synchcola | Jun. 29th, 2006

A neat proof of Lame's formula for the Laplacian.

Let u be a compactly supported test function.

$\int u\,\Delta^2 f\, d\xi = -\int {\partial u \over \partial \xi^i} {\partial f \over \partial \xi^i}d\xi=-\int{ \partial u\over\partial x_a}{\partial x_a\over\partial\xi_i}{\partial x_b\over\partial \xi_i}{\partial f\over\partial x_b} \bigg| {d\xi \over dx}\bigg|dx$

The xⁱ are orthogonal, so ^∂x_a/_{&partξ_i} ^∂x_b/_{∂ξ_i} is g^aa if a = b, and zero otherwise. Also, the Jacobian is √(g¹¹ g²² g³³) = √g.

$-\int {g^{aa} \over \sqrt g} {\partial u \over \partial x_a} {\partial f \over \partial x_a} dx = \int u {\partial \over \partial x_a} \bigg({g^{aa} \over \sqrt g} {\partial f \over \partial x_a}\bigg) dx = \int u \sqrt g {\partial \over \partial x_a} \bigg({g^{aa} \over \sqrt g} {\partial v \over \partial x_a}\bigg) d\xi$

So,

$\Delta^2 f = \sqrt g {\partial \over \partial x_a} \bigg( {g^{aa} \over \sqrt g} {\partial f \over \partial x_a} \bigg)$

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Jun. 29th, 2006

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