## SEIR Epidemic Model with Eventual Vaccination of Newborns
### Basic Reproduction Number and Stability of the Equilibrium Points
So that: (a) either \(S(t)\to(1-q)A/\mu\), \(R(t)\to qA/\mu\) as \(t\to\infty\) and the result is already proven; (b) at least one of \(S(t)\) and \(R(t)\) is unbounded as time increases, which is impossible since this unboundedness would contradict Theorem 1; (c) atleast one of \(S(t)\) and \(R(t)\) exhibits asymptotic oscillations. However, then from the solutions in the proof of Theorem 1 and since \(I(t)\geq 0\) for all and \(t\geq 0\)_as_\(t\to\infty\):  
\[\lim_{t\to\infty}\!\!\left(S(t)-(1-q)A\int_{0}^{t}e^{-\int_{t}^{t} \left(\beta I(s)+\mu\right)ds}d\tau\right)=0\ ;\ \lim_{t\to\infty}\!\!\left(R(t)-\int_{0}^{t}e^{-\mu(t-\tau)} \left(\gamma I(\tau)+qA\right)d\tau\right)=0\] \[\lim_{t\to\infty}\!\!\int_{0}^{t}e^{-\int_{\tau}^{t}\left(\beta I (\sigma)+\mu\right)d\sigma}d\tau=C_{1}>0\ ;\ \lim_{t\to\infty}\!\!\int_{0}^{t}e^{-\mu(t-\tau)} \left(\gamma I(\tau)+qA\right)d\tau=C_{2}\geq 0\]  
therefore, \(S(t)\) and \(R(t)\) cannot oscillate asymptotically so that they converge to finite non-negative limits. Since the disease-free equilibrium point is unique, then \(S(t)\to S^{0}(q)=(1-q)A/\mu\) and \(R(t)\to R^{0}(q)=qA/\mu\) as \(t\to\infty\), and then the disease-free equilibrium point is (not only locally but also globally) asymptotically stable if and only if \(\beta\in[0\,\beta_{c}(q))\). If \(\beta>\beta_{c}(q)\), then the spectral radius of the subsystem of the infective variables \((E,I)\) exceeds unity so that the corresponding sub-matrix of the Jacobian matrix is unstable and then the whole Jacobian matrix at the disease-free equilibrium point is unstable. If \(\beta=\beta_{c}(q)\), then some eigenvalue of the Jacobian matrix of the disease-free equilibrium point is at the stability boundary and the linearized system is critically stable. **Alternative Proof. ** Consider the whole Jacobian matrix at the disease-free equilibrium point:  
\[\mathbf{J}_{EQ^{0}}(q)=\left(\frac{\partial x^{T}}{\partial x}\right)_{EQ^{0} }=\left[ \begin{array}{cccc}-\mu&0&-\frac{\beta(1-q)A}{\mu}&0\\ 0&-(\mu+\sigma)&\frac{\beta(1-q)A}{\mu}&0\\ 0&\sigma&-(\mu+\gamma+\alpha)&0\\ 0&0&\gamma&-\mu\end{array} \right] \tag{12}\]  
where \(x^{T}=(S,E,I,R)\), \(S^{0}(q)=(1-q)A/\mu\), and \(\left(E^{0}\), \(l^{0}\), \(R^{0}\right)=(0\), \(0\), \(0)\). The eigenvalues of Equation (12) are  
\[\lambda_{1,2}=-\mu<0\] \[\lambda_{3,4}=\tfrac{1}{2}\!\left[-(\alpha+\sigma+\gamma+2\mu)\pm \sqrt{\left(\alpha-\sigma\right)^{2}+\gamma(\gamma+2(\alpha-\sigma))+\frac{4 \beta(1-q)\sigma A}{\mu}}\right] \tag{13}\] \[\qquad\qquad=\tfrac{1}{2}\!\left[-(\alpha+\sigma+\gamma+2\mu)\pm \sqrt{\left(\gamma+\alpha-\sigma\right)^{2}+\frac{4\beta(1-q)\sigma A}{\mu}} \right]\]  
Note that \(\lambda_{3}\) and \(\lambda_{4}(<\lambda_{3})\) are real eigenvalues of the above Jacobian matrix. Then, the Jacobian matrix at the disease-free equilibrium point is a stability matrix if and only if \(\lambda_{3}<0\), that is, if and only if  
\[(\alpha+\sigma+\gamma+2\mu)^{2}>(\gamma+\alpha-\sigma)^{2}+\frac{4\beta(1-q) \sigma A}{\mu} \tag{14}\]  
which holds if and only if the transmission rate is small enough to satisfy \(\beta<\beta_{c}(q)\) or, equivalently, if the basic reproduction number Equation (11) is less than one, equivalently \(R_{0}(q)<1\). If the inequality Equation (14), equivalently \(R_{0}(q)<1\), becomes an equality, still equivalent to \(R_{0}(q)=1\) in (11), then \(\lambda_{3}=0\), that is, one eigenvalue of Equation (12) is critically stable and the disease-free equilibrium point is then critically stable. If the inequality in Equation (14) is reversed \((R_{0}(q)>1)\) then the disease-free equilibrium point is unstable. Finally, note that the local asymptotic stability of the disease-free equilibrium point for \(R_{0}(q)<1\) is also global since:  
1. it is the unique attainable equilibrium point since the endemic one is not reachable from Proposition 1.