AP Calculus BC glossary

Logistic differential equation

Also called: Logistic equation

The logistic differential equation, dP/dt = kP times the quantity 1 minus P over L, has two equilibrium solutions: P = 0 and P = L. Their signs settle the rest. A population below L rises toward it, one above L falls toward it, and no solution curve ever crosses L. See logistic growth for the model itself.

\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)

For the model itself and its S-shaped solution curve, see the logistic growth page; what follows is the phase line. Setting $dP/dt=0$ gives the equilibrium solutions $P=0$ and $P=L$ . For $0<P<L$ both factors are positive, so $dP/dt>0$ and $P$ climbs; for $P>L$ the bracket turns negative, so $dP/dt<0$ and $P$ falls. Both arrows point at $L$ , which makes $P=L$ stable and $P=0$ unstable. Since $P=L$ is itself a solution and two solutions of this equation cannot meet, no curve crosses the capacity: a population below it stays below, and one above it stays above.

\frac{d^2P}{dt^2}=\frac{d}{dt}\left[k\left(P-\frac{P^2}{L}\right)\right]=k\left(1-\frac{2P}{L}\right)\frac{dP}{dt}

That product controls the concavity. For a start below the carrying capacity the solution curve is S-shaped and its inflection point is exactly where $P=L/2$ , since $1-2P/L$ changes sign there while $dP/dt$ does not. A population starting above $L$ decreases to $L$ with no inflection point at all: $dP/dt$ and $1-2P/L$ are then both negative, so the second derivative stays positive and the curve is concave up the whole way down. With $k=0.04$ , $L=500$ and $P(0)=600$ it starts at $0.269$ and shrinks toward zero without changing sign.

The mistake

Treating $P=0$ and $P=L$ as the same kind of equilibrium. Both make $dP/dt$ zero, but a population nudged off $P=0$ runs away from it while one nudged off $P=L$ comes back, so $P=0$ is unstable and $P=L$ is stable. Only $P=L$ shows up as the horizontal asymptote of the solution curves.

Appears in: Unit 7: Differential Equations