where $$\underline{\mu}=\frac{\tilde{\delta}\tilde{\gamma} h^{2}_{0}}{4\tilde{M}},~~\tilde{\gamma}=\frac{1}{2}\left[(\frac{\pi}{2})^{2}\frac{d_{1}}{h^{2}_{0}}-b_{1}\right],$$ and $\tilde{\delta}$, $\tilde{M}$ are such that $$\left(\frac{\pi}{2}\right)^{2}\frac{d_{1}}{(1+\tilde{\delta})^{2}h^{2}_{0}}-b_{1}=\tilde{\gamma}$$ and $$u_{0}(x)\leq \tilde{M}\cos\left(\frac{\pi}{2}\frac{x}{h_{0}(1+\tilde{\delta}/2)}\right), \text{ for }x\in [0,h_{0}].$$ Therefore, $h_{\infty}< +\infty$.

The spreading speed {#the speed}
-------------------

If spreading occurs, it is important to estimate the spreading speed of $h(t)$. Following an idea in [@twotaiwanren], one can obtain a rough estimate of the spreading speed as stated in the following theorem.

\[th6.1\] Suppose that $\kappa_1>\kappa_2$ and let $(u,v,h)$ be the solution of . If $h_{\infty}=+\infty$, $u_{0}(x)\leq \kappa_{1}$ in $[0,h_{0})$, $~v_{0}(x)>0$ in $[0,+\infty)$ and $~\displaystyle{\liminf_{x\rightarrow+\infty}v_{0}(x)\geq\kappa_{2}}$, then $$\limsup_{t\rightarrow+\infty}\frac{h(t)}{t}\leq s_{*},$$ where $s_{*}$ is the minimal speed of the traveling waves to the problem related with in the entire space. This estimation of the spreading speed is independent of $\mu$.

However, in the fitness benefit case, we can derive an estimate better than the one in Theorem \[the speed\]. We first recall Proposition 5.1 of  [@dusupandinf].

\[pr4.1\] For any given constants $d_{1}>0$, $b_{1}>0$, $\delta_{1}>0$ and $\beta\in [0,2\sqrt{b_{1}d_{1}})$, the problem $$\label{speed1}\begin{array}{l}
-d_{1}U^{''}+\beta U^{'}=b_{1} U-\delta_{1} U^{2}~ \text{ for }~x\in (0,\infty),\vspace{7 pt}\\
U(0)=0,
\end{array}$$ admits a unique positive solution $U=U_{\beta}$, which depends on $d_{1},b_{1},\delta_{1},\beta$, and satisfies $U_{\beta}(x)\rightarrow \kappa_{1}$ as $x\rightarrow +\infty$. Moreover, $U^{'}(x)>0$ for $x\geq0,$ and for each $\mu>0$, there exists a unique $\beta_{0}=\beta_{0}(\mu,d_{1},b_{1},\delta_{1})\in(0,2\sqrt{b_{1}d_{1}})$ such that $\mu U^{'}_{\beta_{0}}(0)=\beta_{0}$.

