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

Theorem
Let $J[\mathbf{y}]=J[y_1,~y_2,~...,y_n]$ be a functional of the form

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

Let $\mathbf {y}=\mathbf {y}(x)$ which have continuous first derivatives in [a,b] and satisfy the boundary conditions $\mathbf{y}(a)=\mathbf{A}$ and $\mathbf{y}(b)=\mathbf{B}$.

Then a necessary condition for $J[\mathbf{y}]$ to have an extremum (strong or weak) for a given vector function $\mathbf{y}(x)$ is that $\mathbf{y}(x)$ satisfy Euler's equation

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

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


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

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

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

Hence $\mathbf{h}(a)=\mathbf{0}$ and $\mathbf{h}(b)=\mathbf{0}$.

We will start from the increment of a functional:



Using multivariate Taylor's theorem, one can expand $F\left(x,\mathbf{y}+\mathbf{h},\mathbf{y}'+\mathbf{h}'\right)$ with respect to $\mathbf{h}$ and $\mathbf{h}'$:


 * $\displaystyle

F\left(x,\mathbf{y}+\mathbf{h},\mathbf{y}'+\mathbf{h}'\right)=F\left(x,\mathbf{y}+\mathbf{h},\mathbf{y}'+\mathbf{h}'\right)\bigg\rvert_{{h}=f{0},~{h}'={0} }+ \frac{ \partial{F\left(x,{y}+{h},{y}'+{h}'\right)} }{ \partial{{y} } }\bigg\rvert_{{h}={0},~f{h}'=\mathbf{0} } {h} +\frac{ \partial{F\left(x,{y}+{h},f{y}'+{h}'\right)} }{ \partial{{y}'} }\bigg\rvert_{{h}={0},~{h}'={0}} {h}'+\mathcal{O}\left({h}^2, {h}\dot{h}', {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:


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

According to lemma, this implies that


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