Lyapunov function: Difference between revisions

Revision as of 05:59, 20 March 2020

Revision as of 09:32, 5 May 2020

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So let <math>V(x)=x^2</math> on <math>\R </math>. Then,

:<math>\dot V(x) = V'(x) \dot x = 2x\cdot (-x) = -2x^2\leq< 0.</math>

This correctly shows that the above differential equation, <math>x,</math> is asymptotically stable about the origin. Note that using the same Lyapunov candidate one can show that the equilibrium is also globally asymptotically stable.