The logistic equation · Differential Equations

Intuition

A population cannot grow in proportion to itself for ever, because the habitat runs out. The cheapest honest repair is to multiply the growth rate by a factor that falls to zero as the population reaches what the habitat carries. The result is the most-studied equation in this chapter, and everything about it can be read from its phase line before a single integral is attempted.

A room filling with people. While it is nearly empty each new arrival brings friends and the crowd grows in proportion to itself; as the room fills, arrivals slow, and at capacity nobody else comes in. The growth rate is the exponential rate scaled by the fraction of the room still free.

Solutions of $y'=\tfrac45y(1-\tfrac y4)$ . From below the curve is S-shaped, fastest at half the capacity; from above it falls. Both approach the capacity and neither reaches it.

The same equation as a phase line: zero repels, the capacity attracts, and every positive start ends at the capacity. Everything the curves show is already here.

The equation, its equilibria and its curve

The logistic equation is $y'=ry\left(1-\tfrac{y}{K}\right)$ , with the intrinsic growth rate and the carrying capacity. It has equilibria at , unstable, and at , stable. Separation and partial fractions give the solution in closed form, but the phase line already says that every positive start ends at .

y'=ry\Bigl(1-\frac{y}{K}\Bigr)

What the model says

For $y$ small the bracket is near one and the equation is nearly $y'=ry$ : the model begins as exponential growth.
For $y$ near $K$ the bracket is near zero and growth stops. The capacity is where the habitat is full.

Applications

Practice

Two Equilibria, and Which Is Which

The logistic right-hand side vanishes at zero and at the capacity. The derivative test settles both.

\begin{aligned}&f(y)=ry\Bigl(1-\frac{y}{K}\Bigr)\\&f'(0)=r>0,\quad f'(K)=-r<0\end{aligned}

Try it

For $y'=0.5y(1-\tfrac{y}{100})$ , which equilibrium is stable?