Recipe

Find the equilibria of a harvested logistic population

Use when: the equation has the form $\dfrac{dy}{dt}=ry(1-y/K)-h$ — logistic growth with a constant harvest rate $h$ subtracted

Goal: both equilibria (when they exist), which one is stable, and the survival threshold below which the model predicts extinction

Taught in Logistic growth with harvesting · a later section extends this recipe with more cases

Quick recipe

Set the rate to zero and expand into a quadratic
Solve the quadratic and check how many real roots it has
Classify each root's stability from the sign of $f$
Compare the starting population $y_0$ to the unstable threshold $y_1$

1
Set $ry(1-y/K)-h=0$ and multiply out; this is a standard quadratic in $y$ .

Example

For $\dfrac{dy}{dt}=2y(1-y/8)-3$ (so $r=2$ , $K=8$ , $h=3$ ): expanding gives $2y-\dfrac{y^2}{4}-3=0$ ; multiplying by $-4$ to clear the fraction, $y^2-8y+12=0$

Why: an equilibrium is, by definition, any constant $y$ where the rate of change is exactly zero.
2
Apply the quadratic formula (or factor directly, if it factors) to find the roots $y_1<y_2$ , when they exist.

Example

$y^2-8y+12=0$ factors as $(y-2)(y-6)=0$ , giving $y_1=2,\qquad y_2=6$
- the discriminant is positive — two distinct equilibria exist; continue to Step 3.
- the discriminant is negative — no real equilibria exist; the downward parabola never crosses zero, so $f(y)<0$ for every $y$ and the population declines to extinction regardless of where it starts.
- Not yet: the discriminant is exactly zero, so the two equilibria have just merged into a single semistable point at the exact harvest rate called a bifurcation value — covered in Bifurcation of a 1-D autonomous family
Why: the discriminant of this quadratic is exactly what decides whether harvesting at this rate is sustainable at all, not merely how big the two equilibria are.
3
Test a $y$ -value between the two roots and one outside them; $f>0$ between the roots means growth back toward the upper root, $f<0$ outside means decline away from the lower one.

Example

Between the roots, $f(4)=2(4)(1-4/8)-3=4-3=1>0$ . Below $y_1=2$ , $f(1)=2(1)(1-1/8)-3=1.75-3=-1.25<0$ . So $y_2=6$ is stable and $y_1=2$ is unstable.

Why: this is the same sign-of- $f$ phase-line reasoning from earlier in the chapter, applied here to a downward-shifted parabola instead of the plain logistic curve.
4
Check which side of the unstable equilibrium $y_1$ found in Step 3 the starting population $y_0$ sits on — that alone decides the population's long-run fate.

Example

With $y_1=2$ , $y_2=6$ : starting at $y_0=5$ ( $>2$ ) approaches $6$ ; starting at $y_0=1$ ( $<2$ ) declines to extinction.
- $y_0>y_1$ — the population approaches $y_2$ , the harvested carrying capacity.
- $y_0<y_1$ — the population declines to extinction; the model predicts it reaches $0$ in finite time, not just in the limit.
Why: $y_1$ is exactly the boundary Step 3 found between the two directions of monotonic motion — everything on one side is pulled toward $y_2$ , everything on the other is pushed toward $0$ .

Common mistakes

Forgetting the harvest term when expanding — solving the plain logistic equation's equilibria ( $y=0,K$ ) instead of the harvested one.
Assuming the larger root is always unstable; here it's the reverse, because subtracting $h$ flips which side of the parabola sits above zero.
Treating "the population declines" as slow, asymptotic decay like the exponential models elsewhere in the course, rather than reaching zero in finite time.
Treating a starting population just barely above $y_1$ as automatically safe — the model says it survives, but any small negative disturbance can push it back below the threshold.