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\markboth{ Anisotropic conductivities } { Gunther Uhlmann }
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\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
USA-Chile Workshop on Nonlinear Analysis, \newline Electron. J.
Diff. Eqns., Conf. 06, 2001, pp. 303--311. \newline
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)}
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Recent progress in the anisotropic electrical impedance problem
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\thanks{ {\em Mathematics Subject Classifications:} 35R30.
\hfil\break\indent
{\em Key words:} Dirichlet to Neumann map, Electrical Impedance Tomography,
\hfil\break\indent Anisotropic conductivities.
\hfil\break\indent
\copyright 2001 Southwest Texas State University.
\hfil\break\indent Published January 8, 2001.
\hfil\break\indent Partly supported by NSF grant DMS--0070488 } }
\date{}
\author{ Gunther Uhlmann }
\maketitle
\begin{abstract}
We survey some recent progress on the problem of determining an
anisotropic conductivity of a medium by making voltage and current
measurements at the boundary of the medium.
\end{abstract}
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\section{Introduction}
We give more details on open problem 5 stated in \cite{UI} which
was only briefly discussed there for lack of space. We also survey
some recent developments on the same problem.
Let $\Omega\subseteq{\mathbb R}^n$ be a bounded domain with smooth boundary. Let
$\gamma=(\gamma^{ij}(x))$ be the electrical conductivity of $\Omega$ which is
assumed to be a positive definite, smooth, symmetric matrix on
$\overline\Omega$. Muscle tissue in the human body is a prime example of an
anisotropic conductivity since the conductivity in the transverse
direction (for cardiac muscle this is 2.3 mho) is quite different
than in the longitudinal direction (for cardiac muscle this is 6.3
mho).
Under the assumption of no sources or sinks of current in $\Omega$,
the equation for the potential, given a voltage potential $f$ on
$\partial\Omega$, is given by the solution of the Dirichlet problem
\begin{equation}\label{eqn1.1}
\begin{array}{c}
\mathop{\sum}\limits^{n}_{i,j=1}\frac{\partial}{\partial
x_i}\left(\gamma^{ij}\frac{\partial u}{\partial x_j}\right) = 0\hbox{ on }\Omega \\
u\big|_{\partial\Omega} = f.
\end{array}
\end{equation}
The Dirichlet-to-Neumann map (DN) is defined by
\begin{equation}\label{eqn1.2}
\Lambda_\gamma(f)=\sum^n_{i,j=1}\nu^i\gamma^{ij}\frac{\partial u}{\partial x_j}\big|_{\partial\Omega}
\end{equation}
where $\nu=(\nu^1, \ldots, \nu^n)$ denotes the unit outer normal
to $\partial\Omega$ and $u$ is the solution of (\ref{eqn1.1}). $\Lambda_\gamma$ is
also called the \emph{voltage} to \emph{current} map since
$\Lambda_\gamma(f)$ measures the induced current flux at the boundary.
The inverse problem is whether one can determine $\gamma$ by knowing
$\Lambda_\gamma$. Calder\'on proposed this problem in \cite{C}. He worked
as an engineer for YPF (Yacimientos Petroleros Fiscales) in
Argentina and he thought of this problem and his contribution
during that time. This problem arises naturally in geophysical
prospection. In fact the Schlumberger-Doll company was founded in
the early part of the century to find oil using electrical
prospection (see \cite{Z} for an account). We are grateful to
Alberto Gr\"unbaum who convinced Calder\'on to publish his result
in 1980 (personal communication). Paul Malliavin in his lecture at
the conference held at the University of Chicago to honor the
$75^{\mathrm{th}}$ birthday of Calder\'on mentioned that Calder\'on told
him of his inverse result in 1954 (see footnote in page 228 of
\cite{Ch}). More recently this inverse problem has been proposed
as a valuable diagnostic tool in medicine (see for instance
\cite{BB}) and it has been called \emph{electrical impedance
tomography} (EIT). Unfortunately, $\Lambda_\gamma$ doesn't determine $\gamma$
uniquely. This observation is due to L.\ Tartar (see \cite{K-V}
for an account). To see this we define first the Dirichlet
integral associated to a solution of (\ref{eqn1.1}). Let
\begin{equation}\label{eqn1.3}
Q_\gamma(f)=\sum^n_{i,j=1}\int_\Omega\gamma^{ij}(x)\frac{\partial u}{\partial x_i} \frac{\partial
u}{\partial x_j}dx
\end{equation}
with $u$ a solution of (\ref{eqn1.1}).
A standard application of the divergence theorem gives that
\begin{equation}\label{eqn1.4}
Q_\gamma(f)=\int_{\partial\Omega}\Lambda_\gamma(f)fdS,
\end{equation}
where $dS$ denotes surface measure in $\partial\Omega$. In other words,
$\Lambda_\gamma$ is the linear operator associated to the quadratic form
$Q_\gamma$ so that $\Lambda_\gamma$ and $Q_\gamma$ carry the same information.
Let $\psi:\overline\Omega\to\overline\Omega$ be a $C^\infty$ diffeomorphism with
$\psi\big|_{\partial\Omega}= \mathrm{Identity}$. Let $v=u\circ\psi^{-1}$. Then a
straightforward calculation shows that $v$ satisfies
\begin{equation}\label{eqn1.5}
\begin{array}{c} \mathop{\sum}\limits^{n}_{i,j=1}
\frac{\partial}{\partial x_i}\left(\widetilde\gamma_{ij} \frac{\partial v}{\partial
x_j}\right) = 0 \\ v\big|_{\partial\Omega} = f
\end{array}
\end{equation}
where
\begin{equation}\label{eqn1.6}
\widetilde\gamma=\left (\frac{(D\psi)^T\circ\gamma\circ(D\psi)}{|\det
D\psi|}\right )\circ\psi^{-1}=: \psi_*\gamma.
\end{equation}
Here $D\psi$ denotes the (matrix) differential of $\psi$,
$(D\psi)^T$ its transpose and the composition in (\ref{eqn1.6}) is
to be interpreted as composition of matrices.
By making the change of variables $v=u\circ\psi^{-1}$ in the
quadratic form (\ref{eqn1.3}) we see that
\begin{equation}\label{eqn1.7}
Q_{\widetilde\gamma}(f)=Q_\gamma(f)
\end{equation}
and therefore $\Lambda_{\widetilde\gamma}=\Lambda_\gamma$.
We have found a large number of conductivities with the same DN
map: any change of variables of $\Omega$ that leaves the boundary
fixed gives rise to a new conductivity with the same electrical
boundary measurements. The question is then whether this is the
only obstruction to unique identifiability of the conductivity. As
we outline below this is a problem of geometrical nature and we
proceed to state it in invariant form.
\section{Geometric Formulation}
Let $(M,g)$ be a compact Riemannian manifold with boundary. The
Laplace-Beltrami operator associated to the metric $g$ is given in
local coordinates by
\begin{equation}\label{eqn1.8}
\Delta_ g u=\frac{1}{\sqrt{\det g}}\sum^n_{i,j=1}\frac{\partial}{\partial
x_i}\left(\sqrt{\det g} g^{ij} \frac{\partial u}{\partial x_j}\right)
\end{equation}
where $(g^{ij})$ is the inverse of the metric $g$. Let us consider
the Dirichlet problem associated to (\ref{eqn1.1})
\begin{equation}\label{eqn1.9}
\begin{array}{c}
\Delta_ g u = 0\hbox{ on }\Omega \\ u\big|_{\partial\Omega} = f
\end{array}
\end{equation}
We define the DN map in this case by
\begin{equation}\label{eqn1.10}
\Lambda_ g(f)=\sum^n_{i,j=1}\nu^ig^{ij} \frac{\partial u}{\partial x_j}\sqrt{\det
g}\big|_{\partial\Omega}
\end{equation}
where $(\nu^i)=\nu$ is the outer unit normal to $\partial\Omega$. The inverse
problem is to recover $g$ from $\Lambda_g$.
By using a similar argument to the one outlined above we have that
\begin{equation}\label{eqn1.11}
\Lambda_{\psi^\ast g} = \Lambda g
\end{equation}
where $\psi$ is a $C^\infty$ diffeomorphism of $\overline M$ which
is the identity on the boundary. As usual $\psi^\ast g$ denotes
the pull back of the metric $g$ by the diffeomorphism $\psi$.
In the case that $M$ is an open, bounded subset of ${\mathbb R}^n$ with
smooth boundary, it is easy to see that (\cite{L-U}) for $n\ge 3$
\begin{equation}\label{eqn1.12}
\Lambda_g=\Lambda_\gamma
\end{equation}
where
\begin{equation}\label{eqn1.13}
g_{ij}=(\det\gamma^{kl})^{\frac{1}{n-2}}(\gamma^{ij})^{-1}, \quad \gamma^{ij}=(\det
g_{kl})^{1/2}(g_{ij})^{-1}.
\end{equation}
In the two dimensional case (1.12) is not valid. In fact in $n=2$
the Laplace-Beltrami operator is conformally invariant. More
precisely
\[
\Delta_{\alpha g}=\frac{1}{\alpha}\Delta_g
\]
for any function $\alpha$, $\alpha\ne 0$. Therefore we have that for $n=2$
\begin{equation}\label{eqn1.14}
\Lambda_{\alpha(\psi^\ast g)}=\Lambda_{g}
\end{equation}
for any smooth function $\alpha\ne 0$ so that $\alpha\big|_{\partial M}=1$.
Now we give an invariant formulation of the EIT problem in the two
dimensional case. In the Euclidean case a current is a one form
given by
\[
i(x)=\gamma(x)d u(x)
\]
where $u$ is the voltage potential. Then, in two dimensions, the
conductivity $\gamma$ can be viewed as a linear map from 1-forms to
1-forms. Now let $(M,g)$ be a two dimensional Riemannian manifold.
Let $\gamma$ be a positive definite symmetric mapping (with respect to
the inner product defined by the metric $g$) from one forms to one
forms. In this case (\ref{eqn1.1}) takes the form
\begin{equation}\label{eqn1.15}
\begin{array}{c} \delta(\gamma du) = 0\hbox{ in }M \\
u\big|_{\partial M} = f
\end{array}
\end{equation}
where $d$ denotes differentiation and $\delta$ codifferentiation
with respect to the metric $g$.
The DN map is given by the 1-form
\begin{equation}\label{eqn1.16}
\Lambda_{g,\gamma}f = \gamma du\big|_{\partial M}.
\end{equation}
An argument similar to the one outlined above shows that
\begin{equation}\label{eqn1.17}
\Lambda_{g,{\psi_\ast\gamma}}=\Lambda_\gamma
\end{equation}
for every diffeomorphism $\psi:\overline M\to\overline M$ which is
the identity at the boundary. Here $\psi_\ast\gamma$ denotes the
push-forward by the diffeomorphism $\psi$ of the one form $\gamma$. We
remark that Riemannian metrics pullback naturally under smooth
maps and conductivities push-forward naturally under smooth maps.
Now we are in position to state the main conjectures.
\medskip
\noindent{\bf Conjecture A} $(n\ge 3)$.
Let $(M,g)$ be a compact Riemannian manifold with boundary. The
pair $(\partial M,\Lambda_ g)$ determines $(M,g)$ uniquely. Of course
uniquely means up to an isometric copy.
\medskip
\noindent{\bf Conjecture B} $(n=2)$.
Let $(M,g)$ be a compact Riemannian surface. Then the pair $(\partial M,
\Lambda_g)$ determines uniquely the conformal class of $(M,g)$.
Uniquely means again up to an isometric copy.
\medskip
\noindent{\bf Conjecture C} $(n=2)$.
Let $(M,g)$ be a compact Riemannian surface with boundary and $\gamma$
a positive definite symmetric map from one forms to one forms on
$M$. Suppose we know $(M, g, \partial M, \Lambda_{g,\gamma})$ with $\Lambda_{g,\gamma}$
defined as in (\ref{eqn1.16}), then we can recover uniquely $\gamma$.
Uniquely means here up to an isometry which is the identity on the
boundary as in (1.6)
\section{The results}
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A basic result which is used in all the anisotropic results stated
below is the following Lemma proved in \cite{L-U}:
\begin{lemma}\label{lem2.1}
(a) $n\ge 3$. Let $(M,g)$ be a compact Riemannian manifold with
boundary. Then $\Lambda_g$ determines the $C^\infty$-jet of the metric
at the boundary in the following sense. If $g'$ is another
Riemannian metric on $M$ such that $\Lambda_g=\Lambda_{g'}$, then there
exists a diffeomorphism $\varphi:M\to M$, $\varphi\big|_{\partial M}
=\mathrm{Identity}$ such that $g'=\varphi^\ast g$ to infinite order at $\partial
M$.
(b) $n=2$. Let $(M,g)$ be a compact Riemannian manifold with
boundary. Then $\Lambda_g$ determines the conformal class of the
$C^\infty$-jet of the metric at the boundary.
(c) $n=2$. Let $(M,g)$ be a compact Riemannian surface with
boundary. Let $\gamma$ be a positive definite symmetric map from one
forms to one forms. Then the mapping $\Lambda_{g,\gamma}$, as defined in
(\ref{eqn1.6}), determines the $C^\infty$-jet of the map $\gamma$ at
the boundary in the following sense: If $\gamma'$ is another such one
form such that $\Lambda_{g,\gamma}=\Lambda_{g,\gamma'}$. Then there exists a
diffeomorphism $\varphi:M\to M$, $\varphi\big|_{\partial M}=\mathrm{Identity}$ such
that $\gamma'=\varphi_\ast \gamma$ to infinite order at $\partial M$.
\end{lemma}
In other words Lemma \ref{lem2.1} shows that Conjectures A, B, C
above are valid at the boundary. The proof of this result is done
in case a) by showing that $\Lambda_g$ is a pseudodifferential operator
of order 1. Its full symbol, calculated in appropriate
coordinates, determines the $C^\infty$-jet of the metric $g$ at
the boundary. The proofs of b) and c) are similar.
The only case of Conjecture A that has been settled in general is
the isotropic case in Euclidean space. Namely we have in the case
that $M=\Omega$ an open, bounded subset of ${\mathbb R}^n$ with a smooth
boundary and the metric $g$ is given by
\begin{equation}\label{eqn2.1}
g_{ij}=\alpha(x)\delta_{ij},\qquad\alpha>0
\end{equation}
where $\delta_{ij}$ is the Kr\"onecker delta.
Suppose $g^{(1)}$, $g^{(2)}$ are two isotropic Riemannian metrics
\begin{equation}\label{eqn2.2}
g^{(i)}=\alpha^i(x)(\delta_{kl})\qquad i=1,2,\qquad\alpha^i>0.
\end{equation}
Then it is straightforward to show that if $\psi^\ast g_1=g_2$,
$\psi\big|_{\partial\Omega}=\mathrm{Identity}$, then $\psi=\mathrm{Identity}$. So the Conjecture A
in this case is that $g_1=g_2$. This was proven in \cite{S-U}:
\begin{theorem}\label{thm1.1}
Let $\Omega\subseteq{\mathbb R}^n$ $n \ge 3$ be a bounded domain with smooth
boundary. Let $g^{(i)}$, $i=1,2$ be two isotropic Riemannian
manifolds satisfying (\ref{eqn2.2}). Then $\Lambda_{g_1}=\Lambda_{g_2}$
implies $g_1=g_2$.
\end{theorem}
We won't outline the proof here. We mention that a crucial
ingredient in the proof is the construction of complex geometrical
optics solutions of the Laplace-Beltrami operator when the
Riemannian metric is isotropic. More precisely
\begin{lemma}\label{lem2.2}
Let $g$ be an isotropic Riemannian metric as in (\ref{eqn2.1}),
with $\alpha=1$ outside a large ball. Let $\rho\in{\mathbb C}^n$,
$\rho\cdot\rho=0$. Then for $|\rho|$ sufficiently large, there
exist solutions of $\Delta_gu=0$ of the form
\begin{equation}\label{eqn2.3}
u=e^{x\cdot\rho}\alpha^{-\frac{1}{2}}(1+\psi_g(x,\rho))
\end{equation}
with $\psi_g\mathop{\longrightarrow}\limits_{|\rho|\to\infty} 0$
uniformly in compact sets.
\end{lemma}
For more precise statements and a recent survey of other results
using complex geometrical optics solutions, see \cite{UII}.
One of the main difficulties in extending Theorem \ref{thm1.1} to
the general anisotropic case even in the case when $M$ is an open
subset of Euclidean space is to construct an analog of
(\ref{eqn2.3}) for the Laplace-Beltrami operator.
Lassas and the author ([M-U]) proved Conjecture A in the
real-analytic case and Conjecture B in general. Moreover these
results assume that $\Lambda_g$ is measured only on an open subset of
the boundary.
Let $\Gamma$ be an open subset of $\partial M$. we define for $f$, $\mathop{\rm supp}
f\subseteq\Gamma$
\[
\Lambda_{g,\Gamma}(f)=\Lambda_g(f)\big|_{\Gamma}.
\]
The first result of \cite{L-U} is:
\begin{theorem}[$n\ge 3$]\label{thm2.2}
Let $(M,g)$ be a real-analytic compact, connected Riemannian
manifold with boundary. Let $\Gamma\subseteq\partial M$ be real-analytic and
assume that $g$ is real-analytic up to $\Gamma$. Then $(\Lambda_{g,\Gamma}, \partial
M)$ determines uniquely $(M,g)$.
\end{theorem}
Notice that Theorem \ref{thm2.2} doesn't assume any condition on
the topology of the manifold except for connectedness. An earlier
result of \cite{L-U} assumed that $(M,g)$ was strongly convex and
simply connected and $\Gamma=\partial M$.
The second result of \cite{M-U} is the proof of Conjecture B
assuming we only measure the DN map on an open subset of the
boundary.
\begin{theorem}[$n=2$]\label{thm2.3}
Let $(M,g)$ be a compact Riemannian surface with boundary. let
$\Gamma\subseteq\partial M$ be an open subset. Then $(\Lambda_{g,\Gamma}, \partial M)$
determines uniquely the conformal class of $(M,g)$.
\end{theorem}
\noindent{\bf Sketch of proof of Theorems \ref{thm2.2} and
\ref{thm2.3}.} We'll sketch the proof of Theorem \ref{thm2.3}.
Theorem \ref{thm2.2} follows along similar lines. Using Lemma
\ref{lem2.1} we know that $\Lambda_g$ determines $g\big|_{\partial M}$.
We add to $M$ a collar neighborhood to construct $\widetilde
M=M\cup(\partial M\times[0,1])$ with the metric given on $\partial
M\times[0,1]$ by
\[
g\big|_{\partial M\times[0,1]} = g\big|_{\partial M} + ds^2.
\]
With this definition $g\in C^{0,1}(\widetilde M)$. The Green's
kernel is defined by
\[
\begin{array}{c}
\Delta_g h_y = \delta_y\hbox{ in }\widetilde M \\
h_y\big|_{\partial\widetilde M} = 0
\end{array}
\]
It is proven in [M-U] that the DN map determines the Green's
functions in the collared neighborhood. More precisely we have:
\begin{lemma}\label{lem2.3}
$\Lambda_g$ determines $h_y(x),$ $x, y\in\widetilde M-M$.
\end{lemma}
In two dimensions there are special coordinates that change any
Riemannian metric to a conformal multiple of the Euclidean one.
These are called isothermal coordinates ([A]). Given any point
$x\in M$, there exists a coordinate system $(U,\phi)$,
$\phi:U\subseteq M\to{\mathbb R}^2$ so that
\begin{equation}\label{eqn2.4}
g\circ\phi^{-1}=\alpha(x)(\delta_{ij})
\end{equation}
that is, the metric $g$ is isotropic in these coordinates.
Let $V$ be an open neighborhood of $\partial\widetilde M$ so that
$\partial\widetilde M\subseteq V\subseteq\widetilde M$. A fundamental
step in the proof is to show the following result that states,
roughly speaking that we can use the Green's functions based on
points of $V$ as coordinates.
\begin{lemma}\label{lm3.4}
Given any point $x \in \overline M$, there exists a neighborhood
$U$ of $x$ and points $y_1, y_2\in V$ so that
$H_{y_1,y_2}=(h_{y_1}, h_{y_2})$ form a coordinate system on $U$.
\end{lemma}
The next observation is that $H_{y_1, y_2}$ are real-analytic in
isothermal coordinates on $M$. These follow since the Laplacian in
two dimensions is conformally invariant and therefore
\[
\Delta h_y\circ\phi^{-1}=0\hbox{ on }\phi(U)\hbox{ if }y\in\widetilde
M-M
\]
and harmonic functions are real-analytic. Let us take a point
$x\in \overline M$. Then we find a coordinate system $(U,\phi)$
near $x$ and $y_1, y_2\in V$ so that $H_{y_1,y_2}$ is
real-analytic in these coordinates.
Now we continue analytically $h_y, y\in V$ in these coordinates as
much as possible. When this is no longer possible we use Lemma
\ref{lm3.4} to find new points $\widetilde y_1, \widetilde y_2\in
V$ so that $H_{\widetilde y_1, \widetilde y_2}$ is a system of
coordinates and we continue this analytic continuation process
again. This is done in \cite{M-U} using the theory of sheaves. Let
$\cal A$ be the sheaf of sequences of real-analytic maps. We
define an equivalent class $\cal B$ in this sheaf by identifying
elements that are obtained from each other by using real-analytic
diffeomorphisms. Let $p \in \cal B$ be the element corresponding
to the germs of the Green's kernel at a point $x\in \widetilde
M-M$. The isometric copy of the manifold $(M, g)$ is constructed
by taking the path connected component of $\cal B$ containing the
point $p.$
As for Conjecture C the only known result is the case when $M=\Omega$
is an open subset of ${\mathbb R}^n$ with smooth boundary and
$g=(\delta_{ij})=:e$ is the Euclidean metric. More precisely we have
\begin{theorem}[$n=2$]\label{thm2.4}
Let $\Omega\subseteq{\mathbb R}^n$ be a bounded domain with smooth boundary. Let
$\gamma_1, \gamma_2$ be two anisotropic conductivities so that
\[
\Lambda_{e, \gamma_1}=\Lambda_{e, \gamma_2}.
\]
Then there exists $\psi:\overline\Omega\to\overline\Omega$ diffeomorphism
$\psi\big|_{\partial\Omega}=\mathrm{Identity}$ so that
\[
\psi_\ast \gamma_1=\gamma_2.
\]
\end{theorem}
The proof of Theorem \ref{thm2.4} is a combination of the results
of \cite{N} and \cite{S}. In \cite{N} it was proven Theorem
\ref{thm2.4} for isotropic conductivities. Then one uses the
results of \cite{S} to reduce the anisotropic case to the
isotropic one by using the analog of isothermal coordinates in
this case. The result is that given an anisotropic conductivity,
we can find a diffeomorphism $\phi$ so that $\phi_\ast\gamma$ is
isotropic. We end by mentioning that the result of \cite{N} uses
the complex geometrical solutions, (\ref{eqn2.3}) for all complex
frequencies $\rho\in{\mathbb C}^n-0$, $\rho\cdot\rho=0$ (not just large
frequencies). For another construction of these solutions which
allow Lipschitz conductivities see \cite{B-U} (the result of
\cite{N} works for $C^2$ conductivities). Theorem \ref{thm2.4} has
been extended to anisotropic non-linear conductivities in
\cite{Su-U}.
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\noindent{\sc Gunther Uhlmann }\\
Department of Mathematics\\
University of Washington \\
Box 354350\\
Seattle, WA 98195, USA\\
e-mail: gunther@math.washington.edu
\end{document}