that $h_{\infty}>h_0^*$. Then for sufficiently small $\varepsilon>0$, there exists $T>0$ such that $$\displaystyle{h(t)>\frac{\pi}{2}\sqrt{\frac{d_{1}}{\delta_{1}(\kappa_{1}-\kappa_{2})-\varepsilon}}}~\text{ and }~\displaystyle{v(t,x)\leq\kappa_{2}+\frac{\varepsilon}{\delta_{1}}}\text{ for }t\geq T\text{ and }x\in [0,+\infty).$$ Let $\underline{u}$ be the solution of the following problem $$\label{eqth4.3y1}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial \underline{u}}{\partial t}- d_{1}\underline{u}_{xx}=\delta_{1}\underline{u}
\left(\kappa_{1}-\kappa_{2}-\frac{\varepsilon}{\delta_{1}}-\underline{u}\right),}& t>T,~~ 0<x<h(T),\vspace{7 pt}\\
\displaystyle{\underline{u}_{x}(t,0)=0=\underline{u}(t,h(T)), }&t>T,\vspace{7 pt}\\
\underline{u}(T,x)=u(T,x),&0<x<h(T).
\end{array}\right.$$ By the comparison principle, we have $\underline{u}(t,x)\leq u(t,x)$ for all $t\geq T$ and $x\in [0,h(T)]$. Since $h(t)>\frac{\pi}{2}\sqrt{\frac{d_{1}}{\delta_{1}(\kappa_{1}-\kappa_{2})-\varepsilon}}$ for $t>T$ then, by Lemma \[le3.1\], we know that $\displaystyle{\lim_{t\rightarrow+\infty}\underline{u}= \underline{U}>0}$ uniformly in any compact subset of $(0,h(T))$, where $\underline{U}$ is the unique positive solution of $$\label{eqth4.3y2}
\left\{\begin{array}{ll}
\displaystyle{-d_{1}\underline{U}_{xx}=\delta_{1}\underline{U}\left[\kappa_{1}-\kappa_{2}-\frac{\varepsilon}{\delta_{1}}-\underline{U}\right], }& 0<x<h(T),\vspace{7 pt}\\
\underline{U}_{x}(t,0)=0=\underline{U}(t,h(T)).
\end{array}\right.$$ Thus, $$\displaystyle{\liminf_{t\rightarrow+\infty}u(t,x)\geq\lim_{t\rightarrow+\infty}\underline{u}(t,x)=\underline{U}(x)>0},$$ which is a contradiction. Therefore, $h_{\infty}\leq h_0^*$ and this completes the proof.

\[th4.4\] If $h_{0}<h_0^*$, then there exists $\bar{\mu}>0$ such that $h_{\infty}=+\infty$ as $\mu\geq\bar{\mu}$.

Since $\displaystyle{\limsup_{t\rightarrow+\infty}v(t,x)\leq \kappa_{2}+\varepsilon}$ uniformly for $x\in[0,+\infty)$, then there exists $T_{1}>0$ such that $v(t,x)\leq \kappa_{2}$ when $t>T_{1}$. So, $(u,h)$ satisfies $$\label{eq new-add-sharp-criteria1}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial u}{\partial t}\geq d_{1}u_{xx}+\delta_{1}u[\kappa_{1}-\kappa_{2}-u],}& t>T_{1},~~ 0<x<h(t),\vspace{7 pt}\\
h'(t)=-\mu u_{x}(t,h(t)),&t>T_{1},\vspace{7 pt}\\
u_{x}(t,0)= 0, u(t,h(t))= 0,&t>T_{1},\vspace{7 pt}\\
u(T_{1},x)>0, &0<x<h(T_{1}).
\end{array}\right.$$ Note that, $u(T_{1},x)$ depends on $\mu$. So, we consider the following problem.

$$\label{eq new-add-sharp-criteria2}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial \widetilde{u}(t,x)}{\partial t}= d_{1}\widetilde{u}_{xx}+\widetilde{u}(b_{1}-\delta_{1}(\widetilde{u}+\widetilde{v})),}&t>0, ~~ 0<x<h_{0}, \vspace{7 pt}
\\

\displaystyle{\frac{\partial \widetilde{v}(t,x)}{\partial t}=d_{2}\widetilde{v}_{xx}+\widetilde{v}\left(\frac{b_{2}\widetilde{v}}{\widetilde{u}+\widetilde{v}}-\delta_{2}(\widetilde{u}+\widetilde{v})\right),}&t>0, ~~ 0<x<h_{0},\vspace{7 pt}\\
\widetilde{u}_{x}(t,0)=\widetilde{v}_{x}(t,0)=0,&t>0,\vspace{7 pt}\\
\widetilde{u}(t,h_{0})=0,&t>0,\vspace{7 pt}\\
\widetilde{u}(0,x)=u_{0}(x),&0<x<h_{0},\vspace{7 pt}\\
 \widetilde{v}(0,x)=\max\left\{\kappa_{2},\|v_{0}\|_{L^{\infty}([0,+\infty))}\right\},&0<x<h_{0},\vspace{7 pt} 