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I was supposed to solve an exercise and I actually did it, but I had to use a step that I did not comprehend at all.

I know that the fourier transform of $F((-\Delta)f)(k)=|k|^2 F(f)(k).$

But now I had to use that $F((-\Delta+\lambda)^{-1}f)(k)=((|k|^2+\lambda)^{-1})F(f)(k),$ where $\lambda>0$ is a constant.

So I now that the Fourier transform changes differentiation into multiplication and vice versa, but I would have expected that since the Laplacian does not enter this formula directly, but via this $(.+\lambda)^{-1}$ construct, that this should somehow affect the Fourier transform. Could anybody explain to me whether this is actually true and why this holds?

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  • $\begingroup$ I don't think I understand your notation. $k$ is your wavenumber variable, correct? So what does $(-\Delta + k)f$ mean for a function in real space? $\endgroup$
    – Ian
    Commented Nov 24, 2015 at 21:26
  • $\begingroup$ @Ian the $k$ appeared only in the title at the position you mention and now I changed it, but I think the question body makes it clear or do I have another typo? $\endgroup$
    – Zimkovic
    Commented Nov 24, 2015 at 21:28
  • $\begingroup$ Oh, OK, then it is the question that I thought it was. $\endgroup$
    – Ian
    Commented Nov 24, 2015 at 21:30

1 Answer 1

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$$F[f](\xi) = F[(-\Delta+\lambda)((-\Delta+\lambda)^{-1}f)](\xi) = (|\xi|^2+\lambda)F[(-\Delta+\lambda)^{-1}f](\xi),$$ hence $$F[(-\Delta+\lambda)^{-1}f](\xi)=\frac{1}{|\xi|^2+\lambda}F[f](\xi)$$

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