# Difference between Population Growth Models (Differential Equations)

I'm currently studying population growth models in Math class right now and is presented with different equations for different models.

I think I understand that we use $dP/dt = rP$ (where $r$ is the intrinsic growth rate and $P$ is the population) when we have infinite growth.

However, I'm struggling to find the difference between the models:

$dP/dt = rP(1-P/k)$ --> (where $r$ is the intrinsic growth rate, $P$ is the population and $k$ is the carrying capacity)

and

$dP/dt = r(k-P)$ -->(where $r$ is the intrinsic growth rate, $P$ is the population and $k$ is the carrying capacity)

as they both pertain to rate of growth with carrying capacity...

Any explanation/clarification is greatly appreciated!

In the second, $r$ is the intrinsic death rate, growth happens at $rk$ units per time unit.

Note that the first one of the problematic models is quadratic while the second one is linear. Thus the first has two fixed points at $0$ and $k$, while the second only has the one at $k$.

• What does the equilibrium points tell us? i.e., I am still kind of confused as to when to use which. E.g., if I had a population growing at rate r, and wanted to know the population size in x years, which model would be more useful? – Shortytot Jul 10 '18 at 17:00
• Because there is a different number of equilibriums, there is a qualitative difference. The quadratic model is for a closed system where a population develops on its own. The linear model is for a system without internal reproduction that is filled at a constant rate from an external source and that dies off individually, without interaction among the population. – LutzL Jul 10 '18 at 18:32
• @LutzL Indeed, the OP didn't exclude an external source, so that I'm removing my comment. – user539887 Jul 11 '18 at 8:58

The logistic model$$dP/dt = rP(1-P/k)$$ has two equilibrium points namely $P=0$ and $P=k$ with $P=k$ being the stable one.

The $$dP/dt = r(k-P)$$ model has only one equilibrium point which is stable at $P=k$

The behavior at $P<0$ is significantly different for the two models but they behaves similar around $P=k$

• What does the equilibrium points tell us in a population growth model? – Shortytot Jul 10 '18 at 16:58
• It means if you get there you stay there. In our model the equilibrium point $P=k$ is an attractor which means the population will eventually approach the equilibrium point if you start with a positive population. – Mohammad Riazi-Kermani Jul 10 '18 at 17:16

Probably the best way to analyse your models is to plot the r.h.s., let denote it $f(P)$, versus $P$.

In the first case, the population grows a little when close to $0$ (population is too small), then the growth rate increases and attains its maximum at $P=k/2$ (optimal conditions). As $P$ continues to grow, $f(P)$ decreases and reaches $0$ at $P=k$ (due to the competition for resources). This picture agrees pretty well with our intuition.

In contrast to that, in the second model $f(P)$ is maximal when $P=0$ and gradually decreases as $P$ grows. That is to say, the smaller the population, the faster it grows. In this sense, this model gives wrong results for small population sizes.

• You wrote: "this model gives wrong results for small population sizes." But see LutzL's comment under their answer: equations (could) model just different situations, so, without knowing the context, we should not write off some of them as wrong. – user539887 Jul 12 '18 at 11:32
• @user539887, Well, the context is provided in the topic: "Population growth models". I can hardly imagine a population growth model that can be described by the second model. Please correct me if I'm wrong. – Dmitry Jul 12 '18 at 17:18
• In particular, the second model has maximal growth rate when the population is zero! – Dmitry Jul 12 '18 at 17:36
• In my first comment I used an argument that the RHS equal to zero for the zero population density means spontaneous generation. But after seeing LutzL's comment I deleted it. – user539887 Jul 12 '18 at 21:13