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?



Which matrices can be realized as the Dirichlet-to-Neumann map for a given domain?
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 $\lambda=0$.


