Inverse boundary value problem for Maxwell equations with local data

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1 Inverse boundary value problem for Maxwell equations with local data Pedro Caro, Petri la and Mikko Salo February 6, 9 Abstract We prove a uniqueness theorem for an inverse boundary value problem for the Maxwell system with boundary data assumed known only in part of the boundary. We assume that the inaccessible part of the boundary is either part of a plane, or part of a sphere. This work generalizes the results obtained by Isakov [I] for the Schrödinger equation to Maxwell equations. Introduction. Let Ω R 3 be a bounded domain with C, boundary, and let ε, µ, σ be C functions in Ω ε is the permittivity, µ the permeability, and σ the conductivity). We will assume that the coefficients satisfy the positivity conditions γ = ε + iσ/ω, ε >, µ >, σ in Ω..) Let D = i, let ν be the exterior unit normal to Ω, and consider the time-harmonic Maxwell equations for the electric field E and magnetic field H in Ω, { D H + ωγe =,.) D E ωµh =, with the boundary condition ν H = a on Ω..3) This is the magnetic boundary value problem for the Maxwell equations. Here we use to denote the vector product in R 3, and F is the curl of the vector field F. When posed in correct function spaces this problem admits a unique solution E, H) when the angular Supported by the Academy of Finland under CoE project 3476 The first author was also supported by Ministerio de Ciencia e Innovación de España, MTM5-765-C- Department of Mathematics, Universidad Autónoma de Madrid, 849 Madrid, Spain Department of Mathematics and Statistics, P.. Box 68, 4 University of Helsinki, Finland Department of Mathematics and Statistics, P.. Box 68, 4 University of Helsinki, Finland

2 frequency ω > does not belong to a discrete set of resonant frequencies. The impedance map Λγ, µ) is then defined by Λγ, µ) : a ν E Ω. In analogue with the Calderón problem of electrostatics see [C], [N] and [SU]), we consider the inverse problem of determining the electromagnetic parameters γ and µ from the boundary measurement Λγ, µ). Especially, in this work we assume that we can measure the values of Λγ, µ)a) only on a nonempty open subset Γ of Ω, and only for tangential boundary fields a supported in Γ. We further assume that the inaccessible part of the boundary Γ = Ω \ Γ is either part of a sphere, or part of a plane. Similar results were proved by Isakov [I] for the inverse boundary value problem for the Schrödinger operator. ur work extends this method to Maxwell equations. Let us now formulate the main results of this paper in a precise way. Given Ω as above, we define the spaces H DivΩ) = {F H Ω) 3 ; Divν F ) H Ω)}, T H Div Ω) = {a H Ω) 3 ; ν a =, Diva) H Ω)}, where Div denotes the surface divergence on Ω see for example [CP] and [PS] for more information). Provided that µ and γ are coefficients in C Ω) which satisfy.), then for ω outside a discrete set of resonant frequencies, the magnetic boundary value problem.) has a unique solution in HDiv Ω) H Div Ω) for any a T H Div Ω) see [PS]). The impedance map is well defined, and ur first main result is the following. Λγ, µ) : T H Div Ω) T H Div Ω). Theorem. Assume that Ω {R 3 ; x 3 < } is a C, domain. Let Γ = Ω {x 3 = } and Γ = Ω \ Γ. Assume we have two pairs of electromagnetic parameters γ j, µ j ) C 4 Ω) C 4 Ω), j =,, satisfying the positivity conditions.) and the following boundary conditions: γ =, µ = µ up to order one on Γ, and.4) { there exist C 4 extensions of γ j and µ j into R 3 which are.5) invariant under reflection across the plane {x 3 = }. Suppose that ω is not a resonant frequency for γ, µ ) or, µ ). If the impedance maps restricted to Γ coincide, Λγ, µ )a) Γ = Λ, µ )a) Γ for all a T H Div Ω) with suppa) Γ, then the electromagnetic parameters are equal, i.e. γ = and µ = µ in Ω.

3 Before formulating the second main theorem of our paper, we want to comment on the assumptions in Theorem. The condition.4) is not important, and we expect that a suitable boundary determination result would allow to remove this condition completely in the case of C coefficients such a result is given in [JM]). The condition.5) comes from the method of proof, which requires that the Maxwell equations are invariant under reflection across {x 3 = }. The following is a necessary condition for.5) to hold: ) l ) l γ = =, ν ν ) l ) l µ = µ = on Γ for l =, 3..6) ν ν For the sake of definiteness, we have stated Theorem in terms of the impedance map. However, the proof of Theorem extends to the case of restricted Cauchy data sets C Γ γ, µ) = {ν E Γ, ν H Γ ); E, H) H Ω) 3 H Ω) 3 solves.)}. In this setup Ω can be a Lipschitz domain and ω > any number, and the assumption C Γ γ, µ ) = C Γ, µ ) for two pairs γ j, µ j ) satisfying the conditions in Theorem will imply that γ = and µ = µ in Ω. This result allows domains with transversal intersections of Ω and {x 3 = }, in which case.6) is also sufficient for.5). It is well known see for example [MM], [S] and [S] and the references given in these articles) that Maxwell equations may not admit H -solutions even with boundary data in H Ω) if the domain is non convex: the solution may have a finite dimensional singular part near the conical singularity, so the image of the impedance map might be slightly larger than just T H Div Ω). However, all solutions needed in the proof are restrictions to Ω of H -solutions defined in a neighborhood of Ω actually they are defined in the whole R 3 ), and hence it is enough to restrict only to the part of the graph of Λγ, µ) that is actually needed. Let us now formulate our second main theorem: Theorem. Assume Ω B is a C, domain included in an open ball B of R 3 of positive radius, and let Γ = Ω B, Γ B, Γ = Ω \ Γ. Assume we have two pairs of electromagnetic parameters γ j, µ j ) C 4 Ω) C 4 Ω), j =,, satisfying the conditions of Theorem but with respect to B instead of {x 3 = }. If ω is not a resonant frequency and if the impedance maps restricted to Γ coincide, Λγ, µ )a) Γ = Λ, µ )a) Γ for all a T H Div Ω) with suppa) Γ, then the electromagnetic parameters are equal, i.e. γ = and µ = µ in Ω. The unique recovery of electromagnetic parameters from the scattering amplitude was first proven in [CP] under the assumption that the magnetic permeability µ is a constant. The unique recovery of general C -parameters γ and µ from full boundary data was then proved in [PS], and later simplified in [S], and the uniqueness for the inverse scattering 3

4 problem was proved in [Sa]. Boundary determination results were given in [McD] and [JM], and the more general chiral media was studied in [McD]. For a slightly more general approach and more background information, see also the review article [PS]. Note that in this work we need, for technical reasons, two more derivatives of the parameters compared to the full data problems. Inverse problems with partial data for scalar elliptic equations have attracted considerable attention recently. There are two main approaches: the first uses Carleman estimates [BU], [KSU] and the second uses reflection arguments [I]. In the first approach Ω can be any domain but one needs to measure part of Cauchy data in a small set Γ Ω and the other part in a neighborhood of Ω \ Γ. In the second approach it is enough to measure Cauchy data on a fixed small set Γ, but one has the restriction that Ω \ Γ has to be part of a plane or sphere. In this paper, we apply Isakov s reflection method [I] to the Maxwell system to prove Theorems and. As far as we know, these are the first partial data results for an elliptic system which are analogous to [I] an analog of [KSU] for a Dirac system was recently proved in [ST]). The strategy of the proof is similar to [PS] and [S]: we construct special solutions to the Maxwell system, insert them in a suitable integral identity, and recover the coefficients from an asymptotic limit. However, several issues arise when trying to combine this strategy with the reflection method. The integral identities in [PS] and [S] do not seem to go well together with partial data, and the final identity which we use for recovering the coefficients is new. In this identity, many terms survive in the asymptotic limit, and one needs to carefully manipulate these terms to determine the coefficients. Also, there are cross terms involving products of the original and reflected solutions. In [I] these cross terms were handled by the Riemann-Lebesgue lemma. In our case we need to prove a substitute for the Riemann- Lebesgue lemma, and here the C 4 -regularity of the coefficients is used. The structure of the paper is as follows. In the first section we augment the Maxwell equations into a Dirac type elliptic system and prove a crucial factorization result for a second order matrix Schrödinger operator. This follows in principle [S], but with somewhat different notations and some simplifications. In order to keep the article selfcontained, we have included complete proofs. In the second section we construct the complex geometric optics solutions CG solutions for short) needed in the proof. In this section, following Isakov [I], we also perform the reflection across {x 3 = } and analyze the behavior of CG solutions for Maxwell equations under this operation. In the third section we prove the integration by parts formula needed in our argument, and in the fourth section we perform the asymptotical computations needed in our proof. Theorem is then proved in section five, and finally using the Kelvin transform, the proof of Theorem is reduced to Theorem in the last section. Note that with the exception of the last section, we use the vector field formulation for our equations. nly when analyzing the behavior of Maxwell equations under the Kelvin transform does it become useful to revert to a formulation using differential forms. However, we feel that the major part of this work is easier to read when written in vector field notation. 4

5 Acknowledgement: The first author would like to thank Lassi Päivärinta and the Inverse Problems group at the University of Helsinki for their hospitality.. Augmented Maxwell system and reduction to Schrödinger equation. In this section we recall an argument used in [S]. Roughly, the idea is to augment Maxwell equations in such a way that, with a suitable rescaling of the fields, the system can be transformed into a Schrödinger equation. This argument can be written in a slightly more general setting than the one described above, see [PS]. Let U R 3 be an an open set along this section. Consider the time-harmonic Maxwell system in U given by { D H + ωγe =.7) D E + ωµh = where ω > is a fixed angular frequency, γ = ε + iσ/ω and µ, ε, σ C U) satisfy µ, ε >, σ in U. Taking the divergence of both equations in.7) we obtain the additional equations { D γe) = D µh) =. These will be useful in writing the Maxwell equations as an elliptic system. denoting α = log γ and β = log µ we may combine these into four equations D E + Dα E = D E + ωµh = D H + Dβ H = D H + ωγe =. Namely, This gives rise to the 8 8 system D D D D + Dα ωµi 3 Dβ ωγi 3 H E =. Here means that in these positions we may insert anything and would still get the same equations. We wish to consider 8-vectors X t = Φ H t Ψ E t ), where Ψ and Φ are additional scalar fields. We also want the system to be elliptic, in fact, we want the principal part to be a Dirac-type operator. To this end, consider two 4 4 operators acting on 4-vectors Φ H t ) t for instance) D P + D) = D D ) D, P D) = D D ). 5

6 These operators satisfy P + D)P D) = P D)P + D) = I 4, P + D) = P D), P D) = P + D). Motivated by the Pauli-Dirac operator in R 3, we choose the principal part to have the block form ) P P D) = D). P + D) bserve that P D)P D) = I 8 and P D) = P D). The following notations will also be used in the rest of the work. Let us denote ) ) P P b) a, b) =, P ± P+ b) a, b) =,.8) P + a) P a) where a, b C 3 ; and diaga, B) = A B where A and B are 4 4 diagonal C-matrices. When A = λ I 4 and B = λ I 4 with λ, λ C let us write ) λ I diagλ, λ ) = 4. λ I 4 Note that P a, b)p b, a) = diagb b, a a) and P ± a, b)p ± b, a) = diagb b, a a) for any a, b C 3. We shall write for short P a) and P ± a) when b = a, hence P a)p a) = a a)i 8 and P ± a)p ± a) = a a)i 8 for any a C 3. From these last equalities follow the anticommutation formulas ), P a)p b) = P b)p a) + a b)i 8.9) P ± a)p ± b) = P ± b)p ± a) + a b)i 8,.) for a, b C 3. Moreover, the commutation formula P ± a)p b) = P b)p ± a).) holds for a, b C 3. This formula follows easily from P + a)p + b) = P + a)p + b)) t = P b)p a). More generally P ± a, b)p c) = P c)p ± b, a),.) for a, b, c C 3. For the potential there is seemingly considerable freedom in the choice of the entries marked with. However, we later wish to reduce the Maxwell equations into a matrix 6

7 Schrödinger equation with no first order term so a proper choice of the extra entries will be crucial. We will take ωµ Dα V µ,γ = ωµi 3 Dα Dβ ωγ. Dβ ωγi 3 The augmented Maxwell system is then P D) + V µ,γ )X = in U. The solutions of this system for which Φ = Ψ = correspond to the solutions of the original Maxwell system in U. We may write the potential V µ,γ above in the form V µ,γ = ωdiagµ, γ) + P ± Dβ, Dα) + P Dβ, Dα))..3) Next we rescale the X as follows. Let be defined by Then where Y t = Y Φ Y H ) t Y Ψ Y E ) t ) Y = diagµ, γ )X. P D) + V µ,γ )X = diagγ, µ )P D) + W µ,γ )Y W µ,γ = κi 4 P +Dα) P Dβ) κi 4 and κ = ωγµ. Here we used that Hence ) = κi 8 + P ± Dβ, Dα) P D)diagµ, γ )Y ) = diagγ, µ )[P D) P Dβ, Dα)]Y. P D) + V µ,γ )X = P D) + W µ,γ )Y = when X and Y are related as above. This scaling is motivated by the following result. Lemma.. ne has P D) + W µ,γ )P D) W t µ,γ) = I 8 + Q µ,γ P D) W t µ,γ)p D) + W µ,γ ) = I 8 + Q µ,γ 7

8 where the matrix potentials are given by and Q µ,γ = Q µ,γ = with f = x j,x k f) 3 j,k=. Proof: We have α α αi 3 β β βi 3 κ + 4 Dα) )I 4 Dκ Dκ Dκ Dκ κ + 4 Dβ) )I 4 β β βi 3 κ + 4 Dβ) )I 4 Dκ Dκ α α αi 3 κ + 4 Dα) )I 4 P D) + W µ,γ )P D) W t µ,γ) = I 8 + W µ,γ P D) P D)W t µ,γ W µ,γ W t µ,γ. We claim that = W µ,γ P D) P D)W t µ,γ = α α αi 3 β β βi 3 P Dκ). This can be proved by a direct calculation. However, we will use symbol calculus which gives a slightly more elegant proof. In standard left) quantization, since the symbol of P D) is P ξ) and since ξ j P ξ) = ξj W µ,γ = for j =,, 3, we have W µ,γ P D) P D)W t µ,γ = pw µ,γ P ξ) P ξ)w t µ,γ = p 3 ξm P ξ)d xm Wµ,γ) t m= 3 ξm P ξ)d xm Wµ,γ) t = pp D)Wµ,γ). t m= 8

9 The second equality holds because W µ,γ P ξ) P ξ)w t µ,γ = P ± Dβ, Dα)P ξ) P ξ)p ± Dα, Dβ) =, which is true by.). Consequently W µ,γ P D) P D)Wµ,γ t = pp D)Wµ,γ) t = ) P D)P Dα) P Dκ). P + D)P + Dβ) n the other hand κ W µ,γ Wµ,γ t = + 4 Dα) )I 4 P + Dκ) P Dκ) κ + 4 Dβ) )I 4 since Dα + Dβ) = Dlogµγ) = κ Dκ. This shows the desired form of Q µ,γ. It remains to prove that Q µ,γ has the expression given. We have P D) W t µ,γ)p D) + W µ,γ ) = I 8 W t µ,γp D) + P D)W µ,γ W t µ,γw µ,γ. Note that W t µ,γ = W γ,µ. Hence, changing the role of µ and γ we obtain ), Wµ,γP t D) + P D)W µ,γ = W γ,µ P D) + P D)Wγ,µ t = ) P D)P Dβ) + P Dκ). P + D)P + Dα) and ) κ Wµ,γW t µ,γ = W γ,µ Wγ,µ t = + 4 Dβ) )I 4 P + Dκ) P Dκ) κ +. 4 Dα) )I 4 This shows the desired form of Q µ,γ. The crucial point in the above lemma is of course the vanishing of the first order terms. Also, it is worth pointing out that we will need both second order operators given in Lemma. in our work.. Construction of CG solutions. In this section we construct CG solutions for the rescaled Maxwell system and then we produce solutions for the original Maxwell system. In order to get such solutions we apply the Sylvester-Uhlmann method given in [SU]) to the matrix Schrödinger equation obtained in Lemma.. This argument was introduced in [S]. 9

10 For the sake of completeness we recall the Sylvester-Uhlmann estimate whose proof can be found in [SU], [S]. Let q L R 3 ; C) with compact support and < δ <. If ζ C 3 such that ζ ζ = k with ζ large enough, then for any f L δ+ there exists a unique u Hδ solving + ζ D + q)u = f. Moreover u H s C ζ s f δ L, δ+ for s with C a constant independent of ζ and f. The norms are given by f L = + x ) δ fx) dx δ R 3 and u H s = + x ) δ/ u. δ H s R 3 ) Assume that we can extend µ and γ to the whole space so that for some positive constants ε and µ we have γ ε, µ µ C 4 R 3 ), and let Q µ,γ = ω ε µ I 8 + Q µ,γ which is thus compactly supported. Denoting k = ωε µ we are thus led to the equation + k )I 8 + Q µ,γ ) Z =.4) Let θ R 3 any non-zero vector. Consider ζ C 3 such that ζ ζ = k and ζ = τ ζ + θ + τ ) := τ ζ + θτ), where τ is a free parameter controlling the size of ζ, and ζ = η + iη with η, η R 3, η η = and η = η =. Let Z = Z ζ) be a vector which does not depend on x and which is bounded with respect to the parameter τ. Proposition.. Assume that < δ <, and consider ε > such that < δ+ε <. There exists a CG solution of the form for the equation.4) in R 3, such that Z = e iζ x Z + Z + Z r ) D α Z L δ = τ ), D α Z r L δ = τ α +ε) ), for α. Moreover, if we denote then in R 3. R = lim τz M = lim Z, ζ D)I 8 R = Qµ,γ M.5)

11 Proof: Let ζ C 3 be as above, note that e iζ x + k )I 8 + Q µ,γ ) e iζ x Z + Z + Z r ) = = Q µ,γ Z + + ζ D)I 8 + Q µ,γ )Z + Z r ) in R 3. Let χ be a cut-off function which is identically on a neighborhood of {x R 3 : x < ρ} where ρ is so large that Ω { x < ρ}, and define For τ > let χ τ x) = χx/τ) and ζ D) f = e iξ x f π) 3 R ζ 3 ξ dξ. Z = τ χ τ ζ D) Q µ,γ Z ). It is well known that ζ D) : L δ+ L δ is bounded with norm of order ζ [SU]), hence Moreover, differentiating we get Z L δ C τ Q µ,γz L δ+ = τ ). D α Z L δ = τ ), for all α. Here we need the C 4 -regularity of the coefficients. n the other hand, let Z r be the solution of +ζ D)I 8 +Q µ,γ )Z r = +θτ) D)I 8 +Q µ,γ )Z + ζ Dχ τ ) ζ D) Q µ,γ Z ), satisfying D α Z r L C ζ α + θτ) D)I8 + Q µ,γ )Z + ζ Dχ τ ) ζ D) Q µ,γ Z ) δ, L δ+ for α. We then have the estimates Q µ,γ Z L δ+ = τ ), ζ Dχ τ ) ζ D) Q µ,γ Z ) Cτ ε ζ D) Q µ,γ Z ) L δ+ L δ+ε and + θτ) D)I 8 Z L δ+ C ζ D) Q τ µ,γ Z ) + C D ζ D) Q L δ τ µ,γ Z ) +Cτ ε + θτ) D) ζ D) Q µ,γ Z ). L δ+ε L δ

12 Let us explain the second inequality in more detail: consider ε > such that < δ+ε <. Then ζ Dχ τ ) ζ D) Q µ,γ Z ) = + x ) δ+ ζ Dχ τ ) ζ D) Q µ,γ Z ) dx L δ+ R 3 sup { + x ) ε ζ Dχ τ } + x ) δ+ε ζ D) Q µ,γ Z ) dx x suppχ τ ) R 3 Cτ ε) τ ζ D) Q µ,γ Z ). L δ+ε Therefore D α Z r L δ = τ α +ε) ), for α. Finally, R satisfies the equation.5) by the definition of Z and the continuity of ζ D). Considering Z as in Proposition. we have, by Lemma., that is a solution of the rescaled system Note that this solution can be written as Y = P D) W t µ,γ)z P D) + W µ,γ )Y =. Y = e iζ x Y + Y + Y r ).6) where and Y = P ζ)z, Y = P ζ)z W t µ,γz, Y r = P D) W t µ,γ)z + P D + ζ) W t µ,γ)z r,.7) Y L U) = τ), Dα Y L U) = ), Dα Y r L U) = τ α ε ).8) for α and U any bounded open subset of R 3. Recall that our aim is to construct CG solutions for the original Maxwell system. As we mentioned above, Y will produce a solution of the original Maxwell system if Y Φ = Y Ψ =. In order to get this condition we choose Z in a special way. Lemma.. If then we have for τ large enough. P ζ) k)z ) Φ = P ζ) k)z ) Ψ = Y t = Y H ) t Y E ) t )

13 Proof: This proof was given in [S]. Here it is included for the sake of completeness. As we mentioned above Y is solution of so it is also solution of P D) + W µ,γ )Y =, P D) W t µ,γ)p D) + W µ,γ )Y =. From Lemma. we know that Y Φ and Y Ψ are solutions of { + k )Y Φ + q β Y Φ = + k )Y Ψ + q α Y Ψ =,.9) where the potentials q β = β κ + k 4 Dβ), q α = α κ + k 4 Dα) are compactly supported. Adding ±ki 8 one gets Y = e iζ x Ỹ + Ỹ + Y r ) where Ỹ = P ζ) ki 8 )Z and Ỹ = P ζ)z + ki 8 Wµ,γ)Z t. This shows that { + ζ D + q β )Ỹ + Ỹ + Y r ) Φ =,.) + ζ D + q α )Ỹ + Ỹ + Y r ) Ψ =. It is known that these equations have unique solutions in L δ for ζ large enough see [SU]). Since neither Ỹ Φ nor Ỹ Ψ belong to L δ unless Ỹ Φ = Ỹ Ψ =, we choose Z such that Ỹ Φ = P ζ) ki 8 )Z ) Φ =, Ỹ Ψ = P ζ) ki 8 )Z ) Ψ =. In this way we have Ỹ + Ỹ + Y r ) Φ, Ỹ + Ỹ + Y r ) Ψ L δ. Then, the unique solvability for ζ large implies that Y Φ = Y Ψ = for ζ large enough. An explicit choice of Z verifying the condition of last lemma may be done as follows. Let ζ a ζ k ζ Z = kb τ ζ b = ki 3 b τ ζ = ki 3 ζ b τ ζ k ka ki 3 a ζ ki 3 a = ki 8 + ) τ P ± ζ) + P ζ)) m,.) where we denoted m t = b t a t ) for any a, b C 3. The same choice was done in [S]. 3

14 Next we perform the reflection argument given by Isakov in [I]. As in the statement of Theorem, let Ω {x 3 < } be a C, domain. With respect to the standard Cartesian basis e = {e, e, e 3 }, we introduce the reflection x = x, x, x 3 ) x, x, x 3 ) =: ẋx) and the reflected domain Ω = {ẋx) R 3 : x Ω}. Consider = Ω intγ ) Ω. By the assumption.5), we can extend the coefficients µ j and γ j into R 3 as C 4 functions which are even with respect to x 3 for j =,. We still denote the extended coefficients by µ j and γ j. Note that these functions have the proper regularity to construct the CG solutions as above. For the pairs µ, γ ) and µ, ), and for a given vector ξ = ξ, ξ 3 ) R 3 satisfying ξ >, we construct CG solutions Z, Y, X and Z, Y, X in R 3 with complex vectors ζ and ζ, respectively. We choose ζ = ξ + i τ + ξ η + τ + k η 4.) ζ = ξ i τ + ξ η + τ + k η, 4.3) with τ a free parameter controlling the size of ζ and ζ, where η, η R 3 verify η = η =, η η = and η j ξ = for j =,. More precisely, for ξ = ξ, ξ, ξ 3 ) = ξ, ξ 3 ) we choose η = ξ ξ, ξ, ), η = η ξ ξ = ξ ξ ξ ξ 3, ξ ξ 3, ξ ). Note ζ j ζ j = k for j =,. Let us denote and ζ = ζ lim τ = ˇζ = lim ξ τ) = ζ τ ζ) lim ζ τ = η + iη ζ τ = η iη ξ τ) = ζ τ ˇζ). bserve that ξ τ) = ξ + τ ) and ξ τ) = ξ + τ ). Furthermore, if we denote R = M = lim τz lim Z Ř = ˇM = lim τz lim Z 4

15 then the equations ζ D)I 8 R = Q µ,γ M.4) ˇζ D)I 8 Ř = Q µ, ˇM.5) hold in R 3. We now fix another orthonormal basis f = {f, f, f 3 } with With respect to the basis f one has Therefore, ζ = ζ = i f = ξ ξ, ξ, ), f 3 =,, ) = e 3, f = f f 3..6) i τ + ξ 4 τ + ξ 4, ξ It is obvious that for any x, y R 3 ξ =, ξ, ξ 3 ) f, η =,, ) f = f, η = ξ, ξ 3, ξ ) f. τ + k ξ 3 ξ, ξ 3 + τ + k ξ ξ, ξ τ + k ξ 3 ξ, ξ 3 + τ + k ξ ξ ) f ) f. x y = 3 x j y j = j= 3 j= x f j yf j where x = x, x, x 3 ) = x f, x f, x f 3) f and y = y, y, y 3 ) = y f, y f, y f 3 ) f with x 3 = x f 3 and y 3 = y f 3. We next show that the reflection argument allows to construct CG solutions whose tangential magnetic fields ν H on Ω have support contained in Γ. To this end we introduce for j =,, where Ẋ j x) = diag I 4, I 4 )X j ẋx)) I 4 =. 5

16 It is an easy matter to check that Ẋ and Ẋ are solutions of P D) + V µ,γ )Ẋ = P D) + V µ, )Ẋ = in Ω but also in Ω). Therefore, for τ large enough, H = X = X + Ẋ, H = X = X + Ẋ E E are solutions of the same equations and ν H j Γ = for j =,. Let us state the result obtained as the following proposition. Proposition.3. Corresponding to pairs µ, γ ) and µ, ), and for a given vector ξ R 3 satisfying ξ >, there are CG solutions X and X of the augmented Maxwell equations in Ω such that where the terms X j satisfy P D) + V µ,γ )X = P D) + V µ, )X = X = diagµ, γ )Y + diag I 4, I 4 )Y ẋ)) X = diagµ, γ )Y + diag I 4, I 4 )Y ẋ)) and the rescaled fields Y j are given by.6) and.7) for complex vectors ζ j as in.) and.3), with ξ = ζ ζ. Moreover, for τ large enough the fields X j will solve Maxwell equations in Ω and the tangential components of their magnetic fields will vanish on Γ. 3. The orthogonality identity. We assume that ω is not a resonant frequency, so the Maxwell equations.) have a unique solution E, H) for any prescribed value of ν H on Ω as discussed in the introduction. ur uniqueness proofs are based on the following orthogonality relation involving solutions in Ω and impedance maps on Ω. Proposition 3.. Let µ, γ ) and µ, ) be two pairs of coefficients such that ω is not a resonant frequency for either. Let Λ and Λ be the corresponding impedance maps. Then, for any Xj t = ) Hj t Ej t satisfying P D) + V µ,γ )X = P D) + V µ, )X = in Ω one has XV µ,γ V µ, )X dx = i Ω 6 Ω Λ Λ )ν H ) H ds.

17 Proof: If u v) = the integration by parts identity Ω v u dx and u v) Ω = Ω v u ds P D)X X ) = ip ν)x X ) Ω + X P D)X ) holds. Since Ψ j = Φ j = for j =, one has that X V µ, X ) = X V µ, X ), hence Let X t = V µ,γ V µ, )X X ) = V µ,γ X X ) X V µ, X ) = V µ,γ X X ) X V µ, X ) = P D)X X ) + X P D)X ) = ip ν)x X ) Ω = i E ν H ) H ν E ) ds H ) t Ẽt be the solution of the boundary value problem { P D) + V µ, ) X = in Ω ν H = ν H on Ω. Since X on the boundary verifies Ω E ν H ) = E ν H ) we have V µ,γ V µ, )X X ) = i E ν H ) H ν E ) ds. Ω Finally, note that ip ν) X X ) Ω = P D) X X ) X P D)X ) = V µ, X X ) + X V µ, X ) = V µ, X X ) + X V µ, X ) = V µ, V µ, ) X X ) =. Here we used again that Ψ = Φ = Ψ = Φ =. From the last identity one gets i E ν H ) ds = i H ν Ẽ) ds. Therefore Ω V µ,γ V µ, )X X ) = i = i Ω Ω Ω H ν Ẽ) H ν E ) ds H Λ Λ )ν H ) ds, 7

18 which is the desired integral formula. We next plug the solutions constructed in Proposition.3 into the integral formula given in Proposition 3.. Since Λ a = Λ a on Γ for a T H Ω) with suppa) Γ, we have X V µ,γ V µ, )X dx = i Λ Λ )ν H ) H ds Ω Ω = i Λ Λ )ν H ) H ds. Γ Note that ν H Γ = means that the tangential components of H on Γ vanish. n the other hand, Λ Λ )ν H ) only has tangential components. Hence Λ Λ )ν H ) H = on Γ. Since the boundary term in the integral formula vanishes we get the relation X V µ,γ V µ, )X dx =. Following the same notation we write and we thus have Ω Ẏ j x) = diag I 4, I 4 )Y j ẋx)) X = X + Ẋ = diagµ, γ )Y + Ẏ) X = X + Ẋ = diagµ, γ )Y + Ẏ). Since for τ large enough the scalar components vanish, i.e. X Φ j = X Ψ j = for j =,, writing V µ,γ V µ, as in.3) gives where X V µ,γ V µ, )X = X diagωµ µ ), ωγ ))X = Y + Ẏ) diag µ, γ)y + Ẏ) µ = ω µ µ µ µ, γ = ω γ γ. If now we go back to the integral relation, we get = X V µ,γ V µ, )X dx = Y + Ẏ) diag µ, γ)y + Ẏ) dx Ω Ω = Y diag µ, γ)y dx + Ẏ diag µ, γ)ẏ dx + Y diag µ, γ)ẏ dx Ω Ω Ω + Ẏ diag µ, γ)y dx. Ω 8

19 We gather the first two terms and the last two terms cross terms) in different expressions. bserve that Ẏ diag µ, γ)ẏ dx = Y ẋ)) diag I 4, I 4 )diag µ, γ)diag I 4, I 4 )Y ẋ) dx Ω Ω = Y ẋ)) diag µ, γ)y ẋ) dx Ω = Y diag µ, γ)y dx Ω hence Y diag µ, γ)y dx + Ẏ diag µ, γ)ẏ dx = Y diag µ, γ)y dx. Ω Ω With a similar argument the cross terms become Ω Y diag µ, γ)ẏ dx + Therefore, = X V µ,γ V µ, )X dx = Ω Ω Ẏ diag µ, γ)y dx = Ẏ diag µ, γ)y dx. Y diag µ, γ)y dx + Ẏ diag µ, γ)y dx. 3.7) Below we will integrate by parts in many times. Even though may not be a Lipshitz domain, this procedure is easily justified by integrating by parts separately in the C, domains Ω and Ω and by noting that the contribution from {x 3 = } vanishes. Now, the terms on the right hand side of 3.7) may be simplified by writing Y + Ẏ )diag µ, γ)y dx = Y + Ẏ )diag µ, γ)p D) Wµ t,γ )Z dx. 3.8) Note now that P D)diag γ, µ)z ) = diag µ, γ)p D)Z + P D)diag γ, µ))z and = Y + Ẏ )P D)diag γ, µ)z ) dx = P D)Y + Ẏ)) diag γ, µ)z dx i Y + Ẏ )P ν)diag γ, µ)z ds integrating by parts. We know by.4) that µ = µ and γ = on Γ, and this is valid on as well. Since Y + Ẏ is solution for the rescaled system P D) + W µ, )Y + Ẏ) = 9

20 in, we have Y + Ẏ )P D)diag γ, µ)z ) dx = From 3.8) and the above observations we get = Y + Ẏ W µ, Y + Ẏ)) diag γ, µ)z dx. Y + Ẏ )diag µ, γ)y dx = 3.9) )W µ, diag γ, µ) + P D)diag γ, µ)) + diag µ, γ)w t µ,γ )Z dx. It is a straightforward computation to check where W µ, diag γ, µ) + P D)diag γ, µ)) + diag µ, γ)w t µ,γ = κ diag γ, µ) + κ diag µ, γ) + P ± D γ, D µ) + P D γ, D µ), From now on we denote for short µ = ω µ + µ µ µ, γ = ω γ + γ. U := κ diag γ, µ) + κ diag µ, γ) + P ± D γ, D µ) + P D γ, D µ). bserve that these computations could be written separately for Y and Ẏ, giving Y diag µ, γ)y dx = Y UZ dx, Ẏ diag µ, γ)y dx = Ẏ UZ dx. Recall that solutions of the rescaled Maxwell system and the Schrödinger system look respectively like lim Y = e iζ x Y + Y + Y r ), Z = e iζ x Z + Z + Z r ). Using the estimates in Proposition. and in.8, we have Y diag µ, γ)y dx = lim lim lim e iζ ζ ) x Y ) UZ dx e iζ ζ ) x Y ) UZ dx e iζ ζ ) x Y ) UZ dx.

21 Noting that ζ ζ = ξ and taking the limit in 3.7), we obtain = lim e iξ x Y ) UZ dx 3.3) lim e iξ x Y ) UZ dx 3.3) lim e iξ x Y ) UZ dx 3.3) lim Ẏ UZ dx. 3.33) 4. Some technical computations. In this technical section we compute the limit of the terms given in 3.3), 3.3), 3.3) and 3.33). Lemma 4.. The limit in 3.3) is given by lim e iξ x Y ) UZ dx = = k ζ b ) ζ b ) + ζ a ) ζ a )) e iξ x κ + κ ) µ + γ) dx +ω ζ b ) ζ b ) e iξ x µ ) dx + ω ζ a ) ζ a ) e iξ x γ ) dx µ γ +kω e [ ζ iξ x µ γ ) ] b ) ζ a ) D + ζ a ) ζ b ) D dx µ γ +kω e [ ζ iξ x µ γ ) ] b ) ζ a ) D + ζ a ) ζ b ) D dx. 4.34) µ Proof: Remember that with and Z = τ ki 8 + ) P ± ζ ) + P ζ )) m m t = b t a t ), Y = P ζ )Z. A simple commutation in the middle term after writing the definion of U, gives e iξ x Y ) UZ dx = k e iξ x P ζ )Z τ ) Um dx 4.35) + e iξ x Z τ ) P ζ )P ± ζ ) + P ζ ))κ diag µ, γ) + κ diag γ, µ))m dx 4.36) + e iξ x Z τ ) P ζ )P ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx. 4.37)

22 Before computing the limit of 4.35) let us write Y = P ζ )Z = τ It follows from this expression that k lim τ kζ a ζ b )ζ kζ a kζ b ζ a )ζ + kζ b e iξ x P ζ )Z ) Um dx = ) = k e iξ x ζ b ) ζ t ζ a ) ζ t κ diag γ, µ) + κ diag µ, γ)) ) +k e iξ x ζ b ) ζ t ζ a ) ζ t P ± D γ, D µ) + P D γ, D µ)) +kω ] = k e [ ζ iξ x b ) ζ b )κ γ + κ µ) + ζ a ) ζ a )κ µ + κ γ) e [ ζ iξ x µ γ ) ] b ) ζ a ) D + ζ a ) ζ b ) D µ γ To compute the limit of the term in 4.36) observe that one has the relation. 4.38) b a b a dx dx dx dx. 4.39) τ Z ) P ζ )P ± ζ ) + P ζ )) = k τ m P ζ )P ± ζ ) + P ζ )) + 4τ m P ± ζ ) + P ζ ))P ζ )P ± ζ ) + P ζ )). 4.4) Using the commutation formulas.9),.) and.), and the fact that ζ ζ = ξ, we can see that P ± ζ ) + P ζ ))P ζ )P ± ζ ) + P ζ )) = = ζ ζ )P ζ ) P ± ξ)p ± ζ )P ζ ) + ζ ζ )P ± ζ ) P ξ)p ± ζ )P ζ ) +ζ ζ )P ± ζ ) + P ζ )) = P ± ξ)p ± ζ )P ζ ) P ξ)p ± ζ )P ζ ) + τ). 4.4)

23 Having in mind formulas 4.4) and 4.4), we compute the limit of 4.36): lim e iξ x Z τ ) P ζ )P ± ζ ) + P ζ ))κ diag µ, γ) + κ diag γ, µ))m dx = = k e iξ x m P ζ)p ± ζ) + P ζ))κ diag µ, γ) + κ diag γ, µ))m dx e iξ x m 4 P ± ξ) + P ξ))p ± ζ)p ζ)κ diag µ, γ) + κ diag γ, µ))m dx = k e iξ x m P ζ)p ± ζ)κ diag µ, γ) + κ diag γ, µ))m dx ] = k e [ ζ iξ x b ) ζ b )κ µ + κ γ) + ζ a ) ζ a )κ γ + κ µ) dx. 4.4) Here we used P ζ)p ζ) = ζ ζ =, P ± ζ)p ζ) = P ζ)p ± ζ) = ζ ζ ζ ζ and P ± ξ) + P ξ))p ± ζ)p ζ) =. Finally we study the limit of 4.37). Let us write where P ζ ) = τp ζ) + P ξ τ)) P ξ τ)) P ξ) as τ +. We have e iξ x Z τ ) P ζ )P ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx = = e iξ x Z 4τ ) P ξ τ))p ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx + e iξ x Z ) P ζ)p ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx. We split Z into pieces depending of order of τ, ζ a Z = τ ζ b + τ 3 kb ka := τ M + τ M, 4.43)

24 and we set ˇM = lim Z = ˇζ a ˇζ b It follows that lim e iξ x Z τ ) P ζ )P ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx = = e iξ x ˇM 4 P ξ)p ± D γ, D µ) + P D γ, D µ))p ± ζ) + P ζ))m dx + e iξ x M ) P ζ)p ± D γ, D µ) + P D γ, D µ))p ± ζ) + P ζ))m dx + lim e iξ x M ) P ζ)p ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx. τ bserve that for any a C 3 the formula. P a)p ± D γ, D µ) + P D γ, D µ))p ± ζ) + P ζ))m = a Dγ / ζ a = 4ω a Dγ / a Dµ /µ ζ b a Dµ /µ holds. Using that ζ = τ ζ ξ + τ ), the limit is lim e iξ x Z τ ) P ζ )P ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx = = ω e [ ζ iξ x µ γ ) ] b ) ζ b )ξ D) + ζ a ) ζ a )ξ D) dx µ +kω e [ ζ iξ x µ γ ) ] b ) ζ a ) D + ζ a ) ζ b ) D dx µ ω + lim e iξ x µ γ ) ] [ζ τ b )ζ b ) ζ D) + ζ a )ζ a ) ζ D) dx µ In the first term on the right, integration by parts gives ω e [ ζ iξ x µ γ ) ] b ) ζ b )ξ D) + ζ a ) ζ a )ξ D) dx µ = ω e [ ζ iξ x µ γ ) ] b ) ζ b ) ) + ζ a ) ζ a ) ) dx µ +iω e [ ζ iξ x µ γ ) ] b ) ζ b )ν D) + ζ a ) ζ a )ν D) ds. µ 4

25 Since νµ l = νµ l and νγ l = νγ l for l =, on Γ, the boundary term vanishes. n the other hand, for the last term on the right we integrate by parts to obtain e iξ x µ [ µ ) ] ζ D) dx = e iξ x ζ D) dx µ µ [ µ ) ] = e iξ x ζ ξ) dx = µ and e iξ x γ ζ D) dx = = [ γ ) ] e iξ x ζ D) dx = [ γ ) ] e iξ x ζ ξ) dx =. We introduced in order to get rid of the boundary terms in the integrations by parts. Eventually one has lim e iξ x Z τ ) P ζ )P ± D γ, D µ) + P D γ, D µ))p ± ζ ) + P ζ ))m dx = = ω e [ ζ iξ x µ γ ) ] b ) ζ b ) ) + ζ a ) ζ a ) ) dx µ +kω e [ ζ iξ x µ γ ) ] b ) ζ a ) D + ζ a ) ζ b ) D dx. 4.44) µ Putting 4.39), 4.4) and 4.44) together we get the final result. Lemma 4.. The limit in 3.3) is given by = [ +ω e iξ x + +ω lim e iξ x Y ) UZ dx = ] e iξ x κ [ ζ a ) ζ a )κ γ + κ µ) + ζ b ) ζ b )κ µ + κ γ) dx ζ a ) ζ a ) Dγ γ D + ζ b ) ζ b ) Dµ µ ) ] D dx µ µ ] e [ ζ iξ x R E ) ζ a )κ γ + κ µ) + ζ R H ) ζ b )κ µ + κ γ) dx e [ ζ iξ x b ) R E ζ) µ D ζ a ) µ R H ζ) γ ) ] D dx [ µ ω ζ b ) e iξ x ζ D) R ) ] Ψ dx µ [ γ ω ζ a ) e iξ x ζ D) R ) ] Φ dx, 4.45) 5

26 where with R t = RΦ R H ) t RΨ R E ) t ) = Ř Ř = lim τz. Proof: Since Y = P ζ )Z Wµ t, Z we have e iξ x Y ) UZ dx = e iξ x Z ) P ζ )UZ dx 4.46) e iξ x Z) W µ, UZ dx. 4.47) We can write and set M = Then the limit of 4.47) is lim Z = W µ, = κ I 8 P ± Dβ, Dα ) ζ a ζ b ˇM = lim Z = ˇζ a ˇζ b lim e iξ x Z) W µ, UZ dx = = e iξ x κ ˇM U M dx + e iξ x ˇM P ± Dβ, Dα )U M dx = e iξ x κ ˇM κ diag γ, µ) + κ diag µ, γ)) M dx + e iξ x ˇM P ± Dβ, Dα )P ± D γ, D µ) + P D γ, D µ)) M dx ] = e iξ x κ [ ζ a ) ζ a )κ γ + κ µ) + ζ b ) ζ b )κ µ + κ γ) dx [ +ω e iξ x ζ a ) ζ a ) Dγ γ D + ζ b ) ζ b ) Dµ µ ) ] D dx. 4.48) µ µ In the second equality we just wrote down U and we used ˇM P ± D γ, D µ) + P D γ, D µ)) M =. and ˇM P ± Dβ, Dα )κ diag γ, µ) + κ diag µ, γ)) M =. 6

27 We now compute the limit of the right hand side of 4.46): lim e iξ x Z ) P ζ )UZ dx = = e [ ζ iξ x R E ) ζ a )κ γ + κ µ) + ζ R ] H ) ζ b )κ µ + κ γ) dx +ω e [ ζ iξ x b ) R E ζ) µ D ζ a ) µ R H ζ) γ ) ] D dx +ω e [ ζ iξ x b ) R Ψ µ ζ D) + ζ a ) µ R Φ γ ) ] ζ D) dx. Integrating by parts one gets e iξ x µ RΨ ζ D) dx = µ = [ µ ) ] e iξ x RΨ ζ D) dx µ [ µ e iξ x ζ D) R ) ] Ψ dx µ and e iξ x γ RΦ ζ D) dx = = [ γ ) ] e iξ x RΦ ζ D) dx [ γ e iξ x ζ D) R ) ] Φ dx. nce again, we introduced in order to get rid of the boundary terms. Finally lim e iξ x Z ) P ζ )UZ dx = = e [ ζ iξ x R E ) ζ a )κ γ + κ µ) + ζ R ] H ) ζ b )κ µ + κ γ) dx +ω e [ ζ iξ x b ) R E ζ) µ D ζ a ) µ R H ζ) γ ) ] D dx [ µ ω ζ b ) e iξ x ζ D) R ) ] Ψ dx µ [ γ ω ζ a ) e iξ x ζ D) R ) ] Φ dx. 4.49) Putting 4.48) and 4.49) together we get the final formula. 7

28 Lemma 4.3. The limit in 3.3) is given by = +ω lim e iξ x Y ) UZ dx = e [κ iξ x γ + κ µ) ζ b ) ζ R H ) + κ µ + κ γ) ζ a ) ζ R ] E ) dx e [ ζ iξ x a ) ζ R H γ ) D ζ b ) ζ γ R E µ ) ] ) D dx µ [ µ ω ζ b ) e iξ x ζ D) R ) ] Ψ dx µ [ γ ω ζ a ) e iξ x ζ D) R ) ] Φ dx, 4.5) where with R t = RΦ R H ) t RΨ R E ) t ) R = lim τz. Proof: We use the computations done in 4.38) for Y = P ζ )Z and look at the limit lim e iξ x Y ) UZ dx = lim e iξ x P ζ )Z) UZ dx ) = e iξ x ζ b ) ζ t ζ a ) ζ t U R dx. 4.5) We insert the definition of U in 4.5) and we compute ) e iξ x ζ b ) ζ t ζ a ) ζ t U R dx = = e [κ iξ x γ + κ µ) ζ b ) ζ R H ) + κ µ + κ γ) ζ a ) ζ R ] E ) dx ) + e iξ x ζ b ) ζ t ζ a ) ζ t P ± D γ, D µ) + P D γ, D µ)) R dx = e [κ iξ x γ + κ µ) ζ b ) ζ R H ) + κ µ + κ γ) ζ a ) ζ R ] E ) dx [ +ω e iξ x R Ψ µ ζ b ) ζ D) ζ b ) ζ µ R E µ ) ] ) D dx µ [ +ω e iξ x R Φ γ ζ a ) ζ D) + ζ a ) ζ γ R H γ ) ] ) D dx. γ 8

29 Integrating by parts one gets e iξ x µ RΨ ζ D) dx = µ = [ µ ) ] e iξ x RΨ ζ D) dx µ [ µ e iξ x ζ D) R ) ] Ψ dx µ and e iξ x γ RΦ ζ D) dx = = [ γ ) ] e iξ x RΦ ζ D) dx [ γ e iξ x ζ D) R ) ] Φ dx. This gives the final result. The last technical lemma shows that the cross terms do not contribute to the limit. Lemma 4.4. The limit in 3.33) satisfies Proof: Recall that lim Ẏ UZ dx =. 4.5) Ẏ = e iζ ẋ Ẏ + Ẏ + Ẏ r ) Z = e iζ x Z + Z + Z r ), where Ẏ j = diag I 4, I 4 )Yj ẋ) for j =,, r. Note that writing the dot product in terms of the basis f in.6) one has φ τ x) := ζ x ζ ẋx) = ξ x f + τ + k ξ ξ xf 3. Since we chose f 3 = e 3 we have that x f 3 = x 3, hence Ẏ UZ dx = e iφτ Ẏ + Ẏ + Ẏ r ) UZ + Z + Zr ) dx ) = τ + k ξ ξ = τ + k ξ ) ξ ) i τ + k ξ ξ D x3 e iφτ Ẏ + Ẏ + Ẏ r ) UZ + Z + Z r ) dx e iφτ D x3 [Ẏ + Ẏ + Ẏ r ) UZ + Z + Z r )] dx e iφτ ν 3 Ẏ + Ẏ + Ẏ r ) UZ + Z + Z r ) dx. 9

30 The boundary term vanishes because we assumed that l νµ = l νµ and l νγ = l ν for l =, on Γ. Since Ẏ and Z are constant vectors with respect to x we have D x3 [Ẏ + Ẏ + Ẏ r ) UZ + Z + Z r )] = D x3 Ẏ + Ẏ r ) UZ + Z + Z r ) +Ẏ + Ẏ + Ẏ r ) D x3 UZ + Z + Zr ) +Ẏ + Ẏ + Ẏ r ) UD x3 Z + Zr ). Note that from the asymptotic behavior in.8) and in Proposition. one can see that D x3 Ẏ + Ẏ r ) UZ + Z + Zr ) = D x3 Ẏ r ) UZ + ), Ẏ + Ẏ + Ẏ r ) D x3 UZ + Z + Zr ) = Ẏ ) D x3 UZ + ), Ẏ + Ẏ + Ẏ r ) UD x3 Z + Zr ) = Ẏ ) UD x3 Zr + ), in L ); moreover, D x3 Ẏ r ) UZ = τ ε ) L ) n the other hand, we know by 4.38) and 4.43) that Ẏ ) UD x3 Zr = τ ε ). L ) Y ) t = ) ζ b )ζ t ζ a )ζ t + ) := τ τ L ) t + ), Z) t = ζ a ζ b ) + τ ) := τ τ M ) t + τ ) in L ). Therefore lim Note that Ẏ = lim = lim UZ dx = lim + τ + k ξ ) ξ τ + k ξ ) ξ τ τ + k ξ ξ e iφτ Ẏ ) D x3 UZ dx + ) [ e iφτ D x3 Ẏ r ) UZ dx ] e iφτ Ẏ ) UD x3 Zr dx e iφτ Y ẋ)) diag I 4, I 4 )D x3 UZ dx L ) diag I 4, I 4 )D x3 UM = e iφτ L ) diag I 4, I 4 )D x3 UM dx. = ωζ a )ζ a ) ζ D)D x3 γ ωζ b )ζ b ) ζ D)D x3 µ µ. Since the Riemann-Lebesgue lemma implies that µ lim e iφτ D j D x3 dx = lim µ we have proved the result. e iφτ D j D x3 γ dx =, 3

31 5. Proof of Theorem. Assuming the conditions in Theorem, we have proved that the identity in 3.3) 3.33) holds and that the terms in that identity have the limits given in Lemmas 4. to 4.4. In this section we show the equations verified by the coefficients and then give a proof of Theorem. Proposition 5.. Let ξ R 3. With appropriate choices of the constant vectors a, a, b and b one has the following equations: a) If we choose b = b = ˇζ and a = a = ζ [ e iξ x β β ) + ] 4 [Dβ ) Dβ ) ] + κ κ dx =. 5.53) b) If we choose a = a = ˇζ and b = b = ζ [ e iξ x α α ) + ] 4 [Dα ) Dα ) ] + κ κ dx =. 5.54) Proof: In order to get these equations from the previous lemmas we will need the equations.4) and.5). Let us write these equations by components: ζ D) R Φ = α κ + k Dα 4 ) ) ζ a ) ζ D)I 3 RH = ζ b )Dκ ζ D) R Ψ = β κ + k Dβ 5.55) 4 ) ) ζ b ) ζ D)I 3 RE = ζ a )Dκ, ˇζ D)ŘΦ = α κ + k Dα 4 ) )ˇζ a ) ˇζ D)I 3 Ř H = ˇζ b )Dκ ˇζ D)ŘΨ = β κ + k Dβ 5.56) 4 ) )ˇζ b ) ˇζ D)I 3 Ř E = ˇζ a )Dκ. We take complex conjugates in 5.56) in order to get the information required. ζ D) R Φ = α κ + k Dα 4 ) ) ζ a ) ζ D)I 3 RH = ζ b )Dκ ζ D) R Ψ = β κ + k Dβ 4 ) ) ζ b ) ζ D)I 3 RE = ζ a )Dκ. 5.57) These equations are valid in R 3. Multiplying the second and fourth equations of 5.55) and 5.57) by ζ one gets ζ R H = ζ b )κ k) ζ R E = ζ a )κ k) ζ R 5.58) H = ζ b )κ k) ζ R E = ζ a )κ k). 3

32 To see the last identities we used the decay of R j for j =, at infinity, and the fact that ζ R H ζ b )κ, ζ R E ζ a )κ, ζ R H ζ b )κ and ζ R E ζ a )κ are entire functions in the variable ζ x. Let us denote T ) = ω ζ b ) T ) = ω ζ a ) T 3) = ω ζ b ) T 4) = ω ζ a ) [ µ e iξ x ζ D) R ) ] Ψ dx µ [ γ e iξ x ζ D) R ) ] Φ dx γ [ µ e iξ x ζ D) R ) ] Ψ dx µ [ γ e iξ x ζ D) R ) ] Φ dx which are terms from 4.45) and 4.5). Then from the first and third equations in 5.55) and 5.57) one gets ω ζ b ) ζ b ) ω ζ a ) ζ a ) Noting that one can write ω ζ b ) ζ b ) ω ζ a ) ζ a ) where T 5) = ω ζ b ) ζ b ) T 6) = ω ζ a ) ζ a ) T ) + T ) + T 3) + T 4) = [ e iξ x β β ) κ + κ 4 Dβ ) + ] [ 4 Dβ ) µ ) ] µ e iξ x [ α α ) κ + κ 4 Dα ) + 4 Dα ) ] [ γ µ β µ µ µ β ) = D D µ µ µ µ α γ γ γ γ α ) = D D, γ γ T ) + T ) + T 3) + T 4) = T 5) + T 6) [ e iξ x µ µ µ µ ] D D µ µ µ µ β β ) dx e iξ x [ γ D γ γ D γ γ α α ) dx ] dx. ] dx, [ e iξ x κ + κ 4 Dβ ) + ] [ 4 Dβ ) µ ) ] dx µ [ e iξ x κ + κ 4 Dα ) + ] [ 4 Dα ) γ ) ] dx. 3

33 Note that µ µ D µ Denoting we get Dβ µ D µ D = µ µ γ γ γ D D = γ Dα D T 7) = ω ζ b ) ζ b ) T 8) = ω ζ a ) ζ a ) γ µ Dβ µ D µ Dα D e iξ x β β ) dx e iξ x α α ) dx γ. T ) + T ) + T 3) + T 4) = T 5) + T 6) + T 7) + T 8) [ ω ζ b ) ζ b ) e iξ x Dβ µ D µ Dβ µ ] D dx µ [ ω ζ a ) ζ a ) e iξ x Dα γ D Dα γ ] D dx γ ω ζ b ) ζ b ) e iξ x µ ) dx µ ω ζ a ) ζ a ) e iξ x γ ) dx. We consider the next terms from 4.34) and 4.45) T 9) = ω ζ b ) ζ b ) e iξ x µ ) dx µ T ) = ω ζ a ) ζ a ) e iξ x γ ) dx γ T ) = ω e [ ζ iξ x γ a ) ζ a )Dα D + ζ b ) ζ b )Dβ D hence ) ] dx, µ T ) + T ) + T 3) + T 4) + T 9) + T ) + T ) = T 5) + T 6) + T 7) + T 8) [ +ω ζ b ) ζ b ) e iξ x Dβ µ D + µ Dβ µ ] D dx µ [ +ω ζ a ) ζ a ) e iξ x Dα γ D + Dα γ ] D dx. We note that µ µ γ γ D = Dβ Dβ ), D = Dα Dα ) µ µ 33 µ

34 hence T ) + T ) + T 3) + T 4) + T 9) + T ) + T ) = T 5) + T 6) + T 7) + T 8) +ω ζ b ) ζ b ) e iξ x µ [Dβ ) Dβ ) ] dx 4 µ +ω ζ a ) ζ a ) e iξ x γ [Dα ) Dα ) ] dx. 5.59) 4 Considering the terms T ) = e [ ζ iξ x R E ) ζ a )κ γ + κ µ) + ζ R ] H ) ζ b )κ µ + κ γ) dx T 3) = e [κ iξ x γ + κ µ) ζ b ) ζ R H ) + κ µ + κ γ) ζ a ) ζ R ] E ) dx T 4) = k ζ b ) ζ b ) + ζ a ) ζ a )) e iξ x κ + κ ) µ + γ) dx ] T 5) = e iξ x κ [ ζ a ) ζ a )κ γ + κ µ) + ζ b ) ζ b )κ µ + κ γ) dx from 4.34), 4.45) and 4.5); and using 5.58) we have that = + Therefore T ) + T 3) = ] e [ ζ iξ x a ) ζ a )κ k)κ γ + κ µ) + ζ b ) ζ b )κ k)κ µ + κ γ) dx ] e [κ iξ x γ + κ µ) ζ b ) ζ b )κ k) + κ µ + κ γ) ζ a ) ζ a )κ k) dx. T ) + T 3) + T 4) + T 5) = = ζ a ) ζ a ) e iξ x κ κ µ + κ γ) dx + ζ b ) ζ b ) e iξ x κ κ γ + κ µ) dx. 5.6) Now we make the choice b = b = ˇζ and a = a = ζ. The equations 5.55) and 5.57) imply that R j E = for j =,. Considering all the terms from the technical lemmas in Section 4, we see that the only surviving terms in 3.3) 3.33) are those which appear on the left hand sides of 5.59) and 5.6). Thus these computations give [ e iξ x β β ) + µ [Dβ ) Dβ ) ] + κ 4 µ ω κ γ + κ µ) +κ κ + 4 Dβ ) [ 4 Dβ ) µ ) ]] ) dx =. µ 34

35 If we choose a = a = ˇζ and b = b = ζ, we get that R j H = for j =, for the same reason as above. We consider again all the terms from the lemmas in Section 4 and we get [ e iξ x α α ) + γ [Dα ) Dα ) ] + κ 4 ω κ µ + κ γ) +κ κ + 4 Dα ) [ 4 Dα ) γ ) ]] ) dx =. Finally, by a direct computation we see that κ ω κ µ + κ γ) + κ κ ) γ =, κ ω κ γ + κ µ) + κ κ ) µ µ =, hence the result is obtained. The preceding arguments are valid for all ξ R 3 with ξ >, since this was the assumption on ξ in the construction of reflected CG solutions. However, the continuity of the Fourier transform proves the result for all ξ. The above proposition shows that the coefficients satisfy the following equations in : α α ) + 4 Dα + α ) Dα α ) + ω µ γ µ ) =, β β ) + 4 Dβ + β ) Dβ β ) + ω µ γ µ ) =. Let u = γ / and v = µ /µ. The equations become log u) + γ ) Dγ Dlog u) + ω µ γ µ ) =, log v) + µ µ ) Dµ µ Dlog v) + ω µ γ µ ) =. Multiplying the equations by γ and µ µ we obtain Finally, these may be written as D γ Dlog u)) + ω µ γ µ )γ =, D µ µ Dlog v)) + ω µ γ µ )µ µ =. D Du) + ω µ u v )u =, D µ Dv) + ω µ u v )v = in. This is a semilinear elliptic system for u, v C 4 ). The assumptions in Theorem imply that u = v = and ν u = ν v =. The following formulation of the unique continuation property will therefore imply that γ and µ µ in, thus proving Theorem. 35

36 Lemma 5.. Let a, b be non-vanishing complex functions in C ) with positive real part, and let p, q be complex functions in L ). Suppose that u, v C ) satisfy in D adu) + pu v )u =, D bdv) + qu v )v =, and also u = v = and ν u = ν v =. Then u v in. Proof: Let u = + a z and v = + b w. Using the identities we obtain the equations Here p, q L ) are given by D ada z)) = a z + a ) z), a D bdb w)) = b w + b ) w), b u v = ab) zw + b z + a w)uv + ), z + a ) a z + pzw + b z + a w) =, w + b ) b w + qzw + b z + a w) =. p = a pab) uv + )u, q = b qab) uv + )v. Thus, there is C > such that the following differential inequalities for z, w C ) are valid almost everywhere in : zx) C zx) + wx) ), wx) C zx) + wx) ). We also have z = w = and ν z = ν w =. The unique continuation property holds in this setting, which can be seen by applying a scalar Carleman estimate to both z and w for details see [KSaU]). Thus z and w vanish identically in. 6. Proof of Theorem. Here we prove an analog of Theorem for the case where the inaccessible part is part of a sphere. Following the idea in [I], this reduces to the hyperplane case by using the Kelvin transform. We first need to check how the Maxwell equations behave under the Kelvin transform. For this it is convenient to consider E and H as complex valued -forms in Ω that is, a vector field E, E, E 3 ) is identified with the -form E dx + E dx + E 3 dx 3 ). We equip Ω with the Euclidean metric e, and introduce the Hodge star operator e which maps a -form X dx + X dx + X 3 dx 3 to the -form X dx dx 3 + X dx 3 dx + X 3 dx dx. If d is the exterior derivative, the Maxwell equations in form notation are { de = iωµ e H in Ω. dh = iωγ e E 36

37 Assume that is not in Ω. The Kelvin transform is given by xx) = x x, F x) = x x) = x x. If Ω = F Ω), then F is a conformal transformation from Ω, e) onto Ω, e): F e = x 4 e. Let Ẽ = F E, H = F H, µ = F µ, and γ = F γ. The following is the transformation law for the Maxwell equations under the Kelvin transform. Lemma 6.. ne has { de = iωµ e H dh = iωγ e E in Ω { dẽ = iω µ x e H d H = iω γ x e Ẽ in Ω. Proof: We use the following identities valid for k-forms η in a 3-manifold: df η = F dη, F e η) = F ef η, ce η = c 3/ k e η. Here c is any positive smooth function. If de = iωµ e H, then dẽ = df E = F de = F iωµ e H) = iω µ F e H = iω µ ex 4 e H = iω µ x e H. The other direction and the other equation are handled analogously. Next we check how the impedance map transforms. Note that in the form notation, the boundary condition in the Maxwell equations corresponds to fixing the tangential -form e ν H) on Ω, where ν = ν dx + ν dx + ν 3 dx 3 is the outer unit normal written as a -form. Lemma 6.. Let Λ be the impedance map in Ω with coefficients µ, γ), and let Λ be the impedance map in Ω with coefficients µ x, γ x ). If T is a tangential field on Ω, then Λ T ) = F ΛF ) T ). Proof: We take ρ to be a boundary defining function for Ω, that is, ρ is a smooth function R 3 R and Ω = {ρ > }, Ω = {ρ = }. Then ρ = F ρ is a boundary defining function for Ω. The unit normal is the -form ν = d ρ F dρ = = F dρ = F x dρ ) = x F ν, d ρ e F dρ F F ) e dρ F ) e dρ e using that F ) e = x 4 e and ν = dρ/ dρ e. 37

38 Let Ẽ, H) be a solution to the Maxwell system in Ω with e ν H) e Ω = T. Then on Ω one has e ν Ẽ) = x e F ν E) = x F F ) eν E)) = F e ν E)). Similarly e ν H) = F e ν H)), and therefore Λ T ) = e ν Ẽ) = F Λ e ν H)) = F ΛF ) T ). We now assume that µ j, γ j ) are two sets of coefficients satisfying the assumptions of Theorem, and let Λ and Λ be the corresponding impedance maps. Let B be an open ball with Ω B, and suppose that Γ Ω is such that Λ T Γ = Λ T Γ for tangential fields T supported in Γ. Assume that Γ = Ω \ Γ satisfies Γ = Ω B and Γ B, that is, the inaccessible part Γ is part of a sphere. We wish to show that µ = µ and γ = in Ω. To this end, choose coordinates so that B = Bx, ) where x =,, ), and assume that the origin is not in Ω. Let Ω = F Ω) be the image of Ω under the Kelvin transform. From Lemma 6. we obtain that Λ T ) eγ = Λ T ) eγ for tangential fields supported in Γ = F Γ). Since the Kelvin transform maps Γ onto a subset of {x 3 = }, we are in a situation where the inaccessible part of the boundary is part of a hyperplane. Also the other conditions in Theorem are satisfied; in particular,.5) follows from its analog for B which reads { there exist C 4 extensions of γ j and µ j into R 3 which are preserved under the map x F R F x), where F is the Kelvin transform and R is the reflection x, x, x 3 ) x, x, x 3 ). Consequently, Theorem implies that µ x = µ x, γ x = x in Ω. It follows that µ = µ and γ = in Ω, thus proving Theorem. References [BU] [C] Bukhgeim, A. L., Uhlmann, G., Recovering a potential from partial Cauchy data. Comm. PDE 7 ), p Calderón, A. P., n an inverse boundary value problem. Seminar on Numerical Analysis and its Applications to Continuum Physics Río de Janeiro, 98), pp , Soc. Brasil. Mat., Río de Janeiro,

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