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Help with a Helmholtz-like equation

  1. Sep 14, 2010 #1
    I am trying to find the solution to a problem defined as follows:

    (\partial_x^2-K^2)^2 G(\vec{x},\vec{x}')=\delta(\vec{x}'-\vec{x})

    where K is simply a constant and x is three dimensional.

    A \left[ e^{-K(\vec{x}-\vec{x}')}H(\vec{x}-\vec{x}') + e^{K(\vec{x}-\vec{x}')}H(\vec{x}'-\vec{x}) \right]
    + B \left[ (\vec{x}-\vec{x}') e^{-K(\vec{x}-\vec{x}')}H(\vec{x}-\vec{x}') + (\vec{x}'-\vec{x}) e^{K(\vec{x}-\vec{x}')}H(\vec{x}'-\vec{x}) \right]
    Here, H is the step function.

    and I am trying to find A and B such that the first equation is satisfied. Does anyone have any advice on how to proceed?

    My initial guess was to apply [tex] (\partial_x^2-K^2)^2 [/tex] to my proposed solution and then solve for A and B. This gets a bit tricky since once I start taking derivatives I get derivatives of the delta function.
  2. jcsd
  3. Sep 17, 2010 #2
    Hmm, I have never had to deal with problem like that. It seems to me a nonlinear operator.

    But I frequently encountered derivatives of the delta function. I would integrate both sides such that the derivatives of the Dirac Delta is part of the integrand. and then

  4. Sep 18, 2010 #3


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    Why do you think it's nonlinear?

    It's actually linear, I mean we have a fourth order PDE:

    [tex](\partial_x^2-K^2)^2= \partial_x^4 -2\partial_x^2 K^2 +K^4[/tex]
    remember that K is a scalar operator and thus it commutes with any other operator.

    As for the question itself, I guess you know the recipe for finding Green function,
    Do I need to repeat it? (cause I myself kinda keep forgetting).
  5. Sep 18, 2010 #4
    I'm no good with Green's functions, hence the post. However I did come up with a solution:

    First, I know the solution to
    (\partial_x^2 - K^2)G(x,x') = F(x)
    -\int dx' F(x') \frac{e^{-K |x-x'|}}{2 K}

    So we just need to look for the function that produces -1/2K times the exponentials and step functions when operated on by the above.

    Turns out that this produces [tex] A = \frac{1}{4K^3} [/tex] and [tex] B=\frac{1}{4K^2}[/tex]

    Maple spat this answer out earlier, but it is hard to trust maple unless you already know the answer.
  6. Sep 18, 2010 #5
    Do you mean [tex](\frac{\partial^{2}}{\partial x^{2}})^{2}=\frac{\partial^{4}}{\partial x^{4}}[/tex]?
  7. Sep 20, 2010 #6


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    [tex](\frac{\partial^{2}}{\partial x^{2}})^2=(\frac{\partial^2}{\partial x^2})(\frac{\partial^2}{\partial x^2})[/tex].

    Think of it as multiplication of operators.
  8. Sep 20, 2010 #7
    Ok, Thanks. I thought it was the power of the second partial derivative.
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