Nonlinear equations
Objective. Use linearisation and qualitative analysis.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
Nonlinear equations in dynamical systems lack the superposition property of linear ones, so their behaviour near equilibria is studied by linearisation: approximate f(x,y), g(x,y) by their Jacobian at the equilibrium and analyse the resulting linear system. This is valid when the equilibrium is hyperbolic, meaning no eigenvalue of the Jacobian has zero real part, by the Hartman–Grobman theorem. For the pendulum equation x′ = y, y′ = −sin(x), linearising at (0,0) gives y′ = −x, producing a centre, matching the observed small oscillations.
Worked example
For the nonlinear system x′ = y, y′ = −sin(x), find the linearisation at the equilibrium (0,0).
- 1Compute the Jacobian of (y, −sin x) generally
- 2At (0,0), ∂/∂x(−sin x) = −cos(0) = −1
- 3The Jacobian at the origin is [[0,1],[−1,0]]
- 4Write the linearised system explicitly
Answer: x′ = y, y′ = −x
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Start practisingCommon mistake. A common error is trusting the linearisation when an eigenvalue has zero real part, such as a predicted centre, since nonlinear terms can turn it into a spiral. Always check hyperbolicity before relying on the linear classification.
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