Uniqueness Results for Some Inverse Electromagnetic Scattering Problems with Phaseless Far-Field Data

Abstract

Consider three electromagnetic scattering models, namely, electromagnetic scattering by an elastic body, by a chiral medium, and by a cylinder at oblique incidence. We are concerned with the corresponding inverse problems of determining the locations and shapes of the scatterers from phaseless far-field patterns. There are certain essential differences from the usual inverse electromagnetic scattering problems, and some fundamental conclusions need to be proved. First, we show that the phaseless far-field data are invariant under the translation of the scatterers and prove the reciprocity relations of the scattering data. Then, we justify the unique determination of the scatterers by utilizing the reference ball approach and the superpositions of a fixed point source and plane waves as the incident fields. The proofs are based on the reciprocity relations, Green’s formulas, and the analyses of the wave fields in the reference ball.

1. Introduction

Inverse electromagnetic scattering problems arise from many scientific and industrial applications such as radar, nondestructive testing, and medical imaging. They are extensively studied and considerable results are obtained. The uniqueness of inverse problems has a central importance in theoretical studies and numerical algorithms. It is concerned with the unique determination of the unknown objects from the knowledge of the measurable scattering data. In the past few decades, great progress has been achieved for inverse electromagnetic scattering problems from phased measurements (see, for instance, [1,2,3]). However, in many practical situations, only the modulus or intensity of scattering data can be measured. This requires us to study the inverse scattering problems with phaseless data. In this paper, we consider several inverse electromagnetic scattering problems with different physical or structural characteristics and establish some uniqueness results from phaseless far-field data.

Compared with phased inverse scattering problems, inverse scattering problems from phaseless far-field data are more difficult due to the translation invariance. The location of the scatterer cannot be uniquely determined by the phaseless far-field data. Some numerical algorithms have been proposed for shape reconstructions, for example, the hybrid method [4], the nonlinear integral equation method [5], and the Newton method [6]. More recent progress in this direction can be found in [7,8,9,10,11,12] and the references therein. By using superpositions of two plane waves as the incident fields with an interval of frequencies that breaks the translation invariance property of the phaseless far-field pattern, a recursive Newton iteration algorithm in the frequencies is developed to reconstruct both the location and shape of the obstacle from the multi-frequency phaseless far-field data [13]. There have also been extensive studies on the uniqueness issue. It is proved in [14] that the radius of a sound-soft ball centered at the origin can be uniquely determined by a single phaseless far-field datum. For a general smooth, convex, sound-soft obstacle, its shape can be recovered from the modulus of the far-field pattern associated with one incident plane wave [15]. Recently, considerable efforts have been made on the use of uniqueness results to determine general scatterers with the modulus of the far-field pattern. The authors in [16] prove that, under certain a priori assumptions, the obstacle and the index of refraction of an inhomogeneous medium can be uniquely determined from the phaseless far-field patterns generated by infinite sets of superpositions of two plane waves with different directions at a fixed frequency. Later, the a priori assumptions on the obstacle and the index of refraction of the inhomogeneous medium are removed by adding a reference ball to the scattering system [17]. In addition, the reference ball technique is used in [18] to remove the translation invariance, where the uniqueness results are established by using the superpositions of different types of incident waves together with an extra reference ball. The strategy of adding a reference object to the scattering system for phaseless inverse problems has been applied to inverse elastic scattering [19], inverse acoustic–elastic interaction problems [20], inverse cavity scattering [21], and inverse electromagnetic scattering [22]. Based on the techniques mentioned above, some numerical algorithms are proposed to recover both the shape and location of the scattering obstacles from phaseless data (see, for instance, [13,23,24,25]).

We pay special attention to the paper [22], where by adding a known reference ball into the electromagnetic scattering system and using the superpositions of two incident electromagnetic plane waves with different directions and polarizations, the authors prove that the obstacle together with its physical property can be uniquely determined from the modulus of the tangential component of electric far-field measurements at a fixed frequency. The aim of this paper is to extend the uniqueness result to the inverse electromagnetic scattering problems under the following situations:

(1)

The interaction problem between the electromagnetic field and an elastic body;

(2)

Electromagnetic scattering by a chiral medium;

(3)

Electromagnetic scattering by an impedance cylinder at oblique incidence.

The existing results for inverse electromagnetic scattering problems using phaseless data cannot cover our results on the problems under consideration, since the mathematical models are essentially different.

Remark 1.

The reason why we consider these three types of problems is that, in addition to their important applications and the attention of mathematicians, they all originate from the electromagnetic wave scattering model but for scatterers with different physical properties. Moreover, we use the same method to study the related inverse problems.

We combine the reference ball technique with the superpositions of incident plane waves and a fixed point source to uniquely determine the scatterers from the phaseless far-field data. As we show in the subsequent sections, the derivations are not trivial. First, we need to verify whether the far-field patterns have the property of translation invariance. Second, it is necessary to prove the reciprocity principles for far-field patterns under the different situations presented above. Third, the scattering models considered here are different from the classical electromagnetic scattering problems, and some sophisticated modifications are required in the justifications. It is worth noting that electromagnetic scattering from a cylinder at oblique incidence leads to a pair of two-dimensional Helmholtz equations with coupled boundary conditions, and we show that the phaseless far-field of only the electric field is enough to recover the obstacle.

The rest of the paper is organized as follows: In Section 2, we introduce the mathematical model of the interaction problem between the electromagnetic field and an elastic body and formulate the corresponding inverse problem. Then, we study the property of translation invariance and the reciprocity principle for the electromagnetic wave. Consequently, the unique determination of the scatterer from the phaseless far-field data is proved. Since the discussion of the inverse problem for the electromagnetic scattering by a chiral medium is quite similar to the analysis of the first one, we give a brief version in Section 3. Section 4 is devoted to an inverse problem for the electromagnetic scattering problem from a cylinder at oblique incidence.

2. The Interaction Problem between the Electromagnetic Field and an Elastic Body

In this section, we first describe the mathematical model of the interaction problem between the electromagnetic field and an elastic body and then formulate an inverse scattering problem. We derive the property of translation invariance and the reciprocity principle for the electromagnetic wave. Consequently, the unique determination of the scatterer from the phaseless far-field data is proved.

2.1. The Mathematical Formulation

Let $\Omega$ be a bounded domain in ${\mathbb{R}}^3$ with a smooth boundary $\partial \Omega$. Assume that $\Omega$ is occupied by a homogeneous isotropic elastic body with the Lamé constants ${\mu }_i,{\lambda }_i$ and the mass density ${\rho }_i>0$, while the medium in the exterior domain ${\mathbb{R}}^3\setminus \overline{\Omega }$ is homogeneous and isotropic with the positive constants magnetic permeability $\mu$ and electric permittivity $ϵ$. The unit outward normal to $\partial \Omega$ is denoted by $\mathit{n}$. Furthermore, we assume that the system consisting of the electromagnetic field and the elastic body interacts only through the interface $\partial \Omega$.

Consider that the elastic body is illuminated by the plane electromagnetic wave with an incident direction $d\in {\mathbb{O}}^2$ (the unit sphere in ${\mathbb{R}}^3$) and polarization vector $p\in {\mathbb{R}}^3$, namely,

$$\left\{\begin{array}{cc}{\mathit{E}}^i(x,d,p):={\displaystyle \frac{i}{k}}{curl}_x{curl}_{x}p{e}^{ikx·d}=ik(d\times p)\times d{e}^{ikx·d},\hfill & \\ {\mathit{H}}^i(x,d,p):={curl}_{x}p{e}^{ikx·d}=ik(d\times p)e^{ikx·d},\hfill \end{array}\right.$$

(1)

where $d\perp p$ and $k=\omega \sqrt{\epsilon \mu }$ is the wave number with the frequency $\omega$. The total electromagnetic field $(\mathit{E},\mathit{H})$ in ${\mathbb{R}}^3\setminus \overline{\Omega }$ is the sum of the scattered wave $({\mathit{E}}^s,{\mathit{H}}^s)$ and the incident wave $({\mathit{E}}^i,{\mathit{H}}^i)$, i.e., $\mathit{E}={\mathit{E}}^s+{\mathit{E}}^i$, $\mathit{H}={\mathit{H}}^s+{\mathit{H}}^i$, which satisfies the Maxwell equations

$$curl\mathit{E}-ik\mathit{H}=0,curl\mathit{H}+ik\mathit{E}=0\mathrm{in}{\mathbb{R}}^3\setminus \overline{\Omega }.$$

(2)

The elastic displacement $\mathit{u}$ in $\Omega$ satisfies the Navier equation

$${\mu }_i\Delta \mathit{u}+({\lambda }_i+{\mu }_i)\nabla (\nabla ·\mathit{u})+{\rho }_i{\omega }^2\mathit{u}=0\mathrm{in}\Omega .$$

(3)

Voigt’s model leads to the following transmission conditions on the interface [26]:

$$\frac{i}{k}\mathit{n}\times \mathit{H}=T\mathit{u},\mathit{n}\times \mathit{E}=\mathit{n}\times \mathit{u}\mathrm{on}\partial \Omega ,$$

(4)

where T denotes the stress operator given by

$$T\mathit{u}:=2{\mu }_i\mathit{n}·\nabla \mathit{u}+{\lambda }_i\mathit{n}\nabla ·\mathit{u}+{\mu }_i\mathit{n}\times (\nabla \times \mathit{u}).$$

In addition, the scattered field satisfies the Silver–Müller radiation condition

$$\frac{x}{\left|x\right|}\times {\mathit{E}}^s+{\mathit{H}}^s=o\left(\frac{1}{\left|x\right|}\right),\left|x\right|\to \infty$$

(5)

uniformly in all directions $x/\left|x\right|$.

The scattered field $({\mathit{E}}^s,{\mathit{H}}^s)$ has the following asymptotic form:

$$\begin{array}{ccc}\hfill {\mathit{E}}^s(x,d,p)& =& \frac{e^{ik\left|x\right|}}{\left|x\right|}{\mathit{E}}^{\infty }(\widehat{x},d,p)+o\left(\frac{1}{{\left|x\right|}^2}\right)\mathrm{as}\left|x\right|\to \infty ,\hfill \\ \hfill {\mathit{H}}^s(x,d,p)& =& \frac{e^{ik\left|x\right|}}{\left|x\right|}{\mathit{H}}^{\infty }(\widehat{x},d,p)+o\left(\frac{1}{{\left|x\right|}^2}\right)\mathrm{as}\left|x\right|\to \infty ,\hfill \end{array}$$

which uniformly holds in all directions with $\widehat{x}=x/\left|x\right|\in {\mathbb{O}}^2$. Here, ${\mathit{H}}^{\infty }(\widehat{x},d,p)$ and ${\mathit{E}}^{\infty }(\widehat{x},d,p)$ are the far-field patterns of the scattered magnetic and electric fields, respectively. Moreover, it holds that

$${\mathit{H}}^{\infty }(\widehat{x},d,p)=\widehat{x}\times {\mathit{E}}^{\infty }(\widehat{x},d,p),\widehat{x}·{\mathit{H}}^{\infty }(\widehat{x},d,p)=0,\widehat{x}·{\mathit{E}}^{\infty }(\widehat{x},d,p)=0.$$

(6)

The well-posedness of the interaction problem (2)–(5) can be proved by variational methods based on Helmholtz type decomposition techniques of the electromagnetic wave function (see [27,28,29] for details). Moreover, the unique determination of the elastic scatterer from the electromagnetic far-field measurements is shown in [29].

2.2. An Inverse Scattering Problem

Now, we consider an inverse scattering problem, that is, we identify the shape and location of the elastic body from the moduli of the tangential components of electric far-field measurements at a fixed frequency. For this purpose, we add a reference ball $B_1$ as an extra artificial elastic body to the scattering system, such that $B_1\subset {\mathbb{R}}^3\setminus \overline{\Omega }$. The medium in $B_1$ is homogeneous and isotropic with a mass density of ${\rho }_1>0$ and the Lamé constants ${\mu }_1$ and ${\lambda }_1$. Suppose that the wave number k is not an interior transmission eigenvalue in $B_1$, that is, the following interior transmission problem only has a trivial solution:

$$\left\{\begin{array}{cc}curl\widehat{\mathit{E}}-ik\widehat{\mathit{H}}=0,curl\widehat{\mathit{H}}+ik\widehat{\mathit{E}}=0\hfill & \mathrm{in}{B}_1,\hfill \\ {\mu }_1\Delta \widehat{\mathit{u}}+({\lambda }_1+{\mu }_1)\nabla (\nabla ·\widehat{\mathit{u}})+{\rho }_1{\omega }^2\widehat{\mathit{u}}=0\hfill & \mathrm{in}{B}_1,\hfill \\ {\displaystyle \frac{i}{k}}\mathit{n}\times \widehat{\mathit{H}}=T\widehat{\mathit{u}},\mathit{n}\times \widehat{\mathit{E}}=\mathit{n}\times \widehat{\mathit{u}}\hfill & \mathrm{on}\partial B_1.\hfill \end{array}\right.$$

(7)

For the incident electric dipole located at the point $z_0\in {\mathbb{R}}^3\setminus (\overline{\Omega }\cup {\overline{B}}_1)$,

$$\left\{\begin{array}{cc}{\mathit{E}}^i(x,z_0):={\displaystyle \frac{i}{k}}curlcurl{p}_0\Phi (x,z_0),\hfill & x\ne z_0,\hfill \\ {\mathit{H}}^i(x,z_0):=curl{p}_0\Phi (x,z_0),\hfill & x\ne z_0,\hfill \end{array}\right.$$

(8)

where $p_0$ is a constant vector and $\Phi$ is the radiating fundamental solution to the Helmholtz equation given by

$$\Phi (x,z):=\frac{1}{4\pi }\frac{e^{ik|x-z|}}{|x-z|},x,z\in {\mathbb{R}}^3,x\ne z.$$

We denote by $({\mathit{E}}_{\Omega \cup B_1}^s(x,z_0),{\mathit{H}}_{\Omega \cup B_1}^s(x,z_0))$ and $({\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},z_0),{\mathit{H}}_{\Omega \cup B_1}^{\infty }(\widehat{x},z_0))$ the scattered wave and its far-field pattern for the incident electric dipole (8) generated by $\Omega \cup B_1$. The scattered wave and its far-field pattern for the incident plane electromagnetic wave (1) are denoted by $({\mathit{E}}_{\Omega \cup B_1}^s(x,d,p),$ ${\mathit{H}}_{\Omega \cup B_1}^s(x,d,p))$ and $({\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},d,p),$ ${\mathit{H}}_{\Omega \cup B_1}^{\infty }(\widehat{x},d,p))$, respectively.

We take the modulus of the tangential component of the electric far-field pattern on ${\mathbb{O}}^2$ as the measurement data. We introduce the following spherical coordinate transformation:

$${\widehat{z}}_1=sin\phi cos\psi ,{\widehat{z}}_2=sin\phi sin\psi ,{\widehat{z}}_3=cos\phi$$

with $\widehat{z}=({\widehat{z}}_1,{\widehat{z}}_2,{\widehat{z}}_3)\in {\mathbb{O}}^2$ and $(\phi ,\psi )\in [0,\pi ]\times [0,2\pi )$. It gives a one-to-one correspondence between $\widehat{z}\in {\mathbb{O}}^2\setminus \{N,S\}$ and $(\phi ,\psi )$, where $S=(0,0,-1)$ and $N=(0,0,1)$ are the south and north poles of ${\mathbb{O}}^2$, respectively. We introduce

$$e_{\psi }\left(\widehat{z}\right):=(-sin\psi ,cos\psi ,0),e_{\phi }\left(\widehat{z}\right):=(cos\phi cos\psi ,cos\phi sin\psi ,-sin\phi ).$$

Then, $e_{\psi }\left(\widehat{z}\right)$ and $e_{\phi }\left(\widehat{z}\right)$ are two orthogonal tangent vectors of ${\mathbb{O}}^2$ at $\widehat{z}\notin \{N,S\}$.

We now formulate an inverse scattering problem for (2)–(5) using phaseless information, as follows.

Problem 1.

Assume that k is not an interior transmission eigenvalue of the problem (7). Given the phaseless far-field data

$$\left\{\begin{array}{cc}\left\{|e_a\left(\widehat{x}\right)·{\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},z_0)|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\}\right\},\hfill & \\ \left\{|e_a\left(\widehat{x}\right)·{\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},d,p)|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2\mathrm{with}p\perp d\right\},\hfill & \\ \left\{|e_a\left(\widehat{x}\right)·({\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},d,p)+{\mathit{E}}_{\Omega \cup B_1}^{\infty }(\widehat{x},z_0))|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2\mathrm{with}p\perp d\right\}\hfill \end{array}\right.$$

(9)

for $a=\phi ,\psi$, where $z_0\in {\mathbb{R}}^3\setminus (\overline{\Omega }\cup {\overline{B}}_1)$ is a fixed point, identity the shape and location of $\Omega$. To justify the uniqueness of the inverse problem, we first show that the phaseless far-field data have the property of translation invariance for incident plane waves and then present a reciprocity relation for the far-fields.

Lemma 1.

Let ${\Omega }_l:=\{x+l:x\in \Omega \}$ be a shifted domain of Ω with a fixed vector $l\in {\mathbb{R}}^3$ and denote by ${\mathit{E}}_l^{\infty }$ the corresponding electric far-field pattern for the scattering of plane electromagnetic waves by the elastic body Ω. Then, we have

$${\mathit{E}}_l^{\infty }\left(\widehat{x}\right)=e^{ik(d-\widehat{x})·l}{\mathit{E}}^{\infty }\left(\widehat{x}\right),\widehat{x},d\in {\mathbb{O}}^2.$$

(10)

Proof. 

Let $({\mathit{E}}_l^s,{\mathit{H}}_l^s,{\mathit{u}}_l)$ be the unique solution to the problem (2)–(5) with $\Omega$ replaced by ${\Omega }_l$. We want to show that

$$\left\{\begin{array}{cc}{\mathit{E}}_l^s\left(y\right)=e^{ikd·l}{\mathit{E}}^s(y-l),\hfill & y\in {\mathbb{R}}^3\setminus {\overline{\Omega }}_l,\hfill \\ {\mathit{H}}_l^s\left(y\right)=e^{ikd·l}{\mathit{H}}^s(y-l),\hfill & y\in {\mathbb{R}}^3\setminus {\overline{\Omega }}_l,\hfill \\ {\mathit{u}}_l\left(y\right)=e^{ikd·l}\mathit{u}(y-l),\hfill & y\in {\Omega }_l.\hfill \end{array}\right.$$

(11)

It is easy to derive that $({\mathit{E}}_l^s,{\mathit{H}}_l^s,{\mathit{u}}_l)$ expressed by (11) satisfies Equations (2) and (3) and the Silver–Müller radiation condition. Next, we verify the boundary condition (4). In fact, using the definition of the operator T and the boundary conditions satisfied by $(\mathit{E},\mathit{H})$, we have

$$\begin{array}{ccc}& & \frac{i}{k}\mathit{n}\times {\mathit{H}}_l^s\left(y\right)-T\left({\mathit{u}}_l\right)\left(y\right)\hfill \\ & =& e^{ikd·l}(\frac{i}{k}\mathit{n}\left(y\right)\times {\mathit{H}}^s(y-l)-2{\mu }_i\mathit{n}\left(y\right)·\nabla \mathit{u}(y-l)\hfill \\ & & -{\lambda }_i\mathit{n}\left(y\right)\nabla ·\mathit{u}(y-l)-{\mu }_i\mathit{n}\left(y\right)\times (\nabla \times \mathit{u}(y-l)))\hfill \\ & =& e^{ikd·l}\left(\frac{i}{k}\mathit{n}(y-l)\times {\mathit{H}}^s(y-l)-T\left(\mathit{u}\right)(y-l)\right)\hfill \\ & =& -e^{ikd·l}\frac{i}{k}\mathit{n}(y-l)\times {\mathit{H}}^i(y-l)\hfill \\ & =& -e^{ikd·l}\frac{i}{k}\mathit{n}(y-l)\times \left(ik(d\times p)e^{ik(y-l)·d}\right)\hfill \\ & =& -\frac{i}{k}\mathit{n}\left(y\right)\times {\mathit{H}}^i\left(y\right),y\in \partial {\Omega }_l\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \mathit{n}\times {\mathit{E}}_l^s\left(y\right)-\mathit{n}\times {\mathit{u}}_l\left(y\right)\hfill \\ & =& e^{ikd·l}\left(\mathit{n}\left(y\right)\times {\mathit{E}}^s(y-l)-\mathit{n}\left(y\right)\times \mathit{u}(y-l)\right)\hfill \\ & =& e^{ikd·l}\left(\mathit{n}(y-l)\times {\mathit{E}}^s(y-l)-\mathit{n}(y-l)\times \mathit{u}(y-l)\right)\hfill \\ & =& -e^{ikd·l}\mathit{n}(y-l)\times ik(d\times p)\times d{e}^{ik(y-l)·d}\hfill \\ & =& -\mathit{n}\left(y\right)\times {\mathit{E}}^i\left(y\right),y\in \partial {\Omega }_l.\hfill \end{array}$$

Then, from the representation of the far-field pattern, we derive that

$$\begin{array}{ccc}\hfill {\mathit{E}}_l^{\infty }\left(\widehat{x}\right)& =& \frac{ik}{4\pi }\widehat{x}\times {\int }_{\partial {\Omega }_l}\left\{\mathit{n}\left(y\right)\times {\mathit{E}}_l^s\left(y\right)+[\mathit{n}\left(y\right)\times {\mathit{H}}_l^s\left(y\right)]\times \widehat{x}\right\}{e}^{-ik\widehat{x}·y}ds\left(y\right)\hfill \\ & =& e^{ikd·l}\frac{ik}{4\pi }\widehat{x}\times {\int }_{\partial {\Omega }_l}\left\{\mathit{n}(y-l)\times {\mathit{E}}^s(y-l)+[\mathit{n}(y-l)\times {\mathit{H}}^s(y-l)]\times \widehat{x}\right\}{e}^{-ik\widehat{x}·y}ds\left(y\right)\hfill \\ & =& e^{ikd·l}\frac{ik}{4\pi }\widehat{x}\times {\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^s\left(y\right)+[\mathit{n}\left(y\right)\times {\mathit{H}}^s\left(y\right)]\times \widehat{x}\right\}{e}^{-ik\widehat{x}·(y+l)}ds\left(y\right)\hfill \\ & =& e^{ik(d-\widehat{x})·l}\frac{ik}{4\pi }\widehat{x}\times {\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^s\left(y\right)+[\mathit{n}\left(y\right)\times {\mathit{H}}^s\left(y\right)]\times \widehat{x}\right\}{e}^{-ik\widehat{x}·y}ds\left(y\right)\hfill \\ & =& e^{ik(d-\widehat{x})·l}{\mathit{E}}^{\infty }\left(\widehat{x}\right),\widehat{x},d\in {\mathbb{O}}^2.\hfill \end{array}$$

The proof is complete. □

The reciprocity principle of the far-field patterns is stated as follows.

Lemma 2.

The electric far-field pattern for the scattering of plane electromagnetic waves by an elastic body satisfies the reciprocity relation

$$q·{\mathit{E}}^{\infty }(\widehat{x},d,p)=p·{\mathit{E}}^{\infty }(-d,-\widehat{x},q),\widehat{x},d\in {\mathbb{O}}^2,$$

(12)

where $q\in {\mathbb{R}}^3$ is any given vector.

Proof. 

Gauss’s divergence theorem and Maxwell’s equations for the incident fields in $\Omega$ lead to

$${\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^i(y,d,p)·{\mathit{H}}^i(y,-\widehat{x},q)+\mathit{n}\left(y\right)\times {\mathit{H}}^i(y,d,p)·{\mathit{E}}^i(y,-\widehat{x},q)\right\}ds\left(y\right)=0.$$

(13)

By applying Gauss’s divergence theorem, Maxwell’s equations again and the radiation condition for the scattered fields in ${\mathbb{R}}^3\setminus \overline{\Omega }$, we have

$${\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^s(y,d,p)·{\mathit{H}}^s(y,-\widehat{x},q)+\mathit{n}\left(y\right)\times {\mathit{H}}^s(y,d,p)·{\mathit{E}}^s(y,-\widehat{x},q)\right\}ds\left(y\right)=0.$$

(14)

From the Stratton–Chu formula for the scattered fields, we derive that

$$\begin{array}{cc}\hfill & 4\pi q·{\mathit{E}}^{\infty }(\widehat{x},d,p)\hfill \\ \hfill =& {\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^s(y,d,p)·{\mathit{H}}^i(y,-\widehat{x},q)+\mathit{n}\left(y\right)\times {\mathit{H}}^s(y,d,p)·{\mathit{E}}^i(y,-\widehat{x},q)\right\}ds\left(y\right).\hfill \end{array}$$

(15)

By interchanging the roles of d and $-\widehat{x}$, and also the roles of p and q, we obtain

$$\begin{array}{cc}\hfill & 4\pi p·{\mathit{E}}^{\infty }(-d,-\widehat{x},q)\hfill \\ \hfill =& {\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times {\mathit{E}}^s(y,-\widehat{x},q)·{\mathit{H}}^i(y,d,p)+\mathit{n}\left(y\right)\times {\mathit{H}}^s(y,-\widehat{x},q)·{\mathit{E}}^i(y,d,p)\right\}ds\left(y\right).\hfill \end{array}$$

(16)

Then, by direct calculations, we have

$$\begin{array}{cc}\hfill & 4\pi \{q·{\mathit{E}}^{\infty }(\widehat{x},d,p)-p·{\mathit{E}}^{\infty }(-d,-\widehat{x},q)\}\hfill \\ \hfill =& {\int }_{\partial \Omega }\left\{\mathit{n}\left(y\right)\times \mathit{E}(y,d,p)·\mathit{H}(y,-\widehat{x},q)+\mathit{n}\left(y\right)\times \mathit{H}(y,d,p)·\mathit{E}(y,-\widehat{x},q)\right\}ds\left(y\right)\hfill \\ \hfill =& -ik{\int }_{\partial \Omega }\left\{T\mathit{u}(y,d,p)·\mathit{u}(y,-\widehat{x},q)-\mathit{u}(y,d,p)·T\mathit{u}(y,-\widehat{x},q)\right\}ds\left(y\right)\hfill \\ \hfill =& -ik{\int }_{\Omega }\left\{{\Delta }^{*}\mathit{u}(y,d,p)·\mathit{u}(y,-\widehat{x},q)-\mathit{u}(y,d,p)·{\Delta }^{*}\mathit{u}(y,-\widehat{x},q)\right\}ds\left(y\right)\hfill \\ \hfill =& 0,\hfill \end{array}$$

(17)

where the second equality is obtained by the boundary condition (4), the third equality is due to the third Betti formula, and the last equality is deduced by the Navier equation and ${\Delta }^{*}\mathit{u}:={\mu }_i\Delta \mathit{u}+({\lambda }_i+{\mu }_i)\nabla (\nabla ·\mathit{u})$. The proof is complete. □

We are now in a position to justify the unique determination of the scatterer from phaseless far-field data using the strategy in [22]. Differing from [22], we utilize a fixed point source and plane waves as the incident waves. Meanwhile, a reference scatterer is added to the scattering model. The physical properties of the artificial scatterer are analogous to the original one. The introduction of the reference scatterer enables us to recover the phased far-field data, and hence the scatterer can be uniquely determined.

Let ${\Omega }_1$ and ${\Omega }_2$ be two elastic bodies. Denote by $({\mathit{E}}_{{\Omega }_j\cup B_1}^s(x,z_0),{\mathit{H}}_{{\Omega }_j\cup B_1}^s(x,z_0))$ and $({\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},z_0),$ ${\mathit{H}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},z_0))$ the scattered field and the corresponding far-field pattern generated by ${\Omega }_j\cup B_1$ for the incident electric dipole at a fixed point $z_0\in {\mathbb{R}}^3\setminus ({\overline{\Omega }}_j\cup {\overline{B}}_1)$ ($j=1,2$). The scattered field and the corresponding far-field pattern for the incident plane electromagnetic wave $({\mathit{E}}^i(x,d,p),{\mathit{H}}^i(x,d,p))$ are denoted by $({\mathit{E}}_{{\Omega }_j\cup B_1}^s(x,d,p),$ ${\mathit{H}}_{{\Omega }_j\cup B_1}^s(x,d,p))$ and $({\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p),{\mathit{H}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p))$ ($j=1,2$), respectively. The uniqueness result for Problem 1 is stated as follows.

Theorem 1.

Suppose that k is not an interior transmission eigenvalue in $B_1$. Further assume that there is a ball $B_R$ with the center at the origin and the radius R such that ${\overline{\Omega }}_1\cup {\overline{\Omega }}_2\subset B_R$ and ${\overline{B}}_1\cap {\overline{B}}_R=\varnothing$. If the far-field patterns satisfy

$$\left\{\begin{array}{cc}|e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},z_0)|=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},z_0)|,\hfill & \\ |e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)|=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p)|,\hfill & \\ |e_a\left(\widehat{x}\right)·({\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)+{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},z_0))|=|e_a\left(\widehat{x}\right)·({\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p)+{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},z_0))|\hfill \end{array}\right.$$

(18)

for $a=\phi ,\psi$, and all $\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2$ satisfying $d\perp p$, where $z_0\in {\mathbb{R}}^3\setminus ({\overline{\Omega }}_1\cup {\overline{\Omega }}_2\cup {\overline{B}}_1)$ is a fixed point, then it holds that ${\Omega }_1={\Omega }_2$.

Proof. 

Since ${\mathit{E}}^i(x,d,d)=0,{\mathit{H}}^i(x,d,d)=0$, the well-posedness of the interaction problem indicates ${\mathit{E}}^{\infty }(\widehat{x},d,d)=0$. Therefore, the condition $d\perp p$ in the equities (18) is no longer required. We deduce from (18) that

$$\begin{array}{ccc}\hfill & \hfill & \Re \left\{[e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},z_0)]\left[\overline{e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)}\right]\right\}\hfill \\ \hfill & =\hfill & \Re \left\{[e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},z_0)]\left[\overline{e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p)}\right]\right\},\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},d\in {\mathbb{O}}^2,p\in {\mathbb{R}}^3\hfill \end{array}$$

(19)

for $a=\phi ,\psi$. The first two equations in (18) imply

$$\left\{\begin{array}{cc}{e}_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},z_0)=t(\widehat{x},z_0,e_a\left(\widehat{x}\right))e^{i{\beta }_j(\widehat{x},z_0,e_a\left(\widehat{x}\right))},\hfill & \\ e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p)=r(\widehat{x},d,e_a\left(\widehat{x}\right),p)e^{i{\alpha }_j(\widehat{x},d,e_a\left(\widehat{x}\right),p)}\hfill \end{array}\right.$$

(20)

for $j=1,2$, where $t(\widehat{x},z_0,e_a\left(\widehat{x}\right))=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},z_0)|$, $r(\widehat{x},d,e_a\left(\widehat{x}\right),p)=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p)|$, ${\beta }_j(\widehat{x},z_0,e_a\left(\widehat{x}\right))$ and ${\alpha }_j(\widehat{x},d,e_a\left(\widehat{x}\right),p)$ are real-valued functions, $j=1,2$. Then (19) leads to

$$t(\widehat{x},z_0,e_a\left(\widehat{x}\right))r(\widehat{x},d,e_a\left(\widehat{x}\right),p)\Re \left\{e^{i[{\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\alpha }_1(\widehat{x},d,e_a\left(\widehat{x}\right),p)]-i[{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\alpha }_2(\widehat{x},d,e_a\left(\widehat{x}\right),p)]}\right\}=0.$$

(21)

Owing to the fact that $e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p)$ is analytical with respect to $\widehat{x},d,p$ in $({\mathbb{O}}^2\setminus \{N,S\})\times {\mathbb{O}}^2\times {\mathbb{R}}^3$, $j=1,2$, there are some open sets ${\tilde{G}}_1\subset {\mathbb{O}}^2\setminus \{N,S\}$, $G_2\subset {\mathbb{O}}^2$ and $U\subset {\mathbb{R}}^3$ satisfying $r(\widehat{x},d,e_a\left(\widehat{x}\right),p)\ne 0$ for $\widehat{x}\in {\tilde{G}}_1$, $d\in G_2$ and $p\in U$. A similar argument shows that $t(\widehat{x},z_0,e_a\left(\widehat{x}\right))\ne 0$ on an open set $G_1\subset {\tilde{G}}_1$. Thus, we obtain from (21) that

$$cos[{\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\alpha }_1(\widehat{x},d,e_a\left(\widehat{x}\right),p)]=cos[{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\alpha }_2(\widehat{x},d,e_a\left(\widehat{x}\right),p)],$$

(22)

for all $\widehat{x}\in G_1$, $d\in G_2$ and $p\in U$. Consequently, it holds that either

$${\alpha }_1(\widehat{x},d,e_a\left(\widehat{x}\right),p)-{\alpha }_2(\widehat{x},d,e_a\left(\widehat{x}\right),p)={\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right))+2n\pi$$

(23)

or

$${\alpha }_1(\widehat{x},d,e_a\left(\widehat{x}\right),p)+{\alpha }_2(\widehat{x},d,e_a\left(\widehat{x}\right),p)={\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))+{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right))+2n\pi$$

(24)

for some $n\in \mathbb{Z}$ and all $\widehat{x}\in G_1$, $d\in G_2$ and $p\in U$.

We first consider the case (23). For a fixed $z_0$, define

$${\delta }^{\left(a\right)}\left(\widehat{x}\right):={\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))-{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right)),\widehat{x}\in G_1.$$

(25)

Then, we deduce from (20), (23) and (25) that

$$\begin{array}{ccc}\hfill e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)& =& r(\widehat{x},d,e_a\left(\widehat{x}\right),p)e^{i{\alpha }_1(\widehat{x},d,e_a\left(\widehat{x}\right),p)}\hfill \\ & =& r(\widehat{x},d,e_a\left(\widehat{x}\right),p)e^{i{\delta }^{\left(a\right)}\left(\widehat{x}\right)+i{\alpha }_2(\widehat{x},d,e_a\left(\widehat{x}\right),p)}\hfill \\ & =& e^{i{\delta }^{\left(a\right)}\left(\widehat{x}\right)}{e}_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p)\hfill \end{array}$$

for all $\widehat{x}\in G_1$, $d\in G_2$ and $p\in U$. The analyticity of $e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p)$ and ${\delta }^{\left(a\right)}\left(\widehat{x}\right)$ with respect to $d\in {\mathbb{O}}^2$ and $p\in {\mathbb{R}}^3$ indicates that

$$e_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)=e^{i{\delta }^{\left(a\right)}\left(\widehat{x}\right)}{e}_a\left(\widehat{x}\right)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p),\forall \widehat{x}\in G_1,d\in {\mathbb{O}}^2,p\in {\mathbb{R}}^3.$$

(26)

Replacing the variables $\widehat{x}$ by $-d$ and d by $-\widehat{x}$, we have

$$e_a(-d)·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(-d,-\widehat{x},p)=e^{i{\delta }^{\left(a\right)}(-d)}{e}_a(-d)·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(-d,-\widehat{x},p),\forall -d\in G_1,\widehat{x}\in {\mathbb{O}}^2,p\in {\mathbb{R}}^3.$$

The reciprocity relation (12) results in

$$p·{\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,e_a(-d))=e^{i{\delta }^{\left(a\right)}(-d)}p·{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,e_a(-d)),\forall -d\in G_1,\widehat{x}\in {\mathbb{O}}^2,p\in {\mathbb{R}}^3.$$

Due to $e_{\psi }\left(d\right)=-e_{\psi }(-d)$ and $e_{\phi }\left(d\right)=e_{\phi }(-d)$, we conclude that

$${\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))=e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right)),\forall -d\in G_1,\widehat{x}\in {\mathbb{O}}^2.$$

(27)

Then, Rellich’s lemma shows that

$${\mathit{E}}_{{\Omega }_1\cup B_1}^s(x,d,e_a\left(d\right))=e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{E}}_{{\Omega }_2\cup B_1}^s(x,d,e_a\left(d\right)),\forall -d\in G_1,x\in {\mathbb{R}}^3\setminus ({\overline{\Omega }}_1\cup {\overline{\Omega }}_2\cup {\overline{B}}_1),$$

(28)

which immediately leads to

$${\mathit{H}}_{{\Omega }_1\cup B_1}^s(x,d,e_a\left(d\right))=e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{H}}_{{\Omega }_2\cup B_1}^s(x,d,e_a\left(d\right)),\forall -d\in G_1,x\in {\mathbb{R}}^3\setminus ({\overline{\Omega }}_1\cup {\overline{\Omega }}_2\cup {\overline{B}}_1).$$

(29)

On the boundary $\partial B_1$, we have

$$\begin{array}{ccc}& & \frac{i}{k}\mathit{n}\times {\mathit{H}}^i(x,d,e_a\left(d\right))-T{\mathit{u}}_{{\Omega }_j\cup B_1}(x,d,e_a\left(d\right))=-\frac{i}{k}{\mathit{H}}_{{\Omega }_j\cup B_1}^s(x,d,e_a\left(d\right)),\hfill \end{array}$$

(30)

$$\begin{array}{ccc}& & \mathit{n}\times {\mathit{E}}^i(x,d,e_a\left(d\right))-\mathit{n}\times {\mathit{u}}_{{\Omega }_j\cup B_1}(x,d,e_a\left(d\right))=-{\mathit{E}}_{{\Omega }_j\cup B_1}^s(x,d,e_a\left(d\right)),\hfill \end{array}$$

(31)

for $-d\in G_1$, $j=1,2$. From (28)–(31), we deduce that

$$\begin{array}{ccc}\hfill \widehat{\mathit{E}}& :=& (1-e^{i{\delta }^{\left(a\right)}(-d)}){\mathit{E}}^i(x,d,e_a\left(d\right)),x\in B_1,\hfill \\ \hfill \widehat{\mathit{H}}& :=& (1-e^{i{\delta }^{\left(a\right)}(-d)}){\mathit{H}}^i(x,d,e_a\left(d\right)),x\in B_1,\hfill \\ \hfill \widehat{\mathit{u}}& :=& {\mathit{u}}_{{\Omega }_1\cup B_1}(x,d,e_a\left(d\right))-e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{u}}_{{\Omega }_2\cup B_1}(x,d,e_a\left(d\right)),x\in B_1\hfill \end{array}$$

satisfies the problem (7). From the assumption for k, it is known that $\widehat{\mathit{E}}=\widehat{\mathit{H}}=\widehat{\mathit{u}}=0$. Then, we derive that $e^{i{\delta }^{\left(a\right)}(-d)}=1$ for all $-d\in G_1$, since ${\mathit{E}}^i(x,d,e_a\left(d\right))\ne 0$. Thus, (27) yields

$${\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))={\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right)),\forall -d\in G_1,\widehat{x}\in {\mathbb{O}}^2.$$

(32)

The analyticity of ${\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))$ for $d\in {\mathbb{O}}^2\setminus \{N,S\}$, $j=1,2$ shows that

$${\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))={\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right)),\forall d\in {\mathbb{O}}^2\setminus \{N,S\},\widehat{x}\in {\mathbb{O}}^2.$$

(33)

Note that any polarization vector p can be expressed by the linear combination of $e_{\psi }\left(d\right),e_{\phi }\left(d\right)$ and d. We conclude from the fact ${\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,d)=0$ and the analyticity of ${\mathit{E}}_{{\Omega }_j\cup B_1}^{\infty }(\widehat{x},d,p)$ for $d\in {\mathbb{O}}^2$, $j=1,2$ that

$${\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,p)={\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,p),\forall d,\widehat{x}\in {\mathbb{O}}^2,p\in {\mathbb{R}}^3.$$

(34)

From the uniqueness result in [29], we obtain ${\Omega }_1={\Omega }_2$.

Now, we consider the case (24). A similar argument implies that

$${\mathit{E}}_{{\Omega }_1\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))=e^{i{\gamma }^{\left(a\right)}(-d)}\overline{{\mathit{E}}_{{\Omega }_2\cup B_1}^{\infty }(\widehat{x},d,e_a\left(d\right))},\forall -d\in G_1,x\in {\mathbb{O}}^2,$$

(35)

where ${\gamma }^{\left(a\right)}$ is given by

$${\gamma }^{\left(a\right)}\left(\widehat{x}\right):={\beta }_1(\widehat{x},z_0,e_a\left(\widehat{x}\right))+{\beta }_2(\widehat{x},z_0,e_a\left(\widehat{x}\right)),\widehat{x}\in G_1.$$

The same discussion as in Lemma 3.1 of [22] shows that $({\mathit{E}}_{{\Omega }_1\cup B_1}^s(x,d,e_a\left(d\right)),$ ${\mathit{H}}_{{\Omega }_1\cup B_1}^s(x,d,e_a\left(d\right)))$ satisfies Maxwell’s equations in ${\mathbb{R}}^3\setminus {\overline{B}}_R$. Thus, the total electromagnetic field and elastic displacement in $B_1$ satisfy the problem (7). Then, the assumption on k implies

$${\mathit{E}}^i(x,d,e_a\left(d\right))+{\mathit{E}}_{{\Omega }_1\cup B_1}^s(x,d,e_a\left(d\right))=0\mathrm{in}{B}_1,$$

which indicates that the total electric field equals zero in ${\mathbb{R}}^3\setminus {\overline{B}}_R$ owing to the analyticity. This contradicts the fact that ${\mathit{E}}^i(x,d,e_a\left(d\right))\ne 0$. Therefore, (24) does not hold. The proof is complete. □

3. Electromagnetic Scattering by a Chiral Medium

This section deals with the inverse problem of uniquely determining the scatterer from phaseless far-field data for electromagnetic scattering by a chiral medium. The analysis is similar to that presented in Section 2, and we just give a brief version of the argument.

3.1. The Mathematical Formulation

Let $D\in {\mathbb{R}}^3$ be a homogeneous chiral medium with the electric permittivity ${\epsilon }_0>0$, magnetic permeability ${\mu }_0>0$, and chirality admittance ${\beta }_0>0$. Assume that D is a bounded domain with a smooth boundary $\partial D$, and the unit outward normal on $\partial D$ is denoted by $\mathit{n}$. The exterior domain ${\mathbb{R}}^3\setminus \overline{D}$ of D is filled with an achiral homogeneous and isotropic medium with the magnetic permeability $\mu$ and electric permittivity $ϵ$. Consider the scattering of the time-harmonic electromagnetic plane wave $({\mathit{E}}^i,{\mathit{H}}^i)$ by the chiral scatterer D. We still denote the scattered electromagnetic field by $({\mathit{E}}^s,{\mathit{H}}^s)$ and the total electromagnetic field by $(\mathit{E},\mathit{H})$ in ${\mathbb{R}}^3\setminus \overline{D}$. Then, we have

$$curl\mathit{E}-ik\mathit{H}=0,curl\mathit{H}+ik\mathit{E}=0\mathrm{in}{\mathbb{R}}^3\setminus \overline{D},$$

(36)

where $k=\omega \sqrt{\epsilon \mu }$. We denote the electric and magnetic fields in D by ${\mathit{E}}_0$ and ${\mathit{H}}_0$, respectively. They satisfy the Drude–Born–Fedorov system [30]

$$curl{\mathit{E}}_0={\gamma }_0^2{\beta }_0{\mathit{E}}_0+i\frac{{\gamma }_0^2}{k_0}{\mathit{H}}_0,curl{\mathit{H}}_0={\gamma }_0^2{\beta }_0{\mathit{H}}_0-i\frac{{\gamma }_0^2}{k_0}{\mathit{E}}_0\mathrm{in}D,$$

(37)

where $k_0=\omega \sqrt{{\epsilon }_0{\mu }_0}$, ${\gamma }_0=k_0/\sqrt{1-k_0^2{\beta }_0^2}$ and $k_0{\beta }_0<1$.

The continuity of the tangential components of the electric and magnetic fields across the boundary $\partial D$ leads to the following transmission conditions:

$${\epsilon }^{-1/2}\mathit{n}\times \mathit{E}={\epsilon }_0^{-1/2}\mathit{n}\times {\mathit{E}}_0,{\mu }^{-1/2}\mathit{n}\times \mathit{H}={\mu }_0^{-1/2}\mathit{n}\times {\mathit{H}}_0\mathrm{on}\partial D.$$

(38)

In addition, we always assume that the scattered field $({\mathit{E}}^s,{\mathit{H}}^s)$ satisfies the Silver–Müller radiation condition (5).

The solvability of the scattering problem (36)–(38) has been studied in [31,32] under different assumptions. It is shown in [33] that the chiral obstacle is uniquely determined by the far-field patterns of all incident waves with a fixed wave number.

3.2. An Inverse Scattering Problem

We now formulate the inverse problem by using the phaseless electric far-field pattern to identify the chiral scatterer and introduce some notations similar to them for Problem 1. The meanings of the expressions are self-evident.

First, we introduce an extra artificial reference ball $B_2$, which is occupied by a homogeneous and isotropic chiral medium with the electric permittivity ${\epsilon }_2>0$, magnetic permeability ${\mu }_2>0$, and chirality admittance ${\beta }_2>0$ to the scattering system such that $B_2\cap D=\varnothing$. Suppose that the wave number k is not an interior transmission eigenvalue in $B_2$, that is, the following interior transmission problem has only a trivial solution

$$\left\{\begin{array}{cc}curl\tilde{\mathit{E}}-ik\tilde{\mathit{H}}=0,curl\tilde{\mathit{H}}+ik\tilde{\mathit{E}}=0\hfill & \mathrm{in}{B}_2,\hfill \\ curl{\tilde{\mathit{E}}}_0={\gamma }_2^2{\beta }_2{\tilde{\mathit{E}}}_0+i\frac{{\gamma }_2^2}{k_2}{\tilde{\mathit{H}}}_0,curl{\tilde{\mathit{H}}}_0={\gamma }_2^2{\beta }_2{\tilde{\mathit{H}}}_0-i\frac{{\gamma }_2^2}{k_2}{\tilde{\mathit{E}}}_0\hfill & \mathrm{in}{B}_2,\hfill \\ {\epsilon }^{-1/2}\mathit{n}\times \tilde{\mathit{E}}={\epsilon }_2^{-1/2}\mathit{n}\times {\tilde{\mathit{E}}}_0,{\mu }^{-1/2}\mathit{n}\times \tilde{\mathit{H}}={\mu }_2^{-1/2}\mathit{n}\times {\tilde{\mathit{H}}}_0\hfill & \mathrm{on}\partial B_2.\hfill \end{array}\right.$$

(39)

For the incident electric dipole located at the point $z_0\in {\mathbb{R}}^3\setminus (\overline{D}\cup {\overline{B}}_2)$, we use $({\mathit{E}}_{D\cup B_2}^s(x,z_0),{\mathit{H}}_{D\cup B_2}^s(x,z_0))$ and $({\mathit{E}}_{D\cup B_2}^{\infty }(\widehat{x},z_0),{\mathit{H}}_{D\cup B_2}^{\infty }(\widehat{x},z_0))$ to denote the scattered field and the far-field pattern generated by $D\cup B_2$, respectively. The scattered field and far-field pattern for the incident plane waves $({\mathit{E}}^i(x,d,p),{\mathit{H}}^i(x,d,p))$ are denoted by $({\mathit{E}}_{D\cup B_2}^s(x,d,p),$ ${\mathit{H}}_{D\cup B_2}^s(x,d,p))$ and $({\mathit{E}}_{D\cup B_2}^{\infty }(\widehat{x},d,p),{\mathit{H}}_{D\cup B_2}^{\infty }(\widehat{x},d,p))$, respectively. Then, the phaseless inverse scattering problem is formulated as follows.

Problem 2.

Assume that k is not an interior transmission eigenvalue of the problem (39). Given the phaseless far-field data

$$\left\{\begin{array}{cc}\left\{|e_a\left(\widehat{x}\right)·{\mathit{E}}_{D\cup B_2}^{\infty }(\widehat{x},z_0)|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\}\right\},\hfill & \\ \left\{|e_a\left(\widehat{x}\right)·{\mathit{E}}_{D\cup B_2}^{\infty }(\widehat{x},d,p)|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2\mathrm{with}p\perp d\right\},\hfill & \\ \left\{|e_a\left(\widehat{x}\right)·({\mathit{E}}_{D\cup B_2}^{\infty }(\widehat{x},d,p)+E_{D\cup B_2}^{\infty }(\widehat{x},z_0))|:\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2\mathrm{with}p\perp d\right\}\hfill \end{array}\right.$$

(40)

for $a=\psi ,\phi$, where $z_0\in {\mathbb{R}}^3\setminus (\overline{D}\cup {\overline{B}}_2)$ is a fixed point, determine the location and shape of D.

To investigate the above inverse problem, we first give the results for the translation invariance and reciprocity principle.

Lemma 3.

Using the same process as in Lemma 1, one can obtain that the electric far-field pattern for the scattering of plane electromagnetic waves by the chiral scatterer D also satisfies the translation invariance property (10).

Lemma 4.

The reciprocity relation of the electric far-field pattern for the scattering of plane electromagnetic waves by a chiral scatterer also satisfies (12), which was proved in [34].

Next, we study Problem 2 and give the descriptions of some notations. Let $D_1$ and $D_2$ be two chiral scatterers. Denote by $({\mathit{E}}_{D_j\cup B_2}^s(x,z_0),{\mathit{H}}_{D_j\cup B_2}^s(x,z_0))$ and $({\mathit{E}}_{D_j\cup B_2}^{\infty }(\widehat{x},z_0),{\mathit{H}}_{D_j\cup B_2}^{\infty }(\widehat{x},z_0))$ the scattered field and the corresponding far-field pattern generated by $D_j\cup B_2$ for the incident electric dipole at a fixed point $z_0\in {\mathbb{R}}^3\setminus ({\overline{D}}_j\cup {\overline{B}}_2)$ ($j=1,2$). The scattered field and the corresponding far-field pattern for the incident plane electromagnetic wave $({\mathit{E}}^i(x,d,p),{\mathit{H}}^i(x,d,p))$ are denoted by $({\mathit{E}}_{D_j\cup B_2}^s(x,d,p),$ ${\mathit{H}}_{D_j\cup B_2}^s(x,d,p))$ and $({\mathit{E}}_{D_j\cup B_2}^{\infty }(\widehat{x},d,p),{\mathit{H}}_{D_j\cup B_2}^{\infty }(\widehat{x},d,p))$ ($j=1,2$), respectively.

Theorem 2.

Suppose k is not an interior transmission eigenvalue of the problem (39). Further, assume that there is a ball $B_R$ with the center at the origin and the radius R such that ${\overline{D}}_1\cup {\overline{D}}_2\subset B_R$ and ${\overline{B}}_2\cap {\overline{B}}_R=\varnothing$. If the far-field patterns satisfy

$$\left\{\begin{array}{cc}|e_a\left(\widehat{x}\right)·{\mathit{E}}_{D_1\cup B_2}^{\infty }(\widehat{x},z_0)|=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{D_2\cup B_2}^{\infty }(\widehat{x},z_0)|,\hfill & \\ |e_a\left(\widehat{x}\right)·{\mathit{E}}_{D_1\cup B_2}^{\infty }(\widehat{x},d,p)|=|e_a\left(\widehat{x}\right)·{\mathit{E}}_{D_2\cup B_2}^{\infty }(\widehat{x},d,p)|,\hfill & \\ |e_a\left(\widehat{x}\right)·({\mathit{E}}_{D_1\cup B_2}^{\infty }(\widehat{x},d,p)+{\mathit{E}}_{D_1\cup B_2}^{\infty }(\widehat{x},z_0))|=|e_a\left(\widehat{x}\right)·({\mathit{E}}_{D_2\cup B_2}^{\infty }(\widehat{x},d,p)+{\mathit{E}}_{D_2\cup B_2}^{\infty }(\widehat{x},z_0))|\hfill \end{array}\right.$$

(41)

for $a=\psi ,\phi$, and all $\widehat{x}\in {\mathbb{O}}^2\setminus \{N,S\},p\in {\mathbb{R}}^3,d\in {\mathbb{O}}^2$ satisfy $d\perp p$, where $z_0\in {\mathbb{R}}^3\setminus ({\overline{D}}_1\cup {\overline{D}}_2\cup {\overline{B}}_2)$ is a fixed point, then it holds that $D_1=D_2$.

Proof. 

Since the argument is the same as for Theorem 1, we only give some necessary modifications in this situation. The analysis is the same up to the equation (29), in which we use the result for Lemma 4.

For the case (23), it follows that on the boundary $\partial B_2$

$$\begin{array}{ccc}& & {\epsilon }^{-1/2}\mathit{n}\times {\mathit{E}}^i(x,d,e_a\left(d\right))-{\epsilon }_2^{-1/2}\mathit{n}\times {\mathit{E}}_{0,D_j\cup B_2}(x,d,e_a\left(d\right))=-{\epsilon }^{-1/2}{\mathit{E}}_{D_j\cup B_2}^s(x,d,e_a\left(d\right)),\hfill \\ & & {\mu }^{-1/2}\mathit{n}\times {\mathit{H}}^i(x,d,e_a\left(d\right))-{\mu }_2^{-1/2}{\mathit{H}}_{0,D_j\cup B_2}(x,d,e_a\left(d\right))=-{\mu }^{-1/2}{\mathit{H}}_{D_j\cup B_2}^s(x,d,e_a\left(d\right))\hfill \end{array}$$

for $j=1,2$. Let

$$\begin{array}{ccc}\hfill \tilde{\mathit{E}}& :=& (1-e^{i{\delta }^{\left(a\right)}(-d)}){\mathit{E}}^i(x,d,e_a\left(d\right)),x\in B_2,\hfill \\ \hfill \tilde{\mathit{H}}& :=& (1-e^{i{\delta }^{\left(a\right)}(-d)}){\mathit{H}}^i(x,d,e_a\left(d\right)),x\in B_2,\hfill \\ \hfill {\tilde{\mathit{E}}}_0& :=& {\mathit{E}}_{0,D_1\cup B_2}(x,d,e_a\left(d\right))-e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{E}}_{0,D_2\cup B_2}(x,d,e_a\left(d\right)),x\in B_2,\hfill \\ \hfill {\tilde{\mathit{H}}}_0& :=& {\mathit{H}}_{0,D_1\cup B_2}(x,d,e_a\left(d\right))-e^{i{\delta }^{\left(a\right)}(-d)}{\mathit{H}}_{0,D_2\cup B_2}(x,d,e_a\left(d\right)),x\in B_2.\hfill \end{array}$$

Then, they satisfy the problem (39). The assumption of k shows $\tilde{\mathit{E}}=\tilde{\mathit{H}}={\tilde{\mathit{E}}}_0={\tilde{\mathit{H}}}_0=0$ in $B_2$, which indicates $e^{i{\delta }^{\left(a\right)}(-d)}=1$. Following the same discussion as in Theorem 1 and using the uniqueness result presented in [33], we obtain $D_1=D_2$.

When the case (24) holds, the same argument shows that the total electromagnetic fields

$$\begin{array}{c}\hfill {\mathit{E}}_{D_1\cup B_2}(x,d,e_a\left(d\right))={\mathit{E}}^i(x,d,e_a\left(d\right))+{\mathit{E}}_{D_1\cup B_2}^s(x,d,e_a\left(d\right)),\\ \hfill {\mathit{H}}_{D_1\cup B_2}(x,d,e_a\left(d\right))={\mathit{H}}^i(x,d,e_a\left(d\right))+{\mathit{H}}_{D_1\cup B_2}^s(x,d,e_a\left(d\right))\end{array}$$

and $({\mathit{E}}_{0,D_1\cup B_2}(x,d,e_a\left(d\right)),{\mathit{H}}_{0,D_1\cup B_2}(x,d,e_a\left(d\right)))$ in the chiral medium $B_2$ satisfy the problem (39). Then, we have from the assumption on k that

$${\mathit{E}}^i(x,d,e_a\left(d\right))+{\mathit{E}}_{D_1\cup B_2}^s(x,d,e_a\left(d\right))=0\mathrm{in}{B}_2,$$

which implies that the total electric field equals zero in ${\mathbb{R}}^3\setminus {\overline{B}}_R$ due to the analyticity. We, therefore, have a contradiction, since ${\mathit{E}}^i(x,d,e_a\left(d\right))\ne 0$. Thus, (24) does not hold. The proof is complete. □

4. Electromagnetic Scattering by an Impedance Cylinder at Oblique Incidence

In this section, we consider the unique determination of the obstacle from phaseless far-field data of the electromagnetic scattering from a cylinder at oblique incidence. First, the mathematical model of the wave scattering is introduced, and the inverse problem is formulated. Second, the properties of translation invariance and reciprocity principle for the far-field pattern are proved. Third, we present the main result for the unique determination of the obstacle from the phaseless far-field data.

4.1. The Mathematical Formulation

Let us start with the mathematical model and consider the electromagnetic scattering problem of an obliquely incident plane electromagnetic wave by an impedance cylinder [35,36]. Let the cylinder be oriented parallel to the $x_3$-axis and denote by O the cross-section in ${\mathbb{R}}^2$ with a smooth boundary $\partial O$. Assume that the background medium is homogeneous and uniformly dielectric along the $x_3$ direction, which is characterized by the constant electric permittivity $ϵ$ and magnetic permeability $\mu$. For the scattering of the time-harmonic incident plane electromagnetic wave

$$({\mathit{E}}^i,{\mathit{H}}^i)=({\mathit{e}}^i(x_1,x_2),{\mathit{h}}^i(x_1,x_2))e^{-i\omega t-i\sigma x_3},$$

the total electromagnetic field $(\mathit{E},\mathit{H})$ and the scattered field $({\mathit{E}}^s,{\mathit{H}}^s)$ have the representations

$$(\mathit{E},\mathit{H})=(\mathit{e}(x_1,x_2),\mathit{h}(x_1,x_2))e^{-i\omega t-i\sigma x_3},({\mathit{E}}^s,{\mathit{H}}^s)=({\mathit{e}}^s(x_1,x_2),{\mathit{h}}^s(x_1,x_2))e^{-i\omega t-i\sigma x_3},$$

where $\omega$ is the frequency and $\sigma =\omega \sqrt{ϵ\mu }cos\phi =kcos\phi$ with $\phi \in (0,\pi )$ is the incident angle with respect to the negative $x_3$-axis.

Define the $x_3$-component of the incident wave $({\mathit{e}}^i(x_1,x_2),{\mathit{h}}^i(x_1,x_2))$ by $(e^i(x_1,x_2),h^i(x_1,x_2))$, the scattered field $({\mathit{e}}^s(x_1,x_2),{\mathit{h}}^s(x_1,x_2))$ by $(e^s(x_1,x_2),h^s(x_1,x_2))$, and the total field $(\mathit{e}(x_1,x_2),\mathit{h}(x_1,x_2))$ by $(e(x_1,x_2),h(x_1,x_2))$, respectively. By expressing the $x_1$- and $x_2$-components in terms of the $x_3$-components of the corresponding electric and magnetic fields and suppressing $e^{-i\omega t-i\sigma x_3}$, we are led to the following system:

$$\left\{\begin{array}{cc}\Delta e+{\kappa }^{2}e=0,\Delta h+{\kappa }^{2}h=0\hfill & \mathrm{in}{\mathbb{R}}^2\setminus \overline{O},\hfill \\ \mathcal{B}(e,h)=0\hfill & \mathrm{on}\partial O,\hfill \\ {\displaystyle \underset{r\to \infty }{lim}\sqrt{r}(\frac{\partial e^s}{\partial r}-i\kappa e^s)=0,}\hfill & r=\left|x\right|,\hfill \\ {\displaystyle \underset{r\to \infty }{lim}\sqrt{r}(\frac{\partial h^s}{\partial r}-i\kappa h^s)=0,}\hfill & r=\left|x\right|,\hfill \end{array}\right.$$

(42)

where $\kappa =\omega \sqrt{ϵ\mu }sin\phi =ksin\phi$, and $\mathcal{B}(e,h)$ is a coupled oblique boundary condition for e and h, depending on the physical property of the cylinder.

Let $\nu =({\nu }_1,{\nu }_2)$ be the unit outward normal vector and $\tau =(-{\nu }_2,{\nu }_1)$ the unit tangent vector on the boundary $\partial O$. If the cylinder is a perfectly conducting obstacle, then we have the boundary condition

$$\left\{\begin{array}{cc}e=0\hfill & \mathrm{on}\partial O,\hfill \\ {\displaystyle \mu \omega \frac{\partial h}{\partial \nu }+\sigma \frac{\partial e}{\partial \tau }=0}\hfill & \mathrm{on}\partial O\hfill \end{array}\right.$$

(43)

or

$$\left\{\begin{array}{cc}h=0\hfill & \mathrm{on}\partial O,\hfill \\ {\displaystyle ϵ\omega \frac{\partial e}{\partial \nu }-\sigma \frac{\partial h}{\partial \tau }=0}\hfill & \mathrm{on}\partial O.\hfill \end{array}\right.$$

(44)

If the cylinder satisfies the Leontovich impedance boundary condition, one can derive the coupled boundary condition

$$\left\{\begin{array}{cc}{\displaystyle \mu \omega \frac{\partial h}{\partial \nu }+\sigma \frac{\partial e}{\partial \tau }+i\lambda {\kappa }^{2}h=0}\hfill & \mathrm{on}\partial O,\hfill \\ {\displaystyle \lambda ϵ\omega \frac{\partial e}{\partial \nu }-\lambda \sigma \frac{\partial h}{\partial \tau }+i{\kappa }^{2}e=0}\hfill & \mathrm{on}\partial O,\hfill \end{array}\right.$$

(45)

where $\lambda >0$ is the constant impedance coefficient. Note that the boundary conditions (43) and (44) correspond to $\lambda =0$ and $\lambda =\infty$ in (45), respectively.

Let the incident electromagnetic plane wave be a TM polarized plane wave with the incident angle $\phi$, and denote by $\psi$ the polar angle of the incident direction $\tilde{d}$. Then, $\tilde{d}=(sin\phi cos\psi ,sin\phi sin\psi ,-cos\phi )$, the polarization vector $\tilde{p}$ is given by $\tilde{p}=(cos\phi cos\psi ,$ $cos\phi sin\psi ,sin\phi )$, and the incident electric field and magnetic field have the expressions

$${\mathit{e}}^i(x_1,x_2)=\frac{1}{\sqrt{ϵ}}\tilde{p}{e}^{i\kappa (x_{1}cos\psi +x_{2}sin\psi )},{\mathit{h}}^i(x_1,x_2)=\frac{1}{\sqrt{\mu }}{(sin\psi ,-cos\psi ,0)}^{\top }{e}^{i\kappa (x_{1}cos\psi +x_{2}sin\psi )}.$$

Hence, for the TM polarized plane wave, the incident fields $e^i$ and $h^i$ in (42) are given by

$$e^i(x,d)=\frac{sin\phi }{\sqrt{ϵ}}{e}^{i\kappa (x_{1}cos\psi +x_{2}sin\psi )},h^i(x,d)=0,$$

(46)

which can be regarded as a plane wave in ${\mathbb{R}}^2$ with the wave number $\kappa$ and incident direction $d=(cos\psi ,sin\psi )\in \mathbb{S}$ (the unit circle in ${\mathbb{R}}^2$). We denote by $(e^s(x,d),h^s(x,d))$ the scattered field and by $(e^{\infty }(\widehat{x},d),h^{\infty }(\widehat{x},d))$ the corresponding far-field pattern, which have the following asymptotic behaviors:

$$e^s(x,d)=\frac{e^{i\kappa \left|x\right|}}{\sqrt{\left|x\right|}}\left\{e^{\infty }(\widehat{x},d)+o\left(\frac{1}{\left|x\right|}\right)\right\},h^s(x,d)=\frac{e^{i\kappa \left|x\right|}}{\sqrt{\left|x\right|}}\left\{h^{\infty }(\widehat{x},d)+o\left(\frac{1}{\left|x\right|}\right)\right\}$$

which are uniform in all directions $\widehat{x}\in \mathbb{S}$.

It is proved in [36,37] that the problem (42) has a unique solution. The unique determination of the cylinder utilizing only the electric or magnetic far-field data is given in [35]. Moreover, the linear sampling method is established in [38] to reconstruct the scattering object.

4.2. An Inverse Scattering Problem

Next, we describe the inverse problem by using the phaseless far-field pattern $|e^{\infty }|$ to determine the cross-section O. We add an extra artificial reference ball $B_3$ into the scattering system (42) such that $O\cap B_3=\varnothing$ and impose the perfect boundary condition (43) or (44) on $\partial B_3$. The scattered field and its far-field pattern caused by $O\cup B_3$ for the incident plane wave (46) are denoted by $(e_{O\cup B_3}^s(x,d),h_{O\cup B_3}^s(x,d))$ and $(e_{O\cup B_3}^{\infty }(\widehat{x},d),h_{O\cup B_3}^{\infty }(\widehat{x},d))$, respectively. We use $(e_{O\cup B_3}^s(x,z_0),h_{O\cup B_3}^s(x,z_0))$ and $(e_{O\cup B_3}^{\infty }(\widehat{x},z_0),h_{O\cup B_3}^{\infty }(\widehat{x},z_0))$ to represent the scattered field and far-field pattern, respectively, for the incident point source

$$e^i(x,z_0)=\Phi (x,z_0)=\frac{i}{4}{H}_0^{\left(1\right)}\left(\kappa \right|x-z_0\left|\right),h^i(x,z_0)=0,x\in {\mathbb{R}}^2,x\ne z_0$$

with the fixed point $z_0\in {\mathbb{R}}^2\setminus (\overline{O}\cup {\overline{B}}_3)$. Here, $H_0^{\left(1\right)}$ is the Hankel function of the first kind of order zero.

The inverse problem with the phaseless far-field data is introduced as follows.

Problem 3.

Given the phaseless electric far-field data

$$\left\{\begin{array}{c}\left\{|e_{O\cup B_3}^{\infty }(\widehat{x},z_0)|:\widehat{x}\in \mathbb{S}\right\},\hfill \\ \left\{|e_{O\cup B_3}^{\infty }(\widehat{x},d)|:\widehat{x},d\in \mathbb{S}\right\},\hfill \\ \left\{|e_{O\cup B_3}^{\infty }(\widehat{x},d)+e_{O\cup B_3}^{\infty }(\widehat{x},z_0)|:\widehat{x},d\in \mathbb{S}\right\}\hfill \end{array}\right.$$

(47)

for a fixed point $z_0\in {\mathbb{R}}^2\setminus (\overline{O}\cup {\overline{B}}_3)$, identify the shape and location of O.

The translation invariance for the electromagnetic scattering at oblique incidence is given as follows.

Lemma 5.

Let $O_l:=\{x+l:x\in O\}$ be a shifted domain of O with a fixed vector $l\in {\mathbb{R}}^2$, and denote by $e_l^{\infty }$ the corresponding electric far-field pattern for the scattering of the TM polarized incident plane wave at oblique incidence by the cylinder O. Then, we have

$$e_l^{\infty }\left(\widehat{x}\right)=e^{i\kappa (d-\widehat{x})·l}{e}^{\infty }\left(\widehat{x}\right),\widehat{x},d\in \mathbb{S}.$$

(48)

Proof. 

Let $(e_l^s,h_l^s)$ be the unique solution to the problem (42) for ${\Omega }_l$. We claim that

$$\left\{\begin{array}{c}{e}_l^s\left(y\right)=e^{i\kappa d·l}{e}^s(y-l),y\in {\mathbb{R}}^2\setminus {\overline{O}}_l,\\ h_l^s\left(y\right)=e^{i\kappa d·l}{h}^s(y-l),y\in {\mathbb{R}}^2\setminus {\overline{O}}_l.\end{array}\right.$$

(49)

Since $(e_l^s,h_l^s)$ expressed by (49) satisfies the Helmholtz equations in ${\mathbb{R}}^2\setminus {\overline{O}}_l$ and the Sommerfeld radiation condition, we only need to verify the boundary condition (45). Indeed, it holds from the expressions in (49) and the boundary conditions satisfied by $(e^s\left(y\right),h^s\left(y\right))$ that

$$\begin{array}{ccc}& & {\displaystyle \mu \omega \frac{\partial h_l^s}{\partial {\nu }_l}\left(y\right)+\sigma \frac{\partial e_l^s}{\partial {\tau }_l}\left(y\right)+i\lambda {\kappa }^2{h}_l^s\left(y\right)}\hfill \\ & =& {\displaystyle e^{ikd·l}\left(\mu \omega \nu (y-l)·\nabla h^s(y-l)+\sigma \tau (y-l)·\nabla e^s(y-l)+i\lambda {\kappa }^2{h}^s(y-l)\right)}\hfill \\ & =& -e^{ikd·l}\left(\sigma \tau (y-l)·\nabla e^i(y-l)\right)\hfill \\ & =& -e^{ikd·l}\frac{sin\phi }{\sqrt{ϵ}}\left(\sigma \tau (y-l)·\nabla e^{i\kappa (y-l)·d}\right)\hfill \\ & =& -\sigma \frac{\partial e^i}{\partial {\tau }_l}\left(y\right),y\in \partial O_l\hfill \end{array}$$

and

$$\begin{array}{ccc}& & {\displaystyle \lambda ϵ\omega \frac{\partial e_l^s}{\partial {\nu }_l}\left(y\right)-\lambda \sigma \frac{\partial h_l^s}{\partial {\tau }_l}\left(y\right)+i{\kappa }^2{e}_l^s\left(y\right)}\hfill \\ & =& e^{ikd·l}\left(\lambda ϵ\omega \nu (y-l)·\nabla e^s(y-l)-\lambda \sigma \tau (y-l)·\nabla h^s(y-l)+i{\kappa }^2{e}^s(y-l)\right)\hfill \\ & =& -e^{ikd·l}\left(\lambda ϵ\omega \nu (y-l)·\nabla e^i(y-l)+i{\kappa }^2{e}^i(y-l)\right)\hfill \\ & =& -e^{ikd·l}\frac{sin\phi }{\sqrt{ϵ}}\left(\lambda ϵ\omega \nu (y-l)·\nabla e^{i\kappa (y-l)·d}+i{\kappa }^2{e}^{i\kappa (y-l)·d}\right)\hfill \\ & =& {\displaystyle -\lambda ϵ\omega \frac{\partial e^i}{\partial {\nu }_l}\left(y\right)-i{\kappa }^2{e}^i\left(y\right),y\in \partial O_l.}\hfill \end{array}$$

Note that we can obtain from (49) that

$$\frac{\partial e_l^s}{\partial {\nu }_l}\left(y\right)=e^{ikd·l}\frac{\partial e^s}{\partial {\nu }_l}(y-l),y\in \partial O_l.$$

Thus, we have from the representation of the far-field pattern that

$$\begin{array}{ccc}\hfill e_l^{\infty }\left(\widehat{x}\right)& =& \eta {\int }_{\partial O_l}\left\{e_l^s\left(y\right)\frac{\partial e^{-i\kappa \widehat{x}·y}}{\partial {\nu }_l\left(y\right)}-\frac{\partial e_l^s}{\partial {\nu }_l}\left(y\right)e^{-i\kappa \widehat{x}·y}\right\}ds\hfill \\ & =& e^{ikd·l}\eta {\int }_{\partial O_l}\left\{e^s(y-l)\frac{\partial e^{-i\kappa \widehat{x}·y}}{\partial {\nu }_l\left(y\right)}-\frac{\partial e^s}{\partial {\nu }_l}(y-l)e^{-i\kappa \widehat{x}·y}\right\}ds\hfill \\ & =& e^{ikd·l}\eta {\int }_{\partial O}\left\{e^s\left(y\right)\frac{\partial e^{-i\kappa \widehat{x}·(y+l)}}{\partial \nu \left(y\right)}-\frac{\partial e^s}{\partial \nu }\left(y\right)e^{-i\kappa \widehat{x}·(y+l)}\right\}ds\hfill \\ & =& e^{ik(d-\widehat{x})·l}\eta {\int }_{\partial O}\left\{e^s\left(y\right)\frac{\partial e^{-i\kappa \widehat{x}·y}}{\partial \nu \left(y\right)}-\frac{\partial e^s}{\partial \nu }\left(y\right)e^{-i\kappa \widehat{x}·y}\right\}ds\hfill \\ & =& e^{ik(d-\widehat{x})·l}{e}^{\infty }\left(\widehat{x}\right),\widehat{x},d\in \mathbb{S},\hfill \end{array}$$

where $\eta =e^{i\pi /4}/\sqrt{8\pi \kappa }$. The proof is completed. □

We next show the following reciprocity relation in electromagnetic scattering at oblique incidence.

Lemma 6.

The electric far-field pattern for the scattering of the TM polarized incident plane wave by an impedance cylinder at oblique incidence satisfies the reciprocity relation

$$e^{\infty }(\widehat{x},d)=e^{\infty }(-d,-\widehat{x}),\widehat{x},d\in \mathbb{S}.$$

(50)

Proof. 

From Green’s formula and the Helmholtz equation for the incident fields in O, we find

$${\int }_{\partial O}\left\{e^i(·,d)\frac{\partial }{\partial \nu }{e}^i(·,-\widehat{x})-\frac{\partial }{\partial \nu }{e}^i(·,d)e^i(·,-\widehat{x})\right\}ds=0.$$

By Green’s formula, the Helmholtz equation, and the radiation condition for the scattered fields in ${\mathbb{R}}^2\setminus \overline{O}$, we deduce that

$${\int }_{\partial O}\left\{e^s(·,d)\frac{\partial }{\partial \nu }{e}^s(·,-\widehat{x})-\frac{\partial }{\partial \nu }{e}^s(·,d)e^s(·,-\widehat{x})\right\}ds=0.$$

The well-known representation of the far-field reads

$$\frac{1}{\eta }{e}^{\infty }(\widehat{x},d)={\int }_{\partial O}\left\{e^s(·,d)\frac{\partial }{\partial \nu }{e}^i(·,-\widehat{x})-\frac{\partial }{\partial \nu }{e}^s(·,d)e^i(·,-\widehat{x})\right\}ds,$$

and interchanging the roles of d and $-\widehat{x}$ gives

$$\frac{1}{\eta }{e}^{\infty }(-d,-\widehat{x})={\int }_{\partial O}\left\{e^s(·,-\widehat{x})\frac{\partial }{\partial \nu }{e}^i(·,d)-\frac{\partial }{\partial \nu }{e}^s(·,-\widehat{x})e^i(·,d)\right\}ds.$$

The above four equations and the boundary condition (45) show us that

$$\begin{array}{ccc}\hfill & & \frac{1}{\eta }\{e^{\infty }(\widehat{x},d)-e^{\infty }(-d,-\widehat{x})\}\hfill \\ & =& {\int }_{\partial O}\left\{e(·,d)\frac{\partial }{\partial \nu }e(·,-\widehat{x})-\frac{\partial }{\partial \nu }e(·,d)e(·,-\widehat{x})\right\}ds\hfill \\ & =& {\int }_{\partial O}\left\{e(·,d)\left(\frac{\sigma }{ϵ\omega }\frac{\partial }{\partial \tau }{h}^s(·,-\widehat{x})-\frac{i{\kappa }^2}{\lambda ϵ\omega }e(·,-\widehat{x})\right)-\left(\frac{\sigma }{ϵ\omega }\frac{\partial }{\partial \tau }{h}^s(·,d)-\frac{i{\kappa }^2}{\lambda ϵ\omega }e(·,d)\right)e(·,-\widehat{x})\right\}ds\hfill \\ & =& \frac{\sigma }{ϵ\omega }{\int }_{\partial O}\left\{e(·,d)\frac{\partial }{\partial \tau }{h}^s(·,-\widehat{x})-\frac{\partial }{\partial \tau }{h}^s(·,d)e(·,-\widehat{x})\right\}ds\hfill \\ & =& \frac{\sigma }{ϵ\omega }{\int }_{\partial O}\left\{h^s(·,d)\frac{\partial }{\partial \tau }e(·,-\widehat{x})-\frac{\partial }{\partial \tau }e(·,d)h^s(·,-\widehat{x})\right\}ds\hfill \\ & =& \frac{-\mu }{\omega }{\int }_{\partial O}\left\{h^s(·,d)\left(\frac{\partial }{\partial \nu }{h}^s(·,-\widehat{x})+\frac{i\lambda {\kappa }^2}{\mu ϵ}{h}^s(·,-\widehat{x})\right)-\left(\frac{\partial }{\partial \nu }{h}^s(·,d)+\frac{i\lambda {\kappa }^2}{\mu ϵ}{h}^s(·,d)\right)h^s(·,-\widehat{x})\right\}ds\hfill \\ & =& \frac{-\mu }{\omega }{\int }_{\partial O}\left\{h^s(·,d)\frac{\partial }{\partial \nu }{h}^s(·,-\widehat{x})-\frac{\partial }{\partial \nu }{h}^s(·,d)h^s(·,-\widehat{x})\right\}ds\hfill \\ & =& 0,\hfill \end{array}$$

(51)

where the first equality is obtained from the results of last two equations preceding (51), the second and fifth ones are due to the boundary conditions in (45), the fourth equality is owing to the integration by parts, and the last one is derived by Green’s formula, the Helmholtz equation, and the radiation condition. The proof is complete. □

Let $O_1$ and $O_2$ be the cross-sections of two cylinders. The scattered field and its far-field pattern caused by $O_j\cup B_3$ for the incident plane wave (46) are denoted by $(e_{O_j\cup B_3}^s(x,d),h_{O_j\cup B_3}^s(x,d))$ and $(e_{O_j\cup B_3}^{\infty }(\widehat{x},d),h_{O_j\cup B_3}^{\infty }(\widehat{x},d))$ ($j=1,2$), respectively. We denote by $(e_{O_j\cup B_3}^s(x,z_0),h_{O_j\cup B_3}^s(x,z_0))$ and $(e_{O_j\cup B_3}^{\infty }(\widehat{x},z_0),h_{O_j\cup B_3}^{\infty }(\widehat{x},z_0))$, the scattered field and its far-field pattern, respectively, corresponding to the incident point source located at $z_0\in {\mathbb{R}}^2\setminus ({\overline{O}}_j\cup {\overline{B}}_3)$ ($j=1,2$). Then, the main result for the uniqueness is shown by the following theorem.

Theorem 3.

Suppose that κ is neither a Neumann eigenvalue nor a Dirichlet eigenvalue for $B_3$. Further assume that there exists a ball $B_R$ with the center at the origin and the radius R such that ${\overline{O}}_1\cup {\overline{O}}_2\subset B_R$ and ${\overline{B}}_3\cap {\overline{B}}_R=\varnothing$. If the far-field patterns satisfy

$$\left\{\begin{array}{c}|e_{O_1\cup B_3}^{\infty }(\widehat{x},z_0)|=|e_{O_2\cup B_3}^{\infty }(\widehat{x},z_0)|,\hfill \\ |e_{O_1\cup B_3}^{\infty }(\widehat{x},d)|=|e_{O_2\cup B_3}^{\infty }(\widehat{x},d)|,\hfill \\ |e_{O_1\cup B_3}^{\infty }(\widehat{x},d)+e_{O_1\cup B_3}^{\infty }(\widehat{x},z_0)\left)\right|=|e_{O_2\cup B_3}^{\infty }(\widehat{x},d)+e_{O_2\cup B_3}^{\infty }(\widehat{x},z_0)\left)\right|\hfill \end{array}\right.$$

(52)

for all $\widehat{x},d\in \mathbb{S}$ and a fixed point $z_0\in {\mathbb{R}}^2\setminus ({\overline{O}}_1\cup {\overline{O}}_2\cup {\overline{B}}_3)$, then we have $O_1=O_2$.

Proof. 

From the first two equations of (52), we can define $s(\widehat{x},z_0):=\left|e_{O_j\cup B_3}^{\infty }(\widehat{x},z_0)\right|$, $r(\widehat{x},d):=e_{O_j\cup B_3}^{\infty }(\widehat{x},d)$, $j=1,2$. Then, we arrive at

$$\begin{array}{ccc}& & e_{O_j\cup B_3}^{\infty }(\widehat{x},z_0)=s(\widehat{x},z_0)e^{i{\beta }_j(\widehat{x},z_0)},\hfill \\ & & e_{O_j\cup B_3}^{\infty }(\widehat{x},d)=r(\widehat{x},d)e^{i{\alpha }_j(\widehat{x},d)},\hfill \end{array}$$

where ${\beta }_j(\widehat{x},z_0)$ and ${\alpha }_j(\widehat{x},d)$ are real-valued functions, $j=1,2$. The third equation in (52) and the analysis analogous to Theorem 1 yield the existence of the open sets $U_1,U_2\subset \mathbb{S}$ such that

$${\alpha }_1(\widehat{x},d)-{\alpha }_2(\widehat{x},d)={\beta }_1(\widehat{x},z_0)-{\beta }_2(\widehat{x},z_0)+2n\pi$$

(53)

or

$${\alpha }_1(\widehat{x},d)+{\alpha }_2(\widehat{x},d)={\beta }_1(\widehat{x},z_0)+{\beta }_2(\widehat{x},z_0)+2n\pi$$

(54)

for all $\widehat{x}\in U_1$, $d\in U_2$ and $n\in \mathbb{Z}$.

If (53) holds, then the procedure similar to Theorem 1 deduces

$$e_{O_1\cup B_3}^{\infty }(\widehat{x},d)=e^{i\delta (-d)}{e}_{O_2\cup B_3}^{\infty }(\widehat{x},d),\forall -d\in U_1,\widehat{x}\in \mathbb{S},$$

where the reciprocity relation (50) is used, and $\delta \left(\widehat{x}\right):={\beta }_1(\widehat{x},z_0)-{\beta }_2(\widehat{x},z_0),\widehat{x}\in U_1$. Then, Rellich’s lemma shows us that

$$e_{O_1\cup B_3}^s(x,d)=e^{i\delta (-d)}{e}_{O_2\cup B_3}^s(x,d),\forall -d\in U_1,x\in {\mathbb{R}}^2\setminus ({\overline{O}}_1\cup {\overline{O}}_2\cup {\overline{B}}_3),$$

(55)

which leads to

$$\left\{\begin{array}{cc}{e}_{O_1\cup B_3}^s(x,d)=e^{i\delta (-d)}{e}_{O_2\cup B_3}^s(x,d),\hfill & \\ \frac{\partial }{\partial \nu }\left[e_{O_1\cup B_3}^s(x,d)\right]=e^{i\delta (-d)}\frac{\partial }{\partial \nu }\left[e_{O_2\cup B_3}^s(x,d)\right],\hfill & \\ \frac{\partial }{\partial \tau }\left[e_{O_1\cup B_3}^s(x,d)\right]=e^{i\delta (-d)}\frac{\partial }{\partial \tau }\left[e_{O_2\cup B_3}^s(x,d)\right]\hfill \end{array}\right.$$

(56)

for all $-d\in U_1,x\in \partial B_3$.

On the boundary $\partial B_3$, it holds that

$$(1-e^{i\delta (-d)})e^i(x,d)=-[e_{O_1\cup B_3}^s(x,d)-e^{i\delta (-d)}{e}_{O_2\cup B_3}^s(x,d)]=0.$$

If we have the boundary condition (43), then $(1-e^{i\delta (-d)})e^i(x,d)$ satisfies the Helmholtz equation in $B_3$ with the homogeneous Dirichlet boundary condition. The assumption on $\kappa$ indicates that

$$(1-e^{i\delta (-d)})e^i(x,d)=0\mathrm{in}{B}_3,$$

(57)

which leads to $e^{i\delta (-d)}=1$, since $e^i(x,d)\ne 0$.

If we have the boundary condition (44) on $\partial B_3$, then it follows that $\frac{\partial }{\partial \nu }\left[e_{O_j\cup B_3}(x,d)\right]=0$ on $\partial B_3$, since $h_{O_j\cup B_3}^s(x,d)=0$ on $\partial B_3$ implies $\frac{\partial }{\partial \tau }\left[h_{O_j\cup B_3}^s(x,d)\right]=0$ on $\partial B_3$. From (56), we have

$$(1-e^{i\delta (-d)})\frac{\partial }{\partial \nu }\left[e^i(x,d)\right]=-\left\{\frac{\partial }{\partial \nu }\left[e_{O_1\cup B_3}^s(x,d)\right]-e^{i\delta (-d)}\frac{\partial }{\partial \nu }\left[e_{O_2\cup B_3}^s(x,d)\right]\right\}=0.$$

In addition, $(1-e^{i\delta (-d)})\frac{\partial }{\partial \nu }\left[e^i(x,d)\right]$ satisfies the Helmholtz equation in $B_3$. We deduce that this function equals zero by the assumption that $\kappa$ is not a Neumann eigenvalue, which also results in $e^{i\delta (-d)}=1$.

As a result, we obtain for both boundary conditions that

$$e_{O_1\cup B_3}^s(x,d)=e_{O_2\cup B_3}^s(x,d),\forall -d\in U_1,x\in {\mathbb{R}}^2\setminus ({\overline{O}}_1\cup {\overline{O}}_2\cup {\overline{B}}_3).$$

The analyticity of $e_{O_1\cup B_3}^s(x,d)$ implies that the above equality holds for all $d\in \mathbb{S},$ $x\in {\mathbb{R}}^2\setminus ({\overline{O}}_1\cup {\overline{O}}_2\cup {\overline{B}}_3)$. According to the uniqueness result proved in [35] we have $O_1=O_2$. Similarly, we can prove that (54) also does not hold. The proof is complete. □