Necessary Condition for Integral Functional to have Extremum for given function

Theorem
Let $ S $ be a set of real mappings such that:


 * $ S = \{ y \left ( { x } \right ) : \left ( { y : S_1 \subseteq \R \to S_2 \subseteq \R } \right ), \left ( { y \left ( { x } \right ) \in C^1 \left [ { a \,.\,.\, b } \right ] } \right ), \left ( { y \left ( { a } \right ) = A, y \left ( { b } \right ) = B } \right ) \}$

Let $ J \left [ { y } \right ] : S \to S_3 \subseteq \R $ be a functional of the form:


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

Then a necessary condition for $ J \left [ { y } \right ] $ to have an extremum (strong or weak) for a given function $ y \left ( { x } \right ) $ is that $ y \left ( { x } \right )$ 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, 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:


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

Use lemma. Then for any $h(x)$ variation vanishes if


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