The SIR model for epidemics is Susceptible (not yet infected) -> Infected (have disease and are infectious) -> Recovered (recovered from the disease)
and is governed by the system of 3 differential equations$$\begin{split}\frac {dS}{dt}&=-\frac{\beta IS}{N}\\
\frac{dR}{dt}&=\gamma I\\
\frac{dI}{dt}&=\frac{\beta IS}{N}-\gamma I\end{split}$$Determine the final epidemic size, i.e. the total number of people infected
when $S(0)=N$ and there is an infinitesimal initial infection. Express
the fraction infected ($\frac{R(\infty)}{N}$) as a function of $R_0$. Note: it will be an implicit function.
$\beta, \gamma$ are parameters and can be considered constants. It has been a while since I took differential equations – how would one solve for $R(\infty)$? We seek $R(\infty)$ not $I(\infty)$ here because $R$ tells you everyone who has recovered and hence went through the disease phase, but $I$ would just be the current infected.
Best Answer
Note that $S'+I'+R' = 0$, so the equations are linearly dependent, i.e. we have more equations than we need. We limit our study to just the $S'$ and $R'$ equations, eliminating $I$ algbebraically to get
$$S'(t) = -\frac{R_0R'(t)}{N}S(t)$$
where $R_0 = \beta/\gamma$, which has solution
$$ S(t) = S(0)\exp\left(-\frac{R_0(R(t)-R(0))}{N}\right) $$
Taking the limit as $t\to\infty$, while noting that $I\to 0$ and so $S\to N-R$, we get
$$ N - R(\infty) = S(0)\exp\left(-\frac{R_0(R(\infty)-R(0))}{N}\right) $$
Apply the initial conditions $S(0) = N - \varepsilon$ and $R(0) = 0$. Since $\varepsilon$ can be arbitrarily small, we can take the limit as $\varepsilon\to 0$ so it vanishes from the solution
$$N - R(\infty) = N\exp\left(-\frac{R_0R(\infty)}{N}\right)$$
Let's clean it up by naming the quantity of interest $\rho = R(\infty)/N$, and rearranging the equation a bit to get
$$\rho = 1 - \exp(-R_0\rho)$$
Note that this is not explicit, as expected.