# Helmholtz eq. in cylindrical disk (+boundary value)

• Jul 12th 2012, 01:30 PM
marqushogas
Helmholtz eq. in cylindrical disk (+boundary value)
Hi! This is a quite sophisticated problem, but it’s fun and interesting!

Consider the following case: Let’s say we have a 3-dimensional disk with a radius $\displaystyle r_{2}$ and a thickness $\displaystyle d$ (so it actually is a cylinder with a quite short height compared to radius). We’re interested in solving the (complex) vectorfield $\displaystyle E_{z}$ directed in the $\displaystyle \hat{z}$ direction for this disk. The PDE for this field is:

$\displaystyle \nabla^2 E_{z}+k \sigma E_{z}=0$

where $\displaystyle \sigma\geq0$ and $\displaystyle k$ is a pure imaginary number, with a real part 0 and a negative imaginary part. This disk has cylindrical rotation symmetry so $\displaystyle E_{z}$ does not depend on $\displaystyle \phi$. If we choose our cylindrical coordinate system so that the z-axis passes through the center and that z=0 at one of the circular planes of the disk; then the boundary values on the disk are the following:

1. $\displaystyle E_{z}(\rho,z=0)=0$ for all $\displaystyle \rho\in[0,r_{2}]$.
2. $\displaystyle \int_0^{r_{2}}E_{z}(\rho,z’' )\rho\,d\rho=0$ for all $\displaystyle z’' \in(0,d)$.
3. $\displaystyle \int_{r_{1}}^{r_{2}}E_{z}(\rho,z=d)\rho\,d\rho=I/\sigma$, where $\displaystyle 0 \leq r_{1} \leq r_{2}$ is a constant and $\displaystyle I$ is a complex constant.
4. $\displaystyle E_{z}(r’' ,z=d)=0$ for all $\displaystyle r’' \in[r_{0},r_{1}]$, where $\displaystyle r_{0}$ is a constant such that $\displaystyle 0 \leq r_{0} \leq r_{1} \leq r_{2}$.
5. $\displaystyle \int_0^{r_{0}}E_{z}(\rho,z=d)\rhod\rho=I/\sigma$.
6. Obviously $\displaystyle E_z$ must also be finite for all points in the disk.

I would be very thankful for any insight or idea on how to solve this problem (full solutions not necessary acquired!). So if you can help me in any way I owe you a huge amount of thankfulness and respect!
• Jul 13th 2012, 09:08 AM
marqushogas
Re: Helmholtz eq. in cylindrical disk (+boundary value)
Sorry! In boundary value 3. it should be $\displaystyle -I/\sigma$. (but I don't think that changes things too much).