# User:Daviddaved/Layered discretization

The continued fraction approach to the inverse problems can be applied to domains w/inhomogenuous isotropic conductivity in 2D and higher dimensions using layered discretization. The resulting Dirichlet-to-Neumann operator can be written as a ratio of two high order differential operators that satisfy three-term recurrence (similar to the numerators and denominators of functional continued fraction). The fundumental solutions of the differential operators can be directly read from the kernel of the Dirichlet-to-Neumann operator. The conductivity can be then found by a Eucledian type algorithm, reversing the three-term recurrence.

More formally: Let f be the potential and g be the current on a layered domain. Then, on the k'th layer:

$\begin{cases} g_{-1} = 0, \\ g_0 = D\alpha_0 D f_0, \\ f_{k+1}= f_k + g_k/\beta_k,\\ g_{k} = g_{k-1} + D\alpha_k Df_k. \end{cases} \mbox{ In the dual domain, } \begin{cases} f_0 = \frac{\int\beta_0 f_1}{\int\beta_0}, \\ g_0 = \beta_0 (f_1 - \frac{\int\beta_0 f_1}{\int\beta_0}), \\ f_{k+1}= f_k + g_k/\beta_k,\\ g_{k} = g_{k-1} + D\alpha_k Df_k. \end{cases}$

Therefore, for ordinary differential operators F and G,

$\begin{cases} f_k=F_k f_0, \\ g_k=G_k f_0, \\ \Lambda_k = G_k F_k^{-1} = (F_k^*)^{-1}G_k^*, \end{cases}$

since, the Dirichlet-to-Neumann operator is self-adjoint. And

$\begin{cases} F_k^* G_k = G_k^* F_k, \\ G_k^*f_k = F_k^*g_k, \\ G_k = \Lambda_k F_k. \end{cases}$

Exercise (**). Prove that the eigenvalues of the monodromy matrices of the operators F and G are simple, positive and interlace.

Since, the operator G is differential, the support of the function Gf belongs to the closure of the support of the potential f, which allows one to read the solutions of the equation F*g=0 from the Dirichlet-to-Neumann operator and G*f=0 from the dual problem. (Need to find a family of fundamental solutions). The coefficients of the differential operators F, G, F* and G* can be then obtained from the Wronskians of the fundamental solutions.

One also gets from the systems above that,

$\begin{cases} F_0 = Id, \\ G_0 = \alpha_0 D^2+(D\alpha_0)D, \\ F_{k} = \frac{\alpha_0\ldots,\alpha_{k-1}}{\beta_0\ldots\beta_{k-1}}D^{2k}+\ldots, k > 0 \mbox{ and } \\ G_{k} = \frac{\alpha_0\ldots,\alpha_{k}}{\beta_0\ldots\beta{k-1}}D^{2k+2}+\ldots, \end{cases}$

where $\alpha$ and $\beta$ are conductivities of the layers. Therefore, one can find the conductivity of the outmost layer as a ratio of the leading coefficients of the corresponding operators F an G, which allows one to reverse the three-term recurrence and find conductivities of all layers by induction.