IB Maths AI HLCoupled & Second Order Differential EquationsPaper 1 & 2~7 min read
Phase Portraits
A phase portrait is a picture of how x and y evolve together, drawn as solution trajectories in the x–y plane. You don’t plot against time — instead the eigenvalues and eigenvectors of M set the whole shape: straight eigenvector lines for real eigenvalues, spirals or ellipses for complex ones.
📘 What you need to know
Two cases to know: eigenvalues real & distinct, or complex conjugates.
Real eigenvalues: two straight lines through the origin, parallel to the eigenvectors.
Direction: positive eigenvalue → lines point away from the origin; negative → towards it.
Trajectory shape: as t → ∞, parallel to the larger eigenvalue’s line; as t → −∞, parallel to the smaller one. Trajectories never cross an eigenvector line.
Complex eigenvalues: curves orbit the origin — spirals (real part ≠ 0) or ellipses (real part = 0).
Real part sets spiral direction: positive → outward, negative → inward, zero → closed ellipse.
Real, distinct eigenvalues
The portrait always contains the two eigenvector lines through the origin. The origin splits each into trajectories whose direction depends on the sign of the corresponding eigenvalue.
🧠 “Sign sets the flow, size sets the curve”
The sign of each eigenvalue says whether its line flows out (+) or in (−). The relative size shapes the curves: trajectories hug the larger eigenvalue’s line as t → ∞ and the smaller one as t → −∞.
🤔 Why those limiting directions?
The solution is x = Aeλ₁tp1 + Beλ₂tp2. As t → ∞ the term with the larger eigenvalue grows fastest and dominates, so the trajectory lines up with that eigenvector. As t → −∞ the smaller eigenvalue’s term dominates instead. That’s why each trajectory bends from one eigenvector direction to the other.
Eigenvalues
Trajectories
Picture
Both positive
Flow away from origin
Unstable node
Both negative
Flow towards origin
Stable node
One +, one −
In along one line, out along the other
Saddle
Stable node — both eigenvalues negative
For M = (−2, 2; 1, −3): eigenvalues −1 and −4. All trajectories converge on the origin. Since −4 is more negative, far from the origin they run parallel to the y = −x line, and they approach the origin tangent to the y = ½x line.
Complex eigenvalues
Complex conjugate eigenvalues give curves that orbit the origin. The real part decides whether they spiral in, spiral out, or close up into ellipses.
Real part ≠ 0 — spiralRe > 0 spirals away; Re < 0 spirals towards the origin.
Real part = 0 — centrePurely imaginary eigenvalues give closed ellipses orbiting the origin.
Finding the orbit direction: evaluate ẋ = Mx at (1, 0) or (0, 1). It’s clockwise if the motion is downward from (1, 0) or rightward from (0, 1); otherwise counter-clockwise.
🧭 Recipe — drawing a phase portrait
Get the eigenvalues (and eigenvectors) of M.
Real & distinct? Draw the eigenvector lines; arrow them out (λ > 0) or in (λ < 0).
Sketch trajectories: parallel to the smaller-λ line far out, curving to the larger-λ line; never crossing an eigenvector line.
Complex? Decide spiral (Re ≠ 0) or ellipse (Re = 0); find the direction at (1, 0) or (0, 1).
Label the eigenvector lines and arrow the flow.
Worked examples
Examples 1–3 use dxdt = −2x + 2y, dydt = x − 3y, whose matrix has eigenvalues −1 and −4 with eigenvectors (2, 1) and (−1, 1).
WE 1
What type of phase portrait does this system have?
Both eigenvalues are real, distinct, and negative.
λ = −1, −4: both negativea stable node — all trajectories converge on the origin
WE 2
Find the equations of the two eigenvector lines.
Each line passes through the origin parallel to its eigenvector.
(2, 1) → gradient ½ → y = ½x(−1, 1) → gradient −1 → y = −xy = ½x and y = −x
WE 3
Far from the origin, which line are the trajectories parallel to?
As t → −∞ (far out), the smaller eigenvalue’s term dominates.
smaller eigenvalue = −4, eigenvector (−1, 1)parallel to y = −x far from the origin
WE 4
A system has eigenvalues 5 and 3. Which eigenvector does a trajectory align with as t → ∞, and is the node stable?
As t → ∞ the larger eigenvalue dominates; both positive means outward flow.
larger eigenvalue = 5 → align with p1 (its eigenvector)both positive → away from originaligns with the λ = 5 eigenvector; unstable node
WE 5
For ẋ = (1, −2; 1, −1)x the eigenvalues are ±i. Describe the portrait and find its direction.
Purely imaginary → ellipses. Check the velocity at (1, 0).
Re = 0 → closed ellipses (a centre)at (1, 0): ẋ = (1, 1) → moving right and upellipses orbiting counter-clockwise
💡 Top tips
Sign first: positive eigenvalue flows out, negative flows in.
Larger eigenvalue wins as t → ∞; smaller as t → −∞.
Never cross an eigenvector line with a trajectory.
Complex → orbit: spiral if Re ≠ 0, ellipse if Re = 0.
Direction via (1, 0) or (0, 1) — substitute into Mx.
Label the eigenvector lines with their equations.
⚠ Common mistakes
Swapping the limiting lines — larger eigenvalue governs t → ∞, not −∞.
Wrong flow direction from the eigenvalue sign.
Letting a trajectory cross an eigenvector line.
Confusing spiral and ellipse — check whether the real part is zero.
Guessing the orbit direction instead of testing a point.
Next up — Equilibrium Points. A phase portrait shows where every trajectory goes; the origin sits at its centre as a special fixed point. The next topic names what you’ve been seeing: an equilibrium point is where both rates are zero, and the eigenvalue signs you used here decide whether it’s stable or unstable — node, spiral, centre, or saddle.
Need help with Coupled & Second Order Differential Equations?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
