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!

Help on Fredholm Eqn of the First Kind

  1. Jul 20, 2012 #1
    Hello humans,
    Can you offer advice on the following situation?
    [itex] g(s)=\int_{a}^{b}K(t,s)f(t)dt [/itex]
    However I understand that if K can be expressed as
    [itex] K(t,s)=K(t-s) [/itex]
    then we can say
    [itex]f(t)=\mathcal{F}^{-1}\left [ \frac{\mathcal{F}[g(t)]}{\mathcal{F}[K(t)]} \right ] [/itex]
    Where the fancy F is Fourier, natch. Although I am fuzzy on what happened to a and b. Anyway, in my case, my function looks like this:
    [itex] g(s)=\int_{0}^{\infty}t f(t)dt [/itex]
    Can you offer any tips, advice, et cetera?
    Thanks
     
  2. jcsd
  3. Jul 20, 2012 #2
    By the way the idea here is to solve for f, knowing g.
    Thanks
     
  4. Jul 20, 2012 #3

    Mute

    User Avatar
    Homework Helper

    K(t,s) = K(t-s) is not a good enough condition to be able to use the fourier transform method to solve the equation. Your integration limits also have to be [itex]\pm \infty[/itex].

    You could solve the equation numerically using quadrature methods, since then it's basically a linear algebra problem [itex]\mathbf{g} = K\mathbf{f}[/itex].
     
  5. Jul 20, 2012 #4
    Thanks man. I appreciate it.
     
  6. Jul 20, 2012 #5
    followup:
    What if we say
    [itex] t=e^{\omega u/2} [/itex], so [itex] dt=\frac{\omega}{2}e^{\omega u/2}du [/itex]
    and then
    [itex] g=\int_{0}^{\infty}\frac{\omega}{2}e^{\omega u}f(u)du [/itex]
    which means just taking the inverse Laplace transform of g in order to get f?
    Just an idea?
     
  7. Jul 20, 2012 #6

    Mute

    User Avatar
    Homework Helper

    I overlooked the part of your original post where you said your equation was

    [tex]g(s) = \int_0^\infty dt~t f(t).[/tex]

    Unless g(s) happens to be a constant, that is an ill-formed equation. The integration kernel does not depend on s, so the integral is a constant. If g is constant, you can have infinitely many solutions for f.
     
  8. Jul 20, 2012 #7
    Good point, you're right, in the interest of expedience I haven't written it right. The real problem looks like this:
    [itex] g(s)=\int_{0}^{\infty}tf(t,s)dt [/itex]
    I know g, I am trying to find f.
     
    Last edited: Jul 20, 2012
  9. Jul 20, 2012 #8

    Mute

    User Avatar
    Homework Helper

    Hm, well, general solutions to integral equations are hard to come by. You can't apply the Laplace transform as is, for instance, since it will only decouple the integrand if the upper limit is s. You could try, for example, assuming that [itex]f(t,s) = \tilde{f}(s-t)\Theta(s-t)[/itex], where Theta is a Heaviside side function. Then your equation would be of the form

    [tex]g(s) = \int_0^s dt~t \tilde{f}(s-t),[/tex]

    and the Laplace transform would factor the integral for you:

    [tex]\mathcal L g(u) = \frac{\mathcal L f(u)}{u},[/tex]
    which you could then inverse Laplace transform. However, since you have to assume a particular form of the solution here, it might not be general, or perhaps the inverse transform won't exist, etc.

    Another possibility is to consider the Mellin transform. The Mellin transform of a function f(x) is

    [tex]\varphi(y) = \int_0^\infty dx x^{y-1} f(x).[/tex]
    (it's actually related to the Laplace transform by a change of variables). This is almost the form of your equation, if your g were g(s,y=2). If you can choose a suitable generalization of g(s) to some g(s,y), then the solution to

    [tex]g(s,y) = \int_0^\infty dt~t^{y-1} f(t,s)[/tex]
    would simply be the inverse Mellin transform of g:

    [tex]f(t,s) = \frac{1}{2\pi i}\mathcal M^{-1}[g(s,y)](t) = \int_{c-i\infty}^{c+i\infty} dy t^{-y} g(s,y).[/tex]

    (see the wikipedia page for the Mellin transform definition and what the inverse equation means)

    But I'm not sure if you can really get the f(t,s) that you actually want out of that solution.

    Another cause for concern that I have with this approach is that there are effectively infinitely many ways you might generalize g(s) to g(s,y) such that g(s,2) is the g(s) you're starting with. I don't know if that means there are infinitely many solutions, or if some solutions just won't work (e.g., the inverse Mellin transform doesn't exist because your g doesn't satisfy the Mellin inversion theorem requirements), or something else.
     
    Last edited: Jul 20, 2012
  10. Jul 20, 2012 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    My immediate reaction is to get rid of the t factor by change of variable, u = t2, f(t) = 2h(u).
    If g(s) can be written as Ʃans-n, n > 0, then there's a solution h(u, s) = e-usƩanun-1/(n-1)! (the details may be wrong).
    Given any solution h = h(u, s), h + k(u,s) is also a solution iff ∫0k(u,s)du is identically 0.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help on Fredholm Eqn of the First Kind
  1. Logrithmic eqn (Replies: 9)

Loading...