# Two dynamical systems perpendicular

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.

## 1 Answer

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!

• g most certainly could change direction with respect to the orbit. – Paul Apr 11 '17 at 0:10
• @Paul Could you provide an example? – Fimpellizieri Apr 11 '17 at 0:12
• 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 0:15
• 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 7:05
• @Fimpellizieri, to be honest, no I cannot take it from there. – Halinka Apr 11 '17 at 16:35