1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Laplace self-transform

  1. Aug 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Find a function of time f(t) whose Laplace transform is f(s). Other than zero!

    2. Relevant equations

    3. The attempt at a solution

    I've searched the internet and haven't fouind any.
  2. jcsd
  3. Aug 12, 2007 #2
    Neither have I.
    I asked this myself. And read searched other forums. No one knows the answer.

    I believe this is a consequence of the Fredholm Integral Equation, in which, the solutions take trivial form, i.e. only zero.

    Have you ever though about a similar question? Which functions (non-trivial) are its own Fourier Transforms?
    Last edited: Aug 12, 2007
  4. Aug 12, 2007 #3
    Oh, yeah! There are many Fourier self-transforms. Gaussian, for example.
  5. Aug 12, 2007 #4
    The Hankel functions are the eigenfunctions of this Integral equation if you are interested.
  6. Aug 12, 2007 #5
    Thanks for your response, Kummer.

    Yes, I'm aware of the Hankel functions.
    But let's look for the Laplace self-transforms. Do you claim there aren't any?

    Way to go, Tiger!

  7. Aug 12, 2007 #6
    I really do not know, I wish to know myself. I do not think there are, but I cannot prove it, nor have I tried to prove it. I asked some of the professors in the University and I do not have an answer. Next time I go, which is in two weeks, I try again for you.

    Here is your problem stated formally.

    Problem: Does there exist a function [tex]f:[0,\infty)\to \mathbb{R}[/tex] such that [tex]F(s)=\int_0^{\infty} e^{-st}f(t) dt[/tex] is integrable and converges for each [tex]s\geq 0[/tex] with the property that [tex]f(t) = F(s)[/tex].
  8. Aug 13, 2007 #7


    User Avatar
    Science Advisor

    No, f(t) is never equal to F(s).
    You want a function f(t) such that
    [tex]f(s)=\int_0^{\infty} e^{-st}f(t) dt[/tex]
    or, using your F, f(s)= F(s).
  9. Aug 13, 2007 #8
    Yes, I want that function. Other than f(t) identically equal to zero, of course.
    Do you know of such a function? Or how to go about finding one?
    Or how to prove there isn't one?

    Thanks for your interest.
  10. Aug 13, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    How about 1/sqrt(t)? The laplace transform is sqrt(pi)/sqrt(s). It's an eigenfunction, though not with unit eigenvalue...
  11. Aug 14, 2007 #10
    Yes, 1/sqrt(t) is close, and suggests that multiplying it by some other function might do the trick.
  12. Feb 8, 2008 #11
    Kummer, did you ever get an answer to my question about a Laplace Self-Transform?
    You said you were going to ask the profs.

    How about you, Dick, do you know of any?

  13. Feb 8, 2008 #12


    User Avatar
    Science Advisor
    Homework Helper

    No. But then I didn't drive myself all that crazy thinking about the eigenvalue spectrum of the laplace transform either. Is this important?
  14. Feb 8, 2008 #13
    Dick, wouldn't it be important to find a Laplace Self-Transform - something that most
    people say doesn't exist? I think it's very exciting!

  15. Feb 8, 2008 #14


    User Avatar
    Science Advisor
    Homework Helper

    I think the world's gotten along pretty well since August without a Laplace Self-Transform. Can't say I'm all that fired up about it. Why do you say 'most people say doesn't exist'? Has anyone given you a reason?
  16. Feb 8, 2008 #15
    No, they don't give a reason - they just think it doesn't exist - just like you apparently
    do. Look, Dick, as you know, it's not a matter of opinion - either there is such a function or there isn't. I belive there is.

    Thanks for responding in a friendly manner!

  17. Feb 8, 2008 #16


    User Avatar
    Science Advisor
    Homework Helper

    I'm just guessing it doesn't. It's a pure hunch. Now I have to ask, why do you believe it does? I'll look into it again.
  18. Feb 8, 2008 #17


    User Avatar
    Science Advisor
    Homework Helper

    Ok, I've got you something. http://mathforum.org/kb/thread.jspa?forumID=13&threadID=1535681&messageID=5522747 It's a reference to a thread on the sci.math mailing list. It doesn't answer your question but it gives you a lot more eigenfunctions. But none with eigenvalue 1 (near as I can tell). I would suggest you post your question on the sci.math forum. Those people are about as smart as they come, aside from the cranks, and Robert Israel is REALLY SMART. He'll probably pick up on the question and if there is a reason why the laplace transform doesn't have an eigenvalue of 1, he'll know. Try it.
    Last edited: Feb 8, 2008
  19. Feb 9, 2008 #18
    Thanks, Dick, I will!
  20. Feb 9, 2008 #19

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    hmm...two clues here:

    1. Hankel functions are eigenfunctions of the Fourier transform.

    2. [itex]1/\sqrt(t)[/itex] is an eigenfunction of the Laplace transform, but with non-unit eigenvalue.

    This makes me wonder, could the Bessel functions be Laplace self-transforms?
  21. Feb 9, 2008 #20
    In the book "Laplace transforms" by Murray R. Spiegel, is in chapter 1 an exercise on this subject. There are 135 exercises of which I was unable to find 6 of them. One of these 6 is off course the one were it is asked wheter a given function could be it's own transform. :frown:

    Anyway, the exercise asked to prove that

    [tex]L\left[ F(t) \right] = L\left[ at^{-\alpha}+bt^{-\beta \right] =\Lambda \cdot \left(as^{-\alpha}+bs^{-\beta}\right)[/tex]

    In which:

    [tex]\alpha+\beta =1[/tex]


    [tex]\Lambda = \pm\sqrt{\pi csc(\alpha \pi)}[/tex]

    And as a second part it was necessary to answer the question if F(t) was it's own transform and to explain why.

    Maybe this will shed some light on the post.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook