Eigenfunction for Adjoint Operstor

[Pere Callahan]

Hi,

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:

(Af)(x,y)=(\Delta f)(x,y)-f(x,y)+f(0,y)

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

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

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

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

However, the \delta_0(x) makes me think that this is not quite the right approach

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

I further concluded that the "eigenmeasure" for A^* which I am interested in, cannot have a density so it seems necessary to find out how A^* 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 Z(x,y)d^2\lambda in

\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)}

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

Thanks for your help

-Pere

 

[Pere Callahan]

It maybe is a good idea to first consider the problem in only one dimension:

Here the operator is

(Af)(x)=(\partial_x^2f)(x)+f(0)-f(x)

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

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

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...?

Thanks

-Pere

 

[Pere Callahan]

So if nobody replies I do it myself

My last one dimensional DGL is obviously solved for x\neq 0 by

g(x)=A e^{-|x|}

(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..

-Pere

edit:

I solved it again, this time via a Fourier Transform, and I changed my opinion concerning g(x)=A e^{-|x|} 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