Past exam of the short options course of the University of Oxford / 2017 / 9 / a / ii / Solution

At a fixed point the first equation gives $\alpha S_{\ast }{I}_{\ast }=B$. The second gives $I_{\ast }=\alpha S_{\ast }{I}_{\ast }$, so $I_{\ast }=B>0$ andThere is no disease-free fixed point: with $I=0$, susceptibles still increase by $B$ each step.

Set $r=\alpha B>0$. The Jacobian matrix of the update at the fixed point isIts characteristic equation and multipliers areFor $0<r<4$, these are distinct complex conjugates with product one, so both have modulus one. The linearized iterates are bounded and oscillatory: neutral linear stability holds for $0<\alpha B<4$, with no linear asymptotic attraction.

At $r=4$, the repeated multiplier is $-1$, but $M\ne -I$. Writing $M=-I+N$ gives $N\ne 0$, $N^2=0$ and $M^n=(-1{)}^n(I-nN)$, so generic linearized perturbations grow linearly. For $r>4$, a real multiplier has modulus greater than one. Thus the equilibrium is linearly unstable for $\alpha B\ge 4$.

For the unit-modulus range, the nonlinear behavior is not determined by this calculation: unit-modulus multipliers do not determine nonlinear stability.