Math 5110/6830
Homework 5.1 Solutions
1. To find the steady states for Hn and Pn , substitue H ∗ for all H’s and P ∗ for all P’s. Then
H ∗ = kH ∗ e−aP
∗
∗
P ∗ = cH ∗ (1 − e−aP )
From the first equation you can see H ∗ = 0 is a solution, therefore P ∗ = 0 is a solution. To find
the other solution:
∗
1 = ke−aP
1
ln = −aP ∗
k
ln k
P∗ =
a
Now using the second steady state equations we see that H ∗ =
P∗
c(1−e−aP ∗ )
or H ∗ =
ln k
a
−a lnak
c(1−e
)
ln k
ac(1 − k1 )
k ln k
H∗ =
ac(k − 1)
H∗ =
2. To begin, we need to determine the probabilities of a host having exactly one or two or more
encounters. From class, we know that the probability that a host has r encounters is given by
p(r) =
e−aPn · (aPn )r
.
r!
Then the probability of a host having exactly one encounter is aPt e−aPt . Therefore, the probability
that a host has two or more encounters is the probability of having exactly zero encounters and
the probability of having exactly one encounter subtracted from one, i.e. (1 − eaPt − aPt e−aPt ).
So, if f parasitoid progeny will be produced after one encounter and 2f after two or more, only
the second equation of the original model needs to be modified:
Hn+1 = Hn e−aPn
Pn+1 = Hn caPn e−aPn + 2c 1 − e−aPn (1 + aPn ) .
3. (a) The fixed points satisfy
N∗ = e
∗
r 1− NK
N ∗ e−aP
∗
∗
P ∗ = λf N ∗ (1 − e−aP )
Solving this system for fixed points would involve solving a transcendental equation, which
we don’t know how to do. However, we can see that one solution to these equations is
(N ∗ , P ∗ ) = (0, 0). For N ∗ 6= 0,
1=e
∗
r 1− NK
−aP ∗
e
.
The other fixed point can be found by setting P ∗ = 0 in this equation. Then solving for N ∗
gives (N ∗ , P ∗ ) = (K, 0).
1
(b) To determine stability, we need to look at the Jacobian of the system evaluated at the fixed
points.
" N∗ #
∗
∗
r 1− K −aP ∗
r 1− NK −aP ∗
rN
∗
∗
∗
1
−
e
−aN
e
K
J(N , P ) =
∗
∗
λf (1 − e−aP )
aλf N ∗ e−aP
For (N ∗ , P ∗ ) = (0, 0):
er 0
J(0, 0) =
.
0 0
The leading eigenvalue is µ = er , which satisfies |er | = er < 1 for r < 0. Therefore, this
point is stable for r < 0 and unstable otherwise.
For (N ∗ , P ∗ ) = (K, 0):
J(K, 0) =
1 − r −aK
,
0
aλf K
which has eigenvalues µ1 = 1 − r and µ2 = aλf K. For stability, we require that 0 < r < 2
and |aλf K| < 1. This fixed point is therefore unstable when r < 0 or r > 2 and when
|aλf K| > 1.
2
Homework 5.2 Solutions
1. The possible genotypes and corresponding wing colors are as follows:
Genotype
WW
Ww
ww
Wing Color
white
white
black
Mother
W (pn )
w (1 − pn )
2
WW (pn )
Ww (pn (1 − pn ))
W (pn )
Father
w (1 − pn ) Ww (pn (1 − pn )) ww ((1 − pn )2 )
number of W alleles
If pn is the fraction of W alleles in the population, then pn = total
number of alleles . To determine
the total number of alleles in the population, we use the fact that any given genotype in the
population contributes 2 alleles. So the total number of alleles in the n + 1 generation is equal
to 2 times the number of surviving genotypes from the previous generation. Since white-colored
moths can either have a WW or Ww genotype, the total number of these moths that survive
is given by α times the frequency of each genotype: α · (p2n + 2pn (1 − pn )). Similarly, the ww
genotype has survival fraction γ and frequency (1 − pn )2 , thus making the number of black
moths that survive from one generation to the next γ · (1 − pn )2 . Putting it all together, the total
number of alleles in the next generation will be represented by 2·[αp2n +2αpn (1−pn )+γ(1−pn )2 ].
Now, the number of W alleles in the next generation will be given by twice the number of
surviving WW moths plus the number of surviving Ww moths, since there are two W alleles
associated with the former genotype and only one associated with the latter. Note that there is
no contribution from the ww moths. Thus, the number of W alleles in the population will be
2 · αp2n + 1 · 2αpn (1 − pn ). Together, we have
pn+1 =
=
=
=
=
=
2 · αp2n + 1 · 2αpn (1 − pn )
2 · [αp2n + 2αpn (1 − pn ) + γ(1 − pn )2 ]
2αpn
2 · [αp2n + 2αpn (1 − pn ) + γ(1 − pn )2 ]
αpn
2
αpn + 2αpn (1 − pn ) + γ(1 − pn )2
αpn
−αp2n + 2αpn + γ(p2n − 2pn + 1)
αpn
2
(γ − α)pn + 2(α − γ)pn + γ
αpn
.
2
(γ − α)pn − 2(γ − α)pn + γ
(a) The fixed points satisfy
p∗ =
αp∗
,
(γ − α)p∗ 2 − 2(γ − α)p∗ + γ
3
to which p∗1 = 0 is a solution. For p∗ 6= 0 and assuming that α 6= γ, we have
α
1=
2
∗
(γ − α)p − 2(γ − α)p∗ + γ
⇒
0 = (γ − α)p∗ 2 − 2(γ − α)p∗ + (γ − α)
⇒
0 = p∗ 2 − 2p∗ + 1
⇒
p∗2 = 1.
(b) For stability of p∗ , we require |f 0 (p∗ )| < 1, where
αpn
f (pn ) =
.
2
(γ − α)pn − 2(γ − α)pn + γ
Differentiating,
α[(γ − α)p2n − 2(γ − α)pn + γ] − αpn [2(γ − α)pn − 2(γ − α)]
[(γ − α)p2n − 2(γ − α)pn + γ]2
−α(γ − α)p2n + αγ
=
.
[(γ − α)p2n − 2(γ − α)pn + γ]2
f 0 (p∗ ) =
For p∗1 = 0: f 0 (0) = αγ , which implies that p∗ = 0 is stable for α < γ and unstable for α > γ.
For p∗2 = 1: f 0 (1) = 1, which gives us no information about stability. We can do cobwebbing
to verify that this point is stable when α > γ and unstable when α < γ.
(c) If γ = 0.8, then we know that p∗1 = 0 is stable for α < 0.8 and unstable for α > 0.8. In
addition, p∗2 = 1 is unstable for α < 0.8 and stable for α > 0.8.
Bifurcation diagram with α varying
3
2.5
2
p*
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
α
(d) If we start with an environment in which white-colored moths have an advantage, the relationship between the parameters should be α > γ, and the solution will tend toward the
fixed point at 1, favoring the white-colored moths. When the Industrial Revolution takes
place, the black moths have the advantage, suggesting that α < γ now. The solution, after
this turn of events, will now decay toward 0, favoring the black moths. Thus, over time, the
white-colored moths will have the long-lasting disadvantage, and will eventually die out.
4
1
0.9
0.8
0.7
p
n
0.6
0.5
0.4
0.3
0.2
Industrial Revolution
0.1
0
0
5
10
15
20
5
25
n
30
35
40
45
50