7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ This is a naive guess, but does the Poincare-Bendixson theorem say anything useful - maybe it requires better differentiability? $\endgroup$
    – Leo Moos
    Apr 5 at 11:04
  • $\begingroup$ Thinking about it, I don't think that the Poincare-Bendixson helps here -- even in the analytic case. $\endgroup$
    – coudy
    Apr 5 at 12:26
3
$\begingroup$

(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 $\varphi^{-1}$.

$\endgroup$
6
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 6 at 0:37
  • $\begingroup$ Good point! I edited accordingly. $\endgroup$ Apr 6 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
    Apr 6 at 8:52
  • 2
    $\begingroup$ @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. $\endgroup$ Apr 6 at 15:27
  • $\begingroup$ @Wong Indeed, that works, perfect. $\endgroup$
    – coudy
    Apr 6 at 15:32
4
$\begingroup$

Here is my attempt at an answer - what I propose is to use a more topological argument. As far as I can tell this works provided the flow curves are of class $C^0$, because it essentially only uses the fact they are Jordan curves.

Let $(\gamma_\alpha \mid \alpha \in A)$ be the union of flow lines, indexed by some set $A$. Each one of is a compact, continuous loop $\gamma_\alpha: \mathbf{S}^1 \to \mathbf{R}^2$ with $\gamma(1) = 0$, which moreover bounds an open set denoted $\Omega_\alpha \subset \mathbf{R}^2 \setminus \{ 0 \}$ whose closure includes the origin.

Define an equivalence relation on $A$ where $\alpha_1 \sim \alpha_2$ if $\Omega_{\alpha_1} \subset \Omega_{\alpha_2}$ or $\Omega_{\alpha_2} \subset \Omega_{\alpha_1}$. The properties of flow lines imply that this is an equivalence: indeed for any two $\alpha_1,\alpha_2 \in A$ either $\alpha_1 \sim \alpha_2$ or $\Omega_{\alpha_1} \cap \Omega_{\alpha_2} = \emptyset$. Index the equivalence classes by some second set $B$. For each $\beta \in B$ we can form an open set in $\mathbf{R}^2 \setminus \{ 0 \}$: $U_\beta = \cup_{\alpha \in A, [\alpha] = \beta} \Omega_\alpha$. These sets are disjoint, and their union is $\cup_{\beta \in B} U_\beta = \mathbf{R}^2 \setminus \{ 0 \}$. Now, as $\mathbf{R}^2 \setminus \{ 0 \}$ is connected, this means that $\lvert B \rvert = 1$. In terms of the original set $A$, we conclude that any two $\alpha_1,\alpha_2 \in A$ are equivalent, and the open sets $\Omega_\alpha$ form an increasing union.

Next we wish to exploit the fact that $\mathbf{R}^2 \setminus \{ 0 \}$ is not simply connected, whereas the $\Omega_\alpha$ all are. Let $\Gamma: \mathbf{S}^1 \to \mathbf{R}^2 \setminus \{ 0 \}$ be an arbitrary continuous loop that is not null-homotopic in $\mathbf{R}^2 \setminus \{ 0 \}$. This is covered by the open sets $\Omega_\alpha$, from which we may extract a finite subcover, say $\Gamma \subset \Omega_{\alpha_1} \cup \cdots \cup \Omega_{\alpha_N}$ with $\Omega_{\alpha_i} \subset \Omega_{\alpha_{i+1}}$ for $i = 1,\dots,N-1$. As they are increasing, $\Gamma \subset \Omega_{\alpha_N}$. Because $\Omega_{\alpha_N}$ is simply connected, we can continuously deform the loop $\Gamma$ to a point: this contradicts our initial choice for the loop because the deformation is in $\mathbf{R}^2 \setminus \{ 0 \}$.

$\endgroup$
6
  • 1
    $\begingroup$ Nice argument. There is a point that bothers me though. I don't see why the union of the $U_\alpha$ should cover the plane. There may be a trajectory so that close trajectories on different sides wind in opposite directions. The change in direction happens very close to the origin. $\endgroup$
    – coudy
    Apr 5 at 20:28
  • $\begingroup$ I should probably add some details there, as I think the continuous dependence on the initial ways is required. The argument I have in mind would go as follows. By contradiction, assume there exists a point $x \in \mathbf{R}^2 \setminus \{ 0 \}$ which does not belong to any of the $\Omega_\alpha$, and let $\gamma_0$ be the loop containing it. Let next $(x_j \mid j \in \mathbf{N})$ be a sequence of points converging to $x$ as $j \to \infty$, and let $(\gamma_j \mid j \in \mathbf{N})$ be the corresponding sequence of loops. Our aim is to show that for $j$ large enough one has [...] $\endgroup$
    – Leo Moos
    Apr 5 at 23:15
  • $\begingroup$ [...] $\Omega_0 \subset \Omega_j$. (To arrange for this we pick $x_j$ outside of $\Omega_0$). Continuous dependence on the initial conditions means that $\gamma_j \to \gamma_0$ as $j \to \infty$. The convergence need not be particularly strong, although I believe something like $\mathrm{Length}(\gamma_j) \to \mathrm{Length}(\gamma)$ seems necessary. Pick an arbitrary point $y \in \Omega_0$, and let $d_j \in \mathbf{Z}$ be the degree of $\gamma_j$ with respect to this point. Then $d_0 = 1$, and the convergence of the curves implies that $d_j = 1$ for large $j$ enough. This ought to imply [...] $\endgroup$
    – Leo Moos
    Apr 5 at 23:26
  • $\begingroup$ [...] that for large $j$, $\Omega_0 \subset \Omega_j$. But then also $\gamma_0 \setminus \{ 0 \} \subset \Omega_j$, and $x \in \Omega_j$ in particular. This is absurd given how the point was initially chosen. $\endgroup$
    – Leo Moos
    Apr 5 at 23:30
  • $\begingroup$ I don't think that your argument works. It is not true that if $x_j$ converges to $x$, then Length($x_j$) converges to Length($x$). Think about a curve winding around two other curves in succession before reaching the origin. For readability, note that it is better to edit your post instead of answering in comments. $\endgroup$
    – coudy
    Apr 6 at 14:22
1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Is it obvious that $l$ is continuous? I tried to reason like this earlier but got stuck justifying this fact. $\endgroup$ Apr 6 at 0:29
  • 1
    $\begingroup$ @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
    Apr 6 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
    Apr 6 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.