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Helmholtz Differential Equation--Elliptic Cylindrical Coordinates

In Elliptic Cylindrical Coordinates, the Scale Factors are $h_u=h_v=a\sqrt{\sinh^2 u+\sin^2 v}$, $h_z=1$, and the separation functions are $f_1(u)=f_2(v)=f_3(z)=1$, giving a Stäckel Determinant of $S=a^2(\sin^2 v+\sinh^2 u)$. The Helmholtz differential equation is

{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} (1)

Attempt Separation of Variables by writing
\end{displaymath} (2)

then the Helmholtz Differential Equation becomes

{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} (3)

Now divide by $UVZ$ to give

{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} (4)

Separating the $Z$ part,

{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} (5)

{d^2Z\over dz^2}=-(k^2+m^2)Z,
\end{displaymath} (6)

which has the solution
Z(z)= A\cos(\sqrt{k^2+m^2}\,z)+B\sin(\sqrt{k^2+m^2}\,z).
\end{displaymath} (7)

Rewriting (5) gives
\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} (8)

which can be separated into
$\displaystyle {1\over U}{d^2 U\over du^2}-m^2\sinh^2 u$ $\textstyle =$ $\displaystyle c$ (9)
$\displaystyle c+{1\over V}{d^2 V\over dv^2}-m^2\sin^2 v$ $\textstyle =$ $\displaystyle 0,$ (10)

{d^2U\over du^2}-(c+m^2\sinh^2 u)U=0
\end{displaymath} (11)

{d^2V\over dv^2}+(c-m^2\sin^2 v)V=0.
\end{displaymath} (12)

Now use
\sinh^2 u={\textstyle{1\over 2}}[1-\cosh(2u)]
\end{displaymath} (13)

\sin^2 v={\textstyle{1\over 2}}[1-\cos(2v)]
\end{displaymath} (14)

to obtain
{d^2U\over du^2}-\{c+{\textstyle{1\over 2}}m^2[1-\cosh(2u)]\}U=0
\end{displaymath} (15)

{d^2V\over dv^2}+\{c+{\textstyle{1\over 2}}m^2[1-\cos(2v)]\}V=0.
\end{displaymath} (16)

Regrouping gives
{d^2U\over du^2}-[(c+{\textstyle{1\over 2}}m^2)-{\textstyle{1\over 4}}m^22\cosh(2u)]U=0
\end{displaymath} (17)

{d^2V\over dv^2}+[(c+{\textstyle{1\over 2}}m^2)-{\textstyle{1\over 4}}m^22\cos(2v)]V=0.
\end{displaymath} (18)

Let $b\equiv {1\over 2}m^2+c$ and $q\equiv {1\over 4}m^2$, then these become
{d^2U\over du^2}-[b-2q\cosh(2u)]U=0
\end{displaymath} (19)

{d^2V\over dv^2}+[b-2q\cos(2v)]V=0.
\end{displaymath} (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


Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 514 and 657, 1953.

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© 1996-9 Eric W. Weisstein