If $y''(x) +1 = 1/y(x)$ and $y(0)>y'(0)=0$, then for what $L$ does $y(0)=y(L)$? All functions $y(x)$ that satisfy $$y''(x) +1 = \frac{1}{y(x)},\quad y(0)>y'(0)=0$$ seem to have a period $L$ such that $y(L)=y(0)$. Can this be proven, and more importantly, can one calculate $L$ in terms of $y(0)$?
 A: Multiply with $y'$, integrate to get the energy function
$$
E=\frac12y'^2+y-\ln|y|
$$
This is a convex function over the half-plane $y>0$. It has compact level sets without further stationary points except the center at $(y,y')=(1,0)$. The solutions thus rotate along the level curves, the solutions are periodic.
A: I will suppose, without proof, that there is a finite solution $y$ to the equation. In this case, the continuity of the derivative $y'$ is clear by integration.
We first prove that $y$ must have a turning point. By contradiction, suppose $y$ doesn't have a turning point. Suppose further that there is an $\epsilon<1$ and an $x_0$ such that $y\leq \epsilon$ for all $x\geq x_0$. Then,
$$y''\geq\frac{1}{\epsilon}-1>0$$
for all $x\geq x_0$, and so $y$ would become arbitrarally large, a contradiction of $y\leq 1$. Thus, there is a point $x_1>x_0$ and an $M>1$ such that $y\geq M$ for all $x\geq x_1$. But in this case,
$$y''\leq \frac{1}{M}-1<0$$
So $y'$ would have to eventually cross zero, a contradiction of $y$ not having a turning point. Thus, $y$ must have a turning point. 
Now note that our original equation is space-reversal invariant,
$$y''(-x)+1=\frac{1}{y(-x)}$$
Thus, let $\tfrac{L}{2}$ be the first positive turning point of $y$. Then $y(\tfrac{L}{2}-x)=y(\tfrac{L}{2}+x)$ for $0\leq x\leq \tfrac{L}{2}$. Thus $y(L)=y(0)$. Inductively applying this argument shows that $y$ has periodicity $L$.
