\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 125, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/125\hfil Optimal control of an epidemic]
{Optimal control of an epidemic through educational campaigns}

\author[C. Castilho\hfil EJDE-2006/125\hfilneg]
{C\'esar Castilho}  % in alphabetical order

\address{Departamento de Matem\'atica \\
Universidade Federal de Pernambuco\\
Recife, PE CEP 50740-540 Brazil \newline
and The Abdus Salam ICTP \\
Strada Costiera 11 Trieste 34100 Italy}
\email{castilho@dmat.ufpe.br}

\date{}
\thanks{Submitted September 21, 2005. Published October 11, 2006.}
\subjclass[2000]{92D30, 93C15, 34H05}
\keywords{Epidemic; optimal control; educational campaign}

\begin{abstract}
 In this work we study the best strategy for educational campaigns
 during the outbreak of an epidemic. Assuming that the epidemic is
 described by the simplified SIR model and that the total time of
 the campaign is limited due to budget, we consider two possible
 scenarios. In the first scenario we have a campaign oriented to
 decrease the infection rate by stimulating susceptibles to have a
 protective behavior. In the second scenario we have a campaign
 oriented to increase the removal rate by stimulating the infected
 to remove themselves from the infected class. The optimality is
 taken to be to minimize the total number of infected by the end of
 the epidemic outbreak. The technical tool used to determine the
 optimal strategy is the Pontryagin Maximum Principle.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}


\section{Introduction}

 In this work we study the best strategy for educational campaigns
 during the outbreak of an epidemic. We assume that the epidemic is
described by the simplified SIR model \cite{thi} and also assume
that the total time of the campaign is budget limited. Optimality is
measured minimizing the total number of infected at the end of the
optimal outbreak. If we cannot make a campaign during all the
epidemic time, what is the optimal way of using the time we have?
How many campaigns should we make? What should be their intensities?
When should they start? The difficult point is, of course, how to
model the effect of the campaign on the spread of the epidemic. Here
we face two problems: first, the model must be intuitively plausible
and second, it must be mathematically tractable.

 With respect to the first requirement
we will model the campaign effects by reducing the rate at which
the disease is contracted from an average individual \emph{during}
the campaign (called shortly infection rate). We justify this with
an example: suppose during a flu outbreak one starts a campaign
orienting susceptibles to avoid contracting the virus (assuming
some protective behavior, e.g., washing hands, avoiding close
environments, etc.). The effect of the campaign will be that the
probability of a susceptible contracting the virus will decrease.
 The same reasoning applied to a campaign oriented to
the infected (e.g. stimulating quarantine) will be modelled
increasing the rate at which an average individual leaves the
infective rate (called shortly removal rate). With respect to the
second requirement we assume, for mathematical simplicity, that
this reduction (increase) is bounded below (above) and the
campaigns cost are linear on the controls. With those hypotheses
the problem renders itself to analytical treatment and we can
prove the main facts about the optimal campaign. The theorems of
section \ref{C2} reduce the dimension of the optimal problem
allowing a complete numerical study of the problem.

Application of control theory to epidemics is a very large field.
A comprehensive survey of control theory applied to epidemiology
was performed by Wickwire \cite{wick1}. Many different models with
different objective functions have been proposed (see
\cite{gupta,hethe,morton} and more recently
\cite{horst,zaric}). A major difficulty in applying control
theoretic methods to practical epidemiology problems is the
commonly made assumption that one has total knowledge of the state
of the epidemics \cite{dietz}.

\section{Statement of the problem}
\label{statement}

 We denote by $S(t)$, $I(t)$, $R(t)$ the number of susceptible,
infectives and removed in a closed population of size $N$ at time
$t$. We assume the controlled dynamics
\begin{equation} \label{model}
\begin{gathered}
 \dot S=- u_1 S I \, ,\\
 \dot I=u_1 S I - u_2 I \,,\\
 \dot R =u_2 I,
\end{gathered}
\end{equation}
The above models assume a mass-action type interaction ( for more
realistic interactions see \cite{chavez}). We let positive
constants $\beta$ and $\gamma$ denote the infection and removal
rates respectively without the influence of an education campaign.
 Our controls are
$u_1(t) , u_2(t) $ with $u_1(t) \in [ \beta_m , \beta] $ and
$u_2(t) \in [ \gamma , \gamma_M]$ with
$0 <\beta_m$. Observe that $u_1(t)$ and $u_2(t)$ regulate the
goals and efforts of two types of campaigns. For example, if
$u_2(t)=\gamma$ for all $t$ we are controlling only the infection
rate. In this case $u_1(t)=\beta$ will correspond to not having a
campaign affecting the susceptibles and $u_1(t)=\beta_m$ will
correspond to the maximum effort that can be made. The reciprocal
case will be if $u_1(t)=\beta$ for all $t$. In this scenario we
will be controlling only the removal rate $\gamma$. The above
considerations motivate the introduction of the following cost
constraints.
\begin{gather} \label{custo1}
J_1=\int_0^{t^*} [ ( \beta -u_1(t)) + (u_2(t) - \gamma )] I(t) \,dt \,,\\
\label{custo2}
J_2=\int_0^{t^*} ( \beta -u_1(t)) + (u_2(t) - \gamma )\, dt \,,
\end{gather}

In both cases the cost is linear in the controls $u_1$ and $u_2$. In
the first case the cost of the campaign is supposed to be
proportional to the number of infected (if one assumes that the
number of infected is proportional to the number of regions where
the disease occurs and therefore, to the number of regions to be
covered by the campaign, higher the number of infected, higher the
costs) . The second case assumes that the cost is independent of the
number of infected.


 Our goal will be to find the optimal control strategies that
minimize the total number of infected over the course of the
epidemic outbreak (equivalently, that maximize the total number of
susceptibles). In this work, the end of the epidemic outbreak will
be defined as a (very large) time instant $t^*$ for which
$I(t^*) < 1 $ (see remark (\ref{remm}) about the existence of $t^*$).
In other words, $t^*$ is the first time such that $ I(t^*) < 1$.
This is a technicality in order to avoid dealing with a infinite
horizon control problem. Since in the simplified SIR model the
only way to enter in the removed class is from the infected class,
the total number of infected at the end of the epidemics is given
by $  \lim_{t \to \infty} R(t)$. However,
$$
\dot R= \gamma I \,,
$$
and since we always assume $R (0) = 0 $, we obtain that the
total number of infected is given by
$$
\lim_{t \to \infty} R(t) = \int_0^{\infty} \gamma I(t) \,dt \,.
$$

\begin{remark} \label{remm} \rm
We make some remarks that are important for what follows.
\begin{enumerate}
\item Since $S(t)+I(t)+R(t)=N$ we will ignore the last equation of
(\ref{model}).
\item  The set $M= \{ I \ge 0 , S \ge 0, S + I \le N \} $ is an
invariant set for system
(\ref{model}).
\item In the simplified SIR model we always have that
 $  \lim_{t\to \infty} I(t) = 0 $
 (see e.g. \cite{dal}).
\end{enumerate}
 Since we are working only on $M$
and the controls $u_1$ and $u_2$ are bounded and positive, we will
always have that $  \lim_{t \to \infty} I(t) =
0 $ for any control. This establishes the existence of $t^*$.
\end{remark}

 The constant cost constraints $J_1$ and $J_2$ can be imposed
introducing a new variable $w$ to our system. We obtain the two
control systems
\begin{equation}
\label{modelb2}
\begin{gathered}
 \dot S=-u_1 S I ,\\
 \dot I=u_1 S I - u_2 I ,\quad  Y_1=\int_0^{t^*} I(t) \,dt\\
\dot w = \Big( \big( \beta -u_1(t) \big) + \big(u_2(t) -
\gamma \big)\Big) I(t) ,\quad  w(0)=0, \quad w(t^*)=C.
\end{gathered}
\end{equation}
with cost $J_1$, and the system
\begin{equation} \label{modelb}
 \begin{gathered}
 \dot S=-u_1 S I \,,\\
 \dot I=u_1 S I - u_2 I , \quad Y_2=\int_0^{t^*} u_2(t) I(t) \,dt\\
\dot w = \big( \beta -u_1(t) \big) + \big(u_2(t) -
\gamma \big),\quad  w(0)=0, \quad w(t^*)=C.
\end{gathered}
\end{equation}
with cost $J_2$. In both systems we are imposing $J_1=J_2=C$, where
$C$ is a constant.

\begin{remark} \label{custo}\rm
The constant  $C$ is the value of the total amount of campaign effort.
We will assume henceforth that $C$ is such that the controls can not
 be at the maximum effort level during the whole time period.
\end{remark}


 The problems will be referred as problem C1
and C2 respectively. The goal is to find the optimal controls to
(\ref{modelb2}) that minimize $Y_1$ and the optimal controls to
\eqref{modelb} that minimize $Y_2$. We will refer to the first
problem as problem C1 and to the second problem as problem $C2$.
As it will turn out, problem C1 is trivial. We will assume that
 the admissible controls $u_1$ and $u_2$ are measurable locally bounded
 functions.
 Since $u_1$ and $u_2$ appear linearly in our control problems, an
optimal control will in general be a combination of bang-bang
controls and singular controls (see \cite{pontri,ursula}).


\section{Optimality Problem C1}

Problem C1 is such that all the differential equations involved
are multiplied by the positive function $I(t)$. This motivates the
introduction of a new parameter $s$ defined by
$$
s(t) = \int_0^t I(t) dt .
$$
Observing that $   \frac{d}{dt}=I \frac{d}{ds} $ we
obtain for the objective functional that
$$
Y_1 = \int_0^{t^*} I dt = \int_0^{s^*} ds = s^* ,
$$
where $s^* = \int_0^{t^*} I(t) dt $. Therefore, problem C1
write as the minimum time problem
\begin{equation} \label{mintime}
 \begin{gathered}
 S' =-u_1 S ,\\
 I' =u_1 S - u_2 \\
 w' = ( \beta -u_1 ) + (u_2 - \gamma ) ,\quad  w(0)=0, w(s^*)=C ,
\end{gathered}
 \end{equation}
where $' = \frac{d}{ds}$, and $s^* = \int_0^{t^*} I(t) dt $.
 Now we see that in order for the variable $w(s)$ achieve
the value $C$ in the smallest possible time $s$, it suffices that
the derivative $w'$ be the largest possible, therefore it
suffices that $u_1(s)=\beta_m$ and $u_2(s)=\gamma_M$.

\section{Optimality Problem C2} \label{C2}

 Our main tool for the study of the optimality of system
\eqref{modelb} will be the Pontryagin Maximum Principle (PMP)
 \cite{agrachev,pontri}. Let $p_S$, $p_I$ and $p_w$ denote
the adjoint variables to $S$, $I$, and $w$ respectively. The
Hamiltonian for problem $C2$ is
$$
H = p_S ( -u_1 S I ) + p_I ( u_1 S I - u_2 I ) + p_w
[ ( \beta -u_1(t) ) + (u_2(t) - \gamma )]- u_2(t) I .
$$
That we write as
\begin{equation}
\label{pre} H = g + u_1 \phi_1 + u_2 \phi_2 ,
\end{equation}
where
$$
g \equiv p_w ( \beta - \gamma ) , \phi_1 \equiv S I ( p_I - p_S ) - p_w
 , \phi_2 \equiv -I ( p_I + 1 )+p_w.
$$
 The adjoint variables satisfy Hamilton's equations
\begin{equation}
\label{adj} \dot p_S = -\frac{\partial \mathcal{H}}{\partial S}, \quad
\dot p_I = -\frac{\partial \mathcal{H}}{\partial I}, \quad
\dot p_w = -\frac{\partial \mathcal{H}}{\partial w} ,
\end{equation}
that are given by
\begin{equation}\label{adj1}
 \begin{gathered}
 \dot p_S= u_1 I (p_s - p_I ) ,\\
 \dot p_I= u_1 S (p_s - p_I ) + u_2 (p_I +1) ,\\
 \dot p_w =0 .
\end{gathered}
\end{equation}
By the PMP, the optimal controls $u_1(t)$, $u_2(t)$ are the
ones that maximize $\mathcal{H}$ (we are ignoring abnormal controls see
\cite{agrachev}). PMP implies that at optimal trajectories the
following transversality conditions will hold \cite{pontri}
\begin{equation} \label{trans}
p_S(0)=p_I(0)=0 \quad\mbox{and} \quad p_S(t^*)=p_I(t^*)=0.
\end{equation}
This is implied by the boundary
conditions to be satisfied by $w$. The derivatives of the
functions $\phi_1$ and $\phi_2$ along the flow of hamiltonian
dynamical system induced by (\ref{pre}) can be computed using
\eqref{modelb} and (\ref{adj1}). We obtain
\begin{gather} \label{Dphi1}
\dot \phi_1 = u_2 I S ( p_S +1) ,\\
\label{Dphi2}
\dot \phi_2 = -u_1 I S ( p_S +1) .
\end{gather}
 From where it follows that
$$
 u_1 \dot \phi_1 + u_2 \dot \phi_2 =0\, .
$$

\begin{remark} \rm
The existence of the optimal control for problem $C_2$ is given by
an application of Filipov's theorem \cite{agrachev,soueres}:
We observe that the vector field $X$ defined by \eqref{modelb} is
bounded in $M$ and complete ($M$ is compact). Also the controls are
bounded and for each fixed allowed pair $(u_1,u_2)$ the set
$$
\bar X(u_1,u_2)=\Big\{S I  \begin{pmatrix} u_1 \\ -u_1 \\ 0\end{pmatrix}
+ I  \begin{pmatrix} 0 \\ u_2 \\ 0\end{pmatrix}
 , \mbox{ for } S,I \in M \Big\}
$$
is convex, which
 implies that the set $X(u_1,u_2)=\{ X \mbox{for} S , I \in M \}$ is
 convex. To apply directly Filipov's theorem it remains to establish the
 compact support of the vector fields. But this is not necessary by
the boundness and completeness of the vector fields (see
discussion in \cite{agrachev} and \cite{cesari}).
\end{remark}

\subsection{Controlling the infection parameter}
In this subsection we  will control only the infection parameter;
i.e., we will assume $u_2(t)=\gamma$ for all
$ t \ge 0$. The pre-hamiltonian (\ref{pre}) is given by
\begin{equation}
\label{cinf}
 H =\beta p_w - \gamma I ( p_I + 1 ) + u_1(t) \phi_1 .
\end{equation}
We observe that the derivative of the switching function
 $\dot \phi_1=\gamma S I  ( p_S + 1)$ is a continuous function
and its number of zeros is determined only by the behavior of $p_S$ since
$-\gamma I S \ne 0$.

\begin{lemma} \label{lema1}
If $u_2(t)=\gamma$ in the control problem
\eqref{modelb} then there is no open interval where
$\phi_1(t)=\dot \phi_1(t)=0 $.
\end{lemma}

\begin{proof}
Assume there exists an open interval $\mathcal{D}$, where
$\phi_1(t) =\dot \phi_1(t)=0 $ for $t \in \mathcal{D}$. The
derivative of $\phi_1$ being zero implies that
 $p_S=-1$ in $\mathcal{D}$ what implies that $\dot p_S=0$
 and by the first equation of (\ref{adj1}) we have that
 $p_I=p_S=-1$ in $\mathcal{D}$; but equations (\ref{adj1})
 imply that $p_I=p_S=-1$ for all future $t$ ($p_I=p_S=-1$ is
 an equilibrium point for the vector field (\ref{adj1}))
 what contradicts the transversality condition (\ref{trans}).
\end{proof}

\begin{theorem} \label{teori}
If $u_2(t)=\gamma$ in the control problem
\eqref{modelb}, the optimal control $u^*_1(t)$ has at most two
switches.
\end{theorem}

\begin{proof}
First we observe that when $\phi_1 =0$ we have by \eqref{cinf}
that
$$
H - \beta p_w = -\gamma I (p_I + 1).
$$
For latter use we multiply this equation by
$-\frac{S}{\gamma}$ obtaining the equality
\begin{equation} \label{auxx1}
S I (p_I + 1 ) = -\frac{S}{\gamma} ( H - \beta p_w ) .
\end{equation}
When $\phi_1 =0$ we have that $S I ( p_I - p_S)=p_w$.
Solving for $p_S$ and substituting back in
$\dot \phi_1 $, we obtain
$$
\dot \phi_1 = \gamma S I (p_S + 1 )
=\gamma ( S I (p_I + 1 ) - p_w) .
$$
Using (\ref{auxx1}), we obtain that at the zeros of $\phi_1$,
\begin{equation}
\label{auxx2}
 \dot \phi_1 = (\beta p_w - H ) S - p_w .
\end{equation}
 From equation (\ref{auxx2}) we define the function
$$
h =(\beta p_w - H ) S - p_w .
$$
By the first equation of (\ref{model}) we see that $S$ is a strictly
monotone function. Therefore since $p_w$ and $H$ are constant along
the flow we have that $h$ is a monotonic function. (\ref{auxx2})
shows that at the zeros of $\phi_1$, $\dot \phi_1 =h$. Therefore, we
have that at the zeros of the $C^1$ function $\phi_1$, the values of
its derivative $\dot \phi_1 $ is a monotonic function. Therefore
$\dot \phi_1$ can switch signs at most one time. What implies that
$\phi_1$ can have at most two switches of sign (and at most three
zeros).
\end{proof}

\subsection{Controlling the removal parameter}
In this section we will assume that $u_1=\beta$ for all times. The
pre-hamiltonian is
\begin{equation}
\label{cinf2}
 H = -p_w \gamma + \beta S I ( p_I - p_S ) + u_2 \phi_2 .
\end{equation}
We observe that $\dot \phi_2=-\beta S I(p_S + 1 )$ is a continuous function.

\begin{lemma} \label{lema2}
If $u_1(t)=\gamma$ in the control problem
\eqref{modelb} then there is no singular optimal control $u_2(t)$.
\end{lemma}

The proof of the above lemma is similar to the proof of lemma
(\ref{lema1}). Therefore, it is omitted.

\begin{theorem}
If $u_1(t)=\beta$ in the control problem \eqref{modelb},
the optimal control $u_2(t)$ has at most two switches.
\end{theorem}

\begin{proof}
 When $\phi_2=0$ we have that $  p_I - 1 = p_w/I$.
 Since at the zeros of $\phi_2$ we have
 $$
H + \gamma p_w =\beta S I ( p_I - p_S )
$$
it follows that
\begin{equation}
\label{teori2} \dot \phi_2 = H + p_w (\gamma - \beta S) .
\end{equation}
The argument here is the same as the in proof of theorem
(\ref{teori}). The left hand side of (\ref{teori2}) is a monotonic
function. Therefore we have that at the zeros of the $C^1$
function $\phi_2$, $\dot \phi_2 $ can switch signs at most one
time. Therefore $\phi_2$ can have at most two switches of sign
(and at most three zeros).
\end{proof}

\subsection{Controlling the infection and the removal parameters}

In this case we are working in a more complex case. We recall that
$\dot \phi_1= u_2 I S ( p_S +1 ) $ and $\dot
\phi_2= - u_21 I S ( p_S +1 ) $. The
functions $\dot \phi_1$ and $\dot \phi_2$ depend on the controls and
are not necessarily continuous (we are assuming that $u_1(t)$ and
$u_2(t)$ are measurable locally bounded functions). Therefore
$\phi_1$ and $\phi_2$ are not $C^1$ functions and the previous
reasoning does not apply in this case.

\begin{theorem}
Along the optimal solution there is no time instant $\bar t$ for
which $\phi_1(\bar t)=\phi_2(\bar t)=0$.
\end{theorem}

\begin{proof}
At $\bar t$ we would have that $p_S=p_I=-1$ what contradicts the
boundary conditions for $w$ at $t=t^*$. \end{proof}

A corollary of this fact is that $H- g \ne 0$. As in lemma \ref{lema1},
 we can prove the following result.

\begin{theorem}
There is no time interval for which
$\phi_1(t)=\dot \phi_1(t)=0$ and for which $\phi_2(t) =\dot
\phi_2(t)=0$.
\end{theorem}

\begin{theorem}
The two types of campaign, that is, the campaign for reducing
$\beta$ and the campaign for increasing $\gamma$ are either time
disjoint or time nested.
\end{theorem}

The theorem says, for example, that if you start a reducing
infection rate campaign (RIRC), when there is no campaign being
made, then there are only two possibilities: either you start and
finish a increasing removal rate campaign (IRRC) before you finish
the RIRC of you wait until the RIRC is over to start the IRRC.

\begin{proof} We recall that
$$
H = g + u_1 \phi_1 + u_2 \phi_2 .
$$
Since $g$ is constant and $H$ is a first integral for the control system it
follows that the two functions $f_1 \equiv u_1 \phi_1$ and $f_2
\equiv u_2 \phi_2$ add to a constant. The proof is a direct
consequence of this fact. A campaign will start or end at a switch
time, i.e. at a time where some of the functions $\phi_1$ or
$\phi_2$ changes sign. Now let $\alpha \equiv H - g $. Therefore
if $f_1(t_1)$ is zero we have that $f_2 = \alpha$ and vice-versa.
Assume, by way of contradiction, that campaigns are neither disjoint
neither nested. We have two cases to consider a) The number of total
switches is two or b) The number of total switches is greater than
two. (the case of only one switch satisfies the theorem). If we are
in case a) the only situation that does not satisfy the theorem is
the one where each function has one switch and one of the campaigns
(say campaign 2) starts when the other campaign (say campaign 1) is
still on. In this
 case, since there is only one switch left, it follows
that only one of the two will be turned off. As a net result we
will have at least one campaign being made during all epidemic
time what is ruled out by the main hypothesis of the paper: one
can not make campaign for all times (see remark \ref{custo}). In
case b) We have at least three zeros. Now assume, by way of
contradiction, that there are two campaigns that are neither
disjoint or nested. Then there is at least one switching time
$\bar t$ for say $f_2$ that is inside the $f_1$ campaign interval
$I=[t_1,t_2 ]$. Assume without lost of generality that
$\bar t$ is a start and that there is no other switch of $f_2$ in
the interval $\bar I=[\bar t , t_2]$ (intersection
hypothesis) Now at $t_1$ we have $f_1(t_1)=0$ and
$f_2(t_1)=\alpha$. At $t_2$ we also have that $f_1(t_2)=0$ and
$f_2(t_2)=\alpha$. But this impossible, since $f_2$ switches signs
at $\bar t$ and does not switch signs in the interval $\bar I$.
\end{proof}

\section{Controlling an Epidemic} \label{opt}

 In this section we study an example numerically. We will be
controlling only the infection parameter. We assume that the
campaign cost is independent of the number of infected, i.e. We
will be considering the problem $C2$. Since the optimal campaign
has at most two switches it will consist of only one campaign with
maximal effort. Therefore, to determine the optimal campaign, one
must only to determine the time instant when it starts. We call it
the optimal start. The strategy to determine the optimal control
numerically is as follows: For a fixed campaign time $C$ we fix
the susceptible and infective initial values. A grid of $N$
starting campaign times $t_i$, $i=1,..N$ is then specified. The
equations for $S(t)$ and $I(t)$ (the adjoints are not used) are
then integrated $N$ times, one for each campaign starting time
$t_i$. The total number of infected $T_i$ by the end of the
epidemic outbreak is them computed. The optimal start is the $t_i$
that results in the smaller of all $T_i$.

Our goal is to understand how the optimal start depends on the
campaign total time $C$ (we will present only the results for
reducing $\beta$ since the results for increasing $\gamma$ are
equal in nature). The model case is a severe flu epidemic
described in the 4th March 1978 issue of the British Medical
Journal. The parameters for the epidemic were determined by a best
fit numerical technique in \cite{mur}. The values for the
influenza epidemic are $N=763$, $S(0)=762$, $I(0)=1$, $\gamma=2.18
\times 10^{-3}$ and $\beta=0.44036$. Time is measured in days. We
plot the epidemic dynamics in figure 1. The maximum number of
infected occurs at $t=6.49$. This instant is called (according
Bailey \cite{bai}) the central epoch.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.6\textwidth]{fig1}
\end{center}
%\centerline{\includegraphics[width=90mm,height=90mm,angle=270]{fig12.ps}}
\caption{Epidemic dynamics: The number of infected $I$ and the
number of susceptibles $S$. Time is measured in days.}
\end{figure}

 We used a Runge-Kutta Fehlberg 7-8 to integrate the system of equations
with tolerance $10^{-8}$ and step size $h=0.01$. We will take
$\beta_m \equiv 1.08 \times 10 ^{-3}$ what gives a reduction of
$50 \%$of the infection rate. The results obtained are valid for
all ranges of reduction studied. The optimal start can be
determined numerically by a simple search procedure. We partition
the time interval in intervals of length $0.1$. Then we do the
campaign (reducing the infection parameter by $50\% $ ) during
time $C$ for all starting times. In figure 2 we show the number of
infected at the end of the epidemic as a function of the starting
time. Each curve represents different campaign times.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth, height=0.7\textwidth]{fig2}
\end{center}
%\centerline{\includegraphics[width=90mm,height=90mm]{fig2.eps}}
\caption{Number of total infected at the end of epidemics as a
function of the campaign starting time. Different curves represent
different campaigns times $C$.}
\end{figure}

 In figure 3 we show the optimal starting time as a function of the
campaign time. We observe that as the campaign time increases the
starting time decreases until eventually becomes zero.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth, height=0.5\textwidth]{fig3}
\end{center}
%\centerline{\includegraphics[width=90mm,height=90mm]{fig3.eps}}
\caption{Optimal starting time for different values of campaign
values $C$.}
\end{figure}

Figure 4 shows that the optimal campaigns always include the
central epoch. In other words, limited cost campaigns are optimal
around the central epoch for non-controlled epidemics. In the
figure we show in the horizontal axis the campaign duration. The
two solid curves represent the time when the campaign starts
(lower) and the time when the campaign finishes (upper). The
dashed curve shows the central epoch. It is always inside the
campaign duration even for very small times.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth, height=0.5\textwidth]{fig4}
\end{center}
%\centerline{\includegraphics[width=90mm,height=60mm]{fig4.eps}}
\caption{Relative position of the central epoch (dashed line) with
respect to the optimal campaign interval.}
\end{figure}

\subsection*{Conclusions}

In this paper we studied optimal strategies for a limited cost
educational campaign during the outbreak of an epidemic.
Optimality was measured by the minimality of the total number of
infected at the end of the outbreak. Assuming that the effect of
the campaign was to decrease (or increase) infection (removal)
rate we were able to show, using the Pontryagin Maximum Principle,
that the optimal campaign must consist of only one maximum
effort. Numerical simulations, concerning a particular epidemic,
gave us additional information about the optimal start, i.e. the
time to start this maximum effort, in order to minimize our
objective functional. Calling $\tilde t$ the central epoch we
summarize our results in the following: If the campaign cost is
proportional to the number of infected than both campaigns, to
decrease infection rate and to increase removal rate must be done
with maximum intensity at the start of the epidemic. If the
campaign cost is independent of the number of infected and only
one scenario is chosen, then 1) only one maximum effort campaign
should be made, 2) all campaigns should include $\tilde t$. If the
goals of the campaign is both to decrease infection rate
\emph{and} to increase removal rate then campaign for
different scenarios must be nested or disjoint. They should never
start or end at the same time.


\subsection*{Acknowledgments}
It is a great pleasure to thank A. Agrachev for help during the
 preparation of this work. Also I would like to thank
P. D. N. Siriniwasu for many discussions
and S. Lenhart for useful remarks.

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\end{document}