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.

Last modified on 15 April 2013, at 01:09