# Fourier transform of pseudo operator

I try to calculate Fourier transform of $\sqrt{-\Delta_2}\, \Delta_2\phi$ where $\sqrt{-\Delta_2}$ denetos pseudo operator and $\Delta_2$ denotes two-dimensional Laplacian. If we use the double Fourier transform $\displaystyle F(\xi,\eta)=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}f(x,y) e^{-i (x \xi+y \eta)}dx\, dy$, we get Fourier transform of $\Delta_2\phi$ as $-(\xi^2+\eta^2)\phi^{FF}$. But I don't know much about pseudo operator. How to calculate Fourier transform of $\sqrt{-\Delta_2}\, \Delta_2\phi$. Does anyone have any idea about Fourier transform of pseudo operator?

How is your operator $\sqrt{-\Delta}$ defined? Take another look, the answer is probably there. If not, read below (and then read a more comprehensive intro to pseudo-differential operators).

Pseudo-differential operators are defined in terms of the Fourier transform. That is, if $p(x,\xi)$ is an appropriate function of $x,\xi$, you can define an operator $P(x,D)$ by

$$P(x,D)u = \int \int e^{i<x-y,\xi>} p(x,\xi) u(x) dy \ d\xi$$

That is, it's the inverse transform of p times the transform of u. (p is called the 'symbol' of P)

In the case of a differential operator, the symbols are just polynomials in $\xi$. For instance, $\frac{\partial}{\partial x_j}$ has symbol $i \xi_j$, and $\Delta$ has symbol $-(\xi_1^2 + \dots + \xi_n^2)$. Likewise, the operator corresponding to $(\xi_1^2 + \dots + \xi_n^2)^\alpha$ is usually referred to as $(-\Delta)^\alpha$. So $\sqrt{-\Delta}$ works like this:

$$\sqrt{-\Delta}u = \int \int e^{i<x-y,\xi>} \sqrt{\xi_1^2 + \dots + \xi_n^2} u(x) dy \ d\xi$$

• thank you for your advise @BaronVT. I will consider these informations. – Banx Mar 12 '13 at 21:07