Treacherous Euler-Lagrange equation

Question

If I have an Euler-Lagrange equation: $(y')^2 = 2 (1-\cos(y))$ where $y$ is a function of $x$ subjected to boundary conditions $y(x) \to 0$ as $x \to -\infty$ and $y(x) \to 2\pi$ as $x \to +\infty$, how might I find all its solutions?

I can't seem to directly integrate the equation and sub in the conditions... Please help!

Thanks.

Answer 1

Complementary to solutions of joriki and Sivaram you could differentiate your equation once to get

$$2 y''(x) y'(x) = 2 y'(x) \sin y(x)$$

which implies $y''(x) = \sin y(x)$. After substitution $y(x)=\pi - \theta(x)$ this translates into $\theta''(x) = -\sin \theta(x)$ which is the pendulum equation. Your boundary condition require that $\lim\limits_{x\to-\infty} \theta(x) = \pi$ and $\lim\limits_{x\to+\infty} \theta(x) = -\pi$. Hence the solution is not periodic.

This trajectory is described by the Gudermannian function.

Answer 2

$1-\cos(y) = 2 \sin^2 \left( \frac{y}{2} \right)$. Hence, $y'^2 = 4 \sin^2 \left( \frac{y}{2} \right) \implies y' = \pm 2 \sin \left( \frac{y}{2} \right)$

Answer 3

Use $1-\cos y=1-(\cos^2\frac y2-\sin^2\frac y2)=2\sin^2\frac y2$.