it follows from standard elliptic regularity and a diagonal procedure that $(\underline{u}_{L}(x),\bar{v}_{L}(x))$ converges to $(\underline{u}_{\infty}(x),\bar{v}_{\infty}(x))$ uniformly on any compact subset of $[0,+\infty)$, where $(\underline{u}_{\infty},\bar{v}_{\infty})$ satisfies $$\label{eq new-add-a2}
\left\{\begin{array}{ll}
 d_{1}\partial_{xx}\underline{u}_{\infty}+\underline{u}_{\infty}[b_{1}-\delta_{1}(\underline{u}_{\infty}+\bar{v}_{\infty})]=0,& x>0\vspace{7 pt} \\

\displaystyle{d_{2}\partial_{xx}\bar{v}_{\infty}+\bar{v}_{\infty}\left[\frac{b_{2}\bar{v}_{\infty}}{\underline{u}_{\infty}+\bar{v}_{\infty}}-\delta_{2}(\underline{u}_{\infty}+\bar{v}_{\infty})\right]=0,}& x>0\vspace{7 pt}\\

\partial_{x}\underline{u}_{\infty}(0)=\partial_{x}\bar{v}_{\infty }(0)=0, \vspace{7 pt}\\
\displaystyle{\underline{u}_{\infty}(x)\geq\frac{\delta}{2}}, ~~ \bar{v}_{\infty}(x)\leq \kappa_{2}+\delta,&0<x<+\infty.
\end{array}\right.$$ We consider now the following system: $$\label{eq zhbode}
\left\{\begin{array}{ll}
\displaystyle{\frac{d u_{1}}{d t}= u_{1}(b_{1}-\delta_{1}(u_{1}+v_{1})),}&t>0,\vspace{7 pt} \\
\displaystyle{\frac{dv_{1}}{dt}=v_{1}\left(\frac{b_{2}v_{1}}{u_{1}+v_{1}}-\delta_{2}(u_{1}+v_{1})\right),}&t>0,\vspace{7 pt}\\
\displaystyle{u_{1}(0)=\frac{\delta}{2}}, ~~ v_{1}(0)=\kappa_{2}+\delta.
\end{array}\right.$$ Since $\kappa_{1}>\kappa_{2},$ then $(u_{1},v_{1})\rightarrow (\kappa_{1},0)$ as $t\rightarrow +\infty$ (see Lemma 2.2 of [@BOZHENG], for e.g.). Then, the solution $(U,V)$ of the problem $$\label{eq new-add-a3}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial U}{\partial t}=d_{1}U_{xx}+ U(b_{1}-\delta_{1}(U+V)),}&t>0, ~~ x\geq0,\vspace{7 pt}\\
\displaystyle{\frac{\partial V}{\partial t}= d_{2}V_{xx} +V\left(\frac{b_{2}V}{U+V}-\delta_{2}(U+V)\right),}&t>0,~~ x\geq0,\vspace{7 pt}\\
U_{x}(t,0)=V_{x}(t,0)=0,&t>0,\vspace{7 pt}\\
\displaystyle{U(0,x)=\frac{ \delta}{2}}, ~~ V(0,x)= \kappa_{2}+\delta,&x\geq0.
\end{array}\right.$$ satisfies $( U(t,x),V(t,x))\rightarrow (\kappa_{1},0),$ as $t\rightarrow+\infty,$ uniformly in $x\in[0,+\infty)$. By the comparison principle, we have $\underline{u}_{\infty}\geq U$ and $\bar{v}_{\infty}\leq V$ for $t\geq 0$, which immediately yields that $$\lim_{t\rightarrow +\infty}u(t,x)=\kappa_{1}\mbox{ and }\lim_{t\rightarrow +\infty}v(t,x)= 0.$$

The criteria for spreading and vanishing are given in the following theorem.

\[th4.3\] If $\displaystyle{h_{0}\geq\frac{\pi}{2}\sqrt{\frac{d_{1}}{\delta_{1}(\kappa_{1}-\kappa_{2})}}:=h_0^*}$, then $h_{\infty}=+\infty$.

Note that $h(t)$ is nondecreasing. We only need to show that $h_{\infty}<+\infty$ implies $h_{\infty}\leq h_0^*$. It follows from Theorem \[th4.2\] that $h_{\infty}<+\infty$ implies $$\displaystyle{\lim_{t\rightarrow +\infty} \|u(t,\cdot)\|_{C[0, h(t)]}=0}\text{ and }\displaystyle{\lim_{t\rightarrow +\infty}v(t,x)=\kappa_{2}}$$ uniformly in any bounded subset of $[0,+\infty)$. Assume 