Last modified on 5 August 2014, at 11:15

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,


g(\phi) = \int_{\partial\Omega}K(\phi,\theta)f(\theta)d\theta,

where \phi and \theta parametrise the boundary.

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.

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 harmonic function and its harmonic conjugate.

H:u|_{\partial\Omega}\rightarrow v|_{\partial\Omega},


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.

To define discrete Hilbert transform for a planar network, one considers the corresponding graph together w/its dual.