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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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Usually when you define a set to be a connected component, you say it is a connected component of some set. For example, if $A = (-1,0)\cup(1,2)$, then $(-1,0)$ and $(1,2)$ are the connected components of $A$. So when you say $S_\delta$ is a connected component set, it is a connected component of what? Also, is $S_\delta$ meant to be an open set? If it is, your lemma is immediate, and does not need to use the Lyapunov function. Hope these questions help! – froggie Dec 25 '12 at 13:21
S-delta is a connected component of S-epsilon. regarding the openness, from my notes Sdelta is actually compact implying its closed, so im not quite sure about the proof. – d13 Dec 25 '12 at 13:29
What is $S_\epsilon$? – froggie Dec 25 '12 at 13:38
I recommend editing your question to include this definition! It will make the question more clear, and will make it more likely that someone will help you. Good luck! – froggie Dec 25 '12 at 14:16
I'm sorry but the reason you've been waiting for long for an answer is that your question does not make sense. It's a mess of symbols with few relations between them. // What does $B$ have to do with the rest of the problem? // Trying to define $S_\epsilon$, you wrote "{x(t) in S | V(x)<=S}" in a comment but "$\{x \text{ in } S | V(x)\ge \delta \}$" in the question. These are different, and neither version makes sense without a relation between $\epsilon$ and $\delta$. // Your best chance to get an answer would be to provide a reference, or a precise copy of the problem as you saw it. – user53153 Jan 5 '13 at 5:20
up vote 1 down vote accepted

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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i was confused all the way and my professor was confusing me as well. so finally i sat down with him to solve this proof and he did prove it and answering to all my confusions as well. I thank him for giving me this proof! it belongs to him – d13 Jan 9 '13 at 9:41
+1 for taking the right step. Math.SE can be helpful, but it will never replace your professor. – user53153 Jan 10 '13 at 0:15

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