Logistic Growth

Exponential growth cannot continue forever. A bacterial colony eventually runs out of nutrients. A social media platform eventually runs out of new users. A disease eventually runs out of susceptible people. When growth is fast at first but slows as a limit is approached, the pattern is logistic growth — the most realistic model for bounded population dynamics.

The Logistic Model

The logistic growth function is:

$P(t) = \frac{L}{1 + be^{-kt}}$

where:

• $P(t)$ = the quantity at time $t$
• $L$ = the carrying capacity (the maximum sustainable value)
• $b$ = a constant determined by the initial condition
• $k$ = the growth rate constant ($k > 0$)

Key Properties

Initial value: At $t = 0$:

$P(0) = \frac{L}{1 + b}$

So $b = \frac{L}{P(0)} - 1 = \frac{L - P(0)}{P(0)}$.

Long-term behavior: As $t \to \infty$, $e^{-kt} \to 0$, so:

$P(t) \to \frac{L}{1 + 0} = L$

The population approaches the carrying capacity but never exceeds it.

Early behavior: When $P$ is much smaller than $L$, the denominator is large and growth is approximately exponential. The logistic curve looks like exponential growth at the start.

Inflection point: The growth rate switches from increasing to decreasing at $P = \frac{L}{2}$, which occurs at:

$t_{\text{inflection}} = \frac{\ln b}{k}$

At this point, the population is growing fastest. After this, growth slows as the population approaches $L$.

The Shape of the S-Curve

The logistic curve has a distinctive S-shape (also called a sigmoid):

1. Slow start — the population is small and growth is slow in absolute terms
2. Rapid acceleration — growth appears exponential as the population is still far from the carrying capacity
3. Inflection point — maximum growth rate, at $P = L/2$
4. Deceleration — growth slows as resources become scarce
5. Leveling off — the population asymptotically approaches $L$

Logistic vs. Exponential Growth

The key comparison:

Every real-world population eventually transitions from exponential-like growth to logistic behavior when resources become limited.

Building a Logistic Model from Data

To determine $L$, $b$, and $k$ from data, you typically need at least three data points (or two data points plus knowledge of $L$).

When the Carrying Capacity Is Known

If you know $L$ and two data points, you can find $b$ and $k$.

Example 1: A lake can sustain at most 10,000 fish. Currently (at $t = 0$), there are 500 fish. After 3 years, there are 2,000 fish. Find the logistic model.

Find $b$: $P(0) = 500$ and $L = 10000$.

$b = \frac{L - P(0)}{P(0)} = \frac{10000 - 500}{500} = 19$

Find $k$: Use $P(3) = 2000$.

$2000 = \frac{10000}{1 + 19e^{-3k}}$

$1 + 19e^{-3k} = 5$

$19e^{-3k} = 4$

$e^{-3k} = \frac{4}{19} = 0.2105$

$-3k = \ln(0.2105) = -1.558$

$k = 0.5193$

Model: $P(t) = \frac{10000}{1 + 19e^{-0.5193t}}$

Inflection point: $t = \frac{\ln 19}{0.5193} = \frac{2.944}{0.5193} = 5.67$ years, at $P = 5000$ fish.

Example 2: Technology Adoption

A new app has a potential market of 2 million users ($L = 2{,}000{,}000$). At launch, it has 10,000 users. After 6 months, it has 200,000 users.

$b = \frac{2{,}000{,}000 - 10{,}000}{10{,}000} = 199$

$200{,}000 = \frac{2{,}000{,}000}{1 + 199e^{-6k}}$

$1 + 199e^{-6k} = 10$

$199e^{-6k} = 9$

$e^{-6k} = \frac{9}{199} = 0.04523$

$k = \frac{-\ln(0.04523)}{6} = \frac{3.096}{6} = 0.5160$

Model: $P(t) = \frac{2{,}000{,}000}{1 + 199e^{-0.5160t}}$ (months)

When does the app reach 1.5 million users?

$1{,}500{,}000 = \frac{2{,}000{,}000}{1 + 199e^{-0.5160t}}$

$1 + 199e^{-0.5160t} = \frac{4}{3}$

$199e^{-0.5160t} = \frac{1}{3}$

$e^{-0.5160t} = \frac{1}{597}$

$t = \frac{\ln(597)}{0.5160} = \frac{6.392}{0.5160} = 12.4 \text{ months}$

Real-World Application: Epidemic Modeling

The logistic model is foundational in epidemiology. In a population of $N$ susceptible individuals, the number of infected people often follows:

$I(t) = \frac{N}{1 + be^{-kt}}$

where $L = N$ (everyone eventually gets infected in a simple model without interventions).

Example 3: In a school of 800 students, 5 students have the flu on Monday. By Friday (4 days later), 80 students are infected. Model the spread.

$b = \frac{800 - 5}{5} = 159$

$80 = \frac{800}{1 + 159e^{-4k}}$

$1 + 159e^{-4k} = 10$

$159e^{-4k} = 9$

$e^{-4k} = 0.05660$

$k = \frac{-\ln(0.05660)}{4} = \frac{2.872}{4} = 0.7180$

Model: $I(t) = \frac{800}{1 + 159e^{-0.7180t}}$

When will half the school (400) be infected?

$t = \frac{\ln(159)}{0.7180} = \frac{5.069}{0.7180} = 7.06 \text{ days}$

By the following Monday (day 7), roughly half the school is infected. The inflection point — the day with the most new cases — is also around day 7.

Peak daily new infections: At the inflection point, the rate of change is:

$P'(t_{\text{inflection}}) = \frac{kL}{4} = \frac{0.7180 \times 800}{4} = 143.6 \text{ students per day}$

This peak rate is a critical number for school administrators planning responses.

Fitting When the Carrying Capacity Is Unknown

When $L$ is not known in advance, you need at least three data points and must solve a more complex system. A common approach:

1. Plot the data and estimate $L$ from the apparent upper limit
2. Use graphing technology or regression software to fit $L$, $b$, and $k$ simultaneously

For hand calculation, if you have three points $(t_1, P_1)$, $(t_2, P_2)$, $(t_3, P_3)$ with equally spaced times, you can use the formula:

$L = \frac{2P_1 P_2 P_3 - P_2^2(P_1 + P_3)}{P_1 P_3 - P_2^2}$

(provided $P_1 P_3 \neq P_2^2$, which fails only if the data is purely exponential).

Example 4: Data: $P(0) = 100$, $P(5) = 400$, $P(10) = 750$.

$L = \frac{2(100)(400)(750) - 400^2(100 + 750)}{(100)(750) - 400^2}$

$= \frac{60{,}000{,}000 - 160{,}000 \times 850}{75{,}000 - 160{,}000}$

$= \frac{60{,}000{,}000 - 136{,}000{,}000}{-85{,}000}$

$= \frac{-76{,}000{,}000}{-85{,}000} = 894.1$

So $L \approx 894$. Then $b = \frac{894 - 100}{100} = 7.94$ and $k$ can be found from the second data point.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The logistic model $P(t) = \frac{L}{1 + be^{-kt}}$ describes bounded growth with a carrying capacity $L$
• The S-curve has four phases: slow start, rapid acceleration, deceleration, and leveling off
• The inflection point at $P = L/2$ and $t = \frac{\ln b}{k}$ is where growth is fastest
• Unlike exponential growth, logistic growth is sustainable and realistic for populations with limited resources
• To build a model: find $b$ from the initial condition ($b = \frac{L - P_0}{P_0}$), then find $k$ from a second data point
• Applications include population biology, epidemic modeling, technology adoption, and market saturation

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