On Boundary Value Problems in Circle Packing

On Boundary Value Problems in Circle Packing Elias Wegert (with David Bauer and Ken Stephenson) Institute of Applied Analysis TU Bergakademie Freiberg (Germany)

May 19, 2010

E. Wegert (TU Freiberg)

Boundary Value Problems in Circle Packing

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Outline

1

Introduction

2

Riemann-Hilbert Problems for Analytic Functions

3

The Manifold of Circle Packings

4

Circle Packing Riemann-Hilbert Problems

5

Incremental Linear Riemann-Hilbert Problems

6

The Quest for a Hilbert Transform



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Boundary Value Problems in Circle Packing Central Question What are (general) reasonable boundary value problems in circle packing ? Two well-known examples:

maximal packings

prescribed boundary radii

Generalizations: boundary value problems of Riemann and Beurling. E. Wegert (TU Freiberg)


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Circle Packings and Analytic Functions Consider circle packings as discrete models of analytic functions, translate classical boundary value problems into the language of circle packing.

→

A discrete analytic function mimicking a conformal mapping of the unit disc onto an ellipse . The domain packing (left) and the range packing (right) have the same tangency pattern. E. Wegert (TU Freiberg)


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Boundary Value Problems for Analytic Functions

In 1851 Bernhard Riemann published his famous thesis “Grundlagen f¨ ur eine allgemeine Theorie der Functionen einer ver¨anderlichen complexen Gr¨osse”. This paper is well known as the origin of the famous Riemann mapping theorem, but Riemann himself considered conformal mapping just as an example of a more fundamental question. Bernhard Riemanna a

Mathematics



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Riemann’s Well-Posed Boundary Value Problems Which “reasonable” boundary conditions can be imposed on analytic functions?



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Riemann’s Well-Posed Boundary Value Problems Which “reasonable” boundary conditions can be imposed on analytic functions? Riemann considered analytic functions w = u + iv in a domain G and started with a simple example. Es kann also, allgemein zu reden, u am Rande (von G ) als eine ganz willk¨ urliche Function von s gegeben werden, und dadurch ist v u ¨berall mit bestimmt . . .


Generally speaking, u can be prescribed on the boundary (of G ) as an arbitrary function of s, which then also defines v everywhere . . .


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Riemann’s Well-Posed Boundary Value Problems Which “reasonable” boundary conditions can be imposed on analytic functions? Riemann considered analytic functions w = u + iv in a domain G and started with a simple example. Es kann also, allgemein zu reden, u am Rande (von G ) als eine ganz willk¨ urliche Function von s gegeben werden, und dadurch ist v u ¨berall mit bestimmt . . .

Generally speaking, u can be prescribed on the boundary (of G ) as an arbitrary function of s, which then also defines v everywhere . . .

Harmonic continuation of u, conjugate harmonic function v . Leads to Schwarz’ Integral formula. Explicitly for unit disc D Z 1 t +z w (z) = u(t) |dt| + i Im w (0). 2π T t − z



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Nonlinear Riemann-Hilbert Problems (RHP) “Die Bedingungen, welche so eben zur Bestimmung der Function hinreichend und nothwendig befunden worden sind, beziehen sich auf ihren Werth . . . in Begrenzungspunkten, . . . und zwar geben sie f¨ ur jeden Begrenzungspunkt Eine Bedingungsgleichung . . . .”


The conditions which where just found to be sufficient and necessary for determining the function are related to its values . . . at boundary points, . . . namely, they give one equation for each boundary point.


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Nonlinear Riemann-Hilbert Problems (RHP) “Die Bedingungen, welche so eben zur Bestimmung der Function hinreichend und nothwendig befunden worden sind, beziehen sich auf ihren Werth . . . in Begrenzungspunkten, . . . und zwar geben sie f¨ ur jeden Begrenzungspunkt Eine Bedingungsgleichung . . . .”

The conditions which where just found to be sufficient and necessary for determining the function are related to its values . . . at boundary points, . . . namely, they give one equation for each boundary point.

Problem (Riemann-Hilbert Problem) Let f : T × C → R be a given function. Find all functions w which are analytic in the complex unit disc D and continuous on its closure D such that the boundary condition f t, w (t) = 0 is satisfied for all t on the unit circle T. E. Wegert (TU Freiberg)


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Conformal Mapping as Riemann-Hilbert Problem To illustrate his general ideas, Riemann considered conformal mappings of the unit disc D onto a simply connected domain G , “. . . wo also . . . f¨ ur jeden Begrenzungspunkt des Abbildes eine Ortscurve, und zwar f¨ ur alle dieselbe, . . . gegeben ist.”


. . . where . . . for all boundary points of the image a curve is given, namely one and the same for all points.


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Conformal mappings D → G are (special) solutions of the RHP f (w (t)) = 0

for all t ∈ T,

(1)

where f (w ) = 0 describes the boundary of the domain G .



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Conformal mappings D → G are (special) solutions of the RHP f (w (t)) = 0

for all t ∈ T,

(1)

where f (w ) = 0 describes the boundary of the domain G . Besides univalent conformal mappings the Riemann-Hilbert problem (1) also admits constant solutions and branched (conformal) mappings. E. Wegert (TU Freiberg)


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The Modulus Problem Another standard RHP corresponds to f (t, w ) := |w | − 1, where we search for analytic functions in D which are unimodular on the unit circle T. Here the set of solutions (continuous on D) is rather large. It consists of all finite Blaschke products n Y z − zj , w (z) = c 1 − zj z

c ∈ T, zj ∈ D

n = 0, 1, 2, . . . .

j=1



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c ∈ T, zj ∈ D

n = 0, 1, 2, . . . .

j=1

On the other hand there are problems with a single solution or without solution, for example |w (t) − t −1 | = 1,


|w (t) − 2 t −1 | = 1.


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c ∈ T, zj ∈ D

n = 0, 1, 2, . . . .

j=1

On the other hand there are problems with a single solution or without solution, for example |w (t) − t −1 | = 1,

|w (t) − 2 t −1 | = 1.

Both belong to the class of Riemann-Hilbert problems with circular target curves, |w (t) − c(t)| = r (t), which also includes the modulus problem E. Wegert (TU Freiberg)


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Linear Riemann-Hilbert Problems For the well-known linear Riemann-Hilbert problem f is real linear in the second argument, f (t, u + i v ) = a(t) u + b(t) v − c(t), where a, b, c are given real-valued functions with a2 + b 2 6= 0.



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Linear Riemann-Hilbert Problems For the well-known linear Riemann-Hilbert problem f is real linear in the second argument, f (t, u + i v ) = a(t) u + b(t) v − c(t), where a, b, c are given real-valued functions with a2 + b 2 6= 0. In 1920 Fritz Noether discovered that solvability of the problem depends on the “index” κ (winding number about zero) of the “symbol” a + ib.



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Linear Riemann-Hilbert Problems For the well-known linear Riemann-Hilbert problem f is real linear in the second argument, f (t, u + i v ) = a(t) u + b(t) v − c(t), where a, b, c are given real-valued functions with a2 + b 2 6= 0. In 1920 Fritz Noether discovered that solvability of the problem depends on the “index” κ (winding number about zero) of the “symbol” a + ib. In the language of functional analysis, the Riemann-Hilbert operator R : H 2 (D) → L2 (T), w 7→ a Re w + b Im w , maps the Hardy space H 2 (D) into the Lebesgue space L2 (T) and is Fredholm with index 2κ + 1. More precisely dim ker R = max (0, 2κ + 1),


codim im R = max (0, −2κ − 1).


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Some History of Riemann-Hilbert Problems

Problems with special structure: closed solutions, reduction to linear problems (≥ 1941)



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Problems with special structure: closed solutions, reduction to linear problems (≥ 1941) Problems with “small nonlinearities”: perturbations of linear problems and conformal mappings (≥ 1947)



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Problems with special structure: closed solutions, reduction to linear problems (≥ 1941) Problems with “small nonlinearities”: perturbations of linear problems and conformal mappings (≥ 1947) Problems mit “strong nonlinearities”: general nonlinearities with metric restrictions (≥ 1966)



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Problems with special structure: closed solutions, reduction to linear problems (≥ 1941) Problems with “small nonlinearities”: perturbations of linear problems and conformal mappings (≥ 1947) Problems mit “strong nonlinearities”: general nonlinearities with metric restrictions (≥ 1966) Problems with general nonlinearities: topological conditions (≥ 1972)



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Geometric Setting: Back to Bernhard Riemann Riemann himself posed the problem in a geometric language which was practically out of sight for more than a century. Following him we introduce the target curves Mt := {w ∈ C : f (t, w ) = 0},


t ∈ T.


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Geometric Setting: Back to Bernhard Riemann Riemann himself posed the problem in a geometric language which was practically out of sight for more than a century. Following him we introduce the target curves Mt := {w ∈ C : f (t, w ) = 0},

t ∈ T.

Then the boundary condition can be rewritten as w (t) ∈ Mt for all t ∈ T.



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Geometric Setting: the Target Manifold To untangle the family of target curves Mt we lift them from C to T × C. M := {(t, w ) ∈ T×C : f (t, w ) = 0}, is called the target manifold of the problem. The figure shows three colored points t1 , t2 , t3 on T, the associated planes {tj } × C containing the target curves and the target manifold.



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Geometric Setting: the Target Manifold To untangle the family of target curves Mt we lift them from C to T × C. M := {(t, w ) ∈ T×C : f (t, w ) = 0}, is called the target manifold of the problem. The figure shows three colored points t1 , t2 , t3 on T, the associated planes {tj } × C containing the target curves and the target manifold. The red line on M is the graph of the boundary function of a solution w , said to be the trace of w , tr w := {(t, w (t)) : t ∈ T}.



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Admissible Compact Target Manifolds There are several classes of Riemann-Hilbert problems, depending on the global structure and the regularity of the target manifold. Here we consider smooth target manifolds which are built from closed target curves.



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Admissible Compact Target Manifolds There are several classes of Riemann-Hilbert problems, depending on the global structure and the regularity of the target manifold. Here we consider smooth target manifolds which are built from closed target curves. The corresponding RHPs generalize conformal mapping onto smoothly bounded domains.



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Admissible Compact Target Manifolds There are several classes of Riemann-Hilbert problems, depending on the global structure and the regularity of the target manifold. Here we consider smooth target manifolds which are built from closed target curves. The corresponding RHPs generalize conformal mapping onto smoothly bounded domains. A compact target manifold M is said to be admissible, if it has a parametric representation µ ∈ C 1 (T × R) such that M = {(t, µ(t, s)) : t, s ∈ T},



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Admissible Compact Target Manifolds There are several classes of Riemann-Hilbert problems, depending on the global structure and the regularity of the target manifold. Here we consider smooth target manifolds which are built from closed target curves. The corresponding RHPs generalize conformal mapping onto smoothly bounded domains. A compact target manifold M is said to be admissible, if it has a parametric representation µ ∈ C 1 (T × R) such that M = {(t, µ(t, s)) : t, s ∈ T}, (i) for any t ∈ T the mapping s 7→ µ(t, s) is injective,



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Admissible Compact Target Manifolds There are several classes of Riemann-Hilbert problems, depending on the global structure and the regularity of the target manifold. Here we consider smooth target manifolds which are built from closed target curves. The corresponding RHPs generalize conformal mapping onto smoothly bounded domains. A compact target manifold M is said to be admissible, if it has a parametric representation µ ∈ C 1 (T × R) such that M = {(t, µ(t, s)) : t, s ∈ T}, (i) for any t ∈ T the mapping s 7→ µ(t, s) is injective, (ii) there exists a constant C > 1 such that for all s, t ∈ T 0 < 1/C ≤ |∂s µ(t, s)| ≤ C .



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Winding Numbers of Solutions Solutions w can have different winding numbers about the target manifold.



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The pictures show two solutions with winding numbers zero and two, respectively.



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The pictures show two solutions with winding numbers zero and two, respectively. If the origin is contained in the interior of every target curve, this winding number coincides with the number of zeros of w in D. E. Wegert (TU Freiberg)


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Fundamental Solutions of Regular Problems The target manifold (the Riemann-Hilbert problem) is said to be regular if the origin is contained in the interior of every target curve, 0 ∈ int Mt for all t ∈ T.



(2)

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(2)

Theorem (A.I.Shnirel’man, E.W.) If M is a regular admissible target manifold the traces of all solutions without zeros cover the target manifold M in a schlicht manner.



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(2)

Theorem (A.I.Shnirel’man, E.W.) If M is a regular admissible target manifold the traces of all solutions without zeros cover the target manifold M in a schlicht manner. For regular problems the traces of solutions with winding number zero provide the target manifold with a canonical parametrization.



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Arbitrary Solutions of Regular Problems Theorem Assume that M is an admissible regular target manifold, and let n be a nonnegative integer. Then, for arbitrary points t0 ∈ T, w0 ∈ Mt0 , and z1 , . . . , 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 the additional condition w (t0 ) = w0 . For regular problems all traces of solutions have a non-negative winding number about the target manifold.



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Arbitrary Solutions of Regular Problems Theorem Assume that M is an admissible regular target manifold, and let n be a nonnegative integer. Then, for arbitrary points t0 ∈ T, w0 ∈ Mt0 , and z1 , . . . , 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 the additional condition w (t0 ) = w0 . For regular problems all traces of solutions have a non-negative winding number about the target manifold. The family of solutions with winding number n depends on 2n + 1 real parameters. E. Wegert (TU Freiberg)


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Riemann-Hilbert Problems in Circle Packing ? Is there a chance to translate these problem to circle packing ? The conditions which where just found to be sufficient and necessary for determining the function are related to its values . . . at boundary points, . . . namely, they give one (real) equation for each (complex value at a) boundary point . . . . So, roughly speaking, a Riemann-Hilbert problem prescribes “one half” of the boundary data of an analytic function.



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Riemann-Hilbert Problems in Circle Packing ? Is there a chance to translate these problem to circle packing ? The conditions which where just found to be sufficient and necessary for determining the function are related to its values . . . at boundary points, . . . namely, they give one (real) equation for each (complex value at a) boundary point . . . . So, roughly speaking, a Riemann-Hilbert problem prescribes “one half” of the boundary data of an analytic function. Does this fit with circle packing ? Indeed the answer is positive, a finite packing with m boundary circles has m + 3 (real) degrees of freedom. As we shall see, m of these parameters can be directly related to the boundary circles, the remaining 3 are associated with a rigid motion of the packing and can be eliminated by appropriate side conditions.



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The Manifold of Circle Packings



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Circle Packing: Combinatorics In this talk we consider finite circle packings in the Euclidean plane. The skeleton of a packing is its combinatorics, viz. the structure of tangency relations between its circles. The combinatorics of a packing is specified by an abstract simplicial 2-complex K which is a triangulation of an oriented topological surface. We assume that K is a combinatorial closed disc, i.e. finite, simply connected and with nonempty boundary. The vertices, edges and faces of K are denoted by V = {v1 , . . . , vn }, E. Wegert (TU Freiberg)

E = {e1 , . . . , ep },

F = {f1 , . . . , fq }.


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Circle Packings: Definition Recall that we only consider finite circle packings in the Euclidean plane.

Definition (Circle packing) A collection P = {Cv } of circles in C is a circle packing for a complex K , if it satisfies the following: (i) P has a circle Cv associated with each vertex v of K . (ii) Two circles Cu , Cv are externally tangent whenever hu, v i is an edge of K . (iii) If hu, v , w i is a (positively oriented) face of K , then the centers of the circles Cu , Cv , Cw form a positively oriented triangle. We assume that v1 , . . . , vm are the m boundary vertices of K . Circles associated with the boundary vertices of K are termed boundary circles, the others are interior circles. E. Wegert (TU Freiberg)


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Circle Packings: Univalence and Branching A circle packing in which the circles have mutually disjoint interiors is said to be univalent. It is called locally univalent if the chain of neighbors of each interior circle Cj wraps once around Cj . If this winding number is bj + 1 we call bj the branch order of P at Cj .



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The vector (bm+1 , bm+2 , . . . , bn ) is said to be the branch structure of the packing. Branched packings are those with a branch structure b 6= 0.



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The vector (bm+1 , bm+2 , . . . , bn ) is said to be the branch structure of the packing. Branched packings are those with a branch structure b 6= 0. There is a complete description of all “admitted” branch structures for K . E. Wegert (TU Freiberg)


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Rigidity and Flexibility: A Counting Exercise Since K is a combinatorial closed disc with n vertices, p edges, and q faces, we have n + q = p + 1 by Euler’s Theorem. By counting the edges of K in two different ways we obtain 3q = 2p − m, and elimination of q leads to the fundamental relation p = 3n − m − 3. Since a packing with n circles involves 3n real parameters (2n for the centers and n for the radii), and must satisfy p tangency conditions, we expect that 3n − p = m + 3 degrees of freedom remain.



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Rigidity and Flexibility: A Counting Exercise Since K is a combinatorial closed disc with n vertices, p edges, and q faces, we have n + q = p + 1 by Euler’s Theorem. By counting the edges of K in two different ways we obtain 3q = 2p − m, and elimination of q leads to the fundamental relation p = 3n − m − 3. Since a packing with n circles involves 3n real parameters (2n for the centers and n for the radii), and must satisfy p tangency conditions, we expect that 3n − p = m + 3 degrees of freedom remain. Three of these parameters are related to rigid motions (rotation and translation) of the packing in the plane.



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Rigidity and Flexibility: A Counting Exercise Since K is a combinatorial closed disc with n vertices, p edges, and q faces, we have n + q = p + 1 by Euler’s Theorem. By counting the edges of K in two different ways we obtain 3q = 2p − m, and elimination of q leads to the fundamental relation p = 3n − m − 3. Since a packing with n circles involves 3n real parameters (2n for the centers and n for the radii), and must satisfy p tangency conditions, we expect that 3n − p = m + 3 degrees of freedom remain. Three of these parameters are related to rigid motions (rotation and translation) of the packing in the plane. Miraculously, the remaining m parameters can be directly associated with the m boundary circles. E. Wegert (TU Freiberg)


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The Canonical Boundary Value Problem The following theorem shows that the circle packings over K with fixed branch structure can be parametrized by the radii of their boundary circle.

Theorem Let b be an admissible branch structure for the combinatorial closed disc K . Then, for any positive vector r = (r1 , . . . , rm ), there exists a circle packing for K with branch structure b and prescribed boundary radii r1 , . . . , rm . This packing is unique up to rigid motions.



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Theorem Let b be an admissible branch structure for the combinatorial closed disc K . Then, for any positive vector r = (r1 , . . . , rm ), there exists a circle packing for K with branch structure b and prescribed boundary radii r1 , . . . , rm . This packing is unique up to rigid motions. In particular the result holds for locally univalent packings.



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Theorem Let b be an admissible branch structure for the combinatorial closed disc K . Then, for any positive vector r = (r1 , . . . , rm ), there exists a circle packing for K with branch structure b and prescribed boundary radii r1 , . . . , rm . This packing is unique up to rigid motions. In particular the result holds for locally univalent packings. In order to study general boundary value problems we need a stronger version of this result.



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The Manifold of Circle Packings Associating with a circle packing P the radii r := (r1 , . . . , rn ) and the centers z = (z1 , . . . , zn ) of its circles, it can be identified with a point in the ambient space Rn+ × Cn .



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The Manifold of Circle Packings Associating with a circle packing P the radii r := (r1 , . . . , rn ) and the centers z = (z1 , . . . , zn ) of its circles, it can be identified with a point in the ambient space Rn+ × Cn . We denote the subset of all such points (circle packings) in Rn+ × Cn by D, while Db stands for packings with branch structure b.



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Theorem (David Bauer, Ken Stephenson, E.W.) For any fixed combinatorial closed disc K with n vertices and m boundary vertices the set D of circle packings over K is a smooth submanifold of Rn+ × Cn of real dimension m + 3.



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Theorem (David Bauer, Ken Stephenson, E.W.) For any fixed combinatorial closed disc K with n vertices and m boundary vertices the set D of circle packings over K is a smooth submanifold of Rn+ × Cn of real dimension m + 3. The m boundary radii r1 , . . . , rm and three additional “layout parameters” ξ, η, % parametrize Db globally.



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Theorem (David Bauer, Ken Stephenson, E.W.) For any fixed combinatorial closed disc K with n vertices and m boundary vertices the set D of circle packings over K is a smooth submanifold of Rn+ × Cn of real dimension m + 3. The m boundary radii r1 , . . . , rm and three additional “layout parameters” ξ, η, % parametrize Db globally. The manifold D is the disjoint union of the connected components Db with fixed branch structure. The number of components depends on the complex K . E. Wegert (TU Freiberg)


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Global Parametrization of Circle Packings In order to formulate the result more precisely, we denote by zα and zβ the centers of a fixed interior circle (the α-circle) and one of its neighbors (the β-circle) and set ξ + iη := zα ,


eiρ := (zβ − zα )/(|zβ − zα |).


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Global Parametrization of Circle Packings In order to formulate the result more precisely, we denote by zα and zβ the centers of a fixed interior circle (the α-circle) and one of its neighbors (the β-circle) and set ξ + iη := zα ,

eiρ := (zβ − zα )/(|zβ − zα |).

Then, for every admissible branch structure b for K the mapping iρ %b : Db → Rm + × C × T, (r1 , . . . , rn , z1 , . . . , zn ) 7→ (r1 , . . . , rm , ξ + iη, e )

is a diffeomorphism.

Corollary For every admissible branch structure b the mapping %b is a global chart on Db . Its inverse πb is a regular parametrization of Db . E. Wegert (TU Freiberg)


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Ingredients of the Proof: Contact Equations The tangency relations of a circle packing P = (r , z) can be described by a system of contact equations, ωi (r , z) := (xj − xk )2 + (yj − yk )2 − (rj + rk )2 = 0.

(3)

Here ei = hvj , vk i are the edges of K , zj = xj + iyj and zk = xk + iyk for i = 1, . . . , p.



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(3)

Here ei = hvj , vk i are the edges of K , zj = xj + iyj and zk = xk + iyk for i = 1, . . . , p. Attention: Since the contact equations do not reflect the orientation of the triples of mutually tangent circles, the system (3) usually also admits solutions which are not circle packings.



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(3)

Here ei = hvj , vk i are the edges of K , zj = xj + iyj and zk = xk + iyk for i = 1, . . . , p. Attention: Since the contact equations do not reflect the orientation of the triples of mutually tangent circles, the system (3) usually also admits solutions which are not circle packings. The zero set of ω = (ω1 , . . . , ωp ) consists of a finite number of different components, among them are all sets Db consisting of circle packings with branch structure b. E. Wegert (TU Freiberg)


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Ingredients of the Proof: The Jacobian of ω The Jacobian Dω(r , z) of the contact function ω at a point (r , z) is TRB TRI TX TY ∈ Rp×m × Rp×(n−m) × Rp×n × Rp×n .



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Ingredients of the Proof: The Jacobian of ω The Jacobian Dω(r , z) of the contact function ω at a point (r , z) is TRB TRI TX TY ∈ Rp×m × Rp×(n−m) × Rp×n × Rp×n . (r , z) is a circle packing, the reduced matrix If the point of linearization TRI TX TY has (maximal) rank p.



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Ingredients of the Proof: The Jacobian of ω The Jacobian Dω(r , z) of the contact function ω at a point (r , z) is TRB TRI TX TY ∈ Rp×m × Rp×(n−m) × Rp×n × Rp×n . (r , z) is a circle packing, the reduced matrix If the point of linearization TRI TX TY has (maximal) rank p. Then it follows that we get an invertible matrix   TRB TRI TX TY  I 0 0 0  (4) 0 0 N X NY if the last equations involving NX , NY eliminate rigid motion.



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Ingredients of the Proof: The Jacobian of ω The Jacobian Dω(r , z) of the contact function ω at a point (r , z) is TRB TRI TX TY ∈ Rp×m × Rp×(n−m) × Rp×n × Rp×n . (r , z) is a circle packing, the reduced matrix If the point of linearization TRI TX TY has (maximal) rank p. Then it follows that we get an invertible matrix   TRB TRI TX TY  I 0 0 0  (4) 0 0 N X NY if the last equations involving NX , NY eliminate rigid motion. The proof uses two different definitions of angle sums, which coincide for circle packings but not in the ambient space. Another essential ingredient is the discrete maximum principle. The details are technical. E. Wegert (TU Freiberg)


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Circle Packing Riemann-Hilbert Problems



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Circle Packing Riemann-Hilbert Problems Problem Let K be a combinatorial closed disc with m boundary vertices and let M1 , . . . , Mm be a given family of Jordan (target) curves. Find all circle packings P for K such that any boundary circle Bk of P lies in the closure of the domain bounded by the corresponding target curve Mk and meets Mk , Bk ⊂ clos int Mk , Bk ∩ Mk 6= ∅ for k = 1, . . . , m.



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Investigating Circle Packing RHPs Outline of the project:



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Investigating Circle Packing RHPs Outline of the project: Consider packings for a fixed closed combinatorial disc K and find appropriate conditions for the target curves which guarantee that the corresponding problem is solvable.



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Investigating Circle Packing RHPs Outline of the project: Consider packings for a fixed closed combinatorial disc K and find appropriate conditions for the target curves which guarantee that the corresponding problem is solvable. Start with a continuous Riemann-Hilbert problem, discretize it appropriately such that the corresponding circle packing RHP is solvable.



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Investigating Circle Packing RHPs Outline of the project: Consider packings for a fixed closed combinatorial disc K and find appropriate conditions for the target curves which guarantee that the corresponding problem is solvable. Start with a continuous Riemann-Hilbert problem, discretize it appropriately such that the corresponding circle packing RHP is solvable. Show that the solutions of the circle packing problems converge to the solution of the original problem under appropriate refinement of the combinatorics.



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Investigating Circle Packing RHPs Outline of the project: Consider packings for a fixed closed combinatorial disc K and find appropriate conditions for the target curves which guarantee that the corresponding problem is solvable. Start with a continuous Riemann-Hilbert problem, discretize it appropriately such that the corresponding circle packing RHP is solvable. Show that the solutions of the circle packing problems converge to the solution of the original problem under appropriate refinement of the combinatorics. In the above setting the problem is certainly too general to expect “nice” results, there are discretization effects which make the circle packing Riemann-Hilbert problem more involved then its continuous counterpart.



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First Obstacle: No Constants

One obstacle in studying Riemann-Hilbert problems for circle packings comes from the fact that the fundamental class of solutions without zeros is inaccessible in the discrete case.



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One obstacle in studying Riemann-Hilbert problems for circle packings comes from the fact that the fundamental class of solutions without zeros is inaccessible in the discrete case. For the standard example |w (t)| = 1, where Mt = T for all t ∈ T, the set of all solutions consists of the finite Blaschke products. The zero-free solutions are unimodular constants, which have no reasonable counterpart in circle packing.



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One obstacle in studying Riemann-Hilbert problems for circle packings comes from the fact that the fundamental class of solutions without zeros is inaccessible in the discrete case. For the standard example |w (t)| = 1, where Mt = T for all t ∈ T, the set of all solutions consists of the finite Blaschke products. The zero-free solutions are unimodular constants, which have no reasonable counterpart in circle packing. Solution: Consider solutions with winding number one. Think of the Riemann-Hilbert problem as generalization of conformal mapping.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution. But: in general we do not know where the branch points are located.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution. But: in general we do not know where the branch points are located. Idea: Let the packing decide where it needs branch points. Find appropriate indicators.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution. But: in general we do not know where the branch points are located. Idea: Let the packing decide where it needs branch points. Find appropriate indicators. In the classical setting holomorphic functions belong to a linear space and any two functions (univalent or not) are homotopic. But circle packings with different branch structures belong to different components of the manifold D. Changing branch points is a discontinuous operation.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution. But: in general we do not know where the branch points are located. Idea: Let the packing decide where it needs branch points. Find appropriate indicators. In the classical setting holomorphic functions belong to a linear space and any two functions (univalent or not) are homotopic. But circle packings with different branch structures belong to different components of the manifold D. Changing branch points is a discontinuous operation. Idea: “Fractal branching” connects the components D through ambient space.



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Second Obstacle: Branching Even solutions to RHPs with winding number one can have branch points. Since not any branch structure is compatible with a given combinatorics, K must be chosen according to the expected branching of the solution. But: in general we do not know where the branch points are located. Idea: Let the packing decide where it needs branch points. Find appropriate indicators. In the classical setting holomorphic functions belong to a linear space and any two functions (univalent or not) are homotopic. But circle packings with different branch structures belong to different components of the manifold D. Changing branch points is a discontinuous operation. Idea: “Fractal branching” connects the components D through ambient space. For the time being we consider discretized problems where the behaviour of solutions to the continuous problem is known. E. Wegert (TU Freiberg)


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Discretized Riemann-Hilbert Problems: The Setting Let P 0 be a (normalized) maximal packing for K . If tj is the contact point of the boundary circle Bj0 with the unit circle, then we choose the target curve Mtj of the continuous problem as target curve for the corresponding boundary circle Bj in the circle packing Riemann-Hilbert problem.

→

If the boundary circle Bj of the range packing touches its target curve Mtj at one point wj it is natural to consider wj as the value of the discrete solution at tj . E. Wegert (TU Freiberg)


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Test calculations: An Encouraging Example The figures below depict the solution of the Riemann-Hilbert problem |w (t) − 0.2 · t 4 | = 0.8, w (0) = 0, w 0 (0) > 0

[7]

Shown are the range packing modeled on the heptagonal complex K4 , the real and imaginary part of the continuous solution (solid) and the discrete solution (dots), and their traces on the target manifold. E. Wegert (TU Freiberg)


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Test Calculations: Refinement The figure shows the solution of the Riemann-Hilbert problem [6] |w (t) − 0.2 t −4 | = 0.82 discretized with the hexagonal complex K5 .



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The cusps at the boundary of the packing indicate that the discrete solution tries to approximate branch points which emerge at the boundary if 0.82 is replaced by 0.80. E. Wegert (TU Freiberg)


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Why is the Approximation so Good ? First reason: Circle packings provide a conformal approximation.



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Why is the Approximation so Good ? First reason: Circle packings provide a conformal approximation. Second reason: Good approximation of boundary values is automatic.

→



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Why is the Approximation so Good ? First reason: Circle packings provide a conformal approximation. Second reason: Good approximation of boundary values is automatic. Different options boundary values:

→


for

defining

Boundary circle centers


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→

for

defining

Boundary circle centers Contact points



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→

for

defining

Boundary circle centers Contact points (with what ?)



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→

for

defining

Boundary circle centers Contact points (with what ?) Extremal points



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→

for

defining

Boundary circle centers Contact points (with what ?) Extremal points (envelope ?)



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→

for

defining


There are many other local and global constructions. Are there “natural” boundary values of circle packings ?



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→

for

defining


There are many other local and global constructions. Are there “natural” boundary values of circle packings ? Criterion: Boundary values should reflect conformality “best possible”.



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Incremental Linear RHPs



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How to Cage a Circle Packing ? Consider circle packings for complex K normalized to eliminate rigid motion (zα = 0, zβ > 0). How can such packings be deformed ?



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Is a caged packing P rigid when every boundary circle touches the wall ?



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Is a caged packing P rigid when every boundary circle touches the wall ? Can we move the boundary centers along the green lines?



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Is a caged packing P rigid when every boundary circle touches the wall ? Can we move the boundary centers along the green lines? Key: Linearization, study tangent space TP D of D at P. E. Wegert (TU Freiberg)


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Incremental Linear Riemann-Hilbert Problems Intuitive geometric setting gives rise to two types of (incremental) linear Riemann-Hilbert problems in the tangent space TP D of D at P.



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Incremental Linear Riemann-Hilbert Problems Intuitive geometric setting gives rise to two types of (incremental) linear Riemann-Hilbert problems in the tangent space TP D of D at P.

Relevance for nonlinear Riemann-Hilbert Problems: Uniqueness of solution (conformal mapping and RHPs) Homotopy methods for RHPs Numerical methods of Newton type E. Wegert (TU Freiberg)


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Projection on Boundary Centers Can the m parameters of a normalized packing also be associated with its boundary centers ?



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Projection on Boundary Centers Can the m parameters of a normalized packing also be associated with its boundary centers ? Fix complex K , denote by D the manifold of circle packings for K (discrete analytic functions on D). Consider projection on boundary centers ζ : D ⊂ Rn × Cn → Cm , (r1 , . . . , rn , z1 , . . . , zn ) 7→ (z1 , . . . , zm ).

Problem Is is possible to reconstruct a circle packing from its boundary centers ?



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Problem Is is possible to reconstruct a circle packing from its boundary centers ? An effective solution would yield a discrete Cauchy formula.



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Problem Is is possible to reconstruct a circle packing from its boundary centers ? An effective solution would yield a discrete Cauchy formula. Injectivity of the boundary center map ζ on D seems not to be known.



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Problem Is is possible to reconstruct a circle packing from its boundary centers ? An effective solution would yield a discrete Cauchy formula. Injectivity of the boundary center map ζ on D seems not to be known.

Theorem (Ken Stephenson, David Bauer, E.W.) The boundary center map ζ is injective on every component Db . E. Wegert (TU Freiberg)


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Tangent Space of Manifold of Boundary Centers The next result shows that the full flexibility of a circle packing is completely encoded in its boundary centers.



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Theorem (Ken Stephenson, David Bauer, E.W.) For every branch structure b for K the boundary centers of circle packings in Db form a smooth submanifold Cb := ζ(Db ) of Cm with real dimension m + 3 and the boundary center map ζ : Db → Cb is a diffeomorphism.



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Theorem (Ken Stephenson, David Bauer, E.W.) For every branch structure b for K the boundary centers of circle packings in Db form a smooth submanifold Cb := ζ(Db ) of Cm with real dimension m + 3 and the boundary center map ζ : Db → Cb is a diffeomorphism.

Corollary The manifold Cb is globally parametrized by the boundary radii r1 , . . . , rm , and the layout parameters ξ, η, ρ (describing rigid motions). The vectors ∂r1 , . . . , ∂rm , ∂ξ, ∂η, ∂ρ ∈ Cm span the tangent spaces of Cb .



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The Tangent Spaces of the Boundary Center Manifold Each of the three figures visualizes one tangent vector of the tangent space TP C to C at the depicted molecule packing.



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The left one is ∂r1 corresponding to the radius of the shaded circle C1 .



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The left one is ∂r1 corresponding to the radius of the shaded circle C1 . The middle one is ∂ξ corresponding to a horizontal shift of the packing.



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The left one is ∂r1 corresponding to the radius of the shaded circle C1 . The middle one is ∂ξ corresponding to a horizontal shift of the packing. The right one is ∂ρ corresponding to a rotation of the packing.



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Local Frames . . . Fixing layout parameters ξ, η, ρ we get an m dimensional submanifold Cb∗ of Cm . How can the tangent space TP Cb∗ at a packing P be described as a subspace of Cm ?



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Local Frames . . . Fixing layout parameters ξ, η, ρ we get an m dimensional submanifold Cb∗ of Cm . How can the tangent space TP Cb∗ at a packing P be described as a subspace of Cm ? Introduce local frames associated with boundary centers of packing P. Let dxj , dyj be the x, y -coordinate functionals in the plane Cj hosting the center zj . For aj2 + bj2 > 0 the functionals duj := aj dxj + bj dyj dvj := −bj dxj + aj dyj define a basis in Cj .



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Local Frames . . . Fixing layout parameters ξ, η, ρ we get an m dimensional submanifold Cb∗ of Cm . How can the tangent space TP Cb∗ at a packing P be described as a subspace of Cm ? Introduce local frames associated with boundary centers of packing P. Let dxj , dyj be the x, y -coordinate functionals in the plane Cj hosting the center zj . For aj2 + bj2 > 0 the functionals duj := aj dxj + bj dyj dvj := −bj dxj + aj dyj define a basis in Cj . With du := (du1 , . . . , dum ) and dv := (dv1 , . . . , dvm ) any vector w ∈ Cm has coordinates u := hdu, w i ∈ Rm (real part) and v := hdv , w i ∈ Rm (imaginary part). E. Wegert (TU Freiberg)


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...

and Incremental Linear RHPs

Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ?



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...


Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ? The red dot indicates the shift of the u-coordinate of the boundary center.



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...


Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ? So the new center must lie on the green line.



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...


Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ? The pink dot shows a possible position of the new boundary center.



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...


Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ? Projection along the orange line yields the new v coordinate.



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...


Problem: Given the real part hdu, w i of a tangent vector w ∈ TP Cb∗ , find its imaginary part hdv , w i. Intuitively: if we change the u-components of the boundary centers of a packing, what happens with their v -components ? Projection along the orange line yields the new v coordinate.

Problem (Incremental linear Riemann-Hilbert problem) Given c ∈ Rm , find w ∈ TP Cb∗ such that hdu, w i = c. E. Wegert (TU Freiberg)


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Which Local Frames are Appropriate ? Every frame (du, dv ) is generated by a vector g = (γ1 , . . . , γm ) ∈ Cm , hdu, w i = Re (w /g ),

hdv , w i = Im (w /g ).

A frame generated by g is appropriate, if the RHP Re (w /g ) = c is well posed, i.e., it admits a unique solution w ∈ TP C ∗ for every c ∈ Rm .



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hdv , w i = Im (w /g ).

A frame generated by g is appropriate, if the RHP Re (w /g ) = c is well posed, i.e., it admits a unique solution w ∈ TP C ∗ for every c ∈ Rm . Problem: Find explicit geometric conditions for appropriate frames.



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hdv , w i = Im (w /g ).

A frame generated by g is appropriate, if the RHP Re (w /g ) = c is well posed, i.e., it admits a unique solution w ∈ TP C ∗ for every c ∈ Rm . Problem: Find explicit geometric conditions for appropriate frames. What happens if a single component of a frame is rotated ? Let gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ),


j = 1, . . . , m, t ∈ T.


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hdv , w i = Im (w /g ).

A frame generated by g is appropriate, if the RHP Re (w /g ) = c is well posed, i.e., it admits a unique solution w ∈ TP C ∗ for every c ∈ Rm . Problem: Find explicit geometric conditions for appropriate frames. What happens if a single component of a frame is rotated ? Let gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ),

j = 1, . . . , m, t ∈ T.

Lemma For every appropriate g ∈ Cm \ {0} and 0 ≤ j ≤ m the frame generated by gj (t) is appropriate for all t ∈ T except for two values t1 and t2 = −t1 . E. Wegert (TU Freiberg)


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Numerical Experiments: Special Frames The local frames g shown in the figures below are obtained by an optimization procedure. They are special (and perhaps unique) in the sense that they are converted into inappropriate frames gj if an arbitrary single coordinate is rotated by π/2, gj := (γ1 , . . . , γj−1 , iγj , γj+1 , . . . , γm ).



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Tangential Linear Riemann-Hilbert Problems

For the cage problem and more general Riemann-Hilbert problems the conditions on the centers are replaced by tangency conditions. This gives rise to (incremental) tangential Riemann-Hilbert Problems.



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Lemma For every g ∈ Cm \ {0} and j ∈ {1, . . . , m} the tangential Riemann-Hilbert problem for the frame gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ) is well-posed for all t ∈ T with the exception of at most two values t1 and t2 .



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Lemma For every g ∈ Cm \ {0} and j ∈ {1, . . . , m} the tangential Riemann-Hilbert problem for the frame gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ) is well-posed for all t ∈ T with the exception of at most two values t1 and t2 . For the linear RHP involving centers the Jacobian has the form




TRB  0 J= 0


TRI 0 0

TX C NX

 TY S  NY

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Lemma For every g ∈ Cm \ {0} and j ∈ {1, . . . , m} the tangential Riemann-Hilbert problem for the frame gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ) is well-posed for all t ∈ T with the exception of at most two values t1 and t2 . For the linear RHP involving tangents the Jacobian has the form




TRB  I J= 0


TRI 0 0

TX C NX

 TY S  NY

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Lemma For every g ∈ Cm \ {0} and j ∈ {1, . . . , m} the tangential Riemann-Hilbert problem for the frame gj (t) := (γ1 , . . . , γj−1 , tγj , γj+1 , . . . , γm ) is well-posed for all t ∈ T with the exception of at most two values t1 and t2 . If the matrices C and S are replaced by zeros J is invertible.




TRB  I J= 0


TRI 0 0

TX 0 NX

 TY 0  NY

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Special Frames for Tangency Conditions The pictures below illustrate the result of an optimization procedure. Tangency condition at the blue dots generate a special well-posed problem.



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The problem remains well-posed if the tangent at one circle is changed along the green arcs,



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The problem remains well-posed if the tangent at one circle is changed along the green arcs, it gets ill-posed when the red dots are reached.



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The problem remains well-posed if the tangent at one circle is changed along the green arcs, it gets ill-posed when the red dots are reached. The green arcs have angular length greater than π and cover the exterior boundary of the packing. E. Wegert (TU Freiberg)


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The problem remains well-posed if the tangent at one circle is changed along the green arcs, it gets ill-posed when the red dots are reached. The green arcs have angular length greater than π and cover the exterior boundary of the packing. Tangency conditions outperform center conditions. E. Wegert (TU Freiberg)


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A Criterion for Well-Posed Tangential RHPs ? Conjecture: The incremental linear tangential Riemann-Hilbert problem is well-posed if any tangent is attached at a point of the exterior arc of its boundary circle.



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A Criterion for Well-Posed Tangential RHPs ? Conjecture: The incremental linear tangential Riemann-Hilbert problem is well-posed if any tangent is attached at a point of the exterior arc of its boundary circle.

If the green arcs would not completely cover the exterior boundary of the packing we would have a counterexample. E. Wegert (TU Freiberg)


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The Quest for a Hilbert Transform



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A Classical Generalized Hilbert Transform For all appropriate frames g there is a transition operator Hg : u 7→ v , which maps the real part u of a tangent vector to its imaginary part v .



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A Classical Generalized Hilbert Transform For all appropriate frames g there is a transition operator Hg : u 7→ v , which maps the real part u of a tangent vector to its imaginary part v . How does this operator look like in the classical setting ?



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A Classical Generalized Hilbert Transform For all appropriate frames g there is a transition operator Hg : u 7→ v , which maps the real part u of a tangent vector to its imaginary part v . How does this operator look like in the classical setting ? If g : T → C \ {0} is a function such that g (t) defines the u-axis of the local frame we have w = (u + iv ) g with frame coordinates u = Re(w /g ),


v = Im(w /g ).


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v = Im(w /g ).

For g ≡ 1 this is the usual decomposition w = u + iv into real and imaginary part and the mapping u 7→ v is the Hilbert transform, v = Hu.



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v = Im(w /g ).

For g ≡ 1 this is the usual decomposition w = u + iv into real and imaginary part and the mapping u 7→ v is the Hilbert transform, v = Hu. In general w is the solution of a linear Riemann-Hilbert problem.



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v = Im(w /g ).

For g ≡ 1 this is the usual decomposition w = u + iv into real and imaginary part and the mapping u 7→ v is the Hilbert transform, v = Hu. In general w is the solution of a linear Riemann-Hilbert problem. If g has winding number zero, v is given by a generalized Hilbert transform, v = Hg u := µ H(µu), where µ := exp(H arg g ) is a positive function on T. E. Wegert (TU Freiberg)


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Properties of The Hilbert Transform

The classical Hilbert transform has eigenvalues ±i, which is a consequence of the invariance of analytic functions with respect to multiplication by i (rotations by π/2).



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The classical Hilbert transform has eigenvalues ±i, which is a consequence of the invariance of analytic functions with respect to multiplication by i (rotations by π/2). The corresponding eigenfunctions are the analytic and anti-analytic functions, respectively. With t k = eikτ for k ∈ Z we get H : cos kτ 7→ sin kτ,


H : sin kτ 7→ − cos kτ.


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The classical Hilbert transform has eigenvalues ±i, which is a consequence of the invariance of analytic functions with respect to multiplication by i (rotations by π/2). The corresponding eigenfunctions are the analytic and anti-analytic functions, respectively. With t k = eikτ for k ∈ Z we get H : cos kτ 7→ sin kτ,

H : sin kτ 7→ − cos kτ.

The generalized Hilbert transform Hg := µ H(µ . ) has the same eigenvalues ±i. The eigenfunctions are the analytic and anti-analytic functions scaled by multiplication by 1/µ.



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Experiment: Eigenvalues of the Hilbert Transform [6]

[7]

For the maximal packings with combinatorics K36 and K6 the eigenvalues of the Hilbert transform are perfectly real. The figures show the distribution of their imaginary parts (left center condition, right tangency condition).



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Experiment: The transition operator [6]

Test transition operator u 7→ v on maximal (center) packing K36 . Optimal performance: input u(τ ) = cos kτ , output u(t) = sin kτ .



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Experiment: The transition operator [7]

Test transition operator u 7→ v on maximal (center) packing K6 . Optimal performance: input u(τ ) = cos kτ , output u(t) = sin kτ .



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Experiment: Local Frames and Eigenvalues The eigenvalues of the Hilbert transform depend on the local frames.



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Experiment: Local Frames and Eigenvalues The eigenvalues of the Hilbert transform depend on the local frames.

Is there always a special choice where all eigenvalues are imaginary ? E. Wegert (TU Freiberg)


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