1.2 The Logistic Equation

Exponential growth cannot last. A population that doubles every hour eventually runs out of food, space, or patience, and a model that ignores this fact answers questions about the far future with nonsense. The classical fix is to make the per-capita growth rate fall as the population rises. With carrying capacity $8$ (in whatever units the habitat supports) and growth rate $b > 0$ , the logistic equation is

$\frac{dP}{dt} = b\, P \left( 1 - \frac{P}{8} \right).$

Read it before solving it. When $P$ is small the factor $1 - P/8$ is close to $1$ and the equation is nearly $P' = bP$ — exponential growth. When $P$ is near $8$ the factor is near $0$ and growth stalls. Between them, the growth rate $bP(1 - P/8)$ is largest at $P = 4$ , halfway up. And there are two population levels where nothing changes at all: $P = 0$ and $P = 8$ , the equilibria, where the right-hand side vanishes and a population that starts there stays there forever.

The equation can be solved in closed form, and the solutions are

$P(t) = \frac{8}{1 + a e^{-b t}},$

where the constant $a$ encodes the starting value: $P(0) = 8/(1 + a)$ , so large $a$ means a small starting population and $a$ near zero means starting almost at capacity.

A red S-shaped logistic curve rising from its starting value toward an amber horizontal line at height 8; the extinction level 0 lies along the horizontal axis. The slider a moves the starting population between about 0.4 and 7.3, always below 8, and the slider b sharpens or relaxes how fast the curve climbs toward the amber capacity line, which it approaches but never crosses. — Logistic solutions $P(t) = 8/(1 + a e^{-bt})$ with carrying capacity $8$ (the amber line) and the extinction equilibrium $P = 0$ (the horizontal $t$ -axis). The slider $a$ sets the starting population and $b$ sets the growth rate.

Explore

Observe. Set $a = 7$ , $b = 0.8$ , so $P(0) = 1$ . The curve is S-shaped: slow, then fast, then slow. At roughly what height is the curve steepest? Compare with the halfway level $P = 4$ .
Predict. If you increase $a$ to $20$ , the starting population drops to about $0.4$ . Will the curve still reach toward the amber line at $8$ , or level off lower? Decide, then drag the slider.
Verify. With any settings, zoom your attention to the right edge of the window. Does the red curve ever touch or cross the amber line at $8$ ? Explain from the equation why it cannot: what would $dP/dt$ be at the moment of touching?
Observe. Drag $a$ down to $0.1$ , so the population starts near $7.3$ , close to capacity. The S-shape disappears. Which part of the S did this starting value skip?
Predict, then verify. The equilibrium $P = 0$ is also a solution. Judging from the curves with large $a$ , does a tiny population move toward $0$ or away from it? What does that say about the stability of the extinction equilibrium?

Connect

The two equilibria behave differently, and the difference is the most important qualitative fact about the model. Solutions that start just above $0$ flee from it: $P = 0$ is unstable, an equilibrium in the sense of a pencil balanced on its tip. Solutions from every positive starting value climb or descend toward $8$ and flatten against it: $P = 8$ is stable, an equilibrium like a marble at the bottom of a bowl. The sliders let you verify the first fact, and the approach to $8$ from below; the slope field later in this section shows a solution descending onto $8$ from above. Both facts were readable from the equation alone — the sign of $bP(1 - P/8)$ is positive for $0 < P < 8$ and negative for $P > 8$ , so arrows on a vertical $P$ -axis point away from $0$ and toward $8$ . That sign analysis, not the closed formula, is what generalizes to equations nobody can solve.

The Slope Field

Everything above began by solving the equation. The picture below never does. A slope field draws, at every point $(t, P)$ of the plane, a short segment whose slope is $bP(1 - P/8)$ — the equation's instruction at that point, available without any solution formula. A solution curve is then simply a curve that stays tangent to the segments it passes through, and the two equilibria appear as horizontal rivers the flow converges to or splits away from.

A grid of short gray slope marks fills the plane, each tilted to the angle the logistic equation dictates at that point: the rows of marks on the amber lines at heights 8 and 0 lie flat, marks tilt upward in the band between them, most steeply around height 4, and tilt downward above 8. A red solution curve starting at height 1 climbs through the field in an S shape and flattens against the upper amber line, while a blue curve starting at height 10 sinks onto it from above; both stay tangent to every mark they cross. The slider b steepens or relaxes the tilt of the whole field, and the two curves re-thread it accordingly. — The slope field of $dP/dt = bP(1 - P/8)$ , with two solutions threading it: red from $P(0) = 1$ , blue from $P(0) = 10$ . The amber lines mark the equilibria $P = 8$ and $P = 0$ . The segments are flat along both, tilt up between them, and tilt down above $8$ . The slider $b$ re-tilts the whole field.

Observe. Look only at the segments, ignoring the curves. Along the amber line $P = 8$ every segment is flat, and the same is true along $P = 0$ . What does a flat segment say about $dP/dt$ there, and why does that make those two levels equilibria?
Predict. Put a fingertip anywhere in the band $0 < P < 8$ and follow the tilts to the right. Do you end up at $8$ , at $0$ , or somewhere else? Now start above $8$ and do the same.
Verify. The red curve starts at $P(0) = 1$ and the blue at $P(0) = 10$ . Check that each is tangent to every segment it crosses, and that they approach the line $P = 8$ from opposite sides without touching it.
Observe. Within the band $0 < P < 8$ , where do the segments climb most steeply? Compare with your answer from the first Explore list — the halfway level $P = 4$ — and with the fact that the red curve is steepest exactly where it crosses that level.
Predict, then verify. Drag $b$ up toward $2$ . Does the shape of the field change, or only the steepness of its segments? What does that say about which features of this model depend on $b$ and which do not?

The equilibria are exactly the levels where the field goes flat, where the right-hand side is zero: $P = 0$ and $P = 8$ , found with no solution formula involved. The level where the field climbs most steeply is found the same way, one derivative later — where the slope's own derivative in $P$ vanishes.

4

Where the growth rate is largest ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Differentiating $bP(1 - P/8)$ with respect to $P$ gives $b(1 - P/4)$ , and the growth rate peaks where that vanishes. The CAS solves it here: the fastest growth happens at $P = 4$ , halfway to capacity — which is why the red curve is steepest as it crosses the middle of the band, and why the S-shape has its bend exactly there.

The certificate carries one declared hypothesis: $b \neq 0$ . That is not a technicality the engine is being fussy about — if $b$ were zero the equation would hold at every population, there would be no growth to maximize, and " $P = 4$ " would be a wrong answer dressed as a complete one. The section fixed $b > 0$ as the growth rate in its first paragraph, so the hypothesis is the model's own; the engine simply refuses to use it without being told, and the check runs through its full case analysis rather than the generic shortcut.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.