In my class, the professor gave us the following definitions for the $2$D Fourier Transform : $$ \hat{u}(k,\omega)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-i(kx-\omega t)}\ u(x,t)\ dx \ dt $$ $$ u(x,t)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{i(kx-\omega t)}\ \hat{u}(k,\omega)\ dk \ d\omega $$
I want to know what the $2$D Fourier Transform does to partial derivatives and I don't feel as confident in my derivations because of the implications of dealing with double integrals. For example, say I want to find what the $2$D Fourier Transform does to a time derivative $v_t(x,t)$ then by definition I would have: $$ \hat{v}_t(k,\omega)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-i(kx-\omega t)}\ v_t(x,t) \ dx \ dt $$ Just looking at this double integral, my first thought is to switch the order of integration and try to use integration by parts to turn $v_t$ into $v$; however, I am not sure if my function obeys Fubini's Theorem or not. And further, I'm not quite sure how integration by parts would work in this case if it is allowed. Also, I feel I may be well overthinking this. In class, my professor - in one line - would simply start with a PDE such as $$ u_{tt}+\gamma^2u_{xxxx}=0 $$ and then write "take $2$D Fourier Transform" so we have: $$ p(k,\omega)\hat{u}(k,\omega)=0 $$ where $$ p(k,\omega) = \omega^2-\gamma^2k^4=0 $$
This seems to suggest to me that the $2$D Fourier Transform turns partial derivatives with respect to time into a multiplication by $i\omega$ and partial derivatives with respect to space ($x$) into a multiplication by $ik$. I think I should be able to derive these rules from the definition but am stuck on how to treat this double integral.
Any help/hints/advice is much appreciated! Or a reference to any website or paper that goes through these types of derivations would also be incredible.
EDIT: This is my attempt at taking the $2$D Fourier Transform of $v_t$: $$ \hat{v}_t(k,\omega)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}e^{-i(kx-\omega t)}\ v_t(x,t) \ dx \ dt $$ $$ = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-ikx}[\int\limits_{-\infty}^{\infty}e^{i\omega t}v_t(x,t) dt] dx $$ Applying integration by parts on the inner integral and letting the surface term vanish we have: $$ = \frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}(-i\omega)e^{-ikx}e^{i\omega t}v(x,t) dt dx $$ And by definition this implies that we have: $$ =(-i\omega)\hat{u}(k,\omega) $$ My concern is that I'm not quite sure why we would require the surface term to vanish. Would it generally be a condition that $u \rightarrow 0$ as $|t|\rightarrow\infty$? If so why? Would we also require $u \rightarrow 0$ as $|x|\rightarrow\infty$ for the derivation of the $2$D Fourier Transform of the spatial partial derivative $u_x$? If so can someone explain why we require these conditions?