M_0$, $M_1$ and $M_2$ such that for $i=1,2,\dots,N$, the following estimates hold $$|z_i|\leq M_0, \quad |v_i|\leq M_1|z_i|, \quad |\stackrel{\rightarrow}{H_i}| \leq M_2\left(\sum_{j=1}^N v_j^2\right)^{1/2}.$$ Noticing that $z_i$ and $v_i$ have the same sign for $A'_i(s)\geq0$, we get $$v_i z_i=|v_i||z_i|\geq\frac{1}{M_1}v_i^2.$$ By Schwarz's inequality and Young's inequality, we get \begin{eqnarray*} \lefteqn{ \big|\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i \,dx \big| } \\ &\leq&\left(\int_\Omega|\stackrel{\rightarrow}{H_i}|^2 dx\right)^{1/2} \left(\int_\Omega|\nabla T_\lambda z_i|^2 dx\right)^{1/2}\\ &\leq& M_2\sum_{j=1}^N\left(\int_\Omega v_j^2 dx\right)^{1/2} \left(\int_\Omega|\nabla T_\lambda z_i|^2 dx\right)^{1/2}\\ &\leq& \frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+ \frac{N^2 M_1 M_2^2}{4} \int_\Omega(\nabla T_\lambda z_i\nabla T_\lambda z_i)dx\\ &=& \frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+ \frac{N^2 M_1 M_2^2}{4} \int_\Omega(-T_\lambda z_i\Delta T_\lambda z_i)dx\\ &=& \frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+ \frac{N^2 M_1 M_2^2}{4}\int_\Omega\left(-\lambda (T_\lambda z_i)^2 +z_i T_\lambda z_i\right) dx\\ &\leq& \frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+ \frac{N^2 M_1 M_2^2}{4}g_{i\lambda}(t). \end{eqnarray*} Let $$g_\lambda(t)=\sum_{i=1}^N g_{i\lambda}(t).$$ Therefore, \begin{eqnarray*} \lefteqn{g_\lambda(t)}\\ &=& 2\sum_{i=1}^N\Big(\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds -\int_0^t\int_\Omega v_i z_i \,dx\,ds -\int_0^t\int_\Omega\stackrel{\rightarrow}{H_i} \cdot\nabla T_\lambda z_i \,dx\,ds\Big)\\ &\leq& 2\sum_{i=1}^N\Big(\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds -\frac{1}{M_1}\int_0^t\int_\Omega v_i^2 \,dx\,ds\\ && +\frac{1}{N M_1}\sum_{j=1}^N\int_0^t\int_\Omega v_j^2 \,dx\,ds +\frac{N^2 M_1 M_2^2}{4}\int_0^tg_{i\lambda}(s)ds\Big)\\ &\leq& 2\lambda\sum_{i=1}^N\int_0^T\int_\Omega|v_i T_\lambda z_i|\,dx\,ds +\frac{N^2 M_1 M_2^2}{2}\int_0^t g_\lambda(s)ds. \end{eqnarray*} Moreover, it follows that $$g_\lambda(t)\geq0$$ by \begin{eqnarray*} g_{i\lambda}(t) &=&\int_\Omega z_i T_\lambda z_i dx\\ &=& \int_\Omega(-\Delta T_\lambda z_i T_\lambda z_i+ \lambda T_\lambda z_i T_\lambda z_i)dx\\ &=& \int_\Omega(\nabla T_\lambda z_i \nabla T_\lambda z_i+ \lambda T_\lambda z_i T_\lambda z_i)dx\\ &\geq&0\,. \end{eqnarray*} Hence by Gronwall's inequality, we get $$g_\lambda(t)\leq C_1\lambda,$$ where $C_1$ is a constant depending only on $N$, $M_0$, $M_1$, $M_2$, $C_0$, $T$ and the measure of $\Omega$, but independent of $\lambda$ and $t$. So $g_\lambda(t)$ converges to zero on $[0,T]$ uniformly as $\lambda\rightarrow 0$. Noticing that $g_{i\lambda}(t)\geq0$, we get that $g_{i\lambda}(t)$ converges to zero on $[0,T]$ uniformly as $\lambda\rightarrow 0$. Now we prove $$z_i(x,t)=0, \quad \hbox{a.e.}~(x,t)\in Q_T, \quad i=1,2,\dots,N.$$ For any $\varphi\in C^\infty_0(Q_T)$, we have \begin{eqnarray*} \lefteqn{\big|\iint_{Q_T}z_i\varphi \,dx\,dt\big|^2}\\ &=& \left|\iint_{Q_T} (-\Delta T_\lambda z_i +\lambda T_\lambda z_i) \varphi \,dx\,dt\right|^2\\ &=& \left|\iint_{Q_T}(\nabla T_\lambda z_i \nabla\varphi +\lambda\varphi T_\lambda z_i) \,dx\,dt\right|^2\\ &\leq& 2\|\nabla\varphi\|_{L^2(Q_T)}^2\|\nabla T_\lambda z_i\|_{L^2(Q_T)}^2 +2\lambda^2\|\varphi\|_{L^2(Q_T)}^2\|T_\lambda z_i\|_{L^2(Q_T)}^2\\ &\leq& C_2 \left(\iint_{Q_T}|\nabla T_\lambda z_i|^2 \,dx\,dt+ \lambda\iint_{Q_T}(T_\lambda z_i)^2 \,dx\,dt\right)\\ &\leq& C_2\iint_{Q_T} (-\Delta T_\lambda z_i+ \lambda T_\lambda z_i)T_\lambda z_i \,dx\,dt\\ &=& C_2\iint_{Q_T}z_i T_\lambda z_i\,dx\,dt\\ &\leq& C_2 T\sup_{t\in[0,T]}g_{i\lambda}(t)\rightarrow 0, \quad (\lambda\rightarrow 0), \end{eqnarray*} where $C_2=2\|\nabla\varphi\|_{L^2(Q_T)}^2+2\|\varphi\|_{L^2(Q_T)}^2$ independent of $\lambda$. Therefore, $$\iint_{Q_T}z_i\varphi \,dx\,dt=0, \quad \forall \varphi\in C^\infty_0(Q_T).$$ It follows that $$z_i(x,t)=0, \quad \hbox{a.e.}~(x,t)\in Q_T,\quad i=1,2,\dots,N.$$ Thus $$w_i(x,t)=\hat{w}_i(x,t), \quad \hbox{a.e.}~(x,t)\in Q_T, \quad i=1,2,\dots,N,$$ which implies $$u_i(x,t)=\hat{u}_i(x,t), \quad \hbox{a.e.}~(x,t)\in Q_T,\quad i=1,2,\dots,N.$$ The proof is complete. \begin{thebibliography}{10} \bibitem{A1} H.~W. Alt and S. Luckhaus, Quasilinear elliptic--parabolic differential equations, {\it Math. Z.,} {\bf 183} (1983), 311--341. \bibitem{B1} H.~Br\'{e}zis and M.~G.~Crandall, Uniqueness of solutions of the initial value problem for $u_t-\Delta\varphi=0$, {\it J. Math. 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J.,} {\bf 1(2)} (1985), 153--165. \end{thebibliography} \medskip \noindent{\sc Chunpeng Wang \& Jingxue Yin} \\ Department of Mathematics, JiLin University, \\ Changchun, Jilin 130023, People's Republic of China\\ e-mail: yjx@mail.jlu.edu.cn \end{document} 