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doi:10.3934/ipi.2010.4.xx Inverse Problems and Imaging Volume 4, No. 3, 2010, 1–XX RECONSTRUCTING ELECTROMAGNETIC OBSTACLES BY THE ENCLOSURE METHOD Ting Zhou Department of Mathematics, University of Washington, Seattle, WA 98105, USA Abstract. We show that one can determine Perfectly Magnetic Conductor obstacles, Perfectly Electric Conductor obstacles and obstacles satisfying impedance boundary condition, embedded in a known electromagnetic medium, by making electromagnetic measurements at the boundary of the medium. The boundary measurements are encoded in the impedance map that sends the tangential component of the electric field to the tangential component of the magnetic field. We do this by probing the medium with complex geometrical optics solutions to the corresponding Maxwell’s equations and extend the enclosure method to this case. 1. Introduction. We consider in this paper an inverse problem in electromagnetism. We probe a medium in which there is an unknown obstacle with electromagnetic waves. We would like to determine the obstacle from electromagnetic measurements made only at the boundary of the medium. We consider first Perfectly Magnetic Conductor (PMC) obstacles which means that the tangential component of the magnetic field vanishes at the boundary of the obstacle. We also consider Perfectly Electric Conductor (PEC) obstacles which means that the tangential component of the electric field vanishes at the boundary of the obstacle and also obstacles satisfying more general impedance conditions at the boundary. In all these cases we show that under some conditions one can recover the obstacle by making electromagnetic measurements at its boundary. We use the enclosure method [11, 12] of Ikehata that has been applied successfully for the case of electrostatics and acoustics. This method uses as probing waves the so-called complex geometrical optics (CGO) solutions to the conductivity equation, in the case of electrostatics, and Helmholtz equations in the case of acoustics. The case of Maxwell’s equations presents new challenges since it is considerable more difficult in this case to construct CGO solutions. Moreover it is also more difficult to establish the bounds needed for the indicator function in Ikehata’s method. We now describe more precisely the mathematical problem and some historical background. Let Ω ⊂ R3 be a bounded domain with smooth boundary, filled with an isotropic electromagnetic material, characterized by three parameters: the electrical permittivity ε(x), the conductivity σ(x) and magnetic permeability µ(x). In particular, 2000 Mathematics Subject Classification. Primary: 35R30, 35Q61. Key words and phrases. Maxwell’s equations, enclosure methods, complex geometrical optics soltuions. The author is supported by NSF grant 0758357. 1 c 2010 American Institute of Mathematical Sciences 2 Ting Zhou we consider non-dissipative materials for which σ = 0. A PMC obstacle is a subdomain D such that the electric-magnetic field (E, H) satisfies Maxwell’s equations with tangential component of the magnetic field vanishing at the boundary of D, ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D, (1) ν ∧ E|∂Ω = f, ν ∧ H|∂D = 0 where ν is the unit outer normal vector to the boundary ∂Ω ∪ ∂D. A PEC obstacle is such that the tangential component of the electric field ν ∧ E|∂D = 0. Both PMC (λ → 0) and PEC (λ → ∞) conditions are two extreme cases of the general impedance boundary condition iν ∧ H − λ(ν ∧ E) ∧ ν|∂D = 0 where λ ∈ L∞ (∂D) is a complex-valued function, see [9]. Suppose we are given the impedance map which takes the tangential component of the electric field ν ∧ E|∂Ω to the tangential component of the magnetic field ν ∧ H|∂Ω . This map encodes all the possible electromagnetic measurements that can be made at the boundary of the medium. Our goal is to retrieve information of the shape of D from the impedance map. For the case of electrostatics, the well-known Calderón’s problem [6] is to determine the conductivity of a medium by making voltage and current measurements of the boundary. The information is encoded in the Dirichlet-to-Neumann map for the conductivity equation ∇ · (γ∇u) = 0. In [25], Sylvester and Uhlmann constructed complex geometrical optics (CGO) solutions for the conductivity equation and Schrödinger operator ∆ − q and proved unique identifiability of C 2 isotropic conductivities in dimensions n ≥ 3 from the Dirichlet-to-Neumann map. Further developments including improved regularity assumption, 2D problems and partial data problems, were obtained based on this idea. See [27] for a survey of recent developments. As another application of CGO solutions, Ikehata [11, 12] introduced the enclosure method to reconstruct obstacles embedded in an electric conductor (conductivity equations) or acoustic medium (Helmholtz equations). Intuitively, one can enclose the obstacle by hyperplanes. Then one constructs CGO solutions that decay on one side of the hyperplane and grow on the other side. In [16], CGO solutions with nonlinear radial weights are constructed and called complex spherical waves. Using these the enclosure method was generalized to enclose non-convex obstacles, see [15] and [21]. In addition of the conductivity equation and Helmholtz equation, other types of systems are considered. For example, in [28], CGO solutions were constructed for systems with Laplacian as the leading order term, e.g., for the elasticity equations, and used to recover inclusions. For the case of Maxwell’s equations, the uniqueness of parameters in a domain Ω from the impedance map was obtained in [24], [22] and [23]. Particularly, in [23], the Maxwell’s operator was reduced to a matrix Schrödinger operator, and vector CGO solutions were constructed to recover electromagnetic parameters. With these solutions at hand, the enclosure method is applicable using the following steps: one first defines an indicator function Iρ (τ, t) for each direction ρ ∈ S2 ; by adjusting Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 3 t, the hyperplane moves along ρ; for each ρ and t, the asymptotic behavior of Iρ (τ, t) as τ 1 produces the support function of the convex hull of D. One of the main difficulties to determine obstacles in electromagnetic materials is that, unlike the Schrödinger equation, the CGO solutions for Maxwell’s equations don’t behave as small perturbations (w.r.t. τ ) of Calderón’s solutions. However, this can be overcome by carefully choosing appropriate probing fields (see section 3). We will focus on showing the asymptotic behavior of the indicator function of these CGO solutions for large τ. (see section 4) For the reconstruction of the non-convex part of the obstacles, we apply an inversion transformation with respect to a sphere to the domain. Since this transformation maps spheres to planes, by enclosing the image obstacle using hyperplanes, we actually enclose the original obstacle by spheres. The argument relies on the invariance of Maxwell’s equations under this transformation, and a relation between the impedance maps for the original system and the image systems. Throughout the process of reconstruction, one actually constructs the CSW for the original system, which is the pull-back of the CGO solution of the image system by the inversion transformation. See section 5 for more details. In another direction we mention the work of Ammari et al. [2, 3, 4] to determine asymptotically inhomogeneities of small volume embedded in an homogeneous medium, and [1] where efficient numerical algorithms were derived in this context. The rest of the paper is organized as following. In Section 2, we formulate the forward problem for a Perfect Magnetic Conductor (PMC) obstacle and define rigorously the impedance map. Then we construct the appropriate CGO solutions for Maxwell’s equations in Section 3. The main reconstruction scheme for PMC obstacles is developed in Section 4. Section 5 is devoted to introducing the enclosure method to recover the non-convex part of the obstacle based on inversion. In Section 6, we show that the scheme also applies to PEC obstacles and obstacles with impedance boundary conditions. In Section 7, we present our conclusions and propose some open problems related to this paper. We point out that PEC and impedance boundary conditions are more natural boundary conditions in practice. However, this paper presents a detailed proof in the PMC setting, which requires the well-posedness of a mixed boundary value problem for Maxwell’s equations. For completeness, we include a proof in Appendix A. 2. Direct Problems and the main result. Let Ω be a bounded domain in R3 with smooth boundary and connected complement R3 \ Ω. We consider the electric permittivity ε(x), conductivity σ(x) and magnetic permeability µ(x) of the background medium as globally defined functions with the following properties: there are positive constants εm , εM , µm , µM , εc and µc such that for all x ∈ Ω εm ≤ ε(x) ≤ εM , µm ≤ µ(x) ≤ εM , σ(x) = 0 and ε − εc , µ − µc ∈ C03 (Ω). An obstacle D is a Perfect Magnetic Conductor (PMC) if it is a subset of Ω such that Ω \ D is connected. Moreover, the electric field E and the magnetic field H in Ω\D satisfy the following boundary value problem for the time-harmonic Maxwell’s Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 4 Ting Zhou equations (2) ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D, ν ∧ E = f ∈ T H 1/2 (∂Ω) on ∂Ω, and the boundary condition on the interface ∂D is given by (ν ∧ H)|∂D = 0, (3) where ν is the unit outer normal vector on ∂Ω ∪ ∂D. Notations. If F is a function space on ∂Ω, the subspace of all those f ∈ F 3 which 3 are tangent to ∂Ω is denoted by T F . For example, for u ∈ H s (∂Ω) , we have the decomposition u = ut + uν ν, where the tangential component ut = −ν ∧ (ν ∧ u) ∈ T H s (∂Ω) and the normal component uν = u · ν ∈ H s (∂Ω). In addition, we define the weighted L2 space in R3 : Z L2δ = f ∈ L2loc (R3 ) : kf k2L2 = (1 + |x|2 )δ |f (x)|2 dx < ∞ . δ Admissibility. It can be shown (see Appendix A.) that for f ∈ T H 1/2 (∂Ω) and g ∈ T H 1/2 (∂D), the boundary value problem for Maxwell’s equations ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D̄, ν ∧ E|∂Ω = f (4) ν ∧ H|∂D = g, 3 3 has a unique solution (E, H) ∈ H 1 (Ω \ D) × H 1 (Ω \ D) , except for a discrete set of magnetic resonance frequencies {ωn }. It satisfies kEkH 1 (Ω\D) + kHkH 1 (Ω\D) ≤ C kf kT H 1/2 (∂Ω) + kgkT H 1/2 (∂D) . Denote by (E0 , H0 ) the solution of Maxwell’s equations in the domain Ω without an obstacle, namely, the solution of ∇ ∧ E0 = iωµH0 , ∇ ∧ H0 = −iωεE0 , in Ω, ν ∧ E0 |∂Ω = f. Main result. Now we are in the position to define the impedance map for nonresonant frequencies, ΛD (ν ∧ E|∂Ω ) = ν ∧ H|∂Ω . and it can be shown that ΛD : T H 1/2 (∂Ω) → T H 1/2 (∂Ω) is bounded. If ω is a resonance frequency, one can replace the impedance map by the set of Cauchy data Cω = {(ν ∧ E|∂Ω , ν ∧ H|∂Ω ) | (E, H) satisfies (2) and (3)} ⊂ T H 1/2 (∂Ω) × T H 1/2 (∂Ω). Denote by Λ∅ the impedance map for the domain without an obstacle. Then the main result of the presenting work is to show Theorem 2.1. There exists a reconstruction scheme to determine the convex hull of the obstacle D from the impedance map ΛD . Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 5 3. Construction of Complex Geometric Optics (CGO) solutions. In [23], Maxwell’s operator was reduced to an 8 × 8 second order Schrödinger matrix operator by introducing the generalized Sommerfeld potentials. A vector CGO-solution was constructed to simplify the proof in [22]. Similar techniques also appeared in [8] when dealing with the partial data problems for Maxwell’s equations. For completeness, we include the construction of the solution in this work (see [23] for more details). Moreover, to obtain CGO solutions for our use in the enclosure method, a special ”incoming” field will be chosen. Define the scalar fields Φ and Ψ by i i (5) Φ = ∇ · (εE), Ψ = ∇ · (µH). ω ω Under appropriate assumptions on Φ and Ψ, Maxwell’s equations are equivalent to 1 1 1 1 (6) ∇∧E− ∇ Ψ − iωµH = 0, ∇ ∧ H + ∇ Φ + iωεE = 0. ε µ µ ε Moreover, in this case, Φ and Ψ vanish, leading to a solution of Maxwell’s equations. Let X = (ϕ, e, h, ψ)T ∈ (D0 )8 with 1 1 e = ε1/2 E, h = µ1/2 H, ϕ = 1/2 Φ, ψ = 1/2 Ψ. εµ ε µ Then (5) and (6) read (P (i∇) − k + V )X = 0, in Ω where (7) ∇· 0 0 ∇∧ −∇∧ 0 0 ∇· 0 0 ∇ 0 ∇· 0 0 −∇∧ ∇∧ 0 0 ∇· 0 0 D D−1 ∇ 0 0 ∇ P (i∇) = 0 0 and 0 ∇ V = (k − κ)18 + 0 0 are matrix operators and D = diag(µ1/2 , ε1/2 13 , µ1/2 13 , ε1/2 ), κ = ω(εµ)1/2 , k = ω(ε0 µ0 )1/2 . An important property of this operator is that it allows to reduce Maxwell’s equations to a Schrödinger matrix equation by observing that (P (i∇) − k + V )(P (i∇) + k − V T ) = −(∆ + k 2 )18 + Q, where Q = V P (i∇) − P (i∇)V T + k(V + V T ) − V V T is a zeroth-order matrix multiplier. Hence, by writing an ansatz for X, one can define the generalized Sommerfeld potential Y by X = (P (i∇) + k − V T )Y. So it satisfies the Schrödinger equation (8) Inverse Problems and Imaging (−∆ − k 2 + Q)Y = 0. Volume 4, No. 3 (2010), 1–XX 6 Ting Zhou The following CGO-solution is constructed using Faddeev’s kernel. Let ζ ∈ C3 be a vector with ζ · ζ = k 2 . Suppose y0,ζ ∈ C8 is a constant vector with respect to x and bounded with respect to ζ. We refer eix·ζ y0,ζ as the ”incoming” field. Then there exists a unique solution to (8) of the form Yζ (x) = eix·ζ (y0,ζ − vζ (x)) , where vζ (x) ∈ (L2δ+1 )8 satisfying kvζ kL2δ+1 ≤ C/|ζ| for δ ∈ (−1, 0). Moreover, one can show that vζ ∈ (H s (Ω))8 for 0 ≤ s ≤ 2, e.g., see [5], and kvζ (x)kH s (Ω) ≤ C|ζ|s−1 . (9) Lemma 3.1 in [23] states that if we choose y0,ζ such that the first and the last components of (P (ζ)−k)y0,ζ vanish, where P (ζ) is the matrix obtained by replacing i∇ by ζ in (7), then for large |ζ|, Xζ provides the solution to the original Maxwell’s equations. We proceed to provide a more specific choice of y0,ζ such that the CGO solution for Maxwell’s equations has special properties. As in [23], choose y0,ζ = 1 (ζ · a, ka, kb, ζ · b)T , |ζ| where p ζ = −iτ ρ + τ 2 + k 2 ρ⊥ , √ with ρ, ρ⊥ ∈ S2 and ρ · ρ⊥ = 0. τ > 0 is used to control the size of |ζ| = 2τ 2 + k 2 . Then we obtain 0 2 1 −(ζ · a)ζ − kζ ∧ b + k2 a x0,ζ := (P (−ζ) + k)y0,ζ = kζ ∧ a − (ζ · b)ζ + k b |ζ| 0 satisfying the condition in Lemma 3.1 in [23]. Taking τ → ∞, we have ζ 1 → ζ̂ = √ (−iρ + ρ⊥ ). |ζ| 2 In this article, for the enclosure method, we expect different asymptotic behavior of the electric and magnetic fields with respect to the phase growth. Therefore we choose a and b, and correspondingly y0,ζ , such that ζ̂ · b = 1, ζ̂ · a = 0. This is satisfied, for example, by taking a ∈ R3 and b ∈ C3 satisfying a ⊥ ρ, a ⊥ ρ⊥ , b = ζ̂. Given these choices, It’s easy to see that as τ → ∞ η := (x0,ζ )2 → −k ζ̂ ∧ b = ikρ ∧ ρ⊥ = O(1), θ := (x0,ζ )3 = O(τ ). Then Xζ = (P (i∇) + k − V T )Yζ is written in the form √ Xζ = eτ (x·ρ)+i Inverse Problems and Imaging τ 2 +k2 x·ρ⊥ (x0,ζ + rζ (x)) Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 7 where (10) rζ = P (−ζ)vζ + P (i∇)vζ − V T y0,ζ + kvζ − V T vζ satisfying krζ kL2 (Ω) ≤ C for C > 0 independent of ζ. Summing up, we obtain the following Proposition 1. Let ρ, ρ⊥ ∈ S2 with ρ · ρ⊥ = 0. Assume ω is not a resonant frequency. Given θ, η ∈ C3 as above, then for τ > 0 large enough, there exists a 3 3 unique complex geometric optics solution (E0 , H0 ) ∈ H 1 (Ω) ×H 1 (Ω) of Maxwell’s equations ∇ ∧ E0 = iωµH0 ∇ ∧ H0 = −iωεE0 in Ω of the form E0 = ε−1/2 eτ (x·ρ)+i √ τ 2 +k2 x·ρ⊥ (η + R) √ −1/2 τ (x·ρ)+i τ 2 +k2 x·ρ⊥ H0 = µ e (θ + Q). Moreover, we have η = O(1), θ = O(τ ) for τ 1, and R = (rζ )2 , Q = (rζ )3 are bounded in (L2 (Ω))3 for τ 1. For reconstruction, one computes directly the boundary tangential CGO-fields (ν ∧ E0 |∂Ω , ν ∧ H0 |∂Ω ). On the other hand, given unknown background parameters, one can recover tangential CGO fields from the impedance map Λ∅ by solving appropriate boundary integral equations, see [23, 22]. This allows detection of an obstacle by measurements taken before and after its appearance without invading the background area. 4. Reconstruction Scheme. Adding a parameter t > 0 into the CGO-solution in Proposition 1, we use (11) E0 = ε−1/2 eτ (x·ρ−t)+i √ H0 = µ−1/2 eτ (x·ρ−t)+i τ 2 +k2 x·ρ⊥ √ (η + R) τ 2 +k2 x·ρ⊥ (θ + Q) to define an indicator function which physically measures the differences between the energies required to keep the same boundary CGO electric field for the domain Ω with and without the obstacle D. Definition 4.1. For ρ ∈ S2 , τ > 0 and t > 0 we define the indicator function Z Iρ (τ, t) := iω (ν ∧ E0 ) · (ΛD − Λ∅ )(ν ∧ E0 ) ∧ ν dS ∂Ω where E0 is a CGO solution of Maxwell’s equations given by (11). The enclosure method’s aim is to recover the convex hull ch(D) of D by reconstructing the following support function. Definition 4.2. For ρ ∈ S2 , we define the support function of D by hD (ρ) := sup x · ρ. x∈D Then, the reconstruction scheme in Theorem 2.1 is Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 8 Ting Zhou Theorem 4.3. There is a subset Σ ⊂ S2 of measure zero such that when ρ ∈ S2 \ Σ, the support function hD (ρ) can be recovered by hD (ρ) = inf{t ∈ R | lim Iρ (τ, t) = 0}. τ →∞ Moreover, if D is strictly convex, one can reconstruct D. Remark 1. The proof mainly consists of showing the following: lim Iρ (τ, t) = 0 if t > hD (ρ), τ →∞ lim inf Iρ (τ, t) = τ →∞ C>0 ∞ if t = hD (ρ), if t < hD (ρ). Remark 2. Gaussian curvature at a point on the boundary is defined to be the product of the two principal curvatures, which measure how the surface bends by different amounts in different directions at that point. A surface with positive Gaussian curvature at a point is locally convex. In the assumption, the set Σ consists of the directions ρΣ such that some point in {x ∈ ∂D | x · ρΣ = hD (ρΣ )} has Gaussian curvature 0. For a smooth boundary, one can show that the measure of Σ is zero. To show the limits in Remark 1, we need the following equality for Maxwell’s equations. 3 3 Lemma 4.4. Assume (E, H) ∈ H 1 (Ω \ D) × H 1 (Ω \ D) is a solution of (2) and (3) with ν ∧ E|∂Ω = ν ∧ E0 |∂Ω . Denote the reflected solution in Ω \ D by Ẽ = E − E0 and define Z h i I := iω (ν ∧ E0 ) · (ΛD − Λ∅ )(ν ∧ E0 ) ∧ ν dS. ∂Ω Then we have Z (12) I= µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx + Ω\D̄ Z µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx. D Proof. First using integration by parts and by boundary conditions, we have Z µ−1 (∇ ∧ E) · (∇ ∧ Ẽ) − ω 2 εE · (Ẽ)dx Ω\D̄ Z =− Z − ∂Ω iω(ν ∧ H) · (Ẽ)dS = 0. ∂D Adding this to the following equality Z I= (ν ∧ E0 ) · (−iωH + iωH0 )dS ∂Ω Z = −µ−1 (∇ ∧ E0 ) · (∇ ∧ E) + ω 2 εE0 · Edx Ω\D̄ Z Z + µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx + (ν ∧ E0 ) · (−iωH)dS, Ω ∂D since the last term vanishes due to the zero-boundary condition on the interface, we obtain (12). Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 9 Proof of Theorem 4.3. We proceed to show the first limit in Remark 1 (13) lim Iρ (τ, t) = 0 if t > hD (ρ) τ →∞ by proposing an upper bound of the indicator function. 3 The reflected solution Ẽ ∈ H 1 (Ω \ D) satisfies ∇ ∧ (µ−1 ∇ ∧ Ẽ) − sε∇(∇ · εẼ) − ω 2 εẼ = 0 in Ω \ D, ν ∧ Ẽ|∂Ω = 0, ν ∧ (µ−1 ∇ ∧ Ẽ)|∂D = −iων ∧ H0 |∂D ∈ T H 1/2 (∂D). Moreover, we have by (45) (14) kẼk2H 1 (Ω\D) ≤ Ckν ∧ H0 k2H −1/2 (∂D) ≤ C(k∇ ∧ E0 k2L2 (D) + kE0 k2L2 (D) ), where the second inequality is valid since iωhν ∧ H0 , Fi∂D = (µ−1 ∇ ∧ E0 , ∇ ∧ F)D − (iωεE0 , F)D 3 F ∈ H 1 (D) . Therefore, by (12) and (14), we have Iρ (τ, t) ≤ C(kE0 k2L2 (D) + k∇ ∧ E0 k2L2 (D) ) ≤ C(kE0 k2L2 (D) + kH0 k2L2 (D) ). For the CGO-solution (11), we have the following estimates: kE0 k2L2 (D) ≤ Ce2τ (hD (ρ)−t) kη + Rk2L2 (D)3 ∼ e2τ (hD (ρ)−t) (15) kH0 k2L2 (D) ≤ Ce2τ (hD (ρ)−t) kθ + Qk2L2 (D)3 ∼ τ 2 e2τ (hD (ρ)−t) as τ 1. Therefore, we obtain Iρ (τ, t) ≤ Cτ 2 e2τ (hD (ρ)−t) for τ large enough, proving the first limit (13). To show that when t = hD (ρ), (16) lim inf Iρ (τ, hD (ρ)) = C > 0, τ →∞ by (12), it’s sufficient to show the following Lemmas 4.5, 4.6. As for the case when t < hD (ρ), the infinity limit is obtained by observing Iρ (τ, t) = Iρ (τ, hD (ρ))e2τ (hD (ρ)−t) . This completes the proof of the theorem. Lemma 4.5. If t = hD (ρ), then for some positive constant C, Z lim inf µ−1 |∇ ∧ E0 |2 dx = C. τ →∞ D Proof. Thanks to the special choice of H0 in Proposition 1, Z Z −1 2 2 µ |∇ ∧ E0 | dx ≥ CkH0 kL2 (D)3 ≥ C τ 2 e2τ (x·ρ−hD (ρ)) dx ≥ C D D 3 for τ 1. We denote by x0 a point in {x ∈ R | x · ρ = hD (ρ)} ∩ ∂D. The last inequality is due to the accumulation of the integrand in a small cone at x0 . A detailed proof is provided in [13]. Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 10 Ting Zhou Lemma 4.6. If t = hD (ρ), then as τ → ∞, R R ω 2 Ω\D ε|Ẽ|2 dx + D ε|E0 |2 dx R (17) → 0. µ−1 |∇ ∧ E0 |2 dx D Proof. The proof follows a similar scheme in [11] for Helmholtz equations. Noticing that for τ 1 R R ω 2 D ε|E0 |2 dx ε|E0 |2 dx R RD = = O(τ −2 ), −1 |∇ ∧ E |2 dx 2 dx µ µ|H | 0 0 D D it suffices to show ω2 (18) lim R τ →∞ D R Ω\D µ−1 |∇ ε|Ẽ|2 dx ε|Ẽ|2 dx RΩ\D = lim = 0. ∧ E0 |2 dx τ →∞ D µ|H0 |2 dx R To estimate the numerator, we consider the auxiliary boundary value problem: ∇ ∧ P = iωµQ, ∇ ∧ Q = −iωεP − iε ω Ẽ in Ω \ D, ν ∧ P|∂Ω = 0, ν ∧ Q|∂D = 0. This is equivalent to the BVP ∇ ∧ (µ−1 ∇ ∧ P) − sε∇(∇ · εP) − ω 2 εP = εẼ in Ω \ D, (19) ν ∧ P|∂Ω = 0, ν ∧ (∇ ∧ P)|∂D = 0, since ∇ · εP = 0, in Ω \ D. 3 When ω is admissible, (19) is well-posed since εẼ ∈ H 1 (Ω \ D) and divergence free. Moreover, by Proposition 3, one has P ∈ H 2 (Ω \ D)3 satisfying kPkH 2 (Ω\D) ≤ CkẼkL2 (Ω\D) . By the Sobolev embedding theorem, we have |P(x) − P(y)| ≤ C|x − y|1/2 kẼkL2 (Ω\D) for x, y ∈ Ω \ D, sup |P(x)| ≤ CkẼkL2 (Ω\D) . x∈Ω\D Now we are in the position to estimate the numerator in (18) Z Z ε|Ẽ|2 dx = Ẽ · (∇ ∧ (µ−1 ∇ ∧ P) − ω 2 εP)dx Ω\D Ω\D Z = ∇ ∧ (µ−1 ∇ ∧ Ẽ) · P − ω 2 εẼ · Pdx Ω\D (20) Z Z − ν ∧ (µ−1 ∇ ∧ Ẽ) · PdS − ∂Ω Z =− ∂D ν ∧ (µ−1 ∇ ∧ E0 ) · PdS. ∂D Suppose {x ∈ ∂D | x · ρ = hD (ρ)} consists of N points {xj }N j=1 . Without loss of generality, we take N = 2. We decompose D as D1 ∪ D2 ∪ Γ where Dj are two Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 11 disjoint subdomains of D with xj ∈ Γj = ∂Dj \ Γ (j = 1, 2). Then ∂D = Γ1 ∪ Γ2 . Expanding the RHS of (20)at xj , one has Z ε|Ẽ|2 dx Ω\D =iω (Z 2 X j=1 ) Z (P(xj ) − P(x)) · (ν ∧ H0 )dS − Γj P(xj ) · (ν ∧ H0 )dS ∂Dj Z + iω (P(x1 ) − P(x2 )) · (ν ∧ H0 )dS (ΓZ Z 2 X = iω (P(xj ) − P(x)) · (ν ∧ H0 )dS − Γj j=1 ) 2 ω εP(xj ) · E0 dS Dj Z + iω (P(x1 ) − P(x2 )) · (ν ∧ H0 )dS Γ Z Z 2 Z X |x − xj |1/2 |ν ∧ H0 |dS + |E0 |dx + |ν ∧ H0 |dS kẼkL2 (Ω\D) ≤C D Γ j=1 Γj Z 2 Z X τ |x − xj |1/2 eτ (x·ρ−hD (ρ)) dS + eτ (x·ρ−hD (ρ)) dx ≤C Γ D j j=1 Z τ (x·ρ−hD (ρ)) + τe dS kẼkL2 (Ω\D) . Γ This yields kẼk2L2 (Ω\D) dx ≤C 2 X 1/2 τ (x·ρ−hD (ρ)) e dS Γj Z e D |x − xj | τ j=1 + !2 Z τ (x·ρ−hD (ρ)) 2 Z 2 ) τ (x·ρ−hD (ρ)) dx + τe dS . Γ R It’s easy to see that the second term can be absorbed by D µ|H0 |2 dx and the third term decays for τ 1 since points on Γ are a positive distance away from xj . Therefore, we only need to show that Z lim τ |x − xj |1/2 eτ (x·ρ−hD (ρ)) dS = 0, j = 1, 2. τ →∞ Γj This is shown in [11], where nonzero Gaussian curvature of ∂D at xj is required, which is satisfied by the choice of ρ. 5. Complex Spherical Waves and Reconstruction of Non-convex Obstacles. For various applications of CGO solutions, it is natural to consider using CGO solutions with non-linear limiting Carleman weights (LCW). In [16], a full characterization of possible LCWs in Euclidean spaces is given. Among those, to tackle the partial data problems for the Schrödinger equations in [17], a radial weight ln |x − x0 | (x0 ∈ R3 \ Ω) was chosen to construct CGO solutions, known as complex spherical waves (CSW). For Helmholtz equations, they are used in the enclosure method to probe non-convex acoustic obstacles in [21] and [28]. A key ingredient Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 12 Ting Zhou in the construction of CSW are Carleman estimates. Instead, in this work we construct CSW with LCW of the form 2 x − x0 (21) ϕ(x) = R + x0 · ρ |x − x0 |2 for R > 0, x0 ∈ R3 and ρ ∈ R3 \ {0}. The main tool in the construction is the inversion transformation with respect to the sphere S(x0 , R): x − x0 + x0 := y. (22) Tx0 ,R : x 7→ R2 |x − x0 |2 We have that Tx20 ,R = Id. This transformation preserves angles and maps generalized spheres (spheres and hyperplanes) into generalized spheres. Geometrically, fixing a reference circle S(x0 , R), enclosing D with spheres passing through x0 (CSW with LCW ϕ) corresponds to enclosing D̂x0 ,R = Tx0 ,R (D) with hyperplanes that do not pass through x0 , where the reconstruction scheme in Section 4 applies, see Figure 1. Figure 1. Enclosing D with CSW A rigorous argument consists of describing the behavior of the electromagnetic waves in the image domain Ω̂x0 ,R = Tx0 ,R (Ω). To simplify the notation, throughout the following, we drop the subscript (x0 , R). To see more clearly the invariance of Maxwell’s equations under the inversion transformation, we consider electromagnetic fields written as differential 1-forms (Einstein summation notation) E = Ej dxj , H = Hj dxj . The push-forward of E by T in Ω̂, denoted by Ê = T∗ E, is defined by Ê(y) = Êj dy j = (DT −1 )kj (y)Ek (T −1 (y)) dy j , y = T (x). Similarly, for the magnetic field H, we define Ĥ = T∗ H. Suppose (E, H) solves Maxwell’s equations in Ω \ D ∇ ∧ E = iωµH, Inverse Problems and Imaging ∇ ∧ H = −iωεE. Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 13 Then we have (Ê, Ĥ) satisfies in Ω̂ \ D̂ ∇ ∧ Ê = iω µ̂Ĥ, ∇ ∧ Ĥ = −iω ε̂Ê, where the material parameter µ̂ in Ω̂ is defined by (DT )T ◦ µ ◦ (DT ) (23) µ̂ = T∗ µ := ◦ T −1 , |det(DT )| and ε̂ = T∗ ε is defined similarly. Now we are at the point of checking the one-to-one correspondence between boundary measurements. Since the outer normal unit ν̂ of ∂ Ω̂ is the normalization of T∗ ν, and conversely ν is the normalization of (T −1 )∗ ν̂, one has (ν ∧ E)|∂Ω (x) = K1 (x)(ν̂ ∧ Ê)|∂ Ω̂ ◦ T (x), (ν̂ ∧ Ĥ)|∂ Ω̂ (y) = K2 (y)(ν ∧ H)|∂Ω ◦ T −1 (y), where K1 and K2 are defined by det(DT (x))DT −1 (y) , |DT (x)T ν̂(y)| y=T (x) K2 (y) := det(DT −1 (y)) DT (x)T ν̂(y) DT (x) x=T −1 (y) . K1 (x) := Hence, knowing the impedance map ΛD for the original domain, one can compute the map Λ̂D̂ for the image domain by h i (24) Λ̂D̂ (ν̂ ∧ Ê)|∂ Ω̂ = K2 (y) ΛD K1 (x)(ν̂ ∧ Ê)|∂ Ω̂ ◦ T (x) ◦ T −1 (y). We compute explicitly (25) R2 DT −1 (z) = DT (z) = |z − x0 |2 z − x0 I −2 |z − x0 | z − x0 |z − x0 | T ! , and 3 R2 det(DT (z)) = det(DT (z)) = (1 − 2|z − x0 |2 ). |z − x0 |2 According to (25), the push-forward of the isotropic parameters µ̂ and ε̂ given by (23) remain isotropic and satisfy the hypothesis in the previous sections when x0 is outside Ω. Therefore, by Section 3, one can construct CGO solutions with linear phase in Ω̂, denoted by (Ê0 , Ĥ0 ), depending on τ and t. Note that the pull-back of (Ê0 , Ĥ0 ) is a CSW with weight (21) of the original Maxwell system. Compute the impedance map Λ̂D̂ using (24), and define the indicator function for ρ ∈ S2 Z Iˆρ (τ, t) = iω (ν̂ ∧ Ê0 ) · [(Λ̂D̂ − Λ̂∅ )(ν̂ ∧ Ê0 ) ∧ ν̂]dS. −1 ∂ Ω̂ The support function of D̂ hD̂ (ρ) = sup y · ρ, for each ρ ∈ S2 \ Σ y∈D̂ is recovered using Theorem 4.3. Passing back to the original domain by applying the inversion transformation, we recover T (ch(D̂)). The corresponding radial support function of D is x − x0 2 h̃D (ρ) = sup R · ρ + x0 · ρ . |x − x0 |2 x∈D Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 14 Ting Zhou Summing up, we have shown the following theorem. Theorem 5.1. Given x0 ∈ R3 \ Ω and R > 0 such that Ω ⊂ B(x0 , R), there is a zero measure subset Σ of S2 , s.t., when ρ ∈ S2 \ Σ, we have h̃D (ρ) = inf{t ∈ R | lim Iˆρ (τ, t)}. τ →∞ To conclude this section, we give a remark describing the shape we reconstructed, i.e., T (ch(D̂)), using this type of CGO solutions. Suppose x0 ∈ R3 \Ω and R > 0 such that Ω ⊂ Bx0 ,R . In Figure 1, when enclosing D̂, hyperplanes passing through x0 (e.g., l4 , l5 ) have as pre-images (generalized spheres C4 , C5 ) under T themselves. Not only does the cone at x0 , formed by these hyperplanes, narrow the searching area of D, but also its intersection with ∂D separates ∂D into two subsets: the front Fx0 that faces x0 and the back Bx0 = ∂D \ Fx0 (see Figure 2). We observe that in Figure 1, some CSWs (e.g., C1 , C2 ) Figure 2. Front and back of ∂D probe the front Fx0 using their exterior while some (e.g., C3 ) probe the back Bx0 using their interior. Hence in practice, one can expand a sphere passing through x0 , by moving its center along one direction (perpendicular to ρ), until the indicator function is no longer vanishing as τ → ∞, suggesting that the sphere intersects the front Fx0 . In this way, some non-convex part of D from the front side can be recovered. By moving x0 ∈ R3 \ Ω, since the spheres passes x0 , we expect that this reconstruction scheme will recover more non-convex part of the obstacle than the one using CSWs with logarithm weight introduced in [17], whose level surfaces are spheres centered at x0 . 6. Enclosing other obstacles. In this section, we consider reconstruction of other types of obstacles, including a Perfect Electric Conducting (PEC) obstacle and an obstacle with impedance boundary condition. For the acoustic setting, we mention papers [12] and [13] for similar results. In both cases, we suppose our domain Ω, the obstacle D and all the electromagnetic parameters in the background satisfy the same hypothesis as in the PMC case. Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 15 6.1. PEC obstacles. We consider the electromagnetic field (E, H) satisfying Maxwell’s equations (2) with the boundary condition ν ∧ E|∂D = 0. (26) Then the reconstruction scheme Theorem 4.3 is valid simply by a similar equality as Lemma 4.4. 3 3 Lemma 6.1. Suppose (E, H) ∈ H 1 (Ω) × H 1 (Ω) satisfies (2) and (26) with f = ν ∧ E0 |∂Ω . Then we have Z Z (27) −I = µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx + µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx. Ω\D̄ D Therefore, the proof of Theorem 4.3 essentially follows the PMC case, except that when showing R ε|Ẽ|2 dx Ω\D = 0, lim R τ →∞ µ|H0 |2 dx D one needs to apply the Sobolev embedding theorem to the solution of ∇ ∧ P = iωµQ, ∇ ∧ Q = −iωεP − ωi Ẽ in Ω \ D, ν ∧ P|∂Ω = 0, ν ∧ P|∂D = 0. A regularity result for this BVP was studied in the appendix of [16]. 6.2. Obstacles with impedance boundary conditions. In this section, we consider a more general interface condition (iν ∧ H − λ(ν ∧ E) ∧ ν)|∂D = 0, (28) where λ ∈ L∞ (∂D) and Im λ(x) ≥ C > 0. This condition applies when the obstacle D is not perfectly conducting but does not allow the electromagnetic wave to penetrate deeply into D, where λ denotes the electromagnetic impedance of D. First, we state that the following BVP is well-posed by using a similar argument to the one in Appendix A. ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE, in Ω \ D, (29) (iν ∧ H − λ(ν ∧ E) ∧ ν)|∂D = g ∈ T H 1/2 (∂D), ν ∧ E|∂Ω = f ∈ T H 1/2 (∂Ω). Moreover, one has kEkH 1 (Ω\D) + kHkH 1 (Ω\D) ≤ C kf kH 1/2 (∂Ω) + kgkH 1/2 (∂D) . A generalization of the equalities in Lemma 4.4 and Lemma 6.1 can be obtained as following, analogous to Proposition 2.1 in [14], Lemma 6.2. Suppose (E, H) satisfies (29) with g = 0, Inverse Problems and Imaging and f = ν ∧ E0 |∂Ω . Volume 4, No. 3 (2010), 1–XX 16 Ting Zhou Then we have Z Z I= µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx + µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx Ω\D D Z 2 (30) ω −iImλ |(ν ∧ E) ∧ ν| dS + Reλ |(ν ∧ E0 ) ∧ ν|2 − |(ν ∧ Ẽ) ∧ ν|2 ∂D h i − 2Imλ Im (ν ∧ Ẽ) ∧ ν · (ν ∧ E0 ) ∧ ν dS. Proof. First we have with boundary terms Z I= −µ−1 (∇ ∧ E0 ) · (∇ ∧ E) + ω 2 εE0 · Edx Ω\D Z Z + µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx + ω E0 · i(ν ∧ H)dS. Ω ∂D Adding Z µ−1 (∇ ∧ E) · (∇ ∧ Ẽ) − ω 2 εE · Ẽdx − ω 0= Z i(ν ∧ H) · ẼdS, ∂D Ω\D we get Z I= µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx + Z µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx + I1 D Ω\D where the boundary term Z I1 =ω E0 · i(ν ∧ H) + i(ν ∧ H) · E0 − i(ν ∧ H) · EdS Z∂D =ω λ ((ν ∧ E0 ) ∧ ν) · (ν ∧ E) ∧ ν + λ ((ν ∧ E) ∧ ν) · (ν ∧ E0 ) ∧ ν ∂D − λ|(ν ∧ E) ∧ ν|2 dS Z =ω Reλ |(ν ∧ E0 ) ∧ ν|2 − |(ν ∧ Ẽ) ∧ ν|2 − iImλ|(ν ∧ E) ∧ ν|2 ∂D − iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E) ∧ ν + iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E0 ) ∧ ν + iImλ((ν ∧ E) ∧ ν) · (ν ∧ E0 ) ∧ ν − iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E0 ) ∧ νdS. Then we have (30). To deal with the boundary terms, we need the following lemma Lemma 6.3. Let Ω̃ be a bounded domain in Rn with a smooth boundary Γ. Then there exists a positive constant KΩ̃ depending on Ω̃ such that Z (31) |(ν ∧ F) ∧ ν|2 dS = kν ∧ Fk2L2 (Γ) ≤ KΩ̃ kFk2H 1 (Ω̃) , Γ Z (32) Γ |(ν ∧ F) ∧ ν|2 dS ≤ KΩ̃ kFk2H 1 (Ω̃) + −1 kFk2L2 (Ω̃) 3 for all F ∈ H 1 (Ω̃) and ∈ (0, 1). Proof. The proof of (31) is a direct corollary of a standard trace theorem. The proof of (32) is a variation of Theorem 1.5.1.10 in [10] for vector cases. For completeness, we include the proof here. Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 17 Given Γ smooth, one can extend the outer normal unit vector function ν(x) on Γ to a (C 1 (Ω̃))3 function, denoted by νext . Applying integration by parts, we obtain Z Z 2 |(ν ∧ F) ∧ ν| dS = |ν ∧ F|2 dS Γ ZΓ = (νext ∧ F) · (∇ ∧ F) − ∇ ∧ (νext ∧ F) · Fdx ZΩ̃ = (νext ∧ F) · (∇ ∧ F) − (∇ · F)(νext · F) − [(ν · ∇)νext ] · F Ω̃ + (∇ · νext )|F|2 + [(νext · ∇)F] · Fdx Z Z ≤C |∇F||F|dx + |F|2 dx Ω̃ Ω̃ ≤C k∇FkL2 (Ω̃) kFkL2 (Ω̃) + kFk2L2 (Ω̃) . Finally, one has Z |(ν ∧ F) ∧ ν|2 dS ≤ C k∇Fk2L2 (Ω̃) + −1 kFk2L2 (Ω̃) + kFk2L2 (Ω̃) . Γ This implies (32) for ∈ (0, 1), where KΩ̃ is the final general constant C, depending on ν hence on the domain. The reflected solution Ẽ = E − E0 ∈ H 1 (Ω \ D) is a solution of (29) with f = 0 and g = (−iν ∧ H0 + λ(ν ∧ E0 ) ∧ ν)|∂D ∈ T H 1/2 (∂D). Therefore, kẼkH 1 (Ω\D) ≤Ck − iν ∧ H0 + λ(ν ∧ E0 ) ∧ νkH 1/2 (∂D) ≤C(kH0 kH 1 (D) + kE0 kH 1 (D) ). Combined with Lemma 6.2 and Lemma 6.3 one can propose an upper bound |Re(I)| ≤ C kH0 k2H 1 (D) + kE0 k2H 1 (D) , which proves the decaying property (33) lim |Re(Iρ (τ, t))| = 0, τ →∞ for t > hD (ρ). Notice that although the H 1 norm of (E0 , H0 ) introduces one order of τ to the L2 norm, it doesn’t cancel the exponential decay. To show the non-vanishing property at t = hD (ρ), we first present several estimates. Since Ẽ satisfies (37) with f = 0 and g = ω λ(ν ∧ Ẽ) ∧ ν + λ(ν ∧ E0 ) ∧ ν − iν ∧ H0 ∈ T H 1/2 (∂D) ⊂ T H −1/2 (∂D), ∂D using (45), we have kẼkH 1 (Ω\D) ≤C kgkH −1/2 (∂D) n o ≤C k∇ ∧ ẼkL2 (Ω\D) + kẼkL2 (Ω\D) + k∇ ∧ E0 kL2 (D) + kE0 kL2 (D) n o (34) ≤C k∇ ∧ ẼkL2 (Ω\D) + kẼkL2 (Ω\D) + kH0 kL2 (D) + kE0 kL2 (D) Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 18 Ting Zhou 3 The second inequality is valid since for F ∈ H 1 (Ω \ D) , one has Z Z (∇ ∧ Ẽ) · F − Ẽ · (∇ ∧ F)dx, ((ν ∧ Ẽ) ∧ ν) · (ν ∧ F)dS = ∂D Ω\D the same estimate holds for k(ν ∧ E0 ) ∧ νkH −1/2 (∂D) on D and the second part of (14) is true. In addition, by the well-posedness of Maxwell’s equations satisfied by E0 in D with boundary condition ν ∧ µ−1 (∇ ∧ E0 )|∂D = iω(ν ∧ H0 )|∂D ∈ T H −1/2 (∂D), one obtains the estimate (35) kE0 kH 1 (D) ≤ Ckν ∧ H0 kH −1/2 (∂D) ≤ C k∇ ∧ E0 kL2 (D) + kE0 kL2 (D) . Combining Lemma 6.2, Lemma 6.3, (34) and (35), we obtain ! |Re(I)| ≥ min{µ−1 } − L k∇ ∧ Ẽk2L2 (Ω\D) − C kẼk2L2 (Ω\D) Ω\D + min{µ−1 } − L k∇ ∧ E0 k2L2 (D) − C kE0 k2L2 (D) D where L > 0 and C > 0. Choosing small enough such that K := min{µ−1 } − L > 0 Ω and dropping the first term, we obtain the lower bound |Re(I)| ≥ −C kẼk2L2 (Ω\D) + Kk∇ ∧ E0 k2L2 (D) − C kE0 k2L2 (D) . The remainder part of the proof is essentially the same as the PMC case. We propose the auxiliary problem ∇ ∧ (µ−1 ∇ ∧ P) − sε∇(∇ · εP) − ω 2 εP = εẼ in Ω \ D, ν ∧ P|∂Ω = 0, ν ∧ (µ−1 ∇ ∧ P) − ωλ(ν ∧ P) ∧ ν|∂D = 0. Then it can be verified that Z Z 2 ε|Ẽ| dx = − [ν ∧ µ−1 (∇ ∧ E0 )] · P − λ[(ν ∧ E0 ) ∧ ν] · PdS Ω\D ∂D Z Z = − iω (ν ∧ H0 ) · PdS + λ[(ν ∧ E0 ) ∧ ν] · PdS. ∂D ∂D We have shown the first term can be absorbed by kH0 k2L2 (D) for large τ using the regularity of P and Sobolev embedding theorem. It is not hard to see the same holds for the second term since Z Z λ[(ν ∧ E0 ) ∧ ν] · PdS ≤ C eτ (x·ρ−hD (ρ)) dS kẼkL2 (Ω\D) . ∂D ∂D This proves (36) lim inf |Re(Iρ (τ, t))| = τ →∞ C>0 ∞ t = hD (ρ), t < hD (ρ). Summing up, by (33) and (36), we obtain the main reconstruction theorem for obstacles with impedance interface conditions as following Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 19 Theorem 6.4. There is a subset Σ ⊂ S2 of measure zero such that when ρ ∈ S2 \ Σ, the support function hD (ρ) can be recovered by hD (ρ) = inf{t ∈ R | lim |Re (Iρ (τ, t))| = 0}. τ →∞ 7. Conclusion and Open problems. Since the first introduction of the enclosure method in the context of electrostatics, it has been generalized to Helmholtz equations, elasticity equations and some systems with Laplacian as the leading term, the heat equation and so on. In this work, for the first time, we generalized it to Maxwell’s equations. On this direction of research, there are several open problems and further development that we believe can be done in the future. In the acoustic case, [19] recently introduced the enclosure method for recovering penetrable obstacles, namely, coefficients bear a jump discontinuity on the obstacle but allow propagation of waves through the interface. There is loss of elliptic regularity for equations with L∞ coefficients and the Sobolev embedding theorem is not valid. This was tackled by developing some Hölder type of estimates for the equations. It would be interesting to generalize this to the electromagnetic setting. A more general and also much more difficult problem is to explore the method for the full dissipative Maxwell’s equations, in which the conductivity is nonzero. The difficulty is that the complex coefficients will compromise the lower bounds for the indicator function. Another interesting open problem is whether the depth stability analysis in [20] could be extended to Maxwell’s equations. A. Well-posedness of a mixed boundary value problem. Let Ω ⊂ R3 be a bounded domain with smooth boundary and D ⊂ Ω, Consider the boundary value problem for Maxwell’s equations ∇ ∧ E = iωµH ∇ ∧ H = −iωγE + J in Ω \ D, (37) ν ∧ E|∂Ω = f, ν ∧ (µ−1 ∇ ∧ E)|∂D = iων ∧ H|∂D = g. Here µ and γ are complex-valued functions in C k (Ω \ D) with positive real parts and ω ∈ C. Theorem A.1. There is a discrete subset Σ of C such that for ω not in Σ, there 3 3 exists a unique solution (E, H) ∈ H k (Ω \ D) × H k (Ω \ D) of (37) given any f ∈ T H k−1/2 (∂Ω), g ∈ T H k−1/2 (∂D) and J ∈ H k−1 (Ω \ D) with ∇ · J = 0. The solution satisfies (38) kEkH k (Ω\D) + kHkH k (Ω\D) ≤C kf kT H k−1/2 (∂Ω) + kgkT H k−1/2 (∂D) +kJkH k−1 (Ω\D) with C > 0. Here we proceed to prove the theorem by modifying the variational method in [16], [7] and [18]. From the Maxwell’s equations (37), the electric field E satisfies the second order equation ∇ ∧ (µ−1 ∇ ∧ E) − ω 2 γE = iωJ in Ω \ D, and (39) Inverse Problems and Imaging ∇ · γE = 0 in Ω \ D. Volume 4, No. 3 (2010), 1–XX 20 Ting Zhou Therefore, we consider ∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = iωJ in Ω \ D (40) where s is a positive real number. The equation (39) will follow from s∇ · (γ∇(∇ · γE)) + ω 2 ∇ · γE = 0 which is obtained by taking divergence of (40). Denoting the L2 (Ω \ D) inner product by (·, ·) and L2 (Γ) inner products by h·, ·iΓ (where Γ = ∂Ω or ∂D), we define the bilinear form associated with the elliptic system (40): B(E, F) := (µ−1 ∇ ∧ E, ∇ ∧ F) + s(∇ · γE, ∇ · γF) (41) for E, F ∈ X where X = {F ∈ (H 1 (Ω \ D))3 | ν ∧ F|∂Ω = 0, ν · γF|∂D = 0}. By Green’s formulae, we have that B is related to the differential operator P = ∇ ∧ (µ−1 ∇∧) − sγ∇(∇ · γ) by (42) B(E, F) = (P E, F) − hν ∧ (µ−1 ∇ ∧ E), Fi∂(Ω\D) + hs∇ · γE, ν · γFi∂(Ω\D) for E, F ∈ (H 1 (Ω \ D))3 . Then for f˜ ∈ X 0 the weak formulation of the mixed boundary value problem in Ω \ D, P E = f˜ ν ∧ E|∂Ω = 0, ν ∧ (µ−1 ∇ ∧ E)|∂D = 0. is: find E ∈ X such that B(E, F) = (f˜, F) for all F ∈ X. By (42), this implies the natural boundary condition (43) ν ∧ (µ−1 ∇ ∧ E)|∂D = 0, ∇ · γE|∂Ω = 0. To show the theorem, one first has for the homogeneous boundary conditions, Proposition 2. Suppose γ and µ are complex functions in C 1 (Ω \ D) with positive real parts, and let s be a positive real number. There is a discrete set Σs ⊂ C such that if ω is outside this set, then for any f ∈ X 0 there exists a unique solution E ∈ X of (44) ∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = f˜ satisfying kEkH 1 (Ω\D) ≤ Ckf˜kX 0 . Proof. It is sufficient to show that B is bounded and coercive on X. It is clear that B is bounded, |B(E, F)| ≤ CkEkH 1 (Ω\D) kFkH 1 (Ω\D) . To show the coercivity, first we have |B(E, E)| ≥ c k∇ ∧ Ek2L2 (Ω\D) + k∇ · Ek2L2 (Ω\D) − CkEk2L2 (Ω\D) . It can be shown that there is a Poincaré inequality for 1-forms in X similar to that in [26] n o kEk2H 1 (Ω\D) ≤ C kEk2L2 (Ω\D) + k∇ ∧ Ek2L2 (Ω\D) + k∇ · Ek2L2 (Ω\D) , E ∈ X. Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX Reconstructing electromagnetic obstacles 21 We have then B(E, E) ≥ C1 kEk2H 1 (Ω\D) − C2 kEk2L2 (Ω\D) . To show the higher order regularity of solutions, we define for k ≥ 2, X k := F ∈ (H k (Ω \ D))3 | ν ∧ F|∂Ω = ∇ · γF|∂Ω = ν · γF|∂D = ν ∧ (µ−1 ∇ ∧ F)|∂D = 0 . Then we have Proposition 3. Let γ and µ be functions in C k (Ω \ D), k ≥ 2, with positive real parts, and let s > 0. Suppose ω ∈ / Σs , then for any f˜ ∈ (H k−2 (Ω \ D))3 the equation (44) has a unique solution E ∈ X k and kEk k ≤ Ckf˜k k−2 . H (Ω\D) H (Ω\D) This can be proved by the same techniques in Section 5.9 of [26] for the Hodge Laplacian. Proof of Theorem A.1. As in [16], we take Σ to be the set Σ1 in previous propositions, then s can be chosen such that ω ∈ / Σs (for more details see [16]). 3 3 To show uniqueness, suppose (E, H) ∈ H 1 (Ω \ D) ×H 1 (Ω \ D) solves (37) with f = g = 0 and J = 0. One has ∇ ∧ (µ−1 ∇ ∧ E) = ω 2 γE, ∇ · γE = 0. It follows that E ∈ X (by the natural boundary conditions) is a solution of (44) with f˜ = 0, which implies E = H = 0 by Proposition 2. For existence, given f ∈ T H k−1/2 (∂Ω), g ∈ T H k−1/2 (∂D) ⊂ T H k−3/2 (∂D) and J ∈ H k−1 (Ω \ D), we can find E0 ∈ H k (Ω \ D) with ν ∧ E0 |∂Ω = f, ν ∧ (µ−1 ∇ ∧ E0 )|∂D = g such that the extension is bounded, namely kE0 kH k (Ω\D) ≤ C(kf kH k−1/2 (∂Ω) + kgkH k−3/2 (∂D) ) Suppose Ẽ ∈ X k is a solution, given by Proposition 3, of (44) with f˜ = −∇ ∧ (µ−1 ∇ ∧ E0 ) + sγ∇(∇ · γE0 ) + ω 2 γE0 + iωJ ∈ (H k−2 (Ω \ D))3 . Notice that kf˜kH k−2 (Ω\D) ≤C kE0 kH k (Ω\D) + kJkH k−2 (Ω\D) ≤C kE0 kH k (Ω\D) + kJkH k−1 (Ω\D) . Then E = E0 + Ẽ ∈ (H k (Ω \ D))3 satisfies ∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = iωJ. This implies ∇ · γE = 0 by an earlier argument and particular choice of s. If 1 we define H = iωµ ∇ ∧ E ∈ (H k−1 (Ω \ D))3 , then we have (E, H) the solution of Maxwell’s equations. Applying the same argument to H, which satisfies a second order elliptic system ∇ ∧ (γ −1 ∇ ∧ H) − ω 2 µH = ∇ ∧ (γ −1 J) ∈ H k−2 (Ω \ D) Inverse Problems and Imaging Volume 4, No. 3 (2010), 1–XX 22 Ting Zhou by eliminating E from the original Maxwell’s equations. Notice that ∇ · (∇ ∧ (γ −1 J)) = 0. By uniqueness, one has H ∈ (H k (Ω \ D))3 . Finally, the estimate (38) is derived from (45) kEkH k (Ω\D) ≤kE0 kH k (Ω\D) + kẼkH k (Ω\D) ≤C kE0 kH k (Ω\D) + kf˜kH k−2 (Ω\D) ≤C kE0 kH k (Ω\D) + kJkH k−1 (Ω\D) ≤C kf kH k−1/2 (∂Ω) + kgkH k−3/2 (∂D) + kJkH k−1 (Ω\D) ≤C kf kH k−1/2 (∂Ω) + kgkH k−1/2 (∂D) + kJkH k−1 (Ω\D) . 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