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Whittaker Differential Equation


\begin{displaymath}
{d^2u\over dz^2}+{du\over dz}+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)u=0.
\end{displaymath} (1)

Let $u\equiv e^{-z/2} W_{k,m}(z)$, where $W_{k,m}(z)$ denotes a Whittaker Function. Then (1) becomes
${d\over dz}(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')+(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')$
$\mathop{+}\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W = 0.\quad$ (2)
Rearranging,

$({\textstyle{1\over 4}}e^{-z/2}W-{\textstyle{1\over 2}}e^{-z/2}W'-{\textstyle{1\over 2}}e^{-z/2}W'+e^{-z/2}W'')p$
$+(-{\textstyle{1\over 2}}e^{-z/2}W+e^{-z/2}W')+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W = 0\quad$ (3)

\begin{displaymath}
-{\textstyle{1\over 4}}e^{-z/2}W+e^{-z/2}W''+\left({{k\over z}+{{1\over 4}-m^2\over z^2}}\right)e^{-z/2}W=0,
\end{displaymath} (4)

so
\begin{displaymath}
W''+\left({-{1\over 4}+{k\over z}+{{1\over 4}-m^2\over z^2}}\right)W=0,
\end{displaymath} (5)

where $W'\equiv dW/dz$. The solutions are known as Whittaker Functions.


References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 505, 1972.




© 1996-9 Eric W. Weisstein
1999-05-26