Simple Variable End Point Problem

Theorem
Let $ y $ and $ F $ be mappings.

Suppose endpoints of $ y $ lie on two given vertical lines $ x = a $ and $ x = b $.

Suppose $ J $ is a functional of the form


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

and has an extremum for a certain function $ \hat { y } $.

Then $ y $ satisfies the system of equations


 * $ \begin{cases}

& F_y - \frac { \mathrm d }{ \mathrm d x } F_{ y' } = 0 \\ & F_{ y' } \big \rvert_{ x = a } = 0 \\ & F_{ y' } \big \rvert_{ x = b } = 0 \end{cases}$

Proof
From Condition for Differentiable Functional to have Extremum we have


 * $ \displaystyle \delta J \left [ { y; h } \right ] \bigg \rvert_{ y = \hat { y } } = 0 $

For the variation to exist it has to satisfy the requirement for a differentiable functional.

We will start from the increment of a functional:



Using multivariate Taylor's theorem, one can expand $ F \left ( { x, y + h, y' + h' } \right ) $ with respect to $ h $ and $ h' $:


 * $ \displaystyle F \left ( { x, y + h, y' + h' } \right ) = F \left ( { x, y + h, y' + h' } \right ) \bigg \rvert_{ h = 0, ~h' = 0 } + \frac{ \partial { F \left ( { x, y + h, y' + h' } \right ) } }{ \partial { y } }

\bigg \rvert_{ h = 0, ~h' = 0 } h + \frac{ \partial{ F \left ( { x, y + h, y' + h' } \right ) } }{ \partial{ y' } } \bigg \rvert_{ h = 0, ~h' = 0 } h' + \mathcal { O } \left ( { h^2, hh', h'^2 } \right ) $

We can substitute this back into the integral. Note that the first term in the expansion and the negative one in the integral will cancel out.


 * $\displaystyle\Delta J[y;h]=\int_{a}^{b}\left[F(x,y,y')_y h + F(x,y,y')_{y'} h' + \mathcal{O}\left(h^2, hh', h'^2\right)\right]\mathrm{d}{x}$

Terms in $\mathcal{O}\left(h^2,h'^2\right)$ represent terms of order higher than 1 with respect to $h$ and $h'$.

Now, suppose we expand $\int_{a}^{b}\mathcal{O}\left(h^2, hh', h'^2\right)\mathrm{d}{x}$.

Every term in this expansion will be of the form


 * $\displaystyle\int_{a}^{b}A\left(m, n\right)\frac{\partial^{m+n} F\left(x, y, y'\right)}{\partial{y}^m\partial{y'}^n}h^m h'^n \mathrm{d}{x}$

where $m,~n\in\N$ and $m+n\ge 2$

By definition, the integral not counting in $\mathcal{O}(h^2, hh', h'^2)$ is a variation of functional.


 * $\displaystyle \delta J[y;h]=\int_{a}^{b}\left[F_y h+F_{y'}h'\right]\mathrm{d}{x}$

Now, integrate by parts and note that $h(x)$ does not necessarily vanish at the endpoints:

Then, for arbitrary $h(x)$, $J$ has an extremum if


 * $ \begin{cases}

& F_y-\frac{\mathrm{d}{}}{\mathrm{d}{x}}F_{y'}=0 \\ & F_{y'}\big\rvert_{x=a}=0 \\ & F_{y'}\big\rvert_{x=b}=0 \end{cases}$