Math 5110/6830
Homework 8.1 Solutions
1. Including the modifications described, the new systems becomes
dS
dt
dI
dt
= −βIS + γI + bS
= βIS − αI − γI
The I-nullclines for this system, which can by found by setting
dI
dt
= 0, are
I=0
α+γ
S=
β
While the S-nullcline for this system, which can by found by setting
I=
dS
dt
= 0, is
bS
βS − γ
and a vertical asymptote at S = βγ .
b(α+γ)
,
. The non-zero
The I-nullcines and S-nullcine intersect at the fix points: (0, 0) and α+γ
β
αβ
fix point is to the right of the vertical asymptote. To find the stability of these points, we must
first find the Jacobian:
−βI ∗ + b −βS ∗ + γ
.
J(S∗, I∗) =
βI ∗
βS ∗ − α − γ
Notice there is has a horizontal asymptote at I =
b
β
So,
b
γ
.
J(0, 0) =
0 −α − γ
The eigenvalues of this matrix are λ1 = b, λ2 = −α − γ. Because we can assume all parameters
are positive, the fix point (0, 0) is a saddle. The Jacobian for the non-zero steady state is:
#
" −bγ
α
α + γ b(α + γ)
α
J
,
=
.
bγ
0
b + alpha
β
αβ
−bγ
α
which is always negative.
The determinant is αb + bγ which is
α+γ b(α+γ)
always positive. From this we know that
is a stable fix point (either a spiral or a
β , αβ
node). Once we draw the phase plane we will be able to determine that this point is actually a
stable spiral.
The trace of this matrix is
1
b(α+γ)
αβ
0
α+γ
β
2. (a) J describes the population of juveniles while A describes the population of adults. ro A is
the basal reproduction rate, ηµJ is a feedback term, µAJ is a term describing cannibalism
and θ is the rate that juveniles become adults. Finally, δ is the death rate of adults.
(b) See (c)
θJ
A-nullcline is J = δA
θ .
ro+µ(η−1)J . The θ(δ−r
)
o
δ−ro
. Because we cannot
and µ(η−1)
, δµ(η−1)
(c) The J-nullcline is A =
The nullclines intersect at
have a negative population
the fix points (0, 0)
and η < 1, the non-zero fix point only exist if δ < ro . To find the stability of the fix points,
lets look at the Jacobian.
ηµA∗ − µA∗ − θ ro + ηµJ ∗ − µJ ∗
∗
∗
.
J(J , A ) =
θ
−δ
For (0, 0),
−θ ro
J(0, 0) =
.
θ −δ
The trace of this matrix −θ −δ which is negative. The determinant is θ(δ −ro ). Therefore, if
δ > ro , the determinant is positive so (0, 0)
is stable. If δ < ro , the determinant is negative
θ(δ−ro )
δ−ro
so (0, 0) is a saddle. For µ(η−1) , δµ(η−1) ,
J
δ − ro θ(δ − ro )
,
µ(η − 1) δµ(η − 1)
−θro
=
δ
θ
δ
.
−δ
o
The trace is −θr
δ − δ which is negative. The determinant is θ(ro − δ). Since this fix point
only exist if δ < ro , the determinant is positive. Therefore the nonzero steady state is stable
(either a spiral or a node). If this steady state existed for δ > ro , the determinant would be
negative, so the fix point would be a saddle. There are two possible phase planes, δ < ro
and δ > ro .
2
A
δ<ro
J−nullcline
A−nullcline
J
δ>ro
A
J−nullcline
A−nullcline
J
(d) In order for the population to survive there must be a stable nonzero steady state. This
means that δ < ro and η < 1.
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Homework 8.2 Solutions
1. (a) Phase portraits for the system
ẋ = µ − x2 ,
ẏ = −y.
µ=0
µ>0
0
0
√
- µ
√
µ
µ<0
0
0
In the y direction, the behavior tends toward the x-axis.
(b) Nullclines for varying µ:
µ=0
µ>0
µ<0
0
√
- µ
√
µ
0
As µ decreases, the x-nullclines get closer together, coalescing at µ = 0, and then disappear
as µ decreases further. This is the same behavior observed with the fixed points of a saddlenode bifurcation.
2. (a) Nullclines for small a for the system
ẋ = −ax + y,
ẏ =
x2
− by.
1 + x2
The x-nullcline is in red, and the y-nullcline is in green. Note that there are three fixed
points in this case.
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0.1
0.09
0.08
0.07
y
0.06
0.05
0.04
0.03
0.02
0.01
0
0
0.5
1
1.5
2
2.5
x
(b) As a increases, the two nontrivial fixed points get closer and closer together, until they
become one point. As a continues to increase, these fixed points disappear altogether.
a increasing further
a increasing
0.1
0.08
0.08
0.06
0.06
0.04
0.04
0.02
0.02
y
0.1
0
0
0.5
1
x
1.5
0
2
0
0.5
1
x
1.5
2
(c) Based on the fact that the number of points at which the nullclines intersect decreases from
three to two and finally to one, there appears to be a saddle-node bifurcation as
pa increases.
1
±
1 − 4(ab)2
Analytically, the fixed points, when they exist, have x-coordinates x∗ =
.
2ab
∗
∗
∗
Since y = ax , we can plot y for fixed b and varying a to obtain the following bifurcation
diagram:
0.1
0.09
0.08
0.07
y*
0.06
0.05
0.04
0.03
0.02
0.01
0
0
0.02
0.04
0.06
a
5
0.08
0.1
(d) (Extra Credit) We can see from the the differential equations that a is the rate at which x is
degraded. Similarly, x is created at a rate proportional to y. In other words, there must be
mRNA present in order to make proteins. Proteins are used in biological process or degraded
at a rate a. Therefore, mRNA promotes proteins. Similar to proteins, mRNA is degraded
or used up at a rate b. Proteins have a more interesting effect on mRNA. The presence of
proteins does promote messenger RNA to form however this effect is limited. If there is a
lot of protein around the effect of protein on mRNA limits to 1. If there is very little protein
around, the terms limits to zero. Biological, this make senses because there is control over
the production of mRNA (therefore controlling the production of proteins). A large amount
of protein present in the cell is not going to cause the cell to instantly make a large amount
of mRNA.
3. (Extra Credit) The fix points for the system ẋ = r − x − e−x are x = r − e−x . This is a
transcendental equation. Therefore by graphing f (x) = r − x − e−x and seeing where f (x)
crosses the x-axis we can see the fix points.
1.5
r=0
r=.5
r=1
r=1.5
r=2
1
0.5
0
f(x)
−0.5
−1
−1.5
−2
−2.5
−3
−2
−1.5
−1
−0.5
0
0.5
x
1
1.5
2
2.5
3
As you can see, when r < 1 the parabola is below the x-axis for all values of x and there are no
fix points. When r = 1, there is exactly one fix point at x = 0. As r continues to increase, the
parabola shifts up and 2 fix points exist for all r > 1. This is a saddle-node bifurcation with a
bifurcation parameter of r = 1.
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