# User:Daviddaved/Schrodinger equation

The Schrodinger equation provides a link between the local and spectral/global properties of solutions of Laplace-Beltrami equation.

The inverse boundary problem for the Schrodinger equation can be reduced to the Calderon problem due to the identities below that hold for graphs and surfaces. Suppose u on $\Omega$ satisfies the Laplace equation in the domain,

$\Delta_{\gamma}u = \nabla\cdot(\gamma\nabla u) = 0.$

Then

$(\Delta - q)(u\sqrt{\gamma}) = 0,$

where,

$q = \frac{\Delta\sqrt{\gamma}}{\sqrt{\gamma}}.$

For the analog of this system to work on networks, one can define the solution of the Schrodinger equation u on the nodes and the square of the solution on the edges by the following formula:

$\gamma^2(v_l,v_m) = u(v_l)u(v_m).$

Exercise (*). Express the Dirichlet-to-Neumann operator for the Schrodinger equation in terms of the Dirichlet-to-Neumann operator for the corresponding Laplace equation on the network with the same underlying graph.

(Hint). Let

$\Lambda_q = A-B(C+D_q)^{-1}B^T,$

where,

$K = \begin{pmatrix} A & B \\ B^T & C + D_q \end{pmatrix} .$

Then

$\tilde{K} = \begin{pmatrix} A+D_y & BD_x \\ D_x B^T & D_x(C+D_q)D_x \end{pmatrix}$

is the Laplace matrix of the network with

$\Lambda(\tilde{K}) = A + D_y - B D_x (D_x (C+D_q) D_x)^{-1} D_x B^T = \Lambda_q + D_y,$

w/

$x = - (C+D_q)^{-1}B^T1.$

Exercise (**). Reduce the inverse problem for Schrodinger operator to the inverse problem for the Laplace operator on the network w/same underlying graph (w/ possibly signed conductivity).