I am trying to answer the following question (it is not part of an assignement):
I think I have the first question. I am stuck with the 2nd one. I have no idea how to solve it. I looked at several books and I found the following, which makes sense to me, but doesn't use Poincare:
Solution attempt:
Let $V$ be the basin of attraction and $x_{0}\in V$, $x_{0}\neq\tilde{x}$. Then we can find an open ball $B_{\epsilon}(x_{0})\subset V$, since $V$ is open. Fix $s>0$ as in Poincares thm. and let $U:=B_{\epsilon}(x_{0})$. Then by Poincare it follows that there is $x\in B_{\epsilon}(x_{0})$ such that $\Phi_{t}(x)\in B_{\epsilon}(x_{0}) $ for $t>s$.
But since we also have that $x\in V$, we have that
$\lim_{t\to\infty}||\Phi_{t}(x)-\tilde{x}||=0\iff\forall\epsilon>0\exists t_{0}>0:t\geq t_{0}\implies ||\Phi_{t}(x)-\tilde{x}||<\epsilon$.
Consider $0<\epsilon_{1}<\frac{||x_{0}-\tilde{x}||}{2}$.
If $s>t_{0}$, then $\Phi_{t}\in B_{\epsilon}(x_{0})$ according to Poincare. But since $t>s>t_{0}$ we also have that $||\Phi_{t}(x)-\tilde{x}||<\epsilon_{1}$. Thus, $\Phi(x)\in B_{\epsilon}(x_{0})\cap B_{\epsilon_{1}}(\tilde{x})=\emptyset$.Hence, we have a contradiction.
If $t_{0}=s$ we have more or less the same as right above.
If $t_{0}>s$ there are two cases:
i) $t\geq t_{0}>s$, which I think is "clear".
ii) I think this can be solved by choosing s right in the beginning s.t. $s>t_{o}$, where $t_{0}$ is such that $t\geq t_{0}\implies ||\Phi_{t}(x)-\tilde{x}||<\epsilon_{1}$. I think this is possible, because $\tilde{x}$ is attracting.
I would really appreciate any help. Many thanks in advance.
Notation
Theorems used:
