an introduction to the use of integral equation for ... · an introduction to the use of integral...

An introduction to the use of integral equation

for Harmonic Maxwell equation

J.C. NÉDÉLECCMAP, École Polytechnique

Integral for Maxwell

Sommaire1. Introduction

2. The plane waves solutions

3. Fundamental Solution and RadiationConditions

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

CMAP, École PolytechniqueHongKong 2007


10. Approximation for EFIE

11. Single layer for the Laplace Problem

12. Homogeneous and pseudo-homogeneouskernels

13. Pseudo-homogeneous kernels

14. The Helmholtz decomposition

15. The dielectric case



1 – Introduction

Electromagnetic waves are defined by theelectric fieldE and themagneticfield H at each point inR3. An isotropic dielectric mediumis characterizedby the electric permittivityε andthe magnetic permeabilityµ.

The speed of wavesin this dielectric medium is1/√

εµ. We denote byε0,µ0 respectively the permittivity and the permeability of the vacuum, and byc the speed of light in a vacuum, which is

c =1√ε0µ0

. (1)

The relative permittivity and permeability of the medium are defined by

ε = εrε0, εr ≥ 1,

µ = µrµ0, µr ≥ 1.

(2)



In the absence of electric and magnetic charges and currents, the electric

and magnetic fields in a dielectric medium are governed by theMaxwell

system of equations

−ε∂E∂t

+ curlH = 0,

µ∂H∂t

+ curlE = 0.

(3)

We must complete these equations bythe transmission conditionsat

interfaces that separate different dielectric media.The tangential

components of the fieldsE andH are continuous acrossa surfaceΓ of

discontinuity ofε or µ. We denote byn the unit normal toΓ and then these

jump conditionstake the form

[E ∧ n]Γ = 0,

[H ∧ n]Γ = 0.

(4)



Isotropic conducting media are characterized bythe conductivityσ, whichis a real positive number. Maxwell equations in such a mediumare

−ε∂E∂t

+ curlH − σE = 0, (Ampère−Maxwell law),

µ∂H∂t

+ curlE = 0, (Faraday law).

(5)

The interface conditions(4) on dielectric media are unchanged.

We introduce theelectric inductionD and themagnetic inductionB:

D = εE, (6)

B = µH. (7)

From the Maxwell equations, it follows that

divB = 0. (8)



Theharmonic solutionsof Maxwell equations are complex-valued fieldsE

andH 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

),

(9)

satisfy the Maxwell system. They satisfy the system ofHarmonic Maxwell

equations

iωεE + curlH = 0,

−iωµH + curlE = 0.

(10)

It follows from Maxwell equations that

divD = 0. (11)



In the case of a conducting media, we can redefinethe new electricpermittivity asε = ε + iσ/ω and the equation (10) is just similar.

Theperfectly conducting mediumis a common model. The conductor is abounded domain withboundaryΓ. We take the limit whenthe conductivityσ and thusε tends to infinityin the conductor. Then for a fixed pulsationω,the two fieldsE andH tend to zero in this medium. The following interior

limits E ∧ n|Γ is zero, but it is not the case forH ∧ n|Γ. A surfacic electriccurrent appears and also a density of electric charges.From the first part ofthe jump condition(4) the following exterior limit is also zero

E ∧ n|Γ = 0, (12)

while the exterior limitsH ∧ n andE · n are non-zero.We thus only have

to find the solution of (10) and (12) in the infinite exterior dielectricmedium. Aclassical exterior problem ( the PEC problem)is that of aperfectly conducting objectlit by an incident plane wave.



2 – The plane waves solutions

They are specific solutions inR3 of the Harmonic Maxwell equations (10)

for an homogeneous media. Asε andµ are constant inR3, we have

divE = divH = 0; (13)

Using the vectorial identity

∆E = ∇divE − curl curlE (14)

yield thatE andH are also solutions to the wave equations

∆E + k2E = 0,

∆H + k2H = 0, k = ω√

εµ.(15)

The plane waves are associated with a scalar wave equation.



Looking forE, H such that

E(x) = ~e ei(~k.~x),

H(x) = ~h ei(~k.~x), with |~k| = k

(16)

leads to the equations

ωε~e + ~k ∧ ~h = 0,

−ωµ~h + ~k ∧ ~e = 0.(17)

We notice thatthe three vectors~e,~h,~k are mutually orthogonal, and therelations(17) now describe all the plane wave solutions.The vector~k givesthe direction of propagation of the wave. We see thatthe electric and themagnetic fields lie in the plane orthogonal to~k, called thephase plane.These three vectors can be real or complex.Whenk is real, the modulus ofthese solutions is constant.When~k is real and~e is complex of the form~e = ~α1 + i~α2 with (~α1 · ~α2) = 0, we speak of acircular plane wave.



3 – Fundamental Solution and Radiation Conditions

We denote byG 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 equationswith n unknowns and

n equations of the form

au = 0, (19)

wherea is a differential operator with constant coefficients, is defined as a

matrixn × n, denoted byA, such that

aA = δI, (20)

whereδ is the Dirac mass at the originandI is then × n identity matrix.



Theterms ofA 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 a6 × 6 matrix.

Due to the symmetry of the operator, we only need to compute half of this

matrix or equivalently two3 × 3 matrices, solutions of the system

iωεE + curlH = δI, I is the 3 × 3 identity matrix,

−iωµH + curlE = 0.(21)

We obtain the other half of the fundamental solution by exchanging the role

the fieldsE andH, and the parametersε andµ.

Thewave numberk is

k = ω√

εµ. (22)



Theorem 1 The first half of the fundamental solution of the Maxwell

system, solution of(21), is given by

E(x) = iωµG(x)I + iωεD2 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 − iωµD2 G(x).

(24)

The second derivativeD2 should be understood in the sense of

distributions inD′(R3).



Proof

We introduce thescalar potentialV and thevector potentialA

E = ∇V + A,

divA − k2V = 0 ( Lorentz gauge condition).(25)

In the case of the fundamental solution, given by equation(21), thecorresponding quantities are respectively the scalar potential V , which is avector inR

3, and the vector potentialA which is a3 × 3 matrix. Taking thedivergence 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

iωεdiv(δI) = − i

ωε∇δ. (27)



From the gauge condition(25), we can eliminateA and we obtain an

equation forV :

∆V + k2V = − i

ωε∇δ, (28)

which outgoing fundamental solution is

V (x) =i

ωε∇G(x). (29)

Taking the curl in the decomposition(25), it follows that

curlE = curlA. (30)

EliminatingH in (21) yields

curlcurlE − k2E = iωµδI. (31)



From(25), we have

∇divA − k2(E − A) = 0, (32)

Using(32) in connection with(31) and(30) gives

∇divA − curlcurlA + k2A = −iωµδI, (33)

or equivalently

∆A + k2A = −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 quantitiesE andH, andε andµ, without forgetting to changei into−i.



It is important to notice thatthe coefficients of the fundamental matrix arenot integrable functions, nor usual finite parts. In particular,the trace ofD2 G is ∆G, which has a Dirac mass singularity at the origin.

We have used thefundamental solution of the Harmonic Helmholtzequationgiven by(18).

This function satisfies the followingSommerfeld conditionor outgoingwave condition

∣∣∣∣∂u

∂r− iku

∣∣∣∣ ≤c

r2at infinity, (r = |x|). (36)

We twice used thisSommerfeld conditionto select a unique fundamentalsolution for the scalar potential and the vector potential.

We then need to describe precisely the behaviour of the solutions at infinity,which is equivalent todefining the radiation conditions. They appear in theexpression of the fundamental solution.



Theorem 2 The fundamental solution(23) and(24) (Theorem1) satisfies

the usual radiation conditions (r = |x|)∣∣∣∣∂E

∂r− ikE

∣∣∣∣ ≤c

r2, for large r, (37)

∣∣∣∣∂H

∂r− ikH

∣∣∣∣ ≤c


and the followingradiation conditions of Silver-Müller : (n = −→r /r)

∣∣√ε E −√µ H ∧ n

∣∣ ≤ c


∣∣√ε E ∧ n +√

µ H∣∣ ≤ c

r2, for large r. (40)

Any of these four radiation conditions selects a unique fundamental

solution. Moreover, each component ofE andH behaves as1/r at infinity

The proof is based on the expansion at infinity of the functionG and itsderivatives.



A more concise expression of theSilver-Müller radiation conditionsfor the

Maxwell system is (c denotes a generic constant )

|E(x)| ≤ cr , for large r,

|H(x)| ≤ cr , for large r,

∣∣∣√

εE −√µ H ∧

−→rr

∣∣∣ ≤ cr2 , for large r.

(41)

Remark 1 : The plane wave solution do not satisfy the Silver-Müller

radiation conditions (as for example they do not tend to zeroat infinity).

This condition is characteristic of thescattered part of the fieldcreated by

the bounded obstacle.



In the case of an object lit by anincident plane wave, the total fieldtakes theform

E = Einc + Es H = Hinc + Hs (42)

wherethe scattered fieldEs, Hs satifies the Silver-Müller conditions.

Now a formulation of thePEC problemfor a general excitation associatedto a boundary datag, is: FindE andH such that

iωεE + curlH = 0, x ∈ Ωe,

−iωµH + curlE = 0, x ∈ Ωe,

E ∧ n|Γ = g, x ∈ Γ, g a given vector tangent to Γ,

E and H satisfy (41).

(43)

and in the case aboveg = −Einc ∧ n|Γ . (44)



Figure 1: Notations



The interior harmonic Maxwell problemis

iωεE + curlH = 0, x ∈ Ωi,

−iωµE + curlE = 0, x ∈ Ωi,

E ∧ n|Γ = g, x ∈ Γ, g a given vector tangent to Γ.

(45)

Remark 2 : Wheng = −Einc ∧ n|Γ, its solution is just minus the

incoming plane waveE = −Einc, H = −Hinc.

It is thus known in that case.



4 – Elementary differential geometry

A surfaceΓ in R3 is usually defined by a system of chartswhich maps apart of this surface on a bounded squared domain ofR2. We introduce somedifferential operators on such a surfaceΓ. These operators are needed in thestudy of the Maxwell system and its integral representation.

For every pointy in R3, we denote byδ(y) the distance ofy to the surfaceΓ

δ(y) = infx∈Γ |y − x| . (46)

A collection of points whose distance to the surface is less thanε is called atubular neighbourhood of the surfaceΓ. Forε small enough and when thesurface is regular and oriented, any pointy in such a neighbourhoodΓε, hasa unique projectionP(y) on the surface which satisfies

|y − P(y)| = δ(y). (47)



It is easy to check that for a regular surfaceΓ which admitsa tangent plane

at the pointP(y), the liney − P(y) is directed along the normal to the

surfaceΓ at this point. Notice thatthe unit normaln is the gradient of the

functionδ(y) oriented towards the interior or the exterior depending on the

position of the pointy. It holds that

n(P(y)) = ∇δ(y), when y ∈ Ωe

n(P(y)) = −∇δ(y), when y ∈ Ωi.(48)

Any pointy in the tubular neighbourhoodΓε is located usingP(y) and

δ(y), or more precisely we write:

y = P(y) + s n(P(y)), −ε ≤ s ≤ ε,

s = δ(y), when y ∈ Ωe, s = −δ(y), when y ∈ Ωi.(49)



To any functionu defined on the surfaceΓ, we associate thelifting u

defined on the tubular neighbourhoodΓε by

u(y) = u(P(y)). (50)

We continue todenote byn(y) the vector field∇s which is defined onΓε

and has the valuen on the surfaceΓ.

We introduce the familyΓs of parallel surfaces

Γs = y; y = x + sn(x); x ∈ Γ . (51)

The vector fieldn is the field of normals to the surfacesΓs.

We introduce the first family ofdifferential operatorswhich actsonfunctions defined on the surfacesΓ or Γs. Thetangential gradient∇Γu isdefined as

∇Γu = ∇u|Γ . (52)



Thetangential rotational of a functionis defined as

−−→curlΓu = curl(un)|Γ . (53)

The field of normals is a gradient, which implies

curln = 0. (54)

We use the vectorial calculus formula

curl(u−→v ) = ∇u ∧ −→v + ucurl−→v (55)

which yields−−→curlΓu = ∇Γu ∧ n. (56)

Thecurvature operatorRs is

Rs = ∇n. (57)



When we are on the surfaceΓ (i.e., whens = 0), we omit the indexs.

Theorem 3 The curvature operatorRs is a symmetric operator acting in

the tangent plane. We denote by1/R1(s) and1/R2(s) its two eigenvalues,

calledprincipal curvatures. We define itsmean curvatureHs to be half its

trace

Hs =1

2divn =

1

2

(1

R1(s)+

1

R2(s)

). (58)

We define itsGauss curvatureGs to be its determinant

Gs =1

R1R2= det

(Rs|Γ

). (59)

The area element on the surfaceΓs at the pointy = x + sn(x) is related

to the area element on the surfaceΓ at the pointx through the relation

dγs(y) =(1 + 2sH(x) + s2G(x)

)dγ(x). (60)



The normal derivative of the curvature tensorRs is

∂

∂sRs = −R2

s. (61)

Moreover, it holds that

∂

∂sHs = −1

2trace

(R2

s

)= −1

2

(1

(R1(s))2 +

1

(R2(s))2

), (62)

∂

∂sGs = −2HsGs, (63)

[Rs(v ∧ n) − 2Hs(v ∧ n)] ∧ n = Rsv, (for every vector v). (64)



We introduce now afamily of operators acting on tangent vector fieldsv.

We need to define a liftingv of this field in the neigbourhoodΓε. It is

natural to ask it to be tangent to the surfacesΓs.

We transport the tangential vector fieldsin the neighbourhoodΓε using the

following lifting

v(y) = v(x) − sRs(y)v(x). (65)

We introduce the followingfirst order differential operators:

Thesurfacic divergenceof a tangent vector fieldv is

divΓv = divv|Γ. (66)

Thesurfacic rotational of a vector fieldv is

curlΓv = (curlv · n(x))|Γ. (67)



We introduce the followingsecond order differential operators:

Thescalar Laplacianor Laplace-Beltrami operator acting on a functionu is

∆Γu = divΓ∇Γu = −curlΓ−−→curlΓu. (68)

Thevectorial Laplacian or Hodge operator acting on a tangent vectorfield v is

∆Γv = ∇ΓdivΓv −−−→curlΓcurlΓv. (69)

Theorem 4 Letu ∈ C1(Γ) be a function andv ∈(C1(Γ)

)2a tangent

vector field defined on the surfaceΓ. The followingStokes identitieshold:∫

Γ

(∇Γu · v) dγ +

∫

Γ

u divΓvdγ = 0, (70)

∫

Γ

(−−→curlΓu · v

)dγ −

∫

Γ

u curlΓvdγ = 0, (71)



divΓ−−→curlΓu = 0, (72)

curlΓ ∇Γu = 0, (73)

divΓ(v ∧ n) = curlΓv. (74)

Moreover, ifw is a functionC2(Γ), we have:

−∫

Γ

∆Γw udγ =

∫

Γ

(∇Γw · ∇Γu) dγ =

∫

Γ

(−−→curlΓw · −−→curlΓu

)dγ. (75)

If w is a tangent vector field(C2(Γ)

)2, it holds that

−∫

Γ

(∆Γw · v) dγ =

∫

Γ

divΓw divΓvdγ +

∫

Γ

curlΓw curlΓvdγ. (76)



The surfacic operators in the domainΓε and the usual three dimensional

operators inR3 are linked.

Theorem 5 Letu be a function defined onΓε andv a field of vectors inR3

defined onΓε. The following decompositions hold:

∇u = ∇Γsu +∂u

∂sn, (77)

v = (v · n)n + vΓs , vΓs = n ∧ (v ∧ n), (78)

divv = divΓsvΓs + 2Hs(v · n) +∂

∂s(v · n), (79)

curlv = (curlΓsvΓs) n +−−→curlΓs(v · n) + Rs(v ∧ n)

−2Hs(v ∧ n) − ∂∂s

(v ∧ n),

(80)

∆u = ∆Γsu + 2Hs∂u

∂s+

∂2

∂s2u, (81)



∆v = ∆ΓsvΓs + 2Hs∂

∂svΓs +

∂2

∂s2 vΓs

+

[∆Γs(v · n) + 2Hs

∂

∂s(v · n) +

∂2

∂s2 (v · n)

]n

+

[2 (v · ∇ΓsHs) − 2divΓs(Rsv) + 2(v · n)

∂

∂sHs

]n

+ (2HsI − 2Rs)Rsv + 2Rs∇Γ(v · n) + 2(v · n)∇ΓHs.

(82)



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 theDirichlet and the Neumann problems

∆u + k2u = 0, x ∈ Ωi,

u|Γ = ud.(83)

∆u + k2u = 0, x ∈ Ωi,

∂u∂n

∣∣∣Γ= un.

(84)

In the case ofexterior problems, we seek complex-valued solutions. We

need to specify the behavior at infinity and in particular to impose the

radiation condition.



The two model equations are the Dirichlet and the Neumann problems

∆u + k2u = 0, x ∈ Ωe,

u|Γ = ud.(85)

∆u + k2u = 0, x ∈ Ωe,

∂u∂n

∣∣∣Γ= un.

(86)

In both cases, the conditions at infinity are (c is a fixed constant) |u| ≤ c

r , r large,

|∇u| ≤ cr ,

(87)

and theSommerfeld condition∣∣∣∣∂u

∂r− iku

∣∣∣∣ ≤c

r2. (88)



Notice that this last condition implies that the solutions are complex-valued,

althoughk2 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).

Thefundamental solution of the Helmholtz equation which satisfies the

outgoing radiation conditionis

G(x) =1

4π

eikr

r, r2 = x2

1 + x22 + x2

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).



Theorem 6 (Representation theorem)Letu be a function such that (with

possiblyk = 0)

∆u + k2u = 0 in Ωi, (90)

∆u + k2u = 0 in Ωe, (91)

u satisfies the outgoing radiation condition (88) and its tracesuint and

uext, (∂u/∂n)|int and(∂u/∂n)|ext belong toC0(Γ). We define

[u] = uint − uext, (92)[∂u

∂n

]=

∂u

∂n

∣∣∣int

−∂u

∂n

∣∣∣ext

. (93)

For y /∈ Γ, the followingrepresentation formula holds:

u(y) =

∫

Γ

G(x − y)

[∂u

∂n(x)

]dγ(x) −

∫

Γ

∂

∂nx(G(x − y))[u(x)]dγ(x),

(94)



and fory ∈ Γ

uint(y) + uext(y)

2=

∫

Γ

G(x − y)

[∂u

∂n(x)

]dγ(x)

−∫

Γ

∂

∂nx(G(x − y))[u(x)]dγ(x).

(95)

Proof We use the Green formula with on one sideu, and on the other sidethe functionG(· − y). For any domainΩ, whose boundary is∂Ω, and foranyy /∈ Ω, we have

0 =

∫

Ω

(∆u(x) + k2u(x)

)G(x − y)dx

−∫

Ω

(∆G(x − y) + k2G(x − y)

)udx

=

∫

∂Ω

[∂u

∂n(x)G(x − y) − ∂

∂nxG(x − y)u(x)

]dγ(x).

(96)



Let y be in the domainΩi andBε be a ball whose center isy, radiusε and

boundarySε. We also introduce a ballBR whose center isy, radiusR,

chosen large enough to contain the domainΩi, and whose boundary isSR.

We use the above Green formula in the two domains:Ωi − Bε, Ωe ∩ BR.

Adding the two corresponding contributions yields

∫

Γ

G(x − y)

[∂u

∂n(x)

]dγ(x) −

∫

Γ

∂

∂nxG(x − y)[u(x)]dγ(x)

−∫

Sε

G(x − y)∂u

∂n(x)dγ(x) +

∫

Sε

∂

∂nxG(x − y)u(x)dγ(x)

+

∫

SR

G(x − y)∂u

∂n(x)dγ(x) −

∫

SR

∂

∂nxG(x − y)u(x)dγ(x)

= 0.

(97)



The functionu is regular in the ballBε. Thus, whenε tends to zero, the firstintegral onSε is bounded by

∣∣∣∣∫

Sε

G(x − y)∂u

∂n(x)dγ(x)

∣∣∣∣ ≤ εsupx∈Bε

∣∣∣∣∂u

∂n(x)

∣∣∣∣ (98)

and tends to zero. The second integral has the value

∫

Sε

∂

∂nxG(x − y)u(x)dγ(x) = u(y)

∫

Sε

∂

∂nxG(x − y)dγ(x)

+

∫

Sε

∂

∂nx(G(x − y))(u(x) − u(y))dγ(x),

(99)

∂

∂nxG(x − y) =

(ik

4πr− 1

4πr2

)eikr, (100)

∣∣∣∣∫

Sε

∂

∂nx(G(x − y))(u(x) − u(y))dγ(x)

∣∣∣∣

≤ (kε + 1)supx∈Bε|u(x) − u(y)| .

(101)



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 sphereSR. They take the form

−∫

SR

(∂

∂nxG(x − y) − ikG(x − y)

)u(x)dγ(x)

+

∫

SR

G(x − y)

(∂u

∂n− iku(x)

)dγ(x).

(102)

The first of these terms is bounded by∣∣∣∣∫

SR

(∂

∂nxG(x − y) − ikG(x − y)

)u(x)dγ(x)

∣∣∣∣ ≤ supx∈SR|u(x)| .

(103)

ForR large enough, the condition(87) implies that this term tends to zero.



The condition(88) shows that the second term is bounded by∣∣∣∣∫

SR

G(x − y)

(∂u

∂n(x) − iku(x)

)dγ(x)

∣∣∣∣ ≤C

R, (104)

and thus tends to zero whenR tends to infinity.

When the pointy is on the surfaceΓ, the ballBε is split intoΩi ∩ Bε and

Ωe ∩ Bε, which yields two pieces in the corresponding integrals. Whenε

tends to zero, these two pieces are asymptotically separated by the tangent

plane. Thus the associated integrals onSε give rise to the term

(1/2) (uint(y) + uext(y)). We must notice that in this case, the integrands

associated with the surfaceΓ admit a singularity at the pointy. The one

associated with the kernelG(x − y) is singular as1/|x − y| and thus

belongs to the spaceL1(Γ) for x onΓ.



In the expression of the other integrand appears the kernel

∂

∂nxG(x − y) =

eik|x − y|4π|x − y|2

(ik − 1

|x − y|

)((x − y) · nx), (105)

which singularity is equivalent to1/|x − y|, since its numerator

((x − y) · nx) vanishes as|x − y|2 in a neighbourhood ofy, and thus this

kernel belongs to the spaceL1(Γ) for y ∈ Γ. This is no longer true wheny

is not onΓ.



The following integral expression is calleda single layer potential:

u(y) =

∫

Γ

G(x − y)q(x)dγ(x); (106)

The following integral expression is calleda double layer potential:

u(y) =

∫

Γ

∂G

∂nx(x − y)ϕ(x)dγ(x). (107)

These two types of potentials satisfy the Helmholtz equation in Ωi andΩe

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 pointy cross te surfaceΓ. In

the case of the single layer these discontinuity are described in the

following theorem:



Theorem 7 The single layer potential(106) is continuous with respect to

the variabley in R3 and especially when crossing the surfaceΓ where its

value is

u(y) =

∫

Γ

G(x − y)q(x)dγ(x), y ∈ Γ, when q(x) ∈ C0(Γ). (108)

Its tangential derivatives are also continuous. Its normalderivative has a

discontinuity when crossingΓ. Its limit value on both side ofΓ are

∂u

∂n

∣∣∣int

(y) =q(y)

2+

∫

Γ

∂

∂nyG(x − y)q(x)dγ(x), (109)

∂u

∂n

∣∣∣ext

(y) = −q(y)

2+

∫

Γ

∂

∂nyG(x − y)q(x)dγ(x). (110)



Proof

The property(108) results from the integrability of the singularity ofE in a

tubular neighbourhood ofΓ. This singularity is uniformly integrable with

respect to the variabley and thus the classical Lebesgue theorem implies

continuity with respect toy and(108). The theorem6 can be used when the

functionu is represented as a single layer potential. It follows that

[u] = 0 and q =

[∂u

∂n

]. (111)

The gradient of a single layer potential is

∇u(y) =

∫

Γ

∇yG(x − y)q(x)dγ(x). (112)



We use the specific moving system of coordinates introduced above. A

pointy+ in a tubular neighbourhood of the surfaceΓ is determined by its

projectiony0 and the distanceρ to Γ. We introduce the “symmetric” point

y− : y+ = y0 + ρny0

,

y− = y0 − ρny0.

(113)

We compute the following sum:



((∇u (y+) + ∇u (y−)) · ny0)

=

∫

Γ

((∇y+

G (x − y+) + ∇y−G (x − y−)

)· ny0

)q(x)dγ(x)

=

∫

Γ

[eik |x − y+|

4π|x − y+|2(

ik − 1

|x − y+|

)

+ eik |x−y−|4π |x−y−|2

(ik− 1

|x−y−|

)]((y0−x) · ny0

) q(x)dγ(x)

+

∫

Γ

[eik |x − y+|

4π |x − y+|2(

ik − 1

|x − y+|

)

− eik |x − y−|4π |x − y−|2

(ik − 1

|x − y−|

)]ρq(x)dγ(x).

(114)



The classical Lebesgue theorem applies to the first part of the right-handside of(114) since((x − y0) · ny0

) is equivalent to|x − y0|2. Its limit is

2

∫

Γ

∂

∂nyG(x − y)q(x)dγ(x), when ρ tends to 0.

It is more difficult to show that the second part tends to zero with ρ. Let usstudy the more delicate term which is

∫

Γ

(1

|x − y+|3− 1

|x − y−|3

)ρq(x)dγ(x)

=

∫

Γ

(|x−y−|2−|x−y+|2

)(|x−y+|2+|x−y−|2+|x−y+||x−y−|

)

|x − y+|3 |x − y−|3 (|x − y+| + |x − y−|)

× ρq(x)dγ(x).(115)



We can then write

|x − y−|2 − |x − y+|2 = 4ρ ((x − y0) · ny0) . (116)

This integral is bounded by

∣∣∣∣∣

∫

Γ

(1

|x − y+|3− 1

|x − y−|3

)ρq(x)dγ(x)

∣∣∣∣∣

≤ C

∫

Γ

ρ2 |x − y0|2(|x − y0|2 + ρ2

)5/2dγ(x).

(117)

We split this last expression into two parts:

– when|x − y0| ≤ ρ

∫

|x−y0|≤ρ

ρ2 |x − y0|2(|x − y0|2 + ρ2

)5/2dγ ≤ Cρ,



– when|x − y0| ≥ ρ

∫

|x−y0|≥ρ

ρ2 |x − y0|2(|x − y0|2 + ρ2

)5/2dγ

≤ ρε

∫

Γ

|x − y0|4−ε

(|x − y0|2 + ρ2

)5/2dγ ≤ ρε

∫

Γ

dγ

|x − y0|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).



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 andµ are the same for the interior domainΩi and the exterior domainΩe.

Theorem 8 (Representation Theorem)LetΩi be a bounded regular

interior domain, which boundary is the regular surfaceΓ and denote byΩe

the associated exterior domain. The exterior unit normal toΓ is denoted by

n. LetE andH be regular solutions of the Maxwell equations

curlE − iωµH = 0,

curlH + iωεE = 0,inΩi, (118)




curlH + iωεE = 0,inΩe,

E andHsatisfy the radiation

conditions of Silver-Müller.

(119)

Let’s denotej andm the electric and magnetic tangent currentsto the

surfaceΓ

j = Hi ∧ n − He ∧ n

m = Ei ∧ n − Ee ∧ n;(120)

whereHi andHe are respectively the interior and exterior limits of the

fieldH, whileEi andEe are the interior and exterior limits of the fieldE.



Then, the fieldsE and H admit the integral representation

E(y) = iωµ

∫

Γ

G(x − y)j(x)dγ(x)

+ iωε∇

∫

Γ

G(x − y)divΓj(x)dγ(x)

+curl

∫

Γ

G(x − y)m(x)dγ(x), y /∈ Γ.

(121)

H(y) = −iωε

∫

Γ

G(x − y)m(x)dγ(x)

− iωµ∇

∫

Γ

G(x − y)divΓm(x)dγ(x)

+curl

∫

Γ

G(x − y)j(x)dγ(x), y /∈ Γ.

(122)



Threeintegral operatorsappear that are :

Sj(y) =

∫

Γ

G(x − y)j(x)dγ(x), (123)

Tj(y) = ∇∫

Γ

G(x − y)divΓj(x)dγ(x) (124)

Rj(y) = curl

∫

Γ

G(x − y)j(x)dγ(x). (125)

We will need the value on the surfaceG of the fields represent by theintegral representation(121) and(122). These values are associated to therespective jumps of the above operatorsS, T andR.

The operatorsS andT have continuous values acrossG. So is theirtangential derivatives. ButT exhibit a discontinuous normal component andS a discontinuous normal derivative acrossG.

The operatorR have discontinuous tangential values acrossG. But itexhibit a continuous normal component acrossG.



Proof

It is based on the known results for the integral representation of the scalarHelmholtz equation given in Theorems6 and7, and the use of someproperties of distributions.

The fieldsE andH, solutions of(118) and(119), satisfy in the sense ofdistributions inR3 :

curlE − iωµH = mδΓ,

curlH + iωεE = jδΓ.(126)

Thus,E andH are the sum of two contributions, one associated withj, theother associated withm. Let us compute the contribution associated withj,which solves the equation


curlH + iωεE = jδΓ.(127)



It takes the form ofthe sum of a vector potentialA and a scalar potentialV :

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 theLorentz

gauge

divA − k2V = 0. (129)

The potentialsA andV are continuous across the surfaceΓ.

We have

iωεdivE = div (jδΓ) = (divΓj) δΓ, (130)

which, in combination with the gauge condition yields

∆V + k2V = − i

ωεdivΓjδΓ. (131)



Thus, the potentialV is given by a single layer potential for the scalar

Helmholtz equation, with density(idivΓj)/(ωε) :

V (y) =i

ωε

∫

Γ

G(x − y)divΓj(x)dγ(x). (132)

From equation(127), it follows that

∆A + k2A = ∇divA − curlcurlA + k2A

= k2∇V − k2∇V + k2E − curlcurlE

= −iωµjδΓ.

(133)

Thus, the vector potentialA is given by a single layer potential for the

scalar Helmholtz equation , with densityiωµj:

A(y) = iωµ

∫

Γ

G(x − y)j(x)dγ(x).



This give the expression of this part of the fieldE and the similar part ofH

E(y) = iωµ

∫

Γ

G(x − y) j(x) dγ(x)

+i

ωε∇∫

Γ

G(x − y)divΓj(x)dγ(x),

(134)

H(y) = curl

∫

Γ

G(x − y)j(x)dγ(x). (135)

By an argument of symmetry, the part associated withm is

E(y) = curl

∫

Γ

G(x − y) m(x)dγ(x), (136)

H(y) = −iωε

∫

Γ

G(x − y) m(x)dγ(x)

− i

ωµ∇∫

Γ

G(x − y)divΓm(x)dγ(x).

(137)



Theorem 9 Theinterior projectoris the operatorCint which maps the

couple(m, j) to theinterior limit values of the integral representation. It is

given by

Ei ∧ n =m

2− ny ∧ (Rm +

i

ωεT j + iωµSj), (138)

Hi ∧ n =j

2+ ny ∧ (

i

ωµTm + iωεSm − Rj). (139)

Theexterior projectoris the operatorCext which maps the couple(m, j) to

the to theexterior limit values of the integral representation. It is given by

−Ee ∧ n =m

2+ ny ∧ (Rm +

i

ωεT j + iωµSj), (140)

−He ∧ n =j

2− ny ∧ (

i

ωµTm + iωεSm − Rj). (141)

The operatorsCint andCext are projectors.



The expression ofthe interior and exterior valuesare deduced from the

properties of single layer potentials given in Theorems6 and7. (continuity

across the surfaceΓ of the potential and the tangential part of its gradient,

discontinuity of the normal derivative). Besides, using(80), we have

curlu ∧ n =∂

∂nuT −∇T (u · n) + RuT . (142)

Thus, for a single layer potentialu, the jump ofcurlu∧ n is the same as the

jump of∂uT /∂n, whereuT is the tangential component. It follows that

limy→Γ±

(curly

∫

Γ


)∧ ny

= ±j(y)

2+

∫

Γ

(curly(G(x − y)j(x)) ∧ ny) dγ(x).

(143)



We have

curly(G(x − y)j(x)) ∧ ny = (∇yG(x − y) ∧ j(x)) ∧ ny

= ∂G∂ny

(x − y)j(x) −∇yG(x − y) (j(x) · (ny − nx)) .(144)

In order to exhibit the limit of the gradient term, we use

ny ∧∇∫

Γ

G(x − y)ρ(x)dγ(x)

=

∫

Γ

ny ∧∇yG(x − y)ρ(x)dγ(x)

(145)

and thus



ny ∧∇∫

Γ

G(x − y)ρ(x)dγ(x)

=

∫

Γ

((ny − nx) ∧∇yG(x − y)) ρ(x)dγ(x)

+

∫

Γ

−−→curlΓxG(x − y)ρ(x)dγ(x)

=

∫

Γ

((ny − nx) ∧∇yG(x − y)) ρ(x)dγ(x)

−∫

Γ

G(x − y)−−→curlΓρ(x)dγ(x),

(146)

from which we obtain formulas(138), (139), (140) and(141).



7 – Calderon projectors

We introduce theimpedancez =

√µ

ǫand thefrequencyk = ω

√εµ.

We introduce also theoperatorΠΓ which is the projector on the tangent

plane at a point ofΓ. We define theoperatorA which is called the EFIE

operator.

A = k ΠΓS + 1/k ΠΓT (147)

We associated to any operatorL thetwist operatordefined as

Lt = −ny ∧ L (148)

So we consider the following twist operatorsAt, St, Tt andRt.



The Calderon projectorsintroduced in theorem9 are now (in matricial

form)

Cint =

I2 + Rt − i

z At

izAtI2 + Rt

(149)

Cext =

I2 − Rt

iz At

−izAtI2 − Rt

(150)

We thus have the identity (decomposition of the identity)

Cint + Cext = I. (151)

Then, both projectors satisfy

C C = C (152)



which is also equivalent to the two identities

(I

2− C) (

I

2− C) =

I

4. (153)

Cint Cext = Cext Cint = 0 (154)

The two identities (152) (or (153)) are also equivalent to (151) and the

following one

(Cext − Cint) (Cext − Cint) = I. (155)

These identities are equivalent to the two following ones:

At At + Rt Rt =I

4,

At Rt + Rt At = 0.(156)



Two other expressions are the following

(At + Rt) (At + Rt) =I

4

(At − Rt) (At − Rt) =I

4.

(157)

Let B be the rotation or twist operator(which operates in the tangent planeto the surfaceΓ)

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)



8 – The perfect conductor

We examine in this section the specific properties andthe variational

formulation of the integral equationassociated withthe perfect conductor

problem. We have seen that the scattering of a plane wave, by a perfect

conducting object immersed in vacuum, leads to the exteriorMaxwell

problem

curlE − iωµ0H = 0, inΩe,

curlH + iωε0E = 0, inΩe,

E ∧ n|Γ = −Einc ∧ n, on Γ,∣∣√ε0E −√

µ0H ∧ n∣∣ ≤ c

r2 .

(160)



We associate with it, the interior Maxwell problem

curlE − iωµ0H = 0, inΩi,

curlH + iωε0E = 0, inΩi,

E ∧ n|Γ = −Einc ∧ n, on Γ,

(161)

which always admits the solution

E = −Einc,

H = −H inc.(162)

This solution is not unique when the interior problem is not invertible, i.e.,

whenk2 is an eigenvalue of the interior Maxwell problem.



We applythe representation Theorem8 to the solutions of the problems

(160) and(161). We obtain

E(y) = iωµ0

∫

Γ


+i

ωε0∇∫

Γ

G(x − y)divΓj(x)dγ(x),

(163)

H(y) = curl

∫

Γ

G(x − y)j(x)dγ(x), (164)

j = −H inc ∧ n − H ∧ n. (165)

Remark 3 :The tangent fieldj is determined by the boundary condition on

the surfaceΓ, which expression is given by(140).



We have obtained theintegral equation

(Einc ∧ n

)(y) = iωµ0ny ∧

[∫

Γ

G(x − y)j(x)]dγ(x)

+i

ωε0ny ∧

[∇y

∫

Γ

G(x − y)divΓj(x)]dγ(x).

(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 introducea variational formulationfor this equation, where

appears only first derivatives of the unknownj, and yet no finite part or

non-integrable kernels.

First, using the projectorΠΓ, we can rewrite equation(166) as



−ΠΓEinc(y) = iωµ0ΠΓ

[∫

Γ

G(x − y)j(x)]dγ(x)

+i

ωε0∇Γ

[∫

Γ

G(x − y)divΓj(x)]dγ(x).

(167)

We multiply equation(167), written onΓ, by a test vectorjt. Using the

Stokes formula(75) on the surfaceΓ, we obtain

iωµ0

∫

Γ

∫

Γ

G(x − y)(j(x) · jt(y)

)dγ(x)dγ(y)

− iωε0

∫

Γ

∫

Γ

G(x − y)divΓj(x)divΓjt(y)dγ(x)dγ(y)

= −∫

Γ

(Einc · jt

)dγ, for any jt tangent toΓ.

(168)

This variational formulation is calledthe Rumsey principle.



9 – EFIE, MFIE and CFIE equations

There aretwo integral techniques currently useto solve the equations (43)

and (45). The first one is the one introduced above in equation(167). It is

called theElectric Field Integral Equation or EFIE. It has also the following

form using the notation(147) with g = −ΠΓEinc.

Aj =i

zg. (169)

The second one (calledthe MFIE equation) usesthe representation ofH

given by(160)

H(y) = curl

∫

Γ

G(x − y)j(x)dγ(x), y /∈ Γ. (170)

In the case where theinterior conductor domain is not flat, theinterior

value ofH ∧ n (cf remark 2) and then using(139), we can write an



equation for this interior value:

1

2j(y) − n ∧ curl

∫

Γ

G(x − y)j(x)dγ(x) = −H inc ∧ n, y ∈ Γ. (171)

Thus,the MFIE equationis

(I

2+ Rt)j = −H inc ∧ n (172)

The CFIE equationis any convex combination of EFIE and MFIE

((1 − α)A + α(I

2+ Rt))j = (1 − α)

g

z− αHg ∧ n. (173)



10 – Approximation for EFIE

Thevariational formulation for EFIE (called the Rumsey’s principle) leadsto a natural approximation. It is (here we chooseg = −Einc)

k

∫

Γ

∫

Γ

G(x − y)(j(x) · jt(y)

)dγ(x)dγ(y)

−1k

∫

Γ

∫

Γ

G(x − y)divΓj(x)divΓjt(y)dγ(x)dγ(y)

= − i

z

∫

Γ

(Einc · jt

)dγ; for any jt tangent toΓ.

(174)The operatordivΓ is the tangential divergencedefined on the surfaceΓ.

We solve this equation in the Hilbert space

H−1/2div (Γ) =

v ∈ TH−1/2(Γ); divΓv ∈ H−1/2(Γ)



A finite element approximation, which is well adapted to this space,

consists of the well-known 2DRaviart-Thomas mixed finite element(which

leads to stable and convergent approximations).

Let (Th) be a family of triangulations ofΓ and let us denoteS1(Th) the

associated finite element space. Then, there is one unknown for each edge

of (Th) which is the electric flux crossing this edge.This approximation

correspond exactly to the so-called RWG basis for the EFIE. It is

k

∫

Γh

∫

Γh

G(x − y)(jh(x) · jt

h(y))dγ(x)dγ(y)

−1k

∫

Γh

∫

Γh

G(x − y)divΓhjh(x)divΓh

jth(y)dγ(x)dγ(y)

= − i

z

∫

Γh

(Einc · jt

h

)dγ; for any jt

h ∈ S1(Th).

(175)



11 – Single layer for the Laplace Problem

We give and prove in this section some specific properties ofthe single

layer potential and the first kind integral equation associated with the

Laplace equation. The associated representation theorem was introduced in

theorem6 in the specific casek = 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 theprincipal part of

the kernel of the associated integral equations. We will give also a more

precise meaning of the notion of principal part.



Theorem 10 The integral equation associated with thesingle layerpotential

u(y) =1

4π

∫

Γ

1

|x − y|q(x)dγ(x) (176)

for the Dirichlet Laplace problem is

1

4π

∫

Γ

1

|x − y|q(x)dγ(x) = ud(y), y ∈ Γ. (177)

A variational formulation for this integral equation is

1

4π

∫

Γ

∫

Γ

q(x)qt(y)

|x − y| dγ(x)dγ(y)=

∫

Γ

ud(y)qt(y)dγ(y), ∀qt∈H−1/2(Γ). (178)

The corresponding operator, denoted byS, is an isomorphism ofH−1/2(Γ) onto H1/2(Γ) which satisfies the coercivity property

∫

Γ

(Sq)qdγ ≥ α ‖q‖2H−1/2(Γ) , α > 0, ∀ q ∈ H−1/2(Γ). (179)



Proof For q ∈ C0(Γ), the potentialu satisfies the system of equations

∆u = 0, inΩi and Ωe,

[u]Γ = 0,[∂u∂n

]

Γ= q.

(180)

Let W 1(Ωe) denote the Hilbert spaces:

W 1(Ωe) =

u;

u

(1 + r2)1/2∈ L2(Ωe),∇u ∈ L2(Ωe)

.

When the domainΩe is the whole spaceR3, we have

Theorem 11

||v||W 1(R3) ≤ c||∇v||L2(R3), ∀ v ∈ W 1(R3) (181)



A variational formulation for the problem (180) is∫

R3

(∇u · ∇v)dx =

∫

Γ

qvdγ, ∀ v ∈ W 1(R

3). (182)

Theorem11 shows that the bilinear form in the left-hand side of(182), is

coercive on the spaceW 1(R

3). Thus,the Lax-Milgram theorem proves the

existence of a unique solution to this problem whenq is in the space

H−1/2(Γ), as this space is the dual of the trace spaceH1/2(Γ).

Notice that theintegral in(177) is a Lebesgue integralwhenq ∈ L∞(Γ),

but ceases to be so for a less regularq. It is not even a duality when

q ∈ H−1/2(Γ).

From the trace theorem, we know that the trace on the surfaceΓ of the

solution of the problem(180) belongs to the spaceH1/2(Γ). Thus, the

operatorS is continuous fromH−1/2(Γ) into H1/2(Γ). The inverse

operator associated withud ∈ H1/2(Γ), the value ofq.



Both interior and exterior Dirichlet problems admit a unique solution inH1 (Ωi) andW 1 (Ωe). The interior normal derivative is then defined as

∫

Γ

∂u

∂n

∣∣∣int

vdγ =

∫

Ωi

(∇u · ∇v) dx, ∀ v ∈ H1 (Ωi) , (183)

while the exterior normal derivative is defined as∫

Γ

∂u

∂n

∣∣∣ext

vdγ = −∫

Ωe

(∇u · ∇v) dx, ∀ v ∈ W 1 (Ωi) . (184)

The trace theorem proves that both of them are inH−1/2(Γ) and we have

q =∂u

∂n

∣∣∣int

−∂u

∂n

∣∣∣ext

(185)

and satisfies, using the Representation Theorem6 (if q is regular)

Sq = ud. (186)

Thus, the operatorS is a bijective mapping fromH−1/2(Γ) ontoH1/2(Γ).



Moreover, subtracting equalities(183) and(184), we obtain∫

Γ

(Sq)qdγ =

∫

Γ

qudγ =

∫

R3

|∇u|2dx. (187)

The continuity of the trace implies that∫

R3

|∇u|2dγ ≥ β ‖u‖2H1/2(Γ) , β > 0, (188)

while the continuity ofS−1 is equivalent to

‖u‖2H1/2(Γ) ≥ γ ‖q‖2

H−1/2(Γ) , γ > 0. (189)

Combining(188) and(189) yields(179).

Remark 4 :Another meaning of the property(179) is to assert that∫

Γ

∫

Γ

q(x)q(y)

|x − y| dγ(x)dγ(y) (190)

is a scalar product on the spaceH−1/2(Γ).



12 – Homogeneous and pseudo-homogeneous kernels

Let Γ be a bounded regular surface contained inR3. Pseudo homogeneous

operatorsare integral operators of the form

g(y) =

∫

Γ

K(y, x − y)ϕ(x)dγ(x) ; y ∈ Γ. (191)

We establishcontinuity properties of these operators fromHm(Γ) tosimilar Sobolev spacesand any positive integerm. The associatedkernelsK are regular with respect to the variabley and quasi-homogeneous withrespect to the variablex − y.

We establish similar continuity properties forthe family of adjoint operatorswith respect to the scalar product inL2(Γ) which are of the form

g(y) =

∫

Γ

K(x, y − x)ϕ(x)dγ(x), y ∈ Γ. (192)



We examinehomogeneous kernelsK which are the restriction to thesurfaceΓ of kernels defined inR3.

Definition A homogeneous kernelK(y, z) defined inR3 is of class−m,

for an integerm such thatm ≥ 0, when

supy∈R3sup|z|=1

∣∣∣∣∂|α|

∂yα

∂|β|

∂zβK(y, z)

∣∣∣∣ ≤ Cα,β, ∀α and ∀ β. (193)

∂|β|

∂zβK(y, z) is homogeneous of degree − 2

with respect to the variable z, for |β| = m.

(194)

For any planeH whose equation is(h, z) = 0, and anym-uple of vectorsz1, . . . , zm in H,

∫

S1

Dmz K(y, z′) (z1, . . . , zm) dz′ = 0, (195)

whereS1 is the intersection of the sphereS2 with the hyperplaneH.



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 class1 − 2m. Condition(194) is satisfied, since the

partial derivatives of order2m − 1 are odd.

Example 2 : Let α be a multi-index andℓ an integer. A more general

example, which includes the previous one, is

K(x) = (x)α|x|ℓ, (x)α = xα1

1 xα2

2 xα3

3 , |α| = α1 + α2 + α3. (197)

It is of classm = −(|α| + ℓ + 2), for a positivem. Condition(195) is

satisfied whenm is odd and|α| evenor whenm is even and|α| odd, since

in both cases, the derivatives of order−m are odd.



The following theorem gives the continuity properties for these examples:

Theorem 12 The operator associated with the kernelK by either(191) or

(192) is continuous fromHr(Γ) into Hm+r(Γ), for any real positiver,

when the kernelK is of class−m and the surfaceΓ regular enough.



13 – Pseudo-homogeneous kernels

For an integerm such thatm ≥ 0, akernel K(y, z) ispseudo-homogeneous of class−m when it admits for any positive integers an expansion of the form

K(y, z) = Km(y, z) +

ℓ−1∑

j=1

Km+j(y, z) + Km+ℓ(y, z), (198)

whereKm+j is of class−(m + j) for j = 0, ℓ − 1, andl is chosen suchthatKm+ℓ is s times differentiable.

Theorem 13 LetK be a pseudo-homogeneous kernel chosen of class−m.

The associated operators given by(191) or by (192), are continuous from

Hr(Γ) into Hr+m(Γ) for any integerr.

This theorem is a direct consequence of Theorem12 using the expression(198) for s big enough.



We describe now some examples of this situation:

Example 3

Let nx be the unit normal to the surfaceΓ at the pointx. The kernel

K(x, z) =(nx · z)

|z|3 (199)

looks homogeneous of degree−2, but is rather homogeneous of degree−1.

The vectorz which appears in the integral operators(191) and(192) is

eitherx − y or y − z. Whenx andy are close, this vector is asymptotically

tangent. LetP be the projection operator on the surfaceΓ, which is defined

in a tubular neighbourhood of this surface. Then, for az of the formy − x

with y andx on the surfaceΓ, the kernelK takes the form

K∗(x, z) =(nx · (P(x + z) − P(x)))

|z|3 . (200)



A Taylor expansion of the numerator is

P(x + z) = P(x) + DP(x) · z +1

2D2P(x)(z, z) + · · · . (201)

Using the moving frame on the surface, (x = P(x) + sn(P(x))), the

differential ofDP(x) is (I + sR)−1 which acts in the tangent plane to the

surface at the pointP(x), and thus,

(nx · (DP(x) · z)) = 0. (202)

This shows that the operator associated with the kernelK∗ is

pseudo-homogeneous of class−1, the first non-zero term of the Taylor

expansion being homogeneous of degree−1.



Example 4

The single layer potential kernel associated with the Helmholtz equationis

K(z) =eik|z||z| . (203)

A Taylor expansion of the exponential in a neighbourhood of zero is

eik|z| = 1 + ik|z| − k2|z|2 + · · · , (204)

from which it is clear thatthis kernel is pseudo-homogeneous of class−1.

Example 5

The double layer potential kernel associated with the Helmholtz equationis

K(x, z) =∂

∂nx

(eik|x − y||x − y|

). (205)

It is pseudo-homogeneous of class−1 (combining the two previous cases).



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) LetV andW be two Hilbert spaces.

Leta(u, v) be a bilinear form continuous onV × V which satisfies

ℜ [a(u, u)] ≥ α ‖u‖2V − c ‖u‖2

H , α > 0, ∀u ∈ V, (206)

whereH is a Hilbert space containingV . Letb(q, v) be a bilinear form

continuous onW × V which satisfies:

sup‖u‖V =1

|b(q, u)| ≥ β ‖q‖W − c ‖q‖L , β > 0, ∀ q ∈ W, (207)

whereL is a Hilbert space containingW . Consider the following



variational problem, withg1 ∈ V ∗ andg2 ∈ W ∗:

a(u, v) + b(p, v) = (g1, v), ∀ v ∈ V,

b(q, u) = (g2, q), ∀ q ∈ W.(208)

Denote byV0 the kernel of the bilinear form b in V , i.e.,

V0 = u ∈ V, b(q, u) = 0 , ∀ q ∈ W .

Suppose thatthe injection from V0 into H is compactand thattheinjection from W into L is compact. Suppose that there exists an element

ug2∈ V such that

b(q, ug2) = (g2, q), ∀ q ∈ V.

Then, the variational problem(208) satisfiesthe Fredholm alternative,i.e., - either it admits a unique solution inV × W , - or it admits a finite

dimension kernel, and a solution defined up to any element in this kernel,

when the right-hand side(g1, g2) vanishes on any element in this kernel.



Proof Let us first prove that this kernel has finite dimension. From(206)

and(207), any element in this kernel satisfies

‖u‖2V ≤ 1

αℜ [a(u, u)] + c ‖u‖2

H ≤ c ‖u‖2H + c

1

β‖p‖L ‖u‖V ,

or else

‖u‖V + ‖p‖W ≤ c ‖u‖H + c ‖p‖L . (209)

Moreover,u ∈ V0. From the compact injection hypothesis, it follows thatthis linear space has a compact unit ball and thus has finite dimension. Fromthe property ofg2, it is equivalent to consider the case whereg2 is zero.We then introduce, using the Galerkin technique, an approximated solutionfor our problem, denoted(uε, pε), in the quotient space(V × W )/N . Wecontinue to denote this new space byV × W . Notice thatuε belongs toV0.Estimates of this solution are

‖uε‖V + ‖pε‖W ≤ c ‖uε‖H + c ‖pε‖L + c ‖g1‖V ∗ + c ‖g2‖W∗ . (210)



We proceed by contradiction to show the convergence of this sequence:

– either the sequence‖uε‖H + ‖pε‖L is bounded, and then from(210), wededuce that(uε, pε) is bounded inV × W . We then have a weakconvergence inV × W .

– or the sequence‖uε‖H + ‖pε‖L is not bounded, andthenuε = uε/(‖uε‖H + ‖pε‖L) is bounded inV , whilepε = pε/(‖uε‖H + ‖pε‖L) is bounded inW , and their weak limits(u, p)

are such that a(u, v) + b(p, v) = 0, ∀ v ∈ V,

b(q, u) = 0, ∀ q ∈ W.(211)

Thus(u, p) are in the kernel of this problem.

As this kernel is reduced to zero in the quotient space, the quantity‖uε‖H + ‖pε‖L, whose value is1, cannot tend to zero. But, from thecompact injections ofV0 into H and ofW into L, there exists asubsequence which converges strongly. This is contradictory.



TheDirichlet problem for the exterior Helmholtz problem(85) can be

solved using thesingle layer potential(106). This is the following equation

ud(y) =

∫

Γ

G(x − y)q(x)dγ(x); (212)

Using the previous theorem, we obtain thatthis equation has a unique

solution in the spaceH−1/2(Γ) if ud is in the spaceH1/2(Γ), except when

k2 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 fromL2(Γ) into H1(Γ). Its dual operator is also continuous

from L2(Γ) into H1(Γ) and thus by duality it is continuousH−1(Γ) into

L2(Γ). Then by interpolation,this operator is continuousH−1/2(Γ) into

H1/2(Γ).



– Secondly we can expand the kernel using:

eik|z| = 1 + ik|z| − k2|z|2 + · · · , (213)

which gives that

G(x) = 1/|x| + ik − k2|x| + · · · , (214)

The first term is the kernel of the single layer potential for the Laplaceequation. The second one is a constant kernel which result does not dependson the variabley. The rest of the expansion is a pseudo-homogeneouskernel of order−3 and mapsH−1/2(Γ) into H5/2(Γ).

– The integral kernel(212) is the sum of a kernel coercive in the spaceH−1/2(Γ) and a kernel which mapsH−1/2(Γ) into H5/2(Γ). But theinjection ofH5/2(Γ) into H−1/2(Γ) is compact and thus we can used thetheorem(14). The inverse operator is not defined only whenk2 is aneigenvalue of the interior Dirichlet problem for the Laplace equation.



We now return tothe EFIE equation. We look for the vectorj in the form ofits Helmholtz decomposition. If we suppose that the surfaceΓ is simplyconnected, it has the expression

j =−−→curlΓp + ∇Γq. (215)

It follows from the variational equation(168) that

∫

Γ

∫

Γ

G(x − y)∆Γq(x)∆Γqt(y)dγ(x)dγ(y)

− k2

∫

Γ

∫

Γ

G(x − y)(∇Γq(x) · ∇Γqt(y)

)dγ(x)dγ(y)

− k2

∫

Γ

∫

Γ

G(x − y)(−−→curlΓp(x) · ∇Γqt(y)

)dγ(x)dγ(y)

= iωε

∫

Γ

divΓEincT qtdγ,

for any qt inH3/2(Γ).

(216)



−k2

∫

Γ

∫

Γ

G(x − y)(−−→curlΓp(x) · −−→curlΓpt(y)

)dγ(x)dγ(y)

− k2

∫

Γ

∫

Γ

G(x − y)(∇Γq(x) · −−→curlΓpt(y)

)dγ(x)dγ(y)

= −iωε

∫

Γ

curlΓEincT ptdγ,

for any pt inH1/2(Γ).

(217)

These two equations are coupled. The principal part of the first equation is∫

Γ

∫

Γ

G(x − y)∆Γq(x)∆Γqt(y)dγ(x)dγ(y),

which, up to a compact operator, is coercive inH3/2(Γ), due to the

properties of the single layer potential for the Laplace operator.



The principal part of the second equation is∫

Γ

∫

Γ

G(x − y)(−−→curlΓp(x) · −−→curlΓpt(y)

)dγ(x)dγ(y),

which, up to a compact operator, is coercive inH1/2(Γ), 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 surfaceΓ

∫

Γ

∫

Γ

G(x − y)(∇Γq(x) · −−→curlΓp(y)

)dγ(x)dγ(y)

=

∫

Γ

p(y)curlΓ

∫

Γ

G(x − y)∇Γq(x)dγ(x)dγ(y).

(218)



From identity(145), we obtain

∫

Γ

∫

Γ


)dγ(x)dγ(y)

=

∫

Γ

∫

Γ

(((ny−nx) ∧∇yG(x−y)) · ∇Γq(x)) p(y)dγ(x)dγ(y),

(219)which, using the results on pseudo-homogeneous kernels, shows that thisbilinear form is bounded by

∣∣∣∣∫

Γ

∫

Γ


)dγ(x)dγ(y)

∣∣∣∣

≤ c ‖q‖H1/2(Γ) ‖p‖H−1/2(Γ) .

(220)

It follows that the system of equations(216) (217) is of Fredholm type. Itis also equivalent to equation(168), and thus it satisfies the existence anduniqueness properties, whenk2 is not an eigenvalue of the interior problem(161) for the Maxwell equation.



We have prove that the system of equation(216) (217) has a unique

solution except whenk2 is an eigenvalue of the interior problem(161) for

the Maxwell equation. If we plug this solution in the EFIE equation(168),

the solutionj is such that

divΓj = ∆Γq

curlΓj = −∆Γp.(221)

As we know thatp is in H1/2(Γ) andq is in H3/2(Γ), it results that the

vectorj belongs to the space

H−1/2div (Γ) =

v ∈ TH−1/2(Γ); divΓv ∈ H−1/2(Γ)



15 – The dielectric case

Let Ωd be the dielectric, andΩe the exterior complement, with boundaryΓ.

We introduce the following real coefficients (ω is the angular frequency)

ǫd : the electric permittivity ofΩd,

ǫe : the electric permittivityΩe

µd : the magnetic permeabilityΩd

µe : the magnetic permeability ofΩe

kd = ω(µdǫd)12 : the wave number inΩd

ke = ω(µeǫe)12 : the wave number inΩe

zd = (µd

ǫd)

12 : the impedance inΩd

ze = (µe

ǫe)

12 : the impedance inΩe.



The unknows are the electric fieldsEd, Ee and the magnetic fieldsHd, He.

The harmonic Maxwell system is

(Ped)

curlEd − iωµdHd = 0 in Ωd,

curlHd + iωǫdEd = 0 in Ωd,

curlEe − iωµeHe = 0 in Ωe,

curlHe + iωǫeEe = 0 in Ωe,

Ed × n− (Ee + Einc) × n = 0 on Γ,

Hd × n− (He + Hinc) × n = 0 on Γ,

|√ǫeEe −√

µeHe × n| ≤ C|x|2 when |x| → ∞

whereEinc, Hinc is a known plane wave.



We introduce the electricj and the magneticm currents:

jd = Hd × n, je = −He × n

md = Ed × n, me = −Ee × n.(222)

Now, two Green’s functions appear denote byGd andGe.

Separately, the fields inΩd and inΩe admits an integral representation

given by (118) and (119) with their respective currents. Theunknowns are

the currentsje andme. The jump conditions on the boundaryΓ in the

equationPed give the currentsjd andmd.

The Rumsey variational formulation consists in equating the two integral

expressions of the currents on the surfaceΓ.

We denoteby Ad = kd Sd + 1/kd Td, Ae = ke Se + 1/ke Te andRd, Re,

the respectiveEFIE and MFIE operatorsassociated to each domain.



Then, in matrix notation, the system to solve takes the form

A =

izdAkd+ izeAke Rkd

+ Rke

Rkd+ Rke −i(

1

zdAkd

+1

zeAke)

(223)



References

[1] J.C. Nédélec“Acoustics and Electromagnetics Equations, Integral

Representations for Harmonics Problems”; Springer-Verlag, 2001.


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