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

Theorem
Let $\map y x:\R \to \R$ be a real function.

Let $\map F {x, y, y'}:\R^3 \to \R$ be a real function.

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

Suppose $y$ has a continuous first derivative and satisfies Euler's equation:


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

Suppose:


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

Then $\map y x$ has continuous second derivatives.

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 $\Delta x \to 0$:


 * $\ds \lim_{\Delta x \mathop \to 0} \frac {\Delta F_{y'} } {\Delta x} = \lim_{\Delta x \mathop \to 0} \paren {\overline F_{y'x} + \frac {\Delta y} {\Delta x} \overline F_{y' y} + \frac {\Delta y'} {\Delta x} \overline F_{y'y'} }$

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


 * $\ds \lim_{\Delta x \mathop \to 0} \frac {\Delta F_{y'} } {\Delta x} = F_{y' x}$
 * $\ds \lim_{\Delta x \mathop \to 0} \overline F_{y' x} = F_{y' x}$
 * $\ds \lim_{\Delta x \mathop \to 0} \overline F_{y' y} = F_{y' y}$
 * $\ds \lim_{\Delta x \mathop \to 0} \overline F_{y' y} = F_{y'y'}$

Similarly:


 * $\ds \lim_{\Delta x \mathop \to 0} \frac {\Delta y} {\Delta x} = y'$

By Product Rule for Limits of Real Functions, it follows that:


 * $\ds \lim_{\Delta x \mathop \to 0} \frac {\Delta y'} {\Delta x} = y''$

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.