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Could anyone comment on the following ODE problem? Thank you!

Let $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be $C^{1}$ and let $X^{(n)}(t)$ be a sequence of periodic solutions of $$\frac{dX}{dt}(t)=f(X(t)).$$ Assume that $X^{(n)}(0)$ converges and let $X(t)$ be the solution with $X(0)=\lim\limits_{n\rightarrow \infty}X^{(n)}(0)$.

Prove of disprove that $X(t)$ is periodic.

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    I think this could be false based on the following: consider a straight line in $\mathbb{R}^2$, say vertical, and in the left side of the line, put a family of nested cicles oriented clockwise acumulating in this line andon the right side put the same, but counterclockwise. The line will be oriented by this two families and it seems to define a field with the property you have considered, but the solution with initial condition $\lim X^{(n)}(0)$ will be the straight line.2012-04-24
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    Not quite an answer, but may be of interest. Take an ideal pendulum with rigid string/rod (frictionless, etc.). Choose $X^{(n)}(0)$ to correspond to holding the rod $\frac{1}{n}^{\circ}$ from vertical. The solution is obviously periodic. However, the initial states converge to the unstable equilibrium (pendulum 'balanced' vertically). The equilibrium is trivially periodic, however.2012-04-24

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I don't know how to attach a file on the comments, so I post an answer, but just to show what I had in mind:enter image description here

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    I started out looking at the magnetic field for two parallel wires with opposite currents, but the formula has singularities rather than equilibria at the points of interest.2012-04-24
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    Is it like the wires met the plane of the screen at the center of these two most internal cicles I've drawn?2012-04-24
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    Yes. The wires would be coming out of the screen at the equilibrium points of your drawing.2012-04-24
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Actually, I think my pendulum example will work, it just needs a different set of initial conditions.

Choose the initial state $X^{(n)}(0)$ so that the pendulum is at the bottom, but the initial velocity is such that it will end up at $\frac{1}{n}^{\circ}$ from the top. This will be periodic, with ever increasing periods (as a function of $n$).

The initial states will converge, but if the system is started with the converged initial condition, it will approach the top as $t \rightarrow \infty$.

This argument can be made rigorous using some form of Lyapunov/energy function.