# 15.7: 15.7 The rsN model again

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Let’s return to shorter time frames without evolution. Notice that prevalence \(p\) changes with time as the epidemic spreads, but \(\beta\), \(\alpha\), and \(v\) remain constant. So starting with

\[\frac{1}{p}\frac{dp}{dt}\,=\beta(1-v-p)\,-\alpha\]

rearranging terms,

\[\frac{1}{p}\frac{dp}{dt}\,=(\beta(1-v)-\alpha)-\beta\,p\]

and substituting \(r\,=\,\beta(1\,-\,v)\,−\,\alpha,\,s\,=\,−\beta\,,\,and\,N\,=\,p\),

\[\frac{1}{N}\,\frac{dN}{dt}\,=\,r\,+\,sN\]

Voilà, the epidemiological *I* model is revealed to be just the standard model of ecology in another disguise! But now, with mechanisms included (infectivity, virulence, vaccination), deeper conclusions can be reached.

In getting to this standard model, for example, we set s equal to −\(\beta\). Because \(\beta\) is positive, the density dependence term \(s\) is negative, Negative \(s\) implies logistic growth (positive \(s\) is orthologistic), meaning that \(N\) will reach a carrying capacity—an equilibrium, a steady state. But to arrive at the standard model, we set \(N\) equal to \(p\), so since \(N\) reaches an equilibrium in the standard model, the prevalence \(p\) will reach an equilibrium in the \(I\) model.

Without further analysis you can therefore conclude that a disease will not necessarily infect an entire population, but that its prevalence will level out when it reaches a carrying capacity, at the equivalent of −\(r/s\). Substitute backwards \((r\,=\,\beta\,(1\,−\,v)\,−\,\alpha\,and\,s\,=\,−\beta\,)\) and you will find the carrying capacity of the disease:

\[\hat{p} = 1 − \frac{α}{b} − v \]

The small hat atop the \(p\) is just a reminder that this is not the variable prevalence \(p\), but rather the fixed value of the equilibrium prevalence, \(\hat{p}\).

Think about the approach in 15.6, which used \(R_0\). Here is another way to get the result. Since \(v\), the proportion of the population vaccinated, appears in the equation with a minus sign, that means that the greater the proportion vaccinated, the lower the equilibrium prevalence \(\hat{p}\). In fact, setting \(\hat{p}\,=\,0\) and solving for \(v\), when \(v\,=\,1\,−\,\alpha\,/\beta\), the prevalence \(p\) will be zero and the disease will be eradicated. (Actually, for a margin of error, when \(v\,\ge\,1\,−\,\alpha\,/\beta\).