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I was recently reading a paper that used a Fourier decomposition to solve a differential equation. The equation looked like this: $$ \frac{V(t)'}{V(t)} = \Phi_0\,x_0\,(P(t)-Y)\,\exp(-k_0t)+\text{some other terms} $$ The author splits the RHS into three terms, and then takes the Fourier transform of each term individually.

First of all, is the Fourier transform here even valid? Why is it we can talk about the Fourier transform of $\exp(-k_0t)$, when this function is not in $L^2$? I can find the Fourier transform of this function online, but I don't understand why it is defined since it is not in $L^2$. Am I missing something here?

The solution he gave to the term involving simply $(\text{some constants})\cdot\exp(-k_0t)$ was $$ F(\exp(-k_0t))=\frac{\text{some constants}}{k_0^2+4\pi^2 f^2} $$ First of all, this is not what I got. And it is also not what Wolfram Alpha tells me. I thought that perhaps the author is restricting himself to a specific domain (such as $t>0$, which would be reasonable), but I still can't reproduce his results. I have come to the conclusion that there must be some missing assumptions.

I still in the process of earning my masters in mathematics, so I am not sure what kind of wizardy is going on here. Why is it that we can take the Fourier Transform of a function that is not in $L^2$ here? And if there is some asusmption or technique that I am missing, does that assumption or technique allow me to get to the answer given above?

Thanks, Paul

EDIT: Given that $\exp(-k_0t)$, restricted to $t>0$, is in $L^1$, I get the square root of the answer given above. So this seems to be what the author did, though I am still wondering why everything is squared in his solution. Still a general question: does it make sense to restrict ourselves in this way in order to perform a Fourier decomposition? It seems that the author is stretching the Fourier decomposition's purpose in two ways: we are restricting the domain in order to make it in $L^2$, and we are using it on a non-periodic function. Does this have any impact on the robustness/accuracy/usefulness of the Fourier Decomposition approach in this context?

robjohn
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Paul
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  • The Fourier Transform is actually defined on $L^1$ and then extended to $L^2$.

    I assume that the equations are being considered on ${x:x\ge0}$. On this domain, $e^{-k_0t}\in L^1$.

     – robjohn Nov 24 '13 at 15:28
  • @robjohn, thanks for your comment. I had forgotten the way the Fourier transform is actually defined. I still have a lingering "meta" question about the usefullness of the Fourier approach in this context -- I posted it in my edit above. – Paul Nov 24 '13 at 15:38
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    The Fourier transform actually makes sense for Tempered distributions. And $L^1$ and $L^2$ functions are tempered distributions, and their FT as tempered distributions coincide with the standard $L^1$ and/or $L^2$ transforms. – N. S. Nov 24 '13 at 17:07
  • I've modified your question to make it more MathJax friendly. Check to make sure I didn't introduce any errors. – robjohn Nov 24 '13 at 17:11
  • @ N.S. thanks. ok after reading the wikipedia page, I understand how this qualifies as a Tempered Distribution. So I guess my only question left is when it is ok to restrict the domain of the function you are Fourier transforming. If we have full freedom to restrict the domain (to a closed interval for example), couldn't we then ignore the L^1, L^2, or even Tempered Distribuions requirements completely and transform any old function of our choosing, as long as it has a finite integral in our interval? – Paul Nov 24 '13 at 23:59

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$P(t)$ is not a constant, so the term you wrote down is not (some constants)$e^{-k_ot}$, so it's exact Fourier transform depends on the polynomial $P$. Assuming that this polynomial is a constant, so that one is indeed taking the Fourier transform of a constant times $e^{-k_ot}$ on the half-line only, I don't get what either of you get, I get $1\over k_o+i\omega$.

joe johnson
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