\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 70, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2006/70\hfil Quasilinear degenerate parabolic systems]
{Weak solutions for quasilinear degenerate parabolic systems}
\author[Z. Yao, W. Zhou\hfil EJDE-2006/70\hfilneg]
{Zheng'an Yao, Wenshu Zhou}  % in alphabetical order

\address{Zheng'an Yao \newline
 Department of Mathematics, Sun Yat-sen University,
Guangzhou 510275, China}
\email{mcsyao@mail.sysu.edu.cn}

\address{Wenshu Zhou \newline
 Department of Mathematics, Jilin University, Changchun
130012, China} \email{wolfzws@163.com}

\date{}
\thanks{Submitted December 12, 2005. Published July 7, 2006.}
\thanks{Supported by grants: 10171113 and 10471156 from NNSF
 of China and 4009793  from NSF \hfill\break\indent of GuangDong and by 985
 Program, Young Foundation of  Department of Mathemaics
 and \hfill\break\indent  Young Teacher Foundation of Jilin
 University} 
 \subjclass[2000]{35K10, 35K50, 35K55, 35K65} 
 \keywords{Quasilinear degenerate parabolic system; weak solution; 
 \hfill\break\indent existence; uniqueness; asymptotic behavior}

\begin{abstract}
 This paper concerns the initial Dirichlet  boundary-value problem
 for a class of quasilinear degenerate parabolic systems.
 Due to the degeneracies, the problem does not have classical solutions
 in general. Combining the special form of the system, a proper concept
 of  a weak solution is presented, then  the existence and uniqueness
 of weak solutions are proved.  Moreover, the asymptotic behavior
 of weak solutions will also be discussed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction and Results}

This paper concerns the initial Dirichlet boundary-value problem
for the quasilinear degenerate parabolic system
\begin{equation} \label{P}
\begin{gathered}
 u_t=a(u) (\Delta u+\alpha v)\quad\text{in } \Omega_\infty,\\
 v_t=b(v) (\Delta v+\beta u)\quad\text{in } \Omega_\infty,\\
u(x,t)=0,\quad v(x,t)=0,\quad\text{on } \partial \Omega \times (0,\infty),  \\
 u(x,0)=u_0(x),\quad v(x,0)=v_0(x), \quad \text{in }\Omega,\\
 \end{gathered}
\end{equation}
where $\Omega_\infty=\Omega \times (0,+\infty), \Omega$ is a
bounded domain in $\mathbb{R}^N$ with approximately smooth
boundary  $\partial \Omega$, $\alpha$ and $ \beta$ are positive
constants,
  $a(\cdot), b(\cdot) \in \mathcal{K}=\{y(s) \in C^1[0,\infty);
y(0)=0, y'(s)>0, \forall s >0\}$, and
$$
u_0, v_0 \in \mathcal{S}
=\{y(x) \in C(\overline \Omega) \cap H^1(\Omega);
  y(x)\ge 0~\text{on }\overline \Omega, y(x)=0~\text{on }\partial \Omega\}.
$$
This system can be used to describe the development of two groups
in the dynamics of biological groups where $u$ and $v$ are the
densities of the different groups. Similar systems can be found in
\cite{d3,g1,g2,l2,s1,w2}.

The system has been studied in a series of papers, see \cite{d2,w3,d4} and
references therein. For instance, it was proved in \cite{d2} that under
the following assumption conditions:
\begin{itemize}
\item[(H1)] $u_0, v_0 \in C^1(\overline \Omega)$, $u_0>0$, $v_0>0$
in $\Omega$, $u_0=v_0=0$ on $\partial \Omega $;

\item[(H2)]  $\frac{\partial u_0}{\partial \nu}<0$,
$\frac{\partial v_0}{\partial \nu}<0$ on $\partial \Omega$,
where $\nu$ denotes the outward normal to $\partial \Omega$;

\item[(H3)] $a, b \in C[0,\infty) \cap C^1(0,\infty)$ such that $a, b>0
$ in $(0,\infty)$ and $a', b'>0$ in $(0,\infty)$;

\item[(H4)] Either $\mathop{\rm \lim \inf}_{s \to
\infty}\frac{a(s)}{b(s)}>0$ or
      $\mathop{\rm \lim \inf}_{s \to \infty}\frac{b(s)}{a(s)}>0$ holds,

\end{itemize}
 the positive solution of \eqref{P} blows up in
finite time if and only if $\lambda_1^2<\alpha\beta$, $\int^\infty_0
ds/(sa(s))<\infty$ and $\int^\infty_0 ds/(sb(s))<\infty$, where
$\lambda_1$ denotes the first eigenvalue of $-\Delta$ in $\Omega$
with the homogeneous Dirichlet boundary condition. In \cite{w3}, the
author discussed a special case of \eqref{P}: $a(u)=u^{p},
b(u)=u^{q}$ with $p, q \ge 1$, and proved that under the conditions
(H1)-(H2), the positive solutions of \eqref{P} exist globally  if
and only if $\lambda_1^2 \ge \alpha\beta$. For single equation
($a(s)=b(s), \alpha=\beta, u_0=v_0),
 u_t=a(u)(\Delta u+\alpha u)$, we refer to \cite{d1,f1,u1} and references therein.
 In \cite{d1,u1}, for instance, the authors studied the equation with $a(s)=s$ and
   obtained some interesting results.

Since the system is degenerate at points where $u=0$ or $v=0$,
problem \eqref{P} does not always have  classical solutions, and we have
to consider weak solutions. Moreover, we are only interested in the
nonnegative weak solutions.

We remark that, as usual, one may easily define a weak solution
(which, for instance, means a function satisfying the condition (a),
(b) and (d) of the following Definition \ref{def1.1}). However, because of
the special form of this system, such weak solutions may not be
uniquely determined by the initial data. In fact, for single
equation some examples showing the non-uniqueness had been
constructed, see \cite{d1,u1}. So, it is natural
 to ask  how to define a weak solution to guarantee
both uniqueness and existence. One of purposes of this paper is to
give a positive answer to the question. Moreover, the asymptotic
behavior of  solutions will also be discussed. This is the only work
concerning the study of weak solutions to the system, as far as we
know.

Before giving a proper concept of weak solutions, we first define
the support of a nonnegative measurable function $w: \Omega
\to [0, \infty)$:
$$
  \mathop{\rm supp}w
=\overline {\Big\{x \in G ;\lim_{\rho \to 0^+} \frac{\mu(G
\cap B_{\rho}(x))}{\mu(B_{\rho}(x))} >0\Big\}},
 $$
where $G=\{x \in \Omega ; w(x) >0\},~B_{\rho}(x)=\{y \in \Omega ;
|x-y|< \rho\}$, and $\mu (E)$ denotes the Lebesgue measure of a set
$E$ in $\mathbb{R}^N$. It is easy to see that if $w \in C(\Omega)$,
then $\mathop{\rm supp }w =\overline G$.

For  $T>0, \rho>0$ and  $w \in \mathcal{S}$, denote $\Omega_T, \Omega
(w)$ and $\Omega^\rho (w)$ by
\begin{gather*}
\Omega_T=\Omega \times (0,T),\\
 \Omega (w) =\{x\in \Omega; w(x)>0\},\\
\Omega^\rho (w) =\{x\in  \Omega (w); \mbox{\rm dist}(x, \partial
\Omega (w) )>\rho\}.
\end{gather*}


\begin{definition} \label{def1.1} \rm
$(u,v)$ is called a weak solution of \eqref{P},
if for any $T>0$ the following conditions hold:
\begin{itemize}
\item[(a)] $ u, v \ge 0$ a.e. in $\Omega_T$,
$u,v \in  L^{\infty}(\Omega_T)\cap L^2 (0,T;H^1_0(\Omega))$,
 $u_t, v_t\in L^2(\Omega_T)$;

\item[(b)] For any $\varphi, \psi \in C_0^\infty(\Omega_T)$, there holds
 \begin{align*}
  &\iint_{\Omega_T} \Big(-u  \varphi_t+ a(u)\nabla u \nabla\varphi
  +a'(u)|\nabla u|^2 \varphi-\alpha a(u)v\varphi\Big) \,dx\,dt\\
 &+  \iint_{\Omega_T} \Big(-v \psi_t+ b(v)\nabla v \nabla\psi
 +b'(v)|\nabla v|^2 \psi-\beta b(v)u\psi\Big) \,dx\,dt=0;
 \end{align*}

\item[(c)] $\mathop{\rm supp}u(t)=\mathop{\rm supp}u_0$,
 $\mathop{\rm supp}v(t)=\mathop{\rm supp}v_0$,  a.e. in $(0,T)$,
 and for all $\rho>0$ there exist positive constants $c_1=c_1(\rho)$
 and $c_2=c_2(\rho)$  such that
\begin{gather*}
 u \ge c_1 \quad \mbox{\rm a.e. in }\Omega^\rho (u_0) \times (0,T),\\
 v \ge c_2 \quad \mbox{\rm a.e. in }\Omega^\rho (v_0) \times (0,T);
\end{gather*}

\item[(d)] $ (u(t), v(t)) \to (u_0,v_0)$ in $[L^1(\Omega)]^2,$ as
$t \to 0^+$.
\end{itemize}
\end{definition}

The purposes of this paper are to prove the following theorems.

\begin{theorem} \label{thm1.1}
Let $a, b \in \mathcal{K}, \alpha,\beta>0$, and assume  $ u_0,
v_0 \in  \mathcal{S}$. If $\max\{\alpha,\beta\}<\lambda_1$, where
$\lambda_1$ is the same as before,
 then \eqref{P} admits a unique weak solution.
\end{theorem}

\begin{theorem} \label{thm1.2}
Suppose $a, b \in \mathcal{K}$, and there exist positive constants
$\sigma_2,  \rho_2, \rho_1, \rho_0$ and a nonnegative constant
$\sigma_1$ such that $\sigma_2 \ge 1, \sigma_2> \sigma_1 \ge 0,
\rho_2\ge \rho_1$, and
\begin{gather}
\lim_{s \to 0^+}
\frac{a(s)}{s^{\sigma_2}}=\rho_2,\quad
\lim_{s \to +\infty} \frac{a(s)}{s^{\sigma_2}}=\rho_1,\label{e1.1}\\
a(s)\ge  \sigma_2^{-1} s a'(s), \quad  \rho_0 b(s)s^{\sigma_1} \ge
a(s),\quad \forall s\ge 0, \label{e1.2}
\end{gather}
and let  $\alpha,\beta>0, u_0,v_0 \in \mathcal{S}$, and
$A_0\equiv\int_{\Omega}\Big(\frac{u_0^2}{2}+\int_0^{v_0}
\frac{sa(s)}{b(s)}ds\Big)dx>0$. If $\max\{\alpha,\beta\}<\frac{
4}{(2+\sigma_2)^2}\lambda_1$, then there exists a positive constant
$C$ depending only on the known data such that
$$
\int_\Omega \Big(\frac{u^2(x,t)}{2}+\int_0^{v(x,t)}
\frac{sa(s)}{b(s)}ds\Big) dx \le \Big[\frac{1}{Ct+A_0^{-
\sigma_2/2}}\Big]^{2/\sigma_2},
$$
where  $(u,v)$ is the unique weak solution of \eqref{P}.
\end{theorem}

This paper is organized as follows: in next section, we prove
Theorem 1.2. Section 3 is devoted to the proof of Theorem 1.3.

\section{Proof of Theorem 1.2}

\subsection{Proof of existence}
To establish the existence, we use the method of regularization.
For this purpose, we consider for $T>0$,
\begin{equation} \label{Pe}
\begin{gathered}
 u_{\varepsilon t}=a_\varepsilon(u_{\varepsilon})
   (\Delta u_{\varepsilon}+\alpha v_{\varepsilon})\quad \text{in }\Omega_T,\\
v_{\varepsilon t}=b_\varepsilon(v_{\varepsilon})
   (\Delta v_{\varepsilon}+\beta u_{\varepsilon})\quad \text{in }\Omega_T,\\
u_{\varepsilon}(x,t)=v_\varepsilon(x,t)=\varepsilon\quad \text{on }
   \partial \Omega \times (0,T),\\
u_{\varepsilon}(x,0)=u_0(x)+\varepsilon, \quad
v_{\varepsilon}(x,0)=v_0(x)+\varepsilon \quad \text{in } \Omega,
\end{gathered}
\end{equation}
where $\varepsilon \in (0,1)$,
$a_\varepsilon, b_\varepsilon \in C^1(\mathbb{R})$ and
$$
a_\varepsilon(s)=\begin{cases}
a(s),&s \ge \varepsilon,\\
\frac{a^2(s)+a^2(\varepsilon)}{2a(\varepsilon)}, & s<\varepsilon.
\end{cases}\qquad
b_\varepsilon(s)=\begin{cases} b(s),&s \ge \varepsilon,\\
\frac{b^2(s)+a^2(\varepsilon)}{2b(\varepsilon)}, & s<\varepsilon.
\end{cases}
$$

\begin{lemma} \label{lem2.1.1}
Let $\max\{\alpha,\beta\}<\lambda_1$.
If $(u_\varepsilon,v_\varepsilon)$ is a classical solution of
 \eqref{Pe}, then there exists a positive constant $C$
independent of $\varepsilon$ such that
$$
\varepsilon \le u_\varepsilon, v_\varepsilon \le C\quad
\text{in }\Omega_T.\label{e2.1.1}
$$
\end{lemma}

\begin{proof}  First, it is easy to see that the maximal
principle implies the left-hand side of the above claim. It suffices
to show it's right-hand side.

It is well known that  for $\tilde
\lambda=(\max\{\alpha,\beta\}+\lambda_1)/2<\lambda_1$, there
exists a bounded domain $\tilde\Omega$ such that $\tilde\Omega
\supset \overline \Omega$ and $\tilde \lambda$ is the first
eigenvalue of $-\Delta$ in $\tilde \Omega$ with homogeneous
Dirichlet boundary condition \cite{c1}. Denote by $\tilde \phi$
the associated eigenfunction. Then
 $\tilde \phi \in C^2(\Omega) \cap C(\overline \Omega), \tilde \phi>0 $ in $\tilde \Omega$,
 and hence there exists a positive constant
$\delta$ such that $\tilde \phi \ge \delta$ on $\overline \Omega$.
Now choosing a positive constant $\tilde k$ such that
$$
\tilde k \tilde \phi \ge \max_{\overline \Omega}u_0+1
\quad \text{on }\overline \Omega.
$$
Let $w=\tilde k \tilde \phi$. Then we have
\begin{gather*}
w_t-a_\varepsilon(w)(\Delta w+\alpha w)
    a_\varepsilon(w)[\tilde\lambda-\alpha]w>0\quad \text{in }\Omega_T,\\
w_t-b_\varepsilon(w)(\Delta w+\beta w)
    =b_\varepsilon(w)[\tilde\lambda-\beta]w>0\quad \text{in }\Omega_T,
\end{gather*}
hence  it follows from Nagumo's lemma  \cite[pp. 4697]{w1} that
$$
u_\varepsilon, v_\varepsilon \le w \quad \text{in }\Omega_T.
$$
The proof is complete.
\end{proof}

By the standard theory of parabolic equations  \cite[pp. 596]{l1},
 \eqref{Pe} admits a unique classical solution
$(u_\varepsilon,v_\varepsilon)$ satisfying the inequalities of Lemma
2.1. Moreover, the maximal principle implies
\begin{equation}
  u_{\varepsilon_2} \ge u_{\varepsilon_1},\quad
  v_{\varepsilon_2} \ge v_{\varepsilon_1},
  \quad \text{in }\Omega,\quad \hbox{for }\varepsilon_2>\varepsilon_1.
\label{e2.1.2}
\end{equation}
Thus, by Lemma 2.1,  $(u_\varepsilon,v_\varepsilon)$ solves the
 problem
\begin{equation}
\begin{gathered}
 u_{\varepsilon t}=a(u_{\varepsilon})
   (\Delta u_{\varepsilon}+\alpha v_{\varepsilon})\quad\text{in }\Omega_T,\\
v_{\varepsilon t}=b(v_{\varepsilon})
   (\Delta v_{\varepsilon}+\beta u_{\varepsilon})\quad\text{in }\Omega_T,\\
   u_{\varepsilon}(x,t)=v_\varepsilon(x,t)
  = \varepsilon\quad\text{on }\partial \Omega \times (0,T),  \\
u_{\varepsilon}(x,0)=u_0(x)+\varepsilon, \quad
v_{\varepsilon}(x,0)=v_0(x)+\varepsilon\quad\text{in } \Omega,
\end{gathered}\label{e2.1.3}
\end{equation}
In view of  Lemma 2.1  and \eqref{e2.1.2}, one can derive that there
exist nonnegative functions $u, v \in L^\infty(\Omega)$ such that
\begin{equation}
(u_\varepsilon, v_\varepsilon) \to (u,v)\quad
\text{a.e. in $\Omega_T$, as } \varepsilon \to 0.\label{e2.1.4}
\end{equation}
 Next, we shall show that $(u,v)$ is a weak solution of \eqref{P}.
 For this, it suffices to prove that  $(u,v)$ satisfies the conditions
(a)-(d) in Definition \ref{def1.1}.
 Let us first check the condition (a).
To do this, it needs to  establish some basic  estimates on
$u_\varepsilon$ and $v_\varepsilon$.


 \begin{lemma} \label{lem2.1.2}
 For all $\tau \in (0,T)$ and $\varepsilon \in (0,1)$, we have
\begin{itemize}
\item[(1)]
$$\iint_{\Omega_\tau}\frac{u_{\varepsilon t}^2}{a(u_\varepsilon)}\,dx\,dt
 +\int_{\Omega}|\nabla u_\varepsilon(x,\tau)|^2 dx\le C
$$
\item[(2)]
$$
\iint_{\Omega_\tau}\frac{v_{\varepsilon t}^2}{b(u_\varepsilon)}\,dx\,dt+
 \int_{\Omega}|\nabla v_\varepsilon(x,\tau)|^2 dx\le C.
$$
\end{itemize}
 Here $C$ are positive constants independent of $\varepsilon$.
 \end{lemma}

\begin{proof}
 Since the proof is exactly the same for (1) and (2),  we will show the
  validity of (1). Multiplying the first equation of \eqref{e2.1.3} by
  $u_{\varepsilon t}/a(u_\varepsilon)$ and integrating over $\Omega_\tau$
and noticing   $u_{\varepsilon t}=0$ on $\partial \Omega \times (0,T)$,
we obtain
\begin{align*}
&\iint_{\Omega_\tau}\frac{u_{\varepsilon t}^2}{a(u_\varepsilon)}\,dx\,dt\\
 &= \iint_{\Omega_\tau} (\Delta u_\varepsilon +\alpha v_\varepsilon)
 u_{\varepsilon t}\,dx\,dt\\
& =-\iint_{\Omega_\tau}  \nabla u_\varepsilon \nabla u_{\varepsilon t} \,dx\,dt
 +\iint_{\Omega_\tau} \alpha v_\varepsilon u_{\varepsilon t}\,dx\,dt\\
& =-\iint_{\Omega_\tau}\frac{\partial}{\partial t}
   \Big(\frac{|\nabla u_\varepsilon|^2}{2}\Big) \,dx\,dt
  + \iint_{\Omega_\tau} \Big[\alpha v_\varepsilon a(u_\varepsilon)^{1/2}\Big]
       \Big[\frac{u_{\varepsilon t}}{a(u_\varepsilon)^{1/2}}\Big]\,dx\,dt\\
& \leq \frac{1}{2}\int_{\Omega}|\nabla u_0|^2  dx+
   \iint_{\Omega_\tau} \Big[\alpha v_\varepsilon a(u_\varepsilon)^{1/2}\Big]
       \Big[\frac{u_{\varepsilon t}}{a(u_\varepsilon)^{1/2}}\Big]\,dx\,dt.
\end{align*}
By the inequality $a b \le \frac{1}{2}( a^2+b^2 )$, we have,
$$
 \iint_{\Omega_\tau}\frac{u_{\varepsilon t}^2}{a(u_\varepsilon)}dx\, dt
 \leq  \frac{1}{2}\int_{\Omega}|\nabla u_0|^2  dx+
  \frac{\alpha^2}{2} \iint_{\Omega_\tau}  v_\varepsilon^2
  a(u_\varepsilon) \,dx\,dt+\frac{1}{2}
  \iint_{\Omega_\tau}\frac{u_{\varepsilon t}^2}{a(u_\varepsilon)}dx\,  dt,
$$
i.e.
$$
 \iint_{\Omega_\tau}\frac{u_{\varepsilon t}^2}{a(u_\varepsilon)}\,dx\,dt
 \leq  \int_{\Omega}|\nabla u_0|^2  dx+\alpha^2
 \iint_{\Omega_\tau}v_\varepsilon^2 a(u_\varepsilon) \,dx\,dt,
$$
 and then, by Lemma \ref{lem2.1.1}, (1) is proved. This completes the proof.
\end{proof}

 From Lemma 2.1, \eqref{e2.1.4} and Lemma \ref{lem2.1.2}, one may derive that
\begin{equation}
(u_{\varepsilon}, v_\varepsilon) \rightharpoonup
(u,v)\quad \text{in }[H^{1}(\Omega_T)]^2,\quad \mbox{\rm as}~ \varepsilon
\to 0,
\label{e2.1.5}
\end{equation}
where $\rightharpoonup$ denotes the weak convergence, and
$$
0 \leq u,v \in L^{\infty}(\Omega_T)\cap L^2 (0,T;H^1_0(\Omega));\quad
u_t, v_t \in L^2(\Omega_T).
$$
Thus the condition (a) is satisfied. Next, let us check the
condition (b). For this, the following estimates are required.

\begin{lemma} \label{lem2.1.3}
For any $\theta \in (0,1)$, there exist  positive constants $C_1$
and $C_2$ independent of $\varepsilon$ such that
\begin{itemize}
\item[(1)]
$$ \iint_{\Omega_T}   \frac{a'(u_\varepsilon)
   |\nabla u_\varepsilon|^2}{a(u_\varepsilon)^{\theta}} \,dx\,dt\le C_1
$$ provided $a'(0)>0$
\item[(2)]
$$
\iint_{\Omega_T}
  \frac{b'(v_\varepsilon)|\nabla v_\varepsilon|^2}{b(v_\varepsilon)^{\theta}}
  \,dx\,dt\le C_2
$$
provided $b'(0)>0$.
\end{itemize}
  \end{lemma}

\begin{proof}
 Since the proof is exactly the same for
(1) and (2), we shall show the validity of (1).  Given that
$a'(0)>0$ and $\theta \in (0,1)$, we claim that for any $l>0$,
$1/a(s)^{\theta}$ is integrable on $[0,l]$. Indeed, since
$a'(s)>0$ for $s \ge 0$, we have $M=\min_{s \in
[0,l]}a'(s)>0,$ and hence for any $s \in (0, l]$, by mean value
theorem and noticing $a(0)=0$,  there exists $\xi_s \in [0,s]$ such
that $a(s)=a'(\xi_s)s \ge Ms$. Therefore, for $\theta \in (0,1)$, we
have
$$
\int_0^l \frac{1}{a(s)^\theta} ds \le \int_0^l \frac{1}{M^\theta
s^\theta}ds \le \frac{l^{1-\theta}}{M^\theta (1-\theta)}.
$$
This proves the above claim. Now multiplying the first equation of
\eqref{e2.1.3} by
 $a(u_\varepsilon)^{-\theta}$ and integrating $\Omega_T$ and noticing
$\frac{\partial u_\varepsilon}{\partial \nu} \le 0$ on $\partial
\Omega \times (0,T)$, where $\nu$ denotes the unit outward normal
to $\partial \Omega \times (0,T)$, we have
\begin{align*}
& \iint_{\Omega_T}\frac{u_{\varepsilon t}}{a(u_\varepsilon)^\theta}\,dx\,dt\\
 &=\int_{\Omega}\int^{u_\varepsilon(x,T)}_0  \frac{1}{a(s)^\theta} \,ds\,dx
 -\int_{\Omega}\int^{u_0(x)+\varepsilon}_0 \frac{1}{a(s)^\theta} \,ds\,dx
\\
&=\iint_{\Omega_T}a(u_{\varepsilon})^{1-\theta}(\Delta u_\varepsilon
  +\alpha v_\varepsilon )\,dx\,dt\\
&=\iint_{\Omega_T}\Big[\mathop{\rm div}( a(u_{\varepsilon})^{1-\theta}
  \nabla u_\varepsilon)  -(1-\theta)\frac{a'(u_\varepsilon)|\nabla
  u_\varepsilon|^2}{a(u_\varepsilon)^{\theta}}
  +\alpha v_\varepsilon a(u_{\varepsilon})^{1-\theta}\Big]\,dx\,dt\\
&=\int_0^T\int_{\partial\Omega}a(u_{\varepsilon})^{1-\theta}
\frac{\partial u_\varepsilon}{\partial \nu}\,d\sigma\, dt
  -(1-\theta)\iint_{\Omega_T}\frac{a'(u_\varepsilon)|\nabla
  u_\varepsilon|^2}{a(u_\varepsilon)^{\theta}}\,dx\,dt \\
&\quad +\alpha \iint_{\Omega_T}v_\varepsilon a(u_{\varepsilon})^{1-\theta}\,dx\,dt\\
& \le  -(1-\theta)\iint_{\Omega_T}   \frac{a'(u_\varepsilon)|
\nabla u_\varepsilon|^2}{a(u_\varepsilon)^{\theta}}\,dx\,dt
   +\alpha \iint_{\Omega_T}v_\varepsilon a(u_{\varepsilon})^{1-\theta}\,dx\,dt,
\end{align*}
and hence
\begin{align*}
&\iint_{\Omega_T} \frac{a'(u_\varepsilon)
 |\nabla u_\varepsilon|^2}{a(u_\varepsilon)^{\theta}} \,dx\,dt\\
&\le \frac{1}{1-\theta}\int_{\Omega}\int^{u_0(x)+\varepsilon}_0
   \frac{1}{a(s)^\theta} \,ds\,dx
   +\frac{\alpha}{1-\theta} \iint_{\Omega_T}v_\varepsilon
  a(u_{\varepsilon})^{1-\theta}\,dx\,dt,
\end{align*}
 and then, by Lemma \ref{lem2.1.1},  (1) is proved. This completes the proof.
\end{proof}
Denote
$$
\phi_a(s)=\int_0^s a(y) dy,\quad   \phi_b(s)=\int_0^s b(y) dy,\quad \forall
s \ge 0.
$$


\begin{lemma} \label{lem2.1.4}
As $\varepsilon \to 0$, we have
\begin{itemize}
\item[(1)] $\iint_{\Omega_T}|\nabla \phi_a(u_\varepsilon) -\nabla
\phi_a(u)|^2 \,dx\,dt \to 0$;

\item[(2)] $\iint_{\Omega_T}|\nabla \phi_b(v_\varepsilon)
 -\nabla \phi_b(v)|^2 \,dx\,dt \to 0$;

\item[(3)] $\iint_{Q_{u_{\varepsilon}}(c)}|\nabla  u_\varepsilon -\nabla  u
|^2 \,dx\,dt \to 0$;

\item[(4)] $\iint_{Q_{v_\varepsilon}(c)}|\nabla  v_\varepsilon  -\nabla
v |^2 \,dx\,dt \to 0$;
\end{itemize}
where $Q_{u_\varepsilon}(c)=\{(x,t) \in \Omega_T;
u_\varepsilon \ge c>0\}$ and
$Q_{v_\varepsilon}(c)=\{(x,t) \in \Omega_T; v_\varepsilon \ge c>0\}$.
\end{lemma}

\begin{proof} Let us first prove (1).
 Multiplying the first equation of \eqref{e2.1.3} by
$[\phi_a(u_\varepsilon)-\phi_a(u)-\phi_a(\varepsilon)]$
 and integrating over $\Omega_T$, we obtain
\begin{align*}
0&=\iint_{\Omega_T}
u_{\varepsilon t} [\phi_a(u_\varepsilon)-\phi_a(u)-\phi_a(\varepsilon)]\,dx\,dt \\
&\quad +\iint_{\Omega_T}
a(u_\varepsilon)\nabla u_\varepsilon \nabla [\phi_a(u_\varepsilon)-\phi_a(u)
 -\phi_a(\varepsilon)]\,dx\,dt \\
&\quad +\iint_{\Omega_T}
a'(u_\varepsilon)|\nabla u_\varepsilon|^2[\phi_a(u_\varepsilon)-\phi_a(u)
 -\phi_a(\varepsilon)]\,dx\,dt \\
&\quad -\alpha \iint_{\Omega_T}
a(u_\varepsilon)v_\varepsilon[\phi_a(u_\varepsilon)-\phi_a(u)
 -\phi_a(\varepsilon)] \,dx\,dt\\
&=\iint_{\Omega_T}
u_{\varepsilon t} [\phi_a(u_\varepsilon)-\phi_a(u)
-\phi_a(\varepsilon)]\,dx\,dt \\
&\quad +\iint_{\Omega_T} \nabla \phi_a(u_\varepsilon) \nabla [\phi_a(u_\varepsilon)
 -\phi_a(u)]\,dx\,dt \\
&\quad +\iint_{\Omega_T} a'(u_\varepsilon)|\nabla u_\varepsilon|^2[\phi_a
  (u_\varepsilon)-\phi_a(u)-\phi_a(\varepsilon)]\,dx\,dt \\
&\quad -\alpha \iint_{\Omega_T} a(u_\varepsilon)v_\varepsilon[\phi_a(u_\varepsilon)
  -\phi_a(u)-\phi_a(\varepsilon)] \,dx\,dt\,.
\end{align*}
Note that $ u_\varepsilon \ge u$, $\phi_a'(s)\ge 0$, $a'(s) \ge 0$ for all $s \ge 0$,
so the above expression is greater than or equal to
\begin{align*}
&\iint_{\Omega_T} u_{\varepsilon t} [\phi_a(u_\varepsilon)
  -\phi_a(u)-\phi_a(\varepsilon)]\,dx\,dt
+\iint_{\Omega_T} \nabla \phi_a(u_\varepsilon)
  \nabla [\phi_a(u_\varepsilon)-\phi_a(u)]\,dx\,dt \\
&-\phi_a(\varepsilon)\iint_{\Omega_T}a'(u_\varepsilon)
  |\nabla u_\varepsilon|^2\,dx\,dt
-\alpha \iint_{\Omega_T} a(u_\varepsilon)v_\varepsilon
  [\phi_a(u_\varepsilon)-\phi_a(u)-\phi_a(\varepsilon)] \,dx\,dt\\
&=\iint_{\Omega_T}u_{\varepsilon t} [\phi_a(u_\varepsilon)
  -\phi_a(u)-\phi_a(\varepsilon)]\,dx\,dt
  +\iint_{\Omega_T}|\nabla [\phi_a(u_\varepsilon)-\phi_a(u)]|^2 \,dx\,dt \\
&\quad +\iint_{\Omega_T}  \nabla \phi_a(u ) \nabla [\phi_a(u_\varepsilon)
  -\phi_a(u)]\,dx\,dt
  -\phi_a(\varepsilon)\iint_{\Omega_T}a'(u_\varepsilon)
  |\nabla u_\varepsilon|^2\,dx\,dt \\
&\quad -\alpha \iint_{\Omega_T}a(u_\varepsilon)v_\varepsilon
  [\phi_a(u_\varepsilon)-\phi_a(u)-\phi_a(\varepsilon)] \,dx\,dt.
 \end{align*}
Using \eqref{e2.1.4}, \eqref{e2.1.5}, Lemma \ref{lem2.1.1} and Lemma
\ref{lem2.1.2}
and noticing $\phi_a(0)=0$, we have
\begin{align*}
&\iint_{\Omega_T}|\nabla [\phi_a(u_\varepsilon)-\phi_a(u)]|^2 \,dx\,dt  \\
&\le -\iint_{\Omega_T}\! u_{\varepsilon t} [\phi_a(u_\varepsilon)-\phi_a(u)
  -\phi_a(\varepsilon)]\,dx\,dt
  -\iint_{\Omega_T}\! \nabla \phi_a(u)\nabla [\phi_a(u_\varepsilon)-\phi_a(u)]
  \,dx\,dt \\
&\quad +\phi_a(\varepsilon)\iint_{\Omega_T}a'(u_\varepsilon)
 |\nabla u_\varepsilon|^2\,dx\,dt
 +\alpha\iint_{\Omega_T}
a(u_\varepsilon)v_\varepsilon[\phi_a(u_\varepsilon)-\phi_a(u)
  -\phi_a(\varepsilon)] \,dx\,dt\\
& \to   0\quad (\text{as }\varepsilon \to 0).
\end{align*}
Thus (1) is proved. Similarly (2) can be proved.
We shall show (3). Using the equality
$$
a(u_\varepsilon)(\nabla u_\varepsilon-\nabla u) =[\nabla
\phi_a(u_\varepsilon)-\nabla \phi_a(u)]
-[a(u_\varepsilon)-a(u)]\nabla u,
$$
the inequality $(a+b)^2 \le 2(a^2+b^2)$, \eqref{e2.1.4} and (1), we obtain
\begin{align*}
&\iint_{\Omega_T}a(u_\varepsilon)^2|\nabla u_\varepsilon-\nabla u|^2 \,dx\,dt \\
&\le  2\iint_{\Omega_T}|\nabla \phi_a(u_\varepsilon)-\nabla \phi_a(u)|^2 \,dx\,dt
 +2\iint_{\Omega_T}|a(u_\varepsilon)-a(u)|^2|\nabla u|^2 \,dx\,dt\\
&\to 0\quad (\text{as }\varepsilon \to 0),
\end{align*}
and then, by $a'(s) \ge 0$ for all $s \ge 0$, so that (3) is proved.
Similarly (4) can be obtained. The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.1.5}
As $\varepsilon \to 0$, we have
\begin{itemize}
\item[(1)] $\iint_{\Omega_T} |a'(u_\varepsilon)|\nabla
u_\varepsilon|^2-a'(u)|\nabla  u |^2| \,dx\,dt \to 0$;

\item[(2)] $\iint_{\Omega_T} |b'(v_\varepsilon)|\nabla  v_\varepsilon
|^2-b'(v)|\nabla  v |^2| \,dx\,dt \to 0$.
\end{itemize}
\end{lemma}

\begin{proof} Since the proof is exactly the same for
(1) and (2), we will show the validity of (1). For $\rho>0$, let
$\chi^{(\rho)}_\varepsilon $ and $\chi^{(\rho)}$ be the
characteristic functions of $\{(x,t) \in \Omega_T; u_\varepsilon \le
\rho\}$ and $\{(x,t) \in \Omega_T; u \le \rho\}$, respectively. Then
\begin{align*}
 &\iint_{\Omega_T}
 |a'(u_\varepsilon)|\nabla  u_\varepsilon|^2  -a'(u)|\nabla  u |^2 | \,dx\,dt \\
 &\le  \iint_{\Omega_T} a'(u_\varepsilon)||\nabla  u_\varepsilon|^2  - |\nabla  u |^2| \,dx\,dt
  +\iint_{\Omega_T}|a'(u_\varepsilon)-a'(u)||\nabla  u |^2  \,dx\,dt\\
 &\leq  \iint_{\Omega_T} \chi^{(\rho)}_\varepsilon a'(u_\varepsilon)
  |\nabla  u_\varepsilon|^2 \,dx\,dt
 +\iint_{\Omega_T} \chi^{(\rho)}_\varepsilon a'(u_\varepsilon)|\nabla  u |^2
  \,dx\,dt \\
&\quad +\iint_{\Omega_T} (1-\chi^{(\rho)}_\varepsilon ) a'(u_\varepsilon)
 ||\nabla  u_\varepsilon|^2  - |\nabla  u |^2| \,dx\,dt
 +\iint_{\Omega_T}|a'(u_\varepsilon)-a'(u)||\nabla  u |^2  \,dx\,dt\\
& = I_1+I_2+I_3+I_4.
\end{align*}
Clearly, $I_4 \to 0$ as $\varepsilon \to 0$. Since
$u_\varepsilon \ge u$ a.e. in $\Omega$, $\chi_\varepsilon^{(\rho)}
\le \chi^{(\rho)}$ a.e. in $\Omega$. Therefore,
$$
 I_2   \le C\iint_{\Omega_T}\chi^{(\rho)} |\nabla  u |^2 \,dx\,dt \to 0\quad (\rho \to 0).
$$
Next, we estimate $I_1$. If $a'(0)=0$, then
$$
I_1 \le \max_{s \in[0,\rho]}a'(s) \iint_{\Omega_T} |\nabla
u_\varepsilon |^2\,dx\,dt \le C\max_{s \in[0,\rho]}a'(s)
\to 0\quad (\rho \to 0).
$$
If $a'(0)>0$, taking $\theta=1/2$ in Lemma \ref{lem2.1.3} and noticing
 $a'(s)>0$ for $s \ge 0$ and $a(0)=0$, we have
\begin{align*}
 I_1 &=\iint_{\Omega_T} \chi^{(\rho)}_\varepsilon a(u_\varepsilon)^{1/2}
 \frac{a'(u_\varepsilon) |\nabla  u_\varepsilon|^2}{a(u_\varepsilon)^{1/2}}
\,dx\,dt\\
&\le   a(\rho)^{1/2} \iint_{\Omega_T}
 \frac{ a'(u_\varepsilon)|\nabla  u_\varepsilon|^2}{a(u_\varepsilon)^{1/2}}
\,dx\,dt \\
&\le  C a(\rho)^{1/2}
  \to  0\quad  (\text{as }\rho \to 0).
\end{align*}
In any case, we obtain
$$
 I_1  \to 0\quad \text{(as $\rho \to 0$), uniformly in }\varepsilon.
$$
Hence, for any $\delta>0$, we can find a $\rho>0$ sufficiently small
such that
 $I_1+I_2<\delta/2$. For fixed $\rho>0$,  it follows from
Lemma \ref{lem2.1.4} that
$$
 I_3  \le  C\iint_{\Omega_T}(1-\chi^{(\rho)}_\varepsilon )
 ||\nabla  u_\varepsilon|^2  - |\nabla  u |^2| \,dx\,dt
 \to 0\quad (\text{as }\varepsilon \to 0).
$$
Therefore, there exists $\varepsilon_1 \in (0, 1)$ such that
$I_3<\delta/2$   as $\varepsilon<\varepsilon_1$. Consequently, we
obtain
$$
I_1+I_2+I_3<\delta,\quad \forall \varepsilon <\varepsilon_1.
$$
Thus (1) holds. The proof of Lemma \ref{lem2.1.5} is complete.
\end{proof}

 From Lemma \ref{lem2.1.5} it is easy to check that $(u,v)$ satisfies the
condition (b) in Definition \ref{def1.1}.
Finally, we shall show that $(u,v)$ satisfies the condition (c). The
proof can be completed by combining the following two lemmas.

\begin{lemma}
(1) For any $\rho>0$ sufficiently small, there exist positive
constants $c_1=c_1(\rho)$ and $c_2=c_2(\rho)$ such that
\begin{gather*}
u \ge c_1 \quad \text{a.e. in }\Omega^\rho(u_0) \times (0,T),\\
 v \ge c_2 \quad \text{a.e.  in }\Omega^\rho(v_0) \times (0,T);
\end{gather*}
(2) $\mathop{\rm supp}u(t) \supseteq \mathop{\rm supp}u_0$,
$\mathop{\rm supp}v(t) \supseteq \mathop{\rm supp}v_0$,
a.e. in $(0,T)$.
\end{lemma}

\begin{proof}  Note that, in view of the definition of support
of a nonnegative function, the conclusion (1) implies (2). Since the
proof is exactly the same for the first conclusion and the second
conclusion of (1), we will show the validity of the former. It is
easy to see that there exists a positive constant $c=c(\rho)$ such
that
$$
u_0\ge c>0~\text{in }\Omega^\rho (u_0).
$$
Denote by $\lambda_{\rho}$ the first eigenvalue of $-\Delta$ in
$\Omega^{\rho/2}(u_0)$  with the homogeneous Dirichlet boundary
condition  and $\phi_{\rho}$ the associated
 eigenfunction with  $\max_{\overline {\Omega^{\rho/2}(u_0)}}\phi_{\rho}=c$.
Let
$$
\underline u= e^{-\kappa t}\phi_{\rho},\quad (x,t) \in
\Omega^{\rho/2}(u_0)\times (0,T),
$$
where $\kappa=a(\mathop{\rm
sup}_{0<\varepsilon<1}|u_\varepsilon|_\infty)\lambda_{\rho}+1$.Then
$$
\underline u_t-a(u_\varepsilon)\Delta \underline u =e^{-\kappa
t}\phi_{\rho}(-\kappa+a(u_\varepsilon)\lambda_{\rho}) \le 0
~\mbox{in}~\Omega^{\rho/2}(u_0)\times (0,T).
$$
Hence, $u_\varepsilon$ and $\underline u$ are the classical
sup-solution and sub-solution of the equation
$$
w_t-a(u_\varepsilon) \Delta w = 0\quad \text{in }
\Omega^{\rho/2}(u_0) \times (0, T).
$$
On the other hand, obviously we have
$$
u_\varepsilon \ge \underline u\quad \text{on }\partial
\Omega^{\rho/2}(u_0) \times (0,T), \quad  u_\varepsilon(x,0) \ge
\underline u(x,0)\quad \text{in }\Omega^{\rho/2}(u_0).
$$
By the comparison principle, we obtain
$$
u_\varepsilon \ge e^{-\kappa t}\phi_{\rho} \ge e^{-\kappa
T}\min_{\overline {\Omega^\rho(u_0)}}\phi_{\rho}\equiv
c_1(\rho)>0 \quad \text{in }\Omega^\rho(u_0) \times (0,T).
$$
Passing to the limit as $\varepsilon \to 0$, we have
$$
u \ge  c_1(\rho)>0\quad \hbox{a.e. in }\Omega^{\rho}(u_0) \times
(0,T).
$$
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.1.7}
$\mathop{\rm supp}u(t) \subseteq \mathop{\rm supp}u_0$,
$\mathop{\rm supp}v(t) \subseteq \mathop{\rm supp}v_0$,
 a.e. in $(0,T)$.
\end{lemma}

\begin{proof}  Since the proof is exactly the same for the
former and the latter, we will show the validity of the former.
Without loss of generality, we may assume $\mathop{\rm
supp}u_0\subsetneq \overline \Omega$. For any $\delta >0$, let $
\psi (x)=\psi_{\delta }(x)=\inf\big\{\frac{d (x)}{\delta }, 1
\big\}$, where $d (x)=\mathop{\rm dist}(x, \partial \Omega\cup
\mathop{\rm supp}u_0)$. It is well known that the distance functions
$d(x)$ is Lipschitz with the constant 1, and hence  it follows from
Rademacher's theorem  \cite[pp. 49-51]{z1} that $d(x)$ is
differentiable  almost everywhere. Multiplying the first equation of
\eqref{e2.1.3} by $\varphi=\frac{\psi}{a(u_\varepsilon)}$ and
integrating over $\Omega_t$, we have
$$
\int_0^t\int_{\Omega} \left( \frac{u_{\varepsilon
t} \psi}{a(u_\varepsilon)} +  \nabla u_\varepsilon \nabla \psi   -
\alpha  v_\varepsilon \psi \right) dx\,d\tau=0.
$$
By Lemma \ref{lem2.1.1} and Lemma \ref{lem2.1.2}, there exists a positive constant $C$
independent of $\varepsilon$ such that
$$
\int_0^t\int_{\Omega}  \frac{u_{\varepsilon t}
\psi}{a(u_\varepsilon)}dx\,d\tau  \le C,
$$
hence
$$
\int_{\Omega}\Big(\int_\varepsilon^{u_\varepsilon(x,t)}\frac{1}{a(s)}
ds-\int_\varepsilon^{u_0(x)+\varepsilon}\frac{1}{a(s)} ds\Big)
\psi(x) dx \le C.
$$
 Noticing $\psi  u_0 =0$ in $\Omega$, we have
$$
\int_{\Omega}\Big(\int_\varepsilon^{u_0(x)+\varepsilon}\frac{1}{a(s)}
ds\Big) \psi(x) dx=0;
$$
therefore,
$$
\int_{\Omega}\Big(\int_\varepsilon^{u_\varepsilon(x,t)}\frac{1}{a(s)}ds
\Big) \psi(x) dx \leq C.
$$
 By virtue of $u_\varepsilon \ge u $ a.e. in $\Omega_T$, we obtain
 $$
\int_{\{x\in\Omega;\psi(x)=1\}}\int_\varepsilon^{u(x,t)}\frac{1}{a(s)}ds
   dx \leq C.
$$
Hence for any $\sigma \in (0, 1)$ and
$\varepsilon \in (0,\sigma)$ and a.e. $ t \in (0,T)$, we have
 $$
\mu (\{x\in  \{x \in \Omega; \psi=1\} ; u(x,t)>\sigma\})
  \int_\varepsilon^{\sigma} \frac{1}{a(s) }ds   \leq C;
$$
therefore,
$$
 \mu (\{x\in  \{x \in \Omega; \psi=1\} ; u(x,t)>\sigma\})
 \le C\big[\int_\varepsilon^{\sigma} \frac{1}{a(s) }ds\big]^{-1},
$$
where $C$ is a positive constant independent of $\varepsilon$. We
claim that
$$
\lim_{\varepsilon \to 0}\int_\varepsilon^{\sigma}
\frac{1}{a(s) }ds=+\infty.
$$
 Indeed, by the mean value theorem and noticing $a(0)=0$, we derive that
  for any $s \in [0, \sigma]$ there exists $\xi_s \in [0,s]$
 such that $a(s)=a'(\xi_s) s \le Ms$, where
$M=\max_{s \in[0,\sigma]}a'(s) > 0$. Thus
 $$
 \int_\varepsilon^{\sigma}\frac{1}{a(s)}ds
 \ge  \int_\varepsilon^{\sigma}\frac{1}{M s}ds
 =\frac{1}{M}[\ln(\sigma)-\ln(\varepsilon)],
 $$
then passing to the limit as $\varepsilon \to 0$, we prove
the above claim and obtain
$$
\mu (\{x\in \{x \in \Omega;\psi=1\};
u(x,t)>\sigma\})=0\quad\text{a.e.~in}~(0,T),
$$
so that, since $\sigma \in (0,1)$ is arbitrary, we obtain
  $$
 \mu (\{x\in  \{x \in \Omega;\psi=1\} ; u(x,t) >0\})=0\quad\text{a.e.
 in}~(0,T).
$$
Since $\delta >0$ is arbitrary, we conclude that
$$
u(x,t)=0 \quad\text{ a.e. in }(\Omega\setminus \mathop{\rm supp}u_0)
\times (0,T).
$$
This completes the proof of Lemma \ref{lem2.1.7}. Thus the proof of the existence
is complete.
\end{proof}

 \subsection{Proof of uniqueness}
Let $(u_2, v_2)$ and $(u_1, v_1) $ be two weak solutions of
\eqref{P},
 and $E=\mathop{\rm supp}u_0 \cap \mathop{\rm supp}v_0$. It suffices to prove
 that for any $T>0$, $u_2=u_1, v_2=v_1$ a.e. in $\Omega_T$.
\smallskip

\noindent{\bf First Case: $\mu(E)=0$.}
 Without loss of generality, we may
assume that $\mathop{\rm supp}u_0\neq \emptyset$.
 From the definition of weak solutions it follows that $v_2=v_1=0$ a.e.
on $\mathop{\rm supp}u_0$.
Denote by $\lambda_{\rho}$ the first eigenvalue of $-\Delta $ in
$\Omega^\rho (u_0)$ with homogeneous
  Dirichlet  boundary condition and $\phi_{\rho} (x)$ the associated
eigenfunction. Substituting
 $$
 \varphi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta ((u_1-u_2)_+)}{a(u_1)},
 \quad \psi=0,
 $$
  and
 $$
 \varphi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_2)},
\quad \psi=0,
 $$
 in the definition of weak solutions, respectively, where
$\mathop{\rm sign}_\delta(z)=\mathop{\rm sign}(z)\inf\big\{\frac{|z|}{\delta},
1\big\}$
for $\delta>0$, we have for any $t \in (0, T)$
\begin{align*}
&\int_0^t\int_{\Omega} \Big[
\frac{u_{1t}\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_1)}\\
&+\nabla u_1 \nabla (u_1-u_2)_+ \phi_{\rho}
\mathop{\rm sign}{}_\delta'((u_1-u_2)_+)
+\nabla u_1\nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\Big]
dx\,d\tau=0,
\\
&\int_0^t\int_{\Omega} \Big[
\frac{u_{2t}\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_2)}\\
&+\nabla u_2 \nabla (u_1-u_2)_+ \phi_{\rho}
\mathop{\rm sign}{}_\delta'((u_1-u_2)_+)
+\nabla u_2 \nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\Big]
 dx\,d\tau=0,
\end{align*}
and hence
\begin{align*}
&\int_0^t\int_{\Omega} \Big[(f_a(u_1)-f_a(u_2))_t
 \phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\\
&+|\nabla (u_1-u_2)_+|^2\phi_{\rho}\mathop{\rm sign}{}_\delta'((u_1-u_2)_+)\\
&+\nabla(u_1-u_2)\nabla \phi_{\rho}
\mathop{\rm sign}{}_\delta((u_1-u_2)_+) \Big] dx\,d\tau=0,
 \end{align*}
where
$$
f_a(s)=\int_{c_1}^s \frac{1}{a(y)} dy, \quad \quad \forall s >0,
$$
where $c_1$ is the same as that of Definition \ref{def1.1} (note that $c_1$
corresponding to $u_1$ may be different from that corresponding to
$u_2$. Here $c_1$ is minimal between them). Noticing $\mbox{\rm
sgn}_\delta'(z) \ge 0$, we obtain
\begin{align*}
&\int_0^t\int_{\Omega} \Big[(f_a(u_1)-f_a(u_2))_t
\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+) \\
&+\nabla(u_1-u_2)\nabla
\phi_{\rho} \mathop{\rm sign}{}_\delta((u_1-u_2)_+) \Big] dx\,d\tau\le 0,
\end{align*}
Passing to the limit as $\delta \to 0$, we have
 $$
\int_0^t\int_{\Omega} \Big[(f_a(u_1)-f_a(u_2))_t
\phi_{\rho}\mathop{\rm sign} ((u_1-u_2)_+) +\nabla(u_1-u_2)_+\nabla
\phi_{\rho}   \Big] dx\,d\tau\le 0.
$$
Integrating by parts for the second term of the above integral, we
obtain
 $$
\int_0^t\int_{\Omega} \Big[(f_a(u_1)-f_a(u_2))_t
\phi_{\rho}\mathop{\rm sign} ((u_1-u_2)_+)
 +\lambda_{\rho} (u_1-u_2)_+ \phi_{\rho} \Big] dx\,d\tau \le 0,
$$
and then it follows from $\lambda_{\rho}>0$ and $ \phi_{\rho} \ge 0$
that
$$
\int_0^t\int_{\Omega}  (f_a(u_1)-f_a(u_2))_t
\phi_{\rho}\mathop{\rm sign} ((u_1-u_2)_+) dx\,d\tau\le 0.
$$
Since $\mathop{\rm sign} ((u_1-u_2)_+)=\mathop{\rm sign} (f_a(u_1)-f_a(u_2))_+$
a.e. in $\Omega^\rho(u_0)\times (0,T)$, we have
 $$
\int_{\Omega} (f_a(u_1)-f_a(u_2))_+ (x,t) \phi_{\rho}(x)  dx  \le
0,
$$
which implies $(f_a(u_1)-f_a(u_2))_+=0$ a.e. in
$\Omega^\rho(u_0)\times (0,T) $, and hence $u_1 \le u_2$ a.e. in
$\Omega^\rho(u_0)\times (0,T)$, and therefore  $u_1 \le u_2$ a.e. in
$\mathop{\rm supp}u_0 \times (0,T)$. By the condition (c) in
Definition \ref{def1.1}, we derive that $u_1 \le u_2$ a.e. in $\Omega_T$.
Similarly, $u_1 \ge u_2$ a.e. in $\Omega_T$. Thus, $u_1=u_2$ a.e. in
$\Omega_T$.

The same reasoning as those given above  shows that
 $v_1 = v_2$ a.e. in $\Omega_T$. This prove the first case.
\smallskip

\noindent{\bf General Case: $\mu(E)>0$.}
 Denote by $\lambda_{\rho}$ the
first eigenvalue of $-\Delta$ in $E_\rho$ with the homogeneous
Dirichlet boundary condition and $ \phi_{\rho}(x)$ the
 associated eigenfunction,  where  $E_\rho=\{x \in E; \mathop{\rm dist}(x, \partial E)>\rho>0\}$.
Substituting
$$
 \varphi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_1)},
 \quad \psi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)}{b(v_1)},
 $$
 and
  $$
 \varphi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_2)},
 \quad \psi=\frac{\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)}{b(v_2)},
 $$
 in the definition of weak solutions, respectively, we have
\begin{align*}
&\int_0^t\int_{\Omega} \Big[
\frac{u_{1t}\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_1)}
+\nabla u_1 \nabla (u_1-u_2)_+ \phi_{\rho}\mathop{\rm sign}{}_\delta'((u_1-u_2)_+)\\
&+\nabla u_1
\nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)
-\alpha v_1 \phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\Big] dx\,d\tau\\
&+\int_0^t\int_{\Omega}\Big[ \frac{v_{1t}\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)}{a(v_1)}
 +\nabla v_1 \nabla (v_1-v_2)_+\phi_{\rho}\mathop{\rm sign}{}_\delta'((v_1-v_2)_+)\\
& +\nabla v_1 \nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)
 -\beta   u_1 \phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)\Big] dx\,d\tau
=0,
 \end{align*}
\begin{align*}
&\int_0^t\int_{\Omega}
 \Big[\frac{u_{2t}\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)}{a(u_2)}
 +\nabla u_2 \nabla (u_1-u_2)_+ \phi_{\rho}\mathop{\rm sign}{}_\delta'((u_1-u_2)_+)\\
& + \nabla u_2 \nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)
-\alpha v_2\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\Big] dx\,d\tau\\
& +\int_0^t\int_{\Omega} \Big[\frac{v_{2t}\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)}{a(v_2)}
+\nabla v_2 \nabla (v_1-v_2)_+\phi_{\rho}\mathop{\rm sign}{}_\delta'((v_1-v_2)_+)\\
& +\nabla v_2\nabla\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)-\beta u_2
\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)\Big] dx\,d\tau =0,
 \end{align*}
and hence
\begin{align*}
&\int_0^t\int_{\Omega}
\Big[(f_a(u_1)-f_a(u_2))_t\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\\
&+|\nabla (u_1-u_2)_+|^2\phi_{\rho} \mathop{\rm sign}{}_\delta'((u_1-u_2)_+)
 +\nabla (u_1-u_2)  \nabla\phi_{\rho}  \mathop{\rm sign}{}_\delta
((u_1-u_2)_+) \\
& -\alpha  ( v_1-v_2)\phi_{\rho} \mathop{\rm sign}{}_\delta
((u_1-u_2)_+) \Big]  dx\,d\tau\\
& +\int_0^t\int_{\Omega}\Big[(f_b(v_1)-f_b(v_2))_t
  \phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+) \\
& +|\nabla (v_1-v_2)_+|^2\phi_{\rho} \mathop{\rm sign}{}_\delta'((u_1-u_2)_+)
  +\nabla (v_1-v_2)  \nabla \phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+) \\
& -\beta( u_1-u_2)
\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+) \Big] dx\,d\tau =0,
\end{align*}
where $f_a$ is the same as before and $f_b$ is defined by
$$
f_b(s)=\int_{c_2}^s\frac{1}{b(y)} dy,\quad  \forall s>0,
$$
and $c_2$ is the same as that of Definition \ref{def1.1}. Noticing
$\mathop{\rm sign}_\delta'(z) \ge 0$, we obtain from the above
equality
\begin{align*}
&\int_0^t\int_{\Omega}
\Big[(f_a(u_1)-f_a(u_2))_t\phi_{\rho}\mathop{\rm sign}{}_\delta((u_1-u_2)_+)\\
&+\nabla (u_1-u_2)  \nabla\phi_{\rho}  \mathop{\rm sign}{}_\delta ((u_1-u_2)_+)
  -\alpha(v_1-v_2)\phi_{\rho} \mathop{\rm sign}{}_\delta ((u_1-u_2)_+) \Big]
dx\,d\tau\\
& +\int_0^t\int_{\Omega} \Big[(f_b(v_1)-f_b(v_2))_t
 \phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+) \\
& +\nabla (v_1-v_2)  \nabla \phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+)
 - \beta  ( u_1-u_2)
\phi_{\rho}\mathop{\rm sign}{}_\delta((v_1-v_2)_+) \Big] dx\,d\tau \le 0.
 \end{align*}
Passing to the limit as $\delta \to 0$, we have
\begin{align*}
&\int_0^t\int_{\Omega}
\Big[(f_a(u_1)-f_a(u_2))_t\phi_{\rho}\mathop{\rm sign} ((u_1-u_2)_+)
 +\nabla (u_1-u_2)_+  \nabla\phi_{\rho} \\
& - \alpha  ( v_1-v_2)\phi_{\rho} \mathop{\rm sign}  ((u_1-u_2)_+) \Big]  dx\,d\tau\\
& +\int_0^t\int_{\Omega} \Big[(f_b(v_1)-f_b(v_2))_t \phi_{\rho}\mathop{\rm sign}((v_1-v_2)_+)
+\nabla (v_1-v_2)_+  \nabla \phi_{\rho} \\
& - \beta  ( u_1-u_2) \phi_{\rho}\mathop{\rm sign} ((v_1-v_2)_+) \Big] dx\,d\tau \le 0,
 \end{align*}
and hence
\begin{align*}
&\int_0^t\int_{\Omega}
\Big[(f_a(u_1)-f_a(u_2))_t\phi_{\rho}\mathop{\rm sign} ((u_1-u_2)_+)
+\lambda_{\rho}   (u_1-u_2)_+  \phi_{\rho} \\
& -\alpha  ( v_1-v_2)\phi_{\rho} \mathop{\rm sign} ((u_1-u_2)_+) \Big]  dx\,d\tau\\
& +\int_0^t\int_{\Omega} \Big[(f_b(v_1)-f_b(v_2))_t \phi_{\rho}\mathop{\rm sign} ((v_1-v_2)_+)
+\lambda_{\rho}  (v_1-v_2)_+ \phi_{\rho}\\
& -\beta  ( u_1-u_2) \phi_{\rho}\mathop{\rm sign}  ((v_1-v_2)_+) \Big] dx\,d\tau \le 0.
 \end{align*}
 This implies
\begin{align*}
&\int_{\Omega} [(f_a(u_1)-f_a(u_2))_+(x,t)+(f_b(v_1)-f_b(v_2))_+(x,t) ]\phi_{\rho}(x) dx\\
&\le C\int_0^t\int_{\Omega}(| v_1-v_2| +| u_1-u_2|) \phi_{\rho} dx\,d\tau.
 \end{align*}
By the same arguments as the above, we obtain
\begin{align*}
&\int_{\Omega}  [(f_a(u_2)-f_a(u_1))_+(x,t)+(f_b(v_2)-f_b(v_1))_+(x,t)] \phi_{\rho}(x) dx \\
& \le C\int_0^t\int_{\Omega}(| v_1-v_2|+| u_1-u_2|) \phi_{\rho} dx\,d\tau.
 \end{align*}
Thus we have
\begin{equation}
\begin{aligned}
&\int_{\Omega}   [|(f_a(u_2)-f_a(u_1))(x,t)| +|(f_b(v_2)-f_b(v_1))(x,t)|] \phi_{\rho}(x) dx \\
&\le C\int_0^t\int_{\Omega} (| v_1-v_2|+| u_1-u_2|) \phi_{\rho} dx\,d\tau.
\end{aligned}\label{e2.2.1}
\end{equation}
On the other hand, it follows from $a'(s), b'(s) \ge 0$ for
$s \ge 0$ that
\begin{gather*}
|u_1-u_2| \le a(|u_1+u_2|_{L^\infty(\Omega_T)})|f_a(u_2)-f_a(u_1)|
\quad\text{a.e. in }E_\rho \times (0,T),\\
| v_1-v_2|\le
b(|v_1+v_2|_{L^\infty(\Omega_T)})|f_b(v_2)-f_b(v_1)| \quad\text{a.e.
in }E_\rho \times (0,T).
\end{gather*}
Combining this with \eqref{e2.2.1}, we have
\begin{align*}
&\int_{\Omega}   [|(f_a(u_2)-f_a(u_1))(x,t)|
+|(f_b(v_2)-f_b(v_1))(x,t)|] \phi_{\rho}(x) dx \\
&\le C\int_0^t\int_{\Omega}|f_a(u_2)-f_a(u_1)|
+|f_b(v_2)-f_b(v_1)|)\phi_{\rho} dx\,d\tau,
\end{align*}
 and then, by Gronwall's theorem, we obtain
 $$
 |f_a(u_2)-f_a(u_1)| +|f_b(v_2)-f_b(v_1)|=0\quad
\text{a.e. in }E_\rho\times (0,T).
 $$
This shows that
 $$
 u_2=u_1,\quad v_2=v_1,\quad\text{a.e. in}~E_\rho \times (0,T),
 $$
 and hence
 $ u_2=u_1$, $v_2=v_1$, a.e. in $E \times (0,T)$.
 Similar to the proof of the first case, it is not difficult to prove that
 \begin{gather*}
  u_2=u_1\quad\text{a.e. in } (\mathop{\rm supp}u_0-E)\times (0,T),\\
  v_2=v_1\quad\text{a.e. in } (\mathop{\rm supp}v_0-E)\times (0,T).
 \end{gather*}
Combining the above results, we obtain
 \begin{gather*}
 u_2=u_1\quad\text{a.e. in } \mathop{\rm supp}u_0\times (0,T),\\
 v_2=v_1\quad\text{a.e. in } \mathop{\rm supp}v_0\times (0,T).
 \end{gather*}
 This proves the general case  and ends the proof of uniqueness.
%\end{proof}

\section{Proof of  Theorem 1.1 \ref{thm1.2}}

First, we claim that the following inequalities hold:
\begin{gather}
a(u)u^2+a(v)v^2 \ge ( a(u) +  a(v) )uv, \label{e3.1}\\
\rho_2 s^{\sigma_2} \ge a(s) \ge   \rho_1 s^{\sigma_2}\quad
\text{for }s \ge 0, \label{e3.2}\\
\int_0^s a(y)^{1/2} dy \ge \frac{2}{2+\sigma_2} s a(s)^{1/2}\quad \text{for }s
\ge 0, \label{e3.3}\\
\int_0^s a(y)^{1/2} dy \ge \frac{2\rho_1^{1/2}}{2+\sigma_2}
s^{1+\sigma_2/2}\quad \text{for }s \ge 0. \label{e3.4}
\end{gather}
We shall prove \eqref{e3.1}. It follows from $a'(s) \ge 0$ for all
$s \ge 0$ that
$$
[sa(s)]'=a(s)+sa'(s) \ge 0\quad \text{for all } s \ge 0.
$$
This shows that $[ua(u)-va(v)][u-v] \ge 0,$ which implies \eqref{e3.1}.
Let us turn to the proof of \eqref{e3.2}. By virtue of \eqref{e1.1},
it suffices to show that $\big[\frac{a(s)}{s^{\sigma_2}}\big]' \le 0$
for all $s>0$. By \eqref{e1.2}, we immediately obtain
$$
\big[\frac{a(s)}{s^{\sigma_2}}\big]'
=\frac{sa'(s)-\sigma_2a(s)}{s^{\sigma_2+1}} \le 0\quad \mbox{\rm for
all}\quad s > 0,
$$
as asserted. \eqref{e3.4} is an immediate consequence of \eqref{e3.2}.
Finally we shall show \eqref{e3.3}. Let
$$
H(s)=\int_0^s a(y)^{1/2}dy-\frac{2}{2+\sigma_2}sa(s)^{1/2},\quad s\ge0.
$$
Simple calculation shows, by virtue of \eqref{e1.2},  that
\begin{align*}
H'(s)
=&a(s)^{1/2}-\frac{2}{2+\sigma_2}\Big[\frac{1}{2}sa'(s)a(s)^{-1/2}
  +a(s)^{1/2}\Big]\\
=&\frac{a(s)^{-1/2}[\sigma_2 a(s)-a'(s)s]}{2+\sigma_2}
\ge  0
\end{align*}
for all $s>0$. Since $H(0)=0$, we see that $H(s) \ge 0$ for all $s
\ge 0$. This proves \eqref{e3.3}. Thus the above claims hold.

 Now taking $\varphi=u$ and
$\psi=\psi_\varrho=\frac{va(v)}{b(v)+\varrho}$ for $\varrho>0$ as
test functions in Definition \ref{def1.1}, we obtain
\begin{align*}
&\int_\Omega\Big(\frac{u^2(x,t)}{2}+\int_0^{v(x,t)}
\frac{sa(s)}{b(s)+\varrho}ds\Big) dx-
\int_\Omega\Big(\frac{u_0^2}{2}+\int_0^{v_0} \frac{sa(s)}{b(s)+\varrho}ds\Big) d x\\
&= - \int_0^t\int_{\Omega}
\Big[(a(u)+a'(u)u)|\nabla u|^2+\frac{b(v)(a(v)+va'(v))}{b(v)
+\varrho}|\nabla v|^2)\Big]dx\,d\tau\\
&\quad -\int_0^t\int_{\Omega}\frac{\varrho
va(v)b'(v)}{(b(v)+\varrho)^2}|\nabla v|^2dx\,d\tau
 +  \int_0^t\int_{\Omega}\Big(\alpha a(u) +\frac{\beta  a(v)b(v)}{b(v)
+\varrho} \Big)uv \,dx\,d\tau.
\end{align*}
For $\varrho>0$, let
\begin{gather*}
\Phi_\varrho(t)=\int_\Omega\Big(\frac{u^2(x,t)}{2}+\int_0^{v(x,t)}
\frac{sa(s)}{b(s)+\varrho}ds\Big) dx,\\
\Phi (t)=\int_\Omega\Big(\frac{u^2(x,t)}{2}+\int_0^{v(x,t)}
\frac{sa(s)}{b(s) }ds\Big) dx.
\end{gather*}
Then it is easy to see that
$\Phi_\varrho(t) \to \Phi (t)$, $\Phi_\varrho' (t)\to \Phi'(t)$,
in $(0, \infty)$,  as $\varrho\to 0^+$, and
\begin{align*}
\Phi_\varrho'
=&-\int_{\Omega}\Big[(a(u)+a'(u)u)|\nabla u|^2+\frac{b(v)(a(v)+va'(v))}{b(v)+\varrho}|\nabla v|^2\Big]dx \\
&- \int_{\Omega}\frac{\varrho va(v)b'(v)}{(b(v)+\varrho)^2}|\nabla
v|^2dx
 +   \int_{\Omega}  \Big(\alpha a(u) +\frac{\beta  a(v)b(v)}{b(v)+\varrho} \Big)uv  dx \\
\le& -  \int_{\Omega}\Big[ a(u) |\nabla u|^2+\frac{b(v) a(v)
}{b(v)+\varrho}|\nabla v|^2\Big]dx +    \int_{\Omega}\Big(\alpha
a(u) +\frac{\beta  a(v)b(v)}{b(v)+\varrho} \Big)uv dx.
 \end{align*}
Passing to the limit as $\varrho \to 0$ and using \eqref{e3.1}, we
have
\begin{equation}
\begin{aligned}
\Phi'  &\le  -\int_{\Omega}(a(u) |\nabla u|^2+ a(v)|\nabla v|^2)dx
+\max\{\alpha,\beta\}\int_{\Omega}(a(u)u^2 + a(v)v^2 )dx \\
&= -\int_{\Omega}\Big(|\nabla\int_0^{u }{a(s)}^{1/2}ds|^2+|\nabla\int_0^{v }{a(s)}^{1/2}ds|^2\Big)dx \\
\quad &+ \max\{\alpha,\beta\}\int_{\Omega}(a(u)u^2 + a(v)v^2 )dx.
\end{aligned}\label{e3.5}
\end{equation}
In view of  \eqref{e3.3},  we obtain
\begin{gather*}
\int_{\Omega} |\nabla\int_0^{u }{a(s)}^{1/2}ds|^2 dx
 \ge \lambda_1\int_{\Omega} \Big(\int_0^{u }{a(s)}^{1/2}ds\Big)^2 dx
\ge \frac{4\lambda_1}{(2+\sigma_2)^2} \int_{\Omega} a(u)u^2 dx,\\
\int_{\Omega} |\nabla\int_0^{v }{a(s)}^{1/2}ds|^2 dx
 \ge \lambda_1\int_{\Omega} \Big(\int_0^{v }{a(s)}^{1/2}ds\Big)^2 dx
\ge \frac{4\lambda_1}{(2+\sigma_2)^2} \int_{\Omega} a(v)v^2 dx.
\end{gather*}
 Combining this with \eqref{e3.5}, we have
\begin{equation}
\Phi' \le  -\overline\lambda \int_{\Omega}(a(u)u^2+a(v)v^2)dx,\\
\label{e3.6}
\end{equation}
where  $ \overline\lambda= \frac{4\lambda_1}{(2+\sigma_2
)^2}-\max\{\alpha,\beta\}>0$, which, in particular, implies that
$\Phi'(t) \le 0, \forall t>0$, and hence
\begin{equation}
\int_{\Omega}\int_0^{v(x,t)} \frac{sa(s)}{b(s)}\,ds\,dx \le
A_0,\quad \forall t >0.
\label{e3.7}
\end{equation}
By H\"older's inequality  and \eqref{e3.2}, we obtain
\begin{equation}
 \int_{\Omega} u^2dx \le   C \Big(\int_{\Omega} u^{2+\sigma_2}
dx\Big)^{2/(2+\sigma_2)}
\le  C \Big(\int_{\Omega} a(u)u^2 dx\Big)^{2/(2+\sigma_2)}.
\label{e3.8}
\end{equation}
By \eqref{e3.2} and noticing $\sigma_2>\sigma_1$ and using
H\"older's inequality, we have
\begin{equation}
  \begin{aligned}
 \int_{\Omega} \int_0^{v} \frac{sa(s)}{b(s)}ds\, dx
&\le  C  \int_{\Omega} v^{2+\sigma_1} dx\\
&\le  C  \Big(\int_{\Omega} v^{2+\sigma_2} dx\Big)^{(2+\sigma_1)/(2+\sigma_2)} \\
& \le  C    \Big(\int_{\Omega} a(v)v^2
  dx\Big)^{(2+\sigma_1)/(2+\sigma_2)}.
\end{aligned}\label{e3.9}
\end{equation}
Combining  \eqref{e3.6} with \eqref{e3.7}, \eqref{e3.8} and \eqref{e3.9}
and using the inequality $a^r+b^r \ge 2^{1-r}(a+b)^r$ for
$a, b \ge 0, r \ge 1$, we obtain
\begin{align*}
\Phi'(t)\le & -C \Big[\Big(\int_{\Omega}u^2(x,t)
dx\Big)^{(2+\sigma_2)/2}
+\Big(\int_0^{v(x,t)} \frac{sa(s)}{b(s)}ds  dx\Big)^{(2+\sigma_2)/(2+\sigma_1)}\Big]\\
\le & -C \Big[\Big(\int_{\Omega}u^2(x,t) dx\Big)^{(2+\sigma_2)/2}
+\Big(\int_\Omega\int_0^{v(x,t)}\frac{sa(s)}{b(s)}ds dx \Big)^{(2+\sigma_2)/2}\Big]\\
\le & -C  \Big(\int_{\Omega}
\Big(\frac{u^2(x,t)}{2}+\int_0^{v(x,t)} \frac{sa(s)}{b(s)}ds\Big) dx\Big)^{(2+\sigma_2)/2} \\
\le & -C\Phi(t)^{(2+\sigma_2)/2},
\end{align*}
which gives
$$
\Phi(t) \le \Big(\frac{1}{Ct+A_0^{-\sigma_2/2}}\Big)^{2/\sigma_2},
\quad \forall t>0.
 $$
This completes the proof.
%\end{proof}


\subsection*{Acknowledgments} The authors want to thank the anonymous referee
for pointing out some errors  of the original manuscript.

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