I have this question:
What is a separatrix of a equilibrium point of a continuous dynamical system and why it is flow-invariant?
Thanks
Hello and thanks for the answer.
I explain better.
I'm following a first undergraduate course in dynamical systems.
My professor gave me this definition of separatrix curve :
"Let be $\dot X=F(X)$ a planar dynamical system and let be $\hat X$ an equilibrium point.
A differentiable curve $g:I\rightarrow\mathbb R^2$ is a stable separatrix for $\hat X$ if:
1) For every $P\in Im(g)$ the solution with initial condition $P$ exists for every $t\in[0,+\infty)$ and has $\hat X$ for limit as $t$ tends to $ +\infty $ .
2)For every $P\in Im(g)$ exists a neighborhood $U$ of $P$ such that for every $Q\in U-Im(g)$ the solution with initial condition $Q$ does not have $\hat X$ for limit as $t$ tends to $+ \infty$ . "
This is the definition…
Well, it seems to not work…
For example if i consider the system:
$$\begin{cases}
\dot x=-x \\
\dot y=0
\end{cases}
$$
the basin of attraction of $(0,0)$ is the $x$ – axis and it should be even the (image of the) stable separatrix.
Now, according to the definition, even the curve $g:(0,1)\rightarrow \mathbb R^2 $ , with $g(s)=(s,0)$ is a stable separatrix for $(0,0)$ .
Now i also know that the image of the separatrix should be positively invariant…but g is not!
Where i get wrong?
Best Answer
With this definition, the separatrices are exactly the differentiable curves $g:I\to\mathbb R^2$ such that $g(I)\subseteq\mathbb R\times\{0\}$. Thus, $g(I)$ can be any interval of the $x$-axis. There are more restrictive definitions of this notion, though.