Stability of equilibria — stable, unstable, saddle, spiral, centre: Worked examples

Two numbers do all the work (built in the parent, restated so nothing is assumed):

Recall The three numbers we read off

$τ = tr J = f_{x} + g_{y}$ — the sum of eigenvalues ("average growth rate"). $τ < 0$ pulls inward, $τ > 0$ pushes out.
$Δ = det J = f_{x} g_{y} - f_{y} g_{x}$ — the product of eigenvalues. $Δ > 0$ same sign, $Δ < 0$ opposite sign.
$D = τ^{2} - 4Δ$ — the discriminant under the square root in $λ = \frac{τ \pm D}{2}$ . $D < 0$ means complex eigenvalues, i.e. rotation.

See Trace–determinant plane for the map of all these regions at once, Jacobian matrix for where $J$ comes from, and Eigenvalues and eigenvectors for why $e^{λ t}$ appears.

The scenario matrix

Every cell below is a distinct behaviour a fixed point can have. The last column names the example on this page that hits it.

#	$τ$	$Δ$	$D = τ^{2} - 4Δ$	Type	Stability	Example
A	any	$Δ < 0$	$D > 0$ always	Saddle	unstable	Ex 1
B	$τ < 0$	$Δ > 0$	$D > 0$	Stable node	stable	Ex 2
C	$τ > 0$	$Δ > 0$	$D > 0$	Unstable node	unstable	Ex 3
D	$τ < 0$	$Δ > 0$	$D < 0$	Stable spiral	stable	Ex 4
E	$τ > 0$	$Δ > 0$	$D < 0$	Unstable spiral	unstable	Ex 4 (sign flip)
F	$τ = 0$	$Δ > 0$	$D < 0$	Centre	neutral	Ex 5
G	$τ \neq = 0$	$Δ > 0$	$D = 0$	Degenerate / star node	matches sign of $τ$	Ex 6
H	$τ < 0$	$Δ = 0$	$D = τ^{2} > 0$	Line of fixed points	non-isolated	Ex 7
I	nonlinear	—	—	Different type at each equilibrium	mixed	Ex 8 (word problem)
J	exam twist	parameter $a$	—	Type changes as a parameter crosses a threshold	bifurcation	Ex 9

We now walk them in order. Each example makes you forecast first.

Cell A — the saddle

Worked example Example 1 · Cell A (

$Δ < 0$ ) System: $\overset{x}{˙} = x + 3 y, \overset{y}{˙} = 3 x + y$ . Find the equilibrium and classify it.

Forecast: the two variables feed strongly into each other (the cross-coupling $3$ is bigger than the self-terms $1$ ). Guess before reading on: sink, source, or saddle?

Step 1 — locate the equilibrium. Set both velocities to zero: $x + 3 y = 0$ and $3 x + y = 0$ . Why this step? An equilibrium is by definition where $\overset{x}{˙} = \overset{y}{˙} = 0$ ; classification only makes sense at such a point. From the first, $x = - 3 y$ . Substitute into the second: $3 (- 3 y) + y = - 8 y = 0 \Rightarrow y = 0$ , hence $x = 0$ . So the equilibrium is the origin $(0, 0)$ .

Step 2 — build the Jacobian. $f = x + 3 y, g = 3 x + y$ , so $J = (f_{x} g_{x} f_{y} g_{y}) = (1331) .$ Why this step? Near the fixed point the nonlinear flow looks linear; $J$ is that local linear rulebook (here the system is already linear, so $J$ is constant everywhere).

Step 3 — read $τ, Δ$ . $τ = 1 + 1 = 2$ . For the determinant, subtract the off-diagonal product: $Δ = f_{x} g_{y} - f_{y} g_{x} = (1) (1) - (3) (3) = 1 - 9 = - 8 < 0$ . Why this step? The cross terms are what make $Δ$ negative — exactly why "vibes" forecasts fail; you must actually compute $f_{y} g_{x}$ . And $Δ < 0$ forces eigenvalues of opposite sign (their product is negative), which is the definition of a saddle: one direction grows, one decays.

Verify: eigenvalues $λ = \frac{τ \pm τ ^{2} - 4Δ}{2} = \frac{2 \pm 4 + 32}{2} = \frac{2 \pm 6}{2} = {4, - 2}$ . Opposite signs ✔ → saddle, unstable. The mode $e^{4 t}$ blows up, so almost every nudge sends the trajectory flying off.

The figure shows this saddle: the orange line $y = x$ (eigenvalue $λ = 4$ ) is the outward-fleeing direction, the plum line $y = - x$ (eigenvalue $λ = - 2$ ) is the incoming direction, and the teal streamlines sweep in along one and out along the other — the signature "riding-saddle" shape.

Figure — Stability of equilibria — stable, unstable, saddle, spiral, centre

Cells B & C — real nodes

Cells D & E — spirals (with a figure)

Worked example Example 4 · Cells D and E (spirals,

$D < 0$ ) Take the damped-oscillator-style system $\overset{x}{˙} = - x + ω y, \overset{y}{˙} = - ω x - y$ with $ω = 2$ .

Forecast: the antisymmetric $\pm ω$ terms rotate the flow. Diagonal is $- 1$ . Spiral in or out?

Step 1 — Jacobian. $J = (- 1 - 2 2 - 1)$ . Why this step? The skew part ( $+ 2, - 2$ ) is what forces rotation; the diagonal $- 1$ is the decay/growth. Reading them together shows a decaying rotation.

Step 2 — numbers. $τ = - 2$ , $Δ = (- 1) (- 1) - (2) (- 2) = 1 + 4 = 5$ , $D = 4 - 20 = - 16 < 0$ . Why this step? $D < 0$ ⇒ eigenvalues are complex conjugates $α \pm i β$ ; the imaginary part is the circling speed, the real part $α = τ /2$ is the spiral's growth.

Step 3 — eigenvalues. $λ = \frac{- 2 \pm - 16}{2} = - 1 \pm 2 i$ . Real part $- 1 < 0$ . Why this step? $Re λ < 0$ ⇒ amplitude $e^{- t}$ shrinks while phase $e^{2 i t}$ rotates → stable spiral (Cell D).

Sign flip → Cell E. Change diagonals to $+ 1$ : $\overset{x}{˙} = x + 2 y, \overset{y}{˙} = - 2 x + y$ . Then $τ = 2$ , $Δ = 5$ , $D = - 16$ , $λ = 1 \pm 2 i$ , $Re = + 1 > 0$ → unstable spiral. Same rotation, outward growth.

Verify: stable case product $(- 1)^{2} + 2^{2} = 5 = Δ$ ✔ (for $α \pm i β$ , product $= α^{2} + β^{2}$ ), sum $= 2 α = - 2 = τ$ ✔.

The figure puts the two side by side: left, the stable spiral ( $Re λ = - 1$ ) winds inward to the origin; right, the unstable spiral ( $Re λ = + 1$ ) winds outward. The rotation direction is identical in both — only the sign of $τ$ (and hence of $Re λ$ ) decides in-versus-out.

Cell F — the centre (borderline)

Worked example Example 5 · Cell F (

$τ = 0, Δ > 0$ ) $\overset{x}{˙} = 3 y, \overset{y}{˙} = - 3 x$ .

Forecast: pure rotation, no decay term anywhere. Does it spiral or just loop?

Step 1 — Jacobian. $J = (0 - 3 30)$ (equilibrium at the origin, since $\overset{x}{˙} = \overset{y}{˙} = 0$ only when $x = y = 0$ ). Why this step? A purely skew-symmetric $J$ has zero trace — no net inward or outward pull, only rotation.

Step 2 — numbers. $τ = 0$ , $Δ = 0 - (3) (- 3) = 9 > 0$ , $D = 0 - 36 = - 36 < 0$ . Why this step? $τ = 0$ with $Δ > 0$ is the knife-edge between stable and unstable spirals.

Step 3 — eigenvalues. Plug into the formula: $λ = \frac{τ \pm D}{2} = \frac{0 \pm - 36}{2} = \frac{\pm 6 i}{2} = \pm 3 i$ . Purely imaginary. Why this step? No real part ⇒ amplitude neither grows nor shrinks ⇒ closed orbits = centre, neutrally stable.

Verify: $λ_{1} λ_{2} = (3 i) (- 3 i) = 9 = Δ$ ✔, sum $= 0 = τ$ ✔. Centre. Caution: linearisation only guarantees a centre for a linear system; for nonlinear systems a $τ = 0$ point may be a weak spiral (see Linearisation and Hartman–Grobman theorem).

Cell G — degenerate / star node ( $D = 0$ )

Worked example Example 6 · Cell G (repeated eigenvalue)

$\overset{x}{˙} = - 2 x, \overset{y}{˙} = - 2 y$ .

Forecast: both equations decay at the same rate — what shape are the trajectories?

Step 1 — Jacobian. $J = (- 2 0 0 - 2) = - 2 I$ . Why this step? When $J$ is a scalar multiple of the identity, every direction is an eigenvector.

Step 2 — numbers. $τ = - 4$ , $Δ = 4$ , $D = 16 - 16 = 0$ . Why this step? $D = 0$ is the boundary between nodes and spirals — a repeated eigenvalue.

Step 3 — eigenvalue. $λ = \frac{- 4 \pm 0}{2} = - 2$ (double). Why this step? Both modes decay at rate $e^{- 2 t}$ ; because every direction is an eigenvector, trajectories are straight rays into the origin → star node (a special stable node).

Verify: $λ^{2} = 4 = Δ$ ✔, $2 λ = - 4 = τ$ ✔. Stable star node. (If the off-diagonal were nonzero, e.g. $J = (- 2 0 1 - 2)$ , still $D = 0$ but only one eigenvector → an improper/degenerate node, still stable since $λ = - 2 < 0$ .)

Cell H — a whole line of fixed points ( $Δ = 0$ )

Worked example Example 7 · Cell H (non-isolated equilibria)

$\overset{x}{˙} = - x + y, \overset{y}{˙} = x - y$ .

Forecast: the two right-hand sides are exact negatives of each other. What happens to the determinant when the rows of $J$ are proportional?

Step 1 — locate the equilibria. Set $- x + y = 0$ and $x - y = 0$ . Both say the same thing, $y = x$ . So every point on the line $y = x$ is an equilibrium — not just one isolated point. Why this step? When the two zero-velocity conditions coincide, the fixed points form a whole line; classification must acknowledge this is a non-isolated case before touching $J$ .

Step 2 — Jacobian and numbers. $J = (- 1 1 1 - 1)$ , so $τ = - 1 - 1 = - 2$ and $Δ = (- 1) (- 1) - (1) (1) = 1 - 1 = 0$ . Why this step? This lands exactly in Cell H: $τ < 0$ (there is a decaying direction) but $Δ = 0$ (one eigenvalue is exactly zero, reflecting the line of rest points).

Step 3 — eigenvalues. $λ = \frac{- 2 \pm 4 - 0}{2} = \frac{- 2 \pm 2}{2} = {0, - 2}$ . Why this step? The $0$ eigenvalue means no motion along the line $y = x$ ; the $- 2$ eigenvalue pulls every off-line point onto that line. Equilibria are non-isolated and only marginally stable.

Verify: product $0 \cdot (- 2) = 0 = Δ$ ✔, sum $0 + (- 2) = - 2 = τ$ ✔.

Cell I — real-world word problem (nonlinear, mixed types)

Worked example Example 8 · Cell I — competing species

Two species $x, y \geq 0$ obey $\overset{x}{˙} = x (3 - x - 2 y), \overset{y}{˙} = y (2 - x - y)$ . Classify the coexistence equilibrium.

Forecast: competition often means one species wins (saddle between them). Guess: stable coexistence or saddle?

Step 1 — find the interior equilibrium. Need $3 - x - 2 y = 0$ and $2 - x - y = 0$ . Subtract: $(3 - x - 2 y) - (2 - x - y) = 1 - y = 0 \Rightarrow y = 1$ , then $x = 2 - y = 1$ . So $(x^{*}, y^{*}) = (1, 1)$ . Why this step? We want the fixed point where both species survive; a word problem may have several equilibria — pick the one asked for.

Step 2 — Jacobian. With $f = x (3 - x - 2 y) = 3 x - x^{2} - 2 x y$ , $g = y (2 - x - y) = 2 y - x y - y^{2}$ : $J = (3 - 2 x - 2 y - y - 2 x 2 - x - 2 y) .$ At $(1, 1)$ : $J = (3 - 2 - 2 - 1 - 2 2 - 1 - 2) = (- 1 - 1 - 2 - 1)$ . Why this step? Evaluate $J$ at the specific fixed point — the parent warns each equilibrium can differ.

Step 3 — numbers. $τ = - 1 - 1 = - 2$ , $Δ = (- 1) (- 1) - (- 2) (- 1) = 1 - 2 = - 1 < 0$ . Why this step? $Δ < 0$ ⇒ opposite-sign eigenvalues ⇒ saddle: coexistence is unstable — a tiny imbalance drives one species to dominate.

Verify: $λ = \frac{- 2 \pm 4 + 4}{2} = - 1 \pm 2 \approx {0.414, - 2.414}$ ; product $(- 1)^{2} - 2 = - 1 = Δ$ ✔. Saddle — matches the ecological intuition that strong mutual competition rarely allows stable coexistence.

Cell J — exam twist: a bifurcation

Worked example Example 9 · Cell J — parameter crossing a threshold

$\overset{x}{˙} = y, \overset{y}{˙} = - x + a y$ , with a tunable damping/forcing parameter $a$ . Classify the origin for all $a$ .

Forecast: as $a$ increases from negative to positive, the origin must change character. Where does it switch, and to what?

Step 1 — Jacobian. $J = (0 - 1 1 a)$ (origin is always an equilibrium since $\overset{x}{˙} = \overset{y}{˙} = 0$ there).

Step 2 — numbers as functions of $a$ . $τ = a$ , $Δ = (0) (a) - (1) (- 1) = 1 > 0$ always, $D = a^{2} - 4$ . Why this step? $Δ > 0$ fixed means we never get a saddle; only $τ = a$ and $D = a^{2} - 4$ move — so the type is entirely steered by $a$ .

Step 3 — sweep every case.

$a < - 2$ : $τ < 0, D > 0$ → stable node.
$a = - 2$ : $D = 0$ → degenerate stable node (boundary).
$- 2 < a < 0$ : $τ < 0, D < 0$ → stable spiral.
$a = 0$ : $τ = 0, Δ > 0$ → centre (the bifurcation point!).
$0 < a < 2$ : $τ > 0, D < 0$ → unstable spiral.
$a = 2$ : $D = 0$ → degenerate unstable node.
$a > 2$ : $τ > 0, D > 0$ → unstable node. Why this step? Crossing $a = 0$ flips $Re λ$ from negative to positive — a Hopf-type change from stable spiral to unstable spiral through a centre.

Verify (at $a = 1$ ): $λ = \frac{1 \pm 1 - 4}{2} = \frac{1}{2} \pm \frac{3}{2} i$ , so $Re = \frac{1}{2} > 0$ → unstable spiral ✔; product $= \frac{1}{4} + \frac{3}{4} = 1 = Δ$ ✔. Verify (at $a = - 3$ ): $λ = \frac{- 3 \pm 9 - 4}{2} = \frac{- 3 \pm 5}{2} \approx {- 0.382, - 2.618}$ — two real negative roots → stable node ✔; product $\approx (- 0.382) (- 2.618) \approx 1 = Δ$ ✔. Conclusion: the single parameter $a$ walks the origin through five distinct types. Watch the figure: as $a$ slides right, $τ = a$ climbs through zero (stability lost) while $D = a^{2} - 4$ dips below zero between $a = - 2$ and $a = 2$ (spiral region). The dots at $a = - 2, 0, 2$ mark the three boundaries where the type changes.

Consolidation

Recall Which cell does each example live in?

Ex 1 saddle (A) ::: $Δ < 0$ , opposite-sign eigenvalues, always unstable. Ex 2 vs Ex 3 ::: stable node (B) and its time-reversed unstable node (C). Ex 4 ::: stable spiral (D) and, after a sign flip, unstable spiral (E) — $D < 0$ . Ex 5 ::: centre (F) — $τ = 0, Δ > 0$ , purely imaginary eigenvalues. Ex 6 ::: star node (G) — repeated eigenvalue, $D = 0$ . Ex 7 ::: line of fixed points (H) — $Δ = 0$ , a zero eigenvalue. Ex 8 ::: nonlinear ecology (I) — saddle at coexistence; evaluate $J$ per equilibrium. Ex 9 ::: bifurcation (J) — type changes as $a$ crosses $- 2, 0, 2$ .

See the full geography of these cells in the Trace–determinant plane; the physical archetype (Ex 4/5/9) is the Damped harmonic oscillator; and Phase portraits show what each cell looks like as a flow. For a stability proof that survives $τ = 0$ , use Lyapunov stability.