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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.

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  • $\begingroup$ This is a naive guess, but does the Poincare-Bendixson theorem say anything useful - maybe it requires better differentiability? $\endgroup$
    – Leo Moos
    Commented Apr 5, 2021 at 11:04
  • $\begingroup$ Thinking about it, I don't think that the Poincare-Bendixson helps here -- even in the analytic case. $\endgroup$
    – coudy
    Commented Apr 5, 2021 at 12:26

2 Answers 2

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(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 $\varphi^{-1}$.

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    $\begingroup$ technical quibble: if $\lim_{t\to -\infty} \varphi_t (x) = 0$ for all $x$, then $O_t = \mathbb{R}^2$ for all $t$ as you defined it. You probably want $0 < s < t$ in the definition. $\endgroup$
    – Willie Wong
    Commented Apr 6, 2021 at 0:37
  • $\begingroup$ Good point! I edited accordingly. $\endgroup$
    – Martin M. W.
    Commented Apr 6, 2021 at 0:52
  • $\begingroup$ Indeed that works in the $C^1$ case. I am wondering if that argument can be adapted to the $C^0$ setting. $\endgroup$
    – coudy
    Commented Apr 6, 2021 at 8:52
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    $\begingroup$ @coudy: what if you remove the normalization step? The construction of $T$ is the same. Let $\bar{B}$ be the closed ball of radius 1. The argument above shows that the whole space must be contained in the image of $[0,T]\times \bar{B}$, but the latter is compact. $\endgroup$
    – Willie Wong
    Commented Apr 6, 2021 at 15:27
  • $\begingroup$ @Wong Indeed, that works, perfect. $\endgroup$
    – coudy
    Commented Apr 6, 2021 at 15:32
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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.

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    $\begingroup$ Is it obvious that $l$ is continuous? I tried to reason like this earlier but got stuck justifying this fact. $\endgroup$
    – Willie Wong
    Commented Apr 6, 2021 at 0:29
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    $\begingroup$ @Wong The function $l$ is not continuous but it is upper semi-continuous, which is enough to conclude it is bounded. This argument is pretty close to Martin's. $\endgroup$
    – coudy
    Commented Apr 6, 2021 at 14:25
  • $\begingroup$ Well, actually, the upper semi-continuous function is the first return in the open unit ball... which can be used instead. $\endgroup$
    – coudy
    Commented Apr 6, 2021 at 14:48

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