0
$\begingroup$

I am currently working on this paper: http://web.calstatela.edu/faculty/rcooper2/article.pdf. I want to calculate $\phi_\nu$ on page 450. The author states that for $u_{\tau,\epsilon}$ ($\epsilon, \tau>0)$ and $\hat{\phi}_{\nu}= \frac{\hat{u}_{\tau,\epsilon}}{1+\nu(\xi_0^2+\tau^2)}$, $\phi_\nu$ has the following form

\begin{align*} \phi_\nu (x_0,x')= \frac{\pi}{1+\nu\tau^2} \int \limits_{-\infty}^{\infty} e^{-(\frac{1}{\nu}+\tau^2)|x_0-y_0|}u_{\tau,\epsilon}(y_0,x') dy_0. \end{align*}

I tried to calculate the inverse Fourier transform \begin{align*} \phi_\nu (x_0,x')= \frac{1}{2\pi} \int \limits_{-\infty}^{\infty}e^{ix_0\xi_0} \frac{1}{1+\nu(\xi_0^2+\tau^2)} \int \limits_{-\infty}^{\infty} e^{-\xi_0y_0} u_{\tau, \epsilon}(y_0,x') dy_0 d\xi_0\\= \frac{1}{2\pi} \int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{\infty} e^{i\xi_0(x_0-y_0)}\frac{u_{\tau, \epsilon}(y_0,x')}{1+\nu(\xi_0^2+\tau^2)} dy_0 d\xi_0 \end{align*} but I have to admit that I have no clue how to evaluate this integral.

Any help would be greatly appreciated.

$\endgroup$
0
$\begingroup$

Alright, I tried it on my own by using Jordan's Lemma. Could someone take a quick look if this is correct?

First I factorized $1+\nu(\xi_0^2+\tau^2) =\nu (\xi_0+i\sqrt{\frac{1}{\nu}+\tau^2})(\xi_0-i\sqrt{\frac{1}{\nu}+\tau^2})$. Hence $\frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)}$ has two simple poles. As we are only interested in residues satisfying $Im z>0$ we consider $i\sqrt{\frac{1}{\nu}+\tau^2}$.\begin{align*} \int \limits_{\mathbb{R}} \frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)} \mathrm{d}\xi_0= 2\pi i Res(\frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)},i\sqrt{\frac{1}{v}+\tau^2})= 2 \pi i \frac{e^{-\sqrt{\frac{1}{v}+\tau^2}(x_0-y_0}}{2i \nu \sqrt{\frac{1}{v}+\tau^2}} \end{align*} Thus \begin{align*} \phi_\nu (x_0,x') \frac{1}{2\pi \nu \sqrt{\frac{1}{v}+\tau^2}} \int \limits_R e^{-\sqrt{\frac{1}{v}+\tau^2}(x_0-y_0)} u_{\tau,\epsilon}(y_0,x') \mathrm{d}y_0 \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.