Laplace operator/Proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article Laplace operator. This article is currently an experimental vehicle to see how well we can provide proofs and details for a math article without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for some current discussion. This article is "experimental" in the sense that it is a test of one way we may be able to incorporate more detailed proofs in Wikipedia.

1 -div is adjoint to d
2 Laplace-de Rham operator
3 Properties
- 3.1 Proof

[edit] -div is adjoint to d

The claim is made that −div is adjoint to d:

$\int_M df(X) \;\omega = - \int_M f \, \operatorname{div} X \;\omega$

Proof of the above statement:

$\int_M (f\mathrm{div}(X) + X(f)) \omega = \int_M (f\mathcal{L}_X + \mathcal{L}_X(f)) \omega$

$= \int_M \mathcal{L}_X f\omega = \int_M \mathrm{d} \iota_X f\omega = \int_{\partial M} \iota_X f\omega$

If f has compact support, then the last integral vanishes, and we have the desired result.

[edit] Laplace-de Rham operator

One may prove that the Laplace-de Rahm operator is equivalent to the definition of the Laplace-Beltrami operator, when acting on a scalar function f. This proof reads as:

$\Delta f = \mathrm{d}\delta f + \delta\,\mathrm{d}f = \delta\, \mathrm{d}f = \delta \, \partial_i f \, \mathrm{d}x^i$

$= - *\mathrm{d}{*\partial_i f \, \mathrm{d}x^i} = - *\mathrm{d}(\varepsilon_{i J} \sqrt{|g|}\partial^i f \, \mathrm{d}x^J)$

$= - *\varepsilon_{i J} \, \partial_j (\sqrt{|g|}\partial^i f)\, \mathrm{d} x^j \, \mathrm{d}x^J = - * \frac{1}{\sqrt{|g|}} \, \partial_i (\sqrt{|g|}\,\partial^i f) \mathrm{vol}_n$

$= -\frac{1}{\sqrt{|g|}}\, \partial_i (\sqrt{|g|}\,\partial^i f),$

where ω 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) indecies. 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.

[edit] Properties

Given scalar functions f and h, and a real number a, the Laplacian has the property:

$\Delta(fh) = f \, \Delta h + 2 \partial_i f \, \partial^i h + h \, \Delta f.$

[edit] Proof

$\Delta(fh) = \delta\,\mathrm{d}fh = \delta(f\,\mathrm{d}h + h\,\mathrm{d}f) = *\mathrm{d}(f{*\mathrm{d}h}) + *\mathrm{d}(h{*\mathrm{d}f})\;$

$= *(f\,\mathrm{d}*\mathrm{d}h + \mathrm{d}f \wedge *\mathrm{d}h + \mathrm{d}h \wedge *\mathrm{d}f + h\,\mathrm{d}*\mathrm{d}f)$

$= f*\mathrm{d}*\mathrm{d}h + *(\mathrm{d}f \wedge *\mathrm{d}h + \mathrm{d}h \wedge *\mathrm{d}f) + h*\mathrm{d}*\mathrm{d}f$

$= f\, \Delta h$

$+ *(\partial_i f \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j h \, \mathrm{d}x^J + \partial_i h \, \mathrm{d}x^i \wedge \varepsilon_{jJ} \sqrt{|g|} \partial^j f \, \mathrm{d}x^J)$