Helmholtz Differential Equation--Parabolic Cylindrical Coordinates

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

$${1\over u^2+v^2}\left({{\partial^2 F\over \partial u^2} +{\... ...r \partial v^2}}\right)+{\partial^2 F\over\partial z^2}+k^2=0.$$ (1)

Attempt Separation of Variables by writing

$$F(u,v,z)\equiv U(u)V(v)Z(z),$$ (2)

then the Helmholtz Differential Equation becomes

$${1\over u^2+v^2}\left({VZ {d^2U\over du^2}+UZ{d^2V\over dv^2}}\right)+UV{d^2Z\over dz^2}+k^2 UVZ=0.$$ (3)

Divide by $UVZ$,

$${1\over u^2+v^2}\left({{1\over U}{d^2U\over du^2}+{1\over V}{d^2V\over dv^2}}\right)+{1\over Z}{d^2Z\over dz^2}+k^2 = 0.$$ (4)

Separating the $Z$ part,

$${1\over Z}{d^2 Z\over dz^2}=-(k^2+m^2)$$ (5)

$${1\over u^2+v^2}\left({{1\over U}{d^2U\over du^2}+{1\over V}{d^2V\over dv^2}}\right)-k^2=0$$ (6)

$${1\over U}{d^2U\over du^2}+{1\over V}{d^2V\over dv^2}-k^2(u^2+v^2)=0,$$ (7)

so

$${d^2Z\over dz^2}=-(k^2+m^2)Z,$$ (8)

which has solution

$$Z(z)= A\cos(\sqrt{k^2+m^2}\,z)+B\sin(\sqrt{k^2+m^2}\,z),$$ (9)

and

$$\left({{1\over U}{d^2 U\over du^2}-k^2 u^2}\right)+\left({{1\over V}{d^2 V\over dv^2}-k^2 v^2}\right)=0.$$ (10)

This can be separated

$${1\over U}{d^2 U\over du^2}-k^2 u^2 = c$$ (11)

$${1\over V}{d^2 V\over dv^2}-k^2 v^2 = -c,$$ (12)

so

$${d^2U\over du^2}-(c+k^2u^2)U=0$$ (13)

$${d^2V\over dv^2}+(c-k^2v^2)V=0.$$ (14)

These are the Weber Differential Equations, and the solutions are known as Parabolic Cylinder Functions.

See also Parabolic Cylinder Function, Parabolic Cylindrical Coordinates, Weber Differential Equations

References

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 515 and 658, 1953.