Helmholtz eq. in cylindrical disk (+boundary value)

**marqushogas** · July 12th 2012, 01:30 PM

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 $r_{2}$ and a thickness $d$ (so it actually is a cylinder with a quite short height compared to radius). We’re interested in solving the (complex) vectorfield $E_{z}$ directed in the $\hat{z}$ direction for this disk. The PDE for this field is:

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

where $\sigma\geq0$ and $k$ is a pure imaginary number, with a real part 0 and a negative imaginary part. This disk has cylindrical rotation symmetry so $E_{z}$ does not depend on $\phi$ . 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:

$E_{z}(\rho,z=0)=0$ for all $\rho\in[0,r_{2}]$ .
$\int_0^{r_{2}}E_{z}(\rho,z’' )\rho\,d\rho=0$ for all $z’' \in(0,d)$ .
$\int_{r_{1}}^{r_{2}}E_{z}(\rho,z=d)\rho\,d\rho=I/\sigma$ , where $0 \leq r_{1} \leq r_{2}$ is a constant and $I$ is a complex constant.
$E_{z}(r’' ,z=d)=0$ for all $r’' \in[r_{0},r_{1}]$ , where $r_{0}$ is a constant such that $0 \leq r_{0} \leq r_{1} \leq r_{2}$ .
$\int_0^{r_{0}}E_{z}(\rho,z=d)\rhod\rho=I/\sigma$ .
Obviously $E_z$ 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!

**marqushogas** · July 13th 2012, 09:08 AM

Sorry! In boundary value 3. it should be $-I/\sigma$ . (but I don't think that changes things too much).

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