Vanishing First Variational Derivative implies Euler's Equation for Vanishing Variation

Theorem
Let $y(x)$ be a real function such that $y(a)=A$ and $y(b)=B$

Let $J[y]$ be a functional of the form

$\displaystyle J[y]=\int_{a}^{b}F\left(x, y, y'\right)\mathrm{d}{x}$.

Then

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

Proof
The method of finite differences will be used here.

Consider a closed real interval $\left[{{a}\,.\,.\,{b}}\right]$, which is divided in $n+1$ equal parts. Choose its subdivision to be normal:

$a=x_0<x_1<...<x_n<x_{n+1}=b$

such that for $i\in\left\{0, 1, ..., n-1, n\right\}$ we have $x_{i+1}-x_i=\Delta x$.

Approximate the desired function $y$ by a polygonal line with vertices $(x_i, y_i)$ where $i\in\left\{0, 1, ..., n, n+1\right\}$, where $y_i=y(x_i)$.

Hence, the functional $J[y]$ can be approximated by the following sum:

$\displaystyle \mathscr{J}\left(y_1, y_2, ..., y_{n-1}, y_n\right)=\sum_{i=0}^{n}F\left(x_i, y_i, \frac{y_{i+1}-y_i}{\Delta x} \right)\Delta x$

Note that the values $y(x_0)=A$ and $y(x_1)=B$ are fixed, and therefore not varied.

Now, consider a partial derivative of $J$ with respect to $y_k$, where $k\in\left\{1, 2, ..., n-1, n\right\}$.

Since all the functions $y_i$ are independent w.r.t each other, we have $\frac{ \partial{y_m} }{ \partial{y_k} }=\delta_{mk}$, where $\delta_{mk}$ is Kronecker Delta.

Then, the aforementioned sum simplifies to

$\displaystyle \frac{ \partial \mathscr{ J } }{ \partial y_k }= \left[ \frac{\partial{ F } }{ \partial{ y_k } }\left(x_k, y_k, \frac{ y_{ k+1 }-y_k }{ \Delta x } \right)+\frac{ \partial{ F } }{ \partial{ \frac{ y_k-y_{k-1} }{ \Delta x } } } \left(x_{k-1}, y_{k-1}, \frac{ y_{ k }-y_{k-1} }{ \Delta x } \right) \frac{ 1 }{ \Delta x } - \frac{ \partial{ F } }{ \partial{ \frac{ y_{k+1}-y_k }{ \Delta x } } } \left(x_k, y_k, \frac{ y_{ k+1 }-y_k }{ \Delta x } \right) \frac{ 1 }{ \Delta x }\right]\Delta x $

In order to get a variational derivative, the denominator of the LHS has to represent an area.

For this reason, divide everything by $\Delta x$, and take the limit $\Delta\to 0$. Then for all $k\in\left\{1, 2, ..., n-1, n\right\}$

Similarly, for $F(x, y, y')$ we have

Thus,

$\displaystyle \lim_{\Delta x\to 0}\frac{ \partial{\mathscr{J} } }{ \partial{y_k}\Delta x }=F_{y(x_k)}\left(x_k, y(x_k), y'(x_k) \right)-\frac{ \mathrm{ d }{ } }{ \mathrm{ d }{ x } } F_{y'(x_{k-1})} \left(x_{k-1}, y(x_{k-1}), y'(x_{k-1})\right)$

Note that the denominator on the left is an area covered by a rectangle with sides $\Delta x$ and $\partial{y}$, and vanishes as $\Delta x\to 0$.

Finally, since the distance between any two neighbouring points approaches 0 as $\Delta x\to 0$,

the set of all $x_k\in\left[{{a}\,.\,.\,{b}}\right]$ can be treated as continuous, and the index $k$ dropped:

$\displaystyle \lim_{\Delta x\to 0}\frac{ \partial{J } }{ \partial{y}\Delta x }=F_{y(x)}\left(x, y(x), y'(x) \right)-\frac{ \mathrm{ d }{ } }{ \mathrm{ d }{ x } } F_{y'(x)} \left(x, y(x), y'(x)\right)$

The LHS by definition is a variational derivative.

Suppose the LHS vanishes. Then the RHS vanishes as well.