Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Green's Functions

  1. Oct 22, 2007 #1
    [SOLVED] Green's Functions

    1. The problem statement, all variables and given/known data
    The basic equation of the scalar Kirchhoff diffraction theory is:

    [tex]\psi(\vec r_1) = \frac{1}{4\pi}\int_{S_2}\left[\frac{e^{ikr}}{r}\nabla \psi(\vec r_2) - \psi(\vec r_2)\nabla (\frac{e^{ikr}}{r})\right] \cdot d\vec S_2[/tex]

    where [itex]\psi[/itex] satisfies the homogeneous (three-dimensional) Helmholtz equation and [itex]r = |\vec r_1 - \vec r_2|[/itex]. Derive the above equation, assuming that [itex]\vec r_1[/itex] is interior to the closed surface [itex]S_2[/itex]


    2. Relevant equations
    Helmholtz Equation

    [tex](\nabla^2 + k^2)\psi = 0[/tex]

    Modified Helmholtz Equation

    [tex](\nabla^2 - k^2)\psi = 0[/tex]

    Has Green's Function

    [tex]G(\vec r_1, \vec r_2) = \frac{\mathrm{exp}(-k|\vec r_1 - \vec r_2|)}{4\pi|\vec r_1 - \vec r_2|}[/tex]

    Green's Function of three-dimensional Laplacian ([itex]\nabla^2[/itex]) is proportional to

    [tex]G(\vec r_1, \vec r_2) \propto \int \mathrm{exp}(i\vec k\cdot(\vec r_1 - \vec r_2))\frac{d^3k}{k^2}[/tex]

    3. The attempt at a solution

    We have [itex]\psi[/itex] as a solution of the equation

    [tex](\nabla_1{}^2 + k^2)\psi(\vec r_1) = (\nabla_1{}^2 - k^2)\psi(\vec r_1) + 2k^2\psi(\vec r_1) = 0[/tex]

    or

    [tex]\nabla_1{}^2 \psi(\vec r_1) = -k^2\psi(\vec r_1)[/tex]

    Which gives that

    [tex](\nabla_1{}^2 - k^2)\psi(\vec r_1) = -2k^2\psi(\vec r_1) = 2\nabla_1{}^2 \psi(\vec r_1)[/tex]

    So we can write the solution in terms of the Green's function for the modified Helmholtz Equation

    [tex]\psi(\vec r_1) & = & \int_{V_2}G(\vec r_1, \vec r_2)2\nabla_1{}^2\psi(\vec r_1) dV_2[/tex]

    And then I don't know how to take it from there to the solution. It doesn't look like it should be too hard, but I don't know how to do it.

    Any help would be greatly appreciated.

    Thanks in advance,
    Devin
     
  2. jcsd
  3. Oct 23, 2007 #2
    Noone can help?
     
  4. Oct 24, 2007 #3
    Anyone, this is due tomorrow :(
     
  5. Oct 24, 2007 #4
    Ok, I got a little further, still not sure how to finish it though...

    I then apply Green's theorem to the RHS and get

    [tex]\psi(\vec r_1) = 2\int_{S_2}(G(\vec r_1, \vec r_2)\nabla_2\psi(\vec r_2) - \psi(\vec r_2)\nabla_2G(\vec r_2, \vec r_2))dS_2 + 2\int_{V_2}\psi(\vec r_2)\nabla_2{}^2G(\vec r_1, \vec r_2)dV_2[/tex]

    But we know that

    [tex]\nabla_2{}^2G(\vec r_1, \vec r_2) = k^2G(\vec r_1, \vec r_2) - \delta(\vec r_1, \vec r_2)[/tex]

    So the equation becomes

    [tex]3\psi(\vec r_1) = 2\int_{S_2}(G(\vec r_1, \vec r_2)\nabla_2\psi(\vec r_2) - \psi(\vec r_2)\nabla_2G(\vec r_2, \vec r_2))dS_2 + 2k^2\int_{V_2}\psi(\vec r_2)G(\vec r_1, \vec r_2)dV_2[/tex]

    The first term is close to what I need, it is

    [tex]\frac{1}{2\pi}\int_{S_2}\left[\frac{e^{-kr}}{r}\nabla\psi(\vec r_2) - \psi(\vec r_2)\nabla\frac{e^{-kr}}{r}\right]\cdot dS_2[/tex]

    But I don't know how to change that negative to an [itex]i[/itex], or how to get rid of the extra term, or how to get rid of the extra factor of 2/3.
     
  6. Oct 24, 2007 #5
    Ok, for the second term, we have

    [tex] k^2\psi(\vec r_2) = -\nabla_2{}^2\psi(\vec r_2)[/tex]

    And then

    [tex] \int_{V_2}2\nabla_2{}^2\psi(\vec r_2)G(\vec r_1, \vec r_2)dV_2 = \psi(\vec r_1)[/tex]

    So we bring it over to the other side, and divide by 4 to get

    [tex]\psi(\vec r_1) = \frac{1}{8\pi}\int_{S_2}\left[\frac{e^{-kr}}{r}\nabla\psi(\vec r_2) - \psi(\vec r_2)\nabla\frac{e^{-kr}}{r}\right]\cdot d\vec S_2[/tex]

    Which is only off by a factor of 1/2, and has those negatives instead of [itex]i[/itex]'s in the exponent.
     
  7. Oct 25, 2007 #6
    Figured it out, the method I was using was the wrong one.
     
  8. Oct 25, 2007 #7
    I still can't mark my threads as solved...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook