Unique Determination of a Planar Screen in Electromagnetic Inverse Scattering

Abstract

The target of our research is the object being a highly conducting thin plate or a flat screen. We especially focus on the question of when a single measurement uniquely determines an object. By this, we mean that we have one fixed transmitted wave and the resulting scattered field is measured for all directions in the far field. Such measurements are called passive, since there is no need to move the transmitter after its position has been fixed. We show that the far field of a scattered electromagnetic field corresponding to a single incoming plane wave always uniquely determines a bounded super-conductive planar screen. This generalises a previous acoustic result.

1. Introduction

The study of wave scattering from highly conductive objects is rooted in the field of antenna theory. This interest began with a competition initiated by the Prussian Academy in 1879 to demonstrate the existence or non-existence of electromagnetic waves. Maxwell’s theory [1] predicted the existence of such waves 15 years earlier. In 1882, the competition was won by Heinrich Hertz, who constructed a dipole antenna that was able to radiate and measure EM waves, thereby confirming Maxwell’s prediction.

Inverse scattering problems involve determining the properties of an object, such as its shape, size, and composition, from the produced scattered field when it is illuminated with, say, electromagnetic or acoustic waves. A well-studied problem is to determine the shape of an object by analysing the far field of the scattered wave. This is a problem that involves both mathematics and numerical techniques, and it has many practical applications. An overview of this topic can be found in the book [2] by Colton and Kress.

Recently, there has been an increasing number of publications about the inverse scattering problem with fewer measurements. Especially fascinating is the question of when just a single measurement determines an object uniquely. By this, we mean that we have one fixed transmitted wave and the resulting scattered field is measured for all directions in the far field. Such measurements are called passive, since you do not need to move the transmitter after its position has been fixed. This is exactly the target of the research here, the object being a highly conducting thin plate, i.e., a flat screen.

In [3], we considered the problem of fixed frequency acoustic scattering from a sound-soft flat screen. The main result of that paper is that the far field produced by any single incident wave determines the precise shape of the screen, given that it is not anti-symmetric with respect to the plane. Our current work is the generalisation of the result of [3] to Maxwell’s equations. This shape determination problem is known in the literature [2] as Schiffer’s problem. The first uniqueness result for the case of the Dirichlet problem was presented by Schiffer in 1967 [4]. The Schiffer’s uniqueness theorem for the inverse Dirichlet problem assumes a lot of information about the waves as it is using an infinite number of incident frequencies. After Schiffer’s uniqueness result for sound-soft obstacles by countably many incident plane waves [2,4], extensive research in this direction has been conducted. Notable contributions include uniqueness results for general domain [5,6,7,8,9,10,11], polyhedral scatterers [12,13], for the ball or disc [14,15], and for smooth planar curves [16,17,18,19].

Important results on the inverse electromagnetic scattering problem in the TE polarisation case was conducted in [20]. They demonstrated that in the special case of a rectangular penetrable scatterer, it can be uniquely determined just by measuring the electric far-field pattern for a single incoming wave.

In [21], the authors studied uniqueness of an inverse acoustic obstacle scattering problem, demonstrating the unique determination of sound-hard and sound-soft polyhedral scatterers in ${\mathbb{R}}^n$. More precisely, they proved that N far-field measurements corresponding to N incident plane waves given by a fixed wave number and N linearly independent incident directions uniquely determine the obstacle. A few of the uniqueness results in inverse electromagnetic and acoustic obstacle scattering problems were obtained by Liu and Zou [7]. They emphasize recent developments in the unique determination of a general polyhedral scatterer using far-field data corresponding to one or several incident fields. For other recent results in time-harmonic inverse EM-scattering, see the short review by Rainer Kress [6].

A particular case in [22] gives the unique determination of a flat screen by a single incident plane-wave measurement with robin boundary condition. We also wish to mention the article [23], in which the authors demonstrate that the far-field pattern corresponding to one incident plane wave uniquely identifies a sound-soft polyhedral scatterer.

However, they consider the polyhedral sound-soft acoustic problem and not the electromagnetic problem.

In addition, ref. [24] has established a reflection principle for the time-harmonic Maxwell equations. They derive a uniqueness result for the inverse electromagnetic scattering problem for a polyhedral scatterer. The scatterers considered can exhibit a wide variety; for instance, they might comprise a finite number of compact polyhedral shapes along with a finite number of portions from two-dimensional surfaces. Another important work is [25], where the authors consider an obstacle composed of finite solid polyhedra, and they prove that it can be uniquely characterised by the far-field pattern associated with a single incident electromagnetic plane wave. To our knowledge, there is no proof for the unique determination of a planar screen by one far-field pattern without restrictive a priori assumptions such as the assumptioned polyhedron shape. The research in [26] comes very close to ours. There, the obstacle can be any Lipschitz domain provided that its boundary is not an analytic manifold. But that work does not consider screens, which is our priority.

The main motivation for this study comes from antenna theory [27,28,29]. A typical (radar) antenna consists of a configuration of planar screens attached to a common stem, and understanding both the direct and inverse scattering of electromagnetic waves from such structures is a natural and important problem. Our study is the natural first step in analysing the inverse problem of recovering the shape of the antenna using exactly one incoming wave.

The goal of this work is to prove the unique determination of the unknown screen and supporting hyperplane corresponding to a single measurement of the far field. The proof follows from the representation formula for the exterior solution of Maxwell’s equations. The main idea of our paper is to reduce the scattering problem to an integral equation on the screen. Here, the integral operator is the analogue of the double-curl layer potential on the screen, and has as its unknown the jump of the tangential component of the magnetic field. The inverse problem is then solved by showing that, first of all, the incoming plane wave uniquely determines the solution to this integral equation, and secondly, that when the tangential component of the incoming plane wave does not vanish on the screen, the support of the solution is full, i.e., the whole screen. More precisely, our main results state that a single far field corresponding to an incoming plane wave uniquely determines a planar screen, as follows:

Theorem 1.

Let S be a $C^2$–screen contained in a supporting hyperplane L and let

$$E(\theta ;p,q)={\mu }^{1/2}(p\times \theta )e^{ik\langle \theta ,x\rangle },H(\theta ;p,q)={\epsilon }^{1/2}(q\times \theta )e^{ik\langle \theta ,x\rangle }$$

describe the EM-plane wave with wavenumber $k=\omega \sqrt{\epsilon \mu }$, propagation direction θ and polarizations p and q. Let $(e_{\mathrm{sc}},h_{\mathrm{sc}})$ be the electromagnetic wave scattered by S and assume that it does not identically vanish. Then, the non-vanishing far-field pattern of $(e_{\mathrm{sc}},h_{\mathrm{sc}})$ uniquely determines both the supporting hyperplane L and the screen S if neither p or q is parallel to θ.

Remark 1.

As will be clear from the proofs, the scattered field will vanish only if the electric polarization $p\times \theta$ is parallel to the screen.

The plan of this paper is as follows: Analysis of mathematical proof and the concept for the direct scattering problem of EM waves are discussed in Section 1. We start by giving a precise definition of a planar screen and discuss time-harmonic Maxwell’s equations in the exterior of the screen. Representation theorem for the fundamental solution of vector Helmholtz equation is also analysed here. In Section 2, the solution of the inverse problem is presented, including also the unique determination of the supporting hyperplane.

2. Scattering from a Perfectly Conducting Screen

2.1. Formal Definitions

Definition 1.

A planar $C^k$–screen, $k=1,\dots ,\infty$, in ${\mathbb{R}}^3$ is a compact, connected $C^k$–submanifold of an affine hyperplane $L\subset {\mathbb{R}}^3$. The affine hyperplane L is called the supporting hyperplane of S. In the sequel we also fix a globally defined unit normal vector field on S and denote it by ν. Also, the boundary of S as a submanifold of L is denoted by $\partial S$.

Consider the time-harmonic Maxwell’s equations in the exterior of a screen S:

$$\nabla \times E=i\omega \mu H,\nabla \times H=-i\omega \epsilon E\mathrm{in}{\mathbb{R}}^3\setminus S.$$

(1)

Here, the magnetic permeability $\mu$ and the dielectricity $\epsilon$ are known positive constants and we assume also that the angular frequency $\omega >0$ is known.

Given an incident filed $(E_0,H_0)$, i.e., a solution of

$$\nabla \times E_0=i\omega \mu H_0,\nabla \times H_0=-i\omega \epsilon E_0\mathrm{in}{\mathbb{R}}^3,$$

the corresponding scattered field $(E_{sc},H_{sc})$ is (formally) defined by demanding that $(E,H)$, where $E=E_0+E_{sc}$ and $H=H_0+H_{sc}$ satisfy (1), and the scattered field is outgoing in the sense that it satisfies the Silver–Müller radiation conditions,

$$\widehat{r}\times E_{sc}+\sqrt{\frac{\epsilon }{\mu }}{H}_{sc}={o\left(\right|x|}^{-1}),\widehat{r}\times H_{sc}-\sqrt{\frac{\mu }{\epsilon }}{E}_{sc}={o\left(\right|x|}^{-1}),\mathrm{as}\left|x\right|\to \infty .$$

(2)

Here, $\widehat{r}=x/\left|x\right|$. If we further assume that the screen is perfectly conducting, i.e., the total field vanishes on S, this leads to the direct scattering problem for the perfectly conducting screen S: for a given incident field, $(E_0,H_0)$ show that there is a unique scattered field $(E_{sc},H_{sc})$ s.t.

$$\nabla \times E_{sc}=i\omega \mu H_{sc},\nabla \times H_{sc}=-i\omega \epsilon E_{sc}\mathrm{in}{\mathbb{R}}^3\setminus S$$

(3)

satisfying (2) and such that

$$\nu \times (E_{sc}+E_0)=0\mathrm{on}S.$$

(4)

Note that we have not specified in what sense the boundary value (4) holds. This will depend on the spaces where we look for solutions and the availability of suitable trace theorems.

2.2. Representation Theorems

Assume for now that $S\subset {\mathbb{R}}^3$ is a $C^2$ screen. Denote by $C_S^k$ the closed subspace of $C^k({\mathbb{R}}^3\setminus S)$ consisting of those $u\in C^k({\mathbb{R}}^3\setminus S)$ such that u and all its derivatives up to order k have normal limits on S, i.e., for all $\left|\alpha \right|\le k$, there are limits

$$\underset{\delta \to +0}{lim}{\partial }_x^{\alpha }u(x\pm \delta \nu \left(x\right))=u_{\alpha }^{\pm }\left(x\right),x\in S,$$

where $u_{\alpha }^{\pm }\in C\left(S\right)$. Note that we do not assume that limits $u_{\alpha }^{+}$ and $u_{\alpha }^{-}$ coincide on S.

Proposition 1.

Assume $(e,h)\in {\left(C_S^1\right)}^3\times {\left(C_S^1\right)}^3$ solves

$$\nabla \times e=i\omega \mu h,\nabla \times h=-i\omega \epsilon e\mathrm{in}{\mathbb{R}}^3\setminus S,$$

and the Silver–Müller radiation condition

$$\widehat{r}\times e+\sqrt{\frac{\epsilon }{\mu }}h={o\left(\right|x|}^{-1}),\widehat{r}\times h-\sqrt{\frac{\mu }{\epsilon }}e={o\left(\right|x|}^{-1}),\mathit{as}\left|x\right|\to \infty ,\widehat{r}=x/\left|x\right|>0.$$

Then, for all $x\in {\mathbb{R}}^3$,

$$\begin{array}{ccc}\hfill e\left(x\right)& =& \nabla \times {\int }_S\mathsf{\Phi }(x-y)(\nu \times \{e^{+}\left(y\right)-e^{-}\left(y\right)\})ds\left(y\right)\hfill \\ & -& \frac{1}{i\omega \epsilon }{(\nabla \times )}^2{\int }_S\mathsf{\Phi }(x-y)(\nu \times \{h^{+}\left(y\right)-h^{-}\left(y\right)\})ds\left(y\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill h\left(x\right)& =& \nabla \times {\int }_S\mathsf{\Phi }(x-y)(\nu \times \{h^{+}\left(y\right)-h^{-}\left(y\right)\})ds\left(y\right)\hfill \\ & +& \frac{1}{i\omega \mu }{(\nabla \times )}^2{\int }_S\mathsf{\Phi }(x-y)(\nu \times \{e^{+}\left(y\right)-e^{-}\left(y\right)\})ds\left(y\right).\hfill \end{array}$$

Proof. 

For $\delta >0$, let ${𝒪}_{\delta }=\{x\pm t\nu \left(x\right);x\in S,0\le t<\delta \}$ be a collar neighbourhood of S. For sufficiently small $\delta$, this is a bounded, piecewise analytic domain. The standard representation formulas (see for example [30]) give that for all $x\in {\mathbb{R}}^3\setminus {\overline{𝒪}}_{\delta }$, we have

$$e\left(x\right)=\nabla \times {\int }_{\partial {𝒪}_{\delta }}\mathsf{\Phi }(x-y)({\nu }_{\delta }\left(y\right)\times e\left(y\right))ds\left(y\right)-\frac{1}{i\omega \epsilon }{(\nabla \times )}^2{\int }_{\partial {𝒪}_{\delta }}\mathsf{\Phi }(x-y)({\nu }_{\delta }\left(y\right)\times h\left(y\right))ds\left(y\right)$$

and

$$h\left(x\right)=\nabla \times {\int }_{\partial {𝒪}_{\delta }}\mathsf{\Phi }(x-y)({\nu }_{\delta }\left(y\right)\times h\left(y\right))ds\left(y\right)+\frac{1}{i\omega \mu }{(\nabla \times )}^2{\int }_{\partial {𝒪}_{\delta }}\mathsf{\Phi }(x-y)({\nu }_{\delta }\left(y\right)\times e\left(y\right))ds\left(y\right).$$

Here, $\mathsf{\Phi }$ is the outgoing fundamental solution of the Helmholtz operator $\Delta +k^2$ and ${\nu }_{\delta }$ is the exterior unit normal of $\partial {𝒪}_{\delta }$. Then, as $\delta \to +0$,

$$\begin{array}{ccc}\hfill e\left(x\right)& =& \nabla \times {\int }_S\mathsf{\Phi }(x-y)(\nu \times \{e^{+}\left(y\right)-e^{-}\left(y\right)\})ds\left(y\right)\hfill \\ & -& \frac{1}{i\omega \epsilon }{(\nabla \times )}^2{\int }_S\mathsf{\Phi }(x-y)(\nu \times \{h^{+}\left(y\right)-h^{-}\left(y\right)\})ds\left(y\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill h\left(x\right)& =& \nabla \times {\int }_S\mathsf{\Phi }(x-y)(\nu \times \{h^{+}\left(y\right)-h^{-}\left(y\right)\})ds\left(y\right)\hfill \\ & +& \frac{1}{i\omega \mu }{(\nabla \times )}^2{\int }_S\mathsf{\Phi }(x-y)(\nu \times \{e^{+}\left(y\right)-e^{-}\left(y\right)\})ds\left(y\right),\hfill \end{array}$$

as claimed. □

In what follows, we will denote the jumps of a function (or a vector field) u across S by $\left[u\right]$, i.e.,

$$\left[u\right]\left(y\right)=u^{+}\left(y\right)-u^{-}\left(y\right),y\in S.$$

2.3. EM-Plane Waves and Far-Field Patterns

Let $\theta ,p\in S^2$ and denote $q=p\times \theta$. We call the field

$$E(\theta ;p,q)={\mu }^{1/2}(p\times \theta )e^{ik\langle \theta ,x\rangle },H(\theta ;p,q)={\epsilon }^{1/2}(q\times \theta )e^{ik\langle \theta ,x\rangle }$$

the EM-plane wave with wavenumber k, propagation direction θ and polarizations p and q. It is easy to see that these fields satisfy the time-harmonic Maxwell’s equations

$$\nabla \times E(\theta ;p,q)=i\omega \mu H(\theta ;p,q),\nabla \times H(\theta ;p,q)=-i\omega \epsilon E(\theta ;p,q).$$

when $k^2=\epsilon \mu$. Since the scalar components of the scattered electric and magnetic fields are solutions of the Helmholz equation $(\Delta +k^2)u=0$ satisfying the Sommerfeld radiation condition, they have representations

$$E_{}\left(x\right)=\frac{E^{\infty }\left(x\right)}{\left|x\right|}+{o\left(\right|x|}^{-1}),H_{}\left(x\right)=\frac{H^{\infty }\left(x\right)}{\left|x\right|}+{o\left(\right|x|}^{-1})$$

where $E^{\infty }$ and $H^{\infty }$ are the electric and magnetic far-field patterns. If the initial field is the EM-plane wave $\left(E\right(\theta ;p,q),H(\theta ;p,q\left)\right)$, we denote the corresponding far-field patterns by $E^{\infty }(\theta ;p,q)$ and $H^{\infty }(\theta ;p,q)$.

2.4. Relevant Sobolev Spaces

Let (see [31,32]) $L_{\mathrm{loc}}^2({\mathbb{R}}^3\setminus S)$ be the space of measurable functions that are square integrable on compact subsets of ${\mathbb{R}}^3\setminus S$. This becomes a Fréchet space when equipped with semi-norms

$${\parallel f\parallel }_R={\parallel f\parallel }_{L^2({\mathbb{R}}^3\setminus S)\cap B_R\left(0\right))},R>R_0,$$

where $R_0$ is so large that $S\subset B_{R_0}\left(0\right)$. Define also

$$L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)=\{u\in L_{\mathrm{loc}}^2({\mathbb{R}}^3\setminus S);\nabla \times u\in L_{\mathrm{loc}}^2({\mathbb{R}}^3\setminus S)\},$$

$$L_{\mathrm{loc},\mathrm{div}}^2({\mathbb{R}}^3\setminus S)=\{u\in L_{\mathrm{loc}}^2({\mathbb{R}}^3\setminus S);\nabla ·u\in L_{\mathrm{loc}}^2({\mathbb{R}}^3\setminus S)\},$$

and equip these space with semi-norms

$${\parallel f\parallel }_{R,\mathrm{curl}}={(\parallel f\parallel }_R^2{+\parallel \nabla \times f\parallel }_R^2{)}^{1/2}{,\parallel f\parallel }_{R,\mathrm{div}}={(\parallel f\parallel }_R^2{+\parallel \nabla ·f\parallel }_R^2{)}^{1/2}.$$

Also, let

$$T{H}^{-1/2}\left(S\right)=\{u\in H^{-1/2}{\left(S\right)}^3;\langle \nu ,u\rangle =0\},$$

i.e., the space of tangential $H^{-1/2}$ fields on S. We equip this with the norm induced from $H^{-1/2}{\left(S\right)}^3$. With $\mathrm{Div}$ denoting the surface divergence, we also define

$$T{H}_{\mathrm{Div}}^{-1/2}\left(S\right)=\{u\in T{H}^{-1/2}\left(S\right);\mathrm{Div}\left(u\right)\in H^{-1/2}\left(S\right)\}$$

and equip it with the Hilbert norm, defined by

$${\parallel u\parallel }_{T{H}_{\mathrm{Div}}^{-1/2}\left(S\right)}^2={\parallel u\parallel }_{T{H}^{-1/2}\left(S\right)}^2+{\parallel \mathrm{Div}\left(u\right)\parallel }_{H^{-1/2}\left(S\right)}^2.$$

Assume now that $U\in {\mathbb{R}}^3$ is a bounded $C^2$ domain with a connected complement, such that $S\subset \partial U$ is a compact $C^2$-submanifold and fix the unit normal $\nu$ of S so that it extends to a unit exterior normal $\tilde{\nu }$ of U. If $u,\phi \in {\left(C_0^{\infty }\right)}^3$ then the vector Green’s identities give

$${\int }_{\partial U}\langle \tilde{\nu }\times u,\phi \rangle ds={\int }_U\langle \nabla \times u,\phi \rangle -\langle u,\nabla \times \phi \rangle dx,$$

and extending this by density to $\phi \in H^1\left({\mathbb{R}}^3\right)$ and $u\in L_{curl}^2\left(U\right)$ gives the existence of the tangential trace $\tilde{\nu }{ \times u|}_{\partial U}\in T{H}^{-1/2}(\partial U)$. We can argue similarly for the exterior domain. Using this definition, we have well-defined tangential trace maps $t^{\pm }$ from the direction of $\pm \nu$,

$$t^{\pm }:L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)\ni u\mapsto \nu \times u^{\pm }\in T{H}^{-1/2}\left(S\right)$$

Similarly, if $u\in {\left(C_0^{\infty }\right)}^3$ and $\psi \in C_0^{\infty }$, we get from the Divergence Theorem that

$${\int }_U\langle \tilde{\nu },u\rangle \psi ds={\int }_U\langle u,\nabla \psi \rangle +\psi \nabla ·udx,$$

and, using this, we have well-defined normal traces $n^{\pm }$,

$$n^{\pm }:L_{\mathrm{loc},\mathrm{div}}^2({\mathbb{R}}^3\setminus S)\ni u\mapsto \langle \nu ,u^{\pm }\rangle \in H^{-1/2}\left(S\right).$$

Note also that if $u\in L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)$, then $\nabla \times u\in L_{\mathrm{loc},\mathrm{div}}^2({\mathbb{R}}^3\setminus S)$ and

$$\mathrm{Div}(\nu \times u^{\pm })=-\langle \nu ,\nabla \times u^{\pm }\rangle \in H^{-1/2}\left(S\right),$$

i.e., the tangential traces of $L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)$ are in $T{H}_{\mathrm{Div}}^{-1/2}\left(S\right)$. Note also that since extension by zero across a $C^2$ hypersurface is continuous in fractional Sobolev spaces, $H^s$ when $s<1/2$ the space $T{C}_0^{\infty }\left(S\right)$ is dense in $T{H}^s\left(S\right)$. However, this is not necessarily true for the Div spaces, and hence, the closure of $T{C}_0^{\infty }\left(S\right)$ in $T{H}_{\mathrm{Div}}^{-1/2}\left(S\right)$ is denoted by ${\dot{TH}}_{\mathrm{Div}}^{-1/2}\left(S\right)$.

2.5. Layer Potentials in Sobolev spaces

For $x\in {\mathbb{R}}^3\setminus S$ and $u\in C_0^{\infty }{\left(S\right)}^3$ define the (vector) single-layer potential of u by

$$V_{{\mathbb{R}}^3\setminus S}\left(u\right)\left(x\right)={\int }_S\mathsf{\Phi }(x-y)u\left(y\right)ds\left(y\right)$$

and the electromagnetic layer operators by

$$K_{{\mathbb{R}}^3\setminus S}\left(u\right)\left(x\right)=\nabla \times V_{{\mathbb{R}}^3\setminus S}\left(u\right)\left(x\right),$$

and

$$N_{{\mathbb{R}}^3\setminus S}\left(u\right)\left(x\right)={(\nabla \times )}^2{V}_{{\mathbb{R}}^3\setminus S}\left(u\right)\left(x\right).$$

Proposition 2.

Assume that S can be extended to a boundary $\partial U$ for some bounded $C^2$ domain U. Then, the single-layer potential has an extension to a bounded map

$$V_{{\mathbb{R}}^3\setminus S}:H^{-1/2}\left(S\right)\to H_{\mathrm{loc}}^1({\mathbb{R}}^3\setminus S)$$

and $V_{{\mathbb{R}}^3\setminus S}\left(u\right)$ satisfies the Sommerfeld radiation condition. Also, the electromagnetic potentials have extensions into bounded maps

$$K_{{\mathbb{R}}^3\setminus S},N_{{\mathbb{R}}^3\setminus S}:{\dot{TH}}_{\mathrm{Div}}^{-1/2}\left(S\right)\to L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)$$

and $K_{{\mathbb{R}}^3\setminus S}\left(u\right)$ and $N_{{\mathbb{R}}^3\setminus S}\left(u\right)$ satisfy the Sommerfeld radiation conditions for any $u\in {\dot{TH}}_{\mathrm{Div}}{}^{-1/2}\left(S\right)$.

Proof. 

By known continuity properties (see for example [33]), the single-layer potential defines a continuous map $H^{-1/2}\left(U\right)\to H_{\mathrm{loc}}^1({\mathbb{R}}^3\setminus \overline{U})$, and, in fact, $V_{{\mathbb{R}}^3\setminus \overline{U}}\left(\phi \right)$ is continuous across U and the jump in the normal derivative is equal to ${\phi |}_U$. Hence, the claim for V follows since $C_0^{\infty }\left(S\right)$ is dense in $H^{-1/2}\left(S\right)$. This also implies the claims for $K_{{\mathbb{R}}^3\setminus S}$ and $N_{{\mathbb{R}}^3\setminus S}$, since for $u\in T{C}_0^{\infty }\left(S\right)$, we have

$${(\nabla \times )}^2{V}_{{\mathbb{R}}^3\setminus S}\left(u\right)=-\nabla V_{{\mathbb{R}}^3\setminus S}\left(\mathrm{Div}u\right)+k^2{V}_{{\mathbb{R}}^3\setminus S}\left(u\right)$$

and

$${(\nabla \times )}^3{V}_{{\mathbb{R}}^3\setminus S}\left(u\right)=k^2\nabla \times V_{{\mathbb{R}}^3\setminus S}\left(u\right).$$

□

Using this, we can generalise the representation Theorem 1 to weak solutions:

Proposition 3.

Let $S\subset {\mathbb{R}}^3$ be a $C^1$ screen. Assume $(e,h)\in L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)\times L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)$ solves

$$\nabla \times e=i\omega \mu h,\nabla \times h=-i\omega \epsilon e\mathrm{in}{\mathbb{R}}^3\setminus S,$$

and the Silver–Müller radiation condition

$$\widehat{r}\times e+\sqrt{\frac{\epsilon }{\mu }}h={o\left(\right|x|}^{-1}),\widehat{r}\times h-\sqrt{\frac{\mu }{\epsilon }}e={o\left(\right|x|}^{-1}),\mathit{as}\left|x\right|\to \infty ,\widehat{r}=x/\left|x\right|>0.$$

If $\nu \times \left[e\right],\nu \times \left[h\right]\in T{H}_{\mathrm{Div}}^{-1/2}\left(S\right)$, then in ${\mathbb{R}}^3\setminus \overline{S}$,

$$e=K_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[e\right])-\frac{1}{i\omega \epsilon }{N}_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[h\right]),$$

and

$$h=K_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[h\right])+\frac{1}{i\omega \mu }{N}_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[e\right]).$$

Here, ν is the specified unit normal of S.

2.6. Representation Formulas for the Scattered Field

Proposition 4.

Let S be a perfectly conducting $C^2$ screen, and let

$$(E_{sc},H_{sc})\in L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)\times L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^3\setminus S)$$

be the scattered field corresponding to an incoming field $(E_0,H_0)$. Then, in ${\mathbb{R}}^3\setminus \overline{S}$, one has

$$E_{\mathrm{sc}}=-\frac{1}{i\omega \epsilon }{N}_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[H_{\mathrm{sc}}\right])$$

and

$$H_{\mathrm{sc}}=K_{{\mathbb{R}}^3\setminus \overline{S}}(\nu \times \left[H_{\mathrm{sc}}\right]).$$

These fields have the following asymptotic behaviour as $\left|x\right|\to \infty$:

$$E_{\mathrm{sc}}\left(x\right)=-\widehat{x}\times \left(\widehat{x}\times \frac{e^{ik\left|x\right|}}{4\pi i\omega \epsilon \left|x\right|}{\int }_S{e}^{-ik\langle \widehat{x},y\rangle }(\nu \times \left[H_{\mathrm{sc}}\right])\left(y\right)ds\left(y\right)\right)+{O\left(\right|x|}^{-2}),$$

$$H_{\mathrm{sc}}\left(x\right)=\widehat{x}\times \frac{e^{ik\left|x\right|}}{4\pi i\omega \mu \left|x\right|}{\int }_S{e}^{-ik\langle \widehat{x},y\rangle }(\nu \times \left[H_{\mathrm{sc}}\right])\left(y\right)ds\left(y\right)+{O\left(\right|x|}^{-2}),$$

where $\widehat{x}=x/\left|x\right|,x\ne 0$.

Proof. 

Since on a perfectly conducting screen, $\nu \times \left[E_{\mathrm{sc}}\right]=-\nu \times \left[E_0\right]=0$ and the representations of $E_{\mathrm{sc}}$ and $H_{\mathrm{sc}}$ follow from Proposition 1. The asymptotic behaviour is obvious since for $\left|x\right|\to \infty ,y\in S$ and $a\left(y\right)$, a vector field on S,

$${\nabla }_x\times \frac{e^{ik|x-y|}}{|x-y|}a\left(y\right)=\widehat{x}\times \frac{e^{ik\left|x\right|-ik\langle \widehat{x},y\rangle }}{\left|x\right|}a\left(y\right)+{O(|x|}^{-2})$$

$${({\nabla }_x\times )}^2\frac{e^{ik|x-y|}}{|x-y|}a\left(y\right)=\widehat{x}\times (\widehat{x}\times \frac{e^{ik\left|x\right|-ik\langle \widehat{x},y\rangle }}{\left|x\right|}a\left(y\right))+{O\left(\right|x|}^{-2}).$$

□

In view of the above proposition, we can write

$$E_{\mathrm{sc}}\left(x\right)=\frac{e^{ik\left|x\right|}}{4\pi \left|x\right|}{E}^{\infty }\left(\widehat{x}\right)+{O\left(\right|x|}^{-2}),H_{\mathrm{sc}}\left(x\right)=\frac{e^{ik\left|x\right|}}{4\pi \left|x\right|}{H}^{\infty }\left(\widehat{x}\right)+{O\left(\right|x|}^{-2})$$

where the far-field patterns $E^{\infty }$ and $H^{\infty }$ are given by

$$E^{\infty }\left(\widehat{x}\right)=-\widehat{x}\times \left(\widehat{x}\times \frac{1}{i\omega \epsilon }{\int }_S{e}^{-ik\langle \widehat{x},y\rangle }(\nu \times \left[H_{\mathrm{sc}}\right])\left(y\right)ds\left(y\right)\right)$$

$$H^{\infty }\left(\widehat{x}\right)=\widehat{x}\times \frac{1}{i\omega \mu }{\int }_S{e}^{-ik\langle \widehat{x},y\rangle }(\nu \times \left[H_{\mathrm{sc}}\right])\left(y\right)ds\left(y\right).$$

Note also that $\epsilon E^{\infty }\left(\widehat{x}\right)=-\mu \widehat{x}\times H^{\infty }\left(\widehat{x}\right)$, and that the uniqueness of the scattered field follows from the uniqueness of the Dirichlet and Neumann problems for the scalar Helmholtz equation [34].

2.7. Integral Equations for the Scattered Field

Assume now that $U\subset {\mathbb{R}}^3$ is a bounded $C^2$ domain with a connected complement such that $S\subset \partial U$ is a $C^2$ submanifold. Using the usual jump relations (see for example [30,32]), the tangential components of $N_{{\mathbb{R}}^3\setminus \overline{U}}\left(u\right)$ and $N_U\left(u\right)$ are continuous up to $\partial U$ and they have equal traces for all $u\in T{H}_{\mathrm{Div}}^{-1/2}(\partial U)$. Furthermore, for $u\in T{H}_{\mathrm{Div}}^{-1/2}(\partial U)$, one has

$$\nu \times N_{{\mathbb{R}}^3\setminus \overline{U}}\left(u\right){{|}_{\partial U}=\nu \times N_U\left(u\right)|}_{\partial U}=N\left(u\right),$$

where the surface integral operator N is given by

$$N\left(u\right)=\tilde{\nu }\times \nabla S\left(\mathrm{Div}u\right)+k^2\tilde{\nu }\times S\left(u\right).$$

Here, $\tilde{\nu }$ is the exterior unit normal to U, which is assumed to agree with $\nu$ on S, and S is the direct boundary value of the single layer potential, i.e.,

$$S\left(f\right)\left(x\right)={\int }_{\partial U}\mathsf{\Phi }(x-y)f\left(y\right)ds\left(y\right),x\in {\mathbb{R}}^3.$$

Note also that using the trace theorems given in Section 2.5, one has $N:T{H}_{\mathrm{Div}}^{-1/2}(\partial U)\to T{H}_{\mathrm{Div}}^{-1/2}(\partial U)$ continuously and for the restriction to S, one has $N:{\dot{TH}}_{\mathrm{Div}}^{-1/2}\left(S\right)\to {TH}_{\mathrm{Div}}^{-1/2}\left(S\right)$, again continuously.

Assuming now that $(E_{sc},H_{sc})$ is the field scattered by the screen S and $\nu \times \left[H_{sc}\right]\in {\dot{TH}}_{\mathrm{Div}}^{-1/2}\left(S\right)$, we get from Proposition 4 and the continuity results of Section 2.5 that

$$\nu \times E_{sc}=iN(\nu \times \left[H_{sc}\right])/\omega \epsilon ,$$

and since the screen is perfectly conducting, one gets an integral equation for the jump of the tangential component of the magnetic field,

$$-\nu \times E_0=iN(\nu \times \left[H_{sc}\right])/\omega \epsilon ,\nu \times \left[H_{sc}\right]\in {\dot{TH}}_{\mathrm{Div}}^{-1/2}\left(S\right).$$

(5)

Solvability properties of this equation have been considered in [35]. More precisely, it is shown that (5) is uniquely solvable in ${\dot{TH}}^{-1/2}\left(S\right)$.

3. Solution of the Inverse Problem

3.1. Uniqueness When the Supporting Hyperplane Is Known

The following lemma shows that for a planar screen, the tangential density of the far-field pattern is uniquely determined.

Lemma 1.

Assume that ρ is a compactly supported tangential distributional density on a hyperplane L. Let

$${\rho }^{\infty }\left(\widehat{x}\right)=\widehat{x}\times \langle \rho ,exp\{-ik\langle \widehat{x},·\rangle \}\rangle ,\widehat{x}\in {\mathbb{S}}^2.$$

Then the map $\rho \mapsto {\rho }^{\infty }$ is injective.

Proof. 

We may assume that coordinates have been chosen so that L is defined by $\{x;x_3=0\}$. Let $\rho =ad\sigma$ where $a=(a_1,a_2)\in {\epsilon }^{\prime }\left({\mathbb{R}}^2\right)$ and $d\sigma$ is the surface measure on the hyperplane L. Then, ${\rho }^{\infty }=0$ is equivalent to

$$\xi \times ({\widehat{a}}_1\left({\xi }^{\prime }\right),{\widehat{a}}_2\left({\xi }^{\prime }\right),0)=0,\xi =({\xi }^{\prime },(k^2-|{\xi }^{\prime }{|}^2{)}^{1/2}),|{\xi }^{\prime }|<k,$$

and hence, ${\widehat{a}}_1$ and ${\widehat{a}}_2$ vanish in the unit ball of ${\mathbb{R}}^2$ and since they are entire functions, they are identically zero. □

This implies that the far field $E^{\infty }$ (or $H^{\infty }$ for that matter) uniquely determines the density $[\nu \times H_{\mathrm{sc}}]ds$ when the screen S is flat, i.e., included in a hyperplane.

Proposition 5.

Let S be a $C^2$ screen contained in a supporting hyperplane L and let $(e_0,h_0)$ be an electromagnetic plane wave with wave number k with electric and magnetic polarisations p and q. Assume that $\rho \in {\dot{TH}}^{-1/2}\left(S\right)$ solves $-\nu \times e_0=-iN(\nu \times \rho )/\omega \epsilon$ on S. Then, if p or q are not parallel to θ and $\nu \times (p\times \theta )\ne 0$ the density ρ has full support, i.e., $\mathrm{supp}\left(\rho \right)=S$.

Proof. 

Assume coordinates chosen so that the $L=\{x\in {\mathbb{R}}^3;x_3=0\}$. Assume that there is a relatively open $U\subset S$ such that $\rho =0$. Define

$$\tilde{h}=K_{{\mathbb{R}}^3\setminus S}(\nu \times \rho ),\tilde{e}=\frac{i}{\omega \epsilon }\nabla \times \tilde{h}.$$

Then, $\tilde{h},\tilde{e}\in L_{\mathrm{loc},\mathrm{curl}}^2({\mathbb{R}}^2\setminus S)$ and they satisfy Maxwell’s equations

$$\nabla \times \tilde{e}=i\omega \mu \tilde{h},\nabla \times \tilde{h}=-i\omega \epsilon \tilde{e}.$$

The second equation follows from the definition and the first is an immediate consequence of the vector Green’s formulas:

$$\begin{array}{ccc}\hfill \nabla \times \tilde{e}& =& \frac{i}{\omega \epsilon }{(\nabla \times )}^2\tilde{h}=\frac{i}{\omega \epsilon }(\nabla \nabla ·-\Delta )\left(K_{{\mathbb{R}}^3\setminus S}(\nu \times \rho )\right)=\frac{i}{\omega \epsilon }(\nabla \nabla ·-\Delta )(\nabla \times S_{{\mathbb{R}}^3\setminus S}(\nu \times \rho ))\hfill \\ & =& \frac{i}{\omega \epsilon }{k}^2\nabla \times S_{{\mathbb{R}}^3\setminus S}(\nu \times \rho )=-i\omega \mu \nabla \times S_{{\mathbb{R}}^3\setminus S}(\nu \times \rho )=-i\omega \mu \tilde{h}.\hfill \end{array}$$

From the jump relations of the vector potentials, $[\nu \times \tilde{h}]=[\nu \times \tilde{\rho }]=0$ on U and

$$-\tilde{e}=-\frac{i}{\omega \epsilon }\nabla \times \tilde{h}=-\frac{i}{\omega \epsilon }{N}_{{\mathbb{R}}^3\setminus S}(\nu \times \rho )$$

we have

$$\nu \times \tilde{e}{{|}_S=iN(\nu \times \rho )/\omega \epsilon |}_S=\nu \times e_0.$$

Let $E=e_0-\tilde{e}$ and $H=h_0-\tilde{h}$. Then, from the above observations, $(E,H)$ solves (1) and

$${\nu \times E|}_S{=0,[\nu \times H]|}_U=0.$$

Now, let $\widehat{E}$ be an extension of E from the upper half-space $\{x_3>0\}$ to the lower half-space so that it is odd in the tangential component and even in the normal component, i.e.,

$$\widehat{E}(x_1,x_2,-x_3)=(-E_1\left(x\right),-E_2\left(x\right),E_3\left(x\right)),x=(x_1,x_2,x_3),$$

and let $\widehat{H}$ be an extension of H, which is even in the tangential component and odd in the normal component,

$$\widehat{H}(x_1,x_2,-x_3)=(H_1\left(x\right),H_2\left(x\right),-H_3\left(x\right)),x=(x_1,x_2,x_3).$$

Then, a straightforward computation shows that

$$\nabla \times \widehat{E}=i\omega \mu \widehat{H},\nabla \times \widehat{H}=-i\omega \epsilon \widehat{E},x_3\ne 0.$$

Let $V=U\times \mathbb{R}$. Then, since the tangential components of E vanish on S and the tangential components of H are continuous across U, this holds also for the tangential components of $\widehat{E}$ and $\widehat{H}$ across U. Thus, $(\widehat{E},\widehat{H})$ solves

$$\nabla \times \widehat{E}=i\omega \mu \widehat{H},\nabla \times \widehat{H}=-i\omega \epsilon \widehat{E},x\in V,$$

and hence, by unique continuation, $E=\widehat{E}$ and $H=\widehat{H}$ in V and, thus, also in ${\mathbb{R}}^3\setminus S$, since $\nu \times \widehat{H}=\nu \times H$ on U. Hence, we can write

$$e_0=\widehat{E}+\tilde{e},h_0=\widehat{H}+\tilde{h}.$$

(6)

We say that a vector field has parity 1 if the tangential component is even and the normal component is odd with respect to $\{x_3=0\}$, and it has parity $-1$ if the tangential component is odd and the normal component is even. Notice, then, that since $\tilde{h}$ is the EM-double layer of a tangential density, it has parity $-1$. Hence, $\tilde{e}=-i\nabla \times \tilde{h}/\omega \epsilon$ has parity 1. Also, the decomposition of a field as a sum of fields with parity $+1$ and $-1$ is unique. Since $\tilde{e}$ satisfies the Silver–Müller radiation condition, the incoming field $e_0$ must have parity $-1$ and similarly, $h_0$ must have parity 1. Recall that

$$e_0\left(x\right)={\mu }^{1/2}(p\times \theta )e^{ik\langle \theta ,x\rangle },h_0\left(x\right)={\epsilon }^{1/2}(q\times \theta )e^{ik\langle \theta ,x\rangle },q=p\times \theta ,$$

Hence, the parity 1 part of $e_0$ is given by

$${\mu }^{1/2}(q_1{e}_0^{(+)},q_2{e}_0^{(+)},i{q}_3{e}_0^{(-)}),$$

where $e_0^{(+)}\left(x\right)=cos\left(k\langle \theta ,x\rangle \right)$ and $e_0^{(-)}\left(x\right)=sin\left(k\langle \theta ,x\rangle \right)$. This vanishes identically if and only if $q=0$, i.e., $p\times \theta =0$. Similarly, the parity $-1$ part of $h_0$ is

$${\epsilon }^{1/2}(i{(q\times \theta )}_1{e}_0^{(-)},i{(q\times \theta )}_2{e}_0^{(-)},{(q\times \theta )}_3{e}_0^{(+)}),$$

which vanishes identically if and only if $q\times \theta =0$. Since p, q and $\theta$ are unit vectors and $q=p\times \theta$, this is not possible. □

3.2. Unique Determination of a Planar Screen

We show that the supporting hyperplane uniquely determines the far field of a single scattering solution. This, combined with the unique determination results of the previous subsection, then proves Theorem 1.

Proposition 6.

Assume $S_1$ and $S_2$ are two planar screens contained in supporting hyperplanes ${\pi }_1$ and ${\pi }_2$, respectively. Assume $u_1=(e_1,h_2)$ and $u_2=(e_2,h_2)$ are scattering solutions for the screens $S_1$ and $S_2$ corresponding to the same initial field and having equal non-vanishing far fields. Then, ${\pi }_1={\pi }_2$.

Proof. 

Let ${\rho }_1$ and ${\rho }_2$ be the jumps of ${\nu }_1\times h_1$ and ${\nu }_2\times h_2$ across $S_1$ and $S_2$, respectively. Here, ${\nu }_i$ is the specified unit normal to $S_1$. Since $u_1$ and $u_2$ have equal far fields and ${\mathbb{R}}^3\setminus (S_1\cup S_2)$ is connected, we must have $u_1=u_2$ there. Hence, both fields must be smooth across $(S_1\cup S_2)\setminus (S_1\cap S_2)$, i.e., both densities ${\rho }_1$ and ${\rho }_2$ are supported in the intersection $S_1\cap S_2$. If the planes ${\pi }_1$ and ${\pi }_2$ intersect transversally, the jumps are supported on a codimension 2 subspace, and since they belong to ${\dot{\mathrm{TH}}}^{-1/2}(S_1\cup S_2)$, they must vanish (Note that a non-vanishing, compactly supported distribution density on a codimension 2 submanifold of ${\mathbb{R}}^3$ belongs to $H^s$ if and only if $s<-1$. This follows, for example, from estimates at the end of Section 7.1 in [36] by applying these to a suitable dyadic decomposition.) if the intersection is transversal, so the far fields also vanish. □

4. Conclusions

In this article, we proved that a non-vanishing far-field pattern of a single plane wave uniquely determines a planar super-conducting screen. The proof was based on reduction of the scattering problem to a single tangential integral equation on the screen where the unknown is the jump of the tangential component of the scattered magnetic field. We showed that the far field uniquely determines the jump, and that screen is what supports the jump. We plan to generalise this to compact, real-analytic screens in a future work.

However, this will require more advanced techniques. As a possible application, we mention the following problem: Suppose we have an inaccessible array of radars from which we can only obtain distant data. Such information could be, say, whether the array uses classical dipole antennas or more advanced tripole antennas [27]. Our result indicates that such information can, in principle, be obtained with a single measurement.