When can I say , that two curves in the phase space of the following equation never intersect:
$x'=F(x,t)$
Where $x'= \frac{dx}{dt}$ and $F : \mathbb{R}^{3} \to \mathbb{R}^{2}$.
The reason I am asking this is because when our lecturer talked about Poincare map and the Van Der Pol equations , she wanted to prove the following claim:
Poincare map has one fixed point $y_0$, and for any $y \in \mathbb{R}^+$ we have that the limit of $P^ny$ as $n \to \infty$ is $y_0$
Where $P$ is Poincare map , and $P^n$ is poincare map composed wth itself $n$ times.
Before proving the claim she said that " $P $ is monotonic on $\mathbb{R}^+ $ and that we can conclude this frm the fact that curves do not intersect..." (the rest of the sentence is not really important but it was clear why it is true,but I couldn't understand why the curves do not intersect)
I will be happy if you could help me understand
I think I understood (I also talked with some people about this):
Let $x_1 , x_2$ be two solutions to the ode.
$\exists T \in \mathbb{R} $ s.t. $x_1(0) = x_2 (T)$ or $\forall T \in \mathbb{R} x_1(0) \neq x_2 (T)$ . Assuming $F$ is $C^1$ we can conclude from the uniqness theorem that we must have exactly one of the following:
1) The curves in the phase space of $x_1$ and $x_2$ do not intesect.
2) The curves in the phase space of $x_1$ and $x_2$ are the same curve.
If we have $\forall T \in \mathbb{R} x_1(0) \neq x_2 (T)$ , then of course we have (1) , elseways $\exists T \in \mathbb{R} $ s.t. $x_1(0) = x_2 (T)$ , but in this case using uniqness theorem we can conclude that
$\forall t:$ $x_1(t) = x_2(T+t)$
Meaning the curves are equal (as sets)