# 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 $\Omega$ is its kernel. It is a distribution defined on the Cartesian product of the boundary of the domain w/itself, such that if $\Lambda f = g,$ then $g(\phi) = \int_{\partial\Omega}K(\phi,\theta)f(\theta)d\theta,$

where $\phi$ and $\theta$ parametrize the boundary w/arclength measure.

For the case of a half-plane w/unit uniform conductivity one can calculate the kernel explicitly. The kernel is 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 normal derivative at the boundary line.

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 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 C+ the Hilbert transform is given by the following formula: $H_{C^+}f(y) =\frac{1}{\pi} \ \text{p.v.} \int_{-\infty}^{\infty} \frac{f(x)}{y-x}dx.$

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

To define discrete Hilbert transform for a planar network, one needs to consider the network together w/its dual.