NH Modified Helmholtz Equation with Robin Boundary Condition

[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

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

with Robin boundary condition

[itex]\psi - \kappa\hat{\Omega}_n\cdot\vec{\nabla}\psi = 0[/itex]

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

The solution [itex]\psi[/itex] must also vanish at infinity, i.e. [itex]\psi(r\rightarrow\infty,z\rightarrow\pm\infty) = 0[/itex], 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 !

 

[tade]

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

 

[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.

 

[Mandelbroth]

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

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

with Robin boundary condition

[itex]\psi - \kappa\hat{\Omega}_n\cdot\vec{\nabla}\psi = 0[/itex]

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

The solution [itex]\psi[/itex] must also vanish at infinity, i.e. [itex]\psi(r\rightarrow\infty,z\rightarrow\pm\infty) = 0[/itex], 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 !

Yay, radiology!

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

 

[Meconium]

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

 

[Mandelbroth]

I have tried the Green's function approach in Cartesian coordinates, though the Robin boundary condition makes it hard to easily solve.

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

 

[Meconium]

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

 

[Mandelbroth]

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

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.

 

[Meconium]

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

 

[Mandelbroth]

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

You're very welcome.