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Fourier Transforms, Green's function, Helmholtz

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    I've gotten myself mixed up here , appreciate some insights ...

    Using Fourier Transforms, shows that Greens function satisfying the nonhomogeneous Helmholtz eqtn
    $$ \left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2}) \:is\: G(\vec{r_1},\vec{r_2})= \frac{1}{{2\pi}^{3}} \int \frac{e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}}{k^2 - k_0^2} d^3k $$
    2. Relevant equations
    $$ \left(\nabla ^2 +k_0^2 \right) G(\vec{r_1},\vec{r_2})= -\delta (\vec{r_1} -\vec{r_2}) $$
    3. The attempt at a solution
    Taking the Fourier Transforms:
    LHS: $$F\left[ \nabla^2G+k_0^2 G \right] = (- k^2 +k_0^2) \hat{u} $$
    RHS: $$F\left[ -\delta (\vec{r_1} -\vec{r_2}) \right] = - \int 1 e^{-i\vec{k}.(\vec{r_1} -\vec{r_2})} d^3r $$

    $$ \therefore \hat{u} = \int \frac{1}{(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} d^3r = \frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})}$$

    $$ \therefore G(\vec{r_1},\vec{r_2})= F^{-1} \left[\frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} \right] = \frac{1}{{(2\pi)}^{3}} \int_{-\infty}^{\infty}\frac{1}{ik(k^2 - k_0^2)} e^{i\vec{k}.(\vec{r_1} -\vec{r_2})} e^{ik.r}d^3k$$
     
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  3. Oct 13, 2015 #2

    Orodruin

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    First of all, you must decide which variable you are keeping fixed in your Fourier transformation. In this case, the PDE is in terms of ##\vec r_1## and ##\vec r_2## is a constant. You may want to call them ##\vec r## and ##\vec r_0## instead to emphasise this.

    Second, the Fourier transform of the delta function is not an integral - or rather, it is an integral which is trivial to perform. Because of your notation, it is unclear what you mean by the integral you give. The Fourier transform you should be interested in is
    $$
    F[-\delta(\vec r - \vec r')] = -\int e^{-i \vec k\cdot \vec r} \delta(\vec r - \vec r') dV = - e^{-i\vec k \cdot \vec r'}.
    $$
    You should be able to take it from there.
     
  4. Oct 13, 2015 #3
    Awesome, thanks, I clearly need to better understand about the variables in play - may I please get some confirmation of what I have absorbed ...using r1 and r0 as you suggested - and I think r, r1 & r0 are all vectors?

    The formula I have for a 3D FT is $$ F[f(r)] ≡ f ̃(ω)= ∫_{-∞}^∞ f(r) e^{-iω.r} d^3 r $$

    In the above problem (I'm a bit mixed up, sorry) should I be using r or r1?

    Then you use r and r' for the FT of \delta, shouldn't that (for this problem) be $$ F[-\delta (r_1 - r_0)]=- \int e^{-ik.r} \delta (r_{1} - r_{0})] d^3r = e^{-ik.r} $$
    (and I think I want the r in the exponent to be r0?)

    For the inverse transform, I have $$ F^{-1} [\tilde{f} (ω)] ≡ f (r)= \frac{1}{(2\pi)^3} ∫_{-∞}^∞ \tilde{f}(\omega) e^{+iω.r} d^3 \omega $$

    What lead to this doubt was I could see that the (r1 - r0) emerged from the product of the 2 exponents, but I keep mixing up r, r1 & r0 - so would prefer to be certain and not guess. Thanks again, important for me this.
     
  5. Nov 7, 2015 #4
    Hi would really appreciate if someone could briefly check my thoughts above - let me know if they're right, let me know what's wrong ....thanks
     
  6. Nov 8, 2015 #5

    vela

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    Actually, he suggested you use ##\vec{r}## and ##\vec{r}_0## instead of ##\vec{r}_1## and ##\vec{r}_2##, so the original equation becomes
    $$(\nabla^2 + k_0^2)G(\vec{r},\vec{r_0}) = -\delta(\vec{r}-\vec{r}_0).$$ Clearly, ##\vec{r}## is the variable, so multiply both sides by ##e^{-i \vec{k}\cdot \vec{r}}## and integrate to compute the Fourier transforms.

     
  7. Nov 8, 2015 #6
    Thanks Vela, indeed Orodruin did, and I am not seeing my r's clearly, but for the delta function part he used r and r'... should I instead use r and ##r_0##?
     
  8. Nov 8, 2015 #7

    vela

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    Just be consistent. What's the argument of the delta function in
    $$(\nabla^2 + k_0^2)G(\vec{r},\vec{r_0}) = -\delta(\vec{r}-\vec{r}_0)?$$
     
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