differential equations 9. 9.4 models for population growth in this section, we will: investigate...
TRANSCRIPT
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DIFFERENTIAL EQUATIONSDIFFERENTIAL EQUATIONS
9
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9.4Models for
Population Growth
In this section, we will:
Investigate differential equations
used to model population growth.
DIFFERENTIAL EQUATIONS
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One of the models for population growth we
considered in Section 9.1 was based on the
assumption that the population grows at a rate
proportional to the size of the population:
NATURAL GROWTH
dPkP
dt
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Is that a reasonable
assumption?
NATURAL GROWTH
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Suppose we have a population
(of bacteria, for instance) with size
P = 1000.
At a certain time, it is growing at a rate of P’ = 300 bacteria per hour.
NATURAL GROWTH
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Now, let’s take another 1,000 bacteria
of the same type and put them with
the first population.
Each half of the new population was growing at a rate of 300 bacteria per hour.
NATURAL GROWTH
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We would expect the total population
of 2,000 to increase at a rate of 600 bacteria
per hour initially—provided there’s enough
room and nutrition.
So, if we double the size, we double the growth rate.
NATURAL GROWTH
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In general, it seems reasonable that
the growth rate should be proportional
to the size.
NATURAL GROWTH
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In general, if P(t) is the value of a quantity y
at time t and, if the rate of change of P with
respect to t is proportional to its size P(t) at
any time, then
where k is a constant.
This is sometimes called the law of natural growth.
LAW OF NATURAL GROWTH
dPkP
dt
Equation 1
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If k is positive,
the population increases.
If k is negative, it decreases.
LAW OF NATURAL GROWTH
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Equation 1 is a separable differential equation.
Hence, we can solve it by the methods of
Section 9.3:
where A (= ±eC or 0) is an arbitrary constant.
LAW OF NATURAL GROWTH
lnkt C C kt
kt
dPk dt
PP kt C
P e e e
P Ae
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To see the significance of the constant A,
we observe that:
P(0) = Aek·0 = A
Thus, A is the initial value of the function.
LAW OF NATURAL GROWTH
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The solution of the initial-value problem
is:
LAW OF NATURAL GROWTH
0(0)dP
kP P Pdt
Equation 2
0( ) ktP t P e
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Another way of writing Equation 1 is:
This says that the relative growth rate (the growth rate divided by the population size) is constant.
Then, Equation 2 says that a population with constant relative growth rate must grow exponentially.
LAW OF NATURAL GROWTH
1 dPk
P dt
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We can account for emigration
(or “harvesting”) from a population
by modifying Equation 1—as follows.
LAW OF NATURAL GROWTH
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If the rate of emigration is a constant m,
then the rate of change of the population
is modeled by the differential equation
See Exercise 13 for the solution and consequences of Equation 3.
LAW OF NATURAL GROWTH
dPkP m
dt
Equation 3
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As we discussed in Section 9.1, a population
often increases exponentially in its early
stages, but levels off eventually and
approaches its carrying capacity because
of limited resources.
LOGISTIC MODEL
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If P(t) is the size of the population at time t,
we assume that:
This says that the growth rate is initially close to being proportional to size.
In other words, the relative growth rate is almost constant when the population is small.
LOGISTIC MODEL
if is smalldP
kP Pdt
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However, we also want to reflect that
the relative growth rate:
Decreases as the population P increases.
Becomes negative if P ever exceeds its carrying capacity K (the maximum population that the environment is capable of sustaining in the long run).
LOGISTIC MODEL
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The simplest expression for the
relative growth rate that incorporates
these assumptions is:
LOGISTIC MODEL
11
dP Pk
P dt K
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Multiplying by P, we obtain the model
for population growth known as the logistic
differential equation:
LOGISTIC DIFFERENTIAL EQN.
1dP P
kPdt K
Equation 4
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Notice from Equation 4 that:
If P is small compared with K, then P/K is close to 0, and so dP/dt ≈ kP.
If P → K (the population approaches its carrying capacity), then P/K → 1, so dP/dt → 0.
LOGISTIC DIFFERENTIAL EQN.
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From Equation 4, we can deduce
information about whether solutions
increase or decrease directly.
LOGISTIC DIFFERENTIAL EQN.
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If the population P lies between 0 and K,
the right side of the equation is positive.
So, dP/dt > 0 and the population increases.
If the population exceeds the carrying
capacity (P > K), 1 – P/K is negative.
So, dP/dt < 0 and the population decreases.
LOGISTIC DIFFERENTIAL EQN.
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Let’s start our more detailed analysis
of the logistic differential equation by
looking at a direction field.
LOGISTIC DIFFERENTIAL EQN.
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Draw a direction field for the logistic equation
with k = 0.08 and carrying capacity K = 1000.
What can you deduce about the solutions?
LOGISTIC DIFFERENTIAL EQN. Example 1
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In this case, the logistic differential
equation is:
LOGISTIC DIFFERENTIAL EQN.
0.08 11000
dP PP
dt
Example 1
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A direction field for this equation is
shown here.
LOGISTIC DIFFERENTIAL EQN. Example 1
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We show only the first quadrant because:
Negative populations aren’t meaningful. We are interested only in what happens after t = 0.
LOGISTIC DIFFERENTIAL EQN. Example 1
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The logistic equation is autonomous
(dP/dt depends only on P, not on t).
So, the slopes are the same along any horizontal line.
LOGISTIC DIFFERENTIAL EQN. Example 1
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As expected, the slopes are:
Positive for 0 < P < 1000 Negative for P > 1000
LOGISTIC DIFFERENTIAL EQN. Example 1
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The slopes are small when:
P is close to 0 or 1000 (the carrying capacity).
LOGISTIC DIFFERENTIAL EQN. Example 1
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Notice that the solutions move:
Away from the equilibrium solution P = 0 Toward the equilibrium solution P = 1000
LOGISTIC DIFFERENTIAL EQN. Example 1
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Here, we use the direction field to sketch
solution curves with initial populations
P(0) = 100, P(0) = 400, P(0) = 1300
LOGISTIC DIFFERENTIAL EQN. Example 1
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Notice that solution curves that start:
Below P = 1000 are increasing. Above P = 1000 are decreasing.
LOGISTIC DIFFERENTIAL EQN. Example 1
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The slopes are greatest when P ≈ 500.
Thus, the solution curves that start below
P = 1000 have inflection points when P ≈ 500.
In fact, we can prove that all solution curves that start below P = 500 have an inflection point when P is exactly 500.
LOGISTIC DIFFERENTIAL EQN. Example 1
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The logistic equation 4 is separable.
So, we can solve it explicitly using
the method of Section 9.3
LOGISTIC DIFFERENTIAL EQN.
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Since
we have:
LOGISTIC DIFFERENTIAL EQN.
1dP P
kPdt K
(1 / )
dPk dt
P P K
Equation 5
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To evaluate the integral on the left side,
we write:
Using partial fractions (Section 7.4),
we get:
LOGISTIC DIFFERENTIAL EQN.
1
(1 / ) ( )
K
P P K P K P
1 1
( )
K
P K P P K P
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This enables us to rewrite Equation 5:
where A = ±e-C.
LOGISTIC DIFFERENTIAL EQN.
1 1
ln | | ln | |
ln
kt C C kt
kt
dP k dtP K P
P K P kt C
K Pkt C
P
K Pe e e
P
K PAe
P
Equation 6
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Solving Equation 6 for P, we get:
Hence,
LOGISTIC DIFFERENTIAL EQN.
11
1kt
kt
K PAe
P K Ae
1 kt
KP
Ae
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We find the value of A by putting t = 0
in Equation 6.
If t = 0, then P = P0 (the initial population),
so
LOGISTIC DIFFERENTIAL EQN.
00
0
K PAe A
P
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Therefore, the solution to the logistic
equation is:
LOGISTIC DIFFERENTIAL EQN.
0
0
( ) where1 kt
K PKP t A
Ae P
Equation 7
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Using the expression for P(t) in Equation 7,
we see that
which is to be expected.
LOGISTIC DIFFERENTIAL EQN.
lim ( )tP t K
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Write the solution of the initial-value problem
and use it to find the population sizes P(40)
and P(80).
At what time does the population reach 900?
LOGISTIC DIFFERENTIAL EQN.
0.08 1 (0) 1001000
dP PP P
dt
Example 2
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The differential equation is a logistic
equation with:
k = 0.08
Carrying capacity K = 1000
Initial population P0 = 100
LOGISTIC DIFFERENTIAL EQN. Example 2
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Thus, Equation 7 gives the population at
time t as:
Therefore,
LOGISTIC DIFFERENTIAL EQN.
0.08
1000 1000 100( ) where 9
1 100tP t A
Ae
0.08
1000( )
1 9 tP t
e
Example 2
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Hence, the population sizes when t = 40
and 80 are:
LOGISTIC DIFFERENTIAL EQN.
3.2
6.4
1000(40) 731.6
1 9
1000(80) 985.3
1 9
Pe
Pe
Example 2
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The population reaches 900
when:
LOGISTIC DIFFERENTIAL EQN.
0.08
1000900
1 9 te
Example 2
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Solving this equation for t, we get:
So, the population reaches 900 when t is approximately 55.
LOGISTIC DIFFERENTIAL EQN.
0.08 109
0.08 181
181
1 9
0.08 ln ln81
ln8154.9
0.08
t
t
e
e
t
t
Example 2
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As a check on our work, we graph
the population curve and observe where
it intersects the line P = 900.
The cursor indicates that t ≈ 55.
LOGISTIC DIFFERENTIAL EQN. Example 2
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In the 1930s, the biologist G. F. Gause
conducted an experiment with the protozoan
Paramecium and used a logistic equation to
model his data.
COMPARING THE MODELS
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The table gives his daily count of
the population of protozoa.
He estimated the initial relative growth rate to be 0.7944 and the carrying capacity to be 64.
COMPARING THE MODELS
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Find the exponential and logistic models
for Gause’s data.
Compare the predicted values with the
observed values and comment on the fit.
COMPARING THE MODELS Example 3
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Given the relative growth rate k = 0.7944
and the initial population P0 = 2,
the exponential model is:
COMPARING THE MODELS
0.79440( ) 2kt tP t P e e
Example 3
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Gause used the same value of k for
his logistic model.
This is reasonable as P0 = 2 is small compared with the carrying capacity (K = 64).
The equation
shows that the value of k for the logistic model is very close to the value for the exponential model.
0 0
1 2164
t
dPk k
P dt
Example 3COMPARING THE MODELS
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Then, the solution of the logistic equation
in Equation 7 gives
where
So,
0.7944
0
0
64( )
1 1
64 231
2
kt t
KP t
Ae Ae
K PA
P
0.7944
64( )
1 31 tP t
e
Example 3COMPARING THE MODELS
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We use these equations to calculate
the predicted values and compare them
here.
Example 3COMPARING THE MODELS
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Now, let’s compare
the table with this
graph.
Example 3COMPARING THE MODELS
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For the first three
or four days: The exponential model
gives results comparable to those of the more sophisticated logistic model.
Example 3COMPARING THE MODELS
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However, for t ≥ 5: The exponential model
is hopelessly inaccurate.
The logistic model fits the observations reasonably well.
Example 3COMPARING THE MODELS
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Many countries that formerly experienced
exponential growth are now finding that their
rates of population growth are declining and
the logistic model provides a better model.
COMPARING THE MODELS
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The table shows midyear values of B(t),
the population of Belgium, in thousands,
at time t, from 1980 to 2000.
COMPARING THE MODELS
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This figure shows the data points of the table
together with a shifted logistic function
obtained from a calculator with the ability to fit
a logistic function to these points by
regression.
COMPARING THE MODELS
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We see that the logistic model
provides a very good fit.
COMPARING THE MODELS
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The Law of Natural Growth and the logistic
differential equation are not the only equations
that have been proposed to model population
growth.
MODELS FOR POPULATION GROWTH
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In Exercise 18, we look at the Gompertz
growth function.
In Exercises 19 and 20, we investigate
seasonal-growth models.
OTHER MODELS FOR POPULATION GROWTH
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Two of the other models
are modifications of the logistic
model.
OTHER MODELS FOR POPULATION GROWTH
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The differential equation
has been used to model populations that are
subject to harvesting of one sort or another.
Think of a population of fish being caught at a constant rate.
This equation is explored in Exercises 15 and 16.
OTHER MODELS FOR POPULATION GROWTH
1dP P
kP cdt K
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For some species, there is a minimum
population level m below which the species
tends to become extinct.
Adults may not be able to find suitable mates.
OTHER MODELS FOR POPULATION GROWTH
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Such populations have been modeled by
the differential equation
where the extra factor, 1 – m/P,
takes into account the consequences
of a sparse population.
OTHER MODELS FOR POPULATION GROWTH
1 1dP P m
kPdt K P