$h^{'}(t)\rightarrow 0$ as $t\rightarrow\infty$ (see Lemma 3.3 in [@twotaiwanren]). This is a contradiction. Thus, $$\lim_{t\rightarrow +\infty} \|u(t,\cdot)\|_{C[0, h(t)]}=0.$$

Next, we prove that $\displaystyle{\lim_{t \rightarrow +\infty} v (t,x)= \kappa_{2}}$. Having $\displaystyle{\lim_{t\rightarrow +\infty} \|u(t,\cdot)\|_{C[0, h(t)]}=0}$ implies that, for any $\varepsilon \in (0,1)$, there exists $T>0$ such that $0 \leq u(t,x) \leq \varepsilon $ for all $t > T$ and $x\in (0 + \infty)$. Thus, $$\label{eqth4.2y2}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial v}{\partial t}\geq d_{2}v_{xx}+v\left[\frac{b_{2}v}{\varepsilon+v}-\delta_{2}(\varepsilon+v)\right],}&t>T, ~~x>0,\vspace{7 pt}\\
v_{x}(t,0)=0, ~~ v(t,+\infty)\geq 0,&t>T, \vspace{7 pt}\\ v(T,x)>0.
\end{array}\right.$$ By Lemma \[lexin3.4\] and the arbitrariness of $\varepsilon$, we have $\displaystyle{\liminf_{t \rightarrow +\infty}v(t,x)\geq b_2/\delta_2=\kappa_{2}}$ uniformly in any bounded subset of $[0, +\infty)$. This, together with the fact $\displaystyle{\limsup_{t \rightarrow +\infty}v(t,x)\leq \kappa_{2}},$ shows that $\displaystyle{\lim_{t \rightarrow +\infty}v(t,x)=\kappa_{2}}$.

The invasion dynamics {#the fitness benefit}
---------------------

\[th4.1\] Suppose $(u,v,h)$ is the solution of system under conditions . If $h_{\infty}=+\infty$, then $\displaystyle{\lim_{t\rightarrow +\infty}u(t,x)=\kappa_{1}}$ and $\displaystyle{\lim_{t\rightarrow +\infty}v(t,x)=0}$ uniformly in any compact subset of $[0,+\infty)$.

Consider the system $$\label{eq ode1}
\left\{\begin{array}{l}
\tilde{u}^{'}(t)=\delta_{1}\tilde{u}(\kappa_{1}-\tilde{u}),t>0,\vspace{7 pt}\\
\tilde{u}(0)=\|u_{0}\|_{L^{\infty}([0,h_{0}])}.
\end{array}\right.$$ Then, $\displaystyle{\lim_{t\rightarrow+\infty}\tilde{u}(t)=\kappa_{1}}$ and $u(t,x)\leq \tilde{u}(t)$. Consequently, we have $$\limsup_{t\rightarrow+\infty}u(t,x)\leq\kappa_{1}\text{ uniformly for $x\in [0,+\infty)$.}$$ In a similar manner, we can obtain that $$\displaystyle{\limsup_{t\rightarrow+\infty} v(t,x) \leq \kappa_{2}}\text{  uniformly for $x \in [0,+\infty)$.}$$ Since $\kappa_{1}>\kappa_{2}$ then, for $\displaystyle{\delta=\frac{\kappa_{1}-\kappa_{2}}{2}}$, there exists $T_{1}>0$ such that $v(t,x)\leq \kappa_{2}+\delta$ for 