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 $(0,\pi)$?
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3The period for arbitrary amplitude can be written in terms of an elliptic function. This diverges as $\alpha$ approaches $\pi$. – John Barber Jan 22 '18 at 18:15
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1The system can be written as the dynamical system $(x',v') = (v,-\sin(x))$. A relevant theorem for periodic solutions of such a system is Poincaré–Bendixson. – Winther Jan 22 '18 at 18:42
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Related MSE question: https://math.stackexchange.com/questions/1692580/uniqueness-of-a-periodic-solution-for-nonlinear-pendulum – Winther Jan 22 '18 at 18:43
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Is there any 'elementary' way of proving the existence and uniqueness of periodic solutions? – user2471 Jan 22 '18 at 19:41
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Thanks @John Barber for the reference. How would one prove the existence of a periodic solution? – user2471 Jan 22 '18 at 19:49
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There's a constant of motion. See the phase portrait in Figure 5 in the Wikipedia article that @JohnBarber linked to. – Hans Lundmark Jan 23 '18 at 08:06
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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.
Winther
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