2023

1]Department of Mathematics, University of Manchester, Oxford Road,
Manchester, M13 9PL, Greater Manchester, UK

2]Department of Mathematics, University of Manchester, Oxford Road,
Manchester, M13 9PL, Greater Manchester, UK

# Boundary Recovery of Anisotropic Electromagnetic Parameters for the Time
Harmonic Maxwell’s Equations

Sean Holman<EMAIL_ADDRESS>Vasiliki Torega
<EMAIL_ADDRESS>[  [

###### Abstract

This work concerns inverse boundary value problems for the time-harmonic
Maxwell’s equations on differential $1-$forms. We formulate the boundary value
problem on a $3-$dimensional compact and simply connected Riemannian manifold
$M$ with boundary $\partial M$ endowed with a Riemannian metric $g$. Assuming
that the electric permittivity $\varepsilon$ and magnetic permeability $\mu$
are real-valued anisotropic (i.e $(1,1)-$ tensors), we aim to determine
certain metrics induced by these parameters, denoted by $\hat{\varepsilon}$
and $\hat{\mu}$ at $\partial M$. We show that the knowledge of the impedance
and admittance maps determines the tangential entries of $\hat{\varepsilon}$
and $\hat{\mu}$ at $\partial M$ in their boundary normal coordinates, although
the background volume form cannot be determined in such coordinates due to a
non-uniqueness occuring from diffeomorphisms that fix the boundary. Then, we
prove that in some cases, we can also recover the normal components of
$\hat{\mu}$ up to a conformal multiple at $\partial M$ in boundary normal
coordinates for $\hat{\varepsilon}$. Last, we build an inductive proof to show
that if $\hat{\varepsilon}$ and $\hat{\mu}$ are determined at $\partial M$ in
boundary normal coordinates for $\hat{\varepsilon}$, then the same follows for
their normal derivatives of all orders at $\partial M$.

###### keywords:

Maxwell’s equations, Inverse Problems, Anisotropic, Electromagnetic
Parameters, Boundary Normal Coordinates, Pseudodifferential Operators

##  1 Introduction

In this work, we consider a $3-$dimensional compact and simply connected
Riemannian manifold $M$, with smooth boundary $\partial M$, equipped with a
Riemannian metric $g$. In the case of $M$ $\subset$ $\mathbb{R}^{3}$, $g$
could be the Euclidean metric. The coordinate invariant formulation of time
harmonic Maxwell’s equations, in the absence of currents or sources, in $M$ is
provided by

$\begin{split}\ast_{g}\mathrm{d}H=-i\omega\varepsilon
E,\quad\delta_{g}(\varepsilon E)=0,\\\ \ast_{g}\mathrm{d}E=i\omega\mu
H,\quad\delta_{g}(\mu H)=0,\end{split}$ (1)

where $\ast_{g}$ is the Hodge star operator with respect to the metric $g$ and
$\mathrm{d}$ stands for the exterior derivative. The operator $\delta_{g}$
denotes the divergence operator with respect to the metric $g$. The electric
and magnetic fields $E$ and $H$, respectively, are differential $1-$forms. The
electric permittivity $\varepsilon$ and magnetic permeability $\mu$ are
assumed to be real-valued $(1,1)-$tensor fields which are symmetric and
positive definite with respect to the metric $g$. The angular frequency is
represented by $\omega$ and is a real number.

We consider either of the Dirichlet boundary conditions

$\iota^{\ast}H=F\quad\mbox{or}\quad\iota^{\ast}E=G,$ (2)

where $\iota^{\ast}$ stands for the pullback at $\partial M$. Given that the
Dirichlet problem composed of (1) and either boundary condition in (2) is
well-posed, the boundary mappings $\Lambda_{\varepsilon}$ and $\Lambda_{\mu}$,
called the impedance and admittance maps respectively are well defined and are
provided by

$\Lambda_{\varepsilon}(\iota^{\ast}H)=\iota^{\ast}E\quad\mbox{and}\quad\Lambda_{\mu}(\iota^{\ast}E)=\iota^{\ast}H.$

Note that $\Lambda_{\mu}$ is the inverse of $\Lambda_{\varepsilon}$. Assuming
the knowledge of $\Lambda_{\varepsilon}$ and $\Lambda_{\mu}$ we aim to recover
the metrics induced by $\varepsilon$ and $\mu$ on the cotangent bundle of $M$,
denoted by $\varepsilon^{\sharp}$ and $\mu^{\sharp}$, as well as their normal
derivatives at $\partial M$.

###  1.1 Literature Review

The inverse problem of recovering a Riemannian metric using boundary data has
been widely studied for the wave equation. The boundary measurements in this
case are realised by the Dirichlet to Neumann ($DtN$) map. A natural
obstruction to the uniqueness of the reconstruction, also mentioned as
“cloaking", is encountered in this type of inverse problems and is caused by
diffeomorphisms that fix the boundary invisibility , 2danisotropic . The
article lassas2014inverse  studies the inverse problem of reconstructing a
Riemannian manifold through the $DtN$ map for the wave equation. The Dirichlet
boundary conditions represent the sources and they are located in an open
subset of the boundary $S$. The Neumann boundary conditions which represent
the waves produced by the sources in $S$, are being observed in $R$, which is
an open subset of the boundary, with $\overline{S}\cap\overline{R}=\emptyset$.
The method employed to solve this inverse problem uses arguments of the
boundary control method. Applications of the boundary control method are given
in belishev2007recent  which concerns inverse problems in acoustics,
elasticity, electrodynamics, electrical impedance tomography and graph theory.

Well-posedness results for the time harmonic Maxwell’s equations are given in
somersalolinearized  in the case of isotropic electromagnetic parameters and
in Keniganisotropic  in the case of anisotropic electromagnetic parameters
that are conformal multiples. Both the isotropic and anisotropic cases are
considered in costabel1991coercive  where a coercive billinear form for the
decoupled system is constructed. The time domain is also taken into account in
uniquenessanisotropic  and the well-posedness is proved for non-analytic
anisotropic parameters that are conformal multiples of each other. In this
paper, we prove well-posedness for the equations considered by augmenting
Maxwell’s system to an elliptic one. Further details are provided later in
sections 3 and 4.

The linearized inverse boundary value problem of Maxwell’s equations is
studied in somersalolinearized . The electric permittivity, conductivity and
magnetic permeability are assumed to be functions that differ slightly from
given constants and they are recovered through the admittance map. The case of
isotropic (scalar) electromagnetic parameters is also considered in
Mcdowall1997 . In this case the Maxwell’s system can be decoupled to derive a
second order differential equation in terms of the Laplace-Beltrami operator
either with respect to the magnetic field or the electric field. Using the
calculus of Pseudodifferential Operators shubin1987pseudodifferential ,
treves1980introduction  the symbols of the admittance and impedance maps are
determined in terms of a pseudodifferential operator of order $1$ that
factorises the Laplace-Beltrami operator. It is shown that this implies the
boundary recovery of the electric permittivity $\varepsilon$ and magnetic
permeability $\mu$ and their first normal derivatives. This result is extended
in Joshi  where the demonstration of an inductive proof establishes the
boundary determination of the normal derivatives of all orders of the
electromagnetic parameters. This method uses a special class of
Pseudodifferential Operators that depends smoothly on the distance from the
boundary. Similar reasoning is followed in Lionheart  for the inverse problem
of the Laplacian on $k-$forms where the boundary data are represented by the
$DtN$ operator.

A different approach on the inverse problem of time harmonic Maxwell’s
equations for scalar electromagnetic parameters is demonstrated in
Somersalo1994Calderon . The principal symbols of the impedance and admittance
maps are derived using a boundary operator called the Calderon projector for
an augmented system equivalent to the Maxwell’s system. The isotropic inverse
problem for the time harmonic Maxwell’s equations is also studied in
chiralMcDowal  where the medium is a subset of $\mathbb{R}^{3}$ and is assumed
to be chiral. In this method the magnetic permeability is known and the
recovery of the electric permittivity, conductivity and chirality constant
throughout the medium of interest is proven. The augmented system used in
Somersalo1994Calderon  to construct solutions for this system and use complex
geometrical optics, is also used in the present work to establish well-
posedness. The inverse problem for the metrics induced by $\varepsilon$ and
$\mu$ being $(0,2)-$tensors and conformal multiples of each other is studied
in Keniganisotropic . To the best of our knowledge, Keniganisotropic  is the
only anisotropic case regarding the recovery of $\varepsilon$ and $\mu$
studied in the existing literature.

The article Leeconductivities  concerns the anisotropic conductivity problem.
The method employed is the same factorisation method used for the isotropic
case of the time harmonic Maxwell’s equations in Mcdowall1997 . In the case of
the dimension $n$ satisfying $n\geq 3$ the conductivity problem can be reduced
to the problem of recovering a Riemannian metric $g$ using the $DtN$ map for
harmonic functions. The same factorisation method is also applied to the
elasticity problem for transversely isotropic materials in Nakamura1997layer .
The elasticity operator, which is a second order elliptic operator, is
factorised in terms of a pseudodifferential operator of order $1$ making use
of the theory of monic matrix polynomials Lancaster . The aforementioned
operator is used to express the $DtN$ map which yields to the recovery of the
elastic parameters at the boundary. This is the approach that we follow in
this paper in order to calculate the principal symbols of the boundary
mappings.

###  1.2 Summary of Current Work

Now let us describe the contents of the present work. We start in section 2 by
introducing the background material and notation including new metrics
$\hat{\varepsilon}$ and $\hat{\mu}$ which are conformally related to
$\varepsilon^{\sharp}$ and $\mu^{\sharp}$ but also eliminate the dependence of
Maxwell’s equations on $g$. Then follows section 3, where we state our main
results. In order to show that the impedance and admittance maps are well
defined we start in section 4 by illustrating the proof of well-posedness of
the forward problem. To do so, we augment the Maxwell’s system to an $8\times
8$ elliptic system also used in Somersalo1994Calderon  and chiralMcDowal . The
basic step, following taylor1996partial , is to prove a coercivity estimate
for the elliptic operator of the system.

In section 5, we start by defining the impedance and admittance maps. First,
we study two types of non-uniqueness that arise from diffeomorphisms that fix
the boundary. We show that if $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ are two different sets of
metrics connected by such a diffeomorphism $\Phi$ then the boundary data of
the corresponding Dirichlet problems are the same. Next, we examine the
consequences of a diffeomorphism $\Phi$ that fixes the boundary to the
recovery of the determinant of the metric $g$, denoted by $\lvert g\rvert$, in
boundary normal coordinates for $\varepsilon^{\sharp}$ and $\mu^{\sharp}$. We
prove that the Jacobian determinant $\lvert D\Phi\rvert$ of this
diffeomorphism can be arbitrary in boundary coordinates and as a result, the
determinant $\lvert g\rvert$ cannot be uniquely determined in either boundary
normal coordinates for $\varepsilon^{\sharp}$ or $\mu^{\sharp}$. This affects
the boundary recovery of $\varepsilon^{\sharp}$ and $\mu^{\sharp}$ as we show
in section 5.

In section 6 we decouple the system and follow Nakamura1997layer  together
with the theory of monic matrix polynomials Lancaster  to derive the principal
symbols of the factorisation operators denoted by $B(x,D_{\tilde{x}})$ and
$C(x,D_{\tilde{x}})$ representing the normal derivatives in coordinates. Next,
we express the principal symbols of the impedance and admittance maps through
the principal symbols of $B$ and $C$. Proceeding to section 7, we show that
the principal symbols of $\Lambda_{\varepsilon}$ and $\Lambda_{\mu}$ determine
the tangential entries of the metrics $\hat{\varepsilon}$ and $\hat{\mu}$.
Next, investigating the consequences of the boundary recovery on the electric
and magnetic fields, we prove that in some cases we can recover the normal
components of $\hat{\mu}$ in boundary normal coordinates for
$\hat{\varepsilon}$. Under the same choice of coordinates, we show that when
boundary recovery of the metrics $\hat{\varepsilon}$, $\hat{\mu}$ is possible,
the boundary recovery of their normal derivatives of all orders is implied.
This is done by adjusting to our case the inductive proof employed in Joshi
and Lionheart  that uses Riccati equations satisfied by $B(x,D_{\tilde{x}})$,
$C(x,D_{\tilde{x}})$ together with a class of pseudodifferential operators
that allows us to work exclusively with principal symbols at the boundary.

##  2 Background material

Let us introduce some prerequisites from tensor calculus that will be needed
in the following sections. Further details on the topic are given in
lee2013smooth . We will denote by $\tau_{n}^{m}(M)$ the spaces of smooth
$(n,m)-$ complex valued tensor fields on $M$. The Riemmanian $g$ induces a
point-wise billinear form $\langle\cdot,\cdot\rangle_{g}$ on these spaces.
Also, we introduce the space $\Omega^{k}(M)$ that consists of the smooth
complex-valued differential $k-$forms. For $a,b$ $\in$ $\Omega^{k}(M)$ we will
write the point-wise inner product of $a$ with $b$ as $\langle
a,\bar{b}\rangle_{g}$. For $a,b$ $\in$ $\Omega^{1}(M)$ we express $\langle
a,\bar{b}\rangle_{g}$ in a local coordinate chart as

$\langle a,\bar{b}\rangle_{g}=g^{ab}a_{a}\bar{b}_{b},$

which induces the norm $\langle a,\bar{a}\rangle_{g}^{1/2}$. Note carefully
that, in our notation, for complex valued fields the bilinear form $\langle
a,b\rangle_{g}$ is not an inner-product because we do not include the
conjugation. Indeed, in local coordinates it is written as

$\langle a,b\rangle_{g}=g^{ab}a_{a}b_{b}.$

Furthermore, we will write $\lvert a\rvert_{g}^{2}=\langle a,a\rangle_{g}$,
which for complex $a$ does not give a norm.

######  Remark 1.

The purpose of excluding the conjugation from $\langle a,b\rangle_{g}$ and
$\lvert a\rvert_{g}^{2}$ is that in what follows we will make use of the fact
that $\lvert a\rvert_{g}^{2}$ can vanish for complex $a$. When we reach this
point, the reader is reminded with a remark (see Remark 2).

The inner product on $L^{2}(\Omega^{k}(M))$ is given by

$\langle\langle
a,\bar{b}\rangle\rangle_{g}=\int\limits_{M}a\wedge\ast_{g}\bar{b},$

where $a,b$ $\in$ $\Omega^{k}(M)$. In what follows, we will also use inner
products defined as above but with $g$ replaced by metrics induced from the
electromagnetic parameters $\varepsilon$ and $\mu$. Using the Levi-Civita
connection on $(M,g)$, denoted by $\nabla$, we define the spaces
$H^{1}(\Omega^{k}(M))$ as

$H^{1}(\Omega^{k}(M))=\\{a\in L^{2}(\Omega^{k}(M))\mathchar
24635\relax\;\quad\nabla a\in L^{2}(\tau_{k+1}^{0}(M))\\},$

equipped with the norm

$\|a\|_{H^{1}(\Omega^{k}(M))}^{2}=\|a\|_{L^{2}(\Omega^{k}(M))}^{2}+\|\nabla
a\|_{L^{2}(\tau_{k+1}^{0}(M))}^{2}.$

Let $\iota:\partial M\mapsto M$ be the inclusion map at the boundary and
define the pullback $\iota^{\ast}:\Omega^{k}(M)\mapsto\Omega^{k}(\partial M)$
on smooth differential forms. The pullback can be extended by continuity to
$\iota^{\ast}:H^{1}(\Omega^{k}(M))\mapsto H^{1/2}(\Omega^{k}(\partial M))$.
The definition of $H^{1/2}$ and in general fractional Sobolev spaces is given
in section $4.1$ of taylor1996partial , in terms of the Fourier transform.
Moreover, we define the Sobolev space $H_{0}^{1}(\Omega^{k}(M))$ as

$H_{0}^{1}(\Omega^{k}(M))=\\{a\in H^{1}(\Omega^{k}(M))\mathchar
24635\relax\;\quad\iota^{\ast}a=0\\}.$

The $(1,1)-$tensors $\varepsilon$ and $\mu$ are assumed to be smooth and real
valued. We interpret them as point-wise maps on the cotangent bundle
$T^{\ast}(M)$ and assume that they are symmetric and positive definite with
respect to $g$. For $a,b$ $\in$ $T_{x}^{\ast}(M)$ these properties are
described by, respectively,

$\langle a,\varepsilon\bar{b}\rangle_{g}=\langle\varepsilon
a,\bar{b}\rangle_{g},\ \langle a,\mu\bar{b}\rangle_{g}=\langle\mu
a,\bar{b}\rangle_{g},$ $\langle a,\varepsilon\bar{a}\rangle_{g}>0,\quad\langle
a,\mu\bar{a}\rangle_{g}>0\quad\mbox{whenever $a\neq 0$.}$

Due to the above, $\varepsilon$, $\mu$ and their inverses $\varepsilon^{-1}$,
$\mu^{-1}$ induce metrics on the cotangent and tangent bundles $T^{\ast}(M)$
and $T(M)$, respectively. The metrics on $T^{\ast}(M)$ induced by
$\varepsilon$ and $\mu$ are $(2,0)-$tensors and using the $\sharp$ (raising
indices) operator are provided in a local coordinate by

$\varepsilon^{\sharp}=\varepsilon^{ij}=g^{ik}\varepsilon_{k}^{j},\quad\mu^{\sharp}=\mu^{ij}=g^{ik}\mu_{k}^{j}.$

We will write the inverses of $\varepsilon^{\sharp}$ and $\mu^{\sharp}$ using
the $\flat$ (lowering indices) operator as

$(\varepsilon^{-1})_{\flat}=(\varepsilon^{-1})_{ij}=g_{ik}(\varepsilon^{-1})_{j}^{k},\quad(\mu^{-1})_{\flat}=(\mu^{-1})_{ij}=g_{ik}(\mu^{-1})_{j}^{k}.$

Note that we will often use the term metric both for Riemannian metrics, which
are $(0,2)-$tensors and induce a bilinear form on $T(M)$, as well as for
$(2,0)-$tensors and the corresponding bilinear forms on $T^{\ast}(M)$.

###  2.1 Notation and identities

Let us introduce the notation used in this paper. We are working in a set of
coordinates adapted to the boundary $\partial M$ characterised by $x_{3}=0$
within the domain of the coordinate chart. These will be boundary normal
coordinates either for $\varepsilon^{\sharp}$, $\mu^{\sharp}$ or the related
metrics $\hat{\varepsilon}$, $\hat{\mu}$ introduced in (5). Since we are
working on a $3-d$ manifold, the indices of tensor fields $i$ are assumed to
take values from $1$ to $3$. In the case of an index restricted to $i=1,2$, we
will add a tilde as in $\tilde{i}$ . When working with coordinate invariant
expressions we will use the notation $\tilde{\xi}$ for the covector

$\xi_{\tilde{i}}\ dx^{\tilde{i}}=\sum_{\tilde{i}=1}^{2}\xi_{\tilde{i}}\
dx^{\tilde{i}}.$

If $\nu=\ dx^{3}$ is a covector normal to the boundary, we will denote by
$\xi$ a covector that is written in components as

$\xi=\tilde{\xi}+\nu\xi_{3}.$

The determinants of the Riemannian metrics $g$, $(\varepsilon^{-1})_{\flat}$
and $(\mu^{-1})_{\flat}$ in coordinates will be written as

$\lvert
g\rvert:=\det(g),\quad\lvert\varepsilon^{-1}\rvert:=\det((\varepsilon^{-1})_{\flat}),\quad\lvert\mu^{-1}\rvert:=\det((\mu^{-1})_{\flat}).$

The corresponding notation for the determinants of the metrics on
$T^{\ast}(M)$ is

$\lvert
g^{-1}\rvert:=\det(g^{-1}),\quad\lvert\varepsilon\rvert:=\det(\varepsilon^{\sharp}),\quad\lvert\mu\rvert:=\det(\mu^{\sharp}).$

Denoting by $\sigma^{ijk}$ the alternating tensor will use the following
identities for the metric $g^{-1}$

$\lvert g^{-1}\rvert\sigma^{pqr}=\sigma_{ijk}g^{pi}g^{qj}g^{rk},$ (3)

and for the matrix of cofactors of the Riemannian metric $g$

$\sigma^{aqj}\sigma^{dkb}g_{bj}=\lvert g\rvert(g^{ad}g^{qk}-g^{ak}g^{qd}).$
(4)

We define new Riemannian metrics $\hat{\varepsilon}^{-1}$ and $\hat{\mu}^{-1}$
by

$\frac{\hat{\varepsilon}^{-1}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}=\frac{(\varepsilon^{-1})_{\flat}}{\sqrt{\lvert
g\rvert}},\quad\frac{\hat{\mu}^{-1}}{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}=\frac{(\mu^{-1})_{\flat}}{\sqrt{\lvert
g\rvert}},$ (5)

which allow us to write the time harmonic Maxwell’s equations as given in (1)
in the equivalent form

$\begin{split}\ast_{\hat{\varepsilon}}\mathrm{d}H&=-i\omega
E,\quad\delta_{\hat{\varepsilon}}E=0,\\\ \ast_{\hat{\mu}}\mathrm{d}E&=i\omega
H,\quad\delta_{\hat{\mu}}H=0,\\\ \end{split}$ (6)

that is independent of $g$. When working with the metrics (5), we will denote
the associated impedance and admittance maps by $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\mu}}$, respectively. The approach we follow is to work with
two different pairs of metrics $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ and boundary mappings
$\Lambda_{\hat{\varepsilon}},\Lambda_{\hat{\mu}}$ and
$\Lambda_{\hat{\varepsilon}^{\prime}}$, $\Lambda_{\hat{\mu}^{\prime}}$ to
investigate the implications of the assumption
$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$,
$\Lambda_{\hat{\mu}}=\Lambda_{\hat{\mu}^{\prime}}$. We will use the phrases
“boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$" and “boundary normal
coordinates for $\hat{\mu}$/$\hat{\mu}^{\prime}$" to express that we work
separately in boundary normal coordinates for either the metrics
$\hat{\varepsilon}$ and $\hat{\varepsilon}^{\prime}$ or $\hat{\mu}$ and
$\hat{\mu}^{\prime}$. Assume that the representations of $\hat{\varepsilon}$
and $\hat{\varepsilon}^{\prime}$ in boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$ at $x_{3}=0$ are given by,
respectively

$\left(\begin{matrix}\tilde{\varepsilon}&0\\\
0^{T}&1\end{matrix}\right),\quad\left(\begin{matrix}\tilde{\varepsilon}^{\prime}&0\\\
0^{T}&1\end{matrix}\right).$ (7)

Assume further that $\hat{\mu}$ and $\hat{\mu}^{\prime}$ in boundary normal
coordinates for $\hat{\mu}$/$\hat{\mu}^{\prime}$ at $x_{3}=0$ are written as

$\left(\begin{matrix}\tilde{\mu}&0\\\
0^{T}&1\end{matrix}\right),\quad\left(\begin{matrix}\tilde{\mu}^{\prime}&0\\\
0^{T}&1\end{matrix}\right).$ (8)

Note that $\tilde{\varepsilon}$, $\tilde{\varepsilon}^{\prime}$, $\tilde{\mu}$
and $\tilde{\mu}^{\prime}$ define coordinate invariant metrics on $\partial
M$. These are the tangential parts of $\hat{\varepsilon}$,
$\hat{\varepsilon}^{\prime}$, $\hat{\mu}$ and $\hat{\mu}^{\prime}$
respectively.

As mentioned in Section 1, we are going to use the calculus of
Pseudodifferential Operators $(\Psi DOs)$ to determine the principal symbols
of the boundary mappings $\Lambda_{\hat{\varepsilon}}$ and
$\Lambda_{\hat{\mu}}$. Using the Fourier transform and its inverse, $P$ is a
pseudodifferential operator of order $m$ ($P\in\Psi DO^{(m)}$) in local
coordinates if

$P(x,D_{x})u(x)=\frac{1}{(2\pi)^{3}}\int\limits_{\mathbb{R}^{3}}e^{ix\cdot\xi}p(x,\xi)\hat{u}(\xi)\
d\xi,$ (9)

where $D_{x}=-i\partial_{x}$, $\hat{u}(\xi)$ denotes the Fourier transform of
$u(x)$ and $p(x,\xi)$ is called the symbol of $P(x,D_{x})$ and is of order $m$
with respect to the dual variable $\xi$. For details on the definition and
calculus of $\Psi DO$s, which will be used, see shubin1987pseudodifferential ;
treves1980introduction .

##  3 Results

In this section we present the main results included in this paper. The
content of the remainder of the paper contains the proofs for each result.

######  Theorem 1 (Well-posedness of the forward problem.).

For $\omega$ $\in$ $\mathbb{R}$, outside a discrete set, the solution of the
Dirichlet problem

$\begin{split}\ast_{\hat{\varepsilon}}\mathrm{d}H&=-i\omega
E,\quad\delta_{\hat{\varepsilon}}E=0,\\\ \ast_{\hat{\mu}}\mathrm{d}E&=i\omega
H,\quad\delta_{\hat{\mu}}H=0,\end{split}$ $\iota^{\ast}H=F,$

exists, is unique and depends continuously on the data $F$. More particularly,
there exists $\hat{C}>0$ that depends on $\hat{\varepsilon}$, $\hat{\mu}$ and
$\omega$, such that

$\|E\|_{H^{1}(\Omega^{1}(M))}^{2}+\|H\|_{H^{1}(\Omega^{1}(M))}^{2}\leq\hat{C}\|F\|_{H^{1/2}(\Omega^{1}(\partial
M))}^{2}.$

Proof: See Section 4.

######  Theorem 2 (Non-uniqueness.).

If $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ are related by a
diffeomorphism fixing the boundary of $M$ then
$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$.
Additionally, let $g_{\varepsilon}$, $g_{\varepsilon^{\prime}}$, $g_{\mu}$ and
$g_{\mu^{\prime}}$ be the local coordinate representations of the metric $g$
in boundary normal coordinates for
$\varepsilon^{\sharp}$/$\varepsilon^{\prime\sharp}$,
$\mu^{\sharp}$/$\mu^{\prime\sharp}$. There exist pairs of metrics
$(\varepsilon^{\sharp},\mu^{\sharp})$,
$(\varepsilon^{\prime\sharp},\mu^{\prime\sharp})$ not equal to each other such
that $\Lambda_{\epsilon}=\Lambda_{\epsilon^{\prime}}$ and the ratios of the
determinants in the different sets of boundary normal coordinates
$\frac{\lvert g_{\varepsilon}\rvert}{\lvert g_{\varepsilon^{\prime}}\rvert}$,
$\frac{\lvert g_{\mu}\rvert}{\lvert g_{\mu^{\prime}}\rvert}$ are arbitrarily
chosen positive functions in a neighbourhood of the boundary.

Proof: See Section 5.1.

######  Theorem 3 (Boundary recovery of tangential components of the
metrics.).

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
metrics and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated impedance maps. Assuming
that $\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$, then

$\tilde{\varepsilon}=\tilde{\varepsilon}^{\prime},\quad\mbox{at}\ x_{3}=0,$
$\tilde{\mu}=\tilde{\mu}^{\prime},\quad\mbox{at}\ x_{3}=0.$

Proof: See Section 7.1.

######  Theorem 4 (Boundary recovery of the full metrics.).

Let $(\hat{\varepsilon},\hat{\mu})$
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
metrics and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated boundary mappings. Let
us fix boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If
$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$, then at
each point in $\partial M$ least one of the following cases holds:

  1. 1.

The metrics $\tilde{\mu}$ and $\tilde{\varepsilon}$ are multiples and
$\hat{\mu}^{\prime 3\tilde{j}}=c\hat{\mu}^{3\tilde{j}}$ for some
$c\in\mathbb{R}$.

  2. 2.

$\displaystyle\hat{\varepsilon}=\hat{\varepsilon}^{\prime}$ and
$\hat{\mu}=\hat{\mu}^{\prime}$.

Proof: See Section 7.3.

######  Theorem 5 (Boundary recovery of normal derivatives of the metrics.).

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
metrics and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the corresponding boundary mappings.
Let us fix boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If

$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}},$
$\hat{\varepsilon}=\hat{\varepsilon}^{\prime},\quad\hat{\mu}=\hat{\mu}^{\prime}\quad\mbox{at}\
x_{3}=0,$

then,

$\partial_{x_{3}}^{\kappa}\hat{\varepsilon}=\partial_{x_{3}}^{\kappa}\hat{\varepsilon}^{\prime},\quad\partial_{x_{3}}^{\kappa}\hat{\mu}=\partial_{x_{3}}^{\kappa}\hat{\mu}^{\prime}\quad\mbox{at}\
x_{3}=0,$

for any $\kappa\geq 1$.

Proof: See Section 8.

##  4 Sketch of the proof of Theorem 1

Our proof of Theorem 1 relies on augmenting Maxwell’s equations to a larger
elliptic system and proving a coercivity estimate for this system. Since the
methods are fairly standard and well-known, we only give a sketch of this
proof.

To begin, we augment the system (6) to

$PU=-i\left(\begin{matrix}i\omega&0&\delta_{\hat{\mu}}&0\\\
0&i\omega&\ast_{\hat{\varepsilon}}\mathrm{d}&\mathrm{d}\\\
\mathrm{d}&-\ast_{\hat{\mu}}\mathrm{d}&i\omega&0\\\
0&\delta_{\hat{\varepsilon}}&0&i\omega\end{matrix}\right)\left(\begin{matrix}u_{E}\\\
E\\\ H\\\ u_{H}\end{matrix}\right)=\left(\begin{matrix}0\\\ 0\\\ 0\\\
0\end{matrix}\right),$ (10)

where $u_{E}$ and $u_{H}$ are auxiliary $0-$forms. Note that if
$u_{E}=u_{H}=0$, then $E$ and $H$ satisfying (10) also satisfy (6). The
purpose of this step is that $P$ is an elliptic operator, i.e its principal
symbol is invertible for $\xi$ $\in$ $\mathbb{R}^{3}\backslash\\{0\\}$, in
contrast to the operator corresponding to the system (6). The field $U$ is
defined in $\Omega(M)$, where

$\Omega(M)=\Omega^{0}(M)\oplus\Omega^{1}(M)\oplus\Omega^{1}(M)\oplus\Omega^{0}(M)$

and the boundary condition is written in terms of $U$ as

$BU=\left(\begin{matrix}\iota^{\ast}&0&0&0\\\ 0&0&0&0\\\ 0&0&\iota^{\ast}&0\\\
0&0&0&\iota^{\ast}\end{matrix}\right)\left(\begin{matrix}u_{E}\\\ E\\\ H\\\
u_{H}\end{matrix}\right)=\left(\begin{matrix}0\\\ 0\\\ F\\\
0\end{matrix}\right).$ (11)

The spaces $L^{2}(\Omega(M))$ and $H^{1}(\Omega(M))$ are defined using the
inner products and norms given from the product structure

$\begin{split}(U,V)&=\int_{M}\left(\begin{matrix}u_{E}\\\ E\\\ H\\\
u_{H}\end{matrix}\right)\wedge\left(\begin{matrix}\ast_{\hat{\mu}}&0&0&0\\\
0&\ast_{\hat{\varepsilon}}&0&0\\\ 0&0&\ast_{\hat{\mu}}&0\\\
0&0&0&\ast_{\hat{\varepsilon}}\end{matrix}\right)\left(\begin{matrix}\overline{u_{E}}\\\
\overline{E}\\\ \overline{H}\\\
\overline{u_{H}}\end{matrix}\right),\end{split}$ (12)
$\left\|\left(\begin{matrix}u_{E}\\\ E\\\ H\\\
u_{H}\end{matrix}\right)\right\|^{2}=\|u_{E}\|^{2}+\|E\|^{2}+\|H\|^{2}+\|u_{H}\|^{2},$

where the norms are either $L^{2}$ or $H^{1}$. Transforming the boundary value
problem consisting of (10) and (11) to the equivalent one with non-homogeneous
differential equation and homogeneous Dirichlet condition we define the domain
of $P$ to be

$D(P)=\\{U\in H^{1}(\Omega(M))\mathchar 24635\relax\;\quad BU=0\\}.$

The main step in the proof is to prove the following coercivity estimate which
holds for any $U$ $\in$ $D(P)$

$\|U\|_{H^{1}(\Omega(M))}^{2}\leq
C\|PU\|_{L^{2}(\Omega(M))}^{2}+C\|U\|_{L^{2}(\Omega(M))}^{2},$ (13)

for some $C>0$, depending on $\hat{\varepsilon}$, $\hat{\mu}$ and $\omega$. In
(taylor1996partial, , Theorem $11.1$) is shown that the ellipticity of $P$
implies that (13) holds in the interior of $M$. Thus, it remains to show that
the estimate (13) holds in a coordinate chart adapted to the boundary. Using
the method in (taylor1996partial, , Proposition $11.2$), it suffices to show
that $P$ satisfies the hypothesis of regularity upon freezing the
coefficients. The only difference is that we consider the eigenvalues and
eigenvectors of $K_{1}(\tilde{\xi})+K_{0}$ instead of $K_{1}(\tilde{\xi})$ as
denoted in the proof of (taylor1996partial, , Proposition $11.2$) to
accomodate the fact that the eigenvectors of $K_{1}(\tilde{\xi})$ in our case
do not satisfy (taylor1996partial, , Lemma $11.7$). Next, using the symmetry
of $P$ with respect to the the $L^{2}-$product defined in (12) and using the
coercivity estimate we show that $P$ is densely defined self-adjoint with
compact resolvent. Hence, the set of eigenvalues of $P$ is discrete. Finally,
to obtain the result for the original time harmonic Maxwell’s equations we
must show that, for values of $\omega$ outside the spectrum, if
$u_{E}|_{\partial M}=u_{H}|_{\partial M}=0$ then $u_{E}=u_{H}=0$. This is done
by showing that $u_{E}$ and $u_{H}$ satisfy eigenvalue problems which can be
decoupled from the augmented system.

##  5 Boundary mappings and non-uniqueness result

The fact that the boundary value problem with Dirichet condition on $H$ is
well-posed outside a discrete set of $\omega$ implies the corresponding result
for the boundary value problem with Dirichlet condition

$\iota^{\ast}E=G.$

Therefore, we can define the impedance map
$\Lambda_{\hat{\varepsilon}}:\Omega^{1}(\partial M)\mapsto\Omega^{1}(\partial
M)$ realised by

$\Lambda_{\hat{\varepsilon}}(\iota^{\ast}H)=\iota^{\ast}E$

and its inverse $\Lambda_{\hat{\mu}}:\Omega^{1}(\partial
M)\mapsto\Omega^{1}(\partial M)$ given by

$\Lambda_{\hat{\mu}}(\iota^{\ast}E)=\iota^{\ast}H$

which is called the admittance map. Next we study the non-uniqueness implied
by diffeomorphisms $\Phi$ that fix the boundary and thus prove Theorem 2.

###  5.1 Proof of Theorem 2

Let us consider the sets of metrics $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ connected by a
diffeomorphism $\Phi:$ $M\mapsto M$ with $\Phi|_{\partial M}=\mathbb{I}$ as

$\hat{\varepsilon}=\Phi^{\ast}\hat{\varepsilon}^{\prime},\quad\hat{\mu}=\Phi^{\ast}\hat{\mu}^{\prime}.$
(14)

Because of the coordinate invariant form of (6), the action of $\Phi$ on the
magnetic field $H$ is

$H=(D\Phi)^{T}H^{\prime}$

and since $\Phi|_{\partial M}=\mathbb{I}$ we get
$\iota^{\ast}H=\iota^{\ast}H^{\prime}$. The same is also true for the
electrial field: $\iota^{\ast}E=\iota^{\ast}E^{\prime}$. Therefore the
impedance maps corresponding to the $\hat{\varepsilon}$,
$\hat{\varepsilon}^{\prime}$ metrics satisfy

$\Lambda_{\hat{\varepsilon}}(\iota^{\ast}H)=\Lambda_{\hat{\varepsilon}^{\prime}}(\iota^{\ast}H).$

Hence, the metrics $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ lead to the same boundary
data, which implies the same for the metrics
$(\varepsilon^{\sharp},\mu^{\sharp},g)$ and
$(\varepsilon^{\prime\sharp},\mu^{\prime\sharp},g)$ defined through (5).

To complete the proof of Theorem 2 we need the following Lemma.

######  Lemma 1.

Suppose that $\\{x^{j}\\}_{j=1}^{3}$ are a set of local boundary coordinates
on a domain $U\subset M$ and take any positive function $h\in C^{\infty}(M)$
such that $\mathrm{supp}(h-1)\subset U$. Then there exists a diffeomorphism
$\Phi:M\mapsto M$ such that

$\Phi{|}_{\partial M}=\mathbb{I}$

and in the local coordinates the Jacobian determinant satisfies $\lvert
D\Phi\rvert=h$ for all $x^{3}$ sufficiently small.

Proof: We will assume without loss of generality that

$\mathrm{supp}(h-1)\subset\\{(x^{1},x^{2},x^{3})\in[-\epsilon,\epsilon]\times[-\epsilon,\epsilon]\times[0,\epsilon]\\}\subset
U.$

for some $\epsilon>0$. Let $\phi\in C^{\infty}_{c}(\mathbb{R})$ be an even
function satisfying $0\leq\phi\leq 1$, $\mathrm{supp}(\phi)\subset(-1,1)$,
$\phi(s)=1$ for $s$ in a neighourhood of zero and
$\int_{-\infty}^{\infty}\phi(s)\ \mathrm{d}s=1$. Also set

$a=4\ \mathrm{sup}\\{h+1\\}+3,\quad b=\frac{2\epsilon}{a}\left(\
\mathrm{sup}\\{h+1\\}+1\right),\quad
c=\frac{2\epsilon}{a}\mathrm{sup}\\{h+1\\},$
$d(x^{1},x^{2})=\frac{1}{c}\int_{0}^{\infty}(h(x^{1},x^{2},s)-1)\phi\left(\frac{as}{\epsilon}\right)\
\mathrm{d}s.$

Then we define $\Phi$ within $U$ by

$\Phi(x^{1},x^{2},x^{3})=\left(x^{1},x^{2},f(x^{1},x^{2},x^{3})\right),$

with

$f(x^{1},x^{2},x^{3})=\int_{0}^{x^{3}}(h(x^{1},x^{2},s)-1)\phi\left(\frac{as}{\epsilon}\right)-d(x^{1},x^{2})\phi\left(\frac{s-b}{c}\right)+1\
\mathrm{d}s.$ (15)

With this construction, we see first that $f(x^{1},x^{2},x^{3})$ satisfies
$f(x^{1},x^{2},0)=0$ which implies that $\Phi|_{\partial M}=\mathbb{I}$. Also,
noting that the support of the first and second terms in the integrand on the
right side of (15) is contained in $\\{s\in[0,\epsilon]\\}$ we can conclude

$\frac{\partial f}{\partial
x^{3}}(x^{1},x^{2},x^{3})=(h(x^{1},x^{2},x^{3})-1)\phi\left(\frac{ax^{3}}{\epsilon}\right)-d(x^{1},x^{2})\phi\left(\frac{x^{3}-b}{c}\right)+1>0$
(16)

everywhere and $f(x^{1},x^{2},x^{3})=x^{3}$ for
$(x^{1},x^{2},x^{3})\notin[-\epsilon,\epsilon]\times[-\epsilon,\epsilon]\times[0,\epsilon]$.
Thus, $\Phi$ can be extended as the identity outside $U$ to give a
diffeomorphism on $M$ with Jacobian matrix given within $U$ by

$D\Phi=\left(\begin{matrix}1&0&0\\\ 0&1&0\\\ \frac{\partial f}{\partial
x^{1}}&\frac{\partial f}{\partial x^{2}}&\frac{\partial f}{\partial
x^{3}}\end{matrix}\right).$ (17)

From (16), we see that $\frac{\partial f}{\partial x^{3}}=h$ when $x^{3}$ is
sufficiently small, and so the Jacobian determinant is $\lvert D\Phi\rvert=h$
for all such $x^{3}$. This completes the proof.

Now let us proceed to the proof of the second part of Theorem 2 using the
diffeomorphism $\Phi$ constructed in Lemma 1 in (14). Given a set of
coordinates on the boundary, let $\Psi$, $\Psi^{\prime}$ be the corresponding
coordinate transformations from $M$ to boundary normal coordinates for
$\hat{\varepsilon}$ and $\hat{\varepsilon}^{\prime}$, respectively. Let us
denote by $\hat{g}$ and $\hat{g}^{\prime}$ the local representations of the
metric $g$ in boundary normal coordinates for
$\hat{\varepsilon}/\hat{\varepsilon}^{\prime}$. Then, referring to Figure 1
and using the fact that $\Psi=\Psi^{\prime}\circ\Phi$, we deduce that

$\lvert\hat{g}\rvert=\lvert D\Phi\rvert^{-2}\lvert\hat{g}^{\prime}\rvert.$
(18)

Using the result of Lemma 1, that $\lvert D\Phi\rvert$ can be an arbitrary
positive function in a neighbourhood of the boundary, we conclude that the
expression for the volume form of $g$ in boundary normal coordinates for
$\hat{\varepsilon}$ cannot be determined from $\Lambda_{\hat{\varepsilon}}$.

To complete the proof of Theorem 2, let us proceed to the following changes of
coordinates

$\begin{split}\mbox{BNCs for
$\hat{\varepsilon}^{-1}$}\quad\overset{\tilde{\Psi}}{\longrightarrow}\quad\mbox{BNCs
for $(\varepsilon^{-1})_{\flat}$},\\\ \mbox{BNCs for
$\hat{\varepsilon}^{\prime-1}$}\quad\overset{\tilde{\Psi}^{\prime}}{\longrightarrow}\quad\mbox{BNCs
for $(\varepsilon^{\prime-1})_{\flat}$}.\end{split}$

Using the relationship between the $\hat{\varepsilon}^{-1}$ and
$(\varepsilon^{-1})_{\flat}$ metrics given in (5) as well the fact that
$\hat{\varepsilon}$ and $\hat{\varepsilon}^{\prime}$ have the same expression
in their boundary normal coordinates, the Jacobian matrices of
$\tilde{\Psi}^{-1}$ and $(\tilde{\Psi}^{\prime})^{-1}$ are given by,
respectively,

$D\tilde{\Psi}^{-1}=\left(\begin{matrix}1&0&0\\\ 0&1&0\\\
\tilde{\Psi}_{31}&\tilde{\Psi}_{32}&\left(\frac{\lvert\hat{g}\rvert}{\lvert\hat{\varepsilon}^{-1}\rvert}\right)^{-1/4}\end{matrix}\right),\quad
D(\tilde{\Psi}^{\prime})^{-1}=\left(\begin{matrix}1&0&0\\\ 0&1&0\\\
\tilde{\Psi}^{\prime}_{31}&\tilde{\Psi}^{\prime}_{32}&\left(\frac{\lvert\hat{g}^{\prime}\rvert}{\lvert\hat{\varepsilon}^{-1}\rvert}\right)^{-1/4}\end{matrix}\right)$

where the off diagonal entries are undetermined functions which will be zero
at the boundary. Denoting by $g_{\varepsilon}$ and $g_{\varepsilon^{\prime}}$
the metrics $g$ and $g^{\prime}$ in boundary normal coordinates for
$(\varepsilon^{-1})_{\flat}$/$(\varepsilon^{\prime-1})_{\flat}$ we have

$\lvert g_{\varepsilon}\rvert=\lvert
D\tilde{\Psi}\rvert^{-2}\lvert\hat{g}\rvert=\sqrt{\lvert\hat{g}\rvert}\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}\quad\mbox{and}\quad\lvert
g_{\varepsilon^{\prime}}\rvert=\lvert
D\tilde{\Psi}^{\prime}\rvert^{-2}\lvert\hat{g}^{\prime}\rvert=\sqrt{\lvert\hat{g}^{\prime}\rvert}\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}.$

Making use of (18), we deduce

$\lvert g_{\varepsilon}\rvert=\lvert D\Phi\rvert^{-1}\lvert
g_{\varepsilon^{\prime}}\rvert.$

As we showed in Lemma 1, the determinant $\lvert D\Phi\rvert$ can be any
positive function which implies that $\lvert g_{\varepsilon}\rvert/\lvert
g_{\varepsilon^{\prime}}\rvert$ can be chosen arbitrarily in a neighbourhood
of any point on the boundary and is otherwise equal to one in a neighbourhood
of the boundary. By repeating this proof in a neighbourhood of any point on
the boundary and combining the results as a composition we can complete the
proof for $\epsilon^{\sharp}$. Demonstrating an approach symmetric to the
above we can prove the corresponding result for the determinant $\lvert
g\rvert$ in boundary normal coordinates for $\mu^{\sharp}$. This concludes the
proof of Theorem 2.

Figure 1: Changes of coordinates used in proof of Theorem 2. Note that the
upper part of the diagram commutes (i.e. $\Psi=\Psi^{\prime}\circ\Phi$).

##  6 System Decoupling and Factorisation Operators

Aiming to decouple the augmented system given in equation (10) for
$u_{E}=u_{H}=0$, we decompose $U$ as

$U=\left(\begin{matrix}U_{H}\\\ U_{E}\end{matrix}\right),$

where $U_{E}=\left(\begin{matrix}0\\\ E\end{matrix}\right)$ and
$U_{H}=\left(\begin{matrix}H\\\ 0\end{matrix}\right)$ to obtain

$\begin{split}\left(\begin{matrix}-i\delta_{\hat{\mu}}&0\\\
-i\ast_{\hat{\varepsilon}}\mathrm{d}&-i\mathrm{d}\end{matrix}\right)U_{H}&=-\omega
U_{E},\quad\left(\begin{matrix}-i\mathrm{d}&i\ast_{\hat{\mu}}\mathrm{d}\\\
0&-i\delta_{\hat{\varepsilon}}\end{matrix}\right)U_{E}=-\omega
U_{H}.\end{split}$

Eliminating $U_{E}$ from the above system, we derive the following equation in
terms of $H$

$\left(-\mathrm{d}\delta_{\hat{\mu}}+\ast_{\hat{\mu}}\mathrm{d}\ast_{\hat{\varepsilon}}\mathrm{d}\right)H-\omega^{2}H=0.$

which applying the operator $\ast_{\hat{\varepsilon}}\ast_{\hat{\mu}}$ becomes

$L_{H}H=\left(-\ast_{\hat{\varepsilon}}\ast_{\hat{\mu}}\mathrm{d}\delta_{\hat{\mu}}+\ast_{\hat{\varepsilon}}\mathrm{d}\ast_{\hat{\varepsilon}}\mathrm{d}\right)H-\omega^{2}\ast_{\hat{\varepsilon}}\ast_{\hat{\mu}}H=0.$

Or equivalently, in coordinates

$\begin{split}L_{H}H=\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{qa}D_{x^{a}}\left(\hat{\mu}^{kb}D_{x^{b}}-i\frac{\partial_{x^{d}}(\sqrt{\lvert\hat{\mu}^{-1}\rvert}\hat{\mu}^{kd})}{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}\right)H_{k}\\\
-(\hat{\varepsilon}^{-1})_{lq}\frac{\sigma^{ajq}D_{x^{a}}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{bj}\frac{\sigma^{dkb}D_{x^{d}}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}H_{k}-\omega^{2}\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{qk}H_{k}=0.\end{split}$

Following the approach of Nakamura1997layer , the next step is to separate the
differentiation with respect to $x_{3}$ from the differentiation with respect
to $\tilde{x}$ to write the differential operator $L_{H}$ as

$L_{H}=T_{H}(x)D_{x_{3}}^{2}+A_{H}(x,D_{\tilde{x}})D_{x_{3}}+G_{H}(x)D_{x_{3}}+Q_{H}(x,D_{\tilde{x}})+F_{H}(x,D_{\tilde{x}})+R_{H}(x),$

where the symbols of the operators in the representation of $L_{H}$ are given
by

$\begin{split}(T_{H}(x))_{l}^{k}&=\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{q3}\hat{\mu}^{k3}+\left(\delta_{l}^{k}\hat{\varepsilon}^{33}-\delta_{l}^{3}\hat{\varepsilon}^{3k}\right),\\\
\end{split}$ (19)
$\begin{split}(A_{H}(x,\tilde{\xi}))_{l}^{k}&=\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\left(\hat{\mu}^{q3}\hat{\mu}^{k\tilde{a}}\xi_{\tilde{a}}+\hat{\mu}^{q\tilde{a}}\xi_{\tilde{a}}\hat{\mu}^{k3}\right)\\\
&\hskip
85.35826pt+2\delta_{l}^{k}\langle\nu,\tilde{\xi}\rangle_{\hat{\varepsilon}}-\delta_{l}^{3}\xi_{\tilde{a}}\hat{\varepsilon}^{\tilde{a}k}-\delta_{l}^{\tilde{l}}\xi_{\tilde{l}}\hat{\varepsilon}^{3k},\end{split}$
(20)
$\begin{split}(G_{H}(x))_{l}^{k}&=-i\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\left(\hat{\mu}^{qa}\partial_{x^{a}}\hat{\mu}^{k3}+\hat{\mu}^{q3}\frac{\partial_{x^{a}}(\sqrt{\lvert\hat{\mu}^{-1}\rvert}\hat{\mu}^{ka})}{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}\right)\\\
&+i(\hat{\varepsilon}^{-1})_{lq}\frac{\sigma^{ajq}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\partial_{x^{a}}(\hat{\varepsilon}^{-1})_{\tilde{b}j}\frac{\sigma^{3\tilde{k}\tilde{b}}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}},\end{split}$
(21)
$\begin{split}(Q_{H}(x,\tilde{\xi}))_{l}^{k}&=\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{q\tilde{a}}\xi_{\tilde{a}}\hat{\mu}^{k\tilde{b}}\xi_{\tilde{b}}+\left(\delta_{l}^{k}\lvert\tilde{\xi}\rvert_{\hat{\varepsilon}}^{2}-\delta_{l}^{\tilde{l}}\xi_{\tilde{l}}\xi_{\tilde{a}}\hat{\varepsilon}^{\tilde{a}k}\right),\end{split}$
(22)
$\begin{split}(F_{H}(x,\tilde{\xi}))_{l}^{k}&=-i\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\left(\hat{\mu}^{qa}\partial_{x^{a}}\hat{\mu}^{k\tilde{a}}\xi_{\tilde{a}}+\hat{\mu}^{q\tilde{d}}\xi_{\tilde{d}}\frac{\partial_{x^{a}}(\sqrt{\lvert\hat{\mu}^{-1}\rvert}\hat{\mu}^{ka})}{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}\right)\\\
&+i(\hat{\varepsilon}^{-1})_{lq}\frac{\sigma^{ajq}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\partial_{x^{a}}(\hat{\varepsilon}^{-1})_{bj}\frac{\sigma^{\tilde{d}kb}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\xi_{\tilde{d}},\end{split}$
(23)
$(R_{H}(x))_{l}^{k}=-\omega^{2}\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{qk}-(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{qb}\partial_{x^{b}}\frac{\partial_{x^{a}}(\sqrt{\lvert\hat{\mu}^{-1}\rvert}\hat{\mu}^{ka})}{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}.$

Next we show that $L_{H}$ can be factorised using pseudodifferential
operators. We will make use of the principal symbol of $L_{H}$ which is

$\sigma_{p}(L_{H})=M_{H}(\xi_{3})=T_{H}(x)\xi_{3}^{2}+A_{H}(x,\tilde{\xi})\xi_{3}+Q_{H}(x,\tilde{\xi}).$

$M_{H}$ is a matrix polynomial with respect to the variable $\xi_{3}$ in the
sense of Lancaster  and, for fixed $(x,\tilde{\xi})$, the eigenvalues are the
solutions of the sixth order polynomial equation
$\mathrm{det}(M_{H}(\xi_{3}))=0$.

######  Theorem 6.

Suppose that we have some local boundary coordinates $\\{x^{j}\\}_{j=1}^{3}$
and let $\Gamma_{+}(x,\tilde{\xi})\subset\mathbb{C}$ be a contour in
$\mathbb{C}$ enclosing all eigenvlaues of $M_{H}$ with positive imaginary
part. Then, in the domain of the coordinates, there exist pseudodifferential
operators $B(x,D_{\tilde{x}})$ and $G_{0}(x,D_{\tilde{x}})$ of order $1$ and
$0$, respectively, such that the principal symbol of $B$ is given by

$B^{(1)}(x,\tilde{\xi})=\int_{\Gamma^{+}}\xi_{3}M_{H}(\xi_{3})^{-1}\
d\xi_{3}\left(\int_{\Gamma^{+}}M_{H}(\xi_{3})^{-1}\ d\xi_{3}\right)^{-1}$ (24)

and

$L_{H}=\left(\mathbb{I}D_{x^{3}}-B^{\ast}(x,D_{\tilde{x}})+G_{0}(x,D_{\tilde{x}})\right)T_{H}(x)\left(\mathbb{I}D_{x^{3}}-B(x,D_{\tilde{x}})\right)$
(25)

modulo smoothing. Additionally, $B(x,D_{\tilde{x}})$ satisfies the following
Riccati equation

$T_{H}D_{x^{3}}B+(A_{H}+G_{H})B+T_{H}B^{2}+Q_{H}+F_{H}+R_{H}=0$ (26)

and the symbol of $B$ is a smooth function of $\hat{\varepsilon}$ and
$\hat{\mu}$.

Proof: The proof is the same as the proof of (Nakamura1997layer, , Theorem
2.2) which we briefly sketch. First, the Riccati equation (26) can be derived
from (25) and the expression of the operator $L_{H}$ as done in the proof of
(Nakamura1997layer, , Theorem 2.2). Looking at the principal symbol of the
operators on both sides of (26), we see that the principal symbol $B^{(1)}$ of
$B$ must satisfy

$T_{H}(x)(B^{(1)}(x,\tilde{\xi}))^{2}+A_{H}(x,\tilde{\xi})B^{(1)}(x,\tilde{\xi})+Q_{H}(x,\tilde{\xi})=0$

and using (Lancaster, , Theorem 4.2) we see that $B^{(1)}$ given by (24) will
satisfy this equation. The remainder of the symbol of $B$ can then be
determined by looking at an asymptotic symbol expansion of (26) and this
symbol can be asymptotically summed to produce a pseudodifferential operator
that satisfies (26) modulo smoothing. Then the symbol for $G_{0}$ can be
determined from (25) in the same way.

The final statement about smoothness of the symbol of $B$ with respect to the
parameters $\hat{\varepsilon}$ and $\hat{\mu}$ can be proven by first looking
at the principal symbol given by (24). Indeed, given $(x,\tilde{\xi})$ a
single choice of $\Gamma_{+}$ will work also in a neighbourhood of
$(x,\tilde{\xi})$, and so the regularity with respect to the parameters
follows from the regularity of $M_{H}(\xi_{3})^{-1}$ with respect to the same
on the contour $\Gamma_{+}$. The regularity of the rest of the symbol with
respect to $\hat{\varepsilon}$ and $\hat{\mu}$ follows from this and the
construction of the lower order parts of the symbol as given in
Nakamura1997layer .

Next we express the principal symbol $B^{(1)}$ given by (24) more explicitly.
The expression we derive will be used in the calculation of the principal
symbols of $\Lambda_{\hat{\varepsilon}}$ and $\Lambda_{\hat{\mu}}$ later. Our
method uses the theory of monic matrix polynomials Lancaster  to determine the
Jordan pairs of $M_{H}(\xi_{3})$ and then express $B^{(1)}(x,\tilde{\xi})$ in
terms of them. The theory of monic matrix polynomials is applicable to
$M_{H}(\xi_{3})$ as the coefficient $T_{H}(x)$ is invertible.

In order to calculate the Jordan pairs of $M_{H}$, we should find the
eigenvalues of $M_{H}$ and for each eigenvalue $\xi_{3}$ also find covectors
$y_{j}$ such that

$\sum_{p=0}^{j}\frac{1}{p!}\partial_{\xi_{3}}^{p}M_{H}(\xi_{3})y_{j-p}=0,\quad
j=0,\dots k-1$

where $k$ is at most the order of $\xi_{3}$ as a root of the equation
$\det(M_{H}(\xi_{3}))$=0. For each eigenvalue, the covector $y_{0}$ is called
a corresponding eigenvector and the covectors $y_{1},\dots y_{k-1}$ are called
generalised eigenvectors. Writing $M_{H}(\xi_{3})$ in coordinates, we have

$\left(\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(\hat{\varepsilon}^{-1})_{lq}\hat{\mu}^{qa}\xi_{a}\hat{\mu}^{kd}\xi_{d}+\delta_{l}^{k}\lvert\xi\rvert_{\hat{\varepsilon}}^{2}-\xi_{l}\xi_{d}\hat{\varepsilon}^{dk}\right)H_{k}=0,$

where we have made use of the identity (4) for the $\hat{\varepsilon}^{-1}$
metric. Let us now consider the covectors

$\xi,\quad\zeta=\hat{\varepsilon}^{-1}\hat{\mu}\xi,\quad\chi=\ast_{\hat{\varepsilon}}(\xi\wedge\zeta).$
(27)

When these covectors form a basis (see remark 3), we can write $M_{H}$ in the
basis $\\{\chi,\xi,\zeta\\}$ as follows

$[\begin{matrix}\chi&\xi&\zeta\end{matrix}]^{T}\hat{\varepsilon}{M_{H}}[\begin{matrix}\chi&\xi&\zeta\end{matrix}]=\left(\begin{matrix}\lvert\xi\rvert_{\hat{\varepsilon}}^{2}&0&0\\\
0&0&\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\lvert\xi\rvert_{\hat{\mu}}^{2}\\\
0&-\lvert\xi\rvert_{\hat{\mu}}^{2}&\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\lvert\xi\rvert_{\hat{\mu}\hat{\varepsilon}^{-1}\hat{\mu}}^{2}+\lvert\xi\rvert_{\hat{\varepsilon}}^{2}\end{matrix}\right).$

The determinant of the above vanishes for $\xi_{3}$ satisfying

$\lvert\xi\rvert_{\hat{\mu}}^{4}\lvert\xi\rvert_{\hat{\varepsilon}}^{2}=0.$
(28)

######  Remark 2.

Recalling Remark 1, let us point out that the quantities
$\lvert\xi\rvert_{\hat{\mu}}^{2}$ and
$\lvert\xi\rvert_{\hat{\varepsilon}}^{2}$ can vanish for complex $\xi$ as they
are given by the real inner products
$\langle\cdot,\cdot\rangle_{\hat{\varepsilon}}$
$\langle\cdot,\cdot\rangle_{\hat{\mu}}$.

For fixed $\tilde{\xi}$, the solutions $\xi_{3}$ of (28) are conjugate pairs
denoted by ${\xi_{\hat{\mu}}}_{3}$, $\overline{\xi_{\hat{\mu}}}_{3}$ and
${\xi_{\hat{\varepsilon}}}_{3}$, $\overline{\xi_{\hat{\varepsilon}}}_{3}$,
respectively. The values of $\xi_{3}$ with positive imaginary part denoted by
${\xi_{\hat{\mu}}}_{3}$ and ${\xi_{\hat{\varepsilon}}}_{3}$ are given by,
respectively

${\xi_{\hat{\mu}}}_{3}=-\frac{\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}}{\hat{\mu}^{33}}+i\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\sqrt{\hat{\mu}^{33}}},$
(29)
${\xi_{\hat{\varepsilon}}}_{3}=-\frac{\langle\nu,\tilde{\xi}\rangle_{\hat{\varepsilon}}}{\hat{\varepsilon}^{33}}+i\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}{\sqrt{\hat{\varepsilon}^{33}}},$
(30)

where we have made use of the representations of $\hat{\varepsilon}$ and
$\hat{\mu}$ given in equations (7), (8). We notice that
${\xi_{\hat{\mu}}}_{3}$ is a double root of $\det(M_{H}(\xi_{3}))$ and we
expect an eigenvector and a generalised eigenvector associated with this value
of $\xi_{3}$. The eigenvector corresponding to ${\xi_{\hat{\mu}}}_{3}$ will be
denoted by $\xi_{\hat{\mu}}$ and is given by

$\xi_{\hat{\mu}}=\xi|_{{\xi_{3}=\xi_{\hat{\mu}}}_{3}}.$

The generalised eigenvector is given by

$\gamma_{\hat{\mu}}=\nu+m_{\hat{\mu}}\zeta_{\hat{\mu}},$

where

$\zeta_{\hat{\mu}}=\zeta|_{{\xi_{3}=\xi_{\hat{\mu}}}_{3}}\quad\mbox{and}\quad
m_{\hat{\mu}}=-\frac{2\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\langle\nu,\xi_{\hat{\mu}}\rangle_{\hat{\mu}}}{\frac{\sqrt{\lvert\hat{\mu}^{-1}\rvert}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\lvert\xi_{\hat{\mu}}\rvert_{\hat{\mu}\hat{\varepsilon}^{-1}\hat{\mu}}^{2}+\lvert\xi_{\hat{\mu}}\rvert_{\hat{\varepsilon}}^{2}}.$

Proceeding to the eigenvalue ${\xi_{\hat{\varepsilon}}}_{3}$, as it is a
single root of $\det(M_{H}(\xi_{3}))$, it corresponds to the eigenvector
$\chi_{\hat{\varepsilon}}$, defined by

$\chi_{\hat{\varepsilon}}=\chi|_{\xi_{3}={\xi_{\hat{\varepsilon}}}_{3}}.$

Therefore, the Jordan pairs and blocks of the matrix polynomial
$M_{H}(\xi_{3})$ are given by

$Y_{\hat{\varepsilon}}=\chi_{\hat{\varepsilon}},\quad
J_{\hat{\varepsilon}}=[{\xi_{\hat{\varepsilon}}}_{3}],\quad\mbox{and}\quad
Y_{\hat{\mu}}=\left[\begin{matrix}\xi_{\hat{\mu}}&\gamma_{\hat{\mu}}\end{matrix}\right],\quad
J_{\hat{\mu}}=\left(\begin{matrix}{\xi_{\hat{\mu}}}_{3}&1\\\
0&{\xi_{\hat{\mu}}}_{3}\end{matrix}\right).$

Using the Jordan pairs in the above, the principal symbol of the
pseudodifferential operator $B(x,D_{\tilde{x}})$ is expressed as

$B^{(1)}(x,\tilde{\xi})=X_{H}J_{H}X_{H}^{-1},$ (31)

where

$J_{H}=\left(\begin{matrix}{\xi_{\hat{\varepsilon}}}_{3}&0&0\\\
0&{\xi_{\hat{\mu}}}_{3}&\lvert\xi_{\hat{\mu}}\rvert_{\hat{\varepsilon}}\\\
0&0&{\xi_{\hat{\mu}}}_{3}\end{matrix}\right)\quad\mbox{and}\quad
X_{H}=\left[\begin{matrix}\hat{\chi}_{\hat{\varepsilon}}&\hat{\xi}_{\hat{\mu}}&\gamma_{\hat{\mu}}\end{matrix}\right]$

and $\hat{\chi}_{\hat{\varepsilon}}$, $\hat{\xi}_{\hat{\mu}}$ denote that the
covectors $\chi_{\hat{\varepsilon}}$ and $\xi_{\hat{\mu}}$ are normalised in
such a way that $X_{H}$ $\in$ $\Psi DO^{(0)}$. Additionally, $M_{H}(\xi_{3})$
can be factorised in terms of $B^{(1)}(x,\tilde{\xi})$ as

$M_{H}(\xi_{3})=\left(\mathbb{I}\xi_{3}-(B^{\ast})^{(1)}(x,\tilde{\xi})\right)T_{H}(x)\left(\mathbb{I}\xi_{3}-B^{(1)}(x,\tilde{\xi})\right),$
(32)

where in order to derive the above factorisation we use the fact that
$T_{H}^{1/2}B^{(1)}T_{H}^{-1/2}$, $T_{H}^{1/2}(B^{\ast})^{(1)}T_{H}^{-1/2}$
are spectral right and left divisors of
$T_{H}^{-1/2}M_{H}(\xi_{3})T_{H}^{-1/2}$, respectively. Factorising
$M_{H}(\xi_{3})$, we can follow the method of Nakamura1997layer  to obtain the
factorisation of $L_{H}$ provided in (25), where the principal symbol of
$B(x,D_{\tilde{x}})$ is $B^{(1)}(x,\tilde{\xi})$.

######  Remark 3.

We observe that when ${\xi_{\hat{\varepsilon}}}_{3}={\xi_{\hat{\mu}}}_{3}$,
the covectors $\xi$, $\chi$ and $\zeta$ do not constitute a basis and the
representation (31) does not apply. However, using the continuity of the
symbol $B(x,\tilde{\xi})$ with respect to $\hat{\varepsilon}$ and $\hat{\mu}$
the various symbol expressions for the impedance and other maps derived below
are still valid even when
${\xi_{\hat{\varepsilon}}}_{3}={\xi_{\hat{\mu}}}_{3}$.

Next, we state the following result which will be used in the calculation of
the principal symbol of the impedance map. The proof follows the proof of
(Mcdowall1997, , Proposition 2).

######  Theorem 7.

If $H$ is the solution of the Dirichlet problem consisting of (6), then

$D_{x_{3}}H=B(x,D_{\tilde{x}})H,\ \mbox{modulo smoothing},\ \mbox{in a set of
boundary coordinates}\ \mbox{in}\ M.$

The corresponding factorisation operator obtained by decoupling the system in
terms of the electric field $E$ and following similar reasoning to the above,
will be denoted by $C(x,D_{\tilde{x}})$. The pseudodifferential operator
$C(x,D_{\tilde{x}})$ satisfies the Riccati equation

$T_{E}D_{x^{3}}C+(A_{E}+G_{E})C+T_{E}C^{2}+Q_{E}+F_{E}+R_{E}=0,$ (33)

where $T_{E}$, $A_{E}$, $G_{E}$, $Q_{E}$, $F_{E}$, $R_{E}$ are obtained from
$T_{H}$, $A_{H}$, $G_{H}$, $Q_{H}$, $F_{H}$, $R_{H}$ by exchanging
$\hat{\varepsilon}$ with $\hat{\mu}$. The principal symbol of
$C(x,D_{\tilde{x}})$ is provided by

$C^{(1)}(x,\tilde{\xi})=X_{E}J_{E}X_{E}^{-1},$ (34)

where

$J_{E}=\left(\begin{matrix}{\xi_{\hat{\mu}}}_{3}&0&0\\\
0&{\xi_{\hat{\varepsilon}}}_{3}&\lvert\xi_{\hat{\varepsilon}}\rvert_{\hat{\mu}}\\\
0&0&{\xi_{\hat{\varepsilon}}}_{3}\end{matrix}\right)\quad\mbox{and}\quad
X_{E}=\left[\begin{matrix}\hat{\chi}_{\hat{\mu}}&\hat{\xi}_{\hat{\varepsilon}}&\gamma_{\hat{\varepsilon}}\end{matrix}\right]$

and the covectors $\chi_{\hat{\mu}}$, $\xi_{\hat{\varepsilon}}$ and
$\gamma_{\hat{\varepsilon}}$ are obtained from $\chi_{\hat{\varepsilon}}$,
$\xi_{\hat{\mu}}$ and $\gamma_{\hat{\mu}}$ by exchanging $\hat{\mu}$ with
$\hat{\varepsilon}$.

Viewing the solution $H$ of the Dirichlet problem involving equations (6) and
(2) as a pseudodifferential operator acting on the boundary data $F$ and
motivated by the expression of $B^{(1)}(x,\tilde{\xi})$, the principal symbol
of $H$ in the basis
$\\{\chi_{\hat{\varepsilon}},\xi_{\hat{\mu}},\gamma_{\hat{\mu}}\\}$ is written
as

$H_{k}^{(0)}=a_{\hat{\varepsilon}}{{\chi}_{\hat{\varepsilon}}}_{k}+b_{\hat{\varepsilon}}{{\xi}_{\hat{\mu}}}_{k},$
(35)

where

$a_{\hat{\varepsilon}}=\frac{\ast_{\iota^{\ast}\hat{\varepsilon}}(\tilde{\xi}\wedge
F)}{\langle\nu_{\hat{\varepsilon}},\xi_{\hat{\varepsilon}}\rangle_{\hat{\varepsilon}}\lvert\xi_{\hat{\varepsilon}}\rvert_{\hat{\mu}}^{2}},\quad
b_{\hat{\varepsilon}}=\frac{\ast_{\iota^{\ast}\hat{\varepsilon}}(F\wedge\chi_{\hat{\varepsilon}})}{\langle\nu_{\epsilon},\xi_{\hat{\varepsilon}}\rangle_{\hat{\varepsilon}}\lvert\xi_{\hat{\varepsilon}}\rvert_{\hat{\mu}}^{2}}.$
(36)

Here $\ast_{\iota^{\ast}\hat{\varepsilon}}$ denotes the Hodge star operator
with respect to the $\hat{\varepsilon}$ metric pulled back to $\partial M$ and
$\nu_{\hat{\varepsilon}}$ is the normal covector at $\partial M$ with respect
to the $\hat{\varepsilon}$ metric. This decomposition of the principal symbol
of $H$ is implied since it has to vanish on the principal symbol of the
divergence equation
$\xi_{\tilde{a}}\hat{\mu}^{\tilde{a}b}H_{b}^{(0)}+\hat{\mu}^{3a}(B^{(1)}(x,\tilde{\xi}))_{a}^{b}H_{b}^{(0)}=0$.
Similarly, the principal symbol of $E$ seen as a pseudodifferential operator
acting on the boundary data $G$ is given by

$E_{k}^{(0)}=a_{\hat{\mu}}{{\chi}_{\hat{\mu}}}_{k}+b_{\hat{\mu}}{{\xi}_{\hat{\varepsilon}}}_{k},$

where $a_{\hat{\mu}}$ and $b_{\hat{\mu}}$ are obtained from
$a_{\hat{\varepsilon}}$, $b_{\hat{\varepsilon}}$ by exchanging
$\hat{\varepsilon}$ with $\hat{\mu}$ and replacing $F$ by $G$.

##  7 Principal Symbols of Boundary Mappings and Boundary Recovery

######  Theorem 8.

The principal symbol of the impedance map
$\Lambda_{\hat{\varepsilon}}:\iota^{\ast}H\mapsto\iota^{\ast}E$ is given by

$\lambda_{\hat{\varepsilon}}^{(1)}(F)=-\frac{\tilde{\xi}\
\ast_{\iota^{\ast}\hat{\varepsilon}}(\tilde{\xi}\wedge
F)}{\omega\langle\nu_{\hat{\varepsilon}},\xi_{\hat{\varepsilon}}\rangle_{\hat{\varepsilon}}}.$

Proof: Let us consider the Maxwell’s equation

$\ast_{\hat{\varepsilon}}\mathrm{d}H=-i\omega E,$

which taking the pullback and considering the principal symbol in coordinates
gives

$\lambda_{\hat{\varepsilon}}^{(1)}(F)=-\frac{1}{\omega}(\hat{\varepsilon}^{-1})_{\tilde{l}b}\frac{\sigma^{\tilde{a}kb}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}\xi_{\tilde{a}}H_{k}^{(0)}-\frac{1}{\omega}(\hat{\varepsilon}^{-1})_{\tilde{l}\tilde{b}}\frac{\sigma^{3\tilde{a}\tilde{b}}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}(B^{(1)}(x,\tilde{\xi}))_{\tilde{a}}^{k}H_{k}^{(0)}.$

Observing that $\xi_{\hat{\mu}}$ in the decomposition (35) of $H^{(0)}$ is in
the kernel of the right hand-side of the above equation, only the covector
$\chi_{\hat{\varepsilon}}$ contributes to the principal symbol of
$\Lambda_{\hat{\varepsilon}}$ as follows

$\lambda_{\hat{\varepsilon}}^{(1)}(F)=-a_{\hat{\varepsilon}}\frac{1}{\omega}(\hat{\varepsilon}^{-1})_{\tilde{l}b}\frac{\sigma^{akb}}{\sqrt{\lvert\hat{\varepsilon}^{-1}\rvert}}{\xi_{\hat{\varepsilon}}}_{a}{\chi_{\hat{\varepsilon}}}_{k}=-\frac{a_{\hat{\varepsilon}}}{\omega}{\xi}_{\tilde{l}}\lvert\xi_{\hat{\varepsilon}}\rvert_{\hat{\mu}}^{2},$

where we have used the identity (4) for the $\hat{\varepsilon}^{-1}$ metric.
Substituting the expression (36) for $a_{\hat{\varepsilon}}$ we obtain the
result. Working similarly to the above we can calculate the principal symbol
of the admittance map as stated in the following theorem.

######  Theorem 9.

The principal symbol of the admittance map
$\Lambda_{\hat{\mu}}:\iota^{\ast}E\mapsto\iota^{\ast}H$ is given by

$\lambda_{\hat{\mu}}^{(1)}(G)=\frac{\tilde{\xi}\
\ast_{\iota^{\ast}\hat{\mu}}(\tilde{\xi}\wedge
G)}{\omega\langle\nu_{\hat{\mu}},\xi_{\hat{\mu}}\rangle_{\hat{\mu}}}.$

###  7.1 Proof of Theorem 3

From the fact that
$\lambda_{\hat{\varepsilon}}^{(1)}=\lambda_{\hat{\varepsilon}^{\prime}}^{(1)}$
and using Theorem 8, we get

$\sqrt{\lvert\tilde{\varepsilon}^{-1}\rvert}\langle\nu_{\hat{\varepsilon}},\xi_{\hat{\varepsilon}}\rangle_{\hat{\varepsilon}}=\sqrt{\lvert\tilde{\varepsilon}^{\prime-1}\rvert}\langle\nu_{\hat{\varepsilon}^{\prime}},\xi_{\hat{\varepsilon}^{\prime}}\rangle_{\hat{\varepsilon}^{\prime}},\quad\mbox{at}\
x_{3}=0,$

which, from (30), leads to

$\sqrt{\lvert\tilde{\varepsilon}^{-1}\rvert}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}=\sqrt{\lvert\tilde{\varepsilon}^{\prime-1}\rvert}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}^{\prime}},\quad\mbox{at}\
x_{3}=0.$

This implies that

$\tilde{\varepsilon}=\tilde{\varepsilon}^{\prime}\quad\mbox{at}\ x_{3}=0.$
(37)

Working in the same way for
$\lambda_{\hat{\mu}}^{(1)}=\lambda_{\hat{\mu}^{\prime}}^{(1)}$ we reach the
corresponding result for $\tilde{\mu}$ and $\tilde{\mu}^{\prime}$ which
completes the proof of Theorem 3.

The result of Theorem 3 in terms of the metrics
$(\varepsilon^{\sharp},\mu^{\sharp})$,
$(\varepsilon^{\prime\sharp},\mu^{\prime\sharp})$ translates to

$\lvert g\rvert\varepsilon^{\tilde{i}\tilde{j}}=\lvert
g^{\prime}\rvert\varepsilon^{\prime\tilde{i}\tilde{j}},\quad\mbox{at}\
x_{3}=0,$

in boundary normal coordinates for $\varepsilon^{\sharp}$/
$\varepsilon^{\prime\sharp}$ and

$\lvert g\rvert\mu^{\tilde{i}\tilde{j}}=\lvert
g^{\prime}\rvert\mu^{\prime\tilde{i}\tilde{j}},\quad\mbox{at}\ x_{3}=0,$

in boundary normal coordinates for $\mu^{\sharp}$/ $\mu^{\prime\sharp}$.
Theorem 2 implies that the presence of the determinant $\lvert g\rvert$ in the
above equations leads to a non-uniqueness on the boundary recovery of
$\varepsilon^{\sharp}$ and $\mu^{\sharp}$. More specifically, we have that the
tangential components of the metrics $\varepsilon^{\sharp}$, $\mu^{\sharp}$
can be determined up to a conformal factor which is arbitrary.

###  7.2 Further consequences

Now let us investigate the consequences of Theorem 3 to the different pairs of
electromagnetic fields $E$, $H$ and $E^{\prime}$, $H^{\prime}$.

######  Theorem 10.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different metrics and
$\Lambda_{\hat{\varepsilon}}$, $\Lambda_{\hat{\varepsilon}^{\prime}}$, $E$,
$H$, $E^{\prime}$, $H^{\prime}$ the associated impedance maps and
electromagnetic fields, respectively. Assuming that
$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$ and
$\iota^{\ast}H=\iota^{\ast}H^{\prime}$ then

$E=E^{\prime},\quad\mbox{at}\ x_{3}=0,$

in boundary normal coordinates for $\hat{\varepsilon}$/
$\hat{\varepsilon}^{\prime}$ and

$H=H^{\prime},\quad\mbox{at}\ x_{3}=0,$

in boundary normal coordinates for $\hat{\mu}$/$\hat{\mu}^{\prime}$.

Proof: We will work with $E$, $E^{\prime}$ and the result for $H$,
$H^{\prime}$ is derived accordingly. Since
$\iota^{\ast}E=\iota^{\ast}E^{\prime}$, it suffices to prove that
$E_{3}=E_{3}^{\prime}$ at $x_{3}=0$. To start with, let us consider the third
component of the equation

$\ast_{\hat{\varepsilon}}\mathrm{d}H=-i\omega E,\quad\mbox{at}\ x_{3}=0,$

in boundary normal coordinates for $\hat{\varepsilon}$, which is

$E_{3}=\frac{i}{\omega}\frac{\sigma^{\tilde{a}\tilde{k}3}}{\sqrt{\lvert\tilde{\varepsilon}^{-1}\rvert}}\frac{\partial
H_{\tilde{k}}}{\partial x_{\tilde{a}}},\quad\mbox{at}\ x_{3}=0.$

Similarly, the corresponding equation for the ${}^{\prime}-$ parameters in
boundary normal coordinates for $\hat{\varepsilon}^{\prime}$ is given by

$E^{\prime}_{3}=\frac{i}{\omega}\frac{\sigma^{\tilde{a}\tilde{k}3}}{\sqrt{\lvert\tilde{\varepsilon}^{-1}\rvert}}\frac{\partial
H_{\tilde{k}}}{\partial x_{\tilde{a}}},\quad\mbox{at}\ x_{3}=0,$

where we have used the result of Theorem 3 and the assumption
$\iota^{\ast}H=\iota^{\ast}H^{\prime}$. Therefore, $E_{3}=E^{\prime}_{3}$ at
$x_{3}=0$ which completes the proof. Let $\Psi$, $\Psi^{\prime}$ be the
following change of coordinates transformations

$\mbox{BNCs for
}\hat{\varepsilon}\quad\overset{\Psi}{\longrightarrow}\quad\mbox{BNCs for
}\hat{\mu},$ $\mbox{BNCs for
}\hat{\varepsilon}^{\prime}\quad\overset{\Psi^{\prime}}{\longrightarrow}\quad\mbox{BNCs
for }\hat{\mu}^{\prime}.$

The Jacobian matrix of $\Psi$ and its inverse at $x_{3}=0$ are given by

$D\Psi=\left(\begin{matrix}1&0&-\frac{\hat{\mu}^{13}}{\hat{\mu}^{33}}\\\
0&1&-\frac{\hat{\mu}^{23}}{\hat{\mu}^{33}}\\\
0&0&\frac{1}{\sqrt{\hat{\mu}^{33}}}\end{matrix}\right),\quad
D\Psi^{-1}=\left(\begin{matrix}1&0&\frac{\hat{\mu}^{13}}{\sqrt{\hat{\mu}^{33}}}\\\
0&1&\frac{\hat{\mu}^{23}}{\sqrt{\hat{\mu}^{33}}}\\\
0&0&{\sqrt{\hat{\mu}^{33}}}\end{matrix}\right),$ (38)

where the entries of $\hat{\mu}$ are expressed in boundary normal coordinates
for $\hat{\varepsilon}$. Similarly, the Jacobian matrix of $\Psi^{\prime}$ and
its inverse at $x_{3}=0$ are obtained by (38) after substituting the entries
of $\hat{\mu}$ by the entries of $\hat{\mu}^{\prime}$. Using (38) we obtain
the following representations for $\hat{\varepsilon}$ and its determinant in
boundary normal coordinates for $\hat{\mu}$ at $x_{3}=0$

$\hat{\varepsilon}=D\Psi\left(\begin{matrix}\tilde{\varepsilon}&0\\\
0^{T}&1\end{matrix}\right)(D\Psi)^{T}=\left(\begin{matrix}\tilde{\varepsilon}+\frac{\hat{\mu}^{3\tilde{i}}\hat{\mu}^{3\tilde{j}}}{(\hat{\mu}^{33})^{2}}&-\frac{\hat{\mu}^{\tilde{i}3}}{(\hat{\mu}^{33})^{3/2}}\\\
-\frac{\hat{\mu}^{\tilde{j}3}}{(\hat{\mu}^{33})^{3/2}}&\frac{1}{\hat{\mu}^{33}}\end{matrix}\right),\quad\lvert\hat{\varepsilon}\rvert=\frac{\lvert\tilde{\varepsilon}\rvert}{\hat{\mu}^{33}}.$
(39)

Similarly, the metric $\hat{\mu}$ and its determinant in boundary normal
coordinates for $\hat{\varepsilon}$ at $x_{3}=0$ are provided by

$\hat{\mu}=D\Psi^{-1}\left(\begin{matrix}\tilde{\mu}&0\\\
0^{T}&1\end{matrix}\right)(D\Psi^{-1})^{T}=\left(\begin{matrix}\tilde{\mu}+\frac{\hat{\mu}^{3\tilde{i}}\hat{\mu}^{3\tilde{j}}}{\hat{\mu}^{33}}&\hat{\mu}^{\tilde{i}3}\\\
\hat{\mu}^{3\tilde{j}}&\hat{\mu}^{33}\end{matrix}\right),\quad\lvert\hat{\mu}\rvert=\hat{\mu}^{33}\lvert\tilde{\mu}\rvert.$
(40)

The expressions of $\hat{\varepsilon}^{\prime}$ in boundary normal coordinates
for $\hat{\mu}^{\prime}$ and $\hat{\mu}^{\prime}$ in boundary normal
coordinates for $\hat{\varepsilon}^{\prime}$ at $x_{3}=0$ follow from
equations (39) and (40), respectively, after substituting the entries of
$\hat{\mu}$ by the entries of $\hat{\mu}^{\prime}$. Now let us introduce the
notation for the covectors $\xi_{\hat{\mu}}$, $\chi_{\hat{\mu}}$,
$\xi_{\hat{\varepsilon}}$ and $\chi_{\hat{\varepsilon}}$ in boundary normal
coordinates for $\hat{\mu}$ and $\hat{\varepsilon}$ as shown in Table 1 and
then express their normal components in each coordinate system.

Covectors | BNCs for $\hat{\mu}$ | BNCs for $\hat{\varepsilon}$
---|---|---
$\xi_{\hat{\mu}}$ | $\xi_{\hat{\mu},\tilde{\mu}}$ | $\xi_{\hat{\mu},\tilde{\varepsilon}}$
$\chi_{\hat{\mu}}$ | $\chi_{\hat{\mu},\tilde{\mu}}$ | $\chi_{\hat{\mu},\tilde{\varepsilon}}$
$\xi_{\hat{\varepsilon}}$ | $\xi_{\hat{\varepsilon},\tilde{\mu}}$ | $\xi_{\hat{\varepsilon},\tilde{\varepsilon}}$
$\chi_{\hat{\varepsilon}}$ | $\chi_{\hat{\varepsilon},\tilde{\mu}}$ | $\chi_{\hat{\varepsilon},\tilde{\varepsilon}}$
Table 1: Notation for covectors in the different coordinate systems.

######  Lemma 2.

The normal components of the covectors
$\xi_{\hat{\varepsilon},\tilde{\varepsilon}}$, $\xi_{\hat{\mu},\tilde{\mu}}$,
$\xi_{\hat{\mu},\tilde{\varepsilon}}$ and
$\xi_{\hat{\varepsilon},\tilde{\mu}}$ are provided by

${\xi_{\hat{\varepsilon},\tilde{\varepsilon}}}_{3}=i\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}},\quad{\xi_{\hat{\mu},\tilde{\mu}}}_{3}=i\lvert\tilde{\xi}\rvert_{\tilde{\mu}},$
${\xi_{\hat{\mu},\tilde{\varepsilon}}}_{3}=-\frac{\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}}{\hat{\mu}^{33}}+i\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\sqrt{\hat{\mu}^{33}}},\quad{\xi_{\hat{\varepsilon},\tilde{\mu}}}_{3}=\frac{\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}}{\sqrt{\hat{\mu}^{33}}}+i\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\sqrt{\hat{\mu}^{33}}.$

Proof: To prove the above equations it suffices to use equations (29) and (30)
and the representations of each metric in the different coordinate systems.

######  Lemma 3.

The normal components of the covectors
${\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}$,
${\chi_{\hat{\mu}^{\prime},\tilde{\varepsilon}}}_{3}$,
${\chi_{\hat{\varepsilon},\tilde{\mu}}}_{3}$ and
${\chi_{\hat{\varepsilon}^{\prime},\tilde{\mu}}}_{3}$ are expressed as

${\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}=\frac{\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\tilde{\mu}^{\tilde{j}\tilde{b}}\xi_{\tilde{b}}}{\sqrt{\lvert\tilde{\mu}\rvert}\sqrt{\hat{\mu}^{33}}}+i\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\frac{\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{b}}\xi_{\tilde{b}}\hat{\mu}^{\tilde{j}3}}{\sqrt{\lvert\tilde{\mu}\rvert}\hat{\mu}^{33}},$
${\chi_{\hat{\varepsilon},\tilde{\mu}}}_{3}=\frac{\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{b}\tilde{j}}\xi_{\tilde{b}}\tilde{\mu}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\sqrt{\hat{\mu}^{33}}}{\sqrt{\lvert\tilde{{\varepsilon}}\rvert}}-i\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\frac{\sigma_{3\tilde{j}\tilde{i}}\hat{\mu}^{3\tilde{j}}\tilde{\mu}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}}{\sqrt{\lvert\tilde{\varepsilon}\rvert}\sqrt{\hat{\mu}^{33}}}.$

Proof: We use the following expressions of the covectors $\chi_{\hat{\mu}}$
and $\chi_{\hat{\varepsilon}}$ in general coordinates

${\chi_{\hat{\mu}}}_{k}=\frac{\sigma_{kji}\hat{\mu}^{bj}{\xi_{\hat{\mu}}}_{b}\hat{\varepsilon}^{id}{\xi_{\hat{\mu}}}_{d}}{\sqrt{\lvert\hat{\mu}\rvert}},$
${\chi_{\hat{\varepsilon}}}_{k}=\frac{\sigma_{kji}\hat{\varepsilon}^{bj}{\xi_{\hat{\varepsilon}}}_{b}\hat{\mu}^{id}{\xi_{\hat{\varepsilon}}}_{d}}{\sqrt{\lvert\hat{\varepsilon}\rvert}},$

which are derived using the identity (3) for the metrics $\hat{\mu}$ and
$\hat{\varepsilon}$, respectively. The normal components of these covectors in
each coordinate system are calculated using the representations of the metrics
in each coordinate system and Lemma 2. The normal components of the
${}^{\prime}-$ covectors in the different coordinate systems denoted by
${\xi_{\hat{\varepsilon}^{\prime},\tilde{\varepsilon}}}_{3}$,
${\xi_{\hat{\mu}^{\prime},\tilde{\mu}}}_{3}$,
$\xi_{\hat{\mu}^{\prime},\tilde{\varepsilon}}$,
$\xi_{\hat{\varepsilon}^{\prime},\tilde{\mu}}$,
${\chi_{\hat{\varepsilon}^{\prime},\tilde{\mu}}}_{3}$ and
${\chi_{\hat{\mu}^{\prime},\tilde{\varepsilon}}}_{3}$ are obtained from the
expressions derived in Lemma 2 and 3 after substituting the normal components
of $\hat{\mu}$ by the normal components of $\hat{\mu}^{\prime}$.

######  Lemma 4.

Let $(\hat{\varepsilon},\hat{\mu})$
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be 2 different sets of
metrics and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated boundary mappings. If
$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$, then the
following identities hold

$\left({\xi_{\hat{\mu}^{\prime},\tilde{\varepsilon}}}_{3}+{\xi_{\hat{\varepsilon},\tilde{\varepsilon}}}_{3}\right){\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}=\left({\xi_{\hat{\mu},\tilde{\varepsilon}}}_{3}+{\xi_{\hat{\varepsilon},\tilde{\varepsilon}}}_{3}\right){\chi_{\hat{\mu}^{\prime},\tilde{\varepsilon}}}_{3},\quad\mbox{at}\
x_{3}=0,$ (41)
$\left({\xi_{\hat{\varepsilon}^{\prime},\tilde{\mu}}}_{3}+{\xi_{\hat{\mu},\tilde{\mu}}}_{3}\right){\chi_{\hat{\varepsilon},\tilde{\mu}}}_{3}=\left({\xi_{\hat{\varepsilon},\tilde{\mu}}}_{3}+{\xi_{\hat{\mu},\tilde{\mu}}}_{3}\right){\chi_{\hat{\varepsilon}^{\prime},\tilde{\mu}}}_{3},\quad\mbox{at}\
x_{3}=0.$ (42)

Proof: We will show the derivation of the identity (41) as (42) is proven in a
similar way. Let us fix boundary normal coordinates for $\hat{\varepsilon}$.
We aim to write $\tilde{\chi}_{\hat{\mu}}$ as

$\tilde{\chi}_{\hat{\mu}}=a\tilde{\xi}+b\tilde{\xi}_{\tilde{\varepsilon}\perp}$
(43)

for some constants $a$, $b$ to be determined and
$\tilde{\xi}_{\tilde{\varepsilon}\perp}:=\ast_{\tilde{\varepsilon}}\tilde{\xi}$
satisfying

$\langle\tilde{\xi},\tilde{\xi}_{\tilde{\varepsilon}\perp}\rangle_{\tilde{\varepsilon}}=0,\quad\tilde{\xi}\wedge\tilde{\xi}_{\tilde{\varepsilon}\perp}=\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}\
dv_{\tilde{\varepsilon}},$ (44)

where $dv_{\tilde{\varepsilon}}$ denotes the volume form at the boundary with
respect to the $\tilde{\varepsilon}$ metric. Considering the inner product
with respect to the $\tilde{\varepsilon}$ metric of $\tilde{\xi}$ with (43),
we obtain

$a=-\frac{{\xi_{\hat{\mu},\tilde{\varepsilon}}}_{3}{\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}},$

where we have used the fact that
$\langle\chi_{\hat{\mu},\tilde{\varepsilon}},\xi_{\hat{\mu}}\rangle_{\hat{\varepsilon}}=0$.
The wedge product of $\tilde{\xi}$ with (43) results to

$b=-i\frac{\sqrt{\lvert\tilde{\varepsilon}\rvert}\lvert\xi_{\hat{\mu}}\rvert_{\hat{\varepsilon}}^{2}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\sqrt{\lvert\tilde{\mu}\rvert}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}}.$

Now let us write the principal symbol of the normal component of $E$ as

$E_{3}^{(0)}=a_{\hat{\mu}}{\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}+b_{\hat{\mu}}{\xi_{\hat{\varepsilon},\tilde{\varepsilon}}}_{3}.$

Making use of equations (36) as well as equation (43) we arrive at

$E_{3}^{(0)}(G)=\frac{{\chi_{\hat{\mu},\tilde{\varepsilon}}}_{3}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}({\xi_{\hat{\mu},\tilde{\varepsilon}}}_{3}+{\xi_{\hat{\varepsilon},\tilde{\varepsilon}}}_{3})}\ast_{\iota^{*}\hat{\mu}}\left(\tilde{\xi}\wedge
G\right)+\frac{i\sqrt{\lvert\tilde{\varepsilon}\rvert}}{\sqrt{\lvert\tilde{\mu}\rvert}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\ast_{\iota^{*}\hat{\mu}}\left(\tilde{\xi}_{\tilde{\varepsilon}\perp}\wedge
G\right).$

Setting this equal to $E_{3}^{\prime(0)}(G)$ with
$G=\tilde{\xi}_{\tilde{\varepsilon}\perp}$ and using Theorem 10 completes the
proof.

###  7.3 Proof of Theorem 4

Using Lemma 2 and 3, the imaginary parts of (41) and (42) are provided by,
respectively,

$\begin{split}\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)&=\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}\sigma_{3\tilde{i}\tilde{j}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\tilde{\mu}^{\tilde{j}\tilde{b}}\xi_{\tilde{b}}\left(\hat{\mu}^{\prime
33}\sqrt{\hat{\mu}^{33}}-\hat{\mu}^{33}\sqrt{\hat{\mu}^{\prime 33}}\right),\\\
\sigma_{3\tilde{j}\tilde{i}}\tilde{\mu}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)&=\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\sigma_{3\tilde{i}\tilde{j}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\tilde{\mu}^{\tilde{j}\tilde{b}}\xi_{\tilde{b}}\left(\hat{\mu}^{\prime
33}\sqrt{\hat{\mu}^{33}}-\hat{\mu}^{33}\sqrt{\hat{\mu}^{\prime
33}}\right).\end{split}$ (45)

Combining the above displayed formulae we arrive at

$\sigma_{3\tilde{j}\tilde{i}}\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)\left(\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}\tilde{\mu}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}-\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\right)=0.$
(46)

Now we are going distinguish the following two different cases.

  1. 1.

There exists $a>0$ such that

$\tilde{\mu}=a\tilde{\varepsilon}.$ (47)

  2. 2.

The metrics $\tilde{\mu}$ and $\tilde{\varepsilon}$ are not multiples.

Starting with case 1, we have that
$\sigma_{3\tilde{i}\tilde{j}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\tilde{\mu}^{\tilde{j}\tilde{b}}\xi_{\tilde{b}}=0$
and the first equation in (45) reduces to

$\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)=0.$

This implies that if
$\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)\neq
0$, there exists $b\in\mathbb{R}$ such that

$\tilde{\varepsilon}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}=b\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right),$

which cannot hold, since multiplying the above by $\xi_{\tilde{j}}$ leads to
$\lvert\tilde{\xi}\rvert_{\varepsilon}=0$. Therefore,
$\left(\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}\right)=0$
or equivalently

$\hat{\mu}^{\prime 3\tilde{j}}=c\hat{\mu}^{3\tilde{j}},\quad\mbox{at}\
x_{3}=0.$

for $c\in\mathbb{R}$.

It remains to work with case 2, where $\tilde{\mu}$ and $\tilde{\varepsilon}$
are assumed not to be multiples. Under this assumption, equation (46) yields,
at least for all $\tilde{\xi}$ in an open set, to

$\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}=\beta\left(\tilde{\mu}^{\tilde{j}\tilde{a}}\xi_{\tilde{a}}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}-\tilde{\varepsilon}^{\tilde{j}\tilde{a}}\xi_{\tilde{a}}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}\right),$

where $\beta\in\mathbb{R}$. Combining the above with the equations in (45) we
get

$\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}\hat{\mu}^{\prime
3\tilde{j}}-\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}^{\prime}}\hat{\mu}^{3\tilde{j}}=\left(\hat{\mu}^{33}\sqrt{\hat{\mu}^{\prime
33}}-\hat{\mu}^{\prime
33}\sqrt{\hat{\mu}^{33}}\right)\left(\tilde{\mu}^{\tilde{j}\tilde{a}}\xi_{\tilde{a}}\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}-\tilde{\varepsilon}^{\tilde{j}\tilde{a}}\xi_{\tilde{a}}\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\right).$
(48)

Differentiating (48) with respect to $\tilde{\xi}$ results to

$\begin{split}\hat{\mu}^{\prime
3\tilde{j}}\hat{\mu}^{3\tilde{a}}-\hat{\mu}^{3\tilde{j}}\hat{\mu}^{\prime
3\tilde{a}}&=\left(\hat{\mu}^{33}\sqrt{\hat{\mu}^{\prime
33}}-\hat{\mu}^{\prime
33}\sqrt{\hat{\mu}^{33}}\right)\left(\tilde{\mu}^{\tilde{j}\tilde{a}}\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}+\frac{\tilde{\mu}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}\frac{\tilde{\varepsilon}^{\tilde{a}\tilde{b}}\xi_{\tilde{b}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\right.\\\
&\left.-\tilde{\mu}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\frac{\tilde{\mu}^{\tilde{b}\tilde{a}}\xi_{\tilde{b}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{3}}-\tilde{\varepsilon}^{\tilde{j}\tilde{a}}\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}-\frac{\tilde{\varepsilon}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\frac{\tilde{\mu}^{\tilde{a}\tilde{b}}\xi_{\tilde{b}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}+\tilde{\varepsilon}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\frac{\tilde{\varepsilon}^{\tilde{b}\tilde{a}}\xi_{\tilde{b}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{3}}\right).\end{split}$

Let $\eta\in T^{*}_{x}(\partial M)$ be linearly independent from
$\tilde{\xi}$. Multiplying both sides of the above displayed equation by
$\eta$, we get

$\left(\hat{\mu}^{33}\sqrt{\hat{\mu}^{\prime 33}}-\hat{\mu}^{\prime
33}\sqrt{\hat{\mu}^{33}}\right)\left[\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}\left(\lvert\eta\rvert_{\tilde{\mu}}^{2}-\frac{\langle\tilde{\xi},\eta\rangle_{\tilde{\mu}}^{2}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}}\right)-\frac{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}}\left(\lvert\eta\rvert_{\tilde{\varepsilon}}^{2}-\frac{\langle\tilde{\xi},\eta\rangle_{\tilde{\varepsilon}}^{2}}{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{2}}\right)\right]=0.$
(49)

Assuming that the second factor in (49) vanishes and fixing $\eta$ such that

$\eta_{1}=-\xi_{2},\quad\eta_{2}=\xi_{1}$

we obtain

$\frac{\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}^{4}}{\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{4}}=\frac{\lvert\tilde{\varepsilon}\rvert}{\lvert\tilde{\mu}\rvert}.$

The above holds for $\tilde{\xi}$ in an open set, and we observe that the
right hand-side does not depend on $\tilde{\xi}$. Therefore
$\tilde{\varepsilon}$ should be a multiple of $\tilde{\mu}$, which contradicts
to our initial assumption. Hence, the second factor in (49) cannot be zero
which leads to

$\hat{\mu}^{33}=\hat{\mu}^{\prime 33},\quad\mbox{at}\ x_{3}=0.$

Taking this into account and revisiting equation (48) we have, assuming
without loss of generality that $\hat{\mu}^{\tilde{j}3}\neq 0$,

$\hat{\mu}^{\prime\tilde{j}3}=c\hat{\mu}^{\tilde{j}3},\quad\mbox{at}\
x_{3}=0,$

for $c\in\mathbb{R}$. Next, the real part of (41), by virtue of the
relationships between the normal components of $\hat{\mu}$ and
$\hat{\mu}^{\prime}$, becomes

$\begin{split}\left(1-c\right)\langle\tilde{\xi},\nu\rangle_{\hat{\mu}}\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\tilde{\mu}^{\tilde{i}\tilde{b}}\xi_{\tilde{b}}+\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}\left(c-1\right)\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\hat{\mu}^{3\tilde{j}}\\\
=\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\sqrt{\hat{\mu}^{33}}\left(1-c\right)\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\hat{\mu}^{3\tilde{j}},\end{split}$
(50)

which choosing $\tilde{\xi}\neq 0$ such that

$\langle\nu,\tilde{\xi}\rangle_{\hat{\mu}}=0,$ (51)

simplifies to

$\left(c-1\right)\left(\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}+\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\sqrt{\hat{\mu}^{33}}\right)\sigma_{3\tilde{j}\tilde{i}}\tilde{\varepsilon}^{\tilde{i}\tilde{d}}\xi_{\tilde{d}}\hat{\mu}^{3\tilde{j}}=0.$

Since the vectors $\tilde{\varepsilon}^{\tilde{j}\tilde{d}}\xi_{\tilde{d}}$
and $\hat{\mu}^{3{\tilde{j}}}$ are not multiples of each other, we have

$\left(c-1\right)\left(\lvert\tilde{\xi}\rvert_{\tilde{\mu}}^{2}+\lvert\tilde{\xi}\rvert_{\tilde{\varepsilon}}\lvert\tilde{\xi}\rvert_{\tilde{\mu}}\sqrt{\hat{\mu}^{33}}\right)=0.$

Since the second term on the left is positive, we deduce that $c=1$ which
implies

$\hat{\mu}^{3\tilde{j}}=\hat{\mu}^{\prime 3\tilde{j}},\quad\mbox{at}\
x_{3}=0.$

This completes the proof of Theorem 4.

∎

In the following lemma we show that when the different pairs of metrics agree
at $x_{3}=0$ in boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$ then the same holds for the
electromagnetic fields $E$, $H$, $E^{\prime}$, $H^{\prime}$. This will be used
in Section 8.

######  Lemma 5.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
metrics and $E$, $H$, $E^{\prime}$, $H^{\prime}$ be the corresponding electric
and magnetic fields. If $\hat{\varepsilon}=\hat{\varepsilon}^{\prime}$ and
$\hat{\mu}=\hat{\mu}^{\prime}$ at $x_{3}=0$, in boundary normal coordinates
for $\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$, then

$E=E^{\prime},\quad\mbox{at}\ x_{3}=0$ (52)

and

$H=H^{\prime},\quad\mbox{at}\ x_{3}=0.$ (53)

Proof: To prove the above we only need to consider the magnetic fields $H$ and
$H^{\prime}$ as the result for $E$, $E^{\prime}$ is already proven in Lemma
10. Considering the change of coordinates (38) and the fact that
$H=H^{\prime}$ in boundary normal coordinates for
$\hat{\mu}$/$\hat{\mu}^{\prime}$ completes the proof.

##  8 Normal Derivatives of Electromagnetic Parameters

In this section, we apply the methods of Joshi  and Lionheart  to show
inductively that when the metrics
$(\hat{\varepsilon},\hat{\mu})$,$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$
are equal at $x_{3}=0$, then, for any $\kappa>0$,
$\partial_{x_{3}}^{\kappa}\hat{\varepsilon}=\partial_{x_{3}}^{\kappa}\hat{\varepsilon}^{\prime}$,
$\partial_{x_{3}}^{\kappa}\hat{\mu}=\partial_{x_{3}}^{\kappa}\hat{\mu}^{\prime}$
at $x_{3}=0$ which proves Theorem 5.

Let us consider the pseudodifferential operators $B(x,D_{\tilde{x}})$ and
$B^{\prime}(x,D_{\tilde{x}})$ associated with the metrics
$(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ respectively and define
$Y=B^{\prime}-B$ $\in$ $\Psi DO^{(1)}$. Making use of the Riccati equation
(26) and the corresponding Riccati equation for the ${}^{\prime}-$ parameters
we have that $Y$ satisfies

$\begin{split}&T_{H^{\prime}}D_{x^{3}}Y+(A_{H^{\prime}}+G_{H^{\prime}})Y+T_{H^{\prime}}Y^{2}+T_{H^{\prime}}BY+T_{H^{\prime}}YB=\\\
&\hskip
14.22636pt[(A_{H}-A_{H^{\prime}})+(G_{H}-G_{H^{\prime}})+(T_{H}-T_{H^{\prime}})B]B+(R_{H}-R_{H^{\prime}})\\\
&\hskip
14.22636pt+(Q_{H}-Q_{H^{\prime}})+(F_{H}-F_{H^{\prime}})+(T_{H}-T_{H^{\prime}})D_{x^{3}}B.\end{split}$
(54)

Similarly, $J$ $\in$ $\Psi DO^{(1)}$ realised by $J=C^{\prime}-C$, satisfies

$\begin{split}&T_{E^{\prime}}D_{x^{3}}J+(A_{E^{\prime}}+G_{E^{\prime}})J+T_{E^{\prime}}J^{2}+T_{E^{\prime}}CJ+T_{E^{\prime}}JC=\\\
&\hskip
14.22636pt[(A_{E}-A_{E^{\prime}})+(G_{E}-G_{E^{\prime}})+(T_{E}-T_{E^{\prime}})C]C+(R_{E}-R_{E^{\prime}})\\\
&\hskip
14.22636pt+(Q_{E}-Q_{E^{\prime}})+(F_{E}-F_{E^{\prime}})+(T_{E}-T_{E^{\prime}})D_{x^{3}}C.\end{split}$
(55)

Let us now define a class of pseudodifferential operators that depends
smoothly on the normal distance from the boundary with respect to the
$\hat{\varepsilon}$ metric.

######  Definition 1 ($\Psi DO^{(m,p)}$).

We say $P$ $\in\Psi DO^{(m,p)}(\partial M,\mathbb{R}_{+})$ if it is a family
of pseudodifferential operators of order $m$ on $\partial M$, varying smoothly
up to $x_{3}=0$ such that

$P=\sum_{j=0}^{p}x_{3}^{j}P^{(m-p+j)},$

with $P^{(j)}$ $\in$ $\Psi DO^{(m-p+j)}(\partial M)$.

The calculus of $\Psi DO$s as well as the definition immediately implies the
following results which are collected here for later use.

######  Lemma 6 (Calculus for $\Psi DO^{(m,p)}$).

Suppose that $P\in\Psi DO^{(m,p)}(\partial M,\mathbb{R}_{+})$ and
$P^{\prime}\in\Psi DO^{(m^{\prime},p^{\prime})}(\partial M,\mathbb{R}_{+})$.
Then $PP^{\prime}\in\Psi DO^{(m+m^{\prime},p+p^{\prime})}(\partial
M,\mathbb{R}_{+})$. Also, $P\in\Psi DO^{(m+1,p+1)}(\partial M,\mathbb{R}_{+})$
and $x_{3}P\in\Psi DO^{(m,p+1)}(\partial M,\mathbb{R}_{+})$

######  Definition 2 (Principal symbol in $\Psi DO^{(m,p)}$).

For $P\in\Psi DO^{(m,p)}(\partial M,\mathbb{R}_{+})$, we define the vector of
principal symbols of $P=\sum_{j=0}^{p}x_{3}^{j}P^{(m-p+j)}$ at $x_{3}=0$, by
taking

$\sigma_{p,l}(P)=\sigma_{p}\left(\frac{1}{l!}\frac{d^{l}}{dx_{3}^{l}}P\Big{|}_{x_{3}=0}\right),$

where $0\leq l\leq p$. More particularly, we have that

$\sigma_{p}(P)=\left(\sigma_{p}\left(P^{(m-p)}\Big{|}_{x_{3}=0}\right),\sigma_{p}\left(P^{(m-p+1)}\Big{|}_{x_{3}=0}\right),\dots,\sigma_{p}\left(P^{(m)}\Big{|}_{x_{3}=0}\right)\right).$

We notice that when the metrics
$(\hat{\varepsilon},\hat{\mu})$,$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$
are equal at $x_{3}=0$, the principal symbols of $B$ and $B^{\prime}$ agree at
the boundary. Hence $Y$ $\in$ $\Psi DO^{(0)}(\partial M)$, or in terms of the
class of pseudodifferential operators given in Definition 1,
$Y=x_{3}Y^{(1)}+Y^{(0)}$, where, restricted to $\partial M$, $Y^{(1)}\in\Psi
DO^{(1)}(\partial M)$ and $Y^{(0)}\in\Psi DO^{(0)}(\partial M)$. This is
equivalent to $Y\in\Psi DO^{(1,1)}(\partial M,\mathbb{R}^{+})$ and
correspondingly, for the pseudodifferential operator $J$, we have $J\in\Psi
DO^{(1,1)}(\partial M,\mathbb{R}^{+})$. Also, defining
$\tilde{H}=H^{\prime}-H$ we have, by Lemma 5, that $\tilde{H}=0$ at $x_{3}=0$
which allows us to write $\tilde{H}\in x_{3}\Psi DO^{(0,0)}(\partial
M,\mathbb{R}^{+})\subset\Psi DO^{(0,1)}(\partial M,\mathbb{R}^{+})$.
Similarly, writing $\tilde{E}=E^{\prime}-E$, then $\tilde{E}$ is in the same
space.

Aiming to prove Theorem 5 inductively, we calculate the normal derivatives of
order $a$, $a\geq 1$, of $\tilde{H}$ and $\tilde{E}$ at $x_{3}=0$. In Theorems
11 and 12 we express these derivatives in terms of the pseudodifferential
operators $B$, $Y$ and $C$, $J$, respectively. In Theorems 13 and 14 we use
the Maxwell’s and divergence equations to write $D_{x_{3}}^{a}\tilde{H}$ and
$D_{x_{3}}^{a}\tilde{E}$ in terms of the normal derivatives of
$\hat{\varepsilon}$, $\hat{\mu}$, $\hat{\varepsilon}^{\prime}$,
$\hat{\mu}^{\prime}$ in boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$ at $x_{3}=0$. Then, we use
the expressions derived for $D_{x_{3}}^{a}\tilde{H}$ and
$D_{x_{3}}^{a}\tilde{E}$ at $x_{3}=0$ to determine the vectors of principal
symbols of the Riccati equations (54) and (55), in the sense of Definition 2.

######  Theorem 11.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $H$, $H^{\prime}$ the associated magnetic
fields, respectively and let $\tilde{H}=H^{\prime}-H$. Assume that $B$,
$B^{\prime}$ are the factorisation operators corresponding to the magnetic
fields $H$, $H^{\prime}$ and let $\tilde{H}=H^{\prime}-H$, $Y=B^{\prime}-B$.
Let us fix boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If $Y\in\Psi
DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+})$ and $\tilde{H}=0$ at $x_{3}=0$,
then we have $\tilde{H}\in\Psi DO^{(0,\kappa)}(\partial M,\mathbb{R}^{+})$ and
for $1\leq a$

$\sigma_{p}\left(D_{x_{3}}^{a}\tilde{H}\right)=\sum_{j=0}^{a-1}\sum_{n=0}^{j}\left(\begin{matrix}j\\\
n\end{matrix}\right)(B^{(1)})^{a-1-j}\sigma_{p}\left(D_{x_{3}}^{j-n}Y\right)(B^{(1)})^{n}H^{(0)}$
(56)

at $x_{3}=0$.

Proof: Making use of Theorem 7 we have that

$D_{x^{3}}\tilde{H}=YH+B\tilde{H}+Y\tilde{H}$ (57)

modulo smoothing. In the case $\kappa=1$, by taking the principal symbol of
(57) the proof is complete. So assume that $\kappa>1$. Starting from this, we
prove inductively that for $1\leq a\leq\kappa$,

$D_{x^{3}}^{a}\tilde{H}=\sum_{j=0}^{a-1}B^{a-1-j}D_{x_{3}}^{j}(YH)+B^{a}\tilde{H}+\tilde{R}_{a}\tilde{H}+R_{a}$
(58)

where $\tilde{R}_{a}\in\Psi DO^{(a-1,0)}(\partial M,\mathbb{R}^{+})$ and
$R_{a}\in\Psi DO^{(a,\kappa+1)}(\partial M,\mathbb{R}^{+})$. Note that the
case $a=1$ of (58) is already proven from (57) with $\tilde{R}_{1}=0$ and
$R_{1}=Y\tilde{H}$.

Now assume, by induction, that (58) is true for some $1\leq a$. Using Pascal’s
triangle we obtain

$D_{x_{3}}^{j}(YH)=\sum_{n=0}^{j}\left(\begin{matrix}j\\\
n\end{matrix}\right)(D_{x_{3}}^{j-n}Y)B^{n}H,$ (59)

and applying Lemma 6 to each term separately conclude that
$D_{x_{3}}^{j}(YH)\in\Psi DO^{j+1,\kappa}(\partial M,\mathbb{R}^{+})$.
Therefore, applying $D_{x^{3}}$ to (58) gives

$D_{x^{3}}^{a+1}\tilde{H}=\sum_{j=0}^{a-1}B^{a-1-j}D_{x_{3}}^{j+1}(YH)+B^{a}D_{x^{3}}\tilde{H}+\tilde{R}_{a+1}\tilde{H}+R_{a+1}$

where $\tilde{R}_{a+1}\in\Psi DO^{(a,0)}(\partial M,\mathbb{R}^{+})$ and
$R_{a+1}\in\Psi DO^{(a+1,\kappa+1)}(\partial M,\mathbb{R}^{+})$ contain the
derivatives of the corresponding terms from the $a$ step, as well as terms
involving derivatives of $B$. Using (57), rearranging the terms and changing
the remainder $R_{a+1}$, we see that (58) holds for $a+1$. This completes the
induction and so proves (58).

Using (58) and (59) and the fact that $\tilde{H}$ vanishes at the boundary,
the principal symbol of $D^{a}_{x^{3}}\tilde{H}$ at $x_{3}=0$ is given by
(56). This implies that, at $x_{3}=0$, $D^{a}_{x^{3}}\tilde{H}\in\Psi
DO^{(a-\kappa)}(\partial M)$ and so, since this is true for $1\leq
a\leq\kappa$, we get $\tilde{H}\in\Psi DO^{(0,\kappa)}(\partial
M,\mathbb{R}^{+})$.

######  Theorem 12.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $E$, $E^{\prime}$ the associated electric
fields, respectively and let $\tilde{E}=E^{\prime}-E$. Assume that $C$,
$C^{\prime}$ are the factorisation operators corresponding to the magnetic
fields $E$, $E^{\prime}$ and let $J=C^{\prime}-C$. If $J\in\Psi
DO^{(1,\kappa)}(\partial M)$ and $\tilde{E}=0$ at $x_{3}=0$, then have
$\tilde{E}\in\Psi DO^{(0,\kappa)}(\partial M,\mathbb{R}^{+})$ and for $1\leq
a$

$\sigma_{p}(D_{x_{3}}^{a}\tilde{E})=\sum_{j=0}^{a-1}\sum_{n=0}^{j}\left(\begin{matrix}j\\\
n\end{matrix}\right)(C^{(1)})^{a-1-j}\sigma_{p}(D_{x_{3}}^{j-n}J)(C^{(1)})^{n}E^{(0)},$
(60)

at $x_{3}=0$.

Proof: The proof is done in a similar manner to the proof of Theorem 11 and
therefore is omitted.

We now continue to calculate the difference of the normal derivatives of
Maxwell’s and divergence equations for the two sets of parameters.

######  Theorem 13.

Let $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$, $H$, $H^{\prime}$, $E$, $E^{\prime}$
the associated impedance maps and electromagnetic fields, respectively and
$\tilde{H}=H^{\prime}-H$, $\tilde{E}=E^{\prime}-E$. Assume that $B,B^{\prime}$
are the factorisation operators corresponding to the magnetic fields $H$,
$H^{\prime}$ and let $Y=B^{\prime}-B$. Assume further that $C$, $C^{\prime}$
are the factorisation operators corresponding to the electric fields $E$,
$E^{\prime}$ and $J=C^{\prime}-C$. Let us fix boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If, for some $\kappa\geq 1$,

  * •

$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}}$,

  * •

$\hat{\varepsilon}=\hat{\varepsilon}^{\prime}+x_{3}^{\kappa}e_{\varepsilon},\quad\hat{\mu}=\hat{\mu}^{\prime}+x_{3}^{\kappa}e_{\mu},$
where $e_{\varepsilon},e_{\mu}$ are symmetric,

  * •

$\sqrt{|\hat{\varepsilon}^{-1}|}=\sqrt{|\hat{\varepsilon}^{\prime-1}|}+x_{3}^{\kappa}r_{\varepsilon}$,
$\sqrt{|\hat{\mu}^{-1}|}=\sqrt{|\hat{\mu}^{\prime-1}|}+x_{3}^{\kappa}r_{\mu}$,

  * •

$Y$, $J$ $\in$ $\Psi DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+})$,

then,

$\displaystyle\sigma_{p}\left(D_{x_{3}}^{\kappa}\tilde{H}_{\tilde{c}}\right)$
$\displaystyle=$ $\displaystyle
0,\quad\sigma_{p}\left(D_{x_{3}}^{\kappa}\tilde{H}_{3}\right)=(-i)^{\kappa}\kappa!\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)\frac{H_{j}^{(0)}}{\hat{\mu}^{\prime
33}},$ (61)
$\displaystyle\sigma_{p}(D_{x_{3}}^{\kappa+1}\tilde{H}_{\tilde{c}})$
$\displaystyle=$
$\displaystyle(-i)^{\kappa}\kappa!\xi_{\tilde{c}}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)\frac{H_{j}^{(0)}}{\hat{\mu}^{\prime
33}},$ (62) $\displaystyle\hat{\mu}^{\prime
3j}\sigma_{p}\left(D_{x_{3}}^{\kappa+1}\tilde{H}_{j}\right)$ $\displaystyle=$
$\displaystyle(-i)^{\kappa}\kappa!\Bigg{(}-\frac{\hat{\mu}^{\prime\tilde{i}3}\xi_{\tilde{i}}}{\hat{\mu}^{\prime
33}}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)$ $\displaystyle\hskip 28.45274pt+\
e_{\mu}^{\tilde{i}j}\xi_{\tilde{i}}+2e_{\mu}^{3l}(B^{(1)})_{l}^{j}+\frac{r_{\mu}\hat{\mu}^{\prime
3l}}{\sqrt{|\hat{\mu}^{\prime-1}|}}(B^{(1)})_{l}^{j}\Bigg{)}H_{j}^{(0)},\quad\quad$

at $x_{3}=0$ and if $\kappa\geq 2$ then

$\sigma_{p}(D_{x_{3}}^{a}\tilde{H}_{j})=0,\quad 1\leq a\leq\kappa-1.$ (64)

at $x_{3}=0$.

Proof: First note that, by Theorem 11, $\tilde{H}\in\Psi
DO^{(0,\kappa)}(\partial M,\mathbb{R}^{+})$. Let us first consider the
divergence equations for the magnetic fields $H$ and $H^{\prime}$, which are

$-i\delta_{\hat{\mu}}(H)=0,\quad-i\delta_{\hat{\mu}^{\prime}}(H^{\prime})=0.$

Subtracting these equations and writing the result in boundary normal
coordinates for $\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$, we obtain

$\begin{split}\hat{\mu}^{\prime
3j}D_{x_{3}}\tilde{H}_{j}&=-\hat{\mu}^{\prime\tilde{i}j}D_{x_{\tilde{i}}}\tilde{H}_{j}+x_{3}^{\kappa}e_{\mu}^{\tilde{i}j}D_{x_{\tilde{i}}}H_{j}+x_{3}^{\kappa}e_{\mu}^{3j}D_{x_{3}}H_{j}\\\
&\hskip 28.45274pt-i\kappa
x_{3}^{\kappa-1}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)H_{j}+R\end{split}$ (65)

where $R\in\Psi DO^{(0,\kappa)}(\partial M,\mathbb{R}^{+})$ and the other
terms on the right are in $\Psi DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+})$.
Applying $D_{x_{3}}^{a}$ for $a\geq 0$ to (65), we arrive at

$\begin{split}\hat{\mu}^{\prime
3j}D_{x_{3}}^{a+1}\tilde{H}_{j}&=D_{x_{3}}^{a}\Bigg{(}-\hat{\mu}^{\prime\tilde{i}j}D_{x_{\tilde{i}}}\tilde{H}_{j}+x_{3}^{\kappa}e_{\mu}^{\tilde{i}j}D_{x_{\tilde{i}}}H_{j}+x_{3}^{\kappa}e_{\mu}^{3j}D_{x_{3}}H_{j}\\\
&-i\kappa x_{3}^{\kappa-1}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)H_{j}\Bigg{)},\end{split}$ (66)

modulo $\Psi DO^{(a,\kappa)}(\partial M)$. Each of the terms in the right
hand-side of (66) expands as

$\begin{split}D_{x_{3}}^{a}(-\mu^{\prime\tilde{i}j}D_{x_{\tilde{i}}}\tilde{H}_{j})&=-\hat{\mu}^{\prime\tilde{i}j}D_{x_{\tilde{i}}}D_{x_{3}}^{a}\tilde{H}_{j},\\\
D_{x_{3}}^{a}(x_{3}^{\kappa}e_{\mu}^{\tilde{i}j}D_{x_{\tilde{i}}}H_{j})&=\sum_{n=0}^{a}\left(\begin{matrix}a\\\
n\end{matrix}\right)(D_{x_{3}}^{n}x_{3}^{\kappa})e_{\mu}^{\tilde{i}j}D_{x_{\tilde{i}}}D_{x_{3}}^{a-n}H_{j},\\\
D_{x_{3}}^{a}(x_{3}^{\kappa}e_{\mu}^{3j}D_{x_{3}}H_{j})&=\sum_{n=0}^{a}\left(\begin{matrix}a\\\
n\end{matrix}\right)(D_{x_{3}}^{n}x_{3}^{\kappa})e_{\mu}^{3j}D_{x_{3}}^{a+1-n}H_{j},\\\
D_{x_{3}}^{a}\left(-i\kappa
x_{3}^{\kappa-1}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)H_{j}\right)&=\\\
&\hskip-65.44142pt-i\kappa\sum_{n=0}^{a}\left(\begin{matrix}a\\\
n\end{matrix}\right)(D_{x_{3}}^{n}x_{3}^{\kappa-1})\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)D_{x_{3}}^{a-n}H_{j},\end{split}$
(67)

modulo $\Psi DO^{(a,\kappa)}(\partial M)$. Let us consider the case of
$\kappa=1$. The principal symbols of equation (66) for $a=0$ and $a=1$ at
$x_{3}=0$ give respectively

$\begin{split}\hat{\mu}^{\prime
3l}\sigma_{p}(D_{x_{3}}\tilde{H}_{j})H_{j}^{(0)}&=-i\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)H_{j}^{(0)},\\\ \hat{\mu}^{\prime
3j}\sigma_{p}(D_{x_{3}}^{2}\tilde{H}_{j})&=-\hat{\mu}^{\prime\tilde{i}k}\xi_{\tilde{i}}\sigma_{p}(D_{x_{3}}\tilde{H}_{j})\\\
&\hskip
5.69046pt+(-i)\Bigg{(}e_{\mu}^{\tilde{i}k}\xi_{\tilde{i}}+2e_{\mu}^{3j}(B^{(1)})_{j}^{k}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}(B^{(1)})_{j}^{k}\Bigg{)}H_{k}^{(0)}.\end{split}$
(68)

Assuming that $\kappa\geq 2$ and considering the principal symbols of equation
(66) for $1\leq a\leq\kappa$ at $x_{3}=0$ we obtain

$\begin{split}\hat{\mu}^{\prime
3j}\left(D_{x_{3}}^{l}\tilde{H}_{j}\right)&=-\hat{\mu}^{\prime\tilde{i}j}\xi_{\tilde{i}}\sigma_{p}(D_{x_{3}}^{l-1}\tilde{H}_{j}),\quad
1\leq l\leq\kappa-1,\\\ \hat{\mu}^{\prime
3j}\sigma_{p}\left(D_{x_{3}}^{\kappa}\tilde{H}_{j}\right)&=-\hat{\mu}^{\prime\tilde{i}j}\xi_{\tilde{i}}\sigma_{p}(D_{x_{3}}^{\kappa-1}\tilde{H}_{j})+(-i)^{\kappa}\kappa!\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)H_{j}^{(0)},\\\ \hat{\mu}^{\prime
3j}\sigma_{p}\left(D_{x_{3}}^{\kappa+1}\tilde{H}_{j}\right)&=-\hat{\mu}^{\prime\tilde{i}j}\xi_{\tilde{i}}\sigma_{p}(D_{x_{3}}^{\kappa}\tilde{H}_{j})\\\
&\hskip
5.69046pt+(-i)^{\kappa}\kappa!\Bigg{(}e_{\mu}^{\tilde{i}k}\xi_{\tilde{i}}+2e_{\mu}^{3l}(B^{(1)})_{l}^{k}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}(B^{(1)})_{j}^{k}\Bigg{)}H_{k}^{(0)},\end{split}$
(69)

where we have made use of equations (67) for $1\leq a\leq\kappa$ at $x_{3}=0$.

In order to derive the equations regarding the tangential components of
$\tilde{H}$ we start with Maxwell’s equation

$\ast_{\hat{\varepsilon}}\mathrm{d}H=-i\omega E,$

which is written in boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$ as

$\sigma^{lbc}D_{x_{b}}H_{c}=\omega\frac{\hat{\varepsilon}^{lk}}{\sqrt{\lvert\hat{\varepsilon}\rvert}}E_{k}.$

Evaluating the above at either $l=1$ or $l=2$ and taking into account that we
work in boundary normal coordinates for $\hat{\varepsilon}$ implies

$D_{x_{3}}H_{\tilde{c}}=D_{x_{\tilde{c}}}H_{3}+\omega\sigma_{\tilde{a}3\tilde{c}}\frac{\hat{\varepsilon}^{lk}}{\sqrt{\lvert\hat{\varepsilon}\rvert}}E_{\tilde{k}}.$
(70)

Subtracting the corresponding equation for the ${}^{\prime}-$ parameters, we
arrive at

$D_{x_{3}}\tilde{H}_{\tilde{c}}=D_{x_{\tilde{c}}}\tilde{H}_{3}+\omega\sigma_{\tilde{a}3\tilde{c}}\left(\frac{\hat{\varepsilon}^{\prime
lk}}{\sqrt{\lvert\hat{\varepsilon}^{\prime}\rvert}}E^{\prime}_{\tilde{k}}-\frac{\hat{\varepsilon}^{lk}}{\sqrt{\lvert\hat{\varepsilon}\rvert}}E_{\tilde{k}}\right).$

Using Theorem 12 and the hypotheses, we obtain modulo $\Psi
DO^{(0,\kappa)}(\partial M)$

$\begin{split}D_{x_{3}}\tilde{H}_{\tilde{c}}&=D_{x_{\tilde{c}}}\tilde{H}_{3}.\end{split}$

Applying $D_{x_{3}}$ $a$ more times leads to

$\begin{split}D_{x_{3}}^{a+1}\tilde{H}_{\tilde{c}}=D_{x_{\tilde{c}}}D_{x_{3}}^{a}\tilde{H}_{3},\quad\mbox{modulo}\
\Psi DO^{(a,\kappa)}(\partial M)\end{split}$

for $0\leq a\leq\kappa$. The principal symbol of the above equation at
$x_{3}=0$ gives

$\sigma_{p}(D_{x_{3}}^{a+1}\tilde{H}_{\tilde{c}})=\xi_{\tilde{c}}\sigma_{p}(D_{x_{3}}^{a}\tilde{H}_{3}),\quad
0<a\leq\kappa.$ (71)

Combining equations (68) and (69) with (71) completes the proof.

######  Theorem 14.

Assume the same hypotheses as for Theorem 13 and in addition

  * •

$\hat{\varepsilon}=\hat{\varepsilon}^{\prime}+x_{3}^{\kappa+1}e_{\varepsilon}$,

  * •

$\sqrt{|\hat{\varepsilon}^{-1}|}=\sqrt{|\hat{\varepsilon}^{\prime-1}|}+x_{3}^{\kappa+1}r_{\varepsilon}$.

Then

$\displaystyle\sigma_{p}(D_{x^{3}}^{\kappa+1}\tilde{E}_{\tilde{c}})$
$\displaystyle=$ $\displaystyle
0,\quad\sigma_{p}\left(D_{x_{3}}^{{\kappa+1}}\tilde{E}_{3}\right)=(-i)^{\kappa+1}(\kappa+1)!\frac{r_{\varepsilon}}{\sqrt{|\hat{\varepsilon}^{\prime-1}|}}E_{3}^{(0)},$
(72)
$\displaystyle\sigma_{p}\left(D_{x_{3}}^{\kappa+2}\tilde{E}_{\tilde{c}}\right)$
$\displaystyle=$
$\displaystyle(-i)^{\kappa+1}(\kappa+1)!\xi_{\tilde{c}}\frac{r_{\varepsilon}}{\sqrt{|\hat{\varepsilon}^{\prime-1}|}}E_{3}^{(0)},$
(73) $\displaystyle\sigma_{p}\left(D_{x_{3}}^{\kappa+2}\tilde{E}_{3}\right)$
$\displaystyle=$
$\displaystyle(-i)^{\kappa+1}(\kappa+1)!\Bigg{(}e_{\varepsilon}^{\tilde{i}j}\xi_{\tilde{i}}+\frac{r_{\varepsilon}}{\sqrt{|\hat{\varepsilon}^{\prime-1}|}}(C^{(1)})_{3}^{j}\Bigg{)}E_{j}^{(0)},\quad\quad$
(74) $\displaystyle\sigma_{p}(D_{x_{3}}^{a}\tilde{E}_{j})$ $\displaystyle=$
$\displaystyle 0,\quad 1\leq a\leq\kappa$ (75)

at $x_{3}=0$.

Proof: Working as in the proof of Theorem 13 leads to the desired result. Note
that the calculations and results are simpler in this case because of the
choice of boundary normal coordinates.

###  8.1 Proof of Theorem 5

To prove Theorem 5, it suffices to prove Theorems 15, 16 and 17 below. More
specifically, in each inductive step we start with Theorem 15 to prove given
that the normal derivatives of order $\kappa-1\geq 0$ of $\hat{\varepsilon}$
and $\hat{\varepsilon}^{\prime}$ agree at $x_{3}=0$ the same follows for their
normal derivatives of order $\kappa$ at $x_{3}=0$. Then, we proceed to Theorem
16 to prove the corresponding result for the metrics $\hat{\mu}$ and
$\hat{\mu}^{\prime}$. Last, we have to show that as the normal derivatives of
the metrics induced by the electromagnetic parameters agree to a higher order
at $x_{3}=0$ the order of the pseudodifferential operators $Y$ and $J$
decreases at $x_{3}=0$. This is shown in Theorem 17.

In what follows, we will work with specific values of the principal symbols of
$H$ and $E$ by choosing $H^{(0)}=\chi_{\hat{\varepsilon}}$ and
$E^{(0)}=\chi_{\hat{\mu}}$. Recall Section 6 for the definitions of these
covectors. These covectors are not defined when
$\xi_{\hat{\varepsilon}3}=\xi_{\hat{\mu}3}$, but in these cases the
conclusions still follow by density since, as observed in Remark 3, the
symbols are continuous with respect to the parameters.

######  Theorem 15.

Let $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated impedance maps. Assume
that $B$, $B^{\prime}$ are the factorisation operators corresponding to the
magnetic fields $H$, $H^{\prime}$ and let $Y=B^{\prime}-B$. Let us fix
boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If

$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}},$
$\partial_{x_{3}}^{a}\hat{\varepsilon}=\partial_{x_{3}}^{a}\hat{\varepsilon}^{\prime},\quad\partial_{x_{3}}^{a}\hat{\mu}=\partial_{x_{3}}^{a}\hat{\mu}^{\prime},\quad\mbox{at}\
x_{3}=0,$

for $a=0,\dots,\kappa-1$ and

$Y\in\Psi DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+}),$

then

$\partial_{x_{3}}^{\kappa}\hat{\varepsilon}=\partial_{x_{3}}^{\kappa}\hat{\varepsilon}^{\prime},$

at $x_{3}=0$.

Proof: Since $Y\in\Psi DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+})$, both sides
of Riccati equation (54) are in $\Psi DO^{(2,\kappa)}(\partial
M,\mathbb{R}^{+})$ and the vector of principal symbols, in the sense of
Definition 2, is given by

$\begin{split}&A_{H^{\prime}}Y^{(1)}+T_{H^{\prime}}B^{(1)}Y^{(1)}+T_{H^{\prime}}Y^{(1)}B^{(1)}=\\\
&\hskip
14.22636pt\frac{1}{\kappa!}\left(\partial_{x_{3}}^{\kappa}(A_{H}-A_{H^{\prime}})B^{(1)}+\partial_{x_{3}}^{\kappa}(T_{H}-T_{H^{\prime}})(B^{(1)})^{2}+\partial_{x_{3}}^{\kappa}(Q_{H}-Q_{H^{\prime}})\right),\\\
&-\kappa
iT_{H^{\prime}}Y^{(1)}+A_{H^{\prime}}Y^{(0)}+T_{H^{\prime}}B^{(1)}Y^{(0)}+T_{H^{\prime}}Y^{(0)}B^{(1)}=\\\
&\hskip
85.35826pt\frac{1}{(\kappa-1)!}\left(\partial_{x^{3}}^{\kappa-1}(G_{H}-G_{H^{\prime}})B^{(1)}+\partial_{x^{3}}^{\kappa-1}(F_{H}-F_{H^{\prime}})\right),\\\
&-(\kappa-1)iT_{H^{\prime}}Y^{(0)}+A_{H^{\prime}}Y^{(-1)}+T_{H^{\prime}}B^{(1)}Y^{(-1)}+T_{H^{\prime}}Y^{(-1)}B^{(1)}=\\\
&\hskip
213.39566pt\frac{\partial_{x_{3}}^{\kappa-2}}{(\kappa-2)!}(R_{H}-R_{H^{\prime}}),\\\
&-(\kappa-2)iT_{H^{\prime}}Y^{(-1)}+A_{H^{\prime}}Y^{(-2)}+T_{H^{\prime}}B^{(1)}Y^{(-2)}+T_{H^{\prime}}Y^{(-2)}B^{(1)}=0,\\\
&\hskip 142.26378pt\vdots\\\
&-iT_{H^{\prime}}Y^{-(\kappa-2)}+A_{H^{\prime}}Y^{-(\kappa-1)}+T_{H^{\prime}}B^{(1)}Y^{-(\kappa-1)}+T_{H^{\prime}}Y^{-(\kappa-1)}B^{(1)}=0.\end{split}$
(76)

The different sets of electromagnetic parameters are assumed to agree to order
$\kappa-1$ at $x_{3}=0$, which implies that

$\hat{\varepsilon}^{-1}=\hat{\varepsilon}^{\prime-1}+x_{3}^{\kappa}\tilde{e}_{\varepsilon},\quad\hat{\mu}=\hat{\mu}^{\prime}+x_{3}^{\kappa}e_{\mu}.$

To prove the theorem, we aim to prove that $\tilde{e}_{\varepsilon}=0$ at
$x_{3}=0$ beginning by applying the second equation of (76) to
$H^{(0)}=\chi_{\hat{\epsilon}}$. If $\kappa=1$, Lemma 7 is directly satisfied
which implies the desired result. If $\kappa\geq 2$, in order for the
assumption of Lemma 7 for $Y$ to be satisfied we additionally use (56) and
(64) which when combined give

$Y^{(l)}\chi_{\varepsilon}=0,\quad 1-\kappa\leq l\leq-1.$

at $x_{3}=0$. Therefore the assumptions of Lemma 7 are satisfied for any
$\kappa\geq 1$ and the proof is completed.

######  Theorem 16.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated impedance maps. Assume
that $C$, $C^{\prime}$ are the factorisation operators corresponding to the
electric fields $E$, $E^{\prime}$ and let $J=C^{\prime}-C$. Let us fix
boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If

$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}},$
$\partial_{x_{3}}^{a}\hat{\varepsilon}=\partial_{x_{3}}^{a}\hat{\varepsilon}^{\prime},\quad\partial_{x_{3}}^{b}\hat{\mu}=\partial_{x_{3}}^{b}\hat{\mu}^{\prime},\quad\mbox{at}\
x_{3}=0,$

for $a=0,1,2,\dots,\kappa$, $b=0,1,\dots,\kappa-1$ and

$J\in\Psi DO^{(1,\kappa)}(\partial M),$

then

$\partial_{x_{3}}^{\kappa}\hat{\mu}=\partial_{x_{3}}^{\kappa}\hat{\mu}^{\prime},\quad\mbox{at}\
x_{3}=0.$ (77)

Proof: Given that $J$ $\in$ $\Psi DO^{(1,\kappa)}(\partial M)$ the principal
symbol of the Riccati equation (55) at $x_{3}=0$ is given by

$\begin{split}&A_{E}J^{(1)}+T_{E}C^{(1)}J^{(1)}+T_{E}J^{(1)}C^{(1)}=\\\
&\hskip
28.45274pt\frac{1}{\kappa!}\left(\partial_{x_{3}}^{\kappa}(A_{E}-A_{E^{\prime}})C^{(1)}+\partial_{x_{3}}^{\kappa}(T_{E}-T_{E^{\prime}})(C^{(1)})^{2}+\partial_{x_{3}}^{\kappa}(Q_{E}-Q_{E^{\prime}})\right),\\\
&-\kappa
iT_{E}J^{(1)}+A_{E}J^{(0)}+T_{E}C^{(1)}J^{(0)}+T_{E}J^{(0)}C^{(1)}=\\\ &\hskip
93.89418pt\frac{1}{(\kappa-1)!}\left(\partial_{x^{3}}^{\kappa-1}(G_{E}-G_{E^{\prime}})C^{(1)}+\partial_{x^{3}}^{\kappa-1}(F_{E}-F_{E^{\prime}})\right),\\\
&-(\kappa-1)iT_{E}J^{(0)}+A_{E}J^{(-1)}+T_{E}C^{(1)}J^{(-1)}+T_{E}J^{(-1)}C^{(1)}\\\
&\hskip
113.81102pt=\frac{1}{(\kappa-2)!}\partial_{x_{3}}^{\kappa-2}(R_{E}-R_{E^{\prime}}),\\\
&-(\kappa-2)iT_{E}J^{(-1)}+A_{E}J^{(-2)}+T_{E}C^{(1)}J^{(-2)}+T_{E}J^{(-2)}C^{(1)}=0,\\\
&\hskip 142.26378pt\vdots\\\
&-iT_{E}J^{-(\kappa-2)}+A_{E}J^{-(\kappa-1)}+T_{E}C^{(1)}J^{-(\kappa-1)}+T_{E}J^{-(\kappa-1)}C^{(1)}=0.\end{split}$
(78)

According to the assumptions of the theorem, the different sets of metrics
satisfy

$\hat{\varepsilon}=\hat{\varepsilon}^{\prime}+x_{3}^{\kappa+1}e_{\varepsilon},\quad(\hat{\mu}^{-1})=(\hat{\mu}^{\prime-1})+x_{3}^{\kappa}\tilde{e}_{\mu},$

for $e_{\varepsilon}$, $\tilde{e}_{\mu}$ symmetric. Let us fix
$E^{(0)}=\chi_{\hat{\mu}}$. In order for Lemma 8 to be applicable, we will
show that for any $\kappa\geq 1$ we have

$J^{(l)}\chi_{\hat{\mu}}=0,\quad 1-\kappa\leq l\leq 1,$ (79)

at $x_{3}=0$. Equations (60) and (75) show that for any $\kappa\geq 1$

$J^{(l)}\chi_{\hat{\mu}}=0,\quad 1-\kappa\leq l\leq 0\quad\mbox{at}\ x_{3}=0.$

To show that $J^{(1)}$ also vanishes, we consider the second equation in (78),
(19), (21), (23), (60) and (72). Using all of these formulas and taking the
$\hat{\mu}^{\prime}$ inner product with $\nu$, we obtain that $r_{\epsilon}$
appearing in (72) must be zero. This implies, again by (60) and (72), that
$J^{(1)}\chi_{\hat{\mu}}=0$. Hence, we obtain that equation (79) holds for any
$\kappa\geq 1$ which allows us to apply Lemma 8 and the proof is completed.

The following two lemmas are used in the proofs of Theorems 13 and 14 and
include many of the technical calculations.

######  Lemma 7.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters. Let us fix boundary normal coordinates for
$\hat{\varepsilon}$/ $\hat{\varepsilon}^{\prime}$. If

$\hat{\varepsilon}^{-1}=\hat{\varepsilon}^{\prime-1}+x_{3}^{\kappa}\tilde{e}_{\varepsilon},\quad\hat{\mu}=\hat{\mu}^{\prime}+x_{3}^{\kappa}e_{\mu}$
(80)

and $Y$ $\in$ $\Psi DO^{(1,\kappa)}(\partial M)$ such that for any $\kappa\geq
2$

$Y^{(a)}\chi_{\hat{\varepsilon}}=0,\quad 1-\kappa\leq a\leq-1,$

then for any $\kappa\geq 1$, equation

$\begin{split}&\left(T_{H^{\prime}}(-\kappa
iY^{(1)}+B^{(1)}Y^{(0)}+Y^{(0)}B^{(1)})+A_{H^{\prime}}Y^{(0)}\right)\chi_{\hat{\varepsilon}}=\\\
&\hskip
56.9055pt\frac{1}{(\kappa-1)!}\left(\partial_{x^{3}}^{\kappa-1}(G_{H}-G_{H^{\prime}})B^{(1)}+\partial_{x^{3}}^{\kappa-1}(F_{H}-F_{H^{\prime}})\right)\chi_{\hat{\varepsilon}}\end{split}$
(81)

implies that

$\partial_{x_{3}}^{\kappa}\hat{\varepsilon}=\partial_{x_{3}}^{\kappa}\hat{\varepsilon}^{\prime},$

Proof: We first show that for any $\kappa\geq 1$,

$(Y^{(0)})_{\tilde{l}}^{k}{\chi_{\hat{\varepsilon}}}_{k}=0,\quad(Y^{(0)})_{3}^{k}{\chi_{\hat{\varepsilon}}}_{k}=-i\kappa\Bigg{(}\frac{\langle\nu,\chi_{\hat{\varepsilon}}\rangle_{e_{\mu}}}{\hat{\mu}^{\prime
33}}+\frac{r_{\mu}\langle\nu,\chi_{\hat{\varepsilon}}\rangle_{\hat{\mu}^{\prime}}}{\sqrt{|\hat{\mu}^{\prime-1}|}\hat{\mu}^{\prime
33}}\Bigg{)},$ (82)

at $x_{3}=0$. Evaluating equation (56) for $a=\kappa$ at
$H^{(0)}=\chi_{\varepsilon}$ results to

$\sigma_{p}(D_{x_{3}}^{\kappa}\tilde{H})|_{H^{(0)}=\chi_{\hat{\varepsilon}}}=\sum_{j=0}^{\kappa-1}\sum_{n=0}^{j}\left(\begin{matrix}j\\\
n\end{matrix}\right)(B^{(1)})^{\kappa-1-j}\sigma_{p}(D_{x_{3}}^{j-n}Y)(B^{(1)})^{n}\chi_{\hat{\varepsilon}}.$

The assumption of the lemma on $Y$ implies that

$\sigma_{p}(D_{x_{3}}^{\kappa}\tilde{H})|_{H^{(0)}=\chi_{\hat{\varepsilon}}}=\sigma_{p}(D_{x_{3}}^{\kappa-1}Y)\chi_{\hat{\varepsilon}}=(-i)^{\kappa-1}(\kappa-1)!Y^{(0)}\chi_{\hat{\varepsilon}},$

at $x_{3}=0$. Equation (61) then implies (82).

Now let us write (56) for $a=\kappa+1$ evaluated at
$H^{(0)}=\chi_{\hat{\varepsilon}}$, which gives

$\sigma_{p}(D_{x_{3}}^{\kappa+1}\tilde{H})|_{H^{(0)}=\chi_{\hat{\varepsilon}}}=\sum_{j=0}^{\kappa}\sum_{n=0}^{j}\left(\begin{matrix}j\\\
n\end{matrix}\right)(B^{(1)})^{\kappa-j}\sigma_{p}(D_{x_{3}}^{j-n}Y)(B^{(1)})^{n}\chi_{\hat{\varepsilon}}.$

The assumption of the theorem on $Y$ implies that the only nonzero terms in
the above arise for $j=\kappa$, $n=0,1$ and $j=\kappa-1$, $n=0$. Thus, we have

$\sigma_{p}(D_{x_{3}}^{\kappa+1}\tilde{H})|_{H^{(0)}=\chi_{\hat{\varepsilon}}}=(-i)^{\kappa}\kappa!Y^{(1)}\chi_{\hat{\varepsilon}}+(-i)^{\kappa-1}(\kappa-1)!\left(B^{(1)}Y^{(0)}+Y^{(0)}B^{(1)}\right)\chi_{\hat{\varepsilon}}.$

Hence, we observe that the left hand-side of (81) is equal to

$T_{H^{\prime}}\frac{\sigma_{p}(D_{x_{3}}^{\kappa+1}\tilde{H})|_{H^{(0)}=\chi_{\hat{\varepsilon}}}}{(-i)^{\kappa-1}(\kappa-1)!}+A_{H^{\prime}}Y^{(0)}\chi_{\hat{\varepsilon}}.$
(83)

Using equations (19), (20), (62), (13) and (82), the left hand side of (81),
as shown in (83), is equal to

$-i\kappa\frac{\sqrt{|\hat{\mu}^{-1}|}}{\sqrt{|\hat{\varepsilon}^{-1}|}}\left((\hat{\varepsilon}^{\prime-1})_{lq}\hat{\mu}^{q3}\langle\xi_{\hat{\epsilon}},\chi_{\varepsilon}\rangle_{e_{\mu}}+(\hat{\varepsilon}^{\prime-1})_{lq}\hat{\mu}^{qa}\xi_{\hat{\epsilon}a}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)\chi_{\hat{\epsilon}j}\right).$ (84)

Using formulas (21), (23) and (80), the right hand side of (81) is written in
coordinates as

$\begin{split}&-i\kappa\frac{\sqrt{|\hat{\mu}^{-1}|}}{\sqrt{|\hat{\varepsilon}^{-1}|}}\left((\hat{\varepsilon}^{\prime-1})_{lq}\hat{\mu}^{q3}\langle\xi_{\hat{\epsilon}},\chi_{\varepsilon}\rangle_{e_{\mu}}+(\hat{\varepsilon}^{\prime-1})_{lq}\hat{\mu}^{qa}\xi_{\hat{\epsilon}a}\left(e_{\mu}^{3j}+\frac{r_{\mu}\hat{\mu}^{\prime
3j}}{\sqrt{|\hat{\mu}^{\prime-1}|}}\right)\chi_{\hat{\epsilon}j}\right)\\\
&\hskip
85.35826pt+i\kappa\frac{(\hat{\varepsilon}^{\prime-1})_{lq}}{|\hat{\varepsilon}^{\prime-1}|}\sigma^{3\tilde{j}\tilde{q}}(\tilde{e}_{\varepsilon})_{\tilde{b}\tilde{j}}\sigma^{dk\tilde{b}}\xi_{\hat{\varepsilon}d}\chi_{\hat{\varepsilon}k}.\end{split}$
(85)

Equating (84) with (85) we arrive at

$i\kappa\frac{(\hat{\varepsilon}^{\prime-1})_{lq}}{|\hat{\varepsilon}^{\prime-1}|}\sigma^{3\tilde{j}\tilde{q}}(\tilde{e}_{\varepsilon})_{\tilde{b}\tilde{j}}\sigma^{dk\tilde{b}}\xi_{\hat{\varepsilon}d}\chi_{\hat{\varepsilon}k}=0,$

which, substituting a coordinate expression for $\chi_{\hat{\varepsilon}}$ and
cancelling some terms, is simplified to

$(\varepsilon^{-1})_{\tilde{l}\tilde{l^{\prime}}}\sigma^{3\tilde{\beta}\tilde{l^{\prime}}}(\tilde{e}_{\varepsilon})_{\tilde{b}\tilde{\beta}}\varepsilon^{\tilde{b}\tilde{b^{\prime}}}{\xi}_{\tilde{b^{\prime}}}=0.$

This holds for all $\tilde{\xi}$ and $l$ and so is satisfied if and only if
$\tilde{e}_{\varepsilon}=0$. Therefore $\hat{\varepsilon}^{-1}$ and
$\hat{\varepsilon}^{\prime-1}$ agree up to the $\kappa$ derivative, and so the
same is true for $\hat{\varepsilon}$ and $\hat{\varepsilon}^{\prime}$. This
completes the proof.

######  Lemma 8.

Let $(\hat{\varepsilon},\hat{\mu})$,
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters. Let us fix boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If

$\hat{\varepsilon}=\hat{\varepsilon}^{\prime}+x_{3}^{\kappa+1}e_{\varepsilon},\quad(\hat{\mu}^{-1})=(\hat{\mu}^{\prime-1})+x_{3}^{\kappa}\tilde{e}_{\mu}$

and for any $\kappa\geq 1$ the following equations hold at $x_{3}=0$

$\left(\partial_{x^{3}}^{\kappa-1}(G_{E}-G_{E^{\prime}})C^{(1)}+\partial_{x^{3}}^{\kappa-1}(F_{E}-F_{E^{\prime}})\right)\chi_{\hat{\mu}}=0,$
(86)
$\left(\partial_{x_{3}}^{\kappa}(A_{E}-A_{E^{\prime}})C^{(1)}+\partial_{x_{3}}^{\kappa}(Q_{E}-Q_{E^{\prime}})+\partial_{x_{3}}^{\kappa}(T_{E}-T_{E^{\prime}})(C^{(1)})^{2}\right)\chi_{\hat{\mu}}=0,$
(87)

then,

$\partial_{x_{3}}^{\kappa}\hat{\mu}=\partial_{x_{3}}^{\kappa}\hat{\mu}^{\prime},\quad\mbox{at}\
x_{3}=0.$

Proof: Making use of the symbols $G_{E}$, $F_{E}$, which are given by
equations (21) and (23) with $\hat{\varepsilon}$ and $\hat{\mu}$ switched, and
the assumptions of the lemma on the different sets of metrics, equation (86)
implies

$(\hat{\mu}^{-1})_{l\tilde{q}}\sigma^{3\tilde{j}\tilde{q}}(\tilde{e}_{\mu})_{\tilde{j}b}\sigma^{dkb}{\xi_{\hat{\mu}}}_{d}{\chi_{\hat{\mu}}}_{k}=0.$

Substituting the expression of $\chi_{\hat{\mu}}$. the above reduces to

$(\hat{\mu}^{-1})_{l\tilde{q}}\sigma^{3\tilde{j}\tilde{q}}(\tilde{e}_{\mu})_{\tilde{j}b}\hat{\mu}^{bb^{\prime}}{\xi_{\hat{\mu}}}_{b^{\prime}}=0,\quad\mbox{at}\
x_{3}=0.$

Therefore, we obtain

$(\tilde{e}_{\mu})_{\tilde{j}b}=0,\quad\mbox{at}\ x_{3}=0$ (88)

which implies in particular that $(\tilde{e}_{\mu})_{33}$ is the only
remaining nonzero component at $x_{3}=0$. Considering (87) while using the the
symbols $A_{E}$, $Q_{E}$ and $T_{E}$, determined respectively from (20), (22)
and (19), the hypothesis on the metrics as well as (88), we can calculate

$\ast_{\hat{\mu}}\left(\nu\wedge\ast_{\hat{\mu}}(\chi_{\hat{\mu}}\wedge\xi_{\hat{\mu}})\right)\langle\nu,\xi_{\hat{\mu}}\rangle_{\hat{\mu}}(\tilde{e}_{\mu})_{33}=0.$

Explicit calculation in coordinates using the definitions of
$\chi_{\hat{\mu}}$ and $\xi_{\hat{\mu}}$ shows that the coefficient in the
previous formula is not zero, and so we deduce

$(\tilde{e}_{\mu})_{33}=0,\quad\mbox{at}\ x_{3}=0.$

Therefore, we have shown that

$\partial_{x_{3}}^{\kappa}\hat{\mu}^{-1}=\partial_{x_{3}}^{\kappa}\hat{\mu}^{\prime-1},\quad\mbox{at}\
x_{3}=0,$

which completes the proof.

######  Theorem 17.

Let $(\hat{\varepsilon},\hat{\mu})$ and
$(\hat{\varepsilon}^{\prime},\hat{\mu}^{\prime})$ be two different sets of
electromagnetic parameters and $\Lambda_{\hat{\varepsilon}}$,
$\Lambda_{\hat{\varepsilon}^{\prime}}$ the associated impedance maps. Assume
that $B$, $B^{\prime}$ are the factorisation operators corresponding to the
magnetic fields $H$, $H^{\prime}$ and let $Y=B^{\prime}-B$. Assume further
that $C$, $C^{\prime}$ are the factorisation operators corresponding to the
electric fields $E$, $E^{\prime}$ and let $J=C^{\prime}-C$. Let us fix
boundary normal coordinates for
$\hat{\varepsilon}$/$\hat{\varepsilon}^{\prime}$. If

$\Lambda_{\hat{\varepsilon}}=\Lambda_{\hat{\varepsilon}^{\prime}},$
$\partial_{x^{3}}^{a}\hat{\varepsilon}=\partial_{x^{3}}^{a}\hat{\varepsilon}^{\prime},\quad\partial_{x^{3}}^{a}\hat{\mu}=\partial_{x^{3}}^{a}\hat{\mu}^{\prime},\quad\mbox{at}\
x_{3}=0,$

for $a=0,\dots\kappa$ and

$Y,J\in\Psi DO^{(1,\kappa)}(\partial M,\mathbb{R}^{+}),$

then

$Y,J\in\Psi DO^{(1,\kappa+1)}(\partial M,\mathbb{R}^{+}).$

Proof: The case $\kappa=0$ follows from the formula (31) and (34) for $B$,
$B^{\prime}$, $C$ and $C^{\prime}$ since the principal symbols depend only on
the values of the respective parameters at the boundary. For $\kappa\geq 1$,
it suffices to show that the equations

$\left(A_{H}+T_{H}B^{(1)}\right)Z+T_{H}ZB^{(1)}=0,$ (89)
$\left(A_{E}+T_{E}C^{(1)}\right)W+T_{E}WC^{(1)}=0,$ (90)

admit the unique solutions $Z=0$ and $W=0$, respectively. This follows from
the fact that after proving Theorems 15 and 16 the right hand-sides of the
equations in (76) and (78) vanish. We will prove the result for (89) and the
corresponding result for (90) follows in a similar way. Equation (89)
multiplied by $T_{H}^{-1}$ becomes

$\left(T_{H}^{-1}A_{H}+B^{(1)}\right)Z+ZB^{(1)}=0$ (91)

This is a homogeneous Sylvester equation with unknown $Z$. According to
(datta2004numerical, , Theorem 8.2.1), if $T_{H}^{-1}A_{H}+B^{(1)}$ and
$-B^{(1)}$ do not share any eigenvalues equation (91) admits the unique
solution

$Z=0.$

We showed that $B^{(1)}$ satisfies

$T_{H}(B^{(1)})^{2}+A_{H}B^{(1)}+Q_{H}=0.$

Hence we can write $T_{H}^{-1}A_{H}+B^{(1)}$ as

$T_{H}^{-1}A_{H}+B^{(1)}=-T_{H}^{-1}Q_{H}(B^{(1)})^{-1}.$

So we have to investigate if $T_{H}^{-1}Q_{H}(B^{(1)})^{-1}$ and $B^{(1)}$
share any eigenvalues. Revisiting the factorisation of $M_{H}(\xi_{3})$ in
terms of $B^{(1)}$ as given in equation (32) we have

$T_{H}\xi_{3}^{2}+A_{H}\xi_{3}+Q_{H}=\left(\mathbb{I}\xi_{3}-(B^{\ast})^{(1)}\right)T_{H}\left(\mathbb{I}\xi_{3}-B^{(1)}\right).$

Equating the terms that are of order zero with respect to $\xi_{3}$, we have
that $Q_{H}$ is expressed as

$Q_{H}=(B^{\ast})^{(1)}T_{H}B^{(1)}.$

Hence, $T_{H}^{-1}Q_{H}(B^{(1)})^{-1}$ is written as

$T_{H}^{-1}Q_{H}(B^{(1)})^{-1}=T_{H}^{-1}(B^{\ast})^{(1)}T_{H}.$

Since the eigenvalues $\xi_{\hat{\mu}3}$ and $\xi_{\hat{\epsilon}3}$ of
$B^{(1)}$ have nonzero complex component, the eigenvalues
$\overline{\xi_{\hat{\mu}3}}$ and $\overline{\xi_{\hat{\epsilon}3}}$ of
$T_{H}^{-1}(B^{\ast})^{(1)}T_{H}$ are different, which completes the proof.

## Acknowledgements

We would like to thank Bill Lionheart for suggesting the problem and providing
many useful discussions.

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