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

Theorem
Suppose $y=y\left(x\right)$ has a continous first derivative and satisfies Euler's equation

$\displaystyle F_y-\frac{\mathrm{d}{}}{\mathrm{d}{x}}F_{y'}=0$

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

Then $y(x)$ has continuous second derivatives wherever

$\displaystyle F_{y'y'}\left[x, y\left(x\right), y\left(x\right)'\right]\ne 0$

Proof
Consider the difference

Overbar indicates that derivatives are evaluated along certain intermediate curves.

Divide $\displaystyle \Delta F_{y'}$ 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, 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.


 * : $\S 1.4$: The Simplest Variational Problem. Euler's Equation