4.3: Density-limited growth

Last updated
Save as PDF

Page ID: 25438

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

What about outside of the range of the orthologistic model? Think of the same Black-Eyed Susans, not only close enough that pollinators can ﬂit fluently from one to another, but also crowded so that they start to shade one another, and their roots start to compete for water and nutrients. What is a suitable model for this?

The growth rate will again depend on the number of plants, but now more plants will reduce the growth rate. That just means a minus sign on s.

\(\frac{1}{N}\, \frac{∆N}{∆t}\, =\,r\,+\,sN\), s <0

Again, r is the number of offspring each will produce if it is alone in the world, but with s negative, s is the number each plant will be unable to produce for each additional plant that appears in its vicinity.

Suppose we have r = 1 and s = −1/1000, and we start with three plants, so N (0) = 3. Here is the code, with the new negative s in red.

r=1; s=-0.001; dt=1; t=0; N=3; print(N);

while(t<=20)

{ dN=(r+s*N)*N*dt; N=N+dN; t=t+dt; print(N); }

Now, because s is negative, the growth rate \(\frac{1}{N}\, \frac{∆N}{∆t}\) will drop as the population increases, so you might surmise that the rate will eventually reach zero and the population will level off. In fact, it levels off to 1000.

Figure \(\PageIndex{1}\) Logistic growth (green) contrasted with orthologistic growth (red) and exponential growth (blue).

The value at which it levels off is called an “equilibrium,” a value where the dynamical system becomes quiescent and stops changing. In the case of the logistic equation, it is also called the “carrying capacity,” a level at which the environment cannot “carry” any larger population.

But why 1000? What value of \(\frac{1}{N}\, \frac{∆N}{∆t}\) will make the population level off? When ∆N is 0, that means “the change in N is zero.” And that means N stops growing. And when ∆N is zero, the entire term on the left is zero and algebra proceeds as follows.

\(\frac{1}{N}\, \frac{∆N}{∆t}\,=\,r\,+\,sN\)

\(0\,=\,r\,+\,sN\)

\(-sN\,=\,r\)

\(N\,=\frac{-r}{s}\)

So the carrying capacity is −r/s. In Figure 4.3, −r/s = −1/(−0.001) = 1000. Exactly where it ended up!

This is the celebrated “logistic equation,” published in 1838 by Pierre Verhulst. It is commonly written

\(\frac{∆N}{∆t}\,=\,rN(1\,-\frac{N}{K})\)

Notice that when N is equal to K, the factor in parentheses on the right becomes 1 − N/N = 1−1 = 0, so the whole growth term ∆N /∆t becomes zero and the population stops growing. Thus K is carrying capacity, and therefore K = −r /s.

As an exercise, you might want substitute −r /s for K in the equation above, then simplify and see if you get the r + sN formulation.