\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 06, pp. 1--14.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/06\hfil Existence of global solutions]
{Existence of global solutions for a predator-prey
model with cross-diffusion}

\author[S. Xu\hfil EJDE-2008/06\hfilneg]
{Shenghu Xu}

\address{Department of Mathematics, 
Northwest Normal University, 
Lanzhou 730070, China}
\email{xuluck2001@163.com}

\thanks{Submitted October 13, 2007. Published January 12, 2008.}
\subjclass[2000]{35K57, 35B35, 92D25}
\keywords{Cross-diffusion; global solution; gradient estimates; stability}

\begin{abstract}
 In this article, we prove the  existence of global classical solutions
 for a prey-predator model when the space dimension $n<10$. Under
 certain conditions on the coefficients of the reaction functions,
 the convergence of solutions is established for the system with
 large diffusion by constructing a Lyapunov function.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

 To investigate the spatial segregation under
the self and cross  population pressure, Shigesada, Kawasaki and
Teramoto \cite{Shigesada} proposed a competition model in 1979. Then
there have been established many results in the literatures; see for
example  \cite{Lou, Li, Shim1, Choi1, Choi2, Le, Tuoc1, Tuoc2}.
For the cross-diffusion
systems with prey-predator type reaction functions, there are a few
results mainly on the steady-state problems with the elliptic
systems, see \cite{Kuto1, Ahmed, Kuto2, Nakashima, Ryu}.

In this paper, we study the following cross-diffusion system, with
prey-predator type  reactions,
\begin{equation}
\begin{gathered}
  u_{t}-\Delta[(d_1+\alpha_{11}u+\alpha_{12}v)u]=u(a_1-b_1u-c_1v)
  \quad\text{in }\Omega\times[0,\infty),\\
  v_{t}- \Delta[(d_2+\alpha_{21}u+\alpha_{22}v)v]=v(a_2+b_2u-c_2v)
  \quad\text{in }\Omega\times[0,\infty),\\
  \partial_{\eta }u=\partial_{\eta}v=0 \quad\text{on }
  \partial{\Omega}\times[0,\infty),\\
  u(x,0)=u_{0}(x)\geq0, \quad v(x,0)=v_{0}(x)\geq0 \quad\text{in }\Omega,
\end{gathered}\label{e11}
\end{equation}
where $\Omega\subset\mathbb{R}^{n}$ ($n\geq1$) is a bounded domain with
smooth boundary $\partial\Omega$, $\eta$ is the outward unit normal
vector of the boundary $\partial\Omega$, and
$\partial_\eta=\partial/\partial_\eta$. $\alpha_{ij}$ are given nonnegative
constants for $i, j=1,2$. And $d_i, b_i, c_i(i=1,2)$ and $a_1$ are
positive constants only $a_2$ may be non-positive.

In system \eqref{e11}, $u$ and $v$ are nonnegative functions which
represent the population densities of the prey and predator species,
respectively, $d_1$ and $d_2$ are the random diffusion rates of the
two species, $\alpha_{11}$ and $\alpha_{22}$ are self-diffusion
rates, and $\alpha_{12}$ and $\alpha_{21}$ are the so-called
cross-diffusion rates. When $\alpha_{ij}=0$ ($i,j=1,2$), the system is
the well-known Lotka-Volterra prey-predator model. For more details
on the biological background, see references \cite{Shigesada,
Shim2}.

Local existence (in time) of solutions to \eqref{e11} was established by
Amann in a series of important papers \cite{Amann1, Amann2, Amann3}.
His result can be summarized as follows:

\begin{theorem} \label{thmA}
 Suppose that $u_0, v_0$ are in $W_p^1(\Omega)$ for some $p>n$.
Then  \eqref{e11} has a unique non-negative smooth solution
$u(x,t), v(x,t)$ in
\[
C([0,T),W_p^1(\Omega))\bigcap C^\infty((0,T),C^\infty(\Omega))
\]
with maximal existence time $T$.
Moreover, if the solution $(u,v)$ satisfies the estimate
$$
 \sup_{0\leq t\leq T} \|u(.,t)\|_{W_p^1(\Omega)}<\infty
 \quad\text{and}\quad \sup_{0\leq t\leq T}
\|v(.,t)\|_{W_p^1(\Omega)}<\infty ,
$$
then $T=\infty$.
\end{theorem}

However, little is known about global existence of solutions to
\eqref{e11}. In 2006, Shim \cite{Shim2} proved the  existence of
global solutions  to \eqref{e11} in two cases:
Case(A) $n=1$,  $d_1=d_2$ and $\alpha_{11}=\alpha_{22}=0$;
Case(B) $n=1$, $0<\alpha_{21}<8\alpha_{11}$ and
$0<\alpha_{12}<8\alpha_{22}$.

In \cite{Shim3} the author considered the case when $\alpha_{11},
\alpha_{12}, \alpha_{22}>0$ and $\alpha_{21}=0$ for the system
\eqref{e11}, and established the  existence of global solutions with
$n=1$.

We shall prove the  existence of global solutions to the following
system (namely, the system \eqref{e11} for $\alpha_{12}=0$)
\begin{equation} \begin{gathered}
  u_{t}-\Delta[(d_1+\alpha_{11}u)u]=u(a_1-b_1u-c_1v)\quad\text{in }
  \Omega\times[0,\infty),\\
  v_{t}- \Delta[(d_2+\alpha_{21}u+\alpha_{22}v)v]=v(a_2+b_2u-c_2v)
  \quad\text{in }\Omega\times[0,\infty),\\
  \partial_{\eta }u=\partial_{\eta}v=0 \quad\text{on }
  \partial{\Omega}\times[0,\infty),\\
  u(x,0)=u_{0}(x)\geq0,\quad v(x,0)=v_{0}(x)\geq0
  \quad\text{in }\Omega.
\end{gathered}\label{e12}
\end{equation}

This paper draws on ideas from two papers \cite{Choi2} and
\cite{Tuoc2} which deal with cross-diffusion system with competition
type reactions. Duo to the difference in the reaction functions.
Therefore, in order to obtain the $L^p$-estimate of $v$, we have to
estimate the term $uv^p$. We also obtain result on the asymptotic
stability of the global solution to \eqref{e12} if the diffusion
coefficients are large enough by an important Lemma 5.1 from
\cite{Wang}. We summarize our results in the following theorems:

\begin{theorem} \label{thm1.1}
Let $\alpha_{22}>0$ and assume that
$u_0\geq0, v_0\geq0$  satisfy zero Neumann boundary condition and
belong to $C^{2+\lambda}(\overline{\Omega})$ for some $\lambda>0$.
Then \eqref{e12} possesses a unique non-negative solution
$u, v\in C^{2+\lambda,\frac{2+\lambda}{2}}(\overline{\Omega}\times[0,\infty))$
if either
{\rm (i)} $\alpha_{11}=0 $ or
{\rm (ii)} $\alpha_{11}>0$ and $n<10$.
\end{theorem}

\begin{theorem} \label{thm1.2}
 Assume that all conditions in Theorem \ref{thm1.1} are satisfied.
Assume further that
\begin{gather}
-\frac{a_1b_2}{b_1c_2}<\frac{a_2}{c_2}<\frac{a_1}{c_1},\label{e13}\\
4\rho\overline{u}\overline{v}d_1d_2>m^2(\overline{v}\alpha_{21})^2.
\label{e14}
\end{gather}
Then \eqref{e12} has the unique positive equilibrium point
$(\overline{u}, \overline{v})$ which is global asymptotic stable,
where $m$ is the positive constant in Lemma \ref{lem2.1} (independent of
$d_1, d_2$), $\rho=(b_2c_1+2b_1c_2)b_2^{-2}$ and
$$
(\overline{u},\overline{v})=\Big(\frac{a_1c_2-a_2c_1}{b_1c_2+b_2c_1},
\frac{a_2b_1+a_1b_2}{b_1c_2+b_2c_1}\Big).
$$
\end{theorem}

 The paper is organized as follows. In section 2, we
collect some well known results and prove three new lemmas that are
needed in section 3 and section 4. In section 3, we establish
$L^r$-estimates of the solution $v$ of \eqref{e12} and in section 4 we
give a proof of Theorem \ref{thm1.1}. In section 5, we give a proof of
Theorem \ref{thm1.2}.

\section{Preliminaries}

 We list here some notation.
\begin{align*}
&Q_T=\Omega\times[0,T),\\
&\|u\|_{L^{p,q}(Q_T)}=\Big(\int_0^T(\int_{\Omega}|u(x,t)|^p dx
 )^{\frac{q}{p}}dt\Big)^{1/q},
 L^{p}(Q_T):= L^{p,p}(Q_T),\\
&\|u \|_{W_P^{2,1}(Q_T)}:= \|u\|_{L^p(Q_T)}+\|u_t\|_{L^p(Q_T)}+\|\nabla
 u\|_{L^p(Q_T)}+\|\nabla^2u\|_{L^p(Q_T)},\\
&\|u\|_{V_2(Q_T)}:= \sup_{0\leq t\leq T}\|u(.,t)\|_{L^2(\Omega)}
+\|\nabla u(x,t)\|_{L^2(Q_T)}.
\end{align*}
Firstly, we present some  useful lemmas.

\begin{lemma} \label{lem2.1}
 Let $u, v$ be a solution of \eqref{e12} in $[0,T)$. Then
$0\leq u\leq m$ and $v\geq0$ in   $Q_T$,
where $m=\max\{\frac{a_1}{b_1},\|u_0\|_{L^\infty(\Omega)}\}$.
\end{lemma}

\begin{proof}
The first equation  in \eqref{e12} is expressed as
\begin{equation}
u_t=(d_1+2\alpha_{11}u)\Delta u+2\alpha_{11}\nabla u\cdot\nabla
u+u(a_1-b_1u-c_1v),\label{e21}
\end{equation}
and the second equation is written as
\begin{equation}
v_t=(d_2+\alpha_{21}u+2\alpha_{22}v)\Delta v+2(\alpha_{21}\nabla
u+\alpha_{22}\nabla v)\nabla v+v(\alpha_{21}\Delta
u+a_2+b_2u-c_2v).\label{e22}
\end{equation}
Then application of the maximum principle for \eqref{e21} and \eqref{e22}
yields the nonnegative of $u$ and $v$. Applying the maximum
principle to \eqref{e21} again one can also show the boundedness of $u$.
\end{proof}

\begin{lemma} \label{lem2.2}
There exists a positive $C_1(T)$ such that
\begin{gather*}
\sup_{0\leq t\leq T}\|u(.,t)\|_{L^1(\Omega)}<C_1(T),\quad
\sup_{0\leq t\leq T}\|v(.,t)\|_{L^1(\Omega)}<C_1(T),\\
\|u\|_{L^2(Q_T)}<C_1(T), \quad
\|v\|_{L^2(Q_T)}<C_1(T).
\end{gather*}
\end{lemma}

\begin{proof}
 Integrating  the first equation in \eqref{e12} over
the domain $\Omega$, we have
\begin{equation} \label{e23}
\begin{aligned}
  \frac{d}{dt}\int_\Omega u\,dx
&= a_1\int_\Omega u\,dx-b_1\int_\Omega u^2dx
 -c_1\int_\Omega uv dx \\
&\leq a_1\int_\Omega u\,dx-b_1\int_\Omega u^2dx\\
&\leq a_1\int_\Omega u\,dx-\frac{b_1}{|\Omega|}
\Big(\int_\Omega u\,dx\Big)^2,
\end{aligned}
\end{equation}
where we used H\"older's inequality.
Then we have $\|u(.,t)\|_{L^1(\Omega)}\leq M_1'$, where
$M_1'=\max\{\|u_0\|_{L^1(\Omega)}, \frac{a_1}{b_1}|\Omega|\}$.
Furthermore,
\begin{equation}
\sup_{0\leq t\leq T} \|u(.,t)\|_{L^1(\Omega)}<C_1(T). \label{e24}
\end{equation}
Since
\begin{equation}
\frac{d}{dt}\int_\Omega u \,dx\leq a_1\int_\Omega u\,dx
-b_1\int_\Omega u^2dx .\label{e25}
\end{equation}
Integrating \eqref{e25} from $0$ to $T$, we have
\[
\|u\|_{L^2(Q_T)}^2\leq\frac{a_1}{b_1}M_1'|Q_{T}|+\|u_0\|_{L^1(\Omega)}.
\]
Therefore,
\begin{equation}
\|u\|_{L^2(Q_T)}\leq C_1(T) .\label{e26}
\end{equation}
 Now, integrating  the second equation in the system
 \eqref{e12} over the domain $\Omega$ we have
\begin{equation}
\frac{d}{dt}\int_\Omega vdx= a_2\int_\Omega vdx+b_2\int_\Omega
uvdx-c_2\int_\Omega v^2 dx.\label{e27}
\end{equation}
Multiplying \eqref{e23} by
$\frac{b_2}{c_1}$ and adding it to \eqref{e27}, we have
\begin{equation}
\frac{d}{dt}\int_\Omega \big(\frac{b_2}{c_1}u+v\big)dx
\leq \frac{a_1b_2}{c_1}\int_\Omega u\,dx
+|a_2|\int_\Omega v\,dx-\frac{b_1b_2}{c_1}\int_\Omega
u^2dx-c_2\int_\Omega v^2 dx.\label{e28}
\end{equation}
Then
\begin{align*}
&\min\{1,\frac{b_2}{c_1}\} \frac{d}{dt}\int_\Omega
 \big(u+v\big)dx\\
&\leq \max\{\frac{a_1 b_2}{c_1},|a_2|\}\int_\Omega (u +v)dx
  -\min\{\frac{b_1b_2}{c_1},c_2\}\int_\Omega  (u^2+v^2)dx\\
&\leq \max\{\frac{a_1 b_2}{c_1},|a_2|\}\int_\Omega (u +v)dx
 -\frac{1}{2}\min\{\frac{b_1b_2}{c_1},c_2\}\big[\int_\Omega
 (u+v)dx\big]^2.
\end{align*}
Therefore, $\|v(.,t)\|_{L^1(\Omega)}\leq M_2'$, where
$M_2'=\max\{\frac{A_1}{A_2},\|u_0+v_0\|_{L^1(\Omega)}\}$,
\[
A_1=\frac{\max\{\frac{a_1b_2}{c_1},|a_2|\}}{\min\{1,\frac{b_2}{c_1}\}},\quad
A_2=\frac{\min\{\frac{b_1b_2}{c_1},c_2\}}{2\min\{1,\frac{b_2}{c_1}\}}.
\]
Then
\begin{equation}
 \sup_{0\leq t\leq T} \|v(.,t)\|_{L^1(\Omega)}<C_1(T). \label{e29}
\end{equation}
Integrating \eqref{e28} from $0$ to $T$, we have
\[
 c_2\int_{Q_T}v^2\,dx\,dt
 \leq\frac{a_1b_2}{c_1}\int_0^TM_1'dt+|a_2|\int_0^TM_2'dt
  +\frac{b_2}{c_1}\|u_0\|_{L^1(\Omega)}+\|v_0\|_{L^1(\Omega)},
\]
which implies
$\|v\|_{L^2(Q_T)}\leq C_1(T)$. %\label{e210}
\end{proof}

\begin{lemma} \label{lem2.3}
 Let $w_1=(d_1+\alpha_{11}u)u$ .Then there
exists a constant $C_2(T)$, depending on $\|u_0\|_{W_2^1(\Omega)}$
and $\|u_0\|_{L^\infty(\Omega)}$  such that
\begin{equation}
\|w_1\|_{W_{2}^{2,1}(Q_{T})}\leq C_2(T) .\label{e211}
\end{equation}
Furthermore, $\nabla w_1\in V_2(Q_T)$.
\end{lemma}

\begin{proof}
Note that $w_1$ satisfies the equation
\begin{equation}
w_{1t}=(d_1+2\alpha_{11}u)\Delta w_1+n_1+n_2 v,\label{e212}
\end{equation}
where $n_1=u(d_1+2\alpha_{11}u)(a_1-b_1u),
n_2=-c_1(d_1+2\alpha_{11}u)u$ depend on $u$ and are bounded
functions because of Lemma \ref{lem2.1}. Multiplying the above equation by
$-\Delta w_1$ and integration by parts over $\Omega$, we have
\begin{equation}
\frac{1}{2}\frac{d}{dt}\int_\Omega|\nabla
w_1|^2dx=-\int_\Omega(d_1+2\alpha_{11}u)(\Delta
w_1)^2dx-\int_\Omega(n_1+n_2v)\Delta w_1dx.\label{e213}
\end{equation}
Integrating \eqref{e213} from $0$ to $t$, we obtain
\begin{align*}
&\frac{1}{2}\int_\Omega|\nabla w_1(x,t)|^2dx
 -\frac{1}{2}\int_\Omega|\nabla w_1(x,0)|^2dx\\
 &=  -\int_{Q_t}(d_1+2\alpha_{11}u)(\Delta w_1)^2\,dx\,dt
-\int_{Q_t}(n_1+n_2v)\Delta w_1\,dx\,dt\\
&\leq-d_1\int_{Q_t}|\Delta
w_1|^2\,dx\,dt+\int_{Q_t}(n_1+n_2v)\cdot|\Delta w_1|dx\, dt\,.
\end{align*}
By Young's inequality and $\mathrm{H \ddot{o}lder}$'s inequality, we
have
\begin{align*}
& \frac{1}{2}\int_\Omega|\nabla w_1(x,t)|^2dx+d_1\int_{Q_T}|\Delta w_1|^2\,dx\,dt\\
&\le (\|n_1\|_{L^2(Q_T)}+\|n_2v\|_{L^2(Q_T)})\cdot\|\Delta
w_1\|_{L^2(Q_T)}+\frac{1}{2}\int_\Omega|\nabla w_1(x,0)|^2dx\\
&\leq m_1(1+\|v\|_{L^2(Q_T)})\cdot\|\Delta w_1\|_{L^2(Q_T)}
+\frac{1}{2}\int_\Omega|\nabla w_1(x,0)|^2dx\\
&\leq m_1(1+C_1(T))\cdot\|\Delta w_1\|_{L^2(Q_T)}
+\frac{1}{2}\int_\Omega|\nabla w_1(x,0)|^2dx\\
&\leq \frac{d_1}{2}\|\Delta
w_1\|^2_{L^2(Q_T)}+\frac{m_1^2(1+C_1(T))^2}{2d_1}+\frac{1}{2}
\int_\Omega|\nabla w_1(x,0)|^2dx.
\end{align*}
Therefore,
$$
\sup_{0\leq t\leq T} \int_\Omega|\nabla w_1|^2(x,t)dx+d_1
\int_{Q_T}|\Delta w_1|^2\,dx\,dt\leq m_2,
$$
where $m_2$ depends on $\|u_0\|_{W_2^1(\Omega)}$ and
$\|u_0\|_{L^\infty(\Omega)}$. This implies $\nabla w_1\in V_2(Q_T)$.
Since $w_1\in L^2(Q_T)$ we have from the elliptic regularity
estimate \cite[Lemma 2.3]{Lou}
$$
\int_{Q_T}\|(w_1)_{x_ix_j}\|^2\,dx\,dt\leq m_3\quad
 \text{for }  i, j=1,\dots,n.
$$
 From \eqref{e212}, since $n_1, n_2$ and $u$ are bounded and
$v\in L^2(Q_T)$, we have $w_{1t}\in L^2(Q_T)$. Hence,
$w_1\in W^{2,1}_2(Q_T)$.
\end{proof}

Let $a(x,t,\xi)$ be continuous and $(x,\xi)$-differentiable for
$(x,t,\xi)\in \mathrm{Q_T\times R}$. Assume also that $a(x,t,\xi)$
satisfies the following conditions
\begin{itemize}
\item[(i)] There is $d>0$ such that $a(x,t,\xi)\geq d$ and
$a_\xi(x,t,\xi)\geq 0$ for all $(x,t)\in Q_T$ and $\xi$ in
$\mathrm{R}$.

\item[(ii)] There is a continuous function $M$ on $\mathrm{R}$
such that $a(x,t,\xi)\leq M(\xi)$ for all  $(x,t)\in Q_T$.

\item[(iii)] For any bounded measurable function $g$ on $Q_T$,
$|\nabla_{x}a(.,.,g(.,.))|$ is in the space $L^{2p}(Q_T)$.

\end{itemize}

\begin{lemma} \label{lem2.4}
 Assume that $w\in W_p^{2,1}(Q_T)\bigcap
C^{2,1}(\overline{\Omega}\times[0,T))$ is a bounded function
satisfying
$$
w_t\leq a(x,t,w)\Delta w+f(x,t) \quad\text{in} \quad Q_T
$$
with boundary condition $\frac{\partial w}{\partial\nu}\leq0$ on
$\partial_{Q_T}$, where $f\in L^p(Q_T)$. Then, $\nabla w$ is in
$L^{2p}(Q_T)$.
\end{lemma}

The proof of the above lemma can be found in
\cite[Proposition 2.1]{Tuoc2}.

\begin{lemma} \label{lem2.5}
 Let $q>1, \widetilde{q}=2+\frac{4q}{n(q+1)}$,
$\widetilde{\beta}$ in $(0,1)$ and let $C_T>0$ be any number which
may depend on $T$. Then there is a constant $M_1$ depending on $q,
n, \Omega, \widetilde{\beta}$ and $C_T$ such that for any $g$ in
$C([0,T),W_2^1(\Omega))$ with
$(\int_\Omega|g(.,t)|^{\widetilde{\beta}}dx)^{1/\widetilde{\beta}}
\leq C_T$ for all $t\in[0,T]$, we have the  inequality
$$
\|g\|_{L^{\widetilde{q}}(Q_T)}\leq M_1\big\{1
+\big(\sup_{0\leq t\leq T}\|g(.,t)\|_{L^{2q/q+1}(\Omega)}\big)^{4q/n(q+1)
\widetilde{q}}\|\nabla g\|^{2/\widetilde{q}}_{L^2(Q_T)}\big\} .
$$
\end{lemma}

The proof of the above lemma can be found in \cite[Lemmas 2.3, 2.4]{Choi2}.

\section{$L^r$-estimates for $v$}

\begin{lemma} \label{lem3.1}
There exists a constant $C_3(T)$ such that
$\|\nabla u\|_{L^4(Q_T)}\leq C_3(T)$.
\end{lemma}

\begin{proof}
Let $\delta=\alpha_{11}/d_1$, $w_1=(1+\delta u)u$.
By Lemma \ref{lem2.1}, $u$ is bounded. Therefore, $w_1$ is also bounded.
By Lemma \ref{lem2.3}, we have $w_1\in W_2^{2,1}(Q_T)$. Moreover, $w_1$ satisfies
\begin{align*}
w_{1t}
&\leq d_1(1+2\delta u)\Delta w_1+a_1u(1+2\delta u)\\
&= \sqrt{d_1^2+4\delta d_1 w_1}\Delta w_1+a_1u(1+2\delta u).
\end{align*}
By Lemma \ref{lem2.4} with $p=2$, $a(x,t,\xi)=\sqrt{d_1^2+4\delta d_1\xi}$,
$f(x,t)=a_1u(x,t)(1+2\delta u(x,t))$, we obtain the desired result.
\end{proof}

\begin{lemma} \label{lem3.2}
Let $r>2$ and $p_r=\frac{2r}{r-2}$ be two
positive numbers. Assume that $\alpha_{22}>0$ and assume also that
there is a constant $M_{r,T}>0 $ depending only on $r, T,\Omega$ and
the coefficients of \eqref{e12} such that
$$
\|\nabla u\|_{L^r(Q_T)}\leq M_{r,T}.
$$
Then for any $q>1$, there exists a constant $C(r,q,T)>0$
such that
\begin{equation}
\begin{aligned}
&\|v(.,t)\|^q_{L^q(\Omega)}+\|\nabla(v^{q/2})\|^2_{L^2(Q_t)}
 +\|\nabla(v^{(q+1)/2})\|^2_{L^2(Q_t)}\\
&\leq
C(r,q,T)\big(1+\|v\|^{q-1}_{L^{\frac{p_r(q-1)}{2}}(Q_t)}\big).\label{e31}
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
For any constant $q>1$, multiplying the second equation of \eqref{e12}
by $qv^{q-1}$ and using the integration by parts, we obtain
\begin{align*}
&\frac{\partial}{\partial t}\int_{\Omega}v^{q}dx \\
&= q\int_{\Omega}v^{q-1}\nabla\cdot[(d_{2}+\alpha_{21}u+2\alpha_{22}v)\nabla
v+\alpha_{21}v\nabla u]dx
 +q\int_{\Omega}v^{q}(a_{2}+b_{2}u-c_{2}v) dx \\
&= -q(q-1)\int_{\Omega}v^{q-2}(d_{2}+\alpha_{21}u+2\alpha_{22}v)|\nabla
v|^{2} dx
 -\alpha_{21}(q-1)\int_{\Omega}\nabla(v^{q})\cdot\nabla u\,dx \\
&\quad +q\int_{\Omega}v^{q}(a_{2}+b_{2}u-c_{2}v) dx \\
&\le  -q(q-1)d_{2}\int_{\Omega}v^{q-2}|\nabla v|^{2} dx-2\alpha_{22}q(q-1)\int_{\Omega}v^{q-1}|\nabla v|^{2} dx \\
&\quad -\alpha_{21}(q-1)\int_{\Omega}\nabla(v^{q})\cdot\nabla u\,dx
 +q\int_{\Omega}v^{q}(a_{2}+b_{2}u-c_{2}v) dx  \\
&= -\frac{4(q-1)d_{2}}{q}\int_{\Omega}|\nabla(v^{\frac{q}{2}})|^{2}dx
 -\frac{8\alpha_{22}q(q-1)}{(q+1)^{2}}\int_{\Omega}
 |\nabla(v^{\frac{q+1}{2}})|^{2}dx \\
&\quad -\alpha_{21}(q-1)\int_{\Omega}\nabla(v^{q})\cdot\nabla u\,dx
+q\int_{\Omega}v^{q}(a_{2}+b_{2}u-c_{2}v) dx.
%\label{e32)}
\end{align*}
Integrating the above inequality  from $0$ to $t$, we have
\begin{equation}
\begin{aligned}
& \int_{\Omega}v^{q}(x,t)dx+\frac{4(q-1)d_{2}}{q}
\int_{Q_{t}}|\nabla(v^{\frac{q}{2}})|^{2}dx\,dt
 +\frac{8\alpha_{22}q(q-1)}{(q+1)^{2}}
 \int_{Q_{t}}|\nabla(v^{\frac{q+1}{2}})|^{2}\,dx\,dt \\
&\le \int_{\Omega}v^{q}(x,0)dx-\alpha_{21}(q-1)\int_{Q_t}\nabla(v^{q})
 \cdot\nabla u\,dx\,dt+q\int_{Q_t}v^{q}(a_{2}+b_{2}u-c_{2}v) \,dx\,dt.
\end{aligned}  \label{e33}
\end{equation}
By H\"older's inequality, we have
\begin{equation}
\begin{aligned}
&q\int_{Q_t}v^{q}(a_{2}+b_{2}u-c_{2}v) \,dx\,dt\\
&= a_2q\int_{Q_t}v^{q} dx dt-c_2q\int_{Q_t}v^{q+1}dx
 dt+b_2q\int_{Q_t}uv^{q}dx dt\\
&\leq -c_2q\|v\|^{q+1}_{L^{q+1}(Q_t)}+|a_2|q|Q_T|^{\frac{1}{q+1}}\|v\|^{q}_{L^{q+1}(Q_t)}
 +b_2q\int_{Q_t}uv^{q}dx dt\\
&\leq -c_2q\|v\|^{q+1}_{L^{q+1}(Q_t)}+|a_2|q
\big[\varepsilon\|v\|^{q+1}_{L^{q+1}(Q_t)}
 +\varepsilon^{-q}|Q_T|^{\frac{q}{q+1}}\big]
 +b_2q\int_{Q_t}uv^{q}dx dt\\
&\leq B_1+b_2q\int_{Q_t}uv^{q}dx dt,
\end{aligned}\label{e34}
\end{equation}
where $\varepsilon=\frac{c_2}{|a_2|}$, $B_1$ depends on
$T, q, |\Omega|$ and the coefficients of \eqref{e12}.

On the other hand, since that
$\frac{1}{r}+\frac{1}{2}+\frac{1}{p_r}=1$, using the
H\"older's inequality and
Poincar\'e inequality, we have
\begin{equation}
\begin{aligned}
 \int_{Q_t}uv^{q}dx dt
&= \int_{Q_t}u\cdot v^{\frac{q-1}{2}}\cdot
 v^{\frac{q+1}{2}}\,dx\,dt\\
&\leq \|v^{\frac{q-1}{2}}\|_{L^{p_r}(Q_t)}\cdot
  \|v^{\frac{q+1}{2}}\|_{L^{2}(Q_t)}\cdot\|u\|_{L^{r}(Q_t)}\\
&\leq C_4m\|v\|^{(q-1)/2}_{L^{\frac{p_r(q-1)}{2}}(Q_t)}\cdot\|
 \nabla(v^{\frac{q+1}{2}})\|_{L^{2}(Q_t)}.
\end{aligned} \label{e35}
\end{equation}
The substitution \eqref{e35} into \eqref{e34} leads to
\begin{equation}
\begin{aligned}
 q\int_{Q_t}v^{q}(a_{2}+b_{2}u-c_{2}v) dx dt
&\leq B_1+C_5\|v\|^{(q-1)/2}_{L^{\frac{p_r(q-1)}{2}}(Q_t)}
\cdot\|\nabla(v^{\frac{q+1}{2}})\|_{L^{2}(Q_t)}. \label{e36}
\end{aligned}
\end{equation}
Since that $\frac{1}{r}+\frac{1}{2}+\frac{1}{p_r}=1$ and $\nabla u$
is in $L^r(Q_T)$, using the H\"older's inequality,
we have
\begin{align*}
\big|-\int_{Q_t}\nabla(v^{q})\cdot\nabla u\,\,dx\,dt\big|
&= \frac{2q}{q+1}\big|\int_{Q_t}v^{\frac{q-1}{2}}\cdot
 \nabla(v^{\frac{(q+1)}{2}})\cdot\nabla u\,dx\,dt\big|\\
&\leq \frac{2q}{q+1}\|v^{\frac{q-1}{2}}\|_{L^{p_r}(Q_t)}
 \cdot\|\nabla(v^{\frac{q+1}{2}})\|_{L^2(Q_t)}
 \cdot\|\nabla u\|_{L^r(Q_t)}\\
&\leq \frac{2q}{q+1}\|v\|^{\frac{q-1}{2}}_{L^{\frac{p_r(q-1)}{2}}(Q_t)}
 \cdot\|\nabla(v^{\frac{q+1}{2}})\|_{L^2(Q_t)}
 \cdot\|\nabla u\|_{L^r(Q_t)}\\
&\leq \frac{2q}{q+1}M_{r,T}\|v\|^{\frac{q-1}{2}}_{L^{\frac{p_r(q-1)}{2}}
 (Q_t)}\cdot\|\nabla(v^{\frac{q+1}{2}})\|_{L^2(Q_t)}.
%\label{e37}
\end{align*}
The substitution \eqref{e36} and the above inequality into \eqref{e33}
leads to
\begin{equation}
\begin{aligned}
&\int_{\Omega}v^{q}(x,t)dx+\frac{4(q-1)d_{2}}{q}
  \int_{Q_{t}}|\nabla(v^{\frac{q}{2}})|^{2}dx\,dt
 +\frac{8\alpha_{22}q(q-1)}{(q+1)^{2}}\int_{Q_{t}}
 |\nabla(v^{\frac{q+1}{2}})|^{2}\,dx\,dt \\
&\le B_2+C_6\|v\|^{\frac{q-1}{2}}_{L^{\frac{p_r(q-1)}{2}}(Q_t)}
 \cdot\|\nabla(v^{\frac{q+1}{2}})\|_{L^2(Q_t)} \\
&\le B_2+\frac{C_6}{4\varepsilon}\|v\|^{q-1}_{L^{\frac{p_r(q-1)}{2}}
 (Q_t)}+C_6\varepsilon\|\nabla(v^{\frac{q+1}{2}})\|^2_{L^2(Q_t)},
\end{aligned}\label{e38}
\end{equation}
where $B_2>0$ depending on $q, T,\Omega$ coefficients of \eqref{e12} and
initial datal $v_0$. For any $\varepsilon>0$, from \eqref{e38} and by
choosing a sufficiently small $\varepsilon$, such that
$C_6\varepsilon<\frac{8\alpha_{22}q(q-1)}{(q+1)^{2}}$, we get
\eqref{e31}. This completes the proof of the lemma.
\end{proof}

For any number $a$, we denote $a_+=\max\{a,0\}$.

 \begin{proposition} \label{prop3.3}
 Let $\alpha_{22}>0$.
\begin{itemize}
\item[(i)] If $\alpha_{11}>0$, then there is a constant
$C_7(T)>0$ such that
$$
\|v\|_{V_2(Q_T)}\leq C_7(T).
$$
 Moreover, for any constant $r<\frac{4(n+1)}{(n-2)_+}$, there exists
a positive constant $C_{T}$ such that
$$
\|v\|_{L^r(Q_T)}\leq C_{T}.
$$

\item[(ii)] If $\alpha_{11}=0$, then
$$
\|v\|_{L^r(Q_T)}\leq C_{T} \quad\text{for any}\quad  r>1.
$$
\end{itemize}
\end{proposition}

\begin{proof}
(i) Set $w=v^{(q+1)/2}$ so that
$v^q=w^{2q/(q+1)}$ and $v^{q+1}=w^2$. Then
\begin{align*}
 E&\equiv \sup_{0\leq t\leq T} \int_\Omega v^q(x,t)dx
 +\int_{Q_T}|\nabla(v^{(q+1)/2})|^2\,dx\,dt\\
 &= \sup_{0\leq t\leq T} \int_\Omega w^{2q/q+1}dx
 +\int_{Q_T}|\nabla w|^2\,dx\,dt.
\end{align*}
Let $r_0=4$, $p_0=\frac{2r_0}{r_0-2}$. By Lemma \ref{lem3.1}, we see that
$\nabla u$ is in $L^{r_0}(Q_T)$. So, from Lemma \ref{lem3.2}, we have
\begin{equation}
E+\|\nabla(v^{\frac{q}{2}})\|^2_{L^2(Q_T)} \leq
C(r_0,q,T)\Big(1+\|w\|^{\frac{2(q-1)}{q+1}}_{L^{\frac{p_0(q-1)}{q+1}}(Q_T)}
\Big), \label{e39}
\end{equation}
where $C(r_0,q,T)>0$ depending only $T, \Omega$, initial data
$u_0, v_0$ and the coefficients of \eqref{e12}. Since $q>1$, if we restrict
our $q$ so that
\begin{equation}
(np_0-2n-4)q\leq 2n+np_0.\label{e310}
\end{equation}
Then, $\frac{p_0(q-1)}{q+1}\leq\widetilde{q}$, where
$\widetilde{q}=2+\frac{4q}{n(q+1)}$. Therefore, by
H\"older's inequality
\begin{equation}
\|w\|_{L^{\frac{p_0(q-1)}{q+1}}(Q_T)}\leq
C_8(q,T)\|w\|_{L^{\widetilde{q}}(Q_T)},\label{e311}
\end{equation}
where
$C_8(q,T)=|Q_T|^{\frac{q+1}{p_0(q-1)}-\frac{1}{\widetilde{q}}}$.
Setting $\widetilde{\beta}=2/(q+1)\in(0,1)$, by Lemma \ref{lem2.2} we have
\begin{equation}
\|w(.,t)\|_{L^{\widetilde{\beta}}(\Omega)}=\|v(.,t)\|^{\frac{1}{\widetilde{\beta}}}_{L^1(\Omega)}
\leq ( C_1(T))^{\frac{1}{\widetilde{\beta}}},\quad \forall
t\in[0,T).\label{e312}
\end{equation}
Hence, by Lemma \ref{lem2.5} and the definition of $E$, \eqref{e312} yields
\begin{equation}
\|w\|_{L^{p_0(q-1)/q+1}(Q_T)}\leq
C_8(q,T)\|w\|_{L^{\widetilde{q}}(Q_T)}\leq
C_8(q,T)M_1\{1+E^{2/n\widetilde{q}}E^{\frac{1}{\widetilde{q}}}\}.
\label{e313}
\end{equation}
Then \eqref{e39} together with the above inequality,  we can find a
 constant $C_9(q,T)>0$ such that
\begin{equation}
E\leq C_9(q,T)(1+E^\mu E^\nu)\label{e314}
\end{equation}
with
$$
\mu=\frac{4(q-1)}{n\widetilde{q}(q+1)}, \quad
\nu=\frac{2(q-1)}{\widetilde{q}(q+1)}.
$$
Since
\[
\mu+\nu = \frac{2(q-1)}{\widetilde{q}(q+1)}\big[\frac{2}{n}+1\big]
<\frac{1}{\widetilde{q}}\big[\frac{4q}{n(q+2)}+2\big]=1,
\]
it is easy to see from \eqref{e314} that $E$ is bounded. Therefore, from
\eqref{e313} and \eqref{e314} we get $w\in L^{\widetilde{q}}(Q_T)$ which in
turn implies that $v\in  L^{r}(Q_T)$ with
$r=\frac{\widetilde{q}(q+1)}{2}$ for any $q$ satisfying \eqref{e310}.
Now, looking at \eqref{e310}, if $n\leq2$, we have
\begin{equation}
np_0-2n-4=2(n-2)\leq0,\label{e315}
\end{equation}
then \eqref{e310} holds for all $q$. so for $n\leq2$, $v\in L^r(Q_T)$
for all $r>1$. Now, suppose that $n>2$, we see \eqref{e310} is
equivalent to
$$
1<q<q_0:=\frac{2n+np_0}{(np_0-2n-4)}=\frac{3n}{n-2}.
$$
Then, we have
$$
\frac{\widetilde{q}(q+1)}{2}=q+1+\frac{2q}{n}\leq \overline{r}_1:=
q_0+1+\frac{2q_0}{n}
=\frac{4(n+1)}{n-2}.
$$
So, we see that $v$ is in $L^r(Q_T)$ for all
$1<r\leq\overline{r}_1.$ Since \eqref{e310} holds true for $q=2$. So
when $q=2$, we have $E$ is finite. Therefore, from \eqref{e39} and
\eqref{e313}, we see that $\|v\|_{V_2(Q_T)}$ is bounded for any $n$,
this completes the proof of Proposition \ref{prop3.3} when $\alpha_{11}>0$ and
$ r<\frac{4(n+2)}{(n-2)_+}$.

Next, we consider the case $\alpha_{11}=0$.
By H\"older's inequality, we have
\begin{equation}
\begin{aligned}
&q\int_{Q_t}v^{q}(a_{2}+b_{2}u-c_{2}v) \,dx\,dt \\
&= a_2q\int_{Q_t}v^{q} dx dt-c_2q\int_{Q_t}v^{q+1}dx
 dt+b_2q\int_{Q_t}uv^{q}dx dt\\
&\leq -c_2q\|v\|^{q+1}_{L^{q+1}(Q_t)}+|a_2|q|Q_T|^{\frac{1}{q+1}}\|v\|^{q}_{L^{q+1}(Q_t)}\\
&\quad +b_2q\|v\|^{q}_{L^{q+1}(Q_t)}\cdot\|u\|_{L^{q+1}(Q_t)}\\
&\leq -c_2q\|v\|^{q+1}_{L^{q+1}(Q_t)}+|a_2|q|Q_T|^{\frac{1}{q+1}}\|v\|^{q}_{L^{q+1}(Q_t)}
+b_2qm\|v\|^{q}_{L^{q+1}(Q_t)}\\
&\leq -c_2q\|v\|^{q+1}_{L^{q+1}(Q_t)}+q
\varepsilon\|v\|^{q+1}_{L^{q+1}(Q_t)}+B_3\\
&\leq B_3, \hspace{8cm}\label{e316}
\end{aligned}
\end{equation}
where $\varepsilon=c_2$ and $B_3>0$ which depends only on
$T, q, |\Omega|$, $\|u_0\|_{L^{\infty}(\Omega)}$ and the coefficients of
\eqref{e12}.

We can integrate by parts once to obtain from Lemma \ref{lem2.1} and analogue
of \cite[Theorem 9.1, p. 341-342]{Ladyzenskaja} for Neumann boundary condition
\cite[p.351]{Ladyzenskaja}
\begin{equation}
\begin{aligned}
&\big|-\int_{Q_t}\nabla(v^{q})\cdot\nabla u\,\,dx\,dt\big|\\
&= \big|-\int_{Q_t}v^{q}\Delta u\,\,dx\,dt\big|\\
&\leq \|v\|^{q}_{L^{q+1}(Q_T)}\cdot\|\Delta u\|_{L^{q+1}(Q_T)}\\
&\leq C_{10} \|v\|^{q}_{L^{q+1}(Q_T)}
\Big(\|u(a_1-b_1u-c_1v)\|_{L^{q+1}(Q_T)}+\|u_0\|
 _{W^{2-\frac{2}{q+1}}_{q+1}(\Omega)}\Big)\\
&\leq C_{11}\Big(1+\|v\|^{q+1}_{L^{q+1}(Q_T)}\Big).
\end{aligned}\label{e317}
\end{equation}
The substitution of \eqref{e316} and \eqref{e317} into \eqref{e33}
leads to
\begin{equation}
\sup_{0\leq t\leq T} \|v^q(t)\|^q_{L^q(\Omega)}
 +\|\nabla(v^{(q+1)/2})\|_{L^2(Q_T)}^2
 \leq C_{12}\big(1+\|v\|^{q+1}_{L^{q+1}(Q_T)}\big).\label{e318}
\end{equation}
  We introduce $w=v^\frac{q+1}{2}$, then \eqref{e318} leads to
\begin{equation}
E\equiv\sup_{0\leq t\leq T} \|w(t)\|^{\frac{2q}{q+1}}
_{L^{\frac{2q}{q+1}}(\Omega)}
 +\|\nabla w\|^2_{L^2(Q_T)}
 \leq C_{12}\big(1+\|w\|^{2}_{L^{2}(Q_T)}\big). \label{e319}
\end{equation}
 Recall that Lemma \ref{lem2.2} implies $v\in L^2(Q_T)$, so
$\|w\|_{L^{\frac{4}{q+1}}(Q_T)}\leq  C_{13}$.
 Since $\frac{4}{q+1}<2\leq\widetilde{q}$. Then we see from H\"older's
inequality
\begin{equation}
\|w\|^{2}_{L^{2}(Q_T)}\leq\|w\|^{2(1-\lambda)}_{L^{\widetilde{q}}(Q_T)}
\|w\|^{2\lambda}_{L^{\frac{4}{q+1}}(Q_T)}
\leq
C_{13}^{2\lambda}\|w\|^{2(1-\lambda)}_{L^{\widetilde{q}}(Q_T)},\label{e320}
\end{equation}
where $\lambda=(\frac{1}{2}-\frac{1}{\widetilde{q}})/(\frac{q+1}{4}
-\frac{1}{\widetilde{q}})$.
Setting $\widetilde{\beta}=2/(q+1)\in(0,1)$, we have
$\|w(.,t)\|_{L^{\widetilde{\beta}}(\Omega)}
 =\|v(.,t)\|^{\frac{1}{\widetilde{\beta}}}_{L^1(\Omega)}
\leq C_1(T)^{\frac{1}{\widetilde{\beta}}} $
for all $t\in[0,T)$ by
Lemma \ref{lem2.2}. Then it follow from \eqref{e319}, \eqref{e320} and
Lemma \ref{lem2.5} that
\begin{equation}
E\leq C_{14}(1+E^{\alpha})\label{e321}
\end{equation}
with
$$
\alpha=\frac{2(1-\lambda)}{\widetilde{q}}\big(\frac{2}{n}+1\big)<1.
$$
Thus \eqref{e321} implies
$$
\sup_{0\leq t\leq T}\|w(t)\|^{\frac{2q}{q+1}}_{L^{\frac{2q}{q+1}}(\Omega)}\leq
E\leq C_{15}
$$
with some $C_{15}>0$, let $r=q>1$, so that
$\sup_{0\leq t\leq T} \|v(t)\|_{L^{r}(\Omega)} \leq C_{T}$ and the
proof is complete.
\end{proof}

\section{Proof of Theorem \ref{thm1.1}}

The first step of the proof is to show $v$ is in $L^{r}(Q_T)$ for
any $r>1$.

\begin{lemma} \label{lem4.1}
 Let $\alpha_{11}>0$ and suppose that there are
$r_1>\max\{\frac{n+2}{2},3\}$ and a positive constant $C_{r_1,T}$
such that
$$
\|v\|_{L^{r_1}(Q_T)}\leq C_{r_1,T}.
$$
Then, $v$ is in $L^r(Q_T)$ for any $r>1$.
\end{lemma}

\begin{proof}
The proof is almost identical to \cite[Lemma 4.1]{Tuoc2}, but for
completeness we repeat it here.
First, the equation for $u$ can be written in the divergence form as
\begin{equation}
u_t=\nabla\cdot[(d_1+2\alpha_{11}u)\nabla u]+u(a_1-b_1u-c_1v),\label{e41}
\end{equation}
where $d_1+2\alpha_{11}u$ is bounded in $\overline{Q}_T$ by
 Lemma \ref{lem2.1} and $u(a_1-b_1u-c_1v)$ is in $L^{r_1}$
with $r_1>\frac{n+2}{2}$.
Application of the H\"older continuity result in
\cite[Theorem 10.1, p. 204]{Ladyzenskaja} to \eqref{e41} yields
\begin{equation}
u\in C^{\beta,\frac{\beta}{2}}(\overline{Q}_T) \quad\text{with some }
 \beta>0. \label{e42}
\end{equation}
Moreover, we have
\begin{equation}
w_{1t}=(d_1+2\alpha_{11}u)\Delta w_1+f_1, \label{e43}
\end{equation}
where $w_1=(d_1+\alpha_{11}u)u$ is as in the proof of Lemma \ref{lem2.3},
$f_1=(d_1+2\alpha_{11}u)u(a_1-b_1u-c_1v)$. Since $u$ is bounded  and
by the assumption of this Lemma, we see that $f_1$ is in
$L^{r_1}(Q_T)$. From \eqref{e42}, Lemma \ref{lem2.1} and
Proposition \ref{prop3.3},
applying
\cite[Theorem 9.1, pp. 341-342]{Ladyzenskaja} and its remark
\cite[P. 351]{Ladyzenskaja}, we
have
\begin{equation}
w_1\in W^{2,1}_{r_1}(Q_T).\label{e44}
\end{equation}
This implies $\nabla u=\frac{1}{d_1+2\alpha_{11}u}\nabla w_1$ in
$L^{r_1}(Q_T)$. Now, following the proof of Proposition \ref{prop3.3}
 with $r_1$ instead of $r_0$ and $p_1=\frac{2r_1}{r_1-2}$ instead of
$p_0$, we see that either $v$ is in $L^r(Q_T)$ for any $r>1$ or else
$v$ is in $L^{r_2}(Q_T)$ with
$$
r_2:=\frac{(n+1)r_1}{n+2-r_1}.
$$
The later case happens if and only if
$n+2-r_1>0$.

If $v$ is in $L^{r_2}(Q_T)$, we see that $f_1$ is in $L^{r_2}(Q_T)$.
Therefore, applying \cite[Theorem 9.1, p. 341-342]{Ladyzenskaja} and its
remark \cite[p. 351]{Ladyzenskaja} again, we have $\nabla u$ in $L^{r_2}(Q_T)$.
Then we go back
and do the same argument again. Keep doing likes this we will get a
sequence of numbers
\begin{equation}
r_{k+1}:=\frac{(n+1)r_k}{n+2-r_k}.\label{e45}
\end{equation}
We stop and get the conclusion that $v$ is in $L^r(Q_T)$ for any
$r>1$ when
\begin{equation}
n+2-r_k\leq0.\label{e46}
\end{equation}
Since $r_1>3$, from \eqref{e45} we can prove by induction that
$r_k>3,k=1,2,\dots $. Then, we have
\begin{equation}
\frac{r_{k+1}}{r_{k}}=\frac{n+1}{n+2-r_k}\geq\frac{n+1}{n-1}>1.\label{e47}
\end{equation}
Thus, the sequence $r_k$ is strictly increasing. Therefore, there
must be some $k$ such that \eqref{e46} holds. we stop at this $k$ and
conclude that $v$ is in $L^r(Q_T)$ for any $r>1$, namely, there is a
positive constant $C_{16}$ such that $\|v\|_{L^r(Q_T)}\leq C_{16}$,
where $C_{16}>0$ depending on $q, T, \Omega$ and the coefficients of
the system \eqref{e12} but not on $r$.
\end{proof}

So, from Proposition \ref{prop3.3} and Lemma \ref{lem4.1},
we have the following lemma.

\begin{lemma} \label{lem4.2}
 Let $\alpha_{22}>0$ and suppose $\mathrm{(i)}$
$\alpha_{11}=0$ or $\mathrm{(ii)}$ $\alpha_{11}>0$ and $n<10$. Then
there exists $M_2$ such that
$$
\|v\|_{L^r(Q_T)}\leq M_2 \quad\text{for any }  r>1.
$$
Moreover, for any $r>1$, $v$ is in $V_2(Q_T)$.
\end{lemma}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
We give the proof only in case $\alpha_{11}>0$ because the proof for
$\alpha_{11}=0$ is essentially the same.
By Lemma \ref{lem4.2}, $v$ is bounded in $\overline{Q}_T$. From \eqref{e43}, we
have
$$
w_{1t}=(d_1+2\alpha_{11}u)\Delta w_1+f_1,
$$
where $f_1=(d_1+2\alpha_{11}u)u(a_1-b_1u-c_1v)$ is bounded in
$\overline{Q}_T$ by Lemma \ref{lem2.1} and Lemma \ref{lem4.2}, $(d_1+2\alpha_{11}u)\in
C^{\beta,\frac{\beta}{2}}(Q_T)$ by \eqref{e42}. By
\cite[Theorem 9.1, p.341-342]{Ladyzenskaja}, we have
$$
\|w_1\|_{W^{2,1}_r}(Q_T)<M_3 \quad
\text{for }  \frac{n+2}{2}<r<\frac{4(n+1)}{(n-2)_+}.
$$
Hence it follows from \cite[Lemma 3.3, p.80]{Ladyzenskaja} that
\begin{equation}
w_1\in C^{1+\beta^*,\frac{(1+\beta^*)}{2}}(\overline{Q}_T),
 \quad \forall\,  0<\beta^*<1.\label{e48}
\end{equation}
Since $u=\frac{-d_1+\sqrt{d_1^2+4w_1\alpha_{11}}}{2\alpha_{11}}$, it
follow from \eqref{e48} that
\begin{equation}
u\in C^{1+\beta^*,\frac{(1+\beta^*)}{2}}(\overline{Q}_T),
 \quad \forall\,  0<\beta^*<1.\label{e49}
\end{equation}
Next, we rewrite the equation for $v$ in divergence form as
$$
v_t=\nabla\cdot[(d_2+\alpha_{21}u+2\alpha_{22}v)\nabla
v+\alpha_{21}v\nabla u]+f_2(x,t),
$$
where $f_2(x,t)=v(a_2+b_2u-c_2v)$, $u$, $v$ and $\nabla u$ are all
bounded functions because of Lemma \ref{lem2.1}, Lemma \ref{lem4.2}
 and \eqref{e49}. By
\cite[Theorem 10.1, p.204]{Ladyzenskaja}, we have
\begin{equation}
v\in C^{\sigma,\frac{\sigma}{2}}(\overline{Q}_T)
\text{with some}   0<\sigma<1.\label{e410}
\end{equation}
Now, we then return to the equation for $u$ and write it as
\begin{equation}
u_t=(d_1+2\alpha_{11}u)\Delta u+f_3(x,t),\label{e411}
\end{equation}
where $f_3(x,t)=2\alpha_{11}|\nabla u|^2+u(a_1-b_1u-c_1v)\in
C^{\sigma,\frac{\sigma}{2}}(\overline{Q}_T)$ by \eqref{e49} and
\eqref{e410}. Then the Schuader estimate in
\cite[Theorem 5.3, p.320-321]{Ladyzenskaja}
 applied to \eqref{e411} yields
\begin{equation}
u\in C^{2+\sigma^*,\frac{2+\sigma^*}{2}}(\overline{Q}_T)
\quad\text{with }   \sigma^*=\min\{\lambda,\sigma\}.\label{e412}
\end{equation}
Let $w_2=(d_2+\alpha_{21}u+\alpha_{22}v)v$. Then $w_2$ satisfies
\begin{equation}
w_{2t}=(d_2+\alpha_{21}u+2\alpha_{22}v)\Delta
w_2+f_4(x,t),\label{e413}
\end{equation}
where
$f_4(x,t)=(d_2+\alpha_{21}u+2\alpha_{22}v)v(a_2+b_2u-c_2v)
+\alpha_{21}vu_{t}\in C^{ \sigma^*,\frac{\sigma^*}{2}}(\overline{Q}_T)$
by \eqref{e411} and \eqref{e412},
$d_2+\alpha_{21}u+2\alpha_{22}v\in C^{\sigma,\frac{\sigma}{2}}(\overline{Q}_T)$
by \eqref{e49} and \eqref{e410},
by applying the Schuader estimate to the equation \eqref{e413}, we have
\begin{equation}
w_2\in C^{2+\sigma^*,\frac{2+\sigma^*}{2}}(\overline{Q}_T).\label{e414}
\end{equation}
Then
\begin{equation}
v=\frac{-(d_2+\alpha_{21}u)+\sqrt{(d_2+\alpha_{21}u)^2
+4w_2\alpha_{22}}}{2\alpha_{22}}\in
C^{2+ \sigma^*,\frac{2+\sigma^*}{2}}(\overline{Q}_T).\label{e415}
\end{equation}
Now repeat the procedure by making use of \eqref{e412} and \eqref{e415} in
place of \eqref{e49} and \eqref{e410}, we have
\begin{equation}
u, v\in C^{2+ \lambda,\frac{2+\lambda}{2}}(\overline{Q}_T).\label{e416}
\end{equation}
Finally, the estimates \eqref{e412} and \eqref{e415} imply that the
hypotheses of Theorem \ref{thmA} are satisfied. So that $(u,v)$ exists
globally in time. The proof of Theorem \ref{thm1.1} is now complete.
\end{proof}

\section{Stability}

 In this section, we discuss global asymptotic
stability of positive equilibrium point
$(\overline{u},\overline{v})$ for \eqref{e12}, namely to prove
Theorem \ref{thm1.2}.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
Define the  Lyapunov function:
$$
H(u,v)=\int_\Omega\big[(u-\bar{u}-\bar{u}\ln\frac{u}{\bar{u}})
+\rho(v-\bar{v}-\bar{v}\ln\frac{v}{\bar{v}})\big]dx,
$$
where $\rho=(b_2c_1+2b_1c_2)b_2^{-2}$. Obviously, $H(u,v)$ is
nonnegative and $H(u,v)=0$ if and only if
$(u,v)=(\overline{u},\overline{v})$. By Theorem \ref{thm1.1},
$H(u,v)$ is well-posed for $t\geq0$ if $(u,v)$ is positive solution
to system \eqref{e12}. The time derivative of $H(u,v)$ for
system \eqref{e12} satisfies
\begin{align*}
&\frac{d H(u,v)}{dt}\\
&= \int_{\Omega}\big(\frac{u-\bar{u}}{u}u_{t}+\rho\frac{v-\bar{v}}{v}v_{t}
\big)dx \\
&= \int_{\Omega}\{\frac{u-\bar{u}}{u}\nabla\cdot[(d_{1}+2\alpha_{11}u)\nabla
u]+(u-\bar{u})(a_{1}-b_{1}u-c_{1}v) \\
 &\quad +\rho\frac{v-\bar{v}}{v}\nabla\cdot[(d_{2}+\alpha_{21}u+2\alpha_{22}v)\nabla v+\alpha_{21}v\nabla u]
 +\rho(v-\bar{v})(a_{2}+b_{2}u-c_{2}v)\}dx \\
&= -\int_{\Omega}\big[\frac{(d_{1}+2\alpha_{11}u)\bar{u}}{u^{2}}|\nabla u|^{2}
 +\frac{\rho\alpha_{21}\bar{v}}{v}\nabla u\cdot\nabla v
 +\frac{\rho(d_{2}+\alpha_{21}u+2\alpha_{22}v)\bar{v}}{v^{2}}
 |\nabla v|^{2}\big]dx \\
&\quad -\int_{\Omega}[b_{1}(u-\bar{u})^{2}+(c_{1}-\rho b_{2})(u-\bar{u})(v-\bar{v})
 +c_{2}\rho(v-\bar{v})^{2} ]dx.
\end{align*}
The second integrand in the above equality  is positive definite by the
choice of $\rho$. Meanwhile the first integrand is positive semi-definite if
\begin{equation}
4\rho\overline{u}\overline{v}(d_1+2\alpha_{22}u)(d_2+\alpha_{21}u
+2\alpha_{22}v)>u^2(\alpha_{21}\overline{v})^2.\label{e52}
\end{equation}
By the Lemma \ref{lem2.1} and Theorem \ref{thm1.1}, the condition \eqref{e14} implies
\eqref{e52}. Therefore, when  all conditions in Theorem \ref{thm1.2} hold,
there exists positive constant $\delta$ depending on $b_1, b_2, c_1$
and $c_2$ such that
\begin{equation}
\frac{dH(u,v)}{dt}\leq
-\delta\int_{\Omega}[(u-\bar{u})^{2}+(v-\bar{v})^{2}]dx.\label{e53}
\end{equation}
To obtain the uniform convergence of the solution to
\eqref{e12}, we recall the following result which can be find in
\cite{Wang}.

\begin{lemma} \label{lem5.1}
Let $a$ and $b$ positive constant. Assume that
$\varphi,\psi\in C^1[a,+\infty)$, $\psi(t)\geq0$, $\varphi$ is
bounded. If $\varphi'(t)\leq-b\psi(t)$ and $\psi'(t)$ is bounded in
$[a,+\infty)$, then $\lim_{t\to\infty}\psi(t)=0$.
\end{lemma}

Using integration by parts, H\"older's inequality,
Lemma \ref{lem2.1}, and Lemma \ref{lem4.2}, one can easily verify that
$\frac{d}{dt}\int_{\Omega}[(u-\bar{u})^{2}+(v-\bar{v})^{2}]dx$ is
bounded from above. Then from Lemma \ref{lem5.1} and \eqref{e53}, we have
$$
\|u(\cdot,t)-\overline{u}\|_{L^{\infty}(\Omega)}\to 0,\quad
\|v(\cdot,t)-\overline{v}\|_{L^{\infty}(\Omega)}\to0\quad(t\to\infty).
$$
Namely, $(u,v)$ converges uniformly to $(\overline{u},\overline{v})$.
By the fact that $H(u,v)$ is decreasing for $t\geq0$, it is obvious that
$(\overline{u},\overline{v})$ is global asymptotic stable, and the
proof of Theorem \ref{thm1.2} is complete.
\end{proof}

\subsection*{Acknowledgements}
 The author would like to thank professor Sheng-mao Fu
for the encouragement and useful discussions,
also the anonymous referee for the very careful reading
of the original manuscript and the helpful suggestions.


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