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Inverse Fourier Transform of ##1/k^2## in ##\mathbb{R}^N ##

  1. Dec 14, 2014 #1
    1. The problem statement, all variables and given/known data
    This comes up in the context of Poisson's equation
    Solve for ##\mathbf{x} \in \mathbb{R}^n ## $$ \nabla^2 G(\mathbf{x}) = \delta(\mathbf{x})$$
    2. Relevant equations
    $$\int_0^\pi \sin\theta e^{ikr \cos\theta}\mathop{dk} = \int_{-1}^1 e^{ikr \cos\theta}\mathop{d\cos \theta
    }$$
    3. The attempt at a solution
    Attempt by using Fourier Transforms
    $$ \nabla^2 G(\mathbf{x}) = \delta(\mathbf{x}) \Rightarrow -\lVert k\rVert^2 \tilde{G}(\mathbf{k}) = 1$$
    $$ G(\mathbf{x}) = -\mathcal{F}^{-1}\left[ \lVert k\rVert^{-2} \right]$$
    I reasoned that we could use N-dimensional spherical coordinates and be left with an integral integral over one angle in the plane between ##\mathbf{k}## and ##\mathbf{x}## and a radial integral which would have an element ##k^{n-1}\,dk ## where ##n## is the number of dimensions.
    $$G\left(\mathbf{x}\right) = \frac{-1}{(2\pi)^n}\int k^{n-1}F(\phi_1,\dots,\phi_{n-3})\sin(\phi_{n-2})\mathop{dk d\phi_1 \dots d\phi_{n-2}d\phi_{n-1}} \frac{e^{ikr\cos \phi_{n-2}}}{k^2} $$
    with ##u = \cos \phi_{n-2}##
    $$G\left(\mathbf{x}\right) = \frac{-1}{(2\pi)^n}\int F(\phi_1,\dots,\phi_{n-3})\mathop{d\phi_1 \dots d\phi_{n-3}d\phi_{n-1} }\int_{-1}^{1} du\int_0^\infty\mathop{dk} k^{n-3} e^{ikru}$$
    Maybe I'm not thinking clearly but the last two integrals don't seem like they are going to converge to anything nice. I don't know if I've made a mistake or I need to take these integrals in a particular order. I know I can show the Green's function is proportional to ##|| r||^\alpha ## for ##n>2## by using a test function, but the the idea here is to calculate it directly without assuming that much about its form.
     
    Last edited: Dec 14, 2014
  2. jcsd
  3. Dec 14, 2014 #2

    Ray Vickson

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    The last table in
    http://en.wikipedia.org/wiki/Fourier_transform
    seems to have this entry; it has a function whose transform is ##||k||^{-(n-\alpha)}, \; 0 < \alpha < n##, and yours is for ##\alpha = n-2##. Of course, the results are essentially stated without proof.
     
  4. Dec 14, 2014 #3
    Yes, I am seeking to prove that result.
     
  5. Dec 14, 2014 #4

    vela

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    Perhaps you could introduce a convergence factor ##e^{-\lambda k}## and then take the limit as ##\lambda \to 0^+##.
     
  6. Dec 14, 2014 #5
    That will just be equivalent to ignoring the other endpoint which perhaps needs justification. If you proceed this way you should obtain$$ \int_0^\infty k^\eta e^{-ikru} = \frac{\Gamma(\eta + 1)}{(-iru)^{\eta+1}}\;, \eta >-1$$

    $$\eta = n-3 > -1 \Rightarrow n>2 $$

    So now we expect the integral of F() over the variables besides k and u to give us something like $$\frac{\text{Surface Area(k)}}{k^{n-1}}\frac{1}{2} $$
     
    Last edited: Dec 14, 2014
  7. Dec 15, 2014 #6

    Orodruin

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    This should also be equivalent to considering the integral to be over a complex parameter k and closing the contour in the first octant (for positive u). This should reduce the integral to an integral along the positive imaginary axis, where the exponential is ##e^{-kru}##. For negative u you would have to close the contour in the fourth octant instead. Note that I did not try this explicitly, but I think it seems like a reasonable thing to do.

    For ##n = 2##, the integral is not convergent for small k and you have to regulate it by adding something that evaluates to a (in principle infinite) constant. This is related to the 2D Green's function being of the form ##\ln(r/r_0)##, where ##r_0## will depend on your regularisation.
     
    Last edited: Dec 15, 2014
  8. Dec 15, 2014 #7
    I am skeptical because the return path integral will have imaginary bounds. Using this equivalence does not seem to help. An imaginary number will appear either in your integration bounds or in the exponential.Something like
    $$\int _0^\infty f(iz)dz = -i\int_{i\infty}^{i0} f(u) du $$
    is not necessarily that useful.
    Vela's suggestion may simply be the way forward or perhaps there is another approach.
     
  9. Dec 15, 2014 #8

    Orodruin

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    Exactly, you want an additional i to appear in the exponential so that you get rid of the i that is already there. The imaginary units that appear due to k^m are just constants multiplying the integral. In the end it is very similar to what vela is suggesting although fully taking the rotation in the complex plane rather than letting it go to zero.

    The return path is from ##z = i\infty## to ##z = 0## along the imaginary axis. This can be parametrised in terms of the real number ##t## using ##z = it##, giving you a real integral multiplied by i to some power.

    The total integral is
    $$
    0 = \int_C f(z) dz = \int_0^\infty f(z) dz + \int^0_{i\infty} f(z) dz \quad \Rightarrow \quad \int_0^\infty f(z) dz = - \int^0_{i\infty} f(z) dz,
    $$
    where ##C## is the full contour and I have assumed that you can show that the contribution from 0 and the closing contour at infinity go to zero. In the second integral you do the parametrisation ##z = it##, giving you
    $$
    \int_0^\infty f(z) dz = i \int_0^{\infty} f(it) dt.
    $$
    Inserting your ##f(z) = z^m \exp(izur)## should give ##f(it) = i^m t^m \exp(-tur)## upon which the right integral can be solved.
     
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