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:


 * $ \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 all its arguments.

Then $ y \left ( { x } \right ) $ 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.