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How to tell if F(s) is laplace transform?

  1. Nov 15, 2009 #1
    Explain why the functions



    Is it something to do with some basic property of laplace transform?
  2. jcsd
  3. Nov 15, 2009 #2
    Maybe it relates how you take some known Laplace transforms, and make other transforms from them.
  4. Nov 15, 2009 #3
    Because that's what a Laplace transform does. <--Put that into technical terms.
  5. Nov 16, 2009 #4


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    I would be very tempted to answer "How do you these are Laplace transforms" with "You don't. It is quite possible that these functions came from something that has nothing to do with Laplace transforms". But I suspect the question is really "How do you know there are functions that have these Laplace transforms". That, I would probably say, is because they are combinations of well known Laplace transforms- and Laplace transforms of complicated functions are typically combinations of Laplace transforms of simpler functions.
  6. Nov 16, 2009 #5
    The first one is a well-behaved one (simple poles etc.) but for the second, I think you need to come up with a convergence argument such that, we can approximate your function with the squareroot arbitrarily close by rational functions, such that the limit will be your transform.

    I don't remember exactly your answer but I suspect that this has to do with the existence of Pade tables and the related conditions.

    They typically use it in the elasticity or that kind of areas where the Laplace transforms look very unconventional.

    Edit : It should be something related to Tauberian theorems if I am not mistaken
    Last edited: Nov 16, 2009
  7. Nov 16, 2009 #6
    The product of Laplace transforms always has an inverse, given by a convolution product. The Inverse Laplace transform of
    1/sqrt(1+s^2) is the Bessel function of order zero.
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