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In ODE where $x'=f(x(t))$ and $x(0)=x_0$,

Let $S_\delta$= connected component of the set {x in S|$V(x)\leq \delta$} that contains $x_0$. It's also closed.

Lemma: for every $\epsilon>0$ there exists a $\delta>0$ such that:

(1) $B(x_0,\epsilon)$ $\subseteq$ $S_\delta$.

(2)$S_\epsilon$ $\subseteq$ $B(x_0,\delta)$

I want to prove (1) by contradiction using Lyapunov Function $V(x(t))$. My professor asked me to assume that $V'(x)<0$ so that $x_0$ is asymptotically stable and that it's a strictly Lyapunov function ($V(x(t_1))$ $<$ $V(x(t_0))$). He also said $V(x_0)=0$ and that i should fix $\epsilon=1/n$. Because he proved (2) exactly that way!

But I'm still confused as both parts are incomparable. (I don't know about this material outside ODE!)

Thank you for the help!

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    i didn't get the concept right, my professor confused me every time i asked him. but now he actually explained me the answer.2013-01-09

1 Answers 1

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Assume the contradiction: For every $\delta$ $>$ $0$ there exist $\epsilon$ $>$ $0$ such that $B_\epsilon(x_0)$ $\not\subset$ $S_\delta$.

Let $\epsilon=1/n$ then $\exists$ {$x_n$} $\in$ $B_\epsilon(x_0)$ but {$x_n$} $\not\in$ $S_\delta$ $\implies$ $V(x_n$) $\geq$ $\delta$ $>$ $0$ and {$x_n$} is not contained in any of the disconnected components as well.

then $|x_n$ - $x_0$| $<$ $1/n$ $\implies$ $x_n$ converges to $x_0$. Therefore $V(x_n)$ converges to $V(x_0)$ $\implies$ $V(x_n)$=$V(x_0)$=$0$

Therefore a contradiction!

According to my professor.

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    +1 for taking the right step. Math.SE can be helpful, but it will never replace your professor.2013-01-10