# Wave equation and Fourier Transform: conditions for differentiation

Consider the wave equation in one dimension $$u_{tt}-u_{xx}=0$$ together with a Fourier Transform along $$t$$, ie $$\text{FT}[u](x,\omega)=\int_{-\infty}^{+\infty}u(x,t)\exp(-i\omega t)\mathrm{d}t.\tag{1}$$ The above PDE transforms into $$\partial_{xx}\text{FT}[u]+\omega^2\text{FT}[u]=0$$ whose general solution reads $$\text{FT}[u](x,\omega)=A(\omega)\cos\omega x+B(\omega)\sin\omega x\tag{2}$$ which is essentially the Fourier Transform of d'Alembert's solution.

Under which conditions on $$u(x,t)$$ is the classical differentiation of $$\text{FT}[u](x,\omega)$$ with respect to $$x$$ meaningful? When it is meaningful, is $$\partial_x \text{FT}[u]$$ the Fourier Transform of $$u_x(x,t)$$ that is $$\text{FT}[u_x]$$? It is a classical result which is always used when solving PDE via Fourier Transform (and used above in the quantity $$\partial_{xx} FT[u]$$), however I would like to read the exact assumptions on $$u$$. For instance, is this differentiation acceptable when $$u_{xx}(x,t)$$ should be read in the sense of distributions because $$u_x(x,t)$$ is discontinuous?

• A sufficient condition is the uniform convergence of the Fourier Transform of $u_{xx}(x,t)$. – Mark Viola May 25 '20 at 22:57
• @MarkViola Thanks. Do you have a reference on this? – pluton May 26 '20 at 0:40
• Any book on Advanced Calculus should suffice. – Mark Viola May 26 '20 at 1:17
• @MarkViola I've edited my question with the beginning of an answer: is this what you had in mind? Thx – pluton Jun 4 '20 at 19:30
• No. I was not extending the Fourier Transform to include tempered distributions. – Mark Viola Jun 4 '20 at 19:33

A partial answer to the above question is available in the book "Fourier Analysis, by TW Körner, Cambridge University Press, 1988, page 268, Theorem 53.5" (where $$x$$ and $$t$$ should be interchanged to comply with the question):
Let $$g:\mathbb{R}\times\mathbb{R}\to\mathbb{C}$$ be a continuous function such that $$g_2$$ exists and is continuous. Suppose $$\int_{-\infty}^{+\infty}|g(x,t)|\mathrm{d}x$$ and $$\int_{-\infty}^{+\infty}|g_2(x,t)|\mathrm{d}x$$ exist for each $$t$$ and that $$\int_{|x|>R}|g_2(x,t)|\mathrm{d}x\to 0$$ as $$R\to \infty$$ uniformly in $$t$$ on each $$[a,b]$$. Then $$\int_{-\infty}^{+\infty}g(x,t)\mathrm{d}x$$ is differentiable with $$\frac{d}{dt}\int_{-\infty}^{+\infty}g(x,t)\mathrm{d}x=\int_{-\infty}^{+\infty}\frac{\partial g}{\partial t}(x,t)\mathrm{d}x$$
[note by OP] where $$g_2$$ is the first partial derivative of $$g$$ with respect to its second argument.
If we consider for simplicity the left propagating wave, the solution reads $$u(x,t)=T(x+t)$$ where $$T$$ is a distribution. Its Fourier Transform in time is (because of translation) $$\text{FT}[u](x,\omega)=\int_{-\infty}^{+\infty}T(x+t)\exp(-i\omega t)\mathrm{d}t=\exp(i\omega x)\text{FT}[T](\omega) \tag{3}$$ and so $$\partial_x \text{FT}[u]$$ and $$\partial_{xx}\text{FT}[u]$$ are well defined as soon as $$T$$ is a tempered distribution and $$\partial_x \text{FT}[u](x,\omega)=i\omega\text{FT}[u](\omega)\tag{4}$$
Let us now have a look at the Fourier Transform of $$u_x=T_x$$ (in the sense of distributions) \begin{aligned} \text{FT}[u_x](x,\omega)&=\int_{-\infty}^{+\infty}T'(x+t)\exp(-i\omega t)\mathrm{d}t\\ &=\exp(i\omega x)\text{FT}[T'](\omega)=i\omega\exp(i\omega x)\text{FT}[T](\omega)=i\omega\text{FT}[u](x,\omega) \end{aligned} \tag{5} and Equations (5) and (4) are identical.
Conclusion: for the wave equation in 1D with solution $$u(x,t)=T(x+t)$$, the classical differentiation with respect to space of the Fourier Transform in time is legitimate as soon as $$T$$ is a tempered distribution and the following holds: $$\partial_x \text{FT}[u](x,\omega)=\text{FT}[u_x](x,\omega)=i \omega \text{FT}[u](x,\omega)$$ All this is probably obvious and agrees well with (2) :). Same derivations apply for the right propagating wave $$V(x-t)$$.