Famous Theorems of Mathematics/Applied Mathematics

From Wikibooks, open books for an open world
< Famous Theorems of Mathematics
Jump to: navigation, search

-div is adjoint to d[edit]

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.

Laplace-de Rham operator[edit]

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) 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[edit]

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[edit]

\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)
 + 
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.