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When solving the heat equation, one can use the Fourier transform in space to produce an equivalent ODE, which is easy to solve. In many presentations of the topic, it is assumed without proof that the Fourier transform (in space) of the time partial derivative of the solution $u(x,t)$ is equal to the time derivative of the Fourier transform (in space) of $u(x,t)$, i.e. $$ \mathscr{F}\left[\frac{\partial u(x,t)}{\partial t} \right] = \frac{\partial}{\partial t} \mathscr{F}\left[u(x,t) \right], $$ where $\mathscr{F}$ denotes the Fourier transform in the $x$ variable.

Question: How can this be proved rigorously?

It boils down to taking the limit of an integral whose integrand is a difference quotient as $h \rightarrow 0$. We are on the full real line, so don't have compact support for the integrands. Do we need dominated convergence in some form?

This question was addressed here but, though some sufficient conditions on $u(x,t)$ are mentioned in an answer to this post, no proof is given.

Any and all help appreciated!

user759636
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  • That formula is used literally only for smooth solutions, in which case you can use the dominated convergence theorem, as you pointed out. That's no loss of generality since solutions to the heat equations are assumed to be smooth and decaying (there are also unphysical solutions, such as the Tychonoff example, which do not satisfy these properties and, indeed, are not detected by the Fourier transform). – Giuseppe Negro Mar 14 '20 at 23:09

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A useful way to look at this situation is the following: $u(\cdot,t)$ lives in some space $X$ of functions or distributions. It is useful to consider a space on which the fourier transform is continuous, for example $L^1$, $L^2$ or $\mathcal{S}$, because derivatives commute with continuous linear operators.

Therefore a sufficient condition is that $t\mapsto u(\cdot,t)$ is differentiable as a map $(0,\infty) \to X$.

Johannes Hahn
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