Logistic Growth

Q: What is the logistic differential equation?

dP/dt = kP(1 − P/L), where k > 0 is the growth rate for small populations and L is the carrying capacity.

Quick answer

Exponential growth never stops, but real populations run out of space and food. The logistic equation dP/dt = kP(1 − P/L) builds in a carrying capacity L: growth is nearly exponential while P is small and slows to zero as P nears L. Without solving, the equation shows that P approaches L for any positive start, and that the population grows fastest when P = L/2. Separating variables with partial fractions gives the S-shaped solution P = L/(1 + Ae^(−kt)).

What you'll learn

Read the carrying capacity and growth rate from a logistic equation
Find the limiting value and the point of fastest growth without solving
Solve the logistic equation with partial fractions
Use the solution to answer questions in context

Limits to growth

The model $\tfrac{dP}{dt} = kP$ predicts growth without end, which no real population can sustain. The logistic equation adds a brake:

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

$L$ is the carrying capacity, the largest population the environment supports. While $P$ is small, the factor $1 - \tfrac{P}{L}$ is close to $1$ and growth is nearly exponential. As $P$ nears $L$ , that factor shrinks toward $0$ and growth stalls.

P(0) = 1
P(0) = 14
carrying capacity

Logistic growth toward a carrying capacity of 10

Reading the equation without solving it

The equation answers the main questions on its own:

Equilibria. $\tfrac{dP}{dt} = 0$ when $P = 0$ or $P = L$ .
Direction. For $0 < P < L$ the rate is positive, so $P$ grows. For $P > L$ it is negative, so $P$ falls.
Long run. Either way, a positive population approaches $L$ : $\lim_{t \to \infty} P(t) = L$ .

Why growth is fastest at half the capacity

The rate $kP\left(1 - \tfrac{P}{L}\right)$ is a product of two forces pulling in opposite directions. The factor $P$ drives growth: more individuals, more offspring. The factor $1 - \tfrac{P}{L}$ brakes it: less room, fewer resources. As a function of $P$ , the rate is a downward parabola with zeros at $0$ and $L$ , so it peaks halfway between them. Growth is fastest at $P = \tfrac{L}{2}$ , where the push and the brake balance.

That is also where the S-shaped curve changes from concave up to concave down. Differentiating the equation, $\tfrac{d^2P}{dt^2} = k\left(1 - \tfrac{2P}{L}\right)\tfrac{dP}{dt}$ , which changes sign at $P = \tfrac{L}{2}$ .

Solving the equation

Separate the variables and use partial fractions: $\tfrac{L}{P(L - P)} = \tfrac{1}{P} + \tfrac{1}{L - P}$ . Writing $1 - \tfrac{P}{L} = \tfrac{L - P}{L}$ ,

\int \left(\frac{1}{P} + \frac{1}{L - P}\right)dP = \int k\,dt \quad\Longrightarrow\quad \ln\frac{P}{L - P} = kt + C

for $0 < P < L$ . Solve for $P$ :

P(t) = \frac{L}{1 + Ae^{-kt}} \qquad\text{with}\qquad A = \frac{L - P_0}{P_0}

As $t \to \infty$ , $e^{-kt} \to 0$ and $P \to L$ , as the equation predicted.

Worked examples

Common mistakes

Practice problems

For $\tfrac{dP}{dt} = 0.5P\left(1 - \tfrac{P}{200}\right)$ , find the carrying capacity and the population at which growth is fastest.

Answer

$200$ and $100$

Full solution

$L = 200$ , and the fastest growth is at $\tfrac{L}{2}$ .
For $\tfrac{dP}{dt} = 0.2P - 0.0004P^2$ , find the carrying capacity and the fastest growth rate.

Answer

$L = 500$ ; the fastest rate is $25$

Full solution

$0.2P - 0.0004P^2 = 0.2P(1 - 0.002P) = 0.2P\left(1 - \tfrac{P}{500}\right)$ . At $P = 250$ , the rate is $0.2 \cdot 250 \cdot \tfrac{1}{2} = 25$ .
In problem 1, suppose $P(0) = 50$ . Find $\lim_{t \to \infty} P(t)$ .

Answer

$200$

Full solution

Any positive population approaches the carrying capacity.
In problem 1, suppose $P(0) = 300$ . Is the population rising or falling at first, and at what rate?

Answer

Falling, at $75$ per unit of time

Full solution

$\tfrac{dP}{dt} = 0.5 \cdot 300 \cdot \left(1 - \tfrac{300}{200}\right) = 150 \cdot (-0.5) = -75$ .
Solve $\tfrac{dP}{dt} = 0.4P\left(1 - \tfrac{P}{100}\right)$ with $P(0) = 20$ .

Answer

$P = \tfrac{100}{1 + 4e^{-0.4t}}$

Full solution

$L = 100$ , $k = 0.4$ , and $A = \tfrac{100 - 20}{20} = 4$ .
In problem 5, when does the population reach $50$ ?

Answer

$t = \tfrac{\ln 4}{0.4} \approx 3.47$

Full solution

$\tfrac{100}{1 + 4e^{-0.4t}} = 50$ gives $4e^{-0.4t} = 1$ , so $t = \tfrac{\ln 4}{0.4}$ .
In problem 5, find $P(5)$ .

Answer

About $64.9$

Full solution

$P(5) = \tfrac{100}{1 + 4e^{-2}} \approx \tfrac{100}{1.541} \approx 64.9$ .
In problem 5, what is the greatest rate of growth?

Answer

$10$ per unit of time

Full solution

The fastest growth is at $P = 50$ : $0.4 \cdot 50 \cdot \tfrac{1}{2} = 10$ .
Find the equilibrium solutions of $\tfrac{dP}{dt} = 0.1P\left(1 - \tfrac{P}{80}\right)$ . Which one do nearby solutions approach?

Answer

$P = 0$ and $P = 80$ ; solutions near $80$ approach it

Full solution

The rate is $0$ at $P = 0$ and $P = 80$ . Slightly above $0$ the rate is positive, so solutions move away from $0$ . Around $80$ , solutions below rise and solutions above fall, so they all approach $80$ .
For $\tfrac{dP}{dt} = 0.3P\left(1 - \tfrac{P}{600}\right)$ with $P(0) = 100$ , a student says the population grows fastest at the start, because it is growing. What went wrong?

Hint

Compare the rate at $P = 100$ with the rate at $P = 300$ .

Answer

Growing does not mean growing fastest. The rate is largest at $P = 300$ .

Full solution

At $P = 100$ the rate is $0.3 \cdot 100 \cdot \tfrac{5}{6} = 25$ . At $P = 300$ it is $0.3 \cdot 300 \cdot \tfrac{1}{2} = 45$ .

The rate keeps increasing until $P = \tfrac{L}{2} = 300$ , then decreases toward $0$ .

Frequently asked questions

What is the logistic differential equation?

dP/dt = kP(1 − P/L), where k > 0 is the growth rate for small populations and L is the carrying capacity.

What is the carrying capacity?

The value L that the population approaches as t → ∞, from below or from above, for any positive starting value.

When is logistic growth fastest?

When P = L/2. The rate kP(1 − P/L) is a downward parabola in P, and its peak is halfway between its zeros 0 and L.

What is the solution of the logistic equation?

P = L/(1 + Ae^(−kt)), where A = (L − P₀)/P₀ and P₀ is the starting population.

How do I read L from dP/dt = 0.6P − 0.0012P²?

Factor it as 0.6P(1 − 0.002P) = 0.6P(1 − P/500). The carrying capacity is 500.

What to learn next

Calculus with Parametric EquationsSlopes, tangent lines, second derivatives and arc length for curves given by x = f(t) and y = g(t), where dy/dx is dy/dt divided by dx/dt.

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