# The Azimuth Project Logistic equation (Rev #4, changes)

Showing changes from revision #3 to #4: Added | Removed | Changed

## 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, thepopulation 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 equation is

${d x \over d t} = r x\left(1-\frac{x}{K}\right) \, .$

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 a simple equation:

${d x \over d t} = r x$

which describes exponential population growth:

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

When $K$ is finite and positive, the logistic equation 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 solution has

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

It is easy to find the explicit solution of the logistic equation, since it is a first-order separable differential equation. However, instead of doing this, let us consider a special case.

By rescaling the time and population variables (that is, by choosing appropriate units for time and population), we can reduce the general logistic equation to the case where $r = K = 1$:

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

One solution of this is the logistic function:

$x(t) = \frac{e^t}{1 + e^t} \, .$

It looks like this:

This solution goes from $0$ to $1$ as $t$ goes from $-\infty$ to $+\infty$. All other solutions having that behavior are time-translated versions of this one, i.e.:

$x(t) = \frac{e^{t-t_0}}{1 + e^{t-t_0}} \,.$

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

## References

category: ecology