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MathidiaMathematics Academy
Dynamical systems
Lesson 32 of 75
Easy practice

Linear stability

Objective. Classify fixed points by linearisation.

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  2. 2. Worked example
  3. 3. Practice
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Learn: the key idea

Linear stability analysis classifies the behaviour near a fixed point of a nonlinear system by examining the Jacobian matrix of the system evaluated there, since nearby trajectories approximately follow the linearised flow. For ẋ = f(x,y), ẏ = g(x,y) with fixed point (x₀,y₀), compute the Jacobian J of partial derivatives at that point; its eigenvalues determine whether the point is a stable node, saddle, spiral, or centre. Real eigenvalues of the same sign give a node (stable if both negative), opposite signs give a saddle, and complex eigenvalues give a spiral whose stability follows the sign of the real part. Example: for ẋ = −x, ẏ = −2y, the Jacobian is diag(−1,−2) with both eigenvalues negative, so the origin is a stable node.

Worked example

Classify the linear stability of the fixed point at the origin for the system ẋ = −x + y, ẏ = −y.

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Illustrated card showing ?
  1. 1Compute the Jacobian at the origin: J = [[−1, 1], [0, −1]], since the system is already linear.
  2. 2Find eigenvalues by solving det(J − λI) = 0: (−1−λ)² = 0, giving a repeated eigenvalue λ = −1.
  3. 3Both eigenvalues are real, equal, and negative, so trajectories decay towards the origin.
  4. 4This classifies the origin as a stable (degenerate) node.

Answer: Stable node (repeated eigenvalue λ = −1)

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Common mistake. A common error is concluding stability directly from the trace and determinant without checking they come from the correctly evaluated Jacobian at the fixed point, not at an arbitrary point in the plane.

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