Simple Variable End Point Problem

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

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

Suppose $J$ is a functional of the form


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

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


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

For the variation to exist it has to satisfy the requirement for a differential 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}$