A Discrete Hilbert Transform with Circle Packings

A Discrete Hilbert Transform with Circle Packings

Dominik Volland

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Dominik Volland

A Discrete Hilbert Transform with Circle Packings

Dominik VollandGarching near Munich, Germany

BestMasters ISBN 9783658204563 ISBN 9783658204570 (eBook)https://doi.org/10.1007/9783658204570

Library of Congress Control Number: 2017961504

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Acknowledgments

This book originated from a research project of Elias Wegert and FolkmarBornemann together with the author in the TUM elite graduate programTopMath. The work was supported by the DFG-Collaborative ResearchCenter, TRR 109, “Discretization in Geometry and Dynamics”. I would liketo thank both TopMath and DFG for their financial and ideological supportthroughout the project.

I also wish to express my deep gratitude to my supervisors, Elias Wegertand Folkmar Bornemann, for productive personal communication, for theirvaluable support, and also for suggesting me for “Best Masters”.

Finally, I’d like to thank my family, my friends, and my colleagues atTUM for supporting me while I was writing this book.

Contents

1 Introduction 1

2 Auxiliary Material and Notation 3

3 The Continuous Setting 93.1 Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.1 Boundary values of holomorphic functions . . . . . . . 93.1.2 Integral formulas . . . . . . . . . . . . . . . . . . . . . 133.1.3 Fourier series of the boundary functions . . . . . . . . 14

3.2 The Hilbert transform . . . . . . . . . . . . . . . . . . . . . . 173.3 Riemann-Hilbert problems . . . . . . . . . . . . . . . . . . . . 23

3.3.1 Linear Riemann-Hilbert problems . . . . . . . . . . . . 243.3.2 Nonlinear Riemann-Hilbert problems . . . . . . . . . . 25

3.4 Obtaining the Hilbert transform from a Riemann-Hilbertproblem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.4.1 Choice of the problem . . . . . . . . . . . . . . . . . . 263.4.2 Solutions of the problem . . . . . . . . . . . . . . . . . 273.4.3 Counterexamples for u /∈ C1+α . . . . . . . . . . . . . 31

4 Circle Packings 354.1 First examples and ideas . . . . . . . . . . . . . . . . . . . . . 354.2 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.2.1 Complex . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2.2 Circle packing . . . . . . . . . . . . . . . . . . . . . . 42

4.3 Manifold structure . . . . . . . . . . . . . . . . . . . . . . . . 444.3.1 Contact function . . . . . . . . . . . . . . . . . . . . . 454.3.2 Angle sums and branch structures . . . . . . . . . . . 464.3.3 Parametrization of Db . . . . . . . . . . . . . . . . . . 494.3.4 Normalization . . . . . . . . . . . . . . . . . . . . . . . 50

4.4 Discrete harmonic functions on circle packings . . . . . . . . 524.5 Discrete analytic functions . . . . . . . . . . . . . . . . . . . . 554.6 Maximal packings . . . . . . . . . . . . . . . . . . . . . . . . 56

viii Contents

4.7 Some results on discrete analytic functions . . . . . . . . . . . 574.7.1 Discrete maximum principles . . . . . . . . . . . . . . 584.7.2 Approximation of the Riemann Mapping . . . . . . . . 58

5 Discrete Hilbert Transform 615.1 Discrete boundary value problems . . . . . . . . . . . . . . . 62

5.1.1 Definition and examples . . . . . . . . . . . . . . . . . 625.1.2 Linearization of boundary value problems . . . . . . . 67

5.2 Proof of the maximal packing conjecture . . . . . . . . . . . . 685.2.1 The transformed packing . . . . . . . . . . . . . . . . 705.2.2 Differential of ω at the transformed packing . . . . . . 715.2.3 Basis for the kernel of J . . . . . . . . . . . . . . . . . 76

5.3 Discrete Hilbert transform . . . . . . . . . . . . . . . . . . . . 795.3.1 Difficulties of the Schwarz problem . . . . . . . . . . . 805.3.2 Discretization of the nonlinear problem . . . . . . . . 825.3.3 Linearization of the discrete operator . . . . . . . . . . 86

6 Numerical Results and Future Work 896.1 Test computations . . . . . . . . . . . . . . . . . . . . . . . . 896.2 Eigenvalues of the discrete transform . . . . . . . . . . . . . . 916.3 Elimination of constants . . . . . . . . . . . . . . . . . . . . . 946.4 Curvature of the Circle Packing manifold . . . . . . . . . . . 966.5 Local frames . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Bibliography 101

List of Figures

3.1 Construction of a counterexample . . . . . . . . . . . . . . . . 32

4.1 A circle packing . . . . . . . . . . . . . . . . . . . . . . . . . . 354.2 Packings and complexes . . . . . . . . . . . . . . . . . . . . . 364.3 A discrete conformal map . . . . . . . . . . . . . . . . . . . . 374.4 Overlapping circles . . . . . . . . . . . . . . . . . . . . . . . . 384.5 A branched packing . . . . . . . . . . . . . . . . . . . . . . . 384.6 A complex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.7 Complexes and hex refinement . . . . . . . . . . . . . . . . . 414.8 A face and its face circle . . . . . . . . . . . . . . . . . . . . . 434.9 Illustration of the contact equation . . . . . . . . . . . . . . . 454.10 Maximal packings . . . . . . . . . . . . . . . . . . . . . . . . 574.11 Construction of a discrete Riemann map . . . . . . . . . . . . 59

5.1 A discrete circular Riemann-Hilbert problem . . . . . . . . . 665.2 The extended packing . . . . . . . . . . . . . . . . . . . . . . 695.3 Proof of Lemma 5.12 . . . . . . . . . . . . . . . . . . . . . . . 715.4 Freedoms for maximal packings . . . . . . . . . . . . . . . . . 775.5 A difficulty of the Schwarz problem . . . . . . . . . . . . . . . 825.6 A discrete circular Riemann-Hilbert problem . . . . . . . . . 84

6.1 Numerical results for K[7]3 . . . . . . . . . . . . . . . . . . . . 90

6.2 Numerical results for K ′ . . . . . . . . . . . . . . . . . . . . . 906.3 Spectra of H . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.4 Eigenfunctions for K ′ . . . . . . . . . . . . . . . . . . . . . . 926.5 Modified Eigenfunctions for K ′ . . . . . . . . . . . . . . . . . 936.6 Numerical results for H . . . . . . . . . . . . . . . . . . . . . 946.7 Packings solving instances of (DHRHP) . . . . . . . . . . . . 956.8 Results on a different complex . . . . . . . . . . . . . . . . . . 966.9 Nonlinear versus linearized transform . . . . . . . . . . . . . . 97

Abstract

This book deals with the definition of a discrete Hilbert transform. Theclassical Hilbert transform is a bounded linear operator relating the real-and imaginary parts of the boundary values of a holomorphic function.The discretization is based on the theory of circle packings which has beenestablished as a discrete counterpart of complex analysis in the past threedecades.

The Hilbert transform is closely related to Riemann-Hilbert problems. Inparticular, the transform can be computed from the solution of a Riemann-Hilbert problem. We will therefore define the transform based on discreteRiemann-Hilbert problems which have been investigated by Elias Wegertsince 2009.

The definition of the discrete operator requires the regularity of a certainJacobian matrix. In this book, we present a proof of the regularity of thismatrix.

1 Introduction

The mathematical branch of complex analysis deals with the behavior ofholomorphic functions, which are differentiable functions in the complexplane. The theory of functions of one complex variable is quite self-containednowadays. However, during the last three decades, complex analysis hasfound a discrete counterpart in the theory of so-called circle packings.

These packings are configurations of circles that satisfy prescribed tangencyrelations. The idea of using circles to discretize holomorphic functions is dueto the fact that holomorphic functions are locally conformal transformationsof the complex plane wherever their derivative does not vanish. This means,a holomorphic function will map an infinitesimally small circle onto aninfinitesimally small circle again.

Circle packings have turned out to be a very faithful discretization ofholomorphic functions indeed. By now, a wide variety of theorems andconcepts from complex analysis have found counterparts in circle packing.There are still many open problems and the theory of circle packings is onlygrowing. A broad overview over the theory of circle packings can be foundin Stephenson’s book [27].

This book is devoted to one of the numerous concepts of complex analysisand the quest for its discrete counterpart. Namely, we discuss a discreteHilbert transform in the context of Riemann-Hilbert problems.

Roughly speaking, a function holomorphic in the complex unit disc D isdetermined by the real part of its boundary values. The Hilbert transform isa linear operator that relates the real and imaginary parts of the boundaryvalues of a holomorphic function.

Riemann-Hilbert problems are boundary value problems for functionsholomorphic in D. A Riemann-Hilbert problem imposes one real condition toeach boundary value of the solution. The Hilbert operator is closely relatedto these problems. In fact, we will show that the Hilbert transform of a givenfunction can be obtained from the solution of a suitable Riemann-Hilbertproblem.

The German mathematician Elias Wegert, author of several papers [31][32]and a book [30] on Riemann-Hilbert problems, observed that Riemann-Hilbert problems fit nicely into the setting of circle packings. Motivated

© Springer Fachmedien Wiesbaden GmbH 2017D. Volland, A Discrete Hilbert Transform with Circle Packings,BestMasters, https://doi.org/10.1007/978-3-658-20457-0_1

2 1 Introduction

by this, he has outlined the definition of a discrete Hilbert transform. Thestrategy of this definition can be summarized as follows: Find a Riemann-Hilbert problem from which the Hilbert transform can be obtained, discretizethis problem with circle packings, solve it, and extract the discrete transformfrom the solution. In this book, we present the discrete operator obtainedfrom this procedure.

The definition of the discrete operator rests on the regularity of a certainJacobian matrix which arises from the discrete boundary value problem.The regularity of this matrix has been conjectured by Wegert [34]. In thisbook, we give a proof of this conjecture.

The book is organized as follows. Chapter 2 fixes notation conventions andstates a few classical results that will be used throughout the text. In Chapter3 we will be concerned with the continuous theory of the Hilbert operator.We will first introduce the functional analytic basics required to define theHilbert operator and formulate its most important properties. After that,we will define Riemann-Hilbert problems and discuss their connection to theHilbert operator.

In Chapter 4, we will turn to the discrete setting. We will define circlepackings and state their basic properties, with an emphasis on the differentialgeometric structure of circle packings. Chapter 5 will then introduce discreteboundary value problems. We will give some examples, introduce linearizeddiscrete boundary value problems and prove the regularity of the matrixmentioned before. Based on this and the work done in Chapter 3, we willdefine a discrete Hilbert operator.

Finally, Chapter 6 presents some numerical results and outlines questionsrelated to the discrete operator that may be addressed in the future.

2 Auxiliary Material and Notation

This chapter introduces our notation conventions and gives a short overviewover some results that will be used without references throughout the book.

Complex analysis. There are many books covering elementary complexanalysis. Our standard reference is the book of Ahlfors [1].

A domain is an open connected set in C. The domain we are dealing withmost of the time is the unit disc {z ∈ C| |z| < 1} denoted by D. Its boundaryis the unit circle {z ∈ C| |z| = 1} and will be denoted by T. Other discs willbe denoted by Br(z

∗) := {z ∈ C| |z − z∗| < r}. Given a circle C ⊂ C, thedisc bounded by C will be denoted by intC, while extC = C \ (C ∪ intC).

The boundary of a domain U is denoted by ∂U and the closure U ∪ ∂U isdenoted by U . Some caution is required here since the complex conjugate ofa complex number z is denoted by z.

By Log we denote the principal branch of the complex logarithm definedon the slit domain C− := C \ R≤0. By Arg we denote the principal branchof the complex argument function, i.e. Arg(z) = Im Log(z). log resp. argdenotes any branch of these functions if it is not necessary (or possible) tospecify the branch explicitly.

To keep the overview, we have the following convention for variables: zalways refers to a variable ranging on D or sometimes C, while t is a variableranging on T. Arguments of complex numbers are denoted by Greek letters,mostly τ , sometimes also φ or θ.

Mobius transformations. A Mobius transformation is a holomorphic func-tion of the form

T (z) =az + b

cz + d

where a, b, c, d are complex constants such that ad− bc 6= 0. These functionsplay a special role in complex analysis due to their geometric properties. Itis natural to understand them as mappings on the extended complex planeC ∪ {∞} by writing T (∞) = a

c and T (−dc ) =∞ if c 6= 0 (or T (∞) =∞ ifc = 0).


4 2 Auxiliary Material and Notation

Mobius transformations have the following properties (for proofs, seeChapter 3, Section 3 of [1]):

(i) T is a bijective map on the extended complex plane C ∪ {∞}.

(ii) T is holomorphic on C\{T−1(∞)

}and its derivative vanishes nowhere.

(iii) If C ⊂ C is a Mobius circle, i.e. a circle or a line, T (C) is a Mobiuscircle, too.1

(iv) T can be written as a composition of linear functions and the inversionz 7→ 1

z .

(v) Let C ⊂ C be a circle and T−1(∞) /∈ C. Let z1, z2, z3 ∈ C lie incounter-clockwise order on C. Then

• if T−1(∞) ∈ extC, then T (intC) = intT (C), T (extC) = extT (C),and T (z1), T (z2), T (z3) lie in counter-clockwise order on C again.

• if T−1(∞) ∈ intC, then T (intC) = extT (C), T (extC) = intT (C),and T (z1), T (z2), T (z3) lie in clockwise order on C.

If c = 0, let d = 1 w.l.o.g. Then T (z) = az + b is a linear function. If inaddition a ∈ T, then T is an isometry of C and is called a plane rigid motion.Hence, there are 3 real degrees of freedom to construct a plane rigid motion.

Conformal automorphisms. A function f : D → D is bijective and con-formal iff it is a composition of a rotation z 7→ λz for λ ∈ T and a Mobiustransformation of the form

z 7→ z − z0

1− zz0

for some z0 ∈ D. In particular, there are exactly three real degrees of freedomfor constructing such a map. For a proof, see [1] (Chapter 4, 3.4).

Blaschke products. A (finite) Blaschke product of degree n is a functionof the form

B(z) = c

n∏j=1

z − zj1− zjz

1In this context, a line should be thought of as a circle containing the point ∞. Fromthis, it is clear that T maps a circle C to a circle again iff T−1(∞) /∈ C.


for c ∈ T, z1, . . . , zn ∈ D. The set of all finite Blaschke products is denotedby B.

Note that z1, . . . , zn are precisely the zeros of B counted with multiplicity.The perhaps most important property of Blaschke products is that eachB ∈ B satisfies

|B(t)| = 1 ∀t ∈ T

Since all poles of B lie outside D, B is bounded and continuous on D.Blaschke products can be understood as “polynomials in the disc”: A

Blaschke product of degree n is a conformal self-map of D that attains eachvalue in D exactly n times. For details, see [23].

As the term “finite” suggests, one can also consider infinite Blaschkeproducts with an infinite number of zeros in D. These functions are notrelevant for this book, though.

Function spaces. For further details on the spaces introduced here, anytextbook on functional analysis or measure theory can be consulted, e.g.[7]. A function f : T → C is said to be in the Lebesgue space Lp(T) for1 ≤ p <∞ if its normalized Lp-norm

‖f‖p :=

1

2π

2πˆ

0

∣∣f(eiτ )∣∣p dτ

1p

is smaller than ∞. f is said to be in L∞(T) if it is bounded on T \ E for aset E with Lebesgue measure zero and the L∞-norm is defined by

‖f‖∞ := inf{C ∈ R|{t ∈ T|f(t) ≥ C} has Lebesgue measure zero}.

For 1 ≤ p ≤ ∞, Lp(T) is a Banach space w.r.t. the Lp-norm. The spaceL2(T) is a Hilbert space endowed with the scalar product

(f, g)2 :=

1

2π

2πˆ

0

f(eiτ )g(eiτ )dτ

12

.

For a continuous function f : T→ C, we define its modulus of continuity

ωf (θ) := sup|φ−τ |<θ

∣∣f(eiφ)− f(eiτ )∣∣ .

6 2 Auxiliary Material and Notation

f is said to be Holder continuous with Holder exponent α if its Cα-seminorm

|f |Cα := supθ∈[0,π]

ωf (θ)

θα

is finite. For k ∈ N0 and α ∈ [0, 1], we denote by Ck+α(T) the set of allfunctions whose k-th derivative exists and is Holder continuous with Holderexponent α. Endowed with the Ck+α-norm

‖f‖Ck+α :=

k∑j=0

∥∥∥f (j)∥∥∥∞

+∣∣∣f (k)

∣∣∣Cα

,

these spaces are Banach spaces, too.

Matrices and vectors. Let v ∈ Rn. If not explicitly stated otherwise, vjalways denotes the j-th component of v such that v = (v1, . . . , vn)T . We willfrequently drop the transposition for vectors and just write v = (v1, . . . , vn).

Given a differentiable function f : U → Rk for a domain U ⊂ Rn, wedenote its Jacobian evaluated at x ∈ U by Df |x ∈ Rk×n. In context ofmatrices and vectors, the dot · always denotes the usual matrix product.For example, the chain rule of differentiation reads

D(f ◦ g)|x = Df |g(x) ·Dg|x

in our notation.A diagonal matrix with diagonal entries a1, . . . , an will be denoted by

diag(a1, . . . , an). In = diag(1, . . . , 1) denotes the n×n unit matrix. Further-more, 1n = (1, . . . , 1)T ∈ Rn denotes the vector where all entries are equalto 1.

Implicit Function Theorem. We will make use of the well-known ImplicitFunction Theorem. For convenience, we state the theorem here. The formu-lation we are using is a combination of the formulations in [17] (Theorem 5.4in Chapter XVIII) and in [6] (Theorem 7.43).

Theorem (Implicit Function Theorem). Let U ⊂ Rp × Rn be open andF : U → Rn continuously differentiable. Let (x0, y0) ∈ Rp × Rn such thatF (x0, y0) = 0 and the matrix

DyF |(x0,y0)


is regular. Then there are open sets V1 ⊂ Rp and V2 ⊂ Rn with x0 ∈ V1 andy0 ∈ V2 and a continuously differentiable function

g : V1 → V2

with g(x0) = y0 such that the following holds: (x, y) ∈ V1 × V2 satisfiesF (x, y) = 0 if and only if g(x) = y. Moreover, the Jacobian of g at x0 isgiven by

Dg|x0 = −(DyF |(x0,y0)

)−1 ·DxF |(x0,y0).

Manifolds. To avoid the technical machinery of general abstract differenti-able manifolds, we confine ourselves with introducing submanifolds of Rn.For details on manifolds and rigorous definitions of the termini introducedhere, we refer to [16].

A k-dimensional submanifold of Rn is a set M ⊂ Rn which is locallygiven as the zero set of a function F : Rn → Rn−k with maximal rank. Thismeans, given x ∈M , there is an open set U ⊂ Rn with x ∈ U and a functionF : U → Rn−k such that M ∩ U = F−1({0}). DF has maximal rank ateach point on M ∩ U . The kernel of DF |x is the tangent space at x. Thisspace is denoted by TxM and consists of all vectors which are tangent tothe manifold at M .

By the Implicit Function Theorem, M can be parametrized locally: Givenx ∈ M , there are V ⊂ Rk and U ⊂ Rn open with x ∈ M and a func-tion g : V → Rn such that g(V ) = M ∩ U . The function g is called aparametrization of M .

Given U ⊂ Rn open and M ′ := M ∩ U , a diffeomorphism f : M ′ → Rkwith full rank is called a chart on M .

We will mostly work with manifolds consisting of circle packings. In thiscontext, an element of the manifold is usually denoted by P and a tangentialvector is denoted by p.

3 The Continuous Setting

Most of the theory discussed in this chapter will be presented without anyproofs. For proofs, we refer to the respective textbooks. The theory of Hardyspaces can be found in the books of Duren [10], Koosis [14] and Hoffman[13]. A more recent book with an emphasis on representation theorems is byMashreghi [20].

The current state of the art in the theory of Riemann-Hilbert problemsis presented in the books and papers by Wegert. His paper [31] gives acomprehensive overview and proofs for linear problems, while his book [30]contains an extensive study of nonlinear problems including relations tosingular integral equations, applications and numerical methods.

3.1 Hardy spaces

To investigate boundary value problems for functions holomorphic in D,we need appropriate solution spaces. Clearly, the functions in the solutionspaces should have, in some sense, limits at the boundary which allow us tointerpret the boundary value problem.

A simple possible choice for a solution space would be the disc algebraH∞∩C consisting of functions holomorphic in D with a continuous extensionon D. However, there are plenty of holomorphic functions that are lessregular at the boundary but still have a strong structure at the boundary.

We will therefore start with introducing the so-called Hardy spaces, whichwill be seen to constitute suitable solution spaces for linear Riemann-Hilbertproblems. Roughly speaking, these spaces consist of holomorphic functionswhose boundary values lie in the Lebesgue spaces Lp(T). We will also seethat the Lebesgue spaces for 1 < p <∞ are suitable domain spaces for theHilbert operator.

3.1.1 Boundary values of holomorphic functions

We will start off right with the definition of the Hardy spaces.


10 3 The Continuous Setting

Definition 3.1 (Hardy space). Let 1 ≤ p <∞. We define the Hardy spaceHp by

Hp := {w : D→ C|w holomorphic, supr<1

2πˆ

0

∣∣w(reiτ )∣∣p dτ <∞}.

For w ∈ Hp,

‖w‖p = supr<1

1

2π

2πˆ

0

∣∣w(reiτ )∣∣p dτ

1p

Furthermore, we define

H∞ := {w : D→ C|w holomorphic and bounded}

and for w ∈ H∞

‖w‖∞ = supz∈D|w(z)|

The following lemma states that Hardy spaces have convenient functionalanalytic properties at the very least. A proof of this lemma can be found in[10], Chapter 3.2.

Lemma 3.2 (Hardy space norm). For 1 ≤ p ≤ ∞, ‖·‖p is a norm on Hp

and Hp is a Banach space with respect to this norm.

We now discuss the boundary values. Classically, one would define theboundary value of a function defined on D as

limD3z→t

w(z).

However, for a holomorphic function w, this limit does not exist in generaleven if w is bounded. A counterexample is given in [8], Chapter 2.1.

For a weaker concept of boundary values, one could think of using theradial limit

lim(0,1)3r→1

w(rz).

This concept is in fact too weak, though. One can show that there arenon-vanishing holomorphic functions that have vanishing radial limits almosteverywhere on T. An example for such a function has been constructed by

3.1 Hardy spaces 11

Luzin and Privalov and can be found in [24], Chapter IV, §5. This meansthat holomorphic functions are not uniquely determined by their radiallimits, which is problematic when discussing boundary value problems.

Thus, we will need a concept “in the middle” of the above two: a conceptweaker than unrestricted limits but stronger than radial limits. As it turnsout, nontangential boundary values are a suitable choice.

Definition 3.3 (Nontangential Boundary Value). Let t ∈ T and α < π. Wewrite

S(t, α) := {z ∈ D|∣∣arg(1− tz)

∣∣ < α}.Let w be holomorphic on D. wt ∈ C is called the nontangential boundaryvalue of w at t ∈ T if for every α < π we have

limS(t,α)3z→t

w(z) = wt.

We write wt = limz−→

∠tw(z).

The following theorem states that a holomorphic function is indeeduniquely determined by its nontangential boundary values.

Theorem 3.4 (Luzin-Privalov, [14], Chapter III:D.3). Let w be holomorphicon D and A ⊆ T a subset with nonzero Lebesgue measure. Assume thatlimz−→

∠tw(z) = 0 ∀t ∈ A. Then w ≡ 0 holds.

On the other hand, a result of Fatou tells us that all functions in one ofthe Hardy spaces defined above have nontangential boundary values almosteverywhere.

Theorem 3.5 (Fatou, [14], Chapter IV:C.1). Let 1 ≤ p ≤ ∞ and w ∈ Hp.Then lim

z−→∠tw(z) exists for almost every t ∈ T. If we define wb(t) := lim

z−→∠tw(z)

for all these t, the resulting function wb : T → C satisfies wb ∈ Lp(T) and‖wb‖p = ‖w‖p. I.e., the Lp norm of the boundary function coincides withthe Hp norm of the function.

These two theorems exhibit nontangential boundary values as a suitableconcept for holomorphic functions.

Notation 3.6. In the following, we shall simply write w(t) for the nontan-gential boundary value of a function w ∈ Hp, 1 ≤ p ≤ ∞ at t ∈ T. Thus wewill interpret w ∈ Hp as a function w : D→ C which is holomorphic whenrestricted to D.


We close this subsection with defining spaces of holomorphic functionsthat have continuous boundary functions.

Definition 3.7 (Disc algebra). Given k ∈ N0 and α ∈ [0, 1), we define thespaces H∞ ∩ Ck+α = {f ∈ H∞|τ 7→ f(eiτ ) ∈ Ck+α}. In particular, the discalgebra H∞ ∩ C is the set of all bounded holomorphic functions on D whoseboundary function is continuous.

Clearly, H∞ ∩ Ck+α ⊆ H∞ ∩ C ⊂ H∞ for every k ∈ N0 and α ∈ [0, 1).Moreover, Fatou’s theorem implies that H∞ ∩ Ck+α is a Banach space w.r.t.the Ck+α-norm (applied to the boundary values).

Note that there is a subtlety in this definition: A function w ∈ H∞ ∩ Cis continuous when restricted to either T or D but it is not immediatelyclear whether it is continuous on D. In other words: If w ∈ H∞ ∩ C andt ∈ T, does lim

D3z→tw(z) = w(t) hold?1 In fact, this is true for all functions in

H∞ ∩ C, which is a consequence of the following important theorem by thebrothers Riesz.

Theorem 3.8 (Riesz, IV:C.2 and IV:G in [14]). Let f ∈ Hp. For r ≤ 1define fr : T→ R by fr(t) = f(rt). Then

• If p ∈ [1,∞) then limr→1 ‖fr − f1‖p = 0.

• If p =∞ then limr→1 ‖fr − f1‖∞ = 0 iff f is continuous.

Using this theorem, we easily get

Lemma 3.9. Every function in H∞ ∩ C is continuous on D.

Proof. Let f ∈ H∞ and f |T be continuous. Fix z ∈ D and a sequencezk → z. We have to show that f(zk)→ f(z).

If z ∈ D, this is clear, so let t := z ∈ T. For each k ∈ N, write zk = rktkwith rk > 0 and tk ∈ T. Then we have rk → 1 and tk → t, thus, using thenotation of Theorem 3.8,

|f(zk)− f(t)| ≤ |f(zk)− f(tk)|+ |f(tk)− f(t)|≤ ‖frk − f1‖∞ + |f(tk)− f(t)| → 0

due to Theorem 3.8 and the continuity of f on T.

1Fatou’s theorem only tells us that limz−→

∠tw(z) = w(t)!

3.1 Hardy spaces 13

3.1.2 Integral formulas

Elementary complex analysis tells us that the real and imaginary parts of aholomorphic function are harmonic functions which are so-called conjugatesof each other ([1], Chapter 2, Section 1.2). For a function u harmonic onsome domain U ⊆ C and a closed disc D ⊂ U , Poisson’s formula tells usthat the values of u at each point z ∈ D are determined by the values of uon ∂D ([1], Chapter 4, Theorem 22). Combining these results, we obtainthe following: Given a domain U ⊃ D and a function w holomorphic on U ,the values of w in D are determined by the values of Rew on T up to anadditive imaginary constant. We can explicitly recover w from Rew|T usingSchwarz’ formula ([1], ibid.).

This statement can be generalized to functions in the newly introducedHardy spaces. We will start with Poisson’s formula. The theorem forfunctions harmonic on a domain containing D as stated in [1] (Chapter 4,Theorem 22) can be generalized to functions w ∈ Hp as follows.

Theorem 3.10 (Poisson’s formula, [14], Chapter II:B.1). Let 1 ≤ p ≤ ∞and w ∈ Hp. Then for every r ∈ [0, 1), τ ∈ R we have:

w(reiτ ) =1

2π

π

−π

1− r2

1 + r2 − 2r cos(τ − φ)w(eiφ)dφ

This theorem remains valid if we replace Hp by similar spaces of functionsharmonic in D, see the first chapter of [14]. In other words, it holds goodbecause w is a complex linear combination of the harmonic functions Rewand Imw, but it does not use the fact that Rew and Imw are harmonicconjugates of each other.

For holomorphic functions w, we can use the extra information that w isdetermined up to a constant by knowing only its real part. As mentionedabove, we can therefore recover w from the real part of its boundary valuesand Imw(0) by generalizing Schwarz’ formula to the case w ∈ Hp.

Theorem 3.11 (Schwarz’ Formula). Let 1 ≤ p ≤ ∞ and w ∈ Hp withImw(0) = 0. Then for every z ∈ D we have

w(z) =1

2π

π

−π

eiφ + z

eiφ − zRew(eiφ)dφ.


Notation 3.12. For u ∈ Lp(T) we denote the Poisson integral of u by

Pu(reiτ ) :=1

2π

π

−π

1− r2

1 + r2 − 2r cos(τ − φ)u(eiφ)dφ

and the Schwarz integral of u by

Su(z) :=1

2π

π

−π

eiφ + z

eiφ − zu(eiφ)dφ.

Notice that both operators are linear in u. Employing the linearity, theSchwarz operator acts on complex functions as well.

To prove Schwarz’ formula, write u = Rew|T. It suffices to note thatReSu = RePw|T and that ImSu is the harmonic conjugate of ReSu, comparee.g. [14], Chapter I:E.1.

Remark 3.13. It should be noted that the “converse” of the above theoremdoes not hold in general: For 1 < p <∞, u ∈ Lp(T) implies Su ∈ Hp, butfor p = 1 or p =∞, this is wrong in general. For p = 1, a counterexamplecan be constructed using the converse of Zygmund’s L logL Theorem, see[14], Chapter V:C 4. For p =∞ we will see a counterexample later in thecontext of the Hilbert transform.

Now, if we can recover a holomorphic function w from Rew|T, then inparticular we can recover Imw|T from Rew|T. So, there must be a linearoperator H which maps an appropriate function u : T → R to a functionHu : T → R such that u + iHu is the boundary function of a functionholomorphic in D. This is the Hilbert operator we are going to discuss inthe next section.

3.1.3 Fourier series of the boundary functions

WritingHp(T) := {f ∈ Lp(T)|∃f ∈ Hp : f |T = f},

one may ask for a characterization of the space Hp(T) as a subset of Lp(T).This can be done in an elegant way by considering the Fourier coefficients ofthe functions in Hp(T). For details on Fourier series, we refer to [18]. If a

3.1 Hardy spaces 15

function f is holomorphic on a domain D, it can be locally developed in itsTaylor series

f(z) =

∞∑k=0

fk(z − z0)k

at every point z0 ∈ D. The convergence radius of this power series is thenat least sup{r > 0|Br(z0) ⊆ D}. In particular, each function in one of theHardy spaces can be developed in a power series at 0 with a convergenceradius of at least 1.

Now, by formally taking the limit z −→∠t ∈ T, we obtain a series repres-

entation for the boundary function which turns out to be nothing but itsFourier series:

f(eiτ ) =

∞∑k=0

fkeiτk

Note that all negative coefficients vanish.To prove this, we can use Cauchy’s formula ([1], Chapter 4, Lemma 3). If

C is a circle with center 0 and radius r < 1, the Taylor coefficients fk aregiven by

fk =1

2πi

ˆ

C

f(t)

tk+1dt =

1

2πrk+1

2πˆ

0

f(reiτ )

eikτdτ .

Now Theorem 3.8 shows that we can take r → 1 in this formula, thusobtaining

fk =1

2π

2πˆ

0

f(eiτ )e−ikτdτ

which is just the formula for the Fourier coefficients. ([18], Chapter 1)Doing the same computations for negative k and applying Cauchy’s the-

orem ([1], Chapter 4, Theorem 4) yields that indeed all negative Fouriercoefficients of a function f ∈ Hp for some 1 ≤ p ≤ ∞ must vanish. Remark-ably, this is also a sufficient condition for f ∈ Hp to hold.

Theorem 3.14 (Theorem 3.4 in [10]). Let f ∈ Lp(T). Then f ∈ Hp(T) iffall coefficients with negative index in its Fourier series vanish.

For p ∈ (1,∞), the projection P : Lp(T)→ Hp(T) given by

P

∞∑j=−∞

fktk

=

∞∑j=0

fktk


is well defined and continuous. This is the statement of a theorem byM. Riesz ([13], Chapter 9). For p = 2, the monomials tk, k ∈ Z form anorthonormal basis of the Hilbert space L2(T) and P is just the orthogonalprojection onto the closed subspace H2(T). From this, we see that H2(T) isa Hilbert space, too.

It is insightful to study the mapping properties of the integral operators onthe monomial basis. We start with the Poisson integral. For each k ∈ N0, themonomial tk is the boundary function of the bounded holomorphic functionz 7→ zk. Thus, Theorem 3.10 tells us that Ptk(z) = zk. By decomposing thisrelation into its real and imaginary part, we can also find Ptk for negative kand obtain the following formulas:

Ptk(z) =

zk k > 0,1 k = 0,z−k k < 0.

This reflects the fact that, given a function f ∈ Hp(T), Pf is the correspond-ing function in Hp. If the negative Fourier coefficients of f do not vanish,Pf will still be a complex linear combination of the two harmonic functionsRePf and ImPf whose nontangential boundary values coincide with Re fand Im f , but RePf and ImPf are not harmonic conjugates of each otheranymore.

For the Schwarz integral, consider the functions eiτ 7→ cos kτ and eiτ 7→sin kτ . Note that these functions are the real parts of the boundary valuesof zk and −izk, respectively. Using eikτ = cos kτ + i sin kτ , Theorem 3.11therefore gives us

Stk(z) =

2zk k > 0,1 k = 0,0 k < 0.

.

A function u ∈ Lp(T) with Fourier coefficients uk, k ∈ Z, is real-valued ifffor all k ∈ N we have

uk = u−k ∀k ∈ N and Imu0 = 0.

In this case we may rewrite u as

u(eiτ ) = u0 +

∞∑k=1

2 Reuk cos(kτ)− 2 Imuk sin(kτ).

This is the real part of the boundary values of the function

z 7→ u0 +

∞∑k=1

2ukzk

3.2 The Hilbert transform 17

which is precisely Su(z) by the above formula.

Several properties of the integral operators can be proved more easily bymaking use of the monomial basis. As an example, we will derive a propertyof the Schwarz integral we will use later.

If a function u with Fourier coefficients uk for k ∈ Z is differentiable, thecoefficients of its derivative are given by u′k := ikuk which can be shown byusing integration by parts in the formula of the Fourier coefficients (or justby differentiating each summand of the Fourier series). For a holomorphicfunction with Taylor coefficients ak, k ∈ N0, its derivative has coefficientsa′k := (k + 1)ak+1. But combining these results, we see that there is aconnection between S′u and Su:

izS′u(z) = iz

∞∑k=0

2(k + 1)uk+1zk =

∞∑k=1

2ikukzk = Su(z)

Thus, we have proved the following lemma.

Lemma 3.15. Let u ∈ C1(T). Then izS′u(z) = Su(z).

3.2 The Hilbert transform

Let us start with formally deriving the formula of the Hilbert operator.Let w ∈ Hp with Imw(0) = 0 and let z ∈ D with polar form z = reiτ . By

Schwarz’ formula,

Imw(z) =1

2π

π

−π

Imeiφ + z

eiφ − zRew(eiφ)dφ

=1

2π

π

−π

2r sin(τ − φ)

1 + r2 − 2r cos(τ − φ)Rew(eiφ)dφ.


We would like to compute the boundary values of Imw, thus we are interestedin the limit r → 1. If we simply put r = 1 in the above formula, we obtain

Imw(eiτ ) =1

2π

π

−π

sin(τ − φ)

1− cos(τ − φ)Rew(eiφ)dφ

=1

2π

π

−π

cotτ − φ

2Rew(eiφ)dφ.

Note that the cotangent function has non-integrable singularities at integermultiples of π, thus, the improper integral on the right-hand side does ingeneral not exist in the classical sense. We will have to interpret it in aproper way.

It turns out that taking the Cauchy principal value ([1], Chapter 4,Section 5.3) of the integral does the job. We will therefore write

−π

−π

cotτ − φ

2f(φ)dφ := lim

ε→0

τ−εˆ

−π

cotτ − φ

2f(φ)dφ+

π

τ+ε

cotτ − φ

2f(φ)dφ

for τ ∈ (−π, π) and analogously

−π

−π

cotπ − φ

2f(φ)dφ := lim

ε→0

π−εˆ

−π+ε

cotπ − φ

2f(φ)dφ.

Finally, we can use this notation to define the Hilbert transform.

Definition 3.16 (Hilbert transform). Let u : T→ R be given. The Hilberttransform2 Hu : T→ R of u is the function defined by

eiφ 7→ 1

2π−π

−π

cotτ − φ

2u(eiτ )dτ .

wherever the integral on the right-hand side exists.

The next theorem states that the integral appearing in this definitionexists for very general functions.

2In contrast to some other authors, we will use the notation H for the continuoustransform and use H for the discrete transform defined in Section 5.3


Theorem 3.17 ([14], Chapter I:E.3). Let u ∈ L1(T). Then for almost everyφ ∈ R, the principal value integral

1

2π−π

−π

cotτ − φ

2u(eiτ )dτ

exists.

Now, the important question is which spaces are appropriate to make theHilbert transform a bounded operator. The answer to this question wasindicated before and is now given in the following theorem by M. Riesz.

Theorem 3.18 (M. Riesz, [14], Chapter V:B.2). Let u ∈ Lp(T), 1 < p <∞.Then Hu ∈ Lp(T). The Hilbert Operator H : Lp(T)→ Lp(T), u 7→ Hu is abounded linear operator.

It is pretty easy to provide a counterexample showing that Riesz’ theoremis indeed wrong for p = ∞. Consider the function defined by u(t) = 1 ifRe(t) > 0 and u(t) = 0 otherwise3. Obviously ‖u‖∞ = 1. Computationgives

Hu(eiτ ) =1

πln∣∣∣tan

τ

2

∣∣∣ .Thus lim

ε→0

∣∣Hu(eiε)∣∣ =∞, so Hu is unbounded.

Actually, even a continuous function u can have an unbounded Hilberttransform. Such an example will be constructed later in this book.

Riesz’ theorem is also wrong for p = 1. We will discuss a counterexampleshortly.

Now let us have a look at how the Hilbert operator acts on the monomialbasis. We obtain

Htk =

−itk k > 0

0 k = 0itk k < 0

. (3.1)

Thus, given a real-valued function u on T with Fourier coefficients uk, theFourier coefficients wk of u+ iHu are

wk =

2uk k > 0u0 k = 00 k < 0

.

3To avoid confusion, we remind that t is a complex number lying on T and not theargument of such a number.


Hence, u + iHu is the boundary function of an analytic function. Thisindicates that the Hilbert operator indeed does what we expected it to: Itrelates the real and imaginary part of the boundary values of an analyticfunction. Also, we can already see that this function is equal to Su. We thusformulate the following theorem.

Theorem 3.19 (Corollaries 6.7 and 5.17 in [20]).

1. Let 1 < p < ∞, u ∈ Lp(T). Then w = Su satisfies w ∈ Hp and itsboundary values are given by Su(t) = u(t) + iHu(t).

2. Let 1 ≤ p ≤ ∞, w ∈ Hp with Imw(0) = 0. Then HRe(w|T) = Im(w|T).

Let us revisit Remark 3.13. We can now easily construct a counterexamplefor the case p =∞. Take u ∈ L∞(T) such that Hu /∈ L∞(T). If Su ∈ H∞held, then ImSu|T would be bounded, but Hu = HRe(Su|T) = Im(Su|T) isunbounded.

Similarly, we can also construct an example showing that u ∈ L1(T) doesnot imply Hu ∈ L1(T). Take the example from Remark 3.13 of a functionu ∈ L1(T) such that Su /∈ H1. If Hu ∈ L1(T) held, then u+ iHu ∈ H1(T)since this is a L1 function with vanishing Fourier coefficients. Consequently,there is a function w ∈ H1 with w|T = u+ iHu. But in this case, w|T andSu are holomorphic functions whose real parts coincide. Thus, they must beidentical up to a real constant which contradicts Su /∈ H1.

The theorem of M. Riesz is a very powerful statement. However, we willsee that our approach for discretizing the Hilbert operator does not workunder the very general assumption u ∈ Lp. Thus we will state a theorem ofPlemelj and Privalov on Hilbert transforms of Holder continuous functionswhich will provide the foundation for our approach.

Theorem 3.20 (Plemelj-Privalov). Let α ∈ (0, 1) and u ∈ Cα(T). ThenHu ∈ Cα(T) and the Hilbert Operator is bounded as an operator on Cα(T).

Proof. This result was originally proved for boundary values of the Cauchyintegral, see e.g. [22]. It can easily be extended to the Hilbert operator. Inthe following, we want to give a direct proof for the Hilbert operator basedon Mashreghi’s book [20].


Let u ∈ Cα(T) be real-valued. Recall the definition of the modulus ofcontinuity ω. By Theorem 6.12 in [20], there is a constant C > 0 independentof u such that

ωHu(θ) ≤ C

θˆ

0

ωu(φ)

φdφ+ θ

π

θ

ωu(φ)

φ2dφ

≤ C |u|Cα

θˆ

0

1

φ1−α dφ+ θ

π

θ

1

φ2−α dφ

.

Both integrals can be evaluated explicitly and setting C ′ = Cα(1−α) we get

ωHu(θ) ≤ C ′ |u|Cα θα

and thus|Hu|Cα ≤ C

′ |u|Cα .

Now since2π

0

Hu(eiτ )dτ = 0, Hu must have a zero eiτ0 on T. This yields

that for every τ ∈ R∣∣Hu(eiτ )∣∣ =

∣∣Hu(eiτ )−Hu(eiτ0)∣∣ ≤ ωHu(π) ≤ C ′ |u|Cα π

α.

Overall, we get

‖Hu‖Cα = ‖Hu‖C + |Hu|Cα ≤ C′(πα + 1) |u|Cα ≤ C

′(πα + 1) ‖u‖Cα

which shows the boundedness of H.

Note that the Cα-norm of Hu can be estimated from the Cα-seminorm ofu alone. This reflects the fact that H maps constant functions to 0.

The following corollary rephrases the Plemelj-Privalov theorem based onTheorem 3.19.

Corollary 3.21. For every α > 0, the map u 7→ Su is continuous as a mapCα(T)→ H∞ ∩ C.

Proof.

‖Su‖H∞∩C = ‖u+ iHu‖C ≤ ‖u‖Cα + ‖Hu‖Cα ≤ C ‖u‖Cα

for a constant C due to Theorem 3.20.


The following lemma lists some further properties of the Hilbert operator.

Lemma 3.22. As an operator on Lp for 1 < p <∞, H has the followingproperties:

1. If u ∈ C1 and Hu ∈ C1, then H commutes with differentiation with respectto the polar angle, i.e.:

∂τHu = H∂τu

where ∂τu(t) = ddτ u(eiτ )|arg t.

4

2. kerH = {u : T→ R|u constant}.

3. imH = Lp0(T) :=

{v ∈ Lp(T)|

π

−πv(eiφ)dφ = 0

}.

4. As an operator on Lp0(T), H is bijective and H−1 = −H.

Proof. All claims are quite easy to prove when using the basis representationsof the functions involved.

1. This can be verified via straightforward computation using (3.1).

2. Hu vanishes on T iff all of its Fourier coefficients vanish. This is the caseiff all Fourier coefficients of u except for the 0-th vanish, which is thecase for exactly all constant functions.

3. A function v ∈ Hp satisfies

π

−π

v(eiφ)dφ = 0

iff its 0-th Fourier coefficient vanishes. In this case, direct computationconsidering the Fourier coefficients gives HH(−v) = v. Thus, v is theimage of H under the Lp(T)-function −Hv.

4. This follows directly from the proofs of 2. and 3.

4A similar statement holds under weaker assumptions if we replace ∂τ by weak deriv-atives. From this, we can see that H is also bounded as an operator on the Sobolevspaces W p

m(T) for 1 < p < ∞, m ∈ N. We will not pay any further attention to thisfact in this book.

3.3 Riemann-Hilbert problems 23

3.3 Riemann-Hilbert problems

In the following we will be concerned with boundary value problems forholomorphic functions on D. That is, we investigate problems imposingconditions on the values of a function on T. Their solutions are functionsholomorphic in D with nontangential boundary values satisfying these con-ditions. Riemann-Hilbert problems are perhaps the most important class ofthese problems.

This problem class originates from the dissertation of Bernhard Riemannhimself, which is best known for containing the first formulation of the celeb-rated Riemann Mapping Theorem [25]. Riemann used conformal mapping asan example, but he “treated the [general] problem only heuristically” ([31],Chapter 6). David Hilbert was the first person to come up with a solutionoperator at least for linear problems [12], hence the name. A historicaloverview over the development of Riemann-Hilbert problems can be foundin Chapter 6 of [31] or in Chapter 1 of [32].

To understand the general ideas behind Riemann-Hilbert problems, recallthat the space Hp(T) is the space of functions in Lp(T) where the Fouriercoefficients with negative index vanish. Intuitively, this means that we canprescribe “one half” of the boundary function to obtain a unique holomorphicfunction as a solution. For a different perspective on the same fact, onecan also think of Schwarz’ formula: Loosely speaking, we may not prescribethe boundary function of a holomorphic function arbitrarily, but we mayprescribe its real part arbitrarily.

These observations motivate to investigate other boundary value problemswhere “half of the boundary function” is prescribed. I.e., problems thatimpose one real condition on the value of the solution at each point on T.These problems are the so-called Riemann-Hilbert problems.

The most general way to state a Riemann-Hilbert problem is to write itin the form

F (t, w(t)) = 0 ∀t ∈ T

for a prescribed function F : T× C→ R. Under suitable regularity assump-tions, the sets {z ∈ C|F (t, z) = 0} for a given t ∈ T are submanifolds of Cwith real dimension 1, thus constituting one real condition imposed on w att.

Clearly, the solvability of the problem depends crucially on the structureof F . Roughly speaking, the theory of Riemann-Hilbert problems knownto date is based on splitting them into various classes depending on thestructure of their boundary condition.


The following two subsections give a sketchy overview of the solvabilityfor some of the most important classes of Riemann-Hilbert problems.

3.3.1 Linear Riemann-Hilbert problems

If the function F is affine linear in its complex argument, its solution spacemust be an affine linear space. We call the resulting problem a linearRiemann-Hilbert problem. These problems can be written more compactlyin the form

Im(c(t)w(t)) = h(t)

where the function c : T→ C is called the symbol of the problem. If h(t) ≡ 0,then the problem is called homogeneous.

Linear Riemann-Hilbert problems can be solved explicitly if the symbolc is continuous and zero free. In this case, we may assume that |c(t)| ≡ 1by dividing by |c(t)|. Then the symbol can be written as c(t) = eiφ(t) for afunction φ(t) which is continuous on T\{1}. The jump at 1 is then an integermultiple of 2π and tells us the winding number of c about 0. We therefore

call wind c = 12π

(limτ↗2π

φ(eiτ )− limτ↘0

φ(eiτ )

)the index of the problem. The

crucial results are given in the following theorem.

Theorem 3.23 ([31], theorems in Chapter 3). Given 1 < p <∞, let c = eiφ

be a continuous symbol with winding number k and let h ∈ Lp. Consider thelinear Riemann-Hilbert problem

Im(c(t)w(t)) = h(t).

Then the following assertions hold.

1. If k = 0, the set of solutions in Hp has real dimension 1. The boundaryfunctions of the solutions are given by

w0

(µ−H

(c

|w0|

)+ i

(c

|w0|

))where w0 = exp(−Hφ+ iφ) and µ ∈ R is an arbitrary constant.

2. If k > 0, the problem has exactly 2k + 1 linear independent solutions inHp.

3. If k < 0, there are −2k − 1 bounded linear functionals in Lp such thatthe problem is solvable in Hp iff h(t) lies in the kernel of all of thesefunctionals.

3.3 Riemann-Hilbert problems 25

As we see, the Hilbert operator appears in the solution formula. This isan important motivation for the quest for a discrete Hilbert operator: It isthe crucial building block for the explicit solutions of linear Riemann-Hilbertproblems.

3.3.2 Nonlinear Riemann-Hilbert problems

The key to the solvability of nonlinear problems is to approach them geo-metrically. If a problem is given via a function F , we consider the targetmanifold M :=

⋃t∈T

Mt where

Mt := {z ∈ C|F (t, z) = 0}.

This definition allows us to write the boundary condition of a Riemann-Hilbert problem more compactly as w(t) ∈ Mt ∀t ∈ T. The sets Mt arecalled the target curves of the problem.

These manifolds can now be separated into various classes. The first maindistinction is between the classes

• A of compact target manifolds that are diffeomorphic images of T×T and

• B of non-compact target manifolds that are diffeomorphic images of T×R.5

Obviously, manifolds may be part of neither of these classes.Riemann-Hilbert problems with a manifold of type B have a structure

which is very similar to that of linear problems. In fact, they are a general-ization of linear problems with continuously differentiable symbol.

The class A can be split up into three different classes again. A manifoldM ∈ A is a topological torus. Thus, (C× T) \M consists of two connectedcomponents one of which is bounded. We call this component intM andfor each t ∈ T we write intMt := {z ∈ C|(t, z) ∈ intM}. Now we mayask whether there is a holomorphic function wM ∈ H∞ ∩ C that satisfieswM (t) ∈ intMt or at least wM (t) ∈ intMt ∪Mt for all t ∈ T. Based on this,we consider the classes

• R where a function wM ∈ H∞ ∩ C with wM (t) ∈ intMt ∀t ∈ T exists,

• S where such a function does not exist, but where a function wM ∈ H∞∩Cwith wM (t) ∈ intMt ∪Mt ∀t ∈ T exists, and

5The precise definitions have additional requirements and can be found in Wegert’s book[30].


• N where neither of the above holds.

Problems with manifolds in N have no solution. Problems with manifolds inS turn out to have exactly one solution, namely wM itself. Finally, problemswith manifolds in R have an infinite-dimensional set of solutions. In fact,for arbitrary n ∈ N and z1, . . . , zn ∈ D, there is a one-dimensional set ofsolutions w such that the zj are precisely the zeros of wM −w, counted withmultiplicity. Proofs of these statements can be found in Chapter 2 of [30].

Circular Riemann-Hilbert problems. Circular Riemann-Hilbert problemsare a special class of nonlinear Riemann-Hilbert problems where each targetcurve is a circle. I.e., there are functions c : T → C and r : T → R+ suchthat the target curves are given by Mt = {z ∈ C| |z − c(t)| = r(t)}.

This extra information allows to give some more explicit results on thesolution set of the problem which are covered in a paper by Glader andWegert [11]. As it turns out, it suffices to consider problems where r ≡ 1,since the results for this kind of problems can be transferred to the generalcase using a transformation. Assuming that r ≡ 1, it can be shown thatthe regularity of c implies the regularity of solutions in a similar way as inTheorem 3.20: If α ∈ (0, 1) and c ∈ Cα, then the solution lies in H∞ ∩ Cα.

In the special case where c is constant, the problem can be solved explicitlyup to a Schwarz integral. In fact, we will be using such a problem for thediscretization of the Hilbert transform. In this context, we will derive theexplicit solution in the following section and give an extensive proof.

3.4 Obtaining the Hilbert transform from aRiemann-Hilbert problem

As outlined in the introduction, our strategy for the discretization of His to formulate a Riemann-Hilbert problem depending on some functionu ∈ Lp(T) and to obtain Hu from the solution of this problem.

3.4.1 Choice of the problem

First, we have to find a problem which allows us to obtain the Hilberttransform and is also well suited for discretization.

The most evident choice would be the simplest Riemann-Hilbert problem,namely, the Schwarz problem. Given u ∈ Lp(T), we can consider the problem

Rew(t) = u(t) ∀t ∈ T

3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem 27

By Theorem 3.11, the solutions to this problem in Hp are given by Su(z)+ ic.Thence, we know that for a solution w, Hu is equal to Imw|T up to a realconstant.

Unfortunately, the Schwarz problem is not well suited for discretizationwith circle packings. The main reason for this is that, as long as we merelyknow u, it is difficult to make predictions about the critical points of Su.Our discretization, however, requires information about the number andlocation of critical points of the solution. This and other difficulties relatedto the Schwarz problem will be explained in detail in Subsection 5.3.1.

To resolve these difficulties, we are going to consider a circular problemwith a target manifold in the class A. In contrast to the Schwarz problem,this problem has an infinite dimensional set of solutions and we will see thatwe can find one solution among them which has no critical points and whichis suitably normalized in addition.

Therefore, the Riemann-Hilbert problem we will investigate and be con-cerned with throughout the rest of this book is the following:

|w(t)| = expu(t) (HRHP)

All target curves Mt of this problem are circles centered at 0 and withradius expu(t). As mentioned in Subsection 3.3.2, this allows us to solvethe problem explicitly in the disc algebra H∞ ∩ C.

3.4.2 Solutions of the problem

Since our approach uses only basic complex analysis, we may directly state thetheorem characterizing the set of solutions of (HRHP). Basically, the strategyof the proof is to factor out the zeros of a solution using a Blaschke product.The remaining function is then identified by considering its holomorphiclogarithm.

Theorem 3.24. Let α > 0 and u ∈ Cα. Write wu(z) := expSu(z). Then,the solutions of (HRHP) in H∞ ∩ C are exactly the functions of the formz 7→ B(z)wu(z) where B is an arbitrary finite Blaschke product.

Proof. 1. By Theorem 3.19, Su ∈ Hp for every p < ∞ and Su(t) =u(t) + iHu(t) ∀t ∈ T. By Plemelj-Privalov, Hu ∈ C(T), thus Su ∈ H∞ ∩ C.Obviously this also implies wu ∈ H∞ ∩ C.

Let B be a finite Blaschke product. For each t ∈ T we have

|B(t)wu(t)| = |wu(t)| = exp ReSu(t) = expu(t)


Thus z 7→ wu(z)B(z) is a solution of (HRHP) and lies in H∞ ∩ C.2. Conversely, let w ∈ H∞ ∩ C be an arbitrary solution of (HRHP). Then

we have w(t) 6= 0 for all t ∈ T. Since w ∈ H∞ ∩ C, the set of zeros of w cannot have an accumulation point on T.

On the other hand, the set of zeros of w can not have an accumulationpoint in D either since w is holomorphic. Combining these statements, wcan have only finitely many zeros.

Thus, let B be a Blaschke product with precisely the same zeros (including

multiplicity) as w. Then the function w0(z) = w(z)B(z) is zero free, lies in

H∞ ∩ C and is still a solution of (HRHP). Therefore, it has a logarithm gwith exp ◦g = w0 on D (compare [1], Chapter 4, Corollary 2). For each t ∈ T

Re g(t) = log |w0(t)| = log |w(t)| = log expu(t) = u(t).

Consequently, Re g and ReSu are bounded functions which are harmonic inD and continuous on D and whose values on T coincide. Thence, they mustbe identical6. Since they are the real parts of the holomorphic functionsg and Su, we conclude g = Su + ic for some c ∈ R and thus w0 = wu · eic.Since z 7→ eicB(z) is again a Blaschke product, w has the claimed form.

Notation 3.25. Given u ∈ Cα, we will henceforth write wu(z) := expSu(z)for the “default solution” of (HRHP).

How to obtain Hu from a solution of (HRHP) is evident from the proofof Theorem 3.24. This is the content of the following proposition.

Proposition 3.26. Let u ∈ Cα and let w ∈ H∞ ∩ C be a solution of(HRHP), w(z) = wu(z)B(z). Then there is a continuous branch arg of theargument and a constant c such that

Hu(t) = argw(t)

B(t)+ c ∀t ∈ T.

If ‖Hu‖∞ < π, we can choose arg = Arg. In this case, c = 0.

Proof. As in the proof of Theorem 3.24, let logwu be a complex logarithmof the zero-free function wu. Then logwu(z) = Su(z) − ic for some realconstant c. Thus

Hu(t) = ImSu(t) = Im logwu(t) + c = argwu(t) + c = argw(t)

B(t)+ c

6This follows from Poisson’s formula for harmonic functions, compare the remark afterTheorem 3.10.


If ‖Hu‖∞ < π, then ImSu(z) < π holds by the maximum principle forharmonic functions ([1], Chapter 4, Theorem 21). Hence Su(z) = Logwu(z)

and we obtain Hu(t) = Arg w(t)B(t) .

Note that due to Theorem 3.20, ‖Hu‖∞ < π is satisfied if ‖u‖Cα is smallenough.

This theoretical result leaves a practical problem, though: To obtain Hufrom the solution of (HRHP), we need to know the corresponding Blaschkeproduct. Assume we have designed an algorithm which solves a discretizedversion of (HRHP). How do we know which Blaschke product we have todivide by?

Fortunately, this question is easy to answer due to the special structureof circle packings: The algorithm we design will only be able to computediscrete solutions which have no critical points and are suitably normalized.In the next theorem, we will prove that these requirements single out aunique solution in the continuous setting for which the Blaschke product isknown explicitly.

First, we define the normalization we are about to use.

Definition 3.27. A holomorphic function f : D → C is called standardnormalized if f(0) = 0 and f ′(0) > 0.

We can now state the theorem providing the necessary information for thediscretization with circle packings. We will have to strengthen our regularityassumptions on u even more for this theorem to hold.

Theorem 3.28. Let α > 0 and by Corollary 3.21, let C > 0 be a constantsuch that ‖Su‖H∞∩C ≤ C ‖u‖Cα for every u ∈ Cα. Then for every u ∈ Cα+1

with ‖u‖Cα <1C , the problem (HRHP) has a unique standard normalized

solution w which has no critical points in D. This solution is given by

w(z) = zwu(z).

Proof. Clearly, the function w given in the theorem is a solution of (HRHP),has a zero at 0 and its derivative is given by

w′(z) = zw′u(z) + wu(z).

Thus w′(0) = wu(0) > 0 since Su(0) ∈ R by definition of Su.Next, we show that w has no critical points. By definition of wu,

w′(z) = wu(z) (zS′u(z) + 1) .


This expression is zero only if zS′u(z) = −1, therefore, it suffices to show|zS′u(z)| < 1 ∀z ∈ D. By Lemma 3.15, we have izS′u(z) = Su(z). Byassumption, Su is bounded by 1. We conclude that the desired inequalityholds.

Finally, we address uniqueness. Assume that there is a second standardnormalized solution w(z) = wu(z)B(z) meeting the criteria of the theorem.We have

w′(z) = wu(z)(B(z)S′u(z) + B′(z)

).

B must have at least a simple zero at 0. If B has no further zeros, it isa scalar multiple of z 7→ z. In this case, w′(0) > 0 implies B(z) = z, i.e.w = w.

So assume that B has at least two zeros (counted with multiplicity). By a

theorem of Walsh (§5.3.1 in [29]), B′ has at least one zero in D in that case.

We shall show that the same holds for B(z)S′u(z) + B′(z).

First, we derive a lower bound for B′ on T. Let z1, . . . , zn be the zeros ofB with multiplicity (note that z1 = 0). Then by direct computation

B′(z)

B(z)=

n∑j=1

1− |zj |2

(z − zj)(1− zzj).

Now for t ∈ T,

∣∣∣B′(t)∣∣∣ =

∣∣∣∣∣ tB′(t)B(t)

∣∣∣∣∣ =

∣∣∣∣∣∣n∑j=1

1− |zj |2

(1− tzj)(1− tzj)

∣∣∣∣∣∣ =

n∑j=1

1− |zj |2

|1− tzj |2

where the absolute value could be dropped in the last expression since allsummands are real and positive. But since z1 = 0, one of the summands isequal to 1, thus ∣∣∣B′(t)∣∣∣ > 1 ∀t ∈ T.

Now chose r < 1 large enough such that∣∣∣B′(rt)∣∣∣ > 1 ∀t ∈ T still holds and

B′ still has a zero in Br(0). Since ‖S′u‖∞ < 1, we then have∣∣∣B′(rt) + B(rt)S′u(rt)− B′(rt)∣∣∣ =

∣∣∣B(rt)S′u(rt)∣∣∣ ≤ 1 <

∣∣∣B′(rt)∣∣∣for all t ∈ T. By Rouche’s theorem ([1], Chapter 4, Corollary in Section 5.2),

the functions z 7→ B(z)S′u(z) + B′(z) and B′ have the same number ofzeros in Br(0). We conclude that w′ has a zero in D. This contradicts theassumptions on w.


As we can see, the are in fact plenty of u such that the correspondingstandard normalized solution does have critical points. The crucial pointis that if u ∈ C1+α, we can always use the linearity of H to countervailthis problem: We simply multiply u by a sufficiently small positive realnumber to perform our discretization procedure and divide by the samefactor afterwards.

3.4.3 Counterexamples for u /∈ C1+α

Though the assumption u ∈ C1+α is quite restrictive, it is essential for thevalidity of the last theorem. In the following we construct a counterexampleshowing the problem that occurs if we merely assume u ∈ C1. We will alsogive some more counterexamples to indicate problems that occur if we reducethe regularity assumptions even further.

Continuous function with unbounded Hilbert transform. The followingexample was announced in Section 3.2 and will serve as a building block forthe counterexamples constructed afterwards.

Denote by D := {z ∈ D|Re(z) > 0} the upper half of the unit disc andwrite G0 := {x + iy ∈ C|0 < x < (y + 1)e−y}. Let g : D → G0 be thebijective conformal map whose boundary values satisfy f(1) = −i, f(0) = 0and f(−1) =∞.7 Then in particular f((−1, 0)) = iR>0 holds.

Using the symmetry principle ([1], Chapter 4, Theorem 24), extend f ontoD.8 The extension will be a mapping

f : D→ G := {x+ iy ∈ C| |x| < (y + 1)e−y}

which admits a continuous extension to D \ {−1}. Figure 3.1 shows thedomains.

For t ∈ T, let u(t) = −Re f(t). Obviously u admits a continuous extensionat t = −1 by setting u(−1) = 0. Theorem 3.19 then yields Su = −f andHu(t) = ImSu(t) for t ∈ T. Since ImSu is unbounded, the example hasbeen constructed.

7The existence of g follows from the Riemann Mapping Theorem ([1], Chapter 6, The-orem 1)

8This can also be done explicitly by defining f(z) = −f(z)


(a) Unit disc(b) G

Figure 3.1: Construction of a continuous function with unbounded Hilberttransform

Critical points of solutions for u ∈ C1. We will construct a continuouslydifferentiable function u : T→ R such that the solution w(z) = zwλu(z) ofthe problem

|w(t)| = expλu(t) (3.2)

has a critical point for any λ > 0, thus contradicting Theorem 3.28 for α = 0.Let v be the function constructed and denoted by u in the previous

paragraph. Define

u : T→ R, u(eiτ ) =

τ

0

v(eiσ)dσ

Since2π

0

v(σ)dσ = ReSv(0) = 0, this function is well defined and continuously

differentiable. The solutions to (3.2) whose only zero is a simple zero at 0are then given by z 7→ cz exp(Sλu(z)) for arbitrary c ∈ T.

The location and multiplicity of critical points of the solution does notdepend on c, so w.l.o.g. we investigate wλ(z) = z exp(Sλu(z)). We have

w′λ(z) = exp(Sλu(z))(1 + zS′λu(z))

and since S′λu(z) = λSv(z)iz by Lemma 3.15 we obtain

w′λ(z) = exp(Sλu(z))(1 + λif(z))

−1 1

−1

1

D

−1 1

−1

1

2

3

1

2

3

G0


where f is defined as in the first paragraph. Now choose z ∈ (−1, 0) such thatf(z) = i 1

λ . This is possible due to the construction of v. Then w′λ(z) = 0.So, regardless of the choice of λ, the solution always has a critical point inD.

No continuous solutions for u ∈ C. Having in mind the previous example,it is pretty easy to imagine how to construct a continuous function u suchthat (HRHP) has no continuous solution.

Let u be the function constructed in the first paragraph. Let w be acorresponding solution of (HRHP). With the same reasoning as in theproof of Theorem 3.24, we can assume that w is zero-free and conclude thatIm f − Imw is constant, where f is the conformal map constructed in thefirst paragraph. This contradicts the boundedness of w.

This example can be found in the last section of [11].

Singular inner functions for u ∈ H∞. If we allow u to be discontinuousand merely assume its boundedness, the next statement to fail is Proposition3.26. This is due to the existence of so-called singular inner functions.

The function f(z) = z+1z−1 maps D onto the left half plane {Re z < 0}.

The function w(z) = exp f(z) is bounded in D and satisfies |w(t)| = 1for t ∈ T \ {1}. Thus, w is a bounded discontinuous solution of (HRHP)with u ≡ 0. Following Proposition 3.26, we would expect that for eachHu = Im f + c for some constant c. But this is clearly wrong since Hu ≡ 0and Im f is not constant.

No solutions in Hq for u ∈ Lp. Finally, if we even allowed u to be unboun-ded, (HRHP) completely fails to be useful since it may have no solution inany of the Hp spaces, assuming that we understand the problem in the senseof nontangential limits. As an example, consider the function u(eiφ) = 1

p+1√φ

for φ ∈ (0, 2π]. It is straightforward to check that u ∈ Lp.Assume w were a corresponding solution of (HRHP) in Hq for q > 0.

Then,

‖w‖qq =

2πˆ

0

∣∣w(eiφ)∣∣q dφ =

2πˆ

0

exp

(q

p+1√φ

)dφ

But this integral does not exist for any q > 0, which can easily be verified bysubstituting φ = θ−p−1. Thus, there can be no solution w in any of the Hq


spaces, even for q < 19. So, we lack appropriate solution spaces for (HRHP)in this case.

9These spaces were not introduced in this book but can be defined in a way similar tothe other Hp spaces. They are covered e.g. in [10] and [14]

4 Circle Packings

We will now turn to the discrete setting and discuss the theory of circlepackings.

Ken Stephenson’s Book [27] is the standard volume on circle packings todate. Nevertheless, our work only requires few basics and some importanttheorems from this book, since we mainly concentrate on the manifoldstructure of circle packings, which was elaborated in a later paper by Bauer,Stephenson and Wegert [3].

All figures were created using the software PackCircle by Frank Martin [19].This software is based on MATLAB [21] and is interfaced with Stephenson’ssoftware CirclePack [28]. The figures were plotted in MATLAB and exportedusing matlab2tikz [26].

4.1 First examples and ideas

Before introducing circle packings on a formal mathematical level, we willstart with just showing some pictures and present the core ideas on aninformal level.

A circle packing might look like this:

Figure 4.1: A circle packing


36 4 Circle Packings

As one can see, a circle packing is a configuration of circles some of whichare mutually tangent, but which do usually not intersect or overlap.1

A circle packing is given a topological structure by its pattern of tangencies.Figure 4.2a shows the tangency pattern of the circle packing in Figure 4.1. Aswe can see, this is a graph which triangulates a topological closed disc. Thesegraphs are the topological structures we will work with in the following.

Given such a graph, it is not difficult to imagine that there will be severaldifferent circle packings having this tangency pattern. As an example, Figure4.2b shows a second circle packing based on the same graph as the packingin Figure 4.1. Describing the set of all circle packings on a given graph willbe the topic of Section 4.3. It will turn out that circle packings actuallyform a smooth manifold.

(a) The packing form Fig-ure 4.1 with its com-plex

(b) A second packing withthe same complex

Figure 4.2: Packings and complexes

In the context of discretizing complex analysis, circle packings should bethought of as a discretization of a domain in C. Now, assume that we aregiven two circle packings which have the same combinatorics, i.e. the sameunderlying tangency pattern. Thus, we have discretized two complex domainsusing the same combinatorial structure. Now, the idea is to understand thesetwo packings as a map between the two domains. For example, the packings

1Circle packings are based on tangencies. This is not the only approach to discretizecomplex analysis involving circles. E.g., the books of Bobenko and Suris [5] and [4]contain works on discrete complex analysis using patterns of mutually intersectingcircles. In this approach, cross ratios of intersection points and intersection anglesplay the core roles.

4.1 First examples and ideas 37

in Figure 4.3 could be thought of as the discrete version of a conformal mapfrom the unit disc onto a rectangle.

Figure 4.3: A discrete conformal map between D and a rectangle

This idea may take some getting used to for those new to the topic ofcircle packings: Our discrete holomorphic functions are no mappings in theclassical mathematical sense, but they are simply ordered pairs of two circlepackings. However, this notion gives us what we are looking for: a discretestructure relating two objects in the complex plane and identifying circleswith other circles.

Based on this notion, we will translate boundary value problems forholomorphic functions into boundary value problems for circle packings. Wewill give some examples and then formulate a discrete Hilbert transform.

A problem we will face when doing this is the lack of a linear structure incircle packing. Evidently, circle packings are nonlinear objects: they can’tbe “added” like holomorphic functions. This is problematic if we intendto discretize a linear operator in the context of circle packings. Knowingthat circle packings form a manifold, we can use the tangent space of thismanifold as a substitute. This is our main motivation to investigate thestructure of the set of all circle packings in some detail.

Before closing this section, we will show two more pictures which allow usto discover another important aspect of circle packings:


Figure 4.4: Overlapping circles

Look at Figure 4.4. In this snake packing2, circles do, in fact, intersect.Such a packing can not be interpreted as a discrete domain in C anymore,rather, it could be understood as a domain on a Riemann surface. A discretemap having this packing as its image packing would be the discrete versionof a holomorphic function which is not injective.

Nevertheless, this packing exhibits a structure very similar to that of thecircle packings we saw before. However, a circle packing may also look likethat in Figure 4.5.

Figure 4.5: A branched packing

2This packing was constructed by Ken Stephenson as his prototypical example for apacking which is locally, but not globally univalent (see below). It is known as “snakepacking” due to its shape.

4.2 Basic definitions 39

What happens here? In all packings seen before, every interior circle ofthe packing was surrounded by a chain of neighbors winding once around it.In Figure 4.5, the boundary circles wind twice around the shaded centralcircle. The coloring of the packing aids in observing this.

This behavior reminds us of saddle points of analytic functions: Wherethe derivative of a function vanishes, its values are winding multiple timesaround the value at the saddle point which can be seen by considering itsTaylor expansion. Thus, packings like 4.5 help us to discretize holomorphicfunctions whose derivative vanishes.

The packings seen in the first examples were univalent packings. Thesnake packing is a packing which is not univalent, but still locally univalent.The last example is called a branched packing.

While univalent and locally univalent packings can be continuously de-formed into each other, a univalent packing can never be continuouslydeformed into a branched packing3. This will turn out to be important inthe analysis of the manifold of circle packings and also has consequences foralgorithmic aspects.

4.2 Basic definitions

4.2.1 Complex

The topological structures our circle packings are based on are simplical2-complexes which triangulate an orientated closed disc. An example forsuch a complex is shown in Figure 4.6.

For our work, it is not necessary to discuss the combinatorics and topologyof these structures in detail. The details are worked out in the third chapterof Stephenson’s book [27]. We will just gather the properties these complexesexhibit in the following.

We will usually denote a complex by K. A complex consists of a set ofvertices V , a set of undirected edges E and a set of oriented triangular facesF . As can be observed in the example in Figure 4.6, these objects satisfythe following:

• An edge e = {v, w} ∈ E connects two vertices v and w. In this case, thevertices v and w are called adjacent.

3In this context, “deforming” means that there exists a curve of other circle packingsbetween the two packings. “Continuous deforming” means projecting this curve ontothe center or radius of any of the circles yields a continuous map


Figure 4.6: A complex with nodes, edges and faces (slightly shaded). Bound-ary nodes are filled, interior nodes not. Boundary edges are indark gray, while interior edges are in light gray. The black arrowsindicate the orientation of the faces.

• A face f consists of three mutually adjacent vertices u, v, w. It hasa prescribed orientation, which we denote by writing f = 〈u, v, w〉 =〈v, w, u〉 = 〈w, u, v〉. We say that the vertices u, v, w and the edges {u, v},{v, w} and {u,w} are incident to f .

• Every edge is incident to either exactly one or exactly two faces. It iscalled a boundary edge in the first case and an interior edge in the lattercase. If e is an interior edge, then its orientation with respect to its twofaces is different.

• A vertex is called a boundary vertex if it is incident to a boundary edgeand interior vertex otherwise. The boundary vertices v1, . . . , vm can beput in a cyclical order such that every vertex is adjacent to its predecessorand successor.

• Let n = |V | ≥ 3 be the number of vertices and m the number of boundaryvertices. K has Euler characteristic 1, which means that

|V | − |E|+ |F | = 1.

Since all edges except for the boundary edges are part of exactly two facesand all faces are triangles, 2 |E| −m = 3 |F |. Inserting this into the aboveequation yields the fundamental relation

|E| = 3n−m− 3

which is crucial for everything we do in the following.


• For a vertex v, its neighbors can be put into an order v1, . . . , vk suchthat for all j ∈ {1, . . . , k − 1}, vj is adjacent to vj+1. If v is an interiorvertex, the neighbors form a closed chain, i.e. v1 and vk are adjacent, too.{v, v1, . . . , vk} is called the flower of v. Moreover, k = deg v is called thedegree of v.

Standard examples for such complexes are regular polygonal complexes

with k generations. They are denoted by K[n]k . Their construction can

roughly be described as follows: K[n]1 consists of one interior vertex and

n boundary vertices. All boundary vertices are adjacent to the interior

vertex. K[n]k+1 is obtained from K

[n]k by adding vertices in such a way that

all boundary vertices of K[n]k become interior vertices with degree n. The

construction is visualized in Figure 4.7a for n = 7 and k = 1, 2, 3.To obtain a new complex from a given one, one can perform hex refinement.

This means that every face is subdivided into four faces by adding a newvertex on each edge incident to the face. The process is illustrated in Figure4.7b.

(a) Complexes K[7]1 (blue),

K[7]2 (blue and green),

K[7]3 (all colors)

(b) Hex refinement of a face

Figure 4.7: Complexes and hex refinement

Throughout the rest of this chapter, we assume that we are given a fixedcomplex K = (V,E, F ). It is important to note that almost everythingwe define in the following, like manifolds, contact and angle functions, andpackings, depends on K. To keep the notation simple, K will not appear inthe notation we use for these objects, though. We use the following notationsfor K.


Notation 4.1. K has n vertices that we denote by V = {v1, . . . , vn}. m ofthese vertices are boundary vertices. The vertices are arranged in such a waythat {v1, . . . , vm} are the boundary vertices of K in cyclical order. Thus, vjis adjacent to vj+1 for each j = {1, . . . ,m − 1} and vm is adjacent to v1.E = {e1, . . . , e3n−m−3}.

For j ∈ {1, . . . , n}, let d = deg vj . We denote by vj,1, . . . , vj,d the neighborsof vj arranged in such a way that for each k ∈ {1, . . . , d− 1}, vj,k and vj,k+1

are adjacent and 〈vj , vj,k, vj,k+1〉 ∈ F . If vj is an interior vertex, the sameholds for vj,d and vj,1. To simplify indexing, we also set vj,0 := vj,d. Thisnotation is a bit messy due to the double index, but we will use it rarely. Itwill turn out useful to prevent renumbering the vertices when discussing theflower of a vertex.

4.2.2 Circle packing

We are now ready to formally define circle packings.

Definition 4.2 (Circle packing). P = {C1, . . . , Cn} is called a circle packing(or just packing) on K if

(C1) for each j ∈ {1, . . . , n}, Cj is a circle with center zj ∈ C and radiusrj > 0

(C2) for j, k ∈ {1, . . . , n}, Cj are Ck externally tangent if {vj , vk} ∈ E

(C3) for j, k, l ∈ {1, . . . , n}, the centers of the circles Cj , Ck and Cl form apositively oriented triangle if 〈vj , vk, vl〉 ∈ F .

C1, . . . , Cm are called the boundary circles of P while the remaining circlesare called interior circles.

A circle packing is called univalent if for any j 6= k, intCj and intCk aredisjoint.

Remark 4.3.

• We do not allow generalizations of circles in our definition, e.g. Mobiuscircles or points (as circles with radius zero). This is essential for someof the theorems we will discuss in the following section.

• It is also possible to consider circle packings on the sphere or in thehyperbolic plane instead of the euclidean plane C. Since we will be workingwith equations imposed onto the euclidean centers and radii of a circlepacking, we will completely stay in the euclidean setting.


Before continuing, we remark that (C3) can be formulated differently,which will turn out very convenient later. Consider a face f = 〈va, vb, vc〉 ∈ F ,compare Figure 4.8. Let z′a, z′b, z

′c be the contact points of Cb and Cc resp.

Cc and Ca resp. Ca and Cb. These points are not collinear, so let C ′ be theunique circle passing through them. The circle will be called the dual circleof P corresponding to f and will be denoted by Cf henceforth.

z′a

z′b

z′c

za

zb

zc

Figure 4.8: A face and its face circle

As is evident from Figure 4.8, Cf is actually the inscribed circle of thetriangle formed by za, zb and zc.

4 This triangle is positively oriented iff z′a,z′b and z′c lie in counter-clockwise order on Cf . Thus, (C3) is equivalent torequiring that the contact points of Cj , Ck and Cl lie in counter-clockwiseorder on the unique circle passing through them.

Definition 4.2 is purely geometric. This is the usual way to define circlepackings which can be found e.g. in Ken Stephenson’s book [27].

For our purposes it is not sufficient, though. We intend to investigate thestructure of the set of all circle packings on K, therefore, we need a morealgebraic definition.

We thus describe a circle packing by its most natural coordinates: Theradii rj and the centers zj of its circles, j ∈ {1, . . . , n}. Writing zj = xj + iyjwith real xj and yj , we see that a circle packing can be specified by a vector

4This can be verified with elementary geometric arguments. We omit the proof.


of 3n real coordinates, namely, the rj , xj and yj . This motivates to introducethe vector form of a packing.

Definition 4.4 (Vector form). Let P = {C1, . . . , Cn} be a circle packing andfor every j ∈ {1, . . . , n} let zj = xj + iyj be the center of Cj and let rj ∈ R>0

be the radius of Cj . We call (r1, . . . , rn, x1, . . . , xn, y1, . . . , yn)T ∈ Rn>0×R2n

the vector form of P and denote by D the set of all vector forms of packingson K. The vector (r1, . . . , rn)T ∈ Rn>0 is called the label of P.

Henceforth, we will frequently use the term “circle packing” to refer tothe vector form of a circle packing. We will also abuse notation and denotea packing and its vector form by the same variable. I.e., we will speak of thecircle packing P = (r1, . . . , rn, x1, . . . , xn, y1, . . . , yn)T . Sometimes, we willdecompose the vector form into the three vectors r ∈ Rn>0 and x, y ∈ Rnand indicate this by writing P = (r, x, y).

The label of a packing plays a special role in the general theory of circlepackings. This has a simple reason: If two packings have the same label,they can be mapped onto each other with a plane rigid motion. This isthe statement of Stephenson’s Monodromy Theorem (Theorem 5.4 in [27]).When working in the hyperbolic plane or on the sphere, it is convenient towork with the radii alone to avoid using spherical or hyperbolic coordinates.

However, the boundary value problems discussed in this book involve thecenters of a circle packing. For this reason, the label of a packing alone willhardly be used in the following.

4.3 Manifold structure

In this section, we will introduce the prerequisites for linearization in thecontext of circle packings. To do so, we shall now investigate the structureof D, i.e. the set of all circle packings.

The core result of this section states that D is an (m+ 3)-dimensional realsubmanifold of R3n consisting of a finite number of connected components.The paper [3] gives two different proofs of this result: The first one showsthat the set of all packing labels D∗ is a manifold, from which it is easy todeduce that D is a manifold, too. The second one identifies D as the zeroset of the so-called contact function. We will stick to the latter proof in thefollowing, because this proof gives important extra information for boundaryvalue problems.

4.3 Manifold structure 45

4.3.1 Contact function

To find a suitable function which has D as its zero set, we will formulate (C2)and (C3) as nonlinear equations imposed on the vector form of a packing.

For (C2), this yields the contact equations. Given two circles with radiir1, r2 and centers z1, z2, respectively, it is easy to verify that these circleswill be externally tangent iff

|z1 − z2| = r1 + r2.

This is visualized in Figure 4.9.

z1 z2

r1 r2

Figure 4.9: Illustration of the contact equation

Definition 4.5 (Contact function). Let r, x, y ∈ Rn. For j ∈ {1, . . . , 3n−m− 3} let ej = {vk, vl}. We define

ωj(r, x, y) =1

2((xk − xl)2 + (yk − yl)2 − (rk + rl)

2)

and ω = (ω1, . . . , ω3n−m−3). ω : R3n → R3n−m−3 is called the contactfunction of K.

The factor 12 in this definition does not matter since we are only interested

in zeros of ω, but it eliminates the factor 2 appearing when we calculatederivatives of ω.

For (C3), note that three points z1, z2, z3 ∈ C form a positively orientedtriangle iff

Λ(z1, z2, z3) = Im

(z3 − z1

z2 − z1

)> 0.

Since Λ: C3 \ {(z1, z2, z3)|z1 = z2} → R is obviously continuous, thepreimage of (0,∞) under Λ is open in C3 ∼= R6.

With these preparations, we arrive at the following lemma.


Lemma 4.6. There is an open set U ⊃ D, such that

(r, x, y) ∈ D ⇐⇒ ω(r, x, y) = 0 and (r, x, y) ∈ U .

Proof. For a face f = 〈vj , vk, vl〉 ∈ F , write

Uf = {(r, x, y) ∈ R3n|Λ(xj + iyj , xk + iyk, xl + iyl) > 0}

and define U :=⋂f∈F

Uf ∩ (Rn>0 ×R2n). Since the intersection is over finitely

many open sets, U is open. Then, for a given vector (r, x, y) ∈ Rn>0 × R2n,the corresponding configuration of circles satisfies (C2) iff ω(r, x, y) = 0 andit satisfies (C3) iff (r, x, y) ∈ U . This proves the lemma.

To prove that D is a manifold, it remains to show that the contact functionhas maximal rank at every point on D. This is the content of the followingcrucial lemma.

Lemma 4.7 (Lemma 5.7 in [3]). For (r, x, y) ∈ D let M ∈ R3n−m−3×3n−m

be the last 3n−m columns of Dω|(r,x,y) ∈ R3n−m−3×3n. Then, a basis forthe kernel of M is given by 0

1n

0

,

001n

,

0−yx

∈ Rn−m × Rn × Rn.

In particular, M has rank 3n−m− 3, thus Dω|(r,x,y) has full rank, too.

In Section 4.4, we will give a sketch of the proof by Bauer, Stephenson andWegert. Combining the previous lemmas, we have the following theorem.

Theorem 4.8. D is a smooth manifold of dimension m + 3. Its tangentspace coincides with the kernel of Dω.

The importance of Lemma 4.7 goes beyond this theorem. It tells us thatthe contact function is regular at each circle packing and also allows us toconstruct a basis of its kernel. This is important for the linearization ofboundary value problems.

4.3.2 Angle sums and branch structures

The goal of this subsection is a closer analysis of the structure of D. Havingseen that D is a manifold of real dimension m+ 3, we will now investigateits connectedness and how to parametrize the manifold.


In Section 4.1, we saw an example of a branched circle packing. We willstart with introducing angle sum maps that help us to formalize this concept.

By elementary geometry, the angles of a triangle can be computed fromits side lengths. If a triangle has side lengths a, b and c, the angle α betweenthe sides with length b and c is given by

arccosb2 + c2 − a2

2bc.

Given a circle packing P = (r, x, y), we are interested in computing theangles of those triangles formed by three circle centers which correspond tofaces of K. Now, notice that this can be done in two completely independentways. For a face 〈vj , vk, vl〉, the angle α of the triangle zj , zk, zl at zj satisfies

cosα =(rj + rk)2 + (rj + rl)

2 − (rk + rl)2

2(rj + rk)(rj + rl)

=|zj − zk|2 + |zj − zl|2 − |zk − zl|2

2 |zj − zk| |zj − zj |.

The equality of the two expressions is geometrically evident and also followsdirectly from ω(P) = 0. If Λ(zj , zk, zl) > 0, direct computation confirms

that α is actually the argument of the complex numberzl−zjzk−zj , i.e.

zl − zjzk − zj

= reiα (4.1)

for an r > 0. Based on this, we will now define angle sum maps. Intuitively,they assign to each vertex the sum over all angles of faces incident to thisvertex.

Definition 4.9 (Angle sum maps). The function Φ: Rn>0 → Rn,Φ =(Φ1, . . . ,Φn) with

Φk(r) =

deg vk∑j=1

arccos

((rk + rk,j−1)2 + (rk + rk,j)

2 − (rk,j−1 + rk,j)2

2(rk + rk,j−1)(rk + rk,j)

)is called radial angle sum map. Φ = (Φm+1, . . . ,Φn) is called interior radialangle sum map.

Define E := {z ∈ Cn|∃vj , vk ∈ E such that zj = zk}. Then, the functionΨ: Cn \ E → Rn,Ψ = (Ψ1, . . . ,Ψn) with

Ψk(z) =

deg vk∑j=1

arccos

(|zk − zk,j−1|2 + |zk − zk,j |2 − |zk,j−1 − zk,j |2

2 |zk − zk,j−1| |zk − zk,j |

)


is called center angle sum map.For a circle packing P = (r, x, y), we write Φ(P) := Φ(r) and Ψ(P) :=

Ψ(x+ iy). In this case, inserting ω(P) = 0 into the definitions of Ψ and Φimmediately yields that Φ(P) = Ψ(P).

Let z ∈ Cn such that for every face 〈vj , vk, vl〉 ∈ F we have Λ(zj , zk, zl) > 0.Then (4.1) implies that Ψk(z) is a positive integer multiple of 2π for everyk > m. In particular, for every circle packing P,

Φk(P) = Ψk(P) ∈ 2πN for all k > m. (4.2)

Notice, however, that the above observation is false for Φ: Given r ∈ Rn>0,increasing rk decreases Φk(r). In other words, for k > m, Ψk is locallyconstant but Φk is not. This observation is a crucial ingredient for the proofof Lemma 4.7 and also the main reason for us to bother introducing bothangle sum maps, even though they coincide for each circle packing.

The statement in (4.2) actually has a converse. Given a vector r ∈ Rn>0

such that Φk(r) ∈ 2πN, there is always a circle packing with label r whichis unique up to rigid motions. This and (4.2) are the statements of KenStephenson’s Monodromy Theorem, Theorem 5.4 in [27].

It is now time to address the branched packings mentioned in Section 4.1.Having established (4.2), we can divide circle packings into various classesas done in the following definition.

Definition 4.10 (Branch structure). Let P = (r, x, y) be a circle packing.

• bP := 12π Φ(P)− 1n−m = (Φm+1(r)

2π − 1, . . . , Φn(r)2π − 1)T ∈ Nn−m0 is called

the branch structure of P. Note that bP can be computed from the radiialone, so we can also speak of the branch structure of a label.

• P is called locally univalent if bP = 0.

• We denote by Db the set of all packings with branch structure b. Inparticular, D0 is the set of all locally univalent circle packings.

Thus, D may be decomposed as D =⋃

b∈Nn−m0

Db. The following lemma

shows that actually, Db is non-empty only for finitely many b.

Lemma 4.11 (c.f. Definition 11.4, Theorem 11.5 in [27]). Db is non-emptyiff for every closed edge path γ containing g ∈ N edges the following holds:Denoting by Vγ the set of all nodes interior to γ, g > 2

∑vk∈V

bk−m + 2.


Henceforth, any vector b ∈ Nn−m0 that satisfies the assumptions of Lemma4.11 will be called a branch structure.

Remark 4.12. We have introduced branch structures “backwards”. InStephenson’s book, a branch structure is first defined as a vector whichsatisfies the assumptions of Lemma 4.11 ([27], Definition 11.4). In a second

step, it is shown that for each packing P the vector Φ(P)−1n−m is a branchstructure ([27], Lemma 11.5).

4.3.3 Parametrization of DbΦ is a continuous function everywhere on D. Thus, for b1 6= b2, Db1 and Db2can’t be connected.

Intuitively, we expect that Db is connected for a given branch structure b.This is correct, as we will see by finding a global chart on Db.

To do so, we formulate the following theorem which is one of the mostremarkable results in the theory of circle packings. It states that we canalways find a circle packing for arbitrarily prescribed boundary circle radiiand branch structure.

Theorem 4.13 (Combination of Lemma 11.5 and Theorem 11.6 in [27]).Let b be a branch structure and let r1, . . . , rm > 0 be given arbitrarily. Thereis a unique packing label r with branch structure b such that the first mentries of r are given by r1, . . . , rm.

Let P be a circle packing with packing label r whose n-th circle is centeredat 0 and whose (n− 1)-th circle is centered on R+. We henceforth assumethat vn and vn−1 are adjacent. Thus, the n-th and (n − 1)-th circle of apacking can not have the same center.

If P ′ is a second circle packing with packing label r, there is a plane rigidmotion z 7→ λz + ζ that maps P to P ′ by the Monodromy Theorem. Let znand zn−1 be the centers of the n-th resp. (n− 1)-th circle of P ′. Then

ζ = zn, λ =zn−1 − zn|zn−1 − zn|

We conclude that a circle packing is uniquely determined by its branchstructure b, the radii of its boundary circles, zn and zn−1−zn

|zn−1−zn| , where zn and

zn−1 are the centers of its n-th resp. (n− 1)-th circle. From this, we obtaina chart on Db.


Proposition 4.14 (Theorem 5.2 and Corollary 5.3 in [3]). Let b be a branchstructure. The map

πb : Db →Rm>0 × C× T

(r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) 7→(r1, . . . , rm, zn,zn−1 − zn|zn−1 − zn|

)

is a smooth diffeomorphism and a global chart on Db. Its inverse is a regularparametrization of Db. In particular, Db is a connected component of D.(Here, zj = xj + iyj)

Remark 4.15.

• If we allowed circles with radius 0, i.e. circles which degenerate to a point,every circle packing could be continuously transformed into a point. Inparticular, D would be connected in this case.

• The fact that D is not connected is of crucial relevance for algorithms.If we intend to compute a circle packing using Newton’s method or ahomotopy method, we need to know its branch structure in advance, sincecircle packings with different branch structures can not be continuouslytransformed into each other.

This is one of the aspects that explain why we are working with a nonlinearRiemann-Hilbert problem instead of the much simpler Schwarz problem.We will stress this point in some more detail in Subsection 5.3.1.

4.3.4 Normalization

We have learned that the m+ 3 degrees of freedom may be decomposed into3 degrees for a rigid motion and m degrees for the radii of the boundarycircles.

In Chapter 5, we intend to discuss boundary value problems for circlepackings. This means that we will stress the question how the m degrees offreedom can be eliminated. For this, it is helpful to define the set DN ⊂ D0

of vector forms of normalized circle packings where the 3 freedoms for rigidmotions are eliminated.

Definition 4.16. The standard normalization is the function N : R3n →R3 defined by (r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) 7→ (xn, yn, yn−1). A circlepacking P with N(P) = 0 which additionally satisfies xn−1 > 0 is called astandard normalized circle packing.


The circles used for normalization are termed the α-circle resp. β-circleof a packing in circle packing literature. By convention, we have arrangedthe vertices of K in such a way that these are the n-th and (n− 1)-th circle.As stated in the previous subsection, we will confine ourselves to the casewhere the α-circle and the β-circle are adjacent.

Lemma 4.7 immediately gives

Lemma 4.17. The last 3n−m columns of

(DωDN

)form a regular matrix.

Remark 4.18. Note that in fact, N does not globally eliminate all rigidmotions. The rigid motion z 7→ −z maps an N -normalized circle packing toanother N -normalized circle packing. That is why we additionally requirexn−1 > 0 in definition 4.16. Choosing N as standard normalization hasalgorithmic reasons:

1. N still eliminates rigid motions locally: The set of all circle packingsP with bP = 0 and N(P) = 0 consists of two connected components.Namely, one component where xn−1 > 0 and one where xn−1 < 0. Recallthat xn−1 = 0 can not hold for a packing with N(P) = 0. Thus, Newton’smethod or homotopy methods with initial values on or near one of thecomponents will stay near to the same component.

2. Evaluating N and its derivative is extremely cheap due to their simplestructure. A function that eliminates rigid motions completely will makeuse of the complex argument function, thus having a significantly moredifficult structure.

We now define the set we will be concerned with most of the time.

Definition 4.19. DN := {P = (r, x, y) ∈ D|bP = 0, N(P) = 0, xn−1 > 0}denotes the set of all standard normalized locally univalent circle packings.

The following proposition is a summary of everything we have deducedabout DN in this section and is stated here for convenience.

Proposition 4.20. DN is a smooth connected m-dimensional manifold.The map

(r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) 7→ (r1, . . . , rm)

is a global chart on DN and its inverse is a global regular parametrization ofDN . At each P ∈ DN , the tangent space satisfies

TPDN = ker

(DωDN

)∣∣∣∣P

.


There is an open set UN ⊂ Rn>0 ×R2n such that the following holds: p ∈ UNis the vector form of an N -normalized locally univalent circle packing iffN(p) = 0 and ω(p) = 0.

Henceforth, UN will denote an open set that meets the conditions of theproposition.

4.4 Discrete harmonic functions on circlepackings

In this section, we do a little digression from our main topics and discussdiscrete harmonic functions on K. The main motivation for this digressionis to gain insight into the proof of Lemma 4.7. As stated in the previoussection, this lemma originates from [3]. Technically, the proof we will sketchhere is exactly the original proof. However, it is insightful to relate the proofto discrete harmonic functions.

Before we start, a word of caution: The discrete harmonic functionsdiscussed here massively differ from the discrete analytic functions introducedin the following section, so they should not be mixed up.

General discrete harmonic functions appear e.g. in probability theory [2]or in discrete differential geometry [5]. The idea is as follows. Given anundirected graph (of course, we will work on K), a weight is assigned to eachinterior edge by a function ν : E → R>0. It is crucial that the weights arestrictly positive. The functions we discuss are real (or complex) functionson the nodes of K, i.e. functions f : V → R. For such a function, we definethe discrete Laplacian ∆ by

∆f : V → R, (∆f)(v) =∑

{v,w}∈E

ν({v, w})(f(w)− f(v))

A function is called a discrete harmonic function with respect to ν on K if(∆f)(v) = 0 holds for all interior vertices of K. Clearly, constant functionsand linear combinations of discrete harmonic functions are discrete harmonicagain.

Discrete harmonic functions exhibit properties similar to harmonic func-tions in the continuous setting. A very important property is the discretemaximum principle.

Lemma 4.21 (Maximum principle for discrete harmonic functions). Letν be a weight function and let f be a discrete harmonic function w.r.t. ν.

4.4 Discrete harmonic functions on circle packings 53

Then for each interior vertex v ∈ V there is a vertex w ∈ V adjacent to vsuch that f(w) ≥ f(v). In particular, f attains its maximum at a boundaryvertex.

Proof. Assume the contrary, i.e. let v ∈ V be an interior vertex withf(w) < f(v) for all vertices w adjacent to v. Then

(∆f)(v) =∑

{v,w∈E}

ν({v, w})(f(w)− f(v)︸︷︷︸<0

) < 0

contradicting the assumption that f is discrete harmonic.

From this principle, the theorem on existence and uniqueness followseasily.

Theorem 4.22. Let ν be a weight function and let a ∈ Rm. Then there is aunique discrete harmonic function f such that f(vj) = aj ∀j ∈ {1, . . . ,m}.Proof. We start with uniqueness. Let f, g be two such functions. Thenf − g is a discrete harmonic function which vanishes at all boundary vertices.By the maximum principle, f − g vanishes everywhere, which implies thatf ≡ g.

For existence, let x = (f(vm+1), . . . , f(vn)). Rearranging ∆f = 0 showsthat x can be calculated by solving a linear system of equations M ·x = N ·afor two fixed matrices M ∈ R(n−m)×(n−m) and N ∈ R(n−m)×m dependingon the weights. Since solutions to this system are unique, M must be regular.Hence, a solution must exist for each a.

It is quite remarkable that both statements hold for arbitrary weights.The weights ν usually arise from the “context”. E.g., in probability, they

model transition probabilities between various states (i.e. nodes). In circlepacking, the weights we consider have a both a geometric and an algebraicinterpretation. We postpone the geometric interpretation to the end of thissection. Fix a packing label r. Let e = {v, w} ∈ E be an interior edgeincident to the faces vj , vk, va and vk, vj , vb. We define

νP(e) =1

rj + rk

(√rjrkra

rj + rk + ra+

√rjrkrb

rj + rk + rb

)Recall the definition of the radial angle sum function Φ. Differentiating thisfunction at r shows that its partial derivatives can be expressed compactlyusing the weights: One can compute

∂rj,kΦj(r) =1

rj,kν({vj,k, vj})


for every k ∈ {1, . . . , deg vj} and

∂rjΦj(r) =−1

rj

deg vj∑k=1

ν({vj,k, vj}).

Now let u ∈ Rn be a vector in the kernel of DΦ|r. The equation DΦj |r ·u = 0can then be rewritten as

0 = uj∂rjΦj(r) +

deg vj∑k=1

uj,k∂rj,kΦj(r) =

deg vj∑k=1

(uj,krj,k− ujrj

)ν({vj,k, vj})

which just means that the function f : V → R, f(vj) =ujrj

is discrete

harmonic.We have thus proven the following theorem.

Theorem 4.23. Let P = (r, x, y) be a circle packing and let u ∈ Rn. Then

u lies in the kernel of DΦ|r iff the function f : V → R, f(vj) =ujrj

is discrete

harmonic w.r.t. the weight function νP .

Applying Theorem 4.22 to this situation allows us to characterize thekernel of DΦ|r in a straightforward manner.

Corollary 4.24. DΦ|r has full rank n−m. There is a basis u1, . . . , um ofits kernel such that ukj = δjk for j, k ∈ {1, . . . ,m}.

This is a different way to formulate Lemma 4.4 in [3]. We are now in aposition to sketch the proof of Lemma 4.7.

Proof of Lemma 4.7 (Sketch). A simple geometric argument (Lemma 5.4 in[3]) shows that the last 2n columns of Dω|(r,x,y) have rank 2n− 3.

Now, the crucial step is to show that if p = (pr, px, py) ∈ R3n lies in the

kernel of Dω|(r,x,y), then pr must lie in the kernel of DΦ|r (Lemma 4.6 in

[3]). This is done by expressing DΦ|r in terms of derivatives of the centerangle function Ψ, which was noted to be locally constant (recall the remarksafter Definition 4.9!).

With this at hand, the rest of the proof is easy. Assume p = (pr, px, py) isin the kernel of Dω|(r,x,y) and its first m entries are equal to 0. Then pr = 0by Corollary 4.24 and thus (px, py) must lie in the 3-dimensional kernel ofthe last 2n columns of Dω|(r,x,y).

Overall, this shows that the last 3n−m columns of Dω|(r,x,y) have a kernelof dimension 3. The basis given in Lemma 4.7 can be computed using thegeometric argument from the first step.

4.5 Discrete analytic functions 55

The use of discrete harmonic functions emphasizes that the positivity ofthe radii is essential for the proof.

Using Herron’s formula ([9], Section 1.5), it can be shown that√rjrkra

rj + rk + raand

√rjrkrb

rj + rk + rb

are just the radii of the face circles C〈vj ,vk,va〉 resp. C〈vk,vj ,vb〉. This propertycan be used to prove the following beautiful result due to Tomasz Dubejko([27], Theorem 18.3).

Theorem 4.25 (Dubejko). Given a circle packing P = (r, x, y), define thecenter function f : V → C by f(vj) = xj + iyj . Then f is discrete harmonicw.r.t. νP .

4.5 Discrete analytic functions

We have indicated the idea of discrete analytic functions in Section 4.1. Weare now able to formally define these functions.

Definition 4.26 (Discrete analytic function). Let P and P be circle packings

on K and let P be univalent. Together, they define a discrete analyticfunction f on K which we denote by f : P → P.

If in addition P is locally univalent, f : P → P is called discrete conformalmap.

As explained before, we think of P as the discretization of some connecteddomain in C. This is the domain of the function, while P may be thoughtof as the image of the function.

If P = {C1, . . . , Cn} is not locally univalent, let (bP)j > 0. In thiscase, the neighbors of Cm+j wind (bP)j + 1 times around Cm+j . Thisis the discrete analogue of critical points: If a holomorphic function fhas a critical point with multiplicity k at z0, Taylor expansion yields thatf(z) = f(z0) + (z − z0)k+1 +O((z − z0)k+2) as z → z0. Geometrically, thismeans that the values of f near z0 are winding k + 1 times around f(z0).The flowers with branching order k in circle packing imitate this behavior.We therefore think of the circle Cm+j in the packing P as a critical point oforder (bP)j . This justifies the name “discrete conformal map” for functionswhere P is locally univalent.

Moreover, recall Definition 3.27. Let both P and P be standard normalizedcircle packings. The fact that the n-th circle of both packings is centered at


0 exhibits the function f : P → P as discrete version of a function f withf(0) = 0. The fact that the (n− 1)-th circles are centered on the positivereal line should be interpreted as f(ε) > 0 for small ε > 0, i.e. f ′(0) > 0.This justifies using the term “standard normalized” in both definitions 3.27and 4.16.

4.6 Maximal packings

We will now introduce a very important special class of circle packings,namely, the so-called maximal packings.

Definition 4.27 (Maximal packing). A locally univalent circle packing Pis called a maximal packing iff

• each circle C ∈ P satisfies C ⊆ D

• each boundary circle C of P is internally tangent to T.

The following theorem is another core result in the theory of circle packings.

Theorem 4.28 (Proposition 6.1 in [27]). There is a univalent maximal

circle packing P on K. If P and P are two maximal packings on K, thereis a conformal automorphism of D that maps P onto P. In particular,every maximal packing is univalent and there is a unique maximal packingPK ∈ DN .

It should be remarked that the theorem is actually stated slightly differentin Stephenson’s book: It merely states that there is an essentially uniqueunivalent maximal packing. This formulation does not exclude the possibleexistence of packings which satisfy our definition and which are not univalent,but only locally univalent. Rereading Stephenson’s proof however revealsthat he only uses the local univalence of the maximal packing to proveuniqueness. Thus, the theorem holds good in the way stated here.

Theorem 4.28 implies that we could have equivalently defined maximalpackings by requiring univalence instead of just local univalence. This is theusual way to define maximal packings. The advantage of our definition isthat univalent packings are only characterized geometrically, while locallyunivalent packings can be characterized analytically.

The proof given in [27] uses the Poincare model of hyperbolic geometry.For a description of this model, see, for example, [15]. In this setting, circlesinternally tangent to T are circles with infinite hyperbolic radius. It is very

4.7 Some results on discrete analytic functions 57

interesting to note that this fits beautifully with Theorem 4.13: Asking formaximal packings is basically the same as asking for packings with prescribedboundary radii, namely, all radii are prescribed to be equal to ∞.

The left one of the packings seen in Figure 4.3 was a maximal packing on

the complex K[7]3 . Figure 4.10 shows two more maximal packings: one for

the complex of the packing in Figure 4.2a, and one for K[6]3 .

(a) Maximal packing for thecomplex from Figure 4.2a

(b) Maximal packing for K[6]3

Figure 4.10: Maximal packings

In the context of discrete analytic functions, we may understand a maximalpacking as a discretization of D. Consequently, if P is a maximal packing,we understand f : P → P as the discrete version of a holomorphic functionf : D→ C.

The uniqueness of PK makes it our canonical choice for a discretizationof D. In the following, we will often identify a given circle packing P withthe discrete analytic function f : PK → P.

4.7 Some results on discrete analytic functions

To close this chapter, we are going to present two interesting results fromthe theory of discrete analytic functions. We will not need these results later.They are stated here just as a digression to convince the reader that circlepackings are, in fact, a sensible discretization of complex analysis.


4.7.1 Discrete maximum principles

The classical maximum principle for holomorphic functions states the fol-lowing: Given a domain U ⊂ C and a function f holomorphic on U andcontinuous on U , then |f | has a local maximum in U only if f is constant([1], Chapter 4, Theorem 12). If z ∈ U is a local minimum of |f | in U ,then f(z) = 0 or f is constant (apply the maximum principle to z 7→ 1

f(z) ).

Moreover, Re f and Im f can have no local maximum or minimum in U ,since they are harmonic functions ([1], Chapter 4, Theorem 21).

This principle can be recovered in two ways for a discrete analytic functionf : P → P: First, one could interpret the real and imaginary parts of thecenters of P as discrete Re f and Im f . In this case, Dubejko’s Theorem 4.25implies that the discrete Re f and Im f attain their maximum and minimumat boundary circles.

Second, there is a geometrically motivated discrete analogue of |f ′|: Bythe Taylor expansion, |f(z + h)− f(z)| ≈ |f ′(z)|h for small h > 0. Con-sequently, |f ′| measures how much f distorts lengths. A discrete analogue

for this value is the quotient of the radii of two corresponding circles in Pand P.

We therefore define f# : V → R+ by f(vj) =rjρj

, where rj is the radius

of the j-th circle in P and ρj is the radius of the j-th circle in P. Then wehave the following result of Stephenson.

Theorem 4.29 (Theorem 11.12 in [27]). Let f : P → P ′ be a discreteanalytic function. Assume that f# is not constant. Then f# attains itsmaximum at a boundary vertex. If f# attains its minimum at an interiorvertex vj, then P ′ has a branch point at vj.

4.7.2 Approximation of the Riemann Mapping

Theorem 4.28 can be interpreted as a discrete version of the RiemannMapping Theorem from the following point of view. Given an arbitraryunivalent circle Packing P for some given complex K, we will always findthe corresponding maximal packing PK . This means, we can find a discreteconformal map f : PK → P for any packing P. This corresponds to theclassical Riemann Mapping Theorem stating that given any simply connectedproper subdomain G ⊂ C, there is a bijective conformal map f : D→ G. Thismap is called Riemann map and it is unique up to conformal automorphismsof D.

4.7 Some results on discrete analytic functions 59

Based on this, it is actually possible to approximate the classical Riemannmap. For sufficiently small ε > 0, one can construct a circle packing Pε suchthat all of its circles’ centers lie in G and all radii are equal to ε. Figure 4.11suggests how this works. Now, for each Pε, we can compute a correspondingmaximal packing, thus obtaining a discrete conformal map. We can usethis map to compute an approximation fε of the Riemann map which iscontinuous and piecewise affine linear. We are not going to carry out thedetails here and will just state the convergence result.

(a) Discrete G (b) Corresponding maximal packing

Figure 4.11: Construction of a discrete Riemann map

Theorem 4.30 (Rodin-Sullivan, Theorem 19.1 in [27]). For ε > 0 suffi-ciently small, let fε be the map constructed in the way sketched above5. Thenthe fε converge to a bijective conformal map f : D→ G as ε→ 0 where theconvergence is pointwise and uniform on compact subsets of D.

This theorem stands out in the theory of circle packings because it is oneof rather few convergence results up to now. Theorems like the discretemaximum principle indicate that the theory of circle packings provides agood discrete analogue for complex analysis. There are also lots of numericalexperiments that suggest that discrete objects converge to their classical

5Its precise definition can be found in Chapter 19 of [27].


counterparts. Nevertheless, proven convergence theorems are missing inmany cases to date and are a major open problem yet to be addressed. Thisalso holds for the discrete Hilbert transform.

5 Discrete Hilbert Transform

Having introduced circle packings and all technical auxiliary material, weare now in a position to address boundary value problems for circle packings.Throughout this chapter, K will continue to be a fixed complex for whichwe retain the same notation as previously.

In Section 4.3, we deduced that DN is a real manifold of dimension m.Additionally, we stated a theorem that showed that the m degrees of freedomare directly related to the radii of the boundary circles.

The fact that there are exactly m degrees of freedom suggests that itmakes sense to relate each degree to one boundary circle. Now, one mayask whether they may also be related to the centers of the boundary circlesinstead of their radii, or to both centers and radii. This question is thestarting point of this chapter: How can we find suitable conditions on theboundary circles that single out a (locally) unique circle packing? In otherwords: What are well-posed boundary value problems for circle packings?

We can hardly give any concrete answers to this question up to now.However, we will introduce the technique of linearization of the discreteproblems which may be an important tool to answer the question.

We will revisit Riemann-Hilbert problems in this chapter since theseproblems fit very nicely into our discrete setting. Recall that a Riemann-Hilbert problem consisted of imposing one real condition on a holomorphicfunction for each point on the continuous boundary T of D. Accordingly, thediscrete problems impose one real condition on every point of the discreteboundary of K. Consequently, we will introduce discrete Riemann-Hilbertproblems.

Having done this, we will use the discrete problems to finally define adiscrete Hilbert transform.


62 5 Discrete Hilbert Transform

5.1 Discrete boundary value problems

5.1.1 Definition and examples

To begin, let us recall two boundary value problems we have already en-countered in the previous chapter.

Example 5.1 (Fixed radii problem). Given ρ1, . . . , ρm, we ask for a packingP such that the ρj are just the radii of the boundary circles of P.

Theorem 4.13 tells us that this problem has a unique solution on DN .

Example 5.2 (Maximal packing problem). We ask for a maximal packingP on K.

Theorem 4.28 tells us that this problem has the unique solution PK onDN .

We will now formally define discrete boundary value problems in general.

Definition 5.3 (Discrete boundary value problem). Let β : R3n → Rm be afunction with

βj(r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) = βj(rj , xj , yj) ∀j ∈ {1, . . . ,m},

i.e., βj depends only on the variables rj, xj and yj. β defines a discreteboundary value problem. A circle packing P ∈ DN is a solution to theproblem if

β(P) = 0.

Equivalently, a vector P ∈ UN is the vector form of a solution to the problemiff ω(P )

N(P )β(P )

= 0.

It is not difficult to translate the two standard examples for discreteboundary value problems into the newly introduced notation.

• The function β : R3n → Rm defined by

βj(rj , xj , yj) := rj − ρj (FIXED-RAD)

defines the problem of fixed radii for given ρ1, . . . , ρm ∈ R+.

5.1 Discrete boundary value problems 63

• The function

βj(rj , xj , yj) :=1

2(x2j + y2

j − (1− rj)2) (MAX-PACK)

defines the maximal packing problem.

Let P be a circle packing. If P is a zero of (FIXED-RAD), it is clearthat P actually solves the fixed radii problem as stated in Example 5.1.This is not that evident for (MAX-PACK), though. We will thereforegive an elementary proof that every zero of (ω,N, β) in UN with β from(MAX-PACK) corresponds to a maximal packing.

Proposition 5.4. Let β be the function from (MAX-PACK). Then everyzero P ∈ UN of (ω,N, β) is the vector form of an N -normalized maximalpacking.

Proof. Since P ∈ UN and ω(P ) = 0, there is a circle packing P with vectorform P . Since N(P ) = 0, it is N -normalized and again since P ∈ UN , it isalso locally univalent. Denote the circles of P by C1, . . . , Cn

Write P = (r, x, y) and zj = xj + iyj for j ∈ {1, . . . ,m}. We claim thatrj < 1 holds for all j ∈ {1, . . . ,m}.

Assume this was wrong. Fix j such that rj ≥ 1 and let k ∈ {1, . . . ,m}such that vj and vk are adjacent. Then β(P ) = 0 and ω(P ) = 0 imply thatthe following equations hold:

|zj + zk|2 = (rj + rk)2

|zj |2 = (1− rj)2

|zk|2 = (1− rk)2

Since P ∈ UN , rj + rk > 0 holds. By assumption, rj − 1 ≥ 0. Taking squareroots in the first two equations gives

|zj + zk| = rj + rk

|zj | = rj − 1

which impliesrk + 1 = |zj + zk| − |zj | ≤ |zk|

by the triangle inequality. Inserting this into the last equation we obtain

(rk + 1)2 ≤ (rk − 1)2.


This can only be true if rk ≤ 0 which is a contradiction to P ∈ UN . Sorj < 1 for all j ∈ {1, . . . ,m}. Taking square roots in the equation βj(P ) = 0we obtain

|zj |+ rj = 1 (5.1)

for all j ∈ {1, . . . ,m}, implying that all boundary circles are contained in Dand tangent to T.

It remains to show that all interior circles of P are contained in D. Letj ∈ {m+1, . . . , n}. Cj is contained in the closure of the disc D := B|zj |+rj (0)and tangent to ∂D. Since Cj is surrounded by a closed chain of neighbors,there must be vk adjacent to vj such that Ck intersects C \D. In particular,this implies that |zj |+ rj < max{|z| |z ∈ Ck} = |zk|+ rk.

Consequently, the function f : V → R, f(vj) = |zj | + rj can not attainits maximum at an interior vertex. But since f(vj) = 1 ∀j ∈ {1, . . . ,m},this implies that f is bounded by 1, yielding that all circles are contained inD.

Remark 5.5. The last argument is strongly reminiscent of discrete har-monic functions: If f : V 7→ R were discrete harmonic, we could use themaximum principle and get done immediately. Unfortunately, Dubejko’stheorem does not apply to the f appearing in this proof. Nevertheless, thisproof encourages to keep discrete harmonic functions in mind when onewants to prove geometric and algebraic properties of solutions to discreteboundary value problems.

We shall now state three examples for boundary value problems whichare not well-posed. One of them has no solution on DN , another one has nosolution at all, while the solutions of the third one are not locally unique.

Example 5.6 (No solution on DN ). Define

βj(rj , xj , yj) := (xj + 2)2 + y2j − (1− rj)2.

This problem is similar to the maximal packing problem, but in this case, theβj model internal tangencies of the boundary circles with T− 2 instead of T.It is easy to imagine that this does not fit with the requirement that the n-thcircle is centered at 0. In fact, such a packing can’t exist:

Assume P ∈ DN was a solution to the above problem and let zn be thecenter of its n-th circle, i.e. zn = 0. Let P ′ be the packing obtained from Pby adding 2 to the real part of each center. Then P ′ solves (MAX-PACK).By the proof of Proposition 5.4, the absolute value of its circle centers isbounded by one. Hence zn + 2 ≤ 1, which is a contradiction.


Example 5.7 (No solution). Constructing a problem which has no solutionat all is similarly easy. Intuitively speaking, we can just take two adjacentboundary vertices and force their circles to lie in disjoint closed domains. Inthis case, they can not be tangent and satisfy the boundary value problem atthe same time.

So let β define a boundary value problem and assume that

β1(r1, x1, y1) = x1 − r1 − 1

β2(r2, x2, y2) = x2 + r2 + 1

Assume P = {C1, . . . , Cn} is a solution to this problem. Then β(P) = 0implies that C1 ⊂ {z ∈ C|Re(z) ≥ 1} and C2 ⊂ {z ∈ C|Re(z) ≤ −1}. So C1

and C2 can not be tangent, contradicting the assumption that P is a circlepacking.

Example 5.8 (No locally unique solution). Let P be an arbitrary circlepacking and for j ∈ {1, . . . ,m} let ξj + iνj be the center of its j-th boundarycircle. Define

βj(rj , xj , yj) := νjxj − ξjyj.Direct computation gives that if P is a solution to the corresponding problem,then λP is a solution for every λ > 0, too. In particular λP is a solutionfor every λ > 0. Thus, this problem has solutions, but none of them can belocally unique.

Let us now discuss how to discretize a Riemann-Hilbert problem withcircle packings. Let Mt, t ∈ T be the target curves of a Riemann-Hilbertproblem. For j ∈ {1, . . . ,m}, let tj be the contact point of the j-th circleof PK and T. Now, the idea is to ask for a circle packing P = (C1, . . . , Cn)where for each j ∈ {1, . . . ,m}, Cj is tangent to Mtj . This serves the intuitiveidea of the discretization: The circle tangent to T at tj is mapped to a circletangent to Mtj . Analogously, a continuous solution of the original problemwould map the point tj onto the curve Mtj .

A function that fits this idea would be β : R3n → Rm defined by

βj(rj , xj , yj) := min{|xj + iyj − z|2 |z ∈Mtj} − r2j

This is a very general way to express the boundary condition. Depending onthe structure of the Mtj , the functions βj may also lack differentiability. Wewould only use this if we have no additional information about the curvesMtj . In the following we will confine ourselves to defining discrete circularRiemann-Hilbert problems. These are discrete analogues of the problemsdiscussed in Subsection 3.3.2.


Definition 5.9. For each j ∈ {1, . . . , n}, let Mj be a circle with center cjand radius ρj . The corresponding discrete circular Riemann-Hilbert problemis the problem defined by the functions

βj(rj , xj , yj) :=1

2

(|xj + iyj − cj |2 − (ρj − rj)2

).

The circle Mj is called the j-th target circle of the problem.

The equation βj = 0 models an internal tangency between Mj and thej-th circle of the solution. Figure 5.1 shows an example. The maximalpacking is shown in figure 5.1a with colored boundary circles. The solutionis shown in figure 5.1a. For each boundary circle, the corresponding targetcircle is drawn transparently with a dashed line of the same color. Near thetangency, an arc is drawn clearly to emphasize the tangency.

(a) Maximal packing on K[9]1

(b) Solution of the problem

Figure 5.1: A discrete circular Riemann-Hilbert problem

Note that the maximal packing problem is a special case of this problem.Since (HRHP) was a circular Riemann-Hilbert problem, we will also use aproblem of this kind to discretize (HRHP) and define the discrete Hilberttransform.

There are some first investigations on solvability properties of this kind ofproblems which can be found in a paper by Wegert and Bauer [33]. Theypresent the set of solutions for the case where K is a flower.


5.1.2 Linearization of boundary value problems

In general, proving results on existence and uniqueness of solutions to discreteproblems is very difficult. Nevertheless, we can gain some insight into thesequestions via linearization. If a packing P is a solution of a given problem,we can investigate the structure of the matrixDω

DNDβ

∣∣∣∣∣∣P

.

Essentially, this means that we consider the linearized version of the boundaryvalue problem as a problem on the tangent space TPDN . If the above matrixis regular, the Implicit Function Theorem gives us that the solution is locallyunique.

The boundary value problems on the tangent space of the circle packingmanifold were called incremental boundary value problems by Wegert. Wewill define them now.

Definition 5.10 (Incremental boundary value problem). Let P ∈ DN be acircle packing. Let c ∈ Rm and M = (Mjk)jk ∈ Rm×3n, such that all entriesof M are zero except Mjk for j ∈ {1, . . . ,m} and k ∈ {j, n+ j, 2n+ j}. Mand c define an incremental boundary value problem at P. A tangent vectorp ∈ TPDN is a solution to this problem if M · p = c.

The problem is called incremental Riemann-Hilbert problem if in addition,we have (Mj,j)

2 = (Mj,j+n)2 + (Mj,j+2n)2 for all j ∈ {1, . . . ,m}.As indicated before, Proposition 4.20 allows us to rewrite an incremental

boundary value problem defined by M ∈ Rm×3n and c ∈ Rm asDωDNM

∣∣∣∣∣∣P

· p =

00c

.

Thus, an incremental problem is well-posed iff the matrix on the left-handside is regular.

We are mainly interested in incremental problems that arise as linearizationof boundary value problems for circle packing. The linearization of a problemdefined by a differentiable function β at a circle packing P is given asDω

DNDβ

∣∣∣∣∣∣P

· p =

00

−β(P)

. (5.2)


If P is a solution to the problem, the regularity of the matrix gives usinformation about the solution’s local uniqueness. It is straightforward tocheck that the linearization of a discrete circular Riemann-Hilbert problemyields an incremental Riemann-Hilbert problem, which motivates the name.

Let us now have a look at the linearization of the two standard problems.

Proposition 5.11. The linearization matrix of the fixed radii problem isregular for every P.

Proof. By Proposition 4.20, the last 3n−m columns of(DωDN

)have full rank. The first m columns of Dβ equal the m×m unit matrix and

all other entries are equal to zero. Thus,

DωDNDβ

has full rank.

So for the fixed radii problem, the regularity of the matrix is an immediateconsequence of Lemma 4.7.

The linearization of the maximal packing problem is regular, too. This hasbeen conjectured by Wegert and could be proven by the author in 2017. Theproof requires some preparations and is presented in the following section.

5.2 Proof of the maximal packing conjecture

The core idea of the proof is to think of a maximal packing as an extendedcircle packing. After all, the unit circle T is a circle itself and all boundarycircles of PK are internally tangent to it.

Recall that the boundary conditions of (MAX-PACK) are given by

βj(rj , xj , yj) = (xj)2 + (yj)

2 − (1− rj)2.

Note that the structure of these equations has a strong similarity to thecontact equations

(xj − xk)2 + (yj − yk)2 − (rk + rj)2 = 0.

Indeed, if we formally define the (n + 1)-th vertex vn+1 of K and setxn+1 = yn+1 = 0 and rn+1 = −1, then βj(rj , xj , yj) = 0 becomes preciselythe contact equation for the vertices vn+1 and vj .

5.2 Proof of the maximal packing conjecture 69

So we should think of this as a circle packing on a “greater” complex K,which is obtained from K by adding the vertex vn+1. We will postpone the

precise definition of K for a moment.Let PK = (r, x, y) and let ω be the contact function of K. Consider the

extended Packing P = (r1, . . . , rn,−1, x1, . . . , xn, 0, y1, . . . , yn, 0) on K. The

Jacobian of ω at P is a matrix in R(3n−3)×(3n+3). We will henceforth denotethis matrix by J . If the rank of this matrix is maximal, the desired resultcan be deduced with little extra work.

Unfortunately the extended packing P contains a circle whose radius isformally negative. This fact prohibits to copy the proof of Lemma 4.7 andrequires a second idea.

This idea is to make use of a Mobius transformation. Fix a point z∞ ∈ Dlying in the interstice of two adjacent boundary circles of PK and T as shownin figure 5.2a. If we apply the Mobius transformation T (z) = 1

z−z∞ , theimage of PK under T will be a circle packing again.

z∞=T−1(∞)

(a) Maximal packing

0=T (∞)

(b) Transformed extended packing P ′

Figure 5.2: The extended packing

Now what about T? T maps T to a circle, too. But since z∞ ∈ D,the interior of T is mapped to the exterior of T (T). This means that theimages of the boundary circles of PK under T are now externally tangentto T (T). In short, the image of P under T to be seen in figure 5.2b looks

like a completely normal circle packing! In fact we will now define K in an


appropriate way and show that the image of P under T is a circle packingindeed.

5.2.1 The transformed packing

We define V := V ∪ {vn+1}, such that there is a new vertex in the extendedgraph that T will correspond to. This vertex should be adjacent to allboundary vertices of K, therefore E = E ∪ {vn+1, v1} ∪ . . . ∪ {vn+1, vm}.

We will place z∞ in the interstice of T and the 1st and m-th circleof PK . Thus, vn+1, v1 and vm should be the boundary vertices of the

extended complex, as figure 5.2 suggests. We therefore define F = F ∪{〈vn+1, v2, v1〉, 〈vn+1, v3, v2〉, . . . , 〈vn+1, vm, vm−1〉}. These faces make the

edges {v1, v2}, . . . , {vm−1, vm} interior edges of K. Consequently, the verticesv2, . . . , vm−1 become interior vertices.

Lemma 5.12. Let C1, . . . , Cn be the circles of PK . Chose z∞ ∈ D asdescribed above and define T (z) = 1

z−z∞ . Furthermore, write Cn+1 := T.

Then P ′ = {T (C1), . . . , T (Cn+1)} is a circle packing on K.

Proof. We check whether P ′ satisfies the definition of circle packings.(C1) Since z∞ /∈ Cj ∀j ∈ {1, . . . , n + 1}, T (Cj) is a circle for each

j ∈ {1, . . . , n+ 1}.(C2) Let {vj , vk} ∈ E. W.l.o.g. k 6= n+ 1. Since T is continuous, T (Cj)

and T (Ck) are tangent.If j 6= n+ 1, then z∞ does not lie in the interior of either circle. Thus, the

exteriors of Cj resp. Ck are mapped to the exteriors of T (Cj) resp. T (Ck),so they are actually externally tangent.

If j = n+ 1, z∞ lies in intCn+1 but not in intCk. Thus the internaltangency of Cn+1 and Cj results in an external tangency of T (Cn+1) andT (Ck).

In both cases, T (Cj) and T (Ck) are externally tangent.(C3) Let f = 〈va, vb, vc〉 ∈ F . Consider figure 5.3.If vn+1 is not incident to f , f is also a face of K. Hence, the contact

points of the circles Ca, Cb and Cc lie in counter-clockwise order on the facecircle Cf (colored red in figure 5.3). Now z∞ does not lie in intCf ., thus, thecontact points of T (Ca), T (Cb) and T (Cc) lie on T (Cf ) in the same order.

Consequently, (C3) is satisfied for the face f in the packing P ′.Otherwise let f = 〈vn+1, vj+1, vj〉. Let z′ be the contact point of Cj and

Cj+1 and let tj resp. tj+1 be the contact points of Cj resp. Cj+1 with T.From figure 5.3 we can see that tj , tj+1, and z′ lie in counter-clockwise order


on the circle passing through them (colored blue in figure 5.3). We canalso see that z∞ lies outside this circle. Thus, the same argument as beforeapplies, yielding that (C3) is again satisfied for the face f in P ′.

Figure 5.3: Visualization of the argument for (C3) in the proof of Lemma 5.12

Now, for the packing P ′, Lemma 4.7 applies. We conclude that ω(P ′) = 0and Dω|P′ has maximal rank. We will use this to show that Dω|P has fullrank, too. The strategy is as follows: By construction, T−1(z) = 1

z + z∞

maps P ′ to P . We will first introduce the function Υ such that Υ(P ′) is the

image of P ′ under 1z . We then show

Lemma 4.7⇒ Regularity of Dω|P′Theorem 5.15

=========⇒ Regularity of Dω|Υ(P′)Corollary 5.16

=========⇒ Regularity of Dω|P

5.2.2 Differential of ω at the transformed packing

Throughout this subsection, P ′ = (r, x, y), i.e. r, x and y will denote the

components of P ′. To shorten notation, we write n := n+ 1.The first thing to do is examining how a Mobius transformation affects

the radii and centers of P ′. It is clear how a linear function affects a circle’s


center and radius, but this is not trivial for the inversion z 7→ 1z . We can

still deduce a formula for this.Let C be a circle with center z∗ and radius r such that 0 does not lie on

C. Then expanding the equation |z − z∗|2 = r2 yields

(z − z∗)(z − z∗) = r2 ⇔ |z|2 − z∗z − zz∗ + |z∗|2 − r2 = 0.

Since 0 /∈ C, we have z 6= 0 and |z∗|2 − r2 6= 0. Hence, dividing by

(|z∗|2 − r2) |z|2 yields

1

|z∗|2 − r2− 1

z

z∗

|z∗|2 − r2− 1

z

z∗

|z∗|2 − r2+

1

|z|2= 0.

Expanding the first summand as

1

|z∗|2 − r2=|z∗|2 − r2

(|z∗|2 − r2)2=

|z∗|2

(|z∗|2 − r2)2− r2

(|z∗|2 − r2)2

allows us to factorize again to obtain(1

z− z∗

|z∗|2 − r2

)(1

z− z∗

|z∗|2 − r2

)=

(r

|z∗|2 − r2

)2

which is equivalent to∣∣∣∣∣1z − z∗

|z∗|2 − r2

∣∣∣∣∣ =r∣∣∣|z∗|2 − r2

∣∣∣ .Hence, the image of C under the inversion has radius r

||z∗|2−r2| and center

z∗

|z∗|2−r2 .

If now 0 ∈ extC, then |z∗|2 − r2 > 0, so the modulus in the radius can bedropped. If 0 ∈ intC, however, dropping the modulus changes the sign ofthe radius. Note that this fits beautifully with our situation: z∞ lies inside T.We have given T a formally negative radius, but T (T) should have a positive

radius, since P ′ is a “usual” circle packing with only exterior tangencies.This observation motivates to simply ignore the modulus that appears inthe last equation.

We arrive at the following definition.


Definition 5.13. Let UΥ := {(r, x, y) ∈ R3|x2 + y2 − r2 6= 0}. We definethe function Υ: UΥ → R3 by

Υ(r, x, y) =1

x2 + y2 − r2(r, x,−y) (5.3)

To compute the radii and centers of the image of a circle packing underthe map z 7→ 1

z , we extend Υ to R3n in a natural way: For r, x, y ∈ Rn with(rj , xj , yj) ∈ UΥ ∀j ∈ {1, . . . , n} we write

Υ(r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) = (ρ1, . . . , ρn, ξ1, . . . , ξn, ν1, . . . , νn)

where (ρj , ξj , νj) = Υ(rj , xj , yj).In the following lemma we compute the derivatives of Υ.

Lemma 5.14. The differential of Υ at a point (r, x, y) ∈ UΥ is given by theregular matrix

DΥ =1

(x2 + y2 − r2)2

x2 + y2 + r2 −2rx −2ry2rx −r2 − x2 + y2 −2xy−2ry 2xy r2 − x2 + y2

(5.4)

Proof. The formula is obtained by computation. For the regularity, notethat by construction, Υ ◦ Υ is the identity on the open set UΥ. ThusDΥ|Υ(r,x,y) ·DΥ|(r,x,y) = I3, which implies that DΥ|(r,x,y) is regular.

Of course, DΥ is still regular if Υ denotes the extension to R3n.Now, for el = {vj , vk} ∈ E, the derivatives of ωl at P ′ are given by

drj ωl(r, x, y) = −(rj + rk), drk ωl(r, x, y) = −(rk + rj), (5.5)

dxj ωl(r, x, y) = (xj − xk), dxk ωl(r, x, y) = (xk − xj), (5.6)

dyj ωl(r, x, y) = (yj − yk), dyk ωl(r, x, y) = (yk − yj). (5.7)

All other derivatives are equal to 0. Thus, the l-th line of Dω|P′ has thefollowing nonzero entries:

−rj − rk,−rk − rj , xj − xk, xk − xj , yj − yk, yk − yj

located in the columns j, k, j + n, k + n, j + 2n and k + 2n, respectively.


Theorem 5.15. For j = {1, . . . , n}, write aj := x2j + y2

j − r2j . Note that

aj 6= 0 by construction of P ′. Let A ∈ R3n−3×3n−3 be the diagonal matrixwhose l-th entry is given by ajak, where el = {vj , vk}. Then

A ·Dω|Υ(P′) ·DΥ|P′ = Dω|P′ .

In particular, Dω|Υ(P′) and Dω|P′ have the same rank.

Proof. Fix l ∈ {1, . . . , 3n − 3} and let el = {vj , vk}. The l-th line of thematrix on the left-hand side is given by

ajakDωl|Υ(P′) ·DΥ|P′

The non-zero entries of Dωl|Υ(P′) are given by

−(rjaj

+rkak

),−(rjaj

+rkak

),xjaj− xkak,xkak− xjaj,ykak− yjaj

andyjaj− ykak

at positions j, k, n+ j, n+ k, 2n+ j and 2n+ k, respectively.By definition of the extension of Υ, the non-zero entries of the j-th, (n+j)-

th and (2n+ j)-th row of DΥ are at positions j, n+ j, and 2n+ j and givenby formula (5.4). Consequently, the nonzero entries of ajakDωl|Υ(P′) ·DΥ|P′are at positions j, k, n + j, n + k, 2n + j and 2n + k, too. We can nowcompute the matrix product. Since the factor ak

ajappears in all entries, we

will leave this factor out to make the formulas shorter.The entries of ajakDωl|Υ(P′) ·DΥ|P′ divided by ak

ajat positions j, n+ j,

and 2n+ j are given by

−(rjaj

+rkak

)(r2j + x2

j + y2j ) +

(xjaj− xkak

)(2rjxj)−

(yjaj− ykak

)(−2rjyj)

(5.8)(rjaj

+rkak

)(2rjxj) +

(xjaj− xkak

)(−r2

j − x2j + y2

j )−(yjaj− ykak

)(2xjyj)

(5.9)(rjaj

+rkak

)(2rjxj)−

(xjaj− xkak

)(2xjyj) +

(yjaj− ykak

)(r2j − x2

j + y2j ).

(5.10)

The entries at positions k, n + k, and 2n + k are obtained by swapping jand k.


It is now just a matter of rearranging to show that these terms coincidewith the corresponding entries of Dωl|P′ . We will work this out in thefollowing.

The crucial step in all rearrangements is the identity

aj + ak2

= rjrk + xjxk + yjyk, (5.11)

which is a direct consequence of ωl(P ′) = 0. The term (5.8) is equal to

1

a2j

(−(rjak + rkaj)(2x

2j + 2y2

j − aj) + (xjak − xkaj)(2rjxj)

+(yjak − ykaj)(2rjyj))

=1

a2j

(rjajak + rka

2j − 2aj( rkx

2j + rky

2j︸︷︷︸

=rk(aj+r2j )

+rjxjxk + rjyjyk))

=1

aj(rjak + rkaj − 2(rkaj+ rkr

2j + rjxjxk + rjyjyk︸︷︷︸=rj

aj+ak2 by (5.11)

) )

=1

aj(rjak + rkaj − 2rkaj − rjaj − rjak) = −(rj + rk),

which is the corresponding entry of Dωl|P′ as claimed.The term (5.9) is equal to

1

a2j

(−(rjak + rkaj)(2rjxj) + (xjak − xkaj)(−aj − 2r2

j + 2y2j )

−(yjak − ykaj)(2xjyj))

=1

a2j

(−xjajak + xka

2j + 2aj( xkr

2j − xky2

j︸︷︷︸=xk(x2

j−aj)

+xjrjrk + xjyjyk))

=1

aj(−xjak + xkaj + 2(−xkaj+ xjrjrk + x2

jxk + xjyjyk︸︷︷︸=xj

aj+ak2 by (5.11)

) )

=1

aj(−xjak + xkaj − 2xkaj + xjaj + xjak) = xj − xk,

which is again the corresponding entry of Dωl|P′ . For the third term, itsuffices to note that this term is equal to the second term when we switchxj and yj as well as xk and yk.


Corollary 5.16. J has maximal rank 3n− 3.

Proof. By Lemma 4.7 and Lemma 5.12, Dω|P′ has maximal rank. By

Theorem 5.15, Dω|Υ(P′) has full rank, too. By construction, P is obtained

from Υ(P ′) by adding z∞ to all circle centers. From the equations (5.5)-(5.7),

we see that this does not change Dω. Thus DωΥ(P′) = DωP = J , which

proves the corollary.

5.2.3 Basis for the kernel of J

In this subsection, r, x and y denote the components of P , i.e. P = (r, x, y).

The kernel of J has a geometric interpretation. Let us rethink the geometricsituation: In a neighborhood of P, every zero of ω is a configuration ofcircles where the (n+ 1)-th circle is internally tangent to the circles with

indices 1, . . . ,m. Thus, we ask ourselves how to locally transform P into adifferent configuration of this kind.

Obviously, any maximal packing can be extended to a packing on K thesame way PK could. So what we can do is to perturb PK slightly in sucha way that the perturbed packing is still a maximal packing. This gives 3local degrees of freedom which arise from the conformal automorphisms ofD. Recall that conformal automorphisms of D are constituted by rotationsz 7→ λz for λ ∈ T and Mobius transformations of the form

z 7→ z − z0

1− zz0(5.12)

for some arbitrary z0 ∈ D.Now, in the extended setting, the (n+ 1)-th circle is not fixed to be the

unit circle anymore and we are allowed to transform it together with the restof the packing. This can be done in 3 obvious ways: We can scale the entirepacking by some constant factor and we can move it along the x- or y-axis.

These six local freedoms of the equation ω(P) = 0 are visualized infigure 5.4. The black arrows show the direction in which the center ofthe respective circle is being moved. The greyish packing shows a slightlyperturbed extended maximal packing corresponding to the freedom.

Let us translate these freedoms into tangent vectors of the manifold definedby ω(P) = 0 locally at P. From geometric intuition, it is pretty clear thatthe vectors

p1 = (r, x, y), p2 = (0, y,−x),

p3 = (0, 1, 0), p4 = (0, 0, 1)


(a) p1 - Scaling (b) p2 - Rotation

(c) p3 - Movement along x-Axis (d) p4 - Movement along y-Axis

(e) p5 - Automorphism of D (f) p6 - Automorphism of D

Figure 5.4: Visualization of the freedoms for maximal packings


correspond to scaling, rotating, and euclidean motions along the axes, re-spectively.

Determining the vectors corresponding to automorphisms of the form (5.12)requires some work, but they can still be put in closed form. For arbitraryc ∈ D, let Pc be the the packing obtained by applying the transformationT (z) = z−c

1−zc to the circles of PK . Denote by Pc the corresponding extendedpacking.

The map D→ R3n+3, c 7→ Pc can be constructed explicitly with help ofthe function Υ and is differentiable. Since ω(Pc) = 0 for every c ∈ D and

P0 = P, the derivatives dda Pa|0 and d

da Pia|0 are vectors in the kernel of J .They can be computed1 to be given by

p5 = (2rx, r2 + x2 − y2 − 1, 2xy), p6 = (2ry, 2xy, r2 − x2 + y2 − 1)

where multiplication is to be understood elementwise.2

To check the linear independence of p1, . . . , p6 we look at their entries atpositions n+ 1, 2n+ 1, 2n+ 2, 3n+ 1, 3n+ 2, and 3n+ 3. These entriescorrespond to the derivatives with respect to rn+1, xn, xn+1, yn−1, yn, andyn+1. Table 5.1 lists these entries. Recall that xn = yn = 0 and xn−1 6= 0since PK ∈ DN . Moreover, rn < 1 by Proposition 5.4 and rn+1 = −1. Aswe can see, the (3n+ 3)× 6 matrix whose columns are equal to the vectorsp1, . . . , p6 has a regular 6× 6 submatrix.

Table 5.1: Some entries of the vectors p1, . . . , p6 showing their linear independ-ence

rn+1 xn+1 yn+1 xn yn yn−1

p1 −1 ∗ ∗ ∗ ∗ ∗p3 0 1 ∗ ∗ ∗ ∗p4 0 0 1 ∗ ∗ ∗p5 0 0 0 r2

n − 1 ∗ ∗p6 0 0 0 0 r2

n − 1 ∗p2 0 0 0 0 0 −xn−1

Consequently, the vectors p1, . . . , p6 are all linearly independent, so byCorollary 5.16 they must form a basis of the kernel J . With this basis, wecan easily complete the proof.

1This was done using Wolfram Mathematica [35].2For example, rx denotes the vector (r1x1, . . . , rn+1xn+1).

5.3 Discrete Hilbert transform 79

Theorem 5.17. Let β be given as in (MAX-PACK). Then, the matrix

M :=

DωDNDβ

∣∣∣∣∣∣PK

is regular.

Proof. Let p = (pr, px, py) ∈ kerM . Then p ∈ ker DN in particular. Fromthe definition of N we conclude that the entries of p at positions 2n, 3n− 1,and 3n are zero.

Define p := (pr, 0, px, 0, py, 0) ∈ R3n+3. By construction,

(DωDβ

)is the

submatrix of J which is obtained by removing its (n + 1)-th, (2n + 2)-th,

and (3n+ 3)-th columns. Since p ∈ ker

(DωDβ

), this implies that p ∈ ker J .

Hence, p is a linear combination of the vectors p1, . . . , p6 constructed before.Now p has zeros at positions n+ 1, 2n+ 1, 2n+ 2, 3n+ 1, 3n+ 2, and

3n+ 3. Looking at Table 5.1, we see that this is only possible if p = 0. Thusp = 0 too, which proves the claim.

We will retain the notation M for this matrix throughout the rest of thisbook.

5.3 Discrete Hilbert transform

We have now gathered all the necessary prerequisites to introduce the discreteHilbert transform with circle packings.

In the context of discrete boundary value problems, the discrete Hilberttransform may be thought of as a very illustrative case study and also anapplication of Theorem 5.17. Moreover, the need of linearizing a boundaryvalue problem occurs naturally for the Hilbert transform, thus providinganother reason for studying incremental problems beside the investigationof solvability properties.

Since we have already seen how to translate boundary value problems intothe discrete setting, we will make use of this to define the discrete operator.The strategy has been outlined before: Given a vector u ∈ Rm, which isto be understood as a discrete function on T from now on, we will definea suitable discrete boundary value problem and solve it. Motivated by thecontinuous counterpart, we will extract a discrete Hilbert transform of ufrom the solution.


It is remarkable that the Hilbert operator is a purely real object at firstglance. By definition, it is merely a singular integral operator. Nevertheless,its connection to complex analysis allows us to discretize the operator withcircle packings.

5.3.1 Difficulties of the Schwarz problem

Recall that in Subsection 3.4.1 we shortly suggested to use the Schwarzproblem as the foundation for discretization. We mentioned that this problemis not well suited for discretization with circle packings. We are now in aposition to explain the problems that occur in detail.

Critical points and zeros of the solution. Let u ∈ Lp be given for somep > 1. By Theorem 3.11 and Remark 3.13, the solutions of the Schwarzproblem are given by z 7→ Su(z) + ic for arbitrary c ∈ R. The derivative ofany of these solutions is z 7→ S′u(z). As long as we merely know u, we haveno information about the zeros of this function.

Assume now that we have formulated a discrete Schwarz problem. If wehave done this properly, we expect the discrete problem to have similarsolvability properties.

Hence, the discrete problem has one solution P, which is unique up toimaginary constants on the entire manifold Db. And we have no clue onwhich of the connected components it can be found! In other words, weexpect that there is a unique branch structure b such that all solutions ofthe problem lie on Db, but we do not know b.

However, we have stated in Remark 4.15 that knowing b is essentialto be able to compute the packing, at least in the current state of theart. There have been some ideas on writing adaptive algorithms for thecomputation of circle packings that are able to “jump” between variousconnected components [34], but none of these ideas have been pursuedconsequently up to now.

Even if we had a possibility to predict at least the number of critical pointsof Su for a given u, another problem still remains: Assume we knew thatthe solution to the discrete problem has e.g. exactly three branch points.

Consequently, the solution can be found on Db for some b withn−m∑j=1

bj = 3.

In general there are still plenty of different b which satisfy this. So, we wouldalso have to predict the position of the critical points to find the correctbranch structure.


Trivial case. Another problem appears when we consider the trivial caseu = 0.

In this case, the solutions to the Schwarz problem are then just imaginaryconstants. Constant functions do not have reasonable counterparts in circlepacking. Since constant functions map the unit circle to a point in C, thecounterpart in circle packing would be a discrete analytic function where theimage packing degenerates to a point, i.e. where all circles have radius zero.We have forbidden this in the definition of circle packings since statementslike Lemma 4.7 do not hold at degenerate packings.

Though this problem seems to be of minor relevance at the first glance, ityields a major problem. As we have stated in Section 5.1, there are hardlyany results on discrete boundary value problems up to now. In case of thediscrete Hilbert transform, this can be countervailed with a simple idea:We prove the existence of solutions for u in a small neighborhood of 0 bylinearizing the problem at u = 0 and applying the Implicit Function Theorem,as it will be done in Theorem 5.20. To compute the Hilbert transform for uoutside this neighborhood, we can employ the linearity of the transform.

Linearization at a packing that degenerates to a point is not helpful,though, since the derivative of ω is the null matrix at such a packing. Ergo,the linearization matrix (5.2) is not regular. Linearization at a differentpoint is problematic, too: If we can compute Hu in an open set that doesnot contain 0, then we can not use the linearity to compute Hu for any givenu. So, the linearization tactic is very problematic for the Schwarz problem.

Consistent discretization of boundary conditions As an example, considerthe case where u(eiτ ) = cos(τ). In this case, the solutions of the Schwarzproblem are given by z 7→ z + ic. Thus, we would expect that a solution ofthe discrete problem is the discrete identity, i.e. the maximal packing PKitself should solve the problem.

The following figure 5.5 shows an example for a maximal circle packing ona given graph. The boundary circles are shown in different colors. For eachcolor, the figure also shows the corresponding target curve of the Schwarzproblem. I.e., the circle packing shown here should be a solution to thediscrete problem coming from these target curves.

Evidently, it is not clear at all how the boundary condition has to beinterpreted. The red and the cyan circle are tangent to their target curves,while the light green and the indigo target lines pass through the centers ofthe respective circles. Neither of these holds for the remaining four circles.

In more involved examples, we may be facing unsolvable difficulties infinding a consistent way to interpret the boundary condition. If we do


Figure 5.5: A maximal packing and target lines for the Problem Re eiτ =cos(τ)

not care about this, the discrete Hilbert transform will approximate itscontinuous archetype less accurately.

5.3.2 Discretization of the nonlinear problem

We have already discussed the problem we will use instead in the classicalsetting. For convenience, we state it here again.

|w(t)| = expu(t) (HRHP)

Recall that the target lines of this problem are circles centered at 0 withradius expu(t). We should briefly state why (HRHP) does in fact overcomethe difficulties discussed before.

• We have proven that (HRHP) has a unique standard normalized solutionwith no critical points for u ∈ C1+α in a C1+α-neighborhood of 0. Thus,we may expect that the discrete counterpart has a unique solution in DNunder similar circumstances.

• For u = 0, the problem’s standard normalized solution is given by z 7→ z.This function is not only accessible in the discrete setting, but its discretecounterpart is in fact the easiest discrete holomorphic function one may


think of. Thus, the trivial input u = 0 corresponds to the trivial discreteanalytic function PK → PK .

• (HRHP) can be thought of as a generalization of the problem

|w(t)| = 1.

We can therefore discretize it as a generalization of the maximal packingproblem by requiring that the boundary circles of the solution are internallytangent to respective target circles. Thus, we have found a consistent wayto discretize the boundary condition for all u.

The latter of these observations exhibits the discrete version of (HRHP)as a discrete circular Riemann-Hilbert problem. We will now formulate it.Given a vector u ∈ Rm, we consider the discrete problem defined by thefunction βu, which is given by

βuj (r, x, y) =1

2(x2j + y2

j + (expuj − rj)2). (DHRHP)

The entries of u should be understood as values of a function on T evaluatedat the points tj , where tj is the contact point of the j-th circle of PK and T.

Thus, if we have found a standard normalized Packing P that solves(DHRHP), we may understand PK → P as a discrete z expSu(z). Figure5.6 shows an example. The objects are colored and shaded in a similar wayas in Figure 5.1. However, the dashed target circles are not shown in thisfigure, such that we can only see arcs of the target circles.

Let us derive a discrete Hilbert transform from the solution P = (r, x, y).

According to Proposition 3.26, we have Hu(t) = Arg w(t)t in the continuous

case if ‖u‖Cα is small enough. Here, w denotes the standard normalizedsolution.

We need a discrete interpretation for w(tj). Since tj is the contact point ofthe j-th boundary circle of PK with T, it is natural to interpret the contactpoint of the j-th boundary circle of P with its target circle as w(tj). Ifxj + iyj 6= 0, this point is given by

xj + iyj

(1 +

rj|xj + iyj |

). (5.13)

This is easy to see geometrically and straightforward to check algebraically.So, assume that

xj + iyjtj

∈ C− ∀j ∈ {1, . . . ,m}. (5.14)


(a) Maximal packing on K[6]3

(b) Solution of (DHRHP) for uj =310

cos 3tj

Figure 5.6: A discrete circular Riemann-Hilbert problem

The discrete version of the formula from Proposition 3.26 is then given by

vj := Argxj + iyj

tj(5.15)

Finally, we want to eliminate the constant c appearing in Proposition 3.26. Todo so, we normalize the vector v resulting from formula (5.15) by subtractingits mean from all entries. I.e., we apply the linear function

a(v) = v − 1

m

m∑j=1

vj . (5.16)

On a conceptual level, we are now done: We have discretized the procedure tocompute the Hilbert transform we have found in Section 3.4. Reconsideringthe strategy, though, there are several problems that quickly come in sight.

• However faithful discrete analytic functions may mimic their classicalcounterparts, we can not be sure whether the theorems proved in Section3.4 have counterparts in the discrete setting. The problem DHRHP mighthave no solutions at all, no solutions on DN or its solution may not beunique. In fact, it is easy to construct an example where DHRHP can nothave a solution on DN . We will construct this example below. We expectthat such things will not happen if ‖u‖ is small enough, but we have toprove this.


• Proposition 3.26 only tells us that there is a branch of the complexargument function to obtain the Hilbert transform from the solution ofthe boundary value problem. In general, this branch is unknown and wecan not simply use the principal branch Arg. Since formula (5.15) wasderived from Arg, it should not be applied in that case.

Overcoming these problems on the nonlinear level would probably requirean extensive solution theory on discrete circular problems. But there is stilla third problem:

• The resulting transform could be nonlinear.

The boundary value problem β does not depend linearly on u, nor doesv depend linearly on the solution. Most importantly, circle packings arenonlinear objects themselves, as D is clearly a nonlinear manifold.

We have thus discretized a linear operator with a discrete operator whichis likely to be nonlinear3. Evidently, this is not desirable.

For this reason, we will now linearize the discrete operator. Fortunately,this does, in fact, countervail all of our problems: We will prove that(DHRHP) does admit a locally unique solution at least for small ‖u‖. I.e.,there exists a well defined nonlinear discrete Hilbert operator Hnl on somesmall open set U ⊂ Rm with 0 ∈ U and we can define the linear discreteoperator as

H(u) = limλ→0

1

λHnl(λu).

Example 5.18 (No solution on DN ). Assume K = K[m]1 is a flower and

m > 5. The idea is to make one target circle very small, while all the othersare larger and of the same size. Intuitively, this forces the boundary circlesto become too large to wind only once around the interior vertex.

Define u ∈ Rm by u1 = ln(

1−√

22

)and uj = 0 for j > 1. Denote by φ

the function

φ(ra, rb, rc) = arccos

((ra + rb)

2 + (ra + rc)2 − (rb + rc)

2

2(ra + rb)(ra + rc)

)which appears in the definition of the radial angle sum function Φ.

3The transform could actually be linear as a composition of several nonlinear maps.After all, H is a linear operator that can be obtained via a nonlinear boundary valueproblem. However, we can not check this as long as we do not know a solution operatorof β. Indeed, numerical experiments strongly indicate that the procedure above doesin fact yield a nonlinear operator.


Assume P = {C1, . . . , Cm+1} is a locally univalent solution to (DHRHP)

for this u with label r1, . . . , rm+1. The first target circle has radius 1−√

22 .

Now Cm+1 is centered at 0 and externally tangent to C1, which is internally

tangent to the first target circle. Thus rm+1 < 1−√

22 . Since all other target

circles have radius 1, rj >√

22 for j = 2, . . . ,m.

The function φ is decreasing in ra and increasing in rb and rc ([33],Lemma 3.1). We have

φ

(1−√

2

2,

√2

2,

√2

2

)= arccos(0) =

π

2.

Consequently,

Φm+1(P) ≥m−1∑j=2

φ(rm+1, rj , rj+1) > 4φ

(1−√

2

2,

√2

2,

√2

2

)= 4

π

2= 2π.

Thus, the packing is branched at Cm+1, contradicting the assumption that Pis locally univalent.

5.3.3 Linearization of the discrete operator

We will show the existence of a local solution operator in a neighborhood ofu = 0 ∈ Rm. Though we are not able to state this operator explicitly, wecan exploit the fact that we are working in finitely many dimensions andmake use of the Implicit Function Theorem. Throughout the rest of thischapter, we write PK = (r, x, y).

Before starting, we fix the following notation.

Notation 5.19. We denote by R ∈ R3n×m the matrix whose last m linesform a diagonal matrix with entries 1 − r1, . . . , 1 − rm and whose otherentries are all equal to 0.

Theorem 5.20. There is an open connected set U ⊂ Rm with 0 ∈ U and acontinuously differentiable function W : U → R3n such that for each u ∈ U ,W (u) is a standard normalized circle packing which solves the correspondinginstance of (DHRHP) and which satisfies (5.14). The derivative of W at 0is given by

DW |0 = M−1R


Proof. Define the function β : R3n+m → R3n by

β(r, x, y, u) =

ω(r, x, y)N(r, x, y)βu(r, x, y)

.

Then the vector (PK , 0) ∈ R3n+m is a zero of this function. By direct

computation, the first 3n columns of the Jacobian of β at (PK , 0) are justthe matrix M defined in Theorem 5.17. By Theorem 5.17, M is regular. TheImplicit Function Theorem therefore implies the existence of an open setU ⊂ Rm with 0 ∈ U and a continuously differentiable function W : U → R3n

such that (W (u), u) is a zero of β for each u ∈ U .

Using the continuity of W , we may choose U ⊂ U with 0 ∈ U such thatW (U) ⊂ UN . Then by Proposition 4.20, W (u) is a standard normalized

circle packing for each u ∈ U . By definition of β, W (u) solves the boundaryvalue problem corresponding to βu. Again by continuity of W , we canassume that U was chosen so small that W (u) satisfies 5.14 if u ∈ U .

Finally, the Implicit Function Theorem states that the derivative of W at0 is given by

DW |0 = D(r,x,y)β|(PK ,0)

−1·D(u)β|(PK ,0).

Computing the derivatives of β, we see that indeed DW |0 = M−1R.

With the solution operator at hand, we can compute the discrete nonlineartransform from a solution in the manner described before. We thereforeintroduce the function h : R3n → Rm defined by

hj(r1, . . . , rn, x1, . . . , xn, y1, . . . , yn) = Argxj + iyj

tj∈ (−π, π). (5.17)

These preparations put us into a position to formally define the nonlineardiscrete Hilbert operator.

Definition 5.21. Let U ⊂ Rm be a set satisfying the conditions of Theorem5.20 and let W be the corresponding solution operator. Let h be the functiondefined in (5.17) and let a be the function defined in (5.16). The discretenonlinear Hilbert transform defined on K is the function Hnl : U → Rmgiven by Hnl(u) = a(h(W (u))).

Using Arg(z) = Im Log(z), the derivatives of h are

∂xjhj = − yjx2j + y2

j

, ∂yjhj =xj

x2j + y2

j

.


Consequently, the Jacobian of h at PK has the following form:(0n −Y 0n−m X 0n−m

)where X = diag( x1

x21+y21

, . . . , xnx2n+y2n

), Y = diag( y1x21+y21

, . . . , ynx2n+y2n

) and 0n

resp. 0n−m denote the null matrix in Rm×n resp. Rm×(n−m). To shortennotation, we will write L := Dh|PK in the following.

Finally, note that the function a can be written as a(z) = Az whereA ∈ Rm×m is the matrix where all entries are equal to − 1

m except for thediagonal entries, which are all equal to m−1

m . With these formulas, we willnow, finally, define the discrete linear Hilbert transform as linearization ofHnl.

Definition 5.22. The discrete Hilbert transform H defined on K is thelinear map defined by

H(u) = limλ→0

1

λHnl(λu).

Applying Theorem 5.20, we can derive the matrix representation of H:

H(u) = DHnl|0u = ADh|PKDW |0u = ALM−1Ru.

6 Numerical Results and FutureWork

With Theorem 5.17 proven, we now have a well-defined discrete Hilbertoperator for the maximal packing of any given complex. However, this isonly the beginning of most of the work related to the discrete operator. Itis desirable to examine various properties of the Hilbert operator, and, ofcourse, to examine its convergence behavior under refinement of the complex.There are also other questions not directly related to the discrete Hilbertoperator we would like to investigate. In this last chapter, we will show someresults of numerical computations with the discrete operator and shed lighton some of the questions that may be addressed in the near future.

6.1 Test computations

We will start with showing the results of test computations for some differentfunctions and circle packings.

To test the discrete operator, we use functions whose Hilbert transformis known. Given any w ∈ H∞ ∩ C1+α with Imw(0) = 0, we can apply thediscrete transform to u := Rew|T evaluated at the contact points tj . Wethen compare the result to v := Imw|T = Hu.

The functions we use are w1(z) = z and w2(z) = B(z)− C, where B is aBlaschke product with three simple zeros at 0.3, −0.4+0.64i and −0.7−0.2iand C = B(0). We set u1 = Rew1|T and u2 = Rew2|T. Furthermore, wedefine u3 : T → R by u3(t) = 1 if Re t > 0 and u3(t) = 0 otherwise. InSection 3.2, we saw that Hu3(eiτ ) = 1

π ln∣∣tan τ

2

∣∣. It is remarkable that u3 isnot even an element of H∞ ∩ C. Nevertheless, the discrete transform is stilla quite good approximation to the classical transform.

In both Figure 6.1 and Figure 6.2, picture (a) shows the circle packingused to define the discrete transform. Pictures (b), (c) and (d) show theresults for u1, u2 and u3, respectively. The function uj is plotted in greenwhile Huj is plotted in blue. The entries of the discrete Hilbert transformare plotted in red.


90 6 Numerical Results and Future Work

(a) The packing

−π −π2π2

π

−1

1

(b) u1

−π −π2π2

π

−1

1

(c) u2

−π −π2π2

π

−1

1

(d) u3

Figure 6.1: Numerical results for K[7]3 . u in green, Hu in blue, Hu in red.

(a) The packing

−π −π2π2

π

−1

1

(b) u1

−π −π2π2

π

−1

1

(c) u2

−π −π2π2

π

−1

1

(d) u3

Figure 6.2: Numerical results for K′. u in green, Hu in blue, Hu in red

6.2 Eigenvalues of the discrete transform 91

The complex of the packing used in Figure 6.1 is the regular heptagonal

complex with three generations K[7]3 , which was introduced in Subsection

4.2.1. It has 85 vertices, 56 of which are boundary vertices. The complex

of the packing in Figure 6.2 is obtained by hex-refining every face of K[7]4 .

It will be denoted by K ′ and has 778 vertices, 294 of which are boundaryvertices.

We observe a fairly good approximation of the classical Hilbert transformespecially when using K ′. In the following, we will confine ourselves withthese examples. Examining the convergence of the discrete transform andpresenting convergence plots is beyond the scope of this book.

6.2 Eigenvalues of the discrete transform

We expect the discrete Hilbert operator to share or, at least, mimic analyticand algebraic properties of the classical Hilbert operator. This expectation ismotivated by the properties of circles packings and the promising numericalresults.

As an example, we will have a look at the eigenvalues of the discreteoperator. The eigenvalues of the classical Hilbert operatorH : Lp(T)→ Lp(T)for a given p ∈ (1,∞) can be directly deduced from the basis representationin (3.1): They are i, −i and 0 and the respective eigenspaces are given by

Ei = Hp0 (T), E−i = Hp(T)⊥, E0 = {f : T→ C constant}.

Here, Hp0 (T) denotes set of all functions in Hp(T) with zero mean and

Hp(T)⊥ denotes set of all functions in Lp(T) whose nonnegative Fouriercoefficients vanish. In L2(T), this is the orthogonal complement of H2(T),hence the notation.

We expect a similar behavior for the discrete Hilbert transform. Figure6.3 shows the spectra of discrete Hilbert operators for the two maximalpackings seen before.

As we can see, the property of the classical operator does not completelyconfer to the discrete operator. It is noteworthy, though, that the real partof all eigenvalues seems to be equal to 0. In fact, Re(λ) < 10−15 numericallyfor each eigenvalue λ.

To further illustrate the spectra we will have a look at the “eigenfunctions”of H for K ′. Since all entries of H are real, the eigenvalues and correspondingeigenvectors are pairwise complex conjugates, so it suffices to considerσ+(H) := {λ eigenvalue of H| Imλ > 0}. Numerically, no two elements ofσ+(H) have the same modulus and all eigenvalues are simple eigenvalues.


−0.5 0.5

−1

1

(a) K[7]3

−0.5 0.5

−1

1

(b) K′

Figure 6.3: Spectra of H for two different complexes

Let λ1, λ2, λ3, λ4 ∈ σ+(H) such that |λ1| > |λ2| > |λ3| > |λ4| > |λ| for allλ ∈ σ+(H) \ {λ1, λ2, λ3, λ4}. Figure 6.4 shows the real and imaginary partsof the eigenvectors corresponding to λj .

−π −π2π2

π

−1

1

(a) Eigenfunction e1 of λ1

−π −π2π2

π

−1

1

(b) Eigenfunction e2 of λ2

−π −π2π2

π

−1

1

(c) Eigenfunction e3 of λ3

−π −π2π2

π

−1

1

(d) Eigenfunction e4 of λ4

Figure 6.4: Eigenfunctions for K′. Real part in black, imaginary part in gray.

Based on the interpretation of the discrete transform, we expect theeigenfunctions to be linear combinations of discrete versions of zk with k ∈ N.Consequently, their real and imaginary parts should resemble discrete sineand cosine functions.

As we can see in Figure 6.4, the eigenfunctions do indeed look similar

6.2 Eigenvalues of the discrete transform 93

to sine and cosine functions. However, the discrete functions exhibit somekind of quiver. One would expect them to be a better and less “fuzzy”approximation to the trigonometric functions.

Examining the quivering behavior is far beyond the scope of this book.We would nevertheless like to suggest two possible explanations for thisobservation:

1. An experiment suggests that our interpretation of H might be slightlywrong. Given a function u : T → R, let −→u = (u(t1), . . . , u(tm)) andw(z) = Su(z). Recall that the entries of H−→u are interpreted to be thevalues of Imw at the points tj . We suspect that the entries of H−→u shouldpossibly rather be interpreted as values of Imw at the points zj instead.

To explain this, we show a second plot: Assuming that the eigenvalues ek

are discrete versions of zk, we suspect that the entries of ek approximatethe values zkj for j = 1, . . . ,m. For maximal packings, we have that

(1 − rj)tj = zj . In this case, we would obtain tkj ≈ (1 − rj)−kekj . We

will therefore plot the eigenvectors e1, . . . , e4 again, but this time wewill multiply the j-th entry of the k-th vector by (1 − rj)

−k. Themodified vectors ek are shown in Figure 6.5. We can see that the modifiedeigenfunctions look smoother than the functions in Figure 6.4.

−π −π2π2

π

−1

1

(a) Eigenfunction e1 of λ1

−π −π2π2

π

−1

1

(b) Eigenfunction e2 of λ2

−π −π2π2

π

−1

1

(c) Eigenfunction e3 of λ3

−π −π2π2

π

−1

1

(d) Eigenfunction e4 of λ4

Figure 6.5: Modified Eigenfunctions for K′


2. The eigenspace Hp0 (T) of H does not only contain the basis functions tk

for k ∈ N, but also linear combinations of these functions. It is possiblethat the functions seen in Figure 6.4 are actually discrete versions offunctions of the form tk + ctl for some 0 < |c| << 1 and l ∈ N, l >> k,which would result in the fuzzy appearance of the discrete functions.

6.3 Elimination of constants

For the definition of the discrete Hilbert transform, we used the function ato eliminate the constant c appearing in Proposition 3.26.

In this proposition, we remarked that we can choose the branch of theargument function arg to be the principal branch Arg in case ‖u‖ is smallenough. We further noticed that c = 0 in this case.

Recall the construction of the linearized operator. We used the functionh, which is based on the principal branch Arg. To obtain the operator, welinearized the nonlinear operator defined on an arbitrarily small neighborhoodof 0. But consequently, one should expect that we actually do not need thefunction a at all. The elimination of the constant c is already contained inthe ideas our discretization approach is based on.

We will therefore try out what happens if we leave out the function a.I.e., we replace H = ALM−1R by H = LM−1R. In this experiment, thediscretization is based on the complex K ′ from the previous sections again.

Figure 6.6 shows the result of applying H to the vectors u1 and u2, whereu1j = Re tj and u2

j = − Im tj . The expected results are (Hu1)j ≈ Im tj and

(Hu2)j ≈ Re tj , respectively.

−π −π2π2

π

−1

1

(a) u1

−π −π2π2

π

−1

1

(b) u2

Figure 6.6: Numerical results for H applied to the discrete cosine and sine

As we can see, H yields the expected result for u1 but fails for u2. Appar-ently, Hu2 indeed differs from the expected result by a constant.

6.3 Elimination of constants 95

What is the reason for this behavior? Why does our interpretation fail atthis point?

The problem seems to arise from the standard normalization. To illustratethe problem, it is helpful to use the nonlinear transform: For the scaledvectors 1

2u1 and 1

2u2, we plot the standard normalized circle packings P1 resp.

P2 that solve the corresponding instances of (DHRHP). These packingsshould be understood as discrete images of the functions w1(z) = z exp z

2resp. w2(z) = z exp

(−i z2

), so they should (at least roughly) resemble the

shape of the domains w1(D) resp. w2(D). In Figure 6.7, we show P1 resp.P2 together with the curves w1(T) and w2(T) (in black). The n-th and(n− 1)-th circle of the packings are shaded in dark gray resp. light gray.

(a) P1 and w1(T) (b) P2 and w2(T)

Figure 6.7: Packings solving instances of (DHRHP)

While P1 seems to approximate w1(D) fairly well, P2 looks rather like arotated version of w2(D). This is due to the normalization: The dark graycircle is centered at 0, while the light gray circle has its center on the positivereal line. Due to the structure of the complex K ′, this requirement canapparently not be met without inducing the rotation we observe in the figure.This phenomenon eventually results in the constant that was observed inFigure 6.6.

As a consequence, it may be necessary to rethink the normalizationstrategy we applied. Apart from that, we suspect that the behavior may becaused by the structure of the complex. Clearly, the boundary circles of PK′


are significantly smaller than some of its interior circles. This nonuniformdiscretization could be one of the reasons for the observed behavior.

To support this hypothesis, we show another experiment where we useda different complex K ′′. This complex is hexagonal, i.e. every interiornode has exactly 6 neighbors. Figure 6.8 shows PK′′ and the result of thecorresponding operator H being applied to u2.

(a) Maximal packing on K′′

−π −π2π2

π

−1

1

(b) Resulting operator applied to u2

Figure 6.8: Results of H defined on a different complex

As we can see, the constant is significantly smaller in this case. Furtherexperiments with more different complexes have suggested that the constantindeed depends on the complex itself.

Due to these observations, we presume the following: If the discreteoperator H can be proven to converge in some sense to its continuousarchetype H, then the same can be proven for H if suitable assumptions aremade on the complexes involved.

6.4 Curvature of the Circle Packing manifold

Recall that the discrete Hilbert transform is actually the linearization of anonlinear procedure. It was defined as the limit of

1

λHnl (λu) (6.1)

for λ→ 0. Therefore, it could be interesting to evaluate (6.1) for some λ > 0and compare the result to H(u).

As an example, we use the cosine function u(eiτ ) = cos τ and the operator

H based on the complex K[7]3 . In Figure 6.9, the red asterisks show H(u)

6.5 Local frames 97

while the green resp. blue circles show the results of (6.1) evaluated forλ = 0.9 resp. λ = 1

2 . We see that the difference between the results isextremely small. In fact, one has to look very closely to tell the differentvalues apart.

−π −π2π2

π

−1

1

Figure 6.9: Comparison of nonlinear transform (circles) and linearized trans-form (asterisks)

This observation suggests that the curvature of the circle packing manifoldis very small: though circle packings are nonlinear objects, they seem toscale almost linearly.

Interestingly, Stephenson made similar observations before. In his book[27] in Chapter 23, he reports on experiments with curvature flow, wherehe observed that a convex combination of the vector forms of two circlepackings almost behaved like a circle packing itself. It may be interesting toinvestigate these phenomena and the curvature of DN (or, more generally,D) in more detail.

6.5 Local frames

Recall that we have replaced the simple linear Schwarz problem by the non-linear problem (HRHP) due to the problems related with the discretizationof the Schwarz problem.

The reader may already have figured out that there is an alternative tothis. Instead of using a nonlinear problem, we can use a linear problem withindex 1 instead of index 0. As we will see, the advantages of (HRHP) are


also present in this modification, and we have the additional advantage thatthe problem remains linear. In fact, the modified linear problem was thefirst approach suggested and investigated by Wegert [34].

In the following, we give a brief summary of the local frame approach.

The following two core ideas lead to the local frame based problem:

1. Work with a linear Riemann-Hilbert problem of index 1.

2. For u ≡ 0, z 7→ z should be a solution of the problem.

Following these ideas, we investigate Riemann-Hilbert problems of the form

Re(c(t)(w(t)− t)) = u(t) ∀t ∈ T (6.2)

or, when writing them in the classical form,

Re(c(t)w(t)) = u(t) + Re(c(t)t) ∀t ∈ T.

Here, c : T→ C is a continuous, nowhere vanishing function with windingnumber 1, the so called direction function.

Assume that there is a holomorphic function g ∈ H∞ ∩ C with a simplezero at 0 whose boundary function is given by t 7→ 1

c . Then the problem hasthe standard normalized solution

z 7→ (Su(z) + iµ)g(z) + z

If g and Su and their derivatives are small enough, we can apply similararguments as in the proof of Theorem 3.24 to show that this solution has noother zeros and that its derivative vanishes nowhere.

So, let us discretize the problem (6.2). The target curve at tj ∈ T is the

straight line{z ∈ C|Re(c(tj)(z − tj)) = u(tj)

}. For u = 0, this line passes

through tj . In the special case where g(z) ≡ z, the line is tangent to T at tj ,thus it is also tangent to the j-th circle of PK . Motivated by this, we definethe functions

βj(rj , xj , yj) = (xj − Re(tj)) Re c(tj) + (yj − Im(tj)) Im c(tj) + rj − u(tj)

β defines a discrete boundary value problem. If P = (r, x, y) solves thisproblem, we set

vj = (yj − Im(tj)) Re c(tj)− (xj − Re(tj)) Im c(tj)

6.5 Local frames 99

and understand v as the discrete Hilbert transform of u. This is motivatedby the fact that for a solution w of (6.2) we have

Im(c(t)(w(t)− t)) = Hu(t).

As done for (DHRHP), we can linearize the discrete boundary value problemdefined by β. For the special case c(t) = t, note that the linearizations of theabove problem and (DHRHP) actually coincide. Thus, Theorem 5.17 appliesonce more, yielding a well-defined discrete local frame Hilbert operator.

In the local frame setting, we can now vary the function c. This raisestwo questions:

• Which choices of c yield a discrete problem with regular linearization?

• How does the choice of c affect analytic and algebraic properties of theresulting operator?

As we can see, there are way more questions to be answered than answeredquestions regarding the Hilbert transform. It remains to hope that furtherinsight into the discrete Hilbert transform with circle packings will be gainedin the future and that addressing the questions discussed in this last chapterwill further enrich the fruitful theory of circle packings.

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A Discrete Hilbert Transform with Circle Packings

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