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NH Modified Helmholtz Equation with Robin Boundary Condition

  1. Aug 7, 2013 #1
    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 !
     
    Last edited: Aug 7, 2013
  2. jcsd
  3. Aug 7, 2013 #2
    Just curious, is this from what you're studying in college? Which course?
     
  4. Aug 7, 2013 #3
    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.
     
  5. Aug 7, 2013 #4
    Yay, radiology! :tongue:

    I'm assuming your ##D## is the diffusion constant, right, and not some weird differential operator?
     
  6. Aug 7, 2013 #5
    Yeah, it's only the diffusion constant, sorry for not specifying.
     
  7. Aug 8, 2013 #6
    You're sure it's too hard in Cartesian coordinates? Could you show us where it got too difficult for you?
     
  8. Aug 9, 2013 #7
    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]
     
  9. Aug 9, 2013 #8
    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.
     
  10. Aug 9, 2013 #9
    I will try that then. Thanks a lot for the help !
     
  11. Aug 9, 2013 #10
    You're very welcome. :wink:
     
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