Is the following statement true? Linearisation determines stability at every equilibrium

No. ẋ = −y − x³, ẏ = x − y³ at the origin The linearisation is a centre (eigenvalues ±i), but V = x² + y² gives V̇ = −2(x⁴ + y⁴) < 0: asymptotically stable. Non-hyperbolic equilibria escape Hartman–Grobman.

Counterexample

Is this true?

Linearisation determines stability at every equilibrium

No — it is false.

The counterexample

ẋ $= -y - x^{3}$ , ẏ $= x - y^{3}$ at the origin

Why it fails

The linearisation is a centre (eigenvalues ±i), but $V = x^{2} + y^{2}$ gives V̇ $= -2(x^{4} + y^{4}) < 0$ : asymptotically stable. Non-hyperbolic equilibria escape Hartman–Grobman.

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Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Every IVP has a unique solution
A continuous right-hand side gives a unique solution
A locally unique solution exists for all time

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ODE stability

Go deeper· Explained in full in Stability and phase portraits· Asked in CSIR NET December 2024· From Ordinary Differential Equations › Stability and phase portraits

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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