\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,\; 00,\; 00,\; 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 00,\; g(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)0,\; 00,\; 00,\; h(t)0,\; x=h(t),\\ u_x(t,0)=v_x(t,0)=0,\quad t>0,\\ h(0)=b,\quad 0rs$, 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 $pqrs$. 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 $pqrs$. \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_0/2, \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, \; 00, \; 00, \\ 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 $00,\; 00,\\ 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,\; 00, \\ \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,\; 00,\\ 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,\; 00,\\ 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 $00$. 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,\; 00,\; 00,\\ 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 $00$. 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 \[ 00;\quad \overline{v}(0)=\|v_0\|_\infty. \] 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 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 $pq0$ and $pqa_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 $pq h_0$, $(u(t,x),v(t,x))\leq (\mathcal{M}_1,\mathcal{M}_2):=(\overline{u},\overline{v})$. 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 $pqrs$} 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, $\overline{h}'(t^*)h(0)$. 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)