I'm confusing myself over formulating differential equations, and the units of rates of change. I can't quite find other questions that answer this for me either, but please let me know if I have missed anything already posted.
If we have the differential equation:
\begin{equation}
\frac{dP}{dt} = \beta P – \delta P
\end{equation}
where $P$ is the population size, $\beta$ is a birth rate and $\delta$ is the death rate, I understand that $\beta$ and $\delta$ must have the units $t^{-1}$.
Let's say $t$ is in units of months, and everyone in the population dies at rate $\delta$. Can I understand $\delta$ to be equal to $1/(\text{time to die})$. So, for humans, the birth rate would be $1/(\text{months-per-year*average years alive}) = 1/(12*80)$, for example?
But what if we are still in time units of months, but it only takes a couple of weeks for a member of the population to die on average? For instance, if individuals only live for 2 weeks, that's around 50% of a month, and then $\delta = 1/\text{time to die} = 1/0.5 = 2$, which means that the outgoing rate for deaths per month ($\delta P$) will be greater than the number in the population ($2*P$), which to me doesn't make sense: deaths can't be higher than $P$.
Can anyone help me out with what I am missing here?
Best Answer
The value of $\delta P$ corresponds to the total number of deaths per $1$ month. So $\delta$ is the number of deaths per month for one individual.
In the example from your comment, the recovery rate indeed is equal to 1/infection time. So if the infection time is less than $1$, then theoretically more than 100% infected get recovered. The reason it makes sense is that we measure that only on a short period of time, obtaining the derivative.
Let $R(t)$ and $I(t)$ be the number of recovered and infected people. We need to obtain the change of $R$ in a short period of time. We can approximate the number of people recovered during time $\Delta t$ by $\gamma I(t)\cdot\Delta t$ (and as $\Delta t$ is arbitrarily small, $\gamma\Delta t$ will be less than 1). Then $$ R(t+\Delta t) = R(t) + \gamma I(t)\cdot \Delta t, $$ and we get our equation: $\dot{R}(t)=\lim_{\Delta t\to 0}\frac{R(t+\Delta t)-R(t)}{\Delta t}=\gamma I(t)$.