# Analytical Fourier Transforms with convolution rule [closed]

The task is the following, how to extend Mathematica's FourierTransform command in order to be able to analytically deal with non-linear differential equations. To do so it should be able to deal with convolutions and derivatives

One way of dealing with complicated non-linear Partial Differential Equations is to transform them to Fourier space, where the products become convolutions. However the FourierTansform command does not transform products into convolutions. The command

FourierTransform[f[x,y]g[x,y],{x,y},{kx,ky}]

Should result in something like

Convolution[f[qx,qy]g[qx,qy],{qx,qy},{kx,ky}]

I tried to create an extension that (robustly) allows this but failed.

Furthermore, in order to deal with PDE's the extension should also be able to deal with derivatives, such that

FourierTransform[D[f[x,y],x]g[x,y],{x,y},{kx,ky}]

results in

Convolution[qx f[qx,qy]g[qx,qy],{qx,qy},{kx,ky}]

I think such an extension would be extremely useful, but unfortunately FourierTransform is not very good at recognizing derivatives, and can simply not at all recognize convolutions...

any ideas?

• "Any ideas?" Code something, if I t works, well done, if not, post your code and associated question... otherwise, this has a pre-order feel about it. – ciao Jul 21 '15 at 4:28
• You are aware that Convolve[] is built-in? – J. M.'s technical difficulties Jul 21 '15 at 5:22
• In addition to Convolution not being the correct name, your notation for the Fourier transforms should also be different. f[qx, qy] etc. is not the correct form. It would be the (inverse) FourierTransform of f, so the replacement by Convolve would lead to a more complex expression than before. I.e., it's not a simplification, and there's no reason for Mathematica to do it. – Jens Jul 21 '15 at 5:32