# On 2D Inverse Problems/Electrical networks

< On 2D Inverse Problems(Redirected from User:Daviddaved/Electrical networks)An **electrical network** can be defined as a weighted graph with boundary

The weight function defined on the edges of the graph is called **conductivity**. The Laplacian matrix of an electrical network is symmetric. It is often called Kirchhoff matrix. The boundary measurements for inverse problems can be expressed conveniently in terms of the matrix blocks.

The Dirichlet-to-Neumann map is a special type of Poincaré–Steklov operator. On a surface w/boundary it is the **pseudo-differential operator** from the Dirichlet boundary values (potential) to the Neumann boundary values (current) of the harmonic functions. It is well-defined because of uniqueness and existence of the solution of the Dirichlet problem.

**Exercise (*).**Prove that the Dirichlet-to-Neumann map of an electrical network equals to the Schur complement of its Kirchhoff matrix.

The Laplace equation gives the direct connection between the hitting probability of the random walk started at the boundary and the value of a harmonic function at a vertex/point, see [10]. The connection follows from the sum of the geometric series identity applied to the blocks of the Kirchhoff matrix of the network/graph

This is a special case of the convergent geometric Neumann series applied to the diagonally dominated matrix.

The **effective conductivity** b/w two nodes of an electrical network *G* equals to the ratio b/w the total current and the difference of potentials b/w the two nodes. The effective conductivities can be calculated in terms of the Schur complement of the Kirchhoff matrix of the network.

**Exercise (*)**. Prove that the effective conductivities b/w all pairs of boundary nodes of a network *G*. determine the Dirichlet-to-Neumann map of *G*, and vice versa.

**Exercise (***)**. Prove that the effective conductivities b/w all pairs of nodes of a network *G* determine the conductivities of the edges of the network *G*, and vice versa.