Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions

Theorem
Let $J[y_1,~y_2,~...,y_n]$ be a functional of the form

$\displaystyle J[y_1,~y_2,~...,y_n]=\int_{a}^{b}F\left(x,~y_1,~y_2,...,~y_n,~y_1',~y_2',...,~y_n'\right)\mathrm{d}{x}$

Let $y_1,~y_2,~...,y_n$ be functions in $C^1\left[{{a}\,.\,.\,{b}}\right]$, and satisfy boundary conditions $y_i(a)=A_i$ and $y_i(b)=B_i$, for $i=(1,~2,...,~n)$.

Then a necessary condition for $J[y_1,~y_2,~...,y_n]$ to have an extremum (strong or weak) for a given functions $y_1,~y_2,~...,y_n$ is that they satisfy Euler's equations:

$\displaystyle F_{y_i}-\frac{\mathrm{d}{}}{\mathrm{d}{x}}F_{y_i'}=0,~i=(1,~2,~...,~n)$

Proof
From Condition for Differentiable Functional of N Functions to have Extremum we have


 * $\displaystyle\delta J[y_1,~y_2,~...,y_n; h_1,~h_2,~...,h_n]\bigg\rvert_{y_i=\hat{y}_i,~i=(1,~2,~...,~n)}=0$

For the variation to exist it has to satisfy the requirement for a differential functional.

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

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

We will start from the increment of a functional:



Using multivariate Taylor's theorem, one can expand $F\left(x,y_1+h_1,...,y_n+h_n,y_1'+h_1',...,y_n'+h_n'\right)$ with respect to functions $h_i(x)$ and $h_i'(x)$:


 * $\displaystyle

\begin{equation}{\begin{split}  F\left(x,y_i+h_i,y_i'+h_i'\right)=&F\left(x,y_i+h_i,y_i'+h_i'\right)\bigg\rvert_{\begin{array} ~h_i=0,~h_i'=0\\ i=(1,~...,~n) \end{array} }+ \sum_{i=1}^n\frac{ \partial{F\left(x,~y_i+h_i,y_i'+h_i'\right)} }{ \partial{{y_i} } }\bigg\rvert_{h_i=0,~h_i'=0 } {h_i}\\ &+\sum_{i=1}^n\frac{ \partial{F\left(x,~y_i+h_i, y_i'+h_i'\right)} }{ \partial{{y}_i'} }\bigg\rvert_{h_i=0,~h_i'=0 } {h}'+\mathcal{O}\left(h_i h_j, h_i h_j', h_i' h_j'~\text{for}~i,j=(1,~...,~n)\right)\end{split} }\end{equation} $

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.


 * $\displaystyle\Delta J[y;h]=\int_{a}^{b}\sum_{i=1}^n\left[F_{y_i} h_i + F_{y_i'} h_i' + \mathcal{O}\left(h_i h_j, h_i h_j', h_i' h_j'~\text{for}~i,j=(1,~...,~n)\right)\right]\mathrm{d}{x}$

By definition, the integral not counting in $\mathcal{O}\left(h_i h_j, h_i h_j', h_i' h_j'~\text{for}~i,j=(1,~...,~n)\right)$ is a variation of functional:


 * $\displaystyle \delta J[y_i;~h_i]=\int_{a}^{b}\sum_{i=1}^n\left[F_{y_i} h_i+F_{y_i'}h_i'\right]\mathrm{d}{x}$

Apply lemma for each $i=(1,~...,~n)$.

The variation vanishes if for all functions $h_i$ every term containing $h_i$ vanishes independently.

Therefore, we discover a set of Euler's Equations being satisfied simultaneously:
 * $\displaystyle F_{y_i}-\frac{\mathrm{d}{}}{\mathrm{d}{x}}F_{y_i'}=0,~i=(1,~...,~n)$