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 } \left ( { \overline { F }_{ y' x }+\frac{ \Delta y }{ \Delta x }\overline F_{ y' y } + \frac{ \Delta y' }{ \Delta x } \overline F_{ y' y' } } \right) $

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.