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I am having difficulty understanding how epsilon is chosen to prove that a dynamical system is attractive and/or stable. I have taken several analysis modules and was okay at proof writing, well now a year later I am doing 2 more and I seem to have lost my ability a bit!

The fact there is delta and epsilon is perhaps what is confusing me, although I have watched some videos on this and have seen it drawn out - ie an epsilon ball with some smaller boundaries, delta, within it and we want to show that $SI_\delta\subset I_\delta$ where $I_\delta$ is a compact invariant set. Sorry if this description doesnt make sense, hopefully it does or is clear what I am trying to say and any better description would be welcomed.

So anyway, here is a question which I am going to use as an example. It is a continuous time dynamical system. I have the solutions so I don't want any hints, I would like a good well explained explanation of how epsilon is found and how we choose whether delta is equal to epsilon, or half epsilon etc. I want to really understand what I am doing, so that I can write great proofs and not lose any marks in my exam.

Consider $\dot x=0$ on $X=\mathbb R$ with $I=\{0\}$

The solution is given by $x(t)=1$

so $S_tx=x$

We have that $I= \{ 0\}$ is compact as it is bounded and closed.

It is easily seen that it is invariant since $S_tB=B$

i.e $S_t0=0$.

Now this is where I begin to get a little stuck.

I want to first show that $I$ is not attractive - it attracts no points in any neighbourhood of I

it is show by contradiction:

Let W be some neighbourhood of $I$ such that there exists $x\in W:x\ne 0$ and $\epsilon=|x|/2:\forall t_0 \geq 0$, there exists $t=t_0$ such that

$dist(S_tx,I)=|x|=2\epsilon$

So why is $\epsilon=|x|/2$? Have they really just "chosen" this or is there a reason it has been chosen to be this?

Also, why is the distance $|x|$?

Now to show stability;

For all $\epsilon>0$, if we let $\delta=\epsilon$,

we have

$|x|<\delta \implies |S_tx|=|x|<|\delta|=\epsilon \implies$ $I$ is stable.

Why have they chosen $\delta = \epsilon$? Sometimes it can be half epsilon or 2 epsilon. I am confused how this is chosen.

Thanks!

theorems I am using

It would be great if they could be used instead of any alternative ones please so that I do not get confused.

Stability

Let $I$ be a compact and invariant set. I is called stable is for all $\epsilon > 0$ there exists a $\delta>0$ such that $S_tI_\delta \subset I_\epsilon \forall t\geq 0$

Attractivity

Let $I, H \subset X$ and let $I$ be compact and invariant.

$I$ attracts points of H/compact sets $B \subset H$ /bounded sets $B\subset H$ if

$dist(S_tB,I)\rightarrow 0$ as $t \rightarrow \infty$

  • Could you clarify your post a little bit? Some notation appears out of the blue, it's hard to understand what happens. – Evgeny May 03 '15 at 05:54
  • Hey, sorry it's hard to understand! I have added the theorems I am using from class at the bottom. I have noticed the $S_tx$ notation is not really used anywhere, is this what you mean? To be honest, either way, I dont know if it affects how and if you can answer the question, but I would just like to know how epsilon and delta are chosen? – Bernard.Mathews May 03 '15 at 06:44
  • I have just taken this question and solution from my course material, so it is all correct and how I have been taught. However, please let me know what you would like me to clarify and I will try my best :) – Bernard.Mathews May 03 '15 at 06:46
  • Well, $B$ is used without defining it. $S_t$ is understandable because it's an action of the flow (however, many books usually denote it as $f^t(x)$). Also, the line "solution is given by $x(t) = 1$" is confusing, because general solution of $\dot{x}=0$ is $x \equiv {\rm const}$. That were most confusing phrases in your post. – Evgeny May 03 '15 at 09:24
  • Well it was simply taken from my course and copied exactly. Sorry. I wasnt confused about what any of that anyway. I just laid it out to enable it easier for people to help me – Bernard.Mathews May 04 '15 at 02:10

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