Which matrices can be realized as the Dirichlet-to-Neumann map for a given domain?

Question

Consider Poisson equation $\nabla \cdot (\sigma(x)\nabla u)=0$ in a domain $D$, where $\sigma(x)$ is the spatially dependent conductivity. On the boundary we have $n$ electrodes (Dirichlet BC $u=\text{const}$ on each electrode). And the rest of the boundary is insulating material $du/d\vec n=0$ (Neumann BC). The electrodes do not have any contact impedance.

The Dirichlet-to-Neumann map or Poincare–Steklov operator is the map from voltages to total current on electrodes for a given conductivity distribution (and is a linear map according to Ohm's law and Kirchoff's law (and therefore a matrix)). Given an arbitrary $n\times n$ matrix, is there a way to check if there exist a conductivity distribution which has this matrix as a Dirichlet-to-Neumann map?

I am looking for a computationally fast one hopefully, I don't want to compute the conductivity distribution.

Maybe there are some domain independent restrictions? (I know there are because I've seen them before but I can't find it again). But I was hoping you could improve them when you know the domain. Further I also know that the conductivity is bounded above by some constant. Is there a way to find what restriction on the DtN map this gives?

Answer 1

Let $A$ be this matrix. Because of the formula $$\int_D\sigma\nabla u\cdot\nabla v\, dx=\sum_{i,j}a_{ij}U_iI_j,$$ ($U$ for voltages of $u$, $I$ for currents of $v$), we see three necessary conditions:

• the matrix must be symmetric,
• it must be positive semi-definite,
• and $A{\bf1}=0$.

Actually, if $U\ne \mu{\bf1}$, then $u$ is not constant and the choice $v=u$ yields that $\ker A$ is spanned by ${\bf 1}$.

My guess is that every matrix satisfying these restrictions can be such a Dirichlet-to-Neumann map.

Edit. I see another constraint. For the $i$th column is given by the currents $I_j$ when the potential is $1$ on the $i$th electrod and $0$ on the others. By the maximum principle, $0<u<1$ everywhere in $D$. Select a $j\ne i$. Because it achieves its minimum $0$ along the $j$th electrod, the maximum principle tells us that $\nabla u\cdot\vec n$ is $<0$ along it, and therefore $a_{ij}=I_j<0$. To summarize:

the off-diagonal entries of $A$ must be $<0$.

Remark. The facts that $A$ is symmetric, has negative off-diagonal entries, and $A{\bf1}=0$, imply alltogether that $A$ is positive semi-definite and has rank $n-1$. Just apply Perron-Frobenius: the least eigenvalue $\lambda$ of $A$ is simple and is the only one associated with a positive eigenvector. Therefore $\lambda=0$.

Answer 2

In dimensions 3 and higher, and without any constraints on $\sigma$, one can apparently obtain any symmetric matrix $A = (a_{ij})$ such that $a_{ij} < 0$ when $i \ne j$ and $a_{ii} = -\sum_{j \ne i} a_{ij}$, as suggested in Denis Serre's answer. That answer already explains why these conditions are necessary. To see that they are also sufficient, one can approximate electrodes connected by non-crossing wires of given conductance (in the limiting case $\sigma$ is zero outside the wires, and constant in each wire).

In dimension 2, I believe, there are additional geometric constraints (for example, it is not possible to connect the electrodes with non-crossing wires). I bet this has been studied, but I do not know any references.

Answer 3

This question (for two dimensional domains) was answered by Curtis, Ingerman and Morrow, "Circular Graphs and planar Resistor Networks" (1998). Let $a$ be the $n \times n$ response matrix. As already noted, we must have $a_{ij} = a_{ji}$ and $a(1\ 1\ \cdots\ 1)^T=0$. The additional condition is that, if $i_1$, $i_2$, ..., $i_k$ and $j_1$, $j_2$, ..., $j_k$ appear in circular order around the boundary of the network, and $A'$ is the submatrix with rows $i_r$ and columns $j_s$, then $(-1)^k \det A' \geq 0$.