Our result reads:

\[thspeed4\] Assume $\kappa_1>\kappa_2$. If $h_{\infty}=+\infty$, then $$\beta_{0}(\mu,\kappa_{1}-\kappa_{2},d_{1})\leq \liminf_{t\rightarrow+\infty}\frac{h(t)}{t}\leq\limsup_{t\rightarrow+\infty}\frac{h(t)}{t}\leq\beta_{0}(\mu,b_{1}, \delta_{1},d_{1}),$$ where $\beta_0$ is determined by Proposition \[pr4.1\].

Note that $$\label{eqspeed4.6111}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial u}{\partial t}- d_{1}u_{xx}=u[b_{1}-\delta_{1}(u+v)]\leq u(b_{1}-\delta_{1}u),}&t>0, ~~ 0<x<h(t), \vspace{7 pt}\\
u_{x}(t,0)=0,u(t,h(t))=0,&t>0,\vspace{7 pt}\\
h'(t)=-\mu u_{x}(t,h(t)),&t>0,\vspace{7 pt}\\
u(0,x)=u_{0}(x),&x\in[0,h_{0}].
\end{array}\right.$$ Thus, the pair $(u, h)$ is a subsolution to the problem $$\label{eqspeed4.6222}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial \bar{u}}{\partial t}- d_{1}\bar{u}_{xx}= \bar{u}(b_{1}-\delta_{1}\bar{u}),}&t>0, ~~ 0<x<\bar{h}(t), \vspace{7 pt}\\
\bar{u}_{x}(t,0)=0,\bar{u}(t,\bar{h}(t))=0,&t>0,\vspace{7 pt}\\
\bar{h}'(t)=-\mu \bar{u}_{x}(t,\bar{h}(t)),&t>0,\vspace{7 pt}\\
\bar{u}(0,x)=u_{0}(x),\bar{h}(0)=h_{0},&x\in[0,h_{0}].
\end{array}\right.$$ By the comparison principle, $h(t)\leq \bar{h}(t)$ for $t>0$. Theorem 4.2 of [@duyihongshou] yields that $$\displaystyle{\lim_{t\rightarrow+\infty}\frac{\bar{h}(t)}{t}=\beta_{0}(\mu,b_{1}, \delta_{1},d_{1})}.$$ Hence $$\limsup_{t\rightarrow+\infty}\frac{h(t)}{t}\leq\beta_{0}(\mu,b_{1}, \delta_{1},d_{1}).$$

Note that $\displaystyle{\limsup_{t\rightarrow+\infty}v(t,x)\leq \kappa_{2}}$ uniformly for $x\in [0,+\infty)$ and $h_{\infty}=+\infty.$ Then, there exists $T_{\varepsilon}>0$ such that $v(t,x)\leq\kappa_{2}+\varepsilon$ and $$\displaystyle{h(T_{\varepsilon})>\frac{\pi}{2}\sqrt{\frac{d_{1}}{\kappa_{1}-\kappa_{2}-\varepsilon}}}~ \text{  when $t>T_{\varepsilon}$.}$$ Next, we consider the following problem $$\label{eqspeed4.6333}
\left\{\begin{array}{ll}
\displaystyle{\frac{\partial \underline{u}}{\partial t}-d_{1}\frac{\partial^{2}\underline{u}}{\partial x^{2}}= \underline{u}(\kappa_{1}-\kappa_{2}-\varepsilon-\underline{u}),}&t>T_{\varepsilon},~~  0<x<\underline{h}(t),\vspace{7 pt} \\
\underline{u}_{x}(t,0)=0,\underline{u}(t,\underline{h}(t))=0,&t>T_{\varepsilon},\vspace{7 pt}\\
\underline{h}'(t)=-\mu \underline{u}_{x}(t,\underline{h}(t)),&t>T_{\varepsilon},\vspace{7 pt}\\
\underline{u}(T_{\varepsilon},x)=u(T_{\varepsilon},x),&x\in[0,h(T_{\varepsilon})].
\end{array}\right.$$ By the comparison principle, we obtain $h(t)\geq\underline{h}(t)$ for $t>T_{\varepsilon}$. From Theorem \[th4.3\], we know that $\underline{h}(\infty)=+\infty$. Using a similar argument as above, we have $\displaystyle{\lim_{t\rightarrow+\infty}\frac{\underline{h}(t)}{t}=\beta_{0}(\mu,\kappa_{1}-\kappa_{2},d_{1})}$. Therefore, $$\liminf_{t\rightarrow+\infty}\frac{h(t)}{t}\geq\beta_{0}(\mu,\kappa_{1}-\kappa_{2},d_{1}).$$

The free boundary problem with a heterogeneous birth rate {#Section4}
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In this section, we consider the free boundary problem - with the heterogeneous birth rates $b_1(x)$ and $b_2(x)$.

Some useful lemmas
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In this subsection, we first study a related 