Math 5110/6830
Homework 7.1 Solutions
1. (a) The fixed points are P1 = (0, 0), P2 = (0, 3), P3 = (2, 0) and P4 = (1, 1).
(b) The Jacobian for this system, evaluated at the fixed point (x∗ , y ∗ ), is
2 − 2x∗ − y ∗
−x∗
J(x∗ , y ∗ ) =
.
−2y ∗
3 − 2y ∗ − 2x∗
2 0
For P1 : J(0, 0) =
, which has eigenvalues λ1 = 2 and λ2 = 3. Since both eigenvalues
0 3
are positive, this point
node.
is an unstable
−1 0
For P2 : J(0, 3) =
, which has eigenvalues λ1 = −1 and λ2 = −3. Since both
−6 −3
eigenvalues are negative,
thispoint is a stable node.
−2 −2
, which has eigenvalues λ1 = −2 and λ2 = −1. Since both
For P3 : J(2, 0) =
0 −1
eigenvalues are negative,
thispoint is a stable node.
√
√
−1 −1
, which has eigenvalues λ1 = −1 + 2 and λ2 = −1 − 2.
For P4 : J(1, 1) =
−2 −1
Since the eigenvalues are of opposite sign, this point is a saddle.
2. (a) Graphical analysis:
x-nullclines: y = x and y = −x.
y-nullcline: x = 1.
Phase Plane: x-nullclines in red, y-nullcline in green, fixed points in blue
2.5
2
1.5
1
y
0.5
0
−0.5
−1
−1.5
−2
−2.5
−2.5
−2
−1.5
−1
−0.5
0
x
0.5
Fixed points and stability:
The fixed points are P1 = (1, −1) and P2 = (1, 1).
1
1
1.5
2
2.5
The Jacobian for this system, evaluated at the fixed point (x∗ , y ∗ ), is
−2x∗ 2y ∗
∗ ∗
J(x , y ) =
.
1
0
−2 −2
For P1 : J(1, −1) =
, which has eigenvalues λ1 = −1 + i and λ2 = −1 − i. Since
1
0
the eigenvalues are complex
with negative real part, this point is a stable spiral.
√
√
−2 2
For P2 : J(1, 1) =
, which has eigenvalues λ1 = −1 + 3 and λ2 = −1 − 3. Since
1 0
the eigenvalues are of opposite sign, this point is a saddle.
(b) Graphical analysis:
x-nullcline: y = x3 .
y-nullcline: x = y 2 .
Phase Plane: x-nullcline in red, y-nullcline in green, fixed points in blue
2
1.5
y
1
0.5
0
−0.5
−1
−1
−0.5
0
0.5
x
1
1.5
2
Fixed points and stability:
The fixed points are P1 = (0, 0) and P2 = (1, 1).
The Jacobian for this system, evaluated at the fixed point (x∗ , y ∗ ), is
∗2
3x
−1
∗ ∗
J(x , y ) =
.
−1 2y ∗
0 −1
, which has eigenvalues λ1 = −1 and λ2 = 1. Since the
−1 0
eigenvalues are of opposite sign,
this point is a saddle.
√
√
3 −1
For P2 : J(1, 1) =
, which has eigenvalues λ1 = 5+2 5 and λ2 = 5−2 5 . Since the
−1 2
eigenvalues are both positive, this point is an unstable node.
For P1 : J(0, 0) =
2
3. Preliminaries: Because this is a system of species interaction, we are only concerned with x > 0
and y > 0.
Nullclines:
x-nullclines: x = 0 and y = − αβ
y-nullclines: y = 0 and x = − γδ
Fixed points: P1 = (0, 0) and P2 = − γδ , − αβ
Linearization: The Jacobian of this system, evaluated at the fixed point (x∗ , y ∗ ), is
α + βy ∗
βx∗
J(x∗ , y ∗ ) =
.
δy ∗
γ + δx∗
α 0
For P1 : J(0, 0) =
, which has eigenvalues λ1 = α and λ2 = γ.
0 γ
"
#
βγ
0
−
δ , which has eigenvalues λ = √αγ and λ = −√αγ.
For P2 : J − γδ , − αβ =
1
2
αδ
−β
0
(a) For mutualism of two species that cannot survive alone, the parameters take on the following
signs: α < 0, β > 0, γ < 0 and δ > 0. Note that both components of the fixed point P2
are positive here, and so this point makes sense biologically. Therefore, we can still consider
it in our analysis.
Stability:
For P1 : λ1 = α < 0 and λ2 = γ < 0. Since both eigenvalues are negative, this point is a
stable node.
√
√
For P2 : λ1 = αγ > 0 and λ2 = − αγ < 0. Since the eigenvalues are of opposite sign,
this point is a saddle.
Graphical analysis:
Phase space: x-nullclines in red, y-nullclines in green
–
α
β
0
0
–
γ
δ
Behavior for different initial conditions:
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For high x, low y: High x causes a rise in y, but if y is too low to begin with, this will
cause x to decrease. If x decreases substantially, there will be a further decrease in y, and
solutions will tend toward (0, 0). On the other hand, if y is low, but not too low, initially,
high x will increase y enough so that both populations can be sustained, and they will both
increase.
For high x, high y: The mutualistic nature of the system will cause both populations to
increase.
For low x, low y: Here, there’s not enough of either population to sustain the other, and
the trajectory will approach (0, 0).
For low x, high y: High y may cause a rise in x, but if x is too low to begin with, this will
cause y to decrease. As y decreases, so does x, and the populations will approach (0, 0).
However, if x begins low, but not too low, initially, high y will increase x enough so that
both populations can be sustained, and they will both increase.
(b) In a competition model where only one of the populations is self-sustainable, there are two
alternatives: either x is self-sustainable or y is self sustainable. We will consider both cases
separately.
Case 1: x is self-sustainable, and y is not. Then the parameters take on the following signs:
α > 0, β < 0, γ < 0 and δ < 0. Note that here, P2 doesn’t make sense biologically,
and so we will only consider P1 = (0, 0) in our analysis.
Stability:
For P1 : λ1 = α > 0 and λ2 = γ < 0. Since the eigenvalues are of opposite sign, this
point is a saddle.
Graphical analysis: Note that there is only one y-nullcline, since x = − γδ < 0 doesn’t
make sense biologically.
Phase space: x-nullclines in red, y-nullcline in green
–
α
β
0
0
Behavior for different initial conditions:
As a whole, this system favors the x species.
For low x, low y: Since x can sustain itself, the population will increase, causing further
damage to the y population, which will continue to decrease. This being the case, x
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will continue to increase with virtually no competition from y.
For low x, high y: Low x will cause y to decrease. Once y decreases enough, x begins
to thrive in the environment, out-competing y. x will therefore continue to increase.
For high x, low y: x will simply continue to increase, while y declines further.
For high x, high y: Initially, both populations will be hurt by competition. However, x
is hurt less by a decrease in y than y is hurt by a decrease in x. Therefore, y will decline
faster initially, causing x to recover. Once x recovers, y is too low to thrive and will
continue to decrease. Thus, x wins again!
Case 2: y is self-sustainable, and x is not. Then the parameters take on the following signs:
α < 0, β < 0, γ > 0 and δ < 0. Again, P2 doesn’t make sense in this model, so we will
only analyze P1 = (0, 0).
Stability:
For P1 : λ1 = α < 0 and λ2 = γ > 0. Since the eigenvalues are of opposite sign, this
point is a saddle.
Graphical analysis: Note that there is only one x-nullcline, since y = − αβ < 0 doesn’t
make sense biologically.
Phase space: x-nullcline in red, y-nullclines in green
0
0
–
γ
δ
Behavior for different initial conditions:
As a whole, this system favors the y species.
A similar line of reasoning follows in this case as in the first one.
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