Famous Theorems of Mathematics/Applied Mathematics

-div is adjoint to d

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

$\underset{M}{\int }df(X)\omega =-\underset{M}{\int }f\mathrm{div}X\omega$

Proof of the above statement:

$\underset{M}{\int }(f\mathrm{d}\mathrm{i}\mathrm{v}(X)+X(f))\omega =\underset{M}{\int }(f{ℒ}_X+{ℒ}_X(f))\omega$

$=\underset{M}{\int }{ℒ}_{X}f\omega =\underset{M}{\int }\mathrm{d}{\iota }_{X}f\omega =\underset{\partial M}{\int }{\iota }_{X}f\omega$

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:

$\mathrm{\Delta }f=\mathrm{d}\delta f+\delta \mathrm{d}f=\delta \mathrm{d}f=\delta {\partial }_{i}f\mathrm{d}{x}^i$

$=-*\mathrm{d}*{\partial }_{i}f\mathrm{d}{x}^i=-*\mathrm{d}({\epsilon }_{iJ}\sqrt{|g|}{\partial }^{i}f\mathrm{d}{x}^J)$

$=-*{\epsilon }_{iJ}{\partial }_j(\sqrt{|g|}{\partial }^{i}f)\mathrm{d}{x}^j\mathrm{d}{x}^J=-*\frac{1}{\sqrt{|g|}}{\partial }_i(\sqrt{|g|}{\partial }^{i}f){\mathrm{v}\mathrm{o}\mathrm{l}}_n$

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

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) 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.

Properties

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

$\mathrm{\Delta }(fh)=f\mathrm{\Delta }h+2{\partial }_{i}f{\partial }^{i}h+h\mathrm{\Delta }f.$

Proof

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

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

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

$=f\mathrm{\Delta }h$

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

$+h\mathrm{\Delta }f$

$=f\mathrm{\Delta }h+({\partial }_{i}f{\partial }^{i}h+{\partial }_{i}h{\partial }^{i}f)*{\mathrm{v}\mathrm{o}\mathrm{l}}_n+h\mathrm{\Delta }f$

$=f\mathrm{\Delta }h+2{\partial }_{i}f{\partial }^{i}h+h\mathrm{\Delta }f$

where f and h are scalar functions.