Good afternoon!
I have a quick question on limit sets. I am working with a differential equation in $\mathbb{R}^2$ whose flow is given by:
$\varphi(t,x_0) = e^{-t} \begin{bmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{bmatrix}$ $x_0$.
And I am trying to find its « alpha limit set » (ie, the set of limit points of all its negative orbits). However, the solution of my problem says that if $x_0 \ne (0,0)$, then this alpha limit set is the empty set.
I don't understand why. Because, say, take $x_0 = \displaystyle\binom{1}{1}$. Then the flow is given by $\varphi(t,x_0) = e^{-t}\displaystyle\binom{\cos(t) + \sin(t)}{-\sin(t) + \cos(t)}$. We are studying the negative orbits, ie for $t \leq 0$. But in this case, the first component of the flow wiggles between very large positive values and very large negative values, so as to cover all $\mathbb{R}$ as $t \to -\infty$. And since we always have this « alternating » behaviour, then every value in $\mathbb{R}$ is going to be reached infinitely many times. Similarly for the second component.
So the alpha limit set should be $\mathbb{R}^2$, shouldn't it?
I might have got something wrong somewhere, but I can't tell what. Thanks for reading!
Answer was given by Lutz Lehmann in the comment. I was considering the projections of the flow onto the $x$ and $y$ axes. Those can have any real number as a limit point, but the flow altogether cannot.