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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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    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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    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.