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Population growth and the logistic equation

The growth of the earth's population has long been an important issue for humankind. Will the population continue to grow? Or will it perhaps level off at some point, and if so, when?

Introduction

The growth of the earth's population has long been an important issue for humankind. Will the population continue to grow? Or will it perhaps level off at some point, and if so, when? In this section, we look at two ways we may use differential equations to help us address these questions.

Before we begin, let's consider again two important differential equations that we have seen in earlier work this chapter.

Exploration

The earth's population

We will now begin studying the earth's population. In Activity, we consider some recent population data and explore an elementary model.

Our work in Activity shows that that the exponential model is fairly accurate for years relatively close to 2000. However, if we go too far into the future, the model predicts increasingly large rates of change, which causes the population to grow arbitrarily large. This does not make much sense since it is unrealistic to expect that the earth would be able to support such a large population.

The constant \(k\) in the differential equation has an important interpretation. Let's rewrite the differential equation \(\frac{dP}{dt} = kP\) by solving for \(k\), so that we have \[\begin{aligned}\end{aligned}\].

We see that \(k\) is the ratio of the rate of change to the population; in other words, it is the contribution to the rate of change from a single person. We call this the per capita growth rate.

In the exponential model we introduced in Activity, the per capita growth rate is constant. This means that when the population is large, the per capita growth rate is the same as when the population is small. It is natural to think that the per capita growth rate should decrease when the population becomes large, since there will not be enough resources to support so many people. We expect it would be a more realistic model to assume that the per capita growth rate depends on the population \(P\).

In the previous activity, we computed the per capita growth rate in a single year by computing \(k\), the quotient of \(\frac{dP}{dt}\) and \(P\) (which we did for \(t = 0\)). If we return to the data in Table and compute the per capita growth rate over a range of years, we generate the data shown in Figure, which shows how the per capita growth rate is a function of the population, \(P\).

From the data, we see that the per capita growth rate appears to decrease as the population increases. In fact, the points seem to lie very close to a line, which is shown at two different scales in Figure.

Looking at this line carefully, we can find its equation to be \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Solving the logistic differential equation

Since we would like to apply the logistic model in more general situations, we state the logistic equation in its more general form, \[\begin{aligned}\end{aligned}\].

The equilibrium solutions here are \(P=0\) and \(1-\frac PN = 0\), which shows that \(P=N\). The equilibrium at \(P=N\) is called the carrying capacity of the population for it represents the stable population that can be sustained by the environment.

We now solve the logistic equation. The equation is separable, so we separate the variables \[\begin{aligned}\end{aligned}\], and integrate to find that \[\begin{aligned}\end{aligned}\].

To find the antiderivative on the left, we use the partial fraction decomposition \[\begin{aligned}\end{aligned}\].

Now we are ready to integrate, with \[\begin{aligned}\end{aligned}\].

On the left, observe that \(N\) is constant, so we can remove a factor of \(\frac{1}{N}\) and antidifferentiate to find that \[\begin{aligned}\end{aligned}\].

Multiplying both sides of this last equation by \(N\) and using a rule of logarithms, we next find that \[\begin{aligned}\end{aligned}\].

From the definition of the logarithm, replacing \(e^C\) with \(C\), and letting \(C\) absorb the \(\pm\) that arises from the absolute value, we now know that \[\begin{aligned}\end{aligned}\].

The solution to the initial value problem \[\begin{aligned}\end{aligned}\], is \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Summary

If we assume that the rate of growth of a population is proportional to the population, we are led to a model in which the population grows without bound and at a rate that grows without bound.
By assuming that the per capita growth rate decreases as the population grows, we are led to the logistic model of population growth, which predicts that the population will eventually stabilize at the carrying capacity.

Practice (6)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

The logistic equation may be used to model how a rumor spreads through a group of people. Suppose that \(p(t)\) is the fraction of people that have heard the rumor on day \(t\). The equation \[\begin{aligned}\end{aligned}\] describes how \(p\) changes. Suppose initially that one-tenth of the people have heard the rumor; that is, \(p(0) = 0.1\).
1. What happens to \(p(t)\) after a very long time?
2. Determine a formula for the function \(p(t)\).
3. At what time is \(p\) changing most rapidly?
4. How long does it take before 80% of the people have heard the rumor?
Tunjukkan jawapan
1. The equilibrium solutions for this logistic equation are \(p = 0\) and \(p = 1\). Since \(p = 1\) is the carrying capacity and thus a stable equilibrium, \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\).
2. We know that the general solution to the logistic equation \(\frac{dP}{dt} = kP(N-P), \ P(0) = P_0\) is \[\begin{aligned}\end{aligned}\]. In the context of this problem, we know that \(k = 0.2\), \(N = 1\), and \(p(0) = 0.1\) and therefore the solution is \[\begin{aligned}\end{aligned}\].
3. We know that \(p\) is changing most rapidly when its derivative, \(p'\), is largest. Recalling the original differential equation, we know that \[\begin{aligned}\end{aligned}\] which shows that \(p'\) is a quadratic function of \(p\). That quadratic function has zeros at \(p = 0\) and \(p = 1\), so the vertex of this concave down quadratic function lies at \(p = \frac{1}{2}\), and at this value of \(p\) the rate of change of \(p\) is largest. Thus, we solve the equation \(p(t) = 0.5\) to find the time at which the population is increasing fastest. We see that \[\begin{aligned}\end{aligned}\] so \[\begin{aligned}\end{aligned}\] and \(e^{-0.2t} = 1/9\). It follows that \(t = -5 \ln(1/9) \approx 10.986\) days.
4. Since \(p(t)\) tells us the percentage of people who've heard the rumor, we need to solve the equation \(p(t) = 0.8\). Thus we have \[\begin{aligned}\end{aligned}\] so \[\begin{aligned}\end{aligned}\] and \(e^{-0.2t} = 0.25/9\). It follows that \(t = -5 \ln(0.25/9) \approx 17.19\) days.
Suppose that \(b(t)\) measures the number of bacteria living in a colony in a Petri dish, where \(b\) is measured in thousands and \(t\) is measured in days. One day, you measure that there are 6,000 bacteria and the per capita growth rate is 3. A few days later, you measure that there are 9,000 bacteria and the per capita growth rate is 2.
1. Assume that the per capita growth rate \(\frac{db/dt}{b}\) is a linear function of \(b\). Use the measurements to find this function and write a logistic equation to describe \(\frac{db}{dt}\).
2. What is the carrying capacity for the bacteria?
3. At what population is the number of bacteria increasing most rapidly?
4. If there are initially 1,000 bacteria, how long will it take to reach 80% of the carrying capacity?
Tunjukkan jawapan
1. Let \(y\) be the per capita growth rate as a function of \(b\). Using the given data in the problem statement, we know that this linear function passes through the points \((6000,3)\) and \((9000,2)\). The equation of this line is thus \[\begin{aligned}\end{aligned}\] so \(y = (3 + 2) - \frac{1}{3000}b = 5 - \frac{1}{3000}b\). Since the logistic equation comes from taking the product of the population at time \(t\) with its per capita growth rate at that time, we find that \[\begin{aligned}\end{aligned}\]
2. From the form of the logistic equation, we see that the carrying capacity for the bacteria is \(b = 15000\), which is a stable equilibrium for the population.
3. The bacteria is increasing most rapidly when \(\frac{db}{dt}\) is greatest. Since \[\begin{aligned}\end{aligned}\] is a quadratic function of \(b\), \(\frac{db}{dt}\) is maximized at the vertex of this concave down parabola, which occurs halfway between its zeros: \(b = 0\) and \(b = 15000\). Thus, when \(b = 7500\), the population is growing fastest. To find the exact \(t\)-value when this occurs, we'd need to know an initial condition like in the next part of the problem.
4. Using \(b(0) = 1000\), we solve the logistic equation IVP with \[\begin{aligned}\end{aligned}\]. Using \(N = 15000\), \(k = \frac{1}{3000}\), and \(b_0 = 1000\), we know from the standard form of the solution to the logistic equation that \[\begin{aligned}\end{aligned}\]. To find how long it takes the bacteria to reach 80% of the carrying capacity, we solve the equation \[\begin{aligned}\end{aligned}\]. That equation implies that \[\begin{aligned}\end{aligned}\], so \(14e^{-5t} + 1 = 1.2\) and thus \(e^{-5t} = 0.2/14 = 1/70\). It follows that \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.
Suppose that the population of a species of fish is controlled by the logistic equation \[\begin{aligned}\end{aligned}\], where \(P\) is measured in thousands of fish and \(t\) is measured in years.
1. What is the carrying capacity of this population?
2. Suppose that a long time has passed and that the fish population is stable at the carrying capacity. At this time, humans begin harvesting 20% of the fish every year. Modify the differential equation by adding a term to incorporate the harvesting of fish.
3. What is the new carrying capacity?
4. What will the fish population be one year after the harvesting begins?
5. How long will it take for the population to be within 10% of the carrying capacity?
Tunjukkan jawapan
1. From the form of the given logistic equation, we can see the carrying capacity is 10000 fish.
2. In this situation, the initial condition is \(P(0) = 10\), and by removing \(20\)% of the population per year, we subtract \(0.2P\) from the rate that defines the differential equation, thus finding \[\begin{aligned}\end{aligned}\].
3. Expanding and rewriting the differential equation in (b), we see that \[\begin{aligned}\end{aligned}\]. This, too, is a logistic equation, with carrying capacity \(8000\) fish.
4. We can solve this logistic equation in the usual way (keeping in mind the initial condition \(P(0) = 10\), and noting that the carrying capacity is \(8\)): \[\begin{aligned}\end{aligned}\] After one year, the population of fish will thus be \[\begin{aligned}\end{aligned}\] thousand fish.
5. To be within 10% of the carrying capacity, we need \(0.9(8) \le P(t) \le 1.1(8)\). Since the initial condition is larger than the carrying capacity, we want to know when \(P(t) = 1.1(8) = 8.8\). Thus we solve the equation \[\begin{aligned}\end{aligned}\], which implies \(-0.2e^{-0.8t}+1 = 8/8.8 = 10/11\). Thus, \(e^{-0.8t} = 5/11\), so \(t = -1.25 \ln(5/11) \approx 0.986\) years.
The logistic equation may be used to model how a rumor spreads through a group of people. Suppose that \(p(t)\) is the fraction of people that have heard the rumor on day \(t\). The equation \[\begin{aligned}\end{aligned}\] describes how \(p\) changes. Suppose initially that one-tenth of the people have heard the rumor; that is, \(p(0) = 0.1\).
1. What happens to \(p(t)\) after a very long time?
2. Determine a formula for the function \(p(t)\).
3. At what time is \(p\) changing most rapidly?
4. How long does it take before 80% of the people have heard the rumor?
Tunjukkan jawapan
1. The equilibrium solutions for this logistic equation are \(p = 0\) and \(p = 1\). Since \(p = 1\) is the carrying capacity and thus a stable equilibrium, \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\).
2. We know that the general solution to the logistic equation \(\frac{dP}{dt} = kP(N-P), \ P(0) = P_0\) is \[\begin{aligned}\end{aligned}\]. In the context of this problem, we know that \(k = 0.2\), \(N = 1\), and \(p(0) = 0.1\) and therefore the solution is \[\begin{aligned}\end{aligned}\].
3. We know that \(p\) is changing most rapidly when its derivative, \(p'\), is largest. Recalling the original differential equation, we know that \[\begin{aligned}\end{aligned}\] which shows that \(p'\) is a quadratic function of \(p\). That quadratic function has zeros at \(p = 0\) and \(p = 1\), so the vertex of this concave down quadratic function lies at \(p = \frac{1}{2}\), and at this value of \(p\) the rate of change of \(p\) is largest. Thus, we solve the equation \(p(t) = 0.5\) to find the time at which the population is increasing fastest. We see that \[\begin{aligned}\end{aligned}\] so \[\begin{aligned}\end{aligned}\] and \(e^{-0.2t} = 1/9\). It follows that \(t = -5 \ln(1/9) \approx 10.986\) days.
4. Since \(p(t)\) tells us the percentage of people who've heard the rumor, we need to solve the equation \(p(t) = 0.8\). Thus we have \[\begin{aligned}\end{aligned}\] so \[\begin{aligned}\end{aligned}\] and \(e^{-0.2t} = 0.25/9\). It follows that \(t = -5 \ln(0.25/9) \approx 17.19\) days.
Suppose that \(b(t)\) measures the number of bacteria living in a colony in a Petri dish, where \(b\) is measured in thousands and \(t\) is measured in days. One day, you measure that there are 6,000 bacteria and the per capita growth rate is 3. A few days later, you measure that there are 9,000 bacteria and the per capita growth rate is 2.
1. Assume that the per capita growth rate \(\frac{db/dt}{b}\) is a linear function of \(b\). Use the measurements to find this function and write a logistic equation to describe \(\frac{db}{dt}\).
2. What is the carrying capacity for the bacteria?
3. At what population is the number of bacteria increasing most rapidly?
4. If there are initially 1,000 bacteria, how long will it take to reach 80% of the carrying capacity?
Tunjukkan jawapan
1. Let \(y\) be the per capita growth rate as a function of \(b\). Using the given data in the problem statement, we know that this linear function passes through the points \((6000,3)\) and \((9000,2)\). The equation of this line is thus \[\begin{aligned}\end{aligned}\] so \(y = (3 + 2) - \frac{1}{3000}b = 5 - \frac{1}{3000}b\). Since the logistic equation comes from taking the product of the population at time \(t\) with its per capita growth rate at that time, we find that \[\begin{aligned}\end{aligned}\]
2. From the form of the logistic equation, we see that the carrying capacity for the bacteria is \(b = 15000\), which is a stable equilibrium for the population.
3. The bacteria is increasing most rapidly when \(\frac{db}{dt}\) is greatest. Since \[\begin{aligned}\end{aligned}\] is a quadratic function of \(b\), \(\frac{db}{dt}\) is maximized at the vertex of this concave down parabola, which occurs halfway between its zeros: \(b = 0\) and \(b = 15000\). Thus, when \(b = 7500\), the population is growing fastest. To find the exact \(t\)-value when this occurs, we'd need to know an initial condition like in the next part of the problem.
4. Using \(b(0) = 1000\), we solve the logistic equation IVP with \[\begin{aligned}\end{aligned}\]. Using \(N = 15000\), \(k = \frac{1}{3000}\), and \(b_0 = 1000\), we know from the standard form of the solution to the logistic equation that \[\begin{aligned}\end{aligned}\]. To find how long it takes the bacteria to reach 80% of the carrying capacity, we solve the equation \[\begin{aligned}\end{aligned}\]. That equation implies that \[\begin{aligned}\end{aligned}\], so \(14e^{-5t} + 1 = 1.2\) and thus \(e^{-5t} = 0.2/14 = 1/70\). It follows that \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.
Suppose that the population of a species of fish is controlled by the logistic equation \[\begin{aligned}\end{aligned}\], where \(P\) is measured in thousands of fish and \(t\) is measured in years.
1. What is the carrying capacity of this population?
2. Suppose that a long time has passed and that the fish population is stable at the carrying capacity. At this time, humans begin harvesting 20% of the fish every year. Modify the differential equation by adding a term to incorporate the harvesting of fish.
3. What is the new carrying capacity?
4. What will the fish population be one year after the harvesting begins?
5. How long will it take for the population to be within 10% of the carrying capacity?
Tunjukkan jawapan
1. From the form of the given logistic equation, we can see the carrying capacity is 10000 fish.
2. In this situation, the initial condition is \(P(0) = 10\), and by removing \(20\)% of the population per year, we subtract \(0.2P\) from the rate that defines the differential equation, thus finding \[\begin{aligned}\end{aligned}\].
3. Expanding and rewriting the differential equation in (b), we see that \[\begin{aligned}\end{aligned}\]. This, too, is a logistic equation, with carrying capacity \(8000\) fish.
4. We can solve this logistic equation in the usual way (keeping in mind the initial condition \(P(0) = 10\), and noting that the carrying capacity is \(8\)): \[\begin{aligned}\end{aligned}\] After one year, the population of fish will thus be \[\begin{aligned}\end{aligned}\] thousand fish.
5. To be within 10% of the carrying capacity, we need \(0.9(8) \le P(t) \le 1.1(8)\). Since the initial condition is larger than the carrying capacity, we want to know when \(P(t) = 1.1(8) = 8.8\). Thus we solve the equation \[\begin{aligned}\end{aligned}\], which implies \(-0.2e^{-0.8t}+1 = 8/8.8 = 10/11\). Thus, \(e^{-0.8t} = 5/11\), so \(t = -1.25 \ln(5/11) \approx 0.986\) years.

Symbols used here

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

\frac{\partial u}{\partial t},\ \nabla^2 u

partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Population growth and the logistic equation

How can we use differential equations to realistically model the growth of a population?
How can we assess the accuracy of our models?

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

Cubalah sendiri

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

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