Necessary Condition for Integral Functional to have Extremum for given function

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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:

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


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

From Condition for Differentiable Functional to have Extremum we have

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

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:

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

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

$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y'+ h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$

Substitute this back into the integral:

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

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

Suppose we expand $\ds \int_a^b \map \OO {h^2, h h', h'^2} \rd x$.

Every term in this expansion will be of the form:

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

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


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

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

Use lemma.

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

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

$\blacksquare$


Sources

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