The use of Differential Equations allows mathematical models for population to be created. Here are a few common ones. Qualitative Analysis is very useful here, as it shows the state of the model with certain initial populations.
Let p = population at time t. Then dtdp is the rate of population growth.
Remember that both p,t≥0
Malthus (Doomsday) Model §
In the doomsday model, the rate of growth is directly proportional to a a scaling value k, with no other affecting factors:
\frac{dp}{dt} \propto p \implies \frac{dp}{dt} = kp, k > 0$$ ([Separable ODEs](Separable%20ODEs.md))
For an initial population $p_{0}$, the solution is:
p(t) = p_{0}\times e^{kt}
>[!info]- Derivation
>
>If the initial population is $p_{0}$, it implies that $p(0) = p_{0}$ is an [initial condition](Differential%20Equations.md)
>$\frac{dp}{dt} = kp;p(0) = p_{0}$
>$\frac{1}{p} \frac{dp}{dt} = k$ ([Separation Of Variables](Separation%20Of%20Variables))
>$\int \frac{1}{p} \, dp = \int k \, dt$
>$\log(|p|) = kt + c$
>$\log(p_0) = 0 + c$ ($p(0) = p_{0}$)
>$\log(|p|) = kt + \log(p_{0})$
>$\log\left( \frac{p}{p_{0}} \right) = kt$
>$\frac{p}{p_{0}} = e^{kt}$
>$p(t) = p_{0} \times e^{kt}$
The doomsday model has different behaviours for different values of $k$:
* $k > 0$ : unbounded exponential growth
* $k < 0$ : population dies out
* $k = 0$ : population stays constant
It's pretty unrealistic, as you can tell.
##### With harvesting
Remove some of the population at a constant rate.
\frac{dp}{dt} = kp − h, h > 0.
Using Qualitative Analysis, we can see that:
- The equilibrium solution (i.e. when dtdp=0 ) is kh
- The solution is unstable
Logistic Model §
Inspired by Pierre-François Verhulst, the model introduces a ”competition” term, because a larger population will result in more overcrowding, diseases, etc. This competition model helps keep the population semi-stable.
\frac{dp}{dt} = kp \ - \frac{k}{a}p^2 = kp\left( 1-\frac{p}{a} \right), a > 0
>$$
* $kp$ is the birth rate of the population, like the Malthus model
* $-\frac{k}{a}p^2$ is the competition term, with $a$ being the carrying capacity, which is the maximum population an area can sustainably hold.
Using [Qualitative Analysis](Qualitative%20Analysis.md):
* The equilibria solutions are $p = 0, a$
* $p = 0$ is an unstable solution, while $p = a$ is a stable solution, as can be seen in the phase plot:
Fun fact, the logistic model accurately described the population of USA from 1790-1950.
##### With harvesting
Again, removing population at a constant rate
\frac{dp}{dt} = kp \ - \frac{k}{a}p^2 - h= kp\left( 1-\frac{p}{a} \right) - h, a > 0
Solutions for this one aren’t as generalised, it’s better with an example.