Separation of Laplace and Helmholtz in spherical coordinates
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Short personal note by Phil dated 3.26.05. For Laplace it writes the Laplacian as a radial part minus L^2/r^2, uses spherical harmonics Y_nm, and finds radial solutions R_n = A r^n + B r^-(n+1). For Helmholtz it obtains the Bessel-type radial equation, cites Moon and Spencer for the solution in half-integer Bessel functions, and reduces the n = 0 case to e^{±ikr}/r.
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Separation of Laplace and Helmholtz in spherical coordinates PhL 3.26.05
1. Laplace. 2u = 0.
This is a famous and trivial result, but I seem to not have it written down anywhere. We can write the 3D Laplace operator in this manner
2 = r2 - (1/r2) L2θ,φ
from my TK page and memory of what L2 is . We then try a Laplace solution of the form
u(r,θ,φ) = Rn(r) Ynm(θ,φ)
which gives
[ r2 - (1/r2) L2θ,φ] Rn(r) Ynm(θ,φ) = 0
[ r2 - (1/r2)n(n+1)] Rn(r) Ynm(θ,φ) = 0
[ r2 - (1/r2)n(n+1)] Rn(r) = 0
[ 1/r2 ∂r(r2 ∂r) - (1/r2)n(n+1)] Rn(r) = 0
[ ∂r(r2 ∂r) - n(n+1)] Rn(r) = 0
We try a solution of the form rα and find then that
∂r(r2 ∂r)rα = ∂r(r2 α rα-1) = α ∂r rα+1 = α(α+1)rα
We will have a solution if α(α+1) = n(n+1). Thus
α2+ α - n(n+1) = 0
α = [ -1 ± ]/2 = [ -1 ± ]/2
= [ -1 ± ]/2 = [ -1 ± (2n+1)]/2
α+ = [ -1 + 2n+1]/2 = n
α- = [ -1 - 2n-1]/2 = -(n+1)
So, the solutions of α(α+1) = n(n+1) are α = n and α = -n-1 and we have thus found our two radial solutions
Rn(r) = Arn + Br-n-1
2. Helmholtz: (2+k2) u = 0.
We repeat most of the above with k2 sitting in there, so we get to
[ 1/r2 ∂r(r2 ∂r) - (1/r2)n(n+1) + k2] Rn(r) = 0
[ ∂r(r2 ∂r) - n(n+1) + r2k2] Rn(r) = 0
Well, sneaking a peak at page 27 of Moon and Spencer we see the solution is
Rn(r) = r-1/2 [ A Jn+1/2(kr) + B J-n-1/2(kr) ]
If we have an isotropic problem, then n = 0 and we have
R0(r) = r-1/2 [ A J1/2(kr) + B J-1/2(kr) ]
= r-1/2 [ A r-1/2 sin(kr) + B r-1/2 cos(kr) ]
= [ a sin(kr) + b cos(kr) ] / r
= [ α e-ikr + β e-ikr ] / r
and we arrive at the famous result that the solutions for n = 0 are simply e±ikr/r .