An introduction to the use of integral equation. for Harmonic Maxwell equation. J.C. NÉDÉLEC CMAP, École Polytechnique

Transcription

1 An introduction to the use of integral equation for Harmonic Maxwell equation J.C. NÉDÉLEC CMAP, École Polytechnique

2 1. Introduction Sommaire 2. The plane waves solutions 3. Fundamental Solution and Radiation Conditions 4. Elementary differential geometry 5. The single layer for Helmholtz equation 6. Integral Representations 7. Calderon projectors 8. The perfect conductor 9. EFIE, MFIE and CFIE equations

3 10. Approximation for EFIE 11. Single layer for the Laplace Problem 12. Homogeneous and pseudo-homogeneous kernels 13. Pseudo-homogeneous kernels 14. The Helmholtz decomposition 15. The dielectric case

4 1 Introduction Electromagnetic waves are defined by the electric field E and the magnetic field H at each point in R 3. An isotropic dielectric medium is characterized by the electric permittivity ε and the magnetic permeability µ. The speed of waves in this dielectric medium is 1/ εµ. We denote by ε 0, µ 0 respectively the permittivity and the permeability of the vacuum, and by c the speed of light in a vacuum, which is c = 1 ε0 µ 0. (1) The relative permittivity and permeability of the medium are defined by ε = ε r ε 0, ε r 1, µ = µ r µ 0, µ r 1. (2)

5 In the absence of electric and magnetic charges and currents, the electric and magnetic fields in a dielectric medium are governed by the Maxwell system of equations ε E t + curlh = 0, µ H + curle = 0. t We must complete these equations by the transmission conditions at interfaces that separate different dielectric media. The tangential components of the fields E and H are continuous across a surface of discontinuity of ε or µ. We denote by n the unit normal to and then these jump conditions take the form [E n] = 0, (4) [H n] = 0. (3)

6 Isotropic conducting media are characterized by the conductivity σ, which is a real positive number. Maxwell equations in such a medium are ε E t µ H t + curlh σe = 0, + curle = 0, (Ampère Maxwell law), (Faraday law). The interface conditions (4) on dielectric media are unchanged. We introduce the electric induction D and the magnetic induction B: From the Maxwell equations, it follows that (5) D = εe, (6) B = µh. (7) divb = 0. (8)

7 The harmonic solutions of Maxwell equations are complex-valued fields E and H such that the following fields (ω is the pulsation) ( E(t, x) = R E(x)e iωt), H(t, x) = R ( H(x)e iωt), satisfy the Maxwell system. They satisfy the system of Harmonic Maxwell equations iωεe + curlh = 0, (10) iωµh + curle = 0. It follows from Maxwell equations that (9) divd = 0. (11)

8 In the case of a conducting media, we can redefine the new electric permittivity as ε = ε + iσ/ω and the equation (10) is just similar. The perfectly conducting medium is a common model. The conductor is a bounded domain with boundary. We take the limit when the conductivity σ and thus ε tends to infinity in the conductor. Then for a fixed pulsation ω, the two fields E and H tend to zero in this medium. The following interior limits E n is zero, but it is not the case for H n. A surfacic electric current appears and also a density of electric charges. From the first part of the jump condition (4) the following exterior limit is also zero E n = 0, (12) while the exterior limits H n and E n are non-zero. We thus only have to find the solution of (10) and (12) in the infinite exterior dielectric medium. A classical exterior problem ( the PEC problem) is that of a perfectly conducting object lit by an incident plane wave.

9 2 The plane waves solutions They are specific solutions in R 3 of the Harmonic Maxwell equations (10) for an homogeneous media. As ε and µ are constant in R 3, we have Using the vectorial identity dive = divh = 0; (13) E = dive curl curle (14) yield that E and H are also solutions to the wave equations { E + k 2 E = 0, H + k 2 H = 0, k = ω εµ. (15) The plane waves are associated with a scalar wave equation.

10 Looking for E, H such that E(x) = e e i( k. x), H(x) = h e i( (16) k. x), with k = k leads to the equations { ωε e + k h = 0, ωµ h + k e = 0. (17) We notice that the three vectors e, h, k are mutually orthogonal, and the relations (17) now describe all the plane wave solutions. The vector k gives the direction of propagation of the wave. We see that the electric and the magnetic fields lie in the plane orthogonal to k, called the phase plane. These three vectors can be real or complex. When k is real, the modulus of these solutions is constant. When k is real and e is complex of the form e = α 1 + i α 2 with ( α 1 α 2 ) = 0, we speak of a circular plane wave.

11 3 Fundamental Solution and Radiation Conditions We denote by G the fundamental solution of the scalar Helmholtz equation which satisfies the outgoing radiation condition: G(x) = 1 4π eik x x. (18) The fundamental solution for a system of equations with n unknowns and n equations of the form au = 0, (19) where a is a differential operator with constant coefficients, is defined as a matrix n n, denoted by A, such that aa = δi, (20) where δ is the Dirac mass at the origin and I is the n n identity matrix.

12 The terms of A are in general distributions, and this will be the case here, in contrast with the fundamental solution of the Harmonic Helmholtz equation given by (18) which is an integrable function. Thus, the fundamental solution of the Maxwell system is a 6 6 matrix. Due to the symmetry of the operator, we only need to compute half of this matrix or equivalently two 3 3 matrices, solutions of the system { iωεe + curlh = δi, I is the3 3 identity matrix, (21) iωµh + curle = 0. We obtain the other half of the fundamental solution by exchanging the role the fields E and H, and the parameters ε and µ. The wave number k is k = ω εµ. (22)

13 Theorem 1 The first half of the fundamental solution of the Maxwell system, solution of (21), is given by E(x) = iωµg(x)i + i ωε D2 G(x), H(x) = curl(g(x)i). (23) The second half of the fundamental solution of the Maxwell system is given by E(x) = curl(g(x)i), H(x) = iωεg(x)i ωµ i (24) D2 G(x). The second derivative D 2 should be understood in the sense of distributions in D (R 3 ).

14 Proof We introduce the scalar potential V and the vector potential A { E = V + A, diva k 2 V = 0 ( Lorentz gauge condition). (25) In the case of the fundamental solution, given by equation (21), the corresponding quantities are respectively the scalar potential V, which is a vector in R 3, and the vector potential A which is a 3 3 matrix. Taking the divergence in the decomposition (25), it follows that dive = V + diva, (26) which is obtained from the divergence in the first part of equation (21), dive = 1 iωε div(δi) = i δ. (27) ωε

15 From the gauge condition (25), we can eliminate A and we obtain an equation for V : V + k 2 V = i δ, (28) ωε which outgoing fundamental solution is V (x) = i G(x). (29) ωε Taking the curl in the decomposition (25), it follows that Eliminating H in (21) yields curle = curla. (30) curlcurle k 2 E = iωµδi. (31)

16 From (25), we have diva k 2 (E A) = 0, (32) Using (32) in connection with (31) and (30) gives diva curlcurla + k 2 A = iωµδi, (33) or equivalently A + k 2 A = iωµδi. (34) The outgoing fundamental solution of (34) is A(x) = iωµ G(x)I. (35) Thus, we have obtained (23). The expression (24) follows by exchanging the quantities E and H, and ε and µ, without forgetting to change i into i.

17 It is important to notice that the coefficients of the fundamental matrix are not integrable functions, nor usual finite parts. In particular, the trace of D 2 G is G, which has a Dirac mass singularity at the origin. We have used the fundamental solution of the Harmonic Helmholtz equation given by (18). This function satisfies the following Sommerfeld condition or outgoing wave condition u r iku c at infinity, (r = x ). (36) r 2 We twice used this Sommerfeld condition to select a unique fundamental solution for the scalar potential and the vector potential. We then need to describe precisely the behaviour of the solutions at infinity, which is equivalent to defining the radiation conditions. They appear in the expression of the fundamental solution.

18 Theorem 2 The fundamental solution (23) and (24) (Theorem 1) satisfies the usual radiation conditions (r = x ) E r ike c r2, forlarger, (37) H r ikh c r2, forlarger, (38) and the following radiation conditions of Silver-Müller: (n = r /r) ε E µ H n c, r2 forlarge r, (39) ε E n + µ H c, r2 forlarge r. (40) Any of these four radiation conditions selects a unique fundamental solution. Moreover, each component of E and H behaves as 1/r at infinity The proof is based on the expansion at infinity of the function G and its derivatives.

19 A more concise expression of the Silver-Müller radiation conditions for the Maxwell system is (c denotes a generic constant ) E(x) r c, forlarge r,, forlarge r, H(x) r c εe µ H r r c r 2, forlarger. (41) Remark 1 : The plane wave solution do not satisfy the Silver-Müller radiation conditions (as for example they do not tend to zero at infinity). This condition is characteristic of the scattered part of the field created by the bounded obstacle.

20 In the case of an object lit by an incident plane wave, the total field takes the form E = E inc + E s H = H inc + H s (42) where the scattered field E s, H s satifies the Silver-Müller conditions. Now a formulation of the PEC problem for a general excitation associated to a boundary data g, is: Find E and H such that iωεe + curlh = 0, x Ωe, iωµh + curle = 0, x Ωe, E n = g, x, g a given vectortangent to, E and H satisfy (41). and in the case above (43) g = E inc n. (44)

21 Figure 1: Notations

22 The interior harmonic Maxwell problem is iωεe + curlh = 0, x Ωi, iωµe + curle = 0, x Ωi, E n = g, x, g a given vectortangent to. (45) Remark 2 : When g = E inc n, its solution is just minus the incoming plane wave E = E inc, H = H inc. It is thus known in that case.

23 4 Elementary differential geometry A surface in R 3 is usually defined by a system of charts which maps a part of this surface on a bounded squared domain of R 2. We introduce some differential operators on such a surface. These operators are needed in the study of the Maxwell system and its integral representation. For every point y in R 3, we denote by δ(y) the distance of y to the surface δ(y) = inf x y x. (46) A collection of points whose distance to the surface is less than ε is called a tubular neighbourhood of the surface. For ε small enough and when the surface is regular and oriented, any point y in such a neighbourhood ε, has a unique projection P(y) on the surface which satisfies y P(y) = δ(y). (47)

24 It is easy to check that for a regular surface which admits a tangent plane at the point P(y), the line y P(y) is directed along the normal to the surface at this point. Notice that the unit normal n is the gradient of the function δ(y) oriented towards the interior or the exterior depending on the position of the point y. It holds that { n(p(y)) = δ(y), wheny Ωe (48) n(p(y)) = δ(y), wheny Ω i. Any point y in the tubular neighbourhood ε is located using P(y) and δ(y), or more precisely we write: { y = P(y) + s n(p(y)), ε s ε, s = δ(y), wheny Ω e, s = δ(y), wheny Ω i. (49)

25 To any function u defined on the surface, we associate the lifting ũ defined on the tubular neighbourhood ε by ũ(y) = u(p(y)). (50) We continue to denote by n(y) the vector field s which is defined on ε and has the value n on the surface. We introduce the family s of parallel surfaces s = {y; y = x + sn(x); x }. (51) The vector field n is the field of normals to the surfaces s. We introduce the first family of differential operators which acts on functions defined on the surfaces or s. The tangential gradient u is defined as u = ũ. (52)

26 The tangential rotational of a function is defined as The field of normals is a gradient, which implies We use the vectorial calculus formula curl u = curl(ũn). (53) curln = 0. (54) curl(u v ) = u v + ucurl v (55) which yields curl u = u n. (56) The curvature operator R s is R s = n. (57)

27 When we are on the surface (i.e., when s = 0), we omit the index s. Theorem 3 The curvature operator R s is a symmetric operator acting in the tangent plane. We denote by 1/R 1 (s) and 1/R 2 (s) its two eigenvalues, called principal curvatures. We define its mean curvatureh s to be half its trace H s = 1 2 divn = 1 2 We define its Gauss curvature G s to be its determinant ( 1 R 1 (s) + 1 ). (58) R 2 (s) G s = 1 R 1 R 2 = det ( R s ). (59) The area element on the surface s at the point y = x + sn(x) is related to the area element on the surface at the point x through the relation dγ s (y) = ( 1 + 2sH(x) + s 2 G(x) ) dγ(x). (60)

28 The normal derivative of the curvature tensor R s is Moreover, it holds that s H s = 1 2 trace( ) R 2 1 s = 2 s R s = R 2 s. (61) ( 1 (R 1 (s)) (R 2 (s)) 2 ), (62) s G s = 2H s G s, (63) [R s (v n) 2H s (v n)] n = R s v, (forevery vectorv). (64)

29 We introduce now a family of operators acting on tangent vector fields v. We need to define a lifting ṽ of this field in the neigbourhood ε. It is natural to ask it to be tangent to the surfaces s. We transport the tangential vector fields in the neighbourhood ε using the following lifting ṽ(y) = v(x) sr s (y)v(x). (65) We introduce the following first order differential operators: The surfacic divergence of a tangent vector field v is The surfacic rotational of a vector field v is div v = divṽ. (66) curl v = (curlṽ n(x)). (67)

30 We introduce the following second order differential operators: The scalar Laplacian or Laplace-Beltrami operator acting on a function u is u = div u = curl curl u. (68) The vectorial Laplacian or Hodge operator acting on a tangent vector field v is v = div v curl curl v. (69) Theorem 4 Let u C 1 () be a function and v ( C 1 () ) 2 a tangent vector field defined on the surface. The following Stokes identities hold: ( u v) dγ + u div vdγ = 0, (70) ( ) curl u v dγ u curl vdγ = 0, (71)

31 div curl u = 0, (72) curl u = 0, (73) div (v n) = curl v. (74) Moreover, if w is a function C 2 (), we have: w udγ = ( w u)dγ = ( curl w ) curl u dγ. (75) If w is a tangent vector field ( C 2 () ) 2, it holds that ( w v) dγ = div w div vdγ + curl w curl vdγ. (76)

32 The surfacic operators in the domain ε and the usual three dimensional operators in R 3 are linked. Theorem 5 Let u be a function defined on ε and v a field of vectors in R 3 defined on ε. The following decompositions hold: u = s u + u n, (77) s v = (v n)n + v s, v s = n (v n), (78) divv = div s v s + 2H s (v n) + (v n), (79) s curlv = (curl s v s ) n + curl s (v n) + R s (v n) 2H s (v n) s (v n), (80) u u = s u + 2H s s + 2 u, (81) s2

33 v = s v s + 2H s s v s + 2 s 2 v s [ + s (v n) + 2H s s + 2 (v n) + 2 (v n) s ] n [ 2 (v s H s ) 2div s (R s v) + 2(v n) s H s ] n + (2H s I 2R s )R s v + 2R s (v n) + 2(v n) H s. (82)

34 5 The single layer for Helmholtz equation The Helmholtz equation appears in acoustics problems. It described the equation satisfy by the pressure in the case of harmonic solution. The two classical boundary conditions for the interior Helmholtz Equation are the Dirichlet and the Neumann problems { u + k 2 u = 0, x Ω i, (83) u = u d. { u + k 2 u = 0, x Ω i, u (84) n = u n. In the case of exterior problems, we seek complex-valued solutions. We need to specify the behavior at infinity and in particular to impose the radiation condition.

35 The two model equations are the Dirichlet and the Neumann problems { u + k 2 u = 0, x Ω e, u = u d. (85) { u + k 2 u = 0, x Ω e, u (86) n = u n. In both cases, the conditions at infinity are (c is a fixed constant) { u c r, r large, u r c (87), and the Sommerfeld condition u r iku c r. (88) 2

36 Notice that this last condition implies that the solutions are complex-valued, although k 2 is real. The associated operator is not symmetric. This condition is satisfies by the scattered part of the wave and it means that this wave is going to infinity (outgoing). The fundamental solution of the Helmholtz equation which satisfies the outgoing radiation condition is G(x) = 1 4π e ikr r, r2 = x x x 2 3. (89) This function is singular at the origin. It is possible to check that it is integrable and that its trace on a regular surface crossing the origin is still integrable on this surface. We can used this fundamental solution to obtain integral representation of the solutions of the problems (83), (84), (85) and (86).

37 Theorem 6 (Representation theorem) Let u be a function such that (with possibly k = 0) u + k 2 u = 0 in Ω i, (90) u + k 2 u = 0 in Ω e, (91) u satisfies the outgoing radiation condition (88) and its traces u int and u ext, ( u/ n) int and ( u/ n) ext belong to C 0 (). We define [u] = u int u ext, (92) [ ] u = u u. n n int n ext (93) For y /, the following representation formula holds: [ ] u u(y) = G(x y) n (x) dγ(x) (G(x y))[u(x)]dγ(x), n x (94)

38 and for y u int (y) + u ext (y) 2 = G(x y) [ ] u n (x) dγ(x) n x (G(x y))[u(x)]dγ(x). (95) Proof We use the Green formula with on one side u, and on the other side the function G( y). For any domain Ω, whose boundary is Ω, and for any y / Ω, we have ( 0 = u(x) + k 2 u(x) ) G(x y)dx Ω ( G(x y) + k 2 G(x y) ) udx = Ω Ω [ u (x)g(x y) n ] G(x y)u(x) dγ(x). n x (96)

39 Let y be in the domain Ω i and B ε be a ball whose center is y, radius ε and boundary S ε. We also introduce a ball B R whose center is y, radius R, chosen large enough to contain the domain Ω i, and whose boundary is S R. We use the above Green formula in the two domains: Ω i B ε, Ω e B R. Adding the two corresponding contributions yields [ ] u G(x y) n (x) dγ(x) G(x y)[u(x)]dγ(x) n x G(x y) u S ε n (x)dγ(x) + G(x y)u(x)dγ(x) S ε n x (97) + G(x y) u S R n (x)dγ(x) G(x y)u(x)dγ(x) S R n x = 0.

40 The function u is regular in the ball B ε. Thus, when ε tends to zero, the first integral on S ε is bounded by G(x y) u S ε n (x)dγ(x) εsup u x B ε n (x) (98) and tends to zero. The second integral has the value G(x y)u(x)dγ(x) = u(y) G(x y)dγ(x) S ε n x S ε n x (99) + (G(x y))(u(x) u(y))dγ(x), S ε n x ( ik G(x y) = n x 4πr 1 ) e ikr, (100) 4πr 2 (G(x y))(u(x) u(y))dγ(x) S ε n x (101) (kε + 1)sup x Bε u(x) u(y).

41 This last term tends to zero. The first term in the right hand side of (99) admits the limit u(y). Let us now examine the terms on the sphere S R. They take the form ( ) G(x y) ikg(x y) u(x)dγ(x) S R n x ( ) u + G(x y) S R n iku(x) dγ(x). (102) The first of these terms is bounded by ( ) G(x y) ikg(x y) u(x)dγ(x) S R n x sup x S R u(x). (103) For R large enough, the condition (87) implies that this term tends to zero.

42 The condition (88) shows that the second term is bounded by ( ) u G(x y) (x) iku(x) dγ(x) S R n C R, (104) and thus tends to zero when R tends to infinity. When the point y is on the surface, the ball B ε is split into Ω i B ε and Ω e B ε, which yields two pieces in the corresponding integrals. When ε tends to zero, these two pieces are asymptotically separated by the tangent plane. Thus the associated integrals on S ε give rise to the term (1/2) (u int (y) + u ext (y)). We must notice that in this case, the integrands associated with the surface admit a singularity at the point y. The one associated with the kernel G(x y) is singular as 1/ x y and thus belongs to the space L 1 () for x on.

43 In the expression of the other integrand appears the kernel n x G(x y) = eik x y 4π x y 2 ( ik 1 ) ((x y) n x ), (105) x y which singularity is equivalent to 1/ x y, since its numerator ((x y) n x ) vanishes as x y 2 in a neighbourhood of y, and thus this kernel belongs to the space L 1 () for y. This is no longer true when y is not on.

44 The following integral expression is called a single layer potential: u(y) = G(x y)q(x)dγ(x); (106) The following integral expression is called a double layer potential: G u(y) = (x y)ϕ(x)dγ(x). (107) n x These two types of potentials satisfy the Helmholtz equation in Ω i and Ω e and the outgoing radiation condition at infinity. The singularity contained in the kernel implies that these potentials admits some jumps in their values when the current point y cross te surface. In the case of the single layer these discontinuity are described in the following theorem:

45 Theorem 7 The single layer potential (106) is continuous with respect to the variable y in R 3 and especially when crossing the surface where its value is u(y) = G(x y)q(x)dγ(x), y, when q(x) C 0 (). (108) Its tangential derivatives are also continuous. Its normal derivative has a discontinuity when crossing. Its limit value on both side of are u (y) = q(y) n int 2 + G(x y)q(x)dγ(x), (109) n y u (y) = q(y) n ext 2 + G(x y)q(x)dγ(x). (110) n y

46 Proof The property (108) results from the integrability of the singularity of E in a tubular neighbourhood of. This singularity is uniformly integrable with respect to the variable y and thus the classical Lebesgue theorem implies continuity with respect to y and (108). The theorem 6 can be used when the function u is represented as a single layer potential. It follows that [ ] u [u] = 0 and q =. (111) n The gradient of a single layer potential is u(y) = y G(x y)q(x)dγ(x). (112)

47 We use the specific moving system of coordinates introduced above. A point y + in a tubular neighbourhood of the surface is determined by its projection y 0 and the distance ρ to. We introduce the symmetric point y : { y+ = y 0 + ρn y0, (113) y = y 0 ρn y0. We compute the following sum:

48 (( u (y + ) + u (y )) n y0 ) (( = y+ G(x y + ) + y G(x y ) ) ) n y0 q(x)dγ(x) = [ e ik x y + 4π x y eik x y 4π x y 2 [ e ik x y + + 4π x y + 2 eik x y 4π x y 2 ( ik ) 1 x y + )] ((y x y 0 x) n y0 ) q(x)dγ(x) ( ) 1 ik x y + )] ρq(x)dγ(x). ( ik 1 ( ik 1 x y (114)

49 The classical Lebesgue theorem applies to the first part of the right-hand side of (114) since ((x y 0 ) n y0 ) is equivalent to x y 0 2. Its limit is 2 G(x y)q(x)dγ(x), whenρtends to0. n y It is more difficult to show that the second part tends to zero with ρ. Let us study the more delicate term which is ( ) 1 x y x y 3 ρq(x)dγ(x) ( x y 2 x y + 2)( ) x y x y 2 + x y + x y = x y + 3 x y 3 ( x y + + x y ) ρq(x)dγ(x). (115)

50 We can then write x y 2 x y + 2 = 4ρ ((x y 0 ) n y0 ). (116) This integral is bounded by ( ) 1 x y x y 3 ρq(x)dγ(x) C ρ 2 x y 0 2 ( x y ρ 2 ) 5/2 dγ(x). (117) We split this last expression into two parts: when x y 0 ρ x y 0 ρ ρ 2 x y 0 2 ( x y ρ 2 ) 5/2 dγ Cρ,

51 when x y 0 ρ ρ 2 x y 0 2 ) 5/2 dγ x y 0 ρ ( x y ρ 2 ρ ε x y 0 4 ε ( ) 5/2 dγ ρ ε x y ρ 2 dγ x y 0 1+ε, from which it follows that the limit is zero. The other terms can be treated similarly. Equalities (109) and (110) follow from (111) and the limit (114).

52 6 Integral Representations We exhibit the integral representations of the Maxwell equation and give their principal properties. There are obtain when the values of the constants ε and µ are the same for the interior domain Ω i and the exterior domain Ω e. Theorem 8 (Representation Theorem) Let Ω i be a bounded regular interior domain, which boundary is the regular surface and denote by Ω e the associated exterior domain. The exterior unit normal to is denoted by n. Let E and H be regular solutions of the Maxwell equations { curle iωµh = 0, inω i, (118) curlh + iωεe = 0,

53 curle iωµh = 0, curlh + iωεe = 0, E andhsatisfy the radiation conditions of Silver-Müller. inω e, (119) Let s denote j and m the electric and magnetic tangent currents to the surface j = H i n H e n (120) m = E i n E e n; where H i and H e are respectively the interior and exterior limits of the field H, while E i and E e are the interior and exterior limits of the field E.

54 Then, the fields E and H admit the integral representation E(y) = iωµ G(x y)j(x)dγ(x) H(y) = iωε + i ωε i ωµ G(x y)div j(x)dγ(x) +curl G(x y)m(x)dγ(x) +curl G(x y)m(x)dγ(x), y /. G(x y)div m(x)dγ(x) G(x y)j(x)dγ(x), y /. (121) (122)

55 Three integral operators appear that are : Sj(y) = G(x y)j(x)dγ(x), (123) Tj(y) = G(x y)div j(x)dγ(x) (124) Rj(y) = curl G(x y)j(x)dγ(x). (125) We will need the value on the surface G of the fields represent by the integral representation (121) and (122). These values are associated to the respective jumps of the above operators S, T and R. The operators S and T have continuous values across G. So is their tangential derivatives. But T exhibit a discontinuous normal component and S a discontinuous normal derivative across G. The operator R have discontinuous tangential values across G. But it exhibit a continuous normal component across G.

56 Proof It is based on the known results for the integral representation of the scalar Helmholtz equation given in Theorems 6 and 7, and the use of some properties of distributions. The fields E and H, solutions of (118) and (119), satisfy in the sense of distributions in R 3 : { curle iωµh = mδ, (126) curlh + iωεe = jδ. Thus, E and H are the sum of two contributions, one associated with j, the other associated with m. Let us compute the contribution associated with j, which solves the equation { curle iωµh = 0, (127) curlh + iωεe = jδ.

57 It takes the form of the sum of a vector potential A and a scalar potential V : E = A + V. (128) Exactly as in the computation of the fundamental solution, these potentials must be linked by a gauge condition, which we choose to be the Lorentz gauge diva k 2 V = 0. (129) The potentials A and V are continuous across the surface. We have iωεdive = div (jδ ) = (div j) δ, (130) which, in combination with the gauge condition yields V + k 2 V = i ωε div jδ. (131)

58 Thus, the potential V is given by a single layer potential for the scalar Helmholtz equation, with density (idiv j)/(ωε) : V (y) = i G(x y)div j(x)dγ(x). (132) ωε From equation (127), it follows that A + k 2 A = diva curlcurla + k 2 A = k 2 V k 2 V + k 2 E curlcurle = iωµjδ. Thus, the vector potential A is given by a single layer potential for the scalar Helmholtz equation, with density iωµj: A(y) = iωµ G(x y)j(x)dγ(x). (133)

59 This give the expression of this part of the field E and the similar part of H E(y) = iωµ G(x y) j(x) dγ(x) + i (134) ωε G(x y)div j(x)dγ(x), H(y) = curl G(x y)j(x)dγ(x). (135) By an argument of symmetry, the part associated with m is E(y) = curl G(x y) m(x)dγ(x), (136) H(y) = iωε G(x y) m(x)dγ(x) G(x y)div m(x)dγ(x). i ωµ (137)

60 Theorem 9 The interior projector is the operator C int which maps the couple (m, j) to the interior limit values of the integral representation. It is given by E i n = m 2 n y (Rm + i Tj + iωµsj), (138) ωε H i n = j 2 + n y ( i Tm + iωεsm Rj). (139) ωµ The exterior projector is the operator C ext which maps the couple (m, j) to the to the exterior limit values of the integral representation. It is given by E e n = m 2 + n y (Rm + i Tj + iωµsj), (140) ωε H e n = j 2 n y ( i Tm + iωεsm Rj). (141) ωµ The operators C int and C ext are projectors.

61 The expression of the interior and exterior values are deduced from the properties of single layer potentials given in Theorems 6 and 7. (continuity across the surface of the potential and the tangential part of its gradient, discontinuity of the normal derivative). Besides, using (80), we have curlu n = n u T T (u n) + Ru T. (142) Thus, for a single layer potential u, the jump of curlu n is the same as the jump of u T / n, where u T is the tangential component. It follows that ) lim y ± (curl y G(x y)j(x)dγ(x) n y = ± j(y) (143) 2 + (curl y (G(x y)j(x)) n y ) dγ(x).

62 We have { curly (G(x y)j(x)) n y = ( y G(x y) j(x)) n y = n G (x y)j(x) y G(x y) (j(x) (n y n x )). y In order to exhibit the limit of the gradient term, we use (144) and thus n y = G(x y)ρ(x)dγ(x) n y y G(x y)ρ(x)dγ(x) (145)

63 n y = = G(x y)ρ(x)dγ(x) ((n y n x ) y G(x y)) ρ(x)dγ(x) + curl x G(x y)ρ(x)dγ(x) ((n y n x ) y G(x y)) ρ(x)dγ(x) G(x y) curl ρ(x)dγ(x), from which we obtain formulas (138), (139), (140) and (141). (146)

64 We introduce the impedance z = 7 Calderon projectors µ ǫ and the frequency k = ω εµ. We introduce also the operator Π which is the projector on the tangent plane at a point of. We define the operator A which is called the EFIE operator. A = k Π S + 1/k Π T (147) We associated to any operator L the twist operator defined as L t = n y L (148) So we consider the following twist operators A t, S t, T t and R t.

65 The Calderon projectors introduced in theorem 9 are now (in matricial form) C int = I 2 + R t i z A t I iza t 2 + R t (149) C ext = I 2 R t iza t i z A t I 2 R t We thus have the identity (decomposition of the identity) Then, both projectors satisfy (150) C int + C ext = I. (151) C C = C (152)

66 which is also equivalent to the two identities ( I 2 C) (I 2 C) = I 4. (153) C int C ext = C ext C int = 0 (154) The two identities (152) (or (153)) are also equivalent to (151) and the following one (C ext C int ) (C ext C int ) = I. (155) These identities are equivalent to the two following ones: A t A t + R t R t = I 4, A t R t + R t A t = 0. (156)

67 Two other expressions are the following (A t + R t ) (A t + R t ) = I 4 (A t R t ) (A t R t ) = I 4. (157) Let B be the rotation or twist operator (which operates in the tangent plane to the surface ) Bj = n j. (158) Express in term of this operator, the Calderon identities are also B (A + R) B (A + R) = I 4. B (A R) B (A R) = I 4. (159)

68 8 The perfect conductor We examine in this section the specific properties and the variational formulation of the integral equation associated with the perfect conductor problem. We have seen that the scattering of a plane wave, by a perfect conducting object immersed in vacuum, leads to the exterior Maxwell problem curle iωµ 0 H = 0, inω e, curlh + iωε 0 E = 0, inω e, E n = E inc n, on, ε 0 E µ 0 H n c r 2. (160)

69 We associate with it, the interior Maxwell problem curle iωµ 0 H = 0, inω i, curlh + iωε 0 E = 0, inω i, E n = E inc n, on, which always admits the solution { E = E inc, H = H inc. (161) (162) This solution is not unique when the interior problem is not invertible, i.e., when k 2 is an eigenvalue of the interior Maxwell problem.

70 We apply the representation Theorem 8 to the solutions of the problems (160) and (161). We obtain E(y) = iωµ 0 G(x y)j(x)dγ(x) + i (163) G(x y)div j(x)dγ(x), ωε 0 H(y) = curl G(x y)j(x)dγ(x), (164) j = H inc n H n. (165) Remark 3 : The tangent field j is determined by the boundary condition on the surface, which expression is given by (140).

71 We have obtained the integral equation ( E inc n ) [ ] (y) = iωµ 0 n y G(x y)j(x) dγ(x) + i ] n y [ y G(x y)div j(x) dγ(x). ωε 0 (166) In this first expression of the integral equation, only integrable kernels appear. But, derivatives of the integral operator appear, which expression will lead to a non integrable kernel if we derivate under the integral symbol. Instead we introduce a variational formulation for this equation, where appears only first derivatives of the unknown j, and yet no finite part or non-integrable kernels. First, using the projector Π, we can rewrite equation (166) as

72 Π E inc (y) = iωµ 0 Π [ ] G(x y)j(x) dγ(x) + i ] (167) G(x y)div j(x) dγ(x). ωε 0 [ We multiply equation (167), written on, by a test vector j t. Using the Stokes formula (75) on the surface, we obtain iωµ 0 G(x y) ( j(x) j t (y) ) dγ(x)dγ(y) ωε i 0 = G(x y)div j(x)div j t (y)dγ(x)dγ(y) ( E inc j t) dγ, forany j t tangent to. This variational formulation is called the Rumsey principle. (168)

73 9 EFIE, MFIE and CFIE equations There are two integral techniques currently use to solve the equations (43) and (45). The first one is the one introduced above in equation (167). It is called the Electric Field Integral Equation or EFIE. It has also the following form using the notation (147) with g = Π E inc. Aj = i g. (169) z The second one (called the MFIE equation) uses the representation of H given by (160) H(y) = curl G(x y)j(x)dγ(x), y /. (170) In the case where the interior conductor domain is not flat, the interior value of H n (cf remark 2) and then using (139), we can write an

74 equation for this interior value: 1 j(y) n curl G(x y)j(x)dγ(x) = H inc n, y. (171) 2 Thus, the MFIE equation is ( I 2 + R t)j = H inc n (172) The CFIE equation is any convex combination of EFIE and MFIE ((1 α)a + α( I 2 + R t))j = (1 α) g z αh g n. (173)

75 10 Approximation for EFIE The variational formulation for EFIE (called the Rumsey s principle) leads to a natural approximation. It is (here we choose g = E inc ) k G(x y) ( j(x) j t (y) ) dγ(x)dγ(y) 1 G(x y)div k j(x)div j t (y)dγ(x)dγ(y) = i ( E inc j t) dγ; forany j t tangent to. z (174) The operator div is the tangential divergence defined on the surface. We solve this equation in the Hilbert space H 1/2 div () = { } v TH 1/2 (); div v H 1/2 ()

76 A finite element approximation, which is well adapted to this space, consists of the well-known 2D Raviart-Thomas mixed finite element (which leads to stable and convergent approximations). Let (T h ) be a family of triangulations of and let us denote S 1 (T h ) the associated finite element space. Then, there is one unknown for each edge of (T h ) which is the electric flux crossing this edge. This approximation correspond exactly to the so-called RWG basis for the EFIE. It is k G(x y) ( j h (x) jh(y) t ) dγ(x)dγ(y) h h 1 G(x y)div k h j h (x)div h jh(y)dγ(x)dγ(y) t (175) h h = i ( ) E inc jh t dγ; forany j t z h S 1 (T h ). h

77 11 Single layer for the Laplace Problem We give and prove in this section some specific properties of the single layer potential and the first kind integral equation associated with the Laplace equation. The associated representation theorem was introduced in theorem 6 in the specific case k = 0. We introduce a variational formulation. We study and prove a coercivity result for these variational formulation. They will use in the Helmholtz (and also for Maxwell) case, since the kernel associated with the Laplace operator is the principal part of the kernel of the associated integral equations. We will give also a more precise meaning of the notion of principal part.

78 Theorem 10 The integral equation associated with the single layer potential u(y) = 1 1 q(x)dγ(x) (176) 4π x y for the Dirichlet Laplace problem is 1 4π 1 x y q(x)dγ(x) = u d(y), y. (177) A variational formulation for this integral equation is 1 q(x)q t (y) dγ(x)dγ(y) = u d (y)q t (y)dγ(y), q t H 1/2 (). (178) 4π x y The corresponding operator, denoted by S, is an isomorphism of H 1/2 () onto H 1/2 () which satisfies the coercivity property (Sq)qdγ α q 2 H 1/2 (), α > 0, q H 1/2 (). (179)

79 Proof For q C 0 (), the potential u satisfies the system of equations u = 0, inω i and Ω e, [u] = 0, [ ] u = q. n Let W 1 (Ω e ) denote the Hilbert spaces: { } W 1 u (Ω e ) = u; (1 + r 2 ) 1/2 L2 (Ω e ), u L 2 (Ω e ). (180) When the domain Ω e is the whole space R 3, we have Theorem 11 v W 1 (R 3 ) c v L 2 (R 3 ), v W 1 (R 3 ) (181)

80 A variational formulation for the problem (180) is ( u v)dx = qvdγ, v W 1 ( R 3). (182) R 3 Theorem 11 shows that the bilinear form in the left-hand side of (182), is coercive on the space W 1 ( R 3). Thus, the Lax-Milgram theorem proves the existence of a unique solution to this problem when q is in the space H 1/2 (), as this space is the dual of the trace space H 1/2 (). Notice that the integral in (177) is a Lebesgue integral when q L (), but ceases to be so for a less regular q. It is not even a duality when q H 1/2 (). From the trace theorem, we know that the trace on the surface of the solution of the problem (180) belongs to the space H 1/2 (). Thus, the operator S is continuous from H 1/2 () into H 1/2 (). The inverse operator associated with u d H 1/2 (), the value of q.

81 Both interior and exterior Dirichlet problems admit a unique solution in H 1 (Ω i ) and W 1 (Ω e ). The interior normal derivative is then defined as u vdγ = ( u v)dx, v H 1 (Ω i ), (183) n int Ω i while the exterior normal derivative is defined as u vdγ = ( u v) dx, v W 1 (Ω i ). (184) n ext Ω e The trace theorem proves that both of them are in H 1/2 () and we have q = u u (185) n int n ext and satisfies, using the Representation Theorem 6 (if q is regular) Sq = u d. (186) Thus, the operator S is a bijective mapping from H 1/2 () onto H 1/2 ().

82 Moreover, subtracting equalities (183) and (184), we obtain (Sq)qdγ = qudγ = u 2 dx. (187) R 3 The continuity of the trace implies that u 2 dγ β u 2 H 1/2 (), R 3 β > 0, (188) while the continuity of S 1 is equivalent to u 2 H 1/2 () γ q 2 H 1/2 (), γ > 0. (189) Combining (188) and (189) yields (179). Remark 4 : Another meaning of the property (179) is to assert that q(x)q(y) dγ(x)dγ(y) (190) x y is a scalar product on the space H 1/2 ().

83 12 Homogeneous and pseudo-homogeneous kernels Let be a bounded regular surface contained in R 3. Pseudo homogeneous operators are integral operators of the form g(y) = K(y, x y)ϕ(x)dγ(x) ; y. (191) We establish continuity properties of these operators from H m () to similar Sobolev spaces and any positive integer m. The associated kernels K are regular with respect to the variable y and quasi-homogeneous with respect to the variable x y. We establish similar continuity properties for the family of adjoint operators with respect to the scalar product in L 2 () which are of the form g(y) = K(x, y x)ϕ(x)dγ(x), y. (192)

84 We examine homogeneous kernels K which are the restriction to the surface of kernels defined in R 3. Definition A homogeneous kernel K(y, z) defined in R 3 is of class m, for an integer m such that m 0, when α β sup y R 3sup z =1 y α K(y, z) zβ C α,β, α and β. (193) β K(y, z) is homogeneous of degree 2 zβ (194) with respect to the variablez, for β = m. For any plane H whose equation is (h, z) = 0, and any m-uple of vectors z 1,...,z m in H, S 1 D m z K(y, z ) (z 1,...,z m ) dz = 0, (195) where S 1 is the intersection of the sphere S 2 with the hyperplane H.

85 We give some examples of these operators that we will encounter. Example 1: Let m be a positive integer. We consider the kernel K(x) = x 2m 3. (196) This operator is of class 1 2m. Condition (194) is satisfied, since the partial derivatives of order 2m 1 are odd. Example 2 : Let α be a multi-index and l an integer. A more general example, which includes the previous one, is K(x) = (x) α x l, (x) α = x α 1 1 xα 2 2 xα 3 3, α = α 1 + α 2 + α 3. (197) It is of class m = ( α + l + 2), for a positive m. Condition (195) is satisfied when m is odd and α even or when m is even and α odd, since in both cases, the derivatives of order m are odd.

86 The following theorem gives the continuity properties for these examples: Theorem 12 The operator associated with the kernel K by either (191) or (192) is continuous from H r () into H m+r (), for any real positive r, when the kernel K is of class m and the surface regular enough.

87 13 Pseudo-homogeneous kernels For an integer m such that m 0, a kernel K(y, z) is pseudo-homogeneous of class m when it admits for any positive integer s an expansion of the form K(y, z) = K m (y, z) + l 1 j=1 K m+j (y, z) + K m+l (y, z), (198) where K m+j is of class (m + j) for j = 0, l 1, and l is chosen such that K m+l is s times differentiable. Theorem 13 Let K be a pseudo-homogeneous kernel chosen of class m. The associated operators given by (191) or by (192), are continuous from H r () into H r+m () for any integer r. This theorem is a direct consequence of Theorem 12 using the expression (198) for s big enough.

88 We describe now some examples of this situation: Example 3 Let n x be the unit normal to the surface at the point x. The kernel K(x, z) = (n x z) z 3 (199) looks homogeneous of degree 2, but is rather homogeneous of degree 1. The vector z which appears in the integral operators (191) and (192) is either x y or y z. When x and y are close, this vector is asymptotically tangent. Let P be the projection operator on the surface, which is defined in a tubular neighbourhood of this surface. Then, for a z of the form y x with y and x on the surface, the kernel K takes the form K (x, z) = (n x (P(x + z) P(x))) z 3. (200)

89 A Taylor expansion of the numerator is P(x + z) = P(x) + DP(x) z D2 P(x)(z, z) +. (201) Using the moving frame on the surface, (x = P(x) + sn(p(x))), the differential of DP(x) is (I + sr) 1 which acts in the tangent plane to the surface at the point P(x), and thus, (n x (DP(x) z)) = 0. (202) This shows that the operator associated with the kernel K is pseudo-homogeneous of class 1, the first non-zero term of the Taylor expansion being homogeneous of degree 1.

90 Example 4 The single layer potential kernel associated with the Helmholtz equation is K(z) = eik z z. (203) A Taylor expansion of the exponential in a neighbourhood of zero is e ik z = 1 + ik z k 2 z 2 +, (204) from which it is clear that this kernel is pseudo-homogeneous of class 1. Example 5 The double layer potential kernel associated with the Helmholtz equation is ( ) K(x, z) = eik x y. (205) n x x y It is pseudo-homogeneous of class 1 (combining the two previous cases).

91 14 The Helmholtz decomposition We introduce a saddle-point formulation equivalent to the variational formulation (168). It will be more suitable to prove existence and uniqueness, using the following abstract theorem. Theorem 14 (Fredholm alternative) Let V and W be two Hilbert spaces. Let a(u, v) be a bilinear form continuous on V V which satisfies R [a(u, u)] α u 2 V c u 2 H, α > 0, u V, (206) where H is a Hilbert space containing V. Let b(q, v) be a bilinear form continuous on W V which satisfies: sup b(q, u) β q W c q L, β > 0, q W, (207) u V =1 where L is a Hilbert space containing W. Consider the following

92 variational problem, with g 1 V and g 2 W : { a(u, v) + b(p, v) = (g1, v), v V, b(q, u) = (g 2, q), q W. (208) Denote by V 0 the kernel of the bilinear form b in V, i.e., V 0 = {u V, b(q, u) = 0, q W }. Suppose that the injection from V 0 into H is compact and that the injection from W into L is compact. Suppose that there exists an element u g2 V such that b(q, u g2 ) = (g 2, q), q V. Then, the variational problem (208) satisfies the Fredholm alternative, i.e., - either it admits a unique solution in V W, - or it admits a finite dimension kernel, and a solution defined up to any element in this kernel, when the right-hand side (g 1, g 2 ) vanishes on any element in this kernel.

93 Proof Let us first prove that this kernel has finite dimension. From (206) and (207), any element in this kernel satisfies or else u 2 V 1 α R [a(u, u)] + c u 2 H c u 2 H + c 1 β p L u V, u V + p W c u H + c p L. (209) Moreover, u V 0. From the compact injection hypothesis, it follows that this linear space has a compact unit ball and thus has finite dimension. From the property of g 2, it is equivalent to consider the case where g 2 is zero. We then introduce, using the Galerkin technique, an approximated solution for our problem, denoted (u ε, p ε ), in the quotient space (V W)/N. We continue to denote this new space by V W. Notice that u ε belongs to V 0. Estimates of this solution are u ε V + p ε W c u ε H + c p ε L + c g 1 V + c g 2 W. (210)

94 We proceed by contradiction to show the convergence of this sequence: either the sequence u ε H + p ε L is bounded, and then from (210), we deduce that (u ε, p ε ) is bounded in V W. We then have a weak convergence in V W. or the sequence u ε H + p ε L is not bounded, and then ũ ε = u ε /( u ε H + p ε L ) is bounded in V, while p ε = p ε /( u ε H + p ε L ) is bounded in W, and their weak limits (ũ, p) are such that { a(ũ, v) + b( p, v) = 0, v V, (211) b(q, ũ) = 0, q W. Thus (ũ, p) are in the kernel of this problem. As this kernel is reduced to zero in the quotient space, the quantity ũ ε H + p ε L, whose value is 1, cannot tend to zero. But, from the compact injections of V 0 into H and of W into L, there exists a subsequence which converges strongly. This is contradictory.

95 The Dirichlet problem for the exterior Helmholtz problem (85) can be solved using the single layer potential (106). This is the following equation u d (y) = G(x y)q(x)dγ(x); (212) Using the previous theorem, we obtain that this equation has a unique solution in the space H 1/2 () if u d is in the space H 1/2 (), except when k 2 is an eigenvalue of the interior Dirichlet problem for the Laplace equation. The proof is an application of the above results: First, the operator (212) is pseudo-homogeneous of order 1. Thus, it is continuous from L 2 () into H 1 (). Its dual operator is also continuous from L 2 () into H 1 () and thus by duality it is continuous H 1 () into L 2 (). Then by interpolation, this operator is continuous H 1/2 () into H 1/2 ().

96 Secondly we can expand the kernel using: which gives that e ik z = 1 + ik z k 2 z 2 +, (213) G(x) = 1/ x + ik k 2 x +, (214) The first term is the kernel of the single layer potential for the Laplace equation. The second one is a constant kernel which result does not depends on the variable y. The rest of the expansion is a pseudo-homogeneous kernel of order 3 and maps H 1/2 () into H 5/2 (). The integral kernel (212) is the sum of a kernel coercive in the space H 1/2 () and a kernel which maps H 1/2 () into H 5/2 (). But the injection of H 5/2 () into H 1/2 () is compact and thus we can used the theorem (14). The inverse operator is not defined only when k 2 is an eigenvalue of the interior Dirichlet problem for the Laplace equation.

97 We now return to the EFIE equation. We look for the vector j in the form of its Helmholtz decomposition. If we suppose that the surface is simply connected, it has the expression j = curl p + q. (215) It follows from the variational equation (168) that G(x y) q(x) q t (y)dγ(x)dγ(y) k 2 G(x y) ( q(x) q t (y) ) dγ(x)dγ(y) ( ) k 2 G(x y) curl p(x) q t (y) dγ(x)dγ(y) = iωε div ET inc q t dγ, forany q t inh 3/2 (). (216)

98 k 2 G(x y) k 2 G(x y) = iωε curl ET inc p t dγ, forany p t inh 1/2 (). ( curl p(x) ) curl p t (y) dγ(x)dγ(y) ( q(x) ) curl p t (y) dγ(x)dγ(y) (217) These two equations are coupled. The principal part of the first equation is G(x y) q(x) q t (y)dγ(x)dγ(y), which, up to a compact operator, is coercive in H 3/2 (), due to the properties of the single layer potential for the Laplace operator.

99 The principal part of the second equation is ( G(x y) curl p(x) ) curl p t (y) dγ(x)dγ(y), which, up to a compact operator, is coercive in H 1/2 (), due to the properties of the single layer potential for the Laplace operator. We prove now that the coupling terms are also compact operators. We change the integrals using integration by parts on the surface ( G(x y) q(x) ) curl p(y) dγ(x)dγ(y) = p(y)curl G(x y) q(x)dγ(x)dγ(y). (218)

100 From identity (145), we obtain ( G(x y) q(x) ) curl p(y) dγ(x)dγ(y) = (((n y n x ) y G(x y)) q(x)) p(y)dγ(x)dγ(y), (219) which, using the results on pseudo-homogeneous kernels, shows that this bilinear form is bounded by G(x y) ) dγ(x)dγ(y) (220) c q H 1/2 () p H 1/2 (). ( q(x) curl p(y) It follows that the system of equations (216) (217) is of Fredholm type. It is also equivalent to equation (168), and thus it satisfies the existence and uniqueness properties, when k 2 is not an eigenvalue of the interior problem (161) for the Maxwell equation.

101 We have prove that the system of equation (216) (217) has a unique solution except when k 2 is an eigenvalue of the interior problem (161) for the Maxwell equation. If we plug this solution in the EFIE equation (168), the solution j is such that { div j = q (221) curl j = p. As we know that p is in H 1/2 () and q is in H 3/2 (), it results that the vector j belongs to the space { } H 1/2 div () = v TH 1/2 (); div v H 1/2 ()

102 15 The dielectric case Let Ω d be the dielectric, and Ω e the exterior complement, with boundary. We introduce the following real coefficients (ω is the angular frequency) ǫ d : the electric permittivity of Ω d, ǫ e : the electric permittivity Ω e µ d : the magnetic permeability Ω d µ e : the magnetic permeability of Ω e k d = ω(µ d ǫ d ) 1 2 : the wave number in Ω d k e = ω(µ e ǫ e ) 1 2 : the wave number in Ω e z d = ( µ d ǫ d ) 1 2 : the impedance in Ωd z e = ( µ e ǫ e ) 1 2 : the impedance in Ωe.

103 The unknows are the electric fields E d, E e and the magnetic fields H d, H e. The harmonic Maxwell system is curle d iωµ d H d = 0 in Ω d, (P ed ) curlh d + iωǫ d E d = 0 in Ω d, curle e iωµ e H e = 0 in Ω e, curlh e + iωǫ e E e = 0 in Ω e, E d n (E e + E inc ) n = 0 on, H d n (H e + H inc ) n = 0 on, ǫ e E e µ e H e n C x 2 when x where E inc, H inc is a known plane wave.

104 We introduce the electric j and the magnetic m currents: j d = H d n, j e = H e n m d = E d n, m e = E e n. (222) Now, two Green s functions appear denote by G d and G e. Separately, the fields in Ω d and in Ω e admits an integral representation given by (118) and (119) with their respective currents. The unknowns are the currents j e and m e. The jump conditions on the boundary in the equation P ed give the currents j d and m d. The Rumsey variational formulation consists in equating the two integral expressions of the currents on the surface. We denote by A d = k d S d + 1/k d T d, A e = k e S e + 1/k e T e and R d, R e, the respective EFIE and MFIE operators associated to each domain.

105 Then, in matrix notation, the system to solve takes the form à = iz da kd + iz e A ke R kd + R ke R kd + R ke i( 1 A kd + 1 A ke ) z d z e (223)

106 References [1] J.C. Nédélec Acoustics and Electromagnetics Equations, Integral Representations for Harmonics Problems ; Springer-Verlag, 2001.