# 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

$\Lambda ({\tilde {K}})=A+D_{y}-BD_{x}(D_{x}(C+D_{q})D_{x})^{{-1}}D_{x}B^{T}=\Lambda _{q}+D_{y},$

w/

$x=-(C+D_{q})^{{-1}}B^{T}1.$

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).