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| {{Unreferenced|date=December 2007}}
| | Zoologist Stephan from Grand Valley, likes to spend some time kite boarding, property developers [http://Westwoodchristian.org/?option=com_k2&view=itemlist&task=user&id=69149 condo for sale In singapore] singapore and base jumping. Discovered some amazing locales having spent 9 days at Mont-Saint-Michel and its Bay. |
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| {{main|Laplace operator}}
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| ==−div is adjoint to d==
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| The claim is made that −div is adjoint to ''d'':
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| :<math>\int_M df(X) \;\omega = - \int_M f \, \operatorname{div} X \;\omega </math>
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| Proof of the above statement:
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| :<math>\int_M (f\mathrm{div}(X) + X(f)) \omega = \int_M (f\mathcal{L}_X + \mathcal{L}_X(f)) \omega </math>
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| ::<math> = \int_M \mathcal{L}_X f\omega = \int_M \mathrm{d} \iota_X f\omega = \int_{\partial M} \iota_X f\omega</math>
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| If ''f'' has [[compact support]], then the last integral vanishes, and we have the desired result.
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| ==Laplace–de Rham operator== | |
| One may prove that the Laplace–de Rham operator is equivalent to the definition of the Laplace–Beltrami operator, when acting on a scalar function ''f''. This proof reads as:
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| :<math>\Delta f =
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| \mathrm{d}\delta f + \delta\,\mathrm{d}f =
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| \delta\, \mathrm{d}f =
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| \delta \, \partial_i f \, \mathrm{d}x^i</math>
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| ::<math> =
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| - *\mathrm{d}{*\partial_i f \, \mathrm{d}x^i} =
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| - *\mathrm{d}(\varepsilon_{i J} \sqrt{|g|}\partial^i f \, \mathrm{d}x^J)</math>
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| ::<math> =
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| - *\varepsilon_{i J} \, \partial_j
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| (\sqrt{|g|}\partial^i f)\, \mathrm{d} x^j \wedge \mathrm{d}x^J =
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| - * \frac{1}{\sqrt{|g|}} \, \partial_i (\sqrt{|g|}\,\partial^i f) \mathrm{vol}_n</math>
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| ::<math> = -\frac{1}{\sqrt{|g|}}\, \partial_i (\sqrt{|g|}\,\partial^i f),</math>
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| where vol_n; is the [[volume form]] and ε is the completely antisymmetric [[Levi-Civita symbol]]. Note that in the above, the italic lower-case index ''i'' is a single index, whereas the upper-case Roman ''J'' stands for all of the remaining (''n''-1) indices. Notice that the Laplace–de Rham operator is actually minus the Laplace–Beltrami operator; this minus sign follows from the conventional definition of the properties of the [[codifferential]]. Unfortunately, Δ is used to denote both; reader beware.
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| == Properties ==
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| Given scalar functions ''f'' and ''h'', and a real number ''a'', the Laplacian has the property:
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| :<math>\Delta(fh) = f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f.</math>
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| ===Proof===
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| :<math>\Delta(fh) =
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| \delta\,\mathrm{d}fh =
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| \delta(f\,\mathrm{d}h + h\,\mathrm{d}f) =
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| *\mathrm{d}(f{*\mathrm{d}h}) + *\mathrm{d}(h{*\mathrm{d}f})\;</math>
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| :::<math> = *(f\,\mathrm{d}*\mathrm{d}h +
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| \mathrm{d}f \wedge *\mathrm{d}h +
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| \mathrm{d}h \wedge *\mathrm{d}f +
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| h\,\mathrm{d}*\mathrm{d}f)
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| </math>
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| :::<math> =
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| f*\mathrm{d}*\mathrm{d}h +
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| *(\mathrm{d}f \wedge *\mathrm{d}h +
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| \mathrm{d}h \wedge *\mathrm{d}f) +
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| h*\mathrm{d}*\mathrm{d}f</math>
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| :::<math> = f\, \Delta h </math>
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| ::::<math> {} +
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| *(\partial_i f \, \mathrm{d}x^i \wedge
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| \varepsilon_{jJ} \sqrt{|g|} \partial^j h \, \mathrm{d}x^J +
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| \partial_i h \, \mathrm{d}x^i \wedge
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| \varepsilon_{jJ} \sqrt{|g|} \partial^j f \, \mathrm{d}x^J) </math>
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| ::::<math> {} +
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| h \, \Delta f</math>
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| :::<math> = f \, \Delta h +
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| (\partial_i f \, \partial^i h +
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| \partial_i h \, \partial^i f){*\mathrm{vol}_n} +
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| h \, \Delta f </math>
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| :::<math> = f \, \Delta h +
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| 2 \partial_i f \, \partial^i h +
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| h \, \Delta f</math>
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| where ''f'' and ''h'' are scalar functions.
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| {{DEFAULTSORT:Proofs involving the Laplace-Beltrami operator}}
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| [[Category:Article proofs]]
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| [[Category:Differential operators]]
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Zoologist Stephan from Grand Valley, likes to spend some time kite boarding, property developers condo for sale In singapore singapore and base jumping. Discovered some amazing locales having spent 9 days at Mont-Saint-Michel and its Bay.