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I don't understand the sources I found online, so I'm hoping someone can show me how to do the following: I need to put together a 1st-order DE that models some population and includes birth rate, death rate (which is proportional to the size of the population to account for overcrowding), and harvesting. Say, for instance, I'm modelling a population of fish.

The furthest I can get is this:

$$p'(t) = bp(t)-d(t)p(t)-hp(t)$$

where $p(t)$ is the size of the population at time $t$, $b$ is the constant birth rate, $d(t)$ is the death rate, and $h$ is the constant harvest rate.

Now, I've guessed that the death rate is proportional to the difference between the maximum population and the current population, so:

$$d'(t) = (N-p(t))k$$

Where $N$ is the max. pop., and and $k$ is the constant proportion.

So, do I solve this second equation for $d(t)$ and plug it into my first equation, then solve for $p(t)$? In which case, I'd get

$$p'(t) = bp(t) - \left(kNt-k\int p(t)dt + C_0\right)p(t)-hp(t)$$

And that seems far too ugly to be correct.

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If "$d(t)$ is the death rate", and "the death rate is proportional to [such-and-such]", then why did you make $d'(t)$ instead of $d(t)$ proportional to such-and-such? – Rahul Nov 5 '12 at 5:14
@RahulNarain I meant to say, "d(t) is the number of deaths at time t." I've corrected it. Thanks. – Korgan Rivera Nov 5 '12 at 13:19
I don't think that was a correction. In $p'=-dp$, $d$ is a death rate, not a number of deaths. – joriki Nov 5 '12 at 14:21
It was better the first time :) I was suggesting that you set $d(t) = (N-p(t))k$, not change the meaning of $d(t)$. What you've written makes $d(t)$ the total number of deaths till present. Why should $p'(t)$, the rate of change of population, have a term that depends on the total number of deaths that have ever happened? On preview, what joriki said. – Rahul Nov 5 '12 at 14:25
@RahulNarain That makes sense. :) So I have $p'(t) = p(t)(b-Nk-h)-kp(t)^2$. How does that sound? – Korgan Rivera Nov 5 '12 at 16:27

The "standard" model of a density-regulated population is the logistic growth model, also known as the Verhulst equation:

$$\frac{dp}{dt} = rp \left(1-\frac pK\right) = rp - \frac rK p^2,$$

where $r$ is the exponential growth rate of the population at low densities, and $K$ is the effective carrying capacity (i.e. the maximum stable population size) of the population. This differential equation has two fixed points, $p = 0$ and $p = K$, and its general solutions are logistic functions of the form:

$$p(t) = \frac{K p_0 e^{rt}}{K + p_0 (e^{rt} - 1)}.$$

However, the Verhulst equation is not a mechanistic population model, as it does not contain explicit birth and death terms, and there are several different ways to construct mechanistic models which are equivalent to it.

For example, we might assume that the per capita birth rate is fixed but the death rate contains a density-dependent term (e.g. due to conflicts, starvation or diseases), giving a model like:

$$\frac{dp}{dt} = bp - dp - cp^2,$$

where $b$ and $d$ are the baseline per capita birth and death rates and $cp$ is the density-dependent death rate. Alternatively, we might treat the per capita death rate as fixed but the birth rate as proportional to the availability of some resource (e.g. space to grow for plants) of which a constant amount is taken up by each individual, giving us a model like:

$$\frac{dp}{dt} = bp\left(1-\frac pN\right) - dp,$$

where, again, $b$ and $d$ are the baseline birth and death rates at small population sizes, and $N$ is the absolute maximum population size (i.e. the size at which the birth rate drops to zero).

Both of these models can be written in the form of the Verhulst equation: in both cases, $r = b-d$, but in the first case $K = r/c = (b-d)/c$, while in the second $K = rN/b = (1-d/b)N$. Thus, the two models respond differently to changes in the parameter values: in the first model, for instance, increasing the birth rate can increase the equilibrium population size $K$ indefinitely, where as in the second model the population size can never surpass $N$ no matter how large $b$ is.

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