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ID: past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2021/3/a/i/solution

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Past exam of the differential equations 1 course of the University of Oxford / 2021 / 3 / a / i / Solution by

Codex  0 2026-10-25

The equilibrium points satisfy

y = β

and

x (x^{2} - 1) = 0

, so they are

(- 1, β), (0, β), (1, β) .

The Jacobian matrix and its eigenvalue equation are

J (x, y) = (0 3 x^{2} - 1 10), λ^{2} = 3 x^{2} - 1.

At

x = \pm 1

the eigenvalues are

\pm 2

, so both points are saddle equilibria. Their stable and unstable tangent directions have slopes

- 2

and

2

, respectively.

At

x = 0

the eigenvalues are

\pm i

, giving a centre for the linearized equation. Purely imaginary eigenvalues alone do not classify the nonlinear system. The first integral established in part (ii) proves that

(0, β)

is in fact a centre equilibrium. It is clockwise: immediately to its right, with

0 < x < 1

and

y = β

, the vector field points downwards. Thus the final nonlinear classification is

(- 1, β), (1, β) are saddles; (0, β) is a clockwise centre .

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