The Mathematical Theory of Maxwell’s Equations
Andreas Kirsch and Frank Hettlich
Department of Mathematics
Karlsruhe Institute of Technology (KIT)
Karlsruhe, Germany
c May 24, 2014
2
Preface
This book arose from lectures on Maxwell’s equations given by the authors between 2007 and
2013. Graduate students from pure and applied mathematics, physics – including geophysics
– and engineering attended these courses. We observed that the expectations of these groups
of students were quite different: In geophysics expansions of the electromagnetic fields into
spherical (vector-) harmonics inside and outside of balls are of particular interest. Graduate
students from numerical analysis wanted to learn about the variational treatments of interior
boundary value problems including an introduction to Sobolev spaces. A classical approach
in scattering theory – which can be considered as a boundary value problem in the unbounded
exterior of a domain – uses boundary integral equation methods which are particularly
helpful for deriving properties of the far field behaviour of the solution. This approach is, for
polygonal domains or, more generally, Lipschitz domains, also of increasing relevance from
the numerical point of view because the dimension of the region to be discretized is reduced
by one. In our courses we wanted to satisfy all of these wishes and designed an introduction
to Maxwell’s equations which coveres all of these concepts – but restricted ourselves almost
completely (except Section 4.3) to the time-harmonic case or, in other words, to the frequency
domain, 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 of
the Maxwell system (Hertz potentials), and every component of the electric and magnetic
field satisfies an equation of Helmholtz type. Therefore, and also for didactical reasons, we
will consider in each of our approaches first the simpler scalar Helmholtz equation before we
turn to the technically more complicated Maxwell system. In this way one clearly sees the
analogies 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 time
harmonic 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 can
expand the fields inside and outside of D into spherical wave functiuons. First we study the
scalar stationary case; that is, the Laplace equation. We introduce the expansion into scalar
spherical harmonics as the analogon to the Fourier expansion on circles in R2 . This leads
directly to series expansions of solutions of the Laplace equation in spherical coordinates.
The extension to the Helmholtz equation requires the introduction of (spherical) Bessel- and
Hankel functiuons. We derive the most important properties of these special functions in
detail. 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 such
a presentation of both, the scalar and the vectorial case, in the literature. We emphasize that
this chapter is completely self-contained and does not refer to any other chapter – except for
the 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 arbitrary
shape but, in this chapter, with sufficiently smooth boundary ∂D. We present the classical
boundary 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 ourselves
to the case of smooth boundary data (as it is the usual case of the scattering by incident
waves) which allows us to study the setting completely in Hölder spaces and avoids the
notions of “parallel surfaces” and weak forms of the normal derivatives. For the scalar
problem we restrict ourselves to the Neumann boundary condition because our main goal
is the treatment for Maxwell’ equations. Here we believe – as also done in [6] – that the
canonical spaces on the boundary are Hölder spaces where also the surface divergence is
Hölder continuous. In order to prove the necessary properties of the scalar and vector
potentials a careful investigation of the differential geometric properties of the surface ∂D is
needed. Parts of the technical details are moved to the Appendix 6. We emphasize that this
chapter is self countained and does not need any results from other chapters (except from
the appendix).
As an alternative approach for studying boundary value problems for the Helmholtz equation
or 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 and
the homogeneous boundary condition of an ideal conductor. This makes it possible to work
(almost) solely in the Sobolev spaces H01 (D) and H0 (curl, D) of functions with vanishing
boundary traces or tangential boundary traces, respectively. In Section 4.1 we derive the basic properties of these special Sobolev spaces. The characteristic feature is that no regularity
of the boundary and no trace theorems are needed. Probably the biggest difference between
the scalar case of the Helmholtz equation and the vectorial case of Maxwell’s equations is
the fact that H0 (curl, D) is not compactly imbedded in L2 (D, C3 ) in contrast to the space
H01 (D) for the scalar problem. This makes it necessary to introduce the Helmholtz decomposition. The only proof which is beyond the scope of this elementary chapter is the proof that
the subspace of H0 (curl, D) consisting of divergence-free vector fields is compactly imbedded
in L2 (D, C3 ). For this part some regularity of the boundary (e.g. Lipschitz regularity) is
needed. Since the proof of this fact requires more advanced properties of Sobolev spaces it
it transponed to Chapter 5. We note that also this chapter is self contained except of the
beforementioned 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 presented
in Section 4.1. In particular, Sobolev spaces on the boundary ∂D have to be introduced and
the correponding trace operators. Perhaps different from most of the traditional approaches
we first consider the case of the cube (−π, π)3 ⊂ R3 and introduce Sobolev spaces of periodic
functions by the proper decay of the Fourier coefficients. The proofs of imbedding and trace
theorems 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 to
general Lipschitz domains.
We define the scalar and vector potentials in Section 5.2 analogously to the classical case
as in Chapter 3 but have to interpret the boundary integrals as certain dual forms. The
boundary operators are then defined as traces of these potentials. In this way we follow the
classical approach as closely as possible. Our approach is similar but a bit more explicit
than in [16], see also [10]. Once the properties of the potentials and corresponding boundary
operators are known the introduction and investigation of the boundary integral equations is
almost classical. For example, for Lipschitz domains the Dirichlet boundary value problem
for 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 boundary
operator fails to be compact (in contrast to the case of smooth boundaries). Also, the single
layer boundary operator satisfies a Garding’s inequality; that is, can be decomposed into a
coercive and a compart part. Analogously. the Neumann boundary value problem and the
electromagnetic case are treated. In our presentation we try to show the close connection
between the scalar and the vector cases.
Starting perhaps with the pioneering work of Costabel [7] many important contributions to
the study of boundary integral operators in Sobolev spaces for Lipschitz boundaries have
been published. It is impossible for the authors to give an overview on this subject but
instead refer to the monograph [10] and the survey article [4] from which we have learned a
lot. As mentioned above, our approach to introduce the Sobolev spaces, however, is different
from those in, e.g. [1, 2, 3].
In the Appendix 6 we collect results from vector calculus and differential geometry, in particular 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 on
Maxwell’s equations, not even for the time harmonic case or any of the beforementioned subareas. As said before this book arose from – and is intended to be – material for designing
graduate courses on Maxwell’s equations. The students should have some knowledge on vector analysis (curves, surfaces, divergence theorem) and functional analysis (normed spaces,
Hilbert spaces, linear and bounded operators, dual space). The union of the topics covered
in this monograph is certainly far too much for a single course. But it is very well possible
to choose parts of it because the chapters are all independent of each other. For example, in
the summer term 2012 (8 credit points; that is in our place, 16 weeks with 4 hours per week
plus exercises) one of the authors (A.K.) covered Sections 2.1–2.6 of Chapter 2 (only interior
cases), Chapter 3 without all of the proofs of the differential geometric properties of the
surface 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 (spherical
harmonics, Bessel functions) or on Sobolev spaces.
One of the authors wants to dedicate this book to his father, Arnold Kirsch (1922–2013), who
taught him about simplification of problems without falsification (as a concept of teaching
mathematics in high schools). In Chapter 4 of this monograph we have picked up this
concept 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 of
Chapters 3 and 5) are technically rather involved.
Karlsruhe, March 2014
Andreas Kirsch
Frank 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
3.1
3.2
91
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
3.1.4
Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 116
. . . . . . . . . . . . . . . . . . . . . . 112
A Scattering Problem for the Maxwell System . . . . . . . . . . . . . . . . . 124
7
8
CONTENTS
3.3
3.2.1
Representation Theorems . . . . . . . . . . . . . . . . . . . . . . . . 125
3.2.2
Vector Potentials and Boundary Integral Operators . . . . . . . . . . 135
3.2.3
Uniqueness and Existence . . . . . . . . . . . . . . . . . . . . . . . . 141
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
4 The Variational Approach to the Cavity Problem
4.1
4.2
147
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
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
5.1
201
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
6.4
Integral Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
6.5
Surface Gradient and Surface Divergence . . . . . . . . . . . . . . . . . . . . 288
. . . . . . . . . . . . . . . . . 281
Bibliography
292
Index
294
Chapter 1
Introduction
In this introductory chapter we will explain the physical model and derive the boundary value
problems which we will investigate in this monograph. We begin by formulating Maxwell’s
equations in differential form as our starting point. In this monograph we consider exclusively
linear media; that is, the constitutive relations are linear. The restriction to special cases
leads to the analogous equations in electrostatics or magnetostatics and, assuming periodic
time dependence when going into the frequency domain, to time harmonic fields. In the
presence 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 by
introducing 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 Maxwell
equations, which relate five vector fields E, D, H, B, J and the scalar field ρ. In differential
form these read as follows
∂B
+ curlx E = 0
(Faraday’s Law of Induction)
∂t
∂D
− curlx H = −J (Ampere’s Law)
∂t
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 ) and
charge density (in As/m3 ) of the medium.
Here and throughout we use the rationalized MKS-system, i.e. the fields are given with
respect 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 differential
operators 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 and
basic identities are listed in Appendix 6.1 .
The actual equations that govern the behavior of the electromagnetic field were first completely formulated by James Clark Maxwell (1831–1879) in Treatise on Electricity and Magnetism in 1873. It was the ingeneous idea of Maxwell to modify Ampere’s Law which was
known up to that time in the form curl H = J for stationary currents. Furthermore, he
collected 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 derives
from the identity div curl H = 0 the well known equation of continuity which combines the
charge density and the current density.
Conclusion 1.1 Gauss’ Electric Law and Ampere’s Law imply the equation of continuity
∂D
∂ρ
= div
= div curl H − J = − div J .
∂t
∂t
Historically, and more closely connected to the physical situation, the integral forms of
Maxwell’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 a
region Ω0 in R3 where electromagnetic waves propagate. In particular, we require that the
unit 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 boundary
of S at x ∈ ∂S. This vector, lying in the tangent plane of S together with a second vector
n(x) in the tangent plane at x ∈ ∂S and normal to ∂S is oriented such that {τ, n, ν, } form
a mathematically positive system; that is, τ is directed counterclockwise when we look at S
from the positive side, and n(x) is directed to the outside of S. Furthermore, let Ω ∈ R3 be
an open set with boundary ∂Ω and outer unit normal vector ν(x) at x ∈ ∂Ω.
Then Ampere’s law describing the effect of the external and the induced current on the
magnetic field is of the form
Z
Z
Z
d
D · ν ds +
J · ν ds .
(1.1)
H · τ d` =
dt
S
∂S
S
It is named by André Marie Ampère (1775–1836).
Next, Faraday’s law of induction (Michael Faraday, 1791–1867), which is
Z
Z
d
E · τ d` = −
B · ν ds ,
dt
(1.2)
S
∂S
describes how a time-varying magnetic field effects the electric field.
Finally, the equations include Gauss’ Electric Law
Z
Z
D · ν ds =
ρ dx
∂Ω
Ω
(1.3)
1.2. THE CONSTITUTIVE EQUATIONS
11
descibing the sources of the electric displacement, and Gauss’ Magnetic Law
Z
B · ν ds = 0
(1.4)
∂Ω
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 continuous
we can apply the following integral identities due to Stokes and Gauss for surfaces S and
solids Ω lying completely in D.
Z
Z
curl F · ν ds =
F · τ d` (Stokes),
(1.5)
S
∂S
Z
Z
F · ν ds (Gauss),
div F dx =
Ω
(1.6)
∂Ω
see Appendix 6.3. To derive the Maxwell’s equations in differential form we choose F to be
one of the fields H, E, B or D. With these formulas we can eliminate the boundary integrals
in (1.1)–(1.4). We then use the fact that we can vary the surface S and the solid Ω in D
arbitrarily. By equating the integrands we are led to Maxwell’s equations in differential form
as presented in the begining.
With Maxwell’s equations many electromagnetic phenomena became explainable. For instance 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) could
show experimentally the existence of electromagenetic waves, in Karlsruhe, Germany. For
more details on the physical background of Maxwell’s equations we refer to text books as
J.D. Jackson, Classical Electrodynamics [11].
1.2
The Constitutive Equations
In the general setting the equations are not yet complete. Obviously, there are more unknowns 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. In
general, they not only depend on the molecular character but also on macroscopic quantities
as density and temperature of the material. Also, there are time-dependent dependencies
as, e.g., the hysteresis effect, i.e. the fields at time t depend also on the past.
As a first approximation one starts with representations of the form
D = E + 4πP
and B = H − 4πM
12
CHAPTER 1. INTRODUCTION
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 one
can 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 , the
permeability tensor . In this case we call a medium linear .
The special case of an isotropic medium means that polarization and magnetization do not
depend on the directions. Otherwise a medium is called anisotropic. In the isotropic case
dielectricity 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 homogeneous. 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 a
linear 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 is
called the conductivity . If σ = 0 then the material is called a dielectric. In vacuum we have
σ = 0 and ε = ε0 ≈ 8.854 · 10−12 AS/V m, µ = µ0 = 4π · 10−7 V 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 second
order partial differential equations. They serve often as simpler models for electromagnetic
wave propagation. Also in this monograph we will always explain the approaches first for
the simpler scalar wave equation.
Vacuum
Vacuum is a homogeneous, dielectric medium with ε = ε0 , µ = µ0 , and σ = 0, and no charge
distributions and no external currents; that is, ρ = 0 and Je = 0. The law of induction takes
the form
∂H
+ curl E = 0 .
µ0
∂t
1.3. SPECIAL CASES
13
Assuming sufficiently smooth functions a differentiation with respect to time t and an application of Ampere’s Law yields
ε0 µ 0
∂ 2H
+ curl curl H = 0 .
∂t2
√
The term c0 = 1/ ε0 µ0 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 taken
componentwise it follows that the components of H are solutions of the linear wave equation
1 ∂ 2H
− ∆H = 0 .
c20 ∂t2
Analogously, one derives the same equation for the cartesian components of the electric field:
1 ∂ 2E
− ∆E = 0 .
c20 ∂t2
Therefore, a solution of the Maxwell system in vacuum can also be described by a divergence
free solution of one of the two vector valued wave equations and defining the other field by
Amperes 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 region
Ω is called electrostatics. The law of induction reduces to the differential equation
curl E = 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 = div D = − div(ε0 E) = −ε0 ∆u
for the potential u. Thus, the electrostatics is described by the basic elliptic partial differential 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 gener1 1
ated by a point charge, e.g., at the origin. For x ∈ R3 with x 6= 0 the function u(x) = 4π
|x|
is harmonic; that is, satisfies ∆u = 0. Thus by
E(x) = ∇u(x) = −
1 x
,
4π |x|3
x 6= 0 ,
we obtain a stationary solution, the field of an electric monopole .
14
CHAPTER 1. INTRODUCTION
In magnetostatics one considers H being constant in time. For the magnetic field the
situation is different because, by Ampere’s law we have curl H = J . Thus in general
curl H does not vanish. However, according to Gauss’ magnetic law we have
div B = 0 .
From this identity we conclude the existence of a vector potential A : R3 → R3 with
B = − curl A in D. Substituting this into Ampere’s Law yields (for homogeneous media Ω)
after multiplication with µ0 the equation
−µ0 J = curl curl A = ∇ div A − ∆A .
Since curl ∇ = 0 we can add gradients ∇u to A without changing B. We will see later
that we can choose u such that the resulting potential A satisfies div A = 0. This choice of
normalization is called Coulomb gauge named by Charles Augustin de Coulomb (1736–1806).
With this normalization we get also in magnetostatics the Poisson equation
∆A = −µ0 J .
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 time
harmonic fields. Under the assumptions that the fields allow a Fourier transformation in
time we set
Z
E(x, t) eiωt dt ,
E(x; ω) = (Ft E)(x; ω) =
R
Z
H(x; ω) = (Ft H)(x; ω) =
H(x, t) eiωt dt ,
R
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 fields
we denote them by capital Latin letters only. According to Ft (u0 ) = −iωFt u Maxwell’s
equations transform into the time harmonic Maxwell’s equations
−iωB + curl E = 0 ,
iωD + curl H = σE + Je ,
div D = ρ ,
div B = 0 .
Remark: The time harmonic Maxwell system can also be derived from the assumption
that all fields behave periodically with respect to time with the same frequency ω. Then the
complex valued functions E(x, t) = e−iωt E(x), H(x, t) = e−iωt H(x), etc, and their real and
imaginary parts satisfy the time harmonic Maxwell system.
1.3. SPECIAL CASES
15
With the constitutive equations D = εE and B = µH we arrive at
curl E − iωµH
= 0,
(1.7a)
curl H + (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 E
from (1.7a) and (1.7b) which yields
1
curl E + (iωε − σ) E = Je .
(1.8)
curl
iωµ
and
curl
1
curl H
iωε − σ
+ iωµ H = curl
1
Je
iωε − σ
,
(1.9)
respectively. Usually, one writes these equations in a slightly different way by introducing the
constant 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 + i
σ
.
ωε0
Then equations (1.8) and (1.9) take the form
1
curl
curl E − k 2 εc E = iωµ0 Je ,
µr
1
1
2
curl
curl H − k µr H = curl
Je ,
εc
εc
√
with the wave number k = ω ε0 µ0 . We conclude from the second equation that div(µr H) =
0
0 because div curl vanishes. For the electric field we obtain that div(εc E) = − iωµ
div Je =
k2
i
− ωε0 div Je .
In vacuum we have εc = 1, µr = 1 and therefore the equations reduce to
curl curl E − k 2 E = iωµ0 Je ,
(1.10)
curl curl H − k 2 H = curl Je .
(1.11)
Without external current density, Je = 0, we obtain from curl curl = ∇ div −∆ the vector
Helmholtz equations
∆E + k 2 E = 0 and ∆H + k 2 H = 0 .
Obviuously the reduced problems considering E or H are symmetric and we conclude the
following important lemma.
16
CHAPTER 1. INTRODUCTION
1
curl E provides a soluLemma 1.3 A vector field E ∈ C 2 (Ω, C3 ) combined with H := iωµ
0
tion of the time harmonic Maxwell system (1.7a)–(1.7d) for Je = 0 in vacuum if and only if
E is a divergence free solution of the vector Helmholtz equation; that is,
∆E + k 2 E = 0 and
div E = 0 in Ω .
Analougously, a divergence free solution of the vector Helmholtz equation H ∈ C 2 (Ω, C3 )
−1
combined with E := iωε
curl H leads to a solution of Maxwell’s equations in vacuum.
0
The relationship between the Maxwell system and the vector Helmholtz equation remains
true if we consider time harmonic waves in any homogeneous medium because we only have
to substitute µ0 and ε0 by complex valued constants µ and ε, respectively. In this complex
√
valued 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 harmonicP
Maxwell equations (1.10), (1.11) provided
3
3
d is a unit vector in R and p ∈ C 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 fronts
are 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 domain
D. Then x 7→ x · E(x) is a solution of the scalar Helmholtz equation.
Proof: By the vector Helmholtz equation and div E = 0 we obtain
3
3
X
X
∆ x · E(x) =
div ∇ xj Ej (x) =
div Ej (x) e(j) + xj ∇Ej (x)
j=1
j=1
3 X
∂Ej
(x) + xj ∆Ej (x) = 2 div E(x) − k 2 x · E(x)
=
2
∂xj
j=1
= −k 2 x · 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 homogeneous
media 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
1.3. SPECIAL CASES
17
div H = 0 and k 2 div E = −iωµ0 div Je ; that is, div E = −(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 written
as
∆E + k 2 E = −iωµ0 Je +
1
∇ρ ,
ε0
∆H + k 2 H = − curl Je .
(1.12)
(1.13)
Let us consider the magnetic field first and introduce the magnetic Hertz potential : The
equation div H = 0 implies the existence of a vector potential A with H = curl A. Thus
(1.13) takes the form
curl(∆A + k 2 A) = − curl Je
and we obtain
∆A + k 2 A = −Je + ∇ϕ
(1.14)
for some scalar field ϕ. On the other hand, if A and ϕ satisfy (1.14) then
H = curl A and E = −
1
1
(curl H − Je ) = iωµ0 A −
∇(div A − ϕ)
iωε0
iωε0
satisfies the Maxwell system (1.7a)–(1.7d).
Analogously, we can introduce electric Hertz potentials if Je = 0. Because div E = 0 there
exists a vector potential A with E = curl A. Substituting this into (1.12) yields
curl(∆A + k 2 A) = 0
and we obtain
∆A + k 2 A = ∇ϕ
(1.15)
for some scalar field ϕ. On the other hand, if A and ϕ satisfy (1.15) then
E = curl A and H =
1
1
curl E = −iωε0 A +
∇(div A − ϕ)
iωµ0
iωµ0
satifies the Maxwell system (1.7a)–(1.7d). In any case we end up with an inhomogeneous
vector Helmholtz equation.
As a particular example we may take a magnetic Hertz vector A of the form A(x) = u(x) ẑ
with a scalar solution u of the two–dimensional Helmholtz equation and the unit vector
ẑ = (0, 0, 1)> ∈ R3 . Then
>
∂u
∂u
H = curl(uẑ) =
,−
,0
,
∂x2
∂x1
E = iωµ0 ẑ +
1
∇(∂u/∂x3 ) .
−iωε0
18
CHAPTER 1. INTRODUCTION
If u is independent of x3 then E has only a x3 −component, and the vector Helmholtz equation
for A reduces to a scalar Helmholtz equation for the potential u. The situation that E has
only 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 only
one 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 we
consider 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 continuous
with 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 conditions, we apply the law of induction (1.2) to a narrow rectangle-like surface R, containing
the normal n to the surface S and whose long sides C+ and C− are parallel to S and are on
the opposite sides of it, see the following figure.
When we let the height of the narrow sides, AA0 Rand BB 0 , approach zero then C+ and C−
∂
approach a curve C on S, the surface integral ∂t
B · ν ds will vanish in the limit because
R
the field remains finite. Note, that the
normal
ν
is
the
R
R normal to R lying in the tangential
plane of S. Hence, the line integrals C E + · τ d` and C E − · 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 form
Je (x + τ n(x)) = Js (x)δ(τ ) for small
R τ and x ∈ S and withRtangential surface field Js and σ
is finite, then the surface integral R Je · ν ds will tend to C Js · ν d`, and so the boundary
condition is
n × 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 boundary
S. Such a medium is characterized by the fact that the electric field vanishes inside this
medium, 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 = λ n × (E × n) on S
for some non-negative impedance function λ which, under appropriate conditions, may be
considered as an approximation of the transmission conditions. Of course these boundary
conditions occur also in the time harmonic case for the fields which we denote by capital
Latin 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 into
two parts Ω1 and Ω2 . We apply (1.3) first to all of Ω and then to Ω1 and Ω2 separately. The
addition of the last two formulas and the comparison with the first yields that the normal
component 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,1 E 1 − εr,2 E 2 ) = 0 on S
and n · (µr,1 H1 − µr,2 H2 ) = 0 on S .
Conclusion 1.6 The normal components of E and/or H are not continuous at interfaces
where εc and/or µr have jumps.
Finally, we specify the boundary conditions to the E- and H-modes defined above (see page
18). We assume that the surface S is an infinite cylinder in x3 −direction with constant
cross section. Furthermore, we assume that the volume current density J vanishes near the
20
CHAPTER 1. INTRODUCTION
boundary S and that
the surface current densities take the form Js = js ẑ for the E-mode
and Js = js ν × ẑ 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 boundary
condition
∂u
= −js on S ,
[u] = 0 ,
(σ − iωε)
∂ν
and in the H-mode we get
∂u
[u] = js ,
µ
= 0 on S .
∂ν
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 into
account. 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. Now
we look for waves with spherical wave fronts. A direct computation shows that
1 eik|x|
,
4π |x|
Φ(x) =
x ∈ R3 \ {0} ,
is a solution of the Helmholtz equation in R3 \ {0}. It is called the fundamental solution of
the Helmholtz equation at the origin (see Definition 3.1) which is essential as we will see in
Chapter 3. Physically it can be interpretated as the solution generated by a point source at
the origin similar to the electrostatic case (see Example 1.2).
Furthermore, defining
H(x) = curl Φ(x, 0) p
1
curl
=
4π
eik|x|
p
|x|
=
1 eik|x|
∇
×p
4π
|x|
for some a constant vector p ∈ C3 , we obtain from equation (6.5) and the fact that Φ(·, 0)
solves the Helmholtz equation,
curl H(x) = ∇ div Φ(x, 0) p − ∆ Φ(x, 0) p = ∇ div Φ(x, 0) p + k 2 Φ(x, 0) p
and thus by (6.3) curl curl H = k 2 H. Therefore H and E = ωεi 0 curl H constitute a solution
of the time harmonic Maxwell equations in vacuum (see Lemma 1.3).
Computing the gradient
eik|x| x
x
eik|x| x
= ik Φ(x, 0)
−
(1.19)
|x| |x|
|x|
4π|x|2 |x|
√
we obtain by recalling the frequency ω > 0 and wave number k = ω ε0 µ0 that the timedependent magnetic field has the form
1
x
ik
1
−iωt
H(x, t) = H(x) e
=
×p
−
ei(k|x|−ωt)
4π |x|
|x| |x|2
1
∇Φ(x, 0) =
4π
1
ik −
|x|
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| tends
to infinity. This asymptotic is not sufficient in describing a scattered field, since we also
obtain solutions of the Helmholtz equation and the Maxwell equations, respectively, with
this 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 first
case we obtain “outgoing” wave fronts while in the second case where the wave number is
negative we obtain “ingoing” wave fronts. For the scattering of electromagnetic waves the
scattered waves have to be outgoing waves. Thus, it is required to exclude the second ones
by additional conditions which are called radiation conditions.
From (1.19) we observe that the two cases for the Helmholtz equation can be distinguished
by subtracting ikΦ. This motivates a general characterization of radiating solutions u of the
Helmholtz equation by the Sommerfeld radiation condition, which is
x
· ∇u − iku = 0 .
lim |x|
|x|→∞
|x|
Similarly, we find a condition for radiating electromagnetic fields from the behavior of the
Hertz-dipol. Computing
i
curl H(x)
ωε0
ik|x|
i
x
x
1
ik
3x · p x
e
2
=
k
×p ×
+
−
−
p
4πωε0
|x|
|x|
|x|2 |x|
|x| |x|
|x|
E(x) =
we conclude
√
ik|x| 1
x
e
×p
= 0.
|x|→0
|x|→0
ik|x|
|x|
|x|
√
√
Analogously, we obtain lim|x|→0 µ0 H(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
µ0 H(x) × x − |x| ε0 E(x) = 0
(1.20a)
√
√
lim ε0 E(x) × x + µ0 |x|H(x) = lim
µ0
4π
2+
|x|→∞
or
lim
|x|→∞
√
√
ε0 E(x) × x + |x| µ0 H(x) = 0 ,
(1.20b)
are called Silver-Müller radiation condition for time harmonic electromagnetic fields.
Later we will show that these radiation conditions are sufficient for the existence of unique
solutions of scattering problems. Additionally from the representation theorem in Chapter 3
we will prove equivalent formulations and the close relationship of the Sommerfeld and the
Silver-Müller radiation condition. Additionally, the limiting absorption principle will give
another justification for this definition of radiating solutions.
Finally we discuss the energy of scattered waves. In general the energy density of electromagnetic fields in a linear medium is given by 21 (E · D + H · B). Thus, from Maxwell’s
22
CHAPTER 1. INTRODUCTION
equations, the identity (6.9), and the divergence theorem in a region Ω we obtain
Z
Z
∂ 1
E · D + H · B dx
=
E · (curl H − J ) − H · curl E dx
∂t 2 Ω
Ω
Z
=
div(E × H) − E · J dx
Ω
Z
Z
(E × H) · ν ds −
=
∂Ω
E · J dx .
Ω
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 ∂Ω
given by the Poynting vector , E × H, and the electrical work of the fields with the electrical
power J · E.
Let us consider the Poynting theorem in case of time harmonic fields with frequency
ω > 0 in
1
−iωt
vacuum, i.e. J = 0, ε = ε0 , µ = µ0 . Substituting E(x, t) = Re E(x)e
= 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
Z
∂ 1
2
2
ε0 E(x, t) + µ0 H(x, t) dx
∂t 2 Ω
Z
∂ 1
2 −2iωt
2
2 −2iωt
2
=
ε0 Re (E(x) e
) + 2|E(x)| + µ0 Re (H(x) e
) + 2|H| dx
∂t 8 Ω
Z
iω
= −
ε0 Re (E(x)2 e−2iωt ) + µ0 Re (H(x)2 e−2iωt ) dx .
4 Ω
On the other hand we compute the flux term as
Z
Z
Z
1
1
(E × H) · ν ds =
Re E × H · ν ds +
Re E × He−2iωt · ν ds .
2 ∂Ω
2 ∂Ω
∂Ω
By the Poynting Theorem the two integrals coincide for all t. Thus we conclude for the time
independent term
Z
(E × H) · ν ds = 0 .
Re
∂Ω
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 the
ball B(0, R) of radius R we observe conservation of energy in the form
Z
Z
Re
ν · (E × H) ds = Re
ν · (E × H) ds .
|x|=R
∂D
Furthermore, from this identity we obtain
Z
√
2 ε0 µ0 Re
ν · (E × H) ds
∂D
Z
2
2
Z
ε0 |E| + µ0 |H × ν| ds −
=
|x|=R
|x|=R
√
√
| µ0 H × ν − ε0 E|2 dx .
1.5. THE REFERENCE PROBLEMS
23
If we additionally assume radiating fields, the Silver-Müller condition implies
Z
√
µ0 H × ν − √ε0 E 2 ds = 0
lim
R→∞
which yields boundedness of
infinity.
1.5
|x|=R
R
|x|=R
|E|2 ds and, analogously,
R
|x|=R
|H|2 ds as R tends to
The Reference Problems
After this introduction into the mathematical description of electromagnetic waves the aim of
the textbook becomes more obvious. In general we can distinguish (at least) three common
approaches which lead to existence results of boundary value problems for linear partial
differential equations: expanding solutions into spherical wave functions by separation of
variables techniques, reformulation and treatment of a given boundary value problem in
terms of integral equations in Banach spaces of functions, and the reformulation of the
boundary value problem as a variational equation in Hilbert spaces of functions. It is the
aim of this monograph to discuss all of these common methods in the case of time harmonic
Maxwell’s equations.
We already observed the close connection of the Maxwell system with the scalar Helmholtz
equation. Therefore, before we treat the more complicated situation of the Maxwell system
we 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 in
the case of the Helmholtz equation and then we extend it to boundary value problems for
electromagnetic fields.
As we have mentioned already in the preface it is not our aim to present a collection of all
or at least some interesting boundary value problems. Instead, we present the ideas for two
classical 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 region D and some solution E inc and H inc of the unperturbed time harmonic
Maxwell system
curl E inc − iωµ0 H inc = 0 in R3 ,
curl H inc + iωε0 E inc = 0 in R3 ,
the problem is to determine E, H of the Maxwell system
curl E − iωµ0 H = 0 in R3 \ D ,
curl H + iωε0 E = 0 in R3 \ D ,
such that E satisfies the boundary condition
ν × E = 0 on ∂D
24
CHAPTER 1. INTRODUCTION
and both, E and H, can be decomposed into E = E s + E inc and H = H s + H inc in R3 \ D
with some scattered field E s , H s which satisfy the Silver-Müller radiation condition
√
x
√
s
s
lim |x|
µ0 H (x) ×
− ε0 E (x)
= 0
|x|→∞
|x|
√
x
√
s
s
lim |x|
+ µ0 H (x)
ε0 E (x) ×
= 0
|x|→∞
|x|
uniformly with respect to all directions x/|x|.
A
A
A
A
A
A E inc ,
A
A
A
A
A
A
H inc
ν×E =0
D
HH
HH
HH
j
E s, H s
A perfectly conducting cavity
For the second reference problem we consider D ⊆ R3 to be a bounded domain with sufficiently 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 to
determine a solution (E, H) of the time harmonic Maxwell system
curl E − iωµH = 0 in D ,
curl H + (iωε − σ)E = Je
in D ,
(1.21a)
(1.21b)
with the boundary condition
ν × E = 0 on ∂D .
(1.21c)
1.5. THE REFERENCE PROBLEMS
25
µ, ε, σ
ν×E =0
D
-
Of course, for general µ, ε, σ ∈ L∞ (R3 ) we have to give first a correct interpretation of the
differential equations by a so called weak formulation, which will be presented in detail in
Chapter 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 expect
radially symmetric solutions which can be computed by separation of variables in terms
of spherical coordinates. This approach will be worked out in Chapter 2. It will lead us
to a better understanding of electromagnetic waves from its expansion into spherical wave
functions. 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 to
represent electromagnetic waves by integrals over the boundary ∂D of the region D. These
integral representations by boundary potentials are the basis for deriving integral equations
on ∂D. We prefer to choose the “indirect” approach; that is, to search for the solution in
terms of potentials with densities which are determined by the boundary data through a
boundary integral equation. To solve this we will apply the Riesz–Fredholm theory. In this
way we are to prove the existence of unique solutions of the first reference problem for any
perfectly conducting smooth scattering obstacle.
The cavity problem will be investigated in Chapter 4. The treatment by a variational
approach requires the introduction of suitable Sobolev spaces. The Helmholtz decomposition
makes it possible to transfer the ideas of the simpler scalar Helmholtz equation to the Maxwell
system. The Lax–Milgram theorem in Hilbert spaces is the essential tool to establish an
existence result for the second reference problem.
The reason for including Chapter 5 into this monograph is different than for Chapters 3 and
4. While for the latter ones our motivation was the teaching aspect (these chapters arose
from graduate courses) the motivation for Chapter 5 is that we were not able to find such
a thorough presentation of boundary integral methods for Maxwell’s equations on Lipschitz
domains in any textbook. Lipschitz domains, in particular polyedral domains, play obviously
an important role in praxis. We were encouraged by our collegues from the numerical analysis
26
CHAPTER 1. INTRODUCTION
group to include this chapter to have a reference for further studies. The integral equation
methods themselves are not much different from the classical ones on smooth boundaries.
The mapping properties of the boundary operators, however, require a detailed study of
Sobolev spaces on Lipschitz boundaries. In this Chapter 5 we use and combine methods of
Chapters 3 and 4. Therefore, this chapter can not be studied independently of Chapters 3
and 4.
Chapter 2
Expansion into Wave Functions
This chapter, which is totally independent of the remaining parts of this monograph,1 studies
the fact that the solutions of the scalar Helmholtz equation or the vectorial Maxwell system
in balls can be expanded into certain special “wave functions”. We begin by expressing the
Laplacian in spherical coordinates and search for solutions of the scalar Laplace equation
or Helmholtz equation by separation of the (spherical) variables. It will turn out that
the spherical parts are eigensolutions of the Laplace–Beltrami operator while the radial
part solves an equation of Euler type for the Laplace equation and the spherical Bessel
differential equation for the case of the Helmholtz equation. The solutions of these differential
equations will lead to spherical harmonics and spherical Bessel and Hankel functions. We
will investigate these special functions in detail and derive many important properties. The
main goal is to express the solutions of the interior and exterior boundary value problems
as series of these wave functions. As always in this monograph we first present the analysis
for the scalar case of the Laplace equation and the Helmholtz equation before we consider
the 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 in
solutions u of the Laplace equation or Helmholtz equation – and later of the time-harmonic
Maxwell system – which can be separated into a radial part v : R>0 → C and a spherical
part K : S 2 → C; that is,
u(x) = v(r) K(x̂) ,
1
except of the proof of a radiation condition
27
r > 0 , x̂ ∈ S 2 ,
28
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
where S 2 = {x ∈ R3 : |x| = 1} denotes the unit sphere in R3 . Here, r = |x| =
x
∈ S 2 denote the spherical coordinates; that is,
and x̂ = |x|
p
x21 + x22 + x23


r cos ϕ sin θ
x =  r sin ϕ sin θ 
r cos θ
with r ∈ R>0 , ϕ ∈ [0, 2π), θ ∈ [0, π] ,
and x̂ = (cos ϕ sin θ, sin ϕ 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 also
in the differential equations for the magnetic and electric Hertz potentials. Also, solutions
of the full Maxwell system solve the vector Helmholtz equation in particular cases. Thus the
representation of the Laplacian in spherical polar coordinates is of essential importance.
1
1
∂2
∂
∂
1 ∂
2 ∂
+ 2
r
+ 2 2
sin θ
(2.1)
∆ = 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
2 ∂
1
∂2
1
∂
∂
∂2
+
+ 2 2
+ 2
sin θ
.
=
∂r2
r ∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
Definition 2.1 Functions u : Ω → R which satisfy the Laplace equation ∆u = 0 in Ω are
called harmonic functions in Ω. Complex valued functions are harmonic if their real and
imaginary parts are harmonic.
2
∂
∂
∂
2 ∂
(r2 ∂r
) = ∂r
The Laplace operator separates into a radial part, which is r12 ∂r
2 + r ∂r , and the
spherical part which is called the Laplace-Beltrami operator, a differential operator on the
unit sphere.
Definition 2.2 The differential operator ∆S 2 : C 2 (S 2 ) → C(S 2 ) with representation
1 ∂2
1 ∂
∂
∆S 2 =
+
sin θ
.
∂θ
sin2 θ ∂ϕ2 sin θ ∂θ
in spherical coordinates is called spherical Laplace-Beltrami operator on S 2 .
With the notation of the Beltrami operator and Dr =
∂
∂r
we obtain
1
1
2
1
∆ = Dr2 + Dr + 2 ∆S 2 = 2 Dr r2 Dr + 2 ∆S 2 .
r
r
r
r
Assuming that a potential or a component of a time-harmonic electric or magnetic field can
be separated as u(x) = u(rx̂) = v(r)K(x̂) with r = |x| and x̂ = x/|x| ∈ S 2 , a substitution
into the Helmholtz equation ∆u + k 2 u = 0 for some k ∈ C leads to
2
1
0 = ∆u(rx̂) + k 2 u(rx̂) = v 00 (r) K(x̂) + v 0 (r) K(x̂) + 2 v(r) ∆S 2 K(x̂) + k 2 v(r) K(x̂) .
r
r
2.1. SEPARATION IN SPHERICAL COORDINATES
29
Thus we obtain, provided v(r)K(x̂) 6= 0,
v 00 (r) + 2r v 0 (r)
1 ∆S 2 K(x̂)
+ 2
+ k2 = 0 .
v(r)
r K(x̂)
(2.2)
If there is a nontrivial solution of the supposed form, it follows the existence of a constant
λ ∈ C such that v satisfies the ordinary differential equation
r2 v 00 (r) + 2r v 0 (r) +
k 2 r2 + λ v(r) = 0 for r > 0 ,
(2.3)
and K solves the partial differential equation
∆S 2 K(x̂) = λK(x̂) on S 2 .
(2.4)
In the functional analytic language the previous equation describes the problem to determine
eigenfunctions K and corresponding eigenvalues λ of the spherical Laplace-Beltrami operator
∆S 2 . This operator is selfadjoint and non-positive with respect to the L2 (S 2 )−norm, see
Exercise 2.2. Especially, we observe that λ ∈ R<0 . We will explicitely construct a complete
orthonormal system of eigenfunctions although the existence of such a system follows from
functional analytic arguments as well because the resolvent of ∆S 2 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 Laplace
and the Helmholtz equation.
Using the explicit representation of the spherical Laplace-Beltrami operator transforms (2.4)
into
∂
∂
∂2
K(θ, ϕ) + sin θ
sin θ
K(θ, ϕ) = λ sin2 θ K(θ, ϕ)
∂ϕ2
∂θ
∂θ
where we write K(θ, ϕ) for K(x̂). Assuming eigenfunctions of the form K(θ, ϕ) = y1 (ϕ) y2 (θ)
leads to
0
sin θ sin θ y20 (θ)
y100 (ϕ)
+
− λ sin2 θ = 0
(2.5)
y1 (ϕ)
y2 (θ)
provided y1 (ϕ)y2 (θ) 6= 0. If the decomposition is valid there exists a constant µ ∈ C such
that y100 = µy1 . Since we are interested in differentiable solutions u, the function y1 must
be 2π periodic. Using a fundamental system of the linear ordinary differential equation we
conclude that µ = −m2 with m ∈ Z, and the general solution of y100 = µy1 is given by
y1,m (ϕ) = c1 eimϕ + c2 e−imϕ = c̃1 cos(mϕ) + c̃2 sin(mϕ)
with arbitrary constants c1 , c2 ∈ C and c̃1 , c̃2 ∈ C, respectively.
We assume that the reader is familar with the basics of the classical Fourier theory. In
particular, we recall that every function f ∈ L2 (−π, π) allows an expansion in the form
f (t) =
X
m∈Z
fm eimt
30
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
with Fourier coefficients
fm
1
=
2π
Zπ
f (s)e−ims ds ,
m ∈ Z.
−π
The convergence of the series must be understood in the L2 -sense, that is,
2
Zπ M
X
imt f
(t)
−
f
e
dt −→ 0 , N, M → ∞ .
m
−π
m=−N
Thus by the previous result we conclude that the possible functions {y1,m : m ∈ Z} constitute
a complete orthonormal system in L2 (−π, π).
Next we consider the dependence on θ. Using µ = −m2 turns (2.5) into
0
sin θ sin θ y20 (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 associated
Legendre differential equation
0
m2
2
0
(1 − z )w (z) − λ +
w(z) = 0 .
(2.6)
1 − z2
An investigation of this equation will be the main task of the next section leading to the
so called Legendre polynomials and to a complete orthonormal system of spherical surface
harmonics. In particular, we will see that only for λ = −n(n + 1) for n ∈ N0 there exist
smooth 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 Euler
equation
r2 v 00 (r) + 2r v 0 (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 fundamental
set 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 have
to choose v1,n (r) = rn . In the exterior of a ball v2,n (r) = r−(n+1) will lead to solutions which
decay 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 the
substitution z = kr and v(r) = v̂(kr) into the spherical Bessel differential equation
z 2 v̂ 00 (z) + 2z v̂ 0 (z) + z 2 − n(n + 1) v̂(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 to
solutions of the Helmholtz equation and further on also of the Maxwell system.
2.2. LEGENDRE POLYNOMIALS
2.2
31
Legendre Polynomials
Let us consider the case of harmonic functions u, that is ∆u = 0. In the previous section
we saw already that a separation in spherical coordinates leads to solutions of the form
u(x) = rn (c1 eimϕ + c2 e−i m ϕ ) 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; that
is,
d (1 − z 2 ) w0 (z) − λ w(z) = 0 , −1 < z < 1 .
(2.9)
dz
This is a differential equation of Legendre type.
The coefficients vanish at z = ±1. We are going to determine solutions which are continuous
up to the boundary, thus w ∈ C 2 (−1, +1) ∩ C[−1, +1].
Theorem 2.3 (a) For λ = −n(n + 1), n ∈ N ∪ {0}, there exists exactly one solution
w ∈ C 2 (−1, +1) ∩ C[−1, +1] with w(1) = 1. We set Pn = w. The function Pn is a
polynomial of degree n and is called Legendre polynomial of degree n. It satisfies the
differential equation
d (1 − x2 ) Pn0 (x) + n(n + 1) Pn (x) = 0 , −1 ≤ x ≤ 1 .
dx
(b) If there is no n ∈ N ∪ {0} with λ = −n(n + 1) then there is no non-trivial solution of
(2.9) in C 2 (−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) =
∞
X
aj x j
j=0
and substitute this into the differential equation (2.9). A simple calculation shows that the
coefficients have to satisfy the recursion
aj+2 =
j(j + 1) + λ
aj ,
(j + 1)(j + 2)
j = 0, 1, 2, . . . .
(2.10)
The ratio test, applied to
w1 (x) =
∞
X
a2k x
2k
and w2 (x) =
k=0
∞
X
a2k+1 x2k+1
k=0
separately yields that the radius of convergence is one. Therefore, for any a0 , a1 ∈ R the
function 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 reduce
to a finite sum. We study the behaviour of w1 (x) and w2 (x) as x tends to ±1. First we
consider w1 and split w1 in the form
w1 (x) =
kX
0 −1
k=0
a2k x
2k
+
∞
X
k=k0
a2k x2k
32
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
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) for
j = 2k yields
(2k + 1) − λ/2
2k(2k + 1) + λ
+ ln |a2k | = ln 1 −
+ ln |a2k | .
ln |a2(k+1) | = ln
(2k + 1)(2k + 2)
(2k + 1)(k + 1)
Now we use the elementary estimate ln(1 − u) ≥ −u − 2u2 for 0 ≤ u ≤ 1/2 which yields
2
(2k + 1) − λ/2
(2k + 1) − λ/2
−
+ ln |a2k |
ln |a2(k+1) | ≥ −
(2k + 1)(k + 1)
(2k + 1)(k + 1)
≥ −
1
c
− 2 + ln |a2k |
k+1
k
for some c > 0. Therefore, we arrive at the following estimate for ln |a2k |:
ln |a2k | ≥ −
k−1
X
j=k0
k−1
X 1
1
− c
+ ln |a2k0 | for k ≥ k0 .
j+1
j2
j=k
0
Using
k−1
X
j=k0
j+1
Zk
k−1 Z
X
dt
1
dt
≤
=
= ln k − ln k0
j+1
t
t
j=k
0
j
k0
yields
"
#
∞
X
1
ln |a2k | ≥ − ln k + ln k0 − c
+ ln |a2k0 |
2
j
j=k0
|
{z
}
= ĉ
and thus
exp ĉ
for all k ≥ k0 .
k
is positive for all k ≥ k0 and thus
|a2k | ≥
We note that a2k /a2k0
kX
∞
0 −1 a0
a2k
exp ĉ
exp ĉ X 1 2k
exp ĉ
w1 (x)
≥
+
−
x2k +
x ≥ c̃ −
ln(1 − x2 ) .
a2k0
a2k0
a
k
|a
|
|a
|
k
|a
|
2k0
2k0
2k0
2k0
k=1
k=1
From this we observe that w1 (x) → +∞ as x → ±1 or w1 (x) → −∞ as x → ±1 depending
on 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 can
not 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 conclude
that λ = −n(n + 1) for n = m − 2 and, furthermore, that a0 = 0 if n is odd and a1 = 0 if
2.2. LEGENDRE POLYNOMIALS
33
n is even. In particular, w is a polynomial of degree n. We can normalize w by w(1) = 1
because w(1) 6= 0. Indeed, if w(1) = 0 the differential equation
(1 − x2 ) w00 (x) − 2x w0 (x) + n(n + 1) w(x) = 0
for the polynomial w would imply w0 (1) = 0. Differentiating the differential equation would
yield 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) =
Pn
j=0
aj xj have the properties
(a) Pn is even for even n and odd for odd n.
(b) aj+2 =
j(j + 1) − n(n + 1)
aj ,
(j + 1)(j + 2)
j = 0, 1, . . . , n − 2.
Z+1
(c)
Pn (x)Pm (x) dx = 0 for n 6= m.
−1
Proof: Only part (c) has to be shown. We multiply the differential equation (2.9) for Pn
by Pm (x), the differential equation (2.9) for Pm by Pn (x), take the difference and integrate.
This yields
Z+1
0 =
d d 2
0
2
0
(1 − x ) Pn (x) − Pn (x)
(1 − x ) Pm (x) dx
Pm (x)
dx
dx
−1
Z+1
Pn (x) Pm (x) dx .
+ n(n + 1) − m(m + 1)
−1
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 Legendre
polynomials.
Lemma 2.5 For the Legendre polynomials it holds
max |Pn (x)| = 1 for all n = 0, 1, 2, . . . .
−1≤x≤1
Proof:
For fixed n ∈ N we define the function
Φ(x) := Pn (x)2 +
1 − x2
Pn0 (x)2
n(n + 1)
for all x ∈ [−1, +1] .
34
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
We differentiate Φ and have
1 − x2
x
00
0
0
0
P (x) −
P (x)
Φ (x) = 2 Pn (x) Pn (x) +
n(n + 1) n
n(n + 1) n
2 Pn0 (x)
d
2 x Pn0 (x)2
2
0
0
=
n(n + 1) Pn (x) + {(1 − x ) Pn (x)} + x Pn (x) =
.
n(n + 1)
dx
n(n + 1)
|
{z
}
= 0 by the differential equation
Therefore Φ0 > 0 on (0, 1] and Φ0 < 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) ≤ max{Φ(1), Φ(−1)} .
From Φ(1) = Φ(−1) = 1 we conclude that |Pn (x)| ≤ 1 for all x ∈ [−1, +1]. The lemma is
proven by noting that Pn (1) = 1.
2
A useful representation of the Legendre polynomials is a formula first shown by B.O. Rodrigues.
Theorem 2.6 For all n ∈ N0 the Legendre polynomials satisfy the formula of Rodrigues
Pn (x) =
1 dn 2
(x − 1)n ,
2n n! dxn
x ∈ R.
Proof: First we prove that the right hand side solves the Legendre differential equation;
that is, we show that
n+1
d
dn
2
n
2 d
(2.11)
(1 − x ) n+1 (x − 1)
+ n(n + 1) n (x2 − 1)n = 0 .
dx
dx
dx
We observe that both parts are polynomials of degree n. We multiply the first part by xj for
some j ∈ {0, . . . , n}, integrate and use partial integration two times. This yields for j ≥ 1
Z1
Aj :=
Z1
n+1
n+1
d
2 d
2
n
j−1
2 d
x
(1 − x ) n+1 (x − 1) dx = −j x (1 − x ) n+1 (x2 − 1)n dx
dx
dx
dx
j
−1
−1
Z1
= j
(j − 1) xj−2 − (j + 1) xj
dn 2
(x − 1)n dx .
dxn
−1
We note that no boundary contributions occur and, furthermore, that A0 = 0. Now we
use partial integration n times again. No boundary contributions occur either since there is
always at least one factor (x2 − 1) left. For j ≤ n − 1 the integral vanishes because the n-th
derivative of xj and xj−2 vanish for j < n. For j = n we have
n
Z1
An = −n(n + 1) (−1) n!
−1
(x2 − 1)n dx .
2.2. LEGENDRE POLYNOMIALS
35
Analogously, we multiply the second part of (2.11) by xj , integrate, and apply partial integration n-times. This yields
Z1
Bj = n(n + 1)
dn
xj n (x2 − 1)n dx = n(n + 1) (−1)n n!
dx
−1
Z1
(x2 − 1)n dx if j = n
−1
and zero if j < n. This proves that the polynomial
n+1
d
dn
2 d
2
n
(1 − x ) n+1 (x − 1)
+ n(n + 1) n (x2 − 1)n
dx
dx
dx
of degree n is orthogonal in L2 (−1, 1) to all polynomials of degree at most n and, therefore,
has to vanish. Furthermore,
dn dn 2
n
n n
(x + 1) (x − 1) (x − 1) =
dxn
dxn
x=1
x=1
n k
X
n d
dn−k
n
n
(x
+
1)
(x
−
1)
.
=
k dxk
dxn−k
x=1
k=0
x=1
dn−k
n
(x
−
1)
From
= 0 for k ≥ 1 we conclude
dxn−k
x=1
n n dn
dn 2
n
n
=
2
= 2n n!
(x
−
1)
(x
−
1)
n
0
dxn
dx
x=1
x=1
2
which proves the theorem.
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
Z+1
(a)
Pn (x)2 dx =
2
for all n = 0, 1, 2, . . ..
2n + 1
−1
Z+1
(b)
x Pn (x) Pn+1 (x) dx =
2(n + 1)
for all n = 0, 1, 2, . . ..
(2n + 1)(2n + 3)
−1
Proof:
Z+1
(a) We use the representation of Pn by Rodrigues and n partial integrations,
n
dn 2
n d
(x
−
1)
(x2 − 1)n dx = (−1)n
dxn
dxn
−1
Z+1
2
d2n 2
n
(x
−
1)
(x − 1)n dx
dx2n
−1
Z+1
Z+1
2
n
= (−1) (2n)! (x − 1) dx = (2n)! (1 − x2 )n dx .
n
−1
−1
36
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Z+1
It remains to compute In := (1 − x2 )n dx.
−1
We claim:
In = 2
(n!)2
2n(2n − 2) · · · 2
= 2 · 4n
(2n + 1)(2n − 1) · · · 1
(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
Z+1
Z+1
d 1
2
2 n−1
2 n
−
x (1 − x )
dx = In−1 +
x
(1 − x ) dx
dx 2n
−1
= In−1 −
1
In
2n
−1
and thus In =
2n
In−1 .
2n + 1
This proves the representation of In . We arrive at
Z+1
Pn (x)2 dx =
1
2
(n!)2
n
(2n)!
2
·
4
=
.
n
2
(2 n!)
(2n + 1)!
2n + 1
−1
(b) This is proven quite similarily:
Z+1
dn
dxn+1
x n (x2 − 1)n n+1 (x2 − 1)n+1 dx
dx
dx
−1
n
Z+1
n+1
d
n+1
2
n+1 d
2
n
= (−1)
(x − 1)
(x − 1) dx
x
dxn+1 dxn
−1
= (−1)n+1
Z+1


 d2n+1 2

d2n 2
n
(x
−
1)
(x − 1)n 
(x2 − 1)n+1 
x
+
(n
+
1)
 dx
 dx2n+1
2n
dx {z
|
{z
}
|
}
−1
=0
= (n + 1) (−1)n+1 (2n)! In+1 = 2(n + 1)(2n)!
= (n + 1)
=(2n)!
(2n + 2)(2n)(2n − 2) · · · 2
(2n + 3)(2n + 1)(2n − 1) · · · 1
2(2n n!)(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 Legendre
polynomials. We prove only some of these formulas. For the remaining parts we refer
to the exercises.
2.2. LEGENDRE POLYNOMIALS
37
Theorem 2.8 For all x ∈ R and n ∈ N, n ≥ 0, we have
0
0
(a) Pn+1
(x) − Pn−1
(x) = (2n + 1) Pn (x),
(b) (n + 1)Pn+1 (x) = (2n + 1)x Pn (x) − nPn−1 (x),
0
(c) Pn0 (x) = nPn−1 (x) + xPn−1
(x),
0
(x),
(d) xPn0 (x) = nPn (x) + Pn−1
(e) (1 − x2 )Pn0 (x) = (n + 1) x Pn (x) − Pn+1 (x) = −n x Pn (x) − Pn−1 (x) ,
0
0
(x) = Pn (x) + 2x Pn0 (x),
(x) + Pn−1
(f ) Pn+1
(g) (2n + 1)(1 − x2 )Pn0 (x) = n(n + 1) Pn−1 (x) − Pn+1 (x) ,
0
0
(h) n Pn+1
(x) − (2n + 1)x Pn0 (x) + (n + 1)Pn−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 the
formula of Rodrigues,
1
dn+2 2
(x − 1)n+1
2n+1 (n + 1)! dxn+2
1
dn
d
2
n
= n+1
x (x − 1)
2(n + 1)
2
(n + 1)! dxn
dx
0
Pn+1
(x) =
1 dn 2
n
2 2
n−1
(x
−
1)
+
2n
x
(x
−
1)
2n n! dxn
1
dn 2
n
2
n−1
= Pn (x) + n−1
(x
−
1)
+
(x
−
1)
2
(n − 1)! dxn
=
0
= Pn (x) + 2n Pn (x) + Pn−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. This
yields existence of αn , βn ∈ R and qn−3 ∈ Pn−3 such that
Pn+1 (x) = αn xPn (x) + βn Pn−1 (x) + qn−3 .
The orthogonality condition implies that
Z+1
qn−3 Pn+1 dx = 0 ,
Z+1
qn−3 Pn−1 dx = 0 ,
Z+1
x qn−3 (x) Pn (x) dx = 0 ,
−1
−1
−1
38
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
thus
Z+1
qn−3 (x)2 dx = 0 ,
−1
and therefore qn−3 ≡ 0. From 1 = Pn+1 (1) = αn Pn (1) + βn Pn−1 (1) = αn + βn we conclude
that
Pn+1 (x) = αn x Pn (x) + (1 − αn )Pn−1 (x) .
We determine αn by Theorem 2.7:
Z+1
0 =
Pn−1 (x) Pn+1 (x) dx
−1
Z+1
Z+1
= αn x Pn−1 (x) Pn (x) dx + (1 − αn ) Pn−1 (x)2 dx
−1
= αn
thus αn =
2n
2
−
(2n − 1)(2n + 1)
2n − 1
−1
+
2
,
2n − 1
2n + 1
. This proves part (b).
n+1
(c) The definition of Pn yields
1 dn+1 2
(x − 1)n
2n n! dxn+1
1
dn
[x(x2 − 1)n−1 ]
= n−1
2 (n − 1)! dxn
Pn0 (x) =
=
x
n
dn 2
dn−1 2
n−1
(x
−
1)
+
(x − 1)n−1
n−1
n
n−1
n−1
2 (n − 1)! dx
2 (n − 1)! dx
0
= x Pn−1
(x) + n Pn−1 (x) .
(d) Differentiation of (b) and multiplication of (c) for n + 1 instead of n by n + 1 yields
0
0
(x),
(n + 1) Pn+1
(x) = (2n + 1)x Pn0 (x) + (2n + 1)Pn (x) − nPn−1
0
(n + 1) Pn+1
(x) = (n + 1)2 Pn (x) + (n + 1)x Pn0 (x) ,
thus by subtraction
0
0 = (2n + 1) − (n + 1)2 Pn (x) + (2n + 1) − (n + 1) x Pn0 (x) − n Pn−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 solutions
in C 2 (−1, +1) ∩ C[−1, +1] for m 6= 0.
2.2. LEGENDRE POLYNOMIALS
39
Theorem 2.9 The functions
dm
Pn (x) , −1 < x < 1, 0 ≤ m ≤ n ,
dxm
are solutions of the differential equation (2.6) for λ = −n(n + 1); that is,
m2
d
2 d
m
(1 − x ) Pn (x) + n(n + 1) −
Pnm (x) = 0 , −1 < x < 1 ,
2
dx
dx
1−x
Pnm (x) = (1 − x2 )m/2
for all 0 ≤ m ≤ n. The functions Pnm are called associated Legendre functions.
Proof:
We compute
d m
dm
dm+1
Pn (x) = −m x (1 − x2 )m/2 m Pn (x) + (1 − x2 )m/2+1 m+1 Pn (x) ,
dx
dx
dx
d
d
dm
(1 − x2 ) Pnm (x) = −m Pnm (x) + m2 x2 (1 − x2 )m/2−1 m Pn (x)
dx
dx
dx
(1 − x2 )
2 m/2
− m x (1 − x )
dm+1
Pn (x)
dxm+1
− (m + 2) x (1 − x2 )m/2
= −m Pnm (x) +
dm+1
Pn (x) + Pnm+2 (x)
dxm+1
m2 x2 m
(m + 1) 2x m+1
Pn (x) − √
Pn (x)
2
1−x
1 − x2
+ Pnm+2 (x) .
Differentiating equation (2.9) m times yields
m+1
m+2−k
X m + 1 dk
dm
2 d
(1
−
x
)
P
(x)
+
n(n
+
1)
P (x) = 0 .
k
m+2−k n
m n
k
dx
dx
dx
k=0
(2.12)
The sum reduces to three terms only, thus
(1 − x2 )
dm
dm+2
dm+1
P
(x)
−
(m
+
1)
2x
P
(x)
+
n(n
+
1)
−
m(m
+
1)
Pn (x) = 0 .
n
n
dxm+2
dxm+1
dxm
We multiply the identity by (1 − x2 )m/2 and arrive at
Pnm+2 (x) −
(m + 1) 2x m+1
√
Pn (x) + n(n + 1) − m(m + 1) Pnm (x) = 0 .
1 − x2
Combining this with the previous equation by eliminating the terms involving Pnm+1 (x) and
Pnm+2 (x) yields
d
m2 x 2 m
2 d
m
(1 − x ) Pn (x) = −m Pnm (x) +
Pn (x) − n(n + 1) − m(m + 1) Pnm (x)
2
dx
dx
1−x
m2
=
− n(n + 1) Pnm (x)
1 − x2
which proves the theorem.
2
40
2.3
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Expansion into Spherical Harmonics
Now we return to the Laplace equation ∆u = 0 and collect our arguments. We have shown
that for n ∈ N and 0 ≤ m ≤ n the functions
n m
±i m ϕ
hm
n (r, θ, ϕ) = r Pn (cos θ) e
are harmonic functions in all of R3 . Analogously, the functions
v(r, θ, ϕ) = r−n−1 Pnm (cos θ) e±imϕ
are harmonic in R3 \ {0} and decay at infinity.
It is not so obvious that the functions hm
n are polynomials.
±imϕ
n m
with 0 ≤ m ≤ n are homoTheorem 2.10 The functions hm
n (r, θ, ϕ) = r Pn (cos θ) e
n n
geneous polynomials of degree n. The latter property means that hm
n (µx) = µ hm (x) for all
x ∈ 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) =
`
X
aj t2j ,
t ∈ R,
j=0
for some aj . From r = |x| and x3 = r cos θ we write h0n in the form
h0n (x)
n
= |x| Pn
x3
|x|
`
X
x2j
3
2(`−j)
= |x|
a2j x2j
,
a2j 2j =
3 |x|
|x|
j=0
j=0
2`
`
X
and this is obviously a polynomial of degree 2` = n. The same arguments hold for odd
values of n.
Now we show that also the associated functions; that is, for m > 0, are polynomials of degree
n. We write
dm
m
n m
imϕ
n
m
hn (r, θ, ϕ) = r Pn (cos θ) e
= r sin θ m Pn (x)
eimϕ .
dx
x=cos θ
P`
dm
2j−m
and set Q = dx
for some bj .
m Pn . Let again n = 2` be even. Then Q(t) =
j=0 bj t
Furthermore, we use the expressions
q
x3
1
cos θ =
,
sin θ =
x21 + x22 ,
r
r
as well as
cos ϕ =
1 x1
x1
= p 2
,
r sin θ
x1 + x22
x2
sin ϕ = p 2
.
x1 + x22
2.3. EXPANSION INTO SPHERICAL HARMONICS
41
to obtain
hm
n (x)
= r
=
n−m
`
X
(x21
+
x22 )m/2
x x (x + ix )m
3
3
1
2
n−m
= r
Q
(x1 + ix2 )m
Q
2 m/2
2
r (x1 + x2 )
r
bj r2(`−j) x2j−m
(x1 + ix2 )m
3
j=0
which proves the assertion for even n. For odd n one argues analogously. It is clear from the
2
definition that the polynomial hm
n is homogeneous of degree n.
Definition 2.11 Let n ∈ N ∪ {0}.
(a) Homogeneous harmonic polynomials of degree n are called spherical harmonics of order
n.
(b) The functions Kn : S 2 → C, defined as restrictions of spherical harmonics of degree n
to the unit sphere are called spherical surface harmonics of order n.
|m|
n
imϕ
are spherical harmonics of order n
Therefore, the functions hm
n (r, θ, ϕ) = r Pn (cos θ)e
for −n ≤ m ≤ n. Spherical harmonics which do not depend on ϕ are called zonal. Thus, in
case of m = 0 the functions h0n are zonal spherical harmonics.
For any spherical surface harmonic Kn of order n the function
x
n
Hn (x) = |x| Kn
, x ∈ R3 ,
|x|
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 (−x̂) = (−1)n Kn (x̂) for all x̂ ∈ S 2 .
R
(b) Kn (x̂) Km (x̂) ds(x̂) = 0 for all n 6= m.
S2
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
Z Z
∂
∂
Hn Hm − Hm Hn ds =
(Hn ∆Hm − Hm ∆Hn ) dx = 0 .
∂r
∂r
S2
|x|≤1
Setting f (r) = Hn (rx̂) = rn Hn (x̂) for fixed x̂ ∈ S 2 we have that f 0 (1) =
that is,
Z
Z
0 = (m − n)
Hn Hm ds = (m − n)
S2
∂
H (x̂)
∂r n
= n Hn (x̂);
Kn Km ds .
S2
Now we determine the dimension of the space of spherical harmonics for fixed order n.
2
42
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Theorem 2.13 The set of spherical harmonics of order n is a vector space of dimension
2n + 1. In particular, there exists a system {Knm : −n ≤ m ≤ n} of spherical harmonics of
order n such that
Z
1 for m = `,
m
`
Kn Kn ds = δm,` =
0 for m 6= ` ;
S2
that is, {Knm : −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
X
An−j (x1 , x2 ) xj3
x ∈ R3 ,
(2.13)
j=0
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
X
xj3
∆2 An−j (x1 , x2 ) +
j=0
2
n
X
j(j − 1) xj−2
An−j (x1 , x2 ) ,
3
j=2
2
∂
∂
where ∆2 = ∂x
2 + ∂x2 denote the two-dimensional Laplace operator. From ∆2 A0 = ∆2 A1 = 0
1
2
we conclude that
0 =
n−2
X
[∆2 An−j (x1 , x2 ) + (j + 1)(j + 2) An−j−2 (x1 , x2 )] xj3 ,
j=0
and thus by comparing the coefficients
An−j−2 (x1 , x2 ) = −
1
∆2 An−j (x1 , x2 )
(j + 1)(j + 2)
for all (x1 , x2 ) ∈ R2 , j = 0, . . . , n − 2; that is (replace n − j by j)
Aj−2 = −
1
∆2 Aj
(n − j + 1)(n − j + 2)
for j = n, n − 1, . . . , 2 .
(2.14)
Also, one can reverse the arguments: If An and An−1 are homogeneous polynomials of degree
n and n − 1, respectively, then all of the functions Aj defined by (2.14) are homogeneous
polynomials 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
P
ai xi1 x2n−i we note that the dimension of the space of
i=0
all homogeneous polynomials of degree n is just n + 1. Therefore, the dimension of the space
of all spherical harmonics of order n is (n + 1) + n = 2n + 1. Finally, it is well known that
2.3. EXPANSION INTO SPHERICAL HARMONICS
43
any basis of this finite dimensional Euclidian space can be orthogonalized by the method of
Schmidt.
2
Remark: The set {Knm : −n ≤ m ≤ n} is not uniquely determined. Indeed, for any
orthogonal matrix A ∈ R3×3 ; that is, A> A = I, also the set {Knm (Ax) : −n ≤ m ≤ n} is an
orthonormal system. Indeed, the substitution x = Ay yields
Z
Z
m0
m
Knm (Ay) Knm00 (Ay) ds(y) = δn,n0 δm,m0 .
Kn (x) Kn0 (x) ds(x) =
S2
S2
|m|
n
imϕ
m
with
Now we can state that the spherical harmonics hm
n (x) = hn (θ, ϕ) = r Pn (cos θ)e
−n ≤ m ≤ n, which we have determined by separation, constitute such an orthogonal basis
of the space of spherical harmonics of order n.
|m|
n
imϕ
, −n ≤ m ≤ n are spherical
Theorem 2.14 The functions hm
n (r, θ, ϕ) = r Pn (cos θ)e
harmonics of order n. They are mutually orthogonal and, therefore, form an orthogonal basis
of the (2n + 1)-dimensional space of all spherical harmonics of order n.
Proof: We already know that hm
n are spherical harmonics of order n ∈ N. Thus the
theorem follows from
Z
`
hm
n (x) hn (x) dx
Z1 Zπ Z2π
=
0
|x|<1
0
r2n Pn|m| (cos θ) Pn|`| (cos θ) ei(m−`)ϕ r2 sin θ dϕ dθ dr = 0
0
2
for m 6= `.
We want to normalize these functions. First we consider the associated Legendre functions.
Theorem 2.15 The norm of the associated Legendre functions Pnm in L2 (−1, 1) is given by
Z+1
Pnm (x)2 dx =
2
(n + m)!
·
,
2n + 1 (n − m)!
m = 0, . . . , n,
n ∈ N ∪ {0} .
−1
Proof:
The case m = 0 has been proven in Theorem 2.7 already.
For m ≥ 1 partial integration yields
Z+1
Z+1
dm
dm
Pnm (x)2 dx =
(1 − x2 )m m Pn (x) m Pn (x) dx
dx
dx
−1
−1
Z+1
= −
−1
m−1
m
d
d
2 m d
(1 − x )
P
(x)
Pn (x) dx .
n
dx
dxm
dxm−1
(2.15)
44
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Now we differentiate the Legendre differential equation (2.9) (m − 1)-times (see (2.12) for m
replaced by m − 1); that is,
(1 − x2 )
dm−1
dm+1
dm
P
(x)
−
2m
x
P
(x)
+
n(n
+
1)
−
m(m
−
1)
Pn (x) = 0 ,
n
n
dxm+1
dxm
dxm−1
thus, after multiplication by (1 − x2 )m−1 ,
m
m−1
d
2 m d
2 m−1 d
(1 − x )
P
(x)
+
n(n
+
1)
−
m(m
−
1)
(1
−
x
)
Pn (x) = 0 .
n
dx
dxm
dxm−1
Substituting this into (2.15) yields
Z+1
Z+1
Z+1
2
2
m−1
m
Pn (x) dx = (n+m) (n−m+1) Pnm−1 (x)2 dx .
Pn (x) dx = n(n+1)−m(m−1)
−1
−1
−1
This is a recursion formula for
+1
R
Pnm (x)2 dx with respect to m and yields the assertion by
−1
2
using the formula for m = 0.
Now we define the normalized spherical surface harmonics Ynm by
s
(2n + 1) (n − |m|)! |m|
Ynm (θ, ϕ) :=
Pn (cos θ) eimϕ , −n ≤ m ≤ n, n = 0, 1, . . .
4π (n + |m|)!
(2.16)
They form an orthonormal system in L2 (S 2 ). We will identify Ynm (x) with Ynm (θ, ϕ) for
x = (sin θ cos ϕ , sin θ sin ϕ , cos θ)> ∈ S 2 .
From (2.4) with λ = −n(n + 1) we remember that Ynm satisfies the differential equation
1 ∂
∂Ynm (θ, ϕ)
1 ∂ 2 Ynm (θ, ϕ)
+ n(n + 1) Ynm (θ, ϕ) = 0 ;
sin θ
+
2
2
sin θ ∂θ
∂θ
∂ϕ
sin θ
that is,
∆S 2 Ynm + n(n + 1) Ynm = 0
(2.17)
with the Laplace-Beltrami operator ∆S 2 of Definition 2.2. Thus, we see that λ = −n(n + 1)
for n ∈ N are the eigenvalues of ∆S 2 with eigenfunctions Ynm , |m| ≤ n. We mentioned
already (see also Exercise 2.2) that this operator ∆S 2 is selfadjoint and non-negative. In the
setting of abstract functional analysis we note that −∆ is a densely defined and unbounded
operator from L2 (S 2 ) into itself which is selfadjoint. This observation
allows the use of
generalfunctional analytic tools to prove, e.g., that the eigenfunctions Ynm : |m| ≤ n, n =
0, 1, . . . of −∆ form a complete orthonormal system of L2 (S 2 ). We are going to prove this
fact 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.
Z
f (x · y) Ynm (y) ds(y) = λn Ynm (x) , x ∈ S 2 ,
S2
for all n ∈ N and m = −n, . . . , n where λn = 2π
R1
−1
f (t) Pn (t) dt.
2.3. EXPANSION INTO SPHERICAL HARMONICS
45
Proof: We keep x ∈ S 2 fixed and choose an orthogonal matrix A which depends on x
such that x̂ := A−1 x = A> x is the “north pole”; that is, x̂ = (0, 0, 1)> . The transformation
y = Ay 0 yields
Z
Z
Z
m
0
m
0
0
f (x · y) Yn (y) ds(y) =
f (x · Ay ) Yn (Ay ) ds(y ) =
f (x̂ · y) Ynm (Ay) ds(y) .
S2
S2
S2
The function Ynm (Ay) is again a spherical surface harmonic of order n, thus
Ynm (Ay)
n
X
=
ak Ynk (y) ,
y ∈ S2 ,
(2.18)
k=−n
where ak =
R
Ynm (Ay) Yn−k (y) ds(y). Using this and polar coordinates
S2
y = (sin θ cos ϕ , sin θ sin ϕ , cos θ)> ∈ S 2 yields (note that x̂ · y = cos θ)
Z
Z
n
X
m
f (x · y) Yn (y) ds(y) =
ak f (x̂ · y) Ynk (y) ds(y)
S2
k=−n
S2
n
X
s
=
ak
k=−n
Z
π
(2n + 1) (n − |k|)!
·
4π (n + |k|)!
Z
·
0
r
= a0
2π
f (cos θ) Pn|k| (cos θ) eikϕ dϕ sin θ dθ
0
2n + 1
2π
4π
Z
r
π
f (cos θ) Pn (cos θ) sin θ dθ = λn a0
0
2n + 1
4π
where we have used the substitution t = cos θ in the last integral. Now we substitute y = x̂
in (2.18) and have, using Ykn (x̂) = 0 for k 6= 0,
r
n
X
2n + 1
Ynm (x) = Ynm (Ax̂) =
Pn (1) ,
ak Ynk (x̂) = a0
4π | {z }
k=−n
thus
Z
=1
f (x · y) Ynm (y) ds(y) = λn Ynm (x)
S2
2
which proves the theorem.
As a first application of this result we prove the addition formula. A second application will
be the Jacobi–Anger expansion, see Theorem 2.32.
Theorem 2.17 The Legendre polynomials Pn and the spherical surface harmonics defined
in (2.16) satisfy the addition formula
n
X
4π
Pn (x · y) =
Y m (x) Yn−m (y)
2n + 1 m=−n n
for all x, y ∈ S 2 .
46
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Proof: For fixed y ∈ S 2 the function x 7→ Pn (x · y) is a spherical harmonic of order n and,
therefore, has an expansion of the form
Z
n
n
X
X
Pn (x · y) =
Pn (z · y) Y −m (z) ds(z) Ynm (x) =
λn Y −m (y) Ynm (x)
m=−n
S2
m=−n
with
Z
1
4π
2n + 1
−1
by Theorem 2.7 where we used the previously proven Funk–Hecke formula for f = Pn .
λn = 2π
Pn (t) Pn (t) dt =
2
We formulate a simple conclusion as a corollary.
Corollary 2.18
(a)
n
X
|Ynm (x)|2 =
m=−n
r
(b) Ynm (x) ≤
2n + 1
4π
2n + 1
4π
for all x ∈ S 2 and n = 0, 1, . . ..
for all x ∈ S 2 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 harmonics.
Theorem 2.19 The functions Ynm : −n ≤ m ≤ n, n ∈ N ∪ {0} are complete in L2 (S 2 );
that is, every function f ∈ L2 (S 2 ) can be expanded into a generalized Fourier series in the
form
∞ X
n
X
f =
(f, Ynm )L2 (S 2 ) Ynm .
(2.19a)
n=0 m=−n
The series can also be written as
Z
∞
1 X
(2n + 1) f (y) Pn (y · x) ds(y) ,
f (x) =
4π n=0
x ∈ S2 .
(2.19b)
S2
The convergence in (2.19a) and (2.19b) has to be understood in the L2 –sense.
Furthermore, on bounded sets in C 1 (S 2 ) the series converge even uniformly; that is, for every
M > 0 and ε > 0 there exists N0 ∈ N, depending only on M and ε, such that
N
n
X X
(f, Ynm )L2 (S 2 ) Ynm − f n=0 m=−n
∞
N
n
X
X
m
m
= max2 (f, Yn )L2 (S 2 ) Yn (x̂) − f (x̂) ≤ ε
x̂∈S n=0 m=−n
2.3. EXPANSION INTO SPHERICAL HARMONICS
47
for all N ≥ N0 and all f ∈ C 1 (S 2 ) with kf k1,∞ = max{kf k∞ , kf 0 k∞ } ≤ M , and, analogously, for (2.19b).
Here, the space C 1 (S 2 ) consists of those functions f such that (with respect to spherical
coordinates θ and ϕ) the functiuons f , ∂f /∂θ, and sin1 θ ∂f /∂ϕ are continuous and periodic
with respect to ϕ.
Proof: First we prove the second part. Therefore, let f ∈ C 1 (S 2 ) with kf k1,∞ ≤ M . With
the Addition Formula, see Theorem 2.17, we obtain for the partial sum
Z
N
n
N
n
X
X
X
X
m
m
(SN f )(x) =
(f, Yn )L2 (S 2 ) Yn (x) =
f (y)
Ynm (x) Yn−m (y) ds(y)
n=0 m=−n
S2
n=0 m=−n
N Z
X
2n + 1
Pn (x · y) f (y) ds(y) .
=
4π
n=0
S2
0
0
This yields already the equivalence of (2.19a) and (2.19b). With (2n + 1) Pn = Pn+1
− Pn−1
of Theorem 2.8 (set P−1 ≡ 0) this yields
N Z
0
1 X
0
(SN f )(x) =
f (y) Pn+1
(x · y) − Pn−1
(x · y) ds(y)
4π n=0
S2
=
1
4π
Z
f (y) PN0 +1 (x · y) + PN0 (x · y) ds(y) .
S2
Let again z = (0, 0, 1)> be the north pole and choose an orthogonal matrix A (depending on
x) such that Ax = z. Then, by the transformation formula,
Z
1
f (Ay) PN0 +1 (z · y) + PN0 (z · y) ds(y) .
(SN f )(x) =
4π
S2
In spherical polar coordinates y(θ, ϕ) = (sin θ cos ϕ, sin θ sin ϕ, cos θ)> this is, defining F (θ) =
R2π
1
f Ay(θ, ϕ) dϕ,
2π
0
1
(SN f )(x) =
2
Zπ
F (θ) PN0 +1 (cos θ) + PN0 (cos θ) sin θ dθ
0
1
=
2
Z+1
F (arccos t) PN0 +1 (t) + PN0 (t) dt
−1
Z+1
+1
1
1
d
=
F (arccos t) PN +1 (t) + PN (t) −
F (arccos t) PN +1 (t) + PN (t) dt .
2
2
dt
−1
|
{z
}
−1
=F (0)
48
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
We note that partial integration is allowed because t 7→ F (arccos t) is continuously differentiable in (−1, 1) and continuous in [−1, 1]. The value θ = 0 corresponds to the north pole
y = z, thus
Z2π
1
f (Az) dϕ = f (Az) = f (x) ,
F (0) =
2π
0
and therefore for any δ ∈ (0, 1) we obtain
+1
Z
1
d
(SN f )(x) − f (x) =
F
(arccos
t)
P
(t)
+
P
(t)
dt
N
+1
N
2 dt
−1
1
≤
2
−1+δ
Z
+
−1
=
Z1−δ
−1+δ
Z1 d
+
dt F (arccos t) |PN +1 (t)| + |PN (t)| dt
1−δ
1
( I1 + I2 + I3 ) .
2
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 dtd arccos t < 0 we conclude
−1+δ
Z
I1 + I3 ≤ −
Z1
−
−1
|F 0 (arccos t)|
d
arccos t dt
dt
1−δ
≤ kF 0 k∞ 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 ≤ 2ε . Then δ depends only on M and ε.
With this choise of δ we consider I2 and use the inequality of Cauchy-Schwarz.
I2
Z+1
d
≤
max F (arccos t) 1 · |PN +1 (t)| + |PN (t)| dt
−1+δ≤t≤1−δ dt
{z
}−1
|
=:c
√
≤
2 c kPN +1 kL2 + kPN kL2
√
≤ 2 2c
r
2
.
2N + 1
We estimate c by
0
c ≤ kF k∞
d
max arccos t ≤ M
−1+δ≤t≤1−δ dt
d
max arccos t .
−1+δ≤t≤1−δ dt
Now we can choose N0 , depending only on ε and M , such that I2 ≤ 2ε for all N ≥ N0 .
Therefore, |SN (f )(x) − f (x)| ≤ ε for all N ≥ N0 . This proves uniform convergence of the
Fourier series.
2.3. EXPANSION INTO SPHERICAL HARMONICS
49
The first part is proven by an approximation argument. Indeed, we use the general property
of orthonormal systems that SN f is the best approximation of f in the subspace span{Ynm :
|m| ≤ n, n = 0, . . . , N }; that is,
kSN f − f kL2 (S 2 ) ≤ kg − f kL2 (S 2 )
for all g ∈ span{Ynm : |m| ≤ n, n = 0, . . . , N } .
Let now f ∈ L2 (S 2 ) and ε > 0 be given. Since the space C 1 (S 2 ) is dense in L2 (S 2 ) there
exists h ∈ C 1 (S 2 ) such that kh − f kL2 (S 2 ) ≤ ε/2. Therefore,
kSN f − f kL2 (S 2 ) ≤ kSN h − f kL2 (S 2 ) ≤ kSN h − hkL2 (S 2 ) + kh − f kL2 (S 2 )
√
√
ε
4π kSN h − hk∞ + kh − f kL2 (S 2 ) ≤ 4π kSN h − hk∞ + .
≤
2
Since SN h converges uniformly to h we can find N0 ∈ N such that the first part is less than
ε/2 for all N ≥ N0 which ends the proof.
2
As a corollary we can prove completeness of the Legendre polynomials.
p
Corollary 2.20 The polynomials
n + 1/2 Pn : n ∈ N0 form a complete orthonormal
system in L2 (−1, 1); that is, for any f ∈ L2 (−1, 1) there holds
f =
∞
X
f n Pn
with
fn =
n=0
1
n+
2
Z1
f (t) Pn (t) dt , n ∈ N0 .
−1
For f ∈ C 1 [−1, 1] the series converges uniformly.
Proof: The function g(x) = f (x3 ) = f (cos θ) can be considered as a function on the sphere
which is independent of ϕ, thus it is a zonal function. The expansion (2.19a) yields
f (x3 ) =
∞ X
n
X
m
am
n Yn (x)
n=0 m=−n
with
s
am
n
=
(g, Ynm )L2 (S 2 )
=
2n + 1 (n − m)!
4π (n + m)!
Zπ Z2π
0
f (cos θ) Pn|m| (cos θ) eimϕ sin θ dϕ dθ
0

m 6= 0 ,
 0,
π
p
R
=
 (2n + 1)π f (cos θ) Pn (cos θ) sin θ dθ , m = 0 ,
0

m 6= 0 ,
 0,
1
p
R
=
 (2n + 1)π f (t) Pn (t) dt , m = 0 .
−1
Thus,
f (x3 ) =
∞
X
n=0
and the proof is complete.
r
∞
X
4π
0
fn Yn (x) =
fn Pn (x3 )
2n + 1
n=0
2
50
2.4
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Laplace’s Equation in the Interior and Exterior of
a Ball
In the previous two sections we constructed explicitely a complete orthonormal system of
functions in L2 (S 2 ). They play exactly the role of the normalized exponential functions
1
exp(inϕ) : n ∈ Z on the unit circle S 1 , parametrized by x = (cos ϕ, sin ϕ)> , ϕ ∈ [0, 2π];
2π
that is, the classical Fourier expansion functions. It is the aim of this section to expand
solutions of the Laplace equation for balls and solve the corresponding Dirichlet boundary
value problems.
Theorem 2.21 Let u ∈ C 2 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 αnm ∈ C, |m| ≤ n,
n = 0, 1, 2, . . . with
u(rx̂) =
∞ X
n
X
αnm rn Ynm (x̂) ,
0 ≤ r < R , x̂ ∈ S 2 .
(2.20)
n=0 m=−n
The series converges uniformly with all of its derivatives in every closed ball B[0, R0 ] with
R0 < R.
Proof: For every r ∈ (0, R) the function x̂ 7→ u(rx̂) is in C 2 (S 2 ), and, therefore, can be
expanded into a series by Theorem 2.19; that is,
u(rx̂) =
∞ X
n
X
m
um
n (r) Yn (x̂) ,
x̂ ∈ S 2 .
n=0 m=−n
m
The coefficients are given by um
n (r) = u(r, ·), Yn
L2 (S 2 )
.
We show that um
n satisfies a differential equation of Euler type. Using the Laplace equation in spherical coordinates for u, the self-adjoint Laplace-Beltrami operator ∆S 2 , and the
eigenvalue equation (2.17), yields
Z
d
∂
2 d
m
2 ∂u(r, x̂)
r
un (r) =
r
Yn−m (x̂) ds(x̂)
dr
dr
∂r
∂r
S2
Z
= −
∆S 2 u(r, x̂) Yn−m (x̂) ds(x̂)
S2
Z
= −
u(r, x̂) ∆S 2 Yn−m (x̂) ds(x̂)
S2
Z
= n(n + 1)
u(r, x̂) Yn−m (x̂) ds(x̂)
=
n(n + 1) um
n (r) .
S2
m n
The only smooth solution of this Euler differential equation is given by um
n (r) = αn r for
arbitrary αnm . 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. We
fix R0 < R and choose R̂ with R0 < R̂ < R. Multiplying
the representation of u(R̂x̂) with
−q
2
q
Yp (x̂), and integrate over S to obtain u(R̂, ·), Yp L2 (S 2 ) = αpq R̂p which proves uniqueness
of αnm . Furthermore, by the addition formula of Theorem 2.17 we find the representation
n
∞ X
n
X
r
m
u(rx̂) =
u(R̂, ·), Yn L2 (S 2 )
Ynm (x̂)
R̂
n=0 m=−n
=
∞ n Z
X
r
n=0
R̂
n
X
S 2 m=−n
Yn−m (ŷ) Ynm (x̂) u(R̂ŷ) ds(ŷ)
Z
∞ n
X
2n + 1
r
=
Pn (x̂ · ŷ) u(R̂ŷ) ds(ŷ)
4π
R̂
S2
n=0
dj Pn for r ≤ R and x̂ ∈ S . From this representation and the observation that dtj (t) ≤ cj n2j on
0
2
[−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, R0 ] because,
using the Cauchy-Schwarz inequality, it is dominated by the convergent series
0 n
∞
X
R
2`
c ku(R̂, ·)kL2 (S 2 )
(2n + 1) n
R̂
n=0
2
for some c > 0. This ends the proof.
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 (S 2 ) and fR ∈ C 2 (S 2 ) simultanously where we set fR (x̂) = f (Rx̂), x̂ ∈ S 2 , here and in the following.
Theorem 2.22 (a) For given fR ∈ L2 (S 2 ) there exists a unique solution u ∈ C 2 B(0, R)
of ∆u = 0 in B(0, R) with
lim ku(r, ·) − fR kL2 (S 2 ) = 0 .
r→R
The solution is given by the series
u(rx̂) =
∞ X
n
X
(fR , Ynm )L2 (S 2 )
n=0 m=−n
r n
R
Ynm (x̂)
∞
r n Z
1 X
=
f (Rŷ) Pn (x̂ · ŷ) ds(ŷ)
(2n + 1)
4π n=0
R
(2.22a)
(2.22b)
S2
for x = rx̂ ∈ B(0, R). They converge uniformly on every compact ball B[0, R0 ] for any
R0 < R.
(b) If fR ∈ C 2 (S 2 ) there exists a unique solution u ∈ C 2 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].
52
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Proof: First we note that the series coincide by the addition formula as shown in the proof
of Theorem 2.21.
(a) To show uniqueness we assume that u is the difference of two solutions. Then ∆u = 0
in B(0, R) and limr→R ku(r, ·)kL2 (S 2 ) = 0. By the previous theorem u can be representated
as a series in the form (2.20). Let R0 < R and ε > 0 be arbitrary. Choose Rε ∈ [R0 , R) such
that kuε kL2 (S 2 ) ≤ ε where uε (x̂) = u(Rε x̂). Multiplying the representiation of u(Rε x̂) with
Yp−q (x̂) and integrating over S 2 yields (uε , Ypq )L2 (S 2 ) = αpq Rεp , thus
u(rx̂) =
∞ X
n
X
(uε , Ynm )L2 (S 2 )
n=0 m=−n
r
Rε
n
Ynm (x̂) x̂ ∈ S 2 , r ≤ Rε .
Therefore, for r ≤ R0 ,
ku(r, ·)k2L2 (S 2 )
2n
∞ X
n
∞ X
n
X
X
2
r
m
(uε , Ynm )L2 (S 2 ) 2
(uε , Yn )L2 (S 2 )
=
≤
Rε
n=0 m=−n
n=0 m=−n
= kuε k2L2 (S 2 ) ≤ ε2 .
Since this holds for all ε > 0 we conclude that u has to vanish in B(0, R0 ). Since R0 < R
was arbitrary u vanishes in B(0, R).
Uniform convergence of the series and all of its derivates on every ball B[0, R0 ] with R0 < R
is shown as in the proof of the previous theorem. The series for D` u(x) is dominated by the
convergent series
0 n
∞
X
R
2`
c kfR kL2 (S 2 )
(2n + 1) n
.
R
n=0
Therefore, u ∈ C ∞ B(0, R) solves the Laplace equation. Finally, we use the Parseval
identity, applied to the series
u(rx̂) − f (Rx̂) =
∞ X
n
X
n=0 m=−n
(fR , Ynm )L2 (S 2 )
h r n
R
i
− 1 Ynm (x̂) ;
that is,
ku(r, ·) −
fR k2L2 (S 2 )
n
∞ X
i2
X
2 h r n
m
−1 .
=
(fR , Yn )L2 (S 2 )
R
n=0 m=−n
This term
tends to zero
r → R which is shown by standard arguments: For every n the
2 as
2
r n
m
term (fR , Yn )L2 (S 2 )
− 1 tends to zero as r tends to R. Furthermore, it is bounded
R
2
by the summable term (fR , Ynm )L2 (S 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 ∆S 2 we express the expansion coefficient of f as
(fR , Ynm )L2 (S 2 ) = −
1
1
(fR , ∆S 2 Ynm )L2 (S 2 ) = −
(∆S 2 fR , Ynm )L2 (S 2 ) .
n(n + 1)
n(n + 1)
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 any
r ≤ R and any x̂ ∈ S 2 by
∞ X
n
∞ X
n
X
X
n m
1
(fR , Ynm )L2 (S 2 ) r
(∆S 2 fR , Ynm )L2 (S 2 ) |Ynm (x̂)|
|Yn (x̂)| ≤
R
n(n + 1)
n=N m=−n
n=N m=−n
"
∞ X
n
X
(∆S 2 fR , Ynm )L2 (S 2 ) 2
≤
#1/2 "
n=N m=−n
"
= k∆S 2 fR kL2 (S 2 )
∞
X
n=N
1
= √ k∆S 2 fR kL2 (S 2 )
4π
∞ X
n
X
1
|Ynm (x̂)|2
2
2
n
(n
+
1)
n=N m=−n
n
X
1
|Y m (x̂)|2
n2 (n + 1)2 m=−n n
"
∞
X
n=N
2n + 1
2
n (n + 1)2
#1/2
#1/2
#1/2
2
where we again have used part (a) of Corollary 2.18.
To end this section we consider the situation in the exterior of the closed ball B[0, R]. We
are interested in harmonic functions which tend to zero at infinity. The following theorem
corresponds to Theorem 2.21.
Theorem 2.23 Let u ∈ C 2 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
lim u(rx̂) = 0
r→∞
for every x̂ ∈ S 2 .
(2.23)
Then there exist unique coefficients αnm ∈ C, |m| ≤ n, n = 0, 1, 2, . . . with
u(rx̂) =
∞ X
n
X
αnm r−n−1 Ynm (x̂) ,
r > R , x̂ ∈ S 2 .
(2.24)
n=0 m=−n
The series converges uniformly with all of its derivatives outside of every ball B(0, R0 ) with
R0 > R.
The proof is almost the same as the proof of Theorem 2.21. The main difference is that one
d
d
m −n−1
r2 dr
um
has to select the solution um
in Euler’s differential equation dr
n (r) =
n (r) = αn r
m
m
m n
n(n + 1) un (r) instead of un (r) = αn r . 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
to ensure
is needed
uniqueness. Indeed, we observe that
n
n+1
the function u(x) = u(r, ϕ, θ) = (r/R) − (R/r)
Pn (cos θ) is harmonic in the exterior of
B(0, R) and vanishes for r = R.
Again, we study the cases fR ∈ L2 (S 2 ) and fR ∈ C 2 (S 2 ) simultanously where fR (x̂) = f (Rx̂),
x̂ ∈ S 2 , as before.
54
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
2
2
2
3
Theorem
2.24 (a) For3 given fR ∈ L (S ) there exists a unique solution u ∈ C R \
B[0, R] of ∆u = 0 in R \ B[0, R] which satisfies (2.23) and
lim ku(r, ·) − fR kL2 (S 2 ) = 0 .
r→R
The solution is given by the series
u(rx̂) =
∞ X
n
X
(fR , Ynm )L2 (S 2 )
n=0 m=−n
n+1
R
Ynm (x̂)
r
n+1 Z
∞
1 X
R
=
(2n + 1)
f (Rŷ) Pn (x̂ · ŷ) ds(ŷ)
4π n=0
r
(2.26a)
(2.26b)
S2
for x = rx̂ ∈ R3 \ B[0, R]. They converge uniformly for r ≥ R0 for any R0 > R.
(b) If fR ∈ C 2 (S 2 ) there exists a unique solution u ∈ C 2 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 for
r ≥ R.
The proof follows again very closely the proof of the corresponding interior case. We omit
the details and refer again to Exercise 2.7.
2.5
Bessel Functions
The previous investigations for harmonic functions can be applied in electrostatics or magnetostatics (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 Helmholtz
equation 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 important
functions of mathematical physics is subject of the present section.
From the separation u(x) = v(r)K(x̂) in spherical coordinates and the eigenvalues −n(n +1)
of the Beltrami operator we obtained for the radial component v̂(kr) = v(r) with z = kr the
spherical Bessel differential equation
z 2 v̂ 00 (z) + 2z v̂ 0 (z) + z 2 − n(n + 1) v̂(z) = 0 .
(see equation (2.8)). We will investigate this linear differential equation of second order for
arbitrary z ∈ C. For z 6= 0 the differential equation is equivalent to
n(n + 1)
2 0
00
v̂(z) = 0 in C \ {0} .
(2.27)
v̂ (z) + v̂ (z) + 1 −
z
z2
The coefficients of this differential equation are holomorphic in C \ {0} and have poles of first
and second order at 0. As in the case of real z one can show that in every simply connected
domain 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 independent 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
3
P
C. Therefore, if we fix some x0 ∈ Ω there exist αj ∈ C, j = 1, 2, 3, such that
|αj | =
6 0 and
j=1
3
X
j=1
Set w :=
3
P
αj
wj (x0 )
wj0 (x0 )
= 0.
αj wj in Ω. Then w(x0 ) = w0 (x0 ) = 0 and thus from (2.27), also w(j) (x0 ) = 0
j=1
for all j = 0, 1, . . .. Because w is holomorphic in the domain Ω we conclude by the identity
theorem for holomorphic functions that w vanishes in all of Ω. Therefore, {w1 , w2 , w3 } are
linearly dependent.
2
Motivated by the solution v1,n (r) = rn of the corresponding equation (2.7) for the Laplace
equation and λ = −n(n + 1) we make an ansatz for a smooth solution of (2.27) as a power
series in the form
∞
∞
X
X
a` z `+n .
w1 (z) = z n
a` z ` =
`=0
`=0
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
1
a2(`−1) = −
a2(`−1)
(n + 2`)(n + 2` + 1) − n(n + 1)
2` (2n + 2` + 1)
1
1
= − (n + `)
`
(2n + 2`)(2n + 2` + 1)
and thus by induction
a2` =
(2n + 1)! (−1)`
(n + `)!
a0
n!
`! (2n + 2` + 1)!
for all ` ≥ 0 .
Altogether we have that
∞
w1 (z) =
X (−1)`
(2n + 1)!
(n + `)!
a0 z n
z 2` .
n!
`!
(2n
+
2`
+
1)!
`=0
56
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
By the ratio test it is seen that the radius of convergence is infinity. Therefore, this function
w1 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 . Motivated
by the singular solution v2,n (r) = r−n−1 of (2.7) for λ = −n(n + 1) we make an ansatz of the
form
∞
∞
X
X
−n−1
`
w2 (z) = z
a` z =
a` z `−n−1 .
`=0
`=0
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
1
a2(`−1) =
a2(`−1)
(2` − n)(2` − n − 1) − n(n + 1)
(2`) (2n − 2` + 1)
and thus by induction
a2` =
1
1
a0
`! (2n − 2` + 1)(2n − 2` + 3) · · · (2n − 1)
2`
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)!
(2n)!
2 · 4 · · · 2(` − n) 2 · 4 · · · (2n)
= (−1)`−n
(2` − 2n)! (2n)!
2`−n (` − n)! 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)! 2n−` (n − `)!
= n
.
(2n − 2` − 1)!!
2 n! (2n − 2`)!
Therefore, we arrive at the second solution
∞
w2 (z) = a0 z −n−1
n! X 1 (` − n)! 2`
z ,
(2n)! `=0 `! (2` − 2n)!
2.5. BESSEL FUNCTIONS
where we have set
57
(2n − 2`)!
(` − n)!
:= (−1)n−`
(2` − 2n)!
(n − `)!
for ` < n .
The radius of convergence of the series is again infinity. For particular normalizations the
functions w1 and w2 are called spherical Bessel functions.
Definition 2.26 For all z ∈ C the spherical Bessel functions of first and second kind and
order n ∈ N0 are defined by
n
jn (z) = (2z)
∞
X
(−1)`
`=0
`!
(n + `)!
z 2` ,
(2n + 2` + 1)!
z ∈ C,
∞
2 (−1)n+1 X (−1)` (` − n)! 2`
yn (z) =
z ,
(2z)n+1 `=0 `! (2` − 2n)!
where – in the definition of yn – a quantity
(−k)!
(−2k)!
z ∈ C,
for positive integers k is defined by
(2k)!
(−k)!
= (−1)k
,
(−2k)!
k!
k ∈ N.
The functions
h(1)
= jn + i yn ,
n
h(2)
= jn − i yn ,
n
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) yn0 (z) − jn0 (z) yn (z) =
1
.
z2
Proof: We write W (z) for W (jn , yn )(z). We multiply the spherical Bessel differential equation (2.27) for yn by jn and the one for jn by yn and subtract. This yields
yn00 (z) jn (z) − jn00 (z) yn (z) +
2 0
yn (z) jn (z) − jn0 (z) yn (z) = 0 .
z
The first term is just W 0 (z), thus W solves the ordinary differential equation W 0 (z) +
2
W (z) = 0. The general solution is W (z) = cz −2 for some c ∈ C which we determine
z
from the leading coefficient of the Laurant series of W . By the definition of the Bessel
functions we have for fixed n
n!
n!
jn (z) = 2n z n (2n+1)!
1 + O(z 2 ) ,
jn0 (z) = n 2n z n−1 (2n+1)!
1 + O(z 2 )
yn (z) = −2−n z −n−1 (2n)!
1 + O(z 2 ) ,
yn0 (z) = (n + 1) 2−n z −n−2 (2n)!
1 + O(z 2 )
n!
n!
58
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
as z → 0. Therefore,
W (z) = (n + 1) z −2
(2n)!
(2n)!
1 −2
+ nz
1 + O(z 2 ) = 2 1 + O(z 2 )
(2n + 1)!
(2n + 1)!
z
2
which proves that c = 1.
Remark: From this theorem the linear independence of {jn , yn } follows immediately and
(1)
thus also the linear independence of {jn , hn }. 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 the
function cos z/z. Especially, from Rayleigh’s formulas below we note that j0 (z) = sin z/z
(1)
and ih0 (z) = exp(iz)/z.
Theorem 2.28 (Rayleigh’s Formulas)
For any z ∈ C and n ∈ N0 we have
jn (z) =
yn (z) =
h(1)
n (z) =
h(2)
n (z) =
n
1 d
sin z
(−z)
,
z dz
z
n
1 d
cos z
n
,
−(−z)
z dz
z
n
1 d
exp(iz)
n
−i (−z)
,
z dz
z
n
1 d
exp(−iz)
n
i (−z)
.
z dz
z
n
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 expansion
of cos z/z; that is,
∞
X
cos z
(−1)` 2`−1
=
z
,
z
(2`)!
`=0
d n
and observe that the action of z1 dz
on a power of z is given by
n
1 d
z 2`−1 = (2` − 1)(2` − 3) · · · (2` − 2n + 1) z 2`−2n−1 ,
z dz
thus
−(−z)
n
1 d
z dz
n
∞
cos z
(−1)n+1 X (−1)`
=
(2` − 1)(2` − 3) · · · (2` − 2n + 1) z 2` .
z
z n+1 `=0 (2`)!
We discuss the term
q =
(2` − 1)(2` − 3) · · · (2` − 2n + 1)
(2`)!
2.5. BESSEL FUNCTIONS
59
separately for ` ≥ n and ` < n.
For ` ≥ n we have, using again the notation k!! = 1 · 3 · · · k for odd k,
q =
(2`)! 2`−n (` − n)!
1
(` − n)!
(2` − 1)!!
= `
= n
.
(2`)! (2` − 2n − 1)!!
2 `! (2`)! (2` − 2n)!
2 `! (2` − 2n)!
The case ` < n is seen analogously by splitting
(2` − 1)(2` − 3) · · · 1 · (−1) · · · (2` − 2n + 1)
(2` − 1)!! (−1)n−` (2n − 2` − 1)!!
q =
=
(2`)!
(2`)!
= (−1)n−`
2`
(2n − 2`)!
(2`)!
(2n − 2`)!
1
= n (−1)n−`
.
n−`
`! (2`)! 2 (n − `)!
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 the
functions
(1)
m
um
n (rx̂) = hn (kr) Yn (x̂) ,
are called spherical wave functions. They are solutions of ∆u + k 2 u = 0 in R3 \ {0}. The
real part
m
Re um
n (rx̂) = jn (kr) Yn (x̂) ,
satisfies the Helmholtz equation in all of R3 .
The Bessel functions of the first kind jn correspond to the functions rn in the static case, that
is k = 0. For the Helmholtz equation the mappings rx̂ 7→ jn (kr) Ynm (x̂) are the expansion
(1)
functions for solutions inside of balls. Furthermore the functions hn are used to derive
expansions in the exterior of balls. From the definition we observe that they are singular at
the origin of order n. Additionally, we need their asymptotic behavior for r → ∞, which
can be derived from the previous theorem.
Theorem 2.30 For every n ∈ N and z ∈ C we have
exp(iz)
1
(1)
n+1
hn (z) =
(−i)
+ O
for |z| → ∞ ,
z
|z|
d (1)
exp(iz)
1
n
hn (z) =
(−i) + O
for |z| → ∞ ,
dz
z
|z|
uniformly with respect to z/|z|. The corresponding formulas for jn and yn are derived by
taking the real- and imaginary parts, respectively.
60
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Proof: We show by induction that for every n ∈ N0 there exists a polynomial Qn of degree
at most n such that Qn (0) = 1 and
n
exp(iz)
1 d
1
n exp(iz)
= i
.
(2.28)
Qn
n+1
z dz
z
z
z
This would prove the assertions.
For n = 0 this is obvious for the constant polynomial Q0 = 1. Let it be true for n. Then
n+1
1 d
exp(iz)
1 d n exp(iz)
1
=
i
Qn
n+1
z dz
z
z dz
z
z
i
in iz
1
1
1 1
1
1
0
=
e
−
(n
+
1)
−
Q
Q
Q
n
n
n
z
z n+1
z
z n+2
z
z n+1
z z2
1
n+1 exp(iz)
Qn+1
= i
n+2
z
z
with Qn+1 (t) = Qn (t) + i(n + 1) t Qn (t) + it2 Q0n (t). Obviously, Qn+1 is a polynomial of order
(1)
at most n + 1 and Qn+1 (0) = 1. This proves the asymptotic forms of hn and its derivative.
2
For the expansion of solutions of the Helmholtz equation into spherical wave functions also
the asymptotic behaviour with respect to n is necessary.
Theorem 2.31 For every ε > 0 and R > 0 we have the following.
n!
1
1
1
n
n
(2z) 1 + O
z 1+O
jn (z) =
=
for n → ∞ ,
(2n + 1)!
n
(2n + 1)!!
n
(2n)!
1
(2n − 1)!!
1
2
yn (z) = −
1+O
= −
1+O
for n → ∞ ,
n+1
n+1
n! (2z)
n
z
n
(2n)!
2
1
(2n − 1)!!
1
(1)
hn (z) = −i
1+O
= −i
1+O
for n → ∞ ,
n+1
n+1
n! (2z)
n
z
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
∞
∞
X
X
(2n + 1)!
R2` (2n + 1)! (n + `)!
1
R2`
−n
(2z)
j
(z)
−
1
≤
≤
n
n!
`! n! (2n + 2` + 1)!
4(n + 1) `=1 `!
`=1
because for ` ≥ 1
(2n + 1)! (n + `)!
1
(n + 1) · · · (n + `)
1
=
≤
.
(2n + 2` + 1)! n!
(2n + 2) · · · (2n + ` + 1) (2n + ` + 2) · · · (2n + 2` + 1)
4(n + 1)
|
{z
}|
{z
}
≤ 1/(2n+2)
≤ 1/2
2.5. BESSEL FUNCTIONS
61
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·ŷ), x ∈ R3 , of the Helmholtz
equation for some fixed ŷ ∈ S 2 which describes a plane time harmonic electromagnetic field
travelling in the direction ŷ (see example 1.4).
Theorem 2.32 For x̂, ŷ ∈ S 2 and r > 0 there holds the Jacobi–Anger expansion
ikr x̂·ŷ
e
= 4π
∞ X
n
X
n
i
jn (kr) Ynm (x̂) Yn−m (ŷ)
n=0 m=−n
=
∞
X
in (2n + 1) jn (kr) Pn (x̂ · ŷ) .
n=0
For every R > 0 the series converges uniformly with respect to x̂, ŷ ∈ S 2 and 0 ≤ r ≤ R.
Proof: The two representations coincide as seen from the addition formula. Uniform convergence 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 ŷ ∈ S 2 and expand x̂ 7→ eiz x̂·ŷ
into spherical harmonics; that is,
e
iz x̂·ŷ
=
∞ X
n
X
m
am
n (z, ŷ) Yn (x̂)
n=0 m=−n
where
am
n (z, ŷ)
Z
=
S2
eiz x·ŷ Yn−m (x) ds(x) = λn (z) Yn−m (ŷ)
R1
with λn (z) = 2π −1 eizt Pn (t) dt. It remains to show that λn (z) = 4π in jn (z). The function
λn solves the Bessel differential equation. Indeed, using the differential equation (2.9) for
the Legendre polynomial, we conclude that
d 2 0
z λn (z) + z 2 − n(n + 1) λn (z)
dz
Z 1
Z
izt
2 2
= 2π
(1 − t )z + 2izt e Pn (t) dt − n(n + 1)2π
−1
Z
= 2π
eizt Pn (t) dt
−1
1
−1
1
(1 − t2 )z 2 + 2izt eizt Pn (t) dt + 2π
Z
1
−1
eizt
d
(1 − t2 )Pn0 (t) dt
dt
= 0
by two partial integrations of the second integral. Since λn is smooth we conclude that
λn (z) = αn jn (z) for some constant αn ∈ C which we determine by the behavior at the
origin. First, we use the fact that Pn is orthogonal to all polynomials of order less that n.
(2n−1)!
The coefficients of the term tn is given by n! (n−1)!
which can be seen from the recursion
2n−1
formula (b) of Theorem 2.8 by induction (see Excercise 2.5). Therefore,
Z 1
n−1 Z 1
2 n+1
dn
n
n
n n! (n − 1)! 2
2
n (n!) 2
λ
(0)
=
2π
i
t
P
(t)
dt
=
2π
i
P
(t)
dt
=
2π
i
n
n
n
dz n
(2n − 1)!
(2n + 1)!
−1
−1
62
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
where we have used Theorem 2.7. On the other hand, from Definition 2.26 of the Bessel
function we observe that
dn
dn
n!
(n!)2 2n
n
.
j
(0)
=
(2z)
=
n
dz n
(2n + 1)! dz n
(2n + 1)!
z=0
Comparing these two formulas yields αn = 4π in and finishes the proof.
2.6
2
The Helmholtz Equation in the Interior and Exterior of a Ball
Now we are ready to study the series expansion of solutions of the Helmholtz equation
∆u + k 2 u = 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 ∈ C 2 B(0, R) solve the
Helmholtz equation ∆u + k 2 u = 0 in B(0, R). Then there exist unique αnm ∈ C, |m| ≤ n,
n = 0, 1, 2, . . . with
u(rx̂) =
∞ X
n
X
αnm jn (kr) Ynm (x̂) ,
0 ≤ r < R , x̂ ∈ S 2 .
n=0 m=−n
The series converges uniformly with all of its derivatives in every closed ball B[0, R0 ] with
R0 < R.
Proof: We argue as the proof of Theorem 2.21. For every r ∈ (0, R) the function u(rx̂) can
be expanded into a series by Theorem 2.19; that is,
u(rx̂) =
∞ X
n
X
m
um
n (r) Yn (x̂) ,
x̂ ∈ S 2 .
n=0 m=−n
m
m
The coefficients are given by um
n (r) = u(r, ·), Yn L2 (S 2 ) . We show that un satisfies the
spherical Bessel differential equation. Using the Helmholtz equation for u in spherical polar
coordinates, and that the functions Ynm 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
2 2 m
2 d
r
um
n (r) + r k un (r)
dr
dr
Z ∂
2 ∂u(r, x̂)
2 2
=
r
+ r k u(r, x̂) Yn−m (x̂) ds(x̂)
∂r
∂r
S2
Z
= −
∆S 2 u(r, x̂) Yn−m (x̂) ds(x̂)
S2
Z
= −
u(r, x̂) ∆S 2 Yn−m (x̂) ds(x̂)
S2
Z
= n(n + 1)
u(r, x̂) Yn−m (x̂) ds(x̂)
=
n(n + 1) um
n (r) .
S2
After the transformation z = kr it turns into Bessel’s differential equation (2.8) for the
m
m
m
functions um
n . The general solution is given by un (r) = αn jn (kr) + βn yn (kr) for arbitrary
αnm , βnm . Therefore, u has the form
∞ X
n
X
m
u(rx̂) =
αn jn (kr) + βnm yn (kr) Ynm (x̂) ,
x̂ ∈ S 2 , r > 0 .
(2.29)
n=0 m=−n
We are interested in smooth solutions in the ball B(0, R). Therefore, βnm = 0.
To prove uniform convergence we express u(x) in a different form as in the proof of Theorem 2.33. Fix R0 < R and choose R̂ with R0 < R̂ < R. Multiplying
the representation of
2
q
−q
u(R̂x̂) with Yp (x̂), and integrate over S to obtain u(R̂, ·), Yp L2 (S 2 ) = αpq jp (k R̂) which
proves uniqueness of αnm and, furthermore, the representation of u as
u(rx̂) =
∞ X
n
X
u(R̂, ·), Ynm
n=0 m=−n
=
Z
∞
X
jn (kr)
n=0
jn (k R̂)
S2
n
X
jn (kr)
L2 (S 2 )
jn (k R̂)
Ynm (x̂)
Yn−m (ŷ) Ynm (x̂) u(R̂ŷ) ds(ŷ)
m=−n
Z
∞
X
jn (kr) 2n + 1
=
Pn (x̂ · ŷ) u(R̂ŷ) ds(ŷ)
4π
2
j
(k
R̂)
S
n
n=0
for r ≤ R0 and x̂ ∈ S 2 . Here we used again the Addition Formula of Theorem 2.17.
So far, we
exactly the proof of Theorem 2.21. Now we have to use again the
followed
dj Pn estimate dtj (t) ≤ cj n2j 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 any
differential operator D` = ∂ |`| /(∂r`1 ∂θ`2 ∂ϕ`3 ) the series for D` u(x) converges uniformly in
64
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
B[0, R0 ] because it is dominated by the convergent series
0 n
∞
X
R
2`
cku(R̂, ·)kL2 (S 2 )
(2n + 1) n
R̂
n=0
2
with a constant c > 0 depending on R0 . This ends the proof.
Now we transfer the idea of Theorem 2.22 to the interior boundary value problem
∆u + k 2 u = 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 the
unit sphere by setting fR (x̂) = f (Rx̂), x̂ ∈ S 2 .
Theorem 2.34 Assume that k ∈ C with Im k ≥ 0 and R > 0 are such that jn (kR) 6= 0 for
all n ∈ N0 .
(a) For given fR ∈ L2 (S 2 ) there exists a unique solution u ∈ C 2 B(0, R) of ∆u + k 2 u = 0
in B(0, R) with
lim ku(r, ·) − fR kL2 (S 2 ) = 0 .
r→R
The solution is given by
u(rx̂) =
∞ X
n
X
(fR , Ynm )L2 (S 2 )
n=0 m=−n
jn (kr) m
Y (x̂)
jn (kR) n
Z
∞
jn (kr)
1 X
f (Rŷ) Pn (x̂ · ŷ) ds(ŷ) .
(2n + 1)
=
4π n=0
jn (kR)
(2.31a)
(2.31b)
S2
The series converge uniformly on compact subsets of B(0, R).
(b) If fR ∈ C 2 (S 2 ) there exists a unique solution u ∈ C 2 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 proof
of Theorem 2.24. Indeed, let u be a solution of the Helmholtz equation in B(0, R) such that
limr→R ku(r, ·)kL2 (S 2 ) = 0. From Theorem 2.33 we conclude that u can be expanded into
u(rx̂) =
n
∞ X
X
m
am
n jn (kr) Yn (x̂) ,
x̂ ∈ S 2 , r < R .
n=0 m=−n
Let R0 < R and ε > 0 be arbitrary. Choose Rε ∈ (R0 , R] such that kuε kL2 (S 2 ) ≤ ε where
uε (x̂) = u(Rε x̂). Multiplying the representiation of u(Rε x̂) with Yp−q (x̂) and integrating over
S 2 yields (uε , Ypq )L2 (S 2 ) = aqp jp (kRε ), thus
u(rx̂) =
∞ X
n
X
(uε , Ynm )L2 (S 2 )
n=0 m=−n
jn (kr) m
Y (x̂) ,
jn (kRε ) n
x̂ ∈ S 2 , r ≤ Rε .
2.6. THE HELMHOLTZ EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL65
Let now R1 < R0 be arbitrary. Since the asymptotic behavior of the Bessel function (Theorem
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 ,
ku(r, ·)k2L2 (S 2 )
∞ X
n
X
(uε , Ynm )L2 (S 2 ) 2
=
n=0 m=−n
∞ X
n
X
jn (kr) 2
≤ c2
(uε , Ynm )L2 (S 2 ) 2
jn (kRε ) n=0 m=−n
= c2 kuε k2L2 (S 2 ) ≤ c2 ε2 .
This holds for all ε > 0, therefore u has to vanish in B(0, R0 ) \ B(0, R1 ) and thus for |x| < R
because 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 ku(r, .) − fR k2L2 (S 2 ) , since by Theorem 2.31
(kr)
− 1|2 can be uniformly bounded for all n ∈ N and all r ∈ [0, R].
the term | jjnn(kR)
(kr) (b) Also, with the uniform bound on jjnn(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 shows
part (b) of the Theorem.
2
We observe an important difference between the expansion functions rn Ynm (x̂) for the Laplace
equation and jn (kr)Ynm (x̂) 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. From
the asymptotic form of jn (z) for real and positive z (see Theorem 2.30) we conclude that for
every n ∈ N0 and fixed R > 0 there exist infinitely many real and positive wave numbers
kn,1 , kn,2 , . . . with jn (kn,j R) = 0. Therefore, the functions
m
um
n,j (rx̂) = jn (kn,j r) Yn (x̂) ,
n, j ∈ N , |m| ≤ n ,
2
m
solve the homogeneous Dirichlet boundary value problem ∆um
n,j +kn,j un,j = 0 in B[0, R] with
2
um
n,j = 0 on the boundary |x| = R. The values kn,j are called Dirichlet eigenvalues of −∆ in
the ball of radius R. The multiplicity is 2n + 1, and the functions um
n,j , m = −n, . . . , n, are
the 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 the
two linearly independent solutions jn (kr) Ynm (x̂) and yn (kr) Ynm (x̂). 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 member
of 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 with
ε = ε0 , µ = µ0 , and constant σ > 0. We have seen from the previous chapter that in this
case k 2 has to be replaced by k̃ 2 = k 2 εr with εr = 1 + iσ/(ωε0 ). For k̃ itself we take the
66
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
branch with Re k̃ > 0, thus also Im k̃ > 0. From Theorem 2.30 we observe that
1
fn (k̃r) m
m
Yn (x̂) 1 + O
jn (k̃r)Yn (x̂) =
for r → ∞ ,
r
k̃r
where fn (z) = cos z or fn (z) = sin z. In any case we observe that for Im k̃ > 0 these functions
increase exponentially as r tends to infinity. The same holds for the functions yn (k̃r)Ynm (x̂).
(1)
The expansion functions hn (k̃r)Ynm (x̂), however, have the asymptotic forms
exp(ik̃r) m
1
(1)
m
hn (k̃r)Yn (x̂) =
Yn (x̂) 1 + O
for r → ∞ ,
r
k̃r
and these functions decay exponentially at infinity. This is physically plausible since a
(2)
conducting medium is absorbing. We note by passing that the Hankel functions hn of the
second kind are exponentially increasing. Therefore, for conducting media the expansion
(1)
functions hn (k̃r)Ynm (x̂) are the correct ones. In the limiting case σ → 0 the solution should
depend continuously on σ. This makes it plausible to choose these functions also for the
case σ = 0. The requirement that the functions u are bounded for Im k̃ > 0 and depend
continuously on k̃ for Im k̃ → 0 on compact subsets of R3 is called the limiting absorption
principle .
(1)
Comparing hn 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
(1)
u(rx̂) = hn (kr)Ynm (x̂) satisfies the Sommerfeld radiation condition
1
∂u(rx̂)
− ik u(rx̂) = O 2
for r → ∞ ,
(2.32)
∂r
r
uniformly with respect to x̂ ∈ S 2 .
Proof: By Theorem 2.30 we have
d (1)
exp(ikr)
1
(1)
n
n−1
hn (kr) − ik hn (kr) =
(−i) k − ik (−i)
+ O
dr
kr
r
exp(ikr)
1
1
=
O
= O 2
for r → ∞ .
kr
r
r
2
We note that this radiation is necessary only for the case of real values of k. For complex
values 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 every
function defined in the exterior of a bounded domain without making use of the Hankel
functions – and still provides the correct solution. The following theorem is the analog of
Theorem 2.23.
2.6. THE HELMHOLTZ EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL67
Theorem 2.36 Let k ∈ C\{0} with Im k ≥ 0 and u ∈ C 2 R3 \B[0, R] satisfy ∆u+k 2 u = 0
in the exterior of some ball B[0, R]. Furthermore, we assume that u satisfies the Sommerfeld
radiation condition (2.32). Then there exist unique αnm ∈ C, |m| ≤ n, n = 0, 1, 2, . . . with
u(rx̂) =
∞ X
n
X
m
αnm h(1)
n (kr) Yn (x̂) ,
r > R , x̂ ∈ S 2 .
(2.33)
n=0 m=−n
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 um
n (r) satisfies
d m
2
m
u (r) − ik un (r) = O(1/r ). Since only the Hankel functions of the first kind satisfy the
dr n
radiation condition in contrast to the Bessel functions we conclude that βnm = 0 for all n ∈ N0
and |m| ≤ n. We continue with the arguments just as in the proof of Theorem 2.33 and
omit the details.
2
Finally we consider the exterior boundary value problem to determine for given f the complex
valued function u such that
∆u + k 2 u = 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 (S 2 ) there exists a unique solution u ∈ C 2 R3 \B[0, R] of ∆u+k 2 u = 0
with
lim ku(r, ·) − fR kL2 (S 2 ) = 0 ,
r→R
and u satisfies Sommerfeld radiation condition (2.32). The solution is given by
u(rx̂) =
∞ X
n
X
(fR , Ynm )L2 (S 2 )
n=0 m=−n
(1)
hn (kr)
(1)
hn (kR)
Ynm (x̂)
Z
∞
(1)
1 X
hn (kr)
=
f (Rŷ) Pn (x̂ · ŷ) ds(ŷ) .
(2n + 1) (1)
4π n=0
hn (kR)
(2.35a)
(2.35b)
S2
The series converge uniformly on compact subsets of R3 \ B[0, R].
(b) If fR ∈ C 2 (S 2 ) there exists a unique solution u ∈ C 2 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, R0 ) for R0 > R of any
derivative of the series and the validity of the boundary condition follows the well known
(1)
arguments and is omitted. We just mention that hn (kR) never vanishes – in contrast to
jn (kR) – because of the Wronskian of Theorem 2.27.
68
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
The proof that the function u satisfies the radiation condition is more difficult and uses a
result (in the proof of Lemma 2.38) from Chapter 3.2 The problem is that every component
of the series satisfies the radiation condition but the asymptotics of hn (k|x|) as |x| tends to
infinity by Theorem 2.30 does not hold uniformly with respect to n. The idea to overcome
this difficulty is to express u as an integral over a sphere of the form
Z
u(x) =
Φ(x, y) g(y) ds(y) , |x| > R0 ,
(2.36)
|y|=R0
rather than a series, see Lemma 2.38 below. Here, Φ denotes the fundamental solution of
the Helmholtz equation, defined by
Φ(x, y) =
ik (1)
exp(ik|x − y|)
=
h (k|x − y|) ,
4π |x − y|
4π 0
x 6= y .
(2.37)
The fundamental solution satisfies the radiation condition (2.32) uniformly with respect to y
on the compact sphere {y : |y| = R0 } as we show in Lemma 2.39. Therefore, also u satisfies
the 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(rx̂) =
∞ X
n
X
m
αnm h(1)
n (kr) Yn (x̂) ,
r > R , x̂ ∈ S 2 .
(2.38)
n=0 m=−n
We assume that the series converges uniformly on compact subsets of R3 \ B[0, R]. Then
there exists R0 > R and a continuous function g on the sphere with radius R0 such that
Z
u(x) =
Φ(x, y) g(y) ds(y) , |x| > R0 .
|y|=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|)Ynm (ŷ)
and Φ(x, y) inside the ball B(0, R0 ) which yields
Z
jn (kR0 )
Ynm (ŷ)
∂
Φ(x, y) ds(y) − k jn0 (kR0 )
∂ν(y)
Z
Ynm (ŷ) Φ(x, y) ds(y) = 0 .
|y|=R0
|y|=R0
(2.39a)
Second, since v(y) =
satisfies the radiation condition (2.32) we can apply
Green’s representation Theorem 3.6 from the forthcoming Chapter 3 to v in the exterior of
(1)
hn (k|y|) Ynm (ŷ)
2
At least the authors do not know of any proof which uses only the elementary properties of the Hankel
functions derived so far.
2.6. THE HELMHOLTZ EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL69
B(0, R0 ) which yields
m
h(1)
n (k|x|) Yn (x̂)
=
h(1)
n (kR0 )
Z
Ynm (ŷ)
∂
Φ(x, y) ds(y)
∂ν(y)
|y|=R0
d
− k h(1)
(kR0 )
dr n
Z
Ynm (ŷ) Φ(x, y) ds(y) .
(2.39b)
|y|=R0
Equations (2.39a) and (2.39b) are two equations for the integrals. Solving for the second
integral yields
Z
m
(2.40)
Ynm (ŷ) Φ(x, y) ds(y) = ik R02 jn (kR0 ) h(1)
n (k|x|) Yn (x̂)
|y|=R0
where we have used the Wronskian of Theorem 2.27. This holds for all |x| > R0 . Comparing
this with the form (2.38) of u yields
u(x) =
∞ X
n
X
αnm
m
ikR02 h(1)
n (kr) jn (kR0 ) Yn (x̂)
2
ikR
j
(kR
)
0
0 n
n=0 m=−n
∞ X
n
X
Z
=
|y|=R0
gnm Ynm (ŷ) Φ(x, y) ds(y)
n=0 m=−n
for |x| > R0 where
gnm =
αnm
.
ik R02 jn (kR0 )
(2.41)
Let R0 ∈ (R, R0 ). The series (2.38) converges uniformly on {x : |x| = R0 }, thus
∞ X
n
X
m 2 (1)
αn hn (kR0 )2 < ∞
n=0 m=−n
(1)
and, in particular, αnm hn (kR0 ) ≤ c1 for all |m| ≤ n and n ∈ N. Therefore
m
c2
gn ≤ ≤ c3 (2n + 1)
(1)
hn (kR0 )|jn (kR0 )|
R0
R0
n
by Theorem 2.31 which proves uniform convergence of the series
g(R0 x̂) =
∞ X
n
X
gnm Ynm (x̂)
n=0 m=−n
and ends the proof.
2
70
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Lemma 2.39 For any compact set K ⊆ R3 the fundamental solution Φ satisfies the Sommerfeld radiation condition (2.32) uniformly with respect to y ∈ K. More precisely,
exp(ik|x|) −ik x̂·y (2.42a)
1 + O |x|−1 , |x| → ∞ ,
e
Φ(x, y) =
4π|x|
∇x Φ(x, y) = ik x̂
exp(ik|x|) −ik x̂·y e
1 + O |x|−1 ,
4π|x|
|x| → ∞ ,
(2.42b)
uniformly with respect to x̂ = x/|x| ∈ S 2 and y ∈ K.
Proof: We set x = rx̂ with r = |x| and investigate
exp
ikr
|x̂
−
y/r|
−
1
−
|x̂
−
y/r|
exp
−ik
x̂
·
y
4π Φ(rx̂, y) r e−ikr − e−ik x̂·y =
|x̂ − y/r|
= F (1/r; x̂, y)
with
exp ik |x̂ − εy| − 1 /ε − |x̂ − εy| exp −ik x̂ · y
F (ε; x̂, y) =
.
|x̂ − εy|
The function g(ε; x̂, y) = |x̂ − εy| is analytic in a neighborhood of zero with respect to ε and
g(0; x̂, y) = 1 and ∂g(0; x̂, y)/∂ε = −x̂ · y. Therefore, by the rule of l’Hospital we observe
that F (0; x̂, y) = 0, and the assertion of the first part of the lemma follows because also F
is 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 formula
for Bessel functions .
Corollary 2.40 Let k ∈ C \ {0} with Im k ≥ 0. For any x, y ∈ R3 with |x| > |y| we have
∞
ik X
(2n + 1) jn (k|y|) h(1)
Φ(x, y) =
n (k|x|) Pn (x̂ · ŷ) ,
4π n=0
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 ŷ 7→ Φ(x, R0 ŷ) into spherical surface
harmonics; that is,
∞ X Z
X
Φ(x, R0 ŷ) =
Ynm (ẑ) Φ(x, R0 ẑ) ds(ẑ) Ynm (ŷ)
n=0 |m|≤n
|ẑ|=1
∞
1 X X
=
R02 n=0
Z
Yn−m (ẑ) Φ(x, z) ds(z) Ynm (ŷ)
|m|≤n|z|=R
0
= ik
∞ X
X
−m
jn (kR0 ) h(1)
(x̂) Ynm (ŷ)
n (k|x|) Yn
n=0 |m|≤n
∞
ik X
(2n + 1) jn (kR0 ) h(1)
=
n (k|x|) Pn (x̂ · ŷ) ,
4π n=0
2.6. THE HELMHOLTZ EQUATION IN THE INTERIOR AND EXTERIOR OF A BALL71
where we used (2.40) and the addition formula for spherical surface harmonics of Theorem 2.17. The uniform convergence follows from the asymptotic behaviour of jn (t) and
(1)
hn (t) for n → ∞, uniformly with respect to t from compact subsets of R>0 , see Theorem 2.31. Indeed, we have that
1
1
n
(k|y|) 1 + O
for n → ∞ ,
jn (k|y|) =
(2n + 1)!!
n
(2n − 1)!!
1
(1)
hn (k|x|) = −i
for n → ∞ ,
1+O
n+1
(k|x|)
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 n
1
R1
1+O
≤ c
n
R2
2
which ends the proof.
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 boundary
value problem (2.34) satisfies
∞
n
exp(ikr) X X
1
(−i)n+1 m
m
u(rx̂) =
(fR , Yn )L2 (S 2 ) (1)
Yn (x̂) 1 + O 2
(2.43a)
kr
r
h
(kR)
n
n=0 m=−n
Z
∞
exp(ikr) X (2n + 1) (−i)n+1
1
=
f (Rŷ) Pn (x̂ · ŷ) ds(ŷ) 1 + O 2
(2.43b)
(1)
4π kr n=0
r
hn (kR)
2
S
as r tends to infinity, uniformly with respect to x̂ ∈ S 2 .
Proof: We note that (2.43a), (2.43b) follow formally from (2.35a), (2.35b), respectively, by
(1)
using the asymptotics of Theorem 2.30 for hn (kr). The asymptotics, however, are not uniform with respect to n, and we have to use of the representation (2.36) as an integral instead.
(1)
The function g has the expansion coefficients (2.41) with αnm = (fR , Ynm )L2 (S 2 ) /hn (kR); that
is,
(fR , Ynm )L2 (S 2 )
m
.
gn =
(1)
ik R02 hn (kR) jn (kR0 )
Therefore, using the asymptotic form of Φ by Lemma 2.39,
Z
u(x) =
|y|=R0
exp(ikr)
Φ(x, y) g(y) ds(y) =
4π r
Z
e
|y|=R0
−ik x̂·y
g(y) ds(y) 1 + O
1
r2
.
72
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
We compute with the Jacobi-Anger expansion of Theorem 2.32
Z
Z
1
R02
R02
−ik x̂·y
−ikR0 x̂·ŷ
gR , eikR0 x̂· L2 (S 2 )
e
g(y) ds(y) =
e
g(Rŷ) ds(ŷ) =
4π
4π
4π
S2
|y|=R0
=
R02
∞ X
X
gnm (−i)n jn (kR0 ) Ynm (x̂)
n=0 |m|≤n
∞ X
X
1 (fR , Ynm )L2 (S 2 )
=
(−i)n jn (kR0 ) Ynm (x̂)
(1)
ik
hn (kR) jn (kR0 )
n=0 |m|≤n
∞
1 X X (−i)n+1
=
(fR , Ynm )L2 (S 2 ) Ynm (x̂)
(1)
k n=0
|m|≤n hn (kR)
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 asymptotic behaviour of the form
exp(ik|x|) ∞
1
u(x) =
u (x̂) 1 + O 2
, |x| → ∞ ,
r
r
uniformly with respect to x̂ = x/|x| ∈ S 2 . The function u∞ : S 2 → C is called the far field
pattern 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 harmonic
Maxwell system in a homogeneous medium with vanishing external current and charge densities. By the close connection between the Helmholtz equation and the Maxwell system as
discussed in Chapter 1 we are able to use several results from the previous section.
Thus, we consider the boundary value problems
curl E − iωµ0 H = 0 ,
curl H + iωε0 E = 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 and
dielectricity ε as µ0 and ε0 from vacuum. But we can allow the parameters also being
√
complex 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 harmonic
Maxwell’s equations in a domain D ⊆ R3 , if and only if E (or H) is a divergence free solution
of the vector Helmholtz equation; that is,
∆E + k 2 E = 0 and
div E = 0 in D
(see Section 1.3, equation (1.12)). By the following lemma we can construct solutions of
Maxwell’s equations from scalar solutions of the Helmholtz equation.
Lemma 2.42 Let u ∈ C ∞ (D) satisfy ∆u + k 2 u = 0 in a domain D ⊆ R3 . Then
E(x) = curl x u(x)
and
H =
1
curl E ,
iωµ
x ∈ D,
(2.44)
are solutions of the time harmonic Maxwell system
curl E − iωµH = 0 in D
and curl H + iωεE = 0 in D .
(2.45)
1
Furthermore, also Ẽ(x) = curl curl x u(x) and H̃ = iωµ
curl Ẽ 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. We
observe that div E = 0. From curl curl = −∆ + ∇ div and div E = 0 we conclude that
iωµ curl H(x) = curl curl E(x) = −∆E(x) = − curl ∆ x u(x)
= − curl 2 ∇u(x) + x ∆u(x) = k 2 curl x u(x) = k 2 E(x)
which proves the assertion because of k 2 = ω 2 εµ. For the proof of the second part we just
take the curl of the last formula.
2
In view of the expansion results for the Helmholtz equation of the previous section we like
to consider solutions of the Helmholtz equation of the form u(x) = jn (k|x|)Ynm (x̂) where
x
Ynm denotes a spherical surface harmonics of order n ∈ N0 , m = −n, . . . , n and x̂ = |x|
. By
m
m
Lemma 2.42 we see that E(x) = curl(x jn (k|x|)Yn (x̂)) and E(x) = curl curl(x jn (k|x|)Yn (x̂))
are solutions of the Maxwell system. We can also replace the Bessel function by the Hankel
function, if the region D does not contain the origin, which will lead to solutions satisfying
a 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) Ynm (x̂) = jn (kr) GradS 2 Ynm (x̂) × x̂ ,
(2.46a)
and the corresponding function
H(x) =
=
1
curl E(x)
iωµ
1 n(n + 1)
jn (kr) Ynm (x̂) x̂ +
jn (kr) + krjn0 (kr) GradS 2 Ynm (x̂)
iωµr
iωµr
(2.46b)
74
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
satisfy the Maxwell system (2.45). Again, r = |x| and x̂ = x/|x| denote the spherical
coordinates, and GradS 2 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
x̂ × E(x) = jn (kr) GradS 2 Ynm (x̂) ,
1 jn (kr) + krjn0 (kr) x̂ × GradS 2 Ynm (x̂) .
iωµr
(b) Analogously, the vector fields Ẽ(x) = curl curl x jn (kr)Ynm (x̂) and H̃ =
solutions of Maxwell’s equations.
(2.47a)
x̂ × H(x) =
(2.47b)
1
iωµ
curl Ẽ are
(c) The results of (a) and (b) hold in R3 \ {0} if the Bessel function jn is replaced by the
(1)
Hankel function hn .
Proof: (a) From Lemma 2.42 we already know that E(x) = curl x jn (kr)Ynm (x̂) generates
a solution of Maxwell’s equations. We have to prove the second representation.
By using
m
the identity (6.7) and curl x = 0 we find E(x) = ∇ jn (kr)Yn (x̂) × x. Decomposing the
gradient into its tangential part, Grad (see (6.17)), and the radial component, (∂/∂r) x̂,
leads to
m
∂
1
m
jn (k|x|) GradS 2 Yn (x̂) +
jn (k|x|) Yn (x̂) x̂ × x
E(x) =
r
∂r
= jn (kr) GradS 2 Ynm (x̂) × x̂
which proves the second representation in (2.46a).
Now we consider curl E. Analogously to the proof of the previous lemma we use curl curl =
−∆+∇ div and the fact that w(x) = jn (kr)Ynm (x̂) solves the Helmholtz equation. Therefore,
curl E(x) = −∆ xw(x) + ∇ div xw(x)
= −2 ∇w(x) + k 2 x w(x) + 3 ∇w(x) + ∇ x · ∇w(x)
= ∇ jn (kr) + krjn0 (kr) Ynm (x̂) + k 2 x jn (kr)Ynm (x̂)
0
=
jn (kr) + krjn0 (kr) + k 2 r jn (kr) Ynm (x̂) x̂
1
jn (kr) + krjn0 (kr) GradS 2 Ynm (x̂)
r
1
=
2krjn0 (kr) + k 2 r2 jn00 (kr) + k 2 r2 jn (kr) Ynm (x̂) x̂
r
1
+
jn (kr) + krjn0 (kr) GradS 2 Ynm (x̂)
r
n(n + 1)
1
=
jn (kr)Ynm (x̂) x̂ +
jn (kr) + krjn0 (kr) GradS 2 Ynm (x̂)
r
r
+
where we used the Bessel differential equation (2.27) in the last step. Formulas (2.47a) and
(2.47b) follow obviously.
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
75
(1)
2
(c) This is shown as in (a) and (b) by replacing jn by hn .
From the theorem we observe that the tangential components of the boundary values on
a sphere involve the surface gradients GradS 2 Ynm and x̂ × GradS 2 Ynm . We have seen – and
already used several times – that the surface gradient of u ∈ C 1 (S 2 ) is a tangential field,
i.e. ν · GradS 2 u = 0 on S 2 with the unit normal vector ν(x̂) = x̂ on S 2 . Obviously, also
ν × GradS 2 u is tangential. Using the partial integration formula (6.21) we compute
Z
Z
(x̂ × GradS 2 u) · (x̂ × GradS 2 v) ds =
GradS 2 u · (x̂ × GradS 2 v) × x̂ ds
S2
S2
Z
Z
GradS 2 u · GradS 2 v ds = −
=
S2
u ∆S 2 v ds
S2
for u ∈ C 1 (S 2 ) and v ∈ C 2 (S 2 ). Since the spherical harmonics Ynm : −n ≤ m ≤ n, n ∈ N
form an orthonormal system of eigenfunctions of the spherical Laplace-Beltrami operator
∆S 2 with the eigenvalues λ = −n(n + 1), n ∈ N0 , we conclude that the vector fields
1
GradS 2 Ynm (x̂)
Unm (x̂) := p
n(n + 1)
(2.48a)
as well as the vector fields
Vnm (x̂) := x̂ × Unm (x̂) = p
1
n(n + 1)
x̂ × GradS 2 Ynm (x̂)
(2.48b)
are two orthonormal systems of tangential fields on S 2 . Applying partial integration again
and using property (b) of the following lemma we obtain
Z
Z
(x̂ × GradS 2 u) · GradS 2 v ds = −
DivS 2 (x̂ × GradS 2 u) v ds = 0 .
S2
S2
Combining these sets of vector fields Unm and Vnm 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 orthonormal
system in the space of tangential vector fields on S 2 . In preparation of the proof we need
some more properties of the spherical harmonics and the surface differential operators.
Lemma 2.44
(a) DivS 2 (x̂ × GradS 2 u) = 0 for all u ∈ C 2 (S 2 ) where DivS 2 denotes the surface divergence
on the unit sphere (see Example 6.17).
(b) If a tangential field u ∈ Ct1 (S 2 ) = {u ∈ C 1 (S 2 , C3 ) : u · ν = 0 on S 2 } satisfies
DivS 2 u = 0 and
then u = 0 on S 2 .
DivS 2 (ν × u) = 0 on S 2
76
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Proof: (a) We extend u to a C 1 function in a neighborhood of S 2 and apply Corollary 6.20.
This yields DivS 2 (x̂ × GradS 2 u) = DivS 2 (x̂ × ∇u) = −x̂ · curl ∇u = 0 on S 2 .
(b) From DivS 2 (ν × u) = 0 we obtain by partial integration (6.21)
Z
Z
DivS 2 (ν × u) f ds = −
(ν × u) · GradS 2 f ds for every f ∈ C 1 (S 2 ) .
0 =
S2
S2
We express u in spherical coordinates with respect to the basis θ̂, ϕ̂ of the spherical unit
vectors; that is, u = (u · θ̂)θ̂ + (u · ϕ̂)ϕ̂ and thus
ν × u = (u · θ̂)ϕ̂ − (u · ϕ̂)θ̂ .
Let f ∈ C 1 (S 2 ) be arbitrary. The representation of the surface gradient of f on S 2 (see
Example 6.17) implies by partial integration with respect to θ and ϕ, respectively, the
identity
Z 2π Z π
Z
1 ∂f
∂f
θ̂ +
ϕ̂ sin θ dθ dϕ
(ν × u) ·
(ν × u) · GradS 2 f ds =
0 =
∂θ
sin θ ∂ϕ
0
0
S2
Z 2π Z π ∂
∂
=
(u · ϕ̂) sin θ −
(u · θ̂) f dθ dϕ .
∂θ
∂ϕ
0
0
This leads to
∂
∂
(u · ϕ̂) sin θ =
u · θ̂
∂θ
∂ϕ
R
θ
on S 2 . Considering the anti-derivative h(ϕ, θ) = 0 (u · θ̂)(ϕ, θ0 ) dθ0 we have
∂h
= u · θ̂ and
∂θ
Z θ
Z θ
∂h
∂
∂
0
0
(ϕ, θ) =
(u · θ̂)(ϕ, θ ) dθ =
(u · ϕ̂)(ϕ, θ0 ) sin θ0 dθ0 = (u · ϕ̂) sin θ .
∂ϕ
0 ∂ϕ
0 ∂θ
Therefore we conclude that
∂h
1 ∂h
u = (u · θ̂)θ̂ + (u · ϕ̂)ϕ̂ =
θ̂ +
ϕ̂ = GradS 2 h .
∂θ
sin θ ∂ϕ
Finally, we use the condition DivS 2 u = 0 and obtain by partial integration
Z
Z
Z
0 =
h DivS 2 u ds =
u · GradS 2 h ds =
|u|2 ds
S2
which yields u = 0 on S 2 .
S2
S2
2
Remark: From the first part of the proof of part (b) we observe that DivS 2 (ν × u) = 0 on
S 2 implies the existence of a surface potential h. This motivates the common notation of
the surface curl of a tangential field u by Curl u = − Div(ν × u).
The following result is needed as a preparation for the expansion theorem but is also of
independent interest. It proves existence and uniqueness of a solution for the equation
∆S 2 u = f on S 2 where ∆S 2 = DivS 2 GradS 2 denotes again the spherical Laplace-Beltrami
operator (see Definition 2.2).
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
77
R
Theorem 2.45 For every f ∈ C ∞ (SR2 ) with S 2 f ds = 0 there exists aP
unique
solution u ∈
∞ Pn
2
2
2
C (S ) with ∆S 2 u = −f on S and S 2 u ds = 0. If f is given by f = n=1 m=−n fnm Ynm
with fnm = (f, Ynm )L2 (S 2 ) then the solution is given by
Z
∞
1 X 2n + 1
fnm
m
Y (x) =
Pn (x · y) f (y) ds(y) .
u(x) =
n(n + 1) n
4π n=1 n(n + 1) S 2
n=1 m=−n
∞ X
n
X
Both series converge uniformly with all their derivatives. In particular, u ∈ C ∞ (S 2 ).
Proof: Uniqueness is seen by Green’s theorem.
Indeed, if u ∈ C 2 (SR2 ) is the difference of
R
two solutions then ∆S 2 u = 0 and thus 0 = S 2 u DivS 2 GradS 2 u ds =R S 2 | GradS 2 u|2 ds; that
is, GradS 2 u = 0 which implies that u is constant. The requirement S 2 u ds = 0 yields that
u vanishes.
For existence it suffices to show that the series for u converges with all of its derivatives
because ∆S 2 Ynm = −n(n + 1)Ynm . For partial sums we have for any q ∈ N
N
n
X
X
Z X
N
n
X
1
fnm
m
(SN u)(x) =
Yn (x) =
Ynm (x) Yn−m (y) f (y) ds(y)
n(n
+
1)
n(n
+
1)
S 2 m=−n
n=1
n=1 m=−n
Z
N
1 X 2n + 1
Pn (x · y) f (y) ds(y)
=
4π n=1 n(n + 1) S 2
Z
N
(−1)q X
2n + 1
=
∆qS 2 Pn (x · y) f (y) ds(y)
q+1
q+1
4π n=1 n (n + 1)
S2
Z
N
2n + 1
(−1)q X
Pn (x · y) ∆qS 2 f (y) ds(y)
=
4π n=1 nq+1 (n + 1)q+1 S 2
where we used the addition formula of Theorem 2.17 again and ∆S 2 Pn (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 operator
D` of order ` that
Z
N
(−1)q X
2n + 1
Dx` Pn (x · y) ∆qS 2 f (y) ds(y) .
(D SN u)(x) =
q+1
q+1
4π n=1 n (n + 1)
S2
`
By the chain rule and Exercise 2.6 we conclude that |Dx` Pn (x · y)| ≤ cn2` for all n ∈ N and
x, y ∈ S 2 . Therefore,
N
1 X
2n + 1
q+1
4π n=1 n (n + 1)q+1
Z
S2
Dx` Pn (x
·
y) ∆qS 2 f (y) ds(y)
which converges if we choose q ≥ ` + 1.
≤
c k∆qS 2 f k∞
N
X
(2n + 1) n2`
nq+1 (n + 1)q+1
n=1
2
78
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Theorem 2.46 The functions
1
GradS 2 Ynm (x̂)
Unm = p
n(n + 1)
and
Vnm (x̂) = x̂ × Unm (x̂)
for n ∈ N and −n ≤ m ≤ n constitute a complete orthonormal system in
L2t (S 2 ) = u ∈ L2 (S 2 ) : ν · u = 0 on S 2 .
Proof: Let f ∈ Ct∞ (S 2 ) = {u ∈ C ∞ (S 2 , C3 ) : u·ν
= 0 on S 2 }. Then
also DivS 2 f, DivS 2 (ν ×
R
R
∞
2
f ) ∈ C (S ), and by Green’s theorem we have S 2 DivS 2 f ds = S 2 DivS 2 (ν × f ) ds = 0. By
the previous Theorem 2.45 there exist solutions u1 , u2 ∈ C ∞ (S 2 ) with ∆S 2 u1 = DivS 2 f and
∆S 2 u2 = DivS 2 (ν × f ) on S 2 . Their expansions
∞ X
n
X
u1 =
m
am
n Yn
and u2 =
n=1 m=−n
∞ X
n
X
m
bm
n Yn
n=1 m=−n
converge uniformly with all of their derivatives. Set
g = Grad u1 − ν × Grad u2 =
∞ X
n
X
m
m
am
− bm
n Grad Yn
n ν × Grad Yn
n=1 m=−n
=
∞ X
n
X
p
p
m
m
n(n + 1) am
n(n + 1) bm
.
n Un −
n Vn
n=1 m=−n
Then g ∈ Ct∞ (S 2 ) and, by using part (a) of
Lemma 2.44, DivS 2 g = ∆S 2 u1 = DivS 2 f and
DivS 2 (ν × g) = − DivS 2 ν × (ν × GradS 2 u2 ) = DivS 2 GradS 2 u2 = DivS 2 (ν × f ). Application
of part (b) of Lemma 2.44 yields g = f which proves uniform convergence of f into vector
spherical harmonics. The completeness of the set {Unm , Vnm : −n ≤ m ≤ m, n ∈ N} in
2
L2t (S 2 ) follows again by a denseness argument just as in the proof of Theorem 2.19.
As a corollary we formulate an expansion of any vector field A in terms of the normal basis
n
and Vnm .
functions x̂ → Ymn (x̂)x̂ and the tangential basis functions Um
Corollary 2.47 Every vector field A ∈ L2 (S 2 , C3 ) has an expansion of the form
∞ X
X
A(x̂) =
m
m m
m m
am
n Yn (x̂) x̂ + bn Un (x̂) + cn Vn (x̂)
(2.49)
n=0 |m|≤n
where
am
n
bm
n
cm
n
Z
=
S2
Z
=
S2
Z
=
S2
A(x̂) · x̂ Yn−m (x̂) ds(x̂) ,
|m| ≤ n , n ≥ 0 ,
(2.50a)
A(x̂) · Un−m (x̂) ds(x̂) ,
|m| ≤ n , n ≥ 1 , b00 = 0 ,
(2.50b)
A(x̂) · Vn−m (x̂) ds(x̂) ,
|m| ≤ n , n ≥ 1 , c00 = 0 .
(2.50c)
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
79
The convergence is understood in the L2 −sense. Furthermore, Parseval’s equality holds; that
is,
∞ X
X
n 2
|am | + |bnm |2 + |cnm |2 = kAk2L2 (S 2 ,C3 ) .
n=0 |m|≤n
Proof: We decompose the vector field A into
A(x̂) = A(x̂) · x̂ x̂ + x̂ × A(x̂) × x̂ .
The scalar function x̂ 7→ x̂ · A(x̂) and the tangential vector field x̂ 7→ x̂ × A(x̂) × x̂ can
be expanded into basis functions on S 2 according to Theorem 2.19 for the scalar radial part
and the previous Theorem 2.46 for the tangential part.
2
We apply this corollary to the vector field x̂ 7→ E(rx̂) where E satisfies Maxwell’s equation.
Theorem 2.48 Let E, H ∈ C 1 B(0, R), C3 satisfy curl E −iωµH = 0 and curl H +iωεE =
0 in B(0, R). Then there exist unique αnm , βnm ∈ C, |m| ≤ n, n = 0, 1, 2, . . . such that
"
0
∞ X
X
p
rj
(kr)
j
(kr)
n
n
E(x) =
αnm n(n + 1)
Ynm (x̂) x̂ + αnm
Unm (x̂)
r
r
n=1
|m|≤n
+
=
∞ X
X
βnm jn (kr) Vnm (x̂)
1
p
n(n + 1)
−
βnm
n=1 |m|≤n
curl
(2.51a)
h
αnm curl curl x jn (kr) Ynm (x̂)
i
x jn (kr) Ynm (x̂)
(2.51b)
0
∞
rj
(kr)
1 X X h mp
jn (kr) m
n
H(x) = −
Yn (x̂) x̂ + βnm
Unm (x̂)
βn n(n + 1)
iωµ n=1
r
r
|m|≤n
+ k 2 αnm jn (kr) Vnm (x̂)
=
i
∞
h
1
1 X X
p
k 2 αnm curl x jn (kr) Ynm (x̂)
iωµ n=1
n(n + 1)
|m|≤n
i
m
m
− βn curl curl x jn (kr) Yn (x̂)
(2.51c)
(2.51d)
for x = rx̂ and 0 ≤ r < R, x̂ ∈ S 2 . The series converge uniformly in every ball B[0, R0 ] for
R0 < R.
m
m
Proof: Let am
corresponding
n (r), bn (r), and cn (r) be the expansion coefficients of (2.49)
R
1
0
to the vector field given by x̂ 7→ E(rx̂). First we note that a0 (r) = r2 |x|=r x̂ · E(x) ds(x)
vanishes by the divergence theorem because div E = 0.
80
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Second, we consider am
n (r) for n ≥ 1. With
Z
Z
1
m
−m
an (r) =
x̂ · E(rx̂) Yn (x̂) ds(x̂) =
x · E(x) Yn−m (x̂) ds(x̂)
r
2
2
S
S
we observe that ram
n (r) is the expansion coefficient of the solution x 7→ x · E(x) of the scalar
Helmholtz equation, see Lemma 1.5. Thus by Theorem 2.33 we obtain
am
n (r) =
α̃nm
jn (kr)
r
for some α̃nm . Next, the condition div E = 0 in spherical polar coordinates reads
1 ∂ 2
r Er (r, x̂) + DivS 2 Et (r, x̂) = 0
r ∂r
(see (6.20)), and we obtain
1 2 m 0
r an (r)
=
r
Z
1 ∂ 2
r Er (r, x̂) Yn−m (x̂) ds(x̂)
r ∂r
S2
Z
= −
S2
Z
=
S2
DivS 2 Et (r, x̂) Yn−m (x̂) ds(x̂)
E(r, x̂) · GradS 2 Yn−m (x̂) ds(x̂) =
p
n(n + 1) bm
n (r)
by (6.21). Thus, with a common constant αnm we have
p
0
n(n + 1)
m 1
m
m
jn (kr) , bm
an (r) = αn
r jn (kr) .
n (r) = αn
r
r
(2.52)
Finally, we consider cm
n (r). By partial integration it holds
Z
p
m
n(n + 1) cn (r) =
E(r, x̂) · x̂ × GradS 2 Yn−m (x̂) ds(x̂)
S2
Z
E(r, x̂) × x̂ · GradS 2 Yn−m (x̂) ds(x̂)
=
S2
Z
= −
S2
DivS 2 E(r, x̂) × x̂ Yn−m (x̂) ds(x̂)
Z
= −r
S2
Z
= −
S2
x̂ · curl E(rx̂) Yn−m (x̂) ds(x̂)
x · curl E(rx̂) Yn−m (x̂) ds(x̂)
where we have used that DivS 2 E(r, x̂) × x̂ = Div|x|=r E(rx̂) × x̂ = r x̂ · curl E(rx̂) (see
p
Corollary 6.20). Therefore,
n(n + 1) cm
n (r) are the expansion coefficients of the scalar
function ψ(x) = x · curl E(x). Because ψ is a solution of the Helmholtz equation ∆ψ + k 2 ψ =
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
81
m
0, see Lemma 1.5, we again conclude from Theorem 2.33 that cm
n (r) = βn jn (kr) for some
βnm ∈ 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, R0 ] for every R0 < R. As in the scalar case we
will prove another representation of the field E. Fix R0 < R and choose R̂ with R0 < R̂ < R.
The scalar function x̂ 7→ E(R̂x̂) · x̂ is smooth and we obtain from (2.51a) that
Z p
jn (k R̂)
m
E(R̂ŷ) · ŷ Yn−m (ŷ) ds(ŷ) .
=
αn n(n + 1)
R̂
S2
Analogously, we obtain
αnm
jn (k R̂) + k R̂jn0 (k R̂)
R̂
Z
E(R̂ŷ) · Un−m (ŷ) ds(ŷ)
=
S2
= −p
1
n(n + 1)
Z
S2
DivS 2 E(R̂, ŷ) Yn−m (ŷ) ds(ŷ) ,
and
βnm jn (k R̂)
Z
=
S2
E(R̂ŷ) · Vn−m (ŷ) ds(ŷ)
Z
1
= p
n(n + 1)
1
= −p
n(n + 1)
E(R̂, ŷ) · ŷ × GradS 2 Yn−m (ŷ) ds(ŷ)
S2
Z
S2
DivS 2 E(R̂, ŷ) × ŷ Yn−m (ŷ) ds(ŷ) .
Substituting these coefficients into (2.51a) and using the addition formula of Theorem 2.17
yields
"
Z
∞ X
X
R̂ jn (kr)
E(R̂ŷ) · ŷ Yn−m (ŷ) ds(ŷ) Ynm (x̂) x̂
E(rx̂) =
r jn (k R̂) S 2
n=1
|m|≤n
Z
R̂ jn (kr) + krjn0 (kr)
1
DivS 2 E(R̂, ŷ) Yn−m (ŷ) ds(ŷ) Ynm (x̂)
−
GradS 2
r jn (k R̂) + k R̂jn0 (k R̂) n(n + 1)
2
S
#
Z
−m
jn (kr)
1
−
DivS 2 E(R̂, ŷ) × ŷ Yn (ŷ) ds(ŷ) Ynm (x̂)
(2.53)
x̂ × GradS 2
n(n
+
1)
2
jn (k R̂)
S
"
Z
∞
1 X
R̂ jn (kr)
=
(2n + 1)
E(R̂ŷ) · ŷ Pn (x̂ · ŷ) ds(ŷ) x̂
4π n=1
r jn (k R̂) S 2
Z
R̂ jn (kr) + krjn0 (kr)
1
−
GradS 2
DivS 2 E(R̂, ŷ) Pn (x̂ · ŷ) ds(ŷ)
r jn (k R̂) + k R̂jn0 (k R̂) n(n + 1)
S2
#
Z
jn (kr)
1
−
x̂ × GradS 2
DivS 2 E(R̂, ŷ) × ŷ Pn (x̂ · ŷ) ds(ŷ) .
jn (k R̂) n(n + 1)
S2
82
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
This has now the form of the series for the solution of the interior boundary value problem for
the Helmholtz equation. The arguments used there, in the proof of Theorem 2.33, provide
analogously uniform convergence of the series and its derivatives in the closed ball B[0, R0 ].
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 closely
related to the idea from Lemma 2.42 using the scalar expansion functions of the Helmholtz
equation. On the other hand (2.51a) is suitable for treating boundary value problems (see
Theorem 2.49 below) and, additionally, gives some inside into the important relation of the
tangential and the normal component of the electric field on surfaces.
Combining the previous results we obtain the following existence result for the interior
Maxwell problem in a ball; that is, the boundary value problem for Maxwell’s equation in
a ball B(0, R) with boundary values ν × E = fR on |x| = R. We formulate the existence
result again with fR ∈ L2t (S 2 ) where fR (x̂) = f (Rx̂).
√
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 jn0 (kR) 6= 0 for all n ∈ N0 .
1
(a) For given fR ∈ L2t (S 2 ) there exists a unique solution E ∈ C 2 B(0, R) and H = iωµ
curl E
of
curl E − iωµH = 0 and
curl H + iωεE = 0 in B(0, R)
with
lim kν × E(r, ·) − fR kL2 (S 2 ) = 0 .
r→R
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
83
The solution is given by
∞ X
n
X
(fR , Unm )L2 (S 2 )
1
p
E(rx̂) =
curl x jn (kr) Ynm (x̂)
n(n + 1) jn (kR)
n=1 m=−n
(fR , Vnm )L2 (S 2 )
R
m
+ p
curl
curl
xj
(kr)
Y
(x̂)
n
n
n(n + 1) jn (kR) + kR jn0 (kR)
=
∞ X
n
X
(fR , Vnm )L2 (S 2 )
p
n(n + 1)
n=1 m=−n
H(rx̂) =
R
jn (kr) m
Yn (x̂) x̂
jn (kR) + kR jn0 (kR) r
+ (fR , Vnm )L2 (S 2 )
R
jn (kr) + kr jn0 (kr) m
Un (x̂)
jn (kR) + kR jn0 (kR)
r
− (fR , Unm )L2 (S 2 )
jn (kr) m
V (x̂) ,
jn (kR) n
∞
n
1 X X (fR , Unm )L2 (S 2 )
1
p
curl curl x jn (kr) Ynm (x̂)
iωµ n=1 m=−n
n(n + 1) jn (kR)
(fR , V m )L2 (S 2 )
R
+ k2 p n
curl xjn (kr) Ynm (x̂)
0
n(n + 1) jn (kR) + kR jn (kR)
∞
n
p
jn (kr) m
1 X X
1
(fR , Unm )L2 (S 2 ) n(n + 1)
Yn (x̂) x̂
=
iωµ n=1 m=−n
jn (kR) r
+ (fR , Unm )L2 (S 2 )
jn (kr) + kr jn0 (kr) m
1
Un (x̂)
jn (kR)
r
− k 2 (fR , Vnm )L2 (S 2 )
R
jn (kr) Vnm (x̂)
0
jn (kR) + kRjn (kR)
for x = rx̂ ∈ B(0, R). The series converges uniformly on compact subsets of B(0, R).
(b) If fR ∈ Ct2 (S 2 ) the solution of the boundary value problem satisfies E ∈ C 2 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 value
problem for the Helmholtz equation.
Let us repeat the steps. To show uniqueness we assume E is a solution of the homogeneous
boundary value problem, i.e. kν × E(rx̂)kL2 (S 2 ) → 0 for r → R with ν(x̂) = x̂. First we
consider the expansion from Theorem 2.48 for Rε ∈ (R0 , R), where we choose 0 < R0 < R
84
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
and ε > 0 with kν × E(Rε x̂)kL2 (S 2 ) ≤ ε. From (2.53) we obtain
∞ X
X
Rε jn (kr) + kr jn0 (kr)
x̂ × E(rx̂) =
(ν × E(Rε . ), Vnm )L2 (S 2 ) Unm (x̂)
0 (kR )
r
j
(kR
)
+
kR
j
n
ε
ε
ε
n
n=1
|m|≤n
∞ X
X
jn (kr)
−
(ν × E(Rε . ), Unm )L2 (S 2 ) Vnm (x̂) .
jn (kRε )
n=1
|m|≤n
By Parseval’s identity we obtain for 0 < R1 < r < R0 < Rε < R
kν ×
E(r . )k2L2 (S 2 )
∞ X 2
X
Rε 2 jn (kr) + kr jn0 (kr) 2 (ν × E(Rε . ), Vnm ) 2 2 =
L (S )
r jn (kRε ) + kRε j 0 (kRε ) n
n=1 |m|≤n
2
jn (kr) 2 m
+
(ν × E(Rε . ), Un )L2 (S 2 ) jn (kRε )
∞ X 2 2
X
≤ c
(ν × E(Rε . ), Vnm )L2 (S 2 ) + (ν × E(Rε . ), Unm )L2 (S 2 ) 2
n=1 |m|≤n
≤ c2 kν × E(Rε )kL2 (S 2 ) ≤ c2 ε2 .
(2.54)
(kr) Here we have used again as in the proof of 2.34 a uniform bound for all n ∈ N of jjnn(kR
≤c
ε)
0 (kr)
jn (kr)+kr jn
and of Rrε jn (kR
≤ c for r ∈ [R1 , R0 ], where the last estimate follows from
0
ε )+kRε jn (kRε )
Theorem 2.31 by using the identity jn (z) + zjn0 (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 αnm (rjn (kr))0 =
0 and βnm jn (kr) = 0 for all r ∈ (R1 , R0 ) and |m| ≤ n, n ∈ N. Thus it follows αnm = βnm = 0,
and the expansion leads to E = 0 in B(0, R).
Next we show that the function E(rx̂) given in the Theorem satisfies the boundary condition
in L2 sense. For r < R we compare the representation (2.51b) with the given expansion and
define
αnm =
R
(fR , Vnm )L2
jn (kR) + kRjn0 (kR)
and βnm = −
1
(fR , Unm )L2 .
jn (kR)
Thus, we obtain from (2.51a)
∞ X
n
X
αnm
x̂ × E(rx̂) =
(rjn (kr))0 x̂ × Unm (x̂) + βnm jn (kr) x̂ × Vnm (x̂)
r
n=1 m=−n
=
∞ X
n
X
αnm
(rjn (kr))0 Vnm (x̂) − βnm jn (kr) Unm (x̂) .
r
n=1 m=−n
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
85
With the definition of αnm and βnm it follows by Parseval’s identity
kν × E(r . ) − fR kL2
2
∞ X
n
X
jn (kr)
m
2
=
jn (kR) − 1 |(fr , Un )L2 |
n=1 m=−n
2
∞ X
n
X
R jn (kr) + krjn0 (kr)
|(fr , Vnm )L2 |2 .
+
−
1
r jn (kR) + krj 0 (kR)
n
n=1 m=−n
As in Theorem 2.34 we can adapt the arguments
2.22 in the case
of Theorem
of the Lapla jn (kr)
R jn (kr)+krjn0 (kr)
cian, since by Theorem 2.31 the terms jn (kR) − 1 and r jn (kR)+kRj 0 (kR) − 1 are uniformly
n
bounded for r ∈ [0, R] and converge to zero as r tends to R for every n. Thus we conclude
that kν × E(r, ·) − fR kL2 → 0 if r → R.
Furthermore, the second part of the proof of Theorem
2.22
to the electric
can be transfered
jn (kr) R jn (kr)+krjn0 (kr) fields by the uniform bounds on the ratios jn (kR) and r jn (kR)+kRj 0 (kR) and we conclude
n
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 the
first kind in the constructed expansion functions will lead to solutions satisfying a radiation
condition. At this point we find it very convenient to use the scalar Sommerfeld radiation
condition (2.32) for the solutions x · E(x) and x · H(x) of the Helmholtz equation. These
conditions imply in particular that the radial components of the electric and magnetic fields
decay 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 radiation
condition is equivalent to the better known Silver-Müller radiation condition.
Theorem 2.50 Let k ∈ C\{0} with Im k ≥ 0 and E ∈ C 2 R3 \B[0, R] and H =
be solutions of the time harmonic Maxwell equations
curl E − iωµH = 0 in D
and
1
iωµ
curl E
curl H + iωεE = 0 in R3 \ B[0, R] .
Furthermore, let x 7→ x · E(x) and x 7→ x · H(x) satisfy the Sommerfeld radiation condition (2.32). Then there exist unique αnm , βnm ∈ C, |m| ≤ n, n = 0, 1, 2, . . . such that
∞ X mp
X
0 m
αn n(n + 1) (1)
αm
(kr)
Un (x̂)
hn (kr) Ynm (x̂) x̂ + n rh(1)
E(x) =
n
r
r
n=1
|m|≤n
+
=
βnm
m
h(1)
n (kr) Vn (x̂)
(2.55a)
∞ X
X
h
1
m
p
αnm curl curl x h(1)
n (kr) Yn (x̂)
n(n + 1)
n=1 |m|≤n
i
m
(kr)
Y
(x̂)
− βnm curl x h(1)
n
n
(2.55b)
86
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
for x = rx̂ and r > R, x̂ ∈ S 2 . The series converges uniformly with all of its derivatives in
compact subsets of R3 \ B[0, R].
(1)
The field H has the forms (2.51c), (2.51d) with hn replacing jn .
Proof: The proof usesR exactly the same arguments as the proof of Theorem 2.48. First
−m
(x̂) ds(x̂) is the expansion coefficient of the solution
we note that ram
n (r) = S 2 x · E(x) Yn
u(x) = x · E(x) of the Helmholtz equation with respect to {Ynm : |m| ≤ n, n ∈ N}. By
m (1)
m
assumption u is radiating, therefore ram
n (r) = α̃n h (kr) for some coefficient α̃n . Therefore,
(1)
m
m (1)
(2.52) holds for jn replaced by hn . Analogously, cm
n (r) is given by cn (r) = βn hn (kr) for
some βnm because also x 7→ x·curl E(x) is radiating. This yields the form (2.55a) for E(x) and
(1)
from this also (2.55b) by using (2.46a), (2.46b) for hn 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 exterior
Maxwell problem for a ball is uniquely solvable for any k ∈ C with Im k ≥ 0 and every
R > 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 ∈ C 2 R3 \ B[0, R] of
curl H + iωεE = 0 in R3 \ B[0, R]
curl E − iωµH = 0 and
with
lim kν × E(r, ·) − fR kL2 (S 2 ) = 0 ,
r→R
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
∞ X
n
X
(fR , Unm )L2 (S 2 )
1
m
p
curl x h(1)
E(rx̂) =
n (kr) Yn (x̂)
(1)
n(n + 1) hn (kR)
n=1 m=−n
+
=
(1)
(fR , Vnm )L2 (S 2 )
R
m
p
curl
curl
xh
(kr)
Y
(x̂)
n
n
(1) 0
n(n + 1) h(1)
n (kR) + kR hn (kR)
n
∞ X
X
(fR , Vnm )L2 (S 2 )
n=1 m=−n
+
(fR , Vnm )L2 (S 2 )
p
(1)
R
hn (kr) m
n(n + 1) (1)
Yn (x̂) x̂
(1) 0
r
hn (kR) + kR hn (kR)
R
(1)
(1)
−
(fR , Unm )L2 (S 2 )
(1) 0
hn (kr) + kr hn (kr) m
Un (x̂)
(1)
(1) 0
r
hn (kR) + kR hn (kR)
hn (kr)
(1)
hn (kR)
Vnm (x̂) .
The series converges uniformly on compact subsets of R3 \ B[0, R].
(1)
The field H has the forms as in Theorem 2.49 with hn replacing jn .
2.7. EXPANSION OF ELECTROMAGNETIC WAVES
87
(b) If fR ∈ L2t (S 2 ) ∩ C 2 (S 2 ) the solution E is continuous in all of R3 \ B(0, R), and its series
representation 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 boundary
value problem because the arguments are almost the same as in the proof of the cooresponding Theorem 2.49 for the interior boundary value problem. Again, we only mention
(1)
(1)
(1) 0
that hn (kR) and hn (kR) + kRhn (kR) do not vanish for any kR which follows again by
Theorem 2.27. To show the radiation condition we multiply the second representation of
E(x) by x = rx̂ and arrive at
x · E(x) =
∞ X
n
X
(1)
(fR , Vnm )L2 (S 2 )
p
n(n + 1)
n=0 m=−n
Rhn (kr)
(1)
hn (kR)
+
(1) 0
kRhn (kR)
Ynm (x̂) .
This scalar solution of the Helmholtz equation has just the form of (2.38). Application of
Lemma 2.38 and Lemma 2.39 yields that x 7→ x · E(x) satisfies the Sommerfeld radiation
condition.
The corresponding proof for x · H(x) uses the same arguments for the representation of
1
curl E.
2
H = iωµ
For later use we formulate and prove the analog of Lemma 2.38.
Lemma 2.52 Let
E(x) =
∞ X
X
m
m
m
αn curl x h(1)
+ βnm curl curl x h(1)
n (kr) Yn (x̂)
n (kr) Yn (x̂)
n=1 |m|≤n
converge uniformly on compact subsets of R3 \ B[0, R]. Then there exists R0 > R and scalar
continuous functions g and h such that
Z
E(x) = curl
g(y) Φ(x, y) ds(y) x
|y|=R0
+ curl
2
Z
h(y) Φ(x, y) ds(y) x
(2.56a)
|y|=R0
Z
1
1
2
H(x) =
curl E(x) =
curl
g(y) Φ(x, y) ds(y) x
iωµ
iωµ
|y|=R0
Z
+ iωε curl
h(y) Φ(x, y) ds(y) x
(2.56b)
|y|=R0
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 previous
theorem 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
88
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
and substitute the right hand side of (2.40) into the series for E(x). This yields
E(x) =
∞ X
X
n=1 |m|≤n
αnm
curl
ikR02 jn (kR0 )
Z
Ynm (ŷ) Φ(x, y) ds(y) x
|y|=R0
Z
βnm
m
+
curl curl
Yn (ŷ) Φ(x, y) ds(y) x
ikR02 jn (kR0 )
|y|=R0
Z
Z
= curl
g(y) Φ(x, y) ds(y) x + curl curl
h(y) Φ(x, y) ds(y) x
|y|=R0
|y|=R0
with
g(y) =
∞ X
X
n=1 |m|≤n
αnm
Ynm (ŷ) ,
2
ikR0 jn (kR0 )
h(y) =
∞ X
X
n=1 |m|≤n
βnm
Ynm (ŷ) .
2
ikR0 jn (kR0 )
The uniform convergence of the series for g and h is shown in the same way as in the proof
of Lemma 2.38. The form of H is obvious by curl2 = −∆ + ∇ div and k 2 = ω 2 µε.
2
We finish this chapter by the following corollary which states that any pair (E, H) of solutions
of the Maxwell system which satisfy the scalar radiation conditions of Theorems 2.50 and
2.51 satisfy also the Silver–Müller radiation condition (1.20a), (1.20b); or even,
√
and
µ0 H(x) × x̂ −
√
ε0 E(x) × x̂ +
√
ε0 E(x) = O(|x|−2 )
(2.57a)
√
µ0 H(x) = O(|x|−2 )
(2.57b)
uniformly in x/|x| ∈ S 2 . As mentioned above, the scalar radiation conditions of Theorems 2.50 and 2.51 imply in particular that the radial components x̂ · E(x) and x̂ · H(x)
decay faster than O(|x|−1 ) to zero as |x| tends to infinity; that is, E(x) and H(x) are “almost” tangential fields on large spheres. On the other hand, the Silver–Müller radiation
condition implies that the combination H(x) × x − |x|E(x) decays faster than O(|x|−1 ) to
zero 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 ∈ C 2 R3 \B[0, R] and H =
be solutions of the time harmonic Maxwell equations
curl E − iωµH = 0 in D
and
1
iωµ
curl E
curl H + 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–Müller radiation condition (2.57a), (2.57b).
Proof: From Theorem 2.50 and Lemma 2.52 we conclude that a representation of the
form (2.56a) holds. Both terms on the right hand side satisfy the Silver–Müller radiation
condition, see Lemma 3.29 of the next chapter.
2.
2.8. EXERCISES
2.8
89
Exercises
Exercise 2.1 (a) Prove that the Laplacian ∆ is given in spherical polar coordinates by
(2.1).
r cos ϕ
(b) Prove that the Laplacian ∆ is given in planar polar coordinates x =
by
r sin ϕ
1 ∂
∂
1 ∂2
∆ =
r
+
.
r ∂r
∂r
r ∂ϕ2
Exercise 2.2 Show that the Laplace-Beltrami operator ∆S 2 is selfadjoint and non-posivite;
that is,
Z
Z
Z
f (x̂) ∆S 2 f (x̂) ds(x̂) ≤ 0
g(x̂) ∆S 2 f (x̂) ds(x̂) ,
f (x̂) ∆S 2 g(x̂) ds(x̂) =
S2
S2
S2
for all f, g ∈ C 2 (S 2 ).
Exercise 2.3 Construct harmonic functions in R2 or R2 \ {0} by separation of variables in
plane 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 both
sides 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
Pn (t) =
(2n−1)!
;
n! (n−1)! 2n−1
that is,
(2n − 1)!
tn + Qn−1 (t)
n! (n − 1)! 2n−1
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 of
c` > 0 such that
`
d Pn (t) (`)
≤ c` n2` for all n ∈ N .
kPn k∞ = max |t|≤1
dt` Exercise 2.7 Try to prove Theorems 2.23.
Exercise 2.8 Prove the following representations of the derivative by using Rayleigh’s Formulas from Theorem 2.28
n
n+1
fn0 (z) =
fn (z) − fn+1 (z) , fn0 (z) = − fn−1 (z) −
fn (z) , z 6= 0 ,
z
z
(1)
(2)
where fn is any of the functions jn , yn , hn , or hn .
90
CHAPTER 2. EXPANSION INTO WAVE FUNCTIONS
Exercise 2.9 Prove the following asymptotics for the derivatives by modifying the proof of
Theorem 2.31. For any r ∈ N and ε > 0 and R > ε:
dr n
1
dr
1
for n → ∞ ,
jn (z) =
z 1+O
r
r
dz
(2n + 1)!! dz
n
dr (1)
1
dr 1
for n → ∞ ,
hn (z) = −i (2n − 1)!! r n+1 1 + O
r
dz
dz z
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 for
the 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 boundary
value problem we introduce the boundary integral equation method which reformulates the
boundary value problem in terms of an integral equation on the boundary of the region. For
showing existence of a solution of this integral equation we will apply the Riesz–Fredholm
theory. Again, as in the previous chapter, we consider first the simpler scattering problem
for the scalar Helmholtz equation in Section 3.1 before we turn to Maxwell’s equations in
Section 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 +k 2 uinc =
0 in all of R3 , find the total field u ∈ C 2 (R3 \ D) ∩ C 1 (R3 \ D) such that
∆u + k 2 u = 0 in R3 \ D ,
∂u
= 0 on ∂D ,
∂ν
(3.1)
and such that the scattered field us = u − uinc satisfies the Sommerfeld radiation condition
(2.32); that is,
∂us (rx̂)
1
s
for r → ∞ ,
(3.2)
− ik u (rx̂) = O 2
∂r
r
uniformly with respect to x̂ ∈ S 2 . For smooth fields u ∈ C 1 (R3 \ D) and smooth boundaries
the normal derivative is given by ∂u/∂ν = ∇u · ν where ν = ν(x) denotes the exterior unit
normal 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, throughout 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 of
bounded domains Dj such that Dj ∩ D` = ∅ for j 6= `. Furthermore, we assume that the
boundary ∂D is C 2 −smooth (see Definition 6.7 of the appendix) and that the complement
R3 \ D is connected.
Before we investigate uniqueness and existence of a solution of this scattering problem we
study general properties of solutions of the Helmholtz equation ∆u + k 2 u = 0 in bounded
and 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
∆x Φk (x, y) + k 2 Φk (x, y) = 0 for x 6= y .
2
Proof: This is easy to check.
We often suppress the index k; that is, write Φ for Φk . Since the fundamental solution
and/or its derivatives will occur later on as kernels of various integral operators, we begin
with 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
c
K(x, y) ≤
for all x ∈ R3 and y ∈ D with x 6= y .
|x − y|3−β
R
Then the integral D K(x, y) dy exists in the sense of Lebesgue and there exists cβ > 0 with
Z
K(x, y) dy ≤ cβ for all x ∈ R3 and all τ > 0 ,
(3.3a)
D\B(x,τ )
Z
K(x, y) dy ≤ cβ τ β
for all x ∈ R3 and all τ > 0 .
(3.3b)
D∩B(x,τ )
(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
c
K(x, y) ≤
for all x, y ∈ ∂D with x 6= y .
|x − y|2−β
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
Then
R
∂D
93
K(x, y) ds(y) exists and there exists cβ > 0 with
Z
K(x, y) ds(y) ≤ cβ for all x ∈ ∂D and τ > 0 ,
(3.3c)
∂D\B(x,τ )
Z
K(x, y) ds(y) ≤ cβ τ β
for all x ∈ ∂D and τ > 0 .
(3.3d)
∂D∩B(x,τ )
(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−β
for all x, y ∈ ∂D with x 6= y .
Then there exists cβ > 0 with
Z
K(x, y) ds(y) ≤ cβ τ β−1
for all x ∈ ∂D and τ > 0 .
(3.3e)
∂D\B(x,τ )
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 with
respect to x,
Z
D\B(x,τ )
1
dy ≤
|x − y|3−β
=
Z
τ <|y−x|<3R
1
dy = 4π
|x − y|3−β
Z3R
1
r3−β
r2 dr
τ
4π 4π
(3R)β − τ β ≤
(3R)β .
β
β
Second case: |x| > 2R. Then |x − y| ≥ |x| − |y| ≥ R for y ∈ D and thus
Z
Z
1
1 4π 3
1
dy ≤ 3−β
dy = 3−β
R .
3−β
R
R
3
D\B(x,τ ) |x − y|
B(0,R)
This proves (3.3a). For (3.3b) we compute
Z
K(x, y) dy ≤ c
D∩B(x,τ )
Z
|x−y|<τ
1
dy = 4πc
|x − y|3−β
Zτ
1
r3−β
r2 dr =
4πc β
τ .
β
0
(b) We choose a local coordinate system as in Definition 6.7; that is, a covering of ∂D by
cylinders of the form Uj = Rj Cj + z (j) where Rj ∈ R3×3 are rotations and z(j) ∈ R3 and
Cj = B2 (0, αj ) × (−2ρj , 2ρj ), and where
∂D ∩ Uj is expressed as ∂D ∩ Uj = Rj x + 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, a
family of functions φj ∈ C ∞ (R3 ), j = 1, . . . , m, with
• 0 ≤ φj ≤ 1 in R3 ,
94
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
• the support Sj := supp(φj ) is contained in Uj , and
Pm
•
j=1 φj (x) = 1 for all x ∈ ∂D.
S
Then, obviously, ∂D ⊆ m
j=1 Sj . Furthermore, there exists δ > 0 such that |x − y| ≥ δ
S
for all (x, y) ∈ Sj × `6=j (U` \ Uj ) for every j. Indeed, otherwise there would exist some j
S
and a sequence (xk , yk ) ∈ Sj × `6=j (U` \ Uj ) with |xk − yk | → 0. There exist convergent
S
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
Z
Z
m
X
ds(y)
φj (y)
=
ds(y) .
2−β
|x − y|
|x − y|2−β
`=1
∂D
|x−y|>τ
∂D∩U`
|x−y|>τ
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 = Rj u + z (j) and y = Rj v + z (j) for some u, v ∈
B2 (0, αj ) × (−ρj , ρj ) with u3 = ξj (u1 , u2 ) and v3 = ξj (v1 , v2 ). We set ũ = (u1 , u2 ) and
ṽ = (v1 , v2 ) and certainly have an estimate of the form
c1 |x − y| ≤ |ũ − ṽ| ≤ c2 |x − y| for all x, y ∈ Uj ∩ ∂D .
Using polar coordinates with respect to ũ we estimate
Z
Z
q
1
ds(y)
2−β
≤
c
1 + |∇ξj (ṽ)|2 dṽ
2
|x − y|2−β
|ũ − ṽ|2−β
∂D∩Uj
|x−y|>τ
B2 (0,αj )\B2 (ũ,c1 τ )
Z
≤ ĉ
dṽ
= 2π ĉ
|ũ − ṽ|2−β
c1 τ <|ṽ−ũ|<2αj
Z
2αj
c1 τ
r
r2−β
dr ≤ 2π ĉ
(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 is
already determined by its Dirichlet- and Neumann data on the boundary. This theorem is
totally 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 ∈ C 2 (D) ∩ C 1 (D) we have the representation
Z
Z ∂Φ
∂u
2
Φ(x, y) ∆u(y) + k u(y) dy +
u(y)
(x, y) − Φ(x, y) (y) ds(y)
∂ν(y)
∂ν
D
∂D


 −u(x) , x ∈ D ,
− 12 u(x) , x ∈ ∂D ,
=


0,
x 6∈ D .
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
95
Remarks:
• This theorem tells us that, for x ∈ D, any function u can be expressed as a sum of
three potentials:
Z
(S̃ϕ)(x) =
ϕ(y) Φ(x, y) ds(y) , x ∈
/ ∂D ,
(3.4a)
∂D
Z
(D̃ϕ)(x) =
ϕ(y)
∂D
∂Φ
(x, y) ds(y) ,
∂ν(y)
Z
ϕ(y) Φ(x, y) dy ,
(Vϕ)(x) =
x∈
/ ∂D ,
x ∈ R3 ,
(3.4b)
(3.4c)
D
which are called single layer potential, double layer potential, and volume potential,
respectively, with density ϕ. We will investigate these potential in detail in Subsection 3.1.2 below.
• The one-dimensional analogon is (for x ∈ D = (a, b) ⊆ R)
1
u(x) =
2ik
Zb
ik|x−y|
e
00
u (y) + k 2 u(y) dy +
a
b
1
d ik|x−y|
0
ik|x−y|
.
u(y) e
− u (y) e
2ik
dy
a
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 x
x−y
with radius r > 0. For y ∈ ∂B(x, r) the normal vector ν(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) in
the domain Dr := D \ B[x, r]. Then
Z ∂Φ
∂u
u(y)
(x, y) − Φ(x, y) (y) ds(y)
(3.5a)
∂ν(y)
∂ν
∂D
Z
∂u
∂Φ
(x, y) − Φ(x, y) (y) ds(y)
(3.5b)
+
u(y)
∂ν(y)
∂ν
∂B(x,r)
Z
Z
=
u(y) ∆y Φ(x, y) − Φ(x, y) ∆u(y) dy = −
Φ(x, y) ∆u(y) + k 2 u(y) dy ,
Dr
Dr
if one uses the Helmholtz equation for Φ. We compute the integral (3.5b). We observe that
1
y−x
exp(ik|x − y|)
ik −
∇y Φ(x, y) =
4π|x − y|
|x − y| |x − y|
and thus for |y − x| = r:
Φ(x, y) =
exp(ikr)
,
4πr
∂Φ
x−y
exp(ikr)
(x, y) =
· ∇y Φ(x, y) = −
∂ν(y)
r
4πr
1
ik −
.
r
96
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Therefore, we compute the integral (3.5b) as
Z
∂u
∂Φ
(x, y) − Φ(x, y) (y) ds(y)
u(y)
∂ν(y)
∂ν
∂B(x,r)
Z
exp(ikr)
1
∂u
=
u(y)
− ik −
(y) ds
4πr
r
∂ν
|y−x|=r
Z
Z
exp(ikr)
exp(ikr)
∂u
=
(y) ds .
u(y) ds −
ik u(y) +
4πr2
4πr
∂ν
|y−x|=r
|y−x|=r
For r → 0 the first term tends to u(x), because the surface area of ∂B(x, r) is just 4πr2 and
Z
Z
exp(ikr)
exp(ikr)
u(y)
ds
=
u(y)
−
u(x)
ds + exp(ikr) u(x) ,
4πr2
4πr2
|y−x|=r
|y−x|=r
where we obtain by continuity
Z
1
u(y) − u(x) ds ≤
4πr2
|y−x|=r
sup |u(y) − u(x)| → 0 ,
r → 0.
y∈B(x,r)
Similarly the second term tends to zero. Therefore, also the limit of the volume integral
exists 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 the
computation 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 in
all 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 + k 2 u = 0 in D. In this case the function u can be expressed solely as a combination of a
single and a double layer surface potential. This observation is a first hint on a reformulation
of 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 too
strong. For the representation theorem the domain D has to satisfy exactly the assumptions
which 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 ∈ C 2 (D) be a real- or complex valued solution of the Helmholtz equation
in D. Then u is analytic, i.e. one can locally expand u into a power series. That is, for
every z ∈ D there exists r > 0 such that u has the form
u(x) =
X
an (x1 − z1 )n1 (x2 − z2 )n2 (x3 − z3 )n3
n∈N3
where we use the notation N = Z≥0 = {0, 1, 2, . . .}.
for
3
X
j=1
|xj − zj |2 < r2 ,
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
97
Proof: From the previous representation of u(x) as a difference of a single and a double
layer potential and the smoothness of the kernels x 7→ Φ(x, y) and x 7→ ∂Φ(x, y)/∂ν(y) for
x 6= y it follows immediately that u ∈ C ∞ (D). The proof of analyticity is technically not
easy if one avoids methods from complex analysis.1 If one uses these methods then one can
argue as follows: Fix x̂ ∈ D and choose r > 0 such that B[x̂, r] ⊆ D. Define the region
R ⊆ C3 and the function v : R → C by
R = z ∈ C3 : Re z − x̂ < r/2, Im z < r/2 ,
q

qP3

Z expik P3 (z − y )2 2
exp
ik
(z
−
y
)
j
j
j
j
j=1
j=1
∂
∂u
 ds(y)

qP
qP
(y) − u(y)
v(z) =
3
3
∂ν
∂ν(y)
2
2
4π
(z − y )
4π
(z − y )
∂D
j=1
j
j
j=1
j
j
for z ∈ R. Taking the principal
P3 value with2 cut along the negative real axis
P3 of the square
root of the complex number j=1 (zj − yj ) is not a problem because Re j=1 (zj − yj )2 =
P3
2
2
2
2
j=1 (Re zj − yj ) − (Im zj ) = |Re z − y| − |Im z| > 0 because of |Re z − y| ≥ |y − x̂| −
|x̂ − Re z| > r − r/2 = r/2 and |Im z| < r/2. Obviously, the function v is holomorphic in R
and thus (complex) analytic.
2
As a second corollary we can easily prove the following version of Holmgren’s uniqueness
theorem for the Helmholtz equation.
Theorem 3.5 (Holmgren’s uniqueness theorem)
Let D be a domain with C 2 -boundary and u ∈ C 1 (D) ∩ C 2 (D) be a solution of the Helmholtz
equation ∆u + k 2 u = 0 in D. Furthermore, let U be an open set such that U ∩ ∂D 6= ∅. If
u = 0 and ∂u/∂ν = 0 on U ∩ ∂D, then u vanishes in all of D.
Proof: Let z ∈ U ∩ ∂D and B ⊆ U a ball centered at z. Set Γ = D ∩ ∂B. Then
∂(B ∩ D) = Γ ∪ (B ∩ ∂D). The reader should sketch the situation. We define v by
Z ∂Φ
∂u
(x, y) ds(y) , x ∈ B .
v(x) =
Φ(x, y) (y) − u(y)
∂ν
∂ν(y)
Γ
Then v satisfies the Helmholtz equation in B. Application of Green’s representation formula
of Theorem 3.3 to u in B ∩ D yields2
Z
∂Φ
∂u
(x, y) ds(y) = v(x) , x ∈ B ∩ D ,
u(x) =
Φ(x, y) (y) − u(y)
∂ν
∂ν(y)
∂(B∩D)
because the integral vanishes on ∂D ∩ B. By the same theorem we conclude that v(x) = 0
for x ∈ B \ D. Since v is analytic by the previous corollary we conclude that v vanishes in
all 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’s
representation theorem.
1
We refer to [15], Section 2.4, for a proof.
In 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.
2
98
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Theorem 3.6 (Green’s representation theorem in the exterior of D)
Let k ∈ C \ {0} with Im k ≥ 0 and u ∈ C 2 (R3 \ D) ∩ C 1 (R3 \ D) be a solution of the
Helmholtz equation ∆u + k 2 u = 0 in R3 \ D. Furthermore, let u satisfy the Sommerfeld
radiation condition (3.2). Then we have the representation


/ D,
Z  u(x) , x ∈
∂Φ
∂u
1
u(y)
(x, y) − Φ(x, y) (y) ds(y) =
u(x) , x ∈ ∂D ,
2

∂ν(y)
∂ν
∂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’s
representation Theorem 3.3 in the annular region B(0, R) \ D. Noting that ∆u + k 2 u = 0
and that ν(y) for y ∈ ∂D is directed into the interior of B(0, R) \ D yields
Z ∂Φ
∂u
u(x) =
u(y)
(x, y) − Φ(x, y) (y) ds(y)
∂ν(y)
∂ν
∂D
Z
∂Φ
∂u
−
u(y)
(x, y) − Φ(x, y) (y) ds(y) .
∂ν(y)
∂ν
|y|=R
We show that the surface integral over ∂B(0, R) tends to zero as R tends to infinity. We
write this surface integral (for fixed x) as
Z
Z
∂u
∂Φ
(x, y) − ik Φ(x, y) ds(y) −
Φ(x, y)
(y) − ik u(y) ds(y)
IR =
u(y)
∂ν(y)
∂ν
|y|=R
|y|=R
and use the Cauchy-Schwarz inequality:
2
|IR |
Z
2
≤
Z
|u| ds
|y|=R
Z
+
|y|=R
2
∂Φ
∂ν(y) (x, y) − ik Φ(x, y) ds(y)
Φ(x, y)2 ds(y)
|y|=R
2
∂u
ds(y)
−
ik
u
|y|=R ∂ν
Z
From the radiation conditions of Φ(x, ·) for fixed x with respect to y and of u we conclude
that the integrands of the second and forth integral behave as O(1/R4 ) as R → ∞. Since
the surface area of ∂B(0, R) is equal to 4πR2 we conclude that the second and forth integral
tend to zero as R tends to infinity. Furthermore, the integrand of the third integral behaves
2
as O(1/R
) as R → ∞. Therefore, the third integral is bounded. It remains to show that
R
also |y|=R |u|2 ds is bounded. This follows again from the radiation condition. Indeed, from
the radiation condition we have


)
2
Z Z ( 2
Z
∂u
∂u 1
∂u 
ds =
+ |ku|2 ds + 2 Im 
=
−
iku
u
ds .
O
k
2
R
∂r
∂r
∂r
|x|=R
|x|=R
|x|=R
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
99
Green’s theorem, applied in B(0, R) \ D to the function u yields
Z
Z
Z
∂u
∂u
2
u
ds =
u
ds +
|∇u|2 − k |u|2 dx .
∂r
∂ν
|x|=R
∂D
B(0,R)\D
We multiply by k and take the imaginary part. This yields




Z
Z
∂u 
∂u 

Im k
u
ds = Im k u
ds + Im k
∂r
∂ν
|x|=R
∂D
Z
|∇u|2 + |k|2 |u|2 dx
B(0,R)\D


Z
≥ Im k
u
∂u 
ds
∂r
∂D
and thus
Z


2
Z
∂u + |ku|2 ds ≤ −2 Im k u ∂u ds + O 1 .
∂r ∂r
R2
|x|=R
∂D
R
2
This implies, in particular, that |y|=R |u| ds is bounded. Altogether, we have shown that IR
tends 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 combination
of volume and surface potentials. The integral equation method for solving boundary value
problems for the Helmholtz equation and Maxwell’s equations rely heavily on the smoothness
properties of these potentials. This subsection is concerned with the investigation of these
potentials. 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,
eik|x−y|
,
Φ(x, y) =
4π|x − y|
and begin with the volume potential
Z
w(x) =
ϕ(y) Φ(x, y) dy ,
x 6= y ,
x ∈ R3 ,
(3.6)
(3.7)
D
where D ⊆ R3 is any open and bounded set.
Lemma 3.7 Let ϕ ∈ L∞ (D) by any complex–valued function. Then w ∈ C 1 (R3 ) and
Z
∂w
∂Φ
(x) =
ϕ(y)
(x, y) dy , x ∈ R3 , j = 1, 2, 3 .
(3.8)
∂xj
∂x
j
D
100
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Proof: We fix j ∈ {1, 2, 3} and a real valued function η ∈ C 1 (R) with 0 ≤ η(t) ≤ 1 and
η(t) = 0 for t ≤ 1 and η(t) = 1 for t ≥ 2. We set
Z
∂Φ
ϕ(y)
(x, y) dy , x ∈ R3 ,
v(x) =
∂x
j
D
∂Φ
and note that the integral exists by Lemma 3.2 because ∂x
(x,
y)
≤ c0 (|x − y|−2 for some
j
c0 > 0. Furthermore, set
Z
wε (x) =
ϕ(y) Φ(x, y) η |x − y|/ε dy , x ∈ R3 .
D
Then wε ∈ C 1 (R3 ) and
∂wε
(x) =
v(x) −
∂xj
Z
ϕ(y)
|y−x|≤2ε
∂ Φ(x, y) 1 − η |x − y|/ε
dy .
∂xj
Thus, we have
Z
∂Φ
kη 0 k∞ ∂w
ε
v(x) −
Φ(x, y) dy
(x) ≤ kϕk∞
∂xj (x, y) + ε
∂xj
|y−x|≤2ε
Z
1
1
+
dy
≤ c1
2
ε |x − y|
|y−x|≤2ε |x − y|
Z2π Zπ Z2ε = c1
0
0
1
1
+
r2 sin θ dr dθ dϕ = 16π c1 ε .
r2
εr
0
Therefore, ∂wε /∂xj → v uniformly in R3 , which shows w ∈ C 1 (R3 ) and ∂w/∂xj = v.
2
This regularity result is not sufficient in view of second order differential equations. Higher
regularity is obtained for Hölder continuous densities.
Definition 3.8 For a set T ⊆ R3 and α ∈ (0, 1] we define the space C 0,α (T ) of bounded,
uniformly Hölder-continuous functions by
)
(
v(x) − v(y)
C 0,α (T ) := v ∈ C(T ) : v bounded and
sup
< ∞ .
|x − y|α
x,y∈T, x6=y
Note that any Hölder-continuous function is uniformly continuous and, therefore, has a
continuous extension to T . The space C 0,α (T ) is a Banach space with norm
v(x) − v(y)
kvkC 0,α (T ) := supv(x) +
sup
.
(3.9)
|x − y|α
x∈T
x,y∈T, x6=y
| {z }
= kvk∞
Now we can complement the previous Lemma and obtain that volume potentials with Hölder
continuous densities are two times differentiable.
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
101
Theorem 3.9 Let ϕ ∈ C 0,α (D) and let w be the volume potential. Then w ∈ C 2 (D) ∩
C ∞ (R3 \ D) and
−ϕ , in D ,
2
∆w + k w =
0 , in R3 \ D .
Furthermore, if D0 is any C 2 −smooth domain with D ⊆ D0 then
∂ 2w
(x) =
∂xi ∂xj
Z
D0
∂ 2Φ
(x, y) ϕ(y) − ϕ(x) dy − ϕ(x)
∂xi ∂xj
Z
∂D0
∂Φ
(x, y) νj (y) ds(y) (3.10)
∂xi
for x ∈ D and i, j ∈ {1, 2, 3} where we have extended ϕ by zero in D0 \ D. If ϕ = 0 on ∂D
then w ∈ C 2 (R3 ).
Proof: First we note that the volume integral in the last formula exists. Indeed, we fix
x ∈ D and split the region of integration into D0 = D ∪ (D0 \ D). The integral over D0 \ D
exists because
the integrand
is smooth. The integral over D exists again by Lemma 3.2
because ϕ(y) − ϕ(x)∂ 2 Φ/(∂xi ∂xj )(x, y) ≤ c |x − y|α−3 for y ∈ D. The existence of the
surface integral is obvious because x ∈
/ ∂D0 .
We fix i, j ∈ {1, 2, 3} and the same funtion η ∈ C 1 (R) as in the previous lemma. We define
v := ∂w/∂xi ,
Z
u(x) :=
D0
∂ 2Φ
(x, y) ϕ(y) − ϕ(x) dy − ϕ(x)
∂xi ∂xj
Z
∂D0
∂Φ
(x, y) νj (y) ds(y) ,
∂xi
x ∈ D,
and
Z
vε (x) :=
∂Φ
ϕ(y) η |x − y|/ε
(x, y) dy ,
∂xi
D
x ∈ R3 .
Then vε ∈ C 1 (D) and for x ∈ D
∂Φ
∂
ϕ(y)
η |x − y|/ε
(x, y) dy
∂xj
∂xi
D
Z
∂
∂Φ
=
ϕ(y) − ϕ(x)
η |x − y|/ε
(x, y) dy
∂xj
∂xi
D0
Z
∂Φ
∂
+ ϕ(x)
η |x − y|/ε
(x, y) dy
∂xi
D0 ∂xj
Z
∂
∂Φ
=
ϕ(y) − ϕ(x)
η |x − y|/ε
(x, y) dy
∂xj
∂xi
D0
Z
∂Φ
− ϕ(x)
(x, y) νj (y) ds(y)
∂D0 ∂xi
∂vε
(x) =
∂xj
Z
provided 2ε ≤ d(x, ∂D0 ). In the last step we used the Divergence Theorem. Therefore, for
102
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
x ∈ D,
Z
∂Φ
u(x) − ∂vε (x) ≤
ϕ(y) − ϕ(x) ∂
1 − η |x − y|/ε
(x, y) dy
∂xj
∂xj
∂xi
|y−x|≤2ε
Z
kη 0 k∞
1
+
|y − x|α dy
≤ c1
3
2
|y
−
x|
ε
|y
−
x|
|y−x|≤2ε
Z2ε = c2
1
r1−α
(2ε)α
(2ε)1+α
kη 0 k∞ α
r dr = c2
+
≤ c3 ε
+
ε
α
(1 + α) ε
0
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 ∈ C 2 (D) and u = ∂ 2 w/(∂xi ∂xj ).
This proves (3.10). Finally, we fix z ∈ D and set D0 = B(z, R) where R is chosen such that
D ⊆ B(z, R). For x ∈ D we have
Z
2
∆w(x) = −k
Φ(x, y) ϕ(y) − ϕ(x) dy
B(z,R)
Z
− ϕ(x)
|y−z|=R
3
X
xj − y j
yj − zj exp(ikR)
ik − 1/R
ds(y) .
R
4π R
R
j=1
This holds for all x ∈ D. For x = z we conclude
Z
exp
ik|z
−
y|
∆w(z) = −k 2 w(z) + k 2 ϕ(z)
dy − ϕ(z) eikR (1 − ikR) ,
4π
|z
−
y|
B(z,R)
i.e.
"
∆w(z) + k 2 w(z) = −ϕ(z) eikR (1 − ikR) − k 2
Z
B(z,R)
|
{z
= 1
#
exp ik|z − y|
dy
4π |z − y|
}
because
k2
Z
B(z,R)
ZR
exp ik|z − y|
dy = k 2 r eikr dr = −ikR eikR + eikR − 1 .
4π |z − y|
0
If ϕ vanishes on ∂D then the extension of ϕ by zero is in C 0,α (R3 ) (see Exercise 3.5).
Therefore, we can apply the result above to any ball D = D0 = B(0, R) which yields that
the second derivatives of w are continuous in R3 .
2
The following corollary shows that the volume integral is bounded when consider as an
integral operator between suitable spaces.
Corollary 3.10 Let D be C 2 −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 ∈ C 0,α (D). Then there exists c > 0
with
kwkC 1 (R3 ) ≤ c kϕk∞ and kwkC 2 (A) ≤ c kϕkC 0,α (D)
(3.11)
for all ϕ ∈ C 0,α (D).
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
103
Proof: We estimate
1
,
4π|x − y|
∂Φ
1
1
∂xj (x, y) ≤ c1 |x − y|2 + |x − y| ,
2
∂ Φ
1
1
1
∂xi ∂xj (x, y) ≤ c2 |x − y|3 + |x − y|2 + |x − y| .
Φ(x, y) =
Thus for x ∈ R3 we can apply (3.8) and Lemma 3.2 above and obtain
Z
1
w(x) ≤ kϕk∞
dy ≤ c kϕk∞ ,
D 4π|x − y|
Z ∂Φ
∂w
∂xj (x, y) dy ≤ c kϕk∞ ,
∂xj (x) ≤ kϕk∞
D
where the constant c > 0 can be chosen independent of x and ϕ. This proves the first
estimate of (3.11). Let now x ∈ A. If A ⊆ D then there exists δ > 0 with |x − y| ≥ δ for all
x ∈ A and y ∈ ∂D. By (3.10) for D = D0 we have
2
Z 2
Z ∂ w
∂ Φ
∂Φ
α
∂xi ∂xj (x) ≤ kϕkC 0,α (D)
∂xi ∂xj (x, y) |x − y| dy + kϕk∞
∂xi (x, y) ds(y)
D
∂D
Z
Z
dy
ds(y)
+ c4 kϕk∞
≤ c3 kϕkC 0,α (D)
3−α
2
∂D |x − y|
D |x − y|
Z
c4
≤ c5 kϕkC 0,α (D) + 2 kϕk∞
ds
δ
∂D
≤ ckϕkC 0,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
2
Z 2
Z
∂ Φ
∂ w
c
dy ≤ ckϕkC 0,α (D) .
(x, y) dy ≤ kϕk∞ 3
∂xi ∂xj (x) ≤ kϕk∞
δ D
D ∂xi ∂xj
2
We continue with the single layer surface potential ; that is, the function
Z
v(x) = S̃ϕ(x) =
ϕ(y) Φ(x, y) ds(y) , x ∈ R3
(3.12)
∂D
The investigation of this potential requires some elementary facts from differential geometry
which we have collected in Subsection 6.3 of Chapter 6. First we note that for continuous
densities ϕ the integral exists by Lemma 3.2 above even for x ∈ ∂D because Φ has a
singularity of the form Φ(x, y) = 1/(4π|x − y|). Before we prove continuity of v we make
the following general remark.
104
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Remark 3.11 A function v : R3 → C is Hölder continuous if
(a) v is bounded,
(b) v is Hölder 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 thickness
δ as defined in Lemma 6.10; that is,
Hρ := z + tν(z) : z ∈ ∂D , |t| < ρ ,
Uδ := x ∈ R3 : inf |x − z| < δ .
z∈∂D
Proof : Choose δ > 0 such that U3δ ⊆ Hρ .
Ist case: |x1 − x2 | < δ and x1 ∈ U2δ . Then x1 , x2 ∈ U3δ ⊆ Hρ and Hölder continuity follows
from (b).
2nd case: |x1 − x2 | < δ and x1 ∈
/ U2δ . Then x1 , x2 ∈
/ Uδ and thus from (c):
v(x1 ) − v(x2 ) ≤ c|x1 − x2 | ≤ cδ 1−α |x1 − x2 |α .
3rd case: |x1 − x2 | ≥ δ. Then, by (a),
v(x1 ) − v(x2 ) ≤ 2kvk∞ ≤ 2kvk∞ |x1 − x2 |α .
δα
2
Using this remark we show that the single layer potential with continuous density is Hölder
continuous.
Theorem 3.12 The single-layer potential v from (3.12) with continuous density ϕ is uniformly Hölder continuous in all of R3 , and for every α ∈ (0, 1) there exists c > 0 (independent
of ϕ) with
kvkC α (R3 ) ≤ ckϕk∞ .
(3.13)
Proof: We check the
(a), (b), (c) of Remark 3.11. Boundedness follows from
conditions
Lemma 3.2 because Φ(x, y) = 1/(4π|x − y|). For (b) and (c) we write
Z
v(x1 ) − v(x2 ) ≤ kϕk∞ Φ(x1 , y) − Φ(x2 , y) ds(y)
(3.14)
∂D
and estimate
Φ(x1 , y) − Φ(x2 , y) ≤
≤
1
4π
1
ik|x −y|
1 1
e 1 − eik|x2 −y| −
+
|x1 − y| |x2 − y| 4π|x1 − y|
|x1 − x2 |
k |x1 − x2 |
+
4π |x1 − y||x2 − y|
4π |x1 − y|
(3.15)
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
105
because exp(it) − exp(is) ≤ |t − s| for all t, s ∈ R. Now (c) follows because |xj − y| ≥ δ for
xj ∈
/ Uδ .
To show (b); that is, Hölder continuity in Hρ , let x1 , x2 ∈ Hρ , where ρ < ρ0 is chosen as in
Lemma 6.10 such that there is the unique representation
of xj in the form xj = zj + tj ν(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 is
simply estimated by
Z
Z
Z
1
ds(y)
1
ds(y)
Φ(x1 , y) − Φ(x2 , y) ds(y) ≤
+
4π
|x1 − y|
4π
|x2 − y|
Γz1 ,r
Γz1 ,r
1
≤
2π
Z
Γz1 ,r
ds(y)
1
+
|z1 − y|
2π
Γz1 ,r
Z
ds(y)
.
|z2 − y|
Γz2 ,2r
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
Z
ds(y)
≤ c1 ρ
|z − y|
Γz,ρ
for some c1 independent of z and ρ has been proven in (3.3d). Therefore, we have shown
that
Z
Φ(x1 , y) − Φ(x2 , y) ds(y) ≤ c (r + 2r) = 9c |x1 − x2 | ≤ c̃ |x1 − x2 |α
2π
2π
Γz1 ,r
where c̃ 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|/3
and therefore
Φ(x1 , y) − Φ(x2 , y) ≤ 3|x1 − x2 | + k |x1 − x2 | ≤ 3|x1 − x2 | + k |x1 − x2 |
4π |x1 − y|2
4π |x1 − y|
π |z1 − y|2
2π |z1 − y|
because |z1 − y| ≤ 2|x1 − y|. Then we estimate
Z
Φ(x1 , y) − Φ(x2 , y) ds(y)
∂D\Γz1 ,r
|x1 − x2 |
≤
π
Z
3
k
ds(y)
+
|z1 − y|2
2|z1 − y|
∂D\Γz1 ,r
|x1 − x2 |α
=
π
Z
1−α
(r/3)
3
k
+
ds(y)
|z1 − y|2
2|z1 − y|
∂D\Γz1 ,r
|x1 − x2 |α
≤
π 31−α
Z
∂D\Γz1 ,r
3
k
+
ds(y)
|z1 − y|2−(1−α)
2 |z1 − y|1−(1−α)
106
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
because r1−α ≤ |y − z1 |1−α for y ∈ ∂D \ Γz1 ,r . Therefore,
Z
k
1
α
Φ(x1 , y) − Φ(x2 , y) ds(y) ≤
|x1 − x2 | 3 ĉ1−α + ĉ2−α
π 31−α
2
∂D\Γz1 ,r
with the constants ĉβ from Lemma 3.2, part (b). Altogether we have shown the existence of
c > 0 with
v(x1 ) − v(x2 ) ≤ c kϕk∞ |x1 − x2 |α for all x1 , x2 ∈ Hρ0
2
which ends the proof.
In preparation of the investigation of the double layer surface potential we prove an other
auxiliary result which we will be used often.
Lemma 3.13 For ϕ ∈ C 0,α (∂D) and a ∈ C(∂D, C3 ) define
Z
w(x) =
ϕ(y) − ϕ(z) a(y) · ∇y Φ(x, y) ds(y) ,
x ∈ Hρ0 ,
∂D
where x = z + tν(z) ∈ Hρ0 with |t| < ρ0 and z ∈ ∂D. Then the integral exists for x ∈ ∂D
and w is Hölder
continuous in Hρ0 for any exponent β < α. Furthermore, there exists c > 0
with w(x) ≤ ckϕkC 0,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 + tj ν(zj ) ∈ Hρ0 , j = 1, 2, we have to estimate
w(x1 ) − w(x2 )
Z
=
ϕ(y) − ϕ(z1 ) a(y) · ∇y Φ(x1 , y) − ϕ(y) − ϕ(z2 ) a(y) · ∇y Φ(x2 , y) ds(y)
(3.16)
∂D
= ϕ(z2 ) − ϕ(z1 )
Z
a(y) · ∇y Φ(x1 , y) ds(y) +
∂D
Z
+
ϕ(y) − ϕ(z2 ) a(y) · ∇y Φ(x1 , y) − ∇y Φ(x2 , y) ds(y)
∂D
and thus
Z
w(x1 ) − w(x2 ) ≤ ϕ(z2 ) − ϕ(z1 ) ∂D
Z
+ kak∞
a(y) · ∇y Φ(x1 , y) ds(y) +
ϕ(y) − ϕ(z2 ) ∇y Φ(x1 , y) − ∇y Φ(x2 , y) ds(y) (3.17)
∂D
We need the following estimates of ∇y Φ: There exists c > 0 with
c
∇y Φ(x, y) ≤
, x ∈ Hρ0 , y ∈ ∂D, x 6= y ,
|x − y|2
∇y Φ(x1 , y) − ∇y Φ(x2 , y) ≤ c |x1 − x2 | ,
|x1 − y|3
y ∈ ∂D , x` ∈ Hρ0 with 0 < |x1 − y| ≤ 3|x2 − y| .
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
107
Proof of these estimates: The first one is obvious. For the second one we observe that
0
Φ(x, y) = φ(|x − y|) with φ(t) = exp(ikt)/(4πt),
and φ00 (t) =
− 1/t)
thus φ (t) = φ(t)(ik
0
2
2
2
0
2
φ (t)(ik − 1/t) + φ(t)/t = φ(t) (ik − 1/t) + 1/t and therefore φ (t) ≤ c1 /t and φ00 (t) ≤
c2 /t3 for 0 < t ≤ 1. For t ≤ 3s we have
t
t
Z
Z
0
φ (t) − φ0 (s) = φ00 (τ ) dτ ≤ c2 dτ = c2 |t−2 − s−2 |
τ3 2
s
s
c2 |t2 − s2 |
1
c2
1
=
≤
|t − s|
+
2 t 2 s2
2
ts2 t2 s
c2
9
|t − s|
3
≤
|t − s| 3 + 3 = 6c2
.
2
t
t
t3
Setting t = |x1 − y| and s = |x2 − y| and observing that |t − s| ≤ |x1 − x2 | yields the estimate
0
y
−
x
y
−
x
1
2
0
∇y Φ(x1 , y) − ∇y Φ(x2 , y) = φ (|x1 − y|)
−
φ
(|x
−
y|)
2
|y − x1 |
|y − x2 | ≤ φ0 (|x1 − y|) − φ0 (|x2 − y|)
0
y − x1
y − x2 −
+ φ (|x2 − y|) |y − x1 | |y − x2 | ≤ 6 c2
0
|x2 − x1 |
φ (|x2 − y|) |x2 − x1 |
+
2
|y − x1 |3
|y − x1 |
≤ 6 c2
|x2 − x1 |
|x2 − x1 |
+ 2 c1
3
|y − 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 differ
from line to line the integral (in the form (3.16)) over Γz1 ,r is estimated by
Z
ϕ(y) − ϕ(z1 ) a(y) · ∇y Φ(x1 , y) − ϕ(y) − ϕ(z2 ) a(y) · ∇y Φ(x2 , y) ds(y) (3.18)
Γz1 ,r
1
1
α
ds(y)
≤ c
|y − z1 |
+ |y − z2 |
|y − x1 |2
|y − x2 |2
Γz1 ,r
Z
Z
1
1
α
≤ c
|y − z1 |
ds(y) + c
|y − z2 |α
ds(y)
2
|y − z1 |
|y − z2 |2
Γz1 ,r
Γz2 ,2r
Z
α
because Γz1 ,r ⊆ Γz2 ,2r and |xj − y| ≥ |zj − y|/2. Therefore, using (3.3d), this term behaves
as rα = 3α |x1 − x2 |α ≤ c|x1 − x2 |β .
108
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
We finally consider the integral over ∂D \ Γz1 ,r und use the form (3.17):
Z
I := ϕ(z2 ) − ϕ(z1 ) a(y) · ∇y Φ(x1 , y) ds(y) +
∂D\Γz1 ,r
Z
ϕ(y) − ϕ(z2 ) ∇y Φ(x1 , y) − ∇y Φ(x2 , y) ds(y)
+ kak∞
∂D\Γz1 ,r
Z
≤ c
|z2 − z1 |α
1
α |x1 − x2 |
ds(y)
+ |y − z2 |
|x1 − y|2
|x1 − y|3
∂D\Γz1 ,r
Since y ∈ ∂D \ Γz1 ,r we can use the estimates
|x1 − x2 | =
r
1
2
≤ |y − z1 | ≤ |y − x1 | < |y − x1 |
3
3
3
and
|y − z2 | ≤ 2|y − x2 | ≤ 2|y − x1 | + 2|x1 − x2 | ≤ 4|y − x1 | .
Thus, we obtain
Z
I ≤ c
|x1 − x2 |
1
+
|x2 − x1 |
ds(y)
|z1 − y|2 |x1 − y|3−α
α
∂D\Γz1 ,r
β
Z
≤ c|x1 − x2 |
α−β
|x2 − x1 |
|x1 − x2 |1−β
1
+
ds(y)
|z1 − y|2
|z1 − y|3−α
∂D\Γz1 ,r
β
Z
≤ c|x1 − x2 |
1
ds(y) ≤ c ĉα−β |x1 − x2 |β
2−(α−β)
|z1 − y|
∂D\Γz1 ,r
with a constant c > 0 and the constant ĉα−β from Lemma 3.2.
proves the Hölder-continuity of w. The proof of the estimate
x ∈ Hρ0 is simpler and left to the reader.
Next we consider the double layer surface potential
Z
∂Φk
(x, y) ds(y) ,
v(x) = D̃ϕ(x) =
ϕ(y)
∂ν(y)
∂D
This, together with (3.18)
w(x) ≤ vkϕkC 0,β (∂D) for
2
x ∈ R3 \ ∂D ,
(3.19)
for Hölder-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 Hölder-continuous density
ϕ ∈ C 0,α (∂D) can be continuously extended from D to D and from R3 \ D to R3 \ D with
limiting values
Z
1
∂Φk
lim v(x) = − ϕ(x0 ) +
ϕ(y)
(x0 , y) ds(y) , x0 ∈ ∂D ,
(3.20a)
x→x0
2
∂ν(y)
∂D
x∈D
Z
1
∂Φk
lim v(x) = + ϕ(x0 ) +
ϕ(y)
(x0 , y) ds(y) , x0 ∈ ∂D .
(3.20b)
x→x0
2
∂ν(y)
∂D
x∈D
/
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
109
v is Hölder-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
exp(ik|x0 − y|)
ν(y) · (y − x0 )
∂Φk
1
c
ik −
≤
.
∂ν(y) (x0 , y) = 4π|x0 − y|
|x0 − y|
|y − x0 |
|x0 − y|
by Lemma 6.10. Furthermore, v has a decomposition into v = v0 + v1 where
Z
∂Φ0
v0 (x) =
ϕ(y)
(x, y) ds(y) , x ∈ R3 \ ∂D ,
∂ν(y)
∂D
Z
∂(Φk − Φ0 )
(x, y) ds(y) , x ∈ R3 \ ∂D .
v1 (x) =
ϕ(y)
∂ν(y)
∂D
k −Φ0 )
It is easily seen that the kernel K(x, y) = ϕ(y) ∂(Φ∂ν(y)
(x, y) of the integral in the definition
3
3
of v1 is continuous
in R × R and continuously
differentiable with respect to x and bounded
3
on the set (x, y) ∈ R × ∂D : x 6= y . From this it follows that v1 is Hölder-continuous in
all of R3 .
We continue with the analysis of v0 and note that we have again to prove estimates of the
form (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 the
form
Z
Z
∂Φ0
∂Φ0
v0 (x) =
ϕ(y) − ϕ(z)
(x, y) ds(y) + ϕ(z)
(x, y) ds(y) .
∂ν(y)
∂D
∂D ∂ν(y)
|
{z
}
= ṽ0 (x)
The function ṽ0 is Hölder continuous in Uδ/4 with exponent β < α by Lemma 3.13 (take
a(y) = ν(y)). This proves the estimate (a) and (b) of Remark 3.11 for ṽ0 .
Now we consider the decomposition
Z
v0 (x) = ṽ0 (x) + ϕ(z)
∂D
∂Φ0
(x, y) ds(y) .
∂ν(y)
By Green’s representation (Theorem 3.3) for k = 0 and u = 1 we observe that

Z
 −1 , x ∈ D ,
∂Φ0
−1/2 , x ∈ ∂D ,
(x, y) ds(y) =

∂D ∂ν(y)
0,
x∈
/ D.
Therefore,
lim
v0 (x) = ṽ0 (x0 ) − ϕ(x0 ) = v0 (x0 ) +
lim
v0 (x) = ṽ0 (x0 ) = v0 (x0 ) +
x→x0 , x∈D
x→x0 , x∈D
/
1
ϕ(x0 ) .
2
1
1
ϕ(x0 ) − ϕ(x0 ) = − ϕ(x0 ) ,
2
2
110
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
2
This ends the proof.
The next step is an investigation of the derivative of the single layer potential (3.12); that
is,
Z
v(x) =
ϕ(y) Φ(x, y) ds(y) , x ∈ R3 .
∂D
First we show the following auxiliary result.
Lemma 3.15 (a) There exists c > 0 with
Z
∂D\B(x,τ )
∇x Φ(x, y) ds(y) ≤ c for all x ∈ R3 , τ > 0 ,
Z
(b)
lim
Z
τ →0
∂D\B(x,τ )
∇x Φ(x, y) ds(y) =
Z
H(y) Φ(x, y) ds(y) −
∂D
ν(y)
∂Φ
(x, y) ds(y)
∂ν(y)
∂D
>
for x ∈ R3 where H(y) = Div ê1t (y), Div ê2t (y), Div ê3t (y) ∈ R3 and êjt (y) = ν(y) × êj ×
ν(y) , j = 1, 2, 3, the tangential components of the unit vectors êj .
Proof: For any x ∈ R3 we have
Z
∇x Φ(x, y) ds(y) = −
∂D\B(x,τ )
Z
∇y Φ(x, y) ds(y)
∂D\B(x,τ )
Z
= −
Z
Grad y Φ(x, y) ds(y) −
∂D\B(x,τ )
∂Φ
(x, y) ν(y) ds(y)
∂ν(y)
∂D\B(x,τ )
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
Z
Z
a · ∇x Φ(x, y) ds(y) =
∂D\B(x,τ )
Div at (y) Φ(x, y) ds(y)
∂D\B(x,τ )
Z
at (y) · (τ (y) × ν(y)) Φ(x, y) ds(y)
+
∂Γ(x,τ )
Z
+
∂Φ
(x, y) a · ν(y) ds(y) .
∂ν(y)
∂D\B(x,τ )
The first and third integrals converge uniformly with respect to x ∈ ∂D when τ tends to
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
111
zero because the integrands are weakly singular. For the second integral we note that
Z
Z
1 at (y) · ν0 (y) Φ(x, y) ds(y) =
at (y) · (τ (y) × ν(y)) ds(y)
4πτ ∂Γ(x,τ )
∂Γ(x,τ )
Z
1 Div at (y) ds(y)
=
4πτ Γ(x,τ )
and this tends to zero uniformly with respect to x ∈ ∂D when τ tends to zero. The conclusion
follows if we take for a the unit coordinate vectors ê(j) .
2
As an abriviation we write in the following
v(x0 )− = x→x
lim v(x) ,
0
x∈D
v(x0 )+ = x→x
lim v(x) and
0
x∈D
/
∂v
(x0 ) = x→x
lim ν(x0 ) · ∇v(x) and
0
∂ν
−
x∈D
∂v
(x0 ) = x→x
lim ν(x0 ) · ∇v(x) ,
0
∂ν
+
x∈D
/
and obtain the following relations for the traces of the derivatives of the single layer surface
potential.
Theorem 3.16 The derivative of the single layer potential v from (3.12) with Höldercontinuous density ϕ ∈ C 0,α (∂D) can be continuously extended from D to D and from
R3 \ D to R3 \ D. The tangential component is continuous, i.e. Grad v|− = Grad v|+ , and
the limiting values of the normal derivatives are
Z
∂v 1
∂Φ
(x) = ∓ ϕ(x) +
ϕ(y)
(x, y) ds(y) , x ∈ ∂D .
(3.21)
∂ν
2
∂ν(x)
∂D
±
Proof: We note that the integral exists (see proof of Theorem 3.14). First we consider the
density 1, i.e. we set
Z
v1 (x) =
Φ(x, y) ds(y) , x ∈ R3 .
∂D
By the previous lemma we have that
Z
Z
∂Φ
∇v1 (x) = −
ν(y)
(x, y) ds(y) +
H(y) Φ(x, y) ds(y) ,
∂ν(y)
∂D
∂D
>
where again H(y) = Div ê1t (y), Div ê2t (y), Div ê3t (y) ∈ R3 .
x∈
/ ∂D ,
(3.22)
The right–hand side is the sum of a double and a single layer potential. By Theorems 3.12
and 3.14 it has a continuous extension to the boundary from the inside and the outside with
limiting values
Z
Z
1
∂Φ
∇v1 (x)|± = ∓ ν(x) −
ν(y)
(x, y) ds(y) +
H(y) Φ(x, y) ds(y)
2
∂ν(y)
∂D
∂D
Z
1
= ∓ ν(x) +
∇x Φ(x, y) ds(y)
2
∂D
112
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
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 and
the normal derivative jumps, and we have
Z
∂v1 1
∂Φ
(x) = ∓ +
(x, y) ds(y) , x ∈ ∂D .
∂ν
2
∂D ∂ν(x)
±
Now we consider v and have for x = z + tν(z) ∈ Hρ0 \ ∂D (that is, t 6= 0)
Z
∇v(x) =
Z
ϕ(y) ∇x Φ(x, y) ds(y) =
∂D
∇x Φ(x, y) ϕ(y) − ϕ(z) ds(y) + ϕ(z) ∇v1 (x) .
| ∂D
{z
}
= ṽ(x)
Application of Lemma 3.13 yields that ṽ is Hölder continuous in all of Hρ0 with limiting
value
Z
∇x Φ(x, y) ϕ(y) − ϕ(x) ds(y) for x ∈ ∂D .
ṽ(x) =
∂D
For further use we formulate the result derived so far.
Z
1
∇v(x) ± =
∇x Φ(x, y) ϕ(y) − ϕ(x) ds(y) + ϕ(x) ∓ ν(x)
2
∂D
Z
Z
∂Φ
−
ν(y)
(x, y) ds(y) +
H(y) Φ(x, y) ds(y) .
∂ν(y)
∂D
∂D
(3.23)
This proves that the gradient has continuous extensions from the inside and outside of D
and on ∂D we have
Z
Z
∂v 1
∂Φ
∂Φ
(x) = ∓ ϕ(x) + ϕ(x)
(x, y) ds(y) +
(x, y) ϕ(y) − ϕ(x) ds(y)
∂ν
2
∂D ∂ν(x)
∂D ∂ν(x)
±
Z
1
∂Φ
= ∓ ϕ(x) +
ϕ(y)
(x, y) ds(y) ,
2
∂ν(x)
∂D
Z
Grad v(x) ± =
Grad x Φ(x, y) ϕ(y) − ϕ(z) ds(y) + ϕ(z) Grad v1 (x)
(3.24)
∂D
for x ∈ ∂D.
3.1.3
2
Boundary Integral Operators
It is the aim of this subsection to investigate the mapping properties of the traces of the
single and double layer potentials on the boundary ∂D. We start with a general theorem on
boundary integral operators with singular kernels.
Theorem 3.17 Let Λ = (x, y) ∈ ∂D × ∂D : x 6= y and K ∈ C(Λ).
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
(a) Let there exist c > 0 and α ∈ (0, 1) such that
c
K(x, y) ≤
|x − y|2−α
for all (x, y) ∈ Λ ,
K(x1 , y) − K(x2 , y) ≤ c |x1 − x2 |
|x1 − y|3−α
for all (x1 , y), (x2 , y) ∈ Λ
113
(3.25a)
(3.25b)
with |x1 − y| ≥ 3|x1 − x2 | .
Then the operator K1 : C(∂D) → C 0,α (∂D), defined by
Z
(K1 ϕ)(x) =
K(x, y) ϕ(y) ds(y) ,
x ∈ ∂D ,
∂D
is well defined and bounded.
(b) Let there exist c > 0 such that:
K(x, y) ≤
c
|x − y|2
K(x1 , y) − K(x2 , y) ≤ c |x1 − x2 |
|x1 − y|3
Z
∂D\B(x,r)
for all (x, y) ∈ Λ ,
(3.25c)
for all (x1 , y), (x2 , y) ∈ Λ (3.25d)
with |x1 − y| ≥ 3|x1 − x2 | ,
K(x, y) ds(y) ≤ c for all x ∈ ∂D and r > 0 .
Then the operator K2 : C 0,α (∂D) → C 0,α (∂D), defined by
Z
K(x, y) ϕ(y) − ϕ(x) ds(y) ,
(K2 ϕ)(x) =
(3.25e)
x ∈ ∂D ,
∂D
is well defined and bounded.
Proof: (a) We follow the idea of the proof of Theorem 3.12 and write
Z
(K1 ϕ)(x1 ) − (K1 ϕ(x2 ) ≤ kϕk∞
K(x1 , y) − K(x2 , y) ds(y) .
∂D
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
Z
Z
Z
ds(y)
ds(y)
K(x1 , y) − K(x2 , y) ds(y) ≤ c
+ c
2−α
|x1 − y|
|x2 − y|2−α
Γx1 ,r
Γx1 ,r
Γx2 ,2r
≤ c0 rα = (c0 3α ) |x1 − x2 |α
because Γx1 ,r ⊆ Γx2 ,2r . Here we used formula (3.3d) of Lemma 3.2.
114
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
For the integral over ∂D \ Γx1 ,r we note that |y − x1 | ≥ r = 3|x1 − x2 | for y ∈ ∂D \ Γx1 ,r ,
thus
Z
Z
ds(y)
K(x1 , y) − K(x2 , y) ds(y) ≤ c |x1 − x2 |
3−α
∂D\Γx1 ,r
∂D\Γx1 ,r |x1 − y|
≤ c0 |x1 − x2 | rα−1 = c0 3α−1 |x1 − x2 |α
where we used estimate (3.3e). The proof of (K1 ϕ)(x) ≤ ckϕk∞ 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
Z
(K2 ϕ)(x1 ) − (K2 ϕ(x2 ) ≤
K(x1 , y) ϕ(y) − ϕ(x1 ) − K(x2 , y) ϕ(y) − ϕ(x2 ) ds(y)
Γx1 ,r
Z
K(x1 , y) ds(y)
+ ϕ(x2 ) − ϕ(x1 ) ∂D\Γx1 ,r
Z
ϕ(y) − ϕ(x2 ) K(x1 , y) − K(x2 , y) ds(y)
+
∂D\Γx1 ,r
"Z
≤ ckϕkC 0,α (∂D)
Γx1 ,r
ds(y)
+
|y − x1 |2−α
Z
Γx2 ,2r
ds(y)
|y − x2 |2−α
+ ckϕkC 0,α (∂D) |x1 − x2 |α
Z
|x1 − x2 |
|y − x2 |α
+ ckϕkC 0,α (∂D)
ds(y)
|y − x1 |3
∂D\Γx1 ,r

Z
 α
α
≤ ckϕkC 0,α (∂D) r + |x1 − x2 | + |x1 − x2 |
#

ds(y) 

|y − x1 |3−α
∂D\Γx1 ,r
because |y − x2 | ≤ |y − x1 | + |x1 − x2 | = |y − x1 | + r/3 ≤ 2|y − x1 |. The last integral has
been estimated by rα−1 , see (3.3e). This proves that
(K2 ϕ)(x1 ) − (K2 ϕ(x2 ) ≤ ckϕkC 0,α (∂D) |x1 − x2 |α .
The proof of (K2 ϕ)(x) ≤ ckϕkC 0,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 Hölderspaces.
This follows from the previous theorem and the following compactness result.
Lemma 3.18 The imbedding C 0,α (∂D) → C(∂D) is compact for every α ∈ (0, 1).
Proof: We have to prove that the unit ball B = ϕ ∈ C 0,α (∂D) : kϕkC 0,α (∂D) ≤ 1 is
relatively compact, i.e. its closure is compact, in C(∂D). This follows directly by the theorem
of Arcela-Ascoli (see, e.g., [9], Appendix C.7). Indeed, B is equi-continuous because
ϕ(x1 ) − ϕ(x2 ) ≤ kϕkC 0,α (∂D) |x1 − x2 |α ≤ |x1 − x2 |α
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
115
2
for all x1 , x2 ∈ ∂D. Furthermore, B is bounded.
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 from
C 0,α (∂D) into itself for every α ∈ (0, 1).
Proof: The operator K1 is bounded from C(∂D) into C 0,α (∂D) and the imbedding C 0,α (∂D)
into C(∂D) is compact, which implies compactness of K1 : C 0,α (∂D) → C 0,α (∂D).
2
We apply this result to the boundary integral operators which appear in the traces of the
single and double layer potentials of Theorems 3.12, 3.14, and 3.16.
Theorem 3.20 The operators S, D, D0 : C 0,α (∂D) → C 0,α (∂D), defined by
Z
ϕ(y) Φk (x, y) ds(y) , x ∈ ∂D ,
(Sϕ)(x) =
(3.26a)
∂D
Z
(Dϕ)(x) =
ϕ(y)
∂Φk
(x, y) ds(y) ,
∂ν(y)
x ∈ ∂D ,
(3.26b)
ϕ(y)
∂Φk
(x, y) ds(y) ,
∂ν(x)
x ∈ ∂D ,
(3.26c)
∂D
0
Z
(D ϕ)(x) =
∂D
are well defined and compact. Additionally, the operator S is bounded from C 0,α (∂D) into
C 1,α (∂D). Here, C 1,α (∂D) = {u ∈ C 0,α (∂D) : Grad u ∈ C 0,α (∂D, C3 )} equipped with its
canonical 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 6.10
that
1
,
4π|x − y|
∂
(y − x) · ν(y)
1
1
∂ν(y) Φ(x, y) = 4π|x − y| ik − |x − y| |x − y|
c
1 ik −
≤
|x − y|
4π|x − y|
|x − y| Φ(x, y) =
c
c(kd + 1)
k|x − y| + 1 ≤
4π|x − y|
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 ∈ ∂D
such that |x1 −y| ≥ 3|x1 −x2 |. Then, for any t ∈ [0, 1], we conclude that |x1 +t(x2 −x1 )−y| ≥
≤
116
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
|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 | sup ∇x Φ x1 + t(x2 − x1 ), y 0≤t≤1
9 |x1 − x2 |
|x1 − x2 |
≤ c
.
2
4 |x1 − y|2
0≤t≤1 |x1 + t(x2 − x1 ) − y|
≤ c sup
To show the corresponding estimate for the normal derivative of the fundamental solution
we can restrict ourselves to the case k = 0. Indeed, as in the proof of Theorem 3.14 we
observe that ∂(Φk − Φ0 )/∂ν is continuous.
Furthermore, we assume again x1 , x2 , y ∈ ∂D such that |x1 − y| ≥ 3|x1 − x2 |. Then
∂
∂
Φ
(x
,
y)
−
Φ
(x
,
y)
0
1
0
2
∂ν(y)
∂ν(y)
≤
1
4π|x1 − y|3
ν(y) · (y − x1 ) − ν(y) · (y − x2 )
|
{z
}
=ν(y)·(x2 −x1 )
1
+
4π
≤
1
1
ν(y) · (y − x2 )
−
|x1 − y|3 |x2 − y|3 1
(ν(y) − ν(x1 )) · (x2 − x1 ) + ν(x1 ) · (x2 − x1 )
3
4π|x1 − y|
1 1
1
ν(y) · (y − x2 )
+
−
4π |x1 − y|3 |x2 − y|3 Now we use the estimates (a) and (b) of Lemma 6.10 for the first term and the mean value
theorem for the second term. By |x1 + t(x2 − x1 ) − y| ≥ 2|x1 − y|/3 we have
∂
|y − x1 ||x2 − x1 | + |x1 − x2 |2
∂
|y − x2 |2 |x1 − x2 |
Φ
(x
,
y)
−
Φ
(x
,
y)
≤
c
+
c
.
0 2
∂ν(y) 0 1
∂ν(y)
|x1 − y|3
|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 C 0,α (∂D) into C 0,α (∂D, C3 ). But
this 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, uniqueness
can be seen directly from an application of Greens formulas and the radiation condition. But
in scattering theory we are interested in the case of a real and positive wave number k. The
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
117
following lemma is fundamental for proving uniqueness and tells us that a solution of the
Helmholtz equation ∆u + k 2 u = 0 for k > 0 cannot decay faster than 1/|x| as x tends
to infinity. We will give two proofs of this result. The first – and shorter – one uses the
expansion arguments from the previous chapter. In particular, properties of the spherical
Bessel- and Hankel functions are used. The second proof which goes back to the original work
by Rellich (see [22]) avoids the use of these special functions but is far more technical and
also 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 ∈ C 2 (R3 \ B[0, R0 ]) be a solution of the
Helmholtz equation ∆u + k 2 u = 0 for |x| > R0 and wave number k ∈ R>0 such that
Z
|u|2 ds = 0 .
lim
R→∞
|x|=R
Then u vanishes for |x| > R0 .
Proof: The general solution of the Helmholtz equation in the exterior of B(0, R0 ) is given
by (2.29); that is,
∞ X
n
X
m (1)
m
u(rx̂) =
an hn (kr) + bm
j
(kr)
Yn (x̂) ,
n
n
x̂ ∈ S 2 , r > R ,
n=0 m=−n
m
m
:
|m|
≤
n,
n
∈
N
∈
C.
The
spherical
harmonics
Y
form an orthogonal
,
b
for some am
0
n
n
n
system. Therefore, Parseval’s theorem yields
Z
∞ X
n
X
m (1)
2
m
u(rx̂)2 ds(x̂) ,
an hn (kr) + bn jn (kr) =
n=0 m=−n
S2
2
R and from the assumption on u we note that r2 S 2 u(rx̂) ds(x̂) tends to zero as r tends to
infinity. Especially, for every fixed n ∈ N0 and m with |m| ≤ n we conclude that
2
(1)
m
−→ 0
r 2 am
n hn (kr) + bn jn (kr)
m
m
m
m
as r tends to infinity. Defining cm
n = an +bn we can write it as (kr) i an yn (kr)+(kr) cn jn (kr) →
0 for r → ∞. Now we use the asymptotic behaviour of jn (kr) and yn (kr) as r tends to infinity. By Theorem 2.30 we conclude that
ikr
ikr
i am
(−i)n+1 + cm
(−i)n+1 −→ 0 , r → ∞ .
n Im e
n Re e
The term (−i)n+1 can take the values ±1 and ±i. Therefore, depending on n, we have that
m
m
m
i am
n sin(kr) + cn cos(kr) −→ 0 or i an cos(kr) − cn sin(kr) −→ 0 .
m
In any case, am
n and cn have to vanish by taking particular sequences rj → ∞. This shows
m
that also bn = 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 but
needs, however, a stronger assumption on u.
118
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Lemma 3.22 (Rellich’s Lemma, second form) Let u ∈ C 2 (R3 \ B[0, R0 ]) be a solution of
the Helmholtz equation ∆u + k 2 u = 0 for |x| > R0 with wave number k ∈ R>0 such that
Z
|u| ds = 0 and
lim
r→∞
Z
2
|x|=r
lim
r→∞
|x|=r
2
x
ds = 0 .
·
∇u(x)
|x|
Then u vanishes for |x| > R0 .
Proof: The proof, which is taken from the monograph [14] (Section VII.3) is lengthy, and
we will structure it. Without loss of generality we assume that u is real valued because we
can consider real and imaginary parts separately.
1st step: Transforming the integral onto the unit sphere S 2 = {x ∈ R3 : |x| = 1} we conclude
that
Z
u(rx̂)|2 r2 ds(x̂) =
Z
2
∂u
(rx̂) r2 ds(x̂)
|x̂|=1 ∂r
Z
u(x)2 ds(x) and
|x|=r
|x̂|=1
(3.27)
tend to zero as r tends to infinity. We transform the partial differential equation into an
ordinary differential equation (not quite!) for the function v(r, x̂) = r u(r, x̂) with respect to
r. We write v(r) and v 0 (r) and v 00 (r) for v(r, ·) and ∂v(r, ·)/∂r and ∂ 2 v(r, ·)/∂r2 , respectively.
Then (3.27) yields that kv(r)kL2 (S 2 ) → 0 and kv 0 (r)kL2 (S 2 ) → 0 as r → ∞. The latter follows
∂
from ∂r
ru(r, ·, ·) = 1r ru(r, ·) + r ∂u
(r, ·) and the triangle inequality.
∂r
2
∂
1
2 ∂u
2 ∂u
and
We observe that u = r v, thus r ∂r = −v + r ∂v
r
= r ∂r2v , thus
∂r
∂r
∂r
1 ∂
1
2 ∂u
0 = 2
r
(r, θ, φ) + 2 ∆S 2 u(r, θ, φ) + k 2 u(r, θ, φ)
r ∂r
∂r
r
2
1 ∂ v
1
2
=
(r, θ, φ) + k v(r, θ, φ) + 2 ∆S 2 v(r, θ, φ) ,
r ∂r2
r
i.e.
v 00 (r) + k 2 v(r) +
1
∆S 2 v(r) = 0 for r ≥ R0 ,
r2
(3.28)
where again ∆S 2 = Div Grad denotes the Laplace-Beltrami operator on the unit sphere;
that is, in polar coordinates x̂ = (sin θ cos φ, sin θ sin φ, cos θ)>
1 ∂
(∆S 2 w)(θ, φ) =
sin θ ∂θ
∂w
1 ∂ 2w
(θ, φ)
sin θ
(θ, φ) +
∂θ
sin2 θ ∂φ2
for any w ∈ C 2 (S 2 ). It is easily seen either by direct integration or by application of
Theorem 6.11 that ∆S 2 is selfadjoint and negative definite, i.e.
∆S 2 v, w
L2 (S 2 )
= v, ∆S 2 w
L2 (S 2 )
and
∆S 2 v, v
L2 (S 2 )
≤ 0 for all v, w ∈ C 2 (S 2 ) .
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
119
2nd step: We introduce the functions E, vm and F by
E(r) := kv 0 (r)k2L2 (S 2 ) + k 2 kv(r)k2L2 (S 2 ) +
1
∆S 2 v(r), v(r) L2 (S 2 ) ,
2
r
r ≥ R0 ,
vm (r) := rm v(r) ,
r ≥ R0 , m ∈ N ,
m(m + 1) 2c
0
2
2
F (r, m, c) := kvm (r)kL2 (S 2 ) + k +
kvm (r)k2L2 (S 2 )
−
r2
r
+
1
2 vm (r), vm (r)
∆
,
S
L2 (S 2 )
r2
for r ≥ R0 , m ∈ N, c ≥ 0. In the following we write k · k and (·, ·) for k · kL2 (S 2 ) and (·, ·)L2 (S 2 ) ,
respectively. We show:
(a) E satisfies E 0 (r) ≥ 0 for all r ≥ R0 .
(b) The functions vm solve the differential equation
2m 0
m(m + 1)
1
00
2
vm (r) −
vm (r) +
+ k vm (r) + 2 ∆S 2 vm (r) = 0 .
2
r
r
r
(3.29)
(c) For every c > 0 there exist r0 = r0 (c) ≥ R0 and m0 = m0 (c) ∈ N such that
∂ 2
r F (r, m, c) ≥ 0 for all r ≥ r0 , m ≥ m0 .
∂r
(d) Expressed in terms of v the function F has the forms
2
m(m + 1) 2c
m
2
2m 0
−
kv(r)k2
F (r, m, c) = r
v (r) + v(r) + k +
r
r2
r
1
+ 2 ∆S 2 v(r), v(r)
(3.30a)
r
2m
m(2m + 1) 2c
2m
0
2
= r
E(r) +
v(r), v (r) +
−
kv(r)k (. 3.30b)
r
r2
r
Proof of these statements:
(a) We just differentiate E and substitute
the second
derivative from (3.28). Note that
d
d
2
0
0
kv(r)k = 2(v, v ) and dr ∆S 2 v, v = 2 ∆S 2 v, v :
dr
1
2
E 0 (r) = 2 v 0 (r), v 00 (r) + 2k 2 v(r), v 0 (r) − 3 ∆S 2 v(r), v(r) + 2 ∆S 2 v(r), v 0 (r)
r
r
1
1
= 2 v 0 (r) , v 00 (r) + k 2 v(r) + 2 ∆S 2 v(r)
− 3 ∆S 2 v(r), v(r)
r
r
= −
1
∆S 2 v(r), v(r) ≥ 0 .
3
r
120
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
(b) We substitute v(r) = r−m vm (r) into (3.28) and obtain directly (3.29). We omit the
calculation.
00
(c) Again we differentiate r2 F (r, m, c) with respect to r, substitute the form of vm
from
(3.29) and obtain
∂ 2
0
00
0
(r), vm
(r) + 2(k 2 r − c)kvm (r)k2
r F (r, m, c) = 2rkvm
(r)k2 + 2r2 vm
∂r
m(m + 1) 2c
2
2
0
0
2 vm (r), v (r)
+ 2r k +
v
(r),
v
(r)
+
2
∆
−
m
S
m
m
r2
r
0
0
(r), vm (r) + 2(k 2 r − c)kvm (r)k2
(r)k2 − 4cr vm
= · · · = 2r(1 + 2m)kvm
"
2
√
c
0
= 2r 1 + 2m vm (r) − √
vm (r)
1 + 2m
c
c2
2
2
+ k − −
kvm (r)k .
r 1 + 2m
From this the assertion (c) follows if r0 and m0 are chosen such that the bracket (· · · ) is
positive.
(d) The first equation is easy to see by just inserting the form of vm . For the second form
one 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 exists
R1 ≥ R0 such that kv(r)k = 0 for all r ≥ R1 . Assume, on the contrary, that this is not the
case. Then, for every R ≥ R0 there exists r̂ ≥ R such that kv(r̂)k > 0.
We choose the constants ĉ > 0, r0 , m0 , r1 , m1 in the following order:
• Choose ĉ > 0 with k 2 −
2ĉ
R0
> 0.
• Choose r0 = r0 (ĉ) ≥ R0 and m0 = m0 (ĉ) ∈ N according to property (c) above, i.e.
∂
r2 F (r, m, ĉ) ≥ 0 for all r ≥ r0 and m ≥ m0 .
such that ∂r
• Choose r1 > r0 such that kv(r1 )k > 0.
• Choose m1 ≥ m0 such that m1 (m1 + 1)kv(r1 )k2 + ∆S 2 v(r1 ), v(r1 ) > 0.
Then, by (3.30a) and because k 2 − r2ĉ1 ≥ k 2 − R2ĉ0 > 0, it follows that F (r1 , m1 , ĉ) > 0 and thus,
by the monotonicity of r 7→ r2 F (r, m1 , ĉ) that also F (r, m1 , ĉ) > 0 for all r ≥ r1 . Therefore,
from (3.30b) we conclude that, for r ≥ r1 ,
0 < r
−2m1
m1 (2m1 + 1) 2ĉ
−
kv(r)k2
r2
r
m1 d
1 m1 (2m1 + 1)
2
kv(r)k +
− 2ĉ kv(r)k2 .
= E(r) +
r dr
r
r
2m1
F (r, m1 , ĉ) = E(r) +
v(r), v 0 (r) +
r
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
121
1 +1)
− 2ĉ < 0. Finally, choose r̂ ≥ r2 such that
Choose now r2 ≥ r1 such that m1 (2m
r2
d
2
kv(r̂)k ≤ 0. (This is possible because kv(r)k2 → 0 as r → ∞.) We finally have
dr
0 < p := r̂−2m1 F (r̂, m1 , ĉ) ≤ E(r̂) .
By the monotonicity of E we conclude that E(r) ≥ p for all r ≥ r̂. On the other hand, by
the definition of E(r) we have that E(r) ≤ kv 0 (r)k2 + k 2 kv(r)k2 and this tends to zero as r
tends to infinity. This is a contradiction. Therefore, there exists R1 ≥ R0 with v(r) = 0 for
all 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 ∈ C 2 (R3 \
D) ∩ C 1 (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 radiation
condition (3.2). From the radiation condition we conclude that
)
2
Z ( 2
Z
Z ∂u ∂u
∂u
2
2
u
ds
∂r + k |u| ds + 2k Im
∂r − iku ds =
∂r
|x|=R
|x|=R
|x|=R
tends to zero as R tends to infinity. Green’s theorem, applied in BR \ D to the function u
yields that
Z
Z
Z
Z
∂u
∂u
2
2
2
u
ds =
ds +
|∇u| − k |u| dx =
|∇u|2 − k 2 |u|2 dx
u
∂r
∂r
|x|=R
∂D
BR \D
BR \D
because the surface integral over ∂D vanishes by the boundary condition. The volume
integral is real valued. Therefore, its imaginary part vanishes and we conclude that
Z 2
∂u + k 2 |u|2 ds → 0
∂r |x|=R
as R tends to infinity. Rellich’s lemma (in the form Lemma 3.21 or Lemma 3.22) implies that
u vanishes outside of every ball which encloses ∂D. Finally, we note that u is an analytic
function in the exterior of D (see Corollary 3.4). Since the exterior of D is connected we
conclude that u vanishes in R3 \ D.
2
We turn to the question of existence of a solution and choose the integral equation method
for its treatment. We follow the approach of [6], Chapter 3, but prefer to work in the
space C 0,α (∂D) of Hölder continuous functions rather that in the space of merely continuous
functions. This avoids the necessity to introduce the class of continuous functions for which
the normal derivatives exist “in the uniform sense along the normal”.
122
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Let us recall the notion of the single layer surface potential of (3.4a), see also (3.12), and
make the ansatz for the scattered field in the form of a single layer potential; that is,
Z
s
u (x) =
ϕ(y) Φ(x, y) ds(y) , x ∈ R3 \ ∂D ,
(3.31)
∂D
where again Φ(x, y) = exp(ik|x − y|)/(4π|x − y|) denotes the fundamental solution of the
Helmholtz equation, and ϕ ∈ C 0,α (∂D) is some density to be determined. We remark already
here that we will face some difficulties with this ansatz but we begin with this for didactical
reasons. First we note that us solves the Helmholtz equation in the exterior of D and also
the radiation condition. This follows from the corresponding properties of the fundamental
solution Φ(·, y), uniformly with respect to y on the compact surface ∂D. Furthermore, by
Theorems 3.12 and 3.16 the function us and its derivatives can be extended continuously
from the exterior into R3 \ D with limiting values
Z
s
ϕ(y) Φ(x, y) ds(y) = (Sϕ)(x) , x ∈ ∂D ,
(3.32a)
u (x)+ =
∂D
∂us (x)
∂ν
+
1
= − ϕ(x) +
2
Z
ϕ(y)
∂D
∂
Φ(x, y) ds(y)
∂ν(x)
1
= − ϕ(x) + (D0 ϕ)(x) ,
2
x ∈ ∂D ,
(3.32b)
where we used the notations of the boundary integral operators from Theorem 3.20. Therefore, in order that u = uinc + us satisfies the boundary condition ∂u/∂ν = 0 on ∂D the
density ϕ has to satisfy the boundary integral equation
∂uinc
1
− ϕ + D0 ϕ = −
2
∂ν
in C 0,α (∂D) .
(3.33)
By Theorem 3.20 the operator D0 is compact. Therefore, we can apply the Riesz–Fredholm
theory. By Theerem 6.2, existence follows from uniqueness. To prove uniqueness we assume
that ϕ ∈ C 0,α (∂D) satisfies the homogeneous equation − 21 ϕ + D0 ϕ = 0. Define v to be the
single layer potential with density ϕ just as in (3.31), but for arbitrary x ∈
/ ∂D. Then, again
from the jump conditions of the normal derivative of the single layer, ∂v/∂ν + = − 12 ϕ+D0 ϕ =
0. Therefore, v is the solution of the exterior Neumann problem with vanishing boundary
data. 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 D
with 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 ∂D
that ϕ vanishes. However, this is not always the case. Indeed, there are non trivial solutions
in D if, and only if, k 2 is an eigenvalue of −∆ in D with respect to Dirichlet boundary
conditions (see Theorem 2.34).
This is the reason why it is necessary to modify the ansatz (3.31). There are several ways
how to do it, see the discussion in [5], Chapter 3 and 4. We choose a modification which we
have not found in the literature. It avoids the use of double layer potentials. On the other
hand, however, it results in a system of two equations which increases the numerical effort
3.1. A SCATTERING PROBLEM FOR THE HELMHOLTZ EQUATION
123
considerably. We assume for simplicity that D is connected although this is not necessary
as one observes from the following arguments.
We choose an open ball B = B(z, ρ) with boundary Γ such that Γ ⊆ D and such that
k 2 is not an eigenvalue of −∆ inside B with respect to Dirichlet boundary conditions. By
Theorem 2.34 from the previous chapter we observe that we have to choose the radius ρ of
B 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
Z
Z
s
u (x) = (S̃∂D ϕ)(x) + (S̃Γ ψ)(x) =
ϕ(y) Φ(x, y) ds(y) + ψ(y) Φ(x, y) ds(y) ,
∂D
0,α
0,α
where φ ∈ C (∂D) and ψ ∈ C
two boundary integral equations
(3.34)
(Γ) are two densities to be determined from the system of
∂
∂uinc
1
S̃Γ ψ = −
− ϕ + D0 ϕ +
2
∂ν
∂ν
x∈
/ D,
Γ
on ∂D ,
∂
1
+ ik S̃∂D ϕ − ψ + DΓ0 ψ + ik SΓ ψ = 0 on Γ .
∂ν
2
(3.35a)
(3.35b)
The operators SΓ and DΓ0 denote the boundary operators S and D0 , respectively, on the
boundary Γ instead of ∂D. These two equations can be written in matrix form as
inc
1 ϕ
ϕ
∂u /∂ν
D0
∂ S̃Γ /∂ν
−
+
= −
(∂/∂ν + ik)S̃∂D DΓ0 + ik SΓ
2 ψ
ψ
0
in C 0,α (∂D) × C 0,α (Γ). The operators D0 , DΓ0 + ikSΓ , ∂ S̃Γ /∂ν, and (∂/∂ν + ik)S̃∂D are all
compact. Therefore, we can apply the Riez-Fredholm theory to this system. By Theorem 6.2
existence is assured if the homogeneous system admits only the trivial solution ϕ = 0 and
ψ = 0. Therefore, let (ϕ, ψ) ∈ C 0,α (∂D) × C 0,α (Γ) be a solution of the homogeneous system
and 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 integral
equation
we conclude – just in the above case of only one single layer potential – that
∂v/∂ν + = − 21 ϕ + D0 ϕ + ∂ S̃Γ ψ/∂ν = 0. Again, v is a solution of the exterior Neumann
problem with vanishing boundary data. Therefore, by the uniqueness theorem, v vanishes
in the exterior of D. Furthermore, v is continuous in R3 and satisfies also the Helmholtz
equation in D \ B. From the jump conditions on the boundary Γ we conclude that
∂v ∂
1
+ ikv =
+ ik S̃∂D ϕ − ψ + DΓ0 ψ + ik SΓ ψ = 0 on Γ .
∂ν +
∂ν
2
Therefore, v = 0 on ∂D and ∂v/∂ν|+ + ikv = 0 on Γ. Application of Green’s first theorem
in D \ B yields
Z
Z
Z
Z
∂v
∂v 2
2
2
|∇v| − k |v| dx =
v
ds − v
ds = ik |v|2 ds .
∂ν
∂ν +
D\B
∂D
Γ
Γ
124
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Taking the imaginary part yields that v vanishes on Γ and therefore also ∂v/∂ν|+ = 0 on
Γ. Holmgren’s uniqueness Theorem 3.5 implies that v vanishes in all of D \ B. The jump
conditions for the normal derivatives on ∂D yield
∂v ∂v −
= ϕ on ∂D .
0 =
∂ν − ∂ν +
Therefore, v is a single layer potential on Γ with density ψ and vanishes on Γ. The wave
number k 2 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
∂v ∂v 0 =
−
= ψ on Γ .
∂ν − ∂ν +
Therefore, ϕ = 0 on ∂D and ψ = 0 on Γ and we have shown injectivity for the system of
integral equations.
S
If D consists of several components D = M
m=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 for
m = 1, . . . , M .
Application of the Riesz–Fredholm theory to (3.35a) and (3.35b) yields the desired existence
result.
Theorem 3.24 There exists a unique solution u ∈ C 2 (R3 \ D) ∩ C 1 (R3 \ D) of the scattering
problem (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 electromagnetic waves. Here the idea of the integral equation method as shown for the scalar Helmholtz
equation will be extended to the following scattering problem for the Maxwell system.
Given a solution (E inc , H inc ) of the Maxwell system
curl E inc − iωµ0 H inc = 0 ,
curl H inc + iωε0 E inc = 0 in some neighborhood of D ,
determine the total fields E, H ∈ C 1 (R3 \ D, C3 ) ∩ C(R3 \ D, C3 ) such that
curl E − iωµ0 H = 0 and
curl H + iωε0 E = 0 in R3 \ D ,
(3.36a)
E satisfies the boundary condition
ν × E = 0 on ∂D ,
(3.36b)
and the radiating parts E s = E −E inc and H s = H −H inc satisfy the Silver–Müller radiation
conditions
√
x
1
√
s
s
ε0 E (x) − µ0 H (x) ×
= O
,
(3.37a)
|x|
|x|2
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
125
and
√
x
√
µ0 H s (x) + ε0 E s (x) ×
= O
|x|
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 of
the scattering problem in vacuum.
√
Assumption: Let the wave number be given by k = ω ε0 µ0 > 0 with constants ε0 , µ0 > 0
and the obstacle D ⊆ R3 be bounded and C 2 −smooth such that the complement R3 \ D is
connected.
The case of a (homogeneous) conducting medium; that is, σ > 0, can be treated as well
without 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 the
following Exterior Boundary Value Problem:
Given a tangential field f ∈ C 0,α (∂D, C3 ), i.e. ν(x) · f (x) = 0 on ∂D , such that Div f ∈
C 0,α (∂D) determine radiating solutions E s , H s ∈ C 1 (R3 \ D, C3 ) ∩ C(R3 \ D, C3 ); that is,
E s , H s satisfy the radiating conditions (3.37a) and (3.37b), of the system
curl E s − iωµ0 H s = 0 and
curl H s + iωε0 E s = 0 in R3 \ D ,
(3.38a)
such that E s 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 radiation condition. We had made the assumption that the scalar fields x 7→ x · E(x) and
x 7→ x · H(x) – which are solutions of the scalar Helmholtz equation by Lemma 1.5 – satisfy
Sommerfeld’s radiation condition 3.2. Later (in Remark 3.31) we will discuss the equivalence
of both radiation conditions.
3.2.1
Representation Theorems
We have seen in the previous section (Theorem 3.3) that every sufficiently smooth function
u can be written as a sum of a volume potential with density ∆u + k 2 u, a single layer
potential with density ∂u/∂ν, and a double layer potential with potential u. Additionally
we know from Lemma 1.3 that each component of solutions E and H of the homogeneous
Maxwell system in a domain D satisfy the Helmholtz equation. Thus we obtain that these
components are analytic and can be represented by surface potentials. But more appropriate
for Maxwell’s equations is a representation in terms of vector potentials which we will discuss
next.
126
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Theorem 3.25 Let k ∈ C and E ∈ C 1 (D, C3 ) ∩ C(D, C3 ) such that curl E ∈ C(D, C3 ) and
div E ∈ C(D). Then we have for x ∈ D:
Z
Z
Z
2
E(x) = curl Φ(x, y) curl E(y) dy−∇ Φ(x, y) div E(y) dy−k
E(y) Φ(x, y) dy
D
D
Z
D
Z
ν(y) × E(y) Φ(x, y) ds(y) + ∇
− curl
∂D
ν(y) · E(y) Φ(x, y) ds(y) .
∂D
Furthermore, the right–hand side of this equation vanishes for x ∈
/ D and is equal to
for x ∈ ∂D.
1
2
E(x)
Proof: Fix z ∈ D, choose r > 0 such that B[z, r] ⊆ D, and set Dr = D \ B[z, r]. For
x ∈ B(z, r) we set
Z
Z
Z
2
E(y) Φ(x, y) dy
Ir (x) := curl Φ(x, y) curl E(y) dy−∇ Φ(x, y) div E(y) dy−k
Dr
Dr
Z
Dr
ν(y) × E(y) Φ(x, y) ds(y) + ∇
− curl
∂Dr
ν(y) · E(y) Φ(x, y) ds(y)
∂Dr
Z
Z
∇x Φ(x, y) × curl E(y) dy−
=
Z
Dr
∇x Φ(x, y) div E(y) dy−k
Dr
Z
Z
E(y) Φ(x, y) dy
Dr
ν(y) × E(y) Φ(x, y) ds(y) + ∇
− curl
2
∂Dr
Z
ν(y) · E(y) Φ(x, y) ds(y) ,
∂Dr
where we have applied the identity 6.6. We will show that Ir (x) vanishes. Indeed, we can
interchange differentiation and integration and write
Z
Z
Ir (x) = curl
Φ(x, y) curl E(y) dy −
ν(y) × E(y) Φ(x, y) ds(y)
Dr
∂Dr
Z
Z
−∇
Φ(x, y) div E(y) dy −
Dr
− k2
ν(y) · E(y) Φ(x, y) ds(y)
∂Dr
Z
E(y) Φ(x, y) dy
Dr
Z
= curl
curly E Φ(x, ·) − ∇y Φ(x, ·) × E dy −
Z
Dr
∂Dr
Z
−∇
divy E Φ(x, ·) − ∇y Φ(x, ·) · E dy −
Dr
− k2
Z
E Φ(x, ·) dy .
Dr
ν × E Φ(x, ·) ds
Z
∂Dr
ν · E Φ(x, ·) ds
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
Now we use the divergence theorem in the forms:
Z
Z
Z
Z
div F dx =
ν·F ds ,
curl F dx =
Dr
∂Dr
Dr
127
div(0, F3 , −F2 )
..
.
Dr
!
Z
ν×F ds .
dx =
∂Dr
(3.39)
Therefore,
Z
Z
∇y Φ(x, ·) × E dy + ∇
Ir (x) = − curl
Dr
Z
=
∇y Φ(x, ·) · E dy − k
2
Z
Dr
E Φ(x, ·) dy
Dr
∇x E · ∇y Φ(x, ·) − curlx ∇y Φ(x, ·) × E dy − k 2
Dr
Z
E Φ(x, ·) dy
Dr
Z
E ∆y Φ(x, ·) + k 2 Φ(x, ·) dy = 0 .
= −
Dr
Here we used the formula
curlx ∇y Φ(x, ·) × E = −E divx ∇y Φ(x, ·) + (E · ∇x )∇y Φ(x, ·)
= E ∆y Φ(x, ·) + ∇x E · ∇y Φ(x, ·) .
Therefore, Ir (x) = 0 for all x ∈ B(z, r), i.e.
Z
Z
Z
2
0 = ∇x Φ(x, y) × curl E(y) dy− ∇x Φ(x, y) div E(y) dy−k
E(y) Φ(x, y) dy
Dr
Dr
Z
− curl
Dr
ν(y) × E(y) Φ(x, y) ds(y) + ∇
∂D
Z
−
∇x Φ(x, y) × ν(y) × E(y) ds(y) +
|y−z|=r
Z
ν(y) · E(y) Φ(x, y) ds(y)
Z∂D
∇x Φ(x, y) ν(y) · E(y) ds(y) .
|y−z|=r
We set x = z and compute the last two surface integrals explicitely. We recall that
exp(ik|z − y|)
1
z−y
∇z Φ(z, y) =
ik −
4π|z − y|
|z − y| |z − y|
and thus for |z − y| = r:
Z
−
∇z Φ(z, y) × ν(y) × E(y) ds(y) +
|y−z|=r
exp(ikr)
=
4πr
∇z Φ(z, y) ν(y) · E(y) ds(y)
|y−z|=r
1
ik −
r
Z
|y−z|=r
exp(ikr)
=
4πr
Z
1
ik −
r
z−y
|z − y|
|
z−y
z−y
z−y
· E(y) −
×
× E(y) ds(y)
|z − y|
|z − y|
|z − y|
{z
}
= E(y)
Z
E(y) ds(y)
|y−z|=r
ikr
= −e
exp(ikr)
E(z) + ik
4πr
Z
|y−z|=r
exp(ikr)
E(y) ds(y) +
4πr2
Z
|y−z|=r
E(z) − E(y) ds(y) .
128
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
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 expression
vanishes if x ∈
/ D. The formula for x ∈ ∂D follows from the same arguments as in the proof
of Theorem 3.3.
2
This represention holds for any (smooth) vector field E. The relationship with Maxwell’s
equations becomes clear with the following identity.
Lemma 3.26 Let H ∈ C 1 (D, C3 ) ∩ C(D, C3 ) such that curl H ∈ C(D, C3 ), then for x ∈ D
holds
Z
Z
Z
curl Φ(x, y) H(y) dy − Φ(x, y) curl H(y) dy +
ν(y) × H(y) Φ(x, y) ds(y) = 0 .
D
D
∂D
R
Proof: The volume potential D Φ(x, y)H(y) dy is continuously differentiable by Lemma 3.7,
therefore
Z
Z
curl Φ(x, y) H(y) dy −
Φ(x, y) curl H(y) dy
D
D
Z
=
∇x Φ(x, y) × H(y) − Φ(x, y) curl H(y) dy
D
Z
= −
∇y Φ(x, y) × H(y) + Φ(x, y) curl H(y) dy
D
Z
= −
curly Φ(x, y) H(y) dy
D
where we used (6.7). Now we choose r > 0 such that B[x, r] ⊆ D and apply Green’s
formula (6.16a) in Dr = D \ B[x, r]. This yields
Z
Z
ν(y) × H(y) Φ(x, y) ds(y)
curly Φ(x, y) H(y) dy =
Dr
∂Dr
The boundary contribution
Z
ν(y) × H(y) Φ(x, y) ds(y)
|y−x|=r
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 Theorem
3.25 shows that if E and H the Maxwell system the volume integrals annihilate and we
obtain a representation only by surface integrals. This leads to the well known Stratton-Chu
formula.
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
129
Theorem 3.27 (Stratton-Chu formula)
√
Let k = ω ε0 µ0 > 0 and E, H ∈ C 1 (D, C3 ) ∩ C(D, C3 ) satisfy Maxwell’s equations
curl E − iωµ0 H = 0 in D ,
curl H + iωε0 E = 0 in D .
Then we have for x ∈ D:
Z
Z
ν(y) × E(y) Φ(x, y) ds(y) + ∇
E(x) = − curl
∂D
ν(y) · E(y) Φ(x, y) ds(y)
∂D
Z
ν(y) × H(y) Φ(x, y) ds(y)
− iωµ0
∂D
Z
= − curl
ν(y) × E(y) Φ(x, y) ds(y) +
∂D
Z
H(x) = − curl
ν(y) × H(y) Φ(x, y) ds(y) −
∂D
1
curl2
iωε0
Z
1
curl2
iωµ0
ν(y) × H(y) Φ(x, y) ds(y) ,
∂D
Z
ν(y) × E(y) Φ(x, y) ds(y) .
∂D
Proof: The second term in the representation of E in Theorem 3.25 vanishes because of
div E = 0. As mentioned above, subtraction of the identity from Lemma 3.26, multiplied by
iωµ0 , proves the first formula. For the second one we set Dr = D \ B[x, r] for r > 0 such
that B[x, r] ⊆ D and compute
Z
Z
1
ν(y) · E(y) Φ(x, y) ds(y) = −
ν(y) · curl H(y) Φ(x, y) ds(y)
iωε0 ∂Dr
∂Dr
Z
1
= −
iωε0
1
ν(y) · curl H(y) Φ(x, y) ds(y) +
iωε0
| ∂Dr
{z
}
Z
ν(y) · ∇y Φ(x, y) × H(y) ds(y)
∂Dr
= 0 by the divergence theorem
=
1
iωε0
Z
∇y Φ(x, y) · H(y) × ν(y) ds(y) .
∂Dr
R
Now we let r tend to zero. We observe that the integral |y−x|=r ∇y Φ(x, y)· H(y)×ν(y) ds(y)
vanishes because ∇y Φ(x, y) and ν(y) are parallel. Therefore,
Z
Z
1
ν(y) · E(y) Φ(x, y) ds(y) =
∇y Φ(x, y) · H(y) × ν(y) ds(y)
iωε0 ∂D
∂D
Z
1
=
div
Φ(x, y) ν(y) × H(y) ds(y) .
iωε0
∂D
Taking the gradient and using curl curl = ∇ div −∆ yields
Z
Z
1
∇
ν(y) · E(y) Φ(x, y) ds(y) =
curl curl
Φ(x, y) ν(y) × H(y) ds(y)
iωε0
∂D
∂D
2 Z
k
−
Φ(x, y) ν(y) × H(y) ds(y) .
iωε0 ∂D
130
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
This ends the proof for E. The representation for H(x) follows directly by H =
and curl curl curl = − curl ∆.
1
iωµ0
curl E
2
Similar to the scalar case we are also interested in a representation formula in the exterior
of a domain D. As an example of specific solutions to the homogeneous Maxwell equations
which 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
1
1
curl Emd (x) =
curl curl p Φ(x, y) ,
iωµ0
iωµ0
Emd (x) = curl p Φ(x, y) ,
Hmd (x) =
Hed (x) = curl p Φ(x, y) ,
Eed (x) = −
1
1
curl Hed (x) = −
curl curl p Φ(x, y) ,
iωε0
iωε0
for some y ∈ R3 and p ∈ C3 are called magnetic and electric dipols, respectively, at y with
polarization 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, respectively, satisfy the Silver-Müller radiation condition (3.37a), (3.37b); that is,
√
ε0 E(x) −
√
x
µ0 H(x) ×
= O
|x|
1
|x|2
1
|x|2
,
(3.40a)
,
(3.40b)
and
√
µ0 H(x) +
√
x
ε0 E(x) ×
= O
|x|
uniformly with respect to x/|x| ∈ S 2 and (p, y) in compact subsets of C3 × R3 .
(b) The fields x 7→ x · curl p Φ(x, y) and x 7→ x · curl curl p Φ(x, y) satisfies the scalar
Sommerfeld radiation condition (3.2) uniformly with respect to (p, y) in compact subsets
of C3 × R3 .
(c) The fields E(x) = curl x Φ(x, y) and H =
Maxwell system.
1
iωµ0
curl E are radiating solution of the
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
131
Proof: (a) Direct computation yields
ik −
1
|x − y|
x−y
1
= ik Φ(x, y)
×p + O
,
|x − y|
|x|2
Emd (x) = curl p Φ(x, y) = Φ(x, y)
x−y
×p
|x − y|
1
1
curl curl p Φ(x, y) =
(−∆ + ∇ div) p Φ(x, y)
iωµ0
iωµ0
x − y (x − y) · p
1
k2
Φ(x, y) p −
+ O
.
= ··· =
iωµ0
|x − y| |x − y|
|x|2
√
x−y
x
for |x| → ∞. With |x−y|
= |x|
+ O(1/|x|2 ) and k = ω µ0 ε0 the first assertion follows.
Analogously, the second assertion can be proven.
(b) We prove the assertion only for u(x) = x · curl p Φ(x, y) which we write as
x · [(x − y) × p]
1
Φ(x, y)
u(x) = x · ∇x Φ(x, y) × p = ik −
|x − y|
|x − y|
1
p · (y × x)
p · (y × x)
=
ik −
Φ(x, y)
= ik Φ(x, y)
+ O(|x|2 )
|x − y|
|x − y|
|x − y|
Hmd (x) =
and thus by differentiating this expression
x̂·∇u(x) = −k 2 Φ(x, y)
x̂ · (x − y) p · (y × x)
p · (y × x)
+ O(|x|2 ) = −k 2 Φ(x, y)
+ O(|x|2 )
|x − y|
|x − y|
|x − y|
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-Müller radiation
condition. We compute
E(x) = curl x Φ(x, y) = ∇x Φ(x, y) × x
1
x−y
1
x×y
= Φ(x, y) ik −
× x = Φ(x, y) ik −
|x − y| |x − y|
|x − y| |x − y|
1
= ik Φ(x, y) (x̂ × y) + O
and
|x|2
1
x×y
1
x×y
curl E(x) = ∇ Φ(x, y) ik −
×
+ Φ(x, y) ik −
curl
|x − y|
|x − y|
|x − y|
|x − y|
x×y
Φ(x, y)
x×y
1
= ik ∇x Φ(x, y) ×
− ∇x
×
+ O
|x − y|
|x − y| |x − y|
|x|2
x×y
1
+ O
= ik ∇x Φ(x, y) ×
|x − y|
|x|2
132
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
because
1
1
x×y
= ∇x
× (x × y) +
curlx (x × y) = O
curl
|x − y|
|x − y|
|x − y|
and
Φ(x, y)
∇x
= O
|x − y|
1
|x|2
1
|x|
.
.
Finally, we have
x×y
1
x−y
x×y
ik ∇x Φ(x, y) ×
= ik Φ(x, y) ik −
×
|x − y|
|x − y| |x − y| |x − y|
(x − y) · y)
(x − y) · x)
2
= −k Φ(x, y)
x−
y
|x − y|2
|x − y|2
and thus
2
x̂ × curl E(x) = k Φ(x, y) (x̂ × y) + O
√
Substituting the form of H and k = ω ε0 µ0 ends the proof.
1
|x|2
.
2
We note from the lemma that for the electric and magnetic dipols both, the Silver-Müller
radiation condition for the pair (E, H) and the Sommerfeld radiation condition for the scalar
functions e(x) = x·E(x) and h(x) = x·H(x) hold. The following representation theorem will
show 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 ε0 µ0 > 0 and E, H ∈ C 1 (R3 \D, C3 )∩C(R3 \D, C3 ) solutions of the homogeneous
Maxwell’s equations
curl E − iωµ0 H = 0 ,
curl H + iωε0 E = 0
in R3 \ D which satisfy also one of the Silver-Müller radiation conditions (3.40a) or (3.40b).
Then
Z
Z
1
curl
ν(y) × E(y) Φ(x, y) ds(y) −
curl curl
ν(y) × H(y) Φ(x, y) ds(y)
iωε0
∂D
∂D

0,
x ∈ D,

1
E(x)
,
x
∈ ∂D ,
=
 2
E(x) ,
x∈
/ D,
and
Z
ν(y) × H(y) Φ(x, y) ds(y) +
curl
∂D


0,
H(x) ,
=

H(x) ,
1
2
x ∈ D,
x ∈ ∂D ,
x∈
/ D.
1
curl curl
iωµ0
Z
∂D
ν(y) × E(y) Φ(x, y) ds(y)
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
133
Proof: Let us first
condition (3.40a). One applies Theorem 3.27 in
assume the radiation
the region DR = x ∈
/ D : |y| < R for large values of R. Then the assertion follows if one
can show that
Z
Z
1
curl curl
ν(y)×H(y) Φ(x, y) ds(y)
IR := curl
ν(y)×E(y) Φ(x, y) ds(y) −
iωε0
|y|=R
|y|=R
R
tends to zero as R → ∞. To do this we first prove that |y|=R |E|2 ds is bounded with respect
to R. The binomial theorem yields
Z
Z
Z
√
√
2
2
| ε0 E − µ0 H × ν| ds = ε0
|E| ds + µ0
|H × ν|2 ds
|y|=R
|y|=R
|y|=R
√
−2 ε0 µ0 Re
Z
E · (H × ν) ds .
|y|=R
We have by the divergence theorem
Z
Z
Z
E · (H × ν) ds +
E · (H × ν) ds =
|y|=R
∂D
Z
Z
H) · curl E − E · curl H dx
E · (H × ν) ds +
=
∂D
DR
Z
Z
iωµ0 |H|2 − iωε0 |E|2 dx .
E · (H × ν) ds +
=
∂D
DR
This term is purely imaginary, thus
Z
Z
√
√
2
| ε0 E − µ0 H × ν| ds = ε0
|y|=R
div(E × H) dx
DR
Z
2
|E| ds + µ0
|y|=R
√
− 2 ε0 µ0 Re
|H × ν|2 ds
|y|=R
Z
E · (H × ν) ds .
∂D
R
From this the boundedness of |y|=R |E|2 ds follows because the left–hand side tends to zero
by the radiation condition (3.40b).
Now we write IR in the form
Z
1
IR = curl
ν(y) × E(y) Φ(x, y) +
E(y) × ∇y Φ(x, y) ds(y)
ik
|y|=R
r
Z
ε0
1
curl
ν(y) × H(y) +
E(y) × ∇y Φ(x, y) ds(y) .
−
iωε0
µ0
|y|=R
Let us first consider the second term. The bracket (· · · ) tends to zero as 1/R2 by the radiation
condition (3.40a). Taking the curl of the integral results in second order differentiations of
Φ. Since Φ and all derivatives decay as 1/R the total integrand decays as 1/R3 . Therefore,
134
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
this second term tends to zero because the surface area is only 4πR2 .
For the first term we observe that
1
E(y) × ∇y Φ(x, y)
ik
y
1 y−x
1
= E(y) × − Φ(x, y) +
Φ(x, y) ik −
.
R
ik |y − x|
|y − x|
ν(y) × E(y) Φ(x, y) +
For fixed x and arbitrary y with |y| = R the bracket [· · · ] tends to zero of order 1/R2 . The
same holds true for all of the partial derivatives with respect to x. Therefore, the first term
can be estimated by the inequality of Cauchy-Schwartz
sZ
sZ
√ sZ
Z
2
c
c
c
4π
E(y) ds ≤
E(y) ds =
E(y)2 ds
12 ds
2
2
R |y|=R
R
R
|y|=R
|y|=R
|y|=R
and this tends also to zero.
This proves the representation of E by using the first radiation condition (3.40a). The
1
representation of H(x) follows again by computing H = iωµ
curl E. If the second radiation
0
condition (3.40b) is assumed one can argue as before and derive the representation of H
first.
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 radiation
conditions (3.40a) and (3.40b) implies the other one.
(b) The Silver-Müller radiation condition for solutions E, H of the Maxwell system is
equivalent to the Sommerfeld radiation condition (3.2) for every component of E and
H. This follows from the fact that the fundamental solution Φ and every derivative of
Φ satisfies the Sommerfeld radiation condition.
(c) The Silver-Müller radiation condition for solutions E, H of the Maxwell system is
equivalent 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–Müller radiation
condition implies a representation of the form of the previous theorem which implies
the scalar Sommerfeld radiation conditions for e(x) and h(x) by Lemma 3.29, and
the scalar Sommerfeld radiation conditions for e(x) and h(x) imply the representation
(2.56a) which satisfies the Silver–Müller radiation conditions again by Lemma 3.29.
(d) The asymptotic behaviour of Φ yields
1
E(x) = O
|x|
and
H(x) = O
for |x| → ∞ uniformly with respect to all directions x/|x|.
1
|x|
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
135
Sometimes it is convenient to eliminate one of the fields E or H from the Maxwell system
and work with only one of them. If we eliminate H then E solves the second order equation
curl E − k 2 E = 0
(3.41)
√
where again k = ω µ0 ε0 denotes the wave number. If, on the other hand, E satisfies (3.41)
1
E solve the Maxwell system. Indeed, the first Maxwell equation is
then E and H = iωµ
0
satisfied by the definition of H. Also the second Maxwell equation is satisfied because
curl H =
1 k2
1
curl2 E =
∇ div
E
=
E = −iωε0 E .
−
∆E
| {z }
|{z}
iωµ0
iωµ0
iωµ0
2
=0
=−k E
The Silver-Müller radiation condition (3.37a) or (3.37b) turn into
curl E × x̂ − ik E = O |x|−2 , |x| → ∞ ,
(3.42)
uniformly with respect to x̂ = 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 boundary integral equation method. Analogously to the scalar case we have to introduce vector
potentials. Motivated by the Stratton-Chu formulas we have to consider the curl and the
double-curl of the single layer potential
Z
v(x) =
a(y) Φ(x, y) ds(y) , x ∈ R3 ,
(3.43)
∂D
where a ∈ C 0,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 \ ∂D
and 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 Hölder continuous tangential
field a ∈ C 0,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
Z
1
ν(x) × curl v(x) ± = ± a(x) + ν(x) ×
curlx a(y)Φ(x, y) ds(y) , x ∈ ∂D . (3.44)
2
∂D
If, in addition, the surface divergence Div a (see Section 6.5) is continuous, then div v is
continuous in all of R3 with limiting values
Z
div v(x)± =
Φ(x, y) Div a(y) ds(y) , x ∈ ∂D ,
∂D
136
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
and also for the normal component of curl v it holds
ν · curl v + = ν · curl v −
on ∂D .
If, furthermore, Div a ∈ C 0,α (∂D) then curl curl v can be continuously extended from D to
D 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)±
Z 1
∂Φ
2
= ∓ Div a(x) +
(x, y) + k ν(x) · a(y) Φ(x, y) ds(y) , x ∈ ∂D .
Div a(y)
2
∂ν(x)
∂D
Proof: The components of curl v are combinations of partial derivatives of the single layer
potential. Therefore, by Theorem 3.16 the field curl v has continuous extensions to ∂D
from both sides. It remains to show the representation of the tangential components of
these extensions on the boundary. We recall the special neighborhoods Hρ of ∂D from
Subsection 6.3 and write x ∈ Hρ0 in the form x = z + tν(z), z ∈ ∂D, 0 < |t| < ρ0 . Then we
have
Z
ν(z) × curl v(x) =
ν(z) × ∇x Φ(x, y) × a(y) ds(y)
(3.46)
∂D
Z
ν(z) − ν(y) × ∇x Φ(x, y) × a(y) ds(y)
=
∂D
Z
+
a(y)
∂D
∂Φ
(x, y) ds(y) .
∂ν(y)
The first term is continuous in all of R3 by Lemma 3.13, the second in R3 \ ∂D because it is
a double layer potential. The limiting values are
ν(x) × curl v(x)
Z
±
which has the desired form.
=
ν(x) − ν(y) × ∇x Φ(x, y) × a(y) ds(y)
∂D
1
± a(x) +
2
Z
1
= ± a(x) +
2
Z
a(y)
∂D
∂D
∂Φ
(x, y) ds(y)
∂ν(y)
ν(x) × ∇x Φ(x, y) × a(y) ds(y)
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
137
Similar for the normal component we obtain
ν(z) · curl v(x)
Z
Z
=
ν(z) − ν(y) · ∇x Φ(x, y) × a(y) ds(y) +
∂D
Z
=
∂D
a(y) × (ν(z) − ν(y)) · ∇x Φ(x, y) ds(y) −
∂D
Z
=
=
Z
a(y) ν(y) × ∇y Φ(x, y) ds(y)
∂D
a(y) × (ν(z) − ν(y)) · ∇x Φ(x, y) ds(y) −
∂D
Z
ν(y) ∇x Φ(x, y) × a(y) ds(y)
Z
Grad y Φ(x, y) · a(y) × ν(y) ds(y)
∂D
a(y) × (ν(z) − ν(y)) · ∇x Φ(x, y) ds(y) −
∂D
Z
Φ(x, y) Div a(y) × ν(y) ds(y) ,
∂D
where the first integral is continuous at ∂D as above and the second is a single layer potential
and therefore also continuous. Thus taking the limit we obtain
ν · curl v + = ν · curl v − .
For the divergence we write
Z
Z
div v(x) =
a(y) · ∇x Φ(x, y) ds(y) = −
∂D
a(y) · ∇y Φ(x, y) ds(y)
∂D
Z
Z
a(y) · Grad y Φ(x, y) ds(y) =
= −
Φ(x, y) Div a(y) ds(y) .
∂D
∂D
This, again, is a single layer potential and thus continuous.
Finally, because curl curl = ∇ div −∆ and ∆x Φ(x, y) = −k 2 Φ(x, y), we conclude that
Z
Z
2
curl curl v(x) = ∇ div
a(y) Φ(x, y) ds(y) + k
a(y) Φ(x, y) ds(y)
∂D
Z
= ∇
∂D
Φ(x, y) Div a(y) ds(y) + k 2
∂D
Z
a(y) Φ(x, y) ds(y)
∂D
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 integral
operators. It is convenient to not only define the spaces Ct (∂D) and Ct0,α (∂D) of continuous
and Hölder continuous tangential fields, respectively, but also of Hölder continuous tangential
fields such that the surface divergence is also Hölder continuous. Therefore, we define:
Ct (∂D) = a ∈ C(∂D, C3 ) : a(y) · ν(y) = 0 on ∂D ,
Ct0,α (∂D) = Ct (∂D) ∩ C 0,α (∂D, C3 ) ,
0,α
CDiv
(∂D) = a ∈ Ct0,α (∂D) : Div a ∈ C 0,α (∂D) .
We equip Ct (∂D) and Ct0,α (∂D) with the ordinary norms of C(∂D, C3 ) and C 0,α (∂D, C3 ),
0,α
respectively, and CDiv
(∂D) with the norm kakC 0,α (∂D) = kakC 0,α (∂D) + k Div akC 0,α (∂D) . Then
Div
we can prove:
138
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Theorem 3.34
(a) The boundary operator M : Ct (∂D) → Ct0,α (∂D), defined by
Z
(Ma)(x) = ν(x) ×
curlx a(y) Φ(x, y) ds(y) , x ∈ ∂D ,
(3.47)
∂D
is well defined and bounded.
0,α
(b) M is well defined and compact from CDiv
(∂D) into itself.
0,α
0,α
(c) The boundary operator L : CDiv
(∂D) → CDiv
(∂D), defined by
Z
(La)(x) = ν(x) × curl curl
a(y) Φ(x, y) ds(y) ,
x ∈ ∂D ,
(3.48)
∂D
is well defined and bounded. Here, the right–hand side is the trace of curl2 v with v
from (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) = ∇x Φ(x, y) ν(x) − ν(y)
−
∂Φ
(x, y) I .
∂ν(x)
The component gj,` of the first term is given by
gj,` (x, y) =
∂Φ
(x, y) ν` (x) − ν` (y) ,
∂xj
and this satisfies certainly the first assumption of part (a) of Theorem 3.17 for α = 1. For
the second assumption we write
∂Φ
∂Φ
∂Φ
gj,` (x1 , y)−gj,` (x2 , y) = ν` (x1 )−ν` (x2 )
(x1 , y) + ν` (x2 )−ν` (y)
(x1 , y) −
(x2 , y)
∂xj
∂xj
∂xj
and thus by the same arguments as in the proof Lemma 3.13
gj,` (x1 , y) − gj,` (x2 , y) ≤ c |x1 − x2 | + c|x2 − y| |x1 − x2 | ≤ c0 |x1 − x2 | .
|x1 − y|2
|x1 − y|3
|x1 − y|2
This settles the first term of G. For the second term we observe that
∂Φ
∂Φ
(x, y) = −
(x, y) + ν(x) − ν(y) · ∇x Φ(x, y) .
∂ν(x)
∂ν(y)
The first term is just the kernel of the double layer operator treatedP
in the previous theorem.
For the second we can apply the first part again because it is just 3`=1 g`,` (x, y).
0,α
(b) We note that the space CDiv
(∂D) is a subspace of Ct0,α (∂D) with bounded imbedding
and, furthermore, the space Ct0,α (∂D) is compactly imbedded in Ct (∂D) by Lemma 3.18.
0,α
Therefore, M is compact from CDiv
(∂D) into Ct0,α (∂D). It remains to show that Div M is
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
0,α
compact from CDiv
(∂D) into C 0,α (∂D).
0,α
For a ∈ CDiv
(∂D) we define the potential v by
Z
v(x) =
a(y) Φ(x, y) ds(y) ,
139
x ∈ D.
∂D
Then Ma = ν × curl v|− + 21 a by Theorem 3.33. Furthermore, by Theorem 3.33 again we
conclude that for x ∈
/ ∂D
Z
curl curl v(x) = (∇ div −∆)
a(y) Φ(x, y) ds(y)
∂D
Z
= ∇
Div a(y) Φ(x, y) ds(y) + k
2
Z
∂D
a(y) Φ(x, y) ds(y)
∂D
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)|− +
Z
= −
∂D
1
Div a(x)
2
∂Φ
2
Div a(y)
(x, y) + k ν(x) · a(y) Φ(x, y) ds(y)
∂ν(x)
= −D0 (Div a) − k 2 ν · Sa .
0,α
The assertion follows because D0 ◦ Div and ν · S are both compact from CDiv
(∂D) into
0,α
C (∂D).
(c) Define
Z
x∈
/ ∂D .
a(y) Φ(x, y) ds(y) ,
w(x) = curl curl
∂D
Writing again curl2 = ∇ div −∆ yields for x = z + tν(z) ∈ Hρ \ ∂D:
Z
Z
2
w(x) = ∇ div
Φ(x, y) a(y) ds(y) + k
a(y) Φ(x, y) ds(y)
∂D
∂D
Z
∇x Φ(x, y) Div a(y) ds(y) + k
=
∂D
Z
2
Z
a(y) Φ(x, y) ds(y)
∂D
∇x Φ(x, y) Div a(y) − Div a(z) ds(y) + Div a(z)
=
∂D
+k
Z
∇x Φ(x, y) ds(y)
∂D
2
Z
a(y) Φ(x, y) ds(y)
∂D
We use Lemma 3.15 and arrive at
Z
Z
2
w(x) =
∇x Φ(x, y) Div a(y) − Div a(z) ds(y) + k
∂D
a(y) Φ(x, y) ds(y)
∂D
Z
+ Div a(z)
∂D
Z
H(y) Φ(x, y) ds(y) − Div a(z)
∂D
ν(y)
∂Φ
(x, y) ds(y) .
∂ν(y)
140
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Therefore, by Lemma 3.13, Theorems 3.12 and 3.14 the tangential component of w is continuous, thus La is given by
Z
(La)(x) = ν(x) × w(x) = ν(x) ×
∇x Φ(x, y) Div a(y) − Div a(x) ds(y)
∂D
Z
+ Div a(x) ν(x) ×
H(y) Φ(x, y) ds(y)
∂D
Z
∂Φ
(x, y) ds(y) + k 2 ν(x) ×
− Div a(x) ν(x) ×
ν(y)
∂ν(y)
∂D
Z
a(y) Φ(x, y) ds(y) .
∂D
0,α
The boundedness of the last three terms as operators from CDiv
(∂D) into Ct0,α (∂D) follow
from the boundedness of the single and double layer boundary operators S and D. For the
boundedness of the first term we apply part (b) of Theorem 3.17. The assumptions
c
ν(x) × ∇x Φ(x, y) ≤
for all x, y ∈ ∂D , x 6= y , and
|x − y|2
ν(x1 ) × ∇x Φ(x1 , y) − ν(x2 ) × ∇x Φ(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
Z
Z
ν(x) × ∇x Φ(x, y) ds(y) ≤ ∇x Φ(x, y) ds(y) ≤ c
(3.49)
∂D\B(x,r)
∂D\B(x,r)
for all x ∈ ∂D and r > 0 we refer to Lemma 3.15 below. This proves boundedness of
0,α
L from CDiv
(∂D) into Ct0,α (∂D). We consider now the surface divergence Div La. By
Corollary 6.20 and the form of w we conclude that Div La = Div(ν × w) = −ν · curl w. For
x = z + tν(z) ∈ Hρ \ ∂D we compute
Z
Z
3
2
∇x Φ(x, y) × a(y) ds(y)
a(y) Φ(x, y) ds(y) = k
curl w(x) = curl
∂D
∂D
= −k 2
Z
∇y Φ(x, y) × a(y) ds(y)
∂D
= k
2
Z
∇y Φ(x, y) × a(z) − a(y) ds(y) − k 2
Z
∇y Φ(x, y) ds(y) × a(z)
∂D
= k
2
∂D
Z
∇y Φ(x, y) × a(z) − a(y) ds(y) − k 2
Z
Grad y Φ(x, y) ds(y) × a(z)
∂D
∂D
−k
2
Z
ν(y)
∂D
= k
2
Z
∂Φ
(x, y) ds(y) × a(z)
∂ν(y)
∇y Φ(x, y) × a(z) − a(y) ds(y) + k 2
∂D
−k
2
Z
H(y) Φ(x, y) ds(y) × a(z)
∂D
Z
ν(y)
∂D
∂Φ
(x, y) ds(y) × a(z) .
∂ν(y)
The normal component is bounded by the same arguments as above.
2
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
3.2.3
141
Uniqueness and Existence
Since we now have collected the integral operators with their mapping properties
will be applied in case of Maxwell’s equations, we continue with the investigation
scattering problem. As a next step we prove that there exists at most one solution
exterior boundary value problem (3.38a), (3.38b), (3.40a), (3.40b) and therefore also
scattering problem (3.36a), (3.36b),(3.40a), (3.40b).
which
of the
of the
to the
0,α
Theorem 3.35 For every tangential field f ∈ CDiv
(∂D, C3 ) there exists at most one radiat3
3
3
s
s
1
3
ing solution E , H ∈ C (R \ D, C ) ∩ C(R \ D, C ) of the exterior boundary value problem
(3.38a), (3.38b), (3.40a), (3.40b).
Proof: Let Ejs , Hjs for j = 1, 2 be two solutions of the boundary value problem. Then the
difference E s = E1s − E2s and H s = H1s − H2s solve the exterior boundary value problem for
boundary data f = 0.
In the proof of the Stratton-Chu formula (Theorem 3.30) we have derived the following
formula:
Z
Z
Z
√
√
s
s
2
s 2
| ε0 E − µ0 H × ν| ds = ε0
|E | ds + µ0
|H s × ν|2 ds
|y|=R
|y|=R
√
− 2 ε0 µ0 Re
|y|=R
Z
E s · (H s × ν) ds .
∂D
R
R
The integral ∂D E s · (H s × ν) ds = ∂D H s · (ν × E s ) ds vanishes by the boundary condition.
Since the left–hand side tends to zero by the radiation condition we conclude that
R
|E s |2 ds tends to zero as R → ∞. Furthermore, by the Stratton-Chu formula (Theo|y|=R
rem 3.30) in the exterior of D we can represent E s (x) in the form
Z
Z
1
s
s
E (x) = curl
curl curl
ν(y)×E (y) Φ(x, y) ds(y) −
ν(y)×H s (y) Φ(x, y) ds(y)
iωε0
∂D
∂D
for x ∈
/ D. From these two facts we observe that every component u = Ejs satisfies the
R
Helmholtz equation ∆u + k 2 u = 0 in the exterior of D and limR→0 |x|=R |u|ds = 0. Furthermore, we recall that Φ(x, y) satisfies the Sommerfeld radiation condition (3.2) and therefore
also u = Ejs . The triangle inequality in the form |z| ≤ |z−w|+|w|, thus |z|2 ≤ 2|z−w|2 +2|w|2
yields
2
2
Z
Z
Z
x
x
2
u(x)2 ds(x) ,
·
∇u(x)
ds(x)
≤
2
·
∇u(x)
−
iku(x)
ds(x)
+
2k
|x|=R |x|
|x|=R |x|
|x|=R
and this tends to zero as R tends to zero. We are now in the position to apply Rellich’s
Lemma 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 but
rather serves as a preliminary result to motivate a more complicated approach.
142
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Theorem 3.36 Assume that w = 0 is the only solution of the following interior boundary
value problem:
curl curl w − k 2 w = 0 in D ,
ν × curl w = 0 on ∂D .
(3.50)
0,α
Then, for every f ∈ CDiv
(∂D), there exists a unique solution E s , H s ∈ C 1 (R3 \ D, C3 ) ∩
3
3
C(R \ D, C ) of the exterior boundary value problem (3.38a), (3.38b), (3.40a), (3.40b). In
particular, 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
Z
1
s
a(y) Φ(x, y) ds(y) , H s (x) =
curl E s (x) , x ∈
/ D,
(3.51)
E (x) = curl
iωµ
0
∂D
0,α
for a ∈ CDiv
(∂D) which is the unique solution of the boundary integral equation
Z
1
a(x) + ν(x) × curl
a(y) Φ(x, y) ds(y) = f (x) , x ∈ ∂D .
2
∂D
(3.52)
Proof: First we note that, by the jump condition of Theorem 3.33, E s and H s from (3.51)
solve the boundary value problem if a solves (3.52); that is with the operator M,
1
a + Ma = f .
2
Since M is compact we can apply the Riesz–Fredholm theory to this equation; that is,
0,α
existence is assured if uniqueness holds. Therefore, let a ∈ CDiv
(∂D) satisfy 12 a + Ma = 0
on ∂D. Define E s and H s as in (3.51) in all of R3 \ ∂D. Then E s , H s satisfy the Maxwell
system and ν × E s (x) = 1 a + Ma = 0 on ∂D. The uniqueness result of Theorem 3.35
+
2
yields E s = 0 in R3 \ D. Application
of Theorem 3.33 again yields that ν × curl E s is
continuous on ∂D, thus ν × curl E s − = 0 on ∂D. From our assumption we conclude that
E s vanishes also inside of D. Now we apply Theorem 3.33 a third time and have that
a = ν × E s − − ν × E s + = 0. Thus, the homogeneous integral equation admits only a = 0
as a solution and, therefore, there exists a unique solution of the inhomogeneous equation
0,α
for every right hand side f ∈ CDiv
(∂D).
2
The problem of this approach is similarly to the one for scalar case. In general, there exist
eigenvalues of the problem (3.50). To construct an example we apply Lemma 2.42. Thus
we know, if u solves
∆u + k 2 u = 0 in the unit ball B(0, 1)
thescalar Helmholtz equation
2
then v(x) = curl u(x)x solves
curl curl v − k v 2 = 0 in B(0, 1). Furthermore, we have
ν(x) × v(x) = x × ∇u(x) × x = Grad u(x) on S .
Example 3.37 We claim that the field
sin(kr)
w(r, θ, φ) = curl curl sin θ k cos(kr) −
r
r̂
satisfies curl curl w − k 2 w = 0 in B(0, 1) and ν(x) × curl w(x) = k 2 cos θ k cos k − sin k θ̂ on
S 2 = ∂B(0, 1).
3.2. A SCATTERING PROBLEM FOR THE MAXWELL SYSTEM
143
Indeed, we can write w(x) = curl curl u(x)x with
d sin(kr)
cos(kr) sin(kr)
−
= sin θ
.
u(r, θ, φ) = sin θ k
2
r
r
dr r
By direct evaluation of the Laplace operator in spherical coordinates we see that u satisfies
actually is analytic in
the Helmholtz equation ∆u + k 2 u = 0 in all of R3 , because sin(kr)
r
R. By the previous lemma we have that v(x) = curl u(x)x satisfies curl curl v − k 2 v =
0, and
curl curl w− k 2 w = 0. We compute curl w as curl w(x) =
thus also
w satisfies
2
curl ∇ div −∆ u(x)x = k curl u(x)x = k 2 v(x) and thus by the previous lemma ν(x) ×
curl w(x) = k 2 Grad u(x) = k 2 cos θ k cos k − sin k θ̂ on S 2 .
Therefore, if k > 0 is any zero of ψ(k) = k cos k − sin k then the corresponding field w
satisfies (3.50).
Thus again as in the scalar case the insufficent ansatz (3.51) has to be modified. We propose
a modification of the form
Z
s
a(y) Φ(x, y) ds(y)
(3.53a)
E (x) = curl
∂D
Z
+ η curl curl
ν(y) × (Ŝi2 a)(y) Φ(x, y) ds(y) ,
(3.53b)
∂D
Hs =
1
curl E s
iωµ0
(3.53c)
0,α
for some constant η ∈ C, some a ∈ CDiv
(∂D), and where the bounded operator Ŝi :
C 0,α (∂D, C3 ) → C 1,α (∂D, C3 ) is the single layer surface operator with the value k = i, considered componentwise; that is, Ŝi a = (Si a1 , Si a2 , Si a3 )> for a = (a1 , a2 , a3 )> ∈ C 0,α (∂D, C3 ).
We note that Ŝi is bounded from C 0,α (∂D, C3 ) into C 1,α (∂D, C3 ) by Theorem 3.20. There0,α
(∂D, C3 ) into itself. We need the
fore, the operator K : a 7→ ν × Ŝi2 a is compact from CDiv
following additional result for the single layer boundary operator Si for wavenumber k = i.
R
Lemma 3.38 The operator Si is selfadjoint with respect to hϕ, ψi∂D = ∂D ϕ ψ ds; that is
hSi ϕ, ψi∂D = hϕ, Si ψi∂D
for all ϕ, ψ ∈ C 0,α (∂D)
and one-to-one.
Proof: Let ϕ, ψ ∈ C 0,α (∂D) and define u = S̃i ϕ and v = S̃i ψ as the single layer potentials
with 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 decay
exponentially 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
Z
Z
∂v ∂v hSi ϕ, ψi∂D =
(Si ϕ) ψ ds =
u
−
ds
∂ν − ∂ν +
∂D
∂D
Z
Z
Z
∂v
=
(∇u · ∇v + u v) dx +
(∇u · ∇v + u v) dx −
u ds .
∂r
D
B(0,R)\D
|x|=R
144
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
The integral over the sphere {x : |x| = R} tends to zero and thus
Z
hSi ϕ, ψi∂D =
(∇u · ∇v + u v) dx .
R3
This term is symmetric with respect to u and v.
Furthermore, if Si ϕ = 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 Theorem 3.33 that the ansatz (3.53b), (3.53c) solves the exterior boundary value problem if
0,α
a ∈ CDiv
(∂D) solves the equation
1
a + Ma + η LKa = c on ∂D
2
(3.54)
where again Ka = ν × Ŝi2 a. Finally, we can prove the general existence theorem.
0,α
Theorem 3.39 For every f ∈ CDiv
(∂D), there exists a unique radiating solution E s , H s ∈
C 1 (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 solution
for every incident field. The solution has the form of (3.53b), (3.53c) for any η ∈ C \ R and
0,α
some a ∈ CDiv
(∂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 the
composition LK. Therefore, the Riesz–Fredholm theory is applicable to (3.54), in particular
Theorem 6.2.
To show uniqueness, let a be a solution of the homogeneous equation. Then, with the
ansatz (3.53b), (3.53c), we conclude that ν × E s |+ = 0 and thus E s = 0 in R3 \ D by the
uniqueness
result. From the jump conditions
of Theorem 3.33 we conclude
that (note that
R
R
R
curl3 ∂D a(y) Φ(x, y) ds(y) = − curl ∆ ∂D a(y) Φ(x, y) ds(y) = k 2 curl ∂D a(y) Φ(x, y) ds(y))
ν × E s |− = ν × E s |− − ν × E s |+ = −a ,
ν × curl E s |− = ν × curl E s |− − ν × curl E s |+ = −η k 2 Ka .
and thus
Z
s
(ν × curl E |− ) ·
Es
ds = ηk
2
Z
(Ka) · (a × ν) ds
∂D
∂D
= ηk
2
Z
(ν × Ŝi2 a) · (a × ν) ds
∂D
= −ηk
2
Z
Ŝi2 a · a ds
∂D
= −ηk 2
Z
∂D
|Ŝi a|2 ds .
3.3. EXERCISES
145
The left–hand side is real valued by Green’s theorem applied in D. Also the integral on the
right–hand side is real. Because η is not real we conclude that both integrals vanish and
thus also a by the injectivity of Ŝi .
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) = 2ki exp(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
∞ X
n
X
m
−m
jn (k|y|) h(1)
(ŷ)
n (k|x|) Yn (x̂) Yn
n=0 m=−n
=
∞
ik X
(2n + 1) jn (k|y|) h(1)
n (k|x|) Pn (x̂ · ŷ)
4π n=0
where again x̂ = x/|x|, ŷ = 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 interior
Neumann boundary value problem
∆u + k 2 u = 0 in D ,
∂u
= f on ∂D ,
∂ν
(3.55)
is not uniquely solvable.
Exercise 3.4 Show that the interior Neumann boundary value problem (3.55) can be solved
by a single layer ansatz on ∂D provided uniqueness holds. Here D ⊆ R3 is any bounded
domain with sufficiently smooth boundary such that the exterior is connected.
Exercise 3.5 Let ϕ ∈ C 0,α (D) with ϕ = 0 on ∂D. Then the extension of ϕ by zero outside
of D is in C 0,α (R3 ).
The following Exercises 3.6–3.9 study the potential theoretic case; that is, for k = 0, in
R3 . Again, D ⊆ R3 is a bounded domain with sufficiently smooth boundary such that the
exterior is connected.
Exercise 3.6 Show that the exterior Neumann boundary value problem
∆u = 0 in R3 \ D ,
∂
u = f on ∂D ,
∂ν
(3.56a)
has at most one solution u ∈ C 2 (R3 \ D) ∩ C 1 (R3 \ D) which satisfies the decay conditions
u(x) = O(|x|−1 ) ,
uniformly with respect to x/|x| ∈ S 2 .
∇u(x) = O(|x|−2 ) ,
|x| → ∞ ,
(3.56b)
146
CHAPTER 3. SCATTERING FROM A PERFECT CONDUCTOR
Exercise 3.7 Show that for any f ∈ C 0,α (∂D) the exterior Neumann boundary value problem (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 Neumann
boundary value problem; that
R is, (3.56a) in D for f = 0 is only solved by constant functions.
Show that the condition ∂D f (x) ds = 0 is a necessary condition for the inhomogeneous
interior Neumann boundary value problem to be solvable.
Exercise3.9 (a) Show that
boundary operator (3.26b) for k = 0 maps the
R the double layer
0,α
subspace ϕ ∈ C (∂D) : ∂D ϕ(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 ∈
R
C 0,α (∂D) : ∂D ϕ(x)ds = 0 by a single layer ansatz with a density in this subspace.
Chapter 4
The Variational Approach to the
Cavity Problem
In this chapter we want to introduce the reader to a second powerful approach for solving
boundary value problems for the Maxwell system (or more general partial differential equations) which is the basis of, e.g., the Finite Element technique. We introduce this idea for
the 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
curl E − iωµH = 0 in D ,
curl H + (iωε − σ)E = Je
(4.1a)
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 in
L2 (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 curl E, curl H ∈ L2 (D, C3 ).
Assuming scalar valued electric parameter µ, ε, σ and a sufficiently smmoth domain D, we
first multiply (4.1b) by some smooth vector field ψ with vanishing tangential components
on ∂D, then integrate over D, use partial integration and (4.1a). This yields
Z
Z
Je · ψ dx =
curl H · ψ + (iωε − σ)E · ψ dx
D
D
Z
=
H · curl ψ + (iωε − σ)E · ψ dx
D
Z =
D
1
curl E · curl ψ + (iωε − σ)E · ψ
iωµ
dx .
The boundary contribution vanishes because of the boundary condition for ψ. Multiplication
with iω yields
Z Z
1
σ
2
curl E · curl ψ − ω ε + i
E · ψ dx = iω
Je · ψ dx .
(4.2)
ω
D µ
D
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, where
the right–hand side is treated as a linear bounded functional. Considering the left–hand
side as a bilinear form in such a Hilbert space we can hope for an application of the Riesz
representation theorem or, more generally, the Lax-Milgram theorem (see 6.6) to obtain
existence results.
Thus the first crucial step is to find appropriate Hilbert spaces, which in particular incorporates the boundary condition (4.1c) and allows a rigorous treatment of the idea. These
requirements lead to the introduction of Sobolev spaces. We begin with Sobolev spaces
of scalar functions in Subsection 4.1.1 before we continue with vector-valued functions in
Subsection 4.1.2, which will be used in 4.2 for the cavity problem.
4.1
4.1.1
Sobolev Spaces
Basic Properties of Sobolev Spaces of Scalar Functions
We are going to present the definition of the Sobolev space H01 (D) and its basic properties
which are needed to treat the boundary value problem ∆u + k 2 u = f in D and u = 0 on
∂D. The definitions and proofs are all elementary – mainly because we do not need any
smoothness assumptions on the set D ⊆ R3 which is in this section always an open set. Some
of the analogous properties of the space H 1 (D) where no boundary condition is incorporated
are needed for the Helmholtz decomposition of Subsection 4.1.3 and will be postponed to
that subsection.
We assume that the reader is familiar with the following basic function spaces. We have
used some of them before already.
u is k times continuously differentiable in D
k
u:D→C:
,
C (D) =
and all derivatives can be continuously extended to D
C k (D, C3 ) = u : D → C3 : uj ∈ C k (D) for j = 1, 2, 3 ,
C0k (D) = u ∈ C k (D) : supp(u) is compact and supp(u) ⊆ D ,
C0k (D, C3 ) = u ∈ C k (D, C3 ) : uj ∈ C0k (D) for j = 1, 2, 3 ,
Z
p
p
L (D) =
u : D → C : u is Lebesgue-measurable and
|u| dx < ∞ , p ∈ [1, ∞) ,
D
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|p Dq u(x) < ∞ for all p ∈ N, q ∈ N3 .
x∈R3
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 C k (D, C3 )
(for bounded D) and Lp (D, C3 ) for p ∈ [1, ∞) are canonical, the norm in L∞ (D) is given by
kuk∞ := inf c > 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 ∂x2q2 ∂xq33
Definition 4.1 Let D ⊆ R3 be an open set. A function u ∈ L2 (D) possesses a variational
gradient in L2 (D) if there exists F ∈ L2 (D, C3 ) with
Z
Z
u ∇ψ dx = −
F ψ dx for all ψ ∈ C0∞ (D) .
(4.3)
D
D
We write ∇u = F . Using the denseness of C0∞ (R3 ) in L2 (R3 ) (see Lemma 4.9) it is easy to
show that F is unique – if it exists (see Exercise 4.5).
Definition 4.2 We define the Sobolev space H 1 (D) by
H 1 (D) = u ∈ L2 (D) : u possesses a variational gradient ∇u ∈ L2 (D, C3 )
and equip H 1 (D) with the inner product
Z
(u, v)H 1 (D) = (u, v)L2 (D) + (∇u, ∇v)L2 (D) =
u v + ∇u · ∇v dx .
D
Theorem 4.3 The space H 1 (D) is a Hilbert space.
Proof: Only completeness has to be shown. Let (un ) be a Cauchy sequence in H 1 (D).
Then (un ) and (∇un ) are Cauchy sequences in L2 (D) and L2 (D, C3 ), respectively, and thus
convergent: un → u and ∇un → F for some u ∈ L2 (D) and F ∈ L2 (D, C3 ). We show that
F = ∇u: For ψ ∈ C0∞ (D) we conclude that
Z
Z
un ∇ψ dx = −
∇un ψ dx
D
D
by the definition of the variational
gradient. The left–hand side converges to
R
the right–hand side to − D F ψ dx, and thus F = ∇u.
R
D
u ∇ψ dx,
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 C 1 (D) is contained in H 1 (D), and the imbedding is bounded. For u ∈ C 1 (D)
the classical derivatives ∂u/∂xj coincide with the variational derivatives.
150
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
(b) Let u ∈ H 1 (D) and φ ∈ C0∞ (D). Let v be the extension of uφ by zero into R3 . Then
v ∈ H01 (R3 ) and the product rule holds; that is,
φ∇u + u∇φ in D ,
0
in R3 \ D .
∇v =
(4.4)
Proof: (a) For u ∈ C 1 (D) and ψ ∈ C0∞ (D) we note that the extension v of the product uψ
by zero is in C 1 (R3 ). Therefore, choosing a box Q = (−R, R)3 containing D we obtain by
the product rule and the fundamental theorem of calculus
Z
Z
Z
[ψ∇u + u∇ψ] dx =
∇(uψ) dx =
∇v dx = 0 .
D
D
Q
Therefore, the classical derivative is also the variational derivative and thus belongs to
L2 (D, C3 ).
(b) Let ψ ∈ C0∞ (R3 ). Then φψ ∈ C0∞ (D). Let g ∈ L2 (R3 , C3 ) be the right–hand side of
(4.4). Then
Z
Z
[ψ g + v ∇ψ] dx =
[ψφ ∇u + ψu ∇φ + uφ ∇ψ] dx
R3
D
Z
Z
uφ ∇ψ dx
[(ψφ) ∇u + u ∇(ψφ)] dx −
=
D
D
Z
= −
Z
uφ ∇ψ dx = −
v ∇ψ dx
R3
D
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 of
mollifying the given function, the Fourier transform and the convolution of functions. We
just collect and formulate the corresponding results and refer to, e.g., [13], for the proofs.
The Fourier transform F is defined by
Z
1
(Fu)(x) = û(x) =
u(y) e−i x·y dy , x ∈ R3 .
(2π)3/2 R3
F is well defined as an operator from S into itself and also from L1 (R3 ) into the space
Cb (R3 ) of bounded continuous functions on R3 . Furthermore, F is unitary with respect to
the L2 −norm; that is,
(u, v)L2 (R3 ) =
û, v̂
L2 (R3 )
for all u, v ∈ S .
Now we use the following result from the theory of Lebesgue integration.
Lemma 4.5 The space S is dense in L2 (R3 ).
(4.5)
4.1. SOBOLEV SPACES
151
Proof: We only sketch the arguments. First we use without proof that the space of step
functions with compact support are dense in L2 (R3 ). Every step function is a finite linear
combination of functions of the form u(x) = 1 on U and u(x) = 0 outside of U where U is
some open bounded set. We leave the constructive proof that functions of this type can be
approximated 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
Z
u(y) v(x − y) dy ,
(u ∗ v)(x) =
x ∈ R3 .
R3
Obviously by a simple transformation we have u ∗ v = v ∗ u. The Lemma of Young clarifies
the 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
ku ∗ vkLp (R3 ) ≤ kukLp (R3 ) kvkL1 (R3 )
for all u ∈ Lp (R3 ) , v ∈ L1 (R3 ) .
This lemma implies, in particular, that L1 (R3 ) is a commutative algebra with the convolution
as multiplication. The Fourier transform transforms the convolution into the pointwise
multiplication of functions:
F(u ∗ v)(x) = (2π)3/2 û(x) v̂(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
R1
φ(t2 ) t2 dt = 1/(4π) (see Exercise 4.7 for the existence of such a function). Define φδ ∈
0
∞
C (R3 ) by
1
1
2
(4.7)
φδ (x) = 3 φ 2 |x| , x ∈ R3 .
δ
δ
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 = 0
outside 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 kuδ − ukL2 (R3 ) → 0 as δ → 0.
(c) Let u ∈ H 1 (R3 ) and uδ as in part (b). Then uδ ∈ C ∞ (R3 ) ∩ H 1 (R3 ) and kuδ −
ukH 1 (R3 ) → 0 as δ → 0.
152
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Proof: (a) Fix any R > 0 and let |x| < R. Then
Z
(u ∗ φδ )(x) =
u(y) φδ (x − y) dy
|y|≤R+δ
because |x − y| ≥ |y| − |x| ≥ |y| − R ≥ δ for |y| ≥ R + δ and thus φδ (x − y) = 0. The
smoothness of φδ yields that this integral is infinitely often
R differentiable. Furthermore, if u
vanishes outside some compact set K then (u ∗ φδ )(x) = K u(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 that
u ∗ φδ ∈ L2 (R3 ) by the Lemma 4.6 of Young.
(b) Substituting y = δz yields for the Fourier transform (note that dy = δ 3 dz),
(2π)
3/2
1
φ̂δ (x) = 3
δ
Z
φ
R3
Z
1
2
−i x·y
|y| e
dy =
φ |z|2 e−iδ x·z dz = (2π)3/2 φ̂1 (δx) .
2
δ
R3
R
R1
Furthermore, (2π)3/2 φ̂1 (0) = R3 φ |y|2 dy = 4π 0 φ(r2 ) r2 dr = 1 by the normalization of
φ. Therefore, the Fourier transform of uδ is given by
(Fuδ )(x) = e−iδa·x (2π)3/2 û(x) φ̂δ (x)
where we used (4.6) and the translation property. For u ∈ L2 (R3 ) we conclude
2
kuδ − uk2L2 (R3 ) = kFuδ − ûk2L2 (R3 ) = exp(−iδa·)(2π)3/2 φ̂δ − 1 ûL2 (R3 )
Z 2 2
−iδa·x
3/2
=
(2π) φ̂1 (δx) − 1 û(x) dx .
e
R3
The
integrand tends
as δ → 0 for every x and is bounded by the integrable function
2 to zero
3/2
2
(2π) kφ̂1 k∞ + 1 |û(x)| . This proves part (b) by the theorem of dominated convergence.
(c) First we note that ∇(u ∗ φδ ) = ∇u ∗ φδ for u ∈ H 1 (R3 ). Indeed,
Z
Z
∇(u ∗ φδ )(x) =
u(y) ∇x φδ (x − y) dy = −
u(y) ∇y φδ (x − y) dy
R3
R3
Z
∇u(y) φδ (x − y) dy = (∇u ∗ φδ )(x)
=
R3
because, for fixed x, the mapping y 7→ φδ (x − y) is in C0∞ (R3 ). Therefore, again by the
Lemma of Young, ∇uδ ∈ L2 (R3 ); that is uδ ∈ H 1 (R3 ). Application of part (b) yields the
assertion.
2
With this result we can weaken the definition of the variational derivative and prove the
following product rule, compare with Lemma 4.4.
Corollary 4.8 Let D ⊆ R3 be an open and bounded set.
4.1. SOBOLEV SPACES
153
(a) Let u ∈ L2 (D). Then u ∈ H 1 (D) if, and only if, there exists F ∈ L2 (D, C3 ) with
Z
Z
u ∇ψ dx = −
F ψ dx for all ψ ∈ C 1 (D) with compact support in D .
D
D
The field F coincides with ∇u.
(b) Let u ∈ H 1 (D) and v ∈ C 1 (D). Then uv ∈ H 1 (D), and the product rule holds; that is,
∇(uv) = v ∇u + u ∇v .
Proof: (a) Only one direction has to be shown. Let u ∈ H 1 (D) and ψ ∈ C 1 (D) with
compact support K in D. Then the extension by zero into the outside of D belongs to
H 1 (R3 ) by Lemma 4.4. Part (c) of the previous theorem yields convergence of ψδ = ψ ∗ φδ
to ψ in H 1 (D) as δ → 0. Furthermore, supp(ψδ ) ⊆ D for sufficiently small δ > 0; that is,
ψδ ∈ C0∞ (D). Therefore, by the definition of the variational derivative,
Z
[u ∇ψδ + ψδ ∇u] dx = 0
D
for sufficiently small δ > 0. Letting δ tend to zero yields
Z
[u ∇ψ + ψ ∇u] dx = 0
D
which shows the assertion for F = ∇u.
(b) Let ψ ∈ C0∞ (D). Then vψ ∈ C 1 (D) and vanishes in some neighborhood of ∂D. Therefore,
Z
Z
(uv) ∇ψ + (u ∇v + v ∇u) ψ dx =
u ∇(vψ) + (vψ) ∇u dx = 0
D
D
by part (a).
2
Lemma 4.4 implies that for bounded open sets the space C ∞ (D) is contained in H 1 (D). By
the same arguments, the space C0∞ (D) is contained in H 1 (D) even for unbounded sets D.
As a first denseness result we have:
Lemma 4.9 For any open set D ⊆ R3 the space C0∞ (D) is dense in L2 (D).
Proof: First we assume D to be bounded and define the open subsets Dn of D by x ∈ D :
d(x, ∂D) > 1/n where d(x, ∂D) = inf y∈∂D |x − y| denotes the distance of x to the boundary
∂D. Then Dn ⊆ D, and we can find φn ∈ C0∞ (D) with 0 ≤ φn (x) ≤ 1 for all x ∈ D
and φn (x) = 1 for all x ∈ Dn . Let now u ∈ L2 (D). Extend u by zero into all of R3 , then
u ∈ L2 (R3 ). By Theorem 4.7 there exists ψ̃n ∈ C ∞ (R3 ) with kψ̃n −ukL2 (D) ≤ kψ̃n −ukL2 (R3 ) ≤
1/n. The functions ψn = φn ψ̃n are in C0∞ (D) and kψn − ukL2 (D) ≤ kφn (ψ̃n − u)kL2 (D) +
k(φn − 1)ukL2 (D) . The first term is dominated by kψ̃n − ukL2 (D) which converges to zero. For
the second term we apply Lebesgue‘s theorem of dominated convergence. Indeed, writing
154
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
R
k(φn −1)uk2L2 (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)|2
is bounded by the integrable function |u(x)|2 .
Finally, if D is not bounded, we just approximate u ∈ L2 (D) by a function with compact
support and use the previous result for bounded sets to this approximation, compare with
the last part of the proof of Lemma 4.18 below.
2
The space C0∞ (R3 ) of smooth functions with compact support is even dense in H 1 (R3 ), see
Exercise 4.3. For bounded domains, however, this is not the case. Therefore, we add an
other important definition.
Definition 4.10 The space H01 (D) is defined as the closure of C0∞ (D) with respect to
k · kH 1 (D) .
Note that this definition makes sense because C0∞ (D) is a subspace of H 1 (D) by the above
remark. By definition, H01 (D) is a closed subspace of H 1 (D). We understand H01 (D) as
the space of differentiable (in the variational sense) functions u with u = 0 on ∂D. This
interpretation 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 ∈ H 1 (D) the left and the right–hand
sides of the definition (4.3) of the variational derivative; that is,
Z
Z
ψ ∇u dx , ψ ∈ C0∞ (D) ,
u ∇ψ dx and `2 (ψ) := −
`1 (ψ) :=
D
D
are linear functionals from C0∞ (D) into C which are bounded with respect to the norm
k · kH 1 (D) by the Cauchy-Schwarz inequality:
`1 (ψ) ≤ kukL2 (D) k∇ψkL2 (D) ≤ kukL2 (D) kψkH 1 (D)
and analogously for `2 . Therefore, there exist linear and bounded extensions of these functionals to the closure of C0∞ (D) which is H01 (D). Thus the formula of partial integration
holds; that is,
Z
Z
u ∇ψ dx = −
ψ ∇u dx for all u ∈ H 1 (D) and ψ ∈ H01 (D) .
(4.8)
D
D
The next step is to show compactness of the imbedding H01 (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 details
and proofs).
Theorem 4.11 Let Q = (−R, R)3 ⊆ R3 be a bounded cube. For u ∈ L2 (Q) define the
Fourier coefficients un ∈ C by
Z
π
1
un =
u(x) e−i R n·x dx for n ∈ Z3 .
(4.9)
3
(2R)
Q
4.1. SOBOLEV SPACES
155
Then
u(x) =
π
X
un ei R n·x
n∈Z3
in the L2 −sense; that is,
2
Z M
X
π
un ei R n·x dx −→ 0 ,
u(x) −
N, M → ∞ .
n=−N
Q
Furthermore,
(u, v)L2 (Q) = (2R)3
X
un vn
and
kuk2L2 (Q) = (2R)3
X
|un |2 .
n∈Z3
n∈Z3
Therefore,
the space L2 (Q) can be characterized by the space of all functions u such that
P
2
n∈Z3 |un | converges. Because, for sufficiently smooth functions,
π
π X
∇u(x) = i
un n ei R n·x ,
R
3
n∈Z
we observe that i Rπ n un are the Fourier coefficients of ∇u. The requirement ∇u ∈ L2 (D, C3 )
leads to the following definition.
1
(Q) of periodic functions by
Definition 4.12 We define the Sobolev space Hper
(
)
X
1
Hper
(Q) = u ∈ L2 (Q) :
(1 + |n|2 ) |un |2 < ∞
n∈Z3
with inner product
3
1 (Q) = (2R)
(u, v)Hper
X
(1 + |n|2 ) un vn .
n∈Z3
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 H01 (D) belong to
1
Hper
(Q).
Theorem 4.13 Let again D ⊆ R3 be a bounded open set and Q = (−R,R)3 an open box
u on D,
which contains D in its interior. Then the extension operator η̃ : u 7→ ũ =
0 on Q \ D ,
1
1
is linear and bounded from H0 (D) into Hper (Q).
Proof: Let first u ∈ C0∞ (D). Then obviously ũ ∈ C0∞ (Q). We compute its Fourier coefficients ũn as
Z
Z
π
iR
∂ −i π n·x
1
−i R
n·x
ũ(x)
e
ũ(x)
ũn =
dx
=
e R dx
(2R)3
πnj (2R)3
∂xj
Q
= −
iR
πnj (2R)3
Q
Z
Q
π
∂ ũ
iR
(x) e−i R n·x dx = −
vn
∂xj
πnj
156
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
where vn are the Fourier coefficients of v = ∂ ũ/∂xj . Therefore, by Theorem 4.11,
2
2
2 X
X
R2
∂
ũ
R
∂u
R2 2
3
2
3
2
(2R)
nj |ũn | = 2 (2R)
= 2
.
|vn | = 2 π
π ∂xj L2 (Q)
π ∂xj L2 (D)
3
3
n∈Z
n∈Z
Furthermore,
X
(2R)3
|ũn |2 = kũk2L2 (Q) = kuk2L2 (D)
n∈Z3
and thus adding the equations for j = 1, 2, 3 and for ũ,
R2
2
kũkHper
1 + 2 kuk2H 1 (D) .
1 (Q) ≤
π
This holds for all functions u ∈ C0∞ (D). Because C0∞ (D) is dense in H01 (D) we conclude
that
r
R2
1 (Q) ≤
kũkHper
1 + 2 kukH 1 (D) for all u ∈ H01 (D) .
π
This proves boundedness of the extension operator η̃.
2
As a simple application we have the compact imbedding.
Theorem 4.14 Let D ⊆ R3 be a bounded open set. Then the imbedding H01 (D) → L2 (D)
is compact.
Proof: Let again Q = (−R, R)3 ⊆ R3 such that D ⊆ Q. First we show that the imbedding
1
1
(Q) → L2 (Q) by
(Q) → L2 (Q) is compact. Define JN : Hper
J : Hper
X
π
(JN u)(x) :=
un ei R n·x , x ∈ Q , N ∈ N .
|n|≤N
Then JN is obviously
i π n· bounded and
finite dimensional, because the range
R(JN ) = span e R : |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,
k(JN − J)uk2L2 (Q) = (2R)3
X
|un |2 ≤
|n|>N
(2R)3 X
1
(1 + |n|2 ) |un |2 ≤
kuk2Hper
1 (Q) .
2
1+N
1 + N2
|n|>N
1
Therefore, kJN − Jk2Hper
1 (Q)→L2 (Q) ≤ 1+N 2 → 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
η̃
J
R
1
R ◦ J ◦ η̃ : H01 (D) −→ Hper
(Q) −→ L2 (Q) −→ L2 (D)
1
is compact. Here, η̃ : H01 (D) → Hper
(Q) denotes the extension operator from the previous
2
2
theorem and R : L (Q) → L (D) is just the restriction operator.
2
For the space H01 (D) we obtain a useful result which is often called an inequality of Friedrich’s
Type.
4.1. SOBOLEV SPACES
157
Theorem 4.15 For any bounded open set D ⊆ R3 there exists c > 0 with
kukL2 (D) ≤ c k∇ukL2 (D)
for all u ∈ H01 (D) .
Proof: Let again first u ∈ C0∞ (D), extended by zero into R3 . Then, if again D ⊆ Q =
(−R, R)3 ,
Zx1
∂u
(t, x2 , x3 ) dt
u(x) = u(x1 , x2 , x3 ) = u(−R, x2 , x3 ) +
|
{z
}
∂x1
−R
= 0
and thus for x ∈ Q by the inequality of Cauchy–Schwarz
u(x)2 ≤ (x1 + R)
2
2
ZR Zx1 ∂u
∂u
∂x1 (t, x2 , x3 ) dt ≤ 2R ∂x1 (t, x2 , x3 ) dt and
−R
−R
ZR
u(x)2 dx1 ≤ (2R)2
−R
2
ZR ∂u
dt .
(t,
x
,
x
)
2
3
∂x1
−R
Integration with respect to x2 and x3 yields
kuk2L2 (D) = kuk2L2 (Q)
2
ZR ZR ZR ∂u
2
dt dx2 dx3
(t,
x
,
x
)
≤ (2R)
2
3
∂x1
−R −R −R
≤ (2R)2 k∇uk2L2 (Q) = (2R)2 k∇uk2L2 (D) .
By a density argument this holds for all u ∈ H01 (D).
4.1.2
2
Basic Properties of Sobolev Spaces of Vector Valued Functions
We follow the scalar case as closely as possible. We assume here that all functions are
complex valued. First we note that in the formulation (4.1a), (4.1b) not all of the partial
derivatives 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 a
vector field w ∈ L2 (D, C3 ) such that
Z
Z
v · curl ψ dx =
w · ψ dx for all vector fields ψ ∈ C0∞ (D, C3 ) .
D
D
(b) A vector field v ∈ L2 (D, C3 ) possesses a variational divergence in L2 (D) if there exists
a scalar function p ∈ L2 (D) such that
Z
Z
v · ∇ϕ dx = −
p ϕ dx for all scalar functions ϕ ∈ C0∞ (D) .
D
D
158
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
The functions w and p are unique if they exist (compare Exercise 4.5). In view of partial
integration (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) + (curl u, curl v)L2 (D) .
With this inner product the space H(curl, D) is a Hilbert space. The proof follows by the
same 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 C 1 (D, C3 ) is contained in H(curl, D), and the imbedding is bounded. For
u ∈ C 1 (D) the classical curl coincides with the variational curl.
(b) Let u ∈ H(curl, D) and φ ∈ C0∞ (D). Let v be the extension of uφ by zero into R3 .
Then v ∈ H0 (curl, R3 ) and the product rule holds; that is,
φ curl u + ∇φ × u in D ,
curl v =
0
in R3 \ D .
We note that part (a) follows even directly from Lemma 4.4 because H 1 (D, C3 ) ⊆ H(curl, D).
We continue as in the scalar case and prove denseness properties (compare with Exercise 4.3).
Lemma 4.18 C0∞ (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. It
is sufficient to show that
curl(u ∗ φδ ) = (curl u) ∗ φδ
for all u ∈ H(curl, R3 )
with φδ from (4.7). Because supp(φδ ) ⊆ B3 [0, δ] we conclude for |x| < R that
Z
(u ∗ φδ )(x) =
u(y) φδ (x − y) dy
|y|<R+δ
3
and for any fixed vector a ∈ R interchanging differentiation and integration leads to
Z
Z
a · curl(u ∗ φδ )(x) =
a · ∇x φδ (x − y) × u(y) dy = −
a · ∇y φδ (x − y) × u(y) dy
R3
R3
Z
= −
R3
a × ∇y φδ (x − y) · u(y) dy =
Z
curly a φδ (x − y) · u(y) dy
R3
Z
a · curl u(y) φδ (x − y) dy = a · (curl u ∗ φδ )(x)
=
R3
4.1. SOBOLEV SPACES
159
because, for fixed x, the mapping y 7→ a φδ (x−y) is in C0∞ (R3 , C3 ). This holds for all vectors
a, thus curl(u ∗ φδ ) = (curl u) ∗ φδ .
Finally, we have to approximate uδ = u∗φδ by a field with compact support. Let ψ ∈ C0∞ (R3 )
with ψ(x) = 1 for |x| ≤ 1. We define ψR (x) = ψ(x/R) and v R,δ = ψR uδ and have
kv R,δ − uδ k2H(curl,R3 ) = k(ψR − 1)uδ k2L2 (R3 ) + k∇ψR × uδ + (ψR − 1) curl uδ k2L2 (R3 )
≤ k(ψR − 1)uδ k2L2 (R3 ) + 2 k(ψR − 1) curl uδ k2L2 (R3 )
+ 2 k∇ψR × uδ k2L2 (R3 ) .
It is easily seen by the theorem of dominated convergence that all of the terms converge to
zero 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 H01 (D) we define
Definition 4.19 For any open set D ⊆ R3 we define H0 (curl, D) as the closure of C0∞ (D, C3 )
in H(curl, D).
Vector fields v ∈ H0 (curl, D) do not necessarily vanish on ∂D – in contrast to functions
in H01 (D). Only their tangential components vanish. This follows from the trace theorem in Chapter 5, see Theorem 5.21. We finish this short introduction to H(curl, D) and
H0 (curl, D), respectively, by an important observation in view of the following the Helmholtz
decomposition and Maxwell’s equations.
Lemma 4.20 The space ∇H01 (D) = ∇p : p ∈ H01 (D) is a closed subspace of H0 (curl, D).
Proof By the definition of H01 (D) there exists a sequence pj ∈ C0∞ (D) with pj → p in
H 1 (D). It is certainly ∇pj ∈ C0∞ (D, C3 ). Because ∇pj → ∇p in L2 (D, C3 ) and curl ∇pj = 0
we note that (∇pj ) is a Cauchy sequence in H(curl, D) and thus convergent. We conclude
that ∇p ∈ H0 (curl, D).
It remains to show closedness of ∇H01 (D) in L2 (D, C3 ) – and therefore, in H(curl, D). Let
pj ∈ H01 (D) with ∇pj → F in L2 (D, C3 ) for some F ∈ L2 (D, C3 ). Then (∇pj ) is a Cauchy
sequence 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 H 1 (D), which
implies convergence. This shows that F = ∇p for some p ∈ H01 (D) and ends the proof. 2
We recall that H01 (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 compactly
imbedded in L2 (D, C3 ). Therefore, we will decompose the variational problem into two
problems 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 we
discuss in the subsequent subsection.
160
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
4.1.3
The Helmholtz Decomposition
Now we turn to decompositions of vector fields, which are closely related to the Maxwell
system – the Helmholtz decomposition. For a general proof we will apply the Theorem 6.6
of Lax and Milgram.
There are several forms of the Helmholtz decomposition. In this section we need decompositions 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 complex 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) is
symmetric for almost all x. Furthermore, assume the existence of a constant ĉ > 0 such
that Re (z > A(x)z) ≥ ĉ|z|2 for all z ∈ C3 and almost all x ∈ D. Then for the subspaces
H(curl 0, D), VA ⊆ H(curl, D), and H0 (curl 0, D), V0,A ⊆ H0 (curl, D) and Ṽ0,A ⊆ L2 (D, C3 )
defined by
H(curl 0, D) = u ∈ H(curl, D) : curl u = 0 in D ,
(4.10a)
H0 (curl 0, D) = u ∈ H0 (curl, D) : curl u = 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)
Ṽ0,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 Ṽ0,A are closed in L2 (D, C3 ), and L2 (D, C3 ) is the direct sum of Ṽ0,A
and H0 (curl 0, D); that is,
L2 (D, C3 ) = Ṽ0,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 from
H0 (curl, D) onto V0,A is the restriction of the projection operator from L2 (D, C3 ) onto Ṽ0,A .
Proof: The closedness of H(curl 0, D) in H(curl, D) and H0 (curl 0, D) in H0 (curl, D) is obvious. 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
4.1. SOBOLEV SPACES
161
curl un = 0 we observe that (un ) is also a Cauchy sequence in H0 (curl, D) and thus convergent
in H0 (curl, D). This shows that u ∈ H0 (curl, D) and curl u = 0; that is, u ∈ H0 (curl 0, D).
The closedness of Ṽ0,A and V0,A and VA in L2 (D, C3 ) and H0 (curl, D) and H(curl, D), respectively, is easy to see, and we omit this part.
Furthermore it holds Ṽ0,A ∩ H0 (curl 0, D) = {0} because u ∈ Ṽ0,A ∩ H0 (curl 0, D) implies
(Au, u)L2 (D) = 0 and thus, taking the real part, ĉkuk2L2 (D) ≤ Re (Au, u)L2 (D) = 0 which
yields u = 0.
Next, we have to prove that L2 (D, C3 ) ⊆ Ṽ0,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 the
sesquilinear form a : H0 (curl 0, D) × H0 (curl 0, D) → C by
Z
(4.11)
a(ψ, v) = (ψ, Av)L2 (D) =
ψ > Av dx for all v, ψ ∈ H0 (curl 0, D) .
D
Then a is certainly bounded but also coercive in the space H0 (curl 0, D) equipped with the
inner product of H(curl, D). Indeed, we write
Re a(v, v) = Re (v, Av)L2 (D) ≥ ĉ kvk2L2 (D) = ĉ kvk2H(curl,D)
for v ∈ H0 (curl 0, D) because curl v = 0. Therefore, the sesquilinear form a and – for every
fixed 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 given
u ∈ 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 ∈ Ṽ0,A and the decomposition u = u0 + (u − u0 ) ∈ H0 (curl 0, D) + Ṽ0,A has been
shown.
Finally, boundedness of the projection operator u 7→ u − u0 follows from general properties
of direct sums. This proves the result in L2 (D, C3 ). Analogously, the projection operator
P : H0 (curl, D) → V0,A is defined in the same way by P u = 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 ∇H01 (D) = {∇p : p ∈ H01 (D)} is contained in
H0 (curl 0, D). Therefore, V0,A and Ṽ0,A are contained in the spaces
H0 (curl, divA 0, D) := {v ∈ H0 (curl, D) : (Av, ∇ϕ)L2 (D) = 0 for all ϕ ∈ H01 (D)}, (4.12a)
L2 (divA 0, D) := {v ∈ L2 (D, C3 ) : (Av, ∇ϕ)L2 (D) = 0 for all ϕ ∈ H01 (D)}, (4.12b)
respectively, which are the spaces of vector fields with vanishing divergence div(Av) = 0.
The space ∇H01 (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 ∇H01 (D) and the
gradients ∇C(D) of the co-homology space
C(D) = {p ∈ H 1 (D) : ∆p = 0 in D, p constant on every connected component of ∂D} .
Analogously, ∇H 1 (D) is a subspace of H(curl 0, D) – and coincide with it for simply connected domains. Therefore, VA is contained in
H(curl, divA 0, D) = {v ∈ H(curl, D) : (Av, ∇ϕ)L2 (D) = 0 for all ϕ ∈ H 1 (D)}
(4.12c)
162
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
which is the space of vector fields in H(curl, D) with vanishing divergence div(Av) = 0 and
vanishing 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 VµI
and, analogously, Vε for scalar valued ε. In the case σ = 0; that is real-valued and symmetric
and 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 curl u, curl v)L2 (D) + (µu, v)L2 (D) ,
(4.13a)
(u, v)µ,ε = (µ−1 curl u, curl v)L2 (D) + (εu, v)L2 (D) ,
(4.13b)
respectively, and L2 (D, C3 ) = Ṽ0,ε ⊕H0 (curl 0, D) is an orthogonal decomposition with respect
to the inner product (εu, v)L2 (D) .
By the same arguments as in the proof of the previous theorem one can show another kind
of 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) ⊕ ∇H01 (D) ,
H0 (curl, D) = H0 (curl, divA 0, D) ⊕ ∇H01 (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 ). However – and this is the reason for introducing the Helmholtz decomposition – the subspace
V0,A is compactly imbedded in L2 (D, C3 ). We formulate this crucial result in the following
theorem.
Theorem 4.24 Let D be a bounded Lipschitz domain (see Definition 6.7). Then the space
V0,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 (see
Corollary 4.36 below) that also VA is compactly imbedded in L2 (D, C3 ). However, in the
context of this chapter we do not need the compactness of VA .
4.2. THE CAVITY PROBLEM
4.2
4.2.1
163
The Cavity Problem
The Variational Formulation and Existence
Before we investigate the Maxwell system (4.1a)–(4.1c) we consider the variational form
of the following interior boundary value problem for the scalar inhomogeneous Helmholtz
equation, namely
div(a ∇u) + k 2 bu = 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 for
some 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 the
equation by some test function ψ with ψ = 0 on ∂D, integrate over D, and use Green’s first
formula. This yields
Z
Z
2
a ∇u · ∇ψ − k b u ψ dx = − f ψ dx .
D
D
This equation makes perfectly sense in H01 (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 valued
a ∈ L∞ (D) such that a(x) ≥ a0 on D for some constant a0 > 0. A function u ∈ H01 (D) is
called a variational solution (or weak solution) of the boundary value problem (4.14) if
Z
Z
2
a ∇u · ∇ψ − k b u ψ dx = − f ψ dx for all ψ ∈ H01 (D) .
D
D
By Friedrich’s inequality, the first term in this equation defines an inner product in H01 (D).
Lemma 4.26 Let a ∈ L∞ (D) be real valued such that a(x) ≥ a0 on D for some constant
a0 > 0. We define a new inner product in H01 (D) by
(u, v)∗ = (a ∇u, ∇v)L2 (D) ,
u, v ∈ H01 (D) ,
Then H01 (D), (·, ·)∗ is a Hilbert space, and its norm is equivalent to the ordinary norm in
H01 (D); in particular,
s
p
(1 + c2 )kak∞
kuk∗ ≤
kak∞ kukH 1 (D) ≤
kuk∗ for all u ∈ H01 (D)
(4.15)
a0
where c is the constant from Theorem 4.15.
Proof: From the previous theorem we conclude
√
kuk2∗ = k a ∇uk2L2 (D) ≤ kak∞ kuk2H 1 (D) = kak∞ k ∇uk2L2 (D) + kuk2L2 (D)
≤
(1 + c2 ) kak∞ k∇uk2L2 (D) ≤ kak∞
1 + c2
kuk2∗ .
a0
164
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
2
Therefore, we write the variational equation of Definition 4.25 in the form
(u, ψ)∗ − k 2 (bu, ψ)L2 (D) = −(f, ψ)L2 (D)
for all ψ ∈ H01 (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 above
assumptions. Then the boundary value problem (4.14) has a variational solution for exactly
those f ∈ L2 (D) such that (f, v)L2 (D) = 0 for all solutions v ∈ H01 (D) of the corresponding
homogeneous form of (4.14) with b replaced by b; that is,
div(a ∇v) + k 2 bv = 0 in D ,
v = 0 on ∂D .
(4.17)
If, in particular, this boundary value problem (4.17) admits only the trivial solution v = 0
then 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 analysis
to rewrite (4.16) in the form u − k 2 Ku = −f˜ with some compact operator K and apply the
Fredholm alternative of the previous theorem to this equation. To carry out this idea we
note that – for fixed v ∈ L2 (D) – the mapping ψ 7→ (bψ, v)L2 (D) is linear and bounded from
H01 (D), (·, ·)∗ into C. Indeed, by the inequality of Cauchy-Schwarz and Theorem 4.15 we
conclude that there exists c > 0 such that
(bψ, v)L2 (D) ≤ kbk∞ kψkL2 (D) kvkL2 (D) ≤ c kvkL2 (D) kψk∗ for all ψ ∈ H01 (D) .
Therefore, the representation theorem of Riesz assures the existence of a unique gv ∈ H01 (D)
with (bψ, v)L2 (D) = (ψ, gv )∗ for all ψ ∈ H01 (D). We define the operator K̃ : L2 (D) → H01 (D)
by K̃v = gv . Then K̃ satisfies
(K̃v, ψ)∗ = (bv, ψ)L2 (D)
for all v ∈ L2 (D) , ψ ∈ H01 (D) .
K̃ is linear (easy to see) and bounded because kK̃vk2∗ = (K̃v, K̃v)∗ = (bv, K̃v)L2 (D) ≤
kbk∞ kK̃vkL2 (D) kvkL2 (D) ≤ ckK̃vk∗ kvkL2 (D) ; that is, kK̃vk∗ ≤ ckvkL2 (D) . By the same argument there exists f˜ ∈ H01 (D) with (f, ψ)L2 (D) = (f˜, ψ)∗ for all ψ ∈ H01 (D). Therefore, the
equation (4.16) can be written as
(u, ψ)∗ − k 2 (K̃u, ψ)∗ = −(f˜, ψ)∗
for all ψ ∈ H01 (D) .
(4.18)
Because this holds for all such ψ we conclude that this equation is equivalent to
u − k 2 K̃u = −f˜ in H01 (D) .
(4.19)
Because H01 (D) is compactly imbedded in L2 (D) we conclude that the operator K = K̃ ◦
J : H01 (D) → H01 (D) is compact (J denotes the imbedding H01 (D) ⊆ L2 (D)). To study
solvability of this equation we want to apply the Fredholm alternative of Theorem 6.4. We
4.2. THE CAVITY PROBLEM
165
R
choose the bilinear form hu, vi = (u, v)∗ = D a ∇u · ∇v dx in H01 (D). RThen K is self adjoint
because by the definition of K we have hKu, vi = (bu, v)L2 (D) = D b u v dx and this is
symmetric in u and v. Therefore, we know from the Fredholm alternative of Theorem 6.4
that equation (4.19) is solvable if, and only if, the right–hand side f˜ is orthogonal with
respect to h·, ·i to the nullspace of I − k 2 K. Therefore, let v ∈ H01 (D) be a solution of (4.17).
Then v satisfies the homogeneous equation v − k 2 Kv = 0 and hf˜, vi = (f˜, v)∗ = (f, v)L2 (D) .
Therefore, we conclude that hf˜, vi = 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) + k 2 b 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 k 2 ∈ R>0 of
(4.20); that is, there exists a sequence kn > 0 and corresponding eigenfunctions ûn ∈ H01 (D)
with kûn k∗ = 1 such that div(a∇ûn ) + kn b ûn = 0 in D and ûn = 0 on ∂D in the variational
form. Furthermore, kn → ∞ as n → ∞. The sets {ûn : n ∈ N} and {kn ûn : n ∈ N}
form complete orthonormal systems in H01 (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 : H01 (D) → H01 (D) as in the previous theorem. Then K
is compact and self adjoint in H01 (D), (·, ·)∗ because b is real valued. It is also positive
because (Ku, u)∗ = (bu, u)L2 (D) > 0 for u 6= 0. Therefore, the spectral theorem for compact
self adjoint operators in Hilbert spaces (see, e.g,
[12], Section 15.3) implies the existence of a
spectral system (µn , ûn ) of K in H01 (D), (·, ·) ∗ . Furthermore, µn > 0 and µn → 0 as n → ∞
and the set {ûn : n ∈ N} of orthonormal eigenfunctions ûn is complete in H01 (D), (·, ·) ∗ .
Therefore, we have K ûn = µn ûn ; that is, ûn − µ1n K ûn = 0, which is the variational form of
(4.20) for kn = √1µn . The system {kn ûn : n ∈ N} is a system of eigenfunctions as well which
is orthogonal in L2 (D, b dx) because
kn2 (bûn , ûm )L2 (D) = kn2 (K ûn ûm )∗ = (ûn , ûm )∗ = δmn .
This set is also complete in L2 (D). This follows directly from the fact that H01 (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 equations 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
curl E − iωµ H = 0 in D ,
curl H + (iωε − σ) E = Je
in D .
(4.21a)
(4.21b)
166
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
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 this
system; that is,
Z
Z σ
1
2
Je · ψ dx .
(4.22)
curl E · curl ψ − ω ε + i
E · ψ dx = iω
ω
D
D µ
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) and
proceed as in the previous part.
Z
Z
0 =
curl E · ψ − iωµ H · ψ dx =
E · curl ψ − iωµ H · ψ dx
D
Z =
D
D
1
(Je − curl H) · curl ψ − iωµ H · ψ dx .
iωε − σ
We note that the boundary term vanishes because of E ∈ H0 (curl, D). Multiplication with
iω yields
Z
Z 1
1
2
curl H · curl ψ − ω µ H · ψ dx =
Je · curl ψ dx .
(4.23)
D ε + iσ/ω
D ε + iσ/ω
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 Maxwell
system.
Lemma 4.29 Let E ∈ H0 (curl, D) satisfy the variational equation (4.22) for all ψ ∈
H0 (curl, D). Set
1
H =
curl E in D .
iωµ
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
Z h
Z
i
σ
2
iω H · curl ψ − ω ε + i
E · ψ dx = iω
Je · ψ dx
ω
D
D
which is the variational form of iω curl H; that is,
σ
iω curl H = ω 2 ε + i
E + iω Je .
ω
4.2. THE CAVITY PROBLEM
167
2
Division by iω yields (4.21b).
In the following we will discuss the variational equation (4.22) in the space H0 (curl, D). For
the remaining part of this subsection we make the following assumptions on the parameters
µ, 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) + i
σ(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 = εc I) is compactly imbedded in L2 (D, C3 ). By Theorem 4.24 this is the case for
Lipschitz domains D.
Furthermore, it is convenient to define the vector field F ∈ L2 (D, C3 ) by
F =
iω
Je .
εc
Then the variational equation (4.22) takes the form
Z
Z 1
2
εc F · ψ dx
curl E · curl ψ − ω εc E · ψ dx =
D
D µ
(4.25)
for all ψ ∈ H0 (curl, D). To discuss this equation in H0 (curl, D) we will use the Helmholtz
decomposition 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) : (εc u, ψ)L2 (D) = 0 for all ψ ∈ H0 (curl 0, D) .
Analogously, Ṽ0,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 ∈ Ṽ0,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
Z
Z 1
2
curl u · curl ψ − ω εc (u + E0 ) · ψ dx =
εc (f + F0 ) · ψ dx
(4.26)
µ
D
D
for
R all ψ ∈ H0 (curl, D).
R Now we take ψ ∈ H0 (curl 0, D) as test functions. Recalling that
ε
u
·
ψ
dx
=
0
and
ε f · ψ dx = 0 for all ψ ∈ H0 (curl 0, D) yields
D c
D c
Z
Z
2
−ω
εc E0 · ψ dx =
εc F0 · ψ dx for all ψ ∈ H0 (curl 0, D) .
D
D
168
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Rewriting this as
Z
εc [F0 + ω 2 E0 ] · ψ dx = 0 for all ψ ∈ H0 (curl 0, D)
D
and setting ψ = F0 + ω 2 E0 ∈ H0 (curl 0, D) yields E0 = − ω12 F0 .
Therefore, (4.26) reduces to
Z Z
1
2
curl u · curl ψ − ω εc u · ψ dx =
εc f · ψ dx
µ
D
(4.27)
D
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 ∈
Ṽ0,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 and
E0 ∈ H0 (curl 0, D) the Helmholtz decomposition of E. Then E0 = − ω12 F0 and u ∈ V0,εc
solves (4.27) for all ψ ∈ V0,εc .
(b) If u ∈ V0,εc solves (4.27) then E = u −
1
ω2
F0 ∈ H0 (curl, D) solves (4.25).
To show solvability of (4.27) we follow the approach as in the scalar case for the Helmholtz
equation and introduce the equivalent inner product (·, ·)∗ on V0,εc by
Z
1
(v, w)∗ =
curl v · curl w dx , v, w ∈ V0,εc .
(4.28)
µ
D
Then we have the analogue of Lemma 4.26:
Lemma 4.31 The norm kvk∗ =
p
(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 the
existence of a constant c > 0 such that
kvkH(curl,D) ≤ c k curl vkL2 (D)
for all v ∈ V0,εc .
(4.29)
This is sufficient because the estimates k curl vk2L2 (D) ≤ kµk∞ kvk2∗ and kvk∗ ≤
k(1/µ)k∞ kvkH(curl,D) hold obviously.
Assume that the inequality (4.29) is not satisfied. Then there exists a sequence (vn ) in V0,εc
with kvn kH(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 ) by
Theorem 4.24 there exists a subsequence, denoted by the same symbol, which converges
in L2 (D, C3 ). For this subsequence, (vn ) and (curl vn ) are Cauchy sequences in L2 (D, C3 ).
4.2. THE CAVITY PROBLEM
169
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 shows
that v vanishes. This contradicts the fact that kvkH(curl,D) = 1.
2
Now we write (4.27) in the form
(u, ψ)∗ − ω 2 (εc u, ψ)L2 (D) = (εc f, ψ)L2 (D)
for all ψ ∈ V0,εc .
Again, we use the representation theorem of Riesz (Theorem 6.5) to show the existence
of a linear and bounded operator K̃ : L2 (D, C3 ) → V0,εc with (ψ, εc u)L2 (D) = (ψ, K̃u)∗
for all ψ ∈ V0,εc and u ∈ L2 (D, C3 ). We carry out the arguments for the convenience
of the reader although they are completely analogous to the arguments in the proof of
Theorem 4.27. For fixed u ∈ L2 (D, C3 ) the mapping ψ 7→ (ψ, εc u)L2 (D) defines a linear and
bounded functional on V0,εc . Therefore, by the representation theorem of Riesz there exists
a unique g = gu ∈ V0,εc with (ψ, εc u)L2 (D) = (ψ, gu )∗ for all ψ ∈ V0,εc . We set K̃u = gu .
Linearity of K̃ is clear from the uniqueness of the representation gu . Boundedness of K̃ from
L2 (D, C3 ) into V0,εc follows from the estimate
kK̃uk2∗ = (K̃u, K̃u)∗ = (K̃u, εc u)L2 (D) ≤ kεc k∞ kukL2 (D) kK̃ukL2 (D)
≤ c kεc k∞ kukL2 (D) kK̃uk∗
after division by kK̃uk∗ . Because V0,εc is compactly imbedded in L2 (D, C3 ) we conclude
that K = K̃ ◦ J is even compact as an operator from V0,εc into itself, if we denote by
J : V0,εc → L2 (D, C3 ) the compact imbedding operator. Therefore, we can write equation
(4.27) in the form
(u, ψ)∗ − ω 2 (Ku, ψ)∗ = (K̃f, ψ)∗
for all ψ ∈ V0,εc ;
that is,
u − ω 2 Ku = K̃f
in V0,εc .
(4.30)
This is again a Fredholm equation of the second kind for u ∈ V0,ε . In particular, we have
existence once we have uniqueness. The question of uniqueness will be discussed in the next
section below.
For the general question of existence we
again Fredholm’s Theorem 6.4 to X = V0,εc
R apply
1
with bilinear form hu, vi = (u, v)∗ = D µ curl u · curl v dx and operator T = ω 2 K. The
operator K is self adjoint. Indeed, hKu, ψi = (Ku, ψ)∗ = (εc u, ψ)L2 (D) and this is symmetric
in u and ψ. Lemma 4.30 tells us that the variational form (4.25) of the cavity problem is
solvable for F ∈ L2 (D, C3 ) if, and only if, the variational problem (4.27) is solvable in V0,εc
for the part f ∈ Ṽ0,εc of F . By Theorem 6.4 this is solvable for exactly those f ∈ Ṽ0,εc such
that K̃f is orthogonal to the nullspace of I − ω 2 K with respect to h·, ·i. By Lemma 4.30
v ∈ 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 ,
hK̃f, vi = (εc f, v)L2 (D) = εc (F0 + f ), v L2 (D) = (εc F, v)L2 (D) ;
170
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
that is hK̃f, vi vanishes if, and only if, (εc F, v)L2 (D) vanishes. Furthermore we note that v
solves 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) and
H ∈ H(curl, D) for exactly those source terms Je ∈ L2 (D, C3 ) such that (Je , v)L2 (D) = 0
for 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 trivial
solution E = 0, H = 0 then the inhomogeneous boundary value problem (4.21a), (4.21b) has
a 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.
curl E − iωµ H = 0 in D ,
(4.31a)
curl H + iωε E = J
(4.31b)
in D ,
for J = 0.
Definition 4.33 The resolvent set consists of all ω ∈ C for which (4.31a), (4.31b) has a
unique solution (E, H) ∈ H0 (curl, D) × H(curl, D) for all J ∈ L2 (D, C3 ) and such that
J 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 variational
equation (4.27) for u = E ∈ V0,ε (because the right–hand side vanishes); that is,
Z 1
2
curl u · curl ψ − ω εu · ψ dx = 0 for all ψ ∈ V0,ε ,
µ
D
which holds even for all ψ ∈ H0 (curl, D) because u ∈ V0,ε . If u solves this equation then also
its complex conjugate. Therefore, we can assume that u is real valued which we do for the
remaining part of this section for all functions. Then we write the variational equation in
the form
(u, ψ)µ,ε = (ω 2 + 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 h·, ·i = (·, ·)∗ and
the operator K : V0,ε → V0,ε from above,
(u, v)µ,ε = hu, vi + (εu, v)L2 (D) = hu, vi + hKu, vi .
(4.33)
First we consider ω 6= 0. We have seen that ω 6= 0 is in the resolvent set if, and only if, the
operator I − ω 2 K is one-to-one in V0,ε ; that is, 1/ω 2 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 we
take (·, ·)µ,ε instead of just h·, ·i 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 and
eigenvalues 1/ωn2 which converge to zero. Let us now consider ω = 0. From the variational
equation we conclude for ψ = u that curl u = 0. Therefore, ω = 0 is an eigenvalue with the
infinite dimensional eigenspace H0 (curl 0, D). We summarize this in the following theorem.
4.2. THE CAVITY PROBLEM
171
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 valued
functions vn ∈ V0,ε such that
1
curl vn − ωn2 ε vn = 0 in D , ν × vn = 0 on ∂D
curl
µ
in the variational form (4.32); that is,
Z 1
2
curl vn · curl ψ − ωn ε vn · ψ dx = 0 for all ψ ∈ H0 (curl, D) .
D µ
(4.34)
The eigenvalues ωn > 0 have finite multiplicity and tend to infinity as n → ∞. We normalize
vn by kvn kµ,ε = 1 for all n ∈ p
N, where the norm k · kµ,ε had been defined in (4.13b). Then
the sets {vn : n ∈ N} and { 1 + ωn2 vn : n ∈ N} form complete orthonormal systems in
V0,ε , (·, ·)µ,ε and in Ṽ0,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 of
vn ∈ .V0,ε and the definition of V0,ε . From (4.34) we conclude for ψ = vm that
δn,m = (vn , vm )µ,ε = (ωn2 + 1) (εvn , vm )L2 (D)
p
which shows that the set { 1 + ωn2 vn : n ∈ N} forms an orthonormal systems in Ṽ0,ε , (ε·, ·)L2 (D) .
It remains to prove completeness of this system. But this follows from the fact that V0,ε is
dense in Ṽ0,ε . To see this latter denseness result let v ∈ Ṽ0,ε . By the denseness of C0∞ (D, C3 ) in
L2 (D, C3 ) (see Lemma 4.9) there exists a sequence vn ∈ C0∞ (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 ) = Ṽ0,ε + H0 (curl 0) are identical: vn = vn0 + ṽn with vn0 ∈ H0 (curl 0, D) and
ṽn ∈ V0,ε . The boundedness of the projections yields vn0 → 0 in L2 (D, C3 ) and thus ṽn → 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 orthonormal 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 corresponding magnetic fields form a complete orthonormal system in the closed subspace Vµ of
H(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 h·, ·i.
Lemma 4.35 Let ωn > 0 and {vn : n ∈ N} ⊆ V0,ε be the complete orthonormal system of
V0,ε , defined by (4.34). Then wn := ωn1 µ curl 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 the
inner product (·, ·)ε,µ , defined in (4.13a). Furthermore, wn satisfies
Z 1
2
curl wn · curl ψ − ωn µ wn · ψ dx = 0 for all ψ ∈ H(curl, D) .
(4.35)
D ε
172
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Proof: Substituting the definition of wn into (4.34) yields ωn wn , curl ψ L2 (D) =
ωn2 ε vn , ψ L2 (D) for all ψ ∈ H0 (curl, D), that is wn ∈ H(curl, D) and curl wn = ωn εvn by
the definition of the variational curl. Furthermore, wn ∈ Vµ because for φ ∈ H(curl 0, D) we
conclude that ωn (µ wn , φ)L2 (D) = (curl vn , φ)L2 (D) = (vn , curl φ)L2 (D) = 0 by applying Green’s
theorem in the form (6.16b). To show (4.35) let ψ ∈ H(curl, D). Then
ε−1 curl wn , curl ψ L2 (D) = ωn (vn , curl ψ)L2 (D) = ωn (curl vn , ψ)L2 (D) = ωn2 (µ wn , ψ)L2 (D) .
Furthermore, from
(wn , wm )ε,µ =
ε−1 curl wn , curl wm
L2 (D)
= ωn ωm (ε vn , vm )L2 (D) +
=
1
ωn2
+
2
1 + ωn 1 + ωn2
+ (µ wn , wm )L2 (D)
1
µ−1 curl vn , curl vm L2 (D)
ωn ωm
δnm = δnm
we see that the system {wn : n ∈ N} forms an orthonormal set in Vµ . Also completeness in
Vµ 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 curl wn , curl ψ L2 (D) + (µ wn , ψ)L2 (D)
= ωn (vn , curl ψ)L2 (D) +
=
1
ωn +
ωn
1
(curl vn , ψ)L2 (D)
ωn
(vn , curl ψ)L2 (D)
for all n ∈ N. Defining φ = ε−1 curl ψ we note that (ε vn , φ)L2 (D) = 0 for all n ∈ N. The
p
1 + ωn2 vn : n ∈ N in Ṽ0,ε would yield φ = 0 provided
completeness of the system
we had shown that φ ∈ Ṽ0,ε . But this is the case because for ρ ∈ H0 (curl 0, D) we have
(ε φ, ρ)L2 (D) = (curl ψ, ρ)L2 (D) = (ψ, curl ρ)L2 (D) = 0. Therefore, curl ψ = 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
∞
∞
X
X
Tv =
αn wn for v =
αn vn ∈ V0,ε
n=1
because of
kT vk2ε,µ =
n=1
∞
X
αn2 = kvk2µ,ε .
n=1
Then compactness of Vµ follows from the compactness of V0,ε by Theorem 4.24.
2
4.2. THE CAVITY PROBLEM
4.2.2
173
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) with
a(x) ≥ a0 for some a0 > 0 we have uniqueness by Green’s theorem. Indeed, let u ∈ H01 (D)
be a solution of (4.14) in the variational form of Definition 4.25 for f = 0. Substituting
ψ = u into the integral yields
Z
[a |∇u|2 − k 2 b |u|] dx = 0 .
D
R
Taking the imaginary part yields D Im b |u|2 dx = 0 and thus u = 0 because of the assumption 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 same
argument, u vanishes on U . We want to prove that this implies u = 0 in all of D. This property of a differential equation is called the unique continuation property. If u was analytic
this would follow from an analytic continuation argument which is well known from complex
analysis. 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 situation for the boundary value problem (4.21a), (4.21b) for the Maxwell system. As before, we
assume first that the medium is conducting; that is, σ > 0 in D. By the definition (4.24) of
εc this is equivalent to the assumption that the imaginary part of εc is strictly positive on
D. Let E ∈ H0 (curl, D) be a solution of the homogeneous equation. Inserting v = E yields
Z 2
1 2
2
curl E − ω εc |E| dx = 0 ,
µ
D
and thus, taking the imaginary part,
Z
Im εc |E|2 dx = 0
D
from which E = 0 follows because Im εc > 0 on D. Again, if only σ > 0 a.e. on some open
subset 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) and
the 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 spaces
of periodic functions.
For k ∈ N and an open domain D ⊆ R3 the Sobolev space H k (D) is defined as before by
requiring that all partial derivatives up to order k exist in the variational sense and are
L2 −functions (compare with Definition 4.1). The space C ∞ (D) of smooth functions is dense
in H k (D). Again, H0k (D) is the closure of C0∞ (D) with respect to the norm in H k (D).
k
Let Q = (−R, R)3 ⊆ R3 be a cube containing D in its interior. Then we define Hper
(Q) by
(
)
X
k
Hper
(Q) = u ∈ L2 (Q) :
(1 + |n|2 )k |un |2 < ∞
n∈Z3
174
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
with inner product
3
(u, v)Hper
k (Q) = (2R)
X
(1 + |n|2 )k un vn ,
n∈Z3
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):
k
H0k (Q) ,→ Hper
(Q) ,→ H k (Q) .
Proof: First inclusion: We show this by induction with respect to k. For k = 0 there is
k
(Q) and
nothing to show. Assume that the assertion is true for k ≥ 0. Then H0k (Q) ⊆ Hper
there exists c > 0 such that
X
2
(2R)3
(1 + |n|2 )k |un |2 = kuk2Hper
for all u ∈ H0k (Q) .
(4.36)
k (Q) ≤ ckukH k (Q)
n∈Z3
Let u ∈ C0∞ (Q). Then, by partial differentiation (if nj 6= 0),
un
1
=
(2R)3
Z
π
−i R
n·x
u(x) e
1 iR
dx =
(2R)3 π nj
Q
= −
Z
u(x)
∂ −i π n·x
e R dx
∂xj
Q
1 iR
(2R)3 π nj
Z
π
i R (j)
∂u
(x) e−i R n·x dx = −
d ,
∂xj
π nj n
Q
(j)
where dn 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,
kuk2Hper
= (2R)3
k+1
(Q)
X
(1 + |n|2 )k+1 |un |2 = (2R)3
n∈Z3
X
(1 + |n|2 )k |un |2
n∈Z3
3 X
X
2
R2
3
(1 + |n|2 )k d(j)
+ 2 (2R)
n
π
3
j=1
n∈Z
≤ ckuk2H k (Q) + c
3
R2 X
k∂u/∂xj k2H k (Q) ≤ c0 kuk2H k+1 (Q) .
2
π j=1
This proves boundedness of the imbedding with respect to the norm of order k + 1. This
ends the proof of the first inclusion because C0∞ (Q) is dense in H0k+1 (Q).
k
For the second inclusion we truncate the Fourier series of u ∈ Hper
(Q) into
uN (x) =
X
|n|≤N
π
un ei R n·x
4.2. THE CAVITY PROBLEM
175
and compute directly
π j1 +j2 +j3 X
π
∂ j1 +j2 +j3 uN
(x)
=
i
un nj11 nj22 nj33 ei R n·x
j3
j2
j1
R
∂x1 ∂x2 ∂x3
|n|≤N
and thus for j ∈ N3 with j1 + j2 + j3 ≤ k:
j +j +j N 2
∂ 1 2 3u ∂xj1 ∂xj2 ∂xj3 1
2
3
3
≤ (2R)
π 2k X
R
L2 (Q)
3
≤ (2R)
≤
|n|≤N
π 2k X
R
π 2k
R
|un |2 |n1 |2j1 |n2 |2j2 |n3 |2j3
|un |2 |n|2(j1 +j2 +j3 )
|n|≤N
kuN k2Hper
k (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 ∈ H 1 (D) be a solution of the variational equation
Z
Z
∇u · ∇ψ dx =
f ψ dx for all ψ ∈ C0∞ (D) .
D
(4.37)
D
Then u|U ∈ H 2 (U ) and ∆u = −f in U .
(b) Let u ∈ L2 (D) be a solution of the variational equation
Z
Z
u ∆ψ dx = −
f ψ dx for all ψ ∈ C0∞ (D) .
D
(4.38)
D
Then u|U ∈ H 2 (U ) and ∆u = −f in U .
Proof: For both parts we restrict the problem to a periodic problem in a cube by using a
partition of unity. Indeed, let ρ > 0 such that
Sρ < 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
S
j
finitely many open balls B3 (xj , ρ) ⊆ D for xj ∈ U , j = 1, . . . , m, with U ⊆ m
j=1 B3 (x , ρ).
For abbreviation P
we set Bj = B3 (xj , ρ). We choose a partition of unity; that is, ϕj ∈ C0∞ (Bj )
with ϕj ≥ 0 and m
U.
j=1 ϕj (x) = 1 for all x ∈P
1
Let now uj (x) = ϕj (x)u(x), x ∈ D. Then m
j=1 uj = u on U and uj ∈ H0 (D) has support
in Bj .
176
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Proof of (a): For ψ ∈ C0∞ (D) and any j ∈ {1, . . . , m} we have that
Z
Z
∇uj · ∇ψ dx =
ϕj ∇u · ∇ψ + u ∇ϕj · ∇ψ dx
D
D
Z
=
∇u · ∇(ϕj ψ) − ψ ∇u · ∇ϕj − ψ u ∆ϕj − ψ ∇u · ∇ϕj dx
ϕj f − 2 ∇u · ∇ϕj − u ∆ϕj ψ dx
D
Z
=
D
Z
=
gj ψ dx with gj = ϕj f − 2 ∇u · ∇ϕj − u ∆ϕj ∈ L2 (D) .
D
Because the support of ϕj is contained in Bj this equation restricts to
Z
Z
∇uj · ∇ψ dx =
gj ψ dx for all ψ ∈ H01 (D) .
Bj
(4.39)
Bj
Let now R > 0 such that D ⊆ Q = (−R, R)3 . We fix j ∈ {1, . . . , m} and ` ∈ Z and set
ψ` (x) = exp(−i(π/R) ` · x). Because Bj ⊆ D we can find a function ψ̃` ∈ C0∞ (D) with
ψ̃` = ψ` on Bj . Substituting ψ̃` into (4.39) yields
Z
Z
∇uj · ∇ψ` dx =
gj ψ` dx .
(4.40)
Bj
Bj
We can extend the integrals to Q because uj and gj vanish outside of Bj . Now we expand
uj and gj into Fourier series of the form
X
X
π
π
uj (x) =
an ei R n·x and gj (x) =
bn ei R n·x
n∈Z
n∈Z
and substitute this into equation (4.40). This yields
Z
π 2 X
X Z
i(n−`)·x
dx =
bn
ei(n−`)·x dx .
an n · ` e
R n∈Z
Q
Q
n∈Z
From theP
orthogonality of the functions exp(i(π/R)
n·x) we conclude that (π/R)2 a` |`|2 = b` .
P
4
2
2
< ∞; that is, uj ∈ Hper
(Q) ⊆
Because `∈Z |b` |2 < ∞Pwe conclude that
`∈Z |`| |a` | P
m
m
2
2
2
H (Q). Therefore, also j=1 uj ∈ H (Q) and thus u|U = j=1 uj |U ∈ H (U ). This proves
part (a).
Proof of (b): We proceed very similarly and show that u ∈ H 1 (D). Then part (a) applies
and yields the assertion. For ψ ∈ C0∞ (D) and any j ∈ {1, . . . , m} we have that
Z
Z
uj ∆ψ dx =
u ϕj ∆ψ dx
D
D
Z
=
u ∆(ϕj ψ) − 2 ∇ϕj · ∇ψ − ψ ∆ϕj dx
D
Z
=
f ϕj − u ∆ϕj ψ dx − 2
D
Z
gj ψ dx −
D
u ∇ϕj · ∇ψ dx
D
Z
=
Z
Fj · ∇ψ dx
D
4.2. THE CAVITY PROBLEM
177
with gj = f ϕj − 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 C0∞ (D). With the
Fourier series
X
X
X
π
π
π
uj (x) =
bn ei R n·x and Fj (x) =
cn ei R n·x
an ei R n·x and gj (x) =
n∈Z
n∈Z
n∈Z
we conclude that
−
π 2
R
a` |`|2 = b` + i
π
` · c`
R
for all ` ∈ Z ,
P
P
and thus |`|2 |a` | ≤ c |b` | + |`| |c` | which proves that `∈Z (1 + |`|2 ) |a` |2 ≤ c̃ `∈Z |b` |2 +
1
(Q) ⊆ H 1 (Q).
2
|c` |2 < ∞ and thus uj ∈ Hper
Remarks:
(a) If f ∈ H k (D) for some k ∈ N then u|U ∈ H k+2 (U ) by the same arguments, applied
iteratively. Indeed, we have just shown it for k = 0. If it is true for k − 1 and if
f ∈ H k (D) then gj ∈ H k (D) (note that uBj ∈ H0k+1 (Bj ) by assumption of induction!)
and thus uj ∈ H k+2 (D).
(b) This theorem holds without any regularity assumptions on the boundary ∂D. Without
further assumptions on ∂D and the boundary data u|∂D we cannot assure that u ∈
H 2 (D).
The proof of the following fundamental result is taken from [6], Section 8.3. The proof itself
goes back to Müller [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 ∈
H 2 (D) be real valued such that
|∆uj | ≤ c
m
X
|u` | + |∇u` |
in D for j = 1, . . . , m .
(4.41)
`=1
If uj vanish in some open set B ⊆ D for all j = 1, . . . , m, then uj vanish identically in D
for all j = 1, . . . , m.
Proof: Let x0 ∈ B and R ∈ (0, 1) such that B3 [x0 , R] ⊆ D. We show that uj vanishes in
B3 [x0 , R/2]. This is sufficient because for every point x̂ ∈ D we can find finitely many balls
B3 (x` , R` ) ` = 0, . . . , p, with R` ∈ (0, 1), such that B3 (x` , R` /2) ∩ B3 (x`−1 , R`−1 /2) 6= ∅ for
all ` = 1, . . . , p and x̂ ∈ B3 (xp , Rp /2). Then one concludes uj (x̂) = 0 for all j by iteratively
applying the first step.
We choose the coordinate system such that x0 = 0. First we fix j and write u for uj . Let
178
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
ϕ ∈ C0∞ B3 (0, R) with ϕ(x) = 1 for |x| ≤ R/2 und define û, v̂ ∈ H02 (D) by û(x) = ϕ(x)u(x)
and
exp(|x|−n ) û(x) , x 6= 0 ,
v̂(x) =
0,
x = 0,
for some n ∈ N. Note that, indeed, v̂ ∈ H 2 (D) because û vanishes in the neighborhood B
of x0 = 0. Then, with r = |x|,
∆û(x) = exp(−r
−n
) ∆v̂(x) +
2n ∂v̂
n n
(x) + n+2 n − n + 1 v̂(x) .
rn+1 ∂r
r
r
Using the inequality (a + b)2 ≥ 4ab and calling the middle term in the above expression b,
we see that
n
o
2
n n
8n exp(−2r−n ) ∂v̂
(x)
∆v̂(x)
+
−
n
+
1
v̂(x)
.
∆û(x) ≥
rn+1
∂r
rn+2 rn
From now on we drop the argument x. Multiplication with exp(2r−n ) rn+2 and integration
yields
Z
Z
o
∂v̂ n
n n
−n
n+2
2
exp(2r ) r (∆û) dx ≥ 8n r
∆v̂ + n+2 n − n + 1 v̂ dx .
(4.42)
∂r
r
r
D
D
We show that
Z
1
∂v̂
r (x) ∆v̂(x) dx =
∂r
2
D
Z
Z
|∇v̂(x)|2 dx and
(4.43a)
D
1
∂v̂
m−2
v̂(x)
(x)
dx
=
rm
∂r
2
D
Z
v̂(x)2
dx for any integer m .
rm+1
(4.43b)
D
3 P
∂v̂
Indeed, proving the first equation we note that r ∂v̂
=
x
·
∇v̂
and
∇
(x
·
∇v̂)
=
+
e(j) ∂x
∂r
j
j=1
∂v̂
xj ∇ ∂x
and thus
j
2
∇ (x · ∇v̂) · ∇v̂ = |∇v̂| +
3
X
xj ∇
j=1
∂v̂
1
· ∇v̂ = |∇v̂|2 + x · ∇ |∇v̂|2 .
∂xj
2
Since all boundary terms vanish, by partial integration we obtain
Z
Z
Z
∂v̂
∂v̂
1
r
∆v̂ dx = − ∇ r
· ∇v̂ dx = − |∇v̂|2 + x · ∇ |∇v̂|2 dx
∂r
∂r
2
D
D
Z
= −
D
D
1
|∇v̂| dx +
2
2
Z
D
1
div
| {zx} |∇v̂| dx = 2
2
= 3
Z
D
|∇v̂|2 dx .
4.2. THE CAVITY PROBLEM
179
This proves equation (4.43a). For equation (4.43b) we have, using polar coordinates and
partial integration,
Z
Z
1 ∂v̂
dx =
v̂
rm ∂r
D
1 ∂v̂
dx =
v̂
rm ∂r
= −
v̂ ∂v̂ 2
r ds dr
rm ∂r
0 S2
B3 (0,R)
ZR Z
ZR Z
∂
v̂
∂r
1
rm−2
Z
v̂ ds dr = −
0 S2
Z
1
rm−2
v̂
1
dx
r2
D
1 ∂v̂
dx + (m − 2)
v̂
rm ∂r
= −
∂
v̂
∂r
Z
v̂ 2
rm+1
dx
D
D
which proves equation (4.43b). Substituting (4.43a) and (4.43b) for m = 2n+1 and m = n+1
into the inequality (4.42) yields
Z
exp(2r
−n
)r
n+2
Z
2
(∆û) dx ≥ 4n
D
2
v̂ 2
Z
3
|∇v̂| dx + 4n (2n − 1)
D
r2n+2
dx
D
− 4n2 (n − 1)2
Z
v̂ 2
rn+2
dx
D
Z
≥ 4n
2
2
2
Z
|∇v̂| dx + 4n (n + n − 1)
D
v̂ 2
r2n+2
dx
D
where in the last step we used r ≤ R ≤ 1. Now we replace the right hand side again by û
again. With v̂(x) = exp(r−n )û(x) we have
h
i
x
∇v̂(x) = exp(r−n ) ∇û(x) − n r−n−1 û(x)
r
and thus, using |a − b|2 ≥ 12 |a|2 − |b|2 for any vectors a, b ∈ R3 ,
2
2
2
1
−n
2
−2n−2
∇v̂(x) ≥ exp(2r )
∇û(x) − n r
û(x) .
2
180
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Substituting this into the estimate above yields
Z
exp(2r−n ) rn+2 (∆û)2 dx
(4.44)
D
Z
≥ 2n
exp(2r−n ) |∇û|2 dx +
D
2
2
+ 4n (n + n − 1) − 4n
3
Z
exp(2r
−n
)
û2
r2n+2
dx
D
Z
exp(2r
= 2n
−n
2
2
Z
2
) |∇û| dx + 4n (n − 1)
û2
r2n+2
dx
D
D
Z
≥ 2n
exp(2r−n )
exp(2r
−n
2
) |∇û| dx + 2n
D
4
Z
exp(2r
−n
)
û2
rn+2
dx
(4.45)
D
for n ≥ 2. Up to now we have not used the estimate (4.41). We write now uj and ûj for u
and û, respectively. From this estimate, the inequality of Cauchy-Schwarz, and the estimate
(a + b)2 ≤ 2a2 + 2b2 we have the following estimate
m
X
R
∆ûj (x)2 = ∆uj (x)2 ≤ 2m c2
|u` (x)|2 + |∇u` (x)|2
for |x| < .
2
`=1
Therefore, from (4.45) for uj we conclude that
Z
Z
u2j
−n
2
4
−n
2n
exp(2r ) |∇uj | dx + 2n
exp(2r ) 2n+2 dx
r
|x|≤R/2
Z
≤
|x|≤R/2
exp(2r−n ) rn+2 (∆ûj )2 dx
D
2
≤ 2mc
Z
m
X
`=1
exp(2r
−n
)r
n+2
2
|u` | + |∇u` |2 dx +
|x|≤R/2
Z
exp(2r−n )rn+2 (∆ûj )2 dx .
R/2≤|x|≤R
−n
)
for r > 0 and note that ψn is monotonously decreasing. Also,
We set ψn (r) = exp(2r
r2n+2
because r ≤ 1 (and thus rn+2 ≤ r−2n−2 ),
Z
Z
−n
2
4
2n
exp(2r ) |∇uj | dx + 2n
ψn (r) u2j dx
|x|≤R/2
≤ 2mc2
|x|≤R/2


m
X Z
`=1


|x|≤R/2
ψn (r) u2` dx +
Z
|x|≤R/2
exp(2r−n ) |∇u` |2 dx





+ ψn (R/2)k∆ûj k2L2 (D) .
4.2. THE CAVITY PROBLEM
181
Now we sum with respect to j = 1, . . . , m and combine the matching terms. This yields
Z
Z
m
m
X
X
2 2
−n
2
4
2 2
2(n − m c )
exp(2r ) |∇uj | dx + 2(n − m c )
ψn (r) u2j dx
j=1
j=1
|x|≤R/2
≤ ψn (R/2)
m
X
|x|≤R/2
k∆ûj k2L2 (D) .
j=1
2 2
For n ≥ m c we conclude that
4
2 2
2(n − m c ) ψn (R/2)
Z
m
X
j=1
u2j
4
2 2
dx ≤ 2(n − m c )
Z
m
X
j=1
|x|≤R/2
≤ ψn (R/2)
m
X
ψn (r) u2j dx
|x|≤R/2
k∆ûj k2L2 (D)
j=1
and thus
m
Z
m
X
j=1
u2j
|x|≤R/2
X
1
k∆ûj k2L2 (D)
dx ≤
2(n4 − m2 c2 ) j=1
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 boundary
value 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 ∈ C 1 (D) be real valued and a ≥ a0 on D
for some a0 > 0. Furthermore, let b ∈ L∞ (D) be complex valued with Im b ≥ 0 on D and
Im b > 0 a.e. on some open subset U of D. Then the boundary value problem (4.14) has a
unique solution u ∈ H01 (D) for every f ∈ L2 (D).
Proof: By Theorem 4.27 it suffices to prove uniqueness. Let u ∈ H01 (D) be a solution for
f = 0. We have seen above that u vanishes on U , and it remains to show that u vanishes
everywhere on D. In view of Theorem 4.39 we have first to show that u ∈ H 2 (V ) for every
domain V with V ⊆ D.
Let ϕ ∈ C0∞ (D) and define ψ = a1 ϕ on D. Then ψ ∈ C01 (D) and ∇ϕ = a∇ψ + ψ∇a.
Substituting a∇ψ and ψ this into the variational equation yields
Z
1
b
∇u · ∇ϕ − ϕ ∇a · ∇u − k 2 u ϕ dx = 0
a
a
D
R
R
that is, D ∇u · ∇ϕ dx = D f ϕ dx for all ϕ ∈ C0∞ (D) where f = a1 ∇a · ∇u + k 2 ab u. Then
f ∈ L2 (D), and by Theorem 4.38 we conclude that u ∈ H 2 (V ) for any open set with V ⊆ D.
Furthermore, from the differential equation we have the estimate
|∆u(x)| ≤
1
kbk∞
k∇ak∞ |∇u(x)| + k 2
|u(x)| on V .
a0
a0
182
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
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 with
U ⊆ V and V ⊆ D yields that u vanishes in such domains V . This implies that u vanishes
in D and ends the proof.
2
We now turn to the Maxwell case and transform Maxwell’s equations to the vector Helmholtz
equation. 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 ∈ C 1 (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 the
corresponding solution of (4.25). We have shown at the beginning of this subsection that E
1
curl E = 0 in U . By Lemma 4.29, the magnetic field
vanishes on U . Therefore, also H = iωµ
H satisfies
Z 1
2
curl H · curl ψ − ω µ H · ψ dx = 0 for all ψ ∈ H(curl, D) .
(4.46)
εc
D
If we choose ψ = ∇φ for some φ ∈ H01 (D) then we have
Z
H · ∇φ dx = 0 for all φ ∈ H01 (D) ,
D
which is the variational form of div H = 0.
If all of the functions were sufficiently smooth we just would rewrite the equation for H in
the form
1
2
2
0 = εc curl ε−1
× curl H − ω 2 εc µ H ,
c curl H − ω εc µ H = curl H + εc ∇
εc
and thus, because div H = 0 and curl2 = ∇ div −∆,
∆H = εc ∇
1
1
× curl H − ω 2 εc µ H =
curl H × ∇εc − ω 2 εc µ H .
εc
εc
We derive this formula also by the variational equation. Indeed, we set ψ = εc ψ̃ for some
ψ̃ ∈ C0∞ (D, C3 ). Then ψ̃ ∈ H0 (curl, D) and therefore, because curl ψ = εc curl ψ̃ + ∇εc × ψ̃,
Z 1
2
curl H · curl ψ̃ +
curl H · (∇εc × ψ̃) − ω µεc H · ψ̃ dx = 0 for all ψ̃ ∈ C0∞ (D, C3 ) ,
εc
D
which we write as
Z
Z
curl H · curl ψ̃ dx =
D
D
G · ψ̃ dx for all ψ̃ ∈ C0∞ (D, C3 ) ,
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
183
where G = − ε1c curl H × ∇εc + ω 2 µεc H ∈ L2 (D, C3 ). Partial integration yields
Z
Z
Z
Z
Z
2
G · ψ̃ dx = H · curl ψ̃ dx = − H · ∆ψ̃ dx +
H · ∇ div ψ̃ dx = − H · ∆ψ̃ dx .
D
D
D
D
D
R
Here we used the fact that D H · ∇ div ψ̃ dx = 0 because div ψ̃ ∈ H01 (D). This holds for
all ψ̃ ∈ C0∞ (D, C3 ). By the interior regularity result of Theorem 4.38 we conclude that
H ∈ H 2 (U, C3 ) for all domains U with U ⊆ D, and ∆H = −G = ε1c curl H × ∇εc − ω 2 µεc H
in U . Because every component of curl H is a combination of partial derivatives of H` for
` ∈ {1, 2, 3} we conclude the existence of a constant c > 0 such that
3
X
|∆Hj | ≤ c
|∇H` | + |H` | in D for j = 1, 2, 3 .
`=1
Therefore, all of the assumptions of Theorem 4.39 are satisfied and thus H = 0 in all of
U . This implies that also E = 0 in U . Because U is an arbitrary domain with U ⊆ D we
conclude that E = 0 in D.
2
Remarks:
(a) The proof of the theorem can be modified for µ ∈ C 2 (D). Instead of div H = 0 we
have that 0 = div(µH) = ∇µ · H + µ div H, thus
1
2
∇µ · H
curl H = −∆H + ∇ div H = −∆H − ∇
µ
= −∆H − ∇ ∇(ln µ) · H
3 X
∂ ln µ
∂ ln µ
= −∆H −
∇
Hj +
∇Hj
∂x
∂x
j
j
j=1
and this can be treated in the same way. Here we argue classically, but all of the
arguments hold also in the weak case.
(b) The asumption εc ∈ C 1 (D) is very restrictive. One can weaken this assumption to
the requirement that εc is piecewise continuously differentiable. We refer to [17], Section 4.6.
4.3
The Time–Dependent Cavity Problem
The spectral theorem of the previous section allow it to treat the full time–dependent system
of Maxwell’s equations. We begin again with the initial–boundary value problem for the
scalar wave equation in some bounded Lipschitz domain D ⊆ R3 and some interval (0, T ).
1 ∂ 2u
(t,
x)
−
div
a(x)
∇u(t,
x)
= f (t, x) ,
c(x)2 ∂t2
(t, x) ∈ (0, T ) × D ,
(4.47a)
184
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
u(t, x) = 0 for (t, x) ∈ (0, T ) × ∂D ,
(4.47b)
∂u
(0, x) = u1 (x) for x ∈ D .
∂t
We make the following assumptions on the data:
u(0, x) = u0 (x) and
(4.47c)
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 ∈ H01 (D), u1 ∈ L2 (D).
In this section we assume that all functions are real-valued. We set b(x) = 1/c(x)2 for
abbreviation. Then b ∈ L∞ (D) and b(x) ≥ b0 = 1/kck2∞ on D.
The solution has to be understood in a variational sense. To motivate this we multiply the
differential equation (4.47a) by some ψ ∈ C 1 ([0, T ] × D) with ψ(0, x) = ψ(T, x) = 0 for all
x ∈ D and integrate by parts with respect to t and use Green’s first formula with respect to
x. This yields
ZT Z
ZT Z ∂ψ
∂u
(t, x) − a(x) ∇u(t, x) · ∇ψ(t, x) dx dt = −
f (t, x) ψ(t, x) dx dt ,
b(x) (t, x)
∂t
∂t
0
0
D
D
or, using the notation ut = ∂u/∂t and the inner product in L2 (D),
ZT h
i
b ut (t, ·), ψt (t, ·) L2 (D) − a∇u(t, ·), ∇ψ(t, ·) L2 (D) dt = −
0
ZT
f (t, ·), ψ(t, ·) L2 (D) dt .
0
We will require different smoothness properties of u with respect to t and x. This leads to
so called anisotropic function spaces. In particular, the solution has to be differentiable with
respect to t (in a sense to be explained in a moment). It is convenient to consider u to be a
function in t with values u(t) in some function space with respect to x. We have taken this
point of view already when we wrote u(t, ·). To make this idea precise, we recall the notion
of a Frechét-differentiable function for this case.
Definition 4.42 Let V be a normed space (over R) and f : [0, T ] → V a function with
values in V .
(a) The function f is continuous in some t0 ∈ [0, T ] if lim kf (t) − f (t0 )kV = 0. The space
t→t0
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 0 (t0 ) ∈ V if
(t0 )
lim f (t)−f
− f 0 (t0 ) = 0. The space of continuously differentiable functions is
t−t0
t→t0
denoted by C 1 [0, T ; V ].
V
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
185
Remark: If V is a Hilbert space with inner product (·, ·)V then the following
product
1
rule hold. For f, g ∈ C [0, T ; V ] the scalar function h(t) = f (t), g(t) V , t ∈ [0, T ], is
differentiable and
h0 (t) = f 0 (t), g(t) V + f (t), g 0 (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 ; H01 (D) ∩ C 1 0, T ; L2 (D)
(4.48a)
and equip X with the weighted norm
√
√
kukX = max k a ∇u(t)kL2 (D) + max k b u0 (t)kL2 (D) .
(4.48b)
0≤t≤T
0≤t≤T
Lemma 4.43 X is a Banach space. The norm is equivalent to
u 7→ max ku(t)kH 1 (D) + max ku0 (t)kL2 (D) .
0≤t≤T
0≤t≤T
√
Proof: The equivalence is clear√because k a ∇ukL2 (D) is equivalent to the ordinary norm
in H01 (D) by Lemma 4.26 and k b · kL2 (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. Only
the proof of completeness is a little more delicate, but follows the same arguments as in the
proofs of the completeness of C[0, T ] and C 1 [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 , u0 (0) = u1 and
ZT h
i
b u (t), ψ (t) L2 (D) − a∇u(t), ∇ψ(t) L2 (D) dt = −
0
0
0
ZT
f (t, ·), ψ(t) L2 (D) dt
(4.49)
0
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
Therefore, the right–hand side of (4.49) is well defined.
RT
0
kf (t, ·)k2L2 (D) dt = kf kL2 ((0,T )×D) .
The following analysis makes use of the spectral theorem (Theorem 4.28). We recall from
this theorem that there exist eigenvalues kn ∈ R and corresponding eigenfunctions {vn ∈
H01 (D) : n ∈ N} such that
Z
a ∇vn · ∇ϕ − kn2 b vn ϕ dx = 0 for all ϕ ∈ H01 (D) ;
D
that is, using again the notion of the inner product (u, v)∗ =
(vn , ϕ)∗ = kn2 (b vn , ϕ)L2 (D)
R
D
a ∇u · ∇v dx in H01 (D),
for all ϕ ∈ H01 (D) .
(4.50)
186
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Furthermore, the set {vn ∈ H01 (D) : n ∈ N} forms a complete orthonormal system in
H01 (D), (·, ·)∗ , and
{kn vn : n ∈ N} forms a complete orthonormal system in
2
2
L (D), (b·, ·)L (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
u0 = 0,
u solves the problem for
1
u1 = 0, and f = 0. We define cn (t) by cn (t) = u(t), vn ∗ for n ∈ N. Then cn ∈ C [0, T ]. We
choose any ϕ ∈ C 1 [0, T ] such that ϕ(0) = ϕ(T ) = 0 and set ψ(t) = ϕ(t)vm for an arbitrary
m ∈ N. We compute the inner products in (4.49) by using the orthogonality of vn with
respect to (·, ·)∗ and of kn vn with respect to (b ·, ·)L2 (D) . This yields
1 0
c (t) ϕ0 (t) ,
2 m
km
u(t), ψ(t) ∗ = ϕ(t) u(t), vm ∗ = cm (t) ϕ(t) .
b u0 (t), ψ 0 (t) L2 (D) = ϕ0 (t) b u0 (t), vm L2 (D) =
a∇u(t), ∇ψ(t) L2 (D) =
Inserting this into (4.49) yields
Z
T
0
1 0
0
c (t) ϕ (t) − cm (t) ϕ(t) dt = 0
2 m
km
for all such ϕ and all m. Now we use Lemma 4.46 below which yields that cm ∈ C 2 [0, T ] and
c00m (t) + kn2 cm (t) = 0 on [0, T ]. Using the initial conditions cm (0) = 0 and c0m (0) = 0 yields
cm (t) = 0 for all t. This holds for all m ∈ N. The completeness of the system {vn : n ∈ N}
implies that u(t) vanishes for all t.
2
It remains to prove the following lemma which is sometimes called the Fundamental Theorem
of Calculus of Variations.
Lemma 4.46 Let h ∈ C 1 [0, T ] and g ∈ C[0, T ] such that
T
Z
h0 (t) ϕ0 (t) − g(t) ϕ(t) dt = 0 for all ϕ ∈ C 1 [0, T ] with ϕ(0) = ϕ(T ) = 0 .
0
Then h ∈ C 2 [0, T ] and h00 (t) + g(t) = 0 for all t ∈ [0, T ] .
Rt
Proof: Define g̃(t) = 0 g(s)ds for t ∈ [0, T ]. Then g̃ ∈ C 1 [0, T ]. We substitute g into the
variational equation and use partial integration. This yields
Z
0
T
0
h (t) + g̃(t) ϕ0 (t) dt =
| {z }
=: c(t)
Z
0
T
c(t) ϕ0 (t) dt = 0 for all ϕ ∈ C 1 [0, T ] with ϕ(0) = ϕ(T ) = 0 .
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
187
RT
Rt
Let now ρ ∈ C[0, T ] be arbitrary. Define ϕ(t) = 0 ρ(s)ds − Tt 0 ρ(s)ds for t ∈ [0, T ]. Then
ϕ ∈ C 1 [0, T ] with ϕ(0) = ϕ(T ) = 0. Therefore,
Z T
Z T
Z
Z T
1 T
0
0 =
c(t) ϕ (t) dt =
c(t) ρ(t) dt −
ρ(s) ds
c(t) dt
T 0
0
0
0
Z T
Z
1 T
=
ρ(t) c(t) −
c(s) ds dt.
T 0
0
RT
Because this holds for all ρ ∈ C[0, T ] we conclude that c(t) = T1 0 c(s) ds for all t. Therefore,
c is constant and thus h0 = −g̃ + c ∈ C 1 [0, T ] and h00 (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 all
ψ ∈ H01 (D) the scalar function t 7→ b u(t), ψ L2 (D) is twice continuously differentiable and
d2
b
u(t),
ψ
+
L2 (D)
dt2
a ∇u(t), ∇ψ
L2 (D)
=
f (t, ·), ψ
L2 (D)
(4.51)
for all t ∈ [0, T ].
Proof: Let ψ ∈ H01 (D) and ϕ ∈ C 1 [0, T ] with ϕ(0) = ϕ(T ) = 0. We insert ϕψ into (4.49)
which yields
ZT ZT
d
ϕ0 (t)
b u(t), ψ L2 (D) − ϕ(t) a∇u(t), ∇ψ) L2 (D) dt = − ϕ(t) f (t, ·), ψ L2 (D) dt .
dt
0
0
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}.
≤ γn for all t ∈ [0, T ] and n ∈ N where γn > 0 with
Let
c
∈
C[0,
T
]
and
γ
>
0
with
c
(t)
n
n
P∞ n 2
n=1 γn < ∞. Define formally
u(t) =
∞
X
cn (t) hn ,
t ∈ [0, T ] .
n=1
Then u ∈ C[0, T ; H] and kuk2C[0,T ;H] ≤
P∞
n=1
γn2 .
Proof: Let uN be the truncated series; that is,
uN (t) =
N
X
n=1
cn (t) hn ,
t ∈ [0, T ] .
188
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Then uN ∈ C[0, T ; H]. We show that (uN ) is a Cauchy sequence in C[0, T ; H]. Indeed, for
N > M we have
N
N
X
X
2
2
kuN (t) − uM (t)kH =
cn (t) ≤
γn2 .
n=M +1
n=M +1
Taking the maximum yields
kuN − uM k2C[0,T ;H] ≤
N
X
γn2
n=M +1
and this tends to zero as N, M tend to infinity. Therefore, (uN ) is a Cauchy sequence in
C[0, T ; H] and thus convergent.
2
We will apply this result first to the case where H = H01 (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 {kn vn : n ∈ N}. This yields the following
corollary:
Corollary 4.49 Let cn ∈ C 1 [0, T ] and P
γn > 0 with cn (t) ≤ γn and c0n (t) ≤ kn γn for all
2
t ∈ [0, T ] and n ∈ N where γn > 0 with ∞
n=1 γn < ∞. Define formaly
u(t) =
∞
X
cn (t) vn ,
t ∈ [0, T ] .
n=1
Then u ∈ X and kuk2X ≤ 2
P∞
n=1
γn2 .
Proof: Application of the previous lemma to cn in H01 (D), (·, ·)∗ with respect to the
orthonormal system {vn : n ∈ N} yields that u ∈ C[0, T ; H01 (D)] and kuk2C[0,T ;H 1 (D)] ≤
0
P∞ 2
c0n
0
2
2 (D)
.
Then
we
apply
the
lemma
to
u
with
coefficients
γ
in
L
(D),
(b·,
·)
with
L
n=1 n
kn
0
2
respect to the orthonormal system {kn vn : n ∈ N}. This proves that u ∈ C[0, T ; L (D)] and
P
|c0n |
2
ku0 k2C[0,T ;L2 (D)] ≤ ∞
n=1 γn because also 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 ∈ H01 (D) and u1 ∈ L2 (D) there exists a unique solution u of
(4.49) for f = 0 such that u(0) = u0 and u0 (0) = u1 . The solution is given by
∞
X
u(t) =
αn cos(kn t) + βn sin(kn t) vn ,
t ∈ [0, T ] ,
n=1
where αn and βn are the expansion coefficients of u0 ∈ H01 (D) and u1 ∈ L2 (D) with respect
to {vn : n ∈ N} and {kn vn : n ∈ N}, respectively; that is,
u0 =
∞
X
n=1
αn vn ,
u1 =
∞
X
βn kn vn .
n=1
Furthermore, the solution operator (u0 , u1 ) → u is bounded from H01 (D) × L2 (D) into X.
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
189
Proof: To show that u ∈ X we apply the previous corollary with cn (t) = αn cos(kn t) +
βn sin(kn t). The assumptions are obviously satisfied because cn (t)2 ≤ γn2 := 2αn2 + 2βn2 for
all t P
∈ [0,T ] and n
∈ N and, analogously, |c0n (t)| ≤ kn γn for all t ∈ [0, T ] and n ∈ N
2
2
ku k2 + ku1 k2L2 (D,b dx) . Application of Corollary 4.49 yields u ∈ X.
and ∞
n=1 αn + βn =
P∞ 0 ∗
P
Furthermore, u(0) = n=1 αn vn = u0 and u0 (0) = ∞
n=1 kn βn vn = u1 .PIt remains to prove
that 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 {kn vn : n ∈ N} with respect
to (b·, ·)L2 (D) we conclude
ZT h
0
∞ ZT i
X
1 0
0
0
0
bu (t), ψ (t) L2 (D) − u(t), ψ(t) ∗ dt =
c (t)ψn (t) − cn (t)ψn (t) dt ,
kn2 n
n=1
0
where again cn (t) = αn cos(kn t)+βn sin(kn t) are the expansion coefficients of u(t). Let n ∈ N
be fixed. Using partial integration we conclude
ZT 0
ZT 1 0
1 00
0
c (t)ψn (t) − cn (t)ψn (t) dt = −
c (t) + cn (t) ψn (t) dt = 0
kn2 n
kn2 n
0
2
by the special form of cn (t). Therefore, u satisfies (4.49) for f = 0.
Let now f ∈ L2 (0, T ) × D be arbitrary. We construct a particular solution
of the inho
1
2
mogeneous variational equation (4.49). We note that b f ∈ L (0, T ) × D and, therefore,
t 7→ 1b f (t, ·) can be
expanded with respect to the orthonormal system {kn vn : n ∈ N} in
2
L (D), (b·, ·)L2 (D) ; that is,
∞
X
1
f (t, ·) =
fn (t) kn vn
b(·)
n=1
with fn (t) =
f (t, ·), kn vn
L2 (D)
, n ∈ N.
(4.52)
2
Theorem
4.51
Let
f
∈
L
(0,
T
)
×
D
with coefficients fn (t) of (4.52). Then û(t) =
P∞
n=1 an (t) vn , t ∈ [0, T ], with coefficients
Z
1 T
an (t) = −
sin kn |t − s| fn (s) ds , t ∈ [0, T ] , n ∈ N ,
(4.53)
2 0
Furthermore, the operator f → û is bounded from
is a particular solution of (4.49).
L2 (0, T ) × D into X.
Proof: First we note that û ∈ X. Indeed, we observe that an ∈ C 1 [0, T ] and |an (t)|2 ≤
R T
2
RT
1
|f (s)| ds ≤ T4 0 |fn (s)|2 ds for all t ∈ [0, T ] and n ∈ N and analogously |a0n (t)|2 ≤
4 0R n
2T
T
kn
|fn (s)|2 ds for all t ∈ [0, T ] and n ∈ N. Therefore, û ∈ X by Corollary 4.49 and
4
0
Z TX
XZ T
2
2
kûkX ≤ c
|fn (s)| ds = c
|fn (s)|2 ds
0
n
Z
= c
0
0
n
T
kf (s, ·)k2L2 (D) ds = kf k2L2 ((0,T )×D) .
190
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
P∞
Let now ψ ∈ X with ψ(0) = ψ(T ) = 0 and expansion ψ(t) =
n=1 ψn (t)vn . We fix n
and assume first that fn is continuous in [0, T ]. Then it is easy to check that even an ∈
C 2 [0, T ] and an satisfies the differential equation a00n (t) + kn2 an (t) = −kn fn (t). Multiplying
this equation with ψn ∈ C 1 [0, T ], integrating and using integration by parts (note that
ψn (0) = ψn (T ) = 0), yields
Z T
Z T
0
2
0
fn (t) ψn (t) dt .
an (t)ψn (t) − kn an (t) ψn (t) dt = kn
0
0
By a density argument we conclude that this equation holds also if only fn ∈ L2 (0, T ).
Division by kn2 and summing these equations with respect to n yields
Z T
∞ Z T ∞
X
X
1 0
1
0
fn (t) ψn (t) dt ,
a (t)ψn (t) − an (t) ψn (t) dt =
2 n
k
k
n
0
0
n
n=1
n=1
which can again be written as
Z
Z T
0
0
bû (t), ψ (t) L2 (D) − û(t), ψ(t) ∗ dt =
0
0
T
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 ∈ H01 (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 u0 (0) = u1 . The solution is
given by the sum u = û + ũ of the particular solution û constructed in Theorem 4.51 and the
solution ũ of the homogeneous differential equation with initial values ũ(0) = u0 − û(0) and
ũ0 (0) = u1 − û0 (0); that is,
∞ X
a0n (0)
u(t) =
αn − an (0) cos(kn t) + βn −
sin(kn t) + an (t) vn
(4.54)
k
n
n=1
for t ∈ [0, T ] where αn and βn are the expansion coefficients of u0 ∈ H01 (D) and u1 ∈ L2 (D)
with respect to {vn : n ∈ N} and {kn vn : 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 solution
operator (u0 , u1 , f ) 7→ u is bounded from H01 (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 ∈ C 1 [0, T ; L2 (D)] and u0 = 0 and u1 ∈ H01 (D) and u ∈ X be the
solution of (4.49). Then u0 ∈ C 0, T ; H01 (D) and u00 ∈ C 0, T ; L2 (D) ; that is, u0 ∈ X.
1
2
Furthermore, the
mapping
(u
,
f
)
→
7
u
is
bounded
from
H
(D)
×
L
(0,
T
)
×
D
into X̃ =
1
0
1
1
2
2
C 0, T ; H0 (D) ∩ C 0, T ; L (D) equipped with the canonical norm analogously to (4.48b).
The solution satisfies the differential equation pointwise with respect to t; that is
b u00 (t), ψ L2 (D) + a ∇u(t), ∇ψ L2 (D) = f (t, ·), ψ L2 (D)
for all ψ ∈ H01 (D) and all t ∈ [0, T ], compare (4.51).
(4.55)
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
191
Proof: First we consider the particular solution û of Theorem 4.51 and write (4.53) as
Z t
Z T
1
sin(kn (t − s)) fn (s) ds +
sin(kn (s − t)) fn (s) ds
an (t) = −
2 0
t
"
t Z t
1
= −
cos(kn (t − s)) fn0 (s) ds
cos(kn (t − s)) fn (s) −
2kn
0
0
#
T
Z T
cos(kn (s − t)) fn0 (s) ds
− cos(kn (s − t)) fn (s) +
t
t
and
a0n (t)
Z t
Z T
kn
= −
cos(kn (t − s)) fn (s) ds −
cos(kn (s − t)) fn (s) ds
2
0
t
"
t Z t
1
sin(kn (t − s)) fn0 (s) ds
= − − sin(kn (t − s)) fn (s) +
2
0
0
#
T
Z T
− sin(kn (s − t)) fn (s) +
sin(kn (s − t)) fn0 (s) ds
t
t
and thus
c1
c2
kfn k∞ + kfn0 kL2 (0,T ) ≤
kfn kL2 (0,T ) + kfn0 kL2 (0,T ) ,
kn
kn
|a0n (t)| ≤ c2 kfn kL2 (0,T ) + kfn0 kL2 (0,T ) ,
|a00n (t)| = |kn2 an (t) + kn fn (t)| ≤ c3 kn kfn kL2 (0,T ) + kfn0 kL2 (0,T )
|an (t)| ≤
for all t ∈ [0, T ]. Here we used the estimate max0≤t≤T |ϕ(t)|2 ≤ 2 max{T, 1/T } kϕk2L2 (0,T ) +
kϕ0 k2L2 (0,T ) for any function ϕ ∈ C 1 [0, T ] (see Exercise 4.10). Differentiating (4.54) for the
case αn = 0 yields
u0 (t) =
∞
X
kn an (0) sin(kn t) +
kn βn − a0n (0) cos(kn t) + a0n (t) vn ,
n=1
u00 (t) =
∞
X
2
kn an (0) cos(kn t) +
kn a0n (0) − kn2 βn sin(kn t) + a00n (t) vn .
n=1
Now
from the above estimates of an (t), a0n (t), and a00n (t) and the fact that
P∞ we2 observe
2
2
n=1 kn βn = ku1 kH01 (D) that the coefficients of this series satisfy the assumptions of Corollary 4.49. This shows that u0 ∈ X.
2
Equation (4.51) follows from the observation that dtd 2 b u(t), ψ L2 (D) = b u00 (t), ψ L2 (D) be
cause u ∈ C 2 0, T ; L2 (D) .
2
We finish this part with an explicit example which shows that one can not weaken the
assumptions on f for deriving the C 1 −regularity of u.
192
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Example 4.54 Let D = (0, π)3 ⊆ R3 and a = c = 1 and T > 2. The Dirichlet eigenvalues
of −∆ in D are given by kn = |n|, n ∈ N3 , with corresponding eigenfunctions
vn (x) =
3
Y
1
sin(nj xj ) ,
(π/2)3/2 |n| j=1
x ∈ D , n ∈ N3 .
They p
are normalized such that kvn k∗ = k∇vn kL2 (D) = 1 and kvn kL2 (D) = 1/|n|. We set
ρn = |n|2 − |n| for n ∈ N3 and define f by
f (t, x) =
X
fn sin(ρn t) |n| vn (x) ,
t ∈ [0, T ] , x ∈ D ,
n∈N3
where the coefficients fn are such that
P
n
fn2 < ∞. Then
X
kf (t, ·)k2L2 (D) =
fn2 sin2 (ρn t) ,
t ∈ [0, T ] ,
n∈N3
P
2
because {|n|vn : n ∈ N3 } is an orthonormal system in L2 (D). By kf (t, ·)k2L2 (D) ≤
n∈N3 fn
we observe that f ∈ C[0, T ; L2 (D)]. The solution of the initial boundary value problem with
u0 = u1 = 0 is given by
X
ρn
sin(|n|t) vn (x) , t ∈ [0, T ] , x ∈ D ,
u(t, x) =
fn sin(ρn t) −
|n|
3
n∈N
as one checks directly by term-by-term differentiation, which can be made rigorously by
investigating the convergence. The normalization of vn yields
ku(t, ·)k2∗
X
=
fn2
n∈N3
2
ρn
sin(|n|t) ,
sin(ρn t) −
|n|
t ∈ [0, T ] .
If u ∈ C 1 [0, T ; H01 (D)] then
u0 (t) =
X
X
2
fn ρn cos(ρn t) − cos(|n|t) vn and ku0 (t)k2∗ =
fn2 ρ2n cos(ρn t) − cos(|n|t)
n∈N3
n∈N3
for t ∈ [0, T ]. Integration yields
Z
T
0
ku
(t)k2∗
dt =
0
X
fn2 ρ2n
"
=
n∈N3
cos2 (ρn t) + cos2 (|n|t) − 2 cos(ρn t) cos(|n|t) dt
0
n∈N3
X
T
Z
fn2 ρ2n
sin(2ρn T ) sin(2|n|T ) sin (ρn + |n|)T
+
−
T+
4ρn
4|n|
ρn + |n|
sin (|n| − ρn )T
−
|n| − ρn
#
.
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
193
Now we observe that |n| − ρn = 21 + O(1/|n|). Therefore we can choose ε > 0 such that
2 + ε < T and then N so large such that
1
1
1
1
+
+
+
≤ 2 + ε for |n| ≥ N .
4ρn 4|n| ρn + |n| |n| − ρn
This yields
Z
T
ku0 (t)k2∗ dt ≥ (T − 2 − ε)
0
X
fn2 ρ2n
|n|≥N
which implies that f ∈ C 1 [0, T ; L2 (D)].
After the scalar wave equation we consider now the time–dependent Maxwell system; that
is,
∂H
(t, x) = 0 , (t, x) ∈ (0, T ) × D ,
∂t
∂E
curl H(t, x) − ε(x)
(t, x) = Je (t, x) , (t, x) ∈ (0, T ) × D ,
∂t
curl E(t, x) + µ(x)
(4.56a)
(4.56b)
with boundary conditions
ν(x) × E(t, x) = 0 for (t, x) ∈ (0, T ) × ∂D ,
(4.56c)
E(0, x) = e0 (x) and H(0, x) = h0 (x) for x ∈ D .
(4.56d)
and initial conditions
Again we need some conditions on the electrical parameter. Therefore, we make the following
assumptions 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) ∩ C 1 0, T ; L2 (D, C3 ) ,
Xh = C 0, T ; H(curl, D) ∩ C 1 0, T ; L2 (D, C3 ) ,
(4.57a)
(4.57b)
and equip X with the product norm k(E, H)kX = kEk + kHk with
kuk = max ku(t)kH(curl,D) + max ku0 (t)kL2 (D)
0≤t≤T
where u = E or u = H.
0≤t≤T
(4.57c)
194
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
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
curl E(t) + µ H 0 (t) = 0 for all t ∈ [0, T ] ,
(4.58a)
curl H(t) − ε E 0 (t) = Je (t, ·)
(4.58b)
for all t ∈ [0, T ] .
We observe that the solution space is not symmetric with respect to E and H because of the
boundary condition for E. The following analysis makes use of the Helmholtz decompositions
of Theorem 4.21; that is,
L2 (D, C3 ) = H0 (curl 0, D) ⊕ Ṽ0,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 curl ψ = 0 ,
V0,ε = u ∈ H0 (curl, D) : (εu, ψ)L2 (D) = 0 for all ψ ∈ H0 (curl, D) with curl ψ = 0 ,
Ṽ0,ε = u ∈ L2 (D, C3 ) : (εu, ψ)L2 (D) = 0 for all ψ ∈ H0 (curl, D) with curl ψ = 0 .
We recall (see Remark 4.22) that V0,ε and Vµ are just the orthogonal complements of
H0 (curl 0, D) and H(curl 0, D), respectively, with respect to the inner products
(u, v)µ,ε = (µ−1 curl u, curl v)L2 (D) + (ε u, v)L2 (D) ,
(u, v)ε,µ = (ε−1 curl u, curl v)L2 (D) + (µ u, v)L2 (D) ,
respectively, see (4.13a), (4.13b). We note that both norms are equivalent to the ordinary
norm in H(curl, D). Analogously, Ṽ0,ε is the orthogonal complement of H0 (curl 0, D) in
L2 (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 the
existence of an infinite number of eigenvalues ωn > 0 and corresponding eigenfunctions
vn ∈ V0,ε such that
µ−1 curl vn , curl ψ L2 (D) = ωn2 ε vn , ψ L2 (D) for all ψ ∈ H0 (curl, D) .
(4.59)
The functions vn are normalized such that kvn k2µ,ε = kµ−1/2 curl vn k2L2 (D) + kε1/2 vn k2L2 (D) = 1
for all n ∈ N. Then we saw that the system {vn : n ∈ N}
system
pis a complete orthonormal
1 + ωn2 vn : n ∈ N is a complete
in V0,ε with respect to the inner product (·, ·)µ,ε and
orthonormal system in Ṽ0,ε with respect to (εv, w)L2 (D) . Furthermore, we saw in Lemma 4.35
that wn = ωn1 µ curl 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 prove
uniqueness.
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
195
Theorem 4.56 There exists at most one solution of (4.58a), (4.58b) with E(0) = e0 and
H(0) = h0 .
Proof: Let (E, H) be the difference of two solutions. Then (E, H) solves the system for
Je = 0, e0 = 0, and h0 = 0. Let first φ ∈ H(curl 0, D). Then, from (4.58a),
d
d
curl E(t), φ L2 (D) +
µ H(t), φ L2 (D) =
µ H(t), φ L2 (D)
dt
dt
for all t because curl E(t), φ L2 (D) = E(t), curl φ L2 (D) = 0. Therefore, µ H(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) by
wn for some n and have
0 =
0 = curl E(t), wn
= E(t), curl wn
= ωn ε E(t), vn
L2 (D)
L2 (D)
L2 (D)
d
µ H(t), wn L2 (D)
dt
d
µ H(t), wn L2 (D)
+
dt
d
µ H(t), wn L2 (D)
+
dt
+
and analogously
d
ε E(t), vn L2 (D)
dt
d
ε E(t), vn L2 (D)
= H(t), curl vn L2 (D) −
dt
d
ε E(t), vn L2 (D) .
= ωn µ H(t), wn L2 (D) −
dt
Therefore, ε E(t), vn L2 (D) and µ H(t), wn L2 (D) solve a homogeneous linear system of
ordinary differential
equations of
first order with homogeneous initial data. Therefore,
ε E(t), vn L2 (D) and µ H(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 all
t ∈ [0, T ].
2
0 = curl H(t), vn
L2 (D)
−
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 solution
of (4.58a), (4.58b) for Je = 0.
Proof: We decompose the pair (e0 , h0 ) into the sum (e0 , h0 ) = (e00 , h00 ) + (ẽ0 , h̃0 ) with
(e00 , h00 ) ∈ H0 (curl 0, D) × H(curl 0, D) and (ẽ0 , h̃0 ) ∈ V0,ε × Vµ . Then the solution (E, H)
is the sum of the solutions with initial data (e00 , h00 ) and (ẽ0 , h̃0 ), respectively. The solution
196
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
corresponding to initial data (e00 , h00 ) is just the constant (with respect to time t); that is,
E 0 (t) = e00 and H 0 (t) = h00 for all t ∈ [0, T ].
We now construct the solution (Ẽ, H̃) corresponding to initial data (ẽ0 , h̃0 ) ∈ V0,ε × Vµ . We
expand the initial data in the forms
ẽ0 =
∞
X
αn vn ,
h̃0 =
n=1
∞
X
βn w n
n=1
where {vn : n ∈ N} in V0,ε and {wn P
: n ∈ N} in Vµ are the P
orthonormal systems studied
∞
2
2
2
2
above. Then, by Parseval’s identity, n=1 αn = kẽ0 kµ,ε and ∞
n=1 βn = kh̃0 kε,µ . We make
an ansatz in the form
Ẽ(t) =
∞
X
an (t) vn ,
H̃(t) =
n=1
∞
X
bn (t) wn ,
t ∈ [0, T ] ,
n=1
where an , bn ∈ C 1 [0, T ]. Substitution of these series into equations (4.58a), (4.58b) for Je = 0
yields formally
0 =
∞
X
an (t) curl vn + b0n (t) µ wn
=
n=1
0 =
∞
X
∞
X
ωn an (t) + b0n (t) µ wn ,
n=1
bn (t) curl wn − a0n (t) ε vn
n=1
=
∞
X
ωn bn (t) − a0n (t) ε vn .
n=1
The linear independence of the systems {vn : n ∈ N} in V0,ε and {wn : n ∈ N} in Vµ yields
the following system of ordinary differential equations
ωn an (t) + b0n (t) = 0 ,
ωn bn (t) − a0n (t) = 0 ,
t ∈ [0, T ] ,
(4.60)
with initial conditions an (0) = αn , bn (0) = βn . Solving it yields
∞
X
Ẽ(t) =
αn cos(ωn t) + βn sin(ωn t) vn ,
t ∈ [0, T ] ,
n=1
∞
X
H̃(t) =
−αn sin(ωn t) + βn cos(ωn t) wn ,
t ∈ [0, T ] .
n=1
These functions belong to Xe and Xh , respectively. Indeed, by Corollary 4.49 we have to
check uniform estimate of c(t) = αn cos(ωn t) + βn sin(ωn t) and of c(t) = −αn sin(ωn t) +
βn cos(ωn t), respectively, and of there derivatives.
for both cases we have that
P∞Indeed,
2
2
2
2
0
2
|c(t)| ≤ 2αn + 2βn =: γn and |c (t)| ≤ ωn γn and n=1 γn = kẽ0 k2µ,ε + kh̃0 k2ε,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 ) and
some scalar function ρ ∈ C 1 [0, T ].
4.3. THE TIME–DEPENDENT CAVITY PROBLEM
197
Again, we decompose F in the form
F = F 0 + F̃
with F 0 ∈ H0 (curl 0, D) and F̃ ∈ Ṽ0,ε .
A particular solution is obtained as the sum of solutions with right–hand side F 0Rand with F̃ ,
t
respectively. A solution (E 0 , H 0 ) with right–hand side F 0 is simply E 0 (t) = − 0 ρ(s) ds F 0
and H 0 (t) = 0.
Now
consider F̃ ∈ Ṽ0,ε as right–hand side. Then√F̃ has an expansion
the form F̃ =
P∞ we p
P∞ in
2
2 v (convergence in L2 (D)) with k ε F̃ k2
1
+
ω
=
f
.
We make an
f
n n
n=1 n
n=1 n
L2 (D)
ansatz for Ẽ and H̃ in the forms
E(t) =
∞
X
n=1
an (t) vn ,
H̃(t) =
∞
X
bn (t) wn ,
t ∈ [0, T ] ,
n=1
Using the relationship between vn and wn a substitution into the Maxwell system yields,
using the system
p
ωn an (t) + b0n (t) = 0 , ωn bn (t) − a0n (t) = 1 + ωn2 ρ(t) fn . t ∈ [0, T ]
(4.61)
(compare to (4.60)). We can either eliminate an or bn from the system. Because of the
differentiability of ρ we eliminate bn and arrive at
p
1 + ωn2 fn ρ0 (t) , t ∈ [0, T ] .
a00n (t) + ωn2 an (t) =
A solution is given by
p
Z T
1 + ωn2
an (t) =
fn
sin(ωn |t − s|) ρ0 (s) ds , t ∈ [0, T ] .
2 ωn
0
Then an (t) ≤ ckρ0 k∞ |fn | and a0n (t) ≤ c ωn kρ0 k∞ |fn | for all t ∈ [0, T ] and
n∈
N where the
constant c is independent
of n or t. Solving for bn yields the estimates bn (t) ≤ c (kρk∞ +
kρ0 k∞ ) |fn | and b0n (t) ≤ c ωn kρ0 k∞ |fn | for all t ∈ [0, T ] and n ∈ N. Corollary 4.49 yields
(Ẽ, H̃) ∈ X = Xe × Xh and ends the proof.
2
Collecting our results we can formulate the following theorem for the time dependent cavity
problem.
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 ∈ C 1 [0, T ]. Then
there 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 ) ∈ C 1 [0, T ] × L2 (D, C3 )
in the theorem by the assumption (ρ, F ) ∈ C[0, T ] × H0 (curl, D). Then F̃ ∈ V0,ε , and we
solve the system (4.61) for bn and proceed as before.
198
4.4
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Exercises
Exercise 4.1 Let D ⊆ R3 be an open set and u ∈ C(D). Show that the support of u as
defined 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 ; that
is, u(x) = 1 for x ∈ U and u(x) = 0 outside of U . Let φδ ∈ C0∞ (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 C0∞ (R3 ) is dense in H 1 (R3 ). Therefore, H01 (R3 ) coincides with
H 1 (R3 ).
Hint: Use the technique of Lemma 4.18.
Exercise 4.4 Prove that H01 (R3 ) is not compactly imbedded in L2 (R3 ).
Hint: Take a non-vanishing ϕ ∈ C0∞ (R) and some a ∈ R3 , a 6= 0, and discuss the sequence
un (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∗ ∈ D
where a∗ = (a1 , a2 , −a3 )> for any a = (a1 , a2 , a3 )> ∈ C3 . Let D± = {x ∈ D : x3 ≷ 0} and
ϕ ∈ C 1 (D− ) and v ∈ C 1 (D− , C3 ). Extend ϕ and v into D by reflection; that is,
ϕ(x) , x ∈ D− ,
v(x) , x ∈ D− ,
ϕ̃(x) =
ṽ(x) =
∗
+
ϕ(x ) , x ∈ D ,
v ∗ (x∗ ) , x ∈ D+ .
Show that ϕ̃ ∈ H 1 (D) and ṽ ∈ H(curl, D).
Hint: Show that
∇ϕ̃(x) =
∇ϕ(x) ,
x ∈ D− ,
∗ ∗
(∇ϕ) (x ) , x ∈ D+ ,
curl ṽ(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) = 1
for |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 choose
φ 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 H01 (D) is a proper
subspace of H 1 (D).
Hint: Show that functions v with ∆v − v = 0 in D are orthogonal to H01 (D).
4.4. EXERCISES
199
Exercise 4.9 Show that for any v ∈ H01 (D) the extension by zero belongs to H 1 (R3 ); that
is, ṽ ∈ H 1 (R3 ) where ṽ = v in D and ṽ = 0 in R3 \ D.
Exercise 4.10 Prove that |ϕ(t)|2 ≤ 2 max{T, 1/T } kϕk2L2 (0,T ) +kϕ0 k2L2 (0,T ) for all t ∈ [0, T ]
and any function ϕ ∈ C 1 [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 is
this not an obvious fact by our definition of H 1 (D) as the space of restrictions of H 1 (R3 )?
Exercise 4.12 Show that C0∞ (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 compactly
imbedded 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 ) and
H0 (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 of
Variations.
Let u ∈ L1 (0, T ) and v ∈ C[0, T ] such that
Z T
0
ρ (t) u(t) + ρ(t) v(t) dt = 0 for all ρ ∈ C 1 [0, T ] with ρ(T ) = 0 .
0
Then u ∈ C 1 [0, T ] and u(0) = 0 and u0 = v on [0, T ]. If the variational equation holds for
all ρ ∈ C 1 [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 decomposition
by using the results of Chapter 2.
200
CHAPTER 4. THE VARIATIONAL APPROACH TO THE CAVITY PROBLEM
Chapter 5
Boundary Integral Equation Methods
for Lipschitz Domains
For the boundary value problems of Chapters 3 and 4 we made assumptions which are
often not met in applications. Indeed, the classical integral equation methods discussed
in Chapter 3 require smoothness of the boundary ∂D. In case of the cavity problem of
Chapters 4 just a homogeneous boundary condition has been treated. Both restrictions are
connected because if we like to weaken the regularity of the boundary, or if we like to allow
for more general boundary conditions we have to investigate the traces of the functions or
vector fields on the boundary ∂D in detail. Therefore, we continue in Subsections 5.1.1 and
5.1.2 by introducing Sobolev spaces which appear as the range spaces of the trace operators
and prove denseness, trace theorems and compact imbedding results. Finally we use these
results 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 two
dimensional disk of radius α.
Definition 5.1 We call a region D ⊆ R3 to be a Lipschitz domain, if there exists a finite
number of open cylinders Uj of the form Uj = {Rj x + z (j) : x ∈ B2 (0, αj ) × (−2βj , 2βj )}
with z (j) ∈ R3 and rotations Rj ∈ R3×3 and real valued Lipschitz-continuous S
functions ξj ∈
C(B2 [0, αj ]) with |ξj (x1 , x2 )| ≤ βj for all (x1 , x2 ) ∈ B2 [0, αj ] such that ∂D ⊆ m
j=1 Uj and
Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 = ξj (x1 , x2 ) ,
D ∩ Uj = Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 < ξj (x1 , x2 ) ,
Uj \ D = Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 > ξj (x1 , x2 ) .
∂D ∩ Uj =
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 : x21 + x22 < αj2 , |x3 | < 2βj
the cylinders with parameters αj and βj . We can assume without loss of generality that
βj ≥ αj (otherwise split the parameter region into smaller ones). Furthermore, we define
the three-dimensional balls Bj = B3 (0, αj ) ⊆ Cj and introduce the mappings


x1
 + z (j) , x = (x1 , x2 , x3 )> ∈ Bj ,
x2
Ψ̃j (x) = Rj 
ξj (x1 , x2 ) + x3
and their restrictions Ψj to B2 (0, αj ); that is,


x1
 + z (j) ,
x2
Ψj (x̃) = Rj 
ξj (x1 , x2 )
x̃ = (x1 , x2 )> ∈ B2 (0, αj ) .
By Rademacher’s result [21] (see also [9], Section 5.8, and Remark 6.8) we know that Ψj is
differentiable almost everywhere on B2 (0, αj ). This yields
a parametrization
q of ∂D ∩ Uj in
∂Ψj ∂Ψj p
the form y = Ψj (x̃) for x̃ ∈ B2 (0, αj ) with ∂x1 × ∂x2 = 1 + |∇ξj |2 ≤ 1 + L2j where Lj
denotes the Lipschitz constant of ξj .
S
2
0
We set Uj0 = Ψ̃j (Bj ). Then ∂D ⊆ m
j=1 Uj and Bj ∩ (R × {0}) = B2 (0, αj ) × {0}, and
∂D ∩ Uj0 =
Ψ̃j (x) : x ∈ Bj , x3 = 0 = Ψj (x̃) : x̃ ∈ B2 (0, αj ) ,
D ∩ Uj0 = Ψ̃j (x) : x ∈ Bj , x3 < 0 ,
Uj0 \ D = Ψ̃j (x) : x ∈ Bj , x3 > 0 .
Therefore, the mappings Ψ̃j “flatten” the boundary.
We note that the Jacobian Ψ̃0j (x) ∈ R3×3 is given by


1
0
0
0
1
0 
Ψ̃0j (x) = Rj 
∂1 ξj (x̃) ∂2 ξj (x̃) 1
where ∂` ξj = ∂ξj /∂x` for ` = 1, 2 and x̃ = (x1 , x2 ). Therefore, these Jacobians are regular
with constant determinant det Ψ̃0j (x) = 1 and Ψ̃j are isomorphisms from Bj onto Uj0 for every
j = 1, . . . , m.
These parametrizations allow it to transfer the notion of (periodic) Sobolev spaces on two
dimensional planar domains to the boundary ∂D. We begin with Sobolev spaces of scalar
functions in Subsection 5.1.1 and continue with vector-valued functions in Subsection 5.1.2.
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
5.1.1
203
Sobolev Spaces of Scalar Functions
We note first that Green’s formula and the transformation formula hold in the form of
Theorem 6.12 and Remark 6.8, respectively, for Lipschitz domains and sufficiently smooth
functions. This theorem is, e.g. used in the following simple result.
Lemma 5.2 Let D be a bounded Lipschitz domain.
(a) Let u ∈ C 1 (D) with u = 0 on ∂D. Then the extension ũ of u by zero outside of D
yields ũ ∈ H 1 (R3 ) and ∇ũ = ∇u in D and ∇ũ = 0 outside of D.
(b) Let R ∈ R3×3 be an orthogonal matrix and z ∈ R3 . For u ∈ H 1 (D) define v(x) =
u(Rx + z) on V = {x ∈ R3 : Rx + z ∈ D}. Then v ∈ H 1 (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 ∈
C0∞ (R3 ) we have by partial integration:
Z
Z
Z
Z
ũ ∇ψ dx =
u ∇ψ dx = −
ψ ∇u dx +
u ψ ν ds
R3
D
D
∂D
|
{z
}
= 0
Z
Z
ψ ∇u dx = −
= −
ψ g dx .
R3
D
(b) Let ψ ∈ C0∞ (V ). With the substitution y = Rx + z we have
Z
>
R
∇u(y)|y=Rx+z ψ(x) dx
V
= R
Z
>
>
∇u(y) ψ R (y − z) dy = −R
>
= −
u(y) R ∇ψ(x)|x=R> (y−z) dy
D
D
Z
Z
Z
u(y) ∇ψ(x)|x=R> (y−z) dy = −
D
v(x) ∇ψ(x) dx .
V
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 the
space C ∞ (D). Its proof is technically complicated.
Theorem 5.3 Let D be a bounded Lipschitz domain. Then C ∞ (D) is dense in H 1 (D).
Proof: The idea of the proof is to transform the given function u ∈ H 1 (D) onto a cylinder
by using a partition of unity and local coordinates. In a cylinder we construct a smooth
approximation of the transformed function. Finally, we transform back the approximation.
We seperate the proof into four steps.
(i) Let u ∈ H 1 (D) and {Uj , ξj : j = 1, . . . , m} be a local coordinate system. Set U0 = D
204
5 BIE FOR LIPSCHITZ DOMAINS
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, . . . , m
we transform uj by the definition vj (x) = uj (Rj x + z (j) ), x ∈ Cj− , where Cj = B2 (0, αj ) ×
(−2βj , 2βj ) and
Cj− := x = (x̃, x3 ) ∈ Cj : x3 < ξj (x̃)
where again x̃ = (x1 , x2 ). We observe that vj ∈ H 1 (Cj− ) by the chain rule of part (b) of
Lemma 5.2. Furthermore, with
Vj (t) := {x ∈ Cj− : |x̃| > αj − t or x3 < −2βj + t}
we note that there exists δ > 0 with vj (x) = 0 in the neighborhood Vj (2δ) of ∂Cj ∩ Cj−
because the support of ϕj is contained in Uj . The reader should sketch the sets Cj , Cj− , and
Vj (t).
(ii) Let a = (0, 0, L+1)> where L is larger than all of the Lipschitz constants of the functions
ξj and set
Z
ε
vj (x) = (vj ∗ φε )(x − εa) =
vj (y) φε (x − εa − y) dy
Cj− \Vj (2δ)
where φε ∈ C ∞ (R3 ) denotes the mollifier function from (4.7). Its most important properties
had been collected in Theorem 4.7. In particular, we have that vjε ∈ C0∞ (R3 ) and vjε → vj in
L2 (R3 ) as ε tends to zero, the latter property because the zero-extension of vj is in L2 (R3 ).
Next we show that vjε vanishes on Vj (δ) for ε ≤ δ. First we note that the integral in the
definition of vjε is taken over Cj− \Vj (2δ). Therefore, for y ∈ Cj− \Vj (2δ) we have |ỹ| ≤ αj −2δ
and y3 ≥ −2βj + 2δ.
Let first x ∈ Cj− with |x̃| ≥ αj −δ. Then |x−εa−y| ≥ |x̃− ỹ| ≥ |x̃|−|ỹ| ≥ αj −δ −(αj −2δ) =
δ ≥ ε. Therefore, vjε (x) = 0 for these x.
Let second x ∈ Cj− 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, vjε (x) = 0 for these
x.
(iii) We show now that vjε converges to vj even in H 1 (Cj− ). Let ε < δ/(L + 2) and x ∈
Cj− \ Vj (δ). Then
Z
Z
ε
vj (y) ∇x φε (x − εa − y) dy = −
vj (y) ∇y φε (x − εa − y) dy .
∇vj (x) =
Cj−
Cj−
We show that the function y 7→ φε (x − εa − y) belongs to C0∞ (Cj− ). It is sufficient to prove
that the ball B3 (x − εa, ε) is contained in Cj− . Therefore, let y ∈ B3 (x − εa, ε).
Then, first, |ỹ| ≤ |x̃| + |ỹ − x̃| ≤ 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 (x̃) − Lε ≤ ξj (ỹ) + L|ξj (ỹ) − ξj (x̃)| − Lε ≤
ξj (ỹ). This proves that y 7→ φε (x − εa − y) belongs to C0∞ (Cj− ).
By the definition of the weak derivative we have that
Z
ε
∇vj (x) =
∇vj (z) φε (x − εa − z) dz .
(5.1)
Cj−
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
205
For ε < δ/(L + 2) and x ∈ Vj (δ) both sides of (5.1) vanish. Therefore, (5.1) holds for all
x ∈ Cj− and thus (∇vjε )(x) = (gj ∗ φε )(x − εa) for x ∈ Cj− where gj = ∇vj on Cj− and gj = 0
on R3 \ Cj− . By Theorem 4.7 we conclude again that (gj ∗ φε )(x − εa) converges to gj in
L2 (R3 ) and thus ∇vjε |Cj− to ∇vj in L2 (Cj− ). This proves vjε → vj in H 1 (Cj− ) as ε tends to
zero for every j = 1, . . . , m.
(iv) In the last step we transform back the function and set uεj (y) = vjε Rj> (y − z (j) ) for
j = 1, . . . , m and y ∈ Uj ∩ D. We note that uεj ∈ C0∞ (Uj ∩ D) and uεj → ϕj u in H 1 (Uj ∩ D).
This holds for all j = 1, . . . , m. We extend uj by zero into all of D. Finally, we note that
u0 ∈ H 1 (D) with compact support in U0 =PD and thus uε0 = u0 ∗ φε converges to u0 = ϕ0 u
ε
∞
ε
in H 1 (D). To finish the proof we set uε = m
j=0 uj in D and conclude that u ∈ C0 (D) and
P
m
uε converges to j=0 ϕj u = u in H 1 (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 diffeomorphism 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 Jacobians
Ψ0 and (Ψ−1 )0 are essentially bounded; that is, Ψ0 ∈ L∞ (V, R3×3 ) and (Ψ−1 )0 ∈ L∞ (U, R3×3 ).
Then, for u ∈ H 1 (U ) the composition v = u ◦ Ψ belongs to H 1 (V ), and the chain rule holds
in the form
∇v(x) = Ψ0 (x)> ∇u(y)|y=Ψ(x) , x ∈ V .
(5.2)
The mapping u 7→ u ◦ Ψ is bounded from H 1 (U ) into H 1 (V ).
Proof : For u ∈ C ∞ (U ) we have by the transformation formula
Z
Z
2
1
2
dy
|u(y)|2 u Ψ(x) dx =
kvkH 1 (V ) =
0
det Ψ Ψ−1 (y) U
V
≤ c kuk2L2 (U )
and analogously with ∇v(x) = Ψ0 (x)> ∇u(y)|y=Ψ(x) almost everywhere
Z
2
0
2
k∇vkH 1 (V ) ≤ kΨ k∞ ∇u Ψ(x) dx ≤ c kΨ0 k∞ k∇uk2L2 (U ) .
V
Therefore, the mapping u 7→ v = u ◦ Ψ has a bounded extension from H 1 (U ) into H 1 (V ).
Therefore, formula (5.2) extends to all of H 1 (U ).
2
We continue by defining Sobolev spaces of periodic functions as we have done it already in
1
Chapter 4. Definition 4.12 of Hper
(Q3 ) is included in the following definition. For the sake
of 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 scalar
function v : (−π, π)d → C the Fourier coefficients vn ∈ C for n ∈ Zd are defined as
Z
1
vn =
v(y) e−i n·y dy , n ∈ Zd .
(2π)d Q
206
5 BIE FOR LIPSCHITZ DOMAINS
s
Let s ≥ 0 be any real number. The space Hper
(Q) is defined by
(
)
X
s
Hper
(Q) = v ∈ L2 (Q) :
(1 + |n|2 )s |vn |2 < ∞
n∈Z3
with norm
s (Q) = (2π)
kvkHper
3
sX
(1 + |n|2 )s |vn |2 .
n∈Z3
1
We recall from Lemma 4.37 that H01 (Q) ⊆ Hper
(Q) ⊆ H 1 (Q) with bounded inclusions.
1
Therefore, on H01 (Q) the norms of Hper
(Q) and H 1 (Q) are equivalent.
Such a definition of Sobolev spaces of periodic functions is useful to prove, e.g., imbedding
theorems as we did in Chapter 4. We have seen in Theorem 4.14 that H01 (D) is compactly
imbedded 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 < t
s
t
(Q).
(Q) is compactly imbedded in Hper
the space Hper
2
Proof: see Exercise 5.1
The generalization of the definition of Sobolev spaces with respect to any s ∈ R>0 , especially
s = 21 , 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 of
1
(Q3 ) into
trigonometric polynomials P(Q3 ) into P(Q2 ) has a bounded extension from Hper
1/2
Hper (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 :
1/2
1
1
Hper (Q2 ) → Hper
(Q3 ) with γ0 ◦ η = id. In other words, the function u = ηf ∈ Hper
(Q3 )
1/2
coincides with f ∈ Hper (Q2 ) on Q2 × {0}.
1
Proof: Let u ∈P
Hper
(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 x̃ = (x1 , x2 ) and ñ = (n1 , n2 ), respectively. Again | · | denotes the Euclidean norm of
vectors. Then
!
∞
X
X
X
(γ0 u)(x) = u(x̃, 0) =
un ei ñ·x̃ =
vñ ei ñ·x̃
n3 =−∞
ñ∈Z2
|
{z
ñ∈Z2
=: vñ
}
and thus
kγ0 uk2H 1/2 (Q
per
2)
= (2π)3
X
ñ∈Z2
1 + |ñ|2
1/2
|vñ |2 .
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
207
Now we use the following elementary estimate
∞
X
π
1
≤ a
≤ π + 1 for all a ≥ 1
2 + n2
2
a
3
n =−∞
(5.3)
3
which we will show at the end of the proof. Using the inequality of Cauchy–Schwarz and
1/2
the upper estimate for a = 1 + |ñ|2
yields
2
∞
X
1
1/2
1/2 2 1/2
1 + |ñ|2
|vñ |2 = 1 + |ñ|2
u
1
+
|n|
n
1/2 2
1 + |n|
n3 =−∞
∞
X
≤
"
1 + |n|2
|un |2
∞
X
2 1/2
1 + |ñ|
n3 =−∞
n3
∞
X
≤ (π + 1)
1
1 + |ñ|2 + x23
=−∞
#
|un |2 1 + |n|2 .
n3 =−∞
Summation with respect to ñ yields
kγ0 uk2H 1/2 (Q
2)
per
≤ (π + 1) kuk2Hper
1 (Q ) .
3
This holds for all trigonometric polynomials and yields the boundedness of γ0 on all of
1
1
(Q3 ).
(Q3 ) because the space of trigonometric polynomials is dense in Hper
Hper
To showPthe existence of a bounded extension operator η we define the extension ηf for
f (x̃) = ñ∈Z2 fñ eiñ·x̃ by
X
(ηf )(x) =
un ein·x , x ∈ Q3 ,
n∈Z3
where
un
δñ
= fñ
1 + |n|2
"
and δñ =
∞
X
1
1 + |ñ|2 + j 2
j=−∞
#−1
.
1
We first show that ηf ∈ Hper
(Q3 ). We note that
∞
X
|un |2 (1 + |n|2 ) = |fñ |2 δñ2
n3 =−∞
≤
∞
X
p
1
δñ
2
2
p
=
|f
|
1
+
|ñ|
ñ
1 + |n|2
1 + |ñ|2
n3 =−∞
p
2
|fñ |2 1 + |ñ|2
π
by the lower estimate of (5.3). Summing over ñ yields
kηf k2Hper
1 (Q )
3
3
= (2π)
∞
X X
ñ∈Z2 n3 =−∞
=
2
kf k2H 1/2 (Q ) .
2
per
π
|un |2 (1 + |n|2 ) ≤ (2π)3
p
2 X
|fñ |2 1 + |ñ|2
π
2
ñ∈Z
(5.4)
208
5 BIE FOR LIPSCHITZ DOMAINS
1/2
1
Therefore, η : Hper (Q2 ) → Hper
(Q3 ) is well defined and bounded. Furthermore,
(γ0 ηf )(x̃) =
∞
X X
un eiñ·x̃ =
ñ∈Z2 n3 =−∞
X
fñ eiñ·x̃ = f (x̃) .
ñ∈Z2
It remains to show (5.3). We note that
∞
X
∞
X
1
1
1
=
+
2
2
2
2
2
a +j
a
a + j2
j=−∞
j=1
and
N
X
j=1
j
j
ZN
ZN/a
N Z
N Z
X
X
dt
ds
dt
dt
1
1
=
≤
=
=
,
2
2
2
2
2
2
2
2
2
a +j
a
+
j
a
+
t
a
+
t
a
1
+
s
j=1
j=1
j−1
0
j−1
0
where we used the substitution t = as. Analogously,
N
X
j=1
j+1
j+1
N
Z +1
N Z
N Z
X
X
dt
dt
1
1
dt
=
≥
=
=
2
2
2
2
2
2
2
2
a +j
a +j
a +t
a +t
a
j=1
j=1
j
1
j
(NZ
+1)/a
ds
.
1 + s2
1/a
Letting N tend to infinity yields
∞
X
1 π
1
π
− arctan(1/a) ≤
≤
.
2
2
a 2
a +j
2a
j=1
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 H 1/2 (∂D) is given by the following
transformation 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}, corresponding
mappings Ψ̃j from the balls Bj onto Uj0 , and their restrictions Ψj : B2 (0, αj ) → Uj0 ∩ ∂D as
in Definition 5.1.
By Q3 we denote a cube centered at the origin such that all of the balls Bj are contained in
Q3 . Without loss of generality we assume that this is the cube Q3 = (−π, π)3 to make the
notation simpler.
Furthermore, let {φj : j = 1, . . . , m}, be a partition of unity P
on ∂D subordinate to the sets
0
2
2
2
Uj (see Theorem 6.9). For f ∈ L (∂D) we write |f (y)| = m
j=1 φj (y)|f (y)| for y ∈ ∂D,
and thus
Z
m Z
X
2
2
2
f (y) ds =
kf kL2 (∂D) =
φj (y) f (y) ds
∂D
=
j=1
m Z
X
j=1
∂D∩Uj0
2 q
1 + |∇ξj (x)|2 dx
φj Ψj (x) f Ψj (x) B2 (0,αj )
m Z
X
q
f˜j (x)2 1 + |∇ξj (x)|2 dx
=
j=1
Q2
(5.5)
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
209
where
( q
φj Ψj (x) f Ψj (x) , x ∈ B2 (0, αj ) ,
˜
fj (x) :=
(5.6)
0,
x ∈ Q2 \ B2 (0, αj ) .
p
p
1
+
|∇ξ
(x)|
≤
max
1 + k∇ξj k2∞ | we observe
From (5.5) and the estimate 1 ≤
j
j=1,...,m
qP
m
2
˜ 2
that k · kL2 (∂D) is equivalent to
j=1 kfj kL2 (Q2 ) . Therefore, f ∈ L (∂D) if, and only if,
f˜j ∈ L2 (Q2 ) for all j = 1, . . . , m. We extend this definition to define the Sobolev space
H 1/2 (∂D).
Definition 5.8 Let D ⊆ R3 be a Lipschitz domain in the sense of Definition 5.1 with
corresponding local coordinate system {Uj , ξj : j = 1, . . . , m} and corresponding mappings
Ψ̃j from the balls Bj onto Uj0 and their restrictions Ψj : B2 (0, αj ) → Uj0 ∩ ∂D. Furthermore,
let {φj : j = 1, . . . , m}, be a partition of unity on ∂D subordinate to the sets Uj0 (see
Theorem 6.9). Then we define
1/2
H 1/2 (∂D) = f ∈ L2 (∂D) : f˜j ∈ Hper
(Q2 ) for all j = 1, . . . , m
with norm
"
kf kH 1/2 (∂D) =
m
X
#1/2
kf˜j k2H 1/2 (Q
per
2)
,
j=1
where f˜j are given by (5.6), j = 1, . . . , m.
Our definition of the space H 1/2 (∂D) seems to depend on the choice of the local coordinates
ξj and the partition of unity φj . However, we will see in Corollary 5.15 below that the norms
corresponding to two such choices are equivalent.
From now on we assume always that the domain D is a Lipschitz domain in the sense of
Definition 5.1.
We note the following implication of Theorem 5.6.
Corollary 5.9 The space H 1/2 (∂D) is compactly imbedded in L2 (∂D).
Proof: Let (f ` )` be a bounded sequence in H 1/2 (∂D). Then, by definition, (f˜j` )` is bounded
1/2
in Hper (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˜j`k )k of (f˜j` )` which converges
in L2 (Q2 ) for every j = 1, . . . , m as k tends to infinity. By (5.5), applied to the difference
f `k − f `p also (f `k )k converges in L2 (∂D).
2
Definition 5.8 and Theorem 5.7 yield a trace theorem in H 1 (D) for Lipschitz domains.
Theorem 5.10 The trace operator γ0 : C 1 (D) → C(∂D), u 7→ u|∂D , has an extension as
a bounded operator from H 1 (D) to H 1/2 (∂D). Furthermore, γ0 has a bounded right inverse
η : H 1/2 (∂D) → H 1 (D); that is, γ0 (ηf ) = f for all f ∈ H 1/2 (∂D).
210
5 BIE FOR LIPSCHITZ DOMAINS
Proof: Let {Uj , ξj : j = 1, . . . , m} be a local coordinate system of ∂D with corresponding
mappings Ψ̃j from the balls Bj onto Uj0 as in Definition 5.1. Let {φj : j = 1, . . . , m} be a
S
0
1
partition of unity on ∂D subordinate to Uj0 . Set U := m
j=1 Uj and let u ∈ C (D). Set
q
φj Ψ̃j (x) u Ψ̃j (x) , x ∈ Bj−
vj (x) :=
±
±
and extend vj to Q−
3 by zero. Here, Bj = {x ∈ Bj : x3 ≷ 0} and Q3 = {x ∈ Q3 : x3 ≷ 0}.
Furthermore, we extend vj ∈ C(Q−
3 ) into Q3 by even reflection; that is,
vj (x) , x ∈ Q−
3,
ṽj (x) =
∗
vj (x ) , x ∈ Q+
3,
where x∗ = (x1 , x2 , −x3 )> for x = (x1 , x2 , x3 )> ∈ R3 . Then ṽj ∈ H 1 (Q3 ) by Exercise 4.6.
Also, ṽj vanishes in some neighborhood of ∂Q3 . Now we can use the arguments of Lemma 4.37
1
(Q3 ) and that there exists c1 > 0 which is independent of ṽj such that
to show that ṽj ∈ Hper
1 (Q ) ≤ c1 kvj k 1
kṽj kHper
= c1 kvj kH 1 (Bj− ) .
3
H (Q−
3 )
The boundedness of the trace operator of Theorem 5.7 yields
kṽj (·, 0)kHper
≤ c2 kvj kH 1 (Bj− )
1/2
(Q2 )
for all j = 1, . . . , m,
and thus by Definition 5.8
kγ0 uk2H 1/2 (∂D)
=
m
X
kṽj (·, 0)k2H 1/2 (Q
per
2)
j=1
≤
c22
m
X
kvj k2H 1 (B − ) ≤ c kuk2H 1 (D∩U ) ≤ c kuk2H 1 (D) .
j
j=1
This proves boundedness of γ0 .
Now we construct an extension operator η : H 1/2 (∂D) → H 1 (D). For f ∈ H 1/2 (∂D) we
1/2
define f˜j ∈ Hper (Q2 ) for j = 1, . . . , m as in (5.6). By Theorem 5.7 there exist extensions
1
vj ∈ Hper
(Q3 ) of f˜j and the mappings f˜j 7→ vj are bounded for j = 1, . . . , m. Then we set
q
φj (y) vj Ψ̃−1
y ∈ Uj0 .
uj (y) :=
j (y) ,
0
Then uj ∈ H 1 (Uj0 ) for all j and they
Pmvanish in some neighborhood of ∂Uj . We extend uj
3
by zero into all of R and set u = j=1 uj . Then u is an extension of f . Indeed, for fixed
P
y ∈ ∂D let J = j ∈ {1, . . . , m} : y ∈ Uj0 . Then u(y) = j∈J uj (y). Let j ∈ J be fixed;
p
p
that is, y ∈ Uj0 . Then x = Ψ̃−1
(y)
∈
Q
×
{0},
thus
u
(y)
=
φ
(y)
v
(x)
=
φj (y) f˜j (x) =
2
j
j
j
j P
P
φ(y)f (y). Therefore, u(y) = j∈J uj (y) = j∈J f (y) φj (y) = f (y). Furthermore, all the
operations 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
u
S
0
of f into all of R3 (not only into the interior) with support in the neighborhood U = m
U
j=1 j
of ∂D which depends continuously on f .
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
211
A simple conclusion of the trace theorem in combination with the denseness result of Theorem 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
Z
Z
Z
u ∇v dx = −
v ∇u dx +
(γ0 u) (γ0 v) ν ds for all u, v ∈ H 1 (D) .
D
D
∂D
(b) Let Ω ⊆ R3 be a second Lipschitz domain with D ⊆ Ω and let u1 ∈ H 1 (D) and
u2 ∈ H 1 (Ω\D) such that γ0 u1 = γ0 u2 on ∂D. Then the function defined by
u1 (x) ,
x∈D
u(x) =
u2 (x) ,
x ∈ Ω\D
is in H 1 (Ω).
In particular, for u ∈ H 1 (D) with γ0 u = 0 the extension ũ of u by zero outside of D
belongs to H 1 (R3 ).
(c) Let D ⊆ R3 be symmetric with respect to x3 = 0; that is, x ∈ D ⇔ x∗ ∈ D where
z ∗ = (z1 , z2 , −z3 )> for any z = (z1 , z2 , z3 )> ∈ C3 . Let D± = {x ∈ D : x3 ≷ 0} and
ϕ ∈ H 1 (D− ). Extend ϕ into D by reflection; that is,
ϕ(x) , x ∈ D− ,
ϕ̃(x) =
ϕ(x∗ ) , x ∈ D+ ,
Then ϕ̃ ∈ H 1 (D) and the mapping ϕ 7→ ϕ̃ is bounded from H 1 (D− ) into H 1 (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 η̃ : H 1 (D) → H 1 (R3 ) such that η̃|D = u for all
u ∈ H 1 (D) and supp(η̃u) ⊆ U .
Proof: The proof of Theorem 5.10 (see also the remark following that theorem) yields
existence of an operator η from H 1/2 (∂D) into H 1 (R3 ) with γ0 ηf = f on H 1/2 (∂D). Let
φ ∈ C ∞ (R3 ) such that φ = 1 on D and supp(φ) ⊆ U . Define η̃u by
u(x) ,
x ∈ D,
(η̃u)(x) =
φ(x) (ηγ0 u)(x) , x ∈
/ D.
Then γ0 (φ ηγ0 u) = γ0 ηγ0 u = γ0 u, and the previous corollary yields η̃u ∈ H 1 (R3 .
2
As a corollary we have the following imbedding result of Rellich (compare with Theorem 4.14):
212
5 BIE FOR LIPSCHITZ DOMAINS
Theorem 5.13 Let D ⊆ R3 be a bounded Lipschitz domain. Then the imbedding H 1 (D) →
L2 (D) is compact.
Proof: We can just copy the proof of Theorem 4.14 because of the existence of a bounded
extension operator η̃ : H 1 (D) → H 1 (R3 ) of the previous theorem.
2
We recall from Definition 4.10 that the subspace H01 (D) had been defined as the closure of
C0∞ (D) in H 1 (D). The following equivalent characterization will be of essential importance.
Theorem 5.14 The space H01 (D) is the null space N (γ0 ) of the trace operator; that is,
u ∈ H01 (D) if, and only if, γ0 u = 0 on ∂D.
Proof: The inclusion H01 (D) ⊆ N (γ0 ) follows immediately because γ0 u = 0 for all u ∈
C0∞ (D), the boundedness of γ0 and the fact that C0∞ (D) is dense in H01 (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 ∈ H 1 (D) with γ0 u = 0. We extend u by zero into all of R3 . Then u ∈ H 1 (R3 ) by
Corollary 5.11. Let {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 ∈ R3 . Then the support of uj is
contained in Uj ∩D. For j = 1, . . . , m we transform uj by the definition vj (x) = uj (Rj x+z (j) ),
x ∈ R3 , and observe that vj ∈ H 1 (R3 ) by the chain rule of Corollary 5.4, applied to a large
ball containing D. The support of vj is contained in the cylinder Cj = B2 (0, αj )×(−2βj , 2βj ).
Furthermore, with
Cj− := x = (x̃, x3 ) ∈ B2 (0, αj ) × (−2βj , 2βj ) : x3 < ξj (x̃)
and
Vj (t) := {x ∈ Cj− : |x̃| > αj − t or x3 < −2βj + t}
where again x̃ = (x1 , x2 ) we note that there exists δ > 0 with vj (x) = 0 in the neighborhood
Vj (2δ) of ∂Cj ∩ Cj− 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 functions
ξj and set
Z
ε
vj (x) = (vj ∗ φε )(x + εa) =
vj (y) φε (x + εa − y) dy
Cj− \Vj (2δ)
where φε ∈ C ∞ (R3 ) denotes the mollifier function from (4.7). Its most important properties
had been collected in Theorem 4.7. In particular, we have that vjε ∈ C0∞ (R3 ) and vjε → vj in
H 1 (R3 ) as ε tends to zero.
(iii) Next we show that supp(vjε ) ⊆ Cj− for ε < δ/(L + 3). First we note that the integral
in the definition of vjε is taken over Cj− \ Vj (2δ). Therefore, let y ∈ Cj− \ Vj (2δ). Then
|ỹ| ≤ αj − 2δ and y3 ≥ −2βj + 2δ. The boundary of Cj− consists for three parts. We consider
x being in some neighborhood of these parts separately.
For points x ∈ Cj− with |x̃| ≥ αj − δ we have shown vjε (x) = 0 in part (ii) of the proof of
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
213
Theorem 5.3.
Let now x ∈ Cj− 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, vjε (x) = 0
for these x.
Finally, let x ∈ Cj− with x3 ≥ ξj (x̃)−ε and |ỹ−x̃| ≤ ε. Then |x+εa−y| ≥ |x3 +ε(L+2)−y3 | ≥
x3 + ε(L + 2) − y3 ≥ ξj (x̃) − ε + (L + 2)ε − y3 = [ξj (ỹ) − y3 ] − [ξj (ỹ) − ξj (x̃)] + (L + 1)ε ≥
(L+1)ε−L|ỹ− x̃| ≥ ε. Therefore, vjε (x) = 0 also for these x which shows that supp(vjε ) ⊆ Cj−
for ε < δ/(L + 3); that is, vjε ∈ C0∞ (Cj− ).
(iv) In the last step we transform back the function and set uεj (y) = vjε Rj> (y − z (j) ) for
j = 1, . . . , m and y ∈ R3 . We note that uεj ∈ C0∞ (Uj ∩ D) and uεj → ϕj u in H 1 (D). This
holds for all j = 1, . . . , m. Finally, we note that u0 ∈ H 1 (D) with compact support in
the proof we set
U0 = P
D and thus uε0 = u0 ∗ φε converges to u0 = ϕ0 u in H 1 (D). To finishP
m
ε
ε
∞
ε
ε
(D)
and
u
converges
to
in
D
and
conclude
that
u
∈
C
u = m
u
0
j=0 ϕj u = u in
j=0 j
1
H (D).
2
As mentioned before we still have to complete Definition 5.8 by showing that the space
H −1/2 (∂D) does not depend on the choice of the local coordinate system and the partion of
unity.
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 subordinate
to Uj0 .
Proof: First we note that the space C ∞ (D)|∂D = {v|∂D : v ∈ C ∞ (D)} is dense in H 1/2 (∂D).
Indeed, for f ∈ H 1/2 (∂D) we conclude that u = ηf ∈ H 1 (D) can be approximated by a
sequence un ∈ C ∞ (D) with respect to k · kH 1 (D) . But then un |∂D = γ0 un converges to
γ0 u = f in H 1/2 (∂D). Therefore, H 1/2 (∂D) is the completion of C ∞ (D)|∂D with respect to
k·kH 1/2 (∂D) . Furthermore, from the boundedness and surjectivity of γ0 : H 1 (D) → H 1/2 (∂D)
by Theorem 5.14 we observe that the lifted operator [γ0 ] : H 1 (D)/H01 (D) → H 1/2 (∂D) is
a norm-isomorphism from the factor space H 1 (D)/H01 (D) onto H 1/2 (∂D). Therefore, k ·
kH 1/2 (∂D) is equivalent to the norm k[v]k = inf{kv + ψkH 1 (D) : ψ ∈ H01 (D)} in H 1 (D)/H01 (D);
that is, there exist constants c1 , c2 > 0 with c1 k[v]k ≤ kv|∂D kH 1/2 (∂D) ≤ c2 k[v]k for all
v ∈ H 1 (D). The canonical norm in H 1 (D)/H01 (D) depends only on the norm in H 1 (D)
and the subspace H01 (D). Therefore, all of the norms k · kH 1/2 (∂D) originating from different
choices of the coordinate system and partitions of unity are equivalent to each other.
2
1/2
The proof shows
also
that
an
equivalent
norm
in
H
(∂D)
is
given
by
kf
k
=
inf
kukH 1 (D) :
γ0 u = f on ∂D .
Before we turn to the vector valued case we want to discuss the normal derivative ∂u/∂ν
on ∂D which is well defined for u ∈ C 1 (D). It can be considered as the trace of the normal
component of the gradient of u. We denote it by γ1 : C 1 (D) → C(∂D), u 7→ ∂u/∂ν. In
Exercise 5.2 we show for an example that it is not bounded from H 1 (D) into L2 (∂D). (It
214
5 BIE FOR LIPSCHITZ DOMAINS
is not even well-defined as one can show.) However, it defines a bounded operator on the
closed subspace
HD
Z
1
2
1
= u ∈ H (D) :
∇u · ∇ψ − k u ψ dx = 0 for all ψ ∈ H0 (D)
(5.7)
D
of (variational) solutions of the Helmholtz equation into the dual space of H 1/2 (∂D) as we
will see in a moment.
First we recall from Corollary 5.9 that H 1/2 (∂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 H 1/2 (∂D) because |`f (ψ)| ≤ kf kL2 (∂D) kψkL2 (∂D) ≤
c kf kL2 (∂D) kψkH 1/2 (∂D) for all ψ ∈ H 1/2 (∂D). Therefore, if we identify `f with f – as done
by identifying the dual space of L2 (∂D) with itself – then L2 (∂D) can be considered as a
subspace of the dual space H 1/2 (∂D)∗ of H 1/2 (∂D) which we denote by H −1/2 (∂D):
Definition 5.16 Let H −1/2 (∂D) be the dual space of H 1/2 (∂D) equipped with the canonical
norm of a dual space; that is,
k`kH −1/2 (∂D) :=
|h`, ψi∗ |
,
ψ∈H 1/2 (∂D)\{0} kψkH 1/2 (∂D)
sup
` ∈ H −1/2 (∂D) ,
where h`, ψi∗ = `(ψ) denotes the dual form, the evaluation of ` at ψ.
For real valued functions we observe that h`, ψi∗ is just the extension of the L2 −inner product
when we identify `f with f .
Using a local coordinate system and a corresponding partition of unity one can prove that
−1/2
H −1/2 (∂D) can be characterized by the periodic Sobolev space Hper (Q2 ) which is the completion of the space of trigonometric polynomials by the norm
#1/2
"
kvkHper
=
−1/2
(Q2 )
X
(1 + |n|2 )−1/2 |vn |2
.
n
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 ∈ C 2 (D)
with ∆u + k 2 u = 0 in D and ψ ∈ H 1 (D) we have
Z
∂D
∂u
γ0 ψ ds =
∂ν
Z
∇u · ∇ψ − k 2 uψ dx .
D
Since the trace γ0 ψ of ψ is an element of H 1/2 (∂D) the left hand side is a linear functional on
H 1/2 (∂D). The right–hand side is well defined also for u ∈ H 1 (D). Therefore, it is natural
to extend this formula to u, ψ ∈ H 1 (D) and replace the left–hand side by the dual form
h∂u/∂ν, γ0 ψi∗ . This is justified by the following theorem.
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
215
Definition 5.17 (and Theorem) The operator γ1 : HD → H −1/2 (∂D), defined by Green’s
formula; that is,
Z
hγ1 u, ψi∗ =
∇u · ∇ψ̃ − k 2 u ψ̃ dx , ψ ∈ H 1/2 (∂D) ,
(5.8)
D
is well defined and bounded. Here, ψ̃ ∈ H 1 (D) is any extension of ψ into D; that is, γ0 ψ̃ = 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 ψ̃. Indeed, if
1
Theorem5.14, because
ψ̃1 and ψ̃2 are two extensions of ψ then ψ̃R :=
ψ̃1 − ψ̃2 ∈ H0 (D) by
γ0 ψ̃ = 0. With u ∈ HD we conclude that D ∇u · ∇(ψ̃1 − ψ̃2 ) − k 2 u(ψ̃1 − ψ̃2 ) dx = 0. This
shows that the definition of γ1 is independent of the choice of ψ̃.
To show boundedness of γ1 we take ψ̃ = ηψ where η : H 1/2 (∂D) → H 1 (D) denotes the
bounded right inverse of γ0 of Theorem 5.10. Then, by the inequality of Cauchy-Schwarz,
|hγ1 u, ψi∗ | ≤ k∇ukL2 (D) k∇ψ̃kL2 (D) + k 2 kukL2 (D) kψ̃kL2 (D) ≤ max{k 2 , 1}kukH 1 (D) kψ̃kH 1 (D)
= max{k 2 , 1}kukH 1 (D) kηψkH 1 (D) ≤ max{k 2 , 1} kηk kukH 1 (D) kψkH 1/2 (∂D) .
This proves boundedness of γ1 with kγ1 kH −1/2 (∂D) ≤ max{k 2 , 1} kηk kukH 1 (D) .
2
If the region D is of the form D = D1 \ D2 for open sets Dj such that D2 ⊆ D1 then the
boundary ∂D consists of two components ∂D1 and ∂D2 . The spaces H ±1/2 (∂D) can be
written as direct sums H ±1/2 (∂D1 ) × H ±1/2 (∂D2 ), and the trace operators γ0 and γ1 have
two 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 ψ̃ ∈ H 1 (D) which vanish on ∂D3−j .
Remark: Some readers may feel perhaps unhappy with the space HD as the domain of
definition for the trace operator γ1 because it depends on the wave number k. A more
natural way is to define the trace operator on the space
R
R
∃ v ∈ L2 (D) with D ∇u · ∇ψ dx = − D vψ dx
1
1
H (∆, D) := u ∈ H (D) :
for all ψ ∈ H01 (D)
as the space of functions in H 1 (D) for which even ∆u := v ∈ L2 (D). (Note that the
function v is obviously unique.) The space H 1 (∆, D) is equipped with the canonical norm
kukH 1 (∆,D) = kukH 1 (D) + k∆ukL2 (D) . Definition 5.17 should be changed into
Z
hγ1 u, ψi∗ =
∇u · ∇ψ̃ + ψ̃ ∆u dx , ψ ∈ H 1/2 (∂D) .
D
However, we note that HD is a subspace of H 1 (∆, D), and on this subspace the norms
of H 1 (D) and H 1 (∆, D) are equivalent. For the following it is important that we have
boundedness of γ1 in the H 1 (D)−norm, and this is the reason for choosing HD as the
domain of definition.
216
5 BIE FOR LIPSCHITZ DOMAINS
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 space
H(curl, D). It is the aim to prove two trace theorems for this space. To motivate these
traces we consider the integral identity (6.16b); that is,
Z
Z
Z
(v · curl u − u · curl v) dx =
(ν × u) · v ds =
D
∂D
(ν × u) · (ν × v) × ν ds
∂D
for u, v ∈ C 1 (D, C3 ) ∩ C(D, C3 ) and a smooth domain D. Thus, the left–hand side leads
to two possibilities in extending traces of the tangential components of the vector fields
on ∂D namely first, for γt u = ν × u, if we consider v as a test function and, second for
γT v = (ν × v) × ν, if we take u as a test function. Furthermore, taking gradients of the form
v = ∇ϕ for scalar functions ϕ as test functions gives
Z
Z
Z
∇ϕ · curl u dx =
(ν × u) · γT ∇ϕ ds =
(ν × u) · Grad ϕ ds .
D
∂D
∂D
R
From (6.21)) we observe that the right–hand side is just − ∂D Div(ν × u) ϕ ds, see Definition 5.29 below. Thus we observe that the traces of vector fields in H(curl, D) have some
regularity which requires more detailed investigations.
We proceed as for the scalar case and begin with the following simple lemma which corresponds to Lemma 5.2.
Lemma 5.18 Let D be a bounded Lipschitz domain.
(a) Let u ∈ C 1 (D, C3 ) with ν × u = 0 on ∂D. The extension ũ of u by zero outside of D
yields ũ ∈ H(curl, R3 ) and curl ũ = curl u in D and curl ũ = 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) = (det R) R> curl u(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 ∈ C0∞ (U ) and the substitution x = R> (y − z) we have
Z
>
Z
v(x) curl ψ(x) dx =
U
u(Rx + z)> R curl ψ(x) dx
U
Z
= det R
D
u(y)> R curlx ψ(x)|x=R> (y−z) dy .
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
217
Let q (j) , j = 1, 2, 3, be the columns of R. Then
3
3
X
X
(j)
>
>
curly Rψ R (y − z)
= curly
ψj R (y − z) q
=
R∇ψj (x) × q (j)
j=1
j=1
3
X
∂ψj (x) (`)
q × q (j)
=
∂x`
j,`=1
where we have set x = R> (y −z). If det R = 1 then q (1) × q (2) = q (3) and q (2) × q (3) = q (1) and
(3)
(1)
(2)
>
q ×q = q and thus curly Rψ R
= Rcurlx ψ(x)|x=R> (y−z) . If det R = −1 then
(y −z)
>
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
Z
Z
>
v(x) curl ψ(x) dx =
u(y)> curly Rψ R> (y − z) dy
U
D
Z
=
curl u(y)> Rψ R> (y − z) dy
D
Z
= det R
curly u(y)> |y=Rx+z Rψ(x) dx
U
Z
= det R
R> curly u(y)|y=Rx+z
>
ψ(x) dx .
U
This shows that (det R) 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 the
details.
2
As in the scalar case we introduce first some Sobolev spaces of periodic functions. We recall
that every vector valued function v ∈ L2 (Q, Cd ) on a cube Q = (−π, π)d in Rd for d = 2 or
d = 3 has an expansion in the form
X
v(x) =
vn ein·x
n∈Zd
with Fourier coefficients vn ∈ Cd , given by
Z
1
vn =
v(y) e−i n·y dy ,
(2π)d Qd
For d = 3 formal differentiation yields
X
div v(x) = i
n · vn ein·x ,
n∈Z3
curl v(x) = i
n ∈ Zd .
X
n∈Z3
n × vn ein·x .
(5.9)
218
5 BIE FOR LIPSCHITZ DOMAINS
Therefore, curl v ∈ L2 (Q, C3 ) if, and only if,
for the following definition.
P
n∈Z3
|n × vn |2 < ∞. This is the motivation
Definition 5.20 Let again Q3 = (−π, π)3 ⊆ R3 be the cube and Q2 = (−π, π)2 ⊆ R2 be
the square. For any vector function v : Qd → Cd , d = 2 or d = 3, the Fourier coefficients
vn ∈ Cd for n ∈ Zd are defined as
Z
1
v(y) e−i n·y dy , n ∈ Zd .
vn =
d
(2π) Qd
(a) The space Hper (curl, Q3 ) is defined by
(
Hper (curl, Q3 ) =
)
X
v ∈ L2 (Q3 )3 :
|vn |2 + |n × vn |2 < ∞
n∈Z3
with norm
kvkHper (curl,Q3 ) =
sX
|vn |2 + |n × vn |2 .
n∈Z3
s
(b) For any s ∈ R the space Hper
(Div, Q2 ) is defined as the completion of the space


X

T (Q2 , C2 ) :=
vm ei m·x , x ∈ Q2 , vm ∈ C2 , N ∈ N


|m|≤N
of all trigonometric vector polynomials with respect to the norm
sX
s (Div,Q ) =
(1 + |m|2 )s |vm |2 + |m · vm |2 .
kvkHper
2
m∈Z2
s
(c) For any s ∈ R the space Hper
(Curl, Q2 ) is defined as the completion of the space
T (Q2 , C2 ) with respect to the norm
sX
s (Curl,Q ) =
kvkHper
(1 + |m|2 )s |vm |2 + |m × vm |2 ,
2
m∈Z2
where we have set m × a = m1 a2 − m2 a1 for vectors m, a ∈ C2 .
−1/2
Remark: It will be convenient to consider the elements f ∈ Hper (Div, Q2 ) and f ∈
−1/2
Hper (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, we
will 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 .
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
219
Furthermore, let ê = (0, 0, 1)> be the unit normal vector orthogonal to the plane R2 × {0} in
R3 .
−1/2
(a) The trace operator γt : Hper (curl, Q3 ) → Hper (Div, Q2 ), u 7→ ê × u(·, 0) =
>
−u2 (·, 0), u1 (·, 0), 0 , is well-defined and bounded. Furthermore, there exists a bounded
−1/2
right inverse ηt of γt ; that is, a bounded operator ηt : Hper (Div, Q2 ) → Hper (curl, Q3 ) with
γt ◦ ηt = id. In other words, the tangential components ê × u of u = ηt f ∈ Hper (curl, Q3 )
−1/2
coincide with f ∈ Hper (Div, Q2 ) on Q2 × {0}.
−1/2
(b) The trace operator γT : Hper (curl, Q3 ) → Hper (Curl, Q2 ), u 7→ ê × u(·, 0) × ê =
>
u1 (·, 0), u2 (·, 0), 0 , is well-defined and bounded. Furthermore, there exists a bounded
right inverse ηT of γT .
Proof: We restrict ourselves to part (a) and leave the proof of part (b) to the
P reader. iLet
3
3
u ∈ Hper (curl, Q3 ) with Fourier coefficients un ∈ C , n ∈ Z ; that is, u(x) = n∈Z3 un e n·x .
With x̃ = (x1 , x2 ) and ñ = (n1 , n2 ) we observe that
" ∞
#
X X
X
i ñ·x̃
u(x̃, 0) =
un ei ñ·x̃ thus (γt u)(x) =
u⊥
ñ e
ñ∈Z2
n3 =−∞
ñ∈Z2
where
u⊥
ñ
:=
∞
X
ê × un =
n3 =−∞
∞
X
ê × u(ñ,n3 ) ∈ C3 ,
ñ = (n1 , n2 ) ∈ Z2 ,
(5.10)
n3 =−∞
are the Fourier coefficients of γt u. Note that the third component vanishes. We decompose
un in the form
1
1
n
×
(u
×
n)
+
(n · un ) n .
un =
n
|n|2
|n|2
We set for the moment n̂ := n/|n| and vn := un × n and observe that
ê × un =
1
1
ê × (n̂ × vn ) +
(n̂ · un ) (ê × n) ,
|n|
|n|
and thus for fixed ñ 6= (0, 0) by the inequality of Cauchy-Schwarz
⊥ 2
uñ =
∞
2
X
(ê × un )
n3 =−∞
≤
2
2
∞
∞
X
X
1
1
2
ê × (n̂ × vn ) + 2 (n̂ · un ) (ê × n)
n =−∞ |n|
n =−∞ |n|
3
"
≤ 2
3
∞
X
n3
1
|n|2
=−∞
#"
∞
X
n3 =−∞
|vn |2 + |ñ|2
∞
X
n3 =−∞
#
|un |2
(5.11)
220
5 BIE FOR LIPSCHITZ DOMAINS
where we note that |ê × n| = |ñ| does not depend on n3 . We use (5.3) for a = |ñ| with
ñ 6= (0, 0) and arrive at
⊥ 2
uñ ≤ 2(π + 1)
|ñ|
"
∞
X
|n × un |2 + |ñ|2
∞
X
#
|un |2
n3 =−∞
n3 =−∞
provided ñ 6= (0, 0). Now we multiply with (1 + |ñ|2 )−1/2 and sum with respect to ñ 6= (0, 0).
"
X
X
2
≤ 2(π + 1)
(1 + |ñ|2 )−1/2 u⊥
ñ
ñ∈Z2
ñ6=(0,0)
ñ∈Z2
ñ6=(0,0)
≤ 2(π + 1)
#
∞
(1 + |ñ|2 )−1/2 X |n × un |2 + |ñ|2 |un |2
|ñ|
n =−∞
3
X
|n × un |2 + |un |2 .
n∈Z3
Finally, we add the term with ñ = (0, 0). In this case n = n3 ê and thus ê × un =
Therefore,
1
n3
(n × un ).
2
2
∞
X 1
X
X 1 X
X
2
|n × un |2 ,
(ê × un ) = (n × un ) ≤
|n × un | ≤ c1
2
n
n
3
n =−∞
n 6=0 3 n 6=0
n 6=0
n 6=0
3
3
3
with c1 = 2
P∞
1
j=1 j 2 .
3
3
Therefore,
∞
2
∞
X
X
|n × un |2 + 2 |ê × u0 |2 .
ê × un ≤ 2c1
n3 =−∞
n3 =−∞
Adding this to the previous term for ñ 6= (0, 0) yields
X
2
(1 + |ñ|2 )−1/2 |u⊥
ñ | =
ñ∈Z2
2
∞
X
X
2 −1/2 |n × un |2 + |un |2
(1 + |ñ| )
(ê × un ) ≤ c
3
2
X
ñ∈Z
n3 =−∞
n∈Z
(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
X i ñ·x̃
(n1 , n2 , 0)> · u⊥
ñ e
(Div γt u)(x̃, 0) = i
n1 ,n2 ∈Z
3
where again u⊥
ñ ∈ C are the Fourier coefficients of γt u from (5.11) above. For arbitrary
n ∈ Z3 we have from (5.11) with the same notations as above that
(n1 , n2 , 0)> · (ê × un ) = n · (ê × un ) =
1
1
n · ê × (n̂ × vn ) =
(n × ê) · (n̂ × vn )
|n|
|n|
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
221
and thus if (n1 , n2 ) 6= (0, 0)
2
|(n1 , n2 , 0)> · u⊥
ñ |
2
∞
X 1
(n × ê) · (n̂ × vn )
= |n|
n =−∞
3
≤
∞
X
n3
∞
1 X
|n × ê|2 |n̂ × vn |2
2
|n|
=−∞
n =−∞
3
∞
∞
X
π+1 X
≤
|ñ|2 |vn |2 = (π + 1) |ñ|
|un × n|2
|ñ| n =−∞
n =−∞
3
3
where again ñ = (n1 , n2 ). Here we used |n × ê| = |ñ| because the third component of n × ê
vanishes. For ñ = 0 we note that n · (ê × un ) vanishes. Therefore, multiplication with
(1 + |ñ|2 )−1/2 and summation with respect to ñ yields
X
2
(1 + |ñ|2 )−1/2 |(n1 , n2 , 0)> · u⊥
≤ (π + 1)
ñ |
n1 ,n2 ∈Z
X
ñ∈Z2
≤ (π + 1)
X
∞
X
|ñ|
|un × n|2
2
1/2
(1 + |ñ| ) n =−∞
3
|un × n|2 .
n∈Z3
Adding this to (5.12) yields
kγt uk2H −1/2 (Div,Q
per
2)
≤ (c + π + 1)
X
|n × un |2 + |un |2
= (c + π + 1) kuk2Hper (curl,Q3 ) .
n∈Z3
−1/2
Now
a bounded extension operator. Let f ∈ Hper (Q2 ) be given by f (x̃) =
P we construct
iñ·x̃
f
e
with
Fourier coefficients fñ ∈ C3 for which the third components vanish. We
2
ñ
ñ∈Z
define un ∈ C3 for n ∈ Z3 by
δñ
(fñ × an )
un =
1 + |n|2
where again ñ = (n1 , n2 ) and
an


1 2
|n| ê − n3 n , ñ 6= (0, 0) ,
2
=
|ñ|

ê ,
ñ = (0, 0) ,
and δn is given in (5.4) of the proof of Theorem 5.7; that is, δñ =
easily derive the following properties of an :
hP
∞
1
j=−∞ 1+|ñ|2 +j 2
• ê · an = 1 for all n ∈ Z3 ,
• n · an = 0 for all n with ñ 6= (0, 0), and n · an = n3 for all n = n3 ê,
• |an |2 =
|n|2
|ñ|2
for all n with ñ 6= (0, 0) and |an | = 1 for all n = n3 ê.
i−1
. We
222
5 BIE FOR LIPSCHITZ DOMAINS
First we show that (ηt f )(x) = u(x) =
from
∞
∞
X
X
(ê × un ) = δñ
P
n∈Z3
un ein·x defines a right inverse of γt . This follows
1
ê × (fñ × an ) = (ê · an ) fñ = fñ .
2
1
+
|n|
n =−∞
| 3 {z
}
n3 =−∞
= 1
Now we show boundedness of ηt . For ñ 6= (0, 0) we conclude
∞
X
2
|un |
n3 =−∞
∞
X
|an |2
|n|2
2 2 1
≤
= |fñ | δñ 2
(1 + |n|2 )2
|ñ| n =−∞ (1 + |n|2 )2
n3 =−∞
3
"
#
∞
X
δñ
1
2 δñ
= |fñ |2 2 .
δñ
≤ |fñ |
2
2
|ñ|
1 + |n|
|ñ|
n3 =−∞
{z
}
|
|fñ |2 δñ2
∞
X
(5.13)
= 1
Using δñ ≤
2
π
(1 + |ñ|2 )1/2 from estimate (5.3) we arrive at
∞
X
|un |2 ≤
n3 =−∞
1 + |ñ|2
2
4
2 −1/2
2
≤
(1 + |ñ|2 )−1/2 |fñ |2
(1
+
|ñ|
)
|f
|
.
ñ
π
|ñ|2
π
This holds for ñ 6= (0, 0). For ñ = (0, 0) we conclude
∞
X
2
|f0 |2 δ02
|un | ≤
n3 =−∞
∞
X
n3
1
≤ c |f0 |2 .
2 2
(1
+
n
)
3
=−∞
Summing it with respect to ñ yields
X
X
|un |2 ≤ c0
(1 + |ñ|2 )−1/2 |fñ |2 ≤ c0 kf k2H −1/2 (Div,Q2 ) .
ñ∈Z2
n∈Z
Finally, we estimate the norm of curl u(x) = i
n × un =
P
n∈Z3 (n
× un ) ein·x . We have for ñ 6= (0, 0):
δñ
δñ δñ
n × (fñ × an ) =
(n · an ) fñ − (n · fñ ) an = −
(ñ · fñ ) an ,
2
2
1 + |n|
1 + |n|
1 + |n|2
and thus
∞
X
2
|n × un |
2
= |ñ · fñ |
n3 =−∞
δñ2
∞
X
n3
∞
X
|an |2
|n|2
2 2 1
=
|ñ
·
f
|
δ
.
ñ
ñ
(1 + |n|2 )2
|ñ|2 n =−∞ (1 + |n|2 )2
=−∞
3
This expression appeared already in (5.13) above for |fñ |2 instead of |ñ · fñ |2 . For ñ = (0, 0)
δ0
we observe that n × un = 1+n
2 n3 f0 and thus
3
∞
X
n3 =−∞
|n × un |2 = δ02 |f0 |2
∞
X
n3
n23
≤ c |f0 |2 .
2 2
(1
+
n
)
3
=−∞
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
223
Therefore, we arrive at
X
X
(1 + |ñ|2 )−1/2 |n · fñ |2 ≤ c0 kf k2H −1/2 (Div,Q2 ) .
|n × un |2 ≤ c0
n∈Z
ñ∈Z2
2
This ends the proof of part (a). We leave the proof of part (b) to the reader.
−1/2
−1/2
It is the aim to extend the definition of the trace spaces Hper (Div, Q2 ) and Hper (Curl, Q2 )
−1/2
−1/2
to Hper (Div, ∂D) and Hper (Curl, ∂D), respectively, by the local coordinate system of
Definition 5.1. The trace operators map into the tangential plane of the boundary ∂D.
−1/2
−1/2
Therefore, Hper (Div, ∂D) and Hper (Curl, ∂D) are spaces of tangential vector fields. The
transformation u 7→ φj (u ◦ Ψ̃j )|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, Uj0 ) into vector fields of H(curl, Bj ).
We observe that gradients ∇ϕ are special elements of H(curl, Uj0 ) for any ϕ ∈ H 1 (Uj0 ). They
satisfy
∇x (ϕ ◦ Ψ̃j ) = (Ψ̃0j )> (∇y ϕ) ◦ Ψ̃j
where Ψ̃0j ∈ R3×3 denotes again the Jacobian; that is,
0 ∂(Ψ̃j )k (x)
, k, ` ∈ {1, 2, 3} , x ∈ Bj .
Ψ̃j (x) k` =
∂x`
This will provide the correct transformation as we will see shortly. First we assume as above
with loss of generality that the balls Bj are contained in Q3 . We define the subspaces of
L2 (∂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
∂Ψj (x) ∂Ψj (x)
1
×
, y = Ψj (x) ∈ ∂D ∩ Uj0 , where
ν(y) =
ρj (x)
∂x1
∂x2
q
∂Ψj (x) ∂Ψj (x) =
ρj (x) = ×
1 + |∇ξj (x)|2 , x ∈ B2 (0, αj ) .
(5.14)
∂x1
∂x2 Lemma 5.22 Let {Uj , ξj : j = 1, . . . , m} be a local coordinate system of ∂D. Let Ψ̃ : B → U 0
be one of the isomorphism Ψ̃j from the balls Bj onto Uj0 ; that is, we drop the index j. For
u ∈ H(curl, U 0 ∩ D) ∩ C(U 0 ∩ D, C3 ) set
v(x) = Ψ̃0 (x)> u Ψ̃(x) , x ∈ B − ,
(5.15)
where again B − is the lower part of the ball B and Ψ̃0 (x)> ∈ R3×3 is the adjoint of the
Jacobian at x ∈ B. Furthermore, define F (x) ∈ R3×3 as
∂Ψ(x) ∂Ψ(x) ∂Ψ(x) ∂Ψ(x)
F (x) :=
×
, x ∈ B2 (0, αj ) .
(5.16)
∂x1 ∂x2 ∂x1
∂x2
Then, with ê = (0, 0, 1)> ∈ R3 ,
224
5 BIE FOR LIPSCHITZ DOMAINS
(a) v ∈ H(curl, B − ) and
curl v(x) =
−1
Ψ̃0 (x)
curly u(y)|y=Ψ̃(x) ,
x ∈ B− .
(b)
ρ(x) ν(y) × u(y) y=Ψ(x) = F (x) ê × 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)−> ê × v(x, 0) × ê ,
x ∈ B2 (0, αj ) .
Proof: First we note that for any regular matrix M = [a|b|c] ∈ R3×3 with column vectors
a, b, c, the inverse is given by
M −1 = [a|b|c]−1 =
1
1
[b × c|c × a|a × b]> =
[b × c|c × a|a × b]> . (5.17)
(a × b) · c
det M
Applying this to [a|b|c] = Ψ̃0 (x) yields
"
#>
0 −1
∂ Ψ̃(x) ∂ Ψ̃(x) ∂ Ψ̃(x) ∂ Ψ̃(x) ∂ Ψ̃(x) ∂ Ψ̃(x)
Ψ̃ (x)
=
.
×
×
×
∂x2
∂x3 ∂x3
∂x1 ∂x1
∂x2
In the following the subscript k denotes the kth component of the vector function Ψ̃ or v or
u.
(a) We note that the columns of the matrix Ψ̃0 (x)> ∈ R3×3 are just the gradients ∇Ψ̃k (x),
k = 1, 2, 3. Therefore,
3
X
uk Ψ̃(x) ∇Ψ̃k (x) ,
v(x) =
k=1
and
curl v(x) =
3
X
∇uk Ψ̃(x) × ∇Ψ̃k (x)
k=1
=
3
X
3
X
0 >
∇ uk Ψ̃(x) × ∇Ψ̃k (x) =
Ψ̃ (x) ∇y uk (y)y=Ψ̃(x) × ∇Ψ̃k (x)
k=1
k=1
3 X
3
X
∂uk (y) =
∇Ψ̃` (x) × ∇Ψ̃k (x)
∂y` y=Ψ̃(x)
k=1 `=1
= curl u(y) 1 ∇Ψ̃2 (x) × ∇Ψ̃3 (x) + curl u(y) 2 ∇Ψ̃3 (x) × ∇Ψ̃1 (x)
+ curl u(y) 3 ∇Ψ̃1 (x) × ∇Ψ̃2 (x)
= ∇Ψ̃2 (x) × ∇Ψ̃3 (x) | ∇Ψ̃3 (x) × ∇Ψ̃1 (x) | ∇Ψ̃1 (x) × ∇Ψ̃2 (x) curl u(y)y=Ψ̃(x)
=
−1
Ψ̃0 (x)
curly u(y)|y=Ψ̃(x)
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
225
where we applied (5.17) to Ψ̃(x)> .
(b) We write ∂j Ψ for ∂Ψ/∂xj , j = 1, 2, 3, in the following and drop the argument x or
y = Ψ(x). From the definition of v we have v = (u · ∂1 Ψ , u · ∂2 Ψ , u · ∂3 Ψ)> and thus
ê × v = (−u · ∂2 Ψ , u · ∂1 Ψ , 0)> . Furthermore,
F (ê × v) = −(u · ∂2 Ψ) ∂1 Ψ + (u · ∂1 Ψ) ∂2 Ψ = (∂1 Ψ × ∂2 Ψ) × u = ρ (ν × u) .
(c) We have just seen that ρ (ν × u) = −v2 ∂1 Ψ + v1 ∂2 Ψ, thus
ρ2 (ν × u) × ν = −v2 ∂1 Ψ × (∂1 Ψ × ∂2 Ψ) + v1 ∂2 Ψ × (∂1 Ψ × ∂2 Ψ)
= −v2 (∂1 Ψ · ∂2 Ψ) ∂1 Ψ − |∂1 Ψ|2 ∂2 Ψ + v1 |∂2 Ψ|2 ∂1 Ψ − (∂1 Ψ · ∂2 Ψ) ∂2 Ψ



v1
|∂2 Ψ|2
−∂1 Ψ · ∂2 Ψ 0
|∂1 Ψ|2
0   v2 
= F  −∂1 Ψ · ∂2 Ψ
0
0
0
1
{z
}
|
= ρ2 (F > F )−1 = ρ2 F −1 F −>
= ρ2 F −> (ê × v) × ê .
2
This ends the proof.
Now we are able to define the spaces H −1/2 (Div, ∂D) and H −1/2 (Curl, ∂D) and prove the
trace theorems for H(curl, D).
Definition 5.23 Let D ⊆ R3 be a Lipschitz domain with corresponding local coordinate
system (Uj , ξj ) and corresponding isomorphisms Ψ̃j from the balls Bj onto Uj0 and their
restrictions Ψj : B2 (0, αj ) → Uj ∩ ∂D. Furthermore, let {φj : j = 1, . . . , m}, be a partition
of unity on ∂D subordinate to the sets Uj0 and Fj (x) be given by (5.16). We assume without
loss 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 of
−1/2
−1/2
f ∈ L2t (∂D) : f˜jt ∈ Hper
(Div, Q2 )m
and
f ∈ L2t (∂D) : f˜jT ∈ Hper
(Curl, Q2 )m ,
respectively, with respect to the norms
kf kH −1/2 (Div,∂D) =
" m
X
#1/2
kf˜jt k2H −1/2 (Div,Q
per
2)
,
j=1
"
kf kH −1/2 (Curl,∂D) =
m
X
#1/2
kf˜jT k2H −1/2 (Curl,Q
per
2)
,
j=1
where
f˜jt (x) := ρj (x)
q
φj Ψj (x) Fj−1 (x)f Ψj (x) , x ∈ B2 (0, αj ) ,
(5.18a)
f˜jT (x) := ρj (x)
q
φj Ψj (x) Fj> (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.
226
5 BIE FOR LIPSCHITZ DOMAINS
Again, combining the following trace theorem and Theorem 5.25 below one shows exactly
as in Corollary 5.15 that H −1/2 (Div, ∂D) does not depend on the choice of the coordinate
system.
Theorem 5.24 The trace operators γt : H(curl, D) → H −1/2 (Div, ∂D), u 7→ ν × u|∂D and
γT : H(curl, D) → H −1/2 (Curl, ∂D), u 7→ (ν × u|∂D ) × ν are well-defined and bounded and
have 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
kν × ukH −1/2 (Div,∂D) =
" m
X
#1/2
kf˜jt k2H −1/2 (Div,Q
per
2)
j=1
with
f˜jt (x) = ρj (x)
q
Φj (y) Fj−1 (x) ν(y) × u(y) ,
x ∈ B2 (0, αj ) , y = Ψj (x) .
By part (b) of Lemma 5.22 we rewrite this as
q
f˜jt = ê × vj (·, 0) with vj (x) =
Φj (y) Ψ̃0j (x)> u(y)y=Ψ̃j (x) , x ∈ Bj− .
Then vj ∈ H(curl, Bj− ) by part (a) of Lemma 5.22 and vj vanishes in some neighborhood of
∗
>
∂Bj ∩ Bj− , and we extend vj by zero into Q−
3 . Denoting again z = (z1 , z2 , −z3 ) for any
>
3
vector z = (z1 , z2 , z3 ) ∈ C we extend vj into Q3 by
vj (x) , x ∈ Q−
3,
ṽj (x) =
∗ ∗
vj (x ) , x ∈ Q+
3,
The vector field vj fails to be continuous in Q−
3 because of the transformation into the
parameter space. Therefore, we can not directly apply Exercise 4.6 to show that ṽj ∈
(`)
H(curl, Q3 ). However, we can approximate vj by a sequence ψj ∈ C ∞ (Q−
3 ) and study the
(`)
analogous extension to ψ̃j which are in H(curl, Q3 ) by Exercise 4.6. Also, they converge to
(`) ∗
(`) ∗
ṽj in H(curl, Q3 ) because curlx ψ̃j (x∗ ) = − curl ψj (x∗ ) for x ∈ Q+
3 . Also, ṽj vanishes
in some neighborhood of ∂Q3 . Now we show, using partial integration as in Lemma 4.37 (see
Exercise 4.13), that ṽj ∈ Hper (curl, Q3 ) and that there exists c1 > 0 which is independent of
ṽj such that
kṽj kHper (curl,Q3 ) ≤ c1 kvj kH(curl,Q−3 ) = c1 kvj kH(curl,Bj− )
and thus
kf˜jt kHper
= kγt vj kHper
≤ c2 kvj kH(curl,Bj− ) ≤ c3 kukH(curl,Uj0 ) .
−1/2
−1/2
(Div,Q2 )
(Div,Q2 )
Summing these terms yields boundedness of γt . For γT one argues in the same way.
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
227
It remains to construct an extension operator ηt : H −1/2 (Div, ∂D) → H(curl, D). For
−1/2
f ∈ H −1/2 (Div, ∂D) we define f˜jt ∈ Hper (Div, Q2 ) by (5.18a). By Theorem 5.21 there
exist vj ∈ Hper (curl, Q3 ) such that ê × vj = f˜jt and the mappings f˜jt 7→ vj are bounded for
j = 1, . . . , m. We define
q
uj (y) :=
φj (y) Ψ̃0j (x)−> vj (x)x=Ψ̃−1 (y) , y ∈ Uj0 ,
j
P
extend uj by zero into all of R3 and set u = m
u . Then ν × u = f on ∂D. Indeed, for
j=1 j
P
0
fixed y ∈ ∂D let J = j ∈ {1, . . . , m} : y ∈ Uj . Then u(y) = j∈J uj (y). Fix j ∈ J; that
is, y ∈ Uj0 , and x = Ψ̃−1
j (y) ∈ Q2 × {0}. From the definition of uj we observe that
q
q
φj (y) Fj (x) ê × vj (x, 0) =
φj (y) Fj (x) f˜jt (x)
ρj (x) ν(y) × uj (y) =
= 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 . Furthermore, 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 theSproof we have constructed an extension u
0
into all of R3 with support in the neighborhood m
j=1 Uj of ∂D. In particular, this implies
that there exists a bounded extension operator η̃ : H(curl, D) → H(curl, R3 ).
We note that the space H0 (curl, D) has been defined as the closure of C0∞ (D, C3 ) in H(curl, D),
see Definition 4.19. Analogously to Theorem 5.14 it can be characterized as the nullspace of
the 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 ; that
is, u ∈ H0 (curl, D) if, and only if, γt u = 0 on ∂D. The same holds for γT ; that is, u ∈
H0 (curl, D) if, and only if, γT u = 0 on ∂D.
Proof: The inclusion H0 (curl, D) ⊆ N (γt ) follows again immediately from the definition of
H0 (curl, D) as the closure of C ∞ (D, C3 ). For the reverse inclusion we modify the proof of
Theorem 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)∗ defined
as J1 g = λg where λg ∈ H −1/2 (Div, ∂D)∗ is given by
Z
λg (ψ) :=
ũ · curl ψ̃ − ψ̃ · curl ũ dx , g ∈ H −1/2 (Curl, ∂D), ψ ∈ H −1/2 (Div, ∂D) ,
D
228
5 BIE FOR LIPSCHITZ DOMAINS
where ũ, ψ̃ ∈ H(curl, D) are any vector fields with γT ũ = g and γt ψ̃ = 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)∗ defined
as J2 f = µf where µf ∈ H −1/2 (Curl, ∂D)∗ is given by
Z
µf (ϕ) := −
ṽ · curl ϕ̃ − ϕ̃ · curl ṽ dx , f ∈ H −1/2 (Div, ∂D), ϕ ∈ H −1/2 (Curl, ∂D) ,
D
where ṽ, ϕ̃ ∈ H(curl, D) are any vector fields with γt ṽ = f and γT ϕ̃ = ϕ.
(c) It holds λg (f ) = µf (g) =: hf, gi∗ for all f ∈ H −1/2 (Div, ∂D) and g ∈ H −1/2 (Curl, ∂D)
and Green’s formula holds in the form
Z
u · curl v − v · curl u dx for all u, v ∈ H(curl, D) .
(5.19)
hγt v, γT ui∗ =
D
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 depend
on the choices of the extensions. Let ũj , ψ̃j ∈ H(curl, D), j = 1, 2, such that γt ψ̃j = ψ and
γT ũj = g for j = 1, 2. Then
Z
Z
ũ1 · curl ψ̃1 − ψ̃1 · curl ũ1 dx −
ũ2 · curl ψ̃2 − ψ̃2 · curl ũ2 dx
D
D
Z
=
(ũ1 − ũ2 ) · curl ψ̃1 − ψ̃1 · curl(ũ1 − ũ2 ) dx
D
Z
+
ũ2 · curl(ψ̃1 − ψ̃2 ) − (ψ̃1 − ψ̃2 ) · curl ũ2 dx .
D
Because (ũ1 − ũ2 ) ∈ H0 (curl, D) and (ψ̃1 − ψ̃2 ) ∈ H0 (curl, D) by Theorem 5.25 we conclude
that 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
Z
λg (ψ) ≤
|ũ| | curl ψ̃| + |ψ̃| | curl ũ| dx ≤ kũkH(curl,D) kψ̃kH(curl,D)
D
≤ kηT k kgkH(Curl,∂D) kηt k kψkH(Div,∂D)
which proves that λg is bounded and also boundedness of the mapping J1 . Again, the
arguments 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
Z
curl u · curl ψ̃ + u · ψ̃ dx = `(γt ψ̃) for all ψ̃ ∈ H(curl, D)
D
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
229
because the right–hand side is a bounded functional on H(curl, D). Taking ψ̃ ∈ H0 (curl, D)
we note from the definition of the variational curl (Definition 4.16) that curl u ∈ H(curl, D)
and curl2 u = −u in D. Therefore, for ψ ∈ H −1/2 (Div, ∂D) and ψ̃ = ηt ψ we have that
Z
`(ψ) =
curl u · curl ψ̃ − curl2 u · ψ̃ dx
D
which shows that g := γT curl u ∈ H −1/2 (Curl, ∂D) yields λg = `. The proof of surjectivity
of J2 follows again by the same arguments.
(iv) Finally, we prove injectivity of J1 . Let g ∈ H −1/2 (Curl, ∂D) with J1 g = λg = 0. By a general functional analytic argument (see [23], Section IV.6) there exists ` ∈ H −1/2 (Curl, ∂D)∗
with k`k = 1 and `(g) = kgkH −1/2 (Curl,∂D) . By the surjectivity of J2 there exists f ∈
H −1/2 (Div, ∂D) with µf = `. Let ũ = ηT g and ṽ = ηt f . Then, by the definitions of µf
and λg ,
Z
kgkH −1/2 (Curl,∂D) = µf (g) = −
ṽ · curl ũ − ũ · curl ṽ dx = λg (f ) = 0 .
D
Again, injectivity of J2 is proven analogously.
(c) This is obvious by the definition of λg if one takes g = γT u and ψ = γt v. Note that – by
part (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 as
a subspace of H −1/2 (∂D, C3 ). First we note that H 1 (D, C3 ) is boundedly imbedded in
H(curl, D). Therefore, the trace operator γT is well defined and bounded on H 1 (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→ `a for a ∈ H −1/2 (Div, ∂D) where `a ∈ H −1/2 (∂D, C3 ) is
defined by
h`a , ψi∗ = ha, γT ψ̃i∗ ,
a ∈ H −1/2 (Div, ∂D), ψ ∈ H 1/2 (∂D, C3 ) ,
where ψ̃ ∈ H 1 (D, C3 ) is any extension of ψ. Here, h·, ·i∗ denotes the dual form in
hH −1/2 (∂D, C3 ), H 1/2 (∂D, C3 )i and in hH −1/2 (Div, ∂D), H −1/2 (Curl, ∂D)i, respectively, see
part (c) of Theorem 5.26.
Proof: First we show that the definition of `a is independent of the extension. Let ψ̃j ∈
H 1 (D, C3 ), j = 1, 2, be two extensions of ψ. Then ψ̃ := ψ̃1 − ψ̃2 ∈ H01 (D, C3 ) because γ0 ψ̃ = 0
and Theorem 5.14. We choose a sequence φ̃n ∈ C0∞ (D, C3 ) with φ̃n → ψ̃ in H 1 (D, C3 ). Then
γT φ̃n = 0 for all n and γT φ̃n → γT ψ̃ in H −1/2 (Curl, ∂D). This proves γT ψ̃ = 0.
Furthermore, `a ∈ H −1/2 (∂D, C3 ) and the mapping a 7→ `a is bounded because for ψ ∈
H 1/2 (∂D, C3 ) and the extension operator η : H 1/2 (∂D, C3 ) → H 1 (D, C3 ) from the trace
theorem (Theorem 5.10) we conclude that
h`a , ψi∗ = ha, γT ηψi∗ ≤ c kakH −1/2 (Div,∂D) kγT k kηk kψkH 1/2 (∂D) .
230
5 BIE FOR LIPSCHITZ DOMAINS
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 ψ̃n ∈ C ∞ (D, C3 ) with ψ̃n → ηT b in H(curl, D) where
ηT : H −1/2 (Curl, ∂D) → H(curl, D) is the extension operator of Theorem 5.24. This is
possible by the denseness result of Theorem 5.19. Then 0 = ha, γT ψ̃n i∗ → ha, bi∗ . This holds
for all b ∈ H −1/2 (Curl, ∂D) which shows that a has to vanish because H −1/2 (Curl, ∂D) is
the 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
ha, ψi∗ = ha, γT ψ̃i∗ ,
a ∈ H −1/2 (Div, ∂D), ψ ∈ H 1/2 (∂D, C3 ) .
Again, on the left–hand side a is considered as an element of H −1/2 (∂D, C3 ) while on the
right–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 ∈ H 1/2 (∂D) the
dual form ha, ψi∗ ∈ C3 can be defined componentwise. We will use this in the definition of
the vector potentials in Section 5.2 below.
Having in mind the definition of the variational derivative it is not a surprise that we also
define the surface divergence and the surface curl by variational equations; that is, by partial
integration. We take equation (6.21) as a definition. First we note that for ψ̃ ∈ H 1 (D) it
holds that ∇ψ̃ ∈ H(curl, D). Therefore, the traces γT ∇ψ̃ ∈ H −1/2 (Curl, ∂D) and γt ∇ψ̃ ∈
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 divergence Div a ∈ H −1/2 (∂D) and surface curl Curl b ∈ H −1/2 (∂D) are defined as the linear
bounded functionals
hDiv a, ψi∗ = −ha, γT ∇ψ̃i∗ ,
hCurl b, ψi∗ = −hγt ∇ψ̃, bi∗ ,
ψ ∈ H 1/2 (∂D) ,
ψ ∈ H 1/2 (∂D) ,
(5.20a)
(5.20b)
where again ψ̃ ∈ H 1 (D) is any extension of ψ. On the left–hand side, h·, ·i∗ denotes the
dual form in hH −1/2 (∂D), H 1/2 (∂D)i while on the right–hand side it denotes the form in
hH −1/2 (Div, ∂D), H −1/2 (Curl, ∂D)i, see part (c) of Theorem 5.26. We note that the definitions of the surface divergence and surface curl yield Curl b = − Div(ν × b in a variational
sense, see the remark on page ??.
Again, we have to show that this definition is independent on the extension ψ̃. If ψ̃1 and ψ̃2
are two extension then ψ̃ = ψ̃1 − ψ̃2 ∈ H01 (D) and thus ∇ψ̃ ∈ H0 (curl, D) which implies that
γT ∇ψ̃ = 0 and γt ∇ψ̃ = 0.
We note the following applications (compare to Corollary 5.11 for the scalar case and to
Exercise 4.6 for the case of sufficiently smooth functions).
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
231
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 γt u1 = −γt u2 or γT u1 = γT u2 on ∂D.
Then the field
u1 (x) , x ∈ D ,
u(x) =
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 where
z ∗ = (z1 , z2 , −z3 )> for any z = (z1 , z2 , z3 )> ∈ C3 . Let D± = {x ∈ D : x3 ≷ 0} and
v ∈ H(curl, D− ). Extend v into D by even reflection; that is,
v(x) , x ∈ D− ,
ṽ(x) =
v ∗ (x∗ ) , x ∈ D+ ,
Then ṽ ∈ H(curl, D) and the mapping v 7→ ṽ is bounded from H(curl, D− ) into
H(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) ,
x ∈ D− ,
ṽ(x) =
−v ∗ (x∗ ) , x ∈ D+ ,
Then ṽ ∈ H0 (curl, D) and the mapping v 7→ ṽ is bounded from H0 (curl, D− ) into
H0 (curl, D).
Proof: (a) Let γT u1 = γT u2 and ψ ∈ H0 (curl, Ω) arbitrary. Using (5.19) in D for u1 and in
Ω \ D for u2 yields
Z
hγt ψ|− , γT u1 i∗ =
u1 · curl ψ − ψ · curl u1 dx ,
D
Z
hγt ψ|+ , γT u2 , i∗ =
u2 · curl ψ − ψ · curl u2 dx ,
Ω\D
where γt ψ|− and γt ψ|+ denotes the trace from D and Ω \ D on ∂D, respectively. We note
that γt ψ|− = −γt ψ|+ because of the orientation of ν. Now we set v = curl u1 in D and
v = curl u2 in Ω \ D and add both equations. Then v ∈ L2 (Ω, C3 ) and
Z
u · curl ψ − ψ · v dx = 0 for all ψ ∈ H0 (curl, Ω) .
Ω
This is the variational form of v = curl u and proves the lemma for the case that γT u1 = γT u2 .
In the case γt u1 = γt u2 one uses (5.19) with u and v interchanged. Here we note that
γT ψ|− = γT ψ|+ .
(b), (c) These part is proven analogously by validating that
curl v(x) ,
x ∈ D− ,
curl ṽ(x) =
∓(curl v)∗ (x∗ ) , x ∈ D+ ,
232
5 BIE FOR LIPSCHITZ DOMAINS
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 , defined
in (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 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 ∈ B. Then the closed subspace H0 (curl, divA 0, B) ⊆ H0 (curl, B), defined in (4.12a); that
is,
H0 (curl, divA 0, B) = u ∈ H0 (curl, B) : (Au, ∇ϕ)L2 (B) = 0 for all ϕ ∈ H01 (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 (see
Theorem 5.37) by expanding the functions into spherical harmonics.
Let now A be arbitrary. We recall the Helmholtz decompositions from Theorem 4.23 in the
form
H0 (curl, B) = H0 (curl, divA 0, B) ⊕ ∇H01 (B) ,
L2 (B, C3 ) = L2 (divA 0, B) ⊕ ∇H01 (B) ,
(5.21a)
(5.21b)
and note that the projections onto the components are bounded. Let now (un ) be a bounded
sequence in H0 (curl, divA 0, B). Then we decompose un with respect to the Helmholtz decomposition (5.21a) for A(x) = I; that is,
un = vn + ∇pn
with vn ∈ H0 (curl, div 0, B) and pn ∈ H01 (B) .
Because the projections onto the components are bounded we conclude boundedness of the
sequence (vn ) in H0 (curl, div 0, B) with respect to k · kH(curl,B) . Because H0 (curl, div 0, B) is
compactly 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 with
un ∈ H0 (curl, divA 0, B) ⊆ L2 (divA 0, B) and pn ∈ H01 (B) is just the decomposition of vn
in the Helmholtz decomposition (5.21b). Again, by the boundedness of the projections in
L2 (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) is
symmetric for almost all x and there exists ĉ > 0 with Re (z > A(x)z) ≥ ĉ|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 ∈ H01 (D) , (5.22b)
are compact in L2 (D, C3 ). Here, H0 (curl 0, D) denotes the subspace of H0 (curl, D) with
vanishing curl.
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
233
Proof: As noted before (see Remark 4.22) it suffices to consider the subspace H0 (curl, divA 0, D)
because it contains V0,A as a closed subspace. The proof consists of several steps. First we
localize the functions u(`) into U − := U 0 ∩ D = Ψ̃j (B − ), then we transform these functions
into the half ball Bj− in parameter space. We extend them by a suitable reflection into
H0 (curl divM̃ 0, Bj ) for a certain M̃ ∈ L∞ (Bj , C3×3 ). In this space we apply Lemma 5.31 to
prove L2 −convergence of a subsequence which, finally, yields convergence of the corresponding subsequence of (u(`) ).
Let {Uj , ξj : j = 1, . . . , m} be a local coordinate system with corresponding isomorphisms
Ψ̃j from the balls Bj onto Uj0 for j1 , . . . , m. Furthermore, set U00 = D and choose a partition
of unity {φj : j = 0, . . . , m} on D subordinate to the sets Uj0 . Before we consider bounded
sequences we concentrate on one particular element.
Therefore, let u ∈ H0 (curl, divA 0, D) and j ∈ {1, . . . , m} be fixed. The definition of
H0 (curl, divA 0, D) yields for any ϕ̃ ∈ H01 (Uj− ) where again Uj− := Uj0 ∩ D = Ψ̃j (Bj− )
Z
0 =
u(y)> A(y)∇ φj (y)ϕ̃(y) dy
D
Z
=
Uj−
>
u(y) A(y)∇φj (y) ϕ̃(y) dy +
Z
Uj−
φj (y) u(y)> A(y)∇ϕ̃(y) dy
Let now pj ∈ H01 (Uj− ) be defined by
Z
Uj−
Z
>
∇pj (y) A(y)∇ϕ̃(y) dy =
u(y)> A(y)∇φj (y) ϕ̃(y) dy
Uj−
for all ϕ̃ ∈ H01 (Uj− ) .
Substituting ϕ̃ = pj yields the estimate
ĉk∇pj k2L2 (U − ) ≤ c1 kukL2 (D) kpj kL2 (Uj− )
(5.23a)
j
for some c1 > 0 which is independent of u and j. With Friedrich’s inequality we conclude
that this implies
(5.23b)
k∇pj kL2 (Uj− ) ≤ c2 kukL2 (D)
for some c2 > 0 independent of u and j. Defining ũj := φ
R j u + ∇pj we conclude that
ũj ∈ H0 (curl, Uj− ) and even ũj ∈ H0 (curl, divA , Uj− ) because U − ũj (y)> A(y)∇ϕ̃(y) dy = 0
j
for all ϕ̃ ∈ H01 (Uj− ). This holds for all j ∈ {1, . . . , m}. We transform ũj into the parameter
domain as we did it already several times. We define
v(x) = Ψ̃0j (x)> ũj Ψ̃j (x) and M (x) = Ψ̃0j (x)−1 A Ψ̃j (x) Ψ̃0j (x)−> for x ∈ Bj−
where Bj− denotes again the lower part of the ball Bj . Here we dropped the index j for v
and M to keep the presentation simple. Then v ∈ H0 (curl, Bj− ) by Lemma 5.22. Also, we
recall that ∇ϕ(x) = Ψ̃0j (x)> (∇ϕ̃) Ψ̃j (x) for ϕ = ϕ̃ ◦ Ψ̃j and thus
Z
0 =
Uj−
>
ũj (y) A(y)∇ϕ̃(y) dy =
Z
Bj−
v(x)> M (x)∇ϕ(x) dx for all ϕ ∈ H01 (Bj− ) . (5.24)
234
5 BIE FOR LIPSCHITZ DOMAINS
Now we extend v and M into all of the ball Bj by
v(x) ,
x ∈ Bj− ,
ṽ(x) =
∗ ∗
−v (x ) , x ∈ Bj+ ,
and
M`k (x) , x ∈ Bj− , `, k ∈ {1, 2, 3} ,
M`k (x∗ ) , x ∈ Bj+ , `, k ∈ {1, 2} or ` = k = 3 ,
M̃`k (x) =

−M`k (x∗ ) , x ∈ Bj+ , else,


where again z ∗ = (z1 , z2 , −z3 )> for any z = (z1 , z2 , z3 )> ∈ C3 . First we note that M̃ ∈
L∞ (Bj , C3×3 ) and M̃ (x) is symmetric for almost all x ∈ Bj and there exists c1 > 0 with
z > M̃ (x)z ≥ c1 M̃ (x)|z|2 for all z ∈ C3 and x ∈ Bj . Next we show that ṽ ∈ H0 (curl, divM̃ 0, Bj ).
The formula curlx v ∗ (x∗ ) = −(curl v)∗ (x∗ ) yields ṽ ∈ H0 (curl, Bj ). It remains to show that
Z
ṽ(x)> M̃ (x)∇ϕ(x) dx = 0 for all ϕ ∈ H01 (Bj ) .
Bj
Let ϕ ∈ H01 (Bj ). Then the function x 7→ ϕ(x) − ϕ(x∗ ) vanishes for x3 = 0 and is thus in
H01 (Bj− ). Substituting this function into (5.24) and using ∇x ϕ(x∗ ) = (∇ϕ)∗ (x∗ ) yields
Z
v(x)> M (x)∇ ϕ(x) − ϕ(x∗ ) dx
0 =
Bj−
Z
Z
>
v(x) M (x)∇ϕ(x) dx −
=
Bj−
Bj−
Z
Z
>
v(x) M (x)∇ϕ(x) dx −
=
Bj−
v(x)> M (x)∇x ϕ(x∗ ) dx
v(x∗ )> M (x∗ )(∇ϕ)∗ (x) dx .
Bj+
Now a careful elementary calculation shows that
v(x∗ )> M (x∗ )(∇ϕ)∗ (x) = −ṽ(x)> M̃ (x)∇ϕ(x) ,
x ∈ Bj+ ,
which yields
Z
ṽ(x)> M̃ (x)∇ϕ(x) dx = 0 .
Bj
Since this holds for all ϕ ∈
H01 (Bj )
we have shown that ṽ ∈ H0 (curl, divM̃ 0, Bj ).
Now we prove the actual compactness property. Let (u(`) )` be a bounded sequence in
(`)
(`)
H0 (curl, divA 0, D). The scalar functions pj ∈ H01 (Uj− ) and the vector functions ṽj ∈
H0 (curl, divM̃ 0, Bj ) correspond to pj and ṽ above. The estimate (5.23b) implies boundedness
(`)
of (pj )` in H01 (Uj− ). The compact imbedding of H01 (Uj− ) in L2 (Uj− ) yields L2 −convergence
(`)
(`)
(k)
of a subsequence of (pj )` . From estimate (5.23a) appied to the difference pj − pj we
(`)
conclude that this subsequence converges also in H01 (Uj− ). Also, (ũj )` is bounded in
(`)
H0 (curl, divA 0, Uj− ) and thus also (ṽj )` in H0 (curl, divM̃ 0, Bj ) for every j = 1, . . . , m. Now
(`)
we apply Lemma 5.31 which yields convergence of a subsequence of (ṽj )` in L2 (Bj , C3 )
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
235
(`)
which we denote by the same symbol. Then also the sequence (ũj )` converges in L2 (Uj0 , C3 )
for every j = 1, . . . , m. Finally, we consider j = 0; that is, the bounded sequence (φ0 u(`) )`
(`)
in H0 (curl, D). We choose a ball B containing D in its interior and extend u0 = φ0 u(`) by
(`)
(`)
(`)
(`)
zero into B. Then we construct p0 ∈ H01 (B) and ũ0 = u0 − ∇p0 ∈ H0 (curl, divA 0, B)
just as before (but now in B instead of Uj− ). We again have convergence of a further subse(`)
(`)
quence of (p0 )` in L2 (B). Application of Lemma 5.31 to (ũ0 )` yields L2 −convergence of a
(`)
subsequence. Altogether, we have found a common convergent subsequence of (ũj )` for all
(`)
j = 0, 1, . . . , m. Together with the convergence of (pj )` in H01 (Uj− ) this finishes the proof.
2
As a direct consequnce we obtain the imbedding of the subspaces V0,ε and Vµ defined also in
chapter 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 = εI
and 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 and
Helmholtz equation into spherical harmonics for the special case where B is a ball of radius R centered at the origin. We have seen that for L2 −boundary data the series converge
in L2 (D). From Chapter 4 we know that the natural solution space of the boundary value
problems for the Laplace or Helmholtz equation is the Sobolev space H 1 (D). Therefore,
according to the trace theorem, the natural space of boundary data is H 1/2 (∂D). In this
section we will complement the results of Chapter 2 and show convergence in H 1 (D).
In our opinion, expansions into spherical harmionics provide a clear insight into several
properties of Sobolev spaces. For example, the fact that every curl-free vector field has a
potential is seen very explicitly, see Theorem 5.37 below. In the same Theorem we will
show 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 a
characterization of the Sobolev space H 1/2 (∂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 −theory
of Chapter 2 and the H 1 −theory of Chapter 4. The results of this section with the exception
of 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 with
respect to polar coordinates r > 0 and x̂ ∈ S 2 to expand functions into a series of spherical
harmonics 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 ∈
H 1 (B) let um
n (r) ∈ C be the expansion coefficients of x̂ 7→ u(rx̂) with respect to the spherical
236
5 BIE FOR LIPSCHITZ DOMAINS
harmonics; that is,
Z
m
u(r, x̂) Yn−m (x̂) ds(x̂) ,
un (r) =
0 ≤ r < R , |m| ≤ n , n = 0, 1, 2, . . .
(5.25)
S2
Then
N
n
X
X
uN (r, x̂) :=
m
um
n (r) Yn (x̂) ,
0 ≤ r < R , x̂ ∈ S 2 ,
n=0 m=−n
converges to u in H 1 (B) as N tends to infinity.
Proof: First we note that, using (2.17),
Z
1
m
u(r, x̂) DivS 2 GradS 2 Yn−m (x̂) ds(x̂)
un (r) = −
n(n + 1) S 2
Z
1
=
GradS 2 u(r, x̂) · GradS 2 Yn−m (x̂) ds(x̂)
n(n + 1) S 2
Z
1
GradS 2 u(r, x̂) · Un−m (x̂) ds(x̂)
= p
n(n + 1) S 2
p
GradYnm have been defined in Theorem 2.46. Therefore, n(n + 1)um
n (r)
m
are the expansion coefficients of Grad
|m| ≤ n, n ∈ N . From
Sm2 u(r, ·) with respect to Un : m
the orthonormalty of the systems Yn : |m| ≤ n, n ∈ N and {Un : |m| ≤ n, n ∈ N} we
have by Bessel’s inequality
where Unm = √
1
n(n+1)
N
n
X
X
2
|um
n (r)|
Z
≤
n=0 m=−n
N
n
X
X
n(n +
2
1) |um
n (r)|
n=0 m=−n
|u(rx̂)|2 ds(x̂) ,
S2
Z
≤
| GradS 2 u(rx̂)|2 ds(x̂) ,
and
S2
m 2
Z
N
n
X
X
dun
dr (r) ≤
2
∂u
(rx̂) ds(x̂)
S 2 ∂r
n=0 m=−n
for almost all r. By Parseval’s equation we have even equalities for N → ∞.
∂
In spherical polar coordinates the gradient is given by ∇ = ∂r
x̂ + 1r GradS 2 , thus
kuk2H 1 (B) =
ZR Z
0
S2
"
#
2
∂u
1
|u(rx̂)|2 + (rx̂) + 2 | GradS 2 u(rx̂)|2 ds(x̂) r2 dr
∂r
r
ZR "
N
n
X
X
n(n + 1)
2
≥
1+
|um
n (r)| +
2
r
n=0 m=−n
0
m 2 #
dun
r2 dr .
(r)
dr
(5.26)
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
237
The right–hand side converges to kuk2H 1 (B) as N tends to infinity by the theorem of monotoneous convergence. Furthermore, from
p
N
n
X
X
n(n + 1) m
dum
n
∇uN (rx̂) =
(r) Ynm (x̂) x̂ +
un (r) Unm (x̂) , 0 ≤ r < R , x̂ ∈ S 2 ,
dr
r
n=0 m=−n
(5.27)
we conclude that, for M > N ,
m 2 #
Z R"
M
n
X
X
dun
n(n
+
1)
2
r2 dr (5.28)
1
+
+
(r)
kuM − uN k2H 1 (B) =
|um
(r)|
n
dr
2
r
n=N +1 m=−n 0
2
tends to zero as N tends to infinity.
m 0
Remark: From (5.28) we conclude that r 7→ rum
n (r) and r 7→ r(un ) (r) are elements of
L2 (0, R) for all m, n.
In this case of B being a ball the space H 1/2 (∂B) can be characterized by a proper decay of
the 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 fnm ∈ C be the expansion coefficients with respect to the spherical harmonics; that
is,
Z
fnm =
S2
f (Rx̂) Yn−m (x̂) ds(x̂) ,
|m| ≤ n , n = 0, 1, 2, . . .
Then f ∈ H 1/2 (∂B) if, and only if,
"
kf k :=
∞ X
n
X
p
1 + n(n + 1) |fnm |2
#1/2
< ∞.
(5.29)
n=0 m=−n
Proof: For the moment let X be the completion of C ∞ (∂B) with respect to the norm
of (5.29). Note that C ∞ (∂B) is also dense in H 1/2 (∂B) (see Corollary 5.15 for a proof).
Therefore, as in the proof of Corollary 5.15 it suffices to show that the norm k · k of (5.29)
is equivalent to the norm of the factor space H 1 (B)/H01 (B). This is assured if we can prove
Theorems 5.10 and 5.14 for X instead of H 1/2 (∂B). We assume without loss of generality
that R = 1.
To prove boundedness of the trace operator we write u ∈ H 1 (B) in polar coordinates as
u(rx̂) =
n
∞ X
X
m
um
n (r) Yn (x̂)
n=0 m=−n
with coefficients um
n (r) ∈ C and thus
(γ̃0 u)(x̂) =
∞ X
n
X
n=0 m=−n
m
um
n (1) Yn (x̂) .
238
5 BIE FOR LIPSCHITZ DOMAINS
2
for s ∈ [0, 1], using r3 ≤ r2 for r ∈ [0, 1], the inequality of CauchyWe estimate um
n (s)
Schwarz, and 2ab ≤ a2 + b2 , as
Z s
2
d 3 m
3 m
=
s un (s)
r |un (r)|2 dr
dr
0
Z s
d
2 m
2
3
m
3r |un (r)| + 2r Re um
=
u (r) dr
n (r)
dr n
0
s
sZ
2
Z s
Z s
s
d
m
2 2
2 2
um
r2 dr
|un (r)| r dr + 2
|um
≤ 3
(r)
n (r)| r dr
n
dr
0
0
0
Z s
Z s
p
m
2 2
2 2
≤ 3
|un (r)| r dr +
1 + n(n + 1)
|um
n (r)| r dr
0
0
Z s
d m 2 2
1
un (r) r dr .
+ p
1 + n(n + 1) 0 dr
From this and the preceding remark we conclude
that um
n is continous on (0, 1]. We continue
p
with s = 1 and multiply this estimate by 1 + n(n + 1) which yields
Z
Z 1
p
m 2
d m 2 2
1 m
2
2
un (r) r dr
1 + n(n + 1) un (1) ≤ 4 1 + n(n + 1)
|un (r)| r dr +
dr
0
0
and thus
kγ̃0 uk2H 1/2 (∂B)
∞ X
n
X
p
2
2
1 + n(n + 1) |um
=
n (1)| ≤ 4 kukH 1 (B)
n=0 m=−n
by (5.26) for N → ∞. This proves boundedness of γ̃0 from H 1 (B) into X. A right inverse
of γ̃ is given by
∞ X
n
X
(η̃f )(rx̂) =
fnm rn Ynm (x̂) .
n=0 m=−n
It suffices to prove boundedness. From (5.28) for unm (r) = fnm rm we conclude that
∞ X
n
X
n(n + 1) + n2
1
m 2
2
+
|fn |
kη̃f kH 1 (B) =
2n
+
3
2n + 1
n=0 m=−n
≤ c
∞ X
n
X
p
1 + n(n + 1)|fnm |2 = ckf k2 .
n=0 m=−n
This proves the trace theorem for H 1 (B) and X.
To show the analogue of Theorem 5.14 it suffices again to show that the nullspace of γ̃0 is
contained in H01 (B). Let u ∈ H 1 (B) with γ̃0 u = 0; that is, um
n (1) = 0 for all m, n. Let uN
be the truncated sum; that is,
uN (rx̂) =
N
n
X
X
n=0 m=−n
m
um
n (r) Yn (x̂) .
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
239
Then uN is continuous in B \ {0} and converges to u in H 1 (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. Define
vN,ε ∈ H 1 (D) by
N
n
X
X
1 m
un (r) Ynm (x̂) .
vN,ε (rx̂) =
εψ
ε
n=0 m=−n
Now we can argue as we did already several times. By the theorem of dominated convergence
we conclude that vN,ε → uN in H 1 (B) as ε tends to zero. Also, we can approximate vN,ε by
some ṽ ∈ C0∞ (B) because vN,ε vanishes in some neigborhood of S 2 . Altogether this shows
that u can be approximated arbitrarily well in C0∞ (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 ∈
m
m
H(curl, B) let um
n (r), vn (r), wn (r) ∈ C be the expansion coefficients of x̂ 7→ u(rx̂) with
respect to the spherical vector harmonics; that is,
Z
m
u(r, x̂) · x̂ Yn−m (x̂) ds(x̂) , 0 ≤ r < R ,
(5.30a)
un (r) =
S2
vnm (r) =
wnm (r)
Z
S2
Z
=
S2
u(r, x̂) · Un−m (x̂) ds(x̂) ,
0 ≤ r < R,
(5.30b)
u(r, x̂) · Vn−m (x̂) ds(x̂) ,
0 ≤ r < R,
(5.30c)
for |m| ≤ n and n = 0, 1, 2, . . ., where Unm and Vnm have been defined in Theorem 2.46 see
Corollary 2.47. Then
uN (rx̂) :=
N
n
X
X
m
un (r) Ynm (x̂) x̂ + vnm (r) Unm (x̂) + wnm (r) Vnm (x̂) ,
n=0 m=−n
0 ≤ r < R, x̂ ∈ S 2 , 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 are
technically more complicated. First we note from the completeness of the orthonormal
n
systems {Ymn : |m| ≤ n, n ∈ N} in L2 (S 2 ) and {Um
, Vnm : |m| ≤ n, n ∈ N} in L2t (S 2 ) that
∞ X
n
X
m
|un (r)|2 + |vnm (r)|2 + |wnm (r)|2 = ku(r, ·)k2L2 (S 2 ) .
n=0 m=−n
The expansion coefficients of curl u(r, ·) are given by (see Exercise 5.4)
p
Z
n(n + 1) m
−m
curl u(rx̂) · x̂ Yn (x̂) ds(x̂) = −
wn (r) ,
r
S2
Z
0
1
curl u(rx̂) · Un−m (x̂) ds(x̂) = − rwnm (r) ,
r
S2
p
Z
0
n(n + 1) m
1
curl u(rx̂) · Vn−m (x̂) ds(x̂) = −
un (r) +
rvnm (r) .
r
r
S2
240
5 BIE FOR LIPSCHITZ DOMAINS
This yields
∞
n
0 2 p
0 2
1 X X
m
2
m
m
m
n(n + 1) |wn (r)| + rwn (r) + n(n + 1) un (r) − rvn (r) r2 n=0 m=−n
= k curl u(r, ·)k2L2 (S 2 ) .
To compute curl uN we need the formulas (see Exercise 5.4)
m
um
(r)
m
curl un (r) Yn (x̂) x̂ = − n
x̂ × GradS 2 Ynm (x̂) ,
r
0
1
curl vnm (r) GradS 2 Ynm (x̂) = − r vnm (r) x̂ × GradS 2 Ynm (x̂) ,
r
0
1
curl wnm (r) x̂ × GradS 2 Ynm (x̂)
= − r wnm (r) GradS 2 Ynm (x̂)
r
− wnm (r)
n(n + 1) m
Yn (x̂) x̂ .
r
Therefore,
N
n
X
X
p
1
m
curl uN (x) =
− n(n + 1) um
n (r) Vn (x̂) +
r
n=0 m=−n
−
0
r wnm (r)
Unm (x̂)
0
r vnm (r) Vnm (x̂)
i
p
m
m
n(n + 1) wn (r) Yn (x̂) x̂
−
(5.31)
and thus for M > N
kuM −
uN k2H(curl,B)
=
M
X
Z 1
n
X
m
|un (r)|2 + |vnm (r)|2 + |wnm (r)|2 r2 dr
n=N +1 m=−n
+
M
X
0
Z 1
n
X
n=N +1 m=−n
0 2
(n(n + 1) |wnm (r)|2 + rwnm (r) +
0
p
0 2
m
m
+ n(n + 1) un (r) − rvn (r) dr
which 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 funcm
m
m
0
m
0
tions r 7→ rum
n (r), r 7→ rvn (r), r 7→ rwn (r), r 7→ (rwn (r)) , and r 7→ (rvn (r)) are in
L2 (0, R). This yields in particular that vnm and wnm are continuous in (0, R].
As a simple consequence of this result we show that for balls the subspace H(curl 0, B) of
vector fields with vanishing curl coincides with the space ∇H 1 (B) of gradients and also
compactness of the subspaces appearing in the Helmholtz decompositions (see Theorem 4.21
and Remark 4.22) in L2 (B, C3 ).
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
241
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) : curl u = 0 in B}
coincides with the space ∇H 1 (B) = {∇ϕ : ϕ ∈ H 1 (B)}.
R
(b) The space H(curl, div 0, B) = {u ∈ H(curl, B) : B u · ∇ϕ dx = 0 for all ϕ ∈ H 1 (B)}
is compactly imbedded in L2 (B, C3 ).
R
(c) The space H0 (curl, div 0, B) = {u ∈ H0 (curl, B) : B u · ∇ϕ dx = 0 for all ϕ ∈ H01 (B)}
is compactly imbedded in L2 (B, C3 ).
Proof: (a) The inclusion ∇H 1 (B) ⊆ H(curl 0, B) is obvious and holds for any domain. To
show the reverse inclusion let u ∈ H(curl 0, B). Then u has an expansion in the form
u(rx̂) =
∞ X
n
X
m
un (r) Ynm (x̂) x̂ + vnm (r) Unm (x̂) + wnm (r) Vnm (x̂) .
(5.32)
n=0 m=−n
Since curl u = 0 we conclude from (5.31) that
p
n(n + 1) um
wnm (r) = 0 and
n (r) =
0
rvnm (r)
for all r and all n = 0, 1, 2, . . ., |m| ≤ n. Now we set
ϕ(rx̂) =
From
P∞ Pn
n=1
m=−n
∇ϕ(rx̂) =
∞ X
n
X
1
p
r vnm (r) Ynm (x̂) .
n(n + 1
n=1 m=−n
|vnm |2 < ∞ we conclude that ϕ ∈ H 1 (B) and, by (5.27),
N
n
X
X
p
0
1
p
n(n + 1) vnm (r) Unm (x̂)
r vnm (r) Ynm (x̂) +
n(n + 1
n=0 m=−n
N
n
X
X
m
=
un (r) Ynm (x̂) + vnm (r) Unm (x̂) = u(rx̂) .
n=0 m=−n
(b) Without loss of generality let B the unit ball. Every u ∈ H(curl, div 0, B) has an
−m
expansion in the form (5.32). Let ϕ(rx̂) = ρ(r)Y
√ n (x̂) for some n ∈ N and |m| ≤ n and
ρ ∈ C 1 [0, 1]. Then ∇ϕ(rx̂) = ρ0 (r)Yn−m (x̂) x̂ +
Z 1
Z
u(x) · ∇ϕ(x) dx =
0 =
B
0
n(n+1)
r
Un−m (x̂) and thus
p
n(n + 1)
0
m
m
ρ (r) un (r) +
ρ(r) vn (r) r2 dr .
r
(5.33)
This holds for all such ρ. Note that r 7→ rvnm (r) is continuous and r 7→ r2 um
n (r) is in
1
L (0, 1). A modification of the Fundamental Theorem of Calculus
4.46, see Exp
0 (Lemma
m
1
m
2 m
ercise 4.16) yields un ∈ C (0, 1] and un (1) = 0 and r un (r) = n(n + 1) rvnm (r) for
R1 m
P
Pn
2
all r ∈ (0, 1]. From (5.31) we conclude that ∞
n=0
m=−n 0 |qn (r)| dr converges where
242
5 BIE FOR LIPSCHITZ DOMAINS
p
0
m
qnm (r) = n(n + 1) um
n (r) − rvn (r) . We estimate, using these two relationships between
m
m
um
n and vn and qn and integration by parts,
1
Z
p
n(n + 1)
r
2
|vnm (r)|2 dr
Z
=
1
0
r2 um
n (r)
Z
r vnm (r) dr
= −
0
0
1
0
m
r2 um
n (r) r vn (r) dr
0
p
= − n(n + 1)
Z
1
r
2
2
|um
n (r)| dr
Z
1
m
r2 um
n (r) qn (r) dr .
+
0
0
Note that no boundary terms appear because um
n (1) = 0. Therefore,
s
s
Z 1
Z
Z 1
p
2
2
m
m
2
2
m
2
r |un (r)| + |vn (r)| dr ≤
r |un (r)| dr
n(n + 1)
0
0
1
≤
2
Z
1
r2 |qnm (r)|2 dr
0
1
r
2
2
|um
n (r)|
0
1
dr +
2
Z
1
r2 |qnm (r)|2 dr .
0
From this we easily derive the estimate
Z 1
Z
m
1 1 m
2
2
m
2
r |un (r)| + |vn (r)| dr ≤
|qn (r)|2 dr .
n
0
0
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
(IN u)(rx̂) =
N
n
X
X
m
m
m
m
m
um
n (r) Yn (x̂) x̂ + vn (r) Un (x̂) + wn (r) Vn (x̂) .
n=0 m=−n
Then
kIN u − IukL2 (B) =
∞
X
Z
n
X
n=N +1 m=−n
1
2
m
2
m
2
r2 |um
n (r)| + |vn (r)| + |wn (r)| dr
0
Z 1
∞
n
X
X
1
|qnm (r)|2 dr
≤
N + 1 n=N +1 m=−n 0
Z 1
n
∞
X
X
1
+
n(n + 1)|wnm (r)|2 r2 dr .
(N + 2)(N + 1 n=N +1 m=−n 0
This yields kIN u−IukL2 (B) ≤ N1 kukH(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 ∈ C 1 [0, 1] with ρ(1) = 0 in order to have ϕ ∈ H0 (B) by Theorem 5.14. Then
we derive the variational equation
(5.33)
for ρ ∈ C 1 [0, 1] with ρ(1) = 0. Again this shows
p
0
1
2 m
that um
n(n + 1) rvnm (r) for all r ∈ (0, 1] by Exercise 4.16 but
n ∈ C (0, 1] and r un (r) =
5.1. ADVANCED PROPERTIES OF SOBOLEV SPACES
243
m
no longer um
n (1) = 0. Nevertheless, one can continue with the proof because vn (1) = 0 by
the boundary condition ν × u = 0 on ∂B (and Theorem 5.25). We leave the details to the
reader.
2
Now we turn to the characterization of the boundary spaces H −1/2 (Div, ∂B) and H −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. For
m
f ∈ L2t (∂B) let am
n , bn ∈ C be the expansion coefficients with respect to the spherical vectorharmonics; that is,
Z
Z
m
−m
m
f (Rx̂)·Vn−m (x̂) ds(x̂) , |m| ≤ n, n = 0, 1, 2, . . .
f (Rx̂)·Un (x̂) ds(x̂) , bn =
an =
S2
S2
Then f ∈ H −1/2 (Div, ∂B) or f ∈ H −1/2 (Curl, ∂B) if, and only if,
kf k2D
:=
∞ X
n
X
1 + n(n + 1)
1/2
2
|am
n | + 1 + n(n + 1)
−1/2
2
|bm
< ∞,
n|
(5.34)
n=0 m=−n
or
kf k2C
:=
∞ X
n
X
1 + n(n + 1)
−1/2
2
|am
n | + 1 + n(n + 1)
+1/2
2
|bm
< ∞,
n|
(5.35)
n=0 m=−n
respectively.
Proof: Let X be the completion of Ct∞ (∂B) with respect to the norm of (5.34). The
assertion follows as in Corollary 5.15 (see also Theorem 5.35) once we have proven the trace
theorem in X and that H01 (B) coincides with the null space of γ0 ; that is, Theorems 5.24
and 5.25. First we compute kukH(curl,B) . For u of the form
∞ X
n
X
m
un (r) Ynm (x̂) x̂ + vnm (r) Unm (x̂) + wnm (r) Vnm (x̂) ,
u(x) =
n=0 m=−n
we have seen in the previous theorem that
Z R
∞ X
n
X
m
2
kukL2 (B) =
|un (r)|2 + |vnm (r)|2 + |wnm (r)|2 r2 dr ,
n=0 m=−n
k curl uk2L2 (B)
=
Z
∞ X
n
X
n=0 m=−n
(5.36a)
0
0
R
0 2
p
2
m
(r)|
−
r
v
(r)
n(n + 1) |um
n
n
0 2
m
2
m
+ r wn (r) + n(n + 1) |wn (r)| dr .
(5.36b)
Without loss of generality we take R = 1. Then we observe that
∞ X
n X
x̂ × GradS 2 Ynm (x̂)
GradS 2 Ynm (x̂)
m
m
p
x̂ × u(1, x̂) =
vn (1)
− wn (1) p
.
n(n
+
1)
n(n
+
1)
n=0 m=−n
244
5 BIE FOR LIPSCHITZ DOMAINS
Note that vnm and
wnm are continuous
in (0, 1] by the previous remark. Now we have to
2
2
m
m
estimate vn (1) and wn (1) . With 2|a||b| ≤ a2 + b2 we conclude
m 2
vn (1) =
1
Z
2 0
r r vnm (r) dr
0
1
Z
0
r vnm (r)
= 2 Re
vnm (r) r2
0
1
= 2 Re
vnm (r)2 r2 dr
dr +
0
Z
1
Z
0 p
2
m
r vnm (r) − n(n + 1) um
n (r) vn (r) r dr
0
+2
p
Z
1
n(n + 1) Re
2
m
um
n (r) vn (r) r
0
0
2
p
r vnm (r) 0 − n(n + 1) um
n (r) dr +
≤
vnm (r)2 r2 dr
dr +
1
Z
1
Z
0
Z
1
vnm (r)2 r4 dr
0
p
+ n(n + 1)
1
Z
um
2 2
n (r) r
Z 1
p
m 2 2
vn (r) r dr .
dr + (1 + n(n + 1))
0
0
p
We devide by 1 + n(n + 1), observe that r4 ≤ r2 for r ∈ [0, 1] and sum. Comparing this
with (5.36a), (5.36b) yields
∞ X
n
X
1 + n(n + 1)
−1/2 m 2
vn (1) ≤ 2kuk2H(curl,B) .
n=0 m=−n
For wnm (1) we argue analogously:
Z 1
m 2
2 0
wn (1) =
r r wnm (r) dr
0
1
Z
= 2 Re
0
r wnm (r)
wnm (r) r2
0
Z
dr +
1
wnm (r)2 r2 dr
0
Z
1
1
Z 1
p
m 2 4
r wnm (r) 0 2 dr +
wn (r) r dr
1 + n(n + 1)
≤ p
1 + n(n + 1) 0
Z 1
m 2 2
wn (r) r dr ,
+
0
0
and thus
∞ X
n
X
1 + n(n + 1)
1/2 m 2
wn (1) ≤ 2kuk2H(curl,B) .
n=0 m=−n
This proves boundedness of the trace operator H(curl, B) → X. A right inverse is given by
(ηf )(rx̂) =
∞ X
n
X
m m
n
m
bn Un (x̂) − am
n Vn (x̂) r ,
n=0 m=−n
5.2. SURFACE POTENTIALS
245
where anm , bm
n are the Fourier coefficients of f . Boundedness follows from (5.36a), (5.36b) for
m
n
m
m n
un (r) = 0, vnm (r) = bm
n r , and wn (r) = −an r .
Finally, we have to show that the kernel of γt coincides with H0 (curl, B). We follow the
arguments as in the proof of Theorem 5.35 and approximate
uN (rx̂) =
N
n
X
X
m
m
m
m
m
um
(r)
Y
(x̂)
x̂
+
v
(r)
U
(x̂)
+
w
(r)
V
(x̂)
n
n
n
n
n
n
n=0 m=−n
by
∞ X
n X
1−r
ε r m m
m
m
un (r) Yn (x̂) x̂ + ψ vn (r) Un (x̂)
uN,ε (rx̂) =
φ
ε
r
ε
n=0 m=−n
ε r m m
+ ψ wn (r) Vn (x̂)
r
ε
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 vnm and wnm vanish for r = 1. Estimating the difference
kuN − uN,ε kH(curl,B) requires (see (5.36a), (5.36b)) to consider expressions of the form
ε r m 1−r
m
ψ vn (r) − vn (r) and
φ
− 1 um
n (r)
r
ε
ε
and
i
h r
i
0
d h ε r m r
ψ vn (r) − vnm (r)
= ψ 0 vnm (r) − 1 rvnm (r) .
dr
r
ε
ε
These expressions converges to zero pointwise almost everywhere as ε tends to zero. Also
there 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 layer
potentials. First we recall – and rename – the notion of the traces. In this section D is
always a bounded Lipschitz domain with boundary ∂D which separates the interior D from
the exterior R3 \ D. We fix the normal vector ν(x), which exists for almost all points x ∈ ∂D
by the differentiability assumption on the parametrization, and let it direct into the exterior
of D. Then γ0 u|± is the trace of u from the exterior (+) and interior (−), respectively. The
traces of the normal derivatives γ1 u|± for (variational) solutions u of the Helmholtz equation
are defined by
Z
γ1 u|− , ψ ∂D =
∇u · ∇ψ̂ − k 2 uψ̂ dx , ψ ∈ H 1/2 (∂D) , u ∈ HD ,
(5.37a)
D
γ1 u|+ , ψ ∂D = −
Z
R3 \D
∇u · ∇ψ̂ − k 2 uψ̂ dx , ψ ∈ H 1/2 (∂D) , u ∈ HR3 \D , (5.37b)
246
5 BIE FOR LIPSCHITZ DOMAINS
where ψ̂ ∈ H 1 (D) or ψ̂ ∈ H 1 (R3 \D), respectively, are extensions of ψ with bounded support
in the latter case, and
Z
1
2
1
HD =
u ∈ H (D) :
∇u · ∇ψ − k u ψ dx = 0 for all ψ ∈ H0 (D) ,
D
HR3 \D
Z
1
3
=
u ∈ Hloc (R \ D) :
R3 \D
2
1
3
∇u · ∇ψ − k u ψ dx = 0 for all ψ ∈ H0 (R \ D)
denote the spaces of variational solutions of the Helmholtz equation in D and in R3 \ D,
1
respectively, compare with (5.7). Here, Hloc
(R3 \ D) = {u : R3 \ D → C : u|B ∈
1
H (B) for all balls B}.
We denoted the dual form in H −1/2 (∂D), H 1/2 (∂D) by h·, ·i∂D instead of h·, ·i∗ which we
will do from now on. Note, that we have changed the sign in γ1 u|+ because the normal
vector 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 ≥ 0
√
and 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, compare
with 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 ∈ H 1 (D) of the Helmholtz equation; that is, u ∈ HD , we have the
representation
(
Z
−u(x) , x ∈ D ,
∂Φ
(γ0 u)(y)
(x, y) ds(y) − γ1 u, γ0 Φ(x, ·) ∂D =
∂ν(y)
0,
x 6∈ D .
∂D
Proof: First we note that elements of HD are smooth solutions of the Helmholtz equation
in D. Fix x ∈ D. As in the classical case we apply Green’s formula (5.8) to u and Φ(x, ·) in
D \ B[x, ε]:
Z
∂u(y)
γ1 u, γ0 Φ(x, ·) ∂D −
Φ(x, y)
ds(y)
∂ν
|y−x|=ε
Z
=
∇u(y) · ∇y Φ(x, y) − k 2 u(y) Φ(x, y) dy
D\B[x,ε]
Z
=
∇u(y) · ∇y Φ(x, y) + u(y) ∆y Φ(x, y) dy
D\B[x,ε]
Z
∂Φ(x, y)
= −
u(y)
ds(y) +
∂ν(y)
|y−x|=ε
Z
(γ0 u)(y)
∂D
∂Φ(x, y)
ds(y) ;
∂ν(y)
5.2. SURFACE POTENTIALS
247
that is,
Z
∂Φ(x, y)
ds(y)
∂ν(y)
∂D
Z
∂u(y)
∂Φ(x, y)
Φ(x, y)
=
− u(y)
− ds(y) = u(x)
∂ν
∂ν(y)
|y−x|=ε
γ1 u, Φ(x, ·) ∂D −
(γ0 u)(y)
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). Therefore, u is even analytic by Corollary 3.4.
Proof: Green’s representation theorem shows that u can be expressed as a difference of a
double layer potential with density in L2 (∂D) (even in H 1/2 (∂D)) and a single layer potential
with the density ϕ = γ1 u ∈ H −1/2 (∂D). Only the latter
one has
considered. Let A
to be
be a differential operator. It suffices to show that A ϕ, Φ(x, ·) ∂D = ϕ, (Ax Φ)(x, ·) ∂D in
every ball B such that B ⊆ D. Using an argumentby induction it suffices to show thisfor
A = ∂/∂xj and every function Φ̂ ∈ C ∞ B ×(R3 \B) instead of Φ. Set f (x) = ϕ, Φ̂(x, ·) ∂D
for x ∈ B. Choose an open neighborhood U of ∂D such that d := dist(U, B) > 0. Then
Φ̂j := ∂ Φ̂(x, ·)/∂xj ∈ H 1 (U ) and
f (x + hê(j) ) − f (x) − h ϕ, Φ̂j
∂D
= ϕ, Φ̂(x + hê(j) , ·) − Φ̂(x, ·) − h Φ̂j ∂D ≤ c kϕkH −1/2 (∂D) kΦ̂(x + hê(j) , ·) − Φ̂(x, ·) − h Φ̂j kH 1/2 (∂D)
≤ c0 kϕkH −1/2 (∂D) kΦ̂(x + hê(j) , ·) − Φ̂(x, ·) − h Φ̂j kH 1 (U ) .
The differentiabilty of Φ̂ yields that
kΦ̂(x + hê(j) , ·) − Φ̂(x, ·) − h Φ̂j kH 1 (U ) = O(h2 )
which proves the assertion.
2
In the following we drop the symbol γ0 , thus we write write just v for γ0 v. If ∂D is the
interface of two domains then we write v|± and γ1 v|± 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 ∈ H 1/2 (∂D), respectively; that is,
/ ∂D , ϕ ∈ H −1/2 (∂D) ,
(5.39a)
(S̃ϕ)(x) = ϕ, Φ(x, ·) ∂D , x ∈
Z
∂Φ
(D̃ϕ)(x) =
ϕ(y)
(x, y) ds(y) , x ∈
/ ∂D , ϕ ∈ H 1/2 (∂D) ,
(5.39b)
∂ν(y)
∂D
248
5 BIE FOR LIPSCHITZ DOMAINS
compare with (3.4a), (3.4b).
It is the aim to prove that S̃ and D̃ are bounded maps into H 1 (D) and H 1 (B \ D) for any
ball B containing D in its interior.
From the jump conditions of Theorems 3.12 and 3.16 we recall that for smooth domains D
and smooth densities ϕ the single layer u = S̃ϕ solves the transmission problem
∆u + k 2 u = 0 in R3 \ ∂D ,
∂u ∂u u|+ = u|− on ∂D ,
−
= ϕ on ∂D ,
∂ν − ∂ν +
and u satisfies also the radiation condition (3.2). This interpretation of the single layer
potential, together with the trace theorems, will allow us to prove the required properties
of the boundary operators. A problem is that the region of this transmission problem is
all of R3 ; that is, unbounded. Therefore, we will restrict the transmission problem to a
bounded 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 lead
to a compact perturbation of S̃.
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 every
ϕ ∈ H −1/2 (∂D) there exists a unique solution v ∈ H 1 (B) such that
Z
Z
2
v ψ ds = hϕ, ψi∂D for all ψ ∈ H 1 (B) .
(5.40)
∇v · ∇ψ − k v ψ dx − ik
B
∂B
Furthermore, the operator ϕ 7→ v is bounded from H −1/2 (∂D) into H 1 (B).
Proof: We write (5.40) in the form
(v, ψ)H 1 (B) − a(v, ψ) = hϕ, ψi∂D
where a denotes the sequilinear form
Z
Z
2
a(v, ψ) = (k + 1) v ψ dx + ik
B
for all ψ ∈ H 1 (B) ,
v ψ ds ,
v, ψ ∈ H 1 (B) .
∂B
The mapping ` : ψ 7→ hϕ, ψi∂D is a bounded linear functional on H 1 (B) with k`k ≤
c0 kϕkH −1/2 (∂D) because
hϕ, ψi∂D ≤ c kϕkH −1/2 (∂D) kψkH 1/2 (∂D) ≤ c0 kϕkH −1/2 (∂D) kψkH 1 (D)
≤ c0 kϕkH −1/2 (∂D) kψkH 1 (B) .
The boundedness of the sequilinear form a is shown as follows.
a(v, ψ) ≤ (k 2 + 1) kvkL2 (B) kψkL2 (B) + k kvkL2 (∂B) kψkL2 (∂B) .
(5.41)
5.2. SURFACE POTENTIALS
249
Using the boundedness of the trace operator from H 1 (B) into L2 (∂B) we conclude that
a is bounded. The theorem of Riesz (Theorem 6.5) assures the existence of r ∈ H 1 (B)
and a bounded operator K from H 1 (B) into itself such that hϕ, ψi∂D = (r, ψ)H 1 (B) and
a(v, ψ) = (Kv, ψ)H 1 (B) for all v, ψ ∈ H 1 (B). Furthermore, krkH 1 (B) = k`k ≤ c0 kϕkH −1/2 (∂D) .
Then we write (5.40) as v − Kv = r in H 1 (B) and show that K is compact. From (5.41) for
ψ = Kv we note that
kKvk2H 1 (B) = (Kv, Kv)H 1 (B) = a(v, Kv)
≤ (k 2 + 1) kvkL2 (B) kKvkL2 (B) + k kvkL2 (∂B) kKvkL2 (∂B)
≤ c kvkL2 (B) + kvkL2 (∂B) kKvkH 1 (B) ,
thus
kKvkH 1 (B) ≤ c kvkL2 (B) + kvkL2 (∂B) .
Since in a Hilbert space an operator K is compact if, and only if, it maps weakly convergent
sequences into norm-convergent sequences we consider any such sequence vj * 0 weakly in
H 1 (B). From the boundedness of the trace operator γ0 and the compactness of the imbeddings H 1 (B) in L2 (B) and H 1/2 (∂B) in L2 (∂B) we conclude that the right–hand side of the
previous estimate for vj instead of v converges to zero. Therefore, kKvj kH 1 (B) converges to
zero which proves compactness of K.
Therefore, we can apply the Riesz–Fredholm theory to (5.40). To show existence and
boundedness of the solution operator r 7→ v it suffices to prove uniqueness. Therefore,
let v ∈ H 1 (B) be a solution of (5.40) for ϕ = 0. Substituting ψ = kv into (5.40) yields
Z
Z
2
2
2
2
k|∇v| − k|k| |v| dx − i|k|
|v|2 ds = 0 .
B
∂B
Taking the imaginary part and noting that Im k ≥ 0 yields v = 0 on ∂B. Extending v by
zero into the exterior of B yields v ∈ H 1 (R3 ) (see Exercise 4.9) and
Z
∇v · ∇ψ − k 2 v ψ dx = 0 for all ψ ∈ H 1 (R3 ) ;
R3
that is, ∆v + k 2 v = 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 := S̃ϕ in R3 \ ∂D
layer potential with density ϕ ∈
be the single
−1/2
H
(∂D), defined by u(x) = (S̃ϕ)(x) = ϕ, Φ(x, ·) ∂D , x ∈ R3 \ ∂D. Then:
(a) u ∈ C ∞ (R3 \ ∂D) and u satisfies the Helmholtz equation ∆u + k 2 u = 0 in R3 \ ∂D and
Sommerfeld’s radiation condition (3.2) for |x| → ∞; that is,
1
∂us (rx̂)
s
− ik u (rx̂) = O 2
for r → ∞ ,
(5.42)
∂r
r
uniformly with respect to x̂ ∈ S 2 .
250
5 BIE FOR LIPSCHITZ DOMAINS
(b) Let Q be open and bounded such that D ⊆ Q. The operator S̃ is well-defined and
bounded from H −1/2 (∂D) into H 1 (Q).
(c) u|D ∈ HD and u|R3 \D ∈ HR3 \D and γ1 u|− − γ1 u|+ = ϕ.
(d) The traces
S = γ0 S̃ : H −1/2 (∂D) → H 1/2 (∂D) ,
(5.43a)
1
(γ1 S̃|+ + γ1 S̃|− ) : H −1/2 (∂D) → H −1/2 (∂D) ,
2
are well defined and bounded.
D0 =
(5.43b)
(e) With these notations the jump conditions for u = S̃ϕ and ϕ ∈ H −1/2 (∂D) take the
form
1
(5.44)
γ0 u|± = Sϕ , γ1 u|± = ∓ ϕ + D0 ϕ .
2
Proof: (a) follows from Corollary 5.40.
(b) Let B be a ball which contains Q in its interior. We prove the following representation
of S̃ϕ for ϕ ∈ H −1/2 (∂D).
S̃ϕ = v + w in Q ,
(5.45)
where v ∈ H 1 (B) is the solution of (5.40), and w ∈ H 1 (Q) is explicitely given by
Z
∂Φ
v(y)
w(x) =
(x, y) − ik Φ(x, y) ds(y) , x ∈ Q .
∂ν(y)
∂B
Indeed, (5.40) implies that the restrictions satisfy v|D ∈ HD and v|B\D ∈ HB\D if we choose
ψ ∈ H01 (D) and ψ ∈ H01 (B \ D), respectively, and extend them by zero into the remaining
parts of B. Two applications of Green’s representation theorem (Theorem 5.39 in D and in
B \ D) for x ∈ D yield
Z
∂Φ
(x, y) ds(y) + hγ1 v|− , Φ(x, ·)i∂D , x ∈ D ,
v(x) = −
v(y)|−
∂ν(y)
∂D
Z
∂Φ
0 =
v(y)|+
(x, y) ds(y) − hγ1 v|+ , Φ(x, ·)i∂D
∂ν(y)
∂D
Z
∂Φ
−
v(y)
(x, y) ds(y) + hγ1 v, Φ(x, ·)i∂B , x ∈ D ,
∂ν(y)
∂B
where we dropped the symbol γ0 for the trace operator. Adding both equation yields
v(x) = γ1 v|− − γ1 v|+ , Φ(x, ·) ∂D − w̃(x) , x ∈ D ,
(5.46)
where
Z
∂Φ
(x, y) ds(y) − γ1 v, Φ(x, ·) ∂B
∂ν(y)
∂B
Z
Z
∂Φ
=
v(y)
(x, y) ds(y) −
∇v · ∇Φx − k 2 v Φx dx
∂ν(y)
∂B
B\D
Z
Z
∂Φ
=
v(y)
(x, y) ds(y) − ik
v(y) Φx (y) ds = w(x) ,
∂ν(y)
∂B
∂B
w̃(x) :=
v(y)
x ∈ Q.
5.2. SURFACE POTENTIALS
251
Here, Φx ∈ H 1 (B) is chosen such that Φx = Φ(x, ·) on ∂B and Φx = 0 on D. Then Φx is an
extension of Φ(·, x) ∈ H 1/2 (∂B) into B, and the last equation holds by the definition of v.
For fixed x ∈ D we choose Φ̃x ∈ H 1 (B) such that Φ̃x = Φ(x, ·) on ∂D and Φ̃x = 0 on ∂B.
Now we recall the definition (5.37a), (5.37b) of the traces γ1 v|± and rewrite the first term of
(5.46) as
Z
γ1 v|− − γ1 v|+ , Φ(x, ·) ∂D = γ1 v|− − γ1 v|+ , Φ̃x ∂D =
∇v · ∇Φ̃x − k 2 v Φ̃x dx
B
= hϕ, Φ̃x i∂D
= ϕ, Φ(x, ·) ∂D = (S̃ϕ)(x)
by (5.40) for ψ = Φ̃x . Thus v = S̃ϕ − w in D. For x ∈ B \ D we argue exactly in the same
way to show (5.46) which proves (5.45).
Now we observe that v|∂B 7→ w|Q is bounded from L2 (∂B) into H 1 (Q). Furthermore, ϕ 7→ v
is bounded from H −1/2 (∂D) into H 1 (B). Combining this with the boundedness of the trace
operator v 7→ v|∂B yields boundedness of S̃ from H −1/2 (∂D) into H 1 (Q).
(c) From the representation (5.45) it suffices to show that γ1 v|− − γ1 v|+ = ϕ on ∂D because
w is a classical solution of the Helmholtz
R equation in B. By the definitions of γ1 |± and of v
we have that hγ1 v|− − γ1 v|+ , ψi∂D = Q ∇v · ∇ψ̃ − k 2 v ψ̃ dx = hϕ, ψi∂D for all extensions ψ̃
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 D0 .
2
Corollary 5.43 For every ϕ ∈ H −1/2 (∂D) the single layer potential u = S̃ϕ is the only
variational solution of the transmission problem
∂u ∂u 2
3
−
= ϕ on ∂D ,
∆u + k u = 0 in R \ ∂D , u|− = u|+ on ∂D ,
∂ν − ∂ν +
1
and u satisfies the Sommerfeld radiation condition (5.42); that is, u ∈ Hloc
(R3 ) is the unique
radiating solution of
Z
∇u · ∇ψ − k 2 u ψ dx = hϕ, γ0 ψi∂D for all ψ ∈ H 1 (R3 ) with compact support . (5.47)
R3
Proof: The previous theorem implies that u|D ∈ HD and u|R3 \D ∈ HR3 \D and u satisfies the
radiation condition. Let now ψ ∈ H 1 (R3 ) with compact support which is contained in some
ball B. The definitions of γ1 |± yield again
Z
hϕ, ψi∂D = hγ1 u|− − γ1 u|+ , ψi∂D =
∇u · ∇ψ − k 2 u ψ dx .
B
This proves that u = S̃ϕ solves (5.47). To prove uniqueness let ϕ = 0. Then
Z
∇u · ∇ψ − k 2 u ψ dx = 0 for all ψ ∈ H 1 (R3 ) with compact support ;
R3
252
5 BIE FOR LIPSCHITZ DOMAINS
that is, u is a radiating variational solution of the Helmholtz equation in all of R3 . The
regularity result from Corollary 5.40 yields u ∈ C ∞ (R3 ). Also, u satisfies the radiation
condition which implies u = 0 in R3 by Theorem 3.23.
2
The following properties are helpful for using the boundary integral equation method for
solving 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
hϕ, Si φi∂D = hφ, Si ϕi∂D ,
hϕ, Si ϕi∂D ≥ ckϕk2H −1/2 (∂D)
for all φ, ϕ ∈ H −1/2 (∂D) .
In particular, hϕ, Si ϕi∂D is real valued.
(b) Si is an isomorphism from H −1/2 (∂D) onto H 1/2 (∂D).
(c) S − Si is compact from H −1/2 (∂D) onto H 1/2 (∂D) for any k ∈ C with Im k ≥ 0.
Proof: (a) Define u = S̃i ϕ and v = S̃i ψ in R3 \∂D. As shown before, u is a classical solution
of ∆u − u = 0 in R3 \ ∂D and satisfies the radiation condition. From the representation
theorem (Theorem 3.3 for k = i) we observe that u and v and their derivatives decay
exponentially 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) we
conclude that
hϕ, Si ψi∂D = hγ1 u|− − γ1 u|+ , γ0 vi∂D
Z
Z
=
∇u · ∇v + uv dx +
R3 \D
D
Z
=
∇u · ∇v + uv dx +
B(0,R)
Z
=
∇u · ∇(φv) + φuv dx
Z
∇u · ∇(φv) + φuv dx
R<|x|<R+1
∇u · ∇v + uv dx −
Z
v
|x|=R
B(0,R)
∂u
ds
∂ν
where we used the classical Green’s theorem in last step to the smooth functions u and
φv. Now we use the fact that v and ∇u decay exponentially to zero as R tends to infinity.
Therefore we arrive at
Z
hϕ, Si ψi∂D =
∇u · ∇v + uv dx
R3
which is symmetric in u and v. To prove coercivity we choose ψ = ϕ. Then v = u and thus
Z
hϕ, Si ϕi∂D =
|∇u|2 + |u|2 dx = kuk2H 1 (R3 ) .
R3
5.2. SURFACE POTENTIALS
253
Now we recall
from Theorem 5.10 the existence of a bounded extension operator η : H 1/2 (∂D) →
H 1 B(0, R) ; that is γ0 ηψ|± = ψ and ηψ has compact support in B(0, R). Therefore, for
ψ ∈ H 1/2 (∂D) and ψ̃ = ηψ ∈ H 1 B(0, R) we can estimate as above
Z
∇u · ∇ψ̃ + uψ̃ dx
hϕ, ψi∂D = hγ1 u|− − γ1 u|+ , ψi∂D =
R3
≤ kukH 1 (R3 ) kψ̃kH 1 (B(0,R)) ≤ kηk kukH 1 (R3 ) kψkH 1/2 (∂D) ,
and thus kϕkH −1/2 (∂D) =
hϕ, ψi∂D ≤ kηk kukH 1 (R3 ) . Therefore, hϕ, Si ϕi∂D =
sup
kψkH 1/2 (∂D) =1
1
2
kuk2H 1 (R3 ) ≥ kηk
2 kϕkH −1/2 (∂D) . This finishes the proof of part (a).
(b) Injectivity of Si follows immediately from the coerciveness property of part (a). Furthermore, from part (a) we observe that (ϕ, ψ)S := hϕ, Si ψi∂D) defines an inner product
in H −1/2 (∂D) such that its corresponding norm is equivalent to the ordinary norm in
H −1/2 (∂D). Surjectivity of Si is now an immediate consequence of the Riesz Representation Theorem 6.5. Indeed, let f ∈ H 1/2 (∂D). It defines a linear and bounded functional
` on H −1/2 (∂D) by `(ϕ) := hϕ, f i∂D) for ϕ ∈ H −1/2 (∂D). By the theorem of Riesz (Theorem 6.5) there exists ψ ∈ H 1/2 (∂D) such that (ϕ, ψ)S = `(ϕ) for all ϕ ∈ H −1/2 (∂D); that is,
hϕ, Si ψi∂D) = hϕ, f i∂D) for all ϕ ∈ H −1/2 (∂D); that is, Si ψ = f .
(c) Choose open balls Q and B such that D ⊆ Q and Q ⊆ B. We use that facts that S̃ϕ
and S̃i ϕ have representations of the form (5.45); that is,
S̃ϕ = v + w
and S̃i ϕ = vi + wi
in Q
where v, vi ∈ H 1 (B) are the solutions of (5.40) for k and k = i, respectively, and w, wi ∈
H 1 (Q) are explicitely given by
Z
∂Φk
(x, y) − ik Φk (x, y) ds(y) , x ∈ Q ,
w(x) =
v(y)
∂ν(y)
∂B
and analogously for wi (see proof of Theorem 5.42). We note that the mapping ϕ 7→ v is
bounded from H −1/2 (∂D) into H 1 (B). Therefore, by the trace theorem and the compact
imbedding of H 1/2 (∂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) into
H 1/2 (∂D). It remains to consider v − vi . Taking the difference of the equations for v and vi ;
that is,
Z
Z
2
∇v · ∇ψ − k v ψ dx − ik
v ψ ds = hϕ, ψi∂D ,
B
∂B
Z
∇vi · ∇ψ + vi ψ dx +
B
vi ψ ds = hϕ, ψi∂D ,
∂B
yields
Z
Z
∇(v−vi )·∇ψ+(v−vi ) ψ dx +
B
Z
∂B
2
Z
(v−vi ) ψ ds = (k +1)
Z
v ψ dx + (ik+1)
B
v ψ ds
∂B
254
5 BIE FOR LIPSCHITZ DOMAINS
for all ψ ∈ H 1 (B) which is of the form
Z
Z
∇(v − vi ) · ∇ψ + (v − vi ) ψ dx +
B
Z
(v − vi ) ψ ds =
∂B
Z
f ψ dx +
B
g ψ ds (5.48)
∂B
with f ∈ L2 (B) and g ∈ L2 (∂B). As mentioned before, the mapping ϕ 7→ v is bounded from
H −1/2 (∂D) into H 1 (B) and thus compact as a mapping into L2 (B) as well as the mapping
ϕ 7→ v∂B from H −1/2 (∂D) into L2 (∂B). Finally, the mapping (f, g) into the solution v − vi
of (5.48) is bounded from L2 (B) × L2 (∂B) into H 1 (B). This proves compactness of the
mapping ϕ 7→ v − vi from H −1/2 (∂D) into H 1 (B). The trace theorem yields compactness of
S − Si and ends the proof.
2
Remarks 5.45 (i) The proof of part (b) implies in particular that the dual space of H −1/2 (∂D)
can be identified with H 1/2 (∂D); that is, the spaces H 1/2 (∂D) and H −1/2 (∂D) are reflexive
Banach spaces.
(ii) The symmetry of S holds for any k ∈ C \ {0} with Im k ≥ 0. Indeed, we can just copy
the previous proof of part (a) and arrive at
Z
Z
∂u
2
ds
hϕ, Sψi∂D =
∇u · ∇v − k uv dx −
v
B(0,R)
|x|=R ∂ν
and thus
Z
hϕ, Sψi∂D − hψ, Sϕi∂D = −
|x|=R
∂v
∂u
−u
v
∂ν
∂ν
ds .
It suffices to show that the last integral tends to zero as R → ∞. To see this we just write
Z
Z
Z
∂v
∂v
∂u
∂u
−u
− iku ds −
u
− ikv ds ,
ds =
v
v
∂ν
∂ν
∂ν
∂ν
|x|=R
|x|=R
|x|=R
and this tends to zero by Sonmmerfeld’s radiation condition.
We finish this part of the section by formulating the corresponding theorems for the double
layer potential without detailed proofs.
Theorem 5.46 Let Q be open and bounded such that D ⊆ Q. The double layer operator
D̃, defined in (5.39b), is well-defined and bounded from H 1/2 (∂D) into H 1 (D) and into
H 1 (Q \ D).
Furthermore, with u := D̃ϕ in R3 \ ∂D we have that u|D ∈ HD and u|R3 \D ∈ HR3 \D and
γ0 u|+ − γ0 u|− = ϕ and γ1 u|− − γ1 u|+ = 0. In particular, u ∈ C ∞ (R3 \ ∂D) and satisfies the
Helmholtz equation ∆u + k 2 u = 0 in R3 \ ∂D and the Sommerfeld radiation condition. The
traces
T
= γ1 D̃ : H 1/2 (∂D) → H −1/2 (∂D) ,
D =
1
(γ0 D̃|+ + γ0 D̃|− ) : H 1/2 (∂D) → H 1/2 (∂D)
2
(5.49a)
(5.49b)
5.2. SURFACE POTENTIALS
255
are well defined and bounded. Furthermore, we have the jump conditions for u = D̃ϕ and
ϕ ∈ H 1/2 (∂D):
1
γ0 u|± = ± ϕ + Dϕ , γ1 u|± = T ϕ .
(5.50)
2
Proof (only sketch): We follow the proof of Theorem 5.42 and choose a ball B with D ⊆ B
and prove a decomposition of u = D̃ϕ in the form u = v + w where the pair v|D , v|B\D ∈
H 1 (D) × H 1 (B \ D) is the variational solution of the transmission problem
∂v
∆v + k 2 v = 0 in B \ ∂D ,
− ikv = 0 on ∂B ,
∂ν
∂v ∂v =
, v|+ − v|− = ϕ on ∂D .
∂ν +
∂ν −
With this v we define w just as in the proof of Theorem 5.42. The mapping properties of D
and T follow now directly from this representation and the trace theorems. Finally, (5.50)
follows directly from γ0 u|+ − γ0 u|− = 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 with
respect to the special wave number k = i are useful. We leave the proof of the following
result 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 for
the special value k = i. Then:
(a) Ti is symmetric and coercive in the sense that there exists c > 0 with
hTi ϕ, ϕi∂D ≤ −ckϕk2H 1/2 (∂D)
hTi ϕ, φi∂D = hTi φ, ϕi∂D ,
for all φ, ϕ ∈ H 1/2 (∂D) .
In particular, hTi ϕ, ϕi∂D is real valued.
(b) Ti is an isomorphism from H 1/2 (∂D) onto H −1/2 (∂D).
(c) T − Ti is compact from H 1/2 (∂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 k 2 = ω 2 εµ. We recall the
trace 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 exterior
of D and distinguish in the following between the traces from the exterior (+) and interior
(−) by writing γt u|± and γT u|± , respectively. Due to the direction of ν we have the following
forms of Green’s formula 5.19.
Z
hγt v|− , γT u|− i∂D =
u · curl v − v · curl u dx for all u, v ∈ H(curl, D) ,
D
Z
hγt v|+ , γT u|+ i∂D = −
R3 \D
u · curl v − v · curl u dx for all u, v ∈ H(curl, R3 \ D) ,
256
5 BIE FOR LIPSCHITZ DOMAINS
where h·, ·i∂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 h·, ·i∂D instead of h·, ·i∗ .
In the following we want to discuss vector potentials of the form
Z
Z
curl
a(y) Φ(x, y) ds(y) and curl curl
a(y) Φ(x, y) ds(y) for x ∈
/ ∂D ,
∂D
∂D
see Subsection 3.2.2. From the trace theorem we know that in general we have to allow a to be
in H −1/2 (Div, ∂D). Therefore, we have to give meaning to the boundary integral. We recall
from 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 H 1/2 (∂D, C3 ). The dual form is denoted by h·, ·i∂D , see the
beginning of this section. We define the bilinear mapping
h·, ·i∂D : H −1/2 (Div, ∂D) × H 1/2 (∂D) −→ C3
componentwise; that is, ha, ϕi∂D j = haj , ϕi∂D , j = 1, 2, 3, for a ∈ H −1/2 (Div, ∂D) and
ψ ∈ H 1/2 (∂D). We refer to the remark
R following Lemma 5.27. For smooth tangential fields
a and ϕ this is exactly the integral ∂D ϕ(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 ∈ H 1 (D) we have
Z
hγt v, γ0 ψi∂D =
ψ curl v + ∇ψ × v dx .
D
Proof: For any z ∈ C3 we have hγt v, γ0 ψi∂D · z = hγt v, γT (ψz)i∂D which can be seen by
approximating v and ψ by smooth functions. Therefore, by (5.19),
Z
hγt v, γ0 ψi∂D · z = hγt v, γT (ψz)i∂D =
ψ z · curl v − v · curl(ψz) dx
D
Z
=
ψ z · curl v − v · (∇ψ × z) dx = z ·
D
Z
ψ curl v − v × ∇ψ dx .
D
2
Now we are able to generalize the definition of the vector potentials of Subsection 3.2.2 for
any a ∈ H −1/2 (Div, ∂D). We define
v(x) = curl ha, Φ(x, ·)i∂D ,
x ∈ R3 \ ∂D ,
(5.51)
u(x) = curl2 ha, Φ(x, ·)i∂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 the
representation theorem (compare to the Stratton-Chu formula of Theorem 3.27).
5.2. SURFACE POTENTIALS
257
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 curl E − iωµH = 0 and curl H + iωεE = 0 in D
we have the representation
(
E(x) , x ∈ D ,
1
2
curl hγt H, Φ(x, ·)i∂D =
(5.53a)
− curlhγt E, Φ(x, ·)i∂D +
iωε
0,
x 6∈ D .
(b) For any solutions E, H ∈ Hloc (curl, R3 \ D) of curl E − iωµH = 0 and curl H + iωεE = 0
in R3 \ D which satisfy the Silver–Müller radiation condition (3.37a); that is,
√
x
√
= O
ε E(x) − µ H(x) ×
|x|
1
|x|2
|x| → ∞ ,
,
(5.53b)
uniformly with respect to x̂ = x/|x|, we have the representation
curlhγt E, Φ(x, ·)i∂D
1
curl2 hγt H, Φ(x, ·)i∂D =
−
iωε
(
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 of
curl2 u − k 2 u = 0 in D. We fix z ∈ D and choose a ball Br = B(z, r) such that Br ⊆ D. We
apply Green’s formula of Lemma 5.48 in Dr = D \ B[z, r] to E and Φ(x, ·) for any x ∈ Br
and obtain
Z
Z
Φ(x, ·) curl E + ∇y Φ(x, ·) × E dy .
(ν × E) Φ(x, ·) ds =
hγt E, Φ(x, ·)i∂D −
Dr
∂Br
Analogously, we have for H instead of E:
Z
Z
hγt H, Φ(x, ·)i∂D −
(ν × H) Φ(x, ·) ds =
Φ(x, ·) curl H + ∇y Φ(x, ·) × H dy
∂Br
Dr
and thus
Z
Ir (x) := − curl hγt E, Φ(x, ·)i∂D −
(ν × E) Φ(x, ·) ds
∂Br
Z
1
2
curl hγt H, Φ(x, ·)i∂D −
(ν × H) Φ(x, ·) ds
+
iωε
∂Br
Z
= − curl
Φ(x, ·) curl E + ∇y Φ(x, ·) × E dy
Dr
1
+
curl2
iωε
Z
Dr
Φ(x, ·) curl H + ∇y Φ(x, ·) × H dy
258
5 BIE FOR LIPSCHITZ DOMAINS
Now we use that ∇y Φ(x, ·) × E = − curlx Φ(x, ·)E , curl E = iωµH, and curl H = −iωεE.
This yields
Z
Z
2
Φ(x, ·) E dy − iωµ curl
Φ(x, ·) H dy
Ir (x) = curl
Dr
Dr
Z
1
2
3
Φ(x, ·) H dy − curl
Φ(x, ·) E dy
−
curl
iωε
Dr
Dr
Z
Z
1
2
= − iωµ curl
Φ(x, ·) H dy − 2 curl
Φ(x, ·) H dy = 0
k
Dr
Dr
Z
by using curl2 = ∇ div −∆ and the Helmholtz equation for Φ. Therefore,
1
curl2 hγt H, Φ(x, ·)i∂D
− curl[hγt E, Φ(x, ·)i∂D +
iωε
Z
Z
1
2
(ν × E) Φ(x, ·) ds +
(ν × H) Φ(x, ·) ds
= − curl
curl
iωε
∂Br
∂Br
= E(x)
by the classical Stratton-Chu formula of Theorem 3.27. The case x ∈
/ D is treated in the
same way by appling Lemma 5.48 in all of D.
2
Corollary 5.50 For any variational solution E ∈ H(curl, D) of curl2 E − k 2 E = 0 in D we
have the representation
(
E(x) , x ∈ D ,
1
2
− curlhγt E, Φ(x, ·)i∂D − 2 curl hγt curl E, Φ(x, ·)i∂D =
k
0,
x 6∈ D .
1
Proof: We define H = iωµ
curl E 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 the
assertion.
2
For our proof of the boundedness of the vector potentials and the corresponding boundary
operators we need the (unique) solvability of a certain boundary value problem of transmission type – just as in the scalar case. As a familar technique we will use the Helmholtz
decomposition in the form
H(curl, B) = H(curl, div 0, B) ⊕ ∇H 1 (B)
(5.54)
R
in some
ball
B
where
H(curl,
div
0,
B)
=
u
∈
H(curl,
B)
:
u · ∇ϕ dx = 0 for all ϕ ∈
B
H 1 (B) . We have seen in Theorem 5.37 that H(curl, div 0, B) is compactly imbedded in
L2 (B, C3 ).
Theorem 5.51 Let B1 and B2 be two open balls such that B1 ⊆ D and D ⊆ B2 . Let η ∈ C
with Im (ηk) > 0 and Kj : H −1/2 (Div, ∂Bj ) → H −1/2 (Curl, ∂Bj ) for j ∈ {1, 2} be linear
5.2. SURFACE POTENTIALS
259
and compact operators such that hψ, Kj ψi∂Bj are real valued and hψ, Kj ψ, i∂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 following
boundary value problem is uniquely solvable in H(curl, B) where we set B = B2 \ B 1 for
abbreviation.
curl2 v − k 2 v = 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:
Z
2
X
curl v · curl ψ − k 2 v · ψ dx − η
γt ψ, Kj γt v ∂Bj = ha, γT ψi∂D
B
(5.55)
j=1
for all ψ ∈ H(curl, B). The operator a 7→ γt v|∂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 and
ψ = ψ0 + ∇ϕ with v0 , ψ0 ∈ H(curl, div 0, B) and p, ϕ ∈ H 1 (B), the variational equation
(5.55) is equivalent to
Z
B
2
X
2
2
curl v0 · curl ψ0 − k v0 · ψ0 + k ∇p · ∇ϕ dx − η
γt (ψ0 + ∇ϕ), Kj γt (v0 − ∇p) ∂Bj
j=1
= ha, γT (ψ0 + ∇ϕ)i∂D
(5.56)
for all (ψ0 , ϕ) ∈ X := H(curl, div 0, B) × H 1 (B). We equip X with the norm k(ψ0 , ϕ)k2X =
kψ0 k2H(curl,) + k 2 kϕk2H 1 (B) and denote the corresponding inner product by (·, ·)X . Then the
variational equation (5.56) can be written as
(v0 , p), (ψ0 , ϕ) X −
Z
2
X
γt (ψ0 + ∇ϕ), Kj γt (v0 − ∇p) ∂Bj
(k 2 + 1)v0 ψ0 + k 2 p ϕ dx − η
j=1
B
= ha, γT (ψ0 + ∇ϕ)i∂D
for all (ψ0 , ϕ) ∈ X .
(5.57)
The representation theorem of Riesz (Theorem 6.5) guarantees the existence of (g, q) ∈ X
and a bounded operator A from X into itself such that
(g, q), (ψ0 , ϕ) X = ha, γT (ψ0 + ∇ϕ)i∂D
Z
2
A(v0 , p), (ψ0 , ϕ) X =
(k + 1)v0 ψ0 + k 2 p ϕ dx
B
+η
2
X
γt (ψ0 + ∇ϕ), Kj γt (v0 − ∇p) ∂Bj
j=1
for all (v0 , p), (ψ0 , ϕ) ∈ X. Therefore, (5.57) can be written as
(v0 , p) − A(v0 , p) = (g, q) in X .
260
5 BIE FOR LIPSCHITZ DOMAINS
From the estimate
A(v0 , p), (ψ0 , ϕ) ≤ (k 2 + 1)k(v0 , p)kL2 (B)×L2 (B) k(ψ0 , ϕ)kL2 (B)×L2 (B)
X
+ c1
2
X
kKj γt (v0 − ∇p)kH −1/2 (Curl,∂Bj ) kγt (ψ0 + ∇ϕ)kH −1/2 (Div,∂Bj )
j=1
"
≤ c k(v0 , p)kL2 (B)×L2 (B) +
2
X
#
kKj γt (v0 − ∇p)kH −1/2 (Curl,∂Bj ) k(ψ0 , ϕ)kX
j=1
we conclude (set (ψ0 , ϕ) = A(v0 , p)) that
2
X
kA(v0 , p)kX ≤ c k(v0 , p)kL2 (B)×L2 (B) +
kKj γt (v0 − ∇p)kH −1/2 (Curl,∂Bj ) .
j=1
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
P
2 j=1 γt v, Kj γt v ∂Bj = 0 and thus γt v = 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
Z
curl v · curl ψ − k 2 v · ψ dx = 0 for all ψ ∈ H(curl, R3 ) ;
R3
that is, v solves curl2 v − k 2 v = 0 in R3 and vanishes outside of B. The unique continuation
principle yields v = 0 in R3 .
It remains to show that the operator a 7→ γt v|∂D is an isomorphism from H −1/2 (Div, ∂D) onto
itself. This follows immediately from the unique solvability of the following two boundary
value problems for any given b ∈ H −1/2 (Div, ∂D):
curl2 vi − k 2 vi = 0 in D \ B 1 ,
ν × vi = b on ∂D ,
ν × curl vi + η ν × K1 (ν × vi ) = 0 on ∂B1 ,
(5.58a)
(5.58b)
and
curl2 ve − k 2 ve = 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 operators
Kj : H −1/2 (Div, ∂Bj ) → H −1/2 (Curl, ∂Bj ) such that hψ, Kj ψi∂Bj > 0 for all ψ ∈ H −1/2 (Div, ∂Bj ),
ψ 6= 0. Such operators exist. For example, consider the case that ∂Bj = S 2 and the operator
Kj is defined by
∞ X
n
X
m m
1
m
Kj ψ =
an Un + bm
n Vn
1 + n(n + 1)
n=0 m=−n
5.2. SURFACE POTENTIALS
261
m m m m
P
Pn
for ψ = ∞
∈ H −1/2 (Div, S 2 ). This Kj has the desired properties,
n=0
m=−n an Un +bn Vn
see Exercise 5.6.
Now we are able to prove all of the desired properties of the vector potentials and their
behaviours 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 L̃ and M̃, defined by
(L̃a)(x) = curl2 ha, Φ(x, ·)i∂D
(M̃a)(x) = curlha, Φ(x, ·)i∂D
for x ∈ Q ,
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 = M̃a and curl u = L̃a satisfy u|D , curl u|D ∈
H(curl, D) and u|Q\D , curl u|Q\D ∈ H(curl, Q\D) and γt u|− −γt u|+ = a and γt curl u|− −
γt curl u|+ = 0. In particular, u ∈ C ∞ (R3 \ ∂D, C3 ) and u satisfies the equation
curl2 u − k 2 u = 0 in R3 \ ∂D. Furthermore, u and curl u satisfy the Silver–Müller
radiation condition (3.42); that is,
curl u × x̂ − ik u = O |x|−2 , |x| → ∞ ,
uniformly with respect to x̂.
(c) The traces
L = γt L̃
M =
on ∂D ,
(5.59a)
1
(γt M̃|− + γt M̃|+ ) on ∂D ,
2
(5.59b)
are bounded from H −1/2 (Div, ∂D) into itself. With these notations the jump conditions
hold for u = M̃a and a ∈ H −1/2 (Div, ∂D) in the form
1
γt u|± = ∓ a + Ma ,
2
γt curl u|± = La .
(5.60)
(d) L is the sum L = L̂ + K̂ of an isomorphism L̂ from H −1/2 (Div, ∂D) onto itself and a
compact operator K̂.
(e) L̃a can be written as
L̃a = ∇S̃ Div a + k 2 S̃a ,
a ∈ H −1/2 (Div, ∂D) ,
(5.61)
with the single layer S̃a from (5.39a) (in the second occurence taken componentwise).
262
5 BIE FOR LIPSCHITZ DOMAINS
We note that some authors call L̃a the Maxwell single layer and M̃a the Maxwell double
layer. The matrix operator
1
+
M
L
2
C =
1
+M
L
2
is called the electromagnetic Calderon operator.
Proof: (a)–(d)We argue similarily as in the proof of Theorem 5.42. It is sufficient to prove
this for Q being a neighborhood of ∂D. Therefore, let B1 and B2 be two open balls such
that B1 ⊆ D and D ⊆ B2 and let Q be an open set with ∂D ⊆ Q. Choose η ∈ C with
Im (ηk) > 0 and Kj : H −1/2 (Div, ∂Bj ) → H −1/2 (Curl, ∂Bj ) for j ∈ {1, 2} as in the previous
theorem. For a ∈ H −1/2 (Div, ∂D) let v ∈ H(curl, B) be the solution of (5.55). We prove the
following representation of u = L̃f .
L̃a = k 2 (v + w) in B
(5.62)
where
2
X
1
2
j
w(x) =
(−1) curlhγt v, Φ(x, ·)i∂Bj + 2 curl hγt curl v, Φ(x, ·)i∂Bj
k
j=1
for x ∈ B2 \B 1 . To prove this we fix x ∈ D \B 1 and apply Corollary 5.50 of the Stratton-Chu
formula in D \ B 1 and in B2 \ D, respectively, and obtain (note the different signs because
of the orientation of ν)
1
curl2 hγt curl v|− , Φ(x, ·)i∂D
k2
1
+ curlhγt v, Φ(x, ·)i∂B1 + 2 curl2 hγt curl v, Φ(x, ·)i∂B1 ,
k
1
0 = curlhγt v|+ , Φ(x, ·)i∂D + 2 curl2 hγt curl v|+ , Φ(x, ·)i∂D
k
1
− curlhγt v, Φ(x, ·)i∂B2 − 2 curl2 hγt curl v, Φ(x, ·)i∂B2 .
k
v(x) = − curlhγt v|− , Φ(x, ·)i∂D −
Adding both equations and using the transmission condition yields
v(x) =
1
1
curl2 ha, Φ(x, ·)i∂D − w(x) = 2 (L̃a)(x) − w(x) ,
2
k
k
which proves (5.62) for x ∈ D. For x ∈ B2 \ D we argue analogously. Taking the trace in
(5.62) yields
La = k 2 (γt v + γt w) = L̂a + K̂a .
L̂ is an isomorphism by Theorem 5.51, and K̂ is compact by the smoothness of w. Finally,
the properties of M̃ and M follow from the relation curl L̃a = curl2 M̃a = k 2 M̃a, thus
M̃a = curl v + curl w, 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 5.40
for the arguments) yields for x ∈
/ ∂D
L̃a(x) = ∇ divha, Φ(x, ·)i∂D − ∆ha, Φ(x, ·)i∂D
3 X
∂
Φ(x, ·)
+ k 2 ha, Φ(x, ·)i∂D
= ∇
aj ,
∂x
j
∂D
j=1
= ∇ha, ∇x Φ(x, ·)i∂D + k 2 ha, Φ(x, ·)i∂D
= −∇ha, ∇y Φ(x, ·)i∂D + k 2 ha, Φ(x, ·)i∂D .
By the identification of ha, ∇y Φ(x, ·)i∂D with ha, γT ∇y Φ(x, ·)i∂D (see Remark 5.28) and the
definition of the surface divergence (Definition 5.29) we conclude that ha, ∇y Φ(x, ·)i∂D =
−hDiv a, Φ(x, ·)i∂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 value
problems. 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 ∈ H 1/2 (∂D) be
given bounday data.
Interior Dirichlet Problem: Find u ∈ H 1 (D) such that γ0 u = f and ∆u + k 2 u = 0 in D;
that is, in variational form
Z
∇u · ∇ψ − k 2 u ψ dx = 0 for all ψ ∈ H01 (D) .
(5.63a)
D
To formulate the exterior problem we define the local Sobolev space by
1
Hloc
(R3 \ D) := u : R3 \ D → C : u|B ∈ H 1 (B) for all balls B .
1
Exterior Dirichlet Problem: Find u ∈ Hloc
(R3 \ D) such that γ0 u = f and ∆u + k 2 u = 0 in
3
R \ D; that is, in variational form
Z
∇u · ∇ψ − k 2 u ψ dx = 0 for all ψ ∈ H01 (R3 \ D) with compact support, (5.63b)
R3 \D
and u satisfies Sommerfeld’s radiation condition (5.42). Note that u is a smooth solution of
the 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 boundary
value problem (5.63b).
264
5 BIE FOR LIPSCHITZ DOMAINS
(b) The interior Dirichlet boundary value problem (5.63a) has at most one solution if, and
only if, the single layer boundary operator S is one-to-one. More precisely, the null space
N (S) of S is given by all Neumann traces γ1 u of solutions u ∈ H01 (D) of the Helmholtz
equation with vanishing Dirichlet boundary data; that is,
Z
2
1
1
∇u · ∇ψ − k u ψ dx = 0 for all ψ ∈ H0 (D) .
N (S) = γ1 u : u ∈ H0 (D),
D
1
Proof: (a) Let u ∈ Hloc
(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 function
φ ∈ C ∞ (R3 ) such that φ = 1 on B[0, R] und φ = 0 in the exterior of B(0, R +1). Application
of Green’s formula (5.19) in the region B(0, R + 1) \ D to u and ψ = φu yields
Z
Z
∂u
0 = −hγ1 u, ψi∂D +
ψ ds =
∇u · ∇ψ − k 2 u ψ dx
|x|=R+1 ∂ν
B(0,R+1)\D
Z
Z
=
|∇u|2 − k 2 |u|2 dx +
∇u · ∇ψ − k 2 u ψ dx
B(0,R)\D
Z
R<|x|<R+1
=
|∇u|2 − k 2 |u|2 dx −
Z
|x|=R
B(0,R)\D
∂u
u ds .
∂ν
Note that ψ vanishes on ∂D and outside of B(0, R + 1). Now we proceed similarly to the
proof of Theorem 3.23. Indeed,


)
2
Z Z
Z ( 2
∂u
∂u ∂u 
ds =
+ |ku|2 ds − 2 Im 
−
iku
u
ds
k
∂r
∂r ∂r
|x|=R
|x|=R
|x|=R
Z
=
|x|=R
( )
Z
∂u 2
2
2
+ |ku|2 ds + 2 Im k
|∇u|
+
|ku|
dx .
∂r B(0,R)\D
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 R
tends to infinity by the radiation condition. Rellich’s lemma (Lemma 3.21 or 3.22) implies
that u vanishes in the (connected) exterior of D. This proves part (a).
(b) Let first u ∈ H 1 (D) satisfy (5.63a) and γ0 u = 0. Then, by the definition of S̃ and Green’s
representation formula of Theorem 5.39
(S̃γ1 u)(x) = γ1 u, Φ(x, ·) ∂D
Z
∂Φ
= γ1 u, Φ(x, ·) ∂D −
(γ0 u)(y)
(x, y) ds(y)
∂ν(y)
∂D
= u(x) ,
x ∈ D.
Taking the trace yields Sγ1 u = γ0 S̃γ1 u = γ0 u = 0 on ∂D.
5.3. BOUNDARY INTEGRAL EQUATION METHODS
265
Second, let ϕ ∈ H −1/2 (∂D) such that Sϕ = 0. Define u = S̃ϕ in R3 \∂D. Then u|D ∈ H01 (D)
and γ0 u|+ = 0 on ∂D. Therefore, u|R3 \D satisfies the homogeneous exterior boundary value
problem and therefore vanishes by part (a). The jump condition of Theorem 5.42 implies
ϕ = γ1 u|− − γ1 u|+ = γ1 u|− which ends the proof.
2.
Remark: This theorem relates the homogeneous interior Dirichlet problem to the null space
of the boundary operator S. Therefore, k 2 is a Dirichlet eigenvalue of −∆ in the sense of
Subsection 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-toone. Furthermore, the null space N (S) is finite dimensional.
The following theorem studies the question of existence for the case that k 2 is not a Dirichlet
eigenvalue.
Theorem 5.54 Assume in addition to the assumptions at the beginning of this section that
k 2 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 H01 (D) is the trivial one u = 0.
Then there exist (unique) solutions of the exterior and the interior Dirichlet boundary value
problems for every f ∈ H 1/2 (∂D). The solutions can be represented as single layer potentials
in the form
u(x) = (S̃ϕ)(x) = hϕ, Φ(x, ·)i∂D , x ∈
/ ∂D ,
where the density ϕ ∈ H −1/2 (∂D) satisfies Sϕ = f .
Proof: By the mapping properties of S̃ of Theorem 5.42 it suffices to study solvability of the
equation Sϕ = f . Because S is a compact perturbation of the isomorphism Si (the operator
corresponding to k = i) by Theorem 5.42, the well known – and already often used – result
by 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 single
layer potential u solves both, the exterior and the interior boundary value problems with
homogeneous boundary data f = 0. The uniqueness result implies that u vanishes in all of
R3 . The jumps of the normal derivatives (Theorem 5.42 again) yields ϕ = 0.
2
Furthermore we consider the case of k 2 being a Dirichlet eigenvalue of −∆ in D. This
corresponds to the case where S fails to be one-to-one. In this case we have to apply the
Fredholm theory of Theorem 6.4 to the boundary equation Sϕ = f .
Theorem 5.55 The interior and exterior boundary value problems are solvable as single
layer potentials u = S̃ϕ for exactly those f ∈ H 1/2 (∂D) which are orthogonal to all Neumann
traces γ1 v ∈ H −1/2 (∂D) of eigenfunctions v ∈ H01 (D) of −∆ in D; that is,
hγ1 v, f i∂D = 0 for all v ∈ H01 (D) with ∆v + k 2 v = 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 = H 1/2 (∂D), hϕ, f i1 = hϕ, f i∂D for ϕ ∈
266
5 BIE FOR LIPSCHITZ DOMAINS
H −1/2 (∂D), f ∈ H 1/2 (∂D) and hg, ψi2 = hψ, gi∂D for ψ ∈ H −1/2 (∂D), g ∈ H 1/2 (∂D). The
operator S is self adjoint in this dual system, see Remarks 5.45. Application of Theorem 6.4
yields solvability of the equation Sϕ = f for exactly those f ∈ H 1/2 (∂D) with hψ, f i∂D = 0
for all ψ ∈ N (S ∗ ) = N (S). The characterization of the nullspace N (S) in Theorem 5.53
yields the assertion.
2
Remark: For scattering problems by plane waves we have to solve the exterior Dirichlet
problem with boundary data f (x) = − exp(ik θ̂ · x) on ∂D. This boundary data satisfies the
orthogonality condition. Indeed, by Green’s formula (5.8) we conclude for every eigenfunction v ∈ H01 (D) that
Z h
Z
i
∂
ik θ̂·x
2
ik θ̂·x
hγ1 v, exp(ik θ̂·)i∂D =
∇v(x) · ∇e
− k v(x) e
dx =
v(x) eik θ̂·x ds = 0 .
∂ν
D
∂D
Therefore, the scattering problem for plane waves by the obstacle D, can always be solved
by a single layer ansatz.
From the uniqueness result (and Chapter 3) we expect that the exterior boundary value
problem is always uniquely solvable, independently of the wave number. However, this can’t
be done by just one single layer potential. There are several ways to modify the ansatz. In our
context perhaps the simplest possibility is to choose an ansatz as the following combination
of a single and a double layer ansatz
u = S̃ϕ + η D̃Si ϕ in the exterior of D
for some ϕ ∈ H −1/2 (∂D) and some η ∈ C to be chosen in a moment. The boundary operator
Si : H −1/2 (∂D) → H 1/2 (∂D) corresponds to S for k = i. Then u is a solution of the exterior
boundary value problem provided ϕ ∈ H −1/2 (∂D) solves
Sϕ +
η
Si ϕ + η DSi ϕ = f on ∂D .
2
(5.64)
By Theorem 5.44 the operator S is a compact perturbation of the isomorphism Si from
H −1/2 (∂D) onto H 1/2 (∂D) and the operator DSi is bounded. Therefore, for sufficiently
small η0 > 0 also S + η2 Si + η DSi 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 this
choice of η the boundary equation (5.64) is uniquely solvable for every f ∈ H 1/2 (∂D). By
the previous remarks it is sufficient to show that the homogeneous equations admits only the
trivial solution. Therefore, let ϕ ∈ H −1/2 (∂D) be a solution of (5.64) for f = 0 and define
u by u = S̃ϕ + η D̃Si ϕ 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 that
u vanishes in the exterior of D. The jump conditions yield
γ0 u|− = γ0 u|− − γ0 u|+ = −η Si ϕ ,
γ1 u|− = γ1 u|− − γ1 u|+ = ϕ .
Elimination of ϕ yields ηSi γ1 u|− + γ0 u|− = 0. Now we apply Green’s theorem in D; that is,
Z
∇u · ∇ψ − k 2 u ψ dx = hγ1 u|− , γ0 ψi∂D .
D
5.3. BOUNDARY INTEGRAL EQUATION METHODS
267
Substituting ψ = ku yields
Z
k |∇u|2 − k |ku|2 dx = k hγ1 u|, γ0 u|− i∂D = −kη hγ1 u|− , Si γ1 u|− i∂D .
D
Now we note that hγ1 u|− , Si γ1 u|− i∂D is real and nonnegative by Theorem 5.44 and Im (kη) >
0 by assumption. Therefore, taking the imaginary part yields hγ1 u|− , Si γ1 u|− i∂D = 0. The
property of Si from Theorem 5.44 yields γ1 u|− = 0 and thus also ϕ = γ1 u|− − γ1 u|+ = 0.
We formulate the result as a theorem.
Theorem 5.56 The exterior Dirichlet boundary value problem is uniquely solvable for every
f ∈ H 1/2 (∂D). The solution can be expressed in the form
u = S̃ϕ + η D̃Si ϕ
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 of
k. Indeed, one can show by the same arguments as in the proof of Theorem 5.44 that D − Di
is compact. Therefore, we have to choose |η| small enough such that (1 + η/2)Si + ηDi Si is
an isomorphism from H −1/2 (∂D) onto H 1/2 (∂D).
Exactly the same results hold for the interior and exterior Neumann problems. We formulate
the 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 hf, γ0 vi∂D = 0 for all solutions v ∈ H 1 (D) of ∆v + k 2 v = 0 in D and γ1 v = 0 on ∂D.
In this case the solution can be represented as a double layer potential u = D̃ϕ in D with
ϕ ∈ H 1/2 (∂D) satisfying T ϕ = f on ∂D where T : H 1/2 (∂D) → H −1/2 (∂D) denotes the
trace 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 = D̃ϕ + η S̃Ti ϕ in R3 \ D ,
with ϕ ∈ H 1/2 (∂D) for some sufficiently small η ∈ C with Im (ηk) > 0. The density
ϕ ∈ H 1/2 (∂D) satisfies the boundary equation
Tϕ −
η
Ti ϕ + D 0 Ti ϕ = f
2
on ∂D .
We now turn to the electromagnetic case and formulate the boundary value problems. We
assume 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
268
5 BIE FOR LIPSCHITZ DOMAINS
operator γt : H(curl, D) → H −1/2 (∂D) extends the mapping u 7→ ν × u|∂D (and analogously
for exterior domains). Furthermore, let ω > 0, µ > 0, and ε ∈ C with Im ε ≥ 0 and
√
k = ω µε with Re k > 0 and Im k ≥ 0.
The Interior Boundary Value Problem: Find E, H ∈ H(curl, D) such that γt E = f and
curl E − iωµH = 0 in D
and
curl H + iωεE = 0 in D ;
that is, in variational form for the field E (see, e.g., (4.2) or (4.22)).
Z
curl E · curl ψ − k 2 E · ψ dx = 0 for all ψ ∈ H0 (curl, D) .
(5.65a)
(5.65b)
D
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 γt E = f
and
(5.66a)
curl E − iωµH = 0 in R3 \ D and curl H + iωεE = 0 in R3 \ D ;
that is, in variational form for the field E,
Z
curl E · curl ψ − k 2 E · ψ dx = 0 for all ψ ∈ H0 (curl, R3 \ D) with compact support .
R3 \D
(5.66b)
Furthermore, (E, H) have to satisfy the Silver–Müller radiation condition (5.53b); that is,
√
x
1
√
ε E(x) − µ H(x) ×
= O
(5.66c)
|x|
|x|2
uniformly with respect to x/|x| ∈ S 2 . Note that E and H are a smooth solutions of curl2 u −
k 2 u = 0 in the exterior of D.
The question of uniqueness and existence are treated in a very analogous way to the scalar
problems 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, and
only if, the boundary operator L is one-to-one. More precisely, the null space N (L) of L is
given by all traces γt curl u of solutions u ∈ H0 (curl, D) of curl2 u − k 2 u = 0 with vanishing
boundary data γt u; that is,
R curl u · curl ψ − k 2 u · ψ dx = 0
D
N (L) = γt curl u : u ∈ H0 (curl, D),
.
for all ψ ∈ H0 (curl, D)
5.3. BOUNDARY INTEGRAL EQUATION METHODS
Proof: (a) Let
balls such that
on B[0, R] and
and v = H in
∂B(0, R + 1))
269
E, H be a solution of (5.65a)–(5.65b) corresponding to f = 0. Choose again
D ⊆ B(0, R) ⊆ B(0, R + 1) and a function φ ∈ C ∞ (R3 ) such that φ = 1
φ = 0 outside of B(0, R + 1). Green’s formula (5.19) applied to u = φE
the region B(0, R + 1) \ D yields (note that γT u vanishes on ∂D and on
Z
u · (ν × v) ds
0 = −hγt v, γT ui∂D +
|x|=R+1
Z
φE · curl H − H · curl(φE) dx
=
B(0,R+1)\D
Z
= iω
2
2
R<|x|<R+1
= iω
u · curl v − v · curl u dx
ε |E| − µ |H| dx +
B(0,R)\D
Z
Z
ε |E|2 − µ |H|2 dx −
Z
E · (x̂ × H) ds .
|x|=R
B(0,R)\D
We consider again the case Im ε > 0 and Im ε = 0 separately. If Im ε > 0 then the fields
decay exponentially as R tends to infinity. This follows again from the Stratton-Chu formula
of Theorem 5.49. Taking the imaginary part of the previous formula and letting R tend to
infinity yields that E vanishes.
Let now ε be real valued. Taking the real part of the previous formula yields
Z
E · (x̂ × H) ds ≥ 0 .
Re
|x|=R
This yields
√ √
0 ≥ 2 µ ε Re
Z
E · (H × x̂) ds
|x|=R
Z
=
√
√
| εE|2 + | µH × x̂|2 ds −
|x|=R
Z
√
√
| εE − µH × x̂|2 ds .
|x|=R
The Silver–Müller radiation condition
R implies that the second integral tends to zero as R
tends to infinity. This implies that |x|=R |E|2 ds 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 the
exterior of D.
(b) Let first u ∈ H0 (curl, D) with curl2 u − k 2 u = 0 in D. By Corollary 5.50 we conclude for
x ∈ D, because γt u = 0,
(L̃γt curl u)(x) = curl2 hγt curl u, Φ(x, ·)i∂D
= curl2 hγt curl u, Φ(x, ·)i∂D + k 2 curlhγt u, Φ(x, ·)i∂D
= −k 2 u(x) .
Taking the trace yields L(γt curl u) = −k 2 γt u = 0 on ∂D.
270
5 BIE FOR LIPSCHITZ DOMAINS
Second, let a ∈ H −1/2 (Div, ∂D) with La = 0. Define u by u(x) = (L̃a)(x) = curl2 ha, Φ(x, ·)i∂D
for x ∈ R3 \ ∂D. Then u ∈ Hloc (curl, R3 ) solves the exterior and the interior boundary value
problem with homogeneous boundary data γt u = La = 0. The uniqueness result for the exterior problem implies that u vanishes in the exterior. The jump conditions of Theorem 5.52
yield a = γt curl u|− − γt curl u|+ = γt curl u|− . This proves part (b).
2
Theorem 5.60 Assume in addition to the assumptions at the beginning of this section that
k 2 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 one
u = 0. Then there exist (unique) solutions of the exterior and the interior boundary value
problems for every f ∈ H −1/2 (Div, ∂D). The solutions can be represented as boundary layer
potentials in the form
E(x) = (L̃a)(x) = curl2 ha, Φ(x, ·)i∂D ,
H=
1
curl E ,
iωµ
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 boundary 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 the
aim to apply the abstract Fredholm result of Theorem 6.4 and have to find the proper dual
system. The following result will provide the adjoint of L.
Lemma 5.61 (a) The bilinear form h·, ·i on H −1/2 (Div, ∂D) × H −1/2 (Div, ∂D), defined
by ha, bi = ha, ν × bi∂D is well defined and a dual system.
(b) For a, b ∈ H −1/2 (Div, ∂D) we have
hLa, ν × bi∂D = hLb, ν × ai∂D ,
and the adjoint of L with respect to h·, ·i is given by L∗ = −L.
(c) Let Li be the operator L for k = i. Then there exists c > 0 with
hLi a, ai ≥ c kak2H −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
hLa, bi = hDiv b, S Div ai∂D + k 2 hb, Sai∂D
for all a, b ∈ H −1/2 (Div, ∂D)
where S : H −1/2 (∂D) → H 1/2 (∂D) denotes the single layer boundary operator. In
the second occurence it is considered as a bounded operator from H −1/2 (Div, ∂D) into
H −1/2 (Curl, ∂D), see Lemma 5.27.
5.3. BOUNDARY INTEGRAL EQUATION METHODS
271
Proof: (a) On the dense subspace {ν × u|∂D : u ∈ C ∞ (D, C3 )} of H −1/2 (Div, ∂D) the
mapping a 7→ a × ν is expressed as a × ν = γT ηt a 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 γT ηt a has a bounded extension from H −1/2 (Div, ∂D) into
H −1/2 (Curl, ∂D). The bilinear form h·, ·i is non–degenerated because H −1/2 (Curl, ∂D) is the
dual of H −1/2 (Div, ∂D).
(b) Define u, v ∈ Hloc (curl, R3 ) by
u(x) = curl2 ha, Φ(x, ·)i∂D ,
v(x) = curl2 hb, Φ(x, ·)i∂D ,
x∈
/ ∂D .
Then, by Theorem 5.52, γt curl v|− − γt curl v|+ = k 2 b and thus k 2 b × ν = γT curl v|− −
γT curl v|+ . Therefore, by Green’s theorem in the form (5.19), applied in D and in B(0, R)\D,
respectively, and adding the results yields
k 2 hLa, b × νi∂D = hγt u, γT curl v|− i∂D − hγt u, γT curl v|+ i∂D
Z
Z
2
=
[curl v · curl u − u · curl v] dx −
|x|<R
Z
2
Z
[curl v · curl u − k v · u] dx −
=
(ν × u) · curl v ds
|x|=R
|x|<R
(ν × u) · curl v ds .
|x|=R
Changing the roles of a and b and subtracting the results yield
Z
2
2
k hLa, b × νi∂D − k hLb, a × νi∂D =
(ν × v) · curl u − (ν × u) · curl v] ds .
|x|=R
The last term converges to zero as R tends to infinity by the radiation condition. For
a, b ∈ H −1/2 (Div, ∂D) we have with respect to the dual system
hLa, bi = hLa, ν × bi∂D = hLb, ν × ai∂D = −ha, ν × Lbi∂D = −ha, Lbi .
(c) Let now k = i. Taking b = a in part (a) yields
Z
Z
2
2
[| curl u| + |u| ] dx −
hLi a, ai = −hLi a, a × νi∂D =
|x|<R
(ν × u) · curl u ds .
|x|=R
u and curl u decay exponentially to zero as R tends to infinity. Therefore, we arrive at
hLi a, ai = kuk2H(curl,R3 ) .
Finally, the boundedness of the trace operator yields
kakH −1/2 (Div,∂D) = kγt curl u|+ − γt curl u|− kH −1/2 (Div,∂D) ≤ c0 kukH(curl,R3 ) .
(d) Taking the trace in (5.61) yields
La = γt ∇S̃ Div a + k 2 γt S̃a
272
5 BIE FOR LIPSCHITZ DOMAINS
and thus for a, b ∈ H −1/2 (Div, ∂D):
hLa, bi = hLa, ν × bi∂D = hγt ∇S̃ Div a, ν × bi∂D + k 2 hν × Sa, ν × bi∂D
= hb, γT ∇S̃ Div ai∂D + k 2 hb, γT S̃ai∂D = hDiv b, S Div ai∂D + k 2 hb, Sai∂D
by the definition of the surface divergence.
2
Theorem 5.62 The interior and exterior boundary value problems are solvable by a layer
potential u = L̃a for exactly those f ∈ H −1/2 (Div, ∂D) which are orthogonal to all traces
γT curl v ∈ H −1/2 (Curl, ∂D) of eigenfunctions v ∈ H0 (curl, D) of curl2 in D; that is,
hf, γT curl vi∂D = 0 for all v ∈ H0 (curl, D) with curl2 v − k 2 v = 0 in the variational sense.
Proof: We have to discuss solvability of the equation La = f . As mentioned above we want
to apply Theorem 6.4 and define the dual system by X1 = X2 = Y1 = Y2 = H −1/2 (Div, ∂D)
and ha, bi = ha, ν × bi∂D for a, b ∈ H −1/2 (Div, ∂D) as in the previous Lemma. The adjoint
L∗ 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 − k 2 v = 0 in D. The solvability condition hf, gi = 0 reads as 0 = hf, ν ×
γt curl vi∂D = −hf, γT curl vi∂D which proves the theorem.
2
For our reference scattering problem from the introduction (see Section 1.5) we obtain a
final conclusion.
Corollary 5.63 For any Lipschitz domain D ⊆ R3 the scattering problem (3.36a)–(3.37b)
has a unique solution. The scattered field E s can be represented by a layer potential E s = L̃a
where the density a ∈ H −1/2 (Div ∂D) satisfies La = −γt E inc .
Proof: For the scattering problem by a perfect conductor we have to solve the exterior
problem with boundary data f (x) = −γt E inc on ∂D. This boundary data satisfies the
orthogonality condition of the previous theorem. Indeed, by Green’s formula (5.19) we
conclude for every eigenfunction v ∈ H0 (curl, D) that
Z
inc
hγt E , γT curl vi∂D =
curl v · curl E inc − k 2 v E inc dx = hγt v, γT curl E inc i∂D = 0 .
D
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 all
boundary data f , independently of the wave number. By the previous results it is clear that
we have to modify the ansatz u = L̃a. We proceed just as in the scalar case and propose an
ansatz in the form
u = L̃a + η M̃Li a in the exterior of D .
5.3. BOUNDARY INTEGRAL EQUATION METHODS
273
By the jump conditions of Theorem 5.52 the vector field u solves the exterior boundary value
problem for f ∈ H −1/2 (Div, ∂D) if, and only if, a ∈ H −1/2 (Div, ∂D) satisfies the equation
La −
η
Li a + η MLi a = f
2
on ∂D .
(5.67)
As shown in Theorem 5.51 the operator L is the sum of an isomorphism and a compact
operator. Therefore, there exists η0 > 0 such that also L − η2 Li + η MLi 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 = L̃a + η M̃Li a. Then γt u|+ = 0 on
∂D and thus u = 0 in the exterior of D by the uniqueness theorem. The jump conditions
yield
γt u|− = γt u|− − γt u|+ = η Li a ,
γt curl u|− = γt curl u|− − γt curl u|+ = k 2 a
because curl L̃a = k 2 M̃a. Elimination of a yields ηLi γt curl u|− = k 2 γt u|− . Now we apply
Green’s theorem in D; that is,
Z
curl u · curl ψ − k 2 u · ψ dx = hγt ψ, γT curl u|− i∂D .
D
Substituting ψ = ku yields
Z
k | curl u|2 − k |ku|2 dx = −k hγt u|− , ν × γt curl u|− i∂D
D
= −
kη
hLi γt curl u|− , ν × γt curl u|− i∂D .
|k|2
Now we observe that hLi γt curl u|− , ν × γt curl u|− i∂D is nonnegative by Lemma 5.61 and
Im (kη) < 0 by assumption. Therefore, the imaginary part of the right–hand side is nonnegative while the imaginary part of the left–hand side is non-positive. Taking the imaginary
part yields hLi γt curl u|− , ν ×γt curl u|− i∂D = 0 and thus γt curl u|− = 0 by Lemma 5.61. This
ends the proof because k 2 a = γt curl u|− − γt curl u|+ = 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 = L̃a + η M̃Li a
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 of
the 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.
274
5.4
5 BIE FOR LIPSCHITZ DOMAINS
Exercises
Exercise 5.1 Prove Theorem 5.6; that is, compactness of the mapping J : u 7→ u from
s
t
(Q) for s < t.
(Q) into Hper
Hper
P
Hint: Show that the mapping (JN u)(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 H 1 (D) into
L2 (∂D) for D = B2 (0, 1) being the unit disk in R2 by
(a) computing αn for n ∈ N such that kun kH 1 (D) = 1 where un (r, ϕ) = αn rn , and
(b) showing that k∂un /∂rkL2 (∂D) → ∞ as n → ∞.
Exercise 5.3 Prove Corollary 5.11
Hint: 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,
p
Z
n(n + 1) m
curl u(rx̂) · x̂ Yn−m (x̂) ds(x̂) = −
wn (r) ,
r
S2
Z
0
1
curl u(rx̂) · Un−m (x̂) ds(x̂) = − rwnm (r) ,
r
S2
p
Z
0
n(n + 1) m
1
curl u(rx̂) · Vn−m (x̂) ds(x̂) = −
un (r) +
rvnm (r) .
r
r
S2
um (r)
m
curl um
x̂ × GradS 2 Ynm (x̂) ,
= − n
n (r) Yn (x̂) x̂
r
0
1
curl vnm (r) GradS 2 Ynm (x̂) = − r vnm (r) x̂ × GradS 2 Ynm (x̂) ,
r
0
1
curl wnm (r) x̂ × GradS 2 Ynm (x̂)
= − r wnm (r) GradS 2 Ynm (x̂)
r
− wnm (r)
n(n + 1) m
Yn (x̂) x̂ .
r
Exercise 5.5 Let B = B(0, R) be a ball and f ∈ C0∞ (B).
(a) Show that the volume potential
Z
ṽ(x) =
f (y)
B
is in C ∞ (B) and solves ∆ṽ = −f in B.
1
dy ,
4π|x − y|
x∈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 the
form v = ṽ + u.
Exercise 5.6 Define the operator K by
Kψ =
∞ X
n
X
m m
1
m
for
an Un + bm
n Vn
1 + n(n + 1)
n=0 m=−n
∞ X
n
X
m m
m
ψ =
an Un + bm
∈ H −1/2 (Div, S 2 ) .
n Vn
n=0 m=−n
Show that K is well defined and compact from H −1/2 (Div, S 2 ) into H −1/2 (Curl, S 2 ). Show,
furthermore, that hKψ, ψiS 2 is real valued and hKψ, ψiS 2 > 0 for all ψ ∈ H −1/2 (Div, S 2 ),
ψ 6= 0.
Exercise 5.7 Show that the boundary value problem (5.58a), (5.58b) is uniquely solvable
for all b ∈ H −1/2 (Div, ∂D).
Hints: Transform first the boundary value problem to homogeneous boundary data by choosing a proper extension v̂ of b and make the ansatz vi = v̂ + v. Second, use the Helmholtz
decomposition as in the proof of Theorem 5.51.
276
5 BIE FOR LIPSCHITZ DOMAINS
Chapter 6
Appendix
Some notations and basic results are essential throughout the whole monograph. Therefore
we 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 the
common inner product and the cross product by


3
x2 y 3 − x3 y 2
X
x·y =
xj y j
and
x × y =  x3 y 1 − x1 y 3  .
j=1
x1 y 2 − x2 y 1
Frequently we apply the elementary formulas
x · (y × z) = y · (z × x) = z · (x × y)
x × (y × z) = (x · z)y − (x · y)z
(6.1)
(6.2)
Furthermore, we have to fix the notation of some basic differential operators. We will use
the gradient ∇. Additionally we introduce div F = ∇ · F for the divergence of a vector field
F and curl F = ∇ × F for its rotation. For scalar fields we denote the Laplacian operator
by ∆ = div ∇ = ∇ · ∇. Let u : R3 → C and F : R3 → C3 be sufficiently smooth functions
then in a cartesian coordinate system we have




∂F3 ∂F2
∂u
 ∂x1 
 ∂x2 − ∂x3 
 ∂u 
 ∂F
∂F3 




1
curl F = ∇ × F = 
∇u = 
−


 ∂x3
 ∂x2 
∂x1 
 ∂F2 ∂F1 
 ∂u 
−
∂x3
∂x1
∂x2
div F = ∇ · F =
3
X
∂Fj
j=1
∂xj
,
∆u = div ∇u =
3
X
∂ 2u
j=1
277
∂x2j
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 functions
F, G : R3 → C3 we have
curl ∇u = 0
(6.3)
div curl F = 0
(6.4)
curl curl F = ∇ div F − ∆F
(6.5)
div(λF ) = F · ∇λ + λ div F
(6.6)
curl(λF ) = ∇λ × F + λ curl F
(6.7)
∇(F · G) = (F 0 )> G + (G0 )> F
(6.8)
div(F × G) = G · curl F − F · curl G
curl(F × G) = F div G − G div F + F 0 G − G0 F ,
(6.9)
(6.10)
where F 0 (x), G0 (x) ∈ C3×3 are the Jacobian matrices of F and G, respectively, at x; that is,
Fij0 = ∂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 a
vector field and consider cylindrical coordinates


r cos ϕ
x =  r sin ϕ  .
z
With the coordinate unit vectors ẑ = (0, 0, 1)> , r̂ = (cos ϕ, sin ϕ, 0)> , and ϕ̂ = (− sin ϕ, cos ϕ, 0)>
and the representation F = Fr r̂ + Fϕ ϕ̂ + Fz ẑ we obtain
∇f (r, ϕ, z) =
1 ∂f
∂f
∂f
r̂ +
ϕ̂ +
ẑ ,
∂r
r ∂ϕ
∂z
1 ∂(rFr )
1 ∂Fϕ
∂Fz
+
+
,
r ∂r
r ∂ϕ
∂z
1 ∂Fz ∂Fϕ
∂Fr ∂Fz
curl F (r, ϕ, z) =
r̂ +
ϕ̂
−
−
r ∂ϕ
∂z
∂z
∂r
1 ∂(rFϕ ) ∂Fθ
ẑ ,
+
−
r
∂θ
∂ϕ
1 ∂
∂f
1 ∂ 2f
∂ 2f
∆f (r, ϕ, z) =
r
+ 2
+
.
r ∂r
∂r
r ∂ϕ2
∂z 2
div F (r, ϕ, z) =
Essential for the second chapter are spherical coordinates


r sin θ cos ϕ
x =  r sin θ sin ϕ 
r cos θ
(6.11a)
(6.11b)
(6.11c)
(6.11d)
6.2. RESULTS FROM LINEAR FUNCTIONAL ANALYSIS
279
and the coordinate unit vectors r̂ = (sin θ cos ϕ, sin θ sin ϕ, cos θ)> , θ̂ = (cos θ cos ϕ, cos θ sin ϕ, − sin θ)> ,
and ϕ̂ = (− sin ϕ, cos ϕ, 0)> . With F = Fr r̂ + Fθ θ̂ + Fϕ ϕ̂ we have the equations
∇f (r, θ, ϕ) =
1 ∂f
1 ∂f
∂f
r̂ +
θ̂ +
ϕ̂ ,
∂r
r ∂θ
r sin θ ∂ϕ
(6.12a)
1 ∂(r2 Fr )
1 ∂(sin θ Fθ )
1 ∂Fϕ
div F (r, θ, ϕ) = 2
+
+
,
(6.12b)
r
∂r
r sin θ
∂θ
r sin θ ∂ϕ
1
∂(sin θ Fϕ ) ∂Fθ
1
1 ∂Fr ∂(rFϕ )
curl F (r, θ, ϕ) =
−
r̂ +
−
θ̂
r sin θ
∂θ
∂ϕ
r sin θ ∂ϕ
∂r
1 ∂(rFθ ) ∂Fr
+
−
ϕ̂ ,
(6.12c)
r
∂r
∂θ
1 ∂
1
∂
∂f
1
∂ 2f
2 ∂f
.(6.12d)
r
+ 2
sin θ
+ 2 2
∆f (r, θ, ϕ) = 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
6.2
Results from Linear Functional Analysis
We assume that the reader is familiar with the basic facts from linear functional analysis: in
particular with the notions of normes spaces, Banach- and Hilbert spaces, linear, bounded
and 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 linear
and bounded operator; that is, there exists c > 0 with kT xkY ≤ ckxkX for all x ∈ V . Then
there exists a unique extension T̃ : V → Y to the closure V of V ; that is, T̃ x = T x for all
x ∈ V and kT̃ xkY ≤ ckxkX for all x ∈ V . Furthermore, kT̃ k = kT k.
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 differentiable
functions. The proof of boundedness of an operator (e.g. a trace operator) is usually easier
for smooth functions.
The next two theorems are the functional analytic basis of many existence theorems for
boundary value problems.
Theorem 6.2 Let T : X → Y be a linear and bounded operator between the normed spaces
X and Y . Let T be of the form T = A + K such that A is an isomprphism from X onto Y
and K : X → Y is compact. If T is one-to-one then also onto, and T −1 is bounded from Y
onto X. In other words, if the homogeneous equation T x = 0 admits only the trivial solution
x = 0 then the inhomogeneous equation T x = y is uniquely solvalble for all y ∈ Y , and the
solution x depends continuously on y.
By writing T = A(I + A−1 K) it is obvious that it is sufficient to consider the case Y = X
and A = I. For this case the result is proven in, e.g., [12], Chapter 3. It is also a special
casae of Theorem 6.4 below.
280
CHAPTER 6. APPENDIX
The case where I − K fails to be one-to-one is answered by the following theorem which is
called 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 h·, ·i : X × Y → C
is called a dual system, if h·, ·i is nondegenerated; that is, for every x ∈ X, x 6= 0, the linear
form y 7→ hx, yi does not vanish identically and vice versa, for every y ∈ Y , y 6= 0, the linear
form x 7→ hx, yi does not vanish identically.
Let hX1 , Y1 i and hX2 , Y2 i be two dual forms. Two operators T : X1 → X2 and S : Y2 → Y1
are called adjoint to each other if
hT x, yi2 = hx, Syi1
for all x ∈ X1 , y ∈ Y2 .
Theorem 6.4 (Fredholm) Let hXj , Yj ij , j = 1, 2, be two dual systems, T : X1 → X2 a
bounded operator with bounded adjoint operator T ∗ : Y2 → Y1 such that T = T̂ + K and
T ∗ = T̂ ∗ + K ∗ with isomorphisms T̂ and T̂ ∗ and compact operators K and K ∗ . Then the
following hold:
(a) The dimensions of the null spaces of T and T ∗ are finite and coincide; that is, dim N (T ) =
dim N (T ∗ ) < ∞.
(b) The equations T x = u, T ∗ y = v are solvable for exactly those u ∈ X2 and v ∈ Y1 for
which
hu, ψi2 = 0 for all ψ ∈ N (T ∗ ) ⊂ Y2
and
hϕ, vi1 = 0 for all ϕ ∈ N (T ) ⊂ X1 .
Proof: We can easily reduce the problem to the case that X1 = X2 and Y1 = Y2 and
T̂ = T̂ ∗ = id which is of particular importance in itself.. Indeed, the equation T x = u is
equivalent to x + T̂ −1 Kx = T̂ −1 u and the operator K̃ = T̂ −1 K maps X1 into itself. On the
other hand, the equation T ∗ y = v is equivalent to z + K ∗ (T̂ ∗ )−1 z = v for z = T̂ ∗ y. The
operator K ∗ (T̂ ∗ )−1 from Y1 into itself is just the adjoint of K̃. Also, hu, yi2 = hT̂ −1 u, zi1 for
z = T̂ ∗ 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 value
problems, 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 bounded
functional. Then there exists a unique z ∈ X with `(x) = (x, z)X for all x ∈ X. Furthermore,
k`k = kxkX .
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-linear
and bounded and coercive; that is, there exist c1 , c2 > 0 with
a(u, v) ≤ c1 kukX kvkX for all u, v ∈ X ,
Re a(u, u) ≥ c2 kuk2X
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 kukX ≤ k`kX ∗ .
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 rigourously
the notion of domain with a C n −boundary or Lipschitz boundary (see Evans [9]). We denote
by Bj (x, r) := {y ∈ Rj : |y − x| < r} and Bj [x, r] := {y ∈ Rj : |y − x| ≤ r} the open and
closed 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 C n -smooth (that is, D ∈ C n ), if there exists a
finite number of open cylinders Uj of the form Uj = {Rj x+z (j) : x ∈ B2 (0, αj )×(−2βj , 2βj )}
with z (j) ∈ R3 and rotations1 Rj ∈ R3×3 and real valued functions
ξj ∈ C n (B[0, αj ]) with
Sm
|ξj (x1 , x2 )| ≤ βj for all (x1 , x2 ) ∈ B2 [0, αj ] such that ∂D ⊂ j=1 Uj and
Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 = ξj (x1 , x2 ) ,
D ∩ Uj = Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 < ξj (x1 , x2 ) ,
Uj \ D = Rj x + z (j) : (x1 , x2 ) ∈ B2 (0, αj ) , x3 > ξj (x1 , x2 ) .
∂D ∩ Uj =
We call D to be a Lipschitz domain if the functions ξj which describe the boundary locally are
Lipschitz 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 denote
by
Cj = Cj (αj , βj ) = B2 (0, αj )×(−2βj , 2βj ) = x = (x1 , x2 , x3 ) ∈ R3 : x21 +x22 < αj2 , |x3 | < 2βj
the cylinders with parameters αj and βj . We can assume without loss of generality that
βj ≥ αj (otherwise split the parameter region into smaller ones). Furthermore, we set
1
that is, Rj> Rj = I and det Rj = 1
282
CHAPTER 6. APPENDIX
Bj := B3 (0, αj ) ⊂ Cj , j = 1, . . . , m, and introduce the mappings Ψ̃j : Bj → R3 defined by


x1
 + z (j) , x = (x1 , x2 , x3 )> ∈ Bj ,
x2
Ψ̃j (x) = Rj 
ξj (x1 , x2 ) + x3
and its restriction Ψj to B2 (0, αj ); that is,


x1
 + z (j) ,
x2
Ψj (x̃) = Rj 
ξj (x1 , x2 )
x̃ = (x1 , x2 )> ∈ B2 (0, αj ) ,
which
yields
a parametrization of ∂D ∩ Uj in the form y = Ψj (x̃) for x̃ ∈ B2 (0, αj ) with
∂Ψj ∂Ψj p
∂x1 × ∂x2 = 1 + |∇ξj |2 provided the functions ξj are differentiable.
S
0
We set Uj0 = Ψ̃j (Bj ). Then ∂D ⊂ m
j=1 Uj
∂D ∩ Uj0 = Ψ̃j (x) : x ∈ Bj ,
D ∩ Uj0 = Ψ̃j (x) : x ∈ Bj ,
Uj0 \ D = Ψ̃j (x) : x ∈ Bj ,
and Bj ∩ (R2 × {0}) = B2 (0, αj ) × {0}, and
x3 = 0 = Ψj (x̃) : x̃ ∈ B2 (0, αj ) ,
x3 < 0 ,
x3 > 0 .
Therefore, the mappings Ψ̃j “flatten” the boundary. For C 1 −domains D we note that the
Jacobian is given by


1
0
0
0
1
0 ,
Ψ̃0j (x) = Rj 
∂1 ξj (x) ∂2 ξj (x) 1
where ∂` ξj = ∂ξj /∂x` for ` = 1, 2. We note that its determinant is one. The tangential
vectors at y = Ψj (x) ∈ ∂D ∩ Uj are computed as




1
0
∂Ψj
 , ∂Ψj (x) = Rj 
.
0
1
(x) = Rj 
∂x1
∂x2
∂1 ξj (x)
∂2 ξj (x)
They span the tangent plane at y = Ψj (x). The vector


−∂1 ξj (x)
∂Ψj
∂Ψj
(x) ×
(x) = Rj  −∂2 ξj (x) 
∂x1
∂x2
1
is orthogonal to the tangent plane and is directed into the exterior of D. The corresponding
unit vector


∂Ψj
∂Ψj
−∂1 ξj (x)
(x)
×
(x)
1
∂x2
= p
ν(y) = ∂x1
Rj  −∂2 ξj (x) 
2
∂Ψj
∂Ψj
1
+
|∇ξ
(x)|
j
∂x1 (x) × ∂x2 (x)
1
is called the exterior unit normal vector.
6.3. ELEMENTARY FACTS FROM DIFFERENTIAL GEOMETRY
283
Remark 6.8 For Lipschitz domains the functions ξj are merely Lipschitz continuous. Therefore, Ψ̃j and its inverse Ψ̃−1
j , given by


ŷ1

 , ŷ = Rj> (y − z (j) ) , y ∈ Uj0 = Ψ̃j (Bj ) ,
ŷ2
Ψ̃−1
j (y) =
ŷ3 − ξj (ŷ1 , ŷ2 )
are also Lipschitz continuous. A celebrated result of Rademacher [21] (see also [9], Section 5.8) states that every Lipschitz continuous function ξj is differentiable at almost every
point x ∈ B2 (0, αj ) with |∇ξj (x)| ≤ Lj for almost all x ∈ B2 (0, αj ) where Lj is the Lipschitz
constant. Therefore, for Lipschitz domains the exterior unit normal vector ν(x) exists at
almost all points x ∈ ∂D. Furthermore, u ∈ L1 (Uj0 ) if, and only, if u ◦ Ψ̃j ∈ L1 (Bj ) and the
transformation formula holds in the form
Z
Z
u(y) dy =
u Ψ̃j (x) dx .
(6.13)
Uj0
Bj
Application of this result to |u|2 shows that the operator u 7→ u ◦ Ψ̃j is bounded from L2 (Bj )
into L2 (Cj0 ).
R
For such domains and continuous functions f : ∂D → C the surface integral ∂D f 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 KS⊂ R3 be a compact set. For every finite set {Uj : j = 1, . . . , m} of open
domains with
Pm
∞
3
U
there
exist
φ
∈
C
(R
)
with
supp(φ
)
⊂
U
for
all
j
and
K ⊂ m
j
j
j
j=1 φj (y) = 1
j=1 j
for all y ∈ K. We call {φj : j = 1, . . . , m} a partition of unity on K subordinate to
{Uj : j = 1, . . . , m}.
Using a local coordinate system {Uj , ξj : j = 1, . . . , m} of ∂D with corresponding mappings
Ψ̃j from the balls Bj onto Uj0 and their restrictions Ψj : B2 (0, αj ) → Uj0 ∩ ∂D as inR Definition 6.7 and a corresponding partion of unity φj on ∂D with respect to Uj0 we write ∂D f ds
in the form
Z
m Z
m Z
X
X
f ds =
φj f ds =
fj ds
∂D
j=1
∂D∩Uj0
j=1
∂D∩Uj0
with fj (y) = φj (y)f (y). The integral over the surface patch Uj0 ∩ ∂D is given by
Z
Z
q
1 + |∇ξj (x)|2 dx .
fj ds =
fj Ψj (x)
Uj0 ∩∂D
B2 (0,αj )
We collect important properties of the smooth domain D in the following lemma.
Lemma 6.10 Let D ∈ C 2 . Then there exists c0 > 0 such that
(a) ν(y) · (y − z) ≤ c0 |z − y|2 for all y, z ∈ ∂D,
284
CHAPTER 6. APPENDIX
(b) ν(y) − ν(z) ≤ c0 |y − z| for all y, z ∈ ∂D.
(c) Define
Hη :=
z + tν(z) : z ∈ ∂D , |t| < η .
Then there exists η0 > 0 such that for all η ∈ (0, η0 ] and every x ∈ Hη there exist
unique (!) z ∈ ∂D and |t| ≤ η with x = z + tν(z). The set Hη is an open neighborhood
of ∂D for every η ≤ η0 . Furthermore, z − tν(z) ∈ D and z + tν(z) ∈
/ D for 0 < t < η
and 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 : inf z∈∂D |x − z| < δ denotes the strip around ∂D then there exists
δ > 0 with
Uδ ⊂ Hη0 ⊂ Uη0
(6.14)
(d) There exists r0 > 0 such that the surface area of ∂B(z, r) ∩ D for z ∈ ∂D can be
estimated 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 parametrization Ψj : B2 (0, αj ) → ∂D ∩ Uj0 . First, it is easy to see (proof by contradiction) that there
exists δ > 0 with the property that for every pair (z, x) ∈ ∂D × R3 with |z − x| < δ there
exists Uj0 with z, x ∈ Uj0 . 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) δ 2 ≤ diam(D) |y − x|2 .
δ2
δ2
0
Let now |y − x| < δ. Then y, x ∈ Uj for some j. Let x = Ψj (u) and y = Ψj (v). Then
∂Ψj
(u)
∂u1
×
ν(x) = ∂Ψ
∂u1j (u) ×
∂Ψj
(u)
∂u2
∂Ψj
(u)
∂u2
and, by the definition of the derivative,
y − x = Ψj (v) − Ψj (u) =
2
X
k=1
(vk − uk )
∂Ψj
(u) + a(v, u)
∂uk
with a(v, u) ≤ c|u − v|2 for all u, v ∈ Uj0 and some c > 0. Therefore,
2
X
∂Ψj
1
∂Ψ
∂Ψ
j
j
ν(x) · (y − x) ≤ (v
(u)
×
(u)
·
(u)
k − uk ) ∂Ψj
∂Ψj
∂u
∂u
∂u
1
2
k
∂u1 (u) × ∂u2 (u) k=1
|
{z
}
= 0
∂Ψj
1
∂Ψ
j
+ (u)
×
(u)
·
a(v,
u)
∂Ψj
∂Ψj
∂u
∂u
1
2
(u)
×
(u)
∂u1
∂u2
2
−1
≤ c0 |x − y|2 .
≤ c |u − v|2 = c Ψ−1
(x)
−
Ψ
(y)
j
j
6.3. ELEMENTARY FACTS FROM DIFFERENTIAL GEOMETRY
285
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
S
0
(iii) Hη0 ⊂ m
j=1 Uj .
Assume that x ∈ Hη for η ≤ η0 has two representation as x = z1 + t1 ν1 = z2 + t2 ν2 where
we write νj for ν(zj ). Then
1
|z1 − z2 | ,
|z1 − z2 | = (t2 − t1 ) ν2 + t1 (ν2 − ν1 ) ≤ |t1 − t2 | + η c0 |z1 − z2 | ≤ |t1 − t2 | +
16
thus |z1 − z2 | ≤
16
|t
15 1
− t2 | ≤ 2|t1 − t2 |. Furthermore, because ν1 · ν2 ≥ 0,
(ν1 + ν2 ) · (z1 − z2 ) = (ν1 + ν2 ) · (t2 ν2 − t1 ν1 ) = (t2 − t1 ) (ν1 · ν2 + 1) ,
|
{z
}
≥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 | ≥ 1−16c0 η > 0
and thus also z1 = z2 .
Let U 0 be one of the sets Uj0 and Ψ : R2 ⊃ B2 (0, α) → U 0 ∩ ∂D the corresponding bijective
mapping. We define the new mapping F : R2 ⊃ B2 (0, α) × (−η, η) → Hη by
(u, t) ∈ B2 (0, α) × (−η, η) .
For sufficiently small η the mapping F is one-to-one and satisfies det F 0 (u, t) ≥ c̃ > 0 on
B2 (0, α) × (−η, η) for some c̃ > 0. Indeed, this follows from
F (u, t) = Ψ(u) + t ν(u) ,
0
F (u, t) =
>
∂ν
∂Ψ
∂ν
∂Ψ
(u) + t
(u) ,
(u) + t
(u) , ν(u)
∂u1
∂u1
∂u2
∂u2
and the fact that for t = 0 the matrix F 0 (u, 0) has full rank 3. Therefore,
F is a bijective
S
0
mapping 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 for
0 < t < η.
For x = z + tν(z) and y ∈ ∂D we have
2
|x − y|2 = (z − y) + tν(z) ≥ |z − y|2 + 2t(z − y) · ν(z)
≥ |z − y|2 − 2ηc0 |z − y|2
≥
1
|z − y|2
4
because 2ηc0 ≤
3
.
4
286
CHAPTER 6. APPENDIX
Therefore, |z − y| ≤ 2|x − y|. Finally,
2
|x1 − x2 |2 = (z1 − z2 ) + (t1 ν1 − t2 ν2 ) ≥ |z1 − z2 |2 − 2 (z1 − z2 ) · (t1 ν1 − t2 ν2 )
≥ |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
|z1 − z2 |2
4
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 and
arbitrary z ∈ ∂D and σ > 0 we define
Z(σ) = x ∈ ∂B(z, r) : (x − z) · ν(z) ≤ σ
We show that
Z(−2c0 r2 ) ⊂ ∂B(z, r) ∩ D ⊂ Z(+2c0 r2 )
Let x ∈ Z(−2c0 r2 ) have the form x = x0 + tν(x0 ). Then
(x − z) · ν(z) = (x0 − z) · ν(z) + t ν(x0 ) · ν(z) ≤ −2c0 r2 ;
that is,
t ν(x0 ) · ν(z) ≤ −2c0 r2 + (x0 − z) · ν(z) ≤ −2c0 r2 + c0 |x0 − z|2
≤ −2c0 r2 + 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 = 2c0 r2 .
Therefore, the surface area of ∂B(z, r) ∩ D is bounded from below and above by the surface
areas of Z(−2c0 r2 ) and Z(+2c0 r2 ), respectively. Since the surface area of Z(σ) is 2πr(r + σ)
we have
−4πc0 r3 ≤ |∂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 C n (D, C2 )
we denote the space of vector fields F : D → C3 which are n−times continuously differentiable. By C n (D, C3 ) we denote the subspace of C n (D, C3 ) that consists of those functions F
which, together with all derivatives up to order n, have continuous extentions to the closure
D 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 div F ∈ C(D) the
identity
Z
Z
div F (x) dx =
F (x) · ν(x) ds
D
∂D
holds. In particular, the integral on the and left–hand side exists.
Furthermore, application of this formula to F = uve(j) for u, v ∈ C 1 (D) and the j−th unit
vector e(j) yields the formula of partial integration in the form
Z
Z
Z
u ∇v dx = −
v ∇u dx +
u v ν ds .
D
D
∂D
For a proof for Lipschitz domains we refer to [16]. For smooth domains a proof can be
found 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 ∈ C 2 (D) ∩ C 1 (D). Then
Z
Z
∂v
(u ∆v + ∇u · ∇v) dx =
u
ds ,
∂ν
D
∂D
Z Z
∂v
∂u
(u ∆v − ∆u v) dx =
u
−v
ds .
∂ν
∂ν
∂D
D
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. Then
F satisfies the assumption of Theorem 6.11 and div F = u ∆v + ∇u · ∇v.
The second identity is derived by interchanging the roles of u and v in the first identity and
taking 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 ∈ C 1 (D, C3 ) ∩ C(D, C3 )
and let u ∈ C 2 (D) ∩ C 1 (D). Then
Z
Z
curl A dx =
ν × A ds ,
(6.16a)
D
Z
∂D
Z
(B · curl A − A · curl B) dx =
D
(ν × A) · B ds ,
(6.16b)
u (ν · A) ds .
(6.16c)
∂D
Z
Z
(u div A + A · ∇u) dx =
D
∂D
288
CHAPTER 6. APPENDIX
Proof: For the first identity we consider the components separately. For the first one we
have


Z
Z
Z 0
∂A3 ∂A2
−
dx =
div  A3  dx
(curl A)1 dx =
∂x
∂x
2
3
D
D
D
−A2


Z
Z
0
=
ν ·  A3  ds =
(ν × A)1 ds .
∂D
∂D
−A2
For the other components it is proven in the same way.
For the second equation we set F = A × B. Then div F = B · curl A − A · curl B and
ν · F = ν · (A × B) = (ν × A) · B.
For the third identity we set F = uA and have div F = u div A + 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 and
the surface divergence which are differential operators on the boundary ∂D. We assume
throughout this section that D ⊆ R3 is a C 2 −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 C 2 −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) for
x = (x1 , x2 ) ∈ B2 (0, αj ) and define
C 1 (∂D) := f ∈ C(∂D) : (φj f ) ◦ Ψj ∈ C 1 B2 (0, α) for all j = 1, . . . , m ,
C 1 (∂D, C3 ) := F ∈ C(∂D, C3 ) : Fj ∈ C 1 (∂D) for j = 1, 2, 3 ,
Ct (∂D) := F ∈ C(∂D, C3 ) : F · ν = 0 on ∂D ,
Ct1 (∂D) := Ct (∂D) ∩ C 1 (∂D, C3 ) .
There exist several different – but equivalent – approaches to define the surface gradient
and surface divergence. We decided to choose one which uses the ordinary gradient and
divergence, respectively, on a neighborhood of the boundary ∂D. To do this we need to
extend functions and vector fields. We point out that the same technique is used to construct
extension operators for Sobolev spaces, see, e.g., Theorem 4.13.
Lemma 6.15 Let D ⊆ R3 be a C 2 −smooth domain with boundary ∂D.
(a) For every f ∈ C 1 (∂D) there exists f˜ ∈ C 1 (R3 ) with compact support and f˜ = f on
∂D.
6.5. SURFACE GRADIENT AND SURFACE DIVERGENCE
289
(b) For every F ∈ Ct1 (∂D) there exists F̃ ∈ C 1 (R3 , C3 ) with compact support and F̃ = F
on ∂D.
Proof: (a) Using a local coordinate system {Uj , ξj : j = 1, . . . , m} and a corresponding
P
partition of unity {φj : j = 1, . . . , m} on ∂D as in Definition 6.14 we note that f = m
j=1 fj
on ∂D where fj = f φj . The functions fj ◦ Ψj are continuously differentiable functions from
B2 (0, αj ) into C with support in B2 (0, αj ). We extend fj ◦Ψj into the cylinder C̃j = Cj (αj , βj )
by setting gj (x) := ρ(x3 )(fj ◦ Ψj )(x1 , x2 ) for x = (x1 , x2 , x3 ) ∈ C̃j where ρ ∈ C0∞ (−βj , βj )
is such that ρ = 1 in a neighborhood of 0. Then gj : C̃j → C is continuously differentiable,
has compact support, and gj = fj ◦ Ψ̃j on B2 (0, αj ) × {0}. Therefore, f˜j := gj ◦ Ψ̃−1
has
Pj m ˜
3
˜
˜
compact support in Ũj = Ψ̃j (C̃j ). We extend fj by zero into all of R and set f := j=1 fj
S
˜
in R3 . Then f˜ ∈ C 1 (R3 ) with support in m
j=1 Ũj 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 ∈ C 1 (∂D) and f˜ ∈ C 1 (U ) be an extension of f into a neighborhood
U of the boundary ∂D of the domain D ∈ C 2 . Furthermore, let F ∈ Ct1 (∂D) be a tangential
vector field and F̃ ∈ C 1 (U, C3 ) be an extension into U .
(a) The surface gradient of f is defined as the orthogonal projection of ∇f˜ onto the tangent
plane; that is,
∂ f˜
Grad f = ν × (∇f˜ × ν) = ∇f˜ −
ν
∂ν
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
Div F = div F̃ − ν · (F̃ 0 ν)
on ∂D
(6.18)
where F̃ 0 (x) ∈ C3×3 denotes the Jacobian matrix of F̃ at x.
We will see in Lemma 6.19 that the definitions are independent of the choices of the extensions.
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 cos φ, sin θ sin φ, cos θ)> .
Then the surface gradient and surface divergence, respectively, on the sphere ∂D are given
by
Grad f (θ, φ) =
Div F (θ, φ) =
1
∂f
1 ∂f
(θ, φ) θ̂ +
(θ, φ) φ̂ ,
R ∂θ
R sin θ ∂φ
(6.19)
1
∂
1
∂Fφ
sin θ Fθ (θ, φ) +
(θ, φ) ,
R sin θ ∂θ
R sin θ ∂φ
(6.20)
290
CHAPTER 6. APPENDIX
where θ̂ = (cos θ cos φ, cos θ sin φ, − sin θ)> and φ̂ = (− sin φ, cos φ, 0)> are the tangential unit
vectors which span the tangent plane and Fθ , Fφ are the components of F with respect to
these vectors; that is, F = Fθ θ̂ + Fφ φ̂.
In case of calculations in spherical coordinates, x = rx̂ ∈ R3 , often the surface differential
operators 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 S 2 for the differential operators
with respect to the unit sphere. Thus on a sphere of radius r it is
GradS 2 f (r, x̂) = r Grad f (rx̂) ,
and
DivS 2 F (r, x̂) = r Div F (rx̂) ,
where we understand f (r, ·) as a function on the unit sphere on the left side and f as a
function 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| = R
we note that in spherical coordinates it holds
1
∂ 2f
∂
∂f
1
(θ, φ) + 2 2
(θ, φ) .
sin θ
∆∂D f (Rx̂) = Div Grad f (θ, φ) = 2
R sin θ ∂θ
∂θ
R sin θ ∂φ2
Using the corresponding operators for the unit sphere we obtain with the spherical LaplaceBeltrami operator introduced in Definition 2.2 by the different views on the function f that
∆S 2 f (rx̂) = DivS 2 Grad
S 2 f (rx̂)
= r2 ∆∂D f (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 extend
also the vector field ν into a neighborhood U of ∂D such that |ν̃(x)| = 1 on U . This is
possible because by Lemma 6.15 there exists an extension ν̂ ∈ C 1 (R3 , C3 ) of ν. Certainly,
ν̂ 6= 0 in a neighborhood U of ∂D because |ν̂| = 1 on ∂D. Therefore, ν̃ = ν̂/|ν̂| will be the
required extension into U .
Lemma 6.18 Let D ∈ C 2 and F ∈ Ct1 (∂D) and f ∈ C 1 (∂D) with extensions F̃ ∈
C 1 (R3 , C3 ) and f˜ ∈ C 1 (R3 ), respectively. Then:
(a) Div F = ν·curl(ν̃×F̃ ) on ∂D where ν̃ ∈ C 1 (U ) is an extension of ν into a neighborhood
U of ∂D such that |ν̃(x)| = 1 on U .
(b) Let Γ ⊂ ∂D be a relatively open subset2 such that the (relative) boundary C = ∂Γ is
a 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;
2
that 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) is
directed “outwards” of Γ for all x ∈ C. Then
Z
Z
Div F ds =
F · (τ × ν) d` .
Γ
In particular,
R
∂D
C
Div F ds = 0.
(c) Partial integration holds in the following form:
Z
Z
f Div F ds = −
F · Grad f ds .
∂D
(6.21)
∂D
Proof: (a) The product rule for the curl of a vector product (see 6.10) yields
curl(ν̃ × F̃ ) = ν̃ div F̃ − F̃ div ν̃ + ν̃ 0 F̃ − F̃ 0 ν̃
= ν div F̃ − F div ν̃ + ν̃ 0 F − F̃ 0 ν
and thus
ν · curl(ν̃ × F̃ ) = div F̃ − ν| {z
· F} div ν̃ + ν > ν̃ 0 F − ν > F̃ 0 ν .
= 0
>
From ν̃ ν̃ = 1 in U we have by differentiation that ν̃ > ν̃ 0 = 0 in U and thus
ν · curl(ν̃ × F̃ ) = div F̃ − ν > F̃ 0 ν = Div F .
1
3
(b) Let ν̃ and F̃ as inR part (a) and set
R G̃ = ν̃ × F̃ in U . Then G̃ ∈ C (U, C ). ∞By part
(a) we conclude that Γ Div F ds = Γ ν · curl G̃ ds. Choose a sequence G̃n ∈ C (U, C3 )
with
G̃n → G̃ in RC 1 (U, C3 ). Then, by the Theorem of Stokes on Γ, we conclude that
R
ν · curl G̃n ds = C G̃n · τRd`. The convergence
curl G̃n → Rcurl G̃ in C(∂D, C3 ) and G̃n →
Γ
R
ν × F in C(∂D, C3 ) yields Γ Div F ds = C (ν × F ) · τ d` = C F · (τ × ν) d`.
(c) By part (b) it suffices to prove the product rule
Div(f F ) = Grad f · F + f Div F .
(6.22)
Indeed, using the definitions yields
Div(f F ) = div(f˜ F̃ ) − ν · (f˜ F̃ )0 ν
= ∇f˜ · F̃ + f˜ div F̃ − ν > F̃ f˜0 ν − f˜ ν > F̃ 0 ν
>
= Grad f · F + f div F̃ − ν|{z}
F f˜0 ν − f ν > F̃ 0 ν
= 0
= Grad f · F + f Div F .
This ends the proof.
2
Lemma 6.19 The tangential gradient and the tangential divergence depend only on the
values of f and the tangential field F on ∂D, respectively.
292
CHAPTER 6. APPENDIX
Proof Let f˜1 , f˜2 ∈ C 1 (D) be two extensions of f ∈ C 1 (∂D). Then f˜ = f˜1 − f˜2 vanishes on
∂D. The identity (6.21) shows that
Z
ν × (∇f˜ × ν) · F ds = 0 for all F ∈ Ct1 (∂D) .
∂D
The space Ct1 (∂D) is dense in the space L2t (∂D) of all tangential vector fields with L2 −components.
Therefore, ν × (∇f˜ × ν) vanishes which shows that the definition of Grad f is independent
of the extension. Similarly the assertion for the tangential divergence is obtained.
2
Corollary 6.20 Let w ∈ C 2 (D, C3 ) such that w, curl w ∈ C(D, C3 ). Then the surface
divergence of ν × w exists and is given by
Div(ν × w) = −ν · curl w
on ∂D .
(6.23)
Proof: For any ϕ ∈ C 2 (D) we have by the divergence theorem
Z
Z
Z
ϕ ν · curl w ds =
div(ϕ curl w) dx =
∇ϕ · curl w dx
∂D
D
D
Z
Z
ν · (w × ∇ϕ) ds
div(w × ∇ϕ) dx =
=
D
∂D
Z
Z
Div(ν × w) ϕ ds .
(ν × w) · Grad ϕ ds = −
=
∂D
∂D
The assertion follows because the traces ϕ|∂D : ϕ ∈ C 2 (D) are dense in L2 (∂D).
2
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Index
C n −smooth, 281
addition formula
Bessel functions, 70
surface harmonics, 45
adjoint operators, 280
Ampere’s law, 10
anisotropic function space, 184
anisotropic medium, 12
Bessel differential equation, 30, 54
Bessel functions, spherical, 57
boundary operator
D, 254
L, 261
M, 261
T , 254
D0 , 249
S, 249
boundary value problem
Dirichlet, 263
Neumann, 267
vector case, 268
charge density, 9
conductivity, 12
constitutive equations, 11
convolution, 151
current density, 9
dielectric medium, 12
dielectric tensor, 12
Dirichlet eigenvalues of Delta, 65
divergence theorem, 287
dual system, 270, 280
E-mode and M-mode, 18
eigen
-function, 165
-space, 170, 171
-system, 165
-value, 165, 170
electric displacement, 9
electric field, 9
electric monopole, 13
electrostatics, 13
equation of continuity, 10
Euler differential equation, 30
extension operator
η, 209
ηt , ηT , 218, 226
η̃, 211
exterior unit normal vector, 282
far field amplitude, 72
far field pattern, 72
Faraday’s law of induction, 10
Fourier
coefficients, 154
series, 154
transform, 150
Fredholm’s alternative, 280
Fredholm’s theorem, 280
Friedrich’s inequality, 157
fundamental solution, 92, 246
Fundamental Theorem of Calculus of Variations, 186
Funk–Hecke formula, 44
Gauss Theorem, 287
Gauss’ electric law, 10
Gauss’ magnetic law, 11
Green’s formula, 228, 256, 287
vectorial, 287
Green’s Representation Theorem, 246
Hölder-continuous function, 100
Hankel functions, 57
harmonic function, 28
295
296
Hertz dipol, 20
Hertz potential
electric, 17
magnetic, 17
Holmgren’s uniqueness theorem, 97
homogeneous medium, 12
homogeneous polynomial, 40
impedance boundary condition, 19
initial–boundary value problem, 185
interior regularity, 175
isotropic medium, 12
Jacobi–Anger expansion, 61
Laplace equation, 28
Laplace-Beltrami operator, 28, 290
law of induction, 10
Lax – Milgram theorem, 281
Legendre differential equation, 31
associated, 30
Legendre functions, associated, 39
Legendre polynomial, 31
limiting absorption principle, 66
linear medium, 12
Lipschitz domain, 201, 281
local coordinate system, 201, 281
magnetic field, 9
magnetic induction, 9
magnetostatic, 14
Maxwell double layer, 262
Maxwell single layer, 262
MKS-system, 9
Ohm’s law, 12
orthonormal system, 165, 171
partial integration, 154, 211, 287, 291
partition of unity, 175, 204, 212, 283
perfectly conducting medium, 19
permeability tensor, 12
plane time harmonic fields, 16
Poisson equation, 13
potential
double layer D̃ϕ, 95, 247
single layer S̃ϕ, 95, 247
INDEX
vector, 135
vector L̃a, 261
vector M̃a, 261
volume Vϕ, 95
potential theory, 13
Poynting
Theorem, 22
vector, 22
vector, complex, 22
radiation condition
Silver–Müller, 257, 261
Sommerfeld, 249
Rayleigh formulas, 58
resolvent set, 170
Riesz representation theorem, 280
Riesz–Fredholm theorem, 279
Rodrigues formula, 34
Silver-Müller radiation condition, 21
Sobolev spaces
Ṽ0,A , 160
H(curl, D), 158
H(curl 0, D), 160
H 1 (D), 149
1
(Q), 155
Hper
s
Hper (Curl, Q2 ), 218
s
(Div, Q2 ), 218
Hper
s
Hper (Q), 205
H −1/2 (Curl, ∂D), 225
H −1/2 (Div, ∂D), 225
H −1/2 (∂D), 214
H 1/2 (∂D), 209
H0 (curl, D), 159
H0 (curl 0, D), 160
H01 (D), 154
Hper (curl, Q3 ), 218
VA , 160
V0,A , 160
Sommerfeld radiation condition, 21
spectrum, 170
speed of light, 13
spherical harmonics, 41
vector, 75
spherical surface harmonics, 41
spherical wave functions, 59
INDEX
Stratton-Chu formula, 129, 132, 257
support, 148
surface curl, 76, 230
surface divergence, 230, 289
surface gradient, 289
tangential plane, 282
tangential vector, 282
TE-mode and TM-mode, 18
time harmonic Maxwell equations, 14
time–dependent Maxwell system, 193
trace operator
γ0 , 209
γ1 , 215
γt , γT , 218, 226
trace theorem, 206, 209, 218, 226
transformation formula, 283
transmission boundary conditions, 19
unique continuation property, 173, 177
vacuum, 12
variational
curl, 157
divergence, 157
gradient, 149
solution, 163
vector Helmholtz equation, 15
wave equation, 13
wave number, 15, 255
Wronskian of Bessel functions, 57
Young’s inequality, 151
297