Phase plane analysis — trajectories, critical points: Exercises

Two tools we will lean on constantly. Let me restate them in plain words so you never have to guess.

Figure — Phase plane analysis — trajectories, critical points

The picture above is your decision map (the Trace–determinant plane). Every exercise below lands somewhere on it. Keep it open.

Level 1 — Recognition

Exercise 1.1

Given the Jacobian $J = (- 2 0 0 - 5)$ at a critical point, classify the point (type + stability) without solving anything by hand beyond reading the diagonal.

Recall Solution 1.1

WHAT we do: read $τ$ and $Δ$ straight off. $τ = - 2 + (- 5) = - 7, Δ = (- 2) (- 5) - 0 = 10.$ Discriminant $τ^{2} - 4Δ = 49 - 40 = 9 > 0$ → real eigenvalues. Because the matrix is diagonal, the eigenvalues are the diagonal entries: $λ_{1} = - 2, λ_{2} = - 5$ . WHY it looks like this: both eigenvalues are real, same sign, both negative → every direction decays. That is a stable node (sink). Leaves get sucked straight in.

Exercise 1.2

Classify $J = (30 0 - 1)$ .

Recall Solution 1.2

$λ_{1} = 3, λ_{2} = - 1$ (diagonal entries). One grows, one shrinks. $Δ = (3) (- 1) = - 3 < 0$ . WHY: real eigenvalues of opposite sign → saddle, always unstable. The negative determinant alone already screams "saddle" (see the mnemonic: negative det = saddle).

Level 2 — Application

Exercise 2.1

Find all critical points of $\overset{x}{˙} = x + 2 y, \overset{y}{˙} = 2 x + y .$

Recall Solution 2.1

WHAT: set both rates to zero (nothing moves). $x + 2 y = 0, 2 x + y = 0.$ WHY: a critical point is where $f = g = 0$ . From the first, $x = - 2 y$ . Substitute into the second: $2 (- 2 y) + y = - 3 y = 0 \Rightarrow y = 0$ , hence $x = 0$ . Answer: the only critical point is $(0, 0)$ .

Exercise 2.2

For the same system, build the Jacobian, find $τ, Δ$ , the eigenvalues, and classify.

Recall Solution 2.2

WHAT — Jacobian. The right-hand sides are already linear, so the partial derivatives are just the coefficients: $J = (f_{x} g_{x} f_{y} g_{y}) = (1221) .$ WHAT — trace/det. $τ = 1 + 1 = 2$ , $Δ = (1) (1) - (2) (2) = 1 - 4 = - 3$ . WHAT — eigenvalues. Solve $λ^{2} - τ λ + Δ = λ^{2} - 2 λ - 3 = 0$ . $(λ - 3) (λ + 1) = 0 \Rightarrow λ = 3, - 1.$ WHY classify: opposite signs → saddle, unstable. This matches the parent note's Worked Example A exactly.

Exercise 2.3

Critical points and classification for the decoupled system $\overset{x}{˙} = - x, \overset{y}{˙} = - 4 y .$

Recall Solution 2.3

Critical point: $- x = 0, - 4 y = 0 \Rightarrow (0, 0)$ . $J = (- 1 0 0 - 4)$ , so $λ = - 1, - 4$ . Both real, negative, unequal → stable node (sink), called an improper node because the two decay rates differ.

Level 3 — Analysis

Exercise 3.1

The nonlinear system $\overset{x}{˙} = - y + x (1 - x^{2} - y^{2}), \overset{y}{˙} = x + y (1 - x^{2} - y^{2})$ has a critical point at the origin. Linearize there and classify.

Recall Solution 3.1

WHAT — kill the cubics near origin. For tiny $x, y$ , the factor $1 - x^{2} - y^{2} \to 1$ . So the linear part is $\overset{x}{˙} \approx - y + x, \overset{y}{˙} \approx x + y .$ WHY: near the origin, $x^{2} + y^{2}$ is second-order-small — negligible against the linear terms. WHAT — Jacobian. $f = x - y \Rightarrow f_{x} = 1, f_{y} = - 1$ ; $g = x + y \Rightarrow g_{x} = 1, g_{y} = 1$ . $J = (11 - 1 1) .$ WHAT — trace/det/discriminant. $τ = 2$ , $Δ = (1) (1) - (- 1) (1) = 2$ , discriminant $= 4 - 8 = - 4 < 0$ → complex eigenvalues (rotation). $λ = \frac{1}{2} (2 \pm - 4) = 1 \pm i .$ WHY classify: complex with real part $+ 1 > 0$ → unstable spiral (outward). The full nonlinear truth: trajectories spiral outward toward a limit cycle at radius $r = 1$ .

Exercise 3.2

For $J = (0 - 4 10)$ , classify the linearized behaviour and state the caution that applies.

Recall Solution 3.2

$τ = 0 + 0 = 0$ , $Δ = (0) (0) - (1) (- 4) = 4 > 0$ . Discriminant $= 0 - 16 = - 16 < 0$ → complex, and since $τ = 0$ the real part is zero: $λ = \frac{1}{2} (0 \pm - 16) = \pm 2 i .$ Classification: pure imaginary → center (closed loops), neutrally stable in the linear picture. Caution: a center is the borderline case. Nonlinear terms can tip it into a slow spiral, so linearization is inconclusive. To be sure it is a true center you need an energy / Lyapunov argument.

Level 4 — Synthesis

Exercise 4.1

Consider $\overset{x}{˙} = y, \overset{y}{˙} = - x - x^{2}$ . Find all critical points, and classify each by linearization.

Recall Solution 4.1

WHAT — critical points. Set $y = 0$ and $- x - x^{2} = 0$ . $- x (1 + x) = 0 \Rightarrow x = 0$ or $x = - 1$ . With $y = 0$ both times: $critical points: (0, 0) and (- 1, 0) .$ WHAT — general Jacobian. $f = y \Rightarrow f_{x} = 0, f_{y} = 1$ . $g = - x - x^{2} \Rightarrow g_{x} = - 1 - 2 x, g_{y} = 0$ . $J (x, y) = (0 - 1 - 2 x 10) .$

At $(0, 0)$ : $J = (0 - 1 10)$ . $τ = 0, Δ = 1$ , discriminant $= - 4 < 0$ , $λ = \pm i$ → center (linear); borderline, treat with caution.

At $(- 1, 0)$ : $g_{x} = - 1 - 2 (- 1) = 1$ , so $J = (0110)$ . $τ = 0, Δ = (0) (0) - (1) (1) = - 1 < 0$ . $Δ < 0$ → saddle, unstable. (Eigenvalues $λ = \pm 1$ .)

Big picture: the flow has a delicate center at the origin and a saddle to its left — a classic nonlinear phase portrait.

Exercise 4.2

A system has Jacobian $J = (2112)$ at its critical point. Classify it, and state which eigendirections the trajectories align with far from and near the point.

Recall Solution 4.2

$τ = 4, Δ = 4 - 1 = 3$ , discriminant $= 16 - 12 = 4 > 0$ → real. $λ = \frac{1}{2} (4 \pm 2) = 3, 1$ . Both positive → unstable node (source). Eigenvectors. For $λ = 3$ : $(J - 3 I) v = (- 1 1 1 - 1) v = 0 \Rightarrow v = (1, 1)$ . For $λ = 1$ : $(1111) v = 0 \Rightarrow v = (1, - 1)$ . Geometry: trajectories leave the origin. The fast direction $(1, 1)$ (larger $λ = 3$ ) dominates far out; trajectories become tangent to the slow direction $(1, - 1)$ near the origin. That's the tell-tale shape of a node.

Level 5 — Mastery

Exercise 5.1

Design a linear system $\overset{x}{˙} = a x + b y, \overset{y}{˙} = c x + d y$ whose origin is a stable spiral. Give explicit numbers and prove it works.

Recall Solution 5.1

WHAT we need (the target region): stable spiral requires

complex eigenvalues → discriminant $τ^{2} - 4Δ < 0$ ;
decaying rotation → $Re λ < 0$ , i.e. $τ < 0$ . WHY: complex ⇒ rotation; negative real part ⇒ shrinking radius. Construct one. Pick $J = (- 1 2 - 2 - 1)$ , i.e. $a = - 1, b = - 2, c = 2, d = - 1$ . Check: $τ = - 1 + (- 1) = - 2 < 0$ ✓. $Δ = (- 1) (- 1) - (- 2) (2) = 1 + 4 = 5 > 0$ . Discriminant $= 4 - 20 = - 16 < 0$ ✓ complex. $λ = \frac{1}{2} (- 2 \pm - 16) = - 1 \pm 2 i$ . Real part $- 1 < 0$ ✓. Answer: $\overset{x}{˙} = - x - 2 y, \overset{y}{˙} = 2 x - y$ has a stable spiral at the origin. Trajectories spiral inward.

Exercise 5.2

A system's Jacobian at the origin is $J = (1 - 1 α 1)$ with $α$ a real parameter. Find the value(s) of $α$ where the type changes, and name each regime.

Recall Solution 5.2

$τ = 1 + 1 = 2$ (independent of $α$ ), $Δ = (1) (1) - (α) (- 1) = 1 + α$ . Discriminant $= τ^{2} - 4Δ = 4 - 4 (1 + α) = - 4 α$ . WHY track discriminant: its sign flips real ↔ complex — that's where the type changes.

$α < 0$ : discriminant $- 4 α > 0$ → real eigenvalues. Also $Δ = 1 + α$ could be negative if $α < - 1$ .
- $- 1 < α < 0$ : $Δ > 0, τ > 0$ → unstable node.
- $α < - 1$ : $Δ < 0$ → saddle.
- $α = - 1$ : $Δ = 0$ → degenerate (a zero eigenvalue, a line of fixed points).
$α = 0$ : discriminant $= 0$ → repeated eigenvalue $λ = 1$ (borderline node).
$α > 0$ : discriminant $< 0$ → complex, and $τ = 2 > 0$ → unstable spiral. Bifurcation values: $α = 0$ (node ↔ spiral boundary) and $α = - 1$ (saddle ↔ node boundary via $Δ = 0$ ).