= Solution
At the retained order define $K=1+\tau r(I-S)$. Neglecting terms of order $a\tau$ in the incidence correction gives
$$
\dot I=rISK-aI,\qquad \dot S=-rISK.
$$
Dividing and expanding consistently,
$$
\frac{dI}{dS}=-1+\frac{a}{rSK}
=-1+\frac{a}{rS}+O(a\tau,\tau^2).
$$
Thus the <susceptible> count $S_*$ at the first threshold crossing obeys the same retained-order equation as before:
$$
I_*+S_*-\frac ar\log S_*=I_0+S_0-\frac ar\log S_0.
$$
It is important to distinguish completed case transfers at <quarantine> from all <infections> already initiated. Let the <quarantine> time be $T$. Isolation prevents exposures after $T$, but animals exposed during $(T-\tau,T)$ still finish their <incubation period> during $(T,T+\tau)$. Their number, in the printed bookkeeping, is
$$
E_* =\int_{T-\tau}^{T}rI(s)S(s)\,ds
=\tau rI_*S_*+O(\tau^2).
$$
The <infection cases> transferred by time $T$ among the initial $S$ class number $S_0-S_*$. Adding those pending <infection cases> gives the eventual number who catch the <infectious disease>:
$$
\boxed{S_0-S_*+\tau rI_*S_*\quad\text{to the retained order}.}
$$
This adds only <infections> caused by exposures before isolation; it does not allow secondary transmission after <quarantine>. If one asks instead only for the transfers completed by the instant of <quarantine>, the answer is $S_0-S_*$. The distinction is necessary because the model keeps incubating animals in its $S$ <compartment>.
Back to article page
