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Why does the continuity of Lyapunov function matter?

Is there any example where discontinuity of $V$ is problematic even if it does not change the sign but it is strictly decreasing and yet not stable?


From wikipedia:

A Lyapunov function for an autonomous dynamical system

\begin{cases}g:\mathbb {R} ^{n}\to \mathbb {R} ^{n}\\{\dot {y}}=g(y)\end{cases}

an equilibrium point at $y=0$ is a scalar function

$$V: \mathbb{R}^n\to \mathbb{R}$$

that is continuous, has continuous derivatives, is locally positive-definite, and for which $-\nabla V.g$ is also locally positive definite. The condition that $-\nabla V.g$ is locally positive definite is sometimes stated as $\nabla V.g$ is locally negative definite.

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  • $\begingroup$ You might find examples that though the function is decreasing the switching can push the state to higher levels thanks to the dynamics of $g$, I vaguely remember switching system literature has some examples but I can't recall any from the top of my head. Note that Lyapunov function does not stabilize the system, it just points to the dissipation direction of a point in the state space. $\endgroup$
    – percusse
    May 4, 2018 at 12:59
  • $\begingroup$ @percusse, thanks a lot. However, without any example it is hard to judge. $\endgroup$
    – ar2015
    May 5, 2018 at 11:18
  • $\begingroup$ The proof of the Lyapunov theorem proceeds by utilizing level sets of $V$. Without at least piecewise continuity, these level sets may not exist. $\endgroup$
    – ITA
    May 5, 2018 at 16:49
  • $\begingroup$ @ITA, sorry but not convinced. Any example? $\endgroup$
    – ar2015
    May 6, 2018 at 9:34
  • $\begingroup$ Read the proof. $\endgroup$
    – ITA
    May 6, 2018 at 10:57

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