Identify carrying capacity and equilibria

Where does the population stop changing?

The curve shows how fast the number of fish in a lake changes. The horizontal axis is the number of fish \(P\), and the vertical axis is the rate \(\cy{\frac{dP}{dt}}\).

The curve is green where the rate is positive, because the lake gains fish, and red where the rate is negative, because it loses fish.

Move the orange marker to 400 fish. The rate is 60 fish per year, the highest value on the curve.

Move the marker to 0 fish. The rate is 0, because an empty lake stays empty.

Move the marker to 800 fish. The curve crosses the axis there, so the rate is 0 again and the number of fish stays at 800.

Two populations with rate zero

The graph now has no numbers. It shows any population whose rate has this shape, a pattern called logistic growth.

The curve meets the axis at two populations, 0 and \(K\). At each of them the rate is 0, so the population does not change.

A population that does not change is an equilibrium, and the constant function that describes it is an equilibrium solution.

The number \(K\) is the carrying capacity, the largest population that the environment supports in the long run. Move the slider for \(K\). The right-hand crossing moves with it, and the left-hand crossing stays at 0.

The slider for \(k\) sets how fast a small population grows. Move it. The curve becomes taller or shorter, and the crossings stay in place.

Notation

1. The logistic differential equation has this form, where \(P\) is the population, \(t\) is the time, and \(k\) and \(K\) are positive constants.

   \[\frac{dP}{dt} = k\,P\left(1 - \frac{P}{K}\right)\]

2. For a population \(P > 0\), divide both sides by \(P\). The left side is the relative growth rate, the rate divided by the population.

   \[\frac{1}{P}\frac{dP}{dt} = k\left(1 - \frac{P}{K}\right)\]

   If \(P\) is small, the bracket is close to 1, so the constant \(k\) is the relative growth rate of a small population.

3. The bracket falls to 0 as \(P\) rises to \(\cy{K}\), so the relative growth rate reaches 0 there.

4. To find the equilibria, set the rate equal to 0.

   \[kP\left(1 - \frac{P}{K}\right) = 0\]

5. A product is 0 only when one factor is 0. The constant \(k\) is positive, so either \(P = 0\) or the bracket is 0.

   \[P = 0 \qquad\text{or}\qquad 1 - \frac{P}{K} = 0\]

6. The equilibria are \(P = 0\) and \(P = \cy{K}\). The carrying capacity is the positive equilibrium.

When it fails

The expanded equation describes the same curve, but it does not show \(K\). The number in front of \(P^2\) is \(\frac{k}{K}\), not \(K\).

To read \(K\), factor out the first term so that the bracket has the form \(1 - \frac{P}{K}\). For \(0.05P - 0.0002P^2\), this gives \(0.05P\left(1 - \frac{P}{250}\right)\), because \(\frac{0.0002}{0.05} = \frac{1}{250}\), so \(K = 250\).

Example 1: Read the capacity from the equation

A deer population on an island satisfies \(\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{600}\right)\). Find the carrying capacity and the equilibria.

1. Compare the bracket \(1 - \frac{P}{600}\) with \(1 - \frac{P}{K}\). The numbers in the equation are \(k = 0.2\) and \(K = \cy{600}\).

2. The carrying capacity is 600 deer. To find the equilibria, set the rate equal to 0.

   \[0.2\,P\left(1 - \frac{P}{600}\right) = 0\]

3. The factor \(P\) is 0 when \(P = 0\), and the bracket is 0 when \(P = 600\).

   \[\begin{aligned} P &= 0 \\ 1 - \frac{P}{600} = 0 \;&\Rightarrow\; P = 600 \end{aligned}\]

4. The equilibria are \(P = 0\) and \(P = 600\). The curve meets the axis at these two populations.

Example 2: An equation that hides the capacity

A yeast culture satisfies \(\dfrac{dP}{dt} = 0.01P(500 - P)\). Find all equilibria and the carrying capacity.

1. Set the rate equal to 0.

   \[0.01\,P\,(500 - P) = 0\]

2. A tempting wrong move: Divide both sides by \(P\). This leaves \(0.01(500 - P) = 0\), so \(P = 500\) is the only equilibrium.

   \[P = 500\ \text{only}\]

   The curve meets the axis at two populations, not one. Dividing by \(P\) assumes that \(P\) is not 0, so that step hides the solution \(P = 0\). An empty culture stays empty.

3. Do not divide by \(P\). Set each factor equal to 0.

   \[P = 0 \qquad\text{or}\qquad 500 - P = 0\]

4. The equilibria are \(P = 0\) and \(P = 500\). The carrying capacity is the positive one, \(K = \cy{500}\).

Practice

A population satisfies \(\dfrac{1}{P}\dfrac{dP}{dt} = 0.9 - 0.002P\). Find the equilibria and the carrying capacity.

The graph shows the rate, which is \(P\) times the right side of the equation. Move the slider to the population, other than 0, where the orange bar has no height.

Questions

The graph shows the rate \(\dfrac{dP}{dt}\) of the number of fish \(P\) in a reservoir. The equation is \(\dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{1200}\right)\).

1. At which populations does the number of fish stay constant?

   • 0 and 1200
   • 1200 only
   • 600 only
   • 0, 600 and 1200
   Answer

   0 and 1200

2. What is the carrying capacity?

   • 1200
   • 600
   • 75
   • 0
   Answer

   1200

3. A population satisfies \(\dfrac{dP}{dt} = 0.12P - 0.0004P^2\). What is its carrying capacity?

   • 300
   • 0.12
   • 2500
   • 0.0004
   Answer

   300

Summary

You write the equation as \(kP\left(1 - \frac{P}{K}\right)\), read the carrying capacity \(K\), and set the rate equal to 0 to find the equilibria 0 and \(K\).

This works because a product is 0 only when a factor is 0, and the bracket reaches 0 at the capacity.

You use it to find the capacity from an equation in any form and to list every constant solution.

This is one of five skills in topic 7.9 of Which & Why: AP Calculus.