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

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

\Lambda f = g,
g(\phi) = \int_{\partial\Omega}K(\phi,\theta)f(\theta)d\theta,
where \phi 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(\phi,\theta) = k(\phi-\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\Omega}\rightarrow v|_{\partial\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.} \int_{-\infty}^{\infty} \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.