How Much Fishing Is Too Much?

A fish population regrows on its own, fastest at half the ocean's capacity. Catch a fixed amount every year and two balance points appear: one the population returns to, one it falls away from. Fish too hard and both vanish.

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How it works

Equilibrium solutions

In one sentence

An equilibrium of $y' = f(y)$ is a constant solution $y^*$ with $f(y^*) = 0$ ; it is stable if $f$ changes from positive to negative there and unstable if from negative to positive.

General formula

\frac{d{\htmlData{sym=0}{P}}}{dt} = {\htmlData{sym=1}{r}}{\htmlData{sym=0}{P}}\left(1 - \frac{\htmlData{sym=0}{{P}}}{\htmlData{sym=2}{{K}}}\right) - {\htmlData{sym=3}{H}}

Between equilibria the arrows all point one way: up where growth beats the harvest, down where it doesn't. A stable equilibrium has arrows pointing toward it from both sides.

What the symbols mean

$P$
Population
Fish in the stock, in thousands.
$r$
Growth rate
How fast the stock would grow if it were small and unfished.
$K$
Carrying capacity
The most fish the ecosystem can support: 100 thousand here.
$H$
Harvest
Thousands of fish caught per year.

Without fishing the population grows logistically toward the carrying capacity $K$ (here 100 thousand).
A constant harvest $H$ shifts the growth curve down. Equilibria are where growth equals harvest: $rP(1 - P/K) = H$ .
There are two: the upper one is stable (the stock recovers from shocks), the lower one unstable (below it, the stock collapses).
If $H > rK/4$ , growth can never match the harvest: both equilibria vanish and the fishery collapses from any start.

Learn the math

Equilibrium SolutionsDiff Eq · Lesson 11, First Order Differential Equations