Consider $2D$ dynamical systems $X' = F(X)$ where $X=(x,y)$ is a 2-vector, and there is an equilibrium point at the origin. Let $L$ be the set of numbers $x > 0$ such that a limit cycle of the system meets the $x$-axis at $(x,0)$. Is $L$ necessarily closed? More generally what sets $L$ can arise in this way? [The reason I want to know is that I'm teaching dynamical systems.]

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    $\begingroup$ the subject of "Limit periodic set" is fully investigated in the literature about limit cycle theory. see for example "Limit cycles and bifurcation of polynomial vector fields " by Robert Roussarie. $\endgroup$ – Ali Taghavi Feb 18 '14 at 19:23

$L$ need not be closed. Consider a vector field $X$ with the following properties:

  1. $X(x,0)=(0,x)$ for all $x>0$.
  2. For $x\in (0,1)$, the trajectory starting at $(x,0)$ follows a vertical line until it passes through the point $(0,\frac 1{1-x})$, and then travels counter-clockwise around the origin until it returns to $(x,0)$.
  3. The trajectory starting at $(1,0)$ never leaves the vertical line $x=1$.
  4. The trajectory starting at $(x,0)$ for any $x>0$ is non-periodic.

Then $L=[0,1)$ is not closed.

In the other direction, given any closed $L$, it is possible to construct a vector field realising that $L$. Indeed, begin by constructing a map $f\colon [0,\infty)\to [0,\infty)$ such that $L = \{ x \mid f(x)=x \}$, and then construct a flow whose first return map to the positive $x$-axis is the map $f$.

To construct the map, just observe that the complement of $L$ is a countable union of open intervals, and so you can define $f(x) = x$ for all $x\in L$, and then define $f$ on each open interval $(a,b)$ as a "north-south" map, that is, a map of the interval $[a,b]$ that fixes $a$ and $b$ and moves every other point towards $b$.

To construct the flow from the map, just draw circles of radius $x$ for every $x\in L$, and then spirals connecting $(x,0)$ to $(f(x),0)$ for every $x\notin L$.

This still leaves open the question of which non-closed sets can be realised. Clearly there's a topological obstruction to using the procedure at the beginning of my answer to remove lots of boundary points from $L$; indeed, once you remove a single boundary point that way, then the trajectory starting at any larger value of $x$ cannot wind around the origin, and so life becomes a little more complicated...

Edit: Here's a more general construction. Let $X(x,y) = (-y,x)$ so that the flow is along circles centred at the origin. Fix any closed set $E\subset (0,\infty)$ and let $\phi(x,y) = 0$ precisely when $x=0$ and $y\in E$, with $\phi>0$ everywhere else. Then the flow for the vector field $\phi X$ has the property that the set of $x>0$ for which $(x,0)$ is contained in a closed orbit is precisely the set of $x$ not contained in $E$. Thus you can get any open set as your set $L$. Indeed, you can combine the two constructions to get any set that can be written as the intersection of an open set and a closed set. There's probably other stuff you can do too, but this illustrates some of the things that can happen.


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