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Differentiating Integral with Green's function

  1. Nov 19, 2015 #1

    RUber

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    Hello, I am having trouble finding the proper justification for being able to pass the derivative through the integral in the following:
    ## u(x,y) = \frac{\partial}{\partial y} \int_0^\infty\int_{-\infty}^\infty f(x') K_0( \sqrt{ (x - x')^2 + (y-y')^2 } \, dx' dy' ##
    ##K_0## is the Modified Bessel function of the second kind with properties:
    1. ## K_0(z) \approx - \ln z \text{ as } z \to 0 .##
    2. ## \mathop {\lim }\limits_{|z| \to \infty} K_\nu (z) = 0 .##
    3. ##K_\nu(z)## is real and positive for ##\nu > -1## and ## z \in \mathbb{R} >0 .##
    4. ##K_{-\nu}(z) = K_\nu(z) .##
    5. ## \frac{\partial K_\nu (z) }{\partial z } = -\frac 12 \left( K_{\nu -1 }(z) + K_{\nu+1}(z)\right).##

    I have seen it done, but my advisor asked me to justify the step.
    The problem occurs at the singularity ## x= x', y = y'.##

    I also know that ##K_0## has an equivalent representation in the Fourier transform domain of:
    ##c \int_{-\infty}^\infty \frac { e ^{ - \sqrt{ \xi^2+\alpha^2 } |y- y'|}}{\sqrt{ \xi^2+\alpha^2 }} e^{i (x-x') \xi } d\xi##
    for some scaling constant c.
    In a few articles, they discuss the properties of pseudo-differential operators...which may be a hint for justifying the derivative.

    I am really stumped on this one...any insight would be helpful. I feel like there is something intuitive that I am simply missing.
    Thanks.
     
  2. jcsd
  3. Nov 20, 2015 #2
    It looks difficult !
    I don't know the answer, but have you tried this: assuming that you've shown integrability of your double integral on ##\mathbb{R}_+\times \mathbb{R}##, would a polar change of variable transform this complicated double integral to a single, parameter dependent integral ? Then, would the usual theorems on inversion under the integral sign apply ?
     
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