The Logistic Equation

Differential equations can be used to represent the size of a population as it varies over time. We saw this in an earlier chapter in the section on exponential growth and decay, which is the simplest model. A more realistic model includes other factors that affect the growth of the population. In this section, we study the logistic differential equation and see how it applies to the study of population dynamics in the context of biology.

Population Growth and Carrying Capacity

To model population growth using a differential equation, we first need to introduce some variables and relevant terms. The variable

t .

will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem must specify the units used in that particular problem. The variable

P

will represent population. Since the population varies over time, it is understood to be a function of time. Therefore we use the notation

P (t)

represents the instantaneous rate of change of the population as a function of time.

In Exponential Growth and Decay, we studied the exponential growth and decay of populations and radioactive substances. An example of an exponential growth function is

P (t) = P_{0} e^{r t} .

is called the growth rate. [link] shows a graph of

P (t) = 100 e^{0.03 t} .

This differential equation has an interesting interpretation. The left-hand side represents the rate at which the population increases (or decreases). The right-hand side is equal to a positive constant multiplied by the current population. Therefore the differential equation states that the rate at which the population increases is proportional to the population at that point in time. Furthermore, it states that the constant of proportionality never changes.

One problem with this function is its prediction that as time goes on, the population grows without bound. This is unrealistic in a real-world setting. Various factors limit the rate of growth of a particular population, including birth rate, death rate, food supply, predators, and so on. The growth constant

r

usually takes into consideration the birth and death rates but none of the other factors, and it can be interpreted as a net (birth minus death) percent growth rate per unit time. A natural question to ask is whether the population growth rate stays constant, or whether it changes over time. Biologists have found that in many biological systems, the population grows until a certain steady-state population is reached. This possibility is not taken into account with exponential growth. However, the concept of carrying capacity allows for the possibility that in a given area, only a certain number of a given organism or animal can thrive without running into resource issues.

to denote the carrying capacity. The growth rate is represented by the variable

r .

This differential equation can be coupled with the initial condition

P (0) = P_{0}

Suppose that the initial population is small relative to the carrying capacity. Then

\frac{P}{K}

is small, possibly close to zero. Thus, the quantity in parentheses on the right-hand side of [link] is close to

1,

is constant. If the population remains below the carrying capacity, then

\frac{P}{K}

Therefore the right-hand side of [link] is still positive, but the quantity in parentheses gets smaller, and the growth rate decreases as a result. If

P = K

Now suppose that the population starts at a value higher than the carrying capacity. Then

\frac{P}{K} > 1,

Then the right-hand side of [link] is negative, and the population decreases. As long as

P > K,

will get smaller and smaller, but the population approaches the carrying capacity as

t

approaches infinity. This analysis can be represented visually by way of a phase line. A phase line describes the general behavior of a solution to an autonomous differential equation, depending on the initial condition. For the case of a carrying capacity in the logistic equation, the phase line is as shown in [link].

Chapter Opener: Examining the Carrying Capacity of a Deer Population

This is a photograph of a deer. {:}

Let’s consider the population of white-tailed deer (Odocoileus virginianus) in the state of Kentucky. The Kentucky Department of Fish and Wildlife Resources (KDFWR) sets guidelines for hunting and fishing in the state. Before the hunting season of $2004,$

it estimated a population of $900,000$

deer. Johnson notes: “A deer population that has plenty to eat and is not hunted by humans or other predators will double every three years.” (George Johnson, “The Problem of Exploding Deer Populations Has No Attractive Solutions,” January $12, 2001,$

accessed April 9, 2015, http://www.txtwriter.com/onscience/Articles/deerpops.html.) This observation corresponds to a rate of increase $r = \frac{ln (2)}{3} = 0.2311,$

so the approximate growth rate is $23.11 %$

per year. (This assumes that the population grows exponentially, which is reasonable––at least in the short term––with plentiful food supply and no predators.) The KDFWR also reports deer population densities for $32$

counties in Kentucky, the average of which is approximately $27$

deer per square mile. Suppose this is the deer density for the whole state $(39,732$

square miles). The carrying capacity $K$

is $39,732$

square miles times $27$

deer per square mile, or $1,072,764$

deer.

For this application, we have $P_{0} = 900,000, K = 1,072,764,$
and
$r = 0.2311 .$
Substitute these values into [link] and form the initial-value problem.
Solve the initial-value problem from part a.
According to this model, what will be the population in $3$
years? Recall that the doubling time predicted by Johnson for the deer population was
$3$
years. How do these values compare?
Suppose the population managed to reach $1,200,000$
deer. What does the logistic equation predict will happen to the population in this scenario?

The initial value problem is
$\frac{d P}{d t} = 0.2311 P (1 - \frac{P}{1,072,764}), P (0) = 900,000.$
The logistic equation is an autonomous differential equation, so we can use the method of separation of variables.

Step 1: Setting the right-hand side equal to zero gives
$P = 0$
and
$P = 1,072,764 .$
This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change.

Step 2: Rewrite the differential equation and multiply both sides by:

$\begin{matrix} \frac{d P}{d t} & = & 0.2311 P (\frac{1,072,764 - P}{1,072,764}) \\ d P & = & 0.2311 P (\frac{1,072,764 - P}{1,072,764}) d t . \end{matrix}$

Divide both sides by
$P (1,072,764 - P) :$

$\frac{d P}{P (1,072,764 - P)} = \frac{0.2311}{1,072,764} d t .$

Step 3: Integrate both sides of the equation using partial fraction decomposition:

$\begin{matrix} \int \frac{d P}{P (1,072,764 - P)} & = & \int \frac{0.2311}{1,072,764} d t \\ \frac{1}{1,072,764} \int (\frac{1}{P} + \frac{1}{1,072,764 - P}) d P & = & \frac{0.2311 t}{1,072,764} + C \\ \frac{1}{1,072,764} (ln \| P \| - ln \| 1,072,764 - P \|) & = & \frac{0.2311 t}{1,072,764} + C . \end{matrix}$

Step 4: Multiply both sides by
$1,072,764$
and use the quotient rule for logarithms:

$ln \| \frac{P}{1,072,764 - P} \| = 0.2311 t + C_{1} .$

Here
$C_{1} = 1,072,764 C .$
Next exponentiate both sides and eliminate the absolute value:

$\begin{matrix} e^{ln \| \frac{P}{1,072,764 - P} \|} & = & e^{0.2311 t + C_{1}} \\ \| \frac{P}{1,072,764 - P} \| & = & C_{2} e^{0.2311 t} \\ \frac{P}{1,072,764 - P} & = & C_{2} e^{0.2311 t} . \end{matrix}$

Here
$C_{2} = e^{C_{1}}$
but after eliminating the absolute value, it can be negative as well. Now solve for:

$\begin{matrix} P & = & C_{2} e^{0.2311 t} (1,072,764 - P) . \\ P & = & 1,072,764 C_{2} e^{0.2311 t} - C_{2} P e^{0.2311 t} \\ P + C_{2} P e^{0.2311 t} & = & 1,072,764 C_{2} e^{0.2311 t} \\ P (1 + C_{2} e^{0.2311 t}) & = & 1,072,764 C_{2} e^{0.2311 t} \\ P (t) & = & \frac{1,072,764 C_{2} e^{0.2311 t}}{1 + C_{2} e^{0.2311 t}} . \end{matrix}$

Step 5: To determine the value of
$C_{2},$
it is actually easier to go back a couple of steps to where
$C_{2}$
was defined. In particular, use the equation

$\frac{P}{1,072,764 - P} = C_{2} e^{0.2311 t} .$

The initial condition is
$P (0) = 900,000 .$
Replace
$P$
with
$900,000$
and
$t$
with zero:

$\begin{matrix} \frac{P}{1,072,764 - P} & = & C_{2} e^{0.2311 t} \\ \frac{900,000}{1,072,764 - 900,000} & = & C_{2} e^{0.2311 (0)} \\ \frac{900,000}{172,764} & = & C_{2} \\ C_{2} & = & \frac{25,000}{4,799} \approx 5.209 . \end{matrix}$

Therefore

$\begin{matrix} P (t) & = \frac{1,072,764 (\frac{25000}{4799}) e^{0.2311 t}}{1 + (\frac{25000}{4799}) e^{0.2311 t}} \\ = \frac{1,072,764 (25000) e^{0.2311 t}}{4799 + 25000 e^{0.2311 t}} . \end{matrix}$

Dividing the numerator and denominator by
$25,000$
gives

$P (t) = \frac{1,072,764 e^{0.2311 t}}{0.19196 + e^{0.2311 t}} .$

[link] is a graph of this equation.
Using this model we can predict the population in $3$
years.

$P (3) = \frac{1,072,764 e^{0.2311 (3)}}{0.19196 + e^{0.2311 (3)}} \approx 978,830 deer$

This is far short of twice the initial population of
$900,000 .$
Remember that the doubling time is based on the assumption that the growth rate never changes, but the logistic model takes this possibility into account.
If the population reached $1,200,000$
deer, then the new initial-value problem would be

$\frac{d P}{d t} = 0.2311 P (1 - \frac{P}{1,072,764}), P (0) = 1,200,000.$

The general solution to the differential equation would remain the same.

$P (t) = \frac{1,072,764 C_{2} e^{0.2311 t}}{1 + C_{2} e^{0.2311 t}}$

To determine the value of the constant, return to the equation

$\frac{P}{1,072,764 - P} = C_{2} e^{0.2311 t} .$

Substituting the values
$t = 0$
and
$P = 1,200,000,$
you get

$\begin{matrix} C_{2} e^{0.2311 (0)} & = & \frac{1,200,000}{1,072,764 - 1,200,000} \\ C_{2} & = & - \frac{100,000}{10,603} \approx - 9.431 . \end{matrix}$

Therefore

$\begin{matrix} P (t) & = \frac{1,072,764 C_{2} e^{0.2311 t}}{1 + C_{2} e^{0.2311 t}} \\ = \frac{1,072,764 (- \frac{100,000}{10,603}) e^{0.2311 t}}{1 + (- \frac{100,000}{10,603}) e^{0.2311 t}} \\ = - \frac{107,276,400,000 e^{0.2311 t}}{100,000 e^{0.2311 t} - 10,603} \\ \approx \frac{10,117,551 e^{0.2311 t}}{9.43129 e^{0.2311 t} - 1} . \end{matrix}$

This equation is graphed in [link].

Solving the Logistic Differential Equation

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in [link].

as constant solutions. The first solution indicates that when there are no organisms present, the population will never grow. The second solution indicates that when the population starts at the carrying capacity, it will never change.

The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that

Now multiply the numerator and denominator of the right-hand side by

(K - P_{0})

Now that we have the solution to the initial-value problem, we can choose values for

P_{0}, r,

and study the solution curve. For example, in [link] we used the values

r = 0.2311, K = 1,072,764,

This is the same as the original solution. The graph of this solution is shown again in blue in [link], superimposed over the graph of the exponential growth model with initial population

900,000

(appearing in green). The red dashed line represents the carrying capacity, and is a horizontal asymptote for the solution to the logistic equation.

Working under the assumption that the population grows according to the logistic differential equation, this graph predicts that approximately

20

the growth of the population was very close to exponential. The net growth rate at that time would have been around

23.1 %

per year. As time goes on, the two graphs separate. This happens because the population increases, and the logistic differential equation states that the growth rate decreases as the population increases. At the time the population was measured

(2004),

it was close to carrying capacity, and the population was starting to level off.

The solution to the logistic differential equation has a point of inflection. To find this point, set the second derivative equal to zero:

then this quantity is undefined, and the graph does not have a point of inflection. In the logistic graph, the point of inflection can be seen as the point where the graph changes from concave up to concave down. This is where the “leveling off” starts to occur, because the net growth rate becomes slower as the population starts to approach the carrying capacity.

Key Concepts

Key Equations

For the following problems, consider the logistic equation in the form $P' = C P - P^{2} .$

Draw the directional field and find the stability of the equilibria.

C = 3

C = 0

A direction field with arrows pointing to the right. The arrows are horizontal along the x axis. The arrows point down above the x axis and below the x axis. The further away the arrows are from the x axis, the more vertical the lines become.

P = 0

semi-stable

C = −3

Solve the logistic equation for $C = 10$

and an initial condition of $P (0) = 2 .$

P = \frac{10 e^{10 x}}{e^{10 x} + 4}

Solve the logistic equation for $C = −10$

and an initial condition of $P (0) = 2 .$

A population of deer inside a park has a carrying capacity of $200$

and a growth rate of $2 % .$

If the initial population is $50$

deer, what is the population of deer at any given time?

P (t) = \frac{10000 e^{0.02 t}}{150 + 50 e^{0.02 t}}

A population of frogs in a pond has a growth rate of $5 % .$

If the initial population is $1000$

frogs and the carrying capacity is $6000,$

what is the population of frogs at any given time?

[T] Bacteria grow at a rate of $20 %$

per hour in a petri dish. If there is initially one bacterium and a carrying capacity of $1$

million cells, how long does it take to reach $500,000$

cells?

69

hours $5$

minutes

[T] Rabbits in a park have an initial population of $10$

and grow at a rate of $4 %$

per year. If the carrying capacity is $500,$

at what time does the population reach $100$

rabbits?

[T] Two monkeys are placed on an island. After $5$

years, there are $8$

monkeys, and the estimated carrying capacity is $25$

monkeys. When does the population of monkeys reach $16$

monkeys?

7

years $2$

months

[T] A butterfly sanctuary is built that can hold $2000$

butterflies, and $400$

butterflies are initially moved in. If after $2$

months there are now $800$

butterflies, when does the population get to $1500$

butterflies?

The following problems consider the logistic equation with an added term for depletion, either through death or emigration.

[T] The population of trout in a pond is given by $P' = 0.4 P (1 - \frac{P}{10000}) - 400,$

where $400$

trout are caught per year. Use your calculator or computer software to draw a directional field and draw a few sample solutions. What do you expect for the behavior?

A direction field with arrows down for P < 1,000, pointing up for 1,000 < P < 8,500, and pointing down for P > 8,500. Right above P = 8,500, the arrows point down and to the right.

In the preceding problem, what are the stabilities of the equilibria $0 < P_{1} < P_{2} ?$

[T] For the preceding problem, use software to generate a directional field for the value $f = 400 .$

What are the stabilities of the equilibria?

A direction field with arrows pointing down and to the right. Around y = 4,000, the arrows are more horizontal. The further the arrows are from this line, the more vertical the arrows become.

P_{1}

semi-stable

[T] For the preceding problems, use software to generate a directional field for the value $f = 600 .$

What are the stabilities of the equilibria?

[T] For the preceding problems, consider the case where a certain number of fish are added to the pond, or $f = −200 .$

What are the nonnegative equilibria and their stabilities?

A direction field with arrows pointing up for P < 10,000 and arrows pointing down for P > 10,000.

P_{2} > 0

stable

It is more likely that the amount of fishing is governed by the current number of fish present, so instead of a constant number of fish being caught, the rate is proportional to the current number of fish present, with proportionality constant $k,$

P' = 0.4 P (1 - \frac{P}{10000}) - k P .

[T] For the previous fishing problem, draw a directional field assuming $k = 0.1 .$

Draw some solutions that exhibit this behavior. What are the equilibria and what are their stabilities?

[T] Use software or a calculator to draw directional fields for $k = 0.4 .$

What are the nonnegative equilibria and their stabilities?

A direction field with arrows pointing to the right at P = 0. Below 0, the arrows point down and to the right. Above 0, the arrows point down and to the right. The further away from 0, the more vertical the arrows become.

P_{1} = 0

is semi-stable

[T] Use software or a calculator to draw directional fields for $k = 0.6 .$

What are the equilibria and their stabilities?

Solve this equation, assuming a value of $k = 0.05$

and an initial condition of $2000$

fish.

y = \frac{−20}{4 \times 10^{−6} - 0.002 e^{0.01 t}}

Solve this equation, assuming a value of $k = 0.05$

and an initial condition of $5000$

fish.

The following problems add in a minimal threshold value for the species to survive, $T,$

which changes the differential equation to $P' (t) = r P (1 - \frac{P}{K}) (1 - \frac{T}{P}) .$

Draw the directional field of the threshold logistic equation, assuming $K = 10, r = 0.1, T = 2 .$

When does the population survive? When does it go extinct?

A direction field with arrows pointing horizontally to the right along y = 2 and y = 10. For P < 2, the arrows point down and to the right. For 2 < P < 10, the arrows point up and to the right. For P > 10, the arrows point down and to the right. The further the arrows are from 2 and 10, the steeper they become, and the closer they are from 2 and 10, the more horizontal the arrows become.

For the preceding problem, solve the logistic threshold equation, assuming the initial condition $P (0) = P_{0} .$

Bengal tigers in a conservation park have a carrying capacity of $100$

and need a minimum of $10$

to survive. If they grow in population at a rate of $1 %$

per year, with an initial population of $15$

tigers, solve for the number of tigers present.

P (t) = \frac{850 + 500 e^{0.009 t}}{85 + 5 e^{0.009 t}}

A forest containing ring-tailed lemurs in Madagascar has the potential to support $5000$

individuals, and the lemur population grows at a rate of $5 %$

per year. A minimum of $500$

individuals is needed for the lemurs to survive. Given an initial population of $600$

lemurs, solve for the population of lemurs.

The population of mountain lions in Northern Arizona has an estimated carrying capacity of $250$

and grows at a rate of $0.25 %$

per year and there must be $25$

for the population to survive. With an initial population of $30$

mountain lions, how many years will it take to get the mountain lions off the endangered species list (at least $100) ?$

13

years months

The following questions consider the Gompertz equation, a modification for logistic growth, which is often used for modeling cancer growth, specifically the number of tumor cells.

The Gompertz equation is given by $P (t)' = α ln (\frac{K}{P (t)}) P (t) .$

Draw the directional fields for this equation assuming all parameters are positive, and given that $K = 1 .$

Assume that for a population, $K = 1000$

and $α = 0.05 .$

Draw the directional field associated with this differential equation and draw a few solutions. What is the behavior of the population?

A direction field with arrows pointing down and to the right for P < 0, up for 0 < P < 1,000, and down for P > 1,000. The further the arrows are from P = 0 and P = 1,000, the more vertical they become, and the closer they are, the more horizontal they are.

Solve the Gompertz equation for generic $α$

and $K$

and $P (0) = P_{0} .$

[T] The Gompertz equation has been used to model tumor growth in the human body. Starting from one tumor cell on day $1$

and assuming $α = 0.1$

and a carrying capacity of $10$

million cells, how long does it take to reach “detection” stage at $5$

million cells?

31.465

days

[T] It is estimated that the world human population reached $3$

billion people in $1959$

and $6$

billion in $1999 .$

Assuming a carrying capacity of $16$

billion humans, write and solve the differential equation for logistic growth, and determine what year the population reached $7$

billion.

[T] It is estimated that the world human population reached $3$

billion people in $1959$

and $6$

billion in $1999 .$

Assuming a carrying capacity of $16$

billion humans, write and solve the differential equation for Gompertz growth, and determine what year the population reached $7$

billion. Was logistic growth or Gompertz growth more accurate, considering world population reached $7$

billion on October $31, 2011 ?$

September $2008$

Show that the population grows fastest when it reaches half the carrying capacity for the logistic equation $P' = r P (1 - \frac{P}{K}) .$

When does population increase the fastest in the threshold logistic equation $P' (t) = r P (1 - \frac{P}{K}) (1 - \frac{T}{P}) ?$

\frac{K + T}{2}

When does population increase the fastest for the Gompertz equation $P (t)' = α ln (\frac{K}{P (t)}) P (t) ?$

Below is a table of the populations of whooping cranes in the wild from $1940 to 2000 .$

The population rebounded from near extinction after conservation efforts began. The following problems consider applying population models to fit the data. Assume a carrying capacity of $10,000$

cranes. Fit the data assuming years since $1940$

(so your initial population at time $0$

would be $22$

cranes).

*Source:* https://www.savingcranes.org/images/stories/site\_images/conservation/whooping\_crane/pdfs/historic\_wc\_numbers.pdf
Year (years since conservation began)	Whooping Crane Population
$1940 (0)$	$22$
$1950 (10)$	$31$
$1960 (20)$	$36$
$1970 (30)$	$57$
$1980 (40)$	$91$
$1990 (50)$	$159$
$2000 (60)$	$256$

Find the equation and parameter $r$

that best fit the data for the logistic equation.

r = 0.0405

Find the equation and parameters $r$

and $T$

that best fit the data for the threshold logistic equation.

Find the equation and parameter $α$

that best fit the data for the Gompertz equation.

α = 0.0081

Graph all three solutions and the data on the same graph. Which model appears to be most accurate?

Using the three equations found in the previous problems, estimate the population in $2010$

(year $70$

after conservation). The real population measured at that time was $437 .$

Which model is most accurate?

Logistic: $361,$

Threshold: $436,$

Gompertz: $309 .$

The Logistic Equation

Population Growth and Carrying Capacity

Solving the Logistic Differential Equation

Key Concepts

Key Equations

Glossary