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I am given two dynamical systems: $$x'=f(x)$$ $$x'=g(x)$$ where $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ and $g:\mathbb{R}^2\rightarrow \mathbb{R}^2$ are $C^1$ and perpendicular ($\langle f(x),g(x)\rangle=0$ for all $x's$). I am to show that if one of the systems has a (nontrivial) periodic orbit, then the other has a fixed point. Any help with this problem would be appreciated.

Eduardo Longa
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Halinka
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1 Answers1

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Suppose without loss of generality hat $f$ has a nontrivial periodic orbit. Consider the path $\gamma$ that describes the orbit. $\gamma$ is closed because the orbit is periodic; moreover, by the Existence and Uniqueness Theorem for ODEs, $\gamma$ must be simple. It is hence a Jordan curve and we may speak of the interior and exterior of $\gamma$.

EDIT: If $g$ vanishes at some $x\in\gamma$, then $x$ is a trivial fixed point of the system induced by $g$. Assume then that $g$ is nowhere vanishing on $\gamma$.

Because $g$ is continuous, always orthogonal to $f$ and nowhere vanishing on $\gamma$, along $\gamma$ the field $g$ must face either always inwards (towards the interior) or always outwards (towards the exterior).

If $g$ faces always inwards, consider what happens as the time goes to $+\infty$. If $g$ faces always outwards, consider what happens as the time goes to $-\infty$. Do you think you can take it from here?

Hint: Choose a point $p\in\gamma$ and consider its orbit under $g$. Can the orbit intersect $\gamma$ at some point other than $p$? Use the existence and uniqueness theorem!

Fimpellizzeri
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  • g most certainly could change direction with respect to the orbit. – Paul Apr 11 '17 at 00:10
  • @Paul Could you provide an example? – Fimpellizzeri Apr 11 '17 at 00:12
  • What you can say instead is that either g does not change direction on the orbit, or g has a fixed point on the orbit. – Paul Apr 11 '17 at 00:15
  • I've added an edit about the trivial case when $g$ vanishes at some point. This should clear it. – Fimpellizzeri Apr 11 '17 at 00:19
  • If g is nowhere vanishing, it doesn't have a fixed point. – Paul Apr 11 '17 at 00:28
  • Along $\gamma$... – Fimpellizzeri Apr 11 '17 at 01:34
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    This basically leaves untouched the main part of the proof. One can find curious to leave this part to the OP, since it is significantly more difficult than the rest... Note that even if every orbit of a dynamical system crossing $\gamma$ cross it to its interior, they might, nevertheless, contain no fixed point. – Did Apr 11 '17 at 07:05
  • @Did I don't get how your last statement is possible, I would be very glad if you explain how. – Evgeny Apr 11 '17 at 08:08
  • @Evgeny Assume that $$\frac{d\theta}{dr}=\frac{(r-1)^2}{r(2r-1)}$$ then each trajectory issued from some point such that $r>1$ enters the disk $r<1$ then circles counterclockwise infinitely many times in the annulus $\frac12<r<1$, accumulating on the circle $r=\frac12$. Thus, one crosses $\gamma$ ($r=1$) orthogonally and yet, there is no fixed point attached to these trajectories (what happens in the disk $r<\frac12$ being another story). – Did Apr 11 '17 at 11:00
  • @Did Ah, ok, in that sense I agree with you. Because in other sense a "closed curve without contact" is still an evidence that there is an equilibrium point inside of this closed curve. That's where I thought that we are thinking about different setups for statement. – Evgeny Apr 11 '17 at 11:16
  • @Evgeny I see. Indeed, "they" in my first comment refers to "every orbit of a dynamical system crossing γ", not to "every orbit of a dynamical system". – Did Apr 11 '17 at 11:19
  • @Did You're right.., I was trying not to kill a mosquito with a bazooka (say, using Zorn's Lemma or Poincaré-Hopf), but I guess it didn't end up very well. I may try and salvage it later. – Fimpellizzeri Apr 11 '17 at 15:58
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    @Fimpellizieri, to be honest, no I cannot take it from there. – Halinka Apr 11 '17 at 16:35
  • @Halinka do you know Poincaré-Hopf or the Hairy Ball Theorem? – Fimpellizzeri Apr 11 '17 at 16:58
  • Hmmm... Sorry to interrupt again but at this point, throwing around the names of results of the field that may or may not apply is not the most productive endeavour one can imagine. If you can provide precise indications about an approach that does work (and this assumes you are sure it does), then do exactly that, else... – Did Apr 11 '17 at 19:00
  • @Did I won't bother writing an answer using tools that are unfamiliar or otherwise more advanced than what is available to the OP. That said, feel free to do so if you wish. – Fimpellizzeri Apr 11 '17 at 19:04
  • Are you suggesting that I should "writ(e) an answer using tools that are unfamiliar or otherwise more advanced than what is available to the OP"? Odd suggestion. – Did Apr 11 '17 at 19:06
  • @Did I did not suggest that. Merely said you are free to do it if you so wish. – Fimpellizzeri Apr 11 '17 at 20:09
  • @Fimpellizieri, I looked up the hairy ball theorem, and don't find it relevant. In our case, we have a closed orbit, not a ball or sphere, and an orthogonal vector field not tangential. I am sure, I am missing something, but don't know what it is. – Halinka Apr 14 '17 at 00:57
  • @Halinka $\gamma$ is diffeomorphic to the unit circle, while the interior of $\gamma$ is diffeomorphic to the open unit disk (say, via the Riemann mapping theorem), which in turn is diffeomorphic to the open half sphere. If the field $g$ was nowhere vanishing inside $\gamma$, we could push it forward to obtain a nowhere vanishing field inside the open half sphere that is perpendicular to its boundary circle. (continues) – Fimpellizzeri Apr 14 '17 at 03:11
  • (continued) If such a field $F$ on the half sphere were possible, can you think of a way to piece together a nowhere vanishing field $\overline{F}$ on all of the sphere? Hint: The tangent spaces to antipodal points in the sphere are parallel (as planes in $\mathbb{R}^3$). What happens if you extend $F$ by mirroring it across the center of the sphere? – Fimpellizzeri Apr 14 '17 at 03:13