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Introductory materials for the
West Nile Virus module
Gerda de Vries and Cole Zmurchok
Deparment of Mathematical and Statistical Sciences
University of Alberta
Draft: July 9, 2009
Introduction
This document has been developed for teachers to use when introducing their students
to concepts of basic differential equations. Its goal is to provide a introductory resource
for the West Nile Virus Module, which has been developed for CRYSTAL-Alberta. This
document’s intended use is that of an outline for a teacher who wishes to introduce his or
her students to the material in the classroom setting. Alternatively, teachers may opt for
the students to work through the accompanying web-based module. The web-based module
covers the same material and like this document, will equip the students with the necessary
knowledge and tools required to successfully complete the West Nile Virus Module.
This document is divided into four sections, which are outlined below. Solutions to example
questions are given in blue.
Contents
Step 1: Treatment of the differential equation for exponential growth
Step 2: Development and analysis of the differential equation for logistic growth
Step 3: Introduction to a system of differential equations describing an infections disease
Appendices: Methods for the analytical solution to the logistic equation
1
Step 1: Treatment of the differential equation for exponential
growth
Let N (t) be the number of individuals in a population at time t.
Assumptions:
1. The rate of change of N due to births is proportional to N , with proportionality
constant b > 0 (We refer to b as the birth rate).
2. The rate of change of N due to death is proportional to N , with proportionality
constant d > 0 (We refer to d as the death rate).
Then we can write the following differential equation for N (t):
dN
= bN − dN
dt
= (b − d)N
= rN.
Questions that can be posed to students:
• What must be true about N (t)? Can it be positive, zero, negative?
N (t) ≥ 0 for all t. Since it measures the size of a population, it is illogical for
N (t) < 0.
• What is the meaning of:
r > 0?
Here, since r = b − d > 0 ⇒ b > d, the birth rate is greater than the death rate,
and thus, the population is growing. N (t) is increasing.
r = 0?
Here, since r = b − d = 0 ⇒ b = d, the birth rate is equal to the death rate, and
thus, the population is neither growing nor shrinking. N (t) is constant.
r < 0?
Here, since r = b − d < 0 ⇒ b < d, the birth rate is less than death rate, and
thus, the population is shrinking.
2
• What is the sign of
dN
in each of the the cases:
dt
r > 0?
dN
> 0.
dt
r = 0?
dN
= 0.
dt
r<0
dN
< 0.
dt
• What does the sign of
The sign of
dN
imply about N (t)?
dt
dN
alerts us to the behaviour of N (t). For example, when
dt
dN
> 0, then N (t) is increasing;
dt
dN
= 0, then N (t) is constant;
dt
dN
< 0, then N (t) is decreasing.
dt
Solution Method A (“intuitive”, by inspection)
We look for a function N (t) such that when we differentiate it with respect to t we get the
function N back times a factor r.
N (t) = ert works.
Check:
d
dN
= ert (rt)
dt
dt
= rert
= rN,
(by the chain rule for differentiation)
as required.
3
But N (t) = 2ert also works.
Check:
dN
d
= 2 (ert )
dt
dt
= 2rert
(bythechainrulef ordif f erentiation)
= r(2ert )
= rN,
as required.
We can generalize from above, and say that
N (t) = Cert ,
where C is an arbitrary constant (C ∈ R) is a solution to
dN
= rN .
dt
Solution Method B (separation of variables)
Here, we solve the differential equation analytically.
dN
= rN
Z dt
Z
dN
= r dt
N
ln N = rt + C1
N =e
(separation of variables)
(integrate both sides)
(exponentiate both sides)
rt+C1
=e e
(law of exponents)
= Ce
(where C = eC1 )
C1 rt
rt
or N (t) = Cert , as before.
N (t) = Cert describes a family of curves, as C ∈ R.
By specifying the value of C, we pick out one member from the family of curves.
It is customary to specify the value of C via an initial condition, namely the initial value
of N , generally at time t = 0.
4
N(t)
C=2
C=1
C=
1
2
t
Figure 1: Graph of N (t) = Cert for some positive r, and various values of C.
For example, if N (0) = N0 , then we can find C as follows:
N (t) = Cert
gives N (0) = Ce0 = C
but N (0) = N0 as well.
Thus, C = N0 .
We say that
N (t) = N0 ert
is the specific solution of
dN
= rN satisfying the initial condition N (0) = N0 .
dt
Graphs of N (t) = N0 ert and interpretation
Recall the meaning of r, namely r = b − d = birthrate − deathrate.
We have three values of r to consider.
Case r > 0: then b > d, which means that there are more births than deaths. That is, the
dN
> 0 always.
population has unlimited growth. Mathematically,
dt
5
Case r = 0: b = d; the population remains constant.
Case r < 0: b < d; the population goes extinct.
dN
= 0 always.
dt
dN
< 0 always.
dt
N(t)
if r > 0
if r = 0
N0
if r < 0
t
Figure 2: Graphs of N (t) = N0 ert for values of r.
Qualitative analysis of
dN
= rN (phase line analysis)
dt
Here we consider only two cases; when r > 0, and when r < 0.
Case r > 0:
dN
dt
dN
= rN
dt
N
Figure 3: Graph of
dN
= rN with r > 0.
dt
6
dN
If N > 0, then
> 0, so N is increasing, as indicated by the right-facing arrow on the
dt
N -axis.
dN
Moreover, the larger N is, the larger
is. That is, the rate of change of N is increasdt
ing.
We then conclude that the graph of N (t) must look like this:
N(t)
N0
t
Figure 4: Note that the larger N is, the larger the instantaneous slope of N (t).
Case r < 0:
N
dN
= rN
dt
dN
dt
Figure 5: Graph of
dN
= rN with r < 0.
dt
dN
If N < 0, then
< 0, so N is decreasing, as indicated by the left-facing arrow on the
dt
N -axis.
dN
Moreover, the smaller N gets, the smaller
is. In particular, the rate of change of N is
dt
decreasing. It decreases to zero as N → 0.
7
We then conclude that the graph of N (t) must look like this:
N(t)
N0
t
Figure 6: Note that the smaller N is, the smaller the instantaneous slope of N (t). Also
note that the slope approaches zero.
8
Step 2: Development and analysis of the differential equation
for logistic growth
As before, let N (t) be the number of individuals in a population at time t.
In addition, let K be the carrying capacity of the population. That is, if the population is
initially below K, it will increase to K, and if the population is initially above K, it will
decrease to K.
When the population is K, its needs are met precisely by the environment, and can just
be maintained at that level.
A question for students:
• What should be the sign of
dN
when
dt
0 < N < K?
dN
>0
dt
N = K?
dN
=0
dt
N > K?
dN
<0
dt
Return briefly to the exponential growth equation, and take a look at the per-captia growth
dN/dt
.
rate which is defined as
N
dN/dt
dN
= rN , then
= r is the corresponding per-captial growth rate.
If
dt
N
9
The graph of
dN/dt
N
as a function of N looks like this:
dN/dt
N
r
N
Figure 7: The per-capita growth rate is constant for the exponential growth model.
We will now modify this graph, to arrive at the per-capita growth rate for the logistic
growth model, as follows.
First, we need the per capita growth rate to be a decreasing function of N (this will limit
growth when the population is large).
Second, we need the per capita growth rate to satisfy the following:
dN/dt
>0
N
dN/dt
<0
N
when N < K
when N > K
10
The simplest choice is:
dN/dt
N
r
K
Figure 8:
dN/dt
N
N
as a function of N .
Note that this linear function satisfies the conditions imposed above.
Finding the equation of the straight line can be posed as an exercise to students.
The equation is
dN/dt
N
N
=r 1−
.
K
Multiplying by N , we arrive at the following differential equation describing logistic
growth.
dN
N
=r 1−
N.
dt
K
This equation is known as the logistic equation.
N
In the equation, r 1 −
is the per-capita growth rate.
K
11
Another way to look at the logistic equation is as follows:
dN
N
=r 1−
N
dt
K
r
= rN − N 2
K
r 2
When N is small, the second term,
N , this term can be neglected.
K
That is, the population initially undergoes exponential growth.
r 2
As the population increases, the
N term becomes more important,
K
and has a negative effect on the rate of growth of the population (as it
is subtracted). That is, the population growth is limited.
dN
As N → K, the two terms balance, and
= 0. That is, the population
dt
stops growing. It stagnates at N = K.
Now that we have the logistic equation, what can we do? How do we understand the
population dynamics?
We have two options:
1. We can solve the equation with traditional methods. The solution to the logistic
equation is
KN0
where N0 = N (0).
N (t) =
N0 + (K − N0 )e−rt
Then we can graph the solution N (t) as a function of t (after selecting reasonable
values for r, k, and N0 , of course). See the appendices for this method.
2. We can analyse the equation with modern qualitative methods, as follows:
12
We begin by graphing
dN
as a function of N .
dt
x
It may be helpful to ask students to first sketch a graph of y = r 1 −
x, which
K
students should recognize as a parabola opening downwards with x-intercepts at x = 0 and
x = k.
dN
dt
0
Figure 9:
K
dN
dt
N
as a function of N with phase-line analysis.
Note of the following:
dN
= 0. Thus, N stays constant at N = 0.
dt
dN
When 0 < N < K,
> 0. Thus, N is increasing, as denoted by the right-facing
dt
arrows on the N -axis.
dN
= 0. Thus, N stays constant at N = K.
When N = K,
dt
dN
When N > K,
< 0. Thus, N is decreasing, as denoted by the left-facing arrow
dt
on the N -axis.
K
The maximum occurs at N =
(showing this is a good exercise for students).
2
• When N = 0,
•
•
•
•
13
Moreover, we see that:
dN
dt
K
dN
dt
≈0
dN
dt
is maximized
Figure 10:
dN
dt
N
≈0
dN
dt
≈0
dN
as a function of N .
dt
We can thus infer the following qualitative solution N (t) as a function of t:
N(t)
K
K
2
t
Figure 11: Qualitative solution for N (t) as a function of t.
K
Note the inflection point as possible solutions cross the line N (t) = . At this point, the
2
population growth rate is maximized, in other words, the population grows the fastest at
this point.
14
Step 3: Introduction to a system of differential equations describing an infectious disease
Rationale:
• This model can be reduced to the logistic equation, the behaviour of which students
know from Step 2.
• The model introduces students to the concept of dividing the population into a class
of susceptible individuals and a class of infected individuals.
• The model introduces students to the concept of R0 , the basic reproduction number
of the disease.
Let N be the total number of individuals in a population.
We consider the population closed, that is, it is constant. There are no births or deaths,
nor immigration or emigration.
For example, the student body at your school may be considered a closed population (on
the time scale of one year).
Now suppose that one individual in the population falls ill with an infectious disease. How
can we describe the spread of the disease through the population. How many individuals
will fall ill?
To do so, we divide the population into two classes, namely,
Susceptibles: healthy individuals.
Infectives: individuals who have fallen ill with the disease, and are
infectious (capable of infecting susceptibles).
Let S(t) be the number of susceptibles in the population at time t. Similarly, let I(t) be
the number of infectives in the population at time t.
Since the population is closed, we have
S(t) + I(t) = N for all times t.
Initially we have I(0) = 1 and S(0) = N − 1.
A question for students: What can you say about the signs of S(t) and I(t)?
S(t) and I(t) must both be positive as they describe the number of individuals in a population. It is possible for one of them to be 0, as this would indicate that either everyone is
infected or everyone is susceptible.
15
We visualize the movement of individuals between classes as follows:
γI
S
I
βIS
β, γ > 0
Figure 12: Visual representation of the movement of individuals between classes.
The arrow labelled βIS describes the rate at which susceptibles become infected. By
assumption, this rate is proportional to the number of susceptibles and the number of
infectives (I times S measures the number of contacts between susceptibles and infectives),
with proportionality constant β (β measures the infectivity of the disease, or how contagious
the disease is—the larger the value of β, the more contagious the disease).
The arrow labelled γI describes the rate at which infectives recover from the disease and
become susceptible again. By assumption, this rate is proportional to the number of
infectives, with proportionality constant γ (γ measures how quickly infectives recover—the
larger the value of γ, the less time infectives spend in the infective class).
Note that we have ignored immunity to the disease (that is, this model is reasonable for
bacterial infections such as tuberculosis, meningitis, and gonorrhoea).
If we translate the cartoon model (our diagram above, and yes, that is the colloquial name)
into mathematical equations for the unknowns S(t) and I(t), we get
dS
= −βIS + γI
dt
dI
= βIS − γI
dt
Note that the individuals leaving the susceptible class, βIS, turn up in the infective class (as
they are removed—subtracted, from the susceptible class, and added to the infective class).
Similarly, individuals leaving the infective class turn up in the susceptible class.
An exercise for the students is to add the two equations and recall that S + I = N .
Note that we can add the two equations as follows:
dS dI
d(S + I)
dN
+
= −βIS + γI + βIS − γI ⇒
=0⇒
= 0.
dt
dt
dt
dt
16
dN
= 0, consistent with the meaning of N ?
dt
This equation is consistent as we defined our population, N , to be closed—meaning that
it does not change over time.
Is this equation,
We now take advantage of the fact that our population is closed, that is, N is constant,
namely N = S + I.
In particular, if we know I, then we can recover S from S = N − I. Likewise, if we know
S, then we can recover I from I = N − S.
We choose to eliminate S and its corresponding differential equation by writing S = N − 1,
and rewriting the differential equation for I as follows:
dI
= βIS − γI
dt
= βI(N − I) − γI
= βN I − βI 2 − γI
= (βN − γ)I − βI 2
Question: Do you recognize this equation?
The equation is a logistic equation, albeit in disguise!
We can see this by rewriting the equation as follows:
dI
= (βN − γ)I − βI 2
dt
= (βn − γ −βI)I
| {z }
constant
"
= (βN − γ) 1 −
If we let r = βN − γ and K =
βN −γ
β ,
I
βN −γ
β
#
I
then we have
dI
I
=r 1−
I,
dt
K
that is, the logistic equation, as required.
For ease of analysis, we choose to work with the equation in the following form:
dI
= [βN − γ − βI] I.
dt
17
In particular, we will use the qualitative (phase line) analysis we used earlier in the context
of the logistic equation to this differential equation.
dI
We need to graph
versus I. As before, the graph will be a parabola opening downwards
dt
since β > 0. One of the intercepts is at I = 0. The second intercept is determined by
βN − γ − βI = 0
βI = βN − γ
⇒
⇒
I=
βN − γ
β
Note that this intercept may be positive or negative, depending on the sign of βN − γ,
which is determined by the relative sizes of β, γ, and N .
dI
versus I for the two cases
dt
(a) βN − γ > 0
Graphing
(b) βN − γ < 0
can be left as an exercise for students.
dI
dt
dI
dt
I=0
I=
βN −γ
β
I
I=
Figure 13: Case (a): βN − γ > 0.
βN −γ
β
I=0
I
Figure 14: Case (b): βN − γ < 0.
Another exercise for students could be to perform the phase-line analysis for each case, and
draw a graph of the solution I(t) versus t for each case. What are the conclusions?
18
dI
dt
dI
dt
I=0
I=
βN −γ
β
I
I=
Figure 15: Phase-line analysis for (a)
βN −γ
β
I=0
I
Figure 16: Phase-line analysis for (b)
I(t)
I(t)
t
t
Figure 17: Some possible solutions for (a).
Figure 18: A possible solution for (b)
The dotted line in Figure 17 is given by I =
βN − γ
γ
=N− .
β
β
Conclusions:
Case (a): In this case, the number of infectives approaches N − βγ . That is,
there always will be N − βγ infectives within the population.
Case (b): In this case, the number of infectives approaches 0. That is, the
disease will die out (disappear from the population).
Retrospective and introduction to R0 , the disease reproduction
ratio
From the phase-line analysis, we found that the disease is endemic (it persists in the
population) if βN − γ > 0 and that the disease goes extinct if βN − γ < 0.
19
We can rewrite the inequalities as follows:
βN − γ > 0 ⇐⇒ βN > γ ⇐⇒
and βN − γ < 0 ⇐⇒
We let
R0 =
βN
>1
γ
βN
<1
γ
βN
.
γ
R0 is the basic reproduction ratio of the disease.
If R0 > 1, the disease is endemic.
If R0 < 1, the disease goes extinct.
Decomposition of R0 =
βN
:
γ
• βN is the rate at which a single infective introduced into a susceptible population
of size N makes infectious contacts (from the term βIS in the differential equations,
with I = 1 and S = N ).
•
1
is the expected length of time that an infective remains infectious.
γ
Thus, R0 =
βN
is the expected number of infectious contacts made by an infective.
γ
It follows then, if R0 > 1, the disease will take hold and if R0 < 1, the disease will die
out.
Q: How does immunity play into this?
Immunity effectively reduces the (initial) size of the susceptible population.
Let p be the proportion of the population that is immunized (recall 0 ≤ p ≤ 1; the larger
the value of p, the more of the population is immunized).
Then pN is the number of individuals in the population being immunized. Likewise,
N − pN = (1 − p)N is the number of individuals not immunized.
The original R0 value,
βN
, is reduced in this case.
γ
20
The new R0 value, say R̄0 , is
old R0
R̄0 =
(1 − p)
| {z }
z}|{
R0 .
proportion immunized
Q: What proportion of the population should be immunized in order to
eradicate a disease? How does the proportion depend on the R0 value of
that disease?
We need
(1 − p)R0 < 1
⇒ R0 − pR0 < 1
⇒ pR0 > R0 − 1
R0 − 1
1
⇒ p>
=1−
R0
R0
1
= 0.95. In other words, at least 95% of the population
R0
needs to be immunized to eradicate such a disease. We can represent this idea visually:
For example, if R0 = 20, then 1−
p
1
0.95
0.80
0.75
0.66
0.50
12345
20
R0
Figure 19: p, the proportion of population to be immunized, as a function of R0 .
21
Note:
• If R0 < 1, there is no need to immunize.
• The larger R0 , the more of the population needs to immunized.
• Thus, diseases with large R0 , say R0 > 20, are difficult to eradicate since at least 95%
of the population needs to be immunized, and large-scale immunization is difficult to
do.
Possible follow-up questions
1. Will every susceptible individual need to be immunized in order to eradicate the
disease?
2. Look up the term “herd immunity” on Wikipedia. What does it mean?
3. You may be interested to follow the link on the “herd immunity” Wikipedia page to
“The mathematics of mass vaccination.”
4. Look up the R0 value for smallpox and measles. Why do you think global eradication
for smallpox has been accomplished, but not for measles?
22
Appendices
Appendix A: Solving the logistic equation via separation of variables and
partial fractions
dN
r
N
To solve
N =
= r 1−
(K − N ), we recognize that the equation is separadt
K
K
ble:
Z
dN
=
(K − N )N
1
K
Z
K −N
+
1
K
Z
r
dt
K
!
N
dN =
Z
r
dt
K
1
1
r
ln(K − N ) +
ln N = t + C1
K
K
K
ln N − ln(K − N ) = rt + C2
N
ln
= rt + C2
K −N
N
= ert+C2
K −N
−
= ert eC2
N
= C3 ert
K −N
(partial fraction decomposition; see end)
(integrate each term)
(multiply by K; C2 = KC1 )
(law of logarithms)
(exponentiate both sides)
(law of exponents)
(C3 = eC2 )
Now we isolate N :
N = C3 er t(K − N )
= C3 ert K − C − 3ert N
(1 + C3 ert )N = C3 ert K
N=
C3 ert K
1 + C3 ert
(∗)
23
To determine the constant of integration C3 , we apply the initial condition N (0) =
N0 .
C3 K
= N0
N (0) =
1 + C3
An exercise for students could be to show that C3 =
N0
into (∗) to get:
K − N0
N (t) =
=
N0
. Then substitute C3 =
K − N0
N0
rt
K−N0 e K
N0
1 + K−N
ert
0
K
+1
KN0
=
,
N0 + (K − N0 )e−rt
K−N0 −rt
N0 e
as required.
The partial fraction decomposition was obtained as follows:
Let
1
A
B
=
+
(K − N )N
K −N
N
AN + B(K − N )
=
(K − N )N
We need
1 = AN + B(K − N )
= AN + BK − BN
= BK + (A − B)N
Which gives
Thus,
1
K
BK = 1
=⇒
B=
A−B =0
=⇒
A=B=
1/K
1/K
1
=
+
.
(K − N )N
K −N
N
24
1
K
Appendix B: Solving the logistic equation via a Bernoulli transformation
dN
N
To solve
N , we rewrite it as follows:
=r 1−
dt
K
dN
r
= rN − N 2
dt
K
dN
r
− rN = − N 2
dt
K
(1)
and recognize the latter as being in the form of a Bernoulli equation (see below for more
information).
We use the substitution X = N −1 =
1
1
or N = .
N
X
dN
1 dX
=− 2
by the chain rule.
dt
X dt
Substitute into (1) to get
1 dX
1
r 1
−
−r =−
K dt
X
K X2
Then
Multiply by −X 2 to get
dX
r
+ rX =
dt
K
Rearrange to get
dX
r
= −rX +
dt
K
1
= r( − X)
K
(linear equation; looks similar to the exponential growth)
(2)
We introduce another transformation Y (t) = X(t) −
K is a constant). Then
1
1
or X(t) = Y (t) +
(recall that
K
K
dX
dY
=
dt
dt
Substitute into (2) to get
dY
= −rY
dt
We recognize this as an exponential growth equation. We can thus write down the solution
immediately, namely
Y (t) = Y0 e−rt
25
We can now work backwards to retrieve first X(t) and then N (t).
1
Y (t) = Y0 e−rt with Y (t) = X(t) −
gives
K
1
1
e−rt
X(t) −
= X0 −
K
K
1
1
or X(t) =
e−rt
+ X0 −
K
K
Recall that N (t) =
1
from the first transformation, so
X(t)
N (t) =
1
1
K
+
1
N0
−
1
K
e−rt
=
KN0
, as required.
N0 + (K − N0 )e−rt
About Bernoulli Equations:
An equation of the form
dy
+ p(x)y = q(x)y n
dx
is known as a Bernoulli equation, after the mathematician Jakob
Bernoulli.
The substitution v = y 1−n reduces a Bernoulli equation to a linear equation. This substitution method was discovered in 1696 by Gottfried
Leibniz.
In relation to our equation,
dN
r
− rN = − N 2 :
dt
K
In the above, replace y by N and x by t,
dN
+ p(t)N = q(t)N n
dt
r
We have p(t) = −r (constant), and q(t) = − (constant) and n = 2.
K
1
So a substitution X(t) = N (t)1−2 =
should work.
N (t)
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Appendix C: Questions for students about Appendices A & B
dN
N
1. The explicit solution for
N is
=r 1−
dt
K
N (t) =
KN0
.
N0 + (K − N0 )e−rt
Show that
lim N (t) = K
t→∞
and interpret this result in terms of the biological relevance of the problem.
100N0
as a function
N0 + (100 − N0 )e−t
of t, for the following values of the initial condition N0 :
2. Set K = 100, and r = 1. Plot the solution N (t) =
N0 = 0, 1, 25, 75, 99, 100, 101, 150, 200.
Describe the shapes of the curves. Which curves have a point of inflection? Which
ones don’t? Can you tell at which value of N (not t!) the points of inflection occur
if they occur?
3. Find out about the lives and accomplishments of Jakob Bernoulli and Gottfried
Leibniz.
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