Homework 3.1 Solutions
Math 5110/6830
1. (a) Fixed points satisfy:
A∗
B∗
= µ1 A∗ − µ3 A∗ B ∗
= µ2 B ∗ − µ4 A∗ B ∗
Solving this for A∗ and B ∗ gives us:
(A∗ , B ∗ )
(A∗ , B ∗ )
=
(0, 0)
µ2 − 1 µ1 − 1
=
,
µ4
µ3
(b) To determine stability of these points, we need to find a jacobian matrix & look at the determinant
and trace:
µ1 − µ3 B ∗
−µ3 A∗
J(A∗ , B ∗ ) =
−µ4 B ∗
µ2 − µ4 A∗
So, at (0, 0):
J(0, 0)
µ1
0
0
µ2
1.2
0
0
1.3
=
=
For the parameter values given,
T r(J) = µ1 + µ2 = 1.2 + 1.3 = 2.5
Det(J) = µ1 µ2 = (1.2)(1.3) = 1.56
For this point to be stable, we need
|T r(J)| < 1 + Det(J) < 2
Since, |2.5| < 1 + 1.56 = 2.56 > 2, then (0, 0) is unstable. (Note that we could have also looked at
the evals of J). , µ1µ−1
:
At µ2µ−1
4
3
J
µ2 − 1 µ1 − 1
,
µ4
µ3
−µ3 µ2µ−1
4
=
µ1 −1
−µ4 µ3
1
1
−0.15
=
−0.4
1
1
For the parameter values given,
T r(J)
=
1+1=2
µ2 − 1
µ1 − 1
Det(J) = 1 − −µ3
−µ4
µ4
µ3
= 1 − (−0.15)(−0.4) = .94
Since, |2| > 1 + .94, then ( µ2µ−1
, µ1µ−1
) is unstable.
4
3
!
2. (a) The parameter value k is the infectivity rate or the contact rate between susceptible and infected
people.
(b) To find what this system predicts, we should think to look for the stability of the equilibria values.
To find these fixed points:
I∗
= I ∗ + kI ∗ (N − I ∗ )
Solving this for I ∗ we find that the equilibria points are:
I∗
I∗
= 0
= N
To find the stability, let
f (I)
= I + kI(N − I)
Then,
f 0 (I)
= 1 + k(N − I) − kI
= 1 + k(N − 2I)
For I ∗ = 0, |f 0 (0)| = |1 + kN | > 1. Then, I ∗ = 0 is unstable.
For I ∗ = N , |f 0 (0)| = |1 − kN |. For this to be stable:
−1 < 1 − kN
−2 < −kN
2>
kN
<1
<0
>0
Then, I ∗ = N is stable when kN < 2. Therefore, this model predicts that everyone will catch the
disease.
(c) To include people that have recovered from the disease, we need a new variable Rn . Then, we can
also say that the total number of people in the population are N = Hn + In + Rn , where Hn are the
healthy (susceptible) people. We will also need to track these individuals as their population changes
(ie. write an equation for Rn+1 ). To account for the people that have recovered in d days, we need
to subtract a term from the infected equation. Remember that d is NOT a rate at which people are
leaving the infected class. It is also NOT the average time spent in the class. This parameter tells
us exactly how long each individual stays infected. Then, the number of newly infected individuals
on day n + 1 is
kIn (N − In − Rn )
The number of people who move into the recovered class from the infected class on day n are those
that became newly infected d days ago. In mathematics, this is
kIn−d (N − In−d − Rn−d )
The number of people who recover is NOT just In−d because this is the entire population at time
n − d which includes people who became infected at various times NOT just exactly d days ago.
Now, we can write the new system as:
In+1
Rn+1
= In + kIn (N − In − Rn ) − kIn−d (N − In−d − Rn−d )
= Rn + kIn−d (N − In−d − Rn−d )
with
R−d
I−d = ...
I0
= ... = R−1 = R0
Rd
= I−2 = I−1 = 0
= 1
= ... = Rd−1 = 0
= 1
(d) Steady states of the above model:
I∗
R∗
= I ∗ + kI ∗ (N − I ∗ − R∗ ) − kI ∗ (N − I ∗ − R∗ ) = I ∗
= R∗ + kI ∗ (N − I ∗ − R∗ )
Then, we have equilibria values (I ∗ , R∗ ) = (0, 0) (nobody ever gets sick) and (I ∗ , R∗ ) = (0, N )
(everyone gets sick & recovers).
(e) To analyze the stability of these equilibria, look at the Jacobian. By the assumptions of the model,
everyone will recover so (0, N ) should be a stable equilibrium.
Homework 3.2 Solutions
Math 5110/6830
1. (a)
det(A)
a −1
pR
= R
pJ
aJ − 1
= (aR − 1)(aJ − 1) − pR pJ
= (−pJ )(−pR ) − pR pJ = 0
(b)
J
=
aR
pJ
pR
aJ
The eigenvalues are given by (this requires a little algebra and the two conditions: 1 = aR + pJ and
1 = aJ + pR ):
aR − λ
pR pJ
aJ − λ = (λ − aR )(λ − aJ ) − pR pJ
= λ2 − (aR + aJ ) + aR + aJ − 1
= (λ − 1)(λ − (aR + aJ − 1)) = 0
So, the evals are:
λ = 1 and λ = aR + aJ − 1
2. (a) Fixed points must satisfy:
∗
H∗
P∗
= kH ∗ e−aP
∗
= cH ∗ [1 − e−aP ]
The first equation gives us H ∗ = 0 or P ∗ = ln(k)
a .
Substituting H ∗ = 0 into the second eqn we find that
(H ∗ , P ∗ )
Substituting P ∗ =
ln(k)
a
=
(0, 0)
into the second eqn we find that
k ln(k) ln(k)
(H ∗ , P ∗ ) =
,
ac(k − 1)
a
To make sure that P ∗ > 0, we need to require k > 1.
(b) The Jacobian for this system is:
J(H ∗ , P ∗ )
For (H ∗ , P ∗ ) =
k ln(k) ln(k)
ac(k−1) , a
J
=
∗
∗
k ln(k)
c(1−k)
ln(k)
(k−1)
#
ke−aP
−akH ∗ e−aP
∗
−aP ∗
c(1 − e
) acH ∗ e−aP
:
k ln(k) ln(k)
,
ac(k − 1)
a
"
=
1
c(1 − k1 )
For this,
T r(J)
=
Det(J)
=
ln(k)
k−1
ln(k)
k ln(k)
1
−
c 1−
k−1
c(1 − k)
k
ln(k)
ln(k)
+
(k − 1)
k−1
k−1
k ln(k)
k−1
1+
=
=
To satisfy the Jury conditions for stability, we need |T r(J)| < 1 + Det(J) < 2. Checking the second
condition (1 + Det(J) < 2):
k ln(k)
k−1
k ln(k)
k−1
k ln(k)
k ln(k) − k + 1
1+
< 2
< 1
< k−1
< 0
Let f (k) = k ln(k) − k + 1. Since f 0 (k) = 1 + ln(k) − 1 = ln(k) > 0 for k > 1 we see that f (k)
is monotonically increasing function.
Also, f
(1) = 0 so k ln(k) − k + 1 > 0. Then, the 2nd Jury
k ln(k) ln(k)
conditions fails and (H ∗ , P ∗ ) = ac(k−1)
, a
is unstable.
3. (a) The fixed points are given by:
∗
∗
H
= H ∗ er(1− K ) e−aP
∗
= cH ∗ 1 − e−aP
H∗
P∗
Solving this system for fixed points would involve solving a transcendental eqn and we don’t know
how to do this. However, we can see that one solution to these equations is (H ∗ , P ∗ ) = (0, 0). The
other one is found by setting P ∗ = 0, then (H ∗ , P ∗ ) = (K, 0).
(b) The Jacobian for this system is:
"
∗
∗
J(H , P )
=
er(1−
H∗
K
) e−aP ∗ 1 −
∗
c 1 − e−aP
rH ∗
K
−aH ∗ er(1−
H∗
K
) e−aP ∗
acH ∗ e−aP
#
∗
For (H ∗ , P ∗ ) = (0, 0):
J(0, 0)
=
er
0
0
0
Since the leading eigenvalue is λ = er > 1, then (0, 0) is stable for r < 0 and otherwise unstable.
For (H ∗ , P ∗ ) = (K, 0):
1 − r −aK
J(K, 0) =
0
acK
Let’s look at the trace and determinant of this system.
T r(J) = 1 − r + acK
Det(J) = (1 − r)acK
Looking at |T r(J)| < 1 + Det(J) < 2:
|1 − r + acK|
<
for
1 − r + acK
<
−r + acK
<
−r <
1 >
also
−1 − (1 − r)acK
<
−(1 − r)acK
<
racK
<
0 <
r <
1 + (1 − r)acK
1 + (1 − r)acK
(1 − r)acK
−racK
acK
1 − r + acK
2 − r + acK
2 − r + 2acK
(2 − r)(1 + acK)
2
We also need:
1 + (1 − r)acK
(1 − r)acK
< 2
< 1
As long as acK < 1 and r < 2, then this point is stable. Otherwise it’s unstable.
4. (a) Poisson:
0.4
v=1
v=4
v=10
0.35
0.3
p(i)
0.25
0.2
0.15
0.1
0.05
0
0
2
4
6
8
10
i
12
14
16
18
20
(b) Examples of poisson processes:
•
•
•
•
encounters between a predator & prey
seed dispersal
mutations in virus replication or in DNA after radiation
release of the neurotransmitter acetylcholine at a neuromuscular junction
(c) For this new model, we do not need to change the Hn+1 equation since the host population dynamics
will remain unchanged. However, we do need to account for the number of new parasitoids that are
created from the hosts. This value is c for the hosts that only had 1 encounter and 2c for those having
more than one encounter. We have p(0) = e−aPn as the probability of 0 encounters, p(1) = aPn e−aPn
as the probability of 1 encounter, so 1 − p(1) − p(0) is the probability that a host has had more than
1 encounter. Then, the new model becomes:
Hn+1
Pn+1
= kHn e−aPn
= cHn aPn e−aPn + 2cHn 1 − aPn e−aPn − e−aPn