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Helmholtz Differential Equation--Spherical Coordinates

In Spherical Coordinates, the Scale Factors are $h_r=1$, $h_\theta=r\sin\phi$, $h_\phi=r$, and the separation functions are $f_1(r)=r^2$, $f_2(\theta)=1$, $f_3(\phi)=\sin\phi$, giving a Stäckel Determinant of $S=1$. The Laplacian is


\begin{displaymath}
\nabla^2 \equiv {1\over r^2} {\partial\over\partial r}\left(...
...artial\phi}\left({\sin\phi{\partial\over\partial\phi}}\right).
\end{displaymath} (1)

To solve the Helmholtz Differential Equation in Spherical Coordinates, attempt Separation of Variables by writing
\begin{displaymath}
F(r, \theta, \phi) = R(r)\Theta(\theta)\Phi(\phi).
\end{displaymath} (2)

Then the Helmholtz Differential Equation becomes


\begin{displaymath}
{d^2R\over dr^2}\Phi\Theta + {2\over r}{dR\over dr}\Phi\Thet...
...hi }\Theta R + {1\over r^2}{d^2\Phi\over d\phi^2}\Theta R = 0.
\end{displaymath} (3)

Now divide by $R\Theta\Phi$,
$\quad{r^2\sin^2\phi\over\Phi R\Theta}\Phi\Theta{d^2R\over dr^2} + {2\over r}{r^2\sin^2\phi\over\Phi R\Theta}\Phi\Theta{dR\over dr}$
$ + {1\over r^2\sin^2\phi }{r^2\sin^2\phi\over\Phi R\Theta} \Phi R {d^2\Theta\ov...
...hi\over r^2\sin\phi} {r^2\sin^2\phi\over\Phi\Theta R}{d\Phi\over d\phi}\Theta R$
$+ {1\over r^2}{r^2\sin^2\phi \over \Phi R\Theta}{d^2\Phi\over d\phi^2}\Theta R = 0\quad$ (4)


\begin{displaymath}
\left({{r^2\sin^2\phi\over R}{d^2R\over dr^2}+ {2r\sin^2\phi...
...phi} + {\sin^2\phi\over\Phi}{d^2\Phi\over d\phi^2}}\right)= 0.
\end{displaymath} (5)

The solution to the second part of (5) must be sinusoidal, so the differential equation is
\begin{displaymath}
{d^2\Theta\over d\theta^2}{1\over\Theta}= -m^2,
\end{displaymath} (6)

which has solutions which may be defined either as a Complex function with $m=-\infty$, ..., $\infty$
\begin{displaymath}
\Theta(\theta) = A_me^{im\theta},
\end{displaymath} (7)

or as a sum of Real sine and cosine functions with $m=-\infty$, ..., $\infty$
\begin{displaymath}
\Theta(\theta) = S_m\sin(m\theta)+C_m\cos(m\theta).
\end{displaymath} (8)

Plugging (6) back into (7),


\begin{displaymath}
{r^2\over R}{d^2R\over dr^2}+ {2r\over R}{dR\over dr} - {1\o...
...r d\phi }
+ {\sin^2\phi \over \Phi}{d^2\Phi\over d\phi^2}= 0.
\end{displaymath} (9)

The radial part must be equal to a constant
\begin{displaymath}
{r^2\over R}{d^2R\over dr^2}+ {2r\over R}{dR\over dr}= l(l+1)
\end{displaymath} (10)


\begin{displaymath}
r^2 {d^2R\over dr^2}+ 2r {dR\over dr}= l(l+1)R.
\end{displaymath} (11)


But this is the Euler Differential Equation, so we try a series solution of the form

\begin{displaymath}
R = \sum_{n=0}^\infty a_nr^{n+c}.
\end{displaymath} (12)

Then
$r^2\sum_{n=0}^\infty (n+c)(n+c-1)a_nr^{n+c-2}+ 2r \sum_{n=0}^\infty (n+c)a_nr^{n+c-1}$
$ - l(l+1) \sum_{n=0}^\infty a_nr^{n+c}= 0\quad$ (13)
$\sum_{n=0}^\infty (n+c)(n+c-1)a_nr^{n+c}+ 2 \sum_{n=0}^\infty (n+c)a_nr^{n+c}$
$ - l(l+1) \sum_{n=0}^\infty a_nr^{n+c}= 0\quad$ (14)

\begin{displaymath}
\sum_{n=0}^\infty [(n+c)(n+c+1)-l(l+1)]a_nr^{n+c}= 0.
\end{displaymath} (15)

This must hold true for all Powers of $r$. For the $r^c$ term (with $n=0$),
\begin{displaymath}
c(c+1) = l(l+1),
\end{displaymath} (16)

which is true only if $c = l, -l-1$ and all other terms vanish. So $a_n=0$ for $n\not=l$, $-l-1$. Therefore, the solution of the $R$ component is given by
\begin{displaymath}
R_l(r) = A_lr^l+B_lr^{-l-1}.
\end{displaymath} (17)

Plugging (17) back into (9),
\begin{displaymath}
l(l+1)-{m^2\over\sin^2\phi}+ {\cos\phi\over\sin\phi}{1\over\Phi} {d\Phi\over d\phi}+ {1\over\Phi}{d^2\Phi\over d\phi^2}=0
\end{displaymath} (18)


\begin{displaymath}
\Phi''+{\cos\phi\over\sin\phi}\Phi'+\left[{l(l+1)-{m^2\over\sin^2\phi}}\right]\Phi = 0,
\end{displaymath} (19)

which is the associated Legendre Differential Equation for $x = \cos \phi$ and $m = 0$, ..., $l$. The general Complex solution is therefore
$\sum_{l=0}^\infty \sum_{m=-l}^l (A_lr^l+B_lr^{-l-1})P_l^m(\cos \phi)e^{-im\theta}$
$\equiv \sum_{l=0}^\infty \sum_{m=-1}^l(A_lr^l+B_lr^{-l-1})Y_l^m(\theta, \phi),\quad$ (20)
where
\begin{displaymath}
Y_l^m(\theta, \phi) \equiv P_l^m(\cos \phi)e^{-im\theta}
\end{displaymath} (21)

are the (Complex) Spherical Harmonics. The general Real solution is


\begin{displaymath}
\sum_{l=0}^\infty \sum_{m=0}^l (A_lr^l+B_lr^{-l-1})P_l^m(\cos \phi)[S_m\sin(m\theta)+C_m\cos(m\theta)].
\end{displaymath} (22)

Some of the normalization constants of $P_l^m$ can be absorbed by $S_m$ and $C_m$, so this equation may appear in the form

$\sum_{l=0}^\infty \sum_{m=0}^l (A_lr^l+B_lr^{-l-1})P_l^m (\cos \phi)[S_l^m\sin(m\theta)+C_l^m\cos(m\theta)]$
$ \equiv \sum_{l=0}^\infty \sum_{m=0}^l (A_lr^l+B_lr^{-l-1})\times[S_l^mY_l^{m(o)}(\theta, \phi )+C_l^mY_l^{m(e)}(\theta, \phi)],\quad$ (23)
where

\begin{displaymath}
Y_l^{m(o)}(\theta, \phi) \equiv P_l^m(\cos \theta)\sin(m\theta)
\end{displaymath} (24)


\begin{displaymath}
Y_l^{m(e)}(\theta, \phi) \equiv P_l^m(\cos \theta)\cos(m\theta)
\end{displaymath} (25)

are the Even and Odd (real) Spherical Harmonics. If azimuthal symmetry is present, then $\Theta(\theta)$ is constant and the solution of the $\Phi$ component is a Legendre Polynomial $P_l(\cos \phi)$. The general solution is then
\begin{displaymath}
F(r,\phi)=\sum_{l=0}^\infty (A_lr^l+B_lr^{-l-1})P_l(\cos \phi).
\end{displaymath} (26)


Actually, the equation is separable under the more general condition that $k^2$ is of the form

\begin{displaymath}
k^2(r,\theta,\phi)=f(r)+{g(\theta)\over r^2}+{h(\phi)\over r^2\sin\theta}+k'^2.
\end{displaymath} (27)


References

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, p. 514 and 658, 1953.



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© 1996-9 Eric W. Weisstein
1999-05-25