Stability of equilibria — stable, unstable, saddle, spiral, centre: Visual walkthrough

Step 1 — A rule that assigns an arrow to every point

WHAT: we paint an arrow at every point of the plane. WHY: a differential equation is a field of arrows — follow the arrows and you trace a trajectory. PICTURE: below, each blue arrow shows where a particle at that spot would drift next. The lone red dot is special — it has no arrow.

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

Step 2 — Zoom in until the curved field looks straight

Let us name the tiny deviation from the resting point: $u = x - x^{*}, v = y - y^{*} .$

$u, v$ ::: how far we've been nudged, horizontally and vertically. Both are small.

WHAT: we write $f$ and $g$ using their linear approximation (first-order Taylor):

f (x, y) \approx = 0 f (x^{*}, y^{*}) + slope in x f_{x} u + slope in y f_{y} v .

$f_{x}$ ::: partial derivative — "how much $f$ changes if you move a little in $x$ , holding $y$ fixed." It's the steepness of $f$ in the $x$ -direction.
$f_{y}$ ::: same, but for a step in $y$ .
The first term is $0$ because it's an equilibrium — that's what lets the whole thing collapse to something linear.

WHY the derivative and not something else: we want the slope of the arrow-field right at the dot — the derivative is precisely the tool that answers "how fast does the output change per unit nudge?" PICTURE: the curved graph and its tangent line agreeing near the point.

Doing the same for $g$ and stacking both rules gives the linear system.

Step 3 — The Jacobian: the local rulebook in one box

The $2 \times 2$ box $J$ ::: the Jacobian matrix, evaluated at the equilibrium. Read a row across: it says " $\overset{u}{˙}$ is built from $f_{x}$ times $u$ plus $f_{y}$ times $v$ ."
Every entry is a slope of the arrow-field, frozen at the red dot.

WHAT: we packed four numbers into a matrix. WHY: a matrix is exactly the object that stretches and rotates a vector — and stretching/rotating is what the flow does to a tiny nudge. PICTURE: $J$ takes the nudge-arrow $(u, v)$ and outputs the velocity-arrow $(\overset{u}{˙}, \overset{v}{˙})$ — a machine turning position into motion.

Step 4 — Special directions: eigenvectors travel in straight lines

$e^{λ t}$ ::: the exponential — it appears because "rate of change proportional to current size" ( $˙ = λ \cdot$ ) is the equation the exponential solves. That's why this tool and no other.
If $λ > 0$ : $e^{λ t}$ grows → nudge flies out. If $λ < 0$ : it decays → nudge returns.

PICTURE: two special axes; along the green one the arrow shrinks toward the dot, along the coral one it stretches away.

Step 5 — Finding $λ$ with only two numbers: trace and determinant

WHAT: $J e = λ e$ is the same as $(J - λ I) e = 0$ . For a non-zero direction $e$ to be squashed to zero, the matrix $J - λ I$ must be singular, i.e. $det (J - λ I) = 0$ . Multiplying out the $2 \times 2$ determinant:

λ^{2} - τ (f_{x} + g_{y}) λ + Δ (f_{x} g_{y} - f_{y} g_{x}) = 0.

WHY these two numbers: they are exactly the coefficients of the quadratic, so they carry all the information about $λ_{1, 2}$ — you never need the eigenvectors to classify. PICTURE: a single point plotted on the Trace–determinant plane, with $τ$ across and $Δ$ up.

Step 6 — Sign of $Δ$ : same team or opposite team?

WHAT: $Δ = λ_{1} λ_{2}$ . WHY it matters: a product tells you whether the two eigenvalues share a sign.

$Δ < 0$ ::: product negative → one $λ > 0$ , one $λ < 0$ → grows one way, decays the other → saddle (always unstable).
$Δ > 0$ ::: product positive → both same sign → the trace $τ$ then decides which sign.

PICTURE: the saddle — arrows pour in along one axis and blast out along the perpendicular one, like the surface of a mountain pass.

Step 7 — Sign of $D$ : real slide vs complex swirl

WHAT: the root $D$ is real when $D > 0$ and imaginary when $D < 0$ . Here $ω$ (the Greek letter "omega") names the rotation frequency — how many radians per unit time the nudge sweeps around the dot; it is $ω = \frac{1}{2} 4Δ - τ^{2}$ , the size of the imaginary part of $λ$ .

$D > 0$ ::: two distinct real eigenvalues → motion is pure stretch/shrink along two fixed axes → node (or saddle if $Δ < 0$ ).
$D < 0$ ::: $D = i 4Δ - τ^{2}$ brings an imaginary part to $λ$ . Imaginary in the exponent means $e^{iω t}$ — and that is rotation (Euler: $e^{iω t} = cos ω t + i sin ω t$ ). So $D < 0$ → spirals or centre.

WHY the imaginary part = rotation: a real exponent scales; an imaginary exponent spins. The nudge circles the dot instead of walking straight to it.

PICTURE: left, a node (straight radial approach); right, a spiral (rotating approach). The switch is the sign of $D$ .

Step 8 — The knife-edge $D = 0$ : nodes that lose a direction

WHAT: at $D = 0$ you must ask whether the eigenbasis is full. WHY: a defective matrix has no second straight-line direction, so the picture is a curved improper node rather than radial spokes. PICTURE: left, a star node (radial); right, an improper node (all curves funnel tangent to one lonely direction).

Step 9 — Sign of $τ$ and the knife-edge $τ = 0$

WHAT: once $Δ > 0$ fixes "both eigenvalues same sign," the trace picks the sign: $τ = λ_{1} + λ_{2} .$

$τ < 0$ ::: total growth negative → both decay → stable (node or spiral) — arrows spiral/slide inward.
$τ > 0$ ::: total growth positive → both grow → unstable — outward.
$τ = 0$ with $Δ > 0$ ::: eigenvalues are $\pm iω$ , pure imaginary → no growth, no decay, only rotation → centre: perfect closed loops.

PICTURE: three panels — inward spiral ( $τ < 0$ ), closed loops ( $τ = 0$ ), outward spiral ( $τ > 0$ ).

The one-picture summary

Everything above lives on one map: the Trace–determinant plane. Plot $(τ, Δ)$ , and the region you land in is the answer — but remember the boundaries are special:

The parabola $Δ = \frac{τ ^{2}}{4}$ (i.e. $D = 0$ ) separates nodes (below) from spirals (above), and on it live the star / improper nodes (Step 8).
The horizontal axis $Δ = 0$ separates saddles (below) from everything else, and on it sits the non-hyperbolic zero-eigenvalue case (Step 6's edge box).
The vertical axis $τ = 0$ (with $Δ > 0$ ) is the centre knife-edge, splitting stable (left) from unstable (right).

Recall Feynman retelling — say it in plain words

The whole story ::: A rule paints an arrow at every point; where the arrow is zero the system can rest. Nudge it, zoom in until the arrows look straight, and pack the four local slopes into a box called the Jacobian. That box has special directions (eigenvectors) it only stretches by factors $λ$ (eigenvalues); each direction evolves like $e^{λ t}$ . Two summary numbers — the sum $τ$ and product $Δ$ of the $λ$ 's — decide everything: $Δ < 0$ means opposite signs (saddle); $Δ > 0$ means same sign, and then $τ < 0$ pulls inward (stable) while $τ > 0$ pushes out (unstable); the discriminant $D$ decides straight slide (node) versus swirl (spiral), and the razor line $τ = 0$ gives endless loops (centre). The three border curves are their own delicate cases: $Δ = 0$ (a zero eigenvalue, a whole line of near-rest points), $D = 0$ (repeated eigenvalue → star or defective improper node), and $τ = 0$ (a marginal centre).

Recall drill

Recall Quick self-test

Why is $Δ < 0$ always unstable? ::: $Δ = λ_{1} λ_{2} < 0$ forces opposite signs, so one eigenvalue is positive → that mode grows → saddle. What happens exactly on $Δ = 0$ ? ::: One eigenvalue is $0$ and the other is $τ$ → non-hyperbolic, a one-dimensional centre manifold; linearisation can't decide stability. On $D = 0$ , what three node types can occur? ::: Star (proper) node with a full eigenbasis, or a defective improper node with only one eigen-direction (extra $t e^{λ t}$ term). What single sign flips a spiral from stable to unstable? ::: The sign of $τ$ (the trace = total growth rate). Where does the rotation in a spiral come from, algebraically? ::: The imaginary part $ω$ of $λ$ when $D = τ^{2} - 4Δ < 0$ , giving $e^{iω t}$ = rotation. On the trace–determinant map, what curve separates nodes from spirals? ::: The parabola $Δ = τ^{2} /4$ , i.e. $D = 0$ .