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Suppose $\dot{x}=f(x)$ is a dynamical system, with $x$ in $R^n$, and $f:R^n \to R^n$ sufficiently smooth (for example, Lipschitz-continuous).

Assume that $x_e$ is an unstable equilibrium point of the system. Even if $x_e$ is unstable, the union of all trajectories having $x_e$ as a limit point could be larger than just the singleton {$x_e$}. But does this set always have zero measure?

This might be a basic result, but any pointer would be appreciated.

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Title updated to avoid misuse of the term "attracting set". –  Vincenzo Oct 12 '10 at 20:36
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up vote 4 down vote accepted

Take for example the equation

$\dot x = \lambda x$

$\dot y = y^2$

For $\lambda < 0$. The origin is an unstable equilibrium point, however, its stable manifold is the whole lower semiplane (including the $y=0$ axis).

If there is one eigenvalue of real part bigger than $0$ for the derivative in the equilibrium point, it is then true that the set of points converging to $x_e$ has zero measure. See for example here (proposition 4.1).

The proof is simple and based in the existence of center stable manifolds, this implies (toghether with the eigenvalue with positive real part) that the local stable set of the equilibrium has zero measure, after that, it is done since the stable set can be written as a countable union of sets diffeomorphic to this one.

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By the result is true, you mean that ''the union of all .... always have zero measure''. Correct ? –  Denis Serre Oct 12 '10 at 14:16
    
Yes, I mean that the set of points having $x_e$ as limit point in the future has zero measure. I will try to clarify a bit. –  rpotrie Oct 12 '10 at 14:26
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