# Investigate differential equations used to model population growth Contributed by: In this section, we will: Investigate differential equations used to model population growth.
1. 9
DIFFERENTIAL EQUATIONS
2. DIFFERENTIAL EQUATIONS
9.4
Models for
Population Growth
In this section, we will:
Investigate differential equations
used to model population growth.
3. NATURAL GROWTH
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:
dP
kP
dt
4. NATURAL GROWTH
Is that a reasonable
5. NATURAL GROWTH
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.
6. NATURAL GROWTH
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.
7. NATURAL GROWTH
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.
8. NATURAL GROWTH
In general, it seems reasonable that
the growth rate should be proportional
to the size.
9. LAW OF NATURAL GROWTH Equation 1
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 dP
kP
dt
where k is a constant.
 This is sometimes called the law of natural growth.
10. LAW OF NATURAL GROWTH
If k is positive,
the population increases.
If k is negative, it decreases.
11. LAW OF NATURAL GROWTH
Equation 1 is a separable differential equation.
Hence, we can solve it by the methods of
dP
Section 9.3:  P k dt
ln P kt  C
kt C C kt
P e e e
P Ae kt
where A (= ±eC or 0) is an arbitrary constant.
12. LAW OF NATURAL GROWTH
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.
13. LAW OF NATURAL GROWTH Equation 2
The solution of the initial-value problem
dP
kP P (0) P0
dt
kt
is: P(t ) P0 e
14. LAW OF NATURAL GROWTH
Another way of writing Equation 1 is:
1 dP
k
P dt
 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.
15. LAW OF NATURAL GROWTH
We can account for emigration
(or “harvesting”) from a population
by modifying Equation 1—as follows.
16. LAW OF NATURAL GROWTH Equation 3
If the rate of emigration is a constant m,
then the rate of change of the population
is modeled by the differential equation
dP
kP  m
dt
 See Exercise 13 for the solution and consequences
of Equation 3.
17. LOGISTIC MODEL
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.
18. LOGISTIC MODEL
If P(t) is the size of the population at time t,
we assume that:
dP
kP if P is small
dt
 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.
19. LOGISTIC MODEL
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).
20. LOGISTIC MODEL
The simplest expression for the
relative growth rate that incorporates
these assumptions is:
1 dP  P
k  1  
P dt  K
21. LOGISTIC DIFFERENTIAL EQN. Equation 4
Multiplying by P, we obtain the model
for population growth known as the logistic
differential equation:
dP  P
kP  1  
dt  K
22. LOGISTIC DIFFERENTIAL EQN.
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.
23. LOGISTIC DIFFERENTIAL EQN.
From Equation 4, we can deduce
increase or decrease directly.
24. LOGISTIC DIFFERENTIAL EQN.
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.
25. LOGISTIC DIFFERENTIAL EQN.
Let’s start our more detailed analysis
of the logistic differential equation by
looking at a direction field.
26. LOGISTIC DIFFERENTIAL EQN. Example 1
Draw a direction field for the logistic equation
with k = 0.08 and carrying capacity K = 1000.
What can you deduce about the solutions?
27. LOGISTIC DIFFERENTIAL EQN. Example 1
In this case, the logistic differential
equation is:
dP  P 
0.08P  1  
dt  1000 
28. LOGISTIC DIFFERENTIAL EQN. Example 1
A direction field for this equation is
shown here.
29. LOGISTIC DIFFERENTIAL EQN. Example 1
We show only the first quadrant because:
 Negative populations aren’t meaningful.
 We are interested only in what happens after t = 0.
30. LOGISTIC DIFFERENTIAL EQN. Example 1
The logistic equation is autonomous
(dP/dt depends only on P, not on t).
 So, the slopes
are the same
along any
horizontal line.
31. LOGISTIC DIFFERENTIAL EQN. Example 1
As expected, the slopes are:
 Positive for 0 < P < 1000
 Negative for P > 1000
32. LOGISTIC DIFFERENTIAL EQN. Example 1
The slopes are small when:
 P is close to 0 or 1000 (the carrying capacity).
33. LOGISTIC DIFFERENTIAL EQN. Example 1
Notice that the solutions move:
 Away from the equilibrium solution P = 0
 Toward the equilibrium solution P = 1000
34. LOGISTIC DIFFERENTIAL EQN. Example 1
Here, we use the direction field to sketch
solution curves with initial populations
P(0) = 100, P(0) = 400, P(0) = 1300
35. LOGISTIC DIFFERENTIAL EQN. Example 1
Notice that solution curves that start:
 Below P = 1000 are increasing.
 Above P = 1000 are decreasing.
36. LOGISTIC DIFFERENTIAL EQN. Example 1
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.
37. LOGISTIC DIFFERENTIAL EQN.
The logistic equation 4 is separable.
So, we can solve it explicitly using
the method of Section 9.3
38. LOGISTIC DIFFERENTIAL EQN. Equation 5
dP  P
kP  1  
dt  K
we have: dP
P(1  P / K ) k dt
39. LOGISTIC DIFFERENTIAL EQN.
To evaluate the integral on the left side,
we write: 1 K

P (1  P / K ) P( K  P )
Using partial fractions (Section 7.4),
we get:
K 1 1
 
P( K  P) P K  P
40. LOGISTIC DIFFERENTIAL EQN. Equation 6
This enables us to rewrite Equation 5:
1 1 
 P  K  P  dP k dt
ln | P |  ln | K  P |kt  C
K P
ln  kt  C
P
K P  kt  C  C  kt
e e e
P
K P
 Ae  kt where A = ±e-C.
P
41. LOGISTIC DIFFERENTIAL EQN.
Solving Equation 6 for P, we get:
K  kt P 1
 1  Ae    kt
P K 1  Ae
Hence, K
P  kt
1  Ae
42. LOGISTIC DIFFERENTIAL EQN.
We find the value of A by putting t = 0
in Equation 6.
If t = 0, then P = P0 (the initial population),
K  P0 0
 Ae  A
P0
43. LOGISTIC DIFFERENTIAL EQN. Equation 7
Therefore, the solution to the logistic
equation is:
K K  P0
P(t )   kt
where A
1  Ae P0
44. LOGISTIC DIFFERENTIAL EQN.
Using the expression for P(t) in Equation 7,
we see that
lim P(t ) K
t 
which is to be expected.
45. LOGISTIC DIFFERENTIAL EQN. Example 2
Write the solution of the initial-value problem
dP  P 
0.08 P  1   P(0) 100
dt  1000 
and use it to find the population sizes P(40)
and P(80).
At what time does the population reach 900?
46. LOGISTIC DIFFERENTIAL EQN. Example 2
The differential equation is a logistic
equation with:
 k = 0.08
 Carrying capacity K = 1000
 Initial population P0 = 100
47. LOGISTIC DIFFERENTIAL EQN. Example 2
Thus, Equation 7 gives the population at
time t as:
1000 1000  100
P(t )  where A  9
1  Ae  0.08t 100
Therefore, 1000
P(t )   0.08t
1  9e
48. LOGISTIC DIFFERENTIAL EQN. Example 2
Hence, the population sizes when t = 40
and 80 are:
1000
P (40)   3.2
731.6
1  9e
1000
P (80)   6.4
985.3
1  9e
49. LOGISTIC DIFFERENTIAL EQN. Example 2
The population reaches 900
1000
 0.08t
900
1  9e
50. LOGISTIC DIFFERENTIAL EQN. Example 2
Solving this equation for t, we get:
 0.08t
1  9e 109
 0.08t
e  811
 0.08t ln 811  ln 81
ln 81
t 54.9
0.08
 So, the population reaches 900 when t is
approximately 55.
51. LOGISTIC DIFFERENTIAL EQN. Example 2
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.
52. COMPARING THE MODELS
In the 1930s, the biologist G. F. Gause
conducted an experiment with the protozoan
Paramecium and used a logistic equation to
model his data.
53. COMPARING THE MODELS
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.
54. COMPARING THE MODELS Example 3
Find the exponential and logistic models
for Gause’s data.
Compare the predicted values with the
observed values and comment on the fit.
55. COMPARING THE MODELS Example 3
Given the relative growth rate k = 0.7944
and the initial population P0 = 2,
the exponential model is:
kt 0.7944 t
P (t ) P0 e 2e
56. COMPARING THE MODELS Example 3
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 1 dP  2 
k  1   k
P0 dt t 0  64 
shows that the value of k for the logistic model is
very close to the value for the exponential model.
57. COMPARING THE MODELS Example 3
Then, the solution of the logistic equation
in Equation 7 gives
K 64
P (t )   kt
  0.7944 t
1  Ae 1  Ae
where K  P0 64  2
A  31
P0 2
64
So, P (t )   0.7944 t
1  31e
58. COMPARING THE MODELS Example 3
We use these equations to calculate
the predicted values and compare them
59. COMPARING THE MODELS Example 3
Now, let’s compare
the table with this
60. COMPARING THE MODELS Example 3
For the first three
or four days:
 The exponential model
gives results
comparable to those of
the more sophisticated
logistic model.
61. COMPARING THE MODELS Example 3
However, for t ≥ 5:
 The exponential model
is hopelessly
inaccurate.
 The logistic model fits
the observations
reasonably well.
62. COMPARING THE MODELS
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.
63. COMPARING THE MODELS
The table shows midyear values of B(t),
the population of Belgium, in thousands,
at time t, from 1980 to 2000.
64. COMPARING THE MODELS
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
65. COMPARING THE MODELS
We see that the logistic model
provides a very good fit.
66. MODELS FOR POPULATION GROWTH
The Law of Natural Growth and the logistic
differential equation are not the only equations
that have been proposed to model population
67. OTHER MODELS FOR POPULATION GROWTH
In Exercise 18, we look at the Gompertz
growth function.
In Exercises 19 and 20, we investigate
seasonal-growth models.
68. OTHER MODELS FOR POPULATION GROWTH
Two of the other models
are modifications of the logistic
69. OTHER MODELS FOR POPULATION GROWTH
The differential equation dP  P
kP  1    c
dt  K
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.
70. OTHER MODELS FOR POPULATION GROWTH
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.
71. OTHER MODELS FOR POPULATION GROWTH
Such populations have been modeled by
the differential equation
dP  P  m 
kP  1    1  
dt  K  P
where the extra factor, 1 – m/P,
takes into account the consequences
of a sparse population.