Today; when I was doing some Inverse Laplace transformation in the class, I encountered the following problem cited in Zill's book:

The inverse Laplace transformation may be not unique. In Problems 29 and 30 evaluate $\mathscr{L}\{f(t)\}$.

29. $f(t) = \left\{ \begin{array}{ll} 1, & \quad t\ge0,~ t\ne1, t\ne2 \\ 3, & \quad t=1 \\ 4, & \quad t=2 \\ \end{array} \right.$

30. $f(t) = \left\{ \begin{array}{ll} \text{e}^{3t}, & \quad t\ge0,~ t\ne 5, \\ 1, & \quad t=5 \end{array} \right.$

I did them, but I was wondered how to explain the students if they ask me about the criteria for the uniqueness. I think it is rooted in the functional analysis, however, I am weak in this area. Is there any easier way to explain the uniqueness for second year graduate students? Thanks for your time.

  • $\begingroup$ Undergraduate, I mean. :-) $\endgroup$ – mrs May 5 '13 at 15:15
  • $\begingroup$ Babak S: ✿Nice Question✿ $\endgroup$ – Software May 5 '13 at 15:22
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    $\begingroup$ @Babk : for second year undergrads, I would just say that for the purposes of Laplace transforms, Fourier series, Fourier transforms, etc., we consider functions to be equivalent if they differ on only a finite number of points (or a finite number of points per period in the case of Fourier series). When you do the integrals to compute their Laplace transforms, etc., you get the same results. $\endgroup$ – Stefan Smith May 6 '13 at 0:01
  • $\begingroup$ @StefanSmith: Thanks so much for the advices. I'll do that. :-) $\endgroup$ – mrs May 6 '13 at 11:02

The issue here is continuity, since you typically cannot hope to recover the behavior of a function at a point which is not a point of continuity from an integral transform. If you consider continuous functions which decay fast enough (for example, subexponentially decaying functions) then you can prove uniqueness fairly easily by a standard calculation. For example, see here.

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    $\begingroup$ Tanks so much Chris. I got the way. You are a good tutor. :-) $\endgroup$ – mrs May 5 '13 at 15:16

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