On 2D Inverse Problems/Schrodinger equation

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The conductivity equation \Delta_{\gamma}u = \nabla\cdot(\gamma\nabla u) = 0 
is equivalent to Schrodinger equation: (\Delta - q)(u\sqrt{\gamma}) = 0
for potential q = \frac{\Delta\sqrt{\gamma}}{\sqrt{\gamma}}
For an analog of this system on e-networks, one defines the solution of the Schrodinger equation u on the nodes and the square of the conductivity on the edges by the following formula: \gamma^2(a,b) = x(a)x(b).
Exercise (*). Express the DN operator for the Schrodinger equation in terms of the one for the conductivity equation on the same e-network.

(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.

Exercise (**). Reduce the inverse problem for the Schrodinger equation on e-nerwork to the conductivity equation one w/signed conductivity.