4.4 The Logistic Equation

Big idea

Exponential growth comes from the assumption that each individual contributes the same amount of growth regardless of how crowded things are. That assumption is fine for a while and absurd forever: nothing grows without limit in a finite environment. The logistic model keeps the proportional-growth idea but lets the per-capita rate fall off as the population rises, hitting zero at a ceiling called the carrying capacity.

The result is an equation whose right side depends only on the population and not on the clock, so the entire long-run behavior can be read off from the sign of one expression. Where the right side is positive the population climbs; where it is negative the population falls; where it is zero the population stands still. Arrange those signs on a vertical line and you have a phase line, which classifies every possible future without solving anything.

Doing the sign analysis first is worth the minute it costs. It tells you that the carrying capacity attracts every positive starting population, that extinction is a solution but a fragile one, and that a population starting above the ceiling declines toward it rather than crashing. All of that is available before any integration happens, and it is the part you should remember when the algebra fades.

The equation is also separable, so a closed form exists. Partial fractions turn the integral into two logarithms, and unwinding them produces the S-shaped curve that the phase line predicted: slow at first, fastest in the middle, and flattening as it approaches the ceiling. Knowing both the picture and the formula lets each check the other.

Decoder

A population whose per-capita growth rate declines linearly to zero at a fixed ceiling increases toward that ceiling from below and decreases toward it from above, approaching it but never exceeding it from below.

The phrase carrying the content is “per-capita”. Divide the logistic equation by $P$ and the right side becomes $r\left(1 - P/K\right)$, the growth rate per individual. It equals $r$ when the population is negligible and falls steadily to $0$ at $P = K$. Everything else follows from that one linear decline.

Two equilibria appear because the right side is a quadratic in $P$ with roots at $0$ and $K$. Between them the product $rP(1 - P/K)$ is positive, so solutions rise; above $K$ the second factor turns negative, so solutions fall. That makes $K$ stable and $0$ unstable, and stability here means exactly what the arrows say: nearby solutions move toward a stable equilibrium and away from an unstable one.

The mistake to avoid is reading the inflection point as something special about the population size rather than about the growth rate. At $P = K/2$ the population is not at a maximum; the growth rate is. Past that point the population still increases, just less quickly each year, which is what the flattening top of the curve shows.

Definitions and results

The logistic equation. With $r > 0$ the intrinsic growth rate and $K > 0$ the carrying capacity,

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

The factor $1 - P/K$ is the correction that shuts growth off as $P$ approaches $K$.

Autonomous equations. An equation $\frac{dy}{dt} = f(y)$ whose right side does not mention $t$. Its direction field is unchanged by horizontal shifts, so if $y(t)$ is a solution so is $y(t - c)$ for every $c$.

Equilibria. The constant solutions are the roots of $f$. For the logistic equation these are $P = 0$ and $P = K$.

Phase line and stability. Mark the roots of $f$ on a vertical axis and record the sign of $f$ between them. An equilibrium with $f > 0$ below and $f < 0$ above is stable; with the signs reversed it is unstable; with the same sign on both sides it is semi-stable, attracting from one side and repelling from the other.

Closed-form solution. Separating and using $\frac{K}{P(K - P)} = \frac{1}{P} + \frac{1}{K - P}$ gives $\ln\frac{P}{K - P} = rt + C$, and solving for $P$ with $P(0) = P_0$ yields

$$ P(t) = \frac{P_0 K}{P_0 + (K - P_0)e^{-rt}} $$

Long-run behavior. For any $P_0 > 0$, the exponential in the denominator vanishes as $t$ grows and $P(t) \to K$. The approach is from below when $P_0 < K$ and from above when $P_0 > K$.

Inflection point. Differentiating the equation and substituting gives

$$ \frac{d^2P}{dt^2} = r\,\frac{dP}{dt}\left(1 - \frac{2P}{K}\right) $$

so for $0 < P < K$ the concavity changes exactly at $P = K/2$, where the growth rate reaches its largest value $rK/4$.

Comparison with exponential growth. When $P$ is small relative to $K$ the correction factor is close to $1$ and the logistic curve is nearly the exponential $P_0e^{rt}$. The two separate once $P$ becomes a serious fraction of $K$.

Worked examples

Reading the future off the phase line

Take $r = 0.2$ per year and $K = 1000$, so $\frac{dP}{dt} = 0.2\,P\left(1 - \frac{P}{1000}\right)$.

Set the right side to zero: $P = 0$ or $P = 1000$. Now test the sign in each interval. At $P = 100$ the value is $0.2(100)(0.9) = 18 > 0$. At $P = 1200$ it is $0.2(1200)(-0.2) = -48 < 0$. So on $(0, 1000)$ solutions increase and on $(1000, \infty)$ they decrease.

Both arrows point at $P = 1000$, so the carrying capacity is stable. At $P = 0$ the arrow just above points away, so extinction is unstable: any positive population, however small, grows. This settles the long-run behavior for every starting value without a single integration.

Solving an initial-value problem

Keep $r = 0.2$ and $K = 1000$, and start with $P_0 = 100$. The closed form gives

$$ P(t) = \frac{100 \cdot 1000}{100 + 900e^{-0.2t}} = \frac{1000}{1 + 9e^{-0.2t}} $$

Check the endpoints. At $t = 0$ the denominator is $1 + 9 = 10$, so $P(0) = 100$. As $t$ grows the exponential dies and $P(t) \to 1000$, matching the phase line.

When does the population reach half the carrying capacity? Set $P = 500$:

$$ 1 + 9e^{-0.2t} = 2 \quad \Longrightarrow \quad e^{-0.2t} = \frac19 \quad \Longrightarrow \quad t = \frac{\ln 9}{0.2} = 10\ln 3 \approx 11.0 $$

So it takes about $11$ years. For contrast, pure exponential growth at the same rate would reach $500$ when $100e^{0.2t} = 500$, at $t = 5\ln 5 \approx 8.0$ years. Crowding has already cost three years by the halfway point.

Where growth is fastest

Using the same numbers, the growth rate as a function of population is $G(P) = 0.2P\left(1 - P/1000\right)$, a downward parabola in $P$ with roots at $0$ and $1000$. Its maximum is at the midpoint $P = 500$, where

$$ G(500) = 0.2(500)(0.5) = 50 $$

individuals per year. That agrees with the general statement $rK/4 = 0.2(1000)/4 = 50$. Since the previous example put $P = 500$ at about $t = 11$, the curve has its steepest point there and its inflection point there as well, which is why the graph looks like an S rather than an arc.

Classifying equilibria for a different model

Consider $\dfrac{dy}{dt} = y(y - 2)(y - 5)$, an autonomous equation that is not logistic.

The equilibria are $y = 0$, $y = 2$ and $y = 5$. Test one point in each interval. At $y = -1$: $(-1)(-3)(-6) = -18 < 0$. At $y = 1$: $(1)(-1)(-4) = 4 > 0$. At $y = 3$: $(3)(1)(-2) = -6 < 0$. At $y = 6$: $(6)(4)(1) = 24 > 0$.

So the arrows read down, up, down, up from bottom to top. At $y = 0$ the arrows point away on both sides, so it is unstable. At $y = 2$ they point toward it on both sides, so it is stable. At $y = 5$ they point away on both sides, so it is unstable. A solution starting anywhere in $(0, 5)$ approaches $2$; one starting above $5$ runs off upward; one starting below $0$ runs off downward.

Practice

Start with the model itself. These ask you to identify $r$ and $K$ from an equation or from data, to write the closed-form solution, and to answer questions about a population at a given time.

Then the qualitative side. Given an autonomous equation, find the equilibria, build the phase line, classify each equilibrium, and predict the limit of a solution from its starting value.

Quiz

Five items on carrying capacity, phase lines and stability, the closed-form logistic solution, and the point of fastest growth.