# Inverse problems for nonlinear Helmholtz Schrödinger equations and time-
harmonic Maxwell’s equations with partial data

Xuezhu Lu

###### Abstract

We consider Calderón’s inverse boundary value problems for a class of
nonlinear Helmholtz Schrödinger equations and Maxwell’s equations in a bounded
domain in $\mathbb{R}^{n}$. The main method is the higher order linearization
of the Dirichlet-to-Neumann map of the corresponding equations. The local
uniqueness of the linearized partial data Calderón’s inverse problem is
obtained following [11]. The Runge approximation properties and unique
continuation principle allow to extend to global situations. Simultaneous
recovery of some unknown cavity$/$boundary and coefficients are given as some
applications.

###### Contents

  1. 1 Introduction and Main Results
  2. 2 Nonlinear Helmholtz Schrödinger equations
    1. 2.1 Proof of Theorem 1.1
    2. 2.2 Proof of Theorem 1.2
      1. 2.2.1 Local results
      2. 2.2.2 Extend to global results
  3. 3 Applications
    1. 3.1 Simultaneous recovery of cavity and coefficients
    2. 3.2 Simultaneous recovery of boundary and coefficients
  4. 4 Nonlinear Maxwell equations
    1. 4.1 Proof of Theorem
      1. 4.1.1 Local result
      2. 4.1.2 From local to global
    2. 4.2 Proof of Theorem 1.4
      1. 4.2.1 Local result
      2. 4.2.2 From local to global
  5. A Appendix
    1. A.1 Well-posedness of nonlinear Helmholtz Schrödinger equation
    2. A.2 Runge approximation

##  1 Introduction and Main Results

In this paper, we study the partial data inverse boundary value problems for a
class of nonlinear Helmholtz Schrödinger equations and Maxwell’s equations in
a bounded domain.

We first consider the inverse boundary value problems of the following
nonlinear Helmholtz Schrödinger equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u(x)-k^{2}u(x)+q(x)u(x)^{2}=0\nobreak\leavevmode\nobreak\leavevmode&\text{in\leavevmode\nobreak\
}\Omega,\\\
u(x)=f(x)\nobreak\leavevmode\nobreak\leavevmode&\text{on\leavevmode\nobreak\
}\partial\Omega,\end{array}\right.$ (3)

where $f\in C^{s}(\partial\Omega)$, $s>1$ with $s\not\in\mathbb{Z}$. Here,
$\Omega$ is a bounded open set in $\mathbb{R}^{n}$ with $n\geq 2.$ Assume that
the boundary $\partial\Omega$ is smooth.

In Appendix A we show that except for a discrete set of real-valued $k$, the
Dirichlet problem (3) has a unique small solution $u\in W^{2,p}(\Omega)$ for
sufficiently small boundary data $f\in W^{2-\frac{1}{p},p}(\partial\Omega)$,
for $p>n$. It is also shown in [36] that the boundary value problem admits a
unique solution in $C^{s}(\overline{\Omega})$ when
$\|f\|_{C^{s}(\partial\Omega)}$ is small in the case of $k=0$; and then the
full boundary Dirichlet-to-Neumann (DN) map $\Lambda_{q}$ is well defined as
follows:

$\Lambda_{q}:C^{s}(\partial\Omega)\;{\rightarrow}\;C^{s-1}(\partial\Omega),\quad
f\;{\mapsto}\;\nu\cdot\nabla u_{f}|_{\partial\Omega},$

where $u_{f}$ is the solution to (3) and $\nu$ is the unit outer normal vector
to the boundary $\partial\Omega.$

Let $\Gamma_{1}$ and $\Gamma_{2}\subset\partial\Omega$ be two proper open
subsets of the boundary. The $W^{2,p}$ well-posedness guarantees that the
partial boundary Dirichlet-to-Neumann map, denoted by
$\Lambda^{\Gamma_{1},\Gamma_{2}}_{q}$, is well extended to

$\Lambda^{\Gamma_{1},\Gamma_{2}}_{q}(f):=\Lambda_{q}(f)|_{\Gamma_{2}},\leavevmode\nobreak\
\textrm{ for all }f\in W^{2-\frac{1}{p},p}(\partial\Omega)\textrm{ with
}\mathrm{supp}\,(f)\subset\Gamma_{1}.$

We are interested in the unique determination of the nonlinear potential
$q(x)$ in $\Omega$ from the partial DN map
$\Lambda^{\Gamma_{1},\Gamma_{2}}_{q}$. Moreover, we solve an inverse problem
of simultaneously determining the potential and an embedded obstacle or cavity
in $\Omega$ from the partial DN map, similar to that in [37] where the problem
was formulated for the semilinear elliptic equation $\Delta u+a(x,u)=0$.

In the second part of the paper, we consider the inverse boundary value
problems for two types of nonlinear Maxwell’s equations, that arise in
potential applications using electromagnetic waves to explore and image the
nonlinear properties of the media, using measurements of electromagnetic
fields on part of the boundary. The first model of nonlinear electromagnetic
medium we consider is for the Kerr-type, described by the boundary value
problem

$\displaystyle\left\\{\begin{aligned} \nabla\wedge E-i\omega\mu_{0}H&=i\omega
p(x)|H|^{2}H,\\\ \nabla\wedge H+i\omega\varepsilon_{0}E&=-i\omega
q(x)|E|^{2}E,\end{aligned}\right.$ (4)

in $\Omega\subseteq\mathbb{R}^{3},$ with $\nu\wedge E|_{\partial\Omega}=f,$
where $\nu$ is the unit outer normal vector to the boundary $\partial\Omega$.
The electrical permittivity $\varepsilon_{0}$ and the magnetic permeability
$\mu_{0}$ are positive constants and $p(x),q(x)\in C^{1}(\overline{\Omega})$.
Here, with some abuse of notations we still denote by $\Omega$ the domain for
Maxwell’s equations in dimension three instead of arbitrary dimension as
considered in the previous Helmholtz equation.

The above Maxwell’s equations of the Kerr-type are derived from the time-
dependent Maxwell’s equations

$\nabla\wedge\mathcal{E}+\partial_{t}\mathcal{B}=0,\quad\nabla\wedge\mathcal{H}-\partial_{t}\mathcal{D}=0,$

for time harmonic solutions with a fixed frequency $\omega$

$\mathcal{E}(x,t)=E(x)e^{-i\omega t}+\overline{E(x)}e^{i\omega
t},\quad\mathcal{H}(x,t)=H(x,t)e^{-i\omega t}+\overline{H(x)}e^{i\omega t},$

assuming nonlinear constitutive relations of the displacement $\mathcal{D}$
and induction $\mathcal{B}$

$\mathcal{D}=\varepsilon_{0}\mathcal{E}+P_{\textrm{NL}}(\mathcal{E}),\quad\mathcal{B}=\mu_{0}\mathcal{H}+\mathcal{M}_{\textrm{NL}}(\mathcal{H}).$

Here $\mathcal{P}_{\textrm{NL}}$ and $\mathcal{M}_{\textrm{NL}}$ are the
nonlinear polarization and magnetization. In a Kerr-type electromagnetic
medium, usually that of centrosymmetric structure, the nonlinear polarization
is of the form

$\mathcal{P}_{\textrm{NL}}(x,\mathcal{E}(x,t))=\chi_{e}(x,|E|^{2})\mathcal{E}(x,t),$

where $\chi_{e}$ is the scalar susceptibility depending on a memory term such
as the time-average of the intensity of $\mathcal{E}$ 111Note that the time-
averages of the intensities of $\mathcal{E}$ and $\mathcal{H}$ are $2|E|^{2}$
and $2|H|^{2}$ respectively.. It is common to use
$\chi_{e}(x,|E|^{2})=p(x)|E|^{2}$. The reader is refereed to [46, 51] for more
details and other examples of electric nonlinear phenomena. Similarly, the
nonlinearity can be generalized to magnetization by assuming

$\mathcal{M}_{\textrm{NL}}(x,\mathcal{H}(x,t))=\chi_{m}(x,|H|^{2})\mathcal{H}(x,t),\quad\chi_{m}(x,|H|^{2})=q(x)|H|^{2}.$

Such nonlinear magnetizations appear in the study of certain metamaterials
built by combining an array of wires and split-ring resonators embedded into a
Kerr-type dielectric. See [41] for a numerical implementation of this
nonlinear assumption.

The well-posedness of the forward problem of (4) is given in [1] that except
for a discrete set of frequencies, there exists a unique solution $(E,H)\in
W^{1,p}_{\textrm{Div}}(\Omega)\times W^{1,p}_{\textrm{Div}}(\Omega)$ for
sufficiently small $f$ in a $L^{p}$ Sobolev space
$TW^{1-1/p,p}_{\textrm{Div}}(\partial\Omega)$ with $3<p\leq 6$ defined by

$\displaystyle W^{1,p}_{\textrm{Div}}(\Omega)$ $\displaystyle=\\{u\in
W^{1,p}(\Omega;\mathbb{C}^{3}):\textrm{Div}(\nu\wedge u)\in
W^{1-1/p,p}(\partial\Omega)\\},$ $\displaystyle
TW^{1-1/p,p}_{\textrm{Div}}(\partial\Omega)$ $\displaystyle=\\{f\in
W^{1-1/p,p}(\partial\Omega;\mathbb{C}^{3}):\textrm{Div}(f)\in
W^{1-1/p,p}(\partial\Omega)\\},$

where Div is the surface divergence operator on $\partial\Omega$. Thus the
admittance map $\Lambda_{p,q}:$
$TW^{1-1/p,p}_{\textrm{Div}}(\partial\Omega)\rightarrow
TW^{1-1/p,p}_{\textrm{Div}}(\partial\Omega)$ is well defined by

$\Lambda_{p,q}(f):=\nu\wedge H|_{\partial\Omega}.$

For the corresponding partial data inverse problem, we use the partial
admittance map associated to two open non-empty open sets $\Gamma_{1}$ and
$\Gamma_{2}$ of $\partial\Omega$ as

$\Lambda^{\Gamma_{1},\Gamma_{2}}_{p,q}(f):=\Lambda_{p,q}(f)|_{\Gamma_{2}}=\nu\wedge
H|_{\Gamma_{2}}$

for all $f\in C_{c}^{s}(\Gamma_{1};\mathbb{C}^{3})\cap
TW^{1-1/p,p}_{\textrm{Div}}(\partial\Omega)$. We consider the inverse problem
in determining the nonlinear coefficients $p$ and $q$ from this partial
admittance map.

Another important nonlinear electromagnetic behavior of the materials is
second harmonic generation (SHG). Applications of the nonlinear optical
phenomena includes obtaining coherent radiation at a wavelength shorter than
that of the incident laser, through the frequency doubling effect of SHG.
Moreover, in the second-harmonic imaging microscopy (SHIM), a second-harmonic
microscope obtains contrasts from variations in a specimen’s ability to
generate second-harmonic light from the incident light while a conventional
optical microscope obtains its contrast by detecting variations in optical
density, path length, or refractive index of the specimen. The SHIM is also
exploited in imaging flux residues (see the work in Chen Lab at the University
of Michigan). Although nonlinear optical effects are in general very weak, the
significant enhancement of SHG was shown using diffraction gratings or
periodic structures.

We consider an inverse boundary value problem for Maxwell’s equations

$\displaystyle\left\\{\begin{array}[]{ll}\nabla\wedge
E^{\omega}-i\omega\mu_{0}H^{\omega}&=0\\\ \nabla\wedge
H^{\omega}+i\omega\varepsilon_{0}E^{\omega}&=-i\omega\chi^{(2)}\overline{E^{\omega}}\cdot
E^{2\omega}\\\ \nabla\wedge E^{2\omega}-i2\omega\mu_{0}H^{2\omega}&=0\\\
\nabla\wedge
H^{2\omega}+i2\omega\varepsilon_{0}E^{2\omega}&=-i2\omega\chi^{(2)}E^{\omega}\cdot
E^{\omega}\\\ \nu\wedge E^{\omega}|_{\partial\Omega}&=f^{\omega}\\\ \nu\wedge
E^{2\omega}|_{\partial\Omega}&=f^{2\omega}\end{array}\right.$

where $f^{k\omega}\in C^{s}_{c}(\partial\Omega)\cap
TW_{\textrm{Div}}^{1-1/p,p}(\partial\Omega)$ and
$\mathrm{supp}\,f^{k\omega}\subseteq\Gamma_{1}$ for $k=1,2$. This models the
phenomenon when a beam with time-harmonic electric field
$E(t,x)=E(x)e^{-i\omega t}+\textrm{c.c.}$ is incident upon an SHG medium
(e.g., a noncentrosymmetric crystal), new waves are generated at zero
frequency and at frequency $2\omega$, and assuming that the susceptibility
parameter is isotropic $\chi^{(2)}=(\chi^{(2)}_{l})_{l=1}^{3}$. The
corresponding output of the admittance map

$\Lambda_{\chi^{(2)}}^{\Gamma_{1},\Gamma_{2}}(f^{k\omega})=\nu\wedge
H^{k\omega}|_{\Gamma_{2}}$

is measured on $\Gamma_{2}$ with $\Gamma_{1}\cap\Gamma_{2}\not=\emptyset$.
$\chi^{(2)}$ is the second order susceptibility parameter. Then we are solving
the problem of determining the susceptibility $\chi^{(2)}$.

The type of inverse boundary value problems was first formulated by Calderón
in [7] for a proposed imaging method, known as the electrical impedance
tomography, in which one aims to determine the electrical conductivity,
modeled by the function $\sigma(x)$ for $x$ in a body $\Omega$, from the
boundary measurements of the electric voltage and current, formulated using
the DN map $\Lambda_{\sigma}:u|_{\partial\Omega}\mapsto\nu\cdot\sigma\nabla
u|_{\partial\Omega}$ for the linear elliptic equation
$\nabla\cdot(\sigma\nabla u)=0$ in $\Omega$. The first global uniqueness
result was proved by Sylvester and Uhlmann [52] for $C^{2}$-conductivities
$\sigma(x)$ in dimensions $n\geq 3$ by solving the problem of determining an
electric potential $q(x)$ in the Schrödinger equation $(-\Delta+q)u=0$ from
the boundary DN map. Later the regularity condition of conductivities was
relaxed in [5] and [49]. In two dimensions, Nachman in [43] and Astala-
Päivärinta in [4] proved the global uniqueness result for
$C^{2}$-conductivities and for $L^{\infty}$-conductivities, respectively.
Bukhgeim in [6] obtained the uniqueness result for $L^{p}$-potentials of
Schrödinger equations from Cauchy data in two dimenions. Many further results
for global uniqueness of inverse boundary value problems are available in the
literature. We refer readers to [3, 53, 54, 59] for anisotropic conductivities
in two and higher dimensions, [45, 50] for magnetic Schrödinger operators and
[47, 48, 30] for Maxwell’s equations, and [10, 12, 13, 15] in Riemannian
geometries. We also refer readers to the surveys [60, 61] and the references
therein.

The study of the inverse problem for nonlinear elliptic equations goes back to
the 1990’s. The method introduced by Isakov in [20] for the parabolic
equations, shows that the linearization of the nonlinear DN map indeed gives
full information about the DN map of the corresponding linear equation, and
thus the uniqueness results of inverse problems for linear equations could be
applicable. The method was then generalized to recover nonlinear coefficients
for a class of semilinear equations [26, 25, 57] in two and higher dimensions
and for a parabolic systems of semilinear equations in [21]. Also see [22] for
an application of the linearized method to quasilinear equations to recover
nonlinear coefficients from partial data. Furthermore, a second order
linearization was applied to a class of quasilinear equations for the unique
determination of quadratic terms in [27] and determination of anisotropic
conductivities in [56, 58]. Also see [1, 2, 15, 31, 32, 36, 37] for an
application of higher order linearization methods to nonlinear elliptic
equations for the recovery of power-type nonlinear terms. More details on
inverse problems for nonlinear equations can be found in the surveys [60, 61].

In the last few years, the higher order linearization method has become a
powerful tool in dealing with inverse problems for nonlinear hyperbolic
equations including wave equations in [14, 34, 39, 16, 62] and einstein
equations in [40, 33]. In those works, the nonlinearity is proven to be
helpful in determining the information of the coefficients in the operators,
in some situations combined with microlocal analysis of the newly generated
singularities in order to solve the inverse problems whose corresponding
version for linear equations are otherwise still open.

Our first result for the nonlinear Helmholtz Schrödinger equation is given
below.

###### Theorem 1.1

Let $q_{1}$ and $q_{2}\in L^{\infty}(\bar{\Omega}),$ and the Dirichlet-Neumann
map $\Lambda_{q_{j}}$ satisfy that

$\displaystyle\Lambda_{q_{1}}(f)|_{\Gamma_{2}}=\Lambda_{q_{2}}(f)|_{\Gamma_{2}},\leavevmode\nobreak\
\leavevmode\nobreak\ \forall f\in C_{c}^{s}(\Gamma_{1}),$

with $\|f\|_{C_{c}^{s}(\Gamma)}<\delta$, $s>1$ with $s\not\in\mathbb{Z}$,
where $\delta$ is sufficiently small. Then $q_{1}=q_{2}$ in $\Omega.$

###### Remark 1.1

Following Theorem A.1 in appendix, the well-posedness of nonlinear Helmholtz
equation holds for small boundary value $f\in
W^{2-\frac{1}{p},p}(\partial\Omega).$ Thus the Dirichlet-to-Neumann map is
well defined and uniquely determines $q(x)\in L^{\infty}(\Omega)$. The well-
posedness is also given in [36] in Hölder spaces.

We remark here that our result for the nonlinear Helmholtz type Schrödinger
equation (3) is not contained in the cases discussed in [37], where the
linearized equation is of Laplace type. Due to the non-positivity of operator
$-\Delta-k^{2}$, one needs modified density arguments discussed below. On the
other hand, the equation (3) can be viewed as a fixed frequency time-harmonic
equation for the wave operator $\partial_{t}^{2}-\Delta$. Therefore, our
result sheds some light on the inverse boundary problem for wave equations. In
[12], the elliptic DN map of Calderón problem of Schrödinger equation in an
infinite cylinder is reduced to the hyperbolic DN map of a wave equation.
Based on the unique continuation property for the reduced wave equation, the
boundary control method is then applied to recover the uniqueness. Our paper
presents a direct proof of partial data inverse problem for Helmholtz
Schrödinger equation. We also remark a progress in [55], which reduces
Calderón problems to the injectivity of some geodesic X-ray transforms and and
provides a unified approach for inverse boundary problems for Laplace type,
transport and wave equations.

In general, the higher order linearization method gives rise to an integral
identity which involves an integral of some product of solutions to the
linearized equations. The problem is then reduced to an integral geometry
problem after plugging in proper linear solutions. When the linearized
equation is the Laplace equation and the boundary DN-map is given as partial
data $\Lambda^{\Gamma_{1},\Gamma_{2}}$, the solutions are usually chosen to be
harmonic exponentials that vanish on part of the boundary. The density result
of the product of two such exponentials was shown in [11] as an enlightening
step of the study of Calderón problem for Schrödinger equation $-\Delta
u+qu=0$. Their techniques are further exploited in [31] to show the density
result of the set of the product of two gradients of some harmonic functions
with some vanishing boundary values. Such density results were then directly
or indirectly used in proving the uniqueness of inverse boundary problems for
nonlinear elliptic equations in [31, 32, 35, 36, 37]. As a comparison, for the
partial data problems of linear equations, two approaches are mainly proposed:
one is a reflection argument in [23, 8] by assuming that the inaccessible
boundary is either on a plane or on a sphere; the other is to establish the
Carleman estimate as in [30, 10] where the partial data on the boundary are
closely related with the constructed limiting Carleman weight. A combination
of these two approaches is exploited in [28] to extend the uniqueness result
to manifolds. These methods aim at constructing new complex geometrical optics
(CGO) solutions similar to those in [52] for the linear equations and usually
requires some extra assumptions on $\Gamma_{1}$ and $\Gamma_{2}$.

In our paper, following the techniques in [11] we first prove the following
density result in order to obtain the global uniqueness of the coefficient for
a nonlinear Helmholtz Schrödinger equation after higher order linearization.

###### Theorem 1.2

Let $\Omega$ be a bounded open set in $\mathbb{R}^{n}$, $n\geq 2$ with smooth
boundary $\partial\Omega$, and $\Gamma\subset\partial\Omega$ be a proper
nonempty open subset of the boundary. Let $q(x)\in L^{\infty}(\Omega).$
Suppose the cancellation equality

$\int_{\Omega}q(x)v_{1}(x)v_{2}(x)dx=0$ (5)

holds for any smooth solutions $v_{1}(x),v_{2}(x)\in C^{\infty}(\bar{\Omega})$
to the equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v(x)-k^{2}v(x)=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\ }\Omega,\\\
v(x)=0&\text{on\leavevmode\nobreak\
}\partial\Omega\setminus\Gamma.\end{array}\right.$ (8)

Then $q(x)$ vanishes in $\Omega$.

As another application of the density result, our paper also yields the
simultaneous recovery of unknown cavity or boundary and coefficients for the
nonlinear Helmholtz Schrödinger operator $-\Delta u+k^{2}u+q(x)u^{2}$
following the discussion in [37]. In comparison, the solutions to the
linearized equation in our case are no longer harmonic functions, and the
tools available are restricted to the unique continuation principle.

Our main results for the nonlinear Maxwell’s equations are given below.

######  Theorem 1.3 (Kerr-type)

Let $p_{1},p_{2},q_{1}$ and $q_{2}\in L^{\infty}(\bar{\Omega})$ and
$\Gamma_{1}\cap\Gamma_{2}\neq\emptyset$. Suppose

$\displaystyle\Lambda_{p_{1},q_{1}}(f)|_{\Gamma_{2}}=\Lambda_{p_{2},q_{2}}(f)|_{\Gamma_{2}},\leavevmode\nobreak\
\leavevmode\nobreak\ \forall f\in C_{c}^{s}(\Gamma_{1};\mathbb{C}^{3})\cap
TW_{\textrm{Div}}^{1-1/p,p}(\partial\Omega),$

with $\|f\|_{C_{c}^{s}(\Gamma)}<\delta$, $s>1$ with $s\not\in\mathbb{Z}$,
where $\delta$ is sufficiently small. Then $p_{1}=p_{2}$ and $q_{1}=q_{2}$ in
$\Omega.$

We also give the identifiability result of the partial data inverse problem of
a class of nonlinear time-harmonic Maxwell’s system with the second harmonic
generation on $\Omega$

######  Theorem 1.4 (Second Harmonic Generation)

Let $\chi^{(2),1}$ and $\chi^{(2),2}\in(L^{\infty}(\bar{\Omega}))^{3},$ the
admittance map $\Lambda_{\chi^{(2)}}$ satisfy that

$\displaystyle\Lambda_{\chi^{(2),1}}(f^{k\omega})|_{\Gamma_{2}}=\Lambda_{\chi^{(2),2}}(f^{k\omega})|_{\Gamma_{2}},\leavevmode\nobreak\
\leavevmode\nobreak\ \forall f^{k\omega}\in C_{c}^{s}(\Gamma_{1})\cap
TW_{\textrm{Div}}^{1-1/p,p}(\partial\Omega),$

with $\|f^{k\omega}\|_{C_{c}^{s}(\Gamma)}<\delta$, $s>1$ with
$s\not\in\mathbb{Z}$, where $\delta$ is sufficiently small for $k=1,2$. Then
$\chi^{(2),1}=\chi^{(2),2}$ in $\Omega.$

For partial data inverse problems of Maxwell’s equations, the literature
relies on geometrical assumptions on the inaccessible boundary
$\Gamma_{2}^{c}$ to be either on a plane or on a sphere [8] or on the
admissible manifold [9] with $\Gamma_{1}$ to be the global boundary. Our
result applies the high order linearization method and $\Gamma_{1}$ and
$\Gamma_{2}$ could be arbitrary.

We first extend the density result in [11] which states that the set of a
product of harmonic functions with boundary value supported on part of the
boundary of the domain is dense in $L^{2}(\Omega)$ to the set of a product of
solutions to linear Helmholtz equations with boundary value supported on part
of the boundary. Different from calculating symbols of the nonlinear
interaction, we adopt integral identity and follow the microlocal analysis as
in [11] to make the proof simple. Then we apply the result to recover the
coefficients of a class of nonlinear Helmholtz Schrödinger and Maxwell’s
equations with partial data by virtue of the higher order linearization
method. We also give some simultaneous recovery results of coefficients and
unknown cavity as applications.

The paper is organized as follows. In Section 2 we give the proofs of Theorem
1.1 and Theorem 1.2. Additionally, we provide some applications of
simultaneous recovery of nonlinear Helmholtz Schrödinger equations in Section
3. Section 4 is devoted to the identifiability results of Maxwell equations,
i.e., Theorem 4.1 and Theorem 1.4.

##  2 Nonlinear Helmholtz Schrödinger equations

###  2.1 Proof of Theorem 1.1

Following [36, 37], we adopt the higher order linearization of the DN-map to
deal with the nonlinearity. Since the application of the high order
linearization method here is standard, we provide key procedures and refer
readers to [36, 37] for details.

Proof. Let $f=\epsilon_{1}f_{1}+\epsilon_{2}f_{2}$, where
$\epsilon_{1},\epsilon_{2}>0$, and $f_{1}$ and $f_{2}\in
C_{c}^{s}(\Gamma_{1}).$ Consider the parameterized boundary value problem

$\displaystyle\left\\{\begin{array}[]{lll}-\Delta
u_{j}-k^{2}u_{j}+q_{j}(x)u_{j}^{2}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega,\\\
u_{j}=\epsilon_{1}f_{1}+\epsilon_{2}f_{2}&\leavevmode\nobreak\
\text{on\leavevmode\nobreak\ }\partial\Omega,\end{array}\right.$ (11)

with $j=1,2$. It follows from the well-posedness of the nonlinear Helmholtz
equation given in Appendix A.1 that when $k$ is not an eigenvalue of the
Laplace operator, there exist unique solutions
$u_{j}:=u_{j}(x;\epsilon_{1},\epsilon_{2})$ to (3) with $j=1,2$, provided
$\epsilon_{1},\epsilon_{2}$ are sufficiently small. Then the assumption
$\Lambda_{q_{1}}(\epsilon_{1}f_{1}+\epsilon_{2}f_{2})|_{\Gamma_{2}}=\Lambda_{q_{2}}(\epsilon_{1}f_{1}+\epsilon_{2}f_{2})|_{\Gamma_{2}}$
gives

$\partial_{\nu}u_{1}|_{\Gamma_{2}}=\partial_{\nu}u_{2}|_{\Gamma_{2}}.$ (12)

Denote $v_{j}^{(l)}:=\partial_{\epsilon_{l}}u_{j}\big{|}_{\varepsilon_{l}=0}$
for $j=1,2$ and $l=1,2.$ The differentiation of (11) w.r.t. $\epsilon_{l}$ at
$\epsilon_{1}=\epsilon_{2}=0$ and the initial value $u_{j}(x;0,0)=0$ yield

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v_{j}^{(l)}-k^{2}v_{j}^{(l)}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega,\\\
v_{j}^{(l)}=f_{l}&\leavevmode\nobreak\ \text{on\leavevmode\nobreak\
}\partial\Omega,\end{array}\right.$

with $j=1,2$ and $l=1,2.$ $v_{1}^{(l)}=v_{2}^{(l)}$ follows from the
uniqueness of Dirichlet boundary value problem of Helmholtz equation for each
$l$. From now on, denote by

$\displaystyle v^{(l)}:=v_{1}^{(l)}=v_{2}^{(l)},\leavevmode\nobreak\
\text{with\leavevmode\nobreak\ }l=1,2;$

thus $v^{(l)}$ is a solution to

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v^{(l)}-k^{2}v^{(l)}=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\
}\Omega,\\\ v^{(l)}=f_{l}&\leavevmode\nobreak\ \text{on\leavevmode\nobreak\
}\partial\Omega.\end{array}\right.$ (15)

Denote
$w_{j}:=\partial^{2}_{\epsilon_{1}\epsilon_{2}}u_{j}\big{|}_{\varepsilon_{1}=\epsilon_{2}=0}$,
with $j=1,2$. Then the second order linearization of (11) yields

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
w_{j}-k^{2}w_{j}+2q_{j}(x)v^{(1)}v^{(2)}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega,\\\ w_{j}=0&\leavevmode\nobreak\
\text{on\leavevmode\nobreak\ }\partial\Omega.\end{array}\right.$

Moreover, from (12) one has

$\partial_{\nu}w_{1}|_{\Gamma_{2}}=\partial_{\nu}w_{2}|_{\Gamma_{2}}.$ (16)

In order to establish an integral identity we choose some proper solution
$v^{(0)}$ to compensate the unknown information on $\Gamma_{2}$ . Let
$v^{(0)}$ be a solution to

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v^{(0)}-k^{2}v^{(0)}=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\
}\Omega,\\\ v^{(0)}=f_{0}&\text{on\leavevmode\nobreak\
}\Gamma_{2},\end{array}\right.$ (19)

where $f_{0}\in C_{c}^{s}(\Gamma_{2})$ and recall that
$\Gamma_{2}\subseteq\partial\Omega$ is an open nonempty subset. Thus,

$\displaystyle 2\int_{\Omega}(q_{1}(x)-q_{2}(x))v^{(1)}v^{(2)}v^{(0)}dx$
$\displaystyle=$ $\displaystyle
2\int_{\Omega}\Delta(w_{1}-w_{2})v^{(0)}+k^{2}(w_{1}-w_{2})v^{(0)}dx$
$\displaystyle=$ $\displaystyle
2\int_{\Omega}\Delta(w_{1}-w_{2})v^{(0)}-\Delta v^{(0)}(w_{1}-w_{2})dx$
$\displaystyle=$ $\displaystyle
2\left(\int_{\partial\Omega}\frac{\partial}{\partial\nu}v^{(0)}(w_{1}-w_{2})dS-\int_{\partial\Omega}\partial(w_{1}-w_{2})v^{(0)}dS\right)$

vanishes in view of (16) and (19). Hence the integral identity

$\int_{\Omega}(q_{1}(x)-q_{2}(x))v^{(1)}v^{(2)}v^{(0)}dx=0,$ (20)

holds, where $v^{(1)},v^{(2)}$ are solutions to the linearized equation (15)
and $v^{(0)}$ satisfies the boundary value problem (19).

Apply Theorem 1.2 and one has

$(q_{1}(x)-q_{2}(x))v^{(0)}(x)=0$

identically in $\Omega$, where $v^{(0)}$ is a solution to the boundary value
problem (19). By constructing a solution $v^{(0)}$ satisfying
$v^{(0)}(\tilde{x}_{0})\not=0$ for any fixed $\tilde{x}_{0}\in\Omega$, we are
able to show $(q_{1}-q_{2})(\tilde{x}_{0})=0$. In fact, apply Proposition
(A.1) with $L=-\Delta-k^{2}$ and consider $u(x)=e^{-ix\cdot\xi}$ with
$\xi\cdot\xi=k^{2}$ such that $Lu=0$. Moreover, $\xi$ is chosen such that
$u(\tilde{x}_{0})\not=0$. Then Proposition (A.1) yields a solution
$v^{(0)}(x)$ close to $u(x)$ and thus $v^{(0)}(\tilde{x}_{0})\not=0$ by the
continuity. Since $\tilde{x}_{0}$ is arbitrary in $\Omega$, $q_{1}=q_{2}$ in
$\Omega$. This completes the proof of Theorem 1.1. $\Box$

###  2.2 Proof of Theorem 1.2

The proof of Theorem 1.2 consists of three parts. First we establish the new
setting of the local problem and prove local results of Theorem 1.2 in section
2.2.1. Then section 2.2.2 contributes to the extension of local results to
global results is given in section.

####  2.2.1 Local results

###### Proposition 2.1

Let $\Omega$ be a bounded open set in $\mathbb{R}^{n}$, with $n\geq 2$ with
smooth boundary $\partial\Omega$, $x_{0}\in\partial\Omega$ a convex point on
the boundary, and $\Gamma$ be a small neighborhood of $x_{0}$ on the boundary.
Suppose that the cancellation equality (5) holds for any smooth solutions
$v_{1}(x)$ and $v_{2}(x)\in C^{\infty}(\bar{\Omega})$ to the equation (8).
Then there exists $\delta>0$ such that $q(x)$ vanishes on
$B(x_{0},\delta)\cap\Omega$.

Proof. Choose a point $a\in\mathbb{R}^{n}\setminus\bar{\Omega}$ along the
direction of the inward normal vector at $x_{0}\in\partial\Omega$. As $x_{0}$
is convex and $\Omega$ is bounded, suppose that there exists a ball $B(a,r)$
containing $\Omega$ and $\partial B(a,r)\cap\bar{\Omega}=\\{x_{0}\\}$. Then a
transform $\varphi(x)$ consisting of translation and rotation transforms such
that $x_{0}$ is mapped to the origin and $a$ is mapped to
$-e_{1}=(-1,0,\ldots,0)$. Indeed, $\varphi$ can be defined as
$x\in\Omega\mapsto y=\frac{1}{r}\cdot O(x-a)-e_{1},$ where $O$ is an
orthogonal matrix such that $O(x_{0}-a)=|x_{0}-a|\cdot e_{1}=re_{1}.$ Thus
$\tilde{\Omega}=\varphi(\Omega)$ is contained in $B(-e_{1},1).$

Note that the Laplacian $\Delta$ is invariant under $\varphi$ as a combintion
of translation and rotation transform. Then, denote
$\tilde{q}=q\circ\varphi^{-1},$ $\tilde{u}_{1}(y)=u_{1}\circ\varphi^{-1}$ and
$\tilde{u}_{2}(y)=u_{2}\circ\varphi^{-1}$. The cancellation (5) is changed to
$\int_{\tilde{\Omega}}\tilde{q}\tilde{u}_{1}\tilde{u}_{2}dy=0,$ where
$\tilde{u}_{1}$ and $\tilde{u}_{2}$ satisfy $-\Delta u-k^{2}r^{2}u=0$ in
$\tilde{\Omega}$ and vanish on the boundary
$\tilde{\Omega}\setminus\tilde{\Gamma}$.

If $\tilde{q}$ vanishes near $y_{0}=0$, then $q$ also vanishes near $x_{0}$.
Therefore, we start with a new cancellation equality. With some abuse of
notations we omit the tilde sign and still use $x$-coordinate. In our new
setting, denote $x_{0}=0.$ Suppose that $|x+e_{1}|\leq 1$ for $x\in\Omega,$
and
$\partial\Omega\setminus\Gamma\subset\\{x\in\partial\Omega|\leavevmode\nobreak\
x_{1}\leq-2c\\}.$ The following cancellation holds

$\displaystyle\int_{\Omega}qu_{1}u_{2}dx=0,$

for any smooth solutions $u_{1}$ and $u_{2}$ to the equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u-k^{2}r^{2}u=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\ }\Omega,\\\
u=0&\text{on\leavevmode\nobreak\
}\partial\Omega\setminus\Gamma.\end{array}\right.$

The next proof of local results follows the literature [11]. Thus to shorten
the length we only provide some critical procedures and estimates here. For
more details of the calculation we refer to [11].

We first construct $u_{1}$ and $u_{2}$ by adding a correction term
$w_{1}(x,\zeta)$ and $w_{2}(x,\eta)$ to eliminate the value of CGO solutions
of the form $e^{-\frac{i}{h}x\cdot\zeta}$ and $e^{-\frac{i}{h}x\cdot\eta}$ on
the boundary $\Omega\setminus\Gamma$, respectively, where $\zeta$ and $\eta$
are proper complex vectors.

Note that $p(\xi)=\xi^{2}$ is the principal symbol of the Laplacian on
$\mathbb{R}^{n}$ and consider its complexification on $\mathbb{C}^{n}$. Let
$\zeta$ and $\eta$ belong to the set

$p^{-1}(krh)=\\{\zeta\in\mathbb{C}^{n}|\leavevmode\nobreak\
\zeta\cdot\zeta=(krh)^{2}\\}.$

Choose $\zeta_{0}=(i,\sqrt{(krh)^{2}+1},0,\ldots,0)$ and
$\eta_{0}=(i,-\sqrt{(krh)^{2}+1},0,\ldots,0).$ Then $\zeta_{0}$ and $\eta_{0}$
belong to $p^{-1}(krh).$ Since the differential of the map
$s:p^{-1}(krh)\times p^{-1}(krh){\rightarrow}\mathbb{C}^{n},$
$(\zeta,\eta)\mapsto\zeta+\eta$ at $(\zeta_{0},\eta_{0})$ is surjective, any
complex vector $z\in\mathbb{C}^{n}$ with $|z-2aie_{1}|<2a\varepsilon$ can be
decomposed as a sum of the form

$z=\zeta+\eta,\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\zeta,\eta\in p^{-1}(krh),\leavevmode\nobreak\
|\zeta-a\zeta_{0}|<Ca\varepsilon,|\eta-a\eta_{0}|<Ca\varepsilon.$ (21)

In the sequel, we use $C$ to denote different constants independent of $h$ in
the inequalities for convenience.

Take a cutoff function $0\leq\chi\leq 1\in C^{\infty}_{c}(\mathbb{R}^{n})$
which equals 1 on $x_{1}\leq-2c$ containing $\partial\Omega\setminus\Gamma$
with compact support in $x_{1}\leq-c$. There exists a solution
$w_{1}(x,\zeta)$ to the Dirichlet problem

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
w-(kr)^{2}w=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\ }\Omega,\\\
w=-\chi(x)e^{-\frac{i}{h}x\cdot\zeta}&\leavevmode\nobreak\
\text{on\leavevmode\nobreak\ }\partial\Omega.\end{array}\right.$ (24)

Thus, fix $k$ outside a discrete set $\Sigma_{r}$, one has

$\|w_{1}\|_{H^{1}(\Omega)}\leq
C\|\chi(x)e^{-\frac{i}{h}x\cdot\zeta}\|_{H^{\frac{3}{2}}(\partial\Omega)}$

and

$\displaystyle\|w_{1}\|_{H^{1}(\Omega)}\leq
C(1+h^{-1}|\zeta|)^{\frac{3}{2}}e^{-\frac{c}{h}\mathrm{Im}\,\zeta_{1}}e^{\frac{1}{h}|\mathrm{Im}\,\zeta^{\prime}|},\leavevmode\nobreak\
\leavevmode\nobreak\ \text{if\leavevmode\nobreak\ }\mathrm{Im}\,\zeta_{1}\geq
0.$ (25)

Let $u_{1}(x,\zeta)=e^{-\frac{i}{h}x\cdot\zeta}+w_{1}(x,\zeta)$, with
$\zeta\in p^{-1}(krh)$. Then $u_{1}$ is a desired solution and vanishes on
$\partial\Omega\setminus\Gamma$. Similarly, we construct
$u_{2}(x,\eta)=e^{-\frac{i}{h}x\cdot\eta}+w_{2}(x,\eta)$, where $\eta\in
p^{-1}(krh)$ and $w_{2}(x,\eta)$ is a correction term satisfying a similar
estimate to (25) with $\zeta$ replaced by $\eta$. Plug in $u_{1}(x,\zeta)$ and
$u_{2}(x,\eta)$ into the cancellation and combine the estimates of
$\|w_{1}\|_{H^{1}(\Omega)}$ and $\|w_{2}\|_{H^{1}(\Omega)}$. When
$\mathrm{Im}\,\zeta_{1}\geq 0$ and $\mathrm{Im}\,\eta_{1}\geq 0$, the
following estimate

$\displaystyle\left|\int_{\Omega}q(x)e^{-\frac{i}{h}x\cdot(\zeta+\eta)}dx\right|\leq
C\|q\|_{L^{\infty}(\Omega)}$
$\displaystyle(1+h^{-1}|\zeta|)^{\frac{3}{2}}(1+h^{-1}|\eta|)^{\frac{3}{2}}$
$\displaystyle
e^{-\frac{c}{h}\min\\{\mathrm{Im}\,\zeta_{1},\mathrm{Im}\,\eta_{1}\\}}e^{\frac{1}{h}(|\mathrm{Im}\,\zeta^{\prime}|+|\mathrm{Im}\,\eta^{\prime}|)},$

holds. In particular, when $|\zeta-a\zeta_{0}|<Ca\varepsilon$ and
$|\eta-a\eta_{0}|<Ca\varepsilon$ with $\varepsilon\leq\frac{1}{2C}$,

$\displaystyle\left|\int_{\Omega}q(x)e^{-\frac{i}{h}x\cdot(\zeta+\eta)}dx\right|\leq
Ch^{-3}\|q\|_{L^{\infty}(\Omega)}e^{-\frac{ca}{2h}}e^{\frac{2Ca\varepsilon}{h}}.$

Therefore, for all $z\in\mathbb{C}^{n}$ with $|z-2aie_{1}|<2a\varepsilon$, it
follows from the decomposition (21) that

$\displaystyle\left|\int_{\Omega}q(x)e^{-\frac{i}{h}x\cdot z}dx\right|\leq
Ch^{-3}\|q\|_{L^{\infty}(\Omega)}e^{-\frac{ca}{2h}}e^{\frac{2Ca\varepsilon}{h}}.$

This is a similar estimate as in [11], which guarantees an exponential decay
for the F.B.I. transform of $f$. Finally, an application of the water melon
approach as in [11] yields

$q(x)=0,\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
x\in\Omega,\leavevmode\nobreak\ \leavevmode\nobreak\ -\delta\leq x_{1}\leq 0,$

provided $\delta$ small enough. This completes the proof of Proposition 2.1.
$\Box$

####  2.2.2 Extend to global results

In this section, we extend the local result (Proposition 2.1) globally to the
whole $\Omega$. The following lemma is needed.

###### Lemma 2.1

Let $\Omega_{1}\subset\Omega_{2}$ with smooth boundary. Define
$G_{\Omega_{2}}(x,y)$ to be the Green function of the following system

$\left\\{\begin{array}[]{ll}(-\Delta_{y}-k^{2})G_{\Omega_{2}}(x,y)=\delta(x-y)&\leavevmode\nobreak\
\text{in \leavevmode\nobreak\ }\Omega_{2},\\\
G_{\Omega_{2}}(x,y)=0&\leavevmode\nobreak\ \text{on \leavevmode\nobreak\
}\partial\Omega_{2}.\end{array}\right.$

The set

$\mathcal{R}=\left\\{\int_{\Omega_{2}}G_{\Omega_{2}}(x,y)a(y)dy,\leavevmode\nobreak\
\text{for some\leavevmode\nobreak\ }a(y)\in
C^{\infty}(\bar{\Omega}_{2}),\leavevmode\nobreak\
\mathrm{supp}\,a\subset\bar{\Omega}_{2}\setminus\Omega_{1},x\in\Omega_{2}\right\\}$

is dense for $L^{2}(\Omega_{1})$ topology in the space $\mathcal{S}=\\{u(x)\in
C^{\infty}(\bar{\Omega}_{1}):(-\Delta_{x}-k^{2})u=0,u|_{\partial\Omega_{1}\cap\partial\Omega_{2}}=0.\\}$

Proof.  It suffices to show that $\langle v,\mathcal{S}\rangle=0$ for any
$v(x)\in L^{2}(\Omega_{1})$ such that $\langle v,\mathcal{R}\rangle=0$.
Following Fubini Theorem and the assumption, the equality

$\int_{\Omega_{1}}v(x)\int_{\Omega_{1}}G_{\Omega_{2}}(x,y)a(y)dydx=\int_{\Omega_{2}}\int_{\Omega_{1}}v(x)G_{\Omega_{2}}(x,y)dx\leavevmode\nobreak\
a(y)dy=0,$

holds for any $a(y)\in C^{\infty}_{c}(\bar{\Omega}_{2})$. The assumption
$\mathrm{supp}\,a\subset\bar{\Omega}_{2}\setminus\Omega_{1}$ yields

$\int_{\Omega_{1}}v(x)G_{\Omega_{2}}(x,y)dx=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \text{for\leavevmode\nobreak\
}y\in\bar{\Omega}_{2}\setminus\Omega_{1}.$ (26)

Denote $\omega(y)=\int_{\Omega_{1}}v(x)G_{\Omega_{2}}(x,y)dx.$ Then
$\omega(y)=0$ for $y\in\bar{\Omega}_{2}\setminus\Omega_{1}.$ Moreover,
$\omega\in H_{0}^{1}(\Omega_{2})$ in view of
$G_{\Omega_{2}}(x,y)|_{y\in\partial\Omega_{2}}=0$.

Note that $(-\Delta_{y}-k^{2})\omega(y)=v(y)$ for $y\in\Omega_{2}$. Thus, for
any $u\in\mathcal{S},$

$\displaystyle\int_{\Omega_{1}}uvdx$
$\displaystyle=\int_{\Omega_{1}}u(-\Delta_{x}-k^{2})\omega(x)dx=\int_{\Omega_{1}}(\Delta
u\omega-u\Delta\omega)dx$
$\displaystyle=\int_{\partial\Omega_{1}\cap\partial\Omega_{2}}(u\partial_{\nu}\omega-\omega\partial_{\nu}u)dS+\int_{\partial\Omega_{1}\setminus\partial\Omega_{2}}(u\partial_{\nu}\omega-\omega\partial_{\nu}u)dS.$

The right hand side term vanishes by combining
$u|_{\partial\Omega_{1}\cap\partial\Omega_{2}}=0$, $\omega\in
H_{0}^{1}(\Omega_{2})$ and $\omega=0$ in
$\bar{\Omega}_{2}\setminus\Omega_{1}$. Hence $\int_{\Omega_{1}}uvdx=0$ for any
$u\in\mathcal{S}.$ $\hfill\Box$

We are ready to conclude the proof of Theorem 1.2. The approach goes through
as in [11].

Proof of Theorem 1.2 Fix a point $x_{1}\in\Omega$ and let
$\theta:[0,1]{\rightarrow}\bar{\Omega}$ be a $C^{1}$ curve joining
$x_{0}\in\Gamma$ to $x_{1}$ such that $\theta(0)=x_{0}$ and
$\theta^{\prime}(0)$ is the interior normal to $\partial\Omega$ at $x_{0}$ and
$\theta(t)\in\Omega$ for all $t\in(0,1].$ Let
$\Theta_{\epsilon}(t)=\\{x\in\bar{\Omega}|\leavevmode\nobreak\
d(x,\theta([0,t]))\leq\epsilon\\}$ be a closed neighborhood of the curve
$\theta(t),$ $t\in[0,1]$, and set $I=\\{t\in[0,1]|\leavevmode\nobreak\
f=0\text{\leavevmode\nobreak\ a.e. on\leavevmode\nobreak\
}\Theta_{\epsilon}(t)\cap\Omega\\}.$ $I$ is a closed set. From Proposition
2.1, $I$ is non-empty provided that $\epsilon$ is small enough. We would like
to show that $I$ is also open. Thus by connectivity $I=[0,1].$ Then
$x_{1}\not\in\mathrm{supp}\,f$. Since $x_{1}$ is arbitrary, $f=0$ in $\Omega$.
Hence, it suffices to show that $I$ is open.

Let $t\in I$ and $\epsilon$ small enough. Then we can assume that
$\partial\Theta_{\epsilon}(t)\cap\partial\Omega\subset\Gamma,$ and
$\Omega\setminus\Theta_{\epsilon}(t)$ can be smoothed out into an open subset
$\Omega_{1}\subset\Omega$ with smooth boundary such that
$\Omega\setminus\Theta_{\epsilon}(t)\subset\Omega_{1},\leavevmode\nobreak\
\leavevmode\nobreak\
\partial\Omega\setminus\Gamma\subset\partial\Omega\cap\partial\Omega_{1}.$
Furthermore, we augment $\Omega$ by smoothing out the set $\Omega\cup
B(x_{0},\epsilon)$ into an open set $\Omega_{2}$ with smooth boundary such
that
$\partial\Omega\setminus\Gamma\subset\partial\Omega\cap\partial\Omega_{1}\subset\partial\Omega\cap\partial\Omega_{2}.$

Denote by $G_{\Omega_{2}}$ be the fundamental solution to the system

$\left\\{\begin{array}[]{ll}(-\Delta_{y}-k^{2})G_{\Omega_{2}}(x,y)=\delta(x-y)&\leavevmode\nobreak\
\text{in \leavevmode\nobreak\ }\Omega_{2},\\\
G_{\Omega_{2}}(x,y)=0&\leavevmode\nobreak\ \text{on \leavevmode\nobreak\
}\partial\Omega_{2},\end{array}\right.$

and let

$H(x,t)=\int_{\Omega_{1}}f(y)G_{\Omega_{2}}(x,y)G_{\Omega_{2}}(t,y)dy,\leavevmode\nobreak\
\leavevmode\nobreak\ \text{for\leavevmode\nobreak\
}t,x\in\Omega_{2}\setminus\bar{\Omega}_{1}.$

Then $H$ satisfies $-\Delta_{x}H-k^{2}H=0$ $(-\Delta_{t}H-k^{2}H=0)$ when
$x\in\Omega_{2}\setminus\bar{\Omega}_{1}$
$(t\in\Omega_{2}\setminus\bar{\Omega}_{1})$ respectively. Since $f$ vanishes
on $\Omega_{1}\setminus\Omega$,

$H(x,t)=\int_{\Omega_{1}}f(y)G_{\Omega_{2}}(x,y)G_{\Omega_{2}}(t,y)dy=\int_{\Omega}f(y)G_{\Omega_{2}}(x,y)G_{\Omega_{2}}(t,y)dy.$

Thus, $H(x,t)=0$ follows from the assumption, for
$t,x\in\Omega_{2}\setminus\bar{\Omega}.$ Note that $H$ also satisfies
$-(\Delta_{x}+\Delta_{t})H-2k^{2}H=0$, when
$x,t\in\Omega_{2}\setminus\bar{\Omega}_{1}.$ The Unique Continuation Principle
yields $H(x,t)=0\leavevmode\nobreak\ \leavevmode\nobreak\
\text{when\leavevmode\nobreak\ \leavevmode\nobreak\
}t,x\in\bar{\Omega}_{2}\setminus\Omega_{1};$ i.e.,

$\int_{\Omega_{1}}f(y)G_{\Omega_{2}}(x,y)G_{\Omega_{2}}(t,y)dy=0,\leavevmode\nobreak\
\leavevmode\nobreak\ \text{for\leavevmode\nobreak\
}t,x\in\bar{\Omega}_{2}\setminus\Omega_{1}.$

Applying Lemma 2.1, we obtain the equality $\int_{\Omega_{1}}fuv=0,$ where
$u,v\in C^{\infty}(\bar{\Omega}_{1})$ are solutions to $-\Delta u-k^{2}u=0$
and vanish on $\partial\Omega_{1}\cap\partial\Omega_{2}$. Combining
Proposition 2.1, $f$ vanishes on a neighborhood of
$\partial\Omega_{1}\setminus(\partial\Omega_{1}\cap\partial\Omega_{2})$. This
implies that $f$ vanishes on a bigger neighborhood
$\Theta_{\epsilon}(t^{\prime})$ $t^{\prime}>t$ of the curve. Hence $I$ is
open. This completes the proof of Theorem 1.2. $\Box$

##  3 Applications

In this section, we give simultaneous recovery of the partial data inverse
problems for a class of nonlinear Helmholtz Schrödinger equations as an
application of Theorem 1.2.

In some applications, the discontinuity of the medium extends to the cases of
obstacles or cavities embedded in the medium. Mathematically, equations are
satisfied in the medium $\Omega$ minus the obstacle or cavity region $D$ and
subject to certain boundary conditions at $\partial D$. Determining an unknown
obstacle or cavity goes back to Schiffer’s work and is a substantial topic in
inverse scattering theory, for example for sound waves modeled by the linear
wave equation and electromagnetic waves modeled by the linear Maxwell’s
equations. Among many existing methods (such as the probe method and so on),
one method, known as the enclosure method, uses CGO-type solutions to
determine obstacles, cavities and inclusions. The enclosure method was first
introduced by Ikehata in [17, 18] to reconstruct a cavity $D$ inside a
conductive medium $\Omega$ for Schrödinger operator when the surrounding
potential is known a priori. See [19, 24] for more uniqueness results of
obstacle problems when knowing the potential and [42] for a simultaneous
reconstruction of both embedded obstacle and its surrounding potential.

###  3.1 Simultaneous recovery of cavity and coefficients

###### Theorem 3.1

Let $\Omega$ be a bounded open set in $\mathbb{R}^{n}$, $n\geq 2$ with smooth
boundary $\partial\Omega$. Let $D_{1},D_{2}\subset\subset\Omega$ be nonempty
subsets with smooth boundaries such that $\Omega\setminus\bar{D}_{j}$ are
connected. For $j=1,2$ let $q_{j}(x)\in
C^{\infty}(\Omega\setminus\bar{D}_{j})$ and consider the following nonlinear
Helmholtz equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u_{j}-k^{2}u_{j}+q_{j}(x)u_{j}^{2}=0\nobreak\leavevmode\nobreak\leavevmode&\text{in\leavevmode\nobreak\
}\Omega\setminus\bar{D}_{j},\\\
u_{j}=0\nobreak\leavevmode\nobreak\leavevmode&\text{on\leavevmode\nobreak\
}\partial D_{j},\\\
u_{j}=f\nobreak\leavevmode\nobreak\leavevmode&\text{on\leavevmode\nobreak\
}\partial\Omega.\end{array}\right.$ (30)

Assume that the Dirichlet-Neumann map $\Lambda_{q_{j}}^{D_{j}}$ satisfy that

$\displaystyle\Lambda_{q_{1}}^{D_{1}}(f)=\Lambda_{q_{2}}^{D_{2}}(f)\text{on\leavevmode\nobreak\
}\partial\Omega,\leavevmode\nobreak\ \leavevmode\nobreak\ \forall f\in
C^{s}(\partial\Omega)\text{ \leavevmode\nobreak\ is sufficiently small, }$

with $s>1$ with $s\not\in\mathbb{Z}$. Then $D_{1}=D_{2}=D$ and $q_{1}=q_{2}$
in $\Omega\setminus\bar{D}.$

Proof. We apply the standard high order linearization method and and take the
same notations as shown in section 2.1. Let
$f=\epsilon_{1}f_{1}+\epsilon_{2}f_{2}$ for the boundary value of (30), where
$f_{1}$, $f_{2}\in C^{s}(\partial\Omega),$ and $\epsilon_{1},\epsilon_{2}$ are
sufficiently small numbers. Thus,
$v_{j}^{(l)}:=\partial_{\epsilon_{l}}u_{j}\big{|}_{\epsilon_{l}=0}$ satisfy

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v_{j}^{(l)}-k^{2}v_{j}^{(l)}=0&\text{in\leavevmode\nobreak\
}\Omega\setminus\bar{D}_{j},\\\ v_{j}^{(l)}=f_{l}&\text{on\leavevmode\nobreak\
}\partial D_{j},\\\ v_{j}^{(l)}=0&\text{on\leavevmode\nobreak\
}\partial\Omega,\end{array}\right.$ (34)

with $j=1,2$ and $l=1,2.$

Let $G$ be the connected component of
$\Omega\setminus\bar{D}_{1}\cup\bar{D}_{2}$ of which the boundary contains
$\partial\Omega.$ Let $\tilde{v}^{(l)}=v_{1}^{(l)}-v_{2}^{(l)}$ for each $l$
in $G$. Then $\tilde{v}^{(l)}$ satisfies

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\tilde{v}^{(l)}-k^{2}\tilde{v}^{(l)}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }G,\\\
\tilde{v}^{(l)}=\partial_{\nu}\tilde{v}^{(l)}=0&\text{on\leavevmode\nobreak\
}\partial\Omega,\end{array}\right.$

with $l=1,2.$ The unique continuation principle of solutions to Helmholtz
equation yields $\tilde{v}^{(l)}=0$ in $G$. Therefore, for each $l,$
$v_{1}^{(l)}=v_{2}^{(l)}$ in $G$. Hence $v_{2}^{(l)}=0$ on $\partial
D_{1}\setminus D_{2}$ from the continuity.

To recovery the unknown boundary we need to choose $f_{l}$. Let
$U=D_{1}\setminus\bar{D}_{2}$. Then $\partial U=(\partial D_{1}\setminus
D_{2})\cup(\partial D_{2}\cap D_{1})$. Let $\varphi\in H^{1}(U)$ such that

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\varphi-k^{2}\varphi=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }U,\\\ \phi=0&\text{on\leavevmode\nobreak\
}\partial D_{2}\cap D_{1},\\\ \phi=g&\text{on\leavevmode\nobreak\ }\partial
D_{1}\setminus D_{2},\\\ \end{array}\right.$

where $g\in H^{1}(U)$ which has nonvanishing trace supported in $\partial
D_{1}\setminus D_{2}$.

Apply Proposition (A.1) with $U=D_{1}\setminus\bar{D}_{2}$ and
$M=\Omega\setminus D_{2}$. Then
$M\setminus\bar{U}=\Omega\setminus\bar{D}_{1}\cup\bar{D}_{2}$ is connected.
Let $\Gamma=\partial\Omega\subset\partial M$. Hence, there exists some $f_{l}$
with $\mathrm{supp}\,f_{l}\subset\Gamma$ and a corresponding solution
$v_{2}^{(l)}$ to

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v_{2}^{(l)}-k^{2}v_{2}^{(l)}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }M,\\\
v_{2}^{(l)}=f_{l}&\text{on\leavevmode\nobreak\ }\partial M,\end{array}\right.$

such that $v_{2}^{(l)}$ is close to $\phi$ in $H^{1}(U)$ and so is the trace
of $v_{2}^{(l)}$ and $\phi$ on $\partial U$. Note that $v_{2}^{(l)}$ should
vanish on $\partial D_{1}\setminus D_{2}$. This contradicts with the
assumption that the trace of $\phi$ has nonvanishing support $g$ on $\partial
D_{1}\setminus D_{2}$. This shows that $D_{1}=D_{2}.$ From now on, denote by
$D=D_{1}=D_{2}$. It follows that

$\displaystyle v^{(l)}:=v_{1}^{(l)}=v_{2}^{(l)}\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega\setminus\bar{D},$ (35)

for $l=1,2;$ and $v^{(l)}$ is a solution to (34).

To recover the coefficient, we choose $v^{(0)}$ to be a solution to

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v^{(0)}-k^{2}v^{(0)}=0&\leavevmode\nobreak\ \text{in\leavevmode\nobreak\
}\Omega\setminus\bar{D},\\\ v^{(0)}=0&\text{on\leavevmode\nobreak\ }\partial
D,\\\ v_{j}^{(0)}=f_{0}&\text{on\leavevmode\nobreak\
}\partial\Omega,\end{array}\right.$ (39)

where $f_{0}$ is not identically zero on $\partial\Omega.$

Denoted by
$w_{j}=\partial^{2}_{\epsilon_{1}\epsilon_{2}}u_{j}\big{|}_{\epsilon_{1}=\epsilon_{2}=0}.$
It follows from the second order linearization that $w_{j}=0$ on $\partial
D\cup\partial\Omega$. Moreover, the DN map yields

$\partial_{\nu}w_{1}|_{\partial\Omega}=\partial_{\nu}w_{2}|_{\partial\Omega}.$

Thus, similar as in section 2.1, one has the integral identity

$\int_{\Omega\setminus\bar{D}}(q_{1}(x)-q_{2}(x))v^{(1)}v^{(2)}v^{(0)}dx=0,$

where $v^{(1)},v^{(2)}$ satisfy (34) and $v^{(0)}$ satisfies (39). Apply
Theorem 1.2 and one has

$(q_{1}(x)-q_{2}(x))v^{(0)}(x)=0$ (40)

identically in $\Omega\setminus\bar{D}$. Similarly, for any
$\tilde{x}_{0}\in\Omega\setminus\bar{D}$, we apply Proposition (A.1) and
construct $v^{(0)}$ a solution to (39) by choosing some $f_{0}$, with
$v^{(0)}(\tilde{x}_{0})\not=0.$ Hence $(q_{1}-q_{2})(\tilde{x}_{0})=0$. Since
$\tilde{x}_{0}$ is arbitrary in $\Omega\setminus\bar{D}$, then $q_{1}=q_{2}$
in $\Omega\setminus\bar{D}$. $\Box$

###  3.2 Simultaneous recovery of boundary and coefficients

###### Theorem 3.2

Let $\Omega_{j}$ be a bounded open set in $\mathbb{R}^{n}$, $n\geq 2$ with
smooth boundary $\partial\Omega$ for $j=1,2$. Consider
$\Gamma\subset\partial\Omega$ be a proper nonempty open subset of
$\partial\Omega_{1}\cap\partial\Omega_{2}.$ For $j=1,2$ let $q_{j}(x)\in
C^{\infty}(\Omega_{j})$ and consider the following nonlinear Helmholtz
equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u_{j}-k^{2}u_{j}+q_{j}(x)u_{j}^{2}=0&\text{in\leavevmode\nobreak\
}\Omega_{j},\\\ u_{j}=0&\text{on\leavevmode\nobreak\
}\partial\Omega_{j}\setminus\Gamma,\\\ u_{j}=f&\text{on\leavevmode\nobreak\
}\Gamma.\end{array}\right.$ (44)

Assume that the Dirichlet-Neumann map $\Lambda_{q_{j}}^{\Omega_{j}}$ satisfy
that

$\displaystyle\Lambda_{q_{1}}^{\Omega_{1}}(f)=\Lambda_{q_{2}}^{\Omega_{2}}(f)\text{on\leavevmode\nobreak\
}\Gamma,\leavevmode\nobreak\ \leavevmode\nobreak\ \forall f\in
C^{s}_{c}(\Gamma)\text{ \leavevmode\nobreak\ is sufficiently small, }$

with $s>1$ with $s\not\in\mathbb{Z}$. Then $\Omega_{1}=\Omega_{2}=\Omega$ and
$q_{1}=q_{2}$ in $\Omega.$

Proof. We apply the standard high order linearization method and the
procedures and notations as in the proof of Theorem 3.1. Let
$f=\epsilon_{1}f_{1}+\epsilon_{2}f_{2}$. Let
$v_{j}^{(l)}=\frac{\partial}{\partial\epsilon_{l}}\big{|}_{\epsilon_{l}=0}u_{j}$
are solutions to the linearized Helmholtz equation with boundary value
$f_{l}\in C_{c}^{s}(\Gamma),$ for $l=1,2$ and $j=1,2.$ Similarly as in the
proof of Theorem 3.1, the unique continuation principle gives rise to
$v_{1}^{(l)}=v_{2}^{(l)}$ in $G$ for each $l$, where $G$ is a connected
component of $\Omega_{1}\cap\Omega_{2}$ of which the boundary contains
$\Gamma.$ Hence $v_{2}^{(l)}=0$ on $\partial\Omega_{1}\cap\Omega_{2}$ from the
continuity.

To recovery the unknown boundary we need to choose the boundary values $f_{l}$
for $l=1,2$. Let $U=\Omega_{2}\setminus\bar{\Omega}_{1}$. Then $\partial
U=(\partial\Omega_{2}\setminus\Omega_{1})\cup(\partial\Omega_{1}\cap\Omega_{2})$.
Let $\varphi\in H^{1}(U)$ such that

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\varphi-k^{2}\varphi=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }U,\\\ \phi=0&\text{on\leavevmode\nobreak\
}\partial\Omega_{2}\setminus\Omega_{1},\\\ \phi=g&\text{on\leavevmode\nobreak\
}\partial\Omega_{1}\cap\Omega_{2},\\\ \end{array}\right.$

where $g\in H^{1}(U)$ which has nonvanishing trace supported in
$\partial\Omega_{1}\cap\Omega_{2}$.

Apply Proposition (A.1) with $U=\Omega_{2}\setminus\bar{\Omega}_{1}$ and
$M=\Omega_{2}$. Then $M\setminus\bar{U}=\Omega_{1}\cap\Omega_{2}$ is connected
with $\Gamma=\Gamma_{0}\subset\partial M$. Hence, there exists some $f_{l}$
with $\mathrm{supp}\,f_{l}\subset\Gamma$ and a corresponding solution
$v_{2}^{(l)}$ to

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v_{2}^{(l)}-k^{2}v_{2}^{(l)}=0&\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }M,\\\
v_{2}^{(l)}=f_{l}&\text{on\leavevmode\nobreak\ }\partial M\end{array}\right.$

such that $v_{2}^{(l)}$ is close to $\phi$ in $H^{1}(U)$ and so is the trace
of $v_{2}^{(l)}$ and $\phi$ on $\partial U$. Note that $v_{2}^{(l)}$ should
vanish on $\partial\Omega_{1}\cap\Omega_{2}$. This contradicts with the
assumption that the trace of $\phi$ has nonvanishing support $g$ on
$\partial\Omega_{1}\cap\Omega_{2}$. This shows that $\Omega_{1}=\Omega_{2}.$

From now on, denote by $\Omega=\Omega_{1}=\Omega_{2}$. It follows that
$v^{(l)}:=v_{1}^{(l)}=v_{2}^{(l)}\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega,$ for $l=1,2;$ and $v^{(l)}$ are
solutions to the linearized Helmholtz equation with boundary value $f_{l}$
supported on $\Gamma$ for $l=1,2.$ The recovery of the coefficient is then
similar to the proof of Theorem 3.1. By taking the second order linearization
and choosing some $v^{(0)}$ as in (39) with $\partial D$ replaced by
$\partial\Gamma$, one is able to construct an integral identity similar to
(40). An application of Theorem 1.2 and similar arguments complete the proof.
$\Box$

##  4 Nonlinear Maxwell equations

In this section, $\Omega\subseteq\mathbb{R}^{3}$ is a bounded domain with
smooth boundary. The proof of Theorem 4.1 and Theorem 1.4 consist of three
parts. We first adopt the standard procedures of high order linearization of
the admittance map to derive an integral identity. Similarly, we only provide
key procedures of higher order linearization here and refer readers to [36,
37] for more details. Then we follow the literature [11] to give the
uniqueness result locally in a small neighborhood of some point on the
boundary. Lastly, we extend the local result to the global situation.

###  4.1 Proof of Theorem

Proof. Let $f=\epsilon_{1}f_{1}+\epsilon_{2}f_{2}+\epsilon_{3}f_{3}.$
$f_{l}\in C^{\infty}(\partial\Omega)$ with
$\mathrm{supp}\,f_{l}\subseteq\Gamma_{1}$, for $l=1,2,3$. Let
$\epsilon=(\epsilon_{1},\epsilon_{2},\epsilon_{3}).$ Denote

$W_{j}^{(l)}=\partial_{\epsilon_{l}}H_{j}|_{\epsilon=0},\leavevmode\nobreak\
V_{j}^{(l)}=\partial_{\epsilon_{l}}E_{j}|_{\epsilon=0},\leavevmode\nobreak\
W_{j}^{(123)}=\partial_{\epsilon}H_{j}|_{\epsilon=0},\leavevmode\nobreak\
V_{j}^{(123)}=\partial_{\epsilon}E_{j}|_{\epsilon=0},$

for $j=1,2$ and $l=1,2,3.$ Similarly as in section 2.1, one can show that
$V^{(l)}:=V_{1}^{(l)}=V_{2}^{(l)}$ and $W^{(l)}:=W_{1}^{(l)}=W_{2}^{(l)}$
satisfy

$\displaystyle\left\\{\begin{aligned} \nabla\wedge
V^{(l)}-i\omega\mu_{0}W^{(l)}&=0,\\\ \nabla\wedge
W^{(l)}+i\omega\varepsilon_{0}V^{(l)}&=0,\end{aligned}\right.$ (45)

in $\Omega$ with $\nu\wedge V^{(l)}|_{\partial\Omega}=f_{l}$ on the boundary
$\partial\Omega$, for $l=1,2,3.$ By direct computation one can show that
$(E,H)$ of the following form is a solution to (45)

$\displaystyle\begin{split}E&=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\xi}\left(\frac{1}{h^{2}}(\xi\cdot
a)\xi-\frac{k}{h}\xi\times b-k^{2}a\right),\\\
H&=\mu_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\xi}\left(\leavevmode\nobreak\
\frac{1}{h^{2}}(\xi\cdot b)\xi+\frac{k}{h}\xi\times
a-k^{2}b\right),\end{split}$ (46)

where $a,b\in\mathbb{C}^{3}$ are complex vectors to be chosen later; and
$\xi\in\mathbb{C}^{3}$ satisfies $\xi\cdot\xi=k^{2}h^{2}.$

The highest order linearization of the admittance map yields

$\displaystyle\left\\{\begin{aligned} \nabla\wedge
V_{j}^{(123)}-i\omega\mu_{0}W_{j}^{(123)}&=i\omega
p_{j}\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}W_{j}^{(i_{1})}\cdot\overline{W}_{j}^{(i_{2})}\leavevmode\nobreak\
W_{j}^{(i_{3})},\\\ \nabla\wedge
W_{j}^{(123)}-i\omega\varepsilon_{0}V_{j}^{(123)}&=i\omega
q_{j}\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}V_{j}^{(i_{1})}\cdot\overline{V}_{j}^{(i_{2})}\leavevmode\nobreak\
V_{j}^{(i_{3})},\end{aligned}\right.$

in $\Omega$ with $\nu\wedge V_{j}^{(123)}|_{\partial\Omega}=0$. Here,
$\sigma\in S_{3}$ denotes the permutation of the indices. Thus, we obtain the
integral identity

$\displaystyle\int_{\Omega}$
$\displaystyle(p_{1}-p_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}W^{(i_{1})}\cdot\overline{W}^{(i_{2})}\leavevmode\nobreak\
W^{(i_{3})}\cdot\overline{W}^{(0)}dV,$ (47)
$\displaystyle+\int_{\Omega}(q_{1}-q_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}V^{(i_{1})}\cdot\overline{V}^{(i_{2})}\leavevmode\nobreak\
V^{(i_{3})}\cdot\overline{V}^{(0)}dV=0,$ (48)

by choosing $W^{(0)}$ and $V^{(0)}$ satisfying the following conjugate
Maxwell’s equations

$\displaystyle\left\\{\begin{aligned} \nabla\wedge
V^{(0)}-i\omega\mu_{0}W^{(0)}&=0,\\\ \nabla\wedge
W^{(0)}+i\omega\varepsilon_{0}V^{(0)}&=0,\end{aligned}\right.$ (49)

in $\Omega$ with boundary value $\nu\wedge V^{(0)}|_{\partial\Omega}=f_{0}$.
Here $f_{(0)}\in C^{\infty}(\partial\Omega)$ with
$\mathrm{supp}\,f_{0}\subseteq\Gamma_{2}$.

####  4.1.1 Local result

Below we are to prove a local version of Theorem 4.1, i.e., to show
$p_{1}=p_{2}$ and $q_{1}=q_{2}$ locally in a small neighborhood of some
$x_{0}\in\Gamma_{1}$. Without loss of generality, as in section 3, the domain
can be set below the coordinate $\\{x_{1}\leq 0\\}$ by orthogonal
transformation and translation, and $x_{0}=0\in\Gamma_{1}$.

###### Proposition 4.1

Let $\Omega$ be a bounded open set in $\mathbb{R}^{3}$ with smooth boundary
$\partial\Omega$. Let $x_{0}\in\Gamma_{1}\cap\Gamma_{2}\subset\partial\Omega$
be a convex point on the boundary such that
$\Gamma\subset\Gamma_{1}\cap\Gamma_{2}$. Let $\Gamma$ be a small neighborhood
of $x_{0}$ on the boundary. Suppose that the cancellation (47) and (48) hold
for any smooth solutions $(V^{(l)},W^{(l)})$ with $l=1,2,3$, and
$(V^{(0)},W^{(0)})$ to the systems (45) and (49), respectively. Then there
exists $\delta>0$ such that $p_{1}-p_{2}$ and $q_{1}-q_{2}$ vanish on
$B(x_{0},\delta)\cap\Omega$.

Proof. Let $p^{-1}(kh)=\\{\xi\in C^{3}|\xi\cdot\xi=k^{2}h^{2}\\}$,
$\xi_{0}=(i,\sqrt{k^{2}h^{2}+1},0)\in p^{-1}(kh)$ and
$\eta_{0}=(i,-\sqrt{k^{2}h^{2}+1},0)\in p^{-1}(kh)$. Suppose
$|\xi-\xi_{0}|<\epsilon,\xi\in p^{-1}(kh)$, and
$|\eta-\eta_{0}|<\epsilon,\eta\in p^{-1}(kh)$. Then $\mathrm{Im}\,(\xi_{1})>0$
and $\mathrm{Im}\,(\eta_{1})>0$.

Below we construct solutions to the systems (45) and (49), achieved by adding
correction terms to the solutions satisfying the corresponding homogeneous
Maxwell’s equations of the form (46). Choose $b=0$ in (46). Denote

$\begin{split}E_{1}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\xi\cdot
a_{1})\xi-k^{2}a_{1}\right)e^{-\frac{i}{h}x\cdot\xi},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
H_{1}=\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\xi\times
a_{1})e^{-\frac{i}{h}x\cdot\xi},\\\
E_{2}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\xi\cdot
a_{2})\xi-k^{2}a_{2}\right)e^{-\frac{i}{h}x\cdot\xi},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
H_{2}=\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\xi\times
a_{2})e^{-\frac{i}{h}x\cdot\xi},\\\
E_{3}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\xi\cdot
a_{3})\xi-k^{2}a_{3}\right)e^{-\frac{i}{h}x\cdot\xi},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
H_{3}=\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\xi\times
a_{3})e^{-\frac{i}{h}x\cdot\xi},\end{split}$ (50)

and

$\displaystyle
E_{0}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\bar{\eta}\cdot\bar{a}_{0})\bar{\eta}-k^{2}\bar{a}_{0}\right)e^{\frac{i}{h}x\cdot\bar{\eta}},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
H_{0}=-\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\bar{\eta}\times\bar{a}_{0})e^{\frac{i}{h}x\cdot\bar{\eta}},$
(51)

where the constant vectors are defined by

$\displaystyle a_{1}=(1,0,1),\leavevmode\nobreak\ \leavevmode\nobreak\
a_{2}=a_{3}=\bar{\xi}\times\bar{a}_{1},\leavevmode\nobreak\
\leavevmode\nobreak\ a_{0}=\xi\times a_{1}.$ (52)

Take a cutoff function $0\leq\chi\leq 1\in C_{c}^{\infty}$ which equals 1 on
$x_{1}\leq-2c$ containing $\partial\Omega\setminus\Gamma_{1}$ and
$\partial\Omega\setminus\Gamma_{2}$ and is compactly supported in
$x_{1}\leq-c$. There exists $(r_{1},t_{1})$ to the Dirichlet problem

$\displaystyle\left\\{\begin{aligned} \nabla\wedge
r_{1}-i\omega\mu_{0}t_{1}&=0,\\\ \nabla\wedge
t_{1}+i\omega\varepsilon_{0}r_{1}&=0,\end{aligned}\right.$

in $\Omega$ with boundary value $\nu\wedge
r_{1}|_{\partial\Omega}=-\chi(x)E_{1},$ satisfying the following estimates

$\displaystyle\|r_{1}\|_{H^{1}(\Omega)},\|t_{1}\|_{H^{1}(\Omega)}\leq
C(1+h^{-1}|\xi|)^{\frac{3}{2}}e^{-\frac{c}{h}\mathrm{Im}\,\xi_{1}}e^{\frac{1}{h}|\mathrm{Im}\,\xi^{{}^{\prime}}|}\leavevmode\nobreak\
\leavevmode\nobreak\ \text{if\leavevmode\nobreak\ }\mathrm{Im}\,\xi_{1}\geq
0.$ (53)

Likewise, we have $(r_{2},t_{2})$, $(r_{3},t_{3})$ and $(r_{0},t_{0})$ with
similar estimates to (53). Thus,

$\displaystyle V^{(1)}=E_{1}+r_{1},\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ W^{(1)}=H_{1}+t_{1};\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
V^{(2)}=E_{2}+r_{2},\leavevmode\nobreak\ \leavevmode\nobreak\
W^{(2)}=H_{2}+t_{2};$ $\displaystyle V^{(3)}=E_{3}+r_{3},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
W^{(3)}=H_{3}+t_{3};\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ V^{(0)}=E_{0}+r_{0},\leavevmode\nobreak\
\leavevmode\nobreak\ W^{(0)}=H_{0}+t_{0},$

are solutions to the systems (45) and (49), respectively. Now, the derived
cancellation (47) and (48) become

$\displaystyle\int_{\Omega}$
$\displaystyle(p_{1}-p_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(H_{i_{1}}+t_{i_{1}})\cdot(\overline{H_{i_{2}}+t_{i_{2}}})(H_{i_{3}}+t_{i_{3}})\cdot(\overline{H_{0}+t_{0}})dV$
(54)
$\displaystyle+\int_{\Omega}(q_{1}-q_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(E_{i_{1}}+r_{i_{1}})\cdot(\overline{E_{i_{2}}+r_{i_{2}}})(E_{i_{3}}+r_{i_{3}})\cdot(\overline{E_{0}+r_{0}})dV=0.$
(55)

By virtue of (52), one has
$H_{1}\cdot\bar{H}_{2}=H_{1}\cdot\bar{H}_{3}=H_{1}\cdot\bar{H}_{0}=0.$
Therefore, only the items involving some $t_{i}$ survive in the expansion of
the product (54), which are exponentially decaying in view of (53) when
$\mathrm{Im}\,\xi_{1}\geq 0$ and $\mathrm{Im}\,\xi^{\prime}=0$.

Next we calculate the items in the expansion of the product (55). Note that
the items involving some $r_{i}$ also have exponentially decaying properties
from (53) when $\mathrm{Im}\,\xi_{1}\geq 0$ and
$\mathrm{Im}\,\xi^{{}^{\prime}}=0$. We only need to calculate
$(E_{1}\cdot\bar{E}_{2})(E_{3}\cdot\bar{E}_{0}),\leavevmode\nobreak\
\leavevmode\nobreak\
(E_{1}\cdot\bar{E}_{3})(E_{2}\cdot\bar{E}_{0}),\leavevmode\nobreak\
\leavevmode\nobreak\ (E_{2}\cdot\bar{E}_{3})(E_{1}\cdot\bar{E}_{0}).$ Let
$\xi=\xi_{0}+l_{1}$ and $\eta=\eta_{0}+l_{2}$, where
$|l_{1}|,|l_{2}|\leq\epsilon.$

$\displaystyle
h^{8}e^{\frac{i}{h}x\cdot(2\xi-\bar{\xi}+\eta)}(E_{1}\cdot\bar{E}_{2})(E_{3}\cdot\bar{E}_{0})=\bigg{(}(\xi_{0}\cdot
a_{1})\xi_{0}-h^{2}k^{2}a_{1}+((l_{1}\cdot a_{1})\xi+(\xi_{0}\cdot
a_{1})l_{1})\bigg{)}$ $\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
\cdot\bigg{(}(\bar{\xi}_{0}\cdot\bar{a}_{2})\bar{\xi}_{0}-h^{2}k^{2}\bar{a}_{2}+\overline{((l_{1}\cdot
a_{2})\xi+(\xi_{0}\cdot a_{2})l_{1})}\bigg{)}\bigg{(}(\xi_{0}\cdot
a_{3})\xi_{0}-h^{2}k^{2}a_{3}+$ $\displaystyle\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ ((l_{1}\cdot a_{3})\xi+(\xi_{0}\cdot
a_{3})l_{1})\bigg{)}\cdot\bigg{(}(\eta_{0}\cdot
a_{0})\eta_{0}-h^{2}k^{2}a_{0}+((l_{2}\cdot a_{2})\eta+(\eta_{0}\cdot
a_{0})l_{2})\bigg{)}$ $\displaystyle\geq$ $\displaystyle\big{(}(\xi_{0}\cdot
a_{1})(\bar{\xi}_{0}\cdot\bar{a}_{2})(\xi_{0}\cdot\bar{\xi}_{0})+O(\epsilon)+O(h^{2})\big{)}\cdot\big{(}(\xi_{0}\cdot
a_{3})(\eta_{0}\cdot a_{0})(\xi_{0}\cdot\eta_{0})+O(\epsilon)+O(h^{2})\big{)}$
$\displaystyle\geq$
$\displaystyle\big{(}4+24\epsilon|a_{1}||a_{2}||\xi_{0}|^{3}+O(h^{2})\big{)}\big{(}8+2^{5}\epsilon|a_{2}|^{2}|\xi_{0}|^{3}+O(h^{2})\big{)}.$

Therefore,
$h^{8}e^{\frac{i}{h}x\cdot(2\xi-\bar{\xi}+\eta)}(E_{1}\cdot\bar{E}_{2})(E_{3}\cdot\bar{E}_{0})$
is greater than some positive constant, provided $\epsilon$ and $h$ are
sufficiently small. Direct calculations yield similar estimates for
$(E_{1}\cdot\bar{E}_{3})(E_{2}\cdot\bar{E}_{0})$,
$(E_{3}\cdot\bar{E}_{1})(E_{2}\cdot\bar{E}_{0})$,
$(E_{2}\cdot\bar{E}_{3})(E_{1}\cdot\bar{E}_{0})$ and
$(E_{2}\cdot\bar{E}_{3})(E_{1}\cdot\bar{E}_{0})$. Hence, when $\epsilon$ and
$h$ are sufficiently small, one has

$\left|\int_{\Omega}(q_{1}-q_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}E_{i_{1}}\cdot\overline{E_{i_{2}}}\leavevmode\nobreak\
E_{i_{3}}\cdot\overline{E_{i_{0}}}dV\right|\geq\left|\int_{\Omega}(q_{1}-q_{2})h^{-8}e^{-\frac{i}{h}x\cdot(2\xi-\bar{\xi}+\eta)}dV\right|\cdot
4$

and the R.H.S. of the inequality is controlled by items involving at least one
$t_{i}$ or $r_{i}$. Combining the estimate (53) yields

$\displaystyle\left|\int_{\Omega}(q_{1}-q_{2})e^{-\frac{i}{h}x\cdot(2\xi-\bar{\xi}+\eta)}dV\right|\leq
Ch^{8}\|q_{1}-q_{2}\|_{L^{\infty}(\Omega)}$
$\displaystyle(1+h^{-1}|\xi|)^{\frac{9}{2}}(1+h^{-1}|\eta|)^{\frac{3}{2}}$
$\displaystyle
e^{-\frac{c}{h}\min\\{\mathrm{Im}\,\xi_{1},\mathrm{Im}\,\eta_{1}\\}}e^{\frac{1}{h}(3|\mathrm{Im}\,\xi^{\prime}|+|\mathrm{Im}\,\eta^{\prime}|)},$

for $\mathrm{Im}\,\xi_{1}\geq 0$ and $\mathrm{Im}\,\eta_{1}\geq 0$. In
particular, let

$z=2\xi-\bar{\xi}+\eta,\leavevmode\nobreak\ \leavevmode\nobreak\
z_{0}=(4i,0,0).$

When $|\xi-a\zeta_{0}|<Ca\varepsilon$ and $|\eta-a\eta_{0}|<Ca\varepsilon$
with $\varepsilon\leq\frac{1}{2C}$, then $|z-az_{0}|<4Ca\varepsilon$, and

$\displaystyle\left|\int_{\Omega}(q_{1}-q_{2})e^{-\frac{i}{h}x\cdot
z}dx\right|\leq
Ch^{2}\|q_{1}-q_{2}\|_{L^{\infty}(\Omega)}e^{-\frac{ca}{2h}}e^{\frac{4Ca\varepsilon}{h}}.$

Conversely, for any $z\in\mathbb{C}^{3}$, $|z-az_{0}|<4Ca\varepsilon$, $z$ can
be decomposed to

$z=2\xi-\bar{\xi}+\eta,\leavevmode\nobreak\ \leavevmode\nobreak\ \xi,\eta\in
p^{-1}(kh),\leavevmode\nobreak\ \leavevmode\nobreak\
|\xi-a\zeta_{0}|<Ca\varepsilon,|\eta-a\eta_{0}|<Ca\varepsilon.$

Therefore, for all $z\in\mathbb{C}^{n}$, $|z-az_{0}|<4Ca\varepsilon$, the
estimate holds

$\displaystyle\left|\int_{\Omega}(q_{1}-q_{2})e^{-\frac{i}{h}x\cdot
z}dx\right|\leq
Ch^{-3}\|q\|_{L^{\infty}(\Omega)}e^{-\frac{ca}{2h}}e^{\frac{2Ca\varepsilon}{h}}.$

Hence we are able to obtain an exponential decay for the F.B.I. transform of
$T(q_{1}-q_{2})$ in the neighborhood $|z-az_{0}|<4Ca\varepsilon$. Then apply
the water melon approach similarly to show the uniqueness of $q(x)$ in a small
neighborhood of $x_{0}=0$, i.e., there exists some $\delta_{1}>0$ such that

$q_{1}(x)=q_{2}(x),\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
x\in\Omega,\leavevmode\nobreak\ \leavevmode\nobreak\ -\delta_{1}\leq x_{1}\leq
0.$

Now let $a=0$ in (46) and choose $b_{1},b_{2},b_{3},b_{0}$ similar as in (52).
It is easy to see that the expressions of $E$ and $H$ interchange in (50)-(51)
and the similar procedures follow. Hence in the same way we arrive at the
uniqueness of $p(x)$ in a small neighborhood of $x_{0}=0$, i.e., there exists
some $\delta_{2}>0$ such that

$p_{1}(x)=p_{2}(x),\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
x\in\Omega,\leavevmode\nobreak\ \leavevmode\nobreak\ -\delta_{2}\leq x_{1}\leq
0.$

This completes the proof of Proposition 4.1. $\Box$

####  4.1.2 From local to global

To extend to global results we need some preparatory work. Lemma 4.1 describes
a density result involving solutions to Maxwell’s equation as Lemma 4.1.
Remark 4.1 and Remark 4.2 provide some properties of Green functions of
Maxwell’s equations and Helmholtz equation, which will be needed in the
sequel.

###### Lemma 4.1

Let $\Omega_{1}\subset\Omega_{2}\subset\mathbb{R}^{3}$ with smooth boundary
and $\partial\Omega_{1}\cap\partial\Omega_{2}\not=\emptyset$. Let
$\mathcal{R}$ be the set of $\\{(v|_{\Omega_{1}},w|_{\Omega_{1}})\in
C^{\infty}(\Omega_{1},\mathbb{C}^{3})^{2}\\}$ such that

$\left\\{\begin{array}[]{rl}\nabla\wedge v-i\omega\mu_{0}w&=a,\\\ \nabla\wedge
w+i\omega\varepsilon_{0}v&=\bar{a},\end{array}\right.\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega_{2},$

with boundary value $\nu\wedge v|_{\partial\Omega_{2}}=0,$ for some $a\in
C_{c}^{\infty}(\Omega_{2},\mathbb{C}^{3})$ with
$\mathrm{supp}\,a\subset\bar{\Omega}_{2}\setminus\Omega_{1}$. Let
$\mathcal{S}$ be the set of
$\\{(\tilde{v},\tilde{w})\in\mathbb{C}^{\infty}(\Omega_{1},\mathbb{C}^{3})^{2}\\}$
such that

$\left\\{\begin{array}[]{rl}\nabla\wedge\tilde{v}-i\omega\mu_{0}\tilde{w}&=0,\\\
\nabla\wedge\tilde{w}+i\omega\varepsilon_{0}\tilde{v}&=0,\end{array}\right.\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega_{1},$

with boundary value
$\nu\wedge\tilde{v}|_{\partial\Omega_{1}\cap\partial\Omega_{2}}=0.$ Then
$\mathcal{R}$ is dense in the space $\mathcal{S}$ with respect to the
$L^{2}(\Omega_{1},\mathbb{C}^{3})^{2}$ topology.

Proof. Let $g=(g_{1},g_{2})\in L^{2}(\Omega_{1},\mathbb{C}^{3})^{2}$ and $g$
is orthogonal to $\mathcal{R}.$ It suffices to show that $g$ is also
orthogonal to $\mathcal{S}.$

Let $\tilde{g}_{1}$ and $\tilde{g}_{2}$ be the
$L^{2}(\Omega_{2},\mathbb{C}^{3})$ extension of $g_{1}$ and $g_{2}$ by taking
value 0 in $\Omega_{2}\setminus\Omega_{1}$. Suppose $(f,h)$ are the solutions
to the following nonhomogeneous Maxwell’s equations

$\left\\{\begin{array}[]{rl}\nabla\wedge f+i\omega\mu_{0}h&=\tilde{g}_{2},\\\
\nabla\wedge
h-i\omega\varepsilon_{0}f&=\tilde{g}_{1}\end{array}\right.\leavevmode\nobreak\
\text{in\leavevmode\nobreak\ }\Omega_{2},$

with boundary value $\nu\wedge f|_{\partial\Omega_{2}}=0.$ Hence, for every
$(v,w)\in\mathcal{R}$,

$\displaystyle\langle g_{1},v\rangle_{\Omega_{1}}+\langle
g_{2},w\rangle_{\Omega_{1}}=\langle\nabla\wedge
h-i\omega\varepsilon_{0}f,v\rangle_{\Omega_{2}}+\langle\nabla\wedge
f+i\omega\mu_{0}h,w\rangle_{\Omega_{2}}$ $\displaystyle=$
$\displaystyle\int_{\Omega_{2}}h\cdot\overline{\nabla\wedge
v}dV+\int_{\partial\Omega_{2}}\nu\wedge
h\cdot\overline{v}dS-\int_{\Omega_{2}}i\omega\varepsilon_{0}f\cdot\overline{v}dV$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ +\int_{\Omega_{2}}f\cdot\overline{\nabla\wedge
w}dV+\int_{\partial\Omega_{2}}\nu\wedge
f\cdot\overline{w}dS+\int_{\Omega_{2}}i\omega\mu_{0}h\cdot\overline{w}dV$
$\displaystyle=$ $\displaystyle\int_{\Omega_{2}}h\cdot\overline{\nabla\wedge
v-i\omega\mu_{0}w}dV+\int_{\Omega_{2}}f\cdot\overline{\nabla\wedge
w+i\omega\varepsilon_{0}v}dV$ $\displaystyle=$
$\displaystyle\int_{\Omega_{2}}h\cdot\overline{a}dV+\int_{\Omega_{2}}f\cdot
adV$ (56)

vanishes. Suppose $a=c(x)+d(x)i,$ with
$\mathrm{supp}\,c(x),d(x)\subset\bar{\Omega}_{2}\setminus\Omega_{1}$. Then
(56) becomes

$\displaystyle 0$
$\displaystyle=\int_{\Omega_{2}}h\cdot(c(x)-d(x)i)dV+\int_{\Omega_{2}}f\cdot(c(x)+d(x)i)dV$
$\displaystyle=\int_{\Omega_{2}}(f+h)\cdot c(x)dV+\int_{\Omega_{2}}(f-h)\cdot
d(x)idV.$

Since $c(x)$ and $d(x)$ are arbitrary in $L^{2}(\Omega_{2},\mathbb{R}^{3})$,
$f+h$ and $f-h$ vanish on $\bar{\Omega}_{2}\setminus\Omega_{1}$, which implies
$f$ and $h$ vanish on $\bar{\Omega}_{2}\setminus\Omega_{1}.$ Then

$\displaystyle\nu\wedge
f|_{\partial\Omega_{1}\setminus\partial\Omega_{2}}=0\leavevmode\nobreak\
\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\nu\wedge h|_{\partial\Omega_{1}\setminus\partial\Omega_{2}}=0.$ (57)

Combining the assumption $\nu\wedge f|_{\partial\Omega_{2}}=0$, one has

$\displaystyle\nu\wedge f|_{\partial\Omega_{1}}=0.$ (58)

For any $(\tilde{v},\tilde{w})\in\mathcal{S}$,

$\displaystyle\langle g_{1},\tilde{v}\rangle_{\Omega_{1}}+\langle
g_{2},\tilde{w}\rangle_{\Omega_{1}}=\langle\nabla\wedge
h-i\omega\varepsilon_{0}f,\tilde{v}\rangle_{\Omega_{1}}+\langle\nabla\wedge
f+i\omega\mu_{0}h,\tilde{w}\rangle_{\Omega_{1}}$ $\displaystyle=$
$\displaystyle\int_{\Omega_{1}}h\cdot\overline{\nabla\wedge\tilde{v}-i\omega\mu_{0}\tilde{w}}dV+\int_{\Omega_{1}}f\cdot\overline{\nabla\wedge\tilde{w}+i\omega\varepsilon_{0}\tilde{v}}dV$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
\leavevmode\nobreak\ +\int_{\partial\Omega_{1}}\nu\wedge
h\cdot\overline{\tilde{v}}dS+\int_{\partial\Omega_{1}}\nu\wedge
f\cdot\overline{\tilde{w}}dS,$

vanishes by virtue of (57), (58) and
$\nu\wedge\tilde{v}|_{\partial\Omega_{1}\cap\partial\Omega_{2}}=0.$ Therefore,
$g$ is orthogonal to $\mathcal{S}.$ $\Box$

###### Remark 4.1

(i)The fundamental solution
$G^{\mu,\varepsilon}=\left(\begin{array}[]{c}G^{E}\\\
G^{H}\end{array}\right)=\left(\begin{array}[]{cc}E^{(1)}&E^{(2)}\\\
H^{(1)}&H^{(2)}\end{array}\right)$ consists of two parts $(E^{(1)},H^{(1)})$
and $(E^{(2)},H^{(2)})$ satisfying the following Maxwell’s equations,
respectively:

$\left\\{\begin{array}[]{rl}\nabla\wedge
E^{(1)}(x,y)-i\omega\mu_{0}H^{(1)}(x,y)&=\delta(x-y)I_{3},\\\ \nabla\wedge
H^{(1)}(x,y)+i\omega\varepsilon_{0}E^{(1)}(x,y)&=0,\end{array}\right.$

in $\Omega_{2}$ with boundary value $\nu\wedge
E^{(1)}|_{\partial\Omega_{2}}=0;$ and

$\left\\{\begin{array}[]{rl}\nabla\wedge
E^{(2)}(x,y)-i\omega\mu_{0}H^{(2)}(x,y)&=0,\\\ \nabla\wedge
H^{(2)}(x,y)+i\omega\varepsilon_{0}E^{(2)}(x,y)&=\delta(x-y)I_{3},\end{array}\right.$

in $\Omega_{2}$ with boundary value $\nu\wedge
E^{(2)}|_{\partial\Omega_{2}}=0$.

(ii)Similarly, the fundamental solution
$\tilde{G}^{\varepsilon,\mu}=\left(\begin{array}[]{c}\tilde{G}^{E}\\\
\tilde{G}^{H}\end{array}\right)=\left(\begin{array}[]{cc}\tilde{E}^{(1)}&\tilde{E}^{(2)}\\\
\tilde{H}^{(1)}&\tilde{H}^{(2)}\end{array}\right)$ consists of two parts which
satisfy the Maxwell’s equations

$\left\\{\begin{array}[]{rl}\nabla\wedge\tilde{E}^{(1)}(x,y)-i\omega\varepsilon_{0}\tilde{H}^{(1)}(x,y)&=\delta(x-y)I_{3},\\\
\nabla\wedge\tilde{H}^{(1)}(x,y)+i\omega\mu_{0}\tilde{E}^{(1)}(x,y)&=0,\end{array}\right.$

in $\Omega_{2}$ with boundary value
$\nu\wedge\tilde{H}^{(1)}|_{\partial\Omega_{2}}=0;$ and

$\left\\{\begin{array}[]{rl}\nabla\wedge\tilde{E}^{(2)}(x,y)-i\omega\varepsilon_{0}\tilde{H}^{(2)}(x,y)&=0,\\\
\nabla\wedge\tilde{H}^{(2)}(x,y)+i\omega\mu_{0}\tilde{E}^{(2)}(x,y)&=\delta(x-y)I_{3},\end{array}\right.$

in $\Omega_{2}$ with boundary value
$\nu\wedge\tilde{H}^{(2)}|_{\partial\Omega_{2}}=0.$

###### Remark 4.2

If $(v,w)$ satisfies the boundary value problem

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge v-i\omega\mu_{0}w&=a,\\\
\nabla\wedge
w+i\omega\varepsilon_{0}v&=\bar{a},\end{array}\right.\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \text{in\leavevmode\nobreak\
}\Omega_{2},$

with boundary value $\nu\wedge v|_{\partial\Omega_{2}}=0,$ by direct
computation $(\bar{w},\bar{v})$ is a solution to the conjugate Maxwell’s
equations

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge\bar{w}-i\omega\varepsilon_{0}\bar{v}&=a,\\\
\nabla\wedge\bar{v}+i\omega\mu_{0}\bar{w}&=\overline{a},\end{array}\right.\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \text{in\leavevmode\nobreak\
}\Omega_{2},$

with boundary value $\nu\wedge\bar{v}|_{\partial\Omega_{2}}=0.$ This implies
that the conjugate $\bar{w}$ plays a role of the electricity $E$ while the
conjugate $\bar{v}$ works as the magnetism $H$.

Let $G_{\Omega_{2}}(x,y)$ be the matrix of Green function of the Helmholtz
equation
$(-\Delta_{y}-k^{2})G_{\Omega_{2}}(x,y)=\delta(x-y)I_{3},\leavevmode\nobreak\
\text{in \leavevmode\nobreak\ }\Omega_{2},$ where $I_{3}$ is the $3\times 3$
identity matrix.

Proof of Theorem 4.1 Below we show the extension of local result (Proposition
4.1) to the global situation following the same process as in section 2.2.2.

Let $\bar{p}(y)=p_{1}-p_{2}$ and $\bar{q}(y)=q_{1}-q_{2}$. Fix a point
$x_{1}\in\Omega$ and let $\theta:[0,1]{\rightarrow}\bar{\Omega}$ be a $C^{1}$
curve joining $x_{0}\in\Gamma$ to $x_{1}$ such that $\theta(0)=x_{0}$ and
$\theta^{\prime}(0)$ is the interior normal to $\partial\Omega$ at $x_{0}$ and
$\theta(t)\in\Omega$ for all $t\in(0,1].$ Suppose $\Theta_{\epsilon}(t)$ is a
small neighborhood of the curve $\theta([0,t])$ in $\Omega$ and $I$ is the set
of the time where $\bar{p}$ and $\bar{q}$ vanish on $\Theta_{\epsilon}(t)$.
Similarly, $I$ is a closed set and $I$ is also non-empty due to local results,
provided that $\epsilon$ is small enough. It suffices to show that $I$ is
open.

Following the same setting as in section 2.2.2, we have
$\partial\Theta_{\epsilon}(t)\cap\partial\Omega\subset\Gamma,$ and

$\Omega\setminus\Theta_{\epsilon}(t)\subset\Omega_{1},\leavevmode\nobreak\
\leavevmode\nobreak\
\partial\Omega\setminus\Gamma\subset\partial\Omega\cap\partial\Omega_{1};\leavevmode\nobreak\
\leavevmode\nobreak\
\partial\Omega\setminus\Gamma\subset\partial\Omega\cap\partial\Omega_{1}\subset\partial\Omega\cap\partial\Omega_{2}.$

Adopting the notation in Remark 4.1, let

$G_{1}(x_{1},y)=\left(\begin{array}[]{c}G_{1}^{E}(x_{1},y)\\\
G_{1}^{H}(x_{1},y)\end{array}\right),\leavevmode\nobreak\
\tilde{G}_{1}(x_{1},y)=\left(\begin{array}[]{c}\tilde{G}_{1}^{E}(x_{1},y)\\\
\tilde{G}_{1}^{H}(x_{1},y)\end{array}\right),$

for $x_{1},y\in\Omega_{2}$. Similarly, denote $G_{2}(x_{2},y),$
$G_{3}(x_{3},y)$ and $\tilde{G}_{0}(x_{0},y),$ $\tilde{G}_{2}(x_{2},y),$
$\tilde{G}_{3}(x_{3},y)$ for $x_{2},x_{3},x_{0},y\in\Omega_{2}$, respectively.
When $x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\Omega_{1}$, define

$\displaystyle H(x_{1},x_{2},x_{3},x_{0})=$
$\displaystyle\int_{\Omega_{1}}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)dy$
(59) $\displaystyle=$ $\displaystyle\int_{\Omega_{1}}\bar{q}(y)\cdot
G_{1}^{E}(x_{1},y)\otimes\tilde{G}_{2}^{H}(x_{2},y)\cdot
G_{3}^{E}(x_{3},y)\otimes\tilde{G}_{0}^{H}(x_{0},y)$
$\displaystyle+\bar{q}(y)\cdot
G_{2}^{E}(x_{2},y)\otimes\tilde{G}^{H}_{1}(x_{1},y)\cdot
G_{3}^{E}(x_{3},y)\otimes\tilde{G}^{H}_{0}(x_{0},y)$
$\displaystyle+\bar{q}(y)\cdot
G_{1}^{E}(x_{1},y)\otimes\tilde{G}^{H}_{3}(x_{3},y)\cdot
G_{2}^{E}(x_{2},y)\otimes\tilde{G}^{H}_{0}(x_{0},y)$
$\displaystyle+\bar{q}(y)\cdot
G_{3}^{E}(x_{3},y)\otimes\tilde{G}_{1}^{H}(x_{1},y)\cdot
G_{2}^{E}(x_{2},y)\otimes\tilde{G}^{H}_{0}(x_{0},y)$
$\displaystyle+\bar{q}(y)\cdot
G_{2}^{E}(x_{2},y)\otimes\tilde{G}^{H}_{3}(x_{3},y)\cdot
G^{E}_{1}(x_{1},y)\otimes\tilde{G}^{H}_{0}(x_{0},y)$
$\displaystyle+\bar{q}(y)\cdot
G_{3}^{E}(x_{3},y)\otimes\tilde{G}^{H}_{2}(x_{2},y)\cdot
G_{1}^{E}(x_{1},y)\otimes\tilde{G}^{H}_{0}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{1}^{H}(x_{1},y)\otimes\tilde{G}_{2}^{E}(x_{2},y)\cdot
G_{3}^{H}(x_{3},y)\otimes\tilde{G}_{0}^{E}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{2}^{H}(x_{2},y)\otimes\tilde{G}^{E}_{1}(x_{1},y)\cdot
G_{3}^{H}(x_{3},y)\otimes\tilde{G}^{E}_{0}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{1}^{H}(x_{1},y)\otimes\tilde{G}^{E}_{3}(x_{3},y)\cdot
G_{2}^{H}(x_{2},y)\otimes\tilde{G}^{E}_{0}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{3}^{H}(x_{3},y)\otimes\tilde{G}_{1}^{E}(x_{1},y)\cdot
G_{2}^{H}(x_{2},y)\otimes\tilde{G}^{E}_{0}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{2}^{H}(x_{2},y)\otimes\tilde{G}^{E}_{3}(x_{3},y)\cdot
G^{H}_{1}(x_{1},y)\otimes\tilde{G}^{E}_{0}(x_{0},y)$
$\displaystyle+\bar{p}(y)\cdot
G_{3}^{H}(x_{3},y)\otimes\tilde{G}^{E}_{2}(x_{2},y)\cdot
G_{1}^{H}(x_{1},y)\otimes\tilde{G}^{E}_{0}(x_{0},y)dy.$ (60)

Here the tensor sign $\otimes$ represents the tensor product of the matrices
and is also consistent with the tensor of distributions when the matrices are
viewed as distributions in the sense that

$\int(u_{1}(x_{1})\otimes
u_{2}(x_{2}))(\phi_{1}(x_{1})\otimes\phi_{2}(x_{2}))dx_{1}dx_{2}=\int
u_{1}\phi_{1}dx_{1}\int u_{2}\phi_{2}dx_{2},$

for $u_{1},u_{2}\in D^{\prime}(X_{j})$ and $\phi_{j}\in C_{0}^{\infty}(X_{j})$
where $X_{j}$ are open sets of $\mathbb{R}^{3}.$

The term involving $\bar{p}(y)$ plays the role of a magnetic field, while the
term involving $\bar{q}(y)$ serves as an electric field in the coupled
Maxwell’s equations with vanishing tangential component on the boundary.
Moreover, for $x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\Omega_{1}$,

$\displaystyle H(x_{1},x_{2},x_{3},x_{0})$
$\displaystyle=\int_{\Omega_{1}}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)dy$
$\displaystyle=\int_{\Omega}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)dy.$

Note that each entry of $H$ is a sum of terms involving $\bar{p}(y)$ and
$\bar{q}(y)$ and satisfies the Helmholtz equation $(-\Delta_{y}-k^{2})u=0,$
for $x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\bar{\Omega}_{1}.$
Therefore, for $x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\bar{\Omega},$
from the assumption we have

$\int_{\Omega}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)dy=0.$

It follows that $H(x_{1},x_{2},x_{3},x_{0})=0$ when
$x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\bar{\Omega}$.

When $x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}\setminus\bar{\Omega}_{1},$ the
entries of $H$ satisfy the Helmholtz equation $(-\Delta_{y}-k^{2})u=0.$ Hence,
it follows from the Unique Continuation Principle that

$H(x_{1},x_{2},x_{3},x_{0})=0\leavevmode\nobreak\ \leavevmode\nobreak\
\text{when\leavevmode\nobreak\ \leavevmode\nobreak\
}x_{1},x_{2},x_{3},x_{0}\in\bar{\Omega}_{2}\setminus\Omega_{1};$

i.e., $\leavevmode\nobreak\ \leavevmode\nobreak\ \text{for\leavevmode\nobreak\
}x_{1},x_{2},x_{3},x_{0}\in\bar{\Omega}_{2}\setminus\Omega_{1},$

$\displaystyle\int_{\Omega_{1}}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)dy=0.$

Let the left hand side of (60) act on $(a_{1}(x_{1})\otimes
a_{2}(x_{2}),a_{3}(x_{3})\otimes a_{4}(x_{4}))$ for
$x_{1},x_{2},x_{3},x_{0}\in\Omega_{2}$. Here all $a_{i}$ are assumed to have
the forms $a_{i}(x_{i})=(l_{i}(x_{i}),\overline{l_{i}(x_{i})})^{T},$ where
each $l_{i}$ is a smooth vector function on
$C^{\infty}(\Omega_{2},\mathbb{C}^{3})$ with support in
$\bar{\Omega}_{2}\setminus\Omega_{1}$. For each $i$, let

$\displaystyle
v_{i}(y)=\int_{\Omega_{2}}G^{E}_{i}(x_{i},y)a_{i}(x_{i})dx_{i},\leavevmode\nobreak\
\leavevmode\nobreak\
w_{i}(y)=\int_{\Omega_{2}}G^{H}_{i}(x_{i},y)a_{i}(x_{i})dx_{i},$
$\displaystyle\tilde{w}_{i}(y)=\int_{\Omega_{2}}\tilde{G}^{E}_{i}(x_{i},y)a_{i}(x_{i})dx_{i},\leavevmode\nobreak\
\leavevmode\nobreak\
\tilde{v}_{i}(y)=\int_{\Omega_{2}}\tilde{G}^{H}_{i}(x_{i},y)a_{i}(x_{i})dx_{i}.$

Then (60) becomes

$\displaystyle 0=$
$\displaystyle\int_{\Omega_{1}}\bar{q}(y)K^{E}(x_{1},x_{2},x_{3},x_{0},y)\vec{a}+\bar{p}(y)K^{H}(x_{1},x_{2},x_{3},x_{0},y)\vec{a}dy$
$\displaystyle=$
$\displaystyle\int_{\Omega_{1}}\bar{q}(y)(v_{1}\cdot\tilde{v}_{2})\cdot(v_{3}\cdot\tilde{v}_{0})+\bar{q}(y)(v_{2}\cdot\tilde{v}_{1})\cdot(v_{3}\cdot\tilde{v}_{0})$
$\displaystyle+\bar{q}(y)(v_{1}\cdot\tilde{v}_{3})\cdot(v_{2}\cdot\tilde{v}_{0})+\bar{q}(y)(v_{3}\cdot\tilde{v}_{1})\cdot(v_{2}\cdot\tilde{v}_{0})$
$\displaystyle+\bar{q}(y)(v_{2}\cdot\tilde{v}_{3})\cdot(v_{1}\cdot\tilde{v}_{0})+\bar{q}(y)(v_{3}\cdot\tilde{v}_{2})\cdot(v_{1}\cdot\tilde{v}_{0})$
$\displaystyle+\bar{p}(y)(w_{1}\cdot\tilde{w}_{2})\cdot(w_{3}\cdot\tilde{w}_{0})+\bar{p}(y)(w_{2}\cdot\tilde{w}_{1})\cdot(w_{3}\cdot\tilde{w}_{0})$
$\displaystyle+\bar{p}(y)(w_{1}\cdot\tilde{w}_{3})\cdot(w_{2}\cdot\tilde{w}_{0})+\bar{p}(y)(w_{3}\cdot\tilde{w}_{1})\cdot(w_{2}\cdot\tilde{w}_{0})$
$\displaystyle+\bar{p}(y)(w_{2}\cdot\tilde{w}_{3})\cdot(w_{1}\cdot\tilde{w}_{0})+\bar{p}(y)(w_{3}\cdot\tilde{w}_{2})\cdot(w_{1}\cdot\tilde{w}_{0})dy$
(61)

The fundamental solutions of Remark 4.1 imply that $(v_{i},w_{i})$ is a
solution to the inhomogeneous Maxwell’s systems

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge
v_{i}-i\omega\mu_{0}w_{i}&=l_{i},\\\ \nabla\wedge
w_{i}+i\omega\varepsilon_{0}v_{i}&=\overline{l_{i}},\end{array}\right.$ (64)

in $\Omega_{2}$ with boundary value $\nu\wedge v_{i}|_{\partial\Omega_{2}}=0,$
while $(\tilde{v}_{i},\tilde{w}_{i})$ satisfies

$\left\\{\begin{array}[]{rl}\nabla\wedge\tilde{w}_{i}-i\omega\varepsilon_{0}\tilde{v}_{i}&=l_{i},\\\
\nabla\wedge\tilde{v}_{i}+i\omega\mu_{0}\tilde{w}_{i}&=\overline{l_{i}},\end{array}\right.$

in $\Omega_{2}$ with boundary value
$\nu\wedge\tilde{v}_{i}|_{\partial\Omega_{2}}=0.$ Hence, the uniqueness of
solutions gives $\tilde{v}_{i}=\bar{v}_{i}$ and $\tilde{w}_{i}=\bar{w}_{i}$ in
$\Omega_{2},$ for $i=0,1,2,3.$ Therefore, (61) yields

$\displaystyle 0$
$\displaystyle=\int_{\Omega_{1}}(p_{1}-p_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(w_{i_{1}}\cdot\overline{w}_{i_{2}})(w_{i_{3}}\cdot\overline{w}_{0})dy$
$\displaystyle\quad+\int_{\Omega_{1}}(q_{1}-q_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(v_{i_{1}}\cdot\overline{v}_{i_{2}})(v_{i_{3}}\cdot\overline{v}_{0})dy,$
(65)

where $(v_{i},w_{i})$ satisfies $(\ref{value1})$. Since all $a_{i}\in
C^{\infty}(\Omega_{2},\mathbb{C}^{3})$ are supported in
$\bar{\Omega}_{2}\setminus\Omega_{1}$, the set of all the solutions
$(v_{i},w_{i})$ forms the set $\mathcal{R}$. Lemma 4.1 states that
$\mathcal{R}$ is dense in the space $\mathcal{S}$ with respect to the
$L^{2}(\Omega_{1},\mathbb{C}^{3})^{2}$ topology. Thus for any $V_{i}$ and
$W_{i}$ satisfying inhomogeneous Maxwell’s systems on $\Omega_{1}$ with
$\nu\wedge V_{i}$ vanishing on $\partial\Omega_{1}\cap\partial\Omega_{2}$, we
can choose $v_{i}^{(n)}$ and $w_{i}^{(n)}$ converge to $V_{i}$ and $W_{i}$ in
$L^{2}(\Omega_{1},\mathbb{C}^{3})$ respectively.

Note that each $v_{i}^{(n)}$, $w_{i}^{(n)}$ and $V_{i},W_{i}$ are smooth in
$\Omega_{2}$. Then $\|v_{i}^{(n)}\|_{L^{\infty}(\Omega_{2})}$ and
$\|w_{i}^{(n)}\|_{L^{\infty}(\Omega_{2})}$ are bounded for each fixed $i$.
Thus assume

$\|v_{i}^{(n)}\|_{L^{\infty}(\Omega_{2})},\|w_{i}^{(n)}\|_{L^{\infty}(\Omega_{2})},\|V_{i}\|_{L^{\infty}(\Omega_{2})},\|W_{i}\|_{L^{\infty}(\Omega_{2})}\leq
M.$

It is easy to show that
$(v_{1}^{(n)}\cdot\bar{v}_{2}^{(n)})(v_{3}^{(n)}\cdot\bar{v}_{0}^{(n)})$
converges to $(V_{1}\cdot\bar{V}_{2})\leavevmode\nobreak\
(V_{3}\cdot\bar{V}_{0})$ in $L^{1}(\Omega_{1})$ as $n{\rightarrow}\infty$. The
equality (65) extends to

$\displaystyle 0$
$\displaystyle=\int_{\Omega_{1}}(p_{1}-p_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(W_{i_{1}}\cdot\overline{W}_{i_{2}})(W_{i_{3}}\cdot\overline{W}_{0})dV$
$\displaystyle\quad+\int_{\Omega_{1}}(q_{1}-q_{2})\sum_{(i_{1}i_{2}i_{3})=\sigma(123)}(V_{i_{1}}\cdot\overline{V}_{i_{2}})(V_{i_{3}}\cdot\overline{V}_{0})dV,$

where $(V_{i},W_{i})\in C^{\infty}(\bar{\Omega}_{1},\mathbb{C}^{3})^{2}$ is a
solution to inhomogeneous Maxwell’s systems and vanishes on
$\partial\Omega_{1}\cap\partial\Omega_{2}$. Applying Proposition 4.1,
$\bar{p}$ and $\bar{q}$ vanish on a neighborhood of
$\partial\Omega_{1}\setminus(\partial\Omega_{1}\cap\partial\Omega_{2})$. Then
$\bar{p}$ and $\bar{q}$ vanish on a bigger neighborhood
$\Theta_{\epsilon}(t^{\prime})$ $t^{\prime}>t$ of the curve. Hence $I$ is
open. $\Box$

###  4.2 Proof of Theorem 1.4

Below we follow the standard procedures of high order linearization of the
admittance map.

Proof. Let
$f^{k\omega}=\epsilon_{1}f^{k\omega}_{1}+\epsilon_{2}f^{k\omega}_{2},$
$f_{l}^{k\omega}\in C^{\infty}(\partial\Omega)$ with
$\mathrm{supp}\,f_{l}^{k\omega}\subseteq\Gamma_{1}$, for $k=1,2$ and $l=1,2$.
Let $\epsilon=(\epsilon_{1},\epsilon_{2})$. Denote
$(V^{k\omega,j}_{l},W^{k\omega,j}_{l})=(\partial\epsilon_{l}E^{k\omega,j}\big{|}_{\epsilon=0},\partial\epsilon_{l}H^{k\omega,j}\big{|}_{\epsilon=0})$.
Similarly, one can show that
$V_{l}^{k\omega}:=V_{l}^{k\omega,1}=V_{l}^{k\omega,2}$ and
$W_{l}^{k\omega}:=W_{l}^{k\omega,1}=W_{l}^{k\omega,2}$ satisfy

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge
V^{\omega}_{l}-i\omega\mu_{0}W^{\omega}_{l}&=0,\\\ \nabla\wedge
W^{\omega}_{l}+i\omega\varepsilon_{0}V^{\omega}_{l}&=0,\\\ \nabla\wedge
V^{2\omega}_{l}-i2\omega\mu_{0}W^{2\omega}_{l}&=0,\\\ \nabla\wedge
W^{2\omega}_{l}+i2\omega\varepsilon_{0}V^{2\omega}_{l}&=0,\end{array}\right.$
(70)

in $\Omega$ with boundary values $\nu\wedge
V^{k\omega}_{l}|_{\partial\Omega}=f^{k\omega}_{l}$ for $k=1,2$ and $l=1,2.$
The second linearization yields that
$(W_{12}^{k\omega,j},V_{12}^{k\omega,j})=(\partial^{2}_{\epsilon}H^{k\omega,j}|_{\epsilon=0},\partial^{2}_{\epsilon}E^{k\omega,j}|_{\epsilon=0})$
satisfies

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge
V_{12}^{\omega,j}-i\omega\mu_{0}W_{12}^{\omega,j}&=0,\\\ \nabla\wedge
W_{12}^{\omega,j}+i\omega\varepsilon_{0}V_{12}^{\omega,j}&=-i\omega\chi^{(2),j}\big{(}\overline{V_{1}^{\omega}}\cdot
V_{2}^{2\omega}+\overline{V_{2}^{\omega}}\cdot V_{1}^{2\omega}\big{)},\\\
\nabla\wedge V_{12}^{2\omega,j}-i2\omega\mu_{0}W_{12}^{2\omega,j}&=0,\\\
\nabla\wedge
W_{12}^{2\omega,j}+i2\omega\varepsilon_{0}V_{12}^{2\omega,j}&=-i\omega\chi^{(2),j}V^{\omega}_{1}\cdot
V^{\omega}_{2},\end{array}\right.$

in $\Omega$ with boundary values $\nu\wedge
V_{12}^{k\omega,j}|_{\partial\Omega}=0$ with $k=1,2$ and $j=1,2$. We are able
to derive the following integral identities

$\displaystyle\int_{\Omega}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{V_{0}^{\omega}}\leavevmode\nobreak\
\big{(}\overline{V_{1}^{\omega}}\cdot
V_{2}^{2\omega}+\overline{V_{2}^{\omega}}\cdot V_{1}^{2\omega}\big{)}dV=0,$
(71)

and

$\displaystyle\int_{\Omega}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{V_{0}^{2\omega}}\leavevmode\nobreak\
\big{(}V_{1}^{\omega}\cdot V_{2}^{\omega}\big{)}dV=0,$ (72)

by choosing $(V_{0}^{k\omega},W_{0}^{k\omega})$ to be the solution to the
following conjugate Maxwell’s equations

$\displaystyle\left\\{\begin{array}[]{rl}\nabla\wedge
V^{\omega}_{0}-i\omega\mu_{0}W^{\omega}_{0}&=0,\\\ \nabla\wedge
W^{\omega}_{0}+i\omega\varepsilon_{0}V^{\omega}_{0}&=0,\\\ \nabla\wedge
V^{2\omega}_{0}-i2\omega\mu_{0}W^{2\omega}_{0}&=0,\\\ \nabla\wedge
W^{2\omega}_{0}+i2\omega\varepsilon_{0}V^{2\omega}_{0}&=0,\end{array}\right.$
(77)

in $\Omega$ with boundary values $\nu\wedge
V_{0}^{k\omega}|_{\partial\Omega}=f_{0}^{k\omega},$ where $f^{k\omega}_{0}\in
C^{\infty}(\partial\Omega)$ with supports on $\Gamma_{2}$, for each $k$. Next
we construct $V_{1}^{\omega}$, $V_{2}^{\omega}$ and $V_{0}^{2\omega}$
satisfying (70) and (77) to plug in the integral identity (72).

###### Remark 4.3

It is easy to check that $(E,H)$ of the following form

$\displaystyle E$
$\displaystyle=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\xi}\tilde{A}_{\xi},\leavevmode\nobreak\
\leavevmode\nobreak\ \text{satisfying\leavevmode\nobreak\
$\tilde{A}_{\xi}\cdot\xi=0$,}$ (78) $\displaystyle H$
$\displaystyle=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\xi}\tilde{B}_{\xi},\leavevmode\nobreak\
\leavevmode\nobreak\ \text{with\leavevmode\nobreak\
$\tilde{B}_{\xi}=-\frac{1}{kh}\xi\times\tilde{A}_{\xi}$,}$ (79)

satisfies the linearized Maxwell equations by noting that

$\tilde{A}_{\xi}=-\frac{1}{k^{2}h^{2}}\cdot((\xi\cdot\xi)\tilde{A}_{\xi}-(\xi\cdot\tilde{A}_{\xi})\xi)=-\frac{1}{k^{2}h^{2}}\xi\times(\xi\times\tilde{A}_{\xi})=\frac{1}{kh}\xi\times\tilde{B}_{\xi}.$

####  4.2.1 Local result

Below we are to prove a local version of Theorem 1.4, i.e., to show
$\chi^{(2),1}=\chi^{(2),2}$ locally in a small neighborhood of some
$x_{0}\in\Gamma_{1}$. Without loss of generality we assume that the domain is
below the coordinate $\\{x_{1}\leq 0\\}$ by orthogonal transformation and
translation and $x_{0}=0$.

###### Proposition 4.2

Let $\Omega$ be a bounded open set in $\mathbb{R}^{3}$ with smooth boundary
$\partial\Omega$. Let $x_{0}\in\Gamma_{1}\cap\Gamma_{2}\subset\partial\Omega$
be a convex point on the boundary such that
$\Gamma\subset\Gamma_{1}\cap\Gamma_{2}$. Let $\Gamma$ be a small neighborhood
of $x_{0}$ on the boundary. Suppose that the cancellation (72) hold for any
smooth solutions $(V_{1}^{\omega},W_{1}^{\omega})$,
$(V_{2}^{\omega},W_{2}^{\omega})$, and $(V_{0}^{2\omega},W_{0}^{2\omega})$ to
the systems (70) and (77). Then there exists $\delta>0$ such that
$\chi^{(2),1}=\chi^{(2),2}$ on $B(x_{0},\delta)\cap\Omega$.

Proof. (i) Define the set
$p^{-1}(kh)=\\{\xi\in\mathbb{C}^{3}|\xi\cdot\xi=k^{2}h^{2}\\}$ and choose

$\xi_{0}=(i,\sqrt{1+k^{2}h^{2}},0),\leavevmode\nobreak\ \leavevmode\nobreak\
\eta_{0}=(i,0,\sqrt{1+k^{2}h^{2}})\in p^{-1}(kh).$

Suppose $|\xi-\xi_{0}|<\epsilon,\xi\in p^{-1}(kh)$, and
$|\eta-\eta_{0}|<\epsilon,\eta\in p^{-1}(kh)$. Then $\mathrm{Im}\,(\xi_{1})>0$
and $\mathrm{Im}\,(\eta_{1})>0$. Below we construct pairs of solutions as in
subsection 4.1. Here we technically construct
$(E_{0}^{2\omega},H_{0}^{2\omega})$ of the expressions (78) (79) and
$(E_{1}^{\omega},H_{1}^{\omega})$, $(E_{2}^{\omega},H_{2}^{\omega})$ of the
expression (46).

As in Remark 4.3, choose
$E_{0}^{2\omega}=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\varsigma}\tilde{A}_{\varsigma},$
where

$\displaystyle\begin{split}\varsigma&=(\sqrt{2(1-k^{2}h^{2})}i,\sqrt{1+k^{2}h^{2}},\sqrt{1+k^{2}h^{2}})\in
p^{-1}(2kh),\\\
\tilde{A}_{\varsigma}&=\left(-\sqrt{2(1-k^{2}h^{2})}i,-\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}},-\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}}\right),\end{split}$
(80)

Since $\varsigma\cdot\tilde{A}_{\varsigma}=0$, $E_{0}^{2\omega}$ together with
some determined $H_{0}^{2\omega}$ are solutions to the linearized Maxwell’s
equations following Remark 4.3.

Let $b=0$, $a_{1}=a_{2}=(1,0,0)$ in (46). Denote
$H_{1}^{\omega}=\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\xi\times
a_{1})e^{-\frac{i}{h}x\cdot\xi}$,
$H_{2}^{\omega}=\mu_{0}^{-\frac{1}{2}}\frac{k}{h}(\eta\times
a_{2})e^{-\frac{i}{h}x\cdot\eta}$, and

$\begin{split}E_{1}^{\omega}=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\xi}\tilde{A}_{\xi}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\xi\cdot
a_{1})\xi-k^{2}a_{1}\right)e^{-\frac{i}{h}x\cdot\xi},\\\
E_{2}^{\omega}=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\eta}\tilde{A}_{\eta}=\varepsilon_{0}^{-\frac{1}{2}}\left(\frac{1}{h^{2}}(\eta\cdot
a_{2})\xi-k^{2}a_{2}\right)e^{-\frac{i}{h}x\cdot\eta},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \end{split}$ (81)

When $|\xi-\xi_{0}|<\epsilon,$

$\displaystyle\tilde{A}_{\xi}$ $\displaystyle=h^{-2}\left((\xi\cdot
a_{1})\xi-k^{2}h^{2}a_{1}\right)=h^{-2}\xi\times(\xi\times a_{1})$
$\displaystyle=h^{-2}\big{(}\xi_{0}\times(\xi_{0}\times
a_{1})+O(\epsilon)\big{)}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},\sqrt{1+k^{2}h^{2}}i,0)+O(\epsilon)\bigg{)},$

where $O(\epsilon)$ is some terms independent of $x$. Similarly, when
$|\eta-\eta_{0}|<\epsilon,$

$\displaystyle\tilde{A}_{\eta}$ $\displaystyle=h^{-2}\left((\eta\cdot
a_{2})\eta-k^{2}h^{2}a_{2}\right)=h^{-2}\eta\times(\eta\times a_{2})$
$\displaystyle=h^{-2}\big{(}\eta_{0}\times(\eta_{0}\times
a_{2})+O(\epsilon)\big{)}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},0,\sqrt{1+k^{2}h^{2}}i)+O(\epsilon)\bigg{)},$

where $O(\epsilon)$ is some terms independent of $x$. Construct
$(r_{1}^{\omega},t_{1}^{\omega})$, $(r_{2}^{\omega},t_{2}^{\omega})$ and
$(r_{0}^{2\omega},t_{0}^{2\omega})$ with exponentially decaying estimates as
in (53) such that

$\displaystyle
V_{1}^{\omega}=E_{1}^{\omega}+r_{1}^{\omega},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\
W_{1}^{\omega}=H_{1}^{\omega}+t_{1}^{\omega};$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
V_{2}^{\omega}=E_{2}^{\omega}+r_{2}^{\omega},\leavevmode\nobreak\
\leavevmode\nobreak\ W_{2}^{\omega}=H_{2}^{\omega}+t_{2}^{\omega};$
$\displaystyle V_{0}^{2\omega}=E_{0}^{2\omega}+r_{0}^{2\omega},$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
W_{0}^{2\omega}=H_{0}^{2\omega}+t_{0}^{2\omega}.$

are solutions to (70) and (77), respectively. Thus, the integral identity (72)
becomes

$\displaystyle\int_{\Omega}$
$\displaystyle(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{(E_{0}^{2\omega}+r_{0}^{2\omega})}\leavevmode\nobreak\
\leavevmode\nobreak\
(E_{1}^{\omega}+r_{1}^{\omega})\cdot(E_{2}^{\omega}+r_{2}^{\omega})dV=0.$ (82)

Next we calculate the terms in the expansion of the product (82). Note that
the terms involving some $r_{i}$ have exponentially decaying property from
(53) when $\mathrm{Im}\,\xi_{1}\geq 0$ and $\mathrm{Im}\,\xi^{{}^{\prime}}=0$.
As a result,

$\displaystyle\int_{\Omega}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{E_{0}^{2\omega}}\leavevmode\nobreak\
\leavevmode\nobreak\ (E_{1}^{\omega}\cdot
E_{2}^{\omega})dV\text{\leavevmode\nobreak\ consists of terms involving some
$r_{i}^{k\omega}$}.$ (83)

For the L.H.S. of (83), one has

$\displaystyle\left|\int_{\Omega}\varepsilon_{0}^{-\frac{3}{2}}e^{-\frac{i}{h}x\cdot(\xi+\eta-\bar{\varsigma})}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{\tilde{A}_{\varsigma}}\leavevmode\nobreak\
\leavevmode\nobreak\ (\tilde{A}_{\xi}\cdot\tilde{A}_{\eta})dV\right|$
$\displaystyle=$
$\displaystyle\bigg{|}\bigg{(}(-1-k^{2}h^{2},\sqrt{1+k^{2}h^{2}}i,0)\cdot(-1-k^{2}h^{2},0,\sqrt{1+k^{2}h^{2}}i)+O(\epsilon)\bigg{)}\bigg{|}$
$\displaystyle\cdot\bigg{|}\int_{\Omega}h^{-4}\varepsilon_{0}^{-\frac{3}{2}}e^{-\frac{i}{h}x\cdot(\xi+\eta-\bar{\varsigma})}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{\tilde{A}_{\varsigma}}dV\bigg{|}$
$\displaystyle\geq$ $\displaystyle
Ch^{-4}\varepsilon_{0}^{-\frac{3}{2}}\bigg{|}\int_{\Omega}e^{-\frac{i}{h}x\cdot(\xi+\eta-\bar{\varsigma})}(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{\tilde{A}_{\varsigma}}dV\bigg{|},$

when $\epsilon$ and $h$ is sufficiently small. Denote by

$f(x;h)=(\chi^{(2),1}-\chi^{(2),2})\cdot\overline{\tilde{A}_{\varsigma}}.$
(84)

Hence, when $\mathrm{Im}\,\xi_{1}\geq 0$ and $\mathrm{Im}\,\eta_{1}\geq 0$,

$\displaystyle\bigg{|}\int_{\Omega}e^{-\frac{i}{h}x\cdot(\xi+\eta-\bar{\varsigma})}f(x;h)dV\bigg{|}$
$\displaystyle\leq
Ch^{4}\varepsilon_{0}^{\frac{3}{2}}\|\chi^{(2),1}-\chi^{(2),2}\|_{L^{\infty}(\Omega)}(1+h^{-1}|\xi|)^{\frac{3}{2}}(1+h^{-1}|\eta|)^{\frac{3}{2}}$
$\displaystyle(1+h^{-1}|\varsigma|)^{\frac{3}{2}}e^{-\frac{c}{h}\min\\{\mathrm{Im}\,\xi_{1},\mathrm{Im}\,\eta_{1}\\}}e^{\frac{1}{h}(|\mathrm{Im}\,\xi^{{}^{\prime}}|+|\mathrm{Im}\,\eta^{{}^{\prime}}|)}.$

In particular, let

$z=\xi+\eta-\bar{\varsigma},\leavevmode\nobreak\ \leavevmode\nobreak\
z_{0}=((2+\sqrt{2-2k^{2}h^{2}})i,0,0).$

When $|\xi-a\xi_{0}|<Ca\varepsilon$ and $|\eta-a\eta_{0}|<Ca\varepsilon$ with
$\varepsilon\leq\frac{1}{2C}$, $|z-az_{0}|<2Ca\varepsilon,$ and

$\displaystyle\left|\int_{\Omega}f(x;h)e^{-\frac{i}{h}x\cdot z}dx\right|\leq
Ch^{-\frac{1}{2}}\varepsilon_{0}^{\frac{3}{2}}\|\chi^{(2),1}-\chi^{(2),2}\|_{L^{\infty}(\Omega)}e^{-\frac{ca}{2h}}e^{\frac{4Ca\varepsilon}{h}}.$

Conversely, for any $z\in\mathbb{C}^{3}$, $|z-az_{0}|<2Ca\varepsilon$, $z$ can
be decomposed to

$z=\xi+\eta-\bar{\varsigma},\leavevmode\nobreak\ \leavevmode\nobreak\
\xi,\eta\in p^{-1}(kh),\leavevmode\nobreak\ \leavevmode\nobreak\
|\xi-a\zeta_{0}|<Ca\varepsilon,|\eta-a\eta_{0}|<Ca\varepsilon.$

Thus, following the previous procedures, one can show the exponentially
decaying property of the F.B.I. transform $Tf$ for all
$x\in\Omega,\leavevmode\nobreak\ |x_{1}|\leq\delta_{1},$ with $\delta_{1}$
sufficiently small. Hence, in view of (80) (84) and

$\lim_{h{\rightarrow}0}(2\pi
h)^{-\frac{n}{2}}Tf(x)=(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,-1,-1\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \text{in $L^{p}(\Omega)$},$

one has

$\displaystyle(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,-1,-1\right)=0,\leavevmode\nobreak\
\leavevmode\nobreak\ x\in\Omega,-\delta_{1}\leq x_{1}\leq 0.$ (85)

(ii) Now choose

$\xi_{0}=(i,-\sqrt{1+k^{2}h^{2}},0),\leavevmode\nobreak\
\eta_{0}=(i,0,-\sqrt{1+k^{2}h^{2}})\in p^{-1}(kh).$

Suppose $|\xi-\xi_{0}|<\epsilon,\xi\in p^{-1}(kh)$, and
$|\eta-\eta_{0}|<\epsilon,\eta\in p^{-1}(kh)$. Then $\mathrm{Im}\,(\xi_{1})>0$
and $\mathrm{Im}\,(\eta_{1})>0$.

Let
$E_{0}^{2\omega}=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\varsigma}\tilde{A}_{\varsigma}$
as in Remark 4.3, where

$\displaystyle\begin{split}\varsigma&=(\sqrt{2(1-k^{2}h^{2})}i,-\sqrt{1+k^{2}h^{2}},-\sqrt{1+k^{2}h^{2}})\in
p^{-1}(2kh),\\\
\tilde{A}_{\varsigma}&=\left(-\sqrt{2(1-k^{2}h^{2})}i,\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}},\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}}.\right)\end{split}$
(86)

Again $E_{0}^{2\omega}$ with some determined $H_{0}^{2\omega}$ are solutions
to the linearized Maxwell’s equations by noting
$\varsigma\cdot\tilde{A}_{\varsigma}=0$. $E_{1}^{\omega}$ and $E_{2}^{\omega}$
are defined as in (i) of the form (46), where $b=0$, $a_{1}=a_{2}=(1,0,0)$.
Let

$z=\xi+\eta-\bar{\varsigma},\leavevmode\nobreak\ \leavevmode\nobreak\
z_{0}=((2+\sqrt{2-2k^{2}h^{2}})i,0,0).$

Similarly, the amplitudes of $E_{1}^{\omega}$ and $E_{2}^{\omega}$ have the
estimates

$\displaystyle\tilde{A}_{\xi}$
$\displaystyle=h^{-2}\big{(}\xi_{0}\times(\xi_{0}\times
a_{1})+O(\epsilon)\big{)}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},-\sqrt{1+k^{2}h^{2}}i,0)+O(\epsilon)\bigg{)},$

and

$\displaystyle\tilde{A}_{\eta}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},0,-\sqrt{1+k^{2}h^{2}}i)+O(\epsilon)\bigg{)},$

where $O(\epsilon)$ is some terms independent of $x$, when
$|\xi-\xi_{0}|<\epsilon$ and $|\eta-\eta_{0}|<\epsilon,$ respectively.

After going through similar procedures as before, we arrive at an exponential
decaying property of $Tf$, where $f$ is defined in (84). Recall the expression
(86) of $\tilde{A}_{\varsigma}$. Then,

$\lim_{h{\rightarrow}0}(2\pi
h)^{-\frac{n}{2}}Tf(x)=(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,1,1\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \text{in $L^{p}(\Omega)$}.$

Therefore, one has

$\displaystyle(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,1,1\right)=0,\leavevmode\nobreak\
\leavevmode\nobreak\ x\in\Omega,-\delta_{2}\leq x_{1}\leq 0.$ (87)

(iii) We choose

$\xi_{0}=(i,\sqrt{1+k^{2}h^{2}},0),\leavevmode\nobreak\
\eta_{0}=(i,0,-\sqrt{1+k^{2}h^{2}})\in p^{-1}(kh).$

Suppose $|\xi-\xi_{0}|<\epsilon,\xi\in p^{-1}(kh)$, and
$|\eta-\eta_{0}|<\epsilon,\eta\in p^{-1}(kh)$. Then $\mathrm{Im}\,(\xi_{1})>0$
and $\mathrm{Im}\,(\eta_{1})>0$.

Let
$E_{0}^{2\omega}=\varepsilon_{0}^{-\frac{1}{2}}e^{-\frac{i}{h}x\cdot\varsigma}\tilde{A}_{\varsigma},$
where

$\displaystyle\begin{split}\varsigma&=(\sqrt{2(1-k^{2}h^{2})}i,\sqrt{1+k^{2}h^{2}},-\sqrt{1+k^{2}h^{2}})\in
p^{-1}(2kh),\\\
\tilde{A}_{\varsigma}&=\left(-\sqrt{2(1-k^{2}h^{2})}i,-\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}},\frac{(1-k^{2}h^{2})}{\sqrt{1+k^{2}h^{2}}}\right),\end{split}$
(88)

Again $E_{0}^{2\omega}$ together with some determined $H_{0}^{2\omega}$ are
solutions to the linearized Maxwell equations by noting
$\varsigma\cdot\tilde{A}_{\varsigma}=0$ and Remark 4.3.

Similarly, take $a_{1}=a_{2}=(1,0,0).$ When $|\xi-\xi_{0}|<\epsilon,$

$\displaystyle\tilde{A}_{\xi}$
$\displaystyle=h^{-2}\big{(}\xi_{0}\times(\xi_{0}\times
a_{1})+O(\epsilon)\big{)}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},\sqrt{1+k^{2}h^{2}}i,0)+O(\epsilon)\bigg{)},$

where $O(\epsilon)$ is some terms independent of $x$; and when
$|\eta-\eta_{0}|<\epsilon,$

$\displaystyle\tilde{A}_{\eta}$
$\displaystyle=h^{-2}\bigg{(}(-1-k^{2}h^{2},0,-\sqrt{1+k^{2}h^{2}}i)+O(\epsilon)\bigg{)},$

where $O(\epsilon)$ is some terms independent of $x$. Let

$z=\xi+\eta-\bar{\varsigma},\leavevmode\nobreak\ \leavevmode\nobreak\
z_{0}=((2+\sqrt{2-2k^{2}h^{2}})i,0,0).$

Following similar procedures as before and noting the expressions (84) and
(88) for $f$ and $\tilde{A}_{\varsigma}$, we arrive at

$\lim_{h{\rightarrow}0}(2\pi
h)^{-\frac{n}{2}}Tf(x)=(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,-1,1\right)\leavevmode\nobreak\
\leavevmode\nobreak\ \text{in $L^{p}(\Omega)$}.$

It follows that

$\displaystyle(\chi^{(2),1}-\chi^{(2),2})\cdot\left(\sqrt{2}i,-1,1\right)=0,\leavevmode\nobreak\
\leavevmode\nobreak\ x\in\Omega,-\delta_{3}\leq x_{1}\leq 0.$ (89)

In summary of (85) (87) (89), one has

$\displaystyle\chi^{(2),1}-\chi^{(2),2}=0,\leavevmode\nobreak\
\leavevmode\nobreak\ x\in\Omega,-\delta\leq x_{1}\leq 0,$

for some positive $\delta=\min\\{\delta_{1},\delta_{2},\delta_{3}\\}.$ This
completes the proof of Proposition 4.2.1.

####  4.2.2 From local to global

Proof of Theorem 1.4 Below we show the extension of local result (Proposition
4.2.1) to the global situation following the same process as in section 2.2.2
and 4.1.2.

Let $\bar{\chi}(y)=\chi^{(2),1}-\chi^{(2),2}$. Fix a point $x_{1}\in\Omega$
and let $\theta(t)$ be a parameterized curve in $\Omega$ joining
$x_{0}\in\Gamma$ to $x_{1}$ such that $\theta(0)=x_{0}$ and
$\theta^{\prime}(0)$ is the interior normal to $\partial\Omega$ at $x_{0}$.
Suppose $\Theta_{\epsilon}(t)$ is a small neighborhood of the curve
$\theta([0,t])$ in $\Omega$ and $I$ is the set of the time where $\bar{\chi}$
vanishes on $\Theta_{\epsilon}(t)$. The set $\Omega\cup
B(x_{0},\epsilon^{\prime})$ is smoothed out into an open set $\Omega_{2}$ with
smooth boundary such that

$\partial\Omega\setminus\Gamma\subset\partial\Omega\cap\partial\Omega_{1}\subset\partial\Omega\cap\partial\Omega_{2},$

and it suffices to show the set $I$ is open.

Following the notations in Section 4.1.2, define

$\displaystyle
H(x_{1},x_{2},x_{0})=\int_{\Omega_{1}}\bar{\chi}(y)\otimes\tilde{G}^{H}_{0}(x_{0},y)\cdot
G_{1}^{E}(x_{1},y)\otimes G_{2}^{E}(x_{2},y)dy,$

for $x_{1},x_{2},x_{0}\in\Omega_{2}\setminus\Omega_{1}$. Moreover,

$\displaystyle H(x_{1},x_{2},x_{0})$
$\displaystyle=\int_{\Omega_{1}}\bar{\chi}(y)\otimes\tilde{G}^{H}_{0}(x_{0},y)\cdot
G_{1}^{E}(x_{1},y)\otimes G_{2}^{E}(x_{2},y).$ (90)

When $x_{1},x_{2},x_{0}\in\Omega_{2}\setminus\bar{\Omega},$ from the
assumption we have $H(x_{1},x_{2},x_{3},x_{0})=0$, as all the entries of $H$
plays the role of an electric field which satisfies the homogeneous Maxwell’s
equations with vanishing tangential components on the boundary for
$x_{1},x_{2},x_{0}\in\Omega_{2}\setminus\bar{\Omega}_{1}.$

Note that all the entries of $H$ satisfy the Helmholtz equation
$(-\Delta_{y}-k^{2})u=0,$ for
$x_{1},x_{2},x_{0}\in\Omega_{2}\setminus\bar{\Omega}_{1}.$ Hence, from the
unique continuation principle,

$H(x_{1},x_{2},x_{0})=0\leavevmode\nobreak\ \leavevmode\nobreak\
\text{when\leavevmode\nobreak\ \leavevmode\nobreak\
}x_{1},x_{2},x_{0}\in\bar{\Omega}_{2}\setminus\Omega_{1};$

i.e., $\leavevmode\nobreak\ \leavevmode\nobreak\ \text{for\leavevmode\nobreak\
}x_{1},x_{2},x_{3},x_{0}\in\bar{\Omega}_{2}\setminus\Omega_{1},$

$\displaystyle\int_{\Omega_{1}}\bar{\chi}(y)\otimes\tilde{G}^{H}_{0}(x_{0},y)\cdot
G_{1}^{E}(x_{1},y)\otimes G_{2}^{E}(x_{2},y)dy=0.$ (91)

Let the operator at the left hand side of (91) act on

$\left(1\otimes a_{0}(x_{0}),a_{1}(x_{1})\otimes a_{2}(x_{2})\right)$

for $x_{0},x_{1},x_{2}\in\Omega_{2}$. Here all $a_{i}$ are of the forms
$a_{i}(x_{i})=(l_{i}(x_{i}),\overline{l_{i}(x_{i})})^{T},$ and all $l_{i}$ are
smooth vector functions on $C^{\infty}(\Omega_{2},\mathbb{C}^{3})$ with
supports contained in $\bar{\Omega}_{2}\setminus\Omega_{1}$. Denote by

$v_{i}(y)=\int_{\Omega_{2}}G^{E}_{i}(x_{i},y)a_{i}(x_{i})dx_{i},\leavevmode\nobreak\
\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\ \leavevmode\nobreak\
\tilde{v}_{0}(y)=\int_{\Omega_{2}}\tilde{G}^{H}_{0}(x_{0},y)a_{0}(x_{0})dx_{0},$

for $i=1,2$. Thus $v_{i}$ and $w_{i}$ are solutions to the inhomogeneous
Maxwell’s systems (64) and similarly, one has $\tilde{v}_{0}=\bar{v}_{0}$ and
$\tilde{w}_{0}=\bar{w}_{0}$ in $\Omega_{2},$ where $(v_{0},w_{0})$ satisfies
similar Maxwell’s equations (64) with frequency replaced by $2\omega$.
Therefore, (91) becomes

$\displaystyle 0$
$\displaystyle=\int_{\Omega_{1}}\bar{\chi}\cdot\bar{v}_{0}\leavevmode\nobreak\
\leavevmode\nobreak\ v_{1}\cdot v_{2}dy,$

where $(v_{i},w_{i})$ satisfy the system $(\ref{value1})$. Since all $a_{i}\in
C^{\infty}(\Omega_{2},\mathbb{C}^{3})$ with supports contained in
$\bar{\Omega}_{2}\setminus\Omega_{1}$, the solutions $(v_{i},w_{i})$ form the
set $\mathcal{R}$.

Lemma 4.1 states that $\mathcal{R}$ is dense in the space $\mathcal{S}$ with
respect to the $L^{2}(\Omega_{1},\mathbb{C}^{3})^{2}$ topology. In the same
way as in Section 4.1.2, one can show that the equality (65) extends to

$\displaystyle
0=\int_{\Omega_{1}}\bar{\chi}\cdot\bar{V}_{0}\leavevmode\nobreak\
\leavevmode\nobreak\ V_{1}\cdot V_{2}dV$

holds for every $V_{i},W_{i}\in C^{\infty}(\bar{\Omega}_{1},\mathbb{C}^{3})$
which satisfy inhomogeneous Maxwell systems and of which tangential components
vanish on $\partial\Omega_{1}\cap\partial\Omega_{2}$. Hence by virtue of
Proposition 2.1 the cancellation states that $\bar{\chi}$ vanishes on a
neighborhood of
$\partial\Omega_{1}\setminus(\partial\Omega_{1}\cap\partial\Omega_{2})$. This
implies that $\bar{\chi}$ vanishes on a bigger neighborhood
$\Theta_{\epsilon}(t^{\prime})$ $t^{\prime}>t$ of the curve. Hence $I$ is
open.

##  Appendix A Appendix

###  A.1 Well-posedness of nonlinear Helmholtz Schrödinger equation

Let $\Omega\subseteq\mathbb{R}^{n}$ be a bounded domain with smooth boundary.
Consider the boundary value problem for the nonlinear Helmholtz equation with
small boundary value

$\displaystyle\left\\{\begin{array}[]{rl}-\Delta
u-k^{2}u+q(x,u)=0&\text{in\leavevmode\nobreak\ }\Omega,\\\
u=f&\text{on\leavevmode\nobreak\ }\partial\Omega.\end{array}\right.$ (94)

Here $q(x,u)$ is smooth in $u$ and $\partial_{z}^{k}q(x,u_{0})\in
L^{\infty}(\Omega\times\mathbb{R})$ for $k\in\mathbb{N}$. Moreover,
$q(x,0)=\partial_{u}q(x,0)=0.$

###### Theorem A.1

Suppose that $p>\frac{n}{2}$. There exists a discrete subset
$\Sigma\subseteq\mathbb{R}$ such that for every $k\not\in\Sigma$, the boundary
value problem (94) has a unique solution $u(x)$ satisfying

$\|u\|_{W^{2,p}(\Omega)}\leq c\|f\|_{W^{2-\frac{1}{p},p}(\partial\Omega)},$

whenever $\|f\|_{W^{2-\frac{1}{p},p}(\partial\Omega)}<\delta$ is sufficiently
small. Here, $c>0$ is a positive constant depending only on $p,k,\Omega$ and
the coefficients of $q.$

Proof.  Given $F\in L^{p}(\Omega)$, from Fredholm theory, there exists a
discrete subset $\Sigma\subseteq\mathbb{R}$ such that for any
$k\not\in\Sigma$, the Dirichlet boundary problem

$\displaystyle\left\\{\begin{array}[]{rl}-\Delta
u-k^{2}u=F&\text{in\leavevmode\nobreak\ }\Omega,\\\
u=0&\text{on\leavevmode\nobreak\ }\partial\Omega,\end{array}\right.$

has a unique solution $u\in W_{0}^{2,p}(\Omega)$ satisfying
$\|u\|_{W_{0}^{2,p}(\Omega)}\leq c\|F\|_{L^{p}(\Omega)},$ the constant $c$
depending only on $p,k$ and $\Omega$. In other words, for any $k$ outside
$\Sigma$, the operator
$\mathcal{L}=-\Delta-k^{2}:W^{2,p}_{0}(\Omega){\rightarrow}L^{p}(\Omega)$ has
a continuous inverse.

As a result, one can show that, for any $k$ outside $\Sigma$ and $f\in
W^{2-\frac{1}{p},p}(\partial\Omega)$, the linearized equation

$\displaystyle\left\\{\begin{array}[]{rl}-\Delta
u-k^{2}u=0&\text{in\leavevmode\nobreak\ }\Omega,\\\
u=f&\text{on\leavevmode\nobreak\ }\partial\Omega,\end{array}\right.$

of (94) has a solution $u_{0}\in W^{2,p}(\Omega)$ satisfying
$\|u_{0}\|_{W^{2,p}(\Omega)}\leq
c\|f\|_{W^{2-\frac{1}{p},p}(\partial\Omega)}\leq c\delta$. Hence, $u=u_{0}+v$
is a solution to the boundary value problem (94), if $v$ satisfies

$\displaystyle\left\\{\begin{array}[]{cl}-\Delta
v-k^{2}v=-q(x,u_{0}+v)&\text{in\leavevmode\nobreak\ }\Omega,\\\
v=0&\text{on\leavevmode\nobreak\ }\partial\Omega.\end{array}\right.$

Denote by $\mathcal{F}(v)=-q(x,u_{0}+v)$. Since $q(x,0)=\partial_{u}q(x,0)=0,$
then $q(x,u)=q_{r}(x,u)u^{2}$, where
$q_{r}(x,u)=\int_{0}^{1}\partial_{u}^{2}q(x,tu)$ $(1-t)dt$ is bounded.
Furthermore, $q_{r}(x,u)$ is also Lipschitz in $u$ noting
$\partial^{k}_{u}q(x,u)\in L^{\infty}(\Omega\times\mathbb{C})$ for
$k\in\mathbb{N}$. Combining that $W^{2,p}(\Omega)\hookrightarrow
C(\bar{\Omega})$ by Sobolev embedding theorem for $p>\frac{n}{2}$, for $v_{1}$
and $v_{2}$ in $X_{\delta}$ defined as

$X_{\varepsilon}=\\{v\in W^{2,p}(\Omega)|\leavevmode\nobreak\
\|v\|_{W^{2,p}(\Omega)}\leq c\varepsilon\\},$

one has $\|v_{1}\|_{L^{\infty}(\Omega)},\|v_{2}\|_{L^{\infty}(\Omega)}\leq
c\varepsilon,$ and $\|u\|_{L^{\infty}(\Omega)}\leq c\delta,$. Thus,

$\displaystyle\|\mathcal{L}^{-1}\circ\mathcal{F}(v_{1})-\mathcal{L}^{-1}\circ\mathcal{F}(v_{2})\|_{W^{2,p}(\Omega)}$
$\displaystyle\leq$
$\displaystyle\|\mathcal{F}(v_{1})-\mathcal{F}(v_{2})\|_{L^{p}(\Omega)}$
$\displaystyle=$
$\displaystyle\|(q_{r}(x,u_{0}+v_{1})-q_{r}(x,u_{0}+v_{2}))(u_{0}+v_{2})^{2}\|_{L^{p}(\Omega)}$
$\displaystyle+\|q_{r}(x,u_{0}+v_{1})((u_{0}+v_{2})^{2}-(u_{0}+v_{1})^{2})\|_{L^{p}(\Omega)}$
$\displaystyle\leq$ $\displaystyle
c\|v_{1}-v_{2}\|_{L^{p}(\Omega)}\|u_{0}+v_{2}\|^{2}_{L^{\infty}(\Omega)}+c\|2u_{0}+v_{1}+v_{2}\|_{L^{\infty}(\Omega)}\|v_{1}-v_{2}\|_{L^{p}(\Omega)}$
$\displaystyle\leq$ $\displaystyle
c(\delta^{2}+\varepsilon^{2}+\varepsilon\delta+\varepsilon+\delta)\|v_{1}-v_{2}\|_{L^{p}(\Omega)}.$

The constant is independent of $\varepsilon$ and $\delta$. This implies that
$\mathcal{L}^{-1}\circ\mathcal{F}$ is a contraction on $X_{\delta}$ provided
that $\varepsilon$ and $\delta$ are sufficiently small. It follows that there
exists some $v\in X_{\delta}$ such that
$v=\mathcal{L}^{-1}\circ\mathcal{F}(v)$. Furthermore,

$\displaystyle\|v\|_{W^{2,p}(\Omega)}$ $\displaystyle\leq
c\|\mathcal{F}(v)\|_{L^{p}(\Omega)}\leq
c(\|u_{0}\|^{2}_{L^{\infty}(\Omega)}+\|v\|^{2}_{L^{\infty}(\Omega)})$
$\displaystyle\leq
c(\delta\|f\|_{W^{2-\frac{1}{p},p}(\partial\Omega)}+\varepsilon\|v\|_{W^{2,p}(\Omega)}).$

Thus one has
$\|v\|_{W^{2,p}(\Omega)}\leq\|f\|_{W^{2-\frac{1}{p},p}(\partial\Omega)}$
provided $\delta$ and $\varepsilon$ sufficiently small. In summary,
$u=u_{0}+v\in W^{2,p}(\Omega)$ is a solution to the nonlinear Helmholtz
equation with small boundary value satisfying the required estimate. $\Box$

###  A.2 Runge approximation

We need the following Runge approximation property in the previous sections.

Let $M\subset\mathbb{R}^{n},n\geq 2$ be a bounded open set with $C^{\infty}$
boundary. Consider the dual space of $H^{1}(M)$, given by

$\tilde{H}^{-1}(M)=\left\\{f\in H^{-1}(\mathbb{R}^{n})|\leavevmode\nobreak\
\mathrm{supp}\,(f)\subset\bar{M}\right\\}.$

For $f\in\tilde{H}^{-1}(M)$ and $g\in H^{1}(M)$, let
$\tilde{g}=\mathrm{Ext}(g)\in H^{1}(\mathbb{R}^{n})$ be an extension of $g$.

Define the following duality pairing

$(f,g)_{\tilde{H}^{-1}(M),H^{1}(M)}=(f,\tilde{g})_{\tilde{H}^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}=\int_{\mathbb{R}^{n}}f\overline{\tilde{g}}dx.$

###### Proposition A.1

Let $U\subset M$ be a domain with $C^{\infty}$ boundary such that
$M\setminus\bar{U}$ is connected. Assume that $\partial U\cap\partial
M=\bar{V}$, where $V\subset\partial M$ is open with $C^{\infty}$ boundary. Let
$\Gamma$ be a nonempty open subset of $\partial M$, and
$\bar{\Gamma}\cap\bar{V}=\emptyset.$ Then the set

$\mathcal{R}=\\{u|_{U}:\leavevmode\nobreak\ -\Delta
u-k^{2}u=0\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
M,u=g\leavevmode\nobreak\ \text{on}\leavevmode\nobreak\ \partial M;g\in
C_{c}^{\infty}(\Gamma)\\}$

is dense in the space $\mathcal{S}=\\{u\in H^{1}(U):\leavevmode\nobreak\
-\Delta u-k^{2}u=0\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
U,u=0\leavevmode\nobreak\ \text{on}\leavevmode\nobreak\ \partial U\cap\partial
M\\}$ with respect to the $H^{1}(U)$ norm.

Proof. Let $f\in\tilde{H}^{-1}(U)$ be orthogonal to $\mathcal{R}$, i.e., one
has $(f,u|_{U})_{\tilde{H}^{-1}(U),H^{1}(U)}=0$ for any $u|_{U}$ satisfying

$-\Delta u-k^{2}u=0\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
M,u=g\leavevmode\nobreak\ \text{on}\leavevmode\nobreak\ \partial M;g\in
C_{c}^{\infty}(\Gamma).$

From Hahn-Banach theorem it suffices to show that
$(f,v)_{\tilde{H}^{-1}(U),H^{1}(U)}=0$ for any $v\in\mathcal{S}.$

Construct $M_{0}$ with smooth boundary such that $M\subset M_{0}$ and
$\partial M\setminus\bar{\Gamma}\subset\partial M\cap\partial M_{0}\ $. Thus,
$\partial U\cap\partial M\subset\partial U\cap\partial M_{0}$. For
$f\in\tilde{H}^{-1}(U)$, there exists a sequence $\\{f_{j}\\}\subset
C_{0}^{\infty}(U)$ such that $f_{j}{\rightarrow}f$ in $\tilde{H}^{-1}(U).$

Let $w\in H_{0}^{1}(M_{0})$ and $w_{j}\in C^{\infty}(\bar{M_{0}})\cap
H_{0}^{1}(M_{0})$ be the unique solutions to the following Helmholtz equations

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
w-k^{2}w=f\nobreak\leavevmode&\text{in}\leavevmode\nobreak\ M_{0},\\\
w=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial
M_{0},\end{array}\right.$

and

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
w_{j}-k^{2}w_{j}=f_{j}\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M_{0},\\\ w_{j}=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial
M_{0}.\end{array}\right.$

Since $f_{j}{\rightarrow}f$ in $\tilde{H}^{-1}(U),$ the regularity of
solutions to the elliptic equation yields $w_{j}{\rightarrow}w$ in $H^{1}(M).$

Let $\varphi\in H^{1}(M)$ and $\varphi_{j}\in C^{\infty}(\bar{M})\cap
H^{1}(M)$ be the unique solutions to the following Helmholtz equation

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\varphi-k^{2}\varphi=0\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M,\\\ \varphi=w\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial
M.\end{array}\right.$

and

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\varphi_{j}-k^{2}\varphi_{j}=0\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M,\\\ \varphi_{j}=w_{j}\nobreak\leavevmode&\text{on}\leavevmode\nobreak\
\partial M.\end{array}\right.$

Note that $w_{j}{\rightarrow}w$ in $H^{1}(M),$ then
$\varphi_{j}{\rightarrow}\varphi$ in $H^{1}(M).$

Let $\omega_{j}|_{M}=w_{j}-\varphi_{j}$ and $\omega|_{M}=w-\varphi$. Then
$\omega$ and $\omega_{j}$ satisfy

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\omega-k^{2}\omega=f\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M,\\\ w=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial
M.\end{array}\right.$ (97)

and

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\omega_{j}-k^{2}\omega_{j}=f_{j}\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M,\\\ w_{j}=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial
M,\end{array}\right.$

respectively. Thus, for $u\in\mathcal{R},$ one has

$\displaystyle 0$
$\displaystyle=(f,u|_{U})_{\tilde{H}^{-1}(U),H^{1}(U)}=\lim_{j{\rightarrow}\infty}(f_{j},u|_{U})_{\tilde{H}^{-1}(U),H^{1}(U)}=\lim_{j{\rightarrow}\infty}\int_{M}-\Delta
w_{j}u-k^{2}w_{j}udV$
$\displaystyle=\lim_{j{\rightarrow}\infty}\int_{M}(-\Delta
u-k^{2}u)w_{j}dV+\lim_{j{\rightarrow}\infty}\int_{\partial
M}\partial_{\nu}\omega_{j}u-\omega_{j}\partial_{\nu}udS$
$\displaystyle=\lim_{j{\rightarrow}\infty}\int_{\Gamma}\partial_{\nu}\omega_{j}gdS=\int_{\Gamma}\partial_{\nu}\omega
gdS$

Hence, $\partial_{\nu}\omega|_{\Gamma}=0.$ In summary, $\omega$ satisfies

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta\omega-k^{2}\omega=0\nobreak\leavevmode&\text{in}\leavevmode\nobreak\
M\setminus\bar{U},\\\
\omega=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\ \partial M\\\
\partial_{\nu}\omega=0\nobreak\leavevmode&\text{on}\leavevmode\nobreak\
\Gamma\end{array}\right.$

The unique continuation principle implies that

$\omega|_{M\setminus\bar{U}}=0.$

Recall that $\omega\in H^{1}_{0}(M)$,then $\omega|_{\partial U\cap\partial
M}=0$. Hence, $\omega|_{\partial U}=0.$

Thus $\omega$ may be identified with an element in $H^{1}_{0}(U)$. There
exists $\psi_{j}\in C_{0}^{\infty}(U)$ such that $\psi_{j}{\rightarrow}\omega$
in $H^{1}(U)$. Hence
$-\Delta\psi_{j}-k^{2}\psi_{j}{\rightarrow}-\Delta\omega-k^{2}\omega$ in
$H^{-1}(\mathbb{R}^{n}).$

For any $v\in\mathcal{S},$ let $\tilde{v}$ be the extension of $v$ to
$\mathbb{R}^{n}.$

$\displaystyle(-\Delta\omega-k^{2}\omega,\tilde{v})_{H^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}$
$\displaystyle=$
$\displaystyle\lim_{j{\rightarrow}\infty}(-\Delta\psi_{j}-k^{2}\psi_{j},\tilde{v})_{H^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}=\lim_{j{\rightarrow}\infty}\int_{M}(-\Delta\psi_{j}-k^{2}\psi_{j})vdV$
$\displaystyle=$ $\displaystyle\lim_{j{\rightarrow}\infty}\int_{U}(-\Delta
v-k^{2}v)\psi_{j}dV+\lim_{j{\rightarrow}\infty}\int_{\partial
U}\partial_{\nu}\psi_{j}v-\psi_{j}\partial_{\nu}vdS=0.$

Consider $\hat{f}=f-(-\Delta\omega-k^{2}\omega).$ $\hat{f}\in
H^{-1}(\mathbb{R}^{n})$ and $H^{-\frac{1}{2}}(\partial U)$ with
$\mathrm{supp}\,\hat{f}\subseteq\partial U$. From [*],

$\hat{f}=\tilde{f}\otimes\delta_{\partial U},\leavevmode\nobreak\
\leavevmode\nobreak\ \tilde{f}\in H^{-\frac{1}{2}}(\partial U).$

Following (97) and $f\in\tilde{H}^{-1}(U)$ one has
$\mathrm{supp}\,\hat{f}\subseteq\bar{V}=\partial U\cap\partial M.$ Thus
$\mathrm{supp}\,\tilde{f}\subseteq\bar{V}.$ Therefore, there exists
$\tilde{f}_{j}\in C_{0}^{\infty}(V)$ such that
$\tilde{f}_{j}{\rightarrow}\tilde{f}$ in $H^{-\frac{1}{2}}(\partial U).$ Then

$\displaystyle(\hat{f},v|_{U})_{\tilde{H}^{-1}(U),H^{1}(U)}$
$\displaystyle=(\hat{f},\tilde{v})_{H^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}=(\tilde{f},v|_{\partial
U})_{H^{-\frac{1}{2}}(\partial U),H^{\frac{1}{2}}(\partial U)}$
$\displaystyle=$
$\displaystyle\lim_{j{\rightarrow}\infty}(\tilde{f}_{j},v|_{\partial
U})_{H^{-\frac{1}{2}}(\partial U),H^{\frac{1}{2}}(\partial
U)}=\lim_{j{\rightarrow}\infty}\int_{\partial U}\tilde{f}_{j}vdS=0,$

since $\mathrm{supp}\,\tilde{f}_{j}\subseteq V$ and $v|_{\bar{V}}=0.$
Therefore,

$\displaystyle(f,v|_{U})_{\tilde{H}^{-1}(U),H^{1}(U)}$
$\displaystyle=(-\Delta\omega-k^{2}\omega,\tilde{v})_{H^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}+(\hat{f},\tilde{v})_{H^{-1}(\mathbb{R}^{n}),H^{1}(\mathbb{R}^{n})}=0.$

This completes the proof of Proposition A.1. $\Box$

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