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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 249, 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 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/249\hfil Mutualistic models with double fronts]
{Existence of global solutions to a mutualistic model with double fronts}

\author[M. Li, L. Lin \hfil EJDE-2015/249\hfilneg]
{Mei Li, Lin Lin}

\address{Mei Li \newline
School of Mathematical Science,
Nanjing Normal University,
Nanjing 210023, China. \newline
School of Applied Mathematics,
Nanjing University of Finance and Economics,
Nanjing 210023, China}
\email{limei@njue.edu.cn}

\address{Lin Lin \newline
School of Mathematical Science,
Nanjing Normal University,
Nanjing 210023, China}
\email{linlin@njnu.edu.cn}

\thanks{Submitted December 14, 2014. Published September 25, 2015.}
\subjclass[2010]{35R35, 35K60}
\keywords{Mutualistic model; free boundary; grow-up solution; 
\hfill\break\indent global fast solution; global slow solution}

\begin{abstract}
 We study a system of semilinear parabolic equations with two free boundaries
 describing the spreading fronts of the invasive species in a mutualistic
 ecological model. We establish the existence and uniqueness of a local
 classical solution and then study the asymptotic behavior of the free boundary
 problem. The results indicate that two free boundaries tend monotonically to
 finite values at the same time, or to infinite simultaneously. Also
 the free boundary problem admits a global slow solution with unbounded
 free boundaries if the geometric average of the interaction coefficients
 is less than 1, while if it is bigger than 1  there exist the grow-up
 solution and global fast solution with bounded free boundaries.
\end{abstract}

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

\section{Introduction}

Free boundary problems associated with the ecological models have attracted
considerable research attention in the past because of their relevance in
applications. For example, Lin \cite{LIN} introduced the free boundary
in a predator-prey model.  Du and Lou \cite{DL3} considered a two free boundaries
problem with a general nonlinear term.
Wang and Zhao \cite{Wa, WZ} studied the Lotka-Volterra type prey-predator model.
While Lotka-Volterra type competition models had been discussed by Du and
 Lin  \cite{DL1}, and Guo and Wu  \cite{GW}. Some free boundary problems
describing tumor growth had been considered by Tao and Xu \cite{TY, XS}.

For the mutualistic model, Kim and Lin \cite{KLL} studied  the
 free boundary problem
\begin{gather}
u_{t}-d_1 u_{xx}=u(a_1-b_{1}u+c_{1}v),\quad t>0, \; 0<x<h(t),  \nonumber\\
v_{t}-d_2 v_{xx}=v(a_2+b_{2}u-c_{2}v),\quad t>0, \; 0<x<\infty, \nonumber\\
u(t, x)=0,\quad t\geq 0, \;  h(t)<x<\infty,  \nonumber\\
u=0,\quad h'(t)=-\mu \frac{\partial u}{\partial x},\quad  t>0, \;  x=h(t),  \nonumber\\
\frac{\partial u} {\partial x}(t,0)=\frac{\partial v} {\partial x}(t,0)=0,\quad t>0,
\nonumber\\
h(0)=b, \quad (0<b<\infty),\nonumber\\
u(0, x)=u_0(x)\geq 0,\quad  0\leq x\leq b, \nonumber\\
v(0, x)=v_0(x)\geq 0,\quad 0\leq x\leq \infty,
\label{f2}
\end{gather}
and found  blowup  and global solutions.

The condition on the free boundary is
$h'(t)=-\mu u_{x}(t, h(t))$ called the one-phase Stefan condition,
and it was given by Josef Stefan in his papers published in 1989.
Ecologically, it means that the amount of the species flowing across
the free boundary is increasing with respect to the moving length \cite{LIN}.

As for the one-phase Stefan problem for the heat equation with a
superlinear reaction term
\begin{equation}
\begin{gathered}
u_{t}- u_{xx}=u^{1+p},\quad t>0, \; 0<x<h(t), \\
h'(t)=-\frac{\partial u}{\partial x},\quad   t>0, \; x=h(t), \\
\frac{\partial u}{\partial x}(t, 0)=u(0, h(t))=0,\quad t>0,\\
 h(0)=b,  \quad (0<b<\infty),\\
u(0, x)=u_0(x)\geq 0,\quad 0\leq x\leq b,
\end{gathered} \label{f4}
\end{equation}
it was shown in \cite{SP1, SP2} that  all global solutions are
bounded and decay uniformly to $0$ as $t\to\infty$ if the
initial data is small, while if it is big, the
solution will blow up in a finite time. Moreover they showed that
there exist global solutions with slow decay and unbounded free
boundary.

Considering two species mutualistic model proposed by May \cite{May} in 1976,
and the model is described by the following coupled ODE system:
\begin{equation}
\begin{gathered}
\dot{u}(t)=r_1u(1-\frac{u}{K_1+\alpha_1v}), \\
\dot{v}(t)=r_2v(1-\frac{v}{K_2+\alpha_2u}),
\end{gathered} \label{aode1}
\end{equation}
where $r_i, K_i, \alpha_i,(i=1,2)$ are positive constants.
We  deduce that, if $\alpha_1\alpha_2>1$, the solution would
grow up, which means that it becomes infinite as the time goes to infinity,
while if $\alpha_1\alpha_2<1$ there is an unique positive equilibrium
$(\frac{K_1+\alpha_1K_2}{1-\alpha_1\alpha_2},
\frac{K_2+\alpha_2K_1}{1-\alpha_1\alpha_2})$.
Linearization and spectrum analysis shows that the unique positive equilibrium
is locally asymptotically stable, and
it is globally asymptotically stable in the positive quadrant by constructing
the Lyapunov function.

Motivated by the former work, we study the following  mutualistic
model with double fronts,
\begin{equation}
\begin{gathered}
\frac{\partial u}{\partial t}=a\frac{\partial^2u}{\partial x^2}
+r_1u(1-\frac{u}{K_1+\alpha_1v}),\quad t>0, \; g(t)<x<h(t),\\
\frac{\partial v}{\partial t}=b\frac{\partial^2v}{\partial x^2}
 +r_2v(1-\frac{v}{K_2+\alpha_2u}),\quad t>0, \; -\infty<x<\infty,\\
u(t,x)=0,\quad t>0, \; x\leq g(t) \text{ or } x\geq h(t),\\
g(0)=-h_0,\quad  g'(t)=-\mu \frac{\partial u}{\partial x}(t, g(t)), \quad t>0, \\
 h(0)=h_0, \quad h'(t)=-\mu \frac{\partial u}{\partial x}(t, h(t)), \quad t>0,\\
u(0,x)=u_0(x),\quad v(0,x)=v_0(x),\quad -\infty<x<\infty,
\end{gathered} \label{a3}
\end{equation}
where $x=g(t)$ and $x=h(t)$ are the moving left and right boundaries to be
determined, and $h_0$ and $\mu$ are positive constants.
Throughout this paper the initial functions $u_0$ and $v_0$ are nonnegative
and satisfy
\begin{equation}
\begin{gathered}
u_0\in C^2([-h_0, h_0]),\, u_0(\pm h_0)=0,\quad u_0(x)> 0,\quad  x\in (-h_0, h_0), \\
v_0\in C^2(-\infty, \infty)\cap L^\infty (-\infty, \infty), \quad
 v_0(x)=0,\;  x\in (-\infty, -h_0] \cup [h_0, \infty).
\end{gathered} \label{Ae1}
\end{equation}

The paper is organized as follows.
In the next section,  existence and
uniqueness of local solutions for two free boundaries problem \eqref{a3}
is established by using contraction mapping theorem.
Results relating to global slow solution
for $\alpha_1\alpha_2<1$ are presented in
Section 3. In Section 4, the grow-up solution and
global fast solution  for $\alpha_1\alpha_2>1$ are established.

We end this section by recalling two definitions  which will be used
in next sections.

\begin{definition}[\cite{SP1,SP2}] \rm
A solution $(u, v; g, h)$ of \eqref{a3} is said
to be classical if $u\in C([0, T_{\rm max})\times [g(t), h(t)])\cap
C^{1,2}((0, T_{\rm max})\times (g(t), h(t))$,
$v\in C([0, T_{\rm max})\times (-\infty, \infty ))
\cap C^{1,2}((0, T_{\rm max})\times (-\infty, \infty))
\cap C([0, T_{\rm max})\times L^\infty(-\infty, \infty))$
and $h,g\in C^1[0, T_{\rm max})$ with $T_{\rm max}\leq +\infty$ and
satisfy \eqref{a3}, where $T_{\rm max}$ denotes the maximal existing time
of solution.
\end{definition}

\begin{definition}[\cite{BG,SP1,SP2}] \rm
A solution $(u, v; g, h)$ of \eqref{a3} is said to be global
if $T_{\rm max}=+\infty$. If $T_{\rm max}=\infty$ and
$\lim _{t\to T_{max}}(\|u(t, x)\|_{L^{\infty}[g(t), h(t)]}
+\|v(t, x)\|_{L^{\infty}(-\infty, +\infty)})$
  $\to + \infty$, we say that the solution grows up.
If $T_{\rm max}=\infty$ and
$h_\infty:=\lim _{t\to \infty}h(t)<\infty$,
$g_\infty:=\lim _{t\to \infty}g(t)>-\infty$,
the solution is called global fast solution since that the solution decays
uniformly to $0$ at an exponential rate, while If $T_{\rm max}=\infty$ and
$h_\infty=\infty$, $g_\infty=-\infty$, it is called
global slow solution, whose decay rate is at most polynomial.
\end{definition}


\section{Existence and uniqueness}

In this section, we  first present the following local existence and
uniqueness result by the contraction mapping theorem and then give
the property of the double fronts.

\begin{theorem} \label{thm2.1}
For any given $(u_0, v_0)$ satisfying \eqref{Ae1},
and any $\alpha \in (0, 1)$, there exists a $T>0$ such that
 problem \eqref{a3} admits a unique solution
$$
(u, v; g, h)\in C^{ 1+\alpha,(1+\alpha)/2}(D_{T})
\times C^{ 1+\alpha,(1+\alpha)/2}(D^\infty_{T})\times [C^{1+\alpha/2}([0,T])]^2,
$$
moreover,
\begin{equation}
\begin{aligned}
&\|u\|_{C^{1+\alpha,(1+\alpha)/2}({D}_{T})}
 +\|v\|_{C^{1+\alpha,(1+\alpha)/2}({D}_{T})}\\
&+\|g\|_{C^{1+\alpha/2}([0,T])}+\|h\|_{C^{1+\alpha/2}([0,T])}
\leq K,
\end{aligned} \label{b12}
\end{equation}
where $D_{T}=\{(t, x)\in \mathbb{R}^2: t\in [0,T], x\in [g(t), h(t)]\}$,
$D^\infty_{T}=\{(t, x): t\in [0,T], x\in \mathbb{R}\}$,
$K$ and $T$ only depend on $h_0, \alpha, \|u_0\|_{C^{2}([-h_0, h_0])}$,
$\|v_0\|_{C^{2}([-h_0, h_0])}$ and $\|v_0\|_{L^{\infty}(-\infty, \infty)}$.
\end{theorem}

\begin{proof}
As in \cite{ZL}, we first straighten
the double free boundary fronts by making the following change of
variable:
$$
x=\frac{h(t)-g(t)}{2h_0}y+\frac{h(t)+g(t)}{2}.
$$
Now, a straightforward computation yields
\begin{gather*}
\frac{\partial y}{\partial x}=\frac{2h_0}{h(t)-g(t)},\\
\frac{\partial y}{\partial t}=-2h_0\frac{x(h'(t)-g'(t))
+h(t)g'(t)-h'(t)g(t)}{(h(t)-g(t))^2}.
\end{gather*}
If we set
\begin{gather*}
u(t, x)=u(t, \frac{h(t)-g(t)}{2h_0}y+\frac{h(t)+g(t)}{2}):=w(t, y),\\
v(t, x)=v(t, \frac{h(t)-g(t)}{2h_0}y+\frac{h(t)+g(t)}{2}):=z(t, y),
\end{gather*}
then
\begin{gather*}
u_t=w_t-2h_0\frac{x[h'(t)-g'(t)]+h(t)g'(t)-h'(t)g(t)}{[h(t)-g(t)]^2}w_y=w_t-Aw_y,\\
v_t=z_t-2h_0\frac{x[h'(t)-g'(t)]+h(t)g'(t)-h'(t)g(t)}{[h(t)-g(t)]^2}z_y=z_t-Az_y,\\
u_{xx}=Bw_{yy}, \quad  v_{xx}=Bz_{yy},
\end{gather*}
where
\begin{gather*}
A=A(h, g, y)=\frac{y[h'(t)-g'(t)]+h_0[h'(t)+g'(t)]}{h(t)-g(t)},\\
B=B(h,g)=\frac{4h_0^2}{[h(t)-g(t)]^2}.
\end{gather*}
Problem \eqref{a3} can be reduced to
\begin{equation}
\begin{gathered}
w_{t}=Aw_{y}+a Bw_{yy}+r_1w(1-\frac{w}{K_1+\alpha_1z}),\quad
 t>0, \; -h_0<y<h_0, \\
z_{t}=Az_y+b Bw_{yy}+r_2z(1-\frac{z}{K_2+\alpha_2w}),\quad t>0, \;
  -\infty<y<\infty, \\
w=0,\quad h'(t)=-\frac{2h_0\mu}{h(t)-g(t)}\frac{\partial w}{\partial
y},\quad t>0, \; y\geq h_0,\\
w=0,\quad g'(t)=-\frac{2h_0\mu}{h(t)-g(t)}\frac{\partial w}{\partial
y},\quad t>0, \; y\leq -h_0,\\
h(0)=h_0, \quad g(0)=-h_0, \\
w(0, y)=w_0(y):=u_0(y),\quad z(0, y)=z_0(y):=v_0(y),\quad
-\infty\leq y\leq \infty.
\end{gathered}\label{Hb}
\end{equation}
Now the free boundaries $x=h(t)$ and $x=g(t)$ become the fixed lines
$y=h_0$ and $y=-h_0$ respectively, and the equations become more
complex, since the coefficients in the first and second equations of
\eqref{Hb} contain unknown functions $h(t)$,  $g(t)$ and their derivatives.

 The rest of the  proof is by the contraction mapping argument as in \cite{DL, ZL}
with suitable modifications, and we omit the details here.
\end{proof}

To discuss further on \eqref{a3}, we need some preliminary theorems
which will be used in the sequel.
Next we present the monotonicity of the double fronts.

\begin{theorem} \label{thm2.2}
The two free boundaries for problem \eqref{a3} are strictly monotone,
namely, for any solution on $[0, T]$, we have
 $$
h'(t)>0\quad \text{and} \quad g'(t)<0\quad \text{for } 0\leq t\leq T.
$$
\end{theorem}

\begin{proof}
Using the Hopf Lemma to the system of \eqref{a3}, we  immediately deduce that
 $$
u_x(t, h(t))<0,\ u_x(t,g(t))>0 \quad \text{for } 0\leq t\leq T.
$$
Then, combining the above two inequalities with the Stefan conditions
in \eqref{a3}, the result can be obtained.
\end{proof}

The above theorem indicates that $h(t)$ and $g(t)$ are strictly monotone,
and therefore there exists $h_\infty, -g_\infty\in (0, +\infty]$
such that $\lim_{t\to +\infty} \ h(t)=h_\infty$
and $\lim_{t\to +\infty} \ g(t)=g_\infty$.
Thus, we have four possible cases:
(I) $h_\infty=\infty=-g_\infty$,
(II) $h_\infty<\infty, g_\infty>-\infty$,
(III) $h_\infty<\infty, g_\infty=-\infty$ and
(IV) $h_\infty=\infty, g_\infty>-\infty$.
The following theorem shows that the last two cases are unlikely to occur.
It indicates that both $h_\infty$ and $g_\infty$ are finite or infinite
simultaneously.

\begin{theorem} \label{thm2.3}
Let $(u, v; g, h)$ be  a  solution of \eqref{a3} in
$[0, T_{\rm max})\times [g(t), h(t)]$.
Then $g(t)$ and $h(t)$ satisfy
$$
-2h_0<g(t)+h(t)< 2h_0, \quad  t\in [0,T_{\rm max}).
$$
\end{theorem}

\begin{proof}
It follows from continuity that $g(t)+h(t)<2h_0$  for small $t>0$. Define
$$T:= \sup\{s: g(t)+h(t)<2h_0,\ t\in[0,s)\}.$$
We can deduce that $T=T_{\rm max}$ in the following proof by contradiction.
Suppose that $T<T_{\rm max}$, Then we  have
$$
g(t)+h(t)<2h_0,\quad  t\in[0,T),\quad g(T)+h(T)=2h_0.
$$
Hence
\begin{equation}
 g'(T)+h'(T)\geq 0. \label{b121}
\end{equation}
To obtain a contradiction, we define the function
$\mathcal{F}(t,x):=u(t,x)-u(t,-x+2h_0)$
on the region
$$
\Omega' =\{(t, x):  0\leq t\leq T,\;  h_0\leq x\leq h(t)\}.
$$
A straightforward computation yields
$$
\mathcal{F}_{t}=\mathcal{F}_{xx}+c(t,x)\mathcal{F},\quad 0<t\leq T,\; h_0<x<h(t),
$$
with some $c(t,x)\in L^\infty (\Omega')$ and
$$
\mathcal{F}(t, h_0)=0,\ \mathcal{F}(t,h(t))<0,\ 0<t<T.$$
Moreover,
$$
\mathcal{F}(T,h(T))=u(T,h(T))-u(T,-h(T)+2h_0)=u(T,h(T))-u(T,g(T))=0.
$$
Then
\begin{gather*}
\mathcal{F}(t,x)<0,\quad (t,x)\in (0,T]\times (h_0, h(t)),\\
\mathcal{F}_{x}(T,h(T))<0,
\end{gather*}
by applying the strong maximum principle and the Hopf Lemma.
However
$$
\mathcal{F}_x(T,h(T))=u_x(T,h(T))+u_x(T,g(T))=-[g'(T)+h'(T)]/\mu,
$$
namely
$$
g'(T)+h'(T)>0,
$$
which contradicts \eqref{b12}. Therefore
$g(t)+h(t)<2h_0$ for all $0<t<T_{\rm max}$.
Similarly we can prove $g(t)+h(t)>-2h_0$ for all $0<t<T_{\rm max}$.
\end{proof}

Theorem \ref{thm2.1} implies that there exists a $T$ such that the solution exists
in time interval $[0,T]$, and the solution can be further extended to
$[0,T_{\rm max})$  with $T_{\rm max}\leq +\infty$ by Zorn's lemma.
The maximal exist time of the solution
$T_{\rm max}$ depends on a prior estimate with respect to
$\|u\|_{L^\infty}$, $\|v\|_{L^\infty}$ and $g'(t), h'(t)$. Next we show
that if $\|u\|_{L^\infty}<\infty$, then the solution is global.
For this purpose we first provide the following lemma.

\begin{lemma} \label{lem2.4}
Suppose that $\overline{M}:=\|u\|_{L^\infty([0, T]\times[g(t), h(t)])}<\infty$.
Then the solution of the free boundary problem \eqref{a3} satisfies
\begin{gather*}
0\leq v\leq M_2(\overline{M})\quad   \text{for }   0\leq t\leq T,\;
  -\infty\leq x<\infty,\\
 0<-g'(t), h'(t) \leq M_3(\overline{M}) \quad  \text{for } 0\leq t\leq T,
\end{gather*}
where $M_2, M_3$ are independent of $T$.
\end{lemma}

\begin{proof}
Because of  $\overline{M}:=\|u\|_{L^\infty([0, T]\times[g(t), h(t)])}<\infty$,
we obtain
$$
v_t-bv_{xx}\leq r_2v(1-\frac{v}{K_2+\alpha_2\bar{M}})
$$
for $0<t\leq T$, $-\infty<x<\infty$, then we deduce the estimate for $v$  by the
Phragman-Lindelof principle.
Set
$$
\Omega =\{(t, x):  0<t\leq T,  g(t)<x<g(t)+\frac 1M\}
$$
and define an auxiliary function
$$
w(t, x)=\overline{M}[2M(x-g(t))-M^2(x-g(t))^2].
$$
Next, we choose $M$ such that $w(t, x)$ is the supersolution of
$u(t, x)$ in $\Omega$.
Directly computations show that
\begin{gather*}
w_t=-2\overline{M} Mg'(t)\big[1-M(x-g(t))\big]\geq 0,\\
-w_{xx}=2\overline{M} M^2,\\
r_1u(1-\frac{u}{K_1+\alpha_1v})\leq r_1\overline{M}.
\end{gather*}
If $M^2\geq r_1/(2a)$, we have
$$
w_t-aw_{xx}\geq 2a\overline{M} M^2\geq r_1\overline{M}
\geq r_1u(1-\frac{u}{K_1+\alpha_1v}).
$$
On the other hand,
\begin{gather*}
w(t, g(t)+\frac 1M)=\overline{M}\geq u(t, g(t)+\frac 1M),\\
w(t, g(t))=0=u(t, g(t)).
\end{gather*}
Recalling that $u_0(-h_0)=0$ and
$u'_0(-h_0)=-g_1/\mu$ gives that there exists $0<\delta <h_0$ such that
$u_0(x)\leq \frac 34 \overline{M}$ and $|u'_0(x)|\leq |g_1/\mu|+1$ for
$x\in [-h_0, -h_0+\delta]$, we then have $w(0, x)\geq u_0(x)$ in
$[-h_0, -h_0+\frac 1M]$ if
$M\geq \max \{ \frac 1\delta, \frac {|g_1|/\mu+1}{M_1}\}$.
Using the comparison principle yields $u(t,x)\leq w(t, x)$ in $\Omega$.
Noticing that $u(t, g(t))=w(t,g(t))=0$, we have
$$
u_x(t, g(t))\leq w_x(t, g(t))=2M\overline{M}.
$$
Note that the free boundary condition in \eqref{a3} deduces to
$$
0<-g'(t)\leq 2\mu M\overline{M}:= M_3,\quad 0<t\leq T,
$$
where $M_3$ is independent of $T$.
Analogously, we can define
$$
w(t, x)=\overline{M}[2M(h(t)-x)-M^2(h(t)-x)^2]
$$
over the region
$$
\Omega' =\{(t, x):  0<t\leq T,  h(t)-\frac 1M<x<h(t)\},
$$
and derive that $0<h'(t)\leq M_3,\quad 0<t\leq T$.
\end{proof}

\begin{theorem} \label{thm2.5}
Problem \eqref{a3} admits a unique global solution.
\end{theorem}

\begin{proof}
It follows from the uniqueness that there is a number $T_{\rm max}$ such
that $[0, T_{\rm max})$ is the maximal time interval in which the
solution exists. Next we show that $T_{\rm max}=\infty$.
Arguing indirectly, we assume that
$T_{\rm max}<\infty$. It is easy to see that $(le^{r_1t}, le^{r_2t})$
 is the upper solution of the \eqref{a3}, where
\[
l=\max\{\max_{[-h_0, h_0]}u(x,0), \|u(x,0)\|_{L^\infty(-\infty, \infty)}\}.
\]
 We now fix $M>T_{\rm max}$.
Then $u(x,t)$ $\leq le^{r_1M}$ in $[0, T_{\rm max})\times[g(t), h(t)]$.
 By Lemma \ref{lem2.4}, we can find $M_2, M_3$ independent of $T$ such that
\begin{gather*}
0\leq v\leq M_2\quad  \text{for }  0\leq t< T_{\rm max},\;  -\infty\leq x<\infty,\\
0<-g'(t), h'(t) \leq M_3 \quad  \text{for }   0\leq t< T_{\rm max}.
\end{gather*}
It then follows from the proof of Theorem \ref{thm2.1} that there exists
a $\tau>0$ depending only on $M$, $M_2$ and $M_3$ such that the solution
 \eqref{a3} with initial time $T_{\rm max}-\tau/2$ can be extended uniquely to the
time $T_{\rm max}-\tau /2+\tau$. But this contradicts the assumption.
The proof is complete.
\end{proof}


\section{Global bounded solution}
To obtain the existence of a global solution, we first derive a priori
estimate for the solution of \eqref{a3}.

\begin{lemma} \label{lem3.1}
If  $\alpha_1\alpha_2<1$, then the solution of the free boundary problem \eqref{a3}
satisfies
\begin{gather*}
0<u(t, x)\leq C_1\quad \text{for }  0\leq t\leq T, \ g(t)< x< h(t),\\
0\leq v(t, x)\leq C_2\quad \text{for } 0\leq t\leq T,\ -\infty< x<\infty,
\end{gather*}
where $C_i$ is independent of $T$ for $i=1, 2$.
\end{lemma}

\begin{proof}
Firstly we have that $u>0$ in $[g(t), h(t)]\times [0, T]$
and $v\geq 0$ in $(-\infty, \infty)\times [0, T]$ provided that solution exists.

Since the solution is classical in $[0, T]$,  there exists a  $\tilde K(T)$
such that $u(t, x)\leq \alpha_1\tilde K$ and $v(t, x)\leq \tilde K$.
 Next we give the proof for $u(t, x)\leq C_1$ and $v(t, x)\leq C_2$,
where
\begin{gather*}
C_1 :=m \frac{K_1+K_2\alpha_1}{1-\alpha_1\alpha_2} >\max_{[-h_0, h_0]} u_0(x),\\
C_2 :=m \frac{K_1\alpha_2+K_2}{1-\alpha_1\alpha_2}
>\|v_0\|_{L^\infty(-\infty, \infty)}
\end{gather*}
for some $m>1$.

Because  the interval $(-\infty, \infty)$ is unbounded,  maximum principle
does not apply. Next we prove that for any $l>h_0$,
\begin{gather*}
u(t, x)\leq C_1+\alpha_1 \frac {\tilde K[x^2+2\max(a, b) t]}{l^2},\\
v(t, x)\leq C_2+\frac {\tilde K[x^2+2\max(a, b) t]}{l^2}
\end{gather*}
for $0\leq t\leq T$, $ -l\leq x\leq l$.
 Setting
\begin{gather*}
\overline{u}(t, x)= C_1+\alpha_1 \frac {\tilde K[x^2+2\max(a, b) t]}{l^2},\\
\overline{v}(t, x)=C_2+\frac {\tilde K[x^2+2\max(a, b) t]}{l^2},
\end{gather*}
then $(\overline{u}, \overline{v})$ satisfies
\begin{gather*}
\overline{u}_{t}-a\overline{u}_{xx}\geq r_1\overline{u}
 (1-\frac{\overline{u}}{K_{1}+\alpha_1\overline{v}}),\quad 0<t \leq T, \; -l<x<l,  \\
\overline{v}_{t}-b \overline{v}_{xx}\geq r_2\overline{v}
(1-\frac{\overline{v}}{K_{2}+\alpha_2\overline{u}}),\quad  0<t \leq T, \ ;-l<x<l, \\
\overline{u} \geq C_1+\alpha_1 \tilde K>u,\quad
 \overline{v}\geq C_2+\tilde K>v,\quad 0<t\leq T, \; x=\pm l, \\
\overline{u}(0, x)\geq C_1>u_0(x),\quad -l\leq x\leq l \\
\overline{v}(0, x)\geq C_2> v_0(x), \quad -l\leq x\leq l.
\end{gather*}
 It follows  that $ u\leq \overline{u}$ and $v\leq \overline{v}  $  by using
the maximum principle on  $[0, T]\times [-l, l]$.
Now for any fixed $(t_0, x_0)\in [0, T]\times (-\infty, \infty)$,
letting $l$ sufficiently large so that
$(t_0, x_0)\in [0, T]\times [-l, l]$, we deduce from the above proof that
\begin{gather*}
u(t_0, x_0)\leq \overline{u}(t_0, x_0)
= C_1+\alpha_1 \frac {\tilde K[x^2_0+2\max(a, b) t_0]}{l^2},\\
v(t_0, x_0)\leq \overline{v}(t_0, x_0)
=C_2+\frac {\tilde K[x^2_0+2\max(a, b) t_0]}{l^2}.
\end{gather*}
Taking $l\to \infty$ gives the desired estimates.
\end{proof}

Combing Theorem \ref{thm2.5} with Lemma \ref{lem3.1} yields the following result.

\begin{theorem} \label{thm3.2}
If parameters $\alpha_1, \alpha_2$ in double free boundaries
problem \eqref{a3} satisfy  $\alpha_1\alpha_2<1$,
then  \eqref{a3} admits a unique global bounded solution.
\end{theorem}

Next we  discuss the long-time behavior of the free boundary problem \eqref{a3}.
 We first present the slow solution.

\begin{theorem} \label{f5}
If $\alpha_1\alpha_2<1$ and  $h_0>\frac{\pi}{2}\sqrt{a/r_1}$,
the free boundaries of the problem \eqref{a3} satisfy
$h_\infty=\infty$ and $g_\infty=-\infty$.
\end{theorem}

\begin{proof}
Combing Lemma \ref{lem2.4} with Theorem  \ref{thm3.2}, we know that 
the solution is global, 
$x=g(t)$ is monotonic decreasing and $x=h(t)$ is monotonic increasing.
Assuming that $g_\infty>-\infty$  by contradiction, then we have
  $\lim_{t\to +\infty}g'(t)=0$.

On the other hand, the condition $1>a/{r_1}(\frac{\pi}{2h_0})^2 $
implies that $1>\lambda_1$, where $\lambda_1$ denotes the first
eigenvalue of the problem
$$
 -(a/{r_1})\phi''=\lambda \phi \ \ \text{in}\ \ (-h_0, h_0),\quad \phi(\pm h_0)=0.$$
Therefore, for all small $\delta>0$, the first eigenvalue
$\lambda^\delta _1$ of the problem
$$ 
-a\phi''+\delta \phi'=\lambda r_1\phi \quad \text{in } (-h_0, h_0),\quad 
\phi(\pm h_0)=0
$$
satisfies $\lambda^\delta _1<1$. Fix such a $\delta >0$
and consider the problem
\begin{equation}
L_\delta \psi=\psi -\frac{\psi^2}{K_1} \quad \text{in }
 (-h_0,h_0),\quad \psi(\pm h_0)=0,\label{m0}
\end{equation}
where $L_\delta \psi=-(a \psi ''-\delta \psi ')/{r_1}$.  It is well
known \cite[Proposition 3.3]{CC} that  \eqref{m0}
admits a unique positive solution $\psi=\psi_\delta$. By the
moving plane method one easily sees that $\psi(x)$ is symmetric
about $x=0$ with  $\psi ' (x)> 0 $ for $x\in [-h_0, 0)$. Moreover using the
comparison principle, we have $\psi< K_1$ in $[-h_0, h_0]$.
We now set
$$ 
\mathcal{F}(t, x)=\psi \Big(\frac {-h_0 }{g(t)}x \Big),
$$
and directly compute
\[
\mathcal{F}_t -a \mathcal{F}_{xx}
= \frac {h_0 x}{g^2(t)} g'(t) \psi '-a \frac {h_0^2}{g^2(t)} \psi ''
=\frac {h_0^2}{g^2(t)}[ -a \psi ''+ \frac {x g'(t)}{h_0} \psi '].
\]
Note that $g'(t)\to 0$ as $t\to +\infty$, we can choose $T_0>0$
such that $g'(t)>\delta \frac{h_0}{g_\infty}$ for $t\geq T_0$,
then, we obtain
$\frac{xg'(t)}{-h_0}\geq -\delta $ for $t\geq T_0$ and $x\in [g(t), 0]$, 
which leads to
\[
\mathcal{F}_t -a \mathcal{F}_{xx}\leq  \frac {h_0^2}{g^2(t)}(-a\psi ''+\delta
\psi ')
= \frac {h_0^2}{g^2(t)}r_1(\psi -\frac{\psi^2}{K_1}).
\]
Because of  $0\leq \psi< K_1$ and $\frac {-h_0}{g(t)}\leq 1$, we obtain
$$
\mathcal{F}_t- a \mathcal{F}_{xx}
\leq r_1(\psi -\frac{\psi^2}{K_1})= r_1(\mathcal{F} -\frac{\mathcal{F}^2}{K_1})\quad
\text{for } t\geq T_0, \ x\in [g(t), 0].
$$ 

Now we choose $\delta \in (0, 1)$ sufficiently small so that 
$\delta \mathcal{F}(T_0,x)\leq u(T_0, x)$. Then
 $\underline{u}(t, x):=\delta \mathcal{F}(t, x)$
 satisfies
\begin{gather*}
\underline{u}_t-a\underline{u}_{xx}\leq r_1( \underline{u}
-\frac{ \underline{u}^2}{K_1}),\quad t \geq T_0,\; x\in [g(t), 0], \\
\underline{u}(t, g(t))=0, \quad \underline{u}_x(t, 0)=0,  \quad t\geq T_0,\\
\underline{u}(T_0, x)\leq u(T_0, x), \quad g(T_0) \leq x\leq 0.
\end{gather*} 
So we can use the comparison principle to conclude that
$$
\underline{u} (t, x)\leq u(t, x)\quad \text{for }  t\geq T_0, \; x\in [g(t), 0].
$$ 
It follows that
$$
u_x(t, g(t))\geq \underline{u}_x(t, g(t)) =\delta \frac {h_0}{g(t)}
\psi '(h_0)\to \delta \frac {h_0}{g_\infty} \psi '(h_0)>0,
$$ 
which means that $g'(t)\leq -\mu \delta \frac {h_0}{g_\infty} \psi '(h_0)<0$. 
This is a contradiction to the fact that $g'(t)\to 0$ as $t\to \infty$.
This contradiction implies that $g_\infty=-\infty$.
Likewise, we can set
$$
\mathcal{F}(t, x)=\psi \Big(\frac {h_0 }{h(t)}x \Big), \quad x\in [0, h(t)]
$$
to prove that $ h_\infty=+\infty$.
\end{proof}

\section{Global fast solution and grow up solution}

In this section, we discuss the asymptotic behavior of the solution for 
the case $\alpha_1\alpha_2>1$, which is more complicated than that for
the case $\alpha_1\alpha_2<1$. At first, we give the grow-up result.

\begin{theorem} \label{g1} 
Assume that $\alpha_1\alpha_2>1$, then the solution of  \eqref{a3}  with any 
nontrivial nonnegative initial data grows up when $h_0$ is sufficiently large.
\end{theorem}

\begin{proof}
We first show that the solution cannot blow up in any finite time. 
In fact, it has a upper solution $(\overline{u}(t), \overline{v}(t))$ satisfies
\begin{gather*}
\overline{u}_{t}=r_1\overline{u},\quad \overline{v}_{t}=r_2\overline{v},\quad   t>0, \\
\overline{u}(0)=\max_{[-h_0, h_0]}u(x,0)\geq 0, \\
 \overline{v}(0)=\max_{[-h_0, h_0]}v(x,0)\geq 0
\end{gather*}
and the upper solution cannot blow up in finite time.

To prove the solution of  \eqref{a3} grows up, it suffices to compare the 
free boundary problem with the corresponding problem in the fixed domain:
\begin{equation}
\begin{gathered}
u_{t}-a u_{xx}=r_1u(1-\frac{u}{K_1+\alpha_1v}),\quad t>0,\; -h_0<x<h_0, \\
v_{t}-bv_{xx}=r_2v(1-\frac{v}{K_2+\alpha_2u}),\quad t>0, \; -h_0<x<h_0,\\
u(t, -h_0)=v(t, -h_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, \quad -h_0\leq x\leq h_0,\\
\ v(0, x)=v_0(x)\geq 0,\quad -h_0\leq x\leq h_0.
\end{gathered}\label{f13}
\end{equation}

On the other hand, we want to find a lower solution of \eqref{f13} that 
increases exponentially.
Let $(\hat{u}, \hat{v})= (\delta_1w, \delta_2w)$,
where $\delta_i(i=1, 2)$ is some positive constant. Then 
$(\hat{u}, \hat{v})$ is a lower solution of \eqref{f13} if 
$(\delta_1w, \delta_2w)$ satisfies the relations
\begin{equation}
\begin{gathered}
w_{t}-a w_{xx}\leq r_1w(1-\frac{\delta_1w}{K_1+\alpha_1\delta_2w}),\quad
t>0,\; -h_0<x<h_0, \\
w_{t}-bw_{xx}\leq r_2w(1-\frac{\delta_2w}{K_2+\alpha_2\delta_1w}),\quad
 t>0, \; -h_0<x<h_0,\\
w(t, -h_0)=w(t, -h_0)=0, \quad t>0,\\
\delta_1w(0, x)\leq u_0(x), \quad -h_0\leq x\leq h_0,\\
\delta_2w(0, x)\leq v_0(x), \quad -h_0\leq x\leq h_0.
\end{gathered} \label{f16}
\end{equation}
Then \eqref{f16} holds if
\begin{equation}
\begin{gathered}
w_{t}-a w_{xx}\leq \frac{r_1(\alpha_1\delta_2-\delta_1)w}{\alpha_1\delta_2},
\quad t>0,\; -h_0<x<h_0, \\
w_{t}-bw_{xx}\leq \frac{r_2(\alpha_2\delta_1-\delta_2)w}{\alpha_2\delta_1},
\quad  t>0, \; -h_0<x<h_0,\\
w(t, -h_0)=w(t, -h_0)=0,\quad t>0,\\
\delta_1w(0, x)\leq u_0(x), \quad -h_0\leq x\leq h_0,\\
\delta_1w(0, x)\leq v_0(x),\quad -h_0\leq x\leq h_0.
\end{gathered} \label{f17}
\end{equation}
Recall the assumption in the theorem, let $\delta_i>0$ such that
$$\frac{1}{\alpha_2}<\frac{\delta_1}{\delta_2}<\alpha_1$$
and set
$$
D=\max\{\frac{1}{a}, \frac{1}{b}\},\quad
d^*=\min\big\{\frac{r_1(\alpha_1\delta_2-\delta_1)}{\alpha_1\delta_2a}, 
\frac{r_2(\alpha_2\delta_1-\delta_2)}{\alpha_2\delta_1b}\big\}.
$$
Then $d^*>0$ and thus \eqref{f17} holds if
\begin{equation}
\begin{gathered}
Dw_{t}-w_{xx}\leq d^*w,\quad t>0,\; -h_0<x<h_0, \\
w(t, -h_0)=w(t, h_0)=0,\quad t>0,\\
w_0(x)\leq \min\big\{\frac{u_0(x)}{\delta_1}, \frac{v_0(x)}{\delta_2}\big\},\quad
-h_0\leq x\leq h_0.
\end{gathered}\label{f18}
\end{equation}
Let $w(x,t)=\delta e^{\varepsilon t}\cos (\frac{\pi}{2h_0} x)$.
Direct calculations show that if $h_0>\frac \pi 2\frac 1{\sqrt{d^*}}$, 
then we can choose small $\delta$ and $\varepsilon$ such that $w_t \geq 0$ and
\eqref{f18} holds. Therefore the lower solution $(\hat{u}, \hat{v})$ 
increases exponentially, so does the solution of \eqref{a3}.
\end{proof}

Next we introduce a comparison principle for double free boundaries
 $x=h(t)$ and $x=g(t)$,
which can be proved similarly as \cite[Lemma 3.5]{DL}.

\begin{lemma} \label{g3} %\label{lem4.2}
Suppose that $T\in (0,\infty)$, $\overline{h},\overline{g} \in C^1([0,T])$, 
$\overline{u}\in C(\overline D_{1,T}^*)\cap C^{1,2}(D_{1,T}^*)$ and 
$\overline{v}\in C(\overline D_{2,T}^*)\cap C^{1,2}(D_{2,T}^*)$ with
 $D_{1,T}^*=(0, T]\times (\overline{g}(t), \overline{h}(t))$,
$D_{2,T}^*=(0, T]\times (-\infty, +\infty)$, and
\begin{equation}
\begin{gathered}
\overline{u}_{t}-a \overline{u}_{xx}\geq \overline
r_1u(1-\frac{\overline{u}}{K_1+\alpha_1\overline{v}}),\quad
 t>0, \; \overline{g}(t)<x<\overline{h}(t),  \\
\overline{v}_{t}-b \overline{v}_{xx}\geq \overline
r_2v(1-\frac{\overline{v}}{K_2+\alpha_2\overline{u}}),\quad
 t>0, \; -\infty<x<\infty, \\
\overline{u}(t, x)=0,\quad  t>0, \;  -\infty<x<g(t),  \\
\overline{u}(t, x)=0,\quad  t>0, \;  h(t)<x<\infty,  \\
\overline{u}=0,\quad \overline{h}'(t)\geq -\mu \frac{\partial
\overline{u}}{\partial x},\quad  t>0, \;  x=\overline{h}(t),  \\
\overline{u}=0,\quad \overline{g}'(t)\leq -\mu \frac{\partial
\overline{u}}{\partial x},\quad  t>0, \;  x=\overline{g}(t).
\end{gathered} \label{f19}
\end{equation}
If $-h_0\geq \overline{g}(0), h_0\leq \overline{h}(0)$,
$u_0(x)\leq \overline{u}(0,x)$
in $[-h_0,h_0]$ and $v_0(x)\leq \overline{v}(0,x)$
in $(-\infty,+\infty)$, then the solution $(u, v; g, h)$ of the free
 boundary problem \eqref{a3} satisfies
\begin{gather*}
g(t)\geq\overline{g}(t),\quad h(t)\leq\overline{h}(t)\quad \text{in } (0, T], \\
u(t,x)\leq\overline{u}(t,x)\quad  \text{in } [0, T]\times (g(t), h(t)), \\
v(t,x)\leq \overline{v}(t,x)\quad \text{in } [0, T]\times (-\infty,+\infty).
\end{gather*}
\end{lemma}

\begin{remark} \label{g4} \rm
The  $(\overline{u}, \overline{v}; \overline{h}, \overline{g})$ in 
Lemma \ref{g3} is
usually called an upper solution of the problem \eqref{a3}. 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{g3} for lower solutions.
\end{remark}

 In the following theorem, we show existence of a global fast solution.

\begin{theorem}  \label{g5}
If $\alpha_1\alpha_2>1$, then the free boundary problem \eqref{a3}
admits a global fast solution provided that the initial data $u_0$ 
and $h_0$ are suitably small.
Moreover, there exist constant $\beta =r_1/2$ and
 $\eta=\eta(h_0, K_2, a, \mu, \alpha_2)$
such that
$$
\|u\|_\infty\leq \eta e^{-\beta t},\quad t\geq 0.
$$ 
\end{theorem}

\begin{proof}
  As in \cite{RT}, we have only to find a suitable supersolution.
For $t\geq 0$, define
\begin{gather*}
\sigma (t)=2h_0(2-e^{-\gamma t}),\quad \lambda (t)=-\sigma (t),  \quad 
\mathcal{F}(y)=\cos (\frac{\pi}{2}y), \quad -1\leq y\leq 1,\\
\overline{u}(t, x)=\eta e^{-\beta t}\mathcal{F}(x/\sigma (t)), \quad
 t\geq 0,\quad \lambda(t)\leq x\leq \sigma(t),\\
\overline{v}(t, x)=\max\{2K_2, \|v_0(x)\|_{L^\infty(-\infty, +\infty)}\},\quad
 t\geq 0,\; -\infty\leq x\leq \infty,
\end{gather*}
where $\gamma, \beta$ and $\eta>0$ to be determined later.

Straightforward calculations yield
\begin{align*}
& \overline{u}_t-a\overline{u}_{xx}
 -r_1\overline{u}(1-\frac{\overline{u}}{K_1+\alpha_1\overline{v}})\\
&=\eta e^{-\beta t}[-\beta \mathcal{F}-x\sigma '\sigma^{-2}\mathcal{F}'
 -a\sigma^{-2}\mathcal{F}''-r_1\mathcal{F}(1-\frac{\eta e^{-\beta t}
 \mathcal{F}}{K_1+\alpha_1\overline{v}})]\\
&\geq \eta e^{-\beta t}\mathcal{F}[-\beta +(\frac \pi 2)^2
 \frac {a}{16h_0^2}-r_1]
\end{align*}
for all $t>0$  and  $\lambda (t)<x<\sigma (t)$ and
\begin{align*}
\overline{v}_{t}-b\overline{v}_{xx}-r_2\overline{v}
 (1-\frac{\overline{v}}{K_2+\alpha_{2}\overline{u}})
&=\overline{v}(-r_2+\frac{r_2\overline{v}}{K_2+\alpha_2\eta e^{-\beta t} 
 \mathcal{F}})\\
&\geq 2r_2K_2(-1+\frac {2K_2}{K_2+\alpha_2\eta})
\end{align*}
for all $t>0$ and $-\infty<x<\infty$. On the other hand, we can easily deduce
$\sigma'(t)=2\gamma h_0e^{-\gamma t}>0$, 
$-\overline{u}_x(t, \sigma (t))=\frac {\pi}{2}\eta \sigma^{-1}(t)e^{-\beta t}$ and 
$-\overline{u}_x(t, \lambda (t))=\frac {\pi}{2}\eta \lambda^{-1}(t)e^{-\beta t}$. 
Now we set
$$
h=\frac{\pi}{16}\sqrt{\frac{2a}{r_1}},
$$
choosing 
$$0<h_0\leq h,\quad 
\eta =\min \{\frac {K_2}{\alpha_2}, \ \frac {a\pi}{8\mu}(\frac { h_0} {2h})^2\},\quad
\beta=\gamma=(\frac \pi 2)^2\frac{a}{64h^2}=\frac{r_1}{2}.
$$
 It follows that
\begin{gather*}
\overline{u}_{t}-a\overline{u}_{xx}\geq r_1\overline{u}
 (1-\frac{\overline{u}}{K_1+\alpha_{1}\overline{v}}),\quad t>0,\;
 \lambda(t)<x<\sigma(t),  \\
\overline{v}_{t}-b \overline{v}_{xx}\geq r_2\overline{v}
(1-\frac{\overline{v}}{K_2+\alpha_{2}\overline{u}}),\quad t>0, \;
  -\infty<x<\infty, \\
\overline{u}=0,\quad \sigma'(t)> -\mu \frac{\partial \overline{u}}{\partial
x},\quad t>0, \;  x=\sigma(t),\\
\overline{u}=0,\quad \lambda'(t)< -\mu \frac{\partial \overline{u}}{\partial
x},\quad t>0, \;  x=\lambda(t),\\
\sigma (0)=2h_0>h_0, \lambda (0)=-2h_0<-h_0.
\end{gather*} 
Using Lemma \ref{g3}, we can get that $h(t)<\sigma(t)$,  
$g(t)>\lambda(t)$,
and $u(t, x)<\overline{u}(t, x)$, $v(t, x)<\overline{v}(t, x)$ for 
$g(t)\leq x\leq h(t)$ provided  $(u,v)$ exists.  
Therefore $(u, v)$ exists globally and $g_\infty>-\infty$,   
$h_\infty<\infty$.
\end{proof}

\begin{remark} \rm
If $\alpha_1\alpha_2<1$, Theorem \ref{f5} shows that the solution is slow 
for any initial data. If $\alpha_1\alpha_2>1$, 
Theorem \ref{g1} shows that the solution grows up for $h_0$ is sufficiently large. 
Theorem \ref{g5}  implies that the global fast solution is possible 
if the initial data and $h_0$ is suitably small.
\end{remark}

\subsection*{Acknowledgments}
This work is supported  by grants: No. 11171158 from the
NSFC, No. 11KJA110001 from the NSF of Jiangsu Education Committee,
and No. KYLX\_0719 from Project of Graduate Education Innovation of Jiangsu 
Province.


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\end{document}