The logistic growth model is a more realistic population model than exponential growth. It accounts for a carrying capacity $M$ — the maximum population the environment can sustain. Initially, the population grows exponentially, but as it approaches $M$, growth slows and eventually stabilizes.
🎯 In this section you will learn
- The logistic differential equation: $\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)$
- The meaning of $k$ (growth rate) and $M$ (carrying capacity)
- The logistic growth solution: $P(t) = \frac{M}{1 + Ae^{-kt}}$
- How to analyze the logistic model (inflection point, limiting behavior)
📌 The Logistic Equation
$$ \frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right) $$
where $k > 0$ is the growth rate, $M > 0$ is the carrying capacity
💡 Interpreting the Equation
- When $P$ is small, $1 - P/M \approx 1$, so $\frac{dP}{dt} \approx kP$ (exponential growth).
- As $P$ approaches $M$, $1 - P/M \to 0$, so $\frac{dP}{dt} \to 0$ (growth slows).
- The population stabilizes at $P = M$ (carrying capacity).
📌 The Logistic Growth Solution
$$ P(t) = \frac{M}{1 + Ae^{-kt}} $$
where $A = \frac{M - P_0}{P_0}$ and $P_0 = P(0)$ is the initial population
💡 Derivation (Separable DE)
$\frac{dP}{P(1 - P/M)} = k dt$ → Using partial fractions, integrate to get the solution above.
A population grows logistically with carrying capacity $M = 1000$, growth rate $k = 0.2$, and initial population $P_0 = 100$. Find $P(t)$.
①
Compute $A$
$A = \frac{M - P_0}{P_0} = \frac{1000 - 100}{100} = \frac{900}{100} = 9$
②
Write $P(t)$
$P(t) = \frac{1000}{1 + 9e^{-0.2t}}$
For the logistic model $P(t) = \frac{500}{1 + 4e^{-0.3t}}$, find the carrying capacity, initial population, and population after 10 years.
①
Carrying capacity
$M = 500$ (the numerator)
②
Initial population
$P(0) = \frac{500}{1 + 4} = \frac{500}{5} = 100$
③
Population at $t=10$
$P(10) = \frac{500}{1 + 4e^{-3}} \approx \frac{500}{1 + 4(0.0498)} = \frac{500}{1 + 0.1992} = \frac{500}{1.1992} \approx 416.9$
📌 Key Features of Logistic Growth
| Feature | Value | Significance |
|---|
| Carrying capacity | $\lim_{t \to \infty} P(t) = M$ | Population stabilizes at $M$ |
| Initial population | $P(0) = \frac{M}{1+A}$ | Given by $P_0$ |
| Maximum growth rate | Occurs when $P = M/2$ | Inflection point of the logistic curve |
| Growth rate at $P = M/2$ | $\frac{dP}{dt} = \frac{kM}{4}$ | Fastest growth happens at half the carrying capacity |
📌 Inflection Point
The logistic curve has an inflection point at $P = M/2$. This is where the population is growing most rapidly. For $P < M/2$, growth accelerates; for $P > M/2$, growth decelerates.
For $P(t) = \frac{1000}{1 + 9e^{-0.2t}}$, find the time when the population is growing fastest.
①
Fastest growth at $P = M/2$
$M = 1000$, so fastest growth when $P = 500$
②
Solve $P(t) = 500$
$500 = \frac{1000}{1 + 9e^{-0.2t}}$ → $1 + 9e^{-0.2t} = 2$ → $9e^{-0.2t} = 1$ → $e^{-0.2t} = \frac{1}{9}$ → $-0.2t = \ln(1/9) = -\ln 9$ → $t = \frac{\ln 9}{0.2} = 5\ln 9 \approx 5(2.197) \approx 10.99$
⚠️ Common Mistakes
- Confusing $k$ and $r$: In the logistic equation $\frac{dP}{dt} = rP(1 - P/K)$, $r$ is the growth rate, $K$ is the carrying capacity.
- Misidentifying carrying capacity: In $P(t) = \frac{M}{1+Ae^{-kt}}$, the carrying capacity is $M$ (the numerator).
- Forgetting that growth rate is not constant: The logistic model has variable growth rate, unlike exponential.
🔍 Key Takeaways
- Logistic DE: $\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)$.
- Solution: $P(t) = \frac{M}{1 + Ae^{-kt}}$ with $A = \frac{M - P_0}{P_0}$.
- Carrying capacity $M$: the limiting population as $t \to \infty$.
- Fastest growth occurs at $P = M/2$ (inflection point).
- The logistic model is more realistic than exponential for populations with limited resources.
← Back to Module Page