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


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

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

Then $y$ satisfies the system of equations


 * $ \begin{cases}

& F_y-\frac \d {\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


 * $\displaystyle\delta J\sqbrk{y;h}\bigg\rvert_{y=\hat y}=0$

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 \left [ { y; h } \right ] = \int_a^b \left [ { F \left ( { x, y, y' } \right )_y h + F \left ( { x, y, y' } \right )_{ 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 } \left ( { h^2, hh', h'^2 } \right ) $ is a variation of functional.


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

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

Then, for arbitrary $ h \left ( { x } \right ) $, $ 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}$