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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 146, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2015/146\hfil Attraction-repulsion chemotaxis system]
{Asymptotic behavior for small mass in an attraction-repulsion
  chemotaxis system}

\author[Y. Li, K. Lin, C. Mu \hfil EJDE-2015/146\hfilneg]
{Yuhuan Li, Ke Lin, Chunlai Mu}

\address{Yuhuan Li \newline
Department of Mathematics, Sichuan Normal University,
Chengdu 610066,  China}
\email{liyuhuanhuan@163.com}

\address{Ke Lin (corresponding author) \newline
College of Mathematics and Statistics,
Chongqing University, Chongqing 401331, China}
\email{shuxuelk@126.com}

\address{Chunlai Mu \newline
College of Mathematics and Statistics,
Chongqing University, Chongqing 401331, China}
\email{clmu2005@163.com}

\thanks{Submitted April 17, 2015. Published June 6, 2015.}
\subjclass[2010]{35A01, 35B40, 35K55, 92C17}
\keywords{Chemotaxis; attraction-repulsion; boundedness; convergence}

\begin{abstract}
 This article is concerned with the  model
 \begin{gather*}
 u_t=\Delta u-\nabla\cdot(\chi u\nabla v)+\nabla\cdot(\xi u\nabla w),\quad
 x\in \Omega,\; t>0,\\
 0=\Delta v+\alpha u-\beta v,\quad x\in\Omega,\; t>0,\\
 0=\Delta w+\gamma u-\delta w,\quad x\in\Omega,\; t>0
 \end{gather*}
 with homogeneous Neumann boundary conditions in a bounded domain
 $\Omega\subset \mathbb{R}^{n}$ $(n=2,3)$. Under the critical condition
 $\chi \alpha-\xi \gamma=0$, we show that the system  possesses a unique
 global solution that is uniformly bounded in time. Moreover, when $n=2$,
 by  some appropriate smallness conditions on the initial data, we
 assert that this solution converges to
 ($\bar{u}_0$, $\frac{\alpha}{\beta}\bar{u}_0$,
 $\frac{\gamma}{\delta}\bar{u}_0$) exponentially,
 where $\bar{u}_0:=\frac{1}{|\Omega|}\int_{\Omega}u_0$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

Chemotaxis is a phenomenon of the directed movement of cells in response to
the concentration gradient of the chemical which is produced by cells.
A well-known chemotaxis model was  proposed by Keller and Segel \cite{ke}
 in the 1970s, which describes the aggregation of cellular slime
 molds Dictyostelium discoideum. A simple classical Keller-Segel  model
reads as follows
\begin{equation}{\label{1.1}}
\begin{gathered}
u_t=\Delta u- \nabla\cdot(\chi u\nabla v),\quad x\in \Omega, \; t>0,\\
\tau v_t=\Delta v+\alpha u-\beta v,\quad x\in \Omega, \; t>0,\\
\frac{\partial u}{\partial\nu }=\frac{\partial v}{\partial \nu}=0,\quad
x\in\partial\Omega, \; t>0,\\
u(x,0)=u_0(x),\tau v(x,0)=\tau v_0(x),\quad x\in \Omega,
\end{gathered}
\end{equation}
where  $u=u(x,t)$ and $v=v(x,t)$ denote the density of the cells  and
the concentration of the chemical, respectively.
Here $\alpha>0$, $\beta>0$, $\tau=0,1$ are constants, and $\chi>0$
(resp. $\chi<0$) is a constant referred to as the attractive (resp. repulsive)
chemotaxis.

Mathematical study of \eqref{1.1} has been extensively developed
in the past four decades, see [8-10] and the references therein.
In the case $\chi>0$, the outcome in \cite{os} states that a globally bounded
solution of \eqref{1.1} with $\tau=1$ exists when $n=1$.
When $n=2$, it is shown that there exists a critical constant $C$
such that if $\int_{\Omega}u_0<C$, then the solutions of \eqref{1.1}
are bounded \cite{ga,na2} and if $\int_{\Omega}u_0>C$, then
blow-up happens \cite{ho2,na1,se}. When $n\geq 3$, it is insufficient
to rule out blow up in \eqref{1.1} even if $\int_{\Omega}u_0$
is sufficiently small \cite{ca,wi1,wi2}. On the other hand,
the results of repulsive chemotaxis (i.e., $\chi<0$ ) were much less.
For $\tau=0$, it is well known that the solutions of \eqref{1.1}
are uniformly bounded and converge to some stationary solutions
exponentially as time tends to infinity \cite{mo1,mo2}. In \cite{ci}
 the system \eqref{1.1} with $\tau=1$ has been studied based on a Lyapunov function.
It is asserted that \eqref{1.1} possesses a unique classical bounded
solution in two dimensions and a global weak solution exists if $n=3, 4$.

Taking into account attraction and repulsion together, we can get the following
attraction-repulsion system
\begin{equation}\label{1.2}
\begin{gathered}
u_t=\Delta u-\nabla\cdot(\chi u\nabla v)+\nabla\cdot(\xi u\nabla w),
\quad x\in \Omega, \; t>0,\\
\tau v_t=\Delta v+\alpha u-\beta v,\quad x\in\Omega,\; t>0,\\
\tau w_t=\Delta w+\gamma u-\delta w,\quad x\in\Omega, \; t>0,\\
\frac{\partial u}{\partial\nu }
=\frac{\partial v}{\partial \nu}
=\frac{\partial w}{\partial \nu}=0,\quad x\in\partial\Omega, \; t>0,\\
u(x,0)=u_0(x),\; \tau v(x,0)=\tau v_0(x),\; \tau w(x,0)=\tau w_0(x),\quad x\in \Omega
\end{gathered}
\end{equation}
for cell density $u$, concentration of an attractive signal $v$,
 and concentration of a repulsive signal $w$, respectively, where
$\chi,\xi,\alpha,\beta,\gamma$ and $\delta$ are positive and $\tau=0,1$.

Model \eqref{1.2} with $\tau=1$ was proposed in \cite{pa} to describe
the quorum effect in the chemotaxis process, and in \cite{lu}
to describe the aggregation of microglia
in Alzheimer's disease. In the one-dimensional framework, the resulting
variant of \eqref{1.2} with $\tau=1$ was proved to have global solutions
 in \cite{li3}, and large time behavior was obtained in \cite{ji2}
for all $\alpha>0$ and $\beta>0$.
Moreover, the time-periodic solution of \eqref{1.2}
was studied in \cite{li4} for various ranges of parameter values. Since chemical
diffuses faster than cells, it is valuable to consider \eqref{1.2} with $\tau=0$.
Especially in \cite{ta1},
by using the following transformation
\begin{equation} \label{1.3}
s:=\chi v-\xi w,
\end{equation}
Equation \eqref{1.2} can be changed into the general classical Keller-Segel
model \eqref{1.1} for the special case $\beta=\delta$.
Thus under some additional assumptions on the parameters, the global existence,
blow-up, stationary solutions and large-time behavior of \eqref{1.2} with
$\tau=0,1$ were considered in \cite{ta1} by using a number of mathematical
techniques.
But for the case of  $\beta \neq \delta$ in higher dimensions,
it becomes more challenging because there does not exist a Lyapunov functional
for \eqref{1.2}. The first result of this case has  been also found in \cite{ta1},
where global existence was asserted in any bounded domain
$\Omega\subset \mathbb{R}^n(n\geq2)$ if $\chi \alpha-\xi \gamma<0$ and $\tau=0$.
When $\tau=1$, global existence of weak solutions to \eqref{1.2} was
obtained in three dimensions \cite{ji1}. Recently, some further information
on the existence of bounded solutions or on the occurrence of blow-up
has been explored in \cite{ci,li5,li1,li2} in a bounded domain
$\Omega\subset \mathbb{R}^2$.

In this article we focus on \eqref{1.2} with $\tau=0$ for the
cases $\chi \alpha=\xi \gamma$ and  $\beta\not\equiv \delta$.
As for the initial data $u_0$, we may assume that
\begin{equation} \label{1.5}
u_0\in C^{0}(\bar{\Omega}),\quad u_0>0\quad \text{in }\bar{\Omega}.
\end{equation}
To study \eqref{1.2} directly, we turn \eqref{1.2} into the
initial-boundary value problem
\begin{equation} \label{1.6}
\begin{gathered}
u_t=\Delta u-\nabla\cdot( u \nabla s),\quad x\in \Omega,\; t>0,\\
0=\Delta s-\delta s+(\chi \alpha-\xi \gamma)u+\chi(\delta-\beta)v,\quad
 x\in\Omega,\; t>0,\\
0=\Delta v+\alpha u-\beta v,\quad x\in\Omega,; t>0,\\
\frac{\partial u}{\partial\nu }=\frac{\partial s}{\partial \nu}
=\frac{\partial v}{\partial \nu}=0,\quad x\in\partial\Omega,\; t>0,\\
u(x,0)=u_0(x),\quad x\in \Omega\,,
\end{gathered}
\end{equation}
by using the same transformation \eqref{1.3} given in \cite{ta1}.
Firstly, our result involving global existence is stated as follows.

\begin{theorem} \label{thm1.1}
Let $\Omega \subset \mathbb{R}^n(n=2,3)$ be a bounded domain with smooth
boundary $\partial\Omega$ and $\tau=0$.
Assume that
\begin{equation}{\label{1.7}}
\chi \alpha-\xi \gamma=0.
\end{equation}
Then for all  $u_0$ satisfying \eqref{1.5}, \eqref{1.2} possesses a
unique classical solution $(u,v,w)$ which is global in time and uniformly
 bounded in $\Omega\times(0,\infty)$.
\end{theorem}

Secondly, for all positive $\beta$ and $\delta$, inspired by \cite{wi3},
under some suitable smallness on $u_0$, we have the following result.

\begin{theorem} \label{thm1.2}
Let $n=2$, and let $\tau=0$. Suppose that \eqref{1.7} holds.
Given some $u_0$ fulfilling \eqref{1.5}, one can find some $\epsilon_0>0$
such that if
\begin{equation} \label{1.8}
m:=\int_{\Omega}u_0\leq \epsilon
\end{equation}
holds for all $0<\epsilon<\epsilon_0$, then the unique global solution of
\eqref{1.2} satisfies
\begin{equation}{\label{1.9}}
\begin{gathered}
\|u(\cdot,t)-\bar{u}_0\|_{L^{\infty}(\Omega)} \to 0,\\
\|v(\cdot,t)-\frac{\alpha}{\beta}\bar{u}_0\|_{L^{\infty}(\Omega)} \to 0,\\
\|w(\cdot,t)-\frac{\gamma}{\delta}\bar{u}_0\|_{L^{\infty}(\Omega)} \to 0,\\
\end{gathered}
\end{equation}
as $t\to \infty$, where $\bar{u}_0:=\frac{1}{|\Omega|}\int_{\Omega}u_0$.
\end{theorem}

\begin{remark} \label{rmk1.1} \rm
Theorem 1.2 shows that the asymptotic behavior of solutions to \eqref{1.2}
are very similar to the special case $\beta=\delta$ in \cite{ta1}.
Unfortunately, the question of global dynamics for arbitrarily large $m$
has to be left as an open problem here.
\end{remark}

\section{Preliminaries}

 Before proving the main results in this article, we state some basic
and useful properties in this section. We start with the
local-in-time existence of a classical solution to \eqref{1.2}
with $\tau=0$ which has been proved in \cite{ta1}.

\begin{lemma} \label{lem2.1}
For any nonnegative function $u_0\in C^{0}(\bar{\Omega})$,
there exist $T_{\rm max}\in(0,\infty]$  and a unique triple
$(u,v,w)\in C^0(\bar{\Omega}\times{[0,T_{\rm max})})
\cap C^{2,1}( \bar{\Omega}\times(0,T_{\rm max}))$
solving \eqref{1.2} with $\tau=0$ classically.
Moreover, if $T_{\rm max}<\infty$, then
\begin{equation}{\label{2.1}}
\|u(\cdot,t)\|_{L^{\infty}(\Omega)}\to \infty\quad \text{as }t\to  T_{\rm max}.
\end{equation}
\end{lemma}

The following  properties immediately result from an integration of each
equation in \eqref{1.2} with respect to $x\in\bar{\Omega}$, and from
the maximum principle.

\begin{lemma} \label{lem2.2}
Suppose $u_0$ satisfies \eqref{1.5}. Then the solution $(u,v,w)$ of
\eqref{1.2} with $\tau=0$ satisfies
\begin{equation}{\label{2.2}}
\begin{gathered}
\|u(\cdot,t)\|_{L^{1}(\Omega)}=\|u_0\|_{L^{1}(\Omega)}\quad
\text{for all } t\in(0,T_{\rm max}), \\
\|v(\cdot,t)\|_{L^{1}(\Omega)}= \frac{\alpha}{\beta}
\|u_0\|_{L^{1}(\Omega)}\quad \text{for all } t\in(0,T_{\rm max}),\\
\|w(\cdot,t)\|_{L^{1}(\Omega)}=\frac{\gamma}{\delta}
\|u_0\|_{L^{1}(\Omega)}\quad \text{for all }t\in(0,T_{\rm max}).
\end{gathered}
\end{equation}
Moreover,
\begin{equation}{\label{2.3}}
u>0,\quad v>0,\quad w>0\quad \text{in }\bar{\Omega}\times(0,T_{\rm max}).
\end{equation}
\end{lemma}


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

A crucial step towards our boundedness proof will be provided by the following lemma.

\begin{lemma} \label{lem3.1}
Assume that \eqref{1.7} holds, and that $\Omega$ is a bounded domain in
$\mathbb{R}^n$ ($n=2,3$). For any $r> 10/3$, there exists some constant
$C>0$ such that
\begin{equation}{\label{3.1}}
\int_{\Omega}u^{r}(x,t)dx\leq C\quad \text{for all }t\in(0,T_{\rm max}).
\end{equation}
\end{lemma}

\begin{proof} Multiplying $u^{r-1}$ to the first equation in \eqref{1.6}
and integrating by parts, we have
\begin{equation} \label{3.2}
\frac{1}{r}\frac{d}{dt}\int_{\Omega}u^{r}
=-\frac{4(r-1)}{r^{2}}\int_{\Omega}|\nabla u^{r/2}|^2
+\frac{r-1}{r}\int_{\Omega}\nabla u^{r}\cdot\nabla s
\end{equation}
for all  $t\in(0,T_{\rm max})$.
On the other hand, multiplying the second equation in \eqref{1.6} by $u^{r}$,
we get that
\begin{equation}{\label{3.3}}
\begin{split}
\int_{\Omega}\nabla u^{r}\cdot\nabla s
&=-\delta\int_{\Omega}u^{r}s+\chi(\delta-\beta)\int_{\Omega}u^{r}v.\\
&=-\delta\int_{\Omega}u^{r}(\chi v-\xi w)+\chi(\delta-\beta)\int_{\Omega}u^{r}v\\
&=\xi\delta\int_{\Omega}u^{r}w-\chi\beta\int_{\Omega}u^{r}v\quad
\text{for all }t\in(0,T_{\rm max}).
\end{split}
\end{equation}
Noting that $u(x,t)>0$ and $v(x,t)>0$ for all $x\in\bar{\Omega}$ and
$t\in(0,T_{\rm max})$, then combining \eqref{3.2} and \eqref{3.3} yields
\begin{equation}{\label{3.4}}
\frac{d}{dt}\int_{\Omega}u^{r}+\frac{4(r-1)}{r}\int_{\Omega}
 |\nabla u^{r/2}|^2
\leq \xi\delta(r-1)\int_{\Omega}u^{r}w\quad \text{for all }
t\in(0,T_{\rm max}).
\end{equation}
By the Gagliardo-Nirenberg inequality, there exist some constants $C_1>0$
and $C_2>0$ satisfying
\begin{equation}{\label{3.5}}
\begin{split}
\int_{\Omega}u^{\frac{rn+2}{n}}
&=\|u^{r/2}\|^{\frac{2rn+4}{rn}}_{L^{\frac{2rn+4}{rn}}(\Omega)}\\
&\leq C_1 \|\nabla u^{r/2}\|^{\frac{2rn+4}{rn}
  a_1}_{L^2(\Omega)} \|u^{r/2}\|^{\frac{2rn+4}{rn}
 \cdot (1-a_1)}_{L^{\frac{2}{r}}(\Omega)}
+ C_1 \|u^{r/2}\|^{\frac{2rn+4}{rn}}_{L^{\frac{2}{r}}(\Omega)}\\
&\leq C_2 \|\nabla u^{r/2}\|^{2}_{L^2(\Omega)}+C_2\quad
\text{for all } t\in(0,T_{\rm max}),
\end{split}
\end{equation}
where
\begin{equation*}{\label{3.6}}
a_1=\frac{rn}{rn+2}\in(0,1).
\end{equation*}
On the other hand, applying  Young's inequality to \eqref{3.4}, we infer that
\begin{equation}{\label{3.7}}
\frac{d}{dt}\int_{\Omega}u^{r}+\frac{4(r-1)}{r}
 \int_{\Omega}|\nabla u^{r/2}|^2
\leq \frac{r-1}{rC_2}\int_{\Omega}u^{\frac{rn+2}{n}}
 +C_3\int_{\Omega}w^{\frac{rn+2}{2}}
\end{equation}
for all $t\in(0,T_{\rm max})$, where
\begin{equation*} \label{3.8}
C_3=\xi\delta(r-1)\Big(\frac{1}{\xi\delta rC_2}\cdot \frac{rn+2}{rn}\Big)
^{-rn/2}\Big(\frac{rn+2}{2}\Big)^{-1}.
\end{equation*}
To estimate the second term on the right-hand side of \eqref{3.7},
noting that $w$ satisfies
\begin{equation} \label{3.9}
\begin{gathered}
0=\Delta w+\gamma u-\delta w,\quad x\in\Omega, \; t\in(0,T_{\rm max}),\\
\frac{\partial w}{\partial \nu}=0,\quad x\in\partial\Omega, \;t\in(0,T_{\rm max}),
\end{gathered}
\end{equation}
then testing \eqref{3.9} by  $w^{\frac{rn}{2}}$  and applying Young's
inequality again, we immediately obtain
\begin{equation} \label{3.10}
\begin{split}
&\frac{8rn}{(rn+2)^2}\int_{\Omega}|\nabla w^{\frac{rn+2}{4}}|^2
 +\delta \int_{\Omega}w^{\frac{rn+2}{2}}\\
&=\gamma\int_{\Omega}uw^{\frac{rn}{2}}\\
&\leq\frac{\delta(r-1)}{ rC_2C_3}\int_{\Omega}u^{\frac{rn+2}{n}}
+C_{4}\int_{\Omega}w^{\frac{rn(rn+2)}{2(rn-n+2)}},
\end{split}
\end{equation}
where
\[  %\label{3.11}
C_4=\gamma\Big(\frac{\delta(r-1)}{\gamma rC_2C_3} \frac{rn+2}{n}
\Big)^{-\frac{n}{rn-n+2}}\Big(\frac{rn+2}{rn-n+2}\Big)^{-1}.
\]
We use the Gagliardo-Nirenberg inequality to estimate
\begin{equation}{\label{3.12}}
\begin{split}
\int_{\Omega}w^{\frac{rn(rn+2)}{2(rn-n+2)}}
&= \|w^{\frac{rn+2}{4}}\|^{\frac{2rn}{rn-n+2}}_{L^{\frac{2rn}{rn-n+2}}(\Omega)}\\
&\leq C_{5}\|\nabla w^{\frac{rn+2}{4}}\|^{\frac{2rn}{rn-n+2} a_2}_{L^{2}(\Omega)}
\|w^{\frac{rn+2}{4}}\|^{\frac{2rn}{rn-n+2}(1-a_2)}_{L^{\frac{4}{rn+2}}(\Omega)}\\
&\quad +C_5\|w^{\frac{rn+2}{4}}\|^{\frac{2rn}{rn-n+2}}_{L^{\frac{4}{rn+2}}(\Omega)}
\quad \text{for all } t\in(0,T_{\rm max})
\end{split}
\end{equation}
with some constant $C_5>0$ and $a_2$ determined by
\[ %\label{3.13}}
\frac{rn-n+2}{2rn}=\big(\frac{1}{2}-\frac{1}{n}\big) a_2+\frac{rn+2}{4} (1-a_2).
\]
Thus $a_2$ satisfies
\begin{gather*}
a_2=\frac{r^2n^2+2n-4}{(rn^2+4)r}\in(0,1), \\
\frac{2rn}{rn-n+2} a_2=\frac{2rn}{rn-n+2} \frac{r^2n^2+2n-4}{(rn^2+4)r}< 2
\end{gather*}
because $r>\frac{10}{3}$ and $n=2,3$.
By Young's inequality, \eqref{3.12} becomes
\begin{equation}{\label{3.16}}
\begin{split}
\int_{\Omega}w^{\frac{rn(rn+2)}{2(rn-n+2)}}
&\leq C_6\Big(\int_{\Omega}|\nabla w^{\frac{rn+2}{4}}|^2\Big)^{\frac{rn}{rn-n+2}\cdot \frac{r^2n^2+2n-4}{(rn^2+4)r}}+C_6\\
&\leq \epsilon\int_{\Omega}|\nabla w^{\frac{rn+2}{4}}|^2+C_{7}(\epsilon)
\quad \text{for all }t\in(0,T_{\rm max})
\end{split}
\end{equation}
with constants $C_6>0$ and $C_{7}(\epsilon)>0$, where we take
$\epsilon=\frac{4rn}{(rn+2)^2C_4}$.
Inserting \eqref{3.16} into \eqref{3.10}, we find some constant
$C_{8}>0$ satisfying
\begin{equation}{\label{3.17}}
\begin{split}
\int_{\Omega}w^{\frac{rn+2}{2}}
\leq \frac{r-1}{ rC_2C_3}\int_{\Omega}u^{\frac{rn+2}{n}}+C_{8}\quad
\text{for all } t\in(0,T_{\rm max}).
\end{split}
\end{equation}
As a consequence of \eqref{3.17} and \eqref{3.5}, \eqref{3.7} can be turned
into the inequality
\begin{align*} %\label{3.18}
\frac{d}{dt}\int_{\Omega}u^{r}+\frac{4(r-1)}{r}
 \int_{\Omega}|\nabla u^{r/2}|^2
&\leq \frac{2(r-1)}{rC_2}\int_{\Omega}u^{\frac{rn+2}{n}}+C_{9}\\
&\leq \frac{2(r-1)}{rC_2}\Big(C_2\int_{\Omega}|\nabla u^{r/2}|^2+C_2\Big)
+C_{9}
\end{align*}
for all $t\in(0,T_{\rm max})$
with $C_9>0$. Therefore, we can pick $C_{10}>0$ to obtain
\begin{equation}{\label{3.20}}
\frac{d}{dt}\int_{\Omega}u^{r}+\int_{\Omega}u^r
\leq-\frac{2(r-1)}{r}\int_{\Omega}|\nabla u^{r/2}|^2
+\int_{\Omega}u^r+C_{10}
\end{equation}
for all $t\in(0,T_{\rm max})$.
It follows from the Gagliardo-Nirenberg inequality that
\begin{equation}\label{3.21}
\begin{split}
\int_{\Omega}u^r
&=\|u^{r/2}\|^{2}_{L^2(\Omega)}\\
&\leq C_{11}\|\nabla u^{r/2}\|^{2 a_3}_{L^2(\Omega)}
\|u^{r/2}\|^{2 (1-a_3)}_{L^{\frac{2}{r}}(\Omega)}
+C_{11}\|u^{r/2}\|^2_{L^{\frac{2}{r}}(\Omega)}\\
&\leq C_{12}\|\nabla u^{r/2}\|^{2 a_3}_{L^2(\Omega)}+C_{12}
\quad \text{for all } t\in(0,T_{\rm max})
\end{split}
\end{equation}
for some constants $C_{11}>0$ and $C_{12}>0$, where
\[ %\label{3.22}}
a_3=\frac{rn-n}{rn-n+2}\in(0,1).
\]
Inserting \eqref{3.21} in \eqref{3.20} and by Young's inequality,
there exists some constant $C_{13}>0$ such that
\[
\frac{d}{dt}\int_{\Omega}u^{r}+\int_{\Omega}u^r\leq C_{13}\quad
\text{for all } t\in(0,T_{\rm max}),
\]
which leads to
\[
\|u(\cdot,t)\|_{L^{r}(\Omega)}\leq C_{14}\quad \text{for all } t\in(0,T_{\rm max})
\]
with some constant $C_{14}>0$. The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
Since $v$ satisfies
\begin{gather*}
0=\Delta v-\beta v+\alpha u,\quad x\in\Omega,\; t\in(0,T_{\rm max}),\\
\frac{\partial v}{\partial \nu}=0,\quad x\in\partial\Omega,\; t\in(0,T_{\rm max}),
\end{gather*}
then applying the Agmon-Douglis-Nirenberg $L^{r}$ estimates 
\cite{ag1,ag2} on linear elliptic equations with homogeneous Neumann 
boundary condition, there provides some constant $C_1>0$ satisfying
\[
\|v(\cdot,t)\|_{W^{2,r}(\Omega)}
\leq C_{1}\|u(\cdot,t)\|_{L^{r}(\Omega)}\quad
\text{for all }t\in(0,T_{\rm max}).
\]
Now from Lemma \ref{lem3.1} and using the Sobolev embedding:
$W^{2,r}(\Omega)\hookrightarrow
C^1_{B}(\Omega):=\{u\in C^1(\Omega)|Du\in L^{\infty}(\Omega)\}$ if
$r>n$ \cite{gi}, we find
\begin{equation} \label{3.27}
\|\nabla v(\cdot,t)\|_{L^{\infty}(\Omega)}\leq C_2\quad \text{for all }
t\in(0,T_{\rm max})
\end{equation}
with some constant $C_2>0$. Similarly, we can pick some constant $C_3>0$ such that
\begin{equation*}{\label{3.28}}
\begin{split}
\|\nabla w(\cdot,t)\|_{L^{\infty}(\Omega)}\leq C_3\quad\text{for all }
t\in(0,T_{\rm max}).
\end{split}
\end{equation*}
In view of the variation-of-constants formula to the first equation in \eqref{1.2},
we can see that
\begin{align*}
u(\cdot,t)&=e^{t\Delta}u_0-\chi\int^{t}_0e^{(t-\sigma)\Delta}
\nabla\cdot( u(\cdot,\sigma)\nabla v(\cdot,\sigma))d\sigma\\
&\quad+\xi\int^{t}_0e^{(t-\sigma)\Delta}\nabla\cdot(u(\cdot,\sigma)
\nabla w(\cdot,\sigma))d\sigma\\
&=:I_1(\cdot,t)+I_2(\cdot,t)+I_3(\cdot,t)\quad \text{for all }t\in(0,T_{\rm max}).
\end{align*}
As an easy consequence of the smoothing estimates for the Neumann heat semigroup,
we immediately obtain
\[
\|I_1(\cdot,t)\|_{L^{\infty}(\Omega)}\leq \|u_0\|_{L^{\infty}(\Omega)}\quad
\text{for all } t\in(0,T_{\rm max}).
\]
Applying the known  smoothing estimates from \cite{wi2} (see also \cite{ca}),
for some $C_4>0$ we have
\begin{align*} %\label{3.31}
\|I_2(\cdot,t)\|_{L^{\infty}(\Omega)}
&\leq \chi\int^{t}_0\|e^{(t-\sigma)\Delta}\nabla\cdot( u(\cdot,\sigma)
\nabla v(\cdot,\sigma))\|_{L^{\infty}(\Omega)}d\sigma\\
&\leq  C_4\int^{t}_0\left(1+(t-\sigma)^{-\frac{1}{2}
-\frac{n}{2r}}\right)e^{-\lambda_1(t-\sigma)}\| u(\cdot,\sigma)
\nabla v(\cdot,\sigma)\|_{L^{r}(\Omega)}d\sigma
\end{align*}
for all $t\in(0,T_{\rm max})$, where $\lambda_1>0$ denotes the
first eigenvalue of $-\Delta$ in $\Omega$ under Neumann boundary conditions.
For any $r>n$, according to \eqref{3.27} and the boundedness
of $u(\cdot,t)$ in $L^{r}(\Omega)$ asserted by
Lemma \ref{lem3.1}, this yields $C_5>0$ such that
\begin{align*}
&\|I_2(\cdot,t)\|_{L^{\infty}(\Omega)}\\
&\leq  C_4\int^{t}_0\Big(1+(t-\sigma)^{-\frac{1}{2}-\frac{n}{2r}}\Big)
 e^{-\lambda_1(t-\sigma)}\|u(\cdot,\sigma)\|_{L^{r}(\Omega)}
\|\nabla v(\cdot,\sigma)\|_{L^{\infty}(\Omega)}d\sigma\\
&\leq  C_5\int^{t}_0\Big(1+\upsilon^{-\frac{1}{2}-\frac{n}{2r}}\Big)
 e^{-\lambda_1\upsilon}d\upsilon\\
&\leq  C_6\quad \text{for all } t\in(0,T_{\rm max}).
\end{align*}
It is similar to deal with $I_3$, that means
\[
\|I_3(\cdot,t)\|_{L^{\infty}(\Omega)} \leq  C_7\quad \text{for all }
t\in(0,T_{\rm max})
\]
holds with some constant $C_7>0$. Therefore, the maximal existence
time $T_{\rm max}$ of solutions to \eqref{1.2} must be infinite by means
of Lemma \ref{lem2.1} and we finish our proof.
\end{proof}


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

\subsection{A bound for $u$}
To avoid confusion, through this section, we should state that the
constants $c_i$ and $C_i$ $(i=1,2,\ldots)$ are independent of the total mass
$\int_{\Omega}u_0$.

\begin{lemma} \label{lem4.1}
Suppose $\Omega \subset \mathbb{R}^2$ is a bounded domain with smooth boundary
$\partial\Omega$. Then for all $\alpha>0$ and $\beta>0$, the solution $v$ of
\begin{equation}{\label{4.1}}
\begin{gathered}
0=\Delta v-\beta v+\alpha u,\quad x\in\Omega,\\
\frac{\partial v}{\partial \nu}=0,\quad x\in\partial\Omega
\end{gathered}
\end{equation}
satisfies
\begin{gather} \label{4.2}
\|v\|_{L^{p}(\Omega)}\leq  \alpha C_p\|u\|_ {L^{1}(\Omega)}\quad
\text{for all } p\in(1,\infty), \\
\label{4.3}
 \|\nabla v\|_{L^{q}(\Omega)}\leq \alpha C_q\|u\|_{L^{2}(\Omega)}\quad
\text{for all }q\in(1,\infty),
\end{gather}
where $C_p$ (resp. $C_q$) is a positive constant depending on $p$
(resp. $q$).
\end{lemma}

\begin{proof}
From \eqref{4.1},  $v$ can be represented as
\[
 v(x)=\alpha\int_{\Omega}G(x,y)u(y)dy,\quad  \text{a.e. } x\in\Omega,
\]
where $G(x,y)$ is the Green function of $-\Delta+\beta$ in $\Omega$
subject to homogeneous Neumann boundary conditions (see 
\cite{na1,it,ta2}).
Noting that $G(x,y)$ satisfies
\[
|G(x,y)|\leq C\Big(1+\ln \frac{1}{|x-y|}\Big),\quad
|\nabla_x G(x,y)|\leq \frac{C}{|x-y|}\quad
\text{for all }x,y\in\Omega\text{ with }x\neq y
\]
with some constant $C>0$, by means of  Young's inequality for convolutions
we easily arrive at \eqref{4.2}-\eqref{4.3}.
\end{proof}

\begin{lemma} \label{lem4.2}
Assume that the assumptions in Theorem \ref{thm1.2} are satisfied.
Then for all $r>1$ there exists some constant $C>0$ satisfying
\begin{equation}{\label{4.6}}
\limsup_{t\to  \infty}\|u (\cdot,t)\|_{L^{r}(\Omega)}
\leq Cm\big(1+m^{\frac{2}{r}+2}\big),
\end{equation}
where $m:=\int_{\Omega}u_0$.
\end{lemma}

\begin{proof}
In light of the third equation in \eqref{1.6} and the inequality  \eqref{4.2},
for all $p\in(1,\infty)$ we obtain that
\begin{equation}{\label{4.7}}
\|v(\cdot,t)\|_{L^{p}(\Omega)}\leq  C_1m\quad \text{for all } t>0
\end{equation}
with some constant $C_1>0$. Observing that $s$ solves
\begin{gather*} %\label{4.8}
0=\Delta s-\delta s+\chi(\delta-\beta) v,\quad x\in\Omega,\; t>0,\\
\frac{\partial s}{\partial \nu}=0,\quad x\in\partial\Omega,\; t>0,
\end{gather*}
for all $q\in(1,\infty)$ we use \eqref{4.3} and \eqref{4.7} to find some
$C_2>0$ and $C_3>0$ such that
\begin{equation}{\label{4.9}}
\begin{split}
\|\nabla s(\cdot,t)\|_{L^{q}(\Omega)}
&\leq  \chi|\delta-\beta| C_2\cdot\|v(\cdot,t)\|_{L^{2}(\Omega)}\\
&\leq  C_3m\quad \text{for all } t>0.
\end{split}
\end{equation}
Testing the first equation of \eqref{1.6} by $u^{r-1}$ and integrating by parts,
we see that
\begin{equation} \label{4.10}
\frac{d}{dt}\int_{\Omega}u^r
+\int_{\Omega}u^r+\frac{2(r-1)}{r}\int_{\Omega}|\nabla u^{r/2}|^2
\leq\frac{r(r-1)}{2}\int_{\Omega}u^{r}|\nabla s|^2+\int_{\Omega}u^r
\end{equation}
for all $t>0$.
To deal with  the right-hand side of \eqref{4.10}, since
\begin{align*}
\|u\|^{r+1}_{L^{r+1}(\Omega)}
&=\|u^{r/2}\|^{\frac{2(r+1)}{r}}_{L^{\frac{2(r+1)}{r}}(\Omega)}\\
&\leq  C_4 \|\nabla u^{r/2}\|^{2}_{L^2(\Omega)}
 \|u^{r/2}\|^{\frac{2}{r}}_{L^{\frac{2}{r}}(\Omega)}
+ C_4 \|u^{r/2}\|^{\frac{2(r+1)}{r}}_{L^{\frac{2}{r}}(\Omega)}\\
&= C_4 m\|\nabla u^{r/2}\|^{2}_{L^2(\Omega)}+C_4 m^{r+1}
\end{align*}
holds for some constant $C_4>0$, and
\begin{align*}
\|u\|^{r}_{L^{r}(\Omega)}=\|u^{r/2}\|^{2}_{L^{2}(\Omega)}
&\leq  C_5 \|\nabla u^{r/2}\|^{\frac{2(r-1)}{r}}_{L^2(\Omega)}
\|u^{r/2}\|^{\frac{2}{r}}_{L^{\frac{2}{r}}(\Omega)}
+ C_5 \|u^{r/2}\|^{2}_{L^{\frac{2}{r}}(\Omega)}\\
&= C_5m\|\nabla u^{r/2}\|^{\frac{2(r-1)}{r}}_{L^2(\Omega)}
+C_5m^{r}
\end{align*}
holds for $C_5>0$ by means of the Gagliardo-Nirenberg inequality.
Then Young's inequality implies
\begin{align*}
\frac{r(r-1)}{2}\int_{\Omega}u^{r}|\nabla s|^2
&\leq  \frac{r(r-1)}{2}\Big(\epsilon_1\int_{\Omega}u^{r+1}+\epsilon_1^{-r}
\frac{r^{r}}
{(r+1)^{r+1}}\int_{\Omega}|\nabla s|^{2(r+1)}\Big)\\
&\leq \epsilon_1 \frac{r(r-1)}{2}C_4m\|\nabla u^{r/2}\|^{2}_{L^2(\Omega)}+\epsilon_1 \frac{r(r-1)}{2}C_4
m^{r+1}\\
&\quad +\epsilon_1^{-r} \frac{(r-1)}{2}\big(\frac{r}{r+1}\big)^{r+1}
\int_{\Omega}|\nabla s|^{2(r+1)}\quad \text{for all } t>0
\end{align*}
and
\begin{align*}
\int_{\Omega}u^r
&=\|u\|^{r}_{L^{r}(\Omega)}\\
&\leq  C_5m\|\nabla u^{r/2}\|^{\frac{2(r-1)}{r}}_{L^2(\Omega)}
+C_5m^{r}\\
&\leq  \epsilon_2 C_5\|\nabla u^{r/2}\|^{2}_{L^2(\Omega)}
+ \Big(\epsilon_2^{-(r-1)}\cdot \frac{(r-1)^{(r-1)}}{r^r}+1\Big)C_5m^{r}
\quad \text{for all }t>0.
\end{align*}
Taking $\epsilon_1=2r^{-2}C_4^{-1}m^{-1}$ and $\epsilon_2=\frac{r-1}{r}C_5^{-1}$,
inequality \eqref{4.10} becomes
\begin{equation} \label{4.15}
\frac{d}{dt}\int_{\Omega}u^r+\int_{\Omega}u^r
\leq C_6m^r\Big(1+\int_{\Omega}|\nabla s|^{2(r+1)}\Big)\quad \text{for all } t>0.
\end{equation}
Recalling \eqref{4.9}, integrating \eqref{4.15} over $(0,t)$, we find that
\[
\int_{\Omega}u^r\leq e^{-t}\|u_0\|^r_{L^r(\Omega)}
+C_6m^r\Big(1+m^{2(r+1)}\Big)\quad \text{for all }t>0,
\]
which yields \eqref{4.6}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
With Lemma \ref{lem4.2} at hand, the most important step towards  global behavior
of the case $\chi\alpha-\xi\gamma=0$ is to drive a bound for
$U:=u-\overline{u}_0$  in this section (Lemma \ref{lem4.3}). The later will
enforce $\|U(\cdot,t)\|_{L^{\infty}(\Omega)}\to  0$ as $t\to \infty$
under a smallness condition on the initial data $u_0$ by using a fixed-point
type argument (see also \cite{wi3}).
Let us introduce
\begin{gather*}
U(x,t):=u(x,t)-\bar{u}_0,\\
S(x,t):=s(x,t)-\chi\alpha\Big(\frac{1}{\beta}-\frac{1}{\delta}\Big)\bar{u}_0,\\
V(x,t):=v(x,t)-\frac{\alpha}{\beta}\bar{u}_0
\end{gather*}
for all $x\in \bar{\Omega}$ and $t>0$. Then if \eqref{1.7} holds, $(U,S,V)$
solves the  initial-value problem
\begin{equation}{\label{4.18}}
\begin{gathered}
U_t=\Delta U-\nabla\cdot( u \nabla S),\quad x\in \Omega,\; t>0,\\
0=\Delta S-\delta S+\chi(\delta-\beta)V,\quad x\in\Omega,\; t>0,\\
0=\Delta V-\beta V+\alpha U,\quad x\in\Omega,\; t>0,\\
\frac{\partial U}{\partial\nu }=\frac{\partial S}{\partial \nu}
=\frac{\partial V}{\partial \nu}=0,\quad x\in\partial\Omega,\; t>0,\\
U(x,0)=u_0(x)-\bar{u}_0,\quad V(x,0)=v_0(x)-\frac{\alpha}{\beta}\bar{u}_0,\\
S(x,0)=\chi \big(v_0(x)-\frac{\alpha}{\beta}\bar{u}_0\big)
-\xi\big(w_0(x)-\frac{\gamma}{\delta}\bar{u}_0\big),\quad x\in \Omega.
\end{gathered}
\end{equation}
By a straightforward adaptation of the ideas in \cite{li2}, we proceed
to derive  an estimate for $U$ with respect to the norm in $L^{\infty}(\Omega)$.
\end{proof}

\begin{lemma} \label{lem4.3}
Let $n=2$. For some $r>1$, the solution $(U,S,V)$ of \eqref{4.18} satisfies
\begin{equation} \label{4.19}
\limsup_{t\to  \infty}\|U (\cdot,t)\|_{L^{\infty}(\Omega)}
\leq Cm^{2}\big(1+m^{\frac{2}{r}+2}\big).
\end{equation}
\end{lemma}

\begin{proof} 
Since $\nabla S=\nabla s$, we first apply \eqref{4.9}
and Lemma \ref{lem4.2} to pick $t_1=t_1(u,v,w)>0$ such that
\begin{gather} \label{4.20}
\|\nabla S(\cdot,t)\|_{L^{q}(\Omega)}\leq C_1m\quad \text{for all }
t\geq t_1,\; q\in(1,\infty),\\
\label{4.21}
\|u(\cdot,t)\|_{L^{r}(\Omega)}\leq C_2m\big(1+m^{\frac{2}{r}+2}\big)
\quad \text{for all } t\geq t_1,\; r\in(1,\infty),
\end{gather}
where $C_1$ and $C_2$ are positive constant.
 By means of the variation-of-constants formula to the first equation
in \eqref{4.18}, we have
\[
U(\cdot,t)=e^{(t-t_1)\Delta}U(\cdot,t_1)-\int^{t}_{t_1}e^{(t-\sigma)
\Delta}\nabla \cdot\left(u(\cdot,\sigma)\nabla S(\cdot,\sigma)\right)
d\sigma\quad \text{for all } t> t_1.
\]
This in conjunction with some arguments on the asymptotic behavior of the heat
semigroup \cite{li2,wi1} yields \eqref{4.19}
 by using \eqref{4.20}--\eqref{4.21}.
\end{proof}

Now, invoking the upper estimate for $U$ in Lemma \ref{lem4.3},
we can pick $t_2=t_2(u,v,w)>0$ such that
\begin{equation} \label{4.23}
\|U(\cdot,t)\|_{L^{\infty}(\Omega)}
\leq c_1m^{2}(1+m^{\frac{2}{r}+2})\quad \text{for all }t\geq t_2
\end{equation}
with some constant $c_1>0$. With $\epsilon_0>0$ to be specified below,
we fix the total mass $m:=\int_{\Omega}u_0$ small enough such that
$0<m\leq \epsilon$ for  $0<\epsilon<\epsilon_0$.

Suppose that $\epsilon_0$  satisfies
\begin{equation}{\label{4.24}}
2c_1\epsilon_0(1+\epsilon_0^{\frac{2}{r}+2})\leq 1.
\end{equation}
Then \eqref{4.23} implies
\[
\|U(\cdot,t)\|_{L^{\infty}(\Omega)}\leq \frac{1}{2}\epsilon\quad
\text{for all } t\geq t_2.
\]
Now let $\lambda_1>0$ denote the first eigenvalue of $-\Delta$ in $\Omega$
under Neumann boundary conditions, and let some $\kappa$ satisfy
\begin{equation} \label{4.26}
\kappa\in\big(0,\frac{\lambda_1}{2}\big).
\end{equation}
Then since $\|U (\cdot,t)\|_{L^{\infty}(\Omega)}\leq \epsilon/2$ holds
for all $t\geq t_2$, the set
\[
S^*:=\big\{T^*\geq t_2\mid\|U(\cdot,t)\|_{L^{\infty}(\Omega)}
\leq \epsilon e^{-\kappa(t-t_2)}\text{ for all }t\in[t_2,T^*]\big\}
\]
is well-defined.

The following lemma provides $T=\infty$, where $T:=\sup S^*\in(t_2,\infty]$.
Therefore we obtain our goal that  the component $u$ of \eqref{1.2}
actually converges to $\overline{u}_0$, at an exponential rate.


\begin{lemma} \label{lem4.4}
Suppose that $\kappa$ satisfies \eqref{4.26} and that $n=2$.
Then one can find some constant $C>0$ such that
 \begin{equation} \label{4.28}
\|U(\cdot,t)\|_{L^{\infty}(\Omega)}\leq Ce^{-\kappa(t-t_2)}\quad
\text{for all } t> t_2.
\end{equation}
\end{lemma}

\begin{proof}
Given any $p\in(1,\infty)$, since $V$ solves the third equation in \eqref{4.18},
 from \eqref{4.2} we can find some positive $C_1$ and $C_2$ such that
\begin{equation}{\label{4.29}}
\begin{split}
\|V(\cdot,t)\|_{L^{p}(\Omega)}
&\leq   \alpha C_1\cdot\|U(\cdot,t)\|_ {L^{1}(\Omega)}\\
&\leq  \alpha|\Omega|C_1\cdot\|U(\cdot,t)\|_ {L^{\infty}(\Omega)}\\
&\leq  C_2\epsilon e^{-\kappa(t-t_2)}
\quad \text{for all } t\in(t_2,T).
\end{split}
\end{equation}
Moreover, given any $q\in(1,\infty)$, employing the inequality \eqref{4.3},
we can pick some constants $C_3>0$ and $C_4>0$ satisfying
\begin{equation} \label{4.30}
\|\nabla S(\cdot,t)\|_{L^{q}(\Omega)}
\leq \chi|\delta-\beta| C_3\cdot\|V(\cdot,t)\|_{L^{2}(\Omega)}
\leq C_4\epsilon e^{-\kappa(t-t_2)},\quad \forall  t\in(t_2,T).
\end{equation}
Observing that $U=u-\overline{u}_0$, $u$ can be easily controlled as
\[
\|u(\cdot,t)\|_{L^{\infty}(\Omega)}
\leq  \epsilon\Big(e^{-\kappa(t-t_2)}+ \frac{1}{|\Omega|}\Big)\quad
\text{for all }t\in(t_2,T).
\]
In view of \eqref{4.30} and Lemma \ref{lem4.3}, applying some  $L^{p}-L^{q}$
estimates for the Neumann heat semigroup (see \cite[Lemma 1.3]{wi1} or
\cite[Lemma 5.4]{li2}) to the representation of $U(\cdot,t)$,
for some $k>n$ we have
\begin{align*}
&\|U(\cdot,t)\|_{L^{\infty}(\Omega)}\\
&\leq  \|e^{(t-t_2)\Delta}U(\cdot,t_2)\|_{L^{\infty}(\Omega)}
 +\int^{t}_{t_2}\|e^{(t-\sigma)\Delta}\nabla \cdot(u(\cdot,\sigma)
 \nabla S(\cdot,\sigma))\|_{L^{\infty}(\Omega)}d\sigma\\
&\leq  C_5e^{-\lambda_1(t-t_2)}\|U(\cdot,t_2)\|_{L^{\infty}(\Omega)}\\
&\quad +C_5\int^{t}_{t_2}\Big(1+(t-\sigma)^{-\frac{1}{2}-\frac{n}{2k}}\Big)
 e^{-\lambda_1(t-\sigma)}
\|u(\cdot,\sigma)\nabla S(\cdot,\sigma)\|_{L^{k}(\Omega)}d\sigma\\
&\leq  C_6\epsilon^{2}\Big(1+\epsilon^{\frac{2}{r}+2}\Big)e^{-\lambda_1(t-t_2)}\\
&\quad +C_6\epsilon^{2}\int^{t}_{t_2}
\Big(1+(t-\sigma)^{-\frac{1}{2}-\frac{n}{2k}}\Big)e^{-\lambda_1(t-\sigma)}
 \Big(e^{-\kappa(\sigma-t_2)}+e^{-2\kappa(\sigma-t_2)}\Big)d\sigma
\end{align*}
for all $t\in(t_2,T)$. For any $0<\kappa<\frac{\lambda_1}{2}$ and given
some $r>1$, we may use  \cite[Lemma 1.2]{wi1} to find some constant $C_7>0$ such that
\[
\|U(\cdot,t)\|_{L^{\infty}(\Omega)}
\leq C_7\epsilon^{2} \big(1+\epsilon^{\frac{2}{r}+2}\big)e^{-\kappa(t-t_2)}\quad
\text{for all } t\in(t_2,T).
\]
Thus fixing $\epsilon_0>0$  small enough such that
\[
C_7\epsilon_0\Big(1+\epsilon_0^{\frac{2}{r}+2}\Big)<1
\]
and \eqref{4.24}, and in view of the continuity of $U$, we find that
$T=\infty$.
This implies \eqref{4.28} and hence completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
Applying the maximum principle to the second equation in \eqref{1.2} we have
\[
\frac{\alpha}{\beta}\min_{x\in{\bar{\Omega}}} u(x,t)
\leq v(x,t)
\leq \frac{\alpha}{\beta}\max_{x\in{\bar{\Omega}}} u(x,t)
\quad \text{for all }t>0.
\]
In light of Lemma \ref{lem4.4}, there exists $C>0$ satisfying
\[
\|v(\cdot,t)-\frac{\alpha}{\beta}\overline{u}_0\|_{L^{\infty}(\Omega)}
\leq \frac{\alpha}{\beta}\|u(\cdot,t)-\overline{u}_0\|_{L^{\infty}(\Omega)}
\leq Ce^{-\kappa  t}\quad \text{for all }t>0.
\]
The convergence of $w$ can be similarly proved.
\end{proof}

\subsection*{Acknowledgments}
The authors would like to thank the anonymous reviewers for their
valuable suggestions and fruitful comments which  greatly improved this work.
This work is supported by NSF of China (11371384).


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