1.3 Harvesting and Bifurcation

Take last section's pond and start fishing it. If boats remove a fixed quantity per unit time — a quota, not a percentage — the model gains a constant subtraction:

$\frac{d P}{d t} = 2 P (1 - \frac{P}{8}) - a,$

where $a greater than or equal to 0$ is the harvest rate. The question a fishery manager actually asks is not "what is $P (t)$ ?" but "how large can $a$ be before the pond collapses?" That question is about equilibria, and equilibria can be found without solving anything: set the right-hand side to zero.

Expanding, $2 P - P^{2} / 4 - a = 0$ , or $P^{2} - 8 P + 4 a = 0$ , whose roots are

$P_{\pm} = 4 \pm \sqrt{16 - 4 a} .$

Everything interesting is in that square root. For $a less than 4$ there are two equilibria, one above the midpoint $4$ and one below. At $a equals 4$ the root vanishes and the two merge into a single equilibrium at $P equals 4$ . For $a greater than 4$ there is no real root at all: no equilibrium exists, $d P / d t$ is negative for every population, and the fishery drives itself to zero no matter how healthy it started.

The figure below shows the two equilibrium levels as horizontal lines whose heights you control with the harvest slider.

Two horizontal lines in the plane: a solid green line (the stable equilibrium population) and a dashed red line below it (the unstable threshold). Dragging the harvest slider a from 0 toward 4 pulls the green line down from 8 and the red line up from 0 until the two meet at height 4 exactly when a reaches 4, where both equilibria disappear together. — Equilibria of the harvested logistic model. The solid green line is $P_{+} = 4 + \sqrt{16 - 4 a}$ and the dashed red line is $P_{-} = 4 - \sqrt{16 - 4 a}$ . Dragging the harvest rate $a$ toward $4$ squeezes the two together until they collide and disappear.

Explore

Observe. Start with $a equals 0$ (no fishing). Where do the two lines sit? Confirm they are the equilibria $0$ and $8$ you already know from the unharvested logistic equation.
Predict. As $a$ increases, one line must move up and the other down. Which is which, and why? Write down your reasoning from the formula before dragging.
Verify. Drag $a$ slowly from $0$ to $3$ . Note the gap between the lines at $a equals 1$ , at $a equals 2$ , and at $a equals 3$ . Is the gap shrinking at a steady rate, or is it collapsing faster near the end?
Observe. Push $a$ to $3.9$ , then to $3.99$ . Both lines are now nearly at height $4$ . The square root is what makes the last stretch so abrupt — the gap is $2 \sqrt{16 - 4 a}$ , so what happens to the rate of collapse as $a → 4$ ?
Predict, then reason. The green line is stable and the red one unstable. If the pond is sitting comfortably at $P_{+}$ with $a equals 3.9$ and a storm knocks the population down by $1$ , does it recover? Now suppose the harvest is raised to $a equals 4.1$ instead: what happens to a pond sitting at $P equals 4$ ?

Connect

A bifurcation is a qualitative change in the set of equilibria caused by a smooth change in a parameter. Here, nudging $a$ through the critical value $4$ does not merely move the equilibria — it destroys them. Two equilibria of opposite stability approach each other, merge, and annihilate; this pattern is common enough to have a name, the saddle-node (or fold) bifurcation, and it is the mathematical signature of a system that fails suddenly rather than gradually.

The management lesson is sharper than the equations look. Just below the critical rate, the pond still has a stable equilibrium, so the fishery appears to be working — but the unstable equilibrium $P_{-}$ has crept up close underneath it, which means the margin for error has nearly vanished. A bad season that pushes the population below $P_{-}$ sends it to extinction even though the harvest rate never changed. Systems near a saddle-node bifurcation look healthy right up until they do not, and the shrinking gap you watched in step 3 is exactly the shrinking margin.

Note

Coming soon: Euler's method

Everything in this section was equilibrium reasoning, which needs no solution formula — fortunate, because the harvested logistic equation's solutions are unpleasant to write down and most equations have none at all. The general escape is numerical: Euler's method steps forward in small increments, using the equation's own slope at each point to guess the next, and refining the step size until the answer stops changing. The figure for it — a staircase of Euler steps overlaid on a true solution curve, with a slider on the step size to watch the error shrink — needs the trajectory object being added to the XYZ 3D engine, which integrates a differential equation on the fly rather than plotting a formula. It arrives with the same release as the slope field.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.