Necessary Condition for Integral Functional to have Extremum for given Function/Dependent on Nth Derivative of Function

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Theorem

Let $\map F {x, y, z_1, \ldots, z_n}$ be a function in differentiability class $C^2$ with respect to all its variables.

Let $y = \map y x \in C^n\openint a b$ such that:

$\map y a = A_0, \map {y'} a = A_1, \ldots, \map {y^{\paren {n - 1} } } a = A_{n - 1}$

and:

$\map y b = B_0, \map {y'} b = B_1, \ldots, \map {y^{\paren {n - 1} } } b = B_{n - 1}$


Let $J \sqbrk y$ be a functional of the form:

$\ds J \sqbrk y = \int_a^b \map F {x, y, y', \ldots, y^{\paren n} } \rd x$

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:

$F_y - \dfrac \d {\d x} F_{y'} + \dfrac {\d^2} {\d x^2} F_{y''} -\cdots + \paren {-1}^n \dfrac {\d^n} {\d x^n} F_{y^{\paren n} } = 0$


Proof

From Condition for Differentiable Functional to have Extremum we have

$\bigvalueat {\delta J \sqbrk {y h} } {y \mathop = \hat y} = 0$

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

Note that the endpoints of $\map y x$ are fixed.

$\map h x$ is not allowed to change values of $\map y x$ at those points.

Hence the higher derivative $\map {h^{\paren i} } a = 0$ and $\map {h^{\paren i} } b = 0$ for $i \in \openint 1 n$.

We will start from the increment of a functional:

\(\ds \Delta J \sqbrk {y; h}\) \(=\) \(\ds J \sqbrk {y + h} - J \sqbrk y\) definition
\(\ds \) \(=\) \(\ds \int_a^b \map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} \rd x - \int_a^b \map F {x, \ldots, y^{\paren i}, \ldots} \rd x\) where $i \in \set {0, 1, \ldots, n}$
\(\ds \) \(=\) \(\ds \int_a^b \paren {\map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} - \map F {x, \ldots, y^{\paren i}, \ldots} } \rd x\) bringing under the same integral


Using multivariate Taylor's theorem, one can expand $\map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots}$ with respect to $h^{\paren i}$:

$\ds \map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} = \valueat {\map F {x,\ldots,y^{\paren i} + h^{\paren i}, \ldots} } {h^{\paren i} = 0, i \in \openint 0 n} + \valueat {\sum_{i \mathop = 0}^n \frac {\partial \map F {x, \ldots, y^{\paren i} + h^{\paren i}, \ldots} } {\partial y^{\paren i} } } {h^{\paren i} = 0, i \in \openint 0 n} h^{\paren i} + \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$

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.

Hence:

$\ds \Delta J \sqbrk {y; h} = \int_a^b \sum_{i \mathop = 0}^n \paren {\map F {x, \ldots, y^{\paren i}, \ldots}_{y^{\paren i} } h^{\paren i} + \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n} } \rd x$


This article, or a section of it, needs explaining.
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Terms in $\map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$ represent terms of order higher than 1 with respect to $h^{\paren i}$.

Now, suppose we expand $\ds \int_a^b \map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n} \rd x$.


This article, or a section of it, needs explaining.
In particular: Proof that this goes to zero faster than $\size h$, as definition of variation requires
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By definition, the integral not counting in $\map \OO {h^{\paren i} h^{\paren j}, i, j \in \openint 0 n}$ is a variation of functional:


This article, or a section of it, needs explaining.
In particular: The link is to Definition:Differentiable Functional but the label says "variation of functional" which is not defined on that page
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$\ds \delta J \sqbrk {y ; h} = \int_a^b \sum_{i \mathop = 0}^n F_{y^{\paren i} } h^{\paren i} \rd x$

Application of Generalized Integration by Parts, together with boundary values for $h^{\paren i}$ yields:

$\ds \int_a^b h \sum_{i \mathop = 0}^n \paren {-1}^i \frac {\d^i} {\d x^i} F_{y^{\paren i} }\rd x$

From If Definite Integral of a(x)h(x) vanishes for any C^0 h(x) then C^0 a(x) vanishes, then for any $\map h x$ the variation vanishes if:

$F_y - \dfrac \d {\d x} F_{y'} + \dfrac {\d^2} {\d x^2} F_{y''} - \cdots + \paren {-1}^n \dfrac {\d^n} {\d x^n} F_{y^{\paren n} } = 0$

$\blacksquare$


Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: minor refinements, existence of derivatives of F
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Sources

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