Lesson · Separable Equations, Growth and Decay · Shifted and bounded growth

Logistic Growth

A population cannot grow exponentially for ever. The logistic equation y′ = r·y·(1 − y/K) multiplies the exponential rate r·y by a factor that is near 1 for a small population and falls to 0 at the carrying capacity K. Its equilibria are 0 and K, the first repelling and the second attracting; the growth rate is largest at K/2; and Euler's exact steps show a column of fractions whose digits double at every step, so the lab prints them until a stated budget and then says so.

Open the interactive lesson Practice

y′ = r·y·(1 − y/K) equilibria y = 0 and y = K r = 1, K = 4: f(y) = y·(1 − y/4) f′(y) = 1 − y/2 = 0 at y = 2 = K/2 steps from 1: 11/8, 935/512, …

Growth that levels off

Core ideas

What this lesson establishes

Three ideas: the factor that bounds the growth, the two equilibria and which way the rest of the line moves, and where the growth is fastest.

Exponential growth times a bounding factor

The logistic equation is y′ = r·y·(1 − y/K). For y small compared with K the factor 1 − y/K is near 1 and the equation is nearly y′ = r·y. As y approaches K the factor falls to 0 and so does the rate.

Two equilibria, and the direction between them

The right side is zero at y = 0 and at y = K. Between them it is positive, so a population there grows; above K it is negative, so a population there shrinks. Starts anywhere above 0 move towards K, and a start just above 0 moves away from 0: the first equilibrium repels and the second attracts.

The growth is fastest at half the capacity

The rate f(y) = r·y·(1 − y/K) has f′(y) = r·(1 − 2y/K), which is zero at y = K/2. There f is largest, r·K/4, and the curve y(t) is steepest. Before that the curve bends upwards and afterwards downwards.

The material

A population that levels off

The equation, its equilibria read from the sign of the right side, the point of fastest growth from the rate of that rate, and exact Euler steps with the lab's stated limit.

Exponential growth is the right description of a small population with room and food, and it is wrong for a large one, because it has no limit on how big the population can become. Something has to slow the growth as the population fills its surroundings. The simplest correction multiplies the exponential rate by a factor that drops to zero at a ceiling.

Definition · The logistic equation

With a growth rate r > 0 and a carrying capacity K > 0, the logistic equation is y′ = r·y·(1 − y/K).

The carrying capacity is the level at which the growth rate is zero. It is a number the model is given, not one it derives.

Take r = 1 and K = 4, so that f(y) = y·(1 − y/4). Expanded, that is y − y²/4: the exponential term y and a correction that grows with the square of y. Here is the rate at several populations.

y f(y) = y·(1 − y/4) 0 0 1 3/4 2 1 3 3/4 4 0 5 −5/4

The rate rises from 0, peaks at y = 2 with f = 1, and falls to 0 at 4. Above 4 it is negative, so a population above the capacity shrinks back to it. An exponential at y = 3 would be growing at 3, and the logistic is growing at 3/4: the same equation with a quarter of the rate.

Which way the line moves

An equilibrium is a value where f(y) = 0, so that a population placed there does not change. Here those are 0 and 4. Between 0 and 4 the rate is positive and above 4 negative, so every start above 0 heads for 4. The lab labels this 0 unstable; 4 stable, in words the next course defines carefully: nearby starts move away from 0 and towards 4.

Where the growth is fastest

The rate f(y) is largest where its own rate of change is zero. Here f′(y) = 1 − y/2, which is zero at y = 2, half of the capacity. For the curve y(t) this is the steepest point, and the chain rule shows it is also where the curve changes from bending up to bending down, because y″ = f′(y)·y′. The lab's inflection tile prints y = 2.

Exact steps, and the lab's limit

Start at y(0) = 1 with h = 1/2. The first step is 1 + (1/2)·(3/4) = 11/8, a little under the 3/2 that exponential growth would give. The second is 935/512, which is about 1.82617. The right side is quadratic, so the numerator and denominator roughly double their number of digits at every step. The lab prints exact steps until a number has more than 2000 digits and then stops, and says that it has: from this start it prints 11 exact steps, then digit budget. It does not round to carry on.

y′ = y·(1 − y/4), y(0) = 1, h = 1/2 y₁ = 1 + (1/2)·(3/4) = 11/8 y₂ = 11/8 + (1/2)·(231/256) = 935/512 digits roughly double at each step

The population cannot be followed further as exact fractions, and nothing in the equation stopped. The stopping is the lab's, and it is the honest alternative to hiding the growth in the digits by rounding at every step.

Example · Above the capacity

Start at 6. Then f(6) = 6·(1 − 3/2) = −3, so the first step is 6 + (1/2)·(−3) = 9/2: the population falls towards 4, and the lab prints a limit of → 4. The third preset is the same kind of equation with K = 10, whose steepest point is at 5.

Interactive

Autonomous equations

Equilibria, the phase line, and where solutions go

The equilibria of a logistic equation and its first steps

The first tile lists where the right side is zero, the second says what kind of equilibrium each is, and the inflection tile is where the right side is largest. The picture shows the exact Euler polygon from each start, with the equilibria as horizontal lines. The steps tile says when the exact column stopped on the digit budget.

Equation

f(y) in y' = f(y)

Starting values y(0)

Step h

Steps n12

Show

Equilibria—

Types—

f′ at each—

The first start goes—

Exact Euler steps, first start—

Inflection levels—

f is a polynomial in y of degree 4 at most: y^2 - 1, y(1 - y/4), y^3 - 4y^2 + 3y.

Method

Reading a logistic equation

Five moves, done before any step is computed.

Identify r and K

Write the right side as r·y·(1 − y/K). The factor that vanishes gives K; the coefficient of y near 0 gives r.

Find the equilibria

Solve f(y) = 0: y = 0 and y = K.

Read the direction between them

Evaluate f at one point in each region. Positive means y rises, negative that it falls. Mark which equilibria are approached.

Locate the steepest point

Solve f′(y) = 0: y = K/2, where f = r·K/4.

Step it and note the limit

Compute a few exact steps from the starting value and compare each with what exponential growth would have given. Report where the lab stopped.

Equilibria, direction and the steepest point

Score 0 / 0

Choose the answer you can justify.

The logistic equation with K = 4, from y(0) = 1

The equilibria and the steepest point come from the right side alone. The two Euler steps are exact fractions, with h = 1/2.

f(y) = y·(1 − y/4) = y − y²/4 f(y) = 0 at y = 0 and y = 4 f′(y) = 1 − y/2 = 0 at y = 2 = K/2 f(1) = 1·(3/4) = 3/4 y₁ = 1 + (1/2)·(3/4) = 11/8 f(11/8) = (11/8)·(21/32) = 231/256 y₂ = 11/8 + (1/2)·(231/256) = 935/512

The rate at the start is 3/4, and the exponential would have used 1. The difference is the factor 1 − y/4, which has already cost a quarter of the growth at y = 1. The steps rise towards 4 and the lab's picture shows them flattening as they near it.

The value 935/512 ≈ 1.82617 is rounded only to say how big it is; the lab prints the fraction. The next steps double the digits again until the budget, and the equilibria and the steepest point do not depend on any of them.

Common mistakes

Thinking the logistic curve is exponential growth that stops at KThe mistaken model is an exponential that runs until it hits K and is then switched off. The equation has no switch. The factor 1 − y/K acts at every population, trimming the rate smoothly: at y = 1 the rate is 3/4 and not 1, at y = 3 it is 3/4 and not 3. And a switch could not bring a population above K back down, which the equation does, with f(5) = −5/4.

Putting the steepest point at KAt y = K the rate is zero, which is the flattest the curve gets, not the steepest. The rate is largest where f′(y) = 0, which for this family is y = K/2: y = 2 for K = 4, as the inflection tile prints, with f(2) = 1 against f(3) = 3/4.

Reading the lab's stop as the population'sThe exact column stops after eleven steps with a message about the digit budget. That is the lab refusing to keep writing numbers with thousands of digits, not the solution leaving off. The equilibria, their kinds and the steepest point come from the right side and are exact whether or not any step is printed.

Check your understanding

✓

Finish when you can find the equilibria of a logistic equation, say which way the rest of the line moves, and locate its steepest growth.

You should be able to write y′ = r·y·(1 − y/K), solve f(y) = 0, read the direction from the sign of f in each region, solve f′(y) = 0 for the steepest point, compute an exact Euler step, and say what the lab does when the digits run out.

Note: A logistic population has a ceiling, and a harvest takes some of it away at a steady rate. “Harvesting and the Threshold” subtracts a constant from the equation and finds the rate beyond which the population cannot survive.