Phase plane analysis — trajectories, critical points: Question bank

Reminders of the notation used below (all built in the parent):

$\overset{x}{˙} = f (x, y), \overset{y}{˙} = g (x, y)$ — an autonomous system ( $t$ never appears on the right).
A critical point is where $f = g = 0$ (the velocity arrow is zero — no motion).
$J$ = Jacobian, the matrix of first partial derivatives $(f_{x} g_{x} f_{y} g_{y})$ evaluated at the critical point.
$τ = tr J = λ_{1} + λ_{2}$ (sum of eigenvalues), $Δ = det J = λ_{1} λ_{2}$ (product of eigenvalues).
The discriminant $τ^{2} - 4Δ$ decides real vs complex eigenvalues: when $Δ > 0$ , a positive discriminant ( $τ^{2} - 4Δ > 0$ ) gives real eigenvalues (a node), a negative discriminant ( $τ^{2} - 4Δ < 0$ ) gives complex eigenvalues (a spiral).
$r = x^{2} + y^{2}$ is the distance from the origin (radius) — used only in the last edge-case item.

True or false — justify

The trajectory of any solution eventually reaches a critical point.

False. Trajectories can also run off to infinity, or loop forever on a closed orbit (a center or a limit cycle) never reaching a rest point.

Two different trajectories can pass through the same ordinary (non-critical) point.

False. An autonomous field gives exactly one velocity arrow per point, so one direction — two curves through it would need two directions. They can only meet at critical points where the arrow is zero.

A single trajectory can cross itself (form a loop) at an ordinary point.

False for a simple crossing, true only as a closed loop. A trajectory can be a closed curve (periodic orbit), but it cannot cross itself transversally — that would again demand two directions at one point.

det J > 0

the critical point is stable.

False.

det J = λ_{1} λ_{2} > 0

only means the eigenvalues share a sign (node or spiral), not that the sign is negative. Stability is decided by

τ = tr J

: negative

τ

stable, positive

τ

unstable.

det J < 0

the point could be a stable node for the right trace.

False.

det J < 0

forces

λ_{1} λ_{2} < 0

, i.e. real eigenvalues of opposite sign — always a saddle, always unstable, regardless of the trace.

Pure imaginary eigenvalues in the linearization guarantee the nonlinear system has a center.

False. A center is the borderline case

Re λ = 0

; tiny nonlinear terms can push it into a slow spiral. Linearization is inconclusive here — use energy/Lyapunov methods.

Every autonomous 2D system has at least one critical point.

False. For example

\overset{x}{˙} = 1, \overset{y}{˙} = 0

has

f = 1 \neq = 0

everywhere — no rest point at all, so no critical points.

A stable node and a stable spiral have the same long-term fate for nearby trajectories.

True in fate, different in path. Both pull nearby trajectories into the point; the node approaches along straight-ish eigendirections, the spiral winds in because the eigenvalues carry an imaginary part (rotation).

Reversing time (

t \to - t

) turns a stable node into an unstable node.

True. Reversing time negates the velocity field, flipping the sign of every eigenvalue's real part; sinks become sources and vice versa (a saddle stays a saddle, its two signs just swap roles).

The slope relation

\frac{d y}{d x} = \frac{g}{f}

keeps all the information of the original system.

False. It keeps the geometry (shape of curves) but throws away timing/speed — you no longer know how fast the point moves or in which time-direction along the curve.

Spot the error

f (x_{0}, y_{0}) = 0

alone makes

(x_{0}, y_{0})

a critical point."

Error: you need both

f = 0

and

g = 0

. If only

f = 0

then

\overset{y}{˙} = g \neq = 0

, so the point still moves vertically — not at rest.

"The eigenvalues are the roots of

λ^{2} + τ λ + Δ = 0

Error: the sign on the trace term is wrong. The characteristic equation is

λ^{2} - τ λ + Δ = 0

(minus the trace), giving

λ = \frac{1}{2} (τ \pm τ^{2} - 4Δ)

"Discriminant

τ^{2} - 4Δ < 0

means the system is unstable."

Error: a negative discriminant only means the eigenvalues are complex (spiral/center), saying nothing about stability. Stability still comes from the sign of

τ

"We linearize by dropping

x

from

x (1 - x^{2} - y^{2})

because

x

is small."

Error: you only drop the higher-order cubic

- x^{3} - x y^{2}

. The linear term

x \cdot 1 = x

must stay — it belongs in the Jacobian. Dropping it changes the answer entirely.

"Since trajectories can't cross, the phase portrait has no closed curves."

Error: a closed curve doesn't cross another trajectory — it is one continuous trajectory returning to its start. Non-crossing permits loops (centers, limit cycles); it forbids intersections between distinct orbits.

"A saddle is stable in one direction so it's 'half stable'."

Error: any perturbation off the single incoming (stable) direction grows along the outgoing direction, so a saddle is unstable, full stop — almost every nearby trajectory eventually leaves.

Why questions

Why does eliminating

t

give a first-order ODE for the trajectories?

Because

\frac{d y}{d x} = \frac{y ˙}{x ˙} = \frac{g ( x , y )}{f ( x , y )}

relates

y

and

x

directly with a single derivative — one equation, first order — trading two time-ODEs for one shape-ODE.

Why must partial derivatives in

J

be evaluated at the critical point?

Linearization approximates the field by its tangent plane at that specific point; the slopes

f_{x}, f_{y}, g_{x}, g_{y}

change from place to place, so only their values at

(x_{0}, y_{0})

describe the flow near that rest point.

Why do solutions of

\dot{u} = J u

take the form

e^{λ t}

Trying

u = v e^{λ t}

turns the ODE into

J v = λ v

— the eigenvalue equation. So each exponential mode must ride an eigenvector, and

λ

must be an eigenvalue of

J

Why does the sign of

Re λ

decide stability while the imaginary part decides shape?

e^{λ t} = e^{(Re λ) t} e^{i (Im λ) t}

: the real part sets the growth/decay envelope (in vs out), and the imaginary part sets rotation (spiral vs straight-line node).

Why is a center called a "borderline" case?

It sits exactly on the

τ = 0

line of the Trace–determinant plane with

Δ > 0

, right between stable and unstable spirals — any nudge to

τ

off zero tips it decisively one way, so it's structurally fragile.

Why can we usually ignore higher-order terms near a critical point?

Near the point the displacement is tiny, and squares/cubes of a tiny number are far smaller than the number itself, so the linear terms dominate — except in the borderline cases (center, zero eigenvalue) where the small terms decide the tie.

Edge cases

What if

J

has a repeated real eigenvalue (

τ^{2} - 4Δ = 0

This is the boundary between node and spiral — a degenerate/improper node. Its stability is still set by the sign of that repeated

λ

(equivalently the sign of

τ

What if one eigenvalue is exactly zero (

Δ = 0

The linearization is degenerate: there's a whole line of near-equilibria and the flow neither clearly attracts nor repels along it. Linearization is inconclusive; you must keep higher-order terms.

What happens at a critical point when you try to compute

\frac{d y}{d x} = \frac{g}{f}

Both

f = 0

and

g = 0

, giving

\frac{0}{0}

— the slope is genuinely undefined there. That is exactly why multiple trajectories can approach and meet at a critical point.

For

\overset{x}{˙} = - y + x (1 - x^{2} - y^{2}), \overset{y}{˙} = x + y (1 - x^{2} - y^{2})

, is the outward spiral the whole story?

No. The linear part gives an unstable spiral at the origin, but the full nonlinear system has a stable limit cycle on the circle

r = x^{2} + y^{2} = 1

that trajectories spiral toward — linearization only describes the local origin behaviour.

If the field is

\overset{x}{˙} = y, \overset{y}{˙} = - x

(a linear center), can we trust the closed loops?

Yes, here specifically, because the system is exactly linear — there are no higher-order terms to spoil it. The warning about centers applies only when nonlinear terms exist that linearization discarded.

What does time-reversal do to a saddle?

It negates both eigenvalues, so the incoming and outgoing directions swap — but one is still positive and one still negative, so it remains a saddle (just with its "in" and "out" arms exchanged).

Can a trajectory reach a critical point in finite time?

Generally no for a hyperbolic sink: the approach is exponential (

e^{λ t}

with

λ < 0

), so it takes infinite time to actually arrive — it only gets arbitrarily close.

Recall One-line summary of the traps

Two lifetimes of mistakes reduce to: (1) need both $f = 0$ and $g = 0$ ; (2) det gives type, trace gives stability; (3) centers and zero eigenvalues are borderline — linearization can lie.