2D Green's Function - Bessel function equivalence

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RUber
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Homework Statement


This is not a homework problem per se, but I have been working on it for a few days, and cannot make the logical connection, so here it is:
-- The problem is to show that
##\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{-\sqrt{\xi ^2 + \alpha^2 } |y-y'| + i \xi (x-x') }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi= \frac i4 H_0^{(1)}(i \alpha R) = \frac 1{2\pi} K_0(\alpha R) ##
Where ##R=\sqrt{(x-x')^2 + (y-y')^2 } ##.
##\alpha## is the constant wavenumber of a time-harmonic wave.
This is part of the nitty-gritty explanation of the 2D free space Green's function for waves.
Because the answer is a Bessel function, I expect there to be a change to cylindrical coordinates, which I attempt in section 3.

Homework Equations


I have a claim that the equation above is equivalent to
##i \int_{C(\phi)} e ^{ i k r \cos \beta } d\beta ##
with ##C(\phi)## defined by:
## C(\phi) = \left\{ \begin{array}{l l} \displaystyle
x=-|\phi| & y \text{ from } i \infty \text{ to } 0 \\
y=0 & x \text{ from } -|\phi| \text{ to } \pi - |\phi|\\
x= \pi - |\phi| & y \text{ from } 0 \text{ to } -i \infty
\end{array} \right. ##

Referring to Gradshteyn and Ryzhik, \cite{Gradshteyn2000}, this functional form is equivalent to $\rmi \pi H_0^1(kr)$.

The Attempt at a Solution


Attempting to change to polar coordinates centered at (x',y') using ## x = R\cos\theta, y = R\sin\theta## gives:
##\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{-\sqrt{\xi ^2 + \alpha^2 } R\sin\theta + i \xi R\cos\theta }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi ## for ##\theta \in [0, \pi]##
and
##\frac{1}{4\pi} \int_{-\infty}^{\infty} \frac{ e^{\sqrt{\xi ^2 + \alpha^2 } R\sin\theta + i \xi R\cos\theta }}{\sqrt{\xi ^2 + \alpha^2 }} d\xi ## for ##\theta \in [ -\pi,0]##.
The reference I have simply says that with an appropriate subsitition, the equivalence can be seen.
I cannot see what change of variables to make from here. I wonder if perhaps there is a more intuitive form of the Bessel K function that might help me see the connection.
Thank you to anyone who might be able to point me in the right direction.
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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