On 2D Inverse Problems/Kernel of Dirichlet-to-Neumann map

The continuous analog of the matrix representation of a Dirichlet-to-Neumann operator for a domain $\mathrm{\Omega }$ is its kernel. It is a distribution defined on the Cartesian product of the boundary of the domain w/itself, such that if  

$\mathrm{\Lambda }f=g,$ then $g(\varphi )=\underset{\partial \mathrm{\Omega }}{\int }K(\varphi ,\theta )f(\theta )d\theta ,$ where $\varphi$ and $\theta$ parametrize the boundary w/arclength measure.

For the case of the half-plane w/constant unit conductivity the kernel can be calculated explicitly. It's a convolution, because the domain in consideration is shift invariant:

$K(\varphi ,\theta )=k(\varphi -\theta ),$ where k is a distribution on a line. Therefore, the calculation reduces to solving the Dirichlet problem for a ${\delta }_0$-function at the origin and taking outward derivative at the boundary line.

Dirichlet problem for a half-plane

Exercise (**). Complete the calculation of the kernel K for the half-plane to show that $K(x,y)=\frac{-1}{\pi (x-y{)}^2}$ off the diagonal.

Exercise (*). Prove that for rotation invariant domain (disc w/ conductivity depending only on radius) the kernel of Dirichlet-to-Neumann map is a convolution.

The Hilbert transform gives a correspondence between boundary values of a harmonic function and its harmonic conjugate. $H:u{|}_{\partial \mathrm{\Omega }}\to v{|}_{\partial \mathrm{\Omega }},$ where $f(z)=u(z)+iv(z)$ is an analytic function in the domain.

For the case of the complex upper half-plane the Hilbert transform is given by the following formula: $Hf(y)=\frac{1}{\pi }\text{p.v.}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{f(x)}{y-x}dx.$

Exercise (*). Differentiate under the integral sign the formula above to obtain the kernel representation for the Dirichlet-to-Neumann operator for the half plane.

To define discrete Hilbert transform for a planar network, the network together w/its dual is needed.