females.

Let us now recall the origin of system with some details. Let $r_{f}$ and $r_{m}$ denote the number of released female mosquitoes and the number of released males respectively and suppose they were infected with [*Wolbachia*]{}. Also, assume that $r_{f}$ and $r_{m}$ satisfy $$\label{eqintroduce1}
\left\{\begin{array}{ll}
\displaystyle{\frac{{\rm d} r_{f}}{d t}= -\delta_{1}r_{f} T(t)},&t>0,\vspace{7 pt} \\
\displaystyle{\frac{{\rm d} r_{m}}{d t}= -\delta_{1}r_{m} T(t)},&t>0,
\end{array}\right.$$ where $$T(t)=r_{f}+r_{m}+I_{f}+I_{m}+U_{f}+U_{m}$$ denotes the total population size, with $U_{f},$ $U_{m}$, $I_{f}$ and $I_{m}$ standing for the numbers of uninfected reproductive females, uninfected reproductive males, and infected reproductive females and males other than those from releasing, respectively. Let $b_{I}$ (resp. $b_{U}$) be the natural birth rate of the infected (resp. uninfected) mosquitos and $0\leq\delta\leq 1$ be the proportion of mosquitos born female. Then the proportion of mosquitos born male is $1-\delta$. With complete CI (see Table \[liuhua\]) and perfect maternal transmission, we have

$$\label{eqintroduce2}
\left\{\begin{array}{ll}
\displaystyle{\frac{{\rm d} I_{f}}{d t}= \delta b_{I}[I_{f}+r_{f}]-\delta_{1}I_{f} T(t),}&t>0,\vspace{7 pt} \\
\displaystyle{\frac{{\rm d}I_{m}}{d t}= (1-\delta)b_{I}[I_{f}+r_{m}]-\delta_{1}I_{m} T(t),}&t>0,\vspace{7 pt}\\
\displaystyle{\frac{{\rm d} U_{f}}{d t}= \delta b_{U}\left[U_{f}\,\frac{U_{m}}{r_{m}+I_{m}+U_{m}}\right]-\delta_{2}U_{f} T(t),}&t>0, \vspace{7 pt}\\
\displaystyle{\frac{{\rm d} U_{m}}{d t}= (1-\delta)b_{U}\left[U_{f}\frac{U_{m}}{r_{m}+I_{m}+U_{m}}\right]-\delta_{2}U_{m} T(t),}&t>0.
\end{array}\right.$$

\[liuhua\]

    mate             $U_{m}$                   $I_{m}$
  --------- ------------------------- -------------------------
   $U_{f}$   $U_{f}~\hbox{or}~U_{m}$          $\times$
   $I_{f}$   $I_{f}~\hbox{or}~I_{m}$   $I_{f}~\hbox{or}~I_{m}$

  : Strong CI, $\times$ means “no offspring”

One can easily verify 