Limit set

Remarks

Equivalently, $\omega(x) = \bigcap_{T \ge 0} \overline{\set{\varphi_t(x) : t \ge T}} \cap U$, where the bar denotes the closure in $\R^n$. The set $\omega(x)$ is closed in $U$ and invariant. If the forward orbit of $x$ lies in a compact subset of $U$, then $\omega(x)$ is nonempty, compact and connected, and $\varphi_t(x)$ tends to $\omega(x)$ as $t \to \infty$. Example: in polar coordinates, let $r' = r(1 - r^2)$ and $\theta' = 1$ in the plane. For $x \ne 0$ the radius tends to $1$ while the angle keeps turning. So every point of the unit circle is a limit of $\varphi_{t_k}(x)$ for suitable $t_k \to \infty$, and $\omega(x)$ is the unit circle. For $0 < \abs{x} < 1$, the radius tends to $0$ as $t \to -\infty$, so $\alpha(x) = \set{0}$. Non-example: for $x' = 1$ on $\R$, the solution $\varphi_t(x) = x + t$ has no convergent sequence $\varphi_{t_k}(x)$ with $t_k \to \infty$, so $\omega(x)$ is empty. This is possible because the forward orbit is not bounded.