Fourier transform of a differential operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T02:29:59Z http://mathoverflow.net/feeds/question/83019 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83019/fourier-transform-of-a-differential-operator Fourier transform of a differential operator zakk 2011-12-09T01:48:18Z 2012-04-01T21:18:05Z <p>I have a differential operator defined by its Fourier transform:</p> <p>$\left(\alpha k_x^2 + \beta k_y^2 + \gamma k_x k_y \right)^\delta \hspace{20pt} \alpha,\beta,\gamma,\delta \in \mathbb{R}$</p> <p>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.</p> <p>Is my guess correct? If so, how can I prove my guess?</p> <p>Thank you very much!</p> <p>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.</p> http://mathoverflow.net/questions/83019/fourier-transform-of-a-differential-operator/92849#92849 Answer by Bazin for Fourier transform of a differential operator Bazin 2012-04-01T21:18:05Z 2012-04-01T21:18:05Z <p>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 $\gamma^2&lt;\alpha\beta, \alpha >0$, 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 $\gamma^2\ge\alpha\beta$.</p> <p>Bazin.</p>