Fourier transform of a derivative

Question

My PDE book gives the properties for the fourier transform of $u(x,t)$:

$F(\frac{\partial^n}{\partial t^n}u(x,t))=\frac{\partial^n}{\partial t^n}\hat u(\xi,t)$, and

$F(\frac{\partial^n}{\partial x^n}u(x,t))=(i\xi)^n\hat u(\xi,t)$

but I was wondering if these result generalize to higher dimensions. If instead $u(x,y,t)$, would you have:

$F(\frac{\partial^n}{\partial t^n}u(x,y,t))=\frac{\partial^n}{\partial t^n}\hat u(\xi,t)$

$F(\frac{\partial^n}{\partial x^n}u(x,y,t))=(i\xi)^n\hat u(\xi,t)$

$F(\frac{\partial^n}{\partial y^n}u(x,y,t))=(i\xi)^n\hat u(\xi,t)$

And what about mixed derivatives like $F(\frac{\partial^2}{\partial x\partial y}u(x,y,t))$? What if you wanted to take $F(\nabla^4u(x,y,t))$?

I'd appreciate any help.

Answer 1

What happens depends on what variable(s) you are applying the Fourier transform to; if we suppose we are making the Fourier transform with regards to $x \rightsquigarrow \xi$, then if we are using the one-dimensional convention that

$$ \hat{f}(\xi) = \int^{\infty}_{-\infty}f(x)e^{-i\xi x} dx$$

so in higher dimensions we generalise to

$$ \hat{f}(\xi,y,t) = \int^{\infty}_{-\infty}f(x,y,t)e^{-i\xi x} dx $$

it follows that we get that (letting $\mathcal{F}$ represent the Fourier transform operator)

$$\begin{array}{c} \mathcal{F} \left( \frac{\partial^n u(x,y,t)}{\partial t^n} \right) = \frac{\partial^n \hat u(\xi,y,t)}{\partial t^n}\\ \mathcal{F} \left( \frac{\partial^n u(x,y,t)}{\partial x^n} \right) = (i\xi)^n \hat u(\xi,y,t)\\ \mathcal{F} \left( \frac{\partial^n u(x,y,t)}{\partial y^n} \right) = \frac{\partial^n \hat u(\xi,y,t)}{\partial y^n}\\ \end{array} $$

Therefore, for example, if you wanted to find $\mathcal{F}(\partial_{x,y}u(x,y,t))$, we would get that it equals

$$ \frac{\partial}{\partial y} \mathcal{F}\left(\frac{\partial u(x,y,t)}{\partial x}\right) = ix \frac{\partial \hat u(\xi,y,t)}{\partial y}$$

If you end up making Fourier transforms with respect to more than one variable, then just use these rules, but do them a variable at time - so then, if we were to take $x \rightsquigarrow \xi$ and $y \rightsquigarrow \zeta$, we would get that the answer to the above would now be

$$ ix \frac{\partial \hat u(\xi,\zeta,t)}{\partial y} = -xy \hat u(\xi,\zeta,t) $$

Answer 2

I think that as long as everything is well defined (for example, being in the Schwartz space), then the result is what you think it should be. For example, $$F\left(\frac{\partial^2}{\partial x \partial y}\right) u(x,y,t) = (ix)(iy) \hat{u}(x,y,t)$$ (assuming you are doing the transform on $x$ and $y$).

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This is stated in Stein's book on Fourier Analysis in Chapter 6:

Let $f \in \mathcal{S}(\mathbb{R}^d)$. Then $$\left(\frac{\partial}{\partial x}\right)^\alpha f(x) \longrightarrow (i\xi)^\alpha \hat{f}(\xi).$$ (Here, the $x$ and $\xi$ are tuples in $\mathbb{R}^d$, and $\alpha$ is a multi-index. For example, $x=(x_1, x_2)$, and $\alpha=(1,2)$ gives $\left(\frac{\partial}{\partial x}\right)^\alpha:= \frac{\partial^3}{\partial x_1 \partial^2 x_2}$ and $x^\alpha:= x_1 x_2^2$.)