## SEIR Epidemic Model with Eventual Vaccination of Newborns
### Non-Negativity of the Solution  
The solutions of Equation (1) are non-negative for all time for any given non-negative initial condition. This is easily proven by calculating, in a closed analytic form, the unique solution of Equation (1) for any set of non-negative initial conditions. This property is needed for well-posedness of the epidemic model Equation (1). In parallel, the fact that the total population is proven to be bounded for all time guarantees also the boundedness for all time of all the subpopulations under any given arbitrary finite non-negative initial conditions.  
**Theorem 1**.: _The following properties hold for any \(q\in[0\,\ 1]\):_  
1. _The solution trajectory of the epidemic model (_1_) is non-negative for all time for any given non-negative initial conditions._
2. _The solution trajectory of Equation (_1_) is bounded for all time for any given finite non-negative initial conditions and_ \(\underset{t\rightarrow\infty}{\lim\sup}I(t)\leq A\min(1/\alpha\,\ 1/\mu)\)_. As a result, the epidemic model is globally stable for any given arbitrary finite non-negative initial conditions._  
**Proof.** The solution to (1) is  
\[\begin{array}{c}S(t)=e^{-\int_{0}^{t}(\beta I(\tau)+\mu)dt}S(0)+(1-q)A\int_{ 0}^{t}e^{-\int_{t}^{t}(\beta I(s)+\mu)ds}d\tau\\ \\ E(t)=e^{-(\mu+\sigma)t}E(0)+\beta\int_{0}^{t}e^{-(\mu+\sigma)(t-\tau)}S(\tau)I (\tau)d\tau\\ \\ =e^{-(\mu+\sigma)t}E(0)+\beta\int_{0}^{t}e^{-(\mu+\sigma)(t-\tau)}S(\tau) \bigg{[}e^{-(\mu+\gamma+\alpha)\tau}I(0)+\sigma\int_{0}^{\tau}e^{-(\mu+\gamma +\alpha)(\tau-\sigma)}E(s)ds\bigg{]}d\tau\\ \\ I(t)=e^{-(\mu+\gamma+\alpha)t}I(0)+\sigma\int_{0}^{t}e^{-(\mu+\gamma+\alpha)(t -\tau)}E(\tau)d\tau\\ \\ R(t)=e^{-\mu t}R(0)+\int_{0}^{t}e^{-\mu(t-\tau)}(\gamma I(\tau)+qA)d\tau\end{array} \tag{2}\]  
It is obvious that \(S(t)\geq 0\) for all time if \(S(0)\geq 0\) from the above first equation. Then, \(E(t)\geq 0\) for all time from the second equality of the above second equation if \(S(0)\geq 0\), \(E(0)\geq 0\), and \(I(0)\geq 0\) since \(S(t)\geq 0\) for all time. Subsequently, \(I(t)\geq 0\) for all time from the above third equation if \(I(0)\geq 0\), since \(E(t)\geq 0\) for all time. Finally, \(R(t)\geq 0\) for all time from the above fourth equation if \(R(0)\geq 0\), since \(I(t)\geq 0\) for all time. Property (i) has been proved. By summing-up the four subpopulations in Equation (1), one gets that the total population \(N(t)=S(t)+E(t)+I(t)+R(t)\) satisfies the differential Equation (3)  
\[\dot{N}(t)=\dot{S}(t)+\dot{E}(t)+\dot{I}(t)+\dot{R}(t)=-\mu N(t)+A-\alpha I(t) \tag{3}\]  
whose solution is  
\[N(t)=e^{-\mu t}N(0)+\int_{0}^{t}e^{-\mu(t-\tau)}(A-\alpha I(\tau))d\tau\leq e ^{-\mu t}N(0)+\frac{A}{\mu}\leq N(0)+\frac{A}{\mu}<+\infty\]  
so that the total population is bounded for all time. From Property (i), since all the subpopulations are non-negative for all time, it follows that they are also bounded for all time and that \(N(t)\geq 0\) for all time. In addition, \(\underset{t\rightarrow\infty}{\lim\sup}I(t)\leq\underset{t\rightarrow\infty}{ \lim\sup}N(t)\leq A/\mu\) from the above equation. Furthermore \(\underset{t\rightarrow\infty}{\lim\sup}I(t)\leq A/\alpha\) (otherwise, if \(\underset{t\rightarrow\infty}{\lim\sup}I(t)>A/\alpha\) then \(\underset{t\rightarrow\infty}{\lim\sup}N(t)=-\infty\), a contradiction). Therefore, \(\underset{t\rightarrow\infty}{\lim\sup}(t)\leq A\min(1/\alpha\,\ 1/\mu)\). \(\Box\)