EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
EE222 - Problem Set 3 Solutions
February 17, 2016
Problem 1
Problem 1a
The divergence of the vector field is
divf(x) = −1 + a 6= 0
where we used the fact that a 6= 1. Bendixson’s Theorem immediately gives the solution.
Problem 1b
Observe that the unique equilibrium is at (0, 0). Now simply interpret the second equation x2 = x2 whose
solution is x2(t) = x2,0et with initial condition x2,0 ∈ R. Since x2(t) is strictly monotonically increasing
(decreasing) for x2,0 > 0 (< 0), there can be no closed orbits.
Problem 2
Part (i)
Define region M to be the convex hull of the equilibria (−2, 0), (1,±√
3). Then, by the Poincare-Bendixson
Theorem, M contains an equilibrium point and/or a closed orbit. But since (0, 0) is an equilibrium
contained in M , we cannot make any definitive statement about the existence of closed orbits in M .
Part (ii)
Compute the Lie-Derivative of V (x, y) (that is, along its trajectories):
dV
dt=∂V
∂x
∂x
∂t+∂V
∂y
∂y
∂t=
(−x+
1
2(y2 − x2)
)x+ (−y + xy)y = ... = 0
and so V is constant along trajectories of the system.
For a point (ε1, ε2) “close” to (0, 0), V (ε1, ε2) is
V (ε1, ε2) = −1
2(x2 + y2) + h.o.t.
Hence, trajectories close to the origin are circular → closed orbits.
February 17, 2016 1 / 14
EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
Problem 3
Part (i)
We first consider the case
x = ax− bxy
y = −dy + cxy
By letting x(a− by) = 0 and y(cx− d) = 0, the equilibria are (0, 0) and (d/c, a/b). Linearizing the system
gives
Df =
[(a− by) −bxcy (cx− d)
]For the equilibrium at (0, 0), the eigenvalues of the Jacobian are λ1 = a, λ2 = −d, from which it follows
that (0, 0) is a saddle. However, note that (0, 0) is not a proper equilibrium point because we assume the
population of fish to be strictly positive x, y > 0.
For the equilibrium at (d/c, a/b), the eigenvalues of the Jacobian are λ1,2 = ±j√ad. As the Hartman-
Grobman Theorem does not apply here, we need to simulate the system to find that this equilibrium
corresponds to a center, see Figure 1.
Part (ii)
For the more complicated model, the equilibria are at (0, 0), (a/λ, 0), and(bd+aµbc+λµ ,
ac−dλbc+λµ
). The Jacobian
is
Df =
[a− by − 2λx −bx
cy −d+ cx− 2µy
]Evaluating Df at the equilibria yields
• (x∗, y∗) = (0, 0)⇒ saddle
• (x∗, y∗) = (a/λ, 0)⇒ saddle if ac > dλ, stable node otherwise
• (x∗, y∗) =(bd+aµbc+λµ ,
ac−dλbc+λµ
)⇒ stability
Note that the equilibria at (0, 0) and (a/λ, 0) have no physical meaning. Stability of the interior equilibrium
is found by evaluating the Jacobian at the equilibrium:
Df
(bd+ aµ
bc+ λµ,ac− dλbc+ λµ
)=
[−λ bd+aµbc+λµ −b bd+aµbc+λµ
cac−dλbc+λµ −µac−dλbc+λµ
]=:
[−k1 −k2k3 −k4
]where the constants k1, . . . , k4 are strictly greater than zero (for ac > dλ). The characteristic polynomial
is
s2 + (k1 + k4)s2 + k2k3 = 0
February 17, 2016 2 / 14
EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
Figure 1: Problem 3, Simple Model
February 17, 2016 3 / 14
EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
and we see that all coefficients are strictly greater than zero. By the Routh-Hurwitz criterion, the eigen-
values have negative real parts and so this equilibrium is stable. A more precise analysis is possible by
analyzing the discriminant√
(k1 + k4)2 − 4k2k3. See Figure 2 for a phase portrait.
Figure 2: Problem 3, Complicated Model
Problem 4
February 17, 2016 4 / 14
EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
Problem 5
Part (i)
From inspection it follows that the equilibria are at (0,±µ). The Jacobian is just the derivative of x with
respect to x and reads
Df(x∗, µ) = µ2 − 3(x∗)2.
We see that it has a zero eigenvalue at µ = x∗ = 0.
For the case µ 6= 0, do the following case differentiation:
• x∗ = 0→ Df(x∗, µ) = µ2 > 0⇒ unstable for all µ
• x∗ = ±µ→ Df(x∗, µ) = −2µ2 < 0⇒ stable for all µ
See Figure 3 for a plot.
Part (ii)
February 17, 2016 10 / 14
EE222 Spring 2017 - Problem Set 3 Solutions Datong Paul Zhou, [email protected]
Figure 3: Problem 5, Part (i)
February 17, 2016 14 / 14