Logistic growth

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You have likely studied exponential growth and even modeled populations using exponential functions. In this section we'll look at the differential equations that lead to exponential growth models, then refine those models to include some pressure for populations not to grow past a certain limit.

You could think of that limit like the carrying capacity of land: A certain region simply won't support unlimited growth because as one population grows, its resources diminish.

Just as a preview, here's a comparison of exponential and logistic growth curves with some features highlighted (right).

Exponential growth is unchecked growth. Exponential functions aren’t realistic models of population growth and other phenomena.

Exponential growth

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The underlying idea behind exponential growth is that the growth of a population is directly proportional to the number of individuals in it. We model a statement like that as

$$\frac{dn}{dt} = kn$$

The derivative is the change in population (n) with time, and k is a constant that would be a characteristic of the specific population – a proportionality constant. We can separate the variables and integrate,

$$\int \, \frac{dn}{n} = \int \, k \, dt$$

to find the general solution:

$$ln(n) = kt + C$$

These can always be expressed as exponential functions by solving for n(t),

$$e^{ln(n)} = e^{kt + C}$$

and using the laws of exponents to turn the additive constant of integration (C) into a multiplicative one.

"Slope fields"

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One way to display the concept of moving from a general solution to a differential equation to a specific solution is the so-called slope field. A slop field is calculated by plugging (n, t) grid points into the differential equation and plotting the direction of the derivative. The equation doesn't have to be solved to calculate a slope field. On the slope field below, I've drawn the specific solution that would result from our boundary condition (0, 0.5) and the solution that would result from the boundary condition (1, 1.5). Each is just the slope field curve that goes through that point.

The logistic differential equation

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The logistic differential equation recognizes that there is some pressure on a population as it grows past some point, that the presence of other members, competition for resources, &c., can slow down growth. It looks like this:

$$\frac{dn}{dt} = kn (1 - n)$$

Here we've taken the maximum population to be one, which we can change later. The rate of growth (dn/dt) is proportional to both the population (n) and the closeness of the population to its maximum (1-n). The k is the usual proportionality constant. Now let's separate variables and integrate this equation:

$$\int \, \frac{1}{n (1 - n)} \, dn = \int \, kdt$$

The first thing we'll need to do is tackle the integrand on the left. We'll do it by rational decomposition, writing the integrand as

$$\frac{1}{n (1 - n)} = \frac{A}{n} + \frac{B}{(1 - n)}$$

Our goal is to find the A and B that work for this rational function. Multiplying both sides by n(n - 1) gives

$$1 = A(1 - n) + Bn$$

Because this expression has to work for all n, we can set n = 1 to get B = 1, and n = 0 to get A = 1. You should confirm for yourself that those work and that we've successfully converted the integrand into a sum of two fractions. That gives us a new integrated equation:

$$\int \, \frac{dn}{n} + \int \, \frac{dn}{1 - n} = \int \, k\, dt$$

Which is easy to integrate:

$$ln|n| + ln|1 - n| = kt + C$$

Now we can rearrange using the properties of logs:

$$ln \left| \frac{n}{1 - n} \right| = kt + C$$

If we raise each side as a power of e, we get

$$\frac{n}{1 - n} = A e^{kt}$$

where I have let e^kt+C = e^kte^C, and renamed the constant e^C = A. Now we'll do some algebra to solve for n. First multiply both sides by 1 - n:

$$n = A e^{kt} - n A e^{kt}$$

Gather the terms containing n:

$$n + nA e^{kt} = A e^{kt}$$

An example: Spreading a rumor

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One clever example of logistic growth is the spreading of a rumor in a population. Suppose that one person knows a secret, and once a day, anyone who knows the secret can share it with one other person, but without knowing whether that person already knows it. Well, early on, it's unlikely that a teller will run across someone who already knows the secret, but later, when more people know, it's less likely to find a person who doesn't know.

In this example, I assumed we have a group of 20 people, and that person #1 knows the secret to begin with. Then, on each "round," I generated a random number (using a spreadsheet) between 1 and 20, to choose whom to tell the secret next. the results are in the table below.

In round one, 1 told 4. In round two, 1 and 4 told 20 and 6, for a total of four secret-knowers. In round three, those four told four new people to increase the total to 8.

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Transformations of the logistic equation

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Inflection point

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One important point on the logistic curve is the inflection point, the point where the curvature of the graph changes from concave-upward to concave-downward.

The inflection point marks the point where the rate of growth is no longer accelerating, but begins to decelerate.

To find the coordinates of the inflection point, we begin with a logistic function with parameters L, b and k, as shown. We'll rewrite it with a negative exponent so we can easily use the chain rule:

$$f(t) = \frac{L}{1 + b e^{-kt}} = L(1 + be^{-kt})^{-1}$$

The first derivative is

$$ \begin{align} f'(t) &= -L(1 + be^{-kt})^{-2} (-bk \, e^{-kt}) \\[5pt] &= bkL(e^{-kt})(1 + ke^{-kt})^2. \end{align}$$

The second derivative is

$$ \begin{align} f''(t) &= bkL [(-k e^{-kt})(1 + b e^{-kt})^{-2} \\[5pt] &\phantom{000} + e^{-kt}(-2)(1 + b e^{-kt})^{-2}(-kb e^{-kt})] \\[5pt] &= bkL \left[ \frac{-ke^{-kt}}{(1 + be^{-kt})} + \frac{2bk e^{-2kt}}{(1 + be^{-kt})^3} \right] \\[5pt] &= bkL \left[ \frac{2bk e^{-2kt} - ke^{-kt}(1 + be^{-kt})}{(1 + be^{-kt})^3}\right] \\[5pt] \end{align}$$

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