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Why the derivative is negative definite? Shouldn't be negative semi definite? It only depends on the $x$.

CroCo
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  • Roughly speaking semi definite will get stability, definite will get asymptotic stability. So, yes, I would agree that the conclusion is that $x \to 0$. What are the dynamics? – copper.hat Jun 07 '19 at 02:01
  • @copper.hat, true but we need to show the derivative is negative definite to confirm the asymptotic stability. I think we can't show the asymptotic stability if the derivative is negative semi definite. – CroCo Jun 07 '19 at 02:03

1 Answers1

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You're right, this argument is faulty. The standard Lyapunov theorem does not give asymptotic stability in this case.

However, there's a stronger theorem by LaSalle that you can use. Since the set where $\dot V=0$, namely the line $x=0$, contains no complete trajectories except the equilibrium $(0,0)$ (which you see from $\dot x = y - 0$ when $x=0$, so that the vector field is transversal to the line $x=0$ except at the origin), you indeed get asymptotic stability. But there is that extra condition that one needs to check.

Hans Lundmark
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