# Necessary Condition for Integral Functional to have Extremum for given function It has been suggested that this article or section be renamed. One may discuss this suggestion on the talk page.

## Theorem

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

$S = \set {\map y x: \paren {y: S_1 \subseteq \R \to S_2 \subseteq \R}, \paren {\map y x \in C^1 \closedint a b}, \paren {\map y a = A, \map y b = B} }$

Let $J \sqbrk y: S \to S_3 \subseteq \R$ be a functional of the form:

$\displaystyle \int_a^b \map F {x, y, y'} \rd x$

Then a necessary condition for $J \sqbrk y$ to have an extremum (strong or weak) for a given function $\map y x$ is that $\map y x$ satisfy Euler's equation:

$F_y - \dfrac \d {\d x} F_{y'} = 0$

## Proof

$\delta J \sqbrk {y; h} \bigg \rvert_{y = \hat y} = 0$

The variation exists if $J$ is a differentiable functional.

The endpoints of $\map y x$ are fixed.

Hence:

$\map h a = 0$
$\map h b = 0$.

From the definition of increment of a functional:

 $\displaystyle \Delta J \sqbrk {y; h}$ $=$ $\displaystyle J \sqbrk {y + h} - J \sqbrk y$ definition $\displaystyle$ $=$ $\displaystyle \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x$ form of considered functional $\displaystyle$ $=$ $\displaystyle \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x$ bringing under the same integral

Using multivariate Taylor's theorem, expand $\map F {x, y + h, y' + h'}$ with respect to $h$ and $h'$:

$\displaystyle \map F {x, y + h, y' + h'} = \map F {x, y + h, y' + h'} \bigg \rvert_{h = 0, \, h' = 0} + \frac {\partial {\map F {x, y + h, y' + h'} } } {\partial y} \bigg \rvert_{h = 0, \, h' = 0} h + \frac {\partial {\map F {x, y + h, y'+ h'} } } {\partial y'} \bigg \rvert_{h = 0, \, h' = 0} h' + \mathcal O \paren {h^2, h h', h'^2}$

Substitute this back into the integral:

$\displaystyle \Delta J \sqbrk {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \mathcal O \paren {h^2, h h', h'^2} } \rd x$

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

Suppose we expand $\displaystyle \int_a^b \mathcal O \paren {h^2, h h', h'^2} \rd x$.

Every term in this expansion will be of the form:

$\displaystyle \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial y'^n} h^m h'^n \rd x$

where $m, n \in \N: m + n \ge 2$

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

$\displaystyle \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$

Use lemma.

Then for any $\map h x$ variation vanishes if:

$F_y - \dfrac \d {\d x} F_{y'} = 0$

$\blacksquare$