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$x'(t) = y$

$y'(t) = -x + x^3 + y$

$(0,0)$ is a steady state.

According to the solutions, $E(x,y) = V(x) + y^2/2 = x^2/2 - x^4/4 + y^2/2$ is a Lyapunov function.

It then says $E(x,y) > 0$ for all $(x,y) \neq (0,0)$, but that's clearly false if I were to take $x = 4$ and $y = 0$, then I get $8 - 4^3 + 0 = 8 - 4^3 = -56 < 0$

Lemon
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  • Yes, it is saying it is true on an open set minus the steady state is positive. – Lemon Nov 22 '13 at 00:14
  • $(0,0)$ looks like a local attractor, but not a global one. So you won't be able to find a global Lyapunov function. However, you can still find a Lyapunov function on some neighborhood of the origin - for instance, the function $E$. The solution may lack some quantifiers (it should then say "for all $(x, y) \neq (0,0)$ in $U$", where $U$ is some (explicit?) neighborhood of the origin). – D. Thomine Nov 22 '13 at 00:15
  • The solution says we need an open that contains (0,0), but not the other steady states. Why? – Lemon Nov 22 '13 at 02:55

1 Answers1

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As the commenters said, the function is only locally positive. It's easy to see that $$ V(x) = \frac{x^2}{2}\left(1 - \frac{x^2}{2}\right) + \frac{y^2}{2}$$ and so it is positive (at least) in the strip defined by $-\sqrt{2} < x < \sqrt{2}$ (for any $y$). That's enough for local stability.