Logistic Differential Equations

By the end of this lesson, you will be able to analyze and solve logistic growth models, identify equilibria and inflection points.

The Logistic Model

Exponential growth ($dP/dt = kP$) predicts unbounded growth — unrealistic. Logistic growth adds a carrying capacity $K$:

$$\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)$$

Interpretation:

• When $P \ll K$: $(1-P/K) \approx 1$ → approximately exponential growth
• When $P = K$: $dP/dt = 0$ → population is at the carrying capacity (stable equilibrium)
• When $P > K$: $dP/dt < 0$ → population decreases back toward $K$

Equilibria

$P = 0$: unstable equilibrium — any positive population grows away from 0.

$P = K$: stable equilibrium — all solutions with $P > 0$ approach $K$ as $t\to\infty$.

The Inflection Point

The growth rate $dP/dt$ is maximized when the population is at $P = K/2$.

At $P = K/2$: $\frac{d^2P}{dt^2} = 0$ and growth switches from accelerating to decelerating.

Interpretation: the population grows fastest when it's at half the carrying capacity. Beyond $K/2$, growth slows as resources become limited.

Solution of the Logistic DE

The logistic DE is separable. Its general solution is:

$$P(t) = \frac{K}{1 + Ae^{-kt}}$$

where $A = \frac{K - P_0}{P_0}$ and $P_0 = P(0)$.

As $t\to\infty$: $e^{-kt}\to 0$ → $P(t)\to K$ ✓

Example

A population grows logistically with $K = 1000$, $k = 0.3$, and $P(0) = 100$.

$A = \frac{1000-100}{100} = 9$

$P(t) = \frac{1000}{1+9e^{-0.3t}}$

When does the population reach $K/2 = 500$?

$500 = \frac{1000}{1+9e^{-0.3t}}$ → $1+9e^{-0.3t} = 2$ → $9e^{-0.3t} = 1$ → $e^{-0.3t} = 1/9$

$-0.3t = \ln(1/9) = -\ln 9$ → $t = \frac{\ln 9}{0.3} \approx 7.3$ years

What is the population at $t = 10$?

$P(10) = \frac{1000}{1+9e^{-3}} \approx \frac{1000}{1+9(0.0498)} \approx \frac{1000}{1.448} \approx 691$ individuals

Comparing Exponential and Logistic Growth

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