Conditions for C^1 Smooth Solution of Euler's Equation to have Second Derivative

Theorem
Let $y$ be a real function.

Let $y$ have a continuous first derivative and satisfy Euler's equation:


 * $F_y - \dfrac \d {\d x} F_{y'} = 0$

Suppose $F \left({x, y, y'}\right)$ has continuous first and second derivatives all its arguments.

Then $\map y x$ has continuous second derivatives wherever:


 * $F_{y' y'} \sqbrk{x, \map y x, \map y x'} \ne 0$

Proof
Consider the difference

Overbar indicates that derivatives are evaluated along certain intermediate curves.

Divide $\Delta F_{y'} $ by $\Delta x$ and consider the limit $\Delta x\to 0$:

$\displaystyle \lim_{\Delta x\to 0}\frac{\Delta F_{y'} }{\Delta x}=\lim_{\Delta x\to 0}\paren{\overline{F}_{y'x}+\frac{\Delta y}{\Delta x}\overline F_{y'y}+\frac{\Delta y'}{\Delta x}\overline F_{y'y'} }$

Existence of second derivatives and continuity of $F$ is guaranteed by conditions of the theorem:

$\displaystyle\lim_{\Delta x\to 0}\frac{\Delta F_{y'} }{\Delta x}=F_{y'x}$, $\displaystyle\lim_{\Delta x\to 0}\overline F_{y'x}=F_{y'x}$, $\displaystyle\lim_{\Delta x\to 0}\overline F_{y'y}=F_{y'y}$, $\displaystyle\lim_{\Delta x\to 0}\overline F_{y'y}=F_{y'y'}$

Similarly,

$\displaystyle\lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}= y'$

By Product Rule for Limits of Functions, it follows that

$\displaystyle\lim_{\Delta x\to 0}\frac{\Delta y'}{\Delta x}=y''$

Hence $y''$ exists wherever $F_{y' y'} \ne 0$.

Euler's equation and continuity of necessary derivatives of $F$ and $y$ implies that $y''$ is continuous.