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The inverse electromagnetic scattering problem for screens, Inverse Problems 19

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The Inverse Electromagnetic Scattering Problem for Screens
? Department of Mathematical Sciences University of Delaware, Newark, Delaware 19716, USA Abstract. We consider the inverse electromagnetic scattering problem for perfectly conducting screens. We solve this problem by modifying the linear sampling method formulated in [3] for the case of scattering obstacles with nonempty interior. Numerical examples are given for screens in R3 . Keywords: Electromagnetic inverse scattering, linear sampling method, screens.

Fioralba Cakoni,? David Colton?§ and Eric Darrigrand ?

1. Introduction The scattering of electromagnetic waves by a perfect conductor has a venerable history in applied mathematics. In particular, for bounded scattering obstacles with smooth boundary, the mathematical theory of such problems is well developed and is a cornerstone of electromagnetic scattering theory. However, in many problems occurring in practice one encounters thin objects where the thickness is small compared with the wavelength and other characteristic dimensions. Such problems lead to a mathematical model in which the scattering obstacle is an in?nitely thin screen and the mathematical investigation of this type of problem is of relatively recent origin [1], [2]. The main di?culty in this investigation is to establish the compactness of the relevant integral operators in the function spaces appropriate for the mathematical treatment of Maxwell’s equations and the problem was ?rst addressed by Abboud and Starling in 1993 [1]. In recent years the inverse scattering problem for electromagnetic waves has attracted increased attention [8]. However, until now, these investigations have all assumed a priori that the scattering obstacle has a nonempty interior. One reason for this assumption was the fact that in order to understand the inverse problem it is necessary to ?rst have a ?rm mathematical understanding of the direct problem and, as mentioned above, the mathematical basis for scattering by objects with empty interior such as screens was not established until very recently. A second reason for avoiding the inverse electromagnetic scattering problem for screens was that techniques suitable for the numerical solution of inverse electromagnetic scattering problems have only recently been developed (at
§ To whom correspondence should be addressed (colton@math.udel.edu)

The Inverse Electromagnetic Scattering Problem for Screens

2

least for frequencies in the resonance region and with minimal a priori assumptions on the geometry of the scattering obstacles - c.f. [7] and the references contained therein). Since these mathematical reasons for avoiding the inverse electromagnetic scattering problem for screens are no longer valid, it seems to us that the time is appropriate to initiate an investigation of this particular problem in inverse scattering theory. This paper is dedicated to such an e?ort. The plan of our paper is the following. In the next section will summarize the basic existence, uniqueness and regularity results for the direct scattering problem for (orientable) screens and formulate the corresponding inverse problem. We then proceed in Section 3 to study the inverse problem for cylindrical screens assuming TM-polarized incident ?elds. Our approach to this problem is the linear sampling method and in this regard we modify the ideas given in [3] for the case when the scatterer has a nonempty interior. We then turn to the main aim of this paper, the study of the inverse electromagnetic scattering problem for (bounded) screens in R3 , and again base our approach on the linear sampling method as formulated in [3] for obstacles with nonempty interior. We conclude our paper with some numerical examples that illustrate the practicality of our approach. 2. Formulation of the direct and inverse scattering We consider the scattering of a time-harmonic electromagnetic plane wave by a very thin perfectly conducting obstacle in R3 . In particular let Γ be a bounded, simply connected, orientated, smooth open surface in R3 with piecewise smooth boundary curve l which does not intersect itself. We consider Γ as part of the smooth boundary ?D of some bounded connected open set D ? R3 . We denote by ν the normal vector to Γ that coincides with the outward normal vector to ?D. Moreover we denote by ν × E + |Γ , γT E + |Γ and ν · E + |Γ , (ν × E ? |Γ , γT E ? |Γ and ν · E ? |Γ ) the restriction to Γ of the traces ν × E |?D , γT E |?D and ν · E |?D respectively, from the outside (from the inside) of ?D where γT u := ν × (u × ν ) is the tangential component of u. The scattering of electromagnetic waves by the open surface Γ (the screen) leads to the following boundary value problem for the scattered electric ?eld E and magnetic ?eld H: curl H + ikE = 0 γT E = c
±

curl E ? ikH = 0

in in on

R \Γ Γ

R3 \ Γ
3

(1a) (1b) (1c)

where c = ?γT E i := ?ν × (E i × ν )|Γ with E i being the electric ?eld of the electromagnetic plane wave given by i E i (x; d, p) = curl curl p eikd·x = ik (d × p) × d eikd·x k i H (x; d, p) = curl curl p eikd·x = ikd × p eikd·x , (2)

The Inverse Electromagnetic Scattering Problem for Screens

3

where k > 0 is the wave number, d ∈ ? is a unit vector giving the direction of the incident plane wave and p ∈ R3 is the polarization. Here ? denotes the unit sphere in R3 . Moreover the scattered ?eld E , H satis?es the Silver M¨ uller radiation condition
r →∞

lim (H × x ? rE ) = 0

(3)

uniformly in x ? = x/|x|, where r = |x|. From a standard argument (see [1] for details) the Silver-M¨ uller radiation condition implies the following theorem: Theorem 2.1 The scattering problem (1a)-(1c), (3) has at most one solution.
1 Letting Hloc (IR3 ), H 2 (?D) and H ? 2 (?D) denote the usual Sobolev spaces, we recall the de?nition of the following spaces
1 1

L2 t (?D )
?1 ?1

Hloc (curl, R3 \ Γ) :=

:= := :=

3 3 2 3 3 u ∈ (L2 loc (R \ Γ)) : curl u ∈ (Lloc (R \ Γ))

u ∈ (L2 (?D))3 : ν · u = 0 on ?D u ∈ H ? 2 (?D), u ∈ H ? 2 (?D),
1 1 1 1

2 Hdiv (?D) 2 (?D) Hcurl

div?D u ∈ H ? 2 (?D) curl?D u ∈ H ? 2 (?D)

where L2 is the space of square integrable functions de?ned on the indicated domain. To de?ne properly the traces on Γ we need the following spaces H 2 (Γ) := u|Γ : u ∈ H 2 (?D)
1 1 1 ?2 ? ?2 (Γ) and H (Γ) the dual space of Now we denote by H ? 2 (Γ) the dual space of H 1 1 ? 2 ? H 2 (Γ) with L (Γ) as the pivot space. Note that H 2 (Γ) can also be identi?ed with 1 1 ? ?2 H (Γ) := {u ∈ H ? 2 (Γ) : supp u ? Γ}. 1 ?1 2 (Γ) := {u ∈ H 2 (Γ) : supp u ? Γ}. H 1 1

Moreover we de?ne
2 (Γ) := {u ∈ (H ? 2 (Γ))3 , ν · u = 0, div?D u ∈ H ? 2 (Γ)} Hdiv

?1

1

1

? ? 2 (Γ) the dual space of H ? 2 (Γ) in the duality pairing H ? 2 , H ? 2 . Let us denote by H div curl div curl 1 1 ? ? 2 (Γ) and ? ? 2 (Γ))3 , div?D u ∈ H This space contains tangential ?elds u such that u ∈ (H
Γ
3

2 (Γ) := {u ∈ (H ? 2 (Γ))3 , ν · u = 0, curl?D u ∈ (H ? 2 (Γ))3 }. Hcurl 1 1

?1

1

1

1

1

u · grad?D v ds +

div?D u v ds = 0
Γ

for every v ∈ H 2 (Γ). The latter means that the normal trace of u at the edge l is well de?ned and is zero, that is νl · u|l = 0 where νl is the exterior normal vector at the boundary l of Γ (for more details see [1] and [5], p.47). Note also that a function 1 1 ? ? 2 (Γ) can be extended by zero to a function in H ? 2 (?D). u∈H
div div

The forward problem (1a)-(1c), (3) is investigated in detail in [1] and for more complicated geometry in [2]. We recall here some of their results that will be used

The Inverse Electromagnetic Scattering Problem for Screens

4

later on to solve the inverse problem. In terms of the electric ?eld E the scattering problem (1a)-(1c), (3) can be formulated
2 (Γ) ?nd E ∈ Hloc (curl, R3 \ Γ) (we will refer to this problem as (ESP): given c ∈ Hcurl that satis?es

?1

curl curl E ? k 2 E = 0 γT E ± = c
r →∞

in on

R3 \ Γ Γ

(4a) (4b) (4c)

lim (curl E × x ? ikrE ) = 0.

Let [ν × u]|Γ = ν × u+ |Γ ? ν × u? |Γ denote the jump of the tangential component of u across the screen Γ. By applying the vector Green formula with the test function whose support does not intersect Γ one can de?ne the traces [ν × curl E ] on Γ and show the following: ? ? 2 (Γ). Lemma 2.2 If E ∈ Hloc (curl, R3 \ Γ) solves (ESP) then [ν × curl E ]|Γ ∈ H div Furthermore, an application of the vector Green formula and using the continuity of the tangential components of E and curl E across ?D \ Γ shows that the jump [ν × curl E ] of a solution E ∈ Hloc (curl, R3 \ Γ) to (ESP) satis?es γT
Γ
1

[ν × curl E (y )]Φ(x, y ) dsy +
eik|x?y| 4π |x?y |

1 gradx k2

Γ

divΓ [ν × curl E (y )]Φ(x, y ) dsy

= c(x)

where Φ(x, y ) :=

is the fundamental solution of the Helmholtz equation in R3 .
1 1 1

? ? 2 (Γ) onto H ? 2 (Γ) ≡ H ? ? 2 (Γ) Let us denote by A the operator from H div curl div (Aφ)(x) := γT
Γ



de?ned by

φ(y )Φ(x, y ) ds(y ) +

1 gradx k2

divΓ φ(y )Φ(x, y ) ds(y ) .
Γ

Then the jump [ν × curl E ] satis?es the integral equation Aφ = c.
1

(5)

? ? 2 (Γ) solves (5) then a solution E of Conversely, one can easily show that if φ ∈ H div (ESP) is given by: E (x) = =
Γ

1 curlx curlx k2

φ(y )Φ(x, y ) ds(y )
Γ

x ∈ R3 \ Γ divΓ φ(y )Φ(x, y ) ds(y ).

(6)

φ(y )Φ(x, y ) ds(y ) +

1 gradx k2
1

Γ

In [2] (Corollary 3.6) the following lemma it is proved: ? ? 2 (Γ) → H ? 2 (Γ) is an isomorphism. Lemma 2.3 The operator A : H div curl Hence Lemma 2.3 and the previous analysis imply the following: Theorem 2.4 The problem (ESP) has a unique solution E ∈ Hloc (curl, R3 \ Γ). This 1 curl E is the unique solution of the scattering problem (1a)-(1c), (3). E and H = ik
1

The Inverse Electromagnetic Scattering Problem for Screens In [8] it is shown that the scattered ?eld E, H has the asymptotic behavior E (x) = eik|x| |x| E∞ (? x; d, p) + O 1 |x| , H (x) = eik|x| |x| H∞ (? x; d, p) + O 1 |x|

5

as |x| → ∞, where E∞ (· ; d, p) and H∞ (· ; d, p) are de?ned on the unit sphere ? and are the electric far ?eld pattern and the magnetic far ?eld pattern corresponding to the incident direction d and polarization p. Moreover they satisfy [8] H∞ (? x; d, p) for all x ?, d ∈ ? and p ∈ R3 . x ? · H∞ (? x; d, p) = x ? · E∞ (? x; d, p) = 0 =x ? × E∞ (? x; d, p),

The inverse problem we will consider in this paper is to determine Γ from a knowledge of the electric far ?eld pattern for x ?, d ∈ ? (or a subset ?0 ? ? - c.f. [3]) and 3 all polarizations p ∈ R . We will use the linear sampling method to solve the inverse problem. For perfectly conducting screens in the case of TM polarization, in [4], [10]and [11] it is noted that for certain geometries of Γ the set of electric far ?eld patterns is not dense in L2 (?). As will be seen later, this is also the case for an arbitrary polarization in R3 . This causes the linear sampling method to fail for such geometries. To avoid this problem we will adapt here the idea from [3] of using both the electric and magnetic far ?eld patterns as data. More precisely we will investigate the possibility of determining Γ from a knowledge of both E∞ (? x; d, p) and H∞ (? x; d, p × d), for x ?, d ∈ ?0 ? ? and 3 p ∈ R by using the linear sampling method. Note that the electric far ?eld pattern E∞ (? x; d, p) and magnetic far ?eld pattern H∞ (? x; d, p × d) correspond to an incident ?eld propagating in the same direction d but polarized perpendicular to each other. We also note that H∞ can be computed from a knowledge of E∞ . 3. The scalar case To introduce our ideas we ?rst consider the scattering by a perfectly conducting in?nite cylindrical surface parallel to the z-axis. We assume that the incident wave propagates in the direction perpendicular to the cylinder, that is d · nz = 0, where nz is the unit vector on the z axis. Throughout this section we denote by Γ the cross section of this cylindrical surface and again consider it as part of a smooth contour ?D bounding a connected domain D ∈ R2 . Here ν is the unit normal vector to Γ that coincides with the outward normal vector to ?D. The positive and negative traces on Γ are de?ned as in the previous section. If we further assume that the incident electric wave is polarized parallel to the z axis then the electric scattered ?eld has only one component E in the nz direction which satis?es ?E + k 2 E = 0 E =f √ lim r
r →∞ ±

in
±

on Γ , ?E ? ikE = 0 ?r

IR2 \ Γ

(7a) (7b) (7c)

The Inverse Electromagnetic Scattering Problem for Screens

6

where f := ?eikd·x |Γ and the radiation condition (7c) is satis?ed uniformly in x ? = x/r 2 for x ∈ R and r = |x|. Now assume that the incident electric wave is polarized perpendicular to the z axis. Then the magnetic scattered ?eld has also one component H in the nz direction and H satis?es ?H + k 2 H = 0 in IR2 \ Γ ?H ± =h on Γ± , ?ν √ ?H ? ikH = 0 lim r r →∞ ?r
ikd·x 1 1

(8a) (8b) (8c)

where h := ? ?e?ν |Γ . For given f ∈ H 2 (Γ) and h ∈ H ? 2 (Γ) it is proved in [13] (the proof in R2 is the same as in R3 ) that there exists a unique solution to (7a)-(7c) and 1 (8a)-(8c) respectively. In particular these solutions E and H belong to Hloc (R2 \ Γ) and have the asymptotic behavior [8] 1 eikr x; d) + O 3/2 , E (x) = √ E∞ (? r r ikr e 1 H (x) = √ H∞ (? x; d) + O 3/2 , r r (9) (10)

where E∞ (? x; d) and H∞ (? x; d) are the far ?eld patterns corresponding to the Dirichlet problem (7a)-(7c) and the Neumann problem (8a)-(8c) respectively. The inverse scattering problem we are concerned with is to determine the cross section Γ from a knowledge of both E∞ (? x; d) and H∞ (? x; d) for x ? and d on the unit circle ? and ?xed wave number k . To this end we de?ne the combined far ?eld operator F : L2 (?) → L2 (?) by (F g )(? x) := λ
? 2

E∞ (? x; d)g (d)ds(d)+?
?

H∞ (? x; d)g (d)ds(d),

x ? ∈ ?,(11)

where g ∈ L (?), and λ and ? are real numbers not both zero. Note that in contrast to the case of an obstacle with nonempty interior (see [3]) here λ and ? can be of the same sign. We observe that by superposition the combined far ?eld operator (11) is a linear combination of the far ?eld patterns corresponding to the Dirichlet problem (7a)-(7c) and the Neumann problem (8a)-(8c) respectively, corresponding to the Herglotz wave function vg (x) =
?

eikx·d g (d) ds(d),

x ∈ IR2 ,

(12)

with kernel g ∈ L2 (?) as incident ?eld. We want to characterize Γ by the behavior of an approximate solution g of the far ?eld equation (F g ) = Φe ∞, (13)

The Inverse Electromagnetic Scattering Problem for Screens

7

where Φe ∞ is the far ?eld pattern of a suitable scattered ?eld (to be de?ned later). In order to understand better the combined far ?eld operator (11) we de?ne a new operator B as follows.
1 1

De?nition 3.1 The operator B : H 2 (Γ) × H ? 2 (Γ) → L2 (?) maps (f, h) onto λE∞ + ?H∞ where E∞ is the far ?eld pattern of the radiating solution E of (7a)(7c) with boundary data f and H∞ is the far ?eld pattern of the radiating solution H of (8a)-(8c) with boundary data h. If H denotes the mapping of L2 (?) into H 2 (Γ) × H ? 2 (Γ) de?ned by ? ? g (d)eikx·d ds(d) ? ? ? ? Hg (x) := x ∈ Γ, ? ? ? ikx · d ? ds(d) ? ?νx g (d)e
?
1 1

(14)

then by superposition the combined far ?eld operator can be written as ? To characterize the range of B we now consider the compact operator F : H 1 ? 2 (Γ) ?→ L2 (?) de?ned by H ? ?ikx ? ·y dsy + ? β (y ) F (α, β )(? x) = λ α(y )e?ikx e ?·y dsy . ?νy
Γ Γ
1 ? ?2 1 2

(F g ) = ?B(Hg ).

(15)
1 ?2

(Γ) × (16)

? (Γ), the function F (α, β )(? x) is the linear Note that for a given pair (α, β ) ∈ H (Γ) × H ?1 combination of the far ?eld patterns corresponding to γ S (α)(x) and γ ?1 D(β )(x), eiπ/4 where γ = √ and S (α) and D(β ) are the single- and double-layer potentials, 8πk respectively, de?ned by S (α)(x) :=
Γ

α(y )Φ(x, y )dsy ,

D(β )(x) :=
Γ

β (y )

? Φ(x, y )dsy ?νy
1

x ∈ R2 \ Γ.
1

By extending the densities α and β by zero to functions in H 2 (?D) and H ? 2 (?D) respectively, and using the jump relation across ?D for the single- and double- layer 1 1 (R2 \ Γ) (see [12]). potentials, it can be shown that S (α) ∈ Hloc (R2 \ Γ) and D(β ) ∈ Hloc Moreover they satisfy (S α)± |Γ = SΓ α on Γ

? (Dβ )± = TΓ β on Γ, ?ν where the boundary integral operators SΓ and TΓ are de?ned by (SΓ α)(x) :=
Γ

α(y )Φ(x, y )dsy ,

(TΓ β )(x) :=

? ?νx

β (y )
Γ

? Φ(x, y )dsy , ?νy

x ∈ Γ.

Therefore, we have the following relation F (α, β ) = γ ?1 B(SΓ α, TΓ β ) (17)

The Inverse Electromagnetic Scattering Problem for Screens

8

1 1 ?1 ?1 ? ?2 2 (Γ) → H 2 (Γ) are isomorphisms [12], (Γ) → H 2 (Γ) and TΓ : H Note that SΓ : H [13]. 1 ?1 ? ?2 2 (Γ) → (Γ) × H Lemma 3.2 Assume that λ = 0 and ? = 0. Then the operator F : H 2 L (?) de?ned by (16) is injective and has dense range.

Proof. This is Lemma 4.1 of [4] Note that if ? = 0 then the statement of Lemma 3.2 is true under the restriction that does not exist a Herglotz wave function which vanishes on Γ. In particular, by (15) and (17) if the latter is not the case than the set of far ?eld patterns is not dense in L2 (?). The following lemma characterizes the range of F and hence the range of B.

1 ? ?2 (L), Lemma 3.3 For any smooth non intersecting arc L and two functions αL ∈ H 1 L 2 ? 2 (L) we de?ne Φ∞ ∈ L (?) by βL ∈ H

ΦL x) := λ ∞ (?
L

? ·y αL (y )e?ikx dsy + ? L

βL (y )

? ?ikx e ?·y dsy . ?νy

(18)

Then, ΦL x) ∈ Range(F ) if and only if L ? Γ. ∞ (? Proof. This is Lemma 4.3 of [4] Note that since SΓ and TΓ are isomorphisms the range of B and F coincide. Finally we recall an approximation property of the Herglotz wave functions. Lemma 3.4 The range of the operator H : L2 (?) → H 2 (Γ) × H ? 2 (Γ) is dense. Proof. The proof follows as in Theorem 3.1 of [4] setting the surface impedance λ = 0. Note that Lemma 3.4 shows that there exists a Herglotz function vg such that vg |Γ and 1 1 ?vg | approximates f ∈ H 2 (Γ) and h ∈ H ? 2 (Γ), respectively. ?ν Γ By using regularization theory for the ?rst kind ill posed equation
?1 ?1 γ ?1 B(f, h) = F (SΓ f, TΓ h) = ΦL ∞
1 1

and the approximation of (f, h) ∈ H 2 (Γ) × H ? 2 (Γ) by Hg we can now prove exactly in the same way as in Theorem 4.4 of [4] the main result of this section. Let us denote by L the set of open nonintersecting smooth arcs and look for a solution g ∈ L2 (?) of the combined far ?eld equation (F g )(? x) := λ
?

1

1

E∞ (? x; d)g (d)ds(d) + ?
?

H∞ (? x; d)g (d)ds(d) = ?γ ΦL ∞ (19)

for L ∈ L, where

ΦL ∞

is de?ned by (18). Then the following is true:

Theorem 3.5 If F is the combined far ?eld operator (11) corresponding to (7a)-(7c) and (8a)-(8c), and λ = 0 and ? = 0 then we have that

The Inverse Electromagnetic Scattering Problem for Screens
L (i) if L ? Γ then for every ? > 0 there exists a solution g? ∈ L2 (?) of the inequality L F g? + γ ΦL ∞ L2 (?)

9

≤ ?.

L (ii) if L ? Γ then for every ? > 0 and δ > 0 there exists a solution g?,δ ∈ L2 (?) of the inequality L F g?,δ + γ ΦL ∞ L2 (?) ≤ ? + δ

such that
δ →0 L lim g?,δ L2 (?)

=∞

and

δ →0

L lim vg?,δ

H 1 (B R )

=∞

L L is the Herglotz wave function with kernel g where vg?,δ ?,δ .

In practice we need to replace ΦL ∞ in the combined far ?eld equation (19) by a expression independent of L. As we remarked in [4], assuming that there does not exist a Herglotz wave function which vanishes on L (this is not a restriction since L can be arbitrarily chosen), we can conclude from Lemma 4.2 in [4] that the class of potentials of the 1 ? ·y ? ?2 (L) is dense in L2 (?) and hence for numerical form α(y )e?ikx dsy with α ∈ H purposes we ?rst replace ΦL ∞ by an expression of the above form. Next, we note that as L degenerates to a point z with αL an appropriate delta sequence then the above ? ·z ?ikx ? ·z integral approaches e?ikx . Hence, it is reasonable to replace ?γ ΦL ∞ by Φ∞ := γe (the far ?eld pattern of the fundamental solution Φ(x, z )) when numerically solving the combined far ?eld equation (19). 4. The vector case Now we turn our attention to the inverse problem formulated in Section 1 i.e. determining the shape of a smooth open surface Γ in R3 from a knowledge of both E∞ (? x; d, p) and H∞ (? x; d, p × d) = x ? × E∞ (? x; d, p × d), for x ?, d on the unit sphere ? (or 3 a subset ?0 ? ?) and p ∈ R . 2 We again consider the combined far ?eld operator F : L2 t (?) → Lt (?) de?ned by (F g )(? x) := λ
? L

E∞ (? x; d, g (d))ds(d) + ?
?

H∞ (? x; d, g (d) × d)ds(d),

(20)

where g ∈ L2 t (?) and λ and ? are real numbers not both zero. Recall [8] that an electromagnetic Herglotz pair with kernel g ∈ L2 t (?) is de?ned to be a pair of vector ?elds of the form 1 (21) Eg (x) = eikx·d g (d)ds(d), Hg (x) = curl Eg (x). ik
?

Hence by superposition F g is a linear combination of the electric far ?eld pattern corresponding to the electromagnetic Herglotz pair with kernel ikg (d) as incident ?eld and the magnetic far ?eld pattern corresponding to the electromagnetic Herglotz pair ? ∞ where E∞ is the with kernel ikg (d)×d as incident ?eld. More precisely F g = λE∞ +?H

The Inverse Electromagnetic Scattering Problem for Screens

10

?∞ = x ?∞ is the far ?eld pattern of the solution E of (ESP) with c := ?ikγT Eg and H ?×E 1 ? where E ? is the solution of (ESP) with c := γT (curl Eg ). ? = curl E far ?eld pattern of H ik ? follows from the perfectly conducting We note that the boundary condition for E ? expressing the electric ?eld of the Herglotz pair with boundary condition satis?ed by E kernel g (d) × d in terms of the electric ?eld with kernel g (d). The linear sampling method for solving the inverse problem consists in solving the combined far ?eld equation (F g )(? x) = Ee,∞ (22) where Ee,∞ is the electric far ?eld pattern of a suitable electromagnetic ?eld (to be de?ned later). In the following we de?ne an operator, again denoted by B, that gives us more insight into the combined far ?eld operator. De?nition 4.1 The operator B ?∞ where E∞ and λE∞ + ?x ?×E ? of (ESP) with boundary data c E
?1 ?1
2 2 (Γ) × Hcurl (Γ) → L2 ?) onto : Hcurl t (?) maps (c, c ? E∞ are the far ?eld patterns of the solution E and and c ? respectively.

?1

?1

Then by superposition the combined far ?eld operator (20) can be factored as Next we want to characterize the range of B. To do this we introduce a boundary 1 1 ? ? 2 (Γ) × H ? ? 2 (Γ) → L2 integral operator F : H t (?) de?ned by div div F (α, β ) := x ?× λ
1 1

2 2 Let H : L2 t (? → Hcurl (Γ) × Hcurl (Γ) be the operator de?ned by ? ? γT Eg on Γ ? Hg := ? ? i γ curl E on Γ. g k T

(23)

(F g ) = ?ik B(Hg ).

(24)

Γ

? ·y α(y ) e?ikx dsy + ?x ?×

? ·y β (y )e?ikx dsy Γ

×x ? (25)

? ? 2 (Γ). F is clearly a compact operator. Moreover we observe ? ? 2 (Γ) and β ∈ H for α ∈ H div div that ? ·y x ? × α(y ) e?ikx dsy × x ? = (P α)∞
Γ

where the potential P α is given by (P α)(x) = whereas x ?× x ?×

1 curl curl k2

α(y )Φ(x, y )dsy
Γ

x ∈ R3 \ Γ

? ·y α(y ) e?ikx dsy Γ

×x ?=x ? × (P β )∞

where x ? × (P β )∞ is the rotated far ?eld pattern of the potential P β given by (P β )(x) = 1 curl curl k2 β (y )Φ(x, y )dsy
Γ

x ∈ R3 \ Γ.

The Inverse Electromagnetic Scattering Problem for Screens
?1

11

2 (?D), the potentials P α and Since α and β can be extended by zero to functions in Hdiv 3 P β are well de?ned as functions in Hloc (curl, R \ Γ) and satisfy curl curl u ? k 2 u = 0 in R3 \ Γ. Moreover, from the continuity across ?D of the tangetial component of the vector potential P with the layers extended by zero [8], we have

γT (P α)± |Γ = Aα

γT (P β )± |Γ = Aβ 1 gradx k2

on

Γ

where the operator A (c.f. Section 1) is given by (Aφ)(x) := γT
Γ

φ(y )Φ(x, y ) ds(y ) +

divΓ φ(y )Φ(x, y ) ds(y ) .
Γ
1 1

? ? 2 (Γ) into H ? 2 (Γ). Hence From Lemma 2.3 we know that A is an isomorphism from H div curl the operators F and B are related by the relation 1 F (α, β ) = 2 B ((Aα), (Aβ )) (26) k ? 2 (Γ) → L2 ? 2 (Γ) × H Lemma 4.2 The operator F : H t (?) de?ned by (25) is injective div div and has dense range. Proof. Assume that F (α, β ) = 0. We ?rst notice that F (α, β ) coincides with the far ?eld of the surface potential V ∈ Hloc (curl, R3 \ Γ) de?ned by V (x) := λ 1 curl curl k2 i α(y )Φ(x, y )dsy + ? curl k Γ β (y )Φ(x, y )dsy .
Γ ?1 ?1

Hence from Rellich’s lemma V (x) = 0 for x ∈ R3 \ Γ. By using the jump relations across ?D for the vector potentials with densities extended by zero to the whole ?D [8] we obtain that i [ν × V ] |Γ = ?β = 0 k [ν × curl V ]Γ = λα = 0.

2 2 (Γ) × Hcurl (Γ) denotes the dual of the operator F de?ned by Hcurl

This proves injectivity. We now want to characterize the range of F . If F ? : L2 t (?) →
?1 ?1

F (α, β ), g

2 L2 t ,Lt

= (α, β ), F ? g

? ? 2 ,H ? 2 H div curl

1

1

,

(27)

then ([12], page 23) (Range F ) = a Kern F ? where
a

(28) = 0 ?g ? ∈ Kern F ? .

Kern F ? := g ∈ L2 ? t (?) : g, g

2 L2 t ,Lt

By changing the order of integration in (27) one can easily see that ? 1 ? on Γ ? λ k2 γT curlcurl Eg = λγT Eg ? F g := ? ? i γT curlEg on Γ ?k

(29)

The Inverse Electromagnetic Scattering Problem for Screens where Eg by a change of variable is written in the form Eg (y ) =
? ? ·y g (? x)e?ikx ds(? x)

12

y ∈ R3 .

(30)

It su?ces to show that F ? is injective. Let us assume that F ? g = 0. Then from (29) ν × Eg = 0 and ν × curl Eg = 0. From the Stratton-Chu formula and the analyticity of Eg the latter implies that Eg = 0 in R3 and therefore g = 0. Hence the range of F is dense. The proof of the above lemma shows that if ? = 0, then the corresponding operator F has dense range if and only if there does not exist an electromagnetic Herglotz pair such that the tangential component of the electric ?eld vanishes on Γ. In spherical coordinates the spherical wave functions [8] Mn = jn (k |x|) Grad Yn (? x) × x ?, n = 0, 1...

where jn denotes the Bessel function of order n and Yn the spherical harmonic of order n, provide examples of the electric ?eld of an electromagnetic Herglotz pair. Therefore for certain plane surfaces and for spherical surfaces with radius r such that rk is a zero of one of the Bessel functions, for ? = 0 the range of F is not dense and hence by (24) and (26) the set of far ?eld patterns is not dense in L2 t (?). The following lemma will help us to choose the right hand side of the far ?eld equation (22) appropriately. ? ? 2 (S ) we Lemma 4.3 For any smooth open surface S and two functions αS , βS ∈ H div S de?ne E∞ ∈ L2 (?) by t
S E∞ (? x) := x ?× S ? ·y αS (y ) e?ikx dsy + x ?× ? ·y βS (y )e?ikx dsy S
1

×x ?.

(31)

? ? 2 (Γ) it follows directly ? ? 2 (S ) ? H Proof. First assume that S ? Γ. Then since H curl curl S from the de?nition of F that E∞ (? x) ∈ Range(F ). S Now let S ? Γ and assume, on the contrary, that E∞ (? x) ∈ Range(F ), i.e. there exists 1 1 ?2 2 ? (Γ) and β ∈ H ? (Γ) such that α∈H
div div S E∞ (? x) = x ?× Γ ? ·y α(y ) e?ikx dsy + x ?× ? ·y β (y )e?ikx dsy Γ

S Then, E∞ (? x) ∈ Range(F ) if and only if S ? Γ.

1

1

×x ?.

Hence by Rellich’s lemma and the unique continuation principle we have that the potentials i 1 βS (y )Φ(x, y )dsy , αS (y )Φ(x, y )dsy + curl V S (x) := 2 curl curl k k S S 1 i V (x) := 2 curl curl α(y )Φ(x, y )dsy + curl β (y )Φ(x, y )dsy k k Γ Γ

The Inverse Electromagnetic Scattering Problem for Screens

13

/ Γ, and let B? (x0 ) be a small ball with coincide in IR3 \ (Γ ∪ S ). Now let x0 ∈ S , x0 ∈ center at x0 such that B? (x0 ) ∩ Γ = ?. Hence V is analytic in B? (x0 ) while V S has a singularity at x0 which is a contradiction. This proves the lemma. From (26), the above analysis and the fact that A is an isomorphism we have that the following is true:
2 2 (Γ) × Hcurl (Γ) → L2 Theorem 4.4 The operator B : Hcurl t (?) is compact, injective and S has dense range. Moreover the range of B consists of functions E∞ de?ned by (31) for 1 ?2 ? (S ). every S ? Γ and αS , βS ∈ H

?1

?1

div

Finally we need the following approximation property of the electromagnetic Herglotz pairs.
2 2 Lemma 4.5 The trace operator H : L2 t (?) → Hcurl (Γ) × Hcurl (Γ) de?ned by (23) has dense range.

?1

?1

Proof. Let consider Hg de?ned by (23) with Eg written in the form (30). It su?ces to 1 1 ? ? 2 (Γ) → L2 ? ? 2 (Γ) × H show that the dual operator H? : H t (Γ) is injective. In the proof div div ? of Lemma 4.2 it is shown that H coincides with F if we set λ = ? = 1. Therefore H? = F for λ = ? = 1 and hence H? is injective from Lemma 4.2.
2 Note that the result of Lemma 4.5 says that for any pair of functions c, c ? ∈ Hcurl (Γ) there exists an electromagnetic Herglotz pair with kernel g such that γT Eg approximates

?1

i 2 (Γ) norm. c and k γT curl Eg approximates c ? with respect to the Hcurl

?1

We have now all the ingredients to prove the main result of this paper. Let us denote by W the set of smooth open surfaces S and consider the far ?eld equation
S (F g )(? x) = ?ik B(Hg ) = E∞ (? x) ?1 ?1

S ∈ W.

(32)

S From Theorem 4.4 for S ? Γ we have that E∞ (? x) ∈ range B. Therefore we can ?nd a
2 2 solution (cS , c ?S ) ∈ Hcurl (Γ) × Hcurl (Γ) to the equation

i S E . (33) k ∞ Now from Lemma 4.5 for an arbitrarily small ? > 0 we can ?nd a g? ∈ L2 t (?) such that B(cS , c ?S ) =
S (cS , c ?S ) ? Hg?
2 ×H 2 Hcurl curl ?1 ?1

≤ ?.

By the continuity of B we have that
S B(cS , c ?S ) ? B(Hg? ) L2 t (?)

≤ c1 ?.

for some positive constant c1 which implies
S S (F g? )(? x) ? E∞ (? x) L2 t (?)

≤ c1 ?.

The Inverse Electromagnetic Scattering Problem for Screens

14

S Next for S ? Γ, E∞ (? x) does not lies in the range B. But by using a regular regularization technique (e.g. the Morozov discrepancy principle) we can ?nd an approximate solution of (33). In particular for a given δ > 0 let (cρ ?ρ S, c S ) be the regularized solution corresponding to the regularization parameter ρ such that i S ≤δ (34) B(cρ ?ρ S, c S ) ? E∞ L2 t (?) k and ρ→∞

lim (cρ ?ρ S, c S)

2 ×H 2 Hcurl curl

?1

?1

=∞ ≤ ?.

(35)

S (note that ρ → 0 as δ → 0). Hence again from Lemma 4.5 we can ?nd g?,ρ such that S (cρ ?ρ S, c S ) ? Hg?,ρ
2 ×H 2 Hcurl curl ?1 ?1

(36)

By the continuity of B and combining (34) and (36) we have that
S S (F g? , ρ)(? x ) ? E∞ (? x) L2 t (?)

≤ c2 (? + δ ).

Furthermore from (35)
ρ→0 S lim Hg?,ρ
2 ×H 2 Hcurl curl ?1 ?1

= ∞ and

ρ→0

S lim Eg?,ρ

H (curl,BR )

=∞

and
ρ→0 S lim g?,ρ L2 (?)

= ∞.

Summarizing the above analysis we can state the main result of our paper. Theorem 4.6 Assume that Γ is a bounded, simply connected, oriented, smooth open surface. Then if F is the combined far ?eld operator given by (20) corresponding to the scattering problem (1a)-(1c) we have that
S (i) if S ? Γ then for every ? > 0 there exists a solution g? ∈ L2 (?) of the inequality S S F g? ? E∞ L2 (?)

≤ ?.

(ii) if S ? Γ then for every ? > 0 and δ > 0 there exists a solution g L S?, δ ∈ L2 (?) of the inequality S S ? E∞ F g?,δ L2 (?) ≤ ? + δ such that
δ →0 S lim g?,δ L2 (?)

= ∞ and

δ →0

S lim Eg?,δ

H (curl,BR )

=∞

S S where Eg?,ρ is the electric part of the electromagnetic Herglotz pair with kernel g?,ρ . S Again for numerical purposes we want to replace the right hand side E∞ (? x) of the far ?eld equation (32) by an expression independent of S . To this end we observe that S as S degenerates to a point z with αS and βS an appropriate delta sequence then E∞ approaches ik ? ·z ? ·z (? x × q1 ) × x ?e?ikx + (? x × q2 )e?ikx 4π

The Inverse Electromagnetic Scattering Problem for Screens

15

where q1 and q2 are constant vectors. Note that the ?rst term is the electric far ?eld of an electric dipole and the second term is the magnetic far ?eld of an electric dipole. Roughly speaking the screen Γ will now be characterized as the set of points where the L2 t norm of an approximate (regularized) solution of the combined far ?eld equation (F g )(? x) = becomes very large. In ([3], Section 2.3, Section 3.2) it is proved that a Herglotz wave function and a electromagnetic Herglotz pair and their ?rst derivatives can be approximated uniformly on a compact subsets of a disk BR of radius R by a Herglotz wave function and a electromagnetic Herglotz pair respectively, with kernel supported in a subset ?0 ? ?. This new Herglotz wave function and the kernel can now be used in place of g? and vg? in Theorem 3.5, and g? and Eg? in Theorem 4.6. This justi?es the linear sampling method for screens with limited aperture data. We end this section by remarking that for the sake of simplicity we have considered only smooth open surfaces. Our analysis is also valid for open compact Lipschitz manifolds in R3 . For the appropriate analytical frame work see [2]. 5. Numerical examples 5.1. The reconstruction procedure In this section, numerical reconstructions of screens are shown. The numerical algorithm used was de?ned in [7]. We refer the reader to that paper for the details concerning the discretization of (22). For the sake of simplicity, we set ? = 0 in the de?nition of the far ?eld operator (20). Although there is a risk in doing this, we encountered no problems in obtaining the reconstructions shown in the following examples. The discrete solution is computed by using the Tikhonov regularization method where the regularization parameter is chosen by Morosov’s discrepancy principle. All the numerical experiments use synthetic data given by E∞ (? xi , x ?j , eα (? xi )), i = 1 · · · N , j = 1 · · · N , α = 1, 2 where the points (? xi )i=1···N denote the vertices of an evenly distributed mesh on the unit sphere and where (eα (? xi ))α=1,2 is a basis for the plane orthogonal to x ?i . This synthetic data thus requires the solution of 2N forward problems. This is done using a numerical code based on the ultra-weak formulation of Maxwell’s equation introduced in [6]. The meshes were generated by FemLab, and the synthetic data is corrupted using a random noise of maximum level ? (see [7] for details). As explained in [7] the best reconstruction is obtained by combining the use of three independent polarizations q1 , q2 and q3 for the right hand side Ee,∞ . In particular, our algorithm is based on the evaluation of 1 (37) g (·, z, q1 ) ?1 + g (·, z, q2 ) ?1 + g (·, z, q3 ) ?1 G (z ) = 3 ik ? ·z ? ·z (? x × q1 ) × x ?e?ikx + (? x × q2 )e?ikx 4π

The Inverse Electromagnetic Scattering Problem for Screens

16

where q1 = (1, 0, 0), q2 = (0, 1, 0) and q3 = (0, 0, 1). We showed in the previous section that g (·, z, q ) becomes arbitrarily large when z approaches the boundary of the scatterer from inside and remains large when z is outside. Considering a uniform grid in the region where we wish to ?nd the scatterer, the reconstruction is then done by eliminating the zi of the grid where G (zi ) < C max G (zj )
j

(38)

where C < 1 is an appropriately chosen constant (Various ad hoc methods for choosing C are discussed in [7]). In the examples that follow we have simply chosen the value of C that yields the best reconstructions. 5.2. Examples of reconstruction The numerical constructions are done by using 42 uniformly distributed directions of the unit sphere, a noise level ? = 0.01, a uniform grid for z ∈ [?2, 2]3 with mesh step 0.08 and a wave number k = 3. For each example, the picture on the left shows the real object and the picture on the right gives the reconstructed object. 5.2.1. Reconstruction of a disc We consider ?rst a single disc. Figure 1 shows the mesh of the original ?at object on the left and the reconstructed object on the right. Even if the object gained thickness in the reconstruction, the shape of the disc is correct. That result is obtained by linear interpolation using MatLab’s procedure “isosurface” and corresponds to G (z ) = 0.3 max G (zj )
j

This reconstruction is compatible with the criterion (38) with C = 0.3.

2 1.5 1 0.5 0 ?0.5 ?1 ?1.5 ?2 2 1 0 0 ?1 ?2 ?2 ?1 1 2

Figure 1. The disc: Real / Reconstructed object

The Inverse Electromagnetic Scattering Problem for Screens

17

5.2.2. Reconstruction of two parallel squares We also reconstructed two parallel squares (see Figure 2). In this case, C = 0.4. The objects have again gained thickness. Moreover, they are lightly deformed towards each other.

2 1.5 1 0.5 0 ?0.5 ?1 ?1.5 ?2 2 1 0 0 ?1 ?2 ?2 ?1 1 2

Figure 2. Two parallel squares: Real / Reconstructed object

5.2.3. Reconstruction of a curved screen The object considered here is a portion of a sphere, obtained from an eighth of a sphere with a truncation of both corners. The reconstructed object has again gained thickness and the corners have been smoothed out. Figure 3, obtained with C = 0.37, shows that the curvature of the object has been respected.

2 1.5 1 0.5 0 ?0.5 ?1 ?1.5 ?2 2 1 0 ?1 ?2 ?1 ?2 0 2 1

Figure 3. A curved object: Real / Reconstructed object

Acknowledgments The research of F.C. and D.C. was supported in part by grants from the Air Force O?ce of Research.

The Inverse Electromagnetic Scattering Problem for Screens References

18

[1] Abboud T and Starling F, Scattering of an electromagnetic wave by a screen, in Boundary Value Problems and Integral Equations in Nonsmooth Domains, Lecture Notes in Pure and Applied Mathematics, Vol 167 M. Costabel, M. Dauge and S. Nicaise eds., Marcel Decker New York, 1993, 1-17. [2] Bu?a A and Christiansen S H, The electric ?eld equations on Lipschitz screens: de?nitions and numerical approximation, to appear. [3] Cakoni F and Colton D, Combined far ?eld operatots in electromagnetic inverse scattering theory Math. Methods Appl. Sci., to appear. [4] Cakoni F and Colton D, The linear sampling method for cracks, Inverse Problems, to appear. [5] Cessenat M, Mathematical Methods in Electromagnetism, World Sciences, Singapore, 1996. [6] Cessenat O, Application d’une nouvelle formulation variationnelle aux ? equations d’ondes harmoniques. Probl` emes de Helmholtz 2D et de Maxwell 3D. PhD Thesis, Universit? e Paris IX Dauphine, 1996. [7] Colton D, Haddar H and Monk P, The linear sampling method for solving the electromagnetic inverse scattering problem SIAM J. Sci. Com. 24 (2002) 719-731. [8] Colton D and Kress R, Inverse Acoustic and Electromagnetic Scattering Theory, 2nd ed., Springer Verlag, New York, 1998. [9] Colton D and Kress R, On the denseness of Herglotz wave functions and electromagnetic Herglotz pairs in Sobolev spaces., Math. Meth. Appl. Sci., 24 (2001), 1289-1303. [10] Kirsch A and Ritter S, A linear sampling method for inverse scattering from an open arc Inverse Problems, 16 (2000), 89-105. [11] Kress R, Inverse scattering from an open arc, Math. Methods Appl. Sci., 18 (1995), 267-293. [12] McLean W, Strongly Elliptic Systems and Boundary Integral Equations, Cambridge University Press, Cambridge, 2000. [13] Stephan E P, Boundary integral equations for screen problems in R3 Integral Equations and Operator Theory, 10 (1987), 236-257.




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