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Given the following ODE in polar coordinates

$$ \dot{r} = r^2 sin^2(\frac{1}{r}) $$ $$ \dot{\theta} = 1 $$

  1. Show that the origin is Lyapunov Stable

  2. Prove that doesn't exist a Lyapunov function associated with (0,0)

I found a similar question with no answer here: Prove that doesn't exist a Lyapunov function

I can prove the first part, but no clue about the second. I tried looking to the limit sets of the origin, and checking what would happen there with the Lyapunov function, if it existed... but nothing came out of it. Can someone help me?

I'm using the following definition of a Lyapunov function: Given an ODE $\dot{x} =f(x)$ and an equilibrium point $x_0$ (ie, $f(x_0)=0)$, we say that $V:\mathbb{R}^n \rightarrow \mathbb{R}$ is a Lyapunov function if $V(x_0)=0$, $V(x)>0$ if $x \neq x_0$ and $ \langle∇V(x),f(x)\rangle \leq 0$ for all x.

abiessu
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JazzyStudent
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    I guess that for 2. it is implied that the Lyapunov function would have to show asymptotic stability? Namely, one can always construct a positive definite function and call it a "Lyapunov function". – Kwin van der Veen Jun 13 '22 at 02:14
  • If it is implied then i think i can solve the problem with the following argument: If a lyapunov function V existed by continuity, we can find an open set such that V is strictly decreasing, and that would mean that (0,0) is asymptotically stable, that is a contradiction.(the orbits all tend away from the origin) I don't know if this is enough, did i miss something? – JazzyStudent Jun 13 '22 at 17:31
  • I currently don't have time to write a full answer, but instead of trying to prove stability you could try to prove instability using Chetaev's instability theorem. – Kwin van der Veen Jun 13 '22 at 19:57
  • I do not even think that the origin is Lyapunov Stable. Why should it be with $\theta$ going to infinity? – Arastas Jun 14 '22 at 07:19
  • The origin I'm talking is the origin in Cartesian coordinates. – JazzyStudent Jun 14 '22 at 14:07
  • At the next line, you consider $(0,0)$, does it include $\theta$? Or do you work only with the scalar dynamics of $r$? Note that as $r$ has multiple equilibriums, there cannot exist a Lyapunov function for global asymptotic convergence. – Arastas Jun 15 '22 at 09:01

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