1
A harvest lowers the whole rate by H
Taking H per unit time makes the rate of change f(y) − H, where f(y) = r·y·(1 − y/K) is the natural growth. The population grows where the natural growth exceeds the harvest and shrinks where it does not.
Lesson · Separable Equations, Growth and Decay · Shifted and bounded growth
Harvesting and the Threshold
A population harvested at a constant rate H obeys y′ = r·y·(1 − y/K) − H. The equilibria are the roots of a quadratic. Below the threshold H* = r·K/4 there are two, a lower one that repels and an upper one that attracts; at the threshold they merge into one; above it there are none and every start falls. A population that looks sustainable at a given harvest can therefore collapse from a small dip.
y′ = r·y·(1 − y/K) − H
r = 1, K = 4: y² − 4y + 4H = 0
H below 1: two; H = 1: one; H above 1: none
H* = r·K/4, the peak of the growth rate
below the lower equilibrium: collapse
How much a population can give up
Core ideas
What this lesson establishes
Three ideas: what a constant harvest does to the rate, how the quadratic counts the equilibria, and where the threshold comes from.
2
The equilibria solve a quadratic
Setting f(y) − H = 0 gives, for r = 1 and K = 4, y² − 4y + 4H = 0. Its discriminant is 16 − 16·H, so two roots exist for H < 1, one for H = 1, none for H > 1.
3
The threshold is the most the growth ever gives
The natural growth f is largest at y = K/2, where it is r·K/4. A harvest above that is more than the population can replace at any size, and a harvest at exactly that leaves a single, fragile equilibrium.
The material
Two equilibria, one, and none
The harvested equation, the quadratic that counts its equilibria, three harvests of increasing size, and the threshold read off the peak of the growth.
Take the logistic population of the previous lesson, with r = 1 and K = 4, and remove a fixed amount H of it in every unit of time. Fishing and logging are the usual pictures. Left alone the population grows to 4. The question is how large H can be before it no longer does.
Definition · The harvested logistic equation
With a constant harvest rate H ≥ 0, the equation is y′ = r·y·(1 − y/K) − H.
The harvest is taken at the same rate whatever the population is. This is an assumption about the fishery, and the lab computes what follows from it.
r = 1, K = 4:
y′ = y·(1 − y/4) − H
= −y²/4 + y − H
equilibria: y² − 4y + 4H = 0
discriminant: 16 − 16·H = 16·(1 − H)
The discriminant decides the count. For H < 1 it is positive and there are two real roots, for H = 1 it is zero and there is one, and for H > 1 it is negative and there are none. The three presets are one of each.
H = 3/4: y² − 4y + 3 = (y − 1)(y − 3) y = 1, 3
H = 1: y² − 4y + 4 = (y − 2)² y = 2
H = 5/4: y² − 4y + 5, 16 − 20 < 0 none
A harvest of three quarters
The right side is −(y − 1)·(y − 3)/4, which is negative below 1, positive between 1 and 3, and negative above 3. So a population between 1 and 3 grows to 3, a population above 3 falls to 3, and a population below 1 falls further. The lab prints 1 unstable; 3 stable. The harvest has not removed the attracting equilibrium; it has moved it from 4 down to 3, and it has created a repelling one at 1. The slope tile agrees: f′(1) = 1/2 is positive, so nearby populations move away from 1, and f′(3) = −1/2 is negative, so they move towards 3.
The repelling equilibrium is the dangerous one. At y = 1 the natural growth is exactly 3/4, the harvest, so nothing changes. At y = 9/10 the natural growth is 279/400, which is less than 300/400, so the population shrinks, and the less there is the less growth to pay for the harvest. A harvest that the population can replace today may not be one it can replace after a bad year.
At the threshold, and beyond
At H = 1 the right side is −(y − 2)²/4, which is never positive. A population above 2 falls back to 2 and settles. A population below 2 falls away, and so does one that dips just under it. The equilibrium is attracting from one side only, and the lab calls it semistable. At H = 5/4 the right side is negative for every y and every start falls.
The threshold has a general form. The natural growth r·y·(1 − y/K) has its peak at y = K/2, where it equals r·K/4. For a harvest above that peak the right side is negative at every y; for one below it the line H crosses the growth curve twice. So H* = r·K/4, which is 1·4/4 = 1 here.
Example · A population that falls below zero
In the third preset the limit tile reads → −∞: every start falls without bound, and passes zero on the way, since the equation does not know a population cannot be negative. Read a negative value as the population having already gone: the model says the stock is exhausted, and stops describing anything.
Interactive
Autonomous equations
Equilibria, the phase line, and where solutions go
Method
Finding what a harvest does
Five moves, and the threshold is the fourth.
1
Write the natural growth and the harvest
f(y) = r·y·(1 − y/K) and the constant H. The equation is y′ = f(y) − H.
2
Find the equilibria
Solve f(y) = H. For r = 1, K = 4 that is y² − 4y + 4H = 0; use its discriminant first to count the roots.
3
Read the direction in each region
Evaluate f(y) − H at a point in each region between roots. The lower one repels and the upper one attracts.
4
Find the threshold
The peak of f is r·K/4, at y = K/2. A harvest at or above it leaves no attracting equilibrium.
5
Test with a dip
Take a start a little below the lower equilibrium and compute f(y) − H there. If it is negative, a small loss becomes a collapse.
Counting equilibria and finding the threshold
Score 0 / 0Choose the answer you can justify.
r = 1, K = 4, three harvests
The equilibria come from the quadratic, and every figure is an exact fraction.
y′ = y·(1 − y/4) − H
equilibria: y² − 4y + 4H = 0
H = 3/4: (y − 1)(y − 3) y = 1, 3
H = 1: (y − 2)² y = 2
H = 5/4: 16 − 20 < 0 none
threshold H* = r·K/4 = 1·4/4 = 1
f(9/10) − 3/4 = 279/400 − 300/400 = −21/400
The last line is the check on the lower equilibrium: just below 1 the right side is negative, so the population keeps falling. As H rises from 3/4 to 1 the two equilibria, 1 and 3, move together and meet at 2; at any larger harvest nothing is left to hold the population.
The upper equilibrium is 3 at the first harvest, 2 at the second, and absent at the third. It does not fade away smoothly: it goes from 2 to nothing the moment H passes H* = 1.
Common mistakes
1
Believing that any harvest below the natural growth rate is sustainableThe mistaken model is that if the population is growing faster than it is harvested, the harvest is safe. At y = 1 with H = 3/4 the growth exactly matches the harvest, and the equilibrium there repels: at y = 9/10 the growth 279/400 is under the harvest 300/400, and the population falls and keeps falling. A harvest below the peak of the growth is sustainable only for starts above the lower equilibrium.
2
Expecting the population to shrink gradually as the harvest growsThe stable level does come down with H: 4 for no harvest, 3 for H = 3/4, 2 for H = 1. But at H = 1 it meets the repelling equilibrium, past that both are gone, and for H = 5/4 there is nothing left to settle on. There is no small stable population at a slightly larger harvest, so a small increase in H past H* turns a stable stock into a collapse.
3
Treating the negative values as a populationIn the heavy preset the limit is → −∞, so the solution passes below zero. A negative number of fish is not a quantity, and the equation, which knows nothing of the physical problem, carries on past it. The honest reading is that the stock is gone when y reaches 0, and everything after that point is outside the model.
Check your understanding
✓
Finish when you can find the equilibria of a harvested logistic equation, classify them, and compute the threshold harvest.
You should be able to write y′ = f(y) − H, reduce it to a quadratic, count its roots from the discriminant, read which repel and which attract, derive H* = r·K/4 from the peak of the growth, and test a start just below the lower equilibrium.
Note: That finishes the course. Every model in it was one first-order equation with its equilibria read from its right side; the next course takes that reading as its subject and draws the phase line for any equation of the form y′ = f(y).