Dominik Volland studies the construction of a discrete counterpart to the Hilbert transform in the realm of a nonlinear

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*Table of contents : AcknowledgmentsContentsList of FiguresAbstract1 Introduction2 Auxiliary Material and Notation3 The Continuous Setting 3.1 Hardy spaces 3.1.1 Boundary values of holomorphic functions 3.1.2 Integral formulas 3.1.3 Fourier series of the boundary functions 3.2 The Hilbert transform 3.3 Riemann-Hilbert problems 3.3.1 Linear Riemann-Hilbert problems 3.3.2 Nonlinear Riemann-Hilbert problems 3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem 3.4.1 Choice of the problem 3.4.2 Solutions of the problem 3.4.3 Counterexamples for u not in C1+a4Circle Packings 4.1First examples and ideas 4.2Basic definitions 4.2.1Complex 4.2.2Circle packing 4.3Manifold structure 4.3.1Contact function 4.3.2 Angle sums and branch structures 4.3.3Parametrization of Db 4.3.4Normalization 4.4Discrete harmonic functions on circle packings 4.5Discrete analytic functions 4.6Maximal packings 4.7Some results on discrete analytic functions 4.7.1Discrete maximum principles 4.7.2Approximation of the Riemann Mapping5Discrete Hilbert Transform 5.1Discrete boundary value problems 5.1.1Definition and examples 5.1.2Linearization of boundary value problems 5.2Proof of the maximal packing conjecture 5.2.1The transformed packing 5.2.2Differential of w at the transformed packing 5.2.3 Basis for the kernel of J 5.3Discrete Hilbert transform 5.3.1Difficulties of the Schwarz problem 5.3.2Discretization of the nonlinear problem 5.3.3Linearization of the discrete operator6Numerical Results and Future Work 6.1Test computations 6.2Eigenvalues of the discrete transform 6.3Elimination of constants 6.4Curvature of the Circle Packing manifold 6.5Local framesBibliography*

Dominik Volland

A Discrete Hilbert Transform with Circle Packings

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Dominik Volland

A Discrete Hilbert Transform with Circle Packings

Dominik Volland Garching near Munich, Germany

BestMasters ISBN 978-3-658-20456-3 ISBN 978-3-658-20457-0 (eBook) https://doi.org/10.1007/978-3-658-20457-0 Library of Congress Control Number: 2017961504 Springer Spektrum © Springer Fachmedien Wiesbaden GmbH 2017 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Printed on acid-free paper This Springer Spektrum imprint is published by Springer Nature The registered company is Springer Fachmedien Wiesbaden GmbH The registered company address is: Abraham-Lincoln-Str. 46, 65189 Wiesbaden, Germany

Acknowledgments This book originated from a research project of Elias Wegert and Folkmar Bornemann together with the author in the TUM elite graduate program TopMath. The work was supported by the DFG-Collaborative Research Center, TRR 109, “Discretization in Geometry and Dynamics”. I would like to thank both TopMath and DFG for their financial and ideological support throughout the project. I also wish to express my deep gratitude to my supervisors, Elias Wegert and Folkmar Bornemann, for productive personal communication, for their valuable support, and also for suggesting me for “Best Masters”. Finally, I’d like to thank my family, my friends, and my colleagues at TUM for supporting me while I was writing this book.

Contents 1 Introduction

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2 Auxiliary Material and Notation

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3 The Continuous Setting 3.1 Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Boundary values of holomorphic functions . . . . . . 3.1.2 Integral formulas . . . . . . . . . . . . . . . . . . . . 3.1.3 Fourier series of the boundary functions . . . . . . . 3.2 The Hilbert transform . . . . . . . . . . . . . . . . . . . . . 3.3 Riemann-Hilbert problems . . . . . . . . . . . . . . . . . . . 3.3.1 Linear Riemann-Hilbert problems . . . . . . . . . . . 3.3.2 Nonlinear Riemann-Hilbert problems . . . . . . . . . 3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Choice of the problem . . . . . . . . . . . . . . . . . 3.4.2 Solutions of the problem . . . . . . . . . . . . . . . . 3.4.3 Counterexamples for u ∈ / C 1+α . . . . . . . . . . . . 4 Circle Packings 4.1 First examples and ideas . . . . . . . . . . . . . 4.2 Basic definitions . . . . . . . . . . . . . . . . . 4.2.1 Complex . . . . . . . . . . . . . . . . . . 4.2.2 Circle packing . . . . . . . . . . . . . . 4.3 Manifold structure . . . . . . . . . . . . . . . . 4.3.1 Contact function . . . . . . . . . . . . . 4.3.2 Angle sums and branch structures . . . 4.3.3 Parametrization of Db . . . . . . . . . . 4.3.4 Normalization . . . . . . . . . . . . . . . 4.4 Discrete harmonic functions on circle packings 4.5 Discrete analytic functions . . . . . . . . . . . . 4.6 Maximal packings . . . . . . . . . . . . . . . .

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Some results on discrete analytic functions . . . . . . . . . . . 57 4.7.1 Discrete maximum principles . . . . . . . . . . . . . . 58 4.7.2 Approximation of the Riemann Mapping . . . . . . . . 58

5 Discrete Hilbert Transform 5.1 Discrete boundary value problems . . . . . . . . . 5.1.1 Definition and examples . . . . . . . . . . . 5.1.2 Linearization of boundary value problems . 5.2 Proof of the maximal packing conjecture . . . . . . 5.2.1 The transformed packing . . . . . . . . . . 5.2.2 Differential of ω e at the transformed packing 5.2.3 Basis for the kernel of Je . . . . . . . . . . . 5.3 Discrete Hilbert transform . . . . . . . . . . . . . . 5.3.1 Difficulties of the Schwarz problem . . . . . 5.3.2 Discretization of the nonlinear problem . . 5.3.3 Linearization of the discrete operator . . . .

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

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11

A circle packing . . . . . . . . . . . Packings and complexes . . . . . . A discrete conformal map . . . . . Overlapping circles . . . . . . . . . A branched packing . . . . . . . . A complex . . . . . . . . . . . . . . Complexes and hex refinement . . A face and its face circle . . . . . . Illustration of the contact equation Maximal packings . . . . . . . . . Construction of a discrete Riemann

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Abstract This book deals with the definition of a discrete Hilbert transform. The classical Hilbert transform is a bounded linear operator relating the realand imaginary parts of the boundary values of a holomorphic function. The discretization is based on the theory of circle packings which has been established as a discrete counterpart of complex analysis in the past three decades. The Hilbert transform is closely related to Riemann-Hilbert problems. In particular, the transform can be computed from the solution of a RiemannHilbert problem. We will therefore define the transform based on discrete Riemann-Hilbert problems which have been investigated by Elias Wegert since 2009. The definition of the discrete operator requires the regularity of a certain Jacobian matrix. In this book, we present a proof of the regularity of this matrix.

1 Introduction The mathematical branch of complex analysis deals with the behavior of holomorphic functions, which are differentiable functions in the complex plane. The theory of functions of one complex variable is quite self-contained nowadays. However, during the last three decades, complex analysis has found a discrete counterpart in the theory of so-called circle packings. These packings are configurations of circles that satisfy prescribed tangency relations. The idea of using circles to discretize holomorphic functions is due to the fact that holomorphic functions are locally conformal transformations of the complex plane wherever their derivative does not vanish. This means, a holomorphic function will map an infinitesimally small circle onto an infinitesimally small circle again. Circle packings have turned out to be a very faithful discretization of holomorphic functions indeed. By now, a wide variety of theorems and concepts from complex analysis have found counterparts in circle packing. There are still many open problems and the theory of circle packings is only growing. A broad overview over the theory of circle packings can be found in Stephenson’s book [27]. This book is devoted to one of the numerous concepts of complex analysis and the quest for its discrete counterpart. Namely, we discuss a discrete Hilbert transform in the context of Riemann-Hilbert problems. Roughly speaking, a function holomorphic in the complex unit disc D is determined by the real part of its boundary values. The Hilbert transform is a linear operator that relates the real and imaginary parts of the boundary values of a holomorphic function. Riemann-Hilbert problems are boundary value problems for functions holomorphic in D. A Riemann-Hilbert problem imposes one real condition to each boundary value of the solution. The Hilbert operator is closely related to these problems. In fact, we will show that the Hilbert transform of a given function can be obtained from the solution of a suitable Riemann-Hilbert problem. The German mathematician Elias Wegert, author of several papers [31][32] and a book [30] on Riemann-Hilbert problems, observed that RiemannHilbert problems fit nicely into the setting of circle packings. Motivated © Springer Fachmedien Wiesbaden GmbH 2017 D. 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. The strategy of this definition can be summarized as follows: Find a RiemannHilbert problem from which the Hilbert transform can be obtained, discretize this problem with circle packings, solve it, and extract the discrete transform from the solution. In this book, we present the discrete operator obtained from this procedure. The definition of the discrete operator rests on the regularity of a certain Jacobian matrix which arises from the discrete boundary value problem. The regularity of this matrix has been conjectured by Wegert [34]. In this book, we give a proof of this conjecture. The book is organized as follows. Chapter 2 fixes notation conventions and states a few classical results that will be used throughout the text. In Chapter 3 we will be concerned with the continuous theory of the Hilbert operator. We will first introduce the functional analytic basics required to define the Hilbert operator and formulate its most important properties. After that, we will define Riemann-Hilbert problems and discuss their connection to the Hilbert operator. In Chapter 4, we will turn to the discrete setting. We will define circle packings and state their basic properties, with an emphasis on the differential geometric structure of circle packings. Chapter 5 will then introduce discrete boundary value problems. We will give some examples, introduce linearized discrete boundary value problems and prove the regularity of the matrix mentioned before. Based on this and the work done in Chapter 3, we will define a discrete Hilbert operator. Finally, Chapter 6 presents some numerical results and outlines questions related 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 overview over some results that will be used without references throughout the book. Complex analysis. There are many books covering elementary complex analysis. Our standard reference is the book of Ahlfors [1]. A domain is an open connected set in C. The domain we are dealing with most of the time is the unit disc {z ∈ C| |z| < 1} denoted by D. Its boundary is the unit circle {z ∈ C| |z| = 1} and will be denoted by T. Other discs will be denoted by Br (z ∗ ) := {z ∈ C| |z − z ∗ | < r}. Given a circle C ⊂ C, the disc bounded by C will be denoted by int C, while ext C = C \ (C ∪ int C). The boundary of a domain U is denoted by ∂U and the closure U ∪ ∂U is denoted by U . Some caution is required here since the complex conjugate of a complex number z is denoted by z. By Log we denote the principal branch of the complex logarithm defined on the slit domain C− := C \ R≤0 . By Arg we denote the principal branch of the complex argument function, i.e. Arg(z) = Im Log(z). log resp. arg denotes any branch of these functions if it is not necessary (or possible) to specify the branch explicitly. To keep the overview, we have the following convention for variables: z always refers to a variable ranging on D or sometimes C, while t is a variable ranging on T. Arguments of complex numbers are denoted by Greek letters, mostly τ , sometimes also φ or θ. M¨ obius transformations. tion of the form

A M¨obius transformation is a holomorphic func-

az + b cz + d where a, b, c, d are complex constants such that ad − bc 6= 0. These functions play a special role in complex analysis due to their geometric properties. It is natural to understand them as mappings on the extended complex plane C ∪ {∞} by writing T (∞) = ac and T (− dc ) = ∞ if c 6= 0 (or T (∞) = ∞ if c = 0). T (z) =

© Springer Fachmedien Wiesbaden GmbH 2017 D. Volland, A Discrete Hilbert Transform with Circle Packings, BestMasters, https://doi.org/10.1007/978-3-658-20457-0_2

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2 Auxiliary Material and Notation

M¨obius transformations have the following properties (for proofs, see Chapter 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 M¨ obius circle, i.e. a circle or a line, T (C) is a M¨obius circle, too.1 (iv) T can be written as a composition of linear functions and the inversion z 7→ z1 . / C. Let z1 , z2 , z3 ∈ C lie in (v) Let C ⊂ C be a circle and T −1 (∞) ∈ counter-clockwise order on C. Then • if T −1 (∞) ∈ ext C, then T (int C) = int T (C), T (ext C) = ext T (C), and T (z1 ), T (z2 ), T (z3 ) lie in counter-clockwise order on C again. • if T −1 (∞) ∈ int C, then T (int C) = ext T (C), T (ext C) = int T (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 in addition 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 conformal iff it is a composition of a rotation z 7→ λz for λ ∈ T and a M¨obius transformation of the form z 7→

z − z0 1 − zz0

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

A (finite) Blaschke product of degree n is a function

B(z) = c

1 In

n Y z − zj 1 − zj z j=1

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

2 Auxiliary Material and Notation

5

for c ∈ T, z1 , . . . , zn ∈ D. The set of all finite Blaschke products is denoted by B. Note that z1 , . . . , zn are precisely the zeros of B counted with multiplicity. The perhaps most important property of Blaschke products is that each B ∈ 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 each value in D exactly n times. For details, see [23]. As the term “finite” suggests, one can also consider infinite Blaschke products with an infinite number of zeros in D. These functions are not relevant for this book, though. Function spaces. For further details on the spaces introduced here, any textbook 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) for 1 ≤ p < ∞ if its normalized Lp -norm 1 kf kp := 2π

ˆ2π

p1 iτ p f (e ) dτ

0

is smaller than ∞. f is said to be in L∞ (T) if it is bounded on T \ E for a set E with Lebesgue measure zero and the L∞ -norm is defined by kf k∞ := 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 space L2 (T) is a Hilbert space endowed with the scalar product 1 (f, g)2 := 2π

ˆ2π

12 f (eiτ )g(eiτ )dτ .

0

For a continuous function f : T → C, we define its modulus of continuity ωf (θ) := sup f (eiφ ) − f (eiτ ) . |φ−τ | 0, z 1 k = 0, Ptk (z) = −k z k < 0. This reflects the fact that, given a function f ∈ H p (T), Pf is the corresponding function in H p . If the negative Fourier coefficients of f do not vanish, Pf will still be a complex linear combination of the two harmonic functions Re Pf and Im Pf whose nontangential boundary values coincide with Re f and Im f , but Re Pf and Im Pf are not harmonic conjugates of each other anymore. 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 values of z k and −iz k , respectively. Using eikτ = cos kτ + i sin kτ , Theorem 3.11 therefore gives us 2z k k > 0, 1 k = 0, . Stk (z) = 0 k < 0. A function u ∈ Lp (T) with Fourier coefficients uk , k ∈ Z, is real-valued iff for all k ∈ N we have uk = u−k ∀k ∈ N and Im u0 = 0. In this case we may rewrite u as u(eiτ ) = u0 +

∞ X

2 Re uk cos(kτ ) − 2 Im uk sin(kτ ).

k=1

This is the real part of the boundary values of the function z 7→ u0 +

∞ X k=1

2uk z k

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 by making use of the monomial basis. As an example, we will derive a property of the Schwarz integral we will use later. If a function u with Fourier coefficients uk for k ∈ Z is differentiable, the coefficients of its derivative are given by u0k := ikuk which can be shown by using integration by parts in the formula of the Fourier coefficients (or just by differentiating each summand of the Fourier series). For a holomorphic function with Taylor coefficients ak , k ∈ N0 , its derivative has coefficients a0k := (k + 1)ak+1 . But combining these results, we see that there is a connection between Su0 and Su˙ : izSu0 (z) = iz

∞ X

2(k + 1)uk+1 z k =

k=0

∞ X

2ikuk z k = Su˙ (z)

k=1

Thus, we have proved the following lemma. Lemma 3.15. Let u ∈ C 1 (T). Then izSu0 (z) = Su˙ (z).

3.2 The Hilbert transform Let us start with formally deriving the formula of the Hilbert operator. Let w ∈ H p with Im w(0) = 0 and let z ∈ D with polar form z = reiτ . By Schwarz’ formula, 1 Im w(z) = 2π =

1 2π

ˆπ Im −π ˆπ

−π

eiφ + z Re w(eiφ )dφ eiφ − z

2r sin(τ − φ) Re w(eiφ )dφ. 1 + r2 − 2r cos(τ − φ)

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3 The Continuous Setting

We would like to compute the boundary values of Im w, thus we are interested in the limit r → 1. If we simply put r = 1 in the above formula, we obtain 1 Im w(e ) = 2π

ˆπ

sin(τ − φ) Re w(eiφ )dφ 1 − cos(τ − φ)

iτ

=

1 2π

−π ˆπ

cot

τ −φ Re w(eiφ )dφ. 2

−π

Note that the cotangent function has non-integrable singularities at integer multiples of π, thus, the improper integral on the right-hand side does in general not exist in the classical sense. We will have to interpret it in a proper 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 τ −ε ˆ ˆπ ˆπ τ −φ τ −φ τ −φ cot − cot f (φ)dφ := lim f (φ)dφ + cot f (φ)dφ ε→0 2 2 2 −π

−π

τ +ε

for τ ∈ (−π, π) and analogously π−ε ˆ ˆπ π−φ π−φ f (φ)dφ := lim cot f (φ)dφ. − cot ε→0 2 2 −π+ε

−π

Finally, we can use this notation to define the Hilbert transform. Definition 3.16 (Hilbert transform). Let u : T → R be given. The Hilbert transform2 Hu : T → R of u is the function defined by e

iφ

ˆπ τ −φ 1 − cot u(eiτ )dτ . 7→ 2π 2 −π

wherever the integral on the right-hand side exists. The next theorem states that the integral appearing in this definition exists for very general functions. 2 In

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

3.2 The Hilbert transform

19

Theorem 3.17 ([14], Chapter I:E.3). Let u ∈ L1 (T). Then for almost every φ ∈ R, the principal value integral ˆπ 1 τ −φ − cot u(eiτ )dτ 2π 2 −π

exists. Now, the important question is which spaces are appropriate to make the Hilbert transform a bounded operator. The answer to this question was indicated 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 a bounded linear operator. It is pretty easy to provide a counterexample showing that Riesz’ theorem is indeed wrong for p = ∞. Consider the function defined by u(t) = 1 if Re(t) > 0 and u(t) = 0 otherwise3 . Obviously kuk∞ = 1. Computation gives τ 1 Hu(eiτ ) = ln tan . π 2 Thus lim Hu(eiε ) = ∞, so Hu is unbounded. ε→0

Actually, even a continuous function u can have an unbounded Hilbert transform. Such an example will be constructed later in this book. Riesz’ theorem is also wrong for p = 1. We will discuss a counterexample shortly. Now let us have a look at how the Hilbert operator acts on the monomial basis. We obtain −itk k > 0 k 0 k=0 . Ht = (3.1) itk k < 0 Thus, given a real-valued function u on Fourier coefficients wk of u + iHu are 2uk u0 wk = 0 3 To

T with Fourier coefficients uk , the k>0 k=0 . k 0 independent of u such that θ ˆ ˆπ ω (φ) ω (φ) u u ωHu (θ) ≤ C dφ + θ dφ φ φ2 0 θ θ ˆ ˆπ 1 1 ≤ C |u|C α dφ + θ dφ . φ1−α φ2−α 0

θ

Both integrals can be evaluated explicitly and setting C 0 =

C α(1−α)

we get

ωHu (θ) ≤ C 0 |u|C α θα and thus |Hu|C α ≤ C 0 |u|C α . Now since

2π ´

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

0

that for every τ ∈ R Hu(eiτ ) = Hu(eiτ ) − Hu(eiτ0 ) ≤ ωHu (π) ≤ C 0 |u| α π α . C Overall, we get kHukC α = kHukC + |Hu|C α ≤ C 0 (π α + 1) |u|C α ≤ C 0 (π α + 1) kukC α which shows the boundedness of H. Note that the C α -norm of Hu can be estimated from the C α -seminorm of u alone. This reflects the fact that H maps constant functions to 0. The following corollary rephrases the Plemelj-Privalov theorem based on Theorem 3.19. Corollary 3.21. For every α > 0, the map u 7→ Su is continuous as a map C α (T) → H ∞ ∩ C. Proof. kSu kH ∞ ∩C = ku + iHukC ≤ kukC α + kHukC α ≤ C kukC α for a constant C due to Theorem 3.20.

22

3 The Continuous Setting 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 following properties: 1. If u ∈ C 1 and Hu ∈ C 1 , then H commutes with differentiation with respect to the polar angle, i.e.: ∂τ Hu = H∂τ u where ∂τ u(t) =

d iτ 4 dτ u(e )|arg t .

2. ker H = {u : T → R|u constant}. ( ) ´π p p iφ 3. im H = L0 (T) := v ∈ L (T)| v(e )dφ = 0 . −π

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 representations of 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 case iff all Fourier coefficients of u except for the 0-th vanish, which is the case for exactly all constant functions. 3. A function v ∈ H p satisfies ˆπ v(eiφ )dφ = 0 −π

iff its 0-th Fourier coefficient vanishes. In this case, direct computation considering the Fourier coefficients gives HH(−v) = v. Thus, v is the image 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 derivatives. From this, we can see that H is also bounded as an operator on the Sobolev p spaces Wm (T) for 1 < p < ∞, m ∈ N. We will not pay any further attention to this fact 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 for holomorphic functions on D. That is, we investigate problems imposing conditions on the values of a function on T. Their solutions are functions holomorphic in D with nontangential boundary values satisfying these conditions. Riemann-Hilbert problems are perhaps the most important class of these problems. This problem class originates from the dissertation of Bernhard Riemann himself, which is best known for containing the first formulation of the celebrated Riemann Mapping Theorem [25]. Riemann used conformal mapping as an example, but he “treated the [general] problem only heuristically” ([31], Chapter 6). David Hilbert was the first person to come up with a solution operator at least for linear problems [12], hence the name. A historical overview over the development of Riemann-Hilbert problems can be found in Chapter 6 of [31] or in Chapter 1 of [32]. To understand the general ideas behind Riemann-Hilbert problems, recall that the space H p (T) is the space of functions in Lp (T) where the Fourier coefficients with negative index vanish. Intuitively, this means that we can prescribe “one half” of the boundary function to obtain a unique holomorphic function as a solution. For a different perspective on the same fact, one can also think of Schwarz’ formula: Loosely speaking, we may not prescribe the boundary function of a holomorphic function arbitrarily, but we may prescribe its real part arbitrarily. These observations motivate to investigate other boundary value problems where “half of the boundary function” is prescribed. I.e., problems that impose 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 it in the form F (t, w(t)) = 0 ∀t ∈ T for a prescribed function F : T × C → R. Under suitable regularity assumptions, the sets {z ∈ C|F (t, z) = 0} for a given t ∈ T are submanifolds of C with real dimension 1, thus constituting one real condition imposed on w at t. Clearly, the solvability of the problem depends crucially on the structure of F . Roughly speaking, the theory of Riemann-Hilbert problems known to date is based on splitting them into various classes depending on the structure of their boundary condition.

24

3 The Continuous Setting

The following two subsections give a sketchy overview of the solvability for 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 space must be an affine linear space. We call the resulting problem a linear Riemann-Hilbert problem. These problems can be written more compactly in 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 symbol c is continuous and zero free. In this case, we may assume that |c(t)| ≡ 1 by dividing by |c(t)|. Then the symbol can be written as c(t) = eiφ(t) for a function φ(t) which is continuous on T\{1}. The jump at 1 is then an integer multiple of 2π and of c about 0. We therefore tells us the winding number call wind c =

1 2π

lim φ(eiτ ) − lim φ(eiτ )

τ %2π

τ &0

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 the linear Riemann-Hilbert problem Im(c(t)w(t)) = h(t). Then the following assertions hold. 1. If k = 0, the set of solutions in H p has real dimension 1. The boundary functions of the solutions are given by c c w0 µ − H +i |w0 | |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 in H p. 3. If k < 0, there are −2k − 1 bounded linear functionals in Lp such that the problem is solvable in H p iff h(t) lies in the kernel of all of these functionals.

3.3 Riemann-Hilbert problems

25

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

3.3.2 Nonlinear Riemann-Hilbert problems The key to the solvability of nonlinear problems is to approach them geometrically. If a S problem is given via a function F , we consider the target manifold M := Mt where t∈T

Mt := {z ∈ C|F (t, z) = 0}. This definition allows us to write the boundary condition of a RiemannHilbert problem more compactly as w(t) ∈ Mt ∀t ∈ T. The sets Mt are called the target curves of the problem. These manifolds can now be separated into various classes. The first main distinction 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 generalization of linear problems with continuously differentiable symbol. The class A can be split up into three different classes again. A manifold M ∈ A is a topological torus. Thus, (C × T) \ M consists of two connected components one of which is bounded. We call this component int M and for each t ∈ T we write int Mt := {z ∈ C|(t, z) ∈ int M }. Now we may ask whether there is a holomorphic function wM ∈ H ∞ ∩ C that satisfies wM (t) ∈ int Mt or at least wM (t) ∈ int Mt ∪ Mt for all t ∈ T. Based on this, we consider the classes • R where a function wM ∈ H ∞ ∩ C with wM (t) ∈ int Mt ∀t ∈ T exists, • S where such a function does not exist, but where a function wM ∈ H ∞ ∩C with wM (t) ∈ int Mt ∪ Mt ∀t ∈ T exists, and 5 The

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

26

3 The Continuous Setting

• N where neither of the above holds. Problems with manifolds in N have no solution. Problems with manifolds in S turn out to have exactly one solution, namely wM itself. Finally, problems with 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 of solutions w such that the zj are precisely the zeros of wM − w, counted with multiplicity. Proofs of these statements can be found in Chapter 2 of [30]. Circular Riemann-Hilbert problems. Circular Riemann-Hilbert problems are a special class of nonlinear Riemann-Hilbert problems where each target curve is a circle. I.e., there are functions c : T → C and r : T → R+ such that 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 the solution set of the problem which are covered in a paper by Glader and Wegert [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 general case using a transformation. Assuming that r ≡ 1, it can be shown that the regularity of c implies the regularity of solutions in a similar way as in Theorem 3.20: If α ∈ (0, 1) and c ∈ C α , then the solution lies in H ∞ ∩ C α . In the special case where c is constant, the problem can be solved explicitly up to a Schwarz integral. In fact, we will be using such a problem for the discretization of the Hilbert transform. In this context, we will derive the explicit solution in the following section and give an extensive proof.

3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem As outlined in the introduction, our strategy for the discretization of H is to formulate a Riemann-Hilbert problem depending on some function u ∈ 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 Hilbert transform 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 Re w(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 H p are given by Su (z) + ic. Thence, we know that for a solution w, Hu is equal to Im w|T up to a real constant. Unfortunately, the Schwarz problem is not well suited for discretization with circle packings. The main reason for this is that, as long as we merely know u, it is difficult to make predictions about the critical points of Su . Our discretization, however, requires information about the number and location of critical points of the solution. This and other difficulties related to the Schwarz problem will be explained in detail in Subsection 5.3.1. To resolve these difficulties, we are going to consider a circular problem with 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 that we can find one solution among them which has no critical points and which is suitably normalized in addition. Therefore, the Riemann-Hilbert problem we will investigate and be concerned with throughout the rest of this book is the following: |w(t)| = exp u(t)

(HRHP)

All target curves Mt of this problem are circles centered at 0 and with radius exp u(t). As mentioned in Subsection 3.3.2, this allows us to solve the 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 the theorem characterizing the set of solutions of (HRHP). Basically, the strategy of the proof is to factor out the zeros of a solution using a Blaschke product. The remaining function is then identified by considering its holomorphic logarithm. Theorem 3.24. Let α > 0 and u ∈ C α . Write wu (z) := exp Su (z). Then, the solutions of (HRHP) in H ∞ ∩ C are exactly the functions of the form z 7→ B(z)wu (z) where B is an arbitrary finite Blaschke product. Proof. 1. By Theorem 3.19, Su ∈ H p 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 Re Su (t) = exp u(t)

28

3 The Continuous Setting

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 can not have an accumulation point on T. On the other hand, the set of zeros of w can not have an accumulation point in D either since w is holomorphic. Combining these statements, w can have only finitely many zeros. Thus, let B be a Blaschke product with precisely the same zeros (including w(z) multiplicity) as w. Then the function w0 (z) = B(z) is zero free, lies in H ∞ ∩ C and is still a solution of (HRHP). Therefore, it has a logarithm g with 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 exp u(t) = u(t). Consequently, Re g and Re Su are bounded functions which are harmonic in D and continuous on D and whose values on T coincide. Thence, they must be identical6 . Since they are the real parts of the holomorphic functions g and Su , we conclude g = Su + ic for some c ∈ R and thus w0 = wu · eic . Since z 7→ eic B(z) is again a Blaschke product, w has the claimed form. Notation 3.25. Given u ∈ C α , we will henceforth write wu (z) := exp Su (z) for the “default solution” of (HRHP). How to obtain Hu from a solution of (HRHP) is evident from the proof of Theorem 3.24. This is the content of the following proposition. Proposition 3.26. Let u ∈ C α and let w ∈ H ∞ ∩ C be a solution of (HRHP), w(z) = wu (z)B(z). Then there is a continuous branch arg of the argument and a constant c such that Hu(t) = arg

w(t) + c ∀t ∈ T. B(t)

If kHuk∞ < π, we can choose arg = Arg. In this case, c = 0. Proof. As in the proof of Theorem 3.24, let log wu be a complex logarithm of the zero-free function wu . Then log wu (z) = Su (z) − ic for some real constant c. Thus Hu(t) = Im Su (t) = Im log wu (t) + c = arg wu (t) + c = arg 6 This

w(t) +c B(t)

follows from Poisson’s formula for harmonic functions, compare the remark after Theorem 3.10.

3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem

29

If kHuk∞ < π, then Im Su (z) < π holds by the maximum principle for harmonic functions ([1], Chapter 4, Theorem 21). Hence Su (z) = Log wu (z) w(t) and we obtain Hu(t) = Arg B(t) . Note that due to Theorem 3.20, kHuk∞ < π is satisfied if kukC α is small enough. This theoretical result leaves a practical problem, though: To obtain Hu from the solution of (HRHP), we need to know the corresponding Blaschke product. Assume we have designed an algorithm which solves a discretized version of (HRHP). How do we know which Blaschke product we have to divide by? Fortunately, this question is easy to answer due to the special structure of circle packings: The algorithm we design will only be able to compute discrete solutions which have no critical points and are suitably normalized. In the next theorem, we will prove that these requirements single out a unique solution in the continuous setting for which the Blaschke product is known explicitly. First, we define the normalization we are about to use. Definition 3.27. A holomorphic function f : D → C is called standard normalized if f (0) = 0 and f 0 (0) > 0. We can now state the theorem providing the necessary information for the discretization with circle packings. We will have to strengthen our regularity assumptions 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 constant such that kSu kH ∞ ∩C ≤ C kukC α for every u ∈ C α . Then for every u ∈ C α+1 with kuk ˙ C α < C1 , 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 w0 (z) = zwu0 (z) + wu (z). Thus w0 (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 , w0 (z) = wu (z) (zSu0 (z) + 1) .

30

3 The Continuous Setting

This expression is zero only if zSu0 (z) = −1, therefore, it suffices to show |zSu0 (z)| < 1 ∀z ∈ D. By Lemma 3.15, we have izSu0 (z) = Su˙ (z). By assumption, Su˙ is bounded by 1. We conclude that the desired inequality holds. Finally, we address uniqueness. Assume that there is a second standard b normalized solution w(z) b = wu (z)B(z) meeting the criteria of the theorem. We have 0 b b0 w b0 (z) = wu (z) B(z)S u (z) + B (z) . b must have at least a simple zero at 0. If B b has no further zeros, it is B 0 b a scalar multiple of z 7→ z. In this case, w b (0) > 0 implies B(z) = z, i.e. w b = w. b has at least two zeros (counted with multiplicity). By a So assume that B b 0 has at least one zero in D in that case. theorem of Walsh (§5.3.1 in [29]), B 0 b b0 We shall show that the same holds for B(z)S u (z) + B (z). 0 b First, we derive a lower bound for B on T. Let z1 , . . . , zn be the zeros of b with multiplicity (note that z1 = 0). Then by direct computation B n 2 b 0 (z) X 1 − |zj | B = . b (z − zj )(1 − zzj ) B(z) j=1

Now for t ∈ T, n X n 2 2 tB 0 X b 1 − |zj | 1 − |zj | (t) b0 = = B (t) = 2 b B(t) (1 − tzj )(1 − tzj ) |1 − tzj | j=1

j=1

where the absolute value could be dropped in the last expression since all summands are real and positive. But since z1 = 0, one of the summands is equal to 1, thus b0 B (t) > 1 ∀t ∈ T. b0 Now chose r < 1 large enough such that B (rt) > 1 ∀t ∈ T still holds and b 0 still has a zero in Br (0). Since kSu0 k < 1, we then have B ∞ b0 b b0 0 0 b b0 B (rt) + B(rt)S u (rt) − B (rt) = B(rt)Su (rt) ≤ 1 < B (rt) for all t ∈ T. By Rouch´e’s theorem ([1], Chapter 4, Corollary in Section 5.2), 0 b b0 b0 the functions z 7→ B(z)S u (z) + B (z) and B have the same number of 0 zeros in Br (0). We conclude that w b has a zero in D. This contradicts the assumptions on w. b

3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem

31

As we can see, the are in fact plenty of u such that the corresponding standard normalized solution does have critical points. The crucial point is that if u ∈ C 1+α , we can always use the linearity of H to countervail this problem: We simply multiply u by a sufficiently small positive real number to perform our discretization procedure and divide by the same factor afterwards.

3.4.3 Counterexamples for u ∈ / C 1+α Though the assumption u ∈ C 1+α is quite restrictive, it is essential for the validity of the last theorem. In the following we construct a counterexample showing the problem that occurs if we merely assume u ∈ C 1 . We will also give some more counterexamples to indicate problems that occur if we reduce the regularity assumptions even further. Continuous function with unbounded Hilbert transform. The following example was announced in Section 3.2 and will serve as a building block for the counterexamples constructed afterwards. Denote by D := {z ∈ D| Re(z) > 0} the upper half of the unit disc and write G0 := {x + iy ∈ C|0 < x < (y + 1)e−y }. Let g : D → G0 be the bijective conformal map whose boundary values satisfy f (1) = −i, f (0) = 0 and f (−1) = ∞.7 Then in particular f ((−1, 0)) = iR>0 holds. Using the symmetry principle ([1], Chapter 4, Theorem 24), extend f onto D.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 the domains. For t ∈ T, let u(t) = − Re f (t). Obviously u admits a continuous extension at t = −1 by setting u(−1) = 0. Theorem 3.19 then yields Su = −f and Hu(t) = Im Su (t) for t ∈ T. Since Im Su is unbounded, the example has been constructed.

7 The

existence of g follows from the Riemann Mapping Theorem ([1], Chapter 6, Theorem 1) 8 This can also be done explicitly by defining f (z) = −f (z)

32

3 The Continuous Setting 1

3

D

2

−1

1

1

G0 −1

1 −1

−1

(b) G

(a) Unit disc

Figure 3.1: Construction of a continuous function with unbounded Hilbert transform

Critical points of solutions for u ∈ C 1 . We will construct a continuously differentiable function u : T → R such that the solution w(z) = zwλu (z) of the problem |w(t)| = exp λu(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 ˆτ iτ

v(eiσ )dσ

u : T → R, u(e ) = 0

Since

2π ´

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

0

differentiable. The solutions to (3.2) whose only zero is a simple zero at 0 are then given by z 7→ cz exp(Sλu (z)) for arbitrary c ∈ T. The location and multiplicity of critical points of the solution does not depend on c, so w.l.o.g. we investigate wλ (z) = z exp(Sλu (z)). We have 0 wλ0 (z) = exp(Sλu (z))(1 + zSλu (z)) 0 and since Sλu (z) = λ Sviz(z) by Lemma 3.15 we obtain

wλ0 (z) = exp(Sλu (z))(1 + λif (z))

3.4 Obtaining the Hilbert transform from a Riemann-Hilbert problem

33

where f is defined as in the first paragraph. Now choose z ∈ (−1, 0) such that f (z) = i λ1 . This is possible due to the construction of v. Then wλ0 (z) = 0. So, regardless of the choice of λ, the solution always has a critical point in D. 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 such that (HRHP) has no continuous solution. Let u be the function constructed in the first paragraph. Let w be a corresponding solution of (HRHP). With the same reasoning as in the proof of Theorem 3.24, we can assume that w is zero-free and conclude that Im f − Im w is constant, where f is the conformal map constructed in the first 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 discontinuous and merely assume its boundedness, the next statement to fail is Proposition 3.26. This is due to the existence of so-called singular inner functions. z+1 The function f (z) = z−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)| = 1 for t ∈ T \ {1}. Thus, w is a bounded discontinuous solution of (HRHP) with u ≡ 0. Following Proposition 3.26, we would expect that for each Hu = Im f + c for some constant c. But this is clearly wrong since Hu ≡ 0 and Im f is not constant. No solutions in H q for u ∈ Lp . Finally, if we even allowed u to be unbounded, (HRHP) completely fails to be useful since it may have no solution in any of the H p spaces, assuming that we understand the problem in the sense 1√ of nontangential limits. As an example, consider the function u(eiφ ) = p+1 φ p for φ ∈ (0, 2π]. It is straightforward to check that u ∈ L . Assume w were a corresponding solution of (HRHP) in H q for q > 0. Then, ˆ2π ˆ2π q q q iφ √ kwkq = w(e ) dφ = exp p+1 dφ φ 0

0

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

34

3 The Continuous Setting

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

9 These

spaces were not introduced in this book but can be defined in a way similar to the other H p 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 circle packings. Ken Stephenson’s Book [27] is the standard volume on circle packings to date. Nevertheless, our work only requires few basics and some important theorems from this book, since we mainly concentrate on the manifold structure 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’s software CirclePack [28]. The figures were plotted in MATLAB and exported using matlab2tikz [26].

4.1 First examples and ideas Before introducing circle packings on a formal mathematical level, we will start with just showing some pictures and present the core ideas on an informal level. A circle packing might look like this:

Figure 4.1: A circle packing © Springer Fachmedien Wiesbaden GmbH 2017 D. Volland, A Discrete Hilbert Transform with Circle Packings, BestMasters, https://doi.org/10.1007/978-3-658-20457-0_4

36

4 Circle Packings

As one can see, a circle packing is a configuration of circles some of which are 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. As we can see, this is a graph which triangulates a topological closed disc. These graphs 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 several different circle packings having this tangency pattern. As an example, Figure 4.2b shows a second circle packing based on the same graph as the packing in Figure 4.1. Describing the set of all circle packings on a given graph will be the topic of Section 4.3. It will turn out that circle packings actually form a smooth manifold.

(a) The packing form Fig(b) A second packing with ure 4.1 with its comthe same complex plex Figure 4.2: Packings and complexes

In the context of discretizing complex analysis, circle packings should be thought of as a discretization of a domain in C. Now, assume that we are given two circle packings which have the same combinatorics, i.e. the same underlying tangency pattern. Thus, we have discretized two complex domains using the same combinatorial structure. Now, the idea is to understand these two packings as a map between the two domains. For example, the packings 1 Circle

packings are based on tangencies. This is not the only approach to discretize complex analysis involving circles. E.g., the books of Bobenko and Suris [5] and [4] contain works on discrete complex analysis using patterns of mutually intersecting circles. In this approach, cross ratios of intersection points and intersection angles play 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 map from 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 of circle packings: Our discrete holomorphic functions are no mappings in the classical mathematical sense, but they are simply ordered pairs of two circle packings. However, this notion gives us what we are looking for: a discrete structure relating two objects in the complex plane and identifying circles with other circles. Based on this notion, we will translate boundary value problems for holomorphic functions into boundary value problems for circle packings. We will 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 in circle packing. Evidently, circle packings are nonlinear objects: they can’t be “added” like holomorphic functions. This is problematic if we intend to discretize a linear operator in the context of circle packings. Knowing that circle packings form a manifold, we can use the tangent space of this manifold as a substitute. This is our main motivation to investigate the structure of the set of all circle packings in some detail. Before closing this section, we will show two more pictures which allow us to discover another important aspect of circle packings:

38

4 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 discrete map having this packing as its image packing would be the discrete version of a holomorphic function which is not injective. Nevertheless, this packing exhibits a structure very similar to that of the circle packings we saw before. However, a circle packing may also look like that in Figure 4.5.

Figure 4.5: A branched packing

2 This

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

4.2 Basic definitions

39

What happens here? In all packings seen before, every interior circle of the packing was surrounded by a chain of neighbors winding once around it. In Figure 4.5, the boundary circles wind twice around the shaded central circle. The coloring of the packing aids in observing this. This behavior reminds us of saddle points of analytic functions: Where the derivative of a function vanishes, its values are winding multiple times around the value at the saddle point which can be seen by considering its Taylor expansion. Thus, packings like 4.5 help us to discretize holomorphic functions whose derivative vanishes. The packings seen in the first examples were univalent packings. The snake 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 deformed into each other, a univalent packing can never be continuously deformed into a branched packing3 . This will turn out to be important in the analysis of the manifold of circle packings and also has consequences for algorithmic aspects.

4.2 Basic definitions 4.2.1 Complex The topological structures our circle packings are based on are simplical 2-complexes which triangulate an orientated closed disc. An example for such a complex is shown in Figure 4.6. For our work, it is not necessary to discuss the combinatorics and topology of these structures in detail. The details are worked out in the third chapter of Stephenson’s book [27]. We will just gather the properties these complexes exhibit in the following. We will usually denote a complex by K. A complex consists of a set of vertices V , a set of undirected edges E and a set of oriented triangular faces F . As can be observed in the example in Figure 4.6, these objects satisfy the following: • An edge e = {v, w} ∈ E connects two vertices v and w. In this case, the vertices v and w are called adjacent. 3 In

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

40

4 Circle Packings

Figure 4.6: A complex with nodes, edges and faces (slightly shaded). Boundary nodes are filled, interior nodes not. Boundary edges are in dark gray, while interior edges are in light gray. The black arrows indicate the orientation of the faces.

• A face f consists of three mutually adjacent vertices u, v, w. It has a prescribed orientation, which we denote by writing f = hu, v, wi = hv, w, ui = hw, u, vi. 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 is called a boundary edge in the first case and an interior edge in the latter case. If e is an interior edge, then its orientation with respect to its two faces is different. • A vertex is called a boundary vertex if it is incident to a boundary edge and interior vertex otherwise. The boundary vertices v1 , . . . , vm can be put in a cyclical order such that every vertex is adjacent to its predecessor and successor. • Let n = |V | ≥ 3 be the number of vertices and m the number of boundary vertices. 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 faces and all faces are triangles, 2 |E| − m = 3 |F |. Inserting this into the above equation yields the fundamental relation |E| = 3n − m − 3 which is crucial for everything we do in the following.

4.2 Basic definitions

41

• For a vertex v, its neighbors can be put into an order v1 , . . . , vk such that for all j ∈ {1, . . . , k − 1}, vj is adjacent to vj+1 . If v is an interior vertex, 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 the degree of v. Standard examples for such complexes are regular polygonal complexes [n] with k generations. They are denoted by Kk . Their construction can [n] roughly be described as follows: K1 consists of one interior vertex and n boundary vertices. All boundary vertices are adjacent to the interior [n] [n] vertex. Kk+1 is obtained from Kk by adding vertices in such a way that [n]

all boundary vertices of Kk 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 new vertex on each edge incident to the face. The process is illustrated in Figure 4.7b.

[7]

(a) Complexes K1 (blue), [7] K2 (blue and green), [7] K3 (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 fixed complex K = (V, E, F ). It is important to note that almost everything we define in the following, like manifolds, contact and angle functions, and packings, depends on K. To keep the notation simple, K will not appear in the notation we use for these objects, though. We use the following notations for K.

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4 Circle Packings

Notation 4.1. K has n vertices that we denote by V = {v1 , . . . , vn }. m of these vertices are boundary vertices. The vertices are arranged in such a way that {v1 , . . . , vm } are the boundary vertices of K in cyclical order. Thus, vj is 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 neighbors of vj arranged in such a way that for each k ∈ {1, . . . , d − 1}, vj,k and vj,k+1 are adjacent and hvj , vj,k , vj,k+1 i ∈ F . If vj is an interior vertex, the same holds for vj,d and vj,1 . To simplify indexing, we also set vj,0 := vj,d . This notation is a bit messy due to the double index, but we will use it rarely. It will turn out useful to prevent renumbering the vertices when discussing the flower 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 radius rj > 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 a positively oriented triangle if hvj , vk , vl i ∈ F . C1 , . . . , Cm are called the boundary circles of P while the remaining circles are called interior circles. A circle packing is called univalent if for any j 6= k, int Cj and int Ck are disjoint. Remark 4.3. • We do not allow generalizations of circles in our definition, e.g. M¨ obius circles or points (as circles with radius zero). This is essential for some of the theorems we will discuss in the following section. • It is also possible to consider circle packings on the sphere or in the hyperbolic plane instead of the euclidean plane C. Since we will be working with equations imposed onto the euclidean centers and radii of a circle packing, we will completely stay in the euclidean setting.

4.2 Basic definitions

43

Before continuing, we remark that (C3) can be formulated differently, which will turn out very convenient later. Consider a face f = hva , vb , vc i ∈ F , compare Figure 4.8. Let za0 , zb0 , zc0 be the contact points of Cb and Cc resp. Cc and Ca resp. Ca and Cb . These points are not collinear, so let C 0 be the unique circle passing through them. The circle will be called the dual circle of P corresponding to f and will be denoted by Cf henceforth.

za zc0 zb zb0 za0 zc

Figure 4.8: A face and its face circle

As is evident from Figure 4.8, Cf is actually the inscribed circle of the triangle formed by za , zb and zc .4 This triangle is positively oriented iff za0 , zb0 and zc0 lie in counter-clockwise order on Cf . Thus, (C3) is equivalent to requiring that the contact points of Cj , Ck and Cl lie in counter-clockwise order on the unique circle passing through them. Definition 4.2 is purely geometric. This is the usual way to define circle packings which can be found e.g. in Ken Stephenson’s book [27]. For our purposes it is not sufficient, though. We intend to investigate the structure of the set of all circle packings on K, therefore, we need a more algebraic definition. We thus describe a circle packing by its most natural coordinates: The radii rj and the centers zj of its circles, j ∈ {1, . . . , n}. Writing zj = xj + iyj with real xj and yj , we see that a circle packing can be specified by a vector 4 This

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

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4 Circle Packings

of 3n real coordinates, namely, the rj , xj and yj . This motivates to introduce the vector form of a packing. Definition 4.4 (Vector form). Let P = {C1 , . . . , Cn } be a circle packing and for 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 packings on 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 to the vector form of a circle packing. We will also abuse notation and denote a packing and its vector form by the same variable. I.e., we will speak of the circle packing P = (r1 , . . . , rn , x1 , . . . , xn , y1 , . . . , yn )T . Sometimes, we will decompose the vector form into the three vectors r ∈ Rn>0 and x, y ∈ Rn and indicate this by writing P = (r, x, y). The label of a packing plays a special role in the general theory of circle packings. 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 is the statement of Stephenson’s Monodromy Theorem (Theorem 5.4 in [27]). When working in the hyperbolic plane or on the sphere, it is convenient to work with the radii alone to avoid using spherical or hyperbolic coordinates. However, the boundary value problems discussed in this book involve the centers of a circle packing. For this reason, the label of a packing alone will hardly be used in the following.

4.3 Manifold structure In this section, we will introduce the prerequisites for linearization in the context of circle packings. To do so, we shall now investigate the structure of D, i.e. the set of all circle packings. The core result of this section states that D is an (m + 3)-dimensional real submanifold of R3n consisting of a finite number of connected components. The paper [3] gives two different proofs of this result: The first one shows that the set of all packing labels D∗ is a manifold, from which it is easy to deduce that D is a manifold, too. The second one identifies D as the zero set of the so-called contact function. We will stick to the latter proof in the following, because this proof gives important extra information for boundary value 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 radii r1 , r2 and centers z1 , z2 , respectively, it is easy to verify that these circles will 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 ((xk − xl )2 + (yk − yl )2 − (rk + rl )2 ) 2

and ω = (ω1 , . . . , ω3n−m−3 ). ω : R3n → R3n−m−3 is called the contact function 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 calculate derivatives of ω. For (C3), note that three points z1 , z2 , z3 ∈ C form a positively oriented triangle iff z3 − z1 Λ(z1 , z2 , z3 ) = Im > 0. z2 − z1 Since Λ : C3 \ {(z1 , z2 , z3 )|z1 = z2 } → R is obviously continuous, the preimage of (0, ∞) under Λ is open in C3 ∼ = R6 . With these preparations, we arrive at the following lemma.

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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 = hvj , vk , vl i ∈ F , write Uf = {(r, x, y) ∈ R3n |Λ(xj + iyj , xk + iyk , xl + iyl ) > 0} T and define U := Uf ∩ (Rn>0 × R2n ). Since the intersection is over finitely f ∈F

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 and it 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 function has maximal rank at every point on D. This is the content of the following crucial 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 for the kernel of M is given by 0 0 0 1n , 0 , −y ∈ Rn−m × Rn × Rn . 0 1n x 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 and Wegert. Combining the previous lemmas, we have the following theorem. Theorem 4.8. D is a smooth manifold of dimension m + 3. Its tangent space coincides with the kernel of Dω. The importance of Lemma 4.7 goes beyond this theorem. It tells us that the contact function is regular at each circle packing and also allows us to construct a basis of its kernel. This is important for the linearization of boundary value problems.

4.3.2 Angle sums and branch structures The goal of this subsection is a closer analysis of the structure of D. Having seen that D is a manifold of real dimension m + 3, we will now investigate its connectedness and how to parametrize the manifold.

4.3 Manifold structure

47

In Section 4.1, we saw an example of a branched circle packing. We will start with introducing angle sum maps that help us to formalize this concept. By elementary geometry, the angles of a triangle can be computed from its side lengths. If a triangle has side lengths a, b and c, the angle α between the sides with length b and c is given by b2 + c 2 − a 2 . 2bc Given a circle packing P = (r, x, y), we are interested in computing the angles of those triangles formed by three circle centers which correspond to faces of K. Now, notice that this can be done in two completely independent ways. For a face hvj , vk , vl i, the angle α of the triangle zj , zk , zl at zj satisfies arccos

cos α =

(rj + rk )2 + (rj + rl )2 − (rk + rl )2 2(rj + rk )(rj + rl )

=

|zj − zk | + |zj − zl | − |zk − zl | . 2 |zj − zk | |zj − zj |

2

2

2

The equality of the two expressions is geometrically evident and also follows directly from ω(P) = 0. If Λ(zj , zk , zl ) > 0, direct computation confirms z −z that α is actually the argument of the complex number zkl −zjj , i.e. zl − zj = reiα zk − zj

(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 this vertex. Definition 4.9 (Angle sum maps). The function Φ : Rn>0 → Rn , Φ = (Φ1 , . . . , Φn ) with deg Xvk (rk + rk,j−1 )2 + (rk + rk,j )2 − (rk,j−1 + rk,j )2 Φk (r) = arccos 2(rk + rk,j−1 )(rk + rk,j ) j=1 e = (Φm+1 , . . . , Φn ) is called interior radial is called radial angle sum map. Φ angle sum map. Define E := {z ∈ Cn |∃vj , vk ∈ E such that zj = zk }. Then, the function Ψ : Cn \ E → Rn , Ψ = (Ψ1 , . . . , Ψn ) with ! deg 2 2 2 Xvk |zk − zk,j−1 | + |zk − zk,j | − |zk,j−1 − zk,j | Ψk (z) = arccos 2 |zk − zk,j−1 | |zk − zk,j | j=1

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4 Circle Packings

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 Φ immediately yields that Φ(P) = Ψ(P). Let z ∈ Cn such that for every face hvj , vk , vl i ∈ F we have Λ(zj , zk , zl ) > 0. Then (4.1) implies that Ψk (z) is a positive integer multiple of 2π for every k > 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 locally constant but Φk is not. This observation is a crucial ingredient for the proof of Lemma 4.7 and also the main reason for us to bother introducing both angle 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 which is unique up to rigid motions. This and (4.2) are the statements of Ken Stephenson’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 classes as done in the following definition. Definition 4.10 (Branch structure). Let P = (r, x, y) be a circle packing. (r) 1 e • bP := 2π Φ(P) − 1n−m = ( Φm+1 − 1, . . . , Φn2π(r) − 1)T ∈ Nn−m is called 0 2π the branch structure of P. Note that bP can be computed from the radii alone, 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. In particular, D0 is the set of all locally univalent circle packings. S Thus, D may be decomposed as D = Db . The following lemma b∈Nn−m 0

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-empty iff for every closed edge path γ containing g ∈ N edges the P following holds: Denoting by Vγ the set of all nodes interior to γ, g > 2 bk−m + 2. vk ∈V

4.3 Manifold structure

49

Henceforth, any vector b ∈ Nn−m that satisfies the assumptions of Lemma 0 4.11 will be called a branch structure. Remark 4.12. We have introduced branch structures “backwards”. In Stephenson’s book, a branch structure is first defined as a vector which satisfies the assumptions of Lemma 4.11 ([27], Definition 11.4). In a second e step, it is shown that for each packing P the vector Φ(P) − 1n−m is a branch structure ([27], Lemma 11.5).

4.3.3 Parametrization of Db Φ is a continuous function everywhere on D. Thus, for b1 6= b2 , Db1 and Db2 can’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 most remarkable results in the theory of circle packings. It states that we can always find a circle packing for arbitrarily prescribed boundary circle radii and 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. There is a unique packing label r with branch structure b such that the first m entries of r are given by r1 , . . . , rm . Let P be a circle packing with packing label r whose n-th circle is centered at 0 and whose (n − 1)-th circle is centered on R+ . We henceforth assume that vn and vn−1 are adjacent. Thus, the n-th and (n − 1)-th circle of a packing can not have the same center. If P 0 is a second circle packing with packing label r, there is a plane rigid motion z 7→ λz + ζ that maps P to P 0 by the Monodromy Theorem. Let zn and zn−1 be the centers of the n-th resp. (n − 1)-th circle of P 0 . Then ζ = zn ,

λ=

zn−1 − zn |zn−1 − zn |

We conclude that a circle packing is uniquely determined by its branch −zn structure b, the radii of its boundary circles, zn and |zzn−1 , where zn and n−1 −zn | zn−1 are the centers of its n-th resp. (n − 1)-th circle. From this, we obtain a chart on Db .

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Proposition 4.14 (Theorem 5.2 and Corollary 5.3 in [3]). Let b be a branch structure. 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 regular parametrization 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. In particular, 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 a homotopy method, we need to know its branch structure in advance, since circle packings with different branch structures can not be continuously transformed into each other. This is one of the aspects that explain why we are working with a nonlinear Riemann-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 into 3 degrees for a rigid motion and m degrees for the radii of the boundary circles. In Chapter 5, we intend to discuss boundary value problems for circle packings. This means that we will stress the question how the m degrees of freedom 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 rigid motions 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 circle packing P with N (P) = 0 which additionally satisfies xn−1 > 0 is called a standard normalized circle packing.

4.3 Manifold structure

51

The circles used for normalization are termed the α-circle resp. β-circle of a packing in circle packing literature. By convention, we have arranged the 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 case where the α-circle and the β-circle are adjacent. Lemma 4.7 immediately gives Dω Lemma 4.17. The last 3n − m columns of form a regular matrix. DN Remark 4.18. Note that in fact, N does not globally eliminate all rigid motions. The rigid motion z 7→ −z maps an N -normalized circle packing to another N -normalized circle packing. That is why we additionally require xn−1 > 0 in definition 4.16. Choosing N as standard normalization has algorithmic reasons: 1. N still eliminates rigid motions locally: The set of all circle packings P 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. Recall that xn−1 = 0 can not hold for a packing with N (P) = 0. Thus, Newton’s method or homotopy methods with initial values on or near one of the components will stay near to the same component. 2. Evaluating N and its derivative is extremely cheap due to their simple structure. A function that eliminates rigid motions completely will make use of the complex argument function, thus having a significantly more difficult 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 deduced about 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 of DN . At each P ∈ DN , the tangent space satisfies Dω TP DN = ker . DN P

52

4 Circle Packings

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

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

4.4 Discrete harmonic functions on circle packings In this section, we do a little digression from our main topics and discuss discrete harmonic functions on K. The main motivation for this digression is to gain insight into the proof of Lemma 4.7. As stated in the previous section, this lemma originates from [3]. Technically, the proof we will sketch here is exactly the original proof. However, it is insightful to relate the proof to discrete harmonic functions. Before we start, a word of caution: The discrete harmonic functions discussed here massively differ from the discrete analytic functions introduced in 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 an undirected graph (of course, we will work on K), a weight is assigned to each interior edge by a function ν : E → R>0 . It is crucial that the weights are strictly positive. The functions we discuss are real (or complex) functions on the nodes of K, i.e. functions f : V → R. For such a function, we define the discrete Laplacian ∆ by X ∆f : V → R, (∆f )(v) = ν({v, w})(f (w) − f (v)) {v,w}∈E

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 functions and linear combinations of discrete harmonic functions are discrete harmonic again. Discrete harmonic functions exhibit properties similar to harmonic functions in the continuous setting. A very important property is the discrete maximum principle. Lemma 4.21 (Maximum principle for discrete harmonic functions). Let ν 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 v such that f (w) ≥ f (v). In particular, f attains its maximum at a boundary vertex. Proof. Assume the contrary, i.e. let v ∈ V be an interior vertex with f (w) < f (v) for all vertices w adjacent to v. Then X (∆f )(v) = ν({v, w})(f (w) − f (v)) < 0 | {z } {v,w∈E}

0. In this case, the neighbors of Cm+j wind (bP )j + 1 times around Cm+j . This is the discrete analogue of critical points: If a holomorphic function f has a critical point with multiplicity k at z0 , Taylor expansion yields that f (z) = f (z0 ) + (z − z0 )k+1 + O((z − z0 )k+2 ) as z → z0 . Geometrically, this means 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. em+j in the packing P e as a critical point of We therefore think of the circle C order (bP )j . This justifies the name “discrete conformal map” for functions where P is locally univalent. e and P be standard normalized Moreover, recall Definition 3.27. Let both P circle packings. The fact that the n-th circle of both packings is centered at

56

4 Circle Packings

e → P as discrete version of a function f with 0 exhibits the function f : P f (0) = 0. The fact that the (n − 1)-th circles are centered on the positive real line should be interpreted as f (ε) > 0 for small ε > 0, i.e. f 0 (0) > 0. This justifies using the term “standard normalized” in both definitions 3.27 and 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 P is 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 e are two maximal packings on K, there circle packing P on K. If P and P e In particular, is a conformal automorphism of D that maps P onto P. every maximal packing is univalent and there is a unique maximal packing PK ∈ D N . It should be remarked that the theorem is actually stated slightly different in Stephenson’s book: It merely states that there is an essentially unique univalent maximal packing. This formulation does not exclude the possible existence of packings which satisfy our definition and which are not univalent, but only locally univalent. Rereading Stephenson’s proof however reveals that he only uses the local univalence of the maximal packing to prove uniqueness. Thus, the theorem holds good in the way stated here. Theorem 4.28 implies that we could have equivalently defined maximal packings by requiring univalence instead of just local univalence. This is the usual way to define maximal packings. The advantage of our definition is that univalent packings are only characterized geometrically, while locally univalent packings can be characterized analytically. The proof given in [27] uses the Poincar´e model of hyperbolic geometry. For a description of this model, see, for example, [15]. In this setting, circles internally 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 for maximal packings is basically the same as asking for packings with prescribed boundary 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 [7] the complex K3 . Figure 4.10 shows two more maximal packings: one for [6] the complex of the packing in Figure 4.2a, and one for K3 .

(a) Maximal packing for the complex from Figure 4.2a

[6]

(b) Maximal packing for K3

Figure 4.10: Maximal packings

In the context of discrete analytic functions, we may understand a maximal e is a maximal packing, packing as a discretization of D. Consequently, if P e → P as the discrete version of a holomorphic function we understand f : P f : D → C. The uniqueness of PK makes it our canonical choice for a discretization of D. In the following, we will often identify a given circle packing P with the 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 from the 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 circle packings are, in fact, a sensible discretization of complex analysis.

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4.7.1 Discrete maximum principles The classical maximum principle for holomorphic functions states the following: Given a domain U ⊂ C and a function f holomorphic on U and continuous 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 , 1 then f (z) = 0 or f is constant (apply the maximum principle to z 7→ 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 function e → P: First, one could interpret the real and imaginary parts of the f: P centers of P as discrete Re f and Im f . In this case, Dubejko’s Theorem 4.25 implies that the discrete Re f and Im f attain their maximum and minimum at boundary circles. Second, there is a geometrically motivated discrete analogue of |f 0 |: By the Taylor expansion, |f (z + h) − f (z)| ≈ |f 0 (z)| h for small h > 0. Consequently, |f 0 | measures how much f distorts lengths. A discrete analogue e for this value is the quotient of the radii of two corresponding circles in P and P. r We therefore define f # : V → R+ by f (vj ) = ρjj , where rj is the radius e Then we of the j-th circle in P and ρj is the radius of the j-th circle in P. have the following result of Stephenson. Theorem 4.29 (Theorem 11.12 in [27]). Let f : P → P 0 be a discrete analytic function. Assume that f # is not constant. Then f # attains its maximum at a boundary vertex. If f # attains its minimum at an interior vertex vj , then P 0 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 Riemann Mapping Theorem from the following point of view. Given an arbitrary univalent circle Packing P for some given complex K, we will always find the corresponding maximal packing PK . This means, we can find a discrete conformal map f : PK → P for any packing P. This corresponds to the classical Riemann Mapping Theorem stating that given any simply connected proper subdomain G ⊂ C, there is a bijective conformal map f : D → G. This map is called Riemann map and it is unique up to conformal automorphisms of D.

4.7 Some results on discrete analytic functions

59

Based on this, it is actually possible to approximate the classical Riemann map. For sufficiently small ε > 0, one can construct a circle packing Pε such that all of its circles’ centers lie in G and all radii are equal to ε. Figure 4.11 suggests how this works. Now, for each Pε , we can compute a corresponding maximal packing, thus obtaining a discrete conformal map. We can use this map to compute an approximation fε of the Riemann map which is continuous and piecewise affine linear. We are not going to carry out the details 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 sufficiently small, let fε be the map constructed in the way sketched above5 . Then the fε converge to a bijective conformal map f : D → G as ε → 0 where the convergence is pointwise and uniform on compact subsets of D. This theorem stands out in the theory of circle packings because it is one of rather few convergence results up to now. Theorems like the discrete maximum principle indicate that the theory of circle packings provides a good discrete analogue for complex analysis. There are also lots of numerical experiments that suggest that discrete objects converge to their classical 5 Its

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

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

5 Discrete Hilbert Transform Having introduced circle packings and all technical auxiliary material, we are now in a position to address boundary value problems for circle packings. Throughout this chapter, K will continue to be a fixed complex for which we 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 freedom are directly related to the radii of the boundary circles. The fact that there are exactly m degrees of freedom suggests that it makes sense to relate each degree to one boundary circle. Now, one may ask whether they may also be related to the centers of the boundary circles instead of their radii, or to both centers and radii. This question is the starting point of this chapter: How can we find suitable conditions on the boundary circles that single out a (locally) unique circle packing? In other words: 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 discrete problems which may be an important tool to answer the question. We will revisit Riemann-Hilbert problems in this chapter since these problems fit very nicely into our discrete setting. Recall that a RiemannHilbert problem consisted of imposing one real condition on a holomorphic function for each point on the continuous boundary T of D. Accordingly, the discrete problems impose one real condition on every point of the discrete boundary of K. Consequently, we will introduce discrete Riemann-Hilbert problems. Having done this, we will use the discrete problems to finally define a discrete Hilbert transform.

© Springer Fachmedien Wiesbaden GmbH 2017 D. Volland, A Discrete Hilbert Transform with Circle Packings, BestMasters, https://doi.org/10.1007/978-3-658-20457-0_5

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5.1 Discrete boundary value problems 5.1.1 Definition and examples To begin, let us recall two boundary value problems we have already encountered in the previous chapter. Example 5.1 (Fixed radii problem). Given ρ1 , . . . , ρm , we ask for a packing P 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 packing P on K. Theorem 4.28 tells us that this problem has the unique solution PK on DN . We will now formally define discrete boundary value problems in general. Definition 5.3 (Discrete boundary value problem). Let β : R3n → Rm be a function 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 discrete boundary value problem. A circle packing P ∈ DN is a solution to the problem if β(P) = 0. Equivalently, a vector P ∈ UN is the vector form of a solution to the problem iff ω(P ) N (P ) = 0. β(P ) It is not difficult to translate the two standard examples for discrete boundary 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 (x + yj2 − (1 − rj )2 ) 2 j

(MAX-PACK)

defines the maximal packing problem. Let P be a circle packing. If P is a zero of (FIXED-RAD), it is clear that P actually solves the fixed radii problem as stated in Example 5.1. This is not that evident for (MAX-PACK), though. We will therefore give 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 every zero P ∈ UN of (ω, N, β) is the vector form of an N -normalized maximal packing. Proof. Since P ∈ UN and ω(P ) = 0, there is a circle packing P with vector form P . Since N (P ) = 0, it is N -normalized and again since P ∈ UN , it is also 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 that rj < 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 that the following equations hold: 2

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

|zj | = (1 − rj )2 2

|zk | = (1 − rk )2 Since P ∈ UN , rj + rk > 0 holds. By assumption, rj − 1 ≥ 0. Taking square roots in the first two equations gives |zj + zk | = rj + rk |zj | = rj − 1 which implies rk + 1 = |zj + zk | − |zj | ≤ |zk | by the triangle inequality. Inserting this into the last equation we obtain (rk + 1)2 ≤ (rk − 1)2 .

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

(5.1)

for all j ∈ {1, . . . , m}, implying that all boundary circles are contained in D and tangent to T. It remains to show that all interior circles of P are contained in D. Let j ∈ {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 attain its 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 in D. Remark 5.5. The last argument is strongly reminiscent of discrete harmonic functions: If f : V 7→ R were discrete harmonic, we could use the maximum principle and get done immediately. Unfortunately, Dubejko’s theorem does not apply to the f appearing in this proof. Nevertheless, this proof encourages to keep discrete harmonic functions in mind when one wants to prove geometric and algebraic properties of solutions to discrete boundary value problems. We shall now state three examples for boundary value problems which are not well-posed. One of them has no solution on DN , another one has no solution 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 + yj2 − (1 − rj )2 . This problem is similar to the maximal packing problem, but in this case, the βj 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-th circle 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 the center of its n-th circle, i.e. zn = 0. Let P 0 be the packing obtained from P by adding 2 to the real part of each center. Then P 0 solves (MAX-PACK). By the proof of Proposition 5.4, the absolute value of its circle centers is bounded by one. Hence zn + 2 ≤ 1, which is a contradiction.

5.1 Discrete boundary value problems

65

Example 5.7 (No solution). Constructing a problem which has no solution at all is similarly easy. Intuitively speaking, we can just take two adjacent boundary vertices and force their circles to lie in disjoint closed domains. In this case, they can not be tangent and satisfy the boundary value problem at the 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) = 0 implies 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 circle packing. e be an arbitrary circle Example 5.8 (No locally unique solution). Let P packing and for j ∈ {1, . . . , m} let ξj + iνj be the center of its j-th boundary circle. Define βj (rj , xj , yj ) := νj xj − ξj yj . Direct computation gives that if P is a solution to the corresponding problem, e is a solution then λP is a solution for every λ > 0, too. In particular λP for every λ > 0. Thus, this problem has solutions, but none of them can be locally unique. Let us now discuss how to discretize a Riemann-Hilbert problem with circle packings. Let Mt , t ∈ T be the target curves of a Riemann-Hilbert problem. For j ∈ {1, . . . , m}, let tj be the contact point of the j-th circle of 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 intuitive idea of the discretization: The circle tangent to T at tj is mapped to a circle tangent to Mtj . Analogously, a continuous solution of the original problem would map the point tj onto the curve Mtj . A function that fits this idea would be β : R3n → Rm defined by 2

βj (rj , xj , yj ) := min{|xj + iyj − z| |z ∈ Mtj } − rj2 This is a very general way to express the boundary condition. Depending on the structure of the Mtj , the functions βj may also lack differentiability. We would only use this if we have no additional information about the curves Mtj . In the following we will confine ourselves to defining discrete circular Riemann-Hilbert problems. These are discrete analogues of the problems discussed in Subsection 3.3.2.

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Definition 5.9. For each j ∈ {1, . . . , n}, let Mj be a circle with center cj and radius ρj . The corresponding discrete circular Riemann-Hilbert problem is the problem defined by the functions βj (rj , xj , yj ) :=

1 2 |xj + iyj − cj | − (ρj − rj )2 . 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 the j-th circle of the solution. Figure 5.1 shows an example. The maximal packing is shown in figure 5.1a with colored boundary circles. The solution is shown in figure 5.1a. For each boundary circle, the corresponding target circle is drawn transparently with a dashed line of the same color. Near the tangency, an arc is drawn clearly to emphasize the tangency.

[9]

(a) Maximal packing on K1

(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 a problem of this kind to discretize (HRHP) and define the discrete Hilbert transform. There are some first investigations on solvability properties of this kind of problems which can be found in a paper by Wegert and Bauer [33]. They present the set of solutions for the case where K is a flower.

5.1 Discrete boundary value problems

67

5.1.2 Linearization of boundary value problems In general, proving results on existence and uniqueness of solutions to discrete problems is very difficult. Nevertheless, we can gain some insight into these questions via linearization. If a packing P is a solution of a given problem, we can investigate the structure of the matrix Dω DN . Dβ P Essentially, this means that we consider the linearized version of the boundary value problem as a problem on the tangent space TP DN . If the above matrix is regular, the Implicit Function Theorem gives us that the solution is locally unique. The boundary value problems on the tangent space of the circle packing manifold were called incremental boundary value problems by Wegert. We will define them now. Definition 5.10 (Incremental boundary value problem). Let P ∈ DN be a circle packing. Let c ∈ Rm and M = (Mjk )jk ∈ Rm×3n , such that all entries of M are zero except Mjk for j ∈ {1, . . . , m} and k ∈ {j, n + j, 2n + j}. M and c define an incremental boundary value problem at P. A tangent vector p ∈ TP DN 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 as Dω 0 DN · p = 0 . M P c Thus, an incremental problem is well-posed iff the matrix on the left-hand side is regular. We are mainly interested in incremental problems that arise as linearization of boundary value problems for circle packing. The linearization of a problem defined by a differentiable function β at a circle packing P is given as Dω 0 DN · p = 0 . (5.2) Dβ P −β(P)

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If P is a solution to the problem, the regularity of the matrix gives us information about the solution’s local uniqueness. It is straightforward to check that the linearization of a discrete circular Riemann-Hilbert problem yields 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 is regular for every P. Proof. By Proposition 4.20, the last 3n − m columns of Dω DN have full rank. The first m columns of Dβ equalthe m × m unit matrix and Dω all other entries are equal to zero. Thus, DN has full rank. Dβ So for the fixed radii problem, the regularity of the matrix is an immediate consequence of Lemma 4.7. The linearization of the maximal packing problem is regular, too. This has been conjectured by Wegert and could be proven by the author in 2017. The proof 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 extended circle packing. After all, the unit circle T is a circle itself and all boundary circles 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 the contact 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 set xn+1 = yn+1 = 0 and rn+1 = −1, then βj (rj , xj , yj ) = 0 becomes precisely the contact equation for the vertices vn+1 and vj .

5.2 Proof of the maximal packing conjecture

69

e 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 e for a moment. precise definition of K e Consider the Let PK = (r, x, y) and let ω e be the contact function of K. e = (r1 , . . . , rn , −1, x1 , . . . , xn , 0, y1 , . . . , yn , 0) on K. e The extended Packing P (3n−3)×(3n+3) e Jacobian of ω e at P is a matrix in R . We will henceforth denote e If the rank of this matrix is maximal, the desired result this matrix by J. can be deduced with little extra work. e contains a circle whose radius is Unfortunately the extended packing P formally negative. This fact prohibits to copy the proof of Lemma 4.7 and requires a second idea. This idea is to make use of a M¨obius transformation. Fix a point z∞ ∈ D lying in the interstice of two adjacent boundary circles of PK and T as shown 1 in figure 5.2a. If we apply the M¨obius transformation T (z) = z−z , the ∞ image of PK under T will be a circle packing again.

0=T (∞)

z∞ =T −1 (∞)

(a) Maximal packing

e0 (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 the images of the boundary circles of PK under T are now externally tangent e under T to be seen in figure 5.2b looks to T (T). In short, the image of P e in an like a completely normal circle packing! In fact we will now define K

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e under T is a circle packing appropriate way and show that the image of P indeed.

5.2.1 The transformed packing We define Ve := V ∪ {vn+1 }, such that there is a new vertex in the extended graph that T will correspond to. This vertex should be adjacent to all e = E ∪ {vn+1 , v1 } ∪ . . . ∪ {vn+1 , vm }. boundary vertices of K, therefore E We will place z∞ in the interstice of T and the 1st and m-th circle of PK . Thus, vn+1 , v1 and vm should be the boundary vertices of the extended complex, as figure 5.2 suggests. We therefore define Fe = F ∪ {hvn+1 , v2 , v1 i, hvn+1 , v3 , v2 i, . . . , hvn+1 , vm , vm−1 i}. These faces make the e Consequently, the vertices edges {v1 , v2 }, . . . , {vm−1 , vm } interior edges of K. v2 , . . . , vm−1 become interior vertices. Lemma 5.12. Let C1 , . . . , Cn be the circles of PK . Chose z∞ ∈ D as 1 described above and define T (z) = z−z . Furthermore, write Cn+1 := T. ∞ 0 e e Then P = {T (C1 ), . . . , T (Cn+1 )} is a circle packing on K. e0 satisfies the definition of circle packings. Proof. We check whether P (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 int Cn+1 but not in int Ck . Thus the internal tangency of Cn+1 and Cj results in an external tangency of T (Cn+1 ) and T (Ck ). In both cases, T (Cj ) and T (Ck ) are externally tangent. (C3) Let f = hva , vb , vc i ∈ 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 face circle Cf (colored red in figure 5.3). Now z∞ does not lie in int Cf ., thus, the contact points of T (Ca ), T (Cb ) and T (Cc ) lie on T (Cf ) in the same order. e0 . Consequently, (C3) is satisfied for the face f in the packing P 0 Otherwise let f = hvn+1 , vj+1 , vj i. 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 0 lie in counter-clockwise order

5.2 Proof of the maximal packing conjecture

71

on the circle passing through them (colored blue in figure 5.3). We can also see that z∞ lies outside this circle. Thus, the same argument as before e0 . applies, 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

e0 , Lemma 4.7 applies. We conclude that ω e0 ) = 0 Now, for the packing P e (P and De ω |Pe0 has maximal rank. We will use this to show that De ω |Pe has full rank, too. The strategy is as follows: By construction, T −1 (z) = z1 + z∞ e0 to P. e We will first introduce the function Υ such that Υ(P e0 ) is the maps P 1 0 e image of P under z . We then show Theorem 5.15

Lemma 4.7 ⇒ Regularity of De ω |Pe0 =========⇒ Regularity of De ω |Υ(Pe0 ) Corollary 5.16

=========⇒ Regularity of De ω |Pe

5.2.2 Differential of ω e at the transformed packing e0 = (r, x, y), i.e. r, x and y will denote the Throughout this subsection, P e0 . To shorten notation, we write n components of P e := n + 1. The first thing to do is examining how a M¨ obius transformation affects e0 . It is clear how a linear function affects a circle’s the radii and centers of P

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center and radius, but this is not trivial for the inversion z 7→ z1 . 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 2 C. Then expanding the equation |z − z ∗ | = r2 yields 2

2

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

Since 0 ∈ / C, we have z 6= 0 and |z ∗ | − r2 6= 0. Hence, dividing by 2 2 (|z ∗ | − r2 ) |z| yields 1 2 |z ∗ |

− r2

−

1 1 z∗ 1 z∗ − + 2 = 0. 2 2 z |z ∗ | − r2 z |z ∗ | − r2 |z|

Expanding the first summand as 1 2

|z ∗ | − r2

2

=

|z ∗ | − r2 2

(|z ∗ | − r2 )2

=

|z ∗ |

2

2

(|z ∗ | − r2 )2

−

r2 2

(|z ∗ | − r2 )2

allows us to factorize again to obtain z∗ 1 − 2 z |z ∗ | − r2

!

1 z∗ − 2 z |z ∗ | − r2

! =

!2

r 2

|z ∗ | − r2

which is equivalent to 1 z∗ r . − = 2 2 ∗ 2 z ∗ |z | − r |z | − r2 Hence, the image of C under the inversion has radius z∗ ∗ 2

|z | −r 2

r

||z∗ |2 −r2 |

and center

. 2

If now 0 ∈ ext C, then |z ∗ | − r2 > 0, so the modulus in the radius can be dropped. If 0 ∈ int C, however, dropping the modulus changes the sign of the 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 e0 is a “usual” circle packing with only exterior tangencies. radius, since P This observation motivates to simply ignore the modulus that appears in the last equation. We arrive at the following definition.

5.2 Proof of the maximal packing conjecture

73

Definition 5.13. Let UΥ := {(r, x, y) ∈ R3 |x2 + y 2 − r2 6= 0}. We define the function Υ : UΥ → R3 by Υ(r, x, y) =

1 (r, x, −y) x2 + y 2 − r 2

(5.3)

To compute the radii and centers of the image of a circle packing under the map z 7→ z1 , we extend Υ to R3en in a natural way: For r, x, y ∈ Rne with (rj , xj , yj ) ∈ UΥ ∀j ∈ {1, . . . , n e} we write Υ(r1 , . . . , rne , x1 , . . . , xne , y1 , . . . , yne ) = (ρ1 , . . . , ρne , ξ1 , . . . , ξne , ν1 , . . . , νne ) 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 the regular matrix 2 x + y2 + r2 −2rx −2ry 1 2rx −r2 − x2 + y 2 −2xy DΥ = 2 (x2 + y 2 − r2 ) −2ry 2xy r 2 − x2 + y 2 (5.4) Proof. The formula is obtained by computation. For the regularity, note that by construction, Υ ◦ Υ is the identity on the open set UΥ . Thus DΥ|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 R3en . e the derivatives of ω e0 are given by Now, for el = {vj , vk } ∈ E, el at P d rj ω el (r, x, y) = −(rj + rk ),

d rk ω el (r, x, y) = −(rk + rj ),

(5.5)

dxj ω el (r, x, y) = (xj − xk ),

dxk ω el (r, x, y) = (xk − xj ),

(5.6)

d yj ω el (r, x, y) = (yj − yk ),

d yk ω el (r, x, y) = (yk − yj ).

(5.7)

All other derivatives are equal to 0. Thus, the l-th line of De ω |Pe0 has the following nonzero entries: −rj − rk , −rk − rj , xj − xk , xk − xj , yj − yk , yk − yj located in the columns j, k, j + n e, k + n e, j + 2e n and k + 2e n, respectively.

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5 Discrete Hilbert Transform

Theorem 5.15. For j = {1, . . . , n e}, write aj := x2j + yj2 − rj2 . Note that 0 e . Let A ∈ R3n−3×3n−3 be the diagonal matrix aj = 6 0 by construction of P whose l-th entry is given by aj ak , where el = {vj , vk }. Then A · De ω |Υ(Pe0 ) · DΥ|Pe0 = De ω |Pe0 . In particular, De ω |Υ(Pe0 ) and De ω |Pe0 have the same rank. Proof. Fix l ∈ {1, . . . , 3n − 3} and let el = {vj , vk }. The l-th line of the matrix on the left-hand side is given by aj ak De ωl |Υ(Pe0 ) · DΥ|Pe0 The non-zero entries of De ωl |Υ(Pe0 ) are given by −

rj rk + aj ak

,−

rj rk + aj ak

,

xj xk xk x j yk yj yj yk − , − , − and − aj ak ak aj ak aj aj ak

at positions j, k, n e + j, n e + k, 2e n + j and 2e n + k, respectively. By definition of the extension of Υ, the non-zero entries of the j-th, (e n + j)th and (2e n + j)-th row of DΥ are at positions j, n e + j, and 2e n + j and given by formula (5.4). Consequently, the nonzero entries of aj ak De ωl |Υ(Pe0 ) · DΥ|Pe0 are at positions j, k, n e + j, n e + k, 2e n + j and 2e n + k, too. We can now compute the matrix product. Since the factor aakj appears in all entries, we will leave this factor out to make the formulas shorter. The entries of aj ak De ωl |Υ(Pe0 ) · DΥ|Pe0 divided by aakj at positions j, n e + j, and 2e n + j are given by rj rk xj xk yj yk 2 2 2 − + (rj + xj + yj ) + − (2rj xj ) − − (−2rj yj ) aj ak aj ak aj ak (5.8) rk xj xk yj yk rj + (2rj xj ) + − (−rj2 − x2j + yj2 ) − − (2xj yj ) aj ak aj ak aj ak (5.9) rj rk xj xk yj yk + (2rj xj ) − − (2xj yj ) + − (rj2 − x2j + yj2 ). aj ak aj ak aj ak (5.10) The entries at positions k, n e + k, and 2e n + k are obtained by swapping j and k.

5.2 Proof of the maximal packing conjecture

75

It is now just a matter of rearranging to show that these terms coincide with the corresponding entries of De ωl |Pe0 . We will work this out in the following. The crucial step in all rearrangements is the identity aj + ak = r j rk + x j x k + y j y k , 2

(5.11)

e0 ) = 0. The term (5.8) is equal to which is a direct consequence of ωl (P 1 −(rj ak + rk aj )(2x2j + 2yj2 − aj ) + (xj ak − xk aj )(2rj xj ) a2j +(yj ak − yk aj )(2rj yj )) 1 = 2 rj aj ak + rk a2j − 2aj ( rk x2j + rk yj2 +rj xj xk + rj yj yk )) aj {z } | =rk (aj +rj2 )

=

1 (rj ak + rk aj − 2(rk aj + rk rj2 + rj xj xk + rj yj yk ) ) aj {z } | =rj

aj +ak 2

by (5.11)

1 = (rj ak + rk aj − 2rk aj − rj aj − rj ak ) = −(rj + rk ), aj which is the corresponding entry of Dωl |Pe0 as claimed. The term (5.9) is equal to 1 −(rj ak + rk aj )(2rj xj ) + (xj ak − xk aj )(−aj − 2rj2 + 2yj2 ) a2j −(yj ak − yk aj )(2xj yj )) 1 = 2 −xj aj ak + xk a2j + 2aj ( xk rj2 − xk yj2 +xj rj rk + xj yj yk )) aj | {z } =xk (x2j −aj )

=

1 (−xj ak + xk aj + 2(−xk aj + xj rj rk + x2j xk + xj yj yk ) ) aj | {z } =xj

=

aj +ak 2

by (5.11)

1 (−xj ak + xk aj − 2xk aj + xj aj + xj ak ) = xj − xk , aj

which is again the corresponding entry of Dωl |Pe0 . For the third term, it suffices to note that this term is equal to the second term when we switch xj and yj as well as xk and yk .

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5 Discrete Hilbert Transform

Corollary 5.16. Je has maximal rank 3n − 3. ω |Pe0 has maximal rank. By Proof. By Lemma 4.7 and Lemma 5.12, De e is obtained Theorem 5.15, De ω |Υ(Pe0 ) has full rank, too. By construction, P e0 ) by adding z∞ to all circle centers. From the equations (5.5)-(5.7), from Υ(P e which we see that this does not change De ω . Thus De ω e0 = De ω e = J, Υ(P )

P

proves the corollary.

5.2.3 Basis for the kernel of Je e i.e. P e = (r, x, y). In this subsection, r, x and y denote the components of P, e The kernel of J has a geometric interpretation. Let us rethink the geometric e every zero of ω situation: In a neighborhood of P, e is a configuration of circles where the (n + 1)-th circle is internally tangent to the circles with e into a indices 1, . . . , m. Thus, we ask ourselves how to locally transform P different configuration of this kind. e the Obviously, any maximal packing can be extended to a packing on K same way PK could. So what we can do is to perturb PK slightly in such a way that the perturbed packing is still a maximal packing. This gives 3 local degrees of freedom which arise from the conformal automorphisms of D. Recall that conformal automorphisms of D are constituted by rotations z 7→ λz for λ ∈ T and M¨ obius transformations of the form z − z0 (5.12) z 7→ 1 − zz0 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 rest of the packing. This can be done in 3 obvious ways: We can scale the entire packing by some constant factor and we can move it along the x- or y-axis. e = 0 are visualized in These six local freedoms of the equation ω e (P) figure 5.4. The black arrows show the direction in which the center of the respective circle is being moved. The greyish packing shows a slightly perturbed extended maximal packing corresponding to the freedom. Let us translate these freedoms into tangent vectors of the manifold defined e = 0 locally at P. e From geometric intuition, it is pretty clear that by ω e (P) the vectors p1 = (r, x, y),

p2 = (0, y, −x),

p3 = (0, 1, 0),

p4 = (0, 0, 1)

5.2 Proof of the maximal packing conjecture

(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

77

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5 Discrete Hilbert Transform

correspond to scaling, rotating, and euclidean motions along the axes, respectively. Determining the vectors corresponding to automorphisms of the form (5.12) requires some work, but they can still be put in closed form. For arbitrary c ∈ D, let Pc be the the packing obtained by applying the transformation z−c ec the corresponding extended T (z) = 1−zc to the circles of PK . Denote by P packing. ec can be constructed explicitly with help of The map D → R3n+3 , c 7→ P ec ) = 0 for every c ∈ D and the function Υ and is differentiable. Since ω e (P d e d e e e e P0 = P, the derivatives da Pa |0 and da Pia |0 are vectors in the kernel of J. 1 They can be computed to be given by p5 = (2rx, r2 + x2 − y 2 − 1, 2xy),

p6 = (2ry, 2xy, r2 − x2 + y 2 − 1)

where multiplication is to be understood elementwise.2 To check the linear independence of p1 , . . . , p6 we look at their entries at positions n + 1, 2n + 1, 2n + 2, 3n + 1, 3n + 2, and 3n + 3. These entries correspond to the derivatives with respect to rn+1 , xn , xn+1 , yn−1 , yn , and yn+1 . Table 5.1 lists these entries. Recall that xn = yn = 0 and xn−1 6= 0 since PK ∈ DN . Moreover, rn < 1 by Proposition 5.4 and rn+1 = −1. As we can see, the (3n + 3) × 6 matrix whose columns are equal to the vectors p1 , . . . , p6 has a regular 6 × 6 submatrix. Table 5.1: Some entries of the vectors p1 , . . . , p6 showing their linear independence

p1 p3 p4 p5 p6 p2

rn+1 −1 0 0 0 0 0

xn+1 ∗ 1 0 0 0 0

yn+1 ∗ ∗ 1 0 0 0

xn ∗ ∗ ∗ rn2 − 1 0 0

yn ∗ ∗ ∗ ∗ rn2 − 1 0

yn−1 ∗ ∗ ∗ ∗ ∗ −xn−1

Consequently, the vectors p1 , . . . , p6 are all linearly independent, so by e With this basis, we Corollary 5.16 they must form a basis of the kernel J. can easily complete the proof.

1 This 2 For

was done using Wolfram Mathematica [35]. example, rx denotes the vector (r1 x1 , . . . , rn+1 xn+1 ).

5.3 Discrete Hilbert transform

79

Theorem 5.17. Let β be given as in (MAX-PACK). Then, the matrix Dω M := DN Dβ P K

is regular. Proof. Let p = (pr , px , py ) ∈ ker M . Then p ∈ ker DN in particular. From the definition of N we conclude that the entries of p at positions 2n, 3n − 1, and 3n are zero. Dω Define pe := (pr , 0, px , 0, py , 0) ∈ R3n+3 . By construction, is the Dβ submatrix of Je which is obtained by removing its (n + 1)-th, (2n + 2)-th, Dω e and (3n + 3)-th columns. Since p ∈ ker , this implies that pe ∈ ker J. Dβ Hence, pe is a linear combination of the vectors p1 , . . . , p6 constructed before. Now pe 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 pe = 0. Thus p = 0 too, which proves the claim. We will retain the notation M for this matrix throughout the rest of this book.

5.3 Discrete Hilbert transform We have now gathered all the necessary prerequisites to introduce the discrete Hilbert transform with circle packings. In the context of discrete boundary value problems, the discrete Hilbert transform may be thought of as a very illustrative case study and also an application of Theorem 5.17. Moreover, the need of linearizing a boundary value problem occurs naturally for the Hilbert transform, thus providing another reason for studying incremental problems beside the investigation of solvability properties. Since we have already seen how to translate boundary value problems into the 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 is to be understood as a discrete function on T from now on, we will define a suitable discrete boundary value problem and solve it. Motivated by the continuous counterpart, we will extract a discrete Hilbert transform of u from the solution.

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5 Discrete Hilbert Transform

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

5.3.1 Difficulties of the Schwarz problem Recall that in Subsection 3.4.1 we shortly suggested to use the Schwarz problem as the foundation for discretization. We mentioned that this problem is not well suited for discretization with circle packings. We are now in a position to explain the problems that occur in detail. Critical points and zeros of the solution. Let u ∈ Lp be given for some p > 1. By Theorem 3.11 and Remark 3.13, the solutions of the Schwarz problem are given by z 7→ Su (z) + ic for arbitrary c ∈ R. The derivative of any of these solutions is z 7→ Su0 (z). As long as we merely know u, we have no information about the zeros of this function. Assume now that we have formulated a discrete Schwarz problem. If we have done this properly, we expect the discrete problem to have similar solvability properties. Hence, the discrete problem has one solution P, which is unique up to imaginary constants on the entire manifold Db . And we have no clue on which of the connected components it can be found! In other words, we expect that there is a unique branch structure b such that all solutions of the problem lie on Db , but we do not know b. However, we have stated in Remark 4.15 that knowing b is essential to be able to compute the packing, at least in the current state of the art. There have been some ideas on writing adaptive algorithms for the computation of circle packings that are able to “jump” between various connected components [34], but none of these ideas have been pursued consequently up to now. Even if we had a possibility to predict at least the number of critical points of Su for a given u, another problem still remains: Assume we knew that the solution to the discrete problem has e.g. exactly three branch points. n−m P Consequently, the solution can be found on Db for some b with bj = 3. j=1

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

5.3 Discrete Hilbert transform

81

Trivial case. Another problem appears when we consider the trivial case u = 0. In this case, the solutions to the Schwarz problem are then just imaginary constants. Constant functions do not have reasonable counterparts in circle packing. Since constant functions map the unit circle to a point in C, the counterpart in circle packing would be a discrete analytic function where the image packing degenerates to a point, i.e. where all circles have radius zero. We have forbidden this in the definition of circle packings since statements like Lemma 4.7 do not hold at degenerate packings. Though this problem seems to be of minor relevance at the first glance, it yields a major problem. As we have stated in Section 5.1, there are hardly any results on discrete boundary value problems up to now. In case of the discrete Hilbert transform, this can be countervailed with a simple idea: We prove the existence of solutions for u in a small neighborhood of 0 by linearizing 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 u outside 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 different point is problematic, too: If we can compute Hu in an open set that does not contain 0, then we can not use the linearity to compute Hu for any given u. So, the linearization tactic is very problematic for the Schwarz problem. Consistent discretization of boundary conditions As an example, consider the case where u(eiτ ) = cos(τ ). In this case, the solutions of the Schwarz problem are given by z 7→ z + ic. Thus, we would expect that a solution of the discrete problem is the discrete identity, i.e. the maximal packing PK itself should solve the problem. The following figure 5.5 shows an example for a maximal circle packing on a given graph. The boundary circles are shown in different colors. For each color, the figure also shows the corresponding target curve of the Schwarz problem. I.e., the circle packing shown here should be a solution to the discrete problem coming from these target curves. Evidently, it is not clear at all how the boundary condition has to be interpreted. 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 of the respective circles. Neither of these holds for the remaining four circles. In more involved examples, we may be facing unsolvable difficulties in finding a consistent way to interpret the boundary condition. If we do

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5 Discrete Hilbert Transform

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 its continuous archetype less accurately.

5.3.2 Discretization of the nonlinear problem We have already discussed the problem we will use instead in the classical setting. For convenience, we state it here again. |w(t)| = exp u(t)

(HRHP)

Recall that the target lines of this problem are circles centered at 0 with radius exp u(t). We should briefly state why (HRHP) does in fact overcome the difficulties discussed before. • We have proven that (HRHP) has a unique standard normalized solution with no critical points for u ∈ C 1+α in a C 1+α -neighborhood of 0. Thus, we may expect that the discrete counterpart has a unique solution in DN under 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 discrete counterpart is in fact the easiest discrete holomorphic function one may

5.3 Discrete Hilbert transform

83

think of. Thus, the trivial input u = 0 corresponds to the trivial discrete analytic 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 packing problem by requiring that the boundary circles of the solution are internally tangent to respective target circles. Thus, we have found a consistent way to 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 the function β u , which is given by βju (r, x, y) =

1 2 (x + yj2 + (exp uj − rj )2 ). 2 j

(DHRHP)

The entries of u should be understood as values of a function on T evaluated at 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 exp Su (z). Figure 5.6 shows an example. The objects are colored and shaded in a similar way as in Figure 5.1. However, the dashed target circles are not shown in this figure, 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) in the continuous t case if kukC α is small enough. Here, w denotes the standard normalized solution. We need a discrete interpretation for w(tj ). Since tj is the contact point of the j-th boundary circle of PK with T, it is natural to interpret the contact point of the j-th boundary circle of P with its target circle as w(tj ). If xj + iyj 6= 0, this point is given by rj xj + iyj 1 + . (5.13) |xj + iyj | This is easy to see geometrically and straightforward to check algebraically. So, assume that xj + iyj ∈ C− ∀j ∈ {1, . . . , m}. tj

(5.14)

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5 Discrete Hilbert Transform

[6]

(a) Maximal packing on K3

(b) Solution of (DHRHP) for uj = 3 cos 3tj 10

Figure 5.6: A discrete circular Riemann-Hilbert problem

The discrete version of the formula from Proposition 3.26 is then given by vj := Arg

xj + iyj tj

(5.15)

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

a(v) = v −

1 X vj . m j=1

(5.16)

On a conceptual level, we are now done: We have discretized the procedure to compute the Hilbert transform we have found in Section 3.4. Reconsidering the strategy, though, there are several problems that quickly come in sight. • However faithful discrete analytic functions may mimic their classical counterparts, we can not be sure whether the theorems proved in Section 3.4 have counterparts in the discrete setting. The problem DHRHP might have no solutions at all, no solutions on DN or its solution may not be unique. In fact, it is easy to construct an example where DHRHP can not have a solution on DN . We will construct this example below. We expect that such things will not happen if kuk is small enough, but we have to prove this.

5.3 Discrete Hilbert transform

85

• Proposition 3.26 only tells us that there is a branch of the complex argument function to obtain the Hilbert transform from the solution of the boundary value problem. In general, this branch is unknown and we can not simply use the principal branch Arg. Since formula (5.15) was derived from Arg, it should not be applied in that case. Overcoming these problems on the nonlinear level would probably require an extensive solution theory on discrete circular problems. But there is still a third problem: • The resulting transform could be nonlinear. The boundary value problem β does not depend linearly on u, nor does v depend linearly on the solution. Most importantly, circle packings are nonlinear objects themselves, as D is clearly a nonlinear manifold. We have thus discretized a linear operator with a discrete operator which is 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 kuk. I.e., there exists a well defined nonlinear discrete Hilbert operator H nl on some small open set U ⊂ Rm with 0 ∈ U and we can define the linear discrete operator as 1 H(u) = lim H nl (λu). λ→0 λ [m]

Example 5.18 (No solution on DN ). Assume K = K1 is a flower and m > 5. The idea is to make one target circle very small, while all the others are larger and of the same size. Intuitively, this forces the boundary circles to become too large to wind only around the interior vertex. once √ 2 m Define u ∈ R by u1 = ln 1 − 2 and uj = 0 for j > 1. Denote by φ the function (ra + rb )2 + (ra + rc )2 − (rb + rc )2 φ(ra , rb , rc ) = arccos 2(ra + rb )(ra + rc ) which appears in the definition of the radial angle sum function Φ. 3 The

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 value problem. However, we can not check this as long as we do not know a solution operator of β. Indeed, numerical experiments strongly indicate that the procedure above does in fact yield a nonlinear operator.

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5 Discrete Hilbert Transform

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 √ 2 tangent to the first target circle. Thus rm+1 < 1 − 2 . 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 √ √ √ ! π 2 2 2 φ 1− , , = arccos(0) = . 2 2 2 2 Consequently, Φm+1 (P) ≥

m−1 X j=2

√ √ ! π 2 2 2 φ(rm+1 , rj , rj+1 ) > 4φ 1 − , , = 4 = 2π. 2 2 2 2 √

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

5.3.3 Linearization of the discrete operator We will show the existence of a local solution operator in a neighborhood of u = 0 ∈ Rm . Though we are not able to state this operator explicitly, we can exploit the fact that we are working in finitely many dimensions and make use of the Implicit Function Theorem. Throughout the rest of this chapter, 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 lines form a diagonal matrix with entries 1 − r1 , . . . , 1 − rm and whose other entries are all equal to 0. Theorem 5.20. There is an open connected set U ⊂ Rm with 0 ∈ U and a continuously differentiable function W : U → R3n such that for each u ∈ U , W (u) is a standard normalized circle packing which solves the corresponding instance of (DHRHP) and which satisfies (5.14). The derivative of W at 0 is given by DW |0 = M −1 R

5.3 Discrete Hilbert transform

87

Proof. Define the function βe : R3n+m → R3n by ω(r, x, y) e x, y, u) = N (r, x, y) . β(r, β u (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 βe at (PK , 0) are just the matrix M defined in Theorem 5.17. By Theorem 5.17, M is regular. The Implicit Function Theorem therefore implies the existence of an open set e ⊂ Rm with 0 ∈ U e and a continuously differentiable function W : U e → R3n U e e such that (W (u), u) is a zero of β for each u ∈ U . e with 0 ∈ U such that Using the continuity of W , we may choose U ⊂ U W (U ) ⊂ UN . Then by Proposition 4.20, W (u) is a standard normalized e W (u) solves the boundary circle packing for each u ∈ U . By definition of β, value problem corresponding to β u . Again by continuity of W , we can assume 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 at 0 is given by e (P ,0) −1 · D(u) β| e (P ,0) . DW |0 = D(r,x,y) β| K K e we see that indeed DW |0 = M −1 R. Computing the derivatives of β, With the solution operator at hand, we can compute the discrete nonlinear transform from a solution in the manner described before. We therefore introduce the function h : R3n → Rm defined by hj (r1 , . . . , rn , x1 , . . . , xn , y1 , . . . , yn ) = Arg

xj + iyj ∈ (−π, π). tj

(5.17)

These preparations put us into a position to formally define the nonlinear discrete Hilbert operator. Definition 5.21. Let U ⊂ Rm be a set satisfying the conditions of Theorem 5.20 and let W be the corresponding solution operator. Let h be the function defined in (5.17) and let a be the function defined in (5.16). The discrete nonlinear Hilbert transform defined on K is the function H nl : U → Rm given by H nl (u) = a(h(W (u))). Using Arg(z) = Im Log(z), the derivatives of h are yj xj ∂ x j hj = − 2 , ∂ y j hj = 2 . 2 x j + yj xj + yj2

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5 Discrete Hilbert Transform

Consequently, the Jacobian of h at PK has the following form: 0n −Y 0n−m X 0n−m y1 yn xn 1 where X = diag( x2x+y 2 , . . . , x2 +y 2 ), Y = diag( x2 +y 2 , . . . , x2 +y 2 ) and 0n 1

1

n

n

1

1

n

n

resp. 0n−m denote the null matrix in Rm×n resp. Rm×(n−m) . To shorten notation, we will write L := Dh|PK in the following. Finally, note that the function a can be written as a(z) = Az where 1 A ∈ Rm×m is the matrix where all entries are equal to − m except for the m−1 diagonal entries, which are all equal to m . With these formulas, we will now, finally, define the discrete linear Hilbert transform as linearization of H nl . Definition 5.22. The discrete Hilbert transform H defined on K is the linear map defined by 1 nl H (λu). λ→0 λ

H(u) = lim

Applying Theorem 5.20, we can derive the matrix representation of H: H(u) = DH nl |0 u = ADh|PK DW |0 u = ALM −1 Ru.

6 Numerical Results and Future Work With Theorem 5.17 proven, we now have a well-defined discrete Hilbert operator for the maximal packing of any given complex. However, this is only the beginning of most of the work related to the discrete operator. It is desirable to examine various properties of the Hilbert operator, and, of course, to examine its convergence behavior under refinement of the complex. There are also other questions not directly related to the discrete Hilbert operator we would like to investigate. In this last chapter, we will show some results of numerical computations with the discrete operator and shed light on 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 different functions and circle packings. To test the discrete operator, we use functions whose Hilbert transform is known. Given any w ∈ H ∞ ∩ C 1+α with Im w(0) = 0, we can apply the discrete transform to u := Re w|T evaluated at the contact points tj . We then compare the result to v := Im w|T = Hu. The functions we use are w1 (z) = z and w2 (z) = B(z) − C, where B is a Blaschke product with three simple zeros at 0.3, −0.4 + 0.64i and −0.7 − 0.2i and C = B(0). We set u1 = Re w1 |T and u2 = Re w2 |T . Furthermore, we 3 define u3 : T → R by u3 (t) = 1 if Re t > 0 and u (t) = 0 otherwise. 3 In 1 τ 3 iτ Section 3.2, we saw that Hu (e ) = π ln tan 2 . It is remarkable that u is not even an element of H ∞ ∩ C. Nevertheless, the discrete transform is still a quite good approximation to the classical transform. In both Figure 6.1 and Figure 6.2, picture (a) shows the circle packing used to define the discrete transform. Pictures (b), (c) and (d) show the results for u1 , u2 and u3 , respectively. The function uj is plotted in green while Huj is plotted in blue. The entries of the discrete Hilbert transform are plotted in red. © Springer Fachmedien Wiesbaden GmbH 2017 D. Volland, A Discrete Hilbert Transform with Circle Packings, BestMasters, https://doi.org/10.1007/978-3-658-20457-0_6

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6 Numerical Results and Future Work

1

− π2

−π

π

π 2

−1

(b) u

1

(a) The packing 1

− π2

−π

1

π 2

π −π

− π2

−1

π

π 2

−1

(c) u2

(d) u3 [7]

Figure 6.1: Numerical results for K3 . u in green, Hu in blue, Hu in red.

1

− π2

−π

π

π 2

−1

(b) u

1

(a) The packing 1

− π2

−π

1

π 2

π −π

− π2

−1

(c) u

2

π 2

−1

(d) u

3

Figure 6.2: Numerical results for K 0 . u in green, Hu in blue, Hu in red

π

6.2 Eigenvalues of the discrete transform

91

The complex of the packing used in Figure 6.1 is the regular heptagonal [7] complex with three generations K3 , which was introduced in Subsection 4.2.1. It has 85 vertices, 56 of which are boundary vertices. The complex [7] of the packing in Figure 6.2 is obtained by hex-refining every face of K4 . 0 It will be denoted by K and has 778 vertices, 294 of which are boundary vertices. We observe a fairly good approximation of the classical Hilbert transform especially when using K 0 . In the following, we will confine ourselves with these examples. Examining the convergence of the discrete transform and presenting 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 analytic and algebraic properties of the classical Hilbert operator. This expectation is motivated by the properties of circles packings and the promising numerical results. As an example, we will have a look at the eigenvalues of the discrete operator. The eigenvalues of the classical Hilbert operator H : Lp (T) → Lp (T) for a given p ∈ (1, ∞) can be directly deduced from the basis representation in (3.1): They are i, −i and 0 and the respective eigenspaces are given by Ei = H0p (T), E−i = H p (T)⊥ , E0 = {f : T → C constant}. Here, H0p (T) denotes set of all functions in H p (T) with zero mean and H p (T)⊥ denotes set of all functions in Lp (T) whose nonnegative Fourier coefficients vanish. In L2 (T), this is the orthogonal complement of H 2 (T), hence the notation. We expect a similar behavior for the discrete Hilbert transform. Figure 6.3 shows the spectra of discrete Hilbert operators for the two maximal packings seen before. As we can see, the property of the classical operator does not completely confer to the discrete operator. It is noteworthy, though, that the real part of all eigenvalues seems to be equal to 0. In fact, Re(λ) < 10−15 numerically for each eigenvalue λ. To further illustrate the spectra we will have a look at the “eigenfunctions” of H for K 0 . Since all entries of H are real, the eigenvalues and corresponding eigenvectors are pairwise complex conjugates, so it suffices to consider σ + (H) := {λ eigenvalue of H| Im λ > 0}. Numerically, no two elements of σ + (H) have the same modulus and all eigenvalues are simple eigenvalues.

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6 Numerical Results and Future Work

1

1

−0.5

0.5 −0.5

0.5

−1 (a)

−1

[7] K3

(b) K

0

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 all λ ∈ σ + (H) \ {λ1 , λ2 , λ3 , λ4 }. Figure 6.4 shows the real and imaginary parts of the eigenvectors corresponding to λj . 1

−π

1

π 2

− π2

π

−π

−1

−1 1

(b) Eigenfunction e2 of λ2

(a) Eigenfunction e of λ1 1

−π

π

π 2

− π2

1

− π2

π 2

−1

π −π

− π2

π 2

π

−1 3

(c) Eigenfunction e of λ3

(d) Eigenfunction e4 of λ4

Figure 6.4: Eigenfunctions for K 0 . Real part in black, imaginary part in gray.

Based on the interpretation of the discrete transform, we expect the eigenfunctions to be linear combinations of discrete versions of z k with k ∈ N. Consequently, their real and imaginary parts should resemble discrete sine and 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 some kind 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 this observation: 1. An experiment suggests that our interpretation of H might be slightly − wrong. Given a function u : T → R, let → u = (u(t1 ), . . . , u(tm )) and − w(z) = Su (z). Recall that the entries of H → u are interpreted to be the − values of Im w at the points tj . We suspect that the entries of H → u should possibly rather be interpreted as values of Im w at the points zj instead. To explain this, we show a second plot: Assuming that the eigenvalues ek are discrete versions of z k , we suspect that the entries of ek approximate the values zjk for j = 1, . . . , m. For maximal packings, we have that (1 − rj )tj = zj . In this case, we would obtain tkj ≈ (1 − rj )−k ekj . We will therefore plot the eigenvectors e1 , . . . , e4 again, but this time we will multiply the j-th entry of the k-th vector by (1 − rj )−k . The modified vectors ebk are shown in Figure 6.5. We can see that the modified eigenfunctions look smoother than the functions in Figure 6.4. 1

−π

1

− π2

π 2

π

−π

− π2

−1

−1 1

(b) Eigenfunction eb2 of λ2

(a) Eigenfunction eb of λ1 1

−π

π

π 2

1

− π2

π 2

π −π

−1

− π2

π 2

−1 3

(c) Eigenfunction eb of λ3

(d) Eigenfunction eb4 of λ4

Figure 6.5: Modified Eigenfunctions for K 0

π

94

6 Numerical Results and Future Work

2. The eigenspace H0p (T) of H does not only contain the basis functions tk for k ∈ N, but also linear combinations of these functions. It is possible that the functions seen in Figure 6.4 are actually discrete versions of functions of the form tk + ctl for some 0 < |c| > 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 a to eliminate the constant c appearing in Proposition 3.26. In this proposition, we remarked that we can choose the branch of the argument function arg to be the principal branch Arg in case kuk is small enough. We further noticed that c = 0 in this case. Recall the construction of the linearized operator. We used the function h, which is based on the principal branch Arg. To obtain the operator, we linearized the nonlinear operator defined on an arbitrarily small neighborhood of 0. But consequently, one should expect that we actually do not need the function a at all. The elimination of the constant c is already contained in the ideas our discretization approach is based on. We will therefore try out what happens if we leave out the function a. b = LM −1 R. In this experiment, the I.e., we replace H = ALM −1 R by H discretization is based on the complex K 0 from the previous sections again. b to the vectors u1 and u2 , where Figure 6.6 shows the result of applying H 1 2 uj = Re tj and uj = − Im tj . The expected results are (Hu1 )j ≈ Im tj and (Hu2 )j ≈ Re tj , respectively. 1

−π

− π2

1

π 2

π −π

−1

(a) u1

− π2

π 2

π

−1

(b) u2

b applied to the discrete cosine and sine Figure 6.6: Numerical results for H

b yields the expected result for u1 but fails for u2 . ApparAs we can see, H 2 ently, Hu 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 at this point? The problem seems to arise from the standard normalization. To illustrate the problem, it is helpful to use the nonlinear transform: For the scaled vectors 12 u1 and 12 u2 , we plot the standard normalized circle packings P1 resp. P2 that solve the corresponding instances of (DHRHP). These packings should be understood as discrete images of the functions w1 (z) = z exp z2 resp. 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 a rotated version of w2 (D). This is due to the normalization: The dark gray circle is centered at 0, while the light gray circle has its center on the positive real line. Due to the structure of the complex K 0 , this requirement can apparently not be met without inducing the rotation we observe in the figure. This phenomenon eventually results in the constant that was observed in Figure 6.6. As a consequence, it may be necessary to rethink the normalization strategy we applied. Apart from that, we suspect that the behavior may be caused by the structure of the complex. Clearly, the boundary circles of PK 0

96

6 Numerical Results and Future Work

are significantly smaller than some of its interior circles. This nonuniform discretization could be one of the reasons for the observed behavior. To support this hypothesis, we show another experiment where we used a different complex K 00 . This complex is hexagonal, i.e. every interior node has exactly 6 neighbors. Figure 6.8 shows PK 00 and the result of the b being applied to u2 . corresponding operator H

1

−π

− π2

π 2

π

−1 (b) Resulting operator applied to u2 (a) Maximal packing on K

00

b defined on a different complex Figure 6.8: Results of H

As we can see, the constant is significantly smaller in this case. Further experiments with more different complexes have suggested that the constant indeed depends on the complex itself. Due to these observations, we presume the following: If the discrete operator H can be proven to converge in some sense to its continuous b if suitable assumptions are archetype H, then the same can be proven for H made on the complexes involved.

6.4 Curvature of the Circle Packing manifold Recall that the discrete Hilbert transform is actually the linearization of a nonlinear procedure. It was defined as the limit of 1 nl H (λu) λ

(6.1)

for λ → 0. Therefore, it could be interesting to evaluate (6.1) for some λ > 0 and compare the result to H(u). As an example, we use the cosine function u(eiτ ) = cos τ and the operator [7] H based on the complex K3 . 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 for λ = 0.9 resp. λ = 12 . We see that the difference between the results is extremely small. In fact, one has to look very closely to tell the different values apart.

1

−π

− π2

π 2

π

−1 Figure 6.9: Comparison of nonlinear transform (circles) and linearized transform (asterisks)

This observation suggests that the curvature of the circle packing manifold is very small: though circle packings are nonlinear objects, they seem to scale almost linearly. Interestingly, Stephenson made similar observations before. In his book [27] in Chapter 23, he reports on experiments with curvature flow, where he observed that a convex combination of the vector forms of two circle packings almost behaved like a circle packing itself. It may be interesting to investigate 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 nonlinear problem (HRHP) due to the problems related with the discretization of the Schwarz problem. The reader may already have figured out that there is an alternative to this. Instead of using a nonlinear problem, we can use a linear problem with index 1 instead of index 0. As we will see, the advantages of (HRHP) are

98

6 Numerical Results and Future Work

also present in this modification, and we have the additional advantage that the problem remains linear. In fact, the modified linear problem was the first 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 winding number 1, the so called direction function. Assume that there is a holomorphic function g ∈ H ∞ ∩ C with a simple zero at 0 whose boundary function is given by t 7→ 1c . Then the problem has the standard normalized solution z 7→ (Su (z) + iµ)g(z) + z If g and Su and their derivatives are small enough, we can apply similar arguments as in the proof of Theorem 3.24 to show that this solution has no other zeros and that its derivative vanishes nowhere. So, let us discretize the problem (6.2). Theotarget curve at tj ∈ T is the n 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 define the 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 this problem, 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 motivated by 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 problem defined by β. For the special case c(t) = t, note that the linearizations of the above problem and (DHRHP) actually coincide. Thus, Theorem 5.17 applies once more, yielding a well-defined discrete local frame Hilbert operator. In the local frame setting, we can now vary the function c. This raises two questions: • Which choices of c yield a discrete problem with regular linearization? • How does the choice of c affect analytic and algebraic properties of the resulting operator? As we can see, there are way more questions to be answered than answered questions regarding the Hilbert transform. It remains to hope that further insight into the discrete Hilbert transform with circle packings will be gained in the future and that addressing the questions discussed in this last chapter will further enrich the fruitful theory of circle packings.

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