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I know that if a have a strict Lyapunov $L$ function defined in an open set $O$ that contains the equilibrium point $x_{0}$, then that point must be asymptotically stable. Which means that if my solution starts close enough then it will approach $x_{0}$ as $t$ tends to infinity. My question is, if my solution $Y$ starts at any point in $O$, and because $L$ is decreasing over the solution curves then, does it follow that $Y(t)$ will approach $x_{0}$ as $t$ goes to infinity? (Is it true that every point in $O$ is in the basin of attraction of $X_{0}$?)

I know that if $P$ is a positively invariant, closed subset of $O$ with no solutions of $P$ under which $L$ is constant, then $P$ is a subset of the bacin, even if $L$ is not a strict Lyapunov, but just a Lyapunov function.

So, is this theorem only useful for not strict Liapunov functions? (Because if my previous assertion is true then it would be trivial because $O$ is contained in the bacin.)

Thank you!

Rodrigo de Azevedo
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MathUser123
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    The answer to this question follows directly from the proof of the Lyapunov's asymptotic stability theorem. The set $S_{C}= \left{ x:; V(x)\le C \right}$ (or its connectivity component of $x_0$, if it is not connected) such that $\forall x\in S_{C}\setminus {0}; V(x)<0$ is a subset of the domain of attraction – AVK Nov 15 '17 at 11:59
  • And by the way, for the sake of getting relevant results in search, the correct spellings are Lyapunov and basin of attraction – polfosol Nov 15 '17 at 12:37
  • See https://math.stackexchange.com/a/2474778/76258 for a counter-example to your question. – shamisen Mar 03 '23 at 16:44

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