# 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 with respect to 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:

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

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

## Proof

Consider the difference

 $\displaystyle \Delta F_{y'}$ $=$ $\displaystyle F \sqbrk{x +\Delta x,y+\Delta y,y'+\Delta y'} - F \sqbrk{x,y,y'}$ $\displaystyle$ $=$ $\displaystyle \Delta x\overline F_{y'x}+\Delta y\overline F_{y'y}+\Delta y'\overline F_{y'y'}$ Multivariate Mean Value Theorem

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.