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

Theorem
Let $\map y x$ be a real function such that $\map y a = A$ and $\map y b = B$.

Let $J \sqbrk y$ be a functional of the form:


 * $\displaystyle J \sqbrk y = \int_a^b \map F {x, y, y'} \rd x$

Then:


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

Proof
The method of finite differences will be used here.

Consider a closed real interval $\closedint a b$, which is divided in $n + 1$ equal parts.

Choose its subdivision to be normal:


 * $a = x_0 < x_1 < \cdots < x_n < x_{n + 1} = b$

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

Approximate the desired function $y$ by a polygonal line with vertices $\tuple {x_i, y_i}$ where $i \in \set {0, 1, \ldots, n, n + 1}$, where $y_i = \map y {x_i}$.

Hence, the functional $J \sqbrk y$ can be approximated by the following sum:


 * $\displaystyle \map {\mathscr J} {y_1, y_2, \ldots, y_{n - 1}, y_n} = \sum_{i \mathop = 0}^n \map F {x_i, y_i, \frac {y_{i + 1} - y_i} {\Delta x} } \Delta x$

Note that the values $\map y {x_0} = A$ and $\map y {x_1} = B$ are fixed, and therefore not varied.

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

As all the functions $y_i$ are independent each other, we have $\dfrac {\partial y_m} {\partial y_k} = \delta_{m k}$, where $\delta_{m k}$ is the Kronecker Delta.

Then the aforementioned sum simplifies to:


 * $\dfrac {\partial \mathscr J} {\partial y_k} = \paren {\dfrac {\partial F} {\partial y_k} \paren {x_k, y_k, \dfrac {y_{k + 1} - y_k} {\Delta x} } + \dfrac {\partial F} {\partial {\frac {y_k - y_{k - 1} } {\Delta x} } } \paren {x_{k - 1}, y_{k - 1}, \dfrac {y_k - y_{k - 1} } {\Delta x} } \dfrac 1 {\Delta x} - \dfrac {\partial F} {\partial {\frac {y_{k + 1} - y_k} {\Delta x} } } \paren {x_k, y_k, \dfrac {y_{k + 1} - y_k} {\Delta x} } \dfrac 1 {\Delta x} } \Delta x$

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

For this reason, divide everything by $\Delta x$, and take the limit $\Delta \to 0$.

Then for all $k \in \set {1, 2, \dotsc, n - 1, n}$:

Similarly, for $\map F {x, y, y'}$ we have

Thus:


 * $\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\partial \mathscr J} {\partial y_k \Delta x} = \map {F_{\map y {x_k} } } {x_k, \map y {x_k}, \map {y'} {x_k} } - \frac \d {\d x} \map {F_{\map {y'} {x_{k - 1} } } } {x_{k - 1}, \map y {x_{k - 1} }, \map {y'} {x_{k - 1} } }$

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 $x_k \in \closedint a b$ can be treated as continuous, and the index $k$ dropped:


 * $\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\partial J} {\partial y \Delta x} = \map {F_{\map y x} } {x, \map y x, \map {y'} x} - \frac \d {\d x} \map {F_{\map {y'} x} } {x, \map y x, \map {y'} x}$

The by definition is a variational derivative.

Suppose the vanishes.

Then the vanishes as well.