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What can be concluded about an equilibrium point that has finitely many trajectories that converge to it?

Mathematics Asked by abc1455 on November 6, 2021

If you have a dynamical system $dot{x}=f(x)$ and $f(x)$ is smooth , given that $f(x_0)=0$ and only finitely many trajectories converge to $x_0$ . I think if $x_0$ is isolated then the stable manifold is one dimensional since if it was more than one dimensional it would have infinitely many . I also think that the number of trajectories can be at most two .

My questions are :

1) If $x_0$ is isolated , is the stable manifold of $x_0$ only one dimensional and If yes is the smoothness condition needed ?
2) If $x_0$ is not isolated can other conditions be imposed so that the same conclusion can be said ?

Edit : I think I could use the linearization of the system at $x_0$ as $dot{x}=Ax$ , so if the system has one stable eigenvector $v_1$ , since trajectories cannot intersect it will have only two stable trajectories $v_1$ and $-v_1$ , if there is another stable eigenvector then the whole plane containing the two vectors will be stable implying there is an infinite number of trajectories . However , this doesn’t include the cases where the linearization fails or non-isolated points case .

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