# NH Modified Helmholtz Equation with Robin Boundary Condition

1. Aug 7, 2013

### Meconium

Hi,

I am working on a quite difficult, though seemingly simple, non-homogeneous differential equation in cylindrical coordinates. The main equation is the non homogeneous modified Helmholtz Equation

$\nabla^{2}\psi - k^{2}\psi = \frac{-1}{D}\frac{\delta(r-r')\delta(\theta-\theta')\delta(z-z')}{r}$

with Robin boundary condition

$\psi - \kappa\hat{\Omega}_n\cdot\vec{\nabla}\psi = 0$

on $r=a$, the edge of a virtual infinitely long cylinder of radius $r=a$. $\hat{\Omega}_n$ is a vector pointing out of the cylinder.

The solution $\psi$ must also vanish at infinity, i.e. $\psi(r\rightarrow\infty,z\rightarrow\pm\infty) = 0$, to satisfy the Sommerfeld Radiation Condition.

I have tried the Green's function approach in cartesian coordinates, though the Robin boundary condition makes it hard to easily solve. I have also tried it in polar coordinates, but I can't find any reference on how to use Green's function on periodic domains.

This problem arises from the diffusion approximation in biomedical imaging, and a solution would be of great help in my research.

Thanks a lot !

Last edited: Aug 7, 2013
2. Aug 7, 2013

Just curious, is this from what you're studying in college? Which course?

3. Aug 7, 2013

### Meconium

No it's not, I am working in an Optical Radiology Lab, and this problem is well documented in a cartesian semi-infinite medium, where the Robin boundary condition is simply on z = 0 (quite easier, isn't it?). However for a certain application (that I cannot disclose) I need to solve it in cylindrical coordinates.

4. Aug 7, 2013

### Mandelbroth

I'm assuming your $D$ is the diffusion constant, right, and not some weird differential operator?

5. Aug 7, 2013

### Meconium

Yeah, it's only the diffusion constant, sorry for not specifying.

6. Aug 8, 2013

### Mandelbroth

You're sure it's too hard in Cartesian coordinates? Could you show us where it got too difficult for you?

7. Aug 9, 2013

### Meconium

The normal vector $\hat{\Omega}_n$ is directed out of the cylinder, so $\hat{\Omega}_n$ is $\frac{x\vec{i}+y\vec{j}}{\sqrt{x^2+y^2}}$ instead of only $\vec{r}$

8. Aug 9, 2013

### Mandelbroth

However, this hardship is exchanged for a rather unfriendly inner product, which makes the boundary condition more difficult than worth solving.

For Cartesian coordinates, we can just solve using a convolution and attempt to fit the Robin boundary condition.

9. Aug 9, 2013

### Meconium

I will try that then. Thanks a lot for the help !

10. Aug 9, 2013

### Mandelbroth

You're very welcome.