Logistic Growth and Harvesting, differential equations lesson

A fish stock grows quickly while it is small and levels off as food and space run short. If boats then remove a fixed tonnage every year, does the stock settle at a new level, or can it collapse? This lesson solves the logistic model, finds where its natural growth is fastest, and shows how a constant harvest creates a threshold, a critical harvesting rate beyond which no equilibrium remains, and a boundary at zero where the model has to stop.

Key terms

Logistic model

Definition With $t$ as the independent variable and $P' = dP/dt$ , the logistic model for a population $P$ is

P' = rP\left(1 - \frac{P}{K}\right),

with constants $r > 0$ , in units of inverse time, and $K > 0$ , in the units of $P$ . The per-capita rate $P'/P = r(1 - P/K)$ falls linearly in $P$ , from $r$ at $P = 0$ to $0$ at $P = K$ , and is negative above $K$ . The equilibria are $0$ and the carrying capacity $K$ . Since the derivative of the rate with respect to $P$ is $r(1 - 2P/K)$ , which equals $r$ at $0$ and $-r$ at $K$ , the equilibrium $0$ is unstable and $K$ is asymptotically stable. For a start $P_0 > 0$ the solution is

\begin{gathered} P(t) = \frac{K}{1 + Ae^{-rt}}, \\[0.4em] A = \frac{K}{P_0} - 1, \end{gathered}

defined for all $t \ge 0$ . It increases toward $K$ when $0 < P_0 < K$ , decreases toward $K$ when $P_0 > K$ , and stays at $K$ when $P_0 = K$ . The start $P_0 = 0$ gives the zero solution.

In plain words A small population grows almost exponentially at rate $r$ . Crowding lowers the growth per individual in proportion to how full the habitat is, so growth stops at $K$ , and a population above $K$ shrinks back toward it. The carrying capacity is a level built into the model, not a measured law of nature.

Example For $0 < P < K$ , separate the variables with the partial fraction identity

\frac{1}{P\left(1 - P/K\right)} = \frac{1}{P} + \frac{1}{K - P}.

Integrating gives $\ln P - \ln(K - P) = rt + C$ , so $P/(K - P) = e^{C}e^{rt}$ , and solving for $P$ gives the formula above. With $r = \tfrac{1}{5}$ per time unit, $K = 100$ and $P_0 = 20$ , we get $A = 4$ and

P(t) = \frac{100}{1 + 4e^{-t/5}}.

Then $P(0) = 100/5 = 20$ , and both $P'$ and $\tfrac{1}{5}P(1 - P/100)$ equal $80e^{-t/5}/(1 + 4e^{-t/5})^2$ , so the formula solves the problem. As $t \to \infty$ the denominator tends to $1$ and $P(t) \to 100$ .

Maximum natural growth rate

Definition The natural growth rate of the logistic model is $g(P) = rP(1 - P/K)$ , measured in population units per unit time. Completing the square gives

g(P) = \frac{rK}{4} - \frac{r}{K}\left(P - \frac{K}{2}\right)^2,

so the largest value of $g$ is $rK/4$ , reached at $P = K/2$ . Along a logistic curve with $0 < P_0 < K/2$ , growth is fastest when $P$ passes $K/2$ , at the time $t = (\ln A)/r$ , and that point is the inflection point of the S-shaped graph.

In plain words Growth is slow when there are few individuals to reproduce and slow again when the habitat is nearly full. It is fastest in between, at half the carrying capacity.

Example With $r = \tfrac{1}{5}$ and $K = 100$ , the maximum natural growth rate is $rK/4 = 5$ population units per time unit, at $P = 50$ . The curve $P(t) = 100/(1 + 4e^{-t/5})$ reaches $50$ when $4e^{-t/5} = 1$ , at $t = 5\ln 4 \approx 6.93$ , and its slope there is exactly $5$ . Do not confuse this with the per-capita rate, which is largest, close to $r = \tfrac{1}{5}$ , when $P$ is small.

Constant harvesting

Definition Constant harvesting removes a fixed amount $H \ge 0$ per unit time, whatever the size of the population, which gives

P' = rP\left(1 - \frac{P}{K}\right) - H.

The equilibria are the populations where natural growth equals the harvest, $rP(1 - P/K) = H$ . When $0 < H < rK/4$ there are two of them,

P_{\pm} = \frac{K}{2}\left(1 \pm \sqrt{1 - \frac{4H}{rK}}\right),

with $0 < P_- < K/2 < P_+ < K$ . The upper one is asymptotically stable and the lower one is unstable.

In plain words Harvesting shifts the growth parabola down by $H$ . The equilibria sit where the lowered parabola crosses zero, so the stable level drops below $K$ , and a second, unstable equilibrium appears below $K/2$ .

Example Take $r = \tfrac{1}{5}$ , $K = 100$ and $H = \tfrac{16}{5} = 3.2$ . Then $4H/(rK) = 0.64$ and the square root is $0.6$ , so $P_- = 20$ and $P_+ = 80$ . The rate factors as

P' = -\frac{1}{500}(P - 20)(P - 80),

which is negative below $20$ , positive between $20$ and $80$ and negative above $80$ . The arrows of the phase line point toward $80$ from both sides and away from $20$ on both sides.

Threshold equilibrium

Definition A threshold equilibrium is an unstable equilibrium that separates starting values with different long-run outcomes. In the harvested model with $0 < H < rK/4$ , the lower equilibrium $P_-$ is a threshold. Solutions with $P_0 > P_-$ tend to $P_+$ , and solutions with $0 < P_0 < P_-$ decrease and reach $P = 0$ at a finite time. The derivative test agrees, since the rate has derivative $r(1 - 2P/K)$ , which is positive at $P_- < K/2$ and negative at $P_+ > K/2$ .

In plain words A population just above the threshold recovers toward the upper equilibrium, while one just below it declines to zero. The threshold is the edge of collapse, not a safe level to aim for.

Example With $H = \tfrac{16}{5}$ as above, the derivative of the rate is $0.12$ at $20$ and $-0.12$ at $80$ . A start at $30$ rises toward $80$ . A start at $10$ falls, and because the rate increases with $P$ on $[0, 50]$ , it is at most its starting value $-1.4$ for as long as $P \le 10$ . So the population falls by at least $1.4$ per time unit and reaches $0$ before $t = 10/1.4 \approx 7.14$ .

Critical harvesting rate

Definition The critical harvesting rate of the logistic model is $H_c = rK/4$ , the maximum natural growth rate. It is the largest harvest the model can sustain at an equilibrium, often called the maximum sustainable yield. At $H = H_c$ the two equilibria merge at $K/2$ and

P' = -\frac{r}{K}\left(P - \frac{K}{2}\right)^2,

so $K/2$ is semistable. Starts above $K/2$ decrease toward it, and starts below it decrease away from it and reach $0$ at a finite time. For $H > H_c$ there is no equilibrium at all, since $P' \le rK/4 - H < 0$ for every $P$ , and every solution with $P_0 \ge 0$ reaches $0$ by the time $P_0/(H - rK/4)$ .

In plain words Harvesting at the maximum natural growth rate leaves no margin. The only equilibrium sits at half capacity and attracts only from above, so a shortfall below it is never repaired, and harvesting more than the maximum empties the population from every start.

Example With $r = \tfrac{1}{5}$ and $K = 100$ the critical rate is $H_c = 5$ , and at $H = 5$

P' = -\frac{1}{500}(P - 50)^2.

A start at $70$ decreases toward $50$ . A start at $40$ gives $1/(P - 50) = t/500 - 1/10$ , so it reaches $0$ at $t = 40$ . With $H = 6$ the rate is at most $5 - 6 = -1$ , so a start at $80$ reaches $0$ within $80$ time units.

Physical domain

Definition The physical domain of a model is the set of states it is meant to describe. For a population it is $P \ge 0$ , and $P$ itself is a continuous stand-in for a count. The differential equation makes sense for every real $P$ , but a solution is a forecast only while it stays in the physical domain, and stability statements are made in forward time relative to that domain. Without harvesting, $P = 0$ is an equilibrium, so a solution with $P_0 \ge 0$ never becomes negative. With $H > 0$ the rate at $P = 0$ is $-H < 0$ , so solutions cross into negative values. The model must then stop at depletion, the first time $P$ reaches $0$ , or be replaced by a removal rule that depends on the available stock.

In plain words A formula does not know that fish cannot be negative. When a solution reaches zero the population is gone, and whatever the formula says after that moment is not a prediction.

Example With $r = \tfrac{1}{5}$ , $K = 100$ , $H = \tfrac{16}{5}$ and $P_0 = 10$ , partial fractions and separation give

\frac{P - 80}{P - 20} = 7e^{-3t/25}.

The population reaches $0$ when the left side equals $4$ , at $t = \tfrac{25}{3}\ln\tfrac{7}{4} \approx 4.66$ , and the model stops there. Followed past that time, the formula gives negative values that tend to $-\infty$ as $t \to \tfrac{25}{3}\ln 7 \approx 16.2$ , which describes nothing real.

Check yourself

A population follows $P' = 0.5P(1 - P/1000)$ with $t$ in years and $P(0) = 100$ . Find $P(t)$ , the maximum natural growth rate, and the time at which growth is fastest.

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Here $A = 1000/100 - 1 = 9$ , so

P(t) = \frac{1000}{1 + 9e^{-t/2}}.

The maximum natural growth rate is $rK/4 = 0.5 \cdot 1000/4 = 125$ per year, at $P = 500$ . The curve reaches $500$ when $9e^{-t/2} = 1$ , at $t = 2\ln 9 \approx 4.39$ years.

With $r = \tfrac{1}{5}$ and $K = 100$ , a manager sets $H = 4.2$ and plans to hold the population at $30$ , because $30$ is an equilibrium. What goes wrong?

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The rate factors as $P' = -\tfrac{1}{500}(P - 30)(P - 70)$ , so the equilibria are $30$ and $70$ , as the formula $50(1 \pm 0.4)$ also gives. The derivative of the rate, $\tfrac{1}{5}(1 - P/50)$ , is $0.08 > 0$ at $30$ and $-0.08 < 0$ at $70$ . So $30$ is an unstable threshold. A population nudged up to $35$ climbs toward $70$ , and one nudged down to $25$ falls and reaches $0$ at $t = 12.5\ln(27/7) \approx 16.9$ . The stable level for this harvest is $70$ .

Keep $r = \tfrac{1}{5}$ and $K = 100$ but harvest $H = 6$ . Show that the population declines from every start, and find when a population starting at $80$ is depleted.

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Completing the square gives

\begin{aligned} P' &= -\frac{(P - 50)^2 + 500}{500} \\[0.4em] &\le -1, \end{aligned}

so there is no equilibrium and every solution decreases by at least $1$ per time unit. Separating the variables gives

\frac{dP}{(P - 50)^2 + 500} = -\frac{dt}{500}.

An antiderivative of the left side is $\frac{1}{\sqrt{500}}\arctan\frac{P - 50}{\sqrt{500}}$ , and $\sqrt{500} = 10\sqrt{5}$ . Integrating from $P = 80$ down to $P = 0$ gives the depletion time

t = 10\sqrt{5}\,(\alpha + \beta) \approx 46.5,

where $\alpha = \arctan(3/\sqrt{5}) \approx 0.930$ and $\beta = \arctan\sqrt{5} \approx 1.150$ . This agrees with the bound of $80$ time units from the first step.

Logistic Growth and Harvesting

Key terms

Logistic model

Maximum natural growth rate

Constant harvesting

Threshold equilibrium

Critical harvesting rate

Physical domain

Check yourself

How this lesson connects

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