Period of the nonlinear ODE $y''+\cos(y) = 0$

Question

Suppose an equation

$$ y'' + \cos(y) = 0; y(0) = y'(0) = 0 $$

Numerical solution suggest periodic behavior. But how to derive it analytically (a systematic approximate is also acceptable)?

Attempt 1 Write the equation as $y''= - \cos(y)$, together with initial condition, we have $y(0) = 0, y'(0) = 0, y''(0) = -cos(y(0)) = -1$ and so on. So the Taylor expansion approximation is $y = -1/2 x^2 + ...$, but this won't tell the qualitative periodic behavior, not to mention quantitative ones.

Attempt 2 With the familiar trick $y''(x) = dy'/dy \cdot dy/dx = 1/2 dy'^2 /dx$, the equation can be reduced to first order (initial condition used): $y'^2 + 2 sin(y) = 0$. But I don't know how to proceed with this either.

Numerical solution is presented in figure below, the blue curve (get out of plot range around x > 3) is the Taylor series approximation $-1/2x^2$, the yellow and green curves overlap a lot, one is the numerical solution, another is an eyeball fitting with this function ($T\sim 7.4$): $$ \frac{\pi}{2} (\cos(2\pi x/T )-1) $$

Answer 1

This can be thought of physically as motion in a conservative field. The conserved energy is

$$ E = \dfrac{y'^2}{2} + \sin(y) $$

From the initial conditions, the energy $E = 0$. You get periodic orbits where $(y,y')$ goes from $(0,0)$ to $(-\pi/2,-\sqrt{2})$ to $(-\pi, 0)$ to $(\pi/2, \sqrt{2})$ and back to $(0,0)$. By symmetry, the period is $4$ times the time to go from, say, $y=0$ to $-\pi/2$, and that is an elliptic integral: in Maple's notation

$$ 4 \int_0^{\pi/2} \dfrac{dy}{\sqrt{2\sin(y)}} = 4 {\it EllipticK}(\sqrt{2}/2) \approx 7.416298708 $$