# On 2D Inverse Problems/The case of the unit disc

### The 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:

$\Lambda^2=-\Delta_{S^{1}}.$

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?"

### The network settingEdit

To match the functional equation for the Dirichlet-to-Neumann operator of the unit disc with conductivity 1, one needs to find a self-dual layered planar network with rotational symmetry. The Dirichlet-to-Neumann operator for such graph G is equal to:

$\Lambda^2_G = L,$

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

$L = \begin{pmatrix} 2 & -1 & 0 & \ldots & -1 \\ -1 & 2 & -1 & \ldots & 0 \\ 0 & -1 & \ddots & \ddots & \vdots \\ \vdots & \vdots & \ddots & 2 & -1 \\ -1 & 0 & \ldots & -1 & 2 \\ \end{pmatrix} .$
Exercise(*). Prove that the entries of the cofactor matrix of $\Lambda_G$ are ±1 w/the chessboard pattern.

The problem of finding the graph G 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, where n is a natural number, the eigenvalues are 0 with the multiplicity 1 and

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

with the multiplicity 2. The existence and uniqueness of such fraction with n levels follows from our results on layered networks, see [BIMS].

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

Example: the following picture provides the solution for n=8 w/white and black squares representing 1s and -1s.

Correct signs