Past exam of the topics in applied mathematics course of the University of Oxford / 2009 / A / 2 / b

While $q>0$ and $s>0$, division of the equations gives $dq/ds=(1-s)/[(1+\eta )s]$ and $dv/ds=-\eta /(1+\eta )$. Integrating from the initial data yieldsLet $s_{\infty }$ be the final susceptible level. Using $q_{\infty }=0$ givesThis is the smaller positive root selected by decreasing susceptibility; the function $s-\log s$ is decreasing on $(0,1)$, so that selected root is unique. Its upper bound below $s_0$ also follows directly because the added positive term moves the decreasing-branch root left.

Conservation of the total count, or integration of $m_t=q$, then gives the number killed, including those initially infected:The remaining susceptible loss is split between infection and vaccination in fixed ratio $1:\eta$; this explains the factor multiplying $s_0-s_{\infty }$.

Rearrange the final-size equation as $s_{\infty }=s_0\exp [s_{\infty }-s_0-(1+\eta )q_0]$. With fixed $q_0>0,s_0>0$ and $\eta \to \infty$, the selected root is exponentially small:Substitution into the death count yieldsThus strong vaccination suppresses most secondary fatal infections but cannot save the $q_0$ animals already infected, as required by the stated reaction scheme. The limit holds for the fixed positive initial infection assumed here; it need not be uniform if $q_0$ is allowed to shrink with increasing $\eta$.