The Logistic Model

You’ll be able to

Interpret the logistic differential equation and its carrying capacity
Identify the population level of fastest growth
Determine the long-run limit of a logistic solution

Growth that levels off

Exponential growth is unrealistic forever — resources run out. The logistic model corrects this by slowing growth as the population P approaches a carrying capacity M, the maximum the environment can sustain. The logistic differential equation dP/dt = kP(1 − P/M) behaves like exponential growth when P is small (the factor 1 − P/M ≈ 1) but throttles toward zero growth as P → M. The solution is the familiar S-shaped (sigmoidal) curve.

Logistic differential equation

dP/dt = kP(1 − P/M)

M is the carrying capacity. Growth rate is near-exponential for small P and falls to 0 as P → M.

Fastest growth at half capacity

A signature result: the logistic population grows fastest when P = M/2, exactly half the carrying capacity. At that point dP/dt reaches its maximum, and the solution curve has an inflection point where it switches from concave up to concave down. Below M/2 growth accelerates; above M/2 it decelerates. This "half the carrying capacity" fact is one of the most tested single facts in Unit 5.

The long-run limit

As t → ∞, every logistic solution starting with 0 < P < M approaches the carrying capacity M: lim(t→∞) P(t) = M. The carrying capacity is a stable equilibrium — populations below it rise toward it, and populations above it fall toward it. So regardless of the exact initial size, the long-term population settles at M. You can read M straight off the equation as the value that makes the factor (1 − P/M) equal zero.

Worked example

A population follows dP/dt = 0.03P(1 − P/500). Find the carrying capacity, the population of fastest growth, and lim(t→∞) P(t).

1.Compare to dP/dt = kP(1 − P/M): here k = 0.03 and M = 500, so the carrying capacity is M = 500.
2.The population grows fastest at P = M/2 = 500/2 = 250.
3.Since 0 < P < M leads to P → M over time, the long-run limit is 500.
4.At P = 250 the solution has its inflection point (maximum growth rate).

Answer: Carrying capacity M = 500; fastest growth at P = 250 (= M/2); and lim(t→∞) P(t) = 500.

Watch out

The maximum population is the carrying capacity M, but the maximum growth rate occurs at M/2. Confusing "where growth is fastest" (P = M/2) with "the largest the population gets" (P = M) is the classic logistic trap.

Checkpoint

For the logistic equation dP/dt = 0.2P(1 − P/800), at what population is the growth rate the greatest?

Checkpoint

For a logistic model with carrying capacity M = 1000 and initial population P(0) = 50, what is lim(t→∞) P(t)?

On the exam

You can often answer logistic questions without solving the differential equation. Read M directly from the equation as the P that zeroes the (1 − P/M) factor, then use the two facts: fastest growth at M/2, and the long-run limit is M.

Answer the 2 checkpoints as you read.