Stability and phase portraits

Why this is asked: Classify a 2×2 linear system from trace and determinant alone; for nonlinear systems linearise and use Hartman–Grobman — but remember the centre case is the one linearisation cannot decide.

Classification of ẋ = Ax in the plane

Let T = trace $A, D = \det A, \Delta = T^{2} - 4D$ .

Condition	Type	Stability
D < 0	saddle	unstable
$D > 0, \Delta > 0, T < 0$	stable node	asymptotically stable
$D > 0, \Delta > 0, T > 0$	unstable node	unstable
$D > 0, \Delta < 0, T < 0$	stable spiral	asymptotically stable
$D > 0, \Delta < 0, T > 0$	unstable spiral	unstable
D > 0, T = 0	centre	stable, not asymptotically
D = 0	degenerate (a line of equilibria)	—

Summary: asymptotically stable ⇔ T < 0 and D > 0 ⇔ both eigenvalues have negative real part.

Nonlinear systems

Linearise at an equilibrium: J = Df(x*).

Hartman–Grobman: if no eigenvalue of J has zero real part (a hyperbolic equilibrium), the nonlinear phase portrait is locally equivalent to the linear one.
The centre case is undecidable by linearisation: ẋ $= -y - x^{3}$ , ẏ $= x - y^{3}$ has a linear centre but is actually asymptotically stable, while flipping the cubic signs makes it unstable.

Lyapunov's direct method

Find V with V(x*) = 0, V > 0 nearby:

V̇ ≤ 0 ⇒ stable; V̇ < 0 ⇒ asymptotically stable; V̇ > 0 ⇒ unstable.

Works where linearisation fails. For a conservative system, the energy is the natural V.

Key takeaways

T < 0 and D > 0 ⇔ asymptotically stable; D < 0 ⇔ saddle.
Hyperbolic equilibria are settled by linearisation; centres are not.
Lyapunov functions handle the non-hyperbolic cases.

See it move

The trace–determinant planeinteractive

Drag a point across the plane and watch the portrait change: saddle, node, spiral, centre — a map instead of a list of rules.

Unlock interactive visuals

The trap here

“Linearisation determines stability at every equilibrium” — false

ẋ $= -y - x^{3}$ , ẏ $= x - y^{3}$ at the origin

The linearisation is a centre (eigenvalues ±i), but $V = x^{2} + y^{2}$ gives V̇ $= -2(x^{4} + y^{4}) < 0$ : asymptotically stable. Non-hyperbolic equilibria escape Hartman–Grobman.