= 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 gives
$$
v(z)=\frac\eta{1+\eta}[s^*-s(z)],\qquad m(z)=\frac1{1+\eta}\log\frac{s^*}{s(z)}.
$$
Adding the four wave equations yields $q''+c(q+s+v+m)'=0$. Integrate once and use $q'(\infty)=0$ to obtain $q'+c(q+s+v+m)=cs^*$. At minus infinity the assumptions $q=q'=0$ therefore give $s_-+v_-+m_-=s^*$.
Combining these identities shows that $s_-$ satisfies
$$
\boxed{\log(s^*/s_-)=s^*-s_-.}
$$
The nontrivial positive epidemic-wave root has $0<s_-<1<s^*$; the other root $s_-=s^*$ is the disease-free solution. The assumed positive wave consequently requires $s^*>1$; for $s^*\leq1$ 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:
$$
\boxed{s_-+v_-=\frac{s_-+\eta s^*}{1+\eta}.}
$$
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.
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