Simple Variable End Point Problem
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Theorem
Let $y$ and $F$ be mappings.
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Suppose the endpoints of $y$ lie on two given vertical lines $x = a$ and $x = b$.
Suppose $J$ is a functional of the form
- $(1): \quad J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
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:
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\(\ds \Delta J \sqbrk {y; h}\) | \(=\) | \(\ds J \sqbrk {y + h} - J \sqbrk y\) | definition | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x\) | $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x\) |
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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:
- $\ds \Delta J \paren {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$
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Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$.
Now 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 $m, n \in \N$ and $m + n \ge 2$
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By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional.
- $\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$
Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints:
\(\ds \delta J \sqbrk {y; h}\) | \(=\) | \(\ds \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigintlimits {F_{y'} \map h x} {x \mathop = a} {x \mathop = b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigvalueat {F_{y'} } {x \mathop = b} \map h b - \bigvalueat {F_{y'} } {x \mathop = a} \map h a\) |
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}$
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 1.6$: A Simple Variable End Point Problem