2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $k$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Shortytot Jul 10 '18 at 17:00
  • $\begingroup$ 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. $\endgroup$ – LutzL Jul 10 '18 at 18:32
  • $\begingroup$ @LutzL Indeed, the OP didn't exclude an external source, so that I'm removing my comment. $\endgroup$ – user539887 Jul 11 '18 at 8:58
2
$\begingroup$

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 $P=k$

$\endgroup$
  • $\begingroup$ What does the equilibrium points tell us in a population growth model? $\endgroup$ – Shortytot Jul 10 '18 at 16:58
  • $\begingroup$ 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. $\endgroup$ – Mohammad Riazi-Kermani Jul 10 '18 at 17:16
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user539887 Jul 12 '18 at 11:32
  • $\begingroup$ @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. $\endgroup$ – Dmitry Jul 12 '18 at 17:18
  • $\begingroup$ In particular, the second model has maximal growth rate when the population is zero! $\endgroup$ – Dmitry Jul 12 '18 at 17:36
  • $\begingroup$ 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. $\endgroup$ – user539887 Jul 12 '18 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.