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Let $V \in C^1(\mathbb{R}^d)$, and let $S$ denote the solution set for the problem $$ \ddot{x}+\nabla V(x)=0,\quad x(T)-x(0)=0=\dot{x}(T)- \dot{x}(0). $$ When $d=1, T=2\pi$, and $V(x)=\frac12|x|^2$ it is clear that $S$ is the two-dimensional vector space $$ \{\alpha\cos+\beta\sin:\ \alpha,\beta\in \mathbb{R}\} \simeq \mathbb{R}^2. $$ For general $V$ (when the ODE is non-linear), does $S$ have some kind of manifold structure?

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up vote 2 down vote accepted

Let's stick to $d=1$. When $V(x)=-x^2$, the set $S$ consists of one point (the zero function). Maybe this is a zero-dimensional manifold?

Introducing the notation $p=\dot x$, we observe that the trajectories in $(x,p)$ plane follow the level curves of the function $E(x,p)=V(x)+\frac12 p^2$. In general, some of these level curves may be closed while others will not be closed. Among those that are closed, the corresponding periodic solutions may have period $T$ or some other period.

The equation with $V(x)=\frac12 x^2$ (harmonic oscillator) is rather atypical in that all solutions have the same period. This is not the case for the nonlinear pendulum equation, for which $V(x)=-\cos x$. Here is the phase portrait of the nonlinear pendulum (source): Nonlinear pendulum

If $T$ lies in a certain range, there are infinitely many solutions that form a discrete set, namely the closed trajectories around $(2\pi k,0)$. For other values of $T$ (too large, or too small) there are no nonconstant solutions. Of course we always have the constant ones, $x\equiv 2\pi k$, which are periodic with any period.

Conclusion: no, there is no reason to expect a manifold structure. It's more complicated.

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Thank you for your answer. I had a different question in mind though. – albmiz-mth Jan 29 '13 at 11:00

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