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Given the following ODE in polar coordinates

\begin{array}{lcl} \frac{dr}{dt} = r\sin(\frac{1}{r}) \\ \frac{d\theta}{dt} = 1\end{array}

1) Show that the origin $(0,0)$ is Lyapunov stable

2) Prove that doesn't exist Lyapunov function

I could prove the first item, however I'm stuck in the second one. Any hint?

Santos
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  • What definition of Lyapunov function are you using? – user539887 Jun 17 '18 at 18:50
  • Given an ODE $x'=f(x)$ and an equilibrium point $x_0$ (it means $f(x_0) = 0$), we say that $V: \mathbb{R}^n \to \mathbb{R}$ is a Lyapunov function if $V(x_0)=0$, $V(x)>0$ if $x \neq x_0$ and $<\nabla V(x),f(x)> \leq 0$ for all x. – Santos Jun 17 '18 at 18:58
  • I tried, in vain, to prove the nonexistence of Lyapunov function. The origin is surrounded by a countable family of limit cycles (corresponding to $r=\frac{1}{n\pi}$). The limit cycles corresponding to even $n$ are unstable, whereas those corresponding to odd $n$ are stable. So the Lyapunov function on an even cycle must be greater than on both neighboring odd cycles. But I don't see how to obtain any contradiction. Even more, I would be inclined rather to think that there is a Lyapunov function. – user539887 Jun 17 '18 at 20:52
  • @user539887 Well, it could be. But at least this Lyapunov function won't be $C^2$: $C^2$ functions can't have such behaviour in the neighbourhood of critical point. – Evgeny Jun 18 '18 at 05:29
  • @Evgeny If my line of thought is correct, as $V(r\cos{\theta},r\sin{\theta})$ we can take $\tilde{V}(r)$. Then $\tilde{V}$ should have the following properties: $\tilde{V}(0)=0$, $\tilde{V}(r)>0$ for $r>0$, $\tilde{V}'(r)>0$ for $r\in(\frac{1}{(2k+1)\pi},\frac{1}{2k\pi})$ and $\tilde{V}'(r)<0$ for $r\in(\frac{1}{2k\pi},\frac{1}{(2k-1)\pi})$. Why is this incompatible with $V$ being $C^2$? – user539887 Jun 18 '18 at 06:28
  • @user539887 It's also kind of an observation. As you've established, in order to be Lyapunov the function $\tilde{V}$ must have such behaviour on any interval $\left ( \frac{1}{(2k+1)\pi}, \frac{1}{(2k-1)\pi} \right )$ . But is it compatible with $\tilde{V}$ having a strict local minimum at $0$? I think it's not, when function is $C^2$ and $f''(0) > 0$ — the function is strictly U-shaped in some small neighbourhood of $0$. The same holds if higher-order derivatives exist: the first non-zero derivative has even number, and if it's non-zero, then function is U-shaped. – Evgeny Jun 18 '18 at 07:11
  • @Evgeny When the second derivative at zero should be positive, then of course it isn't compatible; but the OP didn't make such an assumption. Incidentally, the second coordinate appears to be a red herring: we have much the same problem when we consider the (non-)existence of a Lyapunov function for $x'=x\sin{\frac{1}{x}}$. – user539887 Jun 18 '18 at 07:59
  • @user539887 The point is that if a function is U-shaped at the origin, it can't be Lyapunov function for this system. Is every strict local minimum U-shaped in some neighbourhood? I think no: while it's true when we have some information about non-zero derivatives, probably there is a continuous counter-example. Can this example be made $C^1$? I don't know, but if it can, then voila, it is a Lyapunov function for $\dot{x} = x \sin{\frac{1}{x}}$. – Evgeny Jun 18 '18 at 09:16
  • @Evgeny Assume that we have a function $V$ such that $V(0)=0$, $V(x)>0$ for $x>0$, $V$ has local mimima, equal to $\frac{1}{(2k-1)^2\pi^2}$ at $x=\frac{1}{(2k-1)\pi}$, $V$ has local maxima, equal to $\frac{1}{(2k-1)^2\pi^2}+\frac{1}{(2k)^3\pi^3}$ at $x=\frac{1}{2k\pi}$. Difference quotients between $x=\frac{1}{2k\pi}$ and $x=\frac{1}{(2k+1)\pi}$ tend to zero as $k\to\infty$, so this seems not to be an obstacle for the existence of such a $C^1$ function. But I don't know. – user539887 Jun 18 '18 at 20:39
  • @user539887 Yeah, I have very similar picture in my mind. I just can't prove right now that such example is $C^1$, that's why I don't know either :) – Evgeny Jun 18 '18 at 21:35
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    Why do you think that there is no Lyapunov function for this particular example? – Artem Oct 26 '19 at 02:54
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    where did this question come from? I mean, is it come from a textbook? – Yaosheng Deng Oct 22 '23 at 04:12

0 Answers0