The Logistic Equation
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 4.4: “The Logistic Equation” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/4-4-the-logistic-equation |
| Textbook used in class | Stewart, Calculus, Section 9.4: “Models for Population Growth” (Examples 1-4) |
Quick Reference
Logistic equation: $$\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right),$$ where $M > 0$ is the carrying capacity and $k > 0$ is the growth rate.
Solution: $$P(t) = \frac{M}{1 + Ae^{-kt}}, \qquad A = \frac{M - P_0}{P_0}.$$
Equilibria: $P = 0$ (unstable) and $P = M$ (stable).
Motivation
Pure exponential growth $dP/dt = kP$ is unrealistic for populations in a limited environment: it predicts $P \to \infty$, but food, space, and other resources cap any real population. The logistic model adds the factor $(1 - P/M)$ to slow growth as $P$ approaches the carrying capacity $M$.
When $P \ll M$ (population far below capacity), $(1 - P/M) \approx 1$ and growth is nearly exponential. When $P \approx M$, $(1-P/M) \approx 0$ and growth nearly stops. The result is an S-shaped curve that starts exponential, bends, and levels off at $M$.
Key Concept: Solving by Separation
The logistic equation is separable. Separate and use partial fractions: $$\frac{dP}{P(1 - P/M)} = k\,dt.$$
Partial fractions: $\dfrac{1}{P(1-P/M)} = \dfrac{1}{P} + \dfrac{1/M}{1-P/M}$.
Integrate both sides: $$\ln|P| - \ln\left|1 - \frac{P}{M}\right| = kt + C \implies \ln\left|\frac{P}{M - P}\right| = kt + C.$$
Exponentiate and solve for $P$: $$\frac{P}{M-P} = Be^{kt} \implies P = \frac{M}{1 + (1/B)e^{-kt}}.$$
Setting $t = 0$ gives $B = P_0/(M - P_0)$, so $A = 1/B = (M - P_0)/P_0$.
Worked Example
A population grows logistically with $M = 1000$, $k = 0.4$, and $P(0) = 100$. Find $P(t)$ and the time at which $P = 500$ (half the carrying capacity).
$A = (1000 - 100)/100 = 9$.
$$P(t) = \frac{1000}{1 + 9e^{-0.4t}}.$$
Set $P = 500$: $$500 = \frac{1000}{1 + 9e^{-0.4t}} \implies 1 + 9e^{-0.4t} = 2 \implies e^{-0.4t} = \frac{1}{9}.$$
$$-0.4t = -\ln 9 \implies t = \frac{\ln 9}{0.4} = \frac{2\ln 3}{0.4} \approx 5.49 \text{ time units.}$$
The inflection point (fastest growth rate) occurs at $P = M/2 = 500$, so this is also the time of maximum growth rate.
the carrying capacity $M$ is the maximum possible population. The carrying capacity $M$ is the equilibrium population, not an absolute ceiling. In the logistic model, if the population starts above $M$ (say, after an unusual event), the equation predicts $dP/dt < 0$, so the population decreases back toward $M$. The model does not forbid $P > M$; it predicts that populations above $M$ will decline and populations below $M$ will grow, both approaching $M$ in the long run. $M$ is stable, not a hard wall.
Leveled Practice
Problem 1. A logistic population has $M = 500$ and $P(0) = 50$. If $P(10) = 200$, find $k$.
Show answer
$A = (500-50)/50 = 9$.
$P(t) = 500/(1 + 9e^{-kt})$.
$P(10) = 200$: $1 + 9e^{-10k} = 2.5$, so $e^{-10k} = 1.5/9 = 1/6$.
$-10k = -\ln 6 \implies k = \frac{\ln 6}{10} \approx 0.179$.