 SIR final removed fraction

For the closed SIR model with $\gamma >0$, the infectious compartment's fraction approaches $i_{\infty }=0$. One proof outline uses the bounded, increasing removed compartment's fraction: since $r(t)-r_0=\gamma {\int }_0^{t}i(u)du$, the total time-integrated $i$ is finite. The bounded derivative $di/dt$ from the SIR model equations prevents indefinitely recurring spikes, so $i(t)$ tends to zero.

For $s_0>0$, the final size relation for an epidemic follows by dividing the SIR model equations for $s$ and $r$ while $i>0$, giving $ds/dr=-(\beta /\gamma )s$. Integration then yieldsThis also follows from the SIR first integral and conservation of population. For the stipulated initial value problem, solveThe bisection method givesAbout $5.88\%$ of the original population remains in the susceptible compartment. The $94.12\%$ is the final removed compartment's fraction, including the initially $1\%$ in the infectious compartment. The newly acquired infection during the modelled period accounts for $s_0-s_{\infty }\approx 0.93120263520$ of the total population, not $1-s_{\infty }$. Dividing by $s_0$ instead gives the proportion of initially susceptible compartment members who acquire infection; its denominator must be stated.

At the SIR infection peak, the flow into the infectious compartment only balances its outflow; it does not stop. Further infection explains why the final susceptible compartment's fraction is well below $1/3$. Also $s_{\infty }=s_0{e}^{-R_0(r_{\infty }-r_0)}>0$ for $s_0>0$ and finite $R_0$: the closed SIR model does not require everybody to acquire infection before its infectious compartment empties in the long-time limit.