varepsilon {\rm e}^{-k(t-t_j)}y_{jt}\sin y_j\\[1mm]
\leq&\kk(2\varepsilon +(\pi/\delta)^2-k\rr) u_j
-4\varepsilon {\rm e}^{-k(t-t_j)}\kk(\frac{\pi-\theta}
{h_\infty-g_\infty}\rr)^2\cos \theta
+\frac{\varepsilon\pi|x-x_0|}{\delta^2}{\rm e}^{-k(t-t_j)}|\sin y_j|\\[1mm]
:=&I(t,x).
  \end{array}$$
Obviously, $2\varepsilon +(\pi/\delta)^2-k<0$ by the third inequality of (\ref{4.10}). Since $|y_j(t,x)|\leq \pi-\theta$ in $\overline\Omega_j$, we can decompose $\Omega_j=D_j\bigcup E_j$ with
\bess
  D_j&=&\kk\{(t,x)\in\Omega_j:\, t_j<t<\tau_j,\
 \pi-2\theta<|y_j(t,x)|<\pi-\theta\rr\},\\[1mm]
 E_j&=&\kk\{(t,x)\in\Omega_j:\, t_j<t<\tau_j,
 \ |y_j(t,x)|\leq\pi-2\theta\rr\}.\eess
It is obvious that $|\sin y_j(t,x)|\leq \sin 2\theta$ in $D_j$,
$\cos y_j(t,x)\geq-\cos 2\theta$ in $E_j$. Note that
$ u_j(t,x)\geq 0$ and $|x-x_0|\leq (h_\infty-g_\infty)/2$ in $\Omega_j$,
in view of (\ref{4.10}), we conclude
 $$ I(t,x)\leq \varepsilon {\rm e}^{-k(t-t_j)}\kk(
 -\dd\frac{3 \pi^2}{(h_\infty-g_\infty)^2}\cos \theta
 +\dd\frac{\pi(h_\infty-g_\infty)}{2\delta^2}\sin 2\theta\rr)<0$$
when $(t,x)\in D_j$, and
 $$I(t,x)\leq \varepsilon {\rm e}^{-k(t-t_j)}\kk(\big(2\varepsilon
 +({\pi}/{\delta})^2-k\big)(\cos\theta-\cos 2\theta)
 +\dd\frac{\pi(h_\infty-g_\infty)}{2\delta^2}\rr)<0$$
when $(t,x)\in E_j$. Therefore, (\ref{4.8}) holds.

\vskip 2pt {\it Step 2: Proof of (\ref{4.5})}. By the comparison principle, $v(t,x)\leq\bar v(t)$ for all $t\in[0,\infty)$ and $x\in\mathbb{R}$, where
 $$\bar v(t)=b{\rm e}^{bt}\kk({\rm e}^{bt}-1
 +\dd\frac{b}{\|v_0\|_\infty}\rr)^{-1},$$
which is the solution of the ODE problem
  $$\bar v'(t)=\bar v(b-\bar v),\ \ t>0;\ \ \bar v(0)=\|v_0\|_\infty.$$
Since $\lim_{t\to\infty}\bar v(t)=b$, it is deduced that
$\limsup_{t\to\infty}v(t,x)\leq\lim_{t\to\infty}\bar v(t)=b$
uniformly for $x\in\mathbb{R}$.

On the other hand, note that (\ref{4.4}) and $u(t,x)\equiv 0$ for $t>0$, $ x\not\in(g(t), h(t))$, we see that for any given $0<\sigma\ll 1$, there exists $T_\sigma>0$ such that $u(t,x)<\sigma$ for $t>T_\sigma$ and $x\in\mathbb{R}$.
For any given $\ep>0$ and $L>0$, let $l_\ep$ be determined by Proposition \ref{pB.1}. Then $v$ satisfies
  $$ \left\{\begin{array}{ll}
  v_t-D v_{xx}\geq v(b-v-c\sigma), \ \ &t>T_\sigma, \ \ -l_\ep<x<l_\ep,\\[1mm]
 v(t,\pm l_\ep)>0,&t\ge T_\sigma,\\[1mm]
 v(T_\sigma,x)>0,&-l_\ep\leq x\leq l_\ep.
 \end{array}\right.$$
Thanks to Proposition \ref{pB.1}, $\liminf_{t\to\infty}v(t,x)\geq (b-c\sigma)-\ep$ uniformly on $[-L,L]$. By the arbitrariness of $\ep$ and $L$, we derive that $\liminf_{t\to\infty}v(t,x)\geq b-c\sigma$ uniformly in the compact subset of $\mathbb{R}$.
Since $\sigma>0$ is arbitrary, it follows that
$\liminf_{t\to\infty} v(t,x)\geq b$ uniformly in any bounded
subset of $\mathbb{R}$. The proof is complete.  \ \ \ \ \fbox{}

\subsection{Spreading case {\rm(}$h_\infty-g_\infty=\infty${\rm)}}

We first provide a proposition which asserts the equivalence of $h_\infty-g_\infty=\infty$ and $g_\infty=-\infty$, $h_\infty=\infty$.

\begin{prop}\label{p4.1} \ If $h_\infty-g_\infty=\infty$, then $g_\infty=-\infty$ and $h_\infty=\infty$.
\end{prop}

{\bf Proof.}  Since $h_\infty-g_\infty=\infty$, there exists $T>0$ such that
$h(T)-g(T)>\pi$. Choose a function $\tilde u_0(x)$ satisfying $\tilde
u_0\in C^2[g(T),h(T)]$, $\tilde u_0(x)\leq u_0(x,T)$ in  $[g(T),h(T)]$, $\tilde
u_0(x)>0$ in $(g(T),h(T))$ and $\tilde u_0(g(T))=\tilde u_0(h(T))=0$. Consider the
following problem
  $$ \left\{\begin{array}{ll}
  \tilde u_t-\tilde u_{xx}=\tilde u(1-\tilde u),\ \ &t>T,\ \ g(T)<x<\tilde
h(t),\\[1mm]
 \tilde u(t, g(T))=0=\tilde u(t,\tilde h(t)),\ \ &t\ge T,\\[1mm]
 \tilde h'(t)=-\mu\tilde u_x(t,\tilde h(t)),\ \ &t\ge T,\\[1mm]
 \tilde h(T)=h(T), \ \tilde u(T,x)=\tilde u_0(x), \ &g(T)\leq x\leq h(T).
 \end{array}\right.$$
By Theorem  2.7 of \cite{KY}, this problem has a unique solution $(\tilde u, \tilde
h)$ and exists for all $t\geq T$, and by Theorem 4.2 of \cite{KY},
$\lim_{t\to\infty}\tilde h(t)=\infty$. In view of Lemma \ref{lm3.2}, it concludes
$h(\infty)=\lim_{t\to\infty} h(t)\geq\lim_{t\to\infty}\tilde h(t)=\infty$. 
Similarly, by Lemma \ref{lm3.3} we can obtain $g(\infty)=-\infty$. \ \ \ \ \fbox{}

\vskip 4pt We deal with the weakly hunting case $b>c$ and $ac<1$ firstly.

\begin{theo}\label{th4.3} Assume that $g_\infty=-\infty$, $h_\infty=\infty$. For the
weakly hunting case $b>c$ and $ac<1$, we have
 \bes
 \lim_{t\to\infty}u(t,x)=\frac{1+ab}{1+ac},\ \ \ \
\lim_{t\to\infty}v(t,x)=\frac{b-c}{1+ac}
 \label{4.11}
 \ees
uniformly in any compact subset of $\mathbb