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Help with contour integration to find green's function of d^3/dx^3

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data
    Given a linear operator [itex]L=\frac{d^3}{dx^3}-1[/itex], show that the fourier transform of the Green's function is [itex]\tilde{G}(k)=\frac{i}{k^3-i}[/itex] and find the three complex poles. Use the Cauchy integral theorem to compute G(x) for x < 0 and x > 0.


    2. Relevant equations



    3. The attempt at a solution

    I found the Fourier transform of the Green's function and solved for the three complex roots. I'm having trouble setting up and carrying out the contour integration though. I think that you should be able to write (k^3 - i ) somehow in terms of the roots, then break the contour integral up into three separate integrals around each contour with terms like (k - ....)(k - ....)(k - ....) in each denominator?

    Any assistance is greatly appreciated!
     
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  3. Dec 16, 2013 #2

    pasmith

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    The cube roots of [itex]i[/itex] are [itex]-i[/itex] and [itex]-ie^{\pm 2\pi i/3}[/itex].
     
  4. Dec 16, 2013 #3
    Thanks for the reply, but I had already solved that
     
  5. Dec 16, 2013 #4

    vela

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    So what's stopping you from doing what the problem says and applying the Cauchy integral theorem to find g(x)?
     
  6. Dec 16, 2013 #5
    I don't know if I'm setting up the integral correctly in a form where a straightforward application of CIF could be done.

    I know about making arcs to form a closed contour but would I just need to evaluate one of these arcs? In this problem for example can I just use the "lower" arc that goes from -x to x and then loops around the pole at -i back to -x? Knowing that the other two poles will have no contribution by the Cauchy integral theorem. Or would I need to sum the contour integrals of the upper and lower arcs?

    So can I just write the lower arc C as this:
    [itex]-\frac{1}{2\pi i}\oint_{C}\frac{e^{ikx}}{k^3-i}dk =-\frac{1}{2\pi i}\oint_{C}\frac{e^{ikx}}{(e^{\frac{i4\pi }{3}}e^{\frac{i2\pi }{3}})}dk[/itex]

    where the denominator was factored like (pole at -i - pole in 1st quadrant)(pole at -i - pole in 2nd quadrant), I can simplify this further but is this the right approach?
     
  7. Dec 16, 2013 #6

    vela

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    The idea is that
    $$\oint_C f(z)\,dz = \int_{-R}^R f(z)\,dz + \int_\text{arc} f(z)\,dz.$$ If you can show the latter integral vanishes as ##R \to \infty## where ##R## is the radius of the arc, you can conclude the integral along the real axis is equal to the contour integral, which you can evaluate using residues. You can close the contour using an arc in the upper half plane or the lower half plane. The one you use depends on the sign of ##x##.
     
  8. Dec 17, 2013 #7
    Then you can't use the Residue Theorem in my opinion.
     
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