User:Daviddaved/The case of the unit disc

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Operator equationEdit

The continuous Dirichlet-to-Neumann operator can be calculated explicitly for certain domains, such as a half-space, a ball and a cylinder and a shell with uniform conductivity 1. For example, for a unit ball in N-dimensions, writing the Laplace equation in spherical coordinates one gets:

\Delta f = r^{1-N}\frac{\partial}{\partial r}\left(r^{N-1}\frac{\partial f}{\partial r}\right) + r^{-2}\Delta_{S^{N-1}}f,

and, therefore, the Dirichlet-to-Neumann operator satisfies the following equation:

\Lambda(\Lambda-(N-2)Id)+\Delta_{S^{N-1}} = 0.

In two-dimensions the equation takes a particularly simple form:


The study of material of this chapter is largely motivated by the question of Professor of Mathematics at the University of Washington Gunther Uhlmann: "Is there a discrete analog of the equation?"

Exercise (**): Prove that for the unit ball the Dirichlet-to-Neumann operator satisfies the quadratic equation above.

Exercise (*): Prove that for the Dirichlet-to-Neumann operator of a half-space of RN with uniform conductivity 1,

\Lambda^2 = -\Delta_{R^{N-1}}.

Network caseEdit

To match the functional equation that the Dirichlet-to-Neumann operator of the unit disc with conductivity 1 satisfies, one would need to look for a self-dual layered planar network with rotational symmetry. The Dirichlet-to-Neumann map for such graph should be equal to:

 \Lambda^2(G) = L,

where -L is equal to the Laplacian on the circle:

L =
 -2     & -1 & 0 & \ldots & -1 \\
 -1     & 2 & -1 & \ldots & 0 \\
 0     & -1 & \ddots & \ddots & \vdots \\
 \vdots    & \vdots & \ddots & 2 & -1 \\
 -1     & 0  & \ldots & -1 & 2 \\

The problem then reduces to finding a Stieltjes continued fraction that is equal to 1 at the non-zero eigenvalues of L. For the (2n+1)-case the eigenvalues are 0 with multiplicity 1 and

 2\sin(\frac{k\pi}{2n+1}), k = 1,2,\ldots n

with multiplicity 2. The existence and uniqueness of such fraction with n floors follows from our results on layered networks.

Exercise (***). Prove that the continued fraction is given by the following formula:

\beta(z) = \cot(\frac{n\pi}{2n+1}) z + \cfrac{1}{\cot(\frac{(n-1)\pi}{2n+1})z + \cfrac{1}{ \ddots + \cfrac{1}{\cot(\frac{\pi}{2n+1}) z} }}.

Exercise 2 (*). Use the previous exercise to prove the trigonometric formula:

\tan(\frac{n\pi}{2n+1}) = 2\sum_k\sin(\frac{k\pi}{2n+1}).

Exercise 3(**). Find the right signs in the following trigonometric formula

\tan(\frac{l\pi}{2n+1}) = 2\sum_k(\pm)\sin(\frac{k\pi}{2n+1}), l = 1,2,\ldots n.