In Elliptic Cylindrical Coordinates, the Scale Factors are
,
, and the separation functions are
, giving a
Stäckel Determinant of
. The Helmholtz differential
equation is
![\begin{displaymath}
{1\over a^2(\sinh^2 u+\sin^2 v)}\left({{\partial^2 F\over \p...
... \partial v^2}}\right)+{\partial^2 F\over \partial z^2}+k^2=0.
\end{displaymath}](h_1021.gif) |
(1) |
Attempt Separation of Variables by writing
![\begin{displaymath}
F(u,v,z)=U(u)V(v)Z(z),
\end{displaymath}](h_1022.gif) |
(2) |
then the Helmholtz Differential Equation becomes
![\begin{displaymath}
{Z\over \sinh^2 u+\sin^2 v}\left({V{d^2 U\over du^2}+U{d^2 V\over dv^2}}\right)+UV{d^2 Z\over dz^2}+k^2 UVZ = 0.
\end{displaymath}](h_1023.gif) |
(3) |
Now divide by
to give
![\begin{displaymath}
{1\over\sinh^2 u+\sin^2 v}\left({{1\over U}{d^2 U\over du^2}...
...}{d^2 V\over dv^2}}\right)+{1\over Z}{d^2 Z\over dz^2}+k^2= 0.
\end{displaymath}](h_1025.gif) |
(4) |
Separating the
part,
![\begin{displaymath}
{1\over Z}{d^2 Z\over dz^2}=-(k^2+m^2){1\over \sinh^2 u+\sin...
...d^2 U\over du^2}+{1\over V}{d^2 V\over dv^2}}\right)= m^2,eqno
\end{displaymath}](h_1026.gif) |
(5) |
so
![\begin{displaymath}
{d^2Z\over dz^2}=-(k^2+m^2)Z,
\end{displaymath}](h_1027.gif) |
(6) |
which has the solution
![\begin{displaymath}
Z(z)= A\cos(\sqrt{k^2+m^2}\,z)+B\sin(\sqrt{k^2+m^2}\,z).
\end{displaymath}](h_1028.gif) |
(7) |
Rewriting (5) gives
![\begin{displaymath}
\left({{1\over U}{d^2 U\over du^2}-m^2\sinh^2 u}\right)+\left({{1\over V}{d^2 V\over dv^2}-m^2\sin^2 v}\right)=0,
\end{displaymath}](h_1029.gif) |
(8) |
which can be separated into
so
![\begin{displaymath}
{d^2U\over du^2}-(c+m^2\sinh^2 u)U=0
\end{displaymath}](h_1033.gif) |
(11) |
![\begin{displaymath}
{d^2V\over dv^2}+(c-m^2\sin^2 v)V=0.
\end{displaymath}](h_1034.gif) |
(12) |
Now use
![\begin{displaymath}
\sinh^2 u={\textstyle{1\over 2}}[1-\cosh(2u)]
\end{displaymath}](h_1035.gif) |
(13) |
![\begin{displaymath}
\sin^2 v={\textstyle{1\over 2}}[1-\cos(2v)]
\end{displaymath}](h_1036.gif) |
(14) |
to obtain
![\begin{displaymath}
{d^2U\over du^2}-\{c+{\textstyle{1\over 2}}m^2[1-\cosh(2u)]\}U=0
\end{displaymath}](h_1037.gif) |
(15) |
![\begin{displaymath}
{d^2V\over dv^2}+\{c+{\textstyle{1\over 2}}m^2[1-\cos(2v)]\}V=0.
\end{displaymath}](h_1038.gif) |
(16) |
Regrouping gives
![\begin{displaymath}
{d^2U\over du^2}-[(c+{\textstyle{1\over 2}}m^2)-{\textstyle{1\over 4}}m^22\cosh(2u)]U=0
\end{displaymath}](h_1039.gif) |
(17) |
![\begin{displaymath}
{d^2V\over dv^2}+[(c+{\textstyle{1\over 2}}m^2)-{\textstyle{1\over 4}}m^22\cos(2v)]V=0.
\end{displaymath}](h_1040.gif) |
(18) |
Let
and
, then these become
![\begin{displaymath}
{d^2U\over du^2}-[b-2q\cosh(2u)]U=0
\end{displaymath}](h_1043.gif) |
(19) |
![\begin{displaymath}
{d^2V\over dv^2}+[b-2q\cos(2v)]V=0.
\end{displaymath}](h_1044.gif) |
(20) |
Here, (20) is the Mathieu Differential Equation and (19) is the modified Mathieu Differential
Equation. These solutions are known as Mathieu Functions.
See also Elliptic Cylindrical Coordinates, Mathieu Differential Equation, Mathieu Function
References
Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York:
McGraw-Hill, pp. 514 and 657, 1953.
© 1996-9 Eric W. Weisstein
1999-05-25