ee222 - problem set 3 solutions - university of california ... · ee222 spring 2017 - problem set 3...

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.

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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

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Figure 1: Problem 3, Simple Model

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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

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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)

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Figure 3: Problem 5, Part (i)

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ee222 - problem set 3 solutions - university of california ... · ee222 spring 2017 - problem set 3...

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