# 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 ${\displaystyle \Omega }$ is its kernel. It is a distribution defined on the Cartesian product of the boundary of the domain w/itself, such that if


${\displaystyle \Lambda f=g,}$ then ${\displaystyle g(\phi )=\int _{\partial \Omega }K(\phi ,\theta )f(\theta )d\theta ,}$ where ${\displaystyle \phi }$ and ${\displaystyle \theta }$ 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:


${\displaystyle K(\phi ,\theta )=k(\phi -\theta ),}$ where k is a distribution on a line. Therefore, the calculation reduces to solving the Dirichlet problem for a ${\displaystyle \delta _{0}}$-function at the origin and taking outward derivative at the boundary line.

Exercise (**). Complete the calculation of the kernel K for the half-plane to show that ${\displaystyle K(x,y)={\frac {-1}{\pi (x-y)^{2}}}}$ 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. ${\displaystyle H:u|_{\partial \Omega }\rightarrow v|_{\partial \Omega },}$ where ${\displaystyle f(z)=u(z)+iv(z)}$ 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: ${\displaystyle Hf(y)={\frac {1}{\pi }}\ {\text{p.v.}}\int _{-\infty }^{\infty }{\frac {f(x)}{y-x}}dx.}$

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.