The Mathematical Theory of Maxwell’s Equations - KIT · are constant (homogeneous medium) or at least continuous. In regions where the vector elds are smooth functions we can apply

The Mathematical Theory of Maxwell’s Equations

Andreas Kirsch and Frank HettlichDepartment of Mathematics

Karlsruhe Institute of Technology (KIT)Karlsruhe, Germany

c© January 14, 2013

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Contents

1 Introduction 7

1.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 The Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Boundary and Radiation Conditions . . . . . . . . . . . . . . . . . . . . . . 15

1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Expansion into Wave Functions 21

2.1 Separation in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.3 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.4 The Boundary Value Problem for the Laplace Equation in a Ball . . . . . . . 46

2.5 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.6 The Boundary Value Problems for the Helmholtz Equation for a Ball . . . . 56

2.7 Expansion of Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . 65

2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

3 Scattering From a Perfect Conductor 77

3.1 A Scattering Problem for the Helmholtz Equation . . . . . . . . . . . . . . . 77

3.1.1 Formulation of the Problem . . . . . . . . . . . . . . . . . . . . . . . 77

3.1.2 Representation Theorems . . . . . . . . . . . . . . . . . . . . . . . . 78

3.1.3 Volume and Surface Potentials . . . . . . . . . . . . . . . . . . . . . . 85

3.1.4 Boundary Integral Operators . . . . . . . . . . . . . . . . . . . . . . 98

3.1.5 Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 102

3.2 A Scattering Problem for the Maxwell System . . . . . . . . . . . . . . . . . 109

3.2.1 Formulation of the Problem . . . . . . . . . . . . . . . . . . . . . . . 109

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4 CONTENTS

3.2.2 Representation Theorems . . . . . . . . . . . . . . . . . . . . . . . . 110

3.2.3 Vector Potentials and Boundary Integral Operators . . . . . . . . . . 118

3.2.4 Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 123

3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

4 The Variational Approach to the Cavity Problem 129

4.1 Formulation of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

4.2 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

4.2.1 Basic Properties of Sobolev Spaces of Scalar Functions . . . . . . . . 129

4.2.2 Basic Properties of Sobolev Spaces of Vector Valued Functions . . . . 136

4.2.3 The Helmholtz Decomposition and Further Results on Sobolev Spaces 139

4.3 The Cavity Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

4.3.1 The Variational Formulation and Existence . . . . . . . . . . . . . . . 151

4.3.2 Uniqueness and Unique Continuation . . . . . . . . . . . . . . . . . . 159

4.4 The Time–Dependent Cavity Problem . . . . . . . . . . . . . . . . . . . . . 169

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

5 Boundary Integral Equation Methods for Lipschitz Domains 173

5.1 Sobolev Spaces on Boundaries . . . . . . . . . . . . . . . . . . . . . . . . . . 173

5.1.1 Boundary-Sobolev Spaces of Scalar Functions . . . . . . . . . . . . . 174

5.1.2 Boundary-Sobolev Spaces of Vector-Valued Functions . . . . . . . . . 182

5.1.3 The Case of a Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

5.2 Surface Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

5.3 Boundary Integral Equation Methods . . . . . . . . . . . . . . . . . . . . . . 221

5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

6 Appendix on Vector Calculus 229

6.1 Table of Differential Operators and Their Properties . . . . . . . . . . . . . . 229

6.2 Elementary Facts from Differential Geometry . . . . . . . . . . . . . . . . . 231

6.3 Integral Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

6.4 Surface Gradient and Surface Divergence . . . . . . . . . . . . . . . . . . . . 238

Bibliography 243

Preface

This book arose from a lecture on Maxwell’s equations given by the authors between ?? and2009.

The emphasis is put on three topics which are clearly structured into Chapters 2, 3, and 4.In each of these chapters we study first the simpler scalar case where we replace the Maxwellsystem by the scalar Helmholtz equation. Then we investigate the time harmonic Maxwell’sequations.

In Chapter 1 we start from the (time dependent) Maxwell system in integral form and derive...

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6 CONTENTS

Chapter 1

Introduction

1.1 Maxwell’s Equations

Electromagnetic wave propagation is described by particular equations relating five vectorfields E , D, H, B, J and the scalar field ρ, where E and D denote the electric field(in V/m) and electric displacement (in As/m2) respectively, while H and B denote themagnetic field (in A/m) and magnetic flux density (in V s/m2 = T =Tesla). Likewise,J and ρ denote the current density (in A/m2) and charge density (in As/m3) of themedium. Here and throughout the lecture we use the rationalized MKS-system, i.e. V ,A, m and s. All fields will be assumed to depend both on the space variable x ∈ R3 and onthe time variable t ∈ R.

The actual equations that govern the behavior of the electromagnetic field, first completelyformulated by Maxwell, may be expressed easily in integral form. Such a formulation has theadvantage of being closely connected to the physical situation. The more familiar differentialform of Maxwell’s equations can be derived very easily from the integral relations as we willsee below.

In order to write these integral relations, we begin by letting S be a connected smooth surfacewith boundary ∂S in the interior of a region Ω0 where electromagnetic waves propagate. Inparticular, we require that the unit normal vector ν(x) for x ∈ S be continuous and directedalways into “one side” of S, which we call the positive side of S. By τ(x) we denote the unitvector tangent to the boundary of S at x ∈ ∂S. This vector, lying in the tangent plane of Stogether with a vector n(x), x ∈ ∂S, normal to ∂S is oriented so as to form a mathematicallypositive system (i.e. τ is directed counterclockwise when we sit on the positive side of S, andn(x) is directed to the outside of S). Furthermore, let Ω ∈ R3 be an open set with boundary∂Ω and outer unit normal vector ν(x) at x ∈ ∂Ω. Then Maxwell’s equations in integral form

7

8 CHAPTER 1. INTRODUCTION

state: ∫∂S

H · τ d` =d

dt

∫S

D · ν ds +

∫S

J · ν ds (Ampere’s Law) (1.1a)

∫∂S

E · τ d` = − d

dt

∫S

B · ν ds (Law of Induction) (1.1b)

∫∂Ω

D · ν ds =

∫∫Ω

ρ dx (Gauss’ Electric Law) (1.1c)

∫∂Ω

B · ν ds = 0 (Gauss’ Magnetic Law) (1.1d)

To derive the Maxwell’s equations in differential form we consider a region Ω0 where µ and εare constant (homogeneous medium) or at least continuous. In regions where the vectorfields are smooth functions we can apply the Stokes and Gauss theorems for surfaces S andsolids Ω lying completely in Ω0:

∫S

curl F · ν ds =

∫∂S

F · τ d` (Stokes), (1.2)

∫∫Ω

div F dx =

∫∂Ω

F · ν ds (Gauss), (1.3)

where F denotes one of the fields H, E , B or D. We recall the differential operators (incartesian coordinates):

div F(x) =3∑j=1

∂Fj∂xj

(x) (divergenz, “Divergenz”)

curl F(x) =

∂F3

∂x2

(x)− ∂F2

∂x3

(x)

∂F1

∂x3

(x)− ∂F3

∂x1

(x)

∂F2

∂x1

(x)− ∂F1

∂x2

(x)

(curl, “Rotation”) .

With these formulas we can eliminate the boundary integrals in (1.1a-1.1d). We then usethe fact that we can vary the surface S and the solid Ω in D arbitrarily. By equating theintegrands we are led to Maxwell’s equations in differential form so that Ampere’s Law,the Law of Induction and Gauss’ Electric and Magnetic Laws, respectively, become:

1.1. MAXWELL’S EQUATIONS 9

∂B∂t

+ curlx E = 0 (Faraday’s Law of Induction)

∂D∂t− curlx H = −J (Ampere’s Law)

divx D = ρ (Gauss’ Electric Law)

divx B = 0 (Gauss’ Magnetic Law)

We note that the differential operators are always taken w.r.t. the spacial variable x (notw.r.t. time t!). Therefore, in the following we often drop the index x.

Physical remarks:

• The law of induction describes how a time-varying magnetic field effects the electricfield.

• Ampere’s Law describes the effect of the current (external and induced) on the mag-netic field.

• Gauss’ Electric Law describes the sources of the electric displacement.

• The forth law states that there are no magnetic currents.

• Maxwell’s equations imply the existence of electromagnetic waves (as ligh, X-rays, etc)in vacuum and explain many electromagnetic phenomena.

• Literature wrt physics: J.D. Jackson, Klassische Elektrodynamik, de Gruyter Verlag

Historical Remark:

• Dates: Andre Marie Ampere (1775–1836), Charles Augustin de Coulomb (1736–1806),Michael Faraday (1791–1867), James Clerk Maxwell (1831–1879)

• It was the ingeneous idea of Maxwell to modify Ampere’s Law which was known up tothat time in the form curl H = J for stationary currents. Furthermore, he collectedthe four equations as a consistent theory to describe the electromagnetic fields. (JamesClerk Maxwell, Treatise on Electricity and Magnetism, 1873).

Conclusion 1.1 Gauss’ Electric Law and Ampere’s Law imply the equation of continuity

∂ρ

∂t= div

∂D∂t

= div(curl H−J

)= − div J

because div curl = 0.


1.2 The Constitutive Equations

In this general setting the equations are not yet consistent (there are more unknowns thanequations). The Constitutive Equations couple them:

D = D(E ,H) and B = B(E ,H)

The electric properties of the material are complicated. In general, they not only depend onthe molecular character but also on macroscopic quantities as density and temperature ofthe material. Also, there are time-dependent dependencies as, e.g., the hysteresis effect, i.e.the fields at time t depend also on the past.

As a first approximation one starts with representations of the form

D = E + 4πP and B = H− 4πM

where P denotes the electric polarization vector and M the magnetization of the material.These can be interpreted as mean values of microscopic effects in the material. Analogously,ρ and J are macroscopic mean values of the free charge and current densities in the medium.

If we ignore ferro-electric and ferro-magnetic media and if the fields are small one can modelthe dependencies by linear equations of the form

D = εE and B = µH

with matrix-valued functions ε : R3 → R3×3 (dielectric tensor), and µ : R3 → R3×3

(permeability tensor). In this case we call the media inhomogenous and anisotropic.

The special case of an isotropic medium means that polarization and magnetization donot depend on the directions. In this case they are just real valued functions, and we have

D = εE and B = µH

with functions ε, µ : R3 → R.

In the simplest case these functions ε and µ are constant. This is the case, e.g., in vacuum.

We indicated already that also ρ and J can depend on the material and the fields. Therefore,we need a further equation. In conducting media the electric field induces a current. In alinear approximation this is described by Ohm’s Law:

J = σE + Je

where Je is the external current density. For isotropic media the function σ : R3 → R iscalled the conductivity.

Remark: If σ = 0 then the material is called dielectric. In vacuum we have σ = 0,ε = ε0 ≈ 8.854 · 10−12AS/V m, µ = µ0 = 4π · 10−7V s/Am. In anisotropic media, also thefunction σ is matrix valued.

1.3. SPECIAL CASES 11

1.3 Special Cases

Vacuum

In regions of vacuum with no charge distributions and (external) currents (that is, ρ = 0and Je = 0) the law of induction takes the form

µ0∂H∂t

+ curl E = 0 .

Differentiation wrt time t and use of Ampere’s Law yields

µ0∂2H∂t2

+1

ε0

curl curl H = 0 ;

that is,

ε0µ0∂2H∂t2

+ curl curl H = 0 .

1/√ε0µ0 has the dimension of velocity and is called the speed of light: c0 = 1/

√ε0µ0.

From curl curl = ∇ div−∆ it follows that the components of H are solutions of the linearwave equation

c20

∂2H∂t2

− ∆H = 0 .

Analogously, one derives the same equation for the electric field:

c20

∂2E∂t2

− ∆E = 0 .

Remark: Heinrich Rudolf Hertz (1857–1894) showed also experimentally the existence ofelectromagenetic waves about 20 years after Maxwell’s paper (in Karlsruhe!).

Electrostatics

If E is in some region Ω constant wrt time t (tha is, in the static case) the law of inductionreduces to

curl E = 0 in Ω .

Therefore, if Ω is simply connected there exists a potential u : Ω→ R with E = −∇u in Ω.Gauss’ Electric Law yields in homogeneous media the Poisson equation

ρ = div D = − div(ε0E) = −ε0∆u

for the potential u. The electrostatics is described by this basic elliptic partial differentialequation ∆u = −ρ/ε0. Mathematically, this is the subject of potential theory.


Magnetostatics

The same technique does not work in magnetostatics because, in general, curl H 6= 0. How-ever, because

div B = 0

we conclude the existence of a vector potential A : R3 → R3 with B = − curl A in D.Substituting this into Ampere’s Law yields (for homogeneous media Ω) after multiplicationwith µ0

−µ0J = curl curl A = ∇ div A − ∆A .

Since curl∇ = 0 we can add gradients ∇u to A without changing B. We will see laterthat we can choose u such that the resulting potential A satisfies div A = 0. This choice ofnormalization is called Coulomb gauge.

With this normalization we also get in the magnetostatic case the Poisson equation

∆A = −µ0J .

We note that in this case the Laplacian has to be taken componentwise.

Time Harmonic Fields

Under the assumptions that the fields allow a Fourier transformation w.r.t. time we set

E(x;ω) = (FtE)(x;ω) =

∫R

E(x, t) eiωt dt ,

H(x;ω) = (FtH)(x;ω) =

∫R

H(x, t) eiωt dt ,

etc. We note that the fields E, H etc are now complex valued, i.e, E(·;ω), H(·;ω) : R3 → C3

(and also the other fields). Although they are vector fields we denote them by capitalLatin letters only. Maxwell’s equations transform into (because Ft(u′) = −iωFtu) the timeharmonic Maxwell’s equations

−iωB + curlE = 0 ,

iωD + curlH = σE + Je ,

divD = ρ ,

divB = 0 .

Remark: The time harmonic Maxwell’s equation can also be derived from the assumptionthat all fields behave periodically w.r.t. time with the same frequency ω. Then the formsE(x, t) = e−iωtE(x), H(x, t) = e−iωtH(x), etc satisfy the time harmonic Maxwell’s equations.

1.3. SPECIAL CASES 13

With the constitutive equations D = εE and B = µH we arrive at

−iωµH + curlE = 0 , (1.4a)

iωεE + curlH = σE + Je , (1.4b)

div(εE) = ρ , (1.4c)

div(µH) = 0 . (1.4d)

Eliminating H or E, respectively, from (1.4a) and (1.4b) yields

curl

(1

iωµcurlE

)+ (iωε− σ)E = Je . (1.5)

and

curl

(1

iωε− σcurlH

)+ iωµH = curl

(1

iωε− σJe

), (1.6)

respectively. Usually, one writes these equations in a slightly different way by introducing theconstant values ε0 > 0 and µ0 > 0 in vacuum and relative values (dimensionless!) µr, εr ∈ Rand εc ∈ C, defined by

µr =µ

µ0

, εr =ε

ε0

, εc = εr + iσ

ωε0

.

Then equations (1.5) and (1.6) take the form

curl

(1

µrcurlE

)− k2εcE = iωµ0Je , (1.7)

curl

(1

εccurlH

)− k2µrH = curl

(1

εcJe

), (1.8)

with the wave number k = ω√ε0µ0. We conclude from this again that div(µrH) = 0

because div curl vanishes. For the electric field we conclude that div(εcE) = − iωµ0

k2 div Je =− iωε0

div Je.

In vacuum we have εc = 1, µr = 1 and thus

curl curlE − k2E = iωµ0Je , (1.9)

curl curlH − k2H = curl Je . (1.10)

For Je = 0 this reduces the vector Helmholtz equations ∆E+k2E = 0 and ∆H+k2H = 0.

### 1. Aequivalenz: Divergenz frei und vector Helmholtz als Lemma,

Also the following observation will be useful.

Lemma 1.2 Is E a divergence free solution of the vector Helmholtz equation in a domainD then x · E is a solution of the scalar Helmholtz equation.


Proof: By the formulae ?? and ?? we obtain

δ(x · E) = div∇(x · E)

= div((E ′)>x+ (x′)E

)= div

((E ′)>x

)+ divE

=3∑i=1

∑j = 13∂

2Ej∂x2

i

xj + 2 div(E)

= ∆A · x = k2x · E .

2 ###

Example 1.3 In the case Je = 0 and in vacuum the fields

E(x) = p eik d·x and H(x) = (p× d) eik d·x

are solutions of the time harmonic Maxwell’s equations (1.9), (1.10) provided d is a unitvector in R3 and p ∈ C3 with p ·d = 01. Such fields are called plane time harmonic fieldswith polarization vector p ∈ C3 and direction d.

We make the assumption εc = 1, µr = 1 for the rest of Section 1.3. Taking the diver-gence of these equations yields divH = 0 and k2 divE = −iωµ0 div Je; that is, divE =−(i/ωε0) div Je. Comparing this to (1.4c) yields the time harmonic version of the equationof continuity

div Je = iωρ .

With the vector identity curl curl = −∆ + div∇ equations (1.10) and (1.9) can be writtenas

∆E + k2E = −iωµ0Je +1

ε0

∇ρ , (1.11)

∆H + k2H = − curl Je . (1.12)

We consider the magnetic field first and introduce magnetic Hertz potentials: divH = 0implies the existence of a vector potential A with H = curlA. Thus (1.12) takes the form

curl(∆A + k2A) = − curl Je

and thus∆A + k2A = −Je + ∇ϕ (1.13)

for some scalar field ϕ. On the other hand, if A and ϕ satisfy (1.13) then

H = curlA and E = − 1

iωε0

(curlH − Je) = iωµ0A −1

iωε0

∇(divA− ϕ)

1We set p · d =∑3

j=1 pjdj even for p ∈ C3

1.4. BOUNDARY AND RADIATION CONDITIONS 15

satisfies the Maxwell system (1.4a)–(1.4d).

Analogously, we can introduce electric Hertz potentials if Je = 0. Since then divE = 0there exists a vector potential A with E = curlA. Substituting this into (1.11) yields

curl(∆A + k2A) = 0

and thus

∆A + k2A = ∇ϕ (1.14)

for some scalar field ϕ. On the other hand, if A and ϕ satisfy (1.14) then

E = curlA and H =1

iωµ0

curlE = −iωε0A +1

iωµ0

∇(divA− ϕ)

satifies the Maxwell system (1.4a)–(1.4d).

As a particular example we take the magnetic Hertz vector A of the form A(x) = u(x) zwith a scalar function u and the uni vector z = (0, 0, 1)> ∈ R3. Then

H = curl(uz) =

(∂u

∂x2

,− ∂u

∂x1

, 0

)>,

E = iωµ0 z +1

−iωε0

∇(∂u/∂x3) .

### TE und TM ausfuehrlicher und Definition formulieren (s. altes Skript, Jackson)

If u is independent of x3 then E has only a x3−component and this mode is called E-mode orTM-mode (for transverse-magnetic). Analogously, the H-mode or TE-mode is defined.In any case, the vector Helmholtz equation for A reduced to the scalar Helmholtz equationfor u. ###

1.4 Boundary and Radiation Conditions

Maxwell’s equations hold only in regions with smooth parameter functions εr, µr and σ. If weconsider a situation in which a surface S separates two homogeneous media from each other,the constitutive parameters ε, µ and σ are no longer continuous but piecewise continuouswith finite jumps on S. While on both sides of S Maxwell’s equations (1.4a)–(1.4d) hold,the presence of these jumps implies that the fields satisfy certain conditions on the surface.

To derive the mathematical form of this behaviour (the boundary conditions) we apply thelaw of induction (1.1b) to a narrow rectangle-like surface R, containing the normal n to thesurface S and whose long sides C+ and C− are parallel to S and are on the opposite sides ofit, see the following figure.


When we let the height of the narrow sides, AA′ and BB′, approach zero then C+ and C−approach a curve C on S, the surface integral ∂

∂t

∫R

B · ν ds will vanish in the limit becausethe field remains finite (note, that the normal ν is the normal to R lying in the tangentialplane of S). Hence, the line integrals

∫C

E+ · τ d` and∫C

E− · τ d` must be equal. Since thecurve C is arbitrary the integrands E+ · τ and E− · τ coincide on every arc C; that is,

n× E+ − n× E− = 0 on S . (1.15)

A similar argument holds for the magnetic field in (1.1a) if the current distribution J =σE + Je remains finite. In this case, the same arguments lead to the boundary condition

n×H+ − n×H− = 0 on S . (1.16)

If, however, the external current distribution is a surface current; that is, if Je is of the formJe(x+ τn(x)) = Js(x)δ(τ) for small τ and x ∈ S and with tangential surface field Js and σis finite, then the surface integral

∫R

Je · ν ds will tend to∫C

Js · ν d`, and so the boundarycondition is

n×H+ − n×H− = Js on S . (1.17)

We will call (1.15) and (1.16) or (1.17) the transmission boundary conditions.

A special and very important case is that of a perfectly conducting medium with bound-ary S. Such a medium is characterized by the fact that the electric field vanishes inside thismedium, and (1.15) reduces to

n× E = 0 on S (1.18)

Another important case is the impedance- or Leontovich boundary condition

n×H = λn× (E × n) on S (1.19)

1.4. BOUNDARY AND RADIATION CONDITIONS 17

which, under appropriate conditions, may be used as an approximation of the transmissionconditions.

The same kind of boundary occur also in the time harmonic case (where we denote the fieldsby capital Latin letters).

Finally, we specify the boundary conditions to the E- and H-modes derived above. ###Wo??? ### We assume that the surface S is an infinite cylinder in x3−direction withconstant cross section. Furthermore, we assume that the volume current density J vanishesnear the boundary S and that the surface current densities take the form Js = jsz for theE-mode and Js = js

(ν× z

)for the H-mode. We use the notation [v] := v|+−v|− for the jump

of the function v at the boundary. Also, we abbreviate (only for this table) σ′ = σ − iωε.We list the boundary conditions in the following table.

Boundary condition E-mode H-mode

transmission [u] = 0 on S ,[µ ∂u∂ν

]= 0 on S ,[

σ′ ∂u∂ν

]= −js on S , [u] = js on S ,

impedance λ k2u+ σ′ ∂u∂ν

= −js on S , k2u− λ iωµ∂u∂ν

= js on S ,

perfect conductor u = 0 on S , ∂u∂ν

= 0 on S .

The situation is different for the normal components. We consider Gauss’ Electric andMagnetic Laws and choose Ω to be a box which is separated by S into two parts Ω1 and Ω2.We apply (1.1c) first to all of Ω and then to Ω1 and Ω2 separately. The addition of the lasttwo formulas and the comparison with the first yields that the normal component D · n hasto be continuous as well as (application of (1.1d)) B ·n. With the constitutive equations onegets

n · (εr,1E1 − εr,2E2) = 0 on S and n · (µr,1H1 − µr,2H2) = 0 on S .

Conclusion 1.4 The normal components of E and/or H are not continuous at interfaceswhere εc and/or µr have jumps.

The Silver-Muller radiation condition

### Physik , Sommerfeld, Silver-Muller ###

Reference Problems

During our course we will consider two classical boundary value problems.


• (Cavity with an ideal conductor as boundary) Let D ⊆ R3 be a bounded domainwith sufficiently smooth boundary ∂D and exterior unit normal vector ν(x) at x ∈ ∂D.Let Je : D → C3 be a vector field. Determine a solution (E,H) of the time harmonicMaxwell system

curlE − iωµH = 0 in D , (1.20a)

curlH + (iωε− σ)E = Je in D , (1.20b)

ν × E = 0 on ∂D . (1.20c)

D

εc, µr

ν × E = 0

• (Scattering by an ideal conductor) Given a bounded region D and some solutionEi and H i of the “unperturbed” time harmonic Maxwell system

curlEi − iµ0Hi = 0 in R3 , curlH i + iε0E

i = 0 in R3 ,

determine E,H of the Maxwell system

curlE − iµ0H = 0 in R3 \D , curlH + iε0E = 0 in R3 \D ,

such that E satisfies the boundary condition ν×E = 0 on ∂D, and E and H have thedecompositions into E = Es + Ei and H = Hs + H i in R3 \ D with some scatteredfield Es, Hs which satisfy the Silver-Muller radiation condition

lim|x|→∞

|x|(Hs(x)× x

|x|−√ε0

µ0

Es(x)

)= 0

lim|x|→∞

|x|(Es(x)× x

|x|+

√µ0

ε0

Hs(x)

)= 0

uniformly with respect to all directions x/|x|.

Remark: For general µr, εc ∈ L∞(R3) we have to give a correct interpretation of thedifferential equations (“variational or weak formulation”) and transmission conditions(“trace theorems”).

1.5. EXERCISES 19

D ν × E = 0

Ei, H i

AAAAAA

AAAAAA

HHHH

HHj Es, Hs

1.5 Exercises


Chapter 2

Expansion into Wave Functions

2.1 Separation in Spherical Coordinates

The starting point of the investigations is the consideration of the boundary value problemsinside or outside the “simple” geometry of a ball in case of a homogeneous medium. Accord-ing to boundary conditions on the sphere we are interested in solutions of the time-harmonicMaxwell equations with components u of E and H, which can be separated by

u(x) = v(r)K(x)

with a radial depending factor v : R>0 → R and a spherical part K : S2 → R, wherer = |x| =

√x2

1 + x22 + x2

3 and x = x|x| ∈ S2 in the unit sphere S2 ⊆ R3. Using spherical

coordinates for x ∈ R3,

x =

r cosϕ sin θr sinϕ sin θr cos θ

with r ∈ R>0, ϕ ∈ [0, 2π), ψ ∈ [0, π] ,

we have x = (cosϕ sin θ, sinϕ sin θ, cos θ)>.

We already saw that the essential differential operator in the modelling of electromangenticwaves is the Laplace operator ∆. It occurs directly in the stationary situation (see page???), in case of the usual polarizations as principal part of the Helmholtz equation and alsoin the full Maxwell system, where the components turned out to be solutions of the vectorHelmholtz equation in particular (see page ???). Thus, a first step must be a representationof the Laplace operator in spherical coordinates.

Lemma 2.1 The Laplace operator in R3 for spherical coordinates (r, ϕ, θ) ∈ R>0× [0, 2π)×[0, π] is given by

∆ =1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin2 θ

∂2

∂ϕ2+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)=

∂2

∂r2+

2

r

∂

∂r+

1

r2 sin2 θ

∂2

∂ϕ2+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

).

21

22 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS

Proof: Let the unit basis vectors of the spherical coordinate system be denoted by

er =1

gr

∂x

∂r, eϕ =

1

gϕ

∂x

∂ϕ, and eθ =

1

gθ

∂x

∂θ

where gr = ‖∂x∂r‖ = 1, gϕ = ‖ ∂x

∂ϕ‖ = r sin θ and gθ = ‖∂x

∂θ‖ = r. With the jacobian of the

transformation,√

detx′>x′ = grgϕgθ = r2 sin θ, and the divergence theorem we obtain∫ r0+εr

r0

∫ ϕ0+εϕ

ϕ0

∫ θ0+εθ

θ0

∆u grgϕgθ d(r, θ, ϕ) =

∫Wε

∆u dx =

∫∂Wε

(∇u)>ν ds

in a domain Wε = x(r, θ, ϕ) ∈ R3 : (r, θ, ϕ) ∈ (r0, r0 + εr) × (θ0, θ0 + εθ) × (ϕ0, ϕ0 + εϕ).Substitution of the unit normal vectors and an application of Fubinis theorem lead to∫ r0+εr

r0

∫ θ0+εθ

θ0

∫ ϕ0+εϕ

ϕ0

∆u grgϕgθ d(r, θ, ϕ)

=

∫ θ0+εθ

θ0

∫ ϕ0+εϕ

ϕ0

(∇u)>e1︸︷︷︸= ∂u∂r

gϕgθ

∣∣∣r0+εr

r0dϕ dθ +

∫ r0+εr

r0

∫ θ0+εθ

θ0

(∇u)>e2︸︷︷︸= 1gϕ

∂u∂ϕ

grgθ

∣∣∣ϕ0+εϕ

ϕ0

dθ dr

+

∫ r0+εr

r0

∫ ϕ0+εϕ

ϕ0

(∇u)>e3︸︷︷︸= 1gθ

∂u∂θ

grgϕ

∣∣∣θ0+εθ

θ0dϕ dr

=

∫ r0+εr

r0

∫ θ0+εθ

θ0

∫ ϕ0+εϕ

ϕ0

∂

∂r(r2 sin θ

∂u

∂r) +

1

sin θ

∂2u

∂ϕ2+

∂

∂θ(sin θ

∂u

∂θ) dϕ dθ dr

Since the identity holds for any εr, εθ, εϕ > 0 and the integrand is a continuous function weconclude the given representation of the Laplace operator. 2

Definition 2.2 Functions u : Ω → R which satisfy the Laplace equation ∆u = 0 in Ωare called harmonic functions in Ω. Complex valued functions are harmonic if real andimaginary parts are harmonic.

The Laplace operator separates into a radial part, which is 1r2

∂∂r

(r2 ∂∂r

) = ∂2

∂r2+ 2

r∂∂r

, and thespherical part which is called the Laplace-Beltrami operator, a differential operator onthe unit sphere.

Definition 2.3 The differential operator ∆S : C2(S2)→ C(S2) with representation

∆S =1

sin2 θ

∂2

∂ϕ2+

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

).

in spherical coordinates is called Laplace-Beltrami operator on S2.

2.1. SEPARATION IN SPHERICAL COORDINATES 23

With the notation of the Beltrami operator and Dr = ∂∂r

we obtain

∆ = D2r +

2

rDr +

1

r2∆S =

1

r2Dr

(r2Dr

)+

1

r2∆S .

### Remark: analoge Zerlegung im Rn, Aufgabe: im R2 (Polarkoordinaten) berechnen###

A substitution of the guess u(x) = u(rx) = v(r)K(x) into the Laplace/Helmholtz equationwith k ∈ C leads to

0 = ∆u(rx) + k2u(rx) = v′′(r)K(x) +2

rv′(r)K(x) +

1

r2v(r) ∆SK(x) + k2v(r)K(x) .

Thus on sets with v(r) 6= 0 and K(x) 6= 0 we obtain

v′′(r) + 2rv′(r)

v(r)+

1

r2

∆SK(x)

K(x)+ k2 = 0 .

If there is a nontrivial solution of the supposed form, it follows the existence of a constant λsuch that v satisfies the differential equation

r2v′′(r) + 2r v′(r) +(k2r2 + λ

)v(r) = 0 (2.1)

and K the equation∆SK(x) = λK(x) . (2.2)

The last equation can be read in a functional analytic contents. The problem of determinefunctions K can be formulated as looking for eigenfunctions K and eigenvalues λ of theLaplace-Beltrami operator ∆S.

In view of Dirichlet boundary conditions like u = f on S2 the idea is to determine a completeset of eigenfunctions of the Laplace-Beltrami operator. Then we can hope to find a solutionfor any given f on S2 by an expansion in terms of eigenfunctions. We observe that thespherical equation (2.2) is independent of k. Thus, results will be useful in any case, theLaplace as well as the Helmholtz equation.

If we consider the expilcit representation of the Laplace-Beltrami operator in spherical co-ordinates the eigenvalue equation (2.2) becomes

∂2

∂ϕ2K + sin θ

∂

∂θ

(sin θ

∂

∂θ

)K = λ sin2 θ K .

Assuming eigenfunctions of the form K(x) = y1(ϕ) y2(θ) leads to

y′′1(ϕ)

y1(ϕ)+

sin θ(sin θ y′2(θ)

)′y2(θ)

− λ sin2 θ = 0 (2.3)

provided y1(ϕ) 6= 0 and y2(θ) 6= 0. If the decomposition is valid there exists a constant µ ∈ Csuch that y′′1 = µy1. Since we are interested in differentiable solutions u, the function y1 must


be 2π periodic. By the fundamental system of the linear ordinary differential equation weconclude that µ = −m2 with m ∈ Z and the general solution is

y1(ϕ) = c1eimϕ + c2e

−imϕ = c1 cos(mϕ) + c2 sin(mϕ)

with arbitrary constants c1, c2 ∈ C and c1, c2 ∈ C, respectively.

We assume that the reader is familar with the classical Fourier theory. In particular withthe fact that every function f ∈ L2(−π, π) allows an expansion in the form

f(t) =1√2π

∑m∈Z

fm eimt

with Fourier coefficients

fm =1√2π

π∫−π

f(s)e−ims ds , m ∈ Z .

The convergence of the series must be understood in the L2-sense, that is,

π∫−π

∣∣∣∣∣f(t)− 1√2π

M∑m=−N

fm eimt

∣∣∣∣∣2

dt −→ 0 , N,M →∞ .

Thus by the previous result we conclude, that the possible functions y1 for m ∈ Z constitutea complete orthonormal system in L2(−π, π).

Next we consider the dependency on θ. Using the result µ = −m2 we get from (2.3) thedifferential equation

sin θ(sin θ y′2(θ)

)′ − (m2 + λ sin2 θ

)y2(θ) = 0

for y2. With the substitution z = cos θ ∈ (−1, 1] and w(z) = y2(θ) we conclude an associatedLegendre differential equation

((1− z2)w′(z)

)′ − (λ+

m2

1− z2

)w(z) = 0 . (2.4)

An investigation of this equation will be the main task of the next section leading to theso called Legendre polynomials and to a complete orthonormal system of spherical surfaceharmonics.

Obviously the radial part of separated solutions given by the ordinary differential equation(2.1) is depending on the wave number k. If k = 0 we obtain the Euler equation

r2v′′(r) + 2r v′(r) + λv(r) = 0 . (2.5)

By the guess v(r) = rµ we compute µ(µ + 1) = −λ, which leads to a fundamental set ofsolutions

v1(r) = r12

+√

14−λ und v2(r) = r

12−√

14−λ .

2.2. LEGENDRE POLYNOMIALS 25

Later we will see that λ = −n(n+ 1) for n ∈ N0 are the eigenvalues of the Laplace-Beltramioperator. Thus v1(r) = rn and v2(r) = r−(n+1). According to non singular solutions of theLaplace equation inside a ball we have to choose v(r) = rn. In the exterior of a ball v2 willlead to solutions with asymptotic behavior |u(x)| → 0 for |x| → ∞.

In case of k 6= 0 we rewrite the differential equation (2.1) by the substitution z = kr andhatv(kr) =

√kr v(r) into

z2v′′(z) + zv′(z) +

(z2 + λ− 1

4

)v(z) = 0

and with the already mentioned eigenvalues λ = −n(n + 1) for n ∈ N we obtain the Besseldifferential equation

z2v′′(z) + zv′(z) +

(z2 −

(n+

1

2

)2)v(z) = 0 . (2.6)

In Section 2.5 we will discuss the Bessel functions, which solve this differential equation.Combining spherical surface harmonics and the corresponding Bessel functions will lead tosolutions of the Helmholtz equation and further on also of Maxwells equations.

### Aufgabe: Separation in kartesischen Koordinaten ###

2.2 Legendre Polynomials

Let us consider the case of harmonic functions u, that is ∆u = 0. We already saw thata separation in spherical coordinates leads to functions of the form u(x) = rµ(c1e

imϕ +c2e−imϕ)y2(θ), where y2(θ) = w(cos θ) is determined by the associated Legendre differential

equation (2.4).

We discuss this ordinary differential equation (2.4) first for the special case m = 0; that is,

d

dz

[(1− z2)w′(z)

]− λw(z) = 0 , −1 < z < 1 . (2.7)

This is a differential equation of Legendre type.

The coefficients vanish at z = ±1. We want to determine solutions which are continuous upto the boundary, thus w ∈ C2(−1,+1) ∩ C[−1,+1].

Theorem 2.4 (a) For λ = −n(n + 1), n ∈ N ∪ 0, there exists exactly one solutionw ∈ C2(−1,+1) ∩ C[−1,+1] with w(1) = 1. We set Pn = w. The function Pn is apolynomial of degree n and is called Legendre polynomial of degree n. It satisfiesthe differential equation

d

dx

[(1− x2)P ′n(x)

]+ n(n+ 1)Pn(x) = 0 , −1 ≤ x ≤ 1 .


(b) If there is no n ∈ N ∪ 0 with λ = −n(n+ 1) then there is no non-trivial solution of(2.7) in C2(−1,+1) ∩ C[−1,+1].

Proof: We make a – first only formal – ansatz for a solution in the form of a power series:

w(x) =∞∑j=0

aj xj

and substitute this into the differential equation (2.7). A simple calculation shows that thecoefficients have to satisfy the recursion

aj+2 =j(j + 1) + λ

(j + 1)(j + 2)aj , j = 0, 1, 2, . . . . (2.8)

The ratio test, applied to

w1(x) =∞∑k=0

a2k x2k and w2(x) =

∞∑k=0

a2k+1 x2k+1

separately yields that the radius of convergence is one. Therefore, for any a0, a1 ∈ R thefunction w is an analytic solution of (2.7) in (−1, 1).

Case 1: There is no n ∈ N∪0 with aj = 0 for all j ≥ n; that is, the series does not reduceto a finite sum. We study the behaviour of w1(x) and w2(x) as x tends to ±1. First weconsider w1 and split w1 in the form

w1(x) =

k0−1∑k=0

a2k x2k +

∞∑k=k0

a2k x2k

where k0 is chosen such that 2k0(2k0+1)+γ > 0 and[(2k0+1)+γ/2

]/[(2k0+1)(k0+1)

]≤ 1/2.

Then a2k does not change its sign anymore for k ≥ k0. Taking the logarithm of (2.8) forj = 2k yields

ln |a2(k+1)| = ln

[2k(2k + 1) + λ

(2k + 1)(2k + 2)

]+ ln |a2k| = ln

[1− (2k + 1) + λ/2

(2k + 1)(k + 1)

]+ ln |a2k| .

Now we use the elementary estimate ln(1− u) ≥ −u− 2u2 for 0 ≤ u ≤ 1/2 which yields

ln |a2(k+1)| ≥ − (2k + 1) + λ/2

(2k + 1)(k + 1)−[

(2k + 1) + λ/2

(2k + 1)(k + 1)

]2

+ ln |a2k|

≥ − 1

k + 1− c

k2+ ln |a2k|

for some c > 0. Therefore, we arrive at the following estimate for ln |a2k|:

ln |a2k| ≥ −k−1∑j=k0

1

j + 1− c

k−1∑j=k0

1

j2+ ln |a2k0| for k ≥ k0 .


Usingk−1∑j=k0

1

j + 1≤

k−1∑j=k0

j+1∫j

dt

t=

k∫k0

dt

t= ln k − ln k0

yields

ln |a2k| ≥ − ln k +

[ln k0 − c

∞∑j=k0

1

j2+ ln |a2k0|

]︸︷︷︸

= c

and thus

|a2k| ≥exp c

kfor all k ≥ k0 .

We note that a2k/a2k0 is positive for all k ≥ k0 and thus

w1(x)

a2k0

≥ a0

a2k0

+

k0−1∑k=1

[a2k

a2k0

− exp c

k a2k0

]x2k +

exp c

|a2k0|

∞∑k=1

1

kx2k = c − exp c

|a2k0|ln(1− x2) .

From this we observe that w1(x)→ +∞ as x→ ±1 or w1(x)→ −∞ as x→ ±1 (dependingon the sign of a2k0). By the same arguments one shows for positive a2k0+1 that w2(x)→ +∞as x → +1 and w2(x) → −∞ as x → −1. For negative a2k0+1 the roles of +∞ and −∞have to be interchanged. In any case, the sum w(x) = w1(x) + w2(x) is not bounded on[−1, 1] which contradicts our requirement on the solution of (2.7). Therefore, this case cannot happen.

Case 2: There is m ∈ N with aj = 0 for all j ≥ m; that is, the series reduces to a finite sum.Let m be the smallest number with this property. From the recursion formula we concludethat λ = −n(n + 1) for n = m − 2 and, furthermore, that a0 = 0 if n is odd and a1 = 0 ifn is even. In particular, w is a polynomial of degree n. We can normalize w by w(1) = 1because w(1) 6= 0. Indeed, if w(1) = 0 the differential equation

(1− x2)w′′(x) − 2xw′(x) + n(n+ 1)w(x) = 0

would imply w′(1) = 0. Differentiating the differential equation would yield w(k)(1) = 0 forall k ∈ N, a contradiction to w 6= 0. 2

In the proof of this theorem we have proven more. We collect the results as a corollary.

Corollary 2.5 The Legendre polynomials Pn(x) =∑n

j=0 ajxj have the properties

(a) Pn is even for even n and odd for odd n.

(b) aj+2 =j(j + 1)− n(n+ 1)

(j + 1)(j + 2)aj , j = 0, 1, . . . , n− 2.

(c)

+1∫−1

Pn(x)Pm(x) dx = 0 for n 6= m.


Proof: Only part (c) has to be shown. We multiply the differential equation (2.7) for Pnby Pm(x), the differential equation (2.7) for Pm by Pn(x), take the difference and integrate.This yields

0 =

+1∫−1

Pm(x)

d

dx

[(1− x2)P ′n(x)

]− Pn(x)

d

dx

[(1− x2)P ′m(x)

]dx

+[n(n+ 1)−m(m+ 1)

] +1∫−1

Pn(x)Pm(x) dx .

The first integral vanishes by partial integration. This proves part (c). 2

Before we return to the Laplace equation we prove some more results for the Legendrepolynomials.

Lemma 2.6

max−1≤x≤1

|Pn(x)| = 1 for all n = 0, 1, 2, . . . .

Proof: For fixed n ∈ N we define the function

Φ(x) := Pn(x)2 +1− x2

n(n+ 1)P ′n(x)2 for all x ∈ [−1,+1] .

We differentiate Φ and have

Φ′(x) = 2P ′n(x)

[Pn(x) +

1− x2

n(n+ 1)P ′′n (x)− x

n(n+ 1)P ′n(x)

]=

2P ′n(x)

n(n+ 1)

[n(n+ 1)Pn(x) +

d

dx(1− x2)P ′n(x)+ xP ′n(x)

]=

2xP ′n(x)2

n(n+ 1).

︸︷︷︸= 0 by the differential equation

Therefore Φ′ > 0 on (0, 1] and Φ′ < 0 on [−1, 0); that is, Φ is monotonously increasing on(0, 1] and monotonously decreasing on [−1, 0). Therefore,

0 ≤ Pn(x)2 ≤ Φ(x) ≤ maxΦ(1),Φ(−1) .

From Φ(1) = Φ(−1) = 1 we conclude that |Pn(x)| ≤ 1 for all x ∈ [−1,+1]. The lemma isproven by noting that Pn(1) = 1. 2


−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Theorem 2.7 (Rodriguez)

For all n ∈ N0 we have

Pn(x) =1

2n n!

dn

dxn(x2 − 1)n , x ∈ R .

Proof: First we prove that the right hand side solves the Legendre differential equation;that is, we show that

d

dx

[(1− x2)

dn+1

dxn+1(x2 − 1)n

]+ n(n+ 1)

dn

dxn(x2 − 1)n = 0 . (2.9)

We observe that both parts are polynomials of degree n. We multiply the first part by xj forsome j ∈ 0, . . . , n, integrate and use partial integration two times. This yields for j ≥ 1

Aj :=

1∫−1

xjd

dx

[(1− x2)

dn+1

dxn+1(x2 − 1)n

]dx = −j

1∫−1

xj−1 (1− x2)dn+1

dxn+1(x2 − 1)n dx

= j

1∫−1

[(j − 1)xj−2 − (j + 1)xj

] dndxn

(x2 − 1)n dx


We note that no boundary contributions occur and, furthermore, that A0 = 0. Now weuse partial integration n times again. No boundary contributions occur either since there isalways at least one factor (x2− 1) left. For j ≤ n− 1 the integral vanishes because the n-thderivative of xj and xj−2 vanish for j < n. For j = n we have

An = −n(n+ 1) (−1)n n!

1∫−1

(x2 − 1)n dx .

Analogously, we multiply the second part of (2.9) by xj, integrate, and apply partial inte-gration n-times. This yields

Bj = n(n+ 1)

1∫−1

xjdn

dxn(x2 − 1)n dx = n(n+ 1) (−1)n n!

1∫−1

(x2 − 1)n dx if j = n

and zero if j < n. This proves that the polynomial

d

dx

[(1− x2)

dn+1

dxn+1(x2 − 1)n

]+ n(n+ 1)

dn

dxn(x2 − 1)n

of degree n is orthogonal in L2(−1, 1) to all polynomials of degree at most n and, therefore,has to vanish. Furthermore,

dn

dxn(x2 − 1)n

∣∣∣∣x=1

=dn

dxn[(x+ 1)n (x− 1)n

]∣∣∣∣x=1

=n∑k=0

(n

k

)dk

dxk(x+ 1)n

∣∣∣∣∣x=1

dn−k

dxn−k(x− 1)n

∣∣∣∣x=1

.

Fromdn−k

dxn−k(x− 1)n

∣∣∣∣x=1

= 0 for k ≥ 1 we conclude

dn

dxn(x2 − 1)n

∣∣∣∣x=1

=

(n

0

)2n

dn

dxn(x− 1)n

∣∣∣∣x=1

= 2n n!

which proves the theorem. 2

As a first application of the formula of Rodriguez we show:

Theorem 2.8 (a)

+1∫−1

Pn(x)2 dx =2

2n+ 1for all n = 0, 1, 2, . . ..

(b)

+1∫−1

xPn(x)Pn+1(x) dx =2(n+ 1)

(2n+ 1)(2n+ 3)for all n = 0, 1, 2, . . ..


Proof: (a) We use the representation of Pn by Rodriguez and n partial integrations,

+1∫−1

dn

dxn(x2 − 1)n

dn

dxn(x2 − 1)n dx = (−1)n

+1∫−1

d2n

dx2n

[(x2 − 1)n

](x2 − 1)n dx

= (−1)n (2n)!

+1∫−1

(x2 − 1)n dx = (2n)!

+1∫−1

(1− x2)n dx .

It remains to compute In :=

+1∫−1

(1− x2)n dx.

We claim: In = 22n(2n− 2) · · · 2

(2n+ 1)(2n− 1) · · · 1= 2 · 4n (n!)2

(2n+ 1)!for all n ∈ N .

The assertion is true for n = 1. Let it be true for n− 1, n ≥ 2. Then

In = In−1 −+1∫−1

x2 (1− x2)n−1 dx = In−1 +

+1∫−1

xd

dx

[1

2n(1− x2)n

]dx

= In−1 −1

2nIn and thus In =

2n

2n+ 1In−1 .

This proves the representation of In. We arrive at

+1∫−1

Pn(x)2 dx =1

(2n n!)2(2n)! 2 · 4n (n!)2

(2n+ 1)!=

2

2n+ 1.


(b) This is proven quite similarily:

+1∫−1

xdn

dxn(x2 − 1)n

dxn+1

dxn+1(x2 − 1)n+1 dx

= (−1)n+1

+1∫−1

(x2 − 1)n+1 dn+1

dxn+1

[xdn

dxn(x2 − 1)n

]dx

= (−1)n+1

+1∫−1

(x2 − 1)n+1

x d2n+1

dx2n+1(x2 − 1)n︸︷︷︸=0

+ (n+ 1)d2n

dx2n(x2 − 1)n︸︷︷︸=(2n)!

dx= (n+ 1) (−1)n+1(2n)! In+1

= 2(n+ 1)(2n)!(2n+ 2)(2n)(2n− 2) · · · 2

(2n+ 3)(2n+ 1)(2n− 1) · · · 1

= (n+ 1)2(2nn!)(2n+1(n+ 1)!)

(2n+ 1)(2n+ 3)

which yields the assertion. 2

The formula of Rodriguez is also useful for proving recursion formulas for the Legendrepolynomials. We prove only some of these formulas.

Theorem 2.9 For all x ∈ R and n ∈ N, n ≥ 0, we have

(a) P ′n+1(x) − P ′n−1(x) = (2n+ 1)Pn(x),

(b) (n+ 1)Pn+1(x) = (2n+ 1)xPn(x) − nPn−1(x),

(c) P ′n(x) = nPn−1(x) + xP ′n−1(x),

(d) xP ′n(x) = nPn(x) + P ′n−1(x),

(e) (1− x2)P ′n(x) = (n+ 1)[xPn(x) − Pn+1(x)

]= −n

[xPn(x) − Pn−1(x)

],

(f) P ′n+1(x) + P ′n−1(x) = Pn(x) + 2xP ′n(x),

(g) (2n+ 1)(1− x2)P ′n(x) = n(n+ 1)[Pn−1(x) − Pn+1(x)

],


(h) nP ′n+1(x) − (2n+ 1)xP ′n(x) + (n+ 1)P ′n−1(x) = 0.

In these formulas we have set P−1 = 0.

Proof: (a) The formula is obvious for n = 0. Let now n ≥ 1. We calculate, using theformula of Rodriguez,

P ′n+1(x) =1

2n+1 (n+ 1)!

dn+2

dxn+2(x2 − 1)n+1 =

1

2n+1 (n+ 1)!

dn

dxn

(2(n+ 1)

d

dx

[x (x2 − 1)n

])=

1

2n n!

dn

dxn[(x2 − 1)n + 2nx2(x2 − 1)n−1

]= Pn(x) +

1

2n−1 (n− 1)!

dn

dxn[(x2 − 1)n + (x2 − 1)n−1

]= Pn(x) + 2nPn(x) + P ′n−1(x)

which proves formula (a).

(b) The orthogonality of the system Pn : n = 0, 1, 2, . . . implies its linear independence.Therefore, P0, . . . , Pn forms a basis of the space Pn of all polynomials of degree ≤ n. Thisyields existence of αn, βn ∈ R and qn−3 ∈ Pn−3 such that

Pn+1(x) = αnxPn(x) + βnPn−1(x) + qn−3 .

The orthogonality condition implies that

+1∫−1

qn−3 Pn+1 dx = 0 ,

+1∫−1

qn−3 Pn−1 dx = 0 ,

+1∫−1

x qn−3(x)Pn(x) dx = 0 ,

thus+1∫−1

qn−3(x)2dx = 0 ,

and therefore qn−3 ≡ 0. From 1 = Pn+1(1) = αnPn(1) + βnPn−1(1) = αn + βn we concludethat

Pn+1(x) = αnxPn(x) + (1− αn)Pn−1(x) .

We determine αn by Theorem 2.8:

0 =

+1∫−1

Pn−1(x)Pn+1(x) dx

= αn

+1∫−1

xPn−1(x)Pn(x) dx + (1− αn)

+1∫−1

Pn−1(x)2 dx

= αn

[2n

(2n− 1)(2n+ 1)− 2

2n− 1

]+

2

2n− 1,


thus αn =2n+ 1

n+ 1. This proves part (b).

(c) The definition of Pn yields

P ′n(x) =1

2nn!

dn+1

dxn+1(x2 − 1)n

=1

2n−1(n− 1)!

dn

dxn[x(x2 − 1)n−1]

=x

2n−1(n− 1)!

dn

dxn(x2 − 1)n−1 +

n

2n−1(n− 1)!

dn−1

dxn−1(x2 − 1)n−1

= xP ′n−1(x) + nPn−1(x) .

(d) Differentiation of (b) and subtraction of (d) for n+ 1 instead of n yields

(n+ 1)P ′n+1(x) = (2n+ 1)xP ′n(x) + (2n+ 1)Pn(x) − nP ′n−1(x),

(n+ 1)P ′n+1(x) = (n+ 1)2Pn(x) + (n+ 1)xP ′n(x) , thus

0 =[(2n+ 1)− (n+ 1)2

]Pn(x) +

[(2n+ 1)− (n+ 1)

]xP ′n(x) − nP ′n−1(x) ;

that is,

0 = −n2Pn(x) + nxP ′n(x) − nP ′n−1(x) .

2

Now we go back to the associated Legendre differential equation (2.4) and determine solutionsin C2(−1,+1) ∩ C[−1,+1] for m 6= 0.

Theorem 2.10 The functions

Pmn (x) = (1− x2)m/2

dm

dxmPn(x) , −1 < x < 1, 0 ≤ m ≤ n , (2.10)

are solutions of the differential equation (2.4) for λ = −n(n+ 1); that is,

d

dx

[(1− x2)

d

dxPmn (x)

]+

(n(n+ 1)− m2

1− x2

)Pmn (x) = 0 , −1 < x < 1 , (2.11)

for all 0 ≤ m ≤ n. The functions Pmn are called associated Legendre functions.

2.3. SPHERICAL HARMONICS 35

Proof: We compute

(1− x2)d

dxPmn (x) = −mx (1− x2)m/2

dm

dxmPn(x) + (1− x2)m/2+1 dm+1

dxm+1Pn(x) ,

d

dx

[(1− x2)

d

dxPmn (x)

]= −mPm

n (x) + m2x2 (1− x2)m/2−1 dm

dxmPn(x)

− mx (1− x2)m/2dm+1

dxm+1Pn(x)

− (m+ 2)x (1− x2)m/2dm+1

dxm+1Pn(x) + Pm+2

n (x)

= −mPmn (x) +

m2 x2

1− x2Pmn (x) − (m+ 1) 2x√

1− x2Pm+1n (x) + Pm+2

n (x) .

Differentiating equation (2.7) m times yields

m+1∑k=0

(m+ 1

k

)dk

dxk(1− x2)

dm+2−k

dxm+2−kPn(x) + n(n+ 1)dm

dxmPn(x) = 0 . (2.12)

The sum reduces to three terms only, thus

(1− x2)dm+2

dxm+2Pn(x) − (m+ 1) 2x

dm+1

dxm+1Pn(x) −

[n(n+ 1)−m(m+ 1)

] dmdxm

Pn(x) = 0 .

We multiply the identity by (1− x2)m/2 and arrive at

Pm+2n (x) − (m+ 1) 2x√

1− x2Pm+1n (x) +

[n(n+ 1)−m(m+ 1)

]Pmn (x) = 0 .

Combining this with the previous equation by eliminating the terms involving Pm+1n (x) and

Pm+2n (x) yields

d

dx

[(1− x2)

d

dxPmn (x)

]= −mPm

n (x) +m2 x2

1− x2Pmn (x) −

[n(n+ 1)−m(m+ 1)

]Pmn (x)

=

(m2

1− x2− n(n+ 1)

)Pmn (x)

which proves the theorem. 2

2.3 Spherical Harmonics

Now we return to the Laplace equation ∆u = 0 and collect our arguments. We have shownthat for n ∈ N and 0 ≤ m ≤ n the functions

hmn (r, θ, ϕ) = c rn Pmn (cos θ) e±imϕ


are harmonic functions in all of R3 for any constant c ∈ C. Analogously, the functions

v(r, θ, ϕ) = c r−n−1 Pmn (cos θ) e±imϕ

are harmonic in R3 \ 0. They decay at infinity.

Another observation on u is that these functions are polynomials.

Theorem 2.11 The functions hmn (r, θ, ϕ) = rnPmn (cos θ) e±imϕ with 0 ≤ m ≤ n are polyno-

mials of degree n.

Proof: First we consider the case m = 0, that is the functions h0n(r, ϕ, θ) = rn Pn(cos θ).

Let n be even, that is, n = 2` for some m. Then

Pn(x) =∑j=0

a2j x2j .

From r = |x| and x3 = r cos θ we write u in the form

h0n(x) = c |x|n Pn

(x3

|x|

)= c |x|2`

∑j=0

a2jx2j

3

|x|2j= c

∑j=0

a2j x2j3 |x|2(m−j) ,

and this is obviously a polynomial of degree 2` = n. The same arguments hold for oddvalues of n.

Now we show that also the associated functions; that is, for m > 0, are polynomials of degreen. We write

hmn (r, θ, ϕ) = rn Pmn (cos θ) eimϕ = rn sinm θ

dm

dxmPn(x)

∣∣∣∣x=cos θ

eimϕ .

and set Q = dm

dxmPn. It holds

cos θ =x3

r, sin θ =

1

r

√x2

1 + x22 ,

as well as

cosϕ =1

r

x1

sin θ=

x1√x2

1 + x22

, sinϕ =x2√x2

1 + x22

.

Therefore,

hmn (x) = rn−m (x21 + x2

2)m/2Q(x3

r

) (x1 + ix2)m

(x21 + x2

2)m/2= rn−mQ

(x3

r

)(x1 + ix2)m .

The assertion follows because Q is a homogeneous polynomial of degree n−m. 2

Furthermore, the functions hmn (x) = c rn Pmn (cos θ) e±imϕ are obviously homogeneous. A

polynomial of degree n ∈ N is called homogeneous, if

u(µx) = µnu(x)

for all x ∈ Rn and µ ∈ R.


Definition 2.12 Let n ∈ N ∪ 0.

(a) Homogeneous harmonic polynomials of degree n are called spherical harmonics oforder n.

(b) The functions Kn : S2 → C, defined as restrictions of spherical harmonics of degree nto the unit sphere are called spherical surface harmonics of order n.

Therefore, the functions hmn (r, θ, ϕ) = rnP|m|n (cos θ)eimϕ are spherical harmonics of order n

for −n ≤ m ≤ n. Spherical harmonics which do not depend on ϕ are called zonal. Thus, incase of m = 0 the functions h0

n are zonal spherical harmonics.

For any surface spherical harmonic Kn of order n the function

Hn(x) = |x|nKn

(x

|x|

), x ∈ R3 ,

is a homogeneous harmonic polynomial; that is, a spherical harmonic. We immediately have

Lemma 2.13 (a) Kn(−x) = (−1)nKn(x) for all x ∈ S2.

(b)∫S2

Kn(x)Km(x) ds(x) = 0 for all n 6= m.

Proof: Part (a) follows immediately since Hn is homogeneous (set µ = −1).

(b) With Green’s second formula in the region x ∈ R3 : |x| < 1 we have∫S2

(Hn

∂

∂rHm −Hm

∂

∂rHn

)ds =

∫|x|≤1

(Hn ∆Hm −Hm ∆Hn) dx = 0 .

Setting f(r) = Hn(rx) = rnHn(x) for fixed x ∈ S2 we have that f ′(1) = ∂∂rHn(x) = nHn(x);

that is,

0 = (m− n)

∫S2

HnHm ds = (m− n)

∫S2

KnKm ds .

2

Now we determine the dimension of the space of spherical harmonics for fixed order n.

Theorem 2.14 The set of spherical harmonics of order n is a vector space of dimension2n + 1. In particular, there exists a system Km

n : −n ≤ m ≤ n of spherical harmonics oforder n such that ∫

S2

Kmn K

`n ds = δm,` =

1 for m = `,0 for m 6= ` ;

that is, Kmn : −n ≤ m ≤ n is an orthonormal basis of this vector space.


Proof: Every homogeneous polynomial of degree n is necessarily of the form

Hn(x) =n∑j=0

An−j(x1, x2)xj3 x ∈ R3 , (2.13)

where An−j are homogeneous polynomials with respect to (x1, x2) of degree n− j. Since Hn

is harmonic it follows that

0 = ∆Hn(x) =n∑j=0

xj3 ∆2An−j(x1, x2) +n∑j=2

j(j − 1)xj−23 An−j(x1, x2) ,

where ∆2 = ∂2

∂x21+ ∂2

∂x22

denote the two-dimensional Laplace operator. From ∆2A0 = ∆2A1 = 0

we conclude that

0 =n−2∑j=0

[∆2An−j(x1, x2) + (j + 1)(j + 2)An−j−2(x1, x2)]xj3 ,

and thus by comparing the coefficients

An−j−2(x1, x2) = − 1

(j + 1)(j + 2)∆2An−j(x1, x2)

for all (x1, x2) ∈ R2, j = 0, . . . , n− 2; that is (replace n− j by j)

Aj−2 = − 1

(n− j + 1)(n− j + 2)∆2Aj for j = n, n− 1, . . . , 2 . (2.14)

Also, one can reverse the arguments: If An and An−1 are homogeneous polynomials of degreen and n − 1, respectively, then all of the functions Aj defined by (2.14) are homogeneouspolynomials of degree j, and Hn is a homogeneous harmonic polynomial of order n.

Therefore, the space of all spherical harmonics of order n is isomorphic to the space(An, An−1) : An, An−1 are homogeneous polynomials of degree n and n− 1, resp.

.

From the representation An(x1, x2) =n∑i=0

ai xi1 x

n−i2 we note that the dimension of the space of

all homogeneous polynomials of degree n is just n+1. Therefore, the dimension of the spaceof all spherical harmonics of order n is (n + 1) + n = 2n + 1. Finally, it is well known thatany basis of this finite dimensional Euclidian space can be orthogonalized by the method ofSchmidt. 2

Remark: The set Kmn : −n ≤ m ≤ n is not uniquely determined. Indeed, for any

orthogonal matrix A ∈ R3×3; that is, A>A = I, also the set Kmn (Ax) : −n ≤ m ≤ n is an

orthonormal system. Indeed, the substitution x = Ay yields∫S2

Kmn (x)Km′

n′ (x) ds(x) =

∫S2

Kmn (Ay)Km′

n′ (Ay) ds(y) = δn,n′δm,m′ .

Now we can state that the spherical harmonics Hmn = crnP

|m|n (cos θ)eimϕ with −n ≤ m ≤ n,

which we have determined by separation, constitute an orthogonal basis of the space ofspherical harmonics of order n.


Theorem 2.15 The functions Hmn (r, θ, ϕ) = rnP

|m|n (cos θ)eimϕ, −n ≤ m ≤ n are spherical

harmonics of order n. They are mutually orthogonal and, therefore, form an orthogonal basisof the (2n+ 1)-dimensional space of all spherical harmonics of order n.

Proof: We already know that Hmn are spherical harmonics of order n ∈ N. Thus the

theorem follows from∫|x|<1

Hmn (x)H`

n(x) dx =

1∫0

π∫0

2π∫0

r2n P |m|n (cos θ)P |`|n (cos θ) ei(m−`)ϕ r2 sin θ dϕ dθ dr = 0

for m 6= `. 2

We want to normalize these functions. First we consider the associated Legendre functions.

Theorem 2.16

+1∫−1

Pmn (x)2 dx =

2

2n+ 1· (n+m)!

(n−m)!, m = 0, . . . , n, n ∈ N ∪ 0 .

Proof: The case m = 0 has been proven in Theorem 2.8 already.

For m ≥ 1 partial integration yields

+1∫−1

Pmn (x)2 dx =

+1∫−1

(1− x2)mdm

dxmPn(x)

dm

dxmPn(x) dx

= −+1∫−1

d

dx

[(1− x2)m

dm

dxmPn(x)

]dm−1

dxm−1Pn(x) dx . (2.15)

Now we differentiate the Legendre differential equation (2.7) (m−1)-times (see (2.12) ###check! ### for m replaced by m− 1); that is,

(1− x2)dm+1

dxm+1Pn(x)− 2mx

dm

dxmPn(x) +

[n(n+ 1)−m(m− 1)

] dm−1

dxm−1Pn(x) = 0 ,

thus, after multiplication by (1− x2)m−1,

d

dx

[(1− x2)m

dm

dxmPn(x)

]+[n(n+ 1)−m(m− 1)

](1− x2)m−1 dm−1

dxm−1Pn(x) = 0 .

Substituting this into (2.12) ### check! ### yields

+1∫−1

Pmn (x)2 dx =

[n(n+1)−m(m−1)

] +1∫−1

Pm−1n (x)2 dx = (n+m) (n−m+1)

+1∫−1

Pm−1n (x)2 dx .


This is a recursion formula for+1∫−1

Pmn (x)2 dx with respect to m and yields the assertion by

using the formula for m = 0. 2

Now we define the normalized spherical surface harmonics Y mn by

Y mn (θ, ϕ) :=

√(2n+ 1) (n− |m|)!

4π (n+ |m|)!P |m|n (cos θ) eimϕ, −n ≤ m ≤ n, n = 0, 1, . . . (2.16)

They form an orthonormal system in L2(S2). We will identify Y mn (x) with Y m

n (θ, ϕ) forx = (sin θ cosϕ , sin θ sinϕ , cos θ)> ∈ S2.

From (2.2) for λ = −n(n+ 1) we remember that Y mn satisfies the differential equation

∆SYmn (θ, ϕ) + n(n+ 1)Y m

n (θ, ϕ)

=1

sin θ

∂

∂θ

[sin θ

∂Y mn (θ, ϕ)

∂θ

]+

1

sin2 θ

∂2Y mn (θ, ϕ)

∂ϕ2+ n(n+ 1)Y m

n (θ, ϕ) = 0 ,

with the Laplace-Beltrami operator ∆S (see page ??). Thus, we now see that λ = −n(n+1)for n ∈ N are the eigenvalues of ∆S with eigenfunctions Y m

n , |m| ≤ n.

The Laplace-Beltrami operator ∆S is symmetric; that is,

(f,∆Sg)L2(S2) = (∆Sf, g)L2(S2) for all f, g ∈ C2(S2) . (2.17)

Indeed, by partial integration we see

(f,∆Sg)L2(S2) =

π∫0

2π∫0

f(θ, ϕ)

1

sin θ

∂

∂θ

[sin θ

∂g(θ, ϕ)

∂θ

]+

1

sin2 θ

∂2g(θ, ϕ)

∂ϕ2

sin θ dϕ dθ

=

2π∫0

π∫0

f(θ, ϕ)∂

∂θ

[sin θ

∂g(θ, ϕ)

∂θ

]dθ dϕ +

π∫0

1

sin θ

2π∫0

f(θ, ϕ)∂2g(θ, ϕ)

∂ϕ2dϕ dθ

= −2π∫

0

π∫0

∂f(θ, ϕ)

∂θsin θ

∂g(θ, ϕ)

∂θdθ dϕ −

π∫0

1

sin θ

2π∫0

∂f(θ, ϕ)

∂ϕ

∂g(θ, ϕ)

∂ϕdϕ dθ

and this is symmetric with respect to f and g. Setting f = g we observe that −∆S is alsonon-negative; that is,

−(f,∆Sf)L2(S2) ≥ 0 for all f ∈ C2(S2) . (2.18)

In the setting of abstract functional analysis we note that −∆ is a (densely defined andunbounded) operator from L2(S2) into itself which is selfadjoint. This observation allowsthe use of general functional analytic tools to prove, e.g., that the eigenfunctions

Y mn :

|m| ≤ n, n = 0, 1, . . .

of −∆ form a complete orthonormal system of L2(S2). We want toprove this fact directly. First we show the following result.


Theorem 2.17 (Addition Formula)

(2n+ 1)Pn(x · y) = 4πn∑

m=−n

Y mn (x)Y −mn (y) for all x, y ∈ S2 .

Proof: Set

K(x, y) :=n∑

m=−n

Y mn (x)Y m

n (y) =n∑

m=−n

Y mn (x)Y −mn (y) , x, y ∈ S2 .

Then ∫S2

K(x, y)Y mn (y) ds(y) = Y m

n (x) , x ∈ S2 .

We show that also 2n+14π

Pn(x · y) solves this equation. We keep x ∈ S2 fixed, set

γ :=

∫S2

Pn(x · y)Y mn (y) ds(y) ,

and choose an orthogonal matrix A (depending on x) such that x := A−1x = A>x is the“north pole”; that is, x = (0, 0, 1)>. The transformation y = Ay′ yields

γ =

∫S2

Pn(x · Ay′)Y mn (Ay′) ds(y′) =

∫S2

Pn(x · y)Y mn (Ay) ds(y) .

The function Y mn (Ay) is again a spherical surface harmonic of order n, thus

Y mn (Ay) =

n∑k=−n

ak Ykn (y) , y ∈ S2 , (2.19)

where ak =∫S2

Y mn (Ay)Y −kn (y) ds(y). Substituting this into the form of γ yields

γ =n∑

k=−n

ak

∫S2

Pn(x · y)Y kn (y) ds(y) .

Using again polar coordinates y = (sin θ cosϕ , sin θ sinϕ , cos θ)> ∈ S2 we observe thatx · y = cos θ, thus

Pn(x · y) = Pn(cos θ) =

√4π

2n+ 1Y 0n (y) ,

and, by the orthogonalty of Y mn ,

γ = a0

√4π

2n+ 1.


We substitute y = x in (2.19) and have, using Y nk (x) = 0 for k 6= 0,

Y mn (x) = Y m

n (Ax) =n∑

k=−n

ak Ykn (x) = a0

√2n+ 1

4πPn(1)︸︷︷︸

=1

,

thus a0 =√

4π2n+1

Y mn (x) and therefore

γ =4π

2n+ 1Y mn (x) .

The difference satisfies therefore∫S2

[K(x, y)− 2n+ 1

4πPn(x · y)

]Y mn (y) ds(y) = 0 for all m = −n, . . . , n,

and this yields the assertion since Y mn : −n ≤ m ≤ n is a basis of all spherical surface

harmonics of order n. 2

We formulate a simple conclusion as a corollary.

Corollary 2.18

(a)n∑

m=−n

|Y mn (x)|2 =

2n+ 1

4πfor all x ∈ S2 and n = 0, 1, . . ..

(b)∣∣Y mn (x)

∣∣ ≤ √2n+ 1

4πfor all x ∈ S2 and n = 0, 1, . . ..

Proof: Part (a) follows immediately from the addition formula for y = x. Part (b) followsdirectly from (a). 2

After this preparation we are able to prove the completeness of the spherical surface har-monics.

Theorem 2.19 The functionsY mn : −n ≤ m ≤ n, n ∈ N ∪ 0

are complete in L2(S2);

that is, every function f ∈ L2(S2) can be expanded into a generalized Fourier series in theform

f =∞∑n=0

n∑m=−n

(f, Y mn )L2(S2) Y

mn . (2.20a)

The series can also be written as

f(x) =1

4π

∞∑n=0

(2n+ 1)

∫S2

f(y)Pn(y · x) ds(y) , x ∈ S2 . (2.20b)


The convergence in (2.20a) and (2.20b) are understood in the L2–sense.

Furthermore, on bounded sets in C1(S2) the series converge even uniformly; that is, for everyM > 0 and ε > 0 there exists N0 ∈ N, depending only on M and ε, such that

∥∥∥∥∥N∑n=0

n∑m=−n

(f, Y mn )L2(S2) Y

mn − f

∥∥∥∥∥∞

= maxx∈S2

∣∣∣∣∣N∑n=0

n∑m=−n

(f, Y mn )L2(S2) Y

mn (x) − f(x)

∣∣∣∣∣ ≤ ε

for all N ≥ N0 and all f ∈ C1(S2) with ‖f‖1,∞ = max‖f‖∞, ‖f ′‖∞ ≤ M , and, analo-gously, for (2.20b). Here, the space C1(S2) consists of those functions f which are continu-ously differentiable in [0, π]× [0, 2π] with respect to the spherical variables θ, ϕ.

Proof: First we prove the second part. Therefore, let f ∈ C1(S2) with ‖f‖1,∞ ≤ M . Withthe addition formula we have for the partial sum

SN(f)(x) =N∑n=0

n∑m=−n

(f, Y mn )L2(S2) Y

mn (x) =

∫S2

f(y)N∑n=0

n∑m=−n

Y mn (x)Y −mn (y) ds(y)

=N∑n=0

∫S2

2n+ 1

4πPn(x · y) f(y) ds(y) .

This yields already the equivalence of (2.20a) and (2.20b). With (2n+ 1)Pn = P ′n+1 − P ′n−1

of Lemma ?? (set P−1 ≡ 0) this yields

SN(f)(x) =1

4π

N∑n=0

∫S2

f(y)[P ′n+1(x · y)− P ′n−1(x · y)

]ds(y)

=1

4π

∫S2

f(y)[P ′N+1(x · y) + P ′N(x · y)

]ds(y) .

Let again x = (0, 0, 1)> be the north pole and the orthogonal matrix A (depending on x)such that Ax = x. Then

SN(f)(x) =1

4π

∫S2

f(Ay)[P ′N+1(x · y) + P ′N(x · y)

]ds(y) .


In spherical polar coordinates this is, defining F (θ) := 12π

2π∫0

(f A)(θ, ϕ) dϕ ,

SN(f) =1

2

π∫0

F (θ)[P ′N+1(cos θ) + P ′N(cos θ)

]sin θ dθ

=1

2

+1∫−1

F (arccos t)[P ′N+1(t) + P ′N(t)

]dt

=1

2F (arccos t)

[PN+1(t) + PN(t)

]∣∣∣∣+1

−1︸︷︷︸=F (0)

− 1

2

+1∫−1

d

dtF (arccos t)

[PN+1(t) + PN(t)

]dt .

We note that partial integration is allowed because t 7→ F (arccos t) is continuously differ-entiable in (−1, 1) and continuous in [−1, 1]. θ = 0 corresponds to the north pole y = x,thus

F (0) =1

2π

2π∫0

f(Ax) dϕ = f(Ax) = f(x) ,

and therefore for arbitrary δ ∈ (0, 1):

∣∣SN(f)(x)− f(x)∣∣ =

1

2

∣∣∣∣∣∣+1∫−1

d

dtF (arccos t)

[PN+1(t) + PN(t)

]dt

∣∣∣∣∣∣≤ 1

2

−1+δ∫−1

+

1−δ∫−1+δ

+

1∫1−δ

∣∣∣∣ ddtF (arccos t)

∣∣∣∣ (|PN+1(t)|+ |PN(t)|)dt

= I1 + I2 + I3 .

We estimate these contributions separately. First we use that∣∣Pn(t)

∣∣ ≤ 1 for all t ∈ [−1, 1]and n ∈ N (see Lemma 2.6).

I1 + I3 ≤ −−1+δ∫−1

−1∫

1−δ

|F ′(arccos t)| ddt

arccos t dt (because ddt

arccos t < 0)

≤ ‖F ′‖∞[π − arccos(−1 + δ) + arccos(1− δ)

]≤ M

[π − arccos(−1 + δ) + arccos(1− δ)

].

Let now ε > 0 be given. Choose δ such that I1 + I3 ≤ ε2. Then δ depends only on M and ε.


With this choise of δ we consider I2 and use the inequality of Cauchy-Schwarz.

I2 ≤ max−1+δ≤t≤1−δ

∣∣∣∣ ddtF (arccos t)

∣∣∣∣︸︷︷︸=:c

+1∫−1

1 ·(|PN+1(t)|+ |PN(t)|

)dt

≤√

2 c(‖PN+1‖L2 + ‖PN‖L2

)≤ 2

√2 c

√2

2N + 1.

We estimate c by

c ≤ ‖F ′‖∞ max−1+δ≤t≤1−δ

∣∣∣∣ ddt arccos t

∣∣∣∣ ≤ M max−1+δ≤t≤1−δ

∣∣∣∣ ddt arccos t

∣∣∣∣ .Now we can choose N0 (only depending on ε and M) such that I2 ≤ ε

2for all N ≥ N0.

Therefore, |SN(f)(x) − f(x)| ≤ ε for all N ≥ N0. This proves uniform convergence of theFourier series.

The first part is proven by an approximation argument. Indeed, we use the fact (which isa general property of orthonormal systems) that SN(f) is the best approximation of f inspanY m

n : |m| ≤ n, n = 0, . . . , N; that is,

‖SN(f)− f‖L2(S2) ≤ ‖g − f‖L2(S2) for all g ∈ spanY mn : |m| ≤ n, n = 0, . . . , N .

Let now f ∈ L2(S2) and ε > 0 be given. Since the space C1(S2) is dense in L2(S2) thereexists h ∈ C1(S2) such that ‖h− f‖L2(S2) ≤ ε/2. Therefore,

‖SN(f)− f‖L2(S2) ≤ ‖SN(h)− f‖L2(S2) ≤ ‖SN(h)− h‖L2(S2) + ‖h− f‖L2(S2)

≤√

4π ‖SN(h)− h‖∞ + ‖h− f‖L2(S2) ≤√

4π ‖SN(h)− h‖∞ +ε

2.

Since SN(h) converges uniformly to h we can find N0 ∈ N such that the first part is less thanε/2 for all N ≥ N0 which ends the proof. 2

As a corollary we prove completeness of the Legendre polynomials.

Corollary 2.20 The polynomials√

n+ 1/2Pn : n ∈ N0

form a complete orthonormal

system in L2(−1, 1); that is for any f ∈ L2(−1, 1) there holds

f =∞∑n=0

fn Pn with fn =

(n+

1

2

) 1∫−1

f(t)Pn(t) dt , n ∈ N0 .

For f ∈ C1[−1, 1] the series converges uniformly.

Proof: The function g(x) = f(x3) = f(cos θ) can be considered as a function on the spherewhich in independent of ϕ, thus it is a zonal function. The expansion (2.20a) yields

f(x3) =∞∑n=0

n∑m=−n

amn Ymn (x)


with

amn = (g, Y mn )L2(S2) =

√2n+ 1

4π

(n−m)!

(n+m)!

π∫0

2π∫0

f(cos θ)P |m|n (cos θ) eimϕ sin θ dϕ dθ

=

0 , m 6= 0 ,√(2n+ 1)π

π∫0

f(cos θ)Pn(cos θ) sin θ dθ , m = 0 ,

=

0 , m 6= 0 ,√

(2n+ 1)π1∫−1

f(t)Pn(t) dt , m = 0 ,

and thus

f(x3) =∞∑n=0

√4π

2n+ 1fn Y

0n (x) =

∞∑n=0

fn Pn(x3)

and ends the proof. 2

2.4 The Boundary Value Problem for the Laplace Equa-

tion in a Ball

### Text ###

Now we consider the Dirichlet problem for the Laplace equation in balls; that is,

∆u = 0 in B(0, R) , u = f on ∂B(0, R) . (2.21)

We study the cases fR ∈ L2(S2) and fR ∈ C2(S2) simultanously where we have set fR(x) =f(Rx), x ∈ S2, here and in the following.

Theorem 2.21 (a) For given fR ∈ L2(S2) there exists a unique solution u ∈ C2(B(0, R)

)of ∆u = 0 in B(0, R) with

limr→R‖u(r, ·)− fR‖L2(S2) = 0 .

The solution is given by the series

u(rx) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

( rR

)nY mn (x) (2.22a)

=1

4π

∞∑n=0

(2n+ 1)( rR

)n ∫S2

f(Ry)Pn(x · y) ds(y) (2.22b)

for x = rx ∈ B(0, R). They converge uniformly on every compact ball B[0, R′] for anyR′ < R.

2.4. THE BOUNDARY VALUE PROBLEM FOR THE LAPLACE EQUATION IN A BALL47

(b) If fR ∈ C2(S2) there exists a unique solution u ∈ C2(B(0, R)

)∩ C

(B[0, R]

)of (2.21)

which is again given by (2.22a), (2.22b). The series converge uniformly on B[0, R].

Proof: First we note that the series coincide by the addition formula as shown in the proofof Theorem 2.19.(a) To show uniqueness we assume that u is the difference of two solutions. Then ∆u = 0in B(0, R) and limr→R ‖u(r, ·)‖L2(S2) = 0. For every r ∈ (0, R) the function x 7→ u(rx) is inC2(S2), and, therefore, can be expanded into a series by Theorem 2.19; that is,

u(rx) =∞∑n=0

n∑m=−n

umn (r)Y mn (x) , x ∈ S2 .

The coefficients are given by umn (r) =(u(r, ·), Y m

n

)L2(S2)

. We show that umn satisfies a differ-

ential equation of Euler type. Using (??) for u, partial integration, and (??), yields

d

dr

(r2 d

dr

)umn (r) =

2π∫0

π∫0

∂

∂r

(r2 ∂u(r, θ, ϕ)

∂r

)Y −mn (θ, ϕ) sin θ dθ dϕ

= −2π∫

0

π∫0

[∂

∂θ

(sin θ

∂u(r, θ, ϕ)

∂θ

)+

1

sin θ

∂2u(r, θ, ϕ)

∂ϕ2

]Y −mn (θ, ϕ) dθ dϕ

= −2π∫

0

π∫0

[∂

∂θ

(sin θ

∂Y −mn (θ, ϕ)

∂θ

)+

1

sin θ

∂2Y −mn (θ, ϕ)

∂ϕ2

]u(r, θ, ϕ) dθ dϕ

= n(n+ 1)

2π∫0

π∫0

Y −mn (θ, ϕ)u(r, θ, ϕ) sin θ dθ dϕ = n(n+ 1)umn (r) .

The only smooth solution of the Euler differential equation is given by umn (r) = amn rn for

arbitrary amn . Therefore, u has the form

u(rx) =∞∑n=0

n∑m=−n

amn rn Y m

n (x) , x ∈ S2 .

Let R0 < R and ε > 0 be arbitrary. Choose Rε ∈ [R0, R) such that ‖uε‖L2(S2) ≤ ε whereuε(x) = u(Rεx). Multiplying the representiation of u(Rεx) with Y −qp (x) and integrating overS2 yields (uε, Y

qp )L2(S2) = aqpR

pε, thus

u(rx) =∞∑n=0

n∑m=−n

(uε, Ymn )L2(S2)

(r

Rε

)nY mn (x) x ∈ S2 , r ≤ Rε .

Therefore, for r ≤ R0,

‖u(r, ·)‖2L2(S2) =

∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2 ( r

Rε

)2n

≤∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2= ‖uε‖2

L2(S2) ≤ ε2 .


Since this holds for all ε > 0 we conclude that u has to vanish in B(0, R0). Since R0 < Rwas arbitrary u vanishes in B(0, R).To show that the series is indeed the solution we show uniform convergence on compactsubsets of B(0, R). Using |Pn(t)| ≤ 1 for |t| ≤ 1 and the inequality of Cauchy-Schwarz, theseries (2.22b) is dominated by the power series

‖f‖L2(S2)√4π

∞∑n=0

(2n+ 1)( rR

)nwith radius of convergence R. In the same way, all of the derivatives converge uniformly aswell. Finally, we use the Parseval identity, applied to the series

u(rx) − f(Rx) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

[( rR

)n− 1]Y mn (x) ;

that is,

‖u(r, ·)− fR‖2L2(S2) =

∞∑n=0

n∑m=−n

∣∣(fR, Y mn )L2(S2)

∣∣2 [( rR

)n− 1]2

.

This term tends to zero as r → R which is shown by standard arguments: For every n the

term∣∣(fR, Y m

n )L2(S2)

∣∣2 [( rR

)n − 1]2

tends to zero as r tends to R. Furthermore, it is bounded

by the summable term∣∣(fR, Y m

n )L2(S2)

∣∣2 (uniformly with respect to r).

(b) It remains to show that the series converges uniformly in B[0, R]. With equation (??),written in the form ∆SY

mn + n(n + 1)Y m

n = 0, and the symmetry of ∆S we express theexpansion coefficient of f as

(fR, Ymn )L2(S2) = − 1

n(n+ 1)(fR,∆SY

mn )L2(S2) = − 1

n(n+ 1)(∆SfR, Y

mn )L2(S2) ,

thus

∞∑n=0

n∑m=−n

∣∣(fR, Y mn )L2(S2)

∣∣ ( rR

)n|Y mn (x)| ≤

∞∑n=0

n∑m=−n

∣∣(∆SfR, Ymn )L2(S2)

∣∣ 1

n(n+ 1)|Y mn (x)|

≤

[∞∑n=0

n∑m=−n

∣∣(∆SfR, Ymn )L2(S2)

∣∣2]1/2 [ ∞∑n=0

n∑m=−n

1

n2(n+ 1)2|Y mn (x)|2

]1/2

= ‖∆SfR‖L2(S2)

[∞∑n=0

1

n2(n+ 1)2

n∑m=−n

|Y mn (x)|2

]1/2

=1√4π‖∆SfR‖L2(S2)

[∞∑n=0

2n+ 1

n2(n+ 1)2

]1/2

where we have also used part (a) of Corollary 2.18. 2

2.5. BESSEL FUNCTIONS 49

To end this section we consider the corresponding exterior boundary value problem.Let again B(0, R) be the open ball with radius R > 0 and center 0 and f : ∂B(), R) −→ C.We want to determine a function u defined in the exterior of B(0, R) such that

∆u = 0 in R3 \B[0, R] , u = f on ∂B(0, R) . (2.23)

We need a further condition to ensure uniqueness. Indeed, we observe that the functionu(x) = u(r, ϕ, θ) =

[(r/R)n− (R/r)n+1

]Pn(cos θ) is harmonic in the exterior of B(0, R) and

vanishes for r = R.

To ensure uniqueness we require that the solution tends zero at infinity; that is

limr→∞

u(rx) = 0 uniforml with respect to x ∈ S2 . (2.24)

Again, we study the cases fR ∈ L2(S2) and fR ∈ C2(S2) simultanously where fR(x) = f(Rx),x ∈ S2, as before.

Theorem 2.22 (a) For given fR ∈ L2(S2) there exists a unique solution u ∈ C2(R3 \

B[0, R])

of ∆u = 0 in R3 \B[0, R] which satisfies (2.24) and

limr→R‖u(r, ·)− fR‖L2(S2) = 0 .

The solution is given by the series

u(rx) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

(R

r

)n+1

Y mn (x) (2.25a)

=1

4π

∞∑n=0

(2n+ 1)

(R

r

)n+1 ∫S2

f(Ry)Pn(x · y) ds(y) (2.25b)

for x = rx ∈ R3 \B[0, R]. They converge uniformly for r ≥ R′ for any R′ > R.

(b) If fR ∈ C2(S2) there exists a unique solution u ∈ C2(R3 \B[0, R]

)∩C

(R3 \B(0, R)

)of

(2.23), (2.24) which is again given by (2.25a), (2.25b). The series converge uniformly forr ≥ R.

The proof is almost the same as for the interior case. In the uniqueness part one has toselect the solution umn (r) = amn r

−n−1 because of the requirement that the solution has todecay at infinity. We omit the further details of the proof. 2

2.5 Bessel Functions

From now on we consider the Helmholtz equation instead of the Laplace equation. Inspherical polar coordinates it takes the form

1

r2

∂

∂r

(r2∂u

∂r

)+

1

r2 sin2 θ

∂2u

∂ϕ2+

1

r2 sin θ

∂

∂θ

(sin θ

∂u

∂θ

)+ k2 u = 0 . (2.26)


We make again an ansatz in the form

u(r, θ, ϕ) = v1(r) v2(ϕ) v3(θ)

and substitute this into (2.26). This yields (after multiplication by r2/u)

r2v′′1(r) + 2r v′1(r)

v1(r)+

v′′2(ϕ)

sin2 θ v2(ϕ)+

sin θ v′′3(θ) + cos θ v′3(θ)

sin θ v3(θ)+ k2r2 = 0 .

This corresponds to (??) for the Laplace equation. By the same arguments one observesthat v2 and v3 satisfies (??) for a constant of the form γ = −n(n + 1), n ∈ N, while (??) isreplaced by

r2v′′1(r) + 2r v′1(r) +[k2r2 − n(n+ 1)

]v1(r) = 0 . (2.27)

Making the substitution v1(r) = w(kr) yields the following spherical Bessel differentialequation for w(z),

z2w′′(z) + 2z w′(z) +(z2 − n(n+ 1)

)w(z) = 0 . (2.28)

We investigate this linear differential equation of second order for arbitrary z ∈ C. For z 6= 0the differential equation is equivalent to

w′′(z) +2

zw′(z) +

(1− n(n+ 1)

z2

)w(z) = 0 in C \ 0 . (2.29)

The coefficients of this differential equation are holomorphic in C\0 and have poles of firstand second order at 0. As in the case of real z one can show that in every simply connecteddomain Ω ⊂ C \ 0 there exist at most two holomorphic solutions of (2.29).

Lemma 2.23 Let Ω ⊂ C \ 0 be a domain. Then there exists at most two holomorphicsolutions of (2.29) in Ω.

Proof: Let wj, j = 1, 2, 3, three solutions. The space C2 is of dimension 2 over the field C.

Therefore, if we fix some x0 ∈ Ω there exist αj ∈ C, j = 1, 2, 3, such that3∑j=1

|αj| 6= 0 and

3∑j=1

αj

(wj(x0)w′j(x0)

)= 0 .

Set w :=3∑j=1

αjwj in Ω. Then w(x0) = w′(x0) = 0 and thus from (2.29), also w(j)(x0) = 0

for all j = 0, 1, . . .. Because w is holomorphic in the domain Ω we conclude by the identitytheorem for holomorphic functions that w vanishes in all of Ω. Therefore, w1, w2, w3 arelinearly dependent. 2

Motivated by the solution v1(r) = rn of the correponding equation (??) for the Laplaceequation and γ = −n(n+ 1) we make an ansatz for a (smooth) solution of (2.29) as a powerseries in the form

w1(z) = zn∞∑`=0

a` z` =

∞∑`=0

a` z`+n .

Substituting the ansatz into (2.29) and comparing the coefficients yields


(a)[(n+ `)(n+ `+ 1)− n(n+ 1)

]a` = 0 for ` = 0, 1, and

(b)[(n+ `)(n+ `+ 1)− n(n+ 1)

]a` + a`−2 = 0 for ` ∈ N, ` ≥ 2.

Therefore, a1 = 0, and from (b) it follows that a` = 0 for all odd `. For even ` we replace `by 2` and arrive at

a2` = − 1

(n+ 2`)(n+ 2`+ 1)− n(n+ 1)a2(`−1) = − 1

2` (2n+ 2`+ 1)a2(`−1)

= −1

`(n+ `)

1

(2n+ 2`)(2n+ 2`+ 1)

and thus by induction

a2` =(2n+ 1)!

n!

(−1)`

`!

(n+ `)!

(2n+ 2`+ 1)!a0 for all ` ≥ 0 .

Altogether we have that

w1(z) =(2n+ 1)!

n!a0 z

n

∞∑`=0

(−1)`

`!

(n+ `)!

(2n+ 2`+ 1)!z2` .

By the ratio test it is easily seen that the radius of convergence ins infinity. Therefore, thisfunction w is a holomorphic solution of the Bessel differential equation in all of C.

Now we determine a second solution of (2.29) which is linearly independent of jn. Motivatedby the singular solution v1(r) = r−n−1 of (??) for γ = −n(n + 1) we make an ansatz of theform

w2(z) = z−n−1

∞∑`=0

a` z` =

∞∑`=0

a` z`−n−1 .

Substituting the ansatz into (2.29) and comparing the coefficients yields

(a)[(`− n)(`− n− 1)− n(n+ 1)

]a` = 0 for ` = 0, 1, and

(b)[(`− n)(`− n− 1)− n(n+ 1)

]a` + a`−2 = 0 for ` ∈ N, ` ≥ 2.

We again set a` = 0 for all odd `. For even ` we replace ` by 2` and arrive at

a2` = − 1

(2`− n)(2`− n− 1)− n(n+ 1)a2(`−1) =

1

(2`) (2n− 2`+ 1)a2(`−1)

and thus by induction

a2` =1

2` `!

1

(2n− 2`+ 1)(2n− 2`+ 3) · · · (2n− 1)a0 for all ` ≥ 1 .


To simplify this expression we look first for ` ≥ n. Then

(2n− 2`+ 1)(2n− 2`+ 3) · · · (2n− 1) = (2n− 2`+ 1)(2n− 2`+ 3) · · · (−1) · 1 · 3 · · · (2n− 1)

= (−1)`−n (2`− 2n− 1)!! (2n− 1)!!

= (−1)`−n(2`− 2n)!

2 · 4 · · · 2(`− n)

(2n)!

2 · 4 · · · (2n)

= (−1)`−n(2`− 2n)!

2`−n(`− n)!

(2n)!

2n 2n!

where we have used the symbol k!! = 1 · 3 · 5 · · · k for any odd k.Now we consider the case ` < n. Analogously to above we have

(2n− 2`+ 1)(2n− 2`+ 3) · · · (2n− 1) =(2n− 1)!!

(2n− 2`− 1)!!=

(2n)!

2n n!

2n−`(n− `)!(2n− 2`)!

.

Therefore, we arrive at the second solution

w2(z) = a0 z−n−1 n!

(2n)!

∞∑`=0

1

`!

(`− n)!

(2`− 2n)!z2` ,

where we have set(`− n)!

(2`− 2n)!:= (−1)n−`

(2n− 2`)!

(n− `)!for ` < n .

The radius of convergence of the series is again infinity. For particular normalizations thefunctions w1 and w2 are called spherical Bessel functions.

Definition 2.24 For all z ∈ C the Bessel and Hankel functions of first and secondkind and order n ∈ N0 are defined by

jn(z) = (2z)n∞∑`=0

(−1)`

`!

(n+ `)!

(2n+ 2`+ 1)!z2` , z ∈ C ,

yn(z) =2 (−1)n+1

(2z)n+1

∞∑`=0

(−1)`

`!

(`− n)!

(2`− 2n)!z2` , z ∈ C ,

h(1)n = jn + i yn ,

h(2)n = jn − i yn ,

where (in the definition of yn) a quanity (−k)!(−2k)!

for positive integers is defined by

(−k)!

(−2k)!= (−1)k

(2k)!

k!, k ∈ N .

For many applications the Wronskian is important.


Theorem 2.25 For all n ∈ N0 and z ∈ C \ 0 we have

W (jn, yn)(z) := jn(z) y′n(z) − j′n(z) yn(z) =1

z2.

Proof: We write W (z) for W (jn, yn)(z). We multiply the spherical Bessel differential equa-tion (2.28) for yn by jn and the one for jn by yn and subtract. This yields

y′′n(z) jn(z)− j′′n(z) yn(z) +2

z

[y′n(z) jn(z)− j′n(z) yn(z)

]= 0 .

The first term is just W ′(z), thus W solves the ordinary differential equation W ′(z) +2zW (z) = 0. The general solution is W (z) = cz−2 for some c ∈ R which we determine

from the leading coefficient of the Laurant series of W . By the definition of the Besselfunctions we have

jn(z) = 2n zn n!(2n+1)!

[1 +O(z2)

], j′n(z) = n 2n zn−1 n!

(2n+1)!

[1 +O(z2)

]yn(z) = −2−n z−n−1 (2n)!

n!

[1 +O(z2)

], y′n(z) = (n+ 1) 2−n z−n−2 (2n)!

n!

[1 +O(z2)

]for z → 0. Therefore,

W (z) =

[(n+ 1) z−2 (2n)!

(2n+ 1)!+ n z−2 (2n)!

(2n+ 1)!

] [1 +O(z2)

]=

1

z2

[1 +O(z2)

]which proves that c = 1. 2

Remark: From this theorem the linear independence of jn, yn follows immediately and

thus also the linear independence of jn, h(1)n . Therefore, they span the solution space of

the differential equation (2.29).

The functions jn and yn are closely related to the functions sin z/z and cos z/z.

Theorem 2.26 (Rayleigh’s Formulas)

For any z ∈ C and n ∈ N0 we have

jn(z) = (−z)n(

1

z

d

dz

)nsin z

z,

yn(z) = −(−z)n(

1

z

d

dz

)ncos z

z,

h(1)n (z) = −i (−z)n

(1

z

d

dz

)nexp(iz)

z,

h(2)n (z) = i (−z)n

(1

z

d

dz

)nexp(−iz)

z.


Proof: We prove only the form for yn. The representation for jn is analogous (even simpler),the ones for the Hankel functions follow imediately. We start with the series expansion ofcos z/z; that is,

cos z

z=

∞∑`=0

(−1)`

(2`)!z2`−1 ,

and observe that the action of(

1zddz

)non a power of z is given by(

1

z

d

dz

)nz2`−1 = (2`− 1)(2`− 3) · · · (2`− 2n+ 1) z2`−2n−1 ,

thus

−(−z)n(

1

z

d

dz

)ncos z

z=

(−1)n+1

zn+1

∞∑`=0

(−1)`

(2`)!, (2`− 1)(2`− 3) · · · (2`− 2n+ 1) z2` .

We discuss the term

q =(2`− 1)(2`− 3) · · · (2`− 2n+ 1)

(2`)!

separately for ` ≥ n and ` < n.For ` ≥ n we have, using again the notation k!! = 1 · 3 · · · k for odd k,

q =(2`− 1)!!

(2`)! (2`− 2n− 1)!!=

(2`)!

2` `!

2`−n (`− n)!

(2`)! (2`− 2n)!=

1

2n `!

(`− n)!

(2`− 2n)!.

The case ` < n is seen analogously by splitting

q =(2`− 1)(2`− 3) · · · 1 · (−1) · · · (2`− 2n+ 1)

(2`)!=

(2`− 1)!! (−1)n−` (2n− 2`− 1)!!

(2`)!

= (−1)n−`(2`)!

2` `! (2`)!

(2n− 2`)!

2n−` (n− `)!=

1

2n `!(−1)n−`

(2n− 2`)!

(n− `)!

This proves the formula for cos z/z. 2

We note that j0(z) = sin z/z and ih(1)0 (z) = exp(iz)/z.

We collect our results in the following definition.

Definition 2.27 For any n ∈ N0 and m ∈ Z with |m| ≤ n and k ∈ C with Im k ≥ 0 thefunctions

umn (rx) = h(1)n (kr)Y m

n (x) ,

are called spherical wave functions. They are solutions of ∆u + k2u = 0 in R3 \ 0.The real part

Re umn (rx) = jn(kr)Y mn (x) ,

satisfies the Helmholtz equation in all of R3.


The Bessel functions of the first kind jn correspond to the functions rn in the static case;that is k = 0, and (rx 7→ jn(kr)Y m

n (x) are the expansion functions for solutions inside of

bounded balls. The functions h(1)n are used to derive expansions in the exterior of balls. From

the definition we observe that they are singular at the origin of order n. The asymptoticform for r →∞ can be derived from the previous theorem.

Theorem 2.28 For every n ∈ N and z ∈ C we have

h(1)n (z) =

exp(iz)

z

[(−i)n+1 + O

(1

|z|

)]for |z| → ∞ ,

d

dzh(1)n (z) =

exp(iz)

z

[(−i)n + O

(1

|z|

)]for |z| → ∞ ,

uniformly with respect to z/|z|.

Proof: We show by induction that for every n ∈ N0 there exists a polynomial Pn of degreeat most n such that Pn(0) = 1 and(

1

z

d

dz

)nexp(iz)

z= in

exp(iz)

zn+1Pn

(1

z

). (2.30)

This would prove the assertions.For n = 0 this is obvious for the constant polynomial P0 = 1. Let it be true for n. Then(

1

z

d

dz

)n+1exp(iz)

z=

1

z

d

dz

[in

exp(iz)

zn+1Pn

(1

z

)]=

in

zeiz[

i

zn+1Pn

(1

z

)− (n+ 1)

1

zn+2Pn

(1

z

)− 1

zn+1P ′n

(1

z

)1

z2

]= in+1 exp(iz)

zn+2Pn+1

(1

z

)with Pn+1(t) = Pn(t) + i(n+ 1) t Pn(t) + it2 P ′n(t). Obviously, Pn+1 is a polynomial of order

at most n+ 1 and Pn+1(0) = 1. This proves the asymptotic forms of h(1)n and its derivative.

2

For the expansion of solutions of the Helmholtz equation into spherical wave functions alsothe asymptotic behaviour with respect to n is necessary.

Theorem 2.29 For every ε > 0 and R > 0

jn(z) =n!

(2n+ 1)!(2z)n

[1 +O

(1

n

)]=

1

(2n+ 1)!!zn[1 +O

(1

n

)]for n→∞ ,

yn(z) = −(2n)!

n!

2

(2z)n+1

[1 +O

(1

n

)]= −(2n− 1)!!

zn+1

[1 +O

(1

n

)]for n→∞ ,

uniformly with respect to |z| ≤ R and uniformly with respect to ε ≤ |z| ≤ R, respectively.Here again, k!! = 1 · 3 · · · k for odd k ∈ N.


Proof: By the definition of jn we have∣∣∣∣(2n+ 1)!

n!(2z)−n jn(z) − 1

∣∣∣∣ ≤ ∞∑`=1

R2`

`!

(2n+ 1)! (n+ `)!

n! (2n+ 2`+ 1)!≤ 1

2n+ 1

∞∑`=1

R2`

`!

because

(2n+ 1)! (n+ `)!

n! (2n+ 2`+ 1)!=

(n+ 1) · · · (n+ `)

(2n+ 2) · · · (2n+ `+ 1)︸︷︷︸≤ 1

· 1

(2n+ `+ 2) · · · (2n+ 2`+ 1)︸︷︷︸≤ 1/(2n+1)` ≤ 1/(2n+1)

.

Analogously, we have for yn:∣∣∣∣ n!

2 (2n)!(2z)n+1 yn(z) + 1

∣∣∣∣ ≤ n−1∑`=1

R2`

`!

n! (2n− 2`)!

(2n)! (n− `)!+

∞∑`=n

R2`

`!

n! (`− n)!

(2n)! (2`− 2n)!

≤ 1

n+ 1

∞∑`=1

R2`

`!,

because

n! (2n− 2`)!

(2n)! (n− `)!=

(n− `+ 1) · · · (2n− 2`)

(n+ 1) · · · (2n)=

1

(n+ 1) · · · (n+ `)︸︷︷︸≤ 1/(n+1)` ≤ 1/n+1

· (n− `+ 1) · · · (2n− 2`)

(n+ `+ 1) · · · (2n)︸︷︷︸≤ 1

for 1 ≤ ` < n and

n! (`− n)!

(2n)! (2`− 2n)!=

n!

(2n)!· (`− n)!

(2`− 2n)!≤ 1

n+ 1

for ` ≥ n. 2

2.6 The Boundary Value Problems for the Helmholtz

Equation for a Ball

Now we are ready to study the series expansion of solutions of the boundary value problemfor the Helmholtz equation ∆u+ k2u = 0 inside and outside of balls. Using the asymptoticproperties of jn we are able to transfer Theorem 2.21 to the interior boundary valueproblem

∆u+ k2u = 0 in B(0, R) , u = f on ∂B(0, R) . (2.31)

We recall that we transform functions on the sphere : |x| = R of radius R onto the unitsphere by setting fR(x) = f(Rx), x ∈ S2.

Theorem 2.30 Assume that k ∈ C with Im k ≥ 0 and R > 0 are such that jn(kR) 6= 0 forall n ∈ N0.

2.6. THE BOUNDARY VALUE PROBLEMS FOR THE HELMHOLTZ EQUATION FOR A BALL57

(a) For given fR ∈ L2(S2) there exists a unique solution u ∈ C2(B(0, R)

)of ∆u + k2u = 0

in B(0, R) with

limr→R‖u(r, ·)− fR‖L2(S2) = 0 .

The solution is given by

u(rx) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

jn(kr)

jn(kR)Y mn (x) (2.32a)

=1

4π

∞∑n=0

(2n+ 1)jn(kr)

jn(kR)

∫S2

f(Ry)Pn(x · y) ds(y) . (2.32b)

The series converge uniformly on compact subsets of B(0, R).

(b) If fR ∈ C2(S2) there exists a unique solution u ∈ C2(B(0, R)

)∩ C

(B[0, R]

)of (2.31)

which is again given by (2.32a), (2.32b). The series converges uniformly on B[0, R].

Proof: We can almost literally copy the proof of the corresponding Theorem 2.21 for theLaplace equation. The ratio (r/R)n has to be replaced by jn(kr)/jn(kR) which asymptoti-cally behaves as (r/R)n. We do not carry out the details but refer to Theorem 2.33 belowfor the corresponding exterior boundary value problem. 2

We observe an important difference between the expansion functions rnY mn (x) for the Laplace

equation and jn(kr)Y mn (x) for the Helmholtz equation. From the asymptotic form of jn(z)

for real and positive z (see Theorem 2.28) we conclude that for every n ∈ N0 there existinfinitely many real and positive wave numbers kn,1, kn,2, . . . with jn(kn,jR) = 0. Therefore,the functions

umn,j(rx) = jn(kn,jr)Ymn (x) , n, j ∈ N , |m| ≤ n ,

solve ∆umn,j + k2n,ju

mn,j = 0 in B[0, R] and umn,j = 0 on the boundary |x| = R. Therefore,

k2n,j are the eigenvalues of −∆ in the ball of radius Rwith respect to Dirichlet boundary

conditions. The multiplicity is 2n + 1, and umn,j = 0, m = −n, . . . , n, are the correspondingeigenfunctions.

In the following theorem we expand the special solution u(rx) = exp(ikr x·y), x = rx ∈ R3,(for some fixed y ∈ S2) of the Helmholtz equation.

Theorem 2.31 (Jacobi-Anger expansion)

eikr x·y = 4π∞∑n=0

n∑m=−n

in jn(kr)Y mn (x)Y −mn (y) =

∞∑n=0

in (2n+ 1) jn(kr)Pn(x · y)

for x, y ∈ S2 and r > 0. For every R > 0 the series converges uniformly with respect tox, y ∈ S2 and 0 ≤ r ≤ R.


Proof: It is sufficient to prove the second formula. Since both sides depend only on theangle between x and y we can assume without loss of generality that y = (0, 0, 1)>. Thenx 7→ exp(ikx · y) = exp(ikx3) = exp(ikr cos θ) is a zonal1 solution of the Helmholtz equationin all of R3 and thus, by the the expansion thorem,

eikr cos θ =∞∑n=0

an jn(kr)Pn(cos θ)

for coefficients an which are given by (see Corollary 2.20)

an jn(kr) =

(n+

1

2

) 1∫−1

eikrt Pn(t) dt =

(n+

1

2

) ∞∑m=0

1

m!(ikr)m

1∫−1

tm Pn(t) dt . (2.33)

We determine an from the asymptotic behaviour as kr tends to zero. First we note that∫ 1

−1tm Pn(t) dt = 0 for all m < n since Pn is orthogonal to the space of polynomials of degree

at most n− 1 (see again Corollary 2.20). Furthermore, from Theorem 2.7 we note that

Pn(t) =1

2n n!

dn

dtn(t2 − 1)n =

1

2n n!

dn

dtn

n∑j=0

(n

j

)(−1)n−j t2j

=1

2n n!(2n)(2n− 1) · · · (n+ 1) tn + Qn−1(t)

for some polynomial Qn−1 of degree at most n − 1. Solving this for tn and using again theorthogonality of the P` yields∫ 1

−1

tn Pn(t) dt =2n(n!)2

(2n)!

∫ 1

−1

Pn(t)2 dt =2n (n!)2

(2n)!

2

2n+ 1.

Therefore, the right hand side of (2.33) behaves as in(kr)n 2n n!(2n)!

+O((kr)n+1

)as kr tends to

zero. The asymptotics of the left hand side of (2.33) is given by the definition of jn andyields an(2kr)n 2n n!

(2n)!+ O

((kr)n+1

). Comparing the coefficient yields an = in(2n + 1) which

ends the proof. 2

We now continue with a boundary value problem for the exterior of a ball. In principle we canexpand the solution of the problem by the spherical wave functions umn (rx) = jn(kr)Y m

n (x) orvmn (rx) = yn(kr)Y m

n (x) – or some combination of them. In order to motivate the physicallycorrect choice we consider the case of a conducting medium with ε = ε0 and µ = µ0 andconstant σ > 0. We have seen from the previous chapter that in this case k2 has to bereplaced by k2 = k2εr with εr = 1+ iσ/(ωε0). For k itself we take the branch with Re k > 0,thus also Im k > 0. From Theorem 2.28 we observe that

jn(kr)Y mn (x) =

fn(kr)

krY mn (x)

[1 + O

(1

r

)]for r →∞ ,

1that is, does not depend on ϕ


where fn(z) = cos z or fn(z) = sin z. In any case we observe that for Im k > 0 these functionsincrease exponentially as r tends to infinity. The same holds for the functions yn(kr)Y m

n (x).

The expansion functions h(1)n (kr)Y m

n (x), however, have the asymptotic forms

h(1)n (kr)Y m

n (x) =exp(ikr)

krY mn (x)

[1 + O

(1

r

)]for r →∞ ,

and these functions decay exponentially at infinity. This is physically plausible since aconducting medium is absorbing! We note by passing that the Hankel functions h

(2)n of the

second kind are exponentially increasing. Therefore, for conducting media the expansionfunctions h

(1)n (kr)Y m

n (x) are the correct ones. In the limiting case σ → 0 the solution shoulddepend continuously on σ. This makes it plausible to choose these functions also for thecase σ = 0. The requirement that the functions u are bounded for Im k > 0 and dependcontinuously on k for Im k → 0 on compact subsets of R3 is called the limiting absorptionprinciple.

Comparing h(1)n with its derivative yields:

Lemma 2.32 Let n ∈ N0 and m ∈ Z with |m| ≤ n. Then u(rx) = h(1)n (kr)Y m

n (x) satisfiesthe Sommerfeld radiation condition

∂u(rx)

∂r− ik u(rx) = O

(1

r2

)for r →∞ , (2.34)

uniformly with respect to x ∈ S2.

Proof: By Theorem 2.28 we have

d

drh(1)n (kr) − ik h(1)

n (kr) =exp(ikr)

kr

[in k − ik in−1 + O

(1

r

)]=

exp(ikr)

krO(

1

r

)= O

(1

r2

)for r →∞ .

2

The importance of this radiation condition lies in the fact that it can be formulated forevery function defined in the exterior of a bounded domain without making use of theHankel functions – and still provides the correct solution.

Now we consider the exterior boundary value problem to determine

∆u+ k2u = 0 for |x| > R , u = f for |x| = R , (2.35)

and u satisfies Sommerfeld radiation condition (2.34).

Theorem 2.33 (a) For given fR ∈ L2(S2) there exists a unique solution u ∈ C2(R3 \

B[0, R])

of ∆u+ k2u = 0 with

limr→R‖u(r, ·)− fR‖L2(S2) = 0 ,


and u satisfies Sommerfeld radiation condition (2.34). The solution is given by

u(rx) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

h(1)n (kr)

h(1)n (kR)

Y mn (x) (2.36a)

=1

4π

∞∑n=0

(2n+ 1)h

(1)n (kr)

h(1)n (kR)

∫S2

f(Ry)Pn(x · y) ds(y) . (2.36b)

The series converge uniformly on compact subsets of R3 \B[0, R].

(b) If fR ∈ C2(S2) there exists a unique solution u ∈ C2(R3 \B[0, R]

)∩C

(R3 \B(0, R)

)of

(2.35) which is again given by (2.36a), (2.36b). The series converges uniformly on B[0, R1]\B(0, R) for every R1 > R.

Proof: (a) The proof of uniqueness follows the lines of the corresponding part in the proofof Theorem 2.22. Indeed, let u satisfy the Helmholtz equation in the exterior of B[0, R] suchthat limr→R ‖u(r, ·)‖L2(S2) = 0. For every r ∈ (0, R) the function u(rx) can be expanded intoa series by Theorem 2.19; that is,

u(rx) =∞∑n=0

n∑m=−n

umn (r)Y mn (x) , x ∈ S2 .

The coefficients are given by umn (r) =(u(r, ·), Y m

n

)L2(S2)

. We show that umn satisfies the

spherical Bessel differential equation. Using the Helmholtz equation for u in spherical polarcoordinates and partial integration yields for r > R

d

dr

(r2 d

dr

)umn (r)+r2k2umn (r) =

2π∫0

π∫0

[∂

∂r

(r2 ∂u(r, θ, ϕ)

∂r

)+ r2k2u(r, θ, ϕ)

]Y −mn (θ, ϕ) sin θ dθ dϕ

= −2π∫

0

π∫0

[∂

∂θ

(sin θ

∂u(r, θ, ϕ)

∂θ

)+

1

sin θ

∂2u(r, θ, ϕ)

∂ϕ2

]Y −mn (θ, ϕ) dθ dϕ

= −2π∫

0

π∫0

[∂

∂θ

(sin θ

∂Y −mn (θ, ϕ)

∂θ

)+

1

sin θ

∂2Y −mn (θ, ϕ)

∂ϕ2

]u(r, θ, ϕ) dθ dϕ

= n(n+ 1)

2π∫0

π∫0

Y −mn (θ, ϕ)u(r, θ, ϕ) sin θ dθ dϕ = n(n+ 1)umn (r) .

The general solution of the spherical Bessel differential equation (after the transformation

z = kr) is given by umn (r) = amn h(1)n (kr) + bmn jn(kr) for arbitrary amn , b

mn . Therefore, u has

the form

u(rx) =∞∑n=0

n∑m=−n

[amn h

(1)n (kr) + bmn jn(kr)

]Y mn (x) , x ∈ S2 , r > R . (2.37)


From the radiation condition for u we conclude that every term umn (r) satisfies ddrumn (r) −

ik umn (r) = O(1/r2). Since only the Hankel functions of the first kind satisfy the radiationcondition in contrast to the Bessel functions we conclude that bmn = 0 for all n ∈ N0 and|m| ≤ n.Let R0 > R and ε > 0 be arbitrary. Choose Rε ∈ (R,R0] such that ‖uε‖L2(S2) ≤ ε whereuε(x) = u(Rεx). Multiplying the representiation of u(Rεx) with Y −qp (x) and integrating over

S2 yields (uε, Yqp )L2(S2) = aqp h

(1)p (kRε), thus

u(rx) =∞∑n=0

n∑m=−n

(uε, Ymn )L2(S2)

h(1)n (kr)

h(1)n (kRε)

Y mn (x) x ∈ S2 , r ≥ Rε .

Let now R1 > R0 be arbitrary. Since the asymptotic behavior of the Hankel function(Theorem 2.29) is uniform with respect to R0 ≤ |z| ≤ R1 there exists a constant c > 0 such

that∣∣h(1)n (kr)/h

(1)n (kRε)

∣∣ ≤ c for all R0 ≤ r ≤ R1. Therefore, for R0 ≤ r ≤ R1,

‖u(r, ·)‖2L2(S2) =

∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2 ∣∣∣∣∣ h(1)n (kr)

h(1)n (kRε)

∣∣∣∣∣2

≤ c2

∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2= c2‖uε‖2

L2(S2) ≤ c2ε2 .

Since this holds for all ε > 0 we conclude that u has to vanish in B(0, R1) \B(0, R0). SinceR1 > R0 > R were arbitrary u vanishes for |x| > R.The proof that the function u satisfies the radiation condition is more difficult and usesa result from Chapter 32 The problem is that every component of the series satisfies theradiation condition but the asymptotics of hn(k|x|) as |x| tends to infinity by Theorem 2.28does not hold uniformly with respect to n. We briefly sketch the idea how to overcome thisdifficulty. We want to express u as an integral over a sphere of the form

u(x) =

∫|y|=R0

Φ(x, y) g(y) ds(y) , |x| > R0 , (2.38)

rather than a series. Here, Φ denotes the fundamental solution of the Helmholtz equation,defined by

Φ(x, y) =exp(ik|x− y|)

4π |x− y|=

ik

4πh

(1)0 (k|x− y|) , x 6= y . (2.39)

Indeed, we fix R0 > R and x with |x| > R0. First, we apply Green’s second identity fromTheorem ?? to the functions w(y) = jn(k|y|)Y m

n (y) and Φ(x, y) inside the ball B(0, R0)which yields

jn(kR0)

∫|y|=R0

Y mn (y)

∂

∂ν(y)Φ(x, y) ds(y) − k j′n(kR0)

∫|y|=R0

Y mn (y) Φ(x, y) ds(y) = 0 .

(2.40a)

2At least the authors do not know of any proof which uses only the elementary results of the Hankelfunctions derived so far.


Second, since v(y) = h(1)n (k|y|)Y m

n (y) satisfies the radiation condition we can apply Green’srepresentation Theorem 3.6 from the forthcoming Chapter 3 to v in the exterior of B(0, R0)which yields

h(1)n (k|x|)Y m

n (x) = h(1)n (kR0)

∫|y|=R0

Y mn (y)

∂

∂ν(y)Φ(x, y) ds(y)

− kd

drh(1)n (kR0)

∫|y|=R0

Y mn (y) Φ(x, y) ds(y) . (2.40b)

Equations (2.40a) and (2.40b) are two equations for the integrals. Solving for the secondintegral yields ∫

|y|=R0

Y mn (y) Φ(x, y) ds(y) = ik R2

0 jn(kR0)h(1)n (k|x|)Y m

n (x) (2.41)

where we have used the Wronskian of Theorem 2.25. This holds for all |x| > R0. Comparingthis with the form (2.36a) of u yields

u(x) =∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

h(1)n (kR) jn(kR0)

h(1)n (kr) jn(kR0)Y m

n (x) =

∫|y|=R0

∞∑n=0

n∑m=−n

gmn Ymn (y) Φ(x, y) ds(y)

for |x| > R0 where

gmn =1

ikR20

(fR, Ymn )L2(S2)

h(1)n (kR) jn(kR0)

. (2.42)

It is easy to see from the asymptotics of h(1)n (kR) and jn(kR0) that

g(R0x) =∞∑n=0

n∑m=−n

gmn Ymn (x) =

1

ikR20

∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

h(1)n (kR) jn(kR0)

Y mn (x)

converges uniformly. Therefore, we have proven the representation of u in the form (2.38)for some continuous g.The fundamental solution satisfies the radiation condition (2.34) uniformly with respect toy on the compact sphere as we show in the following lemma. Therefore, also u satisfies theradiation condition (2.34) and completes the proof. 2

Lemma 2.34 For any compact set K ⊂ R3 the fundamental solution Φ satisfies the radia-tion condition (2.34) uniformly with respect to y ∈ K. More precisely,

Φ(x, y) =exp(ik|x|)

4π|x|[e−ik x·y + O

(|x|−1

)], |x| → ∞ , (2.43a)

∇xΦ(x, y) = ik xexp(ik|x|)

4π|x|[e−ik x·y + O

(|x|−1

)], |x| → ∞ , (2.43b)

uniformly with respect to x = x/|x| ∈ S2 and y ∈ K.


Proof: We set x = rx with r = |x| and investigate

4πΦ(rx, y) r e−ikr−e−ik x·y =exp(ikr[|x− y/r| − 1

])− |x− y/r| exp

(−ik x · y

)|x− y/r|

= F (1/r; x, y)

with

F (ε; x, y) =exp(ik[|x− εy| − 1

]/ε)− |x− εy| exp

(−ik x · y

)|x− εy|

.

The function g1(ε; x, y) = |x − εy| is analytic with respect to ε and g1(0; x, y) = 1 and∂g1(0; x, y)/∂ε = −x · y. Therefore, by the rule of l’Hospital we observe that F (0; x, y) = 0,and the assertion of the first part of the lemma follows since also F is analytic with respectto ε. The second part is proven analogously. 2

From formula (2.41) in the proof of Theorem 2.33 we conclude the following additionformula for Bessel functions.

Corollary 2.35 For any x, y ∈ R3 with |x| > |y| we have

Φ(x, y) =ik

4π

∞∑n=0

(2n+ 1) jn(k|y|)h(1)n (k|x|)Pn(x · y) ,

and the series converges uniformly for |y| ≤ R1 < R2 ≤ |x| ≤ R3 for any R1 < R2 < R3.

Proof: We fix x and R0 < |x| and expand the function y 7→ Φ(x,R0y) into spherical surfaceharmonics; that is,

Φ(x,R0y) =∞∑n=0

∑|m|≤n

∫|z|=1

Y mn (z) Φ(x,R0z) ds(z)Y m

n (y)

=1

R20

∞∑n=0

∑|m|≤n

∫|z|=R0

Y −mn (z) Φ(x, z) ds(z)Y mn (y)

= ik∞∑n=0

∑|m|≤n

jn(kR0)h(1)n (k|x|)Y −mn (x)Y m

n (y)

=ik

4π

∞∑n=0

(2n+ 1) jn(kR0)h(1)n (k|x|)Pn(x · y) ,

where we used (2.41) and the addition formula for spherical surface harmonics of Theo-rem 2.17. The uniform convergence follows from the asymptotic behaviour of jn(t) and

h(1)n (t) for n → ∞, uniformly with respect to t from compact subsets of R>0, see Theo-

rem 2.29. Indeed, we have that

jn(k|y|) =1

(2n+ 1)!!(k|y|)n

[1 +O

(1

n

)]for n→∞ ,

h(1)n (k|x|) = −i (2n− 1)!!

(k|x|)n+1

[1 +O

(1

n

)]for n→∞ ,


uniformly with respect to x and y in the specified regions. Therefore,

∣∣(2n+ 1) jn(kR0)h(1)n (k|x|)

∣∣ =1

k|x|

(|y||x|

)n [1 +O

(1

n

)]≤ c

(R1

R2

)nwhich ends the proof. 2

As a corollary of Lemma 2.34 and the proof of Theorem 2.33 we have:

Corollary 2.36 The solution u of the exterior boundary value problem (2.35) satisfies

u(rx) =exp(ikr)

kr

∞∑n=0

n∑m=−n

(fR, Ymn )L2(S2)

(−i)n+1

h(1)n (kR)

Y mn (x) + O

(1

r2

)(2.44a)

=exp(ikr)

4π kr

∞∑n=0

(2n+ 1) (−i)n+1

h(1)n (kR)

∫S2

f(Ry)Pn(x · y) ds(y) + O(

1

r2

)(2.44b)

as r tends to infinity, uniformly with respect to x ∈ S2.

Proof: We note that (2.44a), (2.44b) follow formally from (2.36a), (2.36b), respectively,

by using the asymptotics of Theorem 2.28 for h(1)n (kr). The asymptotics, however, are not

uniform with respect to n, and we have to use of the representation (2.38) as an integralinstead. The function g has the expansion coefficients (2.42). Therefore, using the asymptoticform of Φ by Lemma 2.34,

u(x) =

∫|y|=R0

Φ(x, y) g(y) ds(y) =exp(ikr)

4π r

∫|y|=R0

e−ik x·y g(y) ds(y) + O(

1

r2

).

We compute with the Jacobi-Anger expansion of Theorem 2.31

1

4π

∫|y|=R0

e−ik x·y g(y) ds(y) =R2

0

4π

∫S2

e−ikR0 x·y g(Ry) ds(y)

=R2

0

4π

(gR, e

ikR0 x·)L2(S2)

=R2

0

4π

∞∑n=0

∑|m|≤n

gmn (−i)n 4π jn(kR0)Y mn (x)

=∞∑n=0

∑|m|≤n

1

ik

(fR, Ymn )L2(S2)

h(1)n (kR) jn(kR0)

(−i)n jn(kR0)Y mn (x)

=∞∑n=0

∑|m|≤n

(−i)n+1

h(1)n (kR)

(fR, Ymn )L2(S2) Y

mn (x)

which proves the first part. For the second representation we use again the addition formula.2

2.7. EXPANSION OF ELECTROMAGNETIC WAVES 65

2.7 Expansion of Electromagnetic Waves

We apply the previous results in case of Maxwell’s equations. The starting point is theobservation that E,H are solutions of the source free time harmonic Maxwell’s equationsin a domain D ⊆ R3, if and only if E (or H) is a divergence free solution of the vectorHelmholtz equation; that is,

∆E + k2E = 0 and divE = 0 in D ,

where H = − 1ik

curlE (see page ### ... chap 1... ###). Thus, by the following Lemmawe can construct solutions of Maxwell’s equations by knowing a scalar valued solution of theHelmholtz equation.

Lemma 2.37 Let u ∈ C3(D) satisfy ∆u+ k2u = 0 in a domain D ⊆ R3. Then

E(x) = curl(xu(x)

)and H(x) =

1

ikcurlE(x)

are solutions of the time harmonic Maxwell equations

curlE − ikH = 0 and curlH + ikE = 0 in D .

Proof : Let u satisfy the Helmholtz equation and define E and H as in the Lemma. Weobserve divE = 0. Furthermore it is

∆E = ∆ curl(xu(x)

)= − curl curl

(∇u(x)× x+ u(x) curlx

)= − curl curl

(∇u(x)× x

)= − curl

(∇u(x) divx︸︷︷︸

=3∇u(x)

−x div∇u(x)︸︷︷︸=∆u(x)=−k2u(x)

+(x · ∇)∇u(x)− (∇u(x) · ∇)x︸︷︷︸=∇u(x)

)

= −k2 curl(xu(x)

)= −k2E ,

where we have used curl(∇u) = 0, equation (6.10), and from (6.9) the identity ∇(x · ∇u) =(x·∇)∇u+∇u, i.e. curl

((x·∇)∇u

)= 0 . We conclude that E is a divergence free solution of

the vector Helmholtz equation and therefore it constitutes a solution of Maxwell’s equations.2

In view of the separation of the Helmholtz equation from the previous section we like toconsider solutions of the Helmholtz equation of the form u(x) = jn(k|x|)Y m

n (x) where Y mn

denotes a spherical surface harmonics of order n ∈ N0, m = −n, . . . , n and x = x|x| . By the

lemma 2.37 we see that by E(x) = curl(xjn(k|x|)Y mn (x)) or E(x) = curl curl(xjn(k|x|)Y m

n (x))we obtain solutions of the Maxwell system. Also we can replace the Besselfunction by theHankel function, if x = 0 ∈ R3 is excluded.


Theorem 2.38 (a) For n ∈ N0 a function E : R3 → R3 defiend by E(x) = curl(x jn(k|x|)Y m

n (x))

with a spherical harmonics Y mn of order n satisfies Maxwell’s equations together with H =

1ik

curlE and it holds

E(x) = jn(k|x|) GradY mn (x)× x . (2.45)

(b) Similar the vector fields curlE(x) = curl curl(x jn(k|x|)Y m

n (x))

are solutions of Maxwell’sequations. For the normal and tangential components with respect to spheres it holds

x · curlE(x) = −n(n+ 1)jn(k|x|)Y mn (x) (2.46)

x× curlE(x) =[ 1

|x|jn(k|x|) + kj′n(k|x|)

]x×GradY m

n (x) (2.47)

(2.48)

with x = x/|x|.

(c) Analogously, the functions E : R3\0 → C3 defined by E(x) = curl(xh

(1)n (k|x|)Y m

n (x))

and curlE are radiating solutions of Maxwell’s equations and satisfy the corresponding iden-tities (2.45)–(2.47) with jn replaced by h

(1)n .

Proof: From Lemma 2.37 we already know that E(x) = curl(x jn(k|x|)Kn(x)

)generates

a solution of Maxwell’s equations. By using the identity (6.7) and curl x = 0 we findE(x) = ∇

(jn(k|x|)Kn(x)

)× x. Decomposing the gradient by the surface gradient Grad

with respect to the sphere of radius |x| and its normal component in the direction ν(x) = xleads to

E(x) =

(Grad (jn(k|x|)Y m

n (x)) +∂

∂ν(jn(k|x|)Y m

n (x)) x

)× x

= jn(k|x|) Grad S2 Y mn (x)× x .

Note that we have used the identity |x|Grad = Grad S2 for the surface gradient on a spherewith radius R = |x| and GradS2 , the surface gradient with respect to the unit sphere. Weavoid different notations of the surface gradient, if the corresponding surface is obvious.

Now we consider the solution curlE and obtain by equation (6.10)

x · curlE(x) = x ·

[jn(k|x|) GradY m

n div(x)− x div (jn(k|x|) GradY mn (x))

+(x · ∇

)(jn(k|x|) GradY m

n (x))−(jn(k|x|) GradY m

n (x) · ∇)x

]= − div(jn(k|x|) Grad Y m

n (x) + x · (jn(k|x|)Y mn (x))′ x

Div(jn(k|x|) Grad Y mn (x)) = −n(n+ 1)jn(k|x|)Y m

n (x) ,

where we have used xtop(x)′ = 12(∇‖x‖2)> = 0.


Furthermore by x · ∇ = ∂∂r

we get

x× curlE(x) = x×

[jn(k|x|) GradY m

n div(x)− x div (jn(k|x|) GradY mn (x))

+(x · ∇

)(jn(k|x|) GradY m

n (x))−(jn(k|x|) GradY m

n (x) · ∇)x

]

=2

|x|jn(k|x|) x×GradY m

n (x) + x×

(3∑j=1

xj∂

∂xj(jn(k|x|) GradY m

n (x))i

)i=1,...,3

− x×

(jn(k|x|)

3∑j=1

(GradY mn (x))j

∂

∂xj(xi|x|

)

)i=1...3

=2


n (x) + x× (kj′n(k|x|)GradY mn (x))

− 1


n (x) + x×

[(3∑j=1

jn(k|x|)(GradY mn )j

xj|x|2

)x

]

=1


n (x) + kj′n(k|x|) x×GradY mn (x) .

Analogously we obtain the representation in case of E(x) = curl(xh

(1)n (k|x|)Y m

n (x)).

Finally, by these representations and the asymptotic behavior of the Hankel functions, seeLemma 2.28, we obtain the Silver-Muller radiation condition,

lim|x|→∞

(H(x)× x− |x|E(x)

)= 0 ,

for the second solution E(x) = curl(xh

(1)n (k|x|)Y m

n (x)), which completes the Theorem. 2

From the theorem we see the tangential boundary values on a sphere of radius R of thesolutions E and curlE from the last theorem are given by

ν×E(x) = jn(kR) GradY mn (x) and ν×curlE(x) =

[ 1

Rjn(kR)+kj′n(kR)

]x×GradY m

n (x)

for x = Rx ∈ R3. Analogously this holds for the radiating fields with the Hankel functions.Thus, in view of solving boundary value problems by expansions w.r.t. spherical harmonicswe consider the surface gradients Grad Y m

n and ν ×Grad Y mn on S2.

By the definition of the surface gradient of u ∈ C1(S2) (see ### Verweis ###) we firstobserve that it is a tangential field, i.e. ν · Gradu = 0 on S2 with the unit normal vectorsν(x) = x on S2 and obviously also x × Gradu is tangential. Using the partial integration


formula (??) we compute∫S2

(x×Gradu) · (x×Grad v) ds =

∫S2

Gradu · ((x×Grad v)× x) ds

=

∫S2

Gradu · Grad v ds =

∫S2

u∆Sv ds

for u ∈ C1(S2) and v ∈ C2(S2). Since the spherical harmonics Y mn for n ∈ N and −n ≤

m ≤ n are an orthonormal system and eigenfunctions of the Laplace-Beltrami operator forthe eigenvalues λ = −n(n+ 1), n ∈ N0, we conclude that the functions

Umn (x) =

1√n(n+ 1)

GradY mn (x)

as well as the functionsV mn (x) = x× Um

n (x)

are two orthonormal systems of tangential fields on S2.

Applying partial integration again we obtain∫S2

(x×Gradu) ·Grad v ds =

∫S2

Div(x×Gradu) v ds = 0 ,

since by Corollary ?? and an arbitrary C1 extension of u it is Div(x × Gradu) = Div(x ×∇u) = −x · curl(∇u) = 0 on S2. We combine both sets. The functions Um

n and V mn are

called vector spherical harmonics. The main step is to show that the vector sphericalharmonics build a complete orthonormal system in the space of tangential vector fields onS2. In preparation of the proof we show the following lemma.

Lemma 2.39 If a tangential field A ∈ C1t (S2) = A ∈ C1(S2) : A · ν = 0 on S2 satisfies

DivA = 0, and Div(ν × A) = 0 on S2 ,

then A = 0.

Proof : From Div(ν × A) = 0 we obtain by partial integration (see equation (??))

0 =

∫S2

Div(ν × A)f ds =

∫S2

(ν × A) ·Grad f ds

for any f ∈ C1(S2). Using the representation A = (A · θ)θ+ (A · ϕ)ϕ for the tangential fieldin spherical coordinates with the basis unit vectors θ, ϕ leads to

ν × A = (A · θ)ϕ− (A · ϕ)θ .

The representation ?? of the surface gradient on S2 implies by partial integration withrespect to θ and ϕ respectively the identity

0 =

∫ 2π

0

∫ π

0

(ν × A)

(∂f

∂θθ +

1

sin θ

∂f

∂ϕϕ

)sin θ dθ dϕ

=

∫ 2π

0

∫ π

0

[∂

∂θ((A · ϕ) sin θ)− ∂

∂ϕ(A · θ)

]f dθ dϕ


for all f ∈ C1(S2). Thus, it holds

∂

∂θ

((A · ϕ) sin θ

)=

∂

∂ϕ

(A · θ

)on S2. Considering an anti-derivative h =

∫(A · ϕ) sin θ dϕ we have

A · θ =∂h

∂θand A · ϕ =

1

sin θ

∂h

∂ϕ.

Therfore it is

A = (A · θ)θ + (A · ϕ)ϕ =∂h

∂θθ +

1

sin θ

∂h

∂ϕϕ = Grad h .

Now, from the condition DivA = 0 we conclude with (??) that

0 =

∫S2

Div(A)h ds =

∫S2

A ·Grad h ds =

∫S2

|Grad h|2 ds .

Thus, we have A = Grad h = 0 on S2. 2

Remark: From the first part of the proof we observe that Div(ν × A) = 0 on S2 impliesthe existence of a surface potential. Thus it becomes obvious, why CurlA = −Div(ν × A)is also called the surface curl of a tangential field A.

Theorem 2.40 The functions Umn = 1√

n(n+1)GradY m

n (x) and V mn (x) = x × Um

n (x) for

n ∈ N and −n ≤ m ≤ n constitute a complete orthonormal system in

L2τ (S

2) =V ∈ L2(S2) : ν · V = 0

.

Proof: Let U ∈ L2τ (S

2) ∩ C3(S2). Then DivU ∈ C2(S2) and the expansion

DivU =∞∑n=1

m∑m=−n

amn Ymn with amn =

∫S2

Div(U)Y mn ds

converges uniformly (see Theorem 2.21). The first coefficient vanishes, since∫S2 DivU ds = 0.

Analogously Div(ν × U) ∈ C2(S2) is given by a uniform expansion

Div(ν × U) =∞∑n=1

m∑m=−n

bmn Ymn with bmn =

∫S2

Div(ν × U)Y mn ds .

Defining

u =∞∑n=1

n∑m=−n

1

n(n+ 1)amn Y

mn and v =

∞∑n=1

n∑m=−n

1

n(n+ 1)bmn Y

mn

on S2 we obtain with the eigenvalue equation Div Grad Y mn = −n(n+ 1)Y m

n that

Div Grad u = DivU and Div Grad v = −Div(ν × U) , (2.49)


if we can take the derivative term by term. Thus, the function V = Grad u + ν × Grad vhas formally an expansion in terms of Um

n and V mn and

DivU = Div V and Div(ν × U) = Div(ν × V ), (2.50)

since Div(ν×Grad u) = −ν ·curl(∇u) = 0 and analogously for v. From the previous Lemmawe conclude U = V . This implies an expansion of U ∈ L2(S2) ∩ C3(S2), if we can showthat the previous thoughts hold rigorously. Finally, by the dense imbedding of this spacein L2

τ (S2) the result follows for any tangential field U ∈ L2

τ (S2) and the Theorem would be

proven.

It remains to show that the above arguments are valid. Therefore, let us consider thedefinitions of u and v. By partial integration (see (??)) we have

n(n+1)amn = n(n+1)

∫S2

DivU Y mn ds =

∫S2

DivU ∆SY mn ds = −

∫S2

Div Grad DivU Y mn ds ,

where we used again the eigenvalues ∆SYmn = Div Grad Y m

n = −n(n + 1)Y mn . We conclude

from Parzeval’s equation

∞∑n=1

n∑m=−n

|n(n+ 1) amn |2 = ‖Div Grad DivU‖2L2 .

Furthermore, we compute with the help of the first part of Corollary 2.18

0 = Div Grad

(n∑

m=−n

|Y mn |2)

= 2n∑

m=−n

|Grad Y mn |2 − 2n(n+ 1)

n∑m=−n

|Y mn |2

= 2n∑

m=−n

|Grad Y mn |2 −

1

2πn(n+ 1)(2n+ 1) ,

i.e.n∑

m=−n

|Grad Y mn |2 =

1

4πn(n+ 1)(2n+ 1) .

The Cauchy-Schwarz inequality leads to(∞∑n=1

n∑m=−n

∣∣∣∣ 1

n(n+ 1)amn Grad Y m

n

∣∣∣∣)2

≤

(∞∑n=1

n∑m=−n

n2(n+ 1)2|amn |2)(

∞∑n=1

n∑m=−n

1

n4(n+ 1)4|Grad Y m

n |2)

≤ ‖Div Grad DivU‖2L2

(∞∑n=1

2n+ 1

4πn3(n+ 1)3

),


which shows uniform convergence of the term by term derivative of u. The same holds forv. Thus, we have u, v ∈ C1(S2) and V = Grad u + ν × Grad v ∈ C(S2) is given by anexpansion in terms of Um

n and V mn .

The last step of the proof is to show that (2.50) is satisfied. Since the expansions of u andGrad u are uniform we obtain∫

S2

Grad u ·Grad f ds =∞∑n=1

m∑m=−n

1

n(n+ 1)amn

∫S2

Grad Y mn ·Grad f ds

=∞∑n=1

m∑m=−n

1

n(n+ 1)amn

∫S2

Div(Grad Y mn ) f ds

=∞∑n=1

m∑m=−n

amn

∫S2

Y mn f ds

=

∫S2

(∞∑n=1

m∑m=−n

amn Ymn

)f ds

=

∫S2

Div(U) f ds =

∫S2

U ·Grad f ds

for all f ∈ C1(S2). Thus in this weak sense we can define Div Grad u = DivU . The sameholds for Div Grad v = −Div(ν × U). With the definition of V we have∫

S2

(U − V ) ·Grad f ds =

∫S2

(U −Grad u) ·Grad f ds = 0

and ∫S2

(ν × U − ν × V ) ·Grad f ds =

∫S2

(ν × U + Grad v) ·Grad f ds = 0 .

Considering the proof of Lemma 2.39 we observe that these weak identities are sufficient toconclude U = V , which completes the proof. 2

As a corollary we formulate an expansion of any vector field E in terms of the normal basisfunctions x→ Y n

m(x)x and the tangential basis functions Unm and V m

n .

Corollary 2.41 Every vector field A ∈ L2(S2)3 has an expansion of the form

A(x) =∞∑n=0

∑|m|≤n

[amn Y

mn (x) x + bmn U

mn (x) + cmn V

mn (x)

](2.51)

where

amn =

∫S2

x · E(x)Y −mn (x) ds(x) , |m| ≤ n , n ≥ 0 , (2.52a)

bmn =

∫S2

E(x) · U−mn (x) ds(x) , |m| ≤ n , n ≥ 1 , b00 = 0 , (2.52b)

cmn =

∫S2

E(x) · V −mn (x) ds(x) , |m| ≤ n , n ≥ 1 , c00 = 0 . (2.52c)


The convergence is understood in the L2−sense. Furthermore, Parseval’s equality holds; thatis,

∞∑n=0

∑|m|≤n

[|anm|2 + |bnm|2 + |cnm|2

]= ‖E‖2

L2(S2)3 .

Proof: We decompose the vector field A into

A(x) =(A(x) · x

)x + x×

(A(x)× x

).

The scalar function x ·A(x) and the tangential vector field x×(A(x)× x

)can be expanded

into basis functions on S2 according to Theorem ?? for x·A(x) and Theorem 2.40 for x×A(x)which proves the result. 2

We apply this corollary to the vector field x 7→ E(rx) where E satisfies Maxwell’s equation.

Theorem 2.42 Let E ∈ C2(B(0, R)

)3satisfy curl2E − k2E = 0 in B(0, R). Then E can

be written as

E(rx) =∞∑n=1

∑|m|≤n

[αmn√n(n+ 1)

rjn(kr)Y m

n (x) x +αmnr

(rjn(kr)

)′Umn (x)

+ (2.53)

betamn jn(kr)V mn (x)

], 0 ≤ r < R , x ∈ S2 , (2.54)

for some αmn , βmn ∈ C. The series converges uniformly in every ball B[0, R′] for R′ < R.

Proof: Let amn (r), bmn (r), cmn (r) be the expansion coefficients of (2.51) corresponding to thevector field given by E(rx). First we note that a0

0(r) = 1√4πr2

∫|x|=r x · E(x) ds(x) vanishes

by the divergence theorem because divE = 0.

Second, we consider amn (r). With

amn (r) =1

r

∫S2

(x · E)Y mn (x) ds(x)

we observe that ramn (r) is the expansion coefficient of the solution x · E(x) of the scalarHelmholtz equation, , see Lemma 1.2. Thus by Theorem ?? we obtain

amn (r) =αmnrjn(kr) .

Next, we use that E is divergence free and prove

1√n(n+ 1)

(r2anm(r)

)′= r bmn (r) . (2.55)


Indeed, the condition divE = 0 in spherical polar coordinates reads

1

r

∂

∂r

(r2Er(r, x)

)+ DivEt(r, x) = 0

and thus

1

r

(r2anm(r)

)′=

∫S2

1

r

∂

∂r

(r2Er(r, x)

)Y −mn (x) ds(x)

= −∫S2

DivEt(r, x)Y −mn (x) ds(x)

=

∫S2

E(r, x) ·Grad Y −mn (x) ds(x) =√n(n+ 1) bmn (r) .

Thus, with a common constant αmn we have

amn (r) = αmn

√n(n+ 1)

rjn(kr) , bmn (r) = αmn

1

r

(r jn(kr)

)′. (2.56)

Finally, we consider cmn (r).

√n(n+ 1) cmn (r) =

∫S2

E(r, x) ·(x×Grad Y −mn (x)

)ds(x)

=

∫S2

(E(r, x)× x

)·Grad Y −mn (x) ds(x)

= −∫S2

Div(E(r, x)× x

)Y −mn (x) ds(x)

= −r∫S2

x · curlE(r, x)Y −mn (x) ds(x)

= −∫S2

x · curlE(r, x)Y −mn (x) ds(x)

where we have used that Div(E(r, x) × x

)= r x · curlE(r, x). Therefore,

√n(n+ 1) cmn (r)

are the expansion coefficients of the scalar function ψ(x) = x · curlE(x). Because ψ is asolution of the Helmholtz equation ∆ψ + k2ψ = 0 we again conclude from Theorem ?? thatcmn (r) = βmn jn(kr) for some βmn ∈ C. 2

### Zusammenhang zum ersten Resultat als Bemerkung: With Theorem ... we havecurl curl(xjn(k|x|)Y m

n (x)) has the normal part and the tangential part ... Thus curl curl() =amn xjn(r)Y m

n + bmn ... ###

Combining the previous results we obtain the following existence result for the interiorMaxwell problem in a ball, that is the boundary value problem for Maxwell’s equation ina ball B(0, R) with boundary values ν × E = f on |x| = R. We formulate the extistenceresult again with fR ∈ L2(S2) with fR(x) = f(Rx).


Theorem 2.43 Assuming that k ∈ C with Im k ≥ 0 and R > 0 are such that jn(kR) 6= 0and jn(kR) 6= −kRj′n(kR) for all n ∈ N0.

(a) For given fR ∈ L2τ (S

2) there exists a unique solution E ∈ C2(B(0, R)

)and H = 1

ikcurlE

ofcurlE − ikH = 0 and curlH + ikE = 0 in B(0, R)

withlimr→R‖ν × E(r·)− fR‖L2(S2) = 0

and the solution is given by

E(rx) =∞∑n=0

n∑m=−n

(fR, Umn )L2(S2)√

n(n+ 1)curl

(r, x

jn(kr)

jn(kR)Y mn (x)

)

+(fR, V

mn )L2(S2)√

n(n+ 1)curl curl

(r, x

rjn(kr) + kj′n(kr)

Rjn(kR) + kj′n(kR)Y mn (x)

).

The series converges uniformly on compact subsets of B(0, R).

(b) If fR ∈ L2τ (S

2) ∩ C2(S2) the solution of the boundary value problem satisfies E ∈C2(B(0, R)

)∩ C

(B(0, R)

)and the series converges uniformly on B(0, R).

Proof:

### knapp mit Verweis auf Theorem 2.44.

Nullstellen jn(kR)+kRj′n(kR) = 0 anders beschreiben, z.B. aus (n+1)jn(kR)+kRj′n(kR) =kRjn−1(kR) dh Schnittpunkte njn(kR) = kRjn−1(kR)? ### 2

### Remark EV ###

In contrast to the interior problem and similar to the Helmholtz equation the exteriorMaxwell problem is uniquely solvable for any R > 0.

Theorem 2.44 Assuming that k ∈ C with Im k ≥ 0 and R > 0.

(a) For given fR ∈ L2τ (S

2) there exists a unique solution E ∈ C2(R3\B(0, R)

)and H =

1ik

curlE of

curlE − ikH = 0 and curlH + ikE = 0 in Rn\B(0, R)

withlimr→R‖ν × E(r, ·)− fR‖L2(S2) = 0 .

The fields E and H satisfies the Sivler-Muller radiation condition (see ### referenz ###)and the solution is given by

E(r, x) =∞∑n=0

n∑m=−n

(fR, Umn )L2(S2)√

n(n+ 1)curl

(r, x

h(1)n (kr)

h(1)n (kR)

Y mn (x)

)

+(fR, V

mn )L2(S2)√

n(n+ 1)curl curl

(r, x

rh(1)n (kr) + kh

(1)n

′(kr)

Rh(1)n (kR) + kh

(1)n

′(kR)

Y mn (x)

).

2.8. EXERCISES 75

The series converges uniformly on compact subsets of R3\B(0, R).

(b) If fR ∈ L2τ (S

2) ∩C2(S2) the solution satisfies E ∈ C2(R3B(0, R)

)∩C

(R3\B(0, R)

)and

its series representation converges uniformly on B(0, R1)\B(0, R) for R1 > R.

Proof:

We follow the same idea as of the proofs of the Theorems ### Referenz 2.20, 2.21, 2.30,2.32 ###.

(i) ### uniqueness ... ###

(ii) ### expansion in L2 ###

From the Wronskian (??) we obtain Im((h

(1)n (z) + z(h

(1)n )′)h

(2)n (z)

)= 1

z. Since the function

h(2)n has no zeros, it is h

(1)n (kR) + kR (h

(1)n )′(kR) 6= 0 for all n ∈ N .

(iii) ### uniformly ... ### 2

### Corollary: jede Lsg in Kugelschale kann durch Entwicklung nach Bessel oder Hankel-funktionen dargestellt werden ###

2.8 Exercises

### Vorschlaege/Ideen:

• im R2, Polarkoordinaten, Laplace separieren

• Separation in kartesischen Koordinaten

• Existenzaussagen im Kreis oder im Quadrat, RR2

• Einige weitere Beweise zu Eigenschaft (e)-(g)

• Aussagen zu Differntialoperatoren auf S2 (oder in chap 1)

• andere Randbedingungen bei Maxwellgln

###


Chapter 3

Scattering From a Perfect Conductor

Version of January 14, 2013

In Section 2.6 of the previous chapter we studied the scattering of a plane wave by balls. Inthis chapter we investigate the same problem for arbitrary shapes. Also, as in the previouschapter, we consider first the simpler scattering problem for the scalar Helmhotz equationin Section 3.1 before we turn to Maxwell’s equations in Section 3.2.

3.1 A Scattering Problem for the Helmholtz Equation

3.1.1 Formulation of the Problem

Throughout this chapter we make the following assumptions on the data:

Assumption: Let the wave number be given by k = ω√ε0µ0 > 0 with constants ε0, µ0 > 0.

Let D ⊂ R3 be a finite union of bounded domains Dj such that Dj ∩ D` = ∅ for j 6= `.Furthermore, we assume that the boundary ∂D is C2−smooth and that the complementR3 \D is connected.

Scattering Problem: Given an incident field ui; that is, a solution ui of the Helmholtzequation ∆ui + k2ui = 0 in all of R3, find the total field u ∈ C2(R3 \D) ∩ C1(R3 \D) suchthat

∆u + k2u = 0 in R3 \D ,∂u

∂ν= 0 on ∂D , (3.1)

and such that the scattered field us = u − ui satisfies the Sommerfeld radiation condition(2.34); that is,

∂us(rx)

∂r− ik us(rx) = O

(1

r2

)for r →∞ , (3.2)

uniformly with respect to x ∈ S2. For smooth fields u ∈ C1(R3 \D) and smooth boundariesthe normal derivative is given by ∂u/∂ν = ∇u · ν where ν = ν(x) denotes the exterior unitnormal vector at x ∈ ∂D.

77

78 CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR

Before we investigate uniqueness and existence of solution of this scattering problem westudy general properties of solutions of the Helmholtz equation ∆u + k2u = 0 in boundedand unbounded domains.

3.1.2 Representation Theorems

We begin with the (really!) fundamental solution of the Helmholtz equation, compare (2.39).

Lemma 3.1 For k ∈ C the function Φk :

(x, y) ∈ R3 × R3 : x 6= y→ C, defined by

Φk(x, y) =eik|x−y|

4π|x− y|, x 6= y ,

is called the fundamental solution of the Helmholtz equation, i.e. it holds that

∆xΦk(x, y) + k2Φk(x, y) = 0 for x 6= y .

Proof: This is easy to check. 2

We often suppress the index k, i.e. write Φ for Φk.

We continue with the existence of certain special improper integrals.

Lemma 3.2 (a) Let K : (x, y) ∈ R3 ×D : x 6= y→ C be continuous. Assume that there

exists c > 0 and β ∈ (0, 1] such that∣∣K(x, y)∣∣ ≤ c

|x− y|3−βfor all x ∈ R3 and y ∈ D with x 6= y .

Then∫∫

DK(x, y) dy exists as an improper integral and there exists cβ > 0 with∫∫

D\B(x,τ)

∣∣K(x, y)∣∣ dy ≤ cβ for all x ∈ R3 and all τ > 0 , (3.3a)∫∫

D∩B(x,τ)

∣∣K(x, y)∣∣ dy ≤ cβ τ

β for all x ∈ R3 and all τ > 0 . (3.3b)

(b) Let K : (x, y) ∈ ∂D × ∂D : x 6= y→ C be continuous. Assume that there exists c > 0

and β ∈ (0, 1] such that∣∣K(x, y)∣∣ ≤ c

|x− y|2−βfor all x, y ∈ ∂D with x 6= y .

Then∫∂DK(x, y) ds(y) exists as an improper integral and there exists cβ > 0 with∫

∂D\B(x,τ)

∣∣K(x, y)∣∣ ds(y) ≤ cβ for all x ∈ ∂D and τ > 0 , (3.3c)∫

∂D∩B(x,τ)

∣∣K(x, y)∣∣ ds(y) ≤ cβ τ

β for all x ∈ ∂D and τ > 0 . (3.3d)

3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION 79

(c) Let K : (x, y) ∈ ∂D × ∂D : x 6= y→ C be continuous. Assume that there exists c > 0

and β ∈ (0, 1) such that∣∣K(x, y)∣∣ ≤ c

|x− y|3−βfor all x, y ∈ ∂D with x 6= y .

Then there exists cβ > 0 with∫∂D\B(x,τ)

∣∣K(x, y)∣∣ ds(y) ≤ cβ τ

β−1 for all x ∈ ∂D and τ > 0 . (3.3e)

Proof: (a) Fix x ∈ R3 and choose R > 0 such that D ⊂ B(0, R).First case: |x| ≤ 2R. Then D ⊂ B(x, 3R) and thus, using spherical polar coordinates w.r.t.x,

∫∫D\B(x,τ)

1

|x− y|3−βdy ≤

∫∫τ<|y−x|<3R

1

|x− y|3−βdy = 4π

3R∫τ

1

r3−β r2 dr

=4π

β

[(3R)β − τβ

]≤ 4π

β(3R)β .

Second case: |x| > 2R. Then |x− y| ≥ |x| − |y| ≥ R for y ∈ D and thus∫∫D\B(x,τ)

1

|x− y|3−βdy ≤ 1

R3−β

∫∫B(0,R)

dy =1

R3−β4π

3R3 .

This proves (3.3a). For (3.3b) we compute∫∫D∩B(x,τ)

∣∣K(x, y)∣∣ dy ≤ c

∫∫|x−y|<τ

1

|x− y|3−βdy = 4πc

τ∫0

1

r3−β r2 dr =

4πc

βrβ .

(b) We choose a local coordinate system as in Definition 6.1; that is, a covering of ∂Dby cylinders of the form Uj = QjCj + zj where Qj ∈ R3×3 are rotations and zj ∈ R3

and Cj = B2(0, ρj) × (−αj, αj), and where ∂D ∩ Uj is expressed as ∂D ∩ Uj =x =

Qju+ zj : (u1, u2) ∈ B2(0, ρj) , u3 = f(u1, u2)

for some (smooth) function fj : B2(0, ρj)→(−αj/2, αj/2). Furthermore, we choose a corresponding partition of unity (see Lemma 6.3);that is, a family of functions φj ∈ C∞(R3), j = 1, . . . ,m, with

• 0 ≤ φj ≤ 1 in R3,

• the support Sj := supp(φj) is contained in Uj, and

•∑m

j=1 φj(x) = 1 for all x ∈ ∂D.

Then, obviously, ∂D ⊂⋃mj=1 Sj. Furthermore, there exists δ > 0 such that |x − y| ≥ δ

for all (x, y) ∈ Sj ×⋃6=j(U` \ Uj) for every j. Indeed, otherwise there would exist some j

and a sequence (xk, yk) ∈ Sj ×⋃`6=j(U` \ Uj) with |xk − yk| → 0. There exist convergent


subsequences xk → z and yk → z. Then z ∈ Sj and z ∈⋃` 6=j(U`\Uj) which is a contradiction

because Sj ⊂ Uj.The integral is decomposed as∫

∂D|x−y|>τ

ds(y)

|x− y|2−β=

m∑`=1

∫∂D∩U`|x−y|>τ

φj(y)

|x− y|2−βds(y) .

We fix x ∈ Uj ∩ ∂D. For y /∈ Uj ∩ ∂D we have that |x− y| ≥ δ. Therefore, the integrals over∂D ∩ U` for ` 6= j are easily estimated.For the integral over ∂D ∩Uj let x = Qju+ zj and y = Qjv + zj for some u, v ∈ B2(0, ρj)×(−αj, αj) with u3 = fj(u1, u2) and v3 = fj(v1, v2). We set u = (u1, u2) and v = (v1, v2) andcertainly have an estimate of the form

c1|x− y| ≤ |u− v| ≤ c2|x− y| for all x, y ∈ Uj ∩ ∂D .

Using polar coordinates w.r.t. u we estimate∫∂D∩Uj|x−y|>τ

ds(y)

|x− y|2−β≤ c2−β

2

∫B2(0,ρj)\B2(u,c1τ)

1

|u− v|2−β√

1 + |∇f(v)|2 dv

≤ c

∫c1τ<|v−u|<2ρj

dv

|u− v|2−β= 2π c

∫ 2ρj

c1τ

r

r2−β dr ≤ 2π c(2ρj)

β

β.

The proofs of (3.3d) and part (c) follow the same lines. 2

The following representation theorem implies that any solution of the Helmholtz equation isalready determined by its Dirichlet- and Neumann data on the boundary. This theorem istotally equivalent to Cauchy’s integral representation formula for holomorphic functions.

Theorem 3.3 (Green’s representation theorem in the interior of D)

For any k ∈ C and u ∈ C2(D) ∩ C1(D) we have the representation∫∫D

Φ(x, y)[∆u(y) + k2u(y)

]dy +

∫∂D

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y) =

=

−u(x) , x ∈ D ,

−12u(x) , x ∈ ∂D ,

0 , x 6∈ D .

The domain integral as well as the surface integral (for x ∈ ∂D) exist as improper integrals.

Remarks:


• This theorem tells us that, for x ∈ D, any function u can be expressed as a sum ofthree potentials:

(Sϕ)(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) , x /∈ ∂D , (3.4a)

(Dϕ)(x) =

∫∂D

ϕ(y)∂Φ

∂ν(y)(x, y) ds(y) , x /∈ ∂D , (3.4b)

(V ϕ)(x) =

∫∫D

ϕ(y) Φ(x, y) dy , x ∈ R3 , (3.4c)

which are called single layer potential, double layer potential, and volume po-tential, respectively, with density ϕ. We will investigate these potential in detail inSubsection 3.1.3 below.

• The one-dimensional analogon is (for x ∈ D = (a, b) ⊂ R)

u(x) =1

2ik

b∫a

eik|x−y|[u′′(y) + k2u(y)

]dy +

1

2ik

[u(y)

d

dyeik|x−y| − u′(y) eik|x−y|

]ba

.

Therefore, the one-dimensional fundamental solution is Φ(x, y) = − exp(ik|x−y|)/(2ik).One should try to prove this representation in the one-dimensional case!

Proof of Theorem 3.3: First we fix x ∈ D and a small closed ball B[x, r] ⊂ D centered at xwith radius r > 0. For y ∈ ∂B(x, r) the normal vector ν(y) = x−y

|y−x| = (x − y)/r is directed

into the interior of B(x, r). We apply Green’s second identity to u und v(y) := Φ(x, y) inthe domain Dr := D \B[x, r]. Then∫

∂D

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y) (3.5a)

+

∫∂B(x,r)

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y) = (3.5b)

=

∫∫Dr

u(y) ∆yΦ(x, y)− Φ(x, y) ∆u(y)

dy = −

∫∫Dr

Φ(x, y)[∆u(y) + k2u(y)

]dy ,

if one uses the Helmholtz equation for Φ. We compute the integral (3.5b). We observe that

∇yΦ(x, y) =exp(ik|x− y|)

4π|x− y|

(ik − 1

|x− y|

)y − x|x− y|

and thus for |y − x| = r:

Φ(x, y) =exp(ikr)

4πr,

∂Φ

∂ν(y)(x, y) =

x− yr· ∇yΦ(x, y) = −exp(ikr)

4πr

(ik − 1

r

).


Therefore, we compute the integral (3.5b) as

Ir(x) :=

∫∂B(x,r)

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y)

=exp(ikr)

4πr

∫|y−x|=r

u(y)

(1

r− ik

)− ∂u

∂ν(y)

ds

=exp(ikr)

4πr2

∫|y−x|=r

u(y) ds − exp(ikr)

4πr

∫|y−x|=r

ik u(y) +

∂u

∂ν(y)

ds .

For r → 0 the first term tends to u(x), the second term to zero since the surface area of∂B(x, r) is just 4πr2. Therefore, also the limit of the volume integral exists as r → 0 andyields the desired formula for x ∈ D.

Let now x ∈ ∂D. Then we procceed in the same way. The domains of integration in(3.5a) and (3.5a) have to be replaced by ∂D \B(x, r) and ∂B(x, r)∩D, respectively. In thecomputation the region y ∈ R3 : |y − x| = r has to be replaced by y ∈ D : |y − x| = r.By Lemma 6.4 its surface area is 2πr2 +O(r3) which gives the factor 1/2 of u(x).

For x 6∈ D the functions u and v = Φ(x, ·) are both solutions of the Helmholtz equation inall of D. Application of Green’s second identity in D yields the assertion. 2

We note that the volume integral vanishes if u is a solution of the Helmholtz equation∆u+ k2u = 0 in D. In this case the function u can be expressed solely as a combination ofa single and a double layer surface potential.

From the proof we observe that our assumptions on the smoothness of the boundary ∂Dare too strong. The domain D has to satisfy exactly the assumptions which are needed forGreen’s theorems to hold.

As a corollary we have:

Conclusion 3.4 Let u ∈ C2(D) be a (real- or complex valued) solution of the Helmholtzequation in D. Then u is analytic, i.e. one can locally expand u into a power series of theform

u(x) =∑n∈N3

anxn11 x

n22 x

n33

where we use the notation N = Z≥0 = 0, 1, 2, . . ..

Proof: From the previous representation of u(x) as a difference of a single and a doublelayer and the smoothness of the kernels x 7→ Φ(x, y) and x 7→ ∂Φ(x, y)/∂ν(y) for x 6= y itfollows immediately that u ∈ C∞(D). The proof of analyticity is technically not easy if oneavoids methods from complex analysis.1 If one uses these methods then one can argue as

1We refer to [?] E. Martensen: Potentialtheorie for a proof.


follows: Fix x ∈ D and choose r > 0 such that B[x, r] ⊂ D. Define the region R ⊂ C3 andthe function v : R→ C by

R =z ∈ C3 :

∣∣Re z − x∣∣ < r/2,

∣∣Im z∣∣ < r/2

,

v(z) =

∫∂D

exp[ik√∑3

j=1(zj − yj)2]

4π√∑3

j=1(zj − yj)2

∂u

∂ν(y)− u(y)

∂

∂ν(y)

exp[ik√∑3

j=1(zj − yj)2]

4π√∑3

j=1(zj − yj)2

ds(y)

for z ∈ R. Taking the square root (principal value, cut along the negative real axis) of thecomplex number

∑3j=1(zj − yj)2 is not a problem since Re

∑3j=1(zj − yj)2 =

∑3j=1(Re zj −

yj)2 − (Im zj)

2 = |Re z − y|2 − |Im z|2 > 0 because of |Re z − y| ≥ |y − x| − |x − Re z| >r − r/2 = r/2 and |Im z| < r/2. Obviously, the function v is holomorphic in R and thus(complex) analytic. 2

As a second corollary we can easily prove the following version of Holmgren’s uniquenesstheorem.

Theorem 3.5 Let D be a domain with C2-boundary and u ∈ C1(D) ∩ C2(D) be a solutionof the Helmholtz equation ∆u+ k2u = 0 in D. Let, furthermore, U be an open set such thatU ∩ ∂D 6= ∅ and u = 0 and ∂u/∂ν = 0 on U ∩ ∂D. Then u vanishes in all of D.

Proof: Let z ∈ U ∩ ∂D and B ⊂ U a ball centered at z. Set Γ = D ∩ ∂B. Then∂(B ∩D) = Γ ∪ (B ∩ ∂D). (The reader should sketch the situation.) We define v by

v(x) =

∫Γ

Φ(x, y)

∂u

∂ν(y)− u(y)

∂Φ

∂ν(y)(x, y)

ds(y) , x ∈ B .

Then v satisfies the Helmholtz equation in B. Application of Green’s representation formulaof Theorem 3.3 to u in B ∩D yields2

u(x) =

∫∂(B∩D)

Φ(x, y)

∂u

∂ν(y)− u(y)

∂Φ

∂ν(y)(x, y)

ds(y) = v(x) , x ∈ B ∩D ,

since the integral vanishes on ∂D ∩ B. By the same theorem we conclude that v(x) = 0 forx ∈ B \D. Since v is analytic by the previous corollary we conclude that v vanishes in all ofB. In particular, u vanishes in B ∩D. Again, u is analytic in D and D is connected, thusalso u vanishes in all of D. 2

For radiating solutions of the Helmholtz equation we have the following version of Green’srepresentation theorem.

Theorem 3.6 (Green’s representation theorem in the exterior of D)

2In this case the region B∩D does not meet the smoothness assumptions of the beginning of this section.The representation theorem still holds by the remark following Theorem 3.3.


Let k ∈ R>0 and u ∈ C2(R3 \ D) ∩ C1(R3 \ D) be a solution of the Helmholtz equation∆u + k2u = 0 in R3 \ D. Furthermore, let u satisfy the Sommerfeld radiation condition(3.2). Then we have the representation

∫∂D

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y) =

u(x) , x /∈ D ,

12u(x) , x ∈ ∂D ,

0 , x ∈ D .

The domain integral as well as the surface integral (for x ∈ ∂D) exists as improper integrals.

Proof: Let first x /∈ D. We choose R > |x| such that D ⊂ B(0, R) and apply Green’srepresentation Theorem 3.3 in the annular region B(0, R) \ D. Noting that ∆u + k2u = 0and that ν(y) for y ∈ ∂D is directed into the interior of B(0, R) \D yields

u(x) =

∫∂D

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y)

−∫|y|=R

u(y)

∂Φ

∂ν(y)(x, y)− Φ(x, y)

∂u

∂ν(y)

ds(y) .

We show that the surface integral over ∂B(0, R) tends to zero as R tends to infinity. Wewrite this surface integral (for fixed x) as

IR =

∫|y|=R

u(y)

[∂Φ

∂ν(y)(x, y)− ikΦ(x, y)

]ds(y) −

∫|y|=R

Φ(x, y)

[∂u

∂ν(y)− ik u(y)

]ds(y)

and use the Cauchy-Schwarz inequality:

|IR|2 ≤∫|y|=R

|u|2ds∫|y|=R

∣∣∣∣ ∂Φ


∣∣∣∣2 ds(y)

+

∫|y|=R

∣∣Φ(x, y)∣∣2ds(y)

∫|y|=R

∣∣∣∣∂u∂ν − ik u∣∣∣∣2 ds(y)

From the radiations of Φ(x, ·) (now for fixed x and with respect to y) and u we concludethat the integrands of the second and forth integral behave as O(1/R4) as R → ∞. Sincethe surface area of ∂B(0, R) is equal to 4πR2 we conclude that second and forth integraltend to zero as R tends to infinity. Furthermore, the integrand of the third integral behavesas O(1/R2) as R → ∞. Therefore, the third integral is bounded. It remains to show thatalso

∫|y|=R |u|

2ds is bounded. This follows again from the radiation condition. Indeed, from

the radiation condition we conclude that∫|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds =

∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + k2|u|2

ds + 2k Im

∫|x|=R

u∂u

∂rds .

Green’s theorem, applied in B(0, R) \D to the function u yields that∫|x|=R

u∂u

∂rds =

∫∂D

u∂u

∂rds +

∫∫B(0,R)\D

[|∇u|2 − k2|u|2

]dx .


The volume integral is real valued. Therefore, its imaginary part vanishes and we concludethat ∫

|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + k2|u|2 ds = −2k Im

∫∂D

u∂u

∂rds + O

(1

R2

).

This implies, in partular, that∫|y|=R |u|

2ds is bounded. Altogether, we have shown that IRtends to zero as R tends to infinity. 2

3.1.3 Volume and Surface Potentials

We have seen in the preceding section that any function can be represented by a combina-tion of volume and surface potentials. The integral equation method for solving Maxwell’sequations rely heavily on the smoothness properties of these potentials. This subsection isconcerned with the investigation of these potentials. The analysis is quite technical and usesthe tools from differential geometry from Subsection ?? of Chapter 1.

We recall the fundamental solution (in this subsection only for k ∈ R, k ≥ 0)

Φ(x, y) =eik|x−y|

4π|x− y|, x 6= y , (3.6)

and begin with the volume potential

w(x) =

∫∫D

ϕ(y) Φ(x, y) dy , x ∈ R3 . (3.7)

Lemma 3.7 Let ϕ : D → C be piecewise continuous.3 Then w ∈ C1(R3) and

∂w

∂xj(x) =

∫∫D

ϕ(y)∂Φ

∂xj(x, y) dy , x ∈ R3 , j = 1, 2, 3 . (3.8)

Proof: We fix j ∈ 1, 2, 3 and a real valued function η ∈ C1(R) with 0 ≤ η(t) ≤ 1 andη(t) = 0 for t ≤ 1 and η(t) = 1 for t ≥ 2. We set

v(x) =

∫∫D

ϕ(y)∂Φ

∂xj(x, y) dy , x ∈ R3 ,

and note that the integal exists as an improper integral by Lemma 3.2 since∣∣∣ ∂Φ∂xj

(x, y)∣∣∣ ≤

c0

(|x− y|−2

)for some c0 > 0. Furthermore, set

wε(x) =

∫∫D

ϕ(y) Φ(x, y) η(|x− y|/ε

)dy , x ∈ R3 .

Then wε ∈ C1(R3) and

v(x)− ∂wε∂xj

(x) =

∫∫|y−x|≤2ε

ϕ(y)∂

∂xj

Φ(x, y)

[1− η

(|x− y|/ε

)]dy ,

3This means: There exist finitely many domains Dj with D = ∪jDj such that ϕ|Djhas a continuous

extension to Dj for every j.


and thus ∣∣∣∣v(x)− ∂wε∂xj

(x)

∣∣∣∣ ≤ ‖ϕ‖∞∫∫|y−x|≤2ε

∣∣∣∣ ∂Φ

∂xj(x, y)

∣∣∣∣+‖η′‖∞ε

∣∣Φ(x, y)∣∣ dy

≤ c1

∫∫|y−x|≤2ε

[1

|x− y|2+

1

ε |x− y|

]dy

= c1

2π∫0

π∫0

2ε∫0

[1

r2+

1

ε r

]r2 sin θ dr dθ dϕ = 16π c1 ε .

Therefore, ∂wε/∂xj → v uniformly in R3. Thus w ∈ C1(R3) and ∂w/∂xj = v. 2

For any set T and α ∈ (0, 1] we define the space Cα(T ) of bounded Holder-continuousfunctions v : T → C by

Cα(T ) :=

v ∈ C(T ) : v bounded and sup

x,y∈T, x 6=y

∣∣v(x)− v(y)∣∣

|x− y|α< ∞

Note that any Holder-continuous function is uniformly continuous and, therefore, has acontinuous extension to T . The space Cα(T ) is a normed space with norm

‖v‖Cα(T ) := supx∈T

∣∣v(x)∣∣︸︷︷︸

= ‖v‖∞

+ supx,y∈T, x 6=y

∣∣v(x)− v(y)∣∣

|x− y|α. (3.9)

Theorem 3.8 Let ϕ ∈ Cα(D) and let w be the volume potential. Then w ∈ C2(D)∩C∞(R3\D) and

∆w + k2w =

−ϕ , in D ,0 , in R3 \D .

Furthermore, if D0 is any C2−smooth domain with D ⊂ D0 then

∂2w

∂xi∂xj(x) =

∫∫D0

∂2Φ

∂xi∂xj(x, y)

[ϕ(y)−ϕ(x)

]dy − ϕ(x)

∫∂D0

∂Φ

∂xi(x, y) νj(y) ds(y) (3.10)

for x ∈ D where we have extended ϕ by zero in D0 \D.

Proof: First we note that the volume integral in the last formula exists. Indeed, we fixx ∈ D and split the region of integration into D0 = D ∪ (D0 \D). The integral over D0 \Dexists because the integrand is smooth. The integral over D exists again by Lemma 3.2since

∣∣ϕ(y)− ϕ(x)∣∣∣∣∂2Φ/(∂xi∂xj)(x, y)

∣∣ ≤ c |x− y|α−3 for y ∈ D. The surface integral is noproblem since x /∈ ∂D0.We fix i, j ∈ 1, 2, 3 and the same funtion η ∈ C1(R) as in the previous lemma. We definev := ∂w/∂xi and

u(x) :=

∫∫D0

∂2Φ

xixj(x, y)

[ϕ(y)− ϕ(x)

]dy − ϕ(x)

∫∂D0

∂Φ

∂xi(x, y) νj(y) ds(y)

vε(x) :=

∫∫D

ϕ(y) η(|x− y|/ε

) ∂Φ

∂xi(x, y) dy , x ∈ R3 .


Then vε ∈ C1(D) and for x ∈ D

∂vε∂xj

(x) =

∫∫D

ϕ(y)∂

∂xj

η(|x− y|/ε

) ∂Φ

∂xi(x, y)

dy

=

∫∫D0

[ϕ(y)− ϕ(x)

] ∂

∂xj

η(|x− y|/ε

) ∂Φ

∂xi(x, y)

dy

+ ϕ(x)

∫∫D0

∂

∂xj

η(|x− y|/ε

) ∂Φ

∂xi(x, y)

dy

=

∫∫D0

[ϕ(y)− ϕ(x)

] ∂

∂xj

η(|x− y|/ε

) ∂Φ

∂xi(x, y)

dy

− ϕ(x)

∫∂D0

∂Φ

∂xi(x, y) νj(y) ds(y)

provided 2ε ≤ d(x, ∂D0). In the last step we used the Divergence Theorem! Therefore,∣∣∣∣u(x)− ∂vε∂xj

(x)

∣∣∣∣ ≤ ∫∫|y−x|≤2ε

∣∣ϕ(y)− ϕ(x)∣∣ ∣∣∣∣ ∂∂xj

(1− η

(|x− y|/ε

)) ∂Φ

∂xi(x, y)

∣∣∣∣ dy≤ c

∫∫|y−x|≤2ε

(1

|y − x|3+‖η′‖∞

ε |y − x|2

)|y − x|α dy

=

2ε∫0

(1

r1−α +‖η′‖∞ε

rα)dr

≤ 4π c

[(2ε)α

α+

(2ε)1+α

(1 + α) ε

]≤ c ε ,

provided 2ε ≤ dist(x, ∂D). Therefore, ∂vε/∂xj → u uniformly on compact subsets of D.Also, vε → v uniformly on compact subsets of D and thus w ∈ C2(D) and u = ∂2w/(∂xi∂xj).This proves (3.10). Finally, setting D0 = B(x,R):

∆w(x) = −k2

∫∫B(x,R)

Φ(x, y)[ϕ(y)− ϕ(x)

]dy

− ϕ(x)3∑j=1

yj − xjR

[exp(ikR)

4π R

(ik − 1/R

) xj − yjR

]4π R2

= −k2w(x) + k2ϕ(x)

∫∫B(x,R)

exp(ik|x− y|

)4π |x− y|

dy − ϕ(x) eikR(1− ikR) ,

i.e.

∆w(x) + k2w(x) = −ϕ(x)

[eikR(1− ikR)− k2

∫∫B(x,R)

exp(ik|x− y|

)4π |x− y|

dy

]︸︷︷︸

= 1


since

k2

∫∫B(x,R)

exp(ik|x− y|

)4π |x− y|

dy = 4π k2

R∫0

exp(ikr)

4π rr2dr = k2

R∫0

r eikrdr

= · · · = −ikR eikR + eikR − 1 .

2

Corollary 3.9 Let D be C2−smooth and A ⊂ R3 be a closed set with A ⊂ D or A ⊂ R3\D.Furthermore, let w be the volume integral with density ϕ ∈ Cα(∂D). Then there exists c > 0with

‖w‖C1(R3) ≤ c ‖ϕ‖∞ and ‖w‖C2(A) ≤ c ‖ϕ‖Cα(D) (3.11)

for all ϕ ∈ Cα(∂D).

Proof: We estimate ∣∣Φ(x, y)∣∣ =

1

4π|x− y|,∣∣∣∣ ∂Φ

∂xj(x, y)

∣∣∣∣ ≤ c1

[1

|x− y|2+

1

|x− y|

],∣∣∣∣ ∂2Φ

∂xi∂xj(x, y)

∣∣∣∣ ≤ c2

[1

|x− y|3+

1

|x− y|2+

1

|x− y|

].

For x ∈ R3 we estimate, using (3.8) and Lemma 3.2 above,

∣∣w(x)∣∣ ≤ ‖ϕ‖∞

∫∫D

1

4π|x− y|dy ≤ c ‖ϕ‖∞ ,∣∣∣∣ ∂w∂xj (x)

∣∣∣∣ ≤ ‖ϕ‖∞∫∫

D

∣∣∣∣ ∂Φ

∂xj(x, y)

∣∣∣∣ dy ≤ c ‖ϕ‖∞ ,

where the constant is independent of x and ϕ. This proves already the first estimate of(3.11). Let now x ∈ A. If A ⊂ D then there exists δ > 0 with |x− y| ≥ δ for all x ∈ A andy ∈ ∂D. By (3.10) for D = D0 we have∣∣∣∣ ∂2w

∂xi∂xj(x)

∣∣∣∣ ≤ ‖ϕ‖Cα(∂D)

∫∫D

∣∣∣∣ ∂2Φ

∂xi∂xj(x, y)

∣∣∣∣ |x− y|α dy + ‖ϕ‖∞∫∂D

∣∣∣∣ ∂Φ

∂xi(x, y)

∣∣∣∣ ds(y)

≤ c3 ‖ϕ‖Cα(∂D)

∫∫D

dy

|x− y|3−α+ c4 ‖ϕ‖∞

∫∂D

ds(y)

|x− y|2

≤ c5 ‖ϕ‖Cα(∂D) +c4

δ2‖ϕ‖∞

∫∂D

ds

≤ c‖ϕ‖Cα(∂D) .


If A ⊂ R3 \D then there exists δ > 0 with |x− y| ≥ δ for all x ∈ A and y ∈ D. Therefore,we can estimate∣∣∣∣ ∂2w

∂xi∂xj(x)

∣∣∣∣ ≤ ‖ϕ‖∞ ∫∫D

∣∣∣∣ ∂2Φ

∂xi∂xj(x, y)

∣∣∣∣ dy ≤ ‖ϕ‖∞ c

δ3

∫∫D

dy .

2

We continue with the single layer surface potential

v(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) , x ∈ R3 . (3.12)

The investigation of this potential requires some elementary facts from differential geometrywhich we have collected in Subsection ?? of Chapter 1.

First we note that for continuous densities ϕ the integral exists (as an improper integral)even for x ∈ ∂D since Φ has a singularity of the form

∣∣Φ(x, y)∣∣ = 1/(4π|x− y|). This follows

from Lemma 3.2 above. Before we prove continuity of v we make the following generalremark.

Remark 3.10 A function v : R3 → C is Holder continuous if

(a) v is bounded,

(b) v is Holder continuous in some Hρ,

(c) v is Lipschitz continuous in R3 \ Uδ for every δ > 0.

Again, Hρ is the strip around ∂D of thickness ρ and Uδ is the neighborhood of ∂D withthickness δ (see Lemma 6.4).

Proof : Choose δ > 0 such that U3δ ⊂ Hρ.Ist case: |x1 − x2| < δ and x1 ∈ U2δ. Then x1, x2 ∈ U3δ ⊂ Hρ and Holder continuity followsfrom (b).2nd case: |x1 − x2| < δ and x1 /∈ U2δ. Then x1, x2 /∈ Uδ and thus from (c):∣∣v(x1)− v(x2)

∣∣ ≤ c|x1 − x2| ≤ cδ1−α|x1 − x2|α .

3rd case: |x1 − x2| ≥ δ. Then, by (a),∣∣v(x1)− v(x2)∣∣ ≤ 2‖v‖∞ ≤

2‖v‖∞δα

|x1 − x2|α .

Theorem 3.11 The single-layer potential v from (3.12) with continuous density ϕ is uni-formly Holder continuous in all of R3, and for every α ∈ (0, 1) there exists c > 0 (independentof ϕ) with

‖v‖Cα(R3) ≤ c‖ϕ‖∞ . (3.13)


Proof: We check the conditions (a), (b), (c) of Remark 3.10. Boundedness follows fromLemma 3.2 since

∣∣Φ(x, y)∣∣ = 1/(4π|x− y|). For (b) and (c) we write∣∣v(x1)− v(x2)

∣∣ ≤ ‖ϕ‖∞ ∫∂D

∣∣Φ(x1, y)− Φ(x2, y)∣∣ ds(y) (3.14)

and estimate∣∣Φ(x1, y)− Φ(x2, y)∣∣ ≤ 1

4π

∣∣∣∣ 1

|x1 − y|− 1

|x2 − y|

∣∣∣∣ +1

4π|x1 − y|∣∣eik|x1−y| − eik|x1−y|

∣∣≤ |x1 − x2|

4π |x1 − y||x2 − y|+

k |x1 − x2|4π |x1 − y|

(3.15)

since∣∣exp(it) − exp(is)

∣∣ ≤ |t − s| for all t, s ∈ R. Now (c) follows since |xj − y| ≥ δ forxj /∈ Uδ.

To show (b); that is, Holder continuity in Hρ0 , let x1, x2 ∈ Hρ0 . We set Γz,r =y ∈ ∂D :

|y − z| < r

and split the domain of integration into Γz1,r and ∂D \ Γz1,r where we setr = 3|x1 − x2|. The integral over Γz1,r is simply estimated by∫

Γz1,r

∣∣Φ(x1, y)− Φ(x2, y)∣∣ ds(y) ≤ 1

4π

∫Γz1,r

ds(y)

|x1 − y|+

1

4π

∫Γz1,r

ds(y)

|x2 − y|

≤ 1

2π

∫Γz1,r

ds(y)

|z1 − y|+

1

2π

∫Γz2,2r

ds(y)

|z2 − y|

where we used the unique representation of xj in the form xj = zj+tjν(zj) with zj ∈ ∂D and|tj| < ρ0 (see Lemma 6.4). By this lemma we conclude that |zj − y| ≤ 2|xj − y| for j = 1, 2and Γz1,r ⊂ Γz2,2r because |y− z2| ≤ |y− z1|+ |z1− z2| ≤ |y− z1|+ 2|x1− x2| ≤ |y− z1|+ r.The estimate ∫

Γz,ρ

ds(y)

|z − y|≤ c1 ρ

for some c1 independent of z and ρ has been proven in (3.3d). Therefore, we have shownthat ∫

Γz1,r

∣∣Φ(x1, y)− Φ(x2, y)∣∣ ds(y) ≤ c

2π(r + 2r) =

9c

2π|x1 − x2| ≤ c |x1 − x2|α

where c is independent of xj.

Now we continue with the integral over ∂D \ Γz1,r. For y ∈ ∂D \ Γz1,r we have 3|x1 − x2| =r ≤ |y− z1| ≤ 2|y− x1|, thus |x2− y| ≥ |x1− y| − |x1− x2| ≥ (1− 2/3) |x1− y| = |x1− y|/3and therefore∣∣Φ(x1, y)− Φ(x2, y)

∣∣ ≤ 3|x1 − x2|4π |x1 − y|2

+k |x1 − x2|4π |x1 − y|

≤ 3|x1 − x2|π |z1 − y|2

+k |x1 − x2|2π |z1 − y|


since |z1 − y| ≤ 2|x1 − y|. Then we estimate∫∂D\Γz1,r

∣∣Φ(x1, y)− Φ(x2, y)∣∣ ds(y)

≤ |x1 − x2|π

∫∂D\Γz1,r

[3

|z1 − y|2+

k

2|z1 − y|

]ds(y)

=|x1 − x2|α

π

∫∂D\Γz1,r

(r/3)1−α[

3

|z1 − y|2+

k

2|z1 − y|

]ds(y)

≤ |x1 − x2|α

π 31−α

∫∂D\Γz1,r

[3

|z1 − y|2−(1−α)+

k

2 |z1 − y|1−(1−α)

]ds(y)

since r1−α ≤ |y − z1|1−α for y ∈ ∂D \ Γz1,r. Therefore,∫∂D\Γz1,r

∣∣Φ(x1, y)− Φ(x2, y)∣∣ ds(y) ≤ 1

π 31−α |x1 − x2|α[3 c1−α +

k

2c2−α

]with the constants cβ from Lemma 3.2, part (b). Altogether we have shown the existence ofc > 0 with ∣∣v(x1)− v(x2)

∣∣ ≤ c ‖ϕ‖∞ |x1 − x2|α for all x1, x2 ∈ Hρ0

which ends the proof. 2

Before we continue with the investigation of the double layer surface potential we prove anauxiliary result which we will use often in the following.

Lemma 3.12 For ϕ ∈ Cα(∂D) and a ∈ C(∂D)3 define

w(x) =

∫∂D

[ϕ(y)− ϕ(z)

]a(y) · ∇yΦ(x, y) ds(y) , x ∈ Hρ0 ,

where x = z+tν(z) ∈ Hρ0 with |t| < ρ0 and z ∈ ∂D. Then the integral exists for x ∈ ∂D as animproper integral and w is Holder continuous in Hρ0 for any exponent β < α. Furthermore,there exists c > 0 with

∣∣w(x)∣∣ ≤ c‖ϕ‖Cβ(∂D) for all x ∈ Hρ0, and the constant c does not

depend on x and ϕ (but may depend on a).

Proof: For x` = z` + t`ν(z`) ∈ Hρ0 , ` = 1, 2, we have to estimate

w(x1)− w(x2)

=

∫∂D

[ϕ(y)− ϕ(z1)

]a(y) · ∇yΦ(x1, y)−

[ϕ(y)− ϕ(z2)

]a(y) · ∇yΦ(x2, y) ds(y) (3.16)

=[ϕ(z2)− ϕ(z1)

] ∫∂D

a(y) · ∇yΦ(x1, y) ds(y) +

+

∫∂D

[ϕ(y)− ϕ(z2)

]a(y) ·

[∇yΦ(x1, y)−∇yΦ(x2, y)

]ds(y)


and thus∣∣w(x1)− w(x2)∣∣ ≤ ∣∣ϕ(z2)− ϕ(z1)

∣∣ ∣∣∣∣∫∂D

a(y) · ∇yΦ(x1, y) ds(y)

∣∣∣∣+

+ ‖a‖∞∫∂D

∣∣ϕ(y)− ϕ(z2)∣∣ ∣∣∇yΦ(x1, y)−∇yΦ(x2, y)

∣∣ ds(y) (3.17)

We need the following estimates of ∇yΦ: There exists c > 0 with

∣∣∇yΦ(x, y)∣∣ ≤ c

|x− y|2, x ∈ Hρ0 , y ∈ ∂D ,

∣∣∇yΦ(x1, y)−∇yΦ(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3

, y ∈ ∂D , x` ∈ Hρ0 with |x1 − y| ≤ 3|x2 − y| .

Proof of these estimates: The first one is obvious. For the second one we observe thatΦ(x, y) = φ(|x − y|) with φ(t) = exp(ikt)/(4πt), thus φ′(t) = φ(t)(ik − 1/t) and φ′′(t) =φ′(t)(ik−1/t)+φ(t)/t2 = φ(t)

[(ik−1/t)2 +1/t2

]and therefore

∣∣φ′(t)∣∣ ≤ c1/t2 and

∣∣φ′′(t)∣∣ ≤c2/t

3 for 0 < t ≤ 1. For t ≤ 3s we have

∣∣φ′(t)− φ′(s)∣∣ =

∣∣∣∣∣∣t∫

s

φ′′(τ) dτ

∣∣∣∣∣∣ ≤ c2

∣∣∣∣∣∣t∫

s

dτ

τ 3

∣∣∣∣∣∣ =c2

2|t−2 − s−2|

=c2

2

|t2 − s2|t2s2

≤ c2

2|t− s|

(1

ts2+

1

t2s

)≤ c2

2|t− s|

(9

t3+

3

t3

)= 6c2

|t− s|t3

.

Setting t = |x1−y| and s = |x2−y| and observing that |t−s| ≤ |x1−x2| yields the estimate

∣∣∇yΦ(x1, y)−∇yΦ(x2, y)∣∣ =

∣∣∣∣φ′(|x1 − y|)y − x1

|y − x1|− φ′(|x2 − y|)

y − x2

|y − x2|

∣∣∣∣≤

∣∣φ′(|x1 − y|)− φ′(|x2 − y|)∣∣

+∣∣φ′(|x2 − y|)

∣∣ ∣∣∣∣ y − x1

|y − x1|− y − x2

|y − x2|

∣∣∣∣≤ 6 c2

|x2 − x1||y − x1|3

+ 2∣∣φ′(|x2 − y|)

∣∣ |x2 − x1||y − x1|

≤ 6 c2|x2 − x1||y − x1|3

+ 2 c1|x2 − x1|

|y − x1||y − x2|2

≤ (6c2 + 18c1)|x2 − x1||y − x1|3

.

This yields the second estimate.


Now we split the region of integration again into Γz1,r and ∂D \ Γz1,r with r = 3|x1 − x2|where again Γz,r = y ∈ ∂D : |y − z| < r. The integral (in the form (3.16)) over Γz1,r isestimated by4∫

Γz1,r

∣∣[ϕ(y)− ϕ(z1)]a(y) · ∇yΦ(x1, y)−

[ϕ(y)− ϕ(z2)

]a(y) · ∇yΦ(x2, y)

∣∣ ds(y) (3.18)

≤ c

∫Γz1,r

[|y − z1|α

1

|y − x1|2+ |y − z2|α

1

|y − x2|2

]ds(y)

≤ c

∫Γz1,r

|y − z1|α1

|y − z1|2ds(y) + c

∫Γz2,2r

|y − z2|α1

|y − z2|2ds(y)

because Γz1,r ⊂ Γz2,2r and |x` − y| ≥ |z` − y|/2. Therefore, using (3.3d), this term behavesas rα = 3α|x1 − x2|α ≤ c|x1 − x2|β.We finally consider the integral over ∂D \ Γz1,r und use the form (3.17):

I :=∣∣ϕ(z2)− ϕ(z1)

∣∣ ∣∣∣∣∣∫∂D\Γz1,r

a(y) · ∇yΦ(x1, y) ds(y)

∣∣∣∣∣+

+ ‖a‖∞∫∂D\Γz1,r

∣∣ϕ(y)− ϕ(z2)∣∣ ∣∣∇yΦ(x1, y)−∇yΦ(x2, y)

∣∣ ds(y)

≤ c

∫∂D\Γz1,r

[|z2 − z1|α

1

|x1 − y|2+ |y − z2|α

|x1 − x2||x1 − y|3

]ds(y)

Since y ∈ ∂D \ Γz1,r we use the estimates

• |x1 − x2| = r3≤ 1

3|y − z1| ≤ 2

3|y − x1| < |y − x1| and

• |y − z2| ≤ 2|y − x2| ≤ 2|y − x1|+ 2|x1 − x2| ≤ 4|y − x1|,

thus

I ≤ c

∫∂D\Γz1,r

[|x2 − x1|α

1

|z1 − y|2+|x1 − x2||x1 − y|3−α

]ds(y)

≤ c′|x1 − x2|β∫

∂D\Γz1,r

[|x2 − x1|α−β

1

|z1 − y|2+|x1 − x2|1−β

|z1 − y|3−α

]ds(y)

≤ c|x1 − x2|β∫

∂D\Γz1,r

1

|z1 − y|2−(α−β)ds(y) ≤ c cα−β|x1 − x2|β

with the constant cα−β from Lemma 3.2. This, together with (3.18) proves the Holder-continuity of w. The proof of the estimate

∣∣w(x)∣∣ ≤ v‖ϕ‖Cβ(∂D) for x ∈ Hρ0 is simpler and

left to the reader. 2

4The constant c is different from line to line.


Now we continue with the double layer surface potential

v(x) =

∫∂D

ϕ(y)∂Φk

∂ν(y)(x, y) ds(y) , x ∈ R3 \ ∂D , (3.19)

for Holder-continuous densities ϕ. Here we indicate the dependence on k ≥ 0 by writing Φk.

Theorem 3.13 The double layer potential v from (3.19) with Holder-continuous densityϕ ∈ Cα(∂D) can be continuously extended from D to D and from R3 \ D to R3 \ D withlimiting values

limx→x0x∈D

v(x) = −1

2ϕ(x0) +

∫∂D

ϕ(y)∂Φk

∂ν(y)(x0, y) ds(y) , x0 ∈ ∂D , (3.20a)

limx→x0

x/∈D

v(x) = +1

2ϕ(x0) +

∫∂D

ϕ(y)∂Φk

∂ν(y)(x0, y) ds(y) , x0 ∈ ∂D . (3.20b)

v is Holder-continuous in D and in R3 \D with exponent β for every β < α. The integralsexist as improper integrals.

Proof: First we note that the integrals exist because for x0, y ∈ ∂D we can estimate∣∣∣∣ ∂Φk

∂ν(y)(x0, y)

∣∣∣∣ =

∣∣∣∣exp(ik|x0 − y|)4π|x0 − y|

(ik − 1

|x0 − y|

)∣∣∣∣∣∣ν(y) · (y − x0)

∣∣|y − x0|

≤ c

|x0 − y|.

by Lemma 6.4. Furthermore, v has a decomposition into v = v0 + v1 where

v0(x) =

∫∂D

ϕ(y)∂Φ0

∂ν(y)(x, y) ds(y) , x ∈ R3 \ ∂D ,

v1(x) =

∫∂D

ϕ(y)∂(Φk − Φ0)

∂ν(y)(x, y) ds(y) , x ∈ R3 \ ∂D .

It is easily seen that the kernel K(x, y) = ϕ(y) ∂(Φk−Φ0)∂ν(y)

(x, y) of the integral in the definition

of v1 is continuous in R3×R3 and continuously differentiable with respect to x and boundedon the set

(x, y) ∈ R3 × ∂D : x 6= y

. From this it follows that v1 is Holder-continuous in

all of R3. Also this is easy to prove.We continue with the analysis of v0 and note that we have again to prove estimates of theform (a) and (b) of Remark 3.10.First, for x ∈ Uδ/4 we have x = z + tν(z) with |t| < δ/4 and z ∈ ∂D. We write v0(x) in theform

v0(x) =

∫∂D

[ϕ(y)− ϕ(z)

] ∂Φ0

∂ν(y)(x, y) ds(y)︸︷︷︸

= v0(x)

+ ϕ(z)

∫∂D

∂Φ0

∂ν(y)(x, y) ds(y) .

The function v0 is Holder continuous in Uδ/4 with exponent β < α by Lemma 3.12 (takea(y) = ν(y)). This proves the estimate (a) and (b) of Remark 3.10 for v0.


Now we go back to the decomposition

v0(x) = v0(x) + ϕ(z)

∫∂D

∂Φ0

∂ν(y)(x, y) ds(y)

By Green’s representation (Theorem 3.3) for k = 0 and u = 1 we observe that

∫∂D

∂Φ0

∂ν(y)(x, y) ds(y) =

−1 , x ∈ D ,−1/2 , x ∈ ∂D ,

0 , x /∈ D .

Therefore,

limx→x0, x∈D

v0(x) = v0(x0)− ϕ(x0) = v0(x0) +1

2ϕ(x0)− ϕ(x0) = −1

2ϕ(x0) ,

limx→x0, x/∈D

v0(x) = v0(x0) = v0(x0) +1

2ϕ(x0) .

This ends the proof. 2

In the following we write

v(x0)∣∣− = lim

x→x0x∈D

v(x) and v(x0)∣∣+

= limx→x0

x/∈D

v(x) and

∂v

∂ν(x0)

∣∣∣∣−

= limx→x0x∈D

ν(x0) · ∇v(x) and∂v

∂ν(x0)

∣∣∣∣+

= limx→x0

x/∈D

ν(x0) · ∇v(x) .

We continue with the derivative of the single layer potential (3.12); that is,

v(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) , x ∈ R3 .

First we show the following auxiliary result.

Lemma 3.14 (a) There exists c > 0 with∣∣∣∣∫∂D\K(x,τ)

∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ c for all x ∈ R3 , τ > 0 ,

(b) limτ→0

∫∂D\K(x,τ)

∇xΦ(x, y) ds(y) =

∫∂D

H(y) Φ(x, y) ds(y)−∫∂D

ν(y)∂Φ


for x ∈ R3 where H(y) =(Div e1

t (y),Div e2t (y),Div e3

t (y))> ∈ R3 and ejt(y) = ν(y) ×

(ej ×

ν(y)), j = 1, 2, 3, the tangential components of the unit vectors ej.


Proof: For any x ∈ R3 we have∫∂D\K(x,τ)

∇xΦ(x, y) ds(y) = −∫

∂D\K(x,τ)

∇yΦ(x, y) ds(y)

= −∫

∂D\K(x,τ)

Grad yΦ(x, y) ds(y)−∫

∂D\K(x,τ)

∂Φ

∂ν(y)(x, y) ν(y) ds(y)

and thus for any fixed vector a ∈ C3 by the previous theorem (here again Γ(x, τ) = ∂D ∩K(x, τ) and at(y) = ν(y)×

(a× ν(y)

))

a ·∫

∂D\K(x,τ)

∇xΦ(x, y) ds(y) =

∫∂D\K(x,τ)

Div at(y) Φ(x, y) ds(y)

+

∫∂Γ(x,τ)

at(y) · ν0(y) Φ(x, y) ds(y)

+

∫∂D\K(x,τ)

∂Φ

∂ν(y)(x, y) a · ν(y) ds(y) .

The first and third integrals converge uniformly with respect to x ∈ ∂D when τ tends tozero because the integrands are weakly singular. For the second integral we note that∣∣∣∣∣∣∣

∫∂Γ(x,τ)

at(y) · ν0(y) Φ(x, y) ds(y)

∣∣∣∣∣∣∣ =1

4πτ

∣∣∣∣∣∣∣∫

∂Γ(x,τ)

at(y) · ν0(y) ds(y)

∣∣∣∣∣∣∣=

1

4πτ

∣∣∣∣∣∣∣∫

Γ(x,τ)

Div at(y) ds(y)

∣∣∣∣∣∣∣and this tends to zero uniformly with respect to x ∈ ∂D when τ tends to zero. The conclusionfollows if we take for a the unit coordinate vectors e(j). 2

Theorem 3.15 The derivative of the single layer potential v from (3.12) with Holder-continuous density ϕ ∈ Cα(∂D) can be continuously extended from D to D and from R3 \Dto R3 \ D. The tangential component is continuous, i.e. Grad v|− = Grad v|+, and thelimiting values of the normal derivatives are

∂v

∂ν(x)

∣∣∣∣±

= ∓1

2ϕ(x) +

∫∂D

ϕ(y)∂Φ

∂ν(x)(x, y) ds(y) , x ∈ ∂D . (3.21)

The integral exists as an improper integral.


Proof: We note that the integral exists (see proof of Theorem 3.13). First we consider thedensity 1, i.e. we set

v1(x) =

∫∂D

Φ(x, y) ds(y) , x ∈ R3 .

By the previous lemma we have that

∇v1(x) = −∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) +

∫∂D

H(y) Φ(x, y) ds(y) , x /∈ ∂D , (3.22)

where again H(y) =(Div e1

t (y),Div e2t (y),Div e3

t (y))> ∈ R3.

The right hand side is the sum of a double and a single layer potential. By Theorems 3.11and 3.13 it has a continuous extension to the boundary from the inside and the outside withlimiting values

∇v1(x)|± = ∓1

2ν(x) −

∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) +

∫∂D

H(y) Φ(x, y) ds(y)

= ∓1

2ν(x) +

∫∂D

∇xΦ(x, y) ds(y)

for x ∈ ∂D. The last integral has to be interpreted as a Cauchy principal value as in part(b) of the previous Lemma3.14. In particular, the tangential component is continuous andthe normal derivatives jumps, and we have

∂v1

∂ν(x)

∣∣∣∣±

= ∓1

2+

∫∂D

∂Φ

∂ν(x)(x, y) ds(y) , x ∈ ∂D .

for x ∈ ∂D. Now we consider v and have for x = z + tν(z) ∈ Hρ0 \ ∂D (that is, t 6= 0)

∇v(x) =

∫∂D

ϕ(y)∇xΦ(x, y) ds(y) =

∫∂D

∇xΦ(x, y)[ϕ(y)− ϕ(z)

]ds(y)︸︷︷︸

= v(x)

+ ϕ(z)∇v1(x) .

Application of Lemma 3.12 yields that v is Holder continuous in all of Hρ0 with limitingvalue

v(x) =

∫∂D

∇xΦ(x, y)[ϕ(y)− ϕ(x)

]ds(y) for x ∈ ∂D .

For further use we formulate the result derived so far.

∇v(x)∣∣± =

∫∂D

∇xΦ(x, y)[ϕ(y)− ϕ(x)

]ds(y) + ϕ(x)

[∓1

2ν(x)

−∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) +

∫∂D

H(y) Φ(x, y) ds(y)

]. (3.23)


This proves that the gradient has continuous extensions from the inside and outside of Dand

∂v

∂ν(x)

∣∣∣∣±

= ∓1

2ϕ(x) + ϕ(x)

∫∂D

∂Φ

∂ν(x)(x, y) ds(y) +

∫∂D

∂Φ

∂ν(x)(x, y)

[ϕ(y)− ϕ(x)

]ds(y)

= ∓1

2ϕ(x) +

∫∂D

ϕ(y)∂Φ

∂ν(x)(x, y) ds(y) ,

Grad v(x)∣∣± =

∫∂D

Grad xΦ(x, y)[ϕ(y)− ϕ(z)

]ds(y) + ϕ(z) Grad v1(x) (3.24)

for x ∈ ∂D. 2

3.1.4 Boundary Integral Operators

It is the aim of this subsection to investigate the mapping properties of the traces of thesingle and double layer potentials on the boundary ∂D. We start with a general theorem onboundary integral operators with singular kernels.

Theorem 3.16 Let Λ =

(x, y) ∈ ∂D × ∂D : x 6= y

and G ∈ C(Λ).

(a) Let there exist c > 0 and α ∈ (0, 1) such that∣∣G(x, y)∣∣ ≤ c

|x− y|2−αfor all (x, y) ∈ Λ , (3.25a)

∣∣G(x1, y)−G(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3−α

for all (x1, y), (x2, y) ∈ Λ (3.25b)

with |x1 − y| ≥ 3|x1 − x2| .

Then the operator K1 : C(∂D)→ Cα(∂D), defined by

(K1ϕ)(x) =

∫∂D

G(x, y)ϕ(y) ds(y) , x ∈ ∂D ,

is well defined and bounded.

(b) Let there exist c > 0 such that:∣∣G(x, y)∣∣ ≤ c

|x− y|2for all (x, y) ∈ Λ , (3.25c)

∣∣G(x1, y)−G(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3

for all (x1, y), (x2, y) ∈ Λ (3.25d)

with |x1 − y| ≥ 3|x1 − x2| ,∣∣∣∣∫∂D\K(x,r)

G(x, y) ds(y)

∣∣∣∣ ≤ c for all x ∈ ∂D and r > 0 . (3.25e)


Then the operator K2 : Cα(∂D)→ Cα(∂D), defined by

(K2ϕ)(x) =

∫∂D

G(x, y)[ϕ(y)− ϕ(x)

]ds(y) , x ∈ ∂D ,


Proof: (a) We follow the idea of the proof of Theorem 3.11 and write

∣∣(K1ϕ)(x1)− (K1ϕ(x2)∣∣ ≤ ‖ϕ‖∞ ∫

∂D

∣∣G(x1, y)−G(x2, y)∣∣ ds(y) .

We split the region of integration again into Γx1,r and ∂D \ Γx1,r where again Γx1,r = y ∈∂D : |y − x1| < r and set r = 3|x1 − x2|. The integral over Γx1,r can be estimated with(3.3d) of Lemma 3.2 by

∫Γx1,r

∣∣G(x1, y)−G(x2, y)∣∣ ds(y) ≤ c

∫Γx1,r

ds(y)

|x1 − y|2−α+ c

∫Γx2,2r

ds(y)

|x2 − y|2−α

≤ c′rα = (c′3α) |x1 − x2|α

because Γx1,r ⊂ Γx2,2r. Here we used formula (3.3d) of Lemma 3.2.

For the integral over ∂D \ Γx1,r we note that |y − x1| ≥ r = 3|x1 − x2| for y ∈ ∂D \ Γx1,r,thus

∫∂D\Γx1,r

∣∣G(x1, y)−G(x2, y)∣∣ ds(y) ≤ c |x1 − x2|

∫∂D\Γx1,r

ds(y)

|x1 − y|3−α

≤ c′|x1 − x2| rα−1 = c′3α−1 |x1 − x2|α

where we used estimate (3.3e). The proof of∣∣(K1ϕ)(x)

∣∣ ≤ c‖ϕ‖∞ is similar (and simpler)and is left to the reader.


For part (b) we follow the ideas of the proof of Lemma 3.12. We write

∣∣(K2ϕ)(x1)− (K2ϕ(x2)∣∣ ≤ ∫

Γx1,r

∣∣G(x1, y)[ϕ(y)− ϕ(x1)

]−G(x2, y)

[ϕ(y)− ϕ(x2)

]∣∣ ds(y)

+∣∣ϕ(x2)− ϕ(x1)

∣∣ ∣∣∣∣∣∫∂D\Γx1,r

G(x1, y) ds(y)

∣∣∣∣∣+

∫∂D\Γx1,r

∣∣ϕ(y)− ϕ(x2)∣∣ ∣∣G(x1, y)−G(x2, y)

∣∣ ds(y)

≤ c‖ϕ‖Cα(∂D)

[∫Γx1,r

ds(y)

|y − x1|2−α+

∫Γx2,2r

ds(y)

|y − x2|2−α

]+ c‖ϕ‖Cα(∂D)|x1 − x2|α

+ c‖ϕ‖Cα(∂D)

∫∂D\Γx1,r

|y − x2|α|x1 − x2||y − x1|3

ds(y)

≤ c‖ϕ‖Cα(∂D)

rα + |x1 − x2|α + |x1 − x2|∫

∂D\Γx1,r

ds(y)

|y − x1|3−α

because |y − x2| ≤ |y − x1| + |x1 − x2| = |y − x1| + r/3 ≤ 2|y − x1|. The last integral hasbeen estimated by rα−1, see (3.3e). This proves that∣∣(K2ϕ)(x1)− (K2ϕ(x2)

∣∣ ≤ c‖ϕ‖Cα(∂D) |x1 − x2|α .

The proof of∣∣(K2ϕ)(x)

∣∣ ≤ c‖ϕ‖Cα(∂D) is again simpler and is left again to the reader. 2

We need compactness properties of boundary operators in Holderspaces. This follows fromthe previous theorem and the compact imbedding of Cα(∂D) in C(∂D).

Lemma 3.17 The imbedding Cα(∂D)→ C(∂D) is compact for every α ∈ (0, 1).

Proof: We have to prove that the unit ball B =ϕ ∈ Cα(∂D) : ‖ϕ‖Cα(∂D) ≤ 1

is

relatively compact5 in C(∂D). This follows directly from the theorem of Arcela-Ascoli (see[3]). Indeed, B is equi-continuous because∣∣ϕ(x1)− ϕ(x2)

∣∣ ≤ ‖ϕ‖Cα(∂D)|x1 − x2|α ≤ |x1 − x2|α

for all x1, x2 ∈ ∂D. Furthermore, B is bounded. 2

Corollary 3.18 Under the assumptions of Theorem 3.16 the operator K1 is compact fromCα(∂D) into itself for every α ∈ (0, 1).

5i.e. its closure is compact


Proof: This follows immediately from the boundedness of K1 from C(∂D) into Cα(∂D) andthe compactness of the imbedding Cα(∂D) into C(∂D). 2

We apply this result to the boundary integral operators which appear in the traces ofthe single and double layer potentials of Theorems 3.11, 3.13, and 3.15.

Theorem 3.19 The operators S,D,D′ : Cα(∂D)→ Cα(∂D), defined by

(Sϕ)(x) =

∫∂D

ϕ(y) Φk(x, y) ds(y) , x ∈ ∂D , (3.26a)

(Dϕ)(x) =

∫∂D

ϕ(y)∂Φk

∂ν(y)(x, y) ds(y) , x ∈ ∂D , (3.26b)

(D′ϕ)(x) =

∫∂D

ϕ(y)∂Φk

∂ν(x)(x, y) ds(y) , x ∈ ∂D , (3.26c)

are well defined and compact. The operator S is bounded from Cα(∂D) into C1,α(∂D)

Proof: We have to check the assumptions (3.25a) and (3.25b) of Theorem 3.16. For x, y ∈∂D we have by the definition of the fundamental solution Φ and part (a) of Lemma 6.4 that∣∣Φ(x, y)

∣∣ =1

4π|x− y|,

∣∣∣∣ ∂

∂ν(y)Φ(x, y)

∣∣∣∣ =1

4π|x− y|

∣∣∣∣ik − 1

|x− y|

∣∣∣∣∣∣(y − x) · ν(y)

∣∣|x− y|

≤ c

4π|x− y|

∣∣∣∣ik − 1

|x− y|

∣∣∣∣ |x− y|≤ c

4π|x− y|[k|x− y|+ 1

]≤ c(kd+ 1)

4π|x− y|

where d = sup|x−y| : x, y ∈ ∂D

. The same stimate holds for ∂Φ(x, y)/∂ν(x). This proves

(3.25a) with α = 1. Furthermore, we will prove (3.25b) with α = 1. Let x1, x2, y ∈ ∂D suchthat |x1 − y| ≥ 3|x1 − x2|. Then, for any t ∈ [0, 1], we conclude that |x1 + t(x2 − x1)− y| ≥|x1 − y| − |x2 − x1| ≥ |x1 − y| − |x1 − y|/3 = 2|x1 − y|/3.First we consider Φ and apply the mean value theorem:∣∣Φ(x1, y)− Φ(x2, y)

∣∣ ≤ |x1 − x2| sup0≤t≤1

∣∣∇xΦ(x1 + t(x2 − x1), y

)∣∣≤ c sup

0≤t≤1

|x1 − x2||x1 + t(x2 − x1)− y|2

≤ c9

4

|x1 − x2||x1 − y|2

.

To show the corresponding estimate for the normal derivative of the fundamental solution wecan restrict ourselves to the case k = 0. Indeed, from the representation Φk(x, y)−Φ0(x, y) =A(|x−y|2

)+|x−y|B

(|x−y|2

)with analytic functions A and B we observe that ∂(Φk−Φ0)/∂ν


is continuous.Let again x1, x2, y ∈ ∂D such that |x1 − y| ≥ 3|x1 − x2|. Then∣∣∣∣ ∂

∂ν(y)Φ0(x1, y)− ∂

∂ν(y)Φ0(x2, y)

∣∣∣∣≤ 1

4π|x1 − y|3∣∣ν(y) · (y − x1)− ν(y) · (y − x2)︸︷︷︸

=ν(y)·(x2−x1)

∣∣+

1

4π

∣∣∣∣ 1

|x1 − y|3− 1

|x2 − y|3

∣∣∣∣ ∣∣ν(y) · (y − x2)∣∣

≤ 1

4π|x1 − y|3[∣∣(ν(y)− ν(x1)) · (x2 − x1)

∣∣+∣∣ν(x1) · (x2 − x1)

∣∣]+

1

4π

∣∣∣∣ 1

|x1 − y|3− 1

|x2 − y|3

∣∣∣∣ ∣∣ν(y) · (y − x2)∣∣

Now we use estimates (a) and (b) of Lemma 6.4 for the first term and the mean valuetheorem for the second term. Using again |x1 + t(x2 − x1)− y| ≥ 2|x1 − y|/3 we have∣∣∣∣ ∂

∂ν(y)Φ0(x1, y)− ∂

∂ν(y)Φ0(x2, y)

∣∣∣∣ ≤ c|y − x1||x2 − x1|+ |x1 − x2|2

|x1 − y|3+ c|y − x2|2|x1 − x2||x1 − y|4

Estimate (3.25b) now follows from the estimates |x1 − x2| ≤ |x1 − y|/3 and |y − x2| ≤|y − x1| + |x1 − x2| ≤ 4|y − x1|/3. The proof for the normal derivative with respect to xfollows the same arguments. Finally, we have to show that Grad S is bounded from Cα(∂D)into Cα(∂D)3. But this follows from the representation (3.24). 2

For later use (in Theorem 3.39) we need the following additional mapping property of thesingle layer boundary operator.

Theorem 3.20 The single layer boundary operator S is bounded from Cα(∂D) into C1,α(∂D).

Proof: ### fehlt, aber einfach! ###

3.1.5 Uniqueness and Existence

Now we come back to the scattering problem (3.1), (3.2) from the beginning of this section.First we study the question of uniqueness. The following lemma is fundamental for provinguniqueness and tells us, that a solution of the Helmholtz equation ∆u + k2u = 0 for realand positive (!) k cannot decay faster than 1/|x| as x tends to infinity. We will give twoproofs of this result. The first - and simpler - one uses the expansion arguments from theprevious chapter. In particular, properties of the spherical Bessel- and Hankel functions areused. The second proof which goes back to the original work by Rellich (see [11]) avoids theuse of these special functions but is far more technical and also needs a stronger assumptionon the field. For completeness, we present both versions. We begin with the first form.


Lemma 3.21 (Rellich’s Lemma, first form) Let u ∈ C2(R3 \ B[0, R0]) be a solution of theHelmholtz equation ∆u+ k2u = 0 for |x| > R0 and wave number k ∈ R>0 such that

limR→∞

∫|x|=R

|u|2 ds = 0 .

Then u vanishes for |x| > R0.

Proof: The general solution of the Helmholtz equation in the exterior of B(0, R0) is givenby (2.37); that is,

u(rx) =∞∑n=0

n∑m=−n

[amn h

(1)n (kr) + bmn jn(kr)

]Y mn (x) , x ∈ S2 , r > R ,

for some amn , bmn ∈ C. The spherical harmonics

Y mn : |m| ≤ n, n ∈ N0

form an orthogonal

system. Therefore, Parceval’s theorem yields

∞∑n=0

n∑m=−n

∣∣amn h(1)n (kr) + bmn jn(kr)

∣∣2 =

∫S2

∣∣u(rx)∣∣2ds(x) ,

and from the assumption on u we note that r2∫S2

∣∣u(rx)∣∣2ds(x) tends to zero as r tends to

infinity. Therefore, for every fixed n ∈ N0 and m with |m| ≤ n we conclude that

r2∣∣amn h(1)

n (kr) + bmn jn(kr)∣∣2 −→ 0

as r tends to infinity. Defining cmn = amn + bmn we can write this as (kr) i amn yn(kr) +(kr) cmn jn(kr) → 0. Now we use the asymptotic behaviour of jn(kr) and yn(kr) as r tendsto infinity. From Theorem 2.28 we conclude that

i amn Im[eikr(−i)n+1

]+ cmn Re

[eikr(−i)n+1

]−→ 0 .

The term (−i)n+1 can take the values ±1 and ±i. Therefore, we have that (depending on n)

i amn sin(kr) + cmn cos(kr) −→ 0 or i amn cos(kr) − cmn sin(kr) −→ 0 .

In any case, amn and cmn have to vanish by taking particular sequences rj →∞. This showsthat also bmn = 0. Since this holds for all n and m we conclude that u vanishes. 2

The second proof avoids the use of the Bessel and Hankel functions but needs, however, astronger assumption on u.

Lemma 3.22 (Rellich’s Lemma, second form) Let u ∈ C2(R3 \ B[0, R0]) be a solution ofthe Helmholtz equation ∆u+ k2u = 0 for |x| > R0 with wave number k ∈ R>0 such that

limr→∞

∫|x|=r|u|2 ds = 0 and lim

r→∞

∫|x|=r

∣∣∣∣ x|x| · ∇u(x)

∣∣∣∣2 ds = 0 .

Then u vanishes for |x| > R0.


Proof:6 The proof is lengthy, and we will structure it. We assume that u is real valued(take real and imaginary parts separately).

1st step: Transforming the integral onto the unit sphere S2 = x ∈ R3 : |x| = 1 we concludethat ∫

|x|=1

∣∣u(rx)|2 r2 ds(x) =

∫|x|=r

∣∣u(x)∣∣2 ds(x) and

∫|x|=1

∣∣∣∣∂u∂r (rx)

∣∣∣∣2 r2 ds(x) (3.27)

tend to zero as r tends to infinity. We transform the partial differential equation into anordinary differential equation (not quite!) for the function v(r, x) = r u(r, x) w.r.t. r. Wewrite v(r) and v′(r) and v′′(r) for v(r, ·) and ∂v(r, ·)/∂r and ∂2v(r, ·)/∂r2, respectively. Then(3.27) yields that ‖v(r)‖L2(S2) → 0 and ‖v′(r)‖L2(S2) → 0 as r →∞. The latter follows from∂∂r

(ru(r, ·, ·)

)= 1

r

(ru(r, ·)

)+ r ∂u

∂r(r, ·) and the triangle inequality.

We observe that u = 1rv, thus r2 ∂u

∂r= −v + r ∂v

∂rand ∂

∂r

(r2 ∂u

∂r

)= r ∂

2vr2

, thus

0 =1

r2

∂

∂r

(r2∂u

∂r(r, θ, φ)

)+

1

r2∆Su(r, θ, φ) + k2u(r, θ, φ)

=1

r

[∂2v

∂r2(r, θ, φ) + k2v(r, θ, φ) +

1

r2∆Sv(r, θ, φ)

],

i.e.

v′′(r) + k2v(r) +1

r2∆Sv(r) = 0 for r ≥ R0 , (3.28)

where again ∆S = Div Grad denotes the Laplace-Beltrami operator; that is, in polar coor-dinates x = (sin θ cosφ, sin θ sinφ, cos θ)>

(∆Sw)(θ, φ) =1

sin θ

∂

∂θ

(sin θ

∂w

∂θ(θ, φ)

)+

1

sin2 θ

∂2w

∂φ2(θ, φ)

for any w ∈ C2(S2). It is easily seen either by direct integration or by application ofTheorem ?? that ∆S is selfadjoint and negative definite, i.e.(

∆Sv, w)L2(S2)

=(v,∆Sw

)L2(S2)

and(∆Sv, v

)L2(S2)

≤ 0 for all v, w ∈ C2(S2) .

2nd step: We introduce the functions E, vm and F by

E(r) := ‖v′(r)‖2L2(S2) + k2‖v(r)‖2

L2(S2) +1

r2

(∆Sv(r), v(r)

)L2(S2)

, r ≥ R0 ,

vm(r) := rmv(r) , r ≥ R0 , m ∈ N ,

F (r,m, c) := ‖v′m(r)‖2L2(S2) +

(k2 +

m(m+ 1)

r2− 2c

r

)‖vm(r)‖2

L2(S2)

+1

r2

(∆Svm(r), vm(r)

)L2(S2)

,

for r ≥ R0, m ∈ N, c ≥ 0. In the following we write ‖·‖ and (·, ·) for ‖·‖L2(S2) and (·, ·)L2(S2),respectively. We show:

6We took the proof from the monograph Partielle Differentialgleichungen zweiter Ordnung by R. Leis


(a) E satisfies E ′(r) ≥ 0 for all r ≥ R0.

(b) The functions vm solve the differential equation

v′′m(r) − 2m

rv′m(r) +

(m(m+ 1)

r2+ k2

)vm(r) +

1

r2∆Svm(r) = 0 . (3.29)

(c) For every c > 0 there exist r0 = r0(c) ≥ R0 and m0 = m0(c) ∈ N such that

∂

∂r

[r2F (r,m, c)

]≥ 0 for all r ≥ r0, m ≥ m0 .

(d) Expressed in terms of v the function F has the forms

F (r,m, c) = r2m

∥∥∥v′(r) +m

rv(r)

∥∥∥2

+

(k2 +

m(m+ 1)

r2− 2c

r

)‖v(r)‖2

+1

r2

(∆Sv(r), v(r)

)(3.30a)

= r2m

E(r) +

2m

r

(v(r), v′(r)

)+

(m(2m+ 1)

r2− 2c

r

)‖v(r)‖2

.(3.30b)

Proof of these statements:(a) We just differentiate E and substitute the second derivative from (3.28). Note thatddr‖v(r)‖2 = 2(v, v′) and d

dr

(∆Sv, v

)= 2(∆Sv, v

′):E ′(r) = 2

(v′(r), v′′(r)

)+ 2k2

(v(r), v′(r)

)− 1

r3

(∆Sv(r), v(r)

)+

2

r2

(∆Sv(r), v′(r)

)= 2

(v′(r) ,

[v′′(r) + k2v(r) +

1

r2∆Sv(r)

])− 1

r3

(∆Sv(r), v(r)

)= − 1

r3

(∆Sv(r), v(r)

)≥ 0 .

(b) We substitute v(r) = r−mvm(r) into (3.28) and obtain directly (3.29). We omit thecalculation.

(c) Again we differentiate r2F (r,m, c) w.r.t. r, substitute the form of v′′m from (3.29) andobtain

∂

∂r

[r2F (r,m, c)

]= 2r‖v′m(r)‖2 + 2r2

(v′m(r), v′′m(r)

)+ 2(k2r − c)‖vm(r)‖2

+ 2r2

(k2 +

m(m+ 1)

r2− 2c

r

)(vm(r), v′m(r)

)+ 2

(∆Svm(r), v′m(r)

)= · · · = 2r(1 + 2m)‖v′m(r)‖2 − 4cr

(v′m(r), vm(r)

)+ 2(k2r − c)‖vm(r)‖2

= 2r

[∥∥∥∥√1 + 2mv′m(r)− c√1 + 2m

vm(r)

∥∥∥∥2

+

(k2 − c

r− c2

1 + 2m

)‖vm(r)‖2

].


From this the assertion (c) follows if r0 and m0 are chosen such that the bracket (· · · ) ispositive.(d) The first equation is easy to see by just inserting the form of vm. For the second formone uses simply the binomial theorem for the first term and the definition of E(r).

3rd step: We begin with the actual proof of the lemma and show first that there existsR1 ≥ R0 such that ‖v(r)‖ = 0 for all r ≥ R1. Assume, on the contrary, that this is not thecase. Then, for every R ≥ R0 there exists r ≥ R such that ‖v(r)‖ > 0.We choose the constants c > 0, r0, m0, r1, m1 in the following order:

• Choose c > 0 with k2 − 2cR0> 0.

• Choose r0 = r0(c) ≥ R0 and m0 = m0(c) ∈ N according to property (c) above, i.e.such that ∂

∂r

[r2F (r,m, c)

]≥ 0 for all r ≥ r0 and m ≥ m0.

• Choose r1 > r0 such that ‖v(r1)‖ > 0.

• Choose m1 ≥ m0 such that m1(m1 + 1)‖v(r1)‖2 +(∆Sv(r1), v(r1)

)> 0.

Then, by (3.30a) and because k2− 2cr1≥ k2− 2c

R0> 0, it follows that F (r1,m1, c) > 0 and thus,

by the monotonicity of r 7→ r2F (r,m1, c) that also F (r,m1, c) > 0 for all r ≥ r1. Therefore,from (3.30b) we conclude that, for r ≥ r1,

0 < r−2m1F (r,m1, c) = E(r) +2m1

r

(v(r), v′(r)

)+

(m1(2m1 + 1)

r2− 2c

r

)‖v(r)‖2

= E(r) +m1

r

d

dr‖v(r)‖2 +

1

r

(m1(2m1 + 1)

r− 2c

)‖v(r)‖2 .

Choose now r2 ≥ r1 such that m1(2m1+1)r2

− 2c < 0. Finally, choose r ≥ r2 such thatddr‖v(r)‖2 ≤ 0. (This is possible because ‖v(r)‖2 → 0 as r →∞.) We finally have

0 < p := r−2m1F (r, m1, c) ≤ E(r) .

By the monotonicity of E we conclude that E(r) ≥ p for all r ≥ r. On the other hand, bythe definition of E(r) we have that E(r) ≤ ‖v′(r)‖2 + k2‖v(r)‖2 and this tends to zero as rtends to infinity. This is a contradiction. Therefore, there exists R1 ≥ R0 with v(r) = 0 forall r ≥ R1 and thus also Re u = 0 for |x| ≥ R1. The same holds for Im u and thus u = 0 for|x| ≥ R1. 2

We can now prove uniqueness of the scattering problem.

Theorem 3.23 For any incident field uinc there exists at most one solution u ∈ C2(R3 \D) ∩ C1(R3 \D) of the scattering problem (3.1), (3.2).

Proof: Let u be the difference of two solutions. Then u satisfies (3.1) and also the radiationcondition (3.2). From the radiation condition we conclude that∫

|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds =

∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + k2|u|2

ds + 2k Im

∫|x|=R

u∂u

∂rds


tends to zero as R tends to infinity. Green’s theorem, applied in BR \D to the function uyields that∫

|x|=R

u∂u

∂rds =

∫∂D

u∂u

∂rds +

∫∫BR\D

[|∇u|2 − k2|u|2

]dx =

∫∫BR\D

[|∇u|2 − k2|u|2

]dx

because the surface integral over ∂D vanishes by the boundary condition. The volumeintegral is real valued. Therefore, its imaginary part vanishes and we conclude that∫

|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + k2|u|2ds

tends to zero as R tends to infinity. Rellich’s lemma (in the form Lemma 3.21 or Lemma 3.22)implies that u vanishes outside of every ball which encloses ∂D. Finally, we note that u isan analytic function in the exterior of D. Since the exterior of D is connected we concludethat u vanishes in R3 \D. 2

We turn to the question of existence and choose the integral equation method for itstreatment. We follow again the approach of [2] but prefer to work in the space Cα(∂D) ofHolder continuous functions rather that in the space of merely continuous functions. Thisavoids the necessity to introduce the class of continuous functions for which the normalderivatives exist “in the uniform sense along the normal”.

We recall the notion of the single layer potentials of (3.4a), see also (3.12), and make theansatz for the scattered field in the form of a single layer potential. We remark already herethat we will face some difficulties with this ansatz. Before we modify the ansatz below wetry the single layer ansatz for the scattered field in the form

us(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) , x ∈ R3 \ ∂D , (3.31)

where again Φ(x, y) = exp(ik|x − y|)/(4π|x − y|) denotes the fundamental solution of theHelmholtz equation, and ϕ ∈ Cα(∂D) is some density to be determined. First we notethat us solves the Helmholtz equation in the exterior of D and also the radiation condition.This follows from the corresponding properties of the fundamental solution Φ(·, y), uniformlywith respect to y on the compact surface ∂D. Furthermore, by Theorems 3.11 and 3.15 thefunction us and its derivatives can be extended continuously (from the exterior) into R3 \Dwith limiting values

us(x)∣∣+

=

∫∂D

ϕ(y) Φ(x, y) ds(y) = (Sϕ)(x) , x ∈ ∂D , (3.32a)

∂us

∂ν(x)

∣∣∣∣+

= −1

2ϕ(x) +

∫∂D

ϕ(y)∂

∂ν(x)Φ(x, y) ds(y)

= −1

2ϕ(x) + (D′ϕ)(x) , x ∈ ∂D , (3.32b)


where we used the notations of the boundary integral operators from Theorem 3.19. There-fore, in order that u = uinc + us satisfies the boundary condition ∂u/∂ν = 0 on ∂D thedensity ϕ has to satisfy the boundary integral equation

−1

2ϕ + D′ϕ = −∂u

inc

∂νin Cα(∂D) . (3.33)

By Theorem 3.19 the operator D′ is compact. Therefore, we can apply the Fredholm theory.In particular, existence follows from uniqueness. To prove uniqueness we assume that ϕ ∈Cα(∂D) satisfies the homogeneous equation −1

2ϕ+D′ϕ = 0. Define v to be the single layer

potential with density ϕ just as in (3.31), but for arbitrary x /∈ ∂D. Then, again fromthe jump contions of the normal derivative of the single layer, ∂v/∂ν

∣∣+

= −12ϕ + D′ϕ = 0.

Therefore, v is the solution of the exterior Neumann problem with vanishing boundarydata. The uniqueness result of Theorem 3.23 yields that v vanishes in the exterior of D.Furthermore, v is continuous in R3, thus v is a solution of the Helmholtz equation in Dwith vanishing boundary data. This is the point where we wish to conclude that v vanishesalso in D. However, this is not always the case. Indeed, this is not the case if, and only if,k2 is an eigenvalue of −∆ in D with respect to Dirichlet boundary conditions. This is thereason why we have to modify the ansatz (3.31). There are several ways how to do it, seethe discussion in [1]. We choose a modification which we have not found in the literature.It avoids the use of double layer potentials. We assume for simplicity that D is connectedalthough this is not necessary as one observes from the following arguments.

We choose an open ball B with boundary Γ such that Γ ⊂ D and such that k2 is not aneigenvalue of −∆ inside B with respect to Dirichlet boundary conditions. By Theorem 2.22from the previous chapter we observe that we have to choose the radius ρ of B such that kρis not a zero of any of the Bessel functions jn.

Now we make an ansatz for us as a sum of two single layer potentials in the form

us(x) = (S∂Dϕ)(x) + (SΓψ)(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) +

∫Γ

ψ(y) Φ(x, y) ds(y) , x /∈ D ,

(3.34)where φ ∈ Cα(∂D) and ψ ∈ Cα(Γ) are two densities to be determined from the system oftwo boundary integral equations

−1

2ϕ + D′ϕ +

∂

∂νSΓψ = −∂u

inc

∂νon ∂D , (3.35a)(

∂

∂ν+ ik

)S∂Dϕ −

1

2ψ + D′Γψ + ik SΓψ = 0 on Γ . (3.35b)

The operators SΓ and D′Γ denote the boundary operators S and D′, respectively, on theboundary Γ instead of ∂D. These two equations can be written in matrix form as

−1

2

(ϕ

ψ

)+

(D′ ∂SΓ/∂ν

(∂/∂ν + ik)S∂D D′Γ + ik SΓ

)(ϕ

ψ

)= −

(∂uinc/∂ν

0

)in Cα(∂D) × Cα(Γ). The operators D′, D′Γ, ∂SΓ/∂ν, and (∂/∂ν + ik)S∂D are all compact.Therefore, we can apply the Fredholm alternative to this system. Existence is assured if

3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM 109

the homogeneous system admits only the trivial solution ϕ = 0 and ψ = 0. Therefore, let(ϕ, ψ) ∈ Cα(∂D) × Cα(Γ) be a solution of the homogeneous system and define the v asthe sum of the single layers with densities ϕ and ψ for all x in R3 \ (∂D ∪ Γ). From thethe jump condition for the normal derivative and the first (homogeneous) integral equationwe conclude – just in the above case of only one single layer potential – that ∂v/∂ν

∣∣+

=

−12ϕ + D′ϕ + ∂SΓψ/∂ν = 0. Again, v is a solution of the exterior Neumann problem with

vanishing boundary data. Therefore, by the uniqueness theorem, v vanishes in the exteriorof D. Furthermore, v is continuous in R3 and satisfies also the Helmholtz equation in D \B.From the jump conditions on the boundary Γ we conclude that

∂v

∂ν

∣∣∣∣+

+ ikv =

(∂

∂ν+ ik

)S∂Dϕ −

1

2ψ + D′Γψ + ik SΓψ = 0 on Γ .

Therefore, v = 0 on ∂D and ∂v/∂ν|+ + ikv = 0 on Γ. Application of Green’s first theoremin D \B yields∫∫

D\B

[|∇v|2 − k2|v|2

]dx =

∫∂D

v∂v

∂νds−

∫Γ

v∂v

∂ν

∣∣∣∣+

ds = ik

∫Γ

|v|2 ds .

Taking the imaginary part yields that v vanishes on Γ and therefore also ∂v/∂ν|+ = 0 onΓ. Holmgren’s uniqueness Theorem 3.5 implies that v vanishes in all of D \ B. The jumpconditions on ∂D yield

0 =∂v

∂ν

∣∣∣∣−− ∂v

∂ν

∣∣∣∣+

= ϕ on ∂D .

Therefore, v is a single layer potential on Γ with density ψ and vanishes on Γ. The wavenumber k2 is not a Dirichlet eigenvalue of −∆ in B by the choice of the radius of B.Therefore, v vanishes also in B. The jump conditions on Γ yield

0 =∂v

∂ν

∣∣∣∣−− ∂v

∂ν

∣∣∣∣+

= ψ on Γ .

Therefore, ϕ = 0 on ∂D and ψ = 0 on Γ.

If D consists of several components D =⋃Mm=1Dm then one has to choose balls Bm in

each of the domains Dm and make an ansatz as a sum of single layers on ∂D and ∂Bm form = 1, . . . ,M .

Application of Fredholm’s alternative yields the following result:

Theorem 3.24 There exists a unique solution u ∈ C2(R3\D)∩C1(R3\D) of the scatteringproblem (3.1), (3.2).

3.2 A Scattering Problem for the Maxwell System

3.2.1 Formulation of the Problem

For this chapter we make the following assumptions on the data:


Assumption: Let the wave number be given by k = ω√ε0µ0 > 0 with constants ε0, µ0 > 0.

Let D ⊂ R3 be bounded and C2−smooth such that the complement R3 \D is connected.

Scattering Problem: Given a solution (Ei, H i) of the Maxwell system

curlEi − iωµ0Hi = 0 , curlH i + iωε0E

i = 0 in some neighborhood of D ,

determine the total fields E,H ∈ C1(R3 \D) ∩ C(R3 \D) such that

curlE − iωµ0H = 0 and curlH + iωε0E = 0 in R3 \D , (3.36a)

E satisfies the boundary condition

ν × E = 0 on ∂D , (3.36b)

and the radiating parts Es = E − Ei and Hs = H −H i satisfy the Silver Muller radiationconditions

√ε0E

s(x) − √µ0Hs(x)× x

|x|= O

(1

|x|2

), (3.37a)

and√µ0H

s(x) +√ε0E

s(x)× x

|x|= O

(1

|x|2

), (3.37b)

uniformly w.r.t. x/|x|. Clearly, after renaming the unknown fields, this is a special case ofthe following problem:

Exterior Boundary Value Problem: Given a tangential field7 c ∈ Cα(∂D)3 such thatDiv c ∈ Cα(∂D) determine radiating8 solutions Es, Hs ∈ C1(R3 \D) ∩ C(R3 \D) of

curlEs − iωµ0Hs = 0 and curlHs + iωε0E

s = 0 in R3 \D , (3.38a)

such that Es satisfies the boundary condition

ν × Es = c on ∂D . (3.38b)

We note again that the assumption on the surface divergence of c is necessary by Corollary ??.

3.2.2 Representation Theorems

We have seen in the previous section (Theorem 3.3) that every sufficiently smooth function ucan be written as a sum of a volume potential with density ∆u+k2u, a single layer potentialwith density ∂u/∂ν, and a double layer potential with potential u. In this subsection weshow the corresponding theorem for vector fields.

7i.e. ν(x) · c(x) = 0 on ∂D8i.e. Es, Hs satisfy the radiating conditions (3.37a) and (3.37b)


Theorem 3.25 Let k ∈ C and E ∈ C1(D)3 ∩C(D)3 such that curlE ∈ C(D)3 and divE ∈C(D). Then we have for x ∈ D:

E(x) = curl

∫∫D

Φ(x, y) curlE(y) dy−∇∫∫D

Φ(x, y) divE(y) dy−k2

∫∫D

E(y) Φ(x, y) dy

− curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) + ∇

∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y)

where the domain integrals exist as improper integrals. The right hand side of this equationvanishes for x /∈ D and is equal to 1

2E(x) for x ∈ ∂D.

Proof: Fix z ∈ D, choose r > 0 such that B[z, r] ⊂ D, and set Dr = D \ B[z, r]. Forx ∈ B(z, r) we set

Ir(x) := curl

∫∫D

Φ(x, y) curlE(y) dy−∇∫∫D

Φ(x, y) divE(y) dy−k2

∫∫D

E(y) Φ(x, y) dy

− curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) + ∇

∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y)

=

∫∫Dr

∇xΦ(x, y)× curlE(y) dy−∫∫Dr

∇xΦ(x, y) divE(y) dy−k2

∫∫Dr

E(y) Φ(x, y) dy

− curl

∫∂Dr

[ν(y)× E(y)

]Φ(x, y) ds(y) + ∇

∫∂Dr

[ν(y) · E(y)

]Φ(x, y) ds(y) .

We show that Ir(x) vanishes. Indeed, we can interchange differentiation and integration andwrite

Ir(x) = curl

[∫∫Dr

Φ(x, y) curlE(y) dy −∫∂Dr

[ν(y)× E(y)

]Φ(x, y) ds(y)

]− ∇

[∫∫Dr

Φ(x, y) divE(y) dy −∫∂Dr

[ν(y) · E(y)

]Φ(x, y) ds(y)

]− k2

∫∫Dr

E(y) Φ(x, y) dy

= curl

[∫∫Dr

curly

[E Φ(x, ·)

]−∇yΦ(x, ·)× E

dy −

∫∂Dr

ν ×[E Φ(x, ·)

]ds

]− ∇

[∫∫Dr

divy

[E Φ(x, ·)

]−∇yΦ(x, ·) · E

dy −

∫∂Dr

ν ·[E Φ(x, ·)

]ds

]− k2

∫∫Dr

E Φ(x, ·) dy .


Now we use the divergence theorem in the forms:∫∫Dr

divF dx =

∫∂Dr

ν · F ds ,∫∫

Dr

curlF dx =

∫∂Dr

ν × F ds .

Therefore,

Ir(x) = − curl

∫∫Dr

∇yΦ(x, ·)× E dy + ∇∫∫

Dr

∇yΦ(x, ·) · E dy − k2

∫∫Dr

E Φ(x, ·) dy

=

∫∫Dr

∇x

[E · ∇yΦ(x, ·)

]− curlx

[∇yΦ(x, ·)× E

]dy − k2

∫∫Dr

E Φ(x, ·) dy

= −∫∫

Dr

E[∆yΦ(x, ·) + k2Φ(x, ·)

]dy = 0 .

Hier we used the formula

curlx[∇yΦ(x, ·)× E

]= −E divx∇yΦ(x, ·) + (E · ∇x)∇yΦ(x, ·)

= E∆yΦ(x, ·) +∇x

[E · ∇yΦ(x, ·)

].

Therefore, Ir(x) = 0 for all x ∈ B(z, r), i.e.

0=

∫∫Dr

∇xΦ(x, y)× curlE(y) dy−∫∫Dr

∇xΦ(x, y) divE(y) dy−k2

∫∫Dr

E(y) Φ(x, y) dy

− curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) + ∇

∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y)

−∫

|y−z|=r

∇xΦ(x, y)×[ν(y)× E(y)

]ds(y) +

∫|y−z|=r

∇xΦ(x, y)[ν(y) · E(y)

]ds(y) .

We set x = z and compute the last two surface integrals explicitely. We recall that

∇zΦ(z, y) =exp(ik|z − y|)

4π|z − y|

(ik − 1

|z − y|

)z − y|z − y|

and thus for |z − y| = r:

−∫

|y−z|=r

∇zΦ(z, y)×[ν(y)× E(y)

]ds(y) +

∫|y−z|=r

∇zΦ(z, y)[ν(y) · E(y)

]ds(y)

=exp(ikr)

4πr

(ik − 1

r

) ∫|y−z|=r

[z − y|z − y|

(z − y|z − y|

· E(y)

)− z − y|z − y|

×(z − y|z − y|

× E(y)

)]︸︷︷︸

= E(y)

ds(y)

=exp(ikr)

4πr

(ik − 1

r

) ∫|y−z|=r

E(y) ds(y)

= −eikr E(z) + ikexp(ikr)

4πr

∫|y−z|=r

E(y) ds(y) +exp(ikr)

4πr2

∫|y−z|=r

[E(z)− E(y)

]ds(y) .


This term converges to −E(z) as r tends to zero. This proves the formula for x ∈ D.The same arguments (replacing Dr by D) which lead to Ir(x) = 0 yield that the expressionvanishes if x /∈ D. The formula for x ∈ ∂D follows from the same arguments as in the proofof Theorem 3.3. 2

Theorem 3.26 (Stratton-Chu formula)Let k = ω

√ε0µ0 > 0 and E,H ∈ C1(D)3 ∩ C(D)3 satisfy Maxwell’s equations

curlE − iωµ0H = 0 in D , curlH + iωε0E = 0 in D .

Then we have for x ∈ D:

E(x) = − curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) + ∇

∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y)

− iωµ0

∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y)

= − curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) +

1

iωε0

curl2∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y) ,

H(x) = − curl

∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y) − 1

iωµ0

curl2∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) .

Proof: The second term in the representation of E in Theorem 3.25 vanishes becausedivE = 0. The first term is rewritten as follows where now Dr = D \B[x, r].∫∫

Dr

∇xΦ(x, y)× curlE(y) dy = iωµ0

∫∫Dr

∇xΦ(x, y)×H(y) dy

= −iωµ0

∫∫Dr

∇yΦ(x, y)×H(y) dy

= −iωµ0

∫∫Dr

curly

[H(y) Φ(x, y)

]− Φ(x, y) curlH(y)

dy

= −iωµ0

∫∫Dr

curly(H(y) Φ(x, y)

)dy + ω2µ0ε0

∫∫Dr

Φ(x, y)E(y) dy

= −iωµ0

∫∂Dr

[ν(y)×H(y)

]Φ(x, y) dy + ω2µ0ε0︸︷︷︸

= k2

∫∫Dr

Φ(x, y)E(y) dy .

This proves the first formula by letting r tend to zero. For the second we continue with:∫∂Dr

[ν(y) · E(y)

]Φ(x, y) ds(y) = − 1

iωε0

∫∂Dr

[ν(y) · curlH(y)

]Φ(x, y) ds(y)

= − 1

iωε0

∫∂Dr

ν(y) · curl[H(y) Φ(x, y)

]ds(y)︸︷︷︸

= 0 by the divergence theorem

+1

iωε0

∫∂Dr

ν(y) ·[∇yΦ(x, y)×H(y)

]ds(y)

=1

iωε0

∫∂Dr

∇yΦ(x, y) ·[H(y)× ν(y)

]ds(y) .


Now we let r tend to zero. We observe that the integral∫|y−x|=r∇yΦ(x, y)·

[H(y)×ν(y)

]ds(y)

vanishes because ∇yΦ(x, y) and ν(y) are parallel. Therefore,∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y) =

1

iωε0

∫∂D

∇yΦ(x, y) ·[H(y)× ν(y)

]ds(y)

=1

iωε0

div

∫∂D

Φ(x, y)[ν(y)×H(y)

]ds(y) .

Taking the gradient and using curl curl = ∇ div−∆ yields

∇∫∂D

[ν(y) · E(y)

]Φ(x, y) ds(y) =

1

iωε0

curl curl

∫∂D


]ds(y)

− k2

iωε0

∫∂D


]ds(y) .

This ends the proof for E. The representation for H(x) follows directly by H = 1iωµ0

curlE

(note that curl curl curl = − curl ∆!) 2

Conclusion 3.27 Solutions E,H of Maxwell’s equations in vacuum D ⊂ R3 are

• analytic in every component

• divergence-free solutions of the vector-Helmholtz equation

On the other side, every divergence-free solution E of the vector-Helmholtz equation is, com-bined with H := 1

iωµ0curlE, a solution of Maxwell’s equations.

Remark: Fields of the form (for some y ∈ R3 and p ∈ C3)

Emd(x) = curl[pΦ(x, y)

], Hmd =

1

iωµ0

curlEmd =1

iωµ0

curl curl[pΦ(x, y)

],

Hed(x) = curl[pΦ(x, y)

], Eed = − 1

iωε0

curlHed = − 1

iωε0

curl curl[pΦ(x, y)

],

are called magnetic and electric dipols, respectively, at y with polarization p. We assumehere, that k is real and positive!

Lemma 3.28 Let k > 0. The electromagnetic fields Emd, Hmd and Eed, Hed of a magneticor electric dipol, respectively, satisfy the Silver-Muller radiation condition

√ε0E(x) − √µ0H(x)× x

|x|= O

(1

|x|2

), (3.39a)

and√µ0H(x) +

√ε0E(x)× x

|x|= O

(1

|x|2

), (3.39b)

uniformly w.r.t. x/|x| and |y| ≤ R for any fixed R > 0.


Proof: Direct computation yields

Emd(x) = curl[pΦ(x, y)

]= Φ(x, y)

(ik − 1

|x− y|

)(x− y|x− y|

× p)

= ikΦ(x, y)

(x− y|x− y|

× p)

+ O(

1

|x|2

),

Hmd(x) =1

iωµ0

curl curl[pΦ(x, y)

]=

1

iωµ0

(−∆ +∇ div)[pΦ(x, y)

]= · · · =

k2

iωµ0

Φ(x, y)

[p− x− y|x− y|

(x− y) · p|x− y|

]+ O

(1

|x|2

).

for |x| → ∞. With x−y|x−y| = x

|x| + O(1/|x|2) and k = ω√µ0ε0 the first assertion follows.

Analogously, the second can be proven. 2

Theorem 3.29 (Stratton-Chu formula in exterior domains)Let k = ω

√ε0µ0 > 0 and E,H ∈ C1(R3 \ D)3 ∩ C(R3 \ D)3 solutions of the homogeneous

Maxwell’s equations

curlE − iωµ0H = 0 , curlH + iωε0E = 0

in R3 \D which satisfy also one of the Silver-Muller radiation conditions (3.39a) or (3.39b).Then

curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) − 1

iωε0

curl curl

∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y) =

=

0 , x ∈ D ,

12E(x) , x ∈ ∂D ,E(x) , x /∈ D ,

and

curl

∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y) +

1

iωµ0

curl curl

∫∂D

[ν(y)× E(y)

]Φ(x, y) ds(y) =

=

0 , x ∈ D ,

12H(x) , x ∈ ∂D ,H(x) , x /∈ D .

Proof: Let us first assume the radiation condition (3.39a). One applies Theorem 3.26 inthe region DR =

x /∈ D : |y| < R

for large values of R. Then the assertion follows if one

can show that

IR := curl

∫|y|=R

[ν(y)×E(y)

]Φ(x, y) ds(y)− 1

iωε0

curl curl

∫|y|=R

[ν(y)×H(y)

]Φ(x, y) ds(y)


tends to zero as R → ∞. To do this we first prove that∫|y|=R |E|

2ds is bounded w.r.t. R.

The binomial theorem yields∫|y|=R

|√ε0E −

√µ0H × ν|2ds = ε0

∫|y|=R

|E|2ds + µ0

∫|y|=R

|H × ν|2ds

−2√ε0µ0 Re

∫|y|=R

E · (H × ν) ds .

We have by the divergence theorem∫|y|=R

E · (H × ν) ds =

∫∂D

E · (H × ν) ds +

∫∫DR

div(E ×H) dx

=

∫∂D

E · (H × ν) ds +

∫∫DR

[H) · curlE − E · curlH

]dx

=

∫∂D

E · (H × ν) ds +

∫∫DR

[iωµ0|H|2 − iωε0|E|2

]dx .

This term is purely imaginary, thus∫|y|=R

|√ε0E −

√µ0H × ν|2ds = ε0

∫|y|=R

|E|2ds + µ0

∫|y|=R

|H × ν|2ds

− 2√ε0µ0 Re

∫∂D

E · (H × ν) ds .

From this the boundedness of∫|y|=R |E|

2ds follows because the left hand side tends to zero

by the radiation condition (3.39b).Now we write IR in the form

IR = curl

∫|y|=R

[ν(y)× E(y)

]Φ(x, y) +

1

ikE(y)×∇yΦ(x, y)

ds(y)

− 1

iωε0

curl

∫|y|=R

([ν(y)×H(y)

]+

√ε0

µ0

E(y)

)×∇yΦ(x, y) ds(y) .

Let us first consider the second term. The bracket (· · · ) tends to zero as 1/R2 by the radiationcondition (3.39a). Taking the curl of the integral results in second order differentiations ofΦ. Since Φ and all derivatives decay as 1/R the total integrand decays as 1/R3. Therefore,this second term tends to zero because the surface area is only 4πR2.For the first term we observe that[

ν(y)× E(y)]

Φ(x, y) +1

ikE(y)×∇yΦ(x, y)

= E(y)×[− yR

Φ(x, y) +1

ik

y − x|y − x|

Φ(x, y)

(ik − 1

|y − x|

)].


For fixed x and arbitrary y with |y| = R the bracket [· · · ] tends to zero of order 1/R2. Thesame holds true for all of the partial derivatives w.r.t. x. Therefore, the first term can beestimated by the inequality of Cauchy-Schwartz

c

R2

∫|y|=R

∣∣E(y)∣∣ ds ≤ c

R2

√∫|y|=R

12 ds

√∫|y|=R

∣∣E(y)∣∣2 ds =

c√

4π

R

√∫|y|=R

∣∣E(y)∣∣2 ds

and this tends also to zero.This proves the representation of E by using the first radiation condition (3.39a). Therepresentation of H(x) follows again by computing H = 1

iωµ0curlE. If the second radiation

condition (3.39b) is assumned one can argue as before and derive the representation of Hfirst. 2

We draw the following conclusions from this result.

Remark 3.30 (a) If E, H are solutions of Maxwell’s equations in vacuum R3 \D then eachof the radiation conditions implies the other one.

(b) The Silver-Muller radiation condition for solutions E, H of the Maxwell system is equiv-alent to the Sommerfeld radiation condition (3.2) for every component of E and H. Thisfollows from the fact that the fundamental solution Φ and every derivative of Φ satisfies theSommerfeld radiation condition.

(c) The asymptotic behaviour of Φ yields

E(x) = O(

1

|x|

)and H(x) = O

(1

|x|

)for |x| → ∞ uniformly w.r.t. all directions x/|x|.

Sometimes it is conveniant to eliminate one of the fields E or H from the Maxwell systemand work with only one of them. If we eliminate H then E solves the second order equation

curlE − k2E = 0 (3.40)

where again k = ω√µ0ε0 denotes the wave number. If, on the other hand, E satisfies (3.40)

then E and H = 1iωµ0

E solve the Maxwell system. Indeed, the first Maxwell equation issatisfied by the definition of H. Also the second Maxwell equation is satisfied because

curlH =1

iωµ0

curl2E =1

iωµ0

[∇ divE︸︷︷︸

=0

− ∆E︸︷︷︸=−k2E

]=

k2

iωµ0

E = −iωε0E .

The Silver-Muller radiation condition (3.37a) or (3.37b) turn into

curlE × x − ik E = O(|x|−2

), |x| → ∞ , (3.41)

uniformly with respect to x = x/|x|.


3.2.3 Vector Potentials and Boundary Integral Operators

In Subsection 3.2.4 we will prove existence of solutions of the scattering problem by a bound-ary integral equation method. Analogously to the scalar case we have to introduce vectorpotentials. Motivated by the Stratton-Chu formulas we have to consider the curl and thedouble-curl of the single layer potential

v(x) =

∫∂D

a(y) Φ(x, y) ds(y) , x ∈ R3 , (3.42)

where a ∈ Cα(∂D)3 is a tangential field, i.e. a(y) · ν(y) = 0 for all y ∈ ∂D.

Lemma 3.31 Let v be defined by (3.42). Then E = curl v satisfies (3.40) in all of R3 \ ∂Dand also the radiation condition (3.41).

The proof follows immediately from Lemma 3.28. 2

In the next theorem we study the behaviour of E at the boundary.

Theorem 3.32 The curl of the potential v from (3.42) with Holder continuous tangentialfield a ∈ Cα(∂D)3 can be continuously extended from D to D and from R3 \ D to R3 \ D.The limiting values of the tangential components are

ν(x)× curl v(x)∣∣± = ±1

2a(x) + ν(x)×

∫∂D

curlx[a(y)Φ(x, y)

]ds(y) , x ∈ ∂D . (3.43)

The integral exists as an improper integral.

If, in addition, the surface divergence Div a (see end of Chapter 1) is continuous, then div vis continuous in all of R3.

If, furthermore, Div a ∈ Cα(∂D) then curl curl v can be continuously extended from D to Dand from R3 \D to R3 \D.

The limiting values are

div v(x)∣∣± =

∫∂D

Φ(x, y) Div a(y) ds(y) , x ∈ ∂D ,

ν × curl curl v∣∣+

= ν × curl curl v∣∣− on ∂D , (3.44)

ν(x) · curl curl v(x)∣∣± = ∓1

2Div a(x)

+

∫∂D

[Div a(y)

∂Φ

∂ν(x)(x, y) + k2ν(x) · a(y) Φ(x, y)

]ds(y) , x ∈ ∂D .

Proof: First we write the second term on the right hand side of (3.43) in the form

ν(x)×∫∂D

curlx[a(y)Φ(x, y)

]ds(y) =

∫∂D

ν(x)×[∇xΦ(x, y)× a(y)

]ds(y)

=

∫∂D

[ν(x)− ν(y)

]×[∇xΦ(x, y)× a(y)

]+

∂Φ

∂ν(y)(x, y) a(y)

ds(y)


and the integrand is weakly singular.The components of curl v are combinations of partial derivatives of the single layer potential.Therefore, by Theorem 3.15 the field curl v has continuous extensions to ∂D from both sides.It remains to show the representation of the tangential components of these extensions onthe boundary. We recall the special neighborhoods Hρ of ∂D from Subsection ?? and writex ∈ Hρ0 in the form x = z + tν(z), z ∈ ∂D, 0 < |t| < ρ0. Then we have

ν(z)× curl v(x) =

∫∂D

ν(z)×[∇xΦ(x, y)× a(y)

]ds(y) (3.45)

=

∫∂D

[ν(z)− ν(y)

]×[∇xΦ(x, y)× a(y)

]ds(y) +

∫∂D

a(y)∂Φ

∂ν(y)(x, y) ds(y) .

The first term is continuous in all of R3 by Lemma 3.12, the second in R3 \ ∂D because it isa double layer potential. The limiting values are

ν(x)× curl v(x)∣∣± =

∫∂D

[ν(x)− ν(y)

]×[∇xΦ(x, y)× a(y)

]ds(y)

±1

2a(x) +

∫∂D

a(y)∂Φ


= ±1

2a(x) +

∫∂D

ν(x)×[∇xΦ(x, y)× a(y)

]ds(y)

which has the desired form. For the divergence we write

div v(x) =

∫∂D

a(y) · ∇xΦ(x, y) ds(y) = −∫∂D

a(y) · ∇yΦ(x, y) ds(y)

= −∫∂D

a(y) ·Grad yΦ(x, y) ds(y) =

∫∂D

Φ(x, y) Div a(y) ds(y)

and this is a single layer potential and thus continuous. Finally, because curl curl = ∇ div−∆and ∆xΦ(x, y) = −k2Φ(x, y), we conclude that

curl curl v(x) = ∇ div

∫∂D

a(y) Φ(x, y) ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

= ∇∫∂D

Φ(x, y) Div a(y) ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

from which the assertion follows by Theorem 3.15. 2

The continuity properties of the derivatives of v give rise to corresponding boundary integraloperators. It is convenient to not only define the spaces Ct(∂D) and Cα

t (∂D) of continuousand Holder continuous tangential fields, respectively, but also of Holder continuous tangentialfields such that the surface divergence is also Holder continuous. Therefore, we define:

Ct(∂D) =a ∈ C(∂D)3 : a(y) · ν(y) = 0 on ∂D

,

Cαt (∂D) = Ct(∂D) ∩ Cα(∂D)3 ,

CαDiv(∂D) =

a ∈ Cα

t (∂D) : Div a ∈ Cα(∂D).


We equip Ct(∂D) and Cαt (∂D) with the ordinary norms of C(∂D)3 and Cα(∂D)3, respec-

tively, and CαDiv(∂D) with the norm ‖a‖CαDiv = ‖a‖Cα + ‖Div a‖Cα . Then we can prove:

Theorem 3.33 (a) The boundary operator M : Ct(∂D)→ Cαt (∂D), defined by

(Ma)(x) = ν(x)×∫∂D

curlx[a(y) Φ(x, y)

]ds(y) , x ∈ ∂D , (3.46)


(b) M is well defined and compact from CαDiv(∂D) into itself.

(c) The boundary operator N : CαDiv(∂D)→ Cα

Div(∂D), defined by

(Na)(x) = ν(x)× curl curl

∫∂D

a(y) Φ(x, y) ds(y) , x ∈ ∂D , (3.47)

is well defined and bounded. Here, the right hand side is the trace of curl2 v with vfrom (3.42), see (3.44).

Proof: (a) We see from (3.45) that the kernel of this integral operator has the form

G(x, y) = ∇xΦ(x, y)[ν(x)− ν(y)

]> − ∂Φ

∂ν(x)(x, y) I .

The component gj,` of the first term is given by

gj,`(x, y) =∂Φ

∂xj(x, y)

[ν`(x)− ν`(y)

],

and this satisfies certainly the first assumption of part (a) of Theorem 3.16 for α = 1. Forthe second assumption we write

gj,`(x1, y)−gj,`(x2, y) =[ν`(x1)−ν`(x2)

] ∂Φ

∂xj(x1, y) +

[ν`(x2)−ν`(y)

] [ ∂Φ

∂xj(x1, y)− ∂Φ

∂xj(x2, y)

]and thus by the same arguments as in the proof Lemma 3.12∣∣gj,`(x1, y)− gj,`(x2, y)

∣∣ ≤ c|x1 − x2||x1 − y|2

+ c|x2 − y||x1 − x2||x1 − y|3

≤ c′|x1 − x2||x1 − y|2

.

This settles the first term of G. For the second term we observe that

∂Φ

∂ν(x)(x, y) = − ∂Φ

∂ν(y)(x, y) +

[ν(x)− ν(y)

]· ∇xΦ(x, y) .

The first term is just the kernel of the double layer operator treated in the previous theorem.For the second we can apply the first part again because it is just

∑3`=1 g`,`(x, y).

(b) We note that the space CαDiv(∂D) is a subspace of Cα

t (∂D) with bounded imbeddingand, furthermore, the space Cα

t (∂D) is compactly imbedded in Ct(∂D) by Lemma 3.17.


Therefore, M is compact from CαDiv(∂D) into Cα

t (∂D). It remains to show that DivM iscompact Cα

Div(∂D) into Cα(∂D).For a ∈ Cα

Div(∂D) we define the potential v by

v(x) =

∫∂D

a(y) Φ(x, y) ds(y) , x ∈ D .

Then Ma = ν × curl v|− + 12a by Theorem 3.32. Furthermore, by Theorem 3.32 again we

conclude that for x /∈ ∂D

curl curl v(x) = (∇ div−∆)

∫∂D

a(y) Φ(x, y) ds(y)

= ∇∫∂D

Div a(y) Φ(x, y) ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

and thus by Corollary ?? and the jump condition of the derivative of the single layer (The-orem 3.15)

Div(Ma)(x) = −ν(x) · curl curl v(x)|− +1

2Div a(x)

= −∫∂D

[Div a(y)

∂Φ

∂ν(x)(x, y) + k2ν(x) · a(y) Φ(x, y)

]ds(y)

= −D′(Div a) − k2 ν · Sa .

The assertion follows because D′ Div and ν · S are both compact from CαDiv(∂D) into

Cα(∂D).

(c) Define

w(x) = curl curl

∫∂D

a(y) Φ(x, y) ds(y) , x /∈ ∂D .

Writing again curl2 = ∇ div−∆ yields for x = z + tν(z) ∈ Hρ \ ∂D:

w(x) = ∇ div

∫∂D

Φ(x, y) a(y) ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

=

∫∂D

∇xΦ(x, y) Div a(y) ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

=

∫∂D

∇xΦ(x, y)[Div a(y)−Div a(z)

]ds(y) + Div a(z)

∫∂D

∇xΦ(x, y) ds(y)

+ k2

∫∂D

a(y) Φ(x, y) ds(y)

We use Lemma 3.14 and arrive at

w(x) =

∫∂D

∇xΦ(x, y)[Div a(y)−Div a(z)

]ds(y) + k2

∫∂D

a(y) Φ(x, y) ds(y)

+ Div a(z)

∫∂D

H(y) Φ(x, y) ds(y) − Div a(z)

∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) .


Therefore, by Lemma 3.12, Theorems 3.11 and 3.13 the tangential component of w is con-tinuous, thus Na is given by9

(Na)(x) = ν(x)× w(x) = ν(x)×∫∂D

∇xΦ(x, y)[Div a(y)−Div a(x)

]ds(y)

+ Div a(x) ν(x)×∫∂D

H(y) Φ(x, y) ds(y)

− Div a(x) ν(x)×∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) + k2 ν(x)×

∫∂D

a(y) Φ(x, y) ds(y) .

The boundedness of the last three terms as operators from CαDiv(∂D) into Cα

t (∂D) followfrom the boundedness of the single and double layer boundary operators S and D. For theboundedness of the first term we apply part (b) of Theorem 3.16. The assumptions∣∣ν(x)×∇xΦ(x, y)

∣∣ ≤ c

|x− y|2for all x, y ∈ ∂D , x 6= y , and

∣∣ν(x1)×∇xΦ(x1, y)− ν(x2)×∇xΦ(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3

for all x1, x2, y ∈ ∂D with |x1−y| ≥ 3|x1−x2| are simple to prove (cf. proof of Lemma 3.12).For the third assumption, namely∣∣∣∣∫

∂D\K(x,r)

ν(x)×∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ ∣∣∣∣∫∂D\K(x,r)

∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ c (3.48)

for all x ∈ ∂D and r > 0 we refer to Lemma 3.14 below. This proves boundedness of N fromCαDiv(∂D) into Cα

t (∂D). We consider now the surface divergence DivN . By Corollary ??and the form of w we conclude that DivNa = Div(ν×w) = −ν · curlw. For x = z+ tν(z) ∈Hρ \ ∂D we compute

curlw(x) = curl3∫∂D

a(y) Φ(x, y) ds(y) = k2

∫∂D

∇xΦ(x, y)× a(y) ds(y)

= −k2

∫∂D

∇yΦ(x, y)× a(y) ds(y)

= k2

∫∂D

∇yΦ(x, y)×[a(z)− a(y)

]ds(y) − k2

∫∂D

∇yΦ(x, y) ds(y) × a(z)

= k2

∫∂D

∇yΦ(x, y)×[a(z)− a(y)

]ds(y) − k2

∫∂D

Grad yΦ(x, y) ds(y) × a(z)

− k2

∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) × a(z)

= k2

∫∂D

∇yΦ(x, y)×[a(z)− a(y)

]ds(y) + k2

∫∂D

H(y) Φ(x, y) ds(y) × a(z)

− k2

∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y) × a(z) .

9This can also serve as a definition!


The normal component is bounded by the same arguments as above. 2

3.2.4 Uniqueness and Existence

First we want to prove that there exists at most one solution of the exterior boundary valueproblem (3.38a), (3.38b),(3.39a), (3.39b) and therefore also to the scattering problem (3.36a),(3.36b),(3.39a), (3.39b).

Theorem 3.34 For every tangential field c ∈ CαDiv(∂D)3 there exists at most one radiating

solution Es, Hs ∈ C1(R3 \D) ∩ C(R3 \D) of the exterior boundary value problem (3.38a),(3.38b), (3.39a), (3.39b).

Proof: Let Esj , H

sj for j = 1, 2 be two solutions of the boundary value problem. Then the

difference Es = Es1 − Es

2 and Hs = Hs1 −Hs

2 solve the exterior boundary value problem forboundary data c = 0.In the proof of the Stratton-Chu formula (Theorem 3.29) we have derived the followingformula:∫

|y|=R|√ε0E

s −√µ0Hs × ν|2ds = ε0

∫|y|=R

|Es|2ds + µ0

∫|y|=R

|Hs × ν|2ds

− 2√ε0µ0 Re

∫∂D

Es · (Hs × ν) ds .

The integral∫∂DEs · (Hs × ν) ds =

∫∂DHs · (ν × Es) ds vanishes by the boundary condi-

tion. Since the left hand side tends to zero by the radiation condition we conclude that∫|y|=R |E

s|2ds tends to zero as R → ∞. Furthermore, by the Stratton-Chu formula (Theo-

rem 3.29) in the exterior of D we can represent Es(x) in the form

Es(x) = curl

∫∂D

[ν(y)×Es(y)

]Φ(x, y) ds(y)− 1

iωε0

curl curl

∫∂D

[ν(y)×Hs(y)

]Φ(x, y) ds(y)

for x /∈ D. From these two facts we observe that every component u = Esj satisfies the

Helmholtz equation ∆u+ k2u = 0 in the exterior of D and limR→0

∫|x|=R |u|ds = 0. Further-

more, we recall that Φ(x, y) satisfies the Sommerfeld radiation condition (3.2) and thereforealso u = Es

j . The triangle inequality in the form |z| ≤ |z−w|+|w|, thus |z|2 ≤ 2|z−w|2+2|w|2yields∫|x|=R

∣∣∣∣ x|x| · ∇u(x)

∣∣∣∣2 ds(x) ≤ 2

∫|x|=R

∣∣∣∣ x|x| · ∇u(x)− iku(x)

∣∣∣∣2 ds(x) + 2k2

∫|x|=R

∣∣u(x)∣∣2ds(x) ,

and this tends to zero as R tends to zero. We are now in the position to apply Rellich’sLemma 3.21 (or Lemma 3.22). This yields u = 0 in the exterior of D and ends the proof. 2

Now we turn to the question of existence of the scattering problem (3.36a), (3.36b), (3.39a),(3.39b) or, more generally, the exterior boundary value problem (3.38a), (3.38b), (3.39a),(3.39b). First we prove an existence result which is not optimal but rather serves as apreliminary result to motivate a more complicated approach..


Theorem 3.35 Assume that w = 0 is the only solution of the following interior boundaryvalue problem:

curl curlw − k2w = 0 in D , ν × curlw = 0 on ∂D . (3.49)

Then, for every c ∈ CαDiv(∂D), there exists a unique solution Es, Hs ∈ C1(R3\D)∩C(R3\D)

of the exterior boundary value problem (3.38a), (3.38b), (3.39a), (3.39b). In particular,under this assumption, the scattering problem (3.36a), (3.36b), (3.39a), (3.39b) has a uniquesolution for every incident field. The solution has the form of

Es(x) = curl

∫∂D

a(y) Φ(x, y) ds(y) , Hs(x) =1

iωµ0

curlEs(x) , x /∈ D , (3.50)

for a ∈ CαDiv(∂D) which is the unique solution of the boundary integral equation

1

2a(x) + ν(x)× curl

∫∂D

a(y) Φ(x, y) ds(y) = c(x) , x ∈ ∂D . (3.51)

Proof: Since M is compact we can apply the Fredholm theory10 to equation (3.51): Ex-istence is assured if uniqueness holds. Therefore, let a ∈ Cα

Div(∂D) be a solution of thehomogeneous boundary integral equation (3.33), i.e. with the operator M

1

2a + Ma = 0 on ∂D .

Define E and H as in (3.50); that is,

E(x) = curl

∫∂D

a(y) Φ(x, y) ds(y) , H =1

iωµ0

curlE , x /∈ ∂D .

Then E, H satisfy the Maxwell system and, by the jump condition of Theorem 3.32 andthe homogeneous integral equation, ν × E(x)

∣∣+

= 12a +Ma = 0 on ∂D. The uniqueness

result of Theorem 3.34 yields E = 0 in R3 \ D. Application of Theorem 3.32 again yieldsthat ν × curlE is continuous on ∂D, thus ν × curlE

∣∣− = 0 on ∂D. From our assumption

we conclude that E vanishes also inside of D. Now we apply Theorem 3.32 a third time andhave that a = ν×E

∣∣−− ν×E

∣∣+

= 0. Therefore, the homogeneous integral equation admitsonly a = 0 as a solution and, therefore, there exists a unique solution of the inhomogeneousequation for every right hand side c ∈ Cα

Div(∂D). 2

In general, there exist eigenvalues of the problem (3.49). First we show:

Lemma 3.36 Let u solve the scalar Helmholtz equation ∆u + k2u = 0 in the unit ballB(0, 1). Then v(x) = curl

[u(x)x

]solves curl curl v− k2v = 0 in B(0, 1) and ν × v = Grad u

on S2 = ∂B(0, 1) where Grad u is again the surface gradient of u.

Proof: From curl curl = ∇ div−∆ we conclude

curl curl v(x) = curl[∇ div

(u(x)x

)−∆

(u(x)x

)]= − curl ∆

(u(x)x

).

10This is actually a result by Riesz.


For any j ∈ 1, 2, 3 we compute

∆(u(x)xj

)=(∆u(x)

)xj + 2∇u(x) · ∇xj = −k2u(x)xj + 2

∂u

∂xj(x)

and thus curl curl v(x) = − curl ∆(u(x)x

)= k2v(x). Furthermore, ν(x) × v(x) = x ×[

∇u(x)× x]

= Grad u(x) on S2. 2

Example 3.37 We claim that the field

w(r, θ, φ) = curl curl

[sin θ

(k cos(kr)− sin(kr)

r

)r

]satisfies curl curlw− k2w = 0 in B(0, 1) and ν(x)× curlw(x) = k2 cos θ

[k cos k− sin k

]θ on

S2 = ∂B(0, 1).

Indeed, we can write w = curl curl(u(x)x

)with

u(r, θ, φ) = sin θ

(k

cos(kr)

r− sin(kr)

r2

)= sin θ

d

dr

sin(kr)

r.

By direct evaluation of the Laplace operator in spherical coordinates we see that u satisfies theHelmholtz equation ∆u+k2u = 0 in all of R3. (Note that sin(kr)

ris analytic in R!) By the pre-

vious lemma we have that v(x) = curl(u(x)x

)satisfies curl curl v− k2v = 0, and thus also w

satisfies curl curlw− k2w = 0. We compute curlw as curlw(x) = curl[∇ div−∆

](u(x)x

)=

k2 curl(u(x)x

)= k2v(x) and thus by the previous lemma ν(x)× curlw(x) = k2 Grad u(x) =

k2 cos θ[k cos k − sin k

]θ on S2.

Therefore, if k > 0 is any zero of ψ(k) = k cos k − sin k then the corresponding field wsatisfies (3.49).

Instead of the insufficent ansatz (3.50) we propose a modified one of the form

Es(x) = curl

∫∂D

a(y) Φ(x, y) ds(y) (3.52a)

+ η curl curl

∫∂D

[ν(y)× (S2

i a)(y)]

Φ(x, y) ds(y) , (3.52b)

Hs =1

iωµ0

curlEs (3.52c)

for some constant η ∈ C, some a ∈ CαDiv(∂D), and where the bounded operator Si :

Cα(∂D)3 → C1,α(∂D)3 is the single layer boundary operator for the value k = i, con-sidered componentwise; that is, Sia = (Sia1, Sia2, Sia3)> for a = (a1, a2, a3)> ∈ Cα(∂D)3.We note that Si is bounded from Cα(∂D)3 into C1,α(∂D)3 by Theorem 3.20. Therefore, theoperator K : a 7→ ν × S2

i a is compact from CαDiv(∂D) into itself. We need the following

additional result for the single layer boundary operator Si for wavenumber k = i.


Lemma 3.38 The operator Si is selfadjoint with respect to 〈ϕ, ψ〉∂D =∫∂Dϕψ ds; that is

〈Siϕ, ψ〉∂D = 〈ϕ, Siψ〉∂D for all ϕ, ψ ∈ Cα(∂D)

and one-to-one.

Proof: Let ϕ, ψ ∈ Cα(∂D) and define u = Siϕ and v = Siψ as the single layer potentialswith densities ϕ and ψ, respectively. Then u and v are solutions of the Helmholtz equation∆u− u = 0 and ∆v − v = 0 in R3 \ ∂D. Furthermore, u and v and their derivatives decayexponentially as |x| tends to infinity. u and v are continuous in all of R3 and ∂v/∂ν|− −∂v/∂ν|+ = ψ. By Green’s formula we have

〈Siϕ, ψ〉∂D =

∫∂D

(Siϕ)ψ ds =

∫∂D

u

(∂v

∂ν

∣∣∣∣−− ∂v

∂ν

∣∣∣∣+

)ds

=

∫∫D

(∇u · ∇v + u v) dx+

∫∫B(0,R)\D

(∇u · ∇v + u v) dx−∫|x|=R

u∂v

∂rds .

The integral over the sphere x : |x| = R tends to zero and thus

〈Siϕ, ψ〉∂D =

∫∫R3

(∇u · ∇v + u v) dx .

This term is symmetric with respect to u and v. Furthermore, if Siϕ = 0 we conclude forψ = ϕ that u vanishes in R3. The jump condition ∂u/∂ν|− − ∂u/∂ν|+ = ϕ implies that ϕvanishes which shows injectivity of Si. 2

Analogously to the beginning of the previous section we observe (with the help of Theo-rem 3.32) that the ansatz (3.52b), (3.52c) solves the exterior boundary value problem ifa ∈ Cα

Div(∂D) solves the equation

1

2a + Ma + ηNKa = c on ∂D (3.53)

where again Ka = ν × S2i a. Finally, we can prove the general existence theorem.

Theorem 3.39 For every c ∈ CαDiv(∂D), there exists a unique radiating solution E,H ∈

C1(R3 \ D) ∩ C(R3 \ D) of the exterior boundary value problem. In particular, under thisassumption, the scattering problem has a unique solution for every incident field. The solu-tion has the form of (3.52b), (3.52c) for any η ∈ C \R and some a ∈ Cα

Div(∂D) which solvesthe boundary equation (3.53).

Proof: We make the ansatz (3.52b), (3.52c) and have to discuss the boundary equation(3.53). The compactness of M and K and the boundedness of N yields compactness of thecomposition NK. Therefore, the Fredholm alternative is applicable to (3.53).To show uniqueness, let a be a solution of the homogeneous equation. Then, with theansatz (3.52b), (3.52c), we conclude that ν × Es|+ = 0 and thus Es = 0 in R3 \ D by the

3.3. EXERCISES 127

uniqueness result. From the jump conditions of Theorem 3.32 we conclude that (note thatcurl3

∫∂Da(y) Φ(x, y) ds(y) = − curl ∆

∫∂Da(y) Φ(x, y) ds(y) = k2 curl

∫∂Da(y) Φ(x, y) ds(y))

ν × Es|− = ν × Es|− − ν × Es|+ = −a ,

ν × curlEs|− = ν × curlEs|− − ν × curlEs|+ = −η k2Ka .

and thus ∫∂D

(ν × curlEs|−) · Es ds = ηk2

∫∂D

(Ka) · (a× ν) ds

= ηk2

∫∂D

(ν × S2i a) · (a× ν) ds

= −ηk2

∫∂D

S2i a · a ds

= −ηk2

∫∂D

|Sia|2 ds .

The left hand side is real valued by Green’s theorem applied in D. Also the integral on theright hand side is real. Because η is not real we conclude that both integrals vanish and thusalso a by the injectivity of Si. 2

3.3 Exercises


Version of January 14, 2013

Chapter 4

The Variational Approach to theCavity Problem

4.1 Formulation of the Problem

The cavity problem has been formulated in the Introduction, see (1.20a)–(1.20c). The prob-lem is to determine vector fields E and H with


curlH + (iωε− σ)E = Je in D , (4.1b)

ν × E = 0 on ∂D . (4.1c)

Here, Je is a given vector field on D which describes the source which we assume to be inL2(D)3. It is natural to search for L2−solutions E and H. But then it follows from (4.1a)and (4.1b) that also curlE, curlH ∈ L2(D)3. Furthermore, the boundary condition (4.1c)has to be incorporated into the space of functions.

These requirements lead to the introduction of Sobolev spaces. We begin with Sobolev spacesof scalar functions in Subsection 4.2.1 before we continue with vector-valued functions inSubsection 4.2.2.

4.2 Sobolev Spaces

4.2.1 Basic Properties of Sobolev Spaces of Scalar Functions

In this section we will present the definition of the Sobolev space H10 (D) and its basic

properties which are needed to treat the boundary value problem ∆u + k2u = f in D andu = 0 on ∂D. The definitions and proves are all elementary – mainly because we do notneed any smoothness assumptions on the set D ⊂ R3 which is in this section always an open

129

130 CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM

set. Some of the analogous properties of the space H1(D) are needed for the Helmholtzdecomposition of Subsection 4.2.3 and will be postponed to this subsection.

We assume that the reader is familiar with the following basic function spaces.1

Ck(D) =

u : D → C :

u is k times continuously differentiable in Dand all derivatives can be continuously extended to D

,

Ck(D)3 =u : D → C3 : uj ∈ Ck(D) for j = 1, 2, 3

,

Ck0 (D) =

u ∈ Ck(D) : supp(u) is compact and supp(u) ⊂ D

,

Ck0 (D)3 =

u ∈ Ck(D)3 : uj ∈ Ck

0 (D) for j = 1, 2, 3,

Lr(D) =

u : D → C : u is Lebesgue-measurable and

∫∫D

|u|rdx <∞,

Lr(D)3 =u : D → C3 : uj ∈ Lr(D) for j = 1, 2, 3

,

S =u ∈ C∞(R3) : sup

x∈R3

[|x|p∣∣Dqu(x)

∣∣] <∞ for all p ∈ N, q ∈ N3.

Here, r ∈ [1,∞), and the support of a measurable function u is defined by

supp(u) =⋂

K ⊂ D : K closed and u(x) = 0 a.e. on D \K.

The differential operator Dq for q = (q1, q2, q3) ∈ N3 is defined by

Dq =∂q1+q2+q3

∂xq11 ∂xq22 ∂x

q33

.

Definition 4.1 Let Ω ⊂ R3 be an open set. A function u ∈ L2(Ω) possesses a variationalgradient in L2(Ω) if there exists F ∈ L2(Ω)3 with∫∫

Ω

u∇ψ dx = −∫∫

Ω

F ψ dx for all ψ ∈ C∞0 (Ω) . (4.2)

We write ∇u = F . It is easy to show (see Exercise 4.2) that F is unique – if it exists.

Definition 4.2 We define the Sobolev space H1(Ω) by

H1(Ω) =u ∈ L2(Ω) : u possesses a variational gradient ∇u ∈ L2(Ω)3

and equip H1(Ω) with the inner product

(u, v)H1(Ω) = (u, v)L2(Ω) + (∇u,∇v)L2(Ω)3 =

∫∫Ω

[u v +∇u · ∇v

]dx .

Theorem 4.3 The space H1(Ω) is a Hilbert space.

1Some of them have been used before already.

4.2. SOBOLEV SPACES 131

Proof: Only completeness has to be shown. Let (un) be a Cauchy sequence in H1(Ω).Then (un) and (∇un) are Cauchy sequences in L2(Ω) and L2(Ω)3, respectively, and thusconvergent: un → u and ∇un → F for some u ∈ L2(Ω) and F ∈ L2(Ω)3. We show thatF = ∇u: For ψ ∈ C∞0 (Ω) we conclude that∫∫

Ω

un∇ψ dx = −∫∫

Ω

∇un ψ dx

by the definition of the variational gradient. The left hand side converges to∫∫

Ωu∇ψ dx,

the right hand side to −∫∫

ΩF ψ dx, and thus F = ∇u. 2

It is the first aim to prove denseness results. The proofs use the technique of mollifying thegiven function and use the Fourier transform and the convolution of functions.The Fourier transform F is defined by

(Fu)(x) = u(x) =1

(2π)3/2

∫∫R3

u(y) e−i x·ydy , x ∈ R3 .

F is well defined as an operator from S into itself and also from L1(R3) into the spaceCb(R3) of bounded continuous functions on R3. Furthermore, F is unitary with respect tothe L2−norm; that is,

(u, v)L2(R3) =(u, v)L2(R3)

for all u, v ∈ S . (4.3)

Therefore, by a general extension theorem from functional analysis, F has an extension toa unitary operator from L2(R3) onto itself; that is, (4.3) holds for all u.v ∈ L2(R3).

The convolution of two functions is defined by

(u ∗ v)(x) =

∫∫R3

u(y) v(x− y) dy , x ∈ R3 .

The Lemma of Young clarifies the mapping properties of the convolution operator.

Lemma 4.4 (Young) The convolution u ∗ v is well defined for u ∈ Lp(R3) and v ∈ L1(R3)for any p ≥ 1. Furthermore, in this case u ∗ v ∈ Lp(R3) and

‖u ∗ v‖Lp(R3) ≤ ‖u‖Lp(R3) ‖v‖L1(R3) for all u ∈ Lp(R3) , v ∈ L1(R3) .

This lemma implies, in particular, that L1(R3) is an algebra with the convolution as multipli-cation. The Fourier transform transforms the convolution into the pointwise multiplicationof functions:

F(u ∗ v)(x) = (2π)3/2 u(x) v(x) for all u ∈ Lp(R3) , v ∈ L1(R3) , p ∈ 1, 2 . (4.4)

The following result to smoothen functions will often be used.


Theorem 4.5 Let φ ∈ C∞(R) with φ(t) = 0 for |t| ≥ 1 and φ(t) > 0 for |t| < 1 and∫ 1

0φ(t2) t2dt = 1/(4π) (see Exercise 4.3). Define φδ ∈ C∞(R3) by

φδ(x) =1

δ3φ

(1

δ2|x|2), x ∈ R3 . (4.5)

Then supp(φδ) ⊂ B3(0, δ) and:

(a) u ∗ φδ ∈ C∞(R3) for all u ∈ L2(R3). Let K ∈ R3 be compact and u = 0 outside of K.Then supp(u ∗ φδ) ⊂ K +B[0, δ] = x+ y : x ∈ K , |y| ≤ δ.

(b) Let u ∈ L2(R3) and a ∈ R3 be a fixed vector. Set uδ(x) = (u ∗ φδ)(x+ δa) for x ∈ R3.Then ‖uδ − u‖L2(R3) → 0 as δ → 0.

(c) Let u ∈ H1(R3) and uδ as in part (b). Then ‖uδ − u‖H1(R3) → 0 as δ → 0.

Proof: (a) Fix any R > 0 and let |x| < R. Then

(u ∗ φδ)(x) =

∫∫|y|≤R+δ

u(y)φδ(x− y) dy

because |x − y| ≥ |y| − |x| ≥ |y| − R ≥ δ for |y| ≥ R + δ and thus φδ(x − y) = 0. Thesmoothness of φδ yields that this integral is infinitely often differentiable. Furthermore,(u ∗ φδ)(x) =

∫∫Ku(y)φδ(x − y) dy which vanishes for x /∈ K + B[0, δ] because |x − y| ≥ δ

for these x and y ∈ K.

(b) Substituting y = δz yields for the Fourier transform (note that dy = δ3dz),

(2π)3/2 φδ(x) =1

δ3

∫∫R3

φ

(1

δ2|y|2)e−i x·y dy =

∫∫R3

φ(|z|2)e−iδ x·z dz = (2π)3/2 φ1(δx) .

Furthermore, (2π)3/2 φ1(0) =∫∫

R3 φ(|y|2)dy = 4π

∫ 1

0φ(r2) r2dr = 1 by the normalization of

φ. Therefore, the Fourier transform of uδ is given by

(Fuδ)(x) = e−iδa·x(2π)3/2u(x) φδ(x)

where we used (4.4) and the translation property. For u ∈ L2(R3) we conclude

‖uδ − u‖2L2(R3) = ‖Fuδ − u‖2

L2(R3) =∥∥[exp(−iδa·)(2π)3/2φδ − 1

]u∥∥2

L2(R3)

=

∫∫R3

∣∣∣e−iδa·x(2π)3/2φ1(δx)− 1∣∣∣2 ∣∣u(x)

∣∣2 dx .The integrand tends to zero as δ → 0 for every x and is bounded by the integrable function‖(2π)3/2φ1 − 1‖2

∞|u(x)|2. This proves part (b) by the theorem of dominated convergence.

(c) First we note that ∇(u ∗ φδ) = ∇u ∗ φδ for u ∈ H1(R3). Indeed,

∇(u ∗ φδ)(x) =

∫∫R3

u(y)∇xφδ(x− y) dy = −∫∫

R3

u(y)∇yφδ(x− y) dy

=

∫∫R3

∇u(y)φδ(x− y) dy = (∇u ∗ φδ)(x)


because, for fixed x, the mapping y 7→ φδ(x−y) is in C∞0 (R3). Application of part (b) yieldsthe assertion. 2

As a direct consequence we have a first denseness result.

Corollary 4.6 The space C∞(R3) is dense in H1(R3).

For the boundary value problem we have to incorporate boundary conditions of Dirichlettype into the space. First we note that the space C∞0 (D) of infinitely often differentiablefunctions with support in D are contained in H1(D). Indeed, these functions u can beextended by zero to u ∈ C∞0 (R3), and partial integration yields that ∇u|D is the variationalgradient of u. We define:

Definition 4.7 The space H10 (D) is defined as the closure of C∞0 (D) with respect to

‖ · ‖H1(D).

By definition, H10 (D) is a closed subspace of H1(D). We understand H1

0 (D) as the space ofdifferentiable (in the variational sense) functions u with u = 0 on ∂D. This interpretationis justified by the following result the proof of which will be postponed to Subsection 4.2.3(Theorem 4.25) because this result is needed only for the motivation of the space H1

0 (D).

Theorem 4.8 Let D be a bounded Lipschitz domain. Thenu ∈ C1(D) : u|∂D = 0

is a

subset of H10 (D).

Remark: Let D be an open bounded set. For fixed u ∈ H1(D) the left and the right handside of the definition (4.2) (for Ω = D) of the variational derivative; that is,

`1(ψ) :=

∫∫D

u∇ψ dx and `2(ψ) := −∫∫

D

ψ∇u dx , ψ ∈ C∞0 (D) ,

are linear functionals from C∞0 (D) into C which are bounded with respect to the norm‖ · ‖H1(D) by the Cauchy-Schwarz inequality:∣∣`1(ψ)

∣∣ ≤ ‖u‖L2(D)‖∇ψ‖L2(D)3 ≤ ‖u‖L2(D)‖ψ‖H1(D)

and analogously for `2. Therefore, there exist linear and bounded extensions of these func-tionals to the closure of C∞0 (D) which is H1

0 (D). Thus the formula of partial integrationholds; that is,∫∫

D

u∇ψ dx = −∫∫

D

ψ∇u dx for all u ∈ H1(D) and ψ ∈ H10 (D) . (4.6)

It is the aim to prove compactness of the imbedding H10 (D)→ L2(D) provided D is bounded.

For this we choose a box Ω of the form Ω = (−R,R)3 ⊂ R3 with D ⊂ Ω. We cite the followingbasic result from the theory of Fourier series.


Theorem 4.9 Let Ω = (−R,R)3 ⊂ R3 be a bounded cube. For u ∈ L2(Ω) define theFourier coefficients un ∈ C by

un =1

(2R)3

∫∫Ω

u(x) e−iπRn·xdx for n ∈ Z3 . (4.7)

Thenu(x) =

∑n∈Z3

un ei πRn·x ,

in the L2−sense; that is,

∫∫Ω

∣∣∣∣∣∣u(x)−∑|n|≤N

un ei πRn·x

∣∣∣∣∣∣2

dx −→ 0 , N →∞ .

Furthermore,

(u, v)L2(Ω) = (2R)3∑n∈Z3

un vn and ‖u‖2L2(Ω) = (2R)3

∑n∈Z3

|un|2 .

Therefore, the space L2(Ω) can be characterized by the space of all functions u such that∑n∈Z3 |un|2 converges. Because, for sufficiently smooth functions,

∇u(x) = iπ

R

∑n∈Z3

un n ei πRn·x ,

we observe that i πRnun are the Fourier coefficients of ∇u. The requirement ∇u ∈ L2(D)3

leads to

Definition 4.10 We define the Sobolev space H1per(Ω) of periodic functions by

H1per(Ω) =

u ∈ L2(Ω) :

∑n∈Z3

(1 + |n|2) |un|2 <∞

with inner product

(u, v)H1per(Ω) = (2R)3

∑n∈Z3

(1 + |n|2)unvn .

Here, un and vn are the Fourier coefficients of u and v, respectively, see (4.7).

The next theorem guarantees that the zero-extensions of functions of H10 (D) belong to

H1per(Ω).

Theorem 4.11 Let again D ⊂ R3 be a bounded open set and Ω = (−R,R)3 an open box

which contains D in its interior. Then the extension operator E : u 7→ u =

u on D,0 on Ω \D ,

is linear and bounded from H10 (D) into H1

per(Ω).


Proof: Let first u ∈ C∞0 (D). We compute the Fourier coefficients un of the zero-extensionu of u:

un =1

(2R)3

∫∫Ω

u(x) e−iπRn·xdx =

iR

πnj(2R)3

∫∫Ω

u(x)∂

∂xje−i

πRn·xdx

= − iR

πnj(2R)3

∫∫Ω

∂u

∂xj(x) e−i

πRn·xdx = − iR

πnjvn

where vn are the Fourier coefficients of v = ∂u/∂xj. Therefore, by Theorem 4.9,

(2R)3∑n∈Z3

n2j |un|2 =

R2

π2(2R)3

∑n∈Z3

|vn|2 =R2

π2

∥∥∥∥ ∂u∂xj∥∥∥∥2

L2(Ω)

=R2

π2

∥∥∥∥ ∂u∂xj∥∥∥∥2

L2(D)

.

Furthermore,

(2R)3∑n∈Z3

|un|2 = ‖u‖2L2(Ω) = ‖u‖2

L2(D)

and thus adding the equations for j = 1, 2, 3 and for u,

‖u‖2H1per(Ω) ≤

(1 +

R2

π2

)‖u‖2

H1(D) .

This holds for all functions u ∈ C∞0 (D). Because C∞0 (D) is dense in H10 (D) we conclude

that

‖u‖H1per(Ω) ≤

√1 +

R2

π2‖u‖H1(D) for all u ∈ H1

0 (D) .

This proves boundedness of the extension operator E. 2

As a simple application we have:

Theorem 4.12 Let D ⊂ R3 be a bounded open set. Then the imbedding H10 (D) → L2(D)

is compact.

Proof: Let again Ω = (−R,R)3 ⊂ R3 such that D ⊂ Ω. First we show that the imbeddingJ : H1

per(Ω)→ L2(Ω) is compact. Define JN : H1per(Ω)→ L2(Ω) by

(JNu)(x) :=∑|n|≤N

un ei πRn·x , x ∈ Ω , N ∈ N .

Then JN is obviously bounded and finite dimensional, because the rangeR(JN) = span

ei

πRn· : |n| ≤ N

is finite dimensional. Therefore, by a well known result

from functional anaysis, JN is compact. Furthermore,

‖(JN − J)u‖2L2(Ω) = (2R)3

∑|n|>N

|un|2 ≤(2R)3

1 +N2

∑|n|>N

(1 + |n|2) |un|2 ≤1

1 +N2‖u‖2

H1per(Ω) .


Therefore, ‖JN − J‖2H1per(Ω)→L2(Ω) ≤

11+N2 → 0 as N tends to infinity and thus, again by a

well known result from functional anaysis, also J is compact.

Now the claim of the theorem follows because the composition

R J E : H10 (D)

E−→ H1per(Ω)

J−→ L2(Ω)R−→ L2(D)

is compact. Here R : L2(Ω)→ L2(D) denotes the restriction operator. 2

The next result is often called an inequality of Friedrich’s Type.

Theorem 4.13 For any bounded open set D ⊂ R3 there exists c > 0 with

‖u‖L2(D) ≤ c ‖∇u‖L2(D)3 for all u ∈ H10 (D) .

Proof: Let again first u ∈ C∞0 (D), extended by zero into R3. Then, if again D ⊂ (−R,R)3,

u(x) = u(x1, x2, x3) = u(−R, x2, x3)︸︷︷︸= 0

+

x1∫−R

∂u

∂x1

(t, x2, x3) dt

and thus for x ∈ Ω by the inequality of Cauchy–Schwarz

∣∣u(x)∣∣2 ≤ (x1 +R)

x1∫−R

∣∣∣∣ ∂u∂x1

(t, x2, x3)

∣∣∣∣2 dt ≤ 2R

R∫−R

∣∣∣∣ ∂u∂x1

(t, x2, x3)

∣∣∣∣2 dt and

R∫−R

∣∣u(x)∣∣2 dx1 ≤ (2R)2

R∫−R

∣∣∣∣ ∂u∂x1

(t, x2, x3)

∣∣∣∣2 dt .Integration with respect to x2 and x3 yields

‖u‖2L2(D) = ‖u‖2

L2(Ω) ≤ (2R)2

R∫−R

R∫−R

R∫−R

∣∣∣∣ ∂u∂x1

(t, x2, x3)

∣∣∣∣2 dt dx2 dx3

≤ (2R)2‖∇u‖2L2(Ω)3 = (2R)2‖∇u‖2

L2(D)3 .

By a density argument this holds for all u ∈ H10 (D). 2

4.2.2 Basic Properties of Sobolev Spaces of Vector Valued Func-tions

We follow the scalar case as closely as possible. We assume here that all functions arecomplex valued. First we note that in the formulation (4.1a), (4.1b) not all of the partialderivatives of the vector field E and H appear but only the curl of E and H.


Definition 4.14 Let Ω ⊂ R3 be an open set.

(a) A vector field v ∈ L2(Ω)3 possesses a variational curl in L2(Ω)3 if there exists a vectorfield w ∈ L2(Ω)3 such that∫∫

Ω

v · curlψ dx =

∫∫Ω

w · ψ dx for all vector fields ψ ∈ C∞0 (Ω)3 .

(b) A vector field v ∈ L2(Ω)3 possesses a variational divergence in L2(Ω) if there exists ascalar function p ∈ L2(Ω) such that∫∫

Ω

v · ∇ϕdx = −∫∫

Ω

pϕ dx for all scalar functions ϕ ∈ C∞0 (Ω) .

The functions w and p are unique if they exist (compare Exercise 4.2). We write again curl vand div v for w and p, respectively.

(c) We define the space H(curl,Ω) by

H(curl,Ω) =u ∈ L2(Ω)3 : u has a variational curl in L2(Ω)3

,

and equip it with the natural inner product

(u, v)H(curl,Ω) = (u, v)L2(Ω)3 + (curlu, curl v)L2(Ω)3 .

With this inner product the space H(curl,Ω) is a Hilbert space. The proof uses the samearguments as in the proof of Theorem 4.3.

In the next result we prove denseness properties as in Corollary 4.6.

Corollary 4.15 C∞(R3)3 is dense in H(curl,R3).

Proof: We apply Theorem 4.5. It is sufficient to show that

curl(u ∗ φδ) = (curlu) ∗ φδ for all u ∈ H(curl,R3)

with φδ from (4.5). Because supp(φδ) ⊂ B[0, δ] we conclude for |x| < R that

(u ∗ φδ)(x) =

∫∫B(0,R+δ)

u(y)φδ(x− y) dy

and thus for any fixed vector a ∈ R3 (interchanging differentiation and integration is allowed!)

a · curl(u ∗ φδ)(x) =

∫∫R3

a ·[∇xφδ(x− y)× u(y)

]dy = −

∫∫R3

a ·[∇yφδ(x− y)× u(y)

]dy

= −∫∫

R3

[a×∇yφδ(x− y)

]· u(y) dy =

∫∫R3

curly[a φδ(x− y)

]· u(y) dy

=

∫∫R3

a · curlu(y)φδ(x− y) dy = a · (curlu ∗ φδ)(x)


because, for fixed x, y 7→ a φδ(x − y) is in C∞0 (R3)3. This holds for all vectors a, thuscurl(u ∗ φδ) = (curlu) ∗ φδ. 2

Analogously to the definition of H10 (D) we define

Definition 4.16 For any open set D ⊂ R3 we define H0(curl, D) as the closure of C∞0 (D)3

in H(curl, D).

Remark: Vector fields v ∈ H0(curl, D) do not necessarily vanish on ∂D – in contrast tofunctions in H1

0 (D). Only their tangential components vanish. This is seen by the followingtwo results:

Lemma 4.17 The space ∇H10 (D) =

∇p : p ∈ H1

0 (D)

is a closed subspace of H0(curl, D).

Therefore2, if p ∈ C1(D) with p = 0 on ∂D then p ∈ H10 (D), thus ∇p ∈ H0(curl, D) but

∇p = (∂p/∂ν)ν does not have to vanish on ∂D.

Proof of the lemma: By the definition of H10 (D) there exists a sequence pj ∈ C∞0 (D) with

pj → p in H1(D). It is certainly ∇pj ∈ C∞0 (D)3. Because ∇pj → ∇p in L2(D)3 andcurl∇pj = 0 we note that (∇pj) is a Cauchy sequence in H(curl, D) and thus convergent.This shows that ∇p ∈ H0(curl, D). It remains to show closedness of ∇H1

0 (D) in L2(D)3 –and thus in H(curl, D). Let pj ∈ H1

0 (D) with ∇pj → F in L2(D)3 for some F ∈ L2(D)2.Then (∇pj) is a Cauchy sequence in L2(D)3. By Friedrichs inequality of Theorem 4.13 weconclude that also (pj) is a Cauchy sequence in L2(D). Therefore, (pj) is a Cauchy sequencein H1(D) and thus convergent. This shows that F = ∇p for some p ∈ H1

0 (D) and ends theproof. 2

The following result corresponds to Theorem 4.8. We transfer again the proof to Subsec-tion 4.2.3, Theorem 4.31.

Theorem 4.18 Let D be a bounded Lipschitz domain. Then:u ∈ C1(D)3 : ν × u = 0 on ∂D

⊂ H0(curl, D) .

We recall that H10 (D) is compactly suppported in L2(D) by Theorem 4.12. This result

was crucial for the solvability of the interior boundary value problem for the Helmholtzequation. As one can easily show (see Exercise 4.8), the space H0(curl, D) fails to be com-pactly imbedded in L2(D)3. Therefore, we will decompose the variational problem into twoproblems which themselves can be treated in the same way as for the Helmholtz equation.The basis of this decomposition is set by the important Helmholtz decomposition which wediscuss in the subsequent subsection.

2if the boundary is sufficiently smooth, see Theorem 4.8


4.2.3 The Helmholtz Decomposition and Further Results on SobolevSpaces

Now we turn to the Helmholtz decomposition. For the proof (in the general formulation) weneed the Theorem of Lax and Milgram:

Theorem 4.19 (Lax and Milgram)Let X be a Hilbert space over C, ` : X → C linear and bounded, a : X×X → C sesqui-linearand bounded and coercive; that is, there exists c1, c2 > 0 with∣∣a(u, v)

∣∣ ≤ c1 ‖u‖X‖v‖X for all u, v ∈ X ,

Re a(u, u) ≥ c2 ‖u‖2X for all u ∈ X .

Then there exists a unique u ∈ X with

a(ψ, u) = `(ψ) for all ψ ∈ X .

Furthermore, there exists c > 0, independent of u, such that ‖u‖X ≤ ‖`‖X∗.

For a proof we refer to the literature (see, e.g. [3]).

Now we can prove the general form of the Helmholtz decomposition. We formulate it inL2(D)3 as well as in H0(curl, D).

Theorem 4.20 Let εc ∈ L∞(D) be complex-valued with Re εc(x) ≥ c > 0 almost everywhereon D. Define the subspaces H0(curl 0, D) and V ⊂ H0(curl, D) and V ⊂ L2(D)3 by

H0(curl 0, D) =u ∈ H0(curl, D) : curlu = 0 in D

, (4.8a)

V =u ∈ H0(curl, D) : (εcu, ψ)L2(D)3 = 0 for all ψ ∈ H0(curl 0, D)

, (4.8b)

V =u ∈ L2(D)3 : (εcu, ψ)L2(D)3 = 0 for all ψ ∈ H0(curl 0, D)

. (4.8c)

Then we have:

(a) H0(curl 0, D) and V are closed in L2(D)3, and L2(D)3 is the direct sum of V andH0(curl 0, D); that is, L2(D)3 = V ⊕ H0(curl 0, D). Furthermore, the correspondingprojection operator P : L2(D)3 → V is bounded.

(b) H0(curl 0, D) and V are closed in H0(curl, D) and H0(curl, D) is the direct sum of Vand H0(curl 0, D); that is, H0(curl, D) = V ⊕ H0(curl 0, D). Furthermore, the corre-sponding projection operator P : H0(curl, D) → V is bounded. P is the restriction ofP ; that is, Pu = P u for all u ∈ H0(curl, D).

Proof: The closedness of H0(curl 0, D) in H0(curl, D) is obvious. The closedness in L3(D)3

is seen as follows. Let un ∈ H0(curl 0, D) converge to some u in L2(D)3. Then (un) is a


Cauchy sequence in L2(D)3. From curlun = 0 we observe that (un) is also a Cauchy se-quence in H0(curl, D) and thus convergent in H0(curl, D). This shows that u ∈ H0(curl, D)and curlu = 0; that is, u ∈ H0(curl 0, D).The closedness of V and V in L2(D)3 and in H(curl, D), respectively, is easy to see andwe omit this part. Also, V ∩ H0(curl 0, D) = 0 because u ∈ V ∩ H0(curl 0, D) implies(εcu, u)L2(D)3 = 0 and thus, taking the real part, c‖u‖2

L2(D)3 ≤ Re (εcu, u)L2(D)3 = 0 whichyields u = 0.Finally, we have to prove that L2(D)3 ⊂ V+H0(curl 0, D) andH0(curl, D) ⊂ V+H0(curl 0, D).Let u ∈ L2(D)3. We define the sesquilinear form a : H0(curl 0, D)×H0(curl 0, D)→ C by

a(ψ, v) = (ψ, εcv)L2(D)3 =

∫∫D

ψ · εc v dx for all v, ψ ∈ H0(curl 0, D) . (4.9)

Then a is certainly bounded but also coercive in the space H0(curl 0, D) equipped with theinner product of H(curl, D). Indeed, we write

Re a(v, v) = Re (εcv, v)L2(D)3 ≥ c ‖v‖2L2(D)3 = c ‖v‖2

H(curl,D)

for v ∈ H0(curl 0, D) because curl v = 0. Therefore, the sesquilinear form a and – forevery fixed u ∈ L2(D)3 – the bounded linear form `(ψ) = (ψ, εcv)L2(D)3 , ψ ∈ H0(curl 0, D),satisfy the assumptions of the Theorem ?? of Lax and Milgram. Therefore, for every givenu ∈ L2(D)3 there exists a unique u0 ∈ H0(curl 0, D) with a(ψ, u0) = `(ψ) for all ψ ∈H0(curl 0, D); that is, (εcu0, ψ)L2(D)3 = (εcu, ψ)L2(D)3 for all ψ ∈ H0(curl 0, D). Therefore, u−u0 ∈ V and the decomposition u = u0+(u−u0) ∈ H0(curl 0, D)+V has been shown. Finally,boundedness of the projection operator P : u 7→ u − u0 follows from general properties ofdirect sums. This proves the result in L2(D)3. Analogously, the operator P : H0(curl, D)→V is defined in the same way by Pu = u− u0 where u0 ∈ H0(curl 0, D) is as before. 2

Remark: If εc is real valued; that is, εc = εr > 0 then the direct sums are orthogonalwith respect to the inner products (εru, v)L2(D)3 and (curlu, curl v)L2(D)3 + (εru, v)L2(D)3 ,

respectively, and P and P , respectively, are the orthogonal projections.

As mentioned before, the space H0(curl, D) is not compactly imbedded in L2(D)3. However– and this is the reason for introducing the Helmholtz decomosition – the subspace V iscompactly imbedded in L2(D)3. We formulate this crucial result in the following theorem.

Theorem 4.21 Let D be a bounded Lipschitz domain (see Definition 6.1 or Definition 4.22below). Then the space V from (4.8b) is compactly imbedded in L2(D)3.

We do not prove this theorem in its full generality but refer to the original reference [?].In the following we want to give a proof of this fact under the assumption that D is aC2−domain. If the reader is not interested in this proof he can just skip this part up to theend of this section and continue directly with Section 4.3.

For the anounced proof we need more sophisticated properties of the Sobolev spaces H1(D)and H0(curl, D) of scalar and vector functions. We have to make smoothness assumptionson the domain and recall Definition 6.1 for the notion of a Lipschitz domain.


Definition 4.22 We call a region D ⊂ R3 to be a Lipschitz domain, if there exists a finitenumber of open cylinders Uj of the form Uj = Qjx + z(j) : x ∈ B2(0, αj) × (−βj, βj)with z(j) ∈ R3 and rotations Qj ∈ R3×3 and real valued Lipschitz-continuous functionsξj ∈ C(B[0, αj]) with |ξj(x1, x2)| ≤ βj/2 for all (x1, x2) ∈ B2[0, αj] such that

∂D ⊂m⋃j=1

Uj ,

∂D ∩ Uj =Qjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 = ξj(x1, x2)

,

D ∩ Uj =Qjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 < ξj(x1, x2)

,

Uj \D =Qjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 > ξj(x1, x2)

.

We call Uj, ξj : j = 1, . . . ,m a local coordinate system of ∂D. For abbreviation wedenote by

Cj = Cj(αj, βj) = B2(0, αj)×(−βj, βj) =x = (x1, x2, x3) ∈ R3 : x2

1+x22 < α2

j , |x3| < βj

the standard cylinders. Furthermore, we define the mappings

Ψj(x) = Qj

x1

x2

ξj(x1, x2) + x3

+ z(j) , x = (x1, x2, x3)> ∈ Cj(αj, βj/2) ,

and its restriction Ψj to B2(0, αj); that is,

Ψj(x) = Qj

x1

x2

ξj(x1, x2)

+ z(j) , x = (x1, x2)> ∈ B2(0, αj) ,

which yields a parametrization of ∂D ∩ Uj in the form y = Ψj(x) for x ∈ B2(0, αj). Themappings Ψj “flatten” the boundary; that is,

∂D ∩ Uj =

Ψj(x) : x ∈ Cj, x3 = 0

=

Ψj(x) : x ∈ B2(0, αj),

D ∩ U ′j =

Ψj(x) : x ∈ C ′j, x3 < 0,

U ′j \D =

Ψj(x) : x ∈ C ′j, x3 > 0,

where C ′j = Cj(αj, βj/2) = B2(αj)× (−βj/2, βj/2) and U ′j = Ψj(C′j).

The proof of the following properties is simple.

Lemma 4.23 Let D be a bounded Lipschitz domain.

(a) Let u ∈ C1(D). Then u ∈ H1(D), and the classical derivatives ∂u/∂xj coincide withthe variational derivatives.


(b) Let u ∈ C1(D) with u = 0 on ∂D. The extension u of u by zero outside of D yieldsu ∈ H1(R3) and ∇u = ∇u in D and ∇u = 0 outside of D.

(c) For u ∈ H1(D) and φ ∈ C∞(D) the product φu ∈ H1(D) and ∇(φu) = u∇φ+ φ∇u.

(d) Let Q ∈ R3×3 be an orthogonal matrix and z ∈ R3. For u ∈ H1(D) define v(x) =u(Qx+ z) on U = x : Qx+ z ∈ D. Then v ∈ H1(U) and ∇v(x) = Q>∇u(Qx+ z).

Proof: (a) follows directly by partial integration (see Theorem 6.5), compare part (b).(b) Set g = ∇u in D and zero outside of D. Then g ∈ L2(R3)3 and for ψ ∈ C∞0 (R3) we haveby partial integration:∫∫

R3

u∇ψ dx =

∫∫D

u∇ψ dx = −∫∫

D

ψ∇u dx +

∫∂D

uψ ν ds︸︷︷︸= 0

= −∫∫

D

ψ∇u dx = −∫∫

R3

ψ g dx .

(c) This follows from the definition of the variational gradient and the product rule. Indeed,let ψ ∈ C∞0 (D). Then ψφ ∈ C∞0 (D) and thus∫∫

D

[u∇φ+ φ∇u]ψ dx =

∫∫D

[u (ψ∇φ) + (φψ)∇u] dx

=

∫∫D

[u (ψ∇φ)−∇(φψ)u] dx =

∫∫D

uφ∇ψ dx .

(d) Let again ψ ∈ C∞0 (D) and let q(j), j = 1, 2, 3, be the columns of the orthogonal matrixQ. We make the substitution y = Qx+ z and have

q(j) ·∫∫

D

∇u(y)|y=Qx+z ψ(x) dx

= q(j) ·∫∫

U

∇u(y)ψ(Q>(y − z)

)dy = −q(j) ·

∫∫U

u(y)Q∇ψ(x)|x=Q>(y−z) dy

= −∫∫

U

u(y)∂ψ(x)

∂xj

∣∣∣∣x=Q>(y−z)

dy = −∫∫

D

v(x)∂ψ(x)

∂xjdx .

This shows that ∂v(x)/∂xj = q(j) · ∇u(y)|y=Qx+z. 2

The proof of the following denseness result is technically complicated.

Theorem 4.24 Let D be a bounded Lipschitz domain. Then C∞(D) is dense in H1(D).

Proof: The proof consists of several steps. Let u ∈ H1(D).(i) Let Uj, ξj : j = 1, . . . ,m be a local coordinate system. Set U0 = D and choose apartition of unity φj : j = 0, . . . ,m on D corresponding to Uj : j = 0, . . . ,m. Set


uj(y) = φj(y)u(y) for j = 0, . . . ,m and y ∈ D. Then the support of uj is contained inUj ∩D. We transform uj by the definition vj(x) = uj(Qjx+ z(j)), x ∈ C−j , where

C−j =x = (x, x3) ∈ B2(0, αj)× (−βj, βj) : x3 < ξj(x)

for j = 1, . . . ,m. Here again x = (x1, x2). Then vj ∈ H1(C−j ) by the previouis lemma and

(because the support of uj is contained in Uj ∩ D) there exists δ > 0 with vj(x) = 0 forall x ∈ C−j \ Cj(αj − 2δ, βj − 2δ) where the cylinder Cj(αj − 2δ, βj − 2δ) is defined as inDefinition 4.22.

(ii) Let a = (0, 0, L+ 1)> where L denotes the (common) Lipschitz constant of the functionsξj. We show that for all j ∈ 1, . . . ,m and all x ∈ C−j ∩ Cj(αj − δ, βj − δ) the ballB3(x − εa, ε) is contained in C−j provided ε < δ/(L + 2). Indeed, for z ∈ B3(x − εa, ε) wehave that |x− z| < ε and |x3 − (L+ 1)ε− z3| < ε where again x = (x1, x2) and z = (z1, z2).Therefore, |z| ≤ |z − x| + |x| < ε + αj − δ ≤ αj. Furthermore, z3 < x3 − (L + 1)ε +ε ≤ ξj(x) − Lε ≤ ξj(z) + |ξj(x) − ξj(z)| − Lε ≤ ξj(z) + L

(|x − z| − ε) ≤ ξj(z). Finally,

z3 > x3 − (L+ 1)ε− ε ≥ −βj + δ − (L+ 2)ε ≥ −βj.

(iii) We extend vj by zero into all of R3 and set vεj (x) = (vj ∗ φε)(x − εa) where againa = (0, 0, L + 1)>. Then vεj ∈ C∞(R3) and vεj → vj in L2(R3). Its support is contained insupp(vj) + B3(εa, ε) ⊂ Cj(αj − δ, βj − δ) for ε < δ/(L + 2). Furthermore, for these ε andx ∈ C−j ∩ Cj(αj − δ, βj − δ) we have

∇vεj (x) =

∫∫C−j

vj(z)∇xφε(x− εa− z) dz = −∫∫

C−j

vj(z)∇zφε(x− εa− z) dz .

The function z 7→ φε(x − εa − z) belongs to C∞0 (C−j ) because is support is contained inB3(x− εa, ε) ⊂ C−j by part (ii). By the definition of the weak derivative we have that

∇vεj (x) =

∫∫C−j

∇vj(z)φε(x− εa− z) dz . (4.10)

For ε < δ/(L+2) and x ∈ C−j \Cj(αj−δ, βj−δ) both sides of (4.10) vanish. Therefore, (4.10)holds for all x ∈ C−j and thus (∇vεj )(x) = (gj ∗φε)(x−εa) for x ∈ C−j where gj = ∇vj on C−jand gj = 0 on R3 \C−j . By Theorem 4.5 we conclude again that (gj ∗φε)(x−εa) converges togj in L2(R3) and thus ∇vεj |C−j to ∇vj in L2(C−j ). This proves vεj → vj in H1(C−j ) for every

j = 1, . . . ,m.

(iv) In this step we transform back the function and set uεj(y) = vεj(Q>j (y − z(j))

)for j =

1, . . . ,m and y ∈ Uj. We note that the support of uεj is contained in Uj. Extending uεj byzero into R3 we have that uεj → φju in H1(D). This holds for all j = 1, . . . ,m. Finally, wenote that u0 ∈ H1(D) with compact support in U0 = D and thus uε0 = u0 ∗ φε converges tou0 = φ0u in H1(D). To finish the proof we set uε =

∑mj=0 u

εj and conclude that uε ∈ C∞(R3)

and uε|D converges to∑m

j=0 φju = u in H1(D). 2

The following theorem has been used as a motivation for introducing the space H10 (D).


Theorem 4.25 Let D be a bounded Lipschitz domain. Thenu ∈ C1(D) : u|∂D = 0

is a

subset of H10 (D).

Proof: We follow very closely the proof of the previous Theorem 4.24. We choose Uj,j = 0, . . . ,m, and Ψj, j = 1, . . . ,m, and a partition of unity φj, j = 0, . . . ,m, on D. Foru ∈ C1(D) with u = 0 on ∂D we define vj ∈ C1(C−j ) for j = 1, . . . ,m as in part (i) of theproof of Theorem 4.24. We extend vj by zero into all of R3 and note that the support ofthis extension of vj is contained in the cylinder Cj = B2(0, αj) × (−βj, βj). This extensionbelongs to H1(R3) as shown in Lemma 4.23. Now we set again a = (0, 0, L+ 1)> as in parts(ii) and (iii) of the proof of Theorem 4.24 and define vεj (x) = (vj ∗ φε)(x + εa) for x ∈ R3.Note the different sign compared to the proof of Theorem 4.24! By Theorem 4.5 the sequencevεj converges to vj in H1(R3) as ε → 0. Furthermore, the support of vεj is contained in Cjfor sufficiently small ε. Now we continue as in the proof of Theorem 4.24 and arrive at thefunctions uε ∈ C∞0 (D) which converge to u in H1(D). 2

As a corollary of these results we state the following chain rules.

Corollary 4.26 Let U,D ⊂ R3 be open sets and Ψ : U → D continuous in U and differen-tiable at almost all points x ∈ U such that the Jacobian Ψ′(x) ∈ R3×3 is essentially bounded;that is, Ψ′ ∈ L∞(U)3×3.

(a) Then, for u ∈ H10 (D) the composition v = u Ψ ∈ H1(U) and the chain rule holds in

the form

∇v(x) = Ψ′(x)>∇u(y)|y=Ψ(x) , x ∈ U . (4.11)

(b) For Lipschitz domains D and u ∈ H1(D) the composition v = u Ψ ∈ H1(U) and thechain rule holds in the form (4.11).

In both cases, the mapping u 7→ u Ψ is bounded from H10 (D) or H1(D), respectively, into

H1(U) and ‖u Ψ‖H1(U) ≤ ‖Ψ′‖∞‖u‖H1(D).

Proof : (a) For u ∈ C∞0 (D) the formula (4.11) holds in all points where Ψ is differentiable.Also, the right hand side of (4.11) is in L2(D)3. Therefore, the mapping u 7→ uΨ is boundedfrom the dense subspace C∞0 (D) of H1

0 (D) into H1(U) and thus has a bounded extension toall of H1

0 (D).Part (b) is proven in the same way. 2

Now turn to the imbedding property of H1(D) into L2(D). We recall from Theorem 4.12that the compactness of the imbeding H1

0 (D) in L2(D) relies on the extension property ofTheorem 4.11. The corresponding result for H1(D) needs D to be a Lipschitz domain.

Theorem 4.27 Let D ⊂ R3 be a bounded Lipschitz domain and R > 0 such that D ⊂B3(0, R). Then there exists a linear and bounded operator E : H1(D) → H1(R3) with(Eu)|D = u and Eu has compact support in B3(0, R) for every u.


Proof: Let Uj, ξj : j = 1, . . . ,m be a local coordinate system of ∂D as in Definition 6.1.We recall the mappings Ψj : Cj → Uj and the sets U ′j = Ψj(C

′j) from Definition 4.22. We

set U ′0 = D and choose a partion of unity φj ∈ C∞(R3), j = 0, . . . ,m, on D with respect toU ′j : j = 0, . . . ,m. For u ∈ H1(D) we set uj(y) = u(y)φj(y) for y ∈ D and j = 0, . . . ,m.

Then uj ∈ H1(U ′j ∩D) with support in U ′j ∩D. For every j = 1, . . . ,m we define vj in C ′j by

vj(x) = uj(Ψj(x1, x2,−|x3|)

), x ∈ C ′j .

The mapping x 7→ Ψj(x1, x2,−|x3|) is continuous and differentiable almost everywhere.Therefore, vj ∈ H1(C ′j) with compact support (see Corollary 4.26). We transform thesefunctions back and set

uj = vj ψ−1j in U ′j , j = 1, . . . ,m, and u0 = u0 = uφ0 in U ′0 = D .

Then uj has compact support in U ′j for j = 0, . . . ,m. Finally, we define Eu =∑m

j=0 uj and

check that Eu is an extension of u. For y ∈ D we set J(y) =j ∈ 0, . . . ,m : y ∈ U ′j

and

thus for fixed j ∈ J(y), j 6= 0, we have y = Ψj(x) for some x ∈ C ′j with x3 < 0. Therefore,uj(y) = vj(x) = uj(y) for j ∈ J(y), j 6= 0. Furthermore, 0 ∈ J(y) and u0(y) = u0(y) andthus

(Eu)(y) =∑j∈J(y)

uj(y) =∑j∈J(y)

uj(y) =∑j∈J(y)

φj(y)u(y) = u(y) .

The mappings u 7→ uj 7→ vj 7→ vj 7→ uj are all linear and bounded in the correspondingH1−norms (chain and product rule). Therefore, the mapping E is linear and also boundedfrom H1(D) into H1(R3). Finally, we choose a function ψ ∈ C∞0

(B3(0, R)

)with ψ = 1 on

D and set (Eu)(x) = ψ(x)(Eu)(x). This operator does the job! 2

As a corollary we have (compare with Theorem 4.12):

Theorem 4.28 Let D ⊂ R3 be a bounded Lipschitz domain. Then the imbedding H1(D)→L2(D) is compact.

The proof follows the same lines as in the proof of Theorem 4.12. 2

Now we turn to the Sobolev space H(curl, D) of vector functions. The following correspondsto Lemma 4.23.

Lemma 4.29 Let D be a bounded Lipschitz domain.

(a) Let u ∈ C1(D)3. Then u ∈ H(curl, D), and the classical curl of u coincides with thevariational curl.

(b) Let u ∈ C1(D)3 with ν × u = 0 on ∂D. The extension u of u by zero outside of Dyields u ∈ H(curl,R3) and curl u = curlu in D and curl u = 0 outside of D.

(c) For u ∈ H(curl, D) and φ ∈ C∞(D) the product φu ∈ H(curl, D) and curl(φu) =∇φ× u+ φ curlu.


(d) Let Q ∈ R3×3 be an orthogonal matrix and z ∈ R3. For u ∈ H(curl, D) define

v(x) = Q>u(Qx+ z) on U = x : Qx+ z ∈ D .

Then v ∈ H(curl, U) and

curl v(x) = (detQ)Q> curlu(y)|y=Qx+z on U .

Proof: The proofs of (a)–(c) are very similar to the corresponding proofs of Lemma 4.23and are omitted.

(d) Let q(j), j = 1, 2, 3, be the columns of Q. Then Qψ(x) =∑3

j=1 ψj(x)q(j) and thus

curly[Qψ](Q>(y − z)

)=

3∑j=1

∇yψj(x)× q(j) =3∑j=1

Q∇xψj(x)× q(j)

=3∑

j,`=1

∂ψj(x)

∂x`q(`) × q(j)

where we have set x = Q>(y − z) = Q−1(y − z). If detQ = 1 then q(1) × q(2) = q(3) andq(2) × q(3) = q(1) and q(3) × q(1) = q(2) and thus

curly[Qψ](Q>(y − z)

)=

(∂ψ2(x)

∂x1

− ∂ψ1(x)

∂x2

)q(3) +

(∂ψ3(x)

∂x2

− ∂ψ2(x)

∂x3

)q(1)

+

(∂ψ1(x)

∂x3

− ∂ψ3(x)

∂x1

)q(2)

= Q curlx ψ(x)|x=Q>(y−z) .

If detQ = 1 then analogous computations yield curly[Qψ](Q>(y−z)

)= −Q curlx ψ(x)|x=Q>(y−z).

Therefore, making the substitution y = Qx+ z twice,∫∫U

Q> curly u(y)|y=Qx+z · ψ(x) dx

=

∫∫U

curly u(y)|y=Qx+z ·Qψ(x) dx =

∫∫D

curly u(y) ·Qψ(Q>(y − z)

)dy

=

∫∫D

u(y) · curly[Qψ](Q>(y − z)

)dy = detQ

∫∫D

u(y) ·Q curlx ψ(x)|x=Q>(y−z) dy

= detQ

∫∫D

Q>u(y) · curlx ψ(x)|x=Q>(y−z) dy = detQ

∫∫U

v(x) · curlψ(x) dx .

This shows that (detQ)Q> curly u(y)|y=Qx+z is the variational curl of v. 2

The following result corresponds to Corollary 4.15 and Theorem 4.24.


Theorem 4.30 If D is a bounded Lipschitz domain then C∞(D)3 is dense in H(curl, D).

Proof: We almost copy the proof of Theorem 4.24.(i) Let Uj, ξj : j = 1, . . . ,m be a local coordinate system. Set U0 = D and choose apartition of unity φj : j = 0, . . . ,m on D corresponding to Uj : j = 0, . . . ,m. Setuj(y) = φj(y)u(y) for j = 0, . . . ,m and y ∈ D. Then the support of uj is contained inUj ∩D. We transform uj by the definition vj(x) = Q>j uj(Qjx+ z(j)), x ∈ C−j , where again

C−j =x = (x, x3) ∈ B2(0, αj)× (−βj, βj) : x3 < ξj(x)

for j = 1, . . . ,m. Here again x = (x1, x2). Then vj ∈ H(curl, C−j ) by the previouis lemma

and (because the support of uj is contained in Uj ∩ D) there exists δ > 0 with vj(x) = 0for all x ∈ C−j \ Cj(αj − 2δ, βj − 2δ) where the cylinder Cj(αj − 2δ, βj − 2δ) is defined as inDefinition 4.22.

(ii) Let a = (0, 0, L+ 1)> where L denotes the (common) Lipschitz constant of the functionsξj. We have shown in part (ii) of the proof of Theorem 4.24 that B3(x+ εa, ε) ⊂ C−j for allx ∈ C−j ∩ Cj(αj − δ, βj − δ) and ε < δ/(L+ 2).

(iii) We extend vj by zero into all of R3 and set vεj (x) = (vj ∗φε)(x−εa). Then vεj ∈ C∞(R3)3

and vεj → vj in L2(R3)3. Its support is contained in supp(vj) +B3(εa, ε) ⊂ Cj(αj − δ, βj − δ)for ε < δ/(L+ 2). Furthermore, for these ε and x ∈ C−j ∩ Cj(αj − δ, βj − δ) and any vectorb ∈ C3 we have

b · curl vεj (x) = b ·∫∫

C−j

∇xφε(x− εa− z)× vj(z) dz

= −∫∫

C−j

[b×∇zφε(x− εa− z)

]· vj(z) dz

=

∫∫C−j

curlz[b φε(x− εa− z)

]· vj(z) dz .

The function z 7→ b φε(x − εa − z) belongs to C∞0 (C−j )3 because is support is contained inB3(x− εa, ε) ⊂ C−j by part (ii). By the definition of the weak curl we have that

b · curl vεj (x) =

∫∫C−j

curl vj(z) ·[b φε(x− εa− z)

]dz

and thus

curl vεj (x) =

∫∫C−j

curl vj(z)φε(x− εa− z) dz (4.12)

For ε < δ/(L + 2) and x ∈ C−j \ Cj(αj − δ, βj − δ) both sides of (4.12) vanish again.Therefore, (4.12) holds for all x ∈ C−j and thus (curl vεj )(x) = (gj ∗ φε)(x − εa) for x ∈ C−jwhere gj = curl vj on C−j and gj = 0 on R3 \ C−j . By Theorem 4.5 we conclude again that(gj ∗ φε)(x− εa) converges to gj in L2(R3)3 and thus curl vεj |C−j to curl vj in L2(C−j )3. This

proves vεj → vj in H(curl, C−j ) for every j = 1, . . . ,m.


(iv) In this step we transform back the function and set uεj(y) = Qjvεj

(Q>j (y − z(j))

)for

j = 1, . . . ,m and y ∈ Uj. We note that the support of uεj is contained in Uj. Extendinguεj by zero into R3 we have that uεj → φju in H(curl, D). This holds for all j = 1, . . . ,m.Finally, we note that u0 ∈ H(curl, D) with compact support in U0 = D and thus uε0 = u0 ∗φεconverges to u0 = φ0u in H(curl, D). Finally we set uε =

∑mj=0 u

εj and conclude that

uε ∈ C∞(R3)3 and uε|D converges to∑m

j=0 φju = u in H(curl, D). 2

The following result illustrates the boundary condition in the definition of H0(curl, D).

Theorem 4.31 Let D be a bounded Lipschitz domain. Then:u ∈ C1(D)3 : ν × u = 0 on ∂D

⊂ H1

0 (D) .

Proof: We follow the proof of Theorem 4.30 just as the proof of Theorem 4.8 followedthe proof of Theorem 4.24. During the proof of Theorem 4.30 we have to show that forany vector valued function vj ∈ C1(B+

3 )3 with vanishing tangential components e3 × vj onB2(0, αj)× 0 its extension by zero into C ′j = B2(0, αj)× (−βj/2, βj/2) defines a functionin H(curl, C−j ). This is a simple exercise, see Exercise ??. 2

Now we are able to prove the desired imbedding property of V from (4.8b) into L2(D)3. Weneed the following auxiliary result.

Lemma 4.32 Let D ∈ C2.

((a) Define the closed subspace W ⊂ H0(curl, D) to be the orthogonal complement (inH0(curl, D)) of ∇H1

0 (D); that is,

W =u ∈ H0(curl, D) : (u,∇ϕ)L2(D)3 = 0 for all ϕ ∈ H1

0 (D).

Then the space

C =u ∈ C1(D)3 ∩ C2(D)3 : div u = 0 in D , ν × u = 0 on ∂D

is dense in W .

(b) The trace operator u 7→ u|∂D from C1(D) into L2(∂D) has an extension to a boundedoperator (trace operator) γ : H1(D)→ L2(D). Furthermore, γ is compact.

Proof: (a) Let u ∈ W . Because W ⊂ H0(curl, D) there exists a sequence vn ∈ C∞0 (D)3 withvn → u in H(curl, D). Set un = Pvn where P : H0(curl, D) → W denotes the projectionoperator corresponding to the orthogonal decomposition H0(curl, D) = W ⊕∇H1

0 (D). Thenun ∈ W and un → Pu = u in H(curl, D). The projection un can be constructed explitely.Indeed, consider the variational equation∫∫

D

∇pn · ∇ψ dx =

∫∫D

vn · ∇ψ dx for all ϕ ∈ H10 (D) ,


for pn ∈ H10 (D). Note that this is exactly the variational equation (4.17) for the case k = 0

and the variational form of

∆pn = div vn in D , pn = 0 on ∂D .

By Theorem 4.37 it has a unique solution pn ∈ H10 (D) Then obviously un = vn−∇pn. Now

we use without proof (“regularity result”, see, e.g. [4]) that pn ∈ C2(D) ∩ C∞(D). Thisrequires sufficient smoothness of the boundary (as in our case). Then∇pn ∈ C1(D)3∩C2(D)3

and thus also un ∈ C. This proves denseness of C in W .

(b) We will prove this in a more detailed context in Chapter 5. 2

Theorem 4.33 Let D ⊂ R3 be open and bounded with D ∈ C2.

(a) The space V from (4.8b) for εc = 1; that is,

V1 =u ∈ H0(curl, D) : (ψ, u)L2(D)3 = 0 for all ψ ∈ H0(curl 0, D)

(4.13)

is boundedly imbedded in H1(D)3.

(b) Let εc ∈ L∞(D) be complex-valued with Re εc(x) ≥ c > 0 almost everywhere on D.Then V from (4.8b) is compactly imbedded in L2(D)3.

Proof: Let first u ∈ C (see Lemma 4.32). Then, by several applications of the various formsof Green’s theorem,

‖ curlu‖2L2(D) =

∫∫D

curlu · curlu dx =

∫∫D

curl2 u · u dx

=

∫∫D

[u · ∇ div u︸︷︷︸

= 0

−3∑j=1

uj ∆uj

]dx

=

∫∫D

3∑j=1

∣∣∇uj∣∣2 dx − ∫∂D

3∑j=1

uj∂uj∂ν

ds .

We study the boundary term.∫∂D

3∑j=1

uj∂uj∂ν

ds =

∫∂D

ν ·3∑j=1

uj∇uj ds

=

∫∂D

ν · (u′)>u ds =

∫∂D

u · (u′ν) ds

=

∫∂D

(u · ν)(ν>u′ν) ds ,


because ν × u vanishes on ∂D. We extend ν to a differentiable vector field ν into a neigh-borhood U of ∂D such that |ν| = 1 on U . From the vanishing of the tangential componentwe have that

0 = Div(ν × (u× u)

)= −ν · curl(u× ν) = ν · curl(ν × u)

= ν ·[ν div u − u div ν + ν ′u − u′ν

]= div u︸︷︷︸

= 0

− (ν · u) div ν + ν>ν ′u − ν>u′ν .

We show that ν>ν ′ vanishes. Indeed∑3

j=1 ν2j = 1, thus

∑3j=1 νj∇νj = 0; that is, (ν ′)>ν = 0.

Therefore, we have shown that ν>u′ν = (ν · u) div ν and thus


3∑j=1

‖∇uj‖2L2(D)3 +

∫∂D

|u · ν|2 div ν ds .

Adding ‖u‖2L2(D)3 =

∑3j=1 ‖uj‖2

L2(D) yields

‖u‖2H1(D)3 ≤ ‖u‖2

H(curl,D) + ‖ div ν‖∞︸︷︷︸=: c

‖γu‖2L2(D)3 (4.14)

where we applied the trace operator γ from Lemma 4.32 componentwise. This holds for allu in the dense subspace C of W by Lemma 4.32 and thus for all u ∈ W .

Now we show the existence of c > 0 such that

‖u‖H1(D)3 ≤ c ‖u‖H(curl,D) for all u ∈ C

by contradiction. If there were no such a constant then there exists a sequence uj ∈ W suchthat ‖uj‖H1(D)3 = 1 for all j and ‖uj‖H(curl,D → 0 as j tends to infinity. From the compactnessof γ : H1(D)3 → L2(∂D)3 we conclude that (for a subsequence) (γuj) converges in L2(D)3.From (4.14) for the difference uj −u` we conclude that (uj) is a Cauchy sequence in H1(D)3

and thus convergent; that is, uj → u in H1(D)3 and ‖u‖H1(D)3 = 1. But this convergence inH1(D)3 implies convergence in H(curl, D), thus u = 0, a contadiction. Thus we have shownthat the space W is boundedly imbedded in H1(D)3. The space V1 is a subspace of W andthus also boundedly imbedded in H1(D)3. This proves part (a).

(b) From part (a) and the compact imbedding of H1(D)3 into L2(D)3 we conclude that V1

is compactly imbedded in L2(D)3. This proves the assertion for the special case εc = 1.Let now εc be arbitrary and (un) a bounded sequence in V . Then we decompose un withrespect to the Helmholtz decomposition H0(curl, D) = V1 ⊕H0(curl 0, D); that is,

un = vn + un with vn = P1un ∈ V1 and un ∈ H0(curl 0, D) .

Because the projection operator P1 is bounded in H0(curl, D) we conclude boundedness ofthe sequence (vn) in V1 w.r.t ‖·‖H(curl,D). Because V1 is compactly imbedded in L2(D)3 thereexists a subsequence (denoted by the same symbol) such that (vn) converges in L2(D)3. Onthe other hand, the form vn = un − un with un ∈ V ⊂ V and un ∈ H0(curl 0, D) is justthe decomposition of vn in the Helmholtz decomposition L2(D)3 = V ⊕ H0(curl 0, D) andtherefore un = P vn. By Theorem 4.20 the operator P is also bounded in L2(D)3, thus (un)converges in L2(D)3. 2

4.3. THE CAVITY PROBLEM 151

4.3 The Cavity Problem

4.3.1 The Variational Formulation and Existence

Before we investigate the Maxwell system (4.1a)–(4.1c) we consider the variational formof the following interior boundary value problem for the scalar inhomogeneous Helmholtzequation, namely

div(a∇u) + k2bu = f in D , u = 0 on ∂D , (4.15)

where k ≥ 0 is real. We assume that a, b ∈ L∞(D) are real valued such that a(x) ≥ a0 andb(x) ≥ b0 for constants a0 > 0, b0 > 0 and f ∈ L2(D).

As for the derivation of the variational equation for Maxwell’s equations we multiply theequation by some test function ψ with ψ = 0 on ∂D, integrate over D, and use Green’s firstformula. This yields ∫∫

D

[a∇u · ∇ψ − k2b uψ

]dx = −

∫∫D

f ψ dx .

This equation makes perfectly sense in H10 (D). We use this as a definition.

Definition 4.34 Let D be a bounded open set and f ∈ L2(D) and real valued a, b ∈ L∞(D)such that a(x) ≥ a0 and b(x) ≥ b0 for constants a0 > 0, b0 > 0. A function u ∈ H1

0 (D) iscalled a variational solution (or weak solution) of the boundary value problem (4.15) if∫∫

D

[a∇u · ∇ψ − k2b uψ

]dx = −

∫∫D

f ψ dx for all ψ ∈ H10 (D) .

By Friedrichs inequality, the first term in this equation defines an inner product in H10 (D).

Lemma 4.35 Let a ∈ L∞(D) such that a(x) ≥ a0 on D for some constant a0 > 0. Wedefine a new inner product in H1

0 (D) by

(u, v)∗ = (a∇u,∇v)L2(D)3 , u, v ∈ H10 (D) ,

Then(H1

0 (D), (·, ·)∗)

is a Hilbert space, and its norm is equivalent to the ordinary norm inH1

0 (D); in particular,

‖u‖∗ ≤√‖a‖∞‖u‖H1(D) ≤

√1 + c2

√‖a‖∞‖u‖∗ for all u ∈ H1

0 (D) (4.16)

where c is the constant from Theorem 4.13.

Proof: From the previous theorem we conclude

‖u‖2∗ = ‖

√a∇u‖2

L2(D) ≤ ‖a‖∞‖u‖2H1(D) = ‖a‖∞‖

[∇u‖2

L2(D)3 + ‖u‖2L2(D)

]≤ (1 + c2) ‖a‖∞‖∇u‖2

L2(D)3 = ‖a‖∞(1 + c2) ‖u‖2∗ .


2

Therefore, we write this variational equation in the form

(ψ, u)∗ − k2(bψ, u)L2(D) = −(ψ, f)L2(D) for all ψ ∈ H10 (D) . (4.17)

The question of existence of solutions of (4.27) is answered by the Fredholm theory. Weformulate the following basic theorem which is a special case of a more general result, see;e.g., [5].

Theorem 4.36 Let X be a normed space (over C) and 〈·, ·〉 : X × X → C a symmetricbilinear form which is bounded and non-degenerated3. Let T : X → X be compact and selfadjoint with respect to 〈·, ·〉. Then the equation

x − Tx = y

is solvable for exactly those y ∈ X which are orthogonal (with respect to 〈·, ·〉) to the nullspaceof I − T .

We are now able to prove the fundamental result on the interior boundary value problem(4.15).

Theorem 4.37 Let again D ⊂ R3 be a bounded open set and a, b ∈ L2(D) satisfy the aboveassumptions. Then the boundary value problem (4.15) has a variational solution for exactlythose f ∈ L2(D) which are orthogonal with respect to the inner product (bu, v)L2(D) to allsolutions of the corresponding homogeneous form of (4.15). If, in particular, the boundaryvalue problem (4.15) for f = 0 admits only the trivial solution u = 0 then the inhomogeneousboundary value problem has a unique solution for every f ∈ L2(D).

Proof: We will use the Riesz representation theorem from functional analysis to rewrite(4.17) in the form u−k2Ku = −Kf with some compact operator K and apply the Fredholmalternative to this equation. To carry out this idea we note that – for fixed v ∈ L2(D) –the mapping ψ 7→ (bψ, v)L2(D) is linear and bounded from

(H1

0 (D), (·, ·)∗)

into C. Indeed,by the inequality of Cauchy-Schwarz and Theorem 4.13 we conclude that there exists c > 0such that∣∣(bψ, v)L2(D)

∣∣ ≤ ‖b‖∞‖ψ‖L2(D)‖v‖L2(D) ≤ c ‖v‖L2(D)‖ψ‖∗ for all ψ ∈ H10 (D) .

Therefore, the representation theorem of Riesz assures the existence of a unique gv ∈ H10 (D)

with (bψ, v)L2(D) = (ψ, gv)∗ for all ψ ∈ H10 (D). We define the operator K : L2(D)→ H1

0 (D)by Kv = gv. Then K is linear (easy to see) and bounded because ‖Kv‖2

∗ = (Kv, gv)∗ =(bKv, v)L2(D) ≤ ‖b‖∞‖Kv‖L2(D)‖v‖L2(D) ≤ c‖Kv‖∗‖v‖L2(D); that is, ‖Kv‖∗ ≤ c‖v‖L2(D).Therefore, the equation (4.17) can be written as

(ψ, u)∗ − k2(ψ, Ku)∗ = −(ψ,Kf)∗ for all ψ ∈ H10 (D) . (4.18)

3that is, for every x ∈ X with x 6= 0, there exists z ∈ X with 〈x, z〉 6= 0


Because this holds for all such ψ we conclude that this equation is equivalent to

u − k2Ku = −Kf in H10 (D) . (4.19)

Because H10 (D) is compactly imbedded in L2(D) we conclude that the operator K = K J :

H10 (D)→ H1

0 (D) is compact (J denotes the imbedding H10 (D) ⊂ L2(D)). Furthermore, the

operator K is self adjount in(H1

0 (D), (·, ·)∗). Indeed,

(Kv, ψ)∗ = (ψ,Kv)∗ = (bψ, v)L2(D) = (bv, ψ)L2(D) = (v, Kψ)∗ , v, ψ ∈ H10 (D) .

Therefore, we know from the Fredholm alternative of Theorem 4.36 that this equation issolvable if, and only if, the right hand side Kf is orthogonal4 to the nullspace of I − k2K.Therefore, let v ∈ H1

0 (D) be a solution of the homogeneous equation v − k2Kv = 0. Thisis the variational form of the boundary value problem (4.15) for f = 0. Then, because(v,Kf)∗ = (bv, f)L2(D), we conclude that (v,Kf)∗ = 0 if, and only if, (bv, f)L2(D) = 0. 2

We can also use the form (4.19) to prove existence and completeness of an eigensystem of−∆. Recall the definition of the inner product (·, ·)∗.

Theorem 4.38 Let D be a bounded open set. Then there exists an infinite number of eigen-values λ ∈ R>0 of (4.15) with respect to Dirichlet boundary conditions; that is, there existsa sequence kn > 0 and corresponding functions un ∈ H1

0 (D) with ‖un‖∗ = 1 such thatdiv(a∇un) + knb un = 0 in D and un = 0 on ∂D in the variational form. Furthermore,kn →∞ as n→∞. The sets un : n ∈ N and knun : n ∈ N form complete orthonormalsystems in

(H1

0 (D), (·, ·)∗)

and(L2(D), (b·, ·)L2(D)

), respectively.

Proof: We define the operator K as in the previous theorem. Then K is compact andself adjoint in

(H1

0 (D), (·, ·)∗). It is also positive because (un, Kun)∗ = (un, un)L2(D) > 0.

Therefore, the spectral theorem for compact self adjoint operators in Hilbert spaces impliesthe existence of a spectral system (µn, un) of K in

(H1

0 (D), (·, ·))∗. Furthermore, µn > 0 and

µn → 0 as n → ∞ and the set un : n ∈ N of orthonormal eigenfunctions un is completein(H1

0 (D), (·, ·))∗. Therefore, we have Kun = µnun; that is, un − 1

µnKun = 0, which is the

homogeneous form of ∆un + λnun = 0 in D and un = 0 on ∂D for kn = 1√µn

. The system

knun : n ∈ N is a system of eigenfunctions as well which are orthogonal in the weightedL2(D)− space because

k2n(bun, um)L2(D) = k2

n

∫∫D

b un um dx =

∫∫D

a∇un · ∇um dx

= (un, um)∗ = δmn .

2

Now we go back to the formulation (4.1a)–(4.1c) of the cavity problem which we recallfor the convenience of the reader. In this section we always assume that D ⊂ R3 is open

4in(H1

0 (D), (·, ·))∗!


and bounded. Given Je ∈ L2(D)3 and ε, µ, σ ∈ L∞(D) determine E ∈ H0(curl, D) andH ∈ H(curl, D) such that


curlH + (iωε− σ)E = Je in D . (4.20b)

The boundary condition (4.1c) is included in the space H0(curl, D) for E.

We show how to eliminate E or H from this system. First, multiply (4.20b) by ψ ∈H0(curl, D), integrate over D, use partial integration and (4.20a). The boundary termvanishes because of ψ ∈ H0(curl, D). This yields∫∫

D

Je · ψ dx =

∫∫D

[curlH · ψ + (iωε− σ)E · ψ

]dx

=

∫∫D

[H · curlψ + (iωε− σ)E · ψ

]dx

=

∫∫D

[1

iωµcurlE · curlψ + (iωε− σ)E · ψ

]dx .

Multiplication with iωµ0 and using k2 = ω2ε0µ0 yields∫∫D

[1

µrcurlE · curlψ − k2

(εr + i

σ

ωε0

)E · ψ

]dx = iωµ0

∫∫D

Je · ψ dx . (4.21)

This variational equation holds for all ψ ∈ H0(curl, D).

Analogously, we can eliminate E. Indeed, now we multiply (4.20a) by ψ ∈ H(curl, D) andproceed as in the previous part.

0 =

∫∫D

[curlE · ψ − iωµH · ψ

]dx

=

∫∫D

[E · curlψ − iωµH · ψ

]dx

=

∫∫D

[1

iωε− σ(Je − curlH) · curlψ − iωµH · ψ

]dx .

We note that the boundary term vanishes because of E ∈ H0(curl, D). Multiplication with−iωε0 yields∫∫

D

[1

εr + iσ/(ωε0)curlH · curlψ − k2µrH · ψ

]dx =

∫∫D

1

εr + iσ/(ωε0)Je · curlψ dx .

(4.22)This holds for all ψ ∈ H(curl, D).

It is easy to see that the variational equations (4.21) and (4.22) are equivalent to the Maxwellsystem.


Lemma 4.39 Let E ∈ H0(curl, D) satisfy the variational equation (4.21) for all ψ ∈H0(curl, D). Set

H =1

iωµcurlE in D .

Then E ∈ H0(curl, D) and H ∈ H(curl, D) satisfy the system (4.20a), (4.20b).

Furthermore, H ∈ H(curl, D) satisfies (4.22) for all ψ ∈ H(curl, D).

The same statement holds for E and H interchanged.

Proof: Let E ∈ H0(curl, D) satisfy (4.21) for all ψ ∈ H0(curl, D). We note that H ∈ L2(D)3.Substituting the definition of H yields∫∫

D

[iωµ0H · curlψ − k2

(εr + i

σ

ωε0

)E · ψ

]dx = iωµ0

∫∫D

Je · ψ dx

which is the variational form of iωµ0 curlH; that is,

iωµ0 curlH = k2

(εr + i

σ

ωε0

)E + iωµ0Je .

Division by iωµ0 yields (4.20b). 2

In the following we will discuss the variational equation (4.21) in the space H0(curl, D) andmake the following assumptions on the parameters µr, εr, and σ (recall Subsection 1.3):

There exists c > 0 such that

(a) µr ∈ L∞(D) is real valued and µr(x) ≥ c almost everywhere,

(b) εr ∈ L∞(D) is real valued and εr(x) ≥ c almost everywhere on D.

(c) σ ∈ L∞(D) is real valued and σ(x) ≥ 0 almost everywhere on D.

For abbreviation we define again the complex relative dielectricity by

εc(x) = εr(x) + iσ(x)

ωε0

, x ∈ D .

Also, it is conveniant to define the vector field F ∈ L2(D)3 by

εcF = iωµ0Je .

Then the variational equation (4.21) takes the form∫∫D

[1

µrcurlE · curlψ − k2εcE · ψ

]dx =

∫∫D

εcF · ψ dx (4.23)


for all ψ ∈ H0(curl, D). To discuss this equation in H0(curl, D) we will use the Helmholtzdecomposition to split this problems into two problems, one in V and one in H0(curl 0, D).(The one in H0(curl 0, D) will be trivial.) Recall that V is given by (4.8b); that is,

V =u ∈ H0(curl, D) : (εcu, ψ)L2(D)3 = 0 for all ψ ∈ H0(curl 0, D)

.

Analogously, V denotes the corresponding subspace in L2(D)3, see (??).

First we use the Helmholtz decomposition to write F as F = F0 + f with F0 ∈ H0(curl 0, D)and f ∈ V . Then we make an ansatz for E in the form E = E0 + u with E0 ∈ H0(curl 0, D)and u ∈ V . Substituting this into (4.23) we get∫∫

D

[1

µrcurlu · curlψ − k2εc(u+ E0) · ψ

]dx =

∫∫D

εc(f + F0) · ψ dx (4.24)

for all ψ ∈ H0(curl, D). Now we take ψ ∈ H0(curl 0, D) as test functions. Recalling that∫∫Dεcu · ψ dx = 0 and

∫∫Dεcf · ψ dx = 0 for all ψ ∈ H0(curl 0, D) yields

−k2

∫∫D

εcE0 · ψ dx =

∫∫D

εcF0 · ψ dx for all ψ ∈ H0(curl 0, D) .

Rewriting this as ∫∫D

εc[F0 + k2E0] · ψ dx = 0 for all ψ ∈ H0(curl 0, D)

and setting ψ = F0 + k2E0 ∈ H0(curl 0, D) yields E0 = − 1k2 F0.

Second, we take ψ ∈ V as test functions. Then (4.24) reduces to∫∫D

[1

µrcurlu · curlψ − k2εcu · ψ

]dx =

∫∫D

εcf · ψ dx (4.25)

for all ψ ∈ V . Before we investigate (4.25) we summarize this splitting in the followinglemma.

Lemma 4.40 Let F ∈ L2(D)3 have the Helmholtz decomposition F = F0 + f with f ∈ Vand F0 ∈ H0(curl 0, D).

(a) Let E ∈ H0(curl, D) be a solution of (4.23) and E = E0 + u with u ∈ V and E0 ∈H0(curl 0, D) the Helmholtz decomposition of E. Then E0 = − 1

k2 F0 and u ∈ V solves(4.25).

(b) If u ∈ V solves (4.25) then E = u− 1k2 F0 ∈ H0(curl, D) solves (4.23).


To show solvability of (4.25) we follow the approach as in the scalar case for the Helmholtzequation and introduce the equivalent inner product (·, ·)∗ on V by

(v, w)∗ =

∫∫D

1

µrcurl v · curlw dx , v, w ∈ V . (4.26)

Then we have the analogue of Corollary ??:

Lemma 4.41 The norm ‖v‖∗ =√

(v, v)∗ ist an equivalent norm on V .

Proof: In contrast to the proof of Corollary ?? we use an indirect argument to show theexistence of a constant c > 0 such that

‖v‖H(curl,D) ≤ c ‖ curl v‖L2(D)3 for all v ∈ V .

(This is sufficient because the estimates ‖ curl v‖2L2(D)3 ≤ ‖µr‖∞‖v‖2

∗ and ‖v‖∗ ≤‖(1/µr)‖∞‖v‖H(curl,D) hold obviously.)Assume that this is not the case. Then there exists a sequence (vn) in V with ‖vn‖H(curl,D) = 1and curl vn → 0 in L2(D)3 as n tends to infinity. The sequence (vn) is therefore bounded inV . Because V is compactly imbedded in L2(D)3 by Theorem 4.33 there exists a subsequence,denoted by the same symbols, which converges in L2(D)3. For this subsequence, (vn) and(curl vn) are Cauchy sequences in L2(D)3. Therefore, (vn) is a Cauchy sequence in V andtherefore convergent: vn → v in H(curl, D) for some v ∈ V and curl v = 0. Therefore,v ∈ V ∩ H0(curl 0, D) = 0 which shows that v vanishes. This contradicts the fact that‖v‖H(curl,D) = 1. 2

Now we write (4.25) in the form

(u, ψ)∗ − k2 (εcu, ψ)L2(D)3 = (εcf, ψ)L2(D)3 for all ψ ∈ V .

Again, we use the representation theorem of Riesz to show the existence of a linear andbounded operator K : L2(D)3 → V with (ψ, εcu)L2(D)3 = (ψ,Ku)∗ for all ψ ∈ V andu ∈ L2(D)3. We carry out the arguments for the convenience of the reader although theyare completely analogous to the arguments in the proof of Theorem 4.37. For fixed u ∈L2(D)3 the mapping ψ 7→ (ψ, εcu)L2(D)3 defines a linear and bounded functional on V .Therefore, by the representation theorem of Riesz there exists a unique g = gu ∈ V with(ψ, εcu)L2(D)3 = (ψ, gu)∗ for all ψ ∈ V . We set Ku = gu. Linearity of K is clear from theuniqueness of the representation gu. Boundedness of K from L2(D)3 into V follows from theestimate

‖Ku‖2∗ = (Ku,Ku)∗ = (Ku, εcu)L2(D)3 ≤ ‖εc‖∞‖u‖L2(D)3‖Ku‖L2(D)3

≤ c ‖εc‖∞‖u‖L2(D)3‖Ku‖∗

after division by ‖Ku‖∗. Because V is compactly imbedded in L2(D)3 we conclude thatK = k J is even compact as an operator from V into itself (here J : V → L2(D)3 denotesthe compact imbedding operator). Therefore, we can write equation (4.25) in the form

(u, ψ)∗ − k2 (Ku, ψ)∗ = (Kf, ψ)∗ for all ψ ∈ V ;


that is,u − k2Ku = Kf in V . (4.27)

This is again a Fredholm equation of the second kind for u ∈ V . In particular, we haveexistence once we have uniqueness. The question of uniqueness will be discussed in the nextsection below.

For the general question of existence we apply again Fredholms Theorem 4.36 to X = V withbilinear form 〈u, v〉 = (u, v)∗ =

∫D

1µr

curlu · curl v dx and operator T = k2K. The operator

K is self adjoint. Indeed, 〈Ku, ψ〉 = (Ku, ψ)∗ = (εcu, ψ)L2(D)3 and this is symmetric in uand ψ. Lemma 4.40 tells us that the variational form (4.23) of the cavity problem is solvablefor F ∈ L2(D)3 if, and only if, the variational problem (4.25) is solvable in V for the partf ∈ V of F . By the Fredholm alternative of Theorem ?? this is solvable for exactly thosef ∈ V such that Kf is orthogonal (with respect to 〈·, ·〉) to the nullspace of I − k2K. ByLemma 4.40 v ∈ V solves (4.25) for f = 0, and only if, v solves (4.23) for F = 0. Also, wenote that

〈Kf, v〉 = (εcf, v)L2(D)3 =(εc(F0 + f), v

)L2(D)3

= (εcF, v)L2(D)3 ;

that is 〈Kf, v〉 vanishes if, and only if, (εcF, v)L2(D)3 vanishes. Further we note that v solves(4.23) if, and only if, v solves (4.23) with εc replaced by εc. We have thus proven:

Theorem 4.42 The cavity problem (4.23) has a solution E ∈ H(curl, D) for exactly thosesource terms εrF ∈ L2(D)3 such that (εrF, v)L2(D) = 0 for all solutions v ∈ H1

0 (D) of thecorresponding homogeneous form of (4.23) with εc replaced by εc; that is,∫∫

D

[1

µrcurlE · curlψ − k2εcE · ψ

]dx = 0 for all ψ ∈ H0(curl, D) .

If, in particular, this boundary value problem admits only the trivial solution E = 0 then theinhomogeneous boundary value problem (4.23) has a unique solution E ∈ H0(curl, D) for allF ∈ L2(D)3.

For the remaining part we assume that εc = εr is real valued; that is σ = 0. Then we canconsider the eigenvalue problem by the same arguments.

Definition 4.43 The resolvent set consists of all k ∈ C for which (4.23) is uniquely solvablein H(curl, D) for all F ∈ L2(D)3 and such that F 7→ E is bounded.

First we consider k 6= 0. We have seen that k 6= 0 is in the resolvent set if, and only if,the operator I − k2K is one-to-one in V ; that is, 1/k2 is in the resolvent set of the compactand selfadjoint operator K : V → V . The spectrum of K consists of 0 and eigenvalues 1/k2

n

which converge to zero. Let us now consider k = 0. The homogeneous form of the variationalform (4.23) takes the form∫∫

D

1

µrcurlE · curlψ dx = 0 for all ψ ∈ H0(curl, D) .


Again, if we decompose E in the form E = E0 + u with u ∈ V and E0 ∈ H0(curl 0, D), thenthis restricts to (E0, ψ)∗ = 0 for all ψ ∈ V which implies that E0 = 0. Therefore, 0 is aneigenvalue with the (infinite dimensional) eigenspace H0(curl 0, D). We summarize this inthe final theorem.

Theorem 4.44 There exists an infinite number of eigenvalues k ∈ R≥0 of (4.23); that is,there exists a sequence kn ≥ 0 for n ∈ N and corresponding functions En ∈ V such that

curl

(1

µrcurlEn

)− k2

nεr En = 0 in D , ν × En = 0 on ∂D

in the variational form (4.23) for Je = 0; that is,∫∫D

[1

µrcurlEn · curlψ − k2

nεrEn · ψ]dx = 0 for all ψ ∈ V .

The positive eigenvalues kn > 0 have finite multiplicity and tend to infinity as n→∞. Theset En : n ∈ N forms a complete orthonormal system in

(V, (·, ·)∗

). The set knEn : n ∈ N

forms a complete orthonormal system in V when we equip V with the weighted L2−innerproduct (εu, v)L2(D)3. Furthermore, k = 0 is also an eigenvalue with infinite dimensionaleigenspace H0(curl 0, D).

4.3.2 Uniqueness and Unique Continuation

For the first uniqueness result we assume that the medium is conducting; that is, σ > 0 inD. By the definition of εc in Subsection 1.3 this is equivalent to the assumption that theimaginary part of εc is strictly positive on D.

Theorem 4.45 Let εc ∈ L∞(D) be complex-valued with Re εc(x) ≥ c > 0 and Im εc(x) > 0almost everywhere on D. Then, for every F ∈ L2(D)3) there exists a unique solution E ∈H(curl, D) of (4.23).

Proof: By Theorem 4.42 it suffices to prove uniqueness of equation (4.23) in H0(curl, D).Let E ∈ H0(curl, D) be a solution of the homogeneous equation. Inserting v = E yields∫∫

D

[1

µr

∣∣curlE∣∣2 − k2εc|E|2

]dx = 0 ,

and thus, taking the imaginary part,∫∫D

Im εc |E|2 dx = 0

from which E = 0 follows because Im εc > 0 on D. 2

Our next goal is to show that uniqueness is assured already if Im εc > 0 on some open partU of D only. The arguments of the previous proof imply that E vanishes on U only, and we


have to prove that from this E = 0 in all of D follows. If E was analytic this would followfrom an analytic continuation argument. However, in general E fails to be analytic.

As a preparation we need an interior regularity result. We can prove it by the familiartechnique of transfering the problem into the Sobolev spaces of periodic functions.For p ∈ N and an open domain D ⊂ R3 the Sobolev space Hp(D) is defined as before byrequiring that all partial derivatives up to order p exist in the variational sense and areL2−functions (compare with Definition 4.1). The space C∞(D) of smooth functions aredense in Hp(D). Again, Hp

0 (D) is the closure of C∞0 (D) with respect to the norm in Hp(D).Let Ω ⊂ R3 be the cube Ω = (−R,R)3. Then we define Hp

per(Ω) by

Hpper(Ω) =

u ∈ L2(Ω) :

∑n∈Z3

(1 + |n|2)p |un|2 <∞

with inner product

(u, v)Hpper(Ω) = (2R)3

∑n∈Z3

(1 + |n|2)p unvn ,

compare with Definition 4.10. Here, un, vn ∈ C are the Fourier coefficients of u and v,respectively, see (4.7). Then it is easy to show:

Lemma 4.46 The following inclusions hold and are bounded (for any p ∈ N):

Hp0 (Ω) → Hp

per(Ω) → Hp(Ω) .

Proof: First inclusion: We show this by induction with respect to p. For p = 0 there isnothing to show. Assume that the assertion is true for p ≥ 0. Then Hp

0 (Ω) ⊂ Hpper(Ω) and

there exists c > 0 such that

(2R)3∑n∈Z3

(1 + |n|2)p |un|2 = ‖u‖2Hpper(Ω) ≤ c‖u‖2

Hp(Ω) for all u ∈ Hp0 (Ω) . (4.28)

Let u ∈ C∞0 (Ω). Then, by partial differentiation (if nj 6= 0),

un =1

(2R)3

∫∫Ω

u(x) e−iπRn·xdx =

1

(2R)3

i R

π nj

∫∫Ω

u(x)∂

∂xje−i

πRn·xdx

= − 1

(2R)3

i R

π nj

∫∫Ω

∂u

∂xj(x) e−i

πRn·xdx = − i R

π njd(j)n ,

where d(j)n are the Fourier coefficients of ∂u/∂xj. Note that the boundary term vanishes. By

assumption of induction we conclude that (4.28) holds for u and for ∂u/∂xj. Therefore,

‖u‖2Hp+1per (Ω)

= (2R)3∑n∈Z3

(1 + |n|2)p+1 |un|2 = (2R)3∑n∈Z3

(1 + |n|2)p |un|2

+R2

π2(2R)3

3∑j=1

∑n∈Z3

(1 + |n|2)p∣∣d(j)n

∣∣2≤ c‖u‖2

Hp(Ω) + cR2

π2

3∑j=1

‖∂u/∂xj‖2Hp(Ω) ≤ c′‖u‖2

Hp+1(Ω) .


This proves boundedness of the imbedding with respect to the norm of order p + 1. Thisends the proof of the first inclusion because C∞0 (Ω) is dense in H1

0 (Ω).

For the second inclusion we truncate the Fourier series of u ∈ Hpper(Ω) into

uN(x) =∑|n|≤N

un ei πRn·x

and compute directly

∂j1+j2+j3uN(x)

∂xj11 ∂xj22 ∂x

j33

=(iπ

R

)j1+j2+j3 ∑|n|≤N

un nj11 n

j22 n

j33 e

i πRn·x

and thus for j ∈ N3 with j1 + j2 + j3 ≤ p:∥∥∥∥ ∂j1+j2+j3uN

∂xj11 ∂xj22 ∂x

j33

∥∥∥∥2

L2(Ω)

≤ (2R)3( πR

)2(j1+j2+j3) ∑|n|≤N

|un|2|n1|2j1 |n2|2j2|n3|2j3

≤ (2R)3( πR

)2(j1+j2+j3) ∑|n|≤N

|un|2|n|2(j1+j2+j3)

≤( πR

)2(j1+j2+j3)

‖uN‖2Hpper(Ω) .

This proves the lemma by letting N tend to infinity. 2

We continue with a regularity result.

Theorem 4.47 (Interior Regularity Property)Let f ∈ L2(D) and U be an open set with U ⊂ D.

(a) Let u ∈ H1(D) be a solution of the variational equation∫∫D

∇u · ∇ψ dx =

∫∫D

f ψ dx for all ψ ∈ C∞0 (D) . (4.29)

Then u|U ∈ H2(U) and ∆u = −f in U .

(b) Let u ∈ L2(D) be a solution of the variational equation∫∫D

u∆ψ dx = −∫∫

D

f ψ dx for all ψ ∈ C∞0 (D) . (4.30)

Then u|U ∈ H2(U) and ∆u = −f in U .

Proof: For both parts we restrict the problem to a periodic problem in a cube by using apartition of unity. Indeed, let ρ > 0 such that ρ < dist(U, ∂D). Then the open balls B(x, ρ)are in D for every x ∈ U . Furthermore, U ⊂

⋃x∈U B(x, ρ). Because U is compact there exist


finitely many open balls B(xj, ρ) ⊂ D for xj ∈ U , j = 1, . . . ,m, with U ⊂⋃mj=1B(xj, ρ). For

abbreviation we set Bj = B(xj, ρ). Again, we choose a partion of unity; that is, ϕj ∈ C∞0 (Bj)with ϕj ≥ 0 and

∑mj=1 ϕj(x) = 1 for all x ∈ U .

Let now uj(x) = ϕj(x)u(x), x ∈ D. Then∑m

j=1 uj = u on U and uj ∈ H10 (D) has support

in Bj.

Proof of (a): For ψ ∈ C∞0 (D) and any j ∈ 1, . . . ,m we have that∫∫D

∇uj · ∇ψ dx =

∫∫D

[ϕj∇u · ∇ψ + u∇ϕj · ∇ψ

]dx

=

∫∫D

[∇u · ∇(ϕj ψ)− ψ∇u · ∇ϕj − ψ u∆ϕj − ψ∇u · ∇ϕj

]dx

=

∫∫D

[ϕj f − 2∇u · ∇ϕj − u∆ϕj

]ψ dx

=

∫∫D

gj ψ dx with gj = ϕj f − 2∇u · ∇ϕj − u∆ϕj ∈ L2(D) .

Because the support of ϕj is contained in Bj this equation restricts to∫∫Bj

∇uj · ∇ψ dx =

∫∫Bj

gj ψ dx for all ψ ∈ H10 (D) . (4.31)

Let now R > 0 such that D ⊂ Ω = (−R,R)3. We fix j ∈ 1, . . . ,m and ` ∈ Z and setψ`(x) = exp(−i(π/R) ` · x). Because Bj ⊂ D we can find a function ψ` ∈ C∞0 (D) withψ` = ψ` on Bj. Substituting ψ` into (4.31) yields∫∫

Bj

∇uj · ∇ψ` dx =

∫∫Bj

gj ψ` dx . (4.32)

We can extend the integrals to Ω because uj and gj vanish outside of Bj. Now we expanduj and gj into Fourier series of the form

uj(x) =∑n∈Z

an ei πRn·x and gj(x) =

∑n∈Z

bn ei πRn·x

and substitute this into equation (4.32). This yields

( πR

)2∑n∈Z

an n · `∫∫

Ω

ei(n−`)·x dx =∑n∈Z

bn

∫∫Ω

ei(n−`)·x dx .

From the orthogonality of the functions exp(i(π/R)n·x) we conclude that (π/R)2a` |`|2 = b`.Because

∑`∈Z |b`|2 <∞ we conclude that

∑`∈Z |`|4|a`|2 <∞; that is, uj ∈ H2

per(Ω) ⊂ H2(Ω).Therefore, also

∑mj=1 uj ∈ H2(Ω) and thus u|U =

∑mj=1 uj|U ∈ H2(U). This proves part (a).


Proof of (b): We proceed very similarly and show that u ∈ H1(D). Then part (a) appliesand yields the assertion. For ψ ∈ C∞0 (D) and any j ∈ 1, . . . ,m we have that∫∫

D

uj ∆ψ dx =

∫∫D

uϕj ∆ψ dx

=

∫∫D

u[∆(ϕjψ)− 2∇ϕj · ∇ψ − ψ∆ϕj

]dx

=

∫∫D

[fϕj − u∆ϕj

]ψ dx − 2

∫∫D

u∇ϕj · ∇ψ dx

=

∫∫D

gj ψ dV −∫∫

D

Fj · ∇ψ dx

with gj = fϕj−u∆ϕj ∈ L2(D) and Fj = 2u∇ϕj ∈ L2(D)3. As in the proof of (a) we observethat the domain of integration is Bj. Therefore, we can again take ψ`(x) = exp(−i(π/R) `·x)for ψ and modify it outside of Bj such that it is in C∞0 (D). With the Fourier series

uj(x) =∑n∈Z

an ei πRn·x and gj(x) =

∑n∈Z

bn ei πRn·x and Fj(x) =

∑n∈Z

cn ei πRn·x

we conclude that

−( πR

)2

a` |`|2 = b` + iπ

R` · c` for all ` ∈ Z ,

and thus

|a`| ≤ c

[1

|`|2|b`| +

1

|`||c`|]

which proves that∑

`∈Z(1 + |`|2) |a`|2 ≤ c∑

`∈Z(|b`|2 + |c`|2

)<∞ and thus uj ∈ H1

per(Ω) ⊂H1(Ω). 2

Remarks:

(a) If f ∈ Hp(D) for some p ∈ N then u|U ∈ Hp+2(U) by the same arguments, appliediteratively. Indeed, we have just shown it for p = 0. If it is true for p − 1 and iff ∈ Hp(D) then gj ∈ Hp(D) (note that uBj ∈ H

p+10 (Bj) by assumption of induction!)

and thus uj ∈ Hp+2(Ω).

(b) This theorem holds without any regularity assumptions on the boundary ∂D. Withoutfurther assumptions on ∂D and the boundary data u|∂D we cannot assure that u ∈H2(D).

The proof of the following fundamental result is an adaption (to the Sobolev space H2(D))of the corresponding proof in [2]. The proof itself goes back to Muller [7] and Protter [9].

Theorem 4.48 (Unique Continuation Principle)


Let D ⊂ R3 be an open domain5 and u1, . . . , um ∈ H2(D) be real valued such that

|∆uj| ≤ c

m∑`=1

|u`|+ |∇u`|

in D for j = 1, . . . ,m . (4.33)

If uj vanish in some open set B ⊂ D for all j = 1, . . . ,m, then uj vanish identically in Dfor all j = 1, . . . ,m.

Proof: Let x0 ∈ B and R ∈ (0, 1) such that B[x0, R] ⊂ D. We show that uj vanishes inB[x0, R/2]. This is sufficient because for every point x ∈ D we can find finitely many ballsB(x`, R`) ` = 0, . . . , p, with R` ∈ (0, 1), such that B(x`, R`/2) ∩ B(x`−1, R`−1/2) 6= ∅ andx ∈ B(xp, Rp/2). Then one concludes uj(x) = 0 for all j by iteratively applying the firststep.We choose the coordinate system such that x0 = 0. First we fix j and write u for uj. Letϕ ∈ C∞0

(B(0, R)

)with ϕ(x) = 1 for |x| ≤ R/2 und define u, v ∈ H2

0 (D) by u(x) = ϕ(x)u(x)and

v(x) =

exp(|x|−n) u(x) , x 6= 0 ,

0 , x = 0 ,

for some n ∈ N. Note that, indeed, v ∈ H2(D) because u vanishes in the neighborhood Bof x0 = 0. Then, with r = |x|,

∆u = exp(−r−n)

∆v +

2n

rn+1

∂v

∂r+

n

rn+2

( nrn− n+ 1

)v

.

Using the inequality (a + b)2 ≥ 4ab and calling the middle term in the above expression b,we see that

(∆u)2 ≥ 8n exp(−2r−n)

rn+1

∂v

∂r

∆v +

n

rn+2

( nrn− n+ 1

)v.

Multiplication with exp(2r−n) rn+2 and integration yields∫∫D

exp(2r−n) rn+2(∆u)2dx ≥ 8n

∫∫D

r∂v

∂r

∆v +

n

rn+2

( nrn− n+ 1

)vdx . (4.34)

We show that ∫∫D

r∂v

∂r∆v dx =

1

2

∫∫D

|∇v|2dx and (4.35a)

∫∫D

1

rmv∂v

∂rdx =

m− 2

2

∫∫D

v2

rm+1dx for any integer m. (4.35b)

Indeed, proving the first equation we note that r ∂v∂r

= x · ∇v and ∇ (x · ∇v) =3∑j=1

e(j) ∂v∂xj

+

xj∇ ∂v∂xj

and thus

∇ (x · ∇v) · ∇v = |∇v|2 +3∑j=1

xj∇∂v

∂xj· ∇v = |∇v|2 +

1

2x · ∇

(|∇v|2

).

5Note that this means that D is connected!


Therefore, by partial integration (note that all boundary terms vanish)∫∫D

r∂v

∂r∆v dx = −

∫∫D

∇(r∂v

∂r

)· ∇v dx = −

∫∫D

|∇v|2 +1

2x · ∇

(|∇v|2

)dx

= −∫∫D

|∇v|2 dx +1

2

∫∫D

div x︸︷︷︸= 3

|∇v|2 dx =1

2

∫∫D

|∇v|2 dx .

This proves equation (4.35a). For equation (4.35b) we have, using polar coordinates andpartial integration,

∫∫D

1

rmv∂v

∂rdx =

∫∫B(0,R)

1

rmv∂v

∂rdx =

R∫0

∫S2

v

rm∂v

∂rr2 ds dr

= −R∫

0

∫S2

v∂

∂r

(1

rm−2v

)ds dr = −

∫∫D

v∂

∂r

(1

rm−2v

)1

r2dx

= −∫∫D

1

rmv∂v

∂rdx + (m− 2)

∫∫D

v2

rm+1dx

which proves equation (4.35b). Substituting (4.35a) and (4.35b) for m = 2n+1 and m = n+1into the inequality (4.34) yields∫∫

D

exp(2r−n) rn+2(∆u)2dx ≥ 4n

∫∫D

|∇v|2 dx + 4n3(2n− 1)

∫∫D

v2

r2n+2dx

− 4n2(n− 1)2

∫∫D

v2

rn+2dx

≥ 4n

∫∫D

|∇v|2 dx + 4n2(n2 + n− 1)

∫∫D

v2

r2n+2dx

where in the last step we used r ≤ R ≤ 1. Now we replace the right hand side again by uagain. With v(x) = exp(r−n)u(x) we have

∇v(x) = exp(r−n)[∇u(x)− n r−n−1x

ru(x)

]and thus, using |a− b|2 ≥ 1

2|a|2 − |b|2 for any vectors a, b ∈ R3,

∣∣∇v(x)∣∣2 ≥ exp(2r−n)

[1

2

∣∣∇u(x)∣∣2 − n2 r−2n−2

∣∣u(x)∣∣2] .


Substituting this into the estimate above yields∫∫D

exp(2r−n) rn+2(∆u)2dx (4.36)

≥ 2n

∫∫D

exp(2r−n) |∇u|2 dx +

+[4n2(n2 + n− 1)− 4n3

] ∫∫D

exp(2r−n)u2

r2n+2dx

= 2n

∫∫D

exp(2r−n) |∇u|2 dx + 4n2(n2 − 1)

∫∫D

exp(2r−n)u2

r2n+2dx

≥ 2n

∫∫D

exp(2r−n) |∇u|2 dx + 2n4

∫∫D

exp(2r−n)u2

rn+2dx (4.37)

for n ≥ 2. Up to now we have not used the estimate (4.33). We write now uj and uj foru and u, respectively. From this estimate and the inequality of Cauchy-Schwarz (and theinequality (α + β)2 ≤ 2α2 + 2β2) we have the following estimate

∣∣∆uj(x)∣∣2 =

∣∣∆uj(x)∣∣2 ≤ 2mc2

m∑`=1

|u`(x)|2 + |∇u`(x)|2

for |x| < R

2.

Therefore, from (4.37) for uj we conclude that

2n

∫∫|x|≤R/2

exp(2r−n) |∇uj|2 dx + 2n4

∫∫|x|≤R/2

exp(2r−n)u2j

r2n+2dx

≤∫∫D

exp(2r−n) rn+2(∆uj)2dx =

∫∫|x|<R/2

+

∫∫R/2<|x|<R

exp(2r−n) rn+2(∆uj)2dx

≤ 2mc2

m∑`=1

∫∫|x|≤R/2

exp(2r−n) rn+2[|u`|2 + |∇u`|2

]dx +

∫∫R/2≤|x|≤R

exp(2r−n)rn+2 (∆uj)2 dx .

We set ψn(r) = exp(2r−n)r2n+2 for r > 0 and note that ψn is monotonously decreasing. Also,

because r ≤ 1 (and thus rn+2 ≤ r−2n−2),

2n

∫∫|x|≤R/2

exp(2r−n) |∇uj|2 dx + 2n4

∫∫|x|≤R/2

ψn(r)u2j dx

≤ 2mc2

m∑`=1

∫∫

|x|≤R/2

ψn(r)u2` dx+

∫∫|x|≤R/2

exp(2r−n) |∇u`|2 dx

+ ψn(R/2)‖∆uj‖2L2(D) .


Now we sum with respect to j = 1, . . . ,m and combine the matching terms. This yields

(2n− 6m2c2)m∑j=1

∫∫|x|≤R/2

exp(2r−n) |∇uj|2 dx+ (2n4 − 6m2c2)m∑j=1

∫∫|x|≤R/2

ψn(r)u2j dx

≤ ψn(R/2)m∑j=1

‖∆uj‖2L2(D) .

For n ≥ 3m2c2 we conclude that

2(n4 − 3m2c2)ψn(R/2)m∑j=1

∫∫|x|≤R/2

u2j dx ≤ 2(n4 − 3m2c2)

m∑j=1

∫∫|x|≤R/2

ψn(r)u2j dx

≤ ψn(R/2)m∑j=1

‖∆uj‖2L2(D)

and thusm∑j=1

∫∫|x|≤R/2

u2j dx ≤

1

2(n4 − 3m2c2)

m∑j=1

‖∆uj‖2L2(D)

The right hand side tends zero as n tends to infinity. This proves uj = 0 in B(0, R/2). 2

The proof of the uniqueness result transforms Maxwell’s equations to the (vector-)Helmholtzequation. We need weaker smoothness conditions on εc if we work with the magnetic field.

Theorem 4.49 Let µr = 1, εc ∈ C1(D) with Im εc > 0 on some open set B ⊂ D. Thenthere exists a unique solution E ∈ H0(curl, D) of the boundary value problem (4.23) for everyF ∈ L2(D)3.

Proof: Again, it suffices to prove uniqueness. Let F = 0 and E ∈ H0(curl, D) be thecorresponding solution of (4.23). As in the proof of Theorem 4.45 we conclude that Evanishes on B. Therefore, also H = 1

iωµ0curlE = 0 in B. By Lemma 4.39, the magnetic

field H satisfies∫∫D

[1

εccurlH · curlψ − k2H · ψ

]dx = 0 for all ψ ∈ H0(curl, D) . (4.38)

If we choose ψ = ∇φ for some φ ∈ H10 (D) then we have∫∫

D

H · ∇φ dx = 0 for all φ ∈ H10 (D) ;

that is, H ∈ V1 which is the variational form of divH = 0.We set now ψ = εcψ for some ψ ∈ C∞0 (D)3. Then ψ ∈ H0(curl, D) and therefore, becausecurlψ = εc curl ψ +∇εc × ψ,∫∫

D

[curlH · curl ψ +

1

εccurlH · (∇εc × ψ)− k2εcH · ψ

]dx = 0 for all ψ ∈ C∞0 (D)3 ,


which we write as∫∫D

curlH · curl ψ dx =

∫∫D

G · ψ dx for all ψ ∈ C∞0 (D)3 ,

where G = − 1εc

curlH ×∇εc + k2εcH ∈ L2(D)3. Partial integration yields∫∫D

G · ψ dx =

∫∫D

H · curl2 ψ dx = −∫∫D

H ·∆ψ dx +

∫∫D

H · ∇ div ψ dx︸︷︷︸= 0 because H∈V1

for all ψ ∈ C∞0 (D)3. By the interior regularity result of Theorem 4.47 we conclude thatH ∈ H2(U)3 for all domains U with U ⊂ D, and ∆H = −G = 1

εccurlH × ∇εc − k2εcH

in U . Because every component of curlH is a combination of partial derivatives of H` for` ∈ 1, 2, 3 we conclude the existence of a constant c > 0 such that

|∆Hj| ≤ c3∑`=1

[|∇H`|+ |H`|

]in D for j = 1, 2, 3 .

Therefore, all of the assumptions of Theorem 4.48 are satisfied and thus H = 0 in all ofU . This implies that also E = 0 in U . Because U is an arbitrary domain with U ⊂ D weconclude that E = 0 in D. 2

Remarks:

(a) The proof of the theorem can be modified that it holds if µr ∈ C2(D). Instead ofdivH = 0 we have6 that 0 = div(µrH) = ∇µr ·H + µr divH, thus

curl2H = −∆H +∇ divH = −∆H −∇(

1

µr∇µr ·H

)= −∆H −∇

[∇(lnµr) ·H

]= −∆H −

3∑j=1

(∇∂ lnµr

∂xjHj +

∂ lnµr∂xj

∇Hj

)and this can be treated in the same way.

(b) The asumption εc ∈ C1(D) is very restrictive. One can weaken this assumption to therequirement that εc is piecewise continuously differentiable. We refer to the monographby Peter Monk.

Finally, we want to comment on the smoothness assumption ∂D ∈ C2. For applicationsand, in particular, for the discretization of the problem by Finite Elements it is advanteous

6Here we argue classically, but all of the arguments hold also in the weak case.

4.4. THE TIME–DEPENDENT CAVITY PROBLEM 169

to allow Lipschitz boundaries (which include polygonal domains). All of the theorems ofthis chapter remain true except part (a) of Theorem 4.33 which holds only if the boundaryis smooth (as in our case) or if D is convex with Lipschitz continuous boundary. However,part (b) is true for all domains with Lipschtiz continuous boundaries – just the way that weproved it via part (a) does not work. Therefore, the main results (Theorems 4.37, 4.38, 4.42,4.44, 4.45, 4.49) hold because their proofs need only part (b) of Theorem 4.33.

###

4.4 The Time–Dependent Cavity Problem

The spectral theorem of the previous section allow it to treat the full time–dependent systemof Maxwells equations. We begin again with the initial–boundary value problem for the scalarwave equation in some bounded Lipschitz domain D ⊂ R3 and some interval (0, T ).

1

c(x)2

∂2u

∂t2(t, x) − div

(a(x)∇u(t, x)

)= f(t, x) , (t, x) ∈ (0, T )×D , (4.39a)

u(t, x) = 0 for (t, x) ∈ (0, T )× ∂D , (4.39b)

u(0, x) = u0(x) and∂u

∂t(0, x) = u1(x) for x ∈ D . (4.39c)

We make the following assumptions on the data:

Assumptions:

• a, c ∈ L∞(D) with a(x) ≥ a0 and c(x) ≥ c0 on D for some a0 > 0 and c0 > 0,

• f ∈ L2((0, T )×D

),

• u0 ∈ H10 (D), u1 ∈ L2(D).

In this section we assume that all functions are real-valued. We set b(x) = 1/c(x)2 forabbreviation. Then b ∈ L∞(D) and b(x) ≥ b0 = 1/‖c‖2

∞ on D.

The solution has to be understood in a variational sense. To motivate this we multiply thedifferential equation (4.39a) by some ψ ∈ C1[(0, T ) ×D] with ψ(0, x) = ψ(T, x) = 0 for allx ∈ D and integrate by parts with respect to t and use Greens first formula with respect tox. This yields

T∫0

∫∫D

[b(x)

∂u

∂t(t, x)

∂ψ

∂t(t, x)− a(x)∇u(t, x) · ∇ψ(t, x)

]dx dt = −

T∫0

∫∫D

f(t, x)ψ(t, x) dx dt ,

or, using the notion (u, v)∗ = (a∇u,∇v)L2(D)3 of the previous section and ut = ∂u/∂t,

T∫0

[(b(x)ut(t, ·), ψt(t, ·)

)L2(D)

−(u(t, ·), ψ(t, ·)

)∗

]dt = −

T∫0

(f(t, ·), ψ(t, ·)

)L2(D)

dt .


We will require different smoothness properties of u with respect to t and x. This leads to“anisotropic” function spaces. In particular, the solution has to be differentiable with respectto t (in a sense to be explained in a moment). It is convenient to consider u to be a functionin t with values u(t) in some function space with respect to x. We have taken this pointof view already when we wrote u(t, ·). To make this idea precise, we recall the notion of a(Frechet-) differentiable function.

Definition 4.50 Let V be a normed space (over R) and f : [0, T ] → V a function withvalues in V .

(a) The function f is continuous in some t0 ∈ [0, T ] if for every ε > 0 there exists δ > 0such that ‖f(t)−f(t0)‖V ≤ ε for all t ∈ [0, T ] with |t−t0| ≤ δ. The space of continuousfunctions on [0, T ] is denoted by C[0, T ;V ].

(b) The function f is differentiable in t0 ∈ [0, T ] with value u′(t0) ∈ V if for every ε > 0there exists δ > 0 such that

∥∥[f(t) − f(t0)]/(t − t0) − u′(t0)

∥∥V≤ ε for all t ∈ [0, T ]

with 0 < |t − t0| ≤ δ. The space of continuously differentiable functions is denoted byC1[0, T ;V ].

We define the solution space X to be

X = C[0, T ;H1

0 (D)]∩ C1

[0, T ;L2(D)

](4.40a)

and equip X with the norm

‖u‖X = max0≤t≤T

‖u(t)‖H1(D) + max0≤t≤T

‖u′(t)‖L2(D) . (4.40b)

Definition 4.51 u is a solution of the initial-boundary value problem (4.39a) – (4.39c) ifu ∈ X and

T∫0

[(b ut(t), ψt(t)

)L2(D)

−(u(t), ψ(t)

)∗

]dt = −

T∫0

(f(t, ·), ψ(t)

)L2(D)

dt (4.41)

for all ψ ∈ X.

We note that t 7→ f(t, ·) is in L2(D) by Fubini’s theorem and∫ T

0‖f(t, ·)‖2

L2(D)dt = ‖f‖L2((0,T )×D).

Therefore, the right hand side of (4.41)) is well defined. ###

4.5 Exercises

Exercise 4.1 Show that C∞0 (R3) is dense in L2(R3) and in H1(R3).Hint:

Exercise 4.2 Show that the vector field F in Definition4.1 is unique (if it exists).Hint: Use Exercise 4.1!

4.5. EXERCISES 171

Exercise 4.3 Construct a function φ ∈ C∞(R) such that φ(t) = 0 for |t| ≥ 1 and φ(t) = 1for |t| ≤ 1/2.Hint:

Exercise 4.4 Prove that for open and bounded D ⊂ R3 the space H10 (D) is a proper

subspace of H1(D).Hint: Show that functions v with ∆v − v = 0 in D are orthogonal to H1

0 (D).

Exercise 4.5 Prove that the spaces H10 (R3) and H1(R3) coincide and that they are not

compactly imbedded in L2(R3).Hint:

Exercise 4.6 In Theorem 4.27 the existence of an extension operator is proven. Why isthis not an obvious fact by our definition of H1(D) as the space of restrictions of H1(R3)?

Exercise 4.7 Show that C∞0 (R3)3 is dense in H(curl,R3).Hint:

Exercise 4.8 Why is, even for bounded sets D, the space H0(curl, D) not compactly imbed-ded in L2(D)3?Hint: Use the Helmholtz decomposition!

Exercise 4.9 Let D = B(0, R) be a ball. Show explicitely the Helmholtz decomposition byusing the results of Chapter ??.

### alternativer Beweis:

Let u ∈ C1(D) such that u is real valued (without loss of generality) and u|∂D = 0. Chooseψ ∈ C∞(R) such that |ψ(t)| ≤ |t| for all t ∈ R and ψ(t) = 0 for |t| ≤ 1 and ψ(t) = t for |t| ≥2. We define uε(x) = ε ψ

(1εu(x)

)for x ∈ D and ε > 0. Then uε(x) = 0 for x with

∣∣u(x)∣∣ ≤ ε

and uε(x) = u(x) for x with∣∣u(x)

∣∣ ≥ 2ε. Therefore, supp(uε) ⊂x ∈ D :

∣∣u(x)∣∣ ≥ ε

.

Because u vanishes on ∂D we conclude thatx ∈ D :

∣∣u(x)∣∣ ≥ ε

⊂ D which proves that

supp(uε) ⊂ D; that is, uε has compact support in D. We extend uε by zero into all of R3.We show that uε → u in H1(D) as ε tends to zero. From ∇uε(x) = ψ′

(1εu(x)

)∇u(x) we

have

‖uε − u‖2L2(D) =

∫∫D

∣∣∣∣ε ψ(1

εu(x)

)− u(x)

∣∣∣∣2 dx , (4.42a)

‖∇uε −∇u‖2L2(D) =

∫∫D

∣∣∣∣ψ′(1

εu(x)

)− 1

∣∣∣∣2 ∣∣∇u(x)∣∣2dx . (4.42b)

We discuss the convergence with the theorem of dominated convergence. For (4.42a) weobserve that the integrand is zero for all x with u(x) = 0 and for all x with u(x) 6= 0 provided2ε ≤

∣∣u(x)∣∣. Also, the integrand is dominated by the integrable function 4|u(x)|2.


For (4.42b) we observe that the integrand is zero for all x with u(x) 6= 0 provided 2ε ≤∣∣u(x)

∣∣.Furthermore, the integrand is zero on the set B =

x ∈ D : u(x) = 0 , ∇u(x) = 0

. Finally,

from a general theorem of measure theory (see Lemma 4.10 below) and the smoothness of theboundary we have that the sets A =

x ∈ D : u(x) = 0 , ∇u(x) 6= 0

and ∂D have measure

zero. This implies pointwise convergence almost everywhere. Furthermore, the integrand isdominated by ‖ψ′ − 1‖2

∞|∇u(x)|2.So far, we approximated u by functions uε ∈ C1

0(D). As the next step we smooth thisfunction; that is, we approximate uε by functions in C∞0 (D). We set uε,δ = uε ∗ φδ with themollifier φδ from Theorem 4.5. Then supp(uε ∗ φδ) ⊂ supp(uε) + B(0, δ). Furthermore, forε > 0 we choose δ(ε) > 0 such that supp(uε) + B(0, δ) ⊂ D and ‖uε ∗ φδ(ε) − uε‖H1(D) ≤‖uε ∗ φδ(ε) − uε‖H1(R3) ≤ ε. Then supp(uε ∗ φδ) ⊂ D. The triangle inequality yields theassertion because

‖uε ∗ φδ(ε) − u‖H1(D) ≤ ‖uε ∗ φδ(ε) − uε‖H1(D) + ‖uε − u‖H1(D) ≤ ε + ‖uε − u‖H1(D) .

2

It remains to show the following result from measure theory.

Lemma 4.10 Let D be an open and bounded set and u ∈ C1(D). Then the set A =x ∈

D : u(x) = 0 , ∇u(x) 6= 0

is of (Lebesgue-) measure zero.

Proof: Define

An =

x ∈ A : d(x, ∂D) ≥ 1

n, |∇u(x)| ≥ 1

n

.

Then An is compact and⋃n∈NAn = A. It is sufficient to show that An has measure zero for

every n.By the implicit function theorem, for every x ∈ An there exists an open box R(x) ⊂ R3

which contains x such that N(x) = x ∈ R(x) : u(x) = 0 is the graph of a function oftwo variables and thus has measure zero by a well known result from measure theory. FromAn ⊂

⋃x∈An R(x) and the compactness of An we conclude that there exist finitely many

x1, . . . , xm ∈ An with An ⊂⋃mj=1R(xj) and thus also An ⊂

⋃mj=1 N(xj). 2 ###

Chapter 5

Boundary Integral Equation Methodsfor Lipschitz Domains

In this chapter we will present an approach for boundary integral equation methods forLipschitz domains. We begin with Sobolev spaces on boundaries of Lipschitz domains andthe corresponding trace operators.

5.1 Sobolev Spaces on Boundaries

We recall Definition 6.1 for the notion of a Lipschitz domain.

Definition 5.1 We call a region D ⊂ R3 to be a Lipschitz domain, if there exists a finitenumber of open cylinders Uj of the form Uj = Qjx + z(j) : x ∈ B2(0, αj) × (−βj, βj)with z(j) ∈ R3 and rotations Qj ∈ R3×3 and real valued Lipschitz-continuous functionsξj ∈ C(B[0, αj]) with |ξj(x1, x2)| ≤ βj/2 for all (x1, x2) ∈ B2[0, αj] such that

∂D ⊂m⋃j=1

Uj ,


,


,


.



1+x22 < α2

j , |x3| < βj

the standard cylinders. Furthermore, we define the mappings

Ψj(x) = Qj

x1

x2

ξj(x1, x2) + x3

+ z(j) , x = (x1, x2, x3)> ∈ Cj(αj, βj/2) ,

173

174 5 BIE FOR LIPSCHITZ DOMAINS


Ψj(x) = Qj

x1

x2

ξj(x1, x2)

+ z(j) , x = (x1, x2)> ∈ B2(0, αj) ,


∂D ∩ Uj =

Ψj(x) : x ∈ Cj, x3 = 0

=

Ψj(x) : x ∈ B2(0, αj),

D ∩ U ′j =

Ψj(x) : x ∈ C ′j, x3 < 0,

U ′j \D =

Ψj(x) : x ∈ C ′j, x3 > 0,

where C ′j = Cj(αj, βj/2) = B2(αj)× (−βj/2, βj/2) and U ′j = Ψj(C′j).

We note that the Jacobian Ψ′j(x) ∈ R3×3 is given by

Ψ′j(x) = Qj

1 0 00 1 0

∂1fj(x) ∂2fj(x) 1

where ∂`fj = ∂fj/∂x` for ` = 1, 2 and x = (x1, x2). Therefore, these Jacobians are regularwith constant determinant det Ψ′j(x) = ±1.

These parametrizations allow it to transfer the notion of (periodic) Sobolev spaces on twodimensional planar domains to ∂D. We begin with Sobolev spaces of scalar function inSubsection 5.1.1 and continue with vector-valued functions in Subsection 5.1.2.

5.1.1 Boundary-Sobolev Spaces of Scalar Functions

We begin by defining Sobolov spaces of periodic functions as we have done it already inChapter 4. The Definition 4.10 of H1

per(Q3) is included in the following definition.

Definition 5.2 Let Q = (−π, π)d ⊂ Rd be the cube in Rd for d = 2 or d = 3. For any scalarfunction v : (−π, π)d → C the Fourier coefficients vn ∈ C for n ∈ Zd are defined as

vn =1

(2π)d

∫∫C

v(y) e−i n·y dy , n ∈ Zd .

Let s ≥ 0 be any real number. The space Hsper(Q) is defined by

Hsper(Q) =

v ∈ L2(Q) :

∑n∈Z3

(1 + |n|2)s |vn|2 <∞

with norm

‖v‖Hsper(Q) =

√∑n∈Z3

(1 + |n|2)s |vn|2 .

5.1. SOBOLEV SPACES ON BOUNDARIES 175

We note that obviously H1per(Q) ⊂ H1(Q) with bounded inclusion. On the other hand,

H10 (Q) ⊂ H1

per(Q) with bounded inclusion. (For both proofs we refer to Excercise ???).Therefore, on H1

0 (Q) the norms of H1per(Q) and H1(Q) are equivalent.

Such a definition of Sobolev spaces of periodic functions is useful to prove; e.g., imbeddingtheorems as we did in Chapter 4. We have seen in Theorem ?? that H1(D) is compactlyimbedded in L2(D). This can be carried over to periodic Sobolev spaces of any order.

Theorem 5.3 Let Q = (−π, π)d ⊂ Rd be the cube in Rd for d = 2 or d = 3. For 0 ≤ s < tthe space H t

per(Q) is compactly imbedded in Hsper(Q).

Proof: see Exercise ?? 2

Also, the proof of the following trace theorem is (rather long but) elementary.

Lemma 5.4 Let again Q3 = (−π, π)3 ⊂ R3 denote the cube in R3 and Q2 = (−π, π)2 ⊂R2 the corresponding square in R2. The trace operator γ : H1

per(Q3) → H1/2per (Q2), u 7→

u|Q2×0, is well-defined and bounded. Furthermore, there exists a bounded right inverse η

of γ; that is, a bounded operator η : H1/2per (Q2) → H1

per(Q3) with γ η = id. In other words,

the function u = ηf ∈ H1per(Q3) coincides with f ∈ H1/2

per (Q2) on Q2 × 0.

Proof: Let u ∈ H1per(Q3) with Fourier coefficients un, n ∈ Z3; that is, u(x) =

∑n∈Z3 un e

i n·x.We truncate the series and define uN(x) =

∑n2

1+n22≤N2

∑|n3|≤N un e

i n·x. For x ∈ R3 and

n ∈ Z3 we set x = (x1, x2) and n = (n1, n2), respectively. Again | · | denotes the Euclideannorm of vectors. Then

(γuN)(x) = uN(x, 0) =∑|n|≤N

(N∑

n3=−N

un

)︸︷︷︸

=: vn

ei n·x =∑|n|≤N

vn ei n·x

and thus

‖γuN‖2

H1/2per (Q2)

=∑|n|≤N

(1 + |n|2

)1/2 |vn|2 .

Now we use the following elementary inclusion

π

2≤ a

∞∑n3=−∞

1

a2 + n23

≤ π + 1 for all a ≥ 1 (5.1)


which we will show at the end of the proof. Using the upper estimate for a =(1 + |n|2

)1/2

and the inequality of Cauchy-Schwarz,

(1 + |n|2

)1/2 |vn|2 =(1 + |n|2

)1/2

∣∣∣∣∣N∑

n3=−N

un(1 + |n|2

)1/2 1(1 + |n|2

)1/2

∣∣∣∣∣2

≤N∑

n3=−N

|un|2(1 + |n|2

) [(1 + |n|2

)1/2N∑

n3=−N

1

1 + |n|2 + x23

]

≤ (π + 1)N∑

n3=−N

|un|2(1 + |n|2

).

Summation with respect to n yields

‖γuN‖2

H1/2per (Q2)

≤ (π + 1) ‖uN‖2H1per(Q3) .

This holds for all N ∈ N. Letting N tend to infinity yields the boundedness of γ.

To show the existence of a bounded extension operator η we define

(ηf)(x) =∑n∈Z3

un ein·x where un = fn

δn1 + |n|2

, x ∈ Q3 ,

and

δn =

[∞∑

j=−∞

1

1 + |n|2 + j2

]−1

. (5.2)

We have to show that ηf ∈ H1per(Q3). We note that

∞∑n3=−∞

|un|2(1 + |n|2) = |fn|2δ2n

∞∑n3=−∞

1

1 + |n|2=(|fn|2

√1 + |n|2

) δn√1 + |n|2

≤ 2

π|fn|2

√1 + |n|2

by the lower estimate of (5.1). Summing over n yields

‖ηf‖2H1per(Q3) =

∑n∈Z2

∞∑n3=−∞

|un|2(1 + |n|2) ≤ 2

π

∑n∈Z2

|fn|2√

1 + |n|2 =2

π‖f‖2

H1/2per (Q2)

.

Therefore, η : H1/2per (Q2)→ H1

per(Q3) is well defined and bounded. Furthermore,

(γηf)(x) =∑n∈Z2

∞∑n3=−∞

un ein·x =

∑n∈Z2

fn ein·x = f(x) .


It remains to show (5.1). We note that

∞∑j=−∞

1

a2 + j2=

1

a2+ 2

∞∑j=1

1

a2 + j2

and

N∑j=1

1

a2 + j2=

N∑j=1

j∫j−1

dt

a2 + j2≤

N∑j=1

j∫j−1

dt

a2 + t2=

N∫0

dt

a2 + t2=

1

a

N/a∫0

ds

1 + s2,

where we used the substitution t = as. Analogously,

N∑j=1

1

a2 + j2=

N∑j=1

j+1∫j

dt

a2 + j2≥

N∑j=1

j+1∫j

dt

a2 + t2=

N+1∫1

dt

a2 + t2=

1

a

(N+1)/a∫1/a

ds

1 + s2.

Letting N tend to infinity yields

1

a

(π2− arctan(1/a)

)≤

∞∑j=1

1

a2 + j2≤ π

2a.

Noting that a ≥ 1 implies arctan(1/a) ≤ arctan(1) = π/4 which yields the estimates (5.1).2

We will need the following corollary.

Corollary 5.5 The trace operator γt : v 7→ v|Q2 from H1(Q+3 ) ∩ C(Q+

3 ) into C(Q2) has abounded extension from H1

0 (Q+3 ) :=

v ∈ H1(Q+

3 ) : supp v ⊂ Q+3 ∪ (Q2×0)

; that is, there

exists c > 0 such that

‖γtv‖H1/2per (Q2)

≤ c ‖v‖H1(Q+3 ) for all v ∈ H1

0 (Q+3 ) . (5.3)

Proof: We extend v ∈ H10 (Q+

3 ) by reflection into Q3, that is, define

v(x1, x2, x3) =

v(x1, x2, x3) , x ∈ Q3 , x3 ∈ (0, π] ,v(x1, x2,−x3) , x ∈ Q3 , x3 ∈ (−π, 0] .

Then v ∈ H1(Q3) with compact support in Q3. Therefore1, v ∈ H10 (Q3) and, by periodic

extension, v ∈ H1per(Q3). By Lemma 5.4 there exists c > 0 such that

‖v(·, 0)‖H

1/2per (Q2)

= ‖v(·, 0)‖H

1/2per (Q2)

≤ c ‖v‖H1per(Q3) .

Now we use that fact that the norms in H1(Q3) and H1per(Q3) are equivalent on H1

0 (Q3).Therefore, there exists c > 0 such that

‖v(·, 0)‖H

1/2per (Q2)

≤ c ‖v‖H1(Q3) =√

2c ‖v‖H1(Q+3 ) .

1see Exercise ??


2

The motivation for the definition of the Sobolev space H1/2(∂D) is given by the followingtransformation of the L2−norm. Let D ⊂ R3 be a regular domain with correspondinglocal coordinate system Uj, ξj : j = 1, . . . ,m. Furthermore, let φj : j = 1, . . . ,m, be apartition of unity on ∂D corresponding to the sets Uj. For f ∈ L2(∂D) we use |f(y)|2 =∑m

j=1 φj(y)|f(y)|2 and thus

‖f‖2L2(∂D) =

∫∂D

∣∣f(y)∣∣2 ds =

m∑j=1

∫∂D∩Uj

φj(y)∣∣f(y)

∣∣2 ds=

m∑j=1

∫Q2

φj(Ψj(x)

)∣∣f(Ψj(x))∣∣2 dx =

m∑j=1

‖fj‖2L2(Q2), , (5.4)

where

fj(x) :=√φj(Ψj(x)

)f(Ψj(x)

), x ∈ Q2 . (5.5)

Therefore, f ∈ L2(∂D) if, and only if, fj ∈ L2(Q2) for all j = 1, . . . ,m.

Definition 5.6 Let D ⊂ R3 be regular in the sense of Definition 5.1 with corresponding localcoordinate system Uj, ξj : j = 1, . . . ,m and Ψj and their restrictions Ψj : Q2 → Uj ∩ ∂D.Furthermore, let φj : j = 1, . . . ,m, be a partition of unity on ∂D corresponding to the setsUj. Then we define

H1/2(∂D) =f ∈ L2(∂D) : fj ∈ H1/2

per (Q2) for all j = 1, . . . ,m

with norm

‖f‖H1/2(∂D) =m∑j=1

[‖fj‖2

H1/2per (Q2)

]1/2

.

Here, fj is given by (5.5).

Our definition of the space H1/2(∂D) seems to depend on the choice of the local coordinatesξj and the partition of unity φj. However, we will see in Corollary 5.10 below that the normscorresponding to two such choices are equivalent.

From now on we assume always that the domain D is regular in the sense of Definition 5.1.

We note the following implication of Theorem 5.3.

Corollary 5.7 The space H1/2(∂D) is compactly imbedded in L2(∂D).

Proof: Let (f `)` be bounded in H1/2(∂D) then, by definition, (f `j )` is bounded in H1/2per (Q2)

for all j = 1, . . . ,m, where f `j = (f `√φj) Ψj. By Theorem 5.3 for s = 0 and t = 1/2

there exists a subsequence (f `kj )k of (f `)` which converges in L2(Q2). By (5.4) also (f `k)kconverges in L2(∂D). 2

Combining Definition 5.6 and Corollary 5.5 we can prove a trace theorem in H1(D).


Theorem 5.8 The trace operator γ : C1(D) → C(∂D), u 7→ u|∂D, has an extension asa bounded operator from H1(D) to H1/2(∂D). Furthermore, γ has a bounded right inverseη : H1/2(∂D)→ H1(D).

Proof: Let Uj, ξj : j = 1, . . . ,m be a local coordinate system of ∂D and φj : j =1, . . . ,m a partition of unity with respect to Uj. Set U :=

⋃mj=1 Uj and let u ∈ C1(D). Set

uj(x) :=√φj(Ψj(x)

)u(Ψj(x)

), x ∈ Q+

3 .

Then uj ∈ H1per(Q

+3 ) ∩ C(Q+

3 ). By Corollary 5.5 there exists c1 > 0 such that∥∥uj(·, 0)∥∥H

1/2per (Q2)

≤ c1‖uj‖H1(Q+3 ) for all j

and thus by Definition 5.6

‖γu‖2H1/2(∂D) =

m∑j=1

‖uj(·, 0)‖2

H1/2per (Q2)

≤ c21

m∑j=1

‖uj‖2H1(Q+

3 )≤ c ‖u‖2

H1(D∩U) ≤ c ‖u‖2H1(D) .

This proves boundedness of γ.

Now we construct an extension operator η : H1/2(∂D) → H1(D). For f ∈ H1/2(∂D) define

fj = (f√φj) Ψj ∈ H

1/2per (Q2) for j = 1, . . . ,m. By Lemma 5.4 there exist extensions

vj ∈ H1per(Q3) of fj and the mappings fj 7→ vj are bounded for j = 1, . . . ,m. Then we set

uj(y) :=√φj(y) vj

(Ψ−1j (y)

), y ∈ Uj ∩D .

Then uj ∈ H10 (Uj ∩D) for all j. We extend uj by zero into all of R3 and set u =

∑mj=1 uj.

Then u is an extension of f . Indeed, for fixed y ∈ ∂D let J =j ∈ 1, . . . ,m : y ∈ Uj

.

Then u(y) =∑

j∈J uj(y). Let j ∈ J ; that is, y ∈ Uj. Then x = Ψ−1j (y) ∈ Q2 × 0,

thus vj(·, 0) = fj = (f√φj) Ψj. Therefore, u(y) =

∑j∈J uj(y) =

∑j∈J√φj(y) vj(x) =∑

j∈J f(y)φj(y) = f(y). Furthermore, all the operations in this constructions are bounded.This proves the theorem. 2

We recall from Definition 4.7 that the subspace H10 (D) had been defined as the closure of

C∞0 (D) in H1(D). The following equivalent characterization is of essential importance inthe following.

Theorem 5.9 The space H10 (D) is the null space ker γ of the trace operator; that is, u ∈

H10 (D) if, and only if, γu = 0 on ∂D.

Proof: The inclusion H10 (D) ⊂ ker γ follows immediately because γu = 0 for all u ∈ C∞0 (D),

the boundedness of γ and the fact that C∞0 (D) is dense in H10 (D) by definition. To show

the reverse inclusion we can proceed exactly as in the proof of Theorem 4.8. 2


Corollary 5.10 Definition 5.6 is independent of the choice of the coordinate system Uj, Ψj :j = 1, . . . ,m and a corresponding partition of unity φj : j = 1, . . . ,m with respect to Uj.

Proof: First we note that the space v∂D : v ∈ C∞(D) is dense in H1/2(∂D). Indeed,for f ∈ H1/2(∂D) we conclude that u = ηf ∈ H1(D) can be approximated by a sequenceun ∈ C∞(D) with respect to ‖ · ‖H1(D). But then un|∂D = γun converges to γu = f in

H1/2(∂D). Therefore, H1/2(∂D) is the completion of v∂D : v ∈ C∞(D) with respect to‖·‖H1/2(∂D). Furthermore, from the boundedness of γ : H1(D)→ H1/2(∂D) and Theorem 5.9

we observe that the lifted operator [γ] : H1(D)/H10 (D)→ H1/2(∂D) is a norm-isomorphism

from the factor space H1(D)/H10 (D) onto H1/2(∂D). Therefore, ‖ · ‖H1/2(∂D) is equivalent to

the norm in H1(D)/H10 (D) which is independent of the choice of the coordinate system. 2

The proof shows also that an equivalent norm in H1/2(∂D) is given by the canonical normof H1(D)/H1

0 (D); that is, ‖f‖ = inf‖u‖H1(D) : γu = f on ∂D

.

Before we turn to the vector valued case we want to discuss the normal derivative ∂u/∂νon ∂D which is well defined for u ∈ C1(D). It can be considered as the trace of the normalcomponent of the gradient of u. We denote it by γ∂ν : C1(D) → C(∂D), u 7→ ∂u/∂ν. InExercise ?? we show that it is not bounded with respect to the norm of H1(D). However, itdefines a bounded operator on the closed subspace

HD =

u ∈ H1(D) :

∫∫D

[∇u · ∇ψ − k2uψ

]dx = 0 for all ψ ∈ H1

0 (D)

(5.6)

of (variational) solutions of the Helmholtz equation into the dual space of H1/2(∂D) as wewill see in a moment.

First we recall from Corollary 5.7 that H1/2(∂D) is a subspace of L2(∂D) with bounded –even compact – inclusion. For any f ∈ L2(∂D) the linear form 〈`f , ψ〉 = (f, ψ)L2(∂D) defines abounded functional onH1/2(∂D) because |`f (ψ)| ≤ ‖f‖L2(∂D)‖ψ‖L2(∂D) ≤ c ‖f‖L2(∂D)‖ψ‖H1/2(∂D)

for all ψ ∈ H1/2(∂D). Therefore, if we identify `f with f – as done by identifying the dualspace of L2(∂D) with itself – then L2(∂D) can be considered as a subspace of the dual spaceH1/2(∂D)∗ of H1/2(∂D) which we denote by H−1/2(∂D):

Definition 5.11 Let H−1/2(∂D) be the dual space of H1/2(∂D) equipped with the canonicalnorm of a dual space; that is,

‖`‖H−1/2(∂D) := supf∈H1/2(∂D)\0

|〈`, f〉|∗‖f‖H1/2(∂D)

, ` ∈ H−1/2(∂D) ,

where 〈`, f〉∗ denote the evaluation of u by ` (dual form).

With the identification of `f with f we observe that 〈`, ψ〉∗ is the extension of the L2−innerproduct.


Using a local coordinate system and a corresponding partion of unity one can prove thatH−1/2(∂D) can be characterized by the periodic Sobolev space H

−1/2per (Q3) which is the com-

pletion of the space of trigonometric polynomials by the norm

‖v‖H−1/2per (Q3)

=

[∑n

(1 + |n|2)−1/2|vn|2]1/2

.

The definition of the trace ∂u/∂ν is motivated by Green’s first theorem: For u ∈ C2(D)with ∆u+ k2u = 0 in D and ψ ∈ H1(D) we have∫

∂D

∂u

∂ν(γψ) ds =

∫∫D

[∇u · ∇ψ − k2uψ

]dx .

Since the trace γψ of ψ is an element of H1/2(∂D) the left hand side is a linear functional onH1/2(∂D). The right hand side is well defined also for u ∈ H1(D). Therefore, it is naturalto extend this formula to u, ψ ∈ H1(D) and replace the left hand side by the dual form〈∂u/∂ν, γψ〉∗. This is justified by the following theorem.

Theorem 5.12 The operator γ∂ν : HD ∩ C1(D) → C(∂D), u 7→ ∂u/∂ν, is bounded as anoperator from HD into H−1/2(∂D). It can be written in the form (Green’s formula)

〈γ∂νu, ψ〉∗ =

∫∫D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) , (5.7)

where ψ ∈ H1(D) is any extension of ψ into D; that is, γψ = ψ – which is possible by thesurjectivity of the trace operator γ, see Theorem 5.8.

Proof: We take (5.7) as the definition of a functional ` on H1/2(∂D); that is, for fixedu ∈ HD we define

〈`, ψ〉∗ :=

∫∫D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) ,

where ψ ∈ H1(D) is any extension of ψ into D. We have to show that this definition isindependent of the choice of ψ. Indeed, if ψ1 and ψ2 are two extensions of ψ then ψ =ψ1− ψ2 ∈ H1

0 (∂D) by Theorem 5.9 because γψ = 0. Therefore,∫D

[∇u · ∇ψ− k2uψ

]dx = 0

because u ∈ HD. This shows that the definition of ` is independent of the choice of ψ. Toshow boundedness of ` we take ψ = ηψ where η : H1/2(∂D)→ H1(D) denotes the boundedright inverse of γ of Theorem 5.8. Then, by the inequality of Cauchy-Schwarz,

|〈`, ψ〉∗| ≤ ‖∇u‖L2(D)‖∇ψ‖L2(D) + k2‖u‖L2(D)‖ψ‖L2(D) ≤ maxk2, 1‖u‖H1(D)‖ψ‖H1(D)

= maxk2, 1‖u‖H1(D)‖ηψ‖H1(D) ≤ maxk2, 1 ‖η‖ ‖u‖H1(D)‖ψ‖H1/2(∂D) .

This proves boundedness of ` and ‖`‖H−1/2(∂D) ≤ maxk2, 1 ‖η‖ ‖u‖H1(D). 2

If the region D is of the form D = D1 \ D2 for open sets Dj such that D2 ⊂ D1 then theboundary ∂D consists of two components ∂D1 and ∂D2. The spaces H±1/2(∂D) can bewritten as direct sums H±1/2(∂D1) × H±1/2(∂D2), and the trace operators γ and γ∂ν havetwo components γ|∂Dj and γ∂ν |∂Dj for j = 1, 2 which are the projections onto ∂Dj. For the

definition of γ|∂Dj or γ∂ν |∂Dj one just takes extensions ψ ∈ H1(D) which vanish on ∂D3−j.


5.1.2 Boundary-Sobolev Spaces of Vector-Valued Functions

Now we turn to Sobolev spaces of vector functions. In Chapter 4 we have studied the spaceH(curl, D). It is the aim to prove two trace theorems for this space. We proceed as for thescalar case and introduce first some Sobolev spaces of periodic functions.

Definition 5.13 Let again Q3 = (−π, π)3 ⊂ R3 be the cube and Q2 = (−π, π)2 ⊂ R2 be thesquare. For any vector function v : (−π, π)d → Cd, d = 2 or d = 3, the Fourier coefficientsvn ∈ Cd for n ∈ Zd are defined as

vn =1

(2π)d

∫∫(−π,π)d

v(y) e−i n·y dy , n ∈ Zd .

(a) The space Hper(curl, Q3) is defined by

Hper(curl, Q3) =

v ∈ L2(Q3)3 :

∑n∈Z3

[|vn|2 + |n× vn|2

]<∞

with norm

‖v‖Hper(curl,Q3) =

√∑n∈Z3

[|vn|2 + |n× vn|2

].

(b) For any s ∈ R the space Hsper(Div, Q2) is defined as the completion of the space

T (Q2)2 :=

∑|m|≤N

am eim·x , x ∈ Q2 , vm ∈ C2 , N ∈ N

of all trigonometric vector polynomials with respect to the norm

‖v‖Hsper(Div,Q2) =

√∑m∈Z2

(1 + |m|2)s[|vm|2 + |m · vm|2

].

(c) For any s ∈ R the space Hsper(Curl, Q2) is defined as the completion of the space T (Q2)2

with respect to the norm

‖v‖Hsper(Curl,Q2) =

√∑m∈Z2

(1 + |m|2)s[|vm|2 + |m× vm|2

],

where we have set m× a = m1a2 −m2a1 for vectors m, a ∈ C2.

It will be convenient to consider the elements f ∈ H−1/2(Div, Q2) and f ∈ H−1/2(Curl, Q2)as vector functions2 from Q2 into C3 with vanishing third component f3. For example, wewill do this identification in the following lemma.

2Actually, these elements are no functions anymore but distributions.


Lemma 5.14 Let again Q3 = (−π, π)3 ⊂ R3 denote the cube in R3 and Q2 = (−π, π)2 ⊂ R2

the corresponding square in R2. Furthermore, let e = (0, 0, 1)> be the unit normal vectororthogonal to the plane R2 × 0 in R3.

(a) The trace operator γt : Hper(curl, Q3)→ H−1/2per (Div, Q2), u 7→ e× u(·, 0) =(

−u2(·, 0), u1(·, 0), 0)>

, is well-defined and bounded. Furthermore, there exists a bounded

right inverse ηt of γt; that is, a bounded operator ηt : H−1/2per (Div, Q2)→ Hper(curl, Q3)

with γt ηt = id. In other words, the tangential components e × u of u = ηtf ∈Hper(curl, Q3) coincides with f ∈ H−1/2

per (Div, Q2) on Q2 × 0.

(b) The trace operator γT : Hper(curl, Q3)→ H−1/2per (Curl, Q2), u 7→

(e× u(·, 0)

)× e =(

u1(·, 0), u2(·, 0), 0)>

, is well-defined and bounded. Furthermore, there exists a boundedright inverse ηT of γT .

Proof: We restrict ourselves to part (a) and leave the proof of part (b) to the reader. Letu ∈ Hper(curl, Q3) with Fourier coefficients un ∈ C3, n ∈ Z3; that is, u(x) =

∑n∈Z3 un e

i n·x.With x = (x1, x2) and n = (n1, n2) we observe that

u(x, 0) =∑n∈Z2

[∞∑

n3=−∞

un

]ei n·x .

Therefore,

u⊥n :=∞∑

n3=−∞

e× un =∞∑

n3=−∞

e× u(n1,n2,n3) ∈ C3 , n = (n1, n2) ∈ Z2 , (5.8)

are the Fourier coefficients of γtu. Note that the third component vanishes. We decomposeun in the form

un =1

|n|2n× (un × n) +

1

|n|2(n · un)n .

We set for the moment n := n/|n| and vn := un × n and observe that

e× un =1

|n|e× (n× vn) +

1

|n|(n · un) (e× n) , (5.9)

and thus for fixed n 6= (0, 0) by the inequality of Cauchy-Schwarz

∣∣u⊥n ∣∣2 =

∣∣∣∣∣∞∑

n3=−∞

(e× un)

∣∣∣∣∣2

≤ 2

∣∣∣∣∣∞∑

n3=−∞

1

|n|e× (n× vn)

∣∣∣∣∣2

+ 2

∣∣∣∣∣∞∑

n3=−∞

1

|n|(n · un) (e× n)

∣∣∣∣∣2

≤ 2

[∞∑

n3=−∞

1

|n|2

][∞∑

n3=−∞

|vn|2 + |n|2∞∑

n3=−∞

|un|2]


where we note that |e × n| = |n| does not depend on n3. We use (5.1) for a = |n| withn 6= (0, 0) and arrive at

∣∣u⊥n ∣∣2 ≤ 2(π + 1)

|n|

[∞∑

n3=−∞

|n× un|2 + |n|2∞∑

n3=−∞

|un|2]

provided n 6= (0, 0). Now we multiply with (1 + |n|2)−1/2 and sum with respect to n 6= (0, 0).

∑n∈Z2

n6=(0,0)

(1 + |n|2)−1/2∣∣u⊥n ∣∣2 ≤ 2(π + 1)

∑n∈Z2

n6=(0,0)

(1 + |n|2)−1/2

|n|

∞∑n3=−∞

[|n× un|2 + |n|2|un|2

]

≤ 2(π + 1)∑n∈Z3

[|n× un|2 + |un|2

].

Finally, we add the term with n = (0, 0). In this case n = n3e and thus e×un = 1n3

(n×un).Therefore,∣∣∣∣∣∑

n3 6=0

(e× un)

∣∣∣∣∣2

=

∣∣∣∣∣∑n3 6=0

1

n3

(n× un)

∣∣∣∣∣2

≤∑n3 6=0

1

n23

∑n3 6=0

|n× un|2 ≤ c1

∞∑n3=−∞

|n× un|2 ,

with c1 = 2∑∞

j=11j2

. Therefore,

∣∣∣∣∣∞∑

n3=−∞

e× un

∣∣∣∣∣2

≤ 2c1

∞∑n3=−∞

|n× un|2 + 2 |e× u0|2 .

Adding this to the previous term for n 6= (0, 0) yields

∑n∈Z2

(1 + |n|2)−1/2|u⊥n |2 =∑n∈Z2

(1 + |n|2)−1/2

∣∣∣∣∣∞∑

n3=−∞

(e× un)

∣∣∣∣∣2

≤ c∑n∈Z3

[|n× un|2 + |un|2

](5.10)

for c = 2(π + 1) + 2c1 + 2.

We study the Fourier coefficients of Div γt by the same arguments. First we note that

(Div γtu)(x, 0) = i∑n∈Z2

(n · u⊥n ) ei n·x

for any extension n of n ∈ N2 where again u⊥n ∈ C3 are the Fourier coefficients of γtu from(5.8) above. We have from (5.9) with the same notations as above that

n · (e× un) = n · (e× un) =1

|n|n ·(e× (n× vn)

)=

1

|n|(n× e) · (n× vn)


and thus for n 6= (0, 0)

|n · u⊥n |2 =

∣∣∣∣∣∞∑

n3=−∞

n · (e× un)

∣∣∣∣∣2

=

∣∣∣∣∣∞∑

n3=−∞

1

|n|(n× e) · (n× vn)

∣∣∣∣∣2

≤∞∑

n3=−∞

1

|n|2∞∑

n3=−∞

|n× e| |n× vn|2

≤ π + 1

|n|

∞∑n3=−∞

|n|2 |vn|2 = (π + 1) |n|∞∑

n3=−∞

|un × n|2 .

Here we used |n × e| ≤ |n| because the third component of n × e vanishes. For n = 0 wenote that n · (e× un) vanishes. Therefore, multiplication with (1 + |n|2)−1/2 and summationwith respect to n yields

∑n∈Z2

(1 + |n|2)−1/2 |n · u⊥n |2 ≤ (π + 1)∑n∈Z2

|n|(1 + |n|2)1/2

∞∑n3=−∞

|un × n|2

≤ (π + 1)∑n∈Z3

|un × n|2 .

Adding this to (5.10) yields

‖γtu‖2

H−1/2per (Div,Q2)

≤ (c+ π + 1)∑n∈Z3

[|n× un|2 + |un|2

]= (c+ π + 1) ‖u‖2

Hper(curl,Q3) .

Now we construct a bounded extension operator. Let f ∈ H−1/2per (Q2) be given by f(x) =∑

n∈Z2 fn ein·x with Fourier coefficients fn ∈ C3 for which the third components vanish. We

define un ∈ C3 for n ∈ Z3 by

un =δn

1 + |n|2(fn × an)

where

an =

1

|n|2[|n|2 e− n3 n

], n 6= (0, 0) ,

e , n = (0, 0) ,

and δn is given in (5.2) of the proof of Lemma 5.4; that is, δn =[∑∞

j=−∞1

1+|n|2+j2

]−1

. We

easily derive the following properties of an:

• e · an = 1 for all n ∈ Z3,

• n · an = 0 for all n with n 6= (0, 0), and n · an = n3 for all n = n3e,

• |an|2 = |n|2|n|2 for all n with n 6= (0, 0) and |an| = 1 for all n = n3e.


First we show that (ηtf)(x) = u(x) =∑

n∈Z3 un ein·x defines a right inverse of γt. This follows

from∞∑

n3=−∞

(e× un) = δn

∞∑n3=−∞

1

1 + |n|2︸︷︷︸= 1

e× (fn × an) = (e · an) fn = fn .

Furthermore, for n 6= (0, 0),

∞∑n3=−∞

|un|2 ≤ |fn|2δ2n

∞∑n3=−∞

|an|2

(1 + |n|2)2= |fn|2 δ2

n

1

|n|2∞∑

n3=−∞

|n|2

(1 + |n|2)2(5.11)

≤ |fn|2δn|n|2

[δn

∞∑n3=−∞

1

1 + |n|2

]︸︷︷︸

= 1

= |fn|2δn|n|2

.

Now we use δn ≤ 2π

(1 + |n|2)1/2 from estimate (5.1) and arrive at

∞∑n3=−∞

|un|2 ≤2

π

[(1 + |n|2)−1/2 |fn|2

] 1 + |n|2

|n|2≤ 4

π

[(1 + |n|2)−1/2 |fn|2

].

This holds for n 6= (0, 0). For n = (0, 0) we conclude

∞∑n3=−∞

|un|2 ≤ |f0|2δ20

∞∑n3=−∞

1

(1 + n23)2≤ c |f0|2 .

Summing this with respect to n yields∑n∈Z

|un|2 ≤ c′∑n∈Z2

(1 + |n|2)−1/2 |fn|2 ≤ c′ ‖f‖2H−1/2(Div,Q2) .

Finally, we estimate the norm of curlu(x) = i∑

n∈Z3(n× un) ein·x. We have for n 6= (0, 0):

n×un =δn

1 + |n|2n×(fn×an) =

δn1 + |n|2

[(n ·an) fn−(n ·fn) an

]= − δn

1 + |n|2(n ·fn) an ,

and thus

∞∑n3=−∞

|n× un|2 = |n · fn|2 δ2n

∞∑n3=−∞

|an|2

(1 + |n|2)2= |n · fn|2 δ2

n

1

|n|2∞∑

n3=−∞

|n|2

(1 + |n|2)2.

This expression appeared already in (5.11) above for |fn|2 instead of |n · fn|2. For n = (0, 0)we observe that n× un = δ0

1+n23n3 f0 and thus

∞∑n3=−∞

|n× un|2 = δ20 |f0|2

∞∑n3=−∞

n23

(1 + n23)2≤ c |f0|2 .


Therefore, we arrive at∑n∈Z

|n× un|2 ≤ c′∑n∈Z2

(1 + |n|2)−1/2 |n · fn|2 ≤ c′ ‖f‖2H−1/2(Div,Q2) .

This ends the proof of part (a). We leave the proof of part (b) to the reader. 2

Theorem 5.15The dual space H

−1/2per (Div, Q2)∗ of H

−1/2per (Div, Q2) can be identified with H

−1/2per (Curl, Q2).

The isomorphism from H−1/2per (Curl, Q2) onto H

−1/2per (Div, Q2)∗ is given by the bounded exten-

sion of g 7→ `g where

〈`g, f〉∗ :=

∫Q2

f(x)·g(x) dx , f ∈ L2t (Q2)∩H−1/2

per (Div, Q2) , g ∈ L2t (Q2)∩H−1/2

per (Curl, Q2) ,

and 〈·, ·〉∗ denotes the dual form. In particular, we have∣∣∣∣∫Q2

f(x) · g(x) dx

∣∣∣∣ ≤ 8π2 ‖f‖H−1/2per (Div,Q2)

‖g‖H−1/2per (Curl,Q2)

(5.12)

for all f ∈ L2t (Q2) ∩H−1/2

per (Div, Q2) and g ∈ L2t (Q2) ∩H−1/2

per (Curl, Q2).

The same statement holds for H−1/2per (Div, Q2) and H

−1/2per (Curl, Q2) interchanged.

Proof: First we show that the mapping g 7→ `g is well-defined and bounded. Let f(x) =∑|m|≤N fm exp(im·x) and g(x) =

∑|m|≤N gm exp(im·x) be trigonometric polynomials. Then∣∣∣∣∫

Q2

f(x) · g(x) dx

∣∣∣∣ = (2π)2

∣∣∣∣∣∣∑|m|≤N

fm · gm

∣∣∣∣∣∣ ≤ (2π)2∑|m|≤N

|fm · gm| .

Using the normalization m = m/|m| we write fm in the form (provided m 6= 0)

fm =1

|m|[(m× fm)×m + (m · fm) m

]and thus

fm · gm =1

|m|[(

(m× fm)×m)· gm + (m · fm) (m · gm)

]=

1

|m|[(m× fm) · (m× gm) + (m · fm) (m · gm)

].

We estimate this by

|fm · gm| ≤1

|m||fm| |m× gm| +

1

|m||m · fm| |gm|

≤ 2√1 + |m|2

[|fm| |m× gm| + |m · fm| |gm|

]≤ 2√

1 + |m|2√|fm|2 + |m · fm|2

√|gm|2 + |m× gm|2


where we used the Cauchy-Schwarz inequality in the last step. This estimate holds also form = 0. Summation and application of the Cauchy-Schwarz inequality again yields∑

|m|≤N

|fm · gm| ≤ 2 ‖f‖H−1/2per (Div,Q2)

‖g‖H−1/2per (Curl,Q2)

.

This yields boundedness of `g for every g ∈ L2t (Q2) ∩H−1/2

per (Curl, Q2) and

‖`g‖H−1/2per (Div,Q2)∗

≤ 8π2‖g‖H−1/2per (Curl,Q2)

.

This shows also boundedness of g 7→ `g from H−1/2per (Curl, Q2) into H

−1/2per (Div, Q2)∗.

It remains to show surjectivity of this mapping g 7→ `g. Let ` ∈ H−1/2per (Div, Q2)∗ be given.

Define gm ∈ C3 by

gm :=[〈`, e1 exp(im·)〉 , 〈`, e2 exp(im·)〉 , 〈`, e2 exp(im·)〉

]>,

where ej denotes the jth coordinate unit vector in C3. We show that g =∑

m gm exp(im·)is in H

−1/2per (Curl, Q2) and ` = `g. Set gN(x) =

∑|m|≤N gm exp(im · x) for N ∈ N and let

f(x) =∑|m|≤N fm exp(im · x) a be any trigonometric polynomial. Then

〈`, f〉∗ =∑|m|≤N

3∑j=1

f jm〈`, ej exp(im·)〉 =∑|m|≤N

fm · gm = 〈`gN , f〉∗ . (5.13)

We choose fm := 1√(1+|m|2)1/2

[gm +m× (gm ×m)

]. Then

〈`, f〉 =∑|m|≤N

fm · gm =∑|m|≤N

1√(1 + |m|2)1/2

[|gm|2 + |m× gm|2

]= ‖gN‖2

H−1/2per (Curl,Q2)

,

and compute ‖f‖H−1/2per (Div,Q2)

. First, we have

|fm|2 =1

1 + |m|2[|gm|2+|m×(gm×m)|2+2|gm×m|2

]=

1

1 + |m|2[|gm|2+(|m|2+2) |gm×m|2

]and second,

|m · fm|2 =|m · gm|2

1 + |m|2=

1

1 + |m|2[|m|2|gm|2 − |gm ×m|2

].

Therefore,

‖f‖2

H−1/2per (Div,Q2)

=∑|m|≤N

1

(1 + |m|2)3/2

[|gm|2 + (|m|2 + 2) |gm ×m|2 + |m|2|gm|2 − |gm ×m|2

]=

∑|m|≤N

1

(1 + |m|2)1/2

[|gm|2 + |gm ×m|2

]= ‖gN‖2


.


We use that ` is bounded and obtain

‖gN‖2


= 〈`, f〉 ≤ ‖`‖H−1/2per (Div,Q2)∗

‖f‖H−1/2per (Div,Q2)

= ‖`‖H−1/2per (Div,Q2)∗

‖gN‖H−1/2per (Curl,Q2)

and thus‖gN‖H−1/2

per (Curl,Q2)≤ ‖`‖

H−1/2per (Div,Q2)∗

.

For N →∞ this proves that g ∈ H−1/2per (Curl, Q2) and, by (5.13), that ` = `g. 2

Analogously to Corollary 5.5 we prove the following implication of Lemma 5.14.

Corollary 5.16 Let H0(curl, Q+3 ) =

v ∈ H(curl, Q+

3 ) : supp v ⊂ Q+3 ∪ (Q2 × 0)

. Then

there exists c > 0 such that

‖γtv‖H−1/2per (Div,Q2)

≤ c ‖v‖H(curl,Q+3 ) , (5.14a)

‖γTv‖H−1/2per (Curl,Q2)

≤ c ‖v‖H(curl,Q+3 ) (5.14b)

for all v ∈ H0(curl, Q+3 ).

Proof: As in the proof of Corollary 5.16 we extend v ∈ H0,∂Ω(curl, Q+3 ) by reflection into

Q3, that is, define

v(x) =

v(x1, x2, x3) , x ∈ Q3, x3 ∈ (0, π) ,v(x1, x2,−x3) , x ∈ Q3, x3 ∈ (−π, 0) .

Then v ∈ H(curl, Q3) with support in Q3. Therefore, v ∈ H0(curl, Q3) and, by periodicextension, v ∈ Hper(curl, Q3). By Lemma 5.14 there exists c > 0 such that∥∥e× v(·, 0)

∥∥H−1/2per (Div,Q2)

=∥∥e× v(·, 0)


≤ c ‖v‖Hper(curl,Q3) .

Now we use that fact that the norms in H(curl, Q3) and Hper(curl, Q3) are equivalent onH0(curl, Q3). Therefore, there exists c > 0 such that∥∥e× v(·, 0)


≤ c1 ‖v‖H(curl,Q3) = c1

√2 ‖v‖H(curl,Q+

3 ) ,

which proves the estimate for e × v(·, 0). The estimate for(e × v(·, 0)

)× e follows by the

same arguments. 2

It is the aim to extend the definition of the trace spaces H−1/2per (Div, Q2) and H

−1/2per (Curl, Q2)

to H−1/2per (Div, ∂D) and H

−1/2per (Curl, ∂D), respectively, by the local coordinate system of

Definition 5.1. The trace operators map into the tangential plane of the boundary ∂D.Therefore, H

−1/2per (Div, ∂D) and H

−1/2per (Curl, ∂D) are spaces of tangential vector fields. The

transformation u 7→ φj(u Ψj)|Q+3

which we used in Definition 5.6 is not adequate in our

case because it does not map vector fields of H(curl, D) into vector fields of H(curl, Q+3 ).


We observe that gradients ∇ϕ are special elements of H(curl, D) for any ϕ ∈ H1(D). Theysatisfy

∇x(ϕ Ψj) = (Ψ′j)>(∇yϕ) Ψj

where Ψ′j ∈ R3×3 denotes again the Jacobian; that is,

[Ψ′j(x)

]k`

=∂(Ψj)k(x)

∂x`, k, ` ∈ 1, 2, 3 .

This will provide the correct transformation as we will see below. First we define the sub-spaces of L2(∂D)3 and L2(Q2)3 of tangential vector fields by

L2t (∂D) :=

f ∈ L2(∂D)3 : ν(y) · f(y) = 0 almost everywhere on ∂D

,

L2t (Q2) :=

g ∈ L2(Q2)3 : g3(y) = 0 almost everywhere on Q2

,

respectively. We note that the tangent plane at y = Ψj(x) is spanned by the vectors∂Ψj(x)/∂x1 and ∂Ψj(x)/∂x2. We recall that the normal vector at y = Ψj(x) for x ∈ Q2 isgiven by

ν(y) =∂Ψj(x)

∂x1

× ∂Ψj(x)

∂x2

, y = Ψj(x) ∈ ∂D ∩ Uj .

Lemma 5.17 Let Uj, ξj : j = 1, . . . ,m be a local coordinate system of ∂D. For u ∈H1(D) ∩ C(D) set

vj(x) = Ψ′j(x)>u(Ψj(x)

), x ∈ Q+

3 , j = 1, . . . , . (5.15)

where Ψ′j(x)> ∈ R3×3 is the adjoint of the Jacobian at x ∈ Q2. Furthermore, define Fj(x) ∈R3×3 as

Fj(x) :=

[∂Ψj(x)

∂x1

∣∣∣∣ ∂Ψj(x)

∂x2

∣∣∣∣ ∂Ψj(x)

∂x1

× ∂Ψj(x)

∂x2

], x ∈ Q2 . (5.16)

Then, with e = (0, 0, 1)> ∈ R3,

(a) vj ∈ H(curl, Q+3 ) and

curl vj(x) =(Ψ′j(x)

)−1curly u(y)|y=Ψj(x) , x ∈ Q+

3 .

(b) (ν(y)× u(y)

)∣∣y=Ψj(x)

= Fj(x)(e× vj(x, 0)

), x ∈ Q2 .

(c) (ν(y)× u(y)

)× ν(y)

∣∣y=Ψj(x)

= Fj(x)−>(e× vj(x, 0)

)× e , x ∈ Q2 .

Proof: First we note that for any regular matrix M = [a|b|c] ∈ R3×3 with column vectorsa, b, c, the inverse is given by

M−1 = [a|b|c]−1 =1

(a× b) · c[b× c|c× a|a× b]> =

1

detM[b× c|c× a|a× b]> .


Applying this to [a|b|c] = Ψ′j(x) yields

[Ψ′j(x)

]−1=

[∂Ψj(x)

∂x2

× ∂Ψj(x)

∂x3

∣∣∣∣∣ ∂Ψj(x)

∂x3

× ∂Ψj(x)

∂x1

∣∣∣∣∣ ∂Ψj(x)

∂x1

× ∂Ψj(x)

∂x2

]>.

In the following proof we drop the index j for abbreviation; that is, write Ψ for Ψj. Thescalar functions Ψk for k = 1, 2, 3 denote now the components of the vector function Ψ.

(a) We note that the columns of the matrix Ψ′(x)> ∈ R3×3 are just the gradients ∇Ψk(x),k = 1, 2, 3. Therefore,

v(x) =3∑

k=1

uk(Ψ(x)

)∇Ψk(x) ,

and

curl v(x) =3∑

k=1

∇uk(Ψ(x)

)×∇Ψk(x)

=3∑

k=1

∇[uk(Ψ(x)

)]×∇Ψk(x) =

3∑k=1

[Ψ′(x)>∇yuk(y)

∣∣y=Ψ(x)

]×∇Ψk(x)

=3∑

k=1

3∑`=1

∂uk(y)

∂y`

∣∣∣∣y=Ψ(x)

[∇Ψ`(x)×∇Ψk(x)

]=

(curlu(y)

)1

[∇Ψ2(x)×∇Ψ3(x)

]+(curlu(y)

)2

[∇Ψ3(x)×∇Ψ1(x)

]+(curlu(y)

)3

[∇Ψ1(x)×∇Ψ2(x)

]=

[∇Ψ2(x)×∇Ψ3(x) | ∇Ψ3(x)×∇Ψ1(x) | ∇Ψ1(x)×∇Ψ2(x)

]curlu(y)

∣∣y=Ψ(x)

=(Ψ′j(x)

)−1curly u(y)|y=Ψj(x) .

(b) We write ∂jΨ for ∂Ψ/∂xj, j = 1, 2, 3, in the following and drop the argument x ory = Ψ(x). From the definition of v we have v = (u · ∂1Ψ , u · ∂2Ψ , u · ∂3Ψ)> and thuse× v = (−u · ∂2Ψ , u · ∂1Ψ , 0)>. Furthermore,

F (e× v) = −(u · ∂2Ψ) ∂1Ψ + (u · ∂1Ψ) ∂2Ψ = (∂1Ψ× ∂2Ψ)× u = α (ν × u) .

(c) We have just seen that α (ν × u) = −v2 ∂1Ψ + v1 ∂2Ψ, thus

α2 (ν × u)× ν = −v2 ∂1Ψ× (∂1Ψ× ∂2Ψ) + v1 ∂2Ψ× (∂1Ψ× ∂2Ψ)

= −v2

[(∂1Ψ · ∂2Ψ) ∂1Ψ− |∂1Ψ|2 ∂2Ψ

]+ v1

[|∂2Ψ|2 ∂1Ψ− (∂1Ψ · ∂2Ψ) ∂2Ψ

]= F

|∂2Ψ|2 −∂1Ψ · ∂2Ψ 0−∂1Ψ · ∂2Ψ |∂1Ψ|2 0

0 0 1

︸︷︷︸

= α2(F>F )−1 = α2F−1F−>

v1

v2

0

= α2 F−>(e× v)× e .



Now we are able to define the spaces H−1/2(Div, ∂D) and H−1/2(Curl, ∂D) and prove thetrace theorems for H(curl, D).

Definition 5.18 Let D ⊂ R3 be piecewise C1 with corresponding local coordinate system(Uj, ξj) and Ψj and their restrictions Ψj : Q2 → Uj ∩ ∂D. Furthermore, let φj : j =1, . . . ,m, be a partition of unity corresponding to the sets Uj and Fj(x) be given by (5.16.

Then we define the spaces H−1/2(Div, ∂D) and H−1/2(Curl, ∂D) as the completion off ∈ L2

t (∂D) : f tj ∈ H−1/2per (Div, Q2)m

and

f ∈ L2

t (∂D) : fTj ∈ H−1/2per (Curl, Q2)m

,

respectively, with respect to the norms

‖f‖H−1/2(Div,∂D) =

[m∑j=1

‖f tj‖2

H−1/2per (Div,Q2)

]1/2

,

‖f‖H−1/2(Curl,∂D) =

[m∑j=1

‖fTj ‖2


]1/2

,

where

f tj (x) :=√φj(Ψj(x)

)F−1j (x)f

(Ψj(x)

), (5.17a)

fTj (x) :=√φj(Ψj(x)

)F>j (x)f

(Ψj(x)

). (5.17b)

Again, combining the following trace theorem and Theorem 5.20 below one shows exactlyas in Corollary 5.10 that H−1/2(Div, ∂D) does not depend on the choice of the coordinatesystem.

Theorem 5.19 The trace operators γt : H(curl, D) → H−1/2(Div, ∂D), u 7→ ν × u|∂D andγT : H(curl, D) → H−1/2(Curl, ∂D), u 7→ (ν × u|∂D) × ν are well-defined and bounded andhave bounded right inverses ηt : H−1/2(Div, ∂D)→ H(curl, D) and ηT : H−1/2(Curl, ∂D)→H(curl, D), respectively.

Proof: Boundedness of the trace operators are seen directly from the Definition 5.18,Lemma 5.17, and Corollary 5.16. Indeed, by definition

‖ν × u‖H−1/2(Div,∂D) =

[m∑j=1

‖f tj‖2

H−1/2per (Div,Q2)

]1/2

with

f tj (x) =√

Φj(y)F−1j (x)

(ν(y)× u(y)

), x ∈ Q2 , y = Ψj(x) .


By Lemma 5.17 we rewrite this as

f tj = e× vj(·, 0) with vj(x) =√

Φj(y) Ψ′j(x)>u(y)∣∣y=Ψj(x)

.

Then vj ∈ H(curl, Q+3 ), and we estimate the norm with Corollary 5.16 as

‖f tj‖H−1/2per (Div,Q2)

≤ c ‖vj‖H(curl,Q+3 ) ≤ c′‖u‖H(curl,D) .

Summing these terms yields boundedness of γt. For γT one argues in the same way.

It remains to construct an extension operator ηt : H−1/2(Div, ∂D) → H(curl, D). For

f ∈ H−1/2(Div, ∂D) we define f tj ∈ H−1/2per (Div, Q2) by (5.17a). By Lemma 5.14 there exist

vj ∈ Hper(curl, Q3) such that e × vj = f tj and the mappings f tj 7→ vj are bounded forj = 1, . . . ,m. We define

uj(y) := Ψ′j(x)−>vj(x)∣∣x=Ψ−1

j (y), y ∈ Uj ∩D .

We extend uj by zero into all of R3 and set u =∑m

j=1

√φj uj. Then ν × u = f on ∂D.

Indeed, for fixed y ∈ ∂D let J =j ∈ 1, . . . ,m : y ∈ Uj

. Then u(y) =

∑j∈J uj(y). Let

j ∈ J ; that is, y ∈ Uj, and x = Ψ−1j (y) ∈ Q2 × 0. From the definition of uj we observe

that vj = (Ψ′j)>(uj Ψj) and thus by Lemma 5.17

ν(y)× uj(y) = Fj(x)(e× vj(x)

)= Fj(x)f tj =

√φj(y) f(y) .

Summation with respect yields ν × u = f . Furthermore, all of the operations in this con-structions of u are bounded. This proves the theorem. 2

We note that the space H0(curl, D) has been defined as the closure of C∞0 (D)3 in H(curl, D),see Definition 4.16. Analogously to Theorem 5.9 it can be characterized as the nullspace ofthe trace operator as the following theorem shows.

Theorem 5.20 The space H0(curl, D) is the null space ker γt of the trace operator γt; thatis, u ∈ H0(curl, D) if, and only if, γtu = 0 on ∂D. The same holds for ηT ; that is, u ∈H0(curl, D) if, and only if, γTu = 0 on ∂D.

Proof: The inclusion H0(curl, D) ⊂ ker γt follows again immediately from the definition ofH0(curl, D) as the closure of C∞(D)3. For the reverse inclusion we can almost copy the proofof Theorem 4.8. The transformation into the parameter space, however, has to be differentaccording to the transformation of the curl. We sketch the formulas. Using again a localcoordinate system Uj, ξj : j = 1, . . . ,m of ∂D and partition of unity φj : j = 1, . . . ,mon ∂D with respect to Uj : j = 1, . . . ,m we define vj(x) :=

√Φj(y) Ψ′j(x)>u(y)

∣∣y=Ψj(x)

for j = 1, . . . ,m on Q+3 . From part (b) of Lemma 5.17) we conclude that γtvj = e × vj

vanishes on Q2 × 0. Now we extend vj by zero to a function vj ∈ H0(curl(Q3), shift thethird component and proceed as in the proof of Theorem 4.8. 2

The following theorem extends Theorem 5.15.


Theorem 5.21 (a) For every g ∈ L2t (∂D) ∩H−1/2(Curl, ∂D) the linear form `g, defined by

〈`g, f〉∗ :=

∫∂D

f(x) · g(x) ds , f ∈ L2t (∂D) ∩H−1/2(Div, ∂D) ,

has a bounded extension to `g ∈ H−1/2(Div, ∂D)∗; that is, the dual space of H−1/2(Div, ∂D).Furthermore, the mapping g 7→ `g from L2

t (∂D) ∩ H−1/2(Curl, ∂D) into H−1/2(Div, ∂D)∗

has an extension to an isomorphism from H−1/2(Curl, ∂D) onto H−1/2(Div, ∂D)∗. In thissense H−1/2(Curl, ∂D) coincides with the dual H−1/2(Div, ∂D)∗ of H−1/2(Div, ∂D). Wewrite 〈g, f〉∂D for 〈`g, f〉∗. In particular,

〈g, f〉∂D =

∫∂D

f(x) · g(x) ds

for all f ∈ L2t (∂D) ∩H−1/2(Div, ∂D) and g ∈ L2

t (∂D) ∩H−1/2(Curl, ∂D).The same statement holds for H−1/2(Div, ∂D) and H−1/2(Curl, ∂D) interchanged.

(b) Green’s formula can be formulated as

〈γTu, γtv〉∂D =

∫∫D

[u · curl v − v · curlu

]dx for all u, v ∈ H(curl, D) . (5.18)

Proof: (a) Let Uj, Ψj : j = 1, . . . ,m be a local coordinate system with restrictionsΨj : Q2 → Uj ∩ ∂D, and φj : j = 1, . . . ,m be a partition of unity on ∂D corresponding tothe sets Uj.

Let g ∈ L2t (∂D) ∩H−1/2(Curl, ∂D) and f ∈ L2

t (∂D) ∩H−1/2(Div, ∂D). Then

〈`g, f〉∂D =

∫∂D

f(x) · g(x) ds =m∑j=1

∫∂D∩Uj

Φj f(x) · g(x) ds =m∑j=1

∫Q2

fj · gj dx ,

where

fj(x) =√

Φj(y)F−1j (x)f(y)|y=Ψ(x) , gj(x) =

√Φj(y)F>j (x)g(y)|y=Ψ(x) .

From (5.12) we conclude that

∣∣〈`g, f〉∂D∣∣ ≤ 8π2

m∑j=1

‖fj‖H−1/2per (Div,Q2)

‖gj‖H−1/2per (Curl,Q2)

≤ 8π2

[m∑j=1

‖fj‖2

H−1/2per (Div,Q2)

]1/2 [ m∑j=1

‖gj‖2


]1/2

= 8π2 ‖f‖H−1/2(Div,∂D) ‖g‖H−1/2(Curl,∂D) .

Therefore, for fixed g ∈ L2t (∂D) ∩ H−1/2(Curl, ∂D) we can extend `g to a bounded linear

functional on H−1/2(Div, ∂D) with

‖`g‖H−1/2(Div,∂D)∗ ≤ 8π2 ‖g‖H−1/2(Curl,∂D) for all g ∈ L2t (∂D) ∩H−1/2(Curl, ∂D) .


Therefore, the mapping g 7→ `g is bounded as well and can be extended to a bounded operatorfrom H−1/2(Curl, ∂D) into H−1/2(Div, ∂D)∗. The proof of injectivity and surjectivity willbe moved to the end of the proof.

(b) Let un, vn ∈ C1(D) such that un → u and vn → v in H(curl, D). Then Green’s formulaholds in the classical form∫∫

D

[un · curl vn − vn · curlun

]dx =

∫∂D

un · (ν × vn) ds = 〈γTun, γtvn〉∂D .

We let n tend to infinity and observe that the left hand side tends to∫∫

D

[u·curl v−v·curlu

]dx

and the right hand side to 〈γTu, γtv〉∂D by part (a). This proves (5.18).Now we show injectivity if the operator g 7→ `g. Let `g = 0 for some g ∈ H−1/2(Curl, ∂D).Choose u ∈ H(curl, D) with γTu = g. From (5.18) we conclude that

0 = 〈`g, γtψ〉∗ = 〈γTu, γtψ〉∂D =

∫∫D

[u · curlψ − ψ · curlu

]dx for all ψ ∈ H(curl, D) .

Taking ψ ∈ H0(curl, D) we note from the variational definition of the curl that curlu ∈H(curl, D) and curl2 u = −u in D. Therefore, we can set ψ = curlu which shows that0 =

∫∫D

[u · curl2 u− | curlu|2

]dx = −

∫∫D

[|u|2 + | curlu|2

]dx; that is, u = 0.

It remains to prove surjectivity of this operator. Let ` ∈ H−1/2(Div, ∂D)∗. By the Theoremof Riesz there exists a unique u ∈ H(curl, D) with∫

D

[curlu · curlψ + u · ψ

]dx = 〈`, γtψ〉∗ for all ψ ∈ H(curl, D) .

Taking again ψ ∈ H0(curl, D) we note that curlu ∈ H(curl, D) and curl2 u = −u in D. ByGreen’s formula (5.18) applied to curlu instead of u we have that

〈`, γtψ〉∗ =

∫D

[curlu · curlψ + u · ψ

]dx = 〈γT curlu, γtψ〉∂D .

Therefore, g := γT curlu ∈ H(Curl, ∂D) yields `g = `.The proof for H−1/2(Div, ∂D) and H−1/2(Curl, ∂D) interchanged follows the same lines. 2

We note the following application.

Lemma 5.22 Let Q3 ⊂ R3 such that D ⊂ Q3 and let u1 ∈ H(curl, D) and u2 ∈ H(curl, Q3\D) such that γtu1 = −γtu2 or γTu1 = γTu2 on ∂D. Then the field

u(x) =

u1(x) , x ∈ D ,u2(x) , x ∈ Q3 \D ,

is in H(curl, Q3).

Proof: Let γTu1 = γTu2 and ψ ∈ H0(curl, Q3) arbitrary. Using (5.18) in D for u1 and inQ3 \D for u2 yields

〈γTu1, γtψ|−〉∂D =

∫∫D

[u1 · curlψ − ψ · curlu1

]dx ,

〈γTu2, γtψ|+〉∂D =

∫∫Q3\D

[u2 · curlψ − ψ · curlu2

]dx ,


where γtψ|− and γtψ|+ denotes the trace from D and Q3 \D on ∂D, respectively. We notethat γtψ|− = −γtψ|+ because of the orientation of ν. Now we set v = curlu1 in D andv = curlu2 in Q3 \D and add both equations. Then v ∈ L2(Q3)3 and∫∫

Q3

[u · curlψ − ψ · v

]dx = 0 for all ψ ∈ H0(curl, Q3) .

This is the variational form of v = curlu and proves the lemma for the case that γTu1 = γTu2.In the case γtu1 = γtu2 one uses (5.18) with u and v interchanged. Here we note thatγTψ|− = γTψ|+. 2

5.1.3 The Case of a Sphere

We consider now the special case where D is a ball of radius R centered at the origin. Wecan use separation of variables with respect to polar coordinates r > 0 and x ∈ S2 to expandfunctions into a series of spherical harmonics. We begin again with spaces of scalar functions.

Theorem 5.23 Let D = B(0, R) be the ball of radius R centered at the origin. For u ∈H1(D) let umn (r) ∈ C be the expansion coefficients of x 7→ u(rx) with respect to the sphericalharmonics; that is,

umn (r) =

∫S2

u(r, x)Y −mn (x) ds(x) , 0 ≤ r < R , |m| ≤ n , n = 0, 1, 2, . . . (5.19)

Then

uN(r, x) :=N∑n=0

n∑m=−n

umn (r)Y mn (x) , 0 ≤ r < R , x ∈ S2 ,

converges to u in H1(D) as N tends to infinity.

Proof: First we note that, using ∇ = ∂∂rx+ 1

rGrad ,

∇uN(r, x) =N∑n=0

n∑m=−n

d

drumn (r)Y m

n (x) x+

√n(n+ 1)

r

Grad Y mn (x)√

n(n+ 1), 0 ≤ r < R , x ∈ S2 ,

and thus by the orthonormalty of the systemsY mn : |m| ≤ n, n = 0, 1, 2, . . .

and

1√n(n+1)

Grad Y mn : |m| ≤ n, n = 0, 1, 2, . . .

‖uM − uN‖2H1(D) =

M∑n=N+1

n∑m=−n

∫ R

0

[|umn (r)|2

(1 +

n(n+ 1)

r2

)+

∣∣∣∣dumndr (r)

∣∣∣∣2]r2 dr (5.20)

for M > N . Bessel’s inequality applied to (5.19) and its derivative with respect to r yieldsconvergence of

∞∑n=0

n∑m=−n

|umn (r)|2 and∞∑n=0

n∑m=−n


∣∣∣∣2 .


Furthermore, using (??) and partial integration in the form of (??) yields√n(n+ 1)umn (r) = − 1√

n(n+ 1)

∫S2

u(r, x) Div Grad Y −mn (x) ds(x)

=

∫S2

Grad u(r, x) · Grad Y −mn (x)√n(n+ 1)

ds(x)

which are the expansion coefficients of Grad u(r, ·) with respect to1√

n(n+1)Grad Y m

n : |m| ≤ n, n = 0, 1, 2, . . .

. Bessel’s inequality yields convergence of

∞∑n=0

n∑m=−n

n(n+ 1) |umn (r)|2 .

which shows convergence of (5.20) as N,M →∞. We note that

‖u‖2H1(D) =

∞∑n=0

n∑m=−n

∫ R

0

[|umn (r)|2

(1 +

n(n+ 1)

r2

)+


∣∣∣∣2]r2 dr . (5.21)

2

In this case of D being a ball the space H1/2(∂D) can be characterized by a proper decay ofthe expansion coeficients with respect to the spherical harmonics.

Theorem 5.24 Let D = B(0, R) be the ball of radius R centered at the origin. For f ∈L2(∂D) let fmn ∈ C be the expansion coefficients with respect to the spherical harmonics; thatis,

fmn =

∫S2

f(Rx)Y −mn (x) ds(x) , |m| ≤ n , n = 0, 1, 2, . . .

Then f ∈ H1/2(∂D) if, and only if,

‖f‖ :=

[∞∑n=0

n∑m=−n

√1 + n(n+ 1) |fmn |2

]1/2

< ∞ . (5.22)

Proof: For the moment let X be the completion of C∞(∂D) with respect to the normof (5.22). Note that C∞(∂D) is also dense in H1/2(∂D) (see Corollary 5.10 for a proof).Therefore, as in the proof of Corollary 5.10 it suffices to show that the norm ‖ · ‖ of (5.22)is equivalent to the norm of the factor space H1(D)/H1

0 (D). This is assured if we can proveTheorems 5.8 and 5.9 for X instead of H1/2(∂D). We assume without loss of generalty thatR = 1.To prove boundedness of the trace operator write u ∈ H1(D) in polar coordinates as

u(rx) =∞∑n=0

n∑m=−n

umn (r)Y mn (x)


with coefficients umn (r) ∈ C and thus

(γu)(x) =∞∑n=0

n∑m=−n

umn (1)Y mn (x) .

We estimate∣∣umn (1)

∣∣2, using r3 ≤ r2 for r ∈ [0, 1], the inequality of Cauchy-Schwartz, and2ab ≤ a2 + b2, as

∣∣umn (1)∣∣2 =

∫ 1

0

d

dr

[r3|umn (r)|2 dr

=

∫ 1

0

[3r2|umn (r)|2 + 2r3 Re

(umn (r)

d

drumn (r)

)]dr

≤ 3

∫ 1

0

|umn (r)|2 r2 dr + 2

√∫ 1

0

|umn (r)|2 r2 dr

√∫ 1

0

∣∣∣∣ ddrumn (r)

∣∣∣∣2 r2 dr

≤ 3

∫ 1

0

|umn (r)|2 r2 dr +√

1 + n(n+ 1)

∫ 1

0

|umn (r)|2 r2 dr

+1√

1 + n(n+ 1)

∫ 1

0


∣∣∣∣2 r2 dr .

Multiplying with√

1 + n(n+ 1) yields

√1 + n(n+ 1)

∣∣umn (1)∣∣2 ≤ 4

(1 + n(n+ 1)

) ∫ 1

0

|umn (r)|2 r2 dr +

∫ 1

0


∣∣∣∣2 r2 dr ,

and thus

‖γu‖2H1/2(∂D) =

∞∑n=0

n∑m=−n

√1 + n(n+ 1) |umn (1)|2 ≤ 4 ‖u‖2

H1(D)

by (5.21). This proves boundedness of γ from H1(D) into X. A right inverse of f is givenby

(ηf)(rx) =∞∑n=0

n∑m=−n

fmn rn Y mn (x) .

It suffices to prove boundedness. From (5.20) for unm(r) = fmn rm we conclude that

‖ηf‖2H1(D) =

∞∑n=0

n∑m=−n

|fmn |2(

1

2n+ 3+n(n+ 1) + n2

2n+ 1

)

≤ c∞∑n=0

n∑m=−n

√1 + n(n+ 1)|fmn |2 = c‖f‖2 .

This proves the trace theorem for H1(D) and X.To show the analogue of Theorem 5.9 it suffices again to show that the nullspace of γ is


contained in H10 (D). Let u ∈ H1(D) with γu = 0; that is, umn (1) = 0 for all m,n. Without

loss of generality we assume that u is real valued. Let uN be the truncated sum; that is,

uN(rx) =N∑n=0

n∑m=−n

umn (r)Y mn (x) .

Then uN → u in H1(D) as N tends to infinity. Furthermore, let ψ ∈ C(R) such thatψ(t) = 0 for |t| ≤ 1 and ψ(t) = t for |t| ≥ 2. Define vN,ε ∈ H1(D) by

vN,ε(rx) =N∑n=0

n∑m=−n

ε ψ

(1

εumn (r)

)Y mn (x) .

Now we can argue as we did already several times. By the theorem of dominated convergencewe conclude that vN,ε → uN in H1(D) as ε tends to zero. Also, we can approximate vN,ε bysome v ∈ C∞0 (D) because vN,ε vanishes in some neigborhood of S2. Altogether this showsthat u can be approximated arbitrarily well in C∞0 (D). This ends the proof. 2

We finish this section by considering the special case where D = B(0, R) is the ball of radiusR. The following theorems correspond to Theorems 5.23 and 5.24.

Theorem 5.25 Let D = B(0, R) be the ball of radius R centered at the origin. For u ∈H(curl, D) let umn (r), vmn (r), wmn (r) ∈ C be the expansion coefficients of x 7→ u(rx) withrespect to the spherical vector harmonics; that is,

umn (r) =

∫S2

u(r, x) · x Y −mn (x) ds(x) , 0 ≤ r < R , (5.23a)

vmn (r) =1√

n(n+ 1)

∫S2

u(r, x) ·Grad Y −mn (x) ds(x) , 0 ≤ r < R , (5.23b)

wmn (r) =1√

n(n+ 1)

∫S2

u(r, x) ·(x×Grad Y −mn (x)

)ds(x) , 0 ≤ r < R , (5.23c)

for |m| ≤ n and n = 0, 1, 2, . . .. Then

uN(r, x) :=N∑n=0

n∑m=−n

[umn (r)Y m

n (x) x +

+vmn (r)√n(n+ 1)

Grad Y mn (x) +

wmn (r)√n(n+ 1)

(x×Grad Y m

n (x))],

0 ≤ r < R, x ∈ S2, converges to u in H(curl, D) as N tends to infinity.


Proof: We compute curluN and use the formulas

curl[umn (r)Y m

n (x) x]

= −umn (r)

r

(x×Grad Y m

n (x)),

curl[vmn (r) Grad Y m

n (x)]

= −1

r

(r vmn (r)

)′ (x×Grad Y m

n (x)),

curl[wmn (r)

(x×Grad Y m

n (x))]

= −1

r

(r wmn (r)

)′Grad Y m

n (x) − wmn (r)n(n+ 1)

rY mn (x) x .

### Hier Herleitung:

curl[umn (r)Y m

n (x) x]

= ∇(umn (r)Y (x)

)× x =

1

rumn (r) Grad Y m

n (x)× x ,

curl[vmn (r) Grad Y m

n (x)]

= curl[r vmn (r)∇Y m

n (x)]

=1

r∇(r vmn (r)

)×Grad Y m

n (x)

=1

r

(r vmn (r)

)′x×Grad Y m

n (x) ,

curl[wmn (r)

(x×Grad Y m

n (x))]

= curl[r wmn (r)

(x×∇Y m

n (x))]

=(r vmn (r)

)′x×

(x×∇Y m

n (x))

+ r vmn (r) curl(x×∇Y mn (x)

= −1

r

(r vmn (r)

)′Grad Y m

n (x) + r vmn (r) curl(x×∇Y mn (x) ,

and

x×∇Y mn (x) =

1

r

∂Y mn

∂θφ − 1

r sin θ

∂Y mn

∂φθ ,

curl(x×∇Y mn (x) =

1

r sin θ

[∂

∂θ

(sin θ

r

∂Y mn

∂θ

)+

1

r sin θ

∂2Y mn

∂φ2

]x

=1

r2

[1

sin θ

∂

∂θ

(sin θ

∂Y mn

∂θ

)+

1

sin2 θ

∂2Y mn

∂φ2

]︸︷︷︸

= Div Grad Ymn (x) = −n(n+1)Ymn (x)

x

= −n(n+ 1)

r2Y mn (x) x .

###

Therefore,

curluN(r, x) =N∑n=0

n∑m=−n

1

r

[−umn (r)

(x×Grad Y m

n (x))

+(r vmn (r)

)′ x×Grad Y mn (x)√

n(n+ 1)

+(r wmn (r)

)′ Grad Y mn (x)√

n(n+ 1)−√n(n+ 1)wmn (r)Y m

n (x) x

]


and

‖ curluN‖2L2(D) =

N∑n=0

n∑m=−n

∫ 1

0

[∣∣∣n(n+ 1)umn (r) +(r vmn (r)

)′∣∣∣2+∣∣(r wmn (r)

)′∣∣2 + n(n+ 1) |wmn (r)|2]dr .

On the other hand, the expansions coefficients of curlu(r, ·) are given by (see Exercise ??)

∫S2

curlu(r, x) · x Y −mn (x) ds(x) = −√n(n+ 1)

rvmn (r) ,∫

S2

curlu(r, x) · Grad Y −mn (x)√n(n+ 1)

ds(x) = −1

r

(rwmn (r)

)′,

∫S2

curlu(r, x) · x×Grad Y −mn (x)√n(n+ 1)

ds(x) = −√n(n+ 1)

rumn (r) +

1

r

(rvmn (r)

)′.

### Hier Herleitung:

∫S2

curlu(r, x) · x Y −mn (x) ds(x) =1

r

π∫0

2π∫0

1

sin θ

[∂

∂θ(sin θ uφ)− ∂uθ

∂φ

]Y −mn sin θ dφ dθ

=1

r

π∫0

2π∫0

[∂

∂θ(sin θ uφ)− ∂uθ

∂φ

]Y −mn dφ dθ

= −1

r

π∫0

2π∫0

[sin θ uφ

∂Y −mn

∂θ− uθ

∂Y −mn

∂φ

]dφ dθ

= −1

r

π∫0

2π∫0

[uφ

∂Y −mn

∂θ− uθ

1

sin θ

∂Y −mn

∂φ

]sin θ dφ dθ

= −1

r

∫S2

u · (x×Grad Y −mn ) ds = −√n(n+ 1)

rvmn (r) ,

202 5 BIE FOR LIPSCHITZ DOMAINS∫S2

curlu(r, x) ·Grad Y −mn (x) ds(x)

=1

r

π∫0

2π∫0

[1

sin θ

∂ur∂φ− ∂

∂r(r uφ)

]∂Y −mn

∂θ+

[∂

∂r(r uθ)−

∂ur∂θ

]1

sin θ

∂Y −mn

∂φ

sin θ dφ dθ

=1

r

∫ π

0

∫ 2π

0

[1

sin θ

∂

∂r(r uθ)

∂Y −mn

∂φ− ∂

∂r(r uφ)

∂Y −mn

∂θ

]sin θ dφ dθ

=1

r

d

dr

[r

∫ π

0

∫ 2π

0

1

sin θuθ∂Y −mn

∂φsin θ dφ dθ

]− 1

r

d

dr

[r

∫ π

0

∫ 2π

0

uφ∂Y −mn

∂θsin θ dφ dθ

]

= −1

r

d

dr

[r

∫S2

u(r, x) ·(x×Grad Y −mn (x)

)ds(x)

]= −

√n(n+ 1)

r

(r wmn (r)

)′,

∫S2

curlu(r, x) ·(x×Grad Y −mn (x)

)ds(x)

=1

r

π∫0

2π∫0

[1

sin θ

∂ur∂φ− ∂

∂r(r uφ)

] [− 1

sin θ

∂Y −mn

∂φ

]+

[∂

∂r(r uθ)−

∂ur∂θ

]∂Y −mn

∂θ

sin θ dφ dθ

= −1

r

π∫0

2π∫0

1

sin2 θ

∂ur∂φ

∂Y −mn

∂φ+∂ur∂θ

∂Y −mn

∂θ

sin θ dφ dθ

+1

r

d

dr

r π∫0

2π∫0

uφ

1

sin θ

∂Y −mn

∂φ+ uθ

∂Y −mn

∂θ

sin θ dφ dθ

= −1

r

∫S2

Grad (u · x) ·Grad Y −mn ds +1

r

d

dr

[r

∫S2

u ·Grad Y −mn ds

]=

1

r

∫S2

(u · x) Div Grad Y −mn ds +1

r

d

dr

[∫S2

u ·Grad Y −mn ds

]= −n(n+ 1)

r

∫S2

(u · x)Y −mn ds +1

r

d

dr

[r

∫S2

u ·Grad Y −mn ds

]

= −n(n+ 1)

rumn (r) +

√n(n+ 1)

r

(r vmn (r)

)′.

###


Bessels inequality applied to u(r, ·) and curlu(r, ·) yields

N∑n=0

n∑m=−n

[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]≤ ‖u(r, ·)‖2

L2(S2) ,

N∑n=0

n∑m=−n

1

r2

[(n(n+ 1) |vmn |2 +

∣∣(rwmn (r))′∣∣2+

+

∣∣∣∣∣√n(n+ 1)

rumn (r)− 1

r

(rvmn (r)

)′∣∣∣∣∣2 ≤ ‖ curlu(r, ·)‖2

L2(S2) .

Multiplication by r2 and integration with respect to r yields

‖uN‖2L2(D) + ‖ curluN‖2

L2(D) ≤ ‖u‖2L2(D) + ‖ curlu‖2

L2(D)

which proves convergence of (uN) in H(curl, D). The limit has to be u. 2

Theorem 5.26 Let D = B(0, R) be the ball of radius R centered at the origin. For f ∈L2t (∂D) let amn , b

mn ∈ C be the expansion coefficients with respect to the spherical vector-

harmonics; that is,

amn =1√

n(n+ 1)

∫S2

f(Rx) ·Grad Y −mn (x) ds(x) ,

bmn =1√

n(n+ 1)

∫S2

f(Rx) · x×(Grad Y −mn (x)

)ds(x) ,

|m| ≤ n and n = 0, 1, 2, . . .. Then f ∈ H−1/2(Div, ∂D) or f ∈ H−1/2(Curl, ∂D) if, and onlyif,

‖f‖2D :=

∞∑n=0

n∑m=−n

[(1 + n(n+ 1)

)1/2 |amn |2 +(1 + n(n+ 1)

)−1/2 |bmn |2 < ∞ , (5.24)

or

‖f‖2C :=

∞∑n=0

n∑m=−n

[(1 + n(n+ 1)

)−1/2 |amn |2 +(1 + n(n+ 1)

)+1/2 |bmn |2 < ∞ , (5.25)

respectively.

Proof: LetX be the completion of C∞t (∂D) with respect to the norm of (5.24). The assertionfollows as in Corollary 5.10 (see also Theorem 5.24) once we have proven Theorems 5.19 and5.20. First we compute ‖u‖H(curl,D). For u of the form

u(r, x) =∞∑n=0

n∑m=−n

[umn (r)Y m

n (x) x +

+ vmn (r)Grad Y m

n (x)√n(n+ 1)

+ wmn (r)x×Grad Y m

n (x)√n(n+ 1)

],


we have seen in the previous theorem that

‖u‖2L2(D) =

∞∑n=0

n∑m=−n

∫ 1

0

[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]r2 dr , (5.26a)


∞∑n=0

n∑m=−n

∫ 1

0

[∣∣∣√n(n+ 1) |umn (r)|2 −(r vmn (r)

)′∣∣∣2+∣∣∣(r wmn (r)

)′∣∣∣2 + n(n+ 1) |wmn (r)|2]dr . (5.26b)

Now we observe that

x× u(1, x) =∞∑n=0

n∑m=−n

[vmn (1)

x×Grad Y mn (x)√

n(n+ 1)− wmn (1)

Grad Y mn (x)√

n(n+ 1)

].

Therefore, we have to estimate∣∣vmn (1)

∣∣2 and∣∣wmn (1)

∣∣2. With 2|a||b| ≤ a2 + b2 we conclude

∣∣vmn (1)∣∣2 =

∫ 1

0

(r∣∣r vmn (r)

∣∣2)′ dr= 2 Re

∫ 1

0

(r vmn (r)

)′vmn (r) r2 dr +

∫ 1

0

∣∣vmn (r)∣∣2 r2 dr

= 2 Re

∫ 1

0

[(r vmn (r)

)′ −√n(n+ 1)umn (r)]vmn (r) r2 dr

+ 2√n(n+ 1) Re

∫ 1

0

umn (r) vmn (r) r2 dr +

∫ 1

0

∣∣vmn (r)∣∣2r2 dr

≤∫ 1

0

∣∣(r vmn (r))′ −√n(n+ 1)umn (r)

∣∣2 dr +

∫ 1

0

∣∣vmn (r)∣∣2r4 dr

+√n(n+ 1)

∫ 1

0

∣∣umn (r)∣∣2r2 dr + (1 +

√n(n+ 1))

∫ 1

0

∣∣vmn (r)∣∣2r2 dr .

We devide by√

1 + n(n+ 1), observe that r4 ≤ r2 for r ∈ [0, 1] and sum. Comparing thiswith (5.26a), (5.26b) yields

∞∑n=0

n∑m=−n

(1 + n(n+ 1)

)−1/2 ∣∣vmn (1)∣∣2 ≤ 2‖u‖2

H(curl,D) .

5.2. SURFACE POTENTIALS 205

For wmn (1) we argue analogously:

∣∣wmn (1)∣∣2 =

∫ 1

0

(r∣∣r wmn (r)

∣∣2)′ dr= 2 Re

∫ 1

0

(r wmn (r)

)′vmn (r) r2 dr +

∫ 1

0

∣∣wmn (r)∣∣2 r2 dr

≤ 1√1 + n(n+ 1)

∫ 1

0

∣∣(r wmn (r))′∣∣2 dr +

√1 + n(n+ 1)

∫ 1

0

∣∣wmn (r)∣∣2r4 dr

+

∫ 1

0

∣∣wmn (r)∣∣2r2 dr ,

and thus∞∑n=0

n∑m=−n

(1 + n(n+ 1)

)1/2 ∣∣wmn (1)∣∣2 ≤ 2‖u‖2

H(curl,D) .

This proves boundedness of the trace operator H(curl, D)→ X. A right inverse is given by

(ηf)(r, x) =∞∑n=0

n∑m=−n

[bmn

Grad Y mn (x)√

n(n+ 1)− amn

x×Grad Y mn (x)√

n(n+ 1)

]rn ,

where anm, bmn are the Fourier coefficients of f . Boundedness follows from (5.26a), (5.26b)for umn (r) = 0, vmn (r) = bmn r

n, and wmn (r) = −amn rn. Also, one shows as in the proof ofTheorem 5.24 that H0(curl, D) is the kernel of the trace operator H(curl, D) → X. For γTone argues analogously. 2

5.2 Surface Potentials

It is the aim of this section to study the mapping properties of the single- and double layerpotentials. First we recall – and rename – the notion of the traces. In this section D isalways a bounded regular domain with boundary ∂D which separates the interior D fromthe exterior R3\D. We fix the normal vector ν(x) (which exists for almost all points x ∈ ∂Dby the differentiability assumption on the parametrization) and let it direct into the exteriorof D. Then γu|± is the trace of u from the exterior (+) and interior (−), respectively. Thetraces γ∂νu|± for (variational) solutions u of the Helmholtz equation are now defined by

⟨γ∂νu|−, ψ

⟩∂D

=

∫∫D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) , u ∈ HD ,

⟨γ∂νu|+, ψ

⟩∂D

= −∫∫

R3\D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) , u ∈ HR3\D ,


where ψ ∈ H1(D) or ψ ∈ H1(R3 \ D), respectively, are extensions of ψ (with compactsupport in the latter case). Here,

HD =

u ∈ H1(D) :

∫D

[∇u · ∇ψ − k2uψ


0 (D)

,

HR3\D =

u ∈ H1

loc(R3 \D) :

∫R3\D

[∇u · ∇ψ − k2uψ


0 (R3 \D)

,

compare with (5.6). By 〈·, ·〉∂D we denote the dual form in⟨H−1/2(∂D), H1/2(∂D)

⟩. Note,

that we have changed the sign in γ∂νu|+ because the normal ν is directed into the interiorof R3 \D.

We begin with the representation theorem for solutions of the Helmholtz equation, comparewith Theorem 3.3.

Theorem 5.27 (Green’s representation theorem)

For any u ∈ HD we have the representation∫∂D

(γu)(y)∂Φ

∂ν(y)(x, y) ds(y) −

⟨γ∂νu, γΦ(x, ·)

⟩∂D

=

−u(x) , x ∈ D ,

0 , x 6∈ D .

Proof: First we note that elements of HD are smooth solutions of the Helmholtz equationin D. Fix x ∈ D. As in the classical case we apply Green’s formula (5.18) to u and Φ(x, ·)in D \B[x, ε]:

⟨γ∂νu, γΦ(x, ·)

⟩∂D−∫|y−x|=ε

Φ(x, y)∂u(y)

∂νds(y)

=

∫∫D\B[x,ε]

[∇u · ∇yΦ(x, ·)− k2uΦ(x, ·)

]dy

=

∫∫D\B[x,ε]

[∇u · ∇yΦ(x, ·) + u∆yΦ(x, ·)

]dy

= −∫|y−x|=ε

u(y)∂Φ(x, y)

∂ν(y)ds(y) +

∫∂D

(γu)(y)∂Φ(x, y)

∂ν(y)ds(y) ;

that is,

⟨γ∂νu,Φ(x, ·)

⟩∂D−∫∂D

(γu)(y)∂Φ(x, y)

∂ν(y)ds(y)

=

∫|y−x|=ε

[Φ(x, y)

∂u(y)

∂ν− u(y)

∂Φ(x, y)

∂ν(y)−]ds(y) = u(x)

where we applied Theorem 3.3 in the last step. For x /∈ D we argue in the same way. 2


In the following we drop the symbol γ; that is, write v for γv. If ∂D is the interface of twodomains then we write v|± and γ∂νv|± to indicate the trace from the interior (−) or exterior(+) of D, compare with the remark at the beginning of this section.

Let now u and v be the single- and double layer potentials with densities ϕ ∈ H−1/2(∂D)and ϕ ∈ H1/2(∂D), respectively; that is,

(Sϕ)(x) =⟨ϕ,Φ(x, ·)

⟩∂D, x /∈ ∂D , ϕ ∈ H−1/2(∂D) , (5.27a)

(Dϕ)(x) =

∫∂D

ϕ(y)∂Φ

∂ν(y)(x, y) ds(y) , x /∈ ∂D , ϕ ∈ H1/2(∂D) , (5.27b)

see (3.4a), (3.4b).

It is the aim to prove that S and D are bounded maps into H1(D) and H1(B \D) for anyball B containing D in its interior.

From the jump conditions of Theorems 3.11 and 3.15 we note that for smooth domains Dand smooth densities ϕ the single layer u solves the transmission problem

∆u+ k2u = 0 in R3 \ ∂D ,

u|+ = u|− on ∂D ,∂u

∂ν

∣∣∣∣−− ∂u

∂ν

∣∣∣∣+

= ϕ on ∂D ,

and u satisfies also the radiation condition (3.2). We consider a related problem in an openball B which contains D in its interior and replace the radiation condition by the boundarycondition ∂u/∂ν − iku = 0 on ∂B. A variational form of this problem is studied in thefollowing theorem.

Lemma 5.28 Let k ∈ C \ 0 with Im k ≥ 0. For every ϕ ∈ H−1/2(∂D) there exists aunique solution v ∈ H1(B) such that∫∫

B

[∇v · ∇ψ − k2v ψ

]dx − ik

∫∂B

v ψ ds = 〈ϕ, ψ〉∂D for all ψ ∈ H1(B) . (5.28)

Furthermore, the operator ϕ 7→ v is bounded from H−1/2(∂D) into H1(B).

Proof: We write (5.28) in the form

(v, ψ)H1(B) − a(v, ψ) = 〈ϕ, ψ〉∂D for all ψ ∈ H1(B) ,

where a denotes the sequilinear form

a(v, ψ) = (k2 + 1)

∫∫B

v ψ dx + ik

∫∂B

v ψ ds , v, ψ ∈ H1(B) .

The mapping ` : ψ 7→ 〈ϕ, ψ〉∂D is a bounded linear functional on H1(B) because∣∣〈ϕ, ψ〉∂D∣∣ ≤ c ‖ϕ‖H−1/2(∂D)‖ψ‖H1/2(∂D) ≤ c′‖ϕ‖H−1/2(∂D)‖ψ‖H1(D)

≤ c′‖ϕ‖H−1/2(∂D)‖ψ‖H1(B) .


Also, ‖`‖ ≤ c′‖ϕ‖H−1/2(∂D). The boundedness of the sequilinear form a is shown as follows.∣∣a(v, ψ)∣∣ ≤ (k2 + 1) ‖v‖L2(B)‖ψ‖L2(B) + k ‖v‖L2(∂B)‖ψ‖L2(∂B) . (5.29)

Using the boundedness of the trace operator from H1(B) into L2(∂B) we conclude that a isbounded. The Theorem of Riesz assures the existence of r ∈ H1(B) and a bounded operatorK from H1(B) into itself such that 〈ϕ, ψ〉∂D = (r, ψ)H1(B) and a(v, ψ) = (Kv, ψ)H1(B) forall v, ψ ∈ H1(B). Furthermore, ‖r‖H1(B) = ‖`‖ ≤ c′‖ϕ‖H−1/2(∂D). Then we write (5.28) asv −Kv = r in H1(B) and show that K is compact. From (5.29) for ψ = Kv we note that

‖Kv‖2H1(B) = (Kv,Kv)H1(B) = a(v,Kv)

≤ (k2 + 1) ‖v‖L2(B)‖Kv‖L2(B) + k ‖v‖L2(∂B)‖Kv‖L2(∂B)

≤ c[‖v‖L2(B) + ‖v‖L2(∂B)

]‖Kv‖H1(B) ,

thus

‖Kv‖2H1(B) ≤ c

[‖v‖L2(B) + ‖v‖L2(∂B)

].

Let now vj 0 weakly in H1(B). From the boundedness of the trace operator γ andthe compactness of the imbeddings H1(B) in L2(B) and H1/2(∂B) in L2(∂B) we concludethat the right hand side of the previous estimate converges to zero. Therefore, ‖Kvj‖H1(B)

converges to zero which proves compactness of K.Therefore, we can apply Fredholm’s alternative to (5.28). To show existence and boundednessof the solution operator r 7→ v it suffices to prove uniqueness. Therefore, let v ∈ H1(B) bea solution of (5.28) for ϕ = 0. Substituting ψ = kv into (5.28) yields∫∫

B

[|∇v|2 − k|k|2|v|2

]dx − i|k|2

∫∂B

|v|2 ds = 0 .

Taking the imaginary part and noting that Im k ≥ 0 yields v = 0. 2

We are now able to prove the following basic properties of the single layer potential.

Theorem 5.29 Let u := Sϕ in R3 \ ∂D be the single layer potential with denity ϕ ∈H−1/2(∂D), defined by u(x) = (Sϕ)(x) =

⟨ϕ,Φ(x, ·)

⟩∂D

, x ∈ R3 \ ∂D. Then:

(a) u ∈ C∞(R3 \ ∂D) and satisfies the Helmholtz equation ∆u + k2u = 0 in R3 \ ∂D andSommerfeld‘s radiation condition (3.2) for |x| → ∞.

(b) Let Q3 be open and bounded such that D ⊂ Q3. The operator S is well-defined andbounded from H−1/2(∂D) into H1(Q3).

(c) u|D ∈ HD and uR3\D ∈ HR3\D and γ∂νu|− − γ∂νu|+ = ϕ.

(d) The trace S = γS on ∂D is bounded from H−1/2(∂D) into H1/2(∂D).


Proof: (a) This follows from the properties of the fundamental solution Φ once we haveshown that (Au)(x) =

⟨ϕ, (AxΦ)(x, ·)

⟩∂D

in every ball B such that B ∩ ∂D = ∅ for everydifferential operator A. Using an argument by induction it suffices to show this for A = ∂/∂xjand every function Φ ∈ C∞

(B × (R3 \ B)

)instead of Φ. Set f(x) =

⟨ϕ, Φ(x, ·)

⟩∂D

forx ∈ B. Choose an open neighborhood U of ∂D such that d := dist(U,B) > 0. ThenΦj := ∂Φ(x, ·)/∂xj ∈ H1(U) and∣∣f(x+ he(j))− f(x)− h

⟨ϕ, Φj

⟩∂D

∣∣=

∣∣⟨ϕ, Φ(x+ he(j), ·)− Φ(x, ·)− h Φj

⟩∂D

∣∣≤ c ‖ϕ‖H−1/2(∂D)‖Φ(x+ he(j), ·)− Φ(x, ·)− h Φj‖H1/2(∂D)

≤ c′ ‖ϕ‖H−1/2(∂D)‖Φ(x+ he(j), ·)− Φ(x, ·)− h Φj‖H1(U) .

The differentiabilty of Φ yields that

‖Φ(x+ he(j), ·)− Φ(x, ·)− h Φj‖H1(U) = O(h2)

which proves the assertion.(b) Let B be a ball which contains Q3 in its interior. We prove the following representationof Sϕ for ϕ ∈ H−1/2(∂D).

Sϕ = v + w in Q3 , (5.30)

where v ∈ H1(B) is the solution of (5.28), and w ∈ H1(Q3) is given by

w(x) =

∫∂B

v(y)

[∂Φ


]ds(y) , x ∈ Q3 .

Indeed, we choose ψ ∈ H10 (D) and ψ ∈ H1

0 (B \ D), respectively, and extend them byzero. Equation (5.28) implies that the restrictions satisfy v|D ∈ HD and v|B\D ∈ HB\D.Application of Green’s representation theorem (Theorem 5.27) for x ∈ D yields

v(x) = −∫∂D

v(y)|−∂Φ

∂ν(y)(x, y) ds(y) + 〈γ∂νv|−,Φ(x, ·)〉∂D , x ∈ D ,

0 =

∫∂D

v(y)|+∂Φ

∂ν(y)(x, y) ds(y) − 〈γ∂νv|+,Φ(x, ·)〉∂D

−∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) + 〈γ∂νv,Φ(x, ·)〉∂B , x ∈ D .

Adding both equation yields

v(x) =⟨γ∂νv|− − γ∂νv|+ , Φ(x, ·)

⟩∂D− w(x) , x ∈ D , (5.31)

where

w(x) :=

∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) −

⟨γ∂νv,Φ(x, ·)

⟩∂B

=

∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) −

∫∫B\D

[∇v · ∇Φx − k2vΦx

]dx

=

∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) − ik

∫∂B

v(y) Φx(y) ds = w(x) , x ∈ Q3 ,


where Φx ∈ H1(B) is chosen such that Φx = Φ(x, ·) on ∂B and Φx = 0 on D. Then Φx isan extension of Φ(·, x) ∈ H1/2(∂B).For fixed x ∈ D we choose Φx ∈ H1(B) such that Φx = Φ(x, ·) on ∂D and Φx = 0 on ∂B.This Φx ∈ H1

0 (B) is an extension of Φ(x, ·) in D and in B \D. Now we recall the definitionof the traces γ∂νv|± and rewrite the first term of (5.31) as⟨γ∂νv|− − γ∂νv|+ , Φ(x, ·)

⟩∂D

=⟨γ∂νv|− − γ∂νv|+ , Φx

⟩∂D

=

∫∫B

[∇v · ∇Φx − k2v Φx

]dx

= 〈ϕ, Φx〉∂D =⟨ϕ,Φ(x, ·)

⟩∂D

= (Sϕ)(x)

by (5.28) for ψ = Φx. Thus v = Sϕ− w in D. For x ∈ B \D we argue exactly in the sameway which proves (5.30).Now we observe that v|∂B → w|Q3 is bounded from L2(∂B) into H1(Q3). Furthermore,ϕ→ v is bounded from H−1/2(∂D) into H1(B). Combing this with the boundedness of thetrace operator v 7→ v|∂B yields boundedness of S from H−1/2(∂D) into H1(Q3).(c) From the representation (5.30) it suffices to show that γ∂νf |− − γ∂νf |+ = 0 on ∂D. Butthis follows directly from the fact that w is a classical solution of the Helmholtz equation inB and thus 〈γ∂νf |− − γ∂νf |+, ψ〉∂D =

∫∫Q3

[∇w · ∇ψ − k2w ψ

]dx = 0 for all extensions ψ of

ψ with compact support.(d) This follows from (b) and the boundedness of the trace operator γ. 2

Corollary 5.30 For every ϕ ∈ H−1/2(∂D) the single layer potential u = Sϕ is the onlyvariational solution of the transmission problem

∆u+ k2u = 0 in R3 \ ∂D , u|− = u|+ on ∂D ,∂u

∂ν

∣∣∣∣−− ∂u

∂ν

∣∣∣∣+

= ϕ on ∂D ;

and u satisfies the Sommerfeld radiation condition (3.2). That is, u ∈ H1loc(R3) is the unique

solution of∫∫R3

[∇u·∇ψ−k2uψ

]dx = 〈ϕ, γψ〉∂D for all ψ ∈ H1(R3) with compact support . (5.32)

Proof: The previous theorem implies that u|D ∈ HD and u <R3\D∈ HR3\D and u satisfies

the radiation condition. Let now ψ ∈ H1(R3) with compact support which is contained insome ball B. The definitions of γ∂ν |± yield

〈γ∂ν |−, γψ〉∂D =

∫∫D

[∇u · ∇ψ − k2uψ

]dx ,

−〈γ∂ν |+, γψ〉∂D =

∫∫B\D

[∇u · ∇ψ − k2uψ

]dx ,

and thus by adding the results

〈ϕ, γψ〉∂D = 〈γ∂νu|− − γ∂νu|+, ψ〉∂D

=

∫∫B

[∇u · ∇ψ − k2uψ

]dx .


This proves that u = Sϕ solves (5.32). 2

The following two properties are essential for using the boundary integral equation methodfor solving the interior exterior boundary value problems.

Theorem 5.31 Let Si be the single layer boundary operator for the special value k = i.Then Si is an isomorphism from H−1/2(∂D) onto H1/2(∂D) and S − Si is compact for anyk ∈ C with Im k ≥ 0.

Proof: To show injectivity let ϕ ∈ H−1/2(∂D) such that Siϕ = 0. Define u = Siϕ in R3\∂D.Then γu|− = γu|+ = 0. As shown before, u is a classical solution of ∆u− u = 0 in R3 \ ∂Dand satisfies the radiation condition. From the representation theorem (Theorem 3.3 fork = i) we observe that u and its derivatives decay exponentially for |x| → ∞. Applicationof Green’s theorem (Theorem??) in D and in B(0, R) \D with ψ = u yields:

0 = 〈γ∂νu|−, γu〉∂D =

∫∫D

[|∇u|2 + |u|2

]dx ,

0 = −〈γ∂νu|+, γu〉∂D =

∫∫B(0,R)\D

[|∇u|2 + |u|2

]dx +

∫|x|=R

∂u

∂νu ds .

The first equation yields u = 0 in D. In the second equation we let R tend to infinity. Thenthe integral over ∂B(0, R) tends to zero, and we obtain that also u = 0 in R3 \D. The jumpconditions yield ϕ = γ∂νu|− − γ∂νu|+ = 0.It remains to show surjectivity of Si. Let f ∈ H1/2(∂D) be given. The idea is to solve theinterior and exterior boundary value problems for ∆u − u = 0 with boundary data f andset ϕ = γ∂νu|− − γ∂νu|+. This ϕ will do the job! To carry out this idea let v1 ∈ H1(D) andv2 ∈ H1(R3 \D such that γv1|− = f and γv2|+ = f . We make the ansatz

u(x) =

v1(x)− u1(x) , x ∈ D ,v2(x)− u2(x) , x /∈ D ,

and determine u1 ∈ H10 (D) and u2 ∈ H1

0 (R3 \D such that∫∫D

[∇u1 · ∇ψ + u1 ψ

]dx =

∫∫D

[∇v1 · ∇ψ + v1 ψ

]dx for all ψ ∈ H1

0 (D) ,∫∫R3\D

[∇u2 · ∇ψ + u2 ψ

]dx =

∫∫R3\D

[∇v2 · ∇ψ + v2 ψ

]dx for all ψ ∈ H1

0 (R3 \D) .

Then u ∈ H1(R3) because u|D ∈ H1(D) and u|R3\D ∈ H1(R3 \D) and γu|− = γ(v1−u1)|− =

f = γu|+. We set ϕ := γ∂νu|− − γ∂νu|+ ∈ H−1/2(∂D). From Theorem 5.12 we concludethat, for ψ ∈ H1(R3),

〈ϕ, ψ〉∂D = 〈γ∂νu|− − γ∂νu|+ , ψ〉 =

∫∫R3

[∇u · ∇ψ + uψ

]dx .

On the other hand, this equation (for ψ with compact support) holds also for Siϕ by Corol-lary 5.30. Taking the difference yields(

Siϕ− u , ψ)H1(R3)

= 0 for all ψ ∈ H1(R3) with compact support.


2

We finish this part of the section by formulating the corresponding theorems for the doublelayer potential without proof.

Theorem 5.32 Let Q3 be open and bounded such that D ⊂ Q3. The double layer operatorD, defined in (5.27b), is well-defined and bounded from H1/2(∂D) into H1(D) and intoH1(Q3 \D).Furthermore, with u := Dϕ in R3 \ ∂D we have that u|D ∈ HD and uR3\D ∈ HR3\D and

γu|+− γu|− = ϕ and γ∂νu|−− γ∂νu|+ = 0. In particular, u ∈ C∞(R3 \ ∂D) and satisfies theHelmholtz equation ∆u+ k2u = 0 in R3 \ ∂D.The trace T = γ∂νD on ∂D is bounded from H1/2(∂D) into H−1/2(∂D).

Theorem 5.33 Let Ti be the boundary operator for the special value k = i. Then T − Ti iscompact from H1/2(∂D) into H−1/2(∂D) and Ti is coercive; that is, there exists c > 0 suchthat

〈Tiϕ, ϕ〉∂D ≥ c ‖ϕ‖2H1/2(∂D) for all ϕ ∈ H1/2(∂D) .

In particular, 〈Tiϕ, ϕ〉∂D is real valued.

We continue with the vector valued case and recall the traces γt : H(curl, D)→ H−1/2(Div, ∂D)and γT : H(curl, D) → H−1/2(Curl, ∂D). As in the scalar case we fix the direction of theunit normal vector to point into the exterior of D and distinguish in the following betweenthe traces from the exterior (+) and interior (−) by writing γtu|± and γTu|±, respectively.Due to the direction of ν we have the following forms of Green’s formula.

〈γTu|−, γtv|−〉∂D =

∫∫D


]dx for all u, v ∈ H(curl, D) ,

〈γTu|+, γtv|+〉∂D = −∫∫

R3\D


]dx for all u, v ∈ H(curl,R3 \D) ,

where 〈·, ·〉∂D denotes the dual form in⟨H−1/2(Div, ∂D), H−1/2(Curl, ∂D)

⟩. In the follow-

ing a degenerated case of this dual form appears, namely a bilinear mapping 〈〈·, ·〉〉∂D :H−1/2(Div, ∂D)×H1/2(∂D)→ C3, defined by the equation

〈〈a, ϕ〉〉∂D · z = 〈a, ϕ γT z〉∂D for all z ∈ C3 (5.33)

and a ∈ H−1/2(Div, ∂D), ϕ ∈ H1/2(∂D). For smooth tangential fields a this is exactly theintegral

∫∂Dϕ(y) a(y) ds(y). We note that ϕγT z ∈ H−1/2(Curl, ∂D) because ϕγT z = γT (ϕz)

for any extension ϕ ∈ H1(D) of ϕ and the fact that ϕz ∈ H(curl, D).For fixed a ∈ H−1/2(Div, ∂D) and ϕ ∈ H1/2(∂D) the right hand defines a linear functionalon C3. Therefore, it has a unique representation in C3 which we denote by 〈〈a, ϕ〉〉∂D ∈ C3.For this mapping we have the following form of Green’s theorem.


Lemma 5.34 For v ∈ H(curl, D) and u ∈ H1(D) we have

〈〈γtv, γu〉〉∂D =

∫∫D

[u curl v +∇u× v

]dx .

Proof: For any z ∈ C3 we have by definition

〈〈γtv, γu〉〉∂D · z = 〈γtv, γT (uz)〉∂D =

∫∫D

[u z · curl v − v · curl(uz)

]dx

=

∫∫D

[u z · curl v − v · (∇u× z)

]dx = z ·

∫∫D

[u curl v − v ×∇u

]dx .

2

Now we are able to generalize the definition of the vector potentials for any a ∈ H−1/2(Div, ∂D).We define

v(x) = curl 〈〈a,Φ(x, ·)〉〉∂D , x ∈ R3 \ ∂D , (5.34)

u(x) = curl2〈〈a,Φ(x, ·)〉〉∂D , x ∈ R3 \ ∂D , (5.35)

where we dropped the symbol γ for the scalar trace operator.

Starting point is the following version of the representation theorem (compare to the Stratton-Chu formula of Theorem 3.26).

Theorem 5.35 (Stratton-Chu formula)

For any solutions v, w ∈ H(curl, D) of curl v − iωµ0w = 0 and curlw + iωε0v = 0 in D wehave the representation

− curl〈〈γtv,Φ(x, ·)〉〉∂D +1

iωε0

curl2〈〈γtw,Φ(x, ·)〉〉∂D =

v(x) , x ∈ D ,

0 , x 6∈ D .(5.36)

Proof: First we note that v and w are smooth solutions of curl2 u − k2u = 0 in D withk2 = ω2µ0ε0. We fix z ∈ D and choose a ball Br = B(z, r) such that Br ⊂ D. We applyGreen’s formula of Lemma 5.34 in Dr = D \ B[z, r] to v and Φ(x, ·) for any x ∈ Br andobtain

〈〈γtv,Φ(x, ·)〉〉∂D −∫∂Br

(ν × v) Φ(x, ·) ds =

∫∫Dr

[Φ(x, ·) curl v +∇yΦ(x, ·)× v

]dy .

Analogously, we have for w instead of v:

〈〈γtw,Φ(x, ·)〉〉∂D −∫∂Br

(ν × w) Φ(x, ·) ds =

∫∫Dr

[Φ(x, ·) curlw +∇yΦ(x, ·)× w

]dy


and thus

Ir(x) := − curl

[〈〈γtv,Φ(x, ·)〉〉∂D −

∫∂Br

(ν × v) Φ(x, ·) ds]

+1

iωε0

curl2[〈〈γtw,Φ(x, ·)〉〉∂D −

∫∂Br

(ν × w) Φ(x, ·) ds]

= − curl

∫∫Dr

[Φ(x, ·) curl v +∇yΦ(x, ·)× v

]dy

+1

iωε0

curl2∫∫

Dr

[Φ(x, ·) curlw +∇yΦ(x, ·)× w

]dy

Now we use that ∇yΦ(x, ·)×v = − curlx(Φ(x, ·)v

)and curl v = iωµ0w and curlw = −iωε0v.

This yields

Ir(x) = curl2∫∫

Dr

Φ(x, ·) v dy − iωµ0 curl

∫∫Dr

Φ(x, ·)w dy

− 1

iωε0

curl3∫∫

Dr

Φ(x, ·)w dy − curl2∫∫

Dr

Φ(x, ·) v dy

= − iωµ0 curl

[∫∫Dr

Φ(x, ·)w dy − 1

k2curl2

∫∫Dr

Φ(x, ·)w dy]

= 0 .

Therefore,

− curl[〈〈γtv,Φ(x, ·)〉〉∂D +1

iωε0

curl2〈〈γtw,Φ(x, ·)〉〉∂D

= − curl

∫∂Br

(ν × v) Φ(x, ·) ds +1

iωε0

curl2∫∂Br

(ν × w) Φ(x, ·) ds

= v(x)

by the classical Stratton-Chu formula of Theorem 3.26. The case x /∈ D is treated in thesame way by appling Lemma 5.34 in all of D. 2

Corollary 5.36 For any variational solution v ∈ H(curl, D) of curl2 v − k2v = 0 in D wehave the representation

− curl〈〈γtv,Φ(x, ·)〉〉∂D −1

k2curl2〈〈γt curl v,Φ(x, ·)〉〉∂D =

v(x) , x ∈ D ,

0 , x 6∈ D .

Proof: We define w = 1iωµ0

curl v and observe that also w ∈ H(curl, D), and v, w satisfy

the assumptions of the previous theorem. Substituting the form of w into (5.36) yields theassertion. 2

For our proof of the boundedness of the vector potentials and the corresponding boundaryoperators we need the (unique) solvability of certain boundary value problems. As a familar


technique we have to use the Helmholtz decomposition and the fact that a certain subspace ofH(curl, D) with divergence-free vector fields is compact in L2(D)3 (compare to Theorem ??where we proved a result of this type. We need this result only for D being of the formB21 = B(0, R2) \B[0, R1] for two balls centered at the origin.

Theorem 5.37 Let B21 = B(0, R2) \ B[0, R1] for two balls with radii R2 > R1 centered atthe origin. Then the subspace

X0(B21) :=

v ∈ H(curl, B21) :

∫∫B21

v · ∇ϕdx = 0 for all ϕ ∈ H1(B21)

is compactly imbedded in L2(B21)3. We note that this is the closed subspace of H(curl, B21)consisting of divengence-free functions with vanishing normal components on the boundary.

Proof: The proof consists of two steps. In the first step we construct a bounded extensioninto the larger region B43 := B(0, R4) \ B[0, R3] where R3 = R2

1/R2 < R1 and and R4 =R2

2/R1 > R2. In the second step we multiply this extension by a suitable mollifier functionwhich defines a bounded extension operator from X0(B21) into H1

per(Q3)3 for some cube Q3

containing B43. The compact imbedding of H1per(Q3)3 into L2(Q3)3 yields the desired result.

Let u ∈ X0(B21) be given by

u(r, x) =∞∑n=0

n∑m=−n

[umn (r)Y m

n (x) x +

+vmn (r)√n(n+ 1)

Grad Y mn (x) +

wmn (r)√n(n+ 1)

(x×Grad Y m

n (x))],

for R1 < r < R2 and x ∈ S2. We construct an extension of the coefficients into the exteriorof B(0, R3) and set

umn (r) =

umn (r) , R1 < r < R2 ,

−R42

r4u(R2

2/r) , R2 < r < R4 ,

−R41

r4u(R2

1/r) , R3 < r < R1 ,

vmn (r) =

vmn (r) , R1 < r < R ,

R42

r4v(R2

2/r) , R2 < r < R4 ,

R41

r4v(R2

1/r) , R3 < r < R1 ,

wmn (r) =

wmn (r) , R1 < r < R ,

R42

r4w(R2

2/r) , R2 < r < R4 ,

R41

r4w(R2

1/r) , R3 < r < R1 ,

and define u by these coefficients. We show that the extension operator u 7→ u is well definedand bounded from X0(B21) into X0(B43).


First we consider the region B(0, R4) \ B[0, R2] and use the substitution s = R22/r. With

ds = −R22/r

2 dr we conclude that∫ R4

R2

∣∣umn (r)∣∣2r2 dr =

1

R22

∫ R2

R1

∣∣umn (s)∣∣2s4 ds ≤

∫ R2

R1

∣∣umn (s)∣∣2s2 ds .

The same formulas hold for vmn and wmn and the region B(0, R1) \B[0, R3]. The formulas forthe curl are a bit more complicated. We recall from Theorem 5.25 that ‖ curl u‖L2(B42) forB42 = B(0, R4) \B[0, R2] is given by

‖ curl u‖2L2(B42) =

∞∑n=0

n∑m=−n

∫ R4

R2

[∣∣∣n(n+ 1) umn (r) +(r vmn (r)

)′∣∣∣2+∣∣(r wmn (r)

)′∣∣2 + n(n+ 1) |wmn (r)|2]dr .

Again, with s = R22/r and ds = −R2

2/r2 dr we conclude that

R4∫R2

∣∣∣n(n+ 1) umn (r) +(r vmn (r)

)′∣∣∣2 dr= R8

2

R4∫R2

∣∣∣∣−n(n+ 1)1

r4umn (R2

2/r) +d

dr

(1

r3vmn (R2

2/r)

)∣∣∣∣2 dr= R8

2

R2∫R1

∣∣∣∣−n(n+ 1)s4

R82

umn (s)− s2

R22

d

ds

(s3

R62

vmn (s)

)∣∣∣∣2 R22

s2ds

=1

R62

R2∫R1

∣∣∣∣n(n+ 1) s3 umn (s) + sd

ds

(s3 vmn (s)

)︸︷︷︸= s2

(s vmn (s)

)′+2s2vmn (s)

∣∣∣∣2ds

=1

R62

R2∫R1

s6∣∣∣n(n+ 1)umn (s) +

(s vmn (s)

)′+ 2 vmn (s)

∣∣∣2 ds≤ 2

R2∫R1

∣∣∣n(n+ 1)umn (s) +(s vmn (s)

)′∣∣∣2 ds +8

R22

R2∫R1

∣∣vmn (s)∣∣2s2 ds .

The other terms are estimated in the same way. Therefore, the extension is bounded. Itremains to show that u ∈ X0(B43). Let ϕ ∈ H1(B43). By expanding ϕ we can assumewithout loss of generality that ϕ is of the form ϕ(r, x) = a(r)Y m

n (x). Then ∇ϕ(r, x) =

a′(r) x Y mn (x) + a(r)

rGrad Y mn(x). Because u ∈ X0(B21) and ϕ|B21 ∈ H1(B21) we conclude


that ∫∫B43

u · ∇ϕdx =

∫ R4

R3

[a′(r) umn (r) +

√n(n+ 1)

ra(r) vmn (r)

]r2 dr

=

∫ R1

R3

+

∫ R4

R2

[a′(r) umn (r) +

√n(n+ 1)

ra(r) vmn (r)

]r2 dr .

We consider again only the integral over (R2, R4). The substitution s = R22/r yields again∫ R4

R2

[a′(r) umn (r) +

√n(n+ 1)

ra(r) vmn (r)

]r2 dr

=

∫ R2

R1

[a′(s)umn (s) +

√n(n+ 1)

sa(s) vmn (s)

]s2 ds

for a(s) = a(R22/s). This term vanishes because ϕ(s, x) = a(s)Y m

n (x) is in H1(B21). Thisends the first step.Now we choose a function ψ ∈ C∞(R3) such that ψ = 1 on B21 and ψ = 0 on R3 \ B43. Wechoose a cube Q3 = (−R,R)3 such that B43 ⊂ Q3 and define the operator E from X0(B21)into H1

per(Q3)3 by the periodic extension of ψu. Then

(Eu)(x) = ψ(x)u(x) =∑n∈Z3

an ei(π/R)n·x , x ∈ Q3 ,

with expansion coefficients

an =1

(2R)3

∫∫Q3

ψ(y) u(y) e−i(π/R)n×ydy =1

(2R)3

∫∫B43

ψ(y) u(y) e−i(π/R)n·ydy , n ∈ Z3 .

From ψu ∈ H(curl, Q3) and Parseval’s equation we conclude that

(2R)3 π2

R2

∑n∈Z3

|n× an|2 =

∫∫Q3

∣∣curl(ψu)∣∣2 ≤ c‖u‖2

H(curl,B43)

for some c which depends only on ψ. Furthermore,

n · an =1

(2R)3

∫∫Q3

ψ(y)n · u(y) e−i(π/R)n·ydy

= iR

π

1

(2R)3

∫∫Q3

ψ(y) u(y) · ∇e−i(π/R)n·ydy

= iR

π

1

(2R)3

∫∫Q3

u(y) · ∇(ψ(y) e−i(π/R)n·y) dy

− iR

π

1

(2R)3

∫∫Q3

u(y) · ∇ψ(y) e−i(π/R)n·ydy

= −i Rπ

1

(2R)3

∫∫B43

u(y) · ∇ψ(y) e−i(π/R)n·ydy


because the first integral vanishes because u ∈ X0(B43) and ψ e−i(π/R)n· ∈ H1(B43). Theseare the Fourier coefficients of −i (R/π) u(y) · ∇ψ(y). Parseval’s equation yields

(2R)3∑n∈Z3

|n · an|2 =R2

π2

∫∫B43

∣∣u(y) · ∇ψ(y)∣∣2 dy ≤ R2

π2‖u‖2

L2(B43)‖∇ψ‖2L2(B43) .

Writing |n|an as |n|an = (n× an)× n+ (n · an)n with n = n/|n| yields∑n∈Z3

|n|2|an|2 ≤ 2∑n∈Z3

|n× an|2 + 2∑n∈Z3

|n · an|2 ≤ c ‖u‖2H(curl,B43)

which proves boundedness of the operator E : u 7→ ψu from X0(B21) into H1per(Q3). 2

Lemma 5.38 Let D contain the origin and let Bj, j = 1, 2, be two balls of radii Rj, centeredat the origin, such that B1 ⊂ D and D ⊂ B2. Furthermore, let η ∈ C such that Im η 6= 0 andKj : H−1/2(Div, ∂Bj) → H−1/2(Curl, ∂Bj) be the operator which multiplies the coefficientsamn , b

mn of f ∈ H−1/2(Div, ∂Bj) by 1/

(1 + n(n + 1)

). Then, for every f ∈ H−1/2(Div, ∂D)

the following boundary value problems are uniquely solvable in H(curl, D \B1) for part (a),in H(curl, B2 \D) for part (b), and in H(curl, D) for part (c).

(a) curl2 v−k2v = 0 in D\B1 , ν×v = f on ∂D , ν×curl v = −η ν×K1(ν×v) on ∂B1 ;

that is, in variational form: γtv = f on ∂D and∫∫D\B1

[curl v · curlψ − k2v · ψ

]dx = η

⟨γtψ,K1(γtv)

⟩∂B1

(5.37a)

for all ψ ∈ H(curl, D \ B1) with γtψ = 0 on ∂D. The solution operator f 7→ v is boundedfrom H−1/2(Div, ∂D) into H(curl, D \B1).

(b) curl2 v−k2v = 0 in B2 \D , ν×v = f on ∂D , ν×curl v = η ν×K2(ν×v) on ∂B2 ;

that is, in variational form: γtv = f on ∂D and∫∫B2\D


]dx = η

⟨γtψ,K2(γtv)

⟩∂B2

(5.37b)

for all ψ ∈ H(curl, B2 \ D) with γtψ = 0 on ∂D. The solution operator f 7→ v is boundedfrom H−1/2(Div, ∂D) into H(curl, B2 \D).

(c) curl2 v − k2v = 0 in B2 \ (B1 ∪ ∂D) , ν × v|− = ν × v|+ on ∂D ,

ν × curl v|+ − ν × curl v|− = f on ∂D ,

ν × curl v = −η ν ×K1(ν × v) on ∂B1 , ν × curl v = η ν ×K2(ν × v) on ∂B2 ;

that is, in variational form:∫∫B2\B1


]dx = 〈f, γTψ〉∂D + η

2∑j=1

⟨γtψ,Kj(γtv)

⟩∂Bj

(5.37c)

for all ψ ∈ H(curl, B2 \ B1). The operator T : f 7→ γtv|∂D is an isomormism fromH−1/2(Div, ∂D) onto itself.


Proof: (a) Set for abbreviation Q3 = D \ B1 and X :=v ∈ H(curl, Q3) : γtv =

0 on ∂D

. We transform the boundary value problem to a problem with homogeneousboundary condition and an inhomogeneous differential equation in the well known way: Forf ∈ H−1/2(Div, ∂D) let f = ηtf ∈ H(curl, Q3) with γtf = f on ∂D and γtf = 0 on ∂B1.The ansatz v = v + f yields the following variational equation for v ∈ X.∫∫

Q3

[curl v ·curlψ−k2v ·ψ

]dx = −η

⟨γtψ,K1(γtv)

⟩∂B1−∫∫

D\B1

[curl f ·curlψ−k2f ·ψ

]dx

(5.38)for all ψ ∈ X. We solve this by the Helmholtz decomposition; that is, we make the ansatzv = v0 + ∇p with p ∈ H1(Q3) and v0 ∈ X0 :=

v ∈ H :

∫∫Q3v · ∇ϕdx = 0 for all ϕ ∈

H1(Q3)

. Then X is the direct sum of X0 and ∇H1(Q3). Substituting ψ = ∇ϕ withϕ ∈ H1(Q3) and ψ = ψ0 ∈ X0 yields the equivalent variational equation

a(v0, p;ψ0, ϕ) := k2

∫∫Q3

∇p · ∇ϕdx +

∫∫Q3

[curl v0 · curlψ0 − k2v0 · ψ0

]dx

+ η⟨γt(ψ0 −∇ϕ), K1γt(v0 +∇p)

⟩∂B1

= −∫∫

Q3

[curl f · curlψ0 − k2f · (ψ0 −∇ϕ)

]dx ,

for all (ψ0, ϕ) ∈ X0 ×H1(Q3). We write a as

a(v0, p;ψ0, ϕ) = k2(p, ϕ)H1(Q3) + (v0, ψ0)H(curl,Q3 + b(v0, p;ψ0, ϕ)

with

b(v0, p;ψ0, ϕ) := −∫∫

Q3

[k2pϕ+ (k2 + 1)v0 ·ψ0

]dx + η

⟨γt(ψ0−∇ϕ), K1γt(v0 +∇p)

⟩∂B1

,

(v0, p), (ψ0, ϕ) ∈ X0×H1(Q3). The representation theorem of Riesz guarantees the existenceof (g0, g1) ∈ X0 ×H1(Q3) and unique bounded operators A0 and K from X0 ×H1(Q3) intoitself such that(

(g0, g1), (ψ0, ϕ))H(curl,Q3)×H1(Q3)

= −∫∫

Q3

[curl f · curlψ0 − k2f · (ψ0 −∇ϕ)

]dx ,

(A0(v0, p), (ψ0, ϕ)

)H(curl,Q3)×H1(Q3)

= k2(p, ϕ)H1(Q3) + (v0, ψ0)H(curl,Q3) and(K(v0, p), (ψ0, ϕ)

)H(curl,Q3)×H1(Q3)

= b(v0, p;ψ0, ϕ)

for all (v0, p), (ψ0, ϕ) ∈ H0 ×H1(Q3). Therefore, (5.38) can be written as

A0(v0, p) + K(v0, p) = (g0, g1) .

It is easily seen that A0 is an isomormism from X0×H1(Q3) onto itself. Using the facts3 thatX0 is compactly imbedded in L2(Q3)3 and H1(Q3) is compactly imbedded in L2(Q3) well-known arguments show that K is compact. We conclude that the variational equation (5.38)

3See Exerecise ??


for v0 ∈ X0 and all ψ ∈ X0 is (uniquely) solvalble provided uniqueness holds. Therefore, letv0 ∈ X0 be such that (5.38) holds for f = 0 and all ψ ∈ X0 – even for all ψ ∈ X because itholds obviously for ψ = ∇ϕ for ϕ ∈ H1(Q3). We set ψ = v0 and take the imaginary part.From the assumption on η we conclude that

⟨γtv0, K1(γtv0)

⟩∂B1

vanishes. The definition ofK1 yields that γtv0 = 0 on ∂B1. Therefore, we can extend v0 by zero into the interior of B1.This extension is in H0(curl, D) and satisfies∫∫

D

[curl v0 · curlψ − k2v0 · ψ

]dx = 0 for all ψ ∈ H0(curl, D) .

Therefore, curl2 v0 − k2v0 = 0 in D and v0 = 0 in B1. The unique continuation principleyields v0 = 0 in D which proves uniqueness. This ends the proof of part (a).

Part (b) is proved in exactly the same way, unique solvability of the boundary value problemof part (c) by very similar arguments. We leave the details to the reader but show surjectivityof the operator T . Let g ∈ H−1/2(Div, ∂D). Solve the boundary value problems (a) in D\B1

and (b) in B2 \D with boundary data g instead of f . This yields v ∈ H(curl, B2 \B1) withγtv = g. Define f = ν × curl v|+ − ν × curl v|−. Then f ∈ H−1/2(Div, ∂D), and v solves (c).This yields Tf = g. 2

Theorem 5.39 Let Q3 be open and bounded such that D ⊂ Q3. The operator N , de-fined by (N f)(x) = u(x) = curl2〈〈f,Φ(x, ·)〉〉∂D, x ∈ Q3, is well-defined and bounded fromH−1/2(Div, ∂D) into H(curl, Q3).Furthermore, we have that curlu|D ∈ H(curl, D) and curlu|Q3\D ∈ H(curl, Q3 \ D) and

γt curlu|− − γt curlu|+ = k2a. In particular, u ∈ C∞(R3 \ ∂D) and satisfies the equationcurl2 u − k2u = 0 in R3 \ ∂D where k2 = ω2µ0ε0. The field u satisfies the Silver-Mullerradiation (3.41); that is,

curlu× x − ik u = O(|x|−2

), |x| → ∞ , (5.39)

uniformly with respect to x = x/|x|.The trace N = γtN on ∂D is bounded from H−1/2(Div, ∂D) into itself and is the sumN = N +K of an isomorphism N from H−1/2(Div, ∂D) onto itself and a compact operatorK.

Proof: We argue similarily as in the proof of Theorem 5.29. Let, without loss of generalty,D contain the origin and let Bj, j = 1, 2, be two balls of radii Rj, centered at the origin,such that B1 ⊂ D and Q3 ⊂ B2. Furthermore, let η ∈ C such that Im η 6= 0 and Kj :H−1/2(Div, ∂Bj)→ H−1/2(Curl, ∂Bj) as in the previous lemma. For f ∈ H−1/2(Div, ∂D) letv ∈ H0(curl, B2 \ B1) be the solution of (5.37c). Then it is sufficient to prove the followingrepresentation of u = N f .

N f = k2 (v + V) in B2 \B1 , (5.40)

where

V(x) = curl〈〈γtv,Φ(x, ·)〉〉∂B1 +1

k2curl2〈〈γt curl v,Φ(x, ·)〉〉∂B1

− curl〈〈γtv,Φ(x, ·)〉〉∂B2 −1

k2curl2〈〈γt curl v,Φ(x, ·)〉〉∂B2 , x ∈ B2 \B1 .

5.3. BOUNDARY INTEGRAL EQUATION METHODS 221

To prove this we fix x ∈ D and apply the Corollary 5.36 of the Stratton-Chu formula inD \B1 and in B2 \D and obtain (note the different sign because of the orientation of ν)

v(x) = − curl〈〈γtv|−,Φ(x, ·)〉〉∂D −1

k2curl2〈〈γt curl v|−,Φ(x, ·)〉〉∂D

+ curl〈〈γtv,Φ(x, ·)〉〉∂B1 +1

k2curl2〈〈γt curl v,Φ(x, ·)〉〉∂B1 ,

0 = curl〈〈γtv|+,Φ(x, ·)〉〉∂D +1

k2curl2〈〈γt curl v|+,Φ(x, ·)〉〉∂D

− curl〈〈γtv,Φ(x, ·)〉〉∂B2 −1

k2curl2〈〈γt curl v,Φ(x, ·)〉〉∂B2 .

Adding both equations and using the transmission condition yields

v(x) =1

k2curl2〈〈f,Φ(x, ·)〉〉∂D + V(x) =

1

k2(N f)(x) + V(x) ,

which proves (5.40) for x ∈ D. For x ∈ B2 \ D we argue analogously. Taking the trace in(5.40) yields

Nf = k2 (γtv + γtV) = Nf + Kf, .

The mapping N : f 7→ k2γtv is an isomorphism from H1/2(Div, ∂D) onto itself by theprevious lemma. Furthermore, the mapping K : f 7→ k2γtV is compact in H1/2(Div, ∂D)because V is a smooth solution of curl2 u− k2u = 0 in B2 \B1. 2

5.3 Boundary Integral Equation Methods

We begin again with the scalar case and formulate the interior and exterior boundary valueproblems and assume that D ⊂ R3 is piecewise C1−smooth bounded domain D with con-nected exterior R3 \D. Furthermore, let f ∈ H1/2(∂D) be given bounday data.

Interior Dirichlet Problem: Find u ∈ H1(D) such that γu = f and ∆u+ k2u = 0 in D;that is, in variational form∫∫

D

[∇u · ∇ψ − k2uψ


0 (D) . (5.41a)

To formulate the exterior problem we define the local Sobolev space by

H1loc(R3 \D) :=

u : R3 \D → C : u|B ∈ H1(B) for all balls B

.

Exterior Dirichlet Problem: Find u ∈ H1loc(R3 \D) such that γu = f and ∆u+ k2u = 0

in R3 \D; that is, in variational form∫∫R3\D

[∇u ·∇ψ−k2uψ


0 (R3 \D) with compact support, (5.41b)

and u satisfies Sommerfeld’s radiation condition (3.2). Note that u is a smooth solution ofthe Helmholtz equation in the exterior of D.

First we consider the question of uniqueness.


Theorem 5.40 (a) There exists at most one solution of the exterior boundary value problem(5.41b).

(b) The interior boundary value problem (5.41a) has at most one solution if, and only if,the single layer boundary operator S is one-to-one. More precisely, the null space kerS of Sis given by all Neumann traces γ∂νu of solutions u ∈ H1

0 (D) of the Helmholtz equation withvanishing Dirichlet boundary data; that is,

kerS =

γ∂νu : u ∈ H1

0 (D),

∫∫D

[∇u · ∇ψ − k2uψ


0 (D)

.

Proof: (a) Let u ∈ H1loc(R3 \ D) be a solution of the exterior boundary value problem for

f = 0. Choose a ball B(0, R) which contains D in its interior and a function φ ∈ C∞(R3)such that φ = 1 on B[0, R] und φ = 0 in the exterior of B(0, R+ 1). Application of Green’sformula (5.18) in the region B(0, R + 1) \D to u and ψ = φu yields4

0 = −〈γδνu, ψ〉∂D +

∫|x|=R+1

∂u

∂νψ ds =

∫∫B(0,R+1)\D

[∇u · ∇ψ − k2uψ

]dx

=

∫∫B(0,R)\D

[|∇u|2 − k2|u|2

]dx +

∫∫R<|x|<R+1

[∇u · ∇(φu)− k2|u|2 φ

]dx

=

∫∫B(0,R)\D

[|∇u|2 − k2|u|2

]dx −

∫|x|=R

∂u

∂νu ds .

Now we can proceed exactly as in the proof of Theorem 3.23. Indeed,∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + k2|u|2

ds =

∫|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds − 2k Im

∫|x|=R

u∂u

∂rds

=

∫|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds

which converges to zero as R tends to infinity by the radiation condition. Rellich’s lemma(Lemma 3.21 or 3.22) implies that u vanishes in the (connected) exterior of D. This provespart (a).

(b) Let first u ∈ H10 (D) satisfy (5.41a). Then, by the definition of S and Green’s represen-

tation formula of Theorem 5.27

(Sγ∂νu)(x) =⟨γ∂νu,Φ(x, ·)

⟩∂D

=⟨γ∂νu,Φ(x, ·)

⟩∂D−∫∂D

(γu)(y)∂Φ


= u(x) , x ∈ D .

4Note that ψ vanishes on ∂D and outside of B(0, R+ 1).


Taking the trace yields Sγ∂νu = γSγ∂νu = γu = 0 on ∂D.

Second, let ϕ ∈ H−1/2(∂D) such that Sϕ = 0. Define u = Sϕ in R3\∂D. Then u|D ∈ H10 (D)

and γ+u = 0 on ∂D. Therefore, u|R3\D satisfies the homogeneous exterior boundary valueproblem and therefore vanishes by part (a). The junp condition of Theorem 5.29 impliesϕ = γ∂νu|− − γ∂νu|+ = γ∂νu|− which ends the proof. 2.

### Remark: finite dimension of null space, discreteness of eigenvalues (analytic Fredholmtheory) ###

The following theorems study the question of existence.

Theorem 5.41 Assume in addition to the assumptions at the beginning of this section thatk2 is not an eigenvalue of −∆ in D with respect to Dirichlet boundary conditions; that is,the only solution of the variational equation (5.41a) in H0(D) is the trivial one u = 0. Thenthere exist (unique) solutions of the exterior and the interior boundary value problems forevery f ∈ H1/2(∂D). The solutions can be represented as single layer potentials in the form

u(x) = (Sϕ)(x) = 〈ϕ,Φ(x, ·)〉∂D , x /∈ ∂D ,

where the density ϕ ∈ H−1/2(∂D) satisfies Sϕ = f .

Proof: By the mapping properties of S of Theorem 5.29 it suffices to study solvability ofthe equation Sϕ = f . Because S is a compact perturbation of the isomorphism Si (theoperator corresponding to k = i) by this Theorem 5.29 a well known – and already oftenused – result by Riesz guarantees surjectivity of this operator S provided injectivity holds.But this is assured by the previous theorem. Indeed, if Sϕ = 0 then the corresponding singlelayer potential u solves both, the exterior and the interior boundary value problems withhomogeneous boundary data f = 0. The uniqueness result implies that u vanishes in all ofR3. The jumps of the normal derivatives (Theorem 5.29 again) yields ϕ = 0. 2

Theorem 5.42 (Fredhom) Let 〈Xj, Yj〉j, j = 1, 2, be two dual systems, T : X1 → X2 a

bounded operator with bounded dual operator T ∗ : Y2 → Y1 such that T = T + K andT ∗ = T ∗ + K∗ with isomorphisms T and T ∗ and compact operators K and K∗. Then thefollowing results hold:

(a) The dimensions of the null spaces of T and T ∗ are finite and coincide; that is, dim kerT =dim ker(T ∗) <∞.

(b) The equations Tx = u, T ∗y = v are solvable for exactly those u ∈ X2 and v ∈ Y1 forwhich

〈u, ψ〉2 = 0 for all ψ ∈ ker(T ∗) ⊂ Y2 and 〈ϕ, v〉1 = 0 for all ϕ ∈ kerT ⊂ X1 .

Proof: ### Rueckfuehrung auf X1 = X2, Y1 = Y2, T = T ∗ = id und Anwendung vonSatz bei Colton/Kress ###


We apply this result to X1 = Y2 = H−1/2(∂D), X2 = Y1 = H1/2(∂D), 〈ϕ, f〉1 = 〈ϕ, f〉∂D,〈g, ψ〉2 = 〈ψ, g〉∂D, and T = T ∗ = S. Then S is indeed self adjoint (to do). Combined withthe uniqueness theorem we conclude:

Theorem 5.43 The interior and exterior boundary value problems are solvable as singlelayer potentials u = Sϕ for those f ∈ H1/2(∂D) which are orthogonal to all Neumann tracesγ∂νv ∈ H−1/2(∂D) of eigenfunctions v ∈ H1

0 (D) of −∆ in D; that is,

〈γ∂νv, f〉∂D = 0 for all v ∈ H10 (D) with ∆v + k2v = 0 in the variational sense.

Proof: ### Application of Fredholm’s theorem ###

Remark: For scattering problems by plane waves we have to solve the exterior Dirichletproblem with boundary data f(x) = − exp(ik θ · x) on ∂D. This boundary data satis-fies the orthogonality condition. Indeed, by Green’s formula (5.18) we conclude for everyeigenfunction v ∈ H1

0 (D) that

〈γ∂νv, exp(ik θ·)〉∂D =

∫∫D

[∇v(x) · ∇eik θ·x − k2v(x) eik θ·x

]dx =

∫∂D

v(x)∂

∂νeik θ·xds = 0 .

Therefore, the scattering problem for plane waves by the obstacle D can always be solvedby a single layer ansatz.

Exactly the same results hold for the interior and exterior Neumann problems when oneuses the double layer potential u = Dϕ for ϕ ∈ H1/2(∂D). The boundary operator S has tobe replaced by T : H1/2(∂D)→ H−1/2(∂D) given in Theorem 5.32.

### Genauer ausfuehren? ###

We now turn to the electromagnetic case and formulate the boundary value problem. Weassume again that D ⊂ R3 is a piecewise C1−smooth bounded domain D with connectedexterior R3 \ D. Furthermore, let f ∈ H−1/2(Div, ∂D) be given bounday data. We recallthat the trace operator γt : H(curl, D)→ H−1/2(∂D) generalizes the mapping u 7→ ν × u|∂D(and analogously for exterior domains).

Interior Boundary Value Problem: Find u ∈ H(curl, D) such that γtu = f and curl2 u−k2u = 0 in D; that is, in variational form∫∫

D

[curlu · curlψ − k2u · ψ

]dx = 0 for all ψ ∈ H0(curl, D) . (5.42a)

As in the scalar case define the local Sobolev space as

Hloc(curl,R3 \D) :=u : R3 \D → C3 : u|B ∈ H(curl, B) for all balls B

.

Exterior Boundary Value Problem: Find u ∈ Hloc(curl,R3 \D) such that γtu = f andcurl2 u− k2u = 0 in R3 \D; that is, in variational form∫∫

R3\D

[curlu ·curlψ−k2u ·ψ

]dx = 0 for all ψ ∈ H0(curl,R3 \D) with compact support,

(5.42b)


and u satisfies the Silver-Muller radiation condition (5.39). Note that u is a smooth solutionof curl2 u− k2u = 0 in the exterior of D.

The question of uniqueness and existence are treated in a very analogous way to the scalarproblems and are subject of the following theorems.

Theorem 5.44 (a) There exists at most one solution of the exterior boundary value problem(5.42b).

(b) The interior boundary value problem (5.42a) has at most one solution if, and only if, theboundary operator N is one-to-one. More precisely, the null space kerN of N is given byall traces γt curlu of solutions u ∈ H0(curl, D) of curl2 u− k2u = 0 with vanishing boundarydata γtu; that is,

kerN =

γt curlu : u ∈ H0(curl, D),

∫∫D

[curlu · curlψ − k2uψ

]dx = 0

for all ψ ∈ H0(curl, D)

.

Proof: (a) Choose again balls such that D ⊂ B(0, R) ⊂ B(0, R + 1) and a function φ ∈C∞(R3) such that φ = 1 on B[0, R] and φ = 0 outside of B(0, R+1). Green’s formula (5.18)applied to u and v = φ curlu in the region B(0, R+ 1) \D yields (note that γtv vanishes on∂D and on ∂B(0, R + 1))

0 = −〈γTu, γt curlu〉∂D +

∫|x|=R+1

u · (ν × curlu ds

=

∫∫B(0,R+1)\D


]dx

=

∫∫B(0,R)\D

[u · curl2 u− | curlu|2

]dx +

∫∫R<|x|<R+1


]dx

=

∫∫B(0,R)\D

[k2|u|2 − | curlu|2

]dx −

∫|x|=R

u · (x× curlu) ds .

The Silver-Muller radiation condition implies that∫|x|=R

| curlu× x|2 + k2|u|2

ds =

∫|x|=R

| curlu× x− iku|2 ds + 2k Im

∫|x|=R

u · (curlu× x) ds

=

∫|x|=R

| curlu× x− iku|2 ds

tends to zero as R tends to infinity. Now we proceed as in the proof of Theorem 3.34 andapply Rellich’s lemma to conclude that u vanishes in the exterior of D.

(b) Let first u ∈ H0(curl, D) with curl2 u− k2u = 0 in D. By Corollary 5.36 we conclude for


x ∈ D(Nγt curlu)(x) = curl2〈〈γt curlu,Φ(x, ·)〉〉∂D

= curl2〈〈γt curlu,Φ(x, ·)〉〉∂D + k2 curl〈〈 γtu︸︷︷︸= 0

,Φ(x, ·)〉〉∂D

= −k2u(x) .

Taking the trace yields N (γt curlu) = γtu = 0 on ∂D.

Second, let ϕ ∈ H−1/2(Div, ∂D) with Nϕ = 0. Define u by u(x) = curl2〈〈ϕ,Φ(x, ·)〉〉∂D forx ∈ R3 \ ∂D. Then u ∈ Hloc(curl,R3) solves the exterior and the interior boundary valueproblem with homogeneous boundary data f = 0. The uniqueness result for the exteriorproblem implies that u vanishes in the exterior. The jump conditions of Theorem 5.39 yieldsϕ = γt curlu|− − γt curlu|+ = γt curlu|−. This proves part (b). 2

Theorem 5.45 Assume in addition to the assumptions at the beginning of this section thatk2 is not an eigenvalue of curl2 in D with respect to the boundary condition ν × u = 0;that is, the only solution of the variational equation (5.42a) in H0(curl, D) is the trivial oneu = 0. Then there exist (unique) solutions of the exterior and the interior boundary valueproblems for every f ∈ H−1/2(Div, ∂D). The solutions can be represented as boundary layerpotentials in the form

u(x) = (Na)(x) = curl2〈〈a,Φ(x, ·)〉〉∂D , x /∈ ∂D ,

where the density a ∈ H−1/2(Div ∂D) satisfies Na = f .

Proof: This is clear from the uniqueness result of both, the interior and the exterior bound-ary value problem and the fact that N is a compact perturbation of an isomophism. 2

We want to study the equation Na = f for the case when N fails to be one-to-one. It isthe aim to apply the abstract Fredholm result of Theorem 5.42 and have to find the properdual form. The following result provides the adjoint of N .

Lemma 5.46 For a, b ∈ CαDiv(∂D) we have

〈Na, b〉∂D = 〈N b, a〉∂D .

Proof: Define u, v ∈ Hloc(curl,R3) by

u(x) = curl2〈〈a,Φ(x, ·)〉〉∂D , v(x) = curl2〈〈b,Φ(x, ·)〉〉∂D , x /∈ ∂D .

Then, by Theorem 5.39, γt curl v|−−γt curl v|+ = k2b and thus b×ν = γT curl v|−−γT curl v|+.Therefore, by Green’s theorem in the form (5.18), applied in D and in B(0, R) \D, respec-tively, and adding the results yields

〈Na, b× ν〉∂D = 〈γta, γT curl v|−〉∂D − 〈γta, γT curl v|+〉∂D

=

∫∫|x|<R

[curl v · curlu− u · curl2 v] dx −∫|x|=R

(ν × u) · curl v ds

=

∫∫|x|<R

[curl v · curlu− k2v · u] dx −∫|x|=R

(ν × u) · curl v ds .

5.4. EXERCISES 227

Changing the roles of a and b and subtracting the results yield

〈Na, b× ν〉∂D − 〈N b, a× ν〉∂D =

∫|x|=R

[(ν × v) · curlu− (ν × u) · curl v] ds .

The last term converges to zero as R tends to infinity by the radiation condition. This endsthe proof. 2

We define the bilinear form 〈·, ·〉 on H−1/2(Div, ∂D) by 〈a, b〉 = 〈a, b× ν〉∂D. Note that thisis well-defined because b × ν ∈ H−1/2(Curl, ∂D) for b ∈ H−1/2(Div, ∂D). Then the adjointN ∗ of N is given by −N because

〈Na, b〉 = 〈Na, b× ν〉∂D = 〈N b, a× ν〉∂D

= 〈ν ×N b, a〉∂D = −〈a, cNb× ν〉∂D = −〈a,N b〉 .

Theorem 5.47 The interior and exterior boundary value problems are solvable by a layerpotential u = Na for exactly those f ∈ H−1/2(Div, ∂D) which are orthogonal to all tracesγT curl v ∈ H−1/2(Curl, ∂D) of eigenfunctions v ∈ H0(D) of curl2 in D; that is,

〈γT curl v, f〉∂D = 0 for all v ∈ H0(curl, D) with curl2 v − k2v = 0 in the variational sense.

Proof: We have to discuss solvability of the equation Na = f . As mentioned above wewant to apply Theorem 5.42 and define X1 = X2 = Y1 = Y2 = H−1/2(Div, ∂D) and 〈a, b〉 =〈a, b × ν〉∂D for a, b ∈ H−1/2(Div, ∂D). The adjoint N ∗ of N is given by −N . Let nowf ∈ H−1/2(Div, ∂D) and g ∈ kerN ; that is, g = γt curl v for some v ∈ H0(curl, D) withcurl2 v−k2v = 0 inD. The solvability condition 〈f, g〉 = 0 reads as 0 = 〈f, γt curl v×ν〉∂D =〈f, γT curl v〉∂D which proves the theorem. 2

5.4 Exercises


Chapter 6

Appendix on Vector Calculus

6.1 Table of Differential Operators and Their Proper-

ties

In this section we collect the most important formulas from vector calculus.

We assume that all functions are sufficiently smooth. In cartesian coordinates:

Operator Application to function

∇ ∇u = (∂u∂x, ∂u∂x2, ∂u∂x3

)>

div = ∇· ∇ · A =3∑j=1

∂Aj∂xj

curl = ∇× ∇× A =

∂A3

∂x2− ∂A2

∂x3∂A1

∂x3− ∂A3

∂x1∂A2

∂x1− ∂A1

∂x2

∆ = div∇ = ∇ · ∇ ∆u =

∑3j=1

∂2u∂x2j

### Formellisten auch auf den Innenseiten des Covers? ###

The following formulas can be obtained from straightforward calculations.

229

230 CHAPTER 6. APPENDIX ON VECTOR CALCULUS

For x, y, z ∈ C3, λ : C3 → C und A,B : C3 → C3 we have

x · (y × z) = y · (z × x) = z · (x× y) (6.1)

x× (y × z) = (x · z)y − (x · y)z (6.2)

curl∇u = 0 (6.3)

div curlA = 0 (6.4)

curl curlA = ∇ divA − ∆A (6.5)

div(λA) = A · ∇λ + λ divA (6.6)

curl(λA) = ∇λ× A + λ curlA (6.7)

∇(A ·B) = (A · ∇)B + (B · ∇)A + A× (curlB) +B × (curlA) (6.8)

div(A×B) = B · curlA − A · curlB (6.9)

curl(A×B) = A divB − B divA + A′B − B′A , (6.10)

where in the last formula A′(x), B′(x) ∈ C3×3 are the Jacobians of A and B, respectively, atx. For completeness we add the expressions of the differential operators ∇, div, curl, and ∆in other coordinate systems. Let f : R3 → C be a scalar function and F : R3 → C3 a vectorfield.

Cylindrical Coordinates

x =

r cosϕr sinϕz

Let z = (0, 0, 1)> and r = (cosϕ, sinϕ, 0)> and ϕ = (− sinϕ, cosϕ, 0)> be the coordinateunit vectors. Let F = Frr + Fϕϕ+ Fz z. Then

∇f(r, ϕ, z) =∂f

∂rr +

1

r

∂f

∂ϕϕ +

∂f

∂zz ,

divF (r, ϕ, z) =1

r

∂(rFr)

∂r+

1

r

∂Fϕ∂ϕ

+∂Fz∂z

,

curlF (r, ϕ, z) =

(1

r

∂Fz∂ϕ− ∂Fϕ

∂z

)r +

(∂Fr∂z− ∂Fz

∂r

)ϕ +

1

r

(∂(rFϕ)

∂θ− ∂Fθ

∂ϕ

)z ,

∆f(r, ϕ, z) =1

r

∂

∂r

(r∂f

∂r

)+

1

r2

∂2f

∂ϕ2+

∂2f

∂z2.

Spherical Coordinates

x =

r sin θ cosϕr sin θ sinϕr cos θ

6.2. ELEMENTARY FACTS FROM DIFFERENTIAL GEOMETRY 231

Let r = (sin θ cosϕ, sin θ sinϕ, cos θ)> and θ = (cos θ cosϕ, cos θ sinϕ,− sin θ)> andϕ = (− sin θ sinϕ, sin θ cosϕ, 0)> be the coordinate unit vectors. Let F = Frr + Fθθ + Fϕϕ.Then

∇f(r, θ, ϕ) =∂f

∂rr +

1

r

∂f

∂θθ +

1

r sin θ

∂f

∂ϕϕ ,

divF (r, θ, ϕ) =1

r2

∂(r2Fr)

∂r+

1

r sin θ

∂(sin θ Fθ)

∂θ+

1

r sin θ

∂Fϕ∂ϕ

,

curlF (r, θ, ϕ) =1

r sin θ

(∂(sin θ Fϕ)

∂θ− ∂Fθ

∂ϕ

)r +

1

r

(1

sin θ

∂Fr∂ϕ− ∂(rFϕ)

∂r

)θ +

+1

r

(∂(rFθ)

∂r− ∂Fr

∂θ

)ϕ ,

∆f(r, θ, ϕ) =1

r2

∂

∂r

(r2∂f

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂f

∂θ

)+

1

r2 sin2 θ

∂2f

∂ϕ2.

6.2 Elementary Facts from Differential Geometry

Before we recall the basic integral identity of Gauss and Green we have to define rigourouslythe notion of domain with a Cn−boundary or Lipschitz boundary (see Evans [3]). We denoteby Bj(x, r) := y ∈ Rj : |y − x| < r and Bj[x, r] := y ∈ Rj : |y − x| ≤ r the open andclosed ball, respectively, of radius r > 0 centered at x in Rj for j = 2 or j = 3.

Definition 6.1 We call a region D ⊂ R3 to be Cn-smooth (that is, D ∈ Cn), if there existsa finite number of open cylinders Uj of the form Uj = Qjx+z(j) : x ∈ B2(0, αj)×(−βj, βj)with z(j) ∈ R3 and rotations1 Qj ∈ R3×3 and real valued functions ξj ∈ Cn(B[0, αj]) with|ξj(x1, x2)| ≤ βj/2 for all (x1, x2) ∈ B2[0, αj] such that

∂D ⊂m⋃j=1

Uj ,


,


,


.

We call D to be a Lipschitz domain if the functions ξj which describe the boundary locallyare Lipschitz continuous; that is, there exists a constant L > 0 such that |ξj(z) − ξj(y)| ≤|z − y| for all z, y ∈ B2[0, αj] and all j = 1, . . . ,m.



1+x22 < α2

j , |x3| < βj

1that is, Q>j Qj = I and detQj = 1


the standard cylinders.

Often we will work with the mappings Ψj : Cj → R3 defined by

Ψj(x) = Qj

x1

x2

ξj(x1, x2) + x3

+ z(j) , x = (x1, x2, x3)> ∈ Cj(αj, βj/2) ,


Ψj(x) = Qj

x1

x2

ξj(x1, x2)

+ z(j) , x = (x1, x2)> ∈ B2(0, αj) ,


∂D ∩ Uj =

Ψj(x) : x ∈ Cj, x3 = 0

=

Ψj(x) : x ∈ B2(0, αj),

D ∩ U ′j =

Ψj(x) : x ∈ C ′j, x3 < 0,

U ′j \D =

Ψj(x) : x ∈ C ′j, x3 > 0,

where C ′j = Cj(αj, βj/2) = B2(αj) × (−βj/2, βj/2) and U ′j = Ψj(C′j). For C1−domains D

we note that the Jacobian is given by

Ψ′j(x) = Qj

1 0 00 1 0

∂1ξj(x) ∂2ξj(x) 1

,

where ∂`ξj = ∂ξj/∂x` for ` = 1, 2. We note that its determinant is one. The tangentialvectors at y = Ψj(x) ∈ ∂D ∩ Uj are computed as

∂Ψj

∂x1

(x) = Qj

10

∂1ξj(x)

,∂Ψj

∂x2

(x) = Qj

01

∂2ξj(x)

.

They span the tangent plane at y = Ψj(x). The vector

∂Ψj

∂x1

(x)× ∂Ψj

∂x2

(x) = Qj

−∂1ξj(x)−∂2ξj(x)

1

is orthogonal to the tangent plane and is directed into the exterior of D. The correspondingunit vector

ν(y) =

∂Ψj∂x1

(x)× ∂Ψj∂x2

(x)∣∣∣∂Ψj∂x1

(x)× ∂Ψj∂x2

(x)∣∣∣ =

1√1 + |∇ξj(x)|2

Qj

−∂1ξj(x)−∂2ξj(x)

1

is called the exterior unit normal vector.


Remark 6.2 For Lipschitz domains the functions ξj are merely Lipschitz continuous. There-fore, Ψj and its inverse Ψ−1

j , given by

Ψ−1j (y) = Q>j

y1 − z(j)1

y2 − z(j)2

y3 − z(j)3 − ξj(y1, y2)

, y ∈ U ′j = Ψj(C′j) ,

are also Lipschitz continuous. A celebrated result of Rademacher [10] (see also [8] states thatevery Lipschitz continuous function ξj is differentiable at almost every point x ∈ B2(0, αj)with |∇ξj(x)| ≤ L for almost all x ∈ B2(0, αj) where L is the Lipschitz constant. Therefore,for Lipschitz domains the exterior unit normal vector ν(x) exists at almost all points x ∈ ∂D.Furthermore, u ∈ L1(U ′j) if, and only, if u Ψj ∈ L1(C ′j) and the transformation formulaholds in the form ∫

U ′j

u(y) dy =

∫C′j

u(Ψj(x)

)dx . (6.11)

Application of this result to |u|2 shows that the operator u 7→ u Ψj is bounded from L2(C ′j)into L2(C ′j).

For such domains and continuous functions f : ∂D → C the surface integral∫∂Df ds

exists. Very often in the following we need the following tool (see, e.g., [3]):

Lemma 6.3 (Partition of Unity)Let K ⊂ R3 be a compact set. For every finite set Uj : j = 1, . . . ,m of open domains withK ⊂

⋃mj=1 Uj there exist φj ∈ C∞(R3) with supp(φj) ⊂ Uj for all j and

∑mj=1 φj(y) = 1 for

all y ∈ K. We call (Uj, φj) a partition of unity on K.

Using a local coordinate system Uj, ξj : j = 1, . . . ,m of ∂D as in Definition 6.1 and acorresponding partion of unity φj on ∂D we write

∫∂Df ds in the form∫

∂D

f ds =m∑j=1

∫∂D∩Uj

φj f ds =m∑j=1

∫∂D∩Uj

fj ds

with fj(y) = φj(y)f(y). the integral over the surface patch Uj ∩ ∂D is given by∫Uj∩∂D

fj ds =

∫B2(0,αj)

fj(Ψj(x)

)√1 + |∇ξj(x)|2 dx .

We collect important properties of the smooth domain D in the following lemma.

Lemma 6.4 Let D ∈ C2. Then there exists c0 > 0 such that

(a)∣∣ν(y) · (y − z)

∣∣ ≤ c0|z − y|2 for all y, z ∈ ∂D,

(b)∣∣ν(y)− ν(z)

∣∣ ≤ c0|y − z| for all y, z ∈ ∂D.


(c) DefineHη :=

z + tν(z) : z ∈ ∂D , |t| < η

.

Then there exists η0 > 0 such that for all η ∈ (0, η0] and every x ∈ Hη there existunique (!) z ∈ ∂D and |t| ≤ η with x = z+ tν(z). The set Hη is an open neighborhoodof ∂D for every η ≤ η0. Furthermore, z − tν(z) ∈ D and z + tν(z) /∈ D for 0 < t < ηand z ∈ ∂D.One can choose η0 such that for all η ≤ η0 the following holds:

• |z − y| ≤ 2|x− y| for all x ∈ Hη and y ∈ ∂D, and

• |z1 − z2| ≤ 2|x1 − x2| for all x1, x2 ∈ Hη.

If Uδ :=x ∈ R3 : infz∈∂D |x − z| < δ

denotes the strip around ∂D then there exists

δ > 0 withUδ ⊂ Hη0 ⊂ Uη0 (6.12)

(d) There exists r0 > 0 such that the surface area of ∂B(z, r) ∩ D for z ∈ ∂D can beestimated by ∣∣|∂B(z, r) ∩D| − 2πr2

∣∣ ≤ 4πc0 r3 for all r ≤ r0 . (6.13)

Proof: We use a local coordinate system Uj, ξj : j = 1, . . . ,m which yields the parametriza-tion Ψj : B2(0, αj) → ∂D ∩ Uj. First, it is easy to see (proof by contradiction) that thereexists δ > 0 with the property that for every pair (z, x) ∈ ∂D × R3 with |z − x| < δ thereexists Uj with z, x ∈ Uj. Let diam(D) = sup

|x1 − x2| : x1, x2 ∈ D

be the diameter of D.

(a) Let x, y ∈ ∂D and assume first that |y − x| ≥ δ. Then∣∣ν(y) · (y − x)∣∣ ≤ |y − x| ≤ diam(D)

δ2δ2 ≤ diam(D)

δ2|y − x|2 .

Let now |y − x| < δ. Then there exists Uj with y, x ∈ Uj. Let x = Ψj(u) and y = Ψj(v).Then

ν(x) =

∂Ψj∂u1

(u)× ∂Ψj∂u2

(u)∣∣∣∂Ψj∂u1

(u)× ∂Ψj∂u2

(u)∣∣∣

and, by the definition of the derivative,

y − x = Ψj(v)−Ψj(u) =2∑

k=1

(vk − uk)∂Ψj

∂uk(u) + a(v, u)

with∣∣a(v, u)

∣∣ ≤ c|u− v|2 for all u, v ∈ Uj and some c > 0. Therefore,

∣∣ν(x) · (y − x)∣∣ ≤ 1∣∣∣∂Ψj

∂u1(u)× ∂Ψj

∂u2(u)∣∣∣

2∑k=1

(vk − uk)∣∣∣∣(∂Ψj

∂u1

(u)× ∂Ψj

∂u2

(u)

)· ∂Ψj

∂uk(u)

∣∣∣∣︸︷︷︸= 0

+1∣∣∣∂Ψj

∂u1(u)× ∂Ψj

∂u2(u)∣∣∣∣∣∣∣(∂Ψj

∂u1

(u)× ∂Ψj

∂u2

(u)

)· a(v, u)

∣∣∣∣≤ c |u− v|2 = c

∣∣Ψ−1j (x)−Ψ−1

j (y)∣∣2 ≤ c0 |x− y|2 .


This proves part (a). The proof of (b) follows analogously from the differentiability of u 7→ ν.

(c) Choose η0 > 0 such that

(i) η0 c0 < 1/16 and

(ii) ν(x1) · ν(x2) ≥ 0 for x1, x2 ∈ ∂D with |x1 − x2| ≤ 2η0 and

(iii) Hη0 ⊂⋃Uj.

Assume that x ∈ Hη for η ≤ η0 has two representation as x = z1 + t1ν1 = z2 + t2ν2 wherewe write νj for ν(zj). Then

|z1−z2| =∣∣(t2− t1) ν2 + t1 (ν2−ν1)

∣∣ ≤ |t1− t2| + η c0|z1−z2| ≤ |t1− t2| +1

16|z1−z2| ,

thus |z1 − z2| ≤ 1615|t1 − t2| ≤ 2|t1 − t2|. Furthermore, because ν1 · ν2 ≥ 0,

(ν1 + ν2) · (z1 − z2) = (ν1 + ν2) · (t2ν2 − t1ν1) = (t2 − t1) (ν1 · ν2 + 1)︸︷︷︸≥1

,

thus|t2 − t1| ≤

∣∣(ν1 + ν2) · (z1 − z2)∣∣ ≤ 2c0|z1 − z2|2 ≤ 8c0|t1 − t2|2 ;

that is, |t2−t1|(1−8c0|t2−t1|

)≤ 0. This yields t1 = t2 because 1−8c0|t2−t1| ≥ 1−16c0η > 0

and thus also z1 = z2.

Let U be one of the sets Uj and Ψ : R2 ⊃ B2(0, α) → U ∩ ∂D the corresponding bijectivemapping. We define the new mapping F : R2 ⊃ B2(0, α)× (−η, η)→ Hη by

F (u, t) = Ψ(u) + t ν(u) , (u, t) ∈ B2(0, α)× (−η, η) .

For sufficiently small η the mapping F is one-to-one and satisfies∣∣detF ′(u, t)

∣∣ ≥ c > 0 onB2(0, α)× (−η, η) for some c > 0. Indeed, this follows from

F ′(u, t) =

(∂Ψ

∂u1

(u) + t∂ν

∂u1

(u) ,∂Ψ

∂u2

(u) + t∂ν

∂u2

(u) , ν(u)

)>and the fact that for t = 0 the matrix F ′(u, 0) has full rank 3. Therefore, F is a bijectivemapping from B2(0, α) × (−η, η) onto U ∩ Hη. Therefore, Hη =

⋃(Hη ∩ Uj) is an open

neighborhood of ∂D. This proves also that x = z − tν(z) ∈ D and x = z + tν(z) /∈ D for0 < t < η.

For x = z + tν(z) and y ∈ ∂D we have

|x− y|2 =∣∣(z − y) + tν(z)

∣∣2 ≥ |z − y|2 + 2t(z − y) · ν(z)

≥ |z − y|2 − 2ηc0|z − y|2

≥ 1

4|z − y|2 because 2ηc0 ≤

3

4.


Therefore, |z − y| ≤ 2|x− y|. Finally,

|x1 − x2|2 =∣∣(z1 − z2) + (t1ν1 − t2ν2)

∣∣2 ≥ |z1 − z2|2 − 2∣∣(z1 − z2) · (t1ν1 − t2ν2)

∣∣≥ |z1 − z2|2 − 2 η

∣∣(z1 − z2) · ν1

∣∣ − 2 η∣∣(z1 − z2) · ν2

∣∣≥ |z1 − z2|2 − 4 η c0 |z1 − z2|2 = (1− 4ηc0) |z1 − z2|2 ≥

1

4|z1 − z2|2

because 1− 4ηc0 ≥ 1/4.The proof of (6.12) is simple and left as an exercise.

(d) Let c0 and η0 as in parts (a) and (c). Choose r0 such that B[z, r] ⊂ Hη0 for all r ≤ r0

(which is possible by (6.12)) and ν(z1) · ν(z2) > 0 for |z1 − z2| ≤ 2r0. For fixed r ≤ r0 andarbitrary z ∈ ∂D and σ > 0 we define

Z(σ) =x ∈ ∂B(z, r) : (x− z) · ν(z) ≤ σ

We show that

Z(−2c0r2) ⊂ ∂B(z, r) ∩D ⊂ Z(+2c0r

2)

Let x ∈ Z(−2c0r2) have the form x = x0 + tν(x0). Then

(x− z) · ν(z) = (x0 − z) · ν(z) + t ν(x0) · ν(z) ≤ −2c0r2 ;

that is,

t ν(x0) · ν(z) ≤ −2c0r2 +

∣∣(x0 − z) · ν(z)∣∣ ≤ −2c0r

2 + c0|x0 − z|2

≤ −2c0r2 + 2c0|x− z|2 = 0 ;

that is, t ≤ 0 because |x0−z| ≤ 2r and thus ν(x0)·ν(z) > 0. This shows x = x0+tν(x0) ∈ D.Analogously, for x = x0 − tν(x0) ∈ ∂B(z, r) ∩D we have t > 0 and thus

(x− z) · ν(z) = (x0 − z) · ν(z)− t ν(x0) · ν(z) ≤ c0|x0 − z|2 ≤ 2c0|x− z|2 = 2c0r2 .

Therefore, the surface area of ∂B(z, r)∩D is bounded from below and above by the surfaceareas of Z(−2c0r

2) and Z(+2c0r2), respectively. Since the surface area of Z(σ) is 2πr(r+σ)

we have−4πc0 r

3 ≤ |∂B(z, r) ∩D| − 2πr2 ≤ 4πc0 r3 .

2

6.3 Integral Identities

Now we can formulate the mentioned integral identities. We do it only in R3. By Cn(D)3 wedenote the space of vector fields F : D → C3 which are n−times continuously differentiable.By Cn(D)3 we denote the subspace of Cn(D)3 that consists of those functions F which,together with all derivatives up to order n, have continuous extentions to the closure D ofD.

6.3. INTEGRAL IDENTITIES 237

Theorem 6.5 (Theorem of Gauss, Divergence Theorem)

Let D ⊂ R3 be a bounded Lipschitz domain. For F ∈ C(D)3 with divF ∈ C(D) the identity∫∫D

divF (x) dx =

∫∂D

F (x) · ν(x) ds

holds. In particular, the integral on the left hand side exists.Furthermore, application of this formula to F = uve(j) for u, v ∈ C1(D) ∩ C(D) and thej−th unit vector e(j) yields the formula of partial integration in the form∫∫

D

u∇v dx = −∫∫

D

v∇u dx +

∫∂D

u v ν ds .

For a proof for Lipschitz domains we refer to [6]. For smooth domains a proof can be foundin [3]. As a conclusion one derives the theorems of Green.

Theorem 6.6 (Green’s first and second theorem)

Let D ⊂ R3 be a bounded Lipschitz domain. Furthermore, let u, v ∈ C2(D) ∩ C1(D). Then∫∫D

(u∆v +∇u · ∇v) dx =

∫∂D

u∂v

∂νds ,∫∫

D

(u∆v −∆u v) dx =

∫∂D

(u∂v

∂ν− v ∂u

∂ν

)ds .

Here, ∂u(x)/∂ν = ν(x) · ∇u(x) denotes the normal derivative of u at x ∈ ∂D.

Proof: The first identity is derived from the divergence theorem be setting F = u∇v. ThenF satisfies the assumption of Theorem 6.5 and divF = u∆v +∇u · ∇v.The second identity is derived by interchanging the roles of u and v in the first identity andtaking the difference of the two formulas. 2

We will also need their vector valued analoga.

Theorem 6.7 (Integral identities for vector fields)Let D ⊂ R3 be a bounded Lipschitz domain. Furthermore, let A,B ∈ C1(D)3 ∩ C(D)3 andlet u ∈ C2(D) ∩ C1(D). Then ∫∫

D

curlAdx =

∫∂D

ν × Ads , (6.14a)∫∫D

(B · curlA− A · curlB) dx =

∫∂D

(ν × A) ·B ds , (6.14b)∫∫D

(u divA+ A · ∇u) dx =

∫∂D

u (ν · A) ds . (6.14c)


Proof: For the first identity we consider the components separately. For the first one wehave ∫∫

D

(curlA)1 dx =

∫∫D

(∂A3

∂x2

− ∂A2

∂x3

)dx =

∫∫D

div

0A3

−A2

dx

=

∫∂D

ν ·

0A3

−A2

ds =

∫∂D

(ν × A)1 ds .

For the other components it is proven in the same way.For the second equation we set F = A × B. Then divF = B · curlA − A · curlB andν · F = ν · (A×B) = (ν × A) ·B.For the third identity we set F = uA and have divF = u divA+A ·∇u and ν ·F = u(ν ·A).2

6.4 Surface Gradient and Surface Divergence

We have to introduce two more notions from differential geometry, the surface gradientand surface divergence which are differential operators on the boundary ∂D. We assumethroughout this section that D ⊆ R3 is a C2−smooth domain in the sense of Definition 6.1.First we define the spaces of differentiable functions and vector fields on ∂D.

Definition 6.8 Let D ⊆ R3 be a C2−smooth domain with boundary ∂D. Let Uj, ξj : j =1, . . . ,m be a local coordinate system and φj : j = 1, . . . ,m be a corresponding partitionof unity on ∂D. We set again Ψj(x) = Qj(x1, x2, ξj(x))> + z(j) for x = (x1, x2) ∈ B2(0, αj)and define

C1(∂D) :=f ∈ C(∂D) : (φjf) Ψj ∈ C1

(B2(0, α)

)for all j = 1, . . . ,m

,

C1(∂D)3 :=F ∈ C(∂D)3 : Fj ∈ C1(∂D) for j = 1, 2, 3

,

Ct(∂D) :=F ∈ C(∂D)3 : F · ν = 0 on ∂D

,

C1t (∂D) := Ct(∂D) ∩ C1(∂D)3 .

There exist several different – but equivalent – approaches to define the surface gradientand surface divergence. We decided to choose one which uses the ordinary gradient anddivergence, respectively, on a neighborhood of the boundary ∂D. To do this we need toextend functions and vector fields. We point out that the same technique is used to constructextension operators for Sobolev spaces, see, e.g., Theorem ??.

Lemma 6.9 Let D ⊆ R3 be a C2−smooth domain with boundary ∂D.

(a) For every f ∈ C1(∂D) there exists f ∈ C1(R3) with compact support and f = f on∂D.

6.4. SURFACE GRADIENT AND SURFACE DIVERGENCE 239

(b) For every F ∈ C1t (∂D) there exists F ∈ C1(R3)3 with compact support and F = F on

∂D.

Proof: (a) Using a local coordinate system Uj, ξj : j = 1, . . . ,m and a correspondingpartition of unity φj : j = 1, . . . ,m on ∂D as in Definition 6.8 we note that f =

∑mj=1 fj

on ∂D where fj = fφj. The functions fj Ψj are continuously differentiable functionsfrom B2(0, αj) into C with support in B2(0, αj). We extend fj Ψj into the cylinder C ′j =Cj(αj, βj/2) by setting gj(x) := ρ(x3)(fj Ψj)(x1, x2) for x = (x1, x2, x3) ∈ C ′j where ρ ∈C∞0 (−βj/2, βj/2) is such that ρ = 1 in a neighborhood of 0. Then gj : C ′j → C is continuously

differentiable, has compact support, and gj = fj Ψj on B2(0, αj) × 0. Therefore, fj :=gj Ψ−1

j has compact support in U ′j = Ψj(C′j). We extend fj by zero into all of R3 and set

f :=∑m

j=1 fj in R3. Then f ∈ C1(R3) with support in⋃mj=1 U

′j such that f = f on ∂D.

The proof of (b) is identical by using the argument for every component. 2

Definition 6.10 Let ϕ ∈ C1(∂D) and ϕ ∈ C1(U) be an extension of ϕ into a neighborhoodU of the boundary ∂D of the domain D ∈ C2. Furthermore, let h ∈ C1

t (∂D) be a tangentialvector field and h ∈ C1(U)3 be an extension into U .

(a) The surface gradient of ϕ is defined as the orthogonal projection of ∇ϕ onto thetangent plane; that is,

Grad ϕ = ν × (∇ϕ× ν) = ∇ϕ − ∂ϕ

∂νν on ∂D , (6.15)

where ν = ν(x) denotes the exterior unit normal vector at x ∈ ∂D.

(b) The surface divergence of h is given by

Div h = div h − ν · (h′ν) on ∂D (6.16)

where h′(x) ∈ C3×3 denotes the Jacobian matrix of h at x.

We will see in Lemma 6.13 that the definitions are independent of the choices of the exten-sions.

Example 6.11 As an example we consider the sphere of radius R > 0; that is, D = B(0, R).We parametrize the boundary of this ball by spherical coordinates

Ψ(θ, φ) = R(sin θ cosφ, sin θ sinφ, cos θ)> .

Then the surface gradient and surface divergence, respectively, on the sphere ∂D are givenby

Grad f(θ, φ) =1

R

∂f

∂θ(θ, φ) θ +

1

R sin θ

∂f

∂φ(θ, φ) φ ,

DivF (θ, φ) =1

R sin θ

∂

∂θ

(sin θ Fθ(θ, φ)

)+

1

R sin θ

∂Fφ∂φ

(θ, φ) ,


where θ = (cos θ cosφ, cos θ sinφ,− sin θ)> and φ = (− sin θ sinφ, sin θ cosφ, 0)> are the tan-gential unit vectors which span the tangent plane and Fθ, Fφ are the components of F w.r.t.

these vectors; that is, F = Fθθ + Fφφ. From this we note that

Div Grad f(θ, φ) =1

R2 sin θ

∂

∂θ

(sin θ

∂f

∂θ(θ, φ)

)+

1

R2 sin2 θ

∂2f

∂φ2(θ, φ) .

This differential operator is called the Laplace-Beltrami operator and will be denoted by∆S = Div Grad .

We collect some important properties in the following lemma. It will be necessary to extendalso the vector field ν into a neighborhood U of ∂D such that |ν(x)| = 1 on U . This ispossible because by Lemma 6.9 there exists an extension ν ∈ C1(R3)3 of ν. Certainly, ν 6= 0in a neighborhood U of ∂D because |ν| = 1 on ∂D. Therefore, ν = ν/|ν| will be the requiredextension into U .

Lemma 6.12 Let D ∈ C2 and h ∈ C1t (∂D) and ϕ ∈ C1(∂D) with extensions h ∈ C1(R3)3

and ϕ ∈ C1(R3), respectively. Then:

(a) Div h = ν ·curl(ν×h) on ∂D where ν ∈ C1(U) is an extension of ν into a neighborhoodU of ∂D such that |ν(x)| = 1 on U .

(b) Let Γ ⊂ ∂D be a relatively open subset2 such that the (relative) boundary C = ∂Γ isa closed curve with continuously differentiable tangential unit vector τ(x) for x ∈ C.The orientation of τ is chosen such that (Γ, C) is mathematically posively orientated;that is, the vector τ(x) × ν(x) (which is a tangential vector to the boundary ∂D) isdirected “outwards” of Γ for all x ∈ C. Then∫

Γ

Div h ds =

∫C

h · (τ × ν) d` .

In particular,∫∂D

Div h ds = 0.

(c) Partial integration holds in the following form:∫∂D

ϕ Div h ds = −∫∂D

h ·Grad ϕds . (6.17)

Proof: (a) The product rule for the curl of a vector product yields

curl(ν × h) = ν div h − h div ν + ν ′h − h′ν

= ν div h − h div ν + ν ′h − h′ν

and thusν · curl(ν × h) = div h − ν · h︸︷︷︸

= 0

div ν + ν>ν ′h − ν>h′ν .

2that is, Γ = ∂D ∩ U for some open set U ⊂ R3

6.4. SURFACE GRADIENT AND SURFACE DIVERGENCE 241

From ν>ν = 1 in U we have by differentiation that ν>ν ′ = 0 in U and thus

ν · curl(ν × h) = div h − ν>h′ν = Div h .

(b) Let ν and h as in part (a) and set g = ν × h in U . Then g ∈ C1(U). By part (a) weconclude that

∫Γ

Div h ds =∫

Γν · curl g ds. Choose a sequence gn ∈ C∞(U) with gn → g in

C1(U). Then, by the Theorem of Stokes on Γ, we conclude that∫

Γν ·curl gn ds =

∫Cgn ·τ d`.

The convergence curl gn → curl g in C(∂D) and gn → ν × h in C(∂D) yields∫

ΓDiv h ds =∫

C(ν × h) · τ d` =

∫Ch · (τ × ν) d`.

(c) By part (b) it suffices to prove the product rule

Div(ϕh) = Grad ϕ · h + ϕ Div h . (6.18)

Indeed, using the definitions yields

Div(ϕh) = div(ϕh) − ν ·((ϕh)′ν

)= ∇ϕ · h + ϕ div h − ν>hϕ′ν − ϕ ν>h′ν

= Grad ϕ · h + ϕ div h − ν>h︸︷︷︸= 0

ϕ′ν − ϕν>h′ν

= Grad ϕ · h + ϕ Div h .


Lemma 6.13 The tangential gradient and the tangential divergence depend only on thevalues of ϕ and the tangential field h on ∂D, respectively.

Proof Let ϕ1, ϕ2 ∈ C1(D) be two extensions of ϕ ∈ C1(∂D). Then ϕ = ϕ1 − ϕ2 vanisheson ∂D. The identity (6.17) shows that∫

∂D

[ν × (∇ϕ× ν)

]· h ds = 0 for all h ∈ C1

t (∂D) .

The space C1t (∂D) is dense in the space L2

t (∂D) of all tangential vector fields with L2−components.Therefore, ν × (∇ϕ× ν) vanishes which shows that the definition of Grad ϕ is independentof the extension. Similarly the assertion for the tangential divergence is obtained. 2

Corollary 6.14 Let w ∈ C2(D)3 such that w, curlw ∈ C(D)3. Then the surface divergenceof ν × w exists and is given by

Div(ν × w) = −ν · curlw on ∂D . (6.19)

Proof: For any ϕ ∈ C2(D) we have by the divergence theorem∫∂D

ϕν · curlw ds =

∫∫D

div(ϕ curlw) dx =

∫∫D

∇ϕ · curlw dx

=

∫∫D

div(w ×∇ϕ) dx =

∫∂D

ν · (w ×∇ϕ) ds

=

∫∂D

(ν × w) ·Grad ϕds = −∫∂D

Div(ν × w)ϕds .


The assertion follows because the tracesϕ|∂D : ϕ ∈ C2(D)

are dense in L2(∂D). 2

This procedure to extend functions defined on the boundary will be used more often in thisbook. The constructions consists always of three steps. First, one localizes and flattens theboundary by using local coordinates and a corresponding partition of unity. Second, oneextends the functions (this part differs from application to application), and, finally, onetransforms the extended parts back.

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The Mathematical Theory of Maxwell’s Equations - KIT · are constant (homogeneous medium) or at least continuous. In regions where the vector elds are smooth functions we can apply