# User:DVD206/The case of the unit disc

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

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

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

${\displaystyle \Lambda (\Lambda -(N-2)Id)+\Delta _{S^{N-1}}=0}$.

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

${\displaystyle \Lambda ^{2}=-\Delta _{S^{1}}.}$

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?"

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,

${\displaystyle \Lambda ^{2}=-\Delta _{R^{N-1}}.}$

### Network case

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:

${\displaystyle \Lambda ^{2}(G)=L,}$

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

${\displaystyle L={\begin{bmatrix}-2&-1&0&\ldots &-1\\-1&2&-1&\ldots &0\\0&-1&\ddots &\ddots &\vdots \\\vdots &\vdots &\ddots &2&-1\\-1&0&\ldots &-1&2\\\end{bmatrix}}.}$

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

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

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:

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

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

${\displaystyle \tan({\frac {n\pi }{2n+1}})=2\sum _{k}\sin({\frac {k\pi }{2n+1}}).}$

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

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