A fishery, a forest or a deer herd grows roughly logistically, $P' = rP\left(1 - \frac{P}{K}\right)$. Harvesting removes some of the population. The phase line shows at once whether a harvest is sustainable, and what happens when it is not.
Key idea
With a constant harvest $H$: $$P' = rP\left(1 - \frac{P}{K}\right) - H.$$ If $0 < H < \frac{rK}{4}$ there are two equilibria: the lower one is an unstable threshold and the upper one is stable. At $H = \frac{rK}{4}$ (the maximum sustainable yield) they merge; for larger $H$ there is no equilibrium and the population collapses.
Why it works
The natural growth $g(P) = rP\left(1 - \frac{P}{K}\right)$ is a downward parabola with roots 0 and $K$ and its peak at $P = \frac{K}{2}$, where $g = \frac{rK}{4}$. The population grows exactly where the parabola lies above the horizontal line at height $H$. The crossings are the equilibria. As $H$ rises, the line climbs, the crossings move together, touch at the peak, and then disappear. A sudden change like this, as a parameter passes a critical value, is called a bifurcation.
Worked example: Two equilibria
A fish population (in thousands) follows $P' = 0.5P\left(1 - \frac{P}{100}\right) - 8$. Find and classify the equilibria.
- Set $0.5P - 0.005P^{2} - 8 = 0$ and multiply by $-200$: $P^{2} - 100P + 1600 = 0$.
- $(P - 20)(P - 80) = 0$, so $P = 20$ or $P = 80$.
- $P'$ $= -0.005(P - 20)(P - 80)$: negative below 20, positive between, negative above 80.
Answer: $P = 20$ is unstable (a collapse threshold); $P = 80$ is stable.
Worked example: The largest sustainable harvest
For the same fishery, what is the largest constant harvest with an equilibrium?
- The peak of $g(P) = 0.5P\left(1 - \frac{P}{100}\right)$ is at $P = 50$.
- $g(50) = 0.5 \cdot 50 \cdot \frac{1}{2} = 12.5$.
Answer: $H = 12.5$ thousand fish per year, with the population held at 50.
Worked example: Constant effort instead
Suppose fishing removes a fixed fraction, $P' = 0.5P\left(1 - \frac{P}{100}\right) - 0.2P$. Find the positive equilibrium and the yield there.
- Factor out $P$: $P\left[0.5\left(1 - \frac{P}{100}\right) - 0.2\right] = 0$.
- $1 - \frac{P}{100} = 0.4$, so $P = 60$.
- Yield $= 0.2 \cdot 60 = 12$ per year.
Answer: $P = 60$ (stable), yield 12. Constant effort never removes the equilibrium entirely, which is one reason it is safer.
Common mistake
Treating the maximum sustainable yield as a safe target. At $H = \frac{rK}{4}$ the single equilibrium is semi-stable: any dip below $\frac{K}{2}$, from a bad year or a counting error, starts a slide to collapse. Another slip is calling the lower equilibrium stable because it is “the smaller population”; check the arrows on both sides.