Stability and phase portraits

In one sentence

Near a hyperbolic equilibrium the linearisation tells you everything, the trace and determinant classify every planar linear system between them, and the centre is the one case where linearisation decides nothing.

Why the exam asks it

Because classifying an equilibrium is a two-number computation, and because the borderline case — purely imaginary eigenvalues — is where the method quietly stops working. An option offering "centre" for a nonlinear system is right about the linearisation and often wrong about the system.

The idea

For an autonomous system $\mathbf{x}' = F(\mathbf{x})$, an equilibrium has $F(\mathbf{x}^*) = 0$. Linearise: $\mathbf{x}' \approx J\,(\mathbf{x}-\mathbf{x}^*)$ with $J$ the Jacobian at $\mathbf{x}^*$.

The planar classification

With $\tau = \operatorname{tr}J$ and $\Delta = \det J$, the eigenvalues satisfy $\lambda^2 - \tau\lambda + \Delta = 0$, so

$$\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}.$$

• $\Delta < 0$: real eigenvalues of opposite sign — saddle, always unstable.
• $\Delta > 0$ and $\tau^2 > 4\Delta$: real, same sign — node, stable if $\tau<0$.
• $\Delta > 0$ and $\tau^2 < 4\Delta$: complex — spiral, stable if $\tau<0$.
• $\Delta > 0$ and $\tau = 0$: purely imaginary — centre.

So the parabola $\tau^2 = 4\Delta$ separates nodes from spirals, and the sign of $\tau$ sets stability. Two numbers classify everything, which is why this is examinable in a single line of work.

Hyperbolic equilibria and Hartman–Grobman

An equilibrium is hyperbolic if no eigenvalue has zero real part.

Hartman–Grobman. Near a hyperbolic equilibrium, the nonlinear system is topologically conjugate to its linearisation.

So for hyperbolic equilibria the linear picture is the true picture: all eigenvalues with negative real part gives asymptotic stability, any eigenvalue with positive real part gives instability.

The centre is the exception

When $\tau = 0$ and $\Delta > 0$ the equilibrium is not hyperbolic, and the linearisation decides nothing. The standard illustration:

$$x' = -y + \alpha x(x^2+y^2), \qquad y' = x + \alpha y(x^2+y^2).$$

The Jacobian at the origin is $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ for every $\alpha$ — a centre by the linear classification. In polar coordinates, though, $r' = \alpha r^3$:

• $\alpha < 0$: spirals inward, asymptotically stable.
• $\alpha = 0$: a genuine centre.
• $\alpha > 0$: spirals outward, unstable.

Same linearisation, three different behaviours. A linear centre is the one case where the nonlinear terms decide, and options exploit it by offering "centre, therefore stable".

Lyapunov functions

When linearisation fails, look for $V$ with $V(\mathbf{x}^*) = 0$, $V > 0$ nearby, and $\dot V = \nabla V \cdot F \le 0$. Then the equilibrium is stable; with $\dot V < 0$ strictly, it is asymptotically stable.

This needs no eigenvalues and handles the non-hyperbolic cases. For a conservative system the energy is usually the right $V$, which is why undamped oscillators give genuine centres.

Where intuition breaks

"The linearisation always determines stability." Only at hyperbolic equilibria. The centre case is decided by the nonlinear terms.

"Purely imaginary eigenvalues mean a centre." For the linear system, yes. For a nonlinear one it may spiral either way, as $\alpha$ above shows.

"$\Delta > 0$ and $\tau < 0$ means a stable node." Stable, but a node only if $\tau^2 > 4\Delta$. Otherwise a stable spiral. Both are stable, so the stability question and the shape question are separate.

"A saddle can be stable." Never — $\Delta<0$ gives eigenvalues of both signs.

"Stable means asymptotically stable." A centre is stable and not asymptotically so: orbits stay near and do not approach. Lyapunov stability and asymptotic stability are different conclusions.

"An equilibrium with a zero eigenvalue is unstable." Undetermined. Zero real part means non-hyperbolic, and the nonlinear terms decide.

"Every planar system settles to an equilibrium." Poincaré–Bendixson allows limit cycles: a bounded orbit avoiding equilibria approaches a periodic orbit. That is a planar phenomenon and fails in three dimensions.

The exam's angle

1. Find the equilibria, then the Jacobian at each. Everything follows from those.
2. Compute $\tau$ and $\Delta$ and read the picture. Sign of $\Delta$ first — negative means saddle and you are done.
3. Check hyperbolicity before trusting the answer. Zero real part means the linearisation is silent.
4. For non-hyperbolic cases, try polar coordinates or a Lyapunov function. The $r' = \alpha r^3$ computation is the model.
5. Separate stability from shape. Node and spiral can both be stable; the question often asks only one.

The night before

• Linearise at each equilibrium; classify by $\tau = \operatorname{tr}J$, $\Delta = \det J$.
• $\Delta<0$ saddle; $\Delta>0$ with $\tau^2>4\Delta$ node, $<4\Delta$ spiral; $\tau=0$, $\Delta>0$ centre. Stability from the sign of $\tau$.
• Hartman–Grobman: at a hyperbolic equilibrium the nonlinear system matches its linearisation.
• The centre is not hyperbolic, and nonlinear terms decide: $r' = \alpha r^3$ gives inward, neutral or outward for the same Jacobian.
• Stable $\ne$ asymptotically stable. A centre is the former only.
• Lyapunov: $V>0$, $\dot V \le 0$ gives stability without eigenvalues.
• Poincaré–Bendixson: in the plane, a bounded orbit avoiding equilibria tends to a limit cycle.