next result provides some bounds on the solutions to system with initial conditions .

\[le2.2\] Let $(u,v,h)$ be a solution of for $t\in [0,T]$ for some $T>0$. Then,

(i) $0<u(t,x)\leq M_{1}$ for all $t\in (0,T]$ and $x\in [0,h(t))$, where $$M_1:=\max\left\{\frac{\|b_{1}\|_{L^{\infty}([0,\infty))}}{\delta_{1}}, \|u_{0}\|_{L^{\infty}([0,h_{0}])}\right\}.$$

(ii) $0<v(t,x)\leq M_{2}$ for all $t\in (0,T]$ and $x\in [0, +\infty)$, where $$M_2:=\max\left\{\frac{\|b_{2}\|_{L^{\infty}([0,\infty))}}{\delta_{2}}, \|v_{0}\|_{L^{\infty}([0,+\infty))}\right\}.$$

(iii) $0< h'(t)\leq \Lambda$ for all $t\in (0,T]$, where $\Lambda> 0$ depends on $\mu$, $d_{1}$, $\|u_{0}\|_{L^{\infty}([0,h_{0}])}$ and $\|u_0'\|_{C[0,h_{0}]}$.

The strong maximum principle yields that $u(t,x)>0$ for all $t\in (0,T]$ and $x\in [0,h(t))$, and $v(t,x)>0$ for all $t\in (0,T]$ and $x\in [0,+\infty)$. Note that $u(t,h(t))=0$ yields that $$u_{x}(t,h(t))<0\text{ for all }t\in(0,T].$$ Thus, $h'(t)>0$ for $t\in (0,T]$. Next, we consider the initial value problem $$\label{le2.204}
\left\{\begin{array}{l}
u'(t)=u(t)(\|b_{1}\|_{L^\infty([0,\infty))}-\delta_{1}u(t)),\mbox{ for } t>0,\vspace{7 pt}\\
u(0)=\|u_{0}\|_{L^{\infty}([0,h_{0}])} .
\end{array}\right.$$ From the comparison principle, we know that $$u(t,x)\leq \max\left\{\frac{\|b_{1}\|_{L^{\infty}([0,\infty))}}{\delta_{1}}, \|u_{0}\|_{L^{\infty}([0,h_{0}])}\right\}.$$ Similarly, we can show that $$v(t,x)\leq \max\left\{\frac{\|b_{2}\|_{L^{\infty}([0,\infty))}}{\delta_{2}}, \|v_{0}\|_{L^{\infty}([0,+\infty))}\right\}.$$ To prove (iii), we first consider the auxiliary function $$\label{le2.205}
\omega_{1}(t,x):=M_{1}\left[2M(h(t)-x)-M^{2}(h(t)-x)^{2}\right]$$ for $t\in[0,T]$ and $x\in[h(t)-M^{-1}, h(t)]$, where $$\displaystyle{M = \max\left\{\frac{1}{h_{0}},\sqrt{\frac{\|b_{1}\|_{L^{\infty}([0,\infty))}}{2 d_{1}}},\frac{\|u_0'\|_{C[0,h_{0}]}}{M_{1}} \right\}}.$$ We have $$\label{le2.206}
\left\{\begin{array}{rl}
\omega_{1t}-d_{1}\omega_{1xx}&\geq 2d_{1}M_{1}M^{2}\geq b_{1} M_{1}\vspace{7 pt}\\
&\geq u[b_{1}-\delta_{1}(u+v)]=u_{t}-d_{1}u_{xx},\vspace{7 pt}\\
\omega_{1}(t,h(t))&=0=u(t,h(t)),\vspace{7 pt}\\
\omega_{1}(t,h(t)-M^{-1})&=M_{1}\geq u(t,h(t)-M^{-1}).
\end{array}\right.$$ We also note that $$u_{0}(x)=-\int^{h_{0}}_{x}u_0'(s)ds\leq (h_{0}-x)\|u_0'\|_{C[0,h_{0}]}$$ and $$\omega_{1}(0,x)=M_{1}M(h_{0}-x)[2-M(h_{0}-x)]\geq M_{1}M(h_{0}-x),
\text{ for }x\in [h_{0}-M^{-1},h_{0}].$$ Thus, $\omega_{1}(0,x)\geq u(0,x)$. 