The Logistic Equation

The model $\frac{dP}{dt} = rP$ predicts endless exponential growth. Real populations run out of food or room and level off. The logistic equation builds that limit in, and you can solve it with separation of variables and partial fractions.

Por qué funciona

Let the growth rate per individual, $\frac{1}{P}\frac{dP}{dt}$, start at $r$ when the population is small and fall in a straight line to 0 when $P$ reaches a carrying capacity $K$: $\frac{1}{P}\frac{dP}{dt} = r\left(1 - \frac{P}{K}\right)$. That gives $$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right).$$ Reading the right side: it is 0 at $P = 0$ and $P = K$ (equilibria); positive between them, so the population rises toward $K$; negative above $K$, so it falls back. As a function of $P$ it is a downward parabola, largest at $P = \frac{K}{2}$: that is where growth is fastest.

Idea clave

$\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ has the solution $$P(t) = \frac{K}{1 + Ae^{-rt}}, \qquad A = \frac{K - P_0}{P_0}.$$ Equilibria: $P = 0$ and $P = K$. Growth is fastest when $P = \frac{K}{2}$.

Where the formula comes from

Separate: $\frac{dP}{P(1 - P/K)} = r\,dt$. Partial fractions give $$\begin{aligned} &\frac{1}{P(1 - P/K)} \\ &= \frac{K}{P(K - P)} \\ &= \frac{1}{P} + \frac{1}{K - P}. \end{aligned}$$ Integrate: $\ln|P| - \ln|K - P| = rt + C$, so $\frac{P}{K - P} = Be^{rt}$. Solving for $P$ gives $P = \frac{K}{1 + Ae^{-rt}}$ with $A = \frac{1}{B}$, and $t = 0$ gives $A = \frac{K - P_0}{P_0}$.

Ejemplo resuelto: Solving a logistic model

Solve $\frac{dP}{dt} = 0.5P\left(1 - \frac{P}{100}\right)$ with $P(0) = 20$.

$K = 100$, $r = 0.5$, and $A = \frac{100 - 20}{20} = 4$.
$P(t) = \frac{100}{1 + 4e^{-0.5t}}$.
Check: $P(0) = \frac{100}{5} = 20$, and $P(t) \to 100$ as $t \to \infty$.

Respuesta: $P(t) = \frac{100}{1 + 4e^{-0.5t}}$

Logistic solutions with $K = 10$: from below they rise in an S-shape and level off at 10; from above they fall to 10. The slopes are steepest along $P = 5$.

Ejemplo resuelto: When is growth fastest?

For the model above, when is the population growing fastest?

Fastest growth happens at $P = \frac{K}{2} = 50$.
Solve $\frac{100}{1 + 4e^{-0.5t}} = 50$: $1 + 4e^{-0.5t} = 2$, so $e^{-0.5t} = \frac{1}{4}$.
$t = \frac{\ln 4}{0.5} = 2\ln 4 \approx 2.77$.

Respuesta: $t = 2\ln 4 \approx 2.77$

Ejemplo resuelto: Reading K from an expanded form

A population follows $\frac{dP}{dt} = 0.03P - 0.0001P^{2}$. Find $r$ and $K$.

Factor out $0.03P$: $$\begin{aligned} &0.03P\left(1 - \frac{0.0001}{0.03}P\right) \\ &= 0.03P\left(1 - \frac{P}{300}\right). \end{aligned}$$
So $r = 0.03$ and $K = 300$.

Respuesta: $r = 0.03$, $K = 300$

Error común

Using $A = \frac{K}{P_0}$ instead of $A = \frac{K - P_0}{P_0}$. With $K = 100$, $P_0 = 20$ that gives $A = 5$ and $P(0) = \frac{100}{6} \ne 20$. Quick check: put $t = 0$ into your formula; you must get $P_0$ back.

Where the formula comes from

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