Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 6C / Solution

Adding the three equations of the SIRS model givessoAt a fixed point, $I(bS-c)=0$. The disease-free equilibrium isFor an endemic equilibrium $I^{\ast }>0$, one has $S^{\ast }=c/b$ and $a{R}^{\ast }=c{I}^{\ast }$. Conservation then givesIt exists with positive infected population exactly whenequivalently when the basic reproduction number $bN/c$ exceeds one.

Eliminating $R=N-S-I$ yieldsAt the endemic equilibrium its Jacobian matrix isIts trace and determinant areTherefore the endemic equilibrium is locally asymptotically stable. It is a stable node whenand a stable focus when the inequality is reversed. Near a node, perturbations are sums of two nonoscillatory decaying exponentials. Near a focus, they execute damped oscillations with decay rate $-T/2$ and angular frequency $\sqrt{4D-T^2}/2$.