Centres · Differential Equations

Intuition

When the eigenvalues are purely imaginary there is turning but no winding: every solution goes round a closed loop for ever, and the origin is a centre. The frictionless spring is the model: its energy is conserved, and the loops are the curves of constant energy. A centre is stable, since nearby solutions stay near, but not asymptotically stable, since they never arrive. It is also fragile: the smallest friction turns it into a stable spiral, and the smallest push into an unstable one.

A marble rolling round the inside of a perfectly smooth bowl at a fixed height. Nothing pulls it down and nothing lifts it, so it goes round the same path for ever; a breath of friction and it slowly spirals to the bottom.

The centre of $x'=y$ , $y'=-4x$ , with eigenvalues $\pm2i$ . Every solution goes clockwise round an ellipse $4x^{2}+y^{2}=\text{constant}$ and returns after the time $\pi$ , whatever its size.

Purely imaginary eigenvalues

If $\operatorname{tr}A=0$ and $\det A>0$ , the eigenvalues are $\pm i\beta$ with $\beta=\sqrt{\det A}$ , and the origin is a centre. Every solution is periodic with period , and its trajectory is an ellipse about the origin. A centre is stable, because each solution stays on its own ellipse, and not asymptotically stable, because none approaches the origin.

\operatorname{tr}A=0,\ \det A>0\ \Longrightarrow\ \lambda=\pm i\sqrt{\det A}

What a centre does

The frictionless spring $x''+\omega_{0}^{2}x=0$ is , , with eigenvalues .

The energy is conserved, and every trajectory closes

x'=y,\ y'=-\omega^{2}x\ \Longrightarrow\ \bigl(\omega^{2}x^{2}+y^{2}\bigr)'=0

Differentiate the quantity along a solution and substitute the two equations: the two terms are equal and opposite, so it never changes, and each solution stays on one ellipse. The only point where the velocity vanishes is the origin, so a solution on an ellipse keeps moving round it in one direction, returns to its start, and from there, by uniqueness, repeats.

Proof steps

E=\omega^{2}x^{2}+y^{2}

Twice the energy of the spring, in these units.

E'=2\omega^{2}xx'+2yy'

Differentiate along a solution.

E'=2\omega^{2}xy+2y\bigl(-\omega^{2}x\bigr)=0

Substitute the equations: the terms cancel.

\omega^{2}x^{2}+y^{2}=E(0)

Each solution stays on one ellipse, and goes round it for ever.

Applications

Practice

Zero Trace, Positive Determinant

When the trace is zero and the determinant positive, the eigenvalues are purely imaginary and every solution goes round a closed loop.

\operatorname{tr}A=0,\ \det A>0\ \Longrightarrow\ \lambda=\pm i\beta

Try it

Which matrix gives a centre?