Proving all solutions of an ODE to be periodic I need to prove that all the solutions of the ODE, with $\alpha \in [0, 1]$, are periodic:
$$x''(t) + x(t) - \sin(\alpha x(t))=0.$$
I just don't know how to approach this problem. I thank you in advance for any help.
 A: If $\alpha = 0$, we get a linear ODE with constant coefficients whose solutions are $x(t) = A\cos(t) + B\sin(t)$ and so periodic. Assume that $\alpha \neq 0$. Performing the substitution $x'(t) = v(x(t))$, we have
$$ x''(t) = v'(x(t))x'(t) = v'(x(t))v(x(t)) $$
and so
$$ v'(x)v(x) + x - \sin(\alpha x) = 0 $$
or
$$ v \, dv = (\sin(\alpha x) - x) \, dx. $$
Integrating, we get
$$ \frac{v^2}{2} + \left( \frac{\cos (\alpha x)}{\alpha} + \frac{x^2}{2} \right) = C. $$
Renaming the constant $C$, we then have
$$ F(x,v) = \alpha(x^2 + v^2) + 2\cos(\alpha x) = C $$
and we found a conserved quantity. What does this mean? Given a solution $x(t)$ with initial conditions $x(t_0) = x_0$ and $x'(t_0) = x'_0$, the solution must satisfy
$$ F(x(t), v(x(t))) = F(x(t), x'(t)) \equiv F(x_0, x'_0) $$
and so $(x(t),x'(t))$ must lie on a fixed level set of $F$. The only critical point of $F$ is at the origin $(0,0)$ as the equation
$$ (\nabla F)(x,v) = 2\alpha(x - \sin(\alpha x), v) = (0, 0) $$
has a unique solution $v = x = 0$ (here we use $0 < \alpha \leq 1$). This critical point is a minimum and all level sets are compact (we can even say that if $C > 2$ then $F^{-1}(C)$ is diffeomorphic to $S^1$) which implies that the solutions exist for all time and must be periodic.
