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Eigenfunction for Adjoint Operstor

  1. Feb 22, 2008 #1

    I am having a question again (I should not work on so many things at the same time).

    I came across an operator A acting on C^2 - functions f:R^2->R:

    [tex](Af)(x,y)=(\Delta f)(x,y)-f(x,y)+f(0,y)[/tex]

    My goal is to find the adjoint operator (acting on measures) with respect to the inner product

    [tex]\left\langle f,\mu\right\rangle=\int_{\mathbb{R}^2}{f(x,y)d\mu(x,y)}[/tex]

    For the case that [tex]\mu[/tex] has a density with respect to lesbesgue measure, i.e
    [tex]d\mu(x,y)=g(x,y)d^2\lambda[/tex] I figured out that the adjoint, acting on [tex]\mu[/tex] and hence on the function g can be written as

    [tex](A^*g)(x,y)=(\Delta g)(x,y)-g(x,y)+\delta_0(x)\int_\mathbb{R}{dt\, g(t,y)}[/tex]

    However, the [tex]\delta_0(x)[/tex] makes me think that this is not quite the right approach:smile:

    My interpreation for this is that [tex]A^*[/tex] acting on a "nice" measure (having a density) gives something which does not have a density.

    I further concluded that the "eigenmeasure" for [tex]A^*[/tex] which I am interested in, cannot have a density so it seems necessary to find out how [tex]A^*[/tex] acts on for example a measure including a point mass at zero. I don't know how this can be done. More specifically, if we only consider the laplace part for now: What is [tex]Z(x,y)d^2\lambda[/tex] in

    [tex]\int_{\mathbb{R}^2}{(\Delta f)(x,y)\delta_0(x)d^2\lambda(x,y)}=\int_{\mathbb{R}^2}{f(x,y)Z(x,y)d^2\lambda(x,y)}[/tex]

    I would take this [tex]Z(x,y)d^2\lambda[/tex] to be [tex](\Delta^*\delta_0(x)d^2\lambda(x,y))(x,y)[/tex] if it happened to exist in some way or another...

    Thanks for your help

    Last edited: Feb 22, 2008
  2. jcsd
  3. Feb 22, 2008 #2
    It maybe is a good idea to first consider the problem in only one dimension:

    Here the operator is


    Again I am interested in an eigenmeasure of the adjoint operator with respect to the same inner product as above (integration). I found

    [tex](A^*g)(x)=(\partial_x^2g)(x)+\delta_0(x)\int_\mathbb{R}{dt\, g(t)}-g(x)[/tex]

    where the function g represents the measure g(x)dx...

    How ever I have a strong feeling that this operator does not have an eigenfunction.. can somebody confirm this...?


    Last edited: Feb 22, 2008
  4. Feb 22, 2008 #3
    So if nobody replies I do it myself:smile:

    My last one dimensional DGL is obviously solved for [tex]x\neq 0[/tex] by

    [tex]g(x)=A e^{-|x|}[/tex]

    (I only want finite measures (in fact probabilty measures) hence integrable functions.

    How ever for x=0 this is not a solution, so something weird might happen there..any suggestions...maybe an Admin could move this into Differential Equations ... if it seems appropriate..



    I solved it again, this time via a Fourier Transform, and I changed my opinion concerning [tex]g(x)=A e^{-|x|}[/tex] not being a solution for x=0....the delta distribution seems to somehow compensate for the non-differentiability at x=0. So I'll try the same method for the original two-dimensional problem and ask again if it does not work out:smile:
    Last edited: Feb 22, 2008
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