Mr. Integral
Math notes/Differential Equations/Systems and Phase Planes

Phase Portraits and Stability

2 MIN READ · PRACTISE AT YOUR OWN PACE   ·   FREE

LESSON 2 OF 5See the unit map ↗

Classify planar linear equilibria and follow time-oriented trajectories.

Builds on Linear Systems and Eigenmodes

The bigger question: How do coupled variables move together?

On this page

The idea

A phase portrait plots state against state, rather than state against time. At each point the vector AxA\mathbf x gives the trajectory’s direction and speed. Arrows are essential: the same geometric curves can describe attraction or repulsion when time is reversed.

Visual guide

VISUAL GUIDEA spiral contracts as time moves forward
For x′ = −x − y, y′ = x − y, the trajectory from (2, 0) is (2e⁻ᵗ cos t, 2e⁻ᵗ sin t). The arrows turn counterclockwise toward the origin. The axes show states x and y, not time.-2.5-2.5-1.25-1.25001.251.252.52.5xy
  • Trajectory
For x′ = −x − y, y′ = x − y, the trajectory from (2, 0) is (2e⁻ᵗ cos t, 2e⁻ᵗ sin t). The arrows turn counterclockwise toward the origin. The axes show states x and y, not time.

Method and assumptions

For a real 2×22\times2 matrix, write trace τ\tau and determinant Δ\Delta. Eigenvalues solve λ2−τλ+Δ=0\lambda^2-\tau\lambda+\Delta=0. Negative determinant gives a saddle. Positive determinant with negative trace gives asymptotic stability; the discriminant distinguishes real node modes from complex spiral modes. Zero real parts or zero eigenvalues require closer analysis.

Worked example: a saddle

For x′=x,y′=−yx'=x,y'=-y, solutions are (aet,be−t)(ae^t,be^{-t}). The vertical axis is stable and the horizontal axis unstable. Off-axis trajectories satisfy xy=abxy=ab. One decaying direction does not make the equilibrium stable.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A planar linear system has real eigenvalues −1 and 2. What type is the origin?

Hint 1 · Find a starting point

One eigenmode decays and one grows.

Hint 2 · Take the next step

Opposite-sign real eigenvalues give opposing time behavior.

Show the reasoning

Answer: An unstable saddle

Trajectories approach along one eigendirection and depart along the other, producing a saddle.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: an attracting spiral

For x′=−x−y,y′=x−yx'=-x-y,y'=x-y, eigenvalues are −1±i-1\pm i. Radius decays as e−te^{-t} while angle increases at one radian per time unit. The point (1,0)(1,0) initially moves up and left, fixing the counterclockwise direction.

Interpreting the result

A center in a linear system has closed orbits and neutral stability, not asymptotic attraction. A nonlinear system with the same linearized eigenvalues can behave differently, so do not extend the classification without checking its hypotheses.

Practice

  1. Classify A=diag⁡(−1,−3)A=\operatorname{diag}(-1,-3).
  2. Classify A=diag⁡(2,−1)A=\operatorname{diag}(2,-1).
  3. What does asymptotic stability add to stability?
Show worked solutions
  1. A stable node.
  2. A saddle because the determinant is negative.
  3. Nearby trajectories also converge to the equilibrium as time increases.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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