The Factorization Method for Maxwell s Equations

The Factorization Method for Maxwell s Equations Andreas Kirsch University of Karlsruhe Department of Mathematics 76128 Karlsruhe Germany December 15, 2004 Abstract: The factorization method can be applied for certain classes of inverse problems where the shape of a domain has to be determined. It constructs a binary criterion which determines whether a given point is inside or outside the domain. The present paper develops the theory of the factorization method for the time harmonic Maxwell system where the support of the contrast of the index of refraction has to be determined from the knowledge of the far field patterns of the scattered fields for plane incident waves. The cases of an absorbing medium and a non-absorbing medium (where in the latter case the dielectricity varies smoothly) are considered. 1 Introduction The factorization method is a relatively new method for solving certain kinds of inverse scattering problems in the frequency domain or problems in impedance tomography. It is designed to determine the shape of the support of the unknown contrast. So far, this method has been successfully applied to many scalar problems, i.e. problems for the scalar Helmholtz equation. We refer to [16, 17, 18, 19, 21] for some of the results. Also, inverse scattering problems for elastic waves have been treated by this method see, e.g., [14, 23]. So far, no corresponding theoretical result is available when the wave propagation is described by the full Maxwell system. This is because of some, probably only technical, assumptions of a crucial functional analytic result (see Theorem 4.2 below. However, experiments with this method for simulated and real data, reported in [8] (see also [9]), indicate the practicability of the method also for the Maxwell system. In this paper we will show first theoretical results for the Maxwell system for two special cases: In the first case we will assume that the scattering medium is absorbing, i.e. the index of refraction is complex and the imaginary part can be bounded from below on by some positive constant. In the second case we will assume the index of refraction to be real-valued but, unfortunately, it has to decay smoothly to one (the background value) at the boundary of. The interesting case when the index jumps to zero is still open. The paper is organized as follows. In Section 2 we formulate the direct scattering, introduce the far field pattern, and formulate the inverse scattering problem. In Section 3 we prove a factorization of the far field operator F in the form F = H T H with some operators H and T. Here, F is the integral operator with the far field pattern as its kernel, and H denotes the adjoint of H. This factorization is central for the following and holds under weak assumptions on the index of refraction. The operators H and H depend explicitly on the domain. Moreover, can be characterized by those points z R 3 for which an explicitly given function ϕ z belongs to the range of H. The factorization F = H T H has now to be used to describe the range of H by F alone (since F is known by the data of the inverse problem). It is this step where functional 1

analytic tools are used. In Sections 4 and 5 we consider the cases of absorbing and non-absorbing media, respectively. We end this introduction with some remarks concerning the difference between the factization method and the linear sampling method. The linear sampling method has beeen introduced by David Colton and the author in [4] for some scalar acoustic scattering problems. It has then been extended to various classes of scattering problems as, eg., [2, 3, 13]. 2 The Direct Scattering Problem In this section we study the direct scattering problem. Let k = ω ε 0 µ 0 > 0 be the wave number with frequency ω, dielectricity ε 0, and permeability µ 0 in vacuum. Furthermore, let ˆθ S 2 = {x R 3 : x = 1} be a unit vector in R 3 and p C 3 such that p ˆθ := 3 j=1 p ˆθ j j = 0. The electromagnetic plane waves of direction ˆθ and polarization p are given by E inc (x) = p e ikˆθ x and H inc (x) = ε0 µ 0 (ˆθ p) e ikˆθ x, x R 3. This plane wave is scattered by a medium with (space dependent) dielectricity ε = ε(x), permeability µ = µ(x), and conductivity σ = σ(x). We assume that ε ε 0, µ µ 0 and σ 0 outside of some bounded domain. The total fields are superpositions of the incident and scattered fields, i.e. E = E inc + E s and H = H inc + H s and satisfy the Maxwell system We define the contrast curl E iωµh = 0 in R 3, (1a) curl H + iωεe = σe in R 3. (1b) q = µε µσ + i 1 µ 0 ε 0 ωµ 0 ε 0 and observe that q vanishes outside of. The refractive index is then given by 1 + q. Furthermore, the scattered field has to satisfy the Silver-Müller radiation condition ( ) µ0 1 H s (x) x x E s (x) = O as x ε 0 x uniformly w.r.t. ˆx = x/ x. In this paper we will always work with the electrical field E only. Eliminating the magnetic field H from the system (1a), (1b) and using the definition of q leads to for the total field while the incident field satisfies Subtracting both equations yields The Silver-Müller radiation condition turns into curl 2 E k 2 (1 + q)e = 0 in R 3 (2) curl 2 E inc k 2 E inc = 0 in R 3. (3) curl 2 E s k 2 (1 + q)e s = k 2 q E inc in R 3. (4) curl E s (x) ˆx ike s (x) = O ( ) 1 x 2, x. (5) 2

It will be necessary to allow more general source terms on the right-hand side of (4). In particular, we will consider the following problem: curl 2 v k 2 (1 + q)v = k 2 qf in R 3 (6) and v satisfies the Silver-Müller radiation condition (5). Here f L 2 (R 3, C 3 ) is an arbitrary vector function with compact support. The solutions v of (6) as well as of (2) and (4) have to be understood in the variational sense, i.e. are sought in the space H loc (curl, R 3 ) = { v : R 3 C 3 : v B H(curl, B) for all balls B R 3} where H(curl, B) is defined as the completion of C 1 (B, C 3 ) with respect to the norm v H(curl,B) := ( v 2 L 2 (B,C 3 ) + curl v 2 L 2 (B,C 3 )) 1/2. We recall that v H loc (curl, R 3 ) is said to be the variational solution of (6) if v satisfies [ curl v curl w k 2 (1 + q) v w ] dx = k 2 R 3 R 3 q f w dx (7) for all w H loc (curl, R 3 ) with compact support. For proving existence of a solution of this variational problem one transforms (7) into a variational equation on a bounded domain as follows: Let B be some ball containing the supports of f and q in its interior. If v H loc (curl, R 3 ) solves (7) then v solves also [ curl v curl w k 2 (1 + q) v w ] dx w Λ(ν v) ds = k 2 q f w dx (8) B for all w H(curl, B). Here, ν = ν(x) denotes the unit normal vector at x B, directed into the exterior of B, and Λ : H 1/2 (Div, B) H 1/2 (Div, B) denotes the exterior Calderon operator which maps λ H 1/2 (Div, B) into (ν curl u) B where u satisfies curl 2 u k 2 u = 0 in the exterior of B, ν u = λ on B and the radiation condition (5). The space H 1/2 (Div, B) is the trace space of H(curl, B) with respect to the traces v ν v. Its dual space is H 1/2 (Curl, B). For their characterizations we refer to [22]. And vice versa, if v H(curl, B) solves (8) then v can be extended into the exterior of B such that v solves (7). We refer to [22] for details. Existence of a variational solution of (8) can be shown under mild conditions on q: Theorem 2.1 Let q L (R 3 ) with compact support in B such that Rq 0 and Iq 0 on R 3. Then (8) has a (unique) solution v H(curl, B) for every f L 2 (R 3, C 3 ) with compact support in B provided the homogeneous equation (i.e. for f 0) admits only the trivial solution v 0. Furthermore, there exists a constant c > 0 (depending only on B, k, and q) with v H(curl,B) c f L2 (B,C 3 ) for all f L 2 (B, C 3 ). For a proof we refer to, e.g., Monk [22]. The assumption of uniqueness is satisfied if, e.g. Iq > 0 on some open and bounded set R 3 and q 0 on R 3 \. ( is then the support of q.) We note that the solution v lies in the space B H(curl 2, B) = { v H(curl, B) : curl v H(curl, B) } (9) B 3

for every ball B and v H(curl 2,B) c f L 2 (B,C 3 ) for all f L 2 (B, C 3 ). Furthermore, it is well known (eg. [5, 22]) that every radiating solution v of (6) has an asymptotic behaviour of the form v(x) = exp(ik x ) ( ) 1 v (ˆx) + O 4π x x 2, x, uniformly w.r.t. ˆx = x/ x. The vector field v is called the far field pattern of v. It is an analytic function on S 2 with respect to ˆx and is a tangential field, i.e. satisfied v (ˆx) ˆx = 0 for all ˆx S 2. In particular, the far field pattern E of E s is defined. Since it depends on ˆθ and p as well, we will also write E = E (ˆx, ˆθ; p). Note again, that it is tangential, i.e. E (ˆx, ˆθ; p) ˆx = 0 for all ˆx S 2 and all ˆθ S 2 and p C 3 with p ˆθ = 0. We are now able to define the inverse problem. Given E (ˆx, ˆθ; p) for all ˆx, ˆθ S 2 and p C 3 with p ˆθ = 0, find the support of q. Because of the linear dependence of E on p it is sufficient to know E only for a basis of three vectors for p. The task of determining only is rather modest since it is well known that one can even reconstruct q uniquely from this set of data, see [6]. However, the proof of uniqueness is non-constructive while we will give an explicit characterization of the characteristic function of which can, e.g., be used for numerical purposes. 3 Factorization of the Far Field Operator For this section we make the general assumption that the problem (8) has a unique solution for every f L 2 (B, C 3 ). In particular, the scattering problem (1a), (1b), (5) has a unique solution for every incident plane wave. Let R 3 be open and bounded such that is the support of q. We introduce the subspace L 2 t (S 2 ) of L 2 (S 2, C 3 ) consisting of all tangential fields, i.e. L 2 t (S 2 ) := { v L 2 (S 2, C 3 ) : v(ˆx) ˆx = 0, ˆx S 2}. The far field operator F : L 2 t (S 2 ) L 2 t (S 2 ) is defined as (F p)(ˆx) := E (ˆx, ) ˆθ; p(ˆθ) ds(ˆθ), ˆx S 2. (10) S 2 It is linear since E depends linearly on the polarization p. We note that F p is the far field pattern of the electrical field which corresponds to the incident field v p (x) = E inc( ) x, ˆθ; p(ˆθ) ds(ˆθ) = p(ˆθ) e ikx ˆθ ds(ˆθ), x R 3. S 2 S 2 p(ˆθ) e ikx ˆθ ds(ˆθ), These entire solutions of curl 2 v k 2 v = 0 are called Herglotz wave functions. We define the operator H : L 2 t (S 2 ) L 2 (, C 3 ) by (Hp)(x) := x, (11) S 2 and call H the Herglotz operator. H is one-to-one as it is easily seen. Recalling the definition of F p we observe that F p = GHp where Gf := v and v is the far field pattern corresponding to the (variational) solution v of (6) with right hand side f. We consider G as an operator from L 2 (, C 3 ) into L 2 t (S 2 ). We will now show that G = H T for some operator T which will give the factorization of F in the form F = H T H. 4

Theorem 3.1 Let F and H be defined by (10) and (11), respectively. Then F = H T H (12) where H : L 2 (, C 3 ) L 2 t (S 2 ) denotes the adjoint of H and T : L 2 (, C 3 ) L 2 (, C 3 ) is given by T f = k 2 q (f + v) and v H loc (curl, R 3 ) solves and the radiation condition (5). curl 2 v k 2 (1 + q)v = k 2 q f in R 3 (13) Proof: Let g C 1 (, C 3 ) (later we will set g = T f). Define w as the radiating solution of { curl 2 w k 2 g in w = 0 in R 3 \. (14) We will show that w = H g. With the fundamental solution of the Helmholtz equation Φ(x, y) = exp(ik x y ) 4π x y x y, we define w by w(x) := g(y) Φ(x, y) dy 1 k 2 g(y) ν(y) Φ(x, y) ds(y) + 1 k 2 div g(y) Φ(x, y) dy, x. We observe that the tangential component ν w is continuous through. Also, curl w = curl g(y) Φ(x, y) dy, and thus also ν curl w is continuous through. Furthermore, for x we have with curl 2 = + div curl 2 w(x) = curl 2 g(y) Φ(x, y) dy = ( + k 2 ) g(y) Φ(x, y) dy + k 2 g(y) Φ(x, y) dy + g(y) x Φ(x, y) dy = k 2 g(y) Φ(x, y) dy g(y) y Φ(x, y) dy = k 2 g(y) Φ(x, y) dy g(y) ν(y) Φ(x, y) ds(y) + div g(y) Φ(x, y) dy = k 2 w(x), 5

where we have used Gauss theorem. For x we have analogously to above curl 2 w(x) = curl 2 g(y) Φ(x, y) dy = ( + k 2 ) g(y) Φ(x, y) dy + k 2 g(y) Φ(x, y) dy + g(y) x Φ(x, y) dy = g(x) + k 2 g(y) Φ(x, y) dy + div g(y) Φ(x, y) dy g(y) ν(y) Φ(x, y) ds(y) = g(x) + k 2 w(x), i.e. we have shown that w is a radiating solution of (14). The uniqueness property yields w = w and, in particular, w(x) = 1 k 2 curl 2 g(y) Φ(x, y) dy, x. Then it is easily seen (cf. [5]) that w (ˆx) = ˆx g(y) e ikˆx y dy ˆx = (H g)(ˆx). (15) This holds for all g C 1 (, C 3 ). Since C 1 (, C 3 ) is dense in L 2 (, C 3 ) we conclude that H g = w even for g L 2 (, C 3 ) where w solves (14) in the variational sense. Now we set g = T f = k 2 q(f + v) for any f L 2 (, C 3 ), thus curl 2 w k 2 w = k 2 q(f + v) = curl 2 v k 2 v in R 3 \. Furthermore, ν (w v) and ν curl (w v) are continuous through and w v satisfies the radiation condition. Therefore, w = v follows by uniqueness of the problem and thus v = w = H T f. On the other hand, from the definition of H we have that v = Gf and thus Gf = H T f for all f L 2 (, C 3 ). Substituting G = H T into F = GH yields the assertion. We note that the operator T is one-to-one. Indeed, T f = 0 implies v+f 0 in, i.e. curl 2 v+k 2 v = 0 in R 3. This yields v 0 in R 3 since ν v and ν curl v are continuous through and v satisfies the radiation condition. Therefore, also f 0. 4 The Factorization Method for Absorbing Media In this section we make the general assumption that there exists q 0 > 0 with Iq(x) q 0 for almost all x. Recall that is the support of q, i.e. q = 0 outside of (the open and bounded set). As we mentioned above (following Theorem 2.1), problem (8) is then uniquely solvable for all f L 2 (, C 3 ). We recall the factorization F = H T H and introduce the operators IF := 1 2i (F F ) and, analogously, IT := 1 2i (T T ). Then IF and IT are self-adjoint and IF = H (IT )H. (16) We show now that IT is an isomorphism from L 2 (, C 3 ) onto itself and, even more, we show that it is coercive. 6

Lemma 4.1 Under the assumptions of this section there exists γ > 0 (depending on q 0 ) with for all f L 2 (, C 3 ). (IT )f, f L 2 (,C 3 ) = I T f, f L 2 (,C 3 ) γ f 2 L 2 (,C 3 ) Proof: Let be f L 2 (, C 3 ). Then T f = k 2 q(f + v) = k 2 qw with w := f + v. Here, v solves (13), i.e. curl 2 v k 2 v = k 2 q(f + v) = k 2 q w in R 3. Then, since f = w v: T f, f L 2 (,C 3 ) = k 2 = k 2 q w 2 dx k 2 q w 2 dx q w v dx (curl 2 v k 2 v) v dx = k 2 q w 2 [ dx curl v 2 k 2 v 2] dx (ν curl v) v ds, where we applied already Green s theorem in. Applying Green s theorem in {x R 3 \ : x < R} and taking the imaginary part yields I T f, f L 2 (,C 3 ) = k 2 Iq w 2 dx I (ˆx curl v) v ds. From the radiation condition x =R lim r (curl v ˆx ikv) = 0 r we conclude that for R I T f, f L 2 (,C 3 ) = k 2 Iq w 2 dx + I ik v 2 ds k 2 q 0 f + v 2 L 2 (,C 3 ) + k S 2 (17a) S 2 v 2 ds. (17b) Finally, we claim that there exists c > 0 with f + v L 2 (,C 3 ) c f L 2 (,C 3 ) for all f L 2 (, C 3 ). Otherwise there existed a sequence {f j } with f j L 2 (,C 3 ) = 1 and f j + v j L 2 (,C 3 ) 0. From Theorem 2.1 we have the well posedness of the system curl 2 v k 2 v = k 2 q(f + v) and conclude that v j 0 in H(curl, ), i.e., v j 0 in L 2 (, C 3 ). Therefore, also f j 0 in L 2 (, C 3 ), a contradiction. The following functional analytic theorem taken from [20] is more general than needed for the present situation. We will, however, apply it a second time for real valued indices in the next section. Theorem 4.2 Let X and Y be Hilbert spaces, B : Y Y, H : Y X, and A : X X be linear and bounded operators with B = H A H. (18) We make the following assumptions: (a) H is one-to-one and compact. 7

(b) RA := 1 2 (A + A ) has the form RA = C + K with some coercive operator C : X X and compact operator K. (c) IA is non-negative on X, i.e. I Aϕ, ϕ 0 for all ϕ X. (d) RA is one-to-one (and thus an isomorphism onto X) or IA is positive on X, i.e. I Aϕ, ϕ > 0 for all ϕ X, ϕ 0. Then the operator B # := RB + IB is also positive, and the ranges of H : X Y and B 1/2 # : Y Y coincide. Furthermore, the operator B 1/2 # L is a bounded isomorphism from X onto Y. We apply this theorem to the factorization (16), i.e., we set B = IF and A = IT. Then A and B themselves are self-adjoint and A is coercive. Therefore, we set C = A and K = 0. All of the other assumptions are also satisfied. Furthermore, RB = B = IF, IB = 0, and B # = IF. Therefore, we conclude that the ranges of (IF ) 1/2 and H coincide, i.e. R ( (IF ) 1/2) = R(H ). (19) The advantage of this relation lies in the fact that we we can explicitly characterize by the range of H. Indeed, recall, that H : L 2 (, C 3 ) L 2 (S 2 ) is given by (H g)(ˆx) = ˆx g(y) e ikˆx y dy ˆx for ˆx S 2. (20) For any z R 3 and fixed p C 3 we define φ z L 2 t (S 2 ) by Then we can prove φ z (ˆx) = ik 4π (ˆx p) ˆx e ikˆx z, ˆx S 2. (21) Theorem 4.3 Let φ z be defined in (21) for z R 3 (where p C 3 is kept fixed). Then z if and only if φ z R(H ). Proof: First, let z, and define v as the electric dipole with moment p and source point z v(x) = 1 ik curl 2 ( ) x p Φ(x, z), x R 3 \ {z}. Its far field pattern is v = φ z (see [5]). Choose an arbitrary vector field ṽ C 2 (, C 3 ) such that ν ṽ = ν v on and ν curl ṽ = ν curl v on and set g := curl 2 ṽ k 2 ṽ in. We show that H g = φ z. First, we recall from the proof of Theorem 3.1 that H g is the far field pattern of the solution w of the problem { curl 2 w k 2 g in, w = 0 in R 3 \. Therefore, curl 2 (w ṽ) k 2 (w ṽ) = 0 in, curl 2 (w v) k 2 (w v) = 0 in R 3 \, and ν (w ṽ) = ν (w v) + and ν curl (w ṽ) = ν curl (w v) +. The uniqueness theorem implies w v in R 3 \, from which H g = w = v = φ z follows. 8

Now we consider the case z and assume, on the contrary, that φ z = H g R(H ) for some g L 2 (, C 3 ). Again, we define v as before and note again, that H g is the far field pattern of the solution w of { curl 2 w k 2 g in w = 0 in R 3 \. From v = φ z = H g = w we conclude that v w on any ball containing {z}. If z this contradicts the fact that w is analytic outside of and v is singular at z. If z this leads also to a contradiction since w L 2 (B \, C 3 ) for any ball B with center z but v L 2 (B \, C 3 ) since v(x) = O( x z 3 ) as x z. The combination of this theorem with the expression (19) yields the main result of this section: Theorem 4.4 Let the assumption of the beginning of the section be satisfied. Furthermore, let φ z be defined in (21) for z R 3 (where again p C 3 is kept fixed). Then z if, and only if, φ z R ( (IF ) 1/2). This result characterizes the support of q by the data of the inverse scattering problem which are collected in the operator IF. In particular, it gives a new direct proof of uniqueness of the inverse scattering problem to determine from E (ˆx, ˆθ, p) for all ˆx, ˆθ S 2 and p C 3 with p ˆθ = 0. A reformulation of this result and the connection to the Linear Sampling Method as described in, e.g., [2] will be given below in Section 7. 5 The Case of Non-absorbing Media For this section we make the general assumption that v 0 is the only solution of curl 2 v k 2 (1 + q)v = 0 in R 3 satisfying the radiation condition (5). Furthermore, we assume that q L (R 3 ) has compact support (which we again denote by ) and is real-valued and positive. Unfortunately, below we will have to make a further smoothness assumption on q. We recall the definition of the operator T from Theorem 3.1 as T f = k 2 q(f + v) and v solves (13). Up to the factor k 2 q the operator T is the sum of the identity operator and a second one which we will show to be compact on a suitable subspace under the before-mentioned smoothness assumption on q. The factor q motivates us to work in the weighted L 2 space L 2 q(, C 3 ) defined as L 2 q(, C 3 ) = { ψ : C 3 : q ψ L 2 (, C 3 ) } and equipped with the inner product ψ, ϕ L 2 q (,C 3 ) = q(x) ψ(x) ϕ(x) dx. First, we note that L 2 (, C 3 ) is imbedded in L 2 q(, C 3 ) with bounded imbedding. This follows from ψ 2 L 2 q (,C3 ) = q(x) ψ(x) 2 dx q ψ(x) 2 dx = q ψ 2 L 2 (,C 3 ). We denote the adjoint of H considered as an operator from L 2 t (S 2 ) into L 2 q(, C 3 ) by H. It is given by H ϕ = H (qϕ), ϕ L 2 q(, C 3 ). Therefore, the factorization (12) takes the form F = H q 1 T H = H T H (22) 9

with T f = k 2 (f + v) and v solves (13). The operator T is well defined and bounded on L 2 q(, C 3 ) since, if f L 2 q(, C 3 ) then the right hand side of (13) is still in L 2 (, C 3 ). It is our aim to apply Theorem 4.2 to the factorization (22). Setting A = T and Cf = k 2 f and K the real part of the operator f k 2 v we have to show that the latter operator is compact. However, this fails to be the case due to the fact that H(curl, ) is not compactly imbedded in L 2 (, C 3 ). To overcome this difficulty we note that it is sufficient to consider T on the closure of the range R(H) of H only. To formulate this space weakly we define H 00 (curl 2, ) = { v H 0 (curl, ) : curl v H 0 (curl, ) } where H 0 (curl, ) denotes the closed subspace of H(curl, ) with vanishing traces ν v and f (curl 2 w k 2 w) dx = 0 X := f L2 q(, C 3 ) : for all w H 00 (curl 2, ) with (23) (curl 2 w k 2 w) L 2 1/q (, C3 ) Then X is a closed subspace of L 2 q(, C 3 ) which contains the range of H. Before we prove that T is compact on X we need the following simple result. Lemma 5.1 Let B be a ball which contains in its interior. The mapping f Λ(ν v) is bounded from L 2 q(, C 3 ) into H s ( B, C 3 ) for all s R. Here, v H loc (curl, R 3 ) denotes the radiating solution of (13). Proof: This mapping can be decomposed in the following way: Choose a ball B such that B and B B. Since f v B is bounded from L 2 q(, C 3 ) into H(curl, B ) we have by the trace theorem that f ν v B is bounded from L 2 q(, C 3 ) into H 1/2 (Div, B ). It is easily seen 1 that the mapping λ ν v B is bounded from H 1/2 (Div, B ) into H s ( B, C 3 ) for all s. Here, v denotes the solution of curl 2 v k 2 v = 0 in the exterior of B with ν v B = λ, satisfying the radiation condition. Therefore, f (ν v) B is bounded from L 2 q(, C 3 ) into H s ( B, C 3 ) for all s R. The form of Λ, given e.g. in [22] implies the assertion. Theorem 5.2 In addition to the assumption at the beginning of this section let q C(R 3 ) with compact support such that q C 1 () and q 1/2 q L β (, R 3 ) for some β > 3. Then the mapping f v is compact from X into L 2 q(, C 3 ). Here, X is defined in (23) and v H loc (curl, R 3 ) denotes the radiating solution of (13). Remark: In the radially symmetric case, i.e. if is the unit ball and q = q(r), r 0, is of the form q(r) = c(1 r) α, 0 r 1, for any α > 0 the assumption is satisfied if 3 < β < 6/(2 α). On the other hand, the example of Section 6 below shows that the imbedding fails to be compact for α = 0, i.e. if q is constant inside of. Before we prove this theorem we collect two auxiliary results in the following lemma. Lemma 5.3 Let q satisfy the assumptions of the previous theorem and let ψ H 1 (). Then (a) ψ q 1/2 q L 2 (, C 3 ) and (b) (qψ) H 00 (curl 2, ). Proof: (a) The space H 1 () is continuously imbedded in L 6 () (cf. [7], p. 131). Therefore, ψ L 6 () and thus ψ q 1/2 q L 2 (, C 3 ) by Hölder s inequality since q 1/2 q L β (, R 3 ) L 3 (, C 3 ). 1 Since B and B are balls one can, e.g., expand the solution into spherical wave functions. 10

(b) Since curl (qψ) = 0 we have to show that (qψ) H 0 (curl, ). From (qψ) = ψ q + q ψ and part (a) we conclude that ψ q = q 1/2( ψ q 1/2 q ) L 2 (, C 3 ) and thus (qψ) L 2 (, C 3 ). It remains to show that the trace ν (qψ) (exists and) vanishes. This is equivalent to the variational equation [ ] v curl (qψ) (qψ) curl v dx = 0 for all v H 1 (, C 3 ), i.e. (qψ) curl v dx = 0 for all v H 1 (, C 3 ). Let v C 2 (, C 3 ) be a smooth test function. Then (qψ) curl v dx = div (q ψ curl v) dx = q ψ (ν curl v) ds = 0 since q C(R 3 ) vanishes on. This proves the lemma. Proof of Theorem 5.2: Again we choose a ball B which contains in its interior. We show that f v is even compact from X into L 2 (, C 3 ). Let f n X with f n L 2 q (,C 3 ) 1 and let v n be the corresponding solution of (13). Then v n H(curl is bounded. Define scalar functions,b) ϕ n H 1 (B) with ϕ n ψ dx = k 2 (1 + q) v n ψ dx for all ψ H 1 (B), B which is the weak form of B Writing this as ϕ n = div [ k 2 (1 + q)v n ] in B, ϕ n ν = k2 (1 + q) ν v n = k 2 ν v n on B. div [ ϕ n k 2 (1 + q) v n ] = 0 in B, ν [ ϕn k 2 (1 + q) v n ] = 0 on B, we note that there exists uniquely determined fields ˆv n with curl ˆv n = k 2 (1 + q) v n ϕ n in B, div ˆv n = 0 in B, ν ˆv n = 0 on B. We note that k 2 (1 + q)v n L 2 (B,C 3 ) and ϕ n L 2 (B,C) are bounded. A classical regularity result (see [11]) yields ˆv n H 1 (B, C 3 ) and ˆv n H 1 (B,C 3 ) c for all n. Since H 1 (B, C 3 ) is compactly imbedded in L 2 (B, C 3 ) there exists a convergent subsequence of {ˆv n } in L 2 (B, C 3 ). We will denote this subsequence again by {ˆv n }. We will now show that { ϕ n } is not only bounded but also contains a convergent subsequence in L 2 (B, C 3 ). Using (13) and div f n = 0 one observes formally that ϕ n = k 2 div (qf n ) = k 2 f n q. 11

This can be made rigorously by substituting w = ψ for any ψ H 1 () into (8): ϕ n ψ dx = k 2 (1 + q) v n ψ dx B Here we used the identity B B = = B B Grad ψ Λ(ν v n ) ds k 2 ψ Div Λ(ν v n ) ds k 2 + k 2 ( q f n ) ψ dx. Grad φ g ds + B φ Div g ds = 0 q f n ψ dx f n (qψ) dx + for all g H 1/2 (Div, B) and all φ H 1/2 ( B), with surface gradient Grad and surface divergence Div. Note that (qψ) L 2 (, C 3 ) by the previous lemma and also ψ q = q 1/2 (ψ q 1/2 q) L 2 (, C 3 ). We will now show that f n (qψ) dx = 0. From Lemma 5.3 we conclude that w := (qψ) H 00 (curl 2, ) and q 1/2 (curl 2 w k 2 w) = k 2 q 1/2 (qψ) = k 2 ψq 1/2 q k 2 q 1/2 ψ L 2 (, C 3 ). Therefore, since f n X we conclude that 0 = f n (curl 2 w k 2 w) dx = k 2 f n (qψ) dx. We have thus shown that ϕ n ψ dx = ψ Div Λ(ν v n )ds + k 2 ( q f n ) ψ dx B B for all ψ H 1 (B) which is the weak form of ϕ n = k 2 q f n in B, ϕ n ν = Div Λ(ν v n) on B. Since q L β (, R 3 ) and f n L 2 (, C 3 ) we observe that q f n L s (, C 3 ) for s = 2β/(β +2) > 6/5 and q f n L s (B) c. Furthermore, by Lemma 5.1, Div Λ(ν v n ) W 1/2,s ( B) c for all n, where we used the Sobolevspaces W k,s ( B) of order of differentiation k and integrability coefficient s. A standard regularity result for solutions of the Poisson equation (see [12], Theorem 2.4.26) yields ϕ n W 2,s (B) and ϕ n W 2,s (B) c for all n. Since W 2,s (B) is compactly imbedded in W 1,2 (B) = H 1 (B) (eg. [1], note that s > 6/5) there exists a convergent subsequence of { } ϕ n in L 2 (B, C 3 ). We will again denote this subsequence by { } ϕ n. Set vnm = v n v m and, analogously, ˆv nm and ϕ nm for n, m N. Then k 2 (1 + q)v nm, v nm = curl ˆv nm, v nm L2 (B,C 3 ) + ϕ nm, v nm L2 (B,C 3 ) L 2 (B,C 3 ) = ˆv nm, curl v nm L 2 (B,C 3 ) + ϕ nm, v nm L 2 (B,C 3 ) since ν ˆv nm = 0 on B. Now, we observe that ˆv nm L 2 (B,C 3 ) 0 and ϕ nm L 2 (B,C 3 ) 0 as n, m tend to infinity. Furthermore, v nm L2 (B,C 3 ) and curl v nm L2 (B,C 3 ) are bounded. Therefore, 12

(1 + q)v nm, v nm L2 (B,C 3 ) 0, n, m, which implies that {v n } is a Cauchy sequence and thus convergent in L 2 (B, C 3 ) and thus in L 2 q(, C 3 ). This ends the proof. As a next step we have to formulate the factorization in the space X rather than in L 2 q(, C 3 ). Therefore, let P : L 2 q(, C 3 ) X be the orthogonal projection operator onto X. Then H = H P since the range of H is contained in X and thus the orthogonal complement X of X is contained in the complement of the range of H which is the nullspace of H. Therefore, we write the factorization (22) in the form F = H P T H. (24) We now check the assumptions of Theorem 4.2 for B = F and A = P T X and Y = L 2 t (S 2 ). We have that A = C + K where C is a multiple of the identity and K a multiple of the compact operator P T. Therefore, parts (a) and (b) of Theorem 4.2 are satisfied. Part (c) follows from (17a) which obviously holds also under the assumptions of this section. Indeed, we have for f X and thus P T f, f L 2 q (,C 3 ) = T f, f L 2 q (,C 3 ) = T f, f L 2 (,C 3 ) I P T f, f L 2 q (,C 3 ) = k v 2 ds 4π from which assumption (c) follows. For assumption (d) we will study under which conditions I P T f, f L 2 q (,C 3 ) is strictly positive. It vanishes if, and only if, v vanishes identically, i.e. v 0 outside of. We recall the weak definition of f X: f (curl 2 w k 2 w) dx = 0, i.e. f [curl 2 w k 2 (1 + q)w ] dx = k 2 S 2 q f w dx (25) for all w H 00 (curl 2, ) with q 1/2 (curl 2 w k 2 w) L 2 (, C 3 ). Furthermore, since v vanishes outside of, (7) reduces to [ curl v curl w k 2 (1 + q) v w ] dx = k 2 q f w dx for all w H(curl, ). From this we first conclude that v H 00 (curl 2, ) and, second, that v [curl 2 w k 2 (1 + q)w ] dx = k 2 q f w dx (26) for all w H 00 (curl 2, ). Adding equations (25) and (26) yields f [curl 2 w k 2 (1 + q)w ] dx = 0 (27) for w H 00 (curl 2, ) with q 1/2 (curl 2 w k 2 w) L 2 (, C 3 ). Here we have set f = f + v. Therefore, f and f satisfy the following equations in the weak sense: f, f L 2 q(, C 3 ), f f H00 (curl 2, ), (28a) curl 2 f k 2 (1 + q) f = 0 in, (28b) 13

curl 2 f k 2 f = 0 in. (28c) This is an eigenvalue problem with respect to the parameter k. Similar transmission eigenvalue problems have been studied before in connection with the Linear Sampling Method and the factorization method (see [5, 2, 13]). At the moment it is not known to the author whether the set of eigenvalues is discrete or not. Application of Theorem 4.2 yields the main theorem of this section. Theorem 5.4 Let the assumption of the beginning of the section be satisfied. Furthermore, assume that k is not an eigenvalue of the problem (28). Let φ z be defined in (21) for z R 3 (where again p C 3 is kept fixed). Then z if, and only if, φ z R ( F 1/2 ) # where F# := RF + IF. Again we note that this implies, in particular, an alternative proof of uniqueness of the inverse problem. Before we reformulate this result in a different way and compare it to the Linear Sampling Method we study the example of being the ball of radius 1 and q being constant. 6 An Example Let be the ball of radius 1 and q constant, real and positive in and zero outside of. We note that the assumption of Theorem 5.2 is certainly not satisfied in this case because q fails to be continuous. We will show by this example that the assertion of Theorem 5.2 does not hold. Set κ = k 1 + q for abbreviation and note that κ > k. First we recall that (weakly) From this and (6) we conclude that Furthermore, v satisfies curl 2 f k 2 f = 0 for x < 1. (29a) curl 2 (v + f) κ 2 (v + f) = 0 for x < 1. (29b) curl 2 v k 2 v = 0 = 0 for x > 1. (29c) Finally, the fields are linked by the transmission conditions ˆx v = ˆx v + and ˆx curl v = ˆx curl v + on. We expand the fields f and v +f and v into spherical wave functions. Therefore, let j n and h (1) n be the spherical Bessel functions and Hankel functions, respectively, of order n, and Yn m be the spherical harmonics of order n and degree m. Define the vector wave functions Mn m and Nn m by M m n (x; k) = curl { x j n (k x ) Y m n (ˆx) } = j n (k x ) Grad Y m n (ˆx) ˆx, N m n (x; k) = curl { x h (1) n (k x ) Y m n (ˆx) } = h (1) n (k x ) Grad Y m n (ˆx) ˆx. Then f and v + f and v can be expanded into the forms f(x) = v(x) + f(x) = v(x) = [ β m n Mn m (x; k) + αn m curl Mn m (x; k) ], x < 1, n=0 m n [ a m n Mn m (x; κ) + b m n curl Mn m (x; κ) ], x < 1, n=0 m n [ c m n Nn m (x; k) + d m n curl Nn m (x; k) ], x > 1. n=0 m n 14

As a particular case we take f n (x) = curl M m n (x; k) for some fixed m Z and arbitrary n m. Then the previous expansions reduce to v n (x) + f n (x) = b n curl M m n (x; κ), x < 1, v n (x) = d n curl N m n (x; k), x > 1. First, we match the transmission condition ˆx v = ˆx v + and use curl M m n (ˆx; k) ˆx = [ j n (k) + k j n(k) ] Grad Y m n (ˆx) ˆx on, and analogously for the inner wave number κ. This yields b n [ jn (κ) + κ j n(κ) ] [ j n (k) + k j n(k) ] = d n [ h (1) n (k) + k (h (1) n ) (k) ]. Now we match ˆx curl v = ˆx curl v + and use curl 2 M m n (x; k) = k 2 M m n (x; k) and This yields Eliminating d n from this system yields ˆx M m n (ˆx; k) = j n (k) Grad Y m n (ˆx) on. b n κ 2 j n (κ) k 2 j n (k) = d n k 2 h (1) n (k). b n { κ 2 j n (κ) [ h (1) n (k) + k (h (1) n ) (k) ] [ j n (κ) + κ j n(κ) ] k 2 h (1) n (k) } = = k 3[ j n (k)h (1) n (k) h (1) n (k) j n(k) ] = ik where we used the Wronskian relation j n (k) h (1) (1) n (k) h n (k) j n(k) = i/(k 2 ). Now we recall the asymptotic behaviour of the Bessel and Hankel functions (cf. [5]): j n (t) = t n [ ] 1 + O(1/n), h (1) n (t) = 1 3 5 (2n + 1) 1 3 5 (2n 1) i t n+1 [ 1 + O(1/n) ] as n tends to infinity. Analogous asymptotic forms hold for the derivatives. Substituting this into the form of b n yields 2k 2 ( ) n k [ ] b n = 1 + O(1/n). κ 2 + k 2 κ Now we compute f n L 2 () and v n L 2 (). We need the following application of Green s theorem: curl Mn m (x; κ) curl Mn m (x; k) dx = = k 2 = k2 S 2 Mn m (x; κ) curl 2 Mn m (x; k) dx + 1 0 = n(n + 1) M m [ n (ˆx; κ) curl M m n (ˆx; k) ˆx ] ds S 2 Mn m (x; κ) Mn m (x; k) dx + M m [ n (ˆx; κ) curl M m n (ˆx; k) ˆx ] ds S 2 j n (κr) j n (kr) r dr + j n (κ) [ j n (k) + k j n(k) ] Grad Y m n (ˆx) ˆx 2 ds(ˆx) k2 1 0 j n (κr) j n (kr) r dr + j n (κ) [ j n (k) + k j n(k) ] 15

since Grad Y m S 2 n (ˆx) ˆx 2 ds(ˆx) = Grad Y m S 2 n (ˆx) 2 ds(ˆx) = n(n + 1). We determine the asymptotic form by curl Mn m (x; κ) curl Mn m (x; k) dx = n(n + 1) [ ] 2 1 3 5 (2n + 1) k2 k n κ n { n(n + 1) k n+2 κ n [ ] 2 1 3 5 (2n + 1) 2n + 2 n 4 [ 1 3 5 (2n 1) ] 2 (kκ)n. r 2n+1 dr + κ n( k n + n k n ) } + (n + 1)κn k n Now we compute the asymptotic forms of f n L 2 () and v n L 2 (): f n 2 L 2 () = curl M m n (x; k) 2 n dx 4 [ 1 3 5 (2n 1) ] 2 k2n and, since v n (x) = b n curl Mn m (x; κ) curl Mn m (x; k), with ρ := (2k 2 )/(k 2 + κ 2 ): v n 2 L 2 () = b 2 n curl M m n (x; κ) 2 dx + curl M m n (x; k) 2 dx 2 Rb n n 4 [ 1 3 5 (2n + 1) ] 2 1 0 curl M m n (x; κ) curl M m n (x; k) dx = (ρ 1) 2 n 4 [ 1 3 5 (2n + 1) ] 2 k2n (ρ 1) 2 f n 2 L 2 (). ] [ρ 2 k2n κ 2n κ2n + k 2n 2ρ kn κ n (kκ)n We note that ρ 1 0. From this relation and the fact that v n and v p are orthogonal in L 2 () for n p we observe that f v cannot be compact. Indeed, we set ˆv n = v n / f n L 2 () and have that for large n p: ˆv n ˆv p 2 L 2 () = ˆv n 2 L 2 () + ˆv p 2 L 2 () 2 (ρ 1) 2. Therefore, the sequence {ˆv n } cannot contain a convergent subsequence although the corresponding sequence { f n / f n L2 ()} is bounded. 7 Conclusion and Additional Remarks In this paper we have shown for two particular cases how the factorization method works (at least theoretically) when the scattering problem is modelled by Maxwell s equations. We would like to reformulate the result of Theorem 5.4 in a different way and compare it with the Linear Sampling Method. 2 First we note that, by Theorem 5.4, a point z R 3 is inside of if and only if the equation F 1/2 # g = φ z (30) 2 The same reformulation holds also for the result of Theorem 4.4 if one sets F # = IF. 16

is solvable in L 2 t (S 2 ). Since the operator F # is positive and self-adjoint there exists a complete ortho-normal system of eigenfunctions ψ n L 2 t (S 2 ) which correspond to positive eigenvalues λ n. Then (30) is solvable if and only if the series n=1 φ z, ψ n 2 λ n (31) converges. Here,, denotes the inner product in L 2 t (S 2 ). Therefore, the sign of the function W (z) = is just the characteristic function of. [ φ z, ψ n 2 ] 1, z R 3, n=1 λ n The Linear Sampling Method, which has been developed by David Colton and the author first for scalar problems (see [4]) studies the (approximate) solvability of the equation F g = φ z (32) instead of (30). Note that F has the form F = RF + iif while F # = RF + IF. It is well known that in the case of Section 5 the operator F is normal. Therefore, the author believes that a result similar to the one in [17] holds, namely that F # can be replaced by (F F ) 1/2 in the characterization of Theorem 5.4 and in equation (30). Therefore, roughly speaking, the difference in the Linear Sampling Method and the Factorization Method is that the first one claims that there is a better approximate solution of equation (32) for points z inside of than for those outside of while the Factorization Method characterizes the points of directly by the solvability of (30). From this remark it should be clear that the Factorization Method is mathematically more satisfactory. However, it should also be mentioned that so far the characterization of by the Factorization Method has been proven only for a few special cases in contrast to the Linear Sampling Method for which many more results are known. For example, an open problem is the characterization of by the Factorization Method for the Maxwell system when is an ideal conductor. Acknowledgements We thank the referees for their useful remarks and suggestions. Special thanks are given to one of them who weakened the original much stronger smoothness assumption on q in Theorem 5.2 to the present one. References [1] Adams, R.A.: Sobolev spaces. Academic Press, New York, 1975. [2] Cakoni, F., Colton, D., and Haddar, H.: The Linear Sampling Method for Anisotropic Media. J. Comp.Appl. Math. 146, 285 299 (2002). [3] Cakoni, F., and Haddar, H.: The Linear Sampling Method for Anisotropic Media: Part II. Preprint, 2002. [4] Colton, D., and Kirsch, A.: A Simple Method for Solving Inverse Scattering Problems in the Resonance Region. Inverse Problems 12 (1996), 383 393. [5] Colton, D., and Kress, R.: Inverse Acoustic and Electromagnetic Scattering Problems. 2nd edition, Springer-Verlag, New-York, 1998. 17

[6] Colton, D., and Päivärinta, L.: The Uniqueness of a Solution to an Inverse Scattering Problem for Electromagnetic Waves. Arch. Rational Mech. Anal. 119 (1992), 59-70. [7] Dautray, R., and Lions, J.-L.: Mathematical Analysis and Numerical Methods for Science and Technology. Volume 2: Functional and Variational Methods. Springer, Berlin, etc., 1988. [8] Fischer, C.: Multistatisches Radar zur Lokalisierung von Objekten im Boden. PhD-Thesis, University of Karlsruhe, 2003. [9] Fischer, C., and Wiesbeck, W.: Multistatic Antenna Configurations and Image Processing for Mine-Detection GPR. Preprint, 2004. [10] Gilbarg, D., and Trudinger, N. S.: Elliptic Partial Differential Equations. 2nd edition, Springer, New-York, 1983. [11] Girault, V., and Raviart, P.A.: Finite Element Methods for Navier-Stokes Equations. Springer, New-York, 1986. [12] Grisvard, P.: Elliptic Problems in Nonsmooth Domains. Pitman, Boston, 1985. [13] Haddar, H.: The Interior Transmission Problem for Anisotropic Maxwell s Equations and its Applications to the Inverse Problem. Preprint, 2002. [14] Hähner, P.: An Inverse Problem in Elastostatics. Inverse Problems 15 (1999), 961 975. [15] Kirsch, A.: Introduction to the Mathematical Theory of Inverse Problems. Springer-Verlag, New-York, 1998. [16] Kirsch, A.: Characterization of the shape of a scattering obstacle using the spectral data of the far field operator. Inverse Problems 14 (1998), 1489 1512. [17] Kirsch, A.: Factorization of the Far Field Operator for the Inhomogeneous Medium Case and an Application in Inverse Scattering Theory. Inverse Problems 15 (1999), 413 429. [18] Kirsch, A.: New Characterizations of Solutions in Inverse Scattering Theory. Applicable Analysis 76 (2000), 319 350. [19] Kirsch, A.: The MUSIC-Algorithm and the Factorization Method in Inverse Scattering Theory for Inhomogeneous Media. Inverse Problems 18 (2002), 1025 1040. [20] Kirsch, A.: The Factorization Method for a Class of Inverse Elliptic Problems. To appear in Math. Nachr. [21] Kirsch, A., and Ritter, S.: A Linear Sampling Method for Inverse Scattering from an Open Arc. Inverse Problems 16 (2000), 89 105. [22] Monk, P.: Finite Element Methods for Maxwell s Equations. Oxford Science Publications. Oxford, 2003. [23] Pelokanos, G., and Sevroglou, V.: The (F F ) 1/4 Method for 2D Penetrable Elastic Bodies. Talk presented at the 6th International Workshop on Mathematical Methods in Scattering Theory and Biomedical Engineering. Tsepelovo, 2003. [24] Rynne, B. P., and Sleeman, B. D.: The Interior Transmission Problem and Inverse Scattering from Inhomogeneous Media. SIAM J. Math. Anal. 22 (1991), 1755 1762. 18