on riemann-hilbert problems in circle packing

Computational Methods and Function TheoryVolume 9 (2009), No. 2, 609–632

On Riemann-Hilbert Problems in Circle Packing

Elias Wegert and David Bauer

(Communicated by Kenneth Stephenson)

Abstract. We propose a discrete counterpart of non-linear boundary valueproblems for holomorphic functions (Riemann-Hilbert problems) in the frame-work of circle packing. For packings with simple combinatorial structure andcircular target curves appropriate solvability conditions are given and the setof all solutions is described. We compare the discrete and the continuoussetting and discuss several discretization effects. In the last section we in-dicate promising directions for further research and report on the results ofsome test calculations which show that solutions of the circle packing problemapproximate the classical solutions surprisingly well.

Keywords. Riemann-Hilbert problems, circle packing, conformal geometry,hyperbolic geometry.

2000 MSC. Primary 30E25; Secondary 52C26, 30C35, 30C80.

1. Introduction

This paper is about the search for a general class of well-posed boundary valueproblems in circle packing. To give the reader a flavor of the topic we avoidall technicalities and start at an informal level, where a circle packing is just acollection of circles (in the complex plane) with specified pattern of tangencies.The basic definitions are given in Section 2, for details we refer to KennethStephenson’s book [8].

The skeleton of a packing is its combinatorics, viz. the structure of tangencyrelations between its circles. Given the combinatorics, radii and centers of thecircles have to be chosen such that the circles fit accordingly.

Two circles of a packing are said to be neighbors if they are supposed to betangent according to the combinatorics. The interior of a packing consists ofthose circles which are surrounded by a chain of neighbors; the other circles form

Received November 19, 2008, in revised form April 6, 2009.Published online June 1, 2009.This research was supported by the Deutsche Forschungsgemeinschaft (DFG) through the grantWE 1704/8-1.

ISSN 1617-9447/$ 2.50 c© 2009 Heldermann Verlag

610 E. Wegert and D. Bauer CMFT

the boundary. Figure 1 depicts two circle packings with shaded boundary circles.In the left picture the combinatorics is shown as a graph. The packing on the

right is modeled on K[7]3 , the heptagonal complex with three generations. Both

packings are univalent, which means that any two disks have disjoint interiors.

Figure 1. Two univalent circle packings.

Kenneth Stephenson’s “snake packing” on the left of Figure 2 is not univalent butstill locally univalent, while the picture on the right shows a branched packing.Here the central circle has branching order one; the chain of its neighbors wrapstwice around it.

Figure 2. A locally univalent and a branched circle packing.

It is clear that the radii of the circles in a packing cannot be chosen arbitrarily butmust satisfy certain compatibility conditions. In order to explore the geometry ofcircle packings it is essential to understand the global structure of these equations.

Comparing the number of equations, which result from the tangency conditions,with the parameters for the centers and radii of the circles, we expect that apacking with n circles and p tangency conditions has 3n− p degrees of freedom.A simple argument using Euler’s theorem shows that 3n− p = m + 3, where mdenotes the number of boundary circles. Three of the parameters can be inter-preted as a rigid motion (rotation and translation) of the packing in the plane,

9 (2009), No. 2 On Riemann-Hilbert Problems in Circle Packing 611

and an optimist might hope that the remaining m parameters can be directlyassociated with the m boundary circles.

Miraculously this is really so — a fundamental result [8, Thm. 11.6] says that,for every fixed combinatorics, there exists an unbranched circle packing witharbitrarily prescribed radii of its boundary circles. This packing is unique up torigid motions.

Another famous theorem pertains to the existence of a maximal packing withprescribed combinatorics, which is a univalent packing where every boundarycircle is internally tangent to the unit circle T.

Figure 3. Two maximal packings with the combinatorics of K[7]3 .

The maximal packing is unique up to conformal automorphisms of the unit disk(see Figure 3) and thus it can be normalized by three appropriate side condi-tions. This again fits with the expectation that m (tangency) conditions for theboundary circles and three additional conditions uniquely determine a packingwith m boundary circles.

If the situation is so fortunate one may ask for more general well-posed boundaryvalue problems — this is the main purpose of the paper.

The problems which we shall introduce and discuss are discrete counterparts ofnon-linear boundary value problems for holomorphic functions, so-called Riemann-Hilbert problems.

Interrelations between circle packing and complex function theory have beenstudied as early as in 1936 by Paul Koebe [4]. The basic idea is rather simple:from a naive point of view, a holomorphic function is a mapping which mapsinfinitely small circles to infinitely small circles and preserves orientation. Ifinfinitely small circles are replaced by “real” circles we end up with circle packingsas models of “discrete analytic functions”.

More precisely, if P is a circle packing laid out in the complex plane and M is amaximal packing with the same combinatorics, then the mapping M �→ P is adiscrete counterpart of a holomorphic function on the unit disk D. The discrete


→

Figure 4. A circle packing as a discrete analytic function on the disk.

analytic function shown in Figure 4 resembles a conformal mapping of the unitdisk onto a square.

Readers of Kenneth Stephenson’s beautiful and inspiring book [8] will be ac-quainted with a number of discrete versions of classical results in function the-ory. It is surprising how faithful the discrete objects mimic their continuouscounterparts even if the discrete structures are rather coarse.

The problem which has been studied most intensively in this context is confor-mal mapping (some important contributions are [3, 6, 9]). However, conformalmapping turns out to be just a special case of a more general problem which wasposed by Riemann, when he was searching for general boundary value problemsfor holomorphic functions. Riemann’s original setting of these problems (see [5,p. 35 ff.] is purely geometric and therefore well suited for translation into thecontext of circle packing.

In the next section we give an outline of some results for the classical problemsand propose an appropriate formulation of circle packing Riemann-Hilbert prob-lems. The simplest problems of this kind are studied in Section 3. Finally, inSection 4, we indicate some promising directions for further research.

2. Riemann-Hilbert problems

To put the problem into the right perspective we summarize some basic factsabout non-linear Riemann-Hilbert problems. For detailed information we referto [12, 13].

A non-linear Riemann-Hilbert problem consists in finding holomorphic functionsw, say in the complex unit disk D, which satisfy the boundary condition

(2.1) f(t, w(t)) = 0,

for all points t on the unit circle T. Here f : T× C → R is a given function.

In order to avoid problems with boundary values, we are looking for solutions inthe disc algebra H∞ ∩ C, which consists of all functions that are holomorphicin D and extend continuously to D.


If, for example, f does not depend on t and f(w) = 0 describes the boundaryof a Jordan domain G, the conformal mappings of D onto G are solutions of theRiemann-Hilbert problem.

To give a second example, all (finite) Blaschke products are solutions of theRiemann-Hilbert Problem (2.1) with f(t, w) := |w| − 1, which shows that thesolution set can be rather large.

Riemann himself posed the problem in a geometric language, which was practi-cally out of sight before it was rediscovered by Alexander Shnirel’man in 1972.Following Riemann we introduce the target curves

Mt := {w ∈ C : f(t, w) = 0}, t ∈ T.

The target curves form the fibers of the target manifold

M := {(t, w) ∈ T× C : f(t, w) = 0},which contains the complete information about the problem. With these notationthe boundary condition (2.1) can be rewritten as w(t) ∈ Mt for all t ∈ T, orsimply as tr w ⊂ M , where the trace of w is defined by

tr w := {(t, w(t)) ∈ T× C : t ∈ T}.Depending on the global geometric structure and the regularity of the targetmanifold several classes of Riemann-Hilbert problems can be distinguished. Herewe restrict ourselves to smooth target manifolds which are built from closedtarget curves. The next definition specifies the assumptions.

A target manifold M is said to be admissible if it can be represented in the form

M = {(t, μ(t, s)) : t, s ∈ T}with a function μ : T

2 → C satisfying the following conditions (i) and (ii):

(i) The parametrization μ is continuously differentiable on T2 and its derivative

∂sμ with respect to the second variable does not vanish.(ii) For each t ∈ T the mapping μt : T → C, s �→ μ(t, s), is injective.

In particular, these conditions guarantee that all target curves Mt are (smooth)Jordan curves. The target manifold (and the corresponding Riemann-Hilbertproblem) is said to be regular if the origin is contained in the interior of eachtarget curve,

0 ∈ int Mt, t ∈ T.

The following result goes back to Alexander Shnirel’man [7], who proved it insomewhat stronger hypotheses. For more details and a proof under the aboveassumptions we refer to [11, 12].

Proposition 2.1. Assume that M is an admissible regular target manifold andlet n be a non-negative integer. Then, for arbitrary points t0 ∈ T, w0 ∈Mt0, andz1, . . . , zn ∈ D there is a unique function w ∈ H∞ ∩ C which has zeros exactly


at z1, . . . , zn, satisfies the boundary condition w(t) ∈ Mt for all t ∈ T, and theadditional condition w(t0) = w0.

Solutions without zeros are the “fundamental” solutions. The traces of thesesolutions cover the target manifold M in a schlicht manner, thus defining acanonical parametrization of M .

Figure 5. Target manifold with traces of solutions without zeros.

One peculiarity in studying Riemann-Hilbert problems for circle packings comesfrom the fact that exactly the class of solutions without zeros, which is most nat-ural in the continuous setting, is inaccessible in the discrete case. This becomesevident by considering the standard example, where Mt = T for all t ∈ T andthe set of solutions consists of the finite Blaschke products. The zero-free solu-tions are unimodular constants, which have no reasonable counterpart in circlepacking.

Before we give an adequate formulation of circle packing Riemann-Hilbert prob-lems we recall some standard definitions (see Stephenson [8]).

Three mutually externally tangent circles A, B, C form a triple (ABC). Wealways assume that triples (ABC) are positively oriented, i.e. the centers of A, Band C form a positively oriented triangle in this order.

Let K be a simplicial 2-complex which is a combinatorial closed disk (whichmeans that it is finite, simply connected, and has non-empty boundary ∂K). Afamily P of circles C1, . . . , Cn is said to be a circle packing for the complex K ifthe following assertions (i), (ii), and (iii) hold:

(i) There is a bijective correspondence between the vertices vk of K and thecircles Ck of P .

(ii) If vk and vm are adjacent in K the associated circles Ck and Cm are exter-nally tangent.

(iii) If the vertices vj, vk, vm form a positively oriented face of K the three asso-ciated circles form a (positively oriented) triple (CjCkCm).


All circles in P which are associated with boundary vertices in K are said to beboundary circles of the packing P .

One fundamental result of circle packing is the Discrete Uniformization Theoremof Koebe [4], Andreev and Thurston. We quote a special case which is relatedto our situation.

Proposition 2.2 ([8, Prop. 6.1]). If K is a combinatorial closed disk, thereexists an essentially unique univalent circle packing PK ⊂ D for K such thatevery boundary circle of PK is internally tangent to T.

Here “essentially unique” means uniqueness up to a conformal automorphismof D. We refer to any such packing as a maximal packing for K (see Figure 3).

In the following we always assume that the boundary circles B1, . . . , Bm of apacking PK are numbered consecutively such that the sequence of correspondingcircles in the maximal packing PK has counter-clockwise orientation.

The circle packing Riemann-Hilbert problem is the following generalization of theuniformization problem.

Problem 2.1 (CRHP). Let K be a combinatorial closed disk with m boundaryvertices and let M1, . . . , Mm be a given family of Jordan (target) curves. Find allcircle packings P for K such that any boundary circle Bk of P lies in the closureof the domain bounded by the corresponding target curve Mk and meets Mk,

Bk ⊂ clos int Mk, Bk ∩Mk �= ∅ for k = 1, . . . , m.

Figure 6 shows three different locally univalent solutions of the same Riemann-Hilbert problem. All target curves are circles with equal radius (in the onlineversion the boundary disks and the associated target curves carry the same color).

Figure 6. Three solutions of a problem with circular target curves.

In the above setting the problem is certainly too general to expect “nice” results.Even if we restrict ourselves to the case of regular problems, where 0 ∈ int Mk forall k, there are discretization effects (some are discussed at the end of the next


section) which make the circle packing Riemann-Hilbert problem more involvedthen its continuous counterpart.

So the first problem consists in finding appropriate conditions which guaranteethat the discrete problems are solvable. Further we are interested in conver-gence of the discrete solutions to the classical solutions under refinement of thediscretization.

Recall that the solution set of classical regular Riemann-Hilbert problems splitsinto different subclasses according to their numbers of zeros in the disk. Thefundamental solutions are those with the smallest possible number of zeros. Inorder to translate this into the language of circle packing we need a substitutefor the concept of zeros, which is problematic in this setting.

Let P be a circle packing and assume that the origin does not lie in clos intBk

for all boundary circles of P . Then the oriented polygon which consecutivelyconnects the centers of B1, B2, . . . , Bm, B1 does not pass through the origin 0,and its (usual) winding number about 0 is called the winding number of P (aboutzero).

Problem 2.2. Find and characterize (for example by an extremal property)the solutions to circle packing Riemann-Hilbert problems which have minimalwinding number.

In particular solutions with winding number one are of special interest, since theyare the natural generalizations of maximal packings. We expect that the familyof solutions with winding number one involves three parameters correspondingto intrinsic rotations and translations of the packing as is shown in Figure 6.Moreover, we conjecture that these solutions are characterized by similar ex-tremal properties as conformal mappings and solutions of classical non-linearRiemann-Hilbert problems.

In analogy to the continuous case we expect that there is at most one solutionwith winding number one having the standard normalization of conformal map-ping: the center of an alpha circle is fixed at the origin and the center of aneighboring (beta) circle lies on the positive real axis.

Note that we could also pose the Riemann-Hilbert problem such that the centersof the boundary circles lie on the target curves. Though this would simplifythe equations for the boundary conditions (and their numerical treatment), weexpect that problems in this setting have worse properties.

3. Problems for chains and flowers

In this section we investigate circle packing Riemann-Hilbert problems for thesimplest combinatorial structures. The objective is to show that even for thesevery coarse packings the problem is reasonable and mimics its continuous coun-terpart, provided the target curves satisfy certain metric restrictions. On the


Figure 7. “Maximal packing” and solution of RHP with centerson target circles.

other hand we indicate some limitations which result from the rough discretiza-tion and the special structure of the packings.

3.1. Fitting triples. We start with some notation. The radius and the center ofa circle A are denoted by r(A) and c(A), respectively. By S(A; B, C) we denotethe closed sector bounded by the two rays emanating from c(A) and passingthrough c(B) and c(C), respectively, such that one can move counter-clockwisefrom c(B) to c(C) through S(A; B, C). The oriented angle at the vertex c(A) ofthat sector is denoted by �BAC.

The following auxiliary result describes how the angles of a triple depend on theradii of its circles. Its proof is elementary.

Lemma 3.1. If (ABC) and (ABC ′) are triples with r(C ′) > r(C), then

�BAC < �BAC ′, �CBA < �C ′BA, �ACB > �AC ′B.

A basic procedure is fitting one circle, say C, of a triple to a prescribed targetcircle T . This is a particular case of Apollonius’ constructions, but since we needsome special facts we formulate and prove the result in some detail.

Lemma 3.2. Let (AB) be a pair of externally tangent circles and assume thatA is contained in the interior of a (target) circle T .

(i) The pair (AB) can be uniquely complemented to a (positively oriented) triple(ABC∗) such that C∗ touches T internally.

(ii) The circle C∗ depends continuously on the centers and radii of A, B, and T .(iii) For any triple (ABC) the circle C does not intersect T if r(C) < r(C∗) and

it intersects T at exactly two points if r(C) > r(C∗).

Proof. (i). We assume that the centers of A and B are located on the negativeand positive real axis, respectively, and that A and B meet at the origin O.


ABT

0

C*

A~

B~

T~ 0

C*~

Figure 8. (a) Figure 8. (b)

Inversion z �→ 1/z sends A and B to straight lines A := {z : Re z = a < 0} and

B := {z : Re z = b > 0}, respectively (see Figures 8 (a) and (b)).

(ABC) is a (positively oriented) triple, if and only if the image of C, C, is a

circle which touches the lines A and B and if its intersection with the imaginary

axis lies strictly in the upper half plane. We call these positions of C admissible.

Since A lies in the interior of T , the image of T , T , is a circle which lies in theopen half-plane Re z > a and intersects the real axis at two points p and q with

a < p < 0 and 0 < q. In particular T intersects the imaginary axis at a pointwith positive imaginary part.

Increasing the radius of C corresponds with shifting C in direction of the negative

imaginary axis. It follows that there exists an admissible position C∗ of C, where

it touches T from the outside at a point S with Re S =: s.

(ii). The statement concerning continuity is easy to verify, taking into account

that C∗ must touch T at an (interior) point of the strip {z : a < Re z < b}.(iii). If C lies “above” C∗, the circles C and T do not meet. If C is in an

admissible position “below” C∗, the interiors of C and T have common points

(for instance on the line Re z = s). Since the intersection of C and A always lies

in the exterior of T , the circles C and T must intersect at exactly two points.

The following result will be used in Section 3.3. Though it is intuitively clear wegive a proof.

Lemma 3.3. Let (AB) be a pair of externally tangent circles contained in theinterior of a (target) circle T . Then there exists a circle T ′ with the followingproperties:

(i) A is contained in the interior of T ′.(ii) B is internally tangent to T ′.(iii) T ′ is contained in the interior of T .


Proof. We assume that A and B satisfy the assumptions made in the proof

of Lemma 3.2. Inversion at O sends T to a circle T which lies in the ver-tical strip {z : a < Re z < b} and intersects the real axis at two points p

and q with a < p < q < b. Then it is clear that there exists a circle T ′ with

Im c(T ′) = Im c(T ), which lies in the strip {z : a < Re z ≤ b}, touches B and

contains T in its interior. Since T ′ intersects the real axis at points p′ and q′

with a < p′ < p < q < q′ ≤ b, the preimage T ′ of T ′ is a circle with the desiredproperties.

Next we assume that two circles B and C of a triple (ABC) are fitted to cor-responding target circles TB and TC , respectively, and prove a rigidity result,which shows that fitted circles B and C cannot be moved without leaving acertain sector. The precise statement is even a bit stronger.

Lemma 3.4. Let (ABC) be a triple such that A lies in the interior of the targetcircles TB and TC, and assume that B and C are internally tangent to targetcircles TB and TC, respectively.

If (AB′C ′) is a triple such that the centers of B′ and C ′ both lie in the closedsector S := S(A; B, C) and B′ ∩ TB �= ∅, C ′ ∩ TC �= ∅, then B = B′ and C = C ′.

Proof. Let gB and gC be the rays through B and C, respectively, which boundthe sector S (see Figure 3.1). A triple (AB′C ′) is said to be admissible if itsatisfies the assumptions of Lemma 3.4.

A

B

C

TB

TC

gC

gB

S

1. Assume, for a moment, that r(B′) ≥ r(B) and r(C ′) ≥ r(C). Then, byLemma 3.1, �B′AC ′ ≥ �BAC. Since B, B′, C, C ′ are in S this is only possibleif �B′AC ′ = �BAC, which, again by Lemma 3.1, gives r(B′) = r(B) andr(C ′) = r(C), and then a little thought shows that B′ = B and C ′ = C.

2. Assume now the complementary case, i.e. without loss of generality, letr(B′) < r(B). We can increase the radius of B′ continuously, always positioningB′ so that (AB′C ′) form a triple, until r(B′) = r(B). It follows from Lemma 3.2


that the enlarged circle B′ intersects the target circle TB at exactly two points.We also observe that the center of B′ cannot leave the sector S: it is clear thatc(B′) cannot cross the ray gC . If it would cross the ray gB then, at this verymoment, r(B′) would still be less than r(B), which would imply that B′ cannotmeet its target circle TB, a contradiction. At the end we get an admissible triple(AB′′C ′) with r(B′′) = r(B) such that B′′ intersects TB twice.

If r(C ′) ≥ r(C) we set C ′′ := C ′. If r(C ′) < r(C) we take the triple (AB′′C ′)and enlarge C ′ according to the above procedure. This yields an admissible triple(AB′′C ′′) with r(B′′) ≥ r(B) and r(C ′′) ≥ r(C) such that B′′ intersects TB twice.By what was shown in the first step, this is impossible and hence the second casecannot occur.

As a corollary we obtain the following lemma, which will be the basis for provingmonotonicity of angle sums of fitted chains.

Lemma 3.5. Let TB and TC be two fixed (target) circles and let (ABC) and(A′B′C ′) be triples with concentric circles A and A′ which lie in the interior ofTB and TC. Assume further that B, B′ and C, C ′ are internally tangent to TB

and TC, respectively.

If the radius of A′ is smaller than the radius of A, the centers of B′ and C ′

cannot both lie in the sector S(A; B, C).

Proof. The result follows immediately from Lemma 3.4 by a homothety of thetriple (A′B′C ′) which maps A′ onto A.

3.2. Fitting chains. We are now going to study circle packing Riemann-Hilbertproblems for the simplest combinatorial structures.

An ensemble C := (C0; B1, . . . , Bm) of circles is called m-chain around the centralcircle C0 (or, if no confusion is possible, simply a chain) if the following holds:

(i) For k = 1, . . . , m, the circles Bk and C0 are externally tangent.(ii) For k = 1, . . . , m− 1, the circles Bk and Bk+1 are externally tangent.(iii) For all k = 1, . . . , m − 1, the oriented angles �BkC0Bk+1 are positive and

less than π.

A chain (C0; B1, . . . , Bm) is said to be closed if the circles Bm and B1 are exter-nally tangent and 0 < �BmC0B1 < π.

A chain (C0; B1, . . . , Bm) is said to be fitted to the target circles T1, . . . , Tm if Bk

is internally tangent to Tk for k = 1, . . . , m.

The Riemann-Hilbert problem for chains consists in finding all closed chains fittedto a given family of target circles. Any such chain is referred to as a solution ofthe problem.


In order to eliminate some degrees of freedom we say that a chain (C0; B1, . . . , Bm)is normalized if C0 is centered at the origin and the center of B1 lies on the pos-itive real axis. Our next goal consists in fitting normalized chains to the targetcircles.

Let C = (C0; B1, . . . , Bm) be an m-chain. We denote by r and r1, . . . , rm theradii of C0 and B1, . . . , Bm, respectively, set rm+1 := r1, and let

(3.1) αk := arccos(r + rk)

2 + (r + rk+1)2 − (rk + rk+1)

2

2(r + rk)(r + rk+1), k = 1, . . . , m.

Given a chain C = (C0; B1, . . . , Bm) the (angular) length L(C) and the angle sumS(C) of C are the numbers

L(C) :=m−1∑k=1

αk, S(C) :=m∑

k=1

αk,

respectively. Since αk is the angle � BkC0Bk+1 formed by the centers of pairwiseexternally tangent circles Bk, C0, and Bk+1 with radii rk, r, and rk+1, respectively,the chain C is closed if and only if its angle sum S(C) is a multiple of 2π. Thenthe (positive) integer

wind C :=S(C)2π

is referred to as the winding number of the closed chain. Closed chains withwinding number one are called flowers.

Recall that we speak of a regular Riemann-Hilbert problem, if the origin lies inthe interior of all target circles, 0 ∈ int Tk (k = 1, . . . , m).

In the continuous setting of Riemann-Hilbert problems all “circles” are infinites-imal and thus the “central circle” (located at the origin) automatically lies inthe interior of all target circles. In the discrete setting this is not guaranteedautomatically, which motivates the following definition.

A solution (C0; B1, . . . , Bm) is said to be regular if its central circle C0 is containedin the interior of all target circles, C0 ⊂ int Tk for all k = 1, . . . , m.

The width r∗ of a regular Riemann-Hilbert problem with target circles T1, . . . , Tn

is the radius r of the largest circle C(r) centered at 0 which fits into all targetcircles,

r∗ := max{r : C(r) ⊂ clos int Tk, k = 1, . . . , m}.Lemma 3.6. Let the problem be regular with width r∗. If r satisfies 0 < r < r∗,there exists exactly one normalized fitted chain with central circle of radius r.The centers and radii of this chain depend continuously on r and on the centersand radii of all target circles.

Proof. 1. If m = 1 we consider the normalized tangent pair (C0; B1). If theradius of B1 is small, B1 lies completely in the interior of T1. Now we let the


radius grow until B1 touches T1. It is clear that B1 depends continuously on rand T1.

2. For m > 1 we assume that the first k circles B1, . . . , Bk with 1 ≤ k < m arealready fitted to their targets. In order to fit the next circle Bk+1 to Tk+1, itsuffices to apply Lemma 3.2 with A := C0, B := Bk, C := Bk+1, and T = Tk+1.Then the result follows by induction.

For all r ∈ (0, r∗) we denote by L(r) and S(r) the length and the angle sum ofthe (unique) normalized fitted chain from Lemma 3.6, respectively.

Lemma 3.7. If m ≥ 2 and the problem is regular, the length L(r) and theangle sum S(r) are continuous and strictly monotone decreasing in r, provided0 < r < r∗.

Proof. The continuity of L and S follows from Lemma 3.6. Monotonicity of Lis shown by induction. For m = 2 the result is a consequence of Lemma 3.5 with(ABC) := (C0B1B2).

Assume that L is strictly decreasing for all regular problems with at most mtarget circles. Let (C0; B1, . . . , Bm+1) be an (m + 1)-chain fitted to the targetcircles T1, . . . , Tm+1 with r := r(C0) and let (C ′

0; B′1, . . . , B

′m+1) be a second fitted

chain such that r′ := r(C0) < r. Further, let Sm := S(C0; Bm, Bm+1) be theclosed sector formed by the two rays (emanating from 0) through the centers ofBm and Bm+1, respectively. If r′ is less than r and r− r′ is sufficiently small, bycontinuity and the assumption on monotonicity made above, the center of B′

m

belongs to the sector Sm. Consequently, by Lemma 3.5 applied to the triples(C0BmBm+1) and (C ′

0B′mB′

m+1), the center of B′m+1 cannot lie in Sm, which

implies that L(r′) > L(r) also for regular (m + 1)-chains.

Finally, monotonicity of S(r) is deduced from monotonicity of L(r) as follows.We choose r and r′ such that r′ < r < r∗ and r − r′ is sufficiently small. LetCm = (C0; B1, . . . , Bm) and C ′m = (C ′

0; B′1, . . . , B

′m) be the normalized chains with

radii r and r′ of the central circles C0 and C ′0, respectively, which are both fitted

to the target circles T1, . . . , Tm.

Then there exists an angle α ∈ [0, 2π) such that rotating a copy of B1 by the angleα about the origin results in a circle Bm+1 so that Cm+1 := (C0; B1, . . . , Bm, Bm+1)is a normalized m+1-chain. If Tm+1 denotes the image of T1 under this rotation R,the chain Cm+1 is fitted to the target circles T1, . . . , Tm, Tm+1.

Next we prove that the image circle B′m+1 of B′

1 under the rotation R mustintersect B′

m. Note that the centers of Bm+1 and of B′m+1 both lie on the same

ray g starting at the origin.

Assume, for a moment, that B′m+1 and B′

m would not intersect each other. Ifthe circles did not even touch each other externally, we could increase the ra-dius of B′

m+1, keeping its center on the ray g, until it touches not only C0

but also B′m from the outside. This results in a triple which we denote by


(C0B′mB′′

m+1). It is easy to see that B′′m+1 must intersect Tm+1. If B′

m+1 touchesB′

m, we set B′′m+1 := B′

m+1 and thus B′′m+1 touches Tm+1. Hence we obtain a

triple (C0B′mB′′

m+1) with B′′m+1 ∩ Tm+1 �= ∅ in every case.

Let S = S(C0; Bm, Bm+1) be the closed sector bounded by the rays from theorigin through the centers of Bm and Bm+1. From continuity and monotonicityof the length for the m-chains (C0; B1, . . . , Bm) and (C ′

0; B′1, . . . , B

′m) we obtain

that the center of B′m lies in S. Since the center of B′′

m+1 also belongs to (theboundary of) S, and B′

m ∩ Tm �= ∅, B′′m+1 ∩ Tm+1 �= ∅, this is in conflict with

Lemma 3.5.

This shows that B′m+1 ∩ B′

m �= ∅, and hence the angle sum of (C ′0; B

′1, . . . , B

′m)

is greater than the length of (C0; B1, . . . , Bm, Bm+1), which is equal to the anglesum of (C0; B1, . . . , Bm).

In order to prove the existence of regular solutions to the Riemann-Hilbert prob-lem it only remains to study the angle sum S(r) of a normalized fitted chain asthe radius r of the central circles approaches the endpoints of the interval (0, r∗).

If r → 0, it follows from the regularity of the problem that all radii rk are boundedfrom below by a positive constant and thus the angles αk defined by (3.1) tendto π. Consequently,

(3.2) limr→+0

S(r) = m π.

Since S(r) is positive and monotone decreasing in r, the finite limit

(3.3) s∗ := limr→r∗−0

S(r) < mπ

exists. This limit is crucial for the existence of regular solutions and we call itthe critical angle sum of the problem.

The critical angle sum s∗ of a given problem can be determined easily if oneadmits chains with degenerate circles Bk having radius zero. Then there existsa unique (possibly degenerate) normalized chain C∗ which is fitted to the targetcircles and has a central circle C0 of radius r∗. The angle sum S(C∗) of this chainis equal to s∗.

In what follows let �x� and x� denote the integers which satisfy x ≤ �x� < x + 1and x− 1 < x� ≤ x, respectively.

Theorem 3.1. A regular circle packing Riemann-Hilbert problem for m-chainshas exactly

M =⌈m

2− 1

⌉−

⌊ s∗

2π

⌋normalized regular solutions. All solutions have distinct winding numbers, namely⌊ s∗

2π

⌋+ 1,

⌊ s∗

2π

⌋+ 2, . . . ,

⌈m

2− 1

⌉.


Proof. The result follows immediately from continuity and monotonicity of theangle sum S(r), the limit relations (3.2), (3.3), and the fact that neither limit isattained by any regular chain.

It is not difficult to find regular problems which have no regular solution, Figure 9shows an example.

Figure 9. A regular RHP and its unique normalized (non-regular) solution.

3.3. Totally regular problems. We now introduce a class of problems whichadmits a complete description of all solutions.

A circle packing Riemann-Hilbert problem is said to be totally regular if itscritical angle sum s∗ is less than 2π.

In order to formulate a simple criterion for total regularity we recall that thewidth r∗ of a regular Riemann-Hilbert problem is the radius of the largest circle,centered at the origin, which fits into all target circles. In addition we define thebounding radius of a problem as the radius d∗ of the smallest circle centered atthe origin which contains all target circles.

Lemma 3.8. If the bounding radius d∗ and the width r∗ of a regular problemsatisfy the inequality

(3.4)d∗

r∗<

1 + sin(π/m)

1− sin(π/m)

the problem is totally regular.

Proof. Note that condition (3.4) is equivalent to

d∗ − r∗

d∗ + r∗< sin

π

m.

We consider the unique (possibly degenerate) chain C∗ = (C∗0 ; B∗

1 , . . . , B∗m) which

is fitted to the target circles and has a central circle C∗0 of radius r∗. The radii r∗k

of B∗k satisfy r∗k ≤ (d∗ − r∗)/2 for all k. Comparing the triple (C0BkBk+1) with


a triple of circles having radii r∗, (d∗ − r∗)/2, (d∗ − r∗)/2, respectively, we inferfrom Lemma 3.1 that the central angle α∗

k satisfies

α∗k ≤ α := 2 arcsin

12(d∗ − r∗)

12(d∗ − r∗) + r∗

= 2 arcsind∗ − r∗

d∗ + r∗<

2π

m.

This remains true if some of the rk are zero, since then α∗k = 0. Thus the critical

angle sum satisfies

s∗ = S(C∗) =m∑

k=1

α∗k < 2π.

Next we prove some auxiliary results, which allows one to describe the full setof solutions (which are a-priori not assumed to be regular) of totally regularproblems. The first lemma sharpens Lemma 3.5 by omitting the assumptionthat A lies in the interior of the target circles.

Lemma 3.9. Let TB and TC be two fixed (target) circles and let (ABC) and(A′B′C ′) be triples with concentric circles A and A′, such that A′ lies in theinterior of TB and TC. Assume further that B, B′ and C, C ′ are internally tangentto TB and TC, respectively. If the radius of A′ is smaller than the radius of A,the centers of B′ and C ′ cannot both lie in the sector S(A; B, C).

Proof. We consider the triple (A′B′′C ′′), which is obtained from (ABC) by ahomothety with center c(A). Since this homothety is a contraction and its centerlies in int TB ∩ int TC the circles B′′ and C ′′ must be contained in the interiors ofTB and TC , respectively.

An application of Lemma 3.3 to the pairs (A′B′′) and (A′C ′′) yields target circlesT ′

B and T ′C in the interior of TB and TC , respectively, where A′ is contained in

the interior of the new target circles, such that B′′, C ′′ are fitted to T ′B and T ′

C ,respectively. Since B′ is fitted to TB and externally tangent to A′, which lies inthe interior of T ′

B, we must have B′ ∩ T ′B �= ∅ and, analogously, C ′ ∩ T ′

C �= ∅.Assume to the contrary that the centers c(B′) and c(C ′) both lie in the sectorS(A; B, C) = S(A′; B′′, C ′′). Then we would have B′ = B′′ and C ′ = C ′′ byLemma 3.4. But B′′ lies in the interior of TB and B′ is fitted to TB, a contradic-tion.

Now it is easy to deduce that the length of any non-regular chain is less than thelength of any regular one.

Lemma 3.10. Let C = (C0; B1, . . . , Bm), C′ = (C ′0; B

′1, . . . , B

′m) be normalized

fitted chains with r := r(C0) ∈ (0, r∗) and r′ := r(C ′0) ≥ r∗. Then we have

L(C) > L(C ′).


Proof. We denote by αk and α′k the central angles of the chains C and C ′,

respectively. If L(C) ≤ L(C′), then there would exist j ∈ {1, . . . , m − 1} suchthat

j−1∑k=1

αk ≥j−1∑k=1

α′k,

j∑k=1

αk ≤j∑

k=1

α′k.

This would imply that the centers c(Bj) and c(Bj+1) both lie in the sectorS(B′

j; C′0, B

′j+1), contradicting Lemma 3.9.

Lemma 3.11. If a problem is totally regular, then all its solutions are regular.

Proof. By Theorem 3.1 we can find a regular solution C = (C0; B1, . . . , Bm)which is a flower. Now suppose, for a moment, that there is a (non-regular)solution C′ = (C ′

0; B′1, . . . , B

′m) with r′ ≥ r∗. We set Tm+1 := T1, Bm+1 := B1,

B′m+1 := B′

1 and consider the normalized chains C = (C0; B1, . . . , Bm, Bm+1)

and C′ = (C ′0; B

′1, . . . , B

′m, B′

m+1) fitted to the target circles T1, . . . , Tm+1. Using

Lemma 3.10 we obtain the contradiction S(C ′) = L(C ′) < L(C) = S(C) = 2π.

The next theorem is an immediate consequence of Theorem 3.1 and Lemma 3.11.

Theorem 3.2. Every totally regular Riemann-Hilbert problem for m-chains hasexactly �m/2− 1� normalized solutions with all possible winding numbers

1, 2, . . . ,⌈m

2− 1

⌉.

In particular there is exactly one normalized flower which solves the problem.

3.4. Discrete versus continuous problems. It is useful to compare the aboveresults for the simplest circle packing Riemann-Hilbert problems with the con-tinuous setting. Recall that in the latter case the regularity condition 0 ∈ int Mt

guarantees the existence of solutions with arbitrary non-negative winding num-bers and that the basic solution class consists of solutions with winding numberzero. Since these solutions are (frequently) inaccessible in the framework of circlepacking we restricted ourselves to solutions with positive winding number. Inorder to make such solutions unique, additional conditions have to be imposed.For instance, solutions with winding number one are uniquely determined by theside conditions w(0) = 0 and w(1) > 0.

In the continuous case all solutions with positive winding number and w(0) = 0are automatically “regular”, in the sense that the infinitesimal “central circle”w(0) lies in the interior of all target circles. The concept of regular solutionswas introduced since this need not be so in the discrete case (see the example inFigure 9).

A normalized chain which is a regular solution with winding number k in thesetting of this paper is the discrete counterpart of a continuous solution with


winding number k normalized by

(3.5) w(0) = w′(0) = . . . = w(k−1)(0) = 0, w(1) > 0.

It is not surprising that solutions with large winding number have a better chanceto exist (see Theorem 3.1). One plausible reason is that chains with large anglesum are more flexible than “short” chains, since the size of the boundary circlescan vary more if the angle sum is large.

Chains are coarse structures of circle packings and can reflect properties of con-tinuous solutions only to a certain extent. In the continuous case there is no strictrelation between the winding number of solutions and their number of branchpoints. In the discrete case of chains both numbers always differ by one andbranching is possible at the central circle only. So chains cannot mimic contin-uous solutions with low winding number and high order of branching, which isreflected in the fact that discrete problems for chains are solved by flowers underrather restrictive assumptions only (for example total regularity). These condi-tions become even stronger when the number m of boundary circles increases(compare Lemma 3.8) since then their radii must be small compared with theradius of the central circle.

The investigation of non-regular solutions is more complicated and requires dif-ferent techniques, since then the fitting chain procedure does in general not giveunique results. Figure 10 shows a situation where two types of bifurcation phe-nomena can be observed. The target circle T1 is a big circle that touches B1 onthe right and is not shown in the picture. If r is decreased the circle B2 shrinksto a point and disappears when the interstice of C0, B1 and B′

2 moves into theinterior of T2. If r is increased both circles B2 and B′

2 disappear at the momentwhen B1 becomes externally tangent to T2.

2

C 0 B 1

T 2

B 2

B’

Figure 10. An almost critical configuration.


Investigations of more general circle packing Riemann-Hilbert problems need abetter understanding of these effects. In particular we are interested in conditionswhich avoid additional complications resulting from discretization.

4. Perspectives

In this section we outline some directions for further research and present firstpromising results of numerical tests (the background of the computations will begiven elsewhere). All problems under consideration have circular target curves.For the corresponding continuous problems we refer to [2].

4.1. Discrete linear problems? The most prominent example of classicalRiemann-Hilbert problems is the linear problem Im

(f(t) w(t)

)= c(t), where

all target curves are straight lines. So the question arises if there are discretecounterparts of linear Riemann-Hilbert problems in circle packing. The obviousanswer seems to be no: circle packings have no linear structure and thus it ismeaningless to speak of linear problems.

The situation improves if we forget about linearity and emphasize that linearproblems involve straight lines. Since circles prefer to live in hyperbolic geometrywe also switch from the Euclidean to the hyperbolic setting.

In the Euclidean plane a straight line can be interpreted in several ways. Firstof all it is a geodesic. A reformulation of linear Riemann-Hilbert problems in thehyperbolic geometry (of the disk) then leads to special circular problems, whereall target curves are circular arcs which are perpendicular to the unit circle.These problems are certainly worth studying, but we expect that they requirerather strong solvability conditions.

Another approach is based on the observation that a straight line has constantcurvature and contains the point at infinity. The corresponding object in hyper-bolic geometry are the horocycles, i.e. circles in D which touch T. For Riemann-Hilbert problems with horocyclic target curves it makes sense to define a discreteindex. The counterpart of linear homogeneous problems are those horocyclicproblems where all target circles pass through a common point, say the origin.

Of course one can also convert classical linear Riemann-Hilbert problems to cir-cular problems by inversion. If none of the target lines meets the origin (whichcan be achieved for problems with non-negative index by an appropriate trans-formation) this results in Riemann-Hilbert problems with target circles throughthe origin. However, in general the original problem is not equivalent to thetransformed problem, since inversion sends zeros of the solution to poles.

In summary we think that the class of circle packing Riemann-Hilbert problemswith horocyclic target curves resembles the classical linear problems in the bestway.


4.2. Incremental circle packing Riemann-Hilbert problems. We brieflymention a conceptually different approach to linear Riemann-Hilbert problems,based on the fact that the discrete analytic functions over a fixed complex Kform a differentiable manifold C of dimension m + 3 (see [1]). The tangent spaceTC of this manifold carries a linear structure and can be interpreted as a (real)m + 3 dimensional linear subspace of C

m−1 × Cα × Cβ. Here Cα and Cβ standfor the coordinate spaces of the centers of two distinguished alpha (interior) andbeta (boundary) circles, and C

m−1 hosts the centers of the remaining m − 1boundary circles.

For a given discrete analytic function w we consider the tangent space TwC of Cat w as a subspace of C

m−1 × Cα × Cβ and impose m linear conditions

aj dxj + bj dyj = cj, j = 1, . . . , m

on the differentials dzj ≡ dxj + i dyj of the boundary circle centers. Togetherwith three appropriate differential side conditions for the centers of the α andthe β circle these m+3 equations define an m−1-dimensional affine subspace Rof C

m−1×Cα×Cβ. The incremental linear Riemann-Hilbert problem for the dif-ferential of a discrete analytic function w then consists in finding the intersectionof R with TwC.

4.3. The role of branch points. In contrast to conformal mappings solutionsof (classical) Riemann-Hilbert problems are typically not univalent; there aresimple examples of the type |w(t)| = r(t) where all solutions (even those withwinding number zero) have branch points.

The treatment of branch points makes a major difference between the continuousand the discrete setting. To demonstrate this we consider a family of classicalRiemann-Hilbert problems depending continuously on some (homotopy) param-eter λ ∈ [0, 1]. Under appropriate assumptions (see [12, Ch. 4]) the solutionsdepend continuously on λ as well. Assume that we have a locally univalent so-lution for λ = 0 and a branched solution for λ = 1. If λ is changed continuouslyfrom 0 to 1, then one or several branch point(s) of the solution will emerge atthe boundary and then move continuously into the interior of the domain D.

A discrete analytic function cannot mimic this behavior. Experiments haveshown that the packing indeed tells us when it needs a branch point at theboundary (see Figure 12). However it is not clear how to move the branchpoint(s) through the interior of the packing afterwards, because a packing withbranching at some circle Cb cannot be continuously deformed into a packing withbranching at a neighbor of Cb. One promising approach to realize moving branchpoints uses the idea of “fractal branching”, introduced by Ken Stephenson andJames Ashe [10], but the crucial point is to understand what are the “drivingforces” of the branch points.


4.4. Discretized circle packing Riemann-Hilbert problems. So far westudied Riemann-Hilbert problems for circle packings with fixed combinatoricswithout requiring (metric) relations between neighboring target circles. Thisadmits rather weird configurations and certainly not all these problems fit intoa reasonable theory.

The situation is usually better if a discrete problem results from discretizationof a continuous problem, since then neighboring target circles will typically beclose to each other.

We briefly describe a discretization scheme for a continuous Riemann-Hilbertproblem with respect to a given complex K.

Since it is not yet clear how branch points can be treated appropriately we assumethat the continuous problem admits a locally univalent solution w with windingnumber one normalized by w(0) = 0 and w′(0) > 0. To set up a correspondingdiscrete problem modeled over a complex K we start with a maximal packing P ′

for K normalized such that the center of its α-circle is at the origin and the centerof a neighboring β circle is on the positive real line. The boundary circles B′

j

of this packing touch the unit circle at contact points tj. Then we choose thetarget curve Mtj of the continuous problem as target curve for the correspondingboundary circle Bj in the circle packing Riemann-Hilbert problem.

→

Figure 11. On the setting of a discretized Riemann-Hilbert problem.

Assume that this problem admits a unique normalized locally univalent solu-tion P . Then any boundary circle Bj of P touches its target curve Mtj at apoint wj. If this point is unique it is natural to consider wj as the value of thediscrete analytic function P ′ �→ P at tj. A visualization of this approach is givenin Figure 11.

Of course the above procedure will not always work, for instance if the structureof the complex K is too coarse. So the next step is to describe appropriateadaptive refinement procedures for the complex K which leads to a sequenceof complexes K1, K2, K3, . . ., such that the discretized circle packing Riemann-Hilbert problems with the complexes Kj admit unique normalized solutions Pj


for all sufficiently large j. The final goal is then to show that the discrete analyticfunctions P ′

j �→ Pj converge to the solution of the continuous Riemann-Hilbertproblem in an appropriate sense.

Figure 12. Solutions of a discretized problem with refined combinatorics.

Figure 12 shows the results of some test calculations for the simple Riemann-Hilbert problem |w(t)− 0.2 · t4| = 0.8 which was discretized with the heptagonal

complexes K[7]3 , K

[7]4 and K

[7]5 . The cusps at the boundary of the last packing

indicate that the discrete solution tries to approximate branch points emergingat the boundary.

The left part of Figure 13 shows the real and imaginary part of the boundaryfunction for the continuous solution (solid lines) and the circle packing (dots)

modeled on K[7]4 . In the right part we see the traces of the continuous (solid line)

and the discrete solution (dots) for the same complex on the target manifold.

−3 −2 −1 0 1 2 3

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure 13. (a) (b)

Isn’t it amazing that such a relatively coarse packing (observe the big circles inthe middle) does approximate the continuous solution so accurately?

Acknowledgement. We would like to thank Kenneth Stephenson for manyinspiring discussions on circle packing and the subject of this paper and thereferee for his very helpful comments.


References

1. D. Bauer, K. Stephenson and E. Wegert, Circle packings as differentiable manifolds, inpreparation.

2. Ch. Glader and E. Wegert, Nonlinear Riemann-Hilbert problems with circular targetcurves, Math. Nachr. 281 No.9 (2008), 1221–1239.

3. Z.-X. He and O. Schramm, On the convergence of circle packings to the Riemann map,Invent. Math. 125 (1996), 285–305.

4. P. Koebe, Kontaktprobleme der konformen Abbildung, Ber. Sachs. Akad. Wiss. Leipzig,Math.-Phys. Klasse 88 (1936), 141–164.

5. B. Riemann, Grundlagen fur eine allgemeine Theorie der Functionen einer veranderlichencomplexen Grosse, in: R. Dedekind and H. Weber (eds.), Gesammelte mathematischeWerke und wissenschaftlicher Nachlaß, Leipzig 1876, 3–47.

6. B. Rodin and D. Sullivan, The convergence of circle packings to the Riemann mapping, J.Differential Geometry 26 (1987), 349–360.

7. A. Shnirel’man, The degree of a quasi-linearlike mapping and the nonlinear Hilbert prob-lem, (in Russian), Mat. Sb. 89 (1972) 3, 366-389; English transl.: Math. USSR Sbornik18 (1974), 373–396.

8. K. Stephenson, Introduction to Circle Packing — The Theory of Discrete Analytic func-tions, Cambridge Univ. Press, Cambridge 2005.

9. , A probabilistic proof of Thurston’s conjecture on circle packings. Rend. Semin.Mat. Fis. Milano 66 (1998), 201–291.

10. K. Stephenson and J. Ashe, Fractal branching, personal communication, University ofTennessee at Knoxville 2007.

11. E. Wegert, Topological methods for strongly nonlinear Riemann-Hilbert problems for holo-morphic functions, Math. Nachr. 134 (1987), 201-230.

12. , Nonlinear Boundary Value Problems for Holomorphic Functions and SingularIntegral Equations, Akademie Verlag, Berlin 1992.

13. , Nonlinear Riemann-Hilbert problems — history and perspectives, in: N. Pa-pamichael et al. (eds.), Proc. 3rd CMFT Conf. 1997 World Scientific, Ser. Approx. Decom-pos. 11 (1999), 583-615.

Elias Wegert E-mail: [email protected]: Tech Univ Bergakademie Freiberg, Institute of Applied Analysis, D-09596 Freiberg,Germany.

David Bauer E-mail: [email protected]: MPI for Mathematics in the Sciences, Inselstrae 22, D-04103 Leipzig, Germany.

on riemann-hilbert problems in circle packing

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