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?

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

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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 $$\partial_x \text{FT}[u](x,\omega)=\text{FT}[u_x](x,\omega)=i \omega \text{FT}[u](x,\omega)$$ All this is probably obvious and agrees well with (2) :). Same derivations apply for the right propagating wave $V(x-t)$.

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