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Given the Hankel function of first kind and zero order

$$ H_0^{(1)}(\omega|x-y|), $$

with $|x-y| \geq C >0$, I would like to calculate its Fourier transform, that is

$$ \int_{-\infty}^{\infty} H_0^{(1)}(\omega|x-y|) e^{-i \omega t} \; d\omega. $$

What I thought was the following:

Given

$$ U_y(x,t) = \frac{H(t-|x-y|)}{2\pi\sqrt{t^2-|x-y|^2}}, $$

where $H$ is the Heaviside function at $0$, we know that $U_y(x,t)$ is the fundamental solution of the 2D transient wave equation [1]:

$$ \begin{cases} (\partial_t^2 - \Delta)U_y(x,t) = \delta_{x=y} \; \delta_{t=0} & (x,t) \in \mathbb{R}^2\times \mathbb{R}, \\ U_y(x,t) = 0 & x \in \mathbb{R}^2, t\ll 0. \end{cases} $$

By applying the Fourier transform to the equation, we should get the Helmholtz equation

$$ (\Delta + \omega^2) \hat{U}(x,\omega) = \delta_{x=y}, $$

which has the Hankel function $H_0^{(1)}(\omega|x-y|)$ as the fundamental solution.

Therefore, I am tempted to write

$$ U_y(x,t) = \int_{-\infty}^{\infty} \hat{U}(x,\omega) e^{-i \omega t} \; d\omega, $$

that is

$$ \frac{H(t-|x-y|)}{2\pi\sqrt{t^2-|x-y|^2}} = \int_{-\infty}^{\infty} H_0^{(1)}(\omega|x-y|) e^{-i \omega t} \; d\omega, $$

but I have the feeling that I am missing something, since no books with tables of Fourier transforms mention it.

[1] Mansur, Carrer, Oyarzun, $Time$-$domain$ $BEM$ $techniques$

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  • $\begingroup$ The idea is correct. Applying the transform gives $(\Delta + \omega^2) \widehat U = -\delta(x - y)$, which has a solution $i H_0^{(1)}(\omega |x - y|)/4$. Note that $i H_0^{(1)}(-\omega |x - y|)/4$ is also a solution, you need additional conditions to determine which one you want. The inverse transform is $1/(2 \pi) \int_{\mathbb R} f(\omega) \, e^{-i t \omega} d\omega$, giving $$\int_{\mathbb R} H_0^{(1)}(\omega) \, e^{-i t \omega} d\omega = \frac {-4 i H(t - 1)} {\sqrt {t^2 - 1}}.$$ The Fourier integrals of $U$ and $\widehat U$ exist as improper integrals as well. $\endgroup$ – Maxim Jun 29 at 15:04
  • $\begingroup$ @Maxim Thank you very much for your comment. I am still thinking about this, and a new problem occured to me. As far as I know, the Hankel function is the solution of the Helmholtz equation when $\omega>0$. Now, taking the Fourier transform for all real values of $\omega$ is correct since the function has a logarithm singularity, but by taking also $\omega = 0$ in the integral am I loosing the fact that the Hankel function is actual the solution of the Helmholtz equation? $\endgroup$ – legars Jul 9 at 16:45
  • $\begingroup$ By itself, the equation does not require $\omega > 0$ (you can verify this easily for the one-dimensional Helmholtz equation). $\widehat U_{-\omega, y}(x)$ satisfies the same equation as $\widehat U_{\omega, y}(x)$. Their normalized linear combinations also satisfy the same equation. We need more information; for instance, we can find the asymptotic behavior of the transform for large $\omega > 0$, which will be $(1 - i) e^{i \omega}/\sqrt {\pi \omega}$. This corresponds to $H_0^{(1)}(\omega)$. $\endgroup$ – Maxim Jul 9 at 18:19
  • $\begingroup$ @Maxim I agree with you. But I have the following doubt: the equation obviously does not require any condition on $\omega$, but the fundamental solution $H_0^{(1)} (\omega)$ is clearly not defined when $\omega =0$. As far as I know, $H_0^{(1)} (\omega)$ is the fundamental solution of $\Delta + \omega^2$ if $\omega \neq 0$. When $\omega=0$, the fundamental solution is $\log|x|$ (up to some constant). Therefore, when I am taking the Fourier transform of $H_0^{(1)} (\omega)$, I am taking also $\omega=0$ in the integral. It makes sense mathematically, $H_0^{(1)} (\omega)$ can be integrated, 1/2 $\endgroup$ – legars Jul 9 at 20:31
  • $\begingroup$ but I think that I am loosing the hypothesis that connects both problems in time- and frequency-domain, since in the limit $H_0^{(1)} (\omega)$ is not going towards the actual solution of the Laplace equation. What is your opinion? Thank you for your patience. 2/2 $\endgroup$ – legars Jul 9 at 20:31

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