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 \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.

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

Hence $ h \left ( { a } \right ) = 0 $ and $ h \left ( { b } \right ) = 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 \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 $

Use lemma. Then for any $ h \left ( { x } \right ) $ variation vanishes if


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