The Azimuth Project
Logistic equation (Rev #12)

Idea

The logistic equation is a simple model of population growth in conditions where there are limited resources. When the population is low it grows in an approximately exponential way. Then, as the effects of limited resources become important, the growth slows, and approaches a limiting value, the equilibrium population or carrying capacity.

The logistic growth model is

{d x \over d t} = {r\over K} (K-x) x \, .

Here $x$ is the population, which is a function of time $t$ . $K$ is the equilibrium population, and $r$ is the growth rate.

Note that in the limit $K \to \infty$ , we get the simpler model:

{d x \over d t} = r x

describing exponential population growth:

x(t) = x_0 e^{r t} \, .

When $K$ is finite and positive, the logistic model describes population growth that is approximately exponential when the population is much less than $K$ , but levels off as the population approaches $K$ . If the population is larger than $K$ , it will decrease. Every positive solution has

lim_{t \to +\infty} x(t) = K \, .

The logistic model can be normalised by rescaling the units of population and time. Define $y\colon = x/K$ and $s\colon = rt$ . The result is

{d y \over d s} = y(1-y) \, .

It is easy to find the explicit solution of the logistic equation, since it is a first-order separable differential equation.

The growing solutions are all time-translated versions of the logistic function (See Wikipedia)

y(s) = {e^s\over 1 + e^s} = {1\over 1 + e^{-s}}

which looks like this:

This function goes from $0$ to $1$ as $t$ goes from $-\infty$ to $+\infty$ . All other growing solutions have the same limiting behavior and are time-translated versions of this one. After rescaling back to the original variables, we have

x(t) = {K\over 1 + e^{-r(t-t_0)}} \, .

There are also decreasing solutions where $x \gt 1$ and solutions (irrelevant to population biology) where $x \lt 0$ decreases explosively to $-\infty$ .

Chaos in the Logistic model

Discrete-time version

When the logistic equation is discretised it displays chaos. It is, in fact, the canonical ground for the study of the period-doubling cascade.

Assume that the Euler formula is used to discretise the logistic growth model, that is,

{\Delta x \over \Delta t} = {r\over K} (K-x) x \, .

Assume $t_{n+1}-t_n = \Delta t$ and $x_n = x(t_n)$ for all $n$ . Then,

x_{n+1} = (1+r\Delta t)x_n - {r\Delta t\over K} x_n^2 \, .

This can be normalised by letting $\lambda = 1+r\Delta t$ :

x_{n+1} = \lambda x_n\Bigl(1 - {1 - 1/\lambda\over K} x_n\Bigr)

and measuring $x$ in units of ${K\over 1-1/\lambda}$ , that is, $x = {Ky\over 1-1/\lambda}$ and one obtains the so-called Logistic Map.

y_{n+1} = \lambda y_n(1 - y_n)\, .

It is known (Wikipedia, again) that, depending on the value of $\lambda$ , this equation has a stable fixed point, multiperiodic stationary states, or displays chaotic behaviour. Note that $\lambda$ depends only on the growth rate and the time step, and not on the carrying capacity.

When the logistic map is interpreted as a numerical approximation to the continuous logistic growth model, the time step $\Delta t$ is largely arbitrary and can be chosen to be less than $1/r$ , so that $1 \lt \lambda \lt 2$ ensuring numerical stability.

However, it is possible to interpret the continuous logistic model as an approximation to a fundamental discrete model. In that case, $\Delta t$ is not arbitrary (representing, for instance, the periodicity with which a species reproduces - usually annually) and neither is $r$ , in which case $\lambda = 1 + r\Delta t$ cannot be tuned. If the population reproduces too fast (high values of lambda), then the sizes of successive generations may be chaotic. If $\lambda \gt 3$ the stationary state fluctuates among at least two states, and if $\lambda \gt 3.57$ (approximately) the dynamics is chaotic.

Continuous version

Here chaos arises as a consequence of delayed feedback.

Rewrite the normalised logistic growth model as

{d x\over x} = (1 - x)d t

and add a delay to the right-hand side, obtaining

{d x(t)\over x(t)} = \bigl[1 - x(t-\Delta)\bigr]d t

This introduces the possibility of overshoot?, as when $x(t)$ reaches the carrying capacity $1$ the earlier value $x(t-\Delta)$ is still away from the carrying capacity and so the system continues in the same direction as it is moving in when it touches 1. In fact, the population trend reverses a time $\Delta$ after reaching the carrying capacity. Depending on the speed with which the $x=1$ level is crossed this may result in growing oscillations as when the constant solution to the discrete version becomes unstable at $\lambda = 3$ .

To see this, write $x = e^y$ :

{d y(t)\over d t} = 1 - e^{y(t-\Delta)}.

Assuming now that $y\approx 0$ ,

{d y(t)\over d t} \approx - y(t-\Delta).

An analysis of this equation is carried out in the article on delayed feedback.

References

category: ecology