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I have to solve

\begin{align} &\frac{\partial h'_{1,1m}(t,r)}{\partial t} + \frac{2}{r} h_{1,1m}(t,r) = 0 \label{beta_0_1}\\ &\frac{\partial^2 h_{1,1m}(t,r)}{\partial t^2} = 0. \label{beta_1-1} \end{align}

The second equation yields

\begin{equation} h_{1,1m}(t,r) = u(r)t + v(r), \end{equation}

Then, the first one gives

\begin{equation} u'(r) + \frac{2}{r} u(r) = 0, \end{equation}

whose solution is $u(r) = 3c/r$, where $c$ in a constant. The function $v(r)$ is entirely arbitrary. The same system, except for $\omega = 0$, does not admit any solution in Fourier domain that is not the null one. Why does it happen?

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    $\begingroup$ Can you say exactly what you mean by "the same system, except for $\omega = 0$, does not admit any solution in Fourier domain"? // Fundamentally isn't your question simply about why there are nice, smooth functions on $\mathbb{R}$ which do not have Fourier transforms expressible as a function? $\endgroup$
    – Willie Wong
    Commented Mar 31 at 9:53
  • $\begingroup$ I mean that if one want to find a solution from the second equation one find either $\omega=0$ or $h_1 = 0$.\\ Yes of course, but I wanted to know why this function is not a smooth function of t, because solving the equation in time domain it looks not clear to me. $\endgroup$
    – AleNekro97
    Commented Mar 31 at 9:59
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    $\begingroup$ you need to allow delta functions when you Fourier transform. $\endgroup$
    – Carlo Beenakker
    Commented Mar 31 at 11:15
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    $\begingroup$ The simpler example is $f' = 0$. Take the Fourier transform you get $i\omega \hat{f}(\omega) = 0$ from which the only "continuous in $\omega$" solution is $\hat{f} \equiv 0$ which corresponds to $f \equiv 0$. but obviously $f \equiv \mathit{const}$ are all solutions. The answer is that the smoothness of the Fourier transform is tied to the decay rate of the function in time domain. But this is basic Fourier theory and this question is not really suitable for this forum. $\endgroup$
    – Willie Wong
    Commented Mar 31 at 12:32
  • $\begingroup$ This is even alluded to on Wikipedia; see the final few sentences on the section on "Differentiation" linked to. $\endgroup$
    – Willie Wong
    Commented Mar 31 at 12:36

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