Simple Variable End Point Problem

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Theorem

Let $y$ and $F$ be mappings.



Suppose the endpoints of $y$ lie on two given vertical lines $x = a$ and $x = b$.

Suppose $J$ is a functional of the form

$(1): \quad J \sqbrk y = \displaystyle \int_a^b \map F {x, y, y'} \rd x$

and has an extremum for a certain function $\hat y$.

Then $y$ satisfies the system of equations

$\begin {cases} F_y - \dfrac \d {\d x} F_{y'} = 0 \\ \bigvalueat {F_{y'} } {x \mathop = a} = 0 \\ \bigvalueat {F_{y'} } {x \mathop = b} = 0 \end {cases}$


Proof

From Condition for Differentiable Functional to have Extremum we have

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

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

We will start from the increment of a functional:



\(\displaystyle \Delta J \sqbrk {y; h}\) \(=\) \(\displaystyle J \sqbrk {y + h} - J \sqbrk y\) definition
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x\) $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x\)



Using multivariate Taylor's theorem, one can 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.

Note that the first term in the expansion and the negative one in the integral will cancel out:

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



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

Now we expand:

$\displaystyle \int_a^b\map \OO {h^2, h h', h'^2} \rd x$

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 $m, n \in \N$ and $m + n \ge 2$



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

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

Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:

\(\displaystyle \delta J \sqbrk {y; h}\) \(=\) \(\displaystyle \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigintlimits {F_{y'} \map h x} {x \mathop = a} {x \mathop = b}\)
\(\displaystyle \) \(=\) \(\displaystyle \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigvalueat {F_{y'} } {x \mathop = b} \map h b - \bigvalueat {F_{y'} } {x \mathop = a} \map h a\)

Then, for arbitrary $\map h x$, $J$ has an extremum if:

$ \begin {cases} F_y - \dfrac \d {\d x} F_{y'} = 0 \\ \bigvalueat {F_{y'} } {x \mathop = a} = 0\\ \bigvalueat {F_{y'} } {x \mathop = b} = 0 \end {cases}$

$\blacksquare$


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