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Integral equation

  1. Jun 23, 2009 #1
    How do you solve an integral equation of the form

    [tex]U(t) = \int_0^\infty U(t-x)K(x)dx[/tex] ?

    where K(x) is a given function.

    One can guess a solution of the form U(t) = e^(a*t), but is there a method for obtaining the general solution?

    edit: for some reason the TeX looks like a black box. Here's the equation

    U(t) = Int_0^infinity U(t-x)K(x)dx
  2. jcsd
  3. Jun 23, 2009 #2
    This is inverstigated under the name of "convolution".

    What do you mean by general solution? It is not very wise to look for a closed formula for arbitrary functions.
  4. Jun 23, 2009 #3


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    Are you sure this even has a solution? If you fourier transform both sides you get

    [tex]\widetilde{U}(\omega) = \widetilde{U}(\omega) \int_0^\infty dx~e^{-i\omega x}K(x)[/tex]

    (provided the functions are nice enough that you can interchange order of integrals, etc), so it looks to me like a solution doesn't exist unless (a) the "half"-fourier transform of K(x) is identically 1 (I'm not sure if that means any function U(x) will satisfy the equation, since if the half-FT of K(x) being one results in the above expression just being an identity, so it doesn't really tell you anything I guess) or (b) there exists a solution U(x) that's not nice such that the convolution theorem (or this semi-convolution theorem) fails and yet still satisfies the equation. I don't know any existence theorems for integral equations, so I don't know if that's possible.
  5. Jun 23, 2009 #4

    Yeah, that's a good point. That seems to restrict the possible functions K that would lead to a solution. And from that result, it does seem that when K has the required property, then any U is a solution. Here are some particular choices of K for which you can guess a solution:

    (i) K(x) = Ax, with A > 0

    Then U(t) = e^(sqrt(A) t) solves the equation.

    (ii) K(x) = Ce^(-bx)

    then U(t) = e^((C-b)t) solves it.

    I don't think either of these choices of K satisfies the "half FT" identity above, though.


    I don't know if equations such as this have a "general solution" like certain differential equations do. But if there is a prescribed method of solving this kind of equation, I was just asking what it is.
  6. Jun 23, 2009 #5


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    I'm not sure, but I think your choice of kernel may have violated the conditions under which the interchange of integration order in my derivation was valid, and hence those kernels fall under category 'b' of what I wrote. The proof of the convolution theorem in the wikipedia article about it requires f and g be in the space L^1, which I think means that the integral of the absolute value of the function over the whole real line is less than infinity, which neither of those kernels satisfies.

    Edit: Of course, that has to be the case: my derivation assumed the Fourier transform of U(t) exists, but the solutions you gave above do not have a fourier transform - hence the derivation I did does not apply to them.
    Last edited: Jun 23, 2009
  7. Jun 23, 2009 #6
    Okay, that makes sense. I haven't thought of a solution for U that has a fourier transform yet. I also haven't thought of a kernel K that satisfies the "half FT equal to 1" condition. But it is still surprising to me to see that if I did find such a K, then any function U with a fourier transform solves the equation. Thanks for your ideas.
  8. Jun 24, 2009 #7
    It occurred to me that K(x) = delta(x) is probably the only function that satisfies

    [tex]\int_0^\infty e^{-i\omega x}K(x)dx = 1[/tex] for all [tex]\omega[/tex],

    in which case the result that any U with a fourier transform solves the integral equation becomes obvious.

    Anyone else having trouble seeing the LaTeX images?
  9. Jun 24, 2009 #8


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    That's kind of what I was thinking before, except that the lower limit of integration is exactly where the delta function "pings" and contributes to the integral, so it's kind of fishy, since if you take the 0 to be "0-", as in the laplace transform, then that works, but if one were to take it to mean "0+", then 0 would just be infinitesimally outside the range of integration and the delta function wouldn't contribute. But, if you just adopt the first definition a delta function works - but note that so would a sum of delta functions with arguments x + x_n, all x_n > 0.

    Also, the LaTeX is working just fine for me.
  10. Jun 24, 2009 #9
    Ah, yes. Since the delta function is centered at the boundary of the integration region, then I probably need to consider the lower limit as 0-. At any rate, the important thing I'm taking away from this is that the interesting solutions to the integral equation will not have a FT (at least not for K that permit a change in the order of integration), because otherwise K would be restricted to delta(x) + some function that is only nonzero for negative x. In which case we learn nothing else about U(t).

    For me, the latex images are black boxes with some specks of white that barely outline the symbols. What's really weird is that I can see the images just fine on old posts. I wonder what has changed?

    thanks again :)
  11. Jun 25, 2009 #10
    LaTeX is not working anymore if you use Internet Explorer 6. This is due to changes made here at PF. A bug in IE6, which was not relevant to the way images were rendered previously, is at play now.
  12. Jun 26, 2009 #11

    Bummer. Thanks for the explanation, though :smile:. Now everyone knows I've got an OLD 'puter.
  13. Jun 26, 2009 #12
    You could try to download the opera web brouser. Opera works even on Windows 95 machines! :smile:
  14. Jun 28, 2009 #13

    Will do. Thanks!
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