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

The operator equation

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. For example, for a unit ball in N-dimensions, writing the Laplace equation in spherical coordinates:

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

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 setting

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

$\Lambda _{G}^{2}=L,$

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 then reduces to calculating a Stieltjes continued fraction equalled 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 one and

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

w/multiplicity two. The existence and uniqueness of such fraction with n levels follow 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