 Solution

For the vaccinated epidemic-wave final size, write $s_{-}=s(-\infty )$. From the travelling-wave equations, $dv/ds=-\eta /(1+\eta )$ and $dm/ds=-1/[(1+\eta )s]$. Using the ahead-of-wave boundary data givesAdding the four wave equations yields $q^{\prime \prime }+c(q+s+v+m{)}^{\prime }=0$. Integrate once and use $q^{\prime }(\infty )=0$ to obtain $q^{\prime }+c(q+s+v+m)=c{s}^{\ast }$. At minus infinity the assumptions $q=q^{\prime }=0$ therefore give $s_{-}+v_{-}+m_{-}=s^{\ast }$.

Combining these identities shows that $s_{-}$ satisfiesThe nontrivial positive epidemic-wave root has $0<s_{-}<1<s^{\ast }$; the other root $s_{-}=s^{\ast }$ is the disease-free solution. The assumed positive wave consequently requires $s^{\ast }>1$; for $s^{\ast }\le 1$ the displayed relation has no distinct smaller root.

At the back of the pulse there are no infectives, so the live biological population consists of susceptible and vaccinated animals:The final-size root is independent of vaccination strength in this travelling-wave problem; vaccination instead changes the split between deaths and survivors for the total biological population lost from susceptibility.