On the periodicity of an non-linear ODE Consider the following equation:
\begin{equation}
  \begin{cases}
    x''+\sin(x)=0 \\
    x(0)=\alpha,\text{ for some }\alpha\in(0,\pi) \\
    x'(0)=0
  \end{cases}
\end{equation}
Is it true that a solution $x\in\mathcal{C}^2([0,\infty),\mathbb{R})$ is necessarily periodic? As the equation represents (in a reduced form) a pendulum with no friction, it seems right to think of periodic solutions, but how would one prove their existence/uniqueness? How would the period $\tau(\alpha)$ behave as a function of $\alpha$ (for instance in the neighborhood of $0$ and $\pi$)? Also what happens to the solutions if $\alpha$ steps outside $(0,\pi)$?   
 A: The 'energy' $E = \frac{\dot{x}^2}{2} - \cos(x)$ of the system is conserved in time since $\dot{E} = \dot{x}[\ddot{x} + \sin(x)] \equiv 0$. This means that the trajectory in the $(x,\dot{x})$ phase-space will be confined to the closed curve $y = \pm \sqrt{2(\cos(x)-\cos(\alpha))}$ with $x \in[-\alpha,\alpha]$ (note that $E = E(t=0) = \frac{0^2}{2} -\cos(\alpha)$).
Consider the phase-space path of the system on this closed curve. If there happens to be two times $t_1$ and $t_2$ where the system is at the same point in phase-space, $(x(t_1),\dot{x}(t_1)) = (x(t_2),\dot{x}(t_2))$, then the evolution will be periodic with period (at most) $T = t_2 - t_1$ (because given initial conditions $(x_0,\dot{x}_0$) the solution is unique).
If the motion is not periodic it means that the trajectory on the closed curve described above has to converge. However if it converge then this must be to a fixpoint $(x_*,\dot{x}_*)$ of the dynamical system $(\dot{x},\dot{y}) = (y,-\sin(x))$. The only fixpoint is $(0,0)$ which does not lie on this curve unless $\alpha = 0$ so this cannot happen.
