Famous Theorems of Mathematics/Applied Mathematics

-div is adjoint to d
The claim is made that &minus;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.

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 &omega; is the volume form and &epsilon; 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, &Delta; is used to denote both; reader beware.

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


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

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


 * $$ = f\, \Delta h $$



\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) $$
 * (\partial_i f \, \mathrm{d}x^i \wedge



h \, \Delta f$$


 * $$ = f \, \Delta h +

(\partial_i f \, \partial^i h + \partial_i h \, \partial^i f){*\mathrm{vol}_n} + h \, \Delta f $$


 * $$ = f \, \Delta h +

2 \partial_i f \, \partial^i h + h \, \Delta f$$

where f and h are scalar functions.