General Variation of Integral Functional/Dependent on N Functions

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Let $J$ be a functional of the form

$\displaystyle J \sqbrk {\ldots y_i \ldots} = \int_{x_0}^{x_1} \map F{x, \ldots y_i \ldots, \ldots y_i' \ldots} \rd x, i = \openint 1 n$


$\displaystyle \delta J=\int_{x_0}^{x_1}\sum_{i=1}^n\paren{F_{y_i}-\frac \d {\d x}F_{y_i'} } \map {h_i} x+\sum_{i=1}^n F_{y_i'}\delta y_i\bigg\rvert_{x=x_0}^{x=x_1}+\paren{F-\sum_{i=1}^n y_i'F_{y_i'} }\delta x\bigg\rvert_{x=x_0}^{x=x_1}$


Let ${y=\map y x, y=\map {y^*} x}$ be smooth real functions.

Let $\map h x=\map {y^*} x-\map y x$

Let the endpoints of the curve $y_i=\map {y_i} x, ~i=\openint 1 n$ be

$P_0=\paren {x_0, ... y_i^0 ...},~ P_1=\paren {x_1, ... y_i^1 ...}$

Let the endpoints of the curve $y_i=y_i^*=\map {y_i} x+\map {h_i} x, ~i=\openint 1 n$ be

$P_0^*=\paren {x_0+\delta x_0, ... y_i^0+\delta y_i^0 ...},~ P_1^*=\paren {x_1+\delta x_1, ... y_i^1+\delta y_i^1 ...}$

Note that the endpoints of both functions may not necessarily match, thus making the functions defined on different intervals.

We choose to extend both curves in such a way, that if there is a difference between original endpoints of intervals at the same end,

then the curve that is not defined in the given interval is extended linearly by drawing a straight line along the tangent of the curve at the point, closest to that interval.

Now both functions $\map {y_i} x$ and $\map {y^*_i} x$ are defined in $\closedint {x_0} {x_1+\delta x_1}$.

The corresponding variation $\delta J$ of $J\sqbrk{...y_i...}$ is defined as the expression which is linear in ${\delta x_0, \delta x_1}$ and ${h_i, h_i', y_i^0, y_i^1}$ for $i=\openint 1 n$,

and which differs from the increment by a quantity of order higher than 1 relative to $\displaystyle\sum_{i=1}^n \map \rho {y_i,y_i^*}$.

Here $\map {\rho} {y_i,y_i^*}$ is defined as

$\map {\rho} {y_i,y_i^*}=$ $\mathrm{max}$$\size {y_i-y_i^*}$ $+\mathrm{max}\size {y_i'-{y_i^*}'}+$$\map {d_2} {P_0, P_0^*}$$+\map {d_2} {P_1, P_1^*}$
\(\displaystyle \Delta J\) \(=\) \(\displaystyle \int_{x_0+\delta x_0}^{x_1+\delta x_1}\map F{x, ...y_i+h_i..., ...y_i'+h_i'...} \rd x-\int_{x_0}^{x_1}\map F {x, ...y_i..., ...y_i'...}\rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{x_0}^{x_1}\sqbrk{\map F { x, ...y_i+h_i..., ...y_i'+h_i'... }-\map F{x, ...y_i..., ...y_i'...} }\rd x+\int_{x_1}^{x_1+\delta x_1}\map F {x, ...y_i+h_i..., ...y_i'+h_i'...}\rd x\)
\(\displaystyle \) \(-\) \(\displaystyle \int_{x_0}^{x_0+\delta x_0}\map F{x, ...y_i+h_i..., ...y_i'+h_i'...} \rd x\)

By using Taylor's theorem, $\Delta J$ can be rewritten as

\(\displaystyle \Delta J\) \(=\) \(\displaystyle \int_{x_0}^{x_1}\sum_{i=1}^n\paren{F_{y_i}h_i+F_{y_i}'h_i'}\rd x+F\big\rvert_{x=x_1}\delta x_1-F \big\rvert_{x=x_0}\delta x_0+\mathcal{O}\paren{\sum_{i=1}^n \map {\rho} {y_i, y_i^*} }\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{x_0}^{x_1}\sum_{i=1}^n\paren{ F_{y_i}-\frac{\d}{\d x}F_{y_i'} } \map {h_i} x \rd x+F\big\rvert_{x=x_1}\delta x_1+\sum_{i=1}^n F_{y_1'}h_i\rvert_{x=x_1}-F\big\rvert_{x=x_0}\delta x_0-\sum_{i=1}^n F_{y_i'}h_i\big\rvert_{x=x_0}+\map {\mathcal{O} } {\sum_{i=1}^n\map {\rho} {y_i, y_i^*} }\)

where $h_i$ terms were integrated by parts.

In the same manner, $\map h x$ can be expanded as

$\displaystyle \map {h_i} {x_0}=\delta y_1^0-\map {y_i'} {x_0}\delta x_0+$$\map {\mathcal{O} } {\map {\rho} {y, y+h} }$
$\map {h_i} {x_1}=\delta y_1^1-\map {y_i'} {x_1}\delta x_1+\map {\mathcal{O} } {\map {\rho} {y,y+h} }$


\(\displaystyle \delta J\) \(=\) \(\displaystyle \int_{x_0}^{x_1}\sum_{i=1}^n\paren{F_{y_i}-\frac{\d}{\d x}F_{y_i'} }\map {h_i} x\rd x+\sum_{i=1}^n F_{y_i'}\big\rvert_{x=x_1}\delta y_i^1+\paren{F-\sum_{i=1}^n y_i' F_{y_i'} }\Bigg\rvert_{x=x_1}\delta x_1-\sum_{i=1}^n F_{y_i'}\big\rvert_{x=x_0}\delta y_i^0-\paren{F-\sum_{i=1}^n y_i' F_{y_i'} }\Bigg\rvert_{x=x_0}\delta x_0\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{x_0}^{x_1}\sum_{i=1}^n\paren{F_{y_i}-\frac{\d}{\d x}F_{y_i'} }\map {h_i} x\rd x+\sum_{i=1}^n F_{y_i'}\delta y_i\big\rvert_{x=x_0}^{x=x_1}+\paren{F-\sum_{i=1}^n y_i' F_{y_i'} }\delta x\Bigg\rvert_{x=x_0}^{x=x_1}\)


$\delta x \rvert_{x = x_j} = \delta x_j$
$\delta y_i \rvert_{x = x_j} = \delta y_i^j, j\in\set{0, 1}$