Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral equation with bounded unknown kernel

  1. Aug 2, 2013 #1
    I need to solve an integral equation of the form

    $$\forall \omega \in [0,1], ~ \int_{\mathbb{R}} K(\omega,y)f(y)dy = \omega$$


    - f is known and positive with $$\int_{\mathbb{R}} f(y)dy = 1$$

    - K: [0,1] x R -> [0,1] is the unknown kernel

    I am looking for a solution other than K(omega,y) = omega. I do not know if such a solution exists, so I am looking either for a solution, or for a proof that K(omega,y) = \omega is the only solution.


    Let me tell you more about my attempts :

    - looking for K(omega,y) = a(omega)b(y) fails because it leads to a(omega) = \omega and b(y) = 1

    - looking for K(omega,y) = a(omega-y) and trying to solve the convolution equation $$\forall \omega \in [0,1], ~ (a \star f)(\omega) = \omega$$ fails (I proved it with Fourier transforms, although I'm not 100% confident about my proof)

    - setting K(omega,y) = omega + H(omega,y) and trying to solve $$\int_{\mathbb{R}} H(\omega,y) f(y) = 0$$ with either H(omega,y) = a(omega)b(y) or H(omega,y) = a(y-omega) fails as well

    There may not be a solution but I hope you can help me find one or prove that it's impossible ! Thank you
  2. jcsd
  3. Aug 3, 2013 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    One way of thinking about it is to assume we can differentiate inside the integral sign with respect to [itex] \omega [/itex]

    [tex] \int_{\mathbb{R}} K'(\omega,y)f(y)dy = 1 [/tex]
    [tex] \int_{\mathbb{R}} K"(\omega,y)f(y)dy = 0 [/tex]

    We can consider [itex] K" [/itex] to be a member of a family of functions of [itex] y [/itex] parameterized by [itex] \omega [/itex] such that each member of the family is orthogonal to [itex] f(y) [/itex] on [itex] \mathbb{R} [/itex].

    Suppose we have an orthogonal basis [itex]\{f_1(y), f_2(y),f_3(y)...\}[/itex] for some space of functions and that [itex] f = f_1 [/itex] Then [itex] \omega f_2 + (1 - \omega) f_3 [/itex] is a family of functions that is orthogonal to [itex] f [/itex].

    I'm just making these suggestion off the top of my head. I haven't tried to work this out with any concrete example.
  4. Aug 3, 2013 #3
    Thank you for your answer, I will also try your approach. Working from K'' may still be difficult since we have a boundedness condition on K.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook