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\begin{document}

\title{Game theory of pre-emptive vaccination before bioterrorism or
  accidental release of smallpox\\\emph{\SUPP}\\\medskip\large\emph{Journal of the Royal Society Interface} (2015)}
\author{{\Large\bfseries Chai Molina and David J.D.~Earn\medskip}\\
Department of Mathematics and Statistics\\McMaster University\\
Hamilton, Ontario, Canada, L8S 4K1}
\date{\today\ @ \thistime}
\maketitle

\newpage
\renewcommand{\contentsname}{APPENDICES}
\tableofcontents

\linenumbers

\newpage

\appendix

\section{Lambert $\W$ function}\label{appx:LambertW}

The standard final size relation, which can be derived from the SIR model \cite{KermMcKe27} and many other epidemic models \cite{MaEarn06}, is
%
\begin{linenomath*}
\begin{equation}\label{eq:finalsizerelation}
Z = 1 - e^{-\R_0 Z} \,.
\end{equation}
\end{linenomath*}
%
Here, $Z$ is the final size ($Z=1-S_\infty$) and $\R_0$ is the basic reproduction number.  $Z$ can be expressed explicitly as a function of $\R_0$ \cite{MaEarn06,Earn+14},
%
\begin{linenomath*}
\begin{equation}\label{eq:finalsizeexplicit}
Z(\R_0) = 1 + \frac{1}{\R_0} \W[-\R_0\,e^{-\R_0}] \,,
\end{equation}
\end{linenomath*}
%
where the Lambert $\W$ function~\cite{Weis,Corl+96} is the inverse function of
%
\begin{linenomath*}
\begin{equation}\label{eq:Winverse}
f(\W) = \W e^{\W} \,.
\end{equation}
\end{linenomath*}
%
Use of the Lambert $\W$ function is critical for our derivations of final size formulae for models we consider here.  $\W(x)$ is real for $x\ge-1/e\simeq-0.368$ and is two-valued for $-1/e< x < 0$.  The upper ``principal'' branch, for which $\W(x)\ge-1$, is denoted $\W_0$ and the lower branch is denoted $\W_{-1}$.  See \Fref{fig:LambertWplot}.

%
\begin{figure}[H]
\begin{center}
\scalebox{0.65}{
\includegraphics{VGfigures/LambertWplot.pdf}
}
\end{center}
\caption{The Lambert $\W$ function, showing the principal branch $\W_0$ and the secondary branch $\W_{-1}$.}
\label{fig:LambertWplot}
\end{figure}
%

\newpage
\section {Interpretation of vaccination effort parameters}\label{appx:phiparam}

The vaccination effort parameters are explained~\Fref{sec:epidemic models}.  ``Fair comparison'' values for these parameters are derived in~\Fref{sec:vaceffort} and listed in~\Fref{tab:vaceffort}.

\subsection {$\phisusc$}\label{appx:phisuscInterpretation}
In~\Fref{sec:vaceffort}, we commented that $\phisusc$ can be considered to be the probability per unit time of a delayer being vaccinated.  To see this, note that the probability ($p_\text{vacc}$) of a susceptible delayer being vaccinated in the time interval $[t,t+\Delta t]$ is the ratio of the number of susceptibles vaccinated in that time interval, $V(t+\Delta t) -V(t)$, to the number of susceptibles present at the beginning of that time interval, $S(t)$. Thus,  
\begin{linenomath*}
\begin{equation*}
p_\text{vacc}=\frac{V(t+\Delta t) -V(t)}{S(t)} = \frac{V(t+\Delta t) -V(t)}{\Delta t} \frac{1}{S(t)}\Delta t\,. 
\end{equation*}
\end{linenomath*}
Since $\lim_{\Delta t \to 0} \frac{V(t+\Delta t) -V(t)}{\Delta t} = \dot V = \phisusc S$, for small $\Delta t$, we have $p_\text{vacc}\approx \phisusc \Delta t$. Thus, $\phisusc$ is the (constant) probability per unit time of a delayer being vaccinated. 
 
\section{Convergence to disease--free equilibrium} \label{appx:dfe}

In this appendix, we show that for all models considered, the epidemic must eventually die out (\emph{i.e.,} the system converges to a disease-free equilibrium).

Consider the $SIRV$ model given by the differential equations
\begin{linenomath*}
\begin{subequations}\label{eq:SIRVgen}
\begin{align}
\dot S &= -\beta SI -\dot V\,,\\
\dot I &= \beta SI -\gamma I\,, \\
\dot R &= \gamma I\,, \label{eq:Rdot} \\
\dot V &= f(t,S,I,R,V)\,,
\end{align}
\end{subequations}
\end{linenomath*}
where $f$ is continuously differentiable and satisfies $f(t,S,I,R,V)\ge0$, $f(t,0,I,R,V) = 0$ whenever $t\ge0$, $S\ge0$, $I\ge0$,$R\ge0$, $V\ge0$. From the fundamental existence and uniqueness theorem \cite{HirsSmal74}, there is a unique solution to \Fref{eq:SIRVgen} for any non-negative initial conditions $S(0)\ge0$, $I(0)\ge0$, $R(0)\ge0$ and $V(0) \ge 0$. Suppose also that $S(0) + I(0) + R(0) + V(0) = 1$.

First, note that $S(t) \ge0$ for all $t\ge0$. To see this, suppose in order to derive a contradiction that $S(T)<0$ for some $T>0$. Then, since $S(t)$ is continuous, there must be some time $0\le \tau<T$ such that $S(\tau) =0$. But because $\dot S \lvert_{S=0}= 0$ it follows that $S(t)=0$ for all $t\ge\tau$, which contradicts $S(T)<0$.

Similarly, it follows that $I(t)\ge0$. Consequently, $R(t)$ is nondecreasing in $t$, and in particular, $R(t)\ge0$. Lastly, $\dot V \ge0$ so $V(t)\ge0$ as well.

Now, $\frac{\mathrm{d}}{\mathrm{d}t} \Big(S(t)+I(t)+R(t)+V(t)\Big)=0$ for all $t$, so $S(0)+I(0)+R(0)+V(0) =1$ implies $S(t)+I(t)+R(t)+V(t) =1$ for all $t$. Consequently, $S(t)$, $I(t)$, $R(t)$ and $V(t)$ each lie in the interval $[0,1]$ for all time.

In addition to being bounded, $S(t)$, $R(t)$ and $V(t)$ are monotonic (their time derivatives are non-positive) and therefore have a limit as $t\to\infty$.  It follows that $I(t)$ also has such a limit ($I= 1 - S - V - R$).  To see that this limit is $0$, suppose instead that $I_\infty = \lim_{t\to\infty} I(t)>0$. Then, $\lim_{t\to\infty} \dot R = \gamma I_\infty$. Thus, there exists a time $t_*$ such that $ \dot R(t)> \gamma I_\infty/2$ for all $t_*<t$. This implies that the proportion in the recovered class increases at least linearly , and must eventually hit $R=1$ (no later than at time $ t_* + 2/(\gamma I_\infty)$) and be greater than 1 thereafter. However, this contradicts the fact that the proportion of the population in any class cannot exceed 1. Thus $I_\infty = \lim_{t\to\infty} I(t) = 0$.

A similar argument can be applied to the constant rate $SEIRV$ model by noting that
\begin{enumerate}
\item If $S(0)\ge0$, $E(0) \ge0$, $I(0)\ge0$, $R(0)\ge0$ and $V(0) \ge 0$, then
$S$, $E$, $I$, $R$ and $V$ remain non-negative for all time.
\item If $S(0) + E(0) + I(0) + R(0) + V(0) = 1$, then $S+E+I+R+V=1$ for all
time.
\item $S(t)\to0$ in finite time in this model. To see this, suppose in order to derive a contradiction, that $S(t)>0$ for all time $t\ge0$. Then, for all $t>\tlag$, $\dot S =\phiconst$, and thus $S(t)\ge \phiconst (t-\tlag)$. Consequently, $S(t)> 1$ for all $t>\tlag + 1/\phiconst$, contradicting the fact that $S(t)\le1$ for all $t$. 
\item Since $S(t)\to0$ in finite time in this model, it follows that after a finite time $\dot E = -\sigma E$, implying that $\lim_{t\to \infty} E = 0$. Since $R$ and $V$ are monotonic and thus have a limit as $t\to\infty$, it follows that
$\lim_{t\to\infty} I=\lim_{t\to\infty} 1-S-E-R-V$ exists as well, and one can
continue as before.
\end{enumerate}
Lastly, a corollary of this convergence to the disease-free equilibrium is that
$S_\infty=\lim_{t\to\infty} S(t)$ is well defined and
\begin{linenomath*}
\begin{equation}
S_\infty < \frac{\gamma}{\beta} = \frac{1}{\R_0}\,. \label{eq:S_infty_<1/R0}
\end{equation}
\end{linenomath*}
To see this, observe that since $S$ is monotonic and bounded, it must converge
to some finite limit within $[0,1]$, so $S_\infty$ is well defined. Next, note
that $I(0) = \alpha (1-p)>0$ and $\lim_{t\to\infty} I(t) = I_\infty = 0$, so
there is some time $t_*$ at which $\dot I(t_*)<0$ and so $S(t_*)<\gamma/\beta$.
But $\dot S\le 0$ and so for any $t_*<t$, we have $S(t)<\gamma/\beta$.
Thus,~\Fref{eq:S_infty_<1/R0} follows because of the monotonicity of $S$.

\section{Calculation of $\pi_1$}\label{appx:pi_1} 

In~\fref{sec:epidemic models}, we stated that as pre-emptive vaccination approaches full coverage ($p\to1$), the probability of a delayer being infected ($\pi_p$) approaches the proportion of susceptibles initially infected in an outbreak ($\alpha$).  Recalling the definition of $\pi_1$ in \Fref{eq:pi_1}, the claim is that for all the models considered
%
\begin{linenomath*}
\begin{equation}\label{eq:pi_1claim}
\pi_1=\lim_{p\to1^-}\pi_p=\alpha \,.
\end{equation}
\end{linenomath*}

To verify \Fref{eq:pi_1claim}, first consider the SIRV models defined in~\Fref{eq:SIRV}.  The proportion of individuals who are eventually removed ($R_\infty$) must be greater than the number initially infected (\Fref{eq:I0}), so
%
\begin{linenomath*}
\begin{equation}\label{eq:}
R_\infty\ge I(0)=\alpha(1-p) \,.
\end{equation}
\end{linenomath*}
%
Thus from \Fref{eq:pi_defn}, we have $\pi_p\ge\alpha$ for any $p\in[0,1)$.  It follows that $\pi_1\ge\alpha$ if the limit exists. 

We now show that $\pi_1\le\alpha$.  The basic reproduction number is $\R_0=\beta/\gamma$ and the effective reproduction number when the outbreak begins is (\Fref{eq:S0})
%
\begin{linenomath*}
\begin{equation}\label{eq:Reff}
\Reff(0) = \R_0 S(0) = \R_0 (1-p)(1-\alpha)\,.
\end{equation}
\end{linenomath*}
%
Thus, if $p>1-1/\R_0$ then $\Reff(0)<1$, and 
%
\begin{linenomath*}
\begin{equation}\label{eq:R0eff0}
\lim_{p\to1^-}\Reff(0)=0 \,.
\end{equation}
\end{linenomath*}
%
To prove that $\pi_1\le\alpha$, we will show that 
%
\begin{linenomath*}
\begin{equation}\label{eq:}
\pi_p \le \frac{\alpha}{1-\Reff(0)} \,, \qquad \text{for all } p>1-\frac1{\R_0}\,.
\end{equation}
\end{linenomath*}
%
\Erefs{eq:Idot} and \eqref{eq:Rdot} imply that $\dot R =\gamma I= \beta S I -\dot I$.   Hence,
\begin{linenomath*}
\begin{equation*}
R_\infty  = \int_{0}^{\infty} \dot R \,\mathrm{d}t 
= \int_{0}^{\infty} \big(\beta S I -\dot I\big) \,\mathrm{d}t \,.
\end{equation*}
\end{linenomath*}
But $S(t)$ decreases monotonically with $t$, and $I_\infty=0$ (\Fref{appx:dfe}), so
\begin{linenomath*}
\begin{equation}\label{eq:Rinftyineq}
R_\infty\le \beta S(0) \int_{0}^{\infty}  I \,\mathrm{d}t + I(0) 
= \Reff(0) \int_{0}^{\infty}  \gamma I \,\mathrm{d}t + I(0) 
= \Reff(0) R_\infty + \alpha(1-p) \,.
\end{equation}
\end{linenomath*}
Thus, 
%
\begin{linenomath*}
\begin{equation}\label{eq:}
\big(1-\Reff(0)\big)R_\infty \le \alpha(1-p) \,,
\end{equation}
\end{linenomath*}
%
and consequently, for any $p>1-1/\R_0$,
%
\begin{linenomath*}
\begin{equation}\label{eq:pi_p_ineq}
\pi_p = \frac{\R_\infty}{1-p}
\le \frac{\alpha}{1-\Reff(0)} \,.
\end{equation}
\end{linenomath*}
%
In the limit $p\to1^-$, \Fref{eq:R0eff0} implies $\pi_1\le\alpha$, as required.

To see that $\pi_1=\alpha$ for the constant rate vaccination (SEIRV) model (\Fref{eq:SEIRV}), we need only note that in this case, 
\begin{linenomath*}
\begin{equation}
R_\infty = \int_{0}^{\infty}\gamma I \,\mathrm{d}t = \int_{0}^{\infty} (\beta S I -\dot I - \dot E) \,\mathrm{d}t \le \R_0S(0) \int_{0}^{\infty}  \gamma I \,\mathrm{d}t + E(0)\,,
\end{equation}
\end{linenomath*}
where $E(0)=\alpha(1-p)$.  Thus, Inequality~\eqref{eq:Rinftyineq} holds for the SEIRV model as well, and the remainder of the proof that $\pi_1=\alpha$ is identical to the argument for SIRV models.

\section{Final size relations, $\pi_p$ and $\psi_p$} \label{appx:finsize+}

\subsection {Vaccination rate $\propto$ disease prevalence} \label {appx:finsize+:vac propto I}
\subsubsection{Final size relations} \label{appx:finsize+:vac propto I:finsize}
A na\"{\i}ve model in which vaccination is proportional to prevalence is 
\begin{subequations}\label{eq:vac propto I}
\begin{linenomath*}
\begin{align} 
\label {eq:vac propto I: S} 
\dot S &= -\beta SI -\phiprev I \\
\dot I &= \beta SI -\gamma I \\
\dot R &= \gamma I \\
\dot V &= \phiprev I\, . \label{eq:vac propto I: V}
\end{align}
\end{linenomath*}
However,~\Fref{eq:vac propto I: S} is not biologically sensible, since if $S=0$ and $I>0$ it follows that $\dot S <0$ and so $S$ attains negative values. Thus, a more realistic model is obtained by replacing the vaccination rate $\phiprev I$ with $\phiprev f(S)I$, where $f$ is a nondecreasing and smooth ``cutoff function" such that $f(S)=1$ except for $0\le S<\delta$, and $f(0)=0$.  Thus,~\Fref{eq:vac propto I: S} is replaced by 
\begin{linenomath*}
\begin{equation} \label{eq:vac propto I: S*}
\tag{\ref{eq:vac propto I: S}$'$}
\dot S = -\beta SI -\phiprev f(S) I \,.
\end{equation} 
\end{linenomath*}
\end{subequations}

For convenience, we also choose $f$ to be an odd function, \emph{i.e.}, $f(-S) = -f(S)$ (however, since negative values of $S$ are not biologically feasible and are unattainable by this model if $S(0)\ge0$, this has no effect on the dynamics of the model for biologically sensible initial conditions).

As $\delta\to0$, $\dot V$ approaches $\phiprev \sgn(S) I$  and~\Fref{eq:vac propto I: S*} approaches 
\begin{linenomath*}
\begin{equation*} 
\dot S = -\beta SI -\phiprev \sgn(S) I \,,
\end{equation*} 
\end{linenomath*}
where
\begin{linenomath*}
\begin{equation*}
\sgn(x) = 
\begin{cases}
-1 & \text{if } x < 0, \\
0 & \text{if } x=0,\\
1 & \text{if } x > 0.
\end{cases}
\end{equation*}
\end{linenomath*}
Thus, a more biologically sensible model where vaccination is proportional to prevalence is:
\begin{subequations}\label{eq:vac propto I**}
\begin{linenomath*}
\begin{align} 
\label {eq:vac propto I: S**}\dot S &= -\beta SI -\phiprev \sgn(S) I \\
\dot I &= \beta SI -\gamma I \\
\dot R &= \gamma I \\
\dot V &= \phiprev \sgn(S)I\, . \label{eq:vac propto I: S**}
\end{align}
\end{linenomath*}
\end{subequations}

In the interior of the biologically meaningful domain, 
\begin{linenomath*}
\begin{equation}
\Delta = \{(S,I,R,V)\lvert S\ge0\,, I\ge0\,, R\ge0\,, V = 1-S+I+R\}\,,
\end{equation} 
\end{linenomath*}
 the phase portrait for this model is similar to that of the original model and the dynamics change only as the hyper-plane $S=0$ is reached. For this reason, we analyze the original model (\Erefs{eq:vac propto I}) and make the necessary corrections to account for~\Fref{eq:vac propto I: S**} afterwards. We denote state-variable solutions to the original model (\Erefs{eq:vac propto I}) with a superscript 1, as in $S^1$, \emph{etc.}

From~\Fref{appx:dfe}, we know that solutions of \Erefs{eq:vac propto I} converge to a disease-free equilibrium.  Thus, we wish to obtain final size relations for this model. We proceed as follows: 

From 
\begin{linenomath*}
\begin{equation}
 \frac{\mathrm{d}R^1}{\mathrm{d}S^1} = -\frac{\gamma}{\beta S^1  +\phiprev}\,,
\end{equation}
\end{linenomath*}
we have
\begin{linenomath*}
\begin{equation} \label{eq:R_S_prevalence} 
 R_\infty^1 =\frac{\gamma}{\beta} \ln\Big(\frac{\beta S(0)+\phiprev}{\beta S_\infty^1 +\phiprev}\Big) \, ,
 \end{equation}
\end{linenomath*}
where a subscript $\infty$ indicates the value of that variable at the end of the epidemic (recall that S(0) also depends on $p$). $S_\infty^1$ is obtained by a similar trick. 
\begin{linenomath*}
\begin{align}
\frac{\mathrm{d}I^1}{\mathrm{d}S^1} &= -\frac{\beta S^1 -\gamma}{\beta S^1+\phiprev}= -1+\frac{\phiprev +\gamma}{\beta S^1+\phiprev}\\
\label{eq:I_S_prevalence_alltime}I(t) - I(0) &= S(0)-S^1(t) + \frac{\phiprev+\gamma}{\beta}\ln\Big(\frac{\beta S^1(t) + \phiprev}{\beta S(0) + \phiprev}\Big) \, .
\end{align}
\end{linenomath*}
As $t\to\infty$, we have 
\begin{linenomath*}
\begin{equation}
\label{eq:I_S_prevalence}I_\infty - I(0) = S(0)-S_\infty^1 + \frac{\phiprev+\gamma}{\beta}\ln\Big(\frac{\beta S_\infty^1 + \phiprev}{\beta S(0) + \phiprev}\Big) \, .
\end{equation}
\end{linenomath*}
Since $I_\infty^1 = 0$ and $S(0)+I(0) = 1-V(0) = 1-p$, 
\begin{linenomath*}
\begin{equation}\label{eq:S_infty_prevalence_implicit}
S_\infty^1 = (1-p)+ \frac{\phiprev+\gamma}{\beta}\ln\left(\frac{\beta S_\infty^1 + \phiprev}{\beta S(0) + \phiprev}\right)=(1-p)-\frac{\phiprev+\gamma}{\gamma}R_\infty^1 \, .
\end{equation}
\end{linenomath*}
Let 
\begin{linenomath*}
\begin{equation}
w(x) = (1-p)+ \frac{\phiprev+\gamma}{\beta}\ln\left(\frac{\beta x + \phiprev}{\beta (1-p)(1-\alpha) + \phiprev}\right) \, .
\end{equation}
\end{linenomath*}
We seek solutions to $S_\infty^1=w(S_\infty^1)$ in the range $S_\infty^1 \in [0,1]$. 
We note that no solutions of~\Fref{eq:vac propto I} cross the $S$-nullcline, $S=-\phiprev/\beta$, and so solutions to~\Fref{eq:S_infty_prevalence_implicit} are in the range $[-\phiprev/\beta,1]$. 

It is possible to use~\Fref{eq:S_infty_prevalence_implicit} to find which initial coverage causes solutions of~\Fref{eq:vac propto I} to hit the $S=0$ hyper-plane, as they will be those for which $S_\infty^1 \le 0$. As long as this does not occur, \Erefs{eq:R_S_prevalence}, \eqref{eq:I_S_prevalence} and \eqref{eq:S_infty_prevalence_implicit} are valid also for the modified system (\Erefs{eq:vac propto I**}). 

From~\Fref{eq:S_infty_prevalence_implicit}, we have
\begin{linenomath*}
\begin{equation*}
S_\infty^1 = -\frac{1}{\beta}\left( \phiprev + (\gamma+ \phiprev) \W_{i} \left( -\frac{\beta S(0)+\phiprev}{\gamma+\phiprev}e^{-\frac{\beta(1-p)+\phiprev}{\gamma+\phiprev}} \right) \right)\,,
\end{equation*} 
\end{linenomath*}
where $i=0$ or $-1$ specifies the branch of the Lambert $\W$ function. To determine which branch of $\W$ gives the correct final size, observe that $S_\infty^1< \gamma/\beta$ for any initial condition (in the biologically meaningful domain). This follows from the following argument: If $S(0) \le \gamma/\beta$, since $S$ is non-increasing and $\dot S<0$ at time $t=0$ (since also $I(0)>0$), we are done. If $S(0)> \gamma/\beta$, we note that $I$ is increasing for any $S$ such that  $S > \gamma/\beta$. But we have seen that $I_\infty=0$. Thus, at some point in time, $S< \gamma/\beta$, and since $S$ is non-increasing, we have $S_\infty^1< \gamma/\beta$. From this we see that it is necessary to use the principal branch, $\W_0$ (rather than $\W_{-1}$, which satisfies $\W_{-1}(x)\le-1$ for all $x$ in its domain of definition, that is $(-1/e,0)$). Thus, 
\begin{linenomath*}
\begin{equation}\label{eq:S_infty_ prevalence_explicit}
S_\infty^1 = -\frac{1}{\beta}\left( \phiprev + (\gamma+ \phiprev) \W_0 \left( -\frac{\beta S(0)+\phiprev}{\gamma+\phiprev}e^{-\frac{\beta(1-p)+\phiprev}{\gamma+\phiprev}} \right) \right)\,.
\end{equation}
\end{linenomath*}

For convenience, we rewrite \Erefs{eq:R_S_prevalence}, \eqref{eq:I_S_prevalence} and \eqref{eq:S_infty_prevalence_implicit} to give: 
\begin{linenomath*}
\begin{equation}\label{eq:ln-s-ratio}
\ln\left(\frac{\beta S(0) + \phiprev}{\beta S_\infty^1 + \phiprev}\right)  =  \frac{\beta}{\phiprev+\gamma}\left(1-p-S_\infty^1 \right) \,,
\end{equation}
\end{linenomath*}
and
\begin{linenomath*}
\begin{equation}
\label{eq:R_infty_prevalence} R_\infty^1  = \frac{\gamma}{\beta}\ln\left(\frac{\beta S(0) + \phiprev}{\beta S_\infty^1 + \phiprev}\right)  =  \frac{\gamma}{\phiprev+\gamma}\left(1-p-S_\infty^1 \right) \,.
\end{equation}
\end{linenomath*}
Thus, 
\begin{linenomath*}
\begin{align}
\label{eq:V_infty_prevalence}
V_\infty^1 & = 1-R_\infty^1 - S_\infty^1  \nonumber \\
& =  1 -  \frac{\gamma}{\phiprev+\gamma}\left(1-p-S_\infty^1 \right) -S_\infty^1 \nonumber \\
& = \frac{\phiprev}{\gamma+ \phiprev} (1-S_\infty^1) + \frac{\gamma}{\gamma+\phiprev}p \,.
\end{align} 
\end{linenomath*}

To see when we can use $S_\infty=S^1_\infty$, $R_\infty=R^1_\infty$ and $V_\infty=V^1_\infty$, it is necessary to find when~\Fref{eq:S_infty_ prevalence_explicit} yields a negative $S_\infty^1$. 
First, we evaluate how $S_\infty$ changes with $p$. 
To find $\partial_p S_\infty^1$, apply $\partial_p:=\frac{\partial}{\partial p}$ to~\Fref{eq:S_infty_prevalence_implicit}, to get (recall that $S(0) = (1-\alpha)(1-p)$, so $\partial_p S(0) =-(1-\alpha)$)
\begin{linenomath*}
\begin{align}
\partial_p S_\infty^1 &=\frac{\gamma+\phiprev}{\beta}\ \left[\frac{\beta(1-\alpha)}{\beta S(0) + \phiprev} +\frac{\beta \partial_p S_\infty^1}{\beta S_\infty^1 + \phiprev}\right] -1 \\
\left(1-\frac{\gamma+\phiprev}{\beta S_\infty^1 + \phiprev}\right)\partial_p S_\infty^1&= \frac{(\gamma+\phiprev)(1-\alpha)}{\beta S(0) + \phiprev} -1 \\
\partial_p S_\infty^1 &= \left(\frac{(\gamma+\phiprev)(1-\alpha)}{\beta S(0) + \phiprev} -1\right)/\left( 1- \frac{\gamma+\phiprev}{\beta S_\infty^1 + \phiprev}\right)\label{eq:prev_partial_p_S_infty}
\end{align}
\end{linenomath*}
Since $S_\infty^1< \gamma/\beta$, it follows that  $1- (\gamma+\phiprev)/(\beta S_\infty^1 + \phiprev)<0$.
Thus, 
\begin{linenomath*}
\begin{equation} 
\sgn(\partial_p S_\infty^1) = \sgn\left[\beta S(0) +\phiprev-(\gamma+\phiprev)(1-\alpha)\right]\,,
\end{equation}
\end{linenomath*}
and since $S(0) = (1-p)(1-\alpha)$, we have: 
\begin{linenomath*}
\begin{equation}
\sgn{\partial_p S_\infty^1}=
\begin{cases}
1 & \text{if } p_m>p\, \\
0 & \text{if } p_m=p,\\
-1 & \text{if } p_m<p \,,
\end{cases}
\end{equation}
\end{linenomath*}
where the local maximum is attained at
\begin{linenomath*}
\begin{equation}
 p=p_m:=1 +\frac{\phiprev-(\gamma+\phiprev)(1-\alpha)}{\beta(1-\alpha)}=1 +\frac{\alpha\phiprev-\gamma(1-\alpha)}{\beta(1-\alpha)}\,.
\end{equation}
\end{linenomath*}
Observe that $p_m\in[0,1]\iff\alpha\phiprev\le\gamma(1-\alpha)$ and $\alpha\phiprev\ge(\gamma-\beta)(1-\alpha)$. However, the second condition, which is necessary to ensure $p_m\ge0$, is trivially satisfied whenever $\beta/\gamma=\mathcal{R}_0\ge1$. Thus, if $\mathcal{R}_0\ge1$, then $p_m\in[0,1]$ \textiff
\begin{linenomath*}
\begin{equation}\label{eq:pmle1}
\alpha\phiprev\le\gamma(1-\alpha)\,.
\end{equation}
\end{linenomath*}

The maximum value of $S_\infty^1$ is thus: 
\begin{linenomath*}
\begin{equation}
\max_{p\in[0,1]}S_\infty^1 = -\frac{1}{\beta}\left( \phiprev + (\gamma+ \phiprev) \W_0 \left( -\frac{\beta S(0)+\phiprev}{\gamma+\phiprev}e^{-\frac{\phiprev-\gamma(1-\alpha)}{(\gamma+\phiprev)(1-\alpha)}} \right) \right)\,.
\end{equation}
\end{linenomath*} 

Now solving for $S_\infty^1=0$ using~\Fref{eq:S_infty_ prevalence_explicit}, we have 
\begin{linenomath*}
\begin{equation} \label{eq:p_zero}
p_0(i)=1 +\frac{\phiprev}{\beta(1-\alpha)} + \frac{\gamma +\phiprev}{\beta} \W_i \left(-\frac{\phiprev}{(1-\alpha) (\gamma +\phiprev )}e^{-\frac{\phiprev }{(1-\alpha) (\gamma +\phiprev )}}\right)
\end{equation}
\end{linenomath*}
for $i=0$ or $-1$.
Note that for $\phiprev>0$,~\Fref{eq:p_zero} gives two values for $p_0$;  we cannot simply cancel out the operation of $\W$ with $xe^x$, since in this case $x=-\frac{\phiprev}{(1-\alpha)(\phiprev+\gamma)}<0$ and $\W$ is not univalued for negative arguments. Instead, we have two possibilities for $\W \left(\frac{\phiprev}{(\alpha -1) (\gamma +\phiprev )}e^{\frac{\phiprev }{(\alpha -1) (\gamma +\phiprev )}}\right)$ corresponding to the two branches, $\W_0$ and $\W_{-1}$. 

If $\alpha\phiprev<(1-\alpha) \gamma$, then $\W_0 \left(-\frac{\phiprev}{(1-\alpha) (\gamma +\phiprev )}e^{-\frac{\phiprev }{(1-\alpha) (\gamma +\phiprev )}}\right)=-\frac{\phiprev }{(1-\alpha) (\gamma +\phiprev )}$, which gives $p_0(0)=1$.  If $\alpha\phiprev\ge(1-\alpha) \gamma$ then similarly $p_0(-1) =1$. This is in agreement with the fact that if $p=1$, $S(0)=0$, and so $S^1_\infty = 0$. 

There are now three cases, which we express as two main cases, the second of which has two subcases:
\begin{itemize}
\item if $p_m\ge1$ (which happens \textiff $\alpha\phiprev\ge(1-\alpha) \gamma$ ), then $S_\infty^1\le0\quad \forall p\in[0,1]$. This follows since if $p_m\ge1$ then $p_0(-1) =1\le p_m\le p_0(0)$. Thus because $S_\infty^1$ is increasing for $p<p_m$, so for $p\in[0,1]$, $S_\infty^1\le S_\infty^1\lvert_{p=1}=0$. In this case, $S_\infty$ is not given by $S_\infty^1$, but is simply $S_\infty=0$. 
\item If $p_m\in(0,1)$ then $S_\infty^1 \lvert_{p_m}>S_\infty^1\lvert_{p=1}=0$ and $p_0(-1)<p_m$.
\begin{itemize}
\item If $0\le p_0(-1)$ (and $p_0(-1)<p_m<1$) then 
\begin{linenomath*}
\begin{equation}
S_\infty=
\begin{cases}
0 & \text{if } p<p_0(-1), \\
S_\infty^1 & \text{if } p_0(-1)\le p\,.
\end{cases}
\end{equation}
\end{linenomath*}
\item If $p_0(-1) \le0$ then $S_\infty^1\ge0\quad \forall p\in[0,1]$ and so $S_\infty=S_\infty^1$ for any $p\in[0,1]$.
\end{itemize} 
\end{itemize}

In all but the very last sub-case, it is also necessary to adjust our formulae for the final sizes of the removed and vaccinated compartments, $R_\infty$ and $V_\infty$, for the values of $p$ for which $S_\infty=0$. Qualitatively, this adjustment is necessary because, when $\delta \to 0^+$,  if a solution reaches $S=0$ in finite time, $S$ remains 0, while $I$ decays exponentially to $0$. However, the solutions of \Erefs{eq:vac propto I} are only identical to those of \Erefs{eq:vac propto I**} so long as $S>0$. Moreover, once $S=0$, $V$ remains constant and all the infectives move into the recovered compartment, which is not the case for solutions of \Erefs{eq:vac propto I}. 

To find formulas for $R_\infty$ and $V_\infty$, fix $p\in[0,1)$ and let $t_0$ be the first time at which no susceptibles remain ($S(t_0) = S^1(t_0) = 0$). Then \Erefs{eq:vac propto I} are valid for any $t< t_0$. We now have  
\begin{align}
R(t) &=\frac{\gamma}{\beta} \ln\left( \frac{\beta S(0)+\phiprev}{ \beta S(t) +\phiprev }\right) \nonumber \\
I(t)  &= I(0)+S(0)-S(t) + \frac{\phiprev+\gamma}{\beta}\ln\left(\frac{\beta S(t) + \phiprev}{\beta S(0) + \phiprev}\right) \nonumber \, ,
\end{align}
in a manner analogous to~\Fref{eq:R_S_prevalence} and~\Fref{eq:I_S_prevalence}.
These equations depend on $S=S(t)$ in a way that is continuous at $S=0$, and so taking $t\to t_0$ is equivalent to taking $S\to0$: 
\begin{align}
 R(t_0) &=\frac{\gamma}{\beta} \ln\left(\frac{\beta S(0)+\phiprev}{\phiprev}\right) \\
I(t_0)  &= I(0)+S(0) + \frac{\phiprev+\gamma}{\beta}\ln\left(\frac{\phiprev}{\beta S(0) + \phiprev}\right) \, .
\end{align}
When $S=0$, $I$ decays exponentially to 0, until all the infectives present at $t_0$ transition into the removed class, $R$.
Thus, we have 
\begin{linenomath*}
\begin{equation}
R_\infty = R(t_0)+I(t_0) = 1-p - \frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)  \, .
\end{equation}
\end{linenomath*}

Next, we know that once $S=0$, $V$ does not change either, since there are no more susceptibles to be vaccinated. Thus, 
\begin{linenomath*}
\begin{equation} 
V_\infty= V(t_0) = 1- R(t_0) -I(t_0) = 1-R_\infty = p + \frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)
\end{equation}
\end{linenomath*}
 
To summarize our results so far, 
\begin{subequations}\label{eq:prev_finsize_summary}
\begin{linenomath*}
\begin{align}
S_\infty &=
\begin{cases}
0 & \text{if } p<p_0 \text{ or } 1 \le p_m, \\
S_\infty^1 & \text{if } p_0\le p\le 1\,,
\end{cases}\\
R_\infty &=
\begin{cases}
1-p-\frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev}S(0)+1\right) & \text{if } p<p_0 \text{ or } 1 \le p_m, \\
\frac{\gamma}{\gamma+\phiprev} \left(1 - p - S_\infty^1\right) & \text{if } p_0\le p\le 1\,,
\end{cases}\\
V_\infty &=
\begin{cases}
p+\frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev}S(0)+1\right) & \text{if } p<p_0 \text{ or } 1 \le p_m, \\
\frac{1}{\gamma+\phiprev} \left(\phiprev(1 - S_\infty^1)+\gamma p\right) & \text{if } p_0\le p\le 1\,,
\end{cases}
\end{align}
\end{linenomath*}
\end{subequations}
where $S^1_\infty$ is given by~\Fref{eq:S_infty_ prevalence_explicit}.
\subsubsection{Qualitative behaviour of $S^1_\infty$ for high vaccine coverage}\label{sec:QualitativeSinftyHighCoverage}

\begin{figure}
\begin{center}
\scalebox{0.8}{
\includegraphics{VGfigures/SinftyPrevPlot.pdf}
}
\end{center}
\caption{The proportion of individuals remaining susceptible at the end of the epidemic, $S_\infty$, as a function of the proportion of the population vaccinated preemptively, $p$, for the model in which vaccination rate is proportional to prevalence.  The line $y(p) = (1-\alpha)(1-p)$ is overlaid in red. We take the proportion of susceptibles initially infected, $\alpha$, to be the estimated value in the left panel ($\alpha=1.72\times10^{-5}$) and a much larger value for comparison in the right panel ($\alpha=0.1$); the remaining model parameters are as in \Fref{tab:estimated}.  See \Fref{sec:QualitativeSinftyHighCoverage}.}
\label{fig:SinftyPrev}
\end{figure}

Qualitatively, observe that for high values of $p$, $S_\infty^1\approx (1-\alpha)(1-p)$ (see \Fref{fig:SinftyPrev}). This is because when $ p>p_d := 1-\frac{\gamma}{\beta(1-\alpha)}$ then $S(0)<\gamma / \beta$, in which case $I$ decays to 0 monotonically (the subscript ``$d$'' denotes decay of $I$). Now in this case, $S$ decreases at least as fast as it does in the vaccination-less $SIR$ model, and so $I$ decreases at least as fast as in the vaccination-less case too. Because $S$ is monotonically decreasing, $I$ decays faster than $I(0)e^{(\beta S(0) -\gamma)t}$. Because $I$ decays at least exponentially, $S$ hardly changes over the course of the epidemic, and so we see an approximately linear decay in $S_\infty^1\approx S(0) =(1-\alpha)(1-p)$ to 0 as we increase $p$. This last phenomenon is related to the herd immunity effect in the standard $SIR$ model: when the entire population is susceptible, in the absence of post-outbreak vaccination ($\phiprev=0$), the critical vaccine coverage which stops the epidemic from taking off is $p=1-\frac{1}{\mathcal{R}_{\text{eff}}}=1-\frac{\gamma}{\beta (1-\alpha)}$. But note, moreover, that the maximal value of $S_\infty^1$ is not attained at $p_d$. Rather, we see that 
\begin{linenomath*}
\begin{equation}
p_m-p_d = \frac{\alpha(\phiprev+\gamma)}{\beta(1-\alpha)}>0 \,.
\end{equation}
\end{linenomath*}
This is because even when $p>p_d$ and $I$ immediately decays to 0 (starting at $t=0$), there are still some susceptibles converted into vaccinated individuals, due to $I>0$. This number of susceptibles lost to vaccination decreases as $p$ is increased, since this decreases $I(0)$ as well. Thus, for $p>p_d$, initially $S_\infty$ increases with $p$. Only when $p>p_m$ does the decrease in $S_\infty$ due to more susceptibles being vaccinated pre-emptively take over. 

\subsubsection{$\pi_p$ decreases and $\psi_p$ increases with $p$ when $p_m\ge 1$}\label{appx:finsize+:pi_p decreases}

Consider the first case above: if $p_m\ge 1$ (see~\Fref{eq:prev_finsize_summary}), then $S_\infty =0$ and we have
\begin{linenomath*}
\begin{align}
V_\infty &=  p + \frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)\nonumber \\ 
\psi_p &= \frac{\phiprev}{\beta(1-p)}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)\nonumber \,.\label{eq:prev_psi_pm_ge1}
\end{align}
\end{linenomath*}
Note that since $S_\infty = 0$, $\pi_p = 1-\psi_p$.
Letting $x=\frac{\beta}{\phiprev}S(0) =  \frac{\beta (1-\alpha)}{\phiprev} (1-p)$,
\begin{linenomath*}
\begin{equation*}
\psi(x)=\psi_{p(x)} = (1-\alpha)\frac{\ln\left(x+1\right)}{x}\,,
\end{equation*}
\end{linenomath*}
so 
\begin{linenomath*}
\begin{align}
\partial_p \psi_p & = -\frac{\beta (1-\alpha)}{\phiprev} \frac{\partial \psi(x)}{\partial x}= -\frac{\beta (1-\alpha)}{\phiprev}\frac{x-(1+x)\ln(1+x)}{x^2(x+1)} \nonumber \\
&= -\frac{\beta (1-\alpha)}{\phiprev}\frac{\frac{\beta}{\phiprev}S(0) -(\frac{\beta}{\phiprev}S(0)+1 )\ln(\frac{\beta}{\phiprev}S(0)+1 )}{\left(\frac{\beta}{\phiprev}S(0)\right)^2(\frac{\beta}{\phiprev}S(0)+1)} \nonumber\\
&= -\frac{\phiprev (1-\alpha)}{\beta}\frac{\beta S(0) -(\beta S(0)+\phiprev )\ln(\frac{\beta}{\phiprev}S(0)+1 )}{\left(S(0)\right)^2(\beta S(0)+\phiprev)} \,.
\end{align}
\end{linenomath*}
Since $x<(x+1)\ln(1+x) \quad \forall x>-1 $, it follows that for $p\in[0,1)$, $\partial_p \psi_p >0$ and consequently, $\partial_p \pi_p <0$.

\subsubsection{Behaviour of $p_0(-1)$ as $\phiprev\to0^+$}
Using~\Fref{eq:p_zero}, one can show that $\lim_{\phiprev\to0^+}p_0(-1)=-\infty$. This follows because $\lim_{\phiprev\to0^-} \W_{-1} \left(x\right)=-\infty$. But because $p_0(-1)\to 1$ as $\phiprev\to \frac{\gamma(1-\alpha}{\alpha}$, it follows that there is some value of $\phiprev$ for which $p_0(-1)=0$. This $\phiprev$ can be found from~\Fref{eq:p_zero}, but the formula is not needed here. 

However, we also note that $\phiprev<\frac{\gamma(1-\alpha}{\alpha}$ (along with $\mathcal{R}_0>1$) implies $p_m\in(0,1)$, so for small $\phiprev$, we have $p_m\in(0,1)$ and $p_0(-1)<0$ (in fact, as $\phiprev\to 0^+$, $p_m\to 1- 1/\mathcal{R}_0$). Thus, for $\phiprev\to 0^+$ (small enough such that $p_0(-1)<0$), $S^1_\infty>0$ for $p\in[0,1)$.

\subsection{Vaccination rate $\propto$ incidence} \label{appx:finsize+:vac propto SI}
The model equations are
\begin{subequations}
\begin{linenomath*}
\begin{align}
\dot S &= -\beta SI -\phiinc SI \\ 
\dot I &= \beta SI -\gamma I \\
\dot R &= \gamma I \\
\dot V &= \phiinc SI\, . \label{eq:dotVinc}
\end{align}
\end{linenomath*}
\end{subequations}

Finding the final sizes for this model is somewhat similar to when vaccination is proportional to prevalence: 
\begin{linenomath*}
\begin{align}
\frac{\mathrm{d}I}{\mathrm{d}S} &= -\frac{\beta SI - \gamma I}{(\beta+\phiinc)SI}=\frac{\gamma}{\beta+\phiinc}\frac{1}{S}-\frac{\beta}{\beta+\phiinc}\\
\label{eq:S_I_incidence}
I(t) -I(0) &= \frac{\gamma}{\beta+\phiinc}\ln\left(\frac{S(t)}{S(0)}\right)-\frac{\beta}{\beta+\phiinc}\left(S(t)-S(0)\right)\\
\label{eq:S_I_infty_incidence}
 -(1-p)\alpha &= \frac{\gamma}{\beta+\phiinc}\ln\left(\frac{S_\infty}{(1-p)(1-\alpha)}\right)-\frac{\beta}{\beta+\phiinc}\left(S_\infty-(1-p)(1-\alpha)\right)\\
\label{eq:S_infty_incidence}  S_\infty&= \frac{\gamma}{\beta}\ln\Big(\frac{S_\infty}{(1-p)(1-\alpha)}\Big) +\Big(1  +\frac{\phiinc}{\beta}\alpha\Big)(1-p)\,,
\end{align}
\end{linenomath*}
where~\Fref{eq:S_I_infty_incidence} is obtained by taking $t\to\infty$ in~\Fref{eq:S_I_incidence}. The solution of~\Fref{eq:S_infty_incidence} is given explicitly by 
\begin{linenomath*}
\begin{equation}
S_\infty = -\frac{\gamma}{\beta}\W_0\left( -\frac{\beta(1-p)(1-\alpha)}{\gamma}e^{-\frac{\beta+\phiinc\alpha}{\gamma}(1-p)} \right)\,,
\end{equation}
\end{linenomath*}
where we take the principle branch of the Lambert function, $\W_0$, because solutions are in the range $S_\infty\in[0,\gamma/\beta]$ (see~\Fref{eq:S_infty_<1/R0} in~\Fref{appx:dfe}). Note also that $S_\infty>0$ \textiff $S(0)>0$ (that is, $p<1$ and $\alpha<1$). Thus, $S=0$ is not attainable in scenarios of interest here. 

To find $V_\infty$, we proceed similarly: 
\begin{linenomath*}
\begin{align}
\dot V &= -\frac{\phiinc}{\beta+\phiinc}\dot S \\
V_\infty - V(0) &=  -\frac{\phiinc}{\beta+\phiinc}(S_\infty - S(0)) \\
V_\infty -p &= \frac{\phiinc}{\beta+\phiinc}( (1-p)(1-\alpha)- S_\infty) \, .
\end{align}
\end{linenomath*}
At the end of the epidemic, $1 = R_\infty + S_\infty + V_\infty$,
hence,
\begin{linenomath*}
\begin{align}
R_\infty &= 1 - p + \frac{\phiinc}{\beta+\phiinc}(S_\infty - (1-p)(1-\alpha)) - S_\infty \nonumber\\
& = 1-p - \frac{\phiinc(1-p)(1-\alpha) +\beta S_\infty}{\beta+\phiinc}\, .
\end{align}
\end{linenomath*}
Thus,
\begin{subequations} \label{eq:pi_psi_inc}
\begin{linenomath*}
\begin{align}
\pi_p & = \frac{R_\infty}{1-p} = 1- \frac{\phiinc}{\beta+\phiinc}(1-\alpha)- \frac{\beta S_\infty}{(\beta+\phiinc)(1-p)}\\
\psi_p &=  \frac{V_\infty-p}{1-p} = \frac{\phiinc}{\beta+ \phiinc}\left((1-\alpha)-\frac{S_\infty}{1-p}\right)  \, . 
\end{align}
\end{linenomath*}
\end{subequations}
Note that whenever $S(0)>0$, there are susceptible individuals left at the end of the epidemic, and so $\pi_p\ne1-\psi_p$. 

Using $\partial_p \left(\frac{S_\infty}{1-p}\right) = \frac{(1-p)\partial_p S_\infty+S_\infty}{(1-p)^2}$ we have
\begin{linenomath*}
\begin{align}  
\partial_p S_\infty & = \frac{\gamma}{\beta}\left(\frac{\partial_p S_\infty}{S_\infty}+ \frac{1}{1-p}\right)  -\left(  1  +\frac{\phiinc}{\beta}\alpha\right) \nonumber\\
\frac{\beta S_\infty-\gamma}{\beta S_\infty}\partial_p S_\infty & =\frac{\gamma}{\beta(1-p)}-1 - \frac{\phiinc}{\beta}\alpha \nonumber\\
\partial_p S_\infty & =\frac{\gamma}{1-p}\frac{S_\infty}{\beta S_\infty-\gamma}-(\phiinc\alpha+\beta)\frac{S_\infty}{\beta S_\infty-\gamma}  \nonumber\\
 & =\frac{\gamma-(\phiinc\alpha+\beta)(1-p)}{(1-p)}\frac{S_\infty}{\beta S_\infty-\gamma} \nonumber\\
(1-p)\partial_p S_\infty +S_\infty & =\frac{(\beta S_\infty -(1-p)(\beta+\phiinc\alpha))S_\infty}{\beta S_\infty-\gamma}\nonumber\\
\partial_p \left(\frac{S_\infty}{1-p}\right) & = \frac{(\beta S_\infty -(1-p)(\beta+\phiinc\alpha))S_\infty}{\beta S_\infty-\gamma}\frac{1}{(1-p)^2} \,. 
\end{align}
\end{linenomath*}
From~\Fref{eq:S_infty_incidence}
\begin{linenomath*}
\begin{equation}
\beta S_\infty - (1-p)(\beta  +\phiinc\alpha)= \gamma\ln\Big(\frac{S_\infty}{S(0)}\Big)<0
\end{equation}
\end{linenomath*}
because $S_\infty<S(0)$ (unless $S(0)$ or $I(0)$ is 0, in which case an outbreak cannot take place). 


Thus $\partial_p \left(\frac{S_\infty}{1-p}\right)>0$ and so 
\begin{linenomath*}
\begin{align}
\partial_p \pi_p&<0 \\
\partial_p\psi_p&<0 \, .
\end{align}
\end{linenomath*}

Note also that $S_\infty$ attains a local maximum (in $p$) at 
\begin{linenomath*}
\begin{equation}
p_m=1-\gamma/(\phiinc\alpha+\beta) \,.
\end{equation}
\end{linenomath*}
$p_m<0$ for $\gamma>\phiinc\alpha+\beta$, in which case $S_\infty$ decreases with $p$ on the interval $[0,1]$. This can only happen when $\gamma>\beta$, that is when $\mathcal{R}_0<1$, implying that for any disease which can spread in the population (with no vaccination), pre-emptive vaccination initially raises, then lowers the proportion of susceptibles remaining at the end of the epidemic.
The maximum level of remaining susceptibles is 
\begin{linenomath*}
\begin{equation}
S_\infty\lvert_{p=p_m} = -\frac{\gamma}{\beta}\W_0\left( -\frac{\beta(1-\alpha)}{(\phiinc\alpha+\beta)e} \right)\,.
\end{equation}
\end{linenomath*}

However, $R_\infty$ is more informative, since susceptibles can be depleted by either infection or vaccination, and so fewer remaining susceptibles does not necessarily imply a larger epidemic, nor does it imply that more individuals were vaccinated. 
However,  
\begin{linenomath*}
\begin{align}
\partial_p R_\infty &=  \partial_p \left((1-p) \pi_p\right)= -\pi_p+(1-p)\partial_p\pi_p<0 \,,
\end{align}
\end{linenomath*}
which shows that increasing pre-emptive vaccine coverage decreases the size of the epidemic, as expected.  

\subsection {Vaccination rate $\propto$ proportion still susceptible}\label{appx:finsize+:vac propto S}
The model equations are
\begin{subequations}
\begin{linenomath*}
\begin{align}
\dot S &= -\beta SI -\phisusc S \\
\dot I &= \beta SI -\gamma I \\
\dot R &= \gamma I \\
\dot V &= \phisusc S\, . 
\end{align}
\end{linenomath*}
\end{subequations}

In this case, a similar strategy to the one we employed for the case where vaccination is proportional to prevalence doesn't quite work. Calculating $S_\infty(I(0))$ is not enough, since we do not know how the remainder of the population is partitioned between the removed and vaccinated classes at the end of the epidemic. The following calculations are also helpful but insufficient:
\begin{linenomath*}
\begin{align}
\frac{\dot I}{I} &= \beta S -\gamma= \frac{\beta}{\phisusc}\dot V - \gamma \\
\frac{\dot S}{S} &= -\beta I -\phisusc=- \frac{\beta}{\gamma}\dot R - \phisusc \, ,
\end{align}
\end{linenomath*}
so
\begin{linenomath*}
\begin{align}
 \label{eq:I-V} \ln\Big(\frac{I(t)}{I(0)}\Big) &=\frac{\beta}{\phisusc}(V(t)-V(0))-\gamma t \\
\label{eq:S-V} \ln\Big(\frac{S(t)}{S(0)}\Big) &=-\frac{\beta}{\gamma}(R(t)-R(0))-\phisusc t \, .
\end{align}
\end{linenomath*}
However, it is not possible to extract $V_\infty$ and $R_\infty$ from here because phase-portrait arguments show that $I$ and $S$ tend to 0 as $t\to\infty$ (this is also implied by \Erefs{eq:I-V} and \eqref{eq:S-V}), thus both sides of these equations diverge as $t \to \infty$.

Nonetheless, a similar method to the one employed in \Fref{appx:finsize+:vac propto SI} yields a relation between $S(t)$ and $I(t)$ from which, using the previous relations, a relation between $R(t)$ and $V(t)$ can be obtained. These will not diverge as $t\to\infty$ (they are bounded), so any divergent components must cancel out. 
\begin{linenomath*}
\begin{align}
\label{eq:dSdI_susceptible}
\frac{\mathrm{d}S}{\mathrm{d}I} &= -\frac{(\beta I +\phisusc) S}{(\beta S -\gamma) I} \\ 
\frac{(\beta S -\gamma)}{S}\mathrm{d}S &=-\frac{(\beta I +\phisusc)}{I}\mathrm{d}I \\
(\beta - \frac{\gamma}{S})\mathrm{d}S &= -(\beta + \frac{\phisusc}{I})\mathrm{d}I \\
\label{eq:S-I_susc} \beta(S(t)-S(0)) - \gamma \ln\Big(\frac{S(t)}{S(0)}\Big) &= -\beta(I(t)-I(0)) - \phisusc \ln\Big(\frac{I(t)}{I(0)}\Big)
\end{align}
\end{linenomath*}
from which we get:
\begin{linenomath*}
\begin{equation}\label{eq:S_susc_explicit}
S(t) = -\frac{\gamma }{\beta} \W\left(-\frac{\beta}{\gamma} e^{\frac{\beta }{\gamma}(I(t)-I(0) -S(0))} (\frac{I(t)}{I(0)})^{\frac{\phisusc }{\gamma }} S(0)\right) \,.
\end{equation}
\end{linenomath*}
By taking the limit $t\to\infty$ in~\Fref{eq:S_susc_explicit}, we see that $S_\infty = 0$, and conseqently, $\psi_p = 1-\pi_p$. 

We now determine under which conditions each of the two branches of the Lambert $\W$ function is used in~\Fref{eq:S_susc_explicit}. First, note that
\begin{linenomath*}
\begin{align}
S(I)  &= -\frac{\gamma}{\beta}\W(z(p))\nonumber \\
z(I) &= -\frac{\beta}{\gamma} e^{\frac{\beta }{\gamma}(I(t)-I(0) -S(0))} \left(\frac{I(t)}{I(0)}\right)^{\frac{\phisusc }{\gamma }} S(0)\nonumber\\
z(I(0)) &= -\frac{\beta}{\gamma}S(0) e^{-\frac{\beta }{\gamma}S(0)} \nonumber\,.
\end{align}
\end{linenomath*}
For $I=I(0)$, we expect to get $\W( -\frac{\beta}{\gamma}S(0)) =  -\frac{\beta}{\gamma}S(0)$ so that $S(I=I(0)) =S(t=0)$. We know that $S(0)>\frac{\gamma}{
\beta}$ which implies that for $t=0$, we must use $\W_1$. Because $S(t)$ monotonically decreases to 0 as $t\to0$, we know that the branch $\W_1$ is used until the peak prevalence is attained (at which time $S=\gamma/\beta$), and then the principal branch $\W_0$ is used. 
Now, 
\begin{linenomath*}
\begin{align}
\frac{\mathrm{d}R}{\mathrm{d}I} &=\frac{\gamma I}{\beta SI-\gamma I}\nonumber\\
& = \frac{\gamma }{\beta S-\gamma}\nonumber\\
& = \frac{-1}{\W\left(-\frac{\beta}{\gamma} e^{\frac{\beta }{\gamma}(I-I(0) -S(0))} (\frac{I}{I(0)})^{\frac{\phisusc }{\gamma }} S(0)\right)+1}\,.
\end{align}
\end{linenomath*}
This can be integrated, to give
\begin{linenomath*}
\begin{align} 
R_\infty&=R(0)+ \int_{I(0)}^{I_\infty} \frac{\gamma }{\beta S-\gamma} \,\mathrm{d}I \label{eq:Rinf_susc}\\
&=R(0) + \int_{I(0)}^{I_\infty} \frac{-1}{\W_i\left(-\frac{\beta}{\gamma} e^{\frac{\beta }{\gamma}(I-I(0) -S(0))} (\frac{I}{I(0)})^{\frac{\phisusc }{\gamma }} S(0)\right)+1}\,\mathrm{d}I \nonumber \\
& = \int_{0}^{I(0)} \frac{1}{\W_i\left(-\frac{\beta}{\gamma} e^{\frac{\beta }{\gamma}(I-I(0) -S(0))} (\frac{I}{I(0)})^{\frac{\phisusc }{\gamma }} S(0)\right)+1}\,\mathrm{d}I\,, 
\end{align}
\end{linenomath*}
where the appropriate branch of $\W_i$ is determined as above, as the integration varible $I$ is varied. Note, however, that the integral in~\Fref{eq:Rinf_susc} is improper, because the integral diverges at the peak prevalence (when $S=\gamma/\beta$). 
\subsection{Instantaneous vaccination of a proportion $\phiinst$ of the population}
\label{appx:finsize+:instvac}
In this case, the disease progresses according to the standard $SIR$ model,
\begin{subequations}
\begin{linenomath*}
\begin{align}
\dot S &= -\beta SI \,,\label{eq:inst_Sdot}\\
\dot I &= (\beta S -\gamma) I \,,\\
\dot R &= \gamma I \,,
\end{align}
\end{linenomath*}
\end{subequations} 
with initial conditions given by
\begin{linenomath*}
\begin{align*}
S(0)&=(1-p)(1-\alpha)(1-\phiinst)\\
I(0)&=(1-p)\alpha\\
R(0)&=0\\
V(0)&=p+\phiinst(1-p)(1-\alpha) \,. 
\end{align*}
\end{linenomath*}
Note that for this model, $S(0)$ is the density of susceptibles after the post-outbreak vaccination response has taken place.

\Fref{eq:inst_Sdot} implies that  
\begin{linenomath*}
\begin{equation}
-\frac{\gamma}{\beta} \frac{\mathrm{d}}{\mathrm{d}t}\ln\left(S\right) = \dot R\,,
\end{equation}
\end{linenomath*}
thus $S_\infty$ satisfies the equation
\begin{linenomath*}
\begin{equation} \label{eq:S_infty_instantaneous}
\frac{\gamma}{\beta} \ln\left( \frac{S(0)}{S_\infty} \right) =R_\infty= 1-V(0) -S_\infty\,,
\end{equation}
\end{linenomath*}
or 
\begin{linenomath*}
\begin{equation}
S_\infty = -\frac{\gamma}{\beta} \W_0 \left( -\frac{\beta}{\gamma}S(0) e^{-\frac{\beta}{\gamma}(1-V(0))} \right)\,.
\end{equation}
\end{linenomath*}
We use the principle branch of the Lambert function in order to obtain solutions satisfying $S_\infty\le S(0)$. Since $-\frac{\beta}{\gamma}S(0) e^{-\frac{\beta}{\gamma}(1-V(0))} > -\frac{\beta}{\gamma}S(0) e^{-\frac{\beta}{\gamma}S(0)}$, and $\W_1$ is monotonically decreasing, $-\frac{\gamma}{\beta}\W_1\left( -\frac{\beta}{\gamma}S(0) e^{-\frac{\beta}{\gamma}(1-V(0))} \right)>S(0) $, which does not correspond to biologically feasible solutions. 
In addition, we have $R_\infty = \frac{\gamma}{\beta}\ln\frac{S(0)}{S_\infty}$.

Since there is no vaccination except during the initial (immediate) response to the outbreak, $V_\infty=V(0) = p+\phiinst(1-p)(1-\alpha)$, and so 
\begin{subequations}
\begin{linenomath*}
\begin{align}
\psi_p &= \frac{V_\infty - p}{1-p} = \phiinst(1-\alpha)\\
\pi_p & = 1- \psi_p - \frac{S_\infty}{1-p} \, .\label{eq:inst_pi_p} 
\end{align}
\end{linenomath*}
\end{subequations}
Using~\Fref{eq:S_infty_instantaneous}, we also have 
\begin{linenomath*}
\begin{equation} \label{eq:pi_p_instantaneous_implicit}
\pi_p =-\frac{\gamma}{\beta(1-p)} \ln\left( \frac{1-\pi_p -\phiinst(1-\alpha)}{(1-\alpha)(1-\phiinst)}\right) 
\end{equation}
\end{linenomath*}
It follows from~\Fref{eq:inst_pi_p} that $\pi_p$ is a decreasing function of $p$: from~\Fref{eq:S_infty_instantaneous} we have 
\begin{linenomath*}
\begin{equation} 
\frac{\gamma}{\beta}\left(-\frac{1}{1-p} -\frac{\partial_p S _\infty}{S_\infty} \right) = -1 + \phiinst(1-\alpha) - \partial_p S_\infty
\end{equation} 
\end{linenomath*}
and so 
\begin{linenomath*}
\begin{equation} \label{eq:partial_p_S_infty_instantaneous}
\partial_p S_\infty  = \left(-1 + \phiinst(1-\alpha) + \frac{\gamma}{\beta}\frac{1}{1-p}\right) \left( \frac{\beta S_\infty}{\beta S_\infty-\gamma}\right)\,.
\end{equation}
\end{linenomath*}
 This gives 
\begin{linenomath*}
\begin{align} 
(1-p)\partial_p S_\infty + S_\infty &=  \frac{\beta S_\infty}{\beta S_\infty-\gamma} \left((1-p)\left(-1 + \phiinst(1-\alpha)  \right) + S_\infty\right)\nonumber \\
&=  \frac{\beta S_\infty}{\gamma-\beta S_\infty} \left(1- V_\infty - S_\infty\right)\,,
\end{align}
\end{linenomath*}
which is positive so long as $R_\infty>0$ (this happens when $S(0)> S_\infty$, which is true whenever $S(0)$ and $I(0)$ are not 0).
 It now follows that $\partial_p \pi_p = -\frac{(1-p)\partial_p S_\infty + S_\infty}{(1-p)^2}<0 $. 
Note that the probability of a delayer being vaccinated post-outbreak ($\psi_p$) is constant.

Lastly, note that
\begin{linenomath*}
\begin{equation}
(1-p)\partial_p S_\infty  = \left(1-V_\infty - \frac{\gamma}{\beta}\right) \left( \frac{\beta S_\infty}{\gamma-\beta S_\infty}\right)\,,
\end{equation}
\end{linenomath*}
which implies that $S_\infty$ increases with $p$ \textiff  $ (1-p)(1 - \phiinst(1-\alpha))\beta>\gamma$, or equivalently, $I(0)+ S(0)=1-V_\infty >\frac{\gamma}{\beta}$. Compare this to the more stringent condition $S(0)>\frac{\gamma}{\beta}$ which ensures that the epidemic  takes off ($I'(0)>0$). 

\section{Maximal vaccination rate for fair comparison of models}
In this section, we find the fair comparison values of the vaccination efforts $\phiinc$ and $\phiprev$ (see \fref{sec:vaceffort}). These are defined as the levels of vaccination effort $\phiinc$ and $\phiprev$ that result in maximal vaccination rates equal to $0.1/\text{day}$ (that is, comparable to \cite{Bauc+03}). 
\subsection{Maximal Vaccination rate when $\dot V = \phiprev I$} \label{appx:phiprevMaxVaccRate}
We begin by finding what the maximal vaccination is when $\dot V = \phiprev I$. Because the vaccination rate, $\dot V$, is maximal when prevalence, $I$, is maximal, we aim to find the maximal prevalence. Now observe that since $\dot I = \left(\beta S -\gamma \right)I$, incidence is maximal when $S= \gamma/\beta$. 
Thus, the peak prevalence is found by substituting $S= \gamma/\beta$, into~\Fref{eq:I_S_prevalence_alltime} to obtain 
\begin{linenomath*}
\begin{equation}
I_{\rm peak} = 1-p -\gamma/\beta+ \frac{\gamma+ \phiprev}{\beta}\ln\Big(\frac{\gamma + \phiprev}{\beta S(0) + \phiprev}\Big)
\end{equation}
\end{linenomath*}
(recall that $I(0) + S(0) = 1-p$).
We now wish to find at which value of $p$ the maximal vaccination rate (over time) is largest. Observe that
\begin{linenomath*}
\begin{align}
\frac{\partial}{\partial p} I_{\rm peak} &= -1 + \frac{(\gamma + \phiprev)(1-\alpha)}{\beta S(0) + \phiprev}\,,\\
\frac{\partial^2}{\partial p^2} I_{\rm peak} &=  \frac{\beta(\gamma + \phiprev)(1-\alpha)^2}{(\beta S(0) + \phiprev)^2}>0 \,,
\end{align}
\end{linenomath*}
which implies that the peak prevalence (and thus the maximal vaccination rate) has a minium in $p$ when 
\begin{linenomath*}
\begin{equation}
p_\text{crit}= \frac{(\beta-\gamma) (1-\alpha)+\phiprev\alpha}{\beta (1-\alpha)}>0
\end{equation}
\end{linenomath*}
($\beta>\gamma$ because $\mathcal{R}_0>1$).

There are now two possibilities: 
\begin{itemize}
\item If $p_\text{crit}\ge1$ (which happens \textiff $\phiprev\alpha\le\gamma (1-\alpha)$) then the maximal vaccination rate is attained when $p=0$.
\item If $p_\text{crit}<1$ (which happens \textiff $\phiprev\alpha>\gamma (1-\alpha)$) then, the maximal vaccination rate must be attained either when $p=0$ or when $p=1$. 
\end{itemize}
Noting that 
\begin{linenomath*}
\begin{align}
I_\text{peak}\lvert_{p=0} &= 1-\gamma/\beta+ \frac{\gamma+ \phiprev}{\beta}\ln\Big(\frac{\gamma + \phiprev}{\beta (1-\alpha) + \phiprev}\Big)\nonumber\\
I_\text{peak}\lvert_{p=1} &= -\gamma/\beta+ \frac{\gamma+ \phiprev}{\beta}\ln\Big(\frac{\gamma + \phiprev}{  \phiprev}\Big)\,, 
\end{align}
\end{linenomath*}
it follows that 
\begin{linenomath*}
\begin{align}
\dot V_{\rm max} =-\phiprev\frac{\gamma}{\beta} +\phiprev\max\bigg\{&1+\frac{\gamma+ \phiprev}{\beta}\ln\Big(\frac{\gamma + \phiprev}{\beta (1-\alpha) + \phiprev}\Big)\,,\nonumber\\ 
&\frac{\gamma+ \phiprev}{\beta}\ln\Big(\frac{\gamma + \phiprev}{  \phiprev}\Big) \bigg\}.
\label{eq:Vdotmaxphiprev}
\end{align}
\end{linenomath*}

We also note that
\begin{linenomath*}
\begin{equation}
\frac{\partial}{\partial \phiprev} I_{\rm peak} = \frac{1}{\beta}\ln\Big(\frac{\gamma + \phiprev}{\beta S(0) + \phiprev}\Big)+ \frac{1}{\beta}\Big(1 - \frac{\gamma+ \phiprev}{\beta S(0) + \phiprev}\Big)\,.
\end{equation}
\end{linenomath*}
Because $1-x+\ln(x)<0$ for any $0<x\ne1$, it follows that  the peak prevalence $I_{\rm peak}$, and thus the peak vaccination rate, decreases with increasing vaccination effort, $\phiprev$ (for any initial coverage, $p$). 
Also, as $\phiprev\to\infty$, we have
\begin{linenomath*}
\begin{align}
I_\text{peak}\lvert_{p=0} &\to  \alpha\nonumber\,,\\
I_\text{peak}\lvert_{p=1} &\to 0\,.\nonumber
\end{align}
\end{linenomath*}

Setting $\dot V_{\rm max}=0.1 /\text{day}$ in \Fref{eq:Vdotmaxphiprev} we can numerically solve for $\phiprev$, with $\alpha$, $\beta$, $\gamma$, as in \Trefs{tab:estimated} and \ref{tab:derived} to obtain $\phiprev \approx 1582/\text{day}$.  

\subsection{Maximal Vaccination rate when $\dot V = \phiinc SI$} \label{appx:phiincMaxVaccRate}
First, we will derrive a formula for the maximal vaccination rate as it depends on the model parameters, $\alpha$, $\beta$, $\gamma$, $\phiinc$. We then use this formula to calculate the appropriate range for $\phiinc$, given the estimates of the other parameters cited in~\Trefs{tab:estimated} and \ref{tab:derived}.

Differentiating~\Fref{eq:dotVinc}, we have 
\begin{linenomath*}
\begin{equation*}
\ddot V=\phiinc\big(\dot S I+ S\dot I\big) = \phiinc SI \big(\beta S - \gamma - (\beta+ \phiinc)I\big)\,.
\end{equation*}
\end{linenomath*}
Thus, critical points of $\dot V$ (excluding those for which $\dot V = 0$) occur when $\beta S - \gamma = (\beta+\phiinc)I$. Using~\Fref{eq:S_I_incidence} and simplifying, this is equivalent to
\begin{linenomath*}
\begin{equation*}
2\beta S =\gamma\ln S + \phiinc(1-p)\alpha+\beta(1-p) +\gamma\Big(1-\ln\big((1-p)(1-\alpha)\big)\Big)\,,
\end{equation*}
\end{linenomath*} 
which has two formal solutions,
\begin{linenomath*}
\begin{equation*}
\hat{S}_k=-\frac{\gamma}{2\beta}\W_k\left(-2\frac{\beta}{\gamma}(1-p)(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}(1-p) -1}\right) \,,
\end{equation*}
\end{linenomath*}
with $k=0$ or $-1$. 

However, it is impossible for $\hat{S}_0$ to be attained by $S(t)$, for all $t\ge0$. To see this, suppose, in order to derive a contradiction, that there is some time $\hat{t}_0\ge0$ such that $S(\hat{t}_0) = \hat{S}_0$. Note that $-1\le\W_0<0$ on the interval $[-1/e,0)$, so $0<\hat{S}_0<\gamma/\beta$.  Because $\hat{S}_0<\gamma/\beta$, we have $\dot{I}(\hat{t}_0)<0$. Since $\dot S <0$ (for all time $t$), it follows that $\ddot V<0$ when $t=\hat{t}_0$, in contradiction to the fact that by definition of $\hat{S}_0$, $\ddot V (\hat{t}_0) =0$. Thus, $S(t)>\hat{S}_0$ is proven\footnote{Consequently, $S_\infty\ge \hat{S}_0$, which is equivalent to a statement about the Lambert $\W$ function: $\W_0(2x/e)\ge2\W_0(x)$, for $-1/e<x<0$.}. It follows that if there is a biologically relevant value of $S$ at which $\ddot V$ changes signs, it must be
\begin{linenomath*}
\begin{equation}\label{eq:Scrit}
\hat{S}_{-1} = -\frac{\gamma}{2\beta}\W_{-1}\left(-2\frac{\beta}{\gamma e}(1-p)(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}(1-p)}\right)\,.
\end{equation}
\end{linenomath*}
Note that $\hat{S}_{-1}>\frac{\gamma}{2\beta}$ because $\W_{-1}(x)<-1\quad \forall x\in[-1/e,0]$ (but it is also possible that $\hat{S}_{-1}>S(0) = (1-p)(1-\alpha)$, which would make this critical point biologically unfeasible). 

Because $S$ decreases with time, we see that $\ddot V$ can change signs at most once for all $t\ge0$. Observe that since $0<S_\infty<\gamma/\beta$, and $\dot V$ decreases when $S\in(0,\gamma/\beta]$, it follows that $\dot V$ eventually (\emph{i.e.,} for large enough $t$) decreases with time. Hence, if $\ddot V(0) \le 0$, then $t=0$ is a maximum of $\dot V$ for $t\ge0$, and if $\ddot V(0) >0$ then $\dot V$ attains its maximum when $S(t) = \hat{S}_{-1}$. The sign of $\ddot V(0)$ is identical to the sign of $\beta S(0) - \gamma - (\beta+ \phiinc)I(0)$, so the maximal vaccination rate is attained at $t=0$ if $(1-p)\big((1-2\alpha)\beta-\alpha\phiinc\big)\le\gamma$, and when $S(t) = \hat{S}_{-1}$ otherwise. Thus, 
\begin{linenomath*}
\begin{equation}
\nu(p) = \max_{t\ge0} \dot V =
\begin{cases}
\phiinc (1-p)^2(1-\alpha)\alpha & \text{if } (1-p)((1-2\alpha)\beta-\alpha\phiinc)\le\gamma \,, \\
 \frac{\phiinc}{\phiinc+\beta} (\beta \hat{S}_{-1} -\gamma)\hat{S}_{-1} & \text{if } (1-p)((1-2\alpha)\beta-\alpha\phiinc)>\gamma \,,
\end{cases}
\end{equation}
\end{linenomath*}
(where, for the second case, we used the fact that $I=\frac{\beta \hat{S}_{-1} -\gamma}{\phiinc+\beta}$ when $S=\hat{S}_{-1}$). 

To maximize $\nu$ over all $p\in [0,1)$ (with $\alpha$, $\beta$, $\gamma$ and $\phiinc$ fixed), we consider the following 3 cases: 
\begin{itemize}
\item First, if $0<{\gamma}< {(1-2\alpha)\beta-\alpha\phiinc}$, then $(1-p)((1-2\alpha)\beta-\alpha\phiinc)\le\gamma$ is equivalent to $\hat{p} = 1-\frac{\gamma}{(1-2\alpha)\beta-\alpha\phiinc}\le p$, and $\hat{p}\in (0,1)$. Hence, for ${p\in[\hat{p},1)}$, ${\nu(p) = \phiinc(1-p)^2(1-\alpha)\alpha}$ is a decreasing function of $p$, and so $\max_{p\in[\hat{p},1)} \nu(p) = \phiinc \Big(\frac{\gamma}{(1-2\alpha)\beta-\alpha\phiinc}\Big)^2(1-\alpha)\alpha $, and  is attained when $p=\hat{p}$. 

When $0\le p<\hat{p}$, we note that because $(\beta x - \gamma)x$ is parabolic with a minimum at $x = \gamma/2\beta$, and $\hat{S}_{-1}\ge\frac{\gamma}{2\beta}$, it follows that 
\begin{linenomath*}
\begin{equation}
\nu(p) =\frac{\phiinc}{\phiinc+\beta} (\beta \hat{S}_{-1} -\gamma)\hat{S}_{-1} 
\end{equation}
\end{linenomath*}
is increasing in $\hat{S}_{-1}$, so $\max_{p\in[0,\hat{p}]} \nu(p)$ is attained on this interval when $\hat{S}_{-1}$ is maximized. Because $-\frac{\gamma}{2\beta}\W_{-1}(x)$ is monotonically increasing, it follows that $\max_{p\in[0,\hat{p}]} \nu(p)$ is maximal when 
\begin{linenomath*}
\begin{equation*}
-2\frac{\beta}{\gamma e}(1-p)(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}(1-p)}
\end{equation*}
\end{linenomath*}
is maximal.
Consequently, we need to maximize $-axe^{-x}$ (with $a>0$), with $x(p) = \frac{\beta+ \phiinc\alpha}{\gamma} (1-p)$, over the interval $0\le p\le \hat{p}$. This corresponds to maximizing $-axe^{-x}$ over $[\frac{\beta+ \phiinc\alpha}{\gamma}, \frac{\beta+ \phiinc\alpha}{(1-2\alpha)\beta-\alpha\phiinc}]\subset[1,\infty)$. Observe that $-axe^{-x}$ has a unique global minimum at $x=1$, and in particular, it is increasing when $x\ge1$. This implies that in the relevant range of $x$, $\hat{S}_{-1}$ increases with $x$, and thus decreases in $p$. It follows that $\hat{S}_{-1}$ is maximal when $p = 0$, and its value is 
\begin{linenomath*}
\begin{equation}\label{eq:Scrit}
\hat{S}_{-1}\lvert_{p=0} = -\frac{\gamma}{2\beta}\W_{-1}\left(-2\frac{\beta}{\gamma e}(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}}\right)\,.
\end{equation}
\end{linenomath*}
Thus, $\max_{p\in[0,1)} \nu(p) =\nu(0)$, and is attained when $p= 0$ (note also that $\nu(p)$ is continuous at $p= \hat{p}$).
\item When $0<{(1-2\alpha)\beta-\alpha\phiinc}\le\gamma$, $(1-p)((1-2\alpha)\beta-\alpha\phiinc)\le\gamma$ is equivalent to $\hat{p} = 1-\frac{\gamma}{(1-2\alpha)\beta-\alpha\phiinc}\le p$, which is satisfied for all $p\in [0,1)$, since $\hat{p}\le0$. Thus, $\nu(p)$ is a decreasing function of $p$ for all $p\in[0,1)$, and thus $\max_{p\in[0,1)} \nu(p)= \phiinc(1-\alpha)\alpha$ is attained at $p=0$.
\item  When ${\alpha\phiinc\ge(1-2\alpha)\beta} $, then $(1-p)((1-2\alpha)\beta-\alpha\phiinc)\le\gamma$ is  always satisfied (since the left hand side is never positive, and $\gamma>0$). Thus, $\nu(p) = \phiinc (1-p)^2(1-\alpha)\alpha$, which decreases with $p$, so $\max_{p\in[0,1)} \nu(p)=\phiinc(1-\alpha)\alpha$, and is attained at $p=0$.
\end{itemize}
Rearranging the conclusions of the preceeding discussion, we see that 
\begin{linenomath*}
\begin{align}\label{eq:Vdotmax}
\max_{p\in[0,1), t\ge0} \dot V =
\begin{cases}
\frac{\phiinc}{\phiinc+\beta} (\beta \hat{S}_{-1} -\gamma)\hat{S}_{-1}\lvert_{p=0} & \text{if } 0\le\alpha\phiinc <(1-2\alpha)\beta -\gamma\,, \\
 \alpha\phiinc(1-\alpha) & \text{if } (1-2\alpha)\beta-\gamma\le \phiinc\alpha \,.
\end{cases}
\end{align}
\end{linenomath*}

When  $ 0\le\alpha\phiinc <(1-2\alpha)\beta -\gamma$, then $\max_{p\in[0,1), t\ge0} \dot V = \frac{\phiinc}{\phiinc+\beta} (\beta \hat{S}_{-1} -\gamma)\hat{S}_{-1}\lvert_{p=0}$, which increases as $S_{-1}\lvert_{p=0}$ increases, as stated earlier. Since $-\W_{-1}(x)$ increases with $x$,  and $-2\frac{\beta}{\gamma e}(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}}$ is an increasing function of $\phiinc$, we conclude that in this range, $\max_{p\in[0,1), t\ge0} \dot V$ increases with $\phiinc$. When $(1-2\alpha)\beta-\gamma\le \phiinc\alpha$, $\max_{p\in[0,1)} \nu(p)$ manifestly increases linearly with $\phiinc$. In all, $\max_{p\in[0,1), t\ge0} \dot V$ is a monotonically increasing function of $\phiinc$. 

Note that at the point separating the two regimes, $\phiinc = \frac{(1-2\alpha)\beta-\gamma}{\alpha}=16570.71/\text{day}$, $\max_{p\in[0,1), t\ge0} \dot V=\big((1-2\alpha)\beta-\gamma\big)(1-\alpha)  = 0.29/\text{day}$ (with parameters as in~\Fref{tab:estimated}).

Finally, to obtain a value of $\phiinc$ that yields a maximal vaccination rate of $\Phi = 0.1 /\text{day}$ (as was estimated in \cite{Bauc+03}), we solve 
\begin{linenomath*}
\begin{equation}
\frac{\phiinc}{\phiinc+\beta} (\beta \hat{S}_{-1} -\gamma)\hat{S}_{-1}\lvert_{p=0}=\Phi\,,
\end{equation}
\end{linenomath*}
 which gives $\hat{S}_{-1}\lvert_{p=0} = \frac{\gamma \pm\sqrt{\gamma^2 +4\beta\Phi(1+\beta/\phiinc)}}{2\beta}$. We take the solution corresponding to the positive sign (the other one gives negative $\hat{S}_{-1}$, which is biologically absurd). Thus,
\begin{linenomath*}
\begin{equation}
\W_{-1}\left(-2\frac{\beta}{\gamma e}(1-\alpha)e^{-\frac{\beta+\phiinc\alpha}{\gamma}}\right)= -1 -\sqrt{1 +4\frac{\beta}{\gamma^2}\Phi(1+\beta/\phiinc)} \, ,
\end{equation}
\end{linenomath*} 
which is equivalent to 
\begin{linenomath*}
\begin{equation*}
\frac{2\beta(1-\alpha)}{\gamma\big(1+ \sqrt{1 +4\frac{\beta}{\gamma^2}\Phi(1+\beta/\phiinc)}\big)}= \exp\Big(\frac{\beta+\phiinc\alpha}{\gamma}-\sqrt{1 +4\frac{\beta}{\gamma^2}\Phi(1+\beta/\phiinc)}\Big) \, ,
\end{equation*}
\end{linenomath*} 
which we solve numerically for $\phiinc$, with parameters as in \Trefs{tab:estimated} and \ref{tab:derived}, to get $\phiinc \approx 5190/\text{day}$.

\section{The individual equilibrium} \label{appx:indeq}
In this section, we show that for each of the five models defined in \Fref{sec:epidemic models}, the game defined in \Fref{sec:game theory} always has a unique convergently stable Nash equilibrium (defined in \Fref{sec:game theory} and abbreviated CSNE).  The proofs given here are constructive, \ie they also provide a method for numerically finding the individual equilbrium ($\pind$).
\subsection{Vaccination rate $\propto$ disease prevalence}\label{appx:indeq:vac propto I}
In this scenario we have 3 cases to examine: 
\begin{enumerate}
\item $p_m\ge1$ \,($\!\iff \alpha\phiprev\ge\gamma(1-\alpha)$)
\item $p_m\le1$ and $p_0\le 0$ 
\item$p_m\le1$ and $p_0 \in (0,1)$
\end{enumerate} 
Recall that in the first case, we have proven (in~\Fref{appx:finsize+:vac propto I}) that $\pi_p$ decreases with $p$. As stated in~\Fref{sec:vac propto I}, we assume that $\pi_p$ decreases with $p$ for the other two cases as well. 

\subsubsection{$p_m\ge1$} \label{appx:ind_eq_prev_S_infty_0}
In this case, we have a unique convergently stable Nash equilibrium (CSNE).
To see this, note that $\psi_p +\pi_p = 1$ and $\partial_p \pi_p<0$. Thus,  
\begin{linenomath*}
\begin{align}
\Delta E &=[\pi_p + (1-\pi_p) r-r/a]a(P-Q) \nonumber\\
& =[\pi_pa(1 -r)-  r(1-a)](P-Q) \nonumber\\
&= a(1 -r)\left[\pi_p - \frac{r (1 - a)}{a(1 - r)}\right](P-Q) \, .
\end{align}
\end{linenomath*}
It is convenient to define
%
\begin{linenomath*}
\begin{equation}\label{eq:oddsratio}
\rho_1 = \frac{r (1 - a)}{a(1 - r)} 
= \left.\frac{r}{1-r} \right/ \frac{a}{1-a} \,,
\end{equation}
\end{linenomath*}
%
which we can interpret as an \term{odds ratio}, namely the odds of a bad outcome from vaccination (compared with infection) relative to the odds of an outbreak occuring.  The odds ratio is well-defined and strictly positive ($\rho_1>0$) because $0<r<1$ and $0<a<1$. Since $\pi_p$ decreases monotonically with $p$, there are three cases: 
\begin{itemize}
\item If $\pi_0 \le\rho_1$ then $\pi_p<\rho_1 \quad \forall p>0$. It follows that $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=0 $. Hence $\pind=0$ is the unique Nash equilibrium. Let $0 \le P < Q$ and fix $\epsilon \in [0,1)$. It follows that $p > 0$, and so $\pi_p - \rho_1 < 0$. Thus $\Delta E >0$ and $\pind$ is convergently stable. 
\item If $\alpha=\pi_1 \ge\rho_1$, then $\pi_p>\rho_1 \quad \forall p<1$.  It follows that $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=1 $. Hence $\pind=1$ is the unique Nash equilibrium. The condition translates to $r(1-a)<\alpha a (1-r)$, or $\rvac  <a \rinf \alpha + a\rvac (1-\alpha)$. Recall that if an outbreak occurs, at the end of the epidemic individuals have either been vaccinated or have contracted the disease. Thus the right hand side is the risk to a vaccinator, and the left hand side is the minimal possible risk to a delayer (assuming no-one is infected after the initial outbreak; if there are secondary infections, then because $\rvac<\rinf$, the delayer's risk can only be increased). Let $1 \ge P > Q$ and fix $\epsilon \in [0,1)$. It follows that $p < 1$, and so $\pi_p - \rho_1 > 0$. Thus $\Delta E >0$ and $\pind$ is convergently stable. 
\item If $\pi_0 > \rho_1 >\pi_1=\alpha $ then there is a unique $\tilde p \in(0,1)$ such that $\pi_p -\rho_1 > 0$ if $p<\tilde p$, $\pi_{\tilde p}=\rho_1$ and $\pi_p-\rho_1 < 0 $ if $p>\tilde p$. Now since for any $\epsilon \in [0,1)$, $Q<P \implies p<P$ and $Q>P \implies p>P$, we have $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=\tilde p $ (for other $P$ take $Q$ between $P$ and $\tilde p$). Thus, the unique Nash equilibrium $\pind$ is the unique solution to $\pi_{\pind} = \rho_1$.  
Fix $\epsilon \in [0,1)$. If $Q<P\le \pind $, $Q \le p < P \le \pind$. Thus $\pi_p - \rho_1 > 0 \implies \Delta E >0$. Similarly, If $Q>P\ge \pind $, $Q \ge p > P \ge \pind$. Thus $\pi_p - \rho_1 < 0 \implies \Delta E >0$. Hence $\pind$ is convergently stable.

Now, to find $\tilde p$:  recall that $1-\pi_p = \frac{\phiprev}{\beta(1-p)}\ln\left(\frac{\beta}{\phiprev}(1-\alpha)(1-p) + 1 \right)$ and thus 
\begin{linenomath*}
\begin{equation} \label{eq:p_tilde_prev}
\tilde p = 1 + \phiprev \frac{ (1-\rho_1) + (1 - \alpha)\W \left(  -\frac{1-\rho_1}{1 - \alpha}e^{-\frac{1-\rho_1}{1 - \alpha}} \right) }{(1 -\alpha) \beta (1-\rho_1)}\,.
\end{equation}
\end{linenomath*}
Again, $\W$ is applied to a negative argument, and it is necessary to determine which branch of $\W$ to use. The principal branch gives $\tilde p = 1$, and $\pi_p\to \alpha$ as $p\to1$, and $\rho_1>\alpha$ by assumption, and so by elimination we must use $\W_{-1}$. Interestingly, $\tilde p$ depends linearly on $\phiprev$. Recall that $\W_{-1}\le-1$, and so $\frac{\partial {\tilde p}}{\partial \phiprev}<0$. 
\end{itemize}
Thus, in all three cases there exists a unique CNSE. 


\subsubsection{$p_m\le1$ and $p_0\le 0$}
In the 2\textsuperscript{nd} case, recall that $\psi_p =  \frac{\phiprev}{\gamma} \pi_p$, and let 
%
\begin{linenomath*}
\begin{equation}\label{eq:rho2}
\rho_2 = \frac{r}{a(1+r\phiprev/\gamma )}\,,
\end{equation}
\end{linenomath*}
%
to obtain
\begin{linenomath*}
\begin{align}
\Delta E &=[\pi_p +  \frac{\phiprev}{\gamma} \pi_p r-r/a]a(P-Q) \nonumber\\
& =[\pi_pa(1 + \frac{\phiprev}{\gamma}r)-  r](P-Q) \nonumber\\
&= a(1 + \frac{\phiprev}{\gamma}r)[\pi_p - \frac{\gamma r}{a(r\phiprev + \gamma )}](P-Q)\nonumber \\ 
&= a(1 + \frac{\phiprev}{\gamma}r)[\pi_p - \rho_2](P-Q)\, .
\end{align}
\end{linenomath*}

Now recall that $\pi_p$ decreases with $p$. So, an identical argument to the one in~\Fref{appx:ind_eq_prev_S_infty_0} also applies here:
\begin{itemize}
\item If $\pi_0\le \rho_2$ then $\pi_p<\rho_2 \quad \forall p>0$. Thus, $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=0 $. Hence $\pind=0$ is the unique Nash equilibrium. Let $0 \le Q < P<1$ and fix $\epsilon \in [0,1)$. It follows that $p >0$, and so $\pi_p - \rho_2 < 0$. Thus $\Delta E >0$ and $\pind$ is convergently stable. 
\item If $\alpha = \pi_1 \ge \rho_2 \ge 0$, then $\pi_p>\rho_2 \quad \forall p<1$.  It follows that $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=1 $. Hence $\pind=1$ is the unique Nash equilibrium. Let $1 \ge P > Q$ and fix $\epsilon \in [0,1)$. It follows that $p < 1$, and so $\pi_p - \rho_2 > 0$. Thus $\Delta E >0$ and $\pind$ is convergently stable. 
\item If $\pi_0 > \rho_2 >\pi_1 = \alpha $ then there is a unique $\tilde p \in(0,1)$ such that $\pi_p -\rho_2 > 0$ if $p<\tilde p$, $\pi_{\tilde p}=\rho_2$ and $\pi_p-\rho_2 < 0 $ if $p>\tilde p$. Now since for any $\epsilon \in [0,1)$, $Q<P \implies p<P$ and $Q>P \implies p>P$, we have $\forall \epsilon\in[0,1) \quad \Delta E>0\; \forall Q\ne P \iff P=\tilde p $. Thus, the unique Nash equilibrium $\pind$ is the unique solution to $\pi_{\pind} = \rho_2$.  
Fix $\epsilon \in [0,1)$. If $Q<P\le \pind $, $Q \le p < P \le \pind$. Thus $\pi_p - \rho_2 > 0 \implies \Delta E >0$. Similarly, If $Q>P\ge \pind$, $Q \ge p > P \ge \pind$. Thus $\pi_p - \rho_2 < 0 \implies \Delta E >0$. Hence $\pind$ is convergently stable.

To find $\tilde p$, recall that $R^1_\infty = (1-p)\pi_p$. Furthermore, from~\Fref{eq:R_infty_prevalence}, we have  
\begin{linenomath*}
\begin{align}
R_\infty^1  &= \frac{\gamma}{\beta}\ln\left(\frac{\beta S(0) + \phiprev}{\beta S_\infty^1 + \phiprev}\right)\nonumber\\
S_\infty^1 &= 1-p- \frac{\phiprev+\gamma}{\gamma} R_\infty^1 \nonumber\\
&\Downarrow\nonumber \\
R_\infty^1  &= \frac{\gamma}{\beta}\ln\left(\frac{\beta S(0) + \phiprev}{\beta (1-p- \frac{\phiprev+\gamma}{\gamma} R_\infty^1) + \phiprev}\right)\nonumber\\
&\Downarrow\nonumber \\
(1-p) \pi_p &= \frac{\gamma}{\beta}\ln\left(\frac{\beta S(0) + \phiprev}{\beta (1-p )\left(1- \frac{\phiprev+\gamma}{\gamma}\pi_p\right) + \phiprev}\right)\nonumber\,.
\end{align}
\end{linenomath*}
Substituting $p=\tilde p$, and using $\pi_{\tilde p} = \rho_2= \frac{\gamma r}{a(r\phiprev + \gamma)}$, we obtain after minor rearrangement 
\begin{linenomath*}
\begin{align}
  \frac{\beta(1-\tilde p)r}{a(r\phiprev + \gamma)} &= \ln\left(\frac{a(\beta (1-\tilde p) (1-\alpha) + \phiprev)(r\phiprev + \gamma)}{\beta (1-\tilde p ) (\gamma(a-r) -(1-a) r \phiprev ) + \phiprev a(r\phiprev + \gamma)}\right)\,.
\end{align}
\end{linenomath*}
However, we have not succeeded in obtaining an analytical solution for the individual equilibrium from this equation. 
\end{itemize}
Thus, in all three cases there exists a unique CSNE. 

\subsubsection{$p_m\le1$ and $p_0 \in (0,1)$}
Since $\pi_p$  decreases with $p$, the argument above shows that there is a unique CSNE in each of the the two intervals $[0,p_0]$ and $[p_0,1]$, denoted $P_{I,1}$ and $P_{I,2}$, respectively. These are the only candidates for Nash equilibria in the interval $[0,1]$: Adding the two sub-intervals together amounts to adding more strategies to the game. Thus, a strategy which was a Nash equilibrium in one of the sub-intervals may not be a Nash equilibrium for the larger strategy set (because players now have a larger strategy set to choose from). However, a strategy which is a Nash equilibrium for $[0,1]$ must be a Nash equilibrium in any sub-interval of $[0,1]$ which contains it. The situation for convergent stability is a bit more subtle, and is considered below. 

Note that when $p\ne p_0$,
\begin{linenomath*}
\begin{equation}
\sgn(\Delta E) = \sgn\big((\pi_p - \rho(p))(P-Q)\big) 
\end{equation}
\end{linenomath*}
 where 
\begin{linenomath*}
\begin{equation}
\rho(p) =
\begin{cases}
\rho_1 & \text{if } p < p_0, \\
\rho_2 & \text{if } p\ge p_0\, .
\end{cases}
\end{equation}
\end{linenomath*}
Note also that
\begin{linenomath*}
\begin{align}
\pi_{p_0} = \rho_1 \iff r=\frac{\gamma a}{\phiprev (1-a) +\gamma} \iff \pi_{p_0} = \rho_2 \,.
\end{align}
\end{linenomath*}
This may seem slightly perplexing at first, but recall that $\pi_{p_0} =\frac{\gamma}{\phiprev+\gamma}$. Thus, if it so happens that $\pi_{p_0} = \rho_1$ or $\pi_{p_0} = \rho_2$, $r$ and $a$ must be related so that in fact $\rho_1=\rho_2$. 

We must now check a number of cases: 

\begin{enumerate}
\item $\pi_0 \le  \rho_1$ and $\pi_{p_0} <\rho_2$: In this case, $\pi_p<\rho(p) \quad \forall p\in (0,1]$, and $P_{I,1}=0$ and $P_{I,2}=p_0$. But if $P=p_0$ and $Q\in[0,p_0)$, $\Delta E<0$ and so $P=p_0$ cannot be a Nash equilibrium. However, if $P=0$, then $\forall \epsilon\in [0,1) \quad \Delta E >0  \; \forall Q\ne P$, and so $P=0$ is a Nash equilibrium. This is trivial for $Q\le p_0$. For $p_0<Q$, we have $\pi_p-\rho(p)<0$ and $P-Q<0$, so $\Delta E>0$ as required. For convergent stability, we only need to check that if $0<P<p_0<Q\le1$, then for any $\epsilon\in[0,1)$ we have $\Delta E>0$. In this case, $p\in [P,Q]$, and $P-Q<0$. Furthermore, $\pi_p-\rho_1<0$ for any $p\le p_0$, and $\pi_p-\rho_2<0$ for any $p\ge p_0$. But since $\pi_{p_0}-\rho_2<0$, $\pi_{p_0} -\rho_1=0$ is impossible and from continuity,  $\pi_p-\rho_1<0$. Thus, $\Delta E (P, Q, \epsilon)>0$ as required for convergent stability. 

\item $\pi_0 \le  \rho_1$ and $\pi_{p_0}>\rho_2>\pi_1 = \alpha$: In this case, $P_{I,1} = 0$ and $P_{I,2}\in(p_0,1)$. Since $P_{I,1} = 0$ is a CSNE in $[0,p_0]$, we know that for any $\epsilon\in[0,1)$ and $0<P<Q\le p_0$, $\Delta E>0$. In particular, for $\epsilon = 0$, and $0<P<Q=p_0$ we get $[\pi_{p_0} + \psi_{p_0} r-r/a]=\frac{\Delta E}{(P-Q)}<0$. But, from $P_{I,2}\in(p_0,1)$ we can similarly get (for $\epsilon = 0$, $p_0=Q<P<P_{I,2}$) $[\pi_{p_0} + \psi_{p_0} r-r/a]a=\frac{\Delta E}{(P-Q)}>0$, a contradiction. 

\item $\pi_0 \le  \rho_1$ and $\alpha=\pi_1 \ge \rho_2>0$: Here, $P_{I,1} = 0$ and $P_{I,2}=1$.  Since $P_{I,1} = 0$ is a CSNE in $[0,p_0]$, $\forall \epsilon\in[0,1)$ and $0\le P<Q\le p_0$, $\Delta E>0$. In particular, for $\epsilon = 0$, and $0<P<Q=p_0$ we get $[\pi_{p_0} + \psi_{p_0} r-r/a]=\frac{\Delta E}{(P-Q)}<0$.
Similarly, since $P_{I,2} = 1$ is a CSNE in $[p_0,1]$, $\forall \epsilon\in[0,1)$ and $p_0\le Q<P\le1$, $\Delta E>0$. In particular, for $\epsilon = 0$, and $p_0=Q<P\le1$ we get $[\pi_{p_0} + \psi_{p_0} r-r/a]=\frac{\Delta E}{(P-Q)}>0$, which is a contradiction.

\item $\rho_2\ge\pi_{p_0} \ge  \rho_1$: In this case, simple algebra gives $\rho_1=\rho_2 =\pi_{p_0}$ and $P_{I,1} = p_0$ and $P_{I,2}=p_0$. Thus, it follows that $p_0$ is the unique CSNE in the interval $[0,1]$. 

\item $\pi_{p_0} >  \rho_1$ and $\alpha=\pi_1 \ge \rho_2$: In a manner analogous to the first case, here, $\pind = 1$ is the unique CSNE.

\item $\pi_{p_0} >  \rho_1$ and $\pi_{p_0}>\rho_2>\pi_1 = \alpha$ : , $P_{I,1} = p_0$ and $P_{I,2}\in(p_0,1)$. $P_{I,1}=p_0$ cannot be a Nash equilibrium since for $p_0<Q<P_{I,2}$, and any $\epsilon\in[0,1)$, $\Delta E <0$ since $P_{I,2}$ is the unique CSNE in $[p_0,1]$. To show that $P_{I,2}$ is a Nash equilibrium, fix $\epsilon\in[0,1)$ and $P=P_{I,2}$ and let $Q<P_{I,2}$. Note that $\pi_p>\rho_1 \; \forall p\in[0,p_0]$ and that  because $\pi_p$ is decreasing and $\pi_{P_{I,2}}=\rho_2$ , $\pi_p>\rho_2 \; \forall p<P_{I,2}$. Thus $\pi_p>\rho(p) \; \forall p<P_{I,2}$ and so in particular, $\Delta E >0$ and $P_{I,2}$ is the unique Nash equilibrium. To see that $P_{I,2}$ is also convergently stable, we must only show that for any $P$ and $Q$, $0\ge Q<p_0<P<P_{I,2}$ and $\epsilon\in[0,1)$, we have $\Delta E >0$. But under these conditions, $p<P_{I,2}$ and so again $\pi_p>\rho(p)$, which implies $\Delta E >0$, as required. 

\item $\pi_0>\rho_1>\pi_{p_0}$ and $\pi_{p_0} <\rho_2$: Now, $P_{I,1}\in(0,p_0)$ and $P_{I,2}=p_0$. Similar to the above case, this implies that $P_{I,1}$ is the unique CSNE (given by~\Fref{eq:p_tilde_prev}).

\item $\pi_0>\rho_1>\pi_{p_0}$ and $\alpha=\pi_1 \ge \rho_2$: Simple algebra shows that this case is impossible:
\begin{linenomath*}
\begin{align}
\rho_1&>\pi_{p_0} = \frac{\gamma}{\gamma+\phiprev} \nonumber \\
r(1-a)(\gamma+\phiprev)&>a(1-r)\gamma \nonumber \\
r((1-a)\phiprev +\gamma)&>a\gamma \nonumber
\end{align}
\end{linenomath*}
but
\begin{linenomath*}
\begin{align}
\rho_2&\le \pi_1<\pi_{p_0} = \frac{\gamma}{\gamma+\phiprev} \nonumber \\
r(\gamma+\phiprev)&<a(\phiprev r + \gamma) \nonumber \\
r((1-a)\phiprev +\gamma)&<a\gamma \nonumber \,.
\end{align}
\end{linenomath*}
\item $\pi_0>\rho_1>\pi_{p_0}$ and $\pi_{p_0}>\rho_2>\pi_1$: 
The reasoning applied to show that the case above is impossible also rules this case out.
\end{enumerate}
We conclude that in all cases there is a unique CSNE, which we denote by $\pind$.

\subsection{Vaccination rate $\propto$ incidence} \label{appx:ind_eq_inc}
Since $\partial_p \psi_p<0$ and $\partial_p \pi_p<0$, $ \partial_p \left(\pi_p + r\psi_p\right)<0$. Thus, an identical argument to the one given in~\Fref{appx:ind_eq_susc} allows us to show that there is always a CSNE for this model. In particular, there are three possibilities:
\begin{itemize}
\item If $\pi_0 + r\psi_0\le r/a$ then $\pind=0$ is a unique CSNE. 
\item If $\alpha = \pi_1 + r\psi_1 \ge r/a$, then $\pind=1$ is a unique CSNE. 
\item If $\pi_0 + r\psi_0 > r/a >\pi_1 + r\psi_1=\alpha$ then there is a unique CSNE, $\pind  \in(0,1)$ such that $\pi_{\pind}+ r\psi_{\pind}=r/a$. To simplify this last condition, we use \Erefs{eq:pi_psi_inc} to obtain 
\begin{linenomath*}
\begin{align}
0=\pi_{\pind} + \psi_{\pind} r-\frac{r}{a}
& = \frac{a\left(\beta + r (1-\alpha)\phiinc \right) -r(\beta+\phiinc)}{a(\beta+\phiinc)} - \frac{\beta+r\phiinc}{\beta+\phiinc}\frac{S_\infty}{1-{\pind}}\nonumber
\end{align}
\end{linenomath*}
which is equivalent to 
\begin{linenomath*}
\begin{equation}\label{eq:inc_Sinfty_at_pind}
\frac{S_\infty}{1-{\pind}} = \frac{a\left(\beta +\alpha \phiinc+  r (1-\alpha)\phiinc \right) -r(\beta+\phiinc)}{ a(\beta+r\phiinc)} \,.
\end{equation} 
\end{linenomath*}
Plugging~\Fref{eq:inc_Sinfty_at_pind} into~\Fref{eq:S_infty_incidence} and rearranging gives the individual equilibrium,
\begin{linenomath*}
\begin{align}
\pind &= 1+ \frac{a\gamma(\beta + r \phiinc)}{r(\beta + \phiinc) (\beta + a\alpha\phiinc)}\nonumber\\ & \times \ln\left(\frac{a\left(\beta +\alpha \phiinc+ r (1-\alpha)\phiinc \right) -r(\beta+\phiinc)}{a(1-\alpha)(\beta + r\phiinc)}\right)\,.
\end{align}
\end{linenomath*}
\end{itemize}

\subsection{Vaccination rate $\propto$ proportion still susceptible}\label{appx:ind_eq_susc}
Recall that for this model, $\psi_p = 1-\pi_p$ (see~\Fref{appx:finsize+:vac propto S}) and that we assume $\pi_p$ decreases with $p$ (as stated in~\Fref{sec:vac_propto_S}). Thus, $\partial_p \left(\pi_p + r\psi_p\right) = \partial_p \left((1-r)\pi_p\right)<0$, and we can infer that:
\begin{itemize}
\item If $\pi_0 + r\psi_0\le r/a$ then $\pi_p + r\psi_p< r/a \quad \forall p>0$. Hence $\pind=0$ is the unique Nash equilibrium. From reasoning similar to that given in~\Fref{appx:ind_eq_prev_S_infty_0}, $\pind$ is convergently stable. 
\item If $\alpha = \pi_1 + r\psi_1 \ge r/a$, then $\pi_p + r\psi_p>r/a \quad \forall p<1$.  It follows that $\pind=1$ is the unique Nash equilibrum and that it is convergently stable. The condition is equivalent to $a \alpha \rinf \ge \rvac$, which is quite intuitive: $a\alpha \rinf$ is a lower bound on the cost for a delayer (attained when there are no new cases after the initial outbreak). If even this lower bound is higher than the cost of vaccinating, then clearly everyone should vaccinate. 
\item If $\pi_0 + r\psi_0 > r/a >\pi_1 + r\psi_1=\alpha$ then there is a unique CSNE, $\pind  \in(0,1)$ such that $\pi_{\pind}+ r\psi_{\pind}=r/a$. 
\end{itemize}

Thus, there is always a CSNE for this model. 

\subsection{Instantaneous vaccination of a proportion $\phiinst$ of the population}\label{appx:ind_eq_inst}
In this case, $\psi_p = \phiinst (1-\alpha)$ and $\pi_p = 1-\psi_p - \frac{S_\infty}{1-p}$. Thus, 
\begin{linenomath*}
\begin{equation}
\Delta E = [\pi_p + \phiinst (1-\alpha) r-r/a]a(P-Q) \, .
\end{equation}
\end{linenomath*}
Letting $\rho = r/a - r\phiinst (1-\alpha)$, since $\pi_p$ decreases monotonically with $p$, we can use an argument similar to those used for the models considered above to show that:
\begin{itemize}
\item if $\pi_0\le \rho$ the $\pind=0$ is the unique CSNE.
\item if $\pi_1  \ge \rho$ then $\pi_p>\rho$ for any $p\in[0,1)$ and so $\pind = 1$ is the unique CSNE. Rearranging the condition $\pi_1 \ge \rho$ gives $a\alpha + ra\phiinst (1-\alpha) \ge r$. This admits a simple biological interpretation: $a\alpha + ra\phiinst (1-\alpha)$ is the relative risk of delaying when the epidemic does not successfully spread (that is, one can only be infected during the initial outbreak). Thus,  if the risks of delaying are greater than vaccinating even if the disease does not spread beyond the cohort initially infected in the outbreak, then it is worthwhile for individuals to vaccinate pre-emptively. 
\item if $\pi_0>\rho>0$ then there is a unique CSNE, $\pind \in (0,1)$ such that $\pi _{\pind }=\rho$. 
In order  to find $\pind$ explicitly, we use ~\Fref{eq:pi_p_instantaneous_implicit} and substitute $\pi_{\pind} =  r/a - r\phiinst (1-\alpha)$:
\begin{linenomath*}
\begin{equation}
 r/a - r\phiinst (1-\alpha)=-\frac{\gamma}{\beta(1-\pind)} \ln\left( \frac{1- r/a  -(1-r)\phiinst(1-\alpha)}{(1-\alpha)(1-\phiinst)}\right) \nonumber \,,
\end{equation}
\end{linenomath*}
and consequently, 
\begin{linenomath*}
\begin{equation}
\pind=1+\frac{a\gamma}{r\beta (1- a\phiinst (1-\alpha))} \ln\left( \frac{ a\left(1-(1-r)\phiinst(1-\alpha)\right)- r}{a(1-\alpha)(1-\phiinst)}\right)\,.
\end{equation}
\end{linenomath*}
\end{itemize}

\subsection{constant rate vaccination} \label{subseq: ind_eq_constant_rate}
As in \cite{Bauc+03}, we assume $\pi_p$ is a decreasing function of $p$. The analysis (performed originally in \cite{Bauc+03}) is then identical to~\Fref{appx:ind_eq_prev_S_infty_0}, implying the existence of a unique CSNE, which we denote by $\pind$.  Using the definition of $\rho_1$ given in \Fref{eq:oddsratio}, we have
\begin{itemize}
\item if $\pi_1\ge\rho_1$ then $\pind = 1$. 
\item if $\pi_1<\rho_1<\pi_0$ then $\pind$ is the unique solution of $\pi_{\pind}=\rho_1$
\item if $\rho_1\ge\pi_0$ then $\pind = 0$. 
\end{itemize}


\section{The group optimum} \label{appx:grpopt}

We have obtained an analytical formula for the group optimum (defined in \Fref{sec:grpopt}), for one sub-case of one of our models. The calculation is given below.

\subsection {Vaccination rate $\propto$ disease prevalence}
We consider the case when $p_m\ge1$. 
Recall that if $p_m\ge1$, then $S_\infty =0$, $\psi_p = \frac{\phiprev}{\beta(1-p)}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right) $, and $\pi_p = 1-\psi_p$ (see~\Fref{appx:finsize+:vac propto I}). Thus, 
\begin{linenomath*}
\begin{align}
C(p) & = r p +(1-p)a(1 - (1-r)\psi_p)\nonumber \\
& = r p +(1-p)a\left(1 - (1-r)\frac{\phiprev}{\beta(1-p)}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)\right) \nonumber \\
& = r p +a\left((1-p) - (1-r)\frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev} S(0)+1\right)\right) \nonumber \\
C(0) &= a(1-(1-r)\frac{\phiprev}{\beta}\ln\left(\frac{\beta}{\phiprev}(1-\alpha) + 1\right) \nonumber \\
C(1) &= r\nonumber \\
C'(p) &= r+a\left(-1-(1-r)\frac{\phiprev}{\beta}\left(-\frac{\beta}{\phiprev}(1-\alpha)\right)\frac{1}{\frac{\beta}{\phiprev} S(0)+1} \right)\nonumber\\
&= r+a\left(\frac{\phiprev(1-\alpha) (1-r)}{\beta S(0)+\phiprev} -1\right)
\end{align}
\end{linenomath*}
Note that $C'(p)$ increases with $p$ since $S(0)$ decreases with $p$ and critical points of $C(p)$ are minima. Thus, if there is a critical point within $[0,1]$, then it is the global minimum; otherwise, the global minimum is at $C(0)$. 
To find critical points, set $C'(p) =0$ or equivalently,
\begin{linenomath*}
\begin{equation}\label{eq:prev_pm_ge1_grp_cost_crit}
 \frac{r}{a}=1-\frac{\phiprev(1-\alpha) (1-r)}{\beta S(0)+\phiprev}
\end{equation} 
\end{linenomath*}
which can only happen if $r<a$ (since $\alpha<1$ and $r<1$), that is, the relative risk (of vaccination \emph{versus} infection) is less than the probability of an outbreak. If $r\ge a$ then $C'(p)\ge0$ throughout $[0,1]$ and the minimal group cost $C(p)$ is attained at $\pgrp=0$. 
This is easily explained: If $\rvac\ge a \rinf$ then the mortality risks from vaccination are no less than those of dying in an outbreak. In this case, vaccinating is not worthwhile for either the individual or the group. 
We now solve~\Fref{eq:prev_pm_ge1_grp_cost_crit} for the initial coverage $p$ at the critical point of the group cost, $C(p)$, assuming $r<a$: 
\begin{linenomath*}
\begin{align}
\beta S(0)&=\frac{\phiprev(1-\alpha) a(1-r)}{a-r} -\phiprev \nonumber \\
1-p&=\frac{\phiprev}{\beta(1-\alpha)}\left(\frac{(1-\alpha) a(1-r)}{a-r} -1\right) \nonumber\\
p&=1-\frac{\phiprev}{\beta(1-\alpha)}\left(\frac{(1-a)r -(1-r)\alpha a}{a-r}\right)\label{eq:prev_pgrp}
\end{align}
\end{linenomath*}

The critical point is attained at $p\ge1$ if and only if $(1-a)r\le(1-r)\alpha a$, which is equivalent to 
\begin{linenomath*}
\begin{equation} \label{eq:prev_grp_opt1}
\rvac\le a(\rinf\alpha + \rvac(1-\alpha))\,,
\end{equation}
\end{linenomath*}
and in this case the group optimum is vaccinating the entire population ($\pgrp=1$). Biologically,~\Fref{eq:prev_grp_opt1} means that more people are expected to die if, in case of an outbreak, all individuals not infected initially are vaccinated (discounted by the outbreak probability, $a$), than the number of people expected to die if the entire population is vaccinated pre-emptively. To see this, note the probability of death due to vaccinating is $\rvac$ (the left hand side of~\Fref{eq:prev_grp_opt1}).  To interpret the right hand side of~\Fref{eq:prev_grp_opt1}, consider an individual who is not vaccinated pre-emptively. If there is an outbreak (represented by the factor $a$), the first term in brackets ($\rinf\alpha$) represents the probability of being in the initially infected cohort ($\alpha$), and then dying due to the disease. The second term ($\rvac(1-\alpha)$), represents the probability of not being in the initially infected cohort, and dying due to the vaccine side effects. Note that because in this scenario $S_\infty=0$, no delayers remain susceptible (they are either infected or vaccinated). Thus, any individual who is not pre-emptively vaccinated, and who is not in the initially infected cohort ($1-\alpha$) has either a probability $\rvac$ of dying due to vaccine side effects, or a probability $\rinf$ of dying due to the disease. But since $\rvac<\rinf$, the term $\rvac(1-\alpha)$ is a lower bound on the probability of death for an individual who is susceptible immediately after the outbreak is seeded (that is, not pre-emptively vaccinated, or in the cohort initially infected at the beginning of the outbreak).

Note that not vaccinating anyone pre-emptively is the group optimum ($\pgrp\le0$) \textiff 
\begin{linenomath*}
\begin{equation}
\beta(1-\alpha)(a-r)\le\phiprev\left((1-a)r -(1-r)\alpha a\right)\,,
\end{equation}
\end{linenomath*}
(but this condition is difficult to interpret biologically). 

Lastly, if  $\phiprev\le\beta\frac{(a-r)(1-\alpha)}{r(1-a(1-\alpha))-a\alpha}$ and  $(1-a)r>(1-r)\alpha a$, then $\pgrp\in[0,1)$ and is given by~\Fref{eq:prev_pgrp}.
It is thus interesting to note that $\pgrp$ depends piece-wise linearly on $\phiprev$. 

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