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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?

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  • $\begingroup$ As another example, consider the equation for recovered individuals from the SIR equation: $dR/dt = \gamma I$. The parameter $\gamma$ is often defined as $1/\text{recovery rate}$ but what if the recovery rate is < than the time units of the model, e.g. if time is in weeks, but recovery takes 2 days, then $\gamma = 1/(2/7) = 3.5$ times the number infected. $\endgroup$
    – user_15
    Commented May 28, 2020 at 12:56
  • $\begingroup$ Thanks @ChristianBlatter. I've changed to Greek letters for the constants. $\endgroup$
    – user_15
    Commented May 28, 2020 at 15:25

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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: $\dot{R}(t)=\lim_{\Delta t\to 0}\frac{R(t+\Delta t)-R(t)}{\Delta t}=\gamma I(t)$.

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  • $\begingroup$ To take an example, look at the description of the $\gamma$ parameter here: cran.r-project.org/web/packages/shinySIR/vignettes/… It's defined in terms of a recovery rate, where $1/\gamma$ corresponds to the average infectious period. But what if the average infectious period is < than the units of time in the model, then $\gamma > 1$, and the number recovered is more than 100% of the infected. Does that make sense? $\endgroup$
    – user_15
    Commented May 28, 2020 at 15:23
  • $\begingroup$ You're right, I'm sorry. I will edit my comment. I now think that to understand why it makes sense, we need to see how the equation is derived. $\endgroup$
    – M_S
    Commented May 28, 2020 at 18:39
  • $\begingroup$ Thanks! That makes sense to me now. $\endgroup$
    – user_15
    Commented May 29, 2020 at 16:06
  • $\begingroup$ Do you mind if I ask a follow-up question? You say that $\delta$ is the "number of deaths per month for one individual" but $\delta$ has units 1/time. If it's the number of deaths per month for one individual, isn't the units 1/time/indiividual = 1/(time * individual)? I'm getting confused when people say units are 1/time for coefficients in DE models, but then interpret them as per-capita rates. Thanks! $\endgroup$
    – user_15
    Commented Jun 19, 2020 at 5:36
  • $\begingroup$ Deaths are also measured in units of individuals (let's say $ind$). So we have $\frac{ind}{time*ind}=1/time$. $\endgroup$
    – M_S
    Commented Jun 19, 2020 at 10:46

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