0
$\begingroup$

I have a differential operator defined by its Fourier transform:

$\left(\alpha k_x^2 + \beta k_y^2 + \gamma k_x k_y \right)^\delta \hspace{20pt} \alpha,\beta,\gamma,\delta \in \mathbb{R}$

I don't know how to do the inverse transform, but I know that it is impossible to compute in the most general case. However I don't need it, as I would only like to demonstrate that this operator, in real space (i.e. after antitrasforming) when applied to a constant term, yields zero. This is just my guess, but seeing how the operator is written, I suppose that after anti-transforming (if that would be possible) then it would be a function of $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and $\frac{\partial ^2}{\partial x \partial y}$, so being null when operating on constants, I suppose.

Is my guess correct? If so, how can I prove my guess?

Thank you very much!

Disclaimer: I am not a mathematician, I am a (wannabe) physicist, so some of my definitions could be inaccurate, I hope the question is clear, though.

Improve this question
$\endgroup$
5
  • $\begingroup$ In general, your operator is not correctly defined because the function written in your question may be very singular. After its multiplication on the Fourier transform of a test function, the result may happen not to be a Fourier transform, at least in the classical case. Therefore the result you ask for is possible only in special cases some of which are given in the book on fractional calculus by Samko, Kilbas, and Marichev. $\endgroup$
    – Anatoly Kochubei
    Commented Dec 9, 2011 at 6:34
  • $\begingroup$ Making sense of this requires the theory of tempered distributions. Assuming everything makes sense, the Fourier transform of the constant function is the dirac distribution, so your guess is correct. $\endgroup$
    – Antoine Levitt
    Commented Dec 9, 2011 at 10:26
  • $\begingroup$ Anatoly: thank you! What if I add some regularities conditions, such as $\delta=\frac{1}{2}$ and a constraint on $\alpha$,$\beta$,$\gamma$ so that what appears under the square root is always $\geq 0$? Would that help? Antoine: thank you for your answer! However, wouldn't the Dirac's distribution have to be convoluted with the operator, as a product in the real space becomes a convolution in reciprocal Fourier space? If I'm correct, then the Dirac's distribution would leave the operator unchanged. $\endgroup$
    – zakk
    Commented Dec 9, 2011 at 10:51
  • $\begingroup$ This is not a multiplication, but an operator application. When you differentiate a function, you can either differentiate it in real space, or multiply by $i\omega$ its Fourier transform. Same applies here. $\endgroup$
    – Antoine Levitt
    Commented Dec 9, 2011 at 11:08
  • $\begingroup$ Also, your operator looks, modulo rotation in the x-y plane, like a fractional laplacian, and there's quite a bit of litterature on that (also lookup sobolev space with non-integer exponent on wikipedia). $\endgroup$
    – Antoine Levitt
    Commented Dec 9, 2011 at 11:55

1 Answer 1

Reset to default
3
$\begingroup$

Let me change slightly your notations and consider the quadratic form in $\mathbb R_{\xi,\eta}^2$ $$ Q(\xi,\eta)=\alpha \xi ^2+2\gamma \xi \eta+\beta \eta^2, $$ where $\alpha, \beta$ are real parameters. This is a Fourier multiplier : $ Fourier\bigl(Q(D_x,D_y)u\bigr)(\xi,\eta)=Q(\xi,\eta)\hat u(\xi,\eta). $ Let us for instance assume that $\gamma^2<\alpha\beta, \alpha >0$, i.e. $Q$ is positive-definite. $Q$ has positive characteristic roots and can be transformed by an orthogonal transformation into $\lambda \xi ^2+\mu \eta^2,\lambda, \mu >0.$ Changing scales leads to $\xi ^2+\eta^2$. This multiplier is the $-$Laplace operator in two dimensions, with the fundamental solution $$ \frac{-1}{2\pi}\ln(\sqrt{x^2+y^2})=E(x,y). $$ So the inverse of the Fourier multiplier $D_x^2+D_y^2$ is the convolution by $E$. Some analogous things can be done when $\gamma^2\ge\alpha\beta$.

Bazin.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .