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I am considering the 1D wave equation with $c=1$ for the sake of simplicity: $$u_{tt}-u_{xx}=0,\quad \forall x\in\mathbb R,\; \forall t\in\mathbb R\tag{1}\label{eq:1}$$ with the following boundary conditions (initial conditions are ignored for now) $$ \begin{align} &u(0,t)=0\tag{2}\label{eq:2}\\ &u(L,t)=0\tag{3}\label{eq:3} \end{align} $$ for some strictly positive $L$. The boundary conditions are just seen as constraints on the sought solution $u(x,t)$ [By this I mean that the solution $u(x,t)$ is still defined for all $x\in\mathbb R$]. I am using the Fourier Transform approach to solve, that is $$\hat{u}(x,\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}u(x,t)\mathrm{e}^{-i\omega t} \mathrm{d}t$$ When plugged into the wave equation, the ODE governing $\hat{u}(x)$ reads $$\omega^2\hat{u}(x,\omega)+\hat{u}_{xx}(x,\omega)=0$$ whose general solution is $$\hat{u}(x,\omega)=A\cos \omega x+B\sin\omega x$$ The Fourier Transform of (2) implies $\hat{u}(0,\omega)=A=0$ while the Fourier Transform of (3) implies $B\sin \omega L=0$, that is $$ \omega_k=k\pi/L,\quad k=1,2,\ldots\tag{4}\label{eq:4}$$ if we decide to discard vanishing solutions. As a consequence: $$\hat{u}(x,\omega)=\sum_{k=1}^{\infty}B_k\sin\omega_k x\tag{5}\label{eq:5}$$ which satisfies (1), (2) and (3).

Question My question is on the inverse transform, whose usual definition is $$u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\hat{u}(x,\omega)\mathrm{e}^{i\omega t}\mathrm{d}\omega,\quad k=1,2,\ldots\tag{6}$$ Since $\omega$ now takes discrete values $\omega_k$ through (5), what is the meaning of the integral in (6) so that the Inverse Fourier Transform makes sense. In other words, through which mathematical argument can we deduce the definition of the discrete Inverse Fourier Transform from the continuous Inverse Fourier Transform?

Possible "wrong" answer

Solution (5) can we expressed as: $$\hat{u}(x,\omega)=\sum_{k=1}^{\infty}B_k \langle\delta_{\omega_k} , \sin\omega x\rangle\tag{7}$$ where $\delta_{\omega_k}$ denotes the usual Dirac distribution at $\omega_k$, that is $\delta_{\omega_k}=\delta(\omega-\omega_k)$. Inserting (7) into (6) yields $$u(x,t)=\frac{1}{\sqrt{2\pi}}\sum_{k=1}^{\infty}B_k\int_{-\infty}^{+\infty}\langle \delta_{\omega_k} , \sin\omega x\,\mathrm{e}^{i\omega t}\rangle \mathrm{d}\omega\tag{8}$$ which becomes (even though the integral itself is not well defined) $$u(x,t)=\sum_{k=1}^{\infty}B_k\sin\omega_k x\,\mathrm{e}^{i\omega_k t}$$ as expected (The $B_k$ have been scaled by a factor $\sqrt{2\pi}$). However (8) is not right, mathematically speaking. I'd be interested in a mathematically sound formulation.

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  • $\begingroup$ Assuming you can pass the Fourier transform inside the summation, you're ultimately trying to take the inverse transform of a nonzero periodic function, namely $\sin(\omega_k x)$. This requires you to define the Fourier transform through distribution theory rather than the Fourier integral, since the Fourier integral does not converge in this situation (not even conditionally). The result of doing that is just Fourier series, which would've been the easier way to look at this problem in the first place. $\endgroup$ – Ian Apr 17 '19 at 13:15
  • $\begingroup$ @Ian yes I know that the the theory of distributions is involved but I am unable to properly write a meaningful answer. I am editing my question with a possible "wrong" answer. $\endgroup$ – pluton Apr 17 '19 at 13:18
  • $\begingroup$ The procedure is very simple, you merely need to figure out what the Fourier transform of a complex exponential is, and then the rest is essentially just algebra. $\endgroup$ – Ian Apr 17 '19 at 13:19
  • $\begingroup$ @Ian I thought is would be fine to proceed with the Dirac $\delta$ distribution, see my edited answer. $\endgroup$ – pluton Apr 17 '19 at 13:29
  • $\begingroup$ The problem is a bit further back. Your $\hat{u}$ is not strictly correct; the point is that the boundary conditions can only be satisfied by a non-identically-zero function if $\omega=\omega_k$. Thus actually your expression for $\hat{u}(x,\omega)$ is not right because it doesn't even involve $\omega$ in the first place; it should have a factor of $\delta(\omega-\omega_k)$ in it so only those $\omega$'s contribute. Then you just need to take the inverse Fourier transform of a Dirac delta. But the cleaner way to proceed would have been to Fourier transform in $x$ in the first place. $\endgroup$ – Ian Apr 17 '19 at 15:02
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Equation (5) is wrong. The correct is $$ \hat{u}(x,\omega) = \sum_{k=1}^{\infty} B_k \sin\omega_k x \, \delta(\omega-\omega_k) \label{new5} \tag{5'} . $$

Then, $$\begin{align} u(x,t) &= \mathcal{F}^{-1}\{ \hat{u}(x,\omega) \} \\ &= \mathcal{F}^{-1}\left\{ \sum_{k=1}^{\infty} B_k \sin\omega_k x \, \delta(\omega-\omega_k) \right\} \\ &= \sum_{k=1}^{\infty} B_k \, \sin\omega_k x \, \mathcal{F}^{-1}\left\{ \delta(\omega-\omega_k) \right\} \\ &= \sum_{k=1}^{\infty} B_k \, \sin\omega_k x \, \frac{1}{\sqrt{2\pi}} \, e^{i\omega_k t} . \end{align}$$

Note:
Equation \eqref{new5} can also be written as $$ \hat{u}(x,\omega) = \sum_{k=1}^{\infty} B_k \sin\omega x \, \delta(\omega-\omega_k) = \sin\omega x \, \sum_{k=1}^{\infty} B_k \delta(\omega-\omega_k) = B(\omega) \sin\omega x \, \sum_{k=1}^{\infty} \delta(\omega-\omega_k) , $$ where $B(\omega_k) = B_k.$ The last sum is called the Dirac comb.

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  • $\begingroup$ Thanks. Your $u(x,t)$ is not a function of $t$? Are you sure about the inverse Fourier Transform? $\endgroup$ – pluton Apr 17 '19 at 18:43
  • $\begingroup$ @pluton. You're right that I had made an error. It has been fixed now. Hopefully I got the factor $\sqrt{2\pi}$ in the right place also. $\endgroup$ – md2perpe Apr 17 '19 at 20:45

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