Homework 6 Solutions
Math 5110/6830
1. (a)
dS/dt
dI/dt
dS
dI
S(I)
−βIS
βIS − kI
−βS
=
βS − k
Z I(S)
=
dI¯
=
β S̄ − k
dS̄
−β S̄
S(0)
I(0)
Z S(I) k
dS̄ = I(S) − I(0)
−1 +
β S̄
S(0)
S(I)
k
−S̄ + ln(S̄)
= I(S) − I(0)
β
S(0)
k
k
= I(S) − I(0)
−S(I) + ln(S(I)) − −S(0) + ln(S(0))
β
β
k
k
−S(I) + ln(S(I)) = I(S) − I(0) − S(0) + ln(S(0))
β
β
Z
(b) Plugging in ICs and terminal conditions:
−S∞ +
k
ln(S∞ )
β
= −S0 +
k
ln(S0 )
β
Clearly, S∞ = S0 is a solution.
(c) Let
f (S)
= −S +
k
ln(S)
β
Note that f (S∞ ) = f (S0 ). Let’s find the maximum:
f 0 (S)
So, the maximum occurs at S =
β
k.
= −1 +
k
βS
Graph of f(S):
Notice that if S0 > βk , then, there is a positive solution S0 6= S∞ such that S∞ <
S0 < βk , then the solution is S0 = S∞ .
k
β
< S0 . But, if
2. (a) Adding in recovery:
dS
dt
dI
dt
= bS − βIS + γI
= βIS − kI − γI
Equilibria points:
0 = bS ∗ − βI ∗ S ∗ + γI ∗ = S ∗ (b − βI ∗ )
0 = βI ∗ S ∗ − kI ∗ − γI ∗ = I ∗ (βS ∗ − k − γ)
So, we have (0, 0) and
k+γ b(k+γ)
β ,
βk
. To determine stability analytically, let’s take the Jacobian:
b − βI ∗
−βS ∗ + γ
∗ ∗
J(S , I ) =
βI ∗
βS ∗ − k − γ
At (0, 0):
J(0, 0)
=
b
γ
0 −k − γ
Recall that an equilibria point is stable if we have both T r(J) < 0 and Det(J) > 0. Then,
= b−k−γ
= −b(k + γ)
T r(J)
Det(J)
Since b > 0, k > 0, and γ > 0, then the determinant is NOT greater than 0. Therefore (0, 0) is
unstable. However, this doesn’t tell us what kind of unstable point we have. For this, we need to
find the eigenvalues of J:
λ1
λ2
= b>0
= −k − γ < 0
So, (0, 0) is a saddle
point.
k+γ b(k+γ)
:
At
β ,
βk
J
k + γ b(k + γ)
,
β
βk
=
− γk
b(k+γ)
k
−k
0
Here,
T r(J)
Det(J)
γ
k
= b(k + γ)
= −
b(k+γ)
Since, T r(J) < 0 and Det(J) > 0, then k+γ
is stable. However, this doesn’t tell us what
β ,
βk
kind of stable point we have. Eigenvalues of J:
p
−bγ ± γ 2 − 4bk 3 − 4bγk 2
λ1,2 =
2k
These are complex, so we are spiraling.
Now, let’s show this with nullclines. The S-nullclines (where dS
dt = 0):
I
S
=
=
0
0
I
=
bS
βS − γ
We’ll be putting S on the x-axis, so we’ll have a → where
bS
Then, we have a ← where I > βS−γ
.
dI
The I-nullclines (where dt = 0):
I
=
S
=
> 0. This occurs where I <
bS
βS−γ .
k+γ
β .
Then,
0
k+γ
β
We’ll be putting I on the y-axis, so we’ll have a ↑ where
we have a ↓ where S >
Phase plane:
dS
dt
dI
dt
> 0. This occurs where S <
k+γ
β .
Just as we found analytically, (0, 0) is a saddle node, and
k+γ b(k+γ)
β ,
βk
is a stable spiral.
(b) Now, let ν = cb where c < 1 so ν < b. Then, adding in reproduction of infected individuals:
dS
dt
dI
dt
= bS − βIS + νI
= βIS − kI
bk
Equilibria points are (0, 0) and βk , β(k−ν)
. For a positive equilibria we need to make the restriction
that k > ν. For stability, let’s find the Jacobian:
b − βI ∗ −βS ∗ + ν
∗ ∗
J(S , I ) =
βI ∗
βS ∗ − k
At (0, 0):
J(0, 0)
=
b ν
0 −k
Theeigenvalues
of this are λ1 = b and λ2 = −k. Then (0, 0) is an unstable saddle point.
k
bk
At β , β(k−ν)
J
bk
k
,
β β(k − ν)
=
bν
− k−ν
bk
k−ν
−k + ν
0
Eigenvalues are complex with negative real part, so this point is a stable spiral.
S-nullclines:
I
=
bS
βS − ν
I-nullclines:
I
=
S
=
0
k
β
Phase Plane:
bk
Therefore, if βk , β(k−ν)
exists, then trajectories will spiral in towards it. And, if it doesn’t exist,
then the population will blow up.
3. (a)
A
J
θJ
δA
µAJ
r(J)A
r0
ηµJ
=
=
=
=
=
=
=
=
no. of adults
no. of juveniles
no. juveniles maturing to adults
no. adults dying
no. people dying from cannibalism
per capita reproduction
initial reproduction rate
feedback response
(b) J-nullclines:
A =
θJ
r0 + µ(η − 1)J
A-nullclines:
A = 0
J = 0
θJ
A =
δ
Phase-plane diagram:
(c) Equilibria values are (0, 0) and
r0 > δ since η < 1. Jacobian:
J(J ∗ , A∗ )
θ(δ−r0 )
δ−r0
µ(η−1) , δµ(η−1)
=
.For the second point to exist, we need to have
−δ
r0 + ηµJ ∗ − µJ ∗
θ
ηµA∗ − µA∗ − θ
At (0, 0):
J(0, 0)
=
−δ
r0
θ
−θ
For this point, T r(J) = −δ − θ, which is less than 0, and Det(J) = θ(δ − r0 ). The determinant is
greater than zero if δ > r0 . So, there are going to be some conditions on stability. Let’s first look at
the other point before
we make conclusions about the stability.
θ(δ−r0 )
δ−r0
At µ(η−1) , δµ(η−1) :
J
δ − r0 θ(δ − r0 )
,
µ(η − 1) δµ(η − 1)
=
−δ
δ
θ
− θrδ0
For this point, T r(J) = −δ − θrδ0 and Det(J) = θ(r0 − δ). So, this is stable if r0 > δ. Now, we can
make some conclusions about the stability of these points.
If δ > r0 , then there is only one equilibria point,
(0, 0), and itsstable. However, if r0 > δ, then the
θ(δ−r0 )
δ−r0
unstable point is (0, 0) and the stable point is µ(η−1)
, δµ(η−1)
.
(d) The population will survive iff r0 > δ and η < 1, otherwise the population will stabilize at 0.