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The Two-Dimensional Inverse Conductivity Problem

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Abstract

In this article, we introduce a process to reconstruct a Riemann surface with boundary equipped with a linked conductivity tensor from its boundary and the Dirichlet–Neumann operator associated with this conductivity. When initial data come from a two- dimensional real Riemannian surface equipped with a conductivity tensor, this process recovers its conductivity structure.

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Notes

  1. We think of a surface with boundary M as a dense open subset of an oriented two-dimensional real manifold with boundary \({\overline{M}}\) whose all connected components are bounded by pure one-dimensional real manifolds ; so the topological boundary bM of M is \({\overline{M}}\backslash M\) ; in the sequel \(\partial M\) is bM equipped with the natural orientation induced by M. A Riemann surface with boundary is a connected complex manifold of dimension 1 which is also a real surface with boundary.

  2. If we fix a point p in \({\overline{M}}\), some coordinates \(\left( x,y\right) \) around p and we set as in [29] \(\left( \xi ,\eta \right) =\left( \mathrm{{d}}y,-\mathrm{{d}}x\right) \) then \(\sigma \left( \mathrm{{d}}x\right) =r\xi +t\eta \) and \(\sigma \left( \mathrm{{d}}y\right) =u\xi +s\eta \), for \(a=a_{x}\mathrm{{d}}x+a_{y}\mathrm{{d}}y\) and \(b=b_{x}\mathrm{{d}}x+b_{y}\mathrm{{d}}y\) in \(T_{p}{\overline{M}}\), \(\sigma _{p}\left( b\right) =\left( b_{x} r+b_{y}u\right) \xi +\left( b_{x}t+b_{y}s\right) \eta \) and

    $$\begin{aligned} a\wedge \sigma _{p}\left( b\right)&=\left( a_{x}\mathrm{{d}}x+a_{y}\mathrm{{d}}y\right) \wedge \left[ \left( b_{x}r+b_{y}u\right) \mathrm{{d}}y-\left( b_{x}t+b_{y}s\right) \mathrm{{d}}x\right] \\&=\left( ra_{x}b_{x}+ua_{x}b_{y}+ta_{y}b_{x}+sb_{x}b_{y}\right) \mathrm{{d}}x\wedge \mathrm{{d}}y. \end{aligned}$$

  3. Since \(B^{-}\cap \delta =\varnothing \), \(B^{-}=\left( B^{-}\cap Y\right) \cup \left( B^{-}\backslash {\overline{Y}}\right) \). \(B^{-}\cap Y\) is an open subset \(B^{-}\) because by construction, \(B^{-} \subset {\mathrm{Reg\,}}{\overline{Y}}\cap {\mathrm{Reg\,}}\overline{Y^{\prime }}\). It is non-empty by hypothesis. Hence \(B^{-}=B^{-}\cap Y\subset Y\).

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Acknowledgements

I would like to thank the referees for their careful reading and suggestions.

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  1. Sorbonne Université, 4 place Jussieu, 75252, Paris Cedex, France

    Vincent Michel

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Correspondence to Vincent Michel.

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In January 2016, my friend Gennadi Henkin, with whom I had worked for more than fifteen years, passed away. This paper, which is on a subject he brought, is dedicated to him. The numerous citations from the articles he authored show the depth of his mathematical thought.

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Michel, V. The Two-Dimensional Inverse Conductivity Problem. J Geom Anal 30, 2776–2842 (2020). https://doi.org/10.1007/s12220-018-00139-2

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