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Spherical Bessel Differential Equation

Take the Helmholtz Differential Equation

\begin{displaymath}
\nabla^2F+k^2 F=0
\end{displaymath} (1)

in Spherical Coordinates. This is just Laplace's Equation in Spherical Coordinates with an additional term,


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

Multiply through by $r^2/R\Phi\Theta$,


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

This equation is separable in $R$. Call the separation constant $n(n+1)$,
\begin{displaymath}
{r^2\over R}{d^2R\over dr^2}+{2r\over R}{dR\over dr}+k^2r^2=n(n+1).
\end{displaymath} (4)

Now multiply through by $R$,
\begin{displaymath}
r^2{d^2R\over dr^2} + 2r{dR\over dr} + [k^2r^2-n(n+1)]R = 0.
\end{displaymath} (5)

This is the Spherical Bessel Differential Equation. It can be transformed by letting $x\equiv kr$, then
\begin{displaymath}
r{dR(r)\over dr} = kr {dR(r)\over k\,dr} = kr {dR(r)\over d(kr)} = x {dR(r)\over dx}.
\end{displaymath} (6)

Similarly,
\begin{displaymath}
r^2 {d^2 R(r)\over dr^2} = x^2 {d^2 R(r)\over dx^2},
\end{displaymath} (7)

so the equation becomes
\begin{displaymath}
x^2{d^2R\over dx^2} + 2x{dR\over dx} + [x^2-n(n+1)]R = 0.
\end{displaymath} (8)

Now look for a solution of the form $R(r)=Z(x)x^{-1/2}$, denoting a derivative with respect to $x$ by a prime,


$\displaystyle R'$ $\textstyle =$ $\displaystyle Z'x^{-1/2}-{\textstyle{1\over 2}}Zx^{-3/2}$ (9)
$\displaystyle R''$ $\textstyle =$ $\displaystyle Z''x^{-1/2}-{\textstyle{1\over 2}}Z'x^{-3/2}-{\textstyle{1\over 2}}Z'x^{-3/2}-{\textstyle{1\over 2}}(-{\textstyle{3\over 2}}) Zx^{-5/2}$  
  $\textstyle =$ $\displaystyle Z''x^{-1/2}-Z'x^{-3/2}+{\textstyle{3\over 4}}Zx^{-5/2},$ (10)

so
$x^2(Z''x^{-1/2}-Z'x^{-3/2}+{\textstyle{3\over 4}}Zx^{-5/2})$
$ +2x(Z'x^{-1/2}-{\textstyle{1\over 2}}Zx^{-3/2}) + [x^2-n(n+1)]Zx^{-1/2}=0$

(11)


\begin{displaymath}
x^2(Z''-Z'x^{-1}+{\textstyle{3\over 4}}Zx^{-2})+2x(Z'-{\textstyle{1\over 2}}Zx^{-1})+[x^2-n(n+1)]Z=0
\end{displaymath} (12)


\begin{displaymath}
x^2Z''+(-x+2x)Z'+[{\textstyle{3\over 4}}-1+x^2-n(n+1)]Z=0
\end{displaymath} (13)


\begin{displaymath}
x^2Z''+xZ'+[x^2-(n^2+n+{\textstyle{1\over 4}})]Z=0
\end{displaymath} (14)


\begin{displaymath}
x^2Z''+xZ'+[x^2-(n+{\textstyle{1\over 2}})^2]Z = 0.
\end{displaymath} (15)

But the solutions to this equation are Bessel Functions of half integral order, so the normalized solutions to the original equation are
\begin{displaymath}
R(r) \equiv A {J_{n+1/2}(kr)\over\sqrt{kr}}+ B {Y_{n+1/2}(kr)\over\sqrt{kr}}
\end{displaymath} (16)

which are known as Spherical Bessel Functions. The two types of solutions are denoted $j_n(x)$ (Spherical Bessel Function of the First Kind) or $n_n(x)$ (Spherical Bessel Function of the Second Kind), and the general solution is written
\begin{displaymath}
R(r) = A'j_n(kr)+B'n_n(kr),
\end{displaymath} (17)

where
$\displaystyle j_n(z)$ $\textstyle \equiv$ $\displaystyle \sqrt{\pi\over 2}{J_{n+1/2}(z)\over\sqrt{z}}$ (18)
$\displaystyle n_n(z)$ $\textstyle \equiv$ $\displaystyle \sqrt{\pi\over 2}{Y_{n+1/2}(z)\over\sqrt{z}}.$ (19)

See also Spherical Bessel Function, Spherical Bessel Function of the First Kind, Spherical Bessel Function of the Second Kind


References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 437, 1972.



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© 1996-9 Eric W. Weisstein
1999-05-26