Logistic Growth and Resource Models, calculus lesson

A few fish released into an empty lake multiply quickly at first. As the lake fills up, food and space run short and the growth slows until the population levels off. Exponential growth cannot describe that slowing down, and the simplest equation that can is the logistic equation.

Key terms

Population growth

Definition In the exponential model a population $P(t)$ grows at a rate proportional to its size,

\frac{dP}{dt} = rP, \qquad P(0) = P_0,

with constant relative growth rate $r$ . The unique solution is $P(t) = P_0 e^{rt}$ . In any model $\frac{1}{P}\frac{dP}{dt}$ is called the per capita growth rate.

In plain words Each individual contributes the same amount of growth no matter how crowded it is. That works only while resources are plentiful.

Example With $P_0 = 100$ and $r = 0.5$ per year, the model predicts $100e^{5} \approx 14{,}841$ after 10 years and about $2.2$ million after 20 years. A real lake would run out of room long before that.

Logistic differential equation

Definition The logistic equation with growth rate $r > 0$ and carrying capacity $K > 0$ is

\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right).

It is separable. For $P(0) = P_0 > 0$ its unique solution is

P(t) = \frac{K}{1 + Ae^{-rt}}, \qquad A = \frac{K - P_0}{P_0},

defined for all $t \ge 0$ . When $0 < P_0 < K$ it is defined for all real $t$ and increases. When $P_0 < \frac{K}{2}$ its graph is S-shaped with the inflection point where $P = \frac{K}{2}$ , the population at which growth is fastest.

In plain words The per capita growth rate $r\left(1 - \frac{P}{K}\right)$ starts near $r$ and drops to zero as the population fills the available room.

Example Take $r = 0.5$ , $K = 1000$ , $P_0 = 100$ . Then $A = 9$ and

P(t) = \frac{1000}{1 + 9e^{-0.5t}}.

$t$ in years	$0$	$2$	$4$	$6$	$10$	$20$
$P(t)$	$100$	$232.0$	$450.9$	$690.6$	$942.8$	$999.6$

The population reaches $500$ when $9e^{-0.5t} = 1$ , that is at $t = 2\ln 9 \approx 4.39$ years.

Carrying capacity

Definition The carrying capacity is the constant $K$ in the logistic equation. It is the positive equilibrium, and for every $P_0 > 0$ the solution satisfies $\lim_{t \to \infty} P(t) = K$ .

In plain words It is the largest population the environment can support over the long run. Below it the population grows, above it the population shrinks.

Example With $K = 1000$ , a lake stocked with $P_0 = 1500$ fish has $A = \frac{1000 - 1500}{1500} = -\frac{1}{3}$ , so

P(t) = \frac{1000}{1 - \frac{1}{3}e^{-0.5t}}.

This starts at $1500$ and decreases toward $1000$ .

Stability of the equilibria

Definition For $g(P) = rP\left(1 - \frac{P}{K}\right)$ the equilibria are $P = 0$ and $P = K$ . Since $g'(P) = r - \frac{2rP}{K}$ , we have $g'(0) = r > 0$ and $g'(K) = -r < 0$ . So $0$ is unstable and $K$ is asymptotically stable.

In plain words A tiny population moves away from zero and grows. A population near capacity is pushed back toward capacity from either side.

Example On the phase line, $g(P) > 0$ for $0 < P < K$ and $g(P) < 0$ for $P > K$ . The arrows point up below $K$ and down above it, both toward $K$ , and away from $0$ on the positive side.

Harvesting

Definition With a constant harvest of $H \ge 0$ individuals per unit time the model becomes

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

Its equilibria solve $\frac{r}{K}P^2 - rP + H = 0$ , so

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

They exist exactly when $H \le \frac{rK}{4}$ . For $H < \frac{rK}{4}$ the larger one $P_+$ is asymptotically stable and the smaller one $P_-$ is unstable. The value $\frac{rK}{4}$ , the largest growth rate $g\left(\frac{K}{2}\right)$ , is the maximum sustainable yield.

In plain words Fishing removes a fixed amount every year. A moderate catch just lowers the level where the population settles. A catch larger than the population’s best possible growth makes it collapse.

Example With $r = 0.5$ and $K = 1000$ the maximum sustainable yield is $125$ per year. For $H = 90$ the number under the root is $1 - \frac{4 \cdot 90}{500} = 0.28$ , and

P_{\pm} = 500\left(1 \pm \sqrt{0.28}\right) \approx 764.6 \text{ and } 235.4.

A population above $235.4$ settles near $764.6$ . One that falls below $235.4$ declines to zero, so the lower equilibrium is a threshold.

Renewable resources

Definition A renewable resource is one that regrows, such as a fish stock or a forest, modeled by $\frac{dP}{dt} = g(P) - h(P)$ with natural growth $g$ and harvest rate $h$ . A yield is sustainable if it is the harvest rate at an asymptotically stable positive equilibrium. With proportional harvesting $h(P) = EP$ and $0 < E < r$ , the logistic model has an asymptotically stable equilibrium $P^*$ and a sustainable yield $Y(E)$ given by

P^* = K\left(1 - \frac{E}{r}\right), \qquad Y(E) = EP^* = EK\left(1 - \frac{E}{r}\right).

In plain words You can take a steady amount forever only if the resource grows back that much each year at the level where it settles.

Example With $r = 0.5$ , $K = 1000$ , the yield $Y(E) = 1000E(1 - 2E)$ is largest when $Y'(E) = 1000 - 4000E = 0$ , so $E = 0.25$ . The population then settles at $P^* = 500$ and the yield is $125$ per year, the same maximum as before.

Model limitations

Definition The logistic model assumes a continuous population with constant $r$ and $K$ , immediate response to crowding, no age structure, no randomness and a per capita growth rate that falls linearly in $P$ . Its predictions are only as good as these assumptions. Solutions are monotone, so the model cannot produce oscillations, and with harvesting it can produce negative values that have no meaning.

In plain words A model is a simplification. Use it for the trend and for what happens near equilibrium, and be suspicious of very small populations, long time spans and conclusions about single animals.

Example With $r = 0.5$ , $K = 1000$ and $H = 130$ , the right side satisfies $\frac{dP}{dt} \le 125 - 130 = -5$ for every $P$ . Starting from $P_0 = 1000$ , the population hits zero within $\frac{1000}{5} = 200$ years. After that the formula would continue to negative values, so the model must be stopped at extinction.

Check yourself

For $\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{400}\right)$ with $P(0) = 50$ , find $P(t)$ and the time when $P = 200$ .

Show the answer

Here $A = \frac{400 - 50}{50} = 7$ , so

P(t) = \frac{400}{1 + 7e^{-0.2t}}.

Setting $P = 200$ gives $7e^{-0.2t} = 1$ , so $t = 5\ln 7 \approx 9.73$ .

What is the largest constant harvest that the population in question 1 can sustain?

Show the answer

The maximum sustainable yield is $\frac{rK}{4} = \frac{0.2 \cdot 400}{4} = 20$ per unit time.

For $\frac{dP}{dt} = P(1 - P) - 0.16$ , find and classify the equilibria.

Show the answer

The equilibria solve $P^2 - P + 0.16 = 0$ , which gives $P = 0.2$ and $P = 0.8$ . With $g(P) = P - P^2 - 0.16$ we have $g'(P) = 1 - 2P$ . Since $g'(0.2) = 0.6 > 0$ , the equilibrium $0.2$ is unstable, and since $g'(0.8) = -0.6 < 0$ , the equilibrium $0.8$ is asymptotically stable.

Logistic Growth and Resource Models

Key terms

Population growth

Logistic differential equation

Carrying capacity

Stability of the equilibria

Harvesting

Renewable resources

Model limitations

Check yourself

How this lesson connects

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