## 2 The \(\mu\)-SEIR Model
### Boundaries
: One gets from Equation (3) via recursive calculations that  
\[I_{i+1}=(1-\gamma)^{i+1}I_{0}+\mu\sum_{j=0}^{i}(1-\gamma)^{i-j}E_{j}\]  
so that \(\lim_{i\to\infty}E_{i}=0\) implies that for any given arbitrary real constant \(\epsilon>0\), there exists a non-negative integer \(m_{0}=m(\epsilon)\) such that \(\sup_{\infty>j\geq m_{0}}E_{j}\leq\epsilon\) and \(lim\sup_{i\to\infty}\left(\sup_{i-m_{0}\leq j\leq i}E_{j}\right)\leq\epsilon\) then one has for any integer \(m\geq m_{0}\) that  
\[I_{k+1}=(1-\gamma)I_{k}+\mu E_{k}\geq 0 \tag{10}\]0 \[I_{k+1}=(1-\gamma)I_{k}+\mu E_{k}\geq 0 \tag{10}\]1 \[I_{k+1}=(1-\gamma)I_{k}+\mu E_{k}\geq 0 \tag{10}\]2  
so that \(\lim_{ \begin{subarray}{c}m\to\infty\\ i-m\to\infty\end{subarray} }I_{i+1}=\mu\epsilon/\gamma\) for any integer \(m\geq m_{0}\). Since \(\epsilon>0\) may be chosen arbitrarily small then \(m_{0}=\lim_{\epsilon\to 0}m_{0}(\epsilon)=\infty\) and so \(m(\geq m_{0})\to\infty\), then \(\lim_{i\to\infty}I_{i}=0\). The first implication has been proved. It remains to be proved that, if \(\lim_{i\to\infty}I_{i}=0\), then \(\lim_{i\to\infty}E_{i}=0\). One has from Equation (2) that:  
\[E_{i+1}=(1-\mu+\beta_{E}S_{i})E_{i}+\beta_{I}S_{i}I_{i}\]Since \(\lim_{i\to\infty}I_{i}=0\) and \(\{S_{i}\}_{i=0}^{\infty}\) is a bounded sequence then \(\lim_{i\to\infty}\beta_{I}I_{i}S_{i}=0\), and then \(\lim_{i\to\infty}(E_{i+1}-(1-\mu+\beta_{E}S_{i})E_{i})=0\). Two cases can arise, namely:  
1. \(\lim_{i\to\infty}E_{i}=0\) and the proof finishes,
2. \(\lim_{i\to\infty}E_{i}\
eq 0\). Then \(\lim_{i\to\infty}E_{i}>0\) so that, from Equation (1), and since \(\lim_{i\to\infty}I_{i}=0\) and \(\frac{1}{2\beta_{E}}\geq E_{i}\geq 0\ \forall i\geq 0\) from Theorem 1, one has that \[\limsup_{i\to\infty}(1-\beta_{I}I_{i}-\beta_{E}E_{i})=\limsup_{i\to\infty}(1- \beta_{E}E_{i})=\left(1-\beta_{E}\lim_{i\to\infty}E_{i}\right)=a\] with \(a\in[\frac{1}{2},1)\), which implies from Equation (1) and Theorem 1 that \(0=\limsup_{i\to\infty}\left(S_{i+1}-a^{i+1}S_{0}\right)=\limsup_{i\to\infty}S _{i+1}\) so that there exist \[\lim_{i\to\infty}S_{i+1}=\limsup_{i\to\infty}S_{i+1}=0\] Then, using again \(E_{i+1}=(1-\mu+\beta_{E}S_{i})E_{i}+\beta_{I}S_{i}I_{i}\), with \(\lim_{i\to\infty}I_{i}=0\) and \(\lim_{i\to\infty}S_{i}=0\) yields \(\lim_{i\to\infty}(E_{i+1}-(1-\mu)E_{i})=0\) which, in turn, implies that: 1. Either \(\lim_{i\to\infty}E_{i}=0\) and the result is proved, 2. or \(\lim_{i\to\infty}E_{i}=E^{*}\
eq 0\) which implies that \(1=1-\mu\), so \(\mu=0\), which contradicts \(\mu\in(0,1]\) and thus, this case cannot occur 3. or \(\frac{E_{i+1}}{E_{i+1}}\to 1-\mu\in[0,1)\). Then, there is a subsequence \(\{E_{i}\}_{i=N}^{\infty}\) of \(\{E_{i}\}_{i=0}^{\infty}\), for some finite integer N, which is non-negative and strictly decreasing so that \(\lim_{i\to\infty}E_{i}=0\) and the result is proved. The proposition has been fully proved. By inspection of Equations (1)-(4), it follows the existence of a, in general, non-unique disease-free equilibrium point. Thus,  
Proposition 4: There is a disease-free equilibrium point \(x_{dfe}=(S_{dfe},0,0,R_{dfe}=N_{0}-S_{dfe})^{T}\) with \(S_{dfe}\), and then \(N_{dfe}\), being dependent on the initial conditions. The demonstration is trivial from Equations (1)-(4) by writing \(E_{i+1}=E_{i}=I_{i+1}=I_{i}=0=E_{dfe}=Idf\) and \(N_{dfe}=S_{dfe}+R_{dfe}\) The following result is concerned with the convergence of the vector sequence solution to the disease-free equilibrium point if the transmission rate of the exposed (E) exceeds that of the infected (I) under a certain minimum lower-bound constraint. This is a reasonable consideration since the exposed subpopulation in the model is also infective and the infectivity along the incubation period is normally of the same magnitude of the infectious subpopulation [49; 50].