Abstract
Tessellations of the Euclidean plane and a circle, a model of the hyperbolic plane, impress by their beauty, elegance and balance equally mathematical amateurs as mathematicians. Besides its impressive figures, esthetic and elegance of related basic formulas affect scientists. The parqueting-reflection method provides besides the tiling related fundamental solutions to complex partial differential equations in the domains of the tessellation.
1. Mathematics and beauty
Longing and aspiration for beauty are implemented into human beings. They do accompany them all their lives for their creative activities and their feelings. Symmetry and balance determine beautiful appearances. They convey aesthetic, calmness, and peace. In mathematics beauty mirrors in like manner as well in geometric figures as in related formulas.
The Pythagorean theorem for right-angled triangles with the catheti a and b and the hypotenuse c in the form 
a2+b2=c2 is easily remembered and very aesthetical (Figure ).
Figure 1. Pythagorean triangle.
For entire triples (a,b,c) this formula leads to Pythagorean numbers, e.g. (3,4,5). As Pierre de Fermat has stated in the 17. century, such entire numbers triple does not exist for any natural k>2, satisfying ak+bk=ck. A deep mathematical proof is only given 30 years ago by Andrew Wiles [Citation1, Citation2].
The Pythagorean theorem with its geometrical background is beautiful, naturally aesthetical, and easily to communicate and to prove. Likewise the statement of the last Fermat theorem is easily to communicate but neither natural nor aesthetical, as negative in its statement and with profound proof. An aesthetical generalization (without geometric background) of the Pythagorean triple for k = 3 rather than for the power 2 is the search for entire quadruples (a,b,c,d), satisfying the relation a3+b3+c3=d3. A possible choice is (3,4,5,6).
2. Parqueting-reflection principle
The development of computer has enabled many people to realize and to enjoy also mathematical insights and results. Figures of fractal curves and of planar and spatial tessellations e.g. are available in internet and became popular also outside mathematics. An example is the work of the Dutch drawer and painter M.C. Escher [Citation3], who without mathematical background inspired from Moorish ornaments has created beautiful tiling figures. Scholars and students groups are experiencing their mastering of computer programming via machine learning with planar and hyperbolic tiling. A planar tiling can be created by continued reflection at circles and lines. Besides creating marvelous tessellations at the same time the reflection process can be utilized to produce useful formulas, which serve for treating differential equations in the domains of the tiling. Moreover, these formulas representing kernel functions for certain fundamental solutions of partial differential equations, do mirror the beauty of the tessellations themselves.
Within the process of creating a general theory of complex partial differential equations of arbitrary order since about three decades mainly model equations are studied. At one hand fundamental kernel functions for the Cauchy–Riemann and the Laplace differential operators are developed explicitly for certain planar domains [Citation4–34]. On the other hand kernel functions are constructed for arbitrary powers of these differential operators and their products [Citation35–61]. The investigation in higher dimensional Euclidean spaces are very challenging and restricted to half spaces and balls [Citation62–67]. Kernel functions for powers of Dirac and Laplace operators are also developed in quaternionic, octonionic and Clifford analysis [Citation68–72,Citation100]. Many young scientists have contributed to the theory as the theses giving reference for [Citation73–94].
This article is reporting on the parqueting reflection process for constructing certain kernel functions for circular domains in the complex plane. From mathematical viewpoint, lines and circles are different samples of the same object. A line is a circle with infinite radius and center at infinity. Reflections at them can mathematically be expressed by related formulas.
A planar domain D, the boundary ∂D of which is built from finitely many arcs from lines and from circles, is a circular polygon in the plane with finitely many boundary arcs from lines or circles. Reflections at these boundary arcs reflect D onto neighboring circular domains, being themselves circular polygons. In this process the fact is important, that reflections preserve (under change of orientation) the set of lines and circles, occasionally called generalized circles. With these circular polygons the reflection process can be repeated. For parqueting of the entire plane such circular polygons are relevant, this continued reflection process of which reveals a set of mutually disjoint polygons the closures of which cover the plane completely. Simple examples are a half plane, the reflection of which at the boundary line leads to the complementary half plane and a disc, the reflection of which at the boundary circle exchanges the inner and the outer discs with one another. Other examples are obviously, half and quarter discs also planes, rectangles, equilateral triangles, parallel strips, certain lenses (digons), half hexagon. A concentric circular ring continued reflected at the outer circle leads to a cover of its outer complement by further concentric rings, while the continued reflection at the inner circle reveals a cover of the inner disc up to the center by further concentric rings. Nevertheless, such exceptional singularities of the covering can still be accepted, but require additional attention [Citation95]. Further examples will be discussed later on.
Two cases appear when tiling the plane. Either finitely many or countably infinite reflections are necessary for the covering. Finite tiling domains are easier to handle than infinite ones. In order to determine certain fundamental solutions for complex partial differential equations in circular polygons D, an arbitrary point z∈D is fixed and its trace in the tiling process is observed. Reflecting a point z of the complex plane (at a line/circle) reveals a point zˆ. It is expressed in dependence of the coefficients of the equation of the generalized circle as a linear transformation in the variable z¯, called inversive transformation. A further reflection leads then all together to a linear transformation in the original variable z, hence a Möbius transformation. Thus, two kinds of points appear in the trace of the point z, either inversive transformations or Möbius transformations in z.
Three kinds of kernel functions are in the focus of this investigation: the Schwarz kernel for the Cauchy–Riemann operator, in particular for analytic functions, and the harmonic Green and Neumann functions for the Poisson equation in the domains of the tiling. They are related to the basic model complex partial differential operators. For the Schwarz kernel the Cauchy–Pompeiu integral representation formula, applied to z∈D and to its trace points, which are all outside D, are properly combined. For the Green function a meromorphic function of the point ζ∈D is constructed, formally having a simple pole at z∈D and at all its trace points being Möbius transformations of z, and having simple zeros at all inversive transformations of z in its trace. This is just achieved by a unique product, say P~(z,ζ). Then G(z,ζ)=log|P~(z,ζ)|2 turns out to be the Green function as well for D as all its reflection domains in the tiling. For the Neumann function likewise a holomorphic function is constructed in ζ∈D, formally having simple zeros in z and in all its trace points, Q~(z,ζ). The Neumann function then turns out to be N(z,ζ)=−log|Q~(z,ζ)|2. As long as a finite parqueting is dealt with these principles work out fine with some modification. Problems appear for infinite parqueting as convergence problems appear, which require attention.
3. Schwarz kernel
An analytic function is essentially, up to an imaginary constant summand, determined by the boundary values of its real part. The reason for this fact is the connection of real and imaginary parts of an analytic function f = u + iv via the Cauchy–Riemann equations. In complex form this Cauchy–Riemann system has the form ∂z¯f=0 with the partial differential operator 2∂z¯=∂x+i∂y. The tool for this result is a transfer of the Cauchy kernel to the Schwarz kernel function. The transformation of the Cauchy integral to the Schwarz integral operator is generally manageable for any domain with a harmonic Green function [Citation96], see also [Citation36, Citation38, Citation47, Citation97]. The Cauchy representation formula for analytic functions is a particular form of the general Cauchy–Pompeiu representation of functions with first-order partial derivatives not necessarily satisfying the Cauchy–Riemann systemn.
Figure 2. Unit disc D.
Cauchy–Pompeiu formula For functions w∈C1(D;C)∩C1(D¯;C) in a regular domain D of the complex plane C the relation 12πi∫∂Dw(ζ)dζζ−z−1π∫Dwζ¯(ζ)dξdηζ−z={w(z),z∈D,0,z∈C∖D¯,is satisfied. A simple modification S(z,ζ)=2ζζ−z−1=ζ+zζ−z of the Cauchy-kernel 1ζ−z in case of the unit disc D=D={|z|<1} reveals the representation formula w(z)=12πi∫|ζ|=1Rew(ζ)S(z,ζ)dζζ+12π∫|ζ|=1Imw(ζ)dζζ−12π∫|ζ|<1[wζ¯(ζ)ζζ+zζ−z+wζ¯(ζ)¯ζ¯1+zζ¯1−zζ¯]dξdη,ζ=ξ+iη,for z∈D. The advantage of the Schwarz kernel for D is ReS(z,ζ)=P(z,ζ)=ζζ−z+ζ¯ζ−z¯−1 for z∈D,ζ∈∂D, which is the Poisson kernel for D.
The Schwarz kernel function can be constructed with the parqueting reflection method for other circular polygons. Some examples are listed here as well as the harmonic Green functions which are also related to the Poisson kernel as essentially the normal derivative of the Green function at the boundary of the domain D.
3.1. Finite parqueting domains
The basic examples are the upper half plane H+={z:0<Imz} and the unit disc D={|z|<1}. Reflecting z∈H+ at the real axis reveals z¯ while reflecting z∈D at the unit circle gives 1z¯.
The analytic Schwarz kernel of
unit disc D (Figure )
S(z,ζ)=ζ+zζ−z=2ζζ−z−1, see e.g. [Citation97]. Its real part ReS(z,ζ)=P(z,ζ)=ζζ−z+ζ¯ζ−z¯−1 is the Poisson kernel for D. The modified Cauchy–Schwarz–Pompeiu representation formula states w(z)=12πi∫∂DRew(ζ)ζ+zζ−zdζζ+iImw(0)−12π∫D(wζ¯(ζ)ζζ+zζ−z+wζ¯(ζ)¯ζ¯1+zζ¯1−zζ¯)dξdη,z∈D.Its polyanalytic generalization is for n∈N w(z)=∑ν=0n−1(−1)ν2πiν!∫∂DRe∂ζ¯νw(ζ)ζ+zζ−z(ζ−z+ζ−z¯)νdζζ+i∑ν=0n−1Im∂ζ¯νw(0)ν!(z+z¯)ν−(−1)n−12π(n−1)!∫D(∂ζ¯nw(ζ)ζζ+zζ−z+∂ζ¯nw(ζ)¯ζ¯1+zζ¯1−zζ¯)(ζ−z+ζ−z¯)ndξdη,z∈D, see [Citation45, Citation46].
upper half plane H+ (Figure )
Figure 3. Upper half plane H+.
The analytic Schwarz kernel and the Poisson kernel are S(z,t)=1i(t−z),P(z,t)=ReS(z,t)=12i(t−z)−12i(t−z¯)=z−z¯2i|t−z|2=y|t−z|2for z=x+iy∈H+, t∈R, see [Citation37, Citation80].
unbounded sector Sπn={0<argz<πn},ω=eiπn,n∈N, (Figure )
has S(z,ζ)=1i∑k=0n−12ζ−ze2kπin,ζ∈∂Sπn,z∈Sπn,as Schwarz kernel, and P(z,ζ)=1i∑k=0n−1(1ζ−ze2kπin−1ζ−z¯e−2kπin),ζ∈∂Sπn,z∈Sπn,as Poisson kernel, see [Citation6].
upper half unit disc D+=D∩H+ (Figure )
the Schwarz kernel is S(z,ζ)={ζ+zζ−z−ζ¯+zζ¯−z,|ζ|=1,0<Imζ,z∈D+,2(1ζ−z−z1−zζ),|Reζ|<1,Imζ=0,z∈D+,see [Citation21].
digon
Figure 4. Unbounded sector for π5.
Figure 5. Upper half unit disc D+.
A circular digon in C∞ is a domain whose boundary is composed of two circular arcs of C∞ with two intersection points. It is a circular polygon with two circular arcs and two vertices.
Figure demonstrates all the four cases of circular digon domains. They are lune domains, circular segment domains, lens domains and cones, respectively corresponding to Figures (a–d).
Figure 6. Four cases of circular digons.
In the last case, as seen in Figure (d), the two boundary rays of a cone are considered to intersect at the corner point and at infinity.
The set Dα,θ:={zz¯−1<0<zz¯sin(α−θ)+zsinθ+z¯sinθ−sin(α+θ)},where 0<α<π, 0<θ≤π, is a circular digon with the two corner points v±:=e±iα and two boundary arcs C0:=∂Dα,θ∩{zz¯sin(α−θ)+zsinθ+z¯sinθ−sin(α+θ)=0},C1:=∂Dα,θ∩{zz¯=1}.C0 and C1 intersect at the corner points v+ and v− with the intersection angle θ. C1 is part of the unit circle {|z|=1}. If α=θ, {zsinθ+z¯sinθ−sin(α+θ)=0} is a straight line and C0 is the straight line segment connecting v+ and v−. In this case, Dα,θ looks like Figure (b).
If α≠θ, C0 is part of the circle whose center is c0:=sinθ/sin(θ−α) and radius is r0:=sinα/|sin(θ−α)|. If α>θ, the domain Dα,θ is a lune domain, while for α<θ, Dα,θ is a lens domain.
Figure depicts the details of Dα,θ in the case of α>θ.
Figure 7. Dα,θ in the case of α>θ.
The Schwarz kernel for Dα,θ is for θ=πn,n∈N S(z,ζ)={∑k=0n−11sinαsin(α−θ)(ζ+z2k)+2sinθζ−z2k,ζ∈C0,z∈Dα,θ,∑k=0n−1ζ+z2kζ−z2k,ζ∈C1,z∈Dα,θ,where z2k=−zsin(α−kθ)+sinkθzsinkθ−in()α+kθ, see [Citation84].
quarter plane Q++={0<Rez,0<Imz} (Figure )
The Schwarz kernel is S(z,ζ)={4zξ2−z2,0<ξ,η=0,z∈Q++,4izη2+z2,0<η,ξ=0,z∈Q++,ζ=ξ+iη, see [Citation4, Citation5].
isosceles orthogonal triangle with the vertices 0, 1, i (Figure )
Figure 8. Quarter plane Q++.
Figure 9. Isosceles orthogonal triangle.
S(z,ζ;z0)=∑m,n∈Z[gm,n(z,ζ)−12[gm,n(z0,ζ)+gm,n(z0¯,ζ)]],gm,n(z,ζ)=1ζ−z−2m−2ni+1ζ+zi−(2m+1)−(2n+1)i+1ζ−zi−(2m+1)−(2n−1)i+1ζ+z−(2m+2)−2ni,m,n∈Z,see [Citation60, Citation61].
circle sector S1,πn={|z|<1,0<argz<πn} (Figure )
has S(z,ζ)={∑k=0n−1(ζ+ze2kπinζ−ze2kπin−ζ¯+ze−2kπinζ¯−ze−2kπin),|ζ|=1,;0<argζ<πn,z∈S1,πn,∑k=0n−1(1ζ−ze2kπin−ze2kπin−zζ),0<|ζ|<1,argζ∈{0,πn},z∈S1,πn,as Schwarz kernel, see [Citation93].
Figure 10. Circle sector for π6.
have commonly S(z,ζ)={[2ζζ−z−1+2ζ(1−mz)ζ(1−mz)+z−m−1],ζ∈∂D∩∂D,z∈D,[2(ζ−m)ζ−z−1+2[ζ−m)(1−mz)(ζ−m)(1−mz)−zr2−1],ζ∈∂D∩∂Dm(r),z∈D,as their Schwarz kernel functions, see [Citation11].
acorn domain (Figure ) D={z∈C||z−(1/2+i)|<1/2,|z−(1+i/2)|<1/2,|z|<1}
has S(z,ζ)={2z−(1+2i)ζ−z+2z−(2+i)(2+2i)zζ−3iζ−3iz−(2−2i)−z(1−2i)zζ−2ζ−2z+(1+2i)−3iz+(2−2i)(2−i)zζ−2ζ−2z+(2+i)+4,z∈∂D∩C1;2z−(2+i)ζ−z−2z−(1+2i)(2+2i)zζ−3iζ−3iz−(2−2i)+3iz+(2−2i)(1−2i)zζ−2ζ−2z+(1+2i)−z(2−i)zζ−2ζ−2z+(2+i)+4,z∈∂D∩C2;2zζ−z+2(3iz+(2−2i))(2+2i)zζ−3iζ−3iz−(2−2i)+2(2z−(1+2i))(1−2i)zζ−2ζ−2z+(1+2i)+2(2z−(2+i))(2−i)zζ−2ζ−2z+(2+i)+4,z∈∂D∩C3.as its Schwarz kernel function, see [Citation28].
3.2. Infinite parqueting domains
The Schwarz kernel for the
circular ring Rr,1={0<r<|z|<1} (Figure )
is S(z,ζ)=ζ+zζ−z+2∑k=1∞(r2kζr2kζ−z+r2kzζ−r2kz),z∈Rr,1,ζ∈∂Rr,1,see [Citation57, Citation91];
upper half concentric ring Rr,1+={0<r<|z|<1,0<Imz} (Figure )
is S(z,ζ)={{ζ+zζ−z−ζ¯+zζ¯−z+2∑n=1∞r2n[ζr2nζ−z−zr2nz−ζ−ζ¯r2nζ¯−z+zr2nz−ζ¯]},|ζ|=1,0<Imζ,z∈R+,−{ζ+zζ−z−ζ¯+zζ¯−z+2∑n=1∞r2n[ζr2nζ−z−zr2nz−ζ−ζ¯r2nζ¯−z+zr2nz−ζ¯]},|ζ|=r,0<Imζ,z∈R+,2{∑n=1∞1t−z−z1−zt+∑n=1∞r2n[1r2nζ−z−zζ(r2nz−ζ)−1ζ(r2n−zζ)+zr2nzζ−1]},ζ∈[−1,r]∪[r,1],z∈R+,see [Citation21];
quarter ring Rr,1++={0<r<|z|<1,0<x=Rez,0<y=Imz} (Figure )
is S(z,ζ)={2[ζ2+z2ζ2−z2−ζ¯2+z2ζ¯2−z2]+4∑k=1∞r4k[ζ2r4kζ2−z2−z2r4kz2−ζ2+z2r4kz2−ζ¯2−ζ¯2r4kζ¯2−z2],|ζ|=1,0<ξ,0<η,−2[ζ2+z2ζ2−z2−ζ¯2+z2ζ¯2−z2]−4∑k=1∞r4k[ζ2r4kζ2−z2−z2r4kz2−ζ2+z2r4kz2−ζi3)2(ζ−ωm,n+1−i3)3−(z+1−i3)3,3(ζ−ωm,n+2)2(ζ−ωm,n+2)3−(z+2)3},and ωm,n=3m+i3n,m+n∈2Z, see [Citation89].
Figure 12. Acorn domain.
4. Green and Neumann functions
Harmonic functions, closely related to analytic functions, play a fundamental role in potential theory. The basic Laplace operator, a second-order partial differential operator, is central partial differential operator in mathematical physics. Integral representation formulas related to the Laplace operator, adjusted to natural boundary conditions, are determined by the harmonic Green and Neumann kernel functions. For certain cyclic planar polygon domains D they can be constructed by the parqueting-reflection method. Also their conformal and their inversive transformation invariance, [Citation17, Citation84], are available for a construction. As function of two variables they are symmetric, harmonic up to a characteristic logarithmic singularity when both variables coincide, but differ in their boundary behavior. While the Green function G1(z,ζ) vanishes at the boundary, the Neumann function N1(z,ζ) has piecewise constant normal derivatives at the boundary, in general these constants differ on different boundary components. Green representation formulas are used for solving the Dirichlet boundary value problem, Neumann functions are needed to solve Neumann boundary value problems. In the Dirichlet problem the boundary values, in the Neumann problem the normal derivatives of the function in question are prescribed. A linear combination of Green and Neumann functions is the Robin function, a real-valued interpolation of Green and Neumann kernels, R1;α,β(z,ζ),α,β∈R,0<α2+β2 satisfying R1;0,β=N1,R1;α,0=G1 [Citation23, Citation26, Citation79].
Figure 13. Circular ring Rr,1.
Figure 14. Upper half circular ring Rr,1+.
Figure 15. Quarter ring Rr,1++.
Definition
For α,β∈R, 0<α2+β2, a real-valued function R1;α,β(z,ζ),z,ζ∈D,z≠ζ, is called Robin function for D, if for any ζ∈D it has the properties
R1;α,β(⋅,ζ) is harmonic in D∖{ζ} and continuously differentiable in D¯∖{ζ},
h(z,ζ)=R1;α,β(z,ζ)+log|ζ−z|2 is harmonic for z∈D,
αR1;α,β(z,ζ)+β∂νzR1;α,β(z,ζ)=βσ(s) for z=z(s)∈∂D, where the density function σ is the real-valued continuous, piecewise constant function of s from the boundary condition for the Neumann function,
β∫∂Dσ(sz)R1;α,β(z,ζ)dsz=0 (normalization).
The respective representation formulas for proper functions are [Citation9, Citation25, Citation39] w(z)=−14π∫∂Dw(ζ)∂νζG1(z,ζ)dsζ−1π∫Dwζζ¯(ζ)G1(z,ζ)dξdηand w(z)=−14π∫∂D{w(ζ)∂νζN1(z,ζ)−∂νw(ζ)N1(z,ζ)}dsζ−1π∫Dwζζ¯(ζ)N1(z,ζ)dξdηrelated to the Dirichlet and the Neumann problems, respectively.
Dirichlet problem. Find for f∈Lp(D;C), 2<p, γ∈C(∂D;C) a solution to wzz¯=finD,w=γon∂D.
Neumann problem. Find for f∈Lp(D;C), 2<p, γ∈C(∂D;C), c∈C a solution to wzz¯=finD,∂νw=γon∂D,12π∫∂Dw(ζ)dsζ=c.
4.1. Finite parqueting domains
Unit disc D={|z|<1} (see Figure )
G1(z,ζ)=log|1−zζ¯ζ−z|2,N1(z,ζ)=−log|(ζ−z)(1−zζ¯)|2with ∂νzN1(z,ζ)=−2 on ∂D, R1;α,β(z,ζ)=log|1−zζ¯ζ−z|2+{2β∑k=1∞(zζ¯)k+(z¯ζ)kα+kβ,−αβ∉N,2β∑k=1,k≠k0∞(zζ¯)k+(z¯ζ)kα+kβ+2[(zζ¯)k0log(zζ¯)+(z¯ζ)k0log(z¯ζ)],k0=−αβ∈N,see [Citation22, Citation26].
Upper (lower) half of D,D±=D∩H±(H+={0<Imz},H−={Imz<0}) (Figure )
G1(z,ζ)=log|(1−z¯ζ)(ζ−z¯)(ζ−z)(1−zζ)|2,N1(z,ζ)=−log|(ζ−z)(ζ−z¯)(1−zζ)(1−z¯ζ)|2,also a modification of N1 N~1(z,ζ)=2log|zζ|2−log|(ζ−z)(ζ−z¯)(1+zζ)(1−z¯ζ)|2,see [Citation21] and also [Citation15].
Right (left) half of D,Dr=D∩{0<Rez}(Dl=D∩{Rez<0}) (Figure )
G1(z,ζ)=log|1−z¯ζζ−zζ+z¯1+zζ|2,N1(z,ζ)=−log|(ζ−z)(1−z¯ζ)(1+zζ)(ζ+z¯)|2,see [Citation14].
Circle sector Dα={z:|z|<1,0<argz<απ},0<α<2 (Figure )
Figure 16. Half hexagon P+.
Figure 17. Halfs of unit disc.
Figure 18. Right (left) half of D,Dr(Dl).
Figure 19. Circle sector D23π.
The Green function for α=1, see before for D+, is log|1−z¯ζζ−zζ−z¯1−zζ|2.Applying the function z1α mapping Dα onto D1=D+, the conformal invariance of the Green function reveals the Green function for Dα as log|1−(z¯ζ)1αζ1α−z1αζ1α−z¯1α1−(zζ)1α|2.This result can be attained also by the parqueting-reflection method.
Let α=mn,0<m<n,m,n∈N, be a rational number and S1=Dα be the respective sector in D. Reflecting S1 at the ray {argζ=mnπ} maps it onto S2=D2α∖Dα¯. At the same time a point z∈S1 is reflected onto exp(2mnπi)z¯∈S2. Continuing consecutively reveals as well the sectors S2k={z:|z|<1,(2k−1)mnπ<argz<2kmnπ},S2k+1={z:|z|<1,2kmnπ<argz<(2k+1)mnπ}as the points z2k=exp(2kmnπi)z¯∈S2k,z2k+1=exp(2kmnπi)z∈S2k+1for 1≤k and z∈S1. In particular, z2n=exp(2mπi)z¯=z¯∈S2n={ζ:|ζ|<1,−mnπ<argζ<0},z2n+1=exp(2mπi)z=z∈S2n+1=S1.The same sectors and points are attained when reflecting S1 at the ray argζ=0 and continuing consecutively clockwise. The covering of the disc D is completed to a covering for the entire plane C. The related parqueting-reflection procedure provides the meromorphic function of ζ P(z,ζ)=∏k=1nζ−exp(2kmnπi)z¯ζ−exp(2kmnπi)zζz¯−exp(2kmnπi)ζz−exp(2kmnπi)=ζn−z¯nζn−zn1−(z¯ζ)n1−(zζ)n.But the parqueting involved is m-fold. This is seen from D¯=⋃k=12nSk¯=⋃k=0n−1(S2k+1¯∪S2(k+1)¯)={z:|z|<1,0≤argz≤2mπ}and reflects to an m-fold covering of C. Applying the m-th root reveals a simple covering and thus log|P(z1m,ζ1m)|2=log|ζmn−z¯mnζmn−zmn1−(z¯ζ)mn1−(zζ)mn|2is the Green function for Dmn.
In case α,0<α<1, is irrational, a rational monotone increasing sequence {rn} approximating α=limn→∞rn, the sequence of the related Green functions log|ζ1rn−z¯1rnζ1rn−z1rn1−(z¯ζ)1rn1−(zζ)1rn|2tends to the Green function log|ζ1α−z¯1αζ1α−z1α1−(z¯ζ)1α1−(zζ)1α|2of the circle section Dα. This is a conclusion of the Wiener principle of domain exhaustions, see [Citation98], p.335. This result can also be extended for Dα with 0<α<2, see [Citation25]. The Neumann function is attained likewise [Citation25].
Almaty apple D={|z|<1,z+z¯<1} (Figure )
Figure 20. Almaty apple.
Parqueting-reflection has to be applied to the Almaty apple D={|z|<1,z+z¯<1} and its complement with respect to the unit disc D∖D¯={|z|<1<z+z¯}. Two reflections have to be executed. Reflecting D at the vertical line z+z¯=1 leads to the image Dˆ={|z−1|<1<z+z¯}. The line z+z¯=1 is reflected at the unit circle onto the circle |z−1|=1. At the same time the set D2={z+z¯<1,|z−1|<1}is mapped onto Dˆ3={1<z+z¯,1<|z−1|}.Similarly, Dˆ2={|z|<1<z+z¯}is mapped onto Dˆ1={|z−1|<1<|z|},and D1={|z|<1<|z−1|}onto D3={1<|z|,z+z¯<1}.The symmetry of the figures involved suggests the reflection at the circle |z−1|=1. The images of the respective sub-domains are needed. The vertical line z+z¯=1 is reflected at |z−1|=1 onto the unit circle |z|=1. The sets D2,Dˆ2,Dˆ1 are reflected at |z−1|=1 onto D1,D3,Dˆ3, respectively.
Remark 4.1
Obviously, these domains Dk,Dˆk,1≤k≤3, provide a parqueting of C. But with respect to the domain D=D1∪D2∪{|z−1|=1,z+z¯<1}, where D2 is reflected at |z−1|=1 onto D1 and vice versa, or at z+z¯=1 onto Dˆ, a double parqueting of C, one parqueting from the reflection at each of the two circles combined with the reflection at the vertical line are needed.
The trace of a point z∈D depends on the location of the point. There are three cases.
z∈D1. Then z1∈Dˆ1,z2∈D3,z3∈D2,z4∈Dˆ3,z5∈Dˆ2.
1.
z∈D2. Then z1∈Dˆ2,z2∈Dˆ3,z3∈D1,z4∈D3,z5∈Dˆ1.
1.
z∈D with |z−1|=1. Then z1∈Dˆ with |z1|=1,z2+z¯2=1 with 1<|z2|,z3=z,z4=z2,z5=z1.
The harmonic Green function for the domains Dk,Dˆk,1≤k≤3, are given by just one expression. As these six domains are related by reflections the Green function for one of them will also serve for the other five. In case ii. the domain D2={z+z¯<1,|z−1|<1} is investigated. Because Dk,Dˆk,1≤k≤3 provide a parqueting of C, the harmonic Green and Neumann functions for D2 are G1(z,ζ)=log|1−zζ¯ζ−zζ−(1−z¯)1−z+zζz¯(1−ζ)+ζ1−(1−z)ζ|2,N1(z,ζ)=−log|(ζ−z)(1−zζ¯)(1−z+zζ)(ζ−(1−z¯))×(1−(1−z)ζ)(z¯(1−ζ)+ζ)ζ¯|2,see [Citation14].
Remark 4.2
The situation is much simpler in case of a half disc, say Dl={|z|<1,z+z¯<0}. With the above notations here the sub-domains D1 and Dˆ1 are both empty. Dl corresponds to D2. The reflection of Dl at the imaginary axis z+z¯=0 is the right half unit disc Dr={|z|<1,0<z+z¯} and C=Dl¯∪Dr¯∪(C∖D) with the unit disc D={|z|<1} is a parqueting of C. z∈Dl is reflected at the imaginary axis onto −z¯∈Dr and both onto 1z¯,−1z outside D¯, as explained before.
Domain bounded by two circular arcs (digon) D={|z|<1,0<|z|2sin(α−θ)+(z+z¯)sinθ−sin(α+θ)},0<α<π,θ=πn,n∈N (see Figure ) G1(z,ζ)=log|P1(z,ζ)|2,N1(z,ζ)=−log|Q1(z,ζ)|2,P1(z,ζ)=1−z¯ζζ−z∏k=1n−1z¯ζsin(α+kθ)−(z¯+ζ)sinkθ−sin(α−kθ)(1+zζ)sinkθ+zsin(α−kθ)−ζsin(α+kθ),Q1(z,ζ)=(ζ−z)(1−z¯ζ)∏k=1n−1[z¯ζsin(α+kθ)−(z¯+ζ)sinkθ−sin(α−kθ)]×[(1+zζ)sinkθ+zsin(α−kθ)−ζsin(α+kθ)], see [Citation31, Citation84].
Hyperbolic half plane, lens D=D∩Dm(r), and lunes D∖Dm(r),Dm(r)∖D with Dm(r)={|z−m|<r},1+r2=m2 (see Figure ) G1(z,ζ)=log|1−zζ¯ζ−zm(ζ¯+z)−(1+zζ¯))ζ+z−m(1+zζ)|2,N1(z,ζ)=−log|(ζ−z)(1−zζ¯)(ζ+z−m(1+zζ))(1+zζ¯−m(ζ¯+z))|2,see [Citation11, Citation25].
Polyharmonic Green–Almansi function for
D
Gn(z,ζ)=|ζ−z|2(n−1)(n−1)!2log|1−zζ¯ζ−z|2−∑μ=1n−11μ|ζ−z|2(n−1−μ)(1−|z|2)μ(1−|ζ|2)μ,z,ζ∈D,z≠ζ,n∈N,see [Citation35, Citation41, Citation99],
4.2. Infinite parqueting domains
Strip S1={z∈C:z=e2iαz¯+2iateiα,0<t<1},0<α<π,a∈R+ (Figure )
G1(z,ζ)=|sinπζ−e2iαz¯2iaeiαsinπζ−z2iaeiα|2,N1(z,ζ)=−log|sinπζ−e2iαz¯2iaeiαsinπζ−z2iaeiα|2,see [Citation12].
Hyperbolic strip D=D∖{D−m1(r1)⋃Dm2(r2)} (Figure )
Dm(r)={|z−m|<r},1+r2=m2,−1<r1−m1<0<m2−r2<1} G1(z,ζ)=log|P(z,ζ)|2,N1(z,ζ)=−log|Q(z,ζ)|2,P(z,ζ)=1−z¯ζζ−zζ−z11−z1¯ζζ−z21−z2¯ζ∏k=1∞1−z4k−1¯ζζ−z4k−11−z4k¯ζζ−z4kζ−z4k+11−z4k+1¯ζζ−z4k+21−z4k+2¯ζ,Q(z,ζ)=(ζ−z)(1−z¯ζ)∏k=1∞ζ−z2k−1ζ+11−z2k−1¯ζz2k−1¯(1+ζ)ζ−z2kζ−11−z2k¯ζz2k¯(1−ζ),α=m1m2+1,β=m1+m2,z1=−m1z¯+1z¯+m1,z2=m2z¯−1z¯−m2,z2k+3=αz2k−1−βα−βz2k−1,z2k+4=αz2k+βα+βz2k,k∈N,z3=αz−βα−βz,z4=αz+βα+βz,[Citation7, Citation10, Citation12].
Degenerate ring D={12<|z−12|,|z|<1}={1<|2z−1|,|z|<1} (Figure )
z2k=(k−1)z−kkz−(k+1),z2k+1=(k+1)z¯−k(k+2)z¯−(k+1),zˆ2k=(k+1)z¯−(k+2)kz¯−(k+1)=1z2k+2¯,zˆ2k+1=(k+3)z−(k+2)(k+2)z−(k+1)=1z2k+3¯,k∈N,G1(z,ζ)=log|P1(z,ζ)|2,N1(z,ζ)=−log|Q1(z,ζ)|2,z,ζ∈D,ζ≠z,P1(z,ζ)=zz¯1−z¯ζζ−zζ+z¯−2zζ+1−2z∏k=0∞ζ−z2k+1ζ−z2k+2ζ−zˆ2k+2ζ−zˆ2k+1,Q1(z,ζ)=zz¯ζ−zζ−11−z¯ζ1−ζzζ+1−2zζ−1ζ+z¯−2ζ−1×∏k=0∞[ζ−z2k+1ζ−1ζ−z2k+2ζ−1ζ−zˆ2k+1ζ−1ζ−zˆ2k+2ζ−1],see [Citation16].
Planar circular rectangle D={0<i(z¯−z),|z∓1|<2<|z−i3|} (Figure )
Figure 21. Strip domain.
Figure 22. Hyperbolic strip domain.
Figure 23. Degenerate ring.
Figure 24. Circular rectangle.
G1(z,ζ)=log|P(z,ζ)|2,P(z,ζ)=1P0(z,ζ)∏k∈N0P2k+1(z,ζ)P2k+2(z,ζ),with Pk(z,ζ)=ζ−zk+ζ−zk+¯1+zk+ζ1+zk+¯ζζ−zk++¯ζ−zk++1+zk++¯ζ1+zk++ζ,k∈N0,where z0+=z,z1+=m1iz¯−1z¯+m1i,m1=3,zk+1+=αkzk−1++βkiαk−βkizk−1+,αk=mkmk+1−1,βk=mk−mk+1,mk+1=m1mk+1m1+mk,zk++=1−zk+¯1+zk+¯,k∈N,see [Citation18–20].
Equilateral triangle T with corner point in −1,1,i3 (Figure )
G1(z,ζ)={log|∏m+n∈2Z(z−ωm,n−1)3−(ζ¯−1)3(z−ωm,n−1)3−(ζ−1)3|2,ζ=ζ¯orζ=−12(1+i3)ζ¯+32(3+i),log|∏m+n∈2Z(z−ωm,n+1)3−(ζ¯+1)3(z−ωm,n+1)3−(ζ+1)3|2,ζ=ζ¯orζ=12(1−i3)ζ¯−32(3−i),for z∈T and with ωm,n=3m+i3n,m+n∈2Z, see [Citation23, Citation24].
Circular ring Rr,1={0<r<|z|<1} (see Figure )
G1(z,ζ)=log|z|2log|ζ|2logr2+log|1−zζζ−z|2+log|∏k=1∞zζ¯−z2kz−r2kζ1−r2kzζ¯ζ−r2kz|2,N1(z,ζ)=log|z|2log|ζ|2logr2−log|(1−zζ¯)(ζ−z)|2−log∏k=1∞|(1−r2kzζ¯)(zζ¯−r2k)(ζ−r2kz)(z−r2kζ)|zζ¯|2|2.R1;α,β(z,ζ)=G1(z,ζ)+2β∑k=−∞,k≠0∞(zζ¯)k+(z¯ζ)k(α+kβ)(1−r2k),ifαβ∉Z,R1;α,β(z,ζ)=G1(z,ζ)+2β∑k=−∞,k≠0,k0∞(zζ¯)k+(z¯ζ)k(α+kβ)(1−r2k)+21−r2k0[(zζ¯)k0log(zζ¯)+(z¯ζ)k0log(z¯ζ)],ifk0=−αβ∈Z,see [Citation26, Citation91].
Upper half circular ring Rr,1+={0<r<|z|<1,0<y=Imz} (see Figure )
G1(z,ζ)=log|1−zζ¯ζ−zζ¯−z1−zζ×∏k=1∞r2kζ¯−zr2kζ−zzζ¯−r2kzζ−r2kζ¯−r2kzζ−r2kz1−r2kzζ¯1−r2kzζ|2,N1(z,ζ)=2log|zζ|2r2−log|(ζ−z)(ζ¯−z)(1−zζ¯)(1−zζ)|2+∑n=1∞[4log|zζ|2−log|(z−r2nζ)(z−r2nζ¯)(zζ−r2n)(zζ¯−r2n)×(ζ−r2nz)(ζ¯−r2nz)(1−r2nzζ)(1−r2nzζ¯)|2],see [Citation21].
Quarter ring Rr,1++={0<rG1(z,ζ)=log|1−zαζ¯αζα−zαζ¯α−zα1−zαζα×∏k=1∞r2kαζ¯α−zαr2kαζα−zαzαζ¯α−r2kαzαζα−r2kαζ¯α−r2kαzαζα−r2kαzα1−r2kαzαζ¯α1−r2kαzαζα|2,see [Citation59].
G1(z,ζ)={log|∏m+n∈2Z(z−ωm,n−2)3−(ζ¯−2)3(z−ωm,n−2)3−(ζ−2)3|2,ζ∈∂1P+∪∂4P+,z∈P+,log|∏m+n∈2Z(z−ωm,n+1−i3)3−(ζ¯+1+i3)3(z−ωm,n+1−i3)3−(ζ+1−i3)3|2,ζ∈∂2P+,z∈P+,log|∏m+n∈2Z(z−ωm,n+2)3−(ζ¯+2)3(z−ωm,n+2)3−(ζ+2)3|2,ζ∈∂3P+∪∂4P+,z∈P+,with the segments ∂kP+ on the respective lines of ∂P+ between the points [2,1+i3],k=1;[1+i3,−1+i3],k=2;[−1+i3,−2],k=3;[−2,2],k=4; and ωm,n=3m+i3n,m+n∈2Z, see [Citation89].
Schweikart triangle on the hyperbolic plane with one right and two zero angles (Figure )
Figure 25. Equilateral triangle.
Figure 26. Ring sector for π5.
Figure 27. All first reflections.
Its sides are the interval from the origin to the point i on the imaginary axis, the interval between i and 1 + i on the parallel to the real axis through the point i and the shorter arc on the circle |z−1|=1 between the origin and 1+i. The sides are described by the 2×2−matrices from the Hermitian space H−={A:A∈GL2(C),A∗=A,detA<0}, where A∗ is the complex conjugate transpose of A, A0=(011;0),A1=(−1110),B0=(0−ii−2),C=(1i−i0),corresponding to the generalized circles z+z¯=0,|z−1|=1,z−z¯=2i,|z−i|=1.
Besides A0−1=A0 also B0−1=(2−ii0),A1−1=(0111),E=(1001)are noted.
Figure 28. Basic circles.
A generalized circle, i.e. a circle or a line, is expressed by the equation azz¯+b¯z+bz¯+c=0 with a,c∈R,b∈C,ac−|b|2<0. It describes for a≠0 the circle |z+ba|2=|b|2−aca2 and for a = 0 the line b¯z+bz¯+c=0. The respective 2×2-matrix from H− is (ab¯bc).Reflecting A0 consecutively at A1 reveals Ak=(−k110),k∈Z,and similarly B0 at B1 Bk=(k−k+i(k−1)−k−i(k−1)2(k−1)),k∈Z.Reflecting a generalized circle B at a generalized circle A gives the generalized circle AB−1A. The four circles A−1,A1,B−1,B0A−1−1B0=A0B−1−1A0 build a circular quadrangle on the hyperbolic plane with four zero angles (Figure 28). Its diagonals are the lines A0 and B0. They divide the quadrangle into four Schweikart triangles. The notation is F−11={A−1,A1,B−1,B0A−1−1B0;A0,B0}.Reflecting F−11 at A1 gives the quadrangle (Figures 29 and ) F11={A1,A3,A2B3−1A2,B3:A2,B2}.Similarly, handling F2k−11 at A2k+1 in the same way provides for k∈N0 F2k+11={A2k+1,A2k+3,Ak+2B3−1Ak+2,Ak+1B−1−1Ak+1;A2k+2,Ak+1B0−1Ak+1}.For attaining the full set F2k+1μ,1≤μ≤3k,k∈N0 of quadrangles, F2k+1μ is reflected at the diagonal Ak−κB0−1Ak−κ of F2(k−κ)−11 resulting in F2k+12⋅3κ−μ+1,1≤μ≤3κ, and of F2k+13κ+μ at the diagonal A2(k−κ) of F2(k−κ)−11 giving F2k+13κ+1−μ+1,1≤μ≤3κ, for any 0≤κ≤k−1,k∈N (Figure ).
Figure 29. Fundamental quadrangle.
Figure 30. Fundamental quadrangle.
Figure 31. First reflections.
The quadrangles F2k+1μ,1≤μ≤3k,k∈N0, determine a parqueting for the lens {|z−1|<1,|z−i|<1}. To complete the parqueting of the hyperbolic plane |z−i|<1 reflection of the F2k+1μ,1≤μ≤3k,k∈N0, at the diagonal B0 and then a final reflection at the diagonal A0 provide the parqueting of the four lenses inside |z−i|<1. F−11 completes the parqueting of the hyperbolic plane and its complementary outside (Figures and ).
Figure 32. Hyperbolic tessellation.
Figure 33. Parqueting of outer circle.
A mathematical description of the parqueting process is possible on the basis of the inversive transformations rk and sk of the reflections at the circles Ak and Bk, respectively. The reflections {rkr0:k∈Z} and {sks0:k∈Z} form cyclic groups of Möbius transformations. Introducing ϱk=rks0rk=(r1r0)ks0(r0r1)kthe relations F2k+13κ+μ=ϱk−κF2k+13κ−μ+1,F2k+12⋅3κ+μ=r2(k−κ)ϱk−κF2k+1μhold for 1≤μ≤3κ,0≤κ≤k−1,k∈N. Introducing triple sets of inversive transformations an explicit but involved description is possible, to describe the F2k+1μ by F2k+11 and finally by F11, see [Citation13].
The trace of a point z∈T during the continued reflections has to be constructed. Reflecting z∈T at the side A1 gives z¯z¯−1. This point reflected at B2 becomes z¯+2iz¯+2i+1. And reflecting this point at A2, or z¯z¯−1 at B2, produces z−2iz−2i−1. These four points build the trace of z∈T in the quadrangle F11 (Figure 34). By the relation F2k+11=rk+2F11 the trace points in F2k+11 turn out to be z(k+1)z+1,z¯(k+1)z¯−1,z−2i(k+1)(z−2i)−1,z¯+2i(k+1)(z¯+2i)+1.These expressions hold also for k = −1, representing the trace in F−11.
Figure 34. First trace points.
Reflecting F2k+11 at the diagonal AkB0−1Ak and F2k+12 at A2k gives the trace points from F2k+12 and F2k+13, respectively, namely {(2−i)z¯+2(2k−(k+1)i)z¯+2k−i,(2−i)z−2(2k−(k+1)i)z−2k+i,(2−i)z¯+4i(2k−(k+1)i)z¯+2+(4k+1)i,(2−i)z−4i(2k−(k+1)i)z−2−(4k+1)i}∈F2k+12,{(2+i)z+2(2k+(k−1)i)z+2k−i,(2+i)z¯−2(2k+(k−1)i)z¯−2k+i,(2+i)z−4i(2k+(k−1)i)z−2−(4k−1)i,(2+i)z¯+4i(2k+(k−1)i)z¯+2+(4k−1)i}∈F2k+13.Following the reflection scheme {F2k+11,…,F2k+13κ}→ϱk−κ→{F2k+12⋅3κ,…,F2k+13κ+1}→r2(k−κ)→{F2k+12⋅3κ+1,…,F2k+13κ+1}for 0≤κ≤k−1,k∈N, the trace points for the quadrangles F2k+1μ,1≤μ≤3k can principally be determined. They are given by the inversive transformations from the Schottky group formed by the basic inversive transformations ![](https://: