I understand what's required in order to prove stability but there are certain parts that seem "out of the blue" to me.
I will give an example as it will portray what I don't understand better.
Consider the discrete-time dynamical system $(\mathbb R,\mathbb N_0, S_t)$ given by the iteration of the map $f: \mathbb R → \mathbb R$ where $f(x)$ = $2x$
Determine the compact and invariant set $I$ & prove the attractivity and stability properties of the set $I$
I understand how to determine the compact and invariant set but when it comes to proving the attracitivity and stability they make choices for $\delta$ and $\epsilon$ that I just don't understand where they got?
$I$ does not attract points in a neighbourhood of $I$. Indeed, for any neighborhood $W$ of $I$ there exists $x ∈ W$, $x \neq 0$ and $\epsilon = 1$ such that for all $t_0 ≥ 0$ there is $t$= max$(t_0,-\frac{ln\lvert x\rvert}{ln2})$$\geq t_0$ with $\lvert S_t(x)\rvert$ = $\lvert 2^t x\rvert$ = $\geq 1$ = $\epsilon$ which is a contradiction.
$I$ is unstable, since there exists $\epsilon = 1$ such that for all $\delta \ge 0$ there exists $t$= max$(-\frac{ln\lvert \frac{\delta}{2}\rvert}{ln2},0)$$\geq 0$ and $x=\frac{\delta}{2} \in I_\delta$ such that $\lvert S_t(x)\rvert$ = $2^t \lvert x\rvert$ = $2^t \frac{\delta}{2} \geq 1$ = $\epsilon$
I can't get my mind around where $t$= max$(t_0,-\frac{ln\lvert x\rvert}{ln2})$ comes from? How do they decide what $\epsilon$ to use? why do they use $x=\frac{\delta}{2}$ why not $x=\frac{\delta}{3}$ or $x=\delta$
I can't put my finger on the inuition behind it as usually you would see an end goal and work towards it but here it seems like the decisions are made after 100's of trial and errors and then the correct one is portrayed as if it would be easy enough to just realise that it should be $x=\frac{\delta}{2}$
I really struggle with these types of proofs so any direction would be massively appreciated!
Thank you very much!
Let's start by solving for the fixed points of the system.
The map is $f(x) = 2x$. A fixed point satisfies $y=2y$. The only solution is zero.
If $x\neq 0$, what happens under the map? $$ x_{t} = f^t(x_0) = 2^t x_0 $$ so that $$ x_t - x_0 = (2^t-1)x_0. $$ Now take any bound $M$. How long until the system started from $x_0$ is further than $M$ from $x_0$? $$ |x_t-x_0| = |(2^t-1)x_0|>M $$ $$ 2^t>\dfrac{M}{|x_0|}+1 $$ $$ t\log(2) > \log \left(\dfrac{M}{|x_0|}+1 \right) $$ $$ t> \dfrac{1}{\log(2)}\log \left( \dfrac{M}{|x_0|}+1 \right). $$ So for any $x_0$ in any nbhd of zero, there is a finite time $t>0$ at which the process escapes any set $[-M,M]$ around 0. The system is unstable.
So that's where a lot of the notation is coming from, like $M = \varepsilon$ or $\log(2)$. They're just making choices about the constants instead of doing it a bit more generally like what I am writing above. I don't know why they are making all choices. Seems a bit arbitary to me. But I don't know what $S_t$ means.
If you think logically about what a dynamical system is doing, stability is mostly about what is happening in a nbhd of a fixed point, like the Hartman Grobman theorem. Are points moving towards, or away from the fixed point in a nbhd? If so, the system is locally stable, otherwise, points are flowing away from the fixed point. You can formalize this with the Jacobian of the system evaluated at the point.
For maps this simple, you might even look at Banach's Fixed Point Theorem. The map here is expansive, rather than a contraction, which is a giveaway that any fixed point is unstable.