\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 122, 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}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/122\hfil Global existence and blowup]
{Global existence and blowup for free boundary problems of
coupled reaction-diffusion systems}

\author[J. Sun, H. Lu, S. Gan, L. Chen \hfil EJDE-2014/122\hfilneg]
{Jianping Sun, Haihua Lu, Shuanglong Gan, Lang Chen} 

\address{Jianping Sun \newline
School of science, Nantong University, Nantong 226007, China}
\email{jpsun@ntu.edu.cn}

\address{Haihua Lu (Corresponding author) \newline
School of science, Nantong University, Nantong 226007, China}
\email{haihualu\_ntu@163.com}

\address{Shuanglong Gan \newline
School of science, Nantong University, Nantong 226007, China}
\email{1102072021@lxy.ntu.edu.cn}

\address{Lang Chen  \newline
School of science, Nantong University, Nantong 226007, China}
\email{814653119@qq.com}

\thanks{Submitted  February 28, 2014. Published May 8, 2014.}
\subjclass[2000]{35K20, 35R35}
\keywords{Free boundary; ecology; interface; existence; blowup}

\begin{abstract}
 This article concerns a free boundary problem for a reaction-diffusion
 system modeling the cooperative interaction of two diffusion biological
 species in one space dimension.
 First we show the existence and uniqueness of a local classical solution,
 then we study the asymptotic behavior of the free boundary problem.
 Our results show that the free boundary problem admits a global solution
 if the inter-specific competitions are strong, while, if the inter-specific
 competitions are weak, there exist the blowup solution and a global fast solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

We consider the  free boundary problem
 \begin{equation} \begin{gathered}
 u_t -d_1 u_{xx} =u(a_1-b_1u^r+v^p), \quad t>0,\; 0<x<h(t), \\
 v_t -d_2v_{xx}=v (a_2-b_2v^s+u^q), \quad t>0,\; 0<x<h(t),\\
 u  =v = 0, \quad t>0,\; x=0,\\
 u  =v = 0,\quad  h'(t)=-\mu(u_x+\rho v_x), \quad t>0,\; x=h(t),\\
 h(0)=h_0,\quad 0<h_0<\infty,\\
  u(x, 0) = u_0(x),\quad v(x, 0) = v_0(x),\quad 0\leq x\leq h_0,\\
\end{gathered} \label{e1.1}
\end{equation}
   where $a_i\geq0, p, q, r, s, b_i, d_i$ $(i = 1, 2)$ and $\mu$ are positive
constants,  $x=h(t)$ is the
free boundary to be determined together with $u(t,x)$ and $ v(t,x)$.
System \eqref{e1.1} is usually referred as the cooperative system. It provides a
simple model to describe, for instance, the cooperative interaction
of two diffusing biological species. $u $ and $v$ represent the
densities of two species, $a_1$ and $a_2$ are their growth rates.
Here, it is assumed that each species finds its subsistence from the
activity of the other one (represented by the reaction terms $v^p$
and $u^q$ ), and disappears by a destruction mechanism,
corresponding for instance to overcrowding or the action of a
predator (represented by the absorption terms $b_1u^r$ and $b_2v^s$).
 For more background for the system, we can refer to \cite{KLL10,LW05}
and references therein.

As we know, the free boundary problems have been used to describe
different types of mathematical models.  For the study of free boundary
problems for some biological models, we refer to, for instance
\cite{CF03,DG11,DG12,GW12,HM03,PZ13} and references cited therein.
Let us recall some work about the blow-up results to the reaction-diffusion
equations or systems with free boundaries. In \cite{ZL12}, Zhang and Lin
investigated the behavior of the positive solution $u(t, x)$ to a
parabolic model with double fronts free boundaries:
  \begin{equation} \begin{gathered}
   u_t -d u_{xx} =u^p, \quad t>0,\; g(t)<x<h(t), \\
   u(t,g(t))=0, \quad g'(t)=\mu u_x(t,g(t),\quad t>0, \\
   u(t,h(t))=0, \quad h'(t)=\mu u_x(t,h(t),\quad t>0,\\
   g(0)=-h_0,\quad h(0)=h_0,\\
   u(0,x) = u_0(x),\quad  v(0,x) = v_0(x),\quad -h_0\leq x\leq h_0.
  \end{gathered} \label{e1.2}
  \end{equation}
The result showed that when $p>1$ blowup occurs if the initial datum
is large enough  and that the solution is global and fast, which
decays uniformly at an exponential rate if the initial datum is
small, while there is a global and slow solution provided that the
initial value is suitably large. In \cite{LLP13}, Ling et al.
studied the global existence and blow-up for a parabolic equation
with a nonlocal source and absorption
  \[
u_t -d u_{xx} =\int_{g(t)}^{h(t)}u^p(t,x)dx-ku^q,\quad  t>0,\; g(t)<x<h(t),
\]
with the same initial and boundary conditions as in \eqref{e1.2}.
As far as the coupled system is concerned,  Kim et al.
\cite{KLL10}  considered the  mutualistic model
\begin{gather*}
    u_t -d_1 u_{xx} =u(a_1-b_1u+c_1v), \quad t>0,\; 0<x<h(t), \\
   v_t -d_2v_{xx}=v (a_2-b_2v+c_2u), \quad t>0,\; 0<x<\infty,\\
   u(t,x)=0, \quad t>0,\; h(t)<x<\infty,\\
  u = 0,\quad h'(t)=-\mu u_x, \quad t>0,\; x=h(t),\\
  u_x(t,0)=v_x(t,0)=0,\quad t>0,\\
  h(0)=b,\quad 0<b<\infty,\\
 u(0, x) = u_0(x)\geq0,\quad  0\leq x\leq b,\\
  v(0, x) = v_0(x)\geq0,\quad 0\leq x\leq \infty.
\end{gather*}
They showed the  existence and uniqueness of a classical local solution
and the asymptotic behavior of the solution.
 And they showed that the free boundary problem
admits a global slow solution if the inter-specific competitions are strong,
 while if the inter-specific
competitions are weak there exist the blowup solution and global fast solution.

As we know, sometimes both species have a tendency to emigrate from
the boundaries to obtain their new habitat; i.e., they will move outward along the
unknown curves (free boundaries) as time increases. It is assumed
that the movement speeds of free boundaries are proportional to the
sum of gradient of these two species, i.e.
 \[
h'(t)=-\mu(u_x+\rho v_x),
\]
which is the well-known Stefan type condition and whose ecological background
can be found in \cite{BDK12}.

 In this article, our interests in studying the long time behavior of
the solution of  \eqref{e1.1} is motivated by previous discussion.
 Differently from above,
 we put zero Dirichlet boundary conditions at the fixed boundary.
 This condition means that the habitat is restricted by a hostile environment
from the left and the species cannot survive on the fixed boundary.
 We will show that if $pq<rs$, the solution of \eqref{e1.1} is global while if
$pq>rs$, there exist a blowup solution and a global fast solution of \eqref{e1.1}.
To this end, we assume that the initial functions $u_0(x)$ and $ v_0(x)$ satisfy
 \begin{equation}
\begin{gathered}
 u_0,v_0\in C^2([0, h_0]),\quad
 u_0(0) =v_0(0) =  u_0(h_0) =v_0(h_0) = 0,\\
 u_0(x),v_0(x) > 0
\quad\text{in } (0,  h_0).
\end{gathered} \label{e1.3}
 \end{equation}
Now let us recall some blowup results of the corresponding problem on a
fixed domain under Dirichlet boundary condition with nonnegative
initial data:
\begin{equation}\begin{gathered}
      u_t=d_1\Delta u+u(a_1-b_1u^r+v^p),\quad    t>0,\; x\in \Omega, \\
      v_t=d_2 \Delta v+v(a_2+u^q-b_2v^s),\quad    t>0,\; x\in \Omega, \\
      u=v=0, \quad t>0,\; x\in \partial\Omega,  \end{gathered}
\label{e1.5}
\end{equation}
where $\Omega\subset \mathbb{R}^N$ is a bounded domain with smooth boundary
$\partial\Omega$. By constructing blowup sub-solution or bounded super-solutions,
Li and Wang \cite{LW05} obtained the optimal conditions on the exponent
 of reaction and absorption terms for the existence or nonexistence of global
solutions. The main results in \cite{LW05} are stated as follows.

\begin{proposition}[{\cite[Theorem 1]{LW05}}] \label{prop1.1}
 If $pq<rs$, then all solutions of \eqref{e1.5} are global and uniformly bounded.
\end{proposition}

 \begin{proposition}[{\cite[Theorem 3]{LW05}}] \label{prop1.2}
 Suppose that $pq>rs$. If $b_1^qb_2^{r_0}<1$ for some $r_0>0$ satisfying
$pq=r_0s$, or $b_1^{s_0}b_2^p<1$ for some $s_0>0$ satisfying $pq=rs_0$,
then all solutions of \eqref{e1.5} blows up in finite time with suitable
initial data.
\end{proposition}

The rest of the paper is organized as follows. In the next section,
local existence and uniqueness of the
free boundary problem are obtained by using the contraction mapping theorem.
In Section 3 a priori estimates will be derived and the global existence will
be given for the case $pq<rs$. Section 4 deals
with the global existence and nonexistence of a classical positive solution
for the case $pq>rs$.


\section{Existence and uniqueness}


In this section, we first prove the  existence and
uniqueness of a local solution using the contraction mapping theorem.

\begin{theorem}\label{thm2.1}
 For any given $(u_0(x),v_0(x))$ satisfying
 \eqref{e1.3} and any $\alpha\in(0, 1)$, there is a $T > 0$ such that
problem  \eqref{e1.1} admits a unique solution
 \[
(u,v,h)\in (C^{\frac{1+\alpha}{2},1+\alpha}(\bar D_T))^2\times
C^{\frac{1+\alpha}{2}}([0,T]).
\]
 Moreover,
   \begin{equation}
   \|u,v\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\bar D_T)}+\|h\|_{
   C^{\frac{1+\alpha}{2}}([0,T])}\leq C,
   \end{equation}
where   $D_T= (0,T]\times(0,h(t))$,
$C$ and $T$ are positive constants only depending on
$h_0, \alpha, \|u_0,v_0\|_{C^2([0,h_0])}$. Here and in the following,
   \[
\|u,v\|_{X}:=\|u \|_{X}+\| v\|_{X}.
\]
\end{theorem}

\begin{proof}
 As in \cite{CF00,DL10}, we first straighten the free boundaries.
 Let $\zeta(y)$   be a function in $C^3(\mathbb{R})$ satisfying
 \begin{gather*}
\zeta(y)=\begin{cases}
1 &\text{if }|y-h_0|<h_0/4, \\
0 &\text{if }  |y-h_0|>h_0/2,
\end{cases} \\
 \zeta'(y)<\frac{6}{h_0},\quad \forall y.
\end{gather*}
Consider the transformation
  \[
(t,y)\mapsto (t,x),\quad\text{where } x=y+\zeta(y)(h(t)-h_0),\;
 0\leq y<\infty.
\]
As long as
$|h(t)-h_0|\leq h_0/8$,
the above transformation is a diffeomorphism from $[0,\infty)$ onto
$[0,\infty)$.
Moreover, it changes the free boundary $x=h(t)$
to the line  $y = h_0$. If we set
 \begin{gather*}
u(t,x)=u(t,y+\zeta(y)(h(t)-h_0))=w(t,y),\\
 v(t,x)=v(t,y+\zeta(y)(h(t)-h_0))=z(t,y),
\end{gather*}
then the free boundary problem \eqref{e1.1} becomes
  \begin{equation}
\begin{gathered}
   w_t - Ad_1w_{yy}-(Bd_1+h'C)w_y =w\left(a_1-b_1w^r+z^p\right), \quad
 t>0, \; 0<y<h_0, \\
  z_t -Ad_2z_{yy}-(Bd_2+h'C)z_y=z\left(a_2+w^q-b_2z^s\right), \quad
 t>0, \; 0<y<h_0,\\
 w(t,0) =z(t,0) =w(t,h_0) =z(t,h_0)= 0,  \quad t>0,  \\
 w(0,y)  =u_0(y),\quad  z(0,y) = v_0(y),\quad 0\leq y\leq h_0,
\end{gathered} \label{e2.2}
\end{equation}
where
\begin{gather*}
 A:=A(h(t),y)=\frac{1}{(1+\zeta'(y)(h(t)-h_0))^2},\\
 B:=B(h(t),y)=-\frac{\zeta''(y)(h(t)-h_0))}{(1+\zeta'(y)(h(t)-h_0))^3},\\
 C:=C(h(t),y)=\frac{\zeta(y)}{1+\zeta'(y)(h(t)-h_0)}.
 \end{gather*}
Denote $h_1=-\mu(u_0'(h_0)+\rho v_0'(h_0))$, and for
$0<T\leq \frac{h_0}{8(1+h_1)}$, define $\Delta_T=[0,T]\times [0,h_0]$,
\begin{gather*}
\begin{aligned}
\mathcal {D}_{1T}=\big\{&w\in C^{\frac{\alpha}{2},\alpha}(\Delta_T):
 w(t,y)\geq0, w(0,y)=u_0(y), w(t,h_0)=0, \\
&\|w-u_0\|_{C^{\frac{\alpha}{2},\alpha}(\Delta_T)}\leq1\big\},
\end{aligned}\\
\mathcal {D}_{2T}=\{h\in C^1([0,T]): h(0 )=h_0, \; h'(0 )=h_1,\;
  \|h'-h_1\|_{C([0,T])}\leq1\}.
  \end{gather*}
It is easily seen that the set
$\mathcal {D}=\mathcal {D}_{1T}\times \mathcal {D}_{2T}$ is a closed convex set in
$C^{\frac{\alpha}{2},\alpha}(\Delta_T)\times C^1([0,T])$.

Next, we shall prove the existence and uniqueness result by using
the contraction mapping theorem. First, we observe that due to our
choice of $T$, for any given $(w, h)\in  \mathcal {D}$, we have
 \[
|h(t)-h_0|\leq T(1+h_1)\leq \frac{h_0}{8}.
\]
Therefore the transformation $(t, y)\to(t, x)$ introduced at the
beginning of the proof is well defined. Applying standard $L^p$
theory and then the Sobolev imbedding theorem, we find that for
any  $(w, h) \in \mathcal {D}$,
the  initial boundary value  problem
\begin{equation} \begin{gathered}
   z_t -Ad_2z_{yy}-(Bd_2+h'C)z_y=z\left(a_2+w^q-b_2z^s\right), \quad
t>0,\; 0<y<h_0,\\
 z(t,0) = z(t,h_0)=0,  \quad t>0,\\
   z(0,y) = v_0(y),\quad 0\leq y\leq h_0,
\end{gathered}
 \end{equation}
admits a unique solution (see \cite{LSU68}) $z\in
C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)$, and
  \[
\|z\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}\leq C_1.
\]
  Moreover, the  initial  boundary value problem
\begin{equation} \begin{gathered}
   \overline{w}_t - Ad_1\overline{w}_{yy}-(Bd_1+h'C)\overline{w}_y =
 w\left(a_1-b_1w^r+z^p\right), \quad t>0,\; 0<y<h_0, \\
 \overline{w}(t,0)  =\overline{w}(t,h_0) =  0,  \quad t>0,  \\
 \overline{w}(0,y)  =u_0(y),\quad 0\leq y\leq h_0,
\end{gathered}
 \end{equation}
admits a unique solution
$\overline{w}\in C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)$, and
 \begin{equation}
\|\overline{w}\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}\leq
 C_2,\label{e2.3}
\end{equation}
where $C_1, C_2$ are two constants depending on $h_0, \alpha, u_0, v_0$.

Defining
  \[
\overline{h}(t)=h_0-\int_0^t \mu( \overline{w}_y (\tau,h_0)
+\rho z_y(\tau,h_0))d\tau,
\]
we have
  \begin{equation}
 \overline{h}'(t)=-\mu( \overline{w} _y  (t,h_0)+\rho z_y(t,h_0)),\quad
  \overline{h}(0)=h_0,\quad    \overline{h}'(0)=h_1,
\label{e2.3p}
\end{equation}
and hence $\overline{h}'\in C^{\alpha/2}([0,t])$ with
  \begin{equation}
\|\overline{h}'\|_{C^{\alpha/2}([0,t])}\leq C_3:=\mu(C_2+\rho
  C_1).\label{e2.4}
\end{equation}
We now define $\mathcal {F}:\mathcal {D} \to
C^{\frac{\alpha}{2},\alpha}(\Delta_ T ) \times C^1([0, T ])$ by
  \[
\mathcal {F}(w,h)=(\overline{w},\overline{h}).
\]
Clearly $(w, h) \in \mathcal {D} $ is a fixed point of $\mathcal{F}$ if and
only if it solves \eqref{e2.2}.

 By \eqref{e2.4}) and \eqref{e2.3}, we have
\begin{gather*}
   \|\overline{h}'-h_1\|_{C([0,T])}
\leq  \|\overline{h}'\|_{C^{\alpha/2}([0,T])T^{\alpha/2}}\leq C_3
  T^{\alpha/2},\\
\begin{aligned}
&\|\overline{w} -u_0\|_{C^{\frac{\alpha}{2},\alpha}(\Delta_T)}\\
&\leq \|\overline{w}\|_{C^{\frac{1+\alpha}{2},0}(\Delta_T)}T^{\frac{1+\alpha}{2}}
  +\|\overline{w}\|_{C^{\frac{1+\alpha}{2},0}(\Delta_T)}T^{\frac{1}{2}}+
  h_0^{1-\alpha}\|\overline{w}_y\|_{C^{\alpha/2,0}(\Delta_T)}T^{\alpha/2}\|\\
&\leq C_2\left(T^{\frac{1+\alpha}{2}}+T^{\frac{1}{2}}
 +h_0^{1-\alpha}T^{\alpha/2}\right).
\end{aligned}
\end{gather*}
Therefore, if we take $T\leq\min\{1, C_3^{-2/\alpha},
[(2+h_0^{1-\alpha})C_1]^{-2/\alpha}\}$, then $\mathcal {F}$ maps
$\mathcal {D}$ into itself.

Next we prove that $\mathcal {F}$ is a contraction mapping on
$\mathcal {D}$ for $T >0$ sufficiently small. Let
$(w_i, h_i) \in \mathcal {D} (i = 1, 2) $ and denote
$(\overline{w}_i, \overline{h}_i) =\mathcal {F}(w_i, h_i)$.
 Then it follows from \eqref{e2.3} and \eqref{e2.4} that
\begin{equation}
  \|\overline{w}_i\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}\leq
 C_2,\quad
  \|\overline{h}_i'\|_{C^{\alpha/2}([0,t])}\leq C_3.
\end{equation}
Setting $U=\overline{w}_1-\overline{w}_2$, $V=z_1-z_2$, we find that
$V(t,y)$ and $U(t,y)$ satisfy
\begin{gather*}
\begin{aligned}
&V_t -A(h_2,y)d_2V_{yy}-(B(h_2,y)d_2+h_2'C(h_2,y))V_y\\
&=[A(h_1,y)-A(h_2,y)]d_2z_{1,yy}
   +[B(h_1,y)-B(h_2,y)]d_2z_{1,y}\\
&\quad +[h_1'C(h_1,y)-h_2'C(h_2,y)]z_{1,y}
 + (a_2-b_2\Phi_2(t,y)+w_1^q)(z_1-z_2)\\
&\quad +z_2\Psi_2(t,y)(w_1-w_2), \quad t>0,\; 0<y<h_0,
\end{aligned}\\
 V(t,0) = V(t,h_0)=0,  \quad t>0,\\
   V(0,y) = 0, \quad 0\leq y\leq h_0,
 \end{gather*}
and
\begin{gather*}
\begin{aligned}
& U_t -A(h_2,y)d_1U_{yy}-(B(h_2,y)+h_2'C(h_2,y))U_y\\
&=  [A(h_1,y)-A(h_2,y)]d_1w_{1,yy}
 +[B(h_1,y)-B(h_2,y)]d_1w_{1,y}\\
&\quad +[h_1'C(h_1,y)-h_2'C(h_2,y)] w_{1,y}
 +  (a_1-b_1\Phi_1(t,y)+z_1^p)(w_1-w_2)\\
&\quad +w_2\Psi_1(t,y)(z_1-z_2), \quad t>0,\;  0<y<h_0,
\end{aligned}\\
 U(t,0) = U(t,h_0)=0,  \quad t>0,\\
   U(0,y) = 0, \quad 0\leq y\leq h_0,
 \end{gather*}
where
\begin{gather*}
\Phi_1(t,y)=\int_0^1(r+1)(\theta w_1+(1-\theta)w_2)^rd\theta,\\
\Phi_2(t,y)=\int_0^1(s+1)(\theta z_1+(1-\theta)z_2)^sd\theta,\\
\Psi_1(t,y)=\int_0^1p(\theta z_1+(1-\theta)z_2)^{p-1}d\theta,\\
\Psi_2(t,y)=\int_0^1q(\theta w_1+(1-\theta)w_2)^{q-1}d\theta.
\end{gather*}
 Using  standard partial differential equation theory \cite{LSU68},
 the $L^p$ estimates for parabolic equations and Sobolev's
imbedding theorem, we obtain
\begin{gather}
  \|{z}_1-
  {z}_2\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}
\leq   C_4(\|w_1-w_2\|_{C(\Delta_T)}+\|h_1-h_2\|_{C^1([0,T])}),\label{e2.5p}\\
\begin{aligned}
&\|\overline{w}_1-
  \overline{w}_2\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}\\
&\leq   C_4(\|w_1-w_2\|_{C(\Delta_T)}+\|h_1-h_2\|_{C^1([0,T])}
 +\|z_1-z_2\|_{C(\Delta_T)}) \\
&\leq   C_5(\|w_1-w_2\|_{C(\Delta_T)}+\|h_1-h_2\|_{C^1([0,T])}).
\end{aligned}\label{e2.6}
\end{gather}
using \eqref{e2.3p}, we have
\begin{equation}
\|\overline{h}'_1-\overline{h}'_2\|_{C^{\alpha/2}([0,T])}\\
\leq \mu(\|\overline{w}_1-  \overline{w}_2\|_{C^{\alpha/2,0}(\Delta_T)}
+\rho\|z_1-  z_2\|_{C^{\frac{\alpha}{2},0}(\Delta_T)}).\label{e2.7}
\end{equation}
Combing \eqref{e2.5p}-\eqref{e2.7}, assuming $T\leq1$, and applying
mean value theorem, we obtain
  \begin{align*}
&\|\overline{w}_1-
  \overline{w}_2\|_{C^{\frac{1+\alpha}{2},1+\alpha}(\Delta_T)}
+ \|\overline{h}'_1-\overline{h}'_2\|_{C^{\alpha/2}([0,T])}\\
&\leq
   C_6(\|w_1-w_2\|_{C(\Delta_T)}+\|h_1-h_2\|_{C^1([0,T])}),
\end{align*}
which implies
 \begin{align*}
\|\overline{w}_1-   \overline{w}_2\|_{C^{\alpha/2,\alpha}(\Delta_T)} 
&\leq     \|w_1-w_2\|_{C^{\frac{1+\alpha}{2},0}(\Delta_T)}T^\frac{1+\alpha}{2}
    +\|w_1-w_2\|_{C^{\frac{1+\alpha}{2},0}(\Delta_T)}T^{1/2}\\
&\quad +h_0^{1-\alpha}\|\overline{w}_{1y}-
  \overline{w}_{2y}\|_{C^{\alpha/2,
  0}(\Delta_T)}T^\frac{\alpha}{2}\\
&\leq (2+h_0^{1-\alpha})T^\frac{\alpha}{2}\|\overline{w}_1-
  \overline{w}_2\|_{C^{\frac{1+\alpha}{2}, 1+\alpha}(\Delta_T)}.
\end{align*}
Hence,   for
 \[
T:=\min\big\{1, (4+2h_0^{1-\alpha})^{-2/\alpha}, C_3^{-2/\alpha},
[(2+h_0^{1-\alpha})C_1]^{-2/\alpha},\frac{h_0}{8(1+h_1)}\big\},
\]
 we have
\begin{align*}
&\|\overline{w}_1-   \overline{w}_2\|_{C^{\alpha/2, \alpha}(\Delta_T)}
+\|h_1-h_2\|_{C^1([0,T])}\\
&\leq     (2+h_0^{1-\alpha})T^\frac{\alpha}{2}\|\overline{w}_1-
  \overline{w}_2\|_{C^{\frac{1+\alpha}{2},  1+\alpha}(\Delta_T)}
+2T^{\alpha/2}\|h_1'-h_2'\|_{C^{\alpha/2}([0,T])}\\
&\leq (2+h_0^{1-\alpha})T^\frac{\alpha}{2}
  C_6(\|w_1-w_2\|_{C(\Delta_T)}+\|h_1-h_2\|_{C^1([0,T])})\\
&\leq \frac{1}{2}(\|\overline{w}_1-
  \overline{w}_2\|_{C^{\alpha/2, \alpha}(\Delta_T)}
  +\|h_1-h_2\|_{C^1([0,T])}).
\end{align*}
This shows that for this $T$, $\mathcal {F}$ is a contraction
mapping on $\mathcal {D}$. It now follows from the contraction
mapping theorem that  $\mathcal {F}$ has a unique fixed point
$(w,h)$ in $\mathcal {D}$. Moreover, by the Schauder estimates, we have
additional regularity for $(w,z, h)$ as a solution of \eqref{e2.2},
namely, $h \in C^{1+\frac{\alpha}{2}}([0, T])$ and
$w,z \in C^{\frac{1+\alpha}{2}, 1+\alpha}((0, T]\times [0,h_0])$,
and \eqref{e2.3}, \eqref{e2.4} hold. In other
words, $(w(t, y),z(t,y), h(t))$  is a unique local classical
solution of the problem \eqref{e2.2}. Hence, $(u,v,h)$ is a unique
classical solution of \eqref{e1.1}.
\end{proof}

\begin{theorem}\label{thm2.2}
 The free boundary for the problem \eqref{e1.1} is strictly monotone increasing;
 i.e., for any solution in $(0,T]$, we have $h'(t) > 0$ for $0< t \leq T$.
\end{theorem}

\begin{proof}
 Firstly, as $u>0$ for $0<x<h(t)$ and $u=0$ at $x=h(t)$, we see that
$u_x(t,h(t))\leq0$ and so $h'(t)\geq0$.
 Since we only know $h\in C^{1+\frac{\alpha}{2}}([0,\infty))$,
 it can not be guaranteed that the domain $(0,\infty)\times[0,h(t)]$ has an
interior sphere property at the right boundary $x=h(t)$.
 hence, the Hopf lemma cannot be used directly to get $h'(t)>0$.
 To solve this, we use a transformation to straighten the free boundary
$x=h(t)$.  Define $y=x/h(t)$ and $w(t,y)=u(t,x), z(t,y)=v(t,x)$.
 A series of detailed calculation asserts that
\begin{gather*}
 w_t-d_1\zeta(t)w_{yy}-\xi(t,y)w_y=w(a_1-b_1w^r+z^p),\quad  t>0,\; 0<y<1,\\
 z_t-d_2\zeta(t)z_{yy}-\xi(t,y)z_y=z(a_2-b_2z^s+w^q),\quad t>0,\; 0<y<1,\\
 w(t,0)=w(t,1)=0,\quad t>0,\\
 w(0,y)=u_0(h_0y),\quad z(0,y)=v_0(h_0y),\quad 0\leq y\leq1,
 \end{gather*}
where $\zeta(t)=h^{-2}(t),\ \xi(t,y)=yh'(t)/{h(t)}$.
 This is an initial and boundary value problem with fixed boundary.
 Since $w>0, z>0$ for $t>0$ and $0<y<1$, by the Hopf lemma,
 we have $w(y,1)<0$, $z(y,1)<0$ for $t>0$.
 This combines with the relation $u_x=h^{-1}(t)w_y$ and $v_x=h^{-1}(t)z_y$
to derive that
 $u_x(t,h(t))<0$ and $v_x(t,h(t))<0$ and so $h'(t)>0$ for $t>0$.
\end{proof}

It is observed that there exists a $T$ such that the solution exists in the
time interval $[0,T]$. The maximal
existing time of the solution $T_{\rm max}$ depends on a prior estimate with respect
 to $\|u\|_{L^\infty}$, $\|v\|_{L^\infty}$ and $h'(t)$.
Next we show that if $\|u\|_{L^\infty}<\infty$ or $\|v\|_{L^\infty}<\infty$,
the solution can be extended. Therefore we first give the following lemma.

\begin{lemma}\label{lem2.1}
Let $(u, v,h)$ be a solution to problem \eqref{e1.1} defined for
 $t \in (0, T_0)$ for some $T_0 \in (0,+\infty]$.
If $M_1:{=}\|u\|_{{L^\infty}([0,T]\times [0,h(t)])}< \infty$, then there exist
constants $  M_2$ and $M_3$ independent of $T_0$ such that
\[
0<v(t,x)\leq M_2(M_1),  \quad  0<h'(t)\leq M_3(M_1)
\]
for $0<t<T_0,\; 0\leq x<h(t)$.
\end{lemma}

\begin{proof}
 By \eqref{e1.1}, we obtain
 \[
v_{t}-d_{2}v_{xx}\leq v(a_2+M_1^q-b_2v^s),\quad     0<t<T_0,\quad 0\leq x<h(t).
\]
 It follows from the comparison principle that $v(t,x)\leq \overline{v}(t)$ for
$t\in(0,T_0)$ and $x\in [0,h(t)]$, where  $\overline{v}(t)$
is a unique solution of the problem
  \[
\frac{d\overline{v}}{dt}=\overline{v}(a_2+M_1^q-b_2\overline{v}^s), \quad t>0;\quad
 \overline{v}(0)=\|v_0\|_\infty.
\]
It is obvious that $\bar v$ is globally bounded.
  Thus we have
  \[
v(t,x)\leq M_2:=\sup_{t\geq0}\overline{v}(t).
\]
 Moreover, by Theorem \ref{thm2.2}, we have $h'(t)>0$ for $t\in(0,T_0)$.
 It remains to show that $h'(t) \leq M_2$ for all $t \in (0, T_0)$
with some $M_2$ independent of $T_0$. To this end, we define
  \[
\Omega_M:=\{(t,x): 0<t<T_0,\ h(t)-1/M<x<h(t)\}
\]
and construct an auxiliary function
 \[
\omega(t,x):=M_1[2M(h(t)-x)-M^2(h(t)-x)^2].
\]
We will choose $M$ so that $\omega(t, x) \geq u(t, x)$ holds over
$\Omega_M$.

Direct calculations show that, for $(t, x) \in \Omega_M$,
\begin{gather*}
w_t =   2M_1Mh'(t)(1 -M(h(t)- x)) \geq 0,\quad
-w_{xx} = 2M_1M^2,\\
u(a_1-b_1u^r+v^p)\leq M_1(a_1+M_2^p).
 \end{gather*}
It follows that
  \[
\omega_t -d_1\omega_{xx }\geq M_1(a_1+M_2^p) \geq u(a_1-b_1u^r+v^p)
\]
if $M^{2}\geq \frac{a_1+M_2^p}{2d_1}$.
On the other hand,
  \[
\omega(t, h(t)-1/M) =M_1 \geq u(t, h(t)-1/M),\quad
\omega(t,h(t)) = 0 = u(t, h(t)).
\]
   Thus, if we can choose $M$ such that
   \begin{equation}
   u_0(x) \leq \omega(0, x)\quad\text{for } x \in [h_0-1/M, h_0],\label{e2.8}
   \end{equation}
then we can apply the maximum principle to $\omega-u$ over $\Omega_M$
to deduce that $u(t,x) \leq \omega(t, x)$ for $(t, x) \in\Omega_M$.
 It would then follow that
  \[
u_x(t, h(t)) \geq \omega _x(t, h(t)) =-2MM_1.
\]
With the same method, we can deduce
  \[
v_x(t, h(t)) \geq \omega _x(t, h(t)) =-2MM_2,
\]
if $M^{2}\geq \frac{a_2+M_1^q}{2d_2}$. Hence, if
$M^2\geq \max\{ \frac{a_1+M_2^p}{2d_1}, \frac{a_2+M_1^q}{2d_2}\}$, we
have
  \[
 h'(t) = -\mu(u_x(t, h(t))+\rho v_x(t, h(t))) \leq M_3:=2M\mu(M_1+\rho M_2).
\]
To complete the proof, we need only find some $M$ such that
\eqref{e2.8} holds. By direct calculation, we obtain
\begin{gather*}
u_0(x)=\int_{h_0}^x u_0'(y)dy\leq \|u_0'\|_{C([0,h_0])}(h_0-x)
\quad\text{on } [h_0-1/M,  h_0],\\
\omega(0,x)=M_1[2M(h_0-x)-M^2(h_0-x)^2]\geq M_1M(h_0-x).
\end{gather*}
Therefore, upon choosing $M\geq \frac{\|u_0'\|_{C([0,h_0])}}{M_1}$,
\eqref{e2.8} follows.
To conclude, we choose
\[
M:=\max\big\{\frac{a_1+M_2^p}{2d_1}, \frac{a_2+M_1^q}{2d_2},
  \frac{\|u_0'\|_{C([0,h_0])}}{M_1},\frac{\|v_0'\|_{C([0,h_0])}}{M_2}\big\},
\]
thus the proof is complete.
\end{proof}

With the same method as in proof of \cite[Theorem 2.3]{DL10}, we can
get the  existence and uniqueness of a global solution for \eqref{e1.1}.

 \begin{theorem}\label{thm2.3} The solution of problem \eqref{e1.1} exists
and is unique, and it can be extended to $[0,T_{\rm max})$
where $T_{\rm max}\leq \infty $. Moreover, if $T_{\rm max} < \infty$, we have
$\limsup_{t\to T_{\rm max}}{\|u,v\|_{L^{\infty}([0,h(t)]\times [0,t])}}=\infty $.
\end{theorem}

\begin{proof}
 It follows from the uniqueness that there is a $T_{\rm max}$ such that
 $[0,T_{\rm max})$ is the maximal time
interval in which the solution exists. In order to prove the present theorem,
it suffices to show that, when
 $T_{\rm max}<\infty$,
 $$
\limsup_{t\to T_{\rm max}}{\|u,v\|_{L^{\infty}([0,t]\times [0,h(t)])}}=\infty .
$$
In what follows we use the contradiction argument.
Assume that $T_{\rm max}< \infty$ and
$\|u\|_{L^{\infty}([0,T_{\rm max})\times[0,h(t)])}< \infty$.
Since $v \leq M_2(M)$ in $[0,h(t)]\times [0,T_{\rm max})$ and
$0 < h'(t) \leq M_3$ in $[0,T_{\rm max})$ by Lemma \ref{lem2.1},
using a bootstrap argument and the Schauder's estimate
yields a priori bound of $\|u(t,x),v(t,x)\|_{C^{1+\alpha}([0,h(t)]} $
for all $t\in [0,T_{\rm max})$. Let the bound be $M_4$.
It follows from the proof of Theorem \ref{thm2.1} that there exists a $\tau > 0$
depending only on $M_1,M_2,M_3$ and
$M_4$ such that the solution of the problem \eqref{e1.1}
 with the initial time $T_{\rm max}-\tau /2$ can be extended
uniquely to the time $T_{\rm max}-\tau /2+\tau $  that contradicts the assumption.
Thus the proof is completed.
\end{proof}


\section{Global solution for the case $pq<rs$}

We first give a comparison principle, whose proof is standard and we omit 
it (see \cite{WX04}).

\begin{lemma}\label{lem3.1a}
Let $a_i(x,t), b_i(x,t), c_i(x,t)$, $(i=1,2)$, be continuous functions in 
$\Omega\times(0,T)$. Assume that
$a_i(x,t),c_i(x,t)\geq0$ in $\Omega\times(0,T)$ and $b_i(x,t),
c_i(x,t)$ are bounded on $\bar{\Omega}\times[0,T_0]$ for any
$T_0<T$. If functions $u_i$ belong to $C^{2,1}(\Omega\times(0,T))\cap
C(\bar{\Omega}\times[0,T])$, $i=1,2$, and
satisfy
 \begin{equation}
\begin{gathered}
 u_{1t}\leq (\geq) a_1\Delta u_1+b_1u_1+c_1u_{2} ,\quad  0<t<T,\; x\in\Omega,\\
 u_{2t}\leq (\geq) a_2\Delta u_2+b_2u_2+c_2u_1 ,
   0<t<T,\quad x\in\Omega,\\
 u_1(0,x)\leq(\geq)0 ,\quad  u_2(0,x)\leq(\geq)0, \quad x\in\Omega ,\\
 u_1(t,x)\leq(\geq)0 , \quad u_2(t,x)\leq(\geq)0 , \quad  0<t<T,\;
 x\in\partial\Omega ,\\
 \end{gathered}\label{e3.1a}
\end{equation}
then
 $$
(u_1,u_2)\leq(\geq)0, \quad \forall (x,t)\in\bar{\Omega}\times[0,T),\; i=1,2.
$$
\end{lemma}

 In order to get the  global existence, we aim to construct a constant 
supersolution of \eqref{e1.1}.
 Now, we state a simple fact without proof.

\begin{lemma}[{\cite[Lemma 1]{LW05}}] \label{lem3.1} 
  If $p,q,r,s>0$ and $pq<rs$, then for any positive constants $A, B$, 
there exist two positive constants $\mathcal{M}_1$ and $\mathcal{M}_2$ such  
that $A\mathcal{M}_1^r\geq\mathcal{M}_2^p$ and $B\mathcal{M}_2^s\geq\mathcal{M}_1^q$. 
In addition, if $(\mathcal{M}_1,\mathcal{M}_2)$ is a solution to this inequalities, 
$(k\mathcal{M}_1,\ell\mathcal{M}_2)$ is a solution to such inequalities for every 
$k, \ell\geq1$ satisfying $k^{q/s}\leq \ell\leq k^{r/p}$.
\end{lemma}

\begin{lemma}\label{lem3.2}
If $pq<rs$, the solution of the free boundary problem \eqref{e1.1} satisfies
\[
0 < u(t,x)<\mathcal{M}_1, \quad 
0<v(x,t)\leq   \mathcal{M}_2 \quad\text{for }  0\leq t \leq T, \;
 0 \leq x < h(t),
\]
where $\mathcal{M}_i$ is independent of $T$ for $i=1,2$.
\end{lemma}

\begin{proof}
Fix $C = \max\{\max_{\bar \Omega}u_0(x), \max_{\bar \Omega}v_0(x)\}$. 
We seek a pair of constant $\mathcal{M}_1$, $\mathcal{M}_2$ such that 
$\mathcal{M}_1, \mathcal{M}_2\geq C$,  and
  \begin{equation}
  b_1\mathcal{M}_1^r>a_1+\mathcal{M}_2^p,\quad
 b_2\mathcal{M}_2^s\geq a_2+\mathcal{M}_1^q.\label{e3.1}
  \end{equation}
To begin with, we choose $(\mathcal{M}_1,\mathcal{M}_2)$ such that
$a_1\leq \mathcal{M}_2^p$ and $a_2\leq \mathcal{M}_1^q$.
Therefore, \eqref{e3.1} holds provided that
 \[
b_1\mathcal{M}_1^r>2\mathcal{M}_2^p,\quad
b_2\mathcal{M}_2^s\geq 2\mathcal{M}_1^q.
\]
 Since $pq <rs$, the existence of suitable $\mathcal{M}_1$ and $\mathcal{M}_2$
is guaranteed by Lemma \ref{lem3.1}.

 Next we prove that
for any $l >h_0$, 
$(u(t,x),v(t,x))\leq (\mathcal{M}_1,\mathcal{M}_2):=(\overline{u},\overline{v})$. 
From the above process, we have $(\overline{u},\overline{v})$ satisfies
\begin{gather*}
\overline{u}_t-d_1\overline{u}_{xx}\geq\overline{u}(a_1-b_{1}\overline{u}^r
 +\overline{v}^p),\quad 0< t \leq T,\; 0< x < l, \\
\overline{v}_t-d_2\overline{v}_{xx}\geq\overline{v}(a_2+\overline{u}^q
 -b_{2}\overline{v}^s),\quad 0< t \leq T,\; 0< x < l, \\
\overline{u}\geq 0,\quad \overline{v}\geq 0, \quad 0< t \leq T,\; x=0,l,\\
\overline{u}(0,x)\geq  u_0(x),\quad \overline{v}(0,x)\geq  v_0(x)\quad
 0 \leq x \leq l.
\end{gather*}

 Set $w=\bar u-u,\ z=\bar v-v$, then we have
 \begin{gather*}
     w_t-d_1w_{xx}\geq(a_1-b_1\Phi_3(t,x)+v^p)w+\bar u\Psi_3(t,x)z,\quad
0< t \leq T,\; 0< x < l, \\
      z_t-d_2z_{xx}\geq(a_2-b_2\Phi_4(t,x)+u^q)w+\bar v\Psi_4(t,x)z,\quad
 0< t \leq T,\; 0< x < l, \\
      w\geq 0,\ z\geq 0, \quad 0< t \leq T,\; x=0,l,\\
      w(0,x)\geq 0,\quad z(0,x)\geq 0, \quad 0 \leq x \leq l,
\end{gather*}
where
\begin{gather*}
\Phi_3(t,x)=\int_0^1(r+1)(\theta \bar u+(1-\theta)u)^rd\theta,\quad
 \Phi_4(t,x)=\int_0^1(s+1)(\theta \bar v+(1-\theta)v)^sd\theta,\\
\Psi_3(t,x)=\int_0^1p(\theta \bar v+(1-\theta)v)^{p-1}d\theta,\quad
 \Psi_4(t,x)=\int_0^1q(\theta \bar u+(1-\theta)u)^{q-1}d\theta.
 \end{gather*}
Using Lemma \ref{lem3.1a} in $[0,T]\times[0,l]$ shows that $u\leq \overline{u}$ 
and $v\leq \overline{v}$. Now for any fixed
$(t_0,x_0)\in [0,T]\times[0,h(t)]$, let $l$ be sufficiently large so that
$(t_0,x_0)\in [0,T]\times[0,l]$,
 and it follows from the above proof that
\begin{gather*}
   u(t_0,x_0)\leq \overline{u}(t_0,x_0)= \mathcal{M}_1,\\
   v(t_0,x_0)\leq \overline{v}(t_0,x_0)= \mathcal{M}_2,
 \end{gather*}
which gives the desired estimates. 
\end{proof}

Combining Theorem \ref{thm2.3} with Lemma \ref{lem3.2} yields the existence
of a global solution.


\begin{theorem}\label{thm3.1}
If $pq<rs$, the free boundary problem \eqref{e1.1} admits a unique global 
classical solution.
\end{theorem}

\section{Global and nonglobal solutions for the case $pq>rs$}

In this section, we consider the asymptotic behavior of the solution for 
the case $pq>rs$. First we give the blowup result.

\begin{theorem} \label{thm4.1}
Assume that $pq>rs$. If $b_1^qb_2^{r_0}<1$ for some $r_0>0$ satisfying $pq=r_0s$, 
or  $b_1^{s_0}b_2^p<1$ for some $s_0>0$ satisfying $pq=rs_0$, then all solutions 
of \eqref{e1.1} blow up in finite time with suitable initial data.
\end{theorem}

\begin{proof} 
To show this, it suffices to compare the free boundary problem with the
corresponding problem in the fixed domain:
\begin{equation}
\begin{gathered}
      u_t-d_1u_{xx}= u(a_1-b_{1}u^r+v^p),\quad  t > 0,\; 0< x < h_0, \\
      v_t-d_2v_{xx}= v(a_2+u^q-b_{2}v^s),\quad  t > 0,\; 0< x < h_0, \\
      u(t,0)=v(t,0)=0,\quad  t > 0,\\
      u(t,h_0)=v(t,h_0)=0,\quad  t > 0,\\
      u(0,x)= u_0(x)\geq 0,\; v(0,x)= v_0(x)\geq 0,\quad  0 \leq x \leq h_0
\end{gathered}
\end{equation}
It follows from Proposition \ref{prop1.2} that the solution blows up if
 $b_1^qb_2^{r_0}<1$ for some $r_0>0$ satisfying $pq=r_0s$, or
 $b_1^{s_0}b_2^p<1$ for some $s_0>0$ satisfying $pq=rs_0$.
 We conclude the result by using comparison principle for the fixed boundary.
\end{proof}

 Now we present a comparison principle for $u,v$ and the free boundary $x = h(t)$  
which can be used to estimate the solution
$(u(t, x), v(t, x))$  and the free boundary   $x = h(t)$.

\begin{lemma}\label{lem4.1}
Suppose that $T \in (0,\infty)$, $\overline{h}\in C^{1}([0,T])$, 
$\overline{u}, \overline{v}\in C(\overline{D}^{*}_{1,T})\cap C^{1,2}(D^{*}_{1,T})$   
with $D^{*}_{1,T}=(0,T]\times (0,\overline{h}(t))$, and
\begin{gather*}
  \overline{u}_t-d_1\overline{u}_{xx}
\geq \overline{u}(a_1-b_{1}\overline{u}^r +\overline{v}^p),\quad  
 t > 0,\; 0< x < \overline{h}(t), \\
\overline{v}_t-d_2\overline{v}_{xx}
\geq \overline{v}(a_2+\overline{u}^q-b_{2}\overline{v}^s),\quad 
 t > 0,\; 0< x < \overline{h}(t), \\
 \bar u,\quad \bar v \geq 0, \quad t>0,\; x=0,\\
 \bar u  =\bar v = 0,\quad \bar h'(t)\geq-\mu(\bar u_x+\rho \bar v_x), \quad
t>0,\quad x=\bar h(t),\\
 \bar u(0, x) \geq u_0(x), \bar v(0, x) \geq v_0(x),\quad 0\leq x\leq h_0.
\end{gather*}
If $h(0)\leq \overline{h}(0)$, 
\begin{gather*}
(\bar u(0,x),\bar v(0,x))\geq(0,0) \quad\text{on }[0,\bar h(0)],\\
(u_0(x),v_0(x)) \leq (\overline{u}(0,x), \overline{v}(0,x))\quad
\text{on }[0,h_0], 
\end{gather*}
then the solution $(u,v,h)$ of the free boundary problem \eqref{e1.1} 
satisfies $h(t)\leq \overline{h}(t)$ in $(0,T]$,
$(u(t,x),v(t,x))\leq (\overline{u}(t,x),\overline{v}(t,x))$ in 
$[0,T]\times (0,h(t))$.
\end{lemma}

\begin{proof}
We first assume that $\overline{h}(0)>h(0)$.
 Then $\overline{h}(t)>h(t)$ for small $t>0$.
 We can derive that $\overline{h}(t)>h(t)$ for all $t\geq0$.
 If this is not true, there exists $t^*>0$ such that $\overline{h}(t^*)=h(t^*)$
 and $\overline{h}(t)>h(t)$ for all $t\in(0,t^*)$.
 Thus, $\overline{h}'(t^*)<h'(t^*)$.
 Recall that $(u_0(x),v_0(x))\leq (\overline{u}(0,x), \overline{v}(0,x))$ on 
$[0,h_0]$,  $u(t^{*},h(t^{*}))=0=\overline{u}(t^{*},\overline{h}(t^{*}))$
 and
 $v(t^{*},h(t^{*}))=0=\overline{v}(t^{*},\overline{h}(t^{*}))$.
 As the proof of Lemma \ref{lem3.2}, and applying Lemma \ref{lem3.1a} for the 
fixed boundary, we can obtain that
 $(u(t,x),v(t,x))\leq (\overline{u}(t,x),\overline{v}(t,x))$
 in $(0,t^{*})\times (0,h(t^*))$
 and
\[
 \frac{\partial}{\partial x}(u-\overline{u}){\bigr |}_{(t^{*},h(t^{*}))} 
\geq 0,\quad
 \frac{\partial}{\partial x}(v-\overline{v}){\bigr |}_{(t^{*},h(t^{*}))} \geq 0,
\]
which shows that
\begin{align*}
h'(t^{*})&=-\mu\Big(\frac{\partial u}{\partial x}(t^{*},h(t^{*}))
 +\rho\frac{\partial v}{\partial x}(t^{*},h(t^{*}))\Big)\\
 &\leq -\mu\Big(\frac{\partial \overline{u}}{\partial x}(t^{*},
 \overline{h}(t^{*}))+\rho \frac{\partial \overline{v}}{\partial x}(t^{*},
 \overline{h}(t^{*}))\Big)\\
&\leq \overline{h}'(t^{*}).
\end{align*}
This leads to a contradiction, which proves that $h(t)< \overline{h}(t)$ 
for $0\leq t \leq T$ in the case $\overline{h}(0)>h(0)$.
The general case can be established through approximation
(we also can refer to \cite[Lemma 5.1]{GW12}).
Since $h(t)\leq \overline{h}(t)$ for $0\leq t \leq T$, we have
$(u(t,x),v(t,x)\leq (\overline{u}(t,x),\overline{v}(t,x))$ in 
$[0,T]\times (0,h(t))$. 
\end{proof}

\begin{remark} \label{rmk4.1}\rm
The pair $(\overline{u},\overline{v},\overline{h})$ in Lemma \ref{lem4.1} 
is usually called an upper solution of  \eqref{e1.1}.
We can define a lower solution by reversing all the inequalities in the 
obvious places. Moreover, one can
easily prove an analogue of Lemma \ref{lem4.1} for lower solutions.
\end{remark}

Next we present some conditions so that the global fast solution is possible.


\begin{theorem} \label{thm4.2}
If $pq>rs$, then the free boundary problem \eqref{e1.1} admits a global 
fast solution, provided the initial data is suitably small and $h_0$ 
is suitably small.
\end{theorem}

\begin{proof}
 It suffices to construct the suitable global supersolution. 
Inspired by \cite{KLL10}, we define
\begin{gather*}
 \sigma(t)=2h_0(2-e^{-\gamma t}),\quad t\geq 0;\\
 V(y)=\cos\bigl(\frac{\pi}{2}y \bigr),\quad 0\leq y \leq 1;\\
 w(t,x)=z(t,x)=\varepsilon e^{-\alpha t}V(\frac{x}{\sigma(t)}\bigr),\quad
 t\geq 0,\; 0\leq x \leq \sigma(t), 
\end{gather*}
where $\gamma,\alpha$ and $\varepsilon > 0$ are to be chosen later.
Direct computation yields
\begin{alignat*}{2}
      &w_t-d_1w_{xx}-w(a_1-b_1w^r+z^p) \\
      &\geq\varepsilon e^{-\alpha t}[-\alpha V+d_1V\sigma^{-2}(\frac{\pi}{2})^2-V(a_1-b_1w^r+z^p)]\\
      &\geq \varepsilon e^{-\alpha t}V\bigl[-\alpha + \bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_1}{16h_0^{2}}-a_1-\varepsilon^p\bigr],
\end{alignat*}
 and
\[
z_t-d_2z_{xx}-z(a_2+z^q-b_2z^s)
\geq  \varepsilon e^{-\alpha t}V
\bigl[-\alpha + \bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_2}{16h_0^{2}}
-a_2-\varepsilon^q\bigr],
\]
for all $t>0$ and $0 < x < \sigma(t)$. On the other hand, we have 
$\sigma'(t)=2\gamma h_0e^{-\gamma t}>0$ and
$-w_x(t,\sigma(t))=-z_x(t,\sigma(t))< 2\varepsilon \sigma^{-1}(t)e^{-\alpha t}$.
Now we choose $h_0$ that satisfies
\begin{gather*}
 a_i\leq \bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_i}{64h^{2}_0},\; i=1,2;\quad
  \alpha =\gamma =\min\big\{ \bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_1}{64h^{2}_0},
\bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_2}{64h^{2}_0}\big\},\\
\varepsilon=\min \Big\{\Big(\bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_1}{64h^{2}_0}\Big)^{1/p},
  \Big(\bigl(\frac{\pi}{2}\bigr)^{2}\frac{d_2}{64h^{2}_0}\Big)^{1/q},
   \frac{8h_0^2 \gamma}{\mu\pi(1+\rho)} \Big\}\,.
\end{gather*}
Then we have
\begin{gather*}
w_t-d_1w_{xx}\geq w(a_1-b_{1}w^r+z^p),\quad t > 0,\; 0< x < \sigma(t), \\
z_t-d_2z_{xx}\geq z(a_2+w^q-b_{2}z^s),\quad t > 0,\; 0< x < \sigma(t), \\
w(t,0)\geq0,\quad  z(t,0)\geq0,  \\
w(t,x)=z(t,x)=0,\quad \sigma'(t)> -\mu (\frac{\partial w}{\partial x}
+\rho\frac{\partial z}{\partial x}),\quad  t > 0,\; x =\sigma(t), \\
 \sigma(0)=2h_0>h_0.
\end{gather*}
By Lemma \ref{lem4.1}, one can show that $h(t)<\sigma(t)$,
as long as the solution exists $u(t,x)<w(t,x)$, $v(t,x)<z(t,x)$
for $ 0\leq x \leq h(t)$. In particular, it follows from Lemma \ref{lem4.1} 
that $(u,v,h)$ exists globally and $\lim_{t\to \infty} h(t)< \infty$.
\end{proof}

\subsection*{Acknowledgments}
This work was supported by natural science fund for colleges
and universities in Jiangsu province (12KJB110018),
by the natural science fund of Nantong University (12Z031, 13040435, 03080719),
and by the college students innovative projects, 2013.
We would like to express our sincere gratitude to the anonymous referees
their valuable suggestions.

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\end{document}
