Necessary Condition for Integral Functional to have Extremum for given function

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

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

defined on the set of functions $y(x)$ which have continuous first derivatives in $[a, b]$ and satisfy the boundary conditions $y(a)=A$ and $y(b)=B$. Then a necessary condition for $J[y]$ to have an extremum (strong or weak) for a given function $y(x)$ is that $y(x)$ satisfy Euler's equation

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

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.

Note that the endpoints of $y(x)$ are fixed. $h(x)$ is not allowed to change values of $y(x)$ at those points.

Hence $h(a)=0$ and $h(b)=0$.

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,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.

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

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

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

According to a theorem

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

According to lemma, this implies that

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