$\int_{\bf R}\frac{\partial u}{\partial t}(x,t)e^{-2\pi ix\xi}~dx=\frac{d}{dt}\int_{\bf R}u(x,t)e^{-2\pi ix\xi}~dx$

Question

What is the sufficient condition such that this following equality holds : $$\int_{\bf R}\frac{\partial u}{\partial t}(x,t)e^{-2\pi ix\xi}~dx=\frac{d}{dt}\int_{\bf R}u(x,t)e^{-2\pi ix\xi}~dx$$

This question is about heat equation which is dealt with Fourier Transform . Any comment or advice will be appreciated . Thanks for considering my request .

Answer 1

Suppose that for each fixed $t$, the function $x\rightarrow u(x,t)$ is $L^{1}({\bf{R}})$.

Suppose further that for each fixed $x$, the function $t\rightarrow u(x,t)$ is $C^{1}({\bf{R}})$ and that for fixed $x$, the function $t\rightarrow\partial_{2}u(x,t)\leq\varphi(x)$ for a $\varphi\in L^{1}({\bf{R}})$.

Now for a fixed $t$, and for small $|h|>0$, we have \begin{align*} \int_{\bf{R}}\dfrac{1}{h}[u(x,t+h)-u(x,t)]e^{-2\pi ix\xi}dx&=\int_{\bf{R}}\partial_{2}u(x,\eta_{x,h,t})e^{-2\pi ix\xi}dx, \end{align*} and we see that \begin{align*} |\partial_{1}u(x,\eta_{x,h,t})|\leq\varphi(x), \end{align*} then by Lebesgue Dominated Convergence Theorem we have \begin{align*} \int_{\bf{R}}\dfrac{1}{h}[u(x,t+h)-u(x,t)]e^{-2\pi ix\xi}dx\rightarrow\int_{\bf{R}}\partial_{2}u(x,t)e^{-2\pi ix\xi}dx, \end{align*} put it another way, \begin{align*} \dfrac{d}{dt}\int_{\bf{R}}u(x,t)e^{-2\pi ix\xi}dx=\int_{\bf{R}}\partial_{2}u(x,t)e^{-2\pi ix\xi}dx. \end{align*}