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The Mathematical Theory of Maxwell’s Equations

Andreas Kirsch and Frank HettlichDepartment of Mathematics

Karlsruhe Institute of Technology (KIT)Karlsruhe, Germany

c© May 24, 2014

Preface

This book arose from lectures on Maxwell’s equations given by the authors between 2007 and2013. Graduate students from pure and applied mathematics, physics – including geophysics– and engineering attended these courses. We observed that the expectations of these groupsof students were quite different: In geophysics expansions of the electromagnetic fields intospherical (vector-) harmonics inside and outside of balls are of particular interest. Graduatestudents from numerical analysis wanted to learn about the variational treatments of interiorboundary value problems including an introduction to Sobolev spaces. A classical approachin scattering theory – which can be considered as a boundary value problem in the unboundedexterior of a domain – uses boundary integral equation methods which are particularlyhelpful for deriving properties of the far field behaviour of the solution. This approach is, forpolygonal domains or, more generally, Lipschitz domains, also of increasing relevance fromthe numerical point of view because the dimension of the region to be discretized is reducedby one. In our courses we wanted to satisfy all of these wishes and designed an introductionto Maxwell’s equations which coveres all of these concepts – but restricted ourselves almostcompletely (except Section 4.3) to the time-harmonic case or, in other words, to the frequencydomain, and to a number of model problems.

The Helmholtz equation is closely related to the Maxwell system (for time-harmonic fields).As we will see, solutions of the scalar Helmholtz equation are used to generate solutions ofthe Maxwell system (Hertz potentials), and every component of the electric and magneticfield satisfies an equation of Helmholtz type. Therefore, and also for didactical reasons, wewill consider in each of our approaches first the simpler scalar Helmholtz equation before weturn to the technically more complicated Maxwell system. In this way one clearly sees theanalogies and differences between the models.

In Chapter 1 we begin by formulating the Maxwell system in differential and integral form.We derive special cases as the E− mode and the H−mode and, in particular, the timeharmonic case. Boundary conditions and radiation conditions complement the models.

In Chapter 2 we study the particular case where the domain D is a ball. In this case we canexpand the fields inside and outside of D into spherical wave functiuons. First we study thescalar stationary case; that is, the Laplace equation. We introduce the expansion into scalarspherical harmonics as the analogon to the Fourier expansion on circles in R2. This leadsdirectly to series expansions of solutions of the Laplace equation in spherical coordinates.The extension to the Helmholtz equation requires the introduction of (spherical) Bessel- andHankel functiuons. We derive the most important properties of these special functions indetail. After these preparations for the scalar Helmholtz equation we extend the analysis

3

4

to the expansion of solutions of Maxwell’s equations with respect to vector wave functions.None of the results of this chapter are new, of course, but we have not been able to find sucha presentation of both, the scalar and the vectorial case, in the literature. We emphasize thatthis chapter is completely self-contained and does not refer to any other chapter – except forthe proof that the series solution of the exterior problem satisfies the radiation condition.

Chapter 3 deals with a particular scattering problem. The scattering object is of arbitraryshape but, in this chapter, with sufficiently smooth boundary ∂D. We present the classicalboundary integral equation method and follow very closely the fundamental monographs[5, 6] by David Colton and Rainer Kress. In contrast to their approach we restrict ourselvesto the case of smooth boundary data (as it is the usual case of the scattering by incidentwaves) which allows us to study the setting completely in Holder spaces and avoids thenotions of “parallel surfaces” and weak forms of the normal derivatives. For the scalarproblem we restrict ourselves to the Neumann boundary condition because our main goalis the treatment for Maxwell’ equations. Here we believe – as also done in [6] – that thecanonical spaces on the boundary are Holder spaces where also the surface divergence isHolder continuous. In order to prove the necessary properties of the scalar and vectorpotentials a careful investigation of the differential geometric properties of the surface ∂D isneeded. Parts of the technical details are moved to the Appendix 6. We emphasize that thischapter is self countained and does not need any results from other chapters (except fromthe appendix).

As an alternative approach for studying boundary value problems for the Helmholtz equationor the Maxwell system we will study the weak or variational solution concept in Chapter 4.We restrict ourselves to the interior boundary value problem with a general source term andthe homogeneous boundary condition of an ideal conductor. This makes it possible to work(almost) solely in the Sobolev spaces H1

0 (D) and H0(curl, D) of functions with vanishingboundary traces or tangential boundary traces, respectively. In Section 4.1 we derive the ba-sic properties of these special Sobolev spaces. The characteristic feature is that no regularityof the boundary and no trace theorems are needed. Probably the biggest difference betweenthe scalar case of the Helmholtz equation and the vectorial case of Maxwell’s equations isthe fact that H0(curl, D) is not compactly imbedded in L2(D,C3) in contrast to the spaceH1

0 (D) for the scalar problem. This makes it necessary to introduce the Helmholtz decompo-sition. The only proof which is beyond the scope of this elementary chapter is the proof thatthe subspace of H0(curl, D) consisting of divergence-free vector fields is compactly imbeddedin L2(D,C3). For this part some regularity of the boundary (e.g. Lipschitz regularity) isneeded. Since the proof of this fact requires more advanced properties of Sobolev spaces itit transponed to Chapter 5. We note that also this chapter is self contained except of thebeforementioned compactness property.

The final Chapter 5 presents the boundary integral equation method for Lipschitz domains.The investigation requires more advanced properties of Sobolev spaces than those presentedin Section 4.1. In particular, Sobolev spaces on the boundary ∂D have to be introduced andthe correponding trace operators. Perhaps different from most of the traditional approacheswe first consider the case of the cube (−π, π)3 ⊂ R3 and introduce Sobolev spaces of periodicfunctions by the proper decay of the Fourier coefficients. The proofs of imbedding and tracetheorems are quite elementary. Then we use, as it is quite common, the partion of unity and

5

local maps to define the Sobolev spaces on the boundary and transfer the trace theorem togeneral Lipschitz domains.We define the scalar and vector potentials in Section 5.2 analogously to the classical caseas in Chapter 3 but have to interpret the boundary integrals as certain dual forms. Theboundary operators are then defined as traces of these potentials. In this way we follow theclassical approach as closely as possible. Our approach is similar but a bit more explicitthan in [16], see also [10]. Once the properties of the potentials and corresponding boundaryoperators are known the introduction and investigation of the boundary integral equations isalmost classical. For example, for Lipschitz domains the Dirichlet boundary value problemfor the scalar Helmholtz equation is solved by a (properly modified) single layer ansatz.This is preferable to a double layer ansatz because the corresponding double layer boundaryoperator fails to be compact (in contrast to the case of smooth boundaries). Also, the singlelayer boundary operator satisfies a Garding’s inequality; that is, can be decomposed into acoercive and a compart part. Analogously. the Neumann boundary value problem and theelectromagnetic case are treated. In our presentation we try to show the close connectionbetween the scalar and the vector cases.Starting perhaps with the pioneering work of Costabel [7] many important contributions tothe study of boundary integral operators in Sobolev spaces for Lipschitz boundaries havebeen published. It is impossible for the authors to give an overview on this subject butinstead refer to the monograph [10] and the survey article [4] from which we have learned alot. As mentioned above, our approach to introduce the Sobolev spaces, however, is differentfrom those in, e.g. [1, 2, 3].

In the Appendix 6 we collect results from vector calculus and differential geometry, in par-ticular various forms of Green’s theorem and the surface gradient and surface divergence for(smooth) functions on (smooth) surfaces.

We want to emphasize that it was not our intention to present a comprehensive work onMaxwell’s equations, not even for the time harmonic case or any of the beforementioned sub-areas. As said before this book arose from – and is intended to be – material for designinggraduate courses on Maxwell’s equations. The students should have some knowledge on vec-tor analysis (curves, surfaces, divergence theorem) and functional analysis (normed spaces,Hilbert spaces, linear and bounded operators, dual space). The union of the topics coveredin this monograph is certainly far too much for a single course. But it is very well possibleto choose parts of it because the chapters are all independent of each other. For example, inthe summer term 2012 (8 credit points; that is in our place, 16 weeks with 4 hours per weekplus exercises) one of the authors (A.K.) covered Sections 2.1–2.6 of Chapter 2 (only interiorcases), Chapter 3 without all of the proofs of the differential geometric properties of thesurface and all of the jump properties of the potentials, and Chapter 4 without Section 4.3.Perhaps these notes can also be useful for designing courses on Special Functions (sphericalharmonics, Bessel functions) or on Sobolev spaces.

One of the authors wants to dedicate this book to his father, Arnold Kirsch (1922–2013), whotaught him about simplification of problems without falsification (as a concept of teachingmathematics in high schools). In Chapter 4 of this monograph we have picked up thisconcept by presenting the ideas for a special case only rather than trying to treat the most

6

general cases. Nevertheless, we admit that other parts of the monograph (in particular ofChapters 3 and 5) are technically rather involved.

Karlsruhe, March 2014 Andreas KirschFrank Hettlich

Contents

Preface 2

1 Introduction 9

1.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2 The Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Boundary and Radiation Conditions . . . . . . . . . . . . . . . . . . . . . . 18

1.5 The Reference Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Expansion into Wave Functions 27

2.1 Separation in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3 Expansion into Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . . . 40

2.4 Laplace’s Equation in the Interior and Exterior of a Ball . . . . . . . . . . . 50

2.5 Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.6 The Helmholtz Equation in the Interior and Exterior of a Ball . . . . . . . . 62

2.7 Expansion of Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . 72

2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

3 Scattering From a Perfect Conductor 91

3.1 A Scattering Problem for the Helmholtz Equation . . . . . . . . . . . . . . . 91

3.1.1 Representation Theorems . . . . . . . . . . . . . . . . . . . . . . . . 92

3.1.2 Volume and Surface Potentials . . . . . . . . . . . . . . . . . . . . . . 99

3.1.3 Boundary Integral Operators . . . . . . . . . . . . . . . . . . . . . . 112

3.1.4 Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 116

3.2 A Scattering Problem for the Maxwell System . . . . . . . . . . . . . . . . . 124

7

8 CONTENTS

3.2.1 Representation Theorems . . . . . . . . . . . . . . . . . . . . . . . . 125

3.2.2 Vector Potentials and Boundary Integral Operators . . . . . . . . . . 135

3.2.3 Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 141

3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

4 The Variational Approach to the Cavity Problem 147

4.1 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

4.1.1 Basic Properties of Sobolev Spaces of Scalar Functions . . . . . . . . 148

4.1.2 Basic Properties of Sobolev Spaces of Vector Valued Functions . . . . 157

4.1.3 The Helmholtz Decomposition . . . . . . . . . . . . . . . . . . . . . . 160

4.2 The Cavity Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

4.2.1 The Variational Formulation and Existence . . . . . . . . . . . . . . . 163

4.2.2 Uniqueness and Unique Continuation . . . . . . . . . . . . . . . . . . 173

4.3 The Time–Dependent Cavity Problem . . . . . . . . . . . . . . . . . . . . . 183

4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

5 Boundary Integral Equation Methods for Lipschitz Domains 201

5.1 Advanced Properties of Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . 201

5.1.1 Sobolev Spaces of Scalar Functions . . . . . . . . . . . . . . . . . . . 203

5.1.2 Sobolev Spaces of Vector-Valued Functions . . . . . . . . . . . . . . . 216

5.1.3 The Case of a Ball Revisited . . . . . . . . . . . . . . . . . . . . . . . 235

5.2 Surface Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

5.3 Boundary Integral Equation Methods . . . . . . . . . . . . . . . . . . . . . . 263

5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

6 Appendix 277

6.1 Table of Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 277

6.2 Results from Linear Functional Analysis . . . . . . . . . . . . . . . . . . . . 279

6.3 Elementary Facts from Differential Geometry . . . . . . . . . . . . . . . . . 281

6.4 Integral Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

6.5 Surface Gradient and Surface Divergence . . . . . . . . . . . . . . . . . . . . 288

Bibliography 292

Index 294

Chapter 1

Introduction

In this introductory chapter we will explain the physical model and derive the boundary valueproblems which we will investigate in this monograph. We begin by formulating Maxwell’sequations in differential form as our starting point. In this monograph we consider exclusivelylinear media; that is, the constitutive relations are linear. The restriction to special casesleads to the analogous equations in electrostatics or magnetostatics and, assuming periodictime dependence when going into the frequency domain, to time harmonic fields. In thepresence of media the fields have to satisfy certain continuity and boundary conditions and,if the region is unbounded, a radiation condition at infinity. We finish this chapter byintroducing two model problems which we will treat in detail in Chapters 3 and 4.

1.1 Maxwell’s Equations

Electromagnetic wave propagation is described by four particular equations, the Maxwellequations, which relate five vector fields E , D, H, B, J and the scalar field ρ. In differentialform these read as follows

∂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) .

The fields E and D denote the electric field (in V/m) and electric displacement (in As/m2)respectively, while H and B denote the magnetic 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) andcharge density (in As/m3) of the medium.

Here and throughout we use the rationalized MKS-system, i.e. the fields are given withrespect to the units Volt (V ), Ampere (A), meter (m), and second (s). All fields depend both

9

10 CHAPTER 1. INTRODUCTION

on the space variable x ∈ R3 and on the time variable t ∈ R. We note that the differentialoperators are always taken with respect to the spacial variable x without indicating this.The definition of the differential operators div and curl, e.g., in cartesian coordinates andbasic identities are listed in Appendix 6.1 .

The actual equations that govern the behavior of the electromagnetic field were first com-pletely formulated by James Clark Maxwell (1831–1879) in Treatise on Electricity and Mag-netism in 1873. It was the ingeneous idea of Maxwell to modify Ampere’s Law which wasknown up to that time in the form curl H = J for stationary currents. Furthermore, hecollected the four equations as a consistent theory to describe electromagnetic phenomena.

As a first observation we note that in domains where the equations are satisfied one derivesfrom the identity div curlH = 0 the well known equation of continuity which combines thecharge density and the current density.

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

∂ρ

∂t= div

∂D∂t

= div(curlH−J

)= − divJ .

Historically, and more closely connected to the physical situation, the integral forms ofMaxwell’s equations should be the starting point. In order to derive these integral relations,we begin by letting S be a connected smooth surface with boundary ∂S in the interior of aregion Ω0 in R3 where electromagnetic waves propagate. In particular, we require that theunit normal vector ν(x) for x ∈ S is continuous and directed always into “one side” of S,which we call the positive side of S. By τ(x) we denote a unit vector tangent to the boundaryof S at x ∈ ∂S. This vector, lying in the tangent plane of S together with a second vectorn(x) in the tangent plane at x ∈ ∂S and normal to ∂S is oriented such that τ, n, ν, forma mathematically positive system; that is, τ is directed counterclockwise when we look at Sfrom the positive side, and n(x) is directed to the outside of S. Furthermore, let Ω ∈ R3 bean open set with boundary ∂Ω and outer unit normal vector ν(x) at x ∈ ∂Ω.

Then Ampere’s law describing the effect of the external and the induced current on themagnetic field is of the form∫

∂S

H · τ d` =d

dt

∫S

D · ν ds +

∫S

J · ν ds . (1.1)

It is named by Andre Marie Ampere (1775–1836).

Next, Faraday’s law of induction (Michael Faraday, 1791–1867), which is∫∂S

E · τ d` = − d

dt

∫S

B · ν ds , (1.2)

describes how a time-varying magnetic field effects the electric field.

Finally, the equations include Gauss’ Electric Law∫∂Ω

D · ν ds =

∫Ω

ρ dx (1.3)

1.2. THE CONSTITUTIVE EQUATIONS 11

descibing the sources of the electric displacement, and Gauss’ Magnetic Law∫∂Ω

B · ν ds = 0 (1.4)

which ensures that there are no magnetic currents. Both named after Carl Friedrich Gauss(1777–1855).

In regions where the vector fields are smooth functions and µ and ε are at least continuouswe can apply the following integral identities due to Stokes and Gauss for surfaces S andsolids Ω lying completely in D.∫

S

curl F · ν ds =

∫∂S

F · τ d` (Stokes), (1.5)

∫Ω

div F dx =

∫∂Ω

F · ν ds (Gauss), (1.6)

see Appendix 6.3. To derive the Maxwell’s equations in differential form we choose F to beone of the fields H, E , B or D. With these formulas we can eliminate the boundary integralsin (1.1)–(1.4). We then use the fact that we can vary the surface S and the solid Ω in Darbitrarily. By equating the integrands we are led to Maxwell’s equations in differential formas presented in the begining.

With Maxwell’s equations many electromagnetic phenomena became explainable. For in-stance they predicted the existence of electromagnetic waves as light or X-rays in vacuum.It took about 20 years after Maxwell’s work when Heinrich Rudolf Hertz (1857–1894) couldshow experimentally the existence of electromagenetic waves, in Karlsruhe, Germany. Formore details on the physical background of Maxwell’s equations we refer to text books asJ.D. Jackson, Classical Electrodynamics [11].

1.2 The Constitutive Equations

In the general setting the equations are not yet complete. Obviously, there are more un-knowns than equations. The Constitutive Equations couple them:

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

The electric properties of the material, which give these relationships are complicated. Ingeneral, they not only depend on the molecular character but also on macroscopic quantitiesas density and temperature of the material. Also, there are time-dependent dependenciesas, 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 relativly small onecan model the dependencies by linear equations of the form

D = εE and B = µH

with matrix-valued functions ε : R3 → R3×3, the dielectric tensor , and µ : R3 → R3×3, thepermeability tensor . In this case we call a medium linear .

The special case of an isotropic medium means that polarization and magnetization do notdepend on the directions. Otherwise a medium is called anisotropic. In the isotropic casedielectricity and permeability can be modeled as just real valued functions, and we have

D = εE and B = µH

with scalar functions ε, µ : R3 → R.

In the simplest case these functions ε and µ are constant and we call such a medium homo-geneous . It 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 relation. 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 . If σ = 0 then the material is called a dielectric. In vacuum we haveσ = 0 and ε = ε0 ≈ 8.854 · 10−12AS/V m, µ = µ0 = 4π · 10−7V s/Am. In anisotropic media,also the function σ is matrix valued.

1.3 Special Cases

Under specific physical assumptions the Maxwell system can be reduced to elliptic secondorder partial differential equations. They serve often as simpler models for electromagneticwave propagation. Also in this monograph we will always explain the approaches first forthe simpler scalar wave equation.

Vacuum

Vacuum is a homogeneous, dielectric medium with ε = ε0, µ = µ0, and σ = 0, and no chargedistributions and no external currents; that is, ρ = 0 and Je = 0. The law of induction takesthe form

µ0∂H∂t

+ curlE = 0 .

1.3. SPECIAL CASES 13

Assuming sufficiently smooth functions a differentiation with respect to time t and an ap-plication of Ampere’s Law yields

ε0 µ0∂2H∂t2

+ curl curlH = 0 .

The term c0 = 1/√ε0µ0 has the dimension of a velocity and is called the speed of light .

From the identity curl curl = ∇ div−∆ where the vector valued Laplace operator ∆ is takencomponentwise it follows that the components of H are solutions of the linear wave equation

1

c20

∂2H∂t2

− ∆H = 0 .

Analogously, one derives the same equation for the cartesian components of the electric field:

1

c20

∂2E∂t2

− ∆E = 0 .

Therefore, a solution of the Maxwell system in vacuum can also be described by a divergencefree solution of one of the two vector valued wave equations and defining the other field byAmperes Law or by Faraday’s Law of Induction, respectively.

Electro- and Magnetostatics

Next we consider the Maxwell system in the case of stationary fields; that is, E , D, H, B,J and ρ are constant with respect to time. For the electric field E this situation in a regionΩ is called electrostatics . The law of induction reduces to the differential equation

curlE = 0 in Ω .

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

ρ = divD = − div(ε0 E) = −ε0 ∆u

for the potential u. Thus, the electrostatics is described by the basic elliptic partial differ-ential equation ∆u = −ρ/ε0. Mathematically, we are led to the field of potential theory .

Example 1.2 The most important example is the spherical symmetric electric field gener-ated by a point charge, e.g., at the origin. For x ∈ R3 with x 6= 0 the function u(x) = 1

4π1|x|

is harmonic; that is, satisfies ∆u = 0. Thus by

E(x) = ∇u(x) = − 1

4π

x

|x|3, x 6= 0 ,

we obtain a stationary solution, the field of an electric monopole .


In magnetostatics one considers H being constant in time. For the magnetic field thesituation is different because, by Ampere’s law we have curlH = J . Thus in generalcurlH does not vanish. However, according to Gauss’ magnetic law we have

divB = 0 .

From this identity we conclude the existence of a vector potential A : R3 → R3 withB = − curlA in D. Substituting this into Ampere’s Law yields (for homogeneous media Ω)after multiplication with µ0 the equation

−µ0J = curl curlA = ∇ divA − ∆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 divA = 0. This choice ofnormalization is called Coulomb gauge named by Charles Augustin de Coulomb (1736–1806).With this normalization we get also in magnetostatics the Poisson equation

∆A = −µ0J .

We note that in this case the Laplacian is vector valued and has to be taken componentwise.

Time Harmonic Fields

For our purpose of considering wave phenomena the most important situtation are timeharmonic fields. Under the assumptions that the fields allow a Fourier transformation intime we set

E(x;ω) = (FtE)(x;ω) =

∫RE(x, t) eiωt dt ,

H(x;ω) = (FtH)(x;ω) =

∫RH(x, t) eiωt dt ,

etc. We note that the fields E, H, etc are now complex valued; that is, E(·;ω), H(·;ω) :R3 → C3 and also all other Fourier transformed fields. Although they are vector fieldswe denote them by capital Latin letters only. According to Ft(u′) = −iωFtu Maxwell’sequations transform into the time harmonic Maxwell’s equations

−iωB + curlE = 0 ,

iωD + curlH = σE + Je ,

divD = ρ ,

divB = 0 .

Remark: The time harmonic Maxwell system can also be derived from the assumptionthat all fields behave periodically with respect to time with the same frequency ω. Then thecomplex valued functions E(x, t) = e−iωtE(x), H(x, t) = e−iωtH(x), etc, and their real andimaginary parts satisfy the time harmonic Maxwell system.


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

curlE − iωµH = 0 , (1.7a)

curlH + (iωε− σ)E = Je , (1.7b)

div(εE) = ρ , (1.7c)

div(µH) = 0 . (1.7d)

Assuming (for simplicity only) additionally an isotropic medium we can eliminate H or Efrom (1.7a) and (1.7b) which yields

curl

(1

iωµcurlE

)+ (iωε− σ)E = Je . (1.8)

and

curl

(1

iωε− σcurlH

)+ iωµH = curl

(1

iωε− σJe

), (1.9)

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

µr =µ

µ0

, εr =ε

ε0

, εc = εr + iσ

ωε0

.

Then equations (1.8) and (1.9) take the form

curl

(1

µrcurlE

)− k2εcE = iωµ0Je ,

curl

(1

εccurlH

)− k2µrH = curl

(1

εcJe

),

with the wave number k = ω√ε0µ0. We conclude from the second equation that div(µrH) =

0 because div curl vanishes. For the electric field we obtain that div(εcE) = − iωµ0k2

div Je =− iωε0

div Je.

In vacuum we have εc = 1, µr = 1 and therefore the equations reduce to

curl curlE − k2E = iωµ0Je , (1.10)

curl curlH − k2H = curl Je . (1.11)

Without external current density, Je = 0, we obtain from curl curl = ∇ div−∆ the vectorHelmholtz equations

∆E + k2E = 0 and ∆H + k2H = 0 .

Obviuously the reduced problems considering E or H are symmetric and we conclude thefollowing important lemma.


Lemma 1.3 A vector field E ∈ C2(Ω,C3) combined with H := 1iωµ0

curlE provides a solu-

tion of the time harmonic Maxwell system (1.7a)–(1.7d) for Je = 0 in vacuum if and only ifE is a divergence free solution of the vector Helmholtz equation; that is,

∆E + k2E = 0 and divE = 0 in Ω .

Analougously, a divergence free solution of the vector Helmholtz equation H ∈ C2(Ω,C3)combined with E := −1

iωε0curlH leads to a solution of Maxwell’s equations in vacuum.

The relationship between the Maxwell system and the vector Helmholtz equation remainstrue if we consider time harmonic waves in any homogeneous medium because we only haveto substitute µ0 and ε0 by complex valued constants µ and ε, respectively. In this complexvalued case the wave number k = ω

√µε is chosen such that Im k ≥ 0. As an example for

solutions of the Maxwell system in a homogeneous medium we consider plane waves.

Example 1.4 In the case of vacuum with Je = 0 a short calculation shows that the fields

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

are solutions of the homogeneous time harmonic Maxwell equations (1.10), (1.11) providedd is a unit vector in R3 and p ∈ C3 with p · d =

∑3j=1 pj dj = 0. Such fields are called plane

time harmonic fields with polarization vector p ∈ C3 and direction d, since its wave frontsare planes perpendicular to d.

Additionally the following observation will be useful.

Lemma 1.5 Let E be a divergence free solution of the vector Helmholtz equation in a domainD. Then x 7→ x · E(x) is a solution of the scalar Helmholtz equation.

Proof: By the vector Helmholtz equation and divE = 0 we obtain

∆(x · E(x)

)=

3∑j=1

div∇(xjEj(x)

)=

3∑j=1

div[Ej(x) e(j) + xj∇Ej(x)

]

=3∑j=1

[2∂Ej∂xj

(x) + xj ∆Ej(x)

]= 2 divE(x) − k2x · E(x)

= −k2x · E(x)

where e(j) denotes the jth cartesian coordinate unit vector. 2

As in the stationary situation also the time harmonic Maxwell equations in homogeneousmedia can be treated with methods from potential theory. We make the assumption εc = 1,µr = 1 and consider (1.10) and (1.11). Taking the divergence of these equations yields


divH = 0 and k2 divE = −iωµ0 div Je; that is, divE = −(i/ωε0) div Je. Comparing this to(1.7c) yields the time harmonic version of the equation of continuity

div Je = iωρ .

With the vector identity curl curl = −∆ + div∇ equations (1.10) and (1.11) can be writtenas

∆E + k2E = −iωµ0Je +1

ε0

∇ρ , (1.12)

∆H + k2H = − curl Je . (1.13)

Let us consider the magnetic field first and introduce the magnetic Hertz potential : Theequation divH = 0 implies the existence of a vector potential A with H = curlA. Thus(1.13) takes the form

curl(∆A + k2A) = − curl Je

and we obtain∆A + k2A = −Je + ∇ϕ (1.14)

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

H = curlA and E = − 1

iωε0

(curlH − Je) = iωµ0A −1

iωε0

∇(divA− ϕ)

satisfies the Maxwell system (1.7a)–(1.7d).

Analogously, we can introduce electric Hertz potentials if Je = 0. Because divE = 0 thereexists a vector potential A with E = curlA. Substituting this into (1.12) yields

curl(∆A + k2A) = 0

and we obtain∆A + k2A = ∇ϕ (1.15)

for some scalar field ϕ. On the other hand, if A and ϕ satisfy (1.15) then

E = curlA and H =1

iωµ0

curlE = −iωε0A +1

iωµ0

∇(divA− ϕ)

satifies the Maxwell system (1.7a)–(1.7d). In any case we end up with an inhomogeneousvector Helmholtz equation.

As a particular example we may take a magnetic Hertz vector A of the form A(x) = u(x) zwith a scalar solution u of the two–dimensional Helmholtz equation and the unit vectorz = (0, 0, 1)> ∈ R3. Then

H = curl(uz) =

(∂u

∂x2

,− ∂u

∂x1

, 0

)>,

E = iωµ0 z +1

−iωε0

∇(∂u/∂x3) .


If u is independent of x3 then E has only a x3−component, and the vector Helmholtz equationfor A reduces to a scalar Helmholtz equation for the potential u. The situation that E hasonly one non zero component is called electric mode, E-mode, or transverse-magnetic mode,TM-mode. Analogously, also the H-mode or TE-mode is consdidered if H consists of onlyone non zero component satisfying the scalar Helmholtz equation.

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 (1.7a)–(1.7d) 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 transmission and boundary condi-tions, we apply the law of induction (1.2) to a narrow rectangle-like surface R, containingthe normal n to the surface S and whose long sides C+ and C− are parallel to S and are onthe opposite sides of it, 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

∫RB · ν ds will vanish in the limit because

the field remains finite. Note, that the normal ν is the normal to R lying in the tangentialplane of S. Hence, the line integrals

∫CE+ · τ d` and

∫CE− · τ d` must be equal. Since the

1.4. BOUNDARY AND RADIATION CONDITIONS 19

curve C is arbitrary the integrands E+ · τ and E− · τ coincide on every arc C; that is,

n× E+ − n× E− = 0 on S . (1.16)

A similar argument holds for the magnetic field in (1.1) 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 . (1.17)

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

∫RJe · ν ds will tend to

∫CJs · ν d`, and so the boundary

condition isn×H+ − n×H− = Js on S . (1.18)

We will call (1.16) and (1.17) or (1.18) transmission boundary conditions.

A special and very important case is that of a perfectly conducting medium with boundaryS. Such a medium is characterized by the fact that the electric field vanishes inside thismedium, and (1.16) reduces to

n× E = 0 on S .

In realistic situations, of course, the exact form of this equation never occurs. Nevertheless,it is a common model for the case of a very large conductivity.

Another important case is the impedance- or Leontovich boundary condition

n×H = λn× (E × n) on S

for some non-negative impedance function λ which, under appropriate conditions, may beconsidered as an approximation of the transmission conditions. Of course these boundaryconditions occur also in the time harmonic case for the fields which we denote by capitalLatin letters.

The situation is different for the normal components E · n and H · n. We consider Gauss’Electric and Magnetic Laws and choose Ω to be a box which is separated by a surface S intotwo parts Ω1 and Ω2. We apply (1.3) first to all of Ω and then to Ω1 and Ω2 separately. Theaddition of the last two formulas and the comparison with the first yields that the normalcomponent D · n has to be continuous. Analogously we obtain B · n to be continuous at S,if we consider (1.4). With the constitutive equations one gets

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

Conclusion 1.6 The normal components of E and/or H are not continuous at interfaceswhere εc and/or µr have jumps.

Finally, we specify the boundary conditions to the E- and H-modes defined above (see page18). We assume that the surface S is an infinite cylinder in x3−direction with constantcross section. Furthermore, we assume that the volume current density J vanishes near the


boundary S and that the surface current densities take the form Js = jsz for the E-modeand 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. Then in the E-mode we obtain the transmission boundarycondition

[u] = 0 ,

[(σ − iωε)∂u

∂ν

]= −js on S ,

and in the H-mode we get

[u] = js ,

[µ∂u

∂ν

]= 0 on S .

In scattering theory the solutions live in the unbounded exterior of a bounded domain D.In these situations the behavior of electromagnetic fields at infinity has to be taken intoaccount. As an example we consider the fields in the case of a Hertz-dipol at the origin.

Example 1.7 In the case of plane waves (see example 1.4) the wave fronts are planes. Nowwe look for waves with spherical wave fronts. A direct computation shows that

Φ(x) =1

4π

eik|x|

|x|, x ∈ R3 \ 0 ,

is a solution of the Helmholtz equation in R3 \ 0. It is called the fundamental solution ofthe Helmholtz equation at the origin (see Definition 3.1) which is essential as we will see inChapter 3. Physically it can be interpretated as the solution generated by a point source atthe origin similar to the electrostatic case (see Example 1.2).

Furthermore, defining

H(x) = curl[Φ(x, 0) p

]=

1

4πcurl

(eik|x|

|x|p

)=

1

4π∇e

ik|x|

|x|× p

for some a constant vector p ∈ C3, we obtain from equation (6.5) and the fact that Φ(·, 0)solves the Helmholtz equation,

curlH(x) = ∇ div[Φ(x, 0) p

]−∆

[Φ(x, 0) p

]= ∇ div

[Φ(x, 0) p

]+ k2 Φ(x, 0) p

and thus by (6.3) curl curlH = k2H. Therefore H and E = iωε0

curlH constitute a solutionof the time harmonic Maxwell equations in vacuum (see Lemma 1.3).

Computing the gradient

∇Φ(x, 0) =1

4π

(ik − 1

|x|

)eik|x|

|x|x

|x|= ikΦ(x, 0)

x

|x|− eik|x|

4π|x|2x

|x|(1.19)

we obtain by recalling the frequency ω > 0 and wave number k = ω√ε0µ0 that the time-

dependent magnetic field has the form

H(x, t) = H(x) e−iωt =1

4π

(x

|x|× p)(

ik

|x|− 1

|x|2

)ei(k|x|−ωt)

of the Hertz-dipol centered at the origin with dipol moment p e−iωt.

1.4. BOUNDARY AND RADIATION CONDITIONS 21

We observe that all of the functions E ,H,Φ of this example decay as 1/|x| as |x| tendsto infinity. This asymptotic is not sufficient in describing a scattered field, since we alsoobtain solutions of the Helmholtz equation and the Maxwell equations, respectively, withthis asymptotic behaviour if we replace in the last example the wave number k by −k.

To distinguish these fields we must consider the factor ei(k|x|−ωt) and ei(−k|x|−ωt). In the firstcase we obtain “outgoing” wave fronts while in the second case where the wave number isnegative we obtain “ingoing” wave fronts. For the scattering of electromagnetic waves thescattered waves have to be outgoing waves. Thus, it is required to exclude the second onesby additional conditions which are called radiation conditions.

From (1.19) we observe that the two cases for the Helmholtz equation can be distinguishedby subtracting ikΦ. This motivates a general characterization of radiating solutions u of theHelmholtz equation by the Sommerfeld radiation condition, which is

lim|x|→∞

|x|(x

|x|· ∇u− iku

)= 0 .

Similarly, we find a condition for radiating electromagnetic fields from the behavior of theHertz-dipol. Computing

E(x) =i

ωε0

curlH(x)

=i

4πωε0

[k2

(x

|x|× p)× x

|x|+

(1

|x|2− ik

|x|

) (3x · p|x|

x

|x|− p)]

eik|x|

|x|

we conclude

lim|x|→0

[√ε0E(x)× x+

√µ0|x|H(x)

]= lim|x|→0

[√µ0

4π

(2 +

1

ik|x|

)(x

|x|× p)eik|x|

|x|

]= 0 .

Analogously, we obtain lim|x|→0

[√µ0H(x) × x − √ε0|x|E(x)

]= 0. We note that both

conditions are not satisfied if we replace k by −k. The radition condition

lim|x|→∞

[√µ0H(x)× x − |x|

√ε0E(x)

]= 0 (1.20a)

orlim|x|→∞

[√ε0E(x)× x + |x| √µ0H(x)

]= 0 , (1.20b)

are called Silver-Muller radiation condition for time harmonic electromagnetic fields.

Later we will show that these radiation conditions are sufficient for the existence of uniquesolutions of scattering problems. Additionally from the representation theorem in Chapter 3we will prove equivalent formulations and the close relationship of the Sommerfeld and theSilver-Muller radiation condition. Additionally, the limiting absorption principle will giveanother justification for this definition of radiating solutions.

Finally we discuss the energy of scattered waves. In general the energy density of electro-magnetic fields in a linear medium is given by 1

2(E · D + H · B). Thus, from Maxwell’s


equations, the identity (6.9), and the divergence theorem in a region Ω we obtain

∂

∂t

(1

2

∫Ω

E ·D + H ·B dx

)=

∫Ω

E · (curlH−J )−H · curlE dx

=

∫Ω

div(E ×H)− E ·J dx

=

∫∂Ω

(E ×H) · ν ds −∫

Ω

E ·J dx .

This conservation law for the energy of electromagnetic fields is called Poynting’s Theorem.Physically, the right hand side is read as the sum of the energy flux through the surface ∂Ωgiven by the Poynting vector , E ×H, and the electrical work of the fields with the electricalpower J · E .

Let us consider the Poynting theorem in case of time harmonic fields with frequency ω > 0 invacuum, i.e. J = 0, ε = ε0, µ = µ0. Substituting E(x, t) = Re

(E(x)e−iωt

)= 1

2

(E(x)e−iωt +

E(x)eiωt)

and H(x, t) = 12

(H(x)e−iωt + H(x)eiωt

)into the left hand side of Poynting’s

theorem lead to

∂

∂t

(1

2

∫Ω

ε0 E(x, t)2 + µ0 H(x, t)2 dx

)=

∂

∂t

(1

8

∫Ω

ε0

[Re (E(x)2e−2iωt) + 2|E(x)|2

]+ µ0

[Re (H(x)2e−2iωt) + 2|H|2

]dx

)= −iω

4

∫Ω

ε0 Re (E(x)2e−2iωt) + µ0 Re (H(x)2e−2iωt) dx .

On the other hand we compute the flux term as∫∂Ω

(E ×H) · ν ds =1

2

∫∂Ω

Re(E ×H

)· ν ds +

1

2

∫∂Ω

Re(E ×He−2iωt

)· ν ds .

By the Poynting Theorem the two integrals coincide for all t. Thus we conclude for the timeindependent term

Re

∫∂Ω

(E ×H) · ν ds = 0 .

The vector field E ×H is called the complex Poynting vector . If Ω has the form Ω = ΩR =B(0, R) \ D for a bounded domain D with sufficiently smooth boundary contained in theball B(0, R) of radius R we observe conservation of energy in the form

Re

∫|x|=R

ν · (E ×H) ds = Re

∫∂D

ν · (E ×H) ds .

Furthermore, from this identity we obtain

2√ε0µ0 Re

∫∂D

ν · (E ×H) ds

=

∫|x|=R

ε0 |E|2 + µ0 |H × ν|2 ds −∫|x|=R

|√µ0H × ν −√ε0E|2 dx .

1.5. THE REFERENCE PROBLEMS 23

If we additionally assume radiating fields, the Silver-Muller condition implies

limR→∞

∫|x|=R

∣∣√µ0H × ν −√ε0E

∣∣2 ds = 0

which yields boundedness of∫|x|=R |E|

2 ds and, analogously,∫|x|=R |H|

2 ds as R tends to

infinity.

1.5 The Reference Problems

After this introduction into the mathematical description of electromagnetic waves the aim ofthe textbook becomes more obvious. In general we can distinguish (at least) three commonapproaches which lead to existence results of boundary value problems for linear partialdifferential equations: expanding solutions into spherical wave functions by separation ofvariables techniques, reformulation and treatment of a given boundary value problem interms of integral equations in Banach spaces of functions, and the reformulation of theboundary value problem as a variational equation in Hilbert spaces of functions. It is theaim of this monograph to discuss all of these common methods in the case of time harmonicMaxwell’s equations.

We already observed the close connection of the Maxwell system with the scalar Helmholtzequation. Therefore, before we treat the more complicated situation of the Maxwell systemwe investigate the methods for this scalar elliptic partial differential equation in detail.Thus, the structure of all following chapters will be similar: we first discuss the technique inthe case of the Helmholtz equation and then we extend it to boundary value problems forelectromagnetic fields.

As we have mentioned already in the preface it is not our aim to present a collection of allor at least some interesting boundary value problems. Instead, we present the ideas for twoclassical reference problems only which we introduce next.

Scattering by a perfect conductor

The first one is the scattering of electromagnetic waves in vacuum by a perfect conductor:Given a bounded regionD and some solution Einc andH inc of the unperturbed time harmonicMaxwell system

curlEinc − iωµ0Hinc = 0 in R3 , curlH inc + iωε0E

inc = 0 in R3 ,

the problem is to determine E,H of the Maxwell system

curlE − iωµ0H = 0 in R3 \D , curlH + iωε0E = 0 in R3 \D ,

such that E satisfies the boundary condition

ν × E = 0 on ∂D


and both, E and H, can be decomposed into E = Es + Einc and H = Hs +H inc in R3 \Dwith some scattered field Es, Hs which satisfy the Silver-Muller radiation condition

lim|x|→∞

|x|(√µ0H

s(x)× x

|x|−√ε0E

s(x)

)= 0

lim|x|→∞

|x|(√ε0E

s(x)× x

|x|+√µ0H

s(x)

)= 0

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

D ν × E = 0

Einc, H inc

AAAAAA

AAAAAA

HHHHHHj Es, Hs

A perfectly conducting cavity

For the second reference problem we consider D ⊆ R3 to be a bounded domain with suffi-ciently smooth boundary ∂D and exterior unit normal vector ν(x) at x ∈ ∂D. Furthermore,functions µ, ε, and σ are given on D and some source Je : D → C3. Then the problem is todetermine a solution (E,H) of the time harmonic Maxwell system

curlE − iωµH = 0 in D , (1.21a)

curlH + (iωε− σ)E = Je in D , (1.21b)

with the boundary condition

ν × E = 0 on ∂D . (1.21c)

1.5. THE REFERENCE PROBLEMS 25

-

D

µ, ε, σ

ν × E = 0

Of course, for general µ, ε, σ ∈ L∞(R3) we have to give first a correct interpretation of thedifferential equations by a so called weak formulation, which will be presented in detail inChapter 4.

Throughout, we will have these two reference problems in mind for the whole presentation.We will start with constant electric parameters inside or outside a ball. Then we can expectradially symmetric solutions which can be computed by separation of variables in termsof spherical coordinates. This approach will be worked out in Chapter 2. It will lead usto a better understanding of electromagnetic waves from its expansion into spherical wavefunctions. In particular, we can see explicitely in which way the boundary data are attained.

In Chapter 3 we will use the fundamental solution of the scalar Helmholtz equation torepresent electromagnetic waves by integrals over the boundary ∂D of the region D. Theseintegral representations by boundary potentials are the basis for deriving integral equationson ∂D. We prefer to choose the “indirect” approach; that is, to search for the solution interms of potentials with densities which are determined by the boundary data through aboundary integral equation. To solve this we will apply the Riesz–Fredholm theory. In thisway we are to prove the existence of unique solutions of the first reference problem for anyperfectly conducting smooth scattering obstacle.

The cavity problem will be investigated in Chapter 4. The treatment by a variationalapproach requires the introduction of suitable Sobolev spaces. The Helmholtz decompositionmakes it possible to transfer the ideas of the simpler scalar Helmholtz equation to the Maxwellsystem. The Lax–Milgram theorem in Hilbert spaces is the essential tool to establish anexistence result for the second reference problem.

The reason for including Chapter 5 into this monograph is different than for Chapters 3 and4. While for the latter ones our motivation was the teaching aspect (these chapters arosefrom graduate courses) the motivation for Chapter 5 is that we were not able to find sucha thorough presentation of boundary integral methods for Maxwell’s equations on Lipschitzdomains in any textbook. Lipschitz domains, in particular polyedral domains, play obviouslyan important role in praxis. We were encouraged by our collegues from the numerical analysis


group to include this chapter to have a reference for further studies. The integral equationmethods themselves are not much different from the classical ones on smooth boundaries.The mapping properties of the boundary operators, however, require a detailed study ofSobolev spaces on Lipschitz boundaries. In this Chapter 5 we use and combine methods ofChapters 3 and 4. Therefore, this chapter can not be studied independently of Chapters 3and 4.

Chapter 2

Expansion into Wave Functions

This chapter, which is totally independent of the remaining parts of this monograph,1 studiesthe fact that the solutions of the scalar Helmholtz equation or the vectorial Maxwell systemin balls can be expanded into certain special “wave functions”. We begin by expressing theLaplacian in spherical coordinates and search for solutions of the scalar Laplace equationor Helmholtz equation by separation of the (spherical) variables. It will turn out thatthe spherical parts are eigensolutions of the Laplace–Beltrami operator while the radialpart solves an equation of Euler type for the Laplace equation and the spherical Besseldifferential equation for the case of the Helmholtz equation. The solutions of these differentialequations will lead to spherical harmonics and spherical Bessel and Hankel functions. Wewill investigate these special functions in detail and derive many important properties. Themain goal is to express the solutions of the interior and exterior boundary value problemsas series of these wave functions. As always in this monograph we first present the analysisfor the scalar case of the Laplace equation and the Helmholtz equation before we considerthe more complicated case of Maxwell’s equations.

2.1 Separation in Spherical Coordinates

The starting point of our investigation is the boundary value problem inside (or outside)the “simple” geometry of a ball in the case of a homogeneous medium. We are interested insolutions u of the Laplace equation or Helmholtz equation – and later of the time-harmonicMaxwell system – which can be separated into a radial part v : R>0 → C and a sphericalpart K : S2 → C; that is,

u(x) = v(r)K(x) , r > 0 , x ∈ S2 ,

1except of the proof of a radiation condition

27

28 CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS

where S2 = x ∈ R3 : |x| = 1 denotes the unit sphere in R3. Here, r = |x| =√x2

1 + x22 + x2

3

and x = x|x| ∈ S

2 denote the spherical coordinates; that is,

x =

r cosϕ sin θr sinϕ sin θr cos θ

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

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

In the previous chapter we have seen already the importance of the differential operator ∆in the modelling of electromagnetic waves. It occurs directly in the stationary cases and alsoin the differential equations for the magnetic and electric Hertz potentials. Also, solutionsof the full Maxwell system solve the vector Helmholtz equation in particular cases. Thus therepresentation of the Laplacian in spherical polar coordinates is of essential importance.

∆ =1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin2 θ

∂2

∂ϕ2+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)(2.1)

=∂2

∂r2+

2

r

∂

∂r+

1

r2 sin2 θ

∂2

∂ϕ2+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

).

Definition 2.1 Functions u : Ω → R which satisfy the Laplace equation ∆u = 0 in Ω arecalled harmonic functions in Ω. Complex valued functions are harmonic if their 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 on theunit sphere.

Definition 2.2 The differential operator ∆S2 : C2(S2)→ C(S2) with representation

∆S2 =1

sin2 θ

∂2

∂ϕ2+

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

).

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

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

we obtain

∆ = D2r +

2

rDr +

1

r2∆S2 =

1

r2Dr

(r2Dr

)+

1

r2∆S2 .

Assuming that a potential or a component of a time-harmonic electric or magnetic field canbe separated as u(x) = u(rx) = v(r)K(x) with r = |x| and x = x/|x| ∈ S2, a substitutioninto the Helmholtz equation ∆u+ k2u = 0 for some k ∈ C leads to

0 = ∆u(rx) + k2u(rx) = v′′(r)K(x) +2

rv′(r)K(x) +

1

r2v(r) ∆S2K(x) + k2v(r)K(x) .

2.1. SEPARATION IN SPHERICAL COORDINATES 29

Thus we obtain, provided v(r)K(x) 6= 0,

v′′(r) + 2rv′(r)

v(r)+

1

r2

∆S2K(x)

K(x)+ k2 = 0 . (2.2)

If there is a nontrivial solution of the supposed form, it follows the existence of a constantλ ∈ C such that v satisfies the ordinary differential equation

r2v′′(r) + 2r v′(r) +(k2r2 + λ

)v(r) = 0 for r > 0 , (2.3)

and K solves the partial differential equation

∆S2K(x) = λK(x) on S2 . (2.4)

In the functional analytic language the previous equation describes the problem to determineeigenfunctions K and corresponding eigenvalues λ of the spherical Laplace-Beltrami operator∆S2 . This operator is selfadjoint and non-positive with respect to the L2(S2)−norm, seeExercise 2.2. Especially, we observe that λ ∈ R<0. We will explicitely construct a completeorthonormal system of eigenfunctions although the existence of such a system follows fromfunctional analytic arguments as well because the resolvent of ∆S2 is compact.

We observe that the parameter k appears in the equation for v only. The spherical equation(2.4) is independent of k, and the system of eigenfunctions will be used for both, the Laplaceand the Helmholtz equation.

Using the explicit representation of the spherical Laplace-Beltrami operator transforms (2.4)into

∂2

∂ϕ2K(θ, ϕ) + sin θ

∂

∂θ

(sin θ

∂

∂θ

)K(θ, ϕ) = λ sin2 θ K(θ, ϕ)

where we write K(θ, ϕ) for K(x). Assuming eigenfunctions of the form K(θ, ϕ) = y1(ϕ) y2(θ)leads to

y′′1(ϕ)

y1(ϕ)+

sin θ(sin θ y′2(θ)

)′y2(θ)

− λ sin2 θ = 0 (2.5)

provided y1(ϕ)y2(θ) 6= 0. If the decomposition is valid there exists a constant µ ∈ C suchthat y′′1 = µy1. Since we are interested in differentiable solutions u, the function y1 mustbe 2π periodic. Using a fundamental system of the linear ordinary differential equation weconclude that µ = −m2 with m ∈ Z, and the general solution of y′′1 = µy1 is given by

y1,m(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 basics of the classical Fourier theory. Inparticular, we recall that every function f ∈ L2(−π, π) allows an expansion in the form

f(t) =∑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)−M∑

m=−N

fm eimt

∣∣∣∣∣2

dt −→ 0 , N,M →∞ .

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

Next we consider the dependence on θ. Using µ = −m2 turns (2.5) into

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 arrive at the associatedLegendre differential equation

((1− z2)w′(z)

)′ − (λ+

m2

1− z2

)w(z) = 0 . (2.6)

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. In particular, we will see that only for λ = −n(n + 1) for n ∈ N0 there existsmooth solutions.

Obviously, the radial part of separated solutions given by the ordinary differential equation(2.3) depends on the wave number k. For k = 0 and λ = −n(n + 1) we obtain the Eulerequation

r2v′′(r) + 2r v′(r) − n(n+ 1) v(r) = 0 . (2.7)

By the ansatz v(r) = rµ we compute µ(µ + 1) = n(n + 1), which leads to the fundamentalset of solutions

v1,n(r) = rn and v2,n(r) = r−(n+1) .

If we are interested in non–singular solutions of the Laplace equation inside a ball we haveto choose v1,n(r) = rn. In the exterior of a ball v2,n(r) = r−(n+1) will lead to solutions whichdecay as |x| tends to infinity.

In the case of k 6= 0 we rewrite the differential equation (2.3) for λ = −n(n + 1) by thesubstitution z = kr and v(r) = v(kr) into the spherical Bessel differential equation

z2 v′′(z) + 2z v′(z) +[z2 − n(n+ 1)

]v(z) = 0 . (2.8)

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 the Maxwell system.

2.2. LEGENDRE POLYNOMIALS 31

2.2 Legendre Polynomials

Let us consider the case of harmonic functions u, that is ∆u = 0. In the previous sectionwe saw already that a separation in spherical coordinates leads to solutions of the formu(x) = rn(c1 e

imϕ + c2 e−imϕ)w(cos θ) where w is determined by the associated Legendre

differential equation (2.6) with λ ∈ R<0.

We discuss this real ordinary differential equation (2.6) first for the special case m = 0; thatis,

d

dz

[(1− z2)w′(z)

]− λw(z) = 0 , −1 < z < 1 . (2.9)

This is a differential equation of Legendre type.

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

Theorem 2.3 (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 satisfies thedifferential 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(2.9) 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 (2.9). 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, . . . . (2.10)

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 (2.9) 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)+λ > 0 and[(2k0+1)−λ/2

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

]≤ 1/2.

Then a2k does not change its sign anymore for k ≥ k0. Taking the logarithm of (2.10) 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 (2.9). 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 if


n 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

for the polynomial w would imply w′(1) = 0. Differentiating the differential equation wouldyield w(k)(1) = 0 for all k ∈ N, a contradiction to w 6= 0. 2

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

Corollary 2.4 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 (2.9) for Pnby Pm(x), the differential equation (2.9) 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 further results for the Legendrepolynomials.

Lemma 2.5 For the Legendre polynomials it holds

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). Thus, we have

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

A useful representation of the Legendre polynomials is a formula first shown by B.O. Ro-drigues.

Theorem 2.6 For all n ∈ N0 the Legendre polynomials satisfy the formula of Rodrigues

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 . (2.11)

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 (2.11) 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 Rodrigues we can compute the norm of Pn in L2(−1, 1).

Theorem 2.7 The Legendre polynomials satisfy

(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 Rodrigues 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 Rodrigues is also useful in proving recursion formulas for the Legendrepolynomials. We prove only some of these formulas. For the remaining parts we referto the exercises.


Theorem 2.8 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 Rodrigues,

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 2.7:

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 multiplication of (c) for n+ 1 instead of n by n+ 1 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 by subtraction

0 =[(2n+ 1)− (n+ 1)2

]Pn(x) +

[(2n+ 1)− (n+ 1)

]xP ′n(x) − nP ′n−1(x)

which yields (d).For the proofs of (e)–(h) we refer to the exercises. 2

Now we go back to the associated Legendre differential equation (2.6) and determine solutionsin C2(−1,+1) ∩ C[−1,+1] for m 6= 0.


Theorem 2.9 The functions

Pmn (x) = (1− x2)m/2

dm

dxmPn(x) , −1 < x < 1, 0 ≤ m ≤ n ,

are solutions of the differential equation (2.6) 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 ,

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

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+2n (x) .

Differentiating equation (2.9) 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)



2.3 Expansion into 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) = rn Pmn (cos θ) e±imϕ

are harmonic functions in all of R3. Analogously, the functions

v(r, θ, ϕ) = r−n−1 Pmn (cos θ) e±imϕ

are harmonic in R3 \ 0 and decay at infinity.

It is not so obvious that the functions hmn are polynomials.

Theorem 2.10 The functions hmn (r, θ, ϕ) = rnPmn (cos θ) e±imϕ with 0 ≤ m ≤ n are homo-

geneous polynomials of degree n. The latter property means that hmn (µx) = µnhnm(x) for allx ∈ Rn and µ ∈ R.

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 `. Then

Pn(t) =∑j=0

aj t2j , t ∈ R ,

for some aj. From r = |x| and x3 = r cos θ we write h0n in the form

h0n(x) = |x|n Pn

(x3

|x|

)= |x|2`

∑j=0

a2jx2j

3

|x|2j=∑j=0

a2j x2j3 |x|2(`−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. Let again n = 2` be even. Then Q(t) =

∑`j=0 bj t

2j−m for some bj.Furthermore, we use the expressions

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

.

2.3. EXPANSION INTO SPHERICAL HARMONICS 41

to obtain

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

=∑j=0

bj r2(`−j) x2j−m

3 (x1 + ix2)m

which proves the assertion for even n. For odd n one argues analogously. It is clear from thedefinition that the polynomial hmn is homogeneous of degree n. 2

Definition 2.11 Let n ∈ N ∪ 0.

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

(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 spherical surface 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 2.12 If Kn denotes a surface spherical harmonic of order n, the following holds.

(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 2.13 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

∂x22denote 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 that


any 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 (x) = hmn (θ, ϕ) = rnP|m|n (cos θ)eimϕ with

−n ≤ m ≤ n, which we have determined by separation, constitute such an orthogonal basisof the space of spherical harmonics of order n.

Theorem 2.14 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 thetheorem 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 2.15 The norm of the associated Legendre functions Pmn in L2(−1, 1) is given by

+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 2.7 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 (2.9) (m− 1)-times (see (2.12) for mreplaced 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 (2.15) 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 (2.4) with λ = −n(n+ 1) we remember that Y mn satisfies the differential equation

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 ;

that is,∆S2Y m

n + n(n+ 1)Y mn = 0 (2.17)

with the Laplace-Beltrami operator ∆S2 of Definition 2.2. Thus, we see that λ = −n(n+ 1)for n ∈ N are the eigenvalues of ∆S2 with eigenfunctions Y m

n , |m| ≤ n. We mentionedalready (see also Exercise 2.2) that this operator ∆S2 is selfadjoint and non-negative. In thesetting of abstract functional analysis we note that −∆ is a densely defined and unboundedoperator from L2(S2) into itself which is selfadjoint. This observation allows the use ofgeneral 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 are going to prove thisfact directly starting with the following result – which is of independent interest.

Theorem 2.16 For any f ∈ L2(−1, 1) the following Funk–Hecke Formula holds.∫S2

f(x · y)Y mn (y) ds(y) = λn Y

mn (x) , x ∈ S2 ,

for all n ∈ N and m = −n, . . . , n where λn = 2π∫ 1

−1f(t)Pn(t) dt.


Proof: We keep x ∈ S2 fixed and choose an orthogonal matrix A which depends on xsuch that x := A−1x = A>x is the “north pole”; that is, x = (0, 0, 1)>. The transformationy = Ay′ yields∫

S2

f(x · y)Y mn (y) ds(y) =

∫S2

f(x · Ay′)Y mn (Ay′) ds(y′) =

∫S2

f(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 , (2.18)

where ak =∫S2

Y mn (Ay)Y −kn (y) ds(y). Using this and polar coordinates

y = (sin θ cosϕ , sin θ sinϕ , cos θ)> ∈ S2 yields (note that x · y = cos θ)∫S2

f(x · y)Y mn (y) ds(y) =

n∑k=−n

ak

∫S2

f(x · y)Y kn (y) ds(y)

=n∑

k=−n

ak

√(2n+ 1) (n− |k|)!

4π (n+ |k|)!·

·∫ π

0

∫ 2π

0

f(cos θ)P |k|n (cos θ) eikϕ dϕ sin θ dθ

= a0

√2n+ 1

4π2π

∫ π

0

f(cos θ)Pn(cos θ) sin θ dθ = λn a0

√2n+ 1

4π

where we have used the substitution t = cos θ in the last integral. Now we substitute y = xin (2.18) and have, using Y n

k (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 ∫S2

f(x · y)Y mn (y) ds(y) = λn Y

mn (x)


As a first application of this result we prove the addition formula. A second application willbe the Jacobi–Anger expansion, see Theorem 2.32.

Theorem 2.17 The Legendre polynomials Pn and the spherical surface harmonics definedin (2.16) satisfy the addition formula

Pn(x · y) =4π

2n+ 1

n∑m=−n

Y mn (x)Y −mn (y) for all x, y ∈ S2 .


Proof: For fixed y ∈ S2 the function x 7→ Pn(x · y) is a spherical harmonic of order n and,therefore, has an expansion of the form

Pn(x · y) =n∑

m=−n

∫S2

Pn(z · y)Y −m(z) ds(z)Y mn (x) =

n∑m=−n

λn Y−m(y)Y m

n (x)

with

λn = 2π

∫ 1

−1

Pn(t)Pn(t) dt =4π

2n+ 1

by Theorem 2.7 where we used the previously proven Funk–Hecke formula for f = Pn. 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 of Theorem 2.17 for y = x.Part (b) follows directly 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 . (2.19a)

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 . (2.19b)

The convergence in (2.19a) and (2.19b) has to be 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 (2.19b).

Here, the space C1(S2) consists of those functions f such that (with respect to sphericalcoordinates θ and ϕ) the functiuons f , ∂f/∂θ, and 1

sin θ∂f/∂ϕ are continuous and periodic

with respect to ϕ.

Proof: First we prove the second part. Therefore, let f ∈ C1(S2) with ‖f‖1,∞ ≤ M . Withthe Addition Formula, see Theorem 2.17, we obtain for the partial sum

(SNf)(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 (2.19a) and (2.19b). With (2n+ 1)Pn = P ′n+1 − P ′n−1

of Theorem 2.8 (set P−1 ≡ 0) this yields

(SNf)(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 z = (0, 0, 1)> be the north pole and choose an orthogonal matrix A (depending onx) such that Ax = z. Then, by the transformation formula,

(SNf)(x) =1

4π

∫S2

f(Ay)[P ′N+1(z · y) + P ′N(z · y)

]ds(y) .

In spherical polar coordinates y(θ, ϕ) = (sin θ cosϕ, sin θ sinϕ, cos θ)> this is, defining F (θ) =

12π

2π∫0

f(Ay(θ, ϕ)

)dϕ,

(SNf)(x) =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 differen-tiable in (−1, 1) and continuous in [−1, 1]. The value θ = 0 corresponds to the north poley = z, thus

F (0) =1

2π

2π∫0

f(Az) dϕ = f(Az) = f(x) ,

and therefore for any δ ∈ (0, 1) we obtain

∣∣(SNf)(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

=1

2( 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 2.5). From d

dtarccos t < 0 we conclude

I1 + I3 ≤ −−1+δ∫−1

−1∫

1−δ

|F ′(arccos t)| ddt

arccos t dt

≤ ‖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, depending only 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 general propertyof orthonormal systems that SNf is the best approximation of f in the subspace spanY m

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

‖SNf − 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,

‖SNf − f‖L2(S2) ≤ ‖SNh− f‖L2(S2) ≤ ‖SNh− h‖L2(S2) + ‖h− f‖L2(S2)

≤√

4π ‖SNh− h‖∞ + ‖h− f‖L2(S2) ≤√

4π ‖SNh− h‖∞ +ε

2.

Since SNh 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 can 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 is independent of ϕ, thus it is a zonal function. The expansion (2.19a) 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 .

Thus,

f(x3) =∞∑n=0

√4π

2n+ 1fn Y

0n (x) =

∞∑n=0

fn Pn(x3)

and the proof is complete. 2


2.4 Laplace’s Equation in the Interior and Exterior of

a Ball

In the previous two sections we constructed explicitely a complete orthonormal system offunctions in L2(S2). They play exactly the role of the normalized exponential functions

12π

exp(inϕ) : n ∈ Z

on the unit circle S1, parametrized by x = (cosϕ, sinϕ)>, ϕ ∈ [0, 2π];that is, the classical Fourier expansion functions. It is the aim of this section to expandsolutions of the Laplace equation for balls and solve the corresponding Dirichlet boundaryvalue problems.

Theorem 2.21 Let u ∈ C2(B(0, R)

)be harmonic in the ball B(0, R); that is, satisfies

the Laplace equation ∆u = 0 in B(0, R). Then there exist unique αmn ∈ C, |m| ≤ n,n = 0, 1, 2, . . . with

u(rx) =∞∑n=0

n∑m=−n

αmn rn Y m

n (x) , 0 ≤ r < R , x ∈ S2 . (2.20)

The series converges uniformly with all of its derivatives in every closed ball B[0, R′] withR′ < R.

Proof: For every r ∈ (0, R) the function x 7→ u(rx) is in C2(S2), and, therefore, can beexpanded 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 differential equation of Euler type. Using the Laplace equa-tion in spherical coordinates for u, the self-adjoint Laplace-Beltrami operator ∆S2 , and theeigenvalue equation (2.17), yields

d

dr

(r2 d

dr

)umn (r) =

∫S2

∂

∂r

(r2 ∂u(r, x)

∂r

)Y −mn (x) ds(x)

= −∫S2

∆S2u(r, x)Y −mn (x) ds(x)

= −∫S2

u(r, x) ∆S2Y −mn (x) ds(x)

= n(n+ 1)

∫S2

u(r, x)Y −mn (x) ds(x) = n(n+ 1)umn (r) .

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

arbitrary αmn . Therefore, u has the desired form (2.20).

2.4. LAPLACE’S EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL 51

It remains to prove uniqueness of the expansion coefficients and uniform convergence. Wefix R′ < R and choose R with R′ < R < R. Multiplying the representation of u(Rx) withY −qp (x), and integrate over S2 to obtain

(u(R, ·), Y q

p

)L2(S2)

= αqp Rp which proves uniqueness

of αmn . Furthermore, by the addition formula of Theorem 2.17 we find the representation

u(rx) =∞∑n=0

n∑m=−n

(u(R, ·), Y m

n

)L2(S2)

(r

R

)nY mn (x)

=∞∑n=0

(r

R

)n ∫S2

n∑m=−n

Y −mn (y)Y mn (x)u(Ry) ds(y)

=∞∑n=0

(r

R

)n2n+ 1

4π

∫S2

Pn(x · y)u(Ry) ds(y)

for r ≤ R′ and x ∈ S2. From this representation and the observation that∣∣∣djPndtj

(t)∣∣∣ ≤ cjn

2j on

[−1, 1] (see Exercise 2.6) we conclude for any differential operator D` = ∂|`|/(∂r`1∂θ`2∂ϕ`3)in spherical coordinates that the series for D`u(x) converges uniformly in B[0, R′] because,using the Cauchy-Schwarz inequality, it is dominated by the convergent series

c ‖u(R, ·)‖L2(S2)

∞∑n=0

(2n+ 1)n2`

(R′

R

)nfor some c > 0. This ends the proof. 2

Now we consider the boundary value problem of Dirichlet type in B(0, R); that is,

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

for given boundary function f . We study the cases fR ∈ L2(S2) and fR ∈ C2(S2) simul-tanously where we set fR(x) = f(Rx), x ∈ S2, here and in the following.

Theorem 2.22 (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.

(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 2.21.(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. By the previous theorem u can be representatedas a series in the form (2.20). Let R′ < R and ε > 0 be arbitrary. Choose Rε ∈ [R′, R) suchthat ‖uε‖L2(S2) ≤ ε where uε(x) = u(Rεx). Multiplying the representiation of u(Rεx) withY −qp (x) and integrating over S2 yields (uε, Y

qp )L2(S2) = αqpR

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 ≤ R′,

‖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, R′). Since R′ < Rwas arbitrary u vanishes in B(0, R).

Uniform convergence of the series and all of its derivates on every ball B[0, R′] with R′ < Ris shown as in the proof of the previous theorem. The series for D`u(x) is dominated by theconvergent series

c ‖fR‖L2(S2)

∞∑n=0

(2n+ 1)n2`

(R′

R

)n.

Therefore, u ∈ C∞(B(0, R)

)solves the Laplace equation. 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 (2.17)and the symmetry of ∆S2 we express the expansion coefficient of f as

(fR, Ymn )L2(S2) = − 1

n(n+ 1)(fR,∆S2Y m

n )L2(S2) = − 1

n(n+ 1)(∆S2fR, Y

mn )L2(S2) .

2.4. LAPLACE’S EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL 53

Thus, uniform convergence follows, since for N ∈ N we can estimate the remainder for anyr ≤ R and any x ∈ S2 by

∞∑n=N

n∑m=−n

∣∣(fR, Y mn )L2(S2)

∣∣ ( rR

)n|Y mn (x)| ≤

∞∑n=N

n∑m=−n

∣∣(∆S2fR, Ymn )L2(S2)

∣∣ 1

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

≤

[∞∑n=N

n∑m=−n

∣∣(∆S2fR, Ymn )L2(S2)

∣∣2]1/2 [ ∞∑n=N

n∑m=−n

1

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

]1/2

= ‖∆S2fR‖L2(S2)

[∞∑n=N

1

n2(n+ 1)2

n∑m=−n

|Y mn (x)|2

]1/2

=1√4π‖∆S2fR‖L2(S2)

[∞∑n=N

2n+ 1

n2(n+ 1)2

]1/2

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

To end this section we consider the situation in the exterior of the closed ball B[0, R]. Weare interested in harmonic functions which tend to zero at infinity. The following theoremcorresponds to Theorem 2.21.

Theorem 2.23 Let u ∈ C2(R3 \ B[0, R]

)harmonic in the exterior of the ball B[0, R]; that

is, satisfies the Laplace equation ∆u = 0 for |x| > R. Furthermore, we assume that

limr→∞

u(rx) = 0 for every x ∈ S2 . (2.23)

Then there exist unique coefficients αmn ∈ C, |m| ≤ n, n = 0, 1, 2, . . . with

u(rx) =∞∑n=0

n∑m=−n

αmn r−n−1 Y m

n (x) , r > R , x ∈ S2 . (2.24)

The series converges uniformly with all of its derivatives outside of every ball B(0, R′) withR′ > R.

The proof is almost the same as the proof of Theorem 2.21. The main difference is that onehas to select the solution umn (r) = αmn r

−n−1 in Euler’s differential equation ddr

(r2 d

dr

)umn (r) =

n(n+ 1)umn (r) instead of umn (r) = αmn rn. We leave the proof to the reader as Exercise 2.7.

Also, the interior boundary value problem has an exterior analog. Let again f : ∂B(0, 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) , (2.25)

and (2.23). This latter condition is needed to ensure uniqueness. Indeed, we observe thatthe function u(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.

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


Theorem 2.24 (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 (2.23) 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) (2.26a)

=1

4π

∞∑n=0

(2n+ 1)

(R

r

)n+1 ∫S2

f(Ry)Pn(x · y) ds(y) (2.26b)

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

(2.25), (2.23) which is again given by (2.26a), (2.26b). The series converge uniformly forr ≥ R.

The proof follows again very closely the proof of the corresponding interior case. We omitthe details and refer again to Exercise 2.7.

2.5 Bessel Functions

The previous investigations for harmonic functions can be applied in electrostatics or mag-netostatics (see section 1.3). But for time harmonic electromagentic fields we will solve,as a next step towards the Maxwell system, the same Dirichlet problem for the Helmholtzequation instead of the Laplace equation. The radial functions rn which appeared in, e.g.,(2.22a), (2.22b) have to be replaced by Bessel functions. The introduction of these importantfunctions of mathematical physics is subject of the present section.

From the separation u(x) = v(r)K(x) in spherical coordinates and the eigenvalues −n(n+1)of the Beltrami operator we obtained for the radial component v(kr) = v(r) with z = kr thespherical Bessel differential equation

z2 v′′(z) + 2z v′(z) +[z2 − n(n+ 1)

]v(z) = 0 .

(see equation (2.8)). We will investigate this linear differential equation of second order forarbitrary z ∈ C. For z 6= 0 the differential equation is equivalent to

v′′(z) +2

zv′(z) +

(1− n(n+ 1)

z2

)v(z) = 0 in C \ 0 . (2.27)

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 linearly independent solutions of (2.27).

2.5. BESSEL FUNCTIONS 55

Lemma 2.25 Let Ω ⊆ C \ 0 be a domain. Then there exist at most two linearly indepen-dent holomorphic solutions of (2.27) in Ω.

Proof: Let wj, j = 1, 2, 3, be 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 (2.27), 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,n(r) = rn of the corresponding equation (2.7) for the Laplaceequation and λ = −n(n + 1) we make an ansatz for a smooth solution of (2.27) as a powerseries in the form

w1(z) = zn∞∑`=0

a` z` =

∞∑`=0

a` z`+n .

Substituting the ansatz into (2.27) 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 seen that the radius of convergence is infinity. Therefore, this functionw1 is a holomorphic solution of the Bessel differential equation in all of C.

Now we determine a second solution of (2.27) which is linearly independent of w1. Motivatedby the singular solution v2,n(r) = r−n−1 of (2.7) for λ = −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 (2.27) 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 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 2.26 For all z ∈ C the spherical Bessel functions of first and second kind andorder 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 ,

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

for positive integers k is defined by

(−k)!

(−2k)!= (−1)k

(2k)!

k!, k ∈ N .

The functions

h(1)n = jn + i yn ,

h(2)n = jn − i yn ,

are called Hankel functions of first and second kind and order n ∈ N0.

For many applications the Wronskian of these functions is important.

Theorem 2.27 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 (2.27) 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 ∈ C which we determine

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

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)

]


as 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 (2.27).

The functions jn and yn are closely related to the sine cardinal function sin z/z and to thefunction cos z/z. Especially, from Rayleigh’s formulas below we note that j0(z) = sin z/z

and ih(1)0 (z) = exp(iz)/z.

Theorem 2.28 (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 power series expansionof cos 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

In the following definition our results for the Helmholtz equation are collected.

Definition 2.29 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. Thereal 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, thatis k = 0. For the Helmholtz equation the mappings rx 7→ jn(kr)Y m

n (x) are the expansion

functions for solutions inside of balls. Furthermore the functions h(1)n are used to derive

expansions in the exterior of balls. From the definition we observe that they are singular atthe origin of order n. Additionally, we need their asymptotic behavior for r → ∞, whichcan be derived from the previous theorem.

Theorem 2.30 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|. The corresponding formulas for jn and yn are derived bytaking the real- and imaginary parts, respectively.


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

1

z

d

dz

)nexp(iz)

z= in

exp(iz)

zn+1Qn

(1

z

). (2.28)

This would prove the assertions.For n = 0 this is obvious for the constant polynomial Q0 = 1. Let it be true for n. Then(

1

z

d

dz

)n+1exp(iz)

z=

1

z

d

dz

[in

exp(iz)

zn+1Qn

(1

z

)]=

in

zeiz[

i

zn+1Qn

(1

z

)− (n+ 1)

1

zn+2Qn

(1

z

)− 1

zn+1Q′n

(1

z

)1

z2

]= in+1 exp(iz)

zn+2Qn+1

(1

z

)with Qn+1(t) = Qn(t) + i(n+ 1) tQn(t) + it2Q′n(t). Obviously, Qn+1 is a polynomial of order

at most n+ 1 and Qn+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 2.31 For every ε > 0 and R > 0 we have the following.

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→∞ ,

h(1)n (z) = −i (2n)!

n!

2

(2z)n+1

[1 +O

(1

n

)]= −i (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: We prove only the first formula. 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

4(n+ 1)

∞∑`=1

R2`

`!

because for ` ≥ 1

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

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

1

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

(n+ 1) · · · (n+ `)

(2n+ `+ 2) · · · (2n+ 2`+ 1)︸︷︷︸≤ 1/2

≤ 1

4(n+ 1).


2

We finish this section with a second application of the Funk–Hecke Formula of Theorem 2.16.We consider the expansion of the special solution u(x) = exp(ikx·y), x ∈ R3, of the Helmholtzequation for some fixed y ∈ S2 which describes a plane time harmonic electromagnetic fieldtravelling in the direction y (see example 1.4).

Theorem 2.32 For x, y ∈ S2 and r > 0 there holds the 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 every R > 0 the series converges uniformly with respect to x, y ∈ S2 and 0 ≤ r ≤ R.

Proof: The two representations coincide as seen from the addition formula. Uniform conver-gence follows from the second representation and the asymptotic behaviour of jn as n→∞because |Pn(t)| ≤ 1 on [−1, 1]. We will prove the first representation and apply the Funk–Hecke formula from Theorem 2.16. Indeed, we fix z := kr and y ∈ S2 and expand x 7→ eiz x·y

into spherical harmonics; that is,

eiz x·y =∞∑n=0

n∑m=−n

amn (z, y)Y mn (x)

where

amn (z, y) =

∫S2

eiz x·y Y −mn (x) ds(x) = λn(z)Y −mn (y)

with λn(z) = 2π∫ 1

−1eizt Pn(t) dt. It remains to show that λn(z) = 4π injn(z). The function

λn solves the Bessel differential equation. Indeed, using the differential equation (2.9) forthe Legendre polynomial, we conclude that

d

dz

(z2λ′n(z)

)+[z2 − n(n+ 1)

]λn(z)

= 2π

∫ 1

−1

[(1− t2)z2 + 2izt

]eizt Pn(t) dt − n(n+ 1)2π

∫ 1

−1

eizt Pn(t) dt

= 2π

∫ 1

−1

[(1− t2)z2 + 2izt

]eizt Pn(t) dt + 2π

∫ 1

−1

eiztd

dt

[(1− t2)P ′n(t)

]dt

= 0

by two partial integrations of the second integral. Since λn is smooth we conclude thatλn(z) = αnjn(z) for some constant αn ∈ C which we determine by the behavior at theorigin. First, we use the fact that Pn is orthogonal to all polynomials of order less that n.The coefficients of the term tn is given by (2n−1)!

n! (n−1)! 2n−1 which can be seen from the recursion

formula (b) of Theorem 2.8 by induction (see Excercise 2.5). Therefore,

dn

dznλn(0) = 2π in

∫ 1

−1

tn Pn(t) dt = 2π inn! (n− 1)! 2n−1

(2n− 1)!

∫ 1

−1

Pn(t)2 dt = 2π in(n!)2 2n+1

(2n+ 1)!


where we have used Theorem 2.7. On the other hand, from Definition 2.26 of the Besselfunction we observe that

dn

dznjn(0) =

n!

(2n+ 1)!

dn

dzn(2z)n

∣∣∣∣z=0

=(n!)2 2n

(2n+ 1)!.

Comparing these two formulas yields αn = 4π in and finishes the proof. 2

2.6 The Helmholtz Equation in the Interior and Exte-

rior of a Ball

Now we are ready to study the series expansion of solutions of the Helmholtz equation∆u + k2u = 0 inside and outside of balls and the corresponding boundary value problems.The first theorem is the analog of Theorem 2.21.

Theorem 2.33 Let k ∈ C \ 0 with Im k ≥ 0 and R > 0 and u ∈ C2(B(0, R)

)solve the

Helmholtz equation ∆u + k2u = 0 in B(0, R). Then there exist unique αmn ∈ C, |m| ≤ n,n = 0, 1, 2, . . . with

u(rx) =∞∑n=0

n∑m=−n

αmn jn(kr)Y mn (x) , 0 ≤ r < R , x ∈ S2 .

The series converges uniformly with all of its derivatives in every closed ball B[0, R′] withR′ < R.

Proof: We argue as the proof of Theorem 2.21. For every r ∈ (0, R) the function u(rx) canbe 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 the

spherical Bessel differential equation. Using the Helmholtz equation for u in spherical polarcoordinates, and that the functions Y m

n are eigenfunctions of the selfadjoint Laplace-Beltrami

2.6. THE HELMHOLTZ EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL63

operator yields for r > R

d

dr

(r2 d

dr

)umn (r) + r2k2umn (r)

=

∫S2

[∂

∂r

(r2 ∂u(r, x)

∂r

)+ r2k2u(r, x)

]Y −mn (x) ds(x)

= −∫S2

∆S2u(r, x)Y −mn (x) ds(x)

= −∫S2

u(r, x) ∆S2Y −mn (x) ds(x)

= n(n+ 1)

∫S2

u(r, x)Y −mn (x) ds(x) = n(n+ 1)umn (r) .

After the transformation z = kr it turns into Bessel’s differential equation (2.8) for thefunctions umn . The general solution is given by umn (r) = αmn jn(kr) + βmn yn(kr) for arbitraryαmn , β

mn . Therefore, u has the form

u(rx) =∞∑n=0

n∑m=−n

[αmn jn(kr) + βmn yn(kr)

]Y mn (x) , x ∈ S2 , r > 0 . (2.29)

We are interested in smooth solutions in the ball B(0, R). Therefore, βmn = 0.

To prove uniform convergence we express u(x) in a different form as in the proof of Theo-rem 2.33. Fix R′ < R and choose R with R′ < R < R. Multiplying the representation ofu(Rx) with Y −qp (x), and integrate over S2 to obtain

(u(R, ·), Y q

p

)L2(S2)

= αqp jp(kR) which

proves uniqueness of αmn and, furthermore, the representation of u as

u(rx) =∞∑n=0

n∑m=−n

(u(R, ·), Y m

n

)L2(S2)

jn(kr)

jn(kR)Y mn (x)

=∞∑n=0

jn(kr)

jn(kR)

∫S2

n∑m=−n

Y −mn (y)Y mn (x)u(Ry) ds(y)

=∞∑n=0

jn(kr)

jn(kR)

2n+ 1

4π

∫S2

Pn(x · y)u(Ry) ds(y)

for r ≤ R′ and x ∈ S2. Here we used again the Addition Formula of Theorem 2.17.So far, we followed exactly the proof of Theorem 2.21. Now we have to use again the

estimate∣∣∣djPndtj

(t)∣∣∣ ≤ cjn

2j on [−1, 1] (see Exercise 2.6) and the asymptotic forms of the

derivatives of the Bessel functions for large orders n (see Exercise 2.9). Therefore, for anydifferential operator D` = ∂|`|/(∂r`1∂θ`2∂ϕ`3) the series for D`u(x) converges uniformly in


B[0, R′] because it is dominated by the convergent series

c‖u(R, ·)‖L2(S2)

∞∑n=0

(2n+ 1)n2`

(R′

R

)nwith a constant c > 0 depending on R′. This ends the proof. 2

Now we transfer the idea of Theorem 2.22 to the interior boundary value problem

∆u+ k2u = 0 in B(0, R) , u = f on ∂B(0, R) . (2.30)

We recall that we transform functions on the sphere x ∈ R3 : |x| = R of radius R onto theunit sphere by setting fR(x) = f(Rx), x ∈ S2.

Theorem 2.34 Assume that k ∈ C with Im k ≥ 0 and R > 0 are such that jn(kR) 6= 0 forall n ∈ N0.

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

)of ∆u + k2u = 0

in B(0, R) withlimr→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) (2.31a)

=1

4π

∞∑n=0

(2n+ 1)jn(kr)

jn(kR)

∫S2

f(Ry)Pn(x · y) ds(y) . (2.31b)

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 (2.30)

which is again given by (2.31a), (2.31b). The series converges uniformly on B[0, R].

Proof: (a) The proof of uniqueness follows the lines of the corresponding part in the proofof Theorem 2.24. Indeed, let u be a solution of the Helmholtz equation in B(0, R) such thatlimr→R ‖u(r, ·)‖L2(S2) = 0. From Theorem 2.33 we conclude that u can be expanded into

u(rx) =∞∑n=0

n∑m=−n

amn jn(kr)Y mn (x) , x ∈ S2 , r < R .

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) = aqp jp(kRε), thus

u(rx) =∞∑n=0

n∑m=−n

(uε, Ymn )L2(S2)

jn(kr)

jn(kRε)Y mn (x) , x ∈ S2 , r ≤ Rε .


Let now R1 < R0 be arbitrary. Since the asymptotic behavior of the Bessel function (Theo-rem 2.31) is uniform with respect to R1 ≤ |z| ≤ R0 there exists a constant c > 0 such that∣∣jn(kr)/jn(kRε)

∣∣ ≤ c for all n ∈ N and R1 ≤ r ≤ R0. Therefore, for R1 ≤ r ≤ R0,

‖u(r, ·)‖2L2(S2) =

∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2 ∣∣∣∣ jn(kr)

jn(kRε)

∣∣∣∣2 ≤ c2

∞∑n=0

n∑m=−n

∣∣(uε, Y mn )L2(S2)

∣∣2= c2‖uε‖2

L2(S2) ≤ c2ε2 .

This holds for all ε > 0, therefore u has to vanish in B(0, R0)\B(0, R1) and thus for |x| < Rbecause R1 < R0 < R was arbitrary u.

It remains to show that the series provides the solution of the boundary value problem.Uniform convergence of the series and its derivatives is proven as in the previous theorem.Therefore u ∈ C∞

(B(0, R)

)and u satisfies the Helmholtz equation. Similar to the proof of

Theorem 2.22 we obtain the boundary condition ‖u(r, .)− fR‖2L2(S2), since by Theorem 2.31

the term | jn(kr)jn(kR)

− 1|2 can be uniformly bounded for all n ∈ N and all r ∈ [0, R].

(b) Also, with the uniform bound on∣∣∣ jn(kr)jn(kR)

∣∣∣ for all n ∈ N and r ∈ [0, R] we can copy the

second part of the proof of Theorem 2.22 in case of the Helmholtz equation which showspart (b) of the Theorem. 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: The additional assumption jn(kR) 6=

0 for all n ∈ N0 is required to obtain a unique solution of the boundary value problem. Fromthe asymptotic form of jn(z) for real and positive z (see Theorem 2.30) we conclude that forevery n ∈ N0 and fixed R > 0 there exist infinitely many real and positive wave numberskn,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 the homogeneous Dirichlet boundary value problem ∆umn,j+k2n,ju

mn,j = 0 in B[0, R] with

umn,j = 0 on the boundary |x| = R. The values k2n,j are called Dirichlet eigenvalues of −∆ in

the ball of radius R. The multiplicity is 2n + 1, and the functions umn,j, m = −n, . . . , n, arethe corresponding eigenfunctions.

We now continue with the solutions of the Helmholtz equation in the exterior of a ball.Following the proof of Theorem 2.33 we derive again at the solution (2.29) involving thetwo linearly independent solutions jn(kr)Y m

n (x) and yn(kr)Y mn (x). Both parts in the terms

are defined for r > 0 – and both tend to zero as r tends to infinity (see Theorem 2.30).Therefore, the requirement u(x) → 0 as |x| → ∞ is not sufficient to pick a unique memberof the solution space. In the introduction we have motivated the physically correct choice.Here, we present a second motivation and consider the case of a conducting medium withε = ε0, µ = µ0, and constant σ > 0. We have seen from the previous chapter that in thiscase k2 has to be replaced 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 2.30 we observe that

jn(kr)Y mn (x) =

fn(kr)

krY mn (x)

[1 + O

(1

r

)]for r →∞ ,

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 2.35 Let k ∈ C \ 0 with Im k ≥ 0 and n ∈ N0 and m ∈ Z with |m| ≤ n. Then

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

n (x) satisfies the Sommerfeld radiation condition

∂u(rx)

∂r− ik u(rx) = O

(1

r2

)for r →∞ , (2.32)

uniformly with respect to x ∈ S2.

Proof: By Theorem 2.30 we have

d

drh(1)n (kr) − ik h(1)

n (kr) =exp(ikr)

kr

[(−i)n k − ik (−i)n−1 + O

(1

r

)]=

exp(ikr)

krO(

1

r

)= O

(1

r2

)for r →∞ .

2

We note that this radiation is necessary only for the case of real values of k. For complexvalues with Im k > 0 we note that u(x) decays even exponentially as |x| tends to infinity.

The importance of this radiation condition lies in the fact that it can be formulated for everyfunction defined in the exterior of a bounded domain without making use of the Hankelfunctions – and still provides the correct solution. The following theorem is the analog ofTheorem 2.23.


Theorem 2.36 Let k ∈ C\0 with Im k ≥ 0 and u ∈ C2(R3\B[0, R]

)satisfy ∆u+k2u = 0

in the exterior of some ball B[0, R]. Furthermore, we assume that u satisfies the Sommerfeldradiation condition (2.32). Then there exist unique αmn ∈ C, |m| ≤ n, n = 0, 1, 2, . . . with

u(rx) =∞∑n=0

n∑m=−n

αmn h(1)n (kr)Y m

n (x) , r > R , x ∈ S2 . (2.33)

The series converges uniformly with all of its derivatives on compact subsets of R3 \B[0, R].

Proof: We follow the arguments of the proof of Theorem 2.33 and arrive at the general form(2.29) of u(x). From the radiation condition for u we conclude that every term umn (r) satisfiesddrumn (r)− ik umn (r) = O(1/r2). Since only the Hankel functions of the first kind satisfy the

radiation condition in contrast to the Bessel functions we conclude that βmn = 0 for all n ∈ N0

and |m| ≤ n. We continue with the arguments just as in the proof of Theorem 2.33 andomit the details. 2

Finally we consider the exterior boundary value problem to determine for given f the complexvalued function u such that

∆u+ k2u = 0 for |x| > R , u = f for |x| = R , (2.34)

and u satisfies Sommerfeld radiation condition (2.32).

Theorem 2.37 Let k ∈ C \ 0 with Im k ≥ 0 and R > 0.

(a) For given fR ∈ L2(S2) there exists a unique solution u ∈ C2(R3\B[0, R]

)of ∆u+k2u = 0

withlimr→R‖u(r, ·)− fR‖L2(S2) = 0 ,

and u satisfies Sommerfeld radiation condition (2.32). 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) (2.35a)

=1

4π

∞∑n=0

(2n+ 1)h

(1)n (kr)

h(1)n (kR)

∫S2

f(Ry)Pn(x · y) ds(y) . (2.35b)

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

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

Proof: The proof of uniform convergence outside of any ball B(0, R′) for R′ > R of anyderivative of the series and the validity of the boundary condition follows the well knownarguments and is omitted. We just mention that h

(1)n (kR) never vanishes – in contrast to

jn(kR) – because of the Wronskian of Theorem 2.27.


The proof that the function u satisfies the radiation condition is more difficult and uses aresult (in the proof of Lemma 2.38) from Chapter 3.2 The problem is that every componentof the series satisfies the radiation condition but the asymptotics of hn(k|x|) as |x| tends toinfinity by Theorem 2.30 does not hold uniformly with respect to n. The idea to overcomethis difficulty is 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 , (2.36)

rather than a series, see Lemma 2.38 below. Here, Φ denotes the fundamental solution ofthe Helmholtz equation, defined by

Φ(x, y) =exp(ik|x− y|)

4π |x− y|=

ik

4πh

(1)0 (k|x− y|) , x 6= y . (2.37)

The fundamental solution satisfies the radiation condition (2.32) uniformly with respect to yon the compact sphere y : |y| = R0 as we show in Lemma 2.39. Therefore, also u satisfiesthe radiation condition (2.32) and completes the proof. 2

Lemma 2.38 Let k ∈ C \ 0 with Im k ≥ 0 and R > 0 and u be given by

u(rx) =∞∑n=0

n∑m=−n

αmn h(1)n (kr)Y m

n (x) , r > R , x ∈ S2 . (2.38)

We assume that the series converges uniformly on compact subsets of R3 \ B[0, R]. Thenthere exists R0 > R and a continuous function g on the sphere with radius R0 such that

u(x) =

∫|y|=R0

Φ(x, y) g(y) ds(y) , |x| > R0 .

Here Φ denotes again the fundamental solution of the Helmholtz equation, given by (2.37).

Proof: Let R0 > R such that jn(kR0) 6= 0 for all n and fix x such that |x| > R0. First,we apply Green’s second identity from Theorem 6.12 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 .

(2.39a)

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

n (y) satisfies the radiation condition (2.32) we can applyGreen’s representation Theorem 3.6 from the forthcoming Chapter 3 to v in the exterior of

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


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) . (2.39b)

Equations (2.39a) and (2.39b) 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) (2.40)

where we have used the Wronskian of Theorem 2.27. This holds for all |x| > R0. Comparingthis with the form (2.38) of u yields

u(x) =∞∑n=0

n∑m=−n

αmnikR2

0jn(kR0)ikR2

0h(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 =αmn

ik R20 jn(kR0)

. (2.41)

Let R′ ∈ (R,R0). The series (2.38) converges uniformly on x : |x| = R′, thus

∞∑n=0

n∑m=−n

∣∣αmn ∣∣2 ∣∣h(1)n (kR′)

∣∣2 < ∞

and, in particular,∣∣αmn ∣∣ ∣∣h(1)

n (kR′)∣∣ ≤ c1 for all |m| ≤ n and n ∈ N. Therefore

∣∣gmn ∣∣ ≤ c2∣∣h(1)n (kR′)

∣∣|jn(kR0)|≤ c3(2n+ 1)

(R′

R0

)nby Theorem 2.31 which proves uniform convergence of the series

g(R0x) =∞∑n=0

n∑m=−n

gmn Ymn (x)

and ends the proof. 2


Lemma 2.39 For any compact set K ⊆ R3 the fundamental solution Φ satisfies the Som-merfeld radiation condition (2.32) uniformly with respect to y ∈ K. More precisely,

Φ(x, y) =exp(ik|x|)

4π|x|e−ik x·y

[1 + O

(|x|−1

)], |x| → ∞ , (2.42a)

∇xΦ(x, y) = ik xexp(ik|x|)

4π|x|e−ik x·y

[1 + O

(|x|−1

)], |x| → ∞ , (2.42b)

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 g(ε; x, y) = |x− εy| is analytic in a neighborhood of zero with respect to ε andg(0; x, y) = 1 and ∂g(0; x, y)/∂ε = −x · y. Therefore, by the rule of l’Hospital we observethat F (0; x, y) = 0, and the assertion of the first part of the lemma follows because also Fis analytic with respect to ε. The second part is proven analogously. 2

From formula (2.40) in the proof of Theorem 2.37 we conclude the following addition formulafor Bessel functions .

Corollary 2.40 Let k ∈ C \ 0 with Im k ≥ 0. 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 (2.40) 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 2.31. 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 2.39 and the proof of Theorem 2.37 we have:

Corollary 2.41 Let k ∈ C \ 0 with Im k ≥ 0. The solution u of the exterior boundaryvalue problem (2.34) 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)

[1 + O

(1

r2

)](2.43a)

=exp(ikr)

4π kr

∞∑n=0

(2n+ 1) (−i)n+1

h(1)n (kR)

∫S2

f(Ry)Pn(x · y) ds(y)

[1 + O

(1

r2

)](2.43b)

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

Proof: We note that (2.43a), (2.43b) follow formally from (2.35a), (2.35b), respectively, by

using the asymptotics of Theorem 2.30 for h(1)n (kr). The asymptotics, however, are not uni-

form with respect to n, and we have to use of the representation (2.36) as an integral instead.

The function g has the expansion coefficients (2.41) with αmn = (fR, Ymn )L2(S2)/h

(1)n (kR); that

is,

gmn =(fR, Y

mn )L2(S2)

ik R20 h

(1)n (kR) jn(kR0)

.

Therefore, using the asymptotic form of Φ by Lemma 2.39,

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)

[1 + O

(1

r2

)].


We compute with the Jacobi-Anger expansion of Theorem 2.32

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)

= R20

∞∑n=0

∑|m|≤n

gmn (−i)n 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)

=1

k

∞∑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

This corollary implies that every radiating solution of the Helmholtz equation has an asymp-totic behaviour of the form

u(x) =exp(ik|x|)

ru∞(x)

[1 +O

(1

r2

)], |x| → ∞ ,

uniformly with respect to x = x/|x| ∈ S2. The function u∞ : S2 → C is called the far fieldpattern or far field amplitude of u and plays an important role in inverse scattering theory.

2.7 Expansion of Electromagnetic Waves

In this section we transfer the results of the previous section to the case of the time harmonicMaxwell system in a homogeneous medium with vanishing external current and charge den-sities. By the close connection between the Helmholtz equation and the Maxwell system asdiscussed in Chapter 1 we are able to use several results from the previous section.

Thus, we consider the boundary value problems

curlE − iωµ0H = 0 , curlH + iωε0E = 0

in the interior or exterior of a ball B(0, R) with boundary condition

ν × E = fR on ∂B(0, R)

where the tangential field fR is given. Throughout we use the constant permability µ anddielectricity ε as µ0 and ε0 from vacuum. But we can allow the parameters also beingcomplex valued. In this case we choose the branch of the wave number k = ω

√εµ with

Im k ≥ 0. Therefore the results will hold for any homogeneous medium.

2.7. EXPANSION OF ELECTROMAGNETIC WAVES 73

The starting point is the observation that E,H are solutions of the source free time harmonicMaxwell’s equations in a domain D ⊆ R3, if and only if E (or H) is a divergence free solutionof the vector Helmholtz equation; that is,

∆E + k2E = 0 and divE = 0 in D

(see Section 1.3, equation (1.12)). By the following lemma we can construct solutions ofMaxwell’s equations from scalar solutions of the Helmholtz equation.

Lemma 2.42 Let u ∈ C∞(D) satisfy ∆u+ k2u = 0 in a domain D ⊆ R3. Then

E(x) = curl(xu(x)

)and H =

1

iωµcurlE , x ∈ D , (2.44)

are solutions of the time harmonic Maxwell system

curlE − iωµH = 0 in D and curlH + iωεE = 0 in D . (2.45)

Furthermore, also E(x) = curl curl(xu(x)

)and H = 1

iωµcurl E are solutions of the time

harmonic Maxwell system (2.45).

Proof : Let u satisfy the Helmholtz equation and define E and H as in the Lemma. Weobserve that divE = 0. From curl curl = −∆ +∇ div and divE = 0 we conclude that

iωµ curlH(x) = curl curlE(x) = −∆E(x) = − curl[∆(xu(x)

)]= − curl

[2∇u(x) + x∆u(x)

]= k2 curl

[xu(x)

]= k2E(x)

which proves the assertion because of k2 = ω2εµ. For the proof of the second part we justtake the curl of the last formula. 2

In view of the expansion results for the Helmholtz equation of the previous section we liketo consider solutions of the Helmholtz equation of the form u(x) = jn(k|x|)Y m

n (x) whereY mn denotes a spherical surface harmonics of order n ∈ N0, m = −n, . . . , n and x = x

|x| . By

Lemma 2.42 we see thatE(x) = curl(x jn(k|x|)Y mn (x)) and E(x) = curl curl(x jn(k|x|)Y m

n (x))are solutions of the Maxwell system. We can also replace the Bessel function by the Hankelfunction, if the region D does not contain the origin, which will lead to solutions satisfyinga radiation condition.

Theorem 2.43 (a) For n ∈ N0 and |m| ≤ n the function E : R3 → R3, defined by

E(x) = curl(x jn(kr)Y m

n (x))

= jn(kr) GradS2Y mn (x)× x , (2.46a)

and the corresponding function

H(x) =1

iωµcurlE(x)

=n(n+ 1)

iωµrjn(kr)Y m

n (x) x +1

iωµr

[jn(kr) + krj′n(kr)

]GradS2Y m

n (x) (2.46b)


satisfy the Maxwell system (2.45). Again, r = |x| and x = x/|x| denote the sphericalcoordinates, and GradS2 is the surface gradient on the unit sphere (see Example 6.17).

For the tangential components with respect to spheres of radius r > 0 it holds

x× E(x) = jn(kr) GradS2Y mn (x) , (2.47a)

x×H(x) =1

iωµr

[jn(kr) + krj′n(kr)

]x×GradS2Y m

n (x) . (2.47b)

(b) Analogously, the vector fields E(x) = curl curl(x jn(kr)Y m

n (x))

and H = 1iωµ

curl E aresolutions of Maxwell’s equations.

(c) The results of (a) and (b) hold in R3 \ 0 if the Bessel function jn is replaced by the

Hankel function h(1)n .

Proof: (a) From Lemma 2.42 we already know that E(x) = curl(x jn(kr)Y m

n (x))

generatesa solution of Maxwell’s equations. We have to prove the second representation. By usingthe identity (6.7) and curl x = 0 we find E(x) = ∇

(jn(kr)Y m

n (x))× x. Decomposing the

gradient into its tangential part, Grad (see (6.17)), and the radial component, (∂/∂r) x,leads to

E(x) =

(1

rjn(k|x|) GradS2Y m

n (x) +∂

∂r

(jn(k|x|)

)Y mn (x) x

)× x

= jn(kr) GradS2Y mn (x)× x

which proves the second representation in (2.46a).

Now we consider curlE. Analogously to the proof of the previous lemma we use curl curl =−∆+∇ div and the fact that w(x) = jn(kr)Y m

n (x) solves the Helmholtz equation. Therefore,

curlE(x) = −∆(xw(x)

)+ ∇ div

(xw(x)

)= −2∇w(x) + k2xw(x) + 3∇w(x) + ∇

(x · ∇w(x)

)= ∇

[(jn(kr) + krj′n(kr)

)Y mn (x)

]+ k2x jn(kr)Y m

n (x)

=[(jn(kr) + krj′n(kr)

)′+ k2r jn(kr)

]Y mn (x) x

+1

r

(jn(kr) + krj′n(kr)

)GradS2Y m

n (x)

=1

r

[2krj′n(kr) + k2r2j′′n(kr) + k2r2 jn(kr)

]Y mn (x) x

+1

r


)GradS2Y m

n (x)

=n(n+ 1)

rjn(kr)Y m

n (x) x +1

r


)GradS2Y m

n (x)

where we used the Bessel differential equation (2.27) in the last step. Formulas (2.47a) and(2.47b) follow obviously.


(c) This is shown as in (a) and (b) by replacing jn by h(1)n . 2

From the theorem we observe that the tangential components of the boundary values ona sphere involve the surface gradients GradS2Y m

n and x × GradS2Y mn . We have seen – and

already used several times – that the surface gradient of u ∈ C1(S2) is a tangential field,i.e. ν · GradS2u = 0 on S2 with the unit normal vector ν(x) = x on S2. Obviously, alsoν ×GradS2u is tangential. Using the partial integration formula (6.21) we compute∫

S2

(x×GradS2u) · (x×GradS2v) ds =

∫S2

GradS2u ·((x×GradS2v)× x

)ds

=

∫S2

GradS2u ·GradS2v ds = −∫S2

u∆S2v ds

for u ∈ C1(S2) and v ∈ C2(S2). Since the spherical harmonicsY mn : −n ≤ m ≤ n, n ∈ N

form an orthonormal system of eigenfunctions of the spherical Laplace-Beltrami operator∆S2 with the eigenvalues λ = −n(n+ 1), n ∈ N0, we conclude that the vector fields

Umn (x) :=

1√n(n+ 1)

GradS2Y mn (x) (2.48a)

as well as the vector fields

V mn (x) := x× Um

n (x) =1√

n(n+ 1)x×GradS2Y m

n (x) (2.48b)

are two orthonormal systems of tangential fields on S2. Applying partial integration againand using property (b) of the following lemma we obtain∫

S2

(x×GradS2u) ·GradS2v ds = −∫S2

DivS2(x×GradS2u) v ds = 0 .

Combining these sets of vector fields Umn and V m

n yields the set of spherical vector harmonics .Actually, one should denote them as spherical vector surface harmonics.

The main task is to show that the spherical vector harmonics form a complete orthonormalsystem in the space of tangential vector fields on S2. In preparation of the proof we needsome more properties of the spherical harmonics and the surface differential operators.

Lemma 2.44

(a) DivS2(x×GradS2u) = 0 for all u ∈ C2(S2) where DivS2 denotes the surface divergenceon the unit sphere (see Example 6.17).

(b) If a tangential field u ∈ C1t (S2) = u ∈ C1(S2,C3) : u · ν = 0 on S2 satisfies

DivS2 u = 0 and DivS2(ν × u) = 0 on S2

then u = 0 on S2.


Proof: (a) We extend u to a C1 function in a neighborhood of S2 and apply Corollary 6.20.This yields DivS2(x×GradS2u) = DivS2(x×∇u) = −x · curl∇u = 0 on S2.

(b) From DivS2(ν × u) = 0 we obtain by partial integration (6.21)

0 =

∫S2

DivS2(ν × u) f ds = −∫S2

(ν × u) ·GradS2f ds for every f ∈ C1(S2) .

We express u in spherical coordinates with respect to the basis θ, ϕ of the spherical unitvectors; that is, u = (u · θ)θ + (u · ϕ)ϕ and thus

ν × u = (u · θ)ϕ − (u · ϕ)θ .

Let f ∈ C1(S2) be arbitrary. The representation of the surface gradient of f on S2 (seeExample 6.17) implies by partial integration with respect to θ and ϕ, respectively, theidentity

0 =

∫S2

(ν × u) ·GradS2f ds =

∫ 2π

0

∫ π

0

(ν × u) ·(∂f

∂θθ +

1

sin θ

∂f

∂ϕϕ

)sin θ dθ dϕ

=

∫ 2π

0

∫ π

0

[∂

∂θ

((u · ϕ) sin θ

)− ∂

∂ϕ(u · θ)

]f dθ dϕ .

This leads to∂

∂θ

((u · ϕ) sin θ

)=

∂

∂ϕ

(u · θ

)on S2. Considering the anti-derivative h(ϕ, θ) =

∫ θ0

(u · θ)(ϕ, θ′) dθ′ we have

∂h

∂θ= u · θ and

∂h

∂ϕ(ϕ, θ) =

∫ θ

0

∂

∂ϕ(u · θ)(ϕ, θ′) dθ′ =

∫ θ

0

∂

∂θ

((u · ϕ)(ϕ, θ′) sin θ′

)dθ′ = (u · ϕ) sin θ .

Therefore we conclude that

u = (u · θ)θ + (u · ϕ)ϕ =∂h

∂θθ +

1

sin θ

∂h

∂ϕϕ = GradS2h .

Finally, we use the condition DivS2 u = 0 and obtain by partial integration

0 =

∫S2

h DivS2 u ds =

∫S2

u ·GradS2h ds =

∫S2

|u|2 ds

which yields u = 0 on S2. 2

Remark: From the first part of the proof of part (b) we observe that DivS2(ν × u) = 0 onS2 implies the existence of a surface potential h. This motivates the common notation ofthe surface curl of a tangential field u by Curlu = −Div(ν × u).

The following result is needed as a preparation for the expansion theorem but is also ofindependent interest. It proves existence and uniqueness of a solution for the equation∆S2u = f on S2 where ∆S2 = DivS2 GradS2 denotes again the spherical Laplace-Beltramioperator (see Definition 2.2).


Theorem 2.45 For every f ∈ C∞(S2) with∫S2 f ds = 0 there exists a unique solution u ∈

C2(S2) with ∆S2u = −f on S2 and∫S2 u ds = 0. If f is given by f =

∑∞n=1

∑nm=−n f

mn Y

mn

with fmn = (f, Y mn )L2(S2) then the solution is given by

u(x) =∞∑n=1

n∑m=−n

fmnn(n+ 1)

Y mn (x) =

1

4π

∞∑n=1

2n+ 1

n(n+ 1)

∫S2

Pn(x · y) f(y) ds(y) .

Both series converge uniformly with all their derivatives. In particular, u ∈ C∞(S2).

Proof: Uniqueness is seen by Green’s theorem. Indeed, if u ∈ C2(S2) is the difference oftwo solutions then ∆S2u = 0 and thus 0 =

∫S2 u DivS2 GradS2u ds =

∫S2 |GradS2u|2ds; that

is, GradS2u = 0 which implies that u is constant. The requirement∫S2 u ds = 0 yields that

u vanishes.

For existence it suffices to show that the series for u converges with all of its derivativesbecause ∆S2Y m

n = −n(n+ 1)Y mn . For partial sums we have for any q ∈ N

(SNu)(x) =N∑n=1

n∑m=−n

fmnn(n+ 1)

Y mn (x) =

N∑n=1

1

n(n+ 1)

∫S2

n∑m=−n

Y mn (x)Y −mn (y) f(y) ds(y)

=1

4π

N∑n=1

2n+ 1

n(n+ 1)

∫S2

Pn(x · y) f(y) ds(y)

=(−1)q

4π

N∑n=1

2n+ 1

nq+1(n+ 1)q+1

∫S2

∆qS2Pn(x · y) f(y) ds(y)

=(−1)q

4π

N∑n=1

2n+ 1

nq+1(n+ 1)q+1

∫S2

Pn(x · y) ∆qS2f(y) ds(y)

where we used the addition formula of Theorem 2.17 again and ∆S2Pn(x · y) = −n(n +1)Pn(x · y) (because x 7→ Pn(x · y) is a spherical surface harmonic) Green’s theorem q times.We consider the extension of this formula to x ∈ R3 and observe for any differential operatorD` of order ` that

(D`SNu)(x) =(−1)q

4π

N∑n=1

2n+ 1

nq+1(n+ 1)q+1

∫S2

D`xPn(x · y) ∆q

S2f(y) ds(y) .

By the chain rule and Exercise 2.6 we conclude that |D`xPn(x · y)| ≤ cn2` for all n ∈ N and

x, y ∈ S2. Therefore,

1

4π

N∑n=1

2n+ 1

nq+1(n+ 1)q+1

∣∣∣∣∫S2

D`xPn(x · y) ∆q

S2f(y) ds(y)

∣∣∣∣ ≤ c ‖∆qS2f‖∞

N∑n=1

(2n+ 1)n2`

nq+1(n+ 1)q+1

which converges if we choose q ≥ `+ 1. 2


Theorem 2.46 The functions

Umn =

1√n(n+ 1)

GradS2Y mn (x) and V m

n (x) = x× Umn (x)

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

L2t (S

2) =u ∈ L2(S2) : ν · u = 0 on S2

.

Proof: Let f ∈ C∞t (S2) = u ∈ C∞(S2,C3) : u ·ν = 0 on S2. Then also DivS2 f, DivS2(ν×f) ∈ C∞(S2), and by Green’s theorem we have

∫S2 DivS2 f ds =

∫S2 DivS2(ν × f) ds = 0. By

the previous Theorem 2.45 there exist solutions u1, u2 ∈ C∞(S2) with ∆S2u1 = DivS2 f and∆S2u2 = DivS2(ν × f) on S2. Their expansions

u1 =∞∑n=1

n∑m=−n

amn Ymn and u2 =

∞∑n=1

n∑m=−n

bmn Ymn

converge uniformly with all of their derivatives. Set

g = Grad u1 − ν ×Grad u2 =∞∑n=1

n∑m=−n

[amn Grad Y m

n − bmn ν ×Grad Y mn

]=

∞∑n=1

n∑m=−n

[√n(n+ 1) amn U

mn −

√n(n+ 1) bmn V

mn

].

Then g ∈ C∞t (S2) and, by using part (a) of Lemma 2.44, DivS2 g = ∆S2u1 = DivS2 f andDivS2(ν × g) = −DivS2

(ν × (ν ×GradS2u2)

)= DivS2 GradS2u2 = DivS2(ν × f). Application

of part (b) of Lemma 2.44 yields g = f which proves uniform convergence of f into vectorspherical harmonics. The completeness of the set Um

n , Vmn : −n ≤ m ≤ m, n ∈ N in

L2t (S

2) follows again by a denseness argument just as in the proof of Theorem 2.19. 2

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

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

n .

Corollary 2.47 Every vector field A ∈ L2(S2,C3) 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)

](2.49)

where

amn =

∫S2

A(x) · x Y −mn (x) ds(x) , |m| ≤ n , n ≥ 0 , (2.50a)

bmn =

∫S2

A(x) · U−mn (x) ds(x) , |m| ≤ n , n ≥ 1 , b00 = 0 , (2.50b)

cmn =

∫S2

A(x) · V −mn (x) ds(x) , |m| ≤ n , n ≥ 1 , c00 = 0 . (2.50c)


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

∞∑n=0

∑|m|≤n

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

]= ‖A‖2

L2(S2,C3) .

Proof: We decompose the vector field A into

A(x) =(A(x) · x

)x + x×

(A(x)× x

).

The scalar function x 7→ x · A(x) and the tangential vector field x 7→ x ×(A(x) × x

)can

be expanded into basis functions on S2 according to Theorem 2.19 for the scalar radial partand the previous Theorem 2.46 for the tangential part. 2

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

Theorem 2.48 Let E,H ∈ C1(B(0, R),C3

)satisfy curlE−iωµH = 0 and curlH+iωεE =

0 in B(0, R). Then there exist unique αmn , βmn ∈ C, |m| ≤ n, n = 0, 1, 2, . . . such that

E(x) =∞∑n=1

∑|m|≤n

[αmn√n(n+ 1)

jn(kr)

rY mn (x) x + αmn

(rjn(kr)

)′r

Umn (x)

+ βmn jn(kr)V mn (x)

](2.51a)

=∞∑n=1

∑|m|≤n

1√n(n+ 1)

[αmn curl curl

(x jn(kr)Y m

n (x))

− βmn curl(x jn(kr)Y m

n (x))]

(2.51b)

H(x) = − 1

iωµ

∞∑n=1

∑|m|≤n

[βmn√n(n+ 1)

jn(kr)

rY mn (x) x + βmn

(rjn(kr)

)′r

Umn (x)

+ k2αmn jn(kr)V mn (x)

](2.51c)

=1

iωµ

∞∑n=1

∑|m|≤n

1√n(n+ 1)

[k2αmn curl

(x jn(kr)Y m

n (x))

− βmn curl curl(x jn(kr)Y m

n (x))]

(2.51d)

for x = rx and 0 ≤ r < R, x ∈ S2. The series converge uniformly in every ball B[0, R′] forR′ < R.

Proof: Let amn (r), bmn (r), and cmn (r) be the expansion coefficients of (2.49) correspondingto the vector field given by x 7→ E(rx). First we note that a0

0(r) = 1r2

∫|x|=r x · E(x) ds(x)

vanishes by the divergence theorem because divE = 0.


Second, we consider amn (r) for n ≥ 1. With

amn (r) =

∫S2

x · E(rx)Y −mn (x) ds(x) =1

r

∫S2

x · E(x)Y −mn (x) ds(x)

we observe that ramn (r) is the expansion coefficient of the solution x 7→ x ·E(x) of the scalarHelmholtz equation, see Lemma 1.5. Thus by Theorem 2.33 we obtain

amn (r) =αmnrjn(kr)

for some αmn . Next, the condition divE = 0 in spherical polar coordinates reads

1

r

∂

∂r

(r2Er(r, x)

)+ DivS2 Et(r, x) = 0

(see (6.20)), and we obtain

1

r

(r2amn (r)

)′=

∫S2

1

r

∂

∂r

(r2Er(r, x)

)Y −mn (x) ds(x)

= −∫S2

DivS2 Et(r, x)Y −mn (x) ds(x)

=

∫S2

E(r, x) ·GradS2Y −mn (x) ds(x) =√n(n+ 1) bmn (r)

by (6.21). Thus, with a common constant αmn we have

amn (r) = αmn

√n(n+ 1)

rjn(kr) , bmn (r) = αmn

1

r

(r jn(kr)

)′. (2.52)

Finally, we consider cmn (r). By partial integration it holds√n(n+ 1) cmn (r) =

∫S2

E(r, x) ·(x×GradS2Y −mn (x)

)ds(x)

=

∫S2

(E(r, x)× x

)·GradS2Y −mn (x) ds(x)

= −∫S2

DivS2

(E(r, x)× x

)Y −mn (x) ds(x)

= −r∫S2

x · curlE(rx)Y −mn (x) ds(x)

= −∫S2

x · curlE(rx)Y −mn (x) ds(x)

where we have used that DivS2

(E(r, x) × x

)= Div|x|=r

(E(rx) × x

)= r x · curlE(rx) (see

Corollary 6.20). Therefore,√n(n+ 1) cmn (r) are the expansion coefficients of the scalar

function ψ(x) = x ·curlE(x). Because ψ is a solution of the Helmholtz equation ∆ψ+k2ψ =


0, see Lemma 1.5, we again conclude from Theorem 2.33 that cmn (r) = βmn jn(kr) for someβmn ∈ C. This proves the first representation (2.51a).

Furthermore, using (2.46a) and (2.46b) yields the second representation (2.51b).

It remains to show uniform convergence in B[0, R′] for every R′ < R. As in the scalar case wewill prove another representation of the field E. Fix R′ < R and choose R with R′ < R < R.The scalar function x 7→ E(Rx) · x is smooth and we obtain from (2.51a) that

αmn√n(n+ 1)

jn(kR)

R=

∫S2

(E(Ry) · y

)Y −mn (y) ds(y) .

Analogously, we obtain

αmnjn(kR) + kRj′n(kR)

R=

∫S2

E(Ry) · U−mn (y) ds(y)

= − 1√n(n+ 1)

∫S2

DivS2 E(R, y)Y −mn (y) ds(y) ,

and

βmn jn(kR) =

∫S2

E(Ry) · V −mn (y) ds(y)

=1√

n(n+ 1)

∫S2

E(R, y) ·(y ×GradS2Y −mn (y)

)ds(y)

= − 1√n(n+ 1)

∫S2

DivS2

(E(R, y)× y

)Y −mn (y)

)ds(y) .

Substituting these coefficients into (2.51a) and using the addition formula of Theorem 2.17yields

E(rx) =∞∑n=1

∑|m|≤n

[R

r

jn(kr)

jn(kR)

∫S2

(E(Ry) · y

)Y −mn (y) ds(y)Y m

n (x) x

− R

r

jn(kr) + krj′n(kr)

jn(kR) + kRj′n(kR)

1

n(n+ 1)GradS2

∫S2

DivS2 E(R, y)Y −mn (y) ds(y)Y mn (x)

− jn(kr)

jn(kR)

1

n(n+ 1)

(x×GradS2

∫S2

DivS2

(E(R, y)× y

)Y −mn (y) ds(y)Y m

n (x)

)](2.53)

=1

4π

∞∑n=1

(2n+ 1)

[R

r

jn(kr)

jn(kR)

∫S2

(E(Ry) · y

)Pn(x · y) ds(y) x

− R

r

jn(kr) + krj′n(kr)

jn(kR) + kRj′n(kR)

1

n(n+ 1)GradS2

∫S2

DivS2 E(R, y)Pn(x · y) ds(y)

− jn(kr)

jn(kR)

1

n(n+ 1)

(x×GradS2

∫S2

DivS2

(E(R, y)× y

)Pn(x · y) ds(y)

)].


This has now the form of the series for the solution of the interior boundary value problem forthe Helmholtz equation. The arguments used there, in the proof of Theorem 2.33, provideanalogously uniform convergence of the series and its derivatives in the closed ball B[0, R′].2

The expansions (2.51a) and (2.51b) as well as (2.51c) and (2.51d) complement each other.The representation (2.51b) shows an expansion into vector wave functions which is closelyrelated to the idea from Lemma 2.42 using the scalar expansion functions of the Helmholtzequation. On the other hand (2.51a) is suitable for treating boundary value problems (seeTheorem 2.49 below) and, additionally, gives some inside into the important relation of thetangential and the normal component of the electric field on surfaces.

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 = fR on |x| = R. We formulate the existenceresult again with fR ∈ L2

t (S2) where fR(x) = f(Rx).

Theorem 2.49 We assume that ε, µ ∈ C with Im k ≥ 0 where again k = ω√εµ and R > 0

are such that jn(kR) 6= 0 and jn(kR) + kR j′n(kR) 6= 0 for all n ∈ N0.

(a) For given fR ∈ L2t (S

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

)and H = 1

iωµcurlE

of

curlE − iωµH = 0 and curlH + iωεE = 0 in B(0, R)

with

limr→R‖ν × E(r, ·)− fR‖L2(S2) = 0 .


The solution is given by

E(rx) =∞∑n=1

n∑m=−n

(fR, Umn )L2(S2)√

n(n+ 1)

1

jn(kR)curl

[x jn(kr)Y m

n (x)]

+(fR, V

mn )L2(S2)√

n(n+ 1)

R

jn(kR) + kR j′n(kR)curl curl

[xjn(kr)Y m

n (x)]

=∞∑n=1

n∑m=−n

(fR, Vmn )L2(S2)

√n(n+ 1)

R

jn(kR) + kR j′n(kR)

jn(kr)

rY mn (x) x

+ (fR, Vmn )L2(S2)

R

jn(kR) + kR j′n(kR)

jn(kr) + kr j′n(kr)

rUmn (x)

− (fR, Umn )L2(S2)

jn(kr)

jn(kR)V mn (x) ,

H(rx) =1

iωµ

∞∑n=1

n∑m=−n

(fR, Umn )L2(S2)√

n(n+ 1)

1

jn(kR)curl curl

[x jn(kr)Y m

n (x)]

+ k2 (fR, Vmn )L2(S2)√

n(n+ 1)

R

jn(kR) + kR j′n(kR)curl

[xjn(kr)Y m

n (x)]

=1

iωµ

∞∑n=1

n∑m=−n

(fR, Umn )L2(S2)

√n(n+ 1)

1

jn(kR)

jn(kr)

rY mn (x) x

+ (fR, Umn )L2(S2)

1

jn(kR)


rUmn (x)

− k2(fR, Vmn )L2(S2)

R

jn(kR) + kRj′n(kR)jn(kr)V m

n (x)

for x = rx ∈ B(0, R). The series converges uniformly on compact subsets of B(0, R).

(b) If fR ∈ C2t (S2) the solution of the boundary value problem satisfies E ∈ C2

(B(0, R)

)∩

C(B[0, R]

)and the series converges uniformly on the closed ball B[0, R].

Proof: The proof follows the arguments as in Theorem 2.37 for the case of a boundary valueproblem for the Helmholtz equation.

Let us repeat the steps. To show uniqueness we assume E is a solution of the homogeneousboundary value problem, i.e. ‖ν × E(rx)‖L2(S2) → 0 for r → R with ν(x) = x. First weconsider the expansion from Theorem 2.48 for Rε ∈ (R0, R), where we choose 0 < R0 < R


and ε > 0 with ‖ν × E(Rεx)‖L2(S2) ≤ ε. From (2.53) we obtain

x× E(rx) =∞∑n=1

∑|m|≤n

Rε

r


jn(kRε) + kRε j′n(kRε)(ν × E(Rε . ), V

mn )L2(S2) U

mn (x)

−∞∑n=1

∑|m|≤n

jn(kr)

jn(kRε)(ν × E(Rε . ), U

mn )L2(S2) V

mn (x) .

By Parseval’s identity we obtain for 0 < R1 < r < R0 < Rε < R

‖ν × E(r . )‖2L2(S2) =

∞∑n=1

∑|m|≤n

∣∣∣∣Rε

r

∣∣∣∣2 ∣∣∣∣ jn(kr) + kr j′n(kr)

jn(kRε) + kRε j′n(kRε)

∣∣∣∣2 ∣∣∣(ν × E(Rε . ), Vmn )L2(S2)

∣∣∣2

+

∣∣∣∣ jn(kr)

jn(kRε)

∣∣∣∣2 ∣∣∣(ν × E(Rε . ), Umn )L2(S2)

∣∣∣2≤ c2

∞∑n=1

∑|m|≤n

∣∣∣(ν × E(Rε . ), Vmn )L2(S2)

∣∣∣2 +∣∣∣(ν × E(Rε . ), U

mn )L2(S2)

∣∣∣2≤ c2‖ν × E(Rε)‖L2(S2) ≤ c2ε2 . (2.54)

Here we have used again as in the proof of 2.34 a uniform bound for all n ∈ N of∣∣∣ jn(kr)jn(kRε)

∣∣∣ ≤ c

and of∣∣∣Rεr jn(kr)+kr j′n(kr)

jn(kRε)+kRε j′n(kRε)

∣∣∣ ≤ c for r ∈ [R1, R0], where the last estimate follows from

Theorem 2.31 by using the identity jn(z) + zj′n(z) = zjn−1(z)− njn(z) (see Excercise 2.8).

Since (2.54) holds for any ε > 0 we obtain by the expansion of Theorem 2.48 that αmn (rjn(kr))′ =0 and βmn jn(kr) = 0 for all r ∈ (R1, R0) and |m| ≤ n, n ∈ N. Thus it follows αmn = βmn = 0,and the expansion leads to E = 0 in B(0, R).

Next we show that the function E(rx) given in the Theorem satisfies the boundary conditionin L2 sense. For r < R we compare the representation (2.51b) with the given expansion anddefine

αmn =R

jn(kR) + kRj′n(kR)(fR, V

mn )L2 and βmn = − 1

jn(kR)(fR, U

mn )L2 .

Thus, we obtain from (2.51a)

x× E(rx) =∞∑n=1

n∑m=−n

αmnr

(rjn(kr))′ x× Umn (x) + βmn jn(kr) x× V m

n (x)

=∞∑n=1

n∑m=−n

αmnr

(rjn(kr))′ V mn (x)− βmn jn(kr)Um

n (x) .


With the definition of αmn and βmn it follows by Parseval’s identity

‖ν × E(r . )− fR‖L2 =∞∑n=1

n∑m=−n

∣∣∣∣ jn(kr)

jn(kR)− 1

∣∣∣∣2 |(fr, Umn )L2|2

+∞∑n=1

n∑m=−n

∣∣∣∣Rr jn(kr) + krj′n(kr)

jn(kR) + krj′n(kR)− 1

∣∣∣∣2 |(fr, V mn )L2|2 .

As in Theorem 2.34 we can adapt the arguments of Theorem 2.22 in the case of the Lapla-

cian, since by Theorem 2.31 the terms∣∣∣ jn(kr)jn(kR)

− 1∣∣∣ and

∣∣∣Rr jn(kr)+krj′n(kr)jn(kR)+kRj′n(kR)

− 1∣∣∣ are uniformly

bounded for r ∈ [0, R] and converge to zero as r tends to R for every n. Thus we concludethat ‖ν × E(r, ·)− fR‖L2 → 0 if r → R.

Furthermore, the second part of the proof of Theorem 2.22 can be transfered to the electric

fields by the uniform bounds on the ratios∣∣∣ jn(kr)jn(kR)

∣∣∣ and∣∣∣Rr jn(kr)+krj′n(kr)

jn(kR)+kRj′n(kR)

∣∣∣ and we conclude

part (b) of the Theorem. 2

The last part in this chapter is devoted to the Maxwell problem in the exterior R3\B(0, R)of a ball. First we observe that replacing the Bessel functions by the Hankel functions of thefirst kind in the constructed expansion functions will lead to solutions satisfying a radiationcondition. At this point we find it very convenient to use the scalar Sommerfeld radiationcondition (2.32) for the solutions x · E(x) and x · H(x) of the Helmholtz equation. Theseconditions imply in particular that the radial components of the electric and magnetic fieldsdecay as 1/r2.; that is, the fields E and H are “almost” tangential fields on large spheres.Later (see Corollary 2.53 and Remark 3.31) we will show that this form of the radiationcondition is equivalent to the better known Silver-Muller radiation condition.

Theorem 2.50 Let k ∈ C\0 with Im k ≥ 0 and E ∈ C2(R3\B[0, R]

)and H = 1

iωµcurlE

be solutions of the time harmonic Maxwell equations

curlE − iωµH = 0 in D and curlH + iωεE = 0 in R3 \B[0, R] .

Furthermore, let x 7→ x · E(x) and x 7→ x · H(x) satisfy the Sommerfeld radiation condi-tion (2.32). Then there exist unique αmn , β

mn ∈ C, |m| ≤ n, n = 0, 1, 2, . . . such that

E(x) =∞∑n=1

∑|m|≤n

[αmn√n(n+ 1)

rh(1)n (kr)Y m

n (x) x +αmnr

(rh(1)

n (kr))′Umn (x)

+ βmn h(1)n (kr)V m

n (x)

](2.55a)

=∞∑n=1

∑|m|≤n

1√n(n+ 1)

[αmn curl curl

(xh(1)

n (kr)Y mn (x)

)− βmn curl

(xh(1)

n (kr)Y mn (x)

)](2.55b)


for x = rx and r > R, x ∈ S2. The series converges uniformly with all of its derivatives incompact subsets of R3 \B[0, R].

The field H has the forms (2.51c), (2.51d) with h(1)n replacing jn.

Proof: The proof uses exactly the same arguments as the proof of Theorem 2.48. Firstwe note that ramn (r) =

∫S2 x · E(x)Y −mn (x) ds(x) is the expansion coefficient of the solution

u(x) = x · E(x) of the Helmholtz equation with respect to Y mn : |m| ≤ n, n ∈ N. By

assumption u is radiating, therefore ramn (r) = αmn h(1)(kr) for some coefficient αmn . Therefore,

(2.52) holds for jn replaced by h(1)n . Analogously, cmn (r) is given by cmn (r) = βmn h

(1)n (kr) for

some βmn because also x 7→ x·curlE(x) is radiating. This yields the form (2.55a) for E(x) and

from this also (2.55b) by using (2.46a), (2.46b) for h(1)n instead of jn. Uniform convergence

of this series is shown as in the proof of Theorem 2.48. 2

In contrast to the interior problem and similar to the Helmholtz equation the exteriorMaxwell problem for a ball is uniquely solvable for any k ∈ C with Im k ≥ 0 and everyR > 0.

Theorem 2.51 Let k ∈ C \ 0 with Im k ≥ 0 and let R > 0.

(a) For given fR ∈ L2t (S

2) there exist unique solutions E,H ∈ C2(R3 \B[0, R]

)of

curlE − iωµH = 0 and curlH + iωεE = 0 in R3 \B[0, R]

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

such that x 7→ x ·E(x) and x 7→ x ·H(x) satisfy the Sommerfeld radiation condition (2.32).The solution is given by

E(rx) =∞∑n=1

n∑m=−n

(fR, Umn )L2(S2)√

n(n+ 1)

1

h(1)n (kR)

curl[xh(1)

n (kr)Y mn (x)

]+

(fR, Vmn )L2(S2)√

n(n+ 1)

R

h(1)n (kR) + kRh

(1)n

′(kR)

curl curl[xh(1)

n (kr)Y mn (x)

]=

∞∑n=1

n∑m=−n

(fR, Vmn )L2(S2)

√n(n+ 1)

R

h(1)n (kR) + kRh

(1)n

′(kR)

h(1)n (kr)

rY mn (x) x

+ (fR, Vmn )L2(S2)

R

h(1)n (kR) + kRh

(1)n

′(kR)

h(1)n (kr) + kr h

(1)n

′(kr)

rUmn (x)

− (fR, Umn )L2(S2)

h(1)n (kr)

h(1)n (kR)

V mn (x) .

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

The field H has the forms as in Theorem 2.49 with h(1)n replacing jn.


(b) If fR ∈ L2t (S

2)∩C2(S2) the solution E is continuous in all of R3 \B(0, R), and its seriesrepresentation converges uniformly on B[0, R1] \B(0, R) for every R1 > R.

Proof: We omit the proofs of uniqueness and the fact that the series solves the boundaryvalue problem because the arguments are almost the same as in the proof of the coore-sponding Theorem 2.49 for the interior boundary value problem. Again, we only mention

that h(1)n (kR) and h

(1)n (kR) + kRh

(1)n

′(kR) do not vanish for any kR which follows again by

Theorem 2.27. To show the radiation condition we multiply the second representation ofE(x) by x = rx and arrive at

x · E(x) =∞∑n=0

n∑m=−n

(fR, Vmn )L2(S2)

√n(n+ 1)

Rh(1)n (kr)

h(1)n (kR) + kRh

(1)n

′(kR)

Y mn (x) .

This scalar solution of the Helmholtz equation has just the form of (2.38). Application ofLemma 2.38 and Lemma 2.39 yields that x 7→ x · E(x) satisfies the Sommerfeld radiationcondition.

The corresponding proof for x · H(x) uses the same arguments for the representation ofH = 1

iωµcurlE. 2

For later use we formulate and prove the analog of Lemma 2.38.

Lemma 2.52 Let

E(x) =∞∑n=1

∑|m|≤n

[αmn curl

(xh(1)

n (kr)Y mn (x)

)+ βmn curl curl

(xh(1)

n (kr)Y mn (x)

)]converge uniformly on compact subsets of R3 \B[0, R]. Then there exists R0 > R and scalarcontinuous functions g and h such that

E(x) = curl

[∫|y|=R0

g(y) Φ(x, y) ds(y)x

]+ curl2

[∫|y|=R0

h(y) Φ(x, y) ds(y)x

](2.56a)

H(x) =1

iωµcurlE(x) =

1

iωµcurl2

[∫|y|=R0


]

+ iωε curl

[∫|y|=R0


](2.56b)

for |x| > R0. Here, Φ denotes again the fundamental solution of the Helmholtz equation,given by (2.37). In particular, the solution of the boundary value problem of the previoustheorem can be represented in this form.

Proof: We follow the proof of Lemma 2.38. Let R0 > R such that jn(kR0) 6= 0 for all n


and substitute the right hand side of (2.40) into the series for E(x). This yields

E(x) =∞∑n=1

∑|m|≤n

αmnikR2

0jn(kR0)curl

[∫|y|=R0

Y mn (y) Φ(x, y) ds(y)x

]

+βmn

ikR20jn(kR0)

curl curl

[∫|y|=R0

Y mn (y) Φ(x, y) ds(y)x

]

= curl

[∫|y|=R0


]+ curl curl

[∫|y|=R0


]with

g(y) =∞∑n=1

∑|m|≤n

αmnikR2

0jn(kR0)Y mn (y) , h(y) =

∞∑n=1

∑|m|≤n

βmnikR2

0jn(kR0)Y mn (y) .

The uniform convergence of the series for g and h is shown in the same way as in the proofof Lemma 2.38. The form of H is obvious by curl2 = −∆ +∇ div and k2 = ω2µε. 2

We finish this chapter by the following corollary which states that any pair (E,H) of solutionsof the Maxwell system which satisfy the scalar radiation conditions of Theorems 2.50 and2.51 satisfy also the Silver–Muller radiation condition (1.20a), (1.20b); or even,

√µ0H(x)× x −

√ε0E(x) = O(|x|−2) (2.57a)

and √ε0E(x)× x +

√µ0H(x) = O(|x|−2) (2.57b)

uniformly in x/|x| ∈ S2. As mentioned above, the scalar radiation conditions of Theo-rems 2.50 and 2.51 imply in particular that the radial components x · E(x) and x · H(x)decay faster than O(|x|−1) to zero as |x| tends to infinity; that is, E(x) and H(x) are “al-most” tangential fields on large spheres. On the other hand, the Silver–Muller radiationcondition implies that the combination H(x) × x − |x|E(x) decays faster than O(|x|−1) tozero as |x| tends to infinity; that is, E(x) and H(x) are “almost” orthogonal to each other.

Corollary 2.53 Let k ∈ C\0 with Im k ≥ 0 and E ∈ C2(R3\B[0, R]

)and H = 1

iωµcurlE

be solutions of the time harmonic Maxwell equations

curlE − iωµH = 0 in D and curlH + iωεE = 0 in R3 \B[0, R]

such that x 7→ x ·E(x) and x 7→ x ·H(x) satisfy the Sommerfeld radiation condition (2.32).Then E and H satisfy the Silver–Muller radiation condition (2.57a), (2.57b).

Proof: From Theorem 2.50 and Lemma 2.52 we conclude that a representation of theform (2.56a) holds. Both terms on the right hand side satisfy the Silver–Muller radiationcondition, see Lemma 3.29 of the next chapter. 2.

2.8. EXERCISES 89

2.8 Exercises

Exercise 2.1 (a) Prove that the Laplacian ∆ is given in spherical polar coordinates by(2.1).

(b) Prove that the Laplacian ∆ is given in planar polar coordinates x =

(r cosϕr sinϕ

)by

∆ =1

r

∂

∂r

(r∂

∂r

)+

1

r

∂2

∂ϕ2.

Exercise 2.2 Show that the Laplace-Beltrami operator ∆S2 is selfadjoint and non-posivite;that is,∫

S2

f(x) ∆S2g(x) ds(x) =

∫S2

g(x) ∆S2f(x) ds(x) ,

∫S2

f(x) ∆S2f(x) ds(x) ≤ 0

for all f, g ∈ C2(S2).

Exercise 2.3 Construct harmonic functions in R2 or R2 \ 0 by separation of variables inplane polar coordinates.

Exercise 2.4 Prove parts (e) – (h) of Lemma 2.8. Hints:(e): Show by using the differential equation for Pn and part (c) that the derivatives of bothsides of the first equation coincide. For the second equation use (b).(f) This follows from (c) and (a).(g) This follows using both equations of (e).(h) This follows from (b) and (a).

Exercise 2.5 Show that the leading coefficient of Pn is given by (2n−1)!n! (n−1)! 2n−1 ; that is,

Pn(t) =(2n− 1)!

n! (n− 1)! 2n−1tn + Qn−1(t)

for some polynomial Qn−1 of order less than n. Hint: Use Theorem 2.8.

Exercise 2.6 Use property (c) of Theorem 2.8 to show for every ` ∈ N the existence ofc` > 0 such that

‖P (`)n ‖∞ = max

|t|≤1

∣∣∣∣d`Pn(t)

dt`

∣∣∣∣ ≤ c` n2` for all n ∈ N .

Exercise 2.7 Try to prove Theorems 2.23.

Exercise 2.8 Prove the following representations of the derivative by using Rayleigh’s For-mulas from Theorem 2.28

f ′n(z) =n

zfn(z) − fn+1(z) , f ′n(z) = − fn−1(z) − n+ 1

zfn(z) , z 6= 0 ,

where fn is any of the functions jn, yn, h(1)n , or h

(2)n .


Exercise 2.9 Prove the following asymptotics for the derivatives by modifying the proof ofTheorem 2.31. For any r ∈ N and ε > 0 and R > ε:

dr

dzrjn(z) =

1

(2n+ 1)!!

dr

dzrzn[1 +O

(1

n

)]for n→∞ ,

dr

dzrh(1)n (z) = −i (2n− 1)!!

dr

dzr1

zn+1

[1 +O

(1

n

)]for n→∞ ,

uniformly with respect to |z| ≤ R and uniformly with respect to ε ≤ |z| ≤ R, respectively.

Exercise 2.10 Try to formulate and prove the corresponding theorems of Section 2.7 forthe boundary condition ν ×H = f on ∂B(0, R).

Chapter 3

Scattering From a Perfect Conductor

In Section 2.6 of the previous chapter we have studied the scattering of plane waves by balls.In this chapter we investigate the same problem for arbitrary shapes. To treat this boundaryvalue problem we introduce the boundary integral equation method which reformulates theboundary value problem in terms of an integral equation on the boundary of the region. Forshowing existence of a solution of this integral equation we will apply the Riesz–Fredholmtheory. Again, as in the previous chapter, we consider first the simpler scattering problemfor the scalar Helmholtz equation in Section 3.1 before we turn to Maxwell’s equations inSection 3.2.

3.1 A Scattering Problem for the Helmholtz Equation

For this first part the sacttering problem we are going to solve is the following one:

Given an incident field uinc; that is, a solution uinc of the Helmholtz equation ∆uinc+k2uinc =0 in all of R3, find the total field u ∈ C2(R3 \D) ∩ C1(R3 \D) such that

∆u + k2u = 0 in R3 \D ,∂u

∂ν= 0 on ∂D , (3.1)

and such that the scattered field us = u− uinc satisfies the Sommerfeld radiation condition(2.32); 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.

In this section we restrict ourselves to the Neumann boundary problem as the model problem.The Dirichlet boundary value problem will be treated in Chapter 5.

Since we want to consider the classical situation of scattering in homogeous media, through-out this chapter we make the following assumptions.

91

92 CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR

Assumption: Let the wave number be given by k = ω√εµ ∈ C with constants ε, µ ∈ C where

we take the branch of the square root such that Im k ≥ 0. Let D ⊆ R3 be a finite union ofbounded domains Dj such that Dj ∩ D` = ∅ for j 6= `. Furthermore, we assume that theboundary ∂D is C2−smooth (see Definition 6.7 of the appendix) and that the complementR3 \D is connected.

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

3.1.1 Representation Theorems

We begin with the (really!) fundamental solution of the Helmholtz equation, compare (2.37).

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; that is, write Φ for Φk. Since the fundamental solutionand/or its derivatives will occur later on as kernels of various integral operators, we beginwith the investigation of certain 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 the integral∫DK(x, y) dy exists in the sense of Lebesgue 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 .

3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION 93

Then∫∂DK(x, y) ds(y) exists 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)

(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 withrespect to 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

βτβ .

(b) We choose a local coordinate system as in Definition 6.7; that is, a covering of ∂D bycylinders of the form Uj = RjCj + z(j) where Rj ∈ R3×3 are rotations and z(j) ∈ R3 andCj = B2(0, αj) × (−2ρj, 2ρj), and where ∂D ∩ Uj is expressed as ∂D ∩ Uj =

Rjx + z(j) :

(x1, x2) ∈ B2(0, αj) , x3 = ξj(x1, x2)

for some (smooth) function ξj : B2(0, αj)→ (−ρj, ρj).Furthermore, we choose a corresponding partition of unity (see Theorem 6.9); that is, afamily 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 = Rju + z(j) and y = Rjv + z(j) for some u, v ∈B2(0, αj) × (−ρj, ρj) with u3 = ξj(u1, u2) and v3 = ξj(v1, v2). We set u = (u1, u2) andv = (v1, v2) and certainly have an estimate of the form

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

Using polar coordinates with respect to u we estimate∫∂D∩Uj|x−y|>τ

ds(y)

|x− y|2−β≤ c2−β

2

∫B2(0,αj)\B2(u,c1τ)

1

|u− v|2−β√

1 + |∇ξj(v)|2 dv

≤ c

∫c1τ<|v−u|<2αj

dv

|u− v|2−β= 2π c

∫ 2αj

c1τ

r

r2−β dr ≤ 2π c(2αj)

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 .


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)

(Vϕ)(x) =

∫D

ϕ(y) Φ(x, y) dy , x ∈ R3 , (3.4c)

which are called single layer potential, double layer potential, and volume potential,respectively, with density ϕ. We will investigate these potential in detail in Subsec-tion 3.1.2 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),see Excercise 3.1.

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∫∂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), because the surface area of ∂B(x, r) is just 4πr2 and

exp(ikr)

4πr2

∫|y−x|=r

u(y) ds =exp(ikr)

4πr2

∫|y−x|=r

[u(y)− u(x)

]ds + exp(ikr)u(x) ,

where we obtain by continuity∣∣∣∣ 1

4πr2

∫|y−x|=r

[u(y)− u(x)

]ds

∣∣∣∣ ≤ supy∈B(x,r)

|u(y)− u(x)| → 0 , r → 0 .

Similarly the second term tends to zero. Therefore, also the limit of the volume integralexists as r → 0 and yields 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 6.10 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 of asingle and a double layer surface potential. This observation is a first hint on a reformulationof the boundary value problem in terms of an integral equation on the surface ∂D.

From the proof we note that our assumptions on the smoothness of the boundary ∂D are toostrong. For the representation theorem the domain D has to satisfy exactly the assumptionswhich are needed for Green’s theorems to hold.

As a corollary from the representation theorem we obtain the following conclusion.

Corollary 3.4 Let u ∈ C2(D) be a real- or complex valued solution of the Helmholtz equationin D. Then u is analytic, i.e. one can locally expand u into a power series. That is, forevery z ∈ D there exists r > 0 such that u has the form

u(x) =∑n∈N3

an(x1 − z1)n1(x2 − z2)n2(x3 − z3)n3 for3∑j=1

|xj − zj|2 < r2 ,

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 potential and the smoothness of the kernels x 7→ Φ(x, y) and x 7→ ∂Φ(x, y)/∂ν(y) forx 6= y it follows immediately that u ∈ C∞(D). The proof of analyticity is technically noteasy if one avoids methods from complex analysis.1 If one uses these methods then one canargue as follows: Fix x ∈ D and choose r > 0 such that B[x, r] ⊆ D. Define the regionR ⊆ C3 and the 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 principal value with cut along the negative real axis of the squareroot of the complex number

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

∑3j=1(zj − yj)2 =∑3

j=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 Rand thus (complex) analytic. 2

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

Theorem 3.5 (Holmgren’s uniqueness theorem)Let D be a domain with C2-boundary and u ∈ C1(D)∩C2(D) be a solution of the Helmholtzequation ∆u + k2u = 0 in D. Furthermore, let U be an open set such that U ∩ ∂D 6= ∅. Ifu = 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 ,

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

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

1We refer to [15], Section 2.4, for a proof.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.


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

Let k ∈ C \ 0 with Im k ≥ 0 and u ∈ C2(R3 \ D) ∩ C1(R3 \ D) be a solution of theHelmholtz equation ∆u + k2u = 0 in R3 \ D. Furthermore, let u satisfy the Sommerfeldradiation 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.

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 radiation conditions of Φ(x, ·) for fixed x with respect to y and of 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 the 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 have

O(

1

R2

)=

∫|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds =

∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + |ku|2

ds + 2 Im

k ∫|x|=R

u∂u

∂rds

.


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

u∂u

∂rds =

∫∂D

u∂u

∂νds +

∫B(0,R)\D

[|∇u|2 − k2|u|2

]dx .

We multiply by k and take the imaginary part. This yields

Im

k ∫|x|=R

u∂u

∂rds

= Im

k ∫∂D

u∂u

∂νds

+ Im k

∫B(0,R)\D

[|∇u|2 + |k|2|u|2

]dx

≥ Im

k ∫∂D

u∂u

∂rds

and thus ∫

|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + |ku|2 ds ≤ −2 Im

k ∫∂D

u∂u

∂rds

+ O(

1

R2

).

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

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

3.1.2 Volume and Surface Potentials

We have seen in the preceding section that any function can be represented by a combinationof volume and surface potentials. The integral equation method for solving boundary valueproblems for the Helmholtz equation and Maxwell’s equations rely heavily on the smoothnessproperties of these potentials. This subsection is concerned with the investigation of thesepotentials. The analysis is quite technical and uses some tools from differential geometry(see Subsection 6.3).

We recall the fundamental solution for k ∈ C; that is,

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)

where D ⊆ R3 is any open and bounded set.

Lemma 3.7 Let ϕ ∈ L∞(D) by any complex–valued function. 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 integral exists by Lemma 3.2 because∣∣∣ ∂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 .

Thus, we have∣∣∣∣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, which shows w ∈ C1(R3) and ∂w/∂xj = v. 2

This regularity result is not sufficient in view of second order differential equations. Higherregularity is obtained for Holder continuous densities.

Definition 3.8 For a set T ⊆ R3 and α ∈ (0, 1] we define the space C0,α(T ) of bounded,uniformly Holder-continuous functions by

C0,α(T ) :=

v ∈ C(T ) : v bounded and sup

x,y∈T, x6=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 C0,α(T ) is a Banach space with norm

‖v‖C0,α(T ) := supx∈T

∣∣v(x)∣∣︸︷︷︸

= ‖v‖∞

+ supx,y∈T, x 6=y

∣∣v(x)− v(y)∣∣

|x− y|α. (3.9)

Now we can complement the previous Lemma and obtain that volume potentials with Holdercontinuous densities are two times differentiable.


Theorem 3.9 Let ϕ ∈ C0,α(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 and i, j ∈ 1, 2, 3 where we have extended ϕ by zero in D0 \D. If ϕ = 0 on ∂Dthen w ∈ C2(R3).

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 3.2because

∣∣ϕ(y) − ϕ(x)∣∣∣∣∂2Φ/(∂xi∂xj)(x, y)

∣∣ ≤ c |x − y|α−3 for y ∈ D. The existence of thesurface integral is obvious because 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,

u(x) :=

∫D0

∂2Φ

∂xi∂xj(x, y)

[ϕ(y)− ϕ(x)

]dy − ϕ(x)

∫∂D0

∂Φ

∂xi(x, y) νj(y) ds(y) , x ∈ D ,

and

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, for


x ∈ D,∣∣∣∣u(x)− ∂vε∂xj

(x)

∣∣∣∣ ≤ ∫|y−x|≤2ε

∣∣ϕ(y)− ϕ(x)∣∣ ∣∣∣∣ ∂∂xj

(1− η

(|x− y|/ε

)) ∂Φ

∂xi(x, y)

∣∣∣∣ dy≤ c1

∫|y−x|≤2ε

(1

|y − x|3+‖η′‖∞

ε |y − x|2

)|y − x|α dy

= c2

2ε∫0

(1

r1−α +‖η′‖∞ε

rα)dr = c2

[(2ε)α

α+

(2ε)1+α

(1 + α) ε

]≤ c3 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, we fix z ∈ D and set D0 = B(z,R) where R is chosen such thatD ⊆ B(z,R). For x ∈ D we have

∆w(x) = −k2

∫B(z,R)

Φ(x, y)[ϕ(y)− ϕ(x)

]dy

− ϕ(x)

∫|y−z|=R

3∑j=1

yj − zjR

[exp(ikR)

4π R

(ik − 1/R

) xj − yjR

]ds(y) .

This holds for all x ∈ D. For x = z we conclude

∆w(z) = −k2w(z) + k2ϕ(z)

∫B(z,R)

exp(ik|z − y|

)4π |z − y|

dy − ϕ(z) eikR(1− ikR) ,

i.e.

∆w(z) + k2w(z) = −ϕ(z)

[eikR(1− ikR)− k2

∫B(z,R)

exp(ik|z − y|

)4π |z − y|

dy

]︸︷︷︸

= 1

because

k2

∫B(z,R)

exp(ik|z − y|

)4π |z − y|

dy = k2

R∫0

r eikrdr = −ikR eikR + eikR − 1 .

If ϕ vanishes on ∂D then the extension of ϕ by zero is in C0,α(R3) (see Exercise 3.5).Therefore, we can apply the result above to any ball D = D0 = B(0, R) which yields thatthe second derivatives of w are continuous in R3. 2

The following corollary shows that the volume integral is bounded when consider as anintegral operator between suitable spaces.

Corollary 3.10 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 ∈ C0,α(D). Then there exists c > 0with

‖w‖C1(R3) ≤ c ‖ϕ‖∞ and ‖w‖C2(A) ≤ c ‖ϕ‖C0,α(D) (3.11)

for all ϕ ∈ C0,α(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|

].

Thus for x ∈ R3 we can apply (3.8) and Lemma 3.2 above and obtain∣∣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 c > 0 can be chosen independent of x and ϕ. This proves the firstestimate of (3.11). Let now x ∈ A. If A ⊆ D then there exists δ > 0 with |x− y| ≥ δ for allx ∈ A and y ∈ ∂D. By (3.10) for D = D0 we have∣∣∣∣ ∂2w

∂xi∂xj(x)

∣∣∣∣ ≤ ‖ϕ‖C0,α(D)

∫D

∣∣∣∣ ∂2Φ

∂xi∂xj(x, y)

∣∣∣∣ |x− y|α dy + ‖ϕ‖∞∫∂D

∣∣∣∣ ∂Φ

∂xi(x, y)

∣∣∣∣ ds(y)

≤ c3 ‖ϕ‖C0,α(D)

∫D

dy

|x− y|3−α+ c4 ‖ϕ‖∞

∫∂D

ds(y)

|x− y|2

≤ c5 ‖ϕ‖C0,α(D) +c4

δ2‖ϕ‖∞

∫∂D

ds

≤ c‖ϕ‖C0,α(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 ≤ c‖ϕ‖C0,α(D) .

2

We continue with the single layer surface potential ; that is, the function

v(x) = Sϕ(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 6.3 of Chapter 6. First we note that for continuousdensities ϕ the integral exists by Lemma 3.2 above even for x ∈ ∂D because Φ has asingularity of the form

∣∣Φ(x, y)∣∣ = 1/(4π|x − y|). Before we prove continuity of v we make

the following general remark.


Remark 3.11 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.

Here, Hρ is a strip around ∂D of thickness ρ and Uδ is the neighborhood of ∂D with thicknessδ as defined in Lemma 6.10; that is,

Hρ :=z + tν(z) : z ∈ ∂D , |t| < ρ

,

Uδ :=x ∈ R3 : inf

z∈∂D|x− z| < δ

.

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 .

2

Using this remark we show that the single layer potential with continuous density is Holdercontinuous.

Theorem 3.12 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 3.11. Boundedness follows fromLemma 3.2 because

∣∣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|x2−y|∣∣

≤ |x1 − x2|4π |x1 − y||x2 − y|

+k |x1 − x2|4π |x1 − y|

(3.15)


because∣∣exp(it)− exp(is)

∣∣ ≤ |t− s| for all t, s ∈ R. Now (c) follows because |xj − y| ≥ δ forxj /∈ Uδ.

To show (b); that is, Holder continuity in Hρ, let x1, x2 ∈ Hρ, where ρ < ρ0 is chosen as inLemma 6.10 such that there is the unique representation of xj in the form xj = zj + tjν(zj)with zj ∈ ∂D and |tj| < ρ0. We set Γz,r =

y ∈ ∂D : |y − z| < r

and split the domain of

integration into Γz1,r and ∂D \ Γz1,r where we set r = 3|x1 − x2|. The integral over Γz1,r issimply 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|.

By Lemma 6.10 we conclude that |zj − y| ≤ 2|xj − y| for j = 1, 2 and 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|

because |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)


because 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

In preparation of the investigation of the double layer surface potential we prove an otherauxiliary result which we will be used often.

Lemma 3.13 For ϕ ∈ C0,α(∂D) and a ∈ C(∂D,C3) 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 ∈ ∂Dand w is Holder continuous in Hρ0 for any exponent β < α. Furthermore, there exists c > 0with

∣∣w(x)∣∣ ≤ c‖ϕ‖C0,β(∂D) for all x ∈ Hρ0, and the constant c does not depend on x and ϕ,

but it may depend on a.

Proof: For xj = zj + tjν(zj) ∈ Hρ0 , j = 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, x 6= y ,

∣∣∇yΦ(x1, y)−∇yΦ(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3

, y ∈ ∂D , x` ∈ Hρ0 with 0 < |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 Γz,r = y ∈ ∂D : |y − z| < r. Let c > 0 denote a generic constant which may differfrom line to line the integral (in the form (3.16)) over Γz1,r is estimated by∫

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 |xj − y| ≥ |zj − 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 can use the estimates

|x1 − x2| =r

3≤ 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, we obtain

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 a constant c > 0 and 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‖C0,β(∂D) for

x ∈ Hρ0 is simpler and left to the reader. 2

Next we consider the double layer surface potential

v(x) = Dϕ(x) =

∫∂D

ϕ(y)∂Φk

∂ν(y)(x, y) ds(y) , x ∈ R3 \ ∂D , (3.19)

for Holder-continuous densities. Here we indicate the dependence on k ≥ 0 by writing Φk.

Theorem 3.14 The double layer potential v from (3.19) with Holder-continuous densityϕ ∈ C0,α(∂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→x0x/∈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.

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 6.10. 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.

We continue with the analysis of v0 and note that we have again to prove estimates of theform (a) and (b) of Remark 3.11.

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 3.13 (takea(y) = ν(y)). This proves the estimate (a) and (b) of Remark 3.11 for v0.

Now we consider 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

The next step is an investigation of the derivative of the single layer potential (3.12); thatis,

v(x) =

∫∂D

ϕ(y) Φ(x, y) ds(y) , x ∈ R3 .

First we show the following auxiliary result.

Lemma 3.15 (a) There exists c > 0 with∣∣∣∣∫∂D\B(x,τ)

∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ c for all x ∈ R3 , τ > 0 ,

(b) limτ→0

∫∂D\B(x,τ)

∇xΦ(x, y) ds(y) =

∫∂D

H(y) Φ(x, y) ds(y)−∫∂D

ν(y)∂Φ

∂ν(y)(x, y) ds(y)

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\B(x,τ)

∇xΦ(x, y) ds(y) = −∫

∂D\B(x,τ)

∇yΦ(x, y) ds(y)

= −∫

∂D\B(x,τ)

Grad yΦ(x, y) ds(y)−∫

∂D\B(x,τ)

∂Φ

∂ν(y)(x, y) ν(y) ds(y)

and thus for any fixed vector a ∈ C3 by the previous theorem, again with Γ(x, τ) = ∂D ∩B(x, τ) and at(y) = ν(y)×

(a× ν(y)

), and the Lemma 6.18 we obtain∫

∂D\B(x,τ)

a · ∇xΦ(x, y) ds(y) =

∫∂D\B(x,τ)

Div at(y) Φ(x, y) ds(y)

+

∫∂Γ(x,τ)

at(y) · (τ(y)× ν(y)) Φ(x, y) ds(y)

+

∫∂D\B(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 to


zero 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) · (τ(y)× ν(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

As an abriviation we write in the following

v(x0)∣∣− = lim

x→x0x∈D

v(x) , v(x0)∣∣+

= limx→x0x/∈D

v(x) and

∂v

∂ν(x0)

∣∣∣∣−

= limx→x0x∈D

ν(x0) · ∇v(x) and∂v

∂ν(x0)

∣∣∣∣+

= limx→x0x/∈D

ν(x0) · ∇v(x) ,

and obtain the following relations for the traces of the derivatives of the single layer surfacepotential.

Theorem 3.16 The derivative of the single layer potential v from (3.12) with Holder-continuous density ϕ ∈ C0,α(∂D) can be continuously extended from D to D and fromR3 \D to R3 \D. The tangential component is continuous, i.e. Grad v|− = Grad v|+, andthe limiting 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)

Proof: We note that the integral exists (see proof of Theorem 3.14). 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 3.12and 3.14 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 Lemma 3.15. In particular, the tangential component is continuous andthe normal derivative jumps, and we have

∂v1

∂ν(x)

∣∣∣∣±

= ∓1

2+

∫∂D

∂Φ

∂ν(x)(x, y) ds(y) , 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 3.13 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 on ∂D we have

∂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

3.1.3 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 3.17 Let Λ =

(x, y) ∈ ∂D × ∂D : x 6= y

and K ∈ C(Λ).


(a) Let there exist c > 0 and α ∈ (0, 1) such that∣∣K(x, y)∣∣ ≤ c

|x− y|2−αfor all (x, y) ∈ Λ , (3.25a)

∣∣K(x1, y)−K(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)→ C0,α(∂D), defined by

(K1ϕ)(x) =

∫∂D

K(x, y)ϕ(y) ds(y) , x ∈ ∂D ,

is well defined and bounded.

(b) Let there exist c > 0 such that:∣∣K(x, y)∣∣ ≤ c

|x− y|2for all (x, y) ∈ Λ , (3.25c)

∣∣K(x1, y)−K(x2, y)∣∣ ≤ c

|x1 − x2||x1 − y|3

for all (x1, y), (x2, y) ∈ Λ (3.25d)

with |x1 − y| ≥ 3|x1 − x2| ,∣∣∣∣∫∂D\B(x,r)

K(x, y) ds(y)

∣∣∣∣ ≤ c for all x ∈ ∂D and r > 0 . (3.25e)

Then the operator K2 : C0,α(∂D)→ C0,α(∂D), defined by

(K2ϕ)(x) =

∫∂D

K(x, y)[ϕ(y)− ϕ(x)

]ds(y) , x ∈ ∂D ,


Proof: (a) We follow the idea of the proof of Theorem 3.12 and write∣∣(K1ϕ)(x1)− (K1ϕ(x2)∣∣ ≤ ‖ϕ‖∞ ∫

∂D

∣∣K(x1, y)−K(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

∣∣K(x1, y)−K(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

∣∣K(x1, y)−K(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, even simpler,and is left to the reader.

For part (b) we follow the ideas of the proof of Lemma 3.13. We write∣∣(K2ϕ)(x1)− (K2ϕ(x2)∣∣ ≤ ∫

Γx1,r

∣∣∣K(x1, y)[ϕ(y)− ϕ(x1)

]−K(x2, y)

[ϕ(y)− ϕ(x2)

]∣∣∣ ds(y)

+∣∣ϕ(x2)− ϕ(x1)

∣∣ ∣∣∣∣∣∫∂D\Γx1,r

K(x1, y) ds(y)

∣∣∣∣∣+

∫∂D\Γx1,r

∣∣ϕ(y)− ϕ(x2)∣∣ ∣∣K(x1, y)−K(x2, y)

∣∣ ds(y)

≤ c‖ϕ‖C0,α(∂D)

[∫Γx1,r

ds(y)

|y − x1|2−α+

∫Γx2,2r

ds(y)

|y − x2|2−α

]+ c‖ϕ‖C0,α(∂D)|x1 − x2|α

+ c‖ϕ‖C0,α(∂D)

∫∂D\Γx1,r

|y − x2|α|x1 − x2||y − x1|3

ds(y)

≤ c‖ϕ‖C0,α(∂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‖C0,α(∂D) |x1 − x2|α .

The proof of∣∣(K2ϕ)(x)

∣∣ ≤ c‖ϕ‖C0,α(∂D) is again simpler and is left to the reader. 2

One of the essential properties of the boundary operators is their compactness in Holderspaces.This follows from the previous theorem and the following compactness result.

Lemma 3.18 The imbedding C0,α(∂D)→ C(∂D) is compact for every α ∈ (0, 1).

Proof: We have to prove that the unit ball B =ϕ ∈ C0,α(∂D) : ‖ϕ‖C0,α(∂D) ≤ 1

is

relatively compact, i.e. its closure is compact, in C(∂D). This follows directly by the theoremof Arcela-Ascoli (see, e.g., [9], Appendix C.7). Indeed, B is equi-continuous because∣∣ϕ(x1)− ϕ(x2)

∣∣ ≤ ‖ϕ‖C0,α(∂D)|x1 − x2|α ≤ |x1 − x2|α


for all x1, x2 ∈ ∂D. Furthermore, B is bounded. 2

Thus from the last theorem we immediatly conclude the following corollary.

Corollary 3.19 Under the assumptions of Theorem 3.17 the operator K1 is compact fromC0,α(∂D) into itself for every α ∈ (0, 1).

Proof: The operator K1 is bounded from C(∂D) into C0,α(∂D) and the imbedding C0,α(∂D)into C(∂D) is compact, which implies compactness of K1 : C0,α(∂D)→ C0,α(∂D). 2

We apply this result to the boundary integral operators which appear in the traces of thesingle and double layer potentials of Theorems 3.12, 3.14, and 3.16.

Theorem 3.20 The operators S,D,D′ : C0,α(∂D)→ C0,α(∂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. Additionally, the operator S is bounded from C0,α(∂D) intoC1,α(∂D). Here, C1,α(∂D) = u ∈ C0,α(∂D) : Grad u ∈ C0,α(∂D,C3) equipped with itscanonical norm.

Proof: We have to check the assumptions (3.25a) and (3.25b) of Theorem 3.17. For x, y ∈∂D we have by the definition of the fundamental solution Φ and part (a) of Lemma 6.10that ∣∣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 estimate 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 ∈ ∂Dsuch that |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 solutionwe can restrict ourselves to the case k = 0. Indeed, as in the proof of Theorem 3.14 weobserve that ∂(Φk − Φ0)/∂ν is continuous.

Furthermore, we assume 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 the estimates (a) and (b) of Lemma 6.10 for the first term and the mean valuetheorem for the second term. By |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) follows from |x1 − x2| ≤ |x1 − y|/3 and |y − x2| ≤ |y − x1| + |x1 − x2| ≤4|y − x1|/3.

The same arguments hold for the normal derivative with respect to x.

Finally, we have to show that Grad S is bounded from C0,α(∂D) into C0,α(∂D,C3). Butthis is given from the representation (3.24). 2

3.1.4 Uniqueness and Existence

Now we come back to the scattering problem (3.1), (3.2) from the beginning of the section.We first study the question of uniqueness. For absorbing media; that is Im k > 0, uniquenesscan be seen directly from an application of Greens formulas and the radiation condition. Butin scattering theory we are interested in the case of a real and positive wave number k. The


following lemma is fundamental for proving uniqueness and tells us that a solution of theHelmholtz equation ∆u + k2u = 0 for k > 0 cannot decay faster than 1/|x| as x tendsto infinity. We will give two proofs of this result. The first – and shorter – one uses theexpansion arguments from the previous chapter. In particular, properties of the sphericalBessel- and Hankel functions are used. The second proof which goes back to the original workby Rellich (see [22]) avoids the use of these special functions but is far more technical andalso needs a stronger assumption on 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 (2.29); 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, Parseval’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. Especially, 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 it as (kr) i amn yn(kr)+(kr) cmn jn(kr)→0 for r → ∞. Now we use the asymptotic behaviour of jn(kr) and yn(kr) as r tends to in-finity. By Theorem 2.30 we conclude that

i amn Im[eikr(−i)n+1

]+ cmn Re

[eikr(−i)n+1

]−→ 0 , r →∞ .

The term (−i)n+1 can take the values ±1 and ±i. Therefore, depending on n, we have that

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 it holds for all n and m we conclude that u vanishes. 2

As mentioned above, the second proof avoids the use of the Bessel and Hankel functions butneeds, however, a stronger 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: The proof, which is taken from the monograph [14] (Section VII.3) is lengthy, andwe will structure it. Without loss of generality we assume that u is real valued because wecan consider 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) with respect tor. We write 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 followsfrom ∂

∂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∆S2u(r, θ, φ) + k2u(r, θ, φ)

=1

r

[∂2v

∂r2(r, θ, φ) + k2v(r, θ, φ) +

1

r2∆S2v(r, θ, φ)

],

i.e.

v′′(r) + k2v(r) +1

r2∆S2v(r) = 0 for r ≥ R0 , (3.28)

where again ∆S2 = Div Grad denotes the Laplace-Beltrami operator on the unit sphere;that is, in polar coordinates x = (sin θ cosφ, sin θ sinφ, cos θ)>

(∆S2w)(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 6.11 that ∆S2 is selfadjoint and negative definite, i.e.(

∆S2v, w)L2(S2)

=(v,∆S2w

)L2(S2)

and(∆S2v, 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

(∆S2v(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

(∆S2vm(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:

(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∆S2vm(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

(∆S2v(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

(∆S2v, v

)= 2(∆S2v, v′

):

E ′(r) = 2(v′(r), v′′(r)

)+ 2k2

(v(r), v′(r)

)− 1

r3

(∆S2v(r), v(r)

)+

2

r2

(∆S2v(r), v′(r)

)= 2

(v′(r) ,

[v′′(r) + k2v(r) +

1

r2∆S2v(r)

])− 1

r3

(∆S2v(r), v(r)

)= − 1

r3

(∆S2v(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) with respect to r, substitute the form of v′′m from(3.29) and obtain

∂

∂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

(∆S2vm(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 +(∆S2v(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

Applying the Rellich Lemma 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→ 0

as R tends to infinity. Rellich’s lemma (in the form Lemma 3.21 or Lemma 3.22) implies thatu vanishes outside of every ball which encloses ∂D. Finally, we note that u is an analyticfunction in the exterior of D (see Corollary 3.4). Since the exterior of D is connected weconclude that u vanishes in R3 \D. 2

We turn to the question of existence of a solution and choose the integral equation methodfor its treatment. We follow the approach of [6], Chapter 3, but prefer to work in thespace C0,α(∂D) of Holder continuous functions rather that in the space of merely continuousfunctions. This avoids the necessity to introduce the class of continuous functions for whichthe normal derivatives exist “in the uniform sense along the normal”.


Let us recall the notion of the single layer surface potential of (3.4a), see also (3.12), andmake the ansatz for the scattered field in the form of a single layer potential; that is,

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 ∈ C0,α(∂D) is some density to be determined. We remark alreadyhere that we will face some difficulties with this ansatz but we begin with this for didacticalreasons. First we note that us solves the Helmholtz equation in the exterior of D and alsothe radiation condition. This follows from the corresponding properties of the fundamentalsolution Φ(·, y), uniformly with respect to y on the compact surface ∂D. Furthermore, byTheorems 3.12 and 3.16 the function us and its derivatives can be extended continuouslyfrom the exterior into R3 \D with 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 3.20. 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 C0,α(∂D) . (3.33)

By Theorem 3.20 the operator D′ is compact. Therefore, we can apply the Riesz–Fredholmtheory. By Theerem 6.2, existence follows from uniqueness. To prove uniqueness we assumethat ϕ ∈ C0,α(∂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, againfrom the jump conditions 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. At this point we wish to conclude that v vanishes also in D,because then we could conclude by the jump of the normal derivatives of the potential at ∂Dthat ϕ vanishes. However, this is not always the case. Indeed, there are non trivial solutionsin D if, and only if, k2 is an eigenvalue of −∆ in D with respect to Dirichlet boundaryconditions (see Theorem 2.34).

This is the reason why it is necessary to modify the ansatz (3.31). There are several wayshow to do it, see the discussion in [5], Chapter 3 and 4. We choose a modification which wehave not found in the literature. It avoids the use of double layer potentials. On the otherhand, however, it results in a system of two equations which increases the numerical effort


considerably. We assume for simplicity that D is connected although this is not necessaryas one observes from the following arguments.

We choose an open ball B = B(z, ρ) with boundary Γ such that Γ ⊆ D and such thatk2 is not an eigenvalue of −∆ inside B with respect to Dirichlet boundary conditions. ByTheorem 2.34 from the previous chapter we observe that we have to choose the radius ρ ofB such that kρ is 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 ∈ C0,α(∂D) and ψ ∈ C0,α(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 C0,α(∂D) × C0,α(Γ). The operators D′, D′Γ + ikSΓ, ∂SΓ/∂ν, and (∂/∂ν + ik)S∂D are allcompact. Therefore, we can apply the Riez-Fredholm theory to this system. By Theorem 6.2existence is assured if the homogeneous system admits only the trivial solution ϕ = 0 andψ = 0. Therefore, let (ϕ, ψ) ∈ C0,α(∂D)×C0,α(Γ) be a solution of the homogeneous systemand define the v as the sum of the single layers with densities ϕ and ψ for all x in R3\(∂D∪Γ).From the the jump condition for the normal derivative and the first homogeneous integralequation we 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 vanishesin the exterior of D. Furthermore, v is continuous in R3 and satisfies also the Helmholtzequation 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 for the normal derivatives 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 and we have shown injectivity for the system ofintegral equations.

If D consists of several components D =⋃Mm=1 Dm 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 the Riesz–Fredholm theory to (3.35a) and (3.35b) yields the desired existenceresult.

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

In the second part of this chapter we focus on our main task, the scattering by electromag-netic waves. Here the idea of the integral equation method as shown for the scalar Helmholtzequation will be extended to the following scattering problem for the Maxwell system.

Given a solution (Einc, H inc) of the Maxwell system

curlEinc − iωµ0Hinc = 0 , curlH inc + iωε0E

inc = 0 in some neighborhood of D ,

determine the total fields E,H ∈ C1(R3 \D,C3) ∩ C(R3 \D,C3) 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−Einc and Hs = H−H inc satisfy the Silver–Muller radiationconditions

√ε0E

s(x) − √µ0Hs(x)× x

|x|= O

(1

|x|2

), (3.37a)

3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM 125

and√µ0H

s(x) +√ε0E

s(x)× x

|x|= O

(1

|x|2

), (3.37b)

uniformly with respect to x/|x|.Again throughout the section we fix some assumptions, in view of a classical formulation ofthe scattering problem in vacuum.

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

and the obstacle D ⊆ R3 be bounded and C2−smooth such that the complement R3 \D isconnected.

The case of a (homogeneous) conducting medium; that is, σ > 0, can be treated as wellwithout any difficulty. In this case we just have k ∈ C with Im k ≥ 0.

Clearly, after renaming the unknown fields, the scattering problem is a special case of thefollowing Exterior Boundary Value Problem:

Given a tangential field f ∈ C0,α(∂D,C3), i.e. ν(x) · f(x) = 0 on ∂D , such that Div f ∈C0,α(∂D) determine radiating solutions Es, Hs ∈ C1(R3 \ D,C3) ∩ C(R3 \ D,C3); that is,Es, Hs satisfy the radiating conditions (3.37a) and (3.37b), of the system

curlEs − iωµ0Hs = 0 and curlHs + iωε0E

s = 0 in R3 \D , (3.38a)

such that Es satisfies the boundary condition

ν × Es = f on ∂D . (3.38b)

We note that the assumption on the surface divergence of f is necessary by Corollary 6.20.

Additionally, we note that in the previous chapter we had assumed a different kind of ra-diation condition. We had made the assumption that the scalar fields x 7→ x · E(x) andx 7→ x ·H(x) – which are solutions of the scalar Helmholtz equation by Lemma 1.5 – satisfySommerfeld’s radiation condition 3.2. Later (in Remark 3.31) we will discuss the equivalenceof both radiation conditions.

3.2.1 Representation Theorems

We have seen in the previous section (Theorem 3.3) that every sufficiently smooth functionu can be written as a sum of a volume potential with density ∆u + k2u, a single layerpotential with density ∂u/∂ν, and a double layer potential with potential u. Additionallywe know from Lemma 1.3 that each component of solutions E and H of the homogeneousMaxwell system in a domain D satisfy the Helmholtz equation. Thus we obtain that thesecomponents are analytic and can be represented by surface potentials. But more appropriatefor Maxwell’s equations is a representation in terms of vector potentials which we will discussnext.


Theorem 3.25 Let k ∈ C and E ∈ C1(D,C3)∩C(D,C3) such that curlE ∈ C(D,C3) anddivE ∈ 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) .

Furthermore, the right–hand side of this equation vanishes for x /∈ D and is equal to 12E(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

∫Dr

Φ(x, y) curlE(y) dy−∇∫Dr

Φ(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)

=

∫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) ,

where we have applied the identity 6.6. We will show that Ir(x) vanishes. Indeed, we caninterchange differentiation and integration and write

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

(div(0, F3,−F2)

...

)dx =

∫∂Dr

ν×F ds .

(3.39)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 .

Here 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) lead to Ir(x) = 0 and 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

This represention holds for any (smooth) vector field E. The relationship with Maxwell’sequations becomes clear with the following identity.

Lemma 3.26 Let H ∈ C1(D,C3) ∩ C(D,C3) such that curlH ∈ C(D,C3), then for x ∈ Dholds

curl

∫D

Φ(x, y)H(y) dy −∫D

Φ(x, y) curlH(y) dy +

∫∂D

[ν(y)×H(y)

]Φ(x, y) ds(y) = 0 .

Proof: The volume potential∫D

Φ(x, y)H(y) dy is continuously differentiable by Lemma 3.7,therefore

curl

∫D

Φ(x, y)H(y) dy −∫D

Φ(x, y) curlH(y) dy

=

∫D

[∇xΦ(x, y)×H(y)− Φ(x, y) curlH(y)

]dy

= −∫D

[∇yΦ(x, y)×H(y) + Φ(x, y) curlH(y)

]dy

= −∫D

curly[Φ(x, y)H(y)

]dy

where we used (6.7). Now we choose r > 0 such that B[x, r] ⊆ D and apply Green’sformula (6.16a) in Dr = D \B[x, r]. This yields∫

Dr

curly[Φ(x, y)H(y)

]dy =

∫∂Dr

[ν(y)×H(y)

]Φ(x, y) ds(y)

The boundary contribution ∫|y−x|=r

[ν(y)×H(y)

]Φ(x, y) ds(y)

tends to zero as r → 0 which yields the desired result. 2

Subtracting the last identity multiplied by iωµ0 from the representation of E in Theorem3.25 shows that if E and H the Maxwell system the volume integrals annihilate and weobtain a representation only by surface integrals. This leads to the well known Stratton-Chuformula.


Theorem 3.27 (Stratton-Chu formula)Let k = ω

√ε0µ0 > 0 and E,H ∈ C1(D,C3) ∩ C(D,C3) 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 because ofdivE = 0. As mentioned above, subtraction of the identity from Lemma 3.26, multiplied byiωµ0, proves the first formula. For the second one we set Dr = D \ B[x, r] for r > 0 suchthat B[x, r] ⊆ D and compute∫

∂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

curlEand curl curl curl = − curl ∆. 2

Similar to the scalar case we are also interested in a representation formula in the exteriorof a domain D. As an example of specific solutions to the homogeneous Maxwell equationswhich exist in R3 except of one source point we already introduced dipoles in Section 1.7.

Definition 3.28 For k real and positive vector fields of the form

Emd(x) = curl[pΦ(x, y)

], Hmd(x) =

1

iωµ0

curlEmd(x) =1

iωµ0

curl curl[pΦ(x, y)

],

Hed(x) = curl[pΦ(x, y)

], Eed(x) = − 1

iωε0

curlHed(x) = − 1

iωε0

curl curl[pΦ(x, y)

],

for some y ∈ R3 and p ∈ C3 are called magnetic and electric dipols, respectively, at y withpolarization p.

As seen from the introduction of Chapter 1 the behaviour of these fields for |x| → ∞motivates the radiation condition. More specific we have the following results.

Lemma 3.29 Let k > 0.

(a) The electromagnetic fields Emd, Hmd and Eed, Hed of a magnetic or electric dipol, re-spectively, satisfy the Silver-Muller radiation condition (3.37a), (3.37b); that is,

√ε0E(x) − √µ0H(x)× x

|x|= O

(1

|x|2

), (3.40a)

and

√µ0H(x) +

√ε0E(x)× x

|x|= O

(1

|x|2

), (3.40b)

uniformly with respect to x/|x| ∈ S2 and (p, y) in compact subsets of C3 × R3.

(b) The fields x 7→ x · curl[pΦ(x, y)

]and x 7→ x · curl curl

[pΦ(x, y)

]satisfies the scalar

Sommerfeld radiation condition (3.2) uniformly with respect to (p, y) in compact subsetsof C3 × R3.

(c) The fields E(x) = curl[xΦ(x, y)

]and H = 1

iωµ0curlE are radiating solution of the

Maxwell system.


Proof: (a) 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 assertion can be proven.

(b) We prove the assertion only for u(x) = x · curl[pΦ(x, y)

]which we write as

u(x) = x ·[∇xΦ(x, y)× p

]=

(ik − 1

|x− y|

)Φ(x, y)

x · [(x− y)× p]|x− y|

=

(ik − 1

|x− y|

)Φ(x, y)

p · (y × x)

|x− y|= ikΦ(x, y)

p · (y × x)

|x− y|+ O(|x|2)

and thus by differentiating this expression

x·∇u(x) = −k2Φ(x, y)x · (x− y)

|x− y|p · (y × x)

|x− y|+O(|x|2) = −k2Φ(x, y)

p · (y × x)

|x− y|+O(|x|2)

which proves the assertion by noting that the O−terms hold uniformly with respect to (p, y)in compact subsets of C3 × R3.

(c) By Lemma 2.42 it suffices to show that E and H satisfy the Silver-Muller radiationcondition. We compute

E(x) = curl[xΦ(x, y)

]= ∇xΦ(x, y)× x

= Φ(x, y)

[ik − 1

|x− y|

]x− y|x− y|

× x = Φ(x, y)

[ik − 1

|x− y|

]x× y|x− y|

= ikΦ(x, y) (x× y) + O(

1

|x|2

)and

curlE(x) = ∇(

Φ(x, y)

[ik − 1

|x− y|

])× x× y|x− y|

+ Φ(x, y)

[ik − 1

|x− y|

]curl

x× y|x− y|

= ik∇xΦ(x, y)× x× y|x− y|

− ∇xΦ(x, y)

|x− y|× x× y|x− y|

+ O(

1

|x|2

)= ik∇xΦ(x, y)× x× y

|x− y|+ O

(1

|x|2

)


because

curlx× y|x− y|

= ∇x1

|x− y|× (x× y) +

1

|x− y|curlx(x× y) = O

(1

|x|

).

and

∇xΦ(x, y)

|x− y|= O

(1

|x|2

).

Finally, we have

ik∇xΦ(x, y)× x× y|x− y|

= ikΦ(x, y)

[ik − 1

|x− y|

]x− y|x− y|

× x× y|x− y|

= −k2Φ(x, y)

[(x− y) · y)

|x− y|2x− (x− y) · x)

|x− y|2y

]and thus

x× curlE(x) = k2Φ(x, y) (x× y) + O(

1

|x|2

).

Substituting the form of H and k = ω√ε0µ0 ends the proof. 2

We note from the lemma that for the electric and magnetic dipols both, the Silver-Mullerradiation condition for the pair (E,H) and the Sommerfeld radiation condition for the scalarfunctions e(x) = x·E(x) and h(x) = x·H(x) hold. The following representation theorem willshow that these two kinds of radiation conditions are indeed equivalent (see Remark 3.31).

Theorem 3.30 (Stratton-Chu formula in exterior domains)Let k = ω

√ε0µ0 > 0 and E,H ∈ C1(R3\D,C3)∩C(R3\D,C3) 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 (3.40a) or (3.40b).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 (3.40a). One applies Theorem 3.27 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 with respect

to 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 (3.40b).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 (3.40a). 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 with respect to x. Therefore, the first termcan be estimated 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 (3.40a). Therepresentation of H(x) follows again by computing H = 1

iωµ0curlE. If the second radiation

condition (3.40b) is assumed one can argue as before and derive the representation of Hfirst. 2

We draw the following conclusions from this result.

Remark 3.31

(a) If E, H are solutions of Maxwell’s equations in R3 \ D then each of the radiationconditions (3.40a) and (3.40b) implies the other one.

(b) The Silver-Muller radiation condition for solutions E, H of the Maxwell system isequivalent to the Sommerfeld radiation condition (3.2) for every component of E andH. This follows from the fact that the fundamental solution Φ and every derivative ofΦ satisfies the Sommerfeld radiation condition.

(c) The Silver-Muller radiation condition for solutions E, H of the Maxwell system isequivalent to the Sommerfeld radiation condition (3.2) for the scalar functions e(x) =x·E(x) and h(x) = x·H(x). This follows from the fact that the Silver–Muller radiationcondition implies a representation of the form of the previous theorem which impliesthe scalar Sommerfeld radiation conditions for e(x) and h(x) by Lemma 3.29, andthe scalar Sommerfeld radiation conditions for e(x) and h(x) imply the representation(2.56a) which satisfies the Silver–Muller radiation conditions again by Lemma 3.29.

(d) The asymptotic behaviour of Φ yields

E(x) = O(

1

|x|

)and H(x) = O

(1

|x|

)for |x| → ∞ uniformly with respect to all directions x/|x|.


Sometimes it is convenient 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 (3.41)

where again k = ω√µ0ε0 denotes the wave number. If, on the other hand, E satisfies (3.41)

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| → ∞ , (3.42)

uniformly with respect to x = x/|x|.

3.2.2 Vector Potentials and Boundary Integral Operators

In Subsection 3.2.3 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 , (3.43)

where a ∈ C0,α(∂D,C3) is a tangential field; that is, a(y) · ν(y) = 0 for all y ∈ ∂D.

Lemma 3.32 Let v be defined by (3.43). Then E = curl v satisfies (3.41) in all of R3 \ ∂Dand also the radiation condition (3.42).

The proof follows immediately from Lemma 3.29. 2

In the next theorem we study the behaviour of E at the boundary.

Theorem 3.33 The curl of the potential v from (3.43) with Holder continuous tangentialfield a ∈ C0,α(∂D,C3) 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 . (3.44)

If, in addition, the surface divergence Div a (see Section 6.5) is continuous, then div v iscontinuous in all of R3 with limiting values

div v(x)∣∣± =

∫∂D

Φ(x, y) Div a(y) ds(y) , x ∈ ∂D ,


and also for the normal component of curl v it holds

ν · curl v∣∣+

= ν · curl v∣∣− on ∂D .

If, furthermore, Div a ∈ C0,α(∂D) then curl curl v can be continuously extended from D toD and from R3 \D to R3 \D.

The limiting values are

ν × curl curl v∣∣+

= ν × curl curl v∣∣− on ∂D , (3.45)

and

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: The components of curl v are combinations of partial derivatives of the single layerpotential. Therefore, by Theorem 3.16 the field curl v has continuous extensions to ∂Dfrom both sides. It remains to show the representation of the tangential components ofthese extensions on the boundary. We recall the special neighborhoods Hρ of ∂D fromSubsection 6.3 and write x ∈ Hρ0 in the form x = z + tν(z), z ∈ ∂D, 0 < |t| < ρ0. Then wehave

ν(z)× curl v(x) =

∫∂D

ν(z)×[∇xΦ(x, y)× a(y)

]ds(y) (3.46)

=

∫∂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 3.13, 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.


Similar for the normal component we obtain

ν(z) · curl v(x)

=

∫∂D

[ν(z)− ν(y)

]·[∇xΦ(x, y)× a(y)

]ds(y) +

∫∂D

ν(y)[∇xΦ(x, y)× a(y)

]ds(y)

=

∫∂D

[a(y)× (ν(z)− ν(y))

]· ∇xΦ(x, y) ds(y)−

∫∂D

a(y)[ν(y)×∇yΦ(x, y)

]ds(y)

=

∫∂D

[a(y)× (ν(z)− ν(y))

]· ∇xΦ(x, y) ds(y)−

∫∂D

Grad yΦ(x, y) ·[a(y)× ν(y)

]ds(y)

=

∫∂D

[a(y)× (ν(z)− ν(y))

]· ∇xΦ(x, y) ds(y)−

∫∂D

Φ(x, y) Div[a(y)× ν(y)

]ds(y) ,

where the first integral is continuous at ∂D as above and the second is a single layer potentialand therefore also continuous. Thus taking the limit we obtain

ν · curl v∣∣+

= ν · curl v∣∣− .

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 y Φ(x, y) ds(y) =

∫∂D

Φ(x, y) Div a(y) ds(y) .

This, again, 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 for the boundary values of curl curl v follows by Theorem 3.16. 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 C0,α

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,C3) : a(y) · ν(y) = 0 on ∂D

,

C0,αt (∂D) = Ct(∂D) ∩ C0,α(∂D,C3) ,

C0,αDiv(∂D) =

a ∈ C0,α

t (∂D) : Div a ∈ C0,α(∂D).

We equip Ct(∂D) and C0,αt (∂D) with the ordinary norms of C(∂D,C3) and C0,α(∂D,C3),

respectively, and C0,αDiv(∂D) with the norm ‖a‖C0,α

Div(∂D) = ‖a‖C0,α(∂D) +‖Div a‖C0,α(∂D). Thenwe can prove:


Theorem 3.34 (a) The boundary operator M : Ct(∂D)→ C0,αt (∂D), defined by

(Ma)(x) = ν(x)×∫∂D

curlx[a(y) Φ(x, y)

]ds(y) , x ∈ ∂D , (3.47)


(b) M is well defined and compact from C0,αDiv(∂D) into itself.

(c) The boundary operator L : C0,αDiv(∂D)→ C0,α

Div(∂D), defined by

(La)(x) = ν(x)× curl curl

∫∂D

a(y) Φ(x, y) ds(y) , x ∈ ∂D , (3.48)

is well defined and bounded. Here, the right–hand side is the trace of curl2 v with vfrom (3.43) which exists by the previous Theorem, see (3.45).

Proof: (a) We see from (3.46) 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 3.17 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 3.13

∣∣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 C0,αDiv(∂D) is a subspace of C0,α

t (∂D) with bounded imbeddingand, furthermore, the space C0,α

t (∂D) is compactly imbedded in Ct(∂D) by Lemma 3.18.Therefore, M is compact from C0,α

Div(∂D) into C0,αt (∂D). It remains to show that DivM is


compact from C0,αDiv(∂D) into C0,α(∂D).

For a ∈ C0,αDiv(∂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 3.33. Furthermore, by Theorem 3.33 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 6.20 and the jump condition of the derivative of the single layer(Theorem 3.16)

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 C0,αDiv(∂D) into

C0,α(∂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 3.15 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 3.13, Theorems 3.12 and 3.14 the tangential component of w is con-tinuous, thus La is given by

(La)(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 C0,αDiv(∂D) into C0,α

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 3.17. 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 3.13).For the third assumption, namely∣∣∣∣∫

∂D\B(x,r)

ν(x)×∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ ∣∣∣∣∫∂D\B(x,r)

∇xΦ(x, y) ds(y)

∣∣∣∣ ≤ c (3.49)

for all x ∈ ∂D and r > 0 we refer to Lemma 3.15 below. This proves boundedness ofL from C0,α

Div(∂D) into C0,αt (∂D). We consider now the surface divergence DivLa. By

Corollary 6.20 and the form of w we conclude that DivLa = Div(ν × w) = −ν · curlw. Forx = 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) .

The normal component is bounded by the same arguments as above. 2


3.2.3 Uniqueness and Existence

Since we now have collected the integral operators with their mapping properties whichwill be applied in case of Maxwell’s equations, we continue with the investigation of thescattering problem. As a next step we prove that there exists at most one solution of theexterior boundary value problem (3.38a), (3.38b), (3.40a), (3.40b) and therefore also to thescattering problem (3.36a), (3.36b),(3.40a), (3.40b).

Theorem 3.35 For every tangential field f ∈ C0,αDiv(∂D,C3) there exists at most one radiat-

ing solution Es, Hs ∈ C1(R3 \D,C3)∩C(R3 \D,C3) of the exterior boundary value problem(3.38a), (3.38b), (3.40a), (3.40b).

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 f = 0.In the proof of the Stratton-Chu formula (Theorem 3.30) 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 3.30) 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

We turn to the question of existence of the scattering problem (3.36a), (3.36b), (3.40a),(3.40b) or, more generally, the exterior boundary value problem (3.38a), (3.38b), (3.40a),(3.40b). as in the scalar case we first prove an existence result which is not optimal butrather serves as a preliminary result to motivate a more complicated approach.


Theorem 3.36 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 . (3.50)

Then, for every f ∈ C0,αDiv(∂D), there exists a unique solution Es, Hs ∈ C1(R3 \ D,C3) ∩

C(R3 \D,C3) of the exterior boundary value problem (3.38a), (3.38b), (3.40a), (3.40b). Inparticular, under this assumption, the scattering problem (3.36a), (3.36b), (3.40a), (3.40b)has a unique solution for every incident field. The solution has the form

Es(x) = curl

∫∂D

a(y) Φ(x, y) ds(y) , Hs(x) =1

iωµ0

curlEs(x) , x /∈ D , (3.51)

for a ∈ C0,αDiv(∂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) = f(x) , x ∈ ∂D . (3.52)

Proof: First we note that, by the jump condition of Theorem 3.33, Es and Hs from (3.51)solve the boundary value problem if a solves (3.52); that is with the operator M,

1

2a + Ma = f .

Since M is compact we can apply the Riesz–Fredholm theory to this equation; that is,existence is assured if uniqueness holds. Therefore, let a ∈ C0,α

Div(∂D) satisfy 12a +Ma = 0

on ∂D. Define Es and Hs as in (3.51) in all of R3 \ ∂D. Then Es, Hs satisfy the Maxwellsystem and ν × Es(x)

∣∣+

= 12a +Ma = 0 on ∂D. The uniqueness result of Theorem 3.35

yields Es = 0 in R3 \ D. Application of Theorem 3.33 again yields that ν × curlEs iscontinuous on ∂D, thus ν × curlEs

∣∣− = 0 on ∂D. From our assumption we conclude that

Es vanishes also inside of D. Now we apply Theorem 3.33 a third time and have thata = ν × Es

∣∣− − ν × E

s∣∣+

= 0. Thus, the homogeneous integral equation admits only a = 0as a solution and, therefore, there exists a unique solution of the inhomogeneous equationfor every right hand side f ∈ C0,α

Div(∂D). 2

The problem of this approach is similarly to the one for scalar case. In general, there existeigenvalues of the problem (3.50). To construct an example we apply Lemma 2.42. Thuswe know, if u solves the scalar Helmholtz equation ∆u + k2u = 0 in the unit ball B(0, 1)then v(x) = curl

[u(x)x

]solves curl curl v − k2v = 0 in B(0, 1). Furthermore, we have

ν(x)× v(x) = x×[∇u(x)× x

]= Grad u(x) on S2.

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(x) = 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 satisfiesthe Helmholtz equation ∆u + k2u = 0 in all of R3, because sin(kr)

ractually is analytic in

R. By the previous 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 (3.50).

Thus again as in the scalar case the insufficent ansatz (3.51) has to be modified. We proposea modification of the form

Es(x) = curl

∫∂D

a(y) Φ(x, y) ds(y) (3.53a)

+ η curl curl

∫∂D

[ν(y)× (S2

i a)(y)]

Φ(x, y) ds(y) , (3.53b)

Hs =1

iωµ0

curlEs (3.53c)

for some constant η ∈ C, some a ∈ C0,αDiv(∂D), and where the bounded operator Si :

C0,α(∂D,C3)→ C1,α(∂D,C3) is the single layer surface operator with the value k = i, consid-ered componentwise; that is, Sia = (Sia1,Sia2,Sia3)> for a = (a1, a2, a3)> ∈ C0,α(∂D,C3).We note that Si is bounded from C0,α(∂D,C3) into C1,α(∂D,C3) by Theorem 3.20. There-fore, the operator K : a 7→ ν × S2

i a is compact from C0,αDiv(∂D,C3) 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 ∈ C0,α(∂D)

and one-to-one.

Proof: Let ϕ, ψ ∈ C0,α(∂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 for ψ = ϕ that u vanishes in R3. The jump condition∂u/∂ν|− − ∂u/∂ν|+ = ϕ implies that ϕ vanishes which shows injectivity of Si. 2

Analogously to the beginning of the previous section we observe with the help of Theo-rem 3.33 that the ansatz (3.53b), (3.53c) solves the exterior boundary value problem ifa ∈ C0,α

Div(∂D) solves the equation

1

2a + Ma + ηLKa = c on ∂D (3.54)

where again Ka = ν × S2i a. Finally, we can prove the general existence theorem.

Theorem 3.39 For every f ∈ C0,αDiv(∂D), there exists a unique radiating solution Es, Hs ∈

C1(R3 \ D) ∩ C(R3 \ D) of the exterior boundary value problem (3.38a), (3.38b), (3.40a),(3.40b). In particular, under this assumption, the scattering problem has a unique solutionfor every incident field. The solution has the form of (3.53b), (3.53c) for any η ∈ C \R andsome a ∈ C0,α

Div(∂D) which is the unique solution of the boundary equation (3.54).

Proof: We make the ansatz (3.53b), (3.53c) and have to discuss the boundary equation(3.54). The compactness of M and K and the boundedness of L yields compactness of thecomposition LK. Therefore, the Riesz–Fredholm theory is applicable to (3.54), in particularTheorem 6.2.To show uniqueness, let a be a solution of the homogeneous equation. Then, with theansatz (3.53b), (3.53c), we conclude that ν × Es|+ = 0 and thus Es = 0 in R3 \ D by theuniqueness result. From the jump conditions of Theorem 3.33 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 .

3.3. EXERCISES 145

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 andthus also a by the injectivity of Si. 2

3.3 Exercises

Exercise 3.1 Prove the one-dimensional form of the representation theorem (Theorem 3.3)with fundamental solution Φ(x, y) = i

2kexp(ik|x− y|) for x, y ∈ R, x 6= y.

Exercise 3.2 Use (2.40) to prove that, for |x| > |y|,

Φ(x, y) = ik

∞∑n=0

n∑m=−n

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

n (x)Y −mn (y)

=ik

4π

∞∑n=0

(2n+ 1) jn(k|y|)h(1)n (k|x|)Pn(x · y)

where again x = x/|x|, y = y/|y|. Show that the series converge uniformly on (x, y) ∈R3 × R3 : |y| ≤ R1 < R2 ≤ |x| ≤ R3 for all R1 < R2 < R3.

Exercise 3.3 Show for the example of the unit ball D = B(0, 1) that in general the interiorNeumann boundary value problem

∆u+ k2u = 0 in D ,∂u

∂ν= f on ∂D , (3.55)

is not uniquely solvable.

Exercise 3.4 Show that the interior Neumann boundary value problem (3.55) can be solvedby a single layer ansatz on ∂D provided uniqueness holds. Here D ⊆ R3 is any boundeddomain with sufficiently smooth boundary such that the exterior is connected.

Exercise 3.5 Let ϕ ∈ C0,α(D) with ϕ = 0 on ∂D. Then the extension of ϕ by zero outsideof D is in C0,α(R3).

The following Exercises 3.6–3.9 study the potential theoretic case; that is, for k = 0, inR3. Again, D ⊆ R3 is a bounded domain with sufficiently smooth boundary such that theexterior is connected.

Exercise 3.6 Show that the exterior Neumann boundary value problem

∆u = 0 in R3 \D ,∂

∂νu = f on ∂D , (3.56a)

has at most one solution u ∈ C2(R3 \D) ∩ C1(R3 \D) which satisfies the decay conditions

u(x) = O(|x|−1) , ∇u(x) = O(|x|−2) , |x| → ∞ , (3.56b)

uniformly with respect to x/|x| ∈ S2.


Exercise 3.7 Show that for any f ∈ C0,α(∂D) the exterior Neumann boundary value prob-lem (3.56a), (3.56b) has a unique solution as a single layer ansatz on ∂D.

Exercise 3.8 Show by using Green’s theorem that the homogeneous interior Neumannboundary value problem; that is, (3.56a) in D for f = 0 is only solved by constant functions.Show that the condition

∫∂Df(x) ds = 0 is a necessary condition for the inhomogeneous

interior Neumann boundary value problem to be solvable.

Exercise 3.9 (a) Show that the double layer boundary operator (3.26b) for k = 0 maps thesubspace

ϕ ∈ C0,α(∂D) :

∫∂Dϕ(x)ds = 0

into itself.

Hint: Study the single layer ansatz with density ϕ in D and make use of the jump conditions.

(b) Prove existence of the interior Neumann boundary value problem for any f ∈ϕ ∈

C0,α(∂D) :∫∂Dϕ(x)ds = 0

by a single layer ansatz with a density in this subspace.

Chapter 4

The Variational Approach to theCavity Problem

In this chapter we want to introduce the reader to a second powerful approach for solvingboundary value problems for the Maxwell system (or more general partial differential equa-tions) which is the basis of, e.g., the Finite Element technique. We introduce this idea forthe cavity problem as our reference problem which has been formulated in the introduction,see (1.21a)–(1.21c). The problem 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,C3). 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,C3).

Assuming scalar valued electric parameter µ, ε, σ and a sufficiently smmoth domain D, wefirst multiply (4.1b) by some smooth vector field ψ with vanishing tangential componentson ∂D, then integrate over D, use partial integration and (4.1a). 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 .

The boundary contribution vanishes because of the boundary condition for ψ. Multiplicationwith iω yields∫

D

[1

µcurlE · curlψ − ω2

(ε+ i

σ

ω

)E · ψ

]dx = iω

∫D

Je · ψ dx . (4.2)

147

148 CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM

Now the idea of the approach is to consider this equation in a Hilbert space setting, wherethe right–hand side is treated as a linear bounded functional. Considering the left–handside as a bilinear form in such a Hilbert space we can hope for an application of the Rieszrepresentation theorem or, more generally, the Lax-Milgram theorem (see 6.6) to obtainexistence results.

Thus the first crucial step is to find appropriate Hilbert spaces, which in particular incor-porates the boundary condition (4.1c) and allows a rigorous treatment of the idea. Theserequirements lead to the introduction of Sobolev spaces. We begin with Sobolev spacesof scalar functions in Subsection 4.1.1 before we continue with vector-valued functions inSubsection 4.1.2, which will be used in 4.2 for the cavity problem.

4.1 Sobolev Spaces

4.1.1 Basic Properties of Sobolev Spaces of Scalar Functions

We are going to 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 and u = 0 on∂D. The definitions and proofs are all elementary – mainly because we do not need anysmoothness assumptions on the set D ⊆ R3 which is in this section always an open set. Someof the analogous properties of the space H1(D) where no boundary condition is incorporatedare needed for the Helmholtz decomposition of Subsection 4.1.3 and will be postponed tothat subsection.

We assume that the reader is familiar with the following basic function spaces. We haveused some of them before already.

Ck(D) =

u : D → C :

u is k times continuously differentiable in Dand all derivatives can be continuously extended to D

,

Ck(D,C3) =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,C3) =

u ∈ Ck(D,C3) : uj ∈ Ck

0 (D) for j = 1, 2, 3,

Lp(D) =

u : D → C : u is Lebesgue-measurable and

∫D

|u|pdx <∞, p ∈ [1,∞) ,

L∞(D) = u : D → C : ∃ c > 0 : |u(x)| ≤ c a.e. ,

Lp(D,C3) =u : D → C3 : uj ∈ Lp(D) for j = 1, 2, 3

, p ∈ [1,∞] .

S =u ∈ C∞(R3) : sup

x∈R3

[|x|p∣∣Dqu(x)

∣∣] <∞ for all p ∈ N, q ∈ N3.

Here, N = 0, 1, 2, . . ., and the support of a measurable function u is defined as

supp(u) =⋂

K ⊆ D : K closed and u(x) = 0 a.e. on D \K.

4.1. SOBOLEV SPACES 149

Note that in general supp(u) ⊆ D, see also Exercise 4.1. The norms in the spaces Ck(D,C3)(for bounded D) and Lp(D,C3) for p ∈ [1,∞) are canonical, the norm in L∞(D) is given by

‖u‖∞ := infc > 0 : |u(x)| ≤ c a.e. on D

.

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 D ⊆ R3 be an open set. A function u ∈ L2(D) possesses a variationalgradient in L2(D) if there exists F ∈ L2(D,C3) with∫

D

u∇ψ dx = −∫D

F ψ dx for all ψ ∈ C∞0 (D) . (4.3)

We write ∇u = F . Using the denseness of C∞0 (R3) in L2(R3) (see Lemma 4.9) it is easy toshow that F is unique – if it exists (see Exercise 4.5).

Definition 4.2 We define the Sobolev space H1(D) by

H1(D) =u ∈ L2(D) : u possesses a variational gradient ∇u ∈ L2(D,C3)

and equip H1(D) with the inner product

(u, v)H1(D) = (u, v)L2(D) + (∇u,∇v)L2(D) =

∫D

[u v +∇u · ∇v

]dx .

Theorem 4.3 The space H1(D) is a Hilbert space.

Proof: Only completeness has to be shown. Let (un) be a Cauchy sequence in H1(D).Then (un) and (∇un) are Cauchy sequences in L2(D) and L2(D,C3), respectively, and thusconvergent: un → u and ∇un → F for some u ∈ L2(D) and F ∈ L2(D,C3). We show thatF = ∇u: For ψ ∈ C∞0 (D) we conclude that∫

D

un∇ψ dx = −∫D

∇un ψ dx

by the definition of the variational gradient. The left–hand side converges to∫Du∇ψ dx,

the right–hand side to −∫DF ψ dx, and thus F = ∇u. 2

The proof of the following simple lemma uses only the definition of the variational derivative.

Lemma 4.4 Let D ⊆ R3 be an open, bounded set.

(a) The space C1(D) is contained in H1(D), and the imbedding is bounded. For u ∈ C1(D)the classical derivatives ∂u/∂xj coincide with the variational derivatives.


(b) Let u ∈ H1(D) and φ ∈ C∞0 (D). Let v be the extension of uφ by zero into R3. Thenv ∈ H1

0 (R3) and the product rule holds; that is,

∇v =

φ∇u+ u∇φ in D ,

0 in R3 \D .(4.4)

Proof: (a) For u ∈ C1(D) and ψ ∈ C∞0 (D) we note that the extension v of the product uψby zero is in C1(R3). Therefore, choosing a box Q = (−R,R)3 containing D we obtain bythe product rule and the fundamental theorem of calculus∫

D

[ψ∇u+ u∇ψ] dx =

∫D

∇(uψ) dx =

∫Q

∇v dx = 0 .

Therefore, the classical derivative is also the variational derivative and thus belongs toL2(D,C3).

(b) Let ψ ∈ C∞0 (R3). Then φψ ∈ C∞0 (D). Let g ∈ L2(R3,C3) be the right–hand side of(4.4). Then∫

R3

[ψ g + v∇ψ] dx =

∫D

[ψφ∇u+ ψu∇φ+ uφ∇ψ] dx

=

∫D

[(ψφ)∇u+ u∇(ψφ)] dx −∫D

uφ∇ψ dx

= −∫D

uφ∇ψ dx = −∫R3

v∇ψ dx

because the first integral vanishes by the definition of the variational derivative of u. 2

We continue by showing some denseness results. The proofs will rely on the technique ofmollifying the given function, the Fourier transform and the convolution of functions. Wejust collect and formulate the corresponding results and refer to, e.g., [13], for the proofs.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 . (4.5)

Now we use the following result from the theory of Lebesgue integration.

Lemma 4.5 The space S is dense in L2(R3).


Proof: We only sketch the arguments. First we use without proof that the space of stepfunctions with compact support are dense in L2(R3). Every step function is a finite linearcombination of functions of the form u(x) = 1 on U and u(x) = 0 outside of U where U issome open bounded set. We leave the constructive proof that functions of this type can beapproximated by smmoth functions to the reader (see Exercise 4.2.) 2

From the denseness of S we obtain by a general extension theorem from functional analysis(see Theorem 6.1) that F has an extension to a unitary operator from L2(R3) onto itself and(4.5) 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 .

Obviously by a simple transformation we have u ∗ v = v ∗ u. The Lemma of Young clarifiesthe mapping properties of the convolution operator.

Lemma 4.6 (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 a commutative algebra with the convolutionas multiplication. The Fourier transform transforms the convolution into the pointwisemultiplication of functions:

F(u ∗ v)(x) = (2π)3/2 u(x) v(x) for all u ∈ Lp(R3) , v ∈ L1(R3) , p ∈ 1, 2 . (4.6)

The following result to smoothen functions will often be used.

Theorem 4.7 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 4.7 for the existence of such a function). Define φδ ∈

C∞(R3) by

φδ(x) =1

δ3φ

(1

δ2|x|2), x ∈ R3 . (4.7)

Then supp(φδ) ⊆ B3[0, δ] and:

(a) u ∗ φδ ∈ C∞(R3) ∩ L2(R3) for all u ∈ L2(R3). Let K ∈ R3 be compact and u = 0outside of K. Then supp(u ∗ φδ) ⊆ K +B3[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 ∈ C∞(R3) ∩ H1(R3) and ‖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, if uvanishes outside some compact set K then (u ∗φδ)(x) =

∫Ku(y)φδ(x− y) dy which vanishes

for x /∈ K + B3[0, δ] because |x − y| ≥ δ for these x and y ∈ K. Finally we note thatu ∗ φδ ∈ L2(R3) by the Lemma 4.6 of Young.

(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 (4.6) 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)3/2‖φ1‖∞ + 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). Therefore, again by theLemma of Young, ∇uδ ∈ L2(R3); that is uδ ∈ H1(R3). Application of part (b) yields theassertion. 2

With this result we can weaken the definition of the variational derivative and prove thefollowing product rule, compare with Lemma 4.4.

Corollary 4.8 Let D ⊆ R3 be an open and bounded set.


(a) Let u ∈ L2(D). Then u ∈ H1(D) if, and only if, there exists F ∈ L2(D,C3) with∫D

u∇ψ dx = −∫D

F ψ dx for all ψ ∈ C1(D) with compact support in D .

The field F coincides with ∇u.

(b) Let u ∈ H1(D) and v ∈ C1(D). Then uv ∈ H1(D), and the product rule holds; that is,

∇(uv) = v∇u + u∇v .

Proof: (a) Only one direction has to be shown. Let u ∈ H1(D) and ψ ∈ C1(D) withcompact support K in D. Then the extension by zero into the outside of D belongs toH1(R3) by Lemma 4.4. Part (c) of the previous theorem yields convergence of ψδ = ψ ∗ φδto ψ in H1(D) as δ → 0. Furthermore, supp(ψδ) ⊆ D for sufficiently small δ > 0; that is,ψδ ∈ C∞0 (D). Therefore, by the definition of the variational derivative,∫

D

[u∇ψδ + ψδ∇u] dx = 0

for sufficiently small δ > 0. Letting δ tend to zero yields∫D

[u∇ψ + ψ∇u] dx = 0

which shows the assertion for F = ∇u.

(b) Let ψ ∈ C∞0 (D). Then vψ ∈ C1(D) and vanishes in some neighborhood of ∂D. Therefore,∫D

[(uv)∇ψ + (u∇v + v∇u)ψ

]dx =

∫D

[u∇(vψ) + (vψ)∇u

]dx = 0

by part (a). 2

Lemma 4.4 implies that for bounded open sets the space C∞(D) is contained in H1(D). Bythe same arguments, the space C∞0 (D) is contained in H1(D) even for unbounded sets D.As a first denseness result we have:

Lemma 4.9 For any open set D ⊆ R3 the space C∞0 (D) is dense in L2(D).

Proof: First we assume D to be bounded and define the open subsets Dn of D byx ∈ D :

d(x, ∂D) > 1/n

where d(x, ∂D) = infy∈∂D |x−y| denotes the distance of x to the boundary

∂D. Then Dn ⊆ D, and we can find φn ∈ C∞0 (D) with 0 ≤ φn(x) ≤ 1 for all x ∈ Dand φn(x) = 1 for all x ∈ Dn. Let now u ∈ L2(D). Extend u by zero into all of R3, thenu ∈ L2(R3). By Theorem 4.7 there exists ψn ∈ C∞(R3) with ‖ψn−u‖L2(D) ≤ ‖ψn−u‖L2(R3) ≤1/n. The functions ψn = φnψn are in C∞0 (D) and ‖ψn − u‖L2(D) ≤ ‖φn(ψn − u)‖L2(D) +

‖(φn−1)u‖L2(D). The first term is dominated by ‖ψn−u‖L2(D) which converges to zero. Forthe second term we apply Lebesgue‘s theorem of dominated convergence. Indeed, writing


‖(φn−1)u‖2L2(D) =

∫D|φn(x)−1|2 |u(x)|2 dx, we observe that every x ∈ D is element of Dn for

sufficiently large n and thus φn(x)− 1 vanishes for these n. Furthermore, |φn(x)− 1|2 |u(x)|2is bounded by the integrable function |u(x)|2.Finally, if D is not bounded, we just approximate u ∈ L2(D) by a function with compactsupport and use the previous result for bounded sets to this approximation, compare withthe last part of the proof of Lemma 4.18 below. 2

The space C∞0 (R3) of smooth functions with compact support is even dense in H1(R3), seeExercise 4.3. For bounded domains, however, this is not the case. Therefore, we add another important definition.

Definition 4.10 The space H10 (D) is defined as the closure of C∞0 (D) with respect to

‖ · ‖H1(D).

Note that this definition makes sense because C∞0 (D) is a subspace of H1(D) by the aboveremark. By definition, H1

0 (D) is a closed subspace of H1(D). We understand H10 (D) as

the space of differentiable (in the variational sense) functions u with u = 0 on ∂D. Thisinterpretation is justified by the trace theorem, introduced in Chapter 5 (see Theorem 5.7).

Remark: Let D be an open bounded set. For fixed u ∈ H1(D) the left and the right–handsides of the definition (4.3) 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) ≤ ‖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) . (4.8)

The next step is to show compactness of the imbedding H10 (D) → L2(D) provided D is

bounded. For the proof we choose a box Q of the form Q = (−R,R)3 ⊆ R3 with D ⊆ Q.We cite the following basic result from the theory of Fourier series (see [13] for more detailsand proofs).

Theorem 4.11 Let Q = (−R,R)3 ⊆ R3 be a bounded cube. For u ∈ L2(Q) define theFourier coefficients un ∈ C by

un =1

(2R)3

∫Q

u(x) e−iπRn·xdx for n ∈ Z3 . (4.9)


Thenu(x) =

∑n∈Z3

un ei πRn·x

in the L2−sense; that is,∫Q

∣∣∣∣∣u(x)−M∑

n=−N

un ei πRn·x

∣∣∣∣∣2

dx −→ 0 , N,M →∞ .

Furthermore,

(u, v)L2(Q) = (2R)3∑n∈Z3

un vn and ‖u‖2L2(Q) = (2R)3

∑n∈Z3

|un|2 .

Therefore, the space L2(Q) 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,C3)

leads to the following definition.

Definition 4.12 We define the Sobolev space H1per(Q) of periodic functions by

H1per(Q) =

u ∈ L2(Q) :

∑n∈Z3

(1 + |n|2) |un|2 <∞

with inner product

(u, v)H1per(Q) = (2R)3

∑n∈Z3

(1 + |n|2)unvn .

Here, un and vn are the Fourier coefficients of u and v, respectively, see (4.9).

The next theorem guarantees that the zero-extensions of functions of H10 (D) belong to

H1per(Q).

Theorem 4.13 Let again D ⊆ R3 be a bounded open set and Q = (−R,R)3 an open box

which contains D in its interior. Then the extension operator η : u 7→ u =

u on D,0 on Q \D ,

is linear and bounded from H10 (D) into H1

per(Q).

Proof: Let first u ∈ C∞0 (D). Then obviously u ∈ C∞0 (Q). We compute its Fourier coeffi-cients un as

un =1

(2R)3

∫Q

u(x) e−iπRn·xdx =

iR

πnj(2R)3

∫Q

u(x)∂

∂xje−i

πRn·xdx

= − iR

πnj(2R)3

∫Q

∂u

∂xj(x) e−i

πRn·xdx = − iR

πnjvn


where vn are the Fourier coefficients of v = ∂u/∂xj. Therefore, by Theorem 4.11,

(2R)3∑n∈Z3

n2j |un|2 =

R2

π2(2R)3

∑n∈Z3

|vn|2 =R2

π2

∥∥∥∥ ∂u∂xj∥∥∥∥2

L2(Q)

=R2

π2

∥∥∥∥ ∂u∂xj∥∥∥∥2

L2(D)

.

Furthermore,

(2R)3∑n∈Z3

|un|2 = ‖u‖2L2(Q) = ‖u‖2

L2(D)

and thus adding the equations for j = 1, 2, 3 and for u,

‖u‖2H1per(Q) ≤

(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(Q) ≤

√1 +

R2

π2‖u‖H1(D) for all u ∈ H1

0 (D) .

This proves boundedness of the extension operator η. 2

As a simple application we have the compact imbedding.

Theorem 4.14 Let D ⊆ R3 be a bounded open set. Then the imbedding H10 (D) → L2(D)

is compact.

Proof: Let again Q = (−R,R)3 ⊆ R3 such that D ⊆ Q. First we show that the imbeddingJ : H1

per(Q)→ L2(Q) is compact. Define JN : H1per(Q)→ L2(Q) by

(JNu)(x) :=∑|n|≤N

un ei πRn·x , x ∈ Q , 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 analysis (see, e.g., [23], Section X.2), JN is compact. Furthermore,

‖(JN −J)u‖2L2(Q) = (2R)3

∑|n|>N

|un|2 ≤(2R)3

1 +N2

∑|n|>N

(1 + |n|2) |un|2 ≤1

1 +N2‖u‖2

H1per(Q) .

Therefore, ‖JN − J‖2H1per(Q)→L2(Q) ≤

11+N2 → 0 as N tends to infinity and thus, again by a

well known result from functional analysis, also J is compact.

Now the claim of the theorem follows because the composition

R J η : H10 (D)

η−→ H1per(Q)

J−→ L2(Q)R−→ L2(D)

is compact. Here, η : H10 (D) → H1

per(Q) denotes the extension operator from the previoustheorem and R : L2(Q)→ L2(D) is just the restriction operator. 2

For the space H10 (D) we obtain a useful result which is often called an inequality of Friedrich’s

Type.


Theorem 4.15 For any bounded open set D ⊆ R3 there exists c > 0 with

‖u‖L2(D) ≤ c ‖∇u‖L2(D) for all u ∈ H10 (D) .

Proof: Let again first u ∈ C∞0 (D), extended by zero into R3. Then, if again D ⊆ Q =(−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 ∈ Q 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(Q) ≤ (2R)2

R∫−R

R∫−R

R∫−R

∣∣∣∣ ∂u∂x1

(t, x2, x3)

∣∣∣∣2 dt dx2 dx3

≤ (2R)2‖∇u‖2L2(Q) = (2R)2‖∇u‖2

L2(D) .

By a density argument this holds for all u ∈ H10 (D). 2

4.1.2 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 4.16 Let D ⊆ R3 be an open set.

(a) A vector field v ∈ L2(D,C3) possesses a variational curl in L2(D,C3) if there exists avector field w ∈ L2(D,C3) such that∫

D

v · curlψ dx =

∫D

w · ψ dx for all vector fields ψ ∈ C∞0 (D,C3) .

(b) A vector field v ∈ L2(D,C3) possesses a variational divergence in L2(D) if there existsa scalar function p ∈ L2(D) such that∫

D

v · ∇ϕdx = −∫D

pϕ dx for all scalar functions ϕ ∈ C∞0 (D) .


The functions w and p are unique if they exist (compare Exercise 4.5). In view of partialintegration (see (6.6) and (6.9)) we write curl v and div v for w and p, respectively.

(c) We define the space H(curl, D) by

H(curl, D) =u ∈ L2(D,C3) : u has a variational curl in L2(D,C3)

,

and equip it with the natural inner product

(u, v)H(curl,D) = (u, v)L2(D) + (curlu, curl v)L2(D) .

With this inner product the space H(curl, D) is a Hilbert space. The proof follows by thesame arguments as in the proof of Theorem 4.3.

The following lemma corresponds to Lemma 4.4 and is proven in exactly the same way.

Lemma 4.17 Let D ⊆ R3 be an open, bounded set.

(a) The space C1(D,C3) is contained in H(curl, D), and the imbedding is bounded. Foru ∈ C1(D) the classical curl coincides with the variational curl.

(b) Let u ∈ H(curl, D) and φ ∈ C∞0 (D). Let v be the extension of uφ by zero into R3.Then v ∈ H0(curl,R3) and the product rule holds; that is,

curl v =

φ curlu+∇φ× u in D ,

0 in R3 \D .

We note that part (a) follows even directly from Lemma 4.4 because H1(D,C3) ⊆ H(curl, D).

We continue as in the scalar case and prove denseness properties (compare with Exercise 4.3).

Lemma 4.18 C∞0 (R3,C3) is dense in H(curl,R3).

Proof: First we show denseness of C∞(R3,C3) in H(curl,R3) by applying Theorem 4.7. Itis sufficient to show that

curl(u ∗ φδ) = (curlu) ∗ φδ for all u ∈ H(curl,R3)

with φδ from (4.7). Because supp(φδ) ⊆ B3[0, δ] we conclude for |x| < R that

(u ∗ φδ)(x) =

∫|y|<R+δ

u(y)φδ(x− y) dy

and for any fixed vector a ∈ R3 interchanging differentiation and integration leads to

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, the mapping y 7→ a φδ(x−y) is in C∞0 (R3,C3). This holds for all vectorsa, thus curl(u ∗ φδ) = (curlu) ∗ φδ.Finally, we have to approximate uδ = u∗φδ by a field with compact support. Let ψ ∈ C∞0 (R3)with ψ(x) = 1 for |x| ≤ 1. We define ψR(x) = ψ(x/R) and vR,δ = ψR u

δ and have

‖vR,δ − uδ‖2H(curl,R3) = ‖(ψR − 1)uδ‖2

L2(R3) + ‖∇ψR × uδ + (ψR − 1) curluδ‖2L2(R3)

≤ ‖(ψR − 1)uδ‖2L2(R3) + 2 ‖(ψR − 1) curluδ‖2

L2(R3)

+ 2 ‖∇ψR × uδ‖2L2(R3) .

It is easily seen by the theorem of dominated convergence that all of the terms converge tozero as R tends to infinity because for every x ∈ R3 there exists R0 with ψR(x) = 1 and∇ψR(x) = 0 for R ≥ R0. 2

Analogously to the definition of H10 (D) we define

Definition 4.19 For any open set D ⊆ R3 we define H0(curl, D) as the closure of C∞0 (D,C3)in H(curl, D).

Vector fields v ∈ H0(curl, D) do not necessarily vanish on ∂D – in contrast to functionsin H1

0 (D). Only their tangential components vanish. This follows from the trace theo-rem in Chapter 5, see Theorem 5.21. We finish this short introduction to H(curl, D) andH0(curl, D), respectively, by an important observation in view of the following the Helmholtzdecomposition and Maxwell’s equations.

Lemma 4.20 The space ∇H10 (D) =

∇p : p ∈ H1

0 (D)

is a closed subspace of H0(curl, D).

Proof 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,C3). Because ∇pj → ∇p in L2(D,C3) and curl∇pj = 0we note that (∇pj) is a Cauchy sequence in H(curl, D) and thus convergent. We concludethat ∇p ∈ H0(curl, D).It remains to show closedness of ∇H1

0 (D) in L2(D,C3) – and therefore, in H(curl, D). Letpj ∈ H1

0 (D) with ∇pj → F in L2(D,C3) for some F ∈ L2(D,C3). Then (∇pj) is a Cauchysequence in L2(D,C3). By Friedrich’s inequality of Theorem 4.15 we conclude that also(pj) is a Cauchy sequence in L2(D). Therefore, (pj) is a Cauchy sequence in H1(D), whichimplies convergence. This shows that F = ∇p for some p ∈ H1

0 (D) and ends the proof. 2

We recall that H10 (D) is compactly imbedded in L2(D) by Theorem 4.14, a result which is

crucial for the solvability of the interior boundary value problem for the Helmholtz equation.As one can easily show (see Exercise 4.14), the space H0(curl, D) fails to be compactlyimbedded in L2(D,C3). 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.


4.1.3 The Helmholtz Decomposition

Now we turn to decompositions of vector fields, which are closely related to the Maxwellsystem – the Helmholtz decomposition. For a general proof we will apply the Theorem 6.6of Lax and Milgram.

There are several forms of the Helmholtz decomposition. In this section we need decomposi-tions only in L2(D,C3) and in H0(curl, D). But for later purposes, however, (see Section 4.3)we prove the corresponding decompositions also in H(curl, D). Also, we consider it for com-plex and matrix-valued coefficients µ and ε which we treat simultaneously by writing A.

Theorem 4.21 Let D ⊆ R3 be open and bounded and A ∈ L∞(D,C3×3) such that A(x) issymmetric for almost all x. Furthermore, assume the existence of a constant c > 0 suchthat Re (z>A(x)z) ≥ c|z|2 for all z ∈ C3 and almost all x ∈ D. Then for the subspacesH(curl 0, D), VA ⊆ H(curl, D), and H0(curl 0, D), V0,A ⊆ H0(curl, D) and V0,A ⊆ L2(D,C3)defined by

H(curl 0, D) =u ∈ H(curl, D) : curlu = 0 in D

, (4.10a)

H0(curl 0, D) =u ∈ H0(curl, D) : curlu = 0 in D

, (4.10b)

VA =u ∈ H(curl, D) : (Au, ψ)L2(D) = 0 for all ψ ∈ H(curl 0, D)

, (4.10c)

V0,A =u ∈ H0(curl, D) : (Au, ψ)L2(D) = 0 for all ψ ∈ H0(curl 0, D)

, (4.10d)

V0,A =u ∈ L2(D,C3) : (Au, ψ)L2(D) = 0 for all ψ ∈ H0(curl 0, D)

, (4.10e)

we have:

(a) H0(curl 0, D) and V0,A are closed in L2(D,C3), and L2(D,C3) is the direct sum of V0,A

and H0(curl 0, D); that is,

L2(D,C3) = V0,A ⊕ H0(curl 0, D) .

(b) H(curl 0, D) and VA are closed in H(curl, D), and

H(curl, D) = VA ⊕ H(curl 0, D) .

(c) H0(curl 0, D) and V0,A are closed in H0(curl, D), and

H0(curl, D) = V0,A ⊕ H0(curl 0, D) .

Furthermore, all of the corresponding projection operators are bounded. The projection fromH0(curl, D) onto V0,A is the restriction of the projection operator from L2(D,C3) onto V0,A.

Proof: The closedness of H(curl 0, D) in H(curl, D) and H0(curl 0, D) in H0(curl, D) is ob-vious. The closedness of H0(curl 0, D) in L2(D,C3) is seen as follows. Let un ∈ H0(curl 0, D)converge to some u in L2(D,C3). Then (un) is a Cauchy sequence in L2(D,C3). From


curlun = 0 we observe that (un) is also a Cauchy sequence in H0(curl, D) and thus convergentin H0(curl, D). This shows that u ∈ H0(curl, D) and curlu = 0; that is, u ∈ H0(curl 0, D).

The closedness of V0,A and V0,A and VA in L2(D,C3) and H0(curl, D) and H(curl, D), re-spectively, is easy to see, and we omit this part.

Furthermore it holds V0,A ∩ H0(curl 0, D) = 0 because u ∈ V0,A ∩ H0(curl 0, D) implies(Au, u)L2(D) = 0 and thus, taking the real part, c‖u‖2

L2(D) ≤ Re (Au, u)L2(D) = 0 whichyields u = 0.

Next, we have to prove that L2(D,C3) ⊆ V0,A + H0(curl 0, D) and H0(curl, D) ⊆ V0,A +H0(curl 0, D) and H(curl, D) ⊆ VA + H(curl 0, D). Let u ∈ L2(D,C3). We define thesesquilinear form a : H0(curl 0, D)×H0(curl 0, D)→ C by

a(ψ, v) = (ψ,Av)L2(D) =

∫D

ψ>Av dx for all v, ψ ∈ H0(curl 0, D) . (4.11)

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 (v, Av)L2(D) ≥ c ‖v‖2L2(D) = c ‖v‖2

H(curl,D)

for v ∈ H0(curl 0, D) because curl v = 0. Therefore, the sesquilinear form a and – for everyfixed u ∈ L2(D,C3) – the bounded linear form `(ψ) = (ψ,Au)L2(D), ψ ∈ H0(curl 0, D),satisfy the assumptions of the Theorem 6.6 of Lax and Milgram. Therefore, for every givenu ∈ L2(D,C3) there exists a unique u0 ∈ H0(curl 0, D) with a(ψ, u0) = `(ψ) for all ψ ∈H0(curl 0, D); that is, (Au0, ψ)L2(D) = (Au, ψ)L2(D) for all ψ ∈ H0(curl 0, D). Therefore,

u − u0 ∈ V0,A and the decomposition u = u0 + (u − u0) ∈ H0(curl 0, D) + V0,A has beenshown.

Finally, boundedness of the projection operator u 7→ u− u0 follows from general propertiesof direct sums. This proves the result in L2(D,C3). Analogously, the projection operatorP : H0(curl, D)→ V0,A is defined in the same way by Pu = u− u0 where u0 ∈ H0(curl 0, D)is as before. The proof of H(curl, D) = VA +H(curl 0, D) follows the same arguments. 2

Remark 4.22 (a) Obviously, the space ∇H10 (D) = ∇p : p ∈ H1

0 (D) is contained inH0(curl 0, D). Therefore, V0,A and V0,A are contained in the spaces

H0(curl, divA 0, D) := v ∈ H0(curl, D) : (Av,∇ϕ)L2(D) = 0 for all ϕ ∈ H10 (D), (4.12a)

L2(divA 0, D) := v ∈ L2(D,C3) : (Av,∇ϕ)L2(D) = 0 for all ϕ ∈ H10 (D), (4.12b)

respectively, which are the spaces of vector fields with vanishing divergence div(Av) = 0.The space ∇H1

0 (D) is a strict subspace of H0(curl 0, D). It can be shown (see, e.g., [8],Section IX.1), that H0(curl 0, D) can be decomposed as a direct sum of ∇H1

0 (D) and thegradients ∇C(D) of the co-homology space

C(D) = p ∈ H1(D) : ∆p = 0 in D, p constant on every connected component of ∂D .

Analogously, ∇H1(D) is a subspace of H(curl 0, D) – and coincide with it for simply con-nected domains. Therefore, VA is contained in

H(curl, divA 0, D) = v ∈ H(curl, D) : (Av,∇ϕ)L2(D) = 0 for all ϕ ∈ H1(D) (4.12c)


which is the space of vector fields in H(curl, D) with vanishing divergence div(Av) = 0 andvanishing co-normal components (Aν) · v on ∂D.

(b) Often, as in our application, µ is a scalar, thus A = µI. In this case we write Vµ for VµIand, analogously, Vε for scalar valued ε. In the case σ = 0; that is real-valued and symmetricand positive definite ε and µ the direct sums

H(curl, D) = Vµ ⊕H(curl 0, D) and H0(curl, D) = V0,ε ⊕H0(curl 0, D)

are orthogonal with respect to the inner products

(u, v)ε,µ = (ε−1 curlu, curl v)L2(D) + (µu, v)L2(D) , (4.13a)

(u, v)µ,ε = (µ−1 curlu, curl v)L2(D) + (εu, v)L2(D) , (4.13b)

respectively, and L2(D,C3) = V0,ε⊕H0(curl 0, D) is an orthogonal decomposition with respectto the inner product (εu, v)L2(D).

By the same arguments as in the proof of the previous theorem one can show another kindof decompositions of L2(D,C3) and H0(curl, D).

Theorem 4.23 Let L2(divA 0, D) and H0(curl, divA 0, D) be defined in (4.12a) and (4.12b),respectively, where D and A are as in Theorem 4.21. Then the decompositions

L2(D,C3) = L2(divA 0, D) ⊕ ∇H10 (D) , H0(curl, D) = H0(curl, divA 0, D) ⊕ ∇H1

0 (D)

hold, and the projections are bounded.

We leave the proof to the reader (Exercise 4.15).

As mentioned before, the space H0(curl, D) is not compactly imbedded in L2(D,C3). How-ever – and this is the reason for introducing the Helmholtz decomposition – the subspaceV0,A is compactly imbedded in L2(D,C3). We formulate this crucial result in the followingtheorem.

Theorem 4.24 Let D be a bounded Lipschitz domain (see Definition 6.7). Then the spaceV0,A from (4.10d) is compactly imbedded in L2(D,C3).

The proof of this important theorem requires more sophisticated properties of Sobolev spaces.Therefore, we postpone it to Section 5.1, Theorem 5.32. From this result it follows (seeCorollary 4.36 below) that also VA is compactly imbedded in L2(D,C3). However, in thecontext of this chapter we do not need the compactness of VA.

4.2. THE CAVITY PROBLEM 163

4.2 The Cavity Problem

4.2.1 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 , (4.14)

where k ≥ 0 is real. We assume that a ∈ L∞(D) is real valued such that a(x) ≥ a0 on D forsome constant a0 > 0, and b ∈ L∞(D), f ∈ L2(D) are allowed to be complex valued.

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 4.25 Let D be a bounded open set and f ∈ L2(D) and b ∈ L∞(D) and real valueda ∈ L∞(D) such that a(x) ≥ a0 on D for some constant a0 > 0. A function u ∈ H1

0 (D) iscalled a variational solution (or weak solution) of the boundary value problem (4.14) if∫

D

[a∇u · ∇ψ − k2b uψ

]dx = −

∫D

f ψ dx for all ψ ∈ H10 (D) .

By Friedrich’s inequality, the first term in this equation defines an inner product in H10 (D).

Lemma 4.26 Let a ∈ L∞(D) be real valued such that a(x) ≥ a0 on D for some constanta0 > 0. We define a new inner product in H1

0 (D) by

(u, v)∗ = (a∇u,∇v)L2(D) , 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‖∞

a0

‖u‖∗ for all u ∈ H10 (D) (4.15)

where c is the constant from Theorem 4.15.

Proof: From the previous theorem we conclude

‖u‖2∗ = ‖

√a∇u‖2

L2(D) ≤ ‖a‖∞‖u‖2H1(D) = ‖a‖∞‖

[∇u‖2

L2(D) + ‖u‖2L2(D)

]≤ (1 + c2) ‖a‖∞‖∇u‖2

L2(D) ≤ ‖a‖∞1 + c2

a0

‖u‖2∗ .


2

Therefore, we write the variational equation of Definition 4.25 in the form

(u, ψ)∗ − k2(bu, ψ)L2(D) = −(f, ψ)L2(D) for all ψ ∈ H10 (D) . (4.16)

The question of existence of solutions of (4.30) is again – as in the previous chapter –answered by the Riesz–Fredholm theory.

Theorem 4.27 Let again D ⊆ R3 be a bounded open set and a, b ∈ L2(D) satisfy the aboveassumptions. Then the boundary value problem (4.14) has a variational solution for exactlythose f ∈ L2(D) such that (f, v)L2(D) = 0 for all solutions v ∈ H1

0 (D) of the corresponding

homogeneous form of (4.14) with b replaced by b; that is,

div(a∇v) + k2bv = 0 in D , v = 0 on ∂D . (4.17)

If, in particular, this boundary value problem (4.17) admits only the trivial solution v = 0then the inhomogeneous boundary value problem has a unique solution for every f ∈ L2(D).

Proof: We will use the Riesz representation theorem (Theorem 6.5) from functional analysisto rewrite (4.16) in the form u− k2Ku = −f with some compact operator K and apply theFredholm alternative of the previous theorem to this equation. To carry out this idea wenote 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 4.15 weconclude that there exists c > 0 such 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 satisfies

(Kv, ψ)∗ = (bv, ψ)L2(D) for all v ∈ L2(D) , ψ ∈ H10 (D) .

K is linear (easy to see) and bounded because ‖Kv‖2∗ = (Kv, Kv)∗ = (bv, Kv)L2(D) ≤

‖b‖∞‖Kv‖L2(D)‖v‖L2(D) ≤ c‖Kv‖∗‖v‖L2(D); that is, ‖Kv‖∗ ≤ c‖v‖L2(D). By the same argu-

ment there exists f ∈ H10 (D) with (f, ψ)L2(D) = (f , ψ)∗ for all ψ ∈ H1

0 (D). Therefore, theequation (4.16) can be written as

(u, ψ)∗ − k2(Ku, ψ)∗ = −(f , ψ)∗ for all ψ ∈ H10 (D) . (4.18)

Because this holds for all such ψ we conclude that this equation is equivalent to

u − k2Ku = −f 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)). To study

solvability of this equation we want to apply the Fredholm alternative of Theorem 6.4. We


choose the bilinear form 〈u, v〉 = (u, v)∗ =∫Da∇u · ∇v dx in H1

0 (D). Then K is self adjointbecause by the definition of K we have 〈Ku, v〉 = (bu, v)L2(D) =

∫Db u v dx and this is

symmetric in u and v. Therefore, we know from the Fredholm alternative of Theorem 6.4that equation (4.19) is solvable if, and only if, the right–hand side f is orthogonal withrespect to 〈·, ·〉 to the nullspace of I−k2K. Therefore, let v ∈ H1

0 (D) be a solution of (4.17).Then v satisfies the homogeneous equation v − k2Kv = 0 and 〈f , v〉 = (f , v)∗ = (f, v)L2(D).

Therefore, we conclude that 〈f , v〉 = 0 if, and only if, (f, v)L2(D) = 0. 2

We can also use the form (4.19) to prove existence and completeness of an eigensystem of

div(a∇v) + k2b v = 0 in D , u = 0 on ∂D , (4.20)

for real valued a, b ∈ L∞(D) such that a(x) ≥ a0 and b(x) ≥ b0 on D for some a0, b0 > 0.This equation has, of course, to be understood in the variational sense of Definition 4.25.Recall the definition of the inner product (u, v)∗ = (a∇u,∇v)L2(D).

Theorem 4.28 Let D be a bounded open set and a, b ∈ L∞(D) with a(x) ≥ a0 and b(x) ≥ b0

on D for some a0, b0 > 0. Then there exists an infinite number of eigenvalues k2 ∈ R>0 of(4.20); that is, there exists a sequence kn > 0 and corresponding eigenfunctions un ∈ H1

0 (D)with ‖un‖∗ = 1 such that div(a∇un) + knb un = 0 in D and un = 0 on ∂D in the variationalform. Furthermore, kn → ∞ as n → ∞. The sets un : n ∈ N and knun : n ∈ Nform complete orthonormal systems in

(H1

0 (D), (·, ·)∗)

and L2(D, b dx), respectively. Here,we denote by L2(D, b dx) the space equipped with the weighted inner product (u, v)L2(D,b dx) =(bu, v)L2(D).

Proof: We define the operator K : H10 (D) → H1

0 (D) as in the previous theorem. Then Kis compact and self adjoint in

(H1

0 (D), (·, ·)∗)

because b is real valued. It is also positivebecause (Ku, u)∗ = (bu, u)L2(D) > 0 for u 6= 0. Therefore, the spectral theorem for compactself adjoint operators in Hilbert spaces (see, e.g, [12], Section 15.3) implies the existence of aspectral 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 complete in(H1

0 (D), (·, ·))∗.

Therefore, we have Kun = µnun; that is, un − 1µnKun = 0, which is the variational form of

(4.20) for kn = 1√µn

. The system knun : n ∈ N is a system of eigenfunctions as well which

is orthogonal in L2(D, b dx) because

k2n(bun, um)L2(D) = k2

n(Kun um)∗ = (un, um)∗ = δmn .

This set is also complete in L2(D). This follows directly from the fact that H10 (D) is dense

in L2(D) (see Lemma 4.9). 2

Now we go back to the formulation (4.1a)–(4.1c) of the cavity problem for Maxwell’s equa-tions which we recall for the convenience of the reader.Given Je ∈ L2(D,C3) and scalar and real valued ε, µ, σ ∈ L∞(D) determine E ∈ H0(curl, D)and H ∈ H(curl, D) such that


curlH + (iωε− σ)E = Je in D . (4.21b)


The boundary condition (4.1c) is included in the space H0(curl, D) for E.

At the beginning of the section we already computed the variational equation (4.2) for thissystem; that is,∫

D

[1

µcurlE · curlψ − ω2

(ε+ i

σ

ω

)E · ψ

]dx = iω

∫D

Je · ψ dx . (4.22)

This variational equation holds for all ψ ∈ H0(curl, D).

Analogously, we can eliminate E. Indeed, now we multiply (4.21a) 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 withiω yields∫

D

[1

ε+ iσ/ωcurlH · curlψ − ω2µH · ψ

]dx =

∫D

1

ε+ iσ/ωJe · curlψ dx . (4.23)

This holds for all ψ ∈ H(curl, D).

It is easy to see that the variational equations (4.22) and (4.23) are equivalent to the Maxwellsystem.

Lemma 4.29 Let E ∈ H0(curl, D) satisfy the variational equation (4.22) 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 (4.21a), (4.21b).

Furthermore, H ∈ H(curl, D) satisfies (4.23) for all ψ ∈ H(curl, D).

The same statement holds for E and H interchanged.

Proof: Let E ∈ H0(curl, D) satisfy (4.22) for all ψ ∈ H0(curl, D). We note that H ∈L2(D,C3). Substituting the definition of H into (4.22) yields∫

D

[iω H · curlψ − ω2

(ε+ i

σ

ω

)E · ψ

]dx = iω

∫D

Je · ψ dx

which is the variational form of iω curlH; that is,

iω curlH = ω2(ε+ i

σ

ω

)E + iω Je .


Division by iω yields (4.21b). 2

In the following we will discuss the variational equation (4.22) in the space H0(curl, D). Forthe remaining part of this subsection we make the following assumptions on the parametersµ, ε, σ, and on D (recall Subsection 1.3):

There exists c > 0 such that

(a) µ ∈ L∞(D) is real valued and µ(x) ≥ c almost everywhere,

(b) ε ∈ L∞(D) is real valued and ε(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 dielectricity by

εc(x) = ε(x) + iσ(x)

ω, x ∈ D . (4.24)

Also, we assume that D ⊆ R3 is open and bounded such that the subspace V0,εc (see (4.10d)for A = εcI) is compactly imbedded in L2(D,C3). By Theorem 4.24 this is the case forLipschitz domains D.Furthermore, it is convenient to define the vector field F ∈ L2(D,C3) by

F =iω

εcJe .

Then the variational equation (4.22) takes the form∫D

[1

µcurlE · curlψ − ω2εcE · ψ

]dx =

∫D

εc F · ψ dx (4.25)

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 V0,εc and one in H0(curl 0, D),where the one in H0(curl 0, D) will be trivial. Recall that V0,εc is given by (4.10d); that is,

V0,εc =u ∈ H0(curl, D) : (εcu, ψ)L2(D) = 0 for all ψ ∈ H0(curl 0, D)

.

Analogously, V0,εc denotes the corresponding subspace in L2(D,C3), see (4.10e).

First we use the Helmholtz decomposition to write F as F = F0 + f with F0 ∈ H0(curl 0, D)and f ∈ V0,εc . Then we make an ansatz for E in the form E = E0 +u with E0 ∈ H0(curl 0, D)and u ∈ V0,εc . Substituting this into (4.25) yields∫

D

[1

µcurlu · curlψ − ω2εc(u+ E0) · ψ

]dx =

∫D

εc(f + F0) · ψ dx (4.26)

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

−ω2

∫D

εcE0 · ψ dx =

∫D

εcF0 · ψ dx for all ψ ∈ H0(curl 0, D) .


Rewriting this as ∫D

εc[F0 + ω2E0] · ψ dx = 0 for all ψ ∈ H0(curl 0, D)

and setting ψ = F0 + ω2E0 ∈ H0(curl 0, D) yields E0 = − 1ω2 F0.

Therefore, (4.26) reduces to∫D

[1

µcurlu · curlψ − ω2εcu · ψ

]dx =

∫D

εcf · ψ dx (4.27)

for all ψ ∈ H0(curl, D). Second, we take ψ ∈ V0,εc as test functions. Before we investigate(4.27) we summarize this splitting in the following lemma.

Lemma 4.30 Let F ∈ L2(D,C3) have the Helmholtz decomposition F = F0 + f with f ∈V0,εc and F0 ∈ H0(curl 0, D).

(a) Let E ∈ H0(curl, D) be a solution of (4.25) and E = E0 + u with u ∈ V0,εc andE0 ∈ H0(curl 0, D) the Helmholtz decomposition of E. Then E0 = − 1

ω2 F0 and u ∈ V0,εc

solves (4.27) for all ψ ∈ V0,εc .

(b) If u ∈ V0,εc solves (4.27) then E = u− 1ω2 F0 ∈ H0(curl, D) solves (4.25).

To show solvability of (4.27) we follow the approach as in the scalar case for the Helmholtzequation and introduce the equivalent inner product (·, ·)∗ on V0,εc by

(v, w)∗ =

∫D

1

µcurl v · curlw dx , v, w ∈ V0,εc . (4.28)

Then we have the analogue of Lemma 4.26:

Lemma 4.31 The norm ‖v‖∗ =√

(v, v)∗ is an equivalent norm on V0,εc.

Proof: In contrast to the proof of Lemma 4.26 we use an indirect argument to show theexistence of a constant c > 0 such that

‖v‖H(curl,D) ≤ c ‖ curl v‖L2(D) for all v ∈ V0,εc . (4.29)

This is sufficient because the estimates ‖ curl v‖2L2(D) ≤ ‖µ‖∞‖v‖2

∗ and ‖v‖∗ ≤‖(1/µ)‖∞‖v‖H(curl,D) hold obviously.

Assume that the inequality (4.29) is not satisfied. Then there exists a sequence (vn) in V0,εc

with ‖vn‖H(curl,D) = 1 and curl vn → 0 in L2(D,C3) as n tends to infinity. The sequence(vn) is therefore bounded in V0,εc . Because V0,εc is compactly imbedded in L2(D,C3) byTheorem 4.24 there exists a subsequence, denoted by the same symbol, which convergesin L2(D,C3). For this subsequence, (vn) and (curl vn) are Cauchy sequences in L2(D,C3).


Therefore, (vn) is a Cauchy sequence in V0,εc and therefore convergent: vn → v in H(curl, D)for some v ∈ V0,εc and curl v = 0. Therefore, v ∈ V0,εc ∩ H0(curl 0, D) = 0 which showsthat v vanishes. This contradicts the fact that ‖v‖H(curl,D) = 1. 2

Now we write (4.27) in the form

(u, ψ)∗ − ω2 (εcu, ψ)L2(D) = (εcf, ψ)L2(D) for all ψ ∈ V0,εc .

Again, we use the representation theorem of Riesz (Theorem 6.5) to show the existenceof a linear and bounded operator K : L2(D,C3) → V0,εc with (ψ, εcu)L2(D) = (ψ, Ku)∗for all ψ ∈ V0,εc and u ∈ L2(D,C3). We carry out the arguments for the convenienceof the reader although they are completely analogous to the arguments in the proof ofTheorem 4.27. For fixed u ∈ L2(D,C3) the mapping ψ 7→ (ψ, εcu)L2(D) defines a linear andbounded functional on V0,εc . Therefore, by the representation theorem of Riesz there existsa unique g = gu ∈ V0,εc with (ψ, εcu)L2(D) = (ψ, gu)∗ for all ψ ∈ V0,εc . We set Ku = gu.

Linearity of K is clear from the uniqueness of the representation gu. Boundedness of K fromL2(D,C3) into V0,εc follows from the estimate

‖Ku‖2∗ = (Ku, Ku)∗ = (Ku, εcu)L2(D) ≤ ‖εc‖∞‖u‖L2(D)‖Ku‖L2(D)

≤ c ‖εc‖∞‖u‖L2(D)‖Ku‖∗

after division by ‖Ku‖∗. Because V0,εc is compactly imbedded in L2(D,C3) we concludethat K = K J is even compact as an operator from V0,εc into itself, if we denote byJ : V0,εc → L2(D,C3) the compact imbedding operator. Therefore, we can write equation(4.27) in the form

(u, ψ)∗ − ω2 (Ku,ψ)∗ = (Kf, ψ)∗ for all ψ ∈ V0,εc ;

that is,

u − ω2Ku = Kf in V0,εc . (4.30)

This is again a Fredholm equation of the second kind for u ∈ V0,ε. 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 Fredholm’s Theorem 6.4 to X = V0,εc

with bilinear form 〈u, v〉 = (u, v)∗ =∫D

1µ

curlu · curl v dx and operator T = ω2K. The

operator K is self adjoint. Indeed, 〈Ku,ψ〉 = (Ku,ψ)∗ = (εcu, ψ)L2(D) and this is symmetricin u and ψ. Lemma 4.30 tells us that the variational form (4.25) of the cavity problem issolvable for F ∈ L2(D,C3) if, and only if, the variational problem (4.27) is solvable in V0,εc

for the part f ∈ V0,εc of F . By Theorem 6.4 this is solvable for exactly those f ∈ V0,εc suchthat Kf is orthogonal to the nullspace of I − ω2K with respect to 〈·, ·〉. By Lemma 4.30v ∈ V0,εc solves (4.27) for f = 0 if, and only if, v solves (4.25) for F = 0. Also we note that,for v ∈ V0,εc ,

〈Kf, v〉 = (εcf, v)L2(D) =(εc(F0 + f), v

)L2(D)

= (εcF, v)L2(D) ;


that is 〈Kf, v〉 vanishes if, and only if, (εcF, v)L2(D) vanishes. Furthermore we note that vsolves the homogeneous form of (4.25) if, and only if, v solves (4.25) with εc replaced by εc.We have thus proven:

Theorem 4.32 The cavity problem (4.21a), (4.21b) has a solution E ∈ H0(curl, D) andH ∈ H(curl, D) for exactly those source terms Je ∈ L2(D,C3) such that (Je, v)L2(D) = 0for all solutions v ∈ H0(curl, D) of the corresponding homogeneous form of (4.21a), (4.21b)with σ replaced by −σ. If, in particular, the boundary value problem admits only the trivialsolution E = 0, H = 0 then the inhomogeneous boundary value problem (4.21a), (4.21b) hasa unique solution (E,H) ∈ H0(curl, D)×H(curl, D) for all Je ∈ L2(D,C3).

For the remaining part of this subsection we assume that σ = 0; that is, εc = ε is real valued.Then we can consider the following eigenvalue problem by the same arguments.


curlH + iωεE = J in D , (4.31b)

for J = 0.

Definition 4.33 The resolvent set consists of all ω ∈ C for which (4.31a), (4.31b) has aunique solution (E,H) ∈ H0(curl, D) × H(curl, D) for all J ∈ L2(D,C3) and such thatJ 7→ (E,H) is bounded from L2(D,C3) into H0(curl, D)×H(curl, D).

We have seen that this eigenvalue problem is equivalent to the corresponding variationalequation (4.27) for u = E ∈ V0,ε (because the right–hand side vanishes); that is,∫

D

[1

µcurlu · curlψ − ω2εu · ψ

]dx = 0 for all ψ ∈ V0,ε ,

which holds even for all ψ ∈ H0(curl, D) because u ∈ V0,ε. If u solves this equation then alsoits complex conjugate. Therefore, we can assume that u is real valued which we do for theremaining part of this section for all functions. Then we write the variational equation inthe form

(u, ψ)µ,ε = (ω2 + 1) (εu, ψ)L2(D) for all ψ ∈ H0(curl, D) , (4.32)

where (·, ·)µ,ε had been defined in (4.13b); that is, using the bilinear form 〈·, ·〉 = (·, ·)∗ andthe operator K : V0,ε → V0,ε from above,

(u, v)µ,ε = 〈u, v〉 + (εu, v)L2(D) = 〈u, v〉 + 〈Ku, v〉 . (4.33)

First we consider ω 6= 0. We have seen that ω 6= 0 is in the resolvent set if, and only if, theoperator I−ω2K is one-to-one in V0,ε; that is, 1/ω2 is in the resolvent set of K : V0,ε → V0,ε.We note that K is also compact and selfadjoint with respect to (·, ·)µ,ε. The reason why wetake (·, ·)µ,ε instead of just 〈·, ·〉 as the inner product will be clear in a moment.The spectrum of K; that is, the complementary set of the resolvent set in C, consists of 0 andeigenvalues 1/ω2

n which converge to zero. Let us now consider ω = 0. From the variationalequation we conclude for ψ = u that curlu = 0. Therefore, ω = 0 is an eigenvalue with theinfinite dimensional eigenspace H0(curl 0, D). We summarize this in the following theorem.


Theorem 4.34 There exists an infinite number of positive eigenvalues ωn ∈ R>0 of (4.31a),(4.31b); that is, there exists a sequence ωn > 0 for n ∈ N and corresponding real valuedfunctions vn ∈ V0,ε such that

curl

(1

µcurl vn

)− ω2

nε vn = 0 in D , ν × vn = 0 on ∂D

in the variational form (4.32); that is,∫D

[1

µcurl vn · curlψ − ω2

nε vn · ψ]dx = 0 for all ψ ∈ H0(curl, D) . (4.34)

The eigenvalues ωn > 0 have finite multiplicity and tend to infinity as n→∞. We normalizevn by ‖vn‖µ,ε = 1 for all n ∈ N, where the norm ‖ · ‖µ,ε had been defined in (4.13b). Thenthe sets vn : n ∈ N and

√1 + ω2

n vn : n ∈ N form complete orthonormal systems in(V0,ε, (·, ·)µ,ε

)and in

(V0,ε, (ε·, ·)L2(D)

), respectively. Furthermore, ω = 0 is also an eigenvalue

with infinite dimensional eigenspace H0(curl 0, D).

Proof: First we note that the variational equation (4.34) has been shown to hold for ψ ∈ V0,ε

only. For ψ ∈ H0(curl 0, D), however, the variational equation holds as well because ofvn ∈ .V0,ε and the definition of V0,ε. From (4.34) we conclude for ψ = vm that

δn,m = (vn, vm)µ,ε = (ω2n + 1) (εvn, vm)L2(D)

which shows that the set √

1 + ω2n vn : n ∈ N forms an orthonormal systems in

(V0,ε, (ε·, ·)L2(D)

).

It remains to prove completeness of this system. But this follows from the fact that V0,ε isdense in V0,ε. To see this latter denseness result let v ∈ V0,ε. By the denseness of C∞0 (D,C3) inL2(D,C3) (see Lemma 4.9) there exists a sequence vn ∈ C∞0 (D,C3) with vn → v in L2(D,C3).The decompositions of vn with respect to the direct sums H0(curl, D) = V0,ε + H0(curl 0)and L2(D,C3) = V0,ε + H0(curl 0) are identical: vn = v0

n + vn with v0n ∈ H0(curl 0, D) and

vn ∈ V0,ε. The boundedness of the projections yields v0n → 0 in L2(D,C3) and thus vn → v

in L2(D,C3) which ends the proof. 2

As a particular result of this theorem we recall that vn : n ∈ N forms a complete orthonor-mal system in the closed subspace V0,ε of H0(curl, D) with respect to the inner product(·, ·)µ,ε. The functions vn correspond to the electric fields E. As we will see now, the corre-sponding magnetic fields form a complete orthonormal system in the closed subspace Vµ ofH(curl, D), defined in (4.10c), with respect to the inner product (·, ·)ε,µ, defined in (4.13a).This symmetry is the reason for using the inner product (·, ·)µ,ε instead of just 〈·, ·〉.

Lemma 4.35 Let ωn > 0 and vn : n ∈ N ⊆ V0,ε be the complete orthonormal system ofV0,ε, defined by (4.34). Then wn := 1

ωn µcurl vn ∈ Vµ for all n, and wn : n ∈ N forms

a complete orthonormal system in the closed subspace Vµ of H(curl, D) with respect to theinner product (·, ·)ε,µ, defined in (4.13a). Furthermore, wn satisfies∫

D

[1

εcurlwn · curlψ − ω2

nµwn · ψ]dx = 0 for all ψ ∈ H(curl, D) . (4.35)


Proof: Substituting the definition of wn into (4.34) yields ωn(wn, curlψ

)L2(D)

=

ω2n

(ε vn, ψ

)L2(D)

for all ψ ∈ H0(curl, D), that is wn ∈ H(curl, D) and curlwn = ωnεvn by

the definition of the variational curl. Furthermore, wn ∈ Vµ because for φ ∈ H(curl 0, D) weconclude that ωn(µwn, φ)L2(D) = (curl vn, φ)L2(D) = (vn, curlφ)L2(D) = 0 by applying Green’stheorem in the form (6.16b). To show (4.35) let ψ ∈ H(curl, D). Then(ε−1 curlwn, curlψ

)L2(D)

= ωn (vn, curlψ)L2(D) = ωn (curl vn, ψ)L2(D) = ω2n (µwn, ψ)L2(D) .

Furthermore, from

(wn, wm)ε,µ =(ε−1 curlwn, curlwm

)L2(D)

+ (µwn, wm)L2(D)

= ωnωm (ε vn, vm)L2(D) +1

ωnωm

(µ−1 curl vn, curl vm

)L2(D)

=

(ω2n

1 + ω2n

+1

1 + ω2n

)δnm = δnm

we see that the system wn : n ∈ N forms an orthonormal set in Vµ. Also completeness inVµ is seen by similar arguments: Let ψ ∈ Vµ such that (wn, ψ)ε,µ = 0 for all all n ∈ N. Then

0 = (wn, ψ)ε,µ =(ε−1 curlwn, curlψ

)L2(D)

+ (µwn, ψ)L2(D)

= ωn (vn, curlψ)L2(D) +1

ωn(curl vn, ψ)L2(D)

=

(ωn +

1

ωn

)(vn, curlψ)L2(D)

for all n ∈ N. Defining φ = ε−1 curlψ we note that (ε vn, φ)L2(D) = 0 for all n ∈ N. The

completeness of the system√

1 + ω2n vn : n ∈ N

in V0,ε would yield φ = 0 provided

we had shown that φ ∈ V0,ε. But this is the case because for ρ ∈ H0(curl 0, D) we have(ε φ, ρ)L2(D) = (curlψ, ρ)L2(D) = (ψ, curl ρ)L2(D) = 0. Therefore, curlψ = 0, thus ψ ∈H(curl 0, D) ∩ Vµ = 0, which proves completeness. 2

As a corollary the analogue of Theorem 4.24 follows.

Corollary 4.36 The closed subspace Vµ is compactly imbedded in L2(D,C3).

Proof: The closed subspaces Vµ and V0,ε are isomorphic. Indeed, the isomophism T : V0,ε →Vµ is given by

Tv =∞∑n=1

αnwn for v =∞∑n=1

αn vn ∈ V0,ε

because of

‖Tv‖2ε,µ =

∞∑n=1

α2n = ‖v‖2

µ,ε .

Then compactness of Vµ follows from the compactness of V0,ε by Theorem 4.24. 2


4.2.2 Uniqueness and Unique Continuation

First we consider again the scalar boundary value problem (4.14). For conducting media;that is, complex b ∈ L∞(D) with Im b > 0 a.e. on D and real valued a ∈ L∞(D) witha(x) ≥ a0 for some a0 > 0 we have uniqueness by Green’s theorem. Indeed, let u ∈ H1

0 (D)be a solution of (4.14) in the variational form of Definition 4.25 for f = 0. Substitutingψ = u into the integral yields ∫

D

[a |∇u|2 − k2b |u|] dx = 0 .

Taking the imaginary part yields∫D

Im b |u|2dx = 0 and thus u = 0 because of the assump-tion on b.If only Im b ≥ 0 on D and Im b > 0 a.e. on some open subset U of D then, by the sameargument, u vanishes on U . We want to prove that this implies u = 0 in all of D. This prop-erty of a differential equation is called the unique continuation property . If u was analyticthis would follow from an analytic continuation argument which is well known from complexanalysis. However, for non-analytic coefficients a and b the solution u fails to be analytic.

Before we turn to the unique continuation property we briefly consider the analogous situa-tion for the boundary value problem (4.21a), (4.21b) for the Maxwell system. As before, weassume first that the medium is conducting; that is, σ > 0 in D. By the definition (4.24) ofεc this is equivalent to the assumption that the imaginary part of εc is strictly positive onD. Let E ∈ H0(curl, D) be a solution of the homogeneous equation. Inserting v = E yields∫

D

[1

µ

∣∣curlE∣∣2 − ω2εc |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. Again, if only σ > 0 a.e. on some opensubset U of D then, by the same argument, E vanishes on U .

Our next goal is to prove the unique continuation property of the scalar equation (4.14) andthe Maxwell system (4.21a), (4.21b). As a preparation we need an interior regularity result.We can prove it by the familiar technique of transfering the problem into the Sobolev spacesof periodic functions.For k ∈ N and an open domain D ⊆ R3 the Sobolev space Hk(D) is defined as before byrequiring that all partial derivatives up to order k exist in the variational sense and areL2−functions (compare with Definition 4.1). The space C∞(D) of smooth functions is densein Hk(D). Again, Hk

0 (D) is the closure of C∞0 (D) with respect to the norm in Hk(D).Let Q = (−R,R)3 ⊆ R3 be a cube containing D in its interior. Then we define Hk

per(Q) by

Hkper(Q) =

u ∈ L2(Q) :

∑n∈Z3

(1 + |n|2)k |un|2 <∞


with inner product

(u, v)Hkper(Q) = (2R)3

∑n∈Z3

(1 + |n|2)k unvn ,

compare with Definition 4.12. Here, un, vn ∈ C are the Fourier coefficients of u and v,respectively, see (4.9). Then it is easy to show:

Lemma 4.37 The following inclusions hold and are bounded (for any k ∈ N):

Hk0 (Q) → Hk

per(Q) → Hk(Q) .

Proof: First inclusion: We show this by induction with respect to k. For k = 0 there isnothing to show. Assume that the assertion is true for k ≥ 0. Then Hk

0 (Q) ⊆ Hkper(Q) and

there exists c > 0 such that

(2R)3∑n∈Z3

(1 + |n|2)k |un|2 = ‖u‖2Hkper(Q) ≤ c‖u‖2

Hk(Q) for all u ∈ Hk0 (Q) . (4.36)

Let u ∈ C∞0 (Q). Then, by partial differentiation (if nj 6= 0),

un =1

(2R)3

∫Q

u(x) e−iπRn·xdx =

1

(2R)3

i R

π nj

∫Q

u(x)∂

∂xje−i

πRn·xdx

= − 1

(2R)3

i R

π nj

∫Q

∂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 (4.36) holds for u and for ∂u/∂xj. Therefore,

‖u‖2Hk+1per (Q)

= (2R)3∑n∈Z3

(1 + |n|2)k+1 |un|2 = (2R)3∑n∈Z3

(1 + |n|2)k |un|2

+R2

π2(2R)3

3∑j=1

∑n∈Z3

(1 + |n|2)k∣∣d(j)n

∣∣2≤ c‖u‖2

Hk(Q) + cR2

π2

3∑j=1

‖∂u/∂xj‖2Hk(Q) ≤ c′‖u‖2

Hk+1(Q) .

This proves boundedness of the imbedding with respect to the norm of order k + 1. Thisends the proof of the first inclusion because C∞0 (Q) is dense in Hk+1

0 (Q).

For the second inclusion we truncate the Fourier series of u ∈ Hkper(Q) into

uN(x) =∑|n|≤N

un ei πRn·x


and compute directly

∂j1+j2+j3uN

∂xj11 ∂xj22 ∂x

j33

(x) =(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 ≤ k:∥∥∥∥ ∂j1+j2+j3uN

∂xj11 ∂xj22 ∂x

j33

∥∥∥∥2

L2(Q)

≤ (2R)3( πR

)2k ∑|n|≤N

|un|2|n1|2j1|n2|2j2|n3|2j3

≤ (2R)3( πR

)2k ∑|n|≤N

|un|2|n|2(j1+j2+j3)

≤( πR

)2k

‖uN‖2Hkper(Q) .

This proves the lemma by letting N tend to infinity. 2

We continue with a regularity result.

Theorem 4.38 (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) . (4.37)

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) . (4.38)

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 B3(x, ρ)are in D for every x ∈ U . Furthermore, U ⊆

⋃x∈U B3(x, ρ). Because U is compact there exist

finitely many open balls B3(xj, ρ) ⊆ D for xj ∈ U , j = 1, . . . ,m, with U ⊆⋃mj=1B3(xj, ρ).

For abbreviation we set Bj = B3(xj, ρ). We choose a partition 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) . (4.39)

Let now R > 0 such that D ⊆ Q = (−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 (4.39) yields∫

Bj

∇uj · ∇ψ` dx =

∫Bj

gj ψ` dx . (4.40)

We can extend the integrals to Q 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 (4.40). This yields( πR

)2∑n∈Z

an n · `∫Q

ei(n−`)·x dx =∑n∈Z

bn

∫Q

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(Q) ⊆H2(Q). Therefore, also

∑mj=1 uj ∈ H2(Q) 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 dx −∫D

Fj · ∇ψ dx


with gj = fϕj − u∆ϕj ∈ L2(D) and Fj = 2u∇ϕj ∈ L2(D,C3). As in the proof of (a)we observe that 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 theFourier 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 |`|2 |a`| ≤ c[|b`| + |`| |c`|

]which proves that

∑`∈Z(1 + |`|2) |a`|2 ≤ c

∑`∈Z(|b`|2 +

|c`|2)<∞ and thus uj ∈ H1

per(Q) ⊆ H1(Q). 2

Remarks:

(a) If f ∈ Hk(D) for some k ∈ N then u|U ∈ Hk+2(U) by the same arguments, appliediteratively. Indeed, we have just shown it for k = 0. If it is true for k − 1 and iff ∈ Hk(D) then gj ∈ Hk(D) (note that uBj ∈ Hk+1

0 (Bj) by assumption of induction!)and thus uj ∈ Hk+2(D).

(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 taken from [6], Section 8.3. The proof itselfgoes back to Muller [18] and Protter [20].

Theorem 4.39 (Unique Continuation Property)

Let D ⊆ R3 be a domain; that is, a nonempty, open and connected set, and u1, . . . , um ∈H2(D) be real valued such that

|∆uj| ≤ c

m∑`=1

|u`|+ |∇u`|

in D for j = 1, . . . ,m . (4.41)

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 B3[x0, R] ⊆ D. We show that uj vanishes inB3[x0, R/2]. This is sufficient because for every point x ∈ D we can find finitely many ballsB3(x`, R`) ` = 0, . . . , p, with R` ∈ (0, 1), such that B3(x`, R`/2) ∩ B3(x`−1, R`−1/2) 6= ∅ forall ` = 1, . . . , p and x ∈ B3(xp, Rp/2). Then one concludes uj(x) = 0 for all j by iterativelyapplying the first step.We choose the coordinate system such that x0 = 0. First we fix j and write u for uj. Let


ϕ ∈ C∞0(B3(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(x) = exp(−r−n)

∆v(x) +

2n

rn+1

∂v

∂r(x) +

n

rn+2

( nrn− n+ 1

)v(x)

.

Using the inequality (a + b)2 ≥ 4ab and calling the middle term in the above expression b,we see that(

∆u(x))2 ≥ 8n exp(−2r−n)

rn+1

∂v

∂r(x)

∆v(x) +n

rn+2

( nrn− n+ 1

)v(x)

.

From now on we drop the argument x. Multiplication with exp(2r−n) rn+2 and integrationyields∫

D

exp(2r−n) rn+2(∆u)2dx ≥ 8n

∫D

r∂v

∂r

∆v +

n

rn+2

( nrn− n+ 1

)vdx . (4.42)

We show that∫D

r∂v

∂r(x) ∆v(x) dx =

1

2

∫D

|∇v(x)|2dx and (4.43a)

∫D

1

rmv(x)

∂v

∂r(x) dx =

m− 2

2

∫D

v(x)2

rm+1dx for any integer m. (4.43b)

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

).

Since all boundary terms vanish, by partial integration we obtain∫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 (4.43a). For equation (4.43b) we have, using polar coordinates andpartial integration,

∫D

1

rmv∂v

∂rdx =

∫B3(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 (4.43b). Substituting (4.43a) and (4.43b) for m = 2n+1 and m = n+1into the inequality (4.42) 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 ≥ 12|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 (4.44)

≥ 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 (4.45)

for n ≥ 2. Up to now we have not used the estimate (4.41). We write now uj and uj for uand u, respectively. From this estimate, the inequality of Cauchy-Schwarz, and the estimate(a+ b)2 ≤ 2a2 + 2b2 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 (4.45) 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

≤ 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

2(n−m2c2)m∑j=1

∫|x|≤R/2

exp(2r−n) |∇uj|2 dx+ 2(n4 −m2c2)m∑j=1

∫|x|≤R/2

ψn(r)u2j dx

≤ ψn(R/2)m∑j=1

‖∆uj‖2L2(D) .

For n ≥ m2c2 we conclude that

2(n4 −m2c2)ψn(R/2)m∑j=1

∫|x|≤R/2

u2j dx ≤ 2(n4 −m2c2)

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 −m2c2)

m∑j=1

‖∆uj‖2L2(D)

The right–hand side tends to zero as n tends to infinity. This proves uj = 0 in B3(0, R/2).2

Now we apply this result to the scalar boundary value problem (4.14) and the boundaryvalue problem (4.21a), (4.21b) for the Maxwell system. We begin with the scalar problem.

Theorem 4.40 Let D ⊆ R3 be a domain and a ∈ C1(D) be real valued and a ≥ a0 on Dfor some a0 > 0. Furthermore, let b ∈ L∞(D) be complex valued with Im b ≥ 0 on D andIm b > 0 a.e. on some open subset U of D. Then the boundary value problem (4.14) has aunique solution u ∈ H1

0 (D) for every f ∈ L2(D).

Proof: By Theorem 4.27 it suffices to prove uniqueness. Let u ∈ H10 (D) be a solution for

f = 0. We have seen above that u vanishes on U , and it remains to show that u vanisheseverywhere on D. In view of Theorem 4.39 we have first to show that u ∈ H2(V ) for everydomain V with V ⊆ D.Let ϕ ∈ C∞0 (D) and define ψ = 1

aϕ on D. Then ψ ∈ C1

0(D) and ∇ϕ = a∇ψ + ψ∇a.Substituting a∇ψ and ψ this into the variational equation yields∫

D

[∇u · ∇ϕ− 1

aϕ∇a · ∇u− k2 b

auϕ]dx = 0

that is,∫D∇u · ∇ϕdx =

∫Df ϕ dx for all ϕ ∈ C∞0 (D) where f = 1

a∇a · ∇u + k2 b

au. Then

f ∈ L2(D), and by Theorem 4.38 we conclude that u ∈ H2(V ) for any open set with V ⊆ D.Furthermore, from the differential equation we have the estimate

|∆u(x)| ≤ 1

a0

‖∇a‖∞ |∇u(x)| + k2 ‖b‖∞a0

|u(x)| on V .


From this we conclude an estimate of the type (4.41) for the real and imaginary parts of u;that is, u1 = Re u and u2 = Im u. Application of Theorem 4.39 to any open domain V withU ⊆ V and V ⊆ D yields that u vanishes in such domains V . This implies that u vanishesin D and ends the proof. 2

We now turn to the Maxwell case and transform Maxwell’s equations to the vector Helmholtzequation. We need weaker smoothness conditions on εc if we work with the magnetic field –provided µ is constant which is the case for many materials. Thus, let us consider this case.

Theorem 4.41 Let µ > 0 constant, εc ∈ C1(D) with Im εc > 0 on some open set U ⊆ D.Then there exists a unique solution E ∈ H0(curl, D) of the boundary value problem (4.25)for every F ∈ L2(D,C3).

Proof: Again, it suffices to prove uniqueness. Let F = 0 and E ∈ H0(curl, D) be thecorresponding solution of (4.25). We have shown at the beginning of this subsection that Evanishes on U . Therefore, also H = 1

iωµcurlE = 0 in U . By Lemma 4.29, the magnetic field

H satisfies∫D

[1

εccurlH · curlψ − ω2µH · ψ

]dx = 0 for all ψ ∈ H(curl, D) . (4.46)

If we choose ψ = ∇φ for some φ ∈ H10 (D) then we have∫

D

H · ∇φ dx = 0 for all φ ∈ H10 (D) ,

which is the variational form of divH = 0.

If all of the functions were sufficiently smooth we just would rewrite the equation for H inthe form

0 = εc curl[ε−1c curlH

]− ω2εcµH = curl2H + εc∇

1

εc× curlH − ω2εcµH ,

and thus, because divH = 0 and curl2 = ∇ div−∆,

∆H = εc∇1

εc× curlH − ω2εcµH =

1

εccurlH ×∇εc − ω2εcµH .

We derive this formula also by the variational equation. Indeed, we set ψ = εcψ for someψ ∈ C∞0 (D,C3). Then ψ ∈ H0(curl, D) and therefore, because curlψ = εc curl ψ +∇εc × ψ,∫D

[curlH · curl ψ +

1

εccurlH · (∇εc × ψ)− ω2µεcH · ψ

]dx = 0 for all ψ ∈ C∞0 (D,C3) ,

which we write as∫D

curlH · curl ψ dx =

∫D

G · ψ dx for all ψ ∈ C∞0 (D,C3) ,

4.3. THE TIME–DEPENDENT CAVITY PROBLEM 183

where G = − 1εc

curlH ×∇εc + ω2µεcH ∈ L2(D,C3). Partial integration yields∫D

G · ψ dx =

∫D

H · curl2 ψ dx = −∫D

H ·∆ψ dx+

∫D

H · ∇ div ψ dx = −∫D

H ·∆ψ dx .

Here we used the fact that∫DH · ∇ div ψ dx = 0 because div ψ ∈ H1

0 (D). This holds for

all ψ ∈ C∞0 (D,C3). By the interior regularity result of Theorem 4.38 we conclude thatH ∈ H2(U,C3) for all domains U with U ⊆ D, and ∆H = −G = 1

εccurlH ×∇εc− ω2µεcH

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 4.39 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 for µ ∈ C2(D). Instead of divH = 0 wehave that 0 = div(µH) = ∇µ ·H + µ divH, thus

curl2H = −∆H +∇ divH = −∆H −∇(

1

µ∇µ ·H

)= −∆H −∇

[∇(lnµ) ·H

]= −∆H −

3∑j=1

(∇∂ lnµ

∂xjHj +

∂ lnµ

∂xj∇Hj

)and this can be treated in the same way. Here we argue classically, but all of thearguments hold also in the weak case.

(b) The asumption εc ∈ C1(D) is very restrictive. One can weaken this assumption tothe requirement that εc is piecewise continuously differentiable. We refer to [17], Sec-tion 4.6.

4.3 The Time–Dependent Cavity Problem

The spectral theorem of the previous section allow it to treat the full time–dependent systemof Maxwell’s equations. We begin again with the initial–boundary value problem for thescalar wave 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 , (4.47a)


u(t, x) = 0 for (t, x) ∈ (0, T )× ∂D , (4.47b)

u(0, x) = u0(x) and∂u

∂t(0, x) = u1(x) for x ∈ D . (4.47c)

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 (4.47a) 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 Green’s 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 notation ut = ∂u/∂t and the inner product in L2(D),

T∫0

[(b ut(t, ·), ψt(t, ·)

)L2(D)

−(a∇u(t, ·),∇ψ(t, ·)

)L2(D)

]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 toso called anisotropic function spaces. In particular, the solution has to be differentiable withrespect to t (in a sense to be explained in a moment). It is convenient to consider u to be afunction in t with values u(t) in some function space with respect to x. We have taken thispoint of view already when we wrote u(t, ·). To make this idea precise, we recall the notionof a Frechet-differentiable function for this case.

Definition 4.42 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 limt→t0‖f(t)− f(t0)‖V = 0. The space

of continuous functions on [0, T ] is denoted by C[0, T ;V ].

(b) The function f is differentiable in t0 ∈ [0, T ] with value f ′(t0) ∈ V if

limt→t0

∥∥∥f(t)−f(t0)t−t0 − f ′(t0)

∥∥∥V

= 0. The space of continuously differentiable functions is

denoted by C1[0, T ;V ].


Remark: If V is a Hilbert space with inner product (·, ·)V then the following productrule hold. For f, g ∈ C1[0, T ;V ] the scalar function h(t) =

(f(t), g(t)

)V

, t ∈ [0, T ], isdifferentiable and

h′(t) =(f ′(t), g(t)

)V

+(f(t), g′(t)

)V, t ∈ [0, T ] .

The proof uses the same arguments as in the case where V = Rn.

We define the solution space X of the initial–boundary value problem to be

X = C[0, T ;H1

0 (D)]∩ C1

[0, T ;L2(D)

](4.48a)

and equip X with the weighted norm

‖u‖X = max0≤t≤T

‖√a∇u(t)‖L2(D) + max

0≤t≤T‖√b u′(t)‖L2(D) . (4.48b)

Lemma 4.43 X is a Banach space. The norm is equivalent to

u 7→ max0≤t≤T

‖u(t)‖H1(D) + max0≤t≤T

‖u′(t)‖L2(D) .

Proof: The equivalence is clear because ‖√a∇u‖L2(D) is equivalent to the ordinary norm

in H10 (D) by Lemma 4.26 and ‖

√b · ‖L2(D) is equivalent to the ordinary norm in L2(D) by

the boundedness of b and 1/b. All properties of a normed space are very easy to see. Onlythe proof of completeness is a little more delicate, but follows the same arguments as in theproofs of the completeness of C[0, T ] and C1[0, T ]. 2

Definition 4.44 u is a (weak) solution of the initial–boundary value problem (4.47a)–(4.47c)if u ∈ X such that u(0) = u0, u′(0) = u1 and

T∫0

[(b u′(t), ψ′(t)

)L2(D)

−(a∇u(t),∇ψ(t)

)L2(D)

]dt = −

T∫0

(f(t, ·), ψ(t)

)L2(D)

dt (4.49)

for all ψ ∈ X with ψ(0) = ψ(T ) = 0.

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 (4.49) is well defined.

The following analysis makes use of the spectral theorem (Theorem 4.28). We recall fromthis theorem that there exist eigenvalues kn ∈ R and corresponding eigenfunctions vn ∈H1

0 (D) : n ∈ N such that∫D

[a∇vn · ∇ϕ− k2

n b vn ϕ]dx = 0 for all ϕ ∈ H1

0 (D) ;

that is, using again the notion of the inner product (u, v)∗ =∫Da∇u · ∇v dx in H1

0 (D),

(vn, ϕ)∗ = k2n(b vn, ϕ)L2(D) for all ϕ ∈ H1

0 (D) . (4.50)


Furthermore, the set vn ∈ H10 (D) : n ∈ N forms a complete orthonormal system in(

H10 (D), (·, ·)∗

), and knvn : n ∈ N forms a complete orthonormal system in(

L2(D), (b·, ·)L2(D)

), see Theorem 4.28.

First we prove uniqueness.

Theorem 4.45 There exists at most one solution u ∈ X of (4.49).

Proof: Let u be the difference of two solutions. Then u solves the problem for u0 = 0,u1 = 0, and f = 0. We define cn(t) by cn(t) =

(u(t), vn

)∗ for n ∈ N. Then cn ∈ C1[0, T ]. We

choose any ϕ ∈ C1[0, T ] such that ϕ(0) = ϕ(T ) = 0 and set ψ(t) = ϕ(t)vm for an arbitrarym ∈ N. We compute the inner products in (4.49) by using the orthogonality of vn withrespect to (·, ·)∗ and of knvn with respect to (b ·, ·)L2(D). This yields

(b u′(t), ψ′(t)

)L2(D)

= ϕ′(t)(b u′(t), vm

)L2(D)

=1

k2m

c′m(t)ϕ′(t) ,(a∇u(t),∇ψ(t)

)L2(D)

=(u(t), ψ(t)

)∗ = ϕ(t)

(u(t), vm

)∗ = cm(t)ϕ(t) .

Inserting this into (4.49) yields∫ T

0

[1

k2m

c′m(t)ϕ′(t)− cm(t)ϕ(t)

]dt = 0

for all such ϕ and all m. Now we use Lemma 4.46 below which yields that cm ∈ C2[0, T ] andc′′m(t) + k2

ncm(t) = 0 on [0, T ]. Using the initial conditions cm(0) = 0 and c′m(0) = 0 yieldscm(t) = 0 for all t. This holds for all m ∈ N. The completeness of the system vn : n ∈ Nimplies that u(t) vanishes for all t. 2

It remains to prove the following lemma which is sometimes called the Fundamental Theoremof Calculus of Variations .

Lemma 4.46 Let h ∈ C1[0, T ] and g ∈ C[0, T ] such that∫ T

0

[h′(t)ϕ′(t)− g(t)ϕ(t)

]dt = 0 for all ϕ ∈ C1[0, T ] with ϕ(0) = ϕ(T ) = 0 .

Then h ∈ C2[0, T ] and h′′(t) + g(t) = 0 for all t ∈ [0, T ] .

Proof: Define g(t) =∫ t

0g(s)ds for t ∈ [0, T ]. Then g ∈ C1[0, T ]. We substitute g into the

variational equation and use partial integration. This yields∫ T

0

[h′(t) + g(t)︸︷︷︸

=: c(t)

]ϕ′(t) dt =

∫ T

0

c(t)ϕ′(t) dt = 0 for all ϕ ∈ C1[0, T ] with ϕ(0) = ϕ(T ) = 0 .


Let now ρ ∈ C[0, T ] be arbitrary. Define ϕ(t) =∫ t

0ρ(s)ds− t

T

∫ T0ρ(s)ds for t ∈ [0, T ]. Then

ϕ ∈ C1[0, T ] with ϕ(0) = ϕ(T ) = 0. Therefore,

0 =

∫ T

0

c(t)ϕ′(t) dt =

∫ T

0

c(t) ρ(t) dt− 1

T

∫ T

0

ρ(s) ds

∫ T

0

c(t) dt

=

∫ T

0

ρ(t)

[c(t)− 1

T

∫ T

0

c(s) ds

]dt.

Because this holds for all ρ ∈ C[0, T ] we conclude that c(t) = 1T

∫ T0c(s) ds for all t. Therefore,

c is constant and thus h′ = −g + c ∈ C1[0, T ] and h′′(t) + g(t) = 0 for all t ∈ [0, T ]. 2

We can draw a second conclusion from this lemma.

Corollary 4.47 Let f ∈ C[0, T ;L2(D)] and u ∈ X be a solution of (4.49). Then, for allψ ∈ H1

0 (D) the scalar function t 7→(b u(t), ψ

)L2(D)

is twice continuously differentiable and

d2

dt2(b u(t), ψ

)L2(D)

+(a∇u(t),∇ψ

)L2(D)

=(f(t, ·), ψ

)L2(D)

(4.51)

for all t ∈ [0, T ].

Proof: Let ψ ∈ H10 (D) and ϕ ∈ C1[0, T ] with ϕ(0) = ϕ(T ) = 0. We insert ϕψ into (4.49)

which yields

T∫0

[ϕ′(t)

d

dt

(b u(t), ψ

)L2(D)

− ϕ(t)(a∇u(t),∇ψ)

)L2(D)

]dt = −

T∫0

ϕ(t)(f(t, ·), ψ

)L2(D)

dt .

Now we apply Lemma 4.46 to h(t) =(b u(t), ψ

)L2(D)

and g(t) =(a∇u(t),∇ψ)

)L2(D)

−(f(t, ·), ψ

)L2(D)

which yields the assertion because h is differentiable and g continuous . 2

The following simple result will be useful for the proofs of existence.

Lemma 4.48 Let H be a Hilbert space with complete orthonormal system hn : n ∈ N.Let cn ∈ C[0, T ] and γn > 0 with

∣∣cn(t)∣∣ ≤ γn for all t ∈ [0, T ] and n ∈ N where γn > 0 with∑∞

n=1 γ2n <∞. Define formally

u(t) =∞∑n=1

cn(t)hn , t ∈ [0, T ] .

Then u ∈ C[0, T ;H] and ‖u‖2C[0,T ;H] ≤

∑∞n=1 γ

2n.

Proof: Let uN be the truncated series; that is,

uN(t) =N∑n=1

cn(t)hn , t ∈ [0, T ] .


Then uN ∈ C[0, T ;H]. We show that (uN) is a Cauchy sequence in C[0, T ;H]. Indeed, forN > M we have

‖uN(t)− uM(t)‖2H =

N∑n=M+1

cn(t)2 ≤N∑

n=M+1

γ2n .

Taking the maximum yields

‖uN − uM‖2C[0,T ;H] ≤

N∑n=M+1

γ2n

and this tends to zero as N,M tend to infinity. Therefore, (uN) is a Cauchy sequence inC[0, T ;H] and thus convergent. 2

We will apply this result first to the case where H = H10 (D) with inner product (·, ·)∗ and

complete orthonormal system vn : n ∈ N and then to H = L2(D) with inner product(b·, ·)L2(D) and complete orthonormal system knvn : n ∈ N. This yields the followingcorollary:

Corollary 4.49 Let cn ∈ C1[0, T ] and γn > 0 with∣∣cn(t)

∣∣ ≤ γn and∣∣c′n(t)

∣∣ ≤ knγn for allt ∈ [0, T ] and n ∈ N where γn > 0 with

∑∞n=1 γ

2n <∞. Define formaly

u(t) =∞∑n=1

cn(t) vn , t ∈ [0, T ] .

Then u ∈ X and ‖u‖2X ≤ 2

∑∞n=1 γ

2n.

Proof: Application of the previous lemma to cn in(H1

0 (D), (·, ·)∗)

with respect to theorthonormal system vn : n ∈ N yields that u ∈ C[0, T ;H1

0 (D)] and ‖u‖2C[0,T ;H1

0 (D)]≤∑∞

n=1 γ2n. Then we apply the lemma to u′ with coefficients c′n

knin(L2(D), (b·, ·)L2(D)

)with

respect to the orthonormal system knvn : n ∈ N. This proves that u′ ∈ C[0, T ;L2(D)] and

‖u′‖2C[0,T ;L2(D)] ≤

∑∞n=1 γ

2n because also |c′n|

kn≤ γn for all t and n. Adding the results yields

the assertion. 2

To prove existence we first consider the case of no source; that is, f = 0.

Theorem 4.50 For every u0 ∈ H10 (D) and u1 ∈ L2(D) there exists a unique solution u of

(4.49) for f = 0 such that u(0) = u0 and u′(0) = u1. The solution is given by

u(t) =∞∑n=1

[αn cos(knt) + βn sin(knt)

]vn , t ∈ [0, T ] ,

where αn and βn are the expansion coefficients of u0 ∈ H10 (D) and u1 ∈ L2(D) with respect

to vn : n ∈ N and knvn : n ∈ N, respectively; that is,

u0 =∞∑n=1

αnvn , u1 =∞∑n=1

βn knvn .

Furthermore, the solution operator (u0, u1)→ u is bounded from H10 (D)× L2(D) into X.


Proof: To show that u ∈ X we apply the previous corollary with cn(t) = αn cos(knt) +βn sin(knt). The assumptions are obviously satisfied because cn(t)2 ≤ γ2

n := 2α2n + 2β2

n forall t ∈ [0, T ] and n ∈ N and, analogously, |c′n(t)| ≤ knγn for all t ∈ [0, T ] and n ∈ Nand

∑∞n=1

[α2n + β2

n

]= ‖u0‖2

∗ + ‖u1‖2L2(D,b dx). Application of Corollary 4.49 yields u ∈ X.

Furthermore, u(0) =∑∞

n=1 αnvn = u0 and u′(0) =∑∞

n=1 knβnvn = u1. It remains to provethat u satisfies (4.49) for f = 0. Let ψ ∈ X and expand ψ(t) as ψ(t) =

∑∞n=1 ψn(t) vn. Using

the orthogonality of vn : n ∈ N with respect to (·, ·)∗ and of knvn : n ∈ N with respectto (b·, ·)L2(D) we conclude

T∫0

[(bu′(t), ψ′(t)

)L2(D)

−(u(t), ψ(t)

)∗

]dt =

∞∑n=1

T∫0

[1

k2n

c′n(t)ψ′n(t)− cn(t)ψn(t)

]dt ,

where again cn(t) = αn cos(knt)+βn sin(knt) are the expansion coefficients of u(t). Let n ∈ Nbe fixed. Using partial integration we conclude

T∫0

[1

k2n

c′n(t)ψ′n(t)− cn(t)ψn(t)

]dt = −

T∫0

[1

k2n

c′′n(t) + cn(t)

]ψn(t) dt = 0

by the special form of cn(t). Therefore, u satisfies (4.49) for f = 0. 2

Let now f ∈ L2((0, T ) × D

)be arbitrary. We construct a particular solution of the inho-

mogeneous variational equation (4.49). We note that 1bf ∈ L2

((0, T ) × D

)and, therefore,

t 7→ 1bf(t, ·) can be expanded with respect to the orthonormal system knvn : n ∈ N in(

L2(D), (b·, ·)L2(D)

); that is,

1

b(·)f(t, ·) =

∞∑n=1

fn(t) knvn with fn(t) =(f(t, ·), knvn

)L2(D)

, n ∈ N . (4.52)

Theorem 4.51 Let f ∈ L2((0, T ) × D

)with coefficients fn(t) of (4.52). Then u(t) =∑∞

n=1 an(t) vn, t ∈ [0, T ], with coefficients

an(t) = −1

2

∫ T

0

sin(kn|t− s|

)fn(s) ds , t ∈ [0, T ] , n ∈ N , (4.53)

is a particular solution of (4.49). Furthermore, the operator f → u is bounded fromL2((0, T )×D

)into X.

Proof: First we note that u ∈ X. Indeed, we observe that an ∈ C1[0, T ] and |an(t)|2 ≤14

[∫ T0|fn(s)| ds

]2 ≤ T4

∫ T0|fn(s)|2 ds for all t ∈ [0, T ] and n ∈ N and analogously |a′n(t)|2 ≤

k2nT4

∫ T0|fn(s)|2 ds for all t ∈ [0, T ] and n ∈ N. Therefore, u ∈ X by Corollary 4.49 and

‖u‖2X ≤ c

∑n

∫ T

0

|fn(s)|2 ds = c

∫ T

0

∑n

|fn(s)|2 ds

= c

∫ T

0

‖f(s, ·)‖2L2(D) ds = ‖f‖2

L2((0,T )×D) .


Let now ψ ∈ X with ψ(0) = ψ(T ) = 0 and expansion ψ(t) =∑∞

n=1 ψn(t)vn. We fix nand assume first that fn is continuous in [0, T ]. Then it is easy to check that even an ∈C2[0, T ] and an satisfies the differential equation a′′n(t) + k2

nan(t) = −knfn(t). Multiplyingthis equation with ψn ∈ C1[0, T ], integrating and using integration by parts (note thatψn(0) = ψn(T ) = 0), yields∫ T

0

[a′n(t)ψ′n(t)− k2

nan(t)ψn(t)]dt = kn

∫ T

0

fn(t)ψn(t) dt .

By a density argument we conclude that this equation holds also if only fn ∈ L2(0, T ).Division by k2

n and summing these equations with respect to n yields

∞∑n=1

∫ T

0

[1

k2n

a′n(t)ψ′n(t)− an(t)ψn(t)

]dt =

∞∑n=1

1

kn

∫ T

0

fn(t)ψn(t) dt ,

which can again be written as∫ T

0

[(bu′(t), ψ′(t)

)L2(D)

−(u(t), ψ(t)

)∗

]dt =

∫ T

0

(f(t), ψ(t)

)L2(D)

dt .

2

As a corollary we have existence and uniqueness for the general inhomogeneous problem.

Corollary 4.52 For every u0 ∈ H10 (D) and u1 ∈ L2(D) and f ∈ L2

((0, T )×D

)there exists

a unique solution u ∈ X of (4.49) such that u(0) = u0 and u′(0) = u1. The solution isgiven by the sum u = u+ u of the particular solution u constructed in Theorem 4.51 and thesolution u of the homogeneous differential equation with initial values u(0) = u0 − u(0) andu′(0) = u1 − u′(0); that is,

u(t) =∞∑n=1

[(αn − an(0)

)cos(knt) +

(βn −

a′n(0)

kn

)sin(knt) + an(t)

]vn (4.54)

for t ∈ [0, T ] where αn and βn are the expansion coefficients of u0 ∈ H10 (D) and u1 ∈ L2(D)

with respect to vn : n ∈ N and knvn : n ∈ N, respectively, and an(t) are defined in (4.53).Furthermore, the solution u ∈ X depends continuously on u0, u1, and f ; that is, the solutionoperator (u0, u1, f) 7→ u is bounded from H1

0 (D)× L2(D)× L2((0, T )×D

)into X.

By the same arguments we can prove the following regularity result.

Theorem 4.53 Let f ∈ C1[0, T ;L2(D)] and u0 = 0 and u1 ∈ H10 (D) and u ∈ X be the

solution of (4.49). Then u′ ∈ C(0, T ;H1

0 (D))

and u′′ ∈ C(0, T ;L2(D)

); that is, u′ ∈ X.

Furthermore, the mapping (u1, f) 7→ u is bounded from H10 (D) × L2

((0, T ) ×D

)into X =

C1(0, T ;H1

0 (D))∩C2

(0, T ;L2(D)

)equipped with the canonical norm analogously to (4.48b).

The solution satisfies the differential equation pointwise with respect to t; that is(b u′′(t), ψ

)L2(D)

+(a∇u(t),∇ψ

)L2(D)

=(f(t, ·), ψ

)L2(D)

(4.55)

for all ψ ∈ H10 (D) and all t ∈ [0, T ], compare (4.51).


Proof: First we consider the particular solution u of Theorem 4.51 and write (4.53) as

an(t) = −1

2

[∫ t

0

sin(kn(t− s)) fn(s) ds +

∫ T

t

sin(kn(s− t)) fn(s) ds

]

= − 1

2kn

[cos(kn(t− s)) fn(s)

∣∣∣∣t0

−∫ t

0

cos(kn(t− s)) f ′n(s) ds

− cos(kn(s− t)) fn(s)

∣∣∣∣Tt

+

∫ T

t

cos(kn(s− t)) f ′n(s) ds

]and

a′n(t) = −kn2

[∫ t

0

cos(kn(t− s)) fn(s) ds −∫ T

t

cos(kn(s− t)) fn(s) ds

]

= −1

2

[− sin(kn(t− s)) fn(s)

∣∣∣∣t0

+

∫ t

0

sin(kn(t− s)) f ′n(s) ds

− sin(kn(s− t)) fn(s)

∣∣∣∣Tt

+

∫ T

t

sin(kn(s− t)) f ′n(s) ds

]and thus

|an(t)| ≤ c1

kn

(‖fn‖∞ + ‖f ′n‖L2(0,T )

)≤ c2

kn

(‖fn‖L2(0,T ) + ‖f ′n‖L2(0,T )

),

|a′n(t)| ≤ c2

(‖fn‖L2(0,T ) + ‖f ′n‖L2(0,T )

),

|a′′n(t)| = |k2nan(t) + knfn(t)| ≤ c3 kn

(‖fn‖L2(0,T ) + ‖f ′n‖L2(0,T )

)for all t ∈ [0, T ]. Here we used the estimate max0≤t≤T |ϕ(t)|2 ≤ 2 maxT, 1/T

(‖ϕ‖2

L2(0,T ) +

‖ϕ′‖2L2(0,T )

)for any function ϕ ∈ C1[0, T ] (see Exercise 4.10). Differentiating (4.54) for the

case αn = 0 yields

u′(t) =∞∑n=1

[knan(0) sin(knt) +

(knβn − a′n(0)

)cos(knt) + a′n(t)

]vn ,

u′′(t) =∞∑n=1

[k2nan(0) cos(knt) +

(kna

′n(0)− k2

nβn)

sin(knt) + a′′n(t)]vn .

Now we observe from the above estimates of an(t), a′n(t), and a′′n(t) and the fact that∑∞n=1 k

2nβ

2n = ‖u1‖2

H10 (D)

that the coefficients of this series satisfy the assumptions of Corol-

lary 4.49. This shows that u′ ∈ X.Equation (4.51) follows from the observation that d2

dt2

(b u(t), ψ

)L2(D)

=(b u′′(t), ψ

)L2(D)

be-

cause u ∈ C2[0, T ;L2(D)

]. 2

We finish this part with an explicit example which shows that one can not weaken theassumptions on f for deriving the C1−regularity of u.


Example 4.54 Let D = (0, π)3 ⊆ R3 and a = c = 1 and T > 2. The Dirichlet eigenvaluesof −∆ in D are given by kn = |n|, n ∈ N3, with corresponding eigenfunctions

vn(x) =1

(π/2)3/2 |n|

3∏j=1

sin(njxj) , x ∈ D , n ∈ N3 .

They are normalized such that ‖vn‖∗ = ‖∇vn‖L2(D) = 1 and ‖vn‖L2(D) = 1/|n|. We set

ρn =√|n|2 − |n| for n ∈ N3 and define f by

f(t, x) =∑n∈N3

fn sin(ρnt) |n| vn(x) , t ∈ [0, T ] , x ∈ D ,

where the coefficients fn are such that∑

n f2n <∞. Then

‖f(t, ·)‖2L2(D) =

∑n∈N3

f 2n sin2(ρnt) , t ∈ [0, T ] ,

because |n|vn : n ∈ N3 is an orthonormal system in L2(D). By ‖f(t, ·)‖2L2(D) ≤

∑n∈N3 f 2

n

we observe that f ∈ C[0, T ;L2(D)]. The solution of the initial boundary value problem withu0 = u1 = 0 is given by

u(t, x) =∑n∈N3

fn

[sin(ρnt)−

ρn|n|

sin(|n|t)]vn(x) , t ∈ [0, T ] , x ∈ D ,

as one checks directly by term-by-term differentiation, which can be made rigorously byinvestigating the convergence. The normalization of vn yields

‖u(t, ·)‖2∗ =

∑n∈N3

f 2n

[sin(ρnt)−

ρn|n|

sin(|n|t)]2

, t ∈ [0, T ] .

If u ∈ C1[0, T ;H10 (D)] then

u′(t) =∑n∈N3

fn ρn[cos(ρnt)− cos(|n|t)

]vn and ‖u′(t)‖2

∗ =∑n∈N3

f 2nρ

2n

[cos(ρnt)− cos(|n|t)

]2for t ∈ [0, T ]. Integration yields∫ T

0

‖u′(t)‖2∗ dt =

∑n∈N3

f 2nρ

2n

∫ T

0

[cos2(ρnt) + cos2(|n|t)− 2 cos(ρnt) cos(|n|t)

]dt

=∑n∈N3

f 2nρ

2n

[T +

sin(2ρnT )

4ρn+

sin(2|n|T )

4|n|−

sin((ρn + |n|)T

)ρn + |n|

−sin((|n| − ρn)T

)|n| − ρn

].


Now we observe that |n| − ρn = 12

+ O(1/|n|). Therefore we can choose ε > 0 such that2 + ε < T and then N so large such that

1

4ρn+

1

4|n|+

1

ρn + |n|+

1

|n| − ρn≤ 2 + ε for |n| ≥ N .

This yields ∫ T

0

‖u′(t)‖2∗ dt ≥ (T − 2− ε)

∑|n|≥N

f 2nρ

2n

which implies that f ∈ C1[0, T ;L2(D)].

After the scalar wave equation we consider now the time–dependent Maxwell system; thatis,

curlE(t, x) + µ(x)∂H

∂t(t, x) = 0 , (t, x) ∈ (0, T )×D , (4.56a)

curlH(t, x) − ε(x)∂E

∂t(t, x) = Je(t, x) , (t, x) ∈ (0, T )×D , (4.56b)

with boundary conditions

ν(x)× E(t, x) = 0 for (t, x) ∈ (0, T )× ∂D , (4.56c)

and initial conditions

E(0, x) = e0(x) and H(0, x) = h0(x) for x ∈ D . (4.56d)

Again we need some conditions on the electrical parameter. Therefore, we make the followingassumptions on the data:

• ε, µ ∈ L∞(D) such that ε(x) ≥ c > 0 and µ(x) ≥ c > 0 on D for some c > 0,

• Je ∈ L2((0, T )×D

),

• e0 ∈ H0(curl, D) and h0 ∈ H(curl, D).

We define the solution space X for the pair (E,H) by X = Xe ×Xh where

Xe = C[0, T ;H0(curl, D)

]∩ C1

[0, T ;L2(D,C3)

], (4.57a)

Xh = C[0, T ;H(curl, D)

]∩ C1

[0, T ;L2(D,C3)

], (4.57b)

and equip X with the product norm ‖(E,H)‖X = ‖E‖+ ‖H‖ with

‖u‖ = max0≤t≤T

‖u(t)‖H(curl,D) + max0≤t≤T

‖u′(t)‖L2(D) (4.57c)

where u = E or u = H.


Definition 4.55 (E,H) is a solution of the initial-boundary value problem (4.56a)–(4.56d)if (E,H) ∈ X such that E(0) = e0, H(0) = h0 and

curlE(t) + µH ′(t) = 0 for all t ∈ [0, T ] , (4.58a)

curlH(t) − εE ′(t) = Je(t, ·) for all t ∈ [0, T ] . (4.58b)

We observe that the solution space is not symmetric with respect to E and H because of theboundary condition for E. The following analysis makes use of the Helmholtz decompositionsof Theorem 4.21; that is,

L2(D,C3) = H0(curl 0, D) ⊕ V0,ε ,

H0(curl, D) = H0(curl 0, D) ⊕ V0,ε ,

H(curl, D) = H(curl 0, D) ⊕ Vµ ,

where the spaces have been defined in (4.10a)–(4.10e) when we take A = εI or A = µI,respectively; that is,

Vµ =u ∈ H(curl, D) : (µu, ψ)L2(D) = 0 for all ψ ∈ H(curl, D) with curlψ = 0

,

V0,ε =u ∈ H0(curl, D) : (εu, ψ)L2(D) = 0 for all ψ ∈ H0(curl, D) with curlψ = 0

,

V0,ε =u ∈ L2(D,C3) : (εu, ψ)L2(D) = 0 for all ψ ∈ H0(curl, D) with curlψ = 0

.

We recall (see Remark 4.22) that V0,ε and Vµ are just the orthogonal complements ofH0(curl 0, D) and H(curl 0, D), respectively, with respect to the inner products

(u, v)µ,ε = (µ−1 curlu, curl v)L2(D) + (ε u, v)L2(D) ,

(u, v)ε,µ = (ε−1 curlu, curl v)L2(D) + (µu, v)L2(D) ,

respectively, see (4.13a), (4.13b). We note that both norms are equivalent to the ordinarynorm in H(curl, D). Analogously, V0,ε is the orthogonal complement of H0(curl 0, D) inL2(D,R3) with respect to the inner product (ε u, v)L2(D).

We recall the spectral theorem (Theorem 4.34) for the Maxwell system. We proved theexistence of an infinite number of eigenvalues ωn > 0 and corresponding eigenfunctionsvn ∈ V0,ε such that(

µ−1 curl vn, curlψ)L2(D)

= ω2n

(ε vn, ψ

)L2(D)

for all ψ ∈ H0(curl, D) . (4.59)

The functions vn are normalized such that ‖vn‖2µ,ε = ‖µ−1/2 curl vn‖2

L2(D) + ‖ε1/2vn‖2L2(D) = 1

for all n ∈ N. Then we saw that the system vn : n ∈ N is a complete orthonormal systemin V0,ε with respect to the inner product (·, ·)µ,ε and

√1 + ω2

n vn : n ∈ N

is a complete

orthonormal system in V0,ε with respect to (εv, w)L2(D). Furthermore, we saw in Lemma 4.35that wn = 1

ωn µcurl vn, n ∈ N, form a complete orthonormal system in Vµ with respect to

the inner product (·, ·)ε,µ.

Now we turn to the investigation of the Maxwell system (4.56a)–(4.56d). First we proveuniqueness.


Theorem 4.56 There exists at most one solution of (4.58a), (4.58b) with E(0) = e0 andH(0) = h0.

Proof: Let (E,H) be the difference of two solutions. Then (E,H) solves the system forJe = 0, e0 = 0, and h0 = 0. Let first φ ∈ H(curl 0, D). Then, from (4.58a),

0 =(curlE(t), φ

)L2(D)

+d

dt

(µH(t), φ

)L2(D)

=d

dt

(µH(t), φ

)L2(D)

for all t because(curlE(t), φ

)L2(D)

=(E(t), curlφ

)L2(D)

= 0. Therefore,(µH(t), φ

)L2(D)

is constant and thus zero for all t because of the initial condition H(0) = 0. Therefore,H(t) ∈ Vµ for all t ∈ [0, T ]. By the same arguments for φ ∈ H0(curl 0, D) and equation(4.58b) one shows that E(t) ∈ V0,ε for all t ∈ [0, T ]. Now we multiply equation (4.58a) bywn for some n and have

0 =(curlE(t), wn

)L2(D)

+d

dt

(µH(t), wn

)L2(D)

=(E(t), curlwn

)L2(D)

+d

dt

(µH(t), wn

)L2(D)

= ωn(εE(t), vn

)L2(D)

+d

dt

(µH(t), wn

)L2(D)

and analogously

0 =(curlH(t), vn

)L2(D)

− d

dt

(εE(t), vn

)L2(D)

=(H(t), curl vn

)L2(D)

− d

dt

(εE(t), vn

)L2(D)

= ωn(µH(t), wn

)L2(D)

− d

dt

(εE(t), vn

)L2(D)

.

Therefore,(εE(t), vn

)L2(D)

and(µH(t), wn

)L2(D)

solve a homogeneous linear system of

ordinary differential equations of first order with homogeneous initial data. Therefore,(εE(t), vn

)L2(D)

and(µH(t), wn

)L2(D)

have to vanish for all t and n ∈ N. The completeness

of vn : n ∈ N in V0,ε and of wn : n ∈ N in Vµ implies that E(t) and H(t) vanish for allt ∈ [0, T ]. 2

Next we show existence for the special case Je = 0.

Theorem 4.57 For all e0 ∈ H0(curl, D) and h0 ∈ H(curl, D) there exists a unique solutionof (4.58a), (4.58b) for Je = 0.

Proof: We decompose the pair (e0, h0) into the sum (e0, h0) = (e00, h

00) + (e0, h0) with

(e00, h

00) ∈ H0(curl 0, D) × H(curl 0, D) and (e0, h0) ∈ V0,ε × Vµ. Then the solution (E,H)

is the sum of the solutions with initial data (e00, h

00) and (e0, h0), respectively. The solution


corresponding to initial data (e00, h

00) is just the constant (with respect to time t); that is,

E0(t) = e00 and H0(t) = h0

0 for all t ∈ [0, T ].We now construct the solution (E, H) corresponding to initial data (e0, h0) ∈ V0,ε × Vµ. Weexpand the initial data in the forms

e0 =∞∑n=1

αn vn , h0 =∞∑n=1

βnwn

where vn : n ∈ N in V0,ε and wn : n ∈ N in Vµ are the orthonormal systems studiedabove. Then, by Parseval’s identity,

∑∞n=1 α

2n = ‖e0‖2

µ,ε and∑∞

n=1 β2n = ‖h0‖2

ε,µ. We makean ansatz in the form

E(t) =∞∑n=1

an(t) vn , H(t) =∞∑n=1

bn(t)wn , t ∈ [0, T ] ,

where an, bn ∈ C1[0, T ]. Substitution of these series into equations (4.58a), (4.58b) for Je = 0yields formally

0 =∞∑n=1

[an(t) curl vn + b′n(t)µwn

]=

∞∑n=1

[ωn an(t) + b′n(t)

]µwn ,

0 =∞∑n=1

[bn(t) curlwn − a′n(t) ε vn

]=

∞∑n=1

[ωn bn(t)− a′n(t)

]ε vn .

The linear independence of the systems vn : n ∈ N in V0,ε and wn : n ∈ N in Vµ yieldsthe following system of ordinary differential equations

ωn an(t) + b′n(t) = 0 , ωn bn(t)− a′n(t) = 0 , t ∈ [0, T ] , (4.60)

with initial conditions an(0) = αn, bn(0) = βn. Solving it yields

E(t) =∞∑n=1

[αn cos(ωnt) + βn sin(ωnt)

]vn , t ∈ [0, T ] ,

H(t) =∞∑n=1

[−αn sin(ωnt) + βn cos(ωnt)

]wn , t ∈ [0, T ] .

These functions belong to Xe and Xh, respectively. Indeed, by Corollary 4.49 we have tocheck uniform estimate of c(t) = αn cos(ωnt) + βn sin(ωnt) and of c(t) = −αn sin(ωnt) +βn cos(ωnt), respectively, and of there derivatives. Indeed, for both cases we have that|c(t)|2 ≤ 2α2

n + 2β2n =: γ2

n and |c′(t)| ≤ ωnγn and∑∞

n=1 γ2n = ‖e0‖2

µ,ε + ‖h0‖2ε,µ. This proves

the theorem. 2

Now we have to determine particular solutions of the inhomogeneous differential equations;that is, for Je 6= 0. We make the following additional assumption on Je:

Assumption: Let Je be of the form Je(t) = ρ(t) ε F , t ∈ [0, T ], for some F ∈ L2(D,C3) andsome scalar function ρ ∈ C1[0, T ].


Again, we decompose F in the form

F = F 0 + F with F 0 ∈ H0(curl 0, D) and F ∈ V0,ε .

A particular solution is obtained as the sum of solutions with right–hand side F 0 and with F ,respectively. A solution (E0, H0) with right–hand side F 0 is simply E0(t) = −

∫ t0ρ(s) dsF 0

and H0(t) = 0.

Now we consider F ∈ V0,ε as right–hand side. Then F has an expansion in the form F =∑∞n=1 fn

√1 + ω2

nvn (convergence in L2(D)) with ‖√ε F‖2

L2(D) =∑∞

n=1 f2n. We make an

ansatz for E and H in the forms

E(t) =∞∑n=1

an(t) vn , H(t) =∞∑n=1

bn(t)wn , t ∈ [0, T ] ,

Using the relationship between vn and wn a substitution into the Maxwell system yields,using the system

ωn an(t) + b′n(t) = 0 , ωn bn(t)− a′n(t) =√

1 + ω2n ρ(t) fn . t ∈ [0, T ] (4.61)

(compare to (4.60)). We can either eliminate an or bn from the system. Because of thedifferentiability of ρ we eliminate bn and arrive at

a′′n(t) + ω2n an(t) =

√1 + ω2

n fn ρ′(t) , t ∈ [0, T ] .

A solution is given by

an(t) =

√1 + ω2

n

2ωnfn

∫ T

0

sin(ωn|t− s|) ρ′(s) ds , t ∈ [0, T ] .

Then∣∣an(t)

∣∣ ≤ c‖ρ′‖∞ |fn| and∣∣a′n(t)

∣∣ ≤ c ωn‖ρ′‖∞ |fn| for all t ∈ [0, T ] and n ∈ N where theconstant c is independent of n or t. Solving for bn yields the estimates

∣∣bn(t)∣∣ ≤ c (‖ρ‖∞ +

‖ρ′‖∞) |fn| and∣∣b′n(t)

∣∣ ≤ c ωn‖ρ′‖∞ |fn| for all t ∈ [0, T ] and n ∈ N. Corollary 4.49 yields

(E, H) ∈ X = Xe ×Xh and ends the proof. 2

Collecting our results we can formulate the following theorem for the time dependent cavityproblem.

Theorem 4.58 Let e0 ∈ H0(curl, D) and h0 ∈ H(curl, D) and Je of the form Je(t) =ρ(t) ε F , t ∈ [0, T ], for some F ∈ L2(D,C3) and some scalar function ρ ∈ C1[0, T ]. Thenthere exists a unique solution (E,H) ∈ X = Xe ×Xh of the system (4.56a), (4.56b).

As a final remark we note that we can replace the assumption (ρ, F ) ∈ C1[0, T ]×L2(D,C3)in the theorem by the assumption (ρ, F ) ∈ C[0, T ] × H0(curl, D). Then F ∈ V0,ε, and wesolve the system (4.61) for bn and proceed as before.


4.4 Exercises

Exercise 4.1 Let D ⊆ R3 be an open set and u ∈ C(D). Show that the support of u asdefined at the beginning of Subsection 4.1.1 can be expressed as the closure of the set S ⊆ D,given by S = x ∈ D : u(x) 6= 0.

Exercise 4.2 Let U ⊂ R3 be open and bounded und u the characteristic function of U ; thatis, u(x) = 1 for x ∈ U and u(x) = 0 outside of U . Let φδ ∈ C∞0 (R3) be as in Theorem 4.7.Show that uk = u ∗ φk converges to u in L2(R3).Hint: Use the theorem of dominated convergence!

Exercise 4.3 Show that C∞0 (R3) is dense in H1(R3). Therefore, H10 (R3) coincides with

H1(R3).Hint: Use the technique of Lemma 4.18.

Exercise 4.4 Prove that H10 (R3) is not compactly imbedded in L2(R3).

Hint: Take a non-vanishing ϕ ∈ C∞0 (R) and some a ∈ R3, a 6= 0, and discuss the sequenceun(x) = ϕ(x+ na), x ∈ R3, n ∈ N.

Exercise 4.5 Show that the vector field F in Definition 4.1 is unique if it exists.Hint: Use Exercise 4.3!

Exercise 4.6 Let D ⊆ R3 be symmetric with respect to x3 = 0; that is, x ∈ D ⇔ x∗ ∈ Dwhere a∗ = (a1, a2,−a3)> for any a = (a1, a2, a3)> ∈ C3. Let D± = x ∈ D : x3 ≷ 0 andϕ ∈ C1(D−) and v ∈ C1(D−,C3). Extend ϕ and v into D by reflection; that is,

ϕ(x) =

ϕ(x) , x ∈ D−,ϕ(x∗) , x ∈ D+,

v(x) =

v(x) , x ∈ D−,v∗(x∗) , x ∈ D+.

Show that ϕ ∈ H1(D) and v ∈ H(curl, D).

Hint: Show that

∇ϕ(x) =

∇ϕ(x) , x ∈ D−,

(∇ϕ)∗(x∗) , x ∈ D+,curl v(x) =

curl v(x) , x ∈ D−,

−(curl v)∗(x∗) , x ∈ D+,

are the gradient and curl, respectively.

Exercise 4.7 Construct a function φ ∈ C∞(R) such that φ(t) = 0 for |t| ≥ 1 and φ(t) = 1for |t| ≤ 1/2.Hint: Define ϕ by ϕ(t) = exp(−1/t) for t > 0 and ϕ(t) = 0 for t ≤ 0. Discuss ϕ and chooseφ as a proper rational combination of ϕ(1− t2) and ϕ(t2 − 1/4).

Exercise 4.8 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).

4.4. EXERCISES 199

Exercise 4.9 Show that for any v ∈ H10 (D) the extension by zero belongs to H1(R3); that

is, v ∈ H1(R3) where v = v in D and v = 0 in R3 \D.

Exercise 4.10 Prove that |ϕ(t)|2 ≤ 2 maxT, 1/T(‖ϕ‖2

L2(0,T )+‖ϕ′‖2L2(0,T )

)for all t ∈ [0, T ]

and any function ϕ ∈ C1[0, T ].Hint: Use the Fundamental Theorem of Calculus of Variations.

Exercise 4.11 In Theorem 4.13 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 4.12 Show that C∞0 (R3,C3) is dense in H(curl,R3).Hint: Compare with Exercise 4.3.

Exercise 4.13 Let Q = (−R,R)3. Show that the following inclusions hold and are bounded:

H0(curl, Q) → Hper(curl, Q) → H(curl, Q) .

Hint: Follow the arguments of the proof of Lemma 4.37.

Exercise 4.14 Why is, even for bounded sets D, the space H0(curl, D) not compactlyimbedded in L2(D,C3)?Hint: Use the Helmholtz decomposition!

Exercise 4.15 Prove Theorem 4.23; that is, show the decompositions of L2(D,C3) andH0(curl, D) in the form

L2(D,C3) = L2(divA 0, D) ⊕ ∇H0(D) , H0(curl, D) = H0(curl, divA 0, D) ⊕ ∇H0(D) .

Hint: Use the arguments as in the proof of Theorem 4.21.

Exercise 4.16 Prove the following version of the Fundamental Theorem of Calculus ofVariations.Let u ∈ L1(0, T ) and v ∈ C[0, T ] such that∫ T

0

[ρ′(t)u(t) + ρ(t) v(t)

]dt = 0 for all ρ ∈ C1[0, T ] with ρ(T ) = 0 .

Then u ∈ C1[0, T ] and u(0) = 0 and u′ = v on [0, T ]. If the variational equation holds forall ρ ∈ C1[0, T ] then also u(T ) = 0.Hint: Use the methods of the proof of Lemma 4.46.

Exercise 4.17 Let D = B3(0, R) be a ball. Show explicitely the Helmholtz decompositionby using the results of Chapter 2.

Chapter 5

Boundary Integral Equation Methodsfor Lipschitz Domains

For the boundary value problems of Chapters 3 and 4 we made assumptions which areoften not met in applications. Indeed, the classical integral equation methods discussedin Chapter 3 require smoothness of the boundary ∂D. In case of the cavity problem ofChapters 4 just a homogeneous boundary condition has been treated. Both restrictions areconnected because if we like to weaken the regularity of the boundary, or if we like to allowfor more general boundary conditions we have to investigate the traces of the functions orvector fields on the boundary ∂D in detail. Therefore, we continue in Subsections 5.1.1 and5.1.2 by introducing Sobolev spaces which appear as the range spaces of the trace operatorsand prove denseness, trace theorems and compact imbedding results. Finally we use theseresults to extend the boundary integral equation methods for Lipschitz domains.

5.1 Advanced Properties of Sobolev Spaces

We recall Definition 6.7 for the notion of a Lipschitz domain. Let B2(0, α) ⊆ R2 be the twodimensional disk of radius α.

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 = Rjx + z(j) : x ∈ B2(0, αj) × (−2βj, 2βj)with z(j) ∈ R3 and rotations Rj ∈ R3×3 and real valued Lipschitz-continuous functions ξj ∈C(B2[0, αj]) with |ξj(x1, x2)| ≤ βj for all (x1, x2) ∈ B2[0, αj] such that ∂D ⊆

⋃mj=1 Uj and

∂D ∩ Uj =Rjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 = ξj(x1, x2)

,

D ∩ Uj =Rjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 < ξj(x1, x2)

,

Uj \D =Rjx+ 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 we denote

201

202 5 BIE FOR LIPSCHITZ DOMAINS

by

Cj = Cj(αj, βj) = B2(0, αj)× (−2βj, 2βj)

=x = (x1, x2, x3) ∈ R3 : x2

1 + x22 < α2

j , |x3| < 2βj

the cylinders with parameters αj and βj. We can assume without loss of generality thatβj ≥ αj (otherwise split the parameter region into smaller ones). Furthermore, we definethe three-dimensional balls Bj = B3(0, αj) ⊆ Cj and introduce the mappings

Ψj(x) = Rj

x1

x2

ξj(x1, x2) + x3

+ z(j) , x = (x1, x2, x3)> ∈ Bj ,

and their restrictions Ψj to B2(0, αj); that is,

Ψj(x) = Rj

x1

x2

ξj(x1, x2)

+ z(j) , x = (x1, x2)> ∈ B2(0, αj) .

By Rademacher’s result [21] (see also [9], Section 5.8, and Remark 6.8) we know that Ψj isdifferentiable almost everywhere on B2(0, αj). This yields a parametrization of ∂D ∩ Uj in

the form y = Ψj(x) for x ∈ B2(0, αj) with∣∣∣∂Ψj∂x1× ∂Ψj

∂x2

∣∣∣ =√

1 + |∇ξj|2 ≤√

1 + L2j where Lj

denotes the Lipschitz constant of ξj.

We set U ′j = Ψj(Bj). Then ∂D ⊆⋃mj=1 U

′j and Bj ∩ (R2 × 0) = B2(0, αj)× 0, and

∂D ∩ U ′j =

Ψj(x) : x ∈ Bj, x3 = 0

=

Ψj(x) : x ∈ B2(0, αj),

D ∩ U ′j =

Ψj(x) : x ∈ Bj, x3 < 0,

U ′j \D =

Ψj(x) : x ∈ Bj, x3 > 0.

Therefore, the mappings Ψj “flatten” the boundary.

We note that the Jacobian Ψ′j(x) ∈ R3×3 is given by

Ψ′j(x) = Rj

1 0 00 1 0

∂1ξj(x) ∂2ξj(x) 1

where ∂`ξj = ∂ξj/∂x` for ` = 1, 2 and x = (x1, x2). Therefore, these Jacobians are regularwith constant determinant det Ψ′j(x) = 1 and Ψj are isomorphisms from Bj onto U ′j for everyj = 1, . . . ,m.

These parametrizations allow it to transfer the notion of (periodic) Sobolev spaces on twodimensional planar domains to the boundary ∂D. We begin with Sobolev spaces of scalarfunctions in Subsection 5.1.1 and continue with vector-valued functions in Subsection 5.1.2.

5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES 203

5.1.1 Sobolev Spaces of Scalar Functions

We note first that Green’s formula and the transformation formula hold in the form ofTheorem 6.12 and Remark 6.8, respectively, for Lipschitz domains and sufficiently smoothfunctions. This theorem is, e.g. used in the following simple result.

Lemma 5.2 Let D be a bounded Lipschitz domain.

(a) Let u ∈ C1(D) with u = 0 on ∂D. Then the extension u of u by zero outside of Dyields u ∈ H1(R3) and ∇u = ∇u in D and ∇u = 0 outside of D.

(b) Let R ∈ R3×3 be an orthogonal matrix and z ∈ R3. For u ∈ H1(D) define v(x) =u(Rx+z) on V = x ∈ R3 : Rx+z ∈ D. Then v ∈ H1(V ) and ∇v(x) = R>∇u(Rx+z).

Proof: (a) Set g = ∇u in D and zero outside of D. Then g ∈ L2(R3,C3), and for ψ ∈C∞0 (R3) we have by 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 .

(b) Let ψ ∈ C∞0 (V ). With the substitution y = Rx+ z we have

R>∫V

∇u(y)|y=Rx+z ψ(x) dx

= R>∫D

∇u(y)ψ(R>(y − z)

)dy = −R>

∫D

u(y)R∇ψ(x)|x=R>(y−z) dy

= −∫D

u(y)∇ψ(x)|x=R>(y−z) dy = −∫V

v(x)∇ψ(x) dx .

This shows that ∇v(x) = R>∇u(y)|y=Rx+z. 2

An important property of Sobolev spaces on Lipschitz domains D is the denseness of thespace C∞(D). Its proof is technically complicated.

Theorem 5.3 Let D be a bounded Lipschitz domain. Then C∞(D) is dense in H1(D).

Proof: The idea of the proof is to transform the given function u ∈ H1(D) onto a cylinderby using a partition of unity and local coordinates. In a cylinder we construct a smoothapproximation of the transformed function. Finally, we transform back the approximation.We seperate the proof into four steps.(i) Let u ∈ H1(D) and Uj, ξj : j = 1, . . . ,m be a local coordinate system. Set U0 = D


and choose a partition of unity ϕj : j = 0, . . . ,m on D subordinate to Uj : j = 0, . . . ,m(see Theorem 6.9). Set uj(y) = ϕj(y)u(y) for j = 0, . . . ,m and y ∈ D. For j = 1, . . . ,mwe transform uj by the definition vj(x) = uj(Rjx + z(j)), x ∈ C−j , where Cj = B2(0, αj) ×(−2βj, 2βj) and

C−j :=x = (x, x3) ∈ Cj : x3 < ξj(x)

where again x = (x1, x2). We observe that vj ∈ H1(C−j ) by the chain rule of part (b) ofLemma 5.2. Furthermore, with

Vj(t) := x ∈ C−j : |x| > αj − t or x3 < −2βj + t

we note that there exists δ > 0 with vj(x) = 0 in the neighborhood Vj(2δ) of ∂Cj ∩ C−jbecause the support of ϕj is contained in Uj. The reader should sketch the sets Cj, C

−j , and

Vj(t).

(ii) Let a = (0, 0, L+1)> where L is larger than all of the Lipschitz constants of the functionsξj and set

vεj (x) = (vj ∗ φε)(x− εa) =

∫C−j \Vj(2δ)

vj(y)φε(x− εa− y) dy

where φε ∈ C∞(R3) denotes the mollifier function from (4.7). Its most important propertieshad been collected in Theorem 4.7. In particular, we have that vεj ∈ C∞0 (R3) and vεj → vj inL2(R3) as ε tends to zero, the latter property because the zero-extension of vj is in L2(R3).

Next we show that vεj vanishes on Vj(δ) for ε ≤ δ. First we note that the integral in thedefinition of vεj is taken over C−j \Vj(2δ). Therefore, for y ∈ C−j \Vj(2δ) we have |y| ≤ αj−2δand y3 ≥ −2βj + 2δ.Let first x ∈ C−j with |x| ≥ αj−δ. Then |x−εa−y| ≥ |x−y| ≥ |x|−|y| ≥ αj−δ−(αj−2δ) =δ ≥ ε. Therefore, vεj (x) = 0 for these x.Let second x ∈ C−j with x3 ≤ −2βj + δ. Then |x − εa − y| ≥ |x3 − ε(L + 1) − y3| ≥y3 − x3 + (L+ 1)ε ≥ −2βj + 2δ + 2βj − δ + (L+ 1)ε ≥ δ ≥ ε. Therefore, vεj (x) = 0 for thesex.

(iii) We show now that vεj converges to vj even in H1(C−j ). Let ε < δ/(L + 2) and x ∈C−j \ Vj(δ). Then

∇vεj (x) =

∫C−j

vj(y)∇xφε(x− εa− y) dy = −∫C−j

vj(y)∇yφε(x− εa− y) dy .

We show that the function y 7→ φε(x− εa− y) belongs to C∞0 (C−j ). It is sufficient to provethat the ball B3(x− εa, ε) is contained in C−j . Therefore, let y ∈ B3(x− εa, ε).Then, first, |y| ≤ |x|+ |y − x| ≤ αj − δ + ε ≤ αj.Second, y3 ≥ x3− (L+ 1)ε− |x3− (L+ 1)ε− y3| ≥ −2βj + δ− (L+ 1)ε− ε ≥ −2βj because(L+ 2)ε ≤ δ.Third, y3 ≤ x3−(L+1)ε+ |y3−x3 +(L+1)ε)| ≤ ξj(x)−Lε ≤ ξj(y)+L|ξj(y)−ξj(x)|−Lε ≤ξj(y). This proves that y 7→ φε(x− εa− y) belongs to C∞0 (C−j ).By the definition of the weak derivative we have that

∇vεj (x) =

∫C−j

∇vj(z)φε(x− εa− z) dz . (5.1)


For ε < δ/(L + 2) and x ∈ Vj(δ) both sides of (5.1) vanish. Therefore, (5.1) holds for allx ∈ C−j and thus (∇vεj )(x) = (gj ∗ φε)(x− εa) for x ∈ C−j where gj = ∇vj on C−j and gj = 0on R3 \ C−j . By Theorem 4.7 we conclude again that (gj ∗ φε)(x − εa) converges to gj inL2(R3) and thus ∇vεj |C−j to ∇vj in L2(C−j ). This proves vεj → vj in H1(C−j ) as ε tends to

zero for every j = 1, . . . ,m.

(iv) In the last step we transform back the function and set uεj(y) = vεj(R>j (y − z(j))

)for

j = 1, . . . ,m and y ∈ Uj ∩D. We note that uεj ∈ C∞0 (Uj ∩D) and uεj → ϕju in H1(Uj ∩D).This holds for all j = 1, . . . ,m. We extend uj by zero into all of D. Finally, we note thatu0 ∈ H1(D) with compact support in U0 = D and thus uε0 = u0 ∗ φε converges to u0 = ϕ0uin H1(D). To finish the proof we set uε =

∑mj=0 u

εj in D and conclude that uε ∈ C∞0 (D) and

uε converges to∑m

j=0 ϕju = u in H1(D). 2

We apply this result and prove the following chain rule (compare with part (b) of Lemma 5.2).

Corollary 5.4 Let U, V ⊆ R3 be bounded Lipschitz domains and Ψ : V → U a diffeomor-phism from V onto U in the sense that Ψ and Ψ−1 are continuous on V and U , respectively,and differentiable at almost all points x ∈ V and y ∈ U , respectively, such that the JacobiansΨ′ and (Ψ−1)′ are essentially bounded; that is, Ψ′ ∈ L∞(V,R3×3) and (Ψ−1)′ ∈ L∞(U,R3×3).Then, for u ∈ H1(U) the composition v = u Ψ belongs to H1(V ), and the chain rule holdsin the form

∇v(x) = Ψ′(x)>∇u(y)|y=Ψ(x) , x ∈ V . (5.2)

The mapping u 7→ u Ψ is bounded from H1(U) into H1(V ).

Proof : For u ∈ C∞(U) we have by the transformation formula

‖v‖2H1(V ) =

∫V

∣∣u(Ψ(x))∣∣2dx =

∫U

|u(y)|2 1∣∣det Ψ′(Ψ−1(y)

)∣∣ dy≤ c ‖u‖2

L2(U)

and analogously with ∇v(x) = Ψ′(x)>∇u(y)|y=Ψ(x) almost everywhere

‖∇v‖2H1(V ) ≤ ‖Ψ′‖∞

∫V

∣∣∇u(Ψ(x))∣∣2dx ≤ c ‖Ψ′‖∞‖∇u‖2

L2(U) .

Therefore, the mapping u 7→ v = u Ψ has a bounded extension from H1(U) into H1(V ).Therefore, formula (5.2) extends to all of H1(U). 2

We continue by defining Sobolev spaces of periodic functions as we have done it already inChapter 4. Definition 4.12 of H1

per(Q3) is included in the following definition. For the sakeof a simpler notation we restrict ourselves to cubes in Rd with edge length 2π.

Definition 5.5 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

∫Q

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) = (2π)3

√∑n∈Z3

(1 + |n|2)s |vn|2 .

We recall from Lemma 4.37 that H10 (Q) ⊆ H1

per(Q) ⊆ H1(Q) with bounded inclusions.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 4.14 that H1

0 (D) is compactlyimbedded in L2(D). This can be carried over to periodic Sobolev spaces of any order.

Theorem 5.6 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 5.1 2

The generalization of the definition of Sobolev spaces with respect to any s ∈ R>0, especiallys = 1

2, is a key ingredient and leads to our first elementary trace theorem.

Theorem 5.7 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 γ0 : u 7→ u|Q2×0 from the spaces oftrigonometric polynomials P(Q3) into P(Q2) has a bounded extension from H1

per(Q3) into

H1/2per (Q2). Here – and in the following – we often identify Q2 × 0 with Q2.

Furthermore, there exists a bounded right inverse η of γ0; that is, a bounded operator η :H

1/2per (Q2) → H1

per(Q3) with γ0 η = id. In other words, the function u = ηf ∈ H1per(Q3)

coincides with f ∈ H1/2per (Q2) on Q2 × 0.

Proof: Let u ∈ H1per(Q3) be a trigonometric polynomial with Fourier coefficients un, n ∈ Z3;

that is, u(x) =∑

n∈Z3 un ei n·x where the coefficients un vanish for all but a finite number of

n. Therefore, all of the series in this proof restrict to finite sums. For x ∈ R3 and n ∈ Z3

we set x = (x1, x2) and n = (n1, n2), respectively. Again | · | denotes the Euclidean norm ofvectors. Then

(γ0u)(x) = u(x, 0) =∑n∈Z2

(∞∑

n3=−∞

un

)︸︷︷︸

=: vn

ei n·x =∑n∈Z2

vn ei n·x

and thus‖γ0u‖2

H1/2per (Q2)

= (2π)3∑n∈Z2

(1 + |n|2

)1/2 |vn|2 .


Now we use the following elementary estimate

π

2≤ a

∞∑n3=−∞

1

a2 + n23

≤ π + 1 for all a ≥ 1 (5.3)

which we will show at the end of the proof. Using the inequality of Cauchy–Schwarz and

the upper estimate for a =(1 + |n|2

)1/2yields

(1 + |n|2

)1/2 |vn|2 =(1 + |n|2

)1/2

∣∣∣∣∣∞∑

n3=−∞

un(1 + |n|2

)1/2 1(1 + |n|2

)1/2

∣∣∣∣∣2

≤∞∑

n3=−∞

|un|2(1 + |n|2

) [(1 + |n|2

)1/2∞∑

n3=−∞

1

1 + |n|2 + x23

]

≤ (π + 1)∞∑

n3=−∞

|un|2(1 + |n|2

).

Summation with respect to n yields

‖γ0u‖2

H1/2per (Q2)

≤ (π + 1) ‖u‖2H1per(Q3) .

This holds for all trigonometric polynomials and yields the boundedness of γ0 on all ofH1per(Q3) because the space of trigonometric polynomials is dense in H1

per(Q3).

To show the existence of a bounded extension operator η we define the extension ηf forf(x) =

∑n∈Z2 fn e

in·x by

(ηf)(x) =∑n∈Z3

un ein·x , x ∈ Q3 ,

where

un = fnδn

1 + |n|2and δn =

[∞∑

j=−∞

1

1 + |n|2 + j2

]−1

. (5.4)

We first 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 (5.3). Summing over n yields

‖ηf‖2H1per(Q3) = (2π)3

∑n∈Z2

∞∑n3=−∞

|un|2(1 + |n|2) ≤ (2π)3 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,

(γ0ηf)(x) =∑n∈Z2

∞∑n3=−∞

un ein·x =

∑n∈Z2

fn ein·x = f(x) .

It remains to show (5.3). 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 (5.3).2

The motivation for the definition of the Sobolev space H1/2(∂D) is given by the followingtransformation of the L2−norm such that the previous lemma can be applied. Let D ⊆ R3

be a Lipschitz domain with local coordinate system Uj, ξj : j = 1, . . . ,m, correspondingmappings Ψj from the balls Bj onto U ′j, and their restrictions Ψj : B2(0, αj) → U ′j ∩ ∂D asin Definition 5.1.By Q3 we denote a cube centered at the origin such that all of the balls Bj are contained inQ3. Without loss of generality we assume that this is the cube Q3 = (−π, π)3 to make thenotation simpler.Furthermore, let φj : j = 1, . . . ,m, be a partition of unity on ∂D subordinate to the setsU ′j (see Theorem 6.9). For f ∈ L2(∂D) we write |f(y)|2 =

∑mj=1 φj(y)|f(y)|2 for y ∈ ∂D,

and thus

‖f‖2L2(∂D) =

∫∂D

∣∣f(y)∣∣2 ds =

m∑j=1

∫∂D∩U ′j

φj(y)∣∣f(y)

∣∣2 ds=

m∑j=1

∫B2(0,αj)

φj(Ψj(x)

)∣∣f(Ψj(x))∣∣2√1 + |∇ξj(x)|2 dx

=m∑j=1

∫Q2

∣∣fj(x)∣∣2√1 + |∇ξj(x)|2 dx (5.5)


where

fj(x) :=

√φj(Ψj(x)

)f(Ψj(x)

), x ∈ B2(0, αj) ,

0 , x ∈ Q2 \B2(0, αj) .(5.6)

From (5.5) and the estimate 1 ≤√

1 + |∇ξj(x)| ≤ maxj=1,...,m

√1 + ‖∇ξj‖2

∞| we observe

that ‖ · ‖L2(∂D) is equivalent to√∑m

j=1 ‖fj‖2L2(Q2). Therefore, f ∈ L2(∂D) if, and only if,

fj ∈ L2(Q2) for all j = 1, . . . ,m. We extend this definition to define the Sobolev spaceH1/2(∂D).

Definition 5.8 Let D ⊆ R3 be a Lipschitz domain in the sense of Definition 5.1 withcorresponding local coordinate system Uj, ξj : j = 1, . . . ,m and corresponding mappingsΨj from the balls Bj onto U ′j and their restrictions Ψj : B2(0, αj)→ U ′j ∩ ∂D. Furthermore,let φj : j = 1, . . . ,m, be a partition of unity on ∂D subordinate to the sets U ′j (seeTheorem 6.9). 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

,

where fj are given by (5.6), j = 1, . . . ,m.

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 5.15 below that the normscorresponding to two such choices are equivalent.

From now on we assume always that the domain D is a Lipschitz domain in the sense ofDefinition 5.1.

We note the following implication of Theorem 5.6.

Corollary 5.9 The space H1/2(∂D) is compactly imbedded in L2(∂D).

Proof: Let (f `)` be a bounded sequence in H1/2(∂D). Then, by definition, (f `j )` is bounded

in H1/2per (Q2) for all j = 1, . . . ,m, where f `j is defined by (5.6) for f ` instead of f . By

Theorem 5.6 for s = 0 and t = 1/2 there exists a subsequence (f `kj )k of (f `j )` which convergesin L2(Q2) for every j = 1, . . . ,m as k tends to infinity. By (5.5), applied to the differencef `k − f `p also (f `k)k converges in L2(∂D). 2

Definition 5.8 and Theorem 5.7 yield a trace theorem in H1(D) for Lipschitz domains.

Theorem 5.10 The trace operator γ0 : C1(D) → C(∂D), u 7→ u|∂D, has an extension asa bounded operator from H1(D) to H1/2(∂D). Furthermore, γ0 has a bounded right inverseη : H1/2(∂D)→ H1(D); that is, γ0(ηf) = f for all f ∈ H1/2(∂D).


Proof: Let Uj, ξj : j = 1, . . . ,m be a local coordinate system of ∂D with correspondingmappings Ψj from the balls Bj onto U ′j as in Definition 5.1. Let φj : j = 1, . . . ,m be a

partition of unity on ∂D subordinate to U ′j. Set U :=⋃mj=1 U

′j and let u ∈ C1(D). Set

vj(x) :=√φj(Ψj(x)

)u(Ψj(x)

), x ∈ B−j

and extend vj to Q−3 by zero. Here, B±j = x ∈ Bj : x3 ≷ 0 and Q±3 = x ∈ Q3 : x3 ≷ 0.Furthermore, we extend vj ∈ C(Q−3 ) into Q3 by even reflection; that is,

vj(x) =

vj(x) , x ∈ Q−3 ,vj(x

∗) , x ∈ Q+3 ,

where x∗ = (x1, x2,−x3)> for x = (x1, x2, x3)> ∈ R3. Then vj ∈ H1(Q3) by Exercise 4.6.Also, vj vanishes in some neighborhood of ∂Q3. Now we can use the arguments of Lemma 4.37to show that vj ∈ H1

per(Q3) and that there exists c1 > 0 which is independent of vj such that

‖vj‖H1per(Q3) ≤ c1‖vj‖H1(Q−3 ) = c1‖vj‖H1(B−j ) .

The boundedness of the trace operator of Theorem 5.7 yields

‖vj(·, 0)‖H

1/2per (Q2)

≤ c2‖vj‖H1(B−j ) for all j = 1, . . . ,m,

and thus by Definition 5.8

‖γ0u‖2H1/2(∂D) =

m∑j=1

‖vj(·, 0)‖2

H1/2per (Q2)

≤ c22

m∑j=1

‖vj‖2H1(B−j )

≤ c ‖u‖2H1(D∩U) ≤ c ‖u‖2

H1(D) .

This proves boundedness of γ0.

Now we construct an extension operator η : H1/2(∂D) → H1(D). For f ∈ H1/2(∂D) we

define fj ∈ H1/2per (Q2) for j = 1, . . . ,m as in (5.6). By Theorem 5.7 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 ∈ U ′j .

Then uj ∈ H1(U ′j) for all j and they vanish in some neighborhood of ∂U ′j. We extend ujby 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 ∈ U ′j

. Then u(y) =

∑j∈J uj(y). Let j ∈ J be fixed;

that is, y ∈ U ′j. Then x = Ψ−1j (y) ∈ Q2 × 0, thus uj(y) =

√φj(y) vj(x) =

√φj(y) fj(x) =

φ(y)f(y). Therefore, u(y) =∑

j∈J uj(y) =∑

j∈J f(y)φj(y) = f(y). Furthermore, all theoperations in these constructions are bounded. This proves the theorem. 2

Remark: We note that in the second part of the proof we have constructed an extension uof f into all of R3 (not only into the interior) with support in the neighborhood U =

⋃mj=1 U

′j

of ∂D which depends continuously on f .


A simple conclusion of the trace theorem in combination with the denseness result of Theo-rem 5.3 is formulated in the following corollary.

Corollary 5.11 Let D ⊆ R3 be a Lipschitz domain.

(a) The formula of partial integration (see Theorem 6.11) holds in the form∫D

u∇v dx = −∫D

v∇u dx +

∫∂D

(γ0u) (γ0v) ν ds for all u, v ∈ H1(D) .

(b) Let Ω ⊆ R3 be a second Lipschitz domain with D ⊆ Ω and let u1 ∈ H1(D) andu2 ∈ H1(Ω\D) such that γ0u1 = γ0u2 on ∂D. Then the function defined by

u(x) =

u1(x) , x ∈ Du2(x) , x ∈ Ω\D

is in H1(Ω).In particular, for u ∈ H1(D) with γ0u = 0 the extension u of u by zero outside of Dbelongs to H1(R3).

(c) Let D ⊆ R3 be symmetric with respect to x3 = 0; that is, x ∈ D ⇔ x∗ ∈ D wherez∗ = (z1, z2,−z3)> for any z = (z1, z2, z3)> ∈ C3. Let D± = x ∈ D : x3 ≷ 0 andϕ ∈ H1(D−). Extend ϕ into D by reflection; that is,

ϕ(x) =

ϕ(x) , x ∈ D−,ϕ(x∗) , x ∈ D+,

Then ϕ ∈ H1(D) and the mapping ϕ 7→ ϕ is bounded from H1(D−) into H1(D).

The proofs are left as Exercise 5.3, see also Exercise 4.6.

Theorem 5.12 Let D be a bounded Lipschitz domain and U an open set such that D ⊆ U .Then there exists a linear bounded operator η : H1(D) → H1(R3) such that η|D = u for allu ∈ H1(D) and supp(ηu) ⊆ U .

Proof: The proof of Theorem 5.10 (see also the remark following that theorem) yieldsexistence of an operator η from H1/2(∂D) into H1(R3) with γ0ηf = f on H1/2(∂D). Letφ ∈ C∞(R3) such that φ = 1 on D and supp(φ) ⊆ U . Define ηu by

(ηu)(x) =

u(x) , x ∈ D ,

φ(x) (ηγ0u)(x) , x /∈ D .

Then γ0(φ ηγ0u) = γ0ηγ0u = γ0u, and the previous corollary yields ηu ∈ H1(R3. 2

As a corollary we have the following imbedding result of Rellich (compare with Theo-rem 4.14):


Theorem 5.13 Let D ⊆ R3 be a bounded Lipschitz domain. Then the imbedding H1(D)→L2(D) is compact.

Proof: We can just copy the proof of Theorem 4.14 because of the existence of a boundedextension operator η : H1(D)→ H1(R3) of the previous theorem. 2

We recall from Definition 4.10 that the subspace H10 (D) had been defined as the closure of

C∞0 (D) in H1(D). The following equivalent characterization will be of essential importance.

Theorem 5.14 The space H10 (D) is the null space N (γ0) of the trace operator; that is,

u ∈ H10 (D) if, and only if, γ0u = 0 on ∂D.

Proof: The inclusion H10 (D) ⊆ N (γ0) follows immediately because γ0u = 0 for all u ∈

C∞0 (D), the boundedness of γ0 and the fact that C∞0 (D) is dense in H10 (D) by definition.

The reverse inclusion is more difficult to show. It follows closely the proof of Theorem 5.3.We separate the proof into four steps.(i) Let u ∈ H1(D) with γ0u = 0. We extend u by zero into all of R3. Then u ∈ H1(R3) byCorollary 5.11. Let Uj, ξj : j = 1, . . . ,m be a local coordinate system. Set U0 = D andchoose a partition of unity ϕj : j = 0, . . . ,m on D subordinate to Uj : j = 0, . . . ,m (seeTheorem 6.9. Set uj(y) = ϕj(y)u(y) for j = 0, . . . ,m and y ∈ R3. Then the support of uj iscontained in Uj∩D. For j = 1, . . . ,m we transform uj by the definition vj(x) = uj(Rjx+z(j)),x ∈ R3, and observe that vj ∈ H1(R3) by the chain rule of Corollary 5.4, applied to a largeball containing D. The support of vj is contained in the cylinder Cj = B2(0, αj)×(−2βj, 2βj).Furthermore, with

C−j :=x = (x, x3) ∈ B2(0, αj)× (−2βj, 2βj) : x3 < ξj(x)

and

Vj(t) := x ∈ C−j : |x| > αj − t or x3 < −2βj + t

where again x = (x1, x2) we note that there exists δ > 0 with vj(x) = 0 in the neighborhood

Vj(2δ) of ∂Cj ∩ C−j because the support of ϕj is contained in Uj.

(ii) Let a = (0, 0, L+2)> where L is larger than all of the Lipschitz constants of the functionsξj and set

vεj (x) = (vj ∗ φε)(x+ εa) =

∫C−j \Vj(2δ)

vj(y)φε(x+ εa− y) dy

where φε ∈ C∞(R3) denotes the mollifier function from (4.7). Its most important propertieshad been collected in Theorem 4.7. In particular, we have that vεj ∈ C∞0 (R3) and vεj → vj inH1(R3) as ε tends to zero.

(iii) Next we show that supp(vεj ) ⊆ C−j for ε < δ/(L + 3). First we note that the integralin the definition of vεj is taken over C−j \ Vj(2δ). Therefore, let y ∈ C−j \ Vj(2δ). Then|y| ≤ αj−2δ and y3 ≥ −2βj +2δ. The boundary of C−j consists for three parts. We considerx being in some neighborhood of these parts separately.For points x ∈ C−j with |x| ≥ αj − δ we have shown vεj (x) = 0 in part (ii) of the proof of


Theorem 5.3.Let now x ∈ C−j with x3 ≤ −2βj + δ. Then |x + εa − y| ≥ |x3 + ε(L + 2) − y3| ≥y3−x3− (L+ 2)ε ≥ −2βj + 2δj + 2βj− δ− (L+ 2)ε = δ− (L+ 2)ε ≥ ε. Therefore, vεj (x) = 0for these x.Finally, let x ∈ C−j with x3 ≥ ξj(x)−ε and |y−x| ≤ ε. Then |x+εa−y| ≥ |x3+ε(L+2)−y3| ≥x3 + ε(L + 2)− y3 ≥ ξj(x)− ε + (L + 2)ε− y3 = [ξj(y)− y3]− [ξj(y)− ξj(x)] + (L + 1)ε ≥(L+1)ε−L|y−x| ≥ ε. Therefore, vεj (x) = 0 also for these x which shows that supp(vεj ) ⊆ C−jfor ε < δ/(L+ 3); that is, vεj ∈ C∞0 (C−j ).

(iv) In the last step we transform back the function and set uεj(y) = vεj(R>j (y − z(j))

)for

j = 1, . . . ,m and y ∈ R3. We note that uεj ∈ C∞0 (Uj ∩ D) and uεj → ϕju in H1(D). Thisholds for all j = 1, . . . ,m. Finally, we note that u0 ∈ H1(D) with compact support inU0 = D and thus uε0 = u0 ∗ φε converges to u0 = ϕ0u in H1(D). To finish the proof we setuε =

∑mj=0 u

εj in D and conclude that uε ∈ C∞0 (D) and uε converges to

∑mj=0 ϕju = u in

H1(D). 2

As mentioned before we still have to complete Definition 5.8 by showing that the spaceH−1/2(∂D) does not depend on the choice of the local coordinate system and the partion ofunity.

Corollary 5.15 Definition 5.8 is independent of the choice of the coordinate system Uj, ξj :j = 1, . . . ,m and a corresponding partition of unity φj : j = 1, . . . ,m on ∂D subordinateto U ′j.

Proof: First we note that the space C∞(D)|∂D = 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 asequence un ∈ C∞(D) with respect to ‖ · ‖H1(D). But then un|∂D = γ0un converges to

γ0u = f in H1/2(∂D). Therefore, H1/2(∂D) is the completion of C∞(D)|∂D with respect to‖·‖H1/2(∂D). Furthermore, from the boundedness and surjectivity of γ0 : H1(D)→ H1/2(∂D)

by Theorem 5.14 we observe that the lifted operator [γ0] : 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 ‖[v]‖ = inf‖v+ψ‖H1(D) : ψ ∈ H10 (D) in H1(D)/H1

0 (D);that is, there exist constants c1, c2 > 0 with c1‖[v]‖ ≤ ‖v|∂D‖H1/2(∂D) ≤ c2‖[v]‖ for allv ∈ H1(D). The canonical norm in H1(D)/H1

0 (D) depends only on the norm in H1(D)and the subspace H1

0 (D). Therefore, all of the norms ‖ · ‖H1/2(∂D) originating from differentchoices of the coordinate system and partitions of unity are equivalent to each other. 2

The proof shows also that an equivalent norm in H1/2(∂D) is given by ‖f‖ = inf‖u‖H1(D) :

γ0u = 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 γ1 : C1(D) → C(∂D), u 7→ ∂u/∂ν. InExercise 5.2 we show for an example that it is not bounded from H1(D) into L2(∂D). (It


is not even well-defined as one can show.) However, it defines a bounded operator on theclosed subspace

HD =

u ∈ H1(D) :

∫D

[∇u · ∇ψ − k2uψ

]dx = 0 for all ψ ∈ H1

0 (D)

(5.7)

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 5.9 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 a bounded linear functional on H1/2(∂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 doneby identifying the dual space of L2(∂D) with itself – then L2(∂D) can be considered as asubspace of the dual space H1/2(∂D)∗ of H1/2(∂D) which we denote by H−1/2(∂D):

Definition 5.16 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) := supψ∈H1/2(∂D)\0

|〈`, ψ〉∗|‖ψ‖H1/2(∂D)

, ` ∈ H−1/2(∂D) ,

where 〈`, ψ〉∗ = `(ψ) denotes the dual form, the evaluation of ` at ψ.

For real valued functions we observe that 〈`, ψ〉∗ is just the extension of the L2−inner productwhen we identify `f with f .

Using a local coordinate system and a corresponding partition of unity one can prove thatH−1/2(∂D) can be characterized by the periodic Sobolev space H

−1/2per (Q2) which is the com-

pletion of the space of trigonometric polynomials by the norm

‖v‖H−1/2per (Q2)

=

[∑n

(1 + |n|2)−1/2|vn|2]1/2

.

We do not need this result and, therefore, omit the details.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

∂νγ0ψ ds =

∫D

[∇u · ∇ψ − k2uψ

]dx .

Since the trace γ0ψ 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, γ0ψ〉∗. This is justified by the following theorem.


Definition 5.17 (and Theorem) The operator γ1 : HD → H−1/2(∂D), defined by Green’sformula; that is,

〈γ1u, ψ〉∗ =

∫D

[∇u · ∇ψ − k2u ψ

]dx , ψ ∈ H1/2(∂D) , (5.8)

is well defined and bounded. Here, ψ ∈ H1(D) is any extension of ψ into D; that is, γ0ψ = ψ– which is possible by the surjectivity of the trace operator γ0, see Theorem 5.10.

Proof: We have to show that this definition is independent of the choice of ψ. Indeed, ifψ1 and ψ2 are two extensions of ψ then ψ := ψ1 − ψ2 ∈ H1

0 (D) by Theorem 5.14, becauseγ0ψ = 0. With u ∈ HD we conclude that

∫D

[∇u · ∇(ψ1 − ψ2)− k2u(ψ1 − ψ2)

]dx = 0. This

shows that the definition of γ1 is independent of the choice of ψ.

To show boundedness of γ1 we take ψ = ηψ where η : H1/2(∂D) → H1(D) denotes thebounded right inverse of γ0 of Theorem 5.10. Then, by the inequality of Cauchy-Schwarz,

|〈γ1u, 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 γ1 with ‖γ1‖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 γ0 and γ1 havetwo components γ0|∂Dj and γ1|∂Dj for j = 1, 2 which are the projections onto ∂Dj. For the

definition of γ0|∂Dj or γ1|∂Dj one just takes extensions ψ ∈ H1(D) which vanish on ∂D3−j.

Remark: Some readers may feel perhaps unhappy with the space HD as the domain ofdefinition for the trace operator γ1 because it depends on the wave number k. A morenatural way is to define the trace operator on the space

H1(∆, D) :=

u ∈ H1(D) :

∃ v ∈ L2(D) with∫D∇u · ∇ψ dx = −

∫Dvψ dx

for all ψ ∈ H10 (D)

as the space of functions in H1(D) for which even ∆u := v ∈ L2(D). (Note that thefunction v is obviously unique.) The space H1(∆, D) is equipped with the canonical norm‖u‖H1(∆,D) = ‖u‖H1(D) + ‖∆u‖L2(D). Definition 5.17 should be changed into

〈γ1u, ψ〉∗ =

∫D

[∇u · ∇ψ + ψ∆u

]dx , ψ ∈ H1/2(∂D) .

However, we note that HD is a subspace of H1(∆, D), and on this subspace the normsof H1(D) and H1(∆, D) are equivalent. For the following it is important that we haveboundedness of γ1 in the H1(D)−norm, and this is the reason for choosing HD as thedomain of definition.


5.1.2 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. To motivate thesetraces we consider the integral identity (6.16b); that is,∫

D

(v · curlu− u · curl v) dx =

∫∂D

(ν × u) · v ds =

∫∂D

(ν × u) ·((ν × v)× ν

)ds

for u, v ∈ C1(D,C3) ∩ C(D,C3) and a smooth domain D. Thus, the left–hand side leadsto two possibilities in extending traces of the tangential components of the vector fieldson ∂D namely first, for γtu = ν × u, if we consider v as a test function and, second forγTv = (ν× v)× ν, if we take u as a test function. Furthermore, taking gradients of the formv = ∇ϕ for scalar functions ϕ as test functions gives∫

D

∇ϕ · curlu dx =

∫∂D

(ν × u) · γT∇ϕds =

∫∂D

(ν × u) ·Grad ϕds .

From (6.21)) we observe that the right–hand side is just −∫∂D

Div(ν × u)ϕds, see Defini-tion 5.29 below. Thus we observe that the traces of vector fields in H(curl, D) have someregularity which requires more detailed investigations.

We proceed as for the scalar case and begin with the following simple lemma which corre-sponds to Lemma 5.2.

Lemma 5.18 Let D be a bounded Lipschitz domain.

(a) Let u ∈ C1(D,C3) 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.

(b) Let R ∈ R3×3 be an orthogonal matrix and z ∈ R3. For u ∈ H(curl, D) define

v(x) = R>u(Rx+ z) on U = x : Rx+ z ∈ D .

Then v ∈ H(curl, U) and

curl v(x) = (detR)R> curlu(y)|y=Rx+z on U .

Proof: The proof of (a) is very similar to the proof of Lemma 5.2 and is omitted.

(b) With ψ ∈ C∞0 (U) and the substitution x = R>(y − z) we have∫U

v(x)> curlψ(x) dx =

∫U

u(Rx+ z)>R curlψ(x) dx

= detR

∫D

u(y)>R curlx ψ(x)|x=R>(y−z)dy .


Let q(j), j = 1, 2, 3, be the columns of R. Then

curly[Rψ(R>(y − z)

)]= curly

3∑j=1

ψj(R>(y − z)

)q(j) =

3∑j=1

R∇ψj(x)× q(j)

=3∑

j,`=1

∂ψj(x)

∂x`q(`) × q(j)

where we have set x = R>(y−z). If detR = 1 then q(1)×q(2) = q(3) and q(2)×q(3) = q(1) andq(3)×q(1) = q(2) and thus curly

[Rψ(R>(y−z)

)]= R curlx ψ(x)|x=R>(y−z). If detR = −1 then

analogous computations yield curly[Rψ(R>(y − z)

)]= −R curlx ψ(x)|x=R>(y−z). Therefore,

making again the substitution y = Rx+ z, yields∫U

v(x)> curlψ(x) dx =

∫D

u(y)> curly[Rψ(R>(y − z)

)]dy

=

∫D

curlu(y)>Rψ(R>(y − z)

)dy

= detR

∫U

curly u(y)>|y=Rx+zRψ(x) dx

= detR

∫U

(R> curly u(y)|y=Rx+z

)>ψ(x) dx .

This shows that (detR)R> curly u(y)|y=Rx+z is the variational curl of v. 2

The following result corresponds to Theorem 5.3.

Theorem 5.19 If D is a bounded Lipschitz domain then C∞(D,C3) is dense in H(curl, D).

Proof: We can almost copy the proof of Theorem 5.3 if we replace ∇ by curl. We omit thedetails. 2

As in the scalar case we introduce first some Sobolev spaces of periodic functions. We recallthat every vector valued function v ∈ L2(Q,Cd) on a cube Q = (−π, π)d in Rd for d = 2 ord = 3 has an expansion in the form

v(x) =∑n∈Zd

vn ein·x

with Fourier coefficients vn ∈ Cd, given by

vn =1

(2π)d

∫Qd

v(y) e−i n·y dy , n ∈ Zd . (5.9)

For d = 3 formal differentiation yields

div v(x) = i∑n∈Z3

n · vn ein·x , curl v(x) = i∑n∈Z3

n× vn ein·x .


Therefore, curl v ∈ L2(Q,C3) if, and only if,∑

n∈Z3 |n × vn|2 < ∞. This is the motivationfor the following definition.

Definition 5.20 Let again Q3 = (−π, π)3 ⊆ R3 be the cube and Q2 = (−π, π)2 ⊆ R2 bethe square. For any vector function v : Qd → Cd, d = 2 or d = 3, the Fourier coefficientsvn ∈ Cd for n ∈ Zd are defined as

vn =1

(2π)d

∫Qd

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,C2) :=

∑|m|≤N

vm 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,C2) 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.

Remark: It will be convenient to consider the elements f ∈ H−1/2per (Div, Q2) and f ∈

H−1/2per (Curl, Q2) as vector valued functions from Q2 into C3 with vanishing third component

f3. Actually, these elements are no functions anymore but distributions. For example, wewill do this identification in the following lemma.

Theorem 5.21 Let again Q3 = (−π, π)3 ⊆ R3 denote the cube in R3 and Q2 = (−π, π)2 ⊆R2 the corresponding square in R2. Often, we will again identify Q2 ⊆ R2 with Q2×0 ⊆ R3.


Furthermore, let e = (0, 0, 1)> be the unit normal vector orthogonal to the plane R2×0 inR3.

(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)

coincide with f ∈ H−1/2per (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 thus (γtu)(x) =

∑n∈Z2

u⊥n ei n·x

where

u⊥n :=∞∑

n3=−∞

e× un =∞∑

n3=−∞

e× u(n,n3) ∈ C3 , n = (n1, n2) ∈ Z2 , (5.10)

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) , (5.11)

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 (5.3) 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

](5.12)

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∑

n1,n2∈Z

[(n1, n2, 0)> · u⊥n

]ei n·x

where again u⊥n ∈ C3 are the Fourier coefficients of γtu from (5.11) above. For arbitraryn ∈ Z3 we have from (5.11) with the same notations as above that

(n1, n2, 0)> · (e× un) = n · (e× un) =1

|n|n ·(e× (n× vn)

)=

1

|n|(n× e) · (n× vn)


and thus if (n1, n2) 6= (0, 0)

|(n1, n2, 0)> · u⊥n |2 =

∣∣∣∣∣∞∑

n3=−∞

1

|n|(n× e) · (n× vn)

∣∣∣∣∣2

≤∞∑

n3=−∞

1

|n|2∞∑

n3=−∞

|n× e|2 |n× vn|2

≤ π + 1

|n|

∞∑n3=−∞

|n|2 |vn|2 = (π + 1) |n|∞∑

n3=−∞

|un × n|2

where again n = (n1, n2). Here we used |n× e| = |n| because the third component of n× evanishes. For n = 0 we note that n · (e × un) vanishes. Therefore, multiplication with(1 + |n|2)−1/2 and summation with respect to n yields

∑n1,n2∈Z

(1 + |n|2)−1/2 |(n1, n2, 0)> · 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 (5.12) 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 again n = (n1, n2) and

an =

1

|n|2[|n|2 e− n3 n

], n 6= (0, 0) ,

e , n = (0, 0) ,

and δn is given in (5.4) of the proof of Theorem 5.7; 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 .

Now we show boundedness of ηt. For n 6= (0, 0) we conclude

∞∑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(5.13)

≤ |fn|2δn|n|2

[δn

∞∑n3=−∞

1

1 + |n|2

]︸︷︷︸

= 1

= |fn|2δn|n|2

.

Using δn ≤ 2π

(1 + |n|2)1/2 from estimate (5.3) we 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 it 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 (5.13) 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

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 have used in Definition 5.8 is not adequate in

our case because it does not map vector fields of H(curl, U ′j) into vector fields of H(curl, Bj).We observe that gradients ∇ϕ are special elements of H(curl, U ′j) for any ϕ ∈ H1(U ′j). 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 , x ∈ Bj .

This will provide the correct transformation as we will see shortly. First we assume as abovewith loss of generality that the balls Bj are contained in Q3. We define the subspaces ofL2(∂D,C3) and L2(Q2,C3) of tangential vector fields by

L2t (∂D) :=

f ∈ L2(∂D,C3) : ν(y) · f(y) = 0 almost everywhere on ∂D

,

L2t (Q2) :=

g ∈ L2(Q2,C3) : 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 ∈ B2(0, αj)is given by

ν(y) =1

ρj(x)

(∂Ψj(x)

∂x1

× ∂Ψj(x)

∂x2

), y = Ψj(x) ∈ ∂D ∩ U ′j , where

ρj(x) =

∣∣∣∣∂Ψj(x)

∂x1

× ∂Ψj(x)

∂x2

∣∣∣∣ =√

1 + |∇ξj(x)|2, x ∈ B2(0, αj) . (5.14)

Lemma 5.22 Let Uj, ξj : j = 1, . . . ,m be a local coordinate system of ∂D. Let Ψ : B → U ′

be one of the isomorphism Ψj from the balls Bj onto U ′j; that is, we drop the index j. For

u ∈ H(curl, U ′ ∩D) ∩ C(U ′ ∩D,C3) set

v(x) = Ψ′(x)>u(Ψ(x)

), x ∈ B− , (5.15)

where again B− is the lower part of the ball B and Ψ′(x)> ∈ R3×3 is the adjoint of theJacobian at x ∈ B. Furthermore, define F (x) ∈ R3×3 as

F (x) :=

[∂Ψ(x)

∂x1

∣∣∣∣ ∂Ψ(x)

∂x2

∣∣∣∣ ∂Ψ(x)

∂x1

× ∂Ψ(x)

∂x2

], x ∈ B2(0, αj) . (5.16)

Then, with e = (0, 0, 1)> ∈ R3,


(a) v ∈ H(curl, B−) and

curl v(x) =(Ψ′(x)

)−1curly u(y)|y=Ψ(x) , x ∈ B− .

(b)ρ(x)

(ν(y)× u(y)

)∣∣y=Ψ(x)

= F (x)(e× v(x, 0)

), x ∈ B2(0, αj) ,

where ρ = ρj is given by (5.14).

(c) (ν(y)× u(y)

)× ν(y)

∣∣y=Ψ(x)

= F (x)−>[(e× v(x, 0)

)× e], x ∈ B2(0, αj) .

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]> . (5.17)

Applying this to [a|b|c] = Ψ′(x) yields

[Ψ′(x)

]−1=

[∂Ψ(x)

∂x2

× ∂Ψ(x)

∂x3

∣∣∣∣∣ ∂Ψ(x)

∂x3

× ∂Ψ(x)

∂x1

∣∣∣∣∣ ∂Ψ(x)

∂x1

× ∂Ψ(x)

∂x2

]>.

In the following the subscript k denotes the kth component of the vector function Ψ or v oru.

(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′(x)

)−1curly u(y)|y=Ψ(x)


where we applied (5.17) to Ψ(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 = ρ (ν × u) .

(c) We have just seen that ρ (ν × u) = −v2 ∂1Ψ + v1 ∂2Ψ, thus

ρ2 (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

︸︷︷︸

= ρ2(F>F )−1 = ρ2F−1F−>

v1

v2

0

= ρ2 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 5.23 Let D ⊆ R3 be a Lipschitz domain with corresponding local coordinatesystem (Uj, ξj) and corresponding isomorphisms Ψj from the balls Bj onto U ′j and theirrestrictions Ψj : B2(0, αj) → Uj ∩ ∂D. Furthermore, let φj : j = 1, . . . ,m, be a partitionof unity on ∂D subordinate to the sets U ′j and Fj(x) be given by (5.16). We assume withoutloss of generality that all of the disks B2(0, αj) are contained in the square Q2 = (−π, π)2.

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

H−1/2per (Curl,Q2)

]1/2

,

where

f tj (x) := ρj(x)√φj(Ψj(x)

)F−1j (x)f

(Ψj(x)

), x ∈ B2(0, αj) , (5.18a)

fTj (x) := ρj(x)√φj(Ψj(x)

)F>j (x)f

(Ψj(x)

), x ∈ B2(0, αj) , (5.18b)

extended by zero into Q2, and Fj and ρj are defined in (5.16) and (5.14), respectively.


Again, combining the following trace theorem and Theorem 5.25 below one shows exactlyas in Corollary 5.15 that H−1/2(Div, ∂D) does not depend on the choice of the coordinatesystem.

Theorem 5.24 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 from the Definition 5.23, Lemma 5.22,and the boundedness of the trace operator of Theorem 5.21. Indeed, for u ∈ H(curl, D) ∩C(D,C3) we have 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) = ρj(x)√

Φj(y)F−1j (x)

(ν(y)× u(y)

), x ∈ B2(0, αj) , y = Ψj(x) .

By part (b) of Lemma 5.22 we rewrite this as

f tj = e× vj(·, 0) with vj(x) =√

Φj(y) Ψ′j(x)>u(y)∣∣y=Ψj(x)

, x ∈ B−j .

Then vj ∈ H(curl, B−j ) by part (a) of Lemma 5.22 and vj vanishes in some neighborhood of

∂Bj ∩ B−j , and we extend vj by zero into Q−3 . Denoting again z∗ = (z1, z2,−z3)> for any

vector z = (z1, z2, z3)> ∈ C3 we extend vj into Q3 by

vj(x) =

vj(x) , x ∈ Q−3 ,v∗j (x

∗) , x ∈ Q+3 ,

The vector field vj fails to be continuous in Q−3 because of the transformation into theparameter space. Therefore, we can not directly apply Exercise 4.6 to show that vj ∈H(curl, Q3). However, we can approximate vj by a sequence ψ

(`)j ∈ C∞(Q−3 ) and study the

analogous extension to ψ(`)j which are in H(curl, Q3) by Exercise 4.6. Also, they converge to

vj in H(curl, Q3) because curlx(ψ

(`)j

)∗(x∗) = −

(curlψ

(`)j

)∗(x∗) for x ∈ Q+

3 . Also, vj vanishesin some neighborhood of ∂Q3. Now we show, using partial integration as in Lemma 4.37 (seeExercise 4.13), that vj ∈ Hper(curl, Q3) and that there exists c1 > 0 which is independent ofvj such that

‖vj‖Hper(curl,Q3) ≤ c1‖vj‖H(curl,Q−3 ) = c1‖vj‖H(curl,B−j )

and thus

‖f tj‖H−1/2per (Div,Q2)

= ‖γtvj‖H−1/2per (Div,Q2)

≤ c2‖vj‖H(curl,B−j ) ≤ c3‖u‖H(curl,U ′j).

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 (5.18a). By Theorem 5.21 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) :=√φj(y) Ψ′j(x)−>vj(x)

∣∣x=Ψ−1

j (y), y ∈ U ′j ,

extend uj by zero into all of R3 and set u =∑m

j=1 uj. Then ν × u = f on ∂D. Indeed, for

fixed y ∈ ∂D let J =j ∈ 1, . . . ,m : y ∈ U ′j

. Then u(y) =

∑j∈J uj(y). Fix j ∈ J ; that

is, y ∈ U ′j, and x = Ψ−1j (y) ∈ Q2 × 0. From the definition of uj we observe that

ρj(x) ν(y)× uj(y) =√φj(y)Fj(x)

(e× vj(x, 0)

)=√φj(y)Fj(x) f tj (x)

= ρj(x)φj(y) f(y) ;

that is, ν(y)× uj(y) = φj(y) f(y). Summation with respect to j yields ν × u = f . Further-more, all of the operations in these constructions of u are bounded. This proves the theorem.2

Remark: We note that in the second part of the proof we have constructed an extension uinto all of R3 with support in the neighborhood

⋃mj=1 U

′j of ∂D. In particular, this implies

that there exists a bounded extension operator η : H(curl, D)→ H(curl,R3).

We note that the spaceH0(curl, D) has been defined as the closure of C∞0 (D,C3) inH(curl, D),see Definition 4.19. Analogously to Theorem 5.14 it can be characterized as the nullspace ofthe trace operator as the following theorem shows.

Theorem 5.25 The space H0(curl, D) is the null space N (γt) of the trace operator γt; thatis, u ∈ H0(curl, D) if, and only if, γtu = 0 on ∂D. The same holds for γT ; that is, u ∈H0(curl, D) if, and only if, γTu = 0 on ∂D.

Proof: The inclusion H0(curl, D) ⊆ N (γt) follows again immediately from the definition ofH0(curl, D) as the closure of C∞(D,C3). For the reverse inclusion we modify the proof ofTheorem 5.14 by, essentially, replacing ∇ by curl. We omit the details. 2

The following theorem gives a precise formulation of the fact that the spaces H−1/2(Div, ∂D)and H−1/2(Curl, ∂D) are dual to each other.

Theorem 5.26(a) The dual space H−1/2(Div, ∂D)∗ of H−1/2(Div, ∂D) is isomorphic to H−1/2(Curl, ∂D).An isomorphism is given by the mapping J1 : H−1/2(Curl, ∂D) → H−1/2(Div, ∂D)∗ definedas J1g = λg where λg ∈ H−1/2(Div, ∂D)∗ is given by

λg(ψ) :=

∫D

[u · curl ψ − ψ · curl u

]dx , g ∈ H−1/2(Curl, ∂D), ψ ∈ H−1/2(Div, ∂D) ,


where u, ψ ∈ H(curl, D) are any vector fields with γT u = g and γtψ = ψ.

(b) The dual space H−1/2(Curl, ∂D)∗ of H−1/2(Curl, ∂D) is isomorphic to H−1/2(Div, ∂D).An isomorphism is given by the mapping J2 : H−1/2(Div, ∂D) → H−1/2(Curl, ∂D)∗ definedas J2f = µf where µf ∈ H−1/2(Curl, ∂D)∗ is given by

µf (ϕ) := −∫D

[v · curl ϕ− ϕ · curl v

]dx , f ∈ H−1/2(Div, ∂D), ϕ ∈ H−1/2(Curl, ∂D) ,

where v, ϕ ∈ H(curl, D) are any vector fields with γtv = f and γT ϕ = ϕ.

(c) It holds λg(f) = µf (g) =: 〈f, g〉∗ for all f ∈ H−1/2(Div, ∂D) and g ∈ H−1/2(Curl, ∂D)and Green’s formula holds in the form

〈γtv, γTu〉∗ =

∫D

[u · curl v − v · curlu

]dx for all u, v ∈ H(curl, D) . (5.19)

Proof: (a), (b) The proof is split into four parts.(i) First we show that λg and µf are well-defined; that is, the right–hand sides do not dependon the choices of the extensions. Let uj, ψj ∈ H(curl, D), j = 1, 2, such that γtψj = ψ andγT uj = g for j = 1, 2. Then∫

D

[u1 · curl ψ1 − ψ1 · curl u1

]dx −

∫D

[u2 · curl ψ2 − ψ2 · curl u2

]dx

=

∫D

[(u1 − u2) · curl ψ1 − ψ1 · curl(u1 − u2)

]dx

+

∫D

[u2 · curl(ψ1 − ψ2)− (ψ1 − ψ2) · curl u2

]dx .

Because (u1 − u2) ∈ H0(curl, D) and (ψ1 − ψ2) ∈ H0(curl, D) by Theorem 5.25 we concludethat both integrals vanish because of the Definition 4.16 of the variational curl. Therefore,λg is well defined. The same arguments show that also µf is well defined.

(ii) λg and µf are bounded because

∣∣λg(ψ)∣∣ ≤ ∫

D

[|u| | curl ψ|+ |ψ| | curl u|

]dx ≤ ‖u‖H(curl,D)‖ψ‖H(curl,D)

≤ ‖ηT‖ ‖g‖H(Curl,∂D) ‖ηt‖ ‖ψ‖H(Div,∂D)

which proves that λg is bounded and also boundedness of the mapping J1. Again, thearguments for µf and J2 are exactly the same.

(iii) We show that J1 is surjective. Let ` ∈ H−1/2(Div, ∂D)∗. By the theorem of Riesz(Theorem 6.5) there exists a unique u ∈ H(curl, D) with∫

D

[curlu · curl ψ + u · ψ

]dx = `(γtψ) for all ψ ∈ H(curl, D)


because the right–hand side is a bounded functional on H(curl, D). Taking ψ ∈ H0(curl, D)we note from the definition of the variational curl (Definition 4.16) that curlu ∈ H(curl, D)and curl2 u = −u in D. Therefore, for ψ ∈ H−1/2(Div, ∂D) and ψ = ηtψ we have that

`(ψ) =

∫D

[curlu · curl ψ − curl2 u · ψ

]dx

which shows that g := γT curlu ∈ H−1/2(Curl, ∂D) yields λg = `. The proof of surjectivityof J2 follows again by the same arguments.

(iv) Finally, we prove injectivity of J1. Let g ∈ H−1/2(Curl, ∂D) with J1g = λg = 0. By a gen-eral functional analytic argument (see [23], Section IV.6) there exists ` ∈ H−1/2(Curl, ∂D)∗

with ‖`‖ = 1 and `(g) = ‖g‖H−1/2(Curl,∂D). By the surjectivity of J2 there exists f ∈H−1/2(Div, ∂D) with µf = `. Let u = ηTg and v = ηtf . Then, by the definitions of µfand λg,

‖g‖H−1/2(Curl,∂D) = µf (g) = −∫D

[v · curl u− u · curl v

]dx = λg(f) = 0 .

Again, injectivity of J2 is proven analogously.

(c) This is obvious by the definition of λg if one takes g = γTu and ψ = γtv. Note that – bypart (a) – any extensions of g and ψ can be taken in the definition of λg. 2

The notation H−1/2(Div, ∂D) indicates the existence of a surface divergence Div u for u ∈H−1/2(Div, ∂D) and that u and Div u belong to H−1/2(∂D,C3) and H−1/2(∂D), respectively.To confirm this we first show that, indeed, the space H−1/2(Div, ∂D) can be considered asa subspace of H−1/2(∂D,C3). First we note that H1(D,C3) is boundedly imbedded inH(curl, D). Therefore, the trace operator γT is well defined and bounded on H1(D,C3).

Lemma 5.27 The space H−1/2(Div, ∂D) can be identified with a subspace of H−1/2(∂D,C3).The identification is given by a 7→ à for a ∈ H−1/2(Div, ∂D) where à ∈ H−1/2(∂D,C3) isdefined by

〈à, ψ〉∗ = 〈a, γT ψ〉∗ , a ∈ H−1/2(Div, ∂D), ψ ∈ H1/2(∂D,C3) ,

where ψ ∈ H1(D,C3) is any extension of ψ. Here, 〈·, ·〉∗ denotes the dual form in〈H−1/2(∂D,C3), H1/2(∂D,C3)〉 and in 〈H−1/2(Div, ∂D), H−1/2(Curl, ∂D)〉, respectively, seepart (c) of Theorem 5.26.

Proof: First we show that the definition of à is independent of the extension. Let ψj ∈H1(D,C3), j = 1, 2, be two extensions of ψ. Then ψ := ψ1−ψ2 ∈ H1

0 (D,C3) because γ0ψ = 0and Theorem 5.14. We choose a sequence φn ∈ C∞0 (D,C3) with φn → ψ in H1(D,C3). ThenγT φn = 0 for all n and γT φn → γT ψ in H−1/2(Curl, ∂D). This proves γT ψ = 0.Furthermore, à ∈ H−1/2(∂D,C3) and the mapping a 7→ à is bounded because for ψ ∈H1/2(∂D,C3) and the extension operator η : H1/2(∂D,C3) → H1(D,C3) from the tracetheorem (Theorem 5.10) we conclude that∣∣〈à, ψ〉∗∣∣ =

∣∣〈a, γTηψ〉∗∣∣ ≤ c ‖a‖H−1/2(Div,∂D) ‖γT‖ ‖η‖ ‖ψ‖H1/2(∂D) .


Finally, the mapping a 7→ `a is also one-to-one. Indeed, let `a = 0 then take any b ∈H−1/2(Curl, ∂D). Choose a sequence ψn ∈ C∞(D,C3) with ψn → ηT b in H(curl, D) whereηT : H−1/2(Curl, ∂D) → H(curl, D) is the extension operator of Theorem 5.24. This ispossible by the denseness result of Theorem 5.19. Then 0 = 〈a, γT ψn〉∗ → 〈a, b〉∗. This holdsfor all b ∈ H−1/2(Curl, ∂D) which shows that a has to vanish because H−1/2(Curl, ∂D) isthe dual space of H−1/2(Div, ∂D). 2

Remark 5.28 In the follwing we always think of this identification when we use the identity

〈a, ψ〉∗ = 〈a, γT ψ〉∗ , a ∈ H−1/2(Div, ∂D), ψ ∈ H1/2(∂D,C3) .

Again, on the left–hand side a is considered as an element of H−1/2(∂D,C3) while on theright–side a is considered as an element of H−1/2(Div, ∂D).This result implies in particular that for a ∈ H−1/2(Div, ∂D) and scalar ψ ∈ H1/2(∂D) thedual form 〈a, ψ〉∗ ∈ C3 can be defined componentwise. We will use this in the definition ofthe vector potentials in Section 5.2 below.

Having in mind the definition of the variational derivative it is not a surprise that we alsodefine the surface divergence and the surface curl by variational equations; that is, by partialintegration. We take equation (6.21) as a definition. First we note that for ψ ∈ H1(D) itholds that ∇ψ ∈ H(curl, D). Therefore, the traces γT∇ψ ∈ H−1/2(Curl, ∂D) and γt∇ψ ∈H−1/2(Div, ∂D) are well defined.

Definition 5.29 Let a ∈ H−1/2(Div, ∂D) and b ∈ H−1/2(Curl, ∂D). Then the surface di-vergence Div a ∈ H−1/2(∂D) and surface curl Curl b ∈ H−1/2(∂D) are defined as the linearbounded functionals

〈Div a, ψ〉∗ = −〈a, γT∇ψ〉∗ , ψ ∈ H1/2(∂D) , (5.20a)

〈Curl b, ψ〉∗ = −〈γt∇ψ, b〉∗ , ψ ∈ H1/2(∂D) , (5.20b)

where again ψ ∈ H1(D) is any extension of ψ. On the left–hand side, 〈·, ·〉∗ denotes thedual form in 〈H−1/2(∂D), H1/2(∂D)〉 while on the right–hand side it denotes the form in〈H−1/2(Div, ∂D), H−1/2(Curl, ∂D)〉, see part (c) of Theorem 5.26. We note that the defini-tions of the surface divergence and surface curl yield Curl b = −Div(ν × b in a variationalsense, see the remark on page ??.

Again, we have to show that this definition is independent on the extension ψ. If ψ1 and ψ2

are two extension then ψ = ψ1− ψ2 ∈ H10 (D) and thus ∇ψ ∈ H0(curl, D) which implies that

γT∇ψ = 0 and γt∇ψ = 0.

We note the following applications (compare to Corollary 5.11 for the scalar case and toExercise 4.6 for the case of sufficiently smooth functions).


Lemma 5.30 (a) Let Ω ⊆ R3 be a Lipschitz domain such that D ⊆ Ω and let u1 ∈H(curl, D) and u2 ∈ H(curl,Ω \ D) such that γtu1 = −γtu2 or γTu1 = γTu2 on ∂D.Then the field

u(x) =

u1(x) , x ∈ D ,u2(x) , x ∈ Ω \D ,

is in H(curl,Ω).

(b) Let D ⊆ R3 be symmetric with respect to x3 = 0; that is, x ∈ D ⇔ x∗ ∈ D wherez∗ = (z1, z2,−z3)> for any z = (z1, z2, z3)> ∈ C3. Let D± = x ∈ D : x3 ≷ 0 andv ∈ H(curl, D−). Extend v into D by even reflection; that is,

v(x) =

v(x) , x ∈ D−,v∗(x∗) , x ∈ D+,

Then v ∈ H(curl, D) and the mapping v 7→ v is bounded from H(curl, D−) intoH(curl, D).

(c) Let D ⊆ R3 as in part (b) and v ∈ H0(curl, D−). Extend v into D by odd reflection;that is,

v(x) =

v(x) , x ∈ D−,−v∗(x∗) , x ∈ D+,

Then v ∈ H0(curl, D) and the mapping v 7→ v is bounded from H0(curl, D−) intoH0(curl, D).

Proof: (a) Let γTu1 = γTu2 and ψ ∈ H0(curl,Ω) arbitrary. Using (5.19) in D for u1 and inΩ \D for u2 yields

〈γtψ|−, γTu1〉∗ =

∫D

[u1 · curlψ − ψ · curlu1

]dx ,

〈γtψ|+, γTu2, 〉∗ =

∫Ω\D

[u2 · curlψ − ψ · curlu2

]dx ,

where γtψ|− and γtψ|+ denotes the trace from D and Ω \ D on ∂D, respectively. We notethat γtψ|− = −γtψ|+ because of the orientation of ν. Now we set v = curlu1 in D andv = curlu2 in Ω \D and add both equations. Then v ∈ L2(Ω,C3) and∫

Ω

[u · curlψ − ψ · v

]dx = 0 for all ψ ∈ H0(curl,Ω) .

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 (5.19) with u and v interchanged. Here we note thatγTψ|− = γTψ|+.

(b), (c) These part is proven analogously by validating that

curl v(x) =

curl v(x) , x ∈ D−,

∓(curl v)∗(x∗) , x ∈ D+,


respectively, see the proof of the Theorem 5.24 and Exercise 4.6. 2

Now we are able to prove the important compactness property of the subspace V0,A, definedin (4.10d). First we consider the case of a ball.

Lemma 5.31 Let B ⊆ R3 be a ball and A ∈ L∞(B,C3×3) such that A(x) is symmetric foralmost all x and there exists c > 0 with Re (z>A(x)z) ≥ c|z|2 for all z ∈ C3 and almost allx ∈ B. Then the closed subspace H0(curl, divA 0, B) ⊆ H0(curl, B), defined in (4.12a); thatis,

H0(curl, divA 0, B) =u ∈ H0(curl, B) : (Au,∇ϕ)L2(B) = 0 for all ϕ ∈ H1

0 (B).

is compactly imbedded in L2(B,C3). For A = I we just write H0(curl, div 0, B).

Proof: For the special case A(x) = I we will prove this result in the next section (seeTheorem 5.37) by expanding the functions into spherical harmonics.Let now A be arbitrary. We recall the Helmholtz decompositions from Theorem 4.23 in theform

H0(curl, B) = H0(curl, divA 0, B) ⊕ ∇H10 (B) , (5.21a)

L2(B,C3) = L2(divA 0, B) ⊕ ∇H10 (B) , (5.21b)

and note that the projections onto the components are bounded. Let now (un) be a boundedsequence in H0(curl, divA 0, B). Then we decompose un with respect to the Helmholtz de-composition (5.21a) for A(x) = I; that is,

un = vn + ∇pn with vn ∈ H0(curl, div 0, B) and pn ∈ H10 (B) .

Because the projections onto the components are bounded we conclude boundedness of thesequence (vn) in H0(curl, div 0, B) with respect to ‖ · ‖H(curl,B). Because H0(curl, div 0, B) iscompactly imbedded in L2(B,C3) there exists a subsequence (denoted by the same symbol)such that (vn) converges in L2(B,C3). On the other hand, the form vn = un − ∇pn withun ∈ H0(curl, divA 0, B) ⊆ L2(divA 0, B) and pn ∈ H1

0 (B) is just the decomposition of vnin the Helmholtz decomposition (5.21b). Again, by the boundedness of the projections inL2(B,C3) we conclude that (un) converges in L2(B,C3) and ends the proof. 2

Now we are able to prove Theorem 4.24 from the previous chapter.

Theorem 5.32 Let D be a Lipschitz domain and let A ∈ L∞(D,C3×3) such that A(x) issymmetric for almost all x and there exists c > 0 with Re (z>A(x)z) ≥ c|z|2 for all z ∈ C3

and almost all x ∈ D. Then the spaces V0,A and H0(curl, divA 0, D), defined by

V0,A =u ∈ H0(curl, D) : (Au, ψ)L2(D) = 0 for all ψ ∈ H0(curl 0, D)

, (5.22a)

H0(curl, divA 0, D) =u ∈ H0(curl, D) : (Au,∇ϕ)L2(D) = 0 for all ϕ ∈ H1

0 (D), (5.22b)

are compact in L2(D,C3). Here, H0(curl 0, D) denotes the subspace of H0(curl, D) withvanishing curl.


Proof: As noted before (see Remark 4.22) it suffices to consider the subspaceH0(curl, divA 0, D)because it contains V0,A as a closed subspace. The proof consists of several steps. First welocalize the functions u(`) into U− := U ′ ∩D = Ψj(B

−), then we transform these functionsinto the half ball B−j in parameter space. We extend them by a suitable reflection into

H0(curl divM 0, Bj) for a certain M ∈ L∞(Bj,C3×3). In this space we apply Lemma 5.31 toprove L2−convergence of a subsequence which, finally, yields convergence of the correspond-ing subsequence of (u(`)).Let Uj, ξj : j = 1, . . . ,m be a local coordinate system with corresponding isomorphismsΨj from the balls Bj onto U ′j for j1, . . . ,m. Furthermore, set U ′0 = D and choose a partition

of unity φj : j = 0, . . . ,m on D subordinate to the sets U ′j. Before we consider boundedsequences we concentrate on one particular element.Therefore, let u ∈ H0(curl, divA 0, D) and j ∈ 1, . . . ,m be fixed. The definition ofH0(curl, divA 0, D) yields for any ϕ ∈ H1

0 (U−j ) where again U−j := U ′j ∩D = Ψj(B−j )

0 =

∫D

u(y)>A(y)∇(φj(y)ϕ(y)

)dy

=

∫U−j

(u(y)>A(y)∇φj(y)

)ϕ(y) dy +

∫U−j

φj(y)u(y)>A(y)∇ϕ(y) dy

Let now pj ∈ H10 (U−j ) be defined by∫

U−j

∇pj(y)>A(y)∇ϕ(y) dy =

∫U−j

(u(y)>A(y)∇φj(y)

)ϕ(y) dy for all ϕ ∈ H1

0 (U−j ) .

Substituting ϕ = pj yields the estimate

c‖∇pj‖2L2(U−j )

≤ c1‖u‖L2(D)‖pj‖L2(U−j ) (5.23a)

for some c1 > 0 which is independent of u and j. With Friedrich’s inequality we concludethat this implies

‖∇pj‖L2(U−j ) ≤ c2‖u‖L2(D) (5.23b)

for some c2 > 0 independent of u and j. Defining uj := φju + ∇pj we conclude thatuj ∈ H0(curl, U−j ) and even uj ∈ H0(curl, divA, U

−j ) because

∫U−j

uj(y)>A(y)∇ϕ(y) dy = 0

for all ϕ ∈ H10 (U−j ). This holds for all j ∈ 1, . . . ,m. We transform uj into the parameter

domain as we did it already several times. We define

v(x) = Ψ′j(x)>uj(Ψj(x)

)and M(x) = Ψ′j(x)−1A

(Ψj(x)

)Ψ′j(x)−> for x ∈ B−j

where B−j denotes again the lower part of the ball Bj. Here we dropped the index j for vand M to keep the presentation simple. Then v ∈ H0(curl, B−j ) by Lemma 5.22. Also, we

recall that ∇ϕ(x) = Ψ′j(x)>(∇ϕ)(Ψj(x)

)for ϕ = ϕ Ψj and thus

0 =

∫U−j

uj(y)>A(y)∇ϕ(y) dy =

∫B−j

v(x)>M(x)∇ϕ(x) dx for all ϕ ∈ H10 (B−j ) . (5.24)


Now we extend v and M into all of the ball Bj by

v(x) =

v(x) , x ∈ B−j ,−v∗(x∗) , x ∈ B+

j ,

and

M`k(x) =

M`k(x) , x ∈ B−j , `, k ∈ 1, 2, 3 ,M`k(x

∗) , x ∈ B+j , `, k ∈ 1, 2 or ` = k = 3 ,

−M`k(x∗) , x ∈ B+

j , else,

where again z∗ = (z1, z2,−z3)> for any z = (z1, z2, z3)> ∈ C3. First we note that M ∈L∞(Bj,C3×3) and M(x) is symmetric for almost all x ∈ Bj and there exists c1 > 0 withz>M(x)z ≥ c1M(x)|z|2 for all z ∈ C3 and x ∈ Bj. Next we show that v ∈ H0(curl, divM 0, Bj).The formula curlx v

∗(x∗) = −(curl v)∗(x∗) yields v ∈ H0(curl, Bj). It remains to show that∫Bj

v(x)>M(x)∇ϕ(x) dx = 0 for all ϕ ∈ H10 (Bj) .

Let ϕ ∈ H10 (Bj). Then the function x 7→ ϕ(x) − ϕ(x∗) vanishes for x3 = 0 and is thus in

H10 (B−j ). Substituting this function into (5.24) and using ∇xϕ(x∗) = (∇ϕ)∗(x∗) yields

0 =

∫B−j

v(x)>M(x)∇(ϕ(x)− ϕ(x∗)

)dx

=

∫B−j

v(x)>M(x)∇ϕ(x) dx −∫B−j

v(x)>M(x)∇xϕ(x∗) dx

=

∫B−j

v(x)>M(x)∇ϕ(x) dx −∫B+j

v(x∗)>M(x∗)(∇ϕ)∗(x) dx .

Now a careful elementary calculation shows that

v(x∗)>M(x∗)(∇ϕ)∗(x) = −v(x)>M(x)∇ϕ(x) , x ∈ B+j ,

which yields ∫Bj

v(x)>M(x)∇ϕ(x) dx = 0 .

Since this holds for all ϕ ∈ H10 (Bj) we have shown that v ∈ H0(curl, divM 0, Bj).

Now we prove the actual compactness property. Let (u(`))` be a bounded sequence in

H0(curl, divA 0, D). The scalar functions p(`)j ∈ H1

0 (U−j ) and the vector functions v(`)j ∈

H0(curl, divM 0, Bj) correspond to pj and v above. The estimate (5.23b) implies boundedness

of (p(`)j )` in H1

0 (U−j ). The compact imbedding of H10 (U−j ) in L2(U−j ) yields L2−convergence

of a subsequence of (p(`)j )`. From estimate (5.23a) appied to the difference p

(`)j − p

(k)j we

conclude that this subsequence converges also in H10 (U−j ). Also, (u

(`)j )` is bounded in

H0(curl, divA 0, U−j ) and thus also (v(`)j )` in H0(curl, divM 0, Bj) for every j = 1, . . . ,m. Now

we apply Lemma 5.31 which yields convergence of a subsequence of (v(`)j )` in L2(Bj,C3)


which we denote by the same symbol. Then also the sequence (u(`)j )` converges in L2(U ′j,C3)

for every j = 1, . . . ,m. Finally, we consider j = 0; that is, the bounded sequence (φ0u(`))`

in H0(curl, D). We choose a ball B containing D in its interior and extend u(`)0 = φ0u

(`) by

zero into B. Then we construct p(`)0 ∈ H1

0 (B) and u(`)0 = u

(`)0 − ∇p

(`)0 ∈ H0(curl, divA 0, B)

just as before (but now in B instead of U−j ). We again have convergence of a further subse-

quence of (p(`)0 )` in L2(B). Application of Lemma 5.31 to (u

(`)0 )` yields L2−convergence of a

subsequence. Altogether, we have found a common convergent subsequence of (u(`)j )` for all

j = 0, 1, . . . ,m. Together with the convergence of (p(`)j )` in H1

0 (U−j ) this finishes the proof.2

As a direct consequnce we obtain the imbedding of the subspaces V0,ε and Vµ defined also inchapter 4 in case of Lipschitz domains.

Corollary 5.33 The closed subspace V0,ε and Vµ, defined in (4.10d) and (4.10c) for A = εIand A = µI, respectively, are compactly imbedded in L2(D,C3).

Proof: For V0,ε this follows from the previous theorem. By Corollary 4.36 the spaces V0,ε

and Vµ are isometric which ends the proof. 2

5.1.3 The Case of a Ball Revisited

In Chapter 2 we have studied the expansion of solutions of the Laplace equation andHelmholtz equation into spherical harmonics for the special case where B is a ball of ra-dius R centered at the origin. We have seen that for L2−boundary data the series convergein L2(D). From Chapter 4 we know that the natural solution space of the boundary valueproblems for the Laplace or Helmholtz equation is the Sobolev space H1(D). Therefore,according to the trace theorem, the natural space of boundary data is H1/2(∂D). In thissection we will complement the results of Chapter 2 and show convergence in H1(D).In our opinion, expansions into spherical harmionics provide a clear insight into severalproperties of Sobolev spaces. For example, the fact that every curl-free vector field has apotential is seen very explicitly, see Theorem 5.37 below. In the same Theorem we willshow the compact imbedding of the spaces which appear in the Helmholtz decomposition(the special case A = I and D being a ball). As an important by-product we will prove acharacterization of the Sobolev space H1/2(∂B) by the decay of the expansion coefficients.We remark that the purpose of this section is mainly to fill the gap between the L2−theoryof Chapter 2 and the H1−theory of Chapter 4. The results of this section with the exceptionof Theorem 5.37 will not be used in the forthcoming sections.

Let B = B(0, R) be the ball of radius R centered at the origin. We separate variables withrespect to polar coordinates r > 0 and x ∈ S2 to expand functions into a series of sphericalharmonics just as we did in Chapter 2. We begin again with spaces of scalar functions.

Theorem 5.34 Let B = B(0, R) be the ball of radius R centered at the origin. For u ∈H1(B) let umn (r) ∈ C be the expansion coefficients of x 7→ u(rx) with respect to the spherical


harmonics; that is,

umn (r) =

∫S2

u(r, x)Y −mn (x) ds(x) , 0 ≤ r < R , |m| ≤ n , n = 0, 1, 2, . . . (5.25)

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(B) as N tends to infinity.

Proof: First we note that, using (2.17),

umn (r) = − 1

n(n+ 1)

∫S2

u(r, x) DivS2 GradS2Y −mn (x) ds(x)

=1

n(n+ 1)

∫S2

GradS2u(r, x) ·GradS2Y −mn (x) ds(x)

=1√

n(n+ 1)

∫S2

GradS2u(r, x) · U−mn (x) ds(x)

where Umn = 1√

n(n+1)GradY m

n have been defined in Theorem 2.46. Therefore,√n(n+ 1)umn (r)

are the expansion coefficients of GradS2u(r, ·) with respect toUmn : |m| ≤ n, n ∈ N

. From

the orthonormalty of the systemsY mn : |m| ≤ n, n ∈ N

and Um

n : |m| ≤ n, n ∈ N wehave by Bessel’s inequality

N∑n=0

n∑m=−n

|umn (r)|2 ≤∫S2

|u(rx)|2ds(x) ,

N∑n=0

n∑m=−n

n(n+ 1) |umn (r)|2 ≤∫S2

|GradS2u(rx)|2ds(x) , and

N∑n=0

n∑m=−n

∣∣∣∣dumndr (r)

∣∣∣∣2 ≤ ∫S2

∣∣∣∣∂u∂r (rx)

∣∣∣∣2 ds(x)

for almost all r. By Parseval’s equation we have even equalities for N →∞.In spherical polar coordinates the gradient is given by ∇ = ∂

∂rx+ 1

rGradS2 , thus

‖u‖2H1(B) =

R∫0

∫S2

[|u(rx)|2 +

∣∣∣∣∂u∂r (rx)

∣∣∣∣2 +1

r2|GradS2u(rx)|2

]ds(x) r2dr

≥N∑n=0

n∑m=−n

R∫0

[(1 +

n(n+ 1)

r2

)|umn (r)|2 +


∣∣∣∣2]r2dr . (5.26)


The right–hand side converges to ‖u‖2H1(B) as N tends to infinity by the theorem of mono-

toneous convergence. Furthermore, from

∇uN(rx) =N∑n=0

n∑m=−n

dumndr

(r)Y mn (x) x+

√n(n+ 1)

rumn (r)Um

n (x) , 0 ≤ r < R , x ∈ S2 ,

(5.27)we conclude that, for M > N ,

‖uM − uN‖2H1(B) =

M∑n=N+1

n∑m=−n

∫ R

0

[|umn (r)|2

(1 +

n(n+ 1)

r2

)+


∣∣∣∣2]r2 dr (5.28)

tends to zero as N tends to infinity. 2

Remark: From (5.28) we conclude that r 7→ rumn (r) and r 7→ r(umn )′(r) are elements ofL2(0, R) for all m,n.

In this case of B being a ball the space H1/2(∂B) can be characterized by a proper decay ofthe expansion coeficients with respect to the spherical harmonics.

Theorem 5.35 Let B = B(0, R) be the ball of radius R centered at the origin. For f ∈L2(∂B) 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(∂B) if, and only if,

‖f‖ :=

[∞∑n=0

n∑m=−n

√1 + n(n+ 1) |fmn |2

]1/2

< ∞ . (5.29)

Proof: For the moment let X be the completion of C∞(∂B) with respect to the normof (5.29). Note that C∞(∂B) is also dense in H1/2(∂B) (see Corollary 5.15 for a proof).Therefore, as in the proof of Corollary 5.15 it suffices to show that the norm ‖ · ‖ of (5.29)is equivalent to the norm of the factor space H1(B)/H1

0 (B). This is assured if we can proveTheorems 5.10 and 5.14 for X instead of H1/2(∂B). We assume without loss of generalitythat R = 1.To prove boundedness of the trace operator we write u ∈ H1(B) in polar coordinates as

u(rx) =∞∑n=0

n∑m=−n

umn (r)Y mn (x)

with coefficients umn (r) ∈ C and thus

(γ0u)(x) =∞∑n=0

n∑m=−n

umn (1)Y mn (x) .


We estimate∣∣umn (s)

∣∣2 for s ∈ [0, 1], using r3 ≤ r2 for r ∈ [0, 1], the inequality of Cauchy-Schwarz, and 2ab ≤ a2 + b2, as

s3∣∣umn (s)

∣∣2 =

∫ s

0

d

dr

[r3|umn (r)|2 dr

=

∫ s

0

[3r2|umn (r)|2 + 2r3 Re

(umn (r)

d

drumn (r)

)]dr

≤ 3

∫ s

0

|umn (r)|2 r2 dr + 2

√∫ s

0

|umn (r)|2 r2 dr

√∫ s

0

∣∣∣∣ ddrumn (r)

∣∣∣∣2 r2 dr

≤ 3

∫ s

0

|umn (r)|2 r2 dr +√

1 + n(n+ 1)

∫ s

0

|umn (r)|2 r2 dr

+1√

1 + n(n+ 1)

∫ s

0


∣∣∣∣2 r2 dr .

From this and the preceding remark we conclude that umn is continous on (0, 1]. We continuewith s = 1 and multiply this estimate by

√1 + n(n+ 1) which 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

‖γ0u‖2H1/2(∂B) =

∞∑n=0

n∑m=−n

√1 + n(n+ 1) |umn (1)|2 ≤ 4 ‖u‖2

H1(B)

by (5.26) for N → ∞. This proves boundedness of γ0 from H1(B) into X. A right inverseof γ is given by

(ηf)(rx) =∞∑n=0

n∑m=−n

fmn rn Y mn (x) .

It suffices to prove boundedness. From (5.28) for unm(r) = fmn rm we conclude that

‖ηf‖2H1(B) =

∞∑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(B) and X.To show the analogue of Theorem 5.14 it suffices again to show that the nullspace of γ0 iscontained in H1

0 (B). Let u ∈ H1(B) with γ0u = 0; that is, umn (1) = 0 for all m,n. Let uNbe the truncated sum; that is,

uN(rx) =N∑n=0

n∑m=−n

umn (r)Y mn (x) .


Then uN is continuous in B \ 0 and converges to u in H1(B) as N tends to infinity.Furthermore, let ψ ∈ C∞(C) such that ψ(z) = 0 for |z| ≤ 1 and ψ(z) = z for |z| ≥ 2. DefinevN,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(B) as ε tends to zero. Also, we can approximate vN,ε bysome v ∈ C∞0 (B) because vN,ε vanishes in some neigborhood of S2. Altogether this showsthat u can be approximated arbitrarily well in C∞0 (B). This ends the proof. 2

We continue by extending the previous Theorems 5.34 and 5.35 to the vector-valued case.

Theorem 5.36 Let B = B(0, R) be the ball of radius R centered at the origin. For u ∈H(curl, B) 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 , (5.30a)

vmn (r) =

∫S2

u(r, x) · U−mn (x) ds(x) , 0 ≤ r < R , (5.30b)

wmn (r) =

∫S2

u(r, x) · V −mn (x) ds(x) , 0 ≤ r < R , (5.30c)

for |m| ≤ n and n = 0, 1, 2, . . ., where Umn and V m

n have been defined in Theorem 2.46 seeCorollary 2.47. Then

uN(rx) :=N∑n=0

n∑m=−n

[umn (r)Y m

n (x) x + vmn (r)Umn (x) + wmn (r)V m

n (x)],

0 ≤ r < R, x ∈ S2, converges to u in H(curl, D) as N tends to infinity.

Proof: We can argue very much as in the proof of Theorem 5.34 but the formulas aretechnically more complicated. First we note from the completeness of the orthonormalsystems Y n

m : |m| ≤ n, n ∈ N in L2(S2) and Unm, V

mn : |m| ≤ n, n ∈ N in L2

t (S2) that

∞∑n=0

n∑m=−n

[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]= ‖u(r, ·)‖2

L2(S2) .

The expansion coefficients of curlu(r, ·) are given by (see Exercise 5.4)∫S2

curlu(rx) · x Y −mn (x) ds(x) = −√n(n+ 1)

rwmn (r) ,∫

S2

curlu(rx) · U−mn (x) ds(x) = −1

r

(rwmn (r)

)′,

∫S2

curlu(rx) · V −mn (x) ds(x) = −√n(n+ 1)

rumn (r) +

1

r

(rvmn (r)

)′.


This yields

1

r2

∞∑n=0

n∑m=−n

[n(n+ 1) |wmn (r)|2 +

∣∣(rwmn (r))′∣∣2 +

∣∣∣√n(n+ 1)umn (r)−(rvmn (r)

)′∣∣∣2]= ‖ curlu(r, ·)‖2

L2(S2) .

To compute curluN we need the formulas (see Exercise 5.4)

curl[umn (r)Y m

n (x) x]

= −umn (r)

r

(x×GradS2Y m

n (x)),

curl[vmn (r) GradS2Y m

n (x)]

= −1

r

(r vmn (r)

)′ (x×GradS2Y m

n (x)),

curl[wmn (r)

(x×GradS2Y m

n (x))]

= −1

r

(r wmn (r)

)′GradS2Y m

n (x)

− wmn (r)n(n+ 1)

rY mn (x) x .

Therefore,

curluN(x) =N∑n=0

n∑m=−n

1

r

[−√n(n+ 1)umn (r)V m

n (x) +(r vmn (r)

)′V mn (x)

−(r wmn (r)

)′Umn (x) −

√n(n+ 1)wmn (r)Y m

n (x) x]

(5.31)

and thus for M > N

‖uM − uN‖2H(curl,B) =

M∑n=N+1

n∑m=−n

∫ 1

0

[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]r2dr

+M∑

n=N+1

n∑m=−n

∫ 1

0

[(n(n+ 1) |wmn (r)|2 +

∣∣(rwmn (r))′∣∣2+

+∣∣∣√n(n+ 1)umn (r)−

(rvmn (r)

)′∣∣∣2] drwhich proves convergence of (uN) in H(curl, B). The limit has to be u. 2

Remark: Analogously to the scalar case we note from the previous equality that the func-tions r 7→ rumn (r), r 7→ rvmn (r), r 7→ rwmn (r), r 7→ (rwmn (r))′, and r 7→ (rvmn (r))′ are inL2(0, R). This yields in particular that vmn and wmn are continuous in (0, R].

As a simple consequence of this result we show that for balls the subspace H(curl 0, B) ofvector fields with vanishing curl coincides with the space ∇H1(B) of gradients and alsocompactness of the subspaces appearing in the Helmholtz decompositions (see Theorem 4.21and Remark 4.22) in L2(B,C3).


Theorem 5.37 Let B be a ball.

(a) The space H(curl 0, B), defined by H(curl 0, B) = u ∈ H(curl, B) : curlu = 0 in Bcoincides with the space ∇H1(B) = ∇ϕ : ϕ ∈ H1(B).

(b) The space H(curl, div 0, B) = u ∈ H(curl, B) :∫Bu · ∇ϕdx = 0 for all ϕ ∈ H1(B)

is compactly imbedded in L2(B,C3).

(c) The space H0(curl, div 0, B) = u ∈ H0(curl, B) :∫Bu · ∇ϕdx = 0 for all ϕ ∈ H1

0 (B)is compactly imbedded in L2(B,C3).

Proof: (a) The inclusion ∇H1(B) ⊆ H(curl 0, B) is obvious and holds for any domain. Toshow the reverse inclusion let u ∈ H(curl 0, B). Then u has an expansion in the form

u(rx) =∞∑n=0

n∑m=−n

[umn (r)Y m


n (x)]. (5.32)

Since curlu = 0 we conclude from (5.31) that

wmn (r) = 0 and√n(n+ 1)umn (r) =

(rvmn (r)

)′for all r and all n = 0, 1, 2, . . ., |m| ≤ n. Now we set

ϕ(rx) =∞∑n=1

n∑m=−n

1√n(n+ 1

r vmn (r)Y mn (x) .

From∑∞

n=1

∑nm=−n |vmn |2 <∞ we conclude that ϕ ∈ H1(B) and, by (5.27),

∇ϕ(rx) =N∑n=0

n∑m=−n

1√n(n+ 1

[(r vmn (r)

)′Y mn (x) +

√n(n+ 1) vmn (r)Um

n (x)]

=N∑n=0

n∑m=−n

[umn (r)Y m

n (x) + vmn (r)Umn (x)

]= u(rx) .

(b) Without loss of generality let B the unit ball. Every u ∈ H(curl, div 0, B) has anexpansion in the form (5.32). Let ϕ(rx) = ρ(r)Y −mn (x) for some n ∈ N and |m| ≤ n and

ρ ∈ C1[0, 1]. Then ∇ϕ(rx) = ρ′(r)Y −mn (x) x+

√n(n+1)

rU−mn (x) and thus

0 =

∫B

u(x) · ∇ϕ(x) dx =

∫ 1

0

[ρ′(r)umn (r) +

√n(n+ 1)

rρ(r) vmn (r)

]r2dr . (5.33)

This holds for all such ρ. Note that r 7→ rvmn (r) is continuous and r 7→ r2umn (r) is inL1(0, 1). A modification of the Fundamental Theorem of Calculus (Lemma 4.46, see Ex-ercise 4.16) yields umn ∈ C1(0, 1] and umn (1) = 0 and

(r2umn (r)

)′=√n(n+ 1) rvmn (r) for

all r ∈ (0, 1]. From (5.31) we conclude that∑∞

n=0

∑nm=−n

∫ 1

0|qmn (r)|2dr converges where


qmn (r) =√n(n+ 1)umn (r) −

(rvmn (r)

)′. We estimate, using these two relationships between

umn and vmn and qmn and integration by parts,

√n(n+ 1)

∫ 1

0

r2|vmn (r)|2dr =

∫ 1

0

(r2umn (r)

)′r vmn (r) dr = −

∫ 1

0

r2umn (r)(r vmn (r)

)′dr

= −√n(n+ 1)

∫ 1

0

r2|umn (r)|2dr +

∫ 1

0

r2umn (r) qmn (r) dr .

Note that no boundary terms appear because umn (1) = 0. Therefore,

√n(n+ 1)

∫ 1

0

r2[|umn (r)|2 + |vmn (r)|2

]dr ≤

√∫ 1

0

r2|umn (r)|2 dr

√∫ 1

0

r2|qmn (r)|2 dr

≤ 1

2

∫ 1

0

r2|umn (r)|2 dr +1

2

∫ 1

0

r2|qmn (r)|2 dr .

From this we easily derive the estimate∫ 1

0

r2[|umn (r)|2 + |vmn (r)|2

]dr ≤ 1

n

∫ 1

0

|qmn (r)|2dr .

This holds for all m and n.Now we prove compactness of the imbedding operator I : H(curl, div 0, B) → L2(B,C3).Define the operator IN : H(curl, div 0, B)→ L2(B,C3) by

(INu)(rx) =N∑n=0

n∑m=−n

[umn (r)Y m


n (x)].

Then

‖INu− Iu‖L2(B) =∞∑

n=N+1

n∑m=−n

∫ 1

0

r2[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]dr

≤ 1

N + 1

∞∑n=N+1

n∑m=−n

∫ 1

0

|qmn (r)|2 dr

+1

(N + 2)(N + 1

∞∑n=N+1

n∑m=−n

∫ 1

0

n(n+ 1)|wmn (r)|2r2 dr .

This yields ‖INu−Iu‖L2(B) ≤ 1N‖u‖H(curl,B); that is, IN converges to I in the operator norm.

IN is compact as a bounded finite dimensional operator, therefore also I is compact.

(c) We can very much follow the arguments of part (b) and only indicate the differences.We choose ρ ∈ C1[0, 1] with ρ(1) = 0 in order to have ϕ ∈ H0(B) by Theorem 5.14. Thenwe derive the variational equation (5.33) for ρ ∈ C1[0, 1] with ρ(1) = 0. Again this showsthat umn ∈ C1(0, 1] and

(r2umn (r)

)′=√n(n+ 1) rvmn (r) for all r ∈ (0, 1] by Exercise 4.16 but


no longer umn (1) = 0. Nevertheless, one can continue with the proof because vmn (1) = 0 bythe boundary condition ν × u = 0 on ∂B (and Theorem 5.25). We leave the details to thereader. 2

Now we turn to the characterization of the boundary spacesH−1/2(Div, ∂B) andH−1/2(Curl, ∂B)by the decay of the expansion coefficients, compare with Theorem 5.35 for the scalar case.

Theorem 5.38 Let again B = B(0, R) be the ball of radius R centered at the origin. Forf ∈ L2

t (∂B) let amn , bmn ∈ C be the expansion coefficients with respect to the spherical vector-

harmonics; that is,

amn =

∫S2

f(Rx)·U−mn (x) ds(x) , bmn =

∫S2

f(Rx)·V −mn (x))ds(x) , |m| ≤ n, n = 0, 1, 2, . . .

Then f ∈ H−1/2(Div, ∂B) or f ∈ H−1/2(Curl, ∂B) if, and only if,

‖f‖2D :=

∞∑n=0

n∑m=−n

[(1 + n(n+ 1)

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

)−1/2 |bmn |2 < ∞ , (5.34)

or

‖f‖2C :=

∞∑n=0

n∑m=−n

[(1 + n(n+ 1)

)−1/2 |amn |2 +(1 + n(n+ 1)

)+1/2 |bmn |2 < ∞ , (5.35)

respectively.

Proof: Let X be the completion of C∞t (∂B) with respect to the norm of (5.34). Theassertion follows as in Corollary 5.15 (see also Theorem 5.35) once we have proven the tracetheorem in X and that H1

0 (B) coincides with the null space of γ0; that is, Theorems 5.24and 5.25. First we compute ‖u‖H(curl,B). For u of the form

u(x) =∞∑n=0

n∑m=−n

[umn (r)Y m


n (x)],

we have seen in the previous theorem that

‖u‖2L2(B) =

∞∑n=0

n∑m=−n

∫ R

0

[|umn (r)|2 + |vmn (r)|2 + |wmn (r)|2

]r2 dr , (5.36a)

‖ curlu‖2L2(B) =

∞∑n=0

n∑m=−n

∫ R

0

[∣∣∣√n(n+ 1) |umn (r)|2 −(r vmn (r)

)′∣∣∣2+∣∣∣(r wmn (r)

)′∣∣∣2 + n(n+ 1) |wmn (r)|2]dr . (5.36b)

Without loss of generality we take R = 1. Then we observe that

x× u(1, x) =∞∑n=0

n∑m=−n

[vmn (1)

x×GradS2Y mn (x)√

n(n+ 1)− wmn (1)

GradS2Y mn (x)√

n(n+ 1)

].


Note that vmn and wmn are continuous in (0, 1] by the previous remark. Now 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 (5.36a), (5.36b) yields

∞∑n=0

n∑m=−n

(1 + n(n+ 1)

)−1/2 ∣∣vmn (1)∣∣2 ≤ 2‖u‖2

H(curl,B) .

For wmn (1) we argue analogously:

∣∣wmn (1)∣∣2 =

∫ 1

0

(r∣∣r wmn (r)

∣∣2)′ dr= 2 Re

∫ 1

0

(r wmn (r)

)′wmn (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,B) .

This proves boundedness of the trace operator H(curl, B)→ X. A right inverse is given by

(ηf)(rx) =∞∑n=0

n∑m=−n

[bmn U

mn (x) − amn V

mn (x)

]rn ,

5.2. SURFACE POTENTIALS 245

where anm, bmn are the Fourier coefficients of f . Boundedness follows from (5.36a), (5.36b) forumn (r) = 0, vmn (r) = bmn r

n, and wmn (r) = −amn rn.Finally, we have to show that the kernel of γt coincides with H0(curl, B). We follow thearguments as in the proof of Theorem 5.35 and approximate

uN(rx) =N∑n=0

n∑m=−n

[umn (r)Y m


n (x)]

by

uN,ε(rx) =∞∑n=0

n∑m=−n

[φ

(1− rε

)umn (r)Y m

n (x) x +ε

rψ(rεvmn (r)

)Umn (x)

+ε

rψ(rεwmn (r)

)V mn (x)

]where again ψ ∈ C∞(C) with ψ(z) = 0 for |z| ≤ 1 and ψ(z) = z for |z| ≥ 2 and φ ∈ C∞(R)with φ(t) = 0 for |t| ≤ 1 and φ(t) = 1 for |t| ≥ 2. Then uN,ε vanishes in some neighborhood of∂D because the continuous functions vmn and wmn vanish for r = 1. Estimating the difference‖uN − uN,ε‖H(curl,B) requires (see (5.36a), (5.36b)) to consider expressions of the form

ε

rψ(rεvmn (r)

)− vmn (r) and

[φ

(1− rε

)− 1

]umn (r)

andd

dr

[r(εrψ(rεvmn (r)

)− vmn (r)

)]=[ψ′(rεvmn (r)

)− 1] (rvmn (r)

)′.

These expressions converges to zero pointwise almost everywhere as ε tends to zero. Alsothere exist integrable bounds. Then we can continue as in the proof of Theorem 5.35.For γT one 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 Lipschitz 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 γ0u|± is the trace of u from the exterior (+) and interior (−), respectively. Thetraces of the normal derivatives γ1u|± for (variational) solutions u of the Helmholtz equationare defined by⟨

γ1u|−, ψ⟩∂D

=

∫D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) , u ∈ HD , (5.37a)

⟨γ1u|+, ψ

⟩∂D

= −∫R3\D

[∇u · ∇ψ − k2uψ

]dx , ψ ∈ H1/2(∂D) , u ∈ HR3\D , (5.37b)


where ψ ∈ H1(D) or ψ ∈ H1(R3\D), respectively, are extensions of ψ with bounded supportin the latter case, and

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)

denote the spaces of variational solutions of the Helmholtz equation in D and in R3 \ D,respectively, compare with (5.7). Here, H1

loc(R3 \ D) = u : R3 \ D → C : u|B ∈H1(B) for all balls B.We denoted the dual form in

⟨H−1/2(∂D), H1/2(∂D)

⟩by 〈·, ·〉∂D instead of 〈·, ·〉∗ which we

will do from now on. Note, that we have changed the sign in γ1u|+ because the normalvector ν is directed into the interior of R3 \D.

In the first part of this section where we study scalar potentials we let k ∈ C with Re k ≥ 0and Im k ≥ 0 be arbitrary. When we consider vector potentials we let k = ω

√εµ for any

constant ε, µ ∈ C.

We begin with the representation theorem for solutions of the Helmholtz equation, comparewith Theorem 3.3.

Theorem 5.39 (Green’s Representation Theorem)

Let Φ(x, y) be the fundamental solution of the Helmholtz equation; that is,

Φ(x, y) =eik|x−y|

4π|x− y|, x 6= y . (5.38)

For any solution u ∈ H1(D) of the Helmholtz equation; that is, u ∈ HD, we have therepresentation∫

∂D

(γ0u)(y)∂Φ

∂ν(y)(x, y) ds(y) −

⟨γ1u, γ0Φ(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 (5.8) to u and Φ(x, ·) inD \B[x, ε]: ⟨

γ1u, γ0Φ(x, ·)⟩∂D−∫|y−x|=ε

Φ(x, y)∂u(y)

∂νds(y)

=

∫D\B[x,ε]

[∇u(y) · ∇yΦ(x, y)− k2u(y) Φ(x, y)

]dy

=

∫D\B[x,ε]

[∇u(y) · ∇yΦ(x, y) + u(y) ∆yΦ(x, y)

]dy

= −∫|y−x|=ε

u(y)∂Φ(x, y)

∂ν(y)ds(y) +

∫∂D

(γ0u)(y)∂Φ(x, y)

∂ν(y)ds(y) ;


that is,

⟨γ1u,Φ(x, ·)

⟩∂D−∫∂D

(γ0u)(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

Corollary 5.40 Every u ∈ HD is infinitely often differentiable; that is, u ∈ C∞(D). There-fore, u is even analytic by Corollary 3.4.

Proof: Green’s representation theorem shows that u can be expressed as a difference of adouble layer potential with density in L2(∂D) (even in H1/2(∂D)) and a single layer potentialwith the density ϕ = γ1u ∈ H−1/2(∂D). Only the latter one has to be considered. Let Abe a differential operator. It suffices to show that A

⟨ϕ,Φ(x, ·)

⟩∂D

=⟨ϕ, (AxΦ)(x, ·)

⟩∂D

in

every ball B such that B ⊆ D. Using an argument by induction it suffices to show this forA = ∂/∂xj and every function Φ ∈ C∞

(B×(R3\B)

)instead of Φ. Set f(x) =

⟨ϕ, Φ(x, ·)

⟩∂D

for x ∈ 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. 2

In the following we drop the symbol γ0, thus we write write just v for γ0v. If ∂D is theinterface of two domains then we write v|± and γ1v|± 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) , (5.39a)

(Dϕ)(x) =

∫∂D

ϕ(y)∂Φ

∂ν(y)(x, y) ds(y) , x /∈ ∂D , ϕ ∈ H1/2(∂D) , (5.39b)


compare with (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 3.12 and 3.16 we recall that for smooth domains Dand smooth densities ϕ the single layer u = Sϕ 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). This interpretation of the single layerpotential, together with the trace theorems, will allow us to prove the required propertiesof the boundary operators. A problem is that the region of this transmission problem isall of R3; that is, unbounded. Therefore, we will restrict the transmission problem to abounded region B containing D in its interior and add the additional boundary condition∂u/∂ν − iku = 0 on ∂B. The solution of this transmission problem in the ball B will leadto a compact perturbation of S.The variational form is studied in the following theorem.

Lemma 5.41 Let k ∈ C \ 0 with Im k ≥ 0 and B be an open ball with D ⊆ B. For everyϕ ∈ H−1/2(∂D) there exists a unique 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) . (5.40)

Furthermore, the operator ϕ 7→ v is bounded from H−1/2(∂D) into H1(B).

Proof: We write (5.40) 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) with ‖`‖ ≤c′‖ϕ‖H−1/2(∂D) 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) .

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) . (5.41)


Using the boundedness of the trace operator from H1(B) into L2(∂B) we conclude thata is bounded. The theorem of Riesz (Theorem 6.5) assures the existence of r ∈ H1(B)and a bounded operator K from H1(B) into itself such that 〈ϕ, ψ〉∂D = (r, ψ)H1(B) anda(v, ψ) = (Kv, ψ)H1(B) for all v, ψ ∈ H1(B). Furthermore, ‖r‖H1(B) = ‖`‖ ≤ c′‖ϕ‖H−1/2(∂D).Then we write (5.40) as v−Kv = r in H1(B) and show that K is compact. From (5.41) forψ = 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‖H1(B) ≤ c

[‖v‖L2(B) + ‖v‖L2(∂B)

].

Since in a Hilbert space an operator K is compact if, and only if, it maps weakly convergentsequences into norm-convergent sequences we consider any such sequence vj 0 weakly inH1(B). From the boundedness of the trace operator γ0 and the compactness of the imbed-dings H1(B) in L2(B) and H1/2(∂B) in L2(∂B) we conclude that the right–hand side of theprevious estimate for vj instead of v converges to zero. Therefore, ‖Kvj‖H1(B) converges tozero which proves compactness of K.Therefore, we can apply the Riesz–Fredholm theory to (5.40). To show existence andboundedness of the solution operator r 7→ v it suffices to prove uniqueness. Therefore,let v ∈ H1(B) be a solution of (5.40) for ϕ = 0. Substituting ψ = kv into (5.40) yields∫

B

[k|∇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 on ∂B. Extending v byzero into the exterior of B yields v ∈ H1(R3) (see Exercise 4.9) and∫

R3

[∇v · ∇ψ − k2v ψ

]dx = 0 for all ψ ∈ H1(R3) ;

that is, ∆v+ k2v = 0 in R3. The regularity result of Corollary 5.40 and unique continuation(see Theorem 4.39) yields v = 0 in R3. 2

We are now able to prove the following basic properties of the single layer potential.

Theorem 5.42 Let u := Sϕ in R3 \ ∂D be the single layer potential with density 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 u satisfies the Helmholtz equation ∆u+ k2u = 0 in R3 \ ∂D andSommerfeld’s radiation condition (3.2) for |x| → ∞; that is,

∂us(rx)

∂r− ik us(rx) = O

(1

r2

)for r →∞ , (5.42)

uniformly with respect to x ∈ S2.


(b) Let Q be open and bounded such that D ⊆ Q. The operator S is well-defined andbounded from H−1/2(∂D) into H1(Q).

(c) u|D ∈ HD and u|R3\D ∈ HR3\D and γ1u|− − γ1u|+ = ϕ.

(d) The traces

S = γ0S : H−1/2(∂D) → H1/2(∂D) , (5.43a)

D′ =1

2(γ1S|+ + γ1S|−) : H−1/2(∂D) → H−1/2(∂D) , (5.43b)

are well defined and bounded.

(e) With these notations the jump conditions for u = Sϕ and ϕ ∈ H−1/2(∂D) take theform

γ0u|± = Sϕ , γ1u|± = ∓ 1

2ϕ + D′ϕ . (5.44)

Proof: (a) follows from Corollary 5.40.(b) Let B be a ball which contains Q in its interior. We prove the following representationof Sϕ for ϕ ∈ H−1/2(∂D).

Sϕ = v + w in Q , (5.45)

where v ∈ H1(B) is the solution of (5.40), and w ∈ H1(Q) is explicitely given by

w(x) =

∫∂B

v(y)

[∂Φ


]ds(y) , x ∈ Q .

Indeed, (5.40) implies that the restrictions satisfy v|D ∈ HD and v|B\D ∈ HB\D if we choose

ψ ∈ H10 (D) and ψ ∈ H1

0 (B \D), respectively, and extend them by zero into the remainingparts of B. Two applications of Green’s representation theorem (Theorem 5.39 in D and inB \D) for x ∈ D yield

v(x) = −∫∂D

v(y)|−∂Φ

∂ν(y)(x, y) ds(y) + 〈γ1v|−,Φ(x, ·)〉∂D , x ∈ D ,

0 =

∫∂D

v(y)|+∂Φ

∂ν(y)(x, y) ds(y) − 〈γ1v|+,Φ(x, ·)〉∂D

−∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) + 〈γ1v,Φ(x, ·)〉∂B , x ∈ D ,

where we dropped the symbol γ0 for the trace operator. Adding both equation yields

v(x) =⟨γ1v|− − γ1v|+ , Φ(x, ·)

⟩∂D− w(x) , x ∈ D , (5.46)

where

w(x) :=

∫∂B

v(y)∂Φ

∂ν(y)(x, y) ds(y) −

⟨γ1v,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 ∈ Q .


Here, Φx ∈ H1(B) is chosen such that Φx = Φ(x, ·) on ∂B and Φx = 0 on D. Then Φx is anextension of Φ(·, x) ∈ H1/2(∂B) into B, and the last equation holds by the definition of v.For fixed x ∈ D we choose Φx ∈ H1(B) such that Φx = Φ(x, ·) on ∂D and Φx = 0 on ∂B.Now we recall the definition (5.37a), (5.37b) of the traces γ1v|± and rewrite the first term of(5.46) as

⟨γ1v|− − γ1v|+ , Φ(x, ·)

⟩∂D

=⟨γ1v|− − γ1v|+ , Φx

⟩∂D

=

∫B

[∇v · ∇Φx − k2v Φx

]dx

= 〈ϕ, Φx〉∂D =⟨ϕ,Φ(x, ·)

⟩∂D

= (Sϕ)(x)

by (5.40) for ψ = Φx. Thus v = Sϕ− w in D. For x ∈ B \D we argue exactly in the sameway to show (5.46) which proves (5.45).Now we observe that v|∂B 7→ w|Q is bounded from L2(∂B) into H1(Q). Furthermore, ϕ 7→ vis bounded from H−1/2(∂D) into H1(B). Combining this with the boundedness of the traceoperator v 7→ v|∂B yields boundedness of S from H−1/2(∂D) into H1(Q).(c) From the representation (5.45) it suffices to show that γ1v|−− γ1v|+ = ϕ on ∂D becausew is a classical solution of the Helmholtz equation in B. By the definitions of γ1|± and of vwe have that 〈γ1v|−− γ1v|+, ψ〉∂D =

∫Q

[∇v · ∇ψ− k2v ψ

]dx = 〈ϕ, ψ〉∂D for all extensions ψ

of ψ with compact support. This proves part (c).(d) This follows from (b) and the boundedness of the trace operator γ0.(e) This follows directly from (c) and the definition of D′. 2

Corollary 5.43 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 (5.42); that is, u ∈ H1loc(R3) is the unique

radiating solution of∫R3

[∇u ·∇ψ−k2uψ

]dx = 〈ϕ, γ0ψ〉∂D for all ψ ∈ H1(R3) with compact support . (5.47)

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 in someball B. The definitions of γ1|± yield again

〈ϕ, ψ〉∂D = 〈γ1u|− − γ1u|+, ψ〉∂D =

∫B

[∇u · ∇ψ − k2uψ

]dx .

This proves that u = Sϕ solves (5.47). To prove uniqueness let ϕ = 0. Then∫R3

[∇u · ∇ψ − k2uψ

]dx = 0 for all ψ ∈ H1(R3) with compact support ;


that is, u is a radiating variational solution of the Helmholtz equation in all of R3. Theregularity result from Corollary 5.40 yields u ∈ C∞(R3). Also, u satisfies the radiationcondition which implies u = 0 in R3 by Theorem 3.23. 2

The following properties are helpful for using the boundary integral equation method forsolving the interior or exterior boundary value problems.

Theorem 5.44 Let Si be the single layer boundary operator for the special value k = i.Then:

(a) Si is symmetric and coercive in the sense that there exists c > 0 with

〈ϕ,Siφ〉∂D = 〈φ,Siϕ〉∂D , 〈ϕ,Siϕ〉∂D ≥ c‖ϕ‖2H−1/2(∂D) for all φ, ϕ ∈ H−1/2(∂D) .

In particular, 〈ϕ,Siϕ〉∂D is real valued.

(b) Si is an isomorphism from H−1/2(∂D) onto H1/2(∂D).

(c) S − Si is compact from H−1/2(∂D) onto H1/2(∂D) for any k ∈ C with Im k ≥ 0.

Proof: (a) Define u = Siϕ and v = Siψ in R3\∂D. As shown before, u is a classical solutionof ∆u − u = 0 in R3 \ ∂D and satisfies the radiation condition. From the representationtheorem (Theorem 3.3 for k = i) we observe that u and v and their derivatives decayexponentially for |x| → ∞. Let φ ∈ C∞(R3) with φ(x) = 1 for |x| ≤ R φ(x) = 0 for|x| ≥ R + 1 where R is chosen such that D ⊆ B(0, R). By the definitions (5.37a), (5.37b)of the normal derivatives (note that u|D, v|D ∈ HD and u|R3\D, v|R3\D ∈ HR3\D for k = i) weconclude that

〈ϕ,Siψ〉∂D = 〈γ1u|− − γ1u|+, γ0v〉∂D

=

∫D

[∇u · ∇v + uv

]dx +

∫R3\D

[∇u · ∇(φv) + φuv

]dx

=

∫B(0,R)

[∇u · ∇v + uv

]dx +

∫R<|x|<R+1

[∇u · ∇(φv) + φuv

]dx

=

∫B(0,R)

[∇u · ∇v + uv

]dx −

∫|x|=R

v∂u

∂νds

where we used the classical Green’s theorem in last step to the smooth functions u andφv. Now we use the fact that v and ∇u decay exponentially to zero as R tends to infinity.Therefore we arrive at

〈ϕ,Siψ〉∂D =

∫R3

[∇u · ∇v + uv

]dx

which is symmetric in u and v. To prove coercivity we choose ψ = ϕ. Then v = u and thus

〈ϕ,Siϕ〉∂D =

∫R3

[|∇u|2 + |u|2

]dx = ‖u‖2

H1(R3) .


Now we recall from Theorem 5.10 the existence of a bounded extension operator η : H1/2(∂D)→H1(B(0, R)

); that is γ0ηψ|± = ψ and ηψ has compact support in B(0, R). Therefore, for

ψ ∈ H1/2(∂D) and ψ = ηψ ∈ H1(B(0, R)

)we can estimate as above

〈ϕ, ψ〉∂D = 〈γ1u|− − γ1u|+, ψ〉∂D =

∫R3

[∇u · ∇ψ + uψ

]dx

≤ ‖u‖H1(R3) ‖ψ‖H1(B(0,R)) ≤ ‖η‖ ‖u‖H1(R3) ‖ψ‖H1/2(∂D) ,

and thus ‖ϕ‖H−1/2(∂D) = sup‖ψ‖

H1/2(∂D)=1

〈ϕ, ψ〉∂D ≤ ‖η‖ ‖u‖H1(R3). Therefore, 〈ϕ, Siϕ〉∂D =

‖u‖2H1(R3) ≥

1‖η‖2 ‖ϕ‖

2H−1/2(∂D)

. This finishes the proof of part (a).

(b) Injectivity of Si follows immediately from the coerciveness property of part (a). Fur-thermore, from part (a) we observe that (ϕ, ψ)S := 〈ϕ,Siψ〉∂D) defines an inner productin H−1/2(∂D) such that its corresponding norm is equivalent to the ordinary norm inH−1/2(∂D). Surjectivity of Si is now an immediate consequence of the Riesz Represen-tation Theorem 6.5. Indeed, let f ∈ H1/2(∂D). It defines a linear and bounded functional` on H−1/2(∂D) by `(ϕ) := 〈ϕ, f〉∂D) for ϕ ∈ H−1/2(∂D). By the theorem of Riesz (Theo-rem 6.5) there exists ψ ∈ H1/2(∂D) such that (ϕ, ψ)S = `(ϕ) for all ϕ ∈ H−1/2(∂D); that is,〈ϕ,Siψ〉∂D) = 〈ϕ, f〉∂D) for all ϕ ∈ H−1/2(∂D); that is, Siψ = f .

(c) Choose open balls Q and B such that D ⊆ Q and Q ⊆ B. We use that facts that Sϕand Siϕ have representations of the form (5.45); that is,

Sϕ = v + w and Siϕ = vi + wi in Q

where v, vi ∈ H1(B) are the solutions of (5.40) for k and k = i, respectively, and w,wi ∈H1(Q) are explicitely given by

w(x) =

∫∂B

v(y)

[∂Φk

∂ν(y)(x, y)− ikΦk(x, y)

]ds(y) , x ∈ Q ,

and analogously for wi (see proof of Theorem 5.42). We note that the mapping ϕ 7→ v isbounded from H−1/2(∂D) into H1(B). Therefore, by the trace theorem and the compactimbedding of H1/2(∂B) into L2(∂B) the mapping ϕ 7→ v|∂B is compact from H−1/2(∂D)into L2(∂B). This proves compactness of the mappings ϕ 7→ w|∂D from H−1/2(∂D) intoH1/2(∂D). It remains to consider v− vi. Taking the difference of the equations for v and vi;that is, ∫

B

[∇v · ∇ψ − k2v ψ

]dx − ik

∫∂B

v ψ ds = 〈ϕ, ψ〉∂D ,∫B

[∇vi · ∇ψ + vi ψ

]dx +

∫∂B

vi ψ ds = 〈ϕ, ψ〉∂D ,

yields∫B

[∇(v−vi)·∇ψ+(v−vi)ψ

]dx +

∫∂B

(v−vi)ψ ds = (k2+1)

∫B

v ψ dx + (ik+1)

∫∂B

v ψ ds


for all ψ ∈ H1(B) which is of the form∫B

[∇(v − vi) · ∇ψ + (v − vi)ψ

]dx +

∫∂B

(v − vi)ψ ds =

∫B

f ψ dx +

∫∂B

g ψ ds (5.48)

with f ∈ L2(B) and g ∈ L2(∂B). As mentioned before, the mapping ϕ 7→ v is bounded fromH−1/2(∂D) into H1(B) and thus compact as a mapping into L2(B) as well as the mappingϕ 7→ v∂B from H−1/2(∂D) into L2(∂B). Finally, the mapping (f, g) into the solution v − viof (5.48) is bounded from L2(B) × L2(∂B) into H1(B). This proves compactness of themapping ϕ 7→ v− vi from H−1/2(∂D) into H1(B). The trace theorem yields compactness ofS − Si and ends the proof. 2

Remarks 5.45 (i) The proof of part (b) implies in particular that the dual space ofH−1/2(∂D)can be identified with H1/2(∂D); that is, the spaces H1/2(∂D) and H−1/2(∂D) are reflexiveBanach spaces.

(ii) The symmetry of S holds for any k ∈ C \ 0 with Im k ≥ 0. Indeed, we can just copythe previous proof of part (a) and arrive at

〈ϕ,Sψ〉∂D =

∫B(0,R)

[∇u · ∇v − k2uv

]dx −

∫|x|=R

v∂u

∂νds

and thus

〈ϕ,Sψ〉∂D − 〈ψ,Sϕ〉∂D = −∫|x|=R

(v∂u

∂ν− u ∂v

∂ν

)ds .

It suffices to show that the last integral tends to zero as R→∞. To see this we just write∫|x|=R

(v∂u

∂ν− u ∂v

∂ν

)ds =

∫|x|=R

v

(∂u

∂ν− iku

)ds −

∫|x|=R

u

(∂v

∂ν− ikv

)ds ,

and this tends to zero by Sonmmerfeld’s radiation condition.

We finish this part of the section by formulating the corresponding theorems for the doublelayer potential without detailed proofs.

Theorem 5.46 Let Q be open and bounded such that D ⊆ Q. The double layer operatorD, defined in (5.39b), is well-defined and bounded from H1/2(∂D) into H1(D) and intoH1(Q \D).Furthermore, with u := Dϕ in R3 \ ∂D we have that u|D ∈ HD and u|R3\D ∈ HR3\D and

γ0u|+ − γ0u|− = ϕ and γ1u|− − γ1u|+ = 0. In particular, u ∈ C∞(R3 \ ∂D) and satisfies theHelmholtz equation ∆u+ k2u = 0 in R3 \ ∂D and the Sommerfeld radiation condition. Thetraces

T = γ1D : H1/2(∂D)→ H−1/2(∂D) , (5.49a)

D =1

2(γ0D|+ + γ0D|−) : H1/2(∂D)→ H1/2(∂D) (5.49b)


are well defined and bounded. Furthermore, we have the jump conditions for u = Dϕ andϕ ∈ H1/2(∂D):

γ0u|± = ± 1

2ϕ + Dϕ , γ1u|± = T ϕ . (5.50)

Proof (only sketch): We follow the proof of Theorem 5.42 and choose a ball B with D ⊆ Band prove a decomposition of u = Dϕ in the form u = v + w where the pair

(v|D, v|B\D

)∈

H1(D)×H1(B \D) is the variational solution of the transmission problem

∆v + k2v = 0 in B \ ∂D ,∂v

∂ν− ikv = 0 on ∂B ,

∂v

∂ν

∣∣∣∣+

=∂v

∂ν

∣∣∣∣−, v|+ − v|− = ϕ on ∂D .

With this v we define w just as in the proof of Theorem 5.42. The mapping properties of Dand T follow now directly from this representation and the trace theorems. Finally, (5.50)follows directly from γ0u|+ − γ0u|− = ϕ and the definition of Dϕ. 2

We note that – in contrast to the case of a smooth boundary – the operator D is, in general,not compact anymore. Similarly to Theorem 5.44 the properties of the operator T withrespect to the special wave number k = i are useful. We leave the proof of the followingresult to the reader because it follows the same arguments as in the proof of Theorem 5.44.

Theorem 5.47 Let Ti be the normal derivative of the double layer boundary operator forthe special value k = i. Then:

(a) Ti is symmetric and coercive in the sense that there exists c > 0 with

〈Tiϕ, φ〉∂D = 〈Tiφ, ϕ〉∂D , 〈Tiϕ, ϕ〉∂D ≤ −c‖ϕ‖2H1/2(∂D) for all φ, ϕ ∈ H1/2(∂D) .

In particular, 〈Tiϕ, ϕ〉∂D is real valued.

(b) Ti is an isomorphism from H1/2(∂D) onto H−1/2(∂D).

(c) T − Ti is compact from H1/2(∂D) into H−1/2(∂D) for any k ∈ C with Im k ≥ 0.

We continue with the vector valued case. Let now ε, µ ∈ C \ 0 be constant and ω > 0.Define the wave number k ∈ C with Re k ≥ 0 and Im k ≥ 0 by k2 = ω2εµ. We recall thetrace operators 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 the unit normal vector to point into the exteriorof D and distinguish in the following between the traces from the exterior (+) and interior(−) by writing γtu|± and γTu|±, respectively. Due to the direction of ν we have the followingforms of Green’s formula 5.19.

〈γtv|−, γTu|−〉∂D =

∫D


]dx for all u, v ∈ H(curl, D) ,

〈γtv|+, γTu|+〉∂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)

⟩, see Theorem 5.26.

Again we changed the notation by writing 〈·, ·〉∂D instead of 〈·, ·〉∗.

In the following we want to discuss vector potentials of the form

curl

∫∂D

a(y) Φ(x, y) ds(y) and curl curl

∫∂D

a(y) Φ(x, y) ds(y) for x /∈ ∂D ,

see Subsection 3.2.2. From the trace theorem we know that in general we have to allow a to bein H−1/2(Div, ∂D). Therefore, we have to give meaning to the boundary integral. We recallfrom Lemma 5.27 that H−1/2(Div, ∂D) can be considered as a subspace of H−1/2(∂D,C3)which is the dual space of H1/2(∂D,C3). The dual form is denoted by 〈·, ·〉∂D, see thebeginning of this section. We define the bilinear mapping

〈·, ·〉∂D : H−1/2(Div, ∂D)×H1/2(∂D) −→ C3

componentwise; that is,(〈a, ϕ〉∂D

)j

= 〈aj, ϕ〉∂D, j = 1, 2, 3, for a ∈ H−1/2(Div, ∂D) and

ψ ∈ H1/2(∂D). We refer to the remark following Lemma 5.27. For smooth tangential fieldsa and ϕ this is exactly the integral

∫∂Dϕ(y) a(y) ds(y). With this bilinear mapping we have

the following form of Green’s theorem.

Lemma 5.48 For v ∈ H(curl, D) and ψ ∈ H1(D) we have

〈γtv, γ0ψ〉∂D =

∫D

[ψ curl v +∇ψ × v

]dx .

Proof: For any z ∈ C3 we have 〈γtv, γ0ψ〉∂D · z = 〈γtv, γT (ψz)〉∂D which can be seen byapproximating v and ψ by smooth functions. Therefore, by (5.19),

〈γtv, γ0ψ〉∂D · z = 〈γtv, γT (ψz)〉∂D =

∫D

[ψ z · curl v − v · curl(ψz)

]dx

=

∫D

[ψ z · curl v − v · (∇ψ × z)

]dx = z ·

∫D

[ψ curl v − v ×∇ψ

]dx .

2

Now we are able to generalize the definition of the vector potentials of Subsection 3.2.2 forany a ∈ H−1/2(Div, ∂D). We define

v(x) = curl 〈a,Φ(x, ·)〉∂D , x ∈ R3 \ ∂D , (5.51)

u(x) = curl2〈a,Φ(x, ·)〉∂D , x ∈ R3 \ ∂D , (5.52)

where we dropped the symbol γ0 for the scalar trace operator.

The starting point for our discussion of the vector potentials is the following version of therepresentation theorem (compare to the Stratton-Chu formula of Theorem 3.27).


Theorem 5.49 (Stratton-Chu formulas)

Let again Φ(x, y) be the fundamental solution from (5.38) wherel k = ω√εµ with Re k ≥ 0

and Im k ≥ 0.(a) For any solutions E,H ∈ H(curl, D) of curlE − iωµH = 0 and curlH + iωεE = 0 in Dwe have the representation

− curl〈γtE,Φ(x, ·)〉∂D +1

iωεcurl2〈γtH,Φ(x, ·)〉∂D =

E(x) , x ∈ D ,

0 , x 6∈ D .(5.53a)

(b) For any solutions E,H ∈ Hloc(curl,R3 \D) of curlE− iωµH = 0 and curlH+ iωεE = 0in R3 \D which satisfy the Silver–Muller radiation condition (3.37a); that is,

√εE(x) − √µH(x)× x

|x|= O

(1

|x|2

), |x| → ∞ , (5.53b)

uniformly with respect to x = x/|x|, we have the representation

curl〈γtE,Φ(x, ·)〉∂D −1

iωεcurl2〈γtH,Φ(x, ·)〉∂D =

0 , x ∈ D ,

E(x) , x 6∈ D .(5.53c)

Proof: We prove only part (a). First we note that E and H are smooth solutions ofcurl2 u− k2u = 0 in D. We fix z ∈ D and choose a ball Br = B(z, r) such that Br ⊆ D. Weapply Green’s formula of Lemma 5.48 in Dr = D \ B[z, r] to E and Φ(x, ·) for any x ∈ Br

and obtain

〈γtE,Φ(x, ·)〉∂D −∫∂Br

(ν × E) Φ(x, ·) ds =

∫Dr

[Φ(x, ·) curlE +∇yΦ(x, ·)× E

]dy .

Analogously, we have for H instead of E:

〈γtH,Φ(x, ·)〉∂D −∫∂Br

(ν ×H) Φ(x, ·) ds =

∫Dr

[Φ(x, ·) curlH +∇yΦ(x, ·)×H

]dy

and thus

Ir(x) := − curl

[〈γtE,Φ(x, ·)〉∂D −

∫∂Br

(ν × E) Φ(x, ·) ds]

+1

iωεcurl2

[〈γtH,Φ(x, ·)〉∂D −

∫∂Br

(ν ×H) Φ(x, ·) ds]

= − curl

∫Dr

[Φ(x, ·) curlE +∇yΦ(x, ·)× E

]dy

+1

iωεcurl2

∫Dr

[Φ(x, ·) curlH +∇yΦ(x, ·)×H

]dy


Now we use that ∇yΦ(x, ·)× E = − curlx(Φ(x, ·)E

), curlE = iωµH, and curlH = −iωεE.

This yields

Ir(x) = curl2∫Dr

Φ(x, ·)E dy − iωµ curl

∫Dr

Φ(x, ·)H dy

− 1

iωεcurl3

∫Dr

Φ(x, ·)H dy − curl2∫Dr

Φ(x, ·)E dy

= − iωµ curl

[∫Dr

Φ(x, ·)H dy − 1

k2curl2

∫Dr

Φ(x, ·)H dy

]= 0

by using curl2 = ∇ div−∆ and the Helmholtz equation for Φ. Therefore,

− curl[〈γtE,Φ(x, ·)〉∂D +1

iωεcurl2〈γtH,Φ(x, ·)〉∂D

= − curl

∫∂Br

(ν × E) Φ(x, ·) ds +1

iωεcurl2

∫∂Br

(ν ×H) Φ(x, ·) ds

= E(x)

by the classical Stratton-Chu formula of Theorem 3.27. The case x /∈ D is treated in thesame way by appling Lemma 5.48 in all of D. 2

Corollary 5.50 For any variational solution E ∈ H(curl, D) of curl2E− k2E = 0 in D wehave the representation

− curl〈γtE,Φ(x, ·)〉∂D −1

k2curl2〈γt curlE,Φ(x, ·)〉∂D =

E(x) , x ∈ D ,

0 , x 6∈ D .

Proof: We define H = 1iωµ

curlE and observe that also H ∈ H(curl, D), and E,H satisfy

the assumptions of the previous theorem. Substituting the form of H into (5.53a) yields theassertion. 2

For our proof of the boundedness of the vector potentials and the corresponding boundaryoperators we need the (unique) solvability of a certain boundary value problem of trans-mission type – just as in the scalar case. As a familar technique we will use the Helmholtzdecomposition in the form

H(curl, B) = H(curl, div 0, B) ⊕ ∇H1(B) (5.54)

in some ball B where H(curl, div 0, B) =u ∈ H(curl, B) :

∫Bu · ∇ϕdx = 0 for all ϕ ∈

H1(B)

. We have seen in Theorem 5.37 that H(curl, div 0, B) is compactly imbedded inL2(B,C3).

Theorem 5.51 Let B1 and B2 be two open balls such that B1 ⊆ D and D ⊆ B2. Let η ∈ Cwith Im (ηk) > 0 and Kj : H−1/2(Div, ∂Bj) → H−1/2(Curl, ∂Bj) for j ∈ 1, 2 be linear


and compact operators such that 〈ψ,Kjψ〉∂Bj are real valued and 〈ψ,Kjψ, 〉∂Bj > 0 for all

ψ ∈ H−1/2(Div, ∂Bj), ψ 6= 0, j = 1, 2. Then, for every a ∈ H−1/2(Div, ∂D) the followingboundary value problem is uniquely solvable in H(curl, B) where we set B = B2 \ B1 forabbreviation.

curl2 v − k2v = 0 in B \ ∂D ,

ν × v|− = ν × v|+ on ∂D , ν × curl v|+ − ν × curl v|− = a on ∂D ,

ν × curl v − η ν ×K2(ν × v) = 0 on ∂B2 , ν × curl v + η ν ×K1(ν × v) = 0 on ∂B1 ;

that is, in variational form:∫B

[curl v · curlψ − k2v · ψ

]dx − η

2∑j=1

⟨γtψ,Kjγtv

⟩∂Bj

= 〈a, γTψ〉∂D (5.55)

for all ψ ∈ H(curl, B). The operator a 7→ γtv|∂D is an isomorphism from H−1/2(Div, ∂D)onto itself.

Proof: We make use of the Helmholtz decomposition (5.54). Setting v = v0 − ∇p andψ = ψ0 + ∇ϕ with v0, ψ0 ∈ H(curl, div 0, B) and p, ϕ ∈ H1(B), the variational equation(5.55) is equivalent to∫

B

[curl v0 · curlψ0 − k2v0 · ψ0 + k2∇p · ∇ϕ

]dx − η

2∑j=1

⟨γt(ψ0 +∇ϕ), Kjγt(v0 −∇p)

⟩∂Bj

= 〈a, γT (ψ0 +∇ϕ)〉∂D (5.56)

for all (ψ0, ϕ) ∈ X := H(curl, div 0, B) ×H1(B). We equip X with the norm ‖(ψ0, ϕ)‖2X =

‖ψ0‖2H(curl,) + k2‖ϕ‖2

H1(B) and denote the corresponding inner product by (·, ·)X . Then the

variational equation (5.56) can be written as

((v0, p), (ψ0, ϕ)

)X−∫B

[(k2 + 1)v0 ψ0 + k2pϕ

]dx − η

2∑j=1


⟩∂Bj

= 〈a, γT (ψ0 +∇ϕ)〉∂D for all (ψ0, ϕ) ∈ X . (5.57)

The representation theorem of Riesz (Theorem 6.5) guarantees the existence of (g, q) ∈ Xand a bounded operator A from X into itself such that(

(g, q), (ψ0, ϕ))X

= 〈a, γT (ψ0 +∇ϕ)〉∂D(A(v0, p), (ψ0, ϕ)

)X

=

∫B

[(k2 + 1)v0 ψ0 + k2pϕ

]dx

+ η

2∑j=1


⟩∂Bj

for all (v0, p), (ψ0, ϕ) ∈ X. Therefore, (5.57) can be written as

(v0, p) − A(v0, p) = (g, q) in X .


From the estimate∣∣(A(v0, p), (ψ0, ϕ))X

∣∣ ≤ (k2 + 1)‖(v0, p)‖L2(B)×L2(B) ‖(ψ0, ϕ)‖L2(B)×L2(B)

+ c1

2∑j=1

‖Kjγt(v0 −∇p)‖H−1/2(Curl,∂Bj)‖γt(ψ0 +∇ϕ)‖H−1/2(Div,∂Bj)

≤ c

[‖(v0, p)‖L2(B)×L2(B) +

2∑j=1

‖Kjγt(v0 −∇p)‖H−1/2(Curl,∂Bj)

]‖(ψ0, ϕ)‖X

we conclude (set (ψ0, ϕ) = A(v0, p)) that

‖A(v0, p)‖X ≤ c[‖(v0, p)‖L2(B)×L2(B) +

2∑j=1

‖Kjγt(v0 −∇p)‖H−1/2(Curl,∂Bj)

].

From this and the compact imbedding of X in L2(B,C3)×L2(B) and the compactness of Kj

we conclude that A is compact from X into itself. Therefore, existence of a solution of (5.57)holds once uniqueness has been shown. To prove uniqueness, let a = 0 and v ∈ H(curl, B)a corresponding solution. Substituting ψ = kv in (5.55) and taking the imginary part yields∑2

j=1

⟨γtv,Kjγtv

⟩∂Bj

= 0 and thus γtv = 0 on ∂B1 ∪ ∂B2. Now we extend v by zero into all

of R3. Then v ∈ H(curl,R3). Furthermore, (5.55) takes the form∫R3

[curl v · curlψ − k2v · ψ

]dx = 0 for all ψ ∈ H(curl,R3) ;

that is, v solves curl2 v− k2v = 0 in R3 and vanishes outside of B. The unique continuationprinciple yields v = 0 in R3.It remains to show that the operator a 7→ γtv|∂D is an isomorphism from H−1/2(Div, ∂D) ontoitself. This follows immediately from the unique solvability of the following two boundaryvalue problems for any given b ∈ H−1/2(Div, ∂D):

curl2 vi − k2vi = 0 in D \B1 , ν × vi = b on ∂D , (5.58a)

ν × curl vi + η ν ×K1(ν × vi) = 0 on ∂B1 , (5.58b)

andcurl2 ve − k2ve = 0 in B2 \D , ν × ve = b on ∂D ,

ν × curl ve − η ν ×K2(ν × ve) = 0 on ∂B2 .

We leave the proof of this fact to the reader (see Exercise 5.7). 2

The previous theorem requires the existence of linear and compact operatorsKj : H−1/2(Div, ∂Bj)→ H−1/2(Curl, ∂Bj) such that 〈ψ,Kjψ〉∂Bj > 0 for all ψ ∈ H−1/2(Div, ∂Bj),ψ 6= 0. Such operators exist. For example, consider the case that ∂Bj = S2 and the operatorKj is defined by

Kjψ =∞∑n=0

n∑m=−n

1

1 + n(n+ 1)

[amn U

mn + bmn V

mn

]


for ψ =∑∞

n=0

∑nm=−n

[amn U

mn +bmn V

mn

]∈ H−1/2(Div, S2). This Kj has the desired properties,

see Exercise 5.6.

Now we are able to prove all of the desired properties of the vector potentials and theirbehaviours at the boundary.

Theorem 5.52 Let k ∈ C\0 with Im k ≥ 0 and Q a bounded domain such that ∂D ⊆ Q.

(a) The operators L and M, defined by

(La)(x) = curl2〈a,Φ(x, ·)〉∂D for x ∈ Q ,

(Ma)(x) = curl〈a,Φ(x, ·)〉∂D for x ∈ Q ,

are well-defined and bounded from H−1/2(Div, ∂D) into H(curl, Q).

(b) For a ∈ H−1/2(Div, ∂D) the fields u = Ma and curlu = La satisfy u|D, curlu|D ∈H(curl, D) and u|Q\D, curlu|Q\D ∈ H(curl, Q\D) and γtu|−−γtu|+ = a and γt curlu|−−γt curlu|+ = 0. In particular, u ∈ C∞(R3 \ ∂D,C3) and u satisfies the equationcurl2 u − k2u = 0 in R3 \ ∂D. Furthermore, u and curlu satisfy the Silver–Mullerradiation condition (3.42); that is,

curlu× x − ik u = O(|x|−2

), |x| → ∞ ,

uniformly with respect to x.

(c) The traces

L = γtL on ∂D , (5.59a)

M =1

2(γtM|− + γtM|+) on ∂D , (5.59b)

are bounded from H−1/2(Div, ∂D) into itself. With these notations the jump conditionshold for u = Ma and a ∈ H−1/2(Div, ∂D) in the form

γtu|± = ∓1

2a + Ma , γt curlu|± = La . (5.60)

(d) L is the sum L = L+ K of an isomorphism L from H−1/2(Div, ∂D) onto itself and acompact operator K.

(e) La can be written as

La = ∇S Div a + k2 Sa , a ∈ H−1/2(Div, ∂D) , (5.61)

with the single layer Sa from (5.39a) (in the second occurence taken componentwise).


We note that some authors call La the Maxwell single layer and Ma the Maxwell doublelayer. The matrix operator

C =

(12

+M LL 1

2+M

)is called the electromagnetic Calderon operator.

Proof: (a)–(d)We argue similarily as in the proof of Theorem 5.42. It is sufficient to provethis for Q being a neighborhood of ∂D. Therefore, let B1 and B2 be two open balls suchthat B1 ⊆ D and D ⊆ B2 and let Q be an open set with ∂D ⊆ Q. Choose η ∈ C withIm (ηk) > 0 and Kj : H−1/2(Div, ∂Bj)→ H−1/2(Curl, ∂Bj) for j ∈ 1, 2 as in the previoustheorem. For a ∈ H−1/2(Div, ∂D) let v ∈ H(curl, B) be the solution of (5.55). We prove thefollowing representation of u = Lf .

La = k2 (v + w) in B (5.62)

where

w(x) =2∑j=1

(−1)j[curl〈γtv,Φ(x, ·)〉∂Bj +

1

k2curl2〈γt curl v,Φ(x, ·)〉∂Bj

]

for x ∈ B2\B1. To prove this we fix x ∈ D\B1 and apply Corollary 5.50 of the Stratton-Chuformula in D \ B1 and in B2 \D, respectively, and obtain (note the different signs becauseof 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〈a,Φ(x, ·)〉∂D − w(x) =

1

k2(La)(x) − w(x) ,

which proves (5.62) for x ∈ D. For x ∈ B2 \ D we argue analogously. Taking the trace in(5.62) yields

La = k2 (γtv + γtw) = La + Ka .

L is an isomorphism by Theorem 5.51, and K is compact by the smoothness of w. Finally,the properties of M and M follow from the relation curl La = curl2 Ma = k2Ma, thusMa = curl v + curlw, and the trace theorem.(e) Using curl curl = ∇ div−∆ and the Helmholtz equation for Φ and the fact that we can

5.3. BOUNDARY INTEGRAL EQUATION METHODS 263

differentiate with respect to the parameter in the dual form (see the proof of Corollary 5.40for the arguments) yields for x /∈ ∂D

La(x) = ∇ div〈a,Φ(x, ·)〉∂D − ∆〈a,Φ(x, ·)〉∂D

= ∇3∑j=1

⟨aj,

∂

∂xjΦ(x, ·)

⟩∂D

+ k2〈a,Φ(x, ·)〉∂D

= ∇〈a,∇xΦ(x, ·)〉∂D + k2〈a,Φ(x, ·)〉∂D

= −∇〈a,∇yΦ(x, ·)〉∂D + k2〈a,Φ(x, ·)〉∂D .

By the identification of 〈a,∇yΦ(x, ·)〉∂D with 〈a, γT∇yΦ(x, ·)〉∂D (see Remark 5.28) and thedefinition of the surface divergence (Definition 5.29) we conclude that 〈a,∇yΦ(x, ·)〉∂D =−〈Div a,Φ(x, ·)〉∂D which proves the representation (5.61). 2

5.3 Boundary Integral Equation Methods

We begin again with the scalar case and formulate the interior and exterior boundary valueproblems. We assume that D ⊆ R3 is a Lipschitz domain in the sense of Definition 5.1.Furthermore, we assume that the exterior R3 \D of D is connected. Let f ∈ H1/2(∂D) begiven bounday data.

Interior Dirichlet Problem: Find u ∈ H1(D) such that γ0u = f and ∆u + k2u = 0 in D;that is, in variational form∫

D

[∇u · ∇ψ − k2uψ


0 (D) . (5.63a)

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 γ0u = f and ∆u+ k2u = 0 in

R3 \D; that is, in variational form∫R3\D

[∇u · ∇ψ − k2uψ


0 (R3 \D) with compact support, (5.63b)

and u satisfies Sommerfeld’s radiation condition (5.42). Note that u is a smooth solution ofthe Helmholtz equation in the exterior of D.

First we consider the question of uniqueness.

Theorem 5.53 (a) There exists at most one solution of the exterior Dirichlet boundaryvalue problem (5.63b).


(b) The interior Dirichlet boundary value problem (5.63a) has at most one solution if, andonly if, the single layer boundary operator S is one-to-one. More precisely, the null spaceN (S) of S is given by all Neumann traces γ1u of solutions u ∈ H1

0 (D) of the Helmholtzequation with vanishing Dirichlet boundary data; that is,

N (S) =

γ1u : u ∈ H1

0 (D),

∫D

[∇u · ∇ψ − k2uψ


0 (D)

.

Proof: (a) Let u ∈ H1loc(R3 \ D) be a solution of the exterior Dirichlet boundary value

problem for f = 0. Choose a ball B(0, R) which contains D in its interior and a functionφ ∈ C∞(R3) such that φ = 1 on B[0, R] und φ = 0 in the exterior of B(0, R+1). Applicationof Green’s formula (5.19) in the region B(0, R + 1) \D to u and ψ = φu yields

0 = −〈γ1u, 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 · ∇ψ − k2uψ

]dx

=

∫B(0,R)\D

[|∇u|2 − k2|u|2

]dx −

∫|x|=R

∂u

∂νu ds .

Note that ψ vanishes on ∂D and outside of B(0, R + 1). Now we proceed similarly to theproof of Theorem 3.23. Indeed,

∫|x|=R

∣∣∣∣∂u∂r − iku∣∣∣∣2 ds =

∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + |ku|2

ds − 2 Im

k ∫|x|=R

u∂u

∂rds

=

∫|x|=R

∣∣∣∣∂u∂r∣∣∣∣2 + |ku|2

ds + 2 Im k

∫B(0,R)\D

[|∇u|2 + |ku|2

]dx .

Now we have to distinguish between two cases. If Im k > 0 then u vanishes in B(0, R) \D.If k is real valued then the left–hand side of the previous equation converge to zero as Rtends to infinity by the radiation condition. Rellich’s lemma (Lemma 3.21 or 3.22) impliesthat u vanishes in the (connected) exterior of D. This proves part (a).

(b) Let first u ∈ H1(D) satisfy (5.63a) and γ0u = 0. Then, by the definition of S and Green’srepresentation formula of Theorem 5.39

(Sγ1u)(x) =⟨γ1u,Φ(x, ·)

⟩∂D

=⟨γ1u,Φ(x, ·)

⟩∂D−∫∂D

(γ0u)(y)∂Φ


= u(x) , x ∈ D .

Taking the trace yields Sγ1u = γ0Sγ1u = γ0u = 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 γ0u|+ = 0 on ∂D. Therefore, u|R3\D satisfies the homogeneous exterior boundary valueproblem and therefore vanishes by part (a). The jump condition of Theorem 5.42 impliesϕ = γ1u|− − γ1u|+ = γ1u|− which ends the proof. 2.

Remark: This theorem relates the homogeneous interior Dirichlet problem to the null spaceof the boundary operator S. Therefore, k2 is a Dirichlet eigenvalue of −∆ in the sense ofSubsection 4.2.1, see Theorem 4.28, if and only if S fails to be one-to-one.In particular, there exist only a countable number of values k for which S fails to be one-to-one. Furthermore, the null space N (S) is finite dimensional.

The following theorem studies the question of existence for the case that k2 is not a Dirichleteigenvalue.

Theorem 5.54 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 (5.63a) in H1

0 (D) is the trivial one u = 0.Then there exist (unique) solutions of the exterior and the interior Dirichlet boundary valueproblems for every f ∈ H1/2(∂D). The solutions can be represented as single layer potentialsin 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 5.42 it suffices to study solvability of theequation Sϕ = f . Because S is a compact perturbation of the isomorphism Si (the operatorcorresponding to k = i) by Theorem 5.42, the well known – and already often used – resultby Riesz (Theorem 6.5) 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 5.42 again) yields ϕ = 0. 2

Furthermore we consider the case of k2 being a Dirichlet eigenvalue of −∆ in D. Thiscorresponds to the case where S fails to be one-to-one. In this case we have to apply theFredholm theory of Theorem 6.4 to the boundary equation Sϕ = f .

Theorem 5.55 The interior and exterior boundary value problems are solvable as singlelayer potentials u = Sϕ for exactly those f ∈ H1/2(∂D) which are orthogonal to all Neumanntraces γ1v ∈ H−1/2(∂D) of eigenfunctions v ∈ H1

0 (D) of −∆ in D; that is,

〈γ1v, f〉∂D = 0 for all v ∈ H10 (D) with ∆v + k2v = 0 in the variational sense.

Proof: We have to study solvability of the equation Sϕ = f and want to apply Theorem 6.4.Therefore, we set X1 = Y2 = H−1/2(∂D), X2 = Y1 = H1/2(∂D), 〈ϕ, f〉1 = 〈ϕ, f〉∂D for ϕ ∈


H−1/2(∂D), f ∈ H1/2(∂D) and 〈g, ψ〉2 = 〈ψ, g〉∂D for ψ ∈ H−1/2(∂D), g ∈ H1/2(∂D). Theoperator S is self adjoint in this dual system, see Remarks 5.45. Application of Theorem 6.4yields solvability of the equation Sϕ = f for exactly those f ∈ H1/2(∂D) with 〈ψ, f〉∂D = 0for all ψ ∈ N (S∗) = N (S). The characterization of the nullspace N (S) in Theorem 5.53yields the assertion. 2

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 satisfies theorthogonality condition. Indeed, by Green’s formula (5.8) we conclude for every eigenfunc-tion v ∈ H1

0 (D) that

〈γ1v, 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.

From the uniqueness result (and Chapter 3) we expect that the exterior boundary valueproblem is always uniquely solvable, independently of the wave number. However, this can’tbe done by just one single layer potential. There are several ways to modify the ansatz. In ourcontext perhaps the simplest possibility is to choose an ansatz as the following combinationof a single and a double layer ansatz

u = Sϕ + η DSiϕ in the exterior of D

for some ϕ ∈ H−1/2(∂D) and some η ∈ C to be chosen in a moment. The boundary operatorSi : H−1/2(∂D)→ H1/2(∂D) corresponds to S for k = i. Then u is a solution of the exteriorboundary value problem provided ϕ ∈ H−1/2(∂D) solves

Sϕ +η

2Siϕ + ηDSiϕ = f on ∂D . (5.64)

By Theorem 5.44 the operator S is a compact perturbation of the isomorphism Si fromH−1/2(∂D) onto H1/2(∂D) and the operator DSi is bounded. Therefore, for sufficientlysmall η0 > 0 also S+ η

2Si +ηDSi is a compact perturbation of an isomorphism for all η ∈ C

with |η| ≤ η0. We choose η ∈ C with |η| ≤ η0 and Im (kη) > 0 and show that for thischoice of η the boundary equation (5.64) is uniquely solvable for every f ∈ H1/2(∂D). Bythe previous remarks it is sufficient to show that the homogeneous equations admits only thetrivial solution. Therefore, let ϕ ∈ H−1/2(∂D) be a solution of (5.64) for f = 0 and defineu by u = Sϕ + η DSiϕ in all of R3 \ ∂D. Then u|+ vanishes by the jump conditions and(5.64). The uniqueness result of the exterior Dirichlet boundary value problem yields thatu vanishes in the exterior of D. The jump conditions yield

γ0u|− = γ0u|− − γ0u|+ = −η Siϕ , γ1u|− = γ1u|− − γ1u|+ = ϕ .

Elimination of ϕ yields ηSiγ1u|− + γ0u|− = 0. Now we apply Green’s theorem in D; that is,∫D

[∇u · ∇ψ − k2uψ

]dx = 〈γ1u|−, γ0ψ〉∂D .


Substituting ψ = ku yields∫D

[k |∇u|2 − k |ku|2

]dx = k 〈γ1u|,γ0u|−〉∂D = −kη 〈γ1u|−,Siγ1u|−〉∂D .

Now we note that 〈γ1u|−,Siγ1u|−〉∂D is real and nonnegative by Theorem 5.44 and Im (kη) >0 by assumption. Therefore, taking the imaginary part yields 〈γ1u|−,Siγ1u|−〉∂D = 0. Theproperty of Si from Theorem 5.44 yields γ1u|− = 0 and thus also ϕ = γ1u|− − γ1u|+ = 0.We formulate the result as a theorem.

Theorem 5.56 The exterior Dirichlet boundary value problem is uniquely solvable for everyf ∈ H1/2(∂D). The solution can be expressed in the form

u = Sϕ + η DSiϕ in R3 \D

for some sufficiciently small η such that Im (kη) > 0 and the density ϕ ∈ H−1/2(∂D) solves(5.64).

We note that for real (and positive) values of k we can choose η with Im η > 0 independent ofk. Indeed, one can show by the same arguments as in the proof of Theorem 5.44 that D−Diis compact. Therefore, we have to choose |η| small enough such that (1 + η/2)Si + ηDiSi isan isomorphism from H−1/2(∂D) onto H1/2(∂D).

Exactly the same results hold for the interior and exterior Neumann problems. We formulatethe results in two theorems but leave the proofs to the reader.

Theorem 5.57 The interior Neumann problem is solvable exactly for those f ∈ H−1/2(∂D)with 〈f, γ0v〉∂D = 0 for all solutions v ∈ H1(D) of ∆v + k2v = 0 in D and γ1v = 0 on ∂D.In this case the solution can be represented as a double layer potential u = Dϕ in D withϕ ∈ H1/2(∂D) satisfying T ϕ = f on ∂D where T : H1/2(∂D) → H−1/2(∂D) denotes thetrace of the normal derivative of the double layer potential, see Theorem 5.46.

Theorem 5.58 The exterior Neumann problem is always uniquely solvable for every f ∈H−1/2(∂D). The solution can be represented in the form

u = Dϕ + ηSTiϕ in R3 \D ,

with ϕ ∈ H1/2(∂D) for some sufficiently small η ∈ C with Im (ηk) > 0. The densityϕ ∈ H1/2(∂D) satisfies the boundary equation

T ϕ − η

2Tiϕ + D′Tiϕ = f on ∂D .

We now turn to the electromagnetic case and formulate the boundary value problems. Weassume again that D ⊆ R3 is a bounded Lipschtiz domain with connected exterior R3 \D.Furthermore, let f ∈ H−1/2(Div, ∂D) be given bounday data. We recall that the trace


operator γt : H(curl, D)→ H−1/2(∂D) extends the mapping u 7→ ν × u|∂D (and analogouslyfor exterior domains). Furthermore, let ω > 0, µ > 0, and ε ∈ C with Im ε ≥ 0 andk = ω

√µε with Re k > 0 and Im k ≥ 0.

The Interior Boundary Value Problem: Find E,H ∈ H(curl, D) such that γtE = f and

curlE − iωµH = 0 in D and curlH + iωεE = 0 in D ; (5.65a)

that is, in variational form for the field E (see, e.g., (4.2) or (4.22)).∫D

[curlE · curlψ − k2E · ψ

]dx = 0 for all ψ ∈ H0(curl, D) . (5.65b)

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

.

The Exterior Boundary Value Problem: Find E,H ∈ Hloc(curl,R3 \D) such that γtE = fand

curlE − iωµH = 0 in R3 \D and curlH + iωεE = 0 in R3 \D ; (5.66a)

that is, in variational form for the field E,∫R3\D

[curlE · curlψ−k2E ·ψ

]dx = 0 for all ψ ∈ H0(curl,R3 \D) with compact support .

(5.66b)Furthermore, (E,H) have to satisfy the Silver–Muller radiation condition (5.53b); that is,

√εE(x) − √µH(x)× x

|x|= O

(1

|x|2

)(5.66c)

uniformly with respect to x/|x| ∈ S2. Note that E and H are a smooth solutions of 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 5.59 (a) There exists at most one solution of the exterior boundary value problem(5.66a)–(5.66c).

(b) The interior boundary value problem (5.65a)–(5.65b) has at most one solution if, andonly if, the boundary operator L is one-to-one. More precisely, the null space N (L) of L isgiven by all traces γt curlu of solutions u ∈ H0(curl, D) of curl2 u− k2u = 0 with vanishingboundary data γtu; that is,

N (L) =

γt curlu : u ∈ H0(curl, D),

∫D

[curlu · curlψ − k2u · ψ

]dx = 0

for all ψ ∈ H0(curl, D)

.


Proof: (a) Let E,H be a solution of (5.65a)–(5.65b) corresponding to f = 0. Choose againballs such that D ⊆ B(0, R) ⊆ B(0, R + 1) and a function φ ∈ C∞(R3) such that φ = 1on B[0, R] and φ = 0 outside of B(0, R + 1). Green’s formula (5.19) applied to u = φEand v = H in the region B(0, R + 1) \ D yields (note that γTu vanishes on ∂D and on∂B(0, R + 1))

0 = −〈γtv, γTu〉∂D +

∫|x|=R+1

u · (ν × v) ds

=

∫B(0,R+1)\D

[φE · curlH −H · curl(φE)

]dx

= iω

∫B(0,R)\D

[ε |E|2 − µ |H|2

]dx +

∫R<|x|<R+1


]dx

= iω

∫B(0,R)\D

[ε |E|2 − µ |H|2

]dx −

∫|x|=R

E · (x×H) ds .

We consider again the case Im ε > 0 and Im ε = 0 separately. If Im ε > 0 then the fieldsdecay exponentially as R tends to infinity. This follows again from the Stratton-Chu formulaof Theorem 5.49. Taking the imaginary part of the previous formula and letting R tend toinfinity yields that E vanishes.Let now ε be real valued. Taking the real part of the previous formula yields

Re

∫|x|=R

E · (x×H) ds ≥ 0 .

This yields

0 ≥ 2√µ√ε Re

∫|x|=R

E · (H × x) ds

=

∫|x|=R

|√εE|2 + |√µH × x|2

ds −

∫|x|=R

|√εE −√µH × x|2ds .

The Silver–Muller radiation condition implies that the second integral tends to zero as Rtends to infinity. This implies that

∫|x|=R |E|

2ds tends to zero. Now we proceed as in the

proof of Theorem 3.35 and apply Rellich’s lemma to conclude that E and H vanish in theexterior of D.

(b) Let first u ∈ H0(curl, D) with curl2 u− k2u = 0 in D. By Corollary 5.50 we conclude forx ∈ D, because γtu = 0,

(Lγt curlu)(x) = curl2〈γt curlu,Φ(x, ·)〉∂D

= curl2〈γt curlu,Φ(x, ·)〉∂D + k2 curl〈γtu,Φ(x, ·)〉∂D

= −k2u(x) .

Taking the trace yields L(γt curlu) = −k2γtu = 0 on ∂D.


Second, let a ∈ H−1/2(Div, ∂D) with La = 0. Define u by u(x) = (La)(x) = curl2〈a,Φ(x, ·)〉∂Dfor x ∈ R3 \ ∂D. Then u ∈ Hloc(curl,R3) solves the exterior and the interior boundary valueproblem with homogeneous boundary data γtu = La = 0. The uniqueness result for the ex-terior problem implies that u vanishes in the exterior. The jump conditions of Theorem 5.52yield a = γt curlu|− − γt curlu|+ = γt curlu|−. This proves part (b). 2

Theorem 5.60 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 (5.65b) 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

E(x) = (La)(x) = curl2〈a,Φ(x, ·)〉∂D , H =1

iωµcurlE , x /∈ ∂D ,

where the density a ∈ H−1/2(Div ∂D) satisfies La = f .

Proof: This is clear from the uniqueness result of both, the interior and the exterior bound-ary value problem and the fact that L is a compact perturbation of an isomophism. 2

We want to study the equation La = f for the case when L fails to be one-to-one. It is theaim to apply the abstract Fredholm result of Theorem 6.4 and have to find the proper dualsystem. The following result will provide the adjoint of L.

Lemma 5.61 (a) The bilinear form 〈·, ·〉 on H−1/2(Div, ∂D) × H−1/2(Div, ∂D), definedby 〈a, b〉 = 〈a, ν × b〉∂D is well defined and a dual system.

(b) For a, b ∈ H−1/2(Div, ∂D) we have

〈La, ν × b〉∂D = 〈Lb, ν × a〉∂D ,

and the adjoint of L with respect to 〈·, ·〉 is given by L∗ = −L.

(c) Let Li be the operator L for k = i. Then there exists c > 0 with

〈Lia, a〉 ≥ c ‖a‖2H−1/2(Div,∂D) for all a ∈ H−1/2(Div, ∂D) .

In particular, the left–hand side is real valued.

(d) L satisfies the variational equation

〈La, b〉 = 〈Div b,S Div a〉∂D + k2〈b,Sa〉∂D for all a, b ∈ H−1/2(Div, ∂D)

where S : H−1/2(∂D) → H1/2(∂D) denotes the single layer boundary operator. Inthe second occurence it is considered as a bounded operator from H−1/2(Div, ∂D) intoH−1/2(Curl, ∂D), see Lemma 5.27.


Proof: (a) On the dense subspace ν × u|∂D : u ∈ C∞(D,C3) of H−1/2(Div, ∂D) themapping a 7→ a × ν is expressed as a × ν = γTηta with the extension operator ηt :H−1/2(Div, ∂D) → H(curl, D) and the trace operator γT : H(curl, D) → H−1/2(Curl, ∂D)from Theorem 5.24. This operator γTηta has a bounded extension from H−1/2(Div, ∂D) intoH−1/2(Curl, ∂D). The bilinear form 〈·, ·〉 is non–degenerated because H−1/2(Curl, ∂D) is thedual of H−1/2(Div, ∂D).

(b) 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 5.52, γt curl v|− − γt curl v|+ = k2 b and thus k2 b × ν = γT curl v|− −γT curl v|+. Therefore, by Green’s theorem in the form (5.19), applied inD and in B(0, R)\D,respectively, and adding the results yields

k2〈La, b× ν〉∂D = 〈γtu, γT curl v|−〉∂D − 〈γtu, 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 .

Changing the roles of a and b and subtracting the results yield

k2〈La, b× ν〉∂D − k2〈Lb, 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. Fora, b ∈ H−1/2(Div, ∂D) we have with respect to the dual system

〈La, b〉 = 〈La, ν × b〉∂D = 〈Lb, ν × a〉∂D = −〈a, ν × Lb〉∂D = −〈a,Lb〉 .

(c) Let now k = i. Taking b = a in part (a) yields

〈Lia, a〉 = −〈Lia, a× ν〉∂D =

∫|x|<R

[| curlu|2 + |u|2] dx −∫|x|=R

(ν × u) · curlu ds .

u and curlu decay exponentially to zero as R tends to infinity. Therefore, we arrive at

〈Lia, a〉 = ‖u‖2H(curl,R3) .

Finally, the boundedness of the trace operator yields

‖a‖H−1/2(Div,∂D) = ‖γt curlu|+ − γt curlu|−‖H−1/2(Div,∂D) ≤ c′ ‖u‖H(curl,R3) .

(d) Taking the trace in (5.61) yields

La = γt∇S Div a + k2γtSa


and thus for a, b ∈ H−1/2(Div, ∂D):

〈La, b〉 = 〈La, ν × b〉∂D = 〈γt∇S Div a, ν × b〉∂D + k2〈ν × Sa, ν × b〉∂D

= 〈b, γT∇S Div a〉∂D + k2〈b, γT Sa〉∂D = 〈Div b,S Div a〉∂D + k2〈b,Sa〉∂D

by the definition of the surface divergence. 2

Theorem 5.62 The interior and exterior boundary value problems are solvable by a layerpotential u = La 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(curl, D) of curl2 in D; that is,

〈f, γT curl v〉∂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 La = f . As mentioned above we wantto apply Theorem 6.4 and define the dual system by X1 = X2 = Y1 = Y2 = H−1/2(Div, ∂D)and 〈a, b〉 = 〈a, ν × b〉∂D for a, b ∈ H−1/2(Div, ∂D) as in the previous Lemma. The adjointL∗ of L is given by −L.Let now f ∈ H−1/2(Div, ∂D) and g ∈ N (L); that is, g = γt curl v for some v ∈ H0(curl, D)with curl2 v − k2v = 0 in D. The solvability condition 〈f, g〉 = 0 reads as 0 = 〈f, ν ×γt curl v〉∂D = −〈f, γT curl v〉∂D which proves the theorem. 2

For our reference scattering problem from the introduction (see Section 1.5) we obtain afinal conclusion.

Corollary 5.63 For any Lipschitz domain D ⊆ R3 the scattering problem (3.36a)–(3.37b)has a unique solution. The scattered field Es can be represented by a layer potential Es = Lawhere the density a ∈ H−1/2(Div ∂D) satisfies La = −γtEinc.

Proof: For the scattering problem by a perfect conductor we have to solve the exteriorproblem with boundary data f(x) = −γtEinc on ∂D. This boundary data satisfies theorthogonality condition of the previous theorem. Indeed, by Green’s formula (5.19) weconclude for every eigenfunction v ∈ H0(curl, D) that

〈γtEinc, γT curl v〉∂D =

∫D

[curl v · curlEinc − k2v Einc

]dx = 〈γtv, γT curlEinc〉∂D = 0 .

Therefore, for this choice of f the boundary equation La = f is solvable. 2

As in the scalar case we expect that the exterior boundary value problem is uniquely for allboundary data f , independently of the wave number. By the previous results it is clear thatwe have to modify the ansatz u = La. We proceed just as in the scalar case and propose anansatz in the form

u = La + η MLia in the exterior of D .


By the jump conditions of Theorem 5.52 the vector field u solves the exterior boundary valueproblem for f ∈ H−1/2(Div, ∂D) if, and only if, a ∈ H−1/2(Div, ∂D) satisfies the equation

La − η

2Lia + ηMLia = f on ∂D . (5.67)

As shown in Theorem 5.51 the operator L is the sum of an isomorphism and a compactoperator. Therefore, there exists η0 > 0 such that also L − η

2Li + ηMLi is a compact

perturbation of an isomorphism for every η ∈ C with |η| ≤ η0. We choose η ∈ C with |η| ≤ η0

and Im (kη) > 0 and have to prove uniqueness of (5.67). Therefore, let a ∈ H−1/2(Div, ∂D)satisfies (5.67) for f = 0 and define u in R3 \ ∂D by u = La + η MLia. Then γtu|+ = 0 on∂D and thus u = 0 in the exterior of D by the uniqueness theorem. The jump conditionsyield

γtu|− = γtu|− − γtu|+ = ηLia , γt curlu|− = γt curlu|− − γt curlu|+ = k2 a

because curl La = k2Ma. Elimination of a yields ηLiγt curlu|− = k2γtu|−. Now we applyGreen’s theorem in D; that is,∫

D

[curlu · curlψ − k2u · ψ

]dx = 〈γtψ, γT curlu|−〉∂D .

Substituting ψ = ku yields∫D

[k | curlu|2 − k |ku|2

]dx = −k 〈γtu|−, ν × γt curlu|−〉∂D

= − kη

|k|2〈Liγt curlu|−, ν × γt curlu|−〉∂D .

Now we observe that 〈Liγt curlu|−, ν × γt curlu|−〉∂D is nonnegative by Lemma 5.61 andIm (kη) < 0 by assumption. Therefore, the imaginary part of the right–hand side is non-negative while the imaginary part of the left–hand side is non-positive. Taking the imaginarypart yields 〈Liγt curlu|−, ν×γt curlu|−〉∂D = 0 and thus γt curlu|− = 0 by Lemma 5.61. Thisends the proof because k2a = γt curlu|− − γt curlu|+ = 0.

We formulate the result as a theorem.

Theorem 5.64 The exterior boundary value problem is uniquely solvable for every f ∈H−1/2(Div, ∂D). The solution can be expressed in the form

u = La + η MLia in R3 \D

for some sufficiciently small η such that Im (kη) > 0 and the density a ∈ H−1/2(Div, ∂D)solves (5.67).

It is not clear to the authors whether or not the parameter η can be chosen independently ofthe wave number k. This is different from the scalar case (see remark following Theorem 5.56)because the difference L − Li fails to be compact.


5.4 Exercises

Exercise 5.1 Prove Theorem 5.6; that is, compactness of the mapping J : u 7→ u fromH tper(Q) into Hs

per(Q) for s < t.Hint: Show that the mapping (JNu)(x) =

∑|n|≤N un exp(in ·x) is compact and converges in

the operator norm to J .

Exercise 5.2 Show that the mapping u 7→ ∂u/∂r fails to be bounded from H1(D) intoL2(∂D) for D = B2(0, 1) being the unit disk in R2 by(a) computing αn for n ∈ N such that ‖un‖H1(D) = 1 where un(r, ϕ) = αnr

n, and(b) showing that ‖∂un/∂r‖L2(∂D) →∞ as n→∞.

Exercise 5.3 Prove Corollary 5.11Hint: Approximate u and v by smooth functions and follow the proof of Lemma 5.2.

Exercise 5.4 Show the formulas from the proof of Theorem 5.4; that is,∫S2

curlu(rx) · x Y −mn (x) ds(x) = −√n(n+ 1)

rwmn (r) ,∫

S2

curlu(rx) · U−mn (x) ds(x) = −1

r

(rwmn (r)

)′,

∫S2

curlu(rx) · V −mn (x) ds(x) = −√n(n+ 1)

rumn (r) +

1

r

(rvmn (r)

)′.

curl[umn (r)Y m

n (x) x]

= −umn (r)

r

(x×GradS2Y m

n (x)),

curl[vmn (r) GradS2Y m

n (x)]

= −1

r

(r vmn (r)

)′ (x×GradS2Y m

n (x)),

curl[wmn (r)

(x×GradS2Y m

n (x))]

= −1

r

(r wmn (r)

)′GradS2Y m

n (x)

− wmn (r)n(n+ 1)

rY mn (x) x .

Exercise 5.5 Let B = B(0, R) be a ball and f ∈ C∞0 (B).

(a) Show that the volume potential

v(x) =

∫B

f(y)1

4π|x− y|dy , x ∈ B ,

is in C∞(B) and solves ∆v = −f in B.

5.4. EXERCISES 275

(b) Prove that there exists a unique solution v ∈ C∞(B) of the boundary value problem∆v = −f in B, v = 0 on ∂B.

Hints: For part (a) use the proof of Theorem 3.9. For part (b) make a proper ansatz in theform v = v + u.

Exercise 5.6 Define the operator K by

Kψ =∞∑n=0

n∑m=−n

1

1 + n(n+ 1)

[amn U

mn + bmn V

mn

]for

ψ =∞∑n=0

n∑m=−n

[amn U

mn + bmn V

mn

]∈ H−1/2(Div, S2) .

Show that K is well defined and compact from H−1/2(Div, S2) into H−1/2(Curl, S2). Show,furthermore, that 〈Kψ,ψ〉S2 is real valued and 〈Kψ,ψ〉S2 > 0 for all ψ ∈ H−1/2(Div, S2),ψ 6= 0.

Exercise 5.7 Show that the boundary value problem (5.58a), (5.58b) is uniquely solvablefor all b ∈ H−1/2(Div, ∂D).Hints: Transform first the boundary value problem to homogeneous boundary data by choos-ing a proper extension v of b and make the ansatz vi = v + v. Second, use the Helmholtzdecomposition as in the proof of Theorem 5.51.

Chapter 6

Appendix

Some notations and basic results are essential throughout the whole monograph. Thereforewe present a brief collection in this appendix for convenience.

6.1 Table of Differential Operators

Maxwell’s equations have to be considered in the eucildian space R3, where we denote thecommon inner product and the cross product by

x · y =3∑j=1

xjyj and x× y =

x2y3 − x3y2

x3y1 − x1y3

x1y2 − x2y1

.

Frequently we apply the elementary formulas

x · (y × z) = y · (z × x) = z · (x× y) (6.1)

x× (y × z) = (x · z)y − (x · y)z (6.2)

Furthermore, we have to fix the notation of some basic differential operators. We will usethe gradient ∇. Additionally we introduce divF = ∇ ·F for the divergence of a vector fieldF and curlF = ∇ × F for its rotation. For scalar fields we denote the Laplacian operatorby ∆ = div∇ = ∇ · ∇. Let u : R3 → C and F : R3 → C3 be sufficiently smooth functionsthen in a cartesian coordinate system we have

∇u =

∂u

∂x1∂u

∂x2∂u

∂x3

curlF = ∇× F =

∂F3

∂x2

− ∂F2

∂x3∂F1

∂x3

− ∂F3

∂x1∂F2

∂x1

− ∂F1

∂x2

divF = ∇ · F =

3∑j=1

∂Fj∂xj

, ∆u = div∇u =3∑j=1

∂2u

∂x2j

277

278 CHAPTER 6. APPENDIX

The following frequently used formulas can be obtained from straightforward calculations.With sufficiently smooth scalar valued functions u, λ : R3 → C and vector valued functionsF,G : R3 → C3 we have

curl∇u = 0 (6.3)

div curlF = 0 (6.4)

curl curlF = ∇ divF − ∆F (6.5)

div(λF ) = F · ∇λ + λ divF (6.6)

curl(λF ) = ∇λ× F + λ curlF (6.7)

∇(F ·G) = (F ′)>G+ (G′)>F (6.8)

div(F ×G) = G · curlF − F · curlG (6.9)

curl(F ×G) = F divG − G divF + F ′G − G′F , (6.10)

where F ′(x), G′(x) ∈ C3×3 are the Jacobian matrices of F and G, respectively, at x; that is,F ′ij = ∂Fi/∂xj.

For completeness we add the expressions of the differential operators ∇, div, curl, and ∆also in other coordinate systems. Let f : R3 → C be a scalar function and F : R3 → C3 avector field and consider cylindrical coordinates

x =

r cosϕr sinϕz

.

With the coordinate unit vectors z = (0, 0, 1)>, r = (cosϕ, sinϕ, 0)>, and ϕ = (− sinϕ, cosϕ, 0)>

and the representation F = Frr + Fϕϕ+ Fz z we obtain

∇f(r, ϕ, z) =∂f

∂rr +

1

r

∂f

∂ϕϕ +

∂f

∂zz , (6.11a)

divF (r, ϕ, z) =1

r

∂(rFr)

∂r+

1

r

∂Fϕ∂ϕ

+∂Fz∂z

, (6.11b)

curlF (r, ϕ, z) =

(1

r

∂Fz∂ϕ− ∂Fϕ

∂z

)r +

(∂Fr∂z− ∂Fz

∂r

)ϕ

+1

r

(∂(rFϕ)

∂θ− ∂Fθ

∂ϕ

)z , (6.11c)

∆f(r, ϕ, z) =1

r

∂

∂r

(r∂f

∂r

)+

1

r2

∂2f

∂ϕ2+

∂2f

∂z2. (6.11d)

Essential for the second chapter are spherical coordinates

x =

r sin θ cosϕr sin θ sinϕr cos θ

6.2. RESULTS FROM LINEAR FUNCTIONAL ANALYSIS 279

and the coordinate unit vectors r = (sin θ cosϕ, sin θ sinϕ, cos θ)>, θ = (cos θ cosϕ, cos θ sinϕ,− sin θ)>,and ϕ = (− sinϕ, cosϕ, 0)>. With F = Frr + Fθθ + Fϕϕ we have the equations

∇f(r, θ, ϕ) =∂f

∂rr +

1

r

∂f

∂θθ +

1

r sin θ

∂f

∂ϕϕ , (6.12a)

divF (r, θ, ϕ) =1

r2

∂(r2Fr)

∂r+

1

r sin θ

∂(sin θ Fθ)

∂θ+

1

r sin θ

∂Fϕ∂ϕ

, (6.12b)

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 , (6.12c)

∆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.(6.12d)

6.2 Results from Linear Functional Analysis

We assume that the reader is familiar with the basic facts from linear functional analysis: inparticular with the notions of normes spaces, Banach- and Hilbert spaces, linear, boundedand compact operators. We list some results which are needed often in this monograph.

Theorem 6.1 Let X, Y be Banach spaces,V ⊂ a linear subspace, and T : V → Y a linearand bounded operator; that is, there exists c > 0 with ‖Tx‖Y ≤ c‖x‖X for all x ∈ V . Thenthere exists a unique extension T : V → Y to the closure V of V ; that is, T x = Tx for allx ∈ V and ‖T x‖Y ≤ c‖x‖X for all x ∈ V . Furthermore, ‖T‖ = ‖T‖.

This theorem is often applied to the case where V is a dense subspace of X. For example,if X is a Sobolev space then V van be taken to be the space of infinitely often differentiablefunctions. The proof of boundedness of an operator (e.g. a trace operator) is usually easierfor smooth functions.The next two theorems are the functional analytic basis of many existence theorems forboundary value problems.

Theorem 6.2 Let T : X → Y be a linear and bounded operator between the normed spacesX and Y . Let T be of the form T = A+K such that A is an isomprphism from X onto Yand K : X → Y is compact. If T is one-to-one then also onto, and T−1 is bounded from Yonto X. In other words, if the homogeneous equation Tx = 0 admits only the trivial solutionx = 0 then the inhomogeneous equation Tx = y is uniquely solvalble for all y ∈ Y , and thesolution x depends continuously on y.

By writing T = A(I + A−1K) it is obvious that it is sufficient to consider the case Y = Xand A = I. For this case the result is proven in, e.g., [12], Chapter 3. It is also a specialcasae of Theorem 6.4 below.


The case where I −K fails to be one-to-one is answered by the following theorem which iscalled Fredholm’s alternative. We need the notions of a dual system and adjoint operators(see [12], Chapter 4).

Definition 6.3 Two normed spaces X, Y , equipped with a bilinear form 〈·, ·〉 : X × Y → Cis called a dual system, if 〈·, ·〉 is nondegenerated; that is, for every x ∈ X, x 6= 0, the linearform y 7→ 〈x, y〉 does not vanish identically and vice versa, for every y ∈ Y , y 6= 0, the linearform x 7→ 〈x, y〉 does not vanish identically.Let 〈X1, Y1〉 and 〈X2, Y2〉 be two dual forms. Two operators T : X1 → X2 and S : Y2 → Y1

are called adjoint to each other if

〈Tx, y〉2 = 〈x, Sy〉1 for all x ∈ X1 , y ∈ Y2 .

Theorem 6.4 (Fredholm) Let 〈Xj, Yj〉j, j = 1, 2, be two dual systems, T : X1 → X2 a

bounded operator with bounded adjoint 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 hold:

(a) The dimensions of the null spaces of T and T ∗ are finite and coincide; that is, dimN (T ) =dimN (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 ∈ N (T ∗) ⊂ Y2 and 〈ϕ, v〉1 = 0 for all ϕ ∈ N (T ) ⊂ X1 .

Proof: We can easily reduce the problem to the case that X1 = X2 and Y1 = Y2 andT = T ∗ = id which is of particular importance in itself.. Indeed, the equation Tx = u isequivalent to x+ T−1Kx = T−1u and the operator K = T−1K maps X1 into itself. On theother hand, the equation T ∗y = v is equivalent to z + K∗(T ∗)−1z = v for z = T ∗y. Theoperator K∗(T ∗)−1 from Y1 into itself is just the adjoint of K. Also, 〈u, y〉2 = 〈T−1u, z〉1 forz = T ∗y ∈ Y1, u ∈ X2. For this special case we refer to Theorem 4.15 of [12]. 2

Also the following two results are used for proving existence of solutions of boundary valueproblems, in particular for those formulated by variational equations.

Theorem 6.5 (Representation theorem of Riesz)Let X be a Hilbert space with inner product (·, ·)X and ` : X → C a linear and boundedfunctional. Then there exists a unique z ∈ X with `(x) = (x, z)X for all x ∈ X. Furthermore,‖`‖ = ‖x‖X .

For a proof we refer to any book on functional analysis as, e.g., [23], Section III.6.

An extension is given by the theorem of Lax–Milgram.

6.3. ELEMENTARY FACTS FROM DIFFERENTIAL GEOMETRY 281

Theorem 6.6 (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 exist 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, e.g., [9], Section 6.2.

6.3 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 [9]). 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 6.7 We call a region D ⊂ R3 to be Cn-smooth (that is, D ∈ Cn), if there exists afinite number of open cylinders Uj of the form Uj = Rjx+z(j) : x ∈ B2(0, αj)×(−2βj, 2βj)with z(j) ∈ R3 and rotations1 Rj ∈ R3×3 and real valued functions ξj ∈ Cn(B[0, αj]) with|ξj(x1, x2)| ≤ βj for all (x1, x2) ∈ B2[0, αj] such that ∂D ⊂

⋃mj=1 Uj and

∂D ∩ Uj =Rjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 = ξj(x1, x2)

,

D ∩ Uj =Rjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 < ξj(x1, x2)

,

Uj \D =Rjx+ z(j) : (x1, x2) ∈ B2(0, αj) , x3 > ξj(x1, x2)

.

We call D to be a Lipschitz domain if the functions ξj which describe the boundary locally areLipschitz continuous; that is, there exist constants Lj > 0 such that |ξj(z)−ξj(y)| ≤ Lj|z−y|for all z, y ∈ B2[0, αj] and all j = 1, . . . ,m.

We call Uj, ξj : j = 1, . . . ,m a local coordinate system of ∂D. For abbreviation we denoteby

Cj = Cj(αj, βj) = B2(0, αj)×(−2βj, 2βj) =x = (x1, x2, x3) ∈ R3 : x2

1+x22 < α2

j , |x3| < 2βj

the cylinders with parameters αj and βj. We can assume without loss of generality thatβj ≥ αj (otherwise split the parameter region into smaller ones). Furthermore, we set

1that is, R>j Rj = I and detRj = 1


Bj := B3(0, αj) ⊂ Cj, j = 1, . . . ,m, and introduce the mappings Ψj : Bj → R3 defined by

Ψj(x) = Rj

x1

x2

ξj(x1, x2) + x3

+ z(j) , x = (x1, x2, x3)> ∈ Bj ,

and its restriction Ψj to B2(0, αj); that is,

Ψj(x) = Rj

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) with∣∣∣∂Ψj∂x1× ∂Ψj

∂x2

∣∣∣ =√

1 + |∇ξj|2 provided the functions ξj are differentiable.

We set U ′j = Ψj(Bj). Then ∂D ⊂⋃mj=1 U

′j and Bj ∩ (R2 × 0) = B2(0, αj)× 0, and

∂D ∩ U ′j =

Ψj(x) : x ∈ Bj, x3 = 0

=

Ψj(x) : x ∈ B2(0, αj),

D ∩ U ′j =

Ψj(x) : x ∈ Bj, x3 < 0,

U ′j \D =

Ψj(x) : x ∈ Bj, x3 > 0.

Therefore, the mappings Ψj “flatten” the boundary. For C1−domains D we note that theJacobian is given by

Ψ′j(x) = Rj

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) = Rj

10

∂1ξj(x)

,∂Ψj

∂x2

(x) = Rj

01

∂2ξj(x)

.

They span the tangent plane at y = Ψj(x). The vector

∂Ψj

∂x1

(x)× ∂Ψj

∂x2

(x) = Rj

−∂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

Rj

−∂1ξj(x)−∂2ξj(x)

1

is called the exterior unit normal vector.


Remark 6.8 For Lipschitz domains the functions ξj are merely Lipschitz continuous. There-fore, Ψj and its inverse Ψ−1

j , given by

Ψ−1j (y) =

y1

y2

y3 − ξj(y1, y2)

, y = R>j (y − z(j)) , y ∈ U ′j = Ψj(Bj) ,

are also Lipschitz continuous. A celebrated result of Rademacher [21] (see also [9], Sec-tion 5.8) states that every Lipschitz continuous function ξj is differentiable at almost everypoint x ∈ B2(0, αj) with |∇ξj(x)| ≤ Lj for almost all x ∈ B2(0, αj) where Lj is the Lipschitzconstant. Therefore, for Lipschitz domains the exterior unit normal vector ν(x) exists atalmost all points x ∈ ∂D. Furthermore, u ∈ L1(U ′j) if, and only, if u Ψj ∈ L1(Bj) and thetransformation formula holds in the form∫

U ′j

u(y) dy =

∫Bj

u(Ψj(x)

)dx . (6.13)

Application of this result to |u|2 shows that the operator u 7→ u Ψj is bounded from L2(Bj)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., [16], Chapter 3):

Theorem 6.9 (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 φj : j = 1, . . . ,m a partition of unity on K subordinate toUj : j = 1, . . . ,m.

Using a local coordinate system Uj, ξj : j = 1, . . . ,m of ∂D with corresponding mappingsΨj from the balls Bj onto U ′j and their restrictions Ψj : B2(0, αj) → U ′j ∩ ∂D as in Defini-tion 6.7 and a corresponding partion of unity φj on ∂D with respect to U ′j we write

∫∂Df ds

in the form ∫∂D

f ds =m∑j=1

∫∂D∩U ′j

φj f ds =m∑j=1

∫∂D∩U ′j

fj ds

with fj(y) = φj(y)f(y). The integral over the surface patch U ′j ∩ ∂D is given by∫U ′j∩∂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 6.10 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 (6.14)

(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 . (6.15)

Proof: We use a local coordinate system Uj, ξj : j = 1, . . . ,m which yields the parametriza-tion Ψj : B2(0, αj) → ∂D ∩ U ′j. 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 U ′j with z, x ∈ U ′j. 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 y, x ∈ U ′j for some j. 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 ∈ U ′j 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 ⊂⋃mj=1 U

′j.

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 U ′j 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 (6.14) 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 (6.14)) 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

6.4 Integral Identities

Now we can formulate the mentioned integral identities. We do it only in R3. By Cn(D,C2)we denote the space of vector fields F : D → C3 which are n−times continuously differen-tiable. By Cn(D,C3) we denote the subspace of Cn(D,C3) that consists of those functions Fwhich, together with all derivatives up to order n, have continuous extentions to the closureD of D.

6.4. INTEGRAL IDENTITIES 287

Theorem 6.11 (Theorem of Gauss, Divergence Theorem)

Let D ⊂ R3 be a bounded Lipschitz domain. For F ∈ C(D,C3) with divF ∈ C(D) theidentity ∫

D

divF (x) dx =

∫∂D

F (x) · ν(x) ds

holds. In particular, the integral on the and left–hand side exists.Furthermore, application of this formula to F = uve(j) for u, v ∈ C1(D) and the j−th unitvector 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 [16]. For smooth domains a proof can befound in [9]. As a conclusion one derives the theorems of Green.

Theorem 6.12 (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 6.11 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 6.13 (Integral identities for vector fields)Let D ⊂ R3 be a bounded Lipschitz domain. Furthermore, let A,B ∈ C1(D,C3) ∩ C(D,C3)and let u ∈ C2(D) ∩ C1(D). Then ∫

D

curlAdx =

∫∂D

ν × Ads , (6.16a)∫D

(B · curlA− A · curlB) dx =

∫∂D

(ν × A) ·B ds , (6.16b)∫D

(u divA+ A · ∇u) dx =

∫∂D

u (ν · A) ds . (6.16c)


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

6.5 Surface Gradient and Surface Divergence

We have to introduce two more notions from differential geometry, the surface gradient andthe 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 6.7.First we define the spaces of differentiable functions and vector fields on ∂D.

Definition 6.14 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 partition of unity on∂D subordinate to Uj : j = 1, . . . ,m. We set again Ψj(x) = Rj(x1, x2, ξj(x))> + z(j) forx = (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,C3) :=F ∈ C(∂D,C3) : Fj ∈ C1(∂D) for j = 1, 2, 3

,

Ct(∂D) :=F ∈ C(∂D,C3) : F · ν = 0 on ∂D

,

C1t (∂D) := Ct(∂D) ∩ C1(∂D,C3) .

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 4.13.

Lemma 6.15 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.

6.5. SURFACE GRADIENT AND SURFACE DIVERGENCE 289

(b) For every F ∈ C1t (∂D) there exists F ∈ C1(R3,C3) 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 6.14 we note that f =

∑mj=1 fj

on ∂D where fj = fφj. The functions fj Ψj are continuously differentiable functions fromB2(0, αj) into C with support in B2(0, αj). We extend fjΨj into the cylinder Cj = Cj(αj, βj)by setting gj(x) := ρ(x3)(fj Ψj)(x1, x2) for x = (x1, x2, x3) ∈ Cj where ρ ∈ C∞0 (−βj, βj)is such that ρ = 1 in a neighborhood of 0. Then gj : Cj → 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 Uj = Ψj(Cj). 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 Uj such that f = f on ∂D.

The proof of (b) is identical by using the argument for every component. 2

Definition 6.16 Let f ∈ C1(∂D) and f ∈ C1(U) be an extension of f into a neighborhoodU of the boundary ∂D of the domain D ∈ C2. Furthermore, let F ∈ C1

t (∂D) be a tangentialvector field and F ∈ C1(U,C3) be an extension into U .

(a) The surface gradient of f is defined as the orthogonal projection of ∇f onto the tangentplane; that is,

Grad f = ν × (∇f × ν) = ∇f − ∂f

∂νν on ∂D , (6.17)

where ν = ν(x) denotes the exterior unit normal vector at x ∈ ∂D.

(b) The surface divergence of F is given by

DivF = div F − ν · (F ′ν) on ∂D (6.18)

where F ′(x) ∈ C3×3 denotes the Jacobian matrix of F at x.

We will see in Lemma 6.19 that the definitions are independent of the choices of the exten-sions.

Example 6.17 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 , (6.19)

DivF (θ, φ) =1

R sin θ

∂

∂θ

(sin θ Fθ(θ, φ)

)+

1

R sin θ

∂Fφ∂φ

(θ, φ) , (6.20)


where θ = (cos θ cosφ, cos θ sinφ,− sin θ)> and φ = (− sinφ, cosφ, 0)> are the tangential unitvectors which span the tangent plane and Fθ, Fφ are the components of F with respect to

these vectors; that is, F = Fθθ + Fφφ.

In case of calculations in spherical coordinates, x = rx ∈ R3, often the surface differentialoperators with respect to the unit sphere are used instead of the opertors on ∂D = x ∈R3 : |x| = r. Therefore, we indicate this by using the index S2 for the differential operatorswith respect to the unit sphere. Thus on a sphere of radius r it is

GradS2f(r, x) = r Grad f(rx) , and DivS2 F (r, x) = r DivF (rx) ,

where we understand f(r, ·) as a function on the unit sphere on the left side and f as afunction on the sphere ∂D on the right–side.

Furthermore, in general the differential operator

∆∂D = Div Grad

on a surface ∂D is called Laplace-Beltrami operator . From the previous example on |x| = Rwe note that in spherical coordinates it holds

∆∂Df(Rx) = Div Grad f(θ, φ) =1

R2 sin θ

∂

∂θ

(sin θ

∂f

∂θ(θ, φ)

)+

1

R2 sin2 θ

∂2f

∂φ2(θ, φ) .

Using the corresponding operators for the unit sphere we obtain with the spherical Laplace-Beltrami operator introduced in Definition 2.2 by the different views on the function f that

∆S2f(rx) = DivS2 Grad S2f(rx) = r2∆∂Df(x) = r2 Div Grad f(x)

where we read the function f on the left as a function on the unit sphere

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 6.15 there exists an extension ν ∈ C1(R3,C3) of ν. Certainly,ν 6= 0 in a neighborhood U of ∂D because |ν| = 1 on ∂D. Therefore, ν = ν/|ν| will be therequired extension into U .

Lemma 6.18 Let D ∈ C2 and F ∈ C1t (∂D) and f ∈ C1(∂D) with extensions F ∈

C1(R3,C3) and f ∈ C1(R3), respectively. Then:

(a) DivF = ν·curl(ν×F ) 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 positively orientated;

2that is, Γ = ∂D ∩ U for some open set U ⊂ R3

6.5. SURFACE GRADIENT AND SURFACE DIVERGENCE 291

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

DivF ds =

∫C

F · (τ × ν) d` .

In particular,∫∂D

DivF ds = 0.

(c) Partial integration holds in the following form:∫∂D

f DivF ds = −∫∂D

F ·Grad f ds . (6.21)

Proof: (a) The product rule for the curl of a vector product (see 6.10) yields

curl(ν × F ) = ν div F − F div ν + ν ′F − F ′ν

= ν div F − F div ν + ν ′F − F ′ν

and thusν · curl(ν × F ) = div F − ν · F︸︷︷︸

= 0

div ν + ν>ν ′F − ν>F ′ν .

From ν>ν = 1 in U we have by differentiation that ν>ν ′ = 0 in U and thus

ν · curl(ν × F ) = div F − ν>F ′ν = DivF .

(b) Let ν and F as in part (a) and set G = ν × F in U . Then G ∈ C1(U,C3). By part(a) we conclude that

∫Γ

DivF ds =∫

Γν · curl G ds. Choose a sequence Gn ∈ C∞(U,C3)

with Gn → G in C1(U,C3). 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,C3) and Gn →

ν × F in C(∂D,C3) yields∫

ΓDivF ds =

∫C

(ν × F ) · τ d` =∫CF · (τ × ν) d`.

(c) By part (b) it suffices to prove the product rule

Div(f F ) = Grad f · F + f DivF . (6.22)

Indeed, using the definitions yields

Div(f F ) = div(f F ) − ν ·((f F )′ν

)= ∇f · F + f div F − ν>F f ′ν − f ν>F ′ν

= Grad f · F + f div F − ν>F︸︷︷︸= 0

f ′ν − f ν>F ′ν

= Grad f · F + f DivF .


Lemma 6.19 The tangential gradient and the tangential divergence depend only on thevalues of f and the tangential field F on ∂D, respectively.


Proof Let f1, f2 ∈ C1(D) be two extensions of f ∈ C1(∂D). Then f = f1 − f2 vanishes on∂D. The identity (6.21) shows that∫

∂D

[ν × (∇f × ν)

]· F ds = 0 for all F ∈ 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 × (∇f × ν) vanishes which shows that the definition of Grad f is independentof the extension. Similarly the assertion for the tangential divergence is obtained. 2

Corollary 6.20 Let w ∈ C2(D,C3) such that w, curlw ∈ C(D,C3). Then the surfacedivergence of ν × w exists and is given by

Div(ν × w) = −ν · curlw on ∂D . (6.23)

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

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[20] M.H. Protter. Unique continuation for elliptic equations. Trans. Amer. Math., 95:81–90,1960.

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[22] F. Rellich. Uber das asymptotische Verhalten von Losungen von ∆u + λu = 0 inunendlichen Gebieten. Jber. Deutsch. Math. Verein., 53:57–65, 1943.

[23] K. Yosida. Functional Analysis. Springer, New York etc, 1978.

Index

Cn−smooth, 281

addition formulaBessel functions, 70surface harmonics, 45

adjoint operators, 280Ampere’s law, 10anisotropic function space, 184anisotropic medium, 12

Bessel differential equation, 30, 54Bessel functions, spherical, 57boundary operatorD, 254L, 261M, 261T , 254D′, 249S, 249

boundary value problemDirichlet, 263Neumann, 267vector case, 268

charge density, 9conductivity, 12constitutive equations, 11convolution, 151current density, 9

dielectric medium, 12dielectric tensor, 12Dirichlet eigenvalues of Delta, 65divergence theorem, 287dual system, 270, 280

E-mode and M-mode, 18eigen

-function, 165-space, 170, 171

-system, 165-value, 165, 170

electric displacement, 9electric field, 9electric monopole, 13electrostatics, 13equation of continuity, 10Euler differential equation, 30extension operator

η, 209ηt, ηT , 218, 226η, 211

exterior unit normal vector, 282

far field amplitude, 72far field pattern, 72Faraday’s law of induction, 10Fourier

coefficients, 154series, 154transform, 150

Fredholm’s alternative, 280Fredholm’s theorem, 280Friedrich’s inequality, 157fundamental solution, 92, 246Fundamental Theorem of Calculus of Varia-

tions, 186Funk–Hecke formula, 44

Gauss Theorem, 287Gauss’ electric law, 10Gauss’ magnetic law, 11Green’s formula, 228, 256, 287

vectorial, 287Green’s Representation Theorem, 246

Holder-continuous function, 100Hankel functions, 57harmonic function, 28

295

296 INDEX

Hertz dipol, 20Hertz potential

electric, 17magnetic, 17

Holmgren’s uniqueness theorem, 97homogeneous medium, 12homogeneous polynomial, 40

impedance boundary condition, 19initial–boundary value problem, 185interior regularity, 175isotropic medium, 12

Jacobi–Anger expansion, 61

Laplace equation, 28Laplace-Beltrami operator, 28, 290law of induction, 10Lax – Milgram theorem, 281Legendre differential equation, 31

associated, 30Legendre functions, associated, 39Legendre polynomial, 31limiting absorption principle, 66linear medium, 12Lipschitz domain, 201, 281local coordinate system, 201, 281

magnetic field, 9magnetic induction, 9magnetostatic, 14Maxwell double layer, 262Maxwell single layer, 262MKS-system, 9

Ohm’s law, 12orthonormal system, 165, 171

partial integration, 154, 211, 287, 291partition of unity, 175, 204, 212, 283perfectly conducting medium, 19permeability tensor, 12plane time harmonic fields, 16Poisson equation, 13potential

double layer Dϕ, 95, 247single layer Sϕ, 95, 247

vector, 135vector La, 261vector Ma, 261volume Vϕ, 95

potential theory, 13Poynting

Theorem, 22vector, 22vector, complex, 22

radiation conditionSilver–Muller, 257, 261Sommerfeld, 249

Rayleigh formulas, 58resolvent set, 170Riesz representation theorem, 280Riesz–Fredholm theorem, 279Rodrigues formula, 34

Silver-Muller radiation condition, 21Sobolev spaces

V0,A, 160H(curl, D), 158H(curl 0, D), 160H1(D), 149H1per(Q), 155

Hsper(Curl, Q2), 218

Hsper(Div, Q2), 218

Hsper(Q), 205

H−1/2(Curl, ∂D), 225H−1/2(Div, ∂D), 225H−1/2(∂D), 214H1/2(∂D), 209H0(curl, D), 159H0(curl 0, D), 160H1

0 (D), 154Hper(curl, Q3), 218VA, 160V0,A, 160

Sommerfeld radiation condition, 21spectrum, 170speed of light, 13spherical harmonics, 41

vector, 75spherical surface harmonics, 41spherical wave functions, 59

INDEX 297

Stratton-Chu formula, 129, 132, 257support, 148surface curl, 76, 230surface divergence, 230, 289surface gradient, 289

tangential plane, 282tangential vector, 282TE-mode and TM-mode, 18time harmonic Maxwell equations, 14time–dependent Maxwell system, 193trace operator

γ0, 209γ1, 215γt, γT , 218, 226

trace theorem, 206, 209, 218, 226transformation formula, 283transmission boundary conditions, 19

unique continuation property, 173, 177

vacuum, 12variational

curl, 157divergence, 157gradient, 149solution, 163

vector Helmholtz equation, 15

wave equation, 13wave number, 15, 255Wronskian of Bessel functions, 57

Young’s inequality, 151

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