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I want to know if it is possible to use a PDE in the function FourierTransform instead of a function.

Consider for example the simple case for the heat equation

Clear["Global`*"]
 uf=
 FourierTransform[
  Derivative[2, 0][u[x, t]] == Derivative[0, 1][u[x, t]], x, k]

I tried to run this but it doesn't work, is there any way I could make this work?

Als

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2 Answers 2

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You can do it in two steps. Let $U(\omega, t)$ be the Fourier transform of $u(x,t)$, and using the relation that $F \left(\frac{\partial u}{\partial t}\right) = \frac{\partial}{\partial t} U(\omega,t)$ then we have (ps. I am using $\omega$ for your $k$ as I find it more clear)

Clear["Global`*"]
uf = FourierTransform[D[u[x,t],{x,2}],x,w]==FourierTransform[D[u[x,t],t],x,w]
uf = uf/.FourierTransform[u[x,t],x,w]->U[w,t]
uf/.FourierTransform[(u^(0,1))[x,t],x,w]->D[U[w,t],t]

Mathematica graphics

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Using ApplySides:

uf = ApplySides[FourierTransform[#, x, w] &, D[u[x, t], {x, 2}] == D[u[x, t], t]]

And you do the following that @Nasser points out.

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