Planar flow with bounded orbits and a single equilibrium point Is there a $C^1$ flow $\varphi_t$ defined on ${\bf R}^2$ with a single fixed point $0$ and such that for all x,
$$\lim_{t\rightarrow +\infty}\varphi_t(x) = 0,$$
$$\lim_{t\rightarrow -\infty}\varphi_t(x) = 0?$$
I think that this should not exist but I can't find a simple argument to rule out the existence of such a flow. Same question for a $C^0$ flow.
 A: (edited to include Willie Wong's idea for $C^0$ case.)
This kind of flow can't exist in any dimension.
Let $S$ be the unit sphere and $B$ be the open unit ball. If the origin is a global attractor for $\varphi$, then $S \subset \bigcup\limits_{t>0}{\varphi_{-t}(B)}$.
By compactness, $S$ is covered by a union of a finite subset of the $\varphi_{-t}(B)$. That implies that there is a constant $T$ such that no point on $S$ (or $\overline{B}$) flows for more than time $T$ outside $\overline{B}$.
Since the image of $[0, T] \times \overline{B}$ under $\varphi$ is compact, it can't cover the whole space, so the origin is not a global attractor for $\varphi^{-1}$.
A: Another proof might be as follows. Consider a circle about the origin. Since the flow is $C^1$, consider the restriction of the flow's continuous vector field to this circle. There must be some point on the circle whose flow line is transverse to the circle (by continuity of the vector field), as otherwise the circle is a closed orbit of $\varphi^t$, a contradiction. So we have a map $T: S^1 \to S^1$, the first return map of the outward pointing vector field. Consider the flow line segment from $p$ to $T(p)$, for any $p \in S^1$ and let $l: S^1 \to [0, \infty)$ denote the length of the flow line segment. Since $S^1$ is compact, $l$ is bounded by some constant $C$. However, this implies that any point outside a ball of radius $C$ about the origin never returns to the origin, a contradiction. Hence, no such flow exists.
This argument might work for the $C^0$ case too, with more work.
