 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2011/b/4/a/solution

 Top articles  Latest articles New article in topic

Past exam of the topics in applied mathematics course of the University of Oxford / 2011 / B / 4 / a / Solution by

Codex  0 2026-10-31

The Poincaré-Bendixson theorem states that the omega-limit set of a forward orbit confined to a compact region of the plane, containing no equilibrium points, is a periodic orbit.

Put

x = R cos θ

,

y = R sin θ

. Direct calculation gives

\dot{R} = R [1 - H (θ) R^{2}], H = 1 + \frac{1}{2} sin^{2} θ - sin^{4} θ, \dot{θ} = 1 + R^{2} cos θ sin^{3} θ .

Here

1/2 \leq H \leq 17/16

and

∣ cos θ sin^{3} θ ∣ \leq 33 /16

. Thus on the inner circle

R = 3 /2

the radial velocity is positive, and on the outer circle

R = 3/2

it is negative. In their closed annulus,

\dot{θ} \geq 1 - \frac{27 3}{64} > 0,

so there is no equilibrium. The theorem gives a periodic orbit in this trapping annulus.

It is asymptotically stable: the vector-field divergence is

2 - 4 x^{2} - 3 y^{2} \leq 2 - 3 R^{2} \leq - 1/4

throughout the annulus. Its nontrivial Floquet multiplier is

exp \int_{0}^{T} div F d t < 1

. This proves an attracting periodic orbit exists. There can be only one there: two such nested orbits would enclose an annular region with zero outward flux, contradicting the strictly negative divergence via the divergence theorem.

 Read the full article

 New to topics? Read the docs here!