some rearrangement, could be written as $$\label{quad}A{I_v^\ast}^2+BI^\ast_v+C=0,$$ where $$\begin{aligned}
A&=a\beta_{vh}\left[  \mu_v+\delta_v+\frac{a}{N_h}\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \, \pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle+\right.\\
      & \left.+\frac{a}{N_h}\frac{\nu_h}{(\mu_h+\nu_h+\eta)}\left(p\pi_h+ \Lambda_h   \right)\bigl\langle\Beta   \, \mbox{\Large $\mid$}   (\textrm{diag}(\alpha)-M)^{-1}e_1  
\bigr\rangle \right]\\
&>0,\end{aligned}$$ $$\begin{aligned}
B&=(\mu_v+\delta_v)\mu_nN_h +\frac{a}{N_h}\pi_h\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle\mu_hN_h\\
      &+\frac{a}{N_h}\frac{\nu_h}{(\mu_h+\nu_h+\eta)}p\pi_h\bigl\langle\Beta   \, \mbox{\Large $\mid$}   (\textrm{diag}(\alpha)-M)^{-1}e_1  
\bigr\rangle \mu_nN_h\\
&-\frac{a^2\beta_{vh}\nu_v}{(\mu_v+\nu_v+\delta_v)}\frac{N_v}{N_h}\pi_h\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle\\
&-\frac{a^2\beta_{vh}\nu_v\nu_h}{(\mu_v+\nu_v+\delta_v)(\mu_h+\nu_h+\eta)}\frac{N_v}{N_h}p\pi_h\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}e_1\bigr\rangle\\
      &-\frac{a^2\beta_{vh}\nu_v\nu_h}{(\mu_v+\nu_v+\delta_v)(\mu_h+\nu_h+\eta)}\frac{N_v}{N_h}\Lambda_h\bigl\langle
      \Beta   \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}e_1\bigr\rangle,\end{aligned}$$ and $$\begin{aligned}
C&=-\frac{a\nu_v}{(\mu_v+\nu_v+\delta_v)}\frac{N_v}{N_h}\pi_h\mu_hN_h \bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, \pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle\\
 &- 
 \frac{a\nu_v N_v}{(\mu_v+\nu_v+\delta_v)} \frac{N_v}{N_h} \frac{\nu_h \mu_hN_h}{(\mu_h+\nu_h+\eta)}p\pi_h\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \,(\textrm{diag}(\alpha)-M)^{-1}e_1  
\bigr\rangle.\end{aligned}$$ Now, we investigate cases for which has non-negative solutions.\
$\bullet$ If $\Beta=0$, then $C=0$ and $B=(\mu_v+\delta_v)\mu_nN_h>0$. Hence, $I_v^*=0$ is the unique solution of the quadratic equation. Thus, the unique equilibrium for System (\[ModelAD2Compact\]) is $(\bar{S}_h, \bar{E}_h,\bar{\mathbf{I}}_h,\bar{S}_v,0,0)$, where $\bar{S}_h=\frac{\Lambda_h}{\mu_h}$, $ \bar E_h=\frac{p\pi_h}{(\mu_h+\nu_h+\eta)}$, $\bar{\mathbf{I}}_h^\ast=\pi_h(\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}+\frac{\nu_hp\pi_h}{(\mu_h+\nu_h+\eta)} (\textrm{diag}(\alpha)-M)^{-1}e_1,
$ and $\bar{S}_v=\frac{\Lambda_v}{\mu_v+\delta_v}$. This proves .\
$\bullet$ If $\Beta\neq0$ and $p\neq0$, then $C<0$ and therefore has a unique solution such that $I_v^*>0$. Thus, from , System (\[ModelAD2Endemic1\]), and using the fact that $(\textrm{diag}(\alpha)-M)^{-1}e_1\gg0$, we deduce .\
$\bullet$ If $\Beta$ and $p$ are such that $\bigl\langle
      \Beta   
 \, \mbox{\Large $\mid$} \, (\textrm{diag}(\alpha)-M)^{-1}\mathbf{p}\bigr\rangle\neq0$, then we also have $C<0$; that is, it exists a unique $I_v^\sharp>0$ of . As in the previous point, this leads to .\
